Objective Mathematics Vol 1
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Janaka PriyalalObjective Mathematics Vol 1
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Janaka PriyalalCOMPLETE STUDY PACK FOR
ENGINEERING
ENTRANCES
OBJECTIVE
MATHEMATICS
Volume 1
COMPLETE STUDY PACK FOR
ENGINEERING
ENTRANCES
OBJECTIVE
MATHEMATICS
Volume 1
Amit M. Agarwal
ARIHANT PRAKASHAN (SERIES), MEERUT
Arihant Prakashan (Series), Meerut
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PREFACE
Engineering offers the most exciting and fulfilling of careers. As a Engineer you can find satisfaction
by serving the society through your knowledge of technology. Although the number of
Engineering colleges imparting quality education and training has significantly increased after
independence in the country, but simultaneous increase in the number of serious aspirants has
made the competition difficult, it is no longer easy to get a seat in a prestigious Engineering
college today.
For success, you require an objective approach of the study. This does not mean you 'prepare'
yourself for just 'objective questions'. Objective Approach means more than that. It could be
defined as that approach through which a student is able to master the concepts of the subject
and also the skills required to tackle the questions asked in different entrances such as JEE Main &
Advanced, as well other regional Engineering entrances. These two-volume books on Mathematics
‘Objective Mathematics (Vol.1 & 2)’ fill the needs of such books in the market in Mathematics and
are borne out of my experience of teaching Mathematics to Engineering aspirants.
The plan of the presentation of the subject matter in the books is as follows
— The whole chapter has been divided under logical topic heads to cover the syllabi of JEE Main &
Advanced and various Engineering entrances in India.
— The Text develops the concepts in an easy going manner, taking the help of the examples from the
day-to-day life.
— Important points of the topics have been highlighted in the text. Under Notes, some extra points
regarding the topics have been given to enrich the students.
— The Solved Examples make the students learn the basic problem solving skills in Mathematics. Very
detailed explanations have been provided to make the students skilled in systematically tackling
the problems.
— The answers / solutions to all the questions have been provided.
— The Objective Questions have been divided according to their types Single correct option, More
than One, Assertion-Reason, Matching Type, Integer Type, Passage Based, etc. which can take the
students to a level required for various Engineering entrances in the present scenario.
— Entrance Corner includes the Previous Years' Questions asked in JEE Main & Advanced and other
various Engineering entrances. At the end of the book, JEE Main & Advanced & Other Regional
Entrances Solved Papers have been given.
I would open-heartedly welcome the suggestions for the further improvements of this book (Vol.1)
from the students and teachers.
Amit M. Agarwal
CONTENTS
1. SETS 1-19 Ÿ Harmonic Progression (HP)
Ÿ Introduction Ÿ Arithmetico-Geometric Progression (AGP)
Ÿ Set Ÿ Some Special Series
Ÿ Notations
Ÿ Representation of Sets 4. COMPLEX NUMBERS 109-181
Ÿ The Real Number System
Ÿ Types of Sets
Ÿ Modulus of a Real Number
Ÿ Venn Diagram
Ÿ Imaginary Number
Ÿ Operations on SetsLaws of Algebra of Sets
Ÿ Complex Number
Ÿ Formulae to Solve Practical Problems on
Union and Intersection of Sets Ÿ Algebra of Complex Numbers
Ÿ Conjugate of a Complex Number
2. FUNDAMENTALS OF RELATION Ÿ Modulus of a Complex Number
AND FUNCTION 20-40 Ÿ Argument (or Amplitude) of a Complex
Ÿ Ordered Pair Number
Ÿ Cartesian Product of Sets Ÿ Various Forms of a Complex Number
Ÿ Properties of Cartesian Product of Sets Ÿ De-Moivre’s Theorem
Ÿ Relation Ÿ Roots of Unity
Ÿ Representation of Relation Ÿ Geometrical Applications of Complex
Ÿ Domain and Range of Relations Numbers
Ÿ Some Particular Types of Relations Ÿ Loci in Complex Plane
Ÿ Inverse Relation Ÿ Logarithm of Complex Numbers
Ÿ Composition of Relations
Ÿ Functions or Mappings 5. INEQUALITIES AND
QUADRATIC EQUATION 182-268
Ÿ Difference between Relation and Function
Ÿ Inequality
Ÿ Domain, Codomain and Range of a Function
Ÿ Generalised Method of Intervals for Solving
Ÿ Equal Functions
Inequalities by Wavy Curve Method
Ÿ Classification of Functions (Line Rule)
Ÿ Algebra of Real Functions Ÿ Absolute Value of a Real Number
Ÿ Composition of Functions Ÿ Logarithms
Ÿ Arithmetico-Geometric Mean Inequality
3. SEQUENCE AND SERIES 41-108
Ÿ Quadratic Equation with Real Coefficients
Ÿ Introduction
Ÿ Formation of a Polynomial Equation from
Ÿ Arithmetic Progression (AP)
Given Roots
Ÿ Geometric Progression (GP)
Ÿ Symmetric Function of the Roots
Ÿ Transformation of Equations 8. BINOMIAL THEOREM 348-417
Ÿ Common Roots Ÿ Binomial Theorem for Positive Integral Index
Ÿ Quadratic Expression and its Graph Ÿ Multinomial Theorem
Ÿ Maximum and Minimum Values of Rational Ÿ Greatest Coefficient
Expression Ÿ Greatest Term
Ÿ Location of the Roots of a Quadratic Equation Ÿ R-f Factor Relation
Ÿ Algebraic Interpretation of Rolle’s Theorem Ÿ Divisibility Problems
Ÿ Condition for Resolution into Linear Factors Ÿ Properties of Binomial Coefficients
Ÿ Some Application of Graphs to Find the Roots Ÿ Binomial Theorem for any Index
of Equations Ÿ Approximation
Ÿ Exponential Series
6. PERMUTATION AND COMBINATION 269-332
Ÿ Logarithmic Series
Ÿ Fundamental Principles of Counting (FPC)
Ÿ Factorial
9. TRIGONOMETRIC FUNCTIONS
Ÿ Exponent of Prime p in Factorial n AND EQUATIONS 418-511
n n
Ÿ Representation of Symbols Pr and Cr Ÿ Introduction
Ÿ Some Basic Arrangements and Selections Ÿ Measure of Angles
Ÿ Summation of Numbers (3 different ways) Ÿ Systems of Measurement of Angles
Ÿ Permutations under Certain Conditions Ÿ Trigonometric Ratios
Ÿ Circular Permutations Ÿ Trigonometric Function
n
Ÿ Geometrical Applications of Cr Ÿ Graph of Trigonometric Functions
Ÿ Selection of One or More Objects Ÿ Trigonometrical Identities
Ÿ Number of Divisors and the Sum of the Ÿ Trigonometric Ratios of Allied Angles
Divisors of a Given Natural Number Ÿ Trigonometrical Ratios of Compound Angles
Ÿ Division of Objects into Groups
Ÿ Trigonometric Ratios of Multiples of an Angle
Ÿ Dearrangements
Ÿ Maximum and Minimum Values of
Ÿ Number of Integral Solutions of Linear Trigonometrical Expressions
Equations and Inequations Ÿ Trigonometric Equations
Ÿ General Solution of Trigonometric Equations
7. MATHEMATICAL INDUCTION 333-347
Ÿ Solution of Trigonometric Inequality
Ÿ Introduction
Ÿ Statement
10. PROPERTIES OF TRIANGLES,
Ÿ Principle of Mathematical Induction
HEIGHTS AND DISTANCES 512-589
Ÿ Algorithm for Mathematical Induction
Ÿ Introduction
Ÿ Types of Problems
Ÿ Relation between the Sides and Angles of
Triangle
Ÿ Trigonometric Ratios of Half Angles of a Ÿ Locus and its Equation
Triangle Ÿ Combined Equation of a Pair of Straight Lines
Ÿ Area of a Triangle Ÿ Bisectors of the Angle between the Lines
Ÿ Conditional Identities Given by a Homogeneous Equation
Ÿ Solution of Triangles Ÿ General Equation of Second Degree
Ÿ Circles Connected with Triangle Ÿ Equations of the Angle Bisectors
Ÿ The Orthocentre and the Pedal Triangle Ÿ Distance between the Pair of Parallel Lines
Ÿ Cyclic Quadrilateral
Ÿ Regular Polygon 13. CIRCLE 695-791
Ÿ Introduction
Ÿ Heights and Distances
Ÿ Standard Equation of a Circle
Ÿ Some Important Properties of Triangles
Ÿ Circle Passing through Three Points
Ÿ Some Properties Related to Circle
Ÿ Position of a Point with respect to a Circle
11. CARTESIAN SYSTEM OF Ÿ Intersection of a Straight Line and a Circle
RECTANGULAR COORDINATES 590-626 Ÿ Equation of Tangent
Ÿ Introduction Ÿ Normal to a Circle
Ÿ Coordinate System Ÿ Pair of Tangents
Ÿ Distance Formulae Ÿ Director Circle
Ÿ Applications of Distance Formula Ÿ Pole and Polar
Ÿ Section Formulae Ÿ Diameter of a Circle
Ÿ Area of a Triangle Ÿ Angle of Intersection of Two Circles
Ÿ Area of a Quadrilateral Ÿ Family of Circles
Ÿ Some Standard Points of a Triangle Ÿ Coaxial System of Circles
Ÿ Locus Ÿ Limiting Points
Ÿ Transformation of Axes
14. PARABOLA 792-853
12. STRAIGHT LINE AND PAIR Ÿ Conic Sections
OF STRAIGHT LINES 627-694 Ÿ Parabola
Ÿ Straight Line Ÿ Other Standard Forms of Parabola
Ÿ Angle between Two Lines Ÿ Position of a Point with respect to a Parabola
Ÿ Point of Intersection of Two Lines Ÿ Intersection of a Line and a Parabola
Ÿ Image of a Point with Respect to a Line Ÿ Equation of Tangent
Ÿ Family of Lines through the Intersection of Ÿ Angle of Intersection of Two Parabolas
Two Given Lines Ÿ Equation of Normal to Parabola
Ÿ Number of Normals and Conormal Points Ÿ Conjugate points
Ÿ Combined Equation of Pair of Tangents Ÿ Conjugate Lines
Ÿ Director Circle Ÿ Diameter
Ÿ Diameter of a Parabola Ÿ Asymptotes
Ÿ Pole and Polar of a Parabola Ÿ Rectangular Hyperbola
Ÿ Lengths of Tangent, Subtangent, Normal Ÿ Tangent to a Rectangular Hyperbola
and Subnormal Ÿ Normals to a Rectangular Hyperbola
15. ELLIPSE 854-918 17. INTRODUCTION TO THREE
Ÿ Introduction DIMENSIONAL (3D) GEOMETRY 972-986
Ÿ Position of a Point with respect to an Ellipse Ÿ Coordinate Axes and Coordinate Planes in
Ÿ Equation of the Chord Three Dimensional Space
Ÿ Intersection of a Line and an Ellipse Ÿ Coordinates of a Point in Space
Ÿ Tangent Ÿ Distance between Two Points
Ÿ Combined Equation of the Pair of Tangents Ÿ Section Formulae
Ÿ Director Circle Ÿ Centroid of a Triangle
Ÿ Normal
18. INTRODUCTION TO LIMITS
Ÿ Number of Normals and Conormal Points
& DERIVATIVES 987-1040
Ÿ Pole and Polar
Ÿ Limits
Ÿ Conjugate Lines
Ÿ Existence of Limit
Ÿ Diameter
Ÿ Algebra of Limits
Ÿ Evaluation of Limits by Using L' Hospital’s Rule
16. HYPERBOLA 919-971
Ÿ Introduction Ÿ Evaluation of Algebraic Limits
Ÿ Conjugate Hyperbola Ÿ Evaluation of Trigonometric Limits
Ÿ Position of a Point with respect to a Ÿ Evaluation of Exponential and Logarithmic
Hyperbola Limits
Ÿ Intersection of a Line and a Hyperbola Ÿ Evaluation of Exponential Limits of the
Form 1¥
Ÿ Tangent to a Hyperbola
Ÿ Sandwich Theorem for Evaluating Limits
Ÿ Director Circle
Ÿ Some Useful Expansions
Ÿ Normals to a Hyperbola
Ÿ Use of Newton-Leibnitz’s Formula in
Ÿ Equation of the Pair of Tangents
Evaluating the Limits
Ÿ Equations of Chord
Ÿ Derivative
Ÿ Pole and Polar
Ÿ Geometrical Meaning of a Derivative 20. STATISTICS 1058-1094
Ÿ Derivative from First Principle Ÿ Measures of Central Tendency
Ÿ Differentiation of Some Important Functions Ÿ Measures of Dispersion
Ÿ Algebra of Derivative of Functions Ÿ Skewness
Ÿ Chain Rule Ÿ Some Results to be Remembered
Ÿ Logarithmic Differentiation Ÿ Correlation Analysis
Ÿ Characteristics of Correlation Coefficient
19. MATHEMATICAL REASONING 1041-1057 Ÿ Regression Analysis
Ÿ Statements or Propositions
Ÿ Properties of Regression Coefficients
Ÿ Use of Venn Diagrams in Checking Truth and
Ÿ Properties of Lines of Regression
Falsity of Statements
Ÿ Truth Table 21. FUNDAMENTALS OF
Ÿ Logical Connectives/Operators PROBABILITY 1095-1132
Ÿ Quantifiers and Quantified Statements Ÿ Introduction
Ÿ Negation of a Quantified Statement Ÿ Some Basic Definitions
Ÿ Logical Equivalence Ÿ Event
Ÿ Negation of a Compound Statement Ÿ Important Events
Ÿ Converse, Inverse and Contrapositive of an Ÿ Algebra of Events
Implication Ÿ Probability
Ÿ Tautologies and Contradictions Ÿ Geometrical Probability
Ÿ Algebra of Statements Ÿ Addition Theorem of Probability
Ÿ Duality Ÿ Independent Events
Ÿ Booley’s Inequality
JEE Advanced Solved Paper 2015 1135-1140
JEE Main & Advanced Solved Papers 2016 1-12
JEE Main & Advanced/ BITSAT/Kerala CEE/ KCET/AP & TS EAMCET/
VIT/MHT CET Solved Papers 2017 1-32
JEE Main & Advanced/ BITSAT/ KCET/AP & TS EAMCET/
VIT/MHT CET Solved Papers 2018 1-35
JEE Main & Advanced/ BITSAT/ AP & TS EAMCET/
MHT CET/WB JEE Solved Papers 2019-20 1-31
1
Sets
Introduction
In our mathematical language, everything in this universe whether living or non-living is Chapter Snapshot
called an object. If we consider a collection of objects given in such a way that it is possible
to tell beyond doubt, whether a given object is in the collection under consideration or not, ● Introduction
then such a collection of objects is called a well-defined collection of objects. ● Set
● Notations
Set ● Representation of Sets
A set is a well-defined collection of objects. By ‘well-defined collection of objects’, it ● Types of Sets
means that we can definitely decide whether a given particular object belongs to a given
● Venn Diagram
collection or not. The objects, elements and members of a set are synonymous terms.
● Operations on Sets
X Example 1. The set of intelligent students in a class is ● Laws of Algebra of Sets
(a) a null set (b) a well-defined collection
● Formulae to Solve Practical
(c) a finite set (d) not a well-defined collection Problems on Union and
Sol. (d) Since, the criterion for determining the intelligence of a student may vary from person to Intersection of Sets
person, so this is not a well-defined collection of objects.
X Example 2. Which of the following is the collection of first five prime numbers?
(a) {1, 2, 3, 5, 7} (b) {2, 3, 5, 7, 11}
(c) {3, 5, 7, 11, 13} (d) {1, 2, 3, 4, 5}
Sol. (b) It is clear that first five prime numbers are {2, 3, 5, 7, 11}.
Notations
Sets are usually denoted by capital letters A, B, C etc., and their elements by small
letters a, b, c, etc.
Let A be any set of objects and let a be a member of A, then we write a ∈ A and read it
as ‘a belongs to A’ or ‘a is an element of A’ or ‘a is member of A’. If a is not an object
of A, then we write a ∉ A and read as ‘a does not belong to A’ or ‘a is not an element of
A’ or ‘a is not a member of A’.
Representation of Sets
A set is often represented in one of the following two forms :
(i) Roster form or Tabular form (ii) Set builder form
1 Roster Form or Tabular Form
In this form, a set is described by listing its
Ø ●
●
The set {0} is not an empty set as it contains the element 0
(zero).
The set {φ} is not a null set. It is a set containing one
element φ.
Objective Mathematics Vol. 1
elements, separated by commas, within braces { }. ● A set which has atleast one element is called a non-empty set.
e.g. The set of months of a year which have thirty
days, in roster form can be described as Singleton Set
{April, June, September, November} A set consisting of only one element is called a
singleton set.
X Example 3. The roster form of set A = {x : x is
a positive integer and divisor of 9} X Example 7. A set A = {x : x ∈ N and 2 < x < 4} is
(a) {3, 6, 9} (b) {1, 5, 9} (c) {1, 3, 9} (d) {2, 3, 9} called
Sol. (c) Since, x is a positive integer and a divisor of 9. So, x (a) null set
can take values 1, 3, 9. (b) infinite set
∴Roster form of the set A is {1, 3, 9}. (c) singleton set
(d) None of the above
Set Builder Form
Sol. (c) {3}; Since, x ∈ N and 2 < x < 4 ⇒ x = 3
In this form, instead of listing all the elements of a
set we describe the set by some special property Finite Set
(properties) satisfied by all of its elements and write it
as A set which is empty or having a definite number
A = {x : P ( x ) holds} = {x | x has the property P ( x )} and of elements is called a finite set.
read it as ‘A is the set of all elements of x such that x X Example 8. A set {x | x ∈ N and 1 ≤ x ≤ 10} is
has the property P’. The symbol ‘:’ or ‘|’ stands for
‘such that’. called
(a) infinite set (b) finite set
X Example 4. If set A = {1, 2, 3, 4}, then it can be (c) singleton set (d) null set
written in set builder form as Sol. (b) Since, elements of the given set is
(a) A = {x : x ∈ N and x ≤ 5} {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(b) A = {x : x ∈ N and x < 5}
(c) A = {x : x ∈ N and 1 < x < 5} Cardinal Number of a Finite Set
(d) A = {x : x ∈ N and x < 4} The number of distinct elements contained in a
finite set A is called its cardinal number and is denoted
Sol. (b) Clearly, A = { x : x ∈ N and x < 5} describes the set
by n( A ).
A = {1, 2, 3, 4}
e.g. If A = {1, 3, 5, 7}, then n( A ) = 4.
Types of Sets X Example 9. The number of elements in a set
{x : x ∈all vowels} is
Empty Set (a) 6 (b) 26
A set which does not contain any element is called (c) 5 (d) 2
an empty set or null set or void set and is denoted by Sol. (c) A = {a, e, i , o, u}
the symbol φ or {}. So, n( A) = 5
X Example 5. The set {x : x ∈ N , 3 < x < 4} is Infinite Set
equal to
A set which is not finite is called an infinite set. In
(a) {3, 4} (b) {4} (c) φ (d) {3}
other words, a set in which the process of counting of
Sol. (c) Since, x ∈ N and 3 < x < 4, so the given set has no elements does not come to an end is called an infinite
element. set.
X Example 6. The set {x : x 2 + 1 = 0 and x ∈ R } is X Example 10. A set of all points in a plane is
equal to called
(a) {−1, 1} (b) {1} (a) empty set (b) finite set
(c) {−1} (d) φ (c) infinite set (d) None of these
Sol. (d) We know that there is no real number x such that Sol. (c) Points in a plane cannot be counted.
2 x2 + 1 = 0.
Equal Sets
Two sets A and B are said to be equal, if every
We write it as B ⊂ A and read it as B is a proper
subset of A. Thus, B is a proper subset of A, if every
element of B is an element of A and there is atleast one
1
Sets
element of A is also an element of B and every element
element in A which is not in B.
of B is also an element of A. Thus, if x ∈ A ⇒ x ∈ B and
y ∈ B ⇒ y ∈ A, then A and B are equal sets and we can Observe that A ⊆ A i.e. every set is a subset of
write A = B . itself, but not a proper subset.
X Example 11. The set A = {x : x ∈ N and X Example 14. The set {1, 2} is the proper
subset of
1 < x < 7} is equal to
(a) {1, 2} (b) {1, 3, 4}
(a) {2, 3, 4, 5, 6} (b) {1, 2, 3, 4, 5, 6}
(c) {1, 2, 3} (d) {{1, 2}, 3, 4}
(c) {2, 3, 4, 5, 6, 7} (d) {1, 2, 3, 4, 5, 6, 7}
Sol. (c)
Sol. (a) Whenever, we have to show that two sets A and B
are equal, show that A ⊆ B and B ⊆ A, then A = B.
Intervals as Subsets of R
Equivalent Sets Let a, b ∈ R and a < b, then
(i) An open interval denoted by ( a, b) is the set of all
Two finite sets A and B are said to be equivalent, if
real numbers such that ( a, b) = {x ∈ R : a < x < b}
n( A ) = n( B ).
(ii) A closed interval denoted by [ a, b] is the set of all
Clearly, equal sets are equivalent, but equivalent real numbers such that [ a, b] = {x ∈ R : a ≤ x ≤ b}
sets need not to be equal.
(iii) Interval closed at one end and open at the other
Equivalence of two sets is denoted by the symbol are given by [ a, b) = {x ∈ R a ≤ x < b}
‘~’. Thus, if A and B are equivalent sets, we write and ( a, b] = {x ∈ R : a < x ≤ b}
A ~ B which is read as ‘A is equivalent to B’.
X Example 15. The subset of R as intervals
X Example 12. The set {1, 2, 3, 4, 5} is {x : x ∈ R , − 12 < x < − 10} is
equivalent to
(a) ( −12, − 10] (b) [ −12, − 10]
(a) {x : x ∈all vowels} (b) {2, 3, 4}
(c) ( −12, − 10) (d) [ −12, − 10)
(c) {1, 2, 4, 5} (d) {1, 2, 3}
Sol. (c) The given subset is belong to open interval.
Sol. (a)
∴ { x : x ∈ R, − 12 < x < − 10} is (−12, − 10).
Subset and Superset X Example 16. Let R be set of points inside a
The set B is said to be subset of set A, if every rectangle of sides a and b ( a, b >1) wth two sides
element of set B is also an element of set A. along the positive direction of X-axis and Y-axis,
Symbolically, we write it as, B ⊆ A or A ⊇ B , where A then
is superset of B. (a) R = {( x, y) : 0 ≤ x ≤ a, 0 ≤ y < b}
(i) B ⊆ A is read as B is contained in A or B is subset (b) R = {( x, y) : 0 ≤ x < a, 0 ≤ y ≤ b}
of A or A is superset of B. (c) R = {( x, y) : 0 ≤ x ≤ a, 0 < y < b}
(ii) A ⊇ B is read as A contains B or B is a subset of A. (d) R = {( x, y) : 0 < x < a, 0 < y < b}
Evidently, if A and B are two sets such that A ∈ B Sol. (d) Since, ( x, y) both lies in first quadrant.
⇒ x ∈ A, then B is subset of A. The symbol ‘⇒’ stands ∴ x, y > 0 and less than side a and b because R lies
for ‘implies’. We read it as ‘x belongs to B’ implies that inside the rectangle.
∴ R = {( x, y) : 0 < x < a, 0 < y < b}
‘x belongs to A’.
X Example 13. The set A = {1, 2, 3} is the subset Power Set
of The set formed by all the subsets of a given set A is
(a) {1, 2, 4, 5} (b) {1, {2, 3}, 4, 5} called the power set of A. It is usually denoted by P ( A ).
(c) {1, 2, 3, 7, 8} (d) {{1, 2}, 3, 4} X Example 17. If set A = {1, 2, 3}, then P ( A ) is
Sol. (c) equal to
(a) {φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}}
Proper Subset (b) {{1}, {2}, {3}, {1, 2}, {2, 3}, {3, 1}}
The set B is said to be a proper subset of set A, if (c) {φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {3, 1}, {1, 2, 3}}
every element of set B is an element of A, whereas (d) {{1}, {2}, {3}, {1, 2, 3}}
every element of A is not an element of B. Sol. (c) 3
1 Some Results on Subsets
(i) Every set is a subset of itself.
Operations on Sets
Now, we introduce some operations on sets to
Objective Mathematics Vol. 1
(ii) The empty set is a subset of every set.
(iii) The total number of subset of finite set containing construct new sets from the given ones.
n elements is 2 n .
i. Union of sets The union of two sets A and
Comparable Sets B, denoted by A ∪ B is the set of all those
elements, each one of which is either in A or
Two sets A and B are said to be comparable, if one
of them is a subset of the other i.e. either A ⊆ B or in B or in both A and B.
B ⊆ A. U
X Example 18. The set {1, 2, 3} is comparable
with
(a) {1, 2, 4, 5} (b) {{1, 2}, 4, 3} A B
(c) {{1, 3}, 2, 4} (d) {1, 2, 3, 4}
Sol. (d) Since, {1, 2, 3} is the subset of {1, 2, 3, 4}.
Thus, A ∪ B = {x : x ∈ A or x ∈ B }
Clearly x ∈A ∪ B
Universal Set ⇒ x ∈A
In any discussion of set theory, there is always a set or x ∈B
that contains all the sets under consideration i.e. it is a and x∉A∪B
superset of each of the given sets. Such a set is called
the universal set and is denoted by U . ⇒ x ∉ A and x ∉ B
In the figure, the shaded part represents, A ∪ B .
X Example 19. Which of the following set is the It is evident that A ⊆ A ∪ B , B ⊆ A ∪ B .
universal set of sets A = {2, 4, 5}, B = {1, 3, 5}?
C = {3, 5, 7, 11}, D = {2, 4, 8, 10} X Example 21. If two sets A = {1, 2, 3} and
(a) {1, 2, 3, 5, 6, 7, 8, 9, 10}
B = {1, 3, 5, 7}, then A ∪ B is equal to
(b) {1, 2, 3, 4, 5, 6, 7, 8, 10}
(c) {1, 2, 3, 4, 5, 7, 8, 10, 11} (a) {1, 2, 3, 7} (b) {1, 2, 3, 5, 7, 8}
(d) {1, 2, 3, 4, 6, 7, 8, 10, 11} (c) {1, 2, 3, 5, 7} (d) {1, 2, 3, 4, 5, 6, 7}
Sol. (c) Sol. (c)
ii. Intersection of sets The intersection of two
Venn Diagram sets A and B, denoted by A ∩ B is the set of all
Venn diagrams are the diagrams which represent elements, common to both A and B.
the relationship between sets. U
X Example 20. The set of natural numbers is a
subset of whole numbers which is a subset of
integers. Then, correct representation of it through
Venn diagram is A B
U U
Thus, A ∩ B = {x : x ∈ A and x ∈ B }
W
(a) N Z (b) W N Z Clearly, x ∈A ∩ B
⇒ x ∈ A and x ∈ B
and x ∉ A ∩ B ⇒ x ∈ A or x ∈ B
U In the figure, the shaded part represents A ∩ B .
It is evident that A ∩ B ⊆ A, A ∩ B ⊆ B .
(c) N ZW (d) None of these
X Example 22. If two sets A = {1, 2, 3, 4} and
B = {2, 4, 5}, then A ∩ B is equal to
Sol. (a) U (a) {1, 3} (b) {2, 4}
W (c) {3, 4} (d) {1, 6}
N Z
Sol. (b) Since, 2 and 4 are the common elements in sets A
and B.
4
iii. Disjoint sets Two sets A and B are said to be
disjoint, if A ∩ B = φ. If A ∩ B ≠ φ, then A and
v. Symmetric difference of two sets The
symmetric difference of two sets A and B is the
1
set ( A − B ) ∪ ( B − A ) and is denoted by A∆B .
Sets
B are said to be intersecting or overlapping
sets. U
U A–B B–A
A B
A B Thus, A∆B = ( A − B ) ∪ ( B − A )
= {x : x ∉ A ∩ B }
The shaded part represents A ∆B .
X Example 23. Given that,
A = {1, 2, 3, 4, 5, 6} X Example 25. If set A = {1, 3, 5, 7, 9} and set
B = {7, 8, 9, 10, 11} B = {2, 3, 5, 7, 11}, then A ∆ B is equal to
C = {6, 8, 10, 12, 14}. (a) {3, 5, 7} (b) {1, 2}
(c) {9, 11} (d) {1, 2, 9, 11}
Which of the following pair of sets are disjoint sets?
(a) A and B Sol. (d) A ∆ B = ( A − B) ∪ (B − A)
(b) B and C = {1, 9} ∪ {2, 11} = {1, 2, 9, 11}
(c) C and A vi. Complement of a set Let U be the universal
(d) None of the above set and A ⊂ U , then the complement of A,
Sol. (a) denoted by A′ or U − A is defined as
A ′ = {x : x ∈U and x ∉ A}
iv. Difference of sets If A and B are two sets, Clearly, x ∈ A ′ ⇔ x ∉ A
then their difference A − B is the set of all
The shaded part represents A′ .
those elements of A which do not belong to B.
A' U
U
A–B A
A B
X Example 26. Given that U = {x : x is a letter in
Thus, A − B = {x : x ∈ A and x ∉ B } English alphabet} and A = {x : x is a vowel}, then
Clearly, x ∈A − B A′ is equal to
(a) φ (b)U
⇒ x ∈ A and x ∉ B (c) {x : x ∈ all consonants} (d) None of these
In the figure, the shaded part represents A − B .
Sol. (c)
Similarly, the difference B − A is the set of all
those elements of B that do not belong to A i.e. vii. Some results on complement The following
U results are the direct consequences of the
definition of the complement of the set.
B–A (a) U ′ = φ
(b) φ′ = {x ∈U : x ∉φ } = U
A B
(c) ( A ′ ) ′ = {x ∈U : x ∉ A ′ } = {x ∈U : x ∈ A} = A
B − A = {x : x ∈ B and x ∉ A} (d) A ∩ A ′ = {x ∈U : x ∈ A} ∩ {x ∈U : x ∉ A} = φ
(e) A ∪ A ′= {x ∈U : x ∈ A} ∪ {x ∈U : x ∉ A} = U
X Example 24. Given that two sets
A = {1, 3, 5, 7, 9} and B = {2, 3, 5, 7, 11}, then A − B is
(a) {1, 9} Laws of Algebra of Sets
(b) {2, 11} In this article, we shall state and prove some
(c) {3, 5, 7} fundamental laws of algebra of sets.
(d) {1, 2, 3, 5, 7, 9, 11} 1. Idempotent laws For any set A, we have
Sol. (a) {1, 9}, since 1, 9 ∉B (i) A ∪ A = A (ii) A ∩ A = A
5
1 2. Identity laws For any set A, we have
φ and U as identity elements for union and
intersection, respectively.
(ii) Let x be an arbitrary element of A ∩ ( B ∩ C ).
Then,
⇒
x ∈ A ∩ (B ∩ C )
x ∈ A and x ∈ ( B ∩ C )
Objective Mathematics Vol. 1
Proof ⇒ x ∈ A and ( x ∈ B and x ∈C )
(i) A ∪ φ = {x : x ∈ A or x ∈φ} = {x : x ∈ A} = A ⇒ ( x ∈ A and x ∈ B ) and x ∈C
(ii) A ∩ U = {x : x ∈ A and x ∈U } = {x : x ∈ A} = A ⇒ x ∈ ( A ∩ B ) and x ∈C
3. Commutative laws For any two sets A and B, ⇒ x ∈( A ∩ B ) ∩ C
we have ∴ A ∩ (B ∩ C ) ⊆ ( A ∩ B ) ∩ C
(i) A ∪ B = B ∪ A (ii) A ∩ B = B ∩ A Similarly, ( A ∩ B ) ∩ C ⊆ A ∩ ( B ∩ C )
i.e. union and intersection are commutative. Hence, A ∩ (B ∩ C ) = ( A ∩ B ) ∩ C
Proof 5. Distributive laws If A, B and C are any three
Recall that two sets X and Y are equal iff X ⊆ Y sets, then
if every element of X belongs to Y .
(i) A ∪ ( B ∩ C ) = ( A ∪ B ) ∩ ( A ∪ C )
(i) Let x be an arbitrary element of A ∪ B . Then,
(ii) A ∩ ( B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C )
x ∈A ∪ B
i.e. union and intersection are distributive over
⇒ x ∈ A or x ∈ B intersection and union, respectively.
⇒ x ∈ B or x ∈ A Proof
⇒ x ∈B ∪ A (i) Let x be an arbitrary element of A ∪ ( B ∩ C ).
∴ A∪B ⊆B ∪ A Then, x ∈ A ∪ (B ∩ C )
Similarly, B ∪ A⊆A∪B ⇒ x ∈ A or x ∈ ( B ∩ C )
Hence, A∪B =B ∪ A ⇒ x ∈A
(ii) Let x be an arbitrary element of A ∩ B . Then, or ( x ∈ B and x ∈C )
x ∈A ∩ B ⇒ ( x ∈ A or x ∈ B )
⇒ x ∈ A and x ∈ B and ( x ∈ A or x ∈C )
⇒ x ∈ B and x ∈ A ⇒ x ∈( A ∪ B )
⇒ x ∈B ∩ A and x ∈( A ∪ C )
A∩B ⊆B ∩ A ⇒ x ∈ (( A ∪ B ) ∩ ( A ∪ C ))
Similarly, B ∩ A⊆A∩B ∴ A ∪ (B ∩ C ) ⊆ ( A ∪ B ) ∩ ( A ∪ C )
Hence, A∩B =B ∩ A Similarly,( A ∪ B ) ∩ ( A ∪ C ) ⊆ A ∪ ( B ∩ C )
4. Associative laws If A, B and C are any three Hence, A ∪ ( B ∩ C ) = ( A ∪ B ) ∩ ( A ∪ C )
sets, then (ii) Let x be an arbitrary element of A ∩ ( B ∪ C ).
(i) ( A ∪ B ) ∪ C = A ∪ ( B ∪ C ) Then, x ∈ A ∩ ( B ∪ C )
(ii) A ∩ ( B ∩ C ) = ( A ∩ B ) ∩ C ⇒ x ∈A
i.e. union and intersection are associative. and x ∈ (B ∪ C )
Proof ⇒ x ∈A
(i) Let x be an arbitrary element of ( A ∪ B ) ∪ C . and ( x ∈ B or x ∈C )
Then, x ∈( A ∪ B ) ∪ C ⇒ ( x ∈ A and x ∈ B )
⇒ x ∈ ( A ∪ B ) or x ∈C or ( x ∈ A and x ∈C )
⇒ ( x ∈ A or x ∈ B ) or x ∈C
⇒ x ∈( A ∩ B )
⇒ x ∈ A or ( x ∈ B or x ∈C )
or x ∈( A ∩ C )
⇒ x ∈ A or x ∈ ( B ∪ C )
⇒ x ∈( A ∩ B ) ∪ ( A ∩ C )
⇒ x ∈ A ∪ (B ∪ C )
∴ A ∩ (B ∪ C ) ⊆ ( A ∩ B ) ∪ ( A ∩ C )
∴ ( A ∪ B ) ∪ C ⊆ A ∪ (B ∪ C )
Similarly, ( A ∩ B ) ∪ ( A ∩ C ) ⊆ A ∩ ( B ∪ C )
Similarly, A ∪ ( B ∪ C ) ⊆ ( A ∪ B ) ∪ C
Hence, ( A ∪ B ) ∪ C = A ∪ (B ∪ C ) Hence, A ∩ (B ∪ C ) = ( A ∩ B ) ∪ ( A ∩ C )
6
6. De-Morgan’s laws If A and B are any two
sets, then
(i) ( A ∪ B ) ′ = A ′ ∩ B ′
Formulae to Solve Practical
Problems on Union and
1
Sets
(ii) ( A ∩ B ) ′ = A ′ ∪ B ′
Intersection of Sets
Proof Let A, B , C be any finite sets and U be the finite
(i) Let x be an arbitrary element of ( A ∪ B ) ′. universal sets, then
Then, x ∈ ( A ∪ B )′ (i) n ( A ∪ B ) = n ( A ) + n ( B ) − n ( A ∩ B )
⇒ x ∉ (A ∪ B) (ii) If ( A ∩ B ) = φ, then n ( A ∪ B ) = n ( A ) + n ( B )
⇒ x∉A (iii) n ( A − B ) = Number of elements which belong to
and x ∉B only A of sets A and B
⇒ x ∈ A′ = n( A ) − n( A ∩ B ) = n( A ∪ B ) − n( B )
and x ∈B ′ (iv) n ( A∆B ) = Number of elements which belong
⇒ x ∈ A′ ∩ B ′ to exactly one of A or B
∴ ( A ∪ B )′ ⊆ A′ ∩ B ′ = n ( A ) + n ( B ) − 2n ( A ∩ B )
Again, let y be an arbitrary element of (v) n ( A ∪ B ∪ C ) = n ( A ) + n ( B ) + n (C )
A ′ ∩ B ′ . Then, y ∈ A ′ ∩ B ′ − n ( A ∩ B ) − n (B ∩ C ) − n ( A ∩ C )
⇒ y ∈ A′ + n (A ∩ B ∩ C)
and y ∈B ′ (vi) Number of elements in exactly two of the sets
⇒ y∉ A A, B , C
and y∉B = n ( A ∩ B ) + n ( B ∩ C ) + n (C ∩ A )
⇒ y∉ A ∩ B − 3n ( A ∩ B ∩ C )
⇒ y ∈ ( A ∪ B )′ (vii) Number of elements in exactly one of the sets
A, B , C
∴ A′ ∩ B ′ ⊆ ( A ∪ B )′
= n ( A ) + n( B ) + n (C ) − 2n ( A ∩ B )
Hence, ( A ∪ B )′ = A′ ∩ B ′
− 2n ( B ∩ C ) − 2n ( A ∩ C ) + 3n ( A ∩ B ∩ C )
(ii) Let x be an arbitrary element of ( A ∩ B ) ′ .
(viii) n( A ′ ∪ B ′ ) = n {( A ∩ B )′ } = n (U ) − n ( A ∩ B )
Then, x ∈ ( A ∩ B )′
(ix) n ( A ′∩ B ′ ) = n {( A ∪ B ) ′ } = n (U ) − n ( A ∪ B )
⇒ x ∉ (A ∩ B)
⇒ x∉A X Example 27. In a group of 50 people,
and x ∉B 35 speak Hindi, 25 speak both English and Hindi
and all the people speak atleast one of two
⇒ x ∈ A′ languages. The number of people who speak
or x ∈B ′ ‘English’, ‘English but not Hindi’ are respectively
⇒ x ∈ A′ ∪ B ′ (a) 40, 15
⇒ ( A ∩ B ) ′ ⊆ A ′∪B ′ (b) 15, 40
Again, let y be an arbitrary element of (c) 35, 15
A ′ ∪ B ′. Then, (d) 25, 15
⇒ y ∈ ( A ′∪B ′ ) Sol. (a) Let H denotes the set of people speaking Hindi and
⇒ y ∈ A′ E denotes the set of people speaking English.
We have, n (H ∪ E ) = 50, n (H) = 35, n(H ∩ E ) = 25
or y ∈B ′
Also, we have
⇒ y ∉ A or y∉B n (H ∪ E ) = n(H) + n(E ) − n (H ∩ E )
⇒ y∉ (A ∩ B) ⇒ 50 = 35 + n(E ) − 25
⇒ n(E ) = 75 − 35 = 40
⇒ y ∈ ( A ∩ B )′ which is the number of people who speak English.
∴ A′ ∪ B ′ ⊆ ( A ∩ B )′ We have to find the people speak only English
i.e. n (E − H) = n (H ∪ E ) − n(H)
Hence, ( A ∩ B ) ′ = A ′ ∪ B ′
= 50 − 35 = 15 7
1 Work Book Exercise
1 Let A = {1, 2, 3}, B = { 3, 4} C = { 4, 5, 6}. Then,
Objective Mathematics Vol. 1
7 Consider the set A of all determinants of order 3
A ∪ (B ∩ C ) is with entries 0 or 1 only. Let B be the subset of A
a {3} b {1, 2, 3, 4} consisting of all determinants with value 1. Let C
c {1, 2, 5, 6} d {1, 2, 3, 4, 5, 6} be the subset of the set of all determinants with
2 If A = { a, b, d , e }, B = {c , d , f , l , m} and value −1. Then,
a C is empty
C = { a, l , m, o}, then C ∩ ( A ∪ B) will be given by b B has as many elements as C
a {a, d , l, m} c A = B∪C
b { b, c, l, m, o} d B has twice as many elements as C
c { a, l, m}
d { a, b, c, d , f, l, m, o} 8 In a town of 840 persons, 450 persons read
Hindi, 300 read English and 200 read both, then
3 If a set contains n elements, then power set will
the number of persons who read neither is
contain
a 210 b 290 c 180 d 260
a n elements b 2 n elements
c n2 elements d None of these 9 Let S = { x : x is a positive multiple of 3 less than
100}, P = { x : x is a prime number less than 20}.
4 If A and B are two sets, then A ∩ ( A ∪ B) equals
Then, n(S ) + n(P ) is
a A b B a 34
c φ d None of these b 41
5 If A = { φ, { φ}}, then the power set of A is c 33
d 30
a A b { φ, { φ}, A}
c { φ, { φ}, {{ φ}}, A} d None of these 10 If A and B are two disjoint sets, then n ( A ∪ B) is
equal to
6 Subtraction of integers is an operation that is
a n ( A ) + n ( B)
a commutative and associative
b not commutative but associative b n ( A ) + n ( B) − n ( A ∩ B)
c neither commutative nor associative c n ( A ) − n ( B)
d commutative but not associative d n ( A ) ⋅ n ( B)
8
WorkedOut Examples
Type 1. Only One Correct Option
Ex 1. If A = {1, 3, 5, 7, 9, 11, 13, 15, 17}, Now,
A ∩ B = {2, 4 , 6, 8, 12, 20} ∩ {3, 6, 9, 12, 15} = {6, 12}
B = {2, 4, ....., 18} and N is the universal set, B ∩ C = {3, 6, 9, 12, 15} ∩ {5, 10, 15, 20} = {15}
then A ′ ∪ {( A ∪ B ) ∩ B ′ } is C ∩ A = {5, 10, 15, 20} ∩ {2, 4 , 6, 8, 12, 20} = {20}
(a) A (b) N and A ∩ B ∩C =φ
(c) B (d) None of these
A B U
Sol. We have, ( A ∪ B ) ∩ B′ = A − ( A ∩ B ) 2 4 6
12
3
Since, A and B are disjoint sets. 8 1 9
20 5
∴ A∩ B =φ 5 10
⇒ ( A ∪ B ) ∩ B′ = A
∴ {( A ∪ B ) ∩ B′ } ∪ A′ = A ∪ A′ = N C
Hence, (b) is the correct answer. Hence, (b) is the correct answer.
Ex 2. Let A, B and C are subsets of universal setU . If Ex 3. Let A and B have 3 and 6 elements,
A = {2, 4, 6, 8, 12, 20}, B = {3, 6, 9, 12, 15}, respectively. What can be minimum number of
C = {5, 10, 15, 20} and U is the set of all whole elements in A ∪ B ?
numbers. Then, the correct Venn diagram is (a) 3 (b) 6
(c) 9 (d) 18
A B U A B U
2 4
6
3 2 4 6 3 Sol. Note that A ∪ B will contain minimum number of
12
8 12
1 9 8 1 9 elements, if A ⊂ B
(a) 20 5 (b) 20 5
So that, n ( A ∪ B) = n ( B) = 6
5 10 5 10 Hence, (b) is the correct answer.
C C
Ex 4. Two finite sets have m and n elements. The
A
2 4 6 3 B U total number of subsets of the first set is 56
8
12
9 more than total number of subsets of second
(c) 20 (d) None of these set. The values of m and n are
15
5 10 (a) 7 and 6 (b) 6 and 3
C (c) 5 and 1 (d) 8 and 7
Sol. Given, A = {2, 4 , 6, 8, 12, 20} Sol. Given, 2m − 2n = 56
B = {3, 6, 9, 12, 15} By hit and trial method, we get m = 6 and n = 3
C = {5, 10, 15, 20} Hence, (b) is the correct answer.
Type 2. More than One Correct Option
Ex 5. Let A and B be any two sets. Which of the Ex 6. If A and B are two sets such that n( A ) = 70,
following are correct? n( B ) = 60, n( A ∪ B ) =110, then
(a) ( A − B ) ∩ B = φ (a) n( A − B ) = 40
(b) ( A − B ) ∪ B = φ (b) n( B − A ) = 40
(c) ( A − B ) ∩ A = A ∩ B ′ (c) n( A ∩ B ) = 20
(d) ( A − B ) ∪ B = A ∪ B (d) n( A − B ) = n( A ∩ B ′ )
Sol. We have, Sol. We have, n( A ) = 70, n( B ) = 60, n( A ∪ B ) = 110
(a) ( A − B ) ∩ B = ( A ∩ B′ ) ∩ B [QA − B = A ∩ B′ ] ∴ n( A ∪ B ) = n( A ) + n( B ) − n( A ∩ B )
= A ∩ ( B′ ∩ B )
⇒ n( A ∩ B ) = n( A ) + n( B ) − n( A ∪ B )
=A∩φ [Q B′ ∩ B = φ]
=φ = 70 + 60 − 110
(b) ( A − B ) ∪ B = ( A ∩ B′ ) ∪ B = 20
= ( A ∪ B) ∩ ( B′ ∪ B) (a) n( A − B ) = n( A ∩ B′ ) = n( A ) − n( A ∩ B )
= ( A ∪ B) ∩ U [Q B ′ ∪ B = U ] = 70 − 20 = 50
= A∪ B (b) n( B − A ) = n( B ∩ A′ )
(c) ( A − B ) ∩ A = ( A ∩ B′ ) ∩ A = A ∩ B ′ = n( B ) − n( A ∩ B )
(d) ( A − B ) ∪ B = A ∪ B = 60 − 20 = 40
Hence, (a), (c) and (d) are the correct answers. Hence, (b), (c) and (d) are the correct answers. 9
1 Type 3. Assertion and Reason Directions (Ex. Nos. 7-9) In the following examples, Ex 8. Statement I If A and B are any two
Objective Mathematics Vol. 1
each example contains Statement I (Assertion) and non-empty sets, then
Statement II (Reason). Each example has 4 choices (a), ( A − B ) ∪ (B − A) = ( A ∪ B ) − ( A ∩ B )
(b), (c) and (d), out of which only one is correct. The Statement II ( A − B ) = A ∩ B ′
choices are and ( A ∩ B ) ∪ C = ( A ∪ C ) ∩ (B ∪ C )
(a) Statement I is true, Statement II is true; Statement II Sol. ( A − B ) ∪ ( B − A ) = ( A ∩ B′ ) ∪ ( B ∩ A′ )
is a correct explanation for Statement I [Q ( A − B ) = A ∩ B′]
(b) Statement I is true, Statement II is true; Statement II = {( A ∩ B′ ) ∪ B} ∩ {( A ∩ B′ ) ∪ A′ } [distributive
is not a correct explanation for Statement I law]
(c) Statement I is true, Statement II is false = {( A ∪ B ) ∩ ( B′ ∪ B )} ∩ {( A ∪ A′ ) ∩ ( B′ ∪ A′ )}
= {( A ∪ B ) ∩ U } ∩ {U ∩ B′∪ A′ )}[U = Universal set]
(d) Statement I is false, Statement II is true = ( A ∪ B ) ∩ ( B′ ∪ A′ ) = ( A ∪ B ) ∩ ( A ∩ B )′
= ( A ∪ B) − ( A ∩ B)
Ex 7. Statement I If A and B are disjoint sets, then
So, both statements are true and Statement II is correct
( A ′ ∪ B ′ ) ′ = φ. explanation of Statement I.
Statement II ( A ∪ B ) ′ = A ′ ∩ B ′ Hence, (a) is the correct answer.
[De-Morgan’s law]
Ex 9. Statement I If A ′ ∪ B = U , then A ⊆ B .
Sol. ( A′ ∪ B′ )′ = ( A′ )′ ∩ ( B′ )′ = A ∩ B = φ, as A and B
Statement II A ⊂ B ↔ B C ⊂ A C
are disjoint.
∴ Statement I is true, Statement II is true. Sol. A′ ∪ B = U , iff B = A [Q A ∪ A′ = U ]
⇒ A=B ⇒ A⊆B
Also, Statement II is a correct explanation of
Also, A⊆B ⇒ BC ⊂ AC
Statement I.
So, both statements are true but Statement II is not the
Hence, (a) is the correct answer. correct explanation of Statement I.
Hence, (b) is the correct answer.
Type 4. Linked Comprehension Based Questions
Passage (Ex. Nos. 10-12) In a town of 10000 10. The number of families which buy newspaper A only
families, it was found that 40% families buy newspaper = n (P ∩ Q′ ∩ R′ ) = n (P ∩ (Q ∪ R )′ )
A, 20% families buy newspaper B and 10% families buy = n(P ) − n [ P ∩ (Q ∪ R )]
newspaper C. 5% families buy A and B, 3% buy B and C [Q n( A ∩ B′ ) = n( A ) − n( A ∩ B )]
and 4% buy A and C. If 2% families buy all the three news = n (P ) − n[ P ∩ Q ] ∪ (P ∩ R )
papers, then = n(P ) − [ n(P ∩ Q ) + n(P ∩ R ) − n(P ∩ Q ∩ R )]
= 4000 − [ 500 + 400 − 200 ] = 3300
Ex 10. The number of families which buy newspaper A
Hence, (a) is the correct answer.
only, is
(a) 3300 (b) 4000 (c) 1400 (d) 2000 11. The number of families which buy newspaper none of
A , B and C
Ex 11. The number of families which buy newspaper
= n(P ∪ Q ∪ R )′
none of A, B and C, is
= n(U ) − n(P ∪ Q ∪ R )
(a) 4400 (b) 4000 (c) 1000 (d) 3300
= n(U ) − [ n(P ) + n(Q ) + n(R ) − n(P ∩ Q )
Ex 12. The number of families which buy newspaper − n(Q ∩ R ) − n(P ∩ R ) + n(P ∩ Q ∩ R )]
B only, is = 10000 − [ 4000 + 2000 + 1000 − 500
(a) 1400 (b) 2500 − 300 − 400 + 200]
(c) 3000 (d) 2000 = 10000 − 6000
= 4000
Sol. (Ex. Nos. 10-12) Let P, Q and R be the sets of families Hence, (b) is the correct answer.
buying newspapers A, B and C respectively. Let U be
the universal set. Then, 12. The number of families which buy newspaper B only
n(U ) = 10000, n(P ) = 40% of 10000 = 4000 = n(P′ ∩ Q ∩ R′ ) = n(Q ∩ P ′ ∩ R ′ )
n(Q ) = 20% of 10000 = 2000, = n[ Q ∩ (P ∪ R )′ ]
n(R ) = 10% of 10000 = 1000 = n(Q ) − n[ Q ∩ (P ∪ R )]
n(P ∩ Q ) = 5% of 10000 = 500 n(Q ) − n [(Q ∩ P ) ∪ (Q ∩ R )]
n(Q ∩ R ) = 3% of 10000 = 300 = n(Q ) − [ n(P ∩ Q ) + n(Q ∩ R ) − n(P ∩ Q ∩ R )]
n(R ∩ P ) = 4% of 10000 = 400 = 2000 − [ 500 + 300 − 200 ]= 2000 − 600 = 1400
n(P ∩ Q ∩ R ) = 2% of 10000 = 200 Hence, (a) is the correct answer.
10
Type 5. Match the Column
Ex 13. Match the following sets for all sets A and B. B. Since, A ⊂ B
1
Sets
∴ A ∩ B′ = φ
Column I Column II
⇒ ( A ∩ B′ )′ = φ′
If A ⊂ B, then A ∩ B′ A∪B
A. p. ⇒ A′ ∪ B = U
B. If A ⊂ B, then A ′ ∪ B q. φ C. ( A − B ) ∪ A = A [Q ( A − B ) ⊂ A]
C. ( A − B) ∪ A r. U D. ( A − B ) ∪ B = ( A ∩ B ′ ) ∪ B
D. ( A − B) ∪ B s. A = ( A ∪ B) ∩ ( B′ ∪ B)
= ( A ∪ B) ∩ U
Sol. A. Since, A ⊂ B, so there is no element in A, which does
not belong to B. =A∪ B
∴ A − B = φ ⇒ A ∩ B′ = φ A → q; B → r, C → s; D → p
Type 6. Single Integer Answer Type Questions
Ex 14. If U = {x : x 5 − 6x 4 + 11x 3 − 6x 2 = 0}, Ex 15. If A = {x : x is an even natural number}
A = {x : x 2 − 5x + 6 = 0} and and B = {x : x is a prime number }, then
B = {x : x 2 − 3x + 2 = 0}, then n( A ∩ B ) ′ n( A ∩ B ) is equal to ________ .
is equal to ________ . Sol. (1) We have,
Sol. (3) We have, U = {x : x 5 − 6x 4 + 11x 3 − 6x 2 = 0} A = {x : x is an even natural number}
= {0, 1, 2, 3}
= {2, 4 , 6, 8, …}
A = {x : x 2 − 5x + 6 = 0} = {2, 3}
and B = {x : x 2 − 3x + 2 = 0} = {1, 2} and B = {x : x is a prime number}
∴ A ∩ B = {2} = {2, 3, 5, 7, 11, …}
Hence, ( A ∩ B )′ = U − ( A ∩ B ) A ∩ B = {2}
= {0, 1, 2, 3} − {2} = {0, 1, 3}
n( A ∩ B ) = 1
∴ n( A ∩ B )′ = 3
11
Target Exercises
Type 1. Only One Correct Option
1. The set {x : x is a positive integer less than 6 and 11. If aN = { an : n ∈ N }, bN = { nb : n ∈ N },
3 − 1 is an even number} in roster form is
x cN = { cN : n ∈ N } and bN ∩ cN = d N , where
(a) {1, 2, 3, 4, 5} (b) {1, 2, 3, 4, 5, 6}
a, b, c ∈ N and b, c are coprime, then
(c) {2, 4, 6} (d) {1, 3, 5} (a) b = cd (b) c = bd
(c) d = bc (d) None of these
2. The set A = {x : x 4 − x 3 − x 2 = 0and x ∈ N }represents
12. If A = { θ : 2cos 2 θ + sin θ < 2} and
(a) a null set (b) a singleton set
π 3π
(c) an infinite set (d) None of these B = θ : ≤ θ ≤ , then A ∩ B is equal to
3. If A = {1, 2, 3}, B = {x ∈ R : x 2 − 2x + 1 = 0}, C = {1, 2, 3} 2 2
π 5π
and D = {x ∈ R : x 3 − 6x 2 + 11x − 6 = 0}, then the (a) θ : < θ <
2 6
equal sets are 3π
(b) θ : π < θ <
(a) A and B (b) A and C 2
(c) A, B and C (d) A, C and D π 5π 3π
(c) θ : < θ < or π < θ <
4. Which of the following is true? 2 6 2
(a) a ∈{{a}, b} (b) {b, c} ⊂ {a, {b, c}} (d) None of the above
(c) {a, b} ⊂ {a,{b, c}} 13. Let A = {x : x is a digit in the number 3591},
Ta rg e t E x e rc is e s
(d) None of these
5. Let P be the set of prime numbers and S = { t : 2 − 1 t B = {x : x ∈ N , x < 10}.
is a prime}. Then, Which of the following is false?
(a) S ⊂ P (b) P ⊂ S (a) A ∩ B = {1, 3, 5, 9} (b) A − B = φ
(c) S = P (d) None of these (c) B − A = {2, 4 , 6, 7, 8} (d) A ∪ B = {1, 2, 3, 5, 9}
6. If U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {2, 4, 6, 8} and 14. Let A and B be two sets, such that A ∪ B = A. Then,
B = {2, 3, 5, 7}, then ( A ∪ B ) ′, ( A ′ ∩ B ′ ), ( A∆B ) is A ∩ B is equal to
equal to (a) φ (b) B
(c) A (d) None of these
(a) {1, 9}, {2, 8}, {3, 4, 5, 6, 7, 8}
(b) {1, 9}, {1, 9}, {3, 4, 5, 6, 7, 8} 15. Let A and B be two sets, then ( A ∪ B )′ ∪ ( A ′ ∩ B ) is
(c) {1, 9}, {1, 9}, {5, 6, 7, 8}
equal to
(d) None of the above
(a) A′ (b) A
7. If A = {x : x is a multiple of 3} and B = { x : x is a (c) B′ (d) None of these
multiple of 5}, then A − B is equal to 16. LetU be the universal set and A ∪ B ∪ C = U. Then,
(a) A ∩ B (b) A ∩ B (c) A ∩ B (d) A ∩ B {( A − B ) ∪ ( B − C ) ∪ (C − A )}′ is equal to
(a) A ∪ B ∪C
8. Which of the following is not correct? (b) A ∪ (B ∩ C )
(a) A ⊆ A′ if and only if A = φ (c) A ∩ B ∩C
(b) A′ ⊆ A if and only if A = X , where X is the universal (d) A ∩ (B ∪ C )
set
(c) If A ∪ B = A ∪ C , then B = C 17. If A and B are two sets, then
(d) B = C if and only if A ∪ B = A ∪ C and A ∩ B = A ∩C ( A − B ) ∪ ( B − A ) ∪ ( A ∩ B ) is equal to
(a) A ∪ B (b) A ∩ B
9. The set ( A ∪ B ∪ C ) ∩ ( A ∩ B ′ ∩ C ′ )′ ∩ C ′ is equal (c) A (d) B′
to 18. If A = { y : y = 2x, x ∈ N }, B = { y : y = 2x − 1, x ∈ N },
(a) B ∩ C ′ (b) A ∩ C
(c) B′ ∩ C ′ (d) None of these then ( A ∩ B )′ is
(a) A (b) B (c) φ (d) U
10. If A ∪ B = A ∪ C and A ∩ B = A ∩ C, then
19. If n( A ) = 4 and n( B ) = 7, then the minimum and
(a) B = C only when A ⊆ B
(b) B = C only when A ⊆ C maximum value of n( A ∪ B ) are, respectively
(c) B = C (a) 4 and 11 (b) 4 and 7
12 (d) None of the above (c) 7 and 11 (d) None of these
20. Let X be the universal set for sets A and B. If
n( A ) = 200, n( B ) = 300 and n( A ∩ B ) = 100, then
n( A ′ ∩ B ′ ) is equal to 300, provided (X) is equal to
23. In a certain town, 25% families own a phone and
15% own a car, 65% families own neither a phone
nor a car. 2000 families own both a car and a phone.
1
Sets
(a) 600 (b) 700 Consider the following statements in this regard :
(c) 800 (d) 900 I. 10% families own both a car and a phone.
21. Suppose A1 , A 2 , ... , A 30 are thirty sets each with five II. 35% families own either a car or a phone.
elements and B1 , B 2 , ... , B n are n sets each with three III. 40000 families live in the town.
30 n Which of the above statements are correct?
elements. Let ∪ A i = ∪ B j = S . Assume that (a) I and II (b) I and III
i=1 j =1
(c) II and III (d) I, II and III
each elements of S belongs to exactly 10 of the A i ’ s
and exactly 9 of B j ’s. The value of n must be 24. 20 teachers of a school either teach Mathematics or
(a) 30 (b) 40 Physics. 12 of them teach Mathematics, while 4 teach
(c) 45 (d) 50 both the subject. Then, the number of teachers
22. If there are three atheletic teams in a school, 21 are in teaching Physics only is
the basketball team, 26 in hockey team and 29 in the (a) 12 (b) 8
(c) 16 (d) None of these
football team. 14 play hockey and basketball, 15 play
hockey and football, 12 play football and basketball 25. A survey shows that 63% of the Americans like
and 8 play all the games. The total number of cheese whereas 76% like apples. If x% of the
members is Americans like both cheese and apples, then we have
(a) 42 (b) 43 (a) x ≥ 39 (b) x ≤ 63
(c) 45 (d) None of these (c) 39 ≤ x ≤ 63 (d) None of these
Type 2. More than One Correct Option
Targ e t E x e rc is e s
26. If A = {1, 5, 7, 9}, B = {2, 5}, then A − B is equal to 28. In a group of 50 students, the number of students
(a) {1, 7, 9} studying French, English, Sanskrit were found to be
(b) {1, 7, 9, 2} as follows
(c) A ∩ B French = 17; English = 13; Sanskrit = 15;
(d) A ∩ B French and English = 09; English and Sanskrit = 4;
French and Sanskrit = 5; English, French and
27. If X ∪ {1, 2} = {1, 2, 3, 5, 9}, then Sanskrit = 3. The number of students who study
(a) the smallest set of X is {3, 5, 9} (a) French only is 6
(b) the smallest set of X is {2, 3, 5, 9} (b) Sanskrit only is 8
(c) the largest set of X is {1, 2, 3, 5, 9} (c) French and Sanskrit but not English is 2
(d) the largest set of X is {2, 3, 4, 9} (d) atleast one of the three language is 30
Type 3. Assertion and Reason
Directions (Q. Nos. 29-31) In the following questions, 29. Statement I If A = {1, 3, 5, 7, 9, 11, 13, 15, 17,
each question contains Statement I (Assertion) and 19, 21}, B = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20} and N
Statement II (Reason). Each question has 4 choices is the universal set, then A ′ ∪ (( A ∪ B ) ∩ B ′ ) = A
(a), (b), (c) and (d) out of which only one is correct. The Statement II ( A ∪ B ) ∩ B ′ is always A.
choices are
30. Statement I If A, B and C are any three non-empty
(a) Statement I is true, Statement II is true; Statement II sets, then A − ( B ∪ C ) = ( A − B ) ∩ ( A − C )
is a correct explanation for Statement I Statement II A ∩ ( B ∆ C ) = ( A ∩ B ) ∆ ( A ∩ C )
(b) Statement I is true, Statement II is true; Statement II
is not a correct explanation for Statement I 31. Statement I If A, B and C any three sets and U is
universal set, such that n(U ) = 700, n( A ) = 200,
(c) Statement I is true, Statement II is false
n( B ) = 300 and n( A ∩ B ) = 100, then
(d) Statement I is false, Statement II is true n( A ′∩ B ′ ) = 300.
Statement II n( A ′ ∩ B ′ ) = n(U ) − n( A ∩ B )
13
1 Type 4. Linked Comprehension Based Questions
Passage (Q. Nos. 32-34) Out of 100 students; 15 33. The number of students only passed in Mathematics
Objective Mathematics Vol. 1
passed in English, 12 passed in Mathematics, 8 in is
Science, 6 in English and Mathematics, 7 in (a) 5 (b) 4
Mathematics and Science, 4 in English and Science, 4 in (c) 3 (d) 2
all the three passed. 34. The number of students only passed in more than one
32. The number of students passed in English and subject is
Mathematics but not in Science is (a) 9 (b) 3
(c) 2 (d) 1
(a) 3 (b) 2 (c) 4 (d) 5
Type 5. Match the Column
35. Match the following sets for all sets A, B and C :
Column I Column II
A. [( A ′ ∪ B′ ) − A ]′ p. A−B
B. [B′ ∪ ( B′ − A )]′ q. A
C. A − ( A ∩ B) r. B
D. ( A − B) ∩ (C − B) s. (A ∩ C) − B
Type 6. Single Integer Answer Type Questions
Ta rg e t E x e rc is e s
36. If A = {a, b, c}, then the number of proper subsets of A is equal to________ .
37. If A = {x ∈ C : x 2 = 1} and B = {x ∈ C : x 4 = 1}, then number of elements in ( A − B ) is ________ .
38. If A and B are two sets containing 3 and 6 elements, respectively. The maximum number of elements in A ∪ B is
equal to ________ .
39. A college awarded 38 medals in football, 15 in basketball and 20 in cricket. If these medals went to a total of 58 men
and only three men got medals in all the three sports, how many received medals in exactly two of three sports?
Entrances Gallery
JEE Advanced/IIT JEE
1. Let P = {θ :sin θ − cos θ = 2 cos θ } and 2. Let S = {1, 2, 3, 4}. The total number of unordered
Q = {θ :sin θ + cos θ = 2 sin θ} be two sets. Then, pairs of disjoint subsets of S is equal to [2010]
(a) P ⊂ Q and Q − T ≠ φ (b) Q ⊄ P [2011] (a) 25 (b) 34
(c) P ⊄ Q (d) P = Q (c) 42 (d) 41
JEE Main/AIEEE
3. Let A and B be two sets containing four and two (a) N (b) Y − X
elements respectively. Then, the number of subsets (c) X (d) Y
of the set A × B, each having atleast three elements, 5. If A, B and C are three sets, such that A ∩ B = A ∩ C
is [2015]
and A ∪ B = A ∪ C, then [2009]
(a) 219 (b) 256 (c) 275 (d) 510
(a) A = C
4. If X = {4 − 3n − 1: n ∈ N }and Y = {9 ( n − 1) : n ∈ N },
n (b) B = C
where N is the set of natural numbers, then X ∪ Y is (c) A ∩ B = φ
(d) A = B
14 equal to [2014]
Other Engineering Entrances
6. In a class of 60 students, 25 students play cricket and 11. If n( A ) = 1000, n( B ) = 500 and if n( A ∩ B ) ≥ 1 and
1
Sets
20 students play tennis and 10 students both the n( A ∪ B ) = p, then [Kerala CEE 2012]
games, then the number of students who play neither (a) 500 ≤ p ≤ 1000 (b) 1001 ≤ p ≤ 1498
is [Karnataka CET 2014] (c) 1000 ≤ p ≤ 1499 (d) 999 ≤ p ≤ 1499
(a) 45 (b) 0 (c) 25 (d) 35 (e) 1000 ≤ p ≤ 1498
7. The set A = { x : 2x + 3 < 7} is equal to the set 12. Out of 64 students, the number of students taking
Mathematics is 45 and number of students taking
[Karnataka CET 2014]
both Mathematics and Biology is 10. Then, the
(a) D = {x : 0 < (x + 5) < 7} (b) B = {x : − 3 < x < 7}
(c) E = {x : − 7 < x < 7} (d) C = {x : 13 < 2x < 4} number of students taking only Biology is
[OJEE 2012]
8. There is a group of 265 persons who like either (a) 18 (b) 19
singing or dancing or painting. In this group 200 like (c) 20 (d) None of these
singing, 110 like dancing and 55 like painting. If 60 13. There are 100 students in a class. In the examination,
persons like both singing and dancing. 30 like both 50 of them failed in Mathematics, 45 failed in
singing and painting and 10 like all three activities, Physics, 40 failed in Biology and 32 failed in exactly
then the number of persons who like only dancing two of the three subjects. Only one student passed in
and painting is [WB JEE 2014] all the subjects. Then, the number of students failing
(a) 10 (b) 20 (c) 30 (d) 40 in all the three subjects is [WB JEE 2012]
2 1 (a) 12 (b) 4
9. Let X n = z = x + iy : z < for all integers n > 1. (c) 2 (d) Cannot be determined
n
∞
14. The set A = {x : x ∈ R , x 2 = 16 and 2x = 6} is equal to
Then, ∩ X n is [WB JEE 2014] [BITSAT 2011]
Targ e t E x e rc is e s
n=1 (a) φ (b) {14, 3, 4} (c) {3} (d) {4}
(a) a singleton set
(b) not a finite set 15. Let A and B be two sets, then ( A ∪ B )′ ∩ ( A ′ ∩ B ) is
(c) an empty set equal to [GGSIPU 2011]
(d) an finite set with more than one element (a) A′ (b) A
10. The shaded region in the figure represents (c) B′ (d) None of these
[Kerala CEE 2014]
16. Let A = {1, 2}, B = {{1}, {2}}, C = {{1, 2}}. Then which
U of the following relation is true? [J&K CET 2011]
A B (a) A = B (b) B ⊆ C
(c) A ∈ C (d) D ⊂ C
17. 25 people employed for programme A, 50 people for
programme B, 10 people for both. So, number of
(a) A ∩ B (b) A ∪ B (c) B − A (d) A − B employee employed for only A is [OJEE 2011]
(e) ( A − B ) ∪ ( B − A ) (a) 15 (b) 20 (c) 35 (d) 40
Answers
Work Book Exercise
1. (b) 2. (c) 3. (b) 4. (a) 5. (c) 6. (c) 7. (b) 8. (b) 9. (b) 10. (a)
Target Exercises
1. (a) 2. (a) 3. (d) 4. (d) 5. (a) 6. (b) 7. (b) 8. (c) 9. (a) 10. (c)
11. (c) 12. (c) 13. (d) 14. (b) 15. (a) 16. (c) 17. (a) 18. (d) 19. (c) 20. (b)
21. (c) 22. (b) 23. (c) 24. (b) 25. (c) 26. (a,d) 27. (a,c) 28. (a,c,d) 29. (c) 30. (b)
31. (c) 32. (b) 33. (c) 34. (a) 35. (*) 36. (7) 37. (0) 38. (9) 39. (9)
*A → q; B → r; C → p; D → s
Entrances Gallery
1. (d) 2. (d) 3. (a) 4. (d) 5. (b) 6. (c) 7. (a) 8. (a) 9. (a) 10. (e)
11. (c) 12. (b) 13. (c) 14. (a) 15. (d) 16. (c) 17. (a) 15
Explanations
Target Exercises
1. Since, 3 x − 1 is an even number for all x ∈ Z + . So, the 7. A − B = A ∩ B is a general result.
given set in roster form is {1, 2, 3, 4, 5}.
8. Options (a) and (b) are trivially true. Option (c) is
2. Given equation can be written as incorrect because if A = {2, 3, 4}, B = { 3}, C = { 4}, then
1± 5 A ∪ B = A ∪ C but B ≠ C. Option (d) is also correct.
x 2 ( x 2 − x − 1) = 0 ⇒ x = 0,
2
9. ( A ∪ B ∪ C ) ∩ ( A ∩ B′ ∩ C′ )′ ∩ C′
∴There is no natural value of x, which satisfies
x4 − x3 − x2 = 0 = ( A ∪ B ∪ C ) ∩ ( A′ ∪ B ∪ C ) ∩ C′
= [( A ∩ A′ ) ∪ (B ∪ C )] ∩ C′
3. A = {1, 2, 3} = (φ ∪ B ∪ C ) ∩ C′
B = { x ∈ R : x 2 − 2 x + 1 = 0} = {1} = (B ∩ C′ ) ∪ (C ∩ C′ )
C = {1, 2, 3} = (B ∩ C′ ) ∪ φ = B ∩ C′
D = { x ∈ R : x 3 − 6 x 2 + 11x − 6 = 0}= {1, 2, 3} 10. Let x ∈ B ⇒ x ∈ A ∪ B ⇒ x ∈ A ∪ C
∴A, C and D are equal. Case I x ∈A
4. a is not an element of {{ a}, b} ∴ x ∈ A ∩ B or x ∈ A ∩ C or x ∈ C
∴ B ⊆C
∴ a ∈{{
/ a}, b}
Case II x ∈C
{ b, c } is the element of { a, { b, c }}
∴ { b, c } ∈ { a, { b, c }} ∴ x ∈ B ⇒ x ∈ C or B ⊆ C
b ∈{ a, b} but b ∈{ / a, { b, c }} Similarly, C⊆B
∴ { a, b} ⊄ { a, { b, c }} ∴ B =C
5. Let x ∈
/ P 11. We have, bN = { nb : n ∈ N}, cN = {cn : n ∈ N}
Ta rg e t E x e rc is e s
⇒ x is a composite number. and dN = {dn : n ∈ N}
Let us now assume that x ∈ S Also, bN ∩ cN = dN
⇒ 2 x − 1 = m [where, m is a prime number] ∴ bc ∈ bN ∩ cN or bc ∈ dN
⇒ 2x = m + 1 ∴ bc = d [as b and c are coprime]
π 3π
which is not true for all composite numbers, say for 12. Let 2 cos θ + sin θ ≤ 2 and ≤ θ ≤
2
x = 4 because 2 4 = 16 which cannot be equal to the 2 2
sum of any prime number m and 1. ⇒ 2 − 2 sin 2 θ + sin θ ≤ 2
Thus, we arrive at a contradiction ⇒ 2 sin 2 θ − sin θ ≥ 0 ⇒ sin θ(2 sin θ − 1) ≥ 0
⇒ x∈ /S π 5π 3π
⇒ ≤θ ≤ or π ≤ θ ≤
So, S ⊂P 2 6 2
6. Given, U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, π 5π 3π
∴ A ∩ B = θ : ≤ θ ≤ or π ≤ θ ≤
A = {2, 4, 6, 8} 2 6 2
and B = {2, 3, 5, 7} 13. We have, A = {1, 3, 5, 9}
Now, A′ = U − A and B = {1, 2, 3, 4, 5, 6, 7, 8, 9}
= {1, 2, 3, 4, 5, 6, 7, 8, 9 } − {2, 4, 6, 8} Now, find A ∪ B, A ∩ B, A − B, B − A
= {1, 3, 5, 7, 9} ∴ Option (d) is false.
and B′ = U − B
= {1, 2, 3, 4, 5, 6, 7, 8, 9} − {2, 3, 5, 7} 14. A ∪ B = A ⇒ B ⊆ A ⇒ A∩B=B
= {1, 4, 6, 8, 9} 15. ( A ∪ B)′ ∪ ( A ′ ∩ B) = ( A′ ∩ B ′ ) ∪ ( A′ ∩ B)
(i) A ∪ B = {2, 4, 6, 8} ∪ {2, 3, 5, 7} = ( A′ ∪ A′ ) ∩ ( A′ ∪ B) ∩ (B′ ∪ A′ ) ∩ (B′ ∪ B)
= {2, 3, 4, 5, 6, 7, 8} = A′ ∩ { A′ ∪ (B ∩ B′ )} ∩ U = A′ ∩ ( A′ ∪ φ ) ∩ U
∴ ( A ∪ B)′ = U − A ∪ B = A′ ∩ A′ ∩ U = A′ ∩ U = A′
= {1, 2, 3, 4, 5, 6, 7, 8, 9}
− {2, 3, 4, 5, 6, 7, 8} 16. ( A − B) ∪ (B − C ) ∪ (C − A) is represented by the
= {1, 9} shaded portion in the figure. The unshaded portion is
(ii) ( A′ ∩ B′ ) = {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9} A ∩ B ∩ C.
A B
= {1, 9}
(iii) Now, A − B = {2, 4, 6, 8} − {2, 3, 5, 7}
= { 4, 6, 8}
and B − A = {2, 3, 5, 7} − {2, 4, 6, 8}
= { 3, 5, 7}
∴ A ∆ B = ( A − B ) ∪ ( B − A)
16 = { 4, 6, 8} ∪ { 3, 5, 7} = { 3, 4, 5, 6, 7, 8} C
∴ {( A − B) ∪ (B − C ) ∪ (C − A)}′ = A ∩ B ∩ C
17. Draw the Venn diagram. From the Venn diagram,
( A − B ) ∪ ( B − A) ∪ ( A ∩ B ) = A ∪ B
18. A = {2, 4, 6, 8, K }, B = {1, 3, 5, 7, K }
24. n(M ∪ P ) = 20, n(M ) = 12, n(M ∩ P ) = 4
Q
⇒
n(M ∪ P ) = n(M ) + n(P ) − n(M ∩ P )
20 = 12 + n(P ) − 4
1
Sets
A∩B=φ ∴ n(P ) = 12
( A ∩ B )′ = U So, the required number
19. n( A ∪ B) is minimum when A ⊆ B. = n(P ) − n(M ∩ P ) = 12 − 4 = 8
In this case, A∪B=B 25. n( X ) = 100, n(C ) = 63, n( A) = 76, n(C ∩ A) = x
and we have n( A ∪ B) = n(B) = 7 Now, n(C ∪ A) = n(C ) + n( A) − n(C ∩ A)
n( A ∪ B) is maximum when A ∩ B = φ. ∴ n(C ∪ A) = 63 + 76 − x = 139 − x
In this case, n( A ∪ B) = n( A) + n(B) ∴ 139 − x ≤ 100 or x ≥ 39
= 4 + 7 = 11 Also, C ∩ A ⊆ C and C ∩ A ⊆ A
20. We have, n( A ∪ B) = n( A) + n(B) − n( A ∩ B) ∴ n(C ∩ A) ⊆ n(C )
∴ n( A ∪ B) = 200 + 300 − 100 = 400 and n(C ∩ A) ⊆ n( A)
Also, n( A ′ ∩ B′ ) = n(( A ∪ B)′ ) = n( X ) − n( A ∪ B) ⇒ n(C ∩ A) ≤ 63
and n(C ∩ A) ≤ 76
⇒ 300 = n( X ) − 400 ⇒ n( X ) = 700
30
⇒ n(C ∩ A) ≤ 63 i.e. x ≤ 63
21. Since, S = ∪ A i and each element of S is in 10 A i’s. ∴ 39 ≤ x ≤ 63
i =1
26. Given, A = {1, 5, 7, 9}, B = {2, 5}
1 30 1
We have, n(S ) = ∑
10 i = 1
n( A i ) =
10
(30 × 5) = 15 A − B = Set of all those elements of A, which do not
belong to B
30
Also, S = ∪ B j and each element of S is in 9 B j’s. = {1, 7, 9}
j =1
Also, A − B = A ∩ B is a general result.
1 n
9∑
We have, n(S ) = n(B j ) Hence, options (a) and (d) are correct.
j =1 27. Given, X ∪ {1, 2} = {1, 2, 3, 5, 9}
1
⇒ 15 = (n × 3) ⇒ n = 45 The smallest set of X = {1, 2, 3, 5, 9} − {1, 2} = { 3, 5, 9}
Targ e t E x e rc is e s
9 The largest set of X = The set which contains atleast
22. Let B, H, F be the sets of members in the basketball elements 3, 5, 9
team, hockey team, football team, respectively. = {1, 2, 3, 5, 9}
∴ n(B) = 21, n(H ) = 26, n(F ) = ,29 28. Given, n(F ) = 17, n(E ) = 13, n(S ) = 15, n(F ∩ E ) = 9,
n(H ∩ B) = 14, n(H ∩ F ) = 15
,
n(E ∩ S ) = 4, n(F ∩ S ) = 5, n(E ∩ F ∩ S ) = 3
n(F ∩ B) = 12, n(B ∩ H ∩ F ) = 8
∴ n(B ∪ H ∪ F ) = n(B) + n(H ) + n(F ) (a) Number of students study only French
− n(B ∩ H ) − n(H ∩ F ) − n(B ∩ F ) = n(F ) − n(F ∩ E ) − n(F ∩ S ) + n(F ∩ E ∩ S )
= 17 − 9 − 5 + 3 = 6
+ n(B ∩ H ∩ F )
= 21 + 26 + 29 − 14 − 15 − 12 + 8 = 43 (b) Number of students study only Sanskrit
= n(S ) − n(S ∩ F ) − n(S ∩ E ) + n(F ∩ E ∩ S )
23. Let X, P, C denote the sets of all families, families owning
= 15 − 5 − 4 + 3 = 9
phone, families owning car, respectively.
(c) n(F ∩ S ∩ E ) = n(F ∩ S ) − n(F ∩ E ∩ S )
Let total number of families be k.
k×5 k k = 5− 3=2
Since, n(P ∩ C ) = = ⇒ 2000 = (d) ∴ n(F ∪ E ∪ S ) = n(F ) + n(E ) + n(S )
100 20 20
⇒ k = 40000 − n(F ∩ E ) − n(E ∩ S ) − n(S ∩ F )
+ n(F ∩ E ∩ S )
∴ Statement III is correct.
2000 = 17 + 13 + 15 − 9 − 4 − 5 + 3 = 30
Now, n(C ∩ P ) = 2000 = × 100% = 5%
40000 29. In Statement I, A ∩ B = φ
∴ Statement I is incorrect. ⇒ ( A ∪ B) ∩ B′ = A
A′ ∪ (( A ∪ B) ∩ B′ ) = A ′ ∪ A = N
Statement I is true and Statement II is false.
Option (c) is true.
n(P) n(C)
= 25% – 5% 5% = 15% – 5%
30. A − (B ∪ C ) = A ∩ (B ∪ C )′ = A ∩ (B′ ∩ C′ )
= 20% = 10% = ( A ∩ B′ ) ∩ ( A ∩ C′ )
= ( A − B) ∩ ( A − C )
Statement I is true.
A ∩ (B ∆ C ) = A ∩ {(B − C ) ∪ (C − B)}
= { A ∩ (B − C )} ∪ { A ∩ (C − B)}
n (P∩C) n (P ∪C) = 65% = {( A ∩ B) − ( A ∩ C )} ∪ {( A ∩ C ) − ( A ∩ B)}
= ( A ∩ B) ∆ ( A ∩ C )
∴ It is clear from the Venn diagram that Statement II is
correct. Statement II is true.
∴ The correct answer is (c). Option (b) is true. 17
1 31. Since, n( A′ ∩ B′ ) = n(U ) − n( A ∪ B)
= n(U ) − { n( A) + n(B) − n( A ∩ B)}
= 700 − {200 + 300 − 100}
C. A − ( A ∩ B) = A ∩ ( A ∩ B)′
= A ∩ ( A′ ∪ B′ )
= ( A ∩ A′ ) ∪ ( A ∩ B′ )
Objective Mathematics Vol. 1
= 300 = φ ∪ ( A ∩ B′ )
∴Statement I is true and Statement II is false. = A ∩ B′ = A − B
Option (c) is true. D. ( A − B) ∩ (C − B)
= ( A ∩ B′ ) ∩ (C ∩ B′ )
Solutions (Q. Nos. 32-34) = ( A ∩ C ) ∩ B′ = ( A ∩ C ) − B
Let U , E, M and S be denote the total number of
students passed in English, passed in Mathematics
36. We know that the total number of proper subsets of a
and passed in Science, respectively. finite set containing n elements is 2 n − 1.
Here, n(U ) = 100, n(E ) = 15, n(M ) = 12, ∴ Number of proper subsets of A
n(S ) = 8, n(E ∩ M ) = 6, n(M ∩ S ) = 7, = 23 − 1 = 7
n(E ∩ S ) = 4 and n(E ∩ M ∩ S ) = 4 37. A = { −1, 1}, B = { −i , i , − 1, 1}
32. The number of students passed in English and A−B=φ
Mathematics but not in Science ∴ n( A − B) = 0
= n(E ∩ M ∩ S )
38. The maximum number of elements in A ∪ B iff
= n(E ∩ M ) − n(E ∩ M ∩ S ) = 6 − 4 = 2
n( A ∩ B) = 0
33. The number of students only passed in Mathematics ∴ n( A ∪ B) = n( A) + n(B) − n( A ∩ B)
= n(M ∩ E ∩ S ) = 3+ 6= 9
= n(M ) − n(M ∩ E ) − n(M ∩ S ) + n(M ∩ E ∩ S )
39. We have, n(F ) = 38, n(B) = 15 ,n(C ) = 20,
= 12 − 6 − 7 + 4 = 16 − 13 = 3
n(F ∪ B ∪ C ) = 58, n(F ∩ B ∩ C ) = 3
34. The number of students only passed in more than one Now, n(F ∪ B ∪ C ) = n(F ) + n(B) + n(C )
subject
− n(F ∩ B) − n(F ∩ C ) − n(B ∩ C )
= n(M ∩ E ) + n(M ∩ S ) + n(S ∩ E ) − 2n (M ∩ E ∩ S )
+ n(F ∩ B ∩ C )
= 6 + 7 + 4 − 2(4) = 17 − 8 = 9
⇒ n(F ∩ B) + n(B ∩ C ) + n(F ∩ C )
Ta rg e t E x e rc is e s
35. A. [( A′ ∪ B′ ) − A]′ = [( A ∩ B)′ ∩ A′ ]′ = 38 + 15 + 20 + 3 − 58 = 18
= ( A ∩ B) ∪ A = A Now, number of men who received medals in exactly
B. [B′ ∪ (B′ − A)]′ = B′ ∪ (B′ ∩ A′ )]′ two of the three sports
= [B′ ∪ (B ∪ A)′ ]′ = n(F ∩ B) + n(B ∩ C ) + n(F ∩ C ) − 3 n (F ∩ B ∩ C )
= [B ∩ (B ∪ A)] = B = 18 − 3 × 3 = 18 − 9 = 9
Entrances Gallery
1. Since, cos θ ( 2 + 1) = sin θ 4. We have, X = { 4n − 3n − 1 : n ∈ N}
⇒ tan θ = 2 + 1 X = { 0, 9, 54, 243, K } [put n = 1, 2, 3, K]
and sin θ ( 2 − 1) = cos θ Y = { 9 (n − 1) : n ∈ N}
1 2 +1 Y = { 0, 9, 18, 27, K } [put n = 1, 2, 3, K]
⇒ tan θ = × = ( 2 + 1)
2 −1 2 +1 It is clear that X ⊂ Y
∴ P =Q ∴ X∪Y=Y
2. Let A ∩ B = φ, A, B ⊂ S 5. Since, A ∩ B = A ∩ C
and A∪ B = A∪C ⇒ B =C
∴Total number of unordered pairs of disjoint subsets of
34 + 1 6. Let the number of students play cricket = C
S = = 41
2 Number of students play tennis = T
and total number of students = S
3. Given, n( A) = 4, n(B) = 2
∴ n(S ) = 60, n(C ) = 25, n(T ) = 20 and n(C ∩ T ) = 10
⇒ n( A × B) = 8 Now, n(C ∪ T ) = n (C ) + n(T ) − n(C ∩ T )
Total number of subsets of set ( A × B) = 2 8 = 25 + 20 − 10 = 35
Number of subsets of set ( A × B) having no element ∴ The number of students who play neither game
(i.e. φ ) = 1 = n(C ∪ T )′ = n(S ) − n(C ∪ T )
Number of subsets of set ( A × B) having one element = 60 − 35 = 25
= 8C1
7. Given, set A = { x :|2 x + 3| < 7}
Number of subsets of set ( A × B) having two elements
= 8C2 Now, |2 x + 3| < 7
⇒ − 7 < 2x + 3 < 7
∴Number of subsets having atleast three elements
⇒ − 7 − 3 < 2x < 7 − 3
= 2 8 − (1 + 8C1 + 8C2 )
⇒ − 10 < 2 x < 4
= 2 8 − 1 − 8 − 28 = 2 8 − 37 ⇒ − 5< x <2
18 = 256 − 37 = 219 ⇒ 0 < ( x + 5) < 7
8. Let D, P, S denote dancing, painting and singing
∴
respectively.
n(D ∪ P ∪ S ) = 265,
12. n(M ) = 45, n(M ∩ B) = 10, n(M ∪ B) = 64
⇒ n(B) = n(M ∪ B) − n(M ) + n (M ∩ B)
= 64 − 45 + 10 = 29
1
Sets
n(S ) = 200, n(D ) = 110, n(P ) = 55 , ⇒ n(only B) = n(B) − n(M ∩ B) = 29 − 10 = 19
n (S ∩ D ) = 60, n(S ∩ P ) = 30
13. n(M ) = 50 = Number of students failed in Mathematics
and n(D ∩ P ∩ S ) = 10
Q n(D ∪ P ∪ S ) = n(D ) + n(P ) + n(S ) − n(D ∩ P ) n(P ) = 45, n(B) = 40
∴ n(M ∩ P ) + n(M ∩ B) + n(P ∩ B)
− n(P ∩ S ) − n(S ∩ D ) + n(D ∩ P ∩ S )
− 3 n (M ∩ P ∩ B) = 32 ...(i)
∴ 265 = 110 + 55 + 200 − n(D ∩ P )
We have, to find n(M ∩ P ∩ B).
− 30 − 60 + 10 Total number of students = 100 [given]
⇒ 265 = 285 − n(D ∩ P ) Clearly, n(M ∪ P ∪ B) = 99
⇒ n(D ∩ P ) = 20 ⇒ n(M ) + n(P ) + n(B) − { n(M ∩ P ) + n(M ∩ B)
∴Number of persons who like dancing and painting + n(P ∩ B)} + n(M ∩ P ∩ B) = 99
= n(D ∩ P ) − n(D ∩ P ∩ S ) ⇒ 50 + 45 + 40 − { 32 + 3 n(M ∩ P ∩ B)}
= 20 − 10 + n(M ∩ P ∩ B) = 99 [from Eq. (i)]
= 10 ⇒ 135 − 32 − 2 n(M ∩ P ∩ B) = 99
⇒ 2 n(M ∩ P ∩ B) = 4
9. Given, Xn = z = x + iy :| z|2 ≤ = x 2 + y 2 ≤
1 1
n n ⇒ n(M ∩ P ∩ B) = 2
1 14. Since, x 2 = 16
∴ X1 = { x 2 + y 2 ≤ 1}, X2 = x 2 + y 2 ≤
2 ⇒ x = ± 4 and 2 x = 6
2 1 ⇒ x=3
X3 = x + y ≤ ,... X∞ = { x + y ≤ 0}
2 2 2
3 Hence, no value of x is satisfied.
∞ ∴ A=φ
∴ ∩ Xn = X1 ∩ X2 ∩ ... ∩ X∞ = { x 2 + y 2 = 0}
n =1 15.
Targ e t E x e rc is e s
∞
Hence, ∩ Xn is a singleton set.
n =1
A A A B A A A B
10. Given, figure clearly represents
( A − B ) ∪ ( B − A)
11. Given, n( A) = 1000, n(B) = 500, n( A ∩ B) ≥ 1, n( A ∪ B) = p (A ∪ B)′ (A′ ∩ B)
Q n( A ∪ B) = n( A) + n(B) − n( A ∩ B) ∴ ( A ∪ B)′ ∩ ( A′ ∩ B) = φ
⇒ 1 ≤ n( A ∩ B) ≤ 500 16. A = {1, 2} ∈ {{1, 2}} = C
Hence, p ≤ 1000 + 500 − 1 = 1499 ∴ A ∈C
and p ≥ 1000 + 500 − 500 = 1000 17. n( A − B) = n( A) − n( A ∩ B)
∴ 1000 ≤ p ≤ 1499 = 25 − 10 = 15
19
2
Fundamentals
of Relation and
Function
Ordered Pair
Two elements a and b listed in a specific order form an ordered pair, denoted by (a, b). In Chapter Snapshot
an ordered pair ( a, b); a is regarded as the first element and b is the second element. It is ● Ordered Pair
evident from the definition that
Cartesian Product of Sets
( a, b) ≠ ( b, a )
●
● Properties of Cartesian Product
Equality of Ordered Pairs of Sets
Two ordered pairs ( a1 , b1 ) and ( a 2 , b2 ) are equal iff ● Relation
a1 = a 2 and b1 = b2 ● Representation of Relation
i.e. ( a1 , b1 ) = ( a 2 , b2 ) ● Domain and Range of Relations
⇒ a1 = a 2 and b1 = b2 ● Some Particular Types of
Thus, it is evident from the definition that, (1, 2) ≠ (2, 1) and (1, 1) ≠ (2, 2). Relations
x 2 5 1 Inverse Relation
Example 1. If + 1, y − = , , find the values of x and y.
●
X
3 3 3 3 ● Composition of Relations
(a) 2 and 1 (b) 2 and 2 (c) −2 and 1 (d)1 and 1 ● Functions or Mappings
Sol. (a) Given, x + 1, y − 2 = 5 , 1 ● Difference between Relation
3 3 3 3
and Function
x 5 2 1
⇒ + 1= and y− =
3 3 3 3 ● Domain, Codomain and Range
⇒
x 5 1
= − and
1 2
y= + of a Function
3 3 1 3 3 ● Equal Functions
x = 2 and y=1
● Classification of Functions
Cartesian Product of Sets ● Algebra of Real Functions
Let A and B be two non-empty sets. The cartesian product of A and B denoted byA × B
● Composition of Functions
is defined as the set of all ordered pairs ( a, b) , where a ∈ A and b ∈ B .
Symbolically, A × B = {( a, b); a ∈ A and b ∈ B }.
If there are three sets A, B , C and a ∈ A, b ∈ B ,
c ∈C , then we form an ordered triplet ( a, b, c). The set
of all ordered triplets ( a, b, c) is called the cartesian
Properties of
Cartesian Product of Sets
2
Fundamentals of Relation and Function
product of these sets A, B and C,
i.e. A × B × C = {( a, b, c) : a ∈ A, b ∈ B , c ∈C } If A, B and C are three sets, then
i. (a) A × ( B ∪ C ) = ( A × B ) ∪ ( A × C )
Number of Elements in the Cartesian
(b) A × ( B ∩ C ) = ( A × B ) ∩ ( A × C )
Product of Two Sets
If A and B are two finite sets, then ii. A × (B − C ) = ( A × B ) − ( A × C )
n ( A × B ) = n ( A) × n (B )
iii. A×B =B × A⇔ A=B
Ø ● If either A or B is an infinite set, then A × B is an infinite set.
● If A , B , C are finite sets, then n(A × B × C) = n(A) × n(B) × n(C) . iv. If A ⊆ B ⇒ A × A ⊆ ( A × B ) ∩ ( B × A )
X Example 2. Let A = {1, 2, 3} and B = {x : x ∈ N , v. IfA ⊆ B ⇒ A × C ⊆ B × C
x is a prime number less than 5}, then n ( A × B ) is
(a) 3 (b) 4 (c) 5 (d) 6 vi. If A ⊆ B and C ⊆ D ⇒ A × C ⊆ B × D
Sol. (d) We have, A = {1, 2, 3}
vii. ( A × B ) ∩ (C × D ) = ( A ∩ C ) × ( B ∩ D )
and B = { x : x ∈ N, x is a prime number less than 5}
= {2, 3}
∴ n ( A) = 3 and n (B) = 2 viii. A × (B ′ ∪ C ′ )′ = ( A × B ) ∩ ( A × C )
⇒ n( A × B) = n( A) × n(B) = 3 × 2 = 6
ix. A × (B ′ ∩ C ′ )′ = ( A × B ) ∪ ( A × C )
Graphical Representation of Cartesian
x. If A and B have n common elements, then
Product of Two Sets A × B and B × A will have n 2 common
Let A and B be two non-empty subsets of R. Then, elements.
the graph of A × B is the set of all points in the plane
represented by the ordered pairs of A × B . X Example 3. If A = {1, 2, 3}, B = {3, 4} and
Let A = {1, 2, 3}, B = {3, 4, 5}, then A × B = {(1, 3),
C = {4, 5, 6}, then A × ( B ∩ C ) is
(1, 4), (1, 5), (2, 3), (2, 4), (2, 5), (3, 3), (3, 4), (3, 5)}
(a) {(1, 3), (2, 3), (3, 3)} (b) {(1, 4), (2, 4), (3, 4)}
Y
(c) {(4, 1), (4, 2), (4, 3)} (d) {(3, 4), (3, 5), (3, 6)}
5
Sol. (b) We have, A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}
4
Now, B ∩ C = {4}
3 ∴ A × (B ∩ C ) = {(1, 4), (2 , 4),(3, 4)}
2
1
X
Relation
0 Let A and B be two non-empty sets. Then, a relation
1 2 3
R from A to B is a subset of A × B .
Diagrammatic Representation of Cartesian Thus, R is a relation from A to B ⇒ R ⊆ A × B . If R is
Product of Two Sets a relation from a non-empty set A to a non-empty set B
In order to represent A × B by an arrow diagram, and if ( a, b) ∈ R , then we write aRb which is read as ‘a is
we first draw Venn diagrams representing sets A and B related to b by the relation R’. If ( a, b) ∉ R , then we write
one opposite to the other as shown in figure. Now, we aRb and we say that ‘a is not related to b by the relation R’.
draw line segments starting from each element of A and X Example 4. Let A = {1, 2, 3} and B = {−1, 0, 1},
terminating to each element of set B. then which of the following is not a relation from A
If A = {x, y, z} and B = {a, b}, then following figure to B?
gives the arrow diagram of A × B . (a) R = {(1, − 1), (2, 0), (3, 1)}
A B (b) R = {(1, − 1), (1, 0), (2, 0), (2, 1)}
x (c) R = {(2, 0), (2, 1), (1, − 1), (3, 1), (3, 0)}
y
a (d) R = {(1, − 1), (2, 0), (1, 3)}
b
z Sol. (d)Q 3 ∉B
∴ {(1, − 1), (2, 0), (1, 3)} is not a relation from A to B. 21
2 X Example 5. Let A = {1, 2, 3}, we define
R1 = {(1, 2), (3, 2), (1, 3)},
Then, R = {(1, 3), (1, 5), (1, 6), (2, 3), (2, 5), (2, 6),
(3, 5), (3, 6)}
Objective Mathematics Vol. 1
R 2 = {(1, 3), (3, 6), (2, 1), (1, 2)}. ∴ Domain of R = {1, 2, 3} , range of R = {3, 5, 6} and
Then, on A codomain of R = {3, 5, 6}
(a) R 1 is relation and R 2 is not Ø Let A and B be two non-empty finite sets having p and q elements
(b) R 1 and R 2 are relations respectively, then the total number of relations from A to B = 2 pq .
(c) R 1 and R 2 are not relations
(d) None of the above X Example 7. Find the domain and range of the
relation R defined by
Sol. (a) R1 ⊆ A × A, so R1 is a relation on A. But
(3, 6) ∉ A × A, so R 2 ⊆/ A × A. Hence, R 2 is not a relation R = {( x, x + 5) : x ∈{0, 1, 2, 3, 4, 5}}
on A. (a) {0, 1, 2, 3, 4, 5}, {5, 6, 7, 8, 9, 10}
(b) {1, 2, 3, 4, 5}, {5, 6, 7, 8, 9, 10}
Representation of Relation (c) {0, 1, 2, 3, 4, 5}, {6, 7, 8, 9, 10}
(d) None of the above
Roster Form Sol. (a) Given, R = {( x, x + 5) : x ∈ {0, 1, 2, 3, 4, 5}}
In this form, a relation is represented by the set of Putting x = 0, 1, 2, 3, 4, 5, we get
all ordered pairs belonging to R. R = {(0, 5), (1, 6), (2, 7 ), (3, 8), (4, 9), (5, 10)}
⇒ Domain = {0, 1, 2, 3, 4, 5}
e.g. The relation R defined on the set of natural
Now, y = x + 5 = 5, 6, 7, 8, 9, 10 [after putting the value of x ]
number as {( a, b) : b is square of a}, then roster form of ⇒ Range = {5, 6, 7, 8, 9, 10}
this relation is
{(1, 1), (2, 4), (3, 9), ( 4, 16), … } X Example 8. If R is a relation on a finite set
having n elements, then the number of relations on
Set Builder Form A is
2
In this form, a relation is represented as (a) 2 n (b) 2 n (c) n 2 (d) n n
R = {( a, b) : a ∈ A, b ∈ B and a, b satisfy the rule Sol. (b) We have, a set having n elements
which associates a and b} i.e. n( A) = n
e.g. In set builder form, the relation and n( A × A) = n( A) × n( A) = n × n = n2
2
× A)
R = {(2, 8), (3, 27 ), (5, 125), ( 7, 343)} is written as ∴Number of relations on A = 2 n( A = 2n
R = {( x, x 3 ) : x is a prime number less than 10}
X Example 6. Let A = {3, 5}, B = {7, 11} and
Some Particular Types of
R = {( a, b) : a + b is odd a ∈ A, b ∈ B }. Relations
Then, R in roster form is
i. Void relation Since, φ ⊂ A × A, it follows
(a) {(3, 7), (5, 11)} (b) {(3, 11), (5, 7)}
that φ is a relation on A, called the empty or
(c) φ (d) None of these
void relation.
Sol. (c) We have, A = {3, 5} and B = {7, 11}. Here, elements
of A and B are odd numbers. Sum of two odd numbers is X Example 9. Consider the relation R on the set
always even. So, there is no element that belongs to R.
A = {2, 4, 6}, R = {( x, y): x − y is odd x, y ∈ A}
Hence, R = φ
Sol. We observe that,| x − y| is always an even number for
every element of A. Therefore, ( x, y) ∉ R, ∀ x, y ∈ A.
Domain and Range of Relations ⇒ R is a void relation on A.
Let R be a relation from A to B. The domain of R is X Example 10. A relation R on a set A is called
the set of all those elements a ∈ A such that ( a, b) ∈ R an empty relation, if
for some b ∈ B . (a) no element of A is related to any element of A
∴ Domain of R = {a ∈ A : ( a, b) ∈ R } (b) every element of A is related to every element of A
and range of R is the set of all those elements b ∈ B (c) some elements of A are related to some element
such that ( a, b) ∈ R for some a ∈ A. of A
∴ Range of R = {b ∈ B : ( a, b) ∈ R } (d) None of the above
Here, B is called the codomain of R. Sol. (a) A relation R on a set A is called an empty relation, if
e.g. Let A = {1, 2, 3} and B = {3, 5, 6} it has no element, i.e. no ordered pair, i.e. no element of A
22 aRb ⇔ a < b is related to any element of A.
ii. Universal relation Since, A × A ⊆ A × A, it
follows that A × A is a relation on A, called the
X Example 13. If A = {1, 2, 3} and
R = {(1, 2), (2, 2), (3, 1), (3, 2)}, then define inverse
2
Fundamentals of Relation and Function
universal relation. relation of R on A.
X Example 11. Consider the relation R on set Sol. R, being a subset of A × A, is a relation on A.
A = {1, 2, 3, 4, 5} defined by Clearly, 1R2; 2 R2; 3R1 and 3R2.
R = {( x, y) :| x − y | ≥ 0, x, y ∈ A}. Domain (R ) = {1, 2, 3}
and range (R ) = {2, 1}
Sol. We observed that,| x − y| ≥ 0, ∀ x, y ∈ A
Also, R −1 = {(2, 1), (2, 2 ), (1, 3), (2, 3)}
⇒ ( x, y) ∈ R, ∀ ( x, y) ∈ A × A
Domain (R −1 ) = {2, 1}
Each element of set A is related to every element of set A
⇒ R= A× A and range (R −1 ) = {1, 2, 3}
⇒ R is universal relation on set A.
iii. Identity relation The relation I A = {( a, a ); Composition of Relations
∀a ∈ A} is called the identity relation on A. Let R ⊆ A × B , S ⊆ B × C be two relations. Then,
composition of the relations R and S denoted by
X Example 12. If A = {1, 2, 3}, then define the
SoR ⊆ A × C and is defined by ( a, c) ∈ ( SoR ) iff ∃ b ∈ B
identity relation on A.
such that ( a, b) ∈ R , ( b, c) ∈ S .
Sol. The identity relation on A is given by
IA = {(1, 1), (2, 2 ), (3, 3)} X Example 14. If A = {1, 2, 3}, B = {a, b, c, d},
C = {α, β, γ }, R ⊆ ( A × B ) = {(1, a ), (1, c), (2, d )},
Inverse Relation S ⊆ ( B × C ) = {( a, α ), ( a, γ ), ( a, β) }, then what is
the composite of relations R and S?
If R is a relation on A, then the relation R −1 on A,
Sol. SoR ⊆ ( A × C ) = {(1, α ), (1, γ ), (1, β )}
defined by R −1 = {( b, a ) : ( a, b) ∈ R } is called an inverse
relation on A. Ø One should be careful in computing the relation RoS. Actually, SoR
starts with R and RoS starts with S.
Clearly, domain ( R −1 ) = range ( R ) In general, SoR ≠ RoS
and range ( R −1 ) = domain ( R ) Also, (SoR)−1 = R−1oS −1, known as reversal rule.
Work Book Exercise 2.1
1 If A and B are two sets, then A × B = B × A, if and a Z, Z b Z +, Z
only if c Z, Z − d None of these
a A⊂ B b B⊂ A
c A=B d None of these
6 Let a relation R be defined by
R = {(4, 5), (1, 4), (4, 6), (7, 6), (3, 7 )}, then RoR is
2 If the set A has 3 elements and the set equal to
B = {1, 3, 4, 5}, then the number of elements in a {(1, 5), (1, 6), (3, 6)}
( A × B) is b {(1, 4), (1, 5), (3, 6)}
a 11 b 12 c 13 d 15 c {(1, 5), (1, 6), (3, 7)}
d {(1, 4), (1, 5), (3, 7)}
3 Let A = {1, 2}, B = {1, 2, 3, 4}, C = { 5, 6} and
D = { 5, 6, 7, 8}
7 If the relation R : A → B, where A = {1, 2, 3, 4} and
B = {1, 3, 5} is defined by R = {( x, y ) : x < y,
Following statements are given below:
x ∈ A, y ∈ B}, then R −1oR is
I. A × (B ∩ C ) = ( A × B) ∩ ( A × C )
a {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}
II. A × C is a subset of B × D. b {(3, 1), (5, 1), (5, 2), (5, 3), (5, 4)}
Which of the above statement is correct? c {(3, 3), (3, 5), (5, 3), (5, 5)}
a Only I b Only II d None of the above
c Both I and II d None of these
8 If R is a relation from set A = {2, 4, 5} to set
4 Let R be a relation from a set A to a set B, then B = {1, 2, 3, 4, 6, 8} defined by xRy ⇔ x divides y,
a R = A∪B b R = A∩B
where x ∈ A, y ∈ B. Then, R −1 is
c R ⊆ A×B d R ⊆ B× A
a {(2, 2), (2, 4), (2, 6), (2, 8), (4, 4), (4, 8)}
5 Let R be the relation on Z defined by b {(2, 2), (1, 4), (1, 2), (1, 5), (2, 4), (4, 4)}
R = {(a, b) : a, b ∈ Z, a − b is an integer}. Find the c {(2, 2), (4, 2), (6, 2), (8, 2), (4, 4), (8, 4)}
d None of the above
23
domain and range of R.
2 Functions or Mappings Let A and B be two non-empty sets. Then, a
then if this line intersects the graph of the expression
in more than one point, then the expression is a relation
else, if it intersects at only one point, the expression is a
Objective Mathematics Vol. 1
function f from set A to B is a rule which associates function. e.g.
elements of set A to elements of set B such that all Y Y
elements of set A are associated to elements of set B in
unique way.
If f associates x ∈ A to y ∈ B , then we say that y is
the image of the element x and denote it by f ( x ) and X'
O
X X'
O
X
write as y = f ( x ). The element x is called the pre-image
or inverse image of y.
Test line
f Test line
A function is denoted by f : A → B or A → B .
Y' Y'
X Example 15. Let A = {1, 2, 3}, B = {2, 3, 4} and (a) Function (b) Relation
f 1 , f 2 and f 3 be three subsets of A × B as given
f 1 = {(1, 2), (2, 3), (3, 4)}
X Example 17. Let A = {1, 2, 3, 4},
f 2 = {(1, 2), (1, 3), (2, 3), (3, 4)} B = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1),
( 4, 5), (2, 11)}, then
f 3 = {(1, 3), (2, 4)}. Then, which is/are function(s)?
(a) f is a relation from A to B
Sol. Obviously, f1 is a function from A to B but f2 and f3 are (b) f is a function from A to B
not functions from A to B, f2 is not a function from A to B
because 1 ∈ A has two images 2 and 3 and f3 is not a
(c) Both (a) and (b)
function from A to B, because 3 ∈ A has no image in B. (d) None of the above
Sol. (a) Since, first elements of the ordered pairs in f belong
X Example 16. If x, y ∈{1, 2, 3, 4}, then which of to A and second elements of the ordered pairs belong to
the following is/are function(s) in the given set? B, so f is relation from A to B.
(a) f 1 = {( x, y) : y = x + 1} Now, 2 has two different images 9 and 11, so f is not a
(b) f 2 = {( x, y) : x + y > 4} function.
(c) f 3 = {( x, y) : y < x}
(d) f 4 = {( x, y) : x + y = 5} Domain, Codomain and Range of
Sol. (d) We have,
f1 = {(1, 2 ), (2, 3), (3, 4)}
a Function
f2 = {(1, 4), (4, 1), (2, 3), (3, 2 ), (2, 4), (4, 2 ), (3, 4), (4, 3)} Let f : A → B . Then, the set A is known as the
f3 = {(2, 1), (3, 1), (4, 1), (3, 2 ), (4, 2 ), (4, 3)} domain of f and the set B is known as the codomain of
f4 = {(1, 4), (2, 3), (3, 2 ), (4, 1)} f . The set of all f -images of elements of A is known as
Here, f4 is a function but f1, f2 and f3 are not functions. the image of A or image set of A under f and is denoted
f1 is not a function because 4 has no image. f2 , f3 are by f ( A ).
not functions because 3, 4 have more than one images.
Thus, f ( A ) = { f ( x ) : x ∈ A} = Range of f
Ø ● There may exist some elements in set B which are not the Clearly, f ( A ) ⊆ B
images of any element in set A.
● To each and every independent element in A, there
corresponds one and only one image in B. X Example 18. Let N be the set of all natural
● Every function is a relation but every relation may or may numbers, then the rule f : N → N such that
not be a function. f ( x ) = x 2 , ∀ x ∈ N . Write domain, range and
● The number of functions from a finite set A into finite set
codomain of the given function.
B = [n(B)][n(A )].
Sol. Given, f( x) = x2
Difference between Relation and f(1) = (1)2 = 1; f(2 ) = 2 2 = 4;
f(3) = 32 = 9 and so on.
Function ∴ Domain (f ) = N
Geometrically, if we draw a Vertical Parallel Line Range (f ) = {1, 4, 9, K}
(VPL), i.e. any line which is parallel to Y -axis (x = a), and codomain (f ) = N
24
X Example 19. The domain of the real function
f (x ) =
1
is
X Example 21. The domain for which the
functions f ( x ) = 2x 2 − 1 and g ( x ) = 1 − 3x are equal, 2
Fundamentals of Relation and Function
4 − x2 is
1
(a) the set of all real numbers (a) {1, 2} (b) {−2, 2} (c) −2, (d) {0, 1}
(b) the set of all positive real numbers 2
(c) (−2, 2) Sol. (c) We have, f( x) = 2 x2 − 1
(d) [−2, 2] and g ( x) = 1 − 3 x
Sol. (c) We have, f( x) = 1 f ( x) = g ( x)
4 − x2 ⇒ 2 x2 − 1 = 1 − 3 x
⇒ 2 x2 + 3 x − 2 = 0
f( x) is defined iff 4 − x > 0
2
⇒ ( x + 2 ) (2 x − 1) = 0
⇒ x2 − 4 < 0 ⇒ x = − 2, 1/2
⇒ ( x + 2) ( x − 2) < 0 Thus, f( x) and g ( x) are equal on the set −2, .
1
⇒ −2 < x < 2 2
∴Domain of f = (−2, 2 )
x2 Classification of Functions
X Example 20. Let f = x, : x ∈ R be a
1 + x
2
Constant Function Y
function from R into R. The range of f is A function which does not
(a) [0, ∞) (b) (0, 1) change as its parameters vary, (0)
(c) [0, 1) (d) [1, ∞) i.e. the function whose rate of X′ X
x2 change is zero. In short, a
Sol. (c) We have, f( x) = constant function is a function Y′
1 + x2
Domain of f = R that always gives or returns the same value.
Range of f : Let f( x) = y, then or
x2 Let k be a constant, then function f ( x ) = k , ∀ x ∈ R
y= ⇒ y + x2 y = x2
1 + x2 is known as constant function.
⇒ x2 =
y
⇒ x=
y Domain of f ( x ) = R and Range of f ( x ) = {k }
1− y 1− y
Clearly, x will take real values iff
y
≥0
Polynomial Function
1− y The function y = f ( x ) = a 0 x n + a1 x n −1 + K + a n ,
y where a 0 , a1 , a 2 ,…,a n are real coefficients and n is a
⇒ ≤ 0 ⇒ 0≤ y< 1
y−1 non-negative integer, is known as a polynomial function.
∴ Range of f = [0, 1) If a 0 ≠ 0, then degree of polynomial function is n.
Domain of f ( x ) = R
Equal Functions and range varies from function to function.
Two functions f and g are said to be equal iff
(i) domain of f = domain of g.
Rational Function
(ii) codomain of f = codomain of g. If P ( x ) and Q ( x ) are polynomial functions,
P (x )
(iii) f ( x ) = g ( x ) for every x belonging to their Q ( x ) ≠ 0, then function f ( x ) = is known as
common domain. Q (x )
rational function.
If two functions f and g are equal,
then we write f = g. Domain of f ( x ) = R − [ x : Q ( x ) = 0]
e.g. Let A = {1, 2}, B = {3, 6} and f : A → B given by and range varies from function to function.
f ( x ) = x 2 + 2 and g : A → B given by g ( x ) = 3x. Then, Irrational Function
we observe that f and g have the same domain and The function containing one or more terms having
codomain. Also, we have non-integral rational powers of x is called irrational
f (1) = 3 = g (1) function.
and f (2) = 6 = g (2) 5x 3/ 2 − 7x 1/ 2
Hence, f =g e.g. y = f (x ) =
x 1/ 2 − 1
∴ f and g are equal functions. Domain varies from function to function.
25
2 Identity Function Function f ( x ) = x,
Y
y=x
Modulus Function
Function y = f ( x ) = | x | is known as modulus
Objective Mathematics Vol. 1
∀ x ∈ R is known as identity function.
function. It is a straight line x, x ≥ 0
passing through origin and X′ X
O y = f (x ) = | x | =
having slope unity.
−x, x < 0
Domain of f ( x ) = R
and range of f ( x ) = R Domain of f ( x ) = x ∈ R
Y′
and range of f ( x ) = [0, ∞ )
Square Root Function Y
The function that associates every positive real
y = x,
number x to + x is called the square root function, i.e. y = –x,
if x > 0
if x < 0
f (x ) = + x .
Y
y = √x X′ X
O
X′ X Y′
O
Y′ Properties of Modulus Function
Range of f ( x ) = [0, ∞ ) i. For any real number x, x 2 = | x |.
Exponential Function ii. If a and b are positive real numbers, then
A function of the form f ( x ) = a x , a is a positive (a) x 2 ≤ a 2 ⇔ | x | ≤ a ⇔ −a ≤ x ≤ a
real number, is an exponential function. The value of
(b) x 2 ≥ a 2 ⇔ | x | ≥ a ⇔ x ≤ − a or x ≥ a
the function depends upon the value of a. For 0 < a < 1,
function is decreasing and for a >1, function is (c) a 2 ≤ x 2 ≤ b 2 ⇔ a ≤ | x | ≤ b
increasing. ⇔ x ∈[ −b, − a ] ∪ [ a, b]
Domain of f ( x ) = R
and range of f ( x ) = [0, ∞ ) iii. | x ± y| ≤| x | +| y|
Y Y
iv. | x + y | ≥ || x | − | y ||
y = ax y = ax
X'
(0, 1)
X X′
(0, 1)
X
Greatest Integer Function
O O
For any real number x, the greatest integer
function [x] is equal to greatest integer less than or
Y′ Y′ equal to x.
0<a<1 a>1
(a) (b) In general, if n is an integer and x is any number
satisfying n ≤ x < n +1, then [ x ] = n. It is also known as
Logarithmic Function integral part function.
Function f ( x ) = log a x, ( x, a > 0) and a ≠1, is e.g. If 2 ≤ x < 3, then [ x ] = 2
known as logarithmic function. Domain = R
Domain of f ( x ) = (0, ∞ ) Range = I
and range of f ( x ) = R Y
Y Y 3
2
1
(1, 0)
X′ X X′ X X′ X
O O (1, 0) –3 –2 –1 0 1 2 3 4
–1
–2
Y′ Y′
0<a<1 a<1 –3
(a) (b)
26 Y′
Properties of Greatest Integer Function
If n is an integer and x is any real number between
n +1, then
X Example 23. The function
1 − x,
f is defined by
x <0
2
Fundamentals of Relation and Function
n and
f ( x ) = 1, x =0
i. [ − n] = − [ n] x + 1, x >0
ii. [ x + n] = [ x ] + n Then, the graph of f ( x ) is
Y Y
iii. [ −x ] = − [ x ] − 1, x is not an integer.
(0, 1)
(0, 1)
iv. [ x + y ] ≥ [ x ] + [ y] (a) X′ X (b) X′ X
O (1, 0) O
v. [ x ] > n ⇒ x ≥ n +1
Y'
Y'
vi. [x ] < n ⇒ x < n
Y Y
vii. [ x + y] = [ x ] + [ y + x − [ x ]] , ∀ x, y ∈ R
(0, 1)
(0, 1)
1 2 n − 1 (1, 0)
viii. [ x ] + x + + x + + ... + x + = [ nx ] , (c) X′ X (d) X′ X
n O (–1, 0) O
n n
n∈N
Y′ Y′
Signum Function 1 − x, x< 0
The function defined by Sol. (b) We have, f( x) = 1, x=0
x + 1, x > 0
−1, x < 0
|x|
f (x ) = = 0, x = 0 Let f( x) = y
x ∴ x + y = 1, x< 0
1, x > 0 y = 1, x=0
is called the signum function. and y − x = 1, x> 0
Y ∴ Graph of this function is
Y
y=1
1
x>0
(0, 1)
X′ X
X′ X
O
y = –1
–1
x<0
Y′
Y′
Domain = R Algebra of Real Functions
Range = {−1, 0, 1}
Let f : X → R and g : X → R be any two real
X Example 22. Which of the following is/are the functions, where X ⊂ R .
polynomial function? (i) Addition of two real functions
I. f ( x ) = x 3 − x 2 + 2, ∀ x ∈ R ( f + g ) (x ) = f (x ) + g (x )
II. f ( x ) = x 4 + 2x, ∀ x ∈ R (ii) Subtraction of two real functions
III. f ( x ) = x 2/ 3 + 2x, ∀ x ∈ R ( f − g ) (x ) = f (x ) − g (x )
(a) I, II (b) II, III (c) I, II, III (d) Only I (iii) Multiplication by a scalar
(α ⋅ f ) ( x ) = α ⋅ f ( x ), α, x ∈ R
Sol. (a) A function f : R → R is said to be polynomial
function if each of x in R, (iv) Multiplication of two real functions
f( x) = a0 + a1 x + a2 x2 + ... + an xn , where x is positive fg ( x ) = f ( x ) ⋅ g ( x )
integer and a0 , a1, a2 , ..., an ∈ R. (v) Quotient of two real functions
∴ I and II are polynomial functions whereas III is not a f f (x )
polynomial function because degree of x in III is not an (x ) = , g (x ) ≠ 0
integer. g g (x ) 27
2 X Example 24. Let f and g be two real
functions defined by f ( x ) =
1
x+4
, x ≠ − 4 and
Sol. (c) We have, f( x) = 2 x + 3 and g( x) = x2 + 7
Now,
⇒
g (f( x)) = (f( x))2 + 7 = 8
(2 x + 3)2 + 7 = 8
Objective Mathematics Vol. 1
g ( x ) = ( x + 4) 3 , then f / g is equal to ⇒ (2 x + 3)2 = 1 ⇒ 2 x + 3 = ± 1
1 1 1 ⇒ 2 x = − 4 and 2 x = − 2 ⇒ x = − 2, x = − 1
(a) ( x + 4) 2 (b) (c) (d)
( x + 4) 4
( x + 4) 3
( x + 4) 2 1 + x
X Example 27. If f (x ) = log and
Sol. (b) We have, f( x) = 1 and g( x) = ( x + 4)3 1 − x
x+ 4
1 3x + x 3
g (x ) = , then f ( g ( x )) is equal to
Now,
f
=
( x + 4)
=
1
, x≠−4 1 + 3x 2
g ( x + 4)3 ( x + 4)4
(a) f (3x ) (b) [ f ( x )]3 (c) 3 f ( x ) (d) − f ( x )
1
X Example 25. Let f (x ) = x, g (x ) = and Sol. (c) We have, f( x) = log 1 + x
x 1 − x
h( x ) = f ( x ) ⋅ g ( x ), then h( x ) =1 3 x + x3
(a) ∀ x ∈ R (b) ∀ x ∈Q and g ( x) =
1 + 3 x2
(c) ∀ x ∈ R − Q (d) ∀ x ∈ R , x ≠ 0
1 + g ( x)
Sol. (d) We have, f( x) = x , ∀ x ∈ R Now, f{(g ( x))} = log
1 − g ( x)
1
and g ( x) = , ∀ x ∈ R − {0}
x 3 x + x3
1 + 1 + 3 x2 + 3 x + x3
Domain of f( x) ⋅ g ( x) = R − {0} 1 + 3 x2
1 = log = log 3
∴ f( x) ⋅ g ( x) = x × = 1, ∀ x ∈ R − {0} 3x + x 3
1 + 3x − 3x − x
x 1 − 2
i.e. x ∈ R, x ≠ 0 1 + 3x
3
1 + x 1 + x
= log = 3 log
Composition of Functions 1−
⇒ f {g ( x)} = 3f( x)
x 1− x
Let A, B and C be three non-empty sets. Let
f : A → B and g : B → C . Since, f : A → B , for each Properties of Composition of Functions
x ∈ A, there exists a unique element g [ f ( x )] of C. Thus,
for each x ∈ A, there is associated a unique element i. The composition of functions is not
g [ f ( x )] of C. Thus, from f and g, we can define a new commutative.
function from A to C. This function is called the i.e. fog ≠ gof
product or composite of f and g, denoted by gof and
defined by ii. The composition of functions is associative.
( gof ) : A → C such that ( gof ) ( x ) = g [ f ( x )] , ∀ x ∈ A. i.e. fo( goh) = ( fog ) oh
X Example 26. If f : R → R and g : R → R are
defined by f ( x ) = 2x + 3 and g ( x ) = x 2 + 7, then
iii. The composition of any function with the
identity function is the function itself.
values of x such that g{ f ( x )} = 8 are
i.e. if f : A → B , then foI A = IB of = f
(a) 1, 2 (b) −1, 2 (c) −1, −2 (d) 1, −2
Work Book Exercise 2.2
1 Find the range of the function f ( x ) = x 2 + 2. a R, [−1, 1] b R − { 3}, {1, − 1}
c RT , R d None of these
a (−2, 2) b [2, ∞ )
c [3, ∞ ) d None of these
4 The domain of the function f ( x ) = (2 − 2 x − x 2 ) is
a x + a− x
2 If the function f ( x ) = , (a > 2 ), then a − 3≤ x≤ 3
2
b ( −1 − 3 ) ≤ x ≤ ( −1 + 3 )
f ( x + y ) + f ( x − y ) is equal to
c −2 ≤ x ≤ 2
a 2 f( x )f( y) b f( x )f( y)
f( x ) d ( −2 − 3 ) ≤ x ≤ ( −2 + 3 )
c d None of these
f( y)
5 If f ( x ) = sin 2 x and the composite function
| x − 3| g { f ( x )} = |sin x |, then the function g( x ) is equal
3 Domain and range of f ( x ) = are,
x−3 to
28 respectively a x−1 b x c x+1 d − x
WorkedOut Examples
Type 1. Only One Correct Option
Ex 1. The composite mapping fog of the maps (x − 2) + (2 + x ), 2≤x≤3
⇒ f (x ) = − (x − 2) + (x + 2), −2 ≤ x < 2
f : R → R , f ( x ) = sin x, g : R → R , g ( x ) = x 2 is − (x − 2) − (x + 2), −3 ≤ x < − 2
sin x
(a) sin x + x 2 (b) (sin x ) 2 (c) sin x 2 (d) x − 2 + 2 + x, 2 ≤ x ≤ 3 2x , 2≤x≤3
x2
= − x + 2 + x + 2, −2 ≤ x < 2 = 4 , −2 ≤ x < 2
Sol. We have, − x + 2 − x − 2, −3 ≤ x < − 2 −2x , −3 ≤ x < − 2
( fog ) (x ) = f (g (x )) = f (x 2 ) = sin x 2
Hence, (a) is the correct answer.
Hence, (c) is the correct answer.
Ex 5. Let A and B be two sets such that n( A × B ) = 6.
Ex 2. If R is a relation ‘<’ from A = {1, 2, 3, 4} to
If three elements of A × B are (3, 2), (7, 5),
B = {1, 3, 5}, i.e. ( a, b) ∈ R iff a < b, then RoR −1 (8, 5), then
is (a) A = {3, 7, 8} (b) B = {2, 5, 7}
(a) {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} (c) B = {3, 5} (d) None of these
(b) {(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)} Sol. Q (3, 2), (7, 5), (8, 5) ∈ A × B
(c) {(3, 3), (3, 5), (5, 3), (5, 5)} ∴ 3, 7, 8 ∈ A and 2, 5 ∈ B
(d) {(3, 3), (3, 4), (4, 5)} Also, n( A × B ) = 6 = 3 × 2
Sol. We have, ∴ A = {3, 7, 8} and B = {2, 5}
R = {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)} Hence, (a) is the correct answer.
∴ R −1 = {(3, 1), (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)} x −1
Now, RoR −1 = {(3, 3), (3, 5), (5, 3), (5, 5)} Ex 6. If f (x ) = , then f (2x ) is equal to
x +1
Hence, (c) is the correct answer. 3 f (x ) + 1 f (x ) + 1
(a) (b)
sin ( π[ x 2 + 1]) f (x ) + 3 f (x ) + 3
Ex 3. The range of the function f (x ) = f (x ) + 3
x 4 +1 (c) (d) None of these
where, [ ] is greatest integer function, is f (x ) + 1
(a) [0, 1] (b) [− 1, 1] x − 1
3 +1
(c) {0} (d) None of these 3 f (x ) + 1 x + 1
Sol. =
Sol. Since, [ x 2 + 1] is an integer. f (x ) + 3 x −1
+3
x+1
sin(π [ x 2 + 1])
∴ sin (π [ x 2 + 1]) = 0 ⇒ f (x ) = =0 3x − 3 + x + 1 4 x − 2
x4 + 1 = = = f (2x )
x − 1 + 3x + 3 4 x + 2
Now, range of f = R f = {0}
Hence, (a) is the correct answer.
Hence, (c) is the correct answer.
Ex 4. The simplified form of Ex 7. Let f : (4, 6) → (6, 8) be a function defined by
x
f ( x ) = | x − 2 | + | 2 − x | , −3 ≤ x ≤ 3 is f ( x ) = x + , where [ ] denotes the greatest
2
2x, 2 ≤ x ≤ 3 −2x, 2≤ x ≤ 3
integer function, then f −1 ( x ) is equal to
(a) 4, −2 ≤ x < 2 (b) 4, −2 ≤ x < 2
x 1
−2x, −3 ≤ x < − 2 2x, −3 ≤ x < − 2 (a) x − (b) − x − 2 (c) x − 2 (d)
2 π
x+
3x, 2 ≤ x ≤ 3 2
(c) 4, −2 ≤ x < 2 (d) None of these
Sol. Since, x ∈ (4 , 6)
−2x, −3 ≤ x < − 2 x
∴ f (x ) = x + =x+2
2
x − 2, x≥2
Sol. Q | x − 2 | = Let f (x ) = y
− (x − 2), x<2 ∴ y=x + 2 ⇒ x = y−2
(x + 2), x ≥ − 2 ∴ f −1 (x ) = x − 2
and |x + 2|= 29
− (x + 2), x < − 2 Hence, (c) is the correct answer.
2 Ex 8. Find the range of f (x ) = 1 + 3 cos 2x.
(a) [2, 3]
Ex 9. Find the domain
f (x ) =
1
x +| x|
.
of the function
Objective Mathematics Vol. 1
(b) [2, 4]
(c) [−2, 4] (a) R (b) R + (c) R − (d) R + R
(d) None of the above 1
Sol. Given, f (x ) = , for the function to be defined,
Sol. f (x ) = 1 + 3 cos 2x x + |x |
Q −1 ≤ cos 2x ≤ 1 the denominator should not be equal to 0.
⇒ −3 ≤ 3 cos 2x ≤ 3 x + |x |> 0 [Qx + | x | Û 0 ]
⇒ 1 − 3 ≤ 1 + 3 cos 2x ≤ 3 + 1 ⇒ |x |> −x
⇒ −2 ≤ f (x ) ≤ 4 It is true for all R + .
[R + means set of all positive real numbers]
∴ Range = [ −2, 4 ]
∴Domain of the given function = R +
Hence, (c) is the correct answer.
Hence, (b) is the correct answer.
Type 2. More than One Correct Option
Ex 10. Let f , g : R → R be defined, respectively by Ex 11. If n( A ) =10 and n(B ) = 5, then
f ( x ) = x +1 and g ( x ) = 2x − 3. Then, (a) n( A × B ) = 50
(a) ( f + g ) ( x ) = 3x − 2 (b) ( f − g ) ( 2) = 6 (b) n( A × B ) = n( B × A )
f (c) number of relations from A to B is 2 10
(c) ( 2) = 3 (d) ( f − g ) ( x ) = 4 − x
g (d) All of the above
Sol. We have, f (x ) = x + 1and g (x ) = 2x − 3 Sol. Given, n( A ) = 10, n( B ) = 5
( f + g ) (x ) = x + 1 + 2x − 3 = 3x − 2 n( A × B ) = n( A ) × n( B )
( f − g ) (x ) = x + 1 − 2x + 3 = 4 − x = 10 × 5 = 50
f x+1
(x ) = n( B × A ) = n( B ) × n( A )
g 2x − 3
= 5 × 10
f 3
(2) = =3 = 50
g 4−3
∴Total number of relations from A to B is 2n(A × B) = 250
( f − g ) (2) = 4 − 2 = 2
Hence, (a), (c) and (d) are the correct answers. Hence, (a) and (b) are the correct answers.
Type 3. Assertion and Reason
Directions (Ex. Nos. 12-13) In the following examples, ∴ R1 is not a relation.
each example contains Statement I (Assertion) and Whereas R2 ⊂ A × A ⇒ R2 is relation.
Statement II (Reason). Each example has 4 choices (a), Hence, (a) is the correct answer.
(b), (c) and (d) out of which only one is correct. The 1
choices are Ex 13. Statement I f (x ) =
, x ≠ 0, 1, then the
1− x
(a) Statement I is true, Statement II is true; Statement II
is a correct explanation for Statement I graph of the function y = f ( f ( f ( x ))), x >1 is a
(b) Statement I is true, Statement II is true; Statement II straight line.
is not a correct explanation for Statement I 1
Statement II If f ( x ) =
(c) Statement I is true, Statement II is false 1− x
(d) Statement I is false, Statement II is true
⇒ f [ f { f ( x )}] = x, ∀ x ∈ R
Ex 12. Let A = {1, 2, 3}, we define 1 1 x −1
R1 = {(1, 3), (1, 4), (2, 1), (2, 2), (2, 3)} and Sol. f { f (x ) = = = , ∀ x ∈ R − {0,1}
1 − f (x ) 1 − 1 x
R 2 = {(1, 1), (1, 2), (1, 3)}. Then, 1− x
Statement I R1 is not a relation and R 2 is 1 1
relation on A. ∴ f [ f { f (x )}] = =
1 − f { f (x )} 1 − x − 1
Statement II If R is relation on A, then x
R ⊆ A × A. = x , ∀ x ∈ R − {0, 1}
Sol. If R1 is a relation on A, then R1 ⊆ A × A ⇒ Statement I is true, Statement II is false.
30 Here, R1 contains (1, 4) ∉ A × A ⇒ R1 ⊆/ A × A Hence, (c) is the correct answer.
Type 4. Linked Comprehension Based Questions
Passage (Ex. Nos. 14-15) The figure shows a relation Sol. It is obvious that the relation R is ‘x is the square of y’.
2
Fundamentals of Relation and Function
R between the sets P and Q. In set builder form, R = {(x , y) : x is the square of
Q y, x ∈ P , y ∈ Q}
P 5 Hence, (c) is the correct answer.
9 3
2 Ex 15. The relation R in roster form is
4 1
–2 (a) {(9, 3), (4, 2), (25, 5)}
25 –3 (b) {(9, −3), (4, −2), (25, −5)}
–5
(c) {( 9, − 3), ( 9, 3), (4, − 2), (4, 2),
( 25, − 5), ( 25, 5)}
Ex 14. The relation R in set builder form is
(d) None of the above
(a) {( x, y ) : x ∈ P , y ∈ Q}
(b) {( x, y ) : x ∈ Q , y ∈ P} Sol. In roster form,
(c) {( x, y ) : x is the square of y, x ∈ P , y ∈ Q} R = {(9, 3), (9, − 3), (4 , 2), (4 , − 2), (25, 5), (25, − 5)}
(d) {( x, y ) : y is the square of x, x ∈ P , y ∈ Q} Hence, (c) is the correct answer.
Type 5. Match the Columns
Ex 16. Let A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}. Ex 17. Let A = {1, 2, 3, 4, ..., 14}. A relation R from A to
Then, match the following in Column I with A is defined by R = {( x, y) :3x − y = 0, where x,
the sets of ordered pairs in Column II. y ∈ A}. Then, match the terms of Column I with
Column I Column II
the terms of Column II and choose the correct
A. A × (B ∩ C ) p. {(1, 4), (2, 4), (3, 4)}
option from the codes given below:
B. ( A × B) ∩ ( A × C ) q. {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), Column I Column II
(2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}
A. In roster form, the relation p. {1, 2, 3, 4}
C. A × (B ∪ C ) r. {(3, 4)} R is
D. ( A × B) ∪ ( A × C ) B. The domain of R is q. {3, 6, 9, 12}
E. ( A ∩ B) × ( B ∩ C )
C. The range of R is r. {1, 2, 3, 4, ..., 14}
Sol. Given, A = {1, 2, 3}, B = {3, 4} and C = {4 , 5, 6} D. The codomain of R is s. {(1, 3), (2, 6), (3, 9), (4, 12)}
A. B ∩ C = {4}
∴ A × ( B ∩ C ) = {(1, 4 ), (2, 4 ), (3, 4 )} Sol. We have, 3x − y = 0 ⇒ y = 3x
For x = 1, y = 3 ∈ A
B. A × B = {(1, 3), (1, 4 ), (2, 3), (2, 4 ), (3, 3), (3, 4 )}
x = 2, y= 6 ∈A
A × C = {(1, 4 ), (1, 5), (1, 6), (2, 4 ), (2, 5), x = 3, y= 9 ∈A
(2, 6), (3, 4 ), (3, 5), (3, 6)} x = 4 , y = 12 ∈ A
∴ ( A × B ) ∩ ( A × C ) = {(1, 4 ), (2, 4 ), (3, 4 )} x = 5 , y = 15 ∉ A
C. B ∪ C = {3, 4 , 5, 6} A. In roster form, R = {(1, 3), (2, 6), (3, 9), (4 , 12)}
∴ A × ( B ∪ C ) = {(1, 3), (1, 4 ), (1, 5), (1, 6), B. Domain of R = Set of first elements of ordered
(2, 3), (2, 4), (2, 5), (2, 6), (3, 3), pairs in R
(3, 4), (3, 5), (3, 6)} = {1, 2, 3, 4}
D. ( A × B ) ∪ ( A × C ) = {(1, 3), (1, 4 ), (1, 5), (1, 6) C. Range of R = Set of second elements of ordered
(2, 3), (2, 4 ), (2, 5), (2, 6), (3, 3), (3, 4 ), (3, 5), (3, 6)} pairs in R
E. A ∩ B = {3}, B ∩ C = {4} = {3, 6, 9, 12}
∴ ( A ∩ B ) × ( B ∩ C ) = {3, 4} D. Codomain of R is the set A.
A → p; B → p; C → q; D → q; E → r A → s; B → p; C → q; D → r
Type 6. Single Integer Answer Type Questions
Ex 18. If n( A ) = 4, n(B ) = 3 and n( A × B × C ) = 24, Ex 19. Let A = {x, y, z} and B = {1, 2}. Then, the
then n(C ) is equal to________. number of relations from A to B is 2 k , then k is
equal to________.
Sol. (2) We have, n( A ) = 4, n( B ) = 3
and n( A × B × C ) = 24 Sol. (6) We have, n( A ) = 3, n( B ) = 2
n ( A × B) = 6
⇒ n( A ) × n( B ) × n(C ) = 24
4 × 3 × n(C ) = 24 Total number of relations from A to B = 2n(A × B) = 26
∴ k =6 31
n(C ) = 2
Target Exercises
Type 1. Only One Correct Option
1. If two sets A and B are having 99 elements in 10. R is a relation from {11, 12, 13} to {8, 10, 12}
common, then the number of elements common to defined by y = x − 3. The relation R −1 is
each of the sets A × B and B × A are (a) {(11, 8), (13, 10)}
(a) 299 (b) 992 (c) 100 (d) 18 (b) {(8, 11), (10, 13)}
(c) {(8, 11), (9, 12), (10, 13)}
2. Let n( A ) = m and n( B ) = n. Then, the total number of (d) None of the above
non-empty relations that can be defined from A to B
11. Let a relation R be defined by
is
R = {( 4, 5), (1, 4 ), ( 4, 6), ( 7, 6), ( 3, 7)}. The relation
(a) mn (b) nm − 1 (c) mn − 1 (d) 2mn − 1
R −1 oR is given by
3. Let X = {1, 2, 3, 4, 5} and Y = {1, 3, 5, 7, 9}. Which of (a) {(1, 1), (4, 4), (7, 4), (4, 7), (7, 7)}
(b) {(1, 1), (4, 4), (4, 7), (7, 4), (7, 7), (3, 3)}
the following is not a relation from X to Y ?
(c) {(1, 5), (1, 6), (3, 6)}
(a) R1 = {(x , y) : y = x + 2, x ∈ X , y ∈ Y} (d) None of the above
(b) R2 = {(1, 1), (2, 1), (3, 3), (4 , 3), (5, 5)}
(c) R3 = {(1, 1), (1, 3), (3, 5), (3, 7), (5, 7)} 12. If R is a relation on the set A = {1, 2, 3, 4, 5, 6, 7, 8, 9}
(d) R4 = {(1, 3), (2, 5), (2, 4 ), (7, 9)}
given by xRy ⇔ y = 3x, then RoR −1 is
Ta rg e t E x e rc is e s
4. A relation R from C to R is defined by xR y iff | x | = y. (a) {(1, 3), (2, 6), (3, 9)}
Which of the following is correct? (b) {(1, 1), (2, 2), (3, 3)}
(a) (2 + 3i ) R 13 (b) 3R (−3) (c) {(3, 3), (6, 6), (9, 9)}
(c) (1 + i ) R 2 (d) i R1 (d) None of the above
5. The relation R defined on the set A = {1, 2, 3, 4, 5} by 13. If R ⊆ A × B and S ⊆ B × C be two relations, then
R = {( x, y ) : | x 2 − y 2 | < 16} is given by ( SoR ) −1 is equal to
(a) {(1, 1) , (2, 1) , (3, 1) , (4, 1) , (2, 3)} (a) S −1 oR −1 (b) RoS
(b) {(2, 2) , (3, 2) , (4, 2) , (2, 4)} (c) R −1 oS −1 (d) None of these
(c) {(3, 3) , (4, 3) , (5, 4) , (3, 4)}
(d) None of the above
14. If R is a relation from a set A to the set B and S is a
6. A relation R is defined in the set Z of integers as relation from B to C, then the relation SoR
follows ( x, y ) ∈ R iff x 2 + y 2 = 9. Which of the (a) is from C to A (b) is from A to C
(c) does not exist (d) None of these
following is false?
(a) R = {(0, 3), (0, − 3), (3, 0), (−3, 0)} 15. Let A = [ − 1, 1], B = [ − 1, 1], C = [ 0, ∞ ). If
(b) Domain of R = {−3, 0, 3}
R1 = {( x, y ) ∈ A × B : x 2 + y 2 = 1}and
(c) Range of R = {− 3, 0, 3}
(d) None of the above R 2 = {( x, y ) ∈ A × C : x 2 + y 2 = 1}, then
(a) R1 defines a function from A into B
7. If R = {( x, y ) : x, y ∈ Z , x 2 + y 2 ≤ 4} is a relation in Z, (b) R2 defines a function from A into C
then domain of R is (c) R1 defines a function from A onto B
(a) {0, 1, 2} (b) {−2, − 1, 0} (d) R2 defines a function from A onto C
(c) {−2, − 1, 0, 1, 2} (d) None of these
x 2 , 0 ≤ x ≤ 3
16. The relation f is defined by f ( x ) =
8. Let R be a relation in N defined by 3x, 3 ≤ x ≤ 10
R = {( x, y ) : x + 2 y = 8}. The range of R is
and the relation g is defined by
(a) {2, 4, 6} (b) {1, 2, 3}
x , 0 ≤ x ≤ 2
2
(c) {1, 2, 3, 4, 6} (d) None of these g (x ) = . Which of the following
9. Let A = {1, 2, 3}, B = {1, 3, 5}. If relation R from A to B 3x, 2 ≤ x ≤ 10
is given by {(1, 3), (2, 5), (3, 3)}, then R −1 is relation is a function?
(a) {(3, 3), (3, 1) , (5, 3)} (a) f
(b) {(1, 3), (2, 5) , (3, 3)} (b) g
(c) {(1, 3), (5, 2) } (c) f,g
32 (d) None of the above (d) None of the above
17. The domain of F ( x ) =
(a) R − {− 1, − 2}
log 2 ( x + 3)
x 2 + 3x + 2
(b) (− 2, ∞ )
is 26. If f ( x ) = x 2 + 1 , then the value of ( fof ) ( x ) is equal
to
(a) x + 2x + 2
4 2
(b) x + 2x − 2
4 2
2
Fundamentals of Relation and Function
(c) R − {− 1, − 2, − 3} (d) (− 3, ∞ ) − {−1 , − 2} (c) x 4 + x 2 + 1 (d) None of these
18. The domain and range of the function f given by 27. If f ( x ) is defined on [0, 1] by the rule
f ( x ) = 2 − | x − 5 | is
x, if x ∈ Q
(a) Domain = R + , Range = (− ∞ , 1] f (x ) = , then for x ∈[ 0, 1]
(b) Domain = R, Range = (− ∞ , 2 ] 1 − x, if x ∉Q
(c) Domain = R, Range = (− ∞ , 2) (a) ( fof ) (x ) = 1 + x (b) ( fof ) (x ) = x
(d) Domain = R + , Range = (− ∞ , 2 ] (c) ( fof ) (x ) = 1 − x (d) None of these
x−2 x3 + 1 x < 0 ( x − 1)1/ 3 , x < 1
19. The range of the function f ( x ) = when x ≠ 2 is 28. If f ( x ) = 2 , g (x ) = ,
2− x x + 1, x ≥ 0 ( x − 1) , x ≥ 1
1/ 2
(a) R (b) R − {1} (c) {− 1} (d) R − {− 1} then ( gof )( x ) is equal to
(a) x, ∀ x ∈ R (b) x − 1, ∀ x ∈ R
20. The range of the function f ( x ) = | x | is (c) x + 1, ∀ x ∈ R (d) None of these
(a) (0, ∞ ) (b) (− ∞ , 0) 29. If f ( x ) = [ x] and g ( x ) = x − [ x] , then which of the
(c) [ 0, ∞ ) (d) None of these following is the zero function?
(a) ( f + g ) (x ) (b) ( fg ) (x )
21. If f and g are two functions defined as f ( x ) = x + 2 , (c) ( f − g ) (x ) (d) ( fog ) (x )
x ≤ 0 ; g ( x ) = 3, x ≥ 0, then the domain of f + g is x
(a) {0} (b) [ 0, ∞ ) (c) (− ∞ , ∞ ) (d) (− ∞ , 0) 30. If f ( x ) = , x ≠ 1, then ( fofo K of ) ( x ) is equal to
x−1 14243
19 times
22. If [ x] 2 − 5[ x] + 6 = 0, where [⋅] denotes the greatest
Targ e t E x e rc is e s
19
integer function, then x x 19x
(a) (b) (c) (d) x
(a) x ∈[ 3, 4 ] (b) x ∈ (2, 3 ] (c) x ∈[ 2, 3 ] (d) x ∈[ 2, 4 ) x −1 x − 1 x −1
1 1 1 1
23. If f ( x ) = 1 − , then f f is 31. Let f x + = x 2 + 2 , x ∈ R − { 0}, then f ( x ) is
x x x x
1 1 x 1 equal to
(a) (b) (c) (d) (a) x 2 (b) x 2 − 1
x 1+ x x −1 x −1
(c) x 2 − 2 , when | x | ≥ 2 (d) None of these
x−1
24. Let f ( x ) = , then f { f ( x )}is 32. Let f :[ 0, 1] → [ 0, 1] and g:[ 0, 1] → [ 0, 1] be two
x+1
1− x
1 1 1 1 functions defined by f (x ) = and
(a) (b) − (c) (d) 1+ x
x x x+1 x −1
g ( x ) = 4x (1 − x ), then ( fog ) ( x ) is equal to
25. Two functions f and g are said to commute, if 8x (1 − x ) 4 (1 − x )
(a) (b)
( fog ) ( x ) = ( gof ) ( x ), ∀ x, then which one of the (1 + x )2 1+ x
following functions are commute? 1 − 4 x + 4 x2
(c) (d) None of these
(a) f (x ) = x 3 , g (x ) = x + 1 1 + 4 x − 4 x2
(b) f (x ) = x , g (x ) = cos x
(c) f (x ) = x m , g (x ) = x n , m ≠ n, m, n ∈ I
33. Let f ( x ) = | x − 1 | , x ∈ R, then
(I is the set of all integers) (a) f (x 2 ) = ( f (x ))2 (b) f (x + y) = f (x ) + f ( y)
(d) f (x ) = x − 1, g (x ) = x 2 + 1 (c) f (| x |) = | f (x ) | (d) None of these
Type 2. More than One Correct Option
34. If A = {a, b, c, d}, B = {1, 2, 3}, which of the following 35. The relation R defined in A = {1, 2, 3} by aRb, if
sets of ordered pairs are not relation from A to B? | a 2 − b 2 | ≤ 5. Which of the following is true?
(a) {(a, 1), (a, 3)} (a) R = {(1, 1), (2, 2), (3, 3), (2, 1), (1, 2),(2, 3), (3, 2)}
(b) {(b, 1), (c, 2), (d , 1)} (b) R −1 = R
(c) {(a, 2), (b, 3), (3, b)} (c) Domain of R = {1, 2, 3}
(d) {(a, 1), (b, 2), (3, c)} (d) Range of R = {5}
33
2 Type 3. Assertion and Reason
Directions (Q. Nos. 36-37) In the following questions, 36. Statement I The domain of the real function f
Objective Mathematics Vol. 1
each question contains Statement I (Assertion) and defined by f ( x ) = x − 1 is R − {1}.
Statement II (Reason). Each question has 4 choices (a),
(b), (c) and (d), out of which only one is correct. The Statement II The range of the function defined by
choices are f ( x ) = x − 1 is [ 0, ∞ ).
(a) Statement I is true, Statement II is true; Statement II 37. Statement I Let f and g be two real functions
is a correct explanation for Statement I given by f = {( 0, 1), ( 2, 0), ( 3, − 4 ), ( 4, 2), ( 5, 1)} and
(b) Statement I is true, Statement II is true; Statement II g = {(1, 0), ( 2, 2), ( 3, − 1), ( 4, 4 ), ( 5, 3)}.
is not a correct explanation for Statement I
Then, domain of f ⋅ g is given by {2, 3, 4, 5}.
(c) Statement I is true, Statement II is false
Statement II Let f and g be two real functions.
(d) Statement I is false, Statement II is true
Then, ( f ⋅ g )( x ) = f {g ( x )}.
Type 4. Linked Comprehension Based Questions
Passage (Q. Nos. 38-39) The figure shows a relation R 38. The domain of the relations R is
between the sets P and Q. (a) {4, 9, 25}
Q (b) {2, 3, 5}
P (c) {−2, − 3, − 5}
5 (d) {−4 , − 9, − 25}
9 3
2 39. The range of the relation R is
4 1 (a) {2, 3, 4}
Ta rg e t E x e rc is e s
–2 (b) {−2, − 3, − 5}
25 –3 (c) {−5, − 3, − 2, 2, 3, 5}
–5 (d) {−5, − 3, − 2, 1, 2, 3, 5}
Type 5. Match the Column
40. Match the terms of Column I with the terms of Column II.
Column I Column II
A. If P = { x : x < 3, x ∈ N} and Q = { x : x ≤ 2, x ∈ N}, then p. {(0, 8), (8, 0), ( 0, − 8), ( −8, 0)}
( P ∪ Q ) × ( P ∩ Q ) is equal to
B. If A = { x : x ∈ W , x < 2}, B = { x : x ∈ N, 1 < x < 5} and q. {( −2, 3), ( −1, 5), (0, 7),
C = { 3, 5}, where W is the set of whole numbers, then (1, 9) , (2, 11)}
A × ( B ∩ C ) is equal to
C. If R = {( x, y) : x and y are integers and { x 2 + y 2 = 64} is r. {(0, 3) , (1, 3)}
a relation, then R in roster form is
D. If R = {( x, y) : y = 2 x + 7, x ∈ Z and −2 ≤ x ≤ 2} is a s. {(1, 1), (1, 2), (2, 1), (2, 2 )}
relation, then R is
Type 6. Single Integer Answer Type Questions
41. Let A and B be two sets having 3 elements in common. If n( A ) = 5 and n( B ) = 4, then n[( A × B ) ∩ ( B × A )] is
equal to _______.
42. Let f = {(1, 1), ( 2, 3), ( 0, − 1), ( −1, − 3)} be a linear function from Z to Z , such that f ( x ) = kx − 1, then k is
equal to _______.
34
Entrances Gallery
AIEEE
1 4. A real valued function f ( x ) satisfies the functional
1. The domain of the function f ( x ) = is
| x | − x [2011] equation f ( x − y ) = f ( x ) f ( y ) − f ( a − x ) f ( a + y ) ,
(a) (0, ∞ ) (b) (−∞ , 0) where a is a given constant and f ( 0) = 1, f ( 2a − x ) is
(c) (−∞ , ∞ ) − {0} (d) (−∞ , ∞ )
equal to [2005]
2. Let for a ≠ a1 ≠ 0, f ( x ) = ax 2 + bx + c, (a) f (− x ) (b) f (a) + f (a − x )
g ( x ) = a1 x 2 + b1 x + c1 and p( x ) = f ( x ) − g ( x ). If (c) f (x ) (d) − f (x )
p( x ) = 0 only for x = − 1 and p( − 2) = 2 , then the 5. If f : R → R satisfies f ( x + y ) = f ( x ) + f ( y ) ,
value of p( 2) is [2011] n
(a) 18 (b) 3 (c) 9 (d) 6 ∀ x, y ∈ R and f (1) = 7, then Σ
r =1
f ( r ) is [2003]
3. Let f : N → Y be a function defined as f ( x ) = 4x + 3 7n
(a)
where Y = { y ∈ N : y = 4x + 3 for some x ∈ N }. 2
If f is invertible, then its inverse is [2008] 7(n + 1)
(b)
y−3 3y + 4 2
(a) g ( y) = (b) g ( y) = (c) 7n(n + 1)
4 3
y+ 3 y+ 3 7n(n + 1)
(c) g ( y) = 4 + (d) g ( y) = (d)
2
4 4
Targ e t E x e rc is e s
Other Engineering Entrances
6. Let A = {1, 2, 3, 4} and R be the relation on A defined x+2
10. If f ( x ) = , then f { f ( x )}is
by {( a, b ) : a, b ∈ A, a × b is an even number}, then 3x − 1 [Kerala CEE 2014]
the range of R is [J&K CET 2014] (a) x (b) −x
1 1
(a) {1, 2, 4} (b) {2, 4} (c) (d) −
(c) {2, 3, 4} (d) {1, 2, 4} x x
(e) 0
( x 2 + 1)
7. The domain of the function f ( x ) = is 11. The range of the function f ( x ) = log e ( 3x 2 + 4 ) is
x 2 − 3x + 3
equal to [Kerala CEE 2012]
[J&K CET 2014]
(a) [loge 2, ∞ ) (b) [loge 3, ∞ )
(a) R − {1, 2}
(b) R − {1, 4} (c) [ 2 loge 3, ∞ ) (d) [ 0, ∞ )
(c) R (e) [ 2 loge 2, ∞ )
(d) R − {1}
12. Let R be the set of real numbers and the functions
8. Let A = {1, 2, 3, 4, 5}. The domain of the relation on A f : R → R and g : R → R be defined by
defined by R = {( x, y ) : y = 2x − 1} is [J&K CET 2014] f ( x ) = x 2 + 2x − 3 and g ( x ) = x + 1. Then, the value
(a) {1, 2, 3} of x for which g ( f ( x )) = f ( g ( x )) is [WB JEE 2012]
(b) {1, 2} (a) −1 (b) 0
(c) {1, 3, 5} (c) 1 (d) 2
(d) {2, 4}
13. Let A = {x, y, z}and B = {a, b, c, d}. Which one of the
9. If f ( x ) = x and g ( x ) = 2x − 3, then domain of
following is not a relation from A to B?
( fog ) ( x ) is [Kerala CEE 2014] [Kerala CEE 2011]
(a) (− ∞ , − 3) (a) {(x , a), (x , c)}
3 (b) {( y, c), ( y, d )}
(b) − ∞ , −
2 (c) {(z, a), (z, d )}
3 (d) {(z, b), ( y, b), (a, d )}
(c) − , 0 (e) {(x , c)}
2
3 14. If f ( x ) = 3 − x, − 4 ≤ x ≤ 4, then the domain of
(d) 0,
2 log e ( f ( x )) is [J&K CET 2011]
3 (a) [ −4 , 4 ] (b) (−∞ , 3 ]
(e) , ∞
2 (c) (−∞ , 3) (d) [ −4 , 3) 35
2 15. Let f = {( 0, − 1), ( −1, 3), ( 2, 3), ( 3, 5)} be a function
from Z to Z defined by f ( x ) = ax + b. Then,
[AMU 2011]
17. Let R be the set of real numbers and the mapping
f : R → R and g : R → R be defined by f ( x ) = 5 − x 2
and g ( x ) = 3x − 4, then the value of ( fog ) ( −1) is
Objective Mathematics Vol. 1
(a) a = 1, b = − 2 [WB JEE 2010]
(b) a = 2, b = 1 (a) − 44 (b) − 54 (c) − 32 (d) − 64
(c) a = 2, b = − 1
(d) a = 1, b = 2 18. Let A = {1, 0, 1, 2}, B = {4, 2, 0, 2} and f , g : A → B be
16. If f ( x ) = x − 1and g ( x ) = ( x + 1) , then ( gof )( x ) is
2 2
functions defined by f (x ) = x 2 − x and
[Kerala CEE 2011] 1
g ( x ) = 2 x − − 1. Then, [AMU 2010]
(a) (x + 1)4 − 1
2
(b) x 4 − 1
(a) f = g
(c) x 4 (b) f = 2g
(d) (x + 1)4 (c) g = 2 f
(e) (x − 1)4 − 1 (d) None of the above
Answers
Work Book Exercise 2.1
1. (c) 2. (b) 3. (c) 4. (c) 5. (a) 6. (a) 7. (c) 8. (c)
Work Book Exercise 2.2
Ta rg e t E x e rc is e s
1. (b) 2. (a) 3. (b) 4. (b) 5. (b)
Target Exercises
1. (b) 2. (d) 3. (d) 4. (d) 5. (d) 6. (d) 7. (c) 8. (b) 9. (d) 10. (b)
11. (b) 12. (b) 13. (c) 14. (b) 15. (d) 16. (a) 17. (d) 18. (b) 19. (c) 20. (c)
21. (a) 22. (d) 23. (c) 24. (b) 25. (c) 26. (a) 27. (b) 28. (a) 29. (d) 30. (a)
31. (c) 32. (c) 33. (d) 34. (c,d) 35. (a,b,c) 36. (d) 37. (c) 38. (a) 39. (c) 40. (*)
41. (9) 42. (2)
* A → s; B → r; C → p; D → q
Entrances Gallery
1. (b) 2. (a) 3. (a) 4. (d) 5. (d) 6. (b) 7. (c) 8. (a) 9. (e) 10. (a)
11. (e) 12. (a) 13. (d) 14. (d) 15. (c) 16. (c) 17. (a) 18. (a)
36
Explanations
Target Exercises
1. n[( A × B) ∩ (B × A)] = n[( A ∩ B) × (B ∩ A)] and R −1 = {(3, 1), (6, 2 ), (9, 3)}
= n( A ∩ B) × n(B ∩ A) ∴ ROR −1 = {(1, 1), (2, 2 ), (3, 3)}
= 99 × 99 = 992 13. SoR is a relation from A to C.
2. Given, n( A) = m and n(B) = n ∴ (SoR )−1 is a relation from C to A.
Now, total number of relations from A to B = 2 mn R −1 is a relation from B to A.
∴Total number of non-empty relations from A to B S −1 is a relation from C to B.
= 2 mn − 1 ∴R −1oS −1 is a relation from C to A.
Let (c , a) ∈ (SoR )−1
3. R1 is a relation from X to Y because R1 ⊆ X × Y. ∴ (a, c ) ∈ SoR
R2 is a relation from X to Y because R2 ⊆ X × Y. ⇒ ∃ b ∈ B : (a, b) ∈ R and (b, c ) ∈ S
R3 is a relation from X to Y because R3 ⊆ X × Y. ⇒ (b, a) ∈ R −1 and (c , b) ∈ S −1
R4 is not a relation from X to Y because (2, 4), (7, 9) ⇒ (c , a) ∈ R −1oS −1
∉ X × Y. ∴ (SoR )−1 ⊆ R −1oS −1
4. | i | = 0 2 + (1)2 = 1 = 1 Conversely, let (c , a) ∈ R −1oS −1
∴ iR1 is correct. ⇒ ∃ b ∈ B : (c , b) ∈ S −1 and (b, a) ∈ R −1
5. We have, R = {( x, y ):| x 2 − y 2| < 16} ⇒ (b, c ) ∈ S and (a, b) ∈ R
⇒ (a, c ) ∈ SoR ⇒ (c , a) ∈ (SoR )−1
∴ R = {(1, 1), (1, 2 ), (1, 3), (1, 4), (2, 1), (2, 2 ), (2, 3),(2, 4),
(3, 1), (3, 2 ), (3, 3), (3, 4), (4, 1), (4, 2 ), ∴ R −1oS −1 ⊆ (SoR )−1
Combining, we get (SoR )−1 = R −1oS −1
Targ e t E x e rc is e s
(4, 3), (4, 4), (4, 5), (5, 4), (5, 5)}
6. x 2 + y 2 = 9 ⇒ y = 9 − x 2 14. Since, R ⊆ A × B and S ⊆ B × C, we have SoR ⊆ A × C
⇒ y = ± 9− x 2 ∴ SoR is a relation from A to C.
∴ R = {(0, 3), (0, − 3), (3, 0 ), (−3, 0 )} 15. R2 = {( x, y ) ∈ A × C : x 2 + y 2 = 1}
Domain of R = { x : ( x, y ) ∈ R} = { 0, 3, − 3} x2 + y2 = 1 ⇒ y = ± 1 − x2
Range of R = { y : ( x, y ) ∈ R} = { 3, − 3, 0} ⇒ y = 1 − x2 [Q y ≥ 0 ]
7. We have, R = {( x, y ) : x, y ∈ Z, x + y ≤ 4} 2 2
∴R2 is a function from A onto C.
R = {(0, 0 ), (0, − 1), (0, 1), (0, − 2 ), (0, 2 ),(−1, 0 ), (1, 0 ), x 2, 0 ≤ x ≤ 3
16. Given, f ( x ) =
(1, 1), (1, − 1), (−1, 1), (−1, − 1), (2, 0 ), (−2, 0 )} 3 x, 3 ≤ x ≤ 10
∴ Domain of R = { x : ( x, y ) ∈ R} = { 0, − 1, 1, − 2, 2} Q f ( x ) = x 2 is well-defined in 0 ≤ x ≤ 3
8. R = {( x, y ) : x + 2 y = 8, x, y ∈ N} and f ( x ) = 3 x is also well-defined in 3 ≤ x ≤ 10
At x = 3, f ( x ) = x 2
8− x
x + 2y = 8 ⇒ y= ⇒ f(3) = 32 = 9
2
R = {(2, 3), (4, 2 ), (6, 1)} At x = 3, f ( x ) = 3 x
⇒ f(3) = 3 × 3 = 9
∴ Range of R = { y : ( x, y ) ∈ R} = {1, 2, 3}
⇒ f ( x ) is defined at x = 3. Hence, f is a function.
9. We have, R = {(1, 3), (2, 5), (3, 3)} x 2, 0 ≤ x ≤ 2
g( x ) =
∴ R −1 = {( y, x ) : ( x, y ) ∈ R} = {(3, 1), (5, 2 ), (3, 3)}
3 x, 2 ≤ x ≤ 10
10. We have, R = {( x, y ) : y = x − 3, x = 11or 12 or 13, Q g( x ) = x is well-defined in 0 ≤ x ≤ 2
2
y = 8 or 10 or 12} and g( x ) = 3 x is also well-defined in 2 ≤ x ≤ 10
= {(11, 8), (13, 10)}
∴ R −1 = {( y, x ) : ( x, y ) ∈ R} 4
2
= {(8, 11), (10, 13)} 6
11. We have, R = {(4, 5), (1, 4), (4, 6), (7, 6), (3, 7 )}
R −1 = {(5, 4), (4, 1), (6, 4), (6, 7 ), (7, 3)} But at x = 2, g( x ) = x 2
∴ R −1oR = {(4, 4), (1, 1), (4, 7 ), (7, 4), (7, 7 ), (3 ,3)} ⇒ g(2 ) = 2 2 = 4
At x = 2, g( x ) = 3 x
12. We have, set A = {1, 2, 3, 4, 5, 6, 7, 8, 9}
⇒ g(2 ) = 3 × 2 = 6
xRy ⇒ y = 3 x Therefore, g( x ) is not defined at x = 2.
∴ R = {(1, 3), (2, 6), (3, 9)} Hence, g is not a function.
37
2 17. We have, F( x ) =
log 2 ( x + 3)
x 2 + 3x + 2
∴ F( x ) is defined, if x + 3 > 0
28. Let x < 0
∴ (gof )( x ) = g{ f ( x )} = g( x 3 + 1)
= [( x 3 + 1) − 1]1/ 3 [Q x < 0 ⇒ x 3 + 1 < 1]
Objective Mathematics Vol. 1
and x 2 + 3x + 2 ≠ 0
= ( x 3 )1/ 3 = x
⇒ F( x ) is defined, if x > − 3
Let x ≥ 0
and x ≠ − 1, − 2
∴ (gof )( x ) = g(f ( x )) = g( x 2 + 1) = [( x 2 + 1) − 1]1/ 2
⇒ Domain of F( x ) = (−3, ∞ ) − { − 1, − 2}
[Q x ≥ 0 ⇒ x 2 + 1 ≥ 1]
18. Given, f ( x ) = 2 −| x − 5|
= ( x 2 )1/ 2 = | x | = x [Q x ≥ 0]
Domain of f ( x ) is defined for all real values of x.
∴ (gof ) ( x ) = x, ∀ x ∈ R
Q | x − 5| ≥ 0 ⇒ − | x − 5| ≤ 0
⇒ 2 − | x − 5| ≤ 2 ⇒ f ( x ) ≤ 2 29. (f + g )( x ) = f ( x ) + g( x ) = [ x ] + ( x − [ x ]) = x ≠ 0
Hence, range of f ( x ) is (−∞, 2 ]. (fg )( x ) = f ( x )⋅ g( x ) = [ x ]( x − [ x ]) ≠ 0
x −2 x −2 (f − g )( x ) = f ( x ) − g( x ) = [ x ] − ( x − [ x ])
19. f ( x ) = =− = − 1, if x ≠ 2
2−x x −2 = 2 [x] − x ≠ 0
∴ Range of f = { −1} (fog )( x ) = f (g( x )) = f ( x − [ x ]) = [ x − [ x ]] = 0
20. f ( x ) =| x | ⇒ R(f ) = [0, ∞) [Q x − [ x ] ∈ [0, 1)]
x
21. D(f + g ) = D(f ) ∩ D(g ) x x −1
(−∞, 0 ] ∩ [0, ∞ ) = { 0} 30. (fof )x = f = =x
x − 1 x
−1
22. Given, [ x ] − 5[ x ] + 6 = 0
2
x − 1
⇒ [ x ]2 − 3[ x ] − 2[ x ] + 6 = 0 x
⇒ (fofof )x = f (fof )( x ) = f ( x ) =
⇒ [ x ]([ x ] − 3) − 2([ x ] − 3) = 0 x −1
⇒ ([ x ] − 3) ([ x ] − 2 ) = 0 x
∴ (fofof…19 times )( x ) =
⇒ [ x ] = 3 or [ x ] = 2 x −1
Ta rg e t E x e rc is e s
⇒ x ∈[3, 4) or x ∈[2, 3) 2
31. f x + = x 2 + 2 = x + − 2
∴ x ∈[2, 4) 1 1 1
x x x
1 1
23. f f = f 1 −
1
= f (1 − x ) = 1 − 1
x (1 / x ) 1− x Let z=x+
x
1− x − 1 x
= = ∴ f ( z) = z2 − 2
1− x x −1
Replacing z by x, we get
x −1 f ( x ) = x 2 − 2, when| x | ≥ 2
−1
x − 1 x+1
24. f { f ( x )}d = f = 1 − g( x ) 1 − 4 ( x )(1 − x )
x + 1 x −1 32. f { g( x )} = =
+1 1 + g( x ) 1 + 4 x(1 − x )
x+1
x − 1 − x − 1 −2 1 1 − 4x + 4x 2
= = =− =
x − 1+ x + 1 2x x 1 + 4x − 4x 2
25. (fog ) ( x ) = (gof ) ( x ) only when 33. We have, f ( x ) =| x − 1|
f( x) = x m (a)| x 2 − 1| = (| x − 1|)2 , which is not true.
and g( x ) = x n (b)| x + y − 1| = | x − 1| + | y − 1|, which is not true.
As f { g( x )} = ( x ) = x
n m nm (c)|| x | − 1| = |(| x − 1|)|, which is not true.
and g{ f ( x )} = ( x m )n = x nm 34. {(a, 1), (a, 3)} ⊆ A × B
∴ f { g( x )} = g{ f ( x )} ∴This is a relation.
{(b, 1), (c , 2 ), (d , 1)} ⊆ A × B
26. (fof )( x ) = f { f ( x )} = f ( x + 1) = ( x + 1) + 1
2 2 2
∴This is a relation.
= x4 + 2 x2 + 2
(3, b) ∉ A × B
27. Let x ∈[0, 1] ∴ (a, 2 ), (b, 3), (3, b) is not a relation from A to B.
Case I x ∈ Q (3, c ) ∈
/ A×B
∴ f( x) = x ∴ {(a, 1), (b, 2 ), (3, c )} is not a relation from A to B.
and (fof ) ( x ) = f { f ( x )} = f ( x ) = x [Q x ∈ Q ] 35. Q R = {(1, 1), (1, 2 ), (2, 1), (2, 2 ), (2, 3), (3, 2 ), (3, 3)}
Case II x ∉ Q
R −1 = {( y, x ) : ( x, y ) ∈ R}
∴ f( x) = 1 − x
and (fof )( x ) = f { f ( x )} = f (1 − x ) = {(1, 1), (2, 1), (1, 2 ), (2, 2 ), (3, 2 ), (2, 3), (3, 3)}= R
= 1 − (1 − x ) = x [Q 1 − x ∉ Q ] Domain of R = { x : ( x, y ) ∈ R} = {1, 2, 3}
38 Now, in both cases, (fof )( x ) = x and range of R = { y : ( x, y ) ∈ R} = {1, 2, 3}
36. We have, f ( x ) = x − 1
f ( x ) is defined, if x − 1 ≥ 0
i.e. x ≥1
39. Range of R = Set of second elements of ordered pairs in R
= { −5, 5, − 3, 3, − 2, 2}
40. A. P = {1, 2}, Q = {1, 2}
2
Fundamentals of Relation and Function
∴ Domain of f = [1, ∞ ) P ∪ Q = {1, 2}, P ∩ Q = {1, 2}
Hence, Statement I is false. ∴ (P ∪ Q ) × (P ∩ Q ) = {(1, 1), (1, 2 ), (2, 1), (2, 2 )}
Let f( x) = y B. A = { 0, 1}, B = {2, 3, 4}, C = { 3, 5}
Then, y = x −1 B ∩ C = { 3}
⇒ y2 = x − 1 A × (B ∩ C ) = {(0, 3), (1, 3)}
⇒ x = y2 + 1 C. R = {(0, 8), (0, − 8), (8, 0 ), (−8, 0 )}
∴ Range of f = [0, ∞ ) [Q x ≥ 1] D. R = {(−2, 3), (−1, 5), (0, 7 ), (1, 9), (2, 11)}
37. Domain of f = { 0, 2, 3, 4, 5} 41. Given, n( A) = 5, n(B) = 4, n( A ∩ B) = 3
Domain of g = {1, 2, 3, 4, 5} n[( A × B) ∩ (B × A)] = n[( A ∩ B) × (B ∩ A)]
Domain of f ⋅ g = Domain of f ∩ Domain of = n( A ∩ B) × n(B ∩ A) = 3 × 3 = 9
g = {2, 3, 4, 5}
42. Since, f ( x ) is a linear function, f ( x ) = mx + c
Hence, Statement I is true.
Also, f (1) = m + c = 1
If f and g are two real functions.
and f (0 ) = c = − 1
Then, (f ⋅ g )( x ) = f ( x )⋅ g( x )
This gives m=2
Hence, Statement II is false. and f( x) = 2 x − 1
38. Domain of R = Set of first elements of ordered pairs in R ∴ k =2
= { 4, 9, 25}
Entrances Gallery
1 ⇒ 1 = 1 − { f (a)} 2 [f(0 ) = 1, given]
Targ e t E x e rc is e s
1. y =
| x| − x ⇒ f (a) = 0
For domain,| x | − x > 0 ∴ f (2 a − x ) = f { a − ( x − a)}
⇒ | x| > x = f (a)f ( x − a) − f (a − a)f ( x )
This is possible, only if x < 0. = 0 − f ( x )⋅ 1 = − f ( x )
∴ x ∈ (−∞, 0 ) n
2. Given, p( x ) = f ( x ) − g( x ) has only one root −1.
5. ∑ f (r ) = f (1) + f (2 ) + f (3) + K + f (n )
r =1
∴ p( x ) = (a − a1 ) x 2 + (b − b1 )x + (c − c1 ) ...(i) = f (1) + 2 f (1) + 3f (1) + K + nf (1)
− (b − b1 ) [since, f ( x + y ) = f ( x ) + f ( y )]
⇒ = −2 …(ii)
(a − a1 ) = (1 + 2 + 3 + K + n ) f (1) = f (1) ∑ n
(c − c1 ) 7 n(n + 1)
and =1 ...(iii) = [Q f(1) = 7, given]
(a − a1 ) 2
Also, p(−2 ) = 2 6. Since, A = {1, 2, 3, 4} and R be a relation on A defined by
⇒ 4 (a − a1 ) − 2 (b − b1 ) + (c − c1 ) = 2 {(a, b); a, b ∈ A, a × b is an even number}.
⇒ 4(a − a1 ) − 4(a − a1 ) + (a − a1 ) = 2 Thus, R be the set of even numbers.
[using Eqs. (ii) and (iii)] ∴ Range of R = {2, 4}
⇒ (a − a1 ) = 2 ( x 2 + 1)
Substituting (a − a1 ) = 2 in Eq. (i) and Eq. (ii), 7. Q f ( x ) =
( x − 3 x + 3)
2
(b − b1 ) = 4, (c − c1 ) = 2
⇒ ( x 2 + 1) is defined for all x ∈ R.
∴ p(2 ) = 4 (a − a1 ) + 2 (b − b1 ) + (c − c1 )
[from Eq. (i)] Also, x 2 − 3 x = − 3 has imaginary roots.
= 8 + 8 + 2 = 18 ∴x 2 − 3 x + 3 is also defined for all real numbers,
3. Since, Y = { y ∈ Y : y = 4 x + 3 for some x ∈ N} i.e. x ∈ R.
∴ Y = {7, 11, K ∞} Hence, domain of the function is R.
Now, y = 4x + 3 8. Since, A = {1, 2, 3, 4, 5}
y−3
⇒ x= ∴Relation from A to A
4 R = {( x, y ) : y = 2 x − 1} = {(1, 1), (2, 3), (3, 5)}
y−3
Inverse of f ( x ) is g( y ) = . ⇒ Domain = {1, 2, 3}
4
9. Given, f ( x ) = x
4. Given, f ( x − y ) = f ( x ) f ( y ) − f (a − x ) f (a + y ) and g( x ) = 2 x − 3
Let x=0=y ∴ (fog ) ( x ) = f { g( x )}
⇒ f (0 ) = { f (0 )} 2 − { f (a)} 2 39
= f (2 x − 3) = 2 x − 3
2 For domain of (fog )( x ),
⇒
2x − 3 ≥ 0 ⇒ 2x ≥ 3
x≥
3
13. {( z, b), ( y, b), (a, d )} is not a relation from A to B because
a ∉ A.
14. Given, f ( x ) = 3 − x, here −4 ≤ x ≤ 4
Objective Mathematics Vol. 1
2
3 To find domain of loge { f ( x )} = loge (3 − x )
∴ x ∈ , ∞
2 For loge { f ( x )} to be defined,
x+2 3− x > 0 ⇒ 3> x
10. Given, f ( x ) = ⇒ x<3
3x − 1
⇒ Domain = [−4, 3)
x+2
+2 15. Here, f (0 ) = 0 + b ⇒ −1 = b
x + 2 3x − 1
∴ f { f ( x )} = f =
3 x − 1 x + 2 and f (−1) = − a + b
3 −1 ⇒ −3 = − a + b
3 x − 1
⇒ −a = − 3 + 1
x + 2 + 6x − 2
= ⇒ a=2
3 x + 6 − (3 x − 1)
16. (gof )( x ) = g{ f ( x )} = g( x 2 − 1) = ( x 2 − 1 + 1)2 = x 4
7x
= =x
7 17. (fog )(−1) = f { g(−1)} = f (−7 ) = 5 − 49 = − 44
11. Given, f ( x ) = loge (3 x + 4)
2
18. We have, A = {1, 0, 1, 2}
Let y = loge (3 x + 4)
2
B = { 4, 2, 0, 2}
⇒ 3x 2 + 4 = e y and f , g : A → B
Now, f( x) = x2 − x
ey − 4
⇒ x= ⇒ f(0 ) = 0 − 0 = 0
3
and f(1) = 1 − 1 = 0
ey − 4 f(2 ) = 2 2 − 2 = 2
Here, ≥0 and
3
Ta rg e t E x e rc is e s
1
⇒ ey ≥ 4 Also, g( x ) = 2 x − − 1
2
⇒ y ≥ 2 loge 2 −1
⇒ g(0 ) = 2 − 1= 1− 1= 0
12. Given, f( x) = x2 + 2 x − 3 2
1
and g( x ) = x + 1 and g(1) = 21 − − 1 = 1 − 1 = 0
2
Q f { g( x )} = g{ f ( x )}
1
⇒ f ( x + 1) = g( x 2 + 2 x − 3) and g(2 ) = 2 2 − − 1
2
⇒ ( x + 1) + 2( x + 1) − 3 = x + 2 x − 3 + 1
2 2
2⋅3
= − 1= 2
⇒ x2 + 2 x + 1 + 2 x − 1 = x2 + 2 x − 2 2
⇒ 2x = −2 Thus, f ( x ) = g( x ), ∀x
∴ x = −1 Hence, f =g
40
3
Sequence and
Series
Introduction
Sequence Sequence is a function whose domain is the set N of natural numbers. Chapter Snapshot
Real Sequence A sequence whose range is a subset of R, is called a real sequence.
● Introduction
Progression It is not necessary that the terms of a sequence always follow a certain
● Arithmetic Progression (AP)
pattern or they are described by some explicit formula for the nth term.
Those sequences whose terms follow certain patterns, are called ● Geometric Progression (GP)
progressions. ● Harmonic Progression (HP)
Series By adding or subtracting the terms of a sequence, we get an expression ● Arithmetico-Geometric
which is called a series. If a1 , a 2 , a 3 , a 4 , ..., a n , ... is a sequence, then Progression (AGP)
the expression a1 + a 2 + a 3 + K + a n + K is the corresponding series. ● Some Special Series
A series is finite or infinite according as corresponding sequence is
finite or infinite, i.e. the number of terms in the corresponding
sequence is finite or infinite.
n
X Example 1. The first 3 terms of the sequence which is defined by a n = ,
n +1
2
are
5 10 1 2 3 2 1
(a) 2, , (b) , , (c) , , 3 (d) None of these
2 3 2 5 10 5 2
n
Sol. (b) On putting n = 1, 2, 3 in an = , we get
n2 + 1
1 1 2 2
a1 = =
, a2 = 2 =
12 + 1
2 2 +1 5
3 3
and a3 = 2 =
3 + 1 10
1 2 3
Hence, the first 3 terms are , and .
2 5 10
Arithmetic Progression (AP)
A sequence is called an arithmetic progression, if the difference of a term and the
previous term is always same, i.e.
a n + 1 − a n = Constant ( d ), ∀ n ∈ N
The constant difference, generally denoted by d is called the common difference.
3 X Example 2. Show that the sequence < a n >
defined by a n = 4n + 5 is an AP. Also, find its
X Example 5. If the sequence < a n > is defined by
a n = 3n + 2, then difference of any two consecutive
terms is
Objective Mathematics Vol. 1
common difference.
Sol. We have, an = 4n + 5 (a) 2 (b) 3 (c) 4 (d) 5
On replacing n by (n + 1,) we get Sol. (b) Since, nth term is of the form An + B , therefore it is
an + 1 = 4(n + 1) + 5 = 4n + 9 an AP and the common difference is coefficient of n i.e. 3.
Now, an + 1 − an = (4 n + 9) − (4 n + 5)
Clearly, an + 1 − an = 4, ∀ n ∈ N iii. If a constant is added to or subtracted from
So, the given sequence is an AP with common each term of an AP, then the resulting
difference 4. sequence is also an AP with the same common
difference.
Properties of an Arithmetic Progression
X Example 6. If a, b and c are in AP, then which
i. If a is the first term and d is the common of the following will also be in AP?
difference of an AP, then its nth term or general
term a n is given by (a) 3 + a, 3 + b, 3 + c
a n = a + ( n − 1) d (b) a − 1, b − 2, c − 3
(c) a + 1, b + 2, c + 3
X Example 3. If the sequence 9, 12, 15, 18, ... is (d) a − k , b − 2k , c − 3k
an AP, then its general term is Sol. (a) a + 3, b + 3 and c + 3 are in AP because each
(a) 3n + 1 (b) 3n + 2 (c) 3n + 4 (d) 3n + 6 term of the given AP is being added by a constant number
Sol. (d) We have, d = (12 − 9) = 3 3.
As the given sequence is an AP with common
difference 3 and first term 9.
iv. If each term of a given AP is multiplied or
∴General term = nth term
divided by a non-zero constant k, then the
= a + (n − 1)d = 9 + (n − 1) 3 = 3n + 6 resulting sequence is also an AP with common
difference kd or d / k respectively, where d is
X Example 4. Let Tr be the rth term of an AP for the common difference of the given AP.
r =1, 2, 3,K , if for some positive integers m, n, we
1 1 X Example 7. If a, b and c are in AP, then
have Tm = and Tn = , then Tmn is equal to
n m prove that
1 1 1 1 1 1 1 1 1
(a) (b) + (c)1 (d) 0 a + , b + , c + are in AP.
mn m n b c c a a b
Sol. (c) Let a be the first term and d be the common Sol. Since, a, b and c are in AP.
difference. On dividing each term by abc, the resulting sequence
1 a b c
∴ Tm = a + (m − 1) d = ...(i) , and are in AP.
n abc abc abc
1 1 1 1
and Tn = a + (n − 1) d = ...(ii) ⇒ , and are in AP.
m bc ac ab
On subtracting Eq. (ii) from Eq. (i), we get ab + bc + ca ab + bc + ca ab + bc + ca
⇒ , , are in AP
1 1 m− n bc ac ab
(m − n)d = − =
n m mn [on multiplying each term by ab + bc + ca]
1 ab + bc + ca ab + bc + ca ab + bc + ca
⇒ d= ⇒ − 1, − 1, −1
mn bc ac ab
Again, Tmn = a + (mn − 1)d are in AP.
= a + (mn − n + n − 1)d [on adding − 1 to each term]
ab + ac ab + bc bc + ca
= a + (n − 1)d + (mn − n)d ⇒ , , are in AP.
1 bc ca ab
= Tn + n(m − 1)
⇒ a + , b + , c + are in AP.
1 1 1 1 1 1
mn
1 (m − 1) b c c a a b
= + =1
m m Hence proved.
v. In a finite AP, the sum of the terms equidistant
ii. A sequence is an AP iff its nth term is of the from the beginning and end is always same and
form An + B , i.e. a linear expression in n. The is equal to the sum of first and last term, i.e.
common difference in such a case is A, i.e. the a k + a n ( k −1) = a1 + a n for all
42 coefficient of n. k = 1, 2, 3, K, ( n − 1).
X Example 8. If an AP is given by 3, a 2 , ..., 20,
then which of the following is the sum of second
X Example 12. Four numbers are in arithmetic
progression. The sum of first and last term is 8 and
the product of both middle terms is 15. The least
3
Sequence and Series
and second last terms?
(a) 20 (b) 3 (c) 23 (d) 22 number of the sequence is
(a) 4 (b) 3 (c) 2 (d) 1
Sol. (c) As we know, ak + an − ( k − 1) = a1 + an
On putting k = 2, we get
Sol. (d) Let A1, A2 , A3 and A4 be four numbers in AP.
a2 + an − 1 = a1 + an = 3 + 20 Given, A1 + A4 = 8 …(i)
= 23 and A2 A3 = 15 ...(ii)
We know, A2 + A3 = A1 + A4
X Example 9. If a1 , a 2 , a 3 , K , a 2n + 1 are in AP, ∴ A2 + A3 = 8 ...(iii)
then ( a 2n + 1 + a1 ) + ( a 2n + a 2 ) + K + ( a n + 2 + a n ) On solving Eqs. (ii) and (iii), we get
is equal to A2 = 3, A3 = 5 or A2 = 5, A3 = 3
(a) 2na n (b) 2na n − 1 Q 2 A2 = A1 + A3 [by property]
⇒ 2 × 3 = A1 + 5 or 2 × 5 = A1 + 3
(c) 2na n + 1 (d) None of these
⇒ A1 = 1 or A1 = 7
Sol. (c) Clearly, a1 + a2 n + 1 = a2 + a2 n = a3 + a2 n −1 Similarly, A4 = 7 or A4 = 1
= K = an + 2 + an Hence, least number of sequence is 1.
and ar = a1 + (r − 1)d
∴ (a2 n + 1 + a1 ) + (a2 n + a2 ) + K + (an + 2 + an ) Selection of Terms in an AP
= n(a2 n + 1 + a1 ) = n(2 a1 + 2 nd ) Sometimes, it is required to select a finite number
= 2 n(a1 + nd ) = 2 nan + 1 of terms in AP. It is always convenient, if we select the
vi. Three numbers a, b and c are in AP iff terms in the following manner:
2b = a + c Selecting Odd Number of Terms of AP
X Example 10. If a, b and c are in AP, then i. Selecting 3 terms of AP
1 1 1 a − d, a, a + d
a + ,b+ and c + are in
bc ca ab
(a) HP (b) AP
ii. Selecting 5 terms of AP
(c) GP (d) None of these a − 2d, a − d, a, a + d, a + 2d
Sol. (b) a + 1 , b + 1 and c + 1 are in AP, if and only if X Example 13. If the sum of three numbers in
bc ca ab
AP is − 3 and their product is 8, then the numbers
b + 1 − a + 1 = c + 1 − b + 1
are
ca bc ab ca
( b − a) (c − b ) (a) 2, 1, 4 (b) 2, −1, − 4
⇒ ( b − a) + = (c − b ) +
abc abc (c) 4, 2, 1 (d) None of these
⇒ (b − a){abc + 1} = (c − b ){abc + 1} Sol. (b) Let the numbers be a − d , a and a + d .
⇒ b− a=c − b Given, sum = − 3
⇒ 2b = a + c ⇒ (a − d ) + a + (a + d ) = − 3
which is true, as given a, b and c are in AP. ⇒ 3a = − 3 ⇒ a = − 1
vii. If the terms of an AP are chosen at regular and product = 8
intervals, then they form an AP. ⇒ ( a − d ) a( a + d ) = 8
⇒ a(a2 − d 2 ) = 8
X Example 11. If T1 , T2 , T3 , T4 , T5 , T6 , T7 ,… are ⇒ (− 1)(1 − d 2 ) = 8
in AP with common difference 2. Then, the ⇒ d2 = 9
difference of any two consecutive terms of the ⇒ d=± 3
sequence T1 , T4 , T7 , ... is If d = 3, then the numbers are − 4, − 1, 2.
(a) 3 (b) 4 (c) 6 (d) 8 If d = − 3, then the numbers are 2, − 1, − 4.
Sol. (c) Clearly, T1, T4 , T7,... are in AP. X Example 14. In a triangle, the lengths of the
Now, T4 − T1 = T1 + 3 × 2 − T1 = 6 two larger sides are 10 and 9, respectively. If the
angles are in AP, then the length of the third side
viii. If a n , a n +1 and a n + 2 are any three consecutive
can be
terms of an AP, then
(a) 91 (b) 3 3
2a n +1 = a n + a n + 2 43
(c) 5 (d) None of these
3 Sol. (d) Since, angles of a triangle are in AP.
∴
⇒
α + (α − δ) + (α + δ) = 180°
α = 60°
Sum of n Terms of an AP
The sum S n of n terms of an AP with first term a
Objective Mathematics Vol. 1
A and common difference d is given by
(α + δ ) n
9 x S n = [2a + ( n − 1) d ]
2
(α – δ ) α n
B
10
C Also, S n = [ a + l]
2
Using cosine law in ∆ACB,
where, l = last term = a + ( n − 1) d
(10)2 + ( x)2 − 92
cos 60° =
2 ⋅ (10) ⋅ ( x) X Example 17. Which of the following is the sum
1 19 + x2 of 20 terms of an AP 1, 4, 7, 10,...?
⇒ =
2 20 x (a) 690 (b) 590
⇒ x2 − 10 x + 19 = 0
(c) 591 (d ) 691
⇒ x = 5± 6
Sol. (b) Let a be the first term and d be the common
Selecting Even Number of Terms of AP difference of the given AP. Then, a = 1and d = 3.
We have to find the sum of 20 terms of the given AP.
i. Selecting 2 terms of AP n
On putting a = 1, d = 3 and n = 20 in S n = [2 a + (n − 1)d ],
a − d, a + d 2
we get
ii. 20
Selecting 4 terms of AP S 20 = [2 × 1 + (20 − 1) × 3]
2
a − 3d, a − d, a + d, a + 3d
= 10 × 59 = 590
Ø ● It should be noted that in case of odd number of terms, the
middle term is a and the common difference is d while in X Example 18. If S r denotes the sum of the first
case of an even number of terms the middle terms are S 3r − S r − 1
a − d , a + d and the common difference is 2d. r terms of an AP, then is equal to
● If sum of the numbers is not given, then the numbers can be S 2r − S 2r − 1
taken as a, a + d , a + 2 d , a + 3 d ,... .
(a) 2r − 1 (b) 2r + 1
X Example 15. If the sum of four numbers of AP (c) 4r + 1 (d) 2r + 3
is 20, then which of the following is first term of S3r − Sr −1
the given AP? Sol. (b)
S2r − S2r −1
(a) 5 (b) 4 (c) 6 (d) 10
3r (r − 1)
[2 a + (3r − 1) d ] − [2 a + (r − 2 )d ]
Sol. (a) Let the four numbers in AP be a − 3d , a − d , a + d , = 2 2
a + 3d . 2r (2 r − 1)
[2 a + (2 r − 1)d ] − [2 a + (2 r − 2 )d ]
Now, a − 3d + a − d + a + d + a + 3d = 20 2 2
⇒ 4 a = 20 ⇒ a = 5 2 a (2 r + 1) + d (8r 2 − 2 )
=
2 a + d(4 r − 2 )
X Example 16. Divide 32 into four parts which (2 r + 1) [2 a + 2(2 r − 1) d ]
= = (2 r + 1)
are in AP such that the ratio of product of extremes [2 a + 2(2 r − 1)d ]
to the product of means is 7 :15.
(a) 2, 6, 10, 14 (b) 3, 9, 15, 18 Ø A sequence is an AP if and only if the sum of its n terms is of the
(c) 4, 12, 20, 28 (d) 5, 15, 25, 35 form An2 + Bn, where A, B are constants. In such a case, the
common difference of the AP is 2A.
Sol. (a) Let the four parts be a − 3d , a − d , a + d and a + 3d .
Q Sum = 32 X Example 19. If the sum of n terms of an AP is
⇒ (a − 3d ) + (a − d ) + (a + d ) + (a + 3d ) = 32 3n 2 + 5n, then the common difference is
⇒ 4 a = 32 ⇒ a = 8
(a − 3d ) (a + 3d ) 7 (a) 2 (b) 3
and =
(a − d ) (a + d ) 15 (c) 5 (d) 6
a2 − 9d 2 7 Sol. (d) Since, S n = 3 n2 + 5 n is of the form An2 + Bn.
⇒ =
a −d
2 2
15 Therefore, the common difference is 2 A = 6.
64 − 9d 2 7
⇒ =
64 − d 2 15 Arithmetic Mean of Two Given Numbers
⇒ 128d = 512
2 Let a and b be two given numbers, then the
⇒ d =4 ⇒ d=±2
2 arithmetic mean of a and b is a number A such that
44 a, A, b are in AP.
Thus, the four parts are 2, 6, 10, 14.
⇒
Now, a, A, b are in AP.
2A = a + b ⇒ A =
a+b
2
X Example 22. The sum of the products of the
numbers ± 1, ± 2, ± 3, ± 4, ± 5 taking two at a time is 3
Sequence and Series
(a)165
Thus, the arithmetic mean of two numbers a and b
a+b (b) − 55
is given by A = . (c) 55
2
(d) None of the above
Arithmetic Mean of n Given Numbers Sol. (b) (1 − 1 + 2 − 2 + ... + 5 − 5)2
Let a, a 2 , ..., a n be n given numbers, then the = 12 + 12 + 2 2 + 2 2 + ... + 52 + 52 + 2S
arithmetic mean of a1 , a 2 , a 3 , ..., a n is given by where, S is the required number.
a + a 2 + a 3 + ... + a n or 0 = 2(12 + 2 2 + 32 + 42 + 52 ) + 2S
A= 1
n − 5(5 + 1) (10 + 1)
∴ 2S = = − 55
6
Insertion of n Arithmetic Means
X Example 23. Observe that 13 = 1, 2 3 = 3 + 5,
between a and b
3 3 = 7 + 9 + 11, 4 3 = 13 + 15 + 17 + 19. Then, n 3 as a
If between two given quantities a and b, we have
to insert n quantities A1 , A2 , ..., An such that a, similar series is
A1 , A2 , ... An , b form an AP, then we say that n( n − 1) ( n + 1) n
A1 , A2 , ..., An are n arithmetic means between a and b. (a) 2 + 1 − 1 + 2 + 1 + 1
2 2
Let d be the common difference of this AP.
( n + 1) n
Clearly, it contains ( n + 2) terms. + ... + 2 + 1 + 2n − 3
∴ b = ( n + 2) th term 2
b−a (b) ( n 2 + n + 1) + ( n 2 + n + 3) + ( n 2 + n + 5)
⇒ b = a + ( n + 1) d ⇒ d =
n +1 + ... + ( n 2 + 3n − 1)
Now, A1 = a + d, A2 = a + 2d, ... and An = a + nd (c) ( n 2 − n + 1) + ( n 2 − n + 3) + ( n 2 − n + 5)
a + b + ... + ( n 2 + n − 1)
Ø Sum of n AM's between a and b is n i.e.
2 (d) None of the above
A1 + A 2 + .... + A n = nA.
Sol. (c) 13 = 1⋅ (1 − 1) + 1, 2 3 = (2 ⋅ 1 + 1) + 5
X Example 20. Three arithmetic means between 33 = (3 ⋅ 2 + 1) + 9 + 11
3 and 19 are 43 = (4 ⋅ 3 + 1) + 15 + 17 + 19, etc.
(a) 7, 11, 15 (b) 7, 10, 13 ∴ n3 = {n ⋅ (n − 1) + 1} + ...
(c) 7, 12, 17 (d) None of these next term being 2 more than the previous
Sol. (a) Let A1, A2 , A3 be three AM’s between 3 and 19. Then, = (n2 − n + 1) + (n2 − n + 3) + ... + (n2 + n − 1)
3, A1, A2 , A3 19 are in AP whose common difference is
19 − 3 X Example 24. Along a road lies an odd number
d= =4
3+1 of stones placed at intervals of 10 m. These stones
∴ A1 = 3 + d = 3 + 4 = 7 have to be assembled around the middle stone. A
A2 = 3 + 2d = 3 + 8 = 11 person can carry only one stone at a time. A man
A3 = 3 + 3d = 3 + 12 = 15 carried out the job starting with the stone in the
Hence, the required AM’s are 7,11,15. middle, carrying stones in succession, thereby
X Example 21. If (2n + r )r, n ∈ N , r ∈ N is covering a distance of 4.8 km. Then, the number of
expressed the sum of k consecutive odd natural stones is
numbers, then k is equal to (a)15 (b) 29
(a) r (b) n (c) r +1 (d) n +1 (c) 31 (d) 35
Sol. (a) (2 n + r )r = (n + r )2 − n2 Sol. (c) Let there be 2 n + 1 stone i.e. n stones on each side
of the middle stone. The man will run 20 m, to pick up the
= {1 + 3 + 5 + ... to (n + r ) terms} first stone and return, 40 m for the second stone and so
− {1 + 3 + 5 + ... to n terms} on. So, he runs (n/2 ) {2 × 20 + (n − 1)20} = 10n(n + 1) m to
as (n + r )2 − n2 = {1 + 3 + 5 + ... (n + r ) terms) pick up the stones on one side and hence 20n(n + 1) m to
− {1 + 3 + 5 + ... n terms) pick up all the stones.
∴ Sum of r consecutive odd natural numbers, ∴ 20n(n + 1) = 4800 or n = 15
k=r Hence, there are 2 n + 1 = 31 stones. 45
3 Work Book Exercise 3.1
1 If the nth term of an AP 5, 8, 11, ... is 320, then n 9 The sum of the first ten terms of an AP is four times
Objective Mathematics Vol. 1
is equal to the sum of the first five terms, then the ratio of the
105 104 106 112 first term to the common difference is
1/2 2 1/4 4
2 If the 5th term of an AP is 11 and the 9th term is
7, then the 14th term is 10 If p, q and r are in AP, then p + r − 8 q is equal
3 3 3
−1 2 1 0 to
3 If p − 1, p + 3 and 3 p − 1 are in AP, then p is equal to − 6 pqr 4 pqr
2 pqr None of these
a 4 b −4 c 2 d −2
4 The 15th term of sequence x −7, x − 2 and x + 3 is 11 Consider an AP with first term a and common
difference d. Let S k denotes the sum of first k
x + 63 x − 63 63 − x 63 + x S
terms. If kx is independent of x, then
5 If the sum of three numbers of an AP is 24, then Sx
middle term is a = 2d a=d
6 8 3 2 2a = d None of these
6 The sum of the integers from 1 to 100 that are 12 If a1, a2 , a3 ,... are in AP such that
divisible by 2 or 5, is a1 + a5 + a10 + a15 + a20 + a24 = 225,
3000 3050
then a1 + a2 + a3 + .... + a23 + a24 is equal to
3600 None of these
909 75 750 900
7 If N which is the set of natural numbers, is
13 If the numbers a, b, c , d , e are in AP, then the
partitioned into subsets S1 = {1},S 2 = {2, 3},
S 3 = { 4, 5, 6}, ... , then the sum of the members in value of a − 4 b + 6 c − 4 d + e is
S 50 is 1 2 0 16c
62255 62525 14 Sum of n terms of the series
65225 62555
1 1 1
3 + 5 + 7 + ... + n terms + + + ... is
8 If = 7, then the value of n 2+ 5 5+ 8 8 + 11
5 + 8 + 11 + ...+ 10 terms
1
is 3n + 2 − 2 ( 3n + 2 − 2 )
3
35 36 37 40 3n + 2 + 2 None of these
Geometric Progression (GP)
A sequence of non-zero numbers is called a X Example 26. Which of the following is a GP?
geometric progression, if the ratio of a term and the (a) 2, 4, 8, 16, ...
term preceding to it is always a constant quantity. 1 −1 1 −1
The constant ratio is called the common ratio of the (b) , , , ,...
9 27 81 243
GP. (c) 0.01, 0.0001, 0.000001,...
In other words, a sequence a1 , a 2 , a 3 , ..., a n , ... is (d) All of the above
called a geometric progression, if
Sol. (d)
an+ 1
= Constant, ∀ n ∈ N (a) We have, a1 = 2 ,
a2 a a
= 2 , 3 = 2, 4 = 2 and so on
an a1 a2 a3
1 a2 −1
(b) We observe, a1 = , = ,
X Example 25. Show that the sequence given by 9 a1 3
a n = 3 (2 n ), ∀ n ∈ N is a GP. Also, find its common a3 −1 a4 −1
= , = and so on.
a2 3 a3 3
ratio. a
(c) We have, a1 = 0.01, 2 = 0.01,
Sol. We have, an = 3 (2 n ) and an + 1 = 3 (2 n + 1 ) a1
a3 a
an + 1 3 (2 n + 1 ) = 0.01, 4 = 0.01 and so on.
Now, = =2 a2 a3
an 3 (2 n ) It is observed that in each case, every term except the
an + 1 first term bears a constant ratio to the term immediately
Clearly, = 2 (constant), ∀ n ∈ N. So, the given
an preceding it.
46 sequence is a GP with common ratio 2. Thus, options (a), (b) and (c) all are in GP.
⇒ 6 + 6r + 6r 2 = 19 r
General Term of a GP
If a is the first term of a GP and r is the common
ratio, then its nth term (general term) t n is given by
⇒
⇒
6r 2 − 13r + 6 = 0
(3r − 2 ) (2 r − 3) = 0
3
Sequence and Series
⇒ r = 3/2 and 2/3
t n = ar n − 1 Hence, on putting the values of a and r, the required
numbers are 8, 12, 18 or 18, 12, 8.
X Example 27. The nth and 10th terms of a GP
5, 25, 125, ..., are respectively X Example 30. If four numbers in GP are such
(a) 5 n − 1 , 5 9 (b) 5 n − 2 , 5 8 that the third number is greater than the first by 9
and the second number is greater than the fourth
(c) 5 n , 510 (d) None of these
by 18, then the numbers are
Sol. (c) Here, a = 5 and r = 5. Thus, the nth term of the (a) 3, 6, 12, 24 (b) − 3, 6, − 12, 24
given GP is given by an = ar n − 1 (c) 3, − 6, 12, − 24 (d) None of these
⇒ an = 5 (5)n − 1 = 5n
Sol. (c) Let the four numbers be a, ar, ar 2 and ar 3 .
Now, for n = 10, we have a10 = 510
It is given that, ar 2 − a = 9
X Example 28. The third term of a geometric and ar − ar 3 = 18
progression is 4. The product of the first five terms ⇒ a (r 2 − 1) = 9 ...(i)
is and ar (1 − r 2 ) = 18 ...(ii)
(a) 4 3 (b) 4 5 On dividing Eq. (ii) by Eq. (i), we get
(c) 4 4
(d) None of these ar (1 − r 2 ) 18
= ⇒r = − 2
a (r 2 − 1) 9
Sol. (b) Here, t 3 = 4 ⇒ ar 2 = 4
On putting r = − 2 in a (r 2 − 1) = 9, we get a = 3
∴Product of first five terms = a. ar. ar 2 . ar 3 . ar 4
Hence, the numbers are 3, − 6, 12, − 24.
= a5 r10 = (ar 2 )5 = 45
Properties of Geometric Progressions
Selection of Terms in GP
Sometimes, it is required to select a finite number i. If all the terms of a GP is multiplied or divided
of terms in GP. It is always convenient, if we select the by the same non-zero constant, then it remains
terms in the following manner: a GP with the same common ratio.
Number of terms Terms Common ratio X Example 31. If 5, 25, 125,... are in GP, then
a 1, 5, 25,... are in
3 , a, ar r
r (a) AP (b) GP
a a
4 , , ar, ar 3 r2 (c) HP (d) None of these
r3 r
a a
, , a, ar, ar 2 Sol. (b)Q5, 25, 125,... are in GP.
5 r
r2 r On dividing each term by 5, we get the following
sequence 1, 5, 25 ,..., which is again a GP as here
If the product of the numbers is not given, then the a a
a1 = 1, 2 = 5, 3 = 5 and so on.
numbers can be taken as a, ar, ar 2 , ar 3 , ... . a1 a2
X Example 29. If the sum of three numbers in
GP is 38 and their product is 1728, then numbers
ii. The reciprocals of the terms of a given GP
form a GP.
are
(a) 8, 12, 18 (b) 8, 16, 32 1 1 1 1
X Example 32. If , 2 , 3 , 4 ,... are in GP,
(c) 18, 12, 8 (d) None of these 2 2 2 2
Sol. (a,c) Let the three numbers be a , a and ar. then 2, 2 2 , 2 3 ,... are in
r (a) AP (b) GP
Then, product = 1728 (c) HP (d) None of these
a. .
⇒ a ar = 1728
r Sol. (b) On reciprocating each term of a GP, we get the
⇒ a3 = 1728 following sequence 2, 2 2 , 2 3 , 2 4 ,... which is again a GP as
⇒ a = 12 a a
here a1 = 2, 2 = 2, 3 = 2 and so on.
Q Sum = 38 a1 a2
+ a + ar = 38 ⇒ a + 1 + r = 38
a 1
∴
r r iii. If each term of a GP is raised to the same
1 + r + r2 power, then the resulting sequence also form
⇒ 12 = 38
r a GP. 47
Example 37. If 3, 3 2 , 3 3 are in GP, then
3 X Example 33. If a, b and c are in GP, then
a 1/ 2 , b1/ 2 , c1/ 2 are in
X
log 3, log 9 and log 27 are in
(a) AP (b) GP
Objective Mathematics Vol. 1
(a) AP (b) GP
(c) HP (d) None of these
(c) HP (d) None of these
Sol. (a) Since, 3, 9 and 27 are in GP.
Sol. (b) Since, a, b and c are in GP.
1/ 2 1/ 2 ∴ 92 = 3 × 27
b b1/ 2 c1/ 2
=
b c c
= ⇒ ⇒ = ⇒ log 92 = log(3 × 27 )
a b a b a1/ 2 b1/ 2
⇒ 2 log 9 = log 3 + log 27
Hence, a1/ 2 , b1/ 2 and c1/ 2 are in GP.
Hence, log 3, log 9 and log 27 are in AP.
iv. In a finite GP, the product of the terms 5c 3b
X Example 38. If log , log and
equidistant from the beginning and the end is a 5c
always same and is equal to the product of the a
first and the last terms. log are in AP, where a, b and c are in GP,
3b
X Example 34. If 2, a, 2 3 , b and 2 5 are in GP, then a, b, c are the lengths of sides of
then ab is (a) an isosceles triangle (b) an equilateral triangle
(a) 32 (b) 64 (c) 128 (d) 55 (c) a scalene triangle (d) None of these
Sol. (b) Since, 2, a, 2 3 , b and 2 5 are in GP.. Sol. (d) Since, log 5c , log 3b and log a are in AP.
a 5c 3b
∴ Product of a and b = Product of first and last terms
= 2 ⋅ 2 5 = 2 6 = 64 ⇒
5c 3b
, and
a
are in GP.
a 5c 3b
v. Three non-zero numbers a, b and c are in GP 3b
2
5c a
⇒ = .
iff b 2 = ac. 5c a 3b
n n n ⇒ 3b = 5c
10
X Example 35. If ∑ n, ⋅ ∑ n 2 and ∑ n 3 Also, b 2 = ac
n =1 3 n =1 n =1 ∴ 9ac = 25c 2
are in GP, then the value of n is ⇒ 9a = 25c
9a
(a) 2 (b) 3 ⇒ = 5c = 3b
5
(c) 4 (d) None of these a b c
n
⇒ = =
10 n 2 n
5 3 9/ 5
∑ n, ⋅ ∑ n and ∑ n are in GP.
3
Sol. (c) Since,
3 n =1 ⇒ b+ c< a
n =1 n =1
Since, sum of two sides is smaller than the third side.
n (n + 1) 10 n (n + 1) (2 n + 1) n (n + 1) 2 2
⇒ , . , are in GP. So, triangle is not formed.
2 3 6 4
10 n (n + 1) (2 n + 1)
. 2 2 2
n (n + 1) n (n + 1)2
2
X Example 39. A circle of radius r is inscribed
⇒ = ⋅
9 .36 2 4 in a square. The mid-points of sides of the square
⇒ 20 (2 n + 1)2 = 81n (n + 1) have been connected by line segment and a new
⇒ n2 + n − 20 = 0 ⇒ n = 4 square resulted. The sides of the resulting square
were also connected by segment so that a new
vi. If the terms of a given GP are chosen at
square was obtained and so on, then the radius of
regular intervals, then the new sequence so
the circle inscribed in the nth square is
formed also form a GP.
1−n
X Example 36. If 1, 5, 25, 125, 625,... forms a (a) 2 2 r
GP, then the sequence consisting odd term of the
given GP, is 3 − 3 n
(a) AP (b) GP (c) HP (d) None of these (b) 2 2 r
Sol. (b) Clearly, the sequence of odd terms of the given GP
is 1, 25, 625 ,..., which is a GP as every term except the −n
first term bears a constant ratio to the term immediately (c) 2 2 r
preceding it.
vii. If a1 , a 2 , a 3 , ... , a n , ... are in GP of non-zero − 5 − 3n
non-negative terms, then log a1 , log a 2 , ..., (d) 2 2 r
48
log a n K are in AP and vice-versa.
n n
Sol. (a) Side of square S1 = 2 r 2r − 1
Side of square S 2 = r 2
Sol. (c) S n =
r
∑
=1 2
r
=
r =1
= n − + 2 + ... + n
1 1 1
1
∑ 1 − 2 r
3
Sequence and Series
S2 2 2 2
S1 1 1
1 − n
S4 2 = n − 1+ 1
=n− 2
S3 1 2n
1−
2
r√2 r
ii. If l is the last term of a GP, then l = ar n − 1 .
a − lr lr − a
r ⇒ Sn = or S n = , r ≠1
2r 1− r r −1
2 −1 2 −1
2r 1
= 2 r
1
= X Example 42. In a GP, if {a n }, a1 = 3, a n = 96
2 2 2
3 −1 2 and S n =189, then the value of r is
Side of square S 3 = 2 r
1
= 2 r
1
and so on. (a) 5 (b) 2 (c) 3 (d) 8
2 2
n −1
Sol. (b) Here, a1 = 3, an = l = 96 and S n = 189
Side of square S n = 2 r
1
a − lr 3 − 96r
2 Q Sn = ⇒ 189 =
∴ Radius = r(2 − 1/ 2 n − 1
) = r(2( 1− n )/ 2
) and so on. 1− r 1− r
Side of square S n = r(2 − 1/ 2 )n − 1 = r[2( 1− n )/ 2 ] ⇒ 189 − 189r = 3 − 96r ⇒ 186 = 93r
⇒ r =2
Sum of n Terms of a GP
iii. If | r| <1, then lim r n = 0. Therefore, the sum S
n→ ∞
i. The sum of n terms of a GP with first term aand
common ratio r is given by of an infinite GP with common ratio r
satisfying | r | <1 is given by
a (1 − r n ) 1 − r n
1 − r , | r| < 1 S = lim S n = lim a ⇒ S=
a
n→ ∞ n→ ∞ 1− r 1− r
Sn = a (r n − 1) Thus, the sum S of an infinite GP with first
r − 1 , | r| > 1
term a and common ratio r ( − 1 < r < 1) is given
an, r =1 by S=
a
1− r
X Example 40. The sum of 7 terms of a GP
3, 6, 12, ... is X Example 43. The sum to infinity of the GP
(a) 200 −5 5 −5
, , , ... , is
(b) 320 4 16 64
(c) 381 (a) 1 (b) 0
(d) None of the above (c) −1 (d) None of these
Sol. (c) Here, a = 3 and r = 2 Sol. (c) The given GP has first term (a) = − 5 and the
4
r 7 − 1 2 7 − 1 −1
∴ S 7 = a = 3 common ratio (r ) = . Also,| r| < 1
r −1 2 −1 4
= 3 (128 − 1) Hence, the sum to infinity is given by
= 381 −5
a 4
S= = = −1
X Example 41. If 1 − r 1 − − 1
n 4
1 + 2 + 2 2 + ... upto r terms
Sn = ∑ , then S n is n
2r 1
r =1 X Example 44. If S k = lim
n→ ∞
∑ (k + 1) i , then
equal to i=0
n
1
(a) 2 n − ( n + 1) (b)1 −
2n
∑ k Sk is equal to
k =1
1
(c) n − 1 + (d) 2 n − 1 n( n +1) n( n −1) n( n + 2) n( n + 3)
2 n (a) (b) (c) (d) 49
2 2 2 2
k+1
3
1 1 1 Sol. (b) Let G1, G2 , G3 , G4 and G5 be five geometric means
Sol. (d) S k = 1 + + + ... = =
k + 1 (k + 1)2
1−
1 k between a = 576 and b = 9.
k+1 Then, 576, G1, G2 , G3 , G4 , G5 , 9 are in GP with common
Objective Mathematics Vol. 1
∴ k Sk = k + 1 ratio r given by
n n 1 1 1
Now, ∑ k Sk = ∑ (k + 1) = 2 + 3 + ... + (n + 1) r =
9 5+1 1 6 1
576
= =
64
Q r = b n + 1
a
k =1 k =1 2
n n(n + 3)
= [2 + (n + 1)] = 1 1
2 2 ∴ G1 = ar = 576 × = 288 , G2 = ar = 576 × = 144
2
2 4
1 1
Geometric Mean of Two Given Numbers G3 = ar 3 = 576 × = 72, G4 = ar 4 = 576 × = 36
8 16
Let a and b be two given numbers. Then, the 1
and G5 = ar 5 = 576 × = 18
geometric mean of a and b is a number G, such that a, 32
G, b are in GP. Hence, 288, 144, 72, 36, 18 are the required geometric
means between 576 and 9.
Now, a, G, b in GP.
G b X Example 46. If a is the AM of b and c and the
⇒ = ⇒ G 2 = ab ⇒ G = ab two geometric means are G1 and G2 , then G13 + G23
a G
This, the geometric mean of a and b is given by is equal to
(a) abc (b) 2abc (c) 3abc (d) 5abc
G = ab.
Sol. (b) It is given that a is the AM of b and c.
b+c
∴ a= ⇒ b + c = 2a ...(i)
Geometric Mean of n Given Numbers 2
Let a1 , a 2 , a 3 , ...., a n be n given numbers, then the Since, G1 and G2 are two geometric means between b
and c.
geometric mean of a1 , a 2 , ..., a n is given by
Therefore, b, G1, G2 , c are in GP with common ratio
G = ( a1 a 2K a n )1/ n 1
r = .
c 3
b
Insertion of n Geometric Means between 1/ 3
G1 = br = b
c
Now, = c1/ 3 b 2 / 3
a and b b
2/ 3
G2 = br 2 = b
c
Let a and b be two given numbers. If n numbers and = b1/ 3 c 2 / 3
b
G1 , G2 , ..., Gn are inserted between a and b such that
the sequence a, G1 , G2 , ..., Gn , b is a GP. Then, the ⇒ G13 = b 2c and G23 = bc 2
numbers G1 , G2 , ..., Gn are known as n geometric ⇒ G13 + G23 = b 2c + bc 2 = bc(b + c )
means (GM’s) between a and b. ⇒ G13 + G23 = 2 abc [from Eq. (i)]
Now, sequence a, G1 , G2 , ..., Gn , b is a GP X Example 47. If the arithmetic mean and
consisting of ( n + 2) terms. Let r be the common ratio of
geometric mean of a and b are A and G
this GP. respectively, then the value of A − G will be
Then, b = ( n + 2)th term = ar n + 1 2
a−b a+b a − b 2ab
1( n + 1) (c)
b b (a) (b) (d)
⇒ rn +1 = ⇒ = 2 2 2 a+b
a a
b
1/ ( n + 1 )
Sol. (c) Arithmetic mean of a and b = A = a + b
∴ G1 = ar = a 2
a Geometric mean of a and b = G = ab
2/ ( n + 1 ) a+ b a + b − 2 ab
b ∴ A−G = − ab =
G2 = ar 2 = a 2 2
a ( a )2 + ( b )2 − 2 a b a − b
2
= =
M M M 2 2
n / ( n + 1)
b
Gn = ar = a
n X Example 48. In the sequence 1, 2, 2, 4, 4, 4,
a 4, 8, 8, 8, 8, 8, 8,…, where n consecutive terms
Ø Product of n GM's between a and b, i.e.
have the value n, then 1025th term is
G1 × G 2 × G3 × ... × G m = (ab)n / 2 = G n (a) 2 9 (b) 210 (c) 211 (d) 2 8
Sol. (b) Let the 1025th term be 2 n . Then,
X Example 45. The five geometric means
1 + 2 + 4 + 8 +…+ 2 n − 1 < 1025 < 1 + 2 + 4 + 8 +…+ 2 n
between 576 and 9 are
∴ 2 n − 1 < 1025 < 2 n + 1 − 1 or 2 n < 1026 < 2 n + 1
(a) 192, 96, 48, 24, 12 (b) 288, 144, 72, 36, 18 ⇒ n = 10
50
(c) 208, 104, 52, 26, 13 (d) None of these Hence, 1025th term is 210 .
Work Book Exercise 3.2
1 The common ratio of a GP having 10th term and 8 If a,2b, 3c are in AP and a, b, c (distinct) are in
3
Sequence and Series
1st term equal to 1536 and − 3, is GP, then the common ratio of a GP is
2 1 −2 2 1 3 1/3 2
2 Any geometric series can be summed till infinite 9 If a GP having an even number of terms and the
number of terms by the formula sum of all terms is five times the sum of terms
a a occupying odd places, then the common ratio
S∞ = ;r∈R S∞ = ; r <1
1− r 1− r will be
a 3 5 4 1/4
S∞ = ; r >1 None of these
1− r
10 For n ≥ 3, the nth roots of unity form
3 If x, 2 x + 2 and 3 x + 3 are in GP, then the 4th an HP an AP
term is a GP None of these
27 − 27 13.5 − 13.5 11 If 1 + a + a2 + a3 + ... + an = (1 + a) (1 + a2 ) (1 + a4 ),
4 If 1, x, y, z and 2 are in GP, then xyz is equal to then n is equal to
3 5 7 9
4 2
8 None of these 12 The rational number 2.357357357 ... is
2355 2379 2355 2379
5 Which term of the sequence
10001 999 999 10001
2, 1, 2 − 1, 4− 1, 8− 1, ... is 128− 1?
13 91/ 3 ⋅ 91/ 9 ⋅ 91/ 27K is equal to
9th 8th 3
9 3 81 81
7th 10th
6 If a1, a2 , K , an are in AP, then 14 If b1, b2 and b3 (b1 > 0 ) are three successive terms
of a GP with common ratio r, then the possible
log a1, log a2 , ..., log an are in
value of r for which the inequality b3 > 4b2 − 3 b1
GP holds, is given by
not GP
4 2 2.5
AP
None of the above 15 If S denotes the sum to infinity and S n denotes
7 If the sum of the first 10 terms of a certain GP is the sum of n terms of the series
1 1 1 1
equal to 244 times the sum of first 5 terms. Then, 1 + + + + K, such that S − S n < , then
the common ratio is 2 4 8 1000
4 7 the least value of n is
2 3 8 9 10 11
Harmonic Progression (HP)
A sequence a1 , a 2 , ..., a n , ... of non-zero numbers is Thus, the harmonic mean H of two numbers a and
called a harmonic progression, if the sequence b is given by
1 1 1 1 2ab
, , , K, , K is an arithmetic progression. H=
a1 a 2 a 3 an a+b
1 1 1
The sequence 1, , , , ... are in HP, because the
3 5 7
Harmonic Mean of n Numbers
sequence 1, 3, 5, 7..., are in AP. If a1 , a 2 , a 3 , K, a n are n non-zero numbers, then
the harmonic mean H of these numbers is given by
Harmonic Mean of Two Given Numbers 1 1 1
+ +K +
If a and b are two non-zero numbers, then the 1 a1 a 2 an
harmonic mean of a and b is a number H such that the =
H n
sequence a, H and b are in HP.
Now, a, H and b are in HP. X Example 49. Which of the following is
1 1 1 1 1
⇒ , and are in AP. harmonic mean of numbers and ?
a H b 3 7
2 1 1 2ab 1 1
⇒ = + ⇒ H= (a)1 (b) (c) (d) 5
H a b a+b 5 10 51
Sol. (b)Q H = 2 ab
3 a+ b
1 1
On putting a = and b = , we get
Properties of Arithmetic, Geometric and Harmonic
Means between Two Given Numbers
Objective Mathematics Vol. 1
3 7 Let A, G and H be arithmetic, geometric and
2
1 1 harmonic means of two positive numbers a and b. Then,
3 7 1 a+b
H= = 2ab
1 1
+ 5 A= , G = ab and H =
3 7 2 a+b
X Example 50. The harmonic mean of the roots i. These three means possess the following
of the equation properties: A ≥G ≥ H
(5 + 2 ) x 2 − ( 4 + 5 ) x + 8 + 2 5 = 0 is X Example 52. The minimum value of
(a) 2 (b) 4 (c) 6 (d) 8 4 x + 41− x , x ∈ R , is
Sol. (b) Let α and β be the roots of given quadratic equation. (a) 2 (b) 4
Then, (c)1 (d) None of these
4+ 5 8+2 5
α +β = and αβ = Sol. (b)QAM ≥ GM for positive numbers.
5+ 2 5+ 2
4
4x +
Let H be the harmonic mean between α and β, then 4x ≥ 4
So, 4x ⋅ =2
2αβ 16 + 4 5 2 4x
H= = =4
α+β 4+ 5 ⇒ 4 x + 41− x ≥ 4
Insertion of n Harmonic Means between X Example 53. If 3, A and 2 are in AP, then
a and b 3+ 2
is greater than or equal to
The n numbers H1 , H 2 , ..., H n are said to be 2
harmonic means between a and b, if a, H1 , H 2 , ..., H n , b (a) 5 (b) 6
1 1 1 1 1 (c) 8 (d) None of these
are in HP, i.e. if , , , ..., , are in AP. Let d
a H1 H 2 Hn b Sol. (b)Q 3, A and 2 are in AP.
be the common difference of this AP. Then, 3+ 2
1 1 ∴ A= is the AM of 3 and 2.
= + ( n + 2 − 1) d 2
b a 3+ 2
Thus, ≥ 3⋅ 2 = 6
1 1 1 a−b 2
⇒ d= − =
n + 1 b a ( n + 1) ab X Example 54. If the product of n positive
1 1 a−b numbers is unity, then their sum is
Thus, = +
H1 a ( n + 1 ) ab (a) a positive integer (b) divisible by n
1
1 1 2( a − b) (c) equal to n + (d) never less than n
⇒ = + n
H 2 a ( n + 1) ab
Sol. (d) Since, product of n positive numbers is unity.
M M
∴ x1 ⋅ x2 ⋅ x3 K xn = 1 ...(i)
1 1 n( a − b) Using AM ≥ GM, we have
= +
H n a ( n + 1) ab x1 + x2 + x3 + K + xn
≥ ( x1 ⋅ x2 ⋅ x3 K xn )1/ n
n
X Example 51. Insert two harmonic means ⇒ x1 + x2 + K + xn ≥ n(1)1/ n
between 3 and 5. Hence, sum of n positive numbers is never less than n.
30 30 45 45
(a) , (b) , ii. A, G, H form a GP i.e. G 2 = AH
11 13 13 11
30 45 X Example 55. If AM of two terms is 9 and HM
(c) , (d) None of these
12 14 is 36, then GM will be
Sol. (b) a = 3, b = 5 and n = 2 (a)18 (b)12
1 1 (3 − 5) 13 45 (c)16 (d) None of these
∴ = + = ⇒ H1 =
H1 3 (2 + 1) (15) 45 13 Sol. (a)Q(AM)(HM) = (GM)2
1 1 2(3 − 5) 11 45
and = + = ⇒ H2 = ∴ 9 ⋅ 36 = (GM)2 ⇒ GM = 18
H2 3 (2 + 1) (15) 45 11
52
iii. If A, G and H are arithmetic, geometric and
harmonic means between three given
Sum of n Terms of an
Arithmetico-Geometric Progression 3
Sequence and Series
numbers a, b and c, then the equation having The sum of n terms of an arithmetico-geometric
a, b, c as its roots, is progression a, ( a + d ) r, ( a + 2d ) r 2 , ( a + 3 d ) r 3 , ... is
3G 3
x 3 − 3A x 2 + x −G3 =0 given by
H Sn =
X Example 56. If α, β and γ are the roots of the a (1 − r n − 1 ) {a + ( n − 1) d}r n
+ d r − , when r ≠ 1
equation x 3 − 6x 2 + 11x − 6 = 0. Then, HM of the 1 − r (1 − r ) 2 1− r
roots is
n
(a) 6 (b) 11 2 [2a + ( n − 1) d ], when r = 1
11 18
(c) (d) Proof Let S n denotes the sum of n terms of the
6 11
Sol. (d) We know that, if α, β and γ are the roots and A, G, H given sequence.
are AM, GM and HM of the roots, then equation is Then, S n = a + (a + d ) r + (a + 2d ) r 2
3
3G
x3 − 3 A x2 + x − G3 = 0 + ... + {a + ( n − 1) d} r n − 1 …(i)
H
Given, x3 − 6 x2 + 11x − 6 = 0 ⇒ rS n = ar + ( a + d ) r 2 + ... + {a + ( n − 2) d} r n − 1
On comparing, we get + {a + ( n − 1) d} r n …(ii)
3A = 6 …(i)
3G 3 On subtracting Eq. (ii) from Eq. (i), we get
= 11 …(ii)
H S n − r S n = a + [ dr + dr 2 + ... + dr n − 1 ]
and G3 = 6 …(iii)
− {a + ( n − 1) d} r n
From Eqs. (ii) and (iii), we get
H=
18 1 − r n − 1
11 ⇒ S n (1 − r ) = a + dr
1− r
Arithmetico-Geometric − {a + ( n − 1) d} r n ; r ≠ 1
1 − r n − 1 [ a + ( n − 1) d ] r n
Progression (AGP) ⇒ Sn =
a
1− r
+ dr −
1− r
; r ≠1
(1 − r ) 2
A sequence of the form
a, ( a + d ) r, ( a + 2d ) r 2 , ..., [ a + ( n − 1) d ]r n − 1 , ... For r =1,
S n = a + ( a + d ) + ( a + 2d ) +…+ [ a + ( n − 1) d ]
is called an arithmetico-geometric progression.
n
= [2a + ( n − 1) d ]
Arithmetico-Geometric Series 2
Let a, ( a + d ) r, ( a + 2d ) r 2 , ( a + 3d ) r 3 , ... be an X Example 58. The sum of 10 terms of the
arithmetico-geometric progression. sequence 1, 6, 27, 108, ... is
Then, a + ( a + d ) r + ( a + 2d ) r 2 + ( a + 3d ) r 3 + K (a) 280481 (b) 280482
is an arithmetico-geometric series. (c) 280484 (d) 280483
Sol. (d) The given sequence can be rewritten as
X Example 57. The 8th term of the sequence 1, (1 + 1) ⋅ 3, (1 + 2 ⋅ 1) ⋅ 32 , (1 + 3 ⋅ 1) ⋅ 33 , ....
1, 4, 12, 32, 80, .... is which is an arithmetico-geometric progression with
(a) 1000 (b) 1200 a = 1, d = 1 and r = 3.
(c) 1024 (d) None of these a 1 − r9 [a + 9d ] r10
Hence, S10 = + dr −
1− r (1 − r )2
1− r
Sol. (c) The given sequence can be rewritten as
1 (1 − 39 ) [1 + 9] 310
1, (1 + 1) ⋅ 2, (1 + 2 ⋅ 1) ⋅ 2 2 , (1 + 3 ⋅ 1) ⋅ 2 3 ,... = + 3 −
−2 (1 − 3)2 (1 − 3)
Clearly, the sequence is an arithmetico-geometric
− 1 (3 − 310 ) 310 ⋅ 10
progression with a = 1, d = 1 and r = 2. = + +
2 4 2
∴ 8th term = (a + 7d ) ⋅ r 7 − 2 + 3 − 310 + 20 ⋅ 310
=
= (1 + 7 ⋅ 1) ⋅ 2 7= 8 ⋅ 2 7 4
1 + 19 ⋅ 310
= 2 3 ⋅ 2 7 = 210 = 1024 = = 280483 53
4
3
1
Sum of an Infinite Arithmetico-Geometric 1 102
= +
Progression 1−
1 1 − 1
2
102
Let r <1, then r n , r n − 1 → 0 as n → ∞ and it can
Objective Mathematics Vol. 1
102
also be shows that n ⋅ r n → 0 as n → ∞. So, from = +
102 104
a (1 − r n −1 ) [ a + ( n − 1) d ] r n 102 − 1 102 (102 − 1)2
Sn = + dr − , we get 102 − 1 + 1 104
1− r (1 − r ) 2 1− r = 102 2
=
(10 − 1) (10 − 1)
2 2 2
a dr 4
Sn → + as n → ∞ ∴ S=
2
×
10
=
2000
1 − r (1 − r ) 2 10 (102 − 1)2 9801
In other words, when r <1, then sum to infinity of Aliter
an arithmetico-geometric series is 2 4 6 8
Let S= + 3 + 5 + 7 + ...
a dr 10 10 10 10
S∞ = + S 2 2 2
1 − r (1 − r ) 2 ∴ S− = + 3 + 5 + ...
102 10 10 10
2
X Example 59. The sum of 99S 2 102 20
⇒ = 10 = × 2 =
0.2 + 0.004 + 0.00006 + 0.0000008 + ... is 100 1− 2
1 10 10 − 1 99
10
200 20 × 100 2000
(a) ⇒ S = =
891 (99)2 9801
2000
(b) X Example 60. Find the value of the expression
9801 n i j
(c)
1000
∑ ∑ ∑ 1.
9801 i =1 j =1 k =1
(d) None of the above n i j n i
Sol. ∑ ∑ ∑1 = ∑ ∑ j
Sol. (b) Sum = 2 + 43 + 65 + 87 + ... i =1 j =1k =1 i =1 j =1
10 10 10 10
1 n 2 n
=
2 2 3 2 S
1 + 2 + 4 + ... =
= ∑ i + ∑ i
10 10 10 10 1 2 i = 1 i =1
Clearly, S1 is an arithmetico-geometric series with a = 1, 1 n(n + 1) (2 n + 1) n(n + 1)
1 = +
d = 1 and r = 2 . 2 6 2
10 n(n + 1)
a dr = [2 n + 1 + 3]
∴ S1 = + n
1 − r (1 − r )2 n(n + 1) (n + 2 )
=
6
Work Book Exercise 3.3
1 The first term of HP having kth term equal to 4 If x < 1, then 1 + 3 x + 5 x 2 + 7 x 3 + ... is equal to
1 1+ x 2 + x
, is a b
a + (k − 1)d (1 − x )2 (1 − x )2
1 1 x
a a b c d k c d None of these
a k (1 − x )2
2 If a, b, c and d are distinct positive real numbers
in HP, then
5 The sum of n terms of the series
2
1 1
a ab > cd b ac > bd 1 + 2 1 + + 3 1 + + ... is
c ad > bc d None of these
n n
2
b n ( n + 1) c n 1 + d n2
3 1
3 The 4th term of the sequence 3, , 1, ... is a n2 + 1
2 n
3
a 6 If a, b, c , d and p are distinct real numbers such
4
b
4 that (a2 + b2 + c 2 ) p2 − 2 (ab + bc + cd )
3 p + (b2 + c 2 + d 2 ) ≤ 0, then a, b, c and d are in
2
c
3 a GP b AP
d None of the above c HP d None of these
54
7 If x, y, z and w are non-zero real numbers such
that ( x + y + z ) ( y + z + w )
2 2 2 2 2 2
≤ ( xy + yz + zw )2 , then x, y, z and w are in
9 If a, b and c are three positive real numbers such
that b + c − a, c + a − b and a + b − c are
positive,then the expression
(b + c − a) (c + a − b) (a + b − c ) − abc is
3
Sequence and Series
a GP b AP a positive b negative
c HP d None of these c non-positive d non-negative
π
8 If 0 ≤ θ < , then cos θ + sec θ is 10 The least value of the expression
2 5 sin x − 1 + 5− sin x − 1 is
a <2 b ≥2
2 1 5
c can have any value d None of these a b c 5 d
5 5 2
Some Special Series First Way
■ Sum of first n natural numbers, i. If the differences T2 − T1 , T3 − T2 and T4 − T3
n( n + 1)
∑ n = 1 + 2 + 3 + ... + n = are in AP, then the nth term is given by
2
Tn = an 2 + bn + c, where a, b and c are
■ Sum of squares of first n natural numbers,
constants. Determine constants a, b and c by
n( n + 1) (2n + 1)
∑ n = 1 + 2 + 3 + ... + n =
2 2 2 2 2
putting n =1, 2, 3 and equating them with
6 values of corresponding terms of the given
■ Sum of cubes of first n natural numbers, series.
2
n( n + 1)
∑ n = 1 + 2 + 3 + ... + n =
3 3 3 3 3
X Example 62. The sum of n terms of the series
2
1 + 5 +12 + 22 + 35 + ... is
X Example 61. The sum of first 12 terms of the n( n 2 + 1) n 2 ( n + 1)
(a) (b)
13 13 + 2 3 1 3 + 2 3 + 3 3 2 2
series + + + ... to n terms is n( n 2 − 1)
1 1+3 1+3 +5 (c) (d) None of these
equal to 4
509 309 Sol. (b) The given series is 1 + 5 + 12 + 22 + ...
(a) (b) The differences of successive terms are 4, 7, 10, 13 which
2 2
are in AP.
409
(c) (d) None of these ∴ Tn = an2 + bn + c
2
On putting n = 1, 2, 3, we get
Sol. (c) Here nth term, Tn = 1 + 2 + 3 + ... + n = ∑ n2
3 3 3 3 3
T1 = a + b + c = 1 ...(i)
1 + 3 + 5 + ... + (2 n − 1) n T2 = 4 a + 2 b + c = 5 ...(ii)
=
1 2 1
n + n+
1 and T3 = 9a + 3b + c = 12 ...(iii)
4 2 4 On solving Eqs. (i), (ii) and (iii), we get
Sum of n terms, S n = ∑ Tn 3 1
a= ,b=− and c = 0
1 1 1 2 2
= ∑ n2 + ∑ n + n
4 2 4 3 1 1
∴ Tn = n2 − n = (3n2 − n)
1 1 1 1 1 2 2 2
= ⋅ ⋅ n(n + 1) (2 n + 1) + ⋅ n(n + 1) + n
4 6 2 2 4 1 3 1
Now, S n = ∑ Tn = ∑ (3n2 − n) = ∑ n2 − ∑ n
n 2 2 2
= (2 n + 9n + 13)
2
3 (n) (n + 1) (2 n + 1) 1 n(n + 1)
24 = −
12 2 6 2 2
So, S12 = [2 ⋅ (12 )2 + 9 × 12 + 13]
24 n2 (n + 1)
=
409 2
=
2
ii. If the difference T2 − T1 , T3 − T2 and T4 − T3
Method of Difference are in GP with common ratio r, then
Sometimes, the nth term of a series cannot be Tn = ar n −1 + bn + c, where a, b and c are
determined by the methods discussed so far. If a series constants.
is such that the difference between successive terms are Determine constants a, b and c by putting
either in AP or in GP, then we determine its nth term by
n =1, 2, 3 and equating with value of
method of difference and then find the sum of the series
corresponding terms of the given series. 55
by using the formulae of ∑ n, ∑ n 2 and ∑ n 3 .
3 X Example 63. The sum of n terms of the series
5 + 7 + 13 + 31 + 85 + ... is
1 1
Second Way
We can find nth term of the series by the following
steps:
Objective Mathematics Vol. 1
(a) (3 n + 8n − 1) (b) (3 n + 7n + 1)
2 2
1 n 1 i. Denote the n th term by Tn and the sum of the
(c) (3 + 5n) (d) (3 n + 7n) series upto n terms by S n .
2 2
Sol. (a) The given series is 5 + 7 + 13 + 31 + 85 + ... ii. Rewrite the given series with each term shifted
The differences of successive terms are 2, 6, 18, 54, which
by one place to the right.
are in GP with common ratio i.e. 3.
∴ Tn = a(3)n −1 + bn + c iii. By subtracting the later series from the
On putting n = 1, 2, 3, we get former, find Tn .
T1 = a + b + c = 5 ...(i)
T2 = 3a + 2 b + c = 7 ...(ii) iv. From Tn , S n can be found by appropriate
and T3 = 9a + 3b + c = 13 ...(iii) summation.
On solving Eqs. (i), (ii) and (iii), we get
a = 1, b = 0 and c = 4
X Example 66. The sum of n terms of the series
∴ Tn = 3n −1 + 4
1 + 2 + 5 + 12 + 25 + 46 + ... is
1
∴ S n = ∑ Tn = ∑ (3n −1 + 4) (a) n ( n − 1) ( n 2 − 3n + 8)
12
3n − 1
= ∑ 3n −1 + ∑ 4 = + 4n 1
3−1 (b) n ( n + 1) ( n 2 − 3n + 8)
24
3n − 1 1
= + 4n = (3n + 8n − 1) 1
2 2 (c) n ( n + 1) ( n 2 − 3n + 8)
12
X Example 64. Find the sum to n terms (d) None of the above
3 + 7 + 13 + 21 + ... . Sol. (c) Let the sum of the series be S n and nth term of the
series be Tn . Then,
Sol. Let S = 3 + 7 + 13 + 21 + ... + Tn ... (i) S n = 1 + 2 + 5 + 12 + 25 + 46 + ... + Tn −1 + Tn …(i)
and S = 3 + 7 + 13 + ... + Tn − 1 + Tn ... (ii) Sn = 1 + 2 + 5 + 12 + 25 + ... + Tn −1 + Tn …(ii)
On subtracting Eq. (ii) from Eq. (i), we get
On subtracting Eq. (ii) from Eq. (i), we get
Tn = 3 + 4 + 6 + 8 + ... + (Tn − Tn − 1 )
0 = 1 + 1 + 3 + 7 + 13 + 21 + ... + (Tn − Tn −1 ) − Tn
n−1
=3+ [8 + (n − 2 )2 ] = 3 + (n − 1) (n + 2 ) ⇒ Tn = 1 + 1 + 3 + 7 + 13 + 21 + ... + t n −1 + t n …(iii)
2
[since, nth term of Tn is t n ]
= n2 + n + 1
⇒ Tn = 1 + 1 + 3 + 7 + 13 + ... + t n −1 + t n …(iv)
Hence, S = Σ(n2 + n + 1)
On subtracting Eq. (iv) from Eq. (iii), we get
= Σn2 + Σn + Σ1
0 = 1 + 0 + 2 + 4 + 6 + 8 + ... + (t n − t n −1 ) − t n
n(n + 1) (2 n + 1) n(n + 1)
= + + n ⇒ t n = 1 + 2 + 4 + 6 + 8 + ... + (n − 1) terms
6 2
n = 1 + [2 + 4 + 6 + 8 + ... + (n − 2 ) terms
= (n2 + 3n + 5) n−2
3 = 1+ [2 ⋅ 2 + (n − 2 − 1) 2 ]
2
X Example 65. Find the sum to n terms = 1 + (n − 2 ) (2 + n − 3)
= n2 − 3n + 3
1 + 4 + 10 + 22 + ... .
∴ Tn = Σt n = Σn2 − 3Σn + 3n
Sol. Let S = 1 + 4 + 10 + 22 + ... + Tn …(i) n(n + 1) (2 n + 1) n (n + 1)
= − 3⋅ + 3n
and S= 1 + 4 + 10 + ... + Tn − 1 + Tn …(ii) 6 2
n
On subtracting Eq. (ii) from Eq. (i), we get = (n2 − 3n + 5)
Tn = 1 + (3 + 6 + 12 + ... + Tn − Tn − 1 ) 3
2 n − 1 − 1 1
∴ S n = Σn3 − Σn2 + 5Σn
⇒ Tn = 1 + 3 3
2 −1 2
1 n (n + 1) 1 n(n + 1) (2 n + 1) 5 n (n + 1)
⇒ Tn = 3 ⋅ 2 n − 1 − 2 = − ⋅ + ⋅
3 2 6 1 3 2
So, S = ΣTn = 3Σ 2 n − 1 − Σ 2 n(n + 1) 2
= ⋅ [n + n − 4n − 2 + 10]
2 n − 1 6×2
= 3⋅ − 2n
2 −1 =
1
⋅ n (n + 1) (n2 − 3n + 8)
56 = 3⋅ 2n − 2n − 3 12
X Example 67. Find the sum to n terms of the
series 1 ⋅ 2 + 2 ⋅ 3 + 3 ⋅ 4 + ... .
Solution of Form (i) Let d be the common
difference of an AP. Then, a n = a1 + ( n − 1) d
Let sum of the series and n th term be denoted by S n
3
Sequence and Series
Sol. Let Tr be the general term of the series.
So, Tr = r(r + 1)
and Tn , respectively. Then,
To express t r = f(r ) − f(r − 1) multiply and divide t r by
1 1
Sn = +
[(r + 2 ) − (r − 1)]. a1 a 2 ... a r a 2 a 3 ... a r +1
r
So, Tr = (r + 1) [(r + 2 ) − (r − 1)] 1
3 +... +
1 a n a n + 1 ... a n + r − 1
= [r(r + 1) (r + 2 ) − (r − 1) r(r + 1)]
3 1
1 Tn = …(i)
Let f(r ) = r(r + 1) (r + 2 ) a n a n +1 a n + 2 ... a n + r − 2 a n + r − 1
3
So, Tr = [f(r ) − f(r − 1)] 1
n Let Vn = …(ii)
a n +1 a n + 2 ... a n + r − 2 a n + r − 1
Now, S= ∑ Tr = T1 + T2 + T3 + ... + Tn
r =1
[avoiding first term for Vn i.e. a n in Tn ]
1
T1 =
[1⋅ 2 ⋅ 3 − 0] 1
3 Vn − 1 =
1 a n a n +1 ... a n + r − 3 a n + r − 2
T2 = [2 ⋅ 3 ⋅ 4 − 1⋅ 2 ⋅ 3]
3 1
1 ∴ Vn − Vn −1 =
T3 = [3 ⋅ 4 ⋅ 5 − 2 ⋅ 3 ⋅ 4] a n + 1 a n + 2 ... a n + r − 2 a n + r −1
3
1
M
1
M
−
Tn = [n(n + 1)(n + 2 ) − (n − 1)n(n + 1)] a n a n +1 ... a n + r − 2
3
a n − a n + r −1
∴
1
S = n(n + 1)(n + 2 ) Vn − Vn −1 =
3 a n a n +1 a n + 2 ... a n + r − 2 a n + r −1
Hence, the sum of series is f(n) − f(0).
Vn − Vn −1 = Tn ( a n − a n + r −1 ) [from Eq. (i)]
X Example 68. Find the sum to n terms of the = Tn {[ a1 + ( n − 1) d ]
series given below − [ a1 + ( n + r − 2) d ]}
4
+
5
+
6
+… = d (1 − r )Tn
1⋅ 2 ⋅ 3 2 ⋅ 3 ⋅ 4 3 ⋅ 4 ⋅ 5 (V − Vn −1 )
∴ Tn = n
Sol. Let Tr = r+ 3 d (1 − r )
r(r + 1)(r + 2 ) 1
⇒ Tn = {Vn − 1 − Vn }
=
1
+
3 d ( r − 1)
(r + 1)(r + 2 ) r(r + 1)(r + 2 )
Putting n =1, 2, 3, 4, ..., n, we get
1 1 3 1 1
= − + − 1
r + 1 r + 2 2 r(r + 1) (r + 1)(r + 2 ) T1 = (V0 − V1 )
d ( r − 1)
1 1 3 1 1
∴ S= − + 2 2 − (n + 1)(n + 2 ) 1
2 n + 2 T2 = (V1 − V2 )
d ( r − 1)
5 1 3
= − 1+
4 n + 2 2(n + 1)
1
T3 = (V2 − V3 )
5 1 d ( r − 1)
= − [2 n + 5]
4 2(n + 1)(n + 2 ) M M M
1
Vn Method Tn = (Vn − 1 − Vn )
d ( r − 1)
To find the sum of the series of the forms
Adding the above equations, we get
1 1 1
i. + T1 + T2 + T3 + L + Tn = (V0 − Vn )
a1 a 2 a 3 ... a r a 2 a 3 ... a r +1 ( r − 1) d
1
+ ... + 1 1
a n a n + 1 ... a n + r − 1 ∴ Sn =
( r − 1) d a1 a 2K a r − 1
ii. a1 a 2 ... a r + a 2 a 3 ... a r +1 +... + a n a n + 1 ... a n + r − 1 1
−
where, a1 , a 2 , a 3 ... a n K are in AP. a n + 1 a n + 2K a n + r − 1 57
3 Hence, the sum of n terms is
Sn =
1
( r − 1)( a 2 − a1 )
⇒ Tn =
1
( r + 1) d
(Vn − Vn − 1 )
Objective Mathematics Vol. 1
Putting n =1, 2, 3, ...., n, we get
1 1 1
− T1 = (V − V0 )
( r + 1) d 1
a1 a 2K a r − 1 a n + 1 a n + 2 K a n + r − 1
1
Corollary If a1 , a 2 , a 3 , ..., a n ,K are in AP, then T2 = (V2 − V1 )
( r + 1) d
For r = 2
1
1 1 1 T3 = (V − V2 )
+ + L+ ( r + 1) d 3
a1 a 2 a 2 a 3 an an+ 1
M M M
1 1 1 1
= − Tn = (Vn − Vn −1 )
( a 2 − a1 ) a1 a n + 1 ( r + 1) d
a n + 1 − a1 Adding the above equations, we get
=
(a 2 − a1 )a1 a n + 1 1
T1 + T2 + T3 + ... + Tn = (V − V0 )
a1 + ( n + 1 − 1) d − a1 ( r + 1) d n
= 1
da1 a n + 1 ∴ Sn =
( r + 1)( a 2 − a1 )
nd n
= =
da1 a n + 1 a1 a n + 1 ( a n a n + 1K a n + r − a 0 a1 a 2K a r )
For r = 3 [Q a 0 = a1 − d ]
Hence, the sum of n terms is
1 1 1
+ +L + 1
a1 a 2 a 3 a 2 a 3 a 4 an an+ 1an+ 2 Sn =
( r + 1)( a 2 − a1 )
1 1 1
= a2 − {a n a n + 1K a n + r − a 0 a1 a 2K a r }
2( a 2 − a1 ) a1 a 2 a n + 1 a n + 2 Corollary If a1 , a 2 , a 3 , ..., a n , K are in AP,
For r = 4 then
1
+
1 For r = 2
a1 a 2 a 3 a 4 a 2 a 3 a 4 a 5 a1 a 2 + a 2 a 3 + ...+ a n a n + 1
1 1
+L + = {a a a − a 0 a1 a 2 }
an an+ 1an+ 2an+ 3 3( a 2 − a1 ) n n + 1 n + 2
1 1 1 For r = 3
= −
3( a 2 − a1 ) a1 a 2 a 3 a n +1 a n + 2 + a n + 3 a1 a 2 a 3 + a 2 a 3 a 4 + ...+ a n a n + 1 a n + 2
1
= {a a a a − a 0 a1 a 2 a 3 }
Solution of Form (ii) Let S n be the sum and 4( a 2 − a1 ) n n + 1 n + 2 n + 3
Tn be the nth term of the series, then
S n = a1 a 2K a r + a 2 a 3K a r + 1 X Example 69. The sum of n terms of the series
+L + a n a n + 1 ... a n + r − 1 4 5 6
+ + + ... is
∴ Tn = a n a n + 1 a n + 2 ... a n + r − 2 a n + r − 1 1⋅ 2 ⋅ 3 2 ⋅ 3 ⋅ 4 3 ⋅ 4 ⋅ 5
Let Vn = a n a n + 1 a n + 2 ... a n + r − 2 a n + r − 2 a n + r n+3
(a)
[taking one extra term in Tn for Vn ] n( n + 1) ( n + 2)
∴ Vn − 1 = a n − 1 a n a n + 1 ... a n + r − 3 a n + r − 2 a n + r −1 n ( n + 1)
(b)
⇒ Vn − Vn − 1 = a n a n + 1 a n + 2 ... n( n + 1) ( n + 2)
a n + r − 1 ( a n + r − a n − 1 ) [from Eq. (i)] 5 2n + 5
= Tn {([ a1 + ( n + r − 1) d ] − [ a1 + ( n − 2) d ])} (c) −
4 2( n + 1) ( n + 2)
= ( r + 1) d Tn
(d) None of the above
58
n+ 3 1
Sol. (c) Here, Tn =
⇒ Tn =
1
n(n + 1) (n + 2 )
+
3
(n + 1) (n + 2 ) n(n + 1) (n + 2 ) ∴
1
=−
1
4
n(n + 1)(n + 2 )
= − (Vn − Vn − 1 )
= −2
Tn
[from Eq. (i)]
3
Sequence and Series
Tn 2
1 1 3 1 1
⇒ Tn = − + − Putting n = 1, 2, 3, ..., n, we get
n + 1 n + 2 2 n (n + 1) (n + 1) (n + 2 ) 1 1
= − (V1 − V0 )
T1 = − +
1 1 3 1 1 T1 2
∴ −
2 3 2 1⋅ 2 2 ⋅ 3 1 1
= − (V2 − V1 )
T2 = − +
1 1 3 1 1 T2 2
+
3 4 2 2 ⋅ 3 3 ⋅ 4 1 1
= − (V3 − V2 )
and so on. T3 2
Now, S = T1 + T2 + T3 + ...
M M
1 1 3 1 1 1 1
⇒ S= − + 2 2 − (n + 1) (n + 2 ) = − (Vn − Vn −1 )
2 n + 2 Tn 2
5 2n + 5 1 1 1 1 1
∴ S= − ∴ + + + ... + = − (Vn − V0 )
4 2 (n + 1) (n + 2 ) T1 T2 T3 Tn 2
n n
1 1 2 2
X
n
Example 70. If ∑ Tr = (n + 1)(n + 2)(n + 3),
⇒ ∑T =−
2 (n + 1)(n + 2 )
−
2
r =1 r
r =1 8
(n + 1)(n + 2 ) − 2
n =
1 2(n + 1)(n + 2 )
then find ∑ .
r =1 Tr n2 + 3n
=
n n −1 2(n + 1)(n + 2 )
Sol. Q Tn = S n − S n − 1, ∑ Tr − ∑ Tr
r =1 r =1
n(n + 1)(n + 2 )(n + 3) (n − 1) n(n + 1)(n + 2 ) Some Important Series
= −
8 8 x x2 x3 xn
n(n + 1)(n + 2 ) (i) e x = 1 + + + +K + +K
= [(n + 3) − (n − 1)] 1! 2! 3! n!
8
n(n + 1)(n + 2 ) n(n + 1)(n + 2 ) x2 x3 x4
= (4) = (ii) log (1 + x ) = x − + −
8 2 2 3 4
1 2
⇒ = …(i) xn
Tn n(n + 1)(n + 2 ) +K + ( − 1) n + 1 +...
2 n
∴ Vn =
(n + 1)(n + 2 ) α α (α − 1) 2
(iii) (1 + x ) α = 1 + x+ x
2 1! 2!
Then, Vn− 1 =
n(n + 1) α (α − 1)K (α − n + 1) n
2 2 + ... + x + ..., ( − 1 < x < 1)
∴ Vn − Vn − 1 = − n!
(n + 1)(n + 2 ) n(n + 1)
Work Book Exercise 3.4
1 The sum of n terms of the series 2 2 + 42 + 62 + ... 3 The sum of n terms of the series
1 1 1
is + + + ... is
n( n + 1) (2 n + 1) 2 n( n + 1) (2 n + 1) 1⋅ 2 2 ⋅ 3 3 ⋅ 4
a b n 1 1
3 3 a b −
n( n + 1) (2 n + 1) n( n + 1) (2 n + 1) ( n + 1)( n + 2 ) n n+1
c d
6 9 n
c d None of these
n+1
2 The sum of n terms of 1⋅ 2 ⋅ 3 + 2 ⋅ 3 ⋅ 4 + ... is
n( n + 1) ( n + 2 ) ( n + 3) 4 The sum of n terms of the series
a 3 5 7
4 + + 2 +... is
2 n( n + 1) ( n + 2 ) ( n + 3) 12 12 + 2 2 1 + 2 2 + 32
b
3 n+1 n
( n + 2 ) ( n + 1) ( n + 3) a b
c n n+ 1
4 6n 6( n − 1)
n( n − 1) ( n − 2 ) ( n − 3) c d
d n+ 1 n
4
59
n n
3
n
1 r
5 If ∑ Tr = (3n − 1), then find the sum of ∑T . 8 The sum of the series ∑ (r + 1)!, where
r =1 r =1 r r =1
n n ! = 1⋅ 2 ⋅ 3 ... n, is
Objective Mathematics Vol. 1
3 1 3 n
a 1 − b ( 3 − 1) n 1
4 3 2 a b 1−
( n + 1)! ( n + 1)!
3 1 1
n
c (1 − 3 n ) d 1 − 1 n +1
4 2 3 c d
( n + 1)! n!
6 The sum of n terms of the series 9 The sum of first n terms of the series
5 + 7 + 13 + 31 + 85 + ... is 13 + 3 × 2 2 + 33 + 3 × 42 + 53 + 3 × 62 + ..., when
1
a n(2 n + 1) n is even, is
2
1 n n+1 3 n 3
b ( 3 + 8n − 1) a ( n + 7 n2 − 3n − 1) b ( n + 7 n2 − 4n + 1)
2 8 8
1 2n n 3
c ( 3 − 2 n) c ( n + 4n2 + 10n + 8) d None of these
2 8
d None of the above
10 The sum of first n terms of the series
360
1 1 2 3
7 The sum of the series ∑ + + + ... is
k + 1 + (k + 1) k 1 + 12 + 14 1 + 2 2 + 2 4 1 + 32 + 34
k = 1 k
is 1 1 1
a 1 − 2 b
a
18
b
15 4 n + n + 1 n2 + n + 1
19 19
1 1 1
c
17
d
16 c d 1 − 2
19 19 n + n2 + n4 2 n + n + 1
60
WorkedOut Examples
Type 1. Only One Correct Option
Ex 1. If S 1 , S 2 and S 3 denote the sum of first n1 , n2 Ex 3. The sum of the products of 2n numbers
and n3 terms respectively of an AP, then ± 1, ± 2, ± 3, ..., ± n taking two at a time is
S1 S S n ( n + 1) n ( n + 1) ( 2n + 1)
( n2 − n3 ) + 2 ( n3 − n1 ) + 3 ( n1 − n2 ) is (a) − (b)
n1 n2 n3 2 6
n ( n + 1) ( 2n + 1)
equal to (c) − (d) None of these
(a) 0 6
(b) 1 Sol. We have, (1 − 1 + 2 − 2 + 3 − 3 + K + n − n)2
(c) S 1 S 2 S 3 = 12 + 12 + 22 + 22 + K + n2 + n2 + 2S
(d) n1 n 2 n 3 where, S is the required sum.
n1 ⇒ 0 = 2(12 + 22 + K + n2 ) + 2S
Sol. We have, S 1 = [ 2a + (n1 − 1)d ]
2 n(n + 1)(2n + 1)
⇒ S = − (12 + 22 + K + n2 ) = −
2S 1 6
⇒ = 2a + (n1 − 1)d
n1 Hence, (c) is the correct answer.
n2
S 2 = [ 2a + (n 2 − 1)d ] Ex 4. The number of terms common to two AP’s
2
2S 2 3, 7, 11, ..., 407 and 2, 9, 16, ..., 709 is
⇒ = 2a + (n 2 − 1)d (a) 21 (b) 28
n2
n3 (c) 14 (d) None of these
and S3 = [ 2a + (n 3 − 1)d ] Sol. By inspection, first common term to both the sequences
2
2S 3 is 23. Second common term = 51
⇒ = 2a + (n 3 − 1)d Third term = 79 and so on.
n3
These numbers form an AP 23, 51, 79, ...
2S 1 2S 2 2S 3
∴ (n 2 − n3 ) + (n 3 − n1 ) + (n1 − n 2 ) Q T15 = 23 + 14 (28) = 23 + 392 = 415 > 407
n1 n2 n3 and T14 = 23 + 13 (28) = 387 < 407
= [ 2a + (n1 − 1)d ](n 2 − n 3 ) + [ 2a + (n 2 − 1)d ] ∴ Number of common terms = 14
(n 3 − n1 ) + [ 2a + (n 3 − 1)d ](n1 − n 2 ) Hence, (c) is the correct answer.
=0 Aliter
Hence, (a) is the correct answer. Let tm = tn
⇒ 3 + (m − 1) 4 = 2 + (n − 1) 7
Ex 2. Let a and b be two given numbers. If A denotes ⇒ 4 m − 1 = 7n − 5
the single AM and S denotes the sum of n AM’s ⇒ 4 (m + 1) = 7n
between a and b, then S / A depends on m+1 n
⇒ = = λ,
(a) n, a, b (b) n, b (c) n, a (d) n 7 4
where, m ≤ 102
Sol. Q A is the single AM between a and b.
and n ≤ 102
a+ b
∴ A= ∴ m = 7λ − 1, n = 4 λ
2 5 2
Let A1 , A2 , K , An be n AM’s between a and b. λ ≤ 14 and λ ≤ 25
7 4
∴ a, A1 , A2 , ..., An , b is an AP with common difference ∴ λ ≤ 14 , number of common terms = 14
b−a Hence, (c) is the correct answer.
d=
n+1
n
Ex 5. The sum of all the numbers of the form n 3
Now, S = A1 + A2 + K + An = ( A1 + An ) which lie between 100 and 10000, is
2
n n (a) 43261
= (a + d + b − d ) = (a + b) (b) 53261
2 2
a + b (c) 63261
∴ S =n = nA (d) None of the above
2
S Sol. The smallest and the largest numbers between 100 and
⇒ =n 10000, which can be written in the form n3, are
A 61
Hence, (d) is the correct answer. 53 = 125 and 213 = 9261
∴ Required sum
3 = 5 + 6 + 7 + ... + 21 =
3 3 3 3
21
∑ n3 − ∑ n
n= 1
4
n= 1
3
Ex 8. If three positive real numbers a, b and c are in
AP such that abc = 4, then the minimum
possible value of b is
Objective Mathematics Vol. 1
1 2 1 2 (a) 23/ 2 (b) 22/ 3 (c) 21/ 3 (d) 25/ 2
= n (n + 1)2 − n (n + 1)2
4 n = 21 4 n= 4 Sol. Let d be the common difference of an AP, then
= 53361 − 100 = 53261 4 = abc = (b − d ) b (b + d )
Hence, (b) is the correct answer. = b (b2 − d 2 )
⇒ b3 = 4 + bd 2 ≥ 4 [Qb > 0, d 2 ≥ 0]
Ex 6. If 1, log y x, log z y and −15 log x z are in AP,
then ⇒ b≥2 2/ 3
−2
(a) z = x
3
(b) x = y Thus, minimum possible value ofb is 22 / 3, that is the
case when d = 0.
(c) z −2 = y (d) None of these
Hence, (b) is the correct answer.
Sol. Let d be the common difference.
Then, log y x = 1 + d ⇒ x = y1 + d Ex 9. If the lengths of sides of a right triangle are in
logz y = 1 + 2d ⇒ y = z1 + 2 d AP, then the sines of the acute angle are
and − 15 logx z = 1 + 3d 3 4 2 1
(a) , (b) ,
⇒ z = x − (1 + 3 d) / 15 5 5 3 3
∴ x = y1 + d = z(1 + 2 d) (1 + d )
5−1 5+1 3 −1 3+1
= x − (1 + d ) (1 + 2 d) (1 + 3 d) / 15 (c) , (d) ,
⇒ (1 + d ) (1 + 2d ) (1 + 3d ) = − 15
2 2 2 2
⇒ 6d 3 + 11d 2 + 6d + 16 = 0 Sol. Let the sides of the triangle be a − d, a, a + d, where
⇒ (d + 2) (6d 2 − d + 8) = 0 a > d > 0. By Pythagoras theorem, we have
⇒ d = −2 (a + d )2 = (a − d )2 + a2
∴ x = y1 + d = y−1 , y = z1 − 4 = z−3 a
⇒ 4 ad = a2 ⇒ d =
4
and x = (z−3 )−1 = z3 a 4
Hence, (a) is the correct answer. We have, sinθ = =
a+ d 5
1 ⋅ 2 ⋅ 3 + 2 ⋅ 3 ⋅ 4 + 3 ⋅ 4 ⋅ 5 +... upto n terms sin (90° − θ ) =
3
Ex 7. L = lim and
n→ ∞ n (1 ⋅ 2 + 2 ⋅ 3 + 3 ⋅ 4 +... upto n terms) 5
Hence, (a) is the correct answer.
is equal to
n
3 1 1
(a)
4
(b)
4
Ex 10. If S n = ∑ t r = 6 n(2n 2 + 9n + 13), then
r =1
1 5 n
(c) (d)
2
n+ 1
4 ∑ t r is equal to
r =1
Sol. Numerator = ∑ (r − 1) r (r + 1) (a)
1
n ( n + 1) (b)
1
n( n + 2)
r=2
n+ 1 n+ 1
2 2
= ∑ (r3 − r) = ∑ (r3 − r) 1
(c) n( n + 3)
1
(d) n( n + 5)
r=2 r=1 2 2
1 1
= (n + 1)2 (n + 2)2 − (n + 1) (n + 2) Sol. We have, tn = S n − S n − 1 , ∀ n ≥ 1
4 2 1
n n ∴ tn = n(2n2 + 9n + 13)
Also, ∑ r (r + 1) = ∑ [(r + 1)2 − (r + 1)] 6
1
r=1
n+ 1
r=1
n+ 1
− (n − 1) {2(n − 1)2 + 9(n − 1) + 13}
6
= ∑ (s2 − s) = ∑ (s2 − s) 1
= [ 2{(n)3 − (n − 1)3} + 9{n2 − (n − 1)2}
s=2 s=1 6
=
1 1
(n + 1)(n + 2)(2n + 3) − (n + 1)(n + 2) + 13 (n − n + 1)]
6 2 1
= [ 6n − 6n + 2 + 9 (2n − 1) + 13 ]
2
Thus, 6
1 1 2
2
1 1 2
2 1
1 + 1 + − 2 1 + 1 + = (6n2 + 12n + 6) = (n + 1)2
4 n n 2n n n 6
L = lim n n
1
n→ ∞ 1 1 2 3 1 1
1 + 1 + 2 + − 1 + 1 +
2 ∴ ∑ tr = ∑ (r + 1) = (n + 1) (n + 2) − 1
6 n n n 2n n n r=1 r=1
2
1 6 3 1
= × = =
n (n + 3)
62 4 2 4 2
Hence, (a) is the correct answer. Hence, (c) is the correct answer.
Ex 11. Let a n be the nth term of an AP. If
1099
∑ a 2r = 10100 and
1099
∑ a 2r − 1 = 10 99 , then the
Ex 14. The sum of infinite terms of a decreasing GP is
equal to the greatest value of the function
f ( x ) = x 3 + 3x − 9 in the interval [− 2, 3] and
3
Sequence and Series
r =1 r =1 the difference between the first two terms is
common difference of an AP is f ′ (0). Then, the common ratio of a GP is
(a) 1 (b) 9 (c) 10 (d) 1099 2
(a) −
Sol. Let d be the common difference of an AP, then 3
4
a 2r = a 2r − 1 + d (b)
1099 1099
3
2
⇒ ∑ a 2r = ∑ (a 2r − 1 + d ) (c)
r=1 r=1 3
99 99 4
(d) −
10 10
⇒ ∑ a 2r = ∑ a 2r − 1 + 1099 d 3
r=1 r=1
Sol. Let the GP be a, ar, ar2, ... (0 < r < 1).
⇒ 10100 = 1099 + 1099 d
According to the question,
10100 − 1099 1099 (10 − 1)
⇒ d= = =9 a
1099 1099 = 33 + 3 ⋅ 3 − 9 …(i)
1− r
Hence, (b) is the correct answer.
f ′ (x ) = 3x 2 + 3 > 0. So f (x ) is monotonically
Ex 12. Let S (n) denotes the sum of first n terms of an increasing and f (3) is the greatest value in [ −2, 3 ]
S (3n) Also, f ′ (0) = 3
AP. Then, the value of is
S (2n) − S ( n) ∴ a − ar = 3 ...(ii)
(a) 3 (b)
1 On solving Eqs. (i) and (ii), i.e. a = 27(1 − r) and
3 2 4
a(1 − r) = 3, we get r = , but r < 1
(c) 9 (d) None of these 3 3
3n Hence, (c) is the correct answer.
Sol. S (3n) = [ 2a + (3n − 1)d ]
2
2n
Ex 15. The value of
S (2n) = [ 2a + (2n − 1)d ] n
1
n
2
∑ a + rx + a + ( r − 1) x
is
S (n) = [ 2a + (n − 1) d ] r =1
2 n
n (a)
∴ S (2n) − S (n) = [ 2{2a + (2n − 1) d} − 2a − (n − 1)d ] a + a + nx
2
n
= [ 2a + (3n − 1)d ] =
S (3n) a + nx + a
2 3 (b)
S (3n) x
∴ =3 n( a + nx − a )
S (2n) − S (n) (c)
Hence, (a) is the correct answer. x
(d) None of the above
Ex 13. The lengths of three unequal edges of a 1
Sol. Let tr =
rectangular solid block are in GP. The volume a + rx + a + (r − 1)x
of the block is 216 cm 3 and the total surface a + rx − a + (r − 1)x
area is 252 cm 2 . The length of the longest edge =
a + rx − a − (r − 1)x
is
1
(a) 12 cm (b) 6 cm (c) 18 cm (d) 3 cm = [ a + rx − a + (r − 1)x ]
x
a 1
Sol. Let the edges be , a and ar, where r > 1. ∴ t1 + t2 + ... + tn = [{ a + x − a} + { a + 2x
r x
According to the question, − a + x } + ... + { a + nx − a + (n − 1)x }]
a
⋅ a ⋅ ar = 216 ⇒ a3 = 216 ⇒ a = 6 1
r = [ a + nx − a ]
x
a a a + nx − a
and 2 ⋅ a + a ⋅ ar + ⋅ ar = 252 =
r r
x ( a + nx + a )
1 7 1
∴ + r + 1= ⇒ r= ,2 n
r 2 2 =
a + a + nx
Since, a = 6 and r = 2, so the longest side = ar = 12 63
Hence, (a) is the correct answer. Hence, (a) is the correct answer.
3 Sol. On taking AM and GM of 7 numbers
n k
Ex 16. If ∑ ∑ m 2 = an 4 + bn 3 + cn 2 + dn + e, then a a b b b
, , , , ,
c c
, , we get
k = 1 m = 1 2 2 3 3 3 2 2
Objective Mathematics Vol. 1
1 1 a b c 1
(a) a = (b) b = 2⋅ + 3⋅ + 2⋅ a 2 b 3 c 2 7
2 3 2 ≥
12 2
7 2 3 2
1
(c) d = (d) e = 1 1
5 3 a2b3c2 7
⇒ ≥
n k n
k (k + 1)(2k + 1) 7 223322
Sol. ∑ ∑ m2 = ∑
k = 1 m = 1
6 37 a2b3c2
k=1 ⇒ ≥ 2 3 2
n 7 7
2 ⋅3 ⋅2
1
6 k∑
= (2k 3 + 3k 2 + k ) 310 ⋅ 24
=1
⇒ a2b3c2 ≤
77
2
1 n(n + 1) 1 n(n + 1)(2n + 1) 1 n(n + 1) ∴ Greatest value of a b c2 3 2
= ⋅ + +
3 2 2 6 6 2 310 ⋅ 24
=
1 1 1
a = Coefficient of n4 = ⋅ = 77
3 4 12 Hence, (a) is the correct answer.
Hence, (a) is the correct answer.
1 1 1
Ex 20. If a, b and c are digits, then the rational number
log + +
5 4
+K represented by 0. cababab..., is
8 16
Ex 17. The value of (0. 2) is 99c + ba 99c + 10a + b
1 (a) (b)
(a) 1 (b) 2 (c) (d) 4 990 99
2 99c + 10a + b
1 (c) (d) None of these
990
1 1 1 1
Sol. We have, + + +K= 4 = = 2−1 Sol. Let R = 0. cababab K
4 8 16 1 2
1− = 0. c + 0.0ab + 0.000ab + K
2
1 1 1 2−1 c ab ab
log + +
54
+ K
1
log
5 = + + +K
∴ (0.2) 8 16
= 10 103 105
5
−1
ab
log 5 2 2 c 103 c ab 102
= (5−1 ) 1/ 2 = 52 log 5 2 = 5log 5 (2) = 22 = 4 = + = + ×
10 1 − 1 10 103 99
Hence, (d) is the correct answer.
10
2
c ab 99c + ab 99c + 10a + b
Ex 18. If x + y + z =1 and x, y and z are positive = + = =
10 990 990 990
numbers such that (1 − x )(1 − y)(1 − z ) ≥ kxyz,
Hence, (c) is the correct answer.
then k is equal to
(a) 2 (b) 4 Aliter
(c) 8 (d) 16 Let R = 0. cababab...
Sol. Q AM ≥ GM ⇒ 10R = c. ababab… …(i)
y+ z and 103 R = cab. ababab... …(ii)
∴ ≥ yz …(i)
2 From Eqs. (i) and (ii), we get
z+ x
≥ zx …(ii) 103 R − 10R = cab − c
2 c ab − c
x+ y ⇒ R=
and ≥ xy …(iii) 990
2
100c + 10a + b − c
On multiplying Eqs. (i), (ii) and (iii), we get ⇒ R=
( y + z)(z + x )(x + y) 990
≥ xyz 99c + 10a + b
8 ⇒ R=
990
or (1 − x )(1 − y)(1 − z) ≥ 8xyz
Hence, (c) is the correct answer.
Hence, (c) is the correct answer.
1 n
Ex 21. If f (n) = + , where [ x ] denotes the
Ex 19. If a + b + c = 3 and a > 0, b > 0, c > 0, then the 2 100
greatest value of a 2 b 3 c 2 is 100
(a)
3 ⋅210 4
(b)
3 ⋅2
9 4 integral part of x, then the value of ∑ f (n) is
n =1
77 77
(a) 50 (b) 51
3 ⋅ 24
8
64 (c) (d) None of these (c) 1 (d) None of these
77
1 1
Sol. f (1) =
M
+
2 100
M
= 0,...,
M
Ex 24. ABCD is a square of length a, a ∈ N , a >1.
Let L1 , L2 , L3 , ... , La − 1 be points on BC such
that BL1 = L1 L2 = L2 L3 = K = 1 and
3
Sequence and Series
1 49 99
f (49) = + = =0 M 1 , M 2 , M 3 , K , M a − 1 be points on CD such
2 100 100
that CM 1 = M 1 M 2 = M 2 M 3 = K = 1. Then,
1 50 1 51 a −1
f (50) = + = 1, f (51) = + = 1,
2 100 2 100 ∑ ( AL2n + Ln M n2 ) is equal to
M M M n =1
1 100 1
f (100) = + =1 (a) a( a − 1) 2
2 100 2
100
1
∴ ∑ f (n) = (10 44
+ 0 + K + 0) + (1 + 1 + K + 1) = 51 (b) a( a − 1)( 4a − 1)
n= 1
2443 1442443 2
49 terms 51terms
1
Hence, (b) is the correct answer. (c) ( a − 1)( 2a − 1)( 4a − 1)
2
Ex 22. Ar , r =1, 2, 3, ..., n are n points on the parabola (d) None of the above
y 2 = 4x in the first quadrant. If Ar = ( x r , yr ), Sol. a
A B
where x1 , x 2 , x 3 , ..., x n are in GP and x1 = 1,
L1
x 2 = 2, then yn is equal to L2
n+1
(a) 2 2 (b) 2n + 1 (c) ( 2 ) n + 1 (d) 2n/ 2
La – 1
Sol. y = 2 x , being in the first quadrant.
D Ma –1 M 2 M2 C
The sequence of x-coordinates is 1, 2, 4, 8, ...
∴The sequence of y -coordinates is AL21 + L1M 12 = (a + 12 ) + {(a − 1)2 + 12}
2
2, 2 2 , 2 4 , 2 8 , ... AL22 + L 2M 22 = (a2 + 22 ) + {(a − 2)2 + 22}
where, the common ratio is 2. M M
n+ 1 AL2a − 1 + La − 1M a2 − 1 = a2 + (a − 1)2 + {12 + (a − 1)2}
∴ yn = 2 ⋅ ( 2 )n − 1 = 2 2 ∴ Required sum
Hence, (a) is the correct answer. = (a − 1)a2 + {12 + 22 + K + (a − 1)2}
+ 2{12 + 22 + K + (a − 1)2}
Ex 23. In the given square, a diagonal is drawn and
(a − 1)a(2a − 1)
parallel line segments joining points on the = (a − 1)a2 + 3 ⋅
6
adjacent sides are drawn on both sides of the
2a − 1 a(a − 1)(4 a − 1)
diagonal. The length of the diagonal is n 2 cm. = a(a − 1) a + =
2 2
If the distance between consecutive line Hence, (b) is the correct answer.
1
segments is cm, then the sum of the lengths
2 Ex 25. ABC is a right angled triangle in which
of all possible line segments and the diagonal is ∠B = 90° and BC = a. If n points L1 , L2 , ..., Ln
on AB are such that AB is divided in n +1equal
parts and L1 M 1 , L2 M 2 , ..., Ln M n are line
n/ √2 segments parallel to BC and M 1 , M 2 , ..., M n
(n – 1) . 1 are on AC, then the sum of the lengths of
√2 L1 M 1 , L2 M 2 , K , Ln M n is
(n – 2). 1 a( n + 1) a( n − 1)
√2 (a) (b)
(a) n( n + 1) 2 cm (b) n cm2 2 2
an
(c) n( n + 2) cm (d) n 2 2 cm (c) (d) None of these
2
Sol. Lengths of line segments on one side of the diagonal are Sol. A
2, 2 2, 3 2, ..., (n − 1) 2.
∴Required sum L1 M1
L2 M2
= 2{ 2 + 2 2 + 3 2 + K + (n − 1) 2} + n 2
= 2 2{1 + 2 + 3 + K + (n − 1)} + n 2
n(n − 1)
= 2 2⋅ + n 2 = n2 2
2 Ln Mn
Hence, (d) is the correct answer. 65
B a C
3
AL1 L1M 1
Q = Ex 28. If x > 1, y > 1 and z > 1 are in GP, then
AB BC
1 L1M 1 1 1 1
∴ = , and are in
1 + ln x 1 + ln y 1 + ln z
Objective Mathematics Vol. 1
n+1 a
a AL 2 L 2M 2 (a) AP (b) GP
⇒ L1M 1 = and =
n+1 AB BC (c) HP (d) None of these
2 L 2M 2
⇒ = Sol. Since, x , y and z are in GP.
n+1 a
∴ y2 = xz
2a
⇒ L 2M 2 = and so on. Taking log on both sides, we get
n+1
2ln y = ln x + ln z
∴ Required sum ⇒ 2 + 2 ln y = (1 + ln x ) + (1 + ln z)
a 2a 3a na
= + + +K+ ⇒ 2(1 + ln y) = (1 + ln x ) + (1 + ln z)
n+1 n+1 n+1 n+1 ∴ 1 + ln x , 1 + ln y and 1 + ln z are in AP.
a a n(n + 1) an 1 1 1
= (1 + 2 + K + n) = ⋅ = ⇒ , and are in HP.
n+1 n+1 2 2 1 + ln x 1 + ln y 1 + ln z
Hence, (c) is the correct answer. Hence, (c) is the correct answer.
Ex 26. If a, b, c and d are in HP, then ab + bc + cd is Ex 29. If a, a1 , a 2 , a 3 , ..., a 2n , b are in AP and
equal to a, g1 , g 2 , ..., g 2n , b are in GP and h is the HM of
(a) ad (b) 2ad a and b, then
(c) 3ad (d) None of these a1 + a 2n a 2 + a 2n − 1 an + an + 1
1 1 1 1 + +K + is
Sol. Since, a, b, c and d are in HP, therefore , , and are g1 g 2n g 2 g 2n − 1 gn gn + 1
a b c d
in AP. equal to
1 1 1 1 1 1 n 2n
∴ − = − = − =k [say] (a) 2nh (b) (c) nh (d)
b a c b d c h h
1 1
⇒ − = k , a − b = kab …(i) Sol. Since, a, a1 , a2 , a3 , K , a2 n, b are in AP.
b a
∴ a + b = a1 + a2n = a2 + a2 n − 1 = K = an + an + 1
1 1
− = k , b − c = kbc …(ii) Also, a, g1 , g2 , ..., g2n , b are in GP.
c b
∴ ab = g1g2 n = g2 g2 n − 1 = K = gn gn + 1
1 1
and − = k , c − d = kcd …(iii) Now, the given expression
d c
a1 + a 2 n a 2 + a 2 n − 1 an + an + 1
On adding Eqs. (i), (ii) and (iii), we get + +K+
1 1 g1g 2 n g2 g2 n − 1 gn gn + 1
− = 3k, a − d = k (ab + bc + cd )
d a n(a + b) 2n 2ab
= = Q h = a + b
or a − d = 3kad, a − d = k (ab + bc + cd ) ab h
∴ 3ad = (ab + bc + cd ) Hence, (d) is the correct answer.
Hence, (c) is the correct answer.
a b Ex 30. If 0.272727..., x and 0.727272... are in HP, then
Ex 27. If a, b and c are in HP, then , and x must be
b+c c+a
c (a) rational (b) integer
are in (c) irrational (d) None of these
a+b
Sol. Let A = 0.272727...
(a) AP (b) GP
⇒ 102 A = 27.2727...
(c) HP (d) None of these
27 3
⇒ A= =
Sol. Q a, b and c are in HP. 99 11
1 1 1 8
∴ , and are in AP. Similarly, 0.727272K =
a b c 11
a+ b+ c a+ b+ c a+ b+ c Since, 0.272727..., x and 0.727272 ... are in HP.
⇒ , , are in AP 3 8
a b c ⇒ , x , are in HP.
b+ c c+ a a+ b 11 11
⇒1+ ,1+ ,1+ are in AP
a b c 3 8
2⋅ ⋅
b+ c c+ a a+ b 48
⇒ , , are in AP. Now, x = 11 11 =
a b c 3 8 121
a b c +
⇒ , , are in HP. 11 11
b+ c c+ a a+ b ∴ x is rational.
66
Hence, (c) is the correct answer. Hence, (a) is the correct answer.
Ex 31. The arithmetic mean, geometric mean and
harmonic mean between two positive real
quantities are themselves in
Sol. Since, a, b and c are in AP.
∴ 2b = a + c
Since, a, mb and c are in GP.
…(i)
3
Sequence and Series
∴ mb = ac ⇒ m2b2 = ac …(ii)
(a) AP (b) GP
(c) HP (d) None of these On dividing Eq. (i) by Eq. (ii), we get
2b a+ c
Sol. Relation between arithmetic mean (A), geometric mean =
m2b2 ac
(G ) and harmonic mean (H ) is given by G 2 = AH or 2 1 1
G = AH . Clearly, A , G and H are in GP. i.e. = +
m2b a c
Hence, (b) is the correct answer.
⇒ a, m2b and c are in HP.
Ex 32. A person purchases 1 kg tomatoes from each of Hence, (a) is the correct answer.
the 4 places at the rate of 1 kg, 2 kg, 3 kg and
4 kg per rupee respectively. On the average, he Ex 35. The AM, HM and GM between two numbers
144
has purchased x kg tomatoes per rupee, then the are , 15 and 12 but not necessarily in this
value of x is 15
(a) 2 (b) 2.5 order. Then, HM, GM and AM respectively are
144 144
(c) 1.92 (d) None of these (a) , 12, 15 (b) , 15, 12
15 15
Sol. Since, we are given rate per rupees, harmonic mean will 144 144
gives the correct answer. (c) 15, 12, (d) 12, 15,
15 15
4 4 × 12
∴ HM = = Sol. Since, A > G > H , i.e. H < G < A
1 1 1 1 12 + 6 + 4 + 3
+ + + 144
1 2 3 4 ∴ Required numbers are , 12, 15, respectively.
48 15
= = 1.92 kg per rupee
25 Hence, (a) is the correct answer.
Hence, (c) is the correct answer.
Ex 36. Given that α, γ are roots of the equation
Ex 33. If m is the root of the equation Ax 2 − 4x + 1 = 0 and β, δ are the roots of the
(1 − ab) x 2 − ( a 2 + b 2 ) x − (1 + ab) = 0 and m equation Bx 2 − 6x + 1 = 0, then values of A and
B such that α, β, γ and δ are in HP, are
harmonic means are inserted between a and b,
(a) A = 3, B = 8 (b) A = − 3, B = 8
then the difference between the last and the
first of the means equals (c) A = 3, B = − 8 (d) None of these
(a) ab ( a − b ) (b)a ( b − a ) 1 1 1 1
Sol. α , β, γ and δ are in HP ⇒ , , and are in AP.
(c) ab ( b − a ) (d) b − a α β γ δ
Let d be the common difference of this AP.
Sol. According to the question,
Now, α and γ are roots of Ax 2 − 4 x + 1 = 0.
(1 − ab)m2 − (a2 + b2 )m − (1 + ab) = 0
α + γ 4/ A 1 1
⇒ m (a2 + b2 ) + (m2 + 1)ab = m2 − 1 …(i) ∴ = =4 ⇒ + =4
αγ 1/ A α γ
Now, H 1 = First HM between a and b 1 1 1
(m + 1)ab (m + 1)ab i.e. + + 2d = 4 ⇒ +d=2 …(i)
= and H m = α α α
a + mb b + ma Since, β and δ are roots of Bx − 6x + 1 = 0.
2
1 1 β + δ 1 1 6/ B
∴ H m − H 1 = (m + 1) ab − ∴ = + = =6
b + ma a + mb βδ β δ 1/ B
(m − 1) (b − a) 1 1
+ d + + 3d = 6
= (m + 1) ab
or
α α
(b + ma)(a + mb) 1
(m2 − 1) ab (b − a) ⇒ + 2d = 3 …(ii)
= α
m (a2 + b2 ) + (m2 + 1) ab From Eqs. (i) and (ii), we get
(m2 − 1) ab (b − a) 1
= 1, d = 1
= [from Eq. (i)]
m2 − 1 α
1 1 1 1
= ab (b − a) ∴ = 1, = 2, = 3 and =4
α β γ δ
Hence, (c) is the correct answer.
1
Since, =A ⇒ A=3
Ex 34. If a, b and c are in AP and a, mb, c are in GP, αγ
1
then a, m 2 b and c are in Also, =B ⇒ B=8
βδ
(a) HP (b) GP Thus, A = 3 and B = 8
(c) AP (d) None of these 67
Hence, (a) is the correct answer.
3 Ex 37. If log (x + z ) + log (x + z − 2 y) = 2 log (x − z ),
then x, y and z are in
Sol. We have, ai = ai − 1 + 1
⇒ ai2 = ai2− 1 + 2ai − 1 + 1
Putting i = 1, 2, 3, ..., n + 1, we get
Objective Mathematics Vol. 1
(a) GP (b) AP
(c) HP (d) None of these a12 = 0
Sol. Given, log(x + z) + log(x + z − 2 y) = 2 log(x − z) a22 = a12 + 2a1 + 1
⇒ log[(x + z − 2 y) (x + z)] = log(x − z)2 a32 = a22 + 2a2 + 1
⇒ (x + z − 2 y) (x + z) = (x − z)2 M M M
an2 = an2 − 1 + 2an − 1 + 1
⇒ x 2 + 2xz + z2 − 2 yz − 2xy = x 2 + z2 − 2xz
⇒ 4 xz = 2 yz + 2xy ⇒ 2xz = y (x + z) an2 + 1 = an2 + 2an + 1
2xz n+ 1 n n
∴ y=
x+z On adding, we get ∑ ai2 = ∑ ai2 + 2 ∑ ai + n
i=1 i=1 i=1
So, x , y and z are in HP. n
Hence, (c) is the correct answer. ⇒ 2 ∑ ai = − n + an2 + 1 ≥ − n
i=1
Ex 38. If H1 , H 2 , ..., H n are n harmonic means a1 + a2 + ... + an 1
⇒ ≥−
between a and b ( ≠ a ), then value of n 2
H1 + a H n + b According to the question,
+ is a + a2 + ... + an
H1 − a H n − b x= 1 ⇒ x≥−
1
n 2
(a) n + 1 (b) n − 1 Hence, (b) is the correct answer.
(c) 2n (d) 2n + 3
Sol. As a, H 1 , H 2 , ..., H n , b are in HP. Ex 40. The sum of first n terms of the series
1 1 1
⇒ , , , ...,
1 1
, are in AP.
12 + 2 ⋅ 2 2 + 3 2 + 2 ⋅ 4 2 + 5 2 + 2 ⋅ 6 2 + ... is
a H1 H2 Hn b n( n +1) 2
Let d be the common difference of an AP, then , when n is even. When n is odd, the
2
1 1
= + (n + 1) d sum is
b a
1 a−b n 2 ( n + 1) n( n + 1) 2
⇒ d= ⋅ (a) (b)
n + 1 ab 2 2
Thus,
1 1
= + d and
1 1
= −d n( n + 1) 2 n( n + 1)
H1 a Hn b (c) (d)
a b 2 2
⇒ = 1 + ad and = 1 − bd
H1 Hn Sol. When n is odd, last term will be n2.
a b Then, the sum is
1+ 1+
H1 + a Hn + b H1 Hn 12 + 2 ⋅ 22 + 32 + 2 ⋅ 4 2 + 52 + 2 ⋅ 62 + ... + 2(n − 1)2 + n2
Now, + = +
H1 − a Hn − b 1 − a 1−
b (n − 1)n2 n(n + 1)2
H1 Hn = + n2 replacing n by n − 1 in
2 2
1 + 1 + ad 1 + 1 − bd n3 − n2 + 2n2
= + =
1 − 1 − ad 1 − 1 + bd 2
2 + ad 2 − bd 2a − abd − 2b − abd n3 + n2 n2 (n + 1)
= + = = =
− ad bd abd 2 2
2[(a − b) − abd ] Hence, (a) is the correct answer.
=
abd
Ex 41. If three positive real numbers a, b, c are in AP
2[(n + 1) dab − abd ]
= = 2n such that abc = 4, then the minimum possible
abd
value of b is
Hence, (c) is the correct answer.
(a) 23/ 2 (b) 22/ 3 (c) 21/ 3 (d) 25/ 2
Ex 39. If a1 = 0 and a1 , a 2 , a 3 , ..., a n are real numbers Sol. Let d be the common difference of an AP, then
such that a i = a i − 1 + 1 for all i, then the AM 4 = abc = (b − d ) b(b + d ) = b(b2 − d 2 )
of the numbers a1 , a 2 , ..., a n has value x, where ⇒ b3 = 4 + bd 2 ≥ 4 [Q b > 0, d 2 ≥ 0 ]
1 1 ⇒ b≥2 2/ 3
(a) x ≤ − (b) x ≥ −
2 2 Thus, minimum possible value of bis 22/ 3, that is the case
1 when d = 0.
68 (c) x < − (d) None of these
2 Hence, (b) is the correct answer.
Type 2. More than One Correct Option
Ex 42. If a, b and c are in HP, then
3
Sequence and Series
a b c Sol. We have,
(a) , , are in HP
b+ c−a c+ a−b a+ b−c 1 1 1 1 1 1
a (n) = 1 + + + + K + + + K +
2 1 1 2 3 4 7 8 15
(b) = +
b b−a b−c 1 1 1
+ K+ n − 1 + n − 1 +K+ n
b b b 2 2 +1 (2 ) − 1
(c) a − , , c − are in GP
2 2 2 2 4 8 2n − 1
a b c <1+
+ + + ... + n − 1 = n
(d) , , are in HP 2 4 8 2
b+ c c+ a a+ b Thus, a (n) < n ⇒ a (2n) < 2n
1 1 1 1 1
Sol. Since, a, b and c are in HP. Now, a(n) = 1 + + + + + ... +
2 3 4 5 8
1 1 1
So, , , are in AP. 1 1 1
a b c + K + n− 1 + K + n − n
a+ b+ c a+ b+ c a+ b+ c 2 +1 2 2
⇒ , , are in AP.
a b c 1 2 4 2n − 1 1
b+ c c+ a a+ b > 1 + + + + K + n − n
⇒ , , are in AP. 2 4 8 2 2
a b c 1 1 1 n
≥ + +K+ =
[on subtracting 1 from each term]
124422444
4 32 2
b+ c c+ a a+ b
⇒ − 1, − 1, − 1 are in AP. n times
a b c n
b + c − a c + a − b a + b− c Thus, a(n) > ⇒ a (2n) > n
⇒ , , are in AP. 2
a b c Hence, (a), (b), (c) and (d) are the correct answers.
a b c
Thus, , , are in HP. Ex 44. If a, b, c are in AP and a 2 , b 2 , c 2 are in HP,
b+ c c+ a a+ b
a b c then
and , , are also in HP.
b+ c−a c+ a−b a+ b−c (a) a = b = c
1
Also we have, b =
2ac (b) a, b and − c are in GP
a+ c 2
(c) a, b and c are in GP
1 1 2b − (a + c)
So, + = 1
(d) − a, b and c are in GP
b − a b − c (b − a) (b − c)
2
2b − (a + c)
= 2
b − b(a + c) + ac Sol. From the given condition,
2a2c2
2b −
2ac 2b = a + c and b2 =
b a2 + c2
= 2
b − b⋅
2 2ac
+ ac a + c 2a2c2
On eliminating b, we get = 2
b 2 a + c2
2 b2 − ac 2 ⇒ (a2 + c2 )2 + 2ac (a2 + c2 ) − 8a2c2 = 0
= . 2 =
b b − ac b ⇒ (a2 + c2 + 4 ac) (a2 + c2 − 2ac) = 0
b b b b2 ⇒ [(a + c)2 + 2ac ] (a − c)2 = 0
Lastly, a − c − = ac − (a + c) +
2 2 2 4 1
⇒ 4 b2 + ac (a − c)2 = 0
b 2ac b2 b2 2
= ac − . + =
2 b 4 4 1
⇒ a − c = 0 or b2 = −ac
b b b 2
∴ a − , , c − are in GP.
2 2 2 If a = c, we get a = b = c and a, b, c are in GP.
Hence, (a), (b), (c) and (d) are the correct answers. 1 1
If b2 = − ac, then either a, b, − c are in GP
2 2
Ex 43. For a positive integer n, let 1
or − a, b, c are in GP.
2
1 1 1 1 Hence, (a), (b), (c) and (d) are the correct answers.
a ( n) = 1 + + + + ... + n . Then,
2 3 4 (2 ) − 1
Ex 45. Let a n = (111… 1), then
n 1424 3
(a) a ( n ) < n (b) a ( n ) > n times
2
(a) a 912 is not prime (b) a 951 is not prime
(c) a ( 2n ) > n (d) a ( 2n ) < 2n 69
(c) a 480 is not prime (d) a 91 is not prime
3 Sol. As a 912 , a 951 and a480 are divisible by 3, none of them is
prime. For a 91 , we have
1
a 91 = (123
1
99...9) = (1091 − 1)
Ex 47. The sum of the series
1 2 1
tan −1 + tan −1 + tan −1
2 9 8
Objective Mathematics Vol. 1
9 9
91 times
2 1
1
= [(107 )13 − 1] + tan −1 + tan −1 + ... is
9 25 18
(107 )13 − 1 107 − 1 1
= (a) tan −1 ( 3) (b) cot −1
10 − 1 10 − 1
7
3
= [(107 )12 + (107 )11 + ...+ 107 + 1] 1
(c) tan −1 (d) None of these
× [106 + 105 + ...+ 10 + 1] 3
⇒ a 91 is not prime. 2
2
Hence, (a), (b), (c) and (d) are the correct answers. Sol. Let Tr = tan −1 , then
r + 1
−1
n n
Ex 46. If a, b, c and d are four unequal positive 2 r+ 2−r
numbers which are in AP, then
Sn = ∑ tan = ∑
r + 2r + 1 r = 1
2
tan −1
1 + r(r + 2)
r=1
1 1 1 1 1 1 1 1 n
+ = + + < +
(a)
a d b c
(b)
a d b c
= ∑ (tan −1 (r + 2) − tan −1 r)
r=1
1 1 1 1 1 1 4 S n = tan −1 (n + 2) + tan −1 (n + 1) − tan −1 2 − tan −1 1
(c) + > + (d) + >
a d b c b c a+d 3n2 + 7n
= tan −1 2
Sol. Let b = a + p, c = a + 2 p, d = a + 3 p n + 9n + 10
1 1 1 1 1
+ + ∴ S ∞ = tan −1 (3) = cot −1
a d = a a + 3p 3
1 1 1 1 Hence, (a) and (b) are the correct answers.
+ +
b c a + p a + 2p
(a + p) (a + 2 p) Ex 48. If a, b and c are three unequal positive
=
a (a + 3 p) quantities in HP, then
a2 + 3ap + 2 p2 (a) a100 + c100 > 2b100 (b) a 3 + c 3 > 2b 3
= >1
a2 + 3ap (c) a 5 + c 5 > 2b 5 (d) a 2 + c 2 > 2b 2
1 1 1 1
⇒ + > + Sol. Since, b is the HM of a and c and their GM = ac
a d b c
1 1 1 1 ∴ GM > HM
∴ + (a + d ) = + (a + a + 3 p) ⇒ ac > b
b c a + p a + 2 p
and AM of a n and c n > GM of an and cn
(2a + 3 p)2
= an + cn
a + 3ap + 2 p2
2
> a nc n
2
p2
=4+ 2 >4 a n + c n > 2( ac )n > 2bn
a + 3ap + 2 p2 Put n = 100, 3, 5, 2
Hence, (c) and (d) are the correct answers. Hence, (a), (b), (c) and (d) are the correct answers.
Type 3. Assertion and Reason
Directions (Ex. Nos. 49-52) In the following Ex 49. Statement I If a + 2b + 3c = 1 and
examples, each example contains Statement I a > 0, b > 0, c > 0, then the greatest value of
(Assertion) and Statement II (Reason ). Each example 1
has 4 choices (a), (b), (c) and (d) out of which only one is a 3 b 2 c is .
5184
correct. The choices are Statement II There exists an AP such that
(a) Statement I is true, Statement II is true; Statement II sum upto its n terms is given by
is a correct explanation for Statement I S n = an 3 + bn 2 + cn + d .
(b) Statement I is true, Statement II is true; Statement II
Sol. We can use AM ≥ GM to have,
is not a correct explanation for Statement I 3a c
(c) Statement I is true, Statement II is false + 2b + 1/ 6
3 1/ 3 a3 2
(d) Statement I is false, Statement II is true ≥ ⋅ b ⋅ 3c
6 27
70
Sol. Let Tk + 1 = ark and T ′ k + 1 = brk
3
6
1 a3b2c 9 1
⇒ ≥ or a3b2 c ≤ 6 ⇒ a3b2c ≤
6 9 6 5184 Q T ″ k + 1 = ark + brk = (a + b)rk
Also, sum upto n terms of an AP is always a quadratic ∴T ″ k + 1 is general term of a GP with common
Sequence and Series
expression in n. Clearly, Statement I is true and ratio r.
Statement II is false. Hence, (a) is the correct answer.
Hence, (c) is the correct answer.
Ex 50. Statement I If a, b and c are non-zero real Ex 52. Statement I If one AM, A and two GM’s, p
numbers such that 3( a 2 + b 2 + c 2 + 1) and q are inserted between any two numbers,
= 2( a + b + c + ab + bc + ca ), then a, b and c are then p 3 + q 3 = 2 Apq.
in AP as well as in GP. Statement II If x, y and z are in GP, then
Statement II A series is an AP as well as a y 2 = xz.
GP, if all the terms in the series are equal and
non-zero. Sol. Statement I a, A and b are in AP.
Sol. We have, 3(a + b + c + 1)
2 2 2 ⇒ 2A = a + b …(i)
Since, a, p, q and b are in GP.
− 2(a + b + c + ab + bc + ca) = 0
∴ pq = ab …(ii)
⇒ (a − 1)2 + (b − 1)2 + (c − 1)2 + (a − b)2 + (b − c)2 Let common ratio of GP be r.
+ (c − a)2 = 0 ∴ b = ar3
⇒ a= b= c=1 1/ 3
b
Hence, (a) is the correct answer. ⇒ r=
a
Ex 51. Statement I Given, 1, 2, 4, 8,... is a GP; Q p = ar
1/ 3
4, 8, 16, 32,... is a GP, then 1 + 4, 2 + 8, 4 + 16, b
⇒ p=a ⇒ p3 = a2b …(iii)
a
8 + 32, ... is also a GP. 2/ 3
b
Statement II If general term of a GP with and q = ar2 ⇒ q = a
a
common ratio r be Tk + 1 and general term of
⇒ q3 = ab2 …(iv)
another GP with common ratio r be T ′ k + 1 ,
From Eqs. (i), (ii), (iii) and (iv), we get
then the series whose general term p3 + q3 = 2 Apq
T ″ k + 1 = Tk + 1 + T ′ k + 1 is also a GP with Statement II is obviously true.
common ratio r. Hence, (b) is the correct answer.
Type 4. Linked Comprehension Based Questions
Passage I (Ex. Nos. 53-55) We know that, if ∴ln E = ( y − z) ln x + (z − x ) ln y + (x − y) ln z
1 1 1 = (q − r) d ln x + (r − p) d ln y + ( p − q) d ln z
a1, a2, K, an are in HP, then , , ..., are in AP and = d [ p (ln z − ln y) + q (ln x − ln z) + r (ln y − ln x )]
a1 a2 an
= d ln r [ p(r − q) + q ( p − r) + r(q − p)] = 0
vice-versa. If a1, a2, ..., an are in AP with common
⇒ E =1
difference d, then for any b (b > 0), the numbers
Hence, (d) is the correct answer.
ba1, ba 2 , ba 3 , ..., ban are in GP with common ratio bd . If
a1, a2, ..., an are positive and in GP with common ratio (r), Ex 54. If a, b, c, d are in GP and a x = b y = c z = d v ,
then for any base b (b > 0), logb a1, logb a2, ..., logb an are then x, y, z, v are in
in AP with common difference logb r. (a) AP (b) GP
(c) HP (d) None of these
Ex 53. If x, y and z are respectively the pth, qth and rth Sol. Let a, b, c, d be in GP with common ratio r then ln a,
terms of an AP as well as of a GP, then ln b, ln c, ln d are in AP with common difference ln r.
x y − z ⋅ y z − x ⋅ z x − y is equal to Also, ax = b y = cz = dv
(a) p (b) q (c) r (d) 1 ⇒ x ln a = y ln b = z ln c = v ln d
⇒ x ln a = y (ln a + ln r) = z (ln a + 2 ln r)
Sol. Let the base be taken as e. Again, let, x , y and z be terms = v (ln a + 3 ln r)
of a GP with common ratio R, then ln x , ln y, ln z are x−z x− y y−v y−z
⇒ = 2⋅ and = 2⋅
terms of AP with common difference ln R. Also, let z y v z
x , y and z be terms of AP with common difference d. 1 1 2 1 1 2
⇒ + = and + =
Now, x − y = ( p − q) d etc., and x z y y v z
ln x − ln y = ( p − q) ln r etc. So, x , y, z and v are in HP.
Let E = x y − z ⋅ yz − x ⋅ zx − y Hence, (c) is the correct answer. 71
3 Ex 55. If p, q and r are in AP, then pth, qth and rth
terms of any GP are in
(a) AP (b) GP (c) HP (d) AGP
Ex 60. The numbers 2 A171 , G52 + 1 and 2 A172 are in
(a) AP
Objective Mathematics Vol. 1
(b) GP
Sol. Since, p, q and r are in AP. (c) HP
∴b p , bq , br are in GP for any b (b > 0) (d) AGP
⇒ b p − 1 , bq − 1 , br − 1 are in GP. Sol. We have, A171 + A172 = − 2 + 1027 = 1025
p −1 q −1 r−1 2 A171 + 2 A172
⇒ ab , ab , ab are in GP for any a (a ≠ 0) ∴ = 1025
2
. So, pth, qth and rth terms of any GP are in GP.
Also, G5 = 1 × 25 = 32
Hence, (b) is the correct answer.
Now, G52 = 1024
Passage II (Ex. Nos. 56-60) Let A1, A2, A3, ..., A m be
i.e. G52 + 1 = 1025
arithmetic means between −2 and 1027 and
G1, G2, G3, ..., Gn be geometric means between 1 and So, 2 A171 , G52 + 1 and 2 A172 are in AP.
1024. Product of geometric means is 2 45 and sum of Hence, (a) is the correct answer.
arithmetic means is 1025 × 171.
Passage III (Ex. Nos. 61-63) There are two sets A
and B each of which consists of three numbers in AP
Ex 56. The value of n is
whose sum is 15 and where D and d are the common
(a) 7 (b) 9 p 7
difference such that D − d = 1. If = , where p and q
(c) 11 (d) None of these q 8
Sol. G1 G2 … Gn = ( 1 × 1024 )n = 25n are the product of the numbers respectively and d > 0, in
the two sets.
∴ 25n = 245 ⇒ n=9
Hence, (b) is the correct answer. Ex 61. The value of p is
(a) 100 (b) 120
Ex 57. The value of m is (c) 105 (d) 110
(a) 340 (b) 342 (c) 344 (d) 346
Sol. Q A1 + A2 + A3 + ... + Am− 1 + Am Ex 62. The value of q is
= 1025 × 171 (a) 100 (b) 120
− 2 + 1027 (c) 105 (d) 110
∴ m = 1025 × 171
2
⇒ m = 342 Ex 63. The value of D + d is
Hence, (b) is the correct answer. (a) 1 (b) 2
(c) 3 (d) 4
Ex 58. The value of G1 + G2 + G3 + K + Gn is Sol. (Ex. Nos. 61-63) Let numbers in set A be a − D, a, a + D
(a) 1022 (b) 2044 and in set B be b − d , b, b + d.
(c) 512 (d) None of these 3a = 3b = 15
⇒ a=b=5
Sol. Q n = 9
1 Now, set A = {5 − D , 5, 5 + D}
∴ r = (1024 )=2 9+ 1 and set B = {5 − d , 5, 5 + d}
where, D=d+1
So, G1 = 2 andr=2
2 (29 − 1) p 5(25 − D 2 ) 7
Now, G1 + G2 + K + Gn = ∴ = =
2−1 q 5(25 − d 2 ) 8
= 1024 − 2 = 1022 ⇒ 25 (8 − 7) = 8 (d + 1)2 − 7d 2
Hence, (a) is the correct answer. ⇒ d = − 17, 1 but d > 0
⇒ d =1
Ex 59. The common difference of the progression
So, numbers in set A are {3, 5, 7} and in set B are{4 , 5, 6}.
A1 , A3 , A5 , ..., Am − 1 is
Now, p = 3 × 5 × 7 = 105
(a) 6 (b) 3 (c) 2 (d) 1
and q = 4 × 5 × 6 = 120
Sol. Common difference of sequence A1 , A2 ,..., Am
Hence, value of D + d = 3
1027 + 2
= =3
342 + 1 61. (c)
∴Common difference of sequence A1 , A3 , A5 ,..., Am− 1 62. (b)
is 6.
Hence, (a) is the correct answer. 63. (c)
72
⇒ 2d 2 − 2d + 3a2 − a = 0
Passage IV (Ex. Nos. 64-66) Four different integers
form an increasing AP. One of these numbers is equal to
the sum of the squares of the other three numbers.
∴
1
d = [1 ± 1 + 2a − 6a2 ] …(i)
2
3
Sequence and Series
Since, d is positive integer.
Ex 64. The smallest number is ∴ 1 + 2a − 6a2 > 0
(a) − 2 (b) 0 (c) −1 (d) 2 ⇒ 6a2 − 2a − 1 < 0
1− 7 1+ 7
Ex 65. The common difference of the four numbers is ⇒ <a<
6 6
(a) 2 (b) 1
Q a is an integer.
(c) 3 (d) 4
∴ a = 0, put in Eq. (i),
Ex 66. The sum of all the four numbers is d = 1 or 0 but d > 0
∴ d =1
(a) 10 (b) 8
Now, the four numbers are − 1, 0, 1, 2 and their sum is 2.
(c) 2 (d) 6
Sol. (Ex. Nos. 64-66) Let four integers be a − d , a, a + d and
64. (c)
a + 2d, where a and d are integers and d > 0. 65. (b)
Q a + 2d = (a − d )2 + a2 + (a + d )2 66. (c)
Type 5. Match the Columns
Ex 67. Match the statements of Column I with values Sum of all two digit numbers divisible by 3
30
of Column II. = (12 + 99) = 15(111)
2
Column I Column II Sum of all two digit numbers divisible by 6
A. Suppose that F ( n + 1) p. 42 15
= (12 + 96)
2 F ( n) + 1 2
= for n = 1, 2, 3, ...
2 = 15 (54 )
and F (1) = 2 . Then, F (101) equals ∴Required sum
B. If a1, a2, a3, ..., a21 are in AP q. 1620 = 45(109) + 15(54 ) − (45) (54 ) − 15 (111) = 1620
and a3 + a5 + a11 + a17 + a19 = 10, A → r; B → p; C → s; D → q
21
then the value of ∑ ai is Ex 68. Match the statements of Column I with values
i =1
C. 10th term of the series r. 52 of Column II.
S = 1 + 5 + 13 + 29 + ... is
D. The sum of all two digit numbers s. 2045 Column I Column II
which are not divisible by 2 or 3, is
A. The arithmetic mean of two positive numbers p. 240
2F (n) + 1 1 is 6 and their geometric mean G and 77
Sol. A. F (n + 1) = = F (n) + harmonic mean H satisfyG 2 + 3H = 48, then
2 2
the product of number is
∴ F (1), F (2), F (3), ... is an AP with common
1 1 B. The sum of the series q. 32
difference . Thus, F (101) = 2 + 100 × = 52 5 11 17
2 2 + + + ... is
12 ⋅ 4 2 42 ⋅ 7 2 7 2 ⋅ 10 2
B. a1 + 2d + a1 + 4 d + a1 + 10d + a1 + 16d
+ a1 + 18d C. If the first two terms of a harmonic r. 1
= 5a1 + 50d = 5(a1 + 10d ) = 10 i.e. a1 + 10d = 2 1 1
progression are and , then the harmonic 3
21
21 2 3
Now, ∑ ai = [ 2a1 + 20d ] = 21 (a1 + 10d ) = 42 mean of the first four terms is
i=1
2
C. S = 1 + 5 + 13 + 29 + K + t10 …(i) Sol. A. a + b = 12
S = 1 + 5 + 13 + K + t9 + t10 …(ii) 6ab
∴ ab + = 48
On subtracting Eq. (ii) from Eq. (i), we get a+ b
t10 = 1 + 4 + 8 + 16 + ... upto 10 terms ab
⇒ ab + = 48
= 1 + (4 + 8 + 16 + ... upto 9 terms) = 2045 2
90 ∴ ab = 32
D. Sum of all two digit numbers = (10 + 99)
2 5 11 17
= (45) (109) B. S = 2 2 + 2 2 + 2 + ...
1 ⋅4 4 ⋅7 7 ⋅ 102
Sum of all two digit numbers divisible by 2
3⋅ 5 3 ⋅ 11 3 ⋅ 17
45 ⇒ 3S = 2 2 + 2 2 + 2 + ...
= (10 + 98) = (45) (54 ) 1 ⋅4 4 ⋅7 7 ⋅ 102
2 73
(4 − 1) ⋅ (4 + 1) (7 − 4 ) (7 + 4 ) ⇒ 3S = 1
3 ⇒ 3S =
12 ⋅ 4 2
+
4 2 ⋅ 72
+
(10 − 7) (10 + 7)
+ ...
⇒ S =
1
3
Objective Mathematics Vol. 1
1 1 1 1
72 ⋅ 102 C. HM of , , ,
2 3 4 5
4 2 − 12 72 − 4 2 102 − 72 4 240
⇒ 3S = + 2 2 + 2 + ... = =
12 ⋅ 4 2 4 ⋅7 7 ⋅ 102 1 1 1 1 77
+ + +
1 1 1 1 1 2 3 4 5
⇒ 3S = 1 − 2 + 2 − 2 + 2 − 2 + ...
4 4 7 7 10 A → q; B → r; C → p
Type 6. Single Integer Answer Type Questions
Ex 69. The largest term of the sequence x4 1
=
1 4 9 16 (x + 1) (x + 1) (x 4 + 1) (x + 1) (x 2 + 1)
2
, , , , ...is T p . Then, the value of 1
503 524 581 692 −
p is __________ . (x + 1) (x 2 + 1) (x 4 + 1)
M M M
n2 2n − 1
Sol. (7) Tn = x
500 + 3n3 n −1
dTn (500 + 3n ) ⋅ 2n − n ⋅ 9n
3 2 2 (x + 1) (x 2 + 1) ... (x 2 + 1)
⇒ = 1
dn (500 + 3n3 )2 = n− 2
n (1000 − 3n3 ) (x + 1) (x 2 + 1) ... (x 2 + 1)
= 1
(500 + 3n3 )2 − n −1
For maximum of Tn , put (x + 1) (x 2 + 1) ... (x 2 + 1)
dTn 1
=0 Sn = 1 − n −1
dn (x + 1) (x 2 + 1) ... (x 2 + 1)
1000
1/ 3
S∞ = 1
⇒ n=
3
1/ 3 Ex 71. If f (x ) = a 0 + a1 x + a 2 x 2 + ... + a n x n + ... and
1000
Now, 6< <7 f (x )
3 = b0 + b1 x + b2 x 2 + ... + bn x n + ...
Hence, T7 is the largest term, so largest term in the given 1− x
49 where, a 0 = 1, b1 = 3 and b10 = k 11 − 1, then k is
sequence is .
1529 _________ .
x x2 Given that a 0 , a1 , a 2 , ... are in GP.
Ex 70. For x >1, evaluate +
x + 1 ( x + 1) ( x 2 + 1) Sol. (2) Q f (x ) = a0 + a1 x + a2 x 2 + ... + an x n + ...
x4 and f (x ) (1 − x )− 1 = b0 + b1x + b2x 2 + ... + bnx n + ... ,
+ + ...
( x + 1) ( x 2 + 1) ( x 4 + 1) ∴ f (x ) {1 + x + x 2 + ... + x n + ...}
= b0 + b1 x + b2 x 2 + ... + bn x n+ ...
x x2
Sol. (1) Let S = +
x + 1 (x + 1) (x 2 + 1) On comparing the coefficient of x n both sides, we get
x4 a0 + a1 + a 2 + ... + a n − 1 + a n = bn
+ ... b0 = a0 and b1 = a0 + a1
(x + 1) (x + 1) (x 4 + 1)
2
Now, we can write ⇒ b0 = 1 and 3 = a0 + a1 ⇒ a1 = 2
x 1 Since, a0 , a1 , a 2, ... are in GP.
=1−
x+1 x+1 So, first term is 1 and common ratio is 2.
x2 1 1 Now, b10 = a0 + a1 + a 2 + ... + a10
= −
(x + 1) (x 2 + 1) x + 1 (x + 1) (x 2 + 1) = 1 + 2 + 22 + ... + 210 = 211 − 1
74
Target Exercises
Type 1. Only One Correct Option
1. The next term of the sequence 1, 5, 14, 30, 55, ... is 1 1 1 1 1 1
12. If , and are in AP, then + −
(a) 91 (b) 85 (c) 90 (d) 95 a b c a b c
1 1 1
2. The next term of the sequence 1, 3, 6, 10, ... is + − is equal to
b c a
(a) 16 (b) 13 (c) 15 (d) 14
4 3 b2 − ac
3. In a certain AP, 5 times the 5th term is equal to (a) − (b)
ac b2 a2 b2 c2
8 times the 8th term, then 13th term is
4 1
(a) − 13 (b) − 12 (c) − 1 (d) 0 (c) − (d) None of these
ac b2
4. If a1 = a 2 = 2, a n = a n − 1 − 1 ( n > 2), then a 5 is 13. If 7th and 13th terms of an AP are 34 and 64
(a) 0 (b) − 2 (c) − 1 (d) 1 respectively, then its 18th term is
(a) 87 (b) 88 (c) 89 (d) 90
5. The second term of an AP is ( x − y ) and the fifth term
is ( x + y ), then its first term is 14. If the roots of the equation x − 12x + 39x − 28 = 0
3 2
1 2 4 5 are in AP, then the common difference will be
(a) x − y (b) x − y (c) x − y (d) x − y
3 3 3 3 (a) ± 1 (b) ± 2 (c) ± 3 (d) ± 4
Targ e t E x e rc is e s
6. If a, b and c are the sides of a ∆ABC are in AP, then 15. Given that a, b and c are in AP. The determinant
C x+1 x+ 2 x+ a
cot is equal to
2 x + 2 x + 3 x + b in its simplest form is equal to
(a) 3 tan
A
(b) 3 tan
B x+3 x+4 x+c
2 2
A B (a) x 3 + 3ax + 7c (b) 0
(c) 3 cot (d) 3 cot
2 2 (c) 15 (d) 10x 2 + 5x + 2c
7. If a, b, c, d , e and f are in AP, then e − c is equal to 1 1 1 1 1 1
16. If a + , b + and c + are in AP,
b c c a a b
(a) 2 (c − a) (b) 2 (d − c)
then
(c) 2 ( f − d ) (d) d − c
1 1 1
(a) , and are in AP (b) a, b and c are in HP
8. If cos x = b, then for what b do the roots of the a b c
equation form an AP? 1 1 1
(c) , and are in GP (d) a, b and c are in AP
1 a b c
(a) − 1 (b)
2 17. If a, b and c are in AP, then ( a + 2b − c ) ( 2b + c − a )
3
(c) (d) None of these ( c + a − b ) equals
2 abc
(a) (b) abc (c) 2abc (d) 4 abc
9. If log 3 2, log 3 ( 2x − 5) and log 3 ( 2x − 7 / 2) are in 2
AP, then x is equal to 18. If a, b and c are in arithmetic progression, then
(a) 2 (b) 3 (c) 4 (d) 2, 3 1 1 1
, and are in
10. If the pth term of an AP is q and the qth term is p, a+ b a+ c b+ c
then the rth term is (a) AP (b) GP
(c) HP (d) None of these
(a) q − p + r (b) p − q + r
1 1 1
(c) p + q + r (d) p + q − r 19. If , and are in AP, then
p + q q+ r r+ p
1 1 1
11. The consecutive terms , , of a (a) p, q and r are in AP
1+ x 1− x 1− x
(b) q2 , p2 and r2 are in AP
series are in
(c) p2 , q2 and r2 are in AP
(a) HP (b) GP
(c) AP (d) AGP (d) p, q and r are in GP 75
3 20. The number of numbers lying between 100 and 500
that are divisible by 7 but not by 21, is
(a) 57 (b) 19
31. The maximum sum of an AP 40, 38, 36, 34, ... is
(a) 390
(c) 460
(b) 420
(d) None of these
Objective Mathematics Vol. 1
(c) 38 (d) None of these
32. A man arranges to pay off a debt of ` 3600 by
21. If the first, second and last terms of an AP are a, b, c 40 annual instalments which are in AP. When 30 of
respectively, then the sum is the instalments are paid, he dies leaving one-third of
(a + b) (a + c − 2b) (b + c) (a + b − 2c) the debt unpaid. The value of 8th instalment is
(a) (b)
2 (b − a) 2 (b − a) (a) ` 35 (b) ` 50
(a + c) (b + c − 2a) (c) ` 65 (d) None of these
(c) (d) None of these
2 (b − a)
33. A polygon has 25 sides, the lengths of which starting
22. The maximum sum of the series from the smallest side are in AP. If the perimeter of
1 2 the polygon is 2100 cm and the length of the largest
20 + 19 + 18 + 18 + K is
3 3 side 20 times that of the smallest, then the length of
(a) 310 (b) 290 the smallest side and the common difference of an
(c) 320 (d) None of these AP are
1 1
23. The sum of all natural numbers less than 200 that are (a) 8 cm, 6 cm (b) 6 cm, 6 cm
3 3
divisible neither by 3 nor by 5, is 1
(c) 8 cm, 5 cm (d) None of these
(a) 10730 (b) 10732 3
(c) 15375 (d) None of these
34. A club consists of members whose ages are in AP,
24. The four numbers are in AP, such that their sum is 50 the common difference being 3 months. If the
and greatest of them is 4 times the least, are youngest member of the club is just 7 yr old and the
(a) 7, 11, 15, 19 (b) 6, 11, 16, 21 sum of the ages of all the members is 250 yr, then the
Ta rg e t E x e rc is e s
(c) 5, 10, 15, 20 (d) None of these number of members in the club is
25. If the sum of four integers in AP is 24 and their (a) 15 (b) 25 (c) 20 (d) 30
product is 945, then the numbers are 35. The sum of the series 2, 5, 8, 11, ... is 60100, then n is
(a) 3, 5, 7, 9 (b) 5, 8, 11, 14 equal to
(c) 4, 8, 12, 16 (d) None of these
(a) 100 (b) 200
26. If the sum of any number of terms in a sequence is (c) 150 (d) 250
always three times the squared number of these 36. If the sum of n terms of an AP is 3n 2 + 5n, then which
terms, then the sequence is
of its terms is 164?
(a) an AP (b) a GP
(c) an HP (d) None of these (a) 26th (b) 27th
(c) 28th (d) None of these
27. The sum of all two digit numbers which when
37. The first and last terms of an AP are 1 and 11. If the
divided by 4, yield unity as remainder, is
sum of its terms is 36, then the number of terms
(a) 1100 (b) 1200 will be
(c) 1210 (d) None of these
(a) 5 (b) 6 (c) 7 (d) 8
28. The sum of numbers of three digits which are
38. Sum of n terms of the series
divisible by 7, is
2 + 8 + 18 + 32 + ... is
(a) 70336 (b) 70331
n (n + 1)
(c) 70330 (d) None of these (a) (b) 2n (n + 1)
2
29. The sum of all odd numbers of four digits which are n (n + 1)
(c) (d) 1
divisible by 9, is 2
(a) 2754000 (b) 2753000
(c) 2752000 (d) None of these 39. The sum of n terms of the series
1 1 1
30. The sum of positive terms of the series + + + ... is
4 1 1+ 3 3+ 5 5+ 7
10 + 9 + 9 + ... is
7 7 2n + 1 −1+ 2n + 1
352 437 (a) (b)
(a) (b) 2 2
7 7 1
852 (c) (d) 2n − 1
(c) (d) None of these 2n + 1
7
76
40. The sum of all two digit odd numbers is
(a) 2475 (b) 2530 (c) 4905 (d) 5049
51. If 4th, 7th and 10th terms of a GP are p, q and r
respectively, then
(a) p2 = q2 + r2 (b) p2 + qr
3
Sequence and Series
41. If the sum of p terms of an AP is q and that of q terms
(c) q2 = pr (d) r2 = p2 + q2
is p, then the sum of p + q terms will be
(a) 0 (b) p − q 52. If 6th term of a GP is 32 and its 8th term is 128, then
(c) p + q (d) − ( p + q) the value of the common ratio is
42. If the sum of n terms of an AP is 3n 2 − n and its (a) −1 (b) 2
common difference is 6, then its first term is (c) 4 (d) − 4
(a) 2 (b) 3 (c) 1 (d) 4 53. If x1 , x 2 , x 3 as well as y1 , y 2 , y 3 are in GP with same
n
43. If S n = nP + ( n − 1) Q, where S n denotes the sum common ratio, then the points ( x1 , y1 ), ( x 2 , y 2 ) and
2 ( x3 , y3 )
of first n terms of an AP, then the common difference (a) lie on a straight line
is (b) lie on an ellipse
(a) P + Q (b) 2P + 3Q (c) lie on a circle
(c) 2Q (d) Q (d) are vertices of a triangle
44. If the sum of first n terms of a series is 5n 2 + 2n, then 54. If a x = b y = c z and a, b, c are in GP, then x, y and z
its second term is are in
27 56
(a) 16 (b) 17 (c) (d) (a) AP (b) GP
14 15 (c) HP (d) None of these
n
45. If the nth term of a series is + y, then sum 55. If p, q, r are in AP and x, y, z are in GP, then
x
x q − r ⋅ y r − p ⋅ z p − q is equal to
of r terms will be
Targ e t E x e rc is e s
(a) 1 (b) 2
r (r + 1) r (r − 1)
(a) + ry (b) (c) − 1 (d) None of these
2x 2x
r (r − 1) r (r + 1)
56. The sum of three numbers in GP is 56. If we subtract
(c) −xy (d) − rx 1, 7, 21 from these numbers in that order, we obtain
2x 2x an AP. Then, three numbers are
46. If S n denotes the sum of n terms of an AP, then (a) 8, 16, 32 (b) 10, 18, 26
S n + 3 − 3S n + 2 + 3S n + 1 − S n is equal to (c) 9, 16, 23 (d) None of these
1 57. If the sides of a right angled triangle are in GP, then
(a) 0 (b) 1 (c) (d) 2
2 the cosine of the greater acute angle is
a + be y b + ce y c + de y 2 1
47. If = = , then a, b, c and d (a)
1+ 5
(b)
1− 5
a − be y
b − ce y
c − de y
1+ 5
are in (c) (d) None of these
2
(a) AP (b) GP
(c) HP (d) None of these 58. The consecutive numbers of a three digit number
form a GP. If we subtract 792 from this number, then
48. In a GP, if ( m + n )th term is p and ( m − n )th term is q,
we get a number consisting of the same digits written
then its mth term is in the reverse order and if we increase the second
1
(a) 0 (b) pq (c) pq (d) ( p + q) digit of the required number by 2, then resulting
2 number forms an AP. The number is
1 π
49. If cosec 2 θ, 2 cot θ, sec θ, 0 < θ < are in GP, then θ (a) 139 (b) 193
2 2 (c) 931 (d) None of these
is equal to 59. Three non-zero numbers a, b and c are in AP. If
π π
(a) (b) increasing a by 1 or increasing c by 2, the numbers
6 4
π are in GP, then b equals
(c) (d) None of these (a) 10 (b) 14
3
(c) 12 (d) 16
50. If 10th term of a GP is 9 and 4th term is 4, then its 7th
term is 60. The sum of the series 0.4 + 0.004 + 0.00004 + ... is
27 56 11 41 40 2
(a) 6 (b) 14 (c) (d) (a) (b) (c) (d)
14 15 25 100 99 5 77
3 61. If x > 0, then the sum of the series
e −x
−e − 2x
+e − 3x
− ... is
69. Let a n be the nth term of a GP of positive numbers.
If
100 100
∑ a 2n = α and ∑ a 2n − 1 = β , such that α ≠ β,
Objective Mathematics Vol. 1
1 1
(a) (b) x n=1 n=1
1 − e− x e −1
then the common ratio is
1 1
(c) (d) α β
1 + e− x 1 + ex (a) (b)
β α
62. The n th term of a GP is 128 and the sum to its n terms α β
(c) (d)
β α
is 255. If its common ratio is 2, then its first term is
(a) 1 (b) 3 70. If f (x ) is a function satisfying
(c) 8 (d) None of these
f ( x + y ) = f ( x ) f ( y ), ∀ x, y ∈ N such that f (1) = 3
n
63. The sum of
( x + 2) n − 1 + ( x + 2) n − 2 ( x + 1) + ( x + 2) n − 3
and ∑ f ( x ) = 120. Then, the value of n is
x =1
( x + 1) 2 + ... + ( x + 1) n − 1 is equal to (a) 4 (b) 5
n− 2
(a) (x + 2) − (x + 1 ) n (c) 6 (d) None of these
(b) (x + 2)n − 1 − (x + 1)n − 1 71. The value of 0.423 is
(c) (x + 2) − (x + 1)
n n
419 419
(a) (b)
(d) None of the above 999 990
423
64. If ( a, b ), ( c, d ), ( e, f ) are the vertices of a triangle (c) (d) None of these
1000
such that a, c, e are in GP with common ratio r and
72. If x = 1 + y + y 2 + ... , then y is
b, d , f are in GP with common ratio s, then the area
x x x −1 1− x
Ta rg e t E x e rc is e s
of the triangle is (a) (b) (c) (d)
ab x −1 1− x x x
(a) (r + 1) (s + 2) (s + r)
2 73. The sum of first four terms of a GP is 12 (1 − 5 ). If
ab
(b) (r − 1) (s − 2) (s − r) the common ratio is − 5, then the first term of a GP
2
ab is
(c) (r − 1) (s − 1) (s − r)
2 (a) 1 (b) 2
ab (c) 3 (d) 4
(d) (r + 1) (s + 1) (s − r)
2
74. The length of a side of a square is a metre. A second
1 square is formed by joining the mid-points of these
65. If sum of all terms of an infinite GP is times the
5 square. Then, a third square is formed by joining the
sum of odd terms. Then, common ratio is mid-points of the second square and so on. Then, the
(a) 2 (b) 3 sum of the area of the squares which carried upto
4 infinity, is
(c) − (d) 5
5 (a) a2 (b) 2a2 (c) 3a2 (d) 4 a2
66. The sum of 10 terms of the series 2 + 6 + 18 + ... 75. The sum of an infinite geometric series is 3. A series
is which is formed by squares of its terms also have
(a) 121 ( 6 + 2) sum equal to 3. Then, the sequence corresponding to
the series will be
(b) 243 ( 3 + 1)
3 3 3 3 1 1 1 1
121 (a) , , , , ... (b) , , , , ...
(c) 2 4 8 16 2 4 8 16
3 −1 1 1 1 1 1 1 1
(c) , , , , ... (d) 1, − , , , ...
(d) 242 ( 3 − 1) 3 9 27 81 3 32 33
49
76. The product of x1/ 2 ⋅ x1/ 4 ⋅ x1/ 8 ... equals
67. If (1.05) 50 = 11.658 , then ∑ (1.05)n equals
n=1 (a) 0 (b) 1 (c) x (d) ∞
(a) 208.34 (b) 212.12 77. Let S 1 , S 2 , ... be squares such that for each n ≥ 1, the
(c) 212.16 (d) 213.16 length of a side of S n equals the length of the
68. If the second term of a GP is 2 and the sum of its diagonal of S n + 1 . If the length of a side of S 1 is
infinite terms is 8, then its first term is 10 cm and the area of S n less than 1 sq cm, then the
1 1
least value of n is
78 (a) (b) (c) 2 (d) 4 (a) 7 (b) 8 (c) 9 (d) 10
2 4
78. Let α , β be the roots of x 2 − x + p = 0 and γ , δ be the
roots of x 2 − 4x + q = 0. If α , β , γ , δ are in GP, then
87. The sum of the series 1 + 2 x + 3x 2 + 4x 3 + ... (where,
x lies between 0 and 1, i.e. 0 < x < 1) is
1 1 1 1
3
Sequence and Series
integral values of p and q respectively are (a) (b) (c) (d)
(a) − 2, − 32 (b) − 2, 3 1+ x 1− x 1 − 2x (1 − x )2
(c) − 6, 3 (d) − 6, − 32
88. The sum of the series
79. Sum of the series 1 + 2 ⋅ 2 + 3 ⋅ 22 + 4 ⋅ 23 + ... + 100 ⋅ 299 is
(1 + x ) + (1 + x + x ) + (1 + x + x + x ) + ... n terms
2 2 3
(a) 100 ⋅ 2100 + 1 (b) 99 ⋅ 2100 + 1
is
(c) 99 ⋅ 299 − 1 (d) 100 ⋅ 2100 − 1
1 x 2 (1 − x n ) 1 x 3 (1 − x n )
(a)
1− x n − 1 − x (b)
1− x n − 1 − x 89. The sum of the series
1 x (1 − x n ) 13 13 + 23 13 + 23 + 33
n− + + + ... upto 16 terms is
1 − x 1 − x
(c) (d) None of these
1 1+ 3 1+ 3 + 5
(a) 246 (b) 346 (c) 446 (d) 546
80. Sum of the series 9 + 99 + 999 + ... upto n terms is
1 2 2 3 3 4
1 . . .
(a) (10n − 9n − 10)
9 90. The sum of 2 3 2 + 32 2 3 + 3 2 3 2 3 + ... upto
1 1 1 +2 1 +2 +3
(b) (10n + 1 − 9n − 10)
9 n term is
1
(c) (10n + 1 + 9n − 10) n−1 n n+1 n+1
(a) (b) (c) (d)
9 2 n+1 n+ 2 n
(d) None of the above
91. The sum of the series
81. Sum of the series 0.5 + 0.55 + 0.555 + ... upto n terms
S = 12 − 22 + 32 − 4 2 + ... − 20022 + 20032 is
Targ e t E x e rc is e s
is
5 2 1 3 1 1 (a) 2007006 (b) 1005004
(a) n − 1 − n
10
(b) 3 − 1 − n
10
(c) 2000506 (d) None of these
9 9 9 9
7 2 1 5 1 1 92. The value of n for which
(c) n − 1 − n
10
(d) n − 1 − n
10 1 1
9 9 9 9 704 + ( 704 ) + ( 704 ) + ... n terms
2 4
82. A number consists of three digits in GP. If the sum of 1 1
extreme digits is greater than twice middle digit by 1 = 1984 − (1984 ) + (1984 ) − ... n upto terms, is
2 4
and the sum of the left hand and middle digit is (a) 5 (b) 3 (c) 4 (d) 10
two-third of the sum of the middle and right hand
digits, then the number is 93. The positive integer n for which
(a) 4, 6, 9 (b) 3, 7, 6 2 ⋅ 22 + 3 ⋅ 23 + 4 ⋅ 24 + ... + n ⋅ 2n = 2n + 10 , is
(c) 4, 6, 8 (d) None of these (a) 510 (b) 511 (c) 512 (d) 513
83. If one geometric mean G and two arithmetic means p 94. If 1 + 2 + 3 + ... + 2003 = ( 2003) ( 4007) ( 334 )
2 2 2 2
and q are inserted between two numbers, then G 2 is
and (1) ( 2003) + ( 2) ( 2002) + ( 3) ( 2001)
equal to
+ ... + ( 2003) (1) = ( 2003) ( 334 ) ( x ),
(a) (3 p − q) (3q − p) (b) (2 p − q) (2q − p)
(c) (4 p − q) (4 q − p) (d) None of these then x equals
(a) 2005 (b) 2004
84. If in a GP of 3n terms, S 1 denotes the sum of the first (c) 2003 (d) 2001
n terms, S 2 denotes the sum of the second block of n 95. If x> 0 and log 2 x + log 2 ( x ) + log 2 ( 4 x )
terms and S 3 denotes the sum of the last n terms, then
S 1 , S 2 and S 3 are in + log 2 ( 8 x ) + log 2 (16 x ) + ... = 4, then x equals
(a) AP (b) GP (a) 2 (b) 3 (c) 4 (d) 5
(c) HP (d) None of these
96. If x1 , x 2 , ... , x n are n non-zero real numbers such that
85. The product of ( 32) ⋅ ( 32)1/ 6 ⋅ ( 32)1/ 36 ⋅ ... is equal to
( x12 + x 22 + ... + x n2 − 1 ) ( x 22 + x 32 + ... + x n2 )
(a) 16 (b) 64 (c) 32 (d) 0
≤ ( x1 x 2 + x 2 x 3 + ... + x n − 1 x n ) 2 ,
4 7 10
86. The sum of the series 1 + + 2 + 3 + ... is
5 5 5 then x1 , x 2 , ... , x n are in
16 11 35 8 (a) AP (b) GP
(a) (b) (c) (d) 79
25 8 16 11 (c) HP (d) None of these
3
n n
∑ r 4 = f ( n ) , then ∑ ( 2r − 1)4 is equal to
104. If the roots of the equation
97. If
r =1 r =1 a ( b − c ) x 2 + b ( c − a ) x + c ( a − b ) = 0 are equal,
(a) f (2n) − 16 f (n) (b) f (2n) − 7 f (n)
Objective Mathematics Vol. 1
then a, b and c are in
(c) f (2n − 1) − 8 f (n) (d) None of these (a) AP (b) GP
(c) HP (d) None of these
n
1
98. The value of lim
n→ ∞
∑ tan − 1 2r 2 is 105. If AM between two numbers is 5 and their GM is 4,
r =1
then the HM will be
π π
(a) (b) 16 14
2 4 (a) (b)
5 5
(c) 1 (d) None of these 11
(c) (d) None of these
99. If 13 = 1, 23 = 3 + 5, 33 = 7 + 9 + 11 5
and 4 3 = 13 + 15 + 17 + 19, 106. If log ( a + c ), log ( c − a ) and log ( a − 2b + c ) are in
3
then n as a similar series is AP, then
n(n − 1) (n + 1)n (a) a, b and c are in AP (b) a2 , b2 and c2 are in AP
(a) 2 + 1 − 1 + 2 2 + 1 + 1
2 (c) a, b and c are in GP (d) a, b and c are in HP
(n + 1)n
+ ... + 2 + 1 + 2n − 3 107. If cos ( x − y ), cos x and cos ( x + y ) are in HP, then
2
y
(b) (n2 + n + 1) + (n2 + n + 3) + (n2 + n + 5) cos x sec is equal to
2
+ ... + (n2 + 3n − 1)
1
(c) (n2 − n + 1) + (n2 − n + 3) + (n2 − n + 5) (a) ± 2 (b) ±
2
+ ... + (n2 + n − 1)
(c) ± 2 (d) None of these
(d) None of the above
Ta rg e t E x e rc is e s
108. log 3 2, log 6 2 and log 12 2 are in
n
r2 − r − 1
100. ∑ is equal to (a) AP (b) GP
r = 1 ( r + 1)!
(c) HP (d) None of these
n
(a) 109. If a, b and c are in HP, then correct statement is
(n + 1)!
−1 (a) a2 + c2 > b2 (b) a2 + c2 > 2b2
(b)
(n + 1)(n − 1)! (c) a2 + c2 < 2b2 (d) a2 + c2 = 2b2
n
(c) −1 110. The AM, GM and HM of two numbers are x, y and z,
(n + 1)!
(d) None of the above respectively. Then, which of the following is true?
(a) z < y < x (b) y < x < z (c) x < y < z (d) z < x < y
101. Solution set for ( 2 + 2 ) x + ( 2 − 2 ) x = 2⋅ 2x / 4
111. If a, b and c are in HP, then the straight line
is
x y 1
(a) {2} + + = 0 always passes through a fixed point
(b) {0} a b c
(c) [0, 2] and that point is
(d) None of the above 1
(a) (− 1, − 2) (b) (− 1, 2) (c) (1, − 2) (d) 1, −
2
102. If x − ( a + b + c ) x + ( ab + bc + ca ) = 0 has
2
imaginary roots, where a, b, c ∈ R + , then 112. Let a1 , a 2 , ... , a10 be in AP and h1 , h2 , ... , h10 be in
a , b and c HP. If a1 = h1 = 2 and a10 = h10 = 3, then a 4 h7 is
(a) can be sides of a triangle (a) 2 (b) 3 (c) 5 (d) 6
(b) can’t be sides of a triangle
113. If x, y and z are in HP, where z > y > x. Then, value
(c) nothing can be said
(d) None of the above of log ( x + z ) + log ( x − 2 y + z ) is
(a) 2log ( y − z) (b) 2log (z − x )
1 1
103. If first three terms of sequence , a, b, are in (c) 4 log (x − z) (d) log (x − z)
16 6
geometric series and the last three terms are in 114. If H1 , H 2 , H 3 , ... , H 2n + 1 are in HP, then
harmonic series, then the values of a and b will be 2n Hi + Hi + 1
1 1 1 ∑ ( − 1)i H is equal to
(a) a = − , b = 1 (b) a = , b = i=1 i − Hi + 1
4 12 9
80 (c) Both (a) and (b) are true (d) None of these (a) 2n − 1 (b) 2n + 1 (c) 2n (d) 2n + 2
Type 2. More than One Correct Option
115. For the AP given by a1 , a 2 , ... , a n , ... , the equation 119. The roots of the equation
3
Sequence and Series
satisfied is x 5 − 40x 4 + Px 3 + Qx 2 + Rx − S = 0 are in GP. If the
(a) a1 + 2a 2 + a3 = 0 sum of their reciprocals is 10, then the value of S can
(b) a1 − 2a 2 + a3 = 0 be equal to
(c) a1 + 3a 2 − 3a 3 − a4 = 0 1
(a) 32 (b) −
(d) a1 − 4 a 2 + 6a 3 − 4 a4 + a5 = 0 32
1
116. If sum of the first three consecutive terms of an AP (c) − 32 (d)
32
is 9 and the sum of their squares is 35. Then, sum to
n terms of the series is 120. Consider an AP a1 , a 2 , ... , a n , ... and the GP
(a) n(n + 1) (b) n 2 b1 , b2 , ... , bn , ... such that a1 = b1 = 1, a 9 = b9 and
9
(c) n(4 − n) (d) n(6 − n)
∑ ar = 369, then
117. If a1 , a 2 , ... , a n are in AP with common difference d, r =1
1 + a1 a 2 1 + a 2 a 3 (a) b6 = 27 (b) b 7 = 27 (c) b 8 = 81 (d) b9 = 81
then cot − 1 + cot − 1
d d 121. If sin β is the geometric mean between sin α and cos α,
1 + a 3 4
a 1 + a n n−1
a then cos 2β is equal to
+ cot − 1 + ... + cot − 1
d d π π
(a) 2 sin 2 − α (b) 2 cos2 − α
4 4
is equal to π π
(c) 2 cos2 + α (d) 2 sin 2 + α
(a) tan −1 a n − tan −1 a1 (b) cot −1 a1 + cot −1 an 4 4
(c) cot −1 a1 − cot −1 a n (d) None of these
122. If the first and ( 2n + 1) th terms of an AP, a GP and an
Targ e t E x e rc is e s
118. If a, b and c are three terms of an AP such that a ≠ b, HP of positive terms are equal and their ( n + 1)th
b−c terms are a, b and c respectively, then
then may be equal to
a−b (a) a = b = c (b) a ≥ b ≥ c
(a) 2 (b) 3 (c) 1 (d) 2 (c) a + c = 2b (d) ac = b2
Type 3. Assertion and Reason
Directions (Q. Nos. 123-129) In the following a b c
124. Statement I If , and are in AP, then
questions, each question contains Statement I a1 b1 c1
(Assertion) and Statement II (Reason). Each question a1 , b1 and c1 are in GP.
has 4 choices (a), (b), (c) and (d) out of which only one is
correct. The choice are Statement II If ax 2 + bx + c = 0 and
a1 x + b1 x + c1 = 0 have a common root and
2
(a) Statement I is true, Statement II is true; Statement II
a b c
is a correct explanation for Statement I , , are in AP, then a1 , b1 and c1 are in GP.
a1 b1 c1
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I 125. Statement I If all terms of a series with positive
(c) Statement I is true, Statement II is false terms are smaller than 10− 5 , then the sum of the
(d) Statement I is false, Statement II is true series upto infinity will be finite.
n
123. If positive numbers x, y and z satisfy x + y + z = 1, Statement II If S n > 5 , then lim S n is infinite.
10 n→ ∞
then
126. Statement I If three positive numbers in GP
(1 − 2x ) + (1 − 2 y ) + (1 − 2z )
Statement I represent sides of a triangle, then the common ratio
3 5−1 5+1
of a GP must lie between and .
≥ (1 − 2x ) (1 − 2 y) (1 − 2 z )1/ 3 2 2
Statement II For any three positive numbers a, b Statement II Three positive real numbers can form
a triangle, if sum of any two numbers is greater than
and c, AM ≥ GM.
the third number.
81
3 127. Statement I There exists an AP, whose three terms
are 2, 3 and 5.
129. Statement I If the sequence {a n } is monotonically
increasing, then the sequence {S n }, where
a + a 2 + ... + a n
Objective Mathematics Vol. 1
Statement II There does not exist distinct real Sn = 1 is monotonically increasing
numbers p, q and r satisfying n
2 = A + ( p − 1) d too.
3 = A + ( q − 1) d and 5 = A + ( r − 1) d Statement II If {a n } is monotonically increasing,
x n then the expression na n+1 − ( a1 + a 2 + ... + a n ) is
128. Statement I lim = 0, ∀ x > 0.
n→ ∞ n! positive.
Statement II Every sequence whose nth term
contains n !in the denominator converges to zero.
Type 4. Linked Comprehension Based Questions
Passage I (Q. Nos. 130-132) If A , G and H are 132. The value of [ 2P − Q ], where [] denotes the greatest
respectively arithmetic, geometric and harmonic means integer function, is
between a and b, both being unequal and positive, then (a) 2 (b) 3
a+b (c) 5 (d) 6
A= ⇒ a + b = 2 A,
2
Passage II (Q. Nos. 133-135) Let the sequence
G = ab ⇒ ab = G 2 a1, a2, a3, ..., an be in GP. If the area bounded by the
2ab
and H= ⇒ G 2 = AH parabolas y 2 = 4an x and y 2 = 4an (an − x ) is A n .
a+b
133. The sequence A1 , A 2 , A 3 , ... , lies in
Ta rg e t E x e rc is e s
From above discussion, we can say that a and b are the
roots of the equation x 2 − 2 Ax + G 2 = 0. (a) AP
(b) GP
Now, quadratic equations x 2 − Px + Q = 0 and (c) HP
a(b − c) x 2 + b(c − a) x + c(a − b) = 0 have both root (d) None of the above
2ac
common and satisfy the relation b = , where a, b
(a + c) 134. For a n = 1, the point of intersection of the curves are
and c are real numbers. 1 4 8 16
(a) , 2 (b) , 2 (c) , 2 (d) , 2
a( b − c ) 2 3 3 3
130. The value of is
b( c − a ) 135. For a n = 1, the area of region bounded by both the
(a) − 2 (b) 2 (c) −1/ 2 (d) 1/ 2
curves is
131. The value of [P], where, [] denotes the greatest 4 2
(a) 4 2 sq units (b) sq units
integer function, is 3
(a) − 2 (b) − 1 (c)
8 2
sq units (d)
16 2
sq units
(c) 2 (d) 1 3 3
Type 5. Match the Columns
136. Let a, b and c be positive integers such that a + b + c = n. Match the statements of Column I with values of Column II.
Column I Column II
a b c 1/ n
A. ( a b c ) is less than or equal to p. ab + bc + ca
n
B. ( a b bcc a )1/ n is less than or equal to q. a2 + b 2 + c 2
n
C. ( ac b ac b )1/ n is less than or equal to r. abc
D. ( a ab bcc )1/ n + ( a b bcc a )1/ n + ( ac b ac b )1/ n is s. n
less than or equal to
82
137. Match the statements of Column I with values of Column II.
Column I Column II
3
Sequence and Series
A. The AM of two positive numbers a and b exceed their GM by 3/2 and the p. α + β 2 = 96
GM exceed their HM by 6/5 such that a + b = α,| a − b| = β, then
B. The AM of two positive numbers a and b exceed their GM by 2 and HM is q. α + β 2 = 74
one-fifth of the greater of a and b such that a + b = α,| a − b| = β, then
C. The HM of two positive numbers a and b is 4, their AM is A and GM G r. α 2 + β = 234
satisfy the relation 2 A + G 2 = 27 . If a + b = α,| a − b| = β, then
s. α 2 + β = 84
2 5
α α α
t. 1 + + + ... + = 364
β β β
Type 6. Single Integer Answer Type Questions
138. If a sequence a1 , a 2 , a 3 , ... , a n of real numbers is 140. Find the HCF of the minimum non-negative values
such that a1 = 0, | a 2 | = | a1 + 1| , | a 3 | = | a 2 + 1| , ..., of a, b and c, given that the equation
| a n | is equal to | a n − 1 + 1| , where the arithmetic mean
of a1 , a 2 , ... , a n cannot be less than − λ /µ, then find x 4 + ax 3 + bx 2 + cx + 1 = 0 has only real roots.
the value of λ + µ.
141. If K is a positive integer such that 36 + K , 300 + K
139. If ( m + 1) th, ( n + 1) th and ( r + 1) th terms of an AP
and 596 + K are the squares of three consecutive
are in GP and m, n, r are in HP, where the ratios of the
common difference to the first term in the AP is terms of an arithmetic progression, then find
− λ / n, then find the value of λ. ( K − 920).
Targ e t E x e rc is e s
Entrances Gallery
JEE Advanced/IIT JEE
1. Let a, b and c be positive integers such that b/a is an 5. Let a1 , a 2 , a 3 ,... , a100 be an arithmetic progression
p
integer. If a, b and c are in geometric progression and
the arithmetic mean of a, b, c is b + 2, then the value with a1 = 3 and S p = ∑ a i , 1≤ p ≤ 100. For any
i=1
a 2 + a − 14
of is __________ . [2014] Sm
a+1 integer n with 1≤ n ≤ 20, let m = 5n. If does not
Sn
23 depend on n, then a 2 is__________ . [2011]
1 + ∑ 2k is
n
2. The value of cot ∑ cot − 1 [2013]
6. The minimum value of the sum of real numbers
n = 1 k = 1
a − 5 , a − 4 , 3a − 3 , 1, a 8 and a10 with a > 0is___. [2011]
23 25
(a) (b)
25 23 7. If S k , k = 1, 2 ... 100, denotes the sum of the infinite
23 24
(c) (d) k −1
24 23 geometric series, whose first term is and the
k!
4n k (k + 1)
common ratio is 1/k. Then, the value of
3. If S n = ∑ ( − 1) 2 k 2 , then S n can take value(s) 1002 100
2 [2013] + ∑ | ( k 2 − 3k + 1) S k is _________ . [2010]
100! k = 1
(a) 1056
(b) 1088
(c) 1120 8. Let a1 , a 2 , a 3 ,... , a11 be real numbers satisfying
(d) 1332 a1 = 15, 27 − 2a 2 > 0 and a k = 2 a k − 1 − a k − 2 for
4. If a1 , a 2 , a 3 ,... are in a harmonic progression with a 2 + a 22 + ...+ a11
2
k = 3, 4,... ,11. If 1 = 90, then the
a1 = 5 and a 20 = 25. Then, least positive integer n for 11
which a n < 0, is a + a 2 + ...+ a11
[2012] value of 1 is _______ . [2010]
(a) 22 (b) 23 (c) 24 (d) 25 11
83
3 JEE Main/AIEEE
9. If m is the AM of two distinct real numbers l and 17. A man saves ` 200 in each of the first three months of
Objective Mathematics Vol. 1
n( l, n > 1) and G1 , G2 and G3 are three geometric his service. In each of the subsequent months, his
savings increases by `40 more than the savings of
means between l and n , then G14 + 2G24 + G34 equals
immediately previous month. His total saving from
[2015]
the start of service will be ` 11040 after [2011]
(a) 4 l 2mn (b) 4 lm2n (c) 4 lmn2 (d) 4 l 2m2n2 (a) 19 months (b) 20 months
(c) 21 months (d) 18 months
10. The sum of first 9 terms of the series
13 13 + 23 13 + 23 + 33 18. A person is to count 4500 currency notes. Let a n
+ + + ... is [2015]
1 1+ 3 1+ 3 + 5 denotes the number of notes he counts in the nth
minute. If a1 = a 2 = ... = a10 = 150 and a10 , a11 ,... are
(a) 71 (b) 96 (c) 142 (d) 192
in AP with common difference −2 , then the time
11. Let α and β be the roots of equation px 2 + qx + r = 0, taken by him to count all notes, is [2010]
1 1 (a) 24 min (b) 34 min (c) 125 min (d) 135 min
p ≠ 0 . If p, q and r are in AP and + = 4, then the
α β 2 6 10 14
19. The sum of the series 1 + + 2 + 3 + 4 +... is
value of |α − β | is [2014] 3 3 3 3 [2009]
61 2 17 34 2 13 (a) 3 (b) 4 (c) 6 (d) 2
(a) (b) (c) (d)
9 9 9 9
20. The first two terms of a geometric progression add
12. If (10) 9 + 2(11)1 (10) 8 + 3(11) 2 (10) 7 + ... + 10(11) 9 upto 12. The sum of the third and fourth terms is 48.
= k(10) 9 , then k is equal to If the terms of the geometric progression are
[2014]
alternately positive and negative, then the first term
121 441
(a) (b) (c) 100 (d) 110 is
Ta rg e t E x e rc is e s
10 100
[2008]
13. Three positive numbers form an increasing GP. If (a) − 4 (b) − 12 (c) 12 (d) 4
the middle term in this GP is doubled, then new 21. In a geometric progression consisting of positive
numbers are in AP. Then, the common ratio of a GP terms, each term equals the sum of the next two
is [2014] terms. Then, the common ratio of this progression
(a) 2 + 3 (b) 3 + 2 equals [2007]
(c) 2 − 3 (d) 2 + 3 1 1 1
(a) (1 − 5 ) (b) 5 (c) 5 (d) ( 5 − 1)
2 2 2
14. The sum of first 20 terms of the sequence 0.7, 0.77,
0.777,… is [2013] 22. Let a1 , a 2 , a 3 ,... be terms of an AP. If
7 7 a1 + a 2 +...+ a p p2 a
(a) (179 − 10− 20 ) (b) (99 − 10− 20 ) = 2 , p ≠ q, then 6 equals
81 9 a1 + a 2 + ...+ a q q a 21
7 7 [2006]
(c) (179 + 10− 20 ) (d) (99 + 10− 20 ) 7 2 11 41
81 9 (a) (b) (c) (d)
2 7 41 11
15. If 100 times the 100th term of an AP with non-zero
common difference equal to 50 times its 50th term, 23. If a1 , a 2 ,... , a n are in HP, then the expression
then the 150th term of this AP is [2012] a1 a 2 + a 2 a 3 +...+ a n− 1 a n is equal to [2006]
(a) − 150 (b) 150 times its 50th term (a) (n − 1)(a1 − an ) (b) na1an
(c) 150 (d) 0 (c) (n − 1)a1an (d) n(a1 − an )
∞ ∞ ∞
16. Statement I The
∑ a n , y = ∑ b n , z = ∑ c n , where a, b, c are
sum of the series
24. If x =
1 + (1 + 2 + 4 ) + ( 4 + 6 + 9) + ( 9 + 12 + 16) n=0 n=0 n=0
+ ...+ ( 361 + 380 + 400) is 8000. in AP and | a| < 1, | b| < 1, | c| < 1, then x, y and z are in
n [2005]
Statement II ∑ [k 3
− ( k − 1) ] = n for any natural
3 3
(a) HP (b) AGP (c) AP (d) GP
k =1
1 1 1
number n. [2012] 25. The sum of the series 1 + + + +...
4 ⋅ 2! 16⋅ 4 ! 64 ⋅ 6!
(a) Statement I is false, Statement II is true
(b) Statement I is true, Statement II is true; Statement II is is [2005]
a correct explanation for Statement I e+1 e−1
(a) (b)
(c) Statement I is true, Statement II is true; Statement II is 2 e 2 e
not a correct explanation for Statement I e+1 e−1
84 (c) (d)
(d) Statement I is true, Statement II is false e e
3
−x
26. Let Tr be the rth term of an AP, whose first term is a 30. If 1, log 3 ( 31 + 2), log 3 ( 4 ⋅ 3x − 1) are in AP.
and common difference is d. If for some positive Then, x equals [2002]
1 1 (b) 1 − log3 4 (c) 1 − log4 3 (d) log4 3
Sequence and Series
integers m, n, m ≠ n, Tm = and Tn = , then a − d (a) log3 4
n m
31. The value of 21/ 4 ⋅ 41/ 8 ⋅ 81/ 16 ... is [2002]
equals [2004]
1 1 1 (a) 1 (b) 2
(a) 0 (b) 1 (c) (d) + 3
mn m n (c) (d) 4
2
27. The sum of the first n terms of the series
32. If term of a GP is 2, then the product of its 9 terms is
n( n + 1) 2
12 + 2⋅ 22 + 32 + 2⋅ 4 2 + 52 + 2⋅ 62 + ... is , [2002]
2 (a) 256
where n is even. When n is odd, then the sum is [2004] (b) 512
3n (n + 1) n2 (n + 1) (c) 1024
(a) (b)
2 2 (d) None of the above
2
n (n + 1)2 n (n + 1) ∞
(log e x ) n
∑
(c) (d)
4 2 33. is equal to [2002]
n=0 n!
1 1 1
28. The sum of the series + + + ... is [2004] (a) loge x (b) x
2! 4 ! 6! (c) logx e (d) None of these
(e2 − 1) (e − 1)2 (e2 − 1) (e2 − 2)
(a) (b) (c) (d) 1 (x − 1) 3 (x − 1) 4
2 2e 2e e (x − 1) − (x − 1) 2 + − + ...
34. e 2 3 4 is equal to [2002]
1 1 1 (a) log (x − 1)
29. The sum of the series − + − ... is [2003]
1⋅ 2 2⋅ 3 3⋅ 4 (b) log x
4 (c) x
(b) loge 2 − 1 (c) loge 2 (d) loge
Targ e t E x e rc is e s
(a) 2 loge 2
e (d) None of the above
Other Engineering Entrances
35. If Vr denotes the sum of the first r terms of an 39. Let α and β denote the cube roots of unity other than
arithmetic progression (AP), whose first term is r and n
302
α
the common difference is ( 2r − 1). Then, the sum of 1 and α ≠ β. If S = ∑ ( − 1)n β , then the value of
V1 + V2 + ... + Vn is [AMU 2014] n=0
1 S is [WB JEE 2014]
(a) n (n + 1) (3n2 − n + 1)
12 (a) either − 2ω or − 2ω 2 (b) either − 2ω or 2ω 2
1 (c) either 2ω or − 2ω 2 (d) either 2ω or 2ω 2
(b) n(n + 1) (3n2 + n + 2)
12
1 40. If α and β are the roots of the equation
(c) n(2n2 − n + 1)
2 x 2 − px + q = 0, then the value of
1
(d) (2n3 − 2n + 3) α 2 + β 2 2 α 3 + β 3 3
3 (α + β ) x − x + x + ...
2 3
36. If a, b and c are positive numbers in a GP, then the is [BITSAT 2014]
roots of the quadratic equation (a) log (1 − px + qx 2 ) (b) log (1 + px − qx 2 )
(log e a ) x 2 − ( 2 log e b ) x + (log e c ) = 0 are
(c) log (1 + px + qx 2 ) (d) None of the above
[WB JEE 2014]
(a) − 1and
loge c
(b) 1 and −
loge c 41. The sum of the series log 4 2 − log 8 2 + log 16 2 − ...
loge a loge a is [BITSAT 2014]
(c) 1 and loga c (d) − 1and logc a (a) e2 (b) loge 2 + 1
37. If f ( x ) = x + 1/ 2 , then the number of real values of x (c) loge 3 − 2 (d) 1 − loge 2
for which the three unequal terms ex
f ( x ), f ( 2 x ), f ( 4x ) are in HP, is [WB JEE 2014] 42. If = B 0 + B1 x + B 2 x 2 + ... + B n x n + ..., then the
1− x
(a) 1 (b) 0 (c) 3 (d) 2
value of B n − B n− 1 is [BITSAT 2014]
38. If p, q, r and s are positive real numbers such that 1
p + q + r + s = 2, then M = ( p + q ) ( r + s ) satisfies (a) 1 (b)
n
the relation [BITSAT 2014] 1
(c) (d) None of these 85
(a) 0 < M ≤ 1 (b) 1 ≤ M ≤ 2 (c) 2 ≤ M ≤ 3 (d) 3 ≤ M ≤ 4 n!
∞ ∞
x 3n− 2 ∞
x 3n−1
3 43. If a = ∑
x 3n
n = 0 ( 3n )!
, b = ∑
n = 1 ( 3n − 2 )!
and c = ∑
n = 1 ( 3n − 1)!
then the value of a 3 + b 3 + c 3 − 3abc is [BITSAT 2014]
, 52. If
3 5
1
Hn = 1+
2
S n = 1+ + + K +
1
+ K + , then
n
2n − 1
is
the value
[AMU 2012]
of
Objective Mathematics Vol. 1
2 3 n
(a) 1 (b) 0 (a) H n + 2n (b) n − 1 + H n (c) H n − 2n (d) 2n − H n
(c) − 1 (d) − 2
53. If AM and HM between two numbers are 27 and 12
44. For every real number x, let respectively, then their GM is [BITSAT 2011]
x 3 7 15
f ( x ) = + x 2 + x 3 + x 4 +… . Then, the (a) 9 (b) 18
1! 2! 3! 4! (c) 24 (d) 36
equation f ( x ) = 0 has [WB JEE 2014]
54. If x, y and z are in geometric progression, then
(a) no real solution
(b) exactly one real solution log x 10, log y 10 and log z 10 are in [UP SEE 2011]
(c) exactly two real solutions (a) AP (b) GP
(d) infinite number of real solutions (c) HP (d) None of these
45. Let denotes the sum
S of the series 2(1 + 22 )
8 21 40 65 55. The value of 4 + 2(1 + 2) log 2 + (log 2) 2
1+ + + + +…, then [WB JEE 2014] 2!
2! 3! 4 ! 5! 2(1 + 23 )
(a) S < 8 (b) S > 12 (c) 8 < S < 12 (d) S = 8 + (log 2) 3 + … is
3! [UP SEE 2011]
46. In an arithmetical progression a1 , a 2 , a 3 ,…, sum (a) 10 (b) 12
(c) log(32 ⋅ 4 2 ) (d) log(22 ⋅ 32 )
( S ) = a12 − a 22 + a 32 − a 42 + ...− a 22k is equal to
2 2+ 4 2+ 4 + 6
[BITSAT 2013] 56. The value of + + + K is
k 2k 1! 2! 3!
(a) (a12 − a22k ) (b) (a22k − a12 ) [BITSAT 2011]
2k − 1 k −1
Ta rg e t E x e rc is e s
k (a) e (b) 2e
(c) (a12 − a22k ) (d) None of these (c) 3e (d) None of these
k +1
57. If the sum to first n terms of an AP 2, 4, 6,… is 240,
47. If log x ax, log x bx and log x cx are in AP, where a, b,
then the value of n is [BITSAT 2010]
c, x belong to (1, ∞ ), then a, b and c are in (a) 14 (b) 15
[UP SEE 2013] (c) 16 (d) 17
(a) AP (b) HP
(c) GP (d) None of these
4
58. If sum of an infinite geometric series is and its 1st
3
48. Six numbers are in AP such that their sum is 3. The 3
first term is 4 times the third term. Then, the fifth term is , then its common ratio is [BITSAT 2010]
4
term is [WB JEE 2012]
7 9 1 7
(a) − 15 (b) − 3 (a) (b) (c) (d)
16 16 9 9
(c) 9 (d) − 4
xn + 1 + yn + 1
49. If log 10 2, log 10 ( 2x − 1) and log 10 ( 2x + 3) are in AP, 59. The value of n for which is the
then x is equal to [AMU 2012]
xn + yn
1 geometric mean of x and y, is [WB JEE 2010]
(a) log2 5 (b) log2 (− 1) (c) log2 (d) log5 2 1 1
5 (a) n = − (b) n = (c) n = 1 (d) n = − 1
2 2
50. 0.2 + 0.22 + 0.222 + ... upto n terms is equal to
60. GM and HM of two numbers are 10 and
[BITSAT 2012]
8 respectively. The numbers are [WB JEE 2010]
2 2 1
(a) − (1 − 10− n ) (b) n − (1 − 10− n ) (a) 5, 20 (b) 4, 25
9 81 9
(c) 2, 50 (d) 1,100
2 1 2
(c) n − (1 − 10− n ) (d) 61. Sum of n terms of the series 13 + 33 + 53 + 73 +... is
9 9 9
[WB JEE 2010]
51. The sum of the series
(a) n2 (2n2 − 1) (b) n3 (n − 1)
1 1⋅ 3 1⋅ 3⋅ 5 1⋅ 3⋅ 5⋅ 7 (c) n3 + 8n + 4 (d) 2n4 + 3n2
1+ + + + +… is [WB JEE 2013]
3 3⋅ 6 3⋅ 6⋅ 9 3⋅ 6⋅ 9⋅ 12
2 4 6
(a) 2 (b) 3 62. The value of + + + K is [WB JEE 2010]
3 1 3! 5! 7!
86 (c)
2
(d)
3 (a) e1/ 2 (b) e−1 (c) e (d) e−1/ 3
Answers
Work Book Exercise 3.1
1. (c) 2. (b) 3. (a) 4. (a) 5. (b) 6. (b) 7. (b) 8. (a) 9. (a) 10. (a)
11. (c) 12. (d) 13. (c) 14. (b)
Work Book Exercise 3.2
1. (c) 2. (b) 3. (d) 4. (d) 5. (a) 6. (b) 7. (d) 8. (c) 9. (c) 10. (c)
11. (c) 12. (c) 13. (b) 14. (a) 15. (d)
Work Book Exercise 3.3
1. (b) 2. (c) 3. (a) 4. (a) 5. (a) 6. (a) 7. (a) 8. (b) 9. (a) 10. (a)
Work Book Exercise 3.4
1. (b) 2. (a) 3. (c) 4. (c) 5. (a) 6. (b) 7. (a) 8. (b) 9. (c) 10. (d)
Target Exercises
1. (a) 2. (c) 3. (d) 4. (c) 5. (d) 6. (a) 7. (b) 8. (a) 9. (b) 10. (d)
11. (c) 12. (a) 13. (c) 14. (c) 15. (b) 16. (d) 17. (d) 18. (a) 19. (b) 20. (c)
21. (c) 22. (a) 23. (b) 24. (c) 25. (a) 26. (a) 27. (c) 28. (a) 29. (a) 30. (c)
31. (b) 32. (c) 33. (a) 34. (b) 35. (b) 36. (b) 37. (b) 38. (c) 39. (b) 40. (a)
Targ e t E x e rc is e s
41. (d) 42. (a) 43. (d) 44. (b) 45. (a) 46. (a) 47. (b) 48. (c) 49. (c) 50. (a)
51. (c) 52. (b) 53. (a) 54. (c) 55. (a) 56. (a) 57. (a) 58. (c) 59. (c) 60. (c)
61. (d) 62. (a) 63. (c) 64. (c) 65. (c) 66. (a) 67. (c) 68. (d) 69. (a) 70. (a)
71. (b) 72. (c) 73. (b) 74. (b) 75. (a) 76. (c) 77. (b) 78. (a) 79. (a) 80. (b)
81. (d) 82. (a) 83. (b) 84. (b) 85. (b) 86. (c) 87. (d) 88. (b) 89. (c) 90. (b)
91. (a) 92. (a) 93. (d) 94. (a) 95. (c) 96. (b) 97. (a) 98. (b) 99. (c) 100. (b)
101 (b) 102. (a) 103. (c) 104. (c) 105. (a) 106. (d) 107. (a) 108. (c) 109. (b) 110. (a)
111. (c) 112. (d) 113. (b) 114. (c) 115. (b,d) 116. (b,d) 117. (a,c) 118. (c,d) 119. (a,c) 120. (b,d)
121. (a,c) 122. (b,d) 123. (b) 124. (a) 125. (d) 126. (a) 127. (d) 128. (c) 129. (a) 130. (c)
131. (c) 132. (b) 133. (b) 134. (a) 135. (b) 136. (* ) 137. (**) 138. (3) 139. (2) 140. (2)
141. (5)
* A → q,r; B → p, q, r; C → p, q, r; D → s
** A → p, r; B → q; C → s, t
Entrances Gallery
1. (4) 2. (b) 3. (a,d) 4. (d) 5. (3) 6. (8) 7. (4) 8. (0) 9. (b) 10. (b)
11. (d) 12. (c) 13. (d) 14. (c) 15. (d) 16. (b) 17. (c) 18. (b) 19. (a) 20. (b)
21. (d) 22. (c) 23. (c) 24. (a) 25. (a) 26. (a) 27. (b) 28. (b) 29. (d) 30. (b)
31. (b) 32. (b) 33. (b) 34. (c) 35. (b) 36. (c) 37. (a) 38. (a) 39. (a) 40. (c)
41. (d) 42. (c) 43. (a) 44. (b) 45. (c) 46. (a) 47. (c) 48. (d) 49. (a) 50. (c)
51. (b) 52. (d) 53. (b) 54. (c) 55. (b) 56. (c) 57. (b) 58. (a) 59. (a) 60. (a)
61. (a) 62. (b)
87
Explanations
Target Exercises
5 = 2 2 + 1, 9. Given, log 3 2, log 3 (2 x − 5) and log 3 2 x − are in AP.
1. We have, 7
14 = 3 + 5,
2 2
30 = 42 + 14, 7
∴ 2 log 3 (2 x − 5) = log 3 2 + log 3 2 x −
2
55 = 52 + 30,
⇒ 2 x = 8 or 2 x = 4
i.e. square of term is added to previous term.
∴ Next term = 62 + 55 = 36 + 55 = 91 ⇒ x = 3 or x = 2
Neglecting x = 2 as (2 x − 5) < 0 for x = 2
2. We have, 3 = 1+ 2 ⇒ x=3
6= 3+ 3
10. Given,Tp = q ⇒a + ( p − 1)d = q ...(i)
10 = 6 + 4
∴ Next term = 10 + 5 = 15 and Tq = p ⇒ a + (q − 1)d = p ...(ii)
On subtracting Eq. (ii) from Eq. (i), we get
3. Given, 5T5 = 8T8
⇒ ( p − q )d = − ( p − q ) ⇒ d = − 1
⇒ 5 (a + 4 d ) = 8(a + 7d )
On putting d = −1 in Eq. (i), we get
∴ 3 a + 36 d = 0
a + ( p − 1) (−1) = q
⇒ a + 12d = 0
⇒ a= p+ q −1
⇒ T13 = 0
∴ Tr = a + (r − 1)d
4. We have, a3 = a2 − 1 = 2 − 1 = 1 = ( p + q − 1) + (r − 1) (−1)
a4 = a3 − 1 = 1 − 1 = 0 = p + q − 1− r + 1= p + q − r
a5 = a4 − 1 = 0 − 1 = − 1 x
Ta rg e t E x e rc is e s
1 1
11. Here, − =
5. Let a and d be the first term and common difference 1− x 1+ x 1− x
respectively of given AP. Then, 1 1 x
T2 = x − y and − =
1− x 1− x 1− x
⇒ a+d = x − y
and T5 = x + y Hence, these terms are in AP.
⇒ a + 4d = x + y 1 1 1
12. Since, , and are in AP.
5 a b c
⇒ a= x − y 1 1 2
3 ∴ + = …(i)
a c b
6. Given a, b and c are in AP.
1 1 1 1 1 1
⇒ 2b = a + c Now, + − + −
a b c b c a
⇒ 3 b = a + b + c = 2s
s(s − c ) 1 1 2 1 1 1 2 1
C
Now, cot = = + − − + − −
2 (s − a)(s − b) a b b a b c b c
[from Eq. (i)]
2 s(2 s − 2c ) 3b(2 s − 2c )
= = 2 1 2 1 4 2 1 1 1
(2 s − 2 a) (2 s − 2 b) (2 s − 2 a) (3b − 2 b) = − − = − + +
a b c b ac b a c b2
3 (s − c ) s −c
= =3 4 2 2 1 4 3
s−a 3 (s − a) = − + 2 = − 2
ac b b b ac b
b(s − c ) (2 s − 2 b) (s − c )
=3 =3 13. Let a and d be the first term and common difference,
3 b (s − a) 2 s(s − a)
respectively of given AP, then
( s − b) ( s − c ) a7 = 34 ⇒ a + 6 d = 34 ...(i)
=3
s(s − a) and a13 = 64 ⇒ a + 12d = 64 ...(ii)
A On solving Eqs. (i) and (ii), we get
= 3 tan
2 a = 4, d = 5
7. Let x be the common difference of AP a, b, c, d, e, f. ∴ a18 = a + 17d = 4 + 85 = 89
Then, e − c = (a + 4 x ) − (a + 2 x ) 14. Let roots of given equation in AP be a − d , a and a + d .
= 2 x = 2 (d − c ) Then, (a − d ) + a + (a + d ) = 12
8. From given options, if we consider b = − 1, cos x = b ⇒ a=4
⇒ cos x = − 1, which is satisfied for x = π, 3π, 5π, ..., Also, (a − d ) a (a + d ) = 28
which form an AP with common difference 2π. For other ⇒ a (a2 − d 2 ) = 28 ⇒ 4 (16 − d 2 ) = 28
88 values of b, the roots of the equation are not in AP. ⇒ 16 − d 2 = 7 ⇒ d 2 = 9 ⇒ d = ± 3
x+1 x+2 x+a
15. x + 2 x + 3 x + b
x+3 x+4 x+c
21. We have, first term = a
∴ T1 = a
and second term = b
3
Sequence and Series
Applying C1 → C1 − C2 ∴ T2 = b
−1 x + 2 x+a Then, common difference
= −1 x + 3 x + b d = T2 − T1 = b − a
−1 x + 4 x +c Also, last term = c ⇒ c = a + (n − 1)d
c −a+d
1 x+2 x+a ⇒ n=
d
= (−1) 1 x + 3 x+b (b + c − 2 a)
⇒ n= [Qd = b − a]
1 x+4 x+c (b − a)
Applying R2 → R2 − R1, R3 → R3 − R1 ∴ Sum of n terms,
1 x+2 x+a n (b + c − 2 a) (a + c )
Sn = (a + l ) =
= (−1) 0 1 b−a 2 2(b − a)
0 2 c −a 22. The given series is arithmetic whose first term is 20 and
= (−1) [c + a − 2 b] = 0 common difference is − .
2
[since, a, b and c are in AP, so 2b = a + c ] 3
As the common difference is negative, the terms will
16. Given a + , b + and c + are in AP.
1 1 1 1 1 1
become negative after some stage. So, the sum is
b c
c a a b maximum, if only positive terms are added.
a(b + c ) b(c + a) c (a + b) 2
⇒ , , are in AP. Now, t n = 20 + (n − 1) − ≥ 0,
bc ca ab 3
⇒ (b − a) (cb + ca + ab) = (c − b) (ac + ab + cb)
if 60 − 2 (n − 1) ≥ 0 ⇒ 62 ≥ 2 n ⇒ 31 ≥ n
⇒ b− a=c − b
So, the first 31 terms are non-negative.
⇒ a, b and c are in AP.
Targ e t E x e rc is e s
∴ Maximum sum,
17. Given a, b and c are in AP. 31 2
S 31 = 2 × 20 + (31 − 1) −
∴ 2b = a + c …(i) 2 3
Now, (a + 2 b − c ) (2 b + c − a) (c + a − b) 31
= { 40 − 20} = 310
= (a + a + c − c ) (a + c + c − a) (2 b − b) [from Eq. (i)] 2
= (2 a) (2c ) (b) = 4 abc 23. Required sum
1 1 1 = (1 + 2 + 3 + K + 199) − (3 + 6 + 9 + K + 198)
18. We know, , and are in AP, if
− (5 + 10 + 15 + K + 195) + (15 + 30 + 45 + K + 195)
a+ b a+ c b+ c
199 66
1 1 1 1 = (1 + 199) − (3 + 198)
− = − 2 2
a+ c a+ b b+ c a+ c 39 13
b− c a− b − (5 + 195) + (15 + 195)
i.e. = 2 2
a+ b b+ c = 199 × 100 − 33 × 201− 39 × 100 + 13 × 105 = 10732
i.e. b − c = a − b, which is true. 24. Let the four numbers in AP be
1 1 1 α − 3 β, α − β, α + β, α + 3 β.
19. Given, , and are in AP.
p+q q + r r+ p Given, α − 3 β + α − β + α + β + α + 3 β = 50
25
⇒
1
−
1
=
1
−
1 ⇒ 4 α = 50 ⇒ α =
q + r p+q r + p q + r 2
and α + 3 β = 4 (α − 3 β )
⇒ p2 − r 2 = q 2 − p2
2 2 2
⇒ 3 α = 15 β
So, q , p and r are in AP. α 25 5
∴ β= = =
20. The numbers between 100 and 500 that are divisible by 5 5×2 2
7 are 105, 112, 119, 126, 133, 140, 147, …, 483, 490, Hence, the four numbers are 5, 10, 15 and 20.
497. 25. Let the four numbers in AP be
Let such numbers be n. α − 3 β, α − β, α + β and α + 3 β.
Then, 497 = 105 + (n − 1) × 7 Given, α − 3 β + α − β + α + β + α + 3 β = 24
⇒ n = 57 ⇒ 4α = 24
The number between 100 and 500 that are divisible by and (α − 3 β ) (α − β ) (α + β ) (α + 3 β ) = 945
21 are 105, 126, 147, ..., 483. ⇒ (α 2 − 9 β 2 ) (α 2 − β 2 ) = 945
Let such numbers be m. ⇒ (36 − 9 β 2 ) (36 − β 2 ) = 945
Then, 483 = 105 + (m − 1) × 21
⇒ β 4 − 40 β 2 + 144 = 105
⇒ m = 19
∴Required number = n − m = 57 − 19 = 38 ⇒ β 4 − 40β 2 + 39 = 0 89
3 ⇒
⇒
β 4 − β 2 − 39 β 2 + 39 = 0
(β − 1) (β − 39) = 0
2 2
Since, numbers are integers.
30. Here, a = 10 and d = −
3
7
3
Then, t n = 10 + (n − 1) −
Objective Mathematics Vol. 1
7
∴ β 2 ≠ 39
Now, t n is positive, if
∴ β2 − 1 = 0
3
⇒ β = ±1 10 + (n − 1) − ≥ 0
7
Hence, the four integers are 3, 5, 7 and 9.
⇒ 70 − 3 (n − 1) ≥ 0
26. Given, S n = 3 n2 1
⇒ 73 ≥ 3 n ⇒ 24 ≥n
∴ Tn = S n − S n −1 3
= 3 n 2 − 3 (n − 1)2 = 6 n − 3 So, first 24 terms are positive.
Now, sum of the positive terms,
On putting n = 1, 2, 3, ..., the sequence is 3, 9, 15, 21, ...
24 −3
which is clearly an AP. S 24 = 2 × 10 + 23 ×
2 7
27. The first two digit number which when divided by 4
69 852
leaves remainder 1 is 4 ⋅ 3 + 1 = 13 and last is = 12 20 − =
7 7
4 ⋅ 24 + 1 = 97.
Thus, we have to find the sum of the series 31. Let nth term be the first negative term.
13 + 17 + 21 + K + 97 Then, nth term, t n < 0
which is an AP. ∴ 40 + (n − 1) (−2 ) < 0 ⇒ 42 − 2 n < 0
∴ 97 = 13 + (n − 1)4 ⇒ 2 n > 42 ⇒ n > 21
⇒ n = 22 ∴ Least value of n = 22
n
and S n = [a + l ] = 1113
[ + 97 ] Hence, first 21 terms of an AP are non-negative.
2 Since, sum will be maximum, if no negative terms is
= 11 × 110 = 1210 taken.
21
Ta rg e t E x e rc is e s
28. The least and the greatest numbers of three digits ∴ Maximum sum, S 21 = [2 ⋅ 40 + (20 ) (−2 )]
divisible by 7 are 105 and 994, respectively. 2
21
So, it is required to find the sum of the series = × 40 = 420
105 + 112 + 119 + K + 994 2
Here, a = 105, d = 7, an = 994 32. Let the first instalment be a and common difference of
Now, an = a + (n − 1)d an AP be d.
⇒ 994 = 105 + (n − 1) × 7 Given, 3600 = Sum of 40 terms
⇒ 994 − 105 = 7(n − 1) 40
⇒ 3600 = [2 a + (40 − 1)d ]
⇒ 889 = 7 (n − 1) 2
889 ⇒ 3600 = 20 [2 a + 39 d ]
⇒ n − 1=
7 ⇒ 180 = 2 a + 39 d …(i)
∴ n − 1 = 127 After 30 instalments, one-third of the debt is unpaid.
⇒ n = 127 + 1 = 128 3600
Hence, = 1200 is unpaid and 2400 is paid.
n 3
∴ Sum = [2 a + (n − 1)d ] 30
2 Now, 2400 = [2 a + (30 − 1)d ]
128 2
= [2 × 105 + (128 − 1) × 7 ]
2 ⇒ 160 = 2 a + 29 d …(ii)
= 64 (210 + 889) On subtracting Eq. (ii) from Eq. (i), we get
= 64 × 1099 20 = 10 d
= 70336 ∴ d =2
From Eq. (i),
29. The odd numbers of four digits which are divisible by 9
180 = 2 a + 39 ⋅ 2
are 1017, 1035, …, 9999.
⇒ 2 a = 180 − 78 = 102
These are in AP with common difference 18.
∴ a = 51
a = 1017, d = 18and l = 9999
Now, the value of 8th instalment
∴ nth term, an = a + (n − 1)d
= a + (8 − 1)d = 51 + 7 ⋅ 2 = ` 65
⇒ 9999 = 1017 + (n − 1) × 18
⇒ 18 n = 9999 − 999 = 9000 33. Let a be the length of the smallest side and d be the
⇒ n = 500 common difference.
n Here, n = 25 and S 25 = 2100
∴ S n = (a1 + an ) n
2 Now, S n = [2 a + (n − 1)d ]
500 2
= (1017 + 9999) 25
2 ⇒ 2100 = [2 a + (25 − 1)d ]
= 250 × 11016 2
90 = 2754000 ⇒ a + 12d = 84 ...(i)
∴
The largest side = 25th side
= a + (25 − 1)d = a + 24 d
a + 24 d = 20 a [given] ...(ii)
40. Two digit odd numbers are 11, 13, ..., 99. These clearly
form an AP with a = 11, d = 2 and n = 45.
∴ Required sum
3
Sequence and Series
On solving Eqs. (i) and (ii), we get n 45
= (a + l ) = (11 + 99) = 2475
1 2 2
a = 8,d = 6
3 41. S p − Sq = 2 a( p − q ) + ( p2 − q 2 − p + q )d = 2(q − p)
n
34. S n = [2 a + (n − 1)d ] ⇒ 2 a + ( p + q − 1)d = − 2
2 (p + q)
1 ∴ S p +q = [2 a + ( p + q − 1)d ]
Here, a = 1st term = 7 yr, d = 3 months = yr 2
4 −2( p + q )
S n = 250 yr =
n 1 2
⇒ 250 = 2 × 7 + (n − 1) × = − (p + q)
2 4
n n + 55 42. Given, S n = 3 n 2 − n
⇒ 250 =
2 4 If t n is the nth term of an AP, then
⇒ 2000 = n 2 + 55 n t n = S n − S n −1
= (3 n 2 − n ) − [3 (n − 1)2 − (n − 1)] = 6 n − 4
⇒ n 2 + 55 n − 2000 = 0
∴ t1 = 6 − 4 = 2
⇒ (n − 25) (n + 80 ) = 0
n
⇒ n = 25 43. We have, S n = [2 p + (n − 1)Q ]
∴ Number of members in the club = 25 2
n By definition, d = Q
35. S n = [2 a + (n − 1)d ]
2 44. Given, S n = 5 n 2 + 2 n
n
⇒ 60100 = [4 + (n − 1)(3)] ⇒ S1 = 5 + 2 = 7
2
n and S 2 = 5 (4) + 2(2 ) = 24
⇒ [3 n + 1] = 60100
Targ e t E x e rc is e s
2 ∴ T2 = S 2 − S1
⇒ n = 200 = 24 − 7 = 17
36. Tn = S n − S n −1 n
45. Given, t n = + y
x
= [3 n + 5 n ] − [3 (n − 1) + 5 (n − 1)]
2 2
1 2
= [3 n 2 + 5 n ] − [3 (n 2 − 2 n + 1) + 5 (n − 1)] ⇒ t1 =
+ y and t 2 = + y
x x
= 6n + 2 ⇒ d = t 2 − t1 =
1
Let Tn = 164 x
Then, 164 = 6n + 2 ⇒ 6 n = 162 ⇒ n = 27 where, d is common difference of given AP.
r
Hence, 164 is 27th term. ∴ S r = [2 a1 + (r − 1)d ]
2
37. Here, a = 1, l = 11and S n = 36 r (r + 1) 1
n = + ry Q a1 = + y
Q Sn = (a + l ) 2x x
2
∴ 36 =
n
(1 + 11) = 6 n ⇒ n = 6 46. S n + 3 − 3 S n + 2 + 3 S n + 1 − S n
2 = (S n + 3 − S n + 2 ) − 2(S n + 2 − S n + 1 ) + (S n + 1 − S n )
38. Given series is 2 + 2 2 + 3 2 + 4 2 + ... = Tn + 3 − 2Tn + 2 + Tn + 1
= 2 [1 + 2 + 3 + 4 + K upto n terms] = (Tn + 3 − Tn + 2 ) − (Tn + 2 − Tn + 1 )
n(n + 1) n(n + 1) =d −d = 0
= 2 (∑ n ) = 2 Q∑ n =
2 2 a + be y b + ce y c + de y
47. = =
n(n + 1) a − be y b − ce y c − de y
∴ Sn =
2 Applying componendo and dividendo rule, we get
1 1 1 b c d
39. Let S n = + + = =
1+ 3 3+ 5 5+ 7 a b c
1 ⇒ a, b, c and d are in GP.
+K+
2n − 1 + 2n + 1 48. Given, Tm+ n = p and Tm− n = q
3 − 1 5 − 3 7 − 5 ⇒ ar m+ n −1 = p …(i)
= + +
2 2 2 and ar m− n −1 = q …(ii)
2 n + 1 − 2 n − 1 On multiplying Eqs. (i) and (ii), we get
+…+ a2 r 2 m− 2 = pq
2
⇒ (ar m−1 )2 = pq ⇒ ar m−1 = pq
2n + 1 − 1
= ⇒ Tm = pq 91
2
3 49. Since,1/ 2 cosec 2θ, 2 cot θ and sec θ are in GP.
∴
1
2
cosec 2θ ⋅ sec θ = 4 cot 2 θ
⇒ a − 2 ar + ar 2 = 8
On subtracting Eq. (ii) from Eq. (i), we get
3 ar = 48 ⇒ a =
16
…(ii)
Objective Mathematics Vol. 1
…(iii)
1 π r
⇒ cos θ = ⇒ θ= 16
2 3 On substituting a = in Eq. (i), we get
r
50. Let a and r be the first term and the common ratio 16
respectively of GP. + 16 + 16 r = 56
r
Then, T10 = 9 ⇒ ar 9 = 9
⇒ 16 r 2 − 40 r + 16 = 0
and T4 = 4 ⇒ ar 3 = 4
⇒ 2r 2 − 5 r + 2 = 0
∴ T7 = ar 6 = (ar 9 ⋅ ar 3 )1/ 2 = (9 × 4)1/ 2 = 6
⇒ (r − 2 ) (2 r − 1) = 0
51. Let A and R be the first term and the common ratio 1
∴ r = 2,
respectively of given GP. Then, 2
T4 = p ⇒ AR 3 = p If r = 2, then from Eq. (iii), a = 8 and the numbers are
T7 = q ⇒ AR 6 = q 8, 16, 32.
1
and T10 = r ⇒ AR 9 = r If r = , then from Eq. (iii), a = 32 and the numbers are
2
Now, ( AR 3 ) ( AR 9 ) = A2 R12 = ( AR 6 )2 ⇒ pr = q 2 32, 16, 8.
52. Let a and r be the first term and the common ratio 57. Let the sides of the right angled triangle be a, ar and ar 2 ,
respectively of GP. out of which ar 2 is the hypotenuse, then r > 1.
Then, a 6 = 32 ⇒ ar 5 = 32 …(i) Now, a2 r 4 = a2 + a2 r 2
and a8 = 128 ⇒ ar = 128 7
…(ii) 1± 5
⇒ r4 − r2 − 1= 0 ⇒ r2 =
On dividing Eq. (ii) by Eq. (i), we get 2
r2 = 4 ⇒ r = ± 2 Q r >1
∴ r2 > 1
Ta rg e t E x e rc is e s
53. Let r be the common ratio of a GP, then 1+ 5
⇒ r2 =
Area formed by three points 2
x1 y1 1 1 1 1 Let ∠C be the greater acute angle.
1 1
= x1r y1r 1 = x1 y1 r r 1 =0 a 1 2
2 2 Now, cos C = 2 = 2 =
x1r 2 y1r 2 1 r2 r2 1 ar r 1+ 5
So, points are collinear, i.e. lie on a straight line. 58. Let the three digits be a, ar and ar 2 .
54. Let a = b =c = k
x y z
According to the hypothesis,
⇒ a = k1/ x , b = k1/ y and c = k1/ z 100 a + 10 ar + ar 2 − 792 = 100 ar 2 + 10 ar + a
Since, a, b and c are in GP. ⇒ a(1 − r 2 ) = 8 ...(i)
∴ b2 = ac ⇒ (k1/ y )2 = k1/ x k1/ z
and a, ar + 2, ar are in AP.
2
2 1 1
⇒ k 2 / y = k1/ x + 1/ z ⇒ = + ∴ 2(ar + 2 ) = a + ar 2
y x z
1 1 1 ⇒ a(r − 2 r + 1) = 4
2
...(ii)
⇒ , and are in AP.
x y z On dividing Eq. (i) by Eq. (ii), we get
⇒ x, y and z are in HP. a(1 − r 2 ) 8
=
55. Let d be the common difference of an AP and R (≠ 0 ) be a(r 2 − 2 r + 1) 4
(1 + r )(1 − r ) r +1
the common ratio of a GP. Then, ⇒ =2 ⇒ =2
q = p + d , r = p + 2d and y = xR, z = xR 2 (r − 1)2 1− r
Q q − r = − d , r − p = 2d and p − q = − d ∴ r = 1/ 3
∴ xq −r
⋅y r −p
⋅ z p − q = x − d ⋅ ( xR )2d ⋅ ( xR 2 )− d From Eq. (i), a = 9
Thus, digits are 9, 3, 1 and so the required number
= ( x − d ⋅ x 2d ⋅ x − d ) (R 2d ⋅ R −2d ) is 931.
= ( x − d + 2d − d )⋅ (R 2d − 2d )
59. Given a, b and c are in AP.
= x 0 ⋅ R0 = 1 × 1 = 1 ⇒ 2b = a + c …(i)
56. Let the three numbers in GP be a, ar and ar 2 . Also, given a + 1, b, c are in GP and a, b, c + 2 are also
in GP.
∴ a + ar + ar 2 = 56 [given]…(i)
b2 = (a + 1)c …(ii)
On subtracting 1, 7, 21 from the numbers, we get
and b2 = a(c + 2 ) …(iii)
a − 1, ar − 7 , ar 2 − 21, which are given to be in AP.
On solving Eqs. (i), (ii) and (iii), we get
∴ (ar − 7 ) − (a − 1) = (ar 2 − 21) − (ar − 7 )
a = 8, b = 12 and c = 16
⇒ ar − a − 6 = ar 2 − ar − 14
92 Hence, b = 12
60. The given series is a GP with a = 0.4.
r=
1
69. Let a and r be the first term and the common ratio
respectively of given GP. Then,
100
3
100 α= ∑
Sequence and Series
a2 n
a 40 a n =1
∴ S = = Q S ∞ = 1 − r
1 − r 99 ar (1 − r 200 ) a(1 − r n )
⇒ α= Q S n = 1 − r
61. The given series is a GP with (1 − r 2 )
− e −2 x a(1 − r 200 )
a = e − x, r = = −e−x ⇒ β=
e−x 1− r2
a e−x 1 α
∴ S = = = [Q − e − x < 1, for x > 0 ] ⇒ =r
1− r 1+ e−x 1+ e x β
128 r − a 256 − a 70. Given, f ( x + y ) = f ( x )f ( y ), ∀ x, y ∈ N
62. We have, = 255 ⇒ = 255 [Qr = 2]
r −1 2 −1
For any x ∈ N, f ( x ) = [f (1)]x = 3 x [Q f(1) = 3]
⇒ 256 − a = 255 ⇒ a = 1 n
63. Clearly,
( x + 2 )n − ( x + 1)n ∴ ∑ f ( x ) = 120
x =1
( x + 2 ) − ( x + 1) n
= ( x + 2 )n − 1 + ( x + 2 )n − 2 ( x + 1) ⇒ ∑ 3 x = 120
n−3 n −1 x =1
+ ( x + 2) ( x + 1) + … + ( x + 1)
2
⇒ 31 + 32 + 33 + K + 3n = 120
∴Required sum = ( x + 2 )n − ( x + 1)n
⇒ 3n − 1 = 80
[Q( x + 2 ) − ( x + 1) = 1]
⇒ 3n = 81 = 34 ⇒ n = 4
64. Here, c = ar , e = ar 2 , d = bs, f = bs2
a b 1 71. 0.423 = 0.4 + 0.023 + 0.00023 + K
1
∴ Area of triangle = c d 1 4 23 23 23 419
2 = + + + +K=
Targ e t E x e rc is e s
e f 1 10 10 3 10 5 10 7 990
1
= ab(s − r )(s − 1)(r − 1)
72. Given, x = 1 + y + y 2 + K
2 1 1
⇒ x= ⇒ 1− y =
65. Let S denotes the sum of all terms and S1 denotes the 1− y x
sum of odd terms. 1 x −1
⇒ y = 1− =
a x x
S1 = (i.e. sum of odd terms)
1− r2 73. Let the GP be a, ar, ar 2 and ar 3 .
1 a 1 a Then, a (1 + r + r 2 + r 3 )
Given, S = ⋅ S1 ⇒ = ⋅
5 1− r 5 1− r2 1 − r 4
1 4 =a×
⇒ 1+ r = ⇒ r =− 1− r
5 5
= 12 (1 − 5 ) [given]
66. The given series is a GP with a = 2, r = 3 > 1 12 (1 + 5 ) (1 − 5 )
=
a(r 10 − 1) 2 [( 3 )10 − 1] 1+ 5
∴ S10 = =
r −1 3 −1 − 48
=
2 1+ 5
= (242 )( 3 + 1) = 121 ( 6 + 2 )
2 a (1 − 25 − 48
⇒ =
49 1+ 5 1+ 5
67. ∑ (105
. )n = 105
. + (105 . )3 + … + (105
. )2 + (105 . )49 ⇒ a=2
n =1
74. Sum of areas of all the squares
105 . )49 − 1]
. [(105
= a2 a2
. −1
105 = a2 + + +K
2 4
= 212 .16 a2
= = 2 a2
68. Let a and r be the first term and the common ratio 1
1−
respectively of given infinite GP. Then, 2
ar = 2 …(i) 75. Let a be the first term and r (| r | < 1) be the common ratio
a
Also, =8 …(ii) of infinite GP. Then,
1− r
a a2
2 = 3 and =3 …(i)
From Eq. (i), r = 1− r 1− r2
a
a2
On putting the value of r in Eq. (ii), we get Also, =9 …(ii)
a=4 (1 − r )2 93
3 On solving Eqs. (i) and (ii), we get
a=
3
2
81. Let S n = 0.5 + 0.55 + 0.555 + K upto n terms
=
5
[0.9 + 0.99 + 0.999 + K upto n terms]
Objective Mathematics Vol. 1
1 9
and r= 5 1 1 1
2 = 1 − + 1 − + 1 − + K
Hence, the sequence corresponding to the series will 9 10 100 1000
3 3 3 5 1 1 1
be , , , K = (1 + 1 + 1 + K ) − + + + K
2 4 8 9 10 10 2 10 3
1
1
2 1 −
1 1 1 1 5 1 10 n 5 1 10 1
+ + + ... 1− = n − ⋅ = n− × 1 −
10 1 − 1 9 10 9 10 n
2
76. Given product = x 2 4 8 =x =x 9
10
77. QLength of a side ofS n = Length of a diagonal ofS n + 1
5 1 1
⇒ Length of a side of S n = 2(Length of a side of S n + 1) = n − 1 − n
9 9 10
Length of a side of S n + 1 1
⇒ = ,∀ n ≥ 1 82. Let the three digits be a, ar and ar 2 .
Length of a side of S n 2
Given, a + ar 2 = 2 ar + 1
⇒ Side of S1, S 2 , K , S n form a GP with common ratio
1 ⇒ a(r − 2 r + 1) = 1
2
and first term 10. ⇒ a(r − 1)2 = 1 …(i)
2
n −1 2
1 Also, given a + ar = (ar + ar ) 2
∴ Side of S n = 10
2 3
10 ⇒ 3 a(1 + r ) = 2 ra (1 + r )
= n −1 ⇒ (1 + r ) (3 − 2 r ) = 0 [Q a ≠ 0 ]
3
2 2 ∴ r = , −1
100 2
⇒ Area of S n = (S n )2 = n − 1
Ta rg e t E x e rc is e s
3 1 1
2 If r = , then from Eq. (i), a = = =4
Area of S n < 1 [given] 2 (r − 1)2 3 2
Q − 1
100 2
⇒ <1
2n − 1 1
If r = − 1, then from Eq. (i), a = , which is not possible,
⇒ 2 n − 1 > 100 4
⇒ n − 1≥ 7 as a is an integer.
3 9
⇒ n≥8 Hence, a = 4, ar = 4 ⋅ = 6 and ar 2 = 4 ⋅ = 9
2 4
78. Let r be the common ratio of the given GP Then, terms of
GP are α , αr αr 2 and αr 3 . 83. Let the two numbers be a and b, then
G = ab or G 2 = ab
According to the question,
Also, p and q are two AM’s between a and b.
α (1 + r ) = 1, α 2 r = p, α r 2 (1 + r ) = 4, α 2 r 5 = q
∴ a, p, q and b are in AP.
On solving, we get ( p, q ) = (− 2, − 32 ) Now, p − a = q − p and q − p = b − q
79. Consider, (1 + x ) + (1 + x + x 2 ) ∴ a = 2 p − q and b = 2q − p
+ (1 + x + x 2 + x 3 ) + K upto n terms Hence, G 2 = ab = (2 p − q ) (2q − p)
1− x 2
1− x 3
1 − x4 84. Let the 3 n terms of a GP be a, ar, ar 2 , …, ar n − 1, ar n ,
= + + + K upto n terms
1− x 1− x 1− x ar n + 1, ar n + 2 , …, ar 2 n − 1, ar 2 n , ar 2 n + 1, ar 2 n + 2 ,…, ar 3 n − 1.
=
1
[(1 + 1 + 1 + K to n terms) Then, S1 = a + ar + ar 2 + K + ar n − 1
1− x a(1 − r n )
− (x 2 + x 3 + x 4 + K upto n terms)] =
1− r
1 x 2 (1 − x n ) S 2 = ar n + ar n + 1 + ar n + 2 + K + ar 2 n − 1
= n−
1− x 1 − x ar n (1 − r n )
=
80. We have, 9 + 99 + 999 + K upto n terms 1− r
= (10 − 1) + (10 2 − 1) + (10 3 − 1) + K upto n terms S 3 = ar 2 n + ar 2 n + 1 + ar 2 n + 2 + K + ar 3 n − 1
= (10 + 10 2 + 10 3 + K upto n terms) − n ar 2 n (1 − r n )
=
10(10 n − 1) 1− r
= −n 2 2 n (1 − r )
n 2
10 − 1 Now, (S 2 ) = a r
2
(1 − r )2
10(10 − 1) n
= −n a(1 − r ) 2 n (1 − r n )
n
9 = ⋅ ar = S1 S 3
1 1− r 1− r
94 = (10 n + 1 − 9n − 10 )
9 Hence, S1, S 2 and S 3 are in GP.
85. (32 )(32 )1/ 6 (32 )1/ 36 K = 32
= 32
1+
1
+
6 36
= (2 ) = 2 = 64
1
+K
11
1−
= 32 6
93. We have, 2 n + 10 = 2 ⋅ 2 2 + 3 ⋅ 2 3 + 4 ⋅ 2 4 + K + n ⋅ 2 n
⇒ 2(2 n + 10 ) = 2 ⋅ 2 3 + 3 ⋅ 2 4 + K + (n − 1)⋅ 2 n + n ⋅ 2 n + 1 3
Sequence and Series
6/ 5 5 6/ 5 6 On subtracting, we get
4 7 10 − 2 n + 10 = 2 ⋅ 2 2 + 2 3 + 2 4 + K + 2 n − n ⋅ 2 n + 1
86. Let S = 1 + + 2 + 3 + K
5 5 8 (2 n − 2 − 1)
5 = 8+ − n ⋅ 2n + 1
1 1 4
⇒ S = + 2 + 3 +K
7 2 −1
5 5 5 5 = 8 + 2 n + 1 − 8 − n ⋅ 2 n + 1 = 2 n + 1 − (n )2 n + 1
On subtracting, we get ⇒ 2 10
= 2n − 2
4 3 3 3
S = 1+ + 2 + 3 + K ⇒ n = 513
5 5 5 5
7 5 35 94. Let S = (1)(2003) + (2 )(2002 ) + (3)(2001) + K + (2003)(1)
⇒ S = × =
4 4 16 and K = 12 + 2 2 + 32 + K + 20032
87. Let S = 1 + 2 x + 3 x + 4 x + K
2 3
On adding, we get
x S = x + 2 x 2 + 3x 3 + K S + K = (2004)[1 + 2 + 3 + K + 2003]
On subtracting, we get ⇒ (2003)(334)( x ) + (2003)(4007 )(334)
(1 − x )S = 1 + x + x 2 + x 3 + K = (2004)(1002 )(2003)
1 1 ⇒ x = 2005
⇒ (1 − x )S = ⇒ S =
1− x (1 − x )2 95. We can write the given equation as
1 1 1 1
88. Let S = 1 + 2 ⋅ 2 + 3 ⋅ 2 + 4 ⋅ 2 + K + 100 ⋅ 2
2 3 99 1+ + + + +K
log 2 x 2 4 8 16 =4
2S = 1⋅ 2 + 2 ⋅ 2 2 + 3 ⋅ 2 3 + 4 ⋅ 2 4
⇒ log 2 ( x 2 ) = 4 ⇒ x 2 = 24 ⇒ x=4
+ K + 99 ⋅ 2 99 + 100 ⋅ 2100
96. We shall make use of the identity
On subtracting, we get
(a12 + a22 + K + am
2
)(b12 + b22 + K + bm2 )
− S = 1 + (1⋅ 2 + 1⋅ 2 2 + 1⋅ 2 3 + K to 99 terms − 100 ⋅ 2100 )
Targ e t E x e rc is e s
− (a1b1 + a2 b2 + K + ambm )2
= 1 + 2(2 99 − 1) − 100 ⋅ 2100
= (a1b2 − a2 b1 ) + (a1b3 − a3 b1 )2 + K+ (am − 1bm − ambm − 1 )2
2
⇒ S = 99 ⋅ 2100 + 1
n 2 (n + 1)2 Thus, ( x12 + x22 + K + xn2 − 1 ) ( x22 + x32 + K + xn2 )
Σn 3
4 − ( x1 x2 + x2 x3 + K + xn −1 xn )2 ≤ 0
89. Here, Tn = =
n n 2
⇒ ( x1 x3 − x2 x2 ) + ( x1 x4 − x3 x3 )2
2
[2 × 1 + (n − 1)2 ]
2 + K + ( xn − 2 xn − xn − 1 xn − 1 )2 ≤ 0
16
1
⇒ S16 = ∑ (k + 1)2 As x1, x2 , …, xn are real, this is possible if and only if
4 k =1
x1 x3 − x22 = x2 x4 − x32 = K = xn − 2 xn − xn2 − 1 = 0
= 446
x1 x2 x3 x
n(n + 1) n(n + 1) ⇒ = = =K= n
x2 x3 x4 xn − 1
90. Tn = 2 ⋅ 2 = 2 4 2
Σ n3 n (n + 1) ⇒ x1, x2 , …, xn are in GP.
4 97. We have,
n 2n n
1 1 1
= = −
n(n + 1) n n + 1 ∑ (2 r − 1)4 = ∑ r4 − ∑ (2 r )4
r =1 r =1 r =1
1 n
∴ Sn = 1 − = = f (2 n ) − 16 f (n )
n+1 n+1
1 2 (2 r + 1) − (2 r − 1)
91. We can write S as 98. We have, = 2 =
2r 2 4r 1 + (2 r + 1)(2 r − 1)
S = (1 − 2 )(1 + 2 ) + (3 − 4)(3 + 4)
1 (2 r + 1) − (2 r − 1)
+ K + (2001 − 2002 )(2001 + 2002 ) + 20032 tan −1 2 = tan −1
2r 1 + (2 r + 1) (2 r − 1)
= − [1 + 2 + 3 + 4 + K + 2002 ] + 20032
1 = tan −1(2 r + 1) − tan −1(2 r − 1)
= − (2002 )(2003) + (2003)2 = 2007006 n n
2 1
92. According to the given condition,
⇒ ∑ tan−1 2 r 2 = ∑ { tan−1 (2 r + 1) − tan−1(2 r − 1)}
r =1 r =1
1 1984 (2 ) (− 1)n = tan −1(2 n + 1) − tan −1(1)
(704)(2 ) 1 − n = 1 − n
2 3 2 π
= tan –1(2 n + 1) −
2112 (−1)n 4
⇒ 128 = n − 1984 n n
−1 1 −1 π
2 2 ⇒ lim ∑ tan 2 = lim tan (2 n + 1) −
If n is odd, then we get 2 n = 32 ⇒ n = 5 n→ ∞ 2r n → ∞ 4
r =1
128 π π π
If n is even, then we get 128 = n ⇒ n = 0 = − =
2 95
2 4 4
3 99. 13 = 1⋅ (1 − 1) + 1
2 3 = (2 ⋅ 1 + 1) + 5
3 = (3 ⋅ 2 + 1) + 9 + 11
3
,
106. Given, log(a + c ), log (c − a) and log (a − 2 b + c ) are
in AP.
⇒ 2 log(c − a) = log (a + c ) + log (a − 2 b + c )
Objective Mathematics Vol. 1
,
⇒ (c − a)2 = (a + c ) (a − 2 b + c )
43 = (4 ⋅ 3 + 1) + 15 + 17 + 19, etc
⇒ (c − a)2 = (a + c )2 − 2 b (a + c )
∴ n 3 = { n ⋅ (n − 1) + 1} + K, in which next term being 2
more than the previous ⇒ 2 b (a + c ) = (a + c )2 − (c − a)2 = 4 ac
∴ n3 = (n2 − n + 1) + (n2 − n + 3) + K + (n2 + n − 1) 2 ac
⇒ b=
r −1 a+c
r
100. ∑ r − r − 1 = −n
n 2 n
∑ r ! − (r + 1)! = (n + 1)! ⇒ a, b and c are in HP.
r = 1 ( r + 1 )! r =1
n 107. Given, cos ( x − y ), cos x and cos ( x + y ) are in HP.
=− 2 cos( x − y ) cos ( x + y )
(n + 1) n ⋅ (n − 1)! ⇒ cos x =
1 cos ( x − y ) + cos ( x + y )
=−
(n + 1)(n − 1)! 2 (cos 2 x − sin 2 y )
⇒ cos x =
2 cos x cos y
101. Applying AM ≥ GM y
Since, AM = GM ⇒ cos x = 1 + cos y = 2 cos 2
2
2
∴ ( 2 + 2 )x = ( 2 − 2 )x i.e. for x = 0 y
⇒ cos 2 x sec 2 = 2
2
[Q AM = GM iff a = b] y
⇒ cos x sec = ± 2
102. As, D < 0 2
⇒ (a + b + c )2 − 4 (ab + bc + ca) < 0 108. log 2 6 = log 2 (3 × 2 ) = log 2 3 + log 2 2 = 1 + log 2 3
⇒ (a − b + c ) < 4 ac
2
and log 2 12 = log 2 (2 2 × 3)
⇒ − 2 ac < a − b + c = log 2 3 + 2 log 2 2
⇒ ( a + c)> b = 2 + log 2 3
Ta rg e t E x e rc is e s
1 Since, log 2 3, 1 + log 2 3, 2 + log 2 3 are in AP.
103. Given , a and b are in GP.
16 ⇒ log 2 3, log 2 6, log 2 12 are in AP.
⇒ a2 =
b
…(i) ⇒ log 3 2, log 6 2, log12 2 are in HP.
16 109. Given a, b and c are in HP ⇒ b is HM of a and c
1 a+c
Also, a, b and are in HP. ⇒ >b [Q AM > HM]
6 2
1 2
2×a× a2 + c 2 a + c
⇒ b= 6 = 2a Now, >
…(ii) 2 2
a+
1 6a + 1
6 a +c
2 2
⇒ > b2
On solving Eqs. (i) and (ii), we get 2
1 1 ⇒ a2 + c 2 > 2 b2
a= , −
12 4
1
110. Q AM> GM> HM
and b = ,1 ∴ x>y>z ⇒ z<y<x
9
104. Since, a(b − c ) + b(c − a) + c (a − b) = 0 111. Given a, b and c are in HP.
1 1 1
∴ x = 1is a root of the given equation. ⇒ , , are in AP.
a b c
Since, both the roots are equal, therefore the other root
2 1 1
is also 1. ⇒ = +
b a c
∴Their sum = 1 + 1 = 2
− b(c − a) 1 2 1
⇒ =2 ⇒ − + =0
a(b − c ) a b c
x y 1
⇒ − bc + ab = 2 ab − 2 ac Hence, the straight line + + = 0 passes through
a b c
⇒ 2ac = b(a + c ) a fixed point (1, − 2 ).
2 ac
∴ b= 112. Given, a10 = 3 ⇒ a1 + 9 d = 3
a+c
Hence, a, b and c are in HP. ⇒ 2 + 9d = 3 [Q a1 = 2 ]
1
105. Let two numbers be a and b. Then, ⇒ d =
a+ b 9
= 5 ⇒ a + b = 10 1 7
2 ∴ a4 = a1 + 3 d = 2 + =
Also, ab = 4 ⇒ ab = 16 3 3
2ab 2(16) 16 1 1
∴HM between a and b = = = h10 = 3 ⇒ =
96 a+ b 10 5 h10 3
⇒
∴
1
D=−
=
1
1
54
+ 6D
118. If a, bandc are pth,qth and rth terms of an AP, then
is always a rational number.
b−c
a−b 3
Sequence and Series
h7 h1
119. Let the roots be a1, a2 , a3 , a4 , a5 , then
1 1
= − 1 1 1 1 1
2 9 + + + + = 10
7 a1 a2 a3 a4 a5
= Σ a1 a2 a3 a4
18 ⇒ = 10
18 a1 a2 a3 a4 a5
⇒ h7 =
7 ⇒ 10S = Σa1 a2 a3 a4
7 18 Since, a1, a2 , a3 , a4 , a5 are in GP, then consider
∴ a4 h7 = × =6
3 7 a1, a2 , a3 , a4 and a5 as a, ar, ar 2 , ar 3 , ar 4
113. Since, x, y and z are in HP. ⇒ 10S = 10 a5 r 10 = a3 r 6 (a + ar + ar 2 + ar 3 + ar 4 )
2 xz ⇒ 10 a5 r 10 = a3 r 6 (40 ) [Q sum of roots = 40]
∴ y=
x+ z
⇒ a2 r 4 = 4
4 xz
⇒ x − 2y + z = x + z − ⇒ ar 2 = ± 2
x+ z
⇒ S = a3 r 6 (4) = (± 2 )3 (4) = ± 32
( x + z )2 − 4 xz
=
x+ z 120. Since, a1, a2 , a3 , …, an , … are in AP and b1, b2 , … bn , …are
( z − x) 2 in GP.
= ⇒ 1, (1 + d ) , …, [1 + (n − 1)d ] … are in AP
x+ z
⇒ 1, r, r 2 , …, r n − 1, … are in GP.
⇒ ( x + z ) ( x − 2 y + z ) = ( z − x )2
Also, a9 = b9 ⇒ 1 + 8 d = r 8
⇒ log( x + z ) + log( x − 2 y + z ) = 2 log( z − x ) 9
1 1 ∴ S 9 = [ 2 + (9 − 1)d ] = 369
− =K 2
Targ e t E x e rc is e s
114. Let
Hi + 1 Hi ⇒ 2 + 8 d = 41 × 2
2n Hi + Hi + 1 ⇒ d = 10
∴ ∑ (− 1)i H ⇒ r 8 = 81
i =1 i − Hi + 1
⇒ r 8 = ( 3 )8
2n
(− 1)i 1 1
=∑ + = 2 n ⇒ r= 3
i =1
K Hi + 1 Hi ∴ b7 = ( 3 )6 = 9 × 3 = 27
115. Let the given AP be a1, a1 + d , a1 + 2d , … and b9 = 81
By substituting the value, we can find that only options 121. Given, sin β = sin α cos α
(b) and (d) are correct. ⇒ sin 2 β = sin α cos α
116. Let the AP be a, a + d , a + 2d , a + 3 d , K Now, cos 2β = 1 − 2 sin 2 β = 1 − 2 sin α cos α
Given that, 3 a + 3d = 9 = (sin α − cos α )2
⇒ a+d =3
π
and a2 + (a + d )2 + (a + 2d )2 = 35 = 2 sin 2 − α
4
⇒ a2 + (3)2 + (3 + d )2 = 35 π
or 2 cos 2 + α
⇒ (3 − d )2 + 32 + (3 + d )2 = 35 4
⇒ d =±2
122. Since, a is AM between 1st and (2 n + 1) th terms, b is GM
and a = 1, 5
n between 1st and (2 n + 1) th terms and c is HM between
∴ S n = [2 a + (n − 1) d ] 1st and (2 n + 1) th terms.
2
⇒ AM ≥ GM ≥ HM and AM × HM = (GM)2
where d = 2, a = 1
⇒ S n = n2 123. If each of x, y and z is less than 1, then Statement I is
When d = − 2, a = 5 , then S n = n(6 − n ) obviously true.
1 + a1a2 Also, 1 − 2 x + 1 − 2 y + 1 − 2 z = 3 − 2( x + y + z ) = 1,
−1 1 + a2 a3
117. cot −1 + cot a − a The sum of the three given numbers is positive also at
2
a − a1 3 2
must one of x, y and z can be more than 1.
1 + a3 a4 1 + an an − 1 1
+ cot −1 −1
+ K + cot a − a If one of x, y and z is more than or equal to , then their
4
a − a 3 n n −1
2
product is less than equal to zero, hence still remains
= cot −1 a1 − cot −1 a2 + cot −1 a2 − cot −1 a3 true.
+ K+ cot −1 an − 1 − cot −1 an Statement II is always true but it does not explain
= cot −1 a1 − cot −1 an = tan −1 an − tan −1 a1 Statement I.
97
3 124. Since, ax 2 + bx + c = 0 and a1 x 2 + b1 x + c1 = 0 have
common root.
(a1 c − ac1 )2 = 4 (ab1 − ba1) (bc1 − b1c ) …(i)
Let other root be α, then
1× α =
c ( a − b)
a(b − c )
Objective Mathematics Vol. 1
2 ac
a b c
If , , are in AP, then c a −
a1 b1 c 1 a + c
= =1
ba1 − ab1 cb1 − c 1b 2 ac
= =k a − c
a1 b1 b1c 1 a+ c
ca1 − ac1 ∴ α =1
and = 2k
a1 c1 Hence, both roots of Eq. (i) are 1, 1.
On putting these values in Eq. (i), we get Then, roots of x 2 − Px + Q = 0 are also 1, 1.
c1a1 = b12 Now, 1 + 1 = P, 1⋅ 1 = Q
125. Statement I is false, since each term of the series ∴ P = 2, Q = 1
1 1 1 1 ∴ [P ] = [2 ] = 2
+ + + + K is smaller than 10 −5 but its
10 6 10 6 10 6 10 6 Now, [2 P − Q ] = [4 − 1] = 3
sum upto infinity is infinity. Q Roots of Eq. (i) are 1, 1, then
n b (c − a)
Statement II is true, since lim is not finite as 1+ 1 = −
n→ ∞ 10 5 a (b − c )
n→ ∞ a (b − c ) 1
⇒ =−
126. Statement I can be proved by taking the intersection of b (c − a) 2
the inequalities. a n/ 2 an
133. An = 2 ∫ 2 an xdx + 2 ∫ 2 an an − xdx
a > 0, ar > 0, ar 2 > 0 0 a n/ 2
at a + ar > ar 2, ar + ar 2 > a, ar 2 + a > ar Y
y 2 = 4an (an – x) y 2 = 4an x
The inequalities follow from reason.
Ta rg e t E x e rc is e s
127. If we could show that Statement II is true, then (an, 0)
Statement I will be false. Indeed if Statement II is false, X′ X
O (an/2, 0)
then
2 − 3 = ( p − q )d ...(i)
and 3 − 5 = (q − r )d ...(ii)
Y′
On dividing Eq. (i) by Eq. (ii) we get
a a
2 − 3 p−q 2 2
3 n/ 2
2
3 n
= = 4 an ⋅ x − 4 an ⋅ (an − x )2
3− 5 q−r 3 3
0 a
⇒ Rational = Irrational n/ 2
8 a a 8 a a
Hence, Statement I is false and Statement II is true. = an n n − an 0 − n n
3 2 2 3 2 2
128. Statement I is true, since for any x > 0, we can choose
xn 2 ⋅ 8 an2 4 2 2
sufficiently larger n such that is small. Statement II is = = an sq units
n! 3⋅ 2 2 3
(n !)2 134. We have, y 2 = 4 x and y 2 = 4 − 4 x
false, since contains n ! in the denominator but
n! ⇒ 8x = 4
diverges to ∞. 1
⇒ x=
129. We can show that Statement I is true and follows from 2
Statement II. Indeed and y=± 2
a1 + a2 + K+ an + an + 1 a1 + a2 + K + an 135. For an = 1,
Sn + 1 − Sn = −
n+1 n 4 2
nan + 1 − (a1 + a2 + K + an ) An = sq units
= 3
n(n + 1)
(an + 1 − a1 ) + (an + 1 − a2 ) + K + (an + 1 − an )
136. aa bbc c = (aa K a times) (bb K b times) (cc K c times)
= >0 Applying AM-GM inequality,
n(n + 1)
(aa bbc c )1/ n
[ Q an + 1 > a1, an + 1 > a2 , K , etc. ]
a + a + K + a times ) + (b + b + K + b times )
Solutions (Q. Nos. 130-132) + (c + c + K + c times )
Given, a(b − c )x 2 + b(c − a)x + c (a − b) = 0 ...(i) ≤
n
Since, a(b − c ) + b(c − a) + c (a − b) = 0, therefore x = 1 a2 + b2 + c 2
is a root of Eq. (i). =
98 n
Similarly, (abbc c a )1/ n ≤
c a b 1/ n
≤
ab + bc + ca a2 + b2 + c 2
n
≤
ac + ba + cb a2 + b2 + c 2
≤
n
138. Let us add one more number an+1 to the given sequence.
The number an+1 is such that | an + 1 | = | an + 1|. On
squaring all the numbers, we have
3
Sequence and Series
Also, (a b c )
n n a12 = 0
So, (aa bbc c )1/ n + (abbc c a )1/ n + (ac bac b )1/ n a22 = a12 + 2 a1 + 1
(a + b + c ) 2
a32 = a22 + 2 a2 + 1
≤ =n
n a42 = a32 + 2 a3 + 1
Further, a<n ⇒ a <aa n
M M M M
∴ (aa bbc c ) < (abc )n ⇒ (aa bbc c )1/ n < (abc ) an2 = an2−1 + 2 an −1 + 1
3 an2+ 1 = an2 + 2 an + 1
137. A. Q A=G + …(i)
2 On adding the above equalities, we get
6
and G=H+ …(ii) a12 + a22 + K + an2 + an2+ 1
5
= a12 + a22 + K + an2
Q G 2 = AH
3 6 + 2(a1 + a2 + K + an ) + n
⇒ G 2 = G + G − [from Eqs. (i) and (ii)] ⇒ 2(a1 + a2 + K + an ) = − n + an2+ 1 ≥ − n
2 5
a1 + a2 + K + an 1 λ
∴ G = 6, A =
15
[from Eq. (i)] ⇒ ≥− =−
2 n 2 µ
Since, a and b are the roots of x − 15 x + 36 = 0.
2 ∴ λ+µ=3
On solving, we get a = 12, b = 3 or a = 3, b = 12 139. Let a be the first term and d be the common difference of
∴ α = 15, β = 9 an AP. Again, let x, y, z be the (m + 1) th, (n + 1) th and
⇒ α + β 2 = 15 + 81 = 96 (r + 1) th terms of an AP. Then, x = a + md , y = a + nd
and z = a + rd .
and α 2 + β = 225 + 9 = 234
Since, x, y and z are in GP.
A=G + 2
Targ e t E x e rc is e s
B. Q …(i) ∴ y 2 = xz i.e. (a + nd )2 = (a + rd ) (a + md )
1
and H= b [if b > a] d r + m − 2n
5 So, =
bA a n 2 − rm
Q G 2 = AH =
5 Now, m, n and r are in HP.
⇒ 5 ab = bA 2 1 1
⇒ = +
⇒ A = 5a n m r
a+ b 2 m+ r
⇒ = 5a ⇒ =
2 n mr
⇒ b = 9a r + m mr
2 − n 2 − n
d 2 n −2 λ
From Eq. (i), 5 a = ab + 2 Hence, = = = =−
⇒ 5a = 3a + 2 a rm rm n n
n n − n n −
∴ a = 1and b = 9 n n
⇒ α = a + b = 10 ⇒ λ =2
and β = | a − b| = 8 140. Let f ( x ) = x 4 + ax 3 + bx 2 + cx + 1
∴ α + β 2 = 10 + 64 = 74
As a, b and c are non-negative, no root of the equation
C. Q H=4 f ( x ) = 0 can be positive. Further as f(0 ) ≠ 0, all the roots
⇒ 2 A + G 2 = 27 of the equation, say x1, x2 , x3 and x4 are negative.
⇒ 2 A + AH = 27 [Q G 2 = AH ] We have,
⇒ 6 A = 27 ∑ x1 = − a, ∑ x1 x2 = b, ∑ x1 x2 x3 = − c and x1 x2 x3 x4 = 1
9 Using AM ≥ GM for positive numbers − x1, − x2 , − x3 and
⇒ A=
2 − x4 , we get
9 a
and G 2 = × 4 = 18 ≥1 ⇒ a≥ 4
2 4
Since, a and b are the roots of x 2 − 9 x + 18 = 0. Using AM ≥ GM for positive numbers x1 x2 , x1 x3 , x1 x4 ,
∴ a = 6, b = 3 or a = 3, b = 6 x2 x3 , x2 x4 and x3 x4 , we get
b
Now, α = a + b = 9, β = | a − b| = 3 ≥1 ⇒ b≥ 6
∴ α 2 + β = 81 + 3 = 84 6
2 3 4 5 Finally, using AM ≥ GM for positive numbers
α α α α α − x1 x2 x3 , − x1 x2 x4 , − x1 x3 x4 and − x2 x3 x4 , we get
and 1+ + + + +
β β β β β c
≥1 ⇒ c ≥ 4
= 1 + 3 + 32 + 33 + 34 + 35 4
= 364 ∴ HCF of { a, b, c } = HCF of {4, 6, 4} = 2 99
3 141. Let the three consecutive terms be
a − d , a, a + d, where d > 0
Then, a2 − 2 ad + d 2 = 36 + K …(i)
Again, subtracting Eq. (iv) from Eq. (v), we get
2d 2 = 32 ⇒ d 2 = 16 ⇒ d = 4
[Qd = − 4, rejected]
Objective Mathematics Vol. 1
a2 = 300 + K …(ii) From Eq. (iv),
and a2 + 2 ad + d 2 = 596 + K …(iii) 4 (2 a − 4) = 264
⇒ 2 a − 4 = 66
On subtracting Eq. (i) from Eq. (ii) we get
⇒ 2 a = 70 ⇒ a = 35
d (2 a − d ) = 264 …(iv)
∴ K = 352 − 300 = 1225 − 300 = 925
On subtracting Eq. (ii) from Eq. (iii), we get
d (2 a + d ) = 296 …(v) ⇒ K − 920 = 5
Entrances Gallery
b c k( k + 1)
1. = = (integer) 4n
2
a b 3. S n = ∑ (− 1) ⋅ k2
b2 k =1
⇒ b2 = ac ⇒ c= ...(i)
a = − (1)2 − 2 2 + 32 + 42 − 52 − 62 + 7 2 + 82 + K
a+ b+c
Also, =b+2 = (32 − 12 ) + (42 − 2 2 ) + (7 2 − 52 ) + (82 − 62 ) + K
3
= 2{(4 + 12 + 20 + ... ) + (6 + 14 + 22 + ... )}
⇒ a+ b+c = 3b + 6 14442444 3 144 42444 3
n terms n terms
⇒ a − 2b + c =6
n n
b2 b2 = 2 {2 × 4 + (n − 1)8} + {2 × 6 + (n − 1)8}
2
⇒ a − 2b + =6 from Eq. (i), c = a 2
a
= 2[n(4 + 4 n − 4) + n(6 + 4 n − 4)]
2 b b2 6
⇒ 1− + 2 = = 2[4n 2 + 4 n 2 + 2 n ] = 4 n(4n + 1)
a a a
Here, 1056 = 32 × 33 , 1088 = 32 × 34 ,
Ta rg e t E x e rc is e s
2
b 6
− 1 = 1120 = 32 × 35 , 1332 = 36 × 37
a a
⇒ a = 6 only Hence, 1056 and 1332 are possible answers.
a2 + a − 14 4. Here, a1 = 5, a20 = 25 for HP
⇒ =4
a+1 ∴
1
=5
a
23 n 1
2. cot ∑ cot −1 1 + ∑ 2 k and = 25
n = 1 k =1 a + 19 d
1 1
23 ⇒ + 19 d =
= cot ∑ cot −1(1 + 2 + 4 + 6 + 8 + ... + 2n ) 5 25
n =1 ⇒ 19 d =
1 1
− =−
4
23 25 5 25
= cot ∑ cot −1{1 + n(n + 1)} −4
∴ d =
n =1 19 × 25
23 1 an < 0 ⇒
1
<0
= cot ∑ tan −1 Q
a + (n − 1)d
n =1 1 + n(n + 1)
1
23 ⇒ + (n − 1)d < 0
n + 1 − n 5
= cot ∑ tan −1
1 4 95
n =1 1 + n(n + 1) ⇒ − (n − 1) < 0 ⇒ (n − 1) >
5 19 × 25 4
23
= cot ∑ [tan −1(n + 1) − tan −1 n ] ⇒ n > 1+
95
⇒ n > 2475 .
n = 1 4
= cot [(tan 2 − tan 1) + (tan 3 − tan −1 2 )
−1 −1 −1 Hence, the least positive value of n is 25.
+ (tan −1 4 − tan −1 3) + K+ (tan −1 24 − tan −1 23)] 5. Given, a1 = 3, m = 5 n and a1, a2 , ..., a100 are in AP.
S m S 5n
= cot{tan −1 24 − tan −1 1} Also, = is independent of n.
Sn Sn
24 − 1 −1 23
= cot tan −1 = cot tan 5n
[2 × 3 + (5 n − 1)d ]
1 + 24 ⋅ (1) 25 Sm
Consider = 2
25 Sn n
= cot cot −1 [2 × 3 + (n − 1)d ]
23 2
5 {(6 − d ) + 5 nd }
=
25 =
100 23 (6 − d ) + nd
For independent of n, put 6 − d = 0
⇒
∴
d =6
a2 = a1 + d = 3 + 6 = 9
Now, G14 + 2G24 + G34 = (lr )4 + 2 (lr 2 )4 + (lr 3 )4
= l 4 × r 4 (1 + 2 r 4 + r 8 )
n n + l
2
3
Sequence and Series
S = l 4 × r 4 (r 4 + 1) = l 4 ×
Also, if d = 0, then m is independent of n. l l
Sn
= ln × 4 lm2 = 4 lm2 n
∴ a2 = 3
6. Using AM ≥ GM, 10. Given, series is
a−5 + a−4 + a−3 + a−3 + a−3 + 1 + a8 + a10 13 13 + 2 3 13 + 2 3 + 33
+ + + ... ∞
8 1 1+ 3 1+ 3 + 5
≥ (a−5 ⋅ a−4 ⋅ a−3 ⋅ a−3 ⋅ a−3 ⋅ 1⋅ a8 ⋅ a10 )1/ 8 Let Tn be the nth term of the given series.
⇒ a−5 + a−4 + 3 a−3 + 1 + a8 + a10 ≥ 8 ⋅ 1 13 + 2 3 + 33 + ... n 3
∴ Tn =
Hence, the minimum value is 8. 1 + 3 + 5 + ... + to n terms
k −1 2
n(n + 1)
7. We have, S k = k ! =
1
1 (k − 1)! 2 (n + 1)2
1− = 2
=
k n 4
1
Now, (k 2 − 3k + 1)S k = {(k − 2 )(k − 1) − 1} ×
(k − 1)!
9
(n + 1)2 1 2
1 1
∴ S9 = ∑ 4
= [(2 + 32 + … + 10 2 ) + 12 − 12 ]
4
n =1
= −
(k − 3)! (k − 1)! 1 10(10 + 1)(20 + 1) 384
= −1 = = 96
4
100
6 4
⇒ ∑|(k 2 − 3k + 1)S k|
k =1 11. Given, α and β are roots of px 2 + qx + r = 0, p ≠ 0.
1 1 −q r
= 1+ 1+ 2 − + ∴ α+β= , αβ = ...(i)
99 ! 98 !
Targ e t E x e rc is e s
p p
100 2 Since, p, q and r are in AP.
= 4−
100 ! ∴ 2q = p + r ...(ii)
100 1 1
100 2 + =4
⇒
100 !
+ ∑|(k 2 − 3k + 1)S k| = 4 Also,
α β
k =1
α+β
⇒ =4
8. Since, ak = 2 ak − 1 − ak − 2 αβ
So, a1, a2 , K a11 are in AP. − q 4r
⇒ α + β = 4αβ ⇒ = [from Eq. (i)]
a2 + a22 + K + a11 2
11a12 + 35 × 11d 2 + 10 × 11a1d p p
∴ 1 =
11 11 ⇒ q = − 4r
= 90 On putting the value of q in Eq. (ii), we get
⇒ 225 + 35 d 2 + 150 d = 90 2(− 4r ) = p + r
⇒ 35 d 2 + 150 d + 135 = 0 ⇒ p = − 9r
9 − q 4r 4r 4
⇒ d = − 3, − Now, α+β= = = =−
7 p p −9r 9
27 r r 1
Given, a2 < and αβ = = =
2 p − 9r − 9
9 16 4 16 + 36
∴ d = − 3 and d ≠ − ∴ (α − β )2 = (α + β )2 − 4αβ = + =
7 81 9 81
a1 + a2 + K + a11 1
⇒ = [30 − 10 × 3] = 0 52 2
11 2 ⇒ (α − β )2 = ⇒ |α − β| = 13
81 9
9. Given, m is the AM of l and n. 12. Given, k ⋅ 10 9 = 10 9 + 2(11)1(10 )8 + 3 (11)2 (10 )7
∴ l + n = 2m …(i)
+ K + 10(11)9
and G1, G2 , G3 are geometric means between l and n.
2 9
∴ l, G1, G2, G3, nare in GP. 11 11 11
⇒ k = 1 + 2 + 3 + K + 10 ...(i)
Let r be the common ratio of this GP. 10 10 10
∴ G1 = lr 11 11 11 11
2
11
9 10
G2 = lr 2 ⇒ k = 1 + 2 + K + 9 + 10
10 10 10 10 10
G3 = lr 3 ...(ii)
n = lr 4 On subtracting Eq. (ii) from Eq. (i), we get
1 2 9 10
n 4 11 11 11 11 11
⇒ r = k 1 − = 1 + + + K + − 10
l 10 10 10 10 10
101
3 ⇒
10 − 11
k
10
=
11 10
1 − 1
10
11
− 10 11
10
10
Now, we can clearly observe the first elements in each
bracket.
In second bracket, the first element is 1 = 12
Objective Mathematics Vol. 1
− 1 In third bracket, the first element is 4 = 2 2
10
In fourth bracket, the first element is 9 = 32
a(r n − 1)
[Q in GP, sum of n terms = , when r > 1] In last bracket, the first element is 361 = 192
r −1
Hence, we can conclude that there are 20 brackets in
11 10 11
10
all.
⇒ − k = 10 10 − 10 − 10
10 10
Also, in each of the bracket, there are 3 terms out of
∴ k = 100 which the first and last terms are perfect squares of
consecutive integers and the middle term is their
13. Let a, ar and ar 2 be in GP (r > 1.) product.
On multiplying middle term by 2, then the numbers Now, the general term of the series is
a, 2ar and ar 2 are in AP. Tr = (r − 1)2 + (r − 1)r + (r )2
⇒ 4 ar = a + ar 2 So, the sum of n terms of the series is
⇒ r − 4r + 1= 0
2 n
4 ± 16 − 4
Sn = ∑ {(r − 1)2 + (r − 1)r + (r )2}
⇒ r= =2 ± 3 r =1
2 n
r 3 − (r − 1)3
⇒ r =2 + 3 [Q GP is increasing] ⇒ Sn = ∑ r − (r − 1)
r = 1
7 77 777
14. Let S = + + + K upto 20 terms n
10 10 2 10 3
79 99 999
⇒ Sn = ∑ { r 3 − (r − 1)3}
r =1
= + + + K upto 20 terms
9 10 100 1000 n
7 1 1
= 1 − + 1 − 2 + 1 − 3
1 Now, let S n = ∑ { k 3 − (k − 1)3}
k =1
9 10 10 10
Ta rg e t E x e rc is e s
On substituting the value of k and rearranging the
+ K upto 20 terms terms, we get
S n = − 0 3 + (13 − 13 ) + K + [(n − 1)3 − (n − 1)3 ] + n 3
7
= [(1 + 1 + K upto 20 terms ) ⇒ S n = n3
9
1 1 1 Since, the number of terms is 20, hence substituting
− + + + K upto 20 terms n = 20, we get
10 10 2 10 3
S 20 = 8000
20
1 −
1 Statement II We have already proved in the
7 1 10 Statement I that
= 20 −
10 1 − 1
n
9
10
Sn = ∑ { k 3 − (k − 1)3} = n 3
k =1
7 1 10 1
20
17. Let the time taken to save ` 11040 be (n + 3) months.
= 20 − × 1 −
9 10 9 10 For first 3 months, he save `s 200 each month.
In (n + 3) months, his total savings is
7 10 −20 7
= 20 − (1 − 10 ) = (180 − 1 + 10 −20 ) n
3 × 200 + [2(240 ) + (n − 1) × 40 ] = 11040
9 9 81 2
7 −20
= (179 + 10 ) ⇒ 600 + 20 n(n + 11) = 11040
81
⇒ 20 n(n + 11) = 10440
15. Here, T100 = a + (100 − 1)d = a + 99 d ⇒ n(n + 11) = 522
T50 = a + (50 − 1)d = a + 49 d ⇒ n 2 + 11n − 522 = 0
T150 = a + (150 − 1)d = a + 149 d
⇒ n 2 + 29n − 18 n − 522 = 0
Now, according to the given condition,
⇒ n(n + 29) − 18 (n + 29) = 0
100 × T100 = 50 × T50
⇒ n = 18
⇒ 100(a + 99 d ) = 50(a + 49 d )
or n = − 29
⇒ 2(a + 99 d ) = (a + 49 d )
∴ n = 18 [neglecting n = − 29]
⇒ a + 149 d = 0
Hence, total time = n + 3
⇒ T150 = 0 = 18 + 3 = 21months
16. Statement I Let S = (1) + (1 + 2 + 4) + (4 + 6 + 9) 18. Number of notes that the person counts in 10 min
+ ... + (361 + 380 + 400 ) = 10 × 150 = 1500
S = (0 + 0 + 1) + (1 + 2 + 4) + (4 + 6 + 9) We have, a10 , a11, a12 , ... are in AP with common
102 + K + (361 + 380 + 400 ) difference − 2.
Let n be the time taken to count remaining 3000 notes,
then
n
[2 × 148 + (n − 1) × (− 2 )] = 3000
∴
⇒
1 1
− =d
a2 a1
a1 − a2 = a1a2d
3
Sequence and Series
2 Similarly, a2 − a3 = a2 a3d
⇒ n 2 − 149 n + 3000 = 0
M M M
⇒ (n − 24)(n − 125) = 0 an − 1 − an = an − 1and
⇒ n = 24,125
On adding all the equations, we get
Hence, total time taken by the person to count all notes
a1 − an = d { a1a2 + a2 a3 + ...+ an − 1an } …(i)
= 10 + 24 = 34 min
1 1
[Q n ≠ 125as a135 = 148 + 124 (−2 ) < 0] Also, = + (n − 1)d
an a1
2 6 10 14
19. Let S = 1 + + 2 + 3 + 4 + ... a − an
3 3 3 3 ⇒ d = 1
2 6 10 14 a1an (n − 1)
⇒ S − 1 = + 2 + 3 + 4 + ... ...(i)
3 3 3 3 On putting the value of d in Eq. (i), we get
S −1 2 6 10 14 a − an
⇒ = 2 + 3 + 4 + 5 +K ...(ii) a1 − an = 1 { a1a2 + a2 a3 + K + an − 1 an }
3 3 3 3 3 a1an (n − 1)
On subtracting Eq. (ii) from Eq. (i), we get ⇒ a1a2 + a2 a3 + ... + an − 1an = a1an (n − 1)
2 2 4 4 4 ∞ ∞ ∞
(S − 1) = + 2 + 3 + 4 + ...
3 3 3
2 2
3
2
3 24. Given that, x = ∑ an, y = ∑ bn, z = ∑ c n
n=0 n=0 n=0
⇒ S − 1= 1+ + 2 + 3 + K
3 3 3 1
⇒ x = 1 + a + a2 + K = ...(i)
2 1− a
⇒ S = 2 + 3 = 2 + 1= 3 Similarly, y=
1
…(ii)
1
1− 1− b
3 1
and z= ...(iii)
20. Since, a + ar = a (1 + r ) = 12 …(i) 1− c
Targ e t E x e rc is e s
and ar 2 + ar 3 = ar 2 (1 + r ) = 48 …(ii) Now, a, b and c are in AP.
From Eqs. (i) and (ii), we get ⇒ − a, − b and − c are in AP.
r2 = 4 ⇒ r = −2 ⇒ 1 − a, 1 − b and 1 − c are also in AP.
On putting the value of r in Eq. (i), we get 1 1 1
⇒ , and are in HP.
a = − 12 1− a 1− b 1− c
21. Since, each term is equal to the sum of the next two ⇒ x, y, z are in HP.
terms. Aliter
∴ ar n − 1 = ar n + ar n + 1 ⇒ 1 = r + r 2 From Eqs. (i), (ii) and (iii),
⇒ r2 + r − 1= 0 1 1 1
x= ,y= and z =
5 −1 − 5 − 1 1− a 1− b 1− c
r= Q r ≠
2 2 x −1 y −1 z −1
⇒ a= ,b= and c =
x y z
a1 + a2 + K + ap p2
22. Since, = Since, a, b and c are in AP.
a1 + a2 + K + aq q2
∴ 2b = a + c
p y − 1 x − 1 z − 1
[2 a1 + ( p − 1)d ] ⇒
2 p2 2 = +
∴ = 2 y x z
q
[2 a1 + (q − 1) d ] q 2 1 1
2 ⇒ 2 − = 1− + 1−
(2 a1 − d ) + pd p y x z
⇒ = 2 1 1
(2 a1 − d ) + qd q ⇒ = +
⇒ (2 a1 − d ) ( p − q ) = 0 y x z
d Hence, x, y and z are in HP.
⇒ a1 = [Q p ≠ q]
2 25. We know that,
d
+ 5d e x + e−x x2 x4 x6
a6 a1 + 5 d 11d 11 = 1+ + + +K
Now, = = 2 = = 2 2! 4! 6!
a21 a1 + 20 d d
+ 20 d 41d 41 1
2 On putting x = in both sides, we get
2
23. Since, a1, a2 , a3 , ..., an are in HP.
e 1/ 2 + e − 1/ 2
2 4
1 1 1 1
1 1 1 1 = 1+ + +K
∴ , , , K , are in AP. 2 2 2 ! 2 4!
a1 a2 a3 an
e +1 1 1 1
Let d be the common difference of AP. ⇒ = 1+ + + +K
2 e 4 ⋅ 2 ! 16 ⋅ 4 ! 64 ⋅ 6 !
103
3 26. Given,
⇒ a + (m − 1)d =
1
n
1
Tm =
...(i)
⇒
⇒
⇒
12 t 2 − 5 t − 3 = 0
(3 t + 1) (4 t − 3) = 0
t =−
1 3
Objective Mathematics Vol. 1
,
n 3 4
1 3
and Tn = ⇒ 3x = [Q 3 x cannot be negative]
m 4
1 3
⇒ a + (n − 1)d = ...(ii) ⇒ log 3 = x
m 4
On solving Eqs. (i) and (ii), we get
⇒ x = 1 − log 3 4
1
a=d = 31. Let S =2 1/ 4
⋅ 41/ 8 ⋅ 81/ 16 ...
mn
∴ a−d = 0 =2 1/ 4
⋅ 2 2 / 8 ⋅ 2 3 / 16 ...
1 2 3
27. Given that the sum of n terms of given series is 1 + + 2 + ...
1
( S1 )
n(n + 1)2 =2 4 2 2
= 24
, if n is even.
2 2 3
where, S1 = 1 +
+ + ...
Let n be odd, i.e. n = 2 m + 1 2 22
Then, S 2 m + 1 = S 2 m + (2 m + 1)th term It is an infinite arithmetico-geometric progression.
(n − 1)n 2 a d ⋅r
= + nth term ∴ S1 = +
2 1 − r (1 − r )2
(n − 1) n 2 1
= + n2 1⋅
2 1 2
= + =2 + 2 = 4
n − 1 + 2 1−
1 1
2
= n2 1 −
2 2 2
(n + 1) n 2 1
= ( 4)
2 ∴ S = 24 =2
Ta rg e t E x e rc is e s
28. We know that, 32. Since, 5th term of a GP = 2
e = 1+ +
1 1
+
1
+
1
+ ... ...(i) ∴ ar 4 = 2 …(i)
1! 2 ! 3 ! 4 ! where, a and r are the first term and common ratio
1 1 1 1
and e−1 = 1− + − + − ... ...(ii) respectively of a GP.
1! 2 ! 3 ! 4 ! Now, required product
On adding Eqs. (i) and (ii), we get = a × ar × ar 2 × ar 3 × ar 4 × ar 5 × ar 6 × ar 7 × ar 8
2 2
e + e −1 = 2 + + + ... = a9 r 36 = (ar 4 )9
2 ! 4!
e2 + 1 2 2 = 2 9 = 512 [from Eq. (i)]
⇒ −2 = + + ... ∞ n 2
e 2 ! 4! (loge x ) loge x (loge x )
e 2 + 1 − 2e 1 1
33. ∑ n !
= 1+
1!
+
2!
+ ...
⇒ =2 + + ... n=0
e 2 ! 4!
= e loge x
=x
(e − 1) 12
1
⇒ = + + ... 1 1
( x − 1) − ( x − 1) 2 + ( x − 1) 3 − ...
2e 2 ! 4! 34. e 2 3 = e log (1 + x − 1)
=x
1 1 1
29. Consider, − + − .... 35. Vr = Sum of first r terms of an AP whose first term is r and
1⋅ 2 2 ⋅ 3 3 ⋅ 4
1 1 1 1 1 common difference(2 r − 1)
= 1 − − − + − −... r
2 2 3 3 4 = [2 r + (r − 1)(2 r − 1)]
1 1 1 2
= 1 − 2 ⋅ + 2 ⋅ − 2 ⋅ + ... r
2 3 4 = [2 r + 2 r 2 − 3r + 1]
2
1 1 1
= 2 1 − + − + ... − 1 r
2 3 4 = [2 r 2 − r + 1]
2
= 2 log (1 + 1) − 1 1
4 = [2 r 3 − r 2 + r ]
= log 2 2 − log e = log 2
e n
1 n n n
30. Since, 1, log 3 31 − x + 2 and log 3 (4 ⋅ 3 x − 1) are in AP. Now, ∑ Vr = 2 2 ∑ r 3 − ∑ r 2 + ∑ r
r =1 r = 1 r =1 r =1
∴ 2 log 3 (31 − x + 2 )1/ 2 = log 3 3 + log 3 (4 ⋅ 3 x − 1) n(n + 1)(2 n + 1) n(n + 1)
2
1 n(n + 1)
⇒ log 3 (31 − x + 2 ) = log 3 3 (4 ⋅ 3 x − 1) = 2 ⋅ − +
2 2 6 2
⇒ 31 − x + 2 = 12 ⋅ 3 x − 3
1 n 2 (n + 1)2 n(n + 1)(2 n + 1) n(n + 1)
3 = − +
∴ + 2 = 12 t − 3 [let 3 x = t ] 2 2 6 2
104 t
=
n(n + 1)
4
n(n + 1) −
2
(2 n + 1)
3
+1
n(n + 1) 3n + 3n − 2 n − 1 + 3
⇒ M ≤1 ⇒ M ≤1
Also, ( p + q ) (r + s) > 0
∴ M>0
[Q p, q, r , s > 0] 3
Sequence and Series
=
4 3 Hence, 0 < M ≤1
n(n + 1) 39. Case I Let α = ω and β = ω 2
= (3n + n + 2 )
2
302 n 302
12 ω
36. Since, a, b and c are in GP.
∴ S = ∑ (− 1)n ω 2 = ∑ (− 1)n (ω 2 )n
n=0 n=0
∴ b2 = ac = 1 − ω + ω − ω + ω 8 − ω10 + ω12
2 4 6
Given equation is + ... + ω 600 − ω 602 + ω 604
(loge a) x 2 − (2 loge b)x + (loge c ) = 0 ...(i)
= 1 − ω 2 + ω − 1 + ω 2 − ω + 1 + ... + 1 − ω 2 + ω
On putting x = 1in Eq. (i), we get
= 0 + ... + 1 − ω 2 + ω
loge a − 2 loge b + loge c = 0
⇒ 2 loge b = loge a + loge c = − ω 2 − ω 2 = − 2ω 2 [Q1 + ω + ω 2 = 0]
⇒ loge b2 = loge ac Case II Let α = ω 2 and β = ω
⇒ b2 = ac, which is true. n n
n ω ω4
302 2 302
Hence, one of the roots of given equation is 1. ∴ S = ∑ ( − 1)
ω
= ∑ (− 1)n 3
ω
Let another root be α. n=0 n=0
302
2 loge b loge b2
∴ Sum of roots, 1 + α =
loge a
=
loge a
= ∑ (− 1)n (ω )n
n=0
loge ac (loge a + loge c ) = 1 − ω + ω2 − ω3 + ω4 − ω5 + ω6
⇒ α= − 1= −1
loge a loge a − ... + ω 300 − ω 301 + ω 302
loge c
= = log a c = 1 − ω + ω − 1 + ω − ω + 1 − ... + 1 − ω + ω 2
2 2
loge a
= 0 + ... + 1 + ω 2 − ω = − ω − ω = − 2ω
Targ e t E x e rc is e s
Hence, roots are 1 and log a c.
1 2x + 1 α 2 + β2 2 α 3 + β3 3
37. Given, f ( x ) = x + = 40. Given, (α + β)x − x + x +K
2 2 2 3
1 4x + 1 1 1
∴ f (2 x ) = 2 x + ⇒ f (2 x ) = = α x − (α x)2 + (α x)3 − ...
2 2 2 3
1
and f (4 x ) = 4 x + 1 1
+ βx − (βx )2 + (βx )3 − ...
2 2 3
8x + 1
⇒ f (4 x ) = = log (1 + αx ) + log (1 + βx )
2
= log [1 + (α + β ) x + αβx 2 ]
Since, f ( x ), f (2 x ) and f (4 x ) are in HP.
1 1 1 Now, α + β = p and αβ = q
∴ , and are in AP.
f ( x ) f (2 x ) f (4 x ) α 2 + β2 2 α 3 + β3 3
Hence, (α + β ) x − x + x +K
1 1 2 3
+
1 f ( x ) f (4 x ) = log (1 + px + qx 2 )
⇒ =
f (2 x ) 2
2 2 41. Given, log 4 2 − log 8 2 + log16 2 − ...
+ 1 1 1
2 2 x + 1 8x + 1 = − + − ...
⇒ = log 2 4 log 2 8 log 2 16
4x + 1 2
10 x + 2 1 1 1
⇒
2
= = − + − ...
4 x + 1 (2 x + 1) (8 x + 1) 2 3 4
1 1 1 1
⇒ (2 x + 1) (8 x + 1) = (5 x + 1) (4 x + 1) = 1 − 1 − + − + − ...
2 3 4 5
⇒ 16 x 2 + 10 x + 1 = 20 x 2 + 9 x + 1
= 1 − loge 2
⇒ 4x 2 − x = 0
42. We have, e x = (1 − x ) (B0 + B1 x + B2 x 2
⇒ x (4 x − 1) = 0
1 + ... + Bn − 1 x n − 1 + Bn x n + .... )
⇒ x= [Q x ≠ 0]
4 By the expansion of e x , we get
Hence, one real value of x for which the three unequal x x2 xn
terms are in HP. 1+ + + ... + + ...
1! 2 ! n!
38. Q AM ≥ GM = (1 − x ) (B0 + B1 x + B2 x 2 + ... + Bn − 1 x n − 1 + Bn x n + ... )
( p + q ) + (r + s)
∴ ≥ ( p + q ) (r + s) On equating the coefficient of x n both sides, we get
2
1
⇒
2
≥ M Bn − Bn − 1 =
n! 105
2
3
∞ ∞
x 3n x 3n − 2 46. Since, a1, a2 , ..., an are in AP, therefore
43. We have, a= ∑ (3 n )! , b = ∑ (3 n − 2 )! a2 − a1 = a3 − a2 = ... = a2 k − a2 k − 1 = d [say]
n=0 n =1
∞ 3n − 1 Now, a12 − a22 = (a1 − a2 ) (a1 + a2 )
Objective Mathematics Vol. 1
x
and c= ∑ (3 n − 1)! = − d (a1 + a2 )
n =1
∞ ∞ ∞
a32 − a42 = − d (a3 + a4 )
x 3n x 3n − 2 x 3n − 1
Now, a + b + c = ∑ + ∑ + ∑ (3 n − 1)! a22k − 1 − a22k = − d (a2 k −1 + a2 k )
n=0
(3 n )! n = 1(3 n − 2 )! n =1
On adding, we get
x2 x3 S = − d (a1 + a2 + ... + a2 k )
=1+ x + + + ... = e x
2! 3! 2 k
= −d (a + a2 k )
ω2 x2 ω3 x3 2 1
a + bω + cω 2 = 1 + ω x + + + ... = e ωx
2! 3! − dk
2 = − dk (a1 + a2 k ) = (a12 − a22k )
and a + bω 2 + cω = e ω x , ω is an imaginary cube root (a1 − a2 k )
of unity. − dk (a12 − a22k )
=
Now, a3 + b3 + c 3 − 3 abc [(a1 − a2 ) + (a2 − a3 ) + (a3 − a4 )
= (a + b + c ) (a + bω + cω 2 ) ( a + bω 2 + cω ) + ...+ (a2 k − 2 − a2 k −1 ) + (a2 k −1 − a2 k )]
2 2 − dk (a12 − a22k ) k
= e x ⋅ e ωx ⋅ e ω x
= e x (1 + ω + ω )
= e 0⋅ x = 1 = = (a12 − a22k )
(− d ) (2 k − 1) 2k − 1
x 3 2 7 3 15 4
44. Given, f ( x ) = + x + x + x + ... 47. Given, log x (ax ), log x (bx )and log x (cx ) are in AP.
1! 2 ! 3! 4!
2 x (2 x )2 (2 x )3 (2 x )4 ⇒ 1 + log x a, 1 + log x b, 1 + log x c are in AP.
= + + +
1! 2! 3! 4! ⇒ log x a, log x b, log x care in AP.
x x2 x3 x4 log a log c log b
+ ... − + + + + ... ∴ + =2
1! 2 ! 3! 4! log x log x log x
2 x (2 x )2 (2 x )3 (2 x )4 ⇒ log a + log c = 2 log b ⇒ ac = b2
Ta rg e t E x e rc is e s
= 1+ + + +
1! 2! 3! 4! 48. Let the six numbers in AP be
x x2 x3 x4
+ ... − 1 + + + + + ... a − 5d , a − 3d , a − d , a + d , a + 3d , a + 5d
1! 2 ! 3! 4! ∴ a − 5d + a − 3d + a − d + a + d + a
⇒ f( x) = e 2 x − e x + 3 d + a + 5 d = 3 [Q sum = 3]
Put f ( x ) = 0, we get ⇒ 6a = 3
1
e2x − e x = 0 ⇒ a=
2
⇒ e x (e x − 1) = 0
Also, T1 = 4T3 , where T1, T3 are respectively first and third
⇒ ex = 0 or e x = 1 terms of an AP.
⇒ x=0 ⇒ a − 5 d = 4 (a − d ) ⇒ d = − 3 a = −
3
Hence, exactly one real solution exists. 2
8 21 40 65 So, the fifth term = a + 3d
45. We have, S = 1 + + + + + ... 1 3 1 9
2 ! 3! 4! 5! = + 3 − = − = − 4
Let S1 = 1 + 8 + 21 + 40 + 65 + .... + Tn ...(i) 2 2 2 2
and S1 = 1 + 8 + 21 + 40 + ...+ Tn − 1 + Tn ...(ii) 49. Since,log10 2, log10 (2 x − 1) and log10 (2 x + 3) are in AP.
On subtracting Eq. (ii) from Eq. (i), we get log10 2 + log10 (2 x + 3)
0 = 1 + 7 + 13 + 19 + 25 + ... − Tn ∴ log10 (2 x − 1) =
2
Tn = 1 + 7 + 13 + 19 + 25 + ... upto n terms ⇒ 2 log10 (2 x − 1) = log10 2 (2 x + 3)
n
= [2(1) + (n − 1) 6] ⇒ log10 (2 x − 1)2 = log10 2 (2 x + 3)
2
= n [1 + 3 (n − 1)] = n (3 n − 2 ) ⇒ (2 x − 1)2 = 2 (2 x + 3)
n(3 n − 2 ) 3n − 2 ⇒ 22 x + 1 − 2 x + 1 = 2 x + 1 + 6
∴ S =Σ =Σ
n! (n − 1)! ⇒ 2 2x
+ 1 − 2 ⋅ 2 x+1 = 6
3n − 3 + 1 ⇒ 22 x − 4⋅ 2 x − 5 = 0
=Σ
(n − 1)! Now, let 2x = t
3 1 ⇒ t − 4t − 5 = 0
2
⇒ S =Σ +Σ = 3e + e = 4e
(n − 2 )! (n − 1)! ⇒ (t − 5) (t + 1) = 0
1 1 ⇒ t = 5 or t = − 1
Qe = 1+ + + ...
1! 2 ! ⇒ 2x = 5
We know that, 2 < e < 3 [neglecting 2 x = − 1as 2 x is always positive]
∴ 8 < 4 e < 12 ⇒ x log 2 2 = log 2 5
106 ⇒ 8 < S < 12 ⇒ x = log 2 5
50. S n = 0.2 + 0.22 + 0.222 + ... upto n terms
= 2 [0.1 + 0.11 + 0.111 + ... upto n terms]
2
= [0.9 + 0.99 + 0.999 + ... upto n terms]
55. 4 + 2 (1 + 2 ) log 2 +
2 (1 + 2 2 ) (log 2 )2
2!
+
2(1 + 2 3 ) (log 2 )3
+ ...
3
Sequence and Series
9 3!
2 (log 2 )2
= [(1 − 0.1) + {1 − (0.1)2 } + {1 − (0.1)3 } = 2 1+ log 2 + + ...
9 2!
+...upto n terms] (2 log 2 )2
2 + 2 1 + 2 log 2 + + ...
= [n − {(0.1) + (0.1) + (0.1) + ... upto n terms}]
2 3
2!
9
2 {1 − (0.1)n } 2 1 = 2 (e log 2 ) + 2 (e 2 log 2 )
= n − (0.1) = n − (1 − 10 − n )
9 1 − 0.1 9 9 = 2 × 2 + 2e log 4
1 1⋅ 3 1⋅ 3 ⋅ 5 1⋅ 3 ⋅ 5 ⋅ 7 = 4 + 2 × 4 = 12
51. 1 + + + + +… 56. Let Tn be the nth term of the given series.
3 3 ⋅ 6 3 ⋅ 6 ⋅ 9 3 ⋅ 6 ⋅ 9 ⋅ 12
1 3 1 3 5 n(n + 1) n + 1 (n − 1) + 2
⋅ 2 ⋅ ⋅ 3 ∴ Tn = = =
1 2 2 2 2 2 2 2 + ... n! (n − 1)! (n − 1)!
= 1+ + +
3 1 ⋅ 2 3 1 ⋅ 2 ⋅ 3 3 =
1
+
2
1 1 (n − 2 )! (n − 1)!
+ 1 ∞ ∞ ∞
1 2 2 2 2 2 1 1
= 1+ . +
2 3 2!
3
∴ S = ∑ Tn = ∑ (n − 2 )! + 2 ∑ (n − 1)!
n =1 n=2 n =1
1 1 1
+ 1 + 2 = e + 2e = 3 e
2 2 2 2 3
+ + ... 57. Given, sum of n terms of an AP = 240
3! 3
−1/ 2 n
2 ∴ [2 × 2 + (n − 1) × 2 ] = 240
= 1 − = 31/ 2 = 3 2
3
Targ e t E x e rc is e s
⇒ n(2 + n − 1) = 240
52. We have, Hn = 1 +
1
+ ... +
1 ⇒ n(n + 1) = 15 × 16
2 n ⇒ n = 15
3 5 2n − 1
Consider, S n = 1 + + + ... + 4 3
2 3 n 58. Given, S ∞ = and a =
3 4
1 1 1
= 1 + 2 − + 2 − + ... + 2 − Let r be the common ratio.
2 3 n a 4
[1 + {2 + 2 + ... + 2}] 1 1 1 ∴ =
= 144424443 − + + ... + 1− r 3
( n − 1) times 2 3 n 4 4 3
⇒ − r=
1 1 1 3 3 4
= [1 + 2 (n − 1)] − 1 + + + ... + +1 16 − 9 4
2 3 n ⇒ = r
1 1 1 12 3
= 2 n − 1 + + + ... + = 2n − Hn 7 4
2 3 n ⇒ = r
12 3
53. Let the numbers be a and b. 7
a+ b 2 ab ⇒ r=
∴ = 27 and = 12 16
2 a+ b
xn + 1 + yn + 1
⇒ a + b = 54 and 2 ab = 12 (a + b) 59. We have, = xy , for some n
xn + yn
⇒ 2 ab = 12 (54)
⇒ ab = 6 (54) = 324 ⇒ x n + 1 + y n + 1 = ( xy )1/ 2 ( x n + y n )
⇒ ab = 18 ⇒ x n ⋅ x + y n ⋅ y = x n ⋅ x1/ 2 y1/ 2 + y n ⋅ x1/ 2 y1/ 2
Thus, GM = 18. ⇒ x n( x − xy ) + y n ( y − xy ) = 0
54. Since, x, y and z are in GP. ⇒ x ⋅ x( x −
n
y) + y ⋅ y n( y − x)= 0
∴ y2 = x z
⇒ ( x − y )( x ⋅ x − y ⋅ y ) = 0
n n
Now, taking log10 on both sides, we get
For x ≠ y, x n
x = yn y
2 log10 y = log10 x + log10 z 1
n+
1 1 1 x 2
⇒ 2 = + ⇒ =1
log y 10 log x 10 log z 10
y
1
⇒
1
,
1
,
1
are in AP. ⇒ n+ =0
log x 10 log y 10 log z 10 2
1
⇒ log x 10, log y 10 and log z 10 are in HP. ⇒ n=−
2
107
3
2
60. Let the numbers be a and b. n(n + 1) n(n + 1)(2 n + 1)
=8 − 12
∴ ab = 10 2 6
⇒ ab = 100 ...(i) n(n + 1)
Objective Mathematics Vol. 1
2 ab +6 −n
and =8 2
a+ b
= 2 n 2 (n + 1)2 − 2 n(n + 1)(2 n + 1) + 3 n(n + 1) − n
200
⇒ =8 = n(n + 1)[2 n(n + 1) − 2 (2 n + 1) + 3] − n
a+ b
= n(n + 1)[2 n 2 + 2 n − 4 n − 2 + 3] − n
⇒ a + b = 25 ...(ii)
= n(n + 1)[2 n 2 − 2 n + 1] − n
On solving Eqs. (i) and (ii), we get
a = 5 and b = 20 = n(n + 1)⋅ 2 n(n − 1) + n(n + 1) − n
or a = 20 and b = 5 = 2 n 2 (n 2 − 1) + n 2 = n 2 (2 n 2 − 1)
61. Tn = (2 n − 1)3 = 8 n 3 − 13 − 3 ⋅ 2 n ⋅ 1(2 n − 1) 62. Here, Tn =
2n
=
1
−
1
= 8 n − 1 − 12 n + 6 n
3 2 (2 n + 1)! 2 n ! (2 n + 1)!
= 8 n 3 − 12 n 2 + 6 n − 1 1 1 1 1
ΣTn = − + − +K
2 ! 3! 4! 5!
∴ S n = ΣTn 1 1 1 1 1
= 8 Σn 3 − 12 Σ n 2 + 6 Σ n − Σ1 = 1− + − + − + K = e −1
1! 2 ! 3 ! 4 ! 5 !
Ta rg e t E x e rc is e s
108
4
Complex
Numbers
The Real Number System
Natural Numbers ( N ) The numbers which are used for counting are known as Chapter Snapshot
natural numbers (also known as set of positive integers),
The Real Number System
i.e. N = {1, 2, 3, …}.
●
● Modulus of a Real Number
Whole Numbers (W ) If ‘0’ is included in the set of natural numbers, then we get
● Imaginary Number
the set of whole numbers, i.e.
Complex Number
W = {0, 1, 2, … } = {N } + {0}.
●
● Algebra of Complex Numbers
Integers (Z or I) If negative natural numbers are included in the set of whole
numbers, then we get set of integers, i.e. ● Conjugate of a Complex
Number
Z or I = {…, −3, −2, −1, 0, 1, 2, 3, …}.
● Modulus of a Complex
p
Rational Numbers (Q ) The numbers which are in the form of , (where p, q ∈ I , Number
q ● Argument (or Amplitude) of a
q ≠ 0), are called as rational numbers. Complex Number
2 1 22 ● Various Forms of a Complex
e.g. , 3, , 0.76, 1.2333, etc.
3 3 7 Number
Irrational Numbers (Q′ ) The numbers which are not rational, i.e. which cannot be ● De-Moivre’s Theorem
p ● Roots of Unity
expressed in form or whose decimal part is
q ● Geometrical Applications of
non-terminating, non-repeating but which may represent Complex Numbers
magnitude of physical quantities are called irrational ● Loci in Complex Plane
numbers. ● Logarithm of Complex
e.g. 2, 51/ 3 , π , e, etc. Numbers
Real Numbers ( R ) The set of rational and irrational numbers is called a set of
real numbers, i.e.
N ⊂W ⊂ Z ⊂Q ⊂ R
4 Ø ●
●
The real number system is totally ordered for any two
numbers a, b ∈R. We must say, either a < b or b < a or b = a.
All real numbers can be represented by points on a straight
line. This line is called as number line.
X Example 3. If f (x ) = | x − 2 | + | x | + | x + 3 | , then
the value of f ( x ) for x ≤ − 3 is
Objective Mathematics Vol. 1
(a) 3x − 1
● Division by zero is meaningless.
● Number zero is neither positive nor negative but it is an even (b) −3x + 1
number. (c) x − 3
● Square of a real number is always non-negative. (d) −(3x + 1)
● An integer is said to be even, if it is divisible by 2, otherwise it
is an odd number. Sol. (d) Q | x − 2| = − ( x − 2 ) = 2 − x for x ≤ − 3
● Number ‘0’ is an additive identity. | x| = − x for x ≤ − 3
● The magnitude of a physical quantity may be expressed as a and | x + 3| = − ( x + 3) for x ≤ −3
real number times, a standard unit.
∴ f ( x) = 2 − x − x − x − 3
● Number ‘1’ is multiplicative identity.
= − 3x − 1
● A positive integer p is called prime, if its only divisors are ± 1
and ± p. = − (3 x + 1)
● Between two real numbers, there lie infinite real numbers.
Infinity (∞) is the concept of the number greater than
●
greatest you can imagine. It is not a number, it is just a Imaginary Number
concept, so we do not associate equality with it.
Square root of a negative real number is imaginary
10 25 number. While solving equation x 2 + 1 = 0, we get
X Example 1. The value of is
∞ x = ± −1 which is imaginary. So, the quantity −1 is
(a)1 (b) 2 (c)10 50 (d) 0 denoted by ‘i’ called ‘iota’. Thus, i = −1.
Sol. (d) As we know, ∞ is the number greater than greatest
we imagine. Also, the value of 1 upon ∞ is tending to zero. e.g. −2, −3, −4, … may expressed as
1025 i 2, i 3, 2i, … .
Hence, = 0.
∞
Integral powers of iota As we have seen
Intervals i = −1, so i 2 = − 1, i 3 = − i and i 4 = 1.
Let a, x, b are real numbers, so that Hence, n ∈ N , i n = i, −1, −i, 1 attains four values
x ∈[ a, b] ⇒ a ≤ x ≤ b, [a, b] is known as the according to the value of n, so
closed interval a, b. i 4n + 1 = i
x ∈( a, b) ⇒ a < x < b, (a, b) is known as the i 4n + 2 = − 1
open interval a, b. i 4n + 3 = − i
x ∈( a, b] ⇒ a < x ≤ b, (a, b] is known as open, i 4n or i 4n + 4 = 1
closed interval a, b.
x ∈[ a, b) ⇒ a ≤ x < b, [a, b) is known as closed, In other words,
open interval a, b. ( −1) n / 2 , if n is an even integer
i =
n
n −1
X Example 2. If a = 3 + 1, b = 2 2, then the value ( −1) 2 i , if n is an odd integer
a
of lies in the interval
b Ø ● i2 = −1 × −1 ≠ 1
(a) ( −1, 0) (b) (0, 1] (c) [0.5, 1.5] (d) (0, 1) ● − a × − b ≠ ab , so for two real numbers a and b but
a ⋅ b = a ⋅ b possible, if both a, b are non-negative.
Sol. (b, c, d) Consider a = 3 + 1 = 0.966 ● i is neither positive, zero nor negative. Due to this reason,
b 2 2 order relations are not defined for imaginary numbers.
Clearly, 0.966 ∈(0, 1,) (0, 1] and [0.5, 1.5].
● The sum of any four consecutive powers of i is zero, i.e.
i 4n + 1 + i 4n + 2 + i 4n + 3 + i 4n + 4 = 0
Modulus of a Real Number X Example 4. The value of i 2014 is
The modulus of a real number x is defined as (a) i
follows: (b) − i
x, when x > 0 (c)1
(d) −1
| x | = 0, when x = 0
−x, when x < 0 Sol. (d) Consider
i 2014 = (i 2 )1007
x − a, when x ≥ a
Ø | x − a| = = (−1)1007 = − 1
110 −(x − a), when x < a
X Example 5.
equal to
(a) − | a |⋅ b
If a < 0, b > 0, then a ⋅ b is
(b) | a |⋅ b ⋅ i
Complex Number
The complex number z = a + ib = ( a, b) can be
4
Complex Numbers
represented by a point P, whose coordinates are referred
(c) | a | b (d) None of these
to rectangular axes XOX ′ and YOY ′, which are called
Sol. (b) As we can only multiply the positive values in square real and imaginary axis respectively. The plane formed
root. by rectangular axes is called argand plane or argand
∴ a b = −|a| b , as a < 0 and b > 0 diagram or complex plane or Gaussian plane.
i.e. −1 ⋅ |a| b = i |a| b = |a|⋅b ⋅ i Y P (a , b )
Imaginary
axis
X Example 6. The value of the sum b
13
∑ (i n + i n + 1 ), where i = −1 is
X′
Real axis O a M
X
n =1
(a) i (b) i −1 (c) − i (d) 0
Y′
Sol. (b) Since, the sum of any four consecutive powers of i,
(i) A number of the form z = x + iy = Re z + i Im z
is zero.
13 13 13 is called a complex number.
∑ (i + i n + 1) = ∑i ∑i
n+1
∴ n n
+ …(i) (ii) Two complex numbers are said to be equal, if
n =1 n =1 n =1
and only if their real parts and imaginary parts
= (i + i 2 + i 3 + K + i 13 ) + (i 2 + i 3 + i 4 + K + i 14 ) are separately equal,
= i −1 [from Eq. (i)] i.e. a + ib = c + id ⇔ a = c and b = d.
(iii) The complex numbers do not possess the
X Example 7. The value of property of order, i.e.
2n (1 + i) 2n x + iy < (or ) > c + id is not defined.
+ , n ∈ I , is equal to (iv) A complex number z is purely real, if Im( z ) = 0
(1 + i) 2n 2n
and said to be purely imaginary, if Re( z ) = 0.
(a) 0 (b) 2 The complex number 0 = 0 + i. 0 is both purely
(c) {1 + ( −1) } ⋅ i
n n
(d) None of these real and purely imaginary.
2n (1 + i )2 n 1
Sol. (c) Here, +
(1 + i ) 2n
2n X Example 9. If z = (−5 i) i , then Im ( z ) is
8
2n (1 + i 2 + 2 i )n
= + equal to
(1 + i 2 + 2 i )n 2n
(a)1 (b) 0
2n (2 i )n 1
= n
+ = n + in (c) −1 (d) None of these
(2 i ) 2n i
in in 1 Sol. (b) Consider z = (−5 i ) 1 i
= 2n + i n = + in = in + 1 8
i (−1)n ( − 1)n
Let us first express z in the format a + ib, then
= i n ⋅ {(−1)n + 1}
−5 2 5 5
z= i = − (−1) = + i 0 ⇒ Im ( z) = 0
X Example 8. If i = −1, then the number of 8 8 8
values of i n + i − n for different n ∈I is
X Example 10. If 4x + i(3x − y) = 3 + i(−6),
where x and y are real numbers, then the value of x
(a) 3 (b) 2 (c) 4 (d) 1 and y respectively are
Sol. (a) As i n + i − n can be written as 3 33
(a) 3, 33 (b) ,
1 i 2n + 1 5 5
x = in + =
in in 3 33
1+ 1 (c) , (d) None of these
If n = 4, x = =2 4 4
1
i +1
2
Sol. (c) We have,
If n = 5, x = =0
i 4 x + i (3 x − y) = 3 + i (−6) …(i)
1+ 1
If n = 6, x = = −2 Equating the real and imaginary parts of Eq. (i), we get
−1 4 x = 3 and 3 x − y = − 6,
i2 + 1 which on solving simultaneously, we get
If n = 7, x = =0
−1 x=
3
and so on. 4
Which shows there exist three different solutions for 33
and y= 111
n ∈ I. 4
4 Algebra of Complex Numbers Multiplication of Complex Numbers
Let z1 = x1 + iy1 and z 2 = x 2 + iy2 be two complex
Addition of Complex Numbers
Objective Mathematics Vol. 1
numbers, then
Let z1 = x1 + iy1 and z 2 = x 2 + iy2 be two complex z1 ⋅ z 2 = ( x1 + iy1 )( x 2 + iy2 )
numbers, then = ( x1 x 2 − y1 y2 ) + i( x1 y2 + x 2 y1 )
z1 + z 2 = x1 + iy1 + x 2 + iy2 ⇒ z1 ⋅ z 2 = [ Re ( z1 ) Re ( z 2 ) − Im ( z1 ) Im ( z 2 )]
= ( x1 + x 2 ) + i( y1 + y2 ) + i [ Re ( z1 ) Im ( z 2 ) + Re ( z 2 )Im ( z1 )]
⇒ Re ( z1 + z 2 ) = Re ( z1 ) + Re ( z 2 )
Properties of Multiplication of Complex Numbers
and Im ( z1 + z 2 ) = Im ( z1 ) + Im ( z 2 )
(a) z1 ⋅ z 2 = z 2 ⋅ z1 [commutative law]
Properties of Addition of Complex Numbers (b) ( z1 ⋅ z 2 ) z 3 = z1 ( z 2 ⋅ z 3 ) [associative law]
(a) z1 + z 2 = z 2 + z1 [commutative law] (c) If z1 ⋅ z 2 = 1 = z 2 ⋅ z1 , then z1 and z 2 are
(b) z1 + ( z 2 + z 3 ) = ( z1 + z 2 ) + z 3 [associative law] multiplicative inverse of each other.
(c) z + 0 = 0 + z (where, 0 = 0 + i 0) (d) (i) z1 ( z 2 + z 3 ) = z1 ⋅ z 2 + z1 ⋅ z 3
[additive identity law] [left distribution law]
(ii) ( z 2 + z 3 ) z1 = z 2 ⋅ z1 + z 3 ⋅ z1
X Example 11. The value of 3(7 + i7) + i(7 + i7) is [right distribution law]
(a)15 + 27 i (b)14 + 28 i (c)14 − 28 i (d)14 + 23 i
X Example 13. The real values of x and y, if
Sol. (b) We have, 3(7 + i 7 ) + i (7 + i 7 ) (1 + i) x − 2i (2 − 3i) y + i
= 21 + 21i + 7 i + 7 i 2 + = i, are respectively
(3 + i) (3 − i)
= 21 + 28 i − 7 [Q i 2 = − 1]
= 14 + 28i (a) 3, −1 (b) 3, 1
(c) −3, 1 (d) −3, −1
Subtraction of Complex Numbers Sol. (a) (1 + i )x − 2 i + (2 − 3i )y + i = i
Let z1 = x1 + iy1 and z 2 = x 2 + iy2 be two complex (3 + i ) (3 − i )
numbers, then ⇒ {(1 + i )x − 2 i }(3 − i ) + {(2 − 3 i )y + i }(3 + i )
z1 − z 2 = ( x1 + iy1 ) − ( x 2 + iy2 ) = i (3 + i )(3 − i )
= ( x1 − x 2 ) + i( y1 − y2 ) ⇒ (1 + i )(3 − i )x − 2 i (3 − i ) + (2 − 3i )(3 + i )y
⇒ Re( z1 − z 2 ) = Re( z1 ) − Re( z 2 ) + i (3 + i ) = 10 i
⇒ (4 + 2 i ) x − 6 i − 2 + (9 − 7 i )y + 3 i − 1 = 10 i
and Im ( z1 − z 2 ) = Im ( z1 ) − Im ( z 2 )
⇒ (4 x − 2 + 9 y − 1) + i (2 x − 6 − 7 y + 3) = 10 i
X Example 12. The value of ⇒ (4 x + 9 y − 3) + i (2 x − 7 y − 3) = 10 i
1 7 1 4 On equating real and imaginary parts on both sides, we
3 + i 3 + 4 + i 3 − − 3 + i is get
4x + 9y = 3 …(i)
5 17 and x − 7 y = 13 …(ii)
(a) − i
3 3 On solving Eqs. (i) and (ii), we get x = 3, y = − 1
17 5
(b) − i X Example 14. The multiplicative inverse of 4 − 3i
3 3
is
17 5
(c) + i 4 3i 4 3i
3 3 (a) − (b) +
25 25 25 25
17 4
(d) − i 4 3i
5 3 (c) + (d) None of these
16 25
Sol. (c) 1 + i 7 + 4 + i 1 + 4 − i Sol. (b) Let z = 4 − 3 i
3 3 3 3
Then, its multiplicative inverse is
= + 4 +
1 4 7 1
+ i + − 1 1 1 1 4 + 3i 4 + 3i
3 3 3 3 = = × =
z 4 − 3 i 4 − 3 i 4 + 3 i 16 − 9 i 2
= + +
1 4 4 7 1 1
+ i + − [Q(a − b )(a + b ) = a2 − b 2 ]
3 1 3 3 3 1
(1 + 12 + 7 + 1 − 3 4 + 3i
=
4)
+ i = [Q i 2 = − 1]
16 + 9
3 3
4 + 3i 4 3i
=
17
+ i
5 = = +
112 3 3 25 25 25
Division of Complex Numbers
Let z1 = x1 + iy1 and z 2 = x 2 + iy2 ( ≠ 0) be two
(iv) Re ( z ) = Re ( z ) =
z−z
z+z
2 4
Complex Numbers
complex numbers, then (v) Im ( z ) =
z1 x1 + iy1 2i
=
z 2 x 2 + iy2 (vi) z1 + z 2 = z1 + z 2
=
1
[( x1 x2 + y1 y2 ) + i ( x2 y1 − x1 y2 )] (vii) z1 − z 2 = z1 − z 2
x22 + y22 (viii) z1 z 2 = z1 z 2
z1 z z
X Example 15. The value of , where z1 = 2 + 3i (ix) 1 = 1 , ( z 2 ≠ 0)
z2
z2 z2
and z 2 = 1 + 2 i, is
8 1 8 1 (x) z1 z 2 + z1 z 2 = 2Re ( z1 z 2 ) = 2Re ( z1 z 2 )
(a) + i (b) − i
5 5 5 5 (xi) z n = ( z ) n = ( z n )
1 8 (xii) If z = f ( z1 ), then z = f ( z1 )
(c) − i (d) None of these
5 5
X Example 16. If z1 = 9 y 2 − 4 − 10 ix and
Sol. (b)Q z1 = 2 + 3i and z2 = 1 + 2 i
1 1− 2 i z 2 = 8 y 2 + 20 i, where z1 = z 2 , then z = x + iy is
∴ z2−1 = =
1 + 2 i (1 + 2 i )(1 − 2 i ) equal to
1− 2 i 1 2 (a) − 2 + 2 i
= = − i
1+ 4 5 5
(b) − 2 ± 2 i
= z1 ⋅ z2 = (2 + 3 i ) − i
z1 −1 1 2
Then,
5 5
(c) − 2 ± i
z2
(d) None of the above
= + + i − + = − i
2 6 4 3 8 1
5 5 5 5 5 5 Sol. (b) Given, z1 = z2
Aliter ⇒ 9 y2 − 4 − 10 i x = 8 y2 + 20 i
Here, x1 = 2, y1 = 3, x2 = 1 and y2 = 2. ⇒ ( y2 − 4) − 10 i ( x + 2 ) = 0
z1 1
∴ = 2 {( x1 x2 + y1 y2 ) + i ( x2 y1 − x1 y2 )} Since, complex number is zero.
z2 x2 + y22 ⇒ y2 − 4 = 0 and x + 2 = 0
1 ∴ y=±2
= 2 {(2 × 1 + 3 × 2 ) + i (3 × 1 − 2 × 2 )}
1 + 22 and x = −2
1 8 1 Thus, z = x + iy = − 2 ± 2 i
= {(2 + 6) + i (3 − 4)} = − i
5 5 5
X Example 17. If (1 + i) z = (1 − i) z, then z is
(a) x (1 − i), x ∈ R
Conjugate of a Complex Number (b) x (1 + i), x ∈ R
Geometrically, the conjugate of z is the reflection x
of point image of z in the real axis. (c) , x ∈R +
1+ i
Y
(d) None of the above
Imaginary axis
z Sol. (a) (1 + i )( x + iy) = (1 − i )( x − iy)
⇒ ( x − y) + i ( x + y) = ( x − y) − i ( x + y)
⇒ x+ y=0
X ∴ z = x − ix = x(1 − i )
O Real axis
z Modulus of a Complex Number
∴ z = x + iy and z = x − iy. Let z = x + iy be any complex number. Then,
e.g. If z = 3 + 4 i, then z = 3 − 4 i. | z | = ( x 2 + y 2 ) is called the modulus of the complex
Properties of Conjugate number z, where modulus | z | represents distance of z
from origin.
z is the mirror image of z along real axis.
e.g. If z = 3 + 2i is a complex number, then
(i) ( z ) = z
| z | = 32 + 22
(ii) z = z ⇔ z is purely real
(iii) z = − z ⇔ z is purely imaginary. = 9 + 4 = 13 113
4 X Example 18. If z is a complex number
satisfying the relation | z + 1| = z + 2(1 + i), then z is
Principal values of the argument are θ, π − θ,
−π + θ, − θ according as the complex number lies on the
Ist, IInd, IIIrd or IVth quadrant.
Objective Mathematics Vol. 1
1
(a) (1 + 4 i)
2 | y|
Here, θ = tan −1 , where z = x + iy
1 |x|
(b) (3 + 4 i)
2 Y
1 arg (z) = π – θ arg (z) = θ
(c) (1 − 4 i)
2
1
(d) (3 − 4 i) θ θ
2 X' θ
X
θ
Sol. (c) Let z = x + iy
∴ | x + iy + 1| = x + iy + 2(1 + i )
arg (z) = – π + θ arg (z) = – θ
⇒ ( x + 1) + y2 = ( x + 2 ) + i ( y + 2 )
2
⇒ ( x + 1)2 + y2 = x + 2 and y+2=0 Y'
⇒ ( x + 1) + 4 = ( x + 2 )
2 2
and y = −2 1 π
e.g. arg (1 + i) = tan −1 =
⇒ 2 x + 5 = 4 x + 4 and y = − 2 1 4
1
⇒ x= and y = − 2 −1 − π
2 arg (1 − i) = tan −1 =
1 1 4
⇒ z = (1 − 4 i )
2 −1 π
arg ( −1 − i) = tan −1 = − π +
X Example 19. If z = 1 + i tan α, where −1 4
3π −1 π
π < α < , then | z | is equal to arg ( −1 + i) = tan −1 = π −
2 1 4
(a) sec α
Ø Argument of 0 is not defined.
(b) −sec α
●
● If z1 = z 2, then | z1| = | z 2| and arg(z1) = arg(z 2)
(c) cosec α ● Argument of purely imaginary number is
(d) None of the above π π
or −
Sol. (b) As z = 1 + i tan α 2 2
∴ | z| = 1 + tan2 α = |sec α|
● Argument of purely real number is 0 or π .
3π
⇒ | z| = − sec α, as π < α <
2 X Example 20. The modulus and argument of the
1+2i
complex number is
Argument (or Amplitude) of a 1−3i
1 3π
Complex Number (a) ,
2 4
Argument of z = θ 1 3π
Argument of z is not unique. General value of (b) ,−
2 4
argument of z is 2nπ + θ.
Y
1 3π
(c) ,
2 4
Imaginary axis
z = (x + iy) (d) None of the above
Sol. (a) Given, z = 1 + 2 i
1 − 3i
θ
X 1 + 2 i 1 + 3i 1 + 3i + 2 i + 6i 2
O Real axis ∴ z= × =
1 − 3i 1 + 3i 12 − (3 i )2
[Q(a + b )(a − b ) = a2 − b 2 ]
Principal Value of Argument =
1 + 5 i + 6(−1)
[Qi 2 = − 1]
1 − 9i 2
The value of θ of the argument, which satisfies the
1 + 5 i − 6 −5 + 5 i −1 + i
inequality −π < θ ≤ π is called the principal value of the = = =
114 1+ 9 10 2
argument.
π π
⇒
∴
z=−
| z| =
1 1
+ i
2 2
2
− 1 + 1
2
[Q|a + ib| = a + b ]
2 2
⇒ tan θ = 1 = tan
4
⇒ θ=
4
Since, the real part of z is negative and imaginary part
4
Complex Numbers
of z is positive, so the point lies in IInd quadrant.
2 2
π 3π
1 1 2 1 1 ∴ arg ( z) = π = θ = π − =
+ == = = 4 4
4 4 4 2 2 1
Hence, modulus =
1 2
Im ( z)
∴ Now, tan θ = 2 Qθ = tan−1 arg( z) =
3π
and
1 Re ( z) 4
−2
Work Book Exercise 4.1
x−3 y−3 10 The diagram shows several numbers in the
1 If + = i , where x, y ∈ R, then
3+ i 3− i complex plane. The circle is the unit circle
a x = 2 and y = − 8 b x = − 2 and y = 8 centered at the origin. One of these numbers is
c x = − 2 and y = − 6 d x = 2 and y = 8 the reciprocal of F, which is
2 What is the real part of (1 + i )50 ? Imaginary axis
25
a 0 b 2 F
c − 2 25 d − 2 50
D
3 The complex number z satisfies z + | z| = 2 + 8 i .
O C Real axis
Then, the value of| z| is
a 10 b 13 c 17 d 23
4 If z + z = 0, then which of the following must be
3
B A
true on the complex plane?
a Re ( z) < 0 b Re ( z) = 0 a A b B c C d D
c Im ( z) = 0 d z4 = 1
11 Identify the incorrect statement.
5 The sequences S = i + 2 i 2 + 3i 3 + K upto 100 a No non-zero complex number z satisfies the
equation z = − 4 z
terms simplifies to, where i = −1
b z = z implies that z is purely real
a 50 (1 − i ) b 25 i
c z = − z implies that z is purely imaginary
c 25 (1 + i ) d 100 (1 − i )
d If z1, z2 are the roots of the quadratic equation
6 Given i = −1, then the value of the sum az2 + bz + c = 0 such that Im( z1 z2 ) ≠ 0, then a, b, c
1 1 1 1 2 must be real numbers
+ + + +
1 + i 1 − i −1 + i −1 − i 1 + i
12 If z = (3 + 7 i )( p + iq ), where p, q ∈ l − { 0}, is
2 2 2
+ + + purely imaginary, then minimum value of| z|2 is
1 − i −1 + i −1 − i
3 3 3 3 a 0 b 58
+ + + + +K c
3364
d 3364
1 + i 1 − i −1 + i −1 − i 3
n n n n
+ + + + is 13 Consider two complex numbers α and β as
1 + i 1 − i −1 + i −1 − i 2 2
a 2n + 2n
2
b 2 in + 2 in
2 a + bi a − bi
α= + , where a, b ∈ R and
c (1 + i ) n2 d None of these a − bi a + bi
z −1
7 Let i = −1. The product of the real part of the β= , where| z| = 1, then
z+1
roots of z 2 − z = 5 − 5 i is
a both α and β are purely real
a − 25 b −6 c −5 d 25 b both α and β are purely imaginary
8 Number of complex numbers z satisfying z 3 = z c α is purely real and β is purely imaginary
d β is purely real and α is purely imaginary
is
a 1 b 2 14 If z is a complex number having the argument θ,
c 4 d 5 π
0 <θ < and satisfying the equality| z − 3 i | = 3.
2
9 Number of real solution of the equation,
z 3 + iz − 1 = 0 is 6
Then, cot θ − is equal to
a zero b one z
c two d three a 1 b −1 c i d −i
115
−1
4 Various Forms of a
Complex Number
Hence, cosθ =
sinθ =
2
3
Objective Mathematics Vol. 1
2
π 2π
⇒ θ= π− =
Polar Form 3 3
Let z = a + ib be any complex number, then by Thus, the required polar form is
2π 2π
taking 8 cos + i sin
3 3
a = r cos θ and b = r sin θ
We have, z = a + ib X Example 22. Let z and w be two non-zero
= r (cos θ + i sin θ ) complex numbers, such that | z | = | w | and
arg( z ) + arg( w) = π. Then, z is equal to
(known as polar form)
(a) w (b) − w (c) − w (d) w
(a, b)
Imaginary axis
Sol. (c) Given| z| = |w| = r and arg(w) = θ
Also, arg ( z) + arg (w) = π
b
r ⇒ arg( z) = π − θ
Now, z = r[cos( π − θ) + i sin( π − θ)]
θ = r[− cos θ + i sin θ]
a Real axis = − r(cos θ + i sin θ) = − w
where, r = z and θ = Principal value of argument Eulerian Form of a Complex Number
of z. We have,
e i θ = cos θ + i sin θ and e − i θ = cos θ − i sin θ.
X Example 21. The polar form of the complex
−16 These two are called Euler’s notations.
number is Let z be any complex number, such that | z | = r and
1+ i 3
arg ( z ) = θ. Then, in polar form, z can be written as
π π 2π 2π
(a) 4 cos + i sin (b) cos + i sin z = r (cos θ + i sin θ )
3 3 3 3
Using Euler’s notations, we have
2π 2π
(c) 8 cos + i sin (d) None of these z = re i θ
3 3
This form of z is known as the Eulerian form.
−16
Sol. (c) Given complex number =
1+ i 3 X Example 23. Express the following complex
−16 (1 − i 3 ) numbers in Eulerian form.
= ×
1 + i 3 (1 − i 3 ) (i) 1 + i (ii) −2 + 2 i (iii) −1 − i 3
−16(1 − i 3 )
= Sol. (i) Given, z = 1 + i .
1 − (i 3 )2
−16 (1 − i 3 ) Then, r = | z| = 12 + 12 = 2
=
1+ 3 1 π
Let θ be the argument of z. Then, tan θ= =1 ⇒ θ =
1 4
= − 4(1 − i 3) iπ
= − 4 + i4 3 So, Eulerian form of z is 2e 4 .
P (– 4, 4√3 ) Y
(ii) Given, z = − 2 + 2 i
Then, r = | z| = (−2 )2 + 2 2 = 2 2
θ
Let θ be the argument of z. Then,
2 3π
X′ X tan θ = = −1 ⇒ θ =
O −2 4
i 3π
So, Eulerian form of z is 2 2e 4 .
(iii) Given, z = −1 − i 3
Y′ Then, r = | z| = (−1)2 + (− 3 )2 = 2
Let − 4 = rcos θ, 4 3 = rsin θ Let θ be the argument of z. Then,
By squaring and adding, we get − 3 2π
tan θ = = 3 ⇒ θ=−
16 + 48 = r 2 (cos 2 θ + sin2 θ) −1 3
2π
−i
116 which gives, r 2 = 64, i.e. r=8 So, Eulerian form of z is 2e 3 .
X Example 24. Complex numbers z1 , z 2 , z 3 are
the vertices A, B , C respectively of an isosceles
xii. | z1 + z 2 | 2 = | z1 | 2 + | z 2 | 2
z
⇔ 1 is purely imaginary.
4
Complex Numbers
right angled triangle with right angle at C. Show
z2
that
( z1 − z 2 ) 2 = 2( z1 − z 3 ) ( z 3 − z 2 ). xiii. | z1 + z2 |2 + | z1 − z2 |2 = 2{| z1 |2 + | z2 |2 }
Sol. In an isosceles ∆ ABC, B(z2)
xiv. | az1 − bz 2 | 2 + | bz1 + az 2 | 2
AC = BC and BC perpendicular
to AC. It means that AC is = ( a 2 + b 2 )(| z1 | 2 + | z 2 | 2 ), where a, b ∈ R .
π
rotated through angle to
2 Ø ● If z is unimodular, then | z | = 1. Now, if f (z) is a unimodular,
occupy the position BC.
C(z3) A(z1) then it is always be expressed as f (z) = cos θ + i sin θ, θ ∈R..
We have,
● Square root of z = a + ib is given by
z2 − z3 iπ / 2 |z| + a |z| − a
=e =i z =± +i , a = Re ( z)
z1 − z3 2 2
⇒ z2 − z3 = + i ( z1 − z3 ) To find the square root of a − ib, replace i by − i in the above
⇒ z22 + z32 − 2 z2 z3 = − ( z12 + z32 − 2 z1 z3 ) result.
⇒ z12 + z22 − 2 z1 z2 = 2 z1 z3 + 2 z2 z3 − 2 z1 z2 − 2 z32
● If x, y ∈R , then
x + iy + x − iy = 2( x 2 + y 2 + x)
= 2( z1 − z3 ) ( z3 − z2 )
⇒ ( z1 − z2 )2 = 2( z1 − z3 ) ( z3 − z2 ) x + iy − x − iy = i 2( x 2 + y 2 − x)
a − ib
Properties of Modulus of Complex Number X Example 25. If x − iy = , then
c − id
i. | z | ≥ 0 ⇒ | z | = 0 iff z = 0 and | z | > 0 iff z ≠ 0 ( x 2 + y 2 ) 2 is equal to
a 2 − b2 a 2 + b2
ii. −| z | ≤ Re ( z ) ≤ | z | and −| z | ≤ Im ( z ) ≤ | z | (a) 2 (b)
c − d2 c2 + d 2
iii. | z | = | z | = |− z | = |− z | a 2 + b2
(c) 2 (d) None of these
c − d2
iv. zz = | z | 2
Sol. (b) Given, x − iy = a − ib
v. | z1 z 2 | = | z1 || z 2 | c − id
1/ 2
In general, | z1 z 2 z 3 K z n | = | z1 || z 2 || z 3 | K | z n | a − ib
⇒ x + i (− y) =
c − id
1
z |z |
= 1 , ( z 2 ≠ 0)
vi. On taking modulus both sides, we get
z 2 | z 2 | a − ib 1/ 2
| x + i (− y)| =
c − id
vii. | z1 ± z 2 | ≤ | z1 | + | z 2 |
1/ 2
In general, | z1 ± z 2 ± z 3 ± K ± z n | a − ib
⇒ x2 + (− y)2 =
≤ | z1 | + | z 2 | + | z 3 | + K + | z n | c − id
[Q| x + iy| = x2 + y2 and| z n| = | z| n ]
viii. | z1 ± z 2 | ≥ || z1 | − | z 2 || 1/ 2
a − ib
⇒ x 2 + y2 =
ix. | z | = | zz |
n n
c − id
On squaring both sides, we get
x. || z1 | − | z 2 || ≤ | z1 + z 2 | ≤ | z1 | + | z 2 | a − ib
x 2 + y2 =
Thus, | z1 | + | z 2 | is the greatest possible value c − id
of | z1 + z 2 | and || z1 | − | z 2 || is the least possible |a − ib| z z1
⇒ x 2 + y2 = Q =
1
value of | z1 + z 2 |. |c − id| z2 z2
a2 + b 2
xi. | z1 ± z 2 | 2 = ( z1 ± z 2 )( z1 ± z 2 ) ⇒ ( x 2 + y2 ) = [Q| x − iy| = x 2 + y2 ]
c2 + d 2
= | z1 | 2 + | z 2 | 2 ± ( z1 z 2 + z1 z 2 ) On squaring both sides, we get
= | z1 | 2 + | z 2 | 2 ± 2Re ( z1 z 2 ) a2 + b 2
( x2 + y2 )2 = 2
c + d2
117
4 X Example 26. For a complex number z, the
minimum value of | z | + | z − 2 | is
X Example 30. The value of − 8 − 6 i is equal to
(a)1 ± 3 i (b) ± (1 − 3 i) (c) ± (1 + 3 i) (d) ± (3 − i)
Objective Mathematics Vol. 1
(a) 1 (b) 2
Sol. (b) Given, −8 − 6i = x + iy = z
(c) 3 (d) None of these | z| + x | z| − x
Then, z=± −i [QIm( z) < 0 ]
Sol. (b) By using | z1| + | z2| ≥ | z1 − z2| 2 2
We have, | z| + | z − 2| ≥ | z − ( z − 2 )| 10 − 8 10 + 8
⇒ z=± −i [Q z = x + y ]
2 2
∴ | z| + | z − 2| ≥ 2 2 2
⇒ z = ± (1 − 3 i )
X Example 27. If | z1 − 1| < 1, | z 2 − 2 | < 2 and
| z 3 − 3 | < 3, then | z1 + z 2 + z 3 | X Example 31. The value of
(a) is less than 6 (b) is more than 3 ( 4 + 3 − 20 )1/ 2 + ( 4 − 3 − 20 )1/ 2 is
(c) is less than 12 (d) lies between 6 and 12 (a) ±6 (b) 0 (c) ± 5 (d) ± 3
Sol. (c)| z1 + z2 + z3| = |( z1 − 1) + ( z2 − 2 ) + ( z3 − 3) + 6|
Sol. (a) We may write, (4 + 3 − 20 ) = (4 + 6 i 5 )
≤ | z1 − 1| + | z2 − 2| + | z3 − 3| + 6 Let (4 + 3 − 20 )1/ 2 = ( x + iy)
< 1 + 2 + 3 + 6 = 12
Then, (4 + 6 i 5 )1/ 2 = ( x + i y)
X Example 28. If α, β are two complex numbers, ⇒ 4 + 6i 5 = ( x2 − y2 ) + (2 xy) i
then | α | 2 + | β | 2 is equal to ⇒ x + y2 = 4 and 2 xy = 6 5
2
1 1 ∴ x2 + y2 = ( x2 − y2 )2 + 4 x2 y2
(a) (| α + β | 2 − | α − β | 2 ) (b) (| α + β | 2 + | α − β | 2 )
2 2 = (16 + 180) = 196 = 14
(c) | α + β | + | α − β |
2 2
(d) None of these On solving the equations x2 = y2 = 14 and x2 − y2 = 4,
we get
Sol. (b)|α + β|2 = (α + β)(α + β) = (α + β)(α + β ) x2 = 9 and y2 = 5
∴ x = ± 3 and y = ± 5
= αα + ββ + αβ + αβ
Since, xy > 0, it follows that x and y are of the same sign.
= |α|2 + |β|2 + αβ + αβ …(i)
∴ x = 3, y = 5 or x = − 3, = − 5
|α − β| = (α − β )(α − β ) = αα + ββ − αβ − αβ
2
So, (4 + 3 − 20 )1/ 2 = (4 + 6 i 5 )1/ 2
= |α|2 + |β|2 − αβ − αβ …(ii) = ± (3 + 5i) …(i)
∴ (4 − 3 −20 )1/ 2 = ± (3 − 5 i) …(ii)
Adding Eqs. (i) and (ii), we get
1 Hence, (4 + 3 − 20 )1/ 2 + (4 − 3 20 )1/ 2 = ± 6
|α|2 + |β|2 = {|α + β|2 + |α − β|2 }
2 [on adding Eqs. (i) and (ii)]
Aliter
X Example 29. If z1 and z 2 are two complex Here, x = 4, y = 6 5
numbers, such that | z1 | < 1 < | z 2 |, then prove that ∴ = (4 + 3 − 20 )1/ 2 + (4 − 3 − 20 )1/ 2
1 − z1 z 2 = 2 { 42 + (6 5 )2 + 4} = 2( 16 + 180 + 4 )
< 1.
z1 − z 2 = 2 (14 + 4) = 36 = ± 6
Sol. Given,| z1| < 1 and| z2| > 1 …(i)
Properties of Arguments
Then, to prove
1 − z1 z2 z1 | z1| i. arg ( z1 z 2 ) = arg ( z1 ) + arg ( z 2 ) + 2kπ
<1 Q =
z1 − z2 z2 | z2| (k = 0 or 1 or −1)
⇒ |1 − z1 z2| < | z1 − z2| …(ii) In general,
On squaring both sides, we get
arg( z1 z 2 z 3 K z n ) = arg( z1 ) + arg( z 2 )
(1 − z1 z2 )(1 − z1 z2 ) < ( z1 − z2 )( z1 − z2 ) (Q| z|2 = zz )
+ arg( z 3 ) +K+ arg( z n ) + 2kπ (k = 0 or 1 or −1)
⇒ 1 − z1 z2 − z1 z2 + z1 z1 z2 z2 < z1 z1 − z1 z2 − z2 z1 + z2 z2
z
⇒ 1 + | z1|2| z2|2 < | z1|2 + | z2|2 ii. arg 1 = arg( z1 ) − arg( z 2 ) + 2kπ
z2 (k = 0 or 1 or −1)
⇒ 1 − | z1|2 − | z2|2 + | z1|2| z2|2 < 0
z
⇒ (1 − | z1|2 )(1 − | z2|2 ) < 0 …(iii) iii. arg = 2 arg( z ) + 2kπ (k = 0 or 1 or −1)
z
which is true by Eq. (i) as| z1| < 1 and| z2| > 1.
∴ (1 − | z1|2 ) > 0
iv. arg( z n ) = n arg( z ) + 2kπ (k = 0 or 1 or −1)
and (1 − | z2|2 ) < 0
z2 z1
∴ Eq. (iii) is true, whenever Eq. (i) is true. v. If arg = θ, then arg = 2kπ − θ, where
1 − z1 z2 z1 z2
⇒ < 1 Hence proved.
118
z1 − z2 k ∈I .
vi. 1
arg( z ) = − arg( z ) = arg
2
Sol. (a) We have, z2 = z1 and z4 = z3
∴
z
z1 z2 =| z1|2
z
and z3 z4 =| z3|2
z1 z2
4
Complex Numbers
Now, arg 1 + arg 2 = arg
vii. If arg( z ) = 0 ⇒ z is real. 4
z z3 z4 z3
| z |2 z 2
viii. arg( z1 z 2 ) = arg( z1 ) − arg( z 2 ) = arg 1 2 = arg 1 = 0
| z3| z
3
π
ix. | z1 + z 2 | = | z1 − z 2 | ⇒ arg( z1 ) − arg( z 2 ) = [Qargument of positive real number is zero]
2
x. | z1 + z 2 | = | z1 | + | z 2 | ⇒ arg( z1 ) = arg( z 2 ) De-Moivre’s Theorem
xi. If | z1 | ≤ 1, | z 2 | ≤ 1, then (a) If n ∈ I (the set of integers), then
(cos θ + i sin θ ) n = cos n θ + i sin n θ.
(a) | z1 − z 2 | 2 ≤ (| z1 | − | z 2 | ) 2 +
(b) If n ∈Q (the set of rational numbers), then
[arg( z1 ) − arg( z 2 )]2 cos n θ + i sin n θ is one of the values of
(b) | z1 + z 2 | 2 ≥ (| z1 | + | z 2 | ) 2 − (cos θ + i sin θ ) n .
[arg( z1 ) − arg( z 2 )]2 Remark
(i) The theorem is also true for (cos θ − i sin θ ), i.e.
xii. | z1 ± z 2 | 2 = | z1 | 2 + | z 2 | 2 ± 2 | z1 | | z 2 | (cos θ − i sin θ ) n = cos n θ − i sin n θ, because
cos (θ 1 − θ 2 ) (cos θ − i sin θ ) n = [cos ( − θ ) + i sin( − θ )]n
= cos( n( − θ ) + i sin ( n( − θ ))
xiii. z1 z 2 + z1 z 2 = 2 | z1 || z 2 |cos (θ 1 − θ 2 ) = cos( −nθ ) + i sin( −nθ )
where, θ 1 = arg( z1 ) and θ 2 = arg( z 2 ) = cos nθ − i sin nθ
1
Ø Proper value of k must be chosen, so that RHS of (i), (ii), (iii) and (ii) = (cos θ + i sin θ ) − 1 = cos θ − i sin θ
cos θ + i sin θ
(iv) lies in (−π, π).
3+i (iii) If z = (cos θ 1 + i sin θ 1 ) (cos θ 2 + i sin θ 2 )
X Example 32. If z = , then the K (cos θ n + i sin θ n )
3−i
Then, z = cos (θ 1 + θ 2 + K + θ n )
fundamental amplitude of z is
π π + i sin(θ 1 + θ 2 + K + θ n )
(a) − (b) (iv) If z = r (cos θ + i sin θ ) and n is a positive integer,
3 3
π then
(c) (d) None of these 2kπ + θ 2kπ + θ
6 z 1/ n = r 1/ n cos + i sin ,
n n
Sol. (b) amp( z) = amp 3 + i
3 −i where k = 0, 1, 2, …, ( n −1)
= amp( 3 + i ) − amp( 3 − i ) Ø ● (sin θ ± i cos θ)n ≠ sin nθ ± i cos nθ
−1 π π π
− tan−1
1
= tan−1 = + = n
3 3 6 6 3 π π
● (sin θ + i cos θ)n = cos − θ + i sin − θ
2 2
X Example 33. The value of amp (iω ) + amp (iω 2 ),
nπ nπ
where i = −1 and ω = 3 1 = non-real, is = cos − nθ + i sin − nθ
2 2
π
(a) 0 (b) ● (cos θ + i sin φ)n ≠ cos nθ + i sin nφ
2
(c) π (d) None of these cos θ + i sin θ π π
X Example 35. If z = , <θ < .
Sol. (c) amp (iω) + amp (iω2 ) = amp (i 2 ⋅ ω3 ) = amp(−1) = π cos θ − i sin θ 4 2
Then, arg ( z ) is
X Example 34. If z1 , z 2 and z 3 , z 4 are two pairs (a) 2θ (b) 2θ − π
of conjugate complex numbers, then (c) π + 2 θ (d) None of these
z z Sol. (a) z = (cos θ + i sin θ)2 = cos 2 θ + i sin 2 θ
arg 1 + arg 2 equals to
z4 z3 where,
π
< 2θ < π
π 3π 2
(a) 0 (b) (c) (d) π Q Clearly, arg ( z) = 2θ Q π < arg( z) < π
2 2 2 119
4 X
the value of x1 x 2 x 3 K ∞ is
π
2
π
Example 36. If x r = cos r + i sin r , then
2 X Example 38. If z =
3 i
2 2
+ +
3 i
5
− ,
2 2
5
Objective Mathematics Vol. 1
then
(a) −1 (b) 1
(a) Re ( z ) = 0 (b) Im ( z ) = 0
(c) 0 (d) None of these
(c) Re ( z ) > 0, Im ( z ) > 0 (d) Re ( z ) > 0, Im ( z ) < 0
Sol. (a)Q xr = cos πr + i sin πr Sol. (b) Given,
2 2 5 5
π π 3 i 3 i
∴ x1 = cos + i sin z= + + −
2 2 2 2 2 2
π π π π
5
π π
5
x2 = cos 2 + i sin 2
= cos + i sin + cos − i sin
2 2
6 6 6 6
M M 5π 5π 5π 5π 5π
= cos + i sin + cos − i sin = 2 cos
π π π π
⇒ x1 x2 x3 … = cos + i sin cos 2 + i sin 2 K
6 6 6 6 6
2 2 2 2 Hence, Im ( z) = 0.
π π π π
= cos + 2 + K + i sin + 2 + K X Example 39. The product of all the values of
2 2 2 2 3/ 4
π π π π
cos + i sin is
= cos 2 + i sin 2 3 3
1
1 − 1 − 1
2 2
(a) −1 (b) 1 (c) 3/2 (d) −1/ 2
3/ 4
= cos π + i sin π = − 1 Sol. (b) Given, cos π + i sin π = [cos π + i sin π ]1/ 4
3
3
(cos θ + i sin θ ) 4 Since, the expression has only 4 different roots, therefore
X Example 37. is equal to on putting n = 0, 1, 2, 3 in
(sin θ + i cos θ ) 5
2 nπ + π 2 nπ + π
(a) cos θ − i sin θ (b) cos 9 θ − i sin 9 θ cos + i sin and multiplying them,
4 4
(c) sin θ − i cos θ (d) sin 9 θ − i cos 9 θ π π 3π 3π
we get cos + i sin cos + i sin
4 4 4 4
(cos θ + i sin θ )4 (cos θ + i sin θ )4
Sol. (d) = cos 5 π + i sin 5 π cos 7 π + i sin 7 π
(sin θ + i cos θ )5
1
5
i 5 sin θ + cos θ 4 4 4 4
i
1 −1 1 −1 −1 1
=
1 1
(cos θ + i sin θ)4 (cos θ + i sin θ)4 + i + i + i −i
= = 2 2 2 2 2 2 2 2
i (cos θ − i sin θ)5 i (cos θ + i sin θ)− 5
1 1 −1 1
1 = − − − = (− 1)(− 1) = 1
= (cos θ + i sinθ)9 = sin 9θ − i cos 9θ 2 2 2 2
i
Work Book Exercise 4.2
1 The maximum and minimum values of| z + 1|, 1 + 2i
3 If = r (cos θ + i sin θ ), then
when| z + 3| ≤ 3 are 2+i
3 4
a (5, 0) b (6, 0) c (7, 1) d (5, 1) a r = 1, θ = tan−1 b r= 5, θ = tan−1
4 3
2 The region represented by inequalities 4
π c r = 1, θ = tan−1 d None of these
arg z ≤ ,| z| ≤ 2, Im ( z ) ≥ 1in the argand 3
3
diagram is given by 4 If z1 and z2 are two non-zero complex numbers,
Y Y such that| z1 + z2| = | z1| + | z2|, then
2 2 arg ( z1 ) − arg ( z2 ) is equal to
π π
1 1 a − π b − c 0 d
2 2
a b
60° π
5 If z = 1 − sin α + i cos α, where α ∈ 0, , then
60°
1 2 X 1 2 X
2
Y Y the modulus and the principal value of the
2 2
argument of z are respectively
π α π α
c 1 d
1 a 2(1 − sin α ), + b 2(1 − sin α ), −
4 2 4 2
60° π α π α
2(1 + sin α ), + d 2(1 + sin α ), −
60°
X X c
1 2 1 2 4 2 4 2
120
4 1
4
8
π π 11 Square root of x 2 +
1
− x − − 6, where
1 + sin + i cos
8 8 is equal to x 2
i x
6 The expression
π π x ∈ R is equal to
1 + sin − i cos
Complex Numbers
8 8 a ± x − + 2 i
1
b ± x − − 2 i
1
a 1 b −1 c i d −i x x
c ± x + + 2 i
d ± x + − 2 i
1 1
2 rπ 2 rπ
7 If zr = cos + i sin , r = 0, 1, 2, 3, 4, …, then x x
5 5
z1 z2 z3 z4 z5 is equal to 12 The minimum value of|1 + z| + |1 − z|, where z is
a −1 b 0 a complex number, is
c 1 d None of these 3
a 2 b
2
π π
8 If zn = cos + i sin , c 1 d 0
(2 n + 1)(2 n + 3) (2 n + 1)(2 n + 3)
π 2π
then lim ( z1 ⋅ z2 ⋅ z3 K zn ) is equal to 13 If arg( z + a) = and arg( z − a) = , a ∈ R + , then
n→ ∞ 6 3
π π π π a z is independent of a b | a| = | z + a|
a cos + i sin b cos + i sin
3 3 6 6 π π
c z = a, c is d z = a, c is
π π 6 3
c cos + i sin d None of these
12 12
14 Let z be a complex number satisfying the
9 If z1, z2 are two complex numbers and a, b are equation ( z 3 + 3)2 = − 16, then| z| has the value
two real numbers, then| az1 − bz2|2 + | bz1 + az2|2 equal to
is equal to a 51/ 2 b 51/ 3 c 5 2/ 3 d 5
a ( a + b )2[| z1|2 + | z2|2 ]
1− i 1− i 1+ i
b ( a + b )[| z1|2 + | z2|2 ] 15 For z1 = 6 , z2 = 6 , z3 = 6 ,
1+ i 3 3+i 3−i
c ( a 2 − b 2 )[| z1|2 + | z2|2 ]
which of the following holds good?
d ( a 2 + b 2 )[| z1|2 + | z2|2 ]
3
a Σ| z1|2 =
10 All real numbers x, which satisfy the inequality 2
b | z1|4 + | z2|4 = | z3|− 8
1 + 4i − 2 − x ≤ 5, where i = −1,x ∈ R are
c Σ| z1|3 + | z2|3 = | z3|− 6
a [ −2, ∞ ) b ( −∞, 2 ]
d | z1|4 + | z2|4 = | z3|8
c [0, ∞ ) d [−2, 0]
Roots of Unity
Cube Roots of Unity Important Identities
Let x= 1 3 (i) + x + 1 = ( x − ω )( x − ω 2 )
x2
⇒ x 3 −1 = 0 (ii) − x + 1 = ( x + ω )( x + ω 2 )
x2
(iii) + xy + y 2 = ( x − yω ) ( x − yω 2 )
x2
⇒ ( x − 1) ( x 2 + x + 1) = 0
(iv) − xy + y 2 = ( x + yω ) + ( x + yω 2 )
x2
−1 + i 3 −1 − i 3
Therefore, x =1, , (v) + y 2 = ( x + iy)( x − iy)
x2
2 2
(vi) + y 3 = ( x + y)( x + yω )( x + yω 2 )
x3
If second root is represented by ω, then third root
(vii) − y 3 = ( x − y) = ( x − yω )( x − yω 2 )
x3
will be ω 2 . Therefore, cube roots of unity are 1, ω, ω 2
and ω, ω 2 are called the imaginary cube roots of unity. (viii) + y 2 + z 2 − xy − yz − zx
x2
= ( x + yω + zω 2 )( x + yω 2 + zω )
Properties or ( xω + yω 2 + z )( xω 2 + yω + z )
0, if r is not a multiple of 3 or ( xω + y + zω 2 ) ( xω 2 + y + zω )
i. 1 + ωr + ω2 r =
3, if r is not a multiple of 3 (ix) x 3 + y 3 + z 3 − 3xyz
= ( x + y + z )( x + ωy + ω 2 y)( x + ωy 2 + ωz )
ii. ω 3 = 1 or ω 3 r = 1
(x) Two points P ( z1 ) and Q ( z 2 ) lie on the same
3r + 1 3r + 2
iii. ω = ω, ω =ω 2 side or opposite side of the line
az + az + b accordingly as az1 + az1 + b and
iv. It always forms an equilateral triangle. az 2 + az 2 + b have same sign or opposite sign.
121
4 X Example 40. If x 2 − x + 1 = 0, then the value of
5
1
2
nth Roots of Unity
Let z =11/ n . Then,
∑ x n + x n is
Objective Mathematics Vol. 1
n =1
z = (cos 0° + i sin 0° )1/ n
(a) 8 z = (cos 2rπ + i sin 2rπ )1/ n , r ∈ Z
2rπ 2rπ
(b) 10 ⇒ z = cos + i sin , r = 0, 1, 2, …, ( n −1)
(c) 12 n n
(d) None of the above [using De-Moivre’s theorem]
i 2rπ
1± 3i
Sol. (a) x2 − x + 1 = 0 ⇒ x= = − ω, − ω2 ⇒ z=e n , r = 0, 1, 2, …, ( n −1)
2
⇒ z = {e i 2π / n }r , r = 0, 1, 2, …, ( n −1)
5
1
+ 2
∴ ∑ x +
2n
n =1 x 2n i 2π
= x2 + 2 + 2 + x4 + 4 + 2 + x6 + 6 + 2
1 1 1 z =αr,α = e n , r = 0, 1, 2, …, ( n −1)
x x x
Thus, nth roots of unity are 1, α, α 2 ,…, α n − 1 , where
+ x + 8 + 2 + x + 10 + 2
8 1 10 1
x x i 2π
2π 2π
= (ω2 + ω4 + ω6 + ω8 + ω10 ) α=e n = cos + i sin
n n
+ 2 + 4 + 6 + 8 + 10 + 10
1 1 1 1 1
ω ω ω ω ω
[Q x = − ω or − ω2 ]
Properties of nth Roots of Unity
= − 1 − 1 + 10 = 8 i. nth roots of unity form a GP with common
100 i 2π
3 3 n
X Example 41. If 3 (x + iy) = + i 49
and ratio e .
2 2
ii. Sum of nth roots of unity is always zero.
x = ky, then k is
(a) −
1
(b) 3
iii. Sum of pth powers of nth roots of unity is
3 zero, if p is not a multiple of n.
1
(c) − 3 (d) − iv. Sum of pth powers of nth roots of unity is n, if
3
p is a multiple of n.
100 100
1 − i 3
Sol. (d) As 349 ( x + iy) = 3 + 3 i = i 3 v. Product of nth roots of unity is ( − 1) n − 1 .
2 2 2
⇒ 349 ( x + iy) = i 100 ⋅ 350 ⋅ (−ω)100 vi. nth roots of unity lie on the unit circle | z | =1
1 i 3
⇒ 3 ( x + iy) = 3 ⋅ ω = 3 ⋅ − +
49 50
50 and divide its circumference into n equal parts.
2 2
3 3 3 X Example 43. If z1 , z 2 , z 3 , …, z n are nth, roots
∴ x + iy = − + i
2 2 of unity, then for k =1, 2, …, n
1
⇒ x=− y (a) | z k | = k | z k + 1 | (b) | z k + 1 | = k | z k |
3
1 (c) | z k + 1 | = | z k | + | z k + 1 | (d) | z k | = | z k + 1 |
∴ k=−
3 Sol. (d) The nth roots of unity are given by
i 2 π ( k − 1)
−n
X Example 42. If z − z + 1 = 0, then z − z 2 n
, zk = e n , k = 1, 2, …, n
where n is a multiple of 3, is i 2 π ( k − 1)
(a) 2( − 1) n ∴ | zk| =
e
n
= 1, ∀ k = 1, 2, …, n
(b) 0
(c) ( − 1) n + 1 ⇒ | zk| = | zk + 1|, ∀ k = 1, 2, …, n
(d) None of the above X Example 44. If n is a positive integer greater
Sol. (b) As z − z + 1 = 0 ⇒
2
z = − ω, − ω 2 than unity and z is a complex number satisfying the
equation z n = ( z + 1) n , then
∴ zn − z− n = (− 1)n ωn − (− 1)− n ω− n (a) Re ( z ) < 0 (b) Re ( z ) > 0
= (− 1)n (ωn − ω− n ) , if n is a multiple of 3. (c) Re ( z ) = 0 (d) None of these
122 = (− 1)n (1 − 1) = 0
4
n
Sol. (a) We have, zn = (1 + z)n ⇒ z = 1 X Example 45. If α is non-real and α = 5 1, then
z + 1
1+ α + α2 + α− 2 − α− 1
z the value of 2 is equal to
Complex Numbers
⇒ = 11/ n
z+1 (a) 4 (b) 2
z
⇒ is nth root of unity. (c) 1 (d) None of these
z+1
z | z|
Sol. (a)Qα 5 = 1
⇒ = 1 ⇒ =1
z + 1 | z + 1| ∴ 1 + α + α2 + α− 2 − α− 1 = 1 + α + α2 + α3 − α4
⇒ | z| = | z + 1| = 1 + α + α 2 + α 3 + α 4 − 2α 4
1
⇒ x+ =0 [taking z = x + iy] 1 − α5
− 2α 4 = 2α 4 = 2 α
4
2 = = 2 × 1= 2
−1 1− α
⇒ x= ⇒ Re( z) < 0
2 ∴ 22 = 4
Work Book Exercise 4.3
1 If ω is a non-real cube root of unity, then the 10 If α is the non-real nth root of unity, then
expression (1 − ω )(1 − ω 2 )(1 + ω 4 )(1 + ω 8 ) is 1 + 3 α + 5 α 2 + K + (2 n − 1) α n − 1 is equal to
equal to 2n n
a b
a 0 b 3 c 1 d 2 1− α 1− α
2 x 3 m + x 3 n − 1 + x 3 r − 2 , where m, n, r ∈ N, is c
n
d None of these
divisible by 2(1 − α )
a x2 − x + 1 b x2 + x + 1
c x2 + x − 1 d None of these
11 −1 − −1 − −1K ∞ is equal to, where ω is the
imaginary cube of root of unity and i = − 1.
3 If ω is a non-real cube root of unity, then a ω or ω 2 b − ω or − ω 2
1 + 2 ω + 3ω2 2 + 3ω + ω2 c 1 + i or 1 − i d − 1 + i or − 1 − i
+ is equal to
2 + 3ω + ω2 3 + ω + 2 ω2
12 If α = e i 2 π / n , then (11 − α )(11 − α 2 ) K (11 − α n − 1 ) is
a −1 b 2ω c 0 d − 2ω
equal to
4 If ( 3 + i )n = ( 3 − i )n , n ∈ N, then least value of 11n − 1 11n − 1 − 1 11n − 1 − 1
a 11n − 1 b c d
n is 10 10 11
a 3 b 4
c 6 d None of these 13 The complex number w satisfying the equation
ω 3 = 8 i and lying in the IInd quadrant on the
5 If x 3 − 1 = 0 has the non-real complex roots α, β, complex plane is
then the value of (1 + 2α + β )3 − (3 + 3 α + 5 β )3 is 3 1
a − 3+ i b − + i
a −7 b 6 2 2
c −5 d 0 c −2 3 + i d − 3 + 2i
6 If ( 3 − i )n = 2 n , n ∈ I, the set of integers, then n is 14 Let z be a complex number satisfying the equation
a multiple of z 6 + z 3 + 1 = 0. If this equation has a root re i θ with
a 6 b 10 c 9 d 12 90 ° < θ < 180 °, then the value of θ is
a 100° b 110° c 160° d 170°
7 If z is a complex number satisfying
z 4 + z 3 + 2 z 2 + z + 1 = 0, then| z| is equal to 15 If ω is an imaginary cube root of unity, then the
value of ( p + q )3 + ( pω + qω 2 )3 + ( pω 2 + qω )3 is
1 3
a b
2 4 a p3 + q 3
c 1 d None of these b 3 ( p3 + q 3 )
c 3 ( p3 + q 3 ) − pq ( p + q )
8 If z is a non-real root of 7 − 1, then
d 3 ( p3 + q 3 ) + pq ( p + q )
z 86 + z175 + z 289 is equal to
a 0 b −1 c 3 d 1 16 If z 2 − z + 1 = 0, then the value of
2 2 2
9 Non-real complex number z satisfying the 1 2 1 3 1
z + + z + 2 + z + 3
equation z 3 + 2 z 2 + 3 z + 2 = 0 are z z z
−1 ± −7 1 + 7i 1 − 7i 1
a b , + K + z 24 + 24 is equal to
2 2 2 z
−1 + 7 i −1 − 7 i a 24 b 32
c − i, , d None of these 123
2 2 c 48 d None of these
4 17 If p = a + bω + cω 2 q = b + cω + aω 2 and
r = c + aω + bω 2 , where a, b, c ≠ 0 and ω is the
complex cube root of unity, then
22 If α, β respectively are the fifth and fourth non-real
roots of unity, then the value of
(1 + α )(1 + β )(1 + α 2 )(1 + β 2 )(1 + β 3 )(1 + α 3 ) is
Objective Mathematics Vol. 1
a p+ q + r = a+ b + c a 0 b (1 + α + α 2 )(1 − β 2 )
b p2 + q 2 + r 2 = a 2 + b 2 + c 2 c (1 + α )(1 + β + β 2 ) d 1
c p2 + q 2 + r 2 = − 2( pq + qr + rp)
d None of the above 23 When the polynomial 5 x 3 + Mx + N is divided by
x 2 + x + 1, the remainder is 0. The value of
18 If a and b are imaginary cube roots of unity, then (M + N ) is equal to
α n + β n is equal to a −3 b 05 c −5 d 15
2nπ 2nπ
a 2 cos b cos 24 If z and w are two complex numbers
3 3
2nπ 2nπ simultaneously satisfying the equations,
c 2 i sin d i sin
3 3 z 3 + w 5 = 0 and z 2 ⋅ w 4 = 1, then
19 If the six solutions of x 6 = − 64 are written in the a z and w both are purely real
b z is purely real and w is purely imaginary
form a + bi , where a and b are real, then the
c w is purely real and z is purely imaginary
product of those solutions with a > 0 is d z and w both are imaginary
a 4 b 8 c 16 d 64
25 Number of ordered pairs (z, ω) of the complex
20 If cos θ + i sin θ is a root of the equation numbers z and ω satisfying the system of
n −1 n−2 equations, z 3 + ω 7 = 0 and z 5 ⋅ ω11 = 1 is
x + a1 x
n
+ a2 x + K + an − 1 x + an = 0, then
n a 7 b 5 c 3 d 2
the value of ∑ ar cos r θ is 26 If 1, z1, z2 , z3 , …, zn − 1 is the nth roots of unity
r =1
a 0
and w is a non-real complex cube root of unity,
n −1
b
c
1
−1
then the product of ∏ (ω − zr ) is cannot be
r =1
d None of the above
equal to
21 If ω is a complex nth root of unity, then a 0 b 1 c −1 d 1+ ω
n
∑ (ar + b)ω r −1
is equal to 27. If Zr , r = 1, 2, 3, …, 50 are the roots of the
50 50
r =1 1
a
n( n + 1) a
b
nb equation ∑ (Z )r = 0, then the value of ∑ Zr −1
r =0 r =1
2 1− n
na is
c d None of these
n−1 a − 85 b − 25 c 25 d 75
Geometrical Applications of Complex Numbers
Basic Concepts in Geometry Sol. (c) Let z1 = − 1 − i and z2 = 2 + 3 i
Distance formula The distance between two Then, required distance = | z2 − z1 |
points P ( z1 ) and Q ( z 2 ) is given by = | 2 + 3i + 1 + i |
PQ = | z 2 − z1 | = | affix of Q − affix of P | =5
Q(z2) Section formulae If R ( z ) divides the line segment
joining P ( z1 ) and Q ( z 2 ) in the ratio m1 : m2 ( m1 , m2 > 0),
P(z1) then
For any complex number z, m z + m2 z1
(i) For internal division, z = 1 2
| z | = | z − 0 | = | z − (0 + i 0)| m1 + m2
Thus, modulus of a complex number z represented m1 z 2 − m2 z1
(ii) For external division, z =
by a point in the argand plane is its distance from origin. m1 − m2
X Example 46. Length of the line segment If R ( z ) is the mid-point of PQ, then affix of R is
joining the points − 1 − i and 2 + 3i is z1 + z 2
(a) −5 (b) 15 (c) 5 (d) 25 2
124
X Example 47. If A, B, C are three points in the
argand plane representing the complex numbers
λz + z 3
X Example 49. Find the general equation of line
joining the points z1 = (1 + i) and z 2 = (1 − i). 4
Complex Numbers
z1 , z 2 , z 3 such that z1 = 2 , where λ ∈R, then Sol. Clearly, the equation of a line is given by
λ +1 z( z1 − z2 ) − z( z1 − z2 ) + z1 z2 − z2 z1 = 0
the distance of A from the line BC is where, z1 = 1 + i and z2 = i − i
(a) λ On substituting the values of z1 and z2 , we get
λ z(1 − i − 1 − i ) − z(1 + i − 1 + i ) + (1 + i )(1 + i )
(b)
λ +1 − (1 − i )(1 − i ) = 0
(c) 1 ⇒ z( − 2 i ) − z (2 i ) + (1 − 1 + 2 i ) − (1 − 1 − 2 i ) = 0
(d) 0 ⇒ − 2 iz − 2 iz + 4 i = 0
⇒ z + z − 2 = 0, which is the required equation.
Sol. (d) As z1 = λz2 + z3
λ+1
Condition of Collinearity
which shows z1 divides z2 , z3 in the ratio of 1: λ.
Three points z1 , z 2 and z 3 are collinear, if
Thus, the points are collinear.
∴Distance of A from line BC is zero. z1 z1 1
z2 z2 1 = 0
X Example 48. Find the relation, if z1 , z 2 , z 3 , z 4
z3 z3 1
are the affixes of the vertices of a parallelogram
taken in order. X Example 50. If z1 , z 2 , z 3 are three complex
Sol. As the diagonals of a parallelogram bisect each other, numbers such that 5 z1 − 13 z 2 + 8 z 3 = 0, then prove
therefore affix of the mid-point of AC is same as the affix of that
the mid-point of BD. z1 z1 1
z1 + z3 z + z4
i.e. = 2
2 2 ` z 2 z 2 1 = 0.
⇒ z1 + z3 = z2 + z4 z3 z3 1
Equation of the Straight Line Sol. 5 z1 − 13 z2 + 8 z3 = 0
5 z1 + 8 z3.
Equation of the Line Passing through the Points z 1 ⇒ = z2
5+ 8
and z 2
⇒ z1, z2 and z3 are collinear.
Let z be any point on the line joining z1 and z 2 , z1 z1 1
then ⇒ z z2 1 = 0 [condition of collinear points]
2
z − z1 z3 z3 1
arg = π or 0
z 2 − z1 Hence proved.
z − z1 Length of Perpendicular
⇒ must be real.
z 2 − z1 The length of perpendicular from a point z1 to
∴ Required equation is az + az + b = 0 is given by
z z 1 az1 + az1 + b
z − z1 z − z1 2a
= ⇒ z1 z1 1 = 0
z 2 − z1 z 2 − z1
z2 z2 1 X Example 51. The length of perpendicular from
⇒ z ( z1 − z 2) − z ( z1 − z 2 ) + z1 z 2 − z 2 z1 = 0 P (2 − 3i) to the line (3 + 4i) z + (3 − 4i) z + 9 = 0 is
equal to
General Equation of a Line 9
(a) 9 (b)
az + az + b = 0, represents a straight line, where b 4
is a real number and a is a complex number. 9
(c) (d) None of these
2
Parametric Equation of a Line
Sol. (c) Let PM be the required length, then
z = z1 + t ( z 2 − z1 ), where t is real parameter |(2 − 3i )(3 + 4i ) + (3 − 4i ) (2 + 3i ) + 9|
PM =
= (1 − t ) z1 + t z 2 , represents the complete line 2 |3 − 4i |
through z1 and z 2 . 45 9
= =
10 2 125
4 Slope of a Line
Slope of the Line Segment Joining Two Points
⇒
⇒
b ( z2 − z1 ) − b ( z2 − z1 ) = 0
b z2 − bz1 − bz2 + bz1 = 0
Adding Eqs. (i) and (ii), we get
…(ii)
Objective Mathematics Vol. 1
If A, B represent complex numbers z1 , z 2 in the 2 (bz1 + bz2 ) = 2c ⇒ bz1 + bz2 = c
argand plane, then the complex slope of AB is defined
by Concept of Rotation
z1 − z 2 In this section, we shall learn about the effect of
z1 − z 2 multiplication of a complex number by e iα which will
Re ( z 2 − z1 ) also be interpreted geometrically.
and real slope is defined by .
Im ( z 2 − z1 ) Complex Number as a Rotating Arrow in the
Slope of Line az + az + b = 0 Argand Plane
The complex slope of the line 1. Let z = r (cos θ + i sin θ) = re iθ be a complex
−a − Coefficient of z number, represented by a point P in the argand
az + az + b = 0 is = plane. Then,
a Coefficient of z
OP = r and ∠XOP = θ
and real slope of the line az + az + b = 0 is Y Q (zeiα)
Re (a ) −i ( a + a )
– = P(z)
Im ( a ) (a − a )
α
X′ θ X
Ø ● If w1 and w 2 are the complex slope of two lines on the argand O x
plane, then the lines are
(a) perpendicular, if w1 + w 2 = 0
(b) parallel, if w1 = w 2
● The equation of a line parallel to the line Y′
az + az + b = 0 is az + az + λ = 0, where λ ∈R .
i ( θ + α)
● The equation of a line perpendicular to the line Now, ze iα = re iθ ⋅ e iα = re
az + az + b = 0 is az − az + iλ = 0, where λ ∈R .
This shows that ze iα is the complex number whose
X Example 52. If a point z1 is the reflection of a modulus is r and argument θ + α. Clearly, ze iα is
point z 2 through the line b z + b z = c, b ≠ 0 in the represented by a point Q in the argand plane such that
argand plane, then b z 2 + b z 1 is equal to OQ = r and ∠XOQ = θ + α.
(a) 4c (b) 2c In other words, to obtain the point representing
(c) c (d) None of these ze iα , we rotate OP through angle α in anti-clockwise
Sol. (c) If P ( z1 ) is the reflection of Q( z2 ) through the line sense. Thus, multiplication by e iα to z rotates the
bz + bz = c in the argand plane. Then, vector OP in anti-clockwise sense through an angle
P(z1)
α and vice-versa. Similarly, multiplication of z with
e − iα will rotate the vector OP in clockwise sense.
Remark Let z1 and Y
z 2 be two complex Q (z2)
R ( z1 2+ z2 ) numbers represented by P(z1)
bz + bz = c points P and Q in the q
X¢ X
argand plane, such that O
iθ
∠POQ = θ. Then, z1 e is
vector of magnitude
Q(z2)
z + z2
| z1 | = OP along OQ and Y¢
R 1 lies on the line.
2 z1 e iθ
z + z2 z1 + z2 = c
is a unit vector
b 1 + b | z1 |
2 2
⇒ b z1 + b z1 + b z2 + b z2 = 2c
z1 e iθ
…(i) along OQ. Consequently, | z 2 |⋅ is a vector of
Since, PQ is perpendicular to the line bz + bz = c. | z1 |
Therefore, magnitude | z 2 | = OQ along OQ.
Slope of PQ + Slope of the line = 0 |z | z
⇒
z2 − z1 − b
+
i.e. z 2 = 2 z1 e i θ ⇒ z 2 = 2 ⋅ z1 e i θ
126 =0 | z1 | z1
z2 − z1 b
X Example 53. The point represented by the
complex number 2 − i is rotated about origin
π
Then,
z 3 − z1 OQ
=
z 2 − z1 OP
(cos α + i sin α )
4
Complex Numbers
CA iα
through an angle in the clockwise direction, the = e
2 BA
new position of point is | z − z1 | iα
(a)1 + 2i (b) −1 − 2i = 3 e
| z 2 − z1 |
(c) 2 + i (d) −1 + 2i Y
Sol. (b) Here, z = 2 − i C (z 3 )
Let z1 be the required complex number.
Q z1 = ze − iπ / 2 B(z2)
π π
z1 = (2 − i ) cos − + i sin −
A(z1)
∴ Q (z3 – z1)
2 2
π π
= (2 − i ) cos – i sin α P (z2 – z1)
2 2
X
= (2 − i ) (0 − i ) O
= − (2 i − i 2 ) = – 2 i − 1
Q ∆OPQ and ∆ABC are congruent,
X Example 54. A particle P starts from the point OQ CA
∴ =
z 0 = 1 + 2i, where i = −1. It moves first horizontally OP BA
away from origin by 5 units and then vertically z − z1
away from origin by 3 units to reach a point z1 . or amp 3 =α
From z1 , the particle moves 2 units in the z 2 − z1
direction of the vector $i + $j and then it moves X Example 55. A man walks a distance of
through an angle π /2 in anti-clockwise direction on 3 units from the origin towards the North-East
a circle with centre at origin to reach a point z 2 . ( N 45° E ) direction. From there, he walks a
The point z 2 is given by distance of 4 units towards the North- West
(a) 6 + 7i (b) −7 + 6i ( N 45° W ) direction to reach a point P. Then, the
(c) 7 + 6i (d) −6 + 7i position of P in the argand plane is
Sol. (d) Imaginary axis
(a) 3e iπ / 4 + 4i (b) (3 − 4i) e iπ / 4
(c) ( 4 + 3i) e iπ / 4 (d) (3 + 4i) e iπ / 4
z2 z'2 (7, 6) Sol. (d) Let OA = 3 units, so that the complex number
1 associated with A is 3 e iπ / 4 . If z is the complex number
)
,2
associated with P, then
(1
90° 3 1
0
5 (6, 2)
z
Y
Real axis P
z2′ = (6 + 2 cos 45° , 5 + 2 sin 45° ) 4
= (7, 6) = 7 + 6i 3ei p/4
By rotation about (0, 0), A
z2 iπ 3
= e iπ / 2 ⇒ z2 = z2 ′ e 2
z2′
p/4
π π
z2 = (7 + 6i ) cos + i sin
X¢ X
O
2 2
= (7 + 6i ) (i ) = − 6 + 7 i Y¢
2. Let z1 , z 2 and z 3 be the vertices of a ∆ABC z − 3 e iπ / 4 4 − iπ / 2
= e
described in anti-clockwise sense. Draw OP and 0 − 3 e iπ / 4 3
OQ parallel and equal to AB and AC, respectively. 4i
=−
Then, point P is z 2 − z1 and Q is z 3 − z1 . If OP is 3
rotated through ∠α in anti-clockwise sense, it ⇒ 3 z − 9 e iπ / 4 = 12 e iπ / 4
coincides with OQ. ⇒ z = (3 + 4i )e iπ / 4
127
4 X Example 56. The complex numbers z1 , z 2 and
z − z3 1 − i 3
z 3 satisfying 1 = are the vertices of a
Applications of Triangle
1. Centroid The centroid of the triangle (in the
Objective Mathematics Vol. 1
z2 − z3 2 argand plane) formed by z1 , z 2 and z 3 is given by
triangle which is 1
(a) of area zero (b) right angled isosceles ( z + z2 + z3 )
3 1
(c) equilateral (d) obtuse angled isosceles
2. Incentre The incentre of the triangle (in the
Sol. (c) z1 − z3 = 1 − i 3
z2
argand plane), formed by z1 , z 2 and z 3 is
z2 − z3 2 az1 + bz 2 + cz 3
(1 − i 3 ) (1 + i 3 ) 1− i23 , where
= = a + b+c
2 (1 + i 3 ) 2 (1 + i 3 )
4 2 a = | z 2 − z 3 |, b = | z 3 − z1 |, c = | z1 − z 2 |
=
2 (1 + i 3 ) (1 + i 3 ) π/3 3. Excentres The excentres of the triangle (in the
z2 − z3 1 + i 3 z3 z1 argand plane), for meet by z1 , z 2 and z 3 are given
⇒ =
z1 − z3 2 by
π π −az1 + bz 2 + cz 3
= cos + i sin
3 3 (i) I 1 =
−a + b + c
z − z3 z − z3 π
⇒ 2 = 1 and arg 2 = az − bz 2 + cz 3
z1 − z3 z1 − z3 3 (ii) I 2 = 1
Hence, the triangle is an equilateral. a −b+c
az + bz 2 − cz 3
Area of Triangle (iii) I 3 = 1
a +b−c
(i) Area of the triangle with vertices z1 , z 2 and z 3 is where, a = | z 2 − z 3 |, b = | z 3 − z1 | and c = | z1 − z 2 |
( z − z 3 ) | z1 | 2
∑ 2 sq unit. 4. Circumcentre The circumcentre of the triangle
4iz1 (in the argand plane), formed by
(ii) The area of triangle whose vertices are z, iz and ∑ z1 z1 ( z 2 − z 3 )
z1 , z 2 , z 3 is given by
1 ∑ z1 ( z 2 − z 3 )
z + iz is | z | 2 .
2 5. Orthocentre The orthocentre of the triangle
(iii) The area of triangle whose vertices are z, ωz and (in the argand plane), formed by z1 , z 2 , z 3 is given
3 by
z + ωz is | z |2.
4 ∑ z12 ( z 2 − z 3 ) + ∑ | z1 | 2 ( z 2 − z 3 )
X Example 57. If the area of the triangle on the ∑ ( z1 z 2 − z1 z 2 )
complex plane formed by the points z, iz and z + iz
is 50 sq units, then | z | is X Example 59. If z1 , z 2 and z 3 are affixes of the
(a) 5 (b) 10 vertices A, B and C, respectively of a ∆ABC having
(c) 15 (d) None of these centroid at G such that z = 0 is the mid-point of AG,
then
Sol. (b) We know that, the area of the triangle formed by the (a) z1 + z 2 + z 3 = 0 (b) z1 + 4 z 2 + z 3 = 0
1 2
points z, iz and z + iz is z .
2 (c) z1 + z 2 + 4 z 3 = 0 (d) 4 z1 + z 2 + z 3 = 0
Sol. (d) The affix of G is z1 + z2 + z3 . Since, z = 0 is the
1
∴ | z|2 = 50
2 3
⇒ | z| = 10 mid-point of AG. Therefore, affix of the mid-point of AG is 0.
z1 + z2 + z3
X Example 58. If the area of the triangle on the + z1
⇒ 3 = 0 ⇒ 4 z1 + z2 + z3 = 0
complex plane formed by complex numbers z, ωz 1+ 1
and z + ωz is 4 3 sq units, then | z | is
(a) 4 (b) 2 (c) 6 (d) 3 X Example 60. The centre of a square ABCD is
at the origin and point A is represented by z1 . The
Sol. (a) The area of the triangle formed by z, ω z and z + ω z
3 2 centroid of ∆BCD is represented by
is | z| . z z
4 (a) 1 (b) − 1
∴
3
| z|2 = 4 3
3 3
4 iz1 iz
128 ⇒ | z| = 4
(c) (d) − 1
3 3
Sol. (b) The affixes of the vertices B, C and D are
iz1 − z1 and − iz1, respectively. Therefore, the affix of the
centroid of ∆BCD is
X Example 62. Prove that the complex numbers
z1 , z 2 and the origin form an equilateral triangle
only, if z12 + z 22 − z1 z 2 = 0.
4
Complex Numbers
iz1 – z1 − iz1 z
=− 1
3 3 Sol. If z1, z2 and z3 form an equilateral triangle, then
D(–iz1) C(–z1) ⇒ z12 + z22 + z32 = z1 z2 + z2 z3 + z3 z1
⇒ z12 + z22 + 02 = z1 z2 + z2 ⋅ 0 + 0 ⋅ z1
⇒ z12 + z22 = z1 z2 ⇒ z12 + z22 − z1 z2 = 0
Hence proved.
O
X Example 63. Let z1 and z 2 be two complex
z z
numbers such that 1 + 2 =1, then
A(z1) B(iz1)
z 2 z1
(a) z1 , z 2 are collinear
X Example 61. Let P (e i θ 1 ), Q (e i θ 2 ) and R (e i θ 3 ) (b) z1 , z 2 and the origin form a right angled triangle
be the vertices of ∆PQR in the argand plane. Then, (c) z1 , z 2 and the origin form an equilateral
the orthocentre of the ∆PQR is triangle
2 (d) None of the above
(a) e i ( θ 1 + θ 2 + θ 3 ) (b) e i ( θ 1 + θ 2 + θ 3 )
3 Sol. (c) We have, z1 + z2 = 1 ⇒ z12 + z2 2 = z1 z2
(c) e i θ 1 + e i θ 2 + e i θ 3
z2 z1
(d) None of these
⇒ z12 + z22 + z32 = z1 z2 + z1 z3 + z2 z3 , where z3 = 0
Sol. (c) We have, |e i θ1| =|e i θ 2| = |e i θ 3| = 1 ⇒ z1, z2 and the origin form an equilateral triangle.
⇒ OP = OQ = OR = 1, where O is the origin.
⇒ Origin O is the circumcentre of ∆PQR. Circle
The affix of the centroid is
1 i θ1
(e + ei θ2 + ei θ3 ) Equation of a Circle
3 (i) The equation of a circle whose centre is at point
Let z be the affix of the orthocentre. Since, centroid having affix z 0 and radius r, is | z − z 0 | = r.
divides the segment joining circumcentre and
orthocentre in the ratio 1 : 2. P(z)
1 1⋅ z + 2 × 0
∴ (e i θ1 + e i θ 2 + e i θ 3 ) =
3 1+ 2 r
i θ1 i θ2 i θ3
⇒ z=e +e +e
C(z0 )
6. Equilateral Triangle
(i) The triangle whose vertices are the points
z1 , z 2 and z 3 on the argand plane, is an
equilateral triangle, if (ii) If the centre of the circle is at origin and radius r,
z12 + z 22 + z 32 = z1 z 2 + z 2 z 3 + z 3 z1 then its equation is | z | = r.
1 1 1 (iii) | z − z 0 | < r represents interior of a circle
or + + =0
z1 − z 2 z 2 − z 3 z 3 − z1 | z − z 0 | = r and | z − z 0 | > r represents exterior of
the circle | z − z 0 | = r.
(ii) If the complex numbers z1 , z 2 and z 3 are the
vertices of an equilateral triangle and z 0 is (iv) General equation of a circle The general
the circumcentre of the triangle, then equation of the circle is zz + az + az + b = 0,
z12 + z 22 + z 32 = 3 z 02 . where a is complex number and b ∈ R .
∴ Centre and radius are −a and | a | 2 − b,
7. Isosceles Triangle
(i) If z1 , z 2 and z 3 are the vertices of a right respectively.
angled isosceles triangle, then (v) Equation of circle in diametric form If end
( z1 − z 2 ) 2 = 2( z1 − z 3 ) ( z 3 − z 2 ) points of diameter represented by A ( z1 ) and B ( z 2 )
and P ( z ) is any point on the circle, then
(ii) If z1 , z 2 and z 3 are the vertices of an ( z − z1 )( z − z 2 ) + ( z − z 2 )( z − z1 ) = 0
isosceles triangle, right angled at z 2 , then
which is required equation of circle in diametric
z12 + z 22 + z 32 = 2 z 2 ( z1 + z 3 ) form.
129
4 X Example 64. A circle whose radius is r and
centre z 0 , then the equation of the circle is
(a) zz − zz 0 − zz 0 + z 0 z 0 = r 2
z − z1
(x) arg = 0 or π
z − z2
Objective Mathematics Vol. 1
⇒ Locus of z is a straight line passing through z1
(b) zz + zz 0 − zz 0 + z 0 z 0 = r 2 and z 2 .
(c) zz − zz 0 + zz 0 − z 0 z 0 = r 2
X Example 66. The complex numbers z = x + iy
(d) None of the above
z − 5i
Sol. (a) Equation of circle| z − z0|2 = r 2 which satisfy the equation = 1, lie on
z + 5i
⇒ ( z − z0 )( z − z0 ) = r 2 ⇒ ( z − z0 )( z − z0 ) = r 2
(a) the X-axis
zz − zz0 − zz0 + z0 z0 = r 2 .
(b) the straight line y = 5
X Example 65. The set of values of k for which (c) a circle passing through the origin
the equation zz + ( −3 + 4i) z − (3 + 4i) z + k = 0 (d) None of the above
represents a circle is
Sol. (a) Given, z − 5 i = 1 ⇒ | z − 5 i| = | z + 5 i|
(a) ( −∞ , 25) (b) (25 , ∞ ) (c) (5, ∞ ) (d) ( −∞, 5) z + 5i
Sol. (a) We have, [if| z − z1| = | z – z2|, then it is a perpendicular
bisector of z1 and z2 ]
zz + (−3 + 4i ) z − (3 + 4i ) z + k = 0
Y
This equation represents a circle with centre
a + (3 − 4i ) and Radius = (−3 − 4 i ) − k = 25 − k
2
(0, 5)
For circle to exist, we must have
25 − k > 0 ⇒ k < 25 X¢
O
X
Hence, the given equation will represent a circle if
k < 25. (0, –5)
Loci in Complex Plane Y¢
If z is a variable point and z1 , z 2 are two fixed ∴ Perpendicular bisector of (0, 5) and (0, − 5) is X-axis.
points in the argand plane, then
(i) | z − z1 | = | z − z 2 | X Example 67. The equation
⇒ Locus of z is the perpendicular bisector of the | z − i | + | z + i | = k, k > 0
line segment joining z1 and z 2 . can represent an ellipse, if k 2 is
(ii) | z − z1 | + | z − z 2 | = k , if | k | > | z1 − z 2 | (a) <1 (b) < 2
⇒ Locus of z is an ellipse. (c) > 4 (d) None of these
(iii) | z − z1 | + | z − z 2 | = | z1 − z 2 | Sol. (c)| z − z1 | + | z – z2 | = k represents ellipse, if
⇒ Locus of z is the line segment joining z1 and z 2 | k | > | z1 − z2|
(iv) | z − z1 | − | z − z 2 | = | z1 − z 2 | Thus,| z − i | + | z − i | = k represents ellipse, if
| k | > |i + i| or | k | > |2 i|
⇒ Locus of z is a straight line joining z1 and z 2
∴ | k | > 2 or k 2 > 4
but z does not lie between z1 and z 2 .
(v) | z − z1 | – | z − z 2 | = k , where k < | z1 − z 2 | X Example 68. The equation | z + i | − | z − i | = k
⇒ Locus of z is a hyperbola.
represents a hyperbola if
(vi) | z − z1 | 2 + | z − z 2 | 2 = | z1 − z 2 | 2 (a) −2 < k < 2 (b) k >2
⇒ Locus of z is a circle with z1 and z 2 as the (c) 0 < k < 2 (d) None of these
extremities of diameter.
Sol. (a)| z − z1| − | z − z2| = k, represents hyperbola,
(vii) | z − z1 | = k | z − z 2 |, ( k ≠ 1) if| k | < |z1 − z2|
⇒ Locus of z is a circle.
Thus, | z + i| − | z − i| = k, represents hyperbola, if
z − z1 | k | < | i + i | or| k | < 2
(viii) arg = α (fixed) ⇒ −2 < k < 2
z − z2
⇒ Locus of z is a segment of circle. X Example 69. If z = 1 − t + i t 2 + t + 2, where t
z − z1
(ix) arg = ± π /2 is a real parameter. The locus of z in the argand
z − z2 plane is
⇒ Locus of z is a circle with z1 and z 2 as the (a) a hyperbola (b) an ellipse
130 vertices of diameter. (c) a straight line (d) None of these
Sol. (a) x + iy = 1 − t + i t 2 + t + 2
⇒ x = 1− t , y = t + t + 2
2
Sol. (c) z − 1 = 1⇒ | z − 1| = | z − i|
⇒
z−i
|( x − 1) + iy| = | x + i ( y − 1)|
4
Complex Numbers
Eliminating t, y2 = t 2 + t + 2
⇒ ( x − 1)2 + y 2 = x2 + ( y − 1)2
= (1 − x)2 + 1 − x + 2
2 ⇒ 2x = 2y
= x − +
3 7 or x− y=0
2 4 which is the equation of a straight line.
2
or y2 − x − =
3 7
2 4
X Example 74. If z = x + iy and
which is a hyperbola.
| z − 2 + i | = | z − 3 − i |, then locus of z is
(a) 2x + 4 y − 5 = 0 (b) 2x − 4 y − 5 = 0
X Example 70. Identify the locus of z, if (c) x + 2 y = 0 (d) x − 2 y + 5 = 0
r2 Sol. (a) | z − 2 + i| = | z − 3 − i|
z=a + , > 0.
z−a ⇒ | ( x − 2 ) + i ( y + 1)| = | ( x − 3) + i ( y − 1)|
2 2
Sol. z = a + r ⇒ z−a=
r ⇒ ( x − 2 )2 + ( y + 1)2 = ( x − 3)2 + ( y − 1)2
z−a z−a
⇒ x 2 + 4 − 4 x + y2 + 1 + 2 y = x 2 + 9 − 6 x + y2 + 1 − 2 y
⇒ ( z − a) ( z − a ) = r 2 ⇒ 2 x + 4y − 5 = 0
⇒ | z − a|2 = r 2
⇒ | z − a| = r X Example 75. If z = z 0 + A ( z − z 0 ), where A is
Hence, locus of z is circle having centre a and radius r. a constant, then prove that locus of z is a straight
line.
X Example 71. If the equation
| z − z1 | 2 + | z − z 2 | 2 = k represents the equation of a Sol. z = z0 + A ( z − z0 )
⇒ Az − z − Az0 + z0 = 0 …(i)
circle, where z1 = 2 + 3 i, z 2 = 4 + 3 i are the extremities
⇒ A z − z − A z0 + z0 = 0 …(ii)
of a diameter, then the value of k is
Adding Eqs. (i) and (ii), we get
1
(a) (b) 4 ( A − 1) z + ( A − 1) z − ( Az0 + Az0 ) + z0 + z0 = 0
4
This is of the form
(c) 2 (d) None of these
az + az + b = 0, where
Sol. (b) As z1 and z2 are the extremities of diameter. a = A − 1 and b = − ( Az0 + Az0 ) + z0 + z0 ∈R
⇒ | z − z1|2 + | z − z2|2 = | z1 − z2|2 Hence, locus of z is a straight line.
⇒ k = | z1 − z2|2 = |2 + 3i − 4 − 3i|2 = | − 2 |2 = 4
X Example 76. Plot the region represented by
Example 72. If | z + 1| = 2 | z – 1,| then the π z + 1 2π
X
≤ arg ≤ in the argand plane.
locus described by the point z in the argand 3 z −1 3
diagram is a Sol. Let us take
(a) straight line (b) circle z + 1 2 π
arg = , clearly z
(c) parabola (d) None of these z − 1 3
Sol. (b)| z + 1| = 2 | z − 1| lies on the minor arc of the (–1, 0) (1, 0)
2π/3
circle passing through (1, 0)
Putting z = x + iy ⇒ x + iy + 1 and (−1 , 0).
= 2| x + iy − 1| π/3
z + 1 π
⇒ | ( x + 1) + iy| = 2 |( x − 1) + iy| Similarly, arg =
z − 1 3
⇒ ( x + 1)2 + y2 = 2 [( x − 1)2 + y2 ]
means that z is lying on the
⇒ x 2 + y2 − 6 x + 1 = 0 major arc of the circle passing through (1, 0) and (−1, 0).
which is the equation of a circle. Now, if we take any point in the region included
between the two arcs, say P1( z1 ), we get
z −1 π z + 1 2 π
X Example 73. The locus of z given by =1 ≤ arg ≤
z−i 3 z − 1 3
is π z + 1 2 π
Thus, ≤ arg ≤ represents the shaded region
(a) a circle (b) an ellipse 3 z − 1 3
(c) a straight line (d) a parabola excluding the points (1, 0) and (−1, 0).
131
4 Some Important Results
i. Four points z1 , z 2 , z 3 and z 4 in anti-clockwise
⇒
⇒
z − z1
arg 2 = π − arg
z3 − z1
α = π −β ⇒ α + β = π
z3
z2
Objective Mathematics Vol. 1
order will be concyclic, if and only if Hence, the points are concyclic.
z − z4 z2 − z3
θ = arg 2 = arg 1
z1 − z 4 z1 − z 3 ii. If z + = a, then the greatest and least
z
z − z4 z2 − z3 a + a2 + 4
⇒ arg 2 −arg = 2nπ , n ∈ I values of z are respectively
1
z − z 4 z1 − z 3 2
z − z 4 z1 − z 3 −a + a + 4 2
⇒ arg 2 = 2nπ and
z1 − z 4 z 2 − z 3 2
z − z 4 z1 − z 3 4
⇒ 2 is real and positive. X Example 79. If z + = a, then find the
z1 − z 4 z 2 − z 3 z
greatest and least values of | z | .
Ø Points z1, z 2, z3 and z4 (not necessarily in order) will be concyclic,
z 2 a z
z − z z − z Sol. + = . Now, put = w
if 2 4 1 3 is positive or negative. 2 z 2 2
z1 − z4 z 2 − z3 1 a
∴ w+ =
w 2
X Example 77. Show that the points 3 + 4i, 3 − 4i,
a a2
−4 + 3i, −4 − 3i are concyclic. + + 4
Now, greatest value of|w| = 2 4 and| z| = 2|w|
Sol. Let z1 = 3 + 4i , z2 = 3 − 4i , z3 = − 4 + 3i 2
z4 = − 4 − 3i a a2
and Greatest value of| z| = + + 4
Then, 2 4
z2 − z4 z1 − z3 (7 − i )(7 + i ) Similarly, we can find least value of
= a a2
1
z − z4 2z − z3 (7 + 7 i )(7 − 7 i ) | z| = − + + 4
50 25 2 4
= =
49 + 49 49
X Example 80. Let z1 and z 2 be two non-real
= a real number
Hence, the given points are concyclic. complex cube roots of unity and
2 2
z − z1 + z − z 2 = λ be the equation of a circle
X Example 78. If z1 , z 2 and z 3 are complex
with z1 , z 2 as ends of a diameter, then the value of
2 1 1 λ is
numbers, such that = + , then show that
z1 z 2 z 3 (a) 4 (b) 3 (c) 2 (d) 2
the points represented by z1 , z 2 and z 3 lie on a 2
circle passing through the origin. Sol. (b) z − ω 2 + z − ω2 − λ could be shows as given
z
Sol. below : z–ω z – ω2
R (z2) 2
⇒λ = ω−ω 2
S(0)
β − 1+ i 3 − 1− i 3
2
= −
2 2
O
α 2
ω – ω2
− 1+ i 3 + 1+ i 3
Q(z1) =
2
P(z3)
= |i 3|2 = 3
2 1 1 1 1 1 1
= + ⇒ − = −
z1 z2
z2 − z1 z1 − z3
z3 z1 z2
z2 − z1
z3
z2
z1
Logarithm of Complex Numbers
⇒ = ⇒ =−
z1 z2 z3 z1 z3 − z1 z3 Let z = α + iβ = re i ( θ + 2nπ )
⇒
z − z1
arg 2
z
= arg − 2 log z = log( re i ( θ + 2nπ ) ) = log r + i(θ + 2nπ )
3
z − z1 z3 = log | z | + i arg z + 2nπi
⇒
z − z1
arg 2
z
= π + arg 2
If we put n = 0, we get principal value of log z.
132 z3 − z1 z3 ∴ Principal value of log z = log| z | + i arg z.
4
− iz
Sol. (a) We know that, cos z = e + e
iz
X Example 81. The value of i i is
2
(a) e −π / 2 (b) e π / 2 2
log e ( 2 − 2
ei 3)
+ e −i log e ( 2 − 3)
(c) e π / 4
Complex Numbers
(d) None of these =
2
iπ / 2
Sol. (a) Let z = i , loge z = loge i = i loge i = i loge e
i i
elog e ( 2 − 3 ) −1
+ elog e ( 2 − 3 )
π π =
= i 2 loge e = − 2
2 2
(2 − 3 )−1 + (2 − 3 )
⇒ z = e −π / 2 =
2
1 1 1 2 + 3
Example 82. If z = i log e (2 − 3 ), then the = + 2 − 3 = + (2 − 3 )
2 2
X
− 3 2 4− 3
value of cos z is 1
= [2 + 3+2− 3]= 2
(a) 2 (b) −2 (c) 2i (d) −2i 2
Work Book Exercise 4.4
z −1 9 The figure formed by four points 1 + 0 i, −1 + 0i,
1 = 1, represents
z+1 25
3 + 4i and on the argand plane is
a a circle b an ellipse −3 − 4i
c a straight line d None of these a a parallelogram but not a rectangle
b a trapezium which is not equilateral
2 | z − 4| <| z − 2|, represents the region given by c a cyclic quadrilateral
a Re ( z) > 0 b Re ( z) < 0 d None of the above
c Re ( z) > 2 d None of these
10 If i = −1 , then define a sequence of complex
3 If 2 z1 − 3 z2 + z3 = 0, then z1, z2 , z3 are number by z1 = 0, zn + 1 = zn 2 + i for n ≥ 1. In the
represented by complex plane, how far from the origin is z111 ?
a three vertices of a triangle a 1 b 2 c 3 d 110
b three collinear points
c three vertices of a rhombus
11 The complex numbers whose real and imaginary
d None of the above parts are integers and satisfy the relation
zz 3 + z 3 z = 350, forms a rectangle on the argand
4 If z = x + iy, such that| z + 1| =| z − 1| and plane, the length of whose diagonal is
z − 1 π a 5 units b 10 units c 15 units d 25 units
amp = , then
z + 1 4 π
12 The locus of z, for arg z = − is
a x= 2 + 1, y = 0 3
x= 0, y = 2 + 1 2π
b a same as the locus of z for arg z =
c x= 0, y = 2 − 1 3
π
d x= 2 − 1, y = 0 b same as the locus of z for arg z =
3
5 If z 8 = ( z − 1)8 , then the roots of this equation are c the part of the straight line 3 x + y = 0 with y < 0,
x> 0
a collinear d the part of the straight line 3 x + y = 0 with y > 0, x < 0
b concyclic
c the vertices of irregular polygon 13 If z1 and z2 are two complex numbers and
d None of the above z1 + z2 π
arg = but| z1 + z2| ≠ | z1 − z2|, then the
6 The equation zz + az + az + b = 0, b ∈ R z1 − z2 2
represents a circle, if figure formed by the points represented by 0, z1,
a | a|2 = b b | a|2 ≥ b
z2 and z1 + z2 is
a a parallelogram but not a rectangle or a rhombus
c | a|2 < b d None of these
b a rectangle but not a square
c a rhombus but not a square
7 If z = (λ + 3) + i 3 − λ2 , where| λ| < 3, then
d a square
locus of z is
a circle b parabola
14 z1 and z2 are two distinct points in an argand
c line d None of these plane. If a| z1| = b| z2|, where a, b ∈ R, then the
az bz
8 If a point P denoting the complex number z point 1 + 2 is a point on the
bz2 az1
moves on the complex plane such that
|Re ( z )| + |Im ( z )| = 1, then the locus of z is a line segment [−2, 2] of the real axis
a a square b a circle b line segment [−2, 2] of the imaginary axis
c two intersecting lines d a line c unit circle| z| = 1
d the line with arg z = tan−1 2 133
4 15 The roots z1, z2 , z3 of the equation
z 3 + 3 αz 2 + 3 βz + γ = 0 correspond to the points
A, B and C on the complex plane. Then, the
23 The straight line (1 + 2 i )z + (2 i − 1)z = 10 i on the
complex plane, has intercept on the imaginary
axis equal to
Objective Mathematics Vol. 1
5 5
triangle is equilateral, if a 5 b c − d −5
2 2
a α =β 2
b α =β 2
c α 2 = 3β 2 d 3α 2 = β 2 24 If the complex number z satisfies the condition
1
| z| ≥ 3, then the least value of z + is equal to
16 If +
z12 + 2 z1 z2 cos θ = 0, then the points
z22 z
represents by z1, z2 and the origin form 5 8
a b
a equilateral triangle b right angled triangle 3 3
c isosceles triangle d None of these 11
c d None of these
3
17 Complex numbers z1, z2 , z3 are the vertices A, B
and C, respectively of an isosceles right angled a b
25. If z1 = , z2 = , z3 = a − bi , for a, b ∈ R and
triangle, right angled at C. Then, ( z1 − z2 )2 is 1− i 2+i
equal to z1 − z2 = 1, then the centroid of the triangle
1
a 2( z1 − z3 )( z3 − z2 ) b ( z1 + z3 )( z3 − z1 ) formed by the points z1, z2 , z3 in the argand
2 plane is given by
c ( z1 − z3 )( z3 − z2 ) d None of these 1 1 1 1
a (1 + 7 i ) b (1 + 7 i ) c (1 − 3i ) d (1 − 3i )
18 A, B and C are points represented by complex 9 3 3 9
numbers z1, z2 and z3 . If the circumcentre of the 26. Let λ ∈ R. If the origin and the non-real roots of
∆ABC is at the origin and the altitude AD of the
2 z 2 + 2 z + λ = 0 form three vertices of an
triangle meets the circumcircle again at P, then
the complex number representing point P is equilateral triangle in the argand plane, then λ is
z1 z2 z3 z2 a 1 b 2 /3
a z= b z=− c 2 d −1
z3 z1
zz z3 z2 27 If a, b, c and u, v, w are complex numbers
c z=− 1 2 d z=
z3 z1 representing the vertices of two triangles, such
that c = (1 − r ) a + rb and w = (1 − r ) u + r v . When r
19 If z1 and z2 are the roots of the equation
is a complex number, then the two triangles
z 2 + pz + q = 0, where the coefficients p and q
a have the same area b are similar
may be complex number. Let A and B represent
c are congruent d None of these
z1 and z2 in the complex plane. If ∠AOB = α ≠ 0
and OA = OB, where O is the origin, then p2 is 28. On the complex Y
equal to plane, ∆OAP and R(z)
α
a 4 q cos 2
b 2q cos α 2 ∆OQR are similar
2 and I(OA) = 1. If the
α
c q cos 2 d None of these points P and Q
4
denote the P(z1)
z + 4 1 complex numbers Q(z2)
20 If Re = , then z is represented by a
2z − i 2 z1 and z2 , then the θ
complex numbers θ
point lying on X
O A (1, 0)
a a circle b an ellipse z denoted by the
c a straight line d None of these point R is given by
z1
21 Suppose z1, z2 and z3 are the vertices of an a z1 z2 b
z2
equilateral triangle inscribed in the circle| z| = 2.
z2 z1 + z2
If z1 = 1 + 3 i and z1, z2 and z3 are in the c d
z1 z2
clockwise sense, then
a z2 = 1 − 3i , z3 = 2 29 Intercept made by the circle zz + αz + αz + r = 0
b z2 = 2 , z3 = 1 − 3i
c z2 = − 1 + 3i , z3 = − 2 on the real axis on complex plane, is
d None of the above a (α + α ) − r b (α + α )2 − 2 r
22 Suppose z1, z2 and z3 are the vertices of an c (α + α )2 + r d (α + α )2 − 4r
equilateral triangle circumscribing in the circle
| z| = 2. If z1 = 1 + 3 i and z1, z2 and z3 are in the 30 The equation of the radical axis of the two circles
anti-clockwise sense, then z2 is represented by the equations,| z − 2| = 3 and
a 1− 3i | z − 2 − 3 i | = 4 on the complex plane is
b 2 a 3y + 1 = 0
c
1
(1 − 3 i ) b 3y − 1 = 0
2 c 2y − 1= 0
d None of the above d None of the above
134
WorkedOut Examples
Type 1. Only One Correct Option
Ex 1. If a = cos α + i sin α, b = cos β + i sin β, α α
tan α − i sin + cos
a b c 2 2
c = cos γ + i sin γ and + + =1, then Ex 4. If is purely
b c a α
1 + 2i sin
cos (α − β) + cos (β − γ ) + cos ( γ − α ) is equal 2
to imaginary, then α is given by
3
(b) −
3 π π
(a) (a) nπ + (b) nπ −
2 2 4 4
(c) 0 (d) 1 π
(c) ( 2n + 1) π (d) 2nπ +
a b c 4
Sol. + + =1
b c a α α
cis α cis β cis γ tan α − i sin + cos
⇒ + + = 1, Sol. Re
2 2
=0
cis β cis γ cis α α
1 + 2i sin
where, cis θ represents cos θ + i sin θ.
2
⇒ cis (α − β ) + cis(β − γ ) + cis( γ − α ) = 1
α α α
Equation real parts of both sides tan α − i sin + cos 1 − 2i sin
2
⇒ Re
2 2
cos (α + β ) + cos (β − γ ) + cos (γ − α ) = 1 =0
α
Hence, (d) is the correct answer. 1 + 4 sin 2
2
Ex 2. The locus of the centre of a circle which α α α
touches the circles | z − z1 | = a and | z − z 2 | = b tan α − 2 sin 2 cos 2 + sin 2
α α α
Re − i tan α ⋅ 2 sin + sin + cos = 0
externally (z, z1 and z 2 are complex numbers) ⇒
will be − 2 2 2
(a) an ellipse (b) a hyperbola α
1 + 4 sin 2
(c) a circle (d) None of these 2
2α
Sol. Let A (z1 ), B (z2 ) be the centres of given circles and P be ⇒ tan α = 2 sin + sin α
the centre of the variable circle which touches given 2
sin α
circles externally, then ⇒ = sin α + 1 − cosα
| AP | = a + r and | BP | = b + r, cosα
⇒ sin α = sin α ⋅ cosα + cosα − cos2 α
where r is the radius of the variable circle.
⇒ sin α (1 − cosα ) = cosα (1 − cosα )
On subtraction, we get
⇒ sin α = cosα , cosα = 1
| AP | − | BP | = a − b
π
⇒ || AP | − | BP || = | a − b | , a constant. ⇒ α = nπ +
4
Hence, locus of P is
or α = 2nπ, n ∈ I
(i) right bisector AB, if a = b.
Hence, (a) is the correct answer.
(ii) a hyperbola, if | a − b | < | AB | = | z2 − z1 |.
(iii) an empty set, if | a − b | > | AB | = | z2 − z1 |. a − ib
(iv) set of all points on line AB except those which lie Ex 5. The expression tan i log reduces to
between A and B, if | a − b | = | AB | ≠ 0. a + ib
Hence, (d) is the correct answer. ab 2ab
(a) (b)
a +b
2 2
a − b2
2
Ex 3. If arg ( z1 ) = arg ( z 2 ), then
ab 2ab
(a) z 2 = kz1−1 , k > 0 (b) z 2 = kz1 , k > 0 (c) (d)
a2 − b2 a2 + b2
(c) | z 2 | = | z1 | (d) None of these
a − ib
z1z1 Sol. Given, tan i log
Sol. z1 = = | z1 |2 z1−1 a + ib
z1
⇒ arg (z1−1 ) = arg (z1 ) = arg (z2 ) Let a + ib = reiθ , r2 = a2 + b2
⇒ z2 = kz1−1 (k > 0) ⇒ a − ib = re− iθ , tan θ =
b
Hence, (a) is the correct answer. a 135
4 ⇒
a − ib
a + ib
a − ib
= e−2 i θ
−2iθ
Therefore,
⇒
z2 = r1[cos (− θ 1 ) + i sin (− θ 1 )]
= r1 (cos θ 1 − i sin θ 1 ) = z1
z2 = (z1 ) = z1 ⇒ | z2 |2 = z1z2
Objective Mathematics Vol. 1
i log = i log(e ) = 2 θ
a + ib Hence, (b) is the correct answer.
a − ib
⇒ tan i log = tan 2θ Ex 8. If z is a complex number, then z 2 + z 2 = 2
a + ib
2 tan θ
represents
= (a) a circle (b) a straight line
1 − tan 2 θ
(c) a hyperbola (d) an ellipse
2b / a 2ab
= =
b2 a2 − b2 Sol. Let z = x + iy, then
1− 2 z2 + z 2 = 2
a
Hence, (b) is the correct answer. ⇒ (x + iy)2 + (x − iy)2 = 2
⇒ x 2 − y2 = 1
Ex 6. If z1 = a + ib and z 2 = c + id are complex
which represents a hyperbola.
numbers such that | z1 | = | z 2 | = 1 and Hence, (c) is the correct answer.
Re ( z1 z 2 ) = 0, then the pair of complex
numbers a + ic = w1 and b + id = w2 satisfies 1 − iα
Ex 9. If = A + iB , then A 2 + B 2 equals
(a) | w1 | ≠ 1 (b) | w 2 | ≠ 1 1 + iα
(c) Re ( w1 w 2 ) = 0 (d) None of these (a) 1 (b) α 2 (c) −1 (d) − α 2
Sol. z1 = a + ib, z2 = c + id 1 − iα 1 + iα
| z1 | = | z2 | = 1 Sol. A + iB = ⇒ A − iB =
1 + iα 1 − iα
⇒ a2 + b2 = c2 + d 2 = 1 (1 − iα )(1 + iα )
w1 = a + ic, w2 = b + id ⇒ ( A + iB ) ( A − iB ) = =1
(1 + iα )(1 − iα )
z1z2 = (a + ib) (c − id )
⇒ A2 + B2 = 1
= (ac + bd ) + i (bc − ad )
As Re (z1z2 ) = 0 Hence, (a) is the correct answer.
⇒ ac + bd = 0 ⇒ ac = − bd
w1w2 = (a + ic) (b − id )
Ex 10. If | z1 | = | z 2 | and arg ( z1 ) + arg ( z 2 ) = π / 2, then
= (ab + cd ) + i (bc − ad ) (a) z1 z 2 is purely real
We have, a2 + b2 = c2 + d 2 (b) z1 z 2 is purely imaginary
⇒ a2 − c2 = d 2 − b2 (c) ( z1 + z 2 ) 2 is purely real
π
⇒ a2 − c2 + 2i ac = d 2 − b2 − 2 ibd [Q as ac = − bd ] (d) arg ( z1−1 ) + arg ( z 2−1 ) =
2
⇒ (a + ic)2 = (d − ib)2
⇒ a + ic = (d − ib) or − d + ib Sol. Let | z1 | = | z2 | = r
⇒ a = d and c = − b or a = − d , b = c ⇒ z1 = r (cosθ + i sin θ )
⇒ c2 + d 2 = b2 + d 2 π π
and z2 = r cos − θ + i sin − θ
2 2
a2 + c2 = a2 + b2
⇒ a2 + c2 = 1, b2 + d 2 = 1 ⇒ z1z2 = r2i, which is purely imaginary.
⇒ | w1 | = | w2 | = 1 z1 + z2 = r [(cosθ + sin θ ) + i (cosθ + sin θ )]
Also, ab + cd = − cd + cd = 0 ⇒ (z1 + z2 )2 = 2r2 ⋅ (cosθ + sin θ )2 ⋅ i
⇒ Re (w1w2 ) = 0 which is purely imaginary.
Hence, (c) is the correct answer. π
Also, arg (z1−1 ) + arg (z2−1 ) = −
2
z1
Ex 7. If = 1 and arg ( z1 z 2 ) = 0, then Hence, (b) is the correct answer.
z2
(a) z1 = z 2 (b) | z 2 |2 = z1 z 2 Ex 11. The value of the expression
(c) z1 z 2 = 1 1 1 1 1
(d) None of these 2 1 + 1 + 2 + 3 2 + 2 + 2
ω ω ω ω
z1
Sol. Let z1 = r1 (cos θ 1 + i sin θ 1 ) , then =1 1 1
z2 + 4 3 + 3 + 2 +… + ( n + 1)
⇒ | z1 | = | z2 | ω ω
⇒ | z1 | = | z2 | = r1 1 1
arg (z1 , z2 ) = 0 n + n + 2 ,
Now, ω ω
⇒ arg(z1 ) + arg (z2 ) = 0
136 ⇒ arg (z2 ) = − θ 1 where ω is an imaginary cube root of unity, is
(a)
n( n 2 + 2)
3
n 2 ( n + 1) 2 + 4n
(b)
n( n 2 − 2)
3
Sol. If z1 is the new complex number, then
| z1 | = | z| + 2 = 2 2
z |z |
Also, 1 = 1 ⋅ ei 3π / 2
4
Complex Numbers
(c) (d) None of these z | z|
4
3π 3π
1 1 1 1 ⇒ z1 = z ⋅ 2 cos + i sin
Sol. tn = (n + 1) n + n + 2 n3 + n2 2 + + 1 2 2
ω ω ω ω
= 2(1 + i )(0 − i ) = − 2i + 2 = 2(1 − i )
1 1
+ n 1 + 2 + + 1 Hence, (d) is the correct answer.
ω ω
= n3 + n2 (ω + ω 2 + 1) + n(ω + ω 2 + 1) + 1 = n3 + 1 1 1
Ex 15. If 2 cos θ = x + and 2 cos φ = y + , then
n n
n2 (n + 1)2 x y
∴ Sn = ∑ tr = ∑ (r3 + 1) = +n x y
r=1 r=1
4 (a) + = 2cos (θ + φ )
y x
Hence, (c) is the correct answer.
1
(b) x m y n + m n = 2cos ( mθ + nφ )
Ex 12. If z1 and z 2 are two complex numbers x y
z + iz 2 z xm yn
satisfying the equation 1 = 1, then 1 (c) += 2cos ( mθ + nφ )
z1 − iz 2 z2 yn xm
is 1
(d) xy + = 2cos (θ − φ )
(a) purely real xy
(b) of unit modulus
1 1
(c) purely imaginary Sol. 2 cosθ = x + , 2 cos φ = y +
x y
(d) None of the above
⇒ x 2 − 2x cos θ + 1 = 0
Sol. (z1 + iz2 )(z1 − iz2 ) = (z1 − iz2 )(z1 + iz2 )
⇒ z1z2 = z1z2 2 cos θ ± 4 cos2 θ − 4
⇒ x=
z1 z1 2
⇒ =
z2 z2 ⇒ x = cosθ ± i sin θ = e± iθ
z
⇒ 1 is purely real. Similarly, y = e± iφ
z2 x y
⇒ + = 2cos (θ − φ )
Hence, (a) is the correct answer. y x
1
Ex 13. If z = − 2 + 2 3i, then z 2n + 2 2n z n + 2 4n may x m yn + m n = 2cos (mθ + nφ )
x y
be equal to xm yn
(a) 22n n
+ m = 2cos (mθ − nφ )
y x
(b) 0 1
(c) 3⋅ 4 2n , n is multiple of 3 xy + = 2cos (θ + φ )
xy
(d) None of the above
Hence, (b) is the correct answer.
Sol. z = − 2 + 2 3i = 4ω
∴z2n + 22n zn + 24n = 4 2nω 2n + 22n ⋅ 4 n ⋅ ω n + 24n Ex 16. The complex numbers z1 and z 2 are such that
= 4 2n[ω 2n + ω n + 1] z1 ≠ z 2 and | z1 | = | z 2 |. If z1 has positive real
part and z 2 has negative imaginary part, then
= 0, if n is not a multiple of 3.
z1 + z 2
= 3 ⋅ 4 2n, if n is a multiple of 3. may be
Hence, (c) is the correct answer.
z1 − z 2
(a) zero
Ex 14. The complex number z = 1 + i is rotated (b) real and positive
through an angle 3π / 2 in anti-clockwise (c) real and negative
direction about the origin and stretched by (d) purely imaginary
additional 2 units, then the new complex Sol. Given, | z1 | = | z2 |
number is Re (z1 ) > 0, Im (z2 ) < 0
(a) − 2 − 2 i z + z2 1 z1 + z2 z1 + z2
Re 1 = +
(b) 2 − 2 i z1 − z2 2 z1 − z2 z1 − z2
(c) 2 − 2 i 1 (z1 + z2 )(z1 − z2 ) + (z1 + z2 )(z1 − z2 )
=
(d) None of the above 2 (z1 − z2 )(z1 − z2 ) 137
4 z1z1 − z1z2 + z2z1 − z2z2 + z1z1 + z1z2
1 − z z − z2z2
= 21
2 | z1 − z2 |2
Sol. Here, z1 (z12 − 3z22 ) = 2
z2 (3z12 − z22 ) = 11
…(i)
…(ii)
Objective Mathematics Vol. 1
On multiplying Eq. (ii) by i and adding it to Eq. (i),
we get
1 2 | z1 |2 − | z2 |2 z13 − 3z22z1 + i (3z12z2 − z23 ) = 2 + 11i
= =0
2 | z1 − z2 |2 ⇒ (z1 + iz2 ) = 2 + 11i …(iii)
z + z2 Again multiplying Eq. (ii) by i and subtracting it from
⇒ 1 is purely imaginary. Eq. (i), we get
z1 − z2
z13 − 3z22z1 − i (3z12z2 − z23 ) = 2 − 11i
Hence, (d) is the correct answer. ⇒ (z1 − iz2 )3 = 2 + 11i …(iv)
z z − z2
Ex 17. If 1 = k , ( z1 , z 2 ≠ 0), then On multiplying Eqs. (iii) and (iv), we get
z1 z + z 2 (z12 + z22 )3 = 4 + 121 ⇒ z12 + z22 = 5
(a) for k ≠ 1, locus z is a straight line Hence, (a) is the correct answer.
(b) for k ∉{1, 0}, z lies on a circle Ex 20. 1 − c 2 = nc − 1 and z = e iθ , then
(c) for k ≠ 0, z represents a point c n
(d) for k ≠ 1, z lies on the perpendicular bisector (1 + nz ) 1 + is equal to
z z 2n z
of the line segment joining 2 and − 2
z1 z1 (a) 1 − ccos θ (b) 1 + 2ccos θ
(c) 1 + ccos θ (d) 1 − 2ccos θ
z − z2
z1z − z2 z1 Sol. Here, 1 − c2 = nc − 1
= k ⇒
Sol. Given, = k ⇒ 1 − c2 = n2c2 − 2nc + 1
z1z + z2 z + z2
z1 c 1
∴ = …(i)
Clearly, if k ≠ 0, 1 then z would lie on a circle. 2n 1 + n2
If k = 1, z would lie on a perpendicular bisector of the c n 1 1
or (1 + nz) 1 + = 2
1 + n2 + n z +
z z 2n z 1+ n z
line segment joining 2 and − 2 .
z1 z1 1
= {1 + n2 + n ⋅ (2 cosθ )}
If k = 0, z represents a point. 1 + n2
Hence, (b) is the correct answer. (1 + n2 ) + 2n cos θ
=
1 + n2
Ex 18. If A ( z1 ), B ( z 2 ) and C ( z 3 ) are the vertices of a
2n
π AB =1+ cos θ
∆ABC in which ∠ABC = and = 2, then 1 + n2
4 BC
the value of z 2 is equal to [using Eq. (i)]
= 1 + ccos θ
(a) z 3 + i( z1 + z 3 )
Hence, (c) is the correct answer.
(b) z 3 − i( z1 − z 3 )
(c) z 3 + i( z1 − z 3 ) Ex 21. Consider an ellipse having its foci at A ( z1 ) and
(d) None of the above B ( z 2 ) in the argand plane. If the eccentricity of
AB
the ellipse is e and it is known that origin is an
Sol. Given, = 2 interior point of the ellipse, then
BC
| z + z2 |
Considering the rotation about B, we get (a) e ∈ 0, 1
z1 − z2 | z1 − z2 | iπ / 4 AB iπ / 4 | z1 | + | z 2 |
= ⋅e = ⋅e
z3 − z2 | z3 − z2 | BC | z − z2 |
(b) e ∈ 0, 1
1 i | z1 | + | z 2 |
= 2 + =1+ i
2 2 | z + z2 |
⇒ z1 − z2 = (1 + i )(z3 − z2 ) (c) e ∈ 0, 1
| z1 | − | z 2 |
⇒ z1 − (1 + i ) z3 = z2 (1 − 1 − i )
⇒ iz2 = − z1 + (1 + i )z3 (d) Can’t be determined
⇒ z2 = iz1 − i (1 + i )z3 Sol. If P (z) is any point on the ellipse. Then, equation of the
∴ z2 = z3 + i (z1 − z3 ) ellipse is
Hence, (c) is the correct answer. | z1 − z2 |
| z − z1 | + | z − z2 | = …(i)
e
Ex 19. If z1 z 2 ∈C , z12 + z 22 ∈ R , z1 ( z12 − 3 z 22 ) = 2 and If we replace z by z1or z2, Eq. (i) becomes | z1 − z2 |.
For P (z) to lie on ellipse, we have
z 2 (3 z12 − z 22 ) = 11, then the value of z12 + z 22 is | z − z2 |
138 | z − z1 | + | z − z2 | < 1
(a) 5 (b) 6 (c) 10 (d) 12 e
It is given that origin is an interior point of the ellipse.
⇒
| z − z2 |
|0 − z1 | + |0 − z2 | < 1
e
By the given conditions, the area of ∆ABC is
1
2
| z1 − z2 | r 4
Complex Numbers
| z1 − z2 | Hence, (b) is the correct answer.
⇒ e ∈ 0,
| z1 | + | z2 |
Ex 25. Locus of z, if
Hence, (b) is the correct answer.
π 3π
Ex 22. If | z − 2 − i | = | z | sin − arg ( z ) , , when | z | ≤ | z − 2 |
then 4
4 arg[ z − (1 + i)] = , is
−π
locus of z is , when | z | > | z − 2 |
(a) a pair of straight lines 4
(b) a circle (a) straight line passing through (2, 0)
(c) a parabola (b) straight line passing through (2, 0), (1, 1)
(d) an ellipse (c) a line segment
(d) a set of two rays
Sol. We have, |(x − 2) + i ( y − 1)| = | z| sin θ
1 1
cosθ −
2 2 Sol. The given equation is written as
where, θ = arg (z) 3π
, when x ≤ 2
1 4
(x − 2) + ( y − 1) =
2
|x − y |
2
arg [ z − (1 + i )] =
2 − π , when x > 2
which is a parabola. 4
Hence, (c) is the correct answer. The locus is a set of two rays.
Ex 23. α 1 , α 2 , α 3 , …, α 100 are all the 100th roots of (0, 2)
unity. The numerical value of
∑ ∑ (α i α j ) is 5
(1, 1)
1 ≤ i < j ≤ 100
(a) 20 (b) 0
(c) ( 20)1/ 20 (d) None of these
(2, 0)
Sol. ∑∑ α 5i α 5j = (α 51 + α 52 + α 53 + α 54 + α 55 + …) 2
1 ≤ i < j ≤ 100
Re (z) = 1
− (α 10
1 + α 2 + α 3 + α 4 + α 5 + …)
10 10 10 10
=0−0=0
100, if r = 100k Hence, (d) is the correct answer.
Because (α 1r + α 2r + … + α 100
r
)=
0 , if r ≠ 100k π
Hence, (b) is the correct answer.
Ex 26. If A = z | arg ( z ) = and
4
Ex 24. The maximum area of the triangle formed by 2π
B = z | arg ( z − 3 − 3i) = , z ∈C . Then,
the complex coordinates z, z1 and z 2 , which 3
satisfy the relations | z − z1 | = | z − z 2 |, n( A ∩ B ) is equal to
z1 + z 2 (a) 1
z − ≤ r, where r >| z1 − z 2 |, is
2 (b) 2
(c) 3
1 1
(a) | z1 − z 2 | 2 (b) | z1 − z 2 | r (d) 0
2 2
1 1 Sol. We can observe that, 3 + 3i ∈ A but ∉ B.
(c) | z1 − z 2 |2 r 2 (d) | z1 − z 2 | r 2 B A
2 2 Y
Sol. A(z)
Y
(3, 3)
B(z1) r X
z 1 +z 2 O
2 C(z2)
O X
∴ n( A ∩ B ) = 0
Hence, (d) is the correct answer.
139
4 Ex 27. Dividing f ( z ) by z − i, we obtain the remainder
i and dividing it by z + i, we get the remainder
1 + i. The remainder upon the division of f ( z )
Sol. f (z) = g (z) (z − i )(z + i ) + az + b; where a, b ∈ C
f (i ) = i ⇒ ai + b = i
f (− i ) = 1 + i ⇒ a(− i ) + b = 1 + i
…(i)
…(ii)
Objective Mathematics Vol. 1
From Eqs. (i) and (ii), we get
by z 2 + 1, is i 1
1 1 a= ,b= + i
(a) ( z + 1) + i (b) ( iz + 1) + i 2 2
2 2 1 1
1 1 Hence, required remainder = az + b = iz + + i
(c) ( iz − 1) + i (d) ( z + i ) + 1 2 2
2 2 Hence, (b) is the correct answer.
Type 2. More than One Correct Option
z + z2 + z3
Ex 28. If z1 = a1 + ib1 and z 2 = a 2 + ib2 are complex (b) 1 = 1 ⇒ | z1 + z2 + z3 |2 = 9
numbers such that | z1 | = 1, | z 2 | = 2 and 3
Re ( z1 z 2 ) = 0, then the pair of complex ⇒ (z1 + z2 + z3 ) (z1 + z2 + z3 ) = 9
ia 4 1 1 4 1 1
numbers ω1 = a1 + 2 and ω 2 = 2b1 + ib2 , ⇒ + + + + =9
2 z1 z2 z3 z1 z2 z3
satisfy (c) ∠QOR = 120°
(a) |ω 1 | = 1 (b) |ω 2 | = 2 Hence, (a), (b), (c) and (d) are the correct answers.
(c) Re (ω 1ω 2 ) = 0 (d) Im (ω 1ω 2 ) = 0
Ex 30. One vertex of the triangle of maximum area
Sol. a12 + b12 = 1, a22 + b22 = 4 and a1a2 = b1b2 that can be inscribed in the curve | z − 2i | = 2, is
a22 + b22 = 4 a12 + 4 b12 2 + 2i, remaining vertices is/are
(a2 + 2ia1 )2 = (2b1 + ib2 )2 ⇒ a2 = ± 2b1 (a) −1 + i( 2 + 3 ) (b) −1 − i( 2 + 3 )
a2 (c) −1 + i( 2 − 3 ) (d) −1 − i( 2 − 3 )
|ω 1 |2 = a1 + a12 + 2 = a12 + b12 = 1
4
⇒ |ω 1 | = 1 and |ω 2 |2 = 4 b12 + b22 = 4 Sol. Clearly, the inscribed triangle is equilateral.
z2
⇒ |ω 2 | = 2
ab
Re (ω 1ω 2 ) = 2a1b1 − 2 2 = 0
2
Im (ω 1ω 2 ) = a1b2 + a2b1 = 2a12 + 2b12 = 2
Hence, (a), (b) and (c) are the correct answers.
z0(2i)
Ex 29. If from a point P representing the complex z3 z1(2 + 2i )
number z1 on the curve | z | = 2, pair of tangents
are drawn to the curve | z | =1, meeting at point 2π 2π
Q ( z 2 ) and R ( z 3 ), then z2 − z0 i z − z0 −i
⇒ =e 3, 3 =e 3
z1 + z 2 + z 3 z1 − z0 z1 − z0
(a) complex number will lie on the
3 ⇒ z2 = − 1 + i (2 + 3)
curve | z | = 1
and z3 = − 1 + i (2 − 3 )
4 1 1 4 1 1
(b) + + + + =9 Hence, options (a) and (c) are the correct answers.
z1 z 2 z 3 z1 z 2 z 3
z 2π Ex 31. Let z be a complex number and a be a real
(c) arg 2 = parameter, such that z 2 + az + a 2 = 0, then
z3 3
(a) locus of z is a pair of straight lines
(d) orthocentre and circumcentre of ∆PQR will (b) locus of z is a circle
coincide 2π
(c) arg ( z ) = ±
Sol. Options (a) and (d) are true as PQR is an equilateral 3
triangle, so orthocentre, circumcentre and centroid will (d) | z | = | a |
coincide.
Y
Sol. z2 + az + a2 = 0
⇒ z = aω, aω 2
Q(z2) P(z1) where, ω is non-real root of cube unity.
⇒ Locus of z is a pair of straight lines
and arg(z) = arg (a) + arg (ω ) or arg (a) + arg (ω 2 )
X¢ X
O 2π
R(z3)
⇒ arg (z) = ±
3
Also, | z | = | a | |ω | or | a | |ω 2 | ⇒ | z | = | a |
140 Y¢ Hence, (a), (c) and (d) are the correct answers.
Type 3. Assertion and Reason
Directions (Ex. Nos. 32-36) In the following 1
Sol. x + = 1 ⇒ x 2 − x + 1 = 0
4
Complex Numbers
examples, each example contains Statement I x
(Assertion) and Statement II (Reason). Each example ∴ x = − ω, − ω2
has 4 choices (a), (b), (c) and (d) out of which only one is 1 1
Now, for x = − ω , p = ω 4000 + =ω + = −1
correct. The choices are ω 4000
ω
(a) Statement I is true, Statement II is true; Statement II Similarly, for x = − ω 2 , p = − 1
is a correct explanation for Statement I For n > 1, 2n = 4 k
n
(b) Statement I is true,Statement II is true; Statement II is ∴ (2)2 = 24k = (16)k = a number with last = 6
not a correct explanation for Statement I ⇒ q = 6 + 1= 7
(c) Statement I is true, Statement II is false Hence, p + q = −1 + 7 = 6
Hence, (d) is the correct answer.
(d) Statement I is false, Statement II is true
Ex 32. Statement I If A ( z1 ), B ( z 2 ), C ( z 3 ) are the Ex 34. Statement I If z1 , z 2 , z 3 are complex
vertices of an equilateral ∆ABC, then numbers represent the points A, B , C such that
2 1 1
z + z 3 − 2 z1 π = + . Then, A, B , C passes through
arg 2 = z1 z 2 z 3
z3 − z2 4
origin.
Statement II If ∠B = α, then Statement II If 2 z 2 = z1 + z 3 , then z1 , z 2 , z 3
z1 − z 2 AB iα z1 − z 2 are concyclic.
= e or arg =α
z 3 − z 2 BC z3 − z2 2 1 1 1 1 1 1
Sol. = + ⇒ − = −
A(z1) z1 z2 z3 z1 z2 z3 z1
z2 − z1 z1 − z3
⇒ =
z1z2 z1z3
z2 − z1 z
⇒ =− 2
z3 − z1 z3
α
z2 − z1 z
B(z2) C(z3) ⇒ arg = arg − 2
z3 − z1 z3
z2 + z3 z −z z z
z2 + z3 − 2z1 − z1 ⇒ arg 2 1 = π − arg 2 = π − arg 2
Sol. Q arg = arg 2 z3 − z1 z3 z3
z3 − z2 z3 − z2
⇒ ∠CAB = π − ∠COB ⇒ ∠CAB + ∠COB = π
A(z1) ⇒ Points O , A , B , C are concyclic.
Hence, (b) is the correct answer.
Ex 35. Statement I 3 + ix 2 y and x 2 + y + 4i are
complex conjugate numbers, then x 2 + y 2 = 4.
B(z2) C(z3) Statement II If sum and product of two
D
z2+z3 complex numbers is real, then they are
2 conjugate complex number.
π Sol. If 3 + ix 2 y and x 2 + y + 4 i are conjugate, then
= [ as AD ⊥ BC ]
2 x 2 y = − 4 and x 2 + y = 3
Hence, (d) is the correct answer.
⇒ x 2 = 4 , y = − 1 ⇒ x 2 + y2 = 5
1 Hence, (d) is the correct answer.
Ex 33. Statement I If x+ =1 and
x Ex 36. Statement I If | z| < 2 − 1, then
p=x 4000
+
1
and q is the digit at unit place | z 2 + 2 z cos α | < 1
x 4000 Statement II | z1 + z 2 | ≤ | z1 | + | z 2 |, also
|cos α| ≤1.
n
in the number 2 2 + 1, n ∈ N and n >1, then the
value of p + q = 8. Sol. | z2 + 2z cos α | < | z2 | + |2z cos α |
< | z|2 + 2| z| |cos α |
Statement II ω, ω 2 are the roots of
1 1 < ( 2 − 1)2 + 2( 2 − 1) < 1
x + = − 1, x 3 + 3 = 2. [Q |cos α | ≤ 1]
x x Hence, (a) is the correct answer.
141
4 Type 4. Linked Comprehension Based Questions
Passage I (Ex. Nos. 37-39) In argand plane, | z | i
4π
Objective Mathematics Vol. 1
Sol. ω 1 = ω 2 e 3
represents the distance of a point z from the origin. In
general, | z1 − z 2 | represents the distance between two ⇒ ω 31 = ω 32 ⇒ω13 ω12 ω22 = ω 32 ω12 ω22
2 2
points z1and z 2 . Also, for a general moving point z in ⇒ ω 1 ω 1 ω22 = ω 2 ω 2 ω12
argand plane, if arg ( z ) = θ, then z = | z | ei θ , where ⇒ω 1 ω12 = ω 2 ω12 , as ω 1 = ω 2
ei θ = cos θ + i sin θ. ∴ ω 2 ω12 = ω 1 ω22
Ex 37. The equation | z − z1 | + | z − z 2 | = 10, if Hence, (b) is the correct answer.
z1 = 3 + 4i and z 2 = − 3 − 4i represents Ex 41. Which of the following must be true?
(a) point circle (b) ordered pair (0, 0) (a) a must be pure imaginary
(c) ellipse (d) None of these (b) β must be pure imaginary
Sol. As z1 − z2 = 6 + 8i = 10 (c) a must be real
∴ z − z1 + z − z2 = z1 − z2 (d) β must be imaginary
⇒ z lies between z1 and z2. Sol. Since, i β is real.
i.e. a line segment. ∴ β is pure imaginary.
Hence, (d) is the correct answer. Hence, (b) is the correct answer.
Ex 38. z − z1 | − | z − z 2 = t, t >10 where t is a real Ex 42. If line (i) makes an angle of 45° with real axis,
parameter, always represents 2α
then (1 + i) − is
(a) ellipse (b) hyperbola α
(c) circle (d) None of these (a) 2 2 (b) 2 2i (c) 2(1 − i ) (d) − 2(1 + i )
Sol. Given equation can represent a hyperbola, since π
α ±i
| z1 − z2 | = 10 Sol. − = e 2 =±i
α
⇒ z lies on hyperbola. 2α
Hence, (b) is the correct answer. ∴ (1 + i ) − = ± 2 (− 1 + i )
α
π
Ex 39. If | z − (3 + 2i)| = z cos − arg ( z ) , then Hence, (c) is the correct answer.
4 Passage III (Ex. Nos. 43-45) Consider ∆ABC in
locus of z is a/an argand plane. Let A(0), B (1) and C (1 + i ) be its vertices
(a) circle (b) parabola and M be the mid-point of CA. Let z be a variable
(c) ellipse (d) hyperbola complex number in the plane. Let us another variable
complex number defined as, u = z 2 + 1.
Sol. Given, | z − (3 + 2i )| = |(x − 3) + i ( y − 2)|
1 1 Ex 43. Locus of u, when z is on BM, is a/an
= |z | cos θ + sin θ , where θ = arg (z)
2 2 (a) circle (b) parabola
1 (c) ellipse (d) hyperbola
⇒ (x − 3)2 + ( y − 2)2 = | x + y|
2
which is a parabola. Ex 44. Axis of locus of u, when z is on BM, is
Hence, (b) is the correct answer. (a) real axis (b) imaginary axis
(c) z + z = 2 (d) z − z = 2i
Passage II (Ex. Nos. 40-42) The complex slope of a
line passing through two points represented by complex Ex 45. Directrix of locus of z, when z is on BM, is
z − z1 (a) real axis (b) imaginary axis
numbers z1 and z 2 is defined by 2 and we shall
z 2 − z1 (c)z + z = 2 (d) z − z = 2i
denote by ω. If z 0 is complex number and c is a real
number, then z 0z + z 0z = 0 represents a straight line. Its Sol. (Ex. Nos. 43-45)
z BM ≡ y − 0 = − 1(x − 1)
complex slope is − 0 . Now ,consider two lines
z0 x + y=1
∴ u − 1 = t + i (1 − t )
α z + αz + i β = 0…(i) and a z + az + b = 0 …(ii)
u = 2t + 2i t (1 − t )
where, α, β and a, b are complex constants and let their
x = 2t and y = 2t (1 − t ), where u = x + iy
complex slopes be denoted by ω1 and ω 2 , respectively.
1
(x − 1)2 = − 2 y − , which is parabola.
Ex 40. If the lines are inclined at an angle of 120° to 2
each other, then Axis is x = 1, i.e. z + z = 2
(a) ω 2 ω1 = ω 1 ω 2 (b) ω 2 ω12 = ω 1 ω 22 Directrix is y = 1, i.e. z − z = 2i
142 (c) ω 12 = ω 22 (d) ω 1 + 2ω 2 = 0 43. (b) 44. (c) 45. (d)
Type 5. Match the Columns
Ex 46. Match the following : ⇒ 75 × 25 = z1 + 1
4
Complex Numbers
∴ z1 − (− 1) = 1875
Column I Column II ⇒ z1 lies on circle.
A. Locus of the point z satisfying the p. circle
equationRe ( z2 ) = Re( z + z), is a/an Ex. 47. If z1 , z 2 , z 3 , z 4 are the roots of the equation
B. Locus of the point z satisfying the q. straight line z 4 + z 3 + z 2 + z + 1 = 0, then
equation| z − z1| + | z − z2| = λ,
λ ∈ R + and λ </ | z1 − z2|, is a/an Column I Column II
4 p. 0
∑
+
C. Locus of the point satisfying the r. ellipse and m ∈ R A. zi4 is equal to
2z − i i =1
equation = m, where
z+1 4
q. 4
∑ zi
5
is equal to
i = −1, is a/an B.
i =1
D. If | z| = 25, then the points s. rectangular 4
r. 1
representing the complex number hyperbola C. ∏( zi + 2 ) is equal to
− 1 + 75 z, is a/an i =1
D. Least value of [| z1 + z2|], where s. 11
Sol. A. Put z = x + iy [ ] represents greatest integer
∴ Re (x + iy)2 = Re (x + iy + x − iy) function, is
x 2 − y2 = 2x
z5 − 1
or x − y2 − 2x = 0
2 Sol. The given equation is = 0, which means that
z−1
Rectangular hyperbola, eccentricity = 2
z1 , z2 , z3 , z4 are four out of five roots of unity except 1.
B. For ellipse, λ > | z1 − z2 |
A. z14 + z24 + z34 + z44 + 14 = 0
and for straight line,
4
λ = | z1 − z2 | ⇒ ∑ zi4 = 1
2z − i i=1
C. Q =m
z+1 B. z15 + z25 + z35 + z45 + 1= 5
i
z− 4
⇒ 2 =
m ⇒ ∑ zi5 = 4
z+1 2 i=1
C. z4 + z3 + z2 + z + 1
For m = 2, = (z − z1 ) (z − z2 ) (z − z3 ) (z − z4 )
1
z− Putting z = − 2 on both the sides, we get
⇒ 2 =1 4
z+1 ∏ (zi + 2) = 11
i=1
i D. | z1 + z2 | = 2 + 2 cos144 ° for minimum
⇒ z − = | z + 1|, i.e. straight line
2
5 −1
D. Given, z = 25 = 2 cos 72° =
2
Let z1 = − 1 + 75 z whose greatest integer is 0.
∴ 75 z = z1 + 1 or 75 z = z1 + 1
A → r; B → q; C → s; D → p
Type 6. Single Integer Answer Type Questions
2i
Ex 48. Consider an equilateral triangle having vertices Sol. (5) A (z1 ) =
at points 3
A
2 iπ 2 − iπ
A e 2 , B e 6 and
3 3
2 − i 5π
C e 6 . If P ( z ) is any point on its
3
O
incircle, then AP + BP + CP
2 2 2
is equal to r
_______ . B C
D 143
4 2 3 1
2
i 2rπ 2rπ
B (z2 ) = − i =1− = 1 − cos − i sin
3 2 2 3 n n
2 2
2rπ 2rπ 2rπ
Objective Mathematics Vol. 1
2 3 i i = 1 − cos + sin = 2 − 2 cos
C (z3 ) = − − = −1− n n n
3 2 2 3
n n
rπ
Radius of incircle of ∆ABC , i.e. r =
1
unit.
∴ ∑ | A1 A2 |2 = ∑ 2 − 2 cos
n
r=1 r=1
3
2π 4π 2(n − 1)π
Hence, any point on incircle i.e. P (z) is = 2 (n − 1) cos + cos + ... + cos
n n n
1 1 1
cos α , sin α i.e. (cosα + i sin α ) = 2 (n − 1) − 2 ⋅ Real part of (α + α + ... + α )
2 n− 1
3 3 3
= 2 (n − 1) − 2 (1)
Solving for | AP |2 + | BP |2 + |CP |2 , we get
[since, {1 + α + α 2 + K + α n− 1 = 0}]
AP 2 + BP 2 + CP 2 = 5 ∴ | A1 A2 |2 + | A1 A3 |2 + ...+ | A1 An |2 = 2n
Ex 49. Let A1 , A2 , ..... An be the vertices of a regular Also, let E = | A1 A2 | | A1 A3 | K | A1 An |
polygon of n sides in a circle of radius unity = |1 − α | |1 − α 2 | |1 − α 3 | ... |1 − α n− 1 |
and = (|1 − α )(1 − α 2 )(1 − α 3 ) K (1 − α n− 1 )|
a = | A1 A2 | 2 + | A1 A3 | 2 + .... + | A1 An | 2 , Since, 1 α ,α 2 ,… , α n− 1 are the roots of zn − 1 = 0.
a zn − 1
b = | A1 A2 | | A1 A3 | ....| A1 An |, then = ____. ⇒ (z − 1) (z − α ) (z − α 2 ) ... (z − α n−1 ) =
b z −1
⇒ (z − α ) (z − α 2 ) (z − α n− 1 ) = zn − 1
Sol. (2) Let us assume that O is the centre of the polygon and
= 1 + z + z2 + … + zn −1
z0 , z1 , … , zn − 1 represent the affixes of A1 , A2 , … , An
such that Substituting z = 1, we have
i 2π (1 − α ) (1 − α 2 )... (1 − α n− 1 ) = n
z0 = 1, z1 = α , z2 = α 2 ,..., zn− 1 = α n− 1, where α = e n
|1 − α | |1 − α 2 |... |1 − α n− 1 | = n
| A1 Ar |2 = |α r − 1|2 = |1 − α r |2 a 2n
Now, Hence, the value of = =2
b n
144
Target Exercises
Type 1. Only One Correct Option
i 592 + i 590 + i 588 + i 586 + i 584 1 + 2i
1. The value of − 1 is 11. The complex number lies in the
i 582
+i 580
+i 578
+i 576
+i 574 1− i
(a) −1 (b) −2 (a) I quadrant (b) II quadrant
(c) −3 (d) −4 (c) III quadrant (d) IV quadrant
1 (1 − i ) 3
2. i 57 + , when simplified has the value 12. The value of is equal to
i 125
1− i3
(a) 0 (b) 2i (a) i (b) − 1 (c) 1 (d) − 2
(c) − 2i (d) 2
2 2
1 + i 1 − i
3. i n + i n + 1 + i n + 2 + i n + 3 is equal to 13. + is equal to
1 − i 1 + i
(a) 1 (b) − 1
(c) 0 (d) None of these (a) 2i (b) − 2i
(c) − 2 (d) 2
4. 1 + i 2 + i 4 + i 6 + K + i 2n is
(1 + i ) 2
(a) positive (b) negative 14. The value of Re is equal to
Targ e t E x e rc is e s
(c) 0 (d) Can’t be determined 3 − i
2 1 1 1 1
1
25 (a) − (b) (c) (d) −
5. The value of i19 + is 5 5 10 10
i
(1+ b + ia )
(b) − 4 (d) − 2
15. If a 2 + b 2 = 1, then is equal to
(a) 4 (c) 2 (1 + b − ia )
6. If n is any positive integer, then the value of (a) 1 (b) 2
i 4n + 1 − i 4n − 1 (c) b + ia (d) a + ib
equals 1 + iz
2 16. If b + ic = (1 + a ) z and a 2 + b 2 + c 2 = 1, then
(a) 1 (b) −1 (c) i (d) −i 1 − iz
equals
7. For a positive integer n, the expression a − ib a − ib a + ib a + ib
n (a) (b) (c) (d)
1 1− c 1+ c 1− c 1+ c
(1 − i ) 1 − equals
n
i
6i −3i 1
(a) 0 (b) 2i n (c) 2n (d) 4 n
17. If 4 3i −1 = x + iy, then
(1 − i ) n 20 3 i
8. If the number is real and positive, then n is
(1 + i ) n − 2 (a) x = 3, y = 1 (b) x = 1, y = 3
(c) x = 0, y = 3 (d) x = 0, y = 0
(a) any integer (b) 2λ
(c) 4 λ + 1 (d) None of these 18. 2i equals
(a) 1 + i (b) 1 − i
9. The smallest positive integer n for which (c) − 2i (d) None of these
n
1 + i
= − 1 is 19. If z is a complex number such that | z | ≠ 0 and
1 − i
Re ( z ) = 0, then
(a) 1 (b) 2 (a) Re (z2 ) = 0 (b) Im (z2 ) = 0
(c) 3 (d) 4
(c) Re (z2 ) = Im (z2 ) (d) None of these
10. The smallest positive number n for which x y
20. If ( x + iy )1/ 3 = a + ib, then + is equal to
(1 + i ) 2n = (1 − i ) 2n is a b
(a) 4 (b) 8 (a) 2 (a2 − b2 ) (b) 4 (a2 − b2 )
(c) 2 (d) 12 (c) 8 (a2 − b2 ) (d) None of these 145
4 21. If 8iz 3 + 12z 2 − 18z + 27i = 0, then
(a) | z | =
3
(b) | z | =
2
(c) | z | = 1 (d) | z | =
3
32. If x + iy =
u + iv
u − iv
, then x 2 + y 2 is equal to
(b) −1
Objective Mathematics Vol. 1
2 3 4 (a) 1
(c) 0 (d) None of these
22. If z = 1 + i, then the multiplicative inverse of z is 2
i i
33. If (1 + i ) (1 + 2i ) (1 + 3i )... (1 + ni ) = α + iβ, then
(a) 1 − i (b) (c) − (d) 2i 2⋅ 5⋅ 10 ... (1 + n 2 ) is equal to
2 2
(a) α − iβ (b) α 2 − β 2
3 3
1 + i 1 − i (c) α 2 + β 2 (d) None of these
23. If − = x + iy, then ( x, y ) is equal to
1 − i 1 + i
34. The principal argument of the complex number
(a) (0, 2) (b) (− 2, 0) (1 + i ) 5 (1 + 3i ) 2
(c) (0, −2) (d) None of these is
−2i ( − 3 + i )
24. The multiplicative inverse of ( 6 + 5i ) 2 is 19π 7π 5π 5π
11 60 11 60 (a) (b) − (c) − (d)
(a) − i (b) − i 12 12 12 12
60 61 61 61
9 60 35. The complex number i + 3 in polar form can be
(c) − i (d) None of these
61 61 written as
1 π π π π
3 + 4i (a) sin + i cos (b) 2 cos + i sin
25. The multiplicative inverse of is 2 6 6 6 6
4 − 5i 1 π π π π
(c) sin + i cos (d) 4 cos + i cos
(a) −
8
+
31
i (b)
8
−
31
i 2 6 6 6 6
25 25 25 25
8 31 36. If z is a complex number, then
(c) − − i (d) None of these
25 25 (a) | z2 | > | z |2 (b) | z2 | = | z |2 (c) | z2 | < | z |2 (d) | z2 | ≥ | z |2
Ta rg e t E x e rc is e s
z
26. If z = x + iy lies in III quadrant, then also lies in 37. If z = x + iy, x and y are real, then | x | + | y | ≤ k | z | ,
z where k is equal to
III quadrant, if (a) 1 (b) 2
(a) x > y > 0 (b) x < y < 0 (c) 3 (d) None of these
(c) y < x < 0 (d) y > x > 0
38. If z is any complex number such that | z + 4 | ≤ 3, then
2− i the least value and greatest value of | z + 1| are
27. The conjugate complex number of is
(1 − 2i ) 2 (a) 1, 6 (b) 0, 6
2 11 2 11 (c) 2, 8 (d) None of these
(a) + i (b) − i
25 25 25 25 39. If z satisfies | z + 1| < | z − 2|, then w = 3z + 2 + i
2 11 2 11 satisfies
(c) − + i (d) − − i
25 25 25 25 (a) | w + 1| < | w − 8| (b) | w + 1| < | w − 7|
(c) | w + w | > 7 (d) | w + 5| < | w − 4 |
2 + i
28. (1 + i ) is equal to 40. For any complex number z, the minimum value of
3 + i
| z | + | z − 1| is
1 1 1 3
(a) − (b) (c) 1 (d) − 1 (a) 1 (b) 0 (c) (d)
2 2 2 2
29. The value of | 2i − − 2i | is 41. The greatest positive argument of complex number
(a) 2 (b) 2 (c) 0 (d) 2 2 satisfying | z − 4| = Re ( z ), is
π 2π
1 − ix (a) (b)
30. If = a − ib and a 2 + b 2 = 1, where a and b are 3 3
1 + ix π π
(c) (d)
real, then x is equal to 2 4
2a 2b
(a) (b) 42. The square roots of −2 + 2 3 i are
(1 + a)2 + b2 (1 + a)2 + b2
(a) ± (1 + 3 i ) (b) ± (1 − 3 i )
2a 2b
(c) (d) (c) ± (− 1 + 3 i ) (d) None of these
(1 + b)2 + a2 (1 + b)2 + a2
43. If z = a + ib, where a > 0, b > 0, then
( 3 + 2i ) 2
31. The modulus of is (a) | z | ≥
1
(a − b) (b) | z | ≥
1
(a + b)
( 4 − 3i ) 2 2
13 11 9 7 1
(a) (b) (c) (d) (c) | z | < (a + b) (d) None of these
146 5 5 5 5 2
44. If for
2
the complex
2
numbers
2
z1
|1 − z1 z 2 | − | z1 − z 2 | = k (1 − | z1 | ) (1 − | z 2 | ),
and
2
z2, 52. If z is any complex number satisfying | z − 1| = 1, then
which of the following is correct?
(a) arg (z − 1) = 2 arg (z)
4
Complex Numbers
then k is equal to (b) 2 arg (z) = 2/ 3 arg (z2 − z)
(a) 1 (b) −1 (c) 2 (d) 4 (c) arg (z − 1) = arg (z + 1)
(d) arg (z) = 2 arg (z + 1)
45. The range of real number α for which the equation
z + α| z − 1| + 2i = 0 has a solution, is 53. If z = − 1, then the principal value of arg ( z 2/ 3 ) is
equal to
5 5 3 3 π 2π 10π
(a) − , (b) − ,
2 2 2 2 (a)
3
(b)
3
or 2π or
3
5 5 5 5π
(c) 0, (d) −∞ , − ∪ , ∞ (c) (d) π
2 2 3
2
z − z1 π
46. If | z | = 2 and locus of 5z − 1 is the circle having 54. If z1 = 8 + 4i, z 2 = 6 + 4i and arg = , then
z − z2 4
radius a and z12 + z 22 − 2z1 z 2 cos θ = 0, then | z1 |: | z 2 |
z satisfies
is equal to
(a) | z − 7 − 4 i | = 1 (b) | z − 7 − 5i | = 2
(a) a : 1 (b) 2a : 1
(c) a : 10 (d) None of these (c) | z − 4 i | = 8 (d) | z − 7i | = 18
47. The maximum value of | z | when z satisfies the 55. If arg ( z ) < 0, then arg ( − z ) − arg (z) is equal to
2 π π
condition z + = 2 is (a) π (b) − π (c) − (d)
z 2 2
(a) 3 − 1 (b) 3 + 1 56. If z is a point on the argand plane such that | z − 1| = 1
(c) 3 (d) 2 + 3 z−2
, then is equal to
Targ e t E x e rc is e s
1 1 z
48. If z1 ≠ z 2 and | z1 + z 2 | = + , then (a) tan (arg z) (b) cot (arg z)
z1 z 2 (c) i tan (arg z) (d) None of these
(a) atleast one of z1 , z2 is unimodular 1 + C + iS
(b) z1 ⋅ z2 is unimodular 57. If C 2 + S 2 = 1, then is equal to
(c) z1 ⋅ z2 is non-unimodular 1 + C − iS
(d) None of the above (a) C + iS (b) C − iS (c) S + iC (d) S − iC
n−1
49. If 1, ω , ω , K , ω
2
are the n, nth roots of unity and 58. The imaginary part of ( z − 1)
z1 , z 2 are any two complex numbers, then (cos α − i sin α) + ( z − 1) −1 × (cos α + i sin α ) is zero,
n −1 if
∑ | z1 + ω k z 2 |2 is equal to (a) | z − 1| = 2 (b) arg (z − 1) = 2α
k =0 (c) arg (z − 1) = α (d) | z | = 1
8
(a) n[| z1 |2 + | z2 |2 ] (b) (n − 1) [| z1 |2 + | z2 |2 ] π π
sin + i cos
(c) (n + 1) [| z1 |2 + | z2 |2 ] (d) None of these 8 8
59. The value of 8
is
50. If z satisfies the equation | z | − z = 1 + 2i, then z is π π
sin − i cos
8 8
equal to
3 3 (a) −1 (b) 0 (c) 1 (d) 2i
(a) + 2i (b) − 2i
2 2 3n
4π 4π 1+ a
3
(c) 2 − i
3
(d) 2 + i 60. If a = cos + i sin , then the value of is
2 2 3 3 2
(−1)n 1
−1 b d (a) (−1)n (b) (c) (d) (−1)n + 1
51. If 3 + i = ( a + ib ) ( c + id ), then tan + tan −1 23n 23n
a c
8 8
is equal to 1 + i 1 − i
π
61. The value of + is equal to
(a) + 2nπ , for some integer n 2 2
2
π (a) 4 (b) 6 (c) 8 (d) 2
(b) − + nπ , for some integer n
3 10
2πk 2πk
π
(c) + nπ , for some integer n
62. The value of ∑ sin
11
− i cos is
11
6 k =1
π (a) 1 (b) −1
(d) − + 2nπ , for some integer n
6 (c) i (d) − i 147
4 63. The value of
10
4 (cos 75° + i sin 75° )
0.4 (cos 30° + i sin 30° )
10
is 72. If ω ( ≠ 1) is cube root of unity satisfying
1
+
a+ω b+ω c+ω
1
+
1
= 2ω 2 and
Objective Mathematics Vol. 1
(a) (1 + i ) (b) (1 − i )
2 2 1 1 1
5 + + = 2 ω, then the value of
(c) (1 + i ) (d) None of these a + ω2 b + ω2 c + ω2
2 1 1 1
+ + is equal to
64. The value of a+1 b+1 c+1
(a) 2 (b) − 2
(cos 2θ − i sin 2θ ) 7 (cos 3θ + i sin 3θ ) −5 (c) − 1 + ω 2 (d) None of these
is
(cos 4θ + i sin 4θ )12 (cos 5θ + i sin 5θ ) −6
(a) cos 33θ + i sin 33θ (b) cos 33θ − i sin 33θ
73. If α and β are the roots of the equation x 2 − 2x + 4 = 0,
(c) cos 47θ + i sin 47θ (d) cos 47θ − i sin 47θ then the value of α 6 + β 6 is
65. (cos 2θ + i sin 2θ ) −5 (cos 3θ − i sin 3θ ) 6 (a) 64 (b) 128
(c) 256 (d) None of these
(sin θ − i cos θ ) 3 is equal to
1
(a) cos 25θ + i sin 25θ 74. If z is any complex number such that z + = 1, then
(b) cos 25θ − i sin 25θ z
(c) sin 25θ + i cos 25θ 1
the value of z 99 + is
(d) sin 25 θ − i cos 25θ z 99
66. The principal value of the arg ( z ) and | z | of the (a) 1 (b) −1
(c) 2 (d) − 2
(1 + cos θ + i sin θ) 5
complex number z = are 75. The value of ( − 1 + −3 ) 62 + ( − 1 − − 3 ) 62 is
(cos θ + i sin θ ) 3
θ θ θ θ (a) 262 (b) 264
(a) − , 32 cos5 (b) , 32 cos5 (c) − 2 62 (d) 0
Ta rg e t E x e rc is e s
2 2 2 2
θ 4 θ
(c) − , 16 cos (d) None of these 76. If ω is a cube root of unity, then
2 2
( 3 + 5ω + 3ω 2 ) 2 ( 3 + 3 ω + 5 ω 2 ) 2 is equal to
67. If z = cos θ + i sin θ, then (a) 4 (b) 0
1 1 (c) − 4 (d) None of these
(a) z + n = 2cos nθ
n
(b) z + n = 2n cos nθ
n
z z 6 6
1 1 1 + i 3 1 − i 3
(c) zn − n = 2n i sin nθ (d) zn − n = (2i )n sin nθ 77. The value of + is
z z 1− i 3 1+ i 3
z 2n − 1 (a) 2 (b) −2 (c) 1 (d) 0
68. If z = cos θ + i sin θ, then , where n is an
z 2n + 1 78. If z1 = 3 + i 3 and z 2 = 3 + i , then the complex
50
integer, is equal to z
number 1 lies in the quadrant number
(a) i cot nθ (b) i tan nθ (c) tan nθ (d) cot nθ z2
(a) I (b) II (c) III (d) IV
69. If a = cos 2α + i sin 2α , b = cos 2β + i sin 2β ,
c = cos 2γ + i sin 2γ and d = cos 2δ + i sin 2δ, then 79. If x = a + b, y = aω + bω 2 , z = aω 2 + bω, then xyz is
1 equal to
abcd + is equal to
abcd (a) (a + b)3 (b) a3 + b3
(a) 2 cos (α + β + γ + δ ) (b) 2cos (α + β + γ + δ) (c) a3 − b3 (d) (a + b)3 + 3ab(a + b)
(c) cos (α + β + γ + δ ) (d) None of these
m
80. If 1, ω and ω 2 are the cube roots of unity, then the
−1 pi + 1 value of (1 + ω ) 3 − (1 + ω 2 ) 3 is
70. e 2mi cot p
⋅ is equal to
pi − 1 (a) 2ω (b) 2 (c) −2 (d) 0
(a) 0 (b) 1 81. If 1, ω and ω 2 are the three cube roots of unity and
(c) − 1 (d) None of these
α , β and γ are the roots of p, p < 0, then for any x, y
8π 8π xα + yβ + zγ
71. If α = cos + i sin , then and z, the expression equals
11 11 xβ + yγ + zα
Re (α + α 2 + α 3 + α 4 + α 5 ) is equal to (a) 1
1 1 (b) ω
(a) (b) −
2 2 (c) ω 2
148 (c) 0 (d) None of these (d) None of the above
82. The value of the expression
1 ( 2 − ω ) ( 2 − ω 2 ) + 2( 3 − ω ) ( 3 − ω 2 ) + .... + ( n − 1)
( n − ω ) ( 2 − ω 2 ), where ω is an imaginary cube root
91. If 1, α , α 2 , K , α n − 1 are the n, nth roots of unity, then
n−1
1
∑ 2 − α i is equal to
4
Complex Numbers
i=1
of unity, is
(n − 2)2n − 1 + 1
n(n + 1)
2
n(n + 1)
2 (a) (b) (n − 2) ⋅ 2n
(a) (b) −n 2n − 1
2 2
2 (n − 2) ⋅ 2n − 1
n(n + 1) (c) (d) None of these
(c) +n (d) None of these 2n − 1
2
1+ i
83. If α and β are the complex cube roots of unity, then 92. The triangle formed by the points 1, and i as
2
α 3 + β 3 + α −2 β −2 is equal to vertices in the argand diagram is a/an
(a) 0 (b) 3 (a) scalene (b) equilateral
(c) − 3 (d) None of these (c) isosceles (d) right angled
84. If ω is a complex cube root of unity, then 93. If the points represented by complex numbers
z1 = a + ib, z 2 = a′ + ib′ and z1 − z 2 are collinear,
(1 − ω + ω 2 ) (1 − ω 2 + ω 4 ) (1 − ω 4 + ω 8 )
then
(1 − ω 8 + ω 16 ) is equal to (a) ab′ + a′ b = 0 (b) ab′ − a′ b = 0
(a) 12 (b) 14 (c) ab + a′ b′ = 0 (d) ab − a′ b′ = 0
(c) 16 (d) None of these
94. If α + iβ = tan −1 z , z = x + iy and α is constant, then
85. If x = ω − ω 2 − 2, then the value of
the locus of z is
x 4 + 3x 3 + 2x 2 − 11 x − 6 is (a) x 2 + y2 + 2x cot 2α = 1 (b) (x 2 + y2 ) cot 2α = 1 + x
(a) 1 (b) − 1 (c) x 2 + y2 + 2 y tan 2α = 1 (d) x 2 + y2 + 2x sin 2α = 1
Targ e t E x e rc is e s
(c) 2 (d) None of these
95. The equation | z + 1 − i | = | z + i − 1| represents
86. If 1, ω and ω 2 are the three cube roots of unity, then
(a) a straight line (b) a circle
(1 + ω ) (1 + ω 2 ) (1 + ω 4 ) (1 + ω 8 ) … to 2n factors is (c) a parabola (d) a hyperbola
equal to
96. If the roots of ( z − 1) n = i ( z + 1) n are plotted in the
(a) 1 (b) − 1
(c) 0 (d) None of these argand plane, they as
(a) on a parabola (b) concyclic
87. If x = a + b + c, y = aα + bβ + c and z = aβ + bα + c (c) collinear (d) the vertices of a triangle
where, α and β are complex cube roots of unity, then
xyz is equal to 97. Locus of the point z satisfying the equation
| iz − 1| + | z − i | = 2 is
(a) 2 (a + b + c )
3 3 3
(b) 2 (a − b − c )
3 3 3
(a) a straight line (b) a circle
(c) a3 + b3 + c3 − 3abc (d) a3 − b3 − c3 (c) an ellipse (d) a pair of straight lines
88. The common roots of the equations 98. For x1 , x 2 , y1 , y 2 , ∈ R , if 0 < x1 , < x 2 , y1 = y 2 and
z 3 + 2z 2 + 2z + 1 = 0 and z 1985 + z 100 + 1 = 0 are 1
z1 = x1 + iy1 , z 2 = x 2 + iy 2 and z 3 = ( z1 + z 2 ),
(a) −1, ω 2
(b) −1, ω 2 then z1 , z 2 , z 3 satisfy
(c) ω , ω 2 (a) | z1 | = | z2 | = | z3 | (b) | z1 | < | z2 | < | z3 |
(c) | z1 | > | z2 | > | z3 | (d) | z1 | < | z3 | < | z2 |
(d) None of the above
99. The equation not representing a circle is given by
89. The values of (16)1/ 4 are
1 + z
(a) ± 2, ± 2i (a) Re =0
1 − z
(b) ± 4 , ± 4 i
(b) zz + iz − iz + 1 = 0
(c) ± 1, ± i z − 1 π
(d) None of the above (c) arg =
z + 1 2
90. If p, q and r are positive integers and ω is an z −1
imaginary cube root of unity and (d) =1
z+1
f ( x ) = x 3 p + x 3q + 1 + x 3r + 2 , then f (ω ) is equal to
(a) ω 100. If z1 , z 2 and z 3 are three complex numbers in AP,
(b) −ω 2 then they lie on
(c) 0 (a) a circle (b) a straight line
(d) None of the above (c) a parabola (d) an ellipse 149
4 101. If the area of the triangle on the argand plane formed
by the complex numbers − z , iz , z − iz is 600 sq units,
then | z | is equal to
110. The locus of the complex number z in an argand
plane satisfying the inequality
| z − 1| + 4 2
Objective Mathematics Vol. 1
log 1/ 2 > 1 where,| z − 1| ≠ is
(a) 10 (b) 20
3| z − 1| − 2 3
(c) 30 (d) None of these
(a) a circle
102. The complex numbers given by 1 − 3i, 4 + 3i and 3 + i (b) interior of a circle
represent the vertices of (c) exterior of a circle
(d) None of the above
(a) a right angled triangle
(b) an isosceles triangle 111. The closest distance of the origin from a curve given
(c) an equilateral triangle
(d) None of the above as az + az + aa = 0 (a is a complex number), is
|a |
(a) 1 (b)
z + 2i 2
103. If Re = 0, then z lies on a circle with centre Re (a) Im (a)
z + 4 (c) (d)
|a | |a |
(a) (− 2, − 1) (b) (− 2, 1)
(c) (2, − 1) (d) (2, 1)
112. Let a be a complex number such that | a| < 1 and
z + 2i z1 , z 2 , … , z n be the vertices of a polygon such that
104. If Im = 0, then z lies on the curve
z + 2 z k = 1 + a + a 2 +… a k , then the vertices of the
(a) x 2 + y2 + 2x + 2 y = 0 polygon lie within the circle
(b) x 2 + y2 − 2x = 0 1
(a) | z − a | =
(c) x + y + 2 = 0 |1 − a |
(d) None of the above 1
(b) | z − 1| =
|1 − a |
105. In the argand plane, all the complex numbers
Ta rg e t E x e rc is e s
1 1
satisfying | z − 4i| + | z + 4i| = 10 lie on (c) z − =
1− a |1 − a |
(a) a straight line (b) a circle
(c) an ellipse (d) a parabola (d) None of these
106. If z = x + iy and a is a real number such that 113. If z1 , z 2 , z 3 and z 4 are the four complex numbers
| z − ai| = | z + ai|, then locus of z is represented by the vertices of a quadrilateral taken in
(a) X -axis order such that z1 − z 4 = z 2 − z 3 and
(b) Y-axis z 4 − z1 π
(c) x = y amp = , then the quadrilateral is a
(d) x 2 + y2 = 1 z 2 − z1 2
(a) square
107. If P represents z = x + iy in the argand plane and (b) rhombus
(c) cyclic quadrilateral
| z − 1|2 + | z + 1|2 = 4, then the locus of P is
(d) None of the above
(a) x 2 + y2 = 2
(b) x 2 + y2 = 1 114. The locus of z satisfying Im ( z 2 ) = 4 is
(c) x 2 + y2 = 4 (a) a circle
(b) a rectangular hyperbola
(d) x + y = 2
(c) a pair of straight lines
108. All the roots of the equation (d) None of the above
a1 z 3 + a 2 z 2 + a 3 z + a 4 = 3, where | a i | ≤ 1, 115. The curve represented by Re ( z 2 ) = 4 is
i = 1, 2, 3, 4, lie outside the circle with centre origin (a) a parabola
and radius (b) an ellipse
(a) 1 (c) a circle
1 (d) a rectangular hyperbola
(b)
3
2 116. A point z is equidistant from three distinct points
(c)
3 z1 , z 2 , z 3 in argand plane. If z , z1 and z 2 are
(d) None of the above z − z1
collinear, then arg 3 will be (z1 , z 2 , z 3 in
109. The region of argand diagram defined by z3 − z2
| z − 1| + | z + 1| ≤ 4 is anti-clockwise sense)
(a) interior of an ellipse π π
(a) (b)
(b) exterior of a circle 2 4
(c) interior and boundary of an ellipse π 2π
150 (c) (d)
(d) None of the above 6 3
Type 2. More than One Correct Option
117. If z1 , z 2 , z 3 and z 4 are roots of the equation 126. Let complex number z of the form x + iy satisfy
4
Complex Numbers
a 0 z 4 + a1 z 3 + a 2 z 2 + a 3 z + a 4 = 0 3z − 6 − 3i π
arg = and | z − 3 + i| = 3. Then, the
where, a 0 , a1 , a 2 , a 3 and a 4 are real, then 2z − 8 − 6i 4
(a) z1 , z2 , z3 , z4 are also roots of equation ordered pairs ( x, y ) are
(b) z1 is equal to atleast one of z1 , z2 , z3 , z4 4 2 4 2
(c) − z1 , − z2 , − z3 , − z4 are also roots of equation (a) 4 − ,1+ (b) 4 + ,1−
5 5 5 5
(d) None of the above
(c) (6, − 1) (d) (0, − 1)
118. If z 3 + ( 3 + 2i ) z + ( − 1 + ia ) = 0 has one real root, 127. If z = ω , ω 2 , where ω is a non-real complex cube root
then the value of a lies in the interval ( a ∈ R ) of unity, are two vertices of an equilateral triangle in
(a) (− 2, 1) (b) (−1, 0) the argand plane, then the third vertex may be
(c) (0, 1) (d) (−2, 3)
represented by
119. The real value of θ for which the expression (a) z = 1 (b) z = 0
i + i cos θ (c) z = − 2 (d) z = − 1
is a real number, is
1 − 2i cos θ 128. If amp ( z1 z 2 ) = 0 and | z1 | = | z 2 | = 1, then
π π (a) z1 + z2 = 0 (b) z1z2 = 1
(a) 2nπ + , n ∈ I (b) 2nπ − , n ∈ I
2 2 (c) z1 = z2 (d) None of these
π π
(c) 2nπ ± , n ∈ I (d) 2nπ ± , n ∈ I
2 4 129. If z is complex number satisfying z + z −1 = 1, then
z n + z − n , n ∈ N has the value
120. If α is a complex constant such that αz + z + a = 0has
(a) 2(−1)n, when n is a multiple of 3
a real root, then
(b) (−1)n − 1, when n is not a multiple of 3
Targ e t E x e rc is e s
(a) α + α = 1
(b) α + α = 0 (c) (−1)n + 1 , when n is a multiple of 3
(c) α + α = − 1 (d) 0, when n is not a multiple of 3
(d) the absolute value of the real root is 1
130. Let z1 , z 2 be two complex numbers represented by
121. If | z1 | = | z 2 | = 1 and arg z1 + arg z 2 = 0, then points on the circle | z | = 1 and | z | = 2 respectively,
(a) z1z2 = 1 then
(b) z1 + z2 = 0
(a) max |2z1 + z2 | = 4 (b) min | z1 − z2 | = 1
(c) z1 = z2
(d) None of the above 1
(c) z2 + ≤3 (d) None of these
z1
122. If | z1 | = 15 and | z 2 − 3 − 4i| = 5, then
(a) | z1 − z2 |min = 5 (b) | z1 − z2 |min = 10 131. ABCD is a square, vertices being taken in the
(c) | z1 − z2 |max = 20 (d) | z1 − z2 |max = 25 anti-clockwise sense. If A represents the complex
123. If | z − (1/ z )| = 1, then number z and the intersection of the diagonals is the
origin, then
1+ 5 5 −1
(a) | z|max = (b) | z|min = (a) B represents the complex number iz
2 2 (b) D represents the complex number iz
5−2 5 −1 (c) B represents the complex number iz
(c) | z|max = (d) | z|min =
2 2 (d) D represents the complex number − iz
124. If z1 and z 2 are two complex numbers ( z1 ≠ z 2 ) 132. If z 0 , z1 represent point P , Q on the locus | z − 1| = 1
π
satisfying | z12 − z 22 | = | z12 + z 22 − 2z1 z 2 | , then and the line segment PQ subtend an angle at the
z z
2
(a) 1 is purely imaginary (b) 1 is purely real point z = 1, then z1 is equal to
z2 z2
i
π (a) 1 + i (z0 − 1) (b)
(c) | arg z1 − arg z2 | = π (d) | arg z1 − arg z2 | = z0 − 1
2
(c) 1 − i (z0 − 1) (d) i (z0 − 1)
125. If | z − 1| = 1, then
133. If | z1 | = | z 2 | = | z 3 | = 1 and z1 , z 2 and z 3 are
(a) arg {(z − 1 − i )/ z} can be equal to − π /4
(b) (z − 2)/ z is purely imaginary number represented by the vertices of an equilateral triangle,
(c) (z − 2)/ z is purely real number then
(d) If arg (z) = θ, where z ≠ 0 and θ is acute, then (a) z1 + z2 + z3 = 0 (b) z1z2z3 = 1
1 − 2 / z = i tan θ (c) z1z2 + z2z3 + z3z1 = 0 (d) None of these
151
4 Type 3. Assertion and Reason
Direction (Q. Nos. 134-144) In the following 139. Statement I If cos (1− i ) = a + ib, where a, b ∈ R and
Objective Mathematics Vol. 1
questions, each question contains Statement I 1 1 1 1
i = −1, then a = e + cos 1, b = e − sin 1.
(Assertion) and Statement II (Reason). Each question 2 e 2 2
has 4 choices (a), (b), (c) and (d) out of which only one is
correct. The choices are Statement II e iθ = cos θ + i sin θ
(a) Statement I is true, Statement II is true; Statement II is
a correct explanation for Statement I.
140. Statement I The product of all values of
(cos α + i sin α ) 3/ 5 is cos 3a + i sin 3a.
(b) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I. Statement II The product of fifth roots of unity is 1.
(c) Statement I is true, Statement II is false.
141. Statement I The locus of the centre of a circle
(d) Statement I is false, Statement II is true. which touches the circles | z − z1 | = a and | z − z 2 | = b
134. Consider z1 = 1, z 2 = 2 and z 3 = 3. externally ( z , z1 and z 2 are complex numbers) will
be hyperbola.
Statement I If z1 + 2z 2 + 3z 3 = 6, then the value
Statement II | z − z1 | − | z − z 2 | < | z 2 − z1 |
of z 2 z 3 + 8z 3 z1 + 27z1 z 2 is 36.
⇒ z lies on hyperbola.
Statement II z1 + z 2 + z 3 ≤ z1 + z 2 + z 3
142. Statement I Consider an ellipse having its foci at
135. Statement I 7 + i > 5 + i A ( z1) and B ( z 2) in the argand plane. If the
Statement II Cancellation laws does not hold true eccentricity of the ellipse is e and it is known that
in complex numbers. origin is an interior point of the ellipse, then
| z + z 2 | | z1 + z 2 |
136. Statement I If z1 and z 2 are two complex numbers e ∈ 1 ,
Ta rg e t E x e rc is e s
z | z1 | + | z 2 | | z1 | + | z 2 |
such that | z1 | = | z 2 | + | z1 − z 2 |, then Im 1 = 0.
z2 Statement II If z 0 is the point interior to curve
Statement II arg ( z ) = 0 ⇒ z is purely real. | z − z1 | + | z − z 2 | = λ
| z 0 − z1 | + | z 0 − z 2 | < λ
1
137. Statement I If z + = 1 ( z ≠ 0 is a complex
z 143. Statement I The equation | z − i| + | z + i| = k , k > 0,
5 +1 can represent an ellipse, if k > 2i.
number), then the maximum value of | z | is .
2
1 Statement II | z − z1 | + | z − z 2 | = k , represents an
Statement II On the locus z + = 1, the farthest
z ellipse, if | k | > | z1 − z 2 |.
5 +1
distance from origin is .
2 144. Statement I The equation zz + az + az + λ = 0 ,
where a is a complex number represents a circle in
138. Statement I The number of complex numbers z
satisfying | z |2 + a | z | + b = 0 ( a, b ∈ R ) is atmost 2. argand plane, if λ is real.
Statement II A quadratic equation in which all the Statement II The radius of the circle
coefficients are non-zero can have atmost two roots. zz + az + az + λ = 0 is aa − λ .
Type 4. Linked Comprehension Based Questions
Passage I (Q. Nos. 145-147) Consider the complex 147. One of the possible argument of complex number
numbers z1 and z 2 satisfying the relation i( z1 / z 2 ) is
2 2 2
z1 + z1 = z1 + z 2 . π π
(a) (b) − (c) 0 (d) None of these
2 2
145. Complex number z1 z 2 is
Passage II (Q. Nos. 148-150) If z satisfy the relation
(a) purely real (b) purely imaginary
(c) zero (d) None of these | z − {(α 2 − 7α + 11) + i }| = 1, α ∈R.
π
Also, arg ( z ) ≥ is satisfied by atleast one z.
146. Complex number z1 / z 2 is 2
(a) purely real
(b) purely imaginary
148. The values of α lies in the interval
7
(c) zero (a) [ 2, 6 ] (b) 1, (c) [ 2, 5 ] (d) [ − 4 , − 2 ]
152 (d) None of the above 2
149. Maximum value of | z − i| is
(a)
7
(b)
4
(c)
5
(d)
9
AB and points A and E lies in the opposite side of line BC.
If A, B and C are represent by the complex numbers 1,
ω and ω 2 respectively.
4
Complex Numbers
2 9 9 4
150. Maximum value of arg ( z ), for which | z − i| is 151. Angle between AC and DE is equal to
π π π π
maximum, is (a) (b) (c) (d)
3 6 4 2
4 9
(a) π − tan −1 (b) π − tan −1
9 4 152. The length of DE is
1 4 3
(c) π − tan −1 (d) π − tan −1 (a) (b) 3 (c) 2 (d) 6
9 5 2
Passage III (Q. Nos. 151-153) On the sides AB and 153. The length of AE is
BC of a ∆ABC, squares are drawn with centres D and E 3− 3 3+ 3
such that points C and D lies on the same side of line AB (a) (b) (c) 3 − 3 (d) 3 + 3
2 2
Type 5. Match the Columns
154. Match the following: 156. Match the following:
Column I Column II Column I Column II
A. z4 − 1 = 0 p. π π
z2 − 1 P. Portions of a z = cos + i sin
A. arg = 0; z ≠ ± i , ± 1 line
8 8
z + 1
2
B. z4 + 1 = 0 q. π π
z = cos − i sin
8 8
B. z − cos −1 cos 12 |−| z − sin−1 sin 12 q. Point of
C. i z4 + 1 = 0 r. π π
intersection of z = cos + i sin
= 8 ( π − 3) hyperbola 4 4
Targ e t E x e rc is e s
D. i z4 − 1 = 0 s. z = cos 0 + i sin 0
C. z2 + k1 = i | z1|2 + k 2; k1 ≠ k 2 ∈ R − { 0} r. Pair of open
rays
157. Match the following:
D. z − 1 − sin−1 1 + z + cos −1 1 − π = 1 s. line segment
3 3 2 Column I ColumnI
A. −1 − −1 − ... ∞ is p. 0
155. For the equation z − 6z + 20 = 0, match the items of
6
B. If zi, i = 1, 2, ..., 6 are the vertices of the q. n
Column I with that of Column II. six sided regular polygon inscribed in a ω−1
6
circle| z| = 2, then ∑ zi
2
Column I Column II is equal to
i =1
A. The number of the roots in the first p. 1 C. If ω k, k ∈ I; 0 ≤ k ≤ n − 1 are the nth r. ω
quadrant can be
roots of unity, then the maximum value
B. The number of the roots in the second q. 2 of ( n!)1/ n ω( n−1)/ 2 is
quadrant can be D. s. ω2
If z1, z2, z3 are the affixes points and A,
C. The number of the roots in the third r. 3 B and C are lying on circle centred at
quadrant can be origin. If the altitude is drawn from
vertex A to base BC, such that it meets
D. The number of the roots in the fourth s. 4 the circumcircle at P( z), then
quadrant can be zz1 + z2 z3 is
Type 6. Single Integer Answer Type Questions
158. If x = a + bi is a complex number such that 161. If z is a complex number satisfying
x 2 = 3 + 4i and x 3 = 2 + 11i, where i = −1, then z + z 3 + 2z 2 + z + 1 = 0, then | z | is equal to _____ .
4
( a + b ) equals _______.
159. If the complex numbers z is simultaneously satisfy 162. ABCD is a rhombus. Its diagonals AC and BD
z − 12 5 z − 4 intersect at the point M and satisfy BD = 2AC. Its
equations = , = 1, then Re ( z ) is ___.
z − 8i 3 z − 8 points D and M represent the complex numbers 1 + i
and 2 − i respectively. Then, the complex number
1 λ
160. If | z | ≥ 3, then the least value of z + is , where λ represented by A is z, has Re ( z ) = {λ 1 , λ 2 }, then the
z 3
value of λ 1 + λ 2 is __________ . 153
is __________ .
Entrances Gallery
JEE Advanced/IIT JEE
1. Match the following : [2014] 6. Let z be a complex number such that the imaginary
2kπ 2kπ part of z is non-zero and a = z 2 + z + 1is real. Then,
Let zk = cos + i sin ; k = 1, 2, ..., 9.
10 10 a cannot take the value [2012]
Column I Column II 1
(a) − 1 (b)
3
A. For each zk , there exists a z j such p. True
zk ⋅ z j = 1 1 3
(c) (d)
2 4
B. There exists k ∈ {1, 2, ..., 9}, such q. False
that z1 ⋅ z = zk has no solution z in the
set of complex numbers
7. If z is any complex number satisfying z − 3 − 2i ≤ 2,
C. 1 − z1 1 − z2 ... 1 − z9 equals r. 1 then the minimum value of 2z − 6 + 5i is________
[2011].
10 π
2i
9
2 kπ 8. Let ω = e 3
and a, b, c, x, y, z be non-zero complex
D. 1 − ∑ cos 10 equals s. 2
k=1 numbers, such that a + b + c = x, a + bω + cω 2 = y
Codes and a + bω 2 + cω = z. Then, the value of
A B C D A B C D 2 2 2
(a) p q s r (b) q p r s x + y + z
is ________ . [2011]
(c) p q r s (d) q p s r 2 2 2
a + b + c
1
2. Let complex numbers α and lie on the circles
α 9. Let z1 andz 2 be two distinct complex numbers and
Ta rg e t E x e rc is e s
( x − x0 ) 2 + ( y − y0 ) 2 = r 2 and z = (1 − t ) z1 + tz 2 for some real number t with
( x − x0 ) + ( y − y0 ) = 4r ,
2 2 2
respectively. If 0 < t <1. If arg (w) denotes the principal argument of a
2 non-zero complex number w, then [2010]
z 0 = x 0 + iy 0 satisfies the equation 2 z 0 = r2 + 2 , (a) z − z1 + z − z2 = z1 − z2
then α equals [2013] (b) arg (z − z1 ) = arg (z − z2 )
1 1 1 1 z − z1 z − z1
(a) (b) (c) (d) (c) =0
2 2 7 3 z2 − z1 z2 − z1
3+i (d) arg (z − z1 ) = arg (z2 − z1 )
3. Let w = and P = {w n : n = 1, 2, 3, ...}. Further
2 10. Match the statement in Column I with those in
H1 = z ∈ C : Re ( z ) >
1 Column II.
and
2
Ø Here, z takes the values in the complex plane and Im z and
− 1
H 2 = z ∈ C : Re ( z ) , where C is the set of all Re z denote, respectively the imaginary part and the real
2 part of z. [2010]
complex numbers. Ifz1 ∈ P ∩ H1 , z 2 ∈ P ∩ H 2 and Column I Column II
O represents the origin, then ∠z1Oz 2 equals [2013]
π π 2π 5π A. The set of point z satisfying p. an ellipse with
(a) (b) (c) (d) 4
2 6 3 6 z − i z = z + i z is eccentricity
5
contained in or equal to
Passage (Q. Nos. 4-5) Let S = S1 ∩ S 2 ∩ S 3
where, S1 = { z ∈ C : z < 4}, B. The set of point z satisfying q. the set of point z
satisfying Im z = 0
z − 1 + 3 i z + 4 + z − 4 = 10 is
S 2 = z ∈ C : Im > 0 contained in or equal to
1− 3 i
C. If ω = 2, then the set of r. the set of point z
and S 3 { z ∈ C : Re z > 0}. satisfying Im ( z) ≤ 1
point z = ω − 1/ω is
4. The area of S equals [2013] contained in or equal to
10π 20π 16π 32π
(a) (b) (c) (d) D. If ω = 1, then the set of s. the set of point z
3 3 3 3 satisfying Re ( z) ≤ 1
point z = ω + 1 / ω is
5. min 1 − 3i − z equals [2013] contained in or equal to
z ∈S
t. the set of point z
2− 3 2+ 3 3− 3 3+ 3 satisfying Im| z| ≤ 5
(a) (b) (c) (d)
2 2 2 2
154
JEE Main/AIEEE
11. A complex number z is said to be unimodular, if 4
19. If z − = 2 , then the maximum value of z is
4
Complex Numbers
| z | = 1. Suppose z1 and z 2 are complex numbers such z
z1 − 2 z 2 equal to [2009]
that is unimodular and z 2 is not (a) 3 + 1 (b) 5 + 1 (d) 2 +
2 − z1 z 2 (c) 2 2
unimodular. Then, the point z1 lies on a [2015] 1
20. The conjugate of a complex number is . Then,
(a) straight line parallel to X -axis i−1
(b) straight line parallel to Y-axis
that complex number is [2008]
(c) circle of radius 2
1 1 1 1
(d) circle of radius 2. (a) − (b) (c) − (d)
i −1 i+1 i+1 i −1
12. If z is a complex number such that z ≥ 2, then the
21. If z + 4 ≤ 3, then the maximum value of z + 1 is
1
minimum value of z + [2014] [2007]
2 (a) 4 (b) 10 (c) 6 (d) 0
5
(a) is equal to 10
2kπ 2kπ
2
(b) lies in the interval (1, 2)
22. The value of ∑ sin 11
+ i cos is
11
[2006]
k =1
5
(c) is strictly greater than (a) 1 (b) − 1 (c) − i (d) i
2
3
(d) is strictly greater than but less than
5 23. If z + z + 1 = 0, where z is complex number, then
2
2 2 2 2 2
1 1
the value of z + + z 2 + 2 + z 3 + 3
1
13. If z is a complex number of unit modulus and z z z
1 + z 2
argument θ, then arg equals 1
Targ e t E x e rc is e s
[2013]
1 + z + ... + z 6 + 6 is [2006]
z
π
(a) − θ (b) −θ (c) θ (d) π − θ (a) 54 (b) 6 (c) 12 (d) 18
2
2
24. If the cube roots of unity are 1, ω, ω , then the roots
z2
14. If z ≠ 1and is real, then the point represented by of the equation ( x − 1) 3 + 8 = 0, are [2005]
z −1
2 2
(a) − 1, 1 + 2ω, 1 + 2ω (b) −1, 1 − 2ω, 1 − 2ω
the complex number z lies [2012]
(a) either on the real axis or on a circle passing through the (c) −1, − 1, − 1 (d) −1, − 1 + 2ω , − 1 − 2ω 2
origin
25. If z1 and z 2 are two non-zero complex numbers such
(b) on a circle with centre at the origin
(c) either on the real axis or on a circle not passing through that z1 + z 2 = z1 + z 2 , then arg ( z1 ) − arg ( z 2 ) is
the origin equal to [2005]
(d) on the imaginary axis π π
(a) − (b) 0 (c) − π (d)
2 2
15. Let α and β be real and z be a complex number. If
z
z 2 + α z + β = 0 has two distinct roots on the line 26. If w = and w = 1, then z lies on [2005]
i
Re ( z ) = 1, then it is necessary that [2012] z−
(a) β ∈ (− 1, 0) (b) β = 1 3
(c) β ∈ (1, ∞ ) (d) β ∈ (0, 1) (a) a parabola (b) a straight line
(c) a circle (d) an ellipse
16. If ω (≠ 1) is a cube root of unity and
27. Let z and w be two complex numbers such that
(1 + ω ) 7 = A + Bω. Then, (A , B) equals [2011]
z + iw = 0 and arg ( zw ) = π. Then, arg ( z ) equals
(a) (1, 1) (b) (1, 0)
(c) (−1, 1) (d) (0, 1) [2004]
π π
17. The number of complex numbers z such that (a) (b)
4 2
z − 1 = | z + 1| = z − i equals [2010] 3π 5π
(c) (d)
(a) 0 (b) 1 4 4
(c) 2 (d) ∞
28. If z = x − iy and z 1/ 3 = p + iq, then
18. If α and β are the roots of the equation x 2 − x + 1 = 0,
x y
+ / ( p + q ) is equal to
2 2
then α 2009 + β 2009 is equal to [2010] [2004]
p q
(a) − 2 (b) − 1
(c) 1 (d) 2 (a) 1 (b) − 1 (c) 2 (d) − 2 155
4 29. If | z 2 − 1| = | z |2 + 1, then z lies on
(a) the real axis
(c) a circle
(b) the imaginary axis
(d) an ellipse
[2004] 1 + i
32. If
1 − i
x
= 1, then [2003]
Objective Mathematics Vol. 1
(a) x = 4 n, where n is any positive integer
30. Let z1 and z 2 be two roots of the equation (b) x = 2n, where n is any positive integer
z 2 + az + b = 0, z being complex. Further, assume (c) x = 4 n + 1, where n is any positive integer
that the origin, z1 and z 2 form an equilateral triangle. (d) x = 2n + 1, where n is any positive integer
Then, [2003]
33. If ω is an imaginary cube root of unity, then
(a) a2 = b (b) a2 = 2b (c) a2 = 3b (d) a2 = 4 b
(1 + ω − ω 2 ) 7 equals [2002]
31. If z and w are two non-zero complex numbers such
π (a) 128 ω
that zw = 1, and arg ( z ) − arg ( w ) = , then z ( w ) is (b) −128 ω
2 (c) 128 ω 2
equal to [2003]
(a) 1 (b) − 1 (c) i (d) − i (d) −128 ω 2
Other Engineering Entrances
34. If α and β are two different complex numbers with 41. If ω, ω 2 are the cube roots of unity, then roots of
β −α equation ( x − 1) 3 + 5 = 0 are [ RPET 2014]
β = 1, then is equal to [Karnataka CET 2014]
1 − αβ (a) −5, − 5 ω, − 5 ω 2
(b) −4 , 1 − 5 ω, 1 − 5 ω 2
(a) 1/2 (b) 0
(c) −1 (d) 1 (c) 6, 1 − 5 ω, 1 + 5 ω 2
(d) None of the above
( 3 + i ) 3 ( 3i + 4 ) 2
35. If z = , then z is equal to 42. The complex number z = x + iy, which satisfies the
( 8 + 6i ) 2 [Kerala CEE 2014]
Ta rg e t E x e rc is e s
z − 3i
(a) 8 (b) 2 (c) 5 equation = 1, lie on [BITSAT 2014]
(d) 4 (e) 10 z + 3i
36. Let w ≠ ± 1 be complex number. If w = 1 and (a) the X -axis
(b) the straight line y = 3
w−1
z= , then Re ( z ) is equal to [Kerala CEE 2014] (c) a circle passing through the origin
w+1 (d) None of the above
1
(a) 1 (b) (c) Re (w) (d) 0 43. If the complex numbers z1 , z 2 and z 3 denote the
w+1
vertices of an isosceles triangle, right angled at z1 ,
(e) w + w
2 2 2
then ( z1 − z 2 ) 2 + ( z1 − z 3 ) 2 is equal to
37. The value of z + z−3 + z−i is minimum, [Kerala CEE 2014]
when z equals [WB JEE 2014] (a) 0 (b) (z2 + z3 )2 (c) 2
2 (e) (z2 − z3 )2
(a) 2 − i (b) 45 + 3i (d) 3
3
i i 44. Let z1 and z 2 be two fixed complex numbers in the
(c) 1 + (d) 1 −
3 3 argand plane and z be an arbitrary point satisfying
π π z − z1 + z − z 2 = 2 z1 − z 2 . Then, the locus of z
38. Convert ( i + 1) / cos − i sin in polar form.
4 4 [J&K CET 2014]
will be [WB JEE 2014]
(a) an ellipse
(a) cos (π /4) + i sin (π / 4 )
(b) a straight line joining z1 and z2
(b) cos (π / 2) − i sin (π / 2)
(c) a parabola
(c) 2 [cos (π /4) + i sin (π /4 )] (d) a bisector of the line segment joining z1 and z2
(d) 2 [cos (π / 2) + i sin (π / 2)]
45. Let z1 be a fixed point on the circle of radius 1
39. Let z1 ≠ z 2 and z1 = z 2 . If z1 has positive real part
z + z1 centred at the origin in the argand plane and z1 ≠ ± 1.
and z 2 has negative imaginary part. Then, 2 Consider an equilateral triangle inscribed in the
z1 − z 2
may be [Manipal 2014] circle with z1 , z 2 and z 3 as the vertices taken in the
(a) 0 (b) real and positive counter clockwise direction. Then, z1 z 2 z 3 is equal
(c) real and negative (d) None of these to [WB JEE 2014]
40. If z = e 2π / 3 , then 1 + z + 3z 2 + 2z 3 + 2z 4 + 3z 5 is (a) z12
equal to [Kerala CEE 2014] (b) z13
(a) − 3eπ i / 3 (b) 3eπ i / 3 (c) 3e2π i / 3 (c) z14
156 (d) − 3e2π i / 3 (e) 0 (d) z1
46. Suppose that z1 , z 2 and z 3 are three vertices of an
equilateral triangle in the argand plane. Let
1
56. If z1 and z 2 are two complex numbers such that
z1 = z 2 + z1 − z 2 , then
z
[Manipal 2012]
z
4
Complex Numbers
α = ( 3 − i ) and β be a non-zero complex (a) Im 1 = 0 (b) Re 1 = 0
2 z2 z2
numbers. The points αz1 + β, αz 2 + β , αz 3 + β will be z z
[WB JEE 2014] (c) Re 1 = Im 1 (d) None of these
z2 z2
(a) the vertices of an equilateral triangle
(b) the vertices of an isosceles triangle
57. If the conjugate of ( x + iy ) (1 − 2i ) is 1 + i, then
(c) collinear
(d) the vertices of a scalene triangle [Karnataka CET 2012]
1− i 1− i
(a) x − iy = (b) x + iy =
47. In the argand plane, the distinct roots of 1 − 2i 1 − 2i
1 + z + z 3 + z 4 = 0 (z is a complex number) 1 1
(c) x = (d) x = −
represent vertices of [WB JEE 2014] 5 5
(a) a square 50
(b) an equilateral triangle 3 3
58. If + i = 3 ( x + iy ), where x and y are real,
25
(c) a rhombus 2 2
(d) a rectangle
then the ordered pair ( x, y ) is [WB JEE 2012]
48. If z1 = 2 2 (1 + i ) and z 2 =1 + i 3, then z12 z 23 is (a) (− 3, 0) (b) (0, 3)
equal to [Kerala CEE 2013] 1 3
(c) (0, − 3) (d) ,
(a) 128i (b) 64i (c) − 64i 2 2
(d) −128i (e) 256
59. If z − z + z + z = 2, then z lies on [AMU 2012]
49.The value of (1 + i ) 3 + (1 − i ) 3 is equal to [J&K CET 2013]
(a) 1 (b) − 2 (c) 0 (d) − 4 (a) a circle (b) a square (c) an ellipse (d) a line
60. If 2x = 3 + 5i, then the value of 2x 3 + 2x 2 − 7x + 72 is
Targ e t E x e rc is e s
50. If z ( 2 − i2 3 ) 2 = i( 3 + i ) 4 , then amplitude of z is
[UP SEE 2013] [MP PET 2011]
−π π (a) 4 (b) − 4
(a) (b) (c) 8 (d) − 8
6 4
π 2 2 2 z1
(c) (d) None of these 61. If z1 + z 2 = z1 + z 2 , then is [MP PET 2011]
6 z2
51. If z1 , z 2 and z 3 are complex numbers such that (a) purely real
(b) purely imaginary
1 1 1
z1 = z 2 = z 3 = + + = 1, then (c) zero of purely imaginary
z1 z 2 z 3 (d) neither real nor imaginary
z1 + z 2 + z 3 is [AMU 2013] 1+ i 3
62. The value of 2
is [Karnataka CET 2011]
(a) 3 (b) 1 1
(c) greater than 3 (d) less than 1 1 +
i + 1
52. If ( 3 i + 1)100 = 299 ( a + ib ), then a 2 + b 2 is equal (a) 20 (b) 9
to [RPET 2013] 5 4
(c) (d)
(a) 4 (b) 3 4 5
(c) 2 (d) 0
63. If z1 and z 2 are two non-zero complex numbers such
53. The value of (1 + 3i ) 4 + (1 − 3i ) 4 is [RPET 2013] z
(a) − 16 (b) 16
that z1 + z 2 = z1 + z 2 , then arg 1 is
z2
(c) 14 (d) − 14
[Kerala CEE 2011]
54. If the fourth roots of unity are z1 , z 2 , z 3 and z 4 , then π
(a) 0 (b) − π (c) −
z12 + z 22 + z 32 + z 42 is equal to [Karnataka CET 2013] 2
π
(a) 0 (b) 2 (d) (e) π
(c) 3 (d) None of these 2
55. Among the complex number z satisfying condition 64. If ω ≠ 1 is a cube root of unity, then the sum of the
z + 1 − i ≤ 1, the number having the least positive series S = 1 + 2ω + 3ω 2 + ... + 3 nω 3n − 1 is
argument is [OJEE 2013] [WB JEE 2011]
3n
(a) 1 − i (a) (b) 3n (ω − 1)
(b) 1 + i ω −1
(c) − i ω− 1
(c) (d) 0
(d) None of the above 3n 157
4 θ
1 + cos 2 − i sin
65.
θ
2
4n
is equal to
70. If
π
π
π
z1 = 2 cos + i sin
4
π
4
and
z 2 = 3 cos + i sin , then z1 z 2 is
[UP SEE 2011]
Objective Mathematics Vol. 1
1 + cos θ + i sin θ
2 2 3 3
[Kerala CEE 2010]
(a) cos nθ − i sin nθ (b) cos nθ + i sin θ (a) 6 (b) 2
(c) cos 2nθ − i sin 2nθ (d) cos 2nθ + i sin 2nθ (c) 6 (d) 3
(e) 2 + 3
− 3+ i 3
66. If x = is a complex number, then the value z z
2 71. If z = r (cos θ + i sin θ ), then the value of + is
of ( x 2 + 3x ) 2 ( x 2 + 3x + 1) is [UP SEE 2011] z z
[Kerala CEE 2010]
9 (a) cos 2 θ (b) 2 cos 2 θ
(a) − (b) 6
8 (c) 2 cos θ (d) 2 sin θ
(c) − 18 (d) 36 (e) 2 sin 2 θ
67. If ( x + iy ) 1/ 3
= 2 + 3i, then 3x + 2 y is equal to 72. If z =
4
, then z is (where, z is complex conjugate
[Kerala CEE 2010] 1− i
(a) − 20 (b) − 60 (c) −120 of z) [WB JEE 2010]
(d) 60 (e) 156
(a) 2(1 + i ) (b) (1 + i )
2 4
68. The modulus of the complex number z such that (c) (d)
1− i 1− i
z + 3 − i = 1 and arg ( z ) = π, is equal to
[Kerala CEE 2010] π
(a) 1 (b) 2 (c) 9 73. If − π < arg ( z ) < − , then arg ( z ) − arg ( − z ) is
2 [WB JEE 2010]
(d) 4 (e) 3
(a) π
69. If z1 , z 2 , .... , z n are complex numbers such that (b) − π
Ta rg e t E x e rc is e s
(c) π /2
z1 = z 2 = ... = z n = 1, then z 2 + z 2 + ... + z n is (d) − π / 2
equal to [Kerala CEE 2010]
cos 30° + i sin 30°
(a) z1z2z3... zn (b) z1 + z2 + ... + zn 74. The value of is equal to
cos 60° − i sin 60° [VITEEE 2010]
1 1 1
(c) + + ... + (d) n (a) i (b) − i
z1 z2 zn 1+ 3i 1 − 3i
(c) (d)
(e) n 2 2
158
Answers
Work Book Exercise 4.1
1. (b) 2. (a) 3. (c) 4. (b) 5. (a) 6. (d) 7. (b) 8. (d) 9. (a) 10. (c)
11. (d) 12. (d) 13. (c) 14. (c)
Work Book Exercise 4.2
1. (a) 2. (b) 3. (a) 4. (c) 5. (a) 6. (b) 7. (c) 8. (b) 9. (d) 10. (d)
11. (a) 12. (a) 13. (d) 14. (b) 15. (b)
Work Book Exercise 4.3
1. (c) 2. (b) 3. (b) 4. (c) 5. (a) 6. (d) 7. (c) 8. (b) 9. (a) 10. (d)
11. (a) 12. (b) 13. (b) 14. (c) 15. (b) 16. (c) 17. (c) 18. (a) 19. (a) 20. (c)
21. (c) 22. (a) 23. (c) 24. (a) 25. (d) 26. (c) 27. (b)
Work Book Exercise 4.4
1. (c) 2. (d) 3. (b) 4. (b) 5. (a) 6. (b) 7. (a) 8. (a) 9. (c) 10. (b)
11. (b) 12. (c) 13. (c) 14. (a) 15. (a) 16. (c) 17. (a) 18. (b) 19. (a) 20. (c)
21. (a) 22. (d) 23. (a) 24. (b) 25. (a) 26. (b) 27. (b) 28. (a) 29. (d) 30. (b)
Target Exercises
1. (b) 2. (a) 3. (c) 4. (d) 5. (b) 6. (c) 7. (c) 8. (c) 9. (b) 10. (c)
11. (b) 12. (d) 13. (c) 14. (a) 15. (c) 16. (d) 17. (d) 18. (a) 19. (b) 20. (b)
21. (a) 22. (c) 23. (c) 24. (d) 25. (c) 26. (c) 27. (d) 28. (c) 29. (a) 30. (b)
Targ e t E x e rc is e s
31. (a) 32. (a) 33. (c) 34. (c) 35. (b) 36. (b) 37. (b) 38. (b) 39. (a) 40. (a)
41. (d) 42. (a) 43. (b) 44. (a) 45. (a) 46. (c) 47. (b) 48. (b) 49. (a) 50. (b)
51. (c) 52. (a) 53. (b) 54. (b) 55. (a) 56. (c) 57. (a) 58. (c) 59. (c) 60. (b)
61. (d) 62. (c) 63. (a) 64. (d) 65. (c) 66. (a) 67. (a) 68. (b) 69. (b) 70. (b)
71. (b) 72. (a) 73. (b) 74. (d) 75. (c) 76. (c) 77. (a) 78. (a) 79. (b) 80. (d)
81. (c) 82. (b) 83. (b) 84. (c) 85. (a) 86. (a) 87. (c) 88. (c) 89. (a) 90. (c)
91. (a) 92. (c) 93. (b) 94. (a) 95. (a) 96. (c) 97. (a) 98. (d) 99. (d) 100. (b)
101. (b) 102. (d) 103. (a) 104. (c) 105. (c) 106. (a) 107. (b) 108. (c) 109. (c) 110. (c)
111. (b) 112. (c) 113. (c) 114. (b) 115. (d) 116. (a) 117. (a,b) 118. (a,b,d) 119. (a,b,c) 120. (a,c,d)
121. (a,c) 122. (a,d) 123. (a,b) 124. (a,d) 125. (a,b,d) 126. (a,b) 127. (a,c) 128. (b,c) 129. (a,b) 130. (a,b,c)
131. (a,d) 132. (a,c) 133. (a,b) 134. (b) 135. (d) 136. (a) 137. (a) 138. (d) 139. (a) 140. (b)
141. (d) 142. (d) 143. (d) 144. (a) 145. (b) 146. (b) 147. (c) 148. (c) 149. (d) 150. (a)
151. (c) 152. (a) 153. (b) 154. (*) 155. (**) 156. (***) 157. (****) 158. (3) 159. (6) 160. (8)
161. (1) 162. (4)
* A → r; B → p; C → q; D → s
** A → p, q; B → p, q; C → p, q; D → p, q
*** A → s; B → r; C → p; D → q
**** A → r,s; B → p; C → q; D → p
Entrances Gallery
1. (c) 2. (c) 3. (c,d) 4. (b) 5. (c) 6. (d) 7. (5) 8. (3) 9. (a,c,d) 10. (*)
11. (c) 12. (b) 13. (c) 14. (a) 15. (c) 16. (a) 17. (b) 18. (c) 19. (b) 20. (c)
21. (c) 22. (c) 23. (c) 24. (b) 25. (b) 26. (b) 27. (c) 28. (d) 29. (b) 30. (c)
31. (d) 32. (a) 33. (d) 34. (d) 35. (b) 36. (d) 37. (c) 38. (d) 39. (a) 40. (a)
41. (b) 42. (a) 43. (a) 44. (a) 45. (b) 46. (a) 47. (b) 48. (d) 49. (d) 50. (a)
51. (b) 52. (a) 53. (a) 54. (a) 55. (d) 56. (a) 57. (b) 58. (d) 59. (a) 60. (a)
61. (b) 62. (d) 63. (a) 64. (a) 65. (c) 66. (c) 67. (c) 68. (e) 69. (c) 70. (c)
71. (b) 72. (d) 73. (a) 74. (a)
* A → q; B → p, t; C → p; D → s,q
159
Explanations
Target Exercises
i 584 (i 8 + i 6 + i 4 + i 2 + 1) i 584 1+ i
2
1− i
2
1. − 1 = 574 − 1 13. We have, +
i (i + i + i + i + 1)
574 8 6 4 2
i 1− i 1+ i
= i 10 − 1 = − 1 − 1 (1 + i )2 (1 − i )2 2i − 2i
= −2 = + = +
(1 − i )2 (1 + i )2 − 2 i 2i
1 1 1
2. i 57 + 125 = (i 4 )14 i + 4 31 = i + = i − i = 0 Q (1 + i )2 = 1 + i 2 + 2 i = 2 i
i (i ) i i
and (1 − i ) = 1 + i − 2 i = − 2 i
2 2
3. We have, i n + i n + 1 + i n + 2 + i n + 3
= −1− 1= −2
= i n (1 + i + i 2 + i 3 )
= i n (1 + i − 1 − i ) [Q i 3 = i 2 ⋅ i = (− 1)⋅ i = − i ] (1 + i )2 2 i 3 + i
14. Re = Re
3 − i 3 + i
= i (0 ) = 0
n
3− i
6i − 2
4. Given expression = 1 + i 2 + i 4 + i 6 + ... + i 2 n = Re
= 1 − 1 + 1 − 1 + ... + (− 1)n 9+ 1
2 6 1
which cannot be determined unless n is known. = Re − + i =−
10 10 5
25 2 2
1
5. We have, i 19 + = i 16 ⋅ i 3 + 24
1 15. Given that, a2 + b2 = 1
i i ⋅i 1 + b + ia (1 + b + ia)(1 + b + ia)
∴ =
1
2 1 + b − ia (1 + b − ia) (1 + b + ia)
= i2 ⋅ i + = (− i − i )2
i (1 + b)2 − a2 + 2 ia (1 + b)
=
Ta rg e t E x e rc is e s
= (− 2 i )2 = 4i 2 = − 4 1 + b2 + 2 b + a2
i 4n + 1 − i 4n − 1 (1 − a2 ) + 2 b + b2 2 ia (1 + b)
6. We have, =
2 2 (1 + b)
i−
1 2 b2 + 2 b + 2 ia (1 + b)
4n − 1
i ⋅i −i ⋅i
4n =
= = i [Q i 4 n = (i 4 )n = (1)n = 1] 2 (1 + b)
2 2 = b + ia
i2 − 1 − 1− 1 − 2 1
= = = = =i 16. We have, b + ic = (1 + a) z
2i 2i 2i −i
b + ic ib − c
n ∴ z= ⇒ iz =
7. (1 − i ) 1 − = (1 − i ) n (1 + i ) n
n 1 1+ a 1+ a
i 1 + iz 1 + a − c + ib
= (1 + 1) n = 2 n ⇒ =
1 − iz 1 + a + c − ib
(1 − i ) n
(1 − i ) (1 − i )
n n−2 (1 + a − c ) + ib (1 + a + c ) + ib
8. = × = ×
(1 + i ) n−2
(1 + i ) n−2
(1 − i ) n − 2 (1 + a + c ) − ib (1 + a + c ) + ib
n −1
a2 + a + iab + ib
= 2 (− i ) n − 1 = 2 (− 1) 2 =
1 + a + ac + c
n −1
This is positive and real, if is even. (a + 1) (a + ib) a + ib
2 = =
n −1 (a + 1) (1 + c ) 1 + c
Let = 2λ
2 17. x + iy = 6 i (3 i 2 + 3) + 3 i (4 i + 20 ) + 1(12 − 60 i )
∴ n = 4λ + 1 = 0 − 12 + 60 i + 12 − 60 i = 0 + 0 i
n n n
1+ i 1+ i 1+ i 1 − 1 + 2i ∴ x = 0, y = 0
9. = × =
1− i 1− i 1+ i 1+ 1 18. 2 i = 1 + i 2 + 2 i = (1 + i )2 = 1 + i
=i n
= − 1, if n = 2
19. Let z = x + iy, where x, y ∈ R
10. (1 + i ) = (1 − i )2 n ⇒ 1 = (− 1)n
2n
∴ x 2 + y 2 ≠ 0 and x = 0
∴The smallest value of n is 2. ∴ z = 0 + iy = iy, y ≠ 0
1 + 2i 1 + 2i 1 + i 1 + 3 i − 2 1 3 ⇒ z2 = − y2
11. = × = =− + i
1− i 1− i 1+ i 2 2 2 ⇒ Im ( z 2 ) = 0
1 + 2i
∴ lies in II quadrant 20. We have, ( x + iy )1/ 3 = a + ib
1− i
On cubing both sides, we get
(1 − i )3 1 − 3 i + 3 i 2 − i 3 − 2 − 2 i x + iy = (a + ib)3
160 12. = = = −2
1− i3 1+ i 1+ i = a3 + 3 a2 (ib) + 3 a (ib)2 + (ib)3
= a3 + 3 a2 bi + 3 ab2 i 2 + i 3 b3
= a3 + 3 a2 bi − 3 ab2 − ib3 [Q i 3 = i 2 ⋅ i = − i ]
= a (a2 − 3 b2 ) + ib (3 a2 − b2 )
∴ Multiplicative inverse of
z= 2 =
z
−
8 31
41
− i
41 = − 8 − 31 i
4
Complex Numbers
z 25 25
∴ x = a (a2 − 3 b2 ) 41
and y = b(3 a2 − b2 ) 8 31
=− − i
x y 25 25
⇒ = a2 − 3 b2 and = 3 a2 − b2
a b 26. Let z = x + iy, where x < 0, y < 0
x y
∴ + = 4 (a2 − b2 ) z x − iy ( x − iy )2 x 2 − y 2 2 xy
a b ∴ = = 2 = − i
z x + iy ( x + y 2 ) x 2 + y 2 x 2 + y 2
21. We have,
z
8 iz 3 + 12 z 2 − 18 z + 27 i = 0 lies in III quadrant, if x 2 − y 2 < 0 and − 2 xy < 0
z
⇒ 4 z (2 iz + 3) + 9 i (2 iz + 3) = 0
2
x < 0, y < 0
⇒ (2 iz + 3) (4 z 2 + 9 i ) = 0 ⇒ − 2 xy < 0
⇒ 2 iz + 3 = 0 or 4z2 + 9 i = 0 Also, x 2 − y 2 < 0, if ( x + y ) ( x − y ) < 0
∴ z =
3 If x − y > 0, if x > y [Q x, y < 0 ⇒ x + y < 0]
2 z
∴ lies in III quadrant, if y < x < 0.
22. Multiplicative inverse of z 2 z
1 1 1 i 2−i
= = = =− 27. Let z =
z 2
(1 + i )2
2 i 2 (1 − 2 i )2
2−i − 2 + i 3 − 4i
1+ i 1− i ∴ z= = ×
23. Since, = i and =−i − 3 − 4i 3 + 4i 3 − 4i
1− i 1+ i
2 11
1+ i
3
1− i
3 =− + i
∴ − = x + iy 25 25
1− i 1+ i ∴ z=−
2 11
− i
Targ e t E x e rc is e s
⇒ i 3 − (− i )3 = x + iy 25 25
⇒ − i + i 3 = x + iy 2 + i 1+ i × 2 + i
28. (1 + i ) =
⇒ − i − i = x + iy 3 + i 3+ i
x + iy = 0 − 2 i 2× 5
Hence, ( x, y ) = (0, − 2 ). = =1
10
24. Let z = (6 + 5i )2 = 36 + 2(6)(5 i ) + 25 i 2
29. 2 i − − 2 i = (1 + i )2 − (1 − i )2
= 36 + 60 i − 25 = 11 + 60 i
= (1 + i ) − (1 − i ) = 2 i
Then, z = 11 − 60 i
∴ 2i − − 2i = 2i = 0 + 4 = 2
and z = (11)2 + (60 )2
= 121 + 3600 1 − ix
30. = a − ib
= 3721 = 61 1 + ix
(1 − ix ) (1 − ix )
Multiplicative inverse of ⇒ = a − ib
z 11 − 60 i 11 60 (1 + ix ) (1 − ix )
z= 2 = = − i
z (61)2
(61)2
(61)2 1 − x 2 − 2 ix a − ib
⇒ =
3 + 4i 3 + 4i 4 + 5i 1 + x2 1
25. Let z = = × 1 − x2 2x
4 − 5i 4 − 5i 4 +5 i ⇒ = a and =b
12 + 15 i + 16 i + 20 i 2 1 + x2 1 + x2
=
16 − 25 i 2 Now, x can be written as
12 + 31 i − 20 2x 2x
= 1 + x2 1 + x2
16 + 25 x= =
8 31 2 1 − x2
= + i +1
41 41 1+ x 2
1 + x2
8 31 b 2b
Then, z=− − i = =
41 41 1 + a 1 + 1 + 2a
2 2 2b 2b
8 31 = =
and z = − + − 1 + (a + b ) + 2 a (1 + a)2 + b2
2 2
41 41
64 + 961 (3 + 2 i )2 3 + 2i
2
= 31. =
(41)2 (4 − 3 i ) 4 − 3i
1025 25 × 41 5 ( 9 + 4 )2 13
= = = = =
(41)2 (41)2 41 ( 16 + 9 ) 5 161
4 32. We have, x + iy =
u + iv
u − iv
We know that, when two complex numbers are equal
…(i) 41.
⇒
Now, z − 4 = Re ( z )
( x − 4) + y 2 = x
2
Objective Mathematics Vol. 1
their conjugates are also equal. Y
u − iv
∴ x − iy = …(ii)
u + iv
On multiplying Eqs. (i) and (ii), we get
u + iv u − iv X′ X
( x + iy ) ( x − iy ) = × O (4, 0)
u − iv u + iv
u 2 − i 2v 2
⇒ x2 + y2 = 2 2 2
u −i v
u 2 + v2 Y′
∴ x + y = 2
2 2
=1
u + v2 ⇒ x − 8 x + 16 + y 2 = x 2
2
⇒ y 2 = 8 ( x − 2)
33. Taking modulus and squaring on both sides, we get
2 2 2 2 2 The given relation represents the part of parabola with
1 + i ⋅ 1 + 2 i ⋅ 1 + 3 i .... 1 + ni = α + i β
focus (4, 0) lying above X-axis and the imaginary axis
⇒ (1 + 1)⋅ (1 + 4)⋅ (1 + 9) ... (1 + n 2 ) = α 2 + β 2 as the directrix. The two tangents from directrix are at
⇒ 2 ⋅ 5 ⋅ 10 ... (1 + n 2 ) = α 2 + β 2 right angle. Hence, the greatest positive argument
π
(1 + i )5 (1 + 3 i )2 of z is .
34. 4
− 2 i (− 3 + i )
2 42. Let a + bi be a square root of − 2 + 2 3 i.
1 2 1 3
5
1 ∴ (a + bi )2 = − 2 + 2 3 i
( 2 )5 + 2 +i
2 2 2 2
= ⇒ a + 2 abi + i 2 b2 = − 2 + 2 3 i
2
3 i
2i 2 − ⇒ (a2 − b2 ) + 2 abi = − 2 + 2 3 i
Ta rg e t E x e rc is e s
2 2 ∴ a2 − b2 = − 2 …(i)
5π 2 π π π 9π
⇒ Argument = + − + = and 2 ab = 2 3 …(ii)
4 3 2 6 12 Now, (a2 + b2 )2 = (a2 − b2 )2 + 4 a2 b2
5π
Therefore, the principal argument is − . = (− 2 )2 + (2 3 )2
12
= 4 + 12 = 16
3 i π π
35. i + 3 = 2 + = 2 cos + i sin ∴ a2 + b2 = 16 = 4 …(iii)
2 2 6 6
On solving Eqs. (i) and (iii), we get
2
36. z 2 = z z = z z = z 2 a2 = 2 or a2 = 1
∴ a=±1
37. For every a ∈ R, a = a2 ⇒ 2 b2 = 6 or b2 = 3
2
∴ a = a2 ∴ b = ± 3.
Now, ( x − y )2 ≥ 0 From Eq. (ii), ab = 3, which is positive.
2 2 Either a = 1, b = 3 or a = − 1, b = − 3
⇒ x + y −2x y ≥ 0
Hence, the two square roots are 1 + 3 i and − 1 − 3 i
⇒ 2x y ≤ x + y
2 2 i.e. ± (1 + 3 i ).
2 2 2 2
⇒ x + y + 2x y ≤2 x + 2y Aliter
z +a z − a
⇒ ( x + y )2 ≤ 2 ( x 2 + y 2 ) z = ± +i
2 2
2
⇒ ( x + y )2 ≤ 2 z 4−2 4 + 2
=± +i = ± (1 + i 3 )
∴ x + y ≤ 2z 2 2
38. z + 1 = ( z + 4) − 3 ≤ z + 4 + − 3 ≤ 3 + 3 = 6 43. As,(a − b)2 ≥ 0, a2 + b2 ≥ 2 ab .... (i)
∴ Greatest value of z + 1 = 6 But z = a +b 2 2
Also, z + 1 ≥ z + 4 − − 3 ≥ 0
∴ 0≤ z+1 ≤6 From Eq. (i), we get| z 2| ≥ 2 ab
∴ | z|2 + a2 + b2 ≥ a2 + b2 + 2 ab
39. z + 1 < z − 2 2 2
⇒ z + z ≥ (a + b)2
⇒ 3z + 3 < 3z − 6 2
⇒ w + 1− i < w − 8 − i ⇒ 2 z ≥ (a + b)2
⇒ w+1< w−8 ⇒ 2 z ≥ a + b, as z is positive
1
40. z + z − 1 ≥ z − z + 1 ≥ 1 ∴ z≥ ( a + b)
162 2
44. We have, 1 − z1 z2 2 − z1 − z2 2
=(1 − z1 z2 ) (1 − z1 z2 ) − ( z1 − z2 ) ( z1 − z2 )
= (1 − z1 z2 ) (1 − z1 z2 ) − ( z1 − z2 ) ( z1 − z2 )
2
[Q z z = z ]
49. Since, 1, ω, ω 2 , ..., ω n − 1 are the nth roots of unity.
∴
n −1
∑ ω k = 0 and
n −1
∑ (ω)k = 0
4
Complex Numbers
k=0 k=0
n −1 n −1
[Q z1 − z2 = z1 − z2 and 1 = 1] 2
=(1 − z1 z2 ) (1 − z1 z2 ) − ( z1 − z2 ) ( z1 − z2 ) [Q ( z1 ) = z1 ]
Now, ∑ z1 + ω z2 =
k
∑ ( z1 + ω k z2 ) { z1 + (ω)k z2 }
k=0 k=0
= 1 − z1 z2 − z1 z2 + z1 z1 z2 z2 − z1 z1 + z1 z2 + z1 z2 − z2 z2 n −1
= 1 + z1
2 2 2 2 2
z2 − z1 − z2 = (1 − z1 ) (1 − z2 )
2 = ∑ [ z1 z1 + z1 z2 (ω )k + z1 z2ω k + z2 z2 (ω )k (ω )k ]
k=0
∴ k =1 n −1 n −1 n −1 n −1
∑ ∑ z1 z2 (ω )k + ∑ z1 z2ω k + ∑
2 2
45. Let z = x + iy = z1 + z2
k=0 k=0 k=0 k=0
We have, z + α z − 1 + 2i = 0 2 2 2 2
= n z1 + 0 + 0 + n z2 = n ( z1 + z2 )
⇒ x + i ( y + 2 ) + α ( x − 1)2 + y 2 = 0
50. Let z = x + iy
On equating real and imaginary parts
∴ z − z = 1 + 2i
y + 2 = 0 ⇒ y = −2
⇒ x 2 + y 2 − ( x + iy ) = 1 + 2 i
and x + α ( x − 1)2 + 4 = 0
⇒ ( x 2 + y 2 − x ) + i (− y ) = 1 + 2 i
∴ x 2 = α 2 ( x 2 − 2 x + 5)
⇒ x2 + y2 − x = 1
or (1 − α 2 )x 2 + 2α 2 x − 5 α 2 = 0
and y = −2
Since, x is real.
∴ D = B 2 − 4 AC ≥ 0 If y = − 2, x2 + 4 − x = 1
⇒ 4 α + 20 α 2 (1 + α 2 ) ≥ 0
4 ⇒ x + 4 = (1 + x )2
2
3
⇒ − 4 α 4 + 5α 2 ≥ 0 ⇒ 2 x = 3 or x =
5 2
⇒ 4 α 2 α 2 − ≤ 0 3
4 ∴ z = x + iy = − 2i
Targ e t E x e rc is e s
2
− 5 5
∴ ≤α ≤ 51. ( 3 + i ) = (a + ib) (c + id )
2 2
On taking argument both sides, we get
46. Given, z = 2 1 b d
tan − 1 = tan − 1 + tan − 1
Let α = 5 z − 1 ⇒ α + 1 = 5 z = 5 (2 ) = 10 3 a c
⇒ α + 1 = 10 −1 b −1 d π
⇒ tan + tan = nπ + , n ∈ Z
∴Locus of α, i.e. 5 z − 1 is the circle having centre at − 1 a c 6
and radius 10. 52. Let z − 1 = e iθ [Q z − 1 = 1]
∴ a = 10
⇒ z = 1 + cos θ + i sin θ
Again, z12 + z22 − 2 z1 z2 cos θ = 0 θ θ θ
2 = 2 cos 2 + 2 i sin cos
z1 z 2 2 2
⇒ − 2 1 cos θ + 1 = 0
z2 z2 θ θ θ
= 2 cos cos + i sin
2 2 2
z1 2 cos θ ± 4 cos θ − 4
2
⇒ = = cos θ ± i sin θ θ 1
z2 2 ∴ arg ( z ) = = arg ( z − 1) or arg ( z − 1) = 2 arg ( z )
2 2
z1 a
⇒ = cos θ ± i sin θ = 1 = 2
53. arg ( z 2 / 3 ) = arg (− 1)
z2 10 3
∴ z1 : z2 = a : 10 2 2 2 2π 10 π
= π , ⋅ 3π , ⋅ 5π = , 2 π,
2 2 2 2 3 3 3 3 3
47. We have, z = z + − ≤ z+ +
z z z z z − z1 ( x + iy ) − (8 + 4i ) ( x − 8) + i ( y − 4)
54. = =
2 2 z − z2 ( x + iy ) − (6 + 4i ) ( x − 6) + i ( y − 4)
⇒ z ≤2 + ⇒ z ≤2 z + 2
z z − z1 π
2 We have, arg =
⇒ z − 2 z + 1≤ 1+ 2 z − z2 4
⇒ ( z − 1)2 ≤ 3 ⇒ − 3 ≤ z − 1 ≤ 3 π
⇒ arg ( z − z1 ) − arg ( z − z2 ) =
⇒ 1− 3 ≤ z ≤ 1+ 3 4
−1 y − 4 −1 y − 4 π
So, the maximum value of z is 1 + 3. ⇒ tan − tan =
x−8 x−6 4
1 1 z + z2 2( y − 4)
48. We have, z1 + z2 = + = 1 ⇒ =1
z1 z2 z1 z2 x + y − 14 x − 8 y + 64
2 2
1 ⇒ x 2 + y 2 − 14 x − 10 y + 72 = 0
⇒ z1 + z2 1 − =0
z1 z2 ⇒ ( x −7 )2 + ( y − 5)2 = ( 2 )2
∴ z − (7 + 5 i ) = 2 163
⇒ z1 z2 = 1 [Q z1 ≠ − z2 ]
4 55. arg (− z ) = π − θ = π + (− θ ) = π + arg ( z ) 8 8
1+ i 1− i
61. We have, +
∴ arg (− z ) − arg ( z ) = π 2 2
8 8
56. Since, z − 1 = 1 π π π π
Objective Mathematics Vol. 1
= cos + i sin + cos − i sin
Let z − 1 = cos θ + i sin θ 4
4
4 4
Then, z − 2 = cos θ + i sin θ − 1 = cos 2 π + i sin 2 π + cos 2 π − i sin 2 π
θ θ θ = 2 cos 2 π = 2 (1) = 2 [using De-Moivre’s theorem]
= − 2 sin 2 + 2 i sin cos
2 2 2 10
2 πk 2 πk
θ θ
= 2 i sin cos + i sin
θ
…(i)
62. We have, ∑ sin 11
− i cos
11
2 2 2 k =1
10
2 πk 2 πk
and z = 1 + cos θ + i sin θ
θ θ θ
= ∑ − i 2 sin 11
− i cos
11
k =1
= 2 cos 2 + 2 i sin cos 2 πk
2 2 2 10
2 πk 2 πk 10 i
θ θ
= 2 cos cos + i sin
θ
…(ii)
=−i ∑ cos 11
+ i sin
11
= −i ∑e 11
2 2 2 k =1 k =1
10 i 2 πk
From Eqs. (i) and (ii), we get = − i ∑ e 11 − 1
z −2 θ θ k = 1
= i tan = i tan (arg z ) Q arg z = , from Eq. (ii)
z 2 2
= − i (Sum of 11th roots of unity − 1)
57. We have, C 2 + S 2 = 1 = − i (0 − 1) = i
Let C = cos θ, S = sin θ 4 (cos 75° + i sin 75° )
63. We have,
1 + C + iS 1 + cos θ + i sin θ 0.4 (cos 30 ° + i sin 30 ° )
∴ =
1 + C − iS 1 + cos θ − i sin θ 10 {cos (75° − 30 ° ) + i sin (75° − 30 ° )}
=
θ θ θ cos 2 30 ° + sin 2 30 °
cos cos + i sin
2 2 2 10
= = 10 (cos 45° + i sin 45° ) = (1 + i )
θ θ θ 2
cos cos − i sin
Ta rg e t E x e rc is e s
2 2 2 64. Using De-Moivre’s theorem, the given expression
2
θ θ (cos θ + i sin θ )− 14 (cos θ + i sin θ)− 15
cos + i sin =
2 2
= = cos θ + i sin θ (cos θ + i sin θ)48 (cos θ + i sin θ )− 30
1
= (cos θ + i sin θ )− 47 = cos 47θ − i sin 47θ
= C + iS
58. Let z − 1 = r (cos θ + i sin θ ) = re iθ 65. We have,
1 sin θ − i cos θ = − i 2 sin θ − i cos θ
∴ Given expression = re iθ ⋅ e − iα + iθ e iα
re = − i (cos θ + i sin θ )
1
= re i( θ − α ) + e −i( θ − α) ∴The given expression, using De-Moivre’s theorem
r
= (− i )3 [cos (− 25 θ) + i sin (− 25 θ )]
Since, imaginary part of given expression is zero,
we have = i [cos 25 θ − i sin 25 θ]
1
r sin (θ − α ) − sin (θ − α ) = 0 = sin 25 θ + i cos 25 θ
r (1 + cos θ + i sin θ)5
r2 − 1= 0 ⇒ r2 = 1 66. We have, z =
(cos θ + i sin θ)3
⇒ r =1 ⇒ z −1 =1 5
sin (θ − α ) = 0 ⇒ (θ − α ) = 0 2 θ θ θ
or 1 + 2 cos − 1 + 2 i sin cos
⇒ θ = α ⇒ arg ( z − 1) = α = 2 2 2
8 cos 3 θ + i sin 3 θ
π π
sin + i cos − 16 θ θ θ
5
8 8 π π
59. = cos + i sin 32 cos 5 cos + i sin
π π
8 8 8 =
2 2 2
sin − i cos cos 3 θ + i sin 3 θ
8 8
= cos 2 π − i sin 2 π = 1 5θ θ 5θ
= 32 cos 5 cos + i sin {cos 3 θ − i sin 3 θ}
1+ a 1 4π 4π 2 2 2
60. = 1 + cos + i sin θ 5 θ 5 θ
2 2 3 3 = 32 cos 5 cos − 3 θ + i sin − 3 θ
1 2π 2π 2π 2 2 2
= . 2 cos cos + i sin
2 3 3 3 θ θ θ
= 32 cos 5 cos − 2 + i sin − 2
1 2π 2π 2
= − cos + i sin
2 3 3 θ
Hence, the modulus and argument of z are 32 cos 5
3n 3n 3n 2
1 + a 2π 2π (− 1)
n
1
∴ = − cos + i sin = θ
2 2 3 3 23n and − , respectively.
164 2
67. We have,
1
=
1
z cos θ + i sin θ
= cos θ − i sin θ
72. We have,
1
1
+
a+ω b+ω c +ω
+
1
1
+
+
1
1
= 2ω 2 =
= 2ω =
2
ω
2
4
Complex Numbers
and
∴ z n = (cos θ + i sin θ )n = cos nθ + i sin nθ , a+ω 2
b+ω 2
c +ω 2
ω
1
and = (cos θ − i sin θ )n = cos nθ − i sin nθ ∴ ω and ω 2 are roots of the equation
zn 1 1 1 2
1 1 + + = .
Hence, z n + n = 2 cos nθ and z n − n = 2 i sin nθ a+ x b+ x c + x x
z z
When x = 1,
68. We have, 1 1 1 2
+ + = =2
z 2 n − 1 (cos θ + i sin θ)2 n − 1 a+1 b+1 c +1 1
=
z 2 n + 1 (cos θ + i sin θ)2 n + 1 1 3
cos 2 nθ + i sin 2 nθ − 1 73. x 2 − 2 x + 4 = 0 ⇒ x = 1 + 3 i = 2 ± i
= 2 2
cos 2 nθ + i sin 2 nθ + 1 = − 2ω − 2ω 2 ⇒ α 6 + β 6 = 128
[using De-Moivre’s theorem] 1
(1 − 2 sin 2 nθ) + 2 i sin nθ cos nθ − 1 74. z + =1
= z
(2 cos 2 nθ − 1) + 2 i sin nθ cos nθ + 1 ⇒ z = −ω or − ω 2
i sin nθ cos nθ + i 2 sin 2 nθ 1 1
= [Q i 2 = − 1] ∴ z 99 + 99 = (− 1) + = −2
cos 2 nθ + i sin nθ cos nθ z (− 1)
i sin nθ (cos nθ + i sin nθ ) 75. (− 1 + − 3 )62 + (− 1 − − 3 )62
= = i tan nθ
cos nθ (cos nθ + i sin nθ ) = 2 62 ω 62 + 2 62 (ω 2 )62
69. We have, abcd = cos (2α + 2β + 2 γ + 2δ ) = 2 62 [(ω 3 )20 ω 2 + (ω 3 )41ω]
+ i sin (2α + 2β + 2 γ + 2δ ) = 2 62 [(ω 2 + ω] = − 2 62
∴ abcd = [cos (2α + 2β + 2 γ + 2δ ) 76. Given expression
Targ e t E x e rc is e s
+ i sin (2α + 2β + 2 γ + 2δ )]1/ 2 = { 3 (1 + ω 2 ) + 5 ω} 2 + { 3 (1 + ω ) + 5 ω 2 } 2
or abcd = cos (α + β + γ + δ ) = (− 3 ω + 5 ω)2 + (− 3 ω 2 + 5 ω 2 )2
+ i sin (α + β + γ + δ ) …(i) = 4ω2 + 4ω = − 4
[using De-Moivre’s theorem] 6
1 ω2 ω
6
− 1+ i 3
∴ = cos (α + β + γ + δ) 77. + 2 = 2, using ω= and
abcd ω ω 2
− i sin (α + β + γ + δ) …(ii) − 1− i 3
On adding Eqs. (i) and (ii), we get ω2 =
1 2
abcd + = 2 cos (α + β + γ + δ) 2
abcd z1 − 3i ω
78. Here, =
70. Let cot − 1 p = θ, then p = cot θ z2 2
25
pi + 1
m
i cot θ + 1
m
z
50 z 2 − 3 iω
25
3
25
2 mi cot −1 p 2 mi θ
∴ e ⋅ =e ⋅ ⇒ 1 = 1 = = − iω
pi − 1 i cot θ − 1 z2 z2 2 2
m m
i (cot θ − i ) cot θ − i 3 3 i
25
= e 2 miθ. = e 2 miθ ⋅ = +
i (cot θ + i ) cot θ + i 2 2 2
m m
cos θ − i sin θ e − iθ z
50
= e 2 miθ ⋅ = e 2 miθ ⋅ iθ ∴ 1 lies in I quadrant.
cos θ + i sin θ e z2
= e 2 miθ (e − 2 iθ )m = e 0 = 1 79. Q xyz = (a + b) (aω + bω 2 ) (aω 2 + bω)
8π
8π 8π = (a + b) (a2 − ab + b2 ) = a3 + b3
71. We have, α = cos + i sin
i
= e 11
11 11
80. (1 + ω )3 − (1 + ω 2 )3 = (− ω 2 )3 − (− ω)3
Re (α + α 2 + α 3 + α 4 + α 5 ) = − ω6 + ω3 = − 1 + 1 = 0
α + α2 + α3 + α4 + α5 + α + α2 + α3 + α4 + α5
= 81. p1/ 3 = (− p)1/ 3 (− 1)1/ 3
2
− 1 + (1 + α + α 2 + α 3 + α 4 + α 5 = − (− p)1/ 3 , − (− p)1/ 3 ω, − (− p)1/ 3 ω 2
= − q, − qω, − qω 2, where q = (− p)1/ 3
+ α + α 2 + α 3 + α 4 + α 5 )
= Let α = − q, β = − qω and γ = − qω 2
2
− 1+ 0 xα + yβ + z γ − q ( x + yω + zω 2 )
= [sum of 11 and 11th roots of unity] ∴ =
2 xβ + yγ + zα − q ( x ω + yω 2 + z )
1 xω 3 + yω 4 + zω 2
=− = = ω2
2 x ω + yω 2 + z 165
4 82. Given expression
=
n
∑ (k − 1)(k − ω ) (k − ω 2 )
88. We have,
⇒
z3 + 2 z2 + 2 z + 1 = 0
( z + 1) ( z 2 + z + 1) = 0
Its roots are − 1, ω and ω 2 . The root z = − 1 does not
Objective Mathematics Vol. 1
k=2
n n satisfy the equation, z1985 + z100 + 1 = 0 but z = ω and
= ∑ (k − 1) (k 2 + k + 1) = ∑ (k 3 − 1) z = ω 2 satisfy it.
k=2 k=2
Hence, ω and ω 2 are the common roots.
= (2 3 + 33 + ... + n 3 ) − (n − 1)
= (Σ n 3 − 13 ) − n + 1 89. We have, (16)1/ 4 = (2 4 )1/ 4 = 2(1)1/ 4
2 = 2 (cos 0 + i sin 0 )1/ 4
n (n + 1)
= −n 1 1
2 = 2 cos (2 kπ + 0 ) + i sin (2 kπ + 0 ), k = 0, 1, 2, 3
4 4
83. Let α = ω and β = ω 2 = 2 × 1, 2 × i , 2 × − 1, 2 × − i , = ± 2, ± 2 i
1 1
Now, α 3 + β 3 + α − 2β − 2 = ω 3 + ω 6 + ⋅ 90. We have, f (ω ) = ω 3 p + ω 3 q + 1 + ω 3 r + 2
ω2 ω4
1 1 = ω 3p + ω 3q ⋅ ω + ω 3r ⋅ ω 2
= ω 3 + (ω 3 )2 + 3 2 = 1 + (1)2 + 2 = 3 [Qω 3 = 1]
(ω ) (1) = (ω 3 ) p + (ω 3 )q ⋅ ω + (ω 3 ) r ⋅ ω 2
84. We have, = 1 + ω + ω2 = 0 [Qω 3 = 1]
(1 − ω + ω 2 ) (1 − ω 2 + ω 4 ) (1 − ω 4 + ω 8 ) (1 − ω 8 + ω16 ) 91. Since 1, α, α 2 , ..., α n − 1 are n and nth roots of unity.
= (1 + ω 2 − ω ) (1 − ω 2 + ω ) (1 − ω + ω 2 ) (1 − ω 2 + ω ) ∴ x n − 1 = ( x − 1) ( x − α) ( x − α 2 ) ...( x − α n − 1 )
[Qω 4 = ω 3 ⋅ ω = ω; ω 8 = (ω 3 )2 ⋅ ω 2 ; ⇒ log ( x n − 1) = log ( x − 1) + log ( x − α )
ω16 = (ω 3 )5 ⋅ ω = ω and ω 3 = 1] + log ( x − α 2 ) + ... + log ( x − α n − 1 )
= (− ω − ω ) (− ω − ω ) (− ω − ω ) (− ω 2 − ω 2 )
2 2
On differentiating both sides w.r.t. x, we get
= (− 2ω ) (− 2ω 2 )(− 2ω ) (− 2ω 2 ) = 16 ⋅ ω 6 nx n − 1 1 1 1 1
= + + + ... +
= 16 (ω 3 )2 xn − 1 x − 1 x − α x − α2 x − αn −1
Ta rg e t E x e rc is e s
= 16 (1)2 = 16 Putting x = 2, we get
x =ω −ω −2 2 n2 n − 1 1 1 1 1
85. We have, = + + + ... +
or x + 2 = ω − ω2 2 −1
n
1 2 − α 2 −α 2
2 − αn −1
n −1
On squaring, x 2 + 4 x + 4 = ω 2 + ω 4 − 2ω 3 n ⋅ 2n − 1 1
∴
2n − 1
− 1= ∑ 2 − αi
= ω 2 + ω 3 ⋅ ω − 2ω 3 i =1
n −1
= ω2 + ω − 2 [Qω 2 = 1] 1 n ⋅ 2 n − 1− 2 n + 1
= − 1− 2
Hence, ∑ 2 − αi = 2n − 1
i =1
=−3
(n − 2 )2 n − 1 + 1
⇒ x 2 + 4x + 7 = 0 =
2n − 1
On dividing x 4 + 3 x 3 + 2 x 2 − 11x − 6 by x 2 + 4 x + 7,
we get 92. The vertices of the triangle are A (1, 0 ), B (1 / 2 , 1 / 2 )
x 4 + 3 x 3 + 2 x 2 − 11x − 6 and C (0, 1.)
= ( x 2 + 4 x + 7 ) ( x 2 − x − 1) + 1 ∴ AB 2 = 2 − 2
= (0 ) ( x 2 − x − 1) + 1 BC 2 = 2 − 2
= 0 + 1=1 AC 2 = 1 + 1 = 2
∴ AB = BC
86. We have,
93. By definition,
(1 + ω ) (1 + ω )2 (1 + ω 4 ) (1 + ω 8 ) ... to 2n factors
z1 = x1 + iy1, z2 = x2 + iy2 and z3 = x3 + iy3 are
= (1 + ω ) (1 + ω 2 ) (1 + ω 3 ⋅ ω ) (1 + ω 6 ⋅ ω 2 ) ... to 2n factors
collinear, if
= (1 + ω ) (1 + ω 2 ) (1 + ω ) (1 + ω 2 ) ... to 2n factors x1 y1 1 a b 1
[Qω 3 = ω 6 = ... = 1] x 2 y2 1 = 0 ⇒ a′ b′ 1=0
= [(1 + ω ) (1 + ω ) ... to n factors] x 3 y3 1 a − a′ b − b′ 1
[(1 + ω 2 ) (1 + ω 2 ) ... to n factors]
⇒ ab′ = a′ b
= (1 + ω ) n (1 + ω 2 ) n = [(1 + ω ) (1 + ω 2 )] n
94. Let z = x + iy, where x, y ∈ R
= (1 + ω + ω 2 + ω 3 ) n = (0 + 1) n = 1
∴ α + iβ = tan − 1 ( x + iy )
[Q1 + ω + ω 2 = 0, ω 3 = 1]
⇒ α − iβ = tan − 1 ( x − iy )
87. We have, α = ω and β = ω 2 ⇒ 2α = tan − 1 ( x + iy ) + tan − 1 ( x − iy )
Then, xyz = (a + b + c ) (aω + bω 2 + c ) (aω 2 + bω + c ) 2x
= tan − 1
= (a + b + c ) (a2 + b2 + c 2 − ab − bc − ca) 1 − x2 − y2
166 = a3 + b3 + c 3 − 3 abc ∴ x 2 + y 2 + 2 x cot 2α = 1
[ x + ( y + 2 ) i ][( x + 2 ) − i y ]
95. z − z1 = z − z2 represents a perpendicular bisector.
96. We have, ( z − 1)n = i ( z + 1) n
=
=
( x + 2 )2 + y 2
( x 2 + y 2 + 2 x + 2 y ) + i (2 x + 2 y + 4)
4
Complex Numbers
n n
∴ z −1 = i z + 1
n n
( x + 2 )2 + y 2
⇒ z −1 = z + 1 z + 2i
⇒ z −1 = z + 1 Since, Im =0 ⇒ x+ y+2=0
2 2
z + 2
⇒ z −1 = z + 1
which represents a straight line.
⇒ ( x − 1)2 + y 2 = ( x + 1)2 + y 2
105. We have,
⇒ 4x = 0 ⇒ x = 0
z − 4 i + z + 4 i = 10
∴The roots lies on the Y-axis.
⇒ z − (0 + 4 i ) + z − (0 − 4 i ) = 10
97. Let z = x + iy, where x, y ∈ R This represents an ellipse.
We have, iz − 1 + z − i = 2 4 i + 4 i < 10 i.e. 10 > 8
⇒ ix − y − 1 + x + iy − i = 2
106. We have, z − a i = z + a i
⇒ (− y − 1)2 + x 2 + x 2 + ( y − 1)2 = 2 2 2
⇒ x + i ( y − a) = x + i ( y + a)
⇒ x 2 = 0 or x = 0 ⇒ x 2 + ( y − a)2 = x 2 + ( y + a)2
∴The locus of z is a straight line. ⇒ 4ay = 0; y = 0, which is X-axis.
98. Clearly, z1 < z3 < z2 , as z3 is mid-point of z1 and z2 . 107. We have, z − 1 2
+ z+1
2
=4
1 + z 1 + z π ⇒
2
( x − 1) + iy + ( x + 1) + iy = 4
2
99. (a) Re = 0 ⇒ arg =
1 − z 1 − z 2 ⇒ ( x − 1)2 + y 2 + ( x + 1)2 + y 2 = 4
This represents a circle. ⇒ 2 ( x 2 + 1) + 2 y 2 = 4
(b) z z + iz − iz + 1 = 0 represent a circle. ∴The locus of P is x + y 2 = 1.
2
z − 1 π 108. We have, a1 z 3 + a2 z 2 + a3 z + a4 = 3
(c) arg =
Targ e t E x e rc is e s
z + 1 2 ⇒ 3 = a1 z 3 + a2 z 2 + a3 z + a4
This represents a circle. ⇒ 3 ≤ a1 z 3 + a2 z 2 + a3 z + a4
z −1
(d) = 1⇒ z − 1 = z + 1 ⇒ 3 ≤ a1 z 3 + a2 z 2 + a3 z + a4
z+1 3 2
⇒ 3≤ z + z + z + 1 [Q ai ≤ 1]
This represents a straight line.
2 3 2 3
z + z3 ⇒ 3 ≤ 1 + z + z + z < 1 + z + z + z + ... ∞
100. We have, z2 = 1 2 3
2 ⇒ 3 < 1 + z + z + z + ... ∞
The point representing z2 divided the line segment 1 1
joining points representing z1 and z3 in the ratio 1 : 1. ⇒ 3< ⇒ 1− z <
1− z 3
∴The points lies on a line. 2
⇒ − z <0
101. Area of the triangle on the argand plane formed by 3
3 2 2
complex numbers − z, iz, z − iz is z . ∴ z >
2 3
3 2
∴ z = 600 ⇒ z = 20 109. We have, z − 1 + z + 1 ≤ 4
2
⇒ ( x − 1)2 + y 2 + ( x + 1)2 + y 2 ≤ 4
102. Let z1 = 1 − 3 i , z2 = 4 + 3 i and z3 = 3 + i . Then, z3
divides the line segment joining z1 and z2 in the ratio 2 : 1 where, z = x + iy
internally. So, the points z1, z2 and z3 are collinear. ⇒ = A+ B≤4 …(i)
103. Let z = x + iy where, A = ( x − 1)2 + y 2
z + 2 i x + iy + 2 i x + i ( y + 2 ) and B = ( x + 1)2 + y 2
Then, = =
z+4 x + iy + 4 ( x + 4) + iy But [( x − 1) + y 2 ] − [( x + 1)2 + y 2 ] = − 4 x
2
[ x + i ( y + 2 )][( x + 4) − iy ] A2 − B 2 = − 4 x
= i.e. …(ii)
( x + 4)2 + y 2 On dividing Eq. (ii) by Eq. (i), we get
( x 2 + 4 x + y 2 + 2 y ) + i (2 x + 4 y + 8)
= ( x − 1)2 + y 2 − ( x + 1)2 + y 2 ≤ − x
( x + 4)2 + y 2
⇒ 2 ( x − 1)2 + y 2 ≤ 4 − x
z + 2i
= 0 ⇒ x + y + 4 x + 2 y = 0,
2 2
Since, Re ⇒ 3 x 2 + 4 y 2 ≤ 12
z + 4
which represents a circle with centre (− 2, − 1). [squaring and simplifying]
x2 y2
104. Let z = x + iy or + ≤1
4 3
z + 2 i x + iy + 2 i x + ( y + 2 )i
Then, = = which represents the interior and boundary of an 167
z+2 x + iy + 2 ( x + 2 ) + iy ellipse.
4 110. We have, log1/ 2
z −1 + 4
3 z − 1 − 2
z−1 + 4 1
1
> 1 = log1/ 2
2
118. Let z = α be a real root. Then,
α 3 + (3 + 2 i )α + (− 1 + ia) = 0
⇒ (α 3 + 3 α − 1) − i (a + 2 α ) = 0
Objective Mathematics Vol. 1
⇒ < <1 ⇒ α 3 + 3α − 1 = 0
3 z −1 −2 2
and α = − a / 2
[Qlog a x is a decreasing function, if a < 1] a3 3a
⇒ z −1 + 4< 3 z −1 −2 ⇒ − − − 1= 0
8 2
⇒ 2 z −1 > 6 ⇒ z −1 > 3 ⇒ a + 12 a + 8 = 0
3
which is an exterior of a circle. Let f (a) = a3 + 12 a + 8
111. The closest distance = length of the perpendicular from ∴ f(− 1) < 0, f(0 ) > 0, f(− 2 ) < 0, f(1) > 0 and f(3) = 0
the origin on the line az + az + aa = 0 119. We have,
2
a (0 ) + a 0 + aa a a 1 + i cos θ (1 + i cos θ ) (1 + 2 i cos θ )
= = = =
2 a 2 a 2 1 − 2 i cos θ (1 − 2 i cos θ ) (1 + 2 i cos θ )
(1 − 2 cos 2 θ ) + i 3 cos θ
1 − ak + 1 =
112. We have, zk = 1 + a + a2 + ... + ak = 1 + 4 cos 2 θ
1− a
(1 + i cos θ )
1 ak + 1 Thus, is a real number, if cos θ = 0
⇒ zk − − (1 − 2 i cos θ )
1− a 1− a π
k+1 ⇒ θ = 2 nπ ±
1 a 1 2
⇒ zk − = < [Q a < 1]
1− a 1− a 1− a where, n is an integer.
∴Vertices of the polygon z1, z2 , ... , zn lie within the circle 120. Let z = c be a real root.
1 1 Then, αc 2 + c + α = 0 …(i)
z− =
1− a 1− a Let α = p + iq
z1 + z3 z2 + z4 Then, ( p + iq )c 2 + c + p − iq = 0
113. We have, z1 − z4 = z2 − z3 or = i.e. the
Ta rg e t E x e rc is e s
2 2 ⇒ pc 2 + c + p = 0
diagonals bisect each other. and qc 2 − q = 0
∴It is a parallelogram.
z − z1 π ⇒ c =±1 [Qq ≠ 0]
Also, amp 4 = From Eq. (i), we get α ± 1 + α = 0
z2 − z1 2 Also, c = 1
⇒ Angle at z1 is a right angle.
121. Let z2 = r2 (cos θ 2 + i sin θ 2 )
∴ It is a rectangle and hence, a cyclic quadrilateral.
⇒ z2 = r2
114. We have, Im ( z 2 ) = 4 Also, arg ( z1 ) + arg ( z2 ) = 0
⇒ Im [( x 2 − y 2 ) + 2 ixy ] = 4 [putting z = x + iy ] ⇒ arg ( z1 ) = − arg ( z2 ) = − θ 2
⇒ 2 xy = 4 or xy = 2, which is a rectangular hyperbola. ∴ z1 = r2 [cos (− θ 2 ) + i sin (− θ 2 )]
115. We have, Re ( z 2 ) = 4 = r2 [cos θ 2 − i sin θ 2 ]
⇒ Re [( x 2 − y 2 ) + 2 ixy ] = 4 [putting z = x + iy ] ⇒ z1 = z2
⇒ x2 − y2 = 4 1
⇒ z1 =
which is a rectangular hyperbola. z2
116. z − z1 = z − z2 = z − z3 ∴ z1 z2 = 1
∴ z must be the mid-point of z1 and z2 and since
122. We have,
z − z1 = z − z2 and z, z1 and z2 are collinear. z1 = 15, z2 − 3 − 4i = 5
⇒ z is circumcentre of ∆ formed by z1, z2 and z3 . z2 – 3 – 4 i = 5
z − z1 π B
∴ arg 3 =± [neglecting − ve value]
z3 − z2 2 A
z1 = 15
117. a0 z 4 + a1 z 3 + a2 z 2 + a3 z + a4 = 0
⇒ a0 z 4 + a1 z 3 + a2 z 2 + a3 z + a4 = 0 O
[taking conjugate on both sides]
⇒ a0 ( z 4 ) + a1( z )3 + a2 ( z )2 + a3 ( z )2 + a3 z + a4 = 0 C
∴z is a root of the equation if z is a root. So, option (a) is Minimum value of z1 − z2 ,
correct.
AB = OB − OA = 15 − 10 = 5
Also, if z1 is real, z1 = z1.
Maximum value of z1 − z2 ,
If z1 is non-real complex, then z1 is also a root because CA = OA + OC
imaginary root occurs in conjugate pairs. So, option (b)
= 10 + 15
168 is also correct.
= 25
123. Given, z −
1
z
=1
1 1
126. Given, z = x + iy
Now,
3 z − 6 − 3 i 3( x + iy ) − 6 − 3 i
=
2 z − 8 − 6 i 2( x + iy ) − 8 − 6 i
4
Complex Numbers
⇒ 1≥ z − ⇒ − 1≤ z − ≤1
z z [3 x − 6 + i (3 y − 3)][(2 x − 8) − i (2 y − 6)]
=
⇒
2
− z ≤ z − 1≤ z [(2 x − 8) + i (2 y − 6)][(2 x − 8) − i (2 y − 6)]
2 6 x 2 − 36 x + 48 + 6 y 2 − 24 y + 18 + i [6 xy
From z − 1 ≥ z , we get
− 6 x − 24 y + 24 − 6 xy + 12 y + 18 x − 36
2
z + z − 1≥ 0 =
(2 x − 8)2 + (2 y − 6)2
− 1+ 5
⇒ z ≥ …(i) 6 x 2 + 6 y 2 − 36 x − 24 y + 66
2 =
2 (2 x − 8)2 + (2 y − 6)2
From z − 1 ≤ z , we get
(12 x − 12 y − 12 )
2
z − z − 1≤ 0 + = a + ib [say]
(2 x − 8)2 + (2 y − 6)2
1− 5 1+ 5 π
⇒ ≤ z ≤ …(ii) Since, arg (a + ib) =
2 2 4
From Eqs. (i) and (ii), we get π b
− 1+ 5 1+ 5 ∴ tan = ⇒ a=b
⇒ ≤ z ≤ 4 a
2 2 ⇒ 6 x + 6 y − 36 x − 24 y + 66 = 12 x − 12 y − 12
2 2
5 −1 1+ 5 ⇒ 6 x 2 + 6 y 2 − 48 x − 12 y + 78 = 8
⇒ z min = , z max =
2 2 ⇒ x 2 + y 2 − 8 x − 2 y + 13 = 0 …(i)
124. z12 − z22 = z12 − z22 − 2 z1 z2 Again, z − 3 − i = 3 ⇒ x + iy − 3 + i = 3
⇒ z1 − z2 z1 + z2 = z1 − z2
2
⇒ ( x − 3)2 + ( y + 1)2 = 9
⇒ z1 + z2 = z1 + z2 ⇒ x 2 + y 2 − 6x + 2 y + 1 = 0 …(ii)
z z On subtracting Eq. (ii) from Eq. (i), W get
z1 + z2 = z1 − z2 ⇒ 1 + 1 − 1 − 1
Targ e t E x e rc is e s
z2 z2 − 2 x − 4 y + 12 = 0
z1 ⇒ x = − 2y + 6 …(iii)
⇒ lies on ⊥ bisector of 1 and − 1. Putting the value of x in Eq. (ii), we get
z2
z1 (− 2 y + 6)2 + y 2 − 6 (− 2 y + 6) + 2 y + 1= 0
⇒ lies on imaginary axis.
z2 ⇒ 5 y 2 − 10 y + 1 = 0
z1 10 ± 4 5
⇒ is purely is imaginary. ∴ y=
z2 10
2
z π = 1±
⇒ arg 1 = ± 5
z2 2 4
π ∴ x = − 2y + 6 = 4 m
∴ arg ( z1 ) − arg( z2 ) = 5
2 4 2
∴ z = x + iy = 4 m + i 1 ±
125. Since, arg (( z − 1 − i ) / z ) is the angle subtended by the 5 5
chord joining the points O and 1 + i at the circumference
Thus, the order pairs ( x, y ) satisfying the given
of the circle z − 1 = 1, so it is equal to − π /4. The line
4 2
joining the points z = 0 and z = 2 + 0 i is the diameter. equations such that x = x + iy and 4 − ,1+
5 5
Y
4 2
P(z) and 4 + ,1− .
5 5
127. Clearly, we have to find it for real z. Let z = x.
θ 2 X
O 1 A Then, x − w = x − w2 = w − w2
2 2
1 3 − 1+ 3 i − 1− 3 i
⇒ x + + = − =3
2 4 2 2
1 3
⇒ x+ =±
z −2 π z −2 2 2
∴ arg =± ⇒ is purely imaginary.
z 2 z−0 ⇒ x = 1, − 2
π 128. amp ( z1 z2 ) = 0 ⇒ amp ( z1 ) + amp ( z2 ) = 0
We have, ∠POA =
2 ∴ amp ( z1 ) = − amp ( z2 ) = amp ( z2 )
2 − z π z − 2 AP z1 = z2 , we get z1 = z2 .
⇒ arg = ⇒ = i Q
0 − z 2 2 OP So, z1 = z2
AP 2
z1 z2 = z2 z2 = z2 = 1, because z2 = 1
Now, in ∆OAP, tan θ = Thus, z − 2 = 0 Also, 169
OP
4 129. z + z − 1 = 1⇒ z 2 − z + 1 = 0
∴ z=
1±
2
3i
= − ω, − ω 2
132. Q z1 − 1 = 1, z0 − 1 = 1
Y z –1 = 1
Objective Mathematics Vol. 1
P (z0)
−n
∴z + zn
= (− ω ) n + (− ω)− n or (− ω 2 ) n + (− ω 2 )− n
1
= (− 1) n ω n + (− 1)− n ⋅ n 1
X
ω C
1
or = (− 1) n ω 2 n + (− 1)− n . 2 n
ω Q (z1)
1
= (− 1) n ω n + n
ω
1 z1 − 1
or = (− 1)n ω 2 n + 2 n ∴ =1
ω z0 − 1
1 π
= (− 1) n ω n + n , because ω 3 n = 1 ∠QCP =
ω 2
= (− 1) n ⋅ (ω n + ω 2 n ) z1 − 1 π π
⇒ amp = ,−
= (− 1) n ⋅ (1 + 1) or (− 1) n ⋅ (− 1) z0 − 1 2 2
Dependening on whether n is a multiple of 3 or not. z1 − 1 ±π ± π
∴ = 1⋅ cos + i sin = i, − i
z0 − 1 2 2
130. 2 z1 + z2 ≤ 2 z1 + z2
∴ z1 − 1 = ( z0 − 1) i , z1 − 1 = − ( z0 − 1) i
= 2 z1 + z2 = 2 × 1 + 2 = 4
133. Given, z1 = z2 = z3
From the figure, z1 − z2 is the least, when 0, z1, z2 are
collinear. A(z1)
z1 – z2 z2
z1
Ta rg e t E x e rc is e s
A
O B
1 1
1 O 1
z =1
B(z2) C(z3)
Then, z1 − z2 = 1 ⇒ The vertices are at equal distances from the origin
z = 0.
1 1 ∴The origin is at the centroid of the equilateral triangle.
Again, z2 + ≤ z2 +
z1 z1 z1 + z2 + z3
∴ =0
1 1 3
=2 + =2 + = 3 ∴ z1 + z2 + z3 = 0
z1 1
with OA as real axis,
131. OB = OD = OA = z z1 = 1, z2 = 1 (cos 120 ° + i sin 120 ° ),
B(z2) A(z) z3 = 1 (cos 120 ° − i sin 120 ° )
1 3 1 3
∴ z1 z2 z3 = 1⋅ − + i − − i
2 2 2 2
2
1
2
3
= − +
2 2
=1
C D (z 1 )
1 8 27
134. z2 z3 + 8 z3 z1 + 27 z1 z2 = z1 z2 z3 + +
π z1 z2 z3
If amp ( z ) = θ, then amp ( z1 ) = θ − ,
2 1 8 27
π = z1 z2 z3 + +
amp ( z2 ) = θ + z1 z2 z3
2
and z = z {cos θ + i sin θ} = z1 z2 z3
z1
+
8 z2
+
27 z3
2 2 2
π π z1 z2 z3
∴ z1 = z cos θ − + i sin θ −
2 2 z1 8 z2 27 z3
= z1 z2 z3 + +
= z {sin θ − i cos θ} 1 4 9
= z (− i ) (cos θ + i sin θ ) = − iz = z1 z2 z3 z1 + 2 z2 + 3 z3
π π = z1 z2 z3 z1 + 2 z2 + 3 z3
z2 = z cos θ + + i sin θ + = iz
170 2 2 = 1⋅ 2 ⋅ 3 ⋅ 6 = 36
135. Statement I is false, since there is no order relation in the
set of complex numbers.
Cancellation laws, a + c > b + c
140. We have,
(cos θ + i sin θ )3 / 5 = (cos 3 θ + i sin 3 θ )1/ 5
= [cos (2 rπ + 3 θ ) + i sin (2 rπ + 3 θ )]1/ 5
4
Complex Numbers
⇒ a > b does not hold true in complex numbers, where, r = 0, 1, 2, 3, 4
therefore Statement II is true. 2 πr + 3 θ
i
136. Given that, arg ( z ) = 0 =e , r = 0, 1, 2, 3, 4
5
⇒ z is purely real. Hence, product of all values of (cos θ + i sin θ )3 / 5
3θ 2π + 3θ 6π + 3θ 8π + 3θ
∴ Statement II is true. i i i i
Also, z1 = z2 + z1 − z2 =e e 4 5
e e 5 5
∴ ( z2 − z1 ) = ( z1 − z2 )
2 2 = e i 3 θ + 4 πi = e 4 iπ e i 3 θ
2 2 = cos 3 θ + i sin 3 θ
⇒ = z1 + z2 − 2 z1 z2 cos (θ1 − θ 2 )
2 2
Also, product of roots of the equation x 5 − 1 = 0 is 1.
= z1 + z2 − 2 z1 z2 Hence, Statement II is true, but it is not a correct
⇒ cos(θ1 − θ 2 ) = 1 ⇒ θ1 − θ 2 = 0 explanation of Statement I.
⇒ arg( z1 ) − arg ( z2 ) = 0 141. Let A( z1 ) and B( z1 ) be the centres of the given circles and
z z1
⇒ arg 1 = 0 ⇒ is purely real. P be the centre of the variable circle, which touches
z2 z2 given circles externally, then
z AP = a + r and BP = b + r
⇒ Im 1 = 0
z2 where, r is the radius of the variable circle.
Hence, Statement I and Statement II are true and On subtraction, we get
Statement II is correct explanation of Statement I. AP − BP = a − b
137. We will show that Statement I is true and follows from ⇒ AP − BP = | a − b is a constant.
Statement II. Hence, locus of P is
Indeed (i) a right bisector of AB, if a = b
Targ e t E x e rc is e s
1 1 1 1 1
z = z+ − ≤ z+ + ≤ 1+ (ii) a hyperbola if a − b < AB = z2 − z1
z z z z z
2 (iii) an empty set, if a − b > AB = z2 − z1
⇒ z − z − 1≤ 0 (iv) set of all points on line AB except those which lie
1 − 5 1 + 5 between A and B, if a − b = AB ≠ 0
⇒ z ∈ ,
2 2 Thus, Statement I is false and Statement II is true.
1 + 5 142. If P( z ) is any point on the ellipse, then equation of the
But as z > 0, we have z ∈ 0,
2 ellipse is
z1 − z2
1+ 5 z − z1 + z − z2 = …(i)
⇒ Maximum value of z is . e
2
For P( z )to lie in ellipse, we have
−a±
a2 − 4 b z1 − z2
138. Statement I is false, since z = , if a2 > 4b z − z1 + z − z2 <
2 e
and z is a positive number c, then z = c
It is given that origin is an interior point of the ellipse.
⇒ z = c (cos θ + i sin θ ) z − z2
⇒ Infinite complex numbers satisfy the given equation. ⇒ 0 − z1 + 0 − z2 < 1
e
Statement II is true. z − z2
A quadratic can have more than two roots, if all ∴ e ∈ 0, 1
z1 + z2
coefficients are zero.
Hence, Statement I is false and Statement II is true.
139. Q e iθ = cos θ + i sin θ
⇒ e − iθ = cos θ − i sin θ 143. As, we know z − z1 + z − z2 = k represents an ellipse,
e iθ + e − iθ if k > z1 − z2 . Thus, z − i + z + i = k represents an
∴ cos θ = ellipse, if k > i + i or k > 2.
2
− (1 + i )
e i(1 − i) + e − i(1− i) e (1 + i) + e ∴ Statement I is false and Statement II is true.
Now, cos (1 − i ) = =
2 2 144. The equation can be rewritten as
e (e i ) + e − 1(e − i ) zz + az + az + aa = aa − λ
=
2 ⇒ ( z + a) ( z + a) = aa − λ
e (cos 1 + i sin 1) + e − 1 (cos 1 − i sin 1) ⇒ z + a = aa − λ
=
2 Since, aa is real, 1 should be real.
1 1 i 1
= e + cos 1 + e − sin 1 aa − λ represents radius of the circle.
2 e 2 e
1 1 1 1 Hence, Statement I and Statement II both are true and
∴ a = e + cos 1, b = e − sin 1
2 e 2 e Statement II is the correct explanation for Statement I.
171
4 Solutions (Q. Nos. 145-147)
Given that,
| z1 + z2|2 = | z1| + | z2|
2 2
152. DE =
( zA − zC )i
1− i
=
1 − ω2
2
=
2
3
Objective Mathematics Vol. 1
⇒| z1| + | z2| + z1 z2 + z1 z2
2 2
= | z1|2 + | z2|2 zB − zC i ω − ω 2 i
153. zE = =
⇒ z1 z2 + z1 z2 =0 …(i) 1− i 1− i
z1 z (ω + ω 2 ) + i (ω − ω 2 ) − 1 − 3
⇒ + 1 =0 [dividing by z2 z2 ] = =
z2 z2 2 2
z1 z1 1+ 3 3 + 3
⇒ + =0 …(ii) ∴ AE = 1 + =
z2 z2 2 2
z 2 − 1
145. From Eq. (i),z2 z2 is purely imaginary. 154. A. arg = 0;z ≠ ± i
z 2 + 1
146. From Eq. (ii), z1 / z2 is purely imaginary z2 − 1 z 2 − 1
= =
147. Also, i ( z1 / z2 ) is purely real. Hence, its possible z2 + 1 z 2 + 1
arguments are 0 and π. ⇒ z − z = 0, z + z = 0, y = 0, x = 0
148. Clearly, according to the least possibility Locus of z is portion of pair of lines xy = 0
α 2 − 7α + 11 ≤ 1 ⇒ α ∈[2, 5] z 2 − 1
Q 2 > 0
149. ( z − i ) − (α 2 − 7α + 11 ) = 1 z + 1
B. Given,
∴ ( z − i ) − (α 2 − 7α + 11) ≥ z − i − α 2 − 7α + 11
z − cos − 1 cos 12 − z − sin − 1 sin 12| = 8 (π − 3)
[using z1 − z2 ≥ z1 − z2 ]
Since, cos − 1 cos 12 − sin − 1 sin 12 = 8 (π − 3)
⇒ α 2 − 7α + 11 + 1 ≥ z − i ∴ Locus of z is portion of a line joining z1 and z2
[since, α 2 − 7α + 11 ≥ − 5/ 4] except the segment between z1 and z2 .
2
∴ z − i ≤ 1 + 5/ 4 or z − i ≤ 9/ 4 C. z 2 − i z1 = k2 − k1
Ta rg e t E x e rc is e s
Y ∴ x 2 − y 2 + 2 ixy − iλ 1 = λ 2
λ1
⇒ x2 − y2 = λ 2 and xy =
2
∴ Locus of z is point of intersection of hyperbola.
1 1 π
D. Given, z − 1 − sin − 1 + z + cos − 1 − =1
3 3 2
X 1 1 π
since, 1 + sin − 1 + cos − 1 − =1
3 3 2
150. AB =
9
and arg ( z ) = π − tan −1 4 ∴ z − z1 + z − z2 = z1 + z2
4 9 ⇒ Locus of z is the segment joining z1 and z2 .
Y
155. There are no real roots of the equation z 6 − 6 z + 20 = 0.
If x + iy is a root, then x − iy is also a root.
Let the roots be x1 ± iy1, x2 ± iy2 , x3 ± iy3
B O A (0, 1) Sum of the roots = 2 ( x1 + x2 + x3 ) = 0
– 9 ,1 – 5 ,1 ⇒ x1 + x2 + x3 = 0
4 4 X ⇒ One of x1, x2 , x3 is negative and other two are
positive or vice-versa.
zB − zA i z − zC i ⇒ The number of roots in each of the quadrant is either
151. zD = , zE = B
1− i 1− i 1 or 2.
A 156. A. z 4 − 1 = 0 ⇒ z 4 = 1 = cos 0 + i sin 0
⇒ z = (cos 0 + i sin 0 )1/ 4
D = cos 0 + i sin 0
B C B. z 4 + 1 = 0 ⇒ z 4 = − 1 = cos π + i sin π
⇒ z = (cos π + i sin π)1/ 4
π π
E = cos + i sin
4 4
π π
C. iz + 1 = 0 ⇒ z = i = cos + i sin
4 4
2 2
1/ 4
π π
∴Angle between AC and DE ⇒ z = cos + i sin
2 2
z − zA z − zA 1 − i π
= arg C − arg C = π π
zE − zD zA − zC i 4 = cos + i sin
172 8 8
D. iz 4 − 1 = 0
⇒
π
z 4 = − i = cos − i sin π /2
2
⇒
⇒
9 (36 + y 2 ) = 25 [36 + ( y − 8)2 ]
y = 17, 8
Thus, the required numbers are z = 6 + 17 i , 6 + 8 i.
4
Complex Numbers
1/ 4
π Hence, the value of Re ( z ) is 6.
⇒ z = cos − i sin π /2
2 1 1 1 1
π π 160. z + ≥ z − = z − = z − as z ≥ 3
= cos − i sin z −z z z
8 8
1 1
157. A. Let z = 1 − z ⇒ z 2 + z + 1 = 0 Let f ( x ) = x − for x ≥ 3,f ' ( x ) = 1 + 2 > 0
x x
⇒ z = ω , ω2 ⇒ f ( x ) is increasing function, so f ( x )min at x = 3 is
B. Here, z1 + z2 + z3 + .... + z6 = 0 ... (i) 1 1 8
z+ = z − =
z2 = z1 e ( i2 π )/ n, if e ( i2π )/ n = α ⇒ z2 = z1 α z min z 3
z =3
Similarly, z3 = z1α 2 , z4 = z2 α 3 , z5 = z2 α 4 , z6 = z1α 5
Hence, λ is equal to 8.
On squaring and adding and then using Eq. (i), we
get 161. Consider
z12 + z22 + z32 + z42 + z52 + z62 = 0 z4 + z3 + 2 z2 + z + 1 = 0
⇒ z + z3 + z2 + z2 + z + 1 = 0
4
n( n + 1) 1/ n
1 + 2 ω + 3 ω 2 + 4 ω 3 + ... + nω n − 1 ⇒ z ( z 2 + z + 1) + ( z 2 + z + 1) = 0
2
C. ≥ n ! ω 2
n
⇒ ( z 2 + z + 1) ( z 2 + 1) = 0
[Q AM ≥ GM] ∴ z = i, − i, ω, ω 2, for each z = 1
Now,
162. Let A be ( x, y ).
E = 1 + 2 ω + 3 ω 2 + 4 ω 3 + ... + nω n − 1 + ....
D(1,1)
ω + 2 ω 2 + 3 ω 3 + ... + (n − 1)ω n − 1 + n ω n
ωE =
(1 − ω)E = 1 + ω + ω 2 + ... + ω n − 1 − n ω n
Targ e t E x e rc is e s
0 − n(1)
⇒ E=
1− ω A M (2, –1) C
n (x, y)
∴ E=
ω −1
D. Let x 2 + y 2 = 1
where, z1 − i , z2 = 1, z3 = − 1 B
∴ z=−i
It is given that, BD = 2 AC
⇒ zz1 + z2 z3 = 1 − 1 = 0
⇒ MD = 2 AM
158. Consider, Also, DM is perpendicular to AM.
x 3 2 + 11 i 3 − 4 i 50 + 25 i
x= = × = ⇒ (1 − 2 )2 + (1 + 1)2 = 4 [( x − 2 )2 + ( y + 1)2 ] …(i)
x2 3 + 4i 3 − 4i 25
y + 1 1+ 1
=2 + i and ⋅ = −1
x − 2 1− 2
∴ a + b = 2 + 1= 3
⇒ 2 ( y + 1) = x − 2
z−4 x − 4 + iy
159. Here, =1 ⇒ =1 With x − 2 = 2( y + 1) , Eq. (i) can be written as
z−8 x − 8 + iy 1
( y + 1)2 =
⇒ ( x − 4)2 + y 2 = ( x − 8)2 + y 2 4
⇒ x=6 1 3
⇒ y =− ,−
z − 12 5 2 2
With x = 6, =
z − 8i 3 ⇒ x = 3, 1
⇒ y 2 − 25 y + 136 = 0 ∴ λ1 + λ 2 = 4
173
4 Entrances Gallery
1. A. zk is 10th root of unity ⇒ zk will also be 10th root of 4. Area of
Objective Mathematics Vol. 1
unity. Take z j as zk . S = S1 ∩ S 2 ∩ S 3 y + √3 x = 0 Y
z
B. z1 ≠ 0, take z = k , we can always find z. π × 42 42 × π
z1 = +
4 6 60°
C. z10 − 1 = ( z − 1)( z − z1 ) K ( z − z9 ) 2 1 1 X′ X
=4 π + 60°
⇒ ( z − z1 )( z − z2 ) K ( z − sz9 ) = 1 4 6 x2 + y2 < 16
+ z + z 2 + K + z 9 , ∀ z ∈ complex number 20 π
=
Put z = 1 3 Y′
(1 − z1 )(1 − z2 ) K (1 − z9 ) = 10 5. Distance of (1,
D. 1 + z1 + z2 + K + z9 = 0 − 3) from y + 3x = 0
⇒ Re (1) + Re ( z1 ) + K + Re ( z9 ) = 0 −3+ 3 ×1 3− 3
⇒ Re (z1) + Re (z2 )|+ K + Re (z9 ) = − 1 > >
2 2
9
2 kπ
∴ 1 − ∑ cos =2 6. Given, is z 2 + z + 1 − a = 0
k =1
10
C(1/α) Clearly, this equation do not have real roots, if
2. OB =|α| Y D<0
B (α) 2r
OC =
1
=
1 ⇒ 1 − 4 (1 − a) < 0
|α| |α| r
⇒ 4a < 3
3
In ∆OBD, ∴ a<
A D 4
| z |2 + |α|2 − r 2
cos θ = 0
2| z0||α | θz 5
0 7. Length AB =
2
In ∆OCD,
Ta rg e t E x e rc is e s
1 Y
| z0|2 + − 4r 2 O X
|α|2
cos θ =
1
2| z0| C (3, 2)
|α|
1
| z0|2 + − 4r 2 X¢ X
| z0|2 + |α|2 − r 2 |α | 2 (0, 0) O B (3, 0)
⇒ =
2| z0||α | 2| z0|
1 A(3, –5/2)
|α |
1
⇒ |α | =
7 Y¢
iπ
3+i ∴ Minimum value = 5
3. w = =e 6
2 8. The expression may not attain integral value for all a, b,c.
If we consider a = b = c , then
B1 A3 x = 3a
y = a (1 + ω + ω 2 ) = 0
B2 π/6 A2 z = a (1 + ω 2 + ω ) = 0
O ∴ | x|2 + | y|2 + | z|2 = 9| a|2
B3 A1
| x|2 + | y|2 + | z|2 9
∴ = =3
x = –1/2 x = 1/2 | a|2 + | b|2 + |c|2 3
2 2 2
nπ
i
As, x + y + z
6
So, wn = e = (a + b + c )(a + b + c ) + (a + b ω + cω 2 )
nπ 1 (a + b ω 2 + c ω ) + (a + bω 2 + cω )(a + b ω + c ω 2 )
Now, for z1, cos >
6 2
nπ 1 = 3 (| a|2 +| b|2 +|c |2 ) + ab (1 + ω + ω 2 ) + ab (1 + ω + ω 2 )
and for z2 , cos <− + bc (1 + ω + ω 2 ) + b c (1 + ω + ω 2 ) + ac (1 + ω + ω 2 )
6 2
Possible position of z1 are A1, A2 , A3 , whereas of z2 are + ac (1 + ω + ω 2 )
B1, B2 , B3 (as shown in the figure). = 3 (| a|2 +| b|2 +|c |2 ) [using 1 + ω + ω 2 = 0 ]
So, possible value of ∠z1Oz2 according to the given | x| +| y| +| z|
2 2 2
2π 5π ∴ =3
options is or . | a|2 +| b|2 +|c |2
3 6
174
9. Given, z = (1 − t )z1 + tz2
⇒
z − z1
z2 − z1
=t
⇒ (| z2|2 − 1)(| z1|2 − 4) = 0
Q
∴
| z2| ≠ 1
| z1| = 2
4
Complex Numbers
Let z1 = x + iy ⇒ x 2 + y 2 = (2 )2
z − z1
⇒ arg =0 ...(i) ∴ Point z1 lies on a circle of radius 2.
z2 − z1
⇒ arg ( z − z1) = arg ( z2 − z1 ) 12. Q | z| ≥ 2
z − z1 z − z1 1 1 1 3
⇒ = ∴ z+
≥ | z| − ≥2− ≥
z2 − z1 z2 − z1 2 2 2 2
z − z1 z − z1 1 3
⇒ =0 Hence, minimum distance between z and − , 0 is .
z2 − z1 z2 − z1 2 2
P(z) 13. Given,| z| = 1, arg z = θ
A(z1) B(z2) ∴ z = e ie
1
AP + PB = AB But z=
z
⇒ | z − z1| + | z − z2| = | z1 − z2|
1+ z
z z ∴ arg = arg( z ) = θ
10. A. −i = + i, z ≠ 0
| z| | z| 1 + 1
z
z
is unimodular complex number
| z| z2
14. Given, a complex number ( z ≠ 1) is purely real.
and lies on perpendicular bisector of i and − i z −1
z
⇒ = ± 1⇒ z = ± 1| z| ⇒ a is real number To find the locus of the complex number z.
| z| z2
Since, ( z ≠ 1) is purely real.
⇒ Im(z) = 0 z −1
B. | z + 4| + | z − 4| = 10
Targ e t E x e rc is e s
z2 z2
z lies on an ellipse, whose focus are (4, 0) and ∴ =
z −1 z −1
(− 4, 0) and length of major axis is 10.
⇒ 2 ae = 8 and 2 a = 10 ⇒ z 2 ( z − 1) = z 2 ( z − 1)
⇒ e = 4/ 5 ⇒ z2 z − z2 = z 2 z − z 2
|Re( z )| ≤ 5. ⇒ zzz − z 2 = zzz − z 2
C. ∴|ω| = 2 ⇒ w = 2(cos θ + i sin θ ) ⇒ z| z|2 − z 2 = z| z|2 − z 2
1
x + iy = 2(cos θ + i sin θ ) − (cos θ − i sin θ ) On rearranging the terms, we get
2
3 5 z| z|2 − z| z|2 = z 2 − z 2
= cos θ + i sin θ ⇒ | z|2 ( z − z ) = ( z − z )( z + z )
2 2
x2 y2 ⇒ | z| ( z − z ) − ( z − z )( z + z ) = 0
2
⇒ + =1
(3 / 2 )2
(5 / 2 )2 ⇒ ( z − z )(| z|2 − ( z + z )) = 0
e2 = 1−
9/ 4 Either ( z − z ) = 0
25/ 4 or [| z|2 − ( z + z )] = 0
9 16
= 1− = Now, z = z
25 25
4 ⇒ Locus of z is real axis and
⇒ e= (| z|2 − ( z + z )) = 0 ⇒ zz − ( z + z ) = 0
5
D. |ω| = 1 Locus of z is a circle passing through the origin.
⇒ x + iy = cos + i sin θ + cos θ − i sin θ Aliter
x + iy = 2 cos θ Put z = x + iy, then
|Re( z )| ≤ 1,|Im( z )| = 0 z2 ( x + iy )2 ( x 2 − y 2 ) + i (2 xy )
= =
11. Given, z2 is not unimodular i.e.| z2| ≠ 1 z − 1 ( x + iy ) − 1 ( x − 1) + iy
z1 − 2 z2 ( x 2 − y 2 ) + i (2 xy ) ( x − 1) − iy
and is unimodular. = ×
2 − z1 z2 ( x − 1) + iy ( x − 1) − iy
z1 − 2 z2 z2
⇒ = 1 ⇒ | z1 − 2 z2|2 = |2 − z1 z2|2 Since, ( z ≠ 1) is purely real, hence its imaginary
2 − z1 z2 z −1
part should be equal to zero.
⇒ ( z1 − 2 z2 )( z1 − 2 z2 ) = (2 − z1 z2 )(2 − z1 z2 )
⇒ ( x 2 − y 2 )(− y ) + (2 xy )( x − 1) = 0
[Q zz = | z|2 ]
⇒ y ( x2 − y2 + 2 x − 2 x2 ) = 0
⇒ | z1|2 + 4| z2|2 − 2 z1 z2 − 2 z1 z2
⇒ y( x 2 + y 2 − 2 x ) = 0
= 4 + | z1|2 | z2|2 − 2 z1 z2 − 2 z1 z2 175
Either y = 0 or x + y 2 − 2 x = 0
2
4 Now, y = 0, locus of z is real axis, and
x 2 + y 2 − 2 x = 0, locus of z is a circle passing through
the origin.
Ø For minimum value of | z |, we take
4 4
z z
z−
− ≤ z −
4 4
+
z z
Objective Mathematics Vol. 1
Locus of z is either real axis or a circle passing through
the origin. from which we get | z| ≥ 5 − 1
15. Let z = x + iy, given Re ( z ) = 1 1
20. Let z =
∴ x = 1⇒ z = 1 + iy i −1
Since, complex roots are conjugate of each other. 1 1 1
∴ z= = =−
∴z = 1 + iy and 1 − iy are two roots of i − 1 − i −1 i +1
z2 + α z + β = 0
21. From the argand Y
Product of roots = β
⇒ (1 + iy )(1 − iy ) = β diagram, maximum
∴ β = 1 + y 2 ≥ 1 ⇒ β ∈ (1, ∞ ) value of| z + 1| is 6.
O
X¢ X
16. Q(1 + ω ) = A + Bω, we know 1 + ω + ω = 0
7 2 Aliter (–7, 0) (–4, 0)(–1, 0)
∴ 1 + ω = − ω2 | z + 1| = | z + 4 − 3|
≤ | z + 4| + |− 3| ≤ 6
⇒ (− ω 2 )7 = A + Bω
Thus, maximum value Y¢
⇒ − ω14 = A + Bω [Q ω14 = ω12 ⋅ ω 2 = ω 2 ]
of| z + 1| is 6.
⇒ − ω 2 = A + Bω 10
2 kπ 2 kπ
⇒
On comparing
1 + ω = A + Bω 22. ∑ sin 11
+ i cos
11
k =1
A = 1, B = 1 10
2 kπ 2 kπ
= i ∑ cos − i sin
17. We have,| z − 1| = | z + 1| = | z − i| 11 11
k =1
10 2 kπ
(0, 1)
= i ∑ e 11
Ta rg e t E x e rc is e s
k = 1
(–1, 0) (1, 0) 10 − 2 kπ
= i ∑ e 11 − 1 = − i
O
k = 0
23. Given, equation is z 2 + z + 1 = 0
z+i ⇒ z = ω, ω 2
2 2 2
1 1 1
Clearly, z is the circumcentre of the triangle formed by Now, z + + z 2 + 2 + z 3 + 3
z z z
the vertices (1, 0), (0, 1) and (− 1, 0), which is unique.
2 2 2
Hence, the number of complex number z is one. 1 1 1
+ z4 + 4 + z5 + 5 + z6 + 6
z z z
18. Since, α and β are roots of the equation x 2 − x + 1 = 0.
−3 2
1± 3i = (ω + ω ) + (ω + ω ) + (ω + ω )
2 2 2 2 3
⇒ α + β = 1, αβ = 1 ⇒ x=
2 + (ω + ω 2 )2 + (ω 2 + ω )2 + (ω 6 + ω −6 )2
1+ 3i 1 − 3i = (− 1) + (− 1) + (1 + 1)2 + (− 1)2 + (− 1)2 + (1 + 1)2
2 2
⇒ x= or
2 2 = 1 + 1 + 4 + 1 + 1 + 4 = 12
⇒ x = −ω or − ω2
24. Given that, ( x − 1)3 + 8 = 0
Thus, α = − ω 2, then β = − ω
⇒ ( x − 1)3 = (− 2 )3
or α = − ω , then β = − ω 2 [where, ω 3 = 1] 3
Hence, α 2009
+ β 2009 = (− ω )2009 + (− ω 2 )2009 x − 1
⇒ =1
−2
= − [(ω 3 )669 ⋅ ω 2 + (ω 3 )1337 ⋅ ω ]
x − 1
= − [ω 2 + ω ] = − (− 1) = 1 ⇒ = (1)
1/ 3
−2
19. | z| = z − +
4 4 x − 1
are 1, ω, ω .
2
Cube roots of
z z −2
4 4 Cube roots of ( x − 1) are − 2, − 2ω and − 2ω 2 .
⇒ | z| ≤ z − +
z | z| ∴Cube roots of x are − 1, 1 − 2ω and 1 − 2ω 2 .
4
⇒ | z| ≤ 2 + 25. Let z1 = x1 + iy1 and z2 = x2 + iy2
| z|
⇒ | z| − 2| z| − 4 ≤ 0
2 Given,| z1 + z2| = | z1| + | z2|
⇒ | z| ≤ 5 + 1 ∴ ( x1 + x2 )2 + ( y1 + y2 )2
176 ∴Maximum value of| z| is 5 + 1 . = x12 + y12 + x22 + y22
On squaring both sides,
= x12 + x22 + 2 x1 x2 + y12 + y22 + 2 y1 y2
= x12 + y12 + x22 + y22 +2 ( x12 + y12 )( x22 + y22 )
and
∴
x
y
q
y
= − 3 p2 + q 2
+ = − 2 p2 − 2q 2
4
Complex Numbers
p q
⇒ x1 x2 + y1 y2 = ( x12 + y12 )( x22 + y22 ) 1 x y
⇒ + = −2
( p2 + q 2 ) p q
Again, squaring both sides, we get
⇒ = x12 x22 + y12 y22 + 2 x1 x2 y1 y2 29. Using the relation, if
= x12 x22 + y12 y22 + x12 y22 + y12 x22 | z1 + z2| = | z1| + | z2|
Then, arg( z1 ) = arg( z2 )
⇒ ( x1 y2 − y1 x2 )2 = 0
Since, | z + (− 1)| = | z 2| + |− 1|
2
y1 y2
⇒ = Then, arg( z 2 ) = arg(− 1)
x1 x2
⇒ 2 arg( z ) = π [Q arg(− 1) = π]
y y π
⇒ tan −1 1 = tan −1 2 ⇒ arg( z ) =
x1 x2 2
⇒ arg ( z1 ) = arg( z2 ) ⇒ z lies on Y-axis (imaginary axis).
∴ arg( z1 ) − arg( z2 ) = 0 30. Since, origin, z1 and z2 are the vertices of an equilateral
Aliter triangle, then
Given, | z1 + z2| = | z1| + | z2| z12 + z22 = z1 z2
= | z1|2 + | z2|2 + 2 Re( z1 z2 ) ⇒ ( z1 + z2 )2 = 3 z1 z2 ...(i)
[squaring both sides] Again, z1, z2 are the roots of the equation
= | z1|2 + | z2|2 + 2| z1|| z2| z 2 + az + b = 0
⇒ Re ( z1 z2 ) = | z1|| z2| Then, z1 + z2 = − a and z1 z2 = b
⇒ | z1|| z2|cos(θ1 − θ 2 ) = | z1|| z2|
On putting these values in Eq. (i), we get
⇒ θ1 − θ 2 = 0 (− a)2 = 3b ⇒ a2 = 3b
Targ e t E x e rc is e s
∴ arg ( z1 ) − arg( z2 ) = 0
z 31. Let z = r1 e iθ ⇒ w = r2 e iφ ⇒ z = r1 e iθ
26. Given, w = and| w| = 1
i Given, | zw| = 1
z−
3 ⇒ | r1 e iθ ⋅ r2 e iφ| = 1
⇒ r1r2 = 1 ...(i)
z i π
⇒ = 1 ⇒ | z| = z − and arg( z ) − arg(w ) =
i 3 2
z−
3 π
⇒ θ−φ= ...(ii)
1 2
⇒ z lies on perpendicular bisector of (0, 0) and 0, .
3 Now, zw = r1 e − i θ ⋅ r2 e i φ
Hence, z lies on a straight line. = r1r2 e − i( θ − φ )
27. Given, z + i w = 0 = 1⋅ e iπ / 2 [from Eqs. (i) and (ii)]
π π
⇒ z = − iw = cos − i sin
⇒ z = iw 2 2
⇒ w = − iz ∴ zw = − i
and arg( zw ) = π 1+ i
x
(1 + i )(1 + i )
x
⇒ arg(− iz 2 ) = π 32. Now, =
1− i (1 − i )(1 + i )
⇒ arg(− i ) + 2 arg( z ) = π x
π π (1 + i )2 1 − 1 + 2 i
x
= =
⇒ − + 2 arg( z ) = π Q arg(− i ) = − 2
2 2 1− i 2
3π 1+ i
x
∴ arg( z ) = ⇒ = (i ) = 1
x
4 [given]
1− i
28. Since, z1/ 3 = p + iq ⇒ (i )x = (i )4 n
∴ z = ( p + iq ) 3
where, n is any positive integer.
= p3 − iq 3 + 3ip2q − 3 pq 2 ∴ x = 4n
Given that, z = x − iy
∴ x − iy = p3 − 3 pq 2 + i (3 p2q − q 3 ) 33. Now, (1 + ω − ω 2 )7 = (− ω 2 − ω 2 )7 [Q1 + ω + ω 2 = 0]
⇒ x = p3 − 3 pq 2 = (− 2ω )2 7
and − y = 3 p2q − q 3 = − 2 7 ⋅ ω14
⇒
x
= p2 − 3 q 2 = − 128 (ω 3 )4 ω 2
p = − 128 ω 2 [Qω 3 = 1]
177
4 34. Given,|β| = 1
∴
β −α
=
β −α
1 − αβ ββ − αβ
[Q1 = ββ]
39. Let z1 = x + iy and z2 = p − iq, where x, q > 0
Given,| z1| = | z2|
⇒ x 2 + y 2 = p2 + q 2
Objective Mathematics Vol. 1
...(i)
z1 + z2 ( x + p) + i ( y − q )
1 β −α 1 ∴ =
= = × 1= 1 [Q| z| = | z|] z1 − z2 ( x − p) + i ( y + q )
|β| β − α (1)
− 2 i ( xq + yp)
= ...(ii)
( 3 + i )3 (3 i + 4)2 ( x − p)2 + ( y + q )2
35. Given, z =
(8 + 6 i )2 If xq + yp ≠ 0, then it is purely imaginary and if
( 3 + i 3 )(3 i + 4)2 x y
∴ | z| = xq + yp = 0 or =− =λ [say]
(8 + 6 i )2 p q
⇒ x = pλ, y = − qλ
|( 3 + i 3 )||(3 i + 4)2| | z1| | z1|
= Q | z | = | z | From Eq. (i), we get
|8 + 6 i|2 2 2 p2 + q 2 = λ2 ( p2 + q 2 )
| 3 + i 3||3 i + 4|2 ⇒ λ2 = 1
= [Q| z n| = | z|n ]
|8 + 6 i|2 ⇒ λ=±1
( 3 + 13 )( 9 + 16 )2 (2 )3 (5)2 10 2 ⋅ 2 For λ = − 1 and z1 ≠ z2 , but| z1| = | z2|
= = = =2 In this case, Eq. (ii) is zero.
( 64 + 36 )2 (10 )2 (10 )2
w −1 40. Given, z = e 2 π / 3
36. Given,|w | = 1and z = ∴ 1 + z + 3z2 + 2 z3 + 2 z4 + 2 z4 + 3z5
w+1
⇒ z(w + 1) = w − 1 ⇒ w( z − 1) = − 1 − z = 1 + e 2 π / 3 + 3 (e 2 π / 3 )2 + 2(e 2 π / 3 )3
1+ z 1+ z + 2(e 2 π / 3 )4 + 3 (e 2 π / 3 )5
⇒ w= ⇒|w | =
1− z 1− z 2π 4π
= 1 + cos + i sin
|1 + z| 3 3
⇒ 1= ⇒ |1 − z| = |1 + z|
Ta rg e t E x e rc is e s
|1 − z| 4π 4π
+ 3 cos + i sin
On squaring both sides, we get 3 3
12 + | z|2 − 2 Re ( z ) = 12 + | z|2 − 2 Re ( z ) 8π 8π
+ 2[cos 2 π + i sin 2 π ] + 2 cos + i sin
3 3
⇒ 4 Re ( z ) = 0
∴ Re ( z ) = 0 10 π 10 π
+ 3 cos + i sin
3 3
37. Let z = x + iy
∴ | z|2 + | z − 3|2 + | z − i|2 = 1 + [cos 120 ° + i sin 120 ° ] + 3 [cos 240 °
+ i sin 240 ° ] + 2[cos 360 ° + i sin 360 ° ]
= | x + iy|2 + |( x − 3) + iy|2 + | x + i ( y − 1)|2
+ 2[cos 480 ° + i sin 480 ° ] + 3 [cos 600 ° + i sin 600 ° ]
= x 2 + y 2 + ( x − 3)2 + y 2 + x 2 + ( y − 1)2 1 3 1 3
= x 2 + y 2 + x 2 − 6x + 9 + y 2 + x 2 + y 2 + 1 − 2 y = 1 + − + i + 3 − − i
2 2 2 2
= 3 x 2 + 3 y 2 − 6 x − 2 y + 10
+ 2[1 + 0 ] + 2[cos 120 ° + i sin 120 ° ]
2 1 1
= 3 ( x 2 − 2 x + 1) + 3 y 2 − y + + 10 − 3 − + 3 [cos 120 ° − i sin 120 ° ]
3 9 3
2 1 3 1 3
1 20 = 3 − 2 − 2 + 2 − + i + 3 − − i
= 3 ( x − 1)2 + 3 y − + 2 2 2 2
3 3
1 −5 3 3 3 3
It is minimum, when x − 1 = 0 and y − =0 = 1 − 3i + − i=− − i
3 2 2 2 2
1 1 3
∴ x = 1 and y = = − 3 + i = − 3 e πi / 3
3
2 2
1
∴ z = 1+ i
3 41. Given, ( x − 1)3 = − 5
1 i ( x − 1) = − 5, − 5ω, − 5 ω 2
2 +
1+ i 2 2 ⇒ x = 1 − 5, 1 − 5 ω, 1 − 5 ω 2
38. =
π π e − iπ / 4 ⇒ x = − 4, 1 − 5 ω, 1 − 5 ω 2
cos − i sin
4 4
z − 3i
π π 42. Given, =1
2 cos + i sin z + 3i
4 4
= ⇒ | z − 3 i| = | z + 3 i|
e − iπ / 4
[if| z − z1| = | z − z2|, then it is perpendicular
= 2e iπ / 4 + iπ / 4 = 2e 2 iπ / 4 = 2e iπ / 2
bisector of z1 and z2 ]
π π
= 2 cos + i sin Hence, perpendicular bisector of (0, 3) and (0, − 3) is
178 2 2 X-axis.
43. ( z3 − z1 ) = ( z2 − z1 )(cos 90 ° + i sin 90 ° )
⇒
⇒
( z3 − z1 ) = i ( z2 − z1 )
( z3 − z1 )2 = − ( z2 − z1 )2
= 8 (1 − 1 + 2 i ) [1 − 3 3 i + 3 3 i (1 + i 3 )]
= 8 (2 i )[1 − 9]
= 16 i (− 8) = − 128 i
4
Complex Numbers
⇒ ( z3 − z1 )2 + ( z2 − z1 )2 = 0 49. 1 + i 3 + 3 i (1 + i ) + 1 − i 3 − 3 i (1 − i )
= 2 + 3i + 3i2 − 3i + 3i2
Aliter
Let z1 = 0, z2 = 1 − i and z3 = 1 + i = 2 + 6i2
Now, ( z1 − z2 )2 + ( z1 − z3 )2 = [0 − (1 − i )]2 + [0 − (1 + i )]2 =2 − 6 = − 4
= (1 − i )2 + (1 + i )2 i ( 3 + i )4 i (2 + 2 3 i )2 i (− 2 + 2 3 i )
50. z = = . =
= 12 + i 2 − 2 i + 12 + i 2 + 2 i 4(1 − 3 i ) 2
4 −2 −2 3i 2(− 1 − 1 3 i )
= 1 − 1 + 1− 1 = 0 i (− 2 + 3 i ) (− 1 + i 3 )
= ×
44. We know that, (− 1 − 3 i ) (− 1 + i 3 )
| z − z1| + | z − z2| = k will represent an ellipse, if i (− 2 − 2 i 3 ) 3 1
| z1 − z2| < k. = = −i
Hence, the equation | z − z1| + | z − z2| = 2| z1 − z2|, 4 2 2
represents an ellipse. π
∴ Amplitude = −
45. Given, z1 ≠ ± 1 6
Since, z2 and z3 can be obtained by rotating vector 51. Q| z1| = | z2| = | z3| = 1
2π 4π ⇒ | z1|2 = | z2|2 = | z3|2 = 1
representing through and , respectively.
3 3
∴ z1 z1 = z2 z2 = z3 z3 = 1
∴ z2 = z1ω 1 1
and z3 = z1ω 2 ⇒ z1 = , z2 =
z1 z2
∴ z1 z2 z3 = z1 × z1ω × z1ω 2 1
and z3 =
= z13ω 3 = z13 [Qω 3 = 1] z3
Targ e t E x e rc is e s
46. Since, z1, z2 and z3 are the vertices of an equilateral 1 1 1
∴ | z1 + z2 + z3| = + +
triangle, therefore z1 z2 z3
| z1 − z2| = | z2 − z3| = 1 =1
= | z3 − z1| = k [say]
1 52. Given, ( 3 i + 1) 100
= 2 99 (a + ib)
Also, α = ( 3 + i) 100
2 3 1 − 1− 3 i
1 1 ∴ 2100 i+ = 2 99 (a + ib) Q ω =
2
⇒ |α| = 3 + 1= × 2 = 1 2 2 2
2 2
⇒ 2(− ω 2 )100 = a + ib
Let A = αz1 + β, B = αz2 + β
and C = αz3 + β ⇒ 2 × ω 200 = a + ib
Now, | AB| = |αz2 + β − (αz1 + β )| ⇒ 2 × (ω 3 )66 × ω 2 = a + ib
= |α ( z2 − z1 )| − 1− 3 i
= |α|| z2 − z1| ⇒ 2 × (1)66 × = a + ib
2
= |1|| z2 − z1| = 1| z2 − z1|
= | z2 − z1| = k ⇒ − 1 − 3i = a + ib
Similarly, BC = CA = k ∴ a2 + b2 = 4
Hence, the points αz1 + β, αz2 + β and αz3 + β are the 53. (1 + 3 i )4 + (1 − 3 i )4 = (− 2ω 2 )4 + (− 2ω )4
vertices of an equilateral triangle.
− 1+ 3i − 1− 3i
47. Given, equation is Q ω = and ω 2 =
2 2
1 + z + z 3 + z 4 = 0.
= 16 ω 8 + 16 ω 4
⇒ (1 + z ) + z 3 (1 + z ) = 0
⇒ (1 + z )(1 + z 3 ) = 1 = 16 (ω 3 )2 ω 2 + 16 (ω 3 )ω
⇒ z = − 1, z 3 = − 1 ⇒ z = − 1 = 16 (ω 2 + ω ) = 16 (− 1) = − 16 [Q 1 + ω + ω 2 = 0]
∴ z = − 1, − ω, − ω 2 54. Q11/ 4 = (cos θ + i sin θ )1/ 4
Hence, roots are in cubic roots of unity. = (cos 2 πr + i sin 2 πr )1/ 4
Hence, these roots are the vertices of an equilateral πr πr
= cos + i sin
triangle. 2 2
where, r = 0, 1, 2, 3
48. Given, z1 = 2 2 (1 + i )
∴ 11/ 4 = 1, i , − 1, − i
and z2 = 1 + i 3
∴Required value = 12 + i 2 + (− 1)2 + (− i )2
∴ z12 z23 = [2 2 (1 + i )]2 [1 + i 3 ]3
= 1− 1+ 1− 1
= 8 (12 + i 2 + 2 i ) [13 + (i 3 )3 ] + 3 ⋅ 1⋅ i 3(1 + i 3 ) =0 179
4 55. From figure, it is clear that argument ≥ 90 °.
Y
59. Locus of z is a circle passing through origin.
60. Given, 2 x = 3 + 5 i
Objective Mathematics Vol. 1
P1 ⇒ 8 x 3 = 27 − 125 i + 27 + 5i − 25 × 9
− 198 + 10 i
⇒ 2 x3 = ...(i)
G 4
P(0, 1) Also, 4 x = 9 − 25 + 30 i
2
(–1,1)
− 16 + 30 i
X' X ⇒ 2 x2 = = − 8 + 15 i ...(ii)
Q 2
P1
∴ 2 x 3 + 2 x 2 − 7 x + 72
Y' − 198 + 10 i
= + (− 8 + 15 i ) − x + 172
4
The given equation | z − (− 1 + i ) ≤ 1| represents the
[form Eqs. (i) and (ii)]
points inside and on the boundary of the circle, centred
at (− 1, 1) and whose radius is 1. It lies in 2nd quadrant. 58 + 70 i − 28 x
=
The point (0, 1,) i.e. i lies on it such that, it has least 4
argument. 58 + 70 i − 14 (3 + 5 i )
∴The required complex number is i. =
4
56. (| z1| − | z2|)2 = | z1 − z2|2 16
= =4
⇒ | z1|2 + | z2|2 − 2| z1|| z2| = | z1|2 + | z2|2 4
− 2| z1|| z2|cos(θ1 − θ 2 ) 61. We have,| z1 + z2|2 = | z1|2 + | z2|2
⇒ cos(θ1 − θ 2 ) = 1 ⇒| z1|2 + | z2|2 + 2Re ( z1 z2 ) = | z1|2 + | z2|2
⇒ θ1 = 2 nπ + θ 2
⇒ Re ( z1 z2 ) = 0
Thus, if z1 = r1(cos θ1 + i sin θ1 )
and z2 = r2 (cos θ1 + i sin θ 2 ) ⇒ z1 z2 + z2 z1 = 0
⇒ z1 = r1(cos θ 2 + i sin θ 2 ) z1 z
⇒ =− 1
Ta rg e t E x e rc is e s
and z2 = r2 (cos θ 2 + i sin θ 2 ) z2 z2
z1 r1 z
⇒ = = − 1
z2 r2 z2
z z1
⇒ Im 1 = 0 ⇒ is purely imaginary.
z2 z2
57. Gievn, ( x + iy )(1 − 2 i ) = 1 + i 1+ i 3 (1 + i 3 )(2 i )
62. Let z = =
⇒ ( x + iy )(1 − 2 i ) = 1 + i i + 2
2
3 + 4i
⇒ ( x − iy )(1 + 2 i ) = 1 + i ...(i) i + 1
1+ i
⇒ x − iy = (1 + i 3 )(2 i )(3 − 4i )
1 + 2i =
25
1+ i 1− i |1 + i 3||2 i||3 − 4i|
⇒ x − iy = = ...(ii) ⇒ | z| =
i + 2i 1 − 2i |25|
1− i 2 ×2 × 5 4
⇒ x + iy = = =
1 − 2i 25 5
50
3 3 63. We have,
58. We have, + i = 325 ( x + iy ) | z1 + z2|2 = (| z1|2 + | z2|2 + 2| z1|| z2|)
2 2
50 ⇒ = | z1|2 + | z2|2 + 2| z1|| z2|cos(θ1 − θ 2 )
3 1
⇒ ( 3 )50 +i = 325 ( x + iy ) = | z1|2 + | z2|2 + 2| z1|| z2|
2 2
50 ⇒ cos(θ1 − θ 2 ) = 1
25 π π ⇒ θ1 = θ 2
⇒ 3 cos + i sin = 325 ( x + iy )
6 6 ⇒ arg ( z1 ) = arg ( z2 )
π π
⇒ cos 50 + i sin 50 = x + iy z
6 6 ∴ arg 1 = 0
π π z2
⇒ cos 25 + i sin 25 = x + iy
3 3 64. Given,
π π
⇒ cos + i sin = x + iy S = 1 + 2 ω + 3 ω 2 + K+ 3 nω 3 n − 1
3 3
∴ Sω = ω + 2ω 2 + 2ω 2 + K+ (3 n − 1)ω 3 n − 1 + 3 n ω 2 n
1 3
⇒ = +i = x + iy ⇒ S(1 − ω ) = 1 + ω + ω 2 + K+ ω 3 n − 1 − 3 nω 3 n = 0 − 3 n
2 2
3n 3n
1 3 ⇒ S =− =
∴ ( x, y ) = , 1− ω ω −1
180 2 2
y
4
4n
2 θ θ θ ⇒ = tan π = 0 ⇒ y = 0 ...(ii)
2 cos 4 − i 2 sin 4 cos 4 x
65. Let z = From Eqs. (i) and (ii), we get
2 cos 2 θ + 2 i sin θ cos θ
Complex Numbers
x = − 3, y = 0
4 4 4
∴ z=−3
4n
θ θ ⇒ | z| = |− 3| = 3
cos 4 − i sin 4
= 69. Given,| z1| = | z2| = K = | zn| = 1
cos θ + i sin θ ⇒ | z1|2 = | z2|2 = K| zn|2 = 1
4 4
⇒ z1 z1 = z2 z2 = K = zn zn = 1
cos nθ − i sin nθ (cos θ + i sin θ )− n 1 1 1
= = ⇒ z1 = , z2 = ,..., zn = ...(i)
cos nθ + i sin nθ (cos θ + i sin θ )n z1 z2 zn
= (cos θ + i sin θ )−2 n = cos 2 nθ − i sin 2 nθ Now, = | z1 + z2 + K + zn|
− 3+ i 3 = | z1 + z2 + K + zn| = | z1 + z2 + K + zn|
66. Given, x =
2 1 1 1
= + +K+ [from Eq. (i)]
− 1+ i 3 z1 z2 zn
⇒ x= − 1= ω − 1
2
∴ ( x 2 + 3 x )2 ( x 2 + 3 x + 1) 70. Q| z1| = 2 and| z2 | = 3
= [(ω − 1)2 + 3 (ω − 1)]2 [(ω − 1)2 + 3 (ω − 1) + 1] ∴ | z1 z2| = | z1|| z2| = 6
iθ
= (ω 2 + ω − 2 )2 (ω 2 + ω − 1) z z re re − i θ
71. + = −i θ + = e i2 θ + e − i2 θ = 2 cos 2 θ
= (− 1 − 2 )2 (− 1 − 1) = − 18 [Q1 + ω + ω 2 = 0] z z re re i θ
4 (1 + i ) 4
67. Given, ( x + iy )1/ 3 = 2 + 3 i 72. z = = 2(1 + i ) =
2 1+ i
⇒ x + iy = (2 + 3 i )3
4
= 8 + 36 i + 54 i 2 + 27 i 3 ∴ z=
1− i
Targ e t E x e rc is e s
= − 46 + 9 i
On equating real and imaginary parts from both sides, 73. If arg ( z ) = − π + θ
we get ⇒ arg ( z ) = π − θ
x = − 46, y = 9 arg (− z ) = − θ
∴ 3 x + 2 y = − 138 + 18 = − 120
∴ arg ( z ) − arg (− z ) = π − θ − (− θ )
68. Let z = x + iy = π −θ + θ = π
∴ | z + 3 − i| = |( x + 3) + i ( y − 1)| = 1 cos 30 ° + i sin 30 °
74.
⇒ ( x + 3) + ( y − 1)2 = 1
2
...(i) cos 60 ° − i sin 60 °
−1 y = (cos 30 ° + i sin 30 ° )(cos 60 ° + i sin 60 ° )
Q arg z = π ⇒ tan =π
x = cos 90 ° + i sin 90 ° = i
181
5
Inequalities and
Quadratic Equation
Inequality
An equation having signs >, <, ≥ or ≤ at the place of sign of equality ( =) is known as Chapter Snapshot
inequality or inequation. ● Inequality
or
Let a and b be two real numbers. If a − b is negative, we say that a is less than b ( a < b)
● Generalised Method of
Intervals for Solving
and if a − b is positive, then a is greater than b ( a > b).
Inequalities by Wavy Curve
Method (Line Rule)
Properties of Inequalities ● Absolute Value of a Real
We shall learn about some elementary properties of inequalities which will be used in Number
the subsequent discussion. ● Logarithms
i. If a > b and b > c, then a > c. Generally, if a1 > a 2 , a 2 > a 3 ,...,a n − 1 > a n , then ● Arithmetico-Geometric Mean
Inequality
a1 > a n
● Quadratic Equation with Real
ii. If a > b, then a ± c > b ± c, ∀c ∈ R Coefficients
a b b a ● Formation of a Polynomial
iii. If a > b, then for m > 0, am > bm, > and for m < 0, bm > am, > Equation from Given Roots
m m m m
● Symmetric Function of the
iv. If a > b > 0, then Roots
1 1 ● Transformation of Equations
(a) a 2 > b 2 (b) | a| > | b | (c) < Common Roots
a b ●
and if a < b < 0 , then ● Quadratic Expression and its
1 1 Graph
(a) a 2 > b 2 (b) | a| > | b | (c) >
a b ● Maximum and Minimum
Values of Rational Expression
v. If a < 0 < b, then ● Location of the Roots of a
(a) a 2 > b 2 , if | a| > | b | (b) a 2 < b 2 , if | a| < | b | Quadratic Equation
● Algebraic Interpretation of
vi. If a < x < b and a, b are positive real numbers, then Rolle’s Theorem
a 2 < x 2 < b2 ● Condition for Resolution into
Linear Factors
vii. If a < x < b and a is negative number and b is positive number, then ● Some Application of Graphs
to Find the Roots of
(a) 0 ≤ x < b , if | b | > | a|
2 2
(b) 0 ≤ x ≤ a , if | a | > | b |
2 2
Equations
a2 < b 2 < c 2
1 1 1
x ∈ b , a ,
and
⇒ 3 a2 < a2 + b 2 + c 2 < 3c 2 …(ii) 5
Inequalities and Quadratic Equation
From Eqs. (i) and (ii), we get
if a and b have same sign
3 a2 a2 + b 2 + c 2 3c 2
1 1 1 < <
viii. If x ∈[ a, b] ⇒ ∈ − ∞, ∪ , ∞ , 3c a+ b+c 3a
x a b
⇒
a2 a2 + b 2 + c 2 c 2
< <
if a and b have opposite c a+ b+c a
signs
a X Example 2. The value of x for which
ix. If > 0, then 12x − 6 < 0, 12 − 3x < 0, is
b
(a) a > 0, if b > 0 (b) a < 0, if b < 0 (a) φ (b) R
(c) R − {0} (d) None of these
x. If a i > bi > 0, where i = 1, 2, 3, …, n, then Sol. (a) 12 x − 6 < 0
a1 a 2 a 3K a n > b1 b2 b3K bn ⇒ 12 x < 6 ⇒ x<
1
2
xi. If a i > bi , where i = 1, 2, 3, …, n, then and 12 − 3 x < 0 …(i)
a1 + a 2 + a 3 + K + a n > b1 + b2 + K + bn ⇒ 12 < 3 x
⇒ 4< x
xii. If 0 < a < 1 and n is a positive rational number, ⇒ x> 4 …(ii)
then So, there is no real number x satisfying both the
inequalities (i) and (ii).
(a) 0 < a n < 1 (b) a − n >1 Hence, the given system of inequalities has no solution.
xiii. If a and b are positive real numbers such that X Example 3. The value of x for which
a < b and if n is any positive rational number, x −3 x −1 x − 2
then −x< − , 2 − x > 2x − 8
4 2 3
(a) a n < b n (b) a − n > b − n 10 10
(c) a 1/ n < b1/ n (a) −1, (b) −1,
3 3
(c) R (d) None of these
xiv. If a >1 and n is any positive rational number,
then Sol. (b) We have, x − 3 − x < x − 1 − x − 2
4 2 3
(a) a n >1 (b) 0 < a − n < 1
⇒ 3 x − 9 − 12 x < 6 x − 6 − 4 x + 8
⇒ −11x < 11 ⇒ − x < 1
xv. If 0 < a < 1 and m, n are positive rational
⇒ x> −1 …(i)
numbers, then and 2 − x> 2x − 8
(a) m > n ⇒ a m < a n ⇒ −3 x > − 10
(b) m < n ⇒ a m > a n ⇒ x<
10
…(ii)
3
xvi. If a >1 and m, n are positive rational numbers, From Eqs. (i) and (ii), the solution of the given system
of inequalities is given by x ∈ −1, .
10
then 3
(a) m > n ⇒ a m > a n
(b) m < n ⇒ a m < a n
Generalised Method of Intervals
X Example 1. If a, b and c are positive real for Solving Inequalities by Wavy
numbers such that a < b < c, then show that
a 2 a 2 + b2 + c2 c2
Curve Method
< < .
c a +b+c a (Line Rule)
Sol. We have, c > b> a> 0 Let F ( x ) = ( x − a1 ) k 1 ( x − a 2 ) k 2 K ( x − a n − 1 ) k n−1
⇒ c >b >a >0
2 2 2
(x − a n ) k n
Now, a< b<c
⇒ 3 a < a + b + c < 3c where, k1 , k 2 , …, k n ∈Z and a1 , a 2 , …, a n are
⇒
1
<
1
<
1
…(i)
fixed real numbers satisfying the condition
3c a + b + c 3 a a1 < a 2 < a 3 < K < a n − 1 < a n 183
For solving F ( x ) > 0 or F ( x ) < 0, consider the following
5 algorithm:
■ We mark the numbers a1 , a 2 ,…, a n on the number
X Example 4. Solve (x − 1)(x − 2)(1 − 2x ) > 0.
Sol. We have, ( x − 1)( x − 2 )(1 − 2 x) > 0
Objective Mathematics Vol. 1
axis and put the plus sign in the interval on the right ⇒ − ( x − 1)( x − 2 )(2 x − 1) > 0
of the largest of these numbers, i.e. on the right of a n . ⇒ ( x − 1)( x − 2 )(2 x − 1) < 0
1
■ Then, we put the plus sign in the interval on the left of On number line mark x = , 1, 2
2
a n , if k n is an even number and the minus sign, if k n is
– + – +
an odd number. In the next interval, we put a sign 1/2 1 2
according to the following rule:
When x > 2, all factors ( x − 1,) (2 x − 1) and ( x − 2 ) are
■ When passing through the point a n − 1 the polynomial positive.
F ( x ) changes sign, if k n − 1 is an odd number.Then, Hence, ( x − 1)( x − 2 )(2 x − 1) > 0 for x > 2.
we consider the next interval and put a sign in it using Now, put positive and negative sign alternatively as
the same rule. shown in figure.
■ Thus, we consider all the intervals. The solution of Hence, solution set of ( x − 1)( x − 2 )(1 − 2 x) > 0 or
the inequality F ( x ) > 0 is the union of all intervals in ( x − 1)( x − 2 ) (2 x − 1) < 0 is −∞, ∪ (1, 2 ).
1
which we have put the plus sign and the solution of 2
the inequality F ( x ) < 0 is the union of all intervals in
2x − 3
which we have put the minus sign. X Example 5. Solve ≥ 3.
3x − 5
Solution of Rational Algebraic Inequation
Sol. 2 x − 3 ≥ 3
If P ( x ) and Q ( x ) are polynomial in x, then the 3x − 5
inequation 2x − 3
⇒ − 3≥ 0
P (x ) P (x ) P (x ) 3x − 5
> 0, < 0, ≥0
Q (x ) Q (x ) Q (x ) 2 x − 3 − 9 x + 15
⇒ ≥0
3x − 5
P (x )
and ≤0 ⇒
−7 x + 12
≥0
Q (x ) 3x − 5
are known as rational algebraic inequations. 7 x − 12
⇒ ≤0
3x − 5
To solve these inequations, we use the sign method
⇒ (7 x − 12 )(3 x − 5) ≤ 0
as explained in the following algorithm :
Sign scheme of (7 x − 12) (3 x − 5) is as follows :
Algorithm + – +
5/3 12/7
Step I Obtain P ( x ) and Q ( x ).
x ∈ ,
5 12
⇒
Step II Factorise P ( x ) and Q ( x ) into linear 3 7
factors.
5 5
Ø x= is not included in the solutions as at x = denominator
Step III Make the coefficient of x positive in all 3 3
factors. becomes zero.
Step IV Obtain critical points by equating all X Example 6. The value of x for which
factors to zero. ( x − 2) 3 ( x − 3) < 0, is
Step V Plot the critical points on the number line. (a) (2, 3) (b) [2, 3) (c) (0, 3) (d) (2, 3]
If there are n critical points, then they
divide the number line into ( n +1) regions. Sol. (a) ( x − 2 )3 ( x − 3) < 0
⇒ ( x − 2 )( x − 3) < 0
Step VI In the right most region the expression
[as ( x − 2 )2 is positive for all real values of x ≠ 2]
P (x )
bears positive sign and in other –∞ + – + ∞
Q (x ) 2 3
regions the expression bears positive and Here, interval is open as x = 2, 3 do not satisfy
negative signs depending on the inequality.
exponents of the factors. i.e. 2 < x < 3 or x ∈(2, 3)
184
Work Book Exercise 5.1 5
Inequalities and Quadratic Equation
1 The value of x satisfying the inequalities 4 If c < d , x 2 + (c + d )x + cd < 0
1
x + ≥ 2 hold (− d , − c ]
x (− d , − c )
( 0, ∞ ) R φ [0, ∞ ) R
φ
x2
2 ≥0 (2 x − 1)( x − 1)4 ( x − 2 )4
x −1 5 ≤0
(1, ∞ ) [1, ∞) ( x − 2 )( x − 4)4
{ 0} ∪ (1, ∞ ) None of these 1, 2
3 ( x − 2 ) ( x − 3) ( x − 4) (1 − x ) ≤ 0
4 3 2 2
R
(1, 3)
φ
( − ∞, 1) ∪ ( 3, ∞ ) 1, 2
( −∞, 1] ∪ [3, ∞ )
2
None of the above
Absolute Value of a Real Forms of the Inequations Containing
Number Absolute Values
Form 1 The inequation of the form
The absolute value of a real number ‘x’ is denoted
x, if x ≥ 0 f (| x| ) < g ( x ) is equivalent to the collection of
by | x| and defined by | x| = systems
−x, if x < 0
f ( x ) < g ( x ), if x ≥ 0
Ø ● (i) | x| is also defined as x 2 f ( −x ) < g ( x ), if x < 0
(ii) If x is positive, then x 2 = x
Form 2 The inequation of the form
| f ( x )| < g ( x ), is equivalent to the systems
(iii) If x is negative, then x2 = − x
f ( x ) < g ( x ), if g ( x ) > 0
e.g. 9 = (−3)2
− f ( x ) < g ( x ), if g ( x ) > 0
= 32 = 3 Form 3 The inequation of the form
or 9 = ±3 | f ( x )| > g ( x ), is equivalent to the systems
● (i) a ≤ | a| f ( x ) > g ( x ), if g ( x ) < 0
(ii) | ab| = | a||b|
− f ( x ) > g ( x ), if g ( x ) < 0
(iii) =
a | a|
b |b| Form 4 The inequation of the form
(iv) | a + b| ≤ | a| + |b| [triangle inequality] | f (| x| )| ≥ g ( x ) is equivalent to the collection of systems
(v) | a − b| ≥ | a| − |b| [triangle inequality] | f ( x )| ≥ g ( x ), if x ≥ 0
(vi) | a + b| = | a| + |b| , iff ab ≥ 0 | f ( −x )| ≥ g ( x ), if x < 0
(vii) | a + b| = | a| − |b| , iff ab ≤ 0 Form 5 The inequation of the form
| f ( x )| ≥ | g ( x )| is equivalent to the collection of
Inequations Containing Absolute Values
system
By definition,
| x| < a ⇒ − a < x < a ( a > 0) f 2 (x ) ≥ g 2 (x )
| x| ≤ a ⇒ −a ≤ x ≤ a Form 6 The inequation of the form
| x| > a ⇒ x < − a and x > a h( x, | f ( x )| ) ≥ g ( x )
| x| ≥ a ⇒ x ≤ − a and x ≥ a is equivalent to the collection of systems
a ≤ | x| ≤ b ⇒ x ∈[ −b, − a ] ∪ [ a, b] h{x, f ( x )} ≥ g ( x ), if f ( x ) ≥ 0
where, a, b > 0
h{x, − f ( x )} ≥ g ( x ), if f ( x ) < 0
185
5 X Example 7. The solution set of |3 − 4x| ≥ 9 is
3
(a) −∞, ∪ [3, ∞ )
X Example 9. Solve the inequation
1 − | x|
+
≥ 1.
Objective Mathematics Vol. 1
2 1 | x| 2
3 Sol. The given inequation is equivalent to the collection of
(b) −∞, − ∪ [3, ∞ )
2 systems.
x 1
(c) ( −∞, 2) ∪ [2, ∞ ) 1 − ≥ , if x ≥ 0 1 ≥ 1 , if x ≥ 0
1 + x 2 |1 + x| 2
(d) None of the above ⇒ ⇒
1 + x ≥ 1 , if x < 0 1 ≥ 1 , if x < 0
Sol. (b) We have, |1 − x| 2
|3 − 4 x| ≥ 9 1 − x 2
⇒ 3 − 4 x ≤ − 9 or 3 − 4 x ≥ 9 1 ≥ 1 , if x ≥ 0
1 + x 2
[since,| x| ≥ a ⇒ x ≤ − a or x ≥ a] ⇒ 1 1
⇒ −4 x ≤ − 12 or −4 x ≥ 6 ≥ , if x < 0
1 − x 2
−3
⇒ x ≥ 3 or x ≤ [dividing both sides by −4] 1 − x ≥ 0, if x ≥ 0 x − 1 ≤ 0, if x ≥ 0
2 1 + x x + 1
−3 ⇒ ⇒
⇒ x ∈ −∞, ∪ [3, ∞ )
2 1 + x ≥ 0, if x < 0 x + 1 ≤ 0, if x < 0
1 − x x − 1
| x + 3| + x x−1
X Example 8. The solution set of > 1 is For ≤ 0, if x ≥ 0, then
x +2 x+1
0≤ x≤ 1 ...(i)
(a) [ −5, − 2] ∪ [ −1, ∞ ) (b) [ −5, − 2) ∪ [ −1, ∞ )
(c) ( −5, − 2) ∪ ( −1, ∞ ) (d) None of these + +
–
Sol. (c) We have, | x + 3| + x > 1
–1 1
x+2 x+1
For ≤ 0, if x < 0
| x + 3| + x x−1
⇒ − 1> 0
x+2 ∴ −1 ≤ x < 0 …(ii)
x+ 3 −2
⇒ >0
x+2 + +
–1 – 1
Case I When x + 3 ≥ 0, i.e. x ≥ − 3
| x + 3| − 2
∴ >0 From Eqs. (i) and (ii), the solution of the given equation
x+2
is x ∈ [−1, 1.]
x+ 3−2
⇒ >0
x+2 X Example 10. The value of x, | x + 3| > |2x − 1| is
x+1 −2 2
⇒ >0
x+2 (a) , 4 (b) − , ∞
3 3
⇒ {x + 1 > 0 and x + 2 > 0} or {x + 1 < 0 and x + 2 < 0}
(c) (0, 1) (d) None of these
⇒ {x > − 1 and x > − 2} or {x < − 1 and x < − 2}
⇒ x > − 1 or x< −2 Sol. (a) Given,| x + 3| > |2 x − 1|
⇒ x ∈ (−1, ∞ ) or x ∈ (−∞, − 2 ) On squaring both sides, we get
⇒ x ∈ (−3, − 2 ) ∪ (−1, ∞ ) [Q x ≥ − 3] …(i) | x + 3|2 > |2 x − 1|2
Case II When x + 3 < 0, i.e. x < − 3 ⇒ {( x + 3) − (2 x − 1)} {( x + 3) + (2 x − 1)} > 0
| x + 3| − 2 ⇒ {(− x + 4)(3 x + 2 )} > 0
∴ >0
x+2 – + –
−x − 3 − 2 –2/3 4
⇒ >0
x+2 −2
⇒ x ∈ , 4
−( x + 5) 3
⇒ >0
x+2
⇒
( x + 5)
<0 Logarithms
x+2 Let there be a number a > 0, ≠1. A number p is
⇒ {x + 5 < 0 and x + 2 > 0} or {x + 5 > 0 and x + 2 < 0} called the logarithm of a number x to the base a, if
⇒ (x < − 5 and x > − 2), which is not possible a p = x and is written as p = log a x. Obviously x must
or {x > − 5 and x < − 2} be positive.
⇒ x ∈ (−5, − 2 ) …(ii)
From Eqs. (i) and (ii), we get Ø ● If x < 0, log a x is imaginary and if x = 0, log a x does not exist .
186 x ∈ (−5, − 2 ) ∪ (−1, ∞ ) ● log a x exists if and only if x, a> 0 and a ≠ 1.
Properties of Logarithms
i. a log a x = x; a ≠ 0, ±1, x > 0 x. If a >1, then
5
Inequalities and Quadratic Equation
(a) log a x > p ⇒ x > a p
ii. a log b x = x log b a ; a > 0, b > 0, ≠1, x > 0
(b) 0 < log a x < p ⇒ 0 < x < a p
iii. log a a =1, log a 1 = 0; a > 0, ≠1
xi. If 0 < a < 1, then
1
iv. log a x = ; x, a > 0, ≠1 (a) log a x > p ⇒ 0 < x < a p
log x a
(b) 0 < log a x < p ⇒ a p < x <1
log b x
v. log a x = log b x ⋅ log a b = ; a, b > 0, ≠1,
log b a X Example 11. The value of x, log e (x − 3) < 1 is
x >0 (a) (0, 3) (b) (0, e) (c) (0, e + 3) (d) (3, 3 + e)
vi. For m, n > 0, a > 0, ≠1 Sol. (d) From definition of logarithms
(a) log a ( m ⋅ n) = log a m + log a n x − 3> 0
m ⇒ x> 3 …(i)
(b) log a = log a m − log a n Also, e > 1 given inequality may written as
n
x − 3 < (e )1 ⇒ x < 3 + e
(c) log a ( m x ) = x log a m ⇒ x ∈ (3, 3 + e )
vii. For x > 0, a > 0, ≠1 X Example 12. The value of x, log 1/ 2 x ≥ log 1/ 3 x
1 is
(a) log a n (x ) = log a x (a) (0, 1] (b) (0, 1)
n
(c) [0, 1) (d) None of these
m
(b) log a n ( x m ) = log a x Sol. (a) Case I When x ≠ 1and x > 0
n
log1/ 2 x ≥ log1/ 3 x
viii. (a) log a x > 0, iff x >1, a >1 or 0 < x < 1, 0 < a < 1 ⇒ log x 2 ≥ log x 3, when x ≠ 1
(b) log a x < 0, iff x >1, 0 < a < 1 or 0 < x < 1, a >1 which is possible, only if 0 < x < 1.
Case II When x = 1
ix. For x > y > 0, log1/ 2 x = log1/ 3 x, i.e. equality holds.
(a) log a x > log a y, if a >1 Combining the above cases,
(b) log a x < log a y, if 0 < a < 1 0 < x ≤ 1 or x ∈(0, 1]
Work Book Exercise 5.2
Solve for x, 7 ( x − 1)2 > 9
1 x+4 >5 ( −∞, − 2 ) ∪ ( 4, ∞ )
( −∞, 1) ( −∞, 9) ( 4, ∞ )
R − [−9, 1] None of these ( −∞, 2 )
2 | x + 2| < 4 ( −∞, − 2 ] ∪ ( 4, ∞ )
( −6, 2 ) ( −6, 0) ( −6, 2 ] ( 0, 2 )
8 ( x + 1)2 < 25
3 log 2 x > 0 ( −6, 4] ( −6, 4)
( 0, ∞ ) ( − ∞ , 0) (1, ∞ ) ( 4, ∞ ) ( 4, 6) ( −4, 6)
9 log x ( x + 7 ) < 0
4 log 4 x > 1
( 0, 1) ( −∞, 1) ( −1, 0) ( −1, 1)
( 0, ∞ ) ( 4, ∞ ) ( −4, ∞ ) ( −∞, 4)
x +6
2
5 log 0. 2 ( x + 5) > 0 10 ≥1
5x
( −5, − 4) ( −5, 4) ( 0, 4) ( 0, 5)
( −∞, − 3)
6 log x 0.5 > 2 ( −∞, − 3) ∪ ( 3, ∞ )
1 R
, 1 ( −∞, 1) ( −1, 0) ( −1, 1)
2 ( −∞, − 3] ∪ [−2, 0) ∪ ( 0, 2 ] ∪ [3, ∞ ) 187
5 Arithmetico-Geometric
Inequality
Mean X Example 15. If the equation
x 4 − 4x 3 + ax 2 + bx + 1 = 0 has four positive roots,
then a, b are
Objective Mathematics Vol. 1
If a1 , a 2 , …, a n are n distinct positive real (a) a = 4, b = 6 (b) a = − 4, b = 6
numbers, then (c) a = 2, b = 3 (d) a = 6, b = − 4
a1 + a 2 + K + a n Sol. (d) Let α, β, γ and δ be four roots of the given equation.
> ( a1 a 2K a n )1/ n
n Then, α + β + γ + δ = 4 and αβγδ = 1
i.e. AM > GM ⇒ AM of α, β, γ, δ = GM of α,β, γ, δ ⇒ α = β = γ = δ
⇒ α = β = γ = δ = 1 [Qα + β + γ + δ = 4, also αβ + αγ
Ø If a1 = a2 = K = an , then AM = GM. Thus, equality occurs when
+ αδ + βγ + βδ + γδ = a, αβγ + αβδ + βγδ + αγδ = − b]
all quantities are equal.
∴ x4 − 4 x3 + ax2 + bx + 1 = x4 − 4 x3 + 6 x2 − 4 x + 1
X Example 13. If a, b and c are three distinct ⇒ a = 6 and b = − 4
positive real numbers, then
1 1 1
Two Important Inequalities
(a) ( a + b + c) + + > 9
a b c i. If a1 , a 2 , ..., a n are n positive distinct real
1 1 1 numbers, then
(b) ( a + b + c) + + < 9
a b c a1m + a 2m + a nm a1 + a 2 +K + a n
m
1 1 1 (a) > ,
(c) ( a + b + c) + + ≥ 9 n n
a b c
if m < 0 or m >1
1 1 1
(d) ( a + b + c) + + ≤ 9 a1m + a 2m + ... + a nm a1 + a 2 + K + a n
m
a b c (b) <
n n
Sol. (a) We have, AM > GM
if 0 < m < 1.
1 1 1
+ + 1/ 3 i.e. the arithmetic mean of the mth power
a+ b+c a b c > 1
∴ > (abc ) and
1/ 3
of n positive quantities is greater than mth
3 3 abc
1 1 1 3 power of their arithmetic mean except
⇒ a + b + c > 3 (abc ) and + + >
1/ 3
a b c (abc )1/ 3 when m is a positive proper fraction known
as mth power mean.
(a + b + c ) + + > 9
1 1 1
⇒
a b c (c) If a1 , a 2 , K, a n and b1 , b2 , K, bn are
rational numbers and m is a rational
Weighted AM and GM Inequalities number, then
If a1 , a 2 , K , a n are n positive real numbers and m1 , b1 a1m + b2 a 2m + K + bn a nm
m2 , …, mn are n positive rational numbers, then b1 + b2 + K + bn
m1 a1 + m2 a 2 + K + mn a n m
b a + b2 a 2 + ... + bn a n
m1 + m2 +... + mn > 1 1 ,
b1 + b2 + K + bn
1
m1 m2 mn m1 + m2 + K + mn if m < 0 or m >1 and
> ( a1 ⋅ a 2 K a n )
b1 a1m + b2 a 2m + K + bn a nm
i.e. weighted AM > weighted GM b1 + b2 +... + bn
m
X Example 14. If a, b and c are distinct positive b a + b2 a2 + K + bn an
integers, then ax b − c + bx c − a + cx a − b is
< 1 1 , if 0 < m < 1
b1 + b2 + K + bn
(a) > 2 ( a + b + c) (b) R
(c) > ( a + b + c) (d) < ( a + b + c) ii. If a1 , a 2 , a 3 , …, a n are distinct positive real
numbers and p, q, r are natural numbers, then
Sol. (c) We have, p+q+r p+q+r
a⋅ x b − c + b⋅ xc − a + c ⋅ x a − b
a1 + a2 + K + a np + q + r
>
a+ b+c n
{( x b −c a
) (x c −a b
) (x a − b c 1/ a + b + c
) } a1p + a 2p +K + a np a1q + a 2q +K + a nq
ax b − c + bx c − a + cx a − b >
⇒ > n n
a+ b+c
{ xa( b − c ) + b( c − a ) + c ( a − b ) 1/ a + b + c a1 + a 2 +K + a n
r r r
b −c c −a a−b
}
188 ⇒ ax + bx + cx > (a + b + c ) n
(a) ≥ ( a12 + a 22 + K + a n2 ) ( b1−2 + b2−2 + K + bn−2 )
X Example 16. If m >1 and n ∈ N , then
(a)
1m + 2 m + 3 m + K + n m n + 1
>
m
(b) ≤ ( a12 + a 22 + K + a n2 ) ( b1−2 + b2−2 + K + bn−2 ) 5
Inequalities and Quadratic Equation
n 2 (c) > ( a12 + a 22 + K + a n2 ) ( b1−2 + b2−2 + K + bn−2 )
m
1m + 2 m + 3 m + K + n m n + 1 (d) < ( a12 + a 22 + K + a n2 ) ( b1−2 + b2−2 + K + bn−2 )
(b) <
n 2
Sol. (b) By Cauchy-Schwartz’s inequality, we have
1m + 2 m + 3 m + K + n m
(c) ≥1 (a1b1−1 + a2 b2−1 + K + an bn−1 )2
n ≤ (a12 + a22 + K + an2 )(b1−2 + b2−2 + K + bn−2 )
1m + 2 m + 3 m + K + n m a
2
(d) ≤1 a a
⇒ 1 + 2 + K+ n
n 1
b b 2 b n
Sol. (a) We know that for m > 1, 1 1 1
AM of mth power > mth power of AM ≤ (a12 + a22 + K + an2 ) 2 + 2 + K + 2
m b1 b2 bn
1m + 2 m + 3m + K + nm 1 + 2 + 3 + K + n
∴ >
n n
1m + 2 m + 3m + K + nm n + 1
m Applications of Inequalities to Find the
⇒ >
n 2 Greatest and Least Values
Some Other Standard Inequalities i. If x1 , x 2 , K , x n are n positive variables such
that x1 + x 2 + K + x n = c (constant), then the
i. Weierstrass Inequality product x1 x 2 K x n is greatest when
(a) If a1 , a 2 , K , a n are n positive real c
numbers, then for n ≥ 2 , x1 = x 2 =K = x n = and the greatest value is
n
(1 + a1 ) (1 + a 2 ) K (1 + a n ) > 1 + a1 + a 2 n
+L + a n c
.
(b) If a1 , a 2 ,K , a n are positive numbers less n
than unity, then
(1 − a1 )(1 − a 2 ) K (1 − a n ) > 1 − a1 − a 2 ii. If x1 , x 2 , K , x n are positive variables such that
−K − a n x1 x 2 K x n = c (constant), then the sum
x1 + x 2 + K + x n is least when
ii. Cauchy-Schwartz Inequality x1 = x 2 = K = x n = c1/ n and the least value of
If a1 , a 2 , K , a n and b1 , b2 , K , bn are 2n real
the sum is n( c1/ n ).
numbers, then
( a1 b1 + a 2 b2 + ... + a n bn ) 2 ≤ ( a12 + a 22 +K + a n2 ) iii. If x1 , x 2 , K, x n are variables and m1 , m2 , K, mn
( b12 + b22 +K + bn2 ) are positive real numbers such that
with the equality holding if and only if x1 + x 2 +… + x n = c (constant), then
m m
a1 a 2 a x1 1 ⋅ x 2 2 … x nmn is greatest, when
= =K = n
b1 b2 bn x1 x x x + x 2 +… + x n
= 2 =… = n = 1
m1 m2 mn m1 + m2 +… + mn
iii. Tchebychef’s Inequality
If a1 , a 2 , K, a n and b1 , b2 , K, bn are real X Example 18. The greatest value of x 2 y 3 when
numbers such that
3x + 4 y = 5, is
a1 ≤ a 2 ≤ a 3 ≤K ≤ a n and b1 ≤ b2 ≤ K ≤ bn ,
3 3 6 8
then (a) (b) (c) (d)
n ( a1 b1 + a 2 b2 + K + a n bn ) 8 16 5 3
≥ ( a1 + a 2 + K + a n ) ( b1 + b2 + K + bn ) Sol. (b) Let P = x2 y3 . Clearly, P is the product of 5 factors
a1 b1 + a 2 b2 + K + a n bn such that two of them are equal to x and the remaining 3
or are equal to y.
n
a1 + a 2 + K + a n b1 + b2 + K + bn Now, 3x + 4y = 5
≥ ⇒ 2 + 3 = 5
3x 4y
n n 2 3
3x 3x 4y 4y 4y
⇒ + + + + =5
X Example 17. If none of b1 , b2 , K , bn is zero, 2 2 3 3 3
5
a a a
2
⇒ 16 x2 y3 ≤ 5 ⇒ x 2 y3 ≤
3
then 1 + 2 + K + n is 3 5 16
1
b b 2 bn or maximum of x2 y3 = 3/16
189
5 Work Book Exercise 5.3
1 If a, b and c are all positive real numbers, which 3 If a and b are two positive quantities whose sum is
Objective Mathematics Vol. 1
one of the following is true? 1 1
λ, then the minimum value of 1 + 1 + is
( b + c )(c + a )( a + b ) ≥ 8abc a b
( b + c )(c + a )( a + b ) < 8abc 1 2 2 1
a c e λ− λ− 1+ 1+
+ + + < 9
b f g
+ λ λ λ λ
e f g a b c
1 1 4 If a2 x 4 + b2 y 4 = c 6 , then the greatest value of xy is
( a + b ) + < 4
a b c3 c3 c3 c3
2 ab 4ab 2 ab 2 ab
2 If a1, a2 , K, an and b1, b2 , K, bn are any two sets of
positive real numbers, then 5 Which of the following is true, where a, b, c > 0?
2
a12 a22 an2 2 ( a 3 + b 3 + c 3 ) ≥ bc( b + c ) + ca(c + a ) + ab( a + b )
+ + ... + a3 + b 3 + c 3 (a + b + c ) ⋅ (a2 + b 2 + c 2 )
b1 b2 bn >
3 9
≤ ( a14 + a24 + K + an4 )( b1−2 + b 2−2 + K + b n−2 ) bc ca ab 1
+ + < (a + b + c )
≥ ( a14 + a24 + K + an4 )( b1−2 + b 2−2 + K + b n−2 ) b+c c+ a a+ b 2
< ( a14 + a24 + K + an4 )( b1−2 + b 2−2 + K + b n−2 ) 2 2 2 1 1 1
+ + < + +
> ( a14 + a24 + K + an4 )( b1−2 + b 2−2 + K + b n−2 ) b+c c+ a a+ b a b c
Quadratic Equation with
Real Coefficients
An equation of the form
ax 2 + bx + c = 0 …(i)
Ø ● If α is a root of the equation f (x) = 0, then the polynomial
where, a ≠ 0, a, b, c ∈ R is called a quadratic f (x) is exactly divisible by (x − α) or (x − α) is a factor of f (x)
equation with real coefficients. and conversely.
● Every equation of nth degree (n≥ 1) has exactly n roots and if
The quantity D = b 2 − 4ac is known as the the equation has more than n roots, it is an identity.
discriminant of the quadratic equation in (i) whose ● If the coefficients of the equation f (x) = 0 are all real and
roots are given by α + iβ is its root, then α − iβ is also a root, i.e. imaginary roots
occur in conjugate pairs.
−b + b 2 − 4ac −b − b 2 − 4ac
α= and β = ● If the coefficients in the equation are all rational and α + β
2a 2a is one of its roots, then α − β is also a root, where α, β ∈ Q
The nature of the roots is as given below: and β is not a perfect square.
● If there is any two real numbers ‘a’ and ‘b’ such that f (a) and
(i) If a quadratic equation in x has more than two f (b) are of opposite signs, then f (x) = 0 must have atleast
roots, then it is an identity in x that is one real root between ‘a’ and ‘b’.
a = b = c = 0. ● Every equation f (x) = 0 of odd degree has atleast one real
(ii) If a =1 and b, c ∈ I and the roots are rational root of a sign opposite to that of its last term.
numbers, then these roots must be integers.
X Example 19. The number of values of a for
(iii) The roots are of the form p ± q ( p, q ∈Q ), iff a, which ( a 2 − 3a + 2) x 2 + ( a 2 − 5a + 6) x + a 2 − 4 = 0
b and c are rational and D is not a perfect square. is an identity in x, is
(iv) The roots are real and distinct, iff D > 0. (a) 0 (b) 2
(v) The roots are real and equal, iff D = 0. (c)1 (d) 3
(vi) The roots are complex with non-zero imaginary Sol. (b) It is an identity in x, if
part, iff D < 0.
a2 − 3 a + 2 = 0, a2 − 5 a + 6 = 0, a2 − 4 = 0
(vii) The roots are rational, iff a, b and c are rational ⇒ a = 1, 2 and a = 2, 3 and a = 2, −2
and D is a perfect square. ∴ Equation is identity, if a = 2
190
Sol. (d) Since, ax2 + x + b = 0 has real roots
X Example 20. If cos θ, sin φ and sin θ are in GP,
then roots of x 2 + 2 cot φ ⋅ x + 1 = 0 are always ⇒
⇒
(1)2 − 4 ab ≥ 0
− 4 ab ≥ − 1
5
Inequalities and Quadratic Equation
(a) equal (b) real
or 4 ab ≤ 1 …(i)
(c) imaginary (d) greater than 1
Now, second equation is x2 − 4 a b x + 1 = 0
Sol. (b) As cos θ, sin φ and sin θ are in GP.
∴ D = 16 ab − 4
⇒ sin2 φ = cos θ ⋅ sin θ
From Eq. (i), D< 0
⇒ cos 2 φ = 1 − sin 2θ …(i)
4cos 2 φ Hence, roots are imaginary.
Also, discriminant, D = 4cot 2 φ − 4 =
sin2 φ
4(1 − sin2 θ )
X Example 25. Let a, b and c be real numbers,
= > 0; as sin 2θ < 1 and sin2 φ > 0 a ≠ 0. If α is a root of a 2 x 2 + bx + c = 0, β is the
sin φ
2
root of a 2 x 2 − bx − c = 0 and 0 < α < β, then the
⇒ Roots are real.
equation a 2 x 2 + 2bx + 2c = 0 has a root γ that
Also, when cos θ = sin θ
always satisfies
⇒ Three numbers are equal which is a special case
α +β β
i.e. the series forms both AP and GP. (a) γ = (b) γ = α +
2 2
X Example 21. If l, m and n are real, l ≠ m, then (c) γ = α (d) α < γ < β
the roots of the equation
Sol. (d) Since, α is a root of a2 x2 + bx + c = 0
( l − m) x 2 − 5( l + m) x − 2( l − m) = 0 are
⇒ a2α 2 + bα + c = 0 …(i)
(a) real and equal (b) complex
and β is a root of a x − bx − c = 0
2 2
(c) real and unequal (d) None of these
⇒ a2β 2 − bβ − c = 0 …(ii)
Sol. (c) Discriminant of the given equation is Let f( x) = a2 x2 + 2 bx + 2c
D = 25(l + m)2 + 8(l − m)2 ∴ f(α ) = a2α 2 + 2 bα + 2c
As, l ≠ m, (l − m)2 > 0. = a2α 2 − 2 a2α 2 = − a2α 2 [from Eq. (i)]
Also, (l + m) ≥ 0
2
and f(β ) = a β + 2 bβ + 2c
2 2
Thus, D> 0 = a2β 2 + 2 a2β 2 = 3a2β 2 [from Eq. (ii)]
Hence, roots are real and unequal.
⇒ f(α ) f(β ) < 0
X Example 22. The number of real solutions of ∴f( x) must have a root lying in the open interval (α, β ).
the equation | x | 2 − 3 | x | + 2 = 0 is ∴ α< γ<β
(a) 4 (b) 1 (c) 3 (d) 2 X Example 26. Find the values of a for which
Sol. (a) Since,| x|2 − 3| x| + 2 = 0 9
4 − ( a − 4)2 + a < 0, ∀ t ∈ (1, 2).
t t
⇒ (| x| − 1)(| x| − 2 ) = 0 ⇒ | x| = 1, 2 4
∴ x = 1, −1, 2, − 2
Hence, four real solutions exist.
Sol. Let 2t = x and f( x) = x2 − (a − 4)x + 9 a
4
We want f( x) < 0, ∀ x ∈ (21, 2 2 )
X Example 23. Let α and β be the roots of the
i.e. ∀ x ∈ (2, 4)
equation ( x − a ) ( x − b) = c, c ≠ 0. Then, the roots of
Since, we want f( x) < 0, ∀ x ∈ (2, 4), one of the f( x) should
the equation ( x − α ) ( x − β) + c = 0 are be less than 2 and the other must be greater than 4.
(a) a, c (b) b , c 128
i.e. f(2 ) < 0 and f(4) < 0, a < − 48 and a > which is not
(c) a, b (d) a + c, b + c 7
possible.
Sol. (c) Given α and β are the roots of ( x − a)( x − b) − c = 0 Hence, no such a exists.
⇒ ( x − a)( x − b ) − c = ( x − α )( x − β )
⇒ ( x − a)( x − b ) = ( x − α )( x − β ) + c
X Example 27. The roots of ax 2 + bx + c = 0,
⇒ a and b are the roots of equation were a ≠ 0 and coefficients are real, are non-real
( x − α) ( x − β) + c = 0 complex and a + c < b. Then,
(a) 4a + c > 2b (b) 4a + c < 2b
X Example 24. If the roots of the equation (c) 4a + c = 2b (d) None of these
ax 2 + x + b = 0 are real, then the roots of the
Sol. (b) Since, ax2 + bx + c = 0 has non-real.
equation x 2 − 4 ab x + 1 = 0 will be Thus, either f( x) = ax2 + bx + c
(a) rational (b) irrational ⇒ f( x) > 0 or f( x) < 0, for all x
191
(c) real (d) imaginary As a + c < b ⇒ f(− 1) < 0
5 ⇒ a − b + c < 0 or a+ c< 0 1
and product of roots =
∴ f( x) < 0, for all x. a −b
2
Y Hence, the quadratic equation
Objective Mathematics Vol. 1
2a 1
x2 − 2 x + 2 = 0
a − b a − b
X′ X ⇒ (a2 − b )x2 − 2 ax + 1 = 0
O
X Example 29. The quadratic equation whose
roots are the AM and HM of the roots of the
equation x 2 + 7x − 1 = 0 is
Y′ (a)14x 2 + 14x − 45 = 0 (b)14x 2 − 14x + 14 = 0
Thus, for all x ∈ R, ax2 + bx + c < 0 (c)14x 2 + 45x − 14 = 0 (d) None of these
or for x = − 2, 4a − 2 b + c < 0
Sol. (c) Here, x2 + 7 x − 1 = 0 has α + β = − 7 and αβ = − 1
⇒ 4a + c < 2 b
QQuadratic equation’s roots are AM and HM of the
roots of the equation x2 + 7 x − 1 = 0
Formation of a Polynomial ⇒ α′ =
α+β
and β ′ =
2αβ
2 α+β
Equation from Given Roots or α′ = −
7
and β′ =
−2 2
=
If α 1 , α 2 , α 3 , … , α n are the roots of an nth degree 2 −7 7
equation, then the equation is ∴Required equation is x2 − (α ′ + β ′ ) x + α ′β ′ = 0
x2 − −
45
x n − S 1 x n − 1 + S 2 x n − 2 − S 3 x n − 3 + ... + ( −1) n S n = 0 ⇒
14
x − 1 = 0
where, S k denotes the sum of the products of roots or 14 x2 + 45 x − 14 = 0
taking k roots at a time.
Particular Cases Symmetric Function
i. Quadratic equation If α and β are the roots of of the Roots
a quadratic equation, then the equation is A function of α and β is said to be a symmetric
x 2 − S1 x + S 2 = 0 function, if it remains unchanged when α and β are
interchanged.
i.e. x 2 − (α + β) x + αβ = 0
e.g. α 2 + β 2 + 2αβ is a symmetric function of α and
ii. Cubic equation If α, β and γ are the roots of a β whereas α 2 − β 2 + 3αβ is not a symmetric function of
cubic equation, then the equation is α and β.
x 3 − S1 x 2 + S 2 x − S 3 = 0 In order to find the value of a symmetric function
i.e. x 3 − (α + β + γ ) x 2 + (αβ + βγ + γα ) x of α and β, express the given function in terms of α + β
− αβγ = 0 and αβ. The following results may be useful:
i. α 2 + β 2 = (α + β) 2 − 2αβ
X Example 28. If a and b are rational and b is
not a perfect square, then the quadratic equation
ii. α 3 + β 3 = (α + β) 3 − 3αβ(α + β)
1
with rational coefficients whose one root is ,
a+ b iii. α 4 + β 4 = (α 3 + β 3 )(α + β) − αβ(α 2 + β 2 )
is
(a) x 2 − 2ax + ( a 2 − b) = 0 iv. α 5 + β 5 = (α 3 + β 3 )(α 2 + β 2 ) − α 2 β 2 (α + β)
(b) ( a 2 − b) x 2 − 2ax + 1 = 0
v. | α − β| = (α + β) 2 − 4αβ
(c) ( a 2 − b) x 2 − 2bx + 1 = 0
(d) None of the above vi. α 2 − β 2 = (α + β)(α − β)
Sol. (b) As irrational roots alway occur in pairs,
1 1 vii. α 3 − β 3 = (α − β)[(α + β) 2 − αβ]
i.e. the roots of given equation are , .
a+ b a− b = (α − β) 3 + 3αβ(α − β)
2a
Thus, sum of roots =
192 a2 − b viii. α 4 − β 4 = (α + β)(α − β)(α 2 + β 2 )
X Example 30. If α and β are the roots of
α β
ax 2 + 2bx + c = 0, then + is equal to
β α
v. To obtain an equation whose roots are
reciprocals of the roots of a given equation is 5
Inequalities and Quadratic Equation
obtained by replacing x by 1/ x in the given
equation.
4b 2 − 2ac 4b 2 − 4ac
(a) (b)
ac ac vi. To transform an equation to another equation
2b 2 − 2ac 2b 2 − 4ac whose roots are negative of the roots of a
(c) (d)
ac ac given equation, replace x by −x.
Sol. (a) Since, α and β are the roots of equation vii. To transform an equation to another equation
ax2 + 2 bx + c = 0 whose roots are square of the roots of a given
−2b c
∴ α+β= and αβ = equation, replace x by x.
a a
α β α 2 + β 2 (α + β )2 − 2αβ viii. To transform an equation to another equation
Now, + = =
β α αβ αβ whose roots are cubes of the roots of a given
α β (4b 2 / a2 ) − 2c / a 4b 2 − 2 ac equation replace x by x 1/ 3 .
⇒ + = =
β α c/a ac
X Example 32. Form an equation whose roots
X Example 31. If α and β are the roots of the are cubes of the roots of equation
equation px 2 + qx + r = 0, then the value of ax 3 + bx 2 + cx + d = 0.
α 3β + β 3α is Sol. Replacing x by x1/ 3 in the given equation, we get
r r ( q − 2 pr )
2
a( x1/ 3 )3 + b( x1/ 3 )2 + c( x1/ 3 ) + d = 0
(a) (b)
p3 p3 ⇒ a x + d = − (b x2 / 3 + cx1/ 3 )
⇒ (a x + d )3 = − (b x2 / 3 + c x1/ 3 )3
r ( q 2 + 2 pr ) ( q 2 − 2 pr )
(c) (d) ⇒ a x + 3 a2 d x2 + 3 ad 2 x + d 3
3 3
p3 p3 = − {b 3 x2 + c 3 x + 3bcx (bx2 / 3 + cx1/ 3 )}
Sol. (b) Since, α and β are the roots of the equation ⇒ a x + 3 a dx + 3 ad 2 x + d 3
3 3 2 2
q r = − {b 3 x2 + c 3 x − 3 bcx (ax + d )}
px2 + qx + r = 0, therefore α + β = − and αβ = .
p p ⇒ a x + x (3 a d − 3 abc + b 3 )
3 3 2 2
∴ α β + β α = αβ(α + β )
3 3 2 2
+ x (3 ad 2 − 3bcd + c 3 ) + d 3 = 0
= (αβ ) [(α + β ) − 2αβ ]
2
which is the required equation.
r q 2 2r r(q 2 − 2 pr )
= 2 − = X Example 33. If the roots of a1 x 2 + b1 x + c1 = 0
p p p p3
are α 1 , β1 and that of a 2 x 2 + b2 x + c2 = 0 are α 2 ,
β 2 such that α 1α 2 = β1β 2 = 1, then
Transformation of Equations a b c
Let α and β be the roots of the equation (a) 1 = 1 = 1
a 2 b2 c2
ax + bx + c = 0, then the equation
2
a b c
(b) 1 = 1 = 1
i. Whose roots are α + k, β + k is obtained by c2 b2 a 2
replacing x by x − k in the given equation. (c) a1 a 2 = b1 b2 = c1 c2
(d) None of the above
ii. Whose roots are α − k, β − k is obtained by
Sol. (b) Roots of the second equation are reciprocal of the
replacing x by x + k in the given equation. first.
1
iii. Whose roots are αk, βk is obtained by As α2 =
α1
multiplying the coefficients of x 2 , x and 1
and β2 =
constant term by k 0 , k 1 , k 2 respectively in the β1
given equation. ∴ Replacing x by
1
in a1 x2 + b1 x + c1 = 0, we get
α β x
iv. Whose roots are , is obtained by c1 x2 + b1 x + a1 = 0 …(i)
k k
and a2 x2 + b2 x + c 2 = 0 [given]…(ii)
multiplying the coefficients of x 2 , x and
Eqs. (i) and (ii) are same equations.
constant term by k 2 , k 1 , k 0 respectively in the c1 b a
∴ = 1 = 1 193
given equation. a2 b2 c 2
5 X Example 34. If α, β and γ are the roots of the
equation x (1 + x 2 ) + x 2 (6 + x ) + 2 = 0, then the
Common Roots
Let a1 x 2 + b1 x + c1 = 0 and a 2 x 2 + b2 x + c2 = 0 be
Objective Mathematics Vol. 1
value of α −1 + β −1 + γ −1 is
two quadratic equations such that a1 , a 2 ≠ 0 and
1 a1 b2 ≠ a 2 b1 .
(a) −3 (b)
2
1 i. When one common root If α is the common
(c) − (d) None of these root of these equations, then
2
( b1 c2 − b2 c1 ) ( a1 b2 − a 2 b1 ) = ( c1 a 2 − c2 a1 ) 2
Sol. (c) Since, 2 x3 + 6 x2 + x + 2 = 0 has roots α, β and γ.
So, 2 x3 + x2 + 6 x + 2 = 0 has roots α −1, β −1 and γ −1
which is the condition for roots of two
quadratic equations to be common.
[writing coefficients in reverse order,
since roots are reciprocal] ii. When two common roots The required
Coefficient of x2 a b c
Hence, sum of the roots = −
Coefficient of x3 condition is 1 = 1 = 1
1 a 2 b2 c2
∴ α −1 + β −1 + γ −1 = −
2
X Example 35. If ax 2 + bx + c = 0 and
Descartes Rule of Signs for the Roots of a bx 2 + cx + a = 0 have a common root and a, b and c
Polynomial are non-zero real numbers, then find the value of
Rule 1 The maximum number of positive real ( a 3 + b 3 + c 3 ) / abc.
roots of a polynomial equation
Sol. Given that, ax2 + bx + c = 0
f ( x ) = a 0 x n + a1 x n − 1 + a 2 x n − 2
and bx2 + cx + a = 0 have a common root. Hence,
+ K + an − 1 x + an = 0
(bc − a2 )2 = (ab − c 2 ) (ac − b 2 )
is the number of changes of the signs of ⇒ b 2c 2 + a4 − 2 a2 bc = a2 bc − ab 3 − ac 3 + b 2c 2
coefficients from positive to negative and
⇒ a4 + ab 3 + ac 3 = 3a2 bc
negative to positive. For instance, in the
a3 + b 3 + c 3
equation x 3 + 3x 2 + 7x − 11 = 0 the signs ⇒ =3
abc
of coefficients are
X Example 36. If the equation x 2 + 5x + 8 = 0
+ + + −
and ax 2 + bx + c = 0; a, b, c ∈ R have a common
As there is just one change of sign, the root, then a : b : c is
number of positive roots of
(a)1 : 5 : 6 (b)1 : 5 : 8
x 3 + 3x 2 + 7x − 11 = 0 is atmost 1. (c) 8 : 5 :1 (d) None of these
Rule 2 The maximum number of negative roots Sol. (b) Since, the quadratic equation x2 + 5 x + 8 = 0 has
of the polynomial equation f ( x ) = 0 is the imaginary roots. So, equation ax2 + bx + c = 0 will have
number of changes from positive to both roots same as the equation x2 + 5 x + 8 has.
negative and negative to positive in the a b c
signs of coefficients of the equation Thus, = = ⇒ a = λ, b = 5λ, c = 8λ
1 5 8
f ( − x ) = 0. Hence, the required ratio is 1 : 5 : 8.
Work Book Exercise 5.4
1 If α and β are the roots of the equation 3 The equations ax 2 + bx + a = 0,
2 x − 3 x − 6 = 0, then equation whose roots are
2
x 3 − 2 x 2 + 2 x − 1 = 0 have two roots in common.
α 2 + 2 , β 2 + 2 is Then, a + b must be equal to
4 x 2 + 49 x + 118 = 0 4 x 2 − 49 x + 118 = 0 1 −1
4 x 2 − 49 x − 118 = 0 x 2 − 49 x + 118 = 0 0 None of these
2 If 2 + i 3 is a root of x 2 + px + q = 0, where 4 If| x 2| + | x| − 2 = 0, then the value of x is equal to
p and q are real, then ( p, q ) is equal to 2
−2
( −4, 7 ) ( 4, − 7 ) ±1
( −7 4) ( 4, 7 ) None of the above
194
5 If| x − 2 | + | x − 9| = 7, then the set of values of x
is
{2, 9} (2, 7) {2} d [2, 9]
11 If α and β are the roots of x 2 + px + q = 0 and α 4 ,
β 4 are the roots of x 2 − r x + s = 0, then the
equation x 2 − 4 qx + 2q 2 − r = 0 has always
5
Inequalities and Quadratic Equation
6 If a, b and c are three distinct positive real two real roots
numbers, then the number of real roots of two positive roots
ax 2 + 2 b| x| + c = 0 is two negative roots
one positive and one negative root
0 1 2 4
7 If sin α, sin β and cos α are in GP, then roots of 12 The real roots of the equation
x + 2 x cot β + 1 = 0 are always
2 | x|3 − 3 x 2 + 3| x| − 2 = 0 are
equal real 0, 2 ±1 ±2 1, 2
imaginary greater than 1
13 If α and β are the roots of the quadratic equation
8 If x 2 + ax + b is an integer for every integer x, 1
6 x 2 − 6 x + 1 = 0, then (a + bα + cα 2 + dα 3 )
then 2
d is always an integer but b need not be an integer 1
b is always an integer but d need not be an integer + (a + bβ + cβ 2 + dβ 3 ) is equal to
Cannot be discussed 2
a and b are always integers 12d + 6 c + 4b + a
12
9 Let a > 0, b > 0 and c > 0. Then, both the roots of 12 a + 6b + 4 c + 9 d
the equation ax 2 + bx + c = 0 1
(12 a + 6b + 4 c + 3 d )
12
are real and negative have negative real part
None of these
are rational numbers None of these
10 If p, q and r are real and p ≠ q, then the roots of 14 Both the roots of the equation
( x − b) ( x − c ) + ( x − a) ( x − c ) + ( x − a) ( x − b) = 0
the equation are always
( p − q ) x 2 + 5 ( p + q )x − 2( p − q ) = 0 are positive negative
real and equal complex real None of these
real and unequal None of these
Quadratic Expression and its Graph
Let a, b and c be real numbers and a ≠ 0. Then, Thus, we may draw the following conclusions:
f ( x ) = ax 2 + bx + c is known as a quadratic expression (i) The parabola will intersect the X -axis in two
or a quadratic polynomial in x. In this section, we shall distinct points iff D > 0
discuss the graph of the quadratic polynomial, i.e. the In this case, the parabola cuts X-axis at (α, 0)
curve whose equation is y = ax 2 + bx + c, a ≠ 0. −b − D −b + D
and (β, 0), where α = and β =
Consider, y = ax 2 + bx + c 2a 2a
For a > 0, we have
4ac − b 2
2
b
= a x + + −D −b
2a 4a 2 y min = at x = , y max → ∞
4a 2a
b
2
D Y
= a x + − 2 y = ax2 + bx + c
2a 4a
2
D b X′ X
⇒ y + = a x + (a,0)
(b,0)
4a 2a
Thus, y = f ( x ) represents a parabola. Y′
The parabola opens upwards or downwards < 0, if α < x <β
accordingly as a > 0 or a < 0
Also, y = f ( x ) is = 0, if x = α, β
When a > 0 When a < 0 > 0, if x < α or x > β
X-axis
For a < 0, we have
a>0
a<0 –D −b
y max = at x = , y min → − ∞
X-axis 4a 2a 195
< 0, if x < α or x > β For a < 0, we have
5 and
y = f ( x ) is = 0, if
> 0, if
x = α, β
α < x <β
y max =
−D
4a
−b
at x = , y min → − ∞
2a
Objective Mathematics Vol. 1
Y and y = f ( x ) < 0, ∀ x
Y
(α,0) (β,0)
X′ X
y = ax2+ bx + c X′ X
Y′ O
(ii) The parabola will just touch the X -axis at one a<0
point iff D = 0 y = ax 2 + bx + c
Y
In this case, the Y′
y = ax 2 + bx + c
parabola touches
Ø ax + bx + c > 0, ∀ x ∈R, iff a> 0 and D < 0 and
2
X-axis at (α, 0), when a>0
ax 2 + bx + c < 0, ∀ x ∈R , iff a< 0 and D < 0.
−b X′ X
α =β = O
2a X Example 37. What is the minimum height of
For a > 0, we have any point on the curve y = x 2 − 4x + 6 above the
y min = 0 Y′ X-axis?
−b
at x = , y max → ∞ Sol. Here, a = 1 > 0 and D = 16 − 4(6) = − 8 < 0
2a Therefore, the parabola remains completely above
> 0, if x ≠ α X-axis and minimum value of any point on the curve is
and y = f ( x ) is −D
= 0, if x = α given by
4a
.
For a < 0, we have −D 8
i.e. ymin = = =2
−b 4a 4
y max = 0 at x = , y min → − ∞
2a Hence, the minimum height is 2.
< 0, if x ≠ α
and y = f (x ) = X Example 38. The roots of ax 2 + bx + c = 0,
0, if x = α where a ≠ 0 and coefficients are real, are non-real
Y
complex and a + c < b. Then,
(a) 4a + c > 2b
(b) 4a + c < 2b
X′ X
O (c) 4a + c = 2b
a<0 (d) None of the above
y = ax2 + bx + c
Y′ Sol. (b) Since, f( x) = ax2 + bx + c = 0 has non-real roots.
(iii) The parabola will not intersect X-axis iff D < 0 Thus, either
In this case, the parabola remains either f( x) > 0 or f( x) < 0, ∀ x
completely above X-axis or completely below As a + c < b ⇒ f(−1) < 0
X-axis according as a > 0 or a < 0. Y
For a > 0, we have
−D −b
y min = at x = , y max → ∞ X′
O
X
4a 2a
and y = f ( x ) > 0, ∀ x
Y
y = ax2 + bx + c
Y¢
a>0
i.e. a − b + c < 0 or a + c < b
X′ X
O ∴ f( x) < 0, ∀ x
Thus, for all x ∈ R; ax2 + bx + c < 0.
Now, for x = − 2; 4 a − 2 b + c < 0
196 Y′ ⇒ 4a + c < 2b
Solution of Quadratic Inequations
Let f ( x ) = ax 2 + bx + c, where a, b, c ∈ R and a ≠ 0.
Step III Obtain the discriminant of the quadratic
equation in Step II.
Step IV Put discriminant ≥ 0 and solve the
5
Inequalities and Quadratic Equation
Then, f ( x ) ≥ 0, f ( x ) < 0, f ( x ) ≤ 0, f ( x ) > 0 are inequation for y. The values of y so
quadratic inequations. The set of all real values of x, obtained determines the set of values
which satisfy an inequation is called its solution set. attained by the given rational expression.
The following example will make it clear:
X Example 40. If x is real, then prove that the
X Example 39. Solve the inequation x 2 − 3x + 4 1
values of 2 lie between and 7.
−x 2 + 3x + 4 < 0. x + 3x + 4 7
Sol. We have,
Sol. Let y = x2 − 3 x + 4. Then,
2
− x + 3x + 4 < 0
2
x + 3x + 4
⇒ x − 3x − 4 > 0
2
x ( y − 1) + 3 x( y + 1) + 4( y − 1) = 0
2
⇒ ( x − 4) ( x + 1) > 0 Now, if D≥ 0
+ – + ⇒ 9( y + 1)2 − 16( y − 1)2 ≥ 0
–1 4 ⇒ − 7 y2 + 50 y − 7 ≥ 0
⇒ x< −1 ⇒ 7 y2 − 50 y + 7 ≤ 0
or x> 4 ⇒ (7 y − 1)( y − 7 ) ≤ 0
⇒ x ∈ (− ∞, − 1) ∪ (4, ∞ ) 1
⇒ ≤ y≤ 7
7
Ø Frequently used inequalities 1
Hence, the given expression lies between and 7.
● (x − a)(x − b) < 0 ⇒ x ∈(a, b) , where a < b 7
(x − a)(x − b) > 0 ⇒ x ∈(−∞ , a) ∪ (b , ∞), where a < b
x 2 −1
●
● x 2 ≤ a2 ⇒ x ∈[− a, a] X Example 41. If f (x ) = for every real
● x 2 ≥ a2 ⇒ x ∈(−∞ , − a] ∪ [a, ∞) x 2 +1
● If ax 2 + bx + c < 0, (a> 0) number x, then the minimum value of f
(a) does not exist because f is unbounded
⇒ x ∈(α , β) , where α and β (α < β) are roots of the equation
ax 2 + bx + c = 0 (b) is not attained even though f is bounded
(c) is equal to 1
● If ax 2 + bx + c > 0, (a> 0) (d) is equal to −1
⇒ x ∈(−∞ , α ) ∪ (β , ∞) , where α and β(α < β) are roots of
Sol. (d) Let y = x2 − 1
2
the equation ax 2 + bx + c = 0.
x +1
⇒ yx2 + y − x2 + 1 = 0
Maximum and Minimum Values ⇒ x2 ( y − 1) + y + 1 = 0
of Rational Expression Now, for real values of x consider D ≥ 0
⇒ − 4( y2 − 1) ≥ 0
Here, we shall find the values attained by a rational ⇒ y2 − 1 ≤ 0
a x 2 + b1 x + c1 ⇒ − 1≤ y ≤ 1
expression of the form 1 2 for real values
a 2 x + b2 x + c2 Thus, minimum value of f( x) is − 1.
of x.
The following algorithm will explain the Location of the Roots of a
procedure: Quadratic Equation
Algorithm Here, we shall discuss various conditions satisfied
by the coefficients of a quadratic equation when the
Step I Equate the given rational expression to y.
location of its roots is given.
Step II Obtain a quadratic equation in x by
Let f ( x ) = ax 2 + bx + c, a, b, c ∈ R , a ≠ 0 and α, β be
simplifying the expression in Step I.
the roots of f ( x ) = 0. Suppose k , k1 , k 2 ∈ R and k1 < k 2 .
197
5 Then, the following hold good :
i. Conditions for a number k (If both the roots
ii. Conditions for a number k (If both the roots
of f ( x ) = 0 are greater than k)
of f ( x ) = 0 are less than k)
Objective Mathematics Vol. 1
a>0
f (k )
a>0 – b
f (k ) 2a
– b α X
k β
2a
α X
β k
– b ,– D
2a 4a
– b ,– D
2a 4a
– b ,– D
2a 4a
– b ,– D
2a 4a
k α β
X
α β k – b
X 2a
f (k )
– b a<0
2a
f(k)
a<0
(a) D ≥ 0 (roots may be equal)
(a) D ≥ 0 (roots may be equal) (b) af ( k ) > 0
(b) af ( k ) > 0 b
(c) k < − , where α ≤ β
b 2a
(c) k > − , where α ≤ β
2a X Example 43. For what values of m ∈ R , both
roots of the equation x 2 − 6mx + 9m 2 − 2m + 2 = 0
X Example 42. Find the values of m for which
exceed 3?
both roots of equation x 2 − mx + 1 = 0 are less than
unity. Sol. Let f( x) = x2 − 6mx + 9m2 − 2 m + 2.
as both roots of f( x) = 0 are greater than 3, we can take
Sol. Let f( x) = x2 − mx + 1, as both roots of f( x) = 0 are less b
D ≥ 0, af(3) > 0 and − > 3.
than 1, we can take D ≥ 0, af(1) > 0 2a
b Case I Consider, D ≥ 0
and − < 1.
2a (−6 m)2 − 4 ⋅ 1⋅ (9m2 − 2 m + 2 ) ≥ 0 ⇒ 8m − 8 ≥ 0
Case I Consider, D ≥ 0 ∴ m ≥ 1 or m ∈ [1, ∞ ) …(i)
(− m)2 − 4 ⋅ 1⋅ 1 ≥ 0 Case II Consider, af (3) > 0
1⋅ (9 − 18m + 9m2 − 2 m + 2 ) > 0
⇒ (m + 2 ) (m − 2 ) ≥ 0
⇒ 9m2 − 20m + 11 > 0
⇒ m ∈ (− ∞, − 2 ] ∪ [2, ∞ ) …(i)
⇒ (9m − 11) (m − 1) > 0
Case II Consider, af (1) > 0
⇒ m − 11 (m − 1) > 0
1⋅ (1 − m + 1) > 0
9
⇒ m− 2 < 0
m ∈ (− ∞, 1) ∪ , ∞
11
⇒ m< 2 ⇒ …(ii)
9
⇒ m ∈ (− ∞, 2 ) …(ii)
b
b Case III Consider, − >3
Case III Consider, − <1 2a
2a
+ +
+ + –
– +1 11/9
–2 2 6m
m > 3 ⇒ m> 1
<1 2
2 ⇒ m ∈ (1, ∞ ) …(iii)
⇒ m< 2
⇒ m ∈ (−∞, 2 ) …(iii) Hence, the values of m satisfying Eqs. (i), (ii) and (iii) at
the same time are m ∈ , ∞ .
11
Hence, the values of m satisfying Eqs. (i), (ii) and (iii) at
9
the same time are m ∈ (− ∞, − 2 ].
198
iii. Conditions for a number k (If k lies between
the roots of f ( x ) = 0)
X Example 45. Find the values of m for which
exactly one root of the equation
x 2 − 2mx + m 2 − 1 = 0 lies in the interval ( − 2, 4).
5
Inequalities and Quadratic Equation
a>0 Sol. Let f( x) = x2 − 2 mx + m2 − 1as exactly one root of
f( x) = 0 lies in the interval, we can take D > 0 and
k
X f(−2 ) f(4) < 0.
α β
f (k) Case I Consider, D > 0
(− 2 m)2 − 4 ⋅ 1⋅ (m2 − 1) > 0
– b ,– D ⇒ 4> 0
2a 4a
∴ m∈R …(i)
– b ,– D Case II Consider, f(−2 ) f(4) < 0
2a 4a
f (k ) +
+ +
X – –
α k β –3 –1 3 5
(4 + 4m + m2 − 1) (16 − 8m + m2 − 1) < 0
a<0 ⇒ (m2 + 4m + 3) (m2 − 8m + 15) < 0
⇒ (m + 1) (m + 3) (m − 3) (m − 5) < 0
(a) D > 0 (b) af ( k ) < 0, where α < β ⇒ (m + 3) (m + 1) (m − 3)(m − 5) < 0
∴ m ∈ (−3, − 1) ∪ (3, 5) ...(ii)
X Example 44. Find all values of p, so that 6 Hence, the values of m satisfying Eqs. (i) and (ii) at the
lies between roots of the equation same time are m ∈ (−3, − 1) ∪ (3, 5).
x 2 + 2 ( p − 3) x + 9 = 0.
Sol. Let f( x) = x2 + 2 ( p − 3) x + 9
v. Conditions for numbers k 1 and k 2 (If both
as 6 lies between the roots of
–
+ + roots of f ( x ) = 0 are confined between
f( x) = 0, we can take D > 0 and 0 6 k1 and k 2 )
af(6) < 0.
Case I Consider, D > 0
{2 ( p − 3)}2 − 4 ⋅ 1⋅ 9 > 0
a>0
⇒ ( p − 3)2 − 9 > 0 ⇒ p( p − 6) > 0
f (k1) f (k2)
⇒ p ∈ (− ∞, 0) ∪ (6, ∞ ) …(i)
Case II Consider, af (6) < 0 X
k1 α β k2
1 ⋅ {36 + 12 ( p − 3) + 9} < 0
3
⇒ 12 p + 9 < 0 ⇒ p + < 0
4 – b,– D
⇒ p ∈ (−∞, − 3/ 4) …(ii) 2a 4a
Hence, the values of p satisfying Eqs. (i) and (ii) at the – b,– D
same time are p ∈ (− ∞, − 3 / 4). 2 a 4a
iv. Conditions for numbers k 1 and k 2 k1 α β k2
(If exactly one root of f ( x ) = 0 lies in the X
interval ( k1 , k 2 )). f (k1) f (k2)
a<0
a>0
f (k1)
(a) D ≥ 0 (roots may be equal)
k2
k1 α β
X (b) af ( k1 ) > 0
f(k2) (c) af ( k 2 ) > 0
b
(d) k1 < − < k 2 , where α ≤ β and k1 < k 2
f (k2) 2a
k1
α β
X X Example 46. Find all values of a for which the
k2
equation 4x 2 − 2x + a = 0 has two roots in the
f (k1)
a<0 interval ( −1, 1).
Sol. Let f( x) = 4 x2 − 2 x + a as both roots of the equation
(a) D > 0 f( x) = 0 lie between (−1, 1,) we can take D ≥ 0, af (−1) > 0 ,
1
(b) f ( k1 ) f ( k 2 ) < 0, where α < β af (1) > 0 and −1 < < 1. 199
4
5 Case I Consider, D ≥ 0
(−2 )2 − 4 ⋅ 4 ⋅ a ≥ 0 ⇒ a ≤
1
4
a ∈ − ∞,
1
X Example 48. If the roots of the equation
x 2 − 2ax + a 2 + a − 3 = 0 are real and less than 3,
then
Objective Mathematics Vol. 1
⇒ …(i)
4 (a) a < 2 (b) 2 ≤ a ≤ 3
Case II Consider, af (−1) > 0 (c) 3a < a ≤ 4 (d) a > 4
4(4 + 2 + a) > 0
⇒ a > − 6 ⇒ a ∈ (−6, ∞ ) …(ii)
Sol. (a) Let f( x) = x2 − 2 ax + a2 + a − 3
Case III Consider, af (1) > 0 Since, both roots are less than 3.
4(4 − 2 + a) > 0 i.e. α < 3, β < 3
⇒ a > − 2 ⇒ a ∈ (−2, ∞ ) …(iii) Sum (S ) = α + β < 6
α+β 2a
Hence, the values of a satisfying Eqs. (i), (ii) and (iii) at ⇒ <3 ⇒ <3
2 2
the same time are a ∈ −2, .
1
4 ⇒ a< 3 …(i)
Again, product (P) = αβ
vi. Conditions for numbers k 1 and k 2 (If k1 and ⇒ P < 9 ⇒ αβ < 9
⇒ a2 + a − 3 < 9
k 2 lie between the roots of f ( x ) = 0)
⇒ a2 + a − 12 < 0
⇒ (a − 3) (a + 4) < 0
a>0
⇒ −4 < a < 3 …(ii)
k1 k2
α X
f (k2) β
f (k1)
f (k1)
f (k2) β α β 3
α
X
k1 k2 Again, D = B − 4 AC ≥ 0
2
⇒ (−2 a)2 − 4 ⋅ 1 (a2 + a − 3) ≥ 0
a< 0
⇒ 4a2 − 4a2 − 4a + 12 ≥ 0
⇒ −4a + 12 ≥ 0
(a) D > 0 ⇒ a≤ 3 …(iii)
(b) af ( k1 ) < 0 Again, af(3) > 0
(c) af ( k 2 ) < 0, where α < β ⇒ 1[(3)2 − 2 a(3) + a2 + a − 3] > 0
⇒ 9 − 6 a + a2 + a − 3 > 0
X Example 47. Find the values of a for which
⇒ a2 − 5 a + 6 > 0
one root of equation ( a − 5) x 2 − 2ax + a − 4 = 0 is
smaller than 1 and the other greater than 2. ⇒ (a − 2 ) (a − 3) > 0
∴ a ∈ (−∞, 2 ) ∪ (3, ∞ ) …(iv)
Sol. Let f( x) = (a − 5)x2 − 2 ax + a − 4(a ≠ 5) as 1 and 2 lie
From Eqs. (i), (ii), (iii) and (iv), we get
between the roots of f( x) = 0, we can take D > 0,
(a − 5)f(1) < 0 and (a − 5) f(2 ) < 0. a ∈ (−4, 2 )
Case I Consider, D > 0
(−2 a)2 − 4 ⋅(a − 5) ⋅ (a − 4) > 0 Algebraic Interpretation of
⇒ 9a − 20 > 0 ⇒ a ∈ (20/ 9, ∞ ) …(i)
Case II Consider, (a − 5) f(1) < 0 Rolle’s Theorem
(a − 5) (a − 5 − 2 a + a − 4) < 0 Let f ( x ) be a polynomial having α and β as its
⇒ (a − 5) (−9) < 0 roots such that α < β. Then, f (α ) = f (β) = 0. Also, a
⇒ a − 5> 0
⇒ a ∈ (5, ∞ ) …(ii)
polynomial function is everywhere continuous and
Case III Consider, (a − 5) f(2 ) < 0
differentiable. Thus, f ( x ) satisfies all the three
conditions of Rolle’s theorem. Consequently, there
+ + exists γ ∈(α, β) such that f ′ ( γ ) = 0, i.e. f ′ ( x ) = 0 at
–
5 24 x = γ. In other words, x = γ is a root of f ′ ( x ) = 0. Thus,
(a − 5) {4(a − 5) − 4a + a − 4} < 0 algebraically Rolle’s theorem can be interpreted as
⇒ (a − 5)(a − 24) < 0 follows:
⇒ a ∈(5, 24) …(iii) Between any two roots of a polynomial f ( x ), there
Hence, the values of a satisfying Eqs. (i), (ii) and (iii) at
200 the same time are a ∈(5, 24). is always a root of its derivative f ′ ( x ).
Example 49. If a + b + c = 0, then the
X
quadratic equation 3ax 2 + 2bx + c = 0 has
(a) atleast one root in (0, 1)
Sol. (a) We have,
Let
x 3 − [ x] = 3 ⇒ x 3 − 3 = [ x]
f( x) = x3 − 3 and g ( x) = [ x]
5
Inequalities and Quadratic Equation
(b) both imaginary roots Y
(c) one root in (1, 2), other in ( −1, 0)
(d) None of the above 3 y = f (x)
Sol. (a) Consider the function f( x) = ax + bx + cx
3 2
2 y = g ( x)
f( x) being a polynomial function is continuous and
differentiable, ∀ x ∈ R 1 (4,1/3 1)
f(0) = 0, f(1) = a + b + c = 0 [given] X′ O X
⇒ f(0) = f(1) –2 –1 1 2 3
–1
Thus, f( x) satisfies all conditions of Rolle’s theorem. So,
there exists a point k ∈(0, 1) such that f ′(k ) = 0. –2
–3
Condition for Resolution into
Linear Factors Y′
Solving the given equations means finding the
The quadratic function x-coordinates of the points of intersection of the curves
ax 2 + 2hxy + by 2 + 2gx + 2 fy + c is resolvable into y = f( x) and y = g ( x).
If 1 < x < 2, we have g ( x) = 1 and f( x) = x3 − 3
linear rational factors iff
At the point of intersection of the two curves, we have
∆ = abc + 2 fgh − af 2
− bg 2 − ch 2 = 0 x3 − 3 = 1 ⇒ x = 41/ 3
a h g Hence, x = 4 is the solution of the equation
1/ 3
x 3 − [ x] = 3
i.e. h b f =0
g f c X Example 52. The number of solutions of the
equation | x | = cos x are
X Example 50. For what values of m can the (a) one (b) two (c) three (d) zero
expression 2x 2 + mxy + 3 y 2 − 5 y − 2 be expressed Sol. (b) Here,| x| = cos x Y
y=–x y=x
as the product of two linear factors? Thus, to find value of x
for which both curves
Sol. Comparing the given equation with y = cos x
have point of
ax2 + 2 hxy + by2 + 2 gx + 2 fy + c = 0, we have X′ X
intersections.
m 5 –π/2 O π/2
a = 2, h = , b = 3, c = − 2, f = − and g = 0 Clearly, there are two
2 2
points of intersection of
The given expression is resolvable into linear factors, if Y′
y =| x| and y = cos x.
abc + 2 hxy − af 2 − bg 2 − ch2 = 0 Hence, there are two real solutions.
25 m2
⇒ −12 − + 2 = 0
2 4
X Example 53. If b > a, then the equation
⇒ m2 = 49 ( x − a )( x − b) − 1 = 0 has
∴ m= ±7 (a) both roots in ( a, b)
(b) both roots in ( −∞, a )
(c) both roots in ( b, + ∞ )
Some Application of Graphs to (d) one root in ( −∞, a ) and the other in ( b, ∞ )
Find the Roots of Equations Sol. (d) Y
y = (x – a)(x – b) – 1
Here, we shall discuss some examples to find the
roots of equations with the help of graphs.
X Example 51. If x 3 − [x ] = 3, where [x ] denotes α a b β
X
the greatest integer less than or equal to x, then
(a) x ∈{41/ 3 } 1
(b) x ∈ R From graph, it is clear that one of the roots of
(c) Cannot be discussed ( x − a) ( x − b ) − 1 = 0 lies in (−∞, a) and other lies in
(b, ∞ ).
201
(d) None of the above
5 Work Book Exercise 5.5
1 The number of values of k for which 9 Let S denotes the set of real values of d for
Objective Mathematics Vol. 1
{ x 2 − (k − 2 ) x + k 2 }{ x 2 + kx + (2 k − 1)} is a which the roots of the equation x 2 − ax − a2 = 0
perfect square, is exceeds d, then S equals
a 1 b 2 a ( −∞, 0)
b −2, −
c 0 d None of these 1
2 If x 2 − 4 x + log1/ 2 a = 0 does not have two distinct 2
c − ,
1 1
real roots, then maximum value of a is
2 4
1 1
d null set
4 16
−
1
None of these 10 x 4 − 4 x − 1 = 0 has
4 a exactly two positive real roots
3 If c > 0 and 4 a + c < 2 b, then a x 2 − bx + c = 0 has b exactly two negative real roots
c exactly two real roots
a root in the interval
d None of the above
a (0, 2) b (2, 4)
c (0, 1) (−2, 0) 11 The roots α and β of the quadratic equation are
ax 2 + bx + c = 0 are real and of opposite sign.
4 Conditions on a and b for which x − ax − b is 2 2
Then, the roots of the equation
less than zero for atleast one positive x are
α ( x − β )2 + β( x − α )2 = 0 are
a a − 3 > 0, b < 0 b a − 3 > 0, b > 0
c a, b ∈ R d Cannot be discussed a positive
b negative
5 The set of possible values of λ for which c real and of opposite sign
x 2 − (λ2 − 5λ + 5)x + (2 λ2 − 3λ − 4) = 0 has roots, d imaginary
whose sum and product are both less than 1, is 12 The graph of curve x 2 = 3 x − y − 2 is
−1, 5 3
(1, 4) a between the lines x = 1 and x =
2 2
1, 5 1, 5 b between the lines x = 1 and x = 2
2
2 c strictly below the line 4 y = 1
d None of the above
6 If ax 2 + bx + 6 = 0 does not have two distinct real
roots, then the least value of 3 a + b is x2 + 5
13 If 0 < a < 5,0 < b < 5 and = x − 2 cos(a + bx )
2 −2
2
1 −1 is satisfied for atleast one real x, then the
greatest value of a + b is
7 If x is real, then the least value of the expression a π
x 2 − 6x + 5 π
is b
x2 + 2 x + 1 2
c 3π
1
−1 − d 4π
2
−
1
None of these
14 If the roots of ax 2 − bx − c = 0 change by the
3 same quantity, then the expression in a, b and c
8 If the quadratic equation that does not change, is
α x 2 + β x + a2 + b2 + c 2 − a b − bc − ca = 0 has b 2 − 4ac
a
imaginary roots, then a2
b − 4c
2(α − β ) + ( a − b )2 + ( b − c )2 + (c − a )2 > 0 b
a
2(α − β ) + ( a − b )2 + ( b − c )2 + (c − a )2 < 0 b 2 + 4ac
c
2(α − β ) + ( a − b )2 + ( b − c )2 + (c − a )2 = 0 a2
None of the above d None of the above
202
WorkedOut Examples
Type 1. Only One Correct Option
Ex 1. If 0x 2 + x + | x | + 1 ≤ 0, then x lies in ⇒ −3x ≥ 6
or 3x ≥ 6
(a) ( 0, ∞ ) (b) ( − ∞ , 0)
⇒ x ≤ −2
(c) R (d) φ or x≥2
Sol. x 2 + x + | x | + 1 ≤ 0 gives two cases due to | x | . ⇒ x ∈ (− ∞ , − 2 ] ∪ [ 2, ∞ )
Case I When x ≥ 0 …(i) Hence, (b) is the correct answer.
x + 2x + 1 ≤ 0 or
2
(x + 1) ≤ 0
2
Ex 3. If a1 , a 2 , ..., a n are positive real numbers whose
⇒ x + 1= 0 product is a fixed real number c, then the
∴ x = − 1 but x ≥ 0 [from Eq. (i)] minimum value of a1 + a 2 + ... + a n − 1 + 2a n is
∴ No value of x ≥ 0.
(a) n ( 2c )1/ n
Case II When x < 0 …(ii)
(b) ( n + 1) c1/ n
x2 + x − x + 1 ≤ 0
(c) 2 nc1/ n
⇒ x2 + 1 ≤ 0
which is not true for any value of x < 0 (d) ( n + 1) ( 2 c )1/ n
∴ No value of x < 0 Sol. We have,
From above two cases no value of x is possible 1
i.e. x ∈ φ. (a + a2 + ... + an −1 + 2an )
n 1
Hence, (d) is the correct answer. ≥ [ a1 a2 ... an −1 (2a n )]1 / n = (2c)1 / n [using AM ≥ GM]
Ex 2. If | x − 1| + | x| + | x + 1| ≥ 6, then x lies in ⇒ a1 + a2 + ... + an − 1 + 2an ≥ n (2c)1 / n
(a) ( − ∞ , 2] (b) ( − ∞ , − 2] ∪ [ 2, ∞ )
Thus, minimum value of a1 + a2 + ... + a n −1 + 2a n is
(c) R (d) φ
n (2c)1 / n .
Sol. | x − 1| + | x | + | x + 1| ≥ 6, gives four cases Hence, (a) is the correct answer.
Case I When x < − 1 …(i)
Then, − (x + 1) − x − (x − 1) ≥ 6 Ex 4. Solution set of
⇒ −x − 1 − x − x + 1 ≥ 6 3
log 3 ( x 2 − 2) < log 3 | x | − 1 is
⇒ −3x ≥ 6 or x ≤ − 2 …(ii) 2
From Eqs. (i) and (ii), x ≤ − 2
(a) ( − 2 , − 1) (b) ( − 2, − 2 )
Case II When −1 ≤ x ≤ 0 …(iii)
(c) ( − 2 , 2) (d) None of these
⇒ x + 1 − x − (x − 1) ≥ 6
⇒ 1− x + 1≥ 6 ⇒ x ≤ −4 …(iv) Sol. We have,
∴ No value of x. [from Eqs. (iii) and (iv)] x 2 − 2 > 0,
3
|x| − 1 > 0
Case III When 0 ≤ x ≤ 1 …(v) 2
3
⇒ x + 1 + x − (x − 1) ≥ 6 and x2 − 2 < |x| − 1
2
⇒ x≥4 …(vi) 2 3
No solution, using Eqs. (v) and (vi). ⇒ x 2 > 2, | x | > and | x |2 − | x | − 1 < 0
3 2
Case IV When x > 1 …(vii) ⇒ | x | > 2 and | x | < 2
⇒ x + 1+ x + x −1≥ 6 ⇒ x ∈ (−2, − 2 ) ∪ ( 2, 2)
⇒ 3x ≥ 6 or x ≥ 2 …(viii)
Hence, (d) is the correct answer.
From Eqs. (vii) and (viii), x ≥ 2
Thus, from above four cases, Ex 5. Solution set of the inequality
x ≤ − 2 or x ≥ 2 1 1
⇒ x ∈ (− ∞ , − 2 ] ∪ [ 2, ∞ ) > is
2 − 1 1 − 2 x− 1
x
Aliter
4
As discussed above, solution exists only when all are (a) (1, ∞ ) (b) 0, log 2
positive or negative: 3
Thus, | x + 1| + | x | + | x − 1| ≥ 6
4
⇒ − (x + 1) − (x ) − (x − 1) ≥ 6 (c) ( −1, ∞ ) (d) 0, log 2 ∪ (1, ∞ )
or (x + 1) + x + (x − 1) ≥ 6 3 203
5 Sol. Put 2x = t, then t > 0. The given inequality becomes
1
>
2
t −1 2 − t
Ex 8. If roots of the equation
x 4 − 8x 3 + bx 2 + cx + 16 = 0
Objective Mathematics Vol. 1
1 2 are positive, then
⇒ − >0
t −1 2− t (a) b = 8 = c (b) c = − 32, b = − 24
2 − t − 2t + 2 (c) b = 24, c = − 32 (d) c = 32, b = 24
⇒ >0
(t − 1) (2 − t ) Sol. If x1 , x2 , x3 and x4 are four roots of given equation, then
4 − 3t x1 + x2 + x3 + x4 = 8 and x1 x2 x3 x4 = 16
⇒ >0
(t − 1) (2 − t ) As AM ≥ GM, we get
(3t − 4 ) 1
⇒ >0 2 = (x1 + x2 + x3 + x4 ) ≥ (x1 x2 x3 x4 )1 / 4 = 2
(t − 1) (t − 2) 4
– + – + i.e. AM = GM
–∞ ∞
1 4 2 Thus, x1 = x2 = x3 = x4 = 2
—
3 ∴ x 4 − 8x 3 + bx 2 + cx + 16 = (x − 2)4
4
⇒ 1< t < or t > 2 ⇒ b = 24 and c = − 32
3 Hence, (c) is the correct answer.
4
⇒ 1 < 2x < or 2x > 2
3 Ex 9. If a, b and c are three positive real numbers
4
⇒ 0 < x < log2 or x > 1 such that a + b ≥ c, then
3
a b c a b c
Hence, (d) is the correct answer. (a) + ≥ (b) + <
1+ a 1+ b 1+ c 1+ a 1+ b 1+ c
Ex 6. Given, n 4 < 10 n for a fixed positive integer a b c
(c) + > (d) None of these
n ≥ 2, then 1+ a 1+ b 1+ c
(a) ( n + 1) 4 < 10n + 1 (b) ( n + 1) 4 > 10n + 1
Sol. We have,
(c) ( n − 1) 4 < 10n + 2 (d) None of these a b a b
+ ≥ +
1+ a 1+ b 1+ a + b 1+ a + b
Sol. We have,
4 4 4 4 a+ b 1
n + 1 1 1 3 = =
= 1 + ≤ 1 + = < 10 [Q n ≥ 2] 1+ a + b 1
n n 2 2 +1
a+ b
⇒ (n + 1)4 ≤ n4 ⋅ 10 < 10n ⋅ 10 = 10n + 1
Since, a + b ≥ c, we get
Hence, (a) is the correct answer. 1 1 1 1 1 1
≤ ⇒1+ ≤1+ ⇒ ≥
Ex 7. If x1 , x 2 , x 3 and x 4 are four positive real a+ b c a+ b c 1
+1
1
+1
1 1 a+ b c
numbers such that x1 + = 4, x 2 + =1, a b 1 1 c
x2 x3 Thus, + ≥ ≥ =
1+ a 1+ b 1 + 1 1 + 1 1+ c
1 1
x 3 + = 4 and x 4 + = 1, then c c
x4 x1 Hence, (a) is the correct answer.
(a) x1 = x 3 and x 2 = x 4 (b) x 2 = x 4 but x1 ≠ x 3
(c) x1 x 2 = 1, x 3 x 4 ≠ 1 (d) x 3 x 4 = 1, x1 x 2 ≠ 1 Ex 10. If a i > 0 for i =1, 2, ..., n and a1 a 2 ... a n = 1, then
Sol. Using AM ≥ GM, minimum value of (1 + a1 ) (1 + a 2 ) ... (1 + a n ) is
1 x 1 x2 (a) 2n/ 2 (b) 2n
x1 + ≥ 2 1 ; x2 + ≥2 (c) 22n (d) 1
x2 x2 x3 x3
1 x3 1 x Sol. We have,
x3 + ≥2 ; x4 + ≥2 4 1
x4 x4 x1 x1 (1 + a1 ) ≥ 1⋅ a1 = a1 [using A.M ≥ G.M.]
2
1 1 1 1 1
⇒ x1 + x2 + x3 + x4 + ≥ 24 (1 + a2 ) ≥ 1⋅ a2 = a2
x2 x3 x4 x1 2
1 1 1 1
M M M
But x1 + x2 + x3 + x4 + = 16 1
(1 + an ) ≥ 1⋅ an = an
x2 x3 x4 x1 2
1 1 1 1 On multiplying above inequalities, we get
∴ x1 = , x2 = , x3 = , x4 =
x2 x3 x4 x1 1
(1 + a1 ) (1 + a2 ) ... (1 + an ) ≥ a1 a2 ... an = 1
1 1 2n
⇒ x1 = 2, x2 = , x3 = 2, x4 = ⇒ (1 + a1 ) (1 + a2 ) ... (1 + an ) ≥ 2n
2 2
204 Hence, (a) is the correct answer. Hence, (b) is the correct answer.
Ex 11. If a, b and c are the sides of a triangle, then
a
+
b
+
c
=
1
4
1
[ 4 a2 + (b + c)2 − 4 a (b + c)]
= [ 2a − (b + c)]2 ≥ 0
5
Inequalities and Quadratic Equation
b+c−a c+a −b a +b−c 4
(a) ≤ 3 (b) ≥ 3 3
Similarly, a2 + b2 + c2 − bc − ca − ab ≥ | c − a|
(c) ≥ 2 (d) ≤ 2 2
3
Sol. As a, b and c are sides of a triangle, b + c − a, and a2 + b2 + c2 − bc − ca − ab ≥ | a − b|
c + a − b, a + b − c > 0. 2
Let x = b + c − a, y = c + a − b and z = a + b − c ⇒ a2 + b2 + c2 − ab − bc − ca
⇒ y + z = 2a, z + x = 2b and x + y = 2c 3
≥ max{| b − c |, | c − a |, | a − b |}
Thus, LHS of the inequality 2
y+ z z+ x x + y Hence, (a) is the correct answer.
= + +
2x 2y 2z
Ex 14. If a, b and c are three distinct positive real
1 y x x z y z
= + + + + + numbers, then the number of real roots of
2 x y z x z y
ax 2 + 2b | x| − c = 0 is
1/ 6
6 y x x z y z (a) 4
≥ ⋅ ⋅ ⋅ ⋅ ⋅ = 3 [Q AM ≥ GM]
2 x y z x z y (b) 2
Hence, (b) is the correct answer. (c) 0
(d) None of the above
Ex 12. If a + b + c = 6, then
Sol. a| x |2 + 2b| x | − c = 0
4a + 1 + 4b + 1 + 4c + 1 −2b ± 4 b2 + 4 ac
∴ |x| =
(a) ≤ 9 (b) ≥ 9 (c) > 9 (d) < 9 2
Sol. By the Cauchy-Schwartz’s inequality, = −b± b2 + ac
( 4 a + 1) + 4b + 1 + 4 c + 1) 2
Since, a, b and c are positive. So, | x | = − b + b2 + ac
≤ (1 + 1 + 1) (4 a + 1 + 4 b + 1 + 4 c + 1)
∴ x has two real values, neglecting
= 3 [ 4 (a + b + c) + 3 ] = (3) (27)
| x | = − b − b2 + ac, as | x | ≥ 0
⇒ 4a + 1 + 4b + 1 + 4c + 1 ≤ 9
Hence, (b) is the correct answer.
Hence, (a) is the correct answer.
Ex 15. If the ratio of the roots of λx 2 + µx + ν = 0 is
Ex 13. If a, b and c ∈ R, then the square root of equal to the ratio of the roots of x 2 + x + 1 = 0,
a 2 + b 2 + c 2 − bc − ca − ab is greater than or then λ, µ and ν are in
equal to (a) AP
3 (b) GP
(a) max{| b − c|, | c − a|, | a − b|}
2 (c) HP
3 (d) None of the above
(b) max{| b − c|, | c − a|, | a − b|}
2 Sol. As ratio of roots for λ x 2 + µ x + v = 0 and
(c) max{| b − c|, | c − a|, | a − b|} x 2 + x + 1 = 0 are equal.
3 α α′
(d) max{| b − c|, | c − a|, | a − b|} ∴ = , where α and β are roots of λx 2 + µx + v = 0
4 β β′
Sol. We have, a2 + b2 + c2 − bc − ca − ab and α ′ , β′ are roots of x 2 + x + 1 = 0
1 Now, ω and ω 2 are the roots of x 2 + x + 1 = 0
= [(b2 + c2 − 2bc) + (c2 + a2 − 2ca) α ω 1
2 ∴ = 2 = ⇒ β = ωα
+ (a2 + b2 − 2ab)] β ω ω
µ ν
1
= [(b − c)2 + (c − a)2 + (a − b)2 ] ≥ 0 and α + β = − , αβ =
2 λ λ
µ ν
3 ⇒ α (1 + ω ) = − , α 2 ω =
Also, a2 + b2 + c2 − bc − ca − ab − (b − c)2 λ λ
4 −µ 2 ν
1 ⇒ − αω 2 = ,α ω =
= [ 4 a2 + 4 b2 + 4 c2 − 4 bc − 4 ca − 4 ab λ λ
4
µ2 ν
−3 (b2 + c2 − 2bc)] ⇒ = or µ 2 = λν
1 λ2 λ
= [ 4 a2 + b2 + c2 + 2bc − 4 a(c + b)] Hence, (b) is the correct answer.
4 205
5 Ex 16. If x 2 − 2r ⋅ pr x + r = 0; r =1, 2, 3 are three
quadratic equations of which each pair has
exactly one root common, then the number of
Thus, min f (x ) > max g (x )
⇒
⇒
2c2 − b2 > b2 + c2
c2 > 2b2
Objective Mathematics Vol. 1
solutions of the triplet ( p1 , p2 , p3 ) is
| c| > 2 | b|
(a) 2 (b) 1 (c) 9 (d) 27
Hence, (d) is the correct answer.
Sol. x 2 − 2 p1 x + 1 = 0 has roots α , β
x 2 − 4 p2 x + 2 = 0 has roots α , γ Ex 18. If a ∈ R and the equation
x − 6 p3 x + 3 = 0 has roots β , γ
2
( a − 2) ( x − [ x ]) 2 + 2( x − [ x ]) + a 2 = 0
⇒α + β = 2 p1 , α + γ = 4 p2 , β + γ = 6 p3 , αβ = 1, (where, [ x ] denotes the greatest integer ≤ x) has
αγ = 2, βγ = 3 no integral solution and has exactly one
∴ γ − β = 4 p2 − 2 p1 and α (γ − β ) = 1 solution in (2, 3), then a lies in the interval
⇒ α=
1 (a) ( − 1, 2) (b) ( 0, 1)
4 p2 − 2 p1 (c) ( − 1, 0) (d) ( 2, 3)
β 1
Also, = ⇒ 2β = γ Sol. Let y = x − [ x ]. Then, equation
γ 2
(a − 2)(x − [ x ])2 + 2 (x − [ x ]) + a2 = 0 …(i)
⇒ β ⋅ 2β = 3
can be written as
3
⇒ β=± f ( y) = (a − 2) y2 + 2 y + a2 = 0 …(ii)
2
As x cannot be an integer,
1 3
∴ ± = 1 …(i) y = x − [x ] ≠ 0
4 p2 − 2 p1 2
When 2 < x < 3, [ x ] = 2
1 1 1 1
Again, + = 2 p1 , + = 2 p2 ⇒ 0 < x − [ x ] < 1 i.e. 0 < y < 1
α β α γ
Since, Eq. (i) has exactly one solution in the interval
1
and = − p1 + p2 + p3 (2, 3), Eq. (ii) has exactly one solution in the interval
γ (0, 1). This is possible, if
1 1 1 f (0) f (1) < 0
∴ + + = p1 + p2 + p3
α β γ For otherwise the Eq. (ii) has either no or two solutions
1 in (0, 1).
⇒ = p1 + p2 − p3
α ⇒ a2 {a − 2 + 2 + a2 } < 0
1 ⇒ a(a + 1) < 0 [Qa2 > 0]
= p1 − p2 + p3
β
⇒ −1 < a < 0 or a ∈ (−1, 0)
1
= − p1 + p2 + p3 Hence, (c) is the correct answer.
γ
γ p − p2 + p3 Ex 19. If a,b and c are all distinct, then
So, =2= 1
β − p1 + p2 + p3 ( x − b)( x − c) ( x − c) ( x − a )
− p1 + p2 + p3
a +b
α ( a − b) ( a − c) ( b − c)( b − a )
and =3=
γ p1 + p2 − p3
( x − a ) ( x − b)
p1 p2 P3
= = = t.
+c − x is equal to
Thus, we get
7 4 9 ( c − a ) ( c − b)
Using this in Eq. (i), we get two values of x. (a) 0 (b) abc
Hence, (a) is the correct answer. (c) a + b + c (d) None of these
Sol. Let
Ex 17. If f (x ) = x 2 + 2bx + 2c 2 and (x − b) (x − c) (x − c) (x − a)
f (x ) = a +b
g ( x ) = − x 2 − 2cx + b 2 are such that min (a − b) (a − c) (b − c) (b − a)
f ( x ) > max g ( x ), then relation between b and c +c
(x − a) (x − b)
−x
is (c − a) (c − b)
b We have, f (a) = 0, f (b) = 0 and f (c) = 0, i.e. f is a
(a) no relation (b) 0 < c <
2 trinomial which has three distinct zeroes but f (x ) is
(c) | c| < 2 | b| (d) | c| > 2 | b| quadratic, so it could not have more than two zeroes.
⇒ f (x ) ≡ 0
Sol. We have, f (x ) = (x + b)2 + 2c2 − b2 Hence,
(x − b) (x − c) (x − a) (x − c) (x − a) (x − b)
⇒ min f (x ) = 2c2 − b2 a +b +c
(a − b) (a − c) (b − a) (b − c) (c − a) (c − b)
and g (x ) = b2 + c2 − (x + c)2
≡ 0, ∀ x ∈ R
206 ⇒ max g (x ) = b2 + c2 Hence, (a) is the correct answer.
Ex 20. If x is real, then the maximum value of
y = 2 ( a − x ) ( x + x 2 + b 2 ) is
Sol. Since, the roots of the given equation are of opposite sign.
∴ Product of the roots < 0
⇒
p ( p − 1)
< 0 ⇒ p ( p − 1) < 0 ⇒ p ∈ (0, 1)
5
Inequalities and Quadratic Equation
(a) a 2 + b 2 (b) a 2 3
Hence, (b) is the correct answer.
(c) a 2 + b 2 (d) a a 2 + b 2
Ex 23. If p, q ∈{1, 2, 3, 4}, then the number of
Sol. Let t = x + x 2 + b2 equations of the form px 2 + qx + 1 = 0 having
1 1 x 2 + b2 − x real roots is
⇒ = =
t x+ x 2 + b2 b2 (a) 15 (b) 9 (c) 7 (d) 8
b2
b 2 Sol. For real roots, we have
∴ t− = 2x and t+ = 2 x 2 + b2 D ≥ 0 ⇒ q2 − 4 p ≥ 0 ⇒ q2 ≥ 4 p
t t
b2 If p = 1, then q2 ≥ 4 p ⇒ q2 ≥ 4 ⇒ q = 2, 3, 4
Thus, 2(a − x )(x + x 2 + b2 ) = 2a − t + (t )
t If p = 2, then q2 ≥ 4 p ⇒ q2 ≥ 8 ⇒ q = 3, 4
= 2at − t 2 + b2 If p = 3, then q2 ≥ 4 p ⇒ q2 ≥ 12 ⇒ q = 4
= a2 + b2 − (a2 − 2at + t 2 ) If p = 4, then q2 ≥ 4 p ⇒ q2 ≥ 16 ⇒ q = 4
= a2 + b2 − (a − t )2 Thus, we see that there are 7 cases.
Hence, (c) is the correct answer.
Therefore, y = 2 (a − x ) x + x 2 + b2 ≤ a2 + b2
Ex 24. If p and q are the roots of the equation
So, the maximum value of y is a2 + b2 . This value is
x 2 + px + q = 0, then
a2 − b2
attained at t = a or x = . (a) p = 1, q = − 2
2a
(b) p = 0, q = 1
Hence, (a) is the correct answer.
(c) p = − 2, q = 0
Ex 21. If a > 0, a ≠1, then the equation (d) p = − 2, q = 1
2 log x a + log ax a + 3 log a 2x a = 0 has Sol. Since, p and q are roots of the equation
(a) exactly one real root x 2 + px + q = 0 ⇒ p + q = − p and pq = q
(b) two real roots pq = q ⇒ q = 0 or p = 1
(c) no real root If q = 0, then p = 0 and if p = 1, then q = − 2
(d) infinite number of real roots Hence, (a) is the correct answer.
Sol. The equation 2 logx a + logax a + 3 loga 2 x a = 0 Ex 25. If x 2 + 6x − 27 > 0 and x 2 − 3x − 4 < 0, then
can be written as
2 log a log a 3 log a (a) x > 3 (b) x < 4
+ + =0 …(i) 7
log x log (ax ) log (a2 x ) (c) 3 < x < 4 (d) x =
2
As a > 0 and a ≠ 1, log a ≠ 0, Eq. (i) can be written as
2
+
1
+
3
=0 Sol. x 2 + 6x − 27 > 0 and x 2 − 3x − 4 < 0
y b + y 2b + y ⇒ (x + 9)(x − 3) > 0 and (x − 4 )(x + 1) < 0
where, b = log a and y = log x ⇒ ∈ (−∞ , − 9) ∪ (3, ∞ ) and (−1 < x < 4 )
⇒ 2 (b + y) (2b + y) + y (2b + y) ⇒ 3<x<4
+ 3 y (b + y) = 0 Hence, (c) is the correct answer.
⇒ 4 b2 + 11by + 6 y2 = 0
Ex 26. If x 2 + ax + b is an integer for every integer x,
− 11b ± 121b − 96b2
4b
2
b then
⇒ y= =− ,−
12 3 2 (a) a is always an integer but b need not be an
4 1 integer
⇒ log x = − log a or − log a
3 2 (b) b is always an integer but a need not be an
⇒ x = a− 4 / 3 , a − 1 / 2 integer
Hence, (b) is the correct answer. (c) a + b is always an integer
(d) Cannot be discussed
Ex 22. The set of values of p for which the roots of the Sol. Let f (x ) = x 2 + ax + b
equation 3x 2 + 2x + p ( p − 1) = 0 are of Clearly, f (0) = b ⇒ b is an integer.
opposite sign, is Now, f (1) = 1 + a + b
(a) ( − ∞ , 0) (b) ( 0, 1) ⇒ a + b is an integer.
(c) (1, ∞ ) (d) ( 0, ∞ ) Hence, (c) is the correct answer. 207
5 Ex 27. Sum of the real roots of the equation
x 2 + 5 | x | + 6 = 0 is equal to
Sol. Since, roots are imaginary, therefore b2 − 4 ac < 0.
The roots α and β are given by
− b + i 4 ac − b2
Objective Mathematics Vol. 1
(a) 5 (b) 10 α=
(c) −5 (d) Does not exist 2a
Sol. Since, x 2 , 5 | x | and 6 are positive, so − b − i 4 ac − b2
and β=
2a
x 2 + 5 | x | + 6 = 0 does not have any real root.
Therefore, |α | = |β |
Therefore, sum does not exist.
b2 4 ac − b2 c
Hence, (d) is the correct answer. Further, |α | = 2
+ =
4a 4 a2 a
Ex 28. The number of real solutions of the equation ⇒ |α | > 1 [Q c > a]
6−x x Hence, (a) is the correct answer.
=2+ is
x −4
2 ( x + 2) Ex 31. The number of real solutions of the equation
(a) two (b) one x
9
= − 3 + x − x is
2
(c) zero (d) None of these
10
6−x x
Sol. =2+ (a) 0 (b) 1
x2 − 4 (x + 2)
(c) 2 (d) None of these
6−x 2x + 4 + x
⇒ = Sol. Let f (x ) = − 3 + x − x 2 . Then, f (x ) < 0, ∀ x, because
x −4
2 (x + 2)
coefficient of x 2 < 0 and D < 0.
6−x 3x + 4
⇒ = Thus, LHS of the given equation is always positive
x2 − 4 x+2 whereas the RHS is always less than zero.
(3x + 4 )(x + 2)(x − 2) The given equation has no solution.
⇒ 6−x=
x+2 Hence, (a) is the correct answer.
⇒ 6 − x = (3x + 4 )(x − 2) [where, x + 2 ≠ 0]
Ex 32. Roots of the equation ax 3 + bx 2 + cx + d = 0
⇒ 3x 2 − x − 14 = 0, x ≠ − 2
remain unchanged by increasing each
7
⇒ x = − 2, , x ≠ − 2 coefficient by one unit, then
3 (a) a = b = c = d ≠ 0 (b) a = b ≠ c = d ≠ 0
∴ Only one solution (c) a ≠ b = c = d ≠ 0 (d) a = b = c ≠ d ≠ 0
Hence, (b) is the correct answer.
Sol. ax 3 + bx 2 + cx + d
Ex 29. For real θ, value of sin θ + cos θ lies in the
2 4
and (a + 1)x 3 + (b + 1) x 2 + (c + 1) x + (d + 1) = 0
interval must be identical.
a+1 b+1 c+1 d +1
3 ⇒ = = =
(a) [1, 2] (b) , 1 a b c d
4 ⇒ a=b=c=d ≠0
1 5 Hence, (a) is the correct answer.
(c) , (d) None of these
4 16
Ex 33. Complete set of values of a such that
Sol. Since, 0 ≤ cos2 θ ≤ 1
x2 − x
⇒ cos4 θ ≤ cos2 θ attains all real values, is
1 − ax
∴ sin 2 θ + cos4 θ ≤ sin 2 θ + cos2 θ ≤ 1 …(i) (a) [1, ∞ ) (b) ( 0, 4]
Also, sin 2 θ + cos4 θ = sin 2 θ + (1 − sin 2 θ )2 (c) ( 0, 1] (d) None of these
= sin 4 θ − sin 2 θ + 1 x2 − x
Sol. y = ⇒ x 2 − x = y − axy
1
2
3 3 1 − ax
= sin 2 θ − + ≥ …(ii)
2 4 4 ⇒ x 2 + x (ay − 1) − y = 0
From Eqs. (i) and (ii), we get Since, x is real ⇒ (ay − 1)2 + 4 y ≥ 0
3
≤ sin 2 θ + cos4 θ ≤ 1 ⇒ a2 y2 + 2 y (2 − a) + 1 ≥ 0, ∀ y ∈ R
4
Hence, (b) is the correct answer. ⇒ a2 > 0, 4 (2 − a)2 − 4 a2 ≤ 0
⇒ 4 + a2 − 4 a − a 2 ≤ 0
Ex 30. If 0 < a < b < c and the roots α and β of the ⇒ a2 > 0, 4 a ≥ 4
equation ax 2 + bx + c = 0 are imaginary, then ⇒ a≥1
(a) | α | = | β | (b) | α | < 1 ⇒ a ∈ [1, ∞ )
208 (c) | b |< 1 (d) None of these Hence, (a) is the correct answer.
Ex 34. If
2
the equations ax 2 − 2bx + c = 0,
bx − 2cx + a = 0 and cx − 2ax + b = 0 all
2
Sol. x 2 − (a − 3) x + a = 0,
D≥0
D = (a − 3)2 − 4 a = a2 − 10a + 9 5
Inequalities and Quadratic Equation
have only positive roots, then
(a) a = b = c (b) a ≠ b ≠ c ⇒ (a − 9) (a − 1) ≥ 0
(c) a ≠ b = c (d) a = b ≠ c ⇒ a ∈ (− ∞ , 1] ∪ [ 9, ∞ )
Case I When both roots are positive.
b c c a a b
Sol. Clearly, > 0, > 0, > 0, > 0, > 0, > 0, D≥0
a a b b c c
⇒ (a − 9) (a − 1) ≥ 0
Thus, a, b and c all have same signs.
⇒ D ≥ 0, a > 0, a > 3
Also, b2 ≥ ac, c2 ≥ ab, a2 ≥ bc ⇒ a ∈ [ 9, ∞ ) …(i)
These last three inequalities can hold simultaneously if Case II When exactly one root is positive.
and only if a = b = c.
⇒ a≤0 …(ii)
Hence, (a) is the correct answer.
Thus, from Eqs. (i) and (ii),
Ex 35. If a ∈ R − and a ≠ − 2, then the equation a ∈ (− ∞ , 0 ] ∪ [ 9, ∞ )
Hence, (c) is the correct answer.
x 2 + a| x| + 1 = 0
(a) cannot have any real root Ex 38. If roots of x 2 − (a − 3)x + a = 0 are such that
(b) must have exactly two real roots both of them is greater than 2, then
(c) must have either exactly two real roots or no (a) a ∈[ 7, 9] (b) a ∈ [ 7, ∞ )
real root (c) a ∈[ 9, 10) (d) a ∈[ 7, 9)
(d) must have either four real roots or no real
root Sol. x 2 − (a − 3)x + a = 0
⇒ D = (a − 3)2 − 4 a
Sol. x 2 + a| x | + 1 = 0
= a2 − 10a + 9 = (a − 1) (a − 9)
⇒ | x |2 + a| x | + 1 = 0
When both roots are greater than 2, then
− a ± a2 − 4 B
|x| = D ≥ 0, f (2) > 0, − >2
2 2A
a < 0 and not equal to −2. a−3
Q ⇒ (a − 1) (a − 9) ≥ 0; 4 − (a − 3) 2 + a > 0; >2
2
−a + a −42
−a − a − 42
⇒ and both are ⇒ a ∈ (− ∞ , 1] ∪ [ 9 , ∞ ); a < 10; a > 7
2 2 a ∈[ 9, 10)
positive.
Hence, (c) is the correct answer.
Thus, there are four roots, if a < − 2, else no real root.
Hence, (d) is the correct answer. Ex 39. If the quadratic equation ax 2 + bx + 6 = 0 does
Ex 36. If x 2 + ax + 1 is a factor of ax 3 + bx + c , then not have real roots and b ∈ R + , then
(a) b + a + a 2 = 0, a = c b2
(a) a > max , b − 6
(b) b − a + a 2 = 0, a = c 24
(c) b + a − a 2 = 0, a = c b2
(b) a < max , b − 6
(d) None of the above 24
Sol. x 2 + ax + 1 must divide ax 3 + bx + c. b2
(c) a > min , b − 6
ax + bx + c
3
Now, 24
x 2 + ax + 1 b2
(b − a + a3 )x + c + a2 (d) a < min , b − 6
= a (x − a) + 24
x 2 + ax + 1
⇒ (b − a + a3 )x + (c + a2 ) = 0, ∀ x Sol. ax 2 + bx + 6 = 0, roots are not real.
⇒ b − a + a3 = 0 and c + a2 = 0 b2
⇒ D < 0 ⇒ b2 − 24 a < 0 ⇒ a >
Hence, (d) is the correct answer. 24
i.e. a is positive. …(i)
Ex 37. If x 2 − (a − 3)x + a = 0 has atleast one positive Also, f (− 1) > 0
root, then ⇒ a−b+ 6>0
(a) a ∈ ( −∞ , 0) ∪ [ 7, 9] ⇒ a>b−6 …(ii)
(b) a ∈ ( − ∞ , 0) ∪ [ 7, ∞ ) b2
⇒ a > max , b − 6 [from Eqs. (i) and (ii)]
(c) a ∈ ( −∞ , 0] ∪ [ 9, ∞ ) 24
(d) None of the above Hence, (a) is the correct answer.
209
5 Ex 40. Consider the equation x 2 + 2x − n = 0, where
n ∈ N and n ∈[5, 100]. Total number of
Ex 43. If 16 − x 2 > | x − a| is to be satisfied by atleast
one negative value of x, then complete set of
Objective Mathematics Vol. 1
different values of n so that the given equation values of a is
has integral roots, is 65 65
(a) − 16, (b) − 8,
(a) 4 (b) 8 (c) 3 (d) 6 4 2
Sol. x 2 + 2x − n = 0 ⇒ (x + 1)2 = n + 1 (c) ( − 16, 16) (d) ( − 8, 8)
⇒ x = −1± n+1 Sol. 16 − x 2 > | x − a|
Thus, n + 1should be a perfect square. ⇒ x 2 − 16 < x − a < 16 − x 2
Since, n ∈[ 5, 100 ] ⇒ n + 1 ∈ [ 6, 100 ] ⇒ x − 16 − x < − a < 16 − x 2 − x
2
Number of perfect squares between 1 and 100 is 10. ⇒ x + 16 − x 2 > a > − 16 + x 2 + x
Thus, n can take 10 − 2, i.e. 8 different values. 2
Hence, (b) is the correct answer. 65 1
⇒ − x − > a > − 16 + x 2 + x; x ≤ 0
4 2
Ex 41. If log 2 (ax 2 + x + a ) ≥ 1, ∀ x ∈ R , then 65
Since, x ∈ R − ⇒ a ∈ − 16,
exhaustive set of values of a is 4
5 5 5 Hence, (a) is the correct answer.
(a) 0, 1 + (b) 1 − ,1+
2 2 2
Ex 44. If x1 , x 2 , x 3 , ..., x n are the roots of
5 5
(c) 0, 1 − (d) 1 + , ∞ x n + ax + b = 0, then the value of
2 2 ( x1 − x 2 ) ( x1 − x 3 ) ( x1 − x 4 ) ... ( x1 − x n ) is
Sol. log2 (ax 2 + x + a) ≥ 1, ∀ x ∈ R (a) nx1 + b (b) n( x1 ) n − 1
⇒ ax 2 + x + a ≥ 2, ∀ x ∈ R (c) n( x1 ) n−1 + a (d) n( x1 ) n−1 + b
⇒ ax 2 + x + (a − 2) ≥ 0, ∀ x ∈ R Sol. x n + x + b = (x − x1 ) (x − x2 ) K (x − xn )
⇒ a > 0, 1 − 4 a (a − 2) ≤ 0 x n + ax + b
⇒ (x − x2 ) (x − x3 ) K (x − xn ) =
⇒ 4 a2 − 8a − 1 ≥ 0 (x − x1 )
5 5 x n + ax + b
⇒ a > 0, a ∈ − ∞ , 1 − ∪ 1+ , ∞ ⇒ (x1 − x2 ) (x1 − x3 ) K (x1 − xn ) = lim
2 2 x→ x1 x − x1
5 = n (x1 )n −1 + a
⇒ a ∈ 1 + , ∞
2 Hence, (c) is the correct answer.
Hence, (d) is the correct answer.
Ex 45. If x 2 + ( a − b) x + (1 − a − b) = 0, where
Ex 42. If x − x + a − 3 < 0 for atleast one negative
2 a, b ∈ R , then the values of a for which
value of x, then complete set of values of ‘a’ is equation has unequal roots for all values of b, is
(a) ( − ∞ , 4 ) given by
(b) ( − ∞ , 2) (a) a < 1 (b) a > 1
(c) ( − ∞ , 3) (c) a ∈ R (d) None of these
(d) ( − ∞ , 1) Sol. For unequal roots, D>0
Sol. The equation x 2 − x + a − 3 = 0 must have atleast one ⇒ (a − b)2 − 4 (1 − a − b) > 0
negative root. ⇒ b2 + b (4 − 2a) + a2 + 4 a − 4 > 0
For real roots, D ≥ 0 ⇒ 1 − 4 (a − 3) ≥ 0 For the above quadratic inequation to be true, ∀ b ∈ R,
13 13 the discriminant of its corresponding equation should
⇒ a≤ ⇒ a ∈ − ∞, ...(i)
4 4 be less than zero.
Both root will be non-negative, if i.e. (4 − 2a)2 − 4 (a2 + 4 a − 4 ) < 0
D ≥ 0, a − 3 ≥ 0 ⇒ − 32a + 32 < 0
⇒
13
a≤ ,a≥3 ⇒ a>1
4 Hence, (b) is the correct answer.
13
⇒ a ∈ 3, ...(ii)
4 Ex 46. Let a, b and c be real numbers with a ≠ 0 and let
Thus, equation will have atleast one negative root, if α, β be the roots of the equation
13 13 ax 2 + bx + c = 0. Then, a 3 x 2 + abcx + c 3 = 0
a ∈ − ∞, ∪ 3, [from Eqs. (i) and (ii)]
4 4 has roots
⇒ a ∈ (− ∞ , 3) (a) α 2β , β 2α (b) α , β 2
210 Hence, (c) is the correct answer. (c) α 2β , βα (d) α 3β , β 3α
Sol. Dividing the equation, a3 x 2 + abcx + c3 = 0 by c2 , we
ax
c
2
ax
get a + b + c = 0
c
Sol. Let f (x ) = ax 2 + bx + c
Since, coefficients are integers and one root is
irrational, then both the roots are irrational.
5
Inequalities and Quadratic Equation
ax ∴ For any λ ∈ Q , f (λ ) ≠ 0
⇒ = α , β are roots
c ⇒ f (λ ) > 0 or f (λ ) < 0 , but f (λ ) ≠ 0
c c
⇒ x = α and x = β ∴ | f (λ )| > 0
a a 2
⇒ x = α 2 β and αβ 2 are roots of above equation. p p
or a +b +c >0
Hence, (a) is the correct answer. q2 q
1
or | ap2 + bpq + cq2 | > 0 ...(i)
Ex 47. The values of ‘a’ for which x 2 + ax q2
+ sin −1 ( x 2 − 4x + 5) + cos −1 ( x 2 − 4x + 5) = 0 As a, b, c, p, q ∈integers
has atleast one solution, is ∴ | ap2 + bpq + cq2 | ≥ 1 ...(ii)
(a) − 2 (b) − 2 + π 1 1
π π ⇒ | f (λ )| = 2 | ap2 + bpq + cq2 | ≥ 2
(c) − (d) − 2 − q q
4 4 [from Eqs. (i) and (ii)]
−1
Sol. Since, sin θ is defined only when Hence, (a) is the correct answer.
−1≤θ ≤1
⇒ − 1 ≤ x2 − 4 x + 5 ≤ 1
Ex 50. The equation x 2 ⋅ 2| x − 3| +4 + 2 x −1
⇒ x 2 − 4 x + 4 ≤ 0 and x2 − 4 x + 6 ≥ 0
= x 2 ⋅ 2 x +1 + 2| x − 3| +2 has
⇒ (x − 2)2 ≤ 0 and (x − 2)2 + 2 ≥ 0
(a) no solution (b) two solutions
1
⇒ x = 2 is only solution (c) x = ± and x ≥ 3 (d) x ≥ 3
Putting x = 2 , we get 2
π Sol. Here, x 2 ⋅ 2|x − 3 |+4 + 2x −1 = x 2 ⋅ 2x +1 + 2|x − 3 |+2
4 + 2a + = 0
2 If x−3≥0
π ⇒ x 2 ⋅ 2x − 3 + 4 + 2 x − 1 = x 2 ⋅ 2 x + 1 + 2 x − 3 + 2
∴ a=−2−
4
⇒ x 2 ⋅ 2x +1 + 2x −1 = x 2 ⋅ 2x +1 + 2x −1
Hence, (d) is the correct answer.
which is always true for x ≥ 3.
Ex 48. The set of values of x satisfying log | sin x | |cos x| Again, if x − 3 < 0
π x 2 ⋅ 2− (x − 3) + 4 + 2x −1 = x 2 ⋅ 2x +1 + 2− (x − 3) + 2
+ log | cos x| |sin x| = 2, when x ∈ 0, , is ⇒ x 2 ⋅ 2 7 − x + 2x − 1 = x 2 ⋅ 2 x + 1 + 2 5 − x
2
π π π π π ⇒ x 2 ⋅ 27 − x − 25 − x = x 2 ⋅ 2x + 1 − 2x − 1
(a) (b) (c) 0, (d) ,
4 3 4 4 2 ⇒ 25 − x (x 2 ⋅ 22 − 1) = 2x −1 (x 2 ⋅ 22 − 1)
1 ⇒ (25 − x − 2x −1 ) (4 x 2 − 1) = 0
Sol. Here, log|sin x | |cos x | + =2
1
log|sin x | |cos x | ⇒ x2 = and 5 − x = x − 1
1 4
⇒ y + = 2, where y = log|sin x | |cos x | ⇒ x = ± 1/ 2 and x = 3
y 1
∴Solution set is x = ± and x ≥ 3
⇒ y2 − 2 y + 1 = 0 2
⇒ y=1 Hence, (c) is the correct answer.
∴ log|sin x | |cos x | = 1
⇒ |cos x | = |sin x | Ex 51. For any real value of θ ≠ π , the value of the
⇒ |tan x | = 1 cos 2 θ − 1
π expression y = is
⇒ x= cos 2 θ + cos θ
4
Hence, (a) is the correct answer. (a) − 1≤ y ≤ 2 (b) y < 0 and y > 2
(c) − 1≤ y ≤ 1 (d) y ≥ 1
Ex 49. If a, b, c ∈ I and ax 2 + bx + c = 0 has irrational cos2 θ − 1
p Sol. Here, y =
root and λ ∈Q = , where f ( x ) = ax 2 + bx + c, cos2 θ + cos θ
q ⇒ ( y − 1)cos2 θ + y cos θ + 1 = 0
then | f ( λ )| 1
1 1 ⇒ cosθ = − 1 and cos θ =
(a) ≥ (b) < 2 1− y
q2 q ∴ −1<
1
<1
1− y 211
(c) Cannot be discussed (d) None of these
5 ⇒
⇒
1
1− y
2− y
+ 1 > 0 and
> 0 and
1
1− y
y
−1< 0
<0
Ex 55. For what values of K ∈ R the expression
2x 2 + K x y + 3 y 2 − 5 y − 2 can be expressed as
Objective Mathematics Vol. 1
1− y 1− y ( a1 x + b1 y + c1 ) ⋅ ( a 2 x + b2 y + c2 )?
⇒ y < 0 and y > 2 (a) −3, − 4 (b) 2, 3
Hence, (b) is the correct answer. (c) 3, 4 (d) 7, − 7
Ex 52. If the roots of the equation x 2 + ax + b = 0 are c Sol. As, ax 2 + 2hx y + by2 + 2g x + 2 f y + c can be
and d, then one of the roots of the equation expressed as product of two linear factors, if
x 2 + (2c + a ) x + (c 2 + ac + b 2 ) = 0 is abc + 2 f g h − af 2 − bg 2 − ch2 = 0
(a) c (b) d − c (c) 2d (d) 2c ⇒ 2x 2 + K x y + 3 y2 − 5 y − 2 can be expressed as
Sol. Here, f (x ) = x 2 + ax + b, then (a1 x + b1 y + c1 ) (a 2 x + b2 y + c2 ) , if
2
f (x + c) = (x + c)2 + a (x + c) + b 5 K 5
2(3) (−2) + 2 − (0) − 2 − − 3 (0)2
2 2 2
= x 2 + (2c + a) x + c2 + ac + b
2
which shows roots of f (x ) are transformed to (x − c), K
− (−2) = 0
i.e. roots of f (x + c) = 0 are c − c and d − c. 2
Thus, x 2 + (2c + a) x + c2 + ac + b2 = 0 has roots 0 ⇒ K 2 − 49 = 0 ⇒ K = 7, − 7
and (d − c). Hence, (d) is the correct answer.
Hence, (b) is the correct answer.
Ex 56. The number of triangles formed by the lines
Ex 53. Number of integers, which satisfy the represented by x 3 − x 2 − x − 2 = 0 and
(16)1/ x xy 2 + 2xy + 4x − 2 y 2 − 4 y − 8 = 0 is
inequality x+ 3 > 1, is equal to
(2 ) (a) 1 (b) 2
(a) infinite (b) 0 (c) 1 (d) 4 (c) 3 (d) None of these
(16)1 / x Sol. Here, x 3 − x 2 − x − 2 = 0
Sol. Here, x+3
>1
2 ⇒ (x − 2) (x 2 + x + 1) = 0 ⇒ x = 2 ...(i)
4/x 4
−x−3
2 and x y + 2x y + 4 x − 2 y − 4 y − 8 = 0
2 2
⇒ > 1 or 2x >1
2x + 3
4
⇒ y2 (x − 2) + 2 y (x − 2) + 4 (x − 2) = 0
−x−3 4
i.e. 2x > 20 ⇒ −x−3>0 ⇒ (x − 2) ( y2 + 2 y + 4 ) = 0
x
⇒ x=2 ...(ii)
(x 2 + 3x − 4 ) − (x + 4 ) (x − 1)
⇒− >0 or >0 Both the equation represent same line. So, number of
x x triangles is zero.
Using number line rule, Hence, (d) is the correct answer.
– + –
+
–4 0 1
Ex 57. If x 2 − (a + b + c) x + (ab + bc + ca ) = 0 has
imaginary roots, where a, b, c ∈ R + , then
⇒ x ∈ (− ∞ , − 4 ) ∪ (0, 1)
Hence, (a) is the correct answer. a , b and c
(a) can be sides of triangle
Ex 54. Number of non-negative integral solution of (b) cannot be side of triangle
equation y 4 + 6xy 2 − 8x = 0 is equal to (c) Nothing can be said
(a) 1 (b) 2 (c) 3 (d) infinite (d) None of the above
y4 Sol. Here, x 2 − (a + b + c) x + (ab + bc + ca) = 0 has
Sol. Here, x = but x is non-negative.
8 − 6 y2 imaginary roots.
⇒D < 0 or (a + b + c)2 − 4 (ab + bc + ca) < 0
y4
So, ≥0 ⇒ (a2 + b2 + c2 − 2ab − 2bc + 2ac) < 4 ac
8 − 6 y2
4 ⇒ (a − b + c)2 < 4 ac
⇒ 8 − 6 y2 > 0 or y2 <
3 ⇒ −2 ac < a − b + c
i.e. y = 0, 1 [as yis non-negative integers] ⇒ (a + c + 2 ac ) > b
For y = 0, we get x = 0
1 ⇒ ( a+ c )2 > b or a+ c> b
For y = 1, we get x = ∉ integer Similarly, b + c> a and a+ b> c
2
∴ Only one non-negative integral solution. ⇒ a , b and c can be sides of triangle.
212 Hence, (a) is the correct answer. Hence, (a) is the correct answer.
Ex 58. The solution set for
( 2 + 2 ) x + ( 2 − 2 ) x = 2 ⋅ 2 x / 4 is
Sol. As x 2 = 2 + 2 [ x ], where LHS is always positive.
Y 5
Inequalities and Quadratic Equation
(a) {2} (b) {0, 2} (c) {0} (d) [0, 2]
Sol. As we know, AM ≥ GM and equality holds only when
values are equal,
X¢ X
( 2+ 2 )x + ( 2 − 2 )x –1 1 2 3
≥ (2x / 2 )1 / 2
2
⇒ ( 2+ 2 ) x + ( 2 − 2 ) x ≥ 2 ⋅ 2x / 4
Here, equality holds only, if
Y¢
( 2+ 2 )x = ( 2 − 2 )x , i.e. x = 0
∴ 2 + 2 [x ] ≥ 0
Hence, (c) is the correct answer. ⇒ [x ] ≥ − 1
If [ x ] = − 1, then
Ex 59. If c < a < b < d, then roots of the equation x2 − 2 = − 2
bx 2 + {1 − b ( c + d )} x + bcd − a = 0 are ⇒ x = 0, not possible
(a) real and distinct in which one lie between c If [ x ] = 0, then
and a ⇒ x2 = 4
(b) real and distinct in which one lie between c ⇒ x = ± 2, not possible
and d If [ x ] = 2, then
(c) real and distinct in which one lie between a x2 = 6
and b ⇒ x=± 6 , only one possible value
(d) complex roots
i.e. x= 6
Sol. Here, f (x ) = bx 2 + {1 − b (c + d )} x + bcd − a Hence, (b) is the correct answer.
or f (x ) = b(x − c) (x − d ) + (x − a)
where, f (c) = (c − a) = − ve Ex 62. The equations x 3 + 5x 2 + px + q = 0 and
f (d ) = (d − a) = + ve x 3 + 7x 2 + px + r = 0 have two roots in
So, one root lies between c and d. common. If their third roots are γ 1 and γ 2
Hence, (b) is the correct answer. respectively, then the ordered pair ( γ 1 , γ 2 ) is
Ex 60. If p and q are odd prime numbers, then which (a) (5, 7) (b) ( − 5, − 7)
of the following statements about the roots of (c) ( − 5, 7) (d) ( 5, − 7)
the equation x 2 + 2 px − 4q = 0 is correct? Sol. Since, two common roots must satisfy
(a) Always integer (x 3 + 5x 2 + px + q) − (x 3 + 7x 2 + px + r) = 0
(b) Always irrational i.e. − 2x 2 + q − r = 0
(c) Always rational but never integer where, sum of common roots = 0
(d) Nothing can be said If x 3 + 5x 2 + px + q = 0 and x 3 + 7x 2 + px + r = 0
Sol. Here, x 2 + 2 px − 4 q = 0 have roots α , β , γ1 and α , β , γ 2 , respectively.
⇒ D = 4 p2 + 16q > 0; as p, q ∈odd prime numbers ∴ α + β + γ1 = − 5 and α + β + γ 2 = − 7, where
α+β=0
If D is perfect square, then
∴ γ1 = − 5 or γ 2 = − 7
p2 + 4 q = d 2
Hence, (b) is the correct answer.
⇒ 4 q = d 2 − p2 ...(i)
But d − p = 8k ; k ∈integer
2 2
...(ii)
Ex 63. Let P (x ) = 0 be the polynomial equation of
least possible degree with rational coefficients
From Eqs. (i) and (ii), q = 2k which is not possible as q
is prime. having 3 7 + 3 49 as a root. Then, the product
∴ D is not a perfect square but positive. of all the roots of P ( x ) = 0 is
Hence, (b) is the correct answer. (a) 56 (b) 42 (c) 343 (d) 7
Ex 61. Number of solutions for x − 2 − 2 [x ] = 0, 2 Sol. If x = 7 + 49
3 3
(where, [ ] denotes the greatest integer ⇒ x 3 = 7 + 49 + 3 ⋅ 3 7 ⋅ 3 49 (3 7 + 3
49 )
function) is ⇒ x = 56 + 21x
3
(a) 0 ⇒ x 3 − 21x − 56 = 0 ; where, P (x ) is least possible
(b) 1 degree 3.
(c) 2 ∴ Product of all roots = 56
(d) infinite Hence, (a) is the correct answer. 213
5 Ex 64. If α and β are the roots of the equation
1
x 2 + px − 2 = 0, where p ∈ R , then the
Sol. If the given inequalities hold, then
a (1 − b) b(1 − c) c(1 − a) >
1
64
... (i)
Objective Mathematics Vol. 1
2p For x > 0, we have
minimum value of α 4 + β 4 is x (1 − x ) = x − x 2
(a) 2 (b) 2 + 2 (c) 2 2 (d) 2 − 2 1 1 1
2 1
2
= − − x ≤ Q − x ≥ 0
Sol. Here, α 4 + β 4 = (α 2 + β 2 )2 − 2α 2 β 2 4 2 4 2
= {(α + β )2 − 2αβ}2 − 2 (αβ )2 1
⇒ a(1 − a) b(1 − b) c(1 − c) ≤ ... (ii)
2
64
1 1 Eqs. (i) and (ii) cannot hold simultaneously.
= p2 + 2 − 4
p 2p Hence, (a) is the correct answer.
1
= p4 + +2 Ex 67. x 4 − 4x − 1 = 0
2 p4
2
(a) all real roots (b) all imaginary roots
1 (c) exactly two real roots (d) None of these
= p2 − + 2+ 2 ≥2+ 2
2 p2
Sol. Let f (x ) = x 4 − 4 x − 1
Thus, minimum value of α 4 + β 4 is 2 + 2. f ′ (x ) = 4 x 3 − 4 = 4 (x − 1)(x 2 + x + 1)
Hence, (b) is the correct answer. So, f (x ) decrease in (−∞ , 1) and increases in (1, ∞ ).
But f (1) = − 4, so f (x ) = 0 has only two real roots in
Ex 65. If α and β are roots of the equation which one will be positive and one will be negative.
x 2 + ax + b = 0, then maximum value of the (as f(0) is negative).
(α − β) 2 Hence, (c) is the correct answer.
expression −( x 2 + ax + b) − will be
4 Ex 68. If an unknown polynomial is divided by (x −1)
a 2 − 4b b 2 − 4a and ( x − 2), we obtain the remainder 2 and 1
(a) (b)
4 4 respectively, then the remainder resulting from
(c) 0 (d) None of these the division of this polynomial by
Sol. Here, y = − (x 2 + ax + b) ( x − 1) ( x − 2) is
(a) 3 (b) −3 (c) 3 − x (d) 3 − 2x
D (a2 − 4 b)
y is maximum i.e. ymax = − =− Sol. Let the polynomial be P (x ), when divided by (x − 1)
4a 4
leaves remainder 2 ⇒ P (1) = 2 and when P (x ) divided
4 b − a2 (α − β )2
∴ ymax = =− by (x − 2) leaves remainder 1 ⇒ P(2) = 1. When P (x ) is
4 4 divided by (x − 1) (x − 2), could be given by
(α − β )2 P (x ) = {q(x )} (x − 1) (x − 2) + r(x )
⇒ Maximum value of − (x 2 + ax + b) − =0
4 where, q (x ) is quotient and r(x ) is remainder.
Hence, (c) is the correct answer. i.e. r (x ) = ax + b
1 1 ⇒ P (x ) = (x − 1) (x − 2) q (x ) + (ax + b) ... (i)
Ex 66. If a, b, c > 0 and a (1 − b) > , b (1 − c) > and ∴ P (1) ⇒ 0 + a + b = 2
4 4
1 and P (2) ⇒ 0 + 2a + b = 1
c(1 − a ) > , then On solving, we get a = − 1, b = 3
4
∴ Remainder = − x + 3
(a) never possible (b) always true
Hence, (c) is the correct answer.
(c) Cannot be discussed (d) None of these
Type 2. More than One Correct Option
Ex 69. If 3 x + 2 − 9 −1/ x > 0, then the interval of x can be Ex 70. If ax 2 − bx + c = 0 has two distinct roots lying
(a) x ∈ ( 0, ∞ ) (b) x ∈ ( 0, 250) in the interval (0, 1), a, b, c ∈ N . Then,
(c) x ∈ R (d) x ∈ ( − 250, 250) (a) log 5 abc = 1
2 x + 2x + 2 2 (b) log 6 abc = 2
Sol. 3x + 2 > 9−1 / x ⇒ x + 2 > − ⇒ >0 (c) log 5 abc = 3
x x
(x + 1)2 + 1 (d) log 6 abc = 4
>0
x b c
Sol. α + β = , αβ = , 0 < αβ < 1
⇒ x>0 a a
214 Hence, (a) and (b) are the correct answers. ⇒ 0 < (1 − α ) ⋅ (1 − β ) < 1
⇒
and 0 < β (1 − β ) ≤
4
1
1
0 < α (1 − α ) ≤ Sol. D1 = b2 − 4 ac < 0, D2 = b2 − 4 ac < 0, as the root is
non-real.
⇒ Both roots will be common.
5
Inequalities and Quadratic Equation
4 a b c
1 ⇒ = = =1
0 < αβ (1 − (α + β ) + αβ ) ≤ c b a
16 ⇒ a=c
c b c 1
0 < 1 − + ≤ Now, b2 − 4 ac < 0
a a a 16
⇒ b2 − 4 a2 ( 4 c2 ) < 0
a2
0 < c (a − b + c) < ⇒ | b| < 2 | a|(2| c |)
16
Hence, (a), (b) and (d) are the correct answers.
⇒ a = b, c = 1
⇒ a2 > 16 Ex 72. Consider the equation x 2 + x − a = 0, a ∈ N .
⇒ a>4 If equation has integral roots, then
⇒ a=5 (a) a = 2
⇒ b=5
(b) a = 6
⇒ min (abc) = 25
(c) a = 12
Hence, (b), (c) and (d) are the correct answers.
(d) a = 20
Ex 71. If ax 2 + bx + c = 0 and cx 2 + bx + a = 0 Sol. Discriminant, D = 1 + 4 a
( a, b, c ∈ R ) have a common non-real root, then ⇒1 + 4 a should be a perfect square.
(a) | b| < 2 | a| As 1 + 4 a is always odd.
(b) | b| < 2 | c| ⇒ 1 + 4 a = (2λ + 1)2 , λ ∈ I +
(c) a = ± c ⇒ a = λ (λ + 1)
(d) a = c Hence, (a), (b), (c) and (d) are the correct answers.
Type 3. Assertion and Reason
Directions (Ex. Nos. 73-76) In the following Statement II The equation ax 2 + bx + c = 0,
examples, each example contains Statement I a, b, c ∈ R has non-real roots, if b 2 − 4ac < 0.
(Assertion) and Statement II (Reason). Each example
has 4 choices (a), (b), (c) and (d) out of which only one is Sol. Statement I
correct. The choices are Let f (x ) = (x − p) (x − r) + λ (x − q) (x − s)
If λ > 0, then f ( p) > 0, f (q) < 0, f (r) < 0 and f (s) > 0
(a) Statement I is true, Statement II is true; Statement II
is a correct explanation for Statement I f (x ) = 0 has one real root between p and q and other
real root between r and s.
(b) Statement I is true, Statement II is true; Statement II
is not a correct explanation for Statement I Statement II Obviously true.
(c) Statement I is true, Statement II is false Hence, (d) is the correct answer.
(d) Statement I is false, Statement II is true Ex 75. Statement I If roots of the equation
Ex 73. Statement I The quadratic equation x 2 − bx + c = 0 are two consecutive integers,
( a − b) x 2 + ( b − c) x + ( c − a ) = 0 has one root then b 2 − 4c < 1.
x =1. Statement II If a, b and c are odd integers,
Statement II If sum of the coefficients in a then the roots of the equation
quadratic equation vanishes, then its one root 4abc x 2 + ( b 2 − 4ac) x − b = 0 are real and
is x =1. distinct.
Sol. Statement I a − b + b − c + c − a = 0 Sol. Statement I Given equation x 2 − bx + c = 0
If quadratic equation ax 2 + bx + c = 0 has one root Let α and β be two roots such that |α − β | = 1.
x = 1, then a + b + c = 0 ⇒ (α + β )2 − 4αβ = 1 ⇒ b2 − 4 c = 1
Sum of coefficients = 0 Statement II Given equation
Hence, (a) is the correct answer. 4 abc x 2 + (b2 − 4 a) x − b = 0
∴ D = (b2 − 4 ac)2 + 16ab2 c
Ex 74. Statement I The equation (x − p) (x − r ) ⇒ D = (b2 + 4 ac)2 > 0
+ λ ( x − q ) ( x − s) = 0, p < q < r < s has So, roots are real and unequal.
non-real roots, if λ > 0. Hence, (d) is the correct answer. 215
5 Ex 76. Statement I If one root is 5 − 2, then the
equation of lowest degree with rational
coefficient is x 4 − 14x 2 + 9 = 0.
Ex 77. Statement I The number of values of a for
which ( a 2 − 3a + 2) x 2 + ( a 2 − 5a + 6) x
+ a 2 − 4 = 0 is an identity in x, is 2.
Objective Mathematics Vol. 1
Statement II For a polynomial equation with Statement II If a = b = c = 0, then equation
rational coefficient irrational roots occur in
pairs. ax 2 + bx + c = 0 is an identity in x.
Sol. Given a root of the equation is Sol. Statement I a2 − 3a + 2 = 0
x= 5− 2 ⇒ a = 1, 2
On squaring both sides, we get a2 − 5a + 6 = 0
x 2 = 5 + 2 − 2 10 ⇒ 7 − x 2 = 2 10 ⇒ a = 2, 3
Again squaring both sides, we get a −4=0
2
49 + x 4 − 14 x 2 = 40 ⇒ x 4 − 14 x 2 + 9 = 0 ⇒ a=± 2
For polynomial equation with rational coefficients a = 2 is the only solution. Hence, Statement I is false.
irrational roots occurs in pairs. Statement II is true by definition.
Hence, (a) is the correct answer. Hence, (d) is the correct answer.
Type 4. Linked Comprehension Based Questions
Passage I (Ex. Nos. 78-80) Let f ( x ) = x 2 + b1x + c1, Passage II (Ex. Nos. 81-83) If roots of the equation
g( x ) = x 2 + b2 x + c2 , real roots of f ( x ) = 0 be α, β and x 4 − 12 x 3 + bx 2 + cx + 81 = 0 are positive.
real roots of g( x ) = 0 be α + δ, β + δ. Least value of f ( x )
1 7 Ex 81. Value of b is
be − . Least value of g( x ) occurs at x = . (a) − 54 (b) 54 (c) 27 (d) − 27
4 2
Ex 78. The least value of g (x ) is Ex 82. Value of c is
(a) −1 (b) −
1 (a) 108 (b) −108 (c) 54 (d) −54
2
1 1 Ex 83. Root of equation 2bx + c = 0 is
(c) − (d) −
4 3 (a) −1/ 2 (b) 1/ 2 (c) 1 (d) − 1
Sol. (Q. Nos. 81-83)
Ex 79. The value of b2 is Let roots of x 4 − 12x 3 + bx 2 + cx + 81 = 0 be
(a) 6 (b) −7 α , β , γ , δ.
(c) 8 (d) 0 ⇒ α + β + γ + δ = 12and αβγδ = 81
α+β+γ+δ
≥ (αβγδ )1 / 4
Ex 80. The roots of g (x ) = 0 are 4
α+β+γ+δ
(a) 3, 4 (b) −3, 4 Since, = 3 and (αβγδ )1 / 4 = 3
4
(c) 3, − 4 (d) −3, − 4
∴ α =β = γ =δ ⇒ α =β = γ =δ = 3
Sol. (Q. Nos. 78-80) Now, b = ∑ αβ = 6 × 9 = 54
(β − α ) = (β + δ ) − (α + δ )(β + α )2 − 4αβ c = − ∑ αβγ = − 4 × 27 = − 108
= [(β + δ ) + (α + δ )]2 − 4 (β + δ ) (α + δ ) (− b1 )2 − 4 c1 ⇒ 2b + c = 0
∴ 1 is a root of equation 2bx + c = 0.
= (− b2 )2 − 4 c2
D1 = D2 ...(i) 81. (b) 82. (b) 83. (c)
D1 1 Passage III (Ex. Nos. 84-86) In the given figure
Least value of f (x ) is − = − ⇒ D1 = 1
4 4 vertices of ∆ABC lie on y = f ( x ) = ax 2 + bx + c. The
and D2 = 1 ∆ABC is right angled isosceles triangle whose
D 1
∴ Least value of g (x )is − 2 = − hypotaneous AC = 4 2 units, then
4 4 Y y = f(x)= ax2 + bx + c
b2 7
Least value of g (x ) occurs at x = − = ⇒ b2 = − 7
2 2
b2 − 4 c2 = D2 ⇒ 49 − 4 c2 = 1
2
48
⇒ = c2 ⇒ c2 = 12
4 A O C X
x − 7x + 12 = 0 ⇒ x = 3, 4
2
216 78. (c) 79. (b) 80. (a) B
Ex 84. y = f (x ) is given by
(a) y =
x 2
−2 2 (b) y =
x
−2
2
Ex 85. Minimum value of y = f (x ) is
(a) 2 2 (b) −2 2 (c) 2 (d) −2 5
Inequalities and Quadratic Equation
2 2 2 x2
Sol. Minimum value of y = at x = 0 is −2 2.
(c) y = x 2 − 8 (d) y = x 2 − 2 2 2 2
Hence, (b) is the correct answer.
Sol. Q AC = 4 2
∴ AB = BC =
4 2
= 4 units
Ex 86. Number of integral value of k for which k /2
2 lies between the roots of f ( x ) = 0, is
and OB = 4 2 − (2 2 )2 = 2 2 (a) 9 (b) 10 (c) 11 (d) 12
∴ A (− 2 2 , 0), B (2 2 , 0), C (0, − 2 2 ) Sol.Q Roots of f (x ) = 0
x2
Q y = ax 2 + bx + c passes through A , B and C, we get i.e. −2 2=0 ⇒ x=±2 2
2 2
x2 ∴ Number of integral value, of k for which k /2 lies in
y= −2 2
2 2 (−2 2 , 2 2 ) is 11.
Hence, (a) is the correct answer. Hence, (c) is the correct answer.
Type 5. Match the Columns
Ex 87. Match the statements of Column I with values From inequality (i),
of Column II. 4 x 2 − (λ − 3) x + 1 > 0
Column I Column II ⇒ (λ − 3)2 − 4 ⋅ 1⋅ 4 ⋅ 1 < 0
A. The equation x 3 − 6 x 2 + 9 x + λ = 0 has p. −3 ⇒ −4 <λ + 2<4
exactly one root in (1, 3), then [λ + 1] is
(where, [ ] denotes the greatest integer ⇒ −1< λ < 7 …(i)
function) From inequality (ii),
x 2 − λx − 2 −2 x 2 + (λ + 2) x + 4 > 0
B. If − 3 < < 2 , ∀ x ∈ R, then [λ ] q.
x + x+1
2
can be (where [ ] denotes the greatest
∴ (l + 2)2 − 4 ⋅ 1⋅ 4 < 0
integer function) ⇒ −4 <λ + 2<4
C. If x 2 + λ x + 1 = 0 and( b − c ) x 2 + (c − a ) r. −1 ⇒ −6<λ <2 …(ii)
x + ( a − b ) = 0 have both the roots
common, then [λ − 1] is (where,[ ]denotes From Eqs. (i) and (ii), we get − 1 < λ < 2
the greatest integer function) ∴ [ λ ] = − 1, 0, 1
D. If N is the number of solutions of the s. 0 C. Q x = 1satisfies
equation| x − | 4 − x || − 2 x = 4, then the
value of − N is (b − c) x 2 + (c − a) x + (a − b) = 0
∴ x = 1 satisfies x 2 + λx + 1 = 0
Sol. A. Let f (x ) = x 3 − 6x 2 + 9x + λ
Then, 1+ λ + 1= 0
∴ f ′ (x ) = 3x 2 − 12x + 9 = 3 (x − 1) (x − 3) ⇒ [ λ − 1] = [ − 2 − 1]
∴ f ′ (x ) = 0 in (1, 3) =−3
+ + D.Q | x 2 − x − 6 | = x + 2
1 – 3 ⇒ | (x − 3) (x + 2) | = x + 2
⇒ | x − 3 | |x + 2 | = x + 2
But f (1) = 4 + λ and f (3) = λ
For f (x ) = 0 to have exactly one root in (1, 3) (x − 3) (x + 2) = (x + 2), x<−2
f (1) and f (3) should have opposite signs ⇒− (x − 3) (x + 2) = (x + 2), −2 ≤ x < 3
∴ f (1) f (3) < 0 (x − 3) (x + 2) = x + 2, x>3
⇒ λ (λ + 4 ) < 0 ⇒ − 3 < λ + 1 < 0
∴ − 3 < λ + 1<1 x = 4, x<−2
⇒ [ λ + 1] = − 3, − 2, − 1, 0 ⇒ x = − 2, 2, −2 ≤ x < 3
x = 4, x>3
x − λx − 2
2
B.Q −3< <2
x2 + x + 1 Hence, x = − 2, 2, 4
⇒ − 3x − 3x − 3 < x − λx − 2 < 2x + 2x + 2
2 2 2 ∴ N =3
[Qx 2 + x + 1 > 0, ∀ x ∈ R] A → p, q, r, s; B → r, s; C → p; D → p
217
5 Ex 88. Match the statements of Column I with the
values of Column II.
A. Number of solutions is 3.
B. 2x − x − 1 = 0
Consider 2x = x + 1
[fig. (a)]
Objective Mathematics Vol. 1
Column I Column II
∴ There are two solutions.
A. Number of real solutions of| x + 1| = e x is p. 2
x = 0, 1 [both are non-negative] [fig. (b)]
B. The number of non-negative real roots of q. 3 C. p + q = α − 2
2 x − x − 1 = 0 is equal to ⇒ pq = − α − 1
C. If p and q are the roots of the quadratic r. 6 ∴ p2 + q2 = (α − 2)2 − 2 (− α − 1)
equation x 2 − (α − 2 ) x − α − 1 = 0, then
= α 2 − 4α + 4 + 2α + 2
minimum value of p2 + q 2 is equal to
= α 2 − 2α + 6 = (α − 1)2 + 5
D. If α and β are the roots of 2 x 2 + 7 x + c = 0 s. 5
7 ∴ Least value of p2 + q2 = 5
and|α 2 − β 2| = , then c is equal to
4 7 c
D. α + β = − , αβ =
2 2
Sol. Y Y y = x+1 7 7
y = |x+1| ∴ = |α 2 − β 2 | = |(α + β )| (α + β )2 − 4αβ =
4 2
y = 2x 49 7 7
y = ex − 2c ⇒ = 49 − 8c
X X 4 4 4
⇒ 49 − 8c = 1 ⇒ c = 6
Fig. (a) Fig. (b) A → q; B → p; C → s; D → r
Type 6. Single Integer Answer Type Questions
1 1 1
Ex 89. If the quadratic polynomial, If β = , then α 2 + 2 = α +
α α α
1
y = (cot α ) x 2 + 2 ( sin α ) x + tan α, 1
2
1
2 ⇒ α+ − 2=α +
α α
α ∈[0, 2 π ] , can take negative values for all
∴ t 2 − t − 2 = 0 ⇒ (t − 2) (t + 1) = 0
5π
x ∈ R and the value of α ∈ , π , then the ⇒ t = 2 or t = − 1
λ If t = 2 , then α = 1 and β = 1. If t = − 1, then
value of λ is . roots are imaginary (ω or ω 2 ).
Sol. (6) b2 − 4 ac < 0 and a < 0 So, the number of quadratic equation is 3.
Hence, cotα < 0 i.e. α ∈ 2nd and 4th quadrants Ex 91. The number of non-zero solutions of the
and 4 sin α − 2 tan α cot α < 0 equation x 2 − 5x − (sgn x) 6 = 0 is .
1
2 sinα < 1 ⇒ sinα < Sol. (1) If x = 0, we have x 2 − 5x = 0 ⇒ x = 0 or 5
2
5π/ 6 π/6 ⇒ No solution. If x > 0, we have x 2 − 5x − 6 = 0
⇒ (x − 6) (x + 1) = 0 ⇒ x = 6 is the solution.
If x < 0, we have x 2 − 5x + 6 = 0
⇒ (x − 3) (x − 2) = 0 ⇒ No solution
∴ Only one solution.
∴ α ∈ 2nd and 4th quadrant
Ex 92. The number of solutions for the equation
sinα ≥ 0 for sinα to exist.
log 4 (2x 2 + x + 1) − log 2 (2x − 1) = 1 is .
∴ α ∈ 2nd quadrant
5π Sol. (1) log4 (2x 2 + x + 1) − log2 (2x − 1) = 1
Hence, α ∈ , π
6
loge (2x 2 + x + 1) loge (2x − 1)
⇒ − =1
Ex 90. Number of quadratic equations with real roots loge (4 ) loge 2
which remain unchanged even after squaring 2x 2 + x + 1
their roots, is . ⇒ loge = loge 4 ⇒ (2x 2 + x + 1)
(2x − 1)2
Sol. (3) αβ = α 2 β 2 ...(i) = 4 (4 x 2 − 4 x + 1) ⇒ 14 x 2 − 17x + 3 = 0
and α + β2 = α + β
2
...(ii) 3
⇒ (14 x − 3) (x − 1) = 0 ⇒ x = ⇒ x =1
So, αβ (1 − αβ ) = 0 ⇒ α = 0 or β = 0 or αβ = 1 14
If α = 0, then from Eq. (ii), β = 0 or β = 1 3
But x = does not lie in the domain of function
⇒ Roots are (0, 0) or (0, 1) 14
218 If β = 0, then α = 0 or β = 1 Hence, x = 1is the only solution.
Target Exercises
Type 1. Only One Correct Option
1. The least integer satisfying 1
12. Solution of 0 < | 3x + 1 | < is
3
27 − x 27 − 9x
49.4 − < 47.4 − is (a) (− 4 / 9, − 2/ 9)
10 10
(b) [ − 4 / 9, − 2/ 9 ]
(a) 2 (b) 3 (c) (− 4 / 9, − 2/ 9) − {− 1/ 3}
(c) 4 (d) None of these (d) [ − 4 / 9, − 2/ 9 ] − {− 1/ 3}
2. Let f ( x ) = ax 2 + bx + c, a ≠ 0. Suppose, f ( −1) < 1, 13. Solution of | x − 1 | ≥ | x − 3 | is
f (1) > − 1 and f ( 3) < − 4. Then, (a) x ≤ 2 (b) x ≥ 2
(a) Cannot be discussed (b) b + 1 > 0 (c) [1, 3 ] (d) None of these
(c) b + 1 < 0 (d) b is positive real
14. Solution of || x | − 1 | < | 1 − x|, x ∈ R is
3. If a and b are positive and unequal, then (a) (− 1, 1) (b) (0, ∞ )
(a) a4 + b4 > a3b + ab3 (b) a4 + b4 < a3b + ab3 (c) (− 1, ∞ ) (d) None of these
(c) a3 + b3 < a2b + ab2 (d) None of these 1
15. Solution of − 2 < 4 is
2 x
Targ e t E x e rc is e s
4. The inequality < 3 is true, when x belongs to
x (a) (− ∞ , − 1/ 2) (b) (1/ 6, ∞ )
(a) [ 2 / 3, ∞ ) (b) (− ∞ , 2 / 3 ] (c) (−1/ 2, 1/ 6) (d) (− ∞ , − 1/ 2) ∪ (1/ 6, ∞ )
(c) (2 / 3, ∞ ) ∪ (− ∞ , 0) (d) None of these 3
16. Solution of 1 + > 2 is
x+4 x
5. < 2 is satisfied when x satisfies
x−3 (a) (0, 3 ] (b) [ − 1, 0)
(a) (−∞ , 3) ∪ (10, ∞ ) (b) (3, 10) (c) (− 1, 0) ∪ (0, 3) (d) None of these
(c) (−∞ , 3) ∪ [10, ∞ ) (d) None of these 17. Solution of | x 2 − 10 | ≤ 6 is
6. Solution of ( x − 1) 2 ( x + 4 ) < 0 is (a) (2, 4) (b) (− 4 , − 2)
(a) (− ∞ , 1) (b) (− ∞ , − 4 ) (c) (− 4 , − 2) ∪ (2, 4 ) (d) [ − 4 , − 2 ] ∪ [ 2, 4 ]
(c) (− 1, 4 ) (d) (1, 4) 1
18. Solution of x + > 2 is
7. Solution of ( 2x + 1) ( x − 3) ( x + 7) < 0 is x
(a) (− ∞ , − 7) ∪ (− 1/ 2, 3) (b) (− ∞ , − 7) ∪ (1/ 2, 3) (a) R − {0} (b) R − {− 1, 0, 1}
(c) (− ∞ , 7) ∪ (− 1/ 2, 3) (d) (− ∞ , − 7) ∪ (3, ∞ ) (c) R − {1} (d) R − {− 1, 1}
x+1 ( x + 1) 2
8. The set of values of x which satisfy the inequations 19. The solution set of + | x + 1| = is
x+2 x | x|
5x + 2 < 3x + 8 and < 4 is
x−1 (a) {x | x ≥ 0} (b) {x | x > 0} ∪ {− 1}
(a) (− ∞ , 1) (b) (2, 3) (c) {− 1, 1} (d) {x | x ≥ 1 or x ≤ − 1}
(c) (− ∞ , 3) (d) (− ∞ , 1) ∪ (2, 3) 20. The set of real values of x satisfying | x − 1 | ≤ 3 and
9. Solution of | 3x + 2 | < 1 is | x − 1 | ≥ 1is
(a) [ − 1, − 1/ 3 ] (b) {− 1/ 3, − 1} (a) [ 2, 4 ] (b) (− ∞ , 2 ] ∪ [ 4 , ∞ )
(c) (− 1, − 1/ 3) (d) None of these (c) [ − 2, 0 ] ∪ [ 2, 4 ] (d) None of these
10. Solution of | 3 − x | = x − 3 is 21. Solution of the inequality x > (1 − x ) is given by
(a) x < 3 (b) x > 3 (c) x ≥ 3 (d) x ≤ 3 (a) ] − ∞ , (− 1 − 5 )/ 2 [
(b) ]( 5 − 1) / 2, ∞[
11. Solution of | 2x − 3 | < | x + 2 | is
(c) ] − ∞ , ( − 1 − 5 ) / 2 [ ∪ ] ( 5 − 1) / 2, ∞[
(a) (− ∞ , 1/ 3) (b) (1/ 3, 5)
(c) (5, ∞ ) (d) (− ∞ , 1 / 3) ∪ (5, ∞ ) (d) ]( 5 − 1) / 2, 1 [
219
5 22. The product of all the solutions of the equation
|( x − 2)|2 − 3 | x − 2 | + 2 = 0 is
33. If log x log y log z = ( y − z ) ( z − x ) ( x − y ) , then
(a) x y ⋅ y z ⋅ z x = 1
Objective Mathematics Vol. 1
(a) 2 (b) − 4 (b) x x ⋅ y y ⋅ z z = 1
(c) 0 (d) None of these
(c) x x ⋅ y y ⋅ z z = 1
23. Which of the following statement(s) is/are correct? (d) None of the above
(a) a > b ⇒ ax > bx , x ≠ 0, (a, b, x ∈ R )
34. The solution set of the inequation
(b) | x | > | y | ⇒ x > y (x , y ∈ R )
1 1 log 1/ 3 ( x 2 + x + 1) + 1> 0 is
(c) a > b ⇒ >
a b (a) (− ∞ , − 2) ∪ (1, + ∞ ) (b) [ − 1, 2 ]
(d) a > b ⇒ a + c > b + c (c) (− 2, 1) (d) (− ∞ , + ∞ )
24. Solution of | x − 1 | + | x − 2 | + | x − 3 | ≥ 6 is 35. Let f ( x ) = log 10 x 2 . The set of all values of x for
(a) [0, 4] (b) (− ∞ , − 2) ∪ [ 4 , ∞ ) which f ( x ) is real, is
(c) (− ∞ , 0 ] ∪ [ 4 , ∞ ) (d) None of these (a) [ − 1, 1] (b) [1, + ∞ )
(c) (− ∞ , − 1] (d) (− ∞ , − 1] ∪ [1, + ∞ )
25. If log 4 5 = a and log 5 6 = b, then log 3 2 is equal to
1 1 36. If x n > x n−1 > K > x 2 > x1 > 1, then the value of
(a) (b)
2a + 1 2b + 1 xKx1
log x1 log x2 log x3 ... log xn x nn− 1 is
1
(c) 2ab + 1 (d)
2ab − 1 (a) 0 (b) 1
(c) 2 (d) None of these
26. If log 5 a ⋅ log a x = 2, then x is equal to 2 2
37. 4 sin x
+ 4 cos x
is equal to
(a) 125 (b) a2
(c) 25 (d) None of these (a) ≤ 4 (b) ≥ 4
(c) ≤ 2 (d) ≥ 2
Ta rg e t E x e rc is e s
27. The number of solutions of log 2 ( x + 5) = 6 − x is
38. If y = 3x − 1 + 3− x −1 , then the least value of y is
(a) 2 (b) 0
(c) 3 (d) None of these (a) 2 (b) 6
2 3
(c) (d)
28. The solution set of log 2 | 4 − 5x | > 2 is 3 2
8
(a) , + ∞
4 8
(b) , b+c c+a a+b
5 5 5 39. Minimum value of + + (for real
a b c
8 positive numbers a, b, c) is
(c) (− ∞ , 0) ∪ , + ∞ (d) None of these
5
(a) 1 (b) 2
x+2 (c) 4 (d) 6
29. The set of real values of x for which log 0. 2 ≤ 1is
x 40. If a, b and c are different positive real numbers such
5 5 that b + c − a, c + a − b and a + b − c are positive,
(a) − ∞ , − ∪ (0, + ∞ ) (b) , + ∞ then ( b + c − a ) ( c + a − b ) ( a + b − c ) − abc is
2 2
(c) (− ∞ , − 2) ∪ (0, + ∞ ) (d) None of these (a) positive (b) negative
(c) non-positive (d) non-negative
30. The number of real values of the parameter k for
which (log 16 x ) 2 − log 16 x + log 16 k = 0, with real 41. If the product of n positive numbers is unity, then
coefficients will have exactly one solution, is their sum is
(a) 2 (b) 1 (a) a negative integer
(c) 4 (d) None of these (b) divisible by n
(c) never less than n
log x a × log a y × log y z 1
31. x is equal to (d) equal to n +
(a) x (b) y n
(c) z (d) None of these 42. For positive numbers a, b and c, the least value of
32. The number of zeroes coming immediately after the 1 1 1
( a 2 + b 2 + c 2 ) 2 + 2 + 2 is
decimal point in the value of ( 5) 25 is (given, a b c
log 10 2 = 0.30103) (a) 3
(a) 16 (b) 9
(b) 17 27
(c)
(c) 18 4
220 (d) None of the above (d) None of the above
43. For positive real numbers a, b and c, which one of the
following holds ?
51. The polynomial ( ax 2 + bx + c ) ( ax 2 − dx − c ), ac ≠ 0,
has 5
Inequalities and Quadratic Equation
(a) a + b + c ≥ bc + ca + ab
2 2 2
(a) four real zeroes
(b) (b + c) (c + a) (a + b) ≤ 8abc (b) atleast two real zeroes
a b c (c) atmost two real zeroes
(c) + + ≤ 3 (d) no real zeroes
b c a
(d) a3 + b3 + c3 ≥ abc 52. If a, b and c are non-zero, unequal rational numbers,
44. For positive real numbers a, b and c, which of the then the roots of the equation
following holds? abc 2 x 2 + ( 3a 2 + b 2 ) cx − 6a 2 − ab + 2b 2 = 0 are
(a) a + b + c > 3 ⇒ a2 + b2 + c2 > 3 (a) rational (b) imaginary
(b) a + b ≤ 12a b − 64
6 6 2 2 (c) irrational (d) None of these
1 1 1 9
(c) a + b + c = α ⇒ + + ≤ 53. The equation ( a + 2) x 2 + ( a − 3) x = 2a − 1, a ≠ − 2
a b c α has roots rational for
(d) None of the above
(a) all rational values of a except a = − 2
45. For positive real numbers a, b and c, such that (b) all real values of a except a = − 2
1
a + b + c = p, which one holds? (c) rational values of a >
2
8 3
(a) ( p − a) ( p − b) ( p − c) ≥ p (d) None of the above
27
(b) ( p − a) ( p − b) ( p − c) ≥ 8abc 54. If the absolute value of the difference of roots of the
(c)
bc ca ab
+ + ≥p equation x 2 + px + 1 = 0 exceeds 3 p , then
a b c
(d) None of the above (a) p < − 1or p > 4 (b) p > 4
(c) − 1 < p < 4 (d) 0 ≤ p < 4
Targ e t E x e rc is e s
46. The minimum value of P = bcx + cay + abz , when 1
xyz = abc, is 55. If is a root of ax 2 + bx + 1 = 0, where a and b
4 − 3i
(a) 3abc (b) 6abc
are real, then
(c) abc (d) 4abc
(a) a = 25, b = − 8 (b) a = 25, b = 8
47. The greatest value of P = a2b3c4, when (c) a = 5, b = 4 (d) None of these
a + b + c = 18, is 56. If a ∈ R , b ∈ R, then the factors of the expression
(a) P = 4 4 ⋅ 64 ⋅ 84 (b) P = 4 2 ⋅ 63 ⋅ 84
a ( x 2 − y 2 ) − bxy are
(c) P = 32 ⋅ 63 ⋅ 84 (d) P = 4 2 ⋅ 62 ⋅ 82
(a) real and different (b) real and identical
48. If x1 , x 2 , ... , x n are any real numbers and n is any (c) complex (d) None of these
positive integer, then 57. The number of positive integral values of k for which
n n
2
n n
2
(16 x 2 + 12 x + 39) + k ( 9x 2 − 2 x + 11) is a perfect
(a) n ∑ xi2 < ∑ xi (b) n ∑ xi2 ≥ ∑ xi
square, is
i=1 i = 1 i=1 i = 1
2
(a) two (b) zero
n n (c) one (d) None of these
(c) ∑ xi2 ≥ n ∑ xi
(d) None of these
i=1 i = 1 58. For the equation 3x 2 + px + 3 = 0, p > 0, if one of the
roots is the square of the other, then p is equal to
49. If x, y and z are positive real numbers such that 1
x + y + z = 2, then (a) (b) 1
3
(a) (2 − x ) (2 − y) (2 − z) ≥ 8xyz 2
(b) x −1 + y−1 + z−1 ≥ 1/ 2 (c) 3 (d)
3
(c) (2 − x ) (2 − y) (2 − z) < 8xyz
(d) None of the above 59. Let α and β be the roots of the quadratic equation
x 2 + px + p 3 = 0 ( p ≠ 0). If (α , β ) is a point on the
50. For non-negative real numbers such that
a1 + a 2 + K + a n = p and q = ∑ a i a j , then parabola y 2 = x, then the roots of the quadratic
i< j equation are
1 1 2 (a) 4 , − 2
(a) q ≤ p2 (b) q > p
2 4 (b) − 4 , − 2
p p2 (c) 4, 2
(c) q < (d) q > (d) − 4 , 2 221
2 2
5 60. If a, b and c are positive real numbers, then the
number of real roots of the equation
ax 2 + b | x | + c = 0 is
69. If the roots of the equation ( a 2 + b 2 ) y 2
− 2 ( ac + bd ) y + c 2 + d 2 = 0 are equal, then
Objective Mathematics Vol. 1
(a) ab = dc (b) ac = bd
(a) 2 (b) 4
a c
(c) 0 (d) None of these (c) ad + bc = 0 (d) =
b d
61. If α and β are the roots of ax 2 + bx + c = 0 and
70. If p, q and r are positive and are in AP, then the roots
α + k , β + k are the roots of px 2 + qx + r = 0, then
of the quadratic equation px 2 + qx + r = 0 are real for
b 2 − 4ac
is equal to r p
q 2 − 4 pr (a) −7 ≥4 3 (b) −7 <4 3
p r
2
p
2
a (c) all p and r (d) no p and r
(a) (b) 1 (c) (d) 0
a p
2x + 1 x 2 + 7 2x + 17
71. The roots of the equation + 2 =
62. If the roots of the equation ax 2 + bx + c = 0 are in the 9 x −7 9
ratio m : n, then are
(a) mnb2 = ac (m + n)2 (b) b2 (m + n) = mn (a) 3, − 3 (b) 5, − 5
(c) m + n = b mn 2
(d) c (mn) = ab (m + n)
2 2 (c) 3 , − 3 (d) 5 , − 5
1 1 1 1 72. If the roots of the equation a ( b − c ) x 2 + b ( c − a ) x
63. If x + = 5, then x 3 + 3 − 5 x 2 + 2 + x +
x x x x + c ( a − b ) = 0 are equal, then a, b and c are in
is equal to (a) AP (b) GP
(c) HP (d) None of these
(a) 0 (b) 5
(c) − 5 (d) 10 73. If α and β are the roots of the equation
Ta rg e t E x e rc is e s
64. The value of 6 + 6 + 6 + K ∞ is x + x α + β = 0, then the values of α and β are
2
(a) α = 1and β = − 1 (b) α = 1and β = − 2
(a) 3 (b) 6
(c) α = 2 and β = 1 (d) α = 2 and β = − 2
(c) − 2 (d) − 4
74. If α and β are the roots of the equation
65. The sum of the roots of the equation x 2 + px + q = 0
8x 2 − 3x + 27 = 0, then the value of
is equal to the sum of their squares, then
1/ 3 1/ 3
(a) p2 − q2 = 0 (b) p2 + q2 = 2q α 2 β 2
+ is
(c) p2 + p = 2q (d) None of these β α
1 1 7
66. The roots of the equation (a) (b) (c) (d) 4
3 4 2
( q − r ) x 2 + ( r − p ) x + ( p − q ) = 0 are
r− p p−q 75. If the expression a 2 ( b 2 − c 2 ) x 2 + b 2 ( c 2 − a 2 ) x
(a) ,1 (b) ,1
q−r q−r + c 2 ( a 2 − b 2 ) is a perfect square, then
q−r r− p (a) a, b and c are in AP (b) a2 , b2 and c2 are in AP
(c) ,1 (d) ,1
p−q p−q 2 2 2
(c) a , b and c are in HP (d) a2 , b2 and c2 are in GP
67. A lad was asked his age by his friend. The lad said, 76. The values of m for which the equation (1+ m ) x 2
‘The number you get when you subtract 25 times my
age from twice the square of my age will be thrice −2 (1+ 3m ) x + (1+ 8m ) = 0 has equal roots, are
your age.’ If the friend’s age is 14, then age of the lad (a) 0, 3 (b) 1 (c) 2 (d) 3
is 77. If α is a root of 4x 2 + 2x − 1 = 0, then the other root
(a) 21 (b) 28 are
(c) 14 (d) 25
(a) 3α 3 − 4α (b) 4α 3 − 3α
2
x − bx λ − 1 (c) 3α + 4α
3
(d) 4α 3 + 3α
68. If the roots of the equation = are such
ax − c λ +1
78. If the roots of the equation
that α + β = 0, then the value of λ is
a−b ( b − c ) x 2 + ( c − a ) x + ( a − b ) = 0 are equal, then
(a) (b) c
a+ b a, b and c are in
1 a+ b (a) AP (b) GP
(c) (d)
222 c a−b (c) HP (d) None of these
79. If the ratio of the roots of l x 2 + nx + n = 0 is p : q,
then
q p l
88. If the sum of the roots of the quadratic equation
ax 2 + bx + c = 0 is equal to the sum of squares of 5
Inequalities and Quadratic Equation
(a) + + =0 a b c
their reciprocals, then , , are in
p q n c a b
p q n (a) AP (b) GP
(b) + + =0
q p l (c) HP (d) None of these
q p l 89. The harmonic mean of the roots of the equation
(c) + + =1
p q n ( 5 + 2 ) x 2 − ( 4 + 5 ) x + ( 8 + 2 5 ) = 0 is
p q n
(d) + + =1 (a) 2 (b) 4
q p l (c) 7 (d) 8
80. The roots of the equation ( x − a ) ( x − b ) = abx 2 are 90. If x 2 − 3x + 2 is a factor of x 4 − px 2 + q, then
always p and q are
(a) real (b) imaginary (a) 2, 3 (b) 4 , − 5
(c) Cannot be discussed (d) None of these (c) 5, 4 (d) 0, 0
81. The equation 91. If 9x − 4 × 3x + 2 + 35 = 0, then the solution set is
x + 3 − 4 x − 1 + x + 8 − 6 x − 1 = 1 has (a) {1, 2} (b) {2, 3} (c) {2, 4} (d) {1, 3}
(a) no solution (b) one solution 1 1 1
(c) two solutions (d) more than two solutions 92. If the roots of the equation + = are
x+ p x+q r
82. If the sum of the roots of the equation equal in magnitude and opposite in sign, then
ax 2 + bx + c = 0 is equal to the sum of the reciprocals (a) p + q = r
of their squares, then bc 2 , ca 2 and ab 2 are in (b) p + q = 2r
Targ e t E x e rc is e s
1 2
(a) AP (b) GP (c) product of roots = ( p + q2 )
2
(c) HP (d) None of these
(d) None of the above
83. If sin θ and cos θ are the roots of the equation x+1 ( x + 1) 2
ax 2 + bx + c, then 93. The solution set of + | x + 1| = is
x |x |
(a) (a − c)2 = b2 − c2 (b) (a − c)2 = b2 + c2
(a) {x | x ≥ 0} (b) {x | x > 0} ∪ {− 1}
(c) (a + c)2 = b2 − c2 (d) (a + c)2 = b2 + c2 (c) {− 1, 1} (d) {x | x ≥ 1 or x ≤ − 1}
84. A car travels 25 km an hour faster than a bus for a 94. { x ∈ R : | x − 2 | = x 2 } is equal to
journey of 500 km. If the bus takes 10 h more than (a) [ − 1, 2 ] (b) [1, 2 ]
the car, then the speeds of the bus and the car are (c) [ − 1, − 2 ] (d) {− 2, 1}
(a) 25 km/h, 40 km/h (b) 25 km/h, 50 km/h
(c) 25 km/ h, 60 km/h (d) None of these 95. Number of solutions of an equation
| x |2 − 3 | x | + 2 = 0 will be
85. If ( ax 2 + bx + c ) y + a ′x 2 + b ′x + c ′ = 0, then the
(a) 4 (b) 1 (c) 3 (d) 2
condition that x may be rational function of y, is
(a) (ac′ − a ′ c)2 = (ab′ − a ′ b) (bc′ − b′ c) 96. In the equation 4 x + 2 = 2x + 3 + 48 , the value of x will
(b) (ab′ − a ′ b)2 = (ac′ − a ′ c) (bc′ − b′ c) be
3
(c) (bc′ − b′ c)2 = (ab′ − a ′ b) (ac′ − a ′ c) (a) − (b) − 2
2
(d) None of the above
(c) − 3 (d) 1
86. If a ( b − c ) x + b ( c − a ) xy + c ( a − b ) y
2 2
is a
97. The real values of x which satisfy the equation
perfect square, then a, b and c are in 2
−3 2
−3
(5 + 2 6 )x + (5 − 2 6 )x = 10 are
(a) AP
(b) GP (a) ± 2 (b) ± 2
(c) HP (c) ± 2, ± 2 (d) 2, 2
(d) None of the above
98. The equation 125x + 45x = 2⋅ ( 27) x has
87. If α and β are the roots of ax 2 + c = bx, then the
(a) no solution
equation ( a + cy ) 2 = b 2 y in y has the roots (b) one solution
(a) α −1 , β −1 (b) α 2 , β 2 (c) two solutions
−1 −1
(c) α β , α β (d) α , β −2 −2 (d) more than two solutions 223
5 99. If α and β are the roots
x − px + q = 0 and α > 0, β > 0, then the value of
2
of the equation 109. The value of x
2 + ( 2 + 1) = ( 5 + 2 2 ) , is
x/ 2 x x/ 2
for which
Objective Mathematics Vol. 1
α 1/ 4 + β 1/ 4 is( p + 6 q + 4q1/ 4 p + 2 q ) k , where (a) 1 (b) 0
(c) 2 (d) None of these
k is equal to
1 1 1 110. If [ x] 2 = [ x + 2], where [ x] is the greatest integer less
(a) 1 (b) (c) (d)
2 3 4 than or equal to x, then x must be such that
(a) x = 2, − 1 (b) [ − 1, 0) ∪ [ 2, 3)
100. The solutions of the equation ( 3 | x | − 3) 2 = | x | + 7 (c) x ∈ [ − 1, 0) (d) None of these
which belongs to the domain of definition of the
function y = x ( x − 3), are given by 111. Number of real roots of the equation
e sin x − e − sin x − 4 = 0 is
1 1 1 1
(a) , − 2 (b) − ,2 (c) ± , ± 2 (d) − , − 2 (a) 1 (b) 0
9 9 9 9
(c) 2 (d) None of these
3x
101. The roots of the equation 2x + 2 ⋅ 3 x −1 = 9 are given by 112. The roots of the equation x x
= x x are
2 (a) 0 and 4 (b) 0 and 1
(a) log2 , − 2 (b) 3, − 3
3 (c) 0, 1 and 4 (d) 1 and 4
(c) − 2, 1 − log2 3 (d) 1 − log2 3, 2
113. The quadratic equation 8sec 2 θ − 6sec θ + 1 = 0 has
102. If f ( x ) = x − [ x], x ( ≠ 0) ∈ R, where [ x] is greatest (a) infinitely many roots (b) exactly two roots
integer less than or equal to x, then the number of (c) exactly four roots (d) no roots
1
solutions of f ( x ) + f = 1is 114. If log 2 x + log x 2 =
10
= log 2 y + log y 2 and x ≠ y,
x
3
then x + y 3 is equal to
Ta rg e t E x e rc is e s
(a) 0 (b) 1 (c) 2 (d) infinite
103. The number of real solutions of the equation 65
(a) 10 (b)
271/ x + 121/ x = 2⋅ 81/ x is 8
37
(a) 1 (b) 2 (c) infinite (d) 0 (c) (d) None of these
6
104. If α and β (α < β ) are the roots of the equation 115. The number of solutions of the equation
x 2 + bx + c = 0, where c < 0 < b, then sin ( e x ) = 5x + 5− x is
(a) 0 < α < β (b) α < 0 < β < | α | (a) 0 (b) 1
(c) α < β < 0 (d) α < 0 < |α | < β (c) 2 (d) infinitely many
105. If x 2 − 2x + sin 2 α = 0, then 116. Let D = a 2 + b 2 + c 2 , where a, b being consecutive
(a) x ∈ [ − 1, 1] (b) x ∈[ 0, 2 ] integers and c = ab, then D is
(c) x ∈ [ − 2, 2 ] (d) x ∈[1, 2 ]
(a) always an even integer
106. The set of values of m for which the roots of the (b) always an odd integer
equation x 2 − ( m + 1) x + m + 4 = 0 are real and (c) always irrational
(d) Can’t be discussed
negative consists of all m such that
(a) − 3 < m ≤ − 1 (b) − 4 < m ≤ − 3 117. The constant term of the quadratic expression
(c) − 3 ≤ m ≤ 5 (d) − 3 ≥ m or m ≥ 5 n
1 1
∑ x − k − 1 x − , as n → ∞ is
k
107. If a ≤ 0 , then the roots of the equation k =2
x 2 − 2a | x − a | − 3a 2 = 0 are (a) − 1
(b) 0
(a) (1 − 2 ) a, ( 6 − 1) a (c) 1
(b) (− 1 + 6 ) a, (1 + 2) a (d) None of the above
(c) (1 + 2 ) a, 0 2
+ 7 x +12)
118. The system y (x = 1 and x + y = 6, y > 0 has
(d) − (1 ± 6 ) a, (1 ± 2) a
(a) no solution (b) one solution
108. For a ≥ 0, the roots of the equation (c) two solutions (d) more than two solutions
x 2 − 2a | x − a | − 3a 2 = 0 are given by 119. The curve y = ( λ + 1) x 2 + 2 intersects the curve
(a) − a ( 6 + 1) (b) a ( 6 + 1) y = λx + 3 in exactly one point, if λ equals
224 (c) a (1 + 2 ), a (− 1 − 6 ) (d) a (1 + 2) (a) {− 2, 2} (b) {1} (c) {− 2} (d) {2}
120. If the expression x 2 + 2 ( a + b + c ) x
+ 3 ( bc + ca + ab ) is a perfect square, then
(a) a = b = c (b) a = ± b = ± c
130. If α and β are the roots of the equation
x 2 + px − 1/ ( 2 p 2 ) = 0, where p ∈ R. Then, the 5
Inequalities and Quadratic Equation
minimum value of α 4 + β 4 is
(c) a = b ≠ c (d) None of these
(a) 2 2 (b) 2 − 2
121. Sum the non-real roots of ( x 2 + x − 2)( x 2 + x − 3) = 12 is (c) 2 (d) 2 + 2
(a) − 1 (b) 1 (c) − 6 (d) 6
131. If x1 and x 2 are the roots of ax 2 + bx + c = 0 and
122. The number of irrational roots of the equation x1 x 2 < 0, then the roots of
4x / ( x 2 + x + 3) + 5x / ( x 2 − 5x + 3) = − 3 / 2 is x1 ( x − x 2 ) 2 + x 2 ( x − x1 ) 2 = 0 are
(a) 4 (b) 0 (c) 1 (d) 2
(a) real and of opposite sign (b) negative
123. If α and β are the roots of the equation x − 2x + 3 = 0. 2 (c) positive (d) non-real
Then, the equation whose roots are 132. The number of integral values of a for which the
P = α 3 − 3α 2 + 5α − 2 and Q = β 3 − β 2 + β + 5, is quadratic equation ( x + a ) ( x + 1991) + 1 = 0 has
(a) x 2 + 3x + 2 = 0 (b) x 2 − 3x − 2 = 0 integral roots, is
(c) x 2 − 3x + 2 = 0 (d) None of these (a) 3 (b) 0
(c) 1 (d) 2
124. If α and β are the roots of the equation
133. The number of real solutions of the equation
2x 2 − 35x + 2 = 0, then the value of ( 9 / 10) x = − 3 + x − x 2 is
( 2α − 35) 3 ( 2 β − 35) 3 is (a) 2 (b) 0
(a) 8 (b) 1 (c) 1 (d) None of these
(c) 64 (d) None of these
134. If x, y, z and t are real numbers
Targ e t E x e rc is e s
125. The sum of values of x satisfying the equation x + y = 9, z + t = 4 and xt − yz = 6. Then, the
2 2 2 2
2 2
−3 −3
( 31 + 8 15 ) x + 1 = ( 32 + 8 15 ) x is greatest value of P = xz is
(a) 3 (b) 0 (a) 2 (b) 3 (c) 4 (d) 6
(c) 2 (d) None of these
135. If x + ax − 3x − ( a + 2) = 0 has real and distinct
2
126. If α and β are the roots of the equation
roots, then minimum value of ( a 2 + 1) / ( a 2 + 2) is
ax 2 + bx + c = 0, then the value of
(a) 1 (b) 0
( aα 2 + c ) / ( aα + b ) + ( aβ 2 + c ) /( aβ + b ) is 1 1
(c) (d)
b (b2 − 2ac) b2 − 4 ac 2 4
(a) (b)
4a 2a 136. The value of the expression x 4 − 8x 3 + 18x 2 − 8x + 2,
b (b2 − 2ac)
(c) (d) None of these when x = 2 + 3, is
a2c
(a) 2 (b) 1 (c) 0 (d) 3
127. If α and β are the roots of the equation x 2 + px + q = 0 n n
137. x 3 + y 3 is divisible by x + y, if
and α 4 , β 4 are the roots of x 2 − rx + q = 0, then the
(a) n is any integer ≥ 0
roots of x 2 − 4qx + 2q 2 − r = 0 are always
(b) n is an odd positive integer
(a) both non-real (b) both positive (c) n is an even positive integer
(c) both negative (d) opposite in sign (d) n is a rational number
128. Suppose A , B and C are defined as A = a 2 b + ab 2 138. If z 0 = α + iβ, i = −1, then the roots of the cubic
− a 2 c − ac 2 , B = b 2 c + bc 2 − a 2 b − ab 2 and C = a 2 c equation x 3 − 2 (1 + α ) x 2 + ( 4α + α 2 + β 2 ) x
+ ac 2 − b 2 c − bc 2 , where a > b > c > 0 and the − 2 (α + β ) = 0 are
2 2
(a) 2, z0 , z0 (b) 1, z0 , − z0
equation Ax 2 + Bx + C = 0 has equal roots, then (c) 2, z0 , − z0 (d) 2, − z0 , z0
a, b and c are in
(a) AP (b) GP (c) HP (d) AGP 139. If α , β and γ are the roots of x 3 + 64 = 0, then the
2 2
129. If α and β are the roots of ax 2 + c = bx, then the α α
equation whose roots are and , is
β γ
equation ( a + cy ) 2 = b 2 y in y has the roots
(a) αβ −1 , α −1β (b) α −2 , β −2 (a) x 2 − 4 x + 16 = 0 (b) x 2 + x + 1 = 0
(c) α −1 , β −1 (d) α 2 , β 2 (c) x 2 + 4 x + 16 = 0 (d) x 2 − x + 1 = 0
225
5 140. If ( x − 1) 3 is a factor of x 4 + ax 3 + bx 2 + cx − 1, then
the other factor is
(a) x − 3 (b) x + 1
149. If f ( x ) = x 4 + 9x 3 + 35x 2 − x + 4, then f ( − 5 + 2 −4 )
is equal to
(a) − 160
Objective Mathematics Vol. 1
(b) 160
(c) x + 2 (d) None of these (c) 0 (d) None of these
141. Let α + iβ , α , β ∈ R be a root of the equation 150. If p ∈ [ − 1, 1] , then the value of x for which
x 3 + qx + r = 0, q, r ∈ R. The cubic equation is 4x 3 − 3x − p = 0 has a root lies in
independent of α and β whose one root is 2α, is 1 1
(a) [0, 1] (b) ,1 (c) 0, (d) [ − 1, 1]
(a) x 3 + qx − r = 0 (b) x 3 + qx + 2r = 0 2 2
(c) x 3 + qx + r = 0 (d) x 3 − qx + r = 0
151. If α , β , γ and σ are the roots of the equation
142. If α , β are the roots of x + px + q = 0 and 2
x 4 + 4x 3 − 6x 2 + 7x − 9 = 0, then the value of
α β (1 + α 2 ) (1 + β 2 ) (1 + γ 2 ) (1 + σ 2 ) is
x 2n + p n x n + q n = 0 and , are the roots of
β α (a) 6 (b) 11
x n + 1 + ( x + 1) n = 0, then n must be (c) 13 (d) 5
(a) even integer (b) odd integer 152. If tan θ1 , tan θ 2 and tan θ 3 are the real roots of the
(c) rational but non-integer (d) None of these
x 3 − ( a + 1) x 2 + ( b − a ) x − b = 0, where
143. The quadratic equation whose roots are cubes of the θ1 + θ 2 + θ 3∈ ( 0, π ), then θ1 + θ 2 + θ 3 is equal to
roots of x 2 + bx + c = 0, is (a) π/2 (b) π/4 (c) 3π/4 (d) π
(a) x 2 + b (b2 − 3c) x − c3 = 0
153. If α , β and γ are the roots of x − x − 1 = 0, then
3 2
(b) x 2 + b (b2 − 3c) x + c3 = 0
(1 + α ) / (1 − α ) + (1 + β ) / (1 − β ) + (1 + γ ) / (1 − γ ) is
(c) x 2 − b (b2 − 3c) x + c3 = 0 equal to
Ta rg e t E x e rc is e s
(d) None of the above (a) − 5 (b) − 6
144. If α and β are the roots of the equation (c) − 7 (d) − 2
ax + bx + c = 0, then the equation whose roots are
2
154. The equation formed by decreasing each root of
1 1 a x 2 + b x + c = 0 by 1 is 2 x 2 + 8 x + 2 = 0, then
and , is
aα + b aβ + b (a) a = − b (b) b = − c
(c) c = − a (d) b = a + c
(a) cax − bx + 1 = 0
2
(b) cax + bx + 1 = 0
2
(c) cax + bx − 1 = 0
2
(d) None of these 155. If the equations ax 2 + bx + c = 0 and x 2 + x + 1 = 0
have a common root, then
145. If x = 2 + 21/ 3 + 22/ 3 , then the value of x 3 − 6x 2 + 6x
(a) a + b + c = 0 (b) a = b = c
is (c) a = b or b = c or c = a (d) a − b + c = 0
(a) 3 (b) 2
(c) 1 (d) None of these 156. If a, b, c ∈ R and the equations ax 2 + bx + c = 0 and
x 3 + 3 x 2 + 3 x + 2 = 0 have two roots in common,
146. If roots of the equation x n − 1 = 0 are
then
1, a1 , a 2 , a 3 , ... , a n−1 , then the value of (1 − a1 )
(a) a = b ≠ c (b) a = b = − c
(1 − a 2 ) (1 − a 3 ) K (1 − a n−1 ) will be
2 n
(c) a = b = c (d) None of these
(a) n (b) n (c) n (d) 0
157. If the equations a x2 + b x + c = 0 and
147. If α and β are the roots of the quadratic
c x + b x + a = 0, a ≠ c have a negative common
2
x 2 − 2cos θx + 1 = 0 , then equation whose roots are
root, then the value of a − b + c is
α n , β n , is (a) 0 (b) 2
(a) x 2 − (2 cos nθ ) x + 1 = 0 (c) 1 (d) None of these
(b) 2x 2 − (2 cos nθ ) x − 1 = 0 158. If the equations x 2 + i x + a = 0 and x 2 − 2 x + ia = 0,
(c) x 2 + (2 cos nθ ) x + 1 = 0
a ≠ 0 have a common root, then
(d) x 2 + (2 cos nθ ) x − 1 = 0 (a) a is real
1
148. If 1, ω , ω , ... , ω
2 n−1
are the nth roots of unity, then (b) a = + i
2
(1 + ω ) (1 + ω ) ... (1 + ω n−1 ) equals
2
1
(c) a = − i
(a) 0, n is even (b) 1, n is even 2
226 (c) n, n is even (d) n2 , n is even (d) the other root is also common
159. If the equations x 2 + 2x + 3λ = 0 and 2x 2 + 3x + 5λ = 0
have a non-zero common root, then λ is equal to
(b) − 1
169. The inequality | 2x − 3 | < 1 is valid when x lies in the
interval 5
Inequalities and Quadratic Equation
(a) 1 (a) (3, 4) (b) (1, 2)
(c) 3 (d) None of these (c) (− 1, 2) (d) (− 4 , 3)
160. If the equations ax 2 + bx + c = 0 and x 2 + 2x + 3 = 0 170. The set of real values of x satisfying | x − 1 | − 1 ≤ 1is
have a common root, then a : b : c is equal to (a) [ − 1, 3 ] (b) [ 0, 2 ]
(a) 2 : 4 : 5 (b) 1 : 3 : 4 (c) [ − 1, 1] (d) None of these
(c) 1 : 2 : 3 (d) None of these
171. If 5x + ( 2 3 ) 2x ≥ 13x , then the solution set for x is
161. If the equations x − px + q = 0 and x − ax + b = 0
2 2
(a) [ 2, + ∞ ] (b) (2, + ∞ )
have a common root and the other root of the second (c) (4 , + ∞ ) (d) None of these
equation is the reciprocal of the other root of the first,
then ( q − b ) 2 is equal to 172. If log 10 x + log 10 y ≥ 2, then the smallest possible
value of x + y is
(a) aq ( p − b)2 (b) bq ( p − a)2
(a) 10 (b) 30
(c) bq ( p − b)2 (d) None of these (c) 20 (d) None of these
162. Let f ( x ) = 1 + 2x + 3x 2 + K + ( n + 1) x n , where n is x2 − x + 1
173. If x ∈ R, then takes values in the interval
even. Then, the number of real roots of the equation x2 + x + 1
f ( x ) = 0 is
1 1
(a) , 3 (b) ,3
3
(a) 0 (b) 1 3
(c) n (d) None of these
(c) (0, 3) (d) None of these
163. The value of x 2 + 2bx + c is positive, if
174. If y = f ( x ) = tan x cot 3x, then
(b) b2 − 4 ac < 0
Targ e t E x e rc is e s
(a) b2 − 4 ac > 0
1
(c) c2 < b (d) b2 < c (a) < y<1
3
1
164. If ax 2 + bx + 10 = 0 does not have two distinct real (b) − ∞ < y < or 3 < y < ∞
3
roots, then the least value of 5a + b is 1
(c) < y<∞
(a) − 3 (b) − 2 3
(c) 3 (d) None of these (d) − ∞ < y < 1
165. If − a 2 x 2 + 2x + 3a 2 > 0, ∀ x ∈ ( 2, 4 ), then the values 1 1
x x
175. The real values of x for which 372 > 1, are
of a lie in the interval 3 3
1 1 2 2 (a) x ∈[ 0, 64 ] (b) x ∈ (0, 64 )
(a) − , (b) − ,
3 3 7 7 (c) x ∈[ 0, 64 ) (d) None of these
4 4 2 2 2 2
(c) − , (d) − , 176. If α and β are the roots of 4x 2 − 16x + λ = 0, λ ∈ R
7 7 13 13
such that 1< α < 2 and 2 < β < 3, then the number of
166. The values of a which make the expression integral solutions of λ is
x 2 − ax + 1 − 2a 2 always positive for real values of x, (a) 5 (b) 6
are (c) 2 (d) 3
2 2 2 2 177. The least integral value of k for which
(a) − <a< (b) − ≤a≤
3 3 3 3 ( k − 2) x 2 + 8x + k + 4 > 0, ∀ x ∈ R, is
2 2
(c) − < a < 1 (d) 0 < a < (a) 5 (b) 4
3 3
(c) 3 (d) None of these
167. If x is integer satisfying x 2 − 6x + 5 ≤ 0 and
178. The values of a for which exactly one root of the
x 2 − 2x > 0, then the number of possible values of x is equation e a x 2 − e 2a x + e a − 1 = 0 lies between 1 and
(a) 3 (b) 4 2 are given by
(c) 2 (d) infinite
(5 − 17 ) 5 + 17
(a) log < a < log
6x 2 − 5x − 3 4 4
168. If < 4, then the least and highest values
x 2 − 2x + 6 (b) 0 < a < 100
5 10
of 4x 2 are (c) log < a < log
4 3
(a) 0, 81 (b) 0, 36 (c) − 10, 3 (d) 10, − 3 (d) None of the above 227
5 179. If the roots of the equation x 2 + 2ax + b = 0 are real
and distinct and they differ by atmost 2 m , then b lies
188. The values of a for which the equation
2x 2 − 2 ( 2a + 1) x + a ( a − 1) = 0 has roots α and β
satisfying the condition α < a < β, are
Objective Mathematics Vol. 1
in the interval
(a) (a2 , a2 + m2 ) (b) (a2 − m2 , a2 ) (a) (− 3, 0) (b) (0, ∞ )
(c) [ a − m , a ]
2 2 2
(d) None of these (c) (− ∞ , − 3) ∪ (0, ∞ ) (d) None of these
189. Let a, b and c be real and ax 2 + bx + c = 0 has two real
180. If a, b, c and d are four consecutive terms of an
c b
increasing AP, then the roots of the equation roots α and β, where α < − 1and β > 1, then 1 + +
( x − a ) ( x − c ) + 2 ( x − b ) ( x − d ) = 0 are a a
(a) non-real complex (b) real and equal (a) < 2 (b) < 1
(c) integers (d) real and distinct (c) < 0 (d) None of these
181.The interval of a for which the equation 190. If 2 a + 3b + 6 c = 0, a, b, c ∈ R, then the quadratic
tan 2 x − ( a − 4 ) tan x + 4 − 2a = 0 has atleast one equation a x 2 + b x + c = 0 has
(a) atleast one root in [0, 1] (b) atleast one root in [2, 3]
solution, ∀ x ∈ [ 0, π / 4], is
(c) atleast one root in (3, 4] (d) None of these
(a) a ∈ (2, 3)
(b) a ∈[ 2, 3 ] 191. Let f ( x ) = ax 2 + bx + c and b, c ∈ R , a ≠ 0, satisfying
(c) a ∈ (1, 4 ) f (1) + f ( 2) = 0. Then, the quadratic equation f ( x ) = 0
(d) a ∈[1, 4 ] has
182. If the expression [ mx − 1 + (1/ x )] is non-negative for (a) no real root (b) 1 and 2 as real roots
all positive real x, then the minimum value of m must (c) two equal roots (d) two distinct real roots
be 192. If x + λy − 2 and x − µy + 1are factors of the expression
(a) − 1/ 2 (b) 0 (c) 1/4 (d) 1/2
6x 2 − xy − y 2 − 6x + 8 y − 12, then
Ta rg e t E x e rc is e s
183. If a, b and c are distinct positive numbers, then the 1 1
(a) λ = ,µ = (b) λ = 2, µ = 3
nature of roots of the equation 1/ ( x − a ) + 1/ ( x − b ) 3 2
+ 1/ ( x − c ) = 1/ x is 1
(c) λ = , µ = −
1
(d) None of these
(a) all real and distinct 3 2
(b) all real and atleast two are distinct
193. If the expression y 2 + 2xy + 2x + my − 3 can be resolved
(c) atleast two real
(d) all non-real into two rational factors, then m must be
(a) any positive real number
184. If a, b, c and d are four consecutive terms of an (b) any negative real number
increasing AP, then the roots of the equation (c) − 2
( x − a ) ( x − c ) + 2 ( x − b ) ( x − d ) = 0 are (d) 3
(a) real and distinct
(b) non-real complex
194. If the expression ax 2 + by 2 + cz 2 + 2ayz + 2bzx +2cxy
(c) real and equal can resolved into rational factors, then
(d) integers (a) a + b + c = 0
(b) a3 + b3 + c3 = 3abc
185. If cos x + sin x − p = 0, p ∈ R has real solutions,
4 2
(c) ab + ac + bc = 0
then (d) None of the above
3
(a) p ≤ 1 (b) ≤ p≤1
4 195. If the equation x 2 + 9 y 2 − 4x + 3 = 0 is satisfied for
3 real values of x and y, then
(c) p ≥ (d) None of these
4 (a) 1 ≤ x ≤ 3 (b) 2 ≤ x ≤ 3
1 2
186. If the roots of x + x + a = 0 exceed a, then
2
(c) − < y < 1 (d) 0 < y <
3 3
(a) 2 < a < 3 (b) a > 3
(c) − 3 < a < − 3 (d) a < − 2 196. If ( mr , 1/ mr ), r = 1, 2, 3, 4 are four pairs of values of x
187. The necessary and sufficient condition for the and y that satisfy the equation
equation (1 − a 2 ) x 2 + 2ax − 1 = 0 to have roots lying x + y + 2gx + 2 fy + c = 0,
2 2
then the value of
in the interval (0, 1) is m1 m2 m3 m4 is
(a) a ≠ 0 (a) 0
(b) a > 0 (b) 1
(c) a < 2 or a > 2 (c) − 1
228 (d) None of the above (d) None of the above
197. The number of values of k for which the equation
x 2 − 3x + k = 0 has two distinct roots lying in the
interval (0, 1), is
199. The number of real solutions of the equation
log 0. 5 x = | x | is 5
Inequalities and Quadratic Equation
(a) 1 (b) 2
(a) 3 (c) 0 (d) None of these
(b) 2
200. If (log 5 x ) 2 + log 5 x < 2, then x belongs to the
(c) infinitely many
(d) no value of k satisfies the requirements interval
1
198. The number of real solutions of the equation e x = x is (a) , 5
25
(a) 1 1 1
(b) ,
(b) 2 5 5
(c) 0
(c) (1, ∞ )
(d) infinite
(d) None of the above
Type 2. More than One Correct Option
2
−15 2
201. The roots of the equation (a + b )x (c) tan (α − β )
1
(d) log + (log β )2
2
−15 α
+ (a − b )x = 2a, where a 2 − b = 1, are
(a) ± 3 (b) ± 4 (c) ± 14 (d) ± 5 207. If the quadratic equation ax 2 + bx + c = 0 ( a > 0) has
sec 2 θ and cosec 2θ as its roots, then which of the
202. If 0 < a < b < c and the roots α , β of the equation following must hold good?
ax 2 + bx + c = 0 are imaginary, then (a) b + c = 0 (b) b2 − 4 ac ≥ 0
(a) | α | = | β | (c) c ≥ 4 a (d) 4 a + b ≥ 0
Targ e t E x e rc is e s
(b) | α | = 1
208. If the roots of the equation x 3 + px 2 + qx – 1 = 0
(c) | β | < 1
(d) None of the above
form an increasing GP, where p and q are real, then
(a) p + q = 0
203. If tan α and tan β are roots of the equation (b) p ∈ (− 3, ∞ )
x 2 + px + q = 0 with p ≠ 0, then (c) one of the roots is unity
(d) one root is smaller than 1 and one root is greater than 1
(a) sin 2 (α + β ) + p sin (α + β )cos (α + β )
+ q cos2 (α + β ) = q 209. Given that α , γ are roots of the equation
p Ax − 4x + 1 = 0 and β , δ are the roots of the equation
2
(b) tan (α + β) =
q−1 of Bx 2 − 6x + 1 = 0 such that α , β , γ and δ are in HP,
(c) cos (α + β ) = − p then
(d) sin (α + β ) = 1 − q
(a) A = 3 (b) A = 4
204. If ( ax + c ) y + ( a ′x + c ′ ) = 0 and x is a rational
2 2 (c) B = 2 (d) B = 8
function of y, then 210. If (1 + k ) tan 2 x − 4 tan x − 1 + k = 0 has real roots
(b) = ′
a c a a
(a) = tan x1 and tan x 2 , then
a′ c′ c c′
(c) ac′ − a′ c = 0 (d) None of these (a) k 2 ≤ 5
2 2
(b) tan (x1 + x2 ) = 2
x x (c) for k = 2, x1 = π / 4
205. The equation + = a ( a − 1) has (d) for k = 1, x1 = 0
x + 1 x − 1
(a) four real roots, if a > 2 211. If the equation x 2 + px + q = 0 has roots u and v,
(b) two real roots, if 1 < a < 2 where p and q are non-zero constants, then
(c) no real root, if a < − 1 (a) qx 2 + px + 1 = 0 has roots 1/u and 1/v
(d) four real roots, if a < − 1 (b) (x + p) (x − q) = 0 has roots u + v and uv
206. If α and β are the roots of the quadratic equation (c) x 2 + p2x + q2 = 0 has roots u2 and v 2
ax 2 + bx + c = 0, then which of the following u
(d) x 2 + px + p = 0 has roots and
v
expression will be the symmetric function of roots v u
α (x − a ) (x − b )
(a) log (b) α 2 β 5 + β 2α 5 212. For real x, the function will assume all
β x−c
real values provided
229
(a) a > b > c
5 (b) b ≤ c ≤ a
(c) a > c > b
(d) a ≤ c ≤ b
(d) µ = 0
215. If x, y ∈ R and 2x 2 + 6xy + 5 y 2 = 1, then
Objective Mathematics Vol. 1
(a) | x | ≤ 5 (b) | x | ≥ 5
213. Let P ( x ) = x 2 + bx + c, where b and c are integers. If
P ( x ) is a factor of both x 4 + 6x 2 + 25 and (c) y ≤ 2
2
(d) y2 ≤ 4
3x 4 + 4x 2 + 28x + 5, then 216. If cos x − y 2 − y − x 2 − 1 ≥ 0, then
(a) P (x ) = 0 has imaginary roots
(a) y ≥ 1 (b) x ∈ R (c) y = 1 (d) x = 0
(b) P (x ) = 0 has roots of opposite sign
(c) P(1) = 4 217. If the following figure shows the graph of
(d) P(1) = 6 f ( x ) = ax 2 + bx + c, then
214. If the equations 4x 2 − x − 1 = 0 and Y
3x 2 + ( λ + µ ) x + λ − µ = 0 have a common root,
then the rational values of λ and µ are
X
−3
(a) λ =
4
(b) λ = 0 (a) ac < 0 (b) bc > 0
3
(c) µ = (c) ab > 0 (d) abc < 0
4
Type 3. Assertion and Reason
Directions (Q. Nos. 218-219) In the following Statement II If ax 2 + bx + c = 0 and
questions, each question contains Statement I
(Assertion) and Statement II (Reason). Each question a1 x + b1 x + c1 = 0 have a common root and
2
Ta rg e t E x e rc is e s
has four choices (a), (b), (c) and (d) out of which only a b c
one is correct. The choices are
, and are in AP, then a1 , b1 and c1 are in
a1 b1 c1
(a) Statement I is true, Statement II is true; Statement II is
a correct explanation for Statement I
GP.
(b) Statement I is true, Statement II is true; Statement II is 219. Statement I If roots of the equation x 2 − bx + c = 0
not a correct explanation for Statement I
are two consecutive integers, then b 2 − 4c = 1.
(c) Statement I is true, Statement II is false
(d) Statement I is false, Statement II is true
Statement II If a, b and c are odd integers, then the
a b c
218. Statement I If , and are in AP, then roots of the equation 4abc x 2 + ( b 2 − 4ac ) x − b = 0
a1 b1 c1
are real and distinct.
a1 , b1 and c1 are in GP.
Type 4. Linked Comprehension Based Questions
Passage I (Q. Nos. 220-222) Let the roots of f ( x ) = x (a) I and II (b) III and IV
(c) II and III (d) I and IV
be α and β, where f ( x ) is quadratic polynomial
ax 2 + bx + c. α and β are also the roots of f (f ( x )) = x . Let 222. If α and β are real and equal, then
the other two roots of f (f ( x )) = x be λ and δ.
(a) λ and δ are imaginary
220. The correct statements are (b) λ and δ are real
(c) λ and δ may be real or imaginary
I. If α and β are real and unequal, then λ and δ are
(d) Cannot say
also real.
II. If α and β are imaginary, then λ and δ are also Passage II (Q. Nos. 223-225) If f ( x ) is a differentiable
imaginary. function wherever it is continuous and f ′(c1) = f ′(c2 ) = 0.
(a) Only I (b) Only II f ′′(c1) ⋅ f ′′(c2 ) < 0, f (c1) = 5, f (c2 ) = 0 and (c1 < c2 ).
(c) both I and II (d) neither I nor II
223. If f (x ) is continuous in [ c1 , c 2 ] and
221. If α + β = λ + δ, then the correct statements are
f ′′ ( c1 ) − f ′′ ( c 2 ) > 0, then minimum number of
I. α and β are real roots of f ′ ( x ) = 0 in [ c1 − 1, c 2 + 1] is
II. λ and δ are real (a) 2 (b) 3
III. α and β cannot be real (c) 4 (d) 5
230
IV. λ and δ cannot be real
224. If f (x ) is continuous in [ c1 , c 2 ]
f ′′ ( c1 ) − f ′′ ( c 2 ) < 0, then minimum number of roots
and (c) 8
228. The roots of g ( x ) = 0 are
(d) 0
5
Inequalities and Quadratic Equation
of f ′ ( x ) = 0 in [ c1 − 1, c 2 + 1] is (a) 3, 4 (b) − 3, 4
(a) 1 (b) 2 (c) 3 (d) 4 (c) 3, − 4 (d) − 3, − 4
225. If f ( x ) is continuous in [ c1 , c 2 ] and Passage IV (Q. Nos. 229-231) Let x 1, x 2, x 3 and x 4
f ′′ ( c1 ) − f ′′ ( c 2 ) > 0, then minimum number of roots be the roots (real or complex) of the equation
of f ( x ) = 0 in [ c1 − 1, c 2 + 1] is x 4 + ax 3 + bx 2 + cx + d = 0 and x 1 + x 2 = x 3 + x 4 and
(a) 2 (b) 3 (c) 4 (d) 5 a, b, c, d ∈ R .
Passage III (Q. Nos. 226-228) Let f ( x ) = x 2 + b1x + c1 229. If a = 2, then the value of b − c is
and g( x ) = x + b2 x + c2 . If real roots of f ( x ) = 0 are α, β
2 (a) − 1 (b) 1 (c) − 2 (d) 2
and real roots of g( x ) = 0 are α + δ, β + δ and least value of 230. If b < 0, then how many different values of a we may
1 7
f ( x ) is − . Then, least value of g( x ) occurs at x = . have?
4 2
(a) 3 (b) 2 (c) 1 (d) 0
226. The least value of g ( x ) is
1
231. If b + c = 1and a ≠ − 2 , then for real values of a, c ∈
(a) − 1 (b) −
2 1
(a) − ∞ , (b) (− ∞ , 3)
1 1 4
(c) − (d) −
4 3 (c) (− ∞ , 1) (d) (− ∞ , 4 )
227. The value of b2 is
(a) 6 (b) −7
Type 5. Match the Columns
Targ e t E x e rc is e s
232. Match the statements of Column I with the values of 233. Match the statements of Column I with the values of
Column II. Column II.
Column I Column II Column I Column II
A. The real values of a for which the p. a≥ 6
A. If the equation x + 2 ( k + 1) x
2
p. 2< k< 4
quadratic equation
2 x 2 − ( a 3 + 8a − 1) x + a 2 − 4a = 0 + ( 9k − 5) = 0 has only negative roots, then
possesses roots of opposite signs are B. If the inequality x 2 − 2 ( 4k − 1) x q. k≥ 6
given by
+ 15k − 2 k − 7 > 0 is valid for all x, then
2
B. If the equation x 2 + 2 ( a + 1) x q. 0≤ a≤ 1
+ 9a − 5 = 0 has only negative roots, C. If x 2 − 2 ( k − 1) x + (2 k + 1) = 0 has both r. k< −1 or
then k> 0
roots positive, then
C. The value of a for which the inequality r. 0< a< 4
x 2 − 2 ( 4a − 1) x + 15a 2 − 2 a − 7 > 0 D. If 2 x 2 − 2 (2 k + 1) x + k ( k + 1) = 0 has one s. k≥ 4
is valid for all x ∈ R, is root less than k and other root greater than k,
then
x2 + 2 x + a s. 2 < a< 4
D. If x ∈ R, then can take all
x 2 + 4 x + 3a E. The graph of the curve x 2 = 3 x − y − 2 is t. 1
real values, if strictly below the line y equal to 4
Type 6. Single Integer Answer Type Questions
234. Let x1 , x 2 , x 3 satisfying the equation the real roots of the equation. If b + c = 2 ( a + 1), then
x 3 − x 2 + βx + γ = 0 are in GP, where ( x1 , x 2 , x 3 > 0), x1 x 2 is equal to ______.
then the maximum value of [β ] + [ γ ] + 2 is ______, 236. If α and β are the roots of x 2 + px − q = 0 and γ , δ are
where [ ] denotes the greatest integer function. the roots of x 2 + px + r = 0, q + r ≠ 0, then
235. Consider the equation x 3 − ax 2 + bx − c = 0 , where (α − γ ) (α − δ )
is equal to ______.
a, b, c ∈ Q ( a ≠ 1). It is given that x1 , x 2 and x1 x 2 are (β − γ ) (β − δ )
231
Entrances Gallery
JEE Advanced/IIT JEE
1. Let a ∈ R and let f : R → R be given by 7. Let ( x 0 , y 0 ) be solution of the following equations
f ( x ) = x 5 − 5x + a, then [2014] (2x )ln 2 = (3 y)ln 3 and 3ln x = 2ln y.
(a) f (x ) has three real roots, if a > 4 Then, x 0 is equal to [2011]
(b) f (x ) has only one real root, if a > 4 1 1
(c) f (x ) has three real roots, if a < − 4 (a) (b)
6 3
(d) f (x ) has three real roots, if − 4 < a < 4 1
(c) (d) 6
2. The quadratic equation p( x ) = 0 with real 2
coefficients has purely imaginary roots. Then, the 8. Let α and β be the roots of x 2 − 6 x − 2 = 0 with
equation p( p( x )) = 0 has [2014]
α > β. If a n = α n − β n for n > 1, then the value of
(a) only purely imaginary roots
(b) all real roots
a10 − 2a 8
is [2011]
(c) two real and two purely imaginary roots 2a 9
(d) neither real nor purely imaginary roots (a) 1 (b) 2 (c) 3 (d) 4
3. The number of points in (− ∞, ∞), for which 9. A value of b for which the equations
x − x sin x − cos x = 0, is
2
[2013] x 2 + bx − 1 = 0
(a) 6 (b) 4 (c) 2 (d) 0 and x 2 + x + b = 0,
Ta rg e t E x e rc is e s
have one root in common, is [2011]
4. If 3x = 4 x − 1 , then x is equal to [2013]
(a) − 2 (b) − i 3 (c) i 5 (d) 2
2 log3 2 2
(a) (b)
2 log3 2 − 1 2 − log2 3 10. Consider the polynomial f ( x ) = 1+ 2x + 3x 2 + 4x 3 .
1 2 log2 3 Let s be the sum of all distinct real roots of f ( x ) and
(c)
1 − log4 3
(d)
2 log2 3 − 1
let t = | s |. The real number s lies in the interval [2010]
1 3
(a) − , 0 (b) − 11, −
5. Let α( a ) and β( a ) be the roots of the equation 4 4
( 3 1 + a − 1) x 2 + ( 1 + a − 1) x + ( 6 1 + a − 1) = 0, 3
(c) − , −
1 1
(d) 0,
4 2 4
where a > − 1. Then, lim α( a ) and lim β ( a ) are
a → 0+ a → 0+ [2012]
5 1 11. Let p and q be real numbers such that p ≠ 0, p 3 ≠ q
(a) − and 1 (b) − and − 1 and p 3 ≠ − q. If α and β are non-zero complex
2 2
7 9 numbers satisfying α + β = − p and α 3 + β 3 = q,
(c) − and 2 (d) − and 3
2 2 α β
then a quadratic equation having and as its
β α
6. The value of
roots, is [2010]
6 + log 3/ 2 K
1 1 1 1 (a) ( p3 + q)x 2 − ( p3 + 2q)x + ( p3 + q) = 0
4− 4− 4−
3 2 3 2 3 2 3 2
(b) ( p3 + q)x 2 − ( p3 − 2q)x + ( p3 + q) = 0
is [2012] (c) ( p3 − q)x 2 − (5 p3 − 2q) x + ( p3 − q) = 0
(a) 4 (b) 3 (c) 1 (d) 0 (d) ( p3 − q)x 2 − (5 p3 + 2q)x + ( p3 − q) = 0
JEE Main/AIEEE
12. Let α and β be the roots of equation x 2 − 6x − 2 = 0. 13. If a ∈ R and the equation
− 3( x − [ x]) 2 + 2( x − [ x]) + a 2 = 0 (where, [x]
If a n = α − β , for n ≥ 1, then the value of
n n
denotes the greatest integer < x) has no integral
a10 − 2a 8
is equal to [2015] solution, then all possible values of a lie in the
2a 9 interval [2014]
(a) 6 (b) −6 (a) (− 1,0) ∪ (0, 1) (b) (1, 2)
232 (c) 3 (d) −3 (c) (− 2, − 1) (d) (− ∞ , − 2) ∪ (2, ∞ )
14. If the equations x 2 + 2x + 3 = 0 and ax 2 + bx + c = 0,
a, b, c ∈ R, have a common root, then a : b : c is [2013]
24. The value of a for which the sum of the squares of
the roots of the equation x 2 − ( a − 2) x − a − 1 = 0
assume the least value, is [2005]
5
Inequalities and Quadratic Equation
(a) 1 : 2 : 3 (b) 3 : 2 : 1 (c) 1 : 3 : 2 (d) 3 : 1 : 2
− sin x (a) 2 (b) 3 (c) 0 (d) 1
15. The equation e sin x
−e − 4 = 0 has [2012]
(a) infinite number of real roots 25. If the roots of the equation x 2 − bx + c = 0 are two
(b) no real roots consecutive integers, then b 2 − 4c is equal to [2005]
(c) exactly one real root
(a) 1 (b) 2 (c) 3 (d) − 2
(d) exactly four real roots
16. Let α , β be real and z be a complex number. If 26. If both the roots of the quadratic equation
z 2 + αz + β = 0 has two distinct roots on the line x 2 − 2 kx + k 2 + k − 5 = 0 are less than 5, then k lies
Re z = 1, then it is necessary that [2011] in the interval [2005]
(a) β ∈ (− 1, 0) (b) β = 1 (a) [4, 5] (b) (− ∞, 4) (c) (6, ∞) (d) (5, 6]
(c) β ∈ (1, ∞ ) (d) β ∈ (0, 1)
27. If the equation
17. Sachin and Rahul attempted to solve a quadratic a n x n + a n − 1 x n − 1 + ... + a1 x = 0,
equation. Sachin made a mistake in writing down the
constant term and ended up in roots (4, 3). Rahul a1 ≠ 0, n > 2 , has a positive root x = α, then the
made a mistake in writing down coefficient of x to equation
get roots (3, 2). The correct roots of equation are na n x n − 1 + ( n − 1)a n − 1 x n − 2 + ... + a1 = 0 has a
[2011]
positive root, which [2005]
(a) − 4 , − 3 (b) 6, 1 (c) 4, 3 (d) − 6, − 1 (a) is equal to α
18. If α and β are the roots of the equation x 2 − x + 1 = 0 (b) is greater than or equal to α
(c) belongs to (0, α )
, then α 2009 + β 2009 is equal to
Targ e t E x e rc is e s
[2010]
(d) is greater than α
(a) − 2 (b) − 1
(c) 1 (d) 2
28. If (1 − p ) is a root of quadratic equation
x 2 + px + (1 − p ) = 0, then its roots are [2004]
19. If the roots of the equation bx 2 + cx + a = 0 are
(a) 0, 1 (b) − 1, 1 (c) 0, − 1 (d) − 1, 2
imaginary, then for all real values of x, the
expression 3b 2 x 2 + 6bcx + 2c 2 is [2009] 29. If one root of the equation x 2 + px + 12 = 0 is 4,
(a) greater than 4ab (b) less than 4ab while the equation x 2 + px + q = 0 has equal roots,
(c) greater than − 4ab (d) less than − 4ab then the value of q is [2004]
49
20. The quadratic equations (a) (b) 12
4
x 2 − 6x + a = 0 and x 2 − cx + 6 = 0 (c) 3 (d) 4
have one root in common. The other roots of the
first and second equations are integers in the ratio 4 30. Let two numbers have arithmetic mean 9 and
: 3. Then, the common root is [2008] geometric mean 4. Then, these numbers are the
(a) 2 (b) 1 (c) 4 (d) 3 roots of the quadratic equation [2004]
(a) x 2 + 18x + 16 = 0 (b) x 2 − 18x + 16 = 0
21. If the difference between the roots of the equation
(c) x + 18x − 16 = 0
2
(d) x 2 − 18x − 16 = 0
x 2 + ax + 1 = 0 is less than 5, then the set of
possible values of a is [2007] 31. Let z1 and z 2 be two roots of the equation
(a) (− 3, 3) (b) (− 3, ∞) z 2 + az + b = 0, z being complex. Further, assume
(c) (3, ∞ ) (d) (− ∞ , − 3)
that the origin, z1 and z 2 form an equilateral
22. If the roots of the quadratic equation x 2 + px + q = 0 triangle. Then, [2003]
are tan 30° and tan 15° respectively, then the value (a) a2 = b (b) a2 = 2b
of 2 + q − p is [2006] (c) a2 = 3b (d) a2 = 4 b
(a) 3 (b) 0 (c) 1 (d) 2
32. If the sum of the roots of the quadratic equation
23. All the values of m for which both roots of the ax 2 + bx + c = 0 is equal to the sum of the squares of
equation x 2 − 2mx + m2 − 1 = 0 are greater than − 2
a b c
but less than 4 lie in the interval [2006] their reciprocals, then , and are in [2003]
c a b
(a) m > 3 (b) − 1 < m < 3
(c) 1 < m < 4 (d) − 2 < m < 0 (a) AP (b) GP (c) HP (d) AGP 233
5 33. The number of the real solutions of the equation
x 2 − 3| x| + 2 = 0 is [2003]
35. If α ≠ β and α 2 = 5α − 3, β 2 = 5β − 3, then the
equation having α /β and β / α as its roots, is [2002]
Objective Mathematics Vol. 1
(a) 2 (b) 4 (c) 1 (d) 3 (a) 3x 2 + 19x + 3 = 0 (b) 3x 2 − 19x + 3 = 0
34. The value of a for which one root of the quadratic (c) 3x − 19x − 3 = 0
2
(d) x 2 − 16x + 1 = 0
equation ( a 2 − 5a + 3)x 2 + ( 3a − 1)x + 2 = 0 is 2
− 7x + 7
36. The number of real roots of 32x = 9 is [2002]
twice as large as the other, is [2003]
(a) 0 (b) 2 (c) 1 (d) 4
(a) 2/ 3 (b) −2/ 3 (c) 1/ 3 (d) −1/ 3
Other Engineering Entrances
37. The number of digits in 20301 44. If a, b and c are positive numbers in a GP, then the
(given, log 10 2 = 0.3010) is [WB JEE 2014] roots of the quadratic equation
(a) 602 (log e a ) x 2 − ( 2 log e b ) x + (log e c ) = 0 are
(b) 301 [WB JEE 2014]
(c) 392 loge c loge c
(d) 391 (a) −1and (b) 1 and −
loge a loge a
38. The number of solutions of the equation (c) 1 and loga c (d) −1and logc a
log 101 log 7 ( x + 7 + x ) = 0 is [WB JEE 2014] 45. If α , β are the roots of ax 2 + bx + c = 0 ( a ≠ 0) and
(a) 3 (b) 7 α + h, β + h are the roots of px 2 + qx + r = 0 ( p ≠ 0),
(c) 9 (d) 49
then the ratio of the squares of their discriminants is
39. The value of a, so that the sum of squares of the [WB JEE 2014]
roots of the equation x 2 − ( a − 2) x − a + 1 = 0 (a) a2 : p2 (b) a : p2 (c) a2 : p (d) a : 2 p
Ta rg e t E x e rc is e s
assume the least value, is [BITSAT 2014]
46. Let f ( x ) = 2x 2 + 5x + 1. If we write f ( x ) is
(a) 2 (b) 0
(c) 3 (d) 1 f ( x ) = [ a ( x + 1)( x − 2) + b( x − 2) ( x − 1)
+ c( x − 1)( x + 1)]
40. If the roots of x 2 − ax + b = 0 are two consecutive
odd integers, then a 2 − 4b is equal to[Kerala CEE 2014] for real numbers a, b and c, then [WB JEE 2014]
(a) 3 (b) 4 (c) 5 (a) there are infinite number of choices for a, b and c
(d) 6 (e) 7 (b) only one choice for a but infinite number of choices for
b and c
41. If α and β are the roots of x 2 − ax + b 2 = 0, then (c) exactly one choice for each of a, b and c
α 2 + β 2 is equal to [Kerala CEE 2014] (d) more than one but finite number of choices for a, b and c
(a) a2 + 2b2 (b) a2 − 2b2 (c) a2 − 2b
47. If α and β are the roots of the equation
(d) a + 2b
2
(e) a − b
2 2
ax 2 + bx + c = 0 and S n = αn + βn,
42. If α and β are the roots of the equation then aS n + 1 + bS n + cS n − 1 is equal to [AMU 2014]
1 1 (a) 0
x 2 + 3x − 4 = 0, then + is equal to
α β [Kerala CEE 2014] (b) abc
−3 3 −4 (c) a + b + c
(a) (b) (c) (d) None of the above
4 4 3
(d)
4
(e)
3 48. If the roots of ax 2 + bx + c = 0 are sin α and cos α for
3 2 some α, then which one of the following is correct?
43. If the roots of the equation x 2 + 2bx + c = 0 are α [RPET 2014]
and β, then b 2 − c is equal to [Kerala CEE 2014] (a) a + b = 2ac
2 2
(α − β )2 (b) b2 − c2 = 2ab
(a)
4 (c) b2 − a2 = 2ac
(b) (α + β )2 − αβ (d) b2 + c2 = 2ab
(c) (α + β )2 + αβ
(α − β )2 49. The value of x such that 32x − 2( 3x + 2 ) + 81 = 0 is
(d) + αβ [Kerala CEE 2014]
2
(a) 1 (b) 2 (c) 3
(α + β )2
(e) + αβ (d) 4 (e) 5
234 2
50. The equation whose roots are the squares of the
roots of the equation 2x 2 + 3x + 1 = 0, is
[Kerala CEE 2014]
59. The number of solutions of the inequation
| x − 2| + | x + 2| < 4 is
(a) 1 (b) 2
[Kerala CEE 2014]
(c) 4
5
Inequalities and Quadratic Equation
(a) 4 x 2 + 5x + 1 = 0 (b) 4 x 2 − x + 1 = 0 (d) 0 (e) 5
(c) 4 x 2 − 5x − 1 = 0 (d) 4 x 2 − 5x + 1 = 0 60. Solve the inequality 2x − 5 ≤ ( 4x − 7)/ 3.
(e) 4 x + 5x − 1 = 0
2
[J&K CET 2014]
(a) x ∈ (− ∞ , 4 ) (b) x ∈ (− ∞ , 4 ]
51. In a ∆ABC, tan A and tan B are the roots of (c) x ∈ (− ∞ , 8 ] (d) x ∈ (− ∞ , − 4 ]
pq( x 2 + 1) = r 2 x. Then, ∆ABC is [WB JEE 2014] 61. Solve the inequality 3x + 2 > − 16 and 2x − 3 ≤ 11 .
(a) a right angled triangle (b) an acute angled triangle [J&K CET 2014]
(c) an obtuse angled triangle (d) an equilateral triangle
(a) (− 6, 7 ] (b) [ − 6, 7) (c) (− 6, 7) (d) [ − 6, 7 ]
52. If α and β are the roots of the quadratic equation 62. The number of real solutions of the equation
x 2 + px + q = 0, then the values of α 3 + β 3 and sin e x = 5x + 5− x is [Karnataka CET 2013]
α 4 + α 2β 2 + β 4 are respectively [WB JEE 2014] (a) 1 (b) 0
(a) 3 pq − p and p − 3 p q + 3q
3 4 2 2 (c) 2 (d) None of these
(b) − p(3q − p2 ) and ( p2 − q) ( p2 + 3q) 63. Let a, b and c be real numbers, a ≠ 0. If α is a root of
(c) pq − 4 and p − q
4 4 a 2 x 2 + bx + c = 0, β is a root of a 2 x 2 − bx − c = 0 and
0 < α < β. Then, the equation a 2 x 2 + 2bx + 2c = 0 has
(d) 3 pq − p3 and ( p2 − q) ( p2 − 3q)
a root γ that always satisfies [UP SEE 2013]
53. The number of solution(s) of the equation (a) γ = α (b) α < β < γ
x + 1 − x − 1 = 4x − 1 is α+β
[WB JEE 2014] (c) α < γ < β (d) γ =
2
Targ e t E x e rc is e s
(a) 2 (b) 0 (c) 3 (d) 1
64. If the arithmetic mean of the roots of a quadratic
54. If 9 x = 12 + 147, then the value of x is equation is 8 and the geometric mean is 5, then the
[RPET 2014] equation is [BITSAT 2013]
(a) −2 (b) 2 (c) 3 (d) −3 (a) x 2 − 16x − 25 = 0 (b) x 2 + 16x − 25 = 0
(c) x − 16x + 25 = 0
2
(d) x 2 − 8x + 5 = 0
55. Solve the equation x 2 − 3x + 1 = 0. [J&K CET 2014]
(a) x = (− 3 ± 2i ) / 2 (b) x = (− 3 ± i ) / 2 65. If the roots of the equation ax 2 + bx + c = 0 are of
(c) x = (− 3 ± i ) (d) x = ( 3 ± i ) / 2 k +1 k+2
the form and , then ( a + b + c ) 2 is equal
k k +1
2π 2π
56. If a = cos + i sin , then the quadratic equation to [Manipal 2013]
7 7
whose roots are α = a + a 2 + a 4 and (a) b2 − 4 ac (b) b2 − 2ac
(c) 2b2 − ac (d) Σa2
β = a 3 + a 5 + a 6 , is [Manipal 2014]
(a) x 2 − x + 2 = 0 (b) x 2 + 2x + 2 = 0 66. If the equations 2ax 2 − 3bx + 4c = 0 and
(c) x 2 + x + 2 = 0 (d) x 2 + x − 2 = 0 3x 2 − 4x + 5 = 0 have a common root, then
( a + b ) / ( b + c ) is equal to ( a, b, c ∈ R )
57. Let p and q be real numbers. If α is the root of
[Kerala CEE 2013]
x 2 + 3 p 2 x + 5q 2 = 0, β is the root of (a) 1/2 (b) 3/35 (c) 34/31
x 2 + 9 p 2 x + 15q 2 = 0 and 0 < α < β, then the (d) 29/23 (e) None of these
equation x 2 + 6 p 2 x + 10q 2 = 0 has a root γ that 67. If log 2 ( 9x −1 + 7) − log 2 ( 3x −1 + 1) = 2 , then the
always satisfies [WB JEE 2014] values of x are [Karnataka CET 2012]
α
(a) γ = + β (b) β < γ (a) 0, 2 (b) 0, 1
4 (c) 1, 4 (d) 1, 2
α
(c) γ = + β (d) α < γ < β 68. If a, b and c are in arithmetic progression, then the
2
roots of the equation ax 2 − 2bx + c = 0 are
58. If the roots of the equation λ 2 + 8λ + µ 2 + 6 µ = 0 [WB JEE 2012]
are real, then µ lies between [RPET 2014] (a) 1 and
c 1
(b) − and −c
(a) −2 and 8 (b) −3 and 6 a a
c c
(c) −8 and 2 (d) −6 and 3 (c) −1and − (d) −2 and − 235
a 2a
77. If one of the roots of 2x 2 − cx + 3 = 0 is 3 and
5 69. The harmonic mean of the roots of the equation
( 5 + 2 ) x 2 − ( 4 + 5 ) x + 8 + 2 5 = 0 is
[Manipal 2012]
another equation 2x 2 − cx + d = 0 has equal roots,
where c and d are real numbers, then d is equal to
Objective Mathematics Vol. 1
(a) 2 (b) 4 (c) 6 (d) 8 [UP SEE 2011]
(a) 3 (b) 49/8 (c) 8/49 (d) −3
70. If α , β and γ are the roots of the equation
( 3 +1+ 3 − 1) 2 ( 3 − 2 )
x 3 + 4x + 2 = 0, then α 3 + β 3 + γ 3 is equal to 78. The value of is
[Karnataka CET 2012] ( 3 + 1) 2 − ( 3 − 1) 2
(a) 2 (b) 6 (c) −2 (d) −6 [UP SEE 2011]
71. If (α + β ) and (α − β ) are the roots of the 3+ 2
(a) (b) 1
equation x 2 + px + q = 0, where α , β , p and q are 3
real, then the roots of the equation 1
(c) 3 (d)
( p 2 − 4q ) ( p 2 x 2 + 4 px ) − 16q = 0 are [WB JEE 2012] 3
1 1 1 1 79. If |2x − 3| < | x + 5|, then x lies in the interval
(a) + and −
α β α β [Kerala CEE 2011]
1 1 1 1 (a) (−3, 5) (b) (5, 9)
(b) + and − −2 2
α β α β (c) , 8 (d) −8,
3 3
1 1 1 1
(c) + and − 2
α β α β (e) −5,
3
(d) ( α + β ) and ( α − β )
80. If α and β are the roots of the equation
72. If the product of the roots of the equation ax 2 + bx + c = 0 , then the equation whose roots are
x 2 − 2 2 kx + 2e 2 log k − 1 = 0 is 31, then the roots of k k
Ta rg e t E x e rc is e s
the equation are real for k, is equal to [AMU 2012] and , is [MP PET 2011]
α β
(a) −4 (b) 1
(c) 4 (d) 0 (a) cx 2 + kbx + k 2a = 0 (b) cx 2 + k 2bx + ka = 0
(c) kcx 2 + bx + k 2a = 0 (d) k 2cx 2 + bx + ka = 0
73. If α and β are the roots of the equation
x 2 − 2x + 4 = 0, then the value of α n + β n will be 81. If 3 ≤ 3t − 18 ≤ 18, then which one of the following is
[BITSAT 2011] correct? [Kerala CEE 2010]
(a) i 2n+ 1 sin (nπ / 3) (a) 15 ≤ 2t + 1 ≤ 20 (b) 8 ≤ t < 12
(b) 2n+ 1 cos (nπ / 3) (c) 8 ≤ t + 1 ≤ 13 (d) 21 ≤ 3t ≤ 24
(c) i 2n− 1 sin (nπ / 3) (e) t ≤ 7 or t ≥ 12
(d) 2n− 1 cos (nπ / 3) x + 11
82. The solution set of the inequation > 0 is
x−3
74. If α , β and γ are the roots of x 3 − 2x + 1 = 0, then the [Kerala CEE 2010]
1 (a) (− ∞ , − 11) ∪ (3, ∞ )
value of Σ is [Karnataka CET 2011]
α + β − γ (b) (− ∞ , − 10) ∪ (2, ∞ )
(c) (− 100, − 11) ∪ (1, ∞ )
(a) −1 / 2 (b) −1 (d) (− 5, 0) ∪ (3, 7)
(c) 0 (d) 1 / 2 (e) (0, 5) ∪ (− 1, 0)
75. The number of integral values of b, for which the 83. In a right angled triangle, the sides are a, b and c,
equation x 2 + bx − 16 has integral roots, is with c as hypotenuse and c − b ≠ 1, c + b ≠ 1 . Then,
[Kerala CEE 2011] the value of
(a) 2 (b) 3 (log c + b a + log c − b a ) / ( 2 log c + b a × log c − b a )
(c) 4 (d) 5 will be [WB JEE 2010]
(e) 6
(a) 2 (b) −1 (c) 1/2 (d) 1
76. If a, b and c are three real numbers such that
84. If α and β are the roots of the equation
a + 2b + 4c = 0. Then, the equation ax 2 + bx + c = 0
ax 2 + bx + c = 0, ( c ≠ 0) , then the equation whose
[WB JEE 2011]
(a) has both the roots complex
1 1
roots are and , is [BITSAT 2010]
(b) has its roots lying within −1 < x < 0 aα + b aβ + b
1 (a) acx 2 − bx + 1 = 0 (b) x 2 − acx + bc + 1 = 0
(c) has one of roots equal to
2
236 (c) acx + bx − 1 = 0
2
(d) x 2 + acx − bc + 11 = 0
(d) has its roots lying within 2 < x < 6
85. If one root of the equation x 2 + px + q = 0 is 2 + 3,
then the value of p and q are, respectively
[VITEEE 2010]
90. The quadratic equation x 2 + 15 | x| + 14 = 0 has
[BITSAT 2010] 5
Inequalities and Quadratic Equation
(a) only positive solutions
(a) − 4 , 1 (b) 4 , − 1 (c) 2, 3 (d) − 2, − 3 (b) only negative solutions
(c) no solution
86. If x 2 + px + q = 0 has the roots α and β, then the
(d) both positive and negative solutions
value of (α − β ) 2 is [VITEEE 2010]
91. The value of a for which the equation
(a) p2 − 4 q (b) ( p2 − 4 q)2
2x 2 + 2 6x + a = 0 has equal roots, is [VITEEE 2010]
(c) p2 + 4 q (d) ( p2 + 4 q)2
(a) 3
87. If a and b are the roots of the equation (b) 4
x 2 + ax + b = 0, a ≠ 0, b ≠ 0 , then the values of a (c) 2
(d) 3
and b are, respectively [Kerala CEE 2010]
(a) 2 and − 2 (b) 2 and −1 (c) 1 and − 2 3 7
92. If + i is a solution of the equation
(d) 1 and 2 (e) −1and 2 2 2
88. If the roots of the equation x 2 − bx + c = 0 are two ax 2 − 6x + b = 0, where a and b are real numbers,
consecutive integers, then b 2 − 4c is equal to then the value of a + b is [Kerala CEE 2010]
[Kerala CEE 2010] (a) 10 (b) 22 (c) 30
(a) −1 (b) 0 (c) 1 (d) 29 (e) 31
(d) 2 (e) 3 93. The roots of the quadratic equation
89. If α and β are the roots of the quadratic equation x 2 − 2 3x − 22 = 0 are [WB JEE 2010]
x 2 + x + 1 = 0, then the equation whose roots are (a) imaginary
α 19 and β 7 , is [WB JEE 2010] (b) real, rational and equal
Targ e t E x e rc is e s
(a) x 2 − x + 1 = 0 (b) x 2 − x − 1 = 0 (c) real, irrational and unequal
(c) x 2 + x − 1 = 0 (d) x 2 + x + 1 = 0 (d) real, rational and unequal
237
Answers
Work Book Exercise 5.1
1. (a) 2. (c) 3. (d) 4. (b) 5. (a)
Work Book Exercise 5.2
1. (c) 2. (a) 3. (d) 4. (b) 5. (a) 6. (a) 7. (a) 8. (b) 9. (a) 10. (d)
Work Book Exercise 5.3
1. (a) 2. (a) 3. (c) 4. (d) 5. (a,b,c,d)
Work Book Exercise 5.4
1. (b) 2. (a) 3. (c) 4. (c) 5. (d) 6. (a) 7. (b) 8. (d) 9. (b) 10. (c)
11. (d) 12. (c) 13. (c) 14. (c)
Work Book Exercise 5.5
1. (a) 2. (b) 3. (a) 4. (c) 5. (d) 6. (b) 7. (c) 8. (a) 9. (d) 10. (c)
11. (c) 12. (c) 13. (c) 14. (c)
Target Exercises
1. (b) 2. (b) 3. (a) 4. (c) 5. (a) 6. (b) 7. (a) 8. (d) 9. (c) 10. (c)
11. (b) 12. (c) 13. (b) 14. (d) 15. (d) 16. (c) 17. (d) 18. (b) 19. (b) 20. (c)
21. (d) 22. (c) 23. (d) 24. (c) 25. (d) 26. (c) 27. (d) 28. (c) 29. (a) 30. (a)
31. (c) 32. (b) 33. (b) 34. (c) 35. (d) 36. (b) 37. (b) 38. (c) 39. (d) 40. (b)
Ta rg e t E x e rc is e s
41. (c) 42. (b) 43. (a) 44. (a) 45. (b) 46. (a) 47. (b) 48. (b) 49. (a) 50. (a)
51. (b) 52. (a) 53. (a) 54. (b) 55. (a) 56. (a) 57. (c) 58. (c) 59. (a) 60. (c)
61. (c) 62. (a) 63. (a) 64. (a) 65. (c) 66. (b) 67. (c) 68. (a) 69. (d) 70. (a)
71. (b) 72. (c) 73. (b) 74. (b) 75. (c) 76. (a) 77. (b) 78. (a) 79. (b) 80. (a)
81. (d) 82. (a) 83. (d) 84. (b) 85. (a) 86. (c) 87. (d) 88. (c) 89. (b) 90. (c)
91. (b) 92. (b) 93. (b) 94. (d) 95. (a) 96. (d) 97. (c) 98. (b) 99. (d) 100. (d)
101. (c) 102. (d) 103. (d) 104. (b) 105. (b) 106. (b) 107. (a) 108. (c) 109. (c) 110. (b)
111. (b) 112. (d) 113. (d) 114. (a) 115. (a) 116. (b) 117. (c) 118. (d) 119. (c) 120. (a)
121. (a) 122. (d) 123. (c) 124. (c) 125. (b) 126. (c) 127. (d) 128. (c) 129. (b) 130. (d)
131. (a) 132. (d) 133. (b) 134. (b) 135. (c) 136. (b) 137. (a) 138. (a) 139. (b) 140. (b)
141. (a) 142. (a) 143. (b) 144. (a) 145. (b) 146. (a) 147. (a) 148. (a) 149. (a) 150. (b)
151. (c) 152. (b) 153. (a) 154. (b) 155. (b) 156. (c) 157. (a) 158. (c) 159. (b) 160. (c)
161. (b) 162. (a) 163. (d) 164. (b) 165. (d) 166. (a) 167. (a) 168. (a) 169. (b) 170. (a)
171. (d) 172. (c) 173. (b) 174. (b) 175. (c) 176. (d) 177. (a) 178. (a) 179. (c) 180. (d)
181. (b) 182. (c) 183. (a) 184. (a) 185. (b) 186. (d) 187. (c) 188. (c) 189. (c) 190. (a)
191. (d) 192. (a) 193. (c) 194. (b) 195. (a) 196. (b) 197. (d) 198. (c) 199. (a) 200. (a)
201. (b,c) 202. (a,b) 203. (a,b) 204. (a,b,c) 205. (a,b,d) 206. (a,b,d) 207. (a,b,c) 208. (a,c,d) 209. (a,d) 210. (a,b,c,d)
211. (a,b) 212. (b,d) 213. (a,c) 214. (a,d) 215. (a,c) 216. (c,d) 217. (a,b,d) 218. (d) 219. (b) 220. (c)
221. (b) 222. (a) 223. (c) 224. (b) 225. (a) 226. (c) 227. (b) 228. (a) 229. (b) 230. (c)
231. (a) 232. (*) 233. (**) 234. (1) 235. (2) 236. (1)
* A → r; B → p; C → s; D → q
** A → q; B → p; C → s; D → r; E → t
Entrances Gallery
1. (b,d) 2. (d) 3. (c) 4. (a,b,c) 5. (b) 6. (a) 7. (c) 8. (c) 9. (b) 10. (c)
11. (b) 12. (c) 13. (a) 14. (a) 15. (d) 16. (c) 17. (b) 18. (c) 19. (c) 20. (a)
21. (a) 22. (a) 23. (b) 24. (d) 25. (a) 26. (b) 27. (c) 28. (c) 29. (a) 30. (b)
31. (c) 32. (c) 33. (b) 34. (a) 35. (b) 36. (b) 37. (c) 38. (c) 39. (d) 40. (b)
41. (b) 42. (b) 43. (a) 44. (c) 45. (a) 46. (c) 47. (a) 48. (c) 49. (b) 50. (d)
51. (a) 52. (d) 53. (b) 54. (c) 55. (d) 56. (c) 57. (d) 58. (c) 59. (d) 60. (a)
61. (a) 62. (d) 63. (c) 64. (c) 65. (a) 66. (c) 67. (d) 68. (a) 69. (b) 70. (d)
71. (a) 72. (c) 73. (b) 74. (b) 75. (a) 76. (c) 77 (b) 78. (b) 79. (c) 80. (a)
81. (c) 82. (a) 83. (d) 84. (a) 85. (a) 86. (a) 87. (c) 88. (c) 89. (d) 90. (c)
238 91. (a) 92. (e) 93. (c)
Explanations
Target Exercises
1. The given inequation reduces to 1
12. Let us first solve|3 x + 1| <
8x 5 3
2< or x>
10 2 1 1
⇒ − < 3x + 1 <
∴ Least integer is 3. 3 3
4 2
2. a − b + c < 1, a + b + c > − 1, 9 a + 3 b + c < − 4 ⇒ − < 3x < −
3 3
⇒ a− b+ c <1 ⇒
4
− <x<−
2
⇒ − 2b < 2 9 9
and −a − b − c < 1
Also, 0 < |3 x + 1| is satisfied by each x except when
or 2b + 2 > 0
3x + 1 = 0
⇒ b + 1> 0 1
⇒ x=−
3. (a4 + b4 ) − (a3 b + ab3 ) = (a4 − a3 b) − (ab3 − b4 ) 3
= a3 (a − b) − b3 (a − b) 4 2 1
∴ Solution is − , − − − .
= (a − b)(a3 − b3 ) 9 9 3
= (a − b)2 (a2 + ab + b2 ) 13. | x − 1| is the distance of x from 1
which is positive as both factors are positive. | x − 3| is the distance of x from 3
Thus, (a4 + b4 ) − (a3 b + ab3 ) > 0 The points x = 2 is equidistant from 1 and 3.
⇒ a4 + b4 > a3 b + ab3 Hence, the solution consists of all x ≥ 2.
4. Using number line rule, 14. || x| − 1| < |1 − x| …(i)
2
Targ e t E x e rc is e s
x ∈ (− ∞, 0 ) ∪ , ∞ Case I x ≥ 0
3
∴ Eq. (i) becomes| x − 1| < x − 1or|1 − x| < 1 − x
5. Using number line rule, which is not satisfied by any x, because
x ∈ (− ∞, 3) ∪ (10, ∞ ) | a| ≥ a, ∀ a ∈ R
6. Using number line rule, Case II − 1≤ x < 0
x ∈ (− ∞, − 4) ∴ Eq. (i) becomes |− 1 − x| < 1 − x
or | x + 1| < 1 − x
7. Using number line rule,
⇒ x + 1 < 1 − x or x < 0
– ve + ve – ve + ve
Thus, Eq. (i) is satisfied for − 1 ≤ x < 0
–7 –—
1 3
2 Case III x < −1
Eq. (i) becomes |− 1 − x| < 1 − x
1
x ∈ (− ∞, − 7 ) ∪ − , 3 ⇒ |1 + x| < 1 − x
2
⇒ − (1 + x ) < 1 − x
x+2 ⇒ − 2 < 0, which is true
8. 5 x + 2 < 3 x + 8 and <4
x −1 So, solution set is (− ∞, 0 ).
⇒ x ∈ (− ∞, 1) ∪ (2, 3) 1 1
15. −2 < 4 ⇒ − 4< −2 < 4
9. |3 x + 2| < 1 ⇔ − 1 < 3 x + 2 < 1 x x
⇔ − 3 < 3x < − 1 ⇒
1
−2 < < 6
1 x
⇔ − 1< x < −
3 1 1
Hence, x ∈ − ∞, − ∪ , ∞ .
10. |3 − x| = x − 3 is true only when x − 3 ≥ 0 2 6
i.e. x ≥ 3 3
16. 1 + >2
11. |2 x − 3| < | x + 2| x
Squaring on both sides, we get 3 3
Case I 1+ >2 ⇒ >1 [clearly x > 0]
(2 x − 3)2 − ( x + 2 )2 < 0 x x
( x − 5)(3 x − 1) < 0 ⇒ 3 > x or x < 3
3 3
+ – + Case II 1 + < − 2 ⇒ <−3 [hence x < 0]
–∞ ∞ x x
1/3 5 ⇒ 3 > − 3x
1 ⇒ − 1 < x or x > − 1
⇒ x ∈ , 5
3 Hence, either 0 < x < 3 or − 1 < x < 0
239
5 17. | x 2 − 10| ≤ 6
⇒
⇒
− 6 < x − 10 ≤ 6
2
4 ≤ x ≤ 16
2
25. Here, 5 = 4a and 6 = 5b
Let
Then,
log 3 2 = x
2 = 3x
Objective Mathematics Vol. 1
⇒ 2 ≤ | x| ≤ 4 Now, 6 = 5b = (4a )b = 4ab
∴ Either 2 ≤ x ≤ 4 or 3 = 2 2 ab − 1
or − 4≤ x ≤ −2 ∴ 2 = (2 2 ab − 1 )x = 2 x( 2 ab − 1)
∴ x ∈ [− 4, − 2 ] ∪ [ 2, 4]
⇒ x(2 ab − 1) = 1
1
18. x + > 2 [clearly x ≠ 0] 26.
log a log x
⋅ =2
x log 5 log a
x2 + 1 x2 + 1 ⇒ log 5 x = 2 ⇒ x = 52 = 25
⇒ >2 ⇒ >2 [Q x 2 + 1 > 0 ]
x | x|
27. x + 5 = 2 6 − x .
⇒ x 2 + 1 > 2| x|
Its graph is as follows
⇒ | x|2 − 2| x| + 1 > 0
2 6–x Y
⇒ (| x| − 1)2 > 0
(x+5)
⇒ | x| ≠ 1 ⇒ x ≠ − 1, 1
∴ x ∈ R − { − 1, 0, 1} O
| x + 1| | x + 1|2
19. + | x + 1| = O
X
| x| | x|
Since,| a| + | b| = | a + b| iff ab ≥ 0 Clearly, there cannot be more than one solution and by
( x + 1) ( x + 1)2 trial, the solution is x = 3.
∴ ⋅ ( x + 1) ≥ 0 ⇒ ≥0
x x 28. |4 − 5 x| > 2 2 = 4
⇒ x > 0 or { − 1} 5x 5x
⇒ − 1 > 1 or − 1< − 1
Ta rg e t E x e rc is e s
20. (− 3 ≤ x − 1 ≤ 3) and (x − 1 ≤ − 1or x − 1 ≥ 1) 4 4
⇒ (− 2 ≤ x ≤ 4) and (x ≤ 0 or x ≥ 2) ∴ x>
8
or x<0
⇒ − 2 ≤ x ≤ 0 or 2 ≤ x ≤ 4 5
8
21. Here, x > 1 − x i.e. greater than zero So, the solution set = (− ∞, 0 ) ∪ , + ∞
5
∴ x2 > 1 − x and 1 − x > 0, x > 0
x+2 x+2 1
⇒ x + x − 1 > 0 and
2
0 < x <1 29. ≥ (0.2 )1 or ≥
x x 5
2
1 5 5 x( x + 2 ) ≥ x 2 or 4 x 2 + 10 x ≥ 0
⇒ x + > and 0 < x < 1
2 4 5
∴ x ≥ 0 or x≤−
5 −1 2
⇒ x> and 0 < x < 1
x+2
2 Also, >0
5 −1 x
⇒ < x <1
2 ⇒ x( x + 2 ) > 0
∴ x < − 2 or x > 0
22. | x − 2|2 − 3| x − 2| + 2 = 0
5
⇒ (| x − 2| − 1)(| x − 2| − 2 ) = 0 ⇒ x = 3, 1, 4, 0 ∴The solution set is − ∞, − ∪ (0, + ∞ ).
2
∴ Product of roots = 0
1 ± 1 − 4 log16 k
23. Here, a > b ⇒ ax > bx, only when x > 0 30. log16 x =
2
and a>b>0
For exactly one solution, 4 log16 k = 1
Option (a) is false
∴ k 4 = 16
| x| > | y| ⇒ x > y > 0,
Option (b) is false Hence, k = 2, − 2, 2 i , − 2 i
1 1 But k is positive and real.
a>b>0 ⇒ <
a b 31. xlog x a ⋅log a y ⋅log y z
Option (c) is false log a log y log z
⋅ ⋅
a > b ⇒ a + c > b + c , for c ∈ R ⇒ x log x log a log y
⇒ xlog x z = z
∴Option (d) is true. 32. log10 x = 25 log10 (5) = 25 {1 − log10 2}
24. For x ≤ 1, the given inequation gives x ≤ 0 and for x ≥ 3, = 25 {1 − 0. 30103}
the given inequation gives x ≥ 4. = 25 { 0.69897}
For 1 < x ≤ 2 or 2 < x < 3, the given inequation gives no = 17.469
solution. ∴ Number of zeroes coming immediately after the
∴ Solution is (− ∞, 0 ] ∪ [4, ∞ ). decimal place is 17.
240
33.
log x log y log z
=
y−z z−x x−y
=
=
x log x + y log y + z log z
On multiplying, we get
1 2
3
1 1
b
1 1
( a + b2 + c 2 ) ⋅ 2 + 2 + 2
3a c
5
Inequalities and Quadratic Equation
1/ 3
0 1
≥ (a2 b2c 2 )1/ 3 2 2 2
⇒ log( x x y y z z ) = 0 ⇒ x x y y z z = 1 a b c
−1 1 1 1
34. log1/ 3 ( x 2 + x + 1) < − 1 = log1/ 3
1 ⇒ (a2 + b2 + c 2 ) 2 + 2 + 2 ≥ 9
3 a b c
−1 ∴ Least value is 9.
1
⇒ x2 + x + 1 <
3 43. Since, AM ≥ GM
⇒ x + x −2 < 0
2 a2 + b2 b2 + c 2
∴ ≥ a2 b2 = ab, ≥ bc
2 2
⇒ x ∈ (− 2, 1)
c +a
2 2
35. log10 x 2 ≥ 0 ⇒ log10 x 2 ≥ log10 1 and ≥ ca
2
⇒ x2 ≥ 1 ⇒ x ≥1 ⇒ x ≤ −1 On adding, we get a + b + c 2 ≥ ab + bc + ca
2 2
⇒ Option (a) holds.
x1
x ...
36. Value = log x1log x2log x3 K log xn − 1 ( xn n−−1 2 log xn xn ) a3 + b3 + c 3
x1 Again, ≥ (a3 b3c 3 )1/ 3
x... 3
= log x1log x2log x3 K log n − 1 xn n−−1 2 ⇒ a3 + b3 + c 3 ≥ 3 abc
= K = log x1 x1 = 1 ⇒ Option (d) does not hold.
4 4 b+c c+a a+ b
37. 4 sin 2 x
+ ≥2 4 sin 2 x
⋅ Next, ≥ bc , ≥ ca, ≥ ab
4sin
2
x
4sin
2
x 2 2 2
b + c c + a a + b
⇒ 4 sin 2 x
+4 cos 2 x
≥4 ⇒ ≥ a bc
2 2 2
2 2 2
x
3 1 ⇒ (b + c )(c + a)(a + b) ≥ 8 abc
38. y = + x , we know AM ≥ GM
Targ e t E x e rc is e s
3 3 ⋅3 ⇒ Option (b) does not hold.
1/ 3
3x 1 1 a b c a b c
+ x x Again, + + ≥ ⋅ ⋅
⇒ 3 3 ⋅3 ≥ 3 . 1 3 b c a b c a
2 3 3x ⋅ 3 a b c
⇒ + + ≥3
2
⇒ 3 x − 1 + 3− x − 1 ≥ b c a
3 ⇒ Option (c) does not hold.
39. Since, AM ≥ GM 44. 3 (a2 + b2 + c 2 ) − (a + b + c )2
a b b c c
+ + + + ≥6 = 2(a2 + b2 + c 2 − bc − ca − ab)
b a c b a
b+c c + a a+ b = (b − c )2 + (c − a)2 + (a − b)2 ≥ 0
⇒ + + ≥6
a b c ⇒ 3(a2 + b2 + c 2 ) ≥ (a + b + c )2 > 9
Hence, the least value is 6. ⇒ a2 + b2 + c 2 > 3
40. Since, AM > GM for different numbers. ⇒ Option (a) holds.
1
(b + c − a) + (c + a − b) Now, a6 + b6 ≥ 12 a2 b2 − 64
∴ > [(b + c − a)(c + a − b)]2
2 If a6 + b6 + 64 ≥ 12 a2 b2
⇒ c > [(b + c − a)(c + a − b)]1/ 2 i.e. if a6 + b6 + 2 6 ≥ 3 ⋅ 2 2 ⋅ a2 b2
Similarly, b > [(b + c − a)(a + b − c )] 1/ 2
a6 + b6 + 2 6
i.e. if ≥ (2 6 a6 b6 )1/ 3 [Q AM ≥ GM]
and a > [(a + b − c )(c + a − b)]1/ 2 3
On multiplying, we get ⇒ Option (b) does not hold.
abc > (b + c − a)(c + a − b)(a + b − c ) Again, AM ≥ HM
⇒ (b + c − a)(c + a − b)(a + b − c ) − abc < 0 a+ b+c 3
∴ ≥
3 1 1 1
41. Here, x1 ⋅ x2 … xn = 1 + +
1 a b c
x1 + x2 + K + xn α 3 1 1 1 9
∴ ≥ ( x1 ⋅ x2 K xn )n ⇒ ≥ ⇒ + + ≥
n 3 1+ 1+ 1 a b c α
⇒ x1 + x2 + K + xn ≥ n a b c
42. Since, AM ≥ GM ⇒ Option (c) does not hold.
a2 + b2 + c 2
∴ ≥ (a2 b2c 2 )1/ 3 45. Using AM ≥ GM, one can show
3 (b + c )(c + a)(a + b) ≥ 8 abc
1/ 3
1 1 1 1 1 ⇒ ( p − a)( p − b)( p − c ) ≥ 8 abc
and 2 + 2 + 2 ≥ 2 2 2
3a b c a b c ⇒ Option (b) holds. 241
( p − a) + ( p − b) + ( p − c )
5 Also,
3
≥ [( p − a)( p − b)( p − c )]1/ 3
52. D = c 2 (3 a2 + b2 )2 − 4 abc 2 (− 6 a2 − ab + 2 b2 )
= c 2 (3 a2 − b2 + 4 ab)2
Objective Mathematics Vol. 1
3 p − (a + b + c ) 53. The sum of the coefficients
⇒ ≥ [( p − a)( p − b)( p − c )]1/ 3
3 = (a + 2 ) + (a − 3) − (2 a − 1) = 0
2p So, x = 1is a root.
⇒ ≥ [( p − a)( p − b)( p − c )]1/ 3 2a − 1
3 ∴ Other root = − = Rational, ∀ a, a ≠ − 2
8 p3 a+2
⇒ ( p − a)( p − b)( p − c ) ≤
27 54. |α − β| > 3p ⇒ (α + β)2 − 4 αβ > 3p
⇒ Option (a) does not hold. ⇒ p2 − 3 p − 4 > 0 and p > 0
1 bc ca bc ca ⇒ ( p − 4)( p + 1) > 0 and p > 0
Again, + ≥ ⋅ , etc.
2 a b a b ⇒ p>4
On adding the inequalities, we get 1 1 1
55. α = ,β = ⇒ αβ =
bc ca ab
+ + ≥a+ b+c = p 4 − 3i 4 + 3i a
a b c or a = 25
⇒ Option (c) does not hold. −b
Also, α+β=
46. AM ≥ GM a
bcx + cay + abz 8 b
⇒ ≥ (a2 b2c 2 ⋅ xyz )1/ 3 ⇒ =−
3 25 25
bcx + cay + abz ≥ 3 xyz ⇒ b=−8
or bcx + acy + abz ≥ 3abc 56. For ax 2 − bxy − ay 2 , D = b2 + 4a2 > 0
a b c 57. (16 + 9 k )x 2 + 2(6 − k )x + (39 + 11k )
2 + 3 + 4 4 1/ 9
2 3 4 a b c
2 3
47. ≥ ⇒ D=0
9 2 3 4 ∴Only one possible integral solution for k.
Ta rg e t E x e rc is e s
a2 b3c 4 p
⇒ 29 ≥ 2 3 4 58. α + α 2 = −
2 ⋅ 3 ⋅ 4 3
3
or a2 b3c 4 ≤ 2 9 ⋅ 2 2 ⋅ 33 ⋅ 44 and α ⋅α 2 = ⇒ α 3 = 1
3
or a2 b3c 4 ≤ 42 ⋅ 63 ⋅ 84
⇒ α = 1, ω, ω 2
⇒ Maximum value of P is 42 ⋅ 63 ⋅ 84 . If α = 1, then p = − 6
2
( x 2 + x22 + K + xn2 ) x1 + x2 + K + xn If α = ω, then p = 3
48. Here, 1 ≥
n n But p>0 ⇒ p≠−6
2 Hence, p=3
n n
⇒ n∑ xi2 ≥ ∑ xi 59. α + β = − p, αβ = p3 and β 2 = α
i =1 i =1 Hence, the roots are 4 and − 2.
49. x + y + z = 2 60. Since, a, b and c are positive, therefore
Now, (2 − x )(2 − y )(2 − z ) = ( y + z )( z + x )( x + y ) ax 2 + b| x| + c > 0
We know, ( y + z ) ≥ 2 yz , ( x + z ) ≥ 2 xz , ( x + y ) ≥ 2 xy Hence, the equation ax 2 + b| x| + c = 0 has no real root.
∴ ( y + z )( z + x )( x + y ) ≥ 8 xyz b c
61. α + β = − , αβ =
⇒ (2 − x ) (2 − y ) (2 − z ) ≥ 8 xyz a a
x −1 + y −1 + z −1 x + y + z − 1 ⇒ [{(α + k ) + ( β + k )} 2 − 4 (α + k )( β + k )]
Also, ≥
3 3 = (α + β )2 − 4 αβ
2
−1 q r b2 4 c
2 ⇒ −4 = 2 −
⇒ x −1 + y −1 + z −1 ≥ 3 ⋅ p 2
p a a
3
q 2 − 4 pr b2 − 4 ac
⇒ x −1 + y −1 + z −1 ≥
9 ⇒ =
2 p2 a2
2
n n b2 − 4 ac a
50. p = (a1 + a2 + K + an ) = ∑
2 2
ai2 + 2 ∑ ai aj ⇒ =
q 2 − 4 pr p
i =1 i< j
α m
n
1 2 62. Since, =
⇒ p2 − 2q = ∑ ai2 ≥ 0 ⇒ q ≤ p β n
i =1
2 α +β m+ n
⇒ =
51. D = b − 4ac, D′ = d + 4 ac
2 2 α −β m− n
⇒ D + D′ = b2 + d 2 > 0 (α + β )2 (m + n )2
⇒ =
(α + β ) − 4 αβ (m + n )2 − 4 mn
2
242 ∴Atleast one of D, D′ is positive.
4 αβ
⇒
⇒
c/a
=
=
4 mn
(α + β )2 (m + n )2
mn
70. Given, p, q and r are in AP ⇒ 2q = p + r
The roots of px 2 + qx + r = 0 are real, if
p + r
2
5
Inequalities and Quadratic Equation
2
b /a 2
(m + n )2 q 2 − 4 pr ≥ 0 ⇒ − 4 pr ≥ 0
2
⇒ ac (m + n )2 = mnb2 2
r r r
∴ − 14 + 1 ≥ 0 ⇒ −7 ≥ 4 3
63. x 3 + 3 − 5 x 2 + 2 + x +
1 1 1
p p p
x x x
3 2 2 x + 1 x 2 + 7 2 x + 17
1 1 1 1 71. Given, + 2 =
= x + − 3 x + − 5 x + − 2 + x + 9 x −7 9
x x x x
3 2 ⇒ 7 x 2 = 175
1 1 1
= x + − 5 x + − 2 x + + 10 ⇒ x 2 = 25
x x x
⇒ x=±5
= 125 − 125 − 2(5) + (10 )
=0 72. x = 1, satisfy a(b − c )x + b(c − a)x + c (a − b) = 0
2
∴ α =β =1
64. Let x = 6 + 6 + 6 + K ∞ = 6 + x 2 ac
⇒ c (a − b) = a(b − c ) ⇒ b =
Squaring on both sides, we get a+c
x2 = 6 + x Hence, a, b and c are in HP.
⇒ x = − 2, 3 73. Since, α and β are roots of equation x 2 + x α + β = 0.
But x cannot be negative. ∴ α + β = − α , αβ = β ⇒ α = 1
Hence, x = 3 [Qβ = 0 does not satisfy the equation]
65. α + β = − p, αβ = q ∴ 1+ β = − 1
Given, α + β =α + β
2 2
⇒ β = −2
= (α + β )2 − 2αβ 74. α and β are roots of 8 x 2 − 3 x + 27 = 0.
Targ e t E x e rc is e s
⇒ − p = p2 − 2q ⇒ α+β=
3
, αβ =
27
⇒ p + p = 2q
2 8 8
1 1 1 1
66. Here, q − r + r − p + p − q = 0 α 2 3 β 2 3 (α 3 )3 + (β 3 )3
∴ + =
p−q β α (αβ )1/ 3
∴Its roots are 1, .
q−r α+β 3/ 8
= =
67. 2 x 2 − 25 x = 3 (14) (αβ)1/ 3 (27 / 8)1/ 3
3/ 8 1
⇒ 2 x 2 − 25 x − 42 = 0 = =
3/ 2 4
⇒ (2 x + 3)( x − 14) = 0
⇒
3
x = − , 14 75. Since, f(1) = 0 and f ( x ) is perfect square.
2 ∴ α =β =1
But x cannot be negative. ⇒ αβ = 1
Hence, x = 14 or c 2 (a2 − b2 ) = a2 (b2 − c 2 )
x 2 − bx λ − 1 ⇒ b2 (a2 + c 2 ) = 2 a2c 2
68. Given, =
ax − c λ+1 2 a2c 2
⇒ b2 =
⇒ (λ + 1)( x − bx ) = (λ − 1)(ax − c )
2
a2 + c 2
⇒ (λ + 1)x 2 − [λb + b + λa − a]x + (λ − 1)c = 0 Hence, a2 , b2 and c 2 are in HP.
Since, α+β=0 76. We have,(1 + m)x 2 − 2(1 + 3 m)x + (1 + 8 m) = 0 ...(i)
λb + b + λa − a
⇒ =0 Let D be the discriminant of Eq. (i). Roots of Eq. (i) will
λ+1
be equal, if D = 0
⇒ λ(a + b) = a − b ⇒ m2 − 3 m = 0 ⇒ m(m − 3) = 0
a−b
⇒ λ= ∴ m = 0, 3
a+ b
77. Let the other root be β.
69. Discriminant = 0 2 1
⇒ 4(ac + bd )2 − 4(a2 + b2 )(c 2 + d 2 ) = 0
Then, α+β=− =−
4 2
⇒ 2 abcd − a2d 2 − b2c 2 = 0 ⇒
1
β = − −α ...(i)
⇒ (ad − bc )2 = 0 2
and 4 α 2 + 2α − 1 = 0
⇒ ad = bc
a c Now, 4 α 3 − 3 α = α (4 α 2 − 3) = α (1 − 2 α − 3)
⇒ = 243
b d [Q 4 α 2 + 2 α − 1 = 0]
5 = − 2α 2 − 2α = −
=−
1
(1 − 2α ) − 2α
1
2
(4 α 2 ) − 2α
[Q 4 α 2 = 1 − 2α ]
⇒
b a2
− =
a
b2 2 c
−
c2
a = b − 2 ca
2
c2
Objective Mathematics Vol. 1
2 a2
1 ⇒ 2ca2 = bc 2 + ab2
= − −α =β [from Eq. (i)]
2
Hence, 4 α 3 − 3 α is the other root. Hence, bc 2 , ca2 and ab2 are in AP.
78. Since, the roots are equal. 83. Since, sin θ and cos θ are the roots of the equation
∴ B − 4 AC = 0
2 ax 2 + bx + c = 0.
b c
⇒ (c − a)2 − 4(a − b)(b − c ) = 0 ∴ sin θ + cos θ = − and sin θ cos θ =
a a
⇒ c 2 + a2 − 2ca − 4ab + 4ac + 4b2 − 4bc = 0 Now, (sin θ + cos θ )2 = 1 + 2 sin θ cos θ
⇒ c + a + 2 ac + 4b − 4b(c + a) = 0
2 2 2
b2 2c a + 2 c
∴ = 1+ =
⇒ (c + a)2 + (2 b)2 − 2 ⋅ 2 b(c + a) = 0 a2 a a
⇒ [(c + a) − (2 b)]2 = 0 ⇒ b2 = a(a + 2c ) = a2 + 2 ac
⇒ c + a − 2b = 0 ⇒ b + c2
2
= a2 + 2 ac + c 2 = (a + c )2
⇒ 2b = a + c Hence, (a + c )2 = b2 + c 2
Hence, a, b and c are in AP. 84. Let the speed of the bus = x km/h, then the speed of the
Aliter car = ( x + 25) km/h
As x = 1satisfy given equation. Time taken by the bus to cover 500 km =
500
h
∴ α = 1is the root x
⇒ α =β =1 and time taken by the car to cover 500 km =
500
h
a−b ( x + 25)
∴ αβ = 1 =
b−c 500 500 50 50
Given, = + 10 ⇒ = +1
( x + 25) ( x + 25)
Ta rg e t E x e rc is e s
⇒ 2b = a + c x x
50 50
79. Let the roots be α and β. ⇒ − =1
−n n α p x ( x + 25)
Then, α + β = , αβ = and = 50( x + 25 − x )
l l β q ⇒ =1
x( x + 25)
p q n α β
Now, + + = + + αβ ⇒ 1250 = x 2 + 25 x
q p l β α
n n ⇒ x 2 + 25 x − 1250 = 0
− +
α + β + αβ ⇒ x + 50 x − 25 x − 1250 = 0
2
= = l l =0
αβ n ⇒ ( x + 50 )( x − 25) = 0
l Q x ≠ − 50 or x = 25
80. The given equation can be written as Hence, the speed of the bus = 25 km/h and speed of
(ab − 1)x 2 + (a + b)x − ab = 0 the car = (25 + 25) = 50 km/h.
Here, D = (a + b)2 + 4ab(ab − 1) 85. The given equation can be written as
(ay + a′ )x 2 + (by + b′ )x + (cy + c ′ ) = 0
= (a + b) − 4ab + 4(ab)
2 2
The condition that x may be rational function of y is,
= (a − b)2 + (2 ab)2 ≥ 0, ∀ a, b
(by + b′ )2 − 4(ay + a′ )(cy + c ′ ) is a perfect square
∴Roots are always real.
i.e. (b2 − 4ac )y 2 + (2 bb′ − 4ac ′ − 4a′ c )y + b′ 2
81. Put x − 1 = t ⇒ x − 1 = t 2 ⇒ x = t 2 + 1 − 4a′ c ′ is a perfect square ⇒ D = 0
The given equation reduces to i.e. 4 (bb′ − 2 ac ′ − 2 a′ c )2 − 4(b2 − 4ac )(b′ 2 − 4a′ c ′ ) = 0
t 2 + 1 + 3 − 4 t + t 2 + 1 + 8 − 6 t = 1, where t ≥ 0 or (ac ′ + a′ c )2 − 4aa′ cc ′ = abb′ c + a′ bb′ c
− a′ bb′ c − a′ c ′ b2 − acb′ 2
⇒|t − 2| + |t − 3| = 1, where t ≥ 0.
⇒ (ac ′ − a′ c ) = (ab′ − a′ b)(bc ′ − b′ c )
2
This equation will be satisfied, if 2 ≤ t ≤ 3
∴ 2 ≤ x − 1 ≤ 3 or 5 ≤ x ≤ 10 86. The corresponding quadratic equation of the given
Hence, the given equation is satisfied for all values of expression is
lying in [5, 10]. a(b − c )x 2 + b(c − a)xy + c (a − b)y 2 = 0
2
82. Let α and β be the roots of the given equation. x x
⇒ a(b − c ) + b(c − a) + c (a − b) = 0
b c y y
Then, α + β = − and αβ = x
a a Let =X
1 1 α 2 + β 2 (α + β )2 − 2αβ y
Given, α + β = 2 + 2 = =
α β (αβ )2 (αβ )2 Then, a(b − c )X 2 + b(c − a)X + c (a − b) = 0
244
The given expression will be perfect square, if the
discriminant of the corresponding equation of the
given expression is zero.
92. We have,
⇒
1
+
1
=
x+ p x+q r
1
r ( x + q + x + p) = ( x + p)( x + q )
5
Inequalities and Quadratic Equation
∴ b2 (c − a)2 − 4a(b − c )⋅ c (a − b) = 0 or x 2 + x( p + q − 2 r ) + pq − r ( p + q ) = 0
⇒ (bc + ab − 2 ac )2 = 0
Let its roots be α and − α.
⇒ b(a + c ) = 2 ac Then, α + (− α ) = − ( p + q − 2 r )
2 ac
⇒ b= ⇒ p + q = 2r
a+c Also, product = α (− α ) = pq − r ( p + q )
Hence, a, b and c are in HP. ( p + q )2 1
= pq − = − ( p2 + q 2 )
87. ax 2 − bx + c = 0 2 2
b | x + 1| | x + 1|2
⇒ α+β= 93. + | x + 1| =
a | x| | x|
c
and αβ =
a 1 ( x + 1)
1 ⇒| x + 1| + 1− =0
Clearly, y= 2 | x| | x|
x
1 b ∴ | x + 1| = 0 ⇒ 1 + | x | − | x + 1| = 0
⇒ a⋅ − +c =0
y y | x + 1| = 0 ⇒ x + 1 > 0 and x≠0
⇒ a − b y + cy = 0 ∴ x = − 1, x > 0
⇒ b2 y = (a + cy )2 Aliter
1 1 ( x + 1)2
⇒ (a + cy )2 = b2 y has roots 2 , 2 . >0
α β x
1 1 ⇒ x > 0 ∪ { − 1}
88. α + β = +
α 2 β2 94. Given,| x − 2| = x 2
b b − 2 ac
2
Targ e t E x e rc is e s
⇒ −= On taking x = 1and x = − 2, the equation is satisfied.
a c2
Hence, its solution is { − 2, 1}.
⇒ − bc = ab − 2 a2c
2 2
b c 2a 95. Given equation is| x|2 − 3| x| + 2 = 0
⇒ + =
c a b ⇒ (| x| − 1) (| x| − 2 ) = 0
c a b Either| x| = 1 or | x| = 2
⇒ , and are in AP.
a b c ⇒ x = ± 1, x = ± 2
a b c Hence, the given equation has four solutions.
i.e. , and are in HP.
c a b 96. Put 2 x = y, the equation becomes 16 y 2 = 8 y + 48
4+ 5 8+ 2 5 ⇒ 2 y2 − y − 6 = 0
89. α + β = and αβ =
5+ 2 5+ 2 ⇒ ( y − 2 ) (2 y + 3) = 0
2αβ 3
HM = ⇒ y = 2, −
α+β 2
16 + 4 5 But 2 x cannot be negative.
= =4
4+ 5 Hence, 2 x = 2 ⇒ x = 1
90. Since, x 2 − 3 x + 2 1
97. 5 − 2 6 =
5+ 2 6
i.e. ( x − 2 )( x − 1) is a factor of x 4 − px 2 + q.
∴ x = 2, x = 1are roots of x 4 − px 2 + q = 0 ∴Given equation
2 2
−3 −3
⇒ 16 − 4 p + q = 0 and 1 − p + q = 0 (5 + 2 6 )x + (5 − 2 6 )x = 10
On solving these equations, we get 1 2
Becomes y + = 10, where y = (5 + 2 6 )x − 3
p = 5, q = 4 y
91. (3 x )2 − 36 ⋅ 3 x + 243 = 0 ⇒ y 2 − 10 y + 1 = 0
Put 3 = y, the equation becomes
x ⇒ y = 5± 2 6
2
−3
y 2 − 36 y + 243 = 0 ⇒ (5 + 2 6 )x = (5 + 2 6 )1
⇒ ( y − 9)( y − 27 ) = 0
2
−3
⇒ (5 + 2 6 )x = (5 + 2 6 )−1
⇒ y = 9, 27
⇒ x − 3=1
2
⇒ 3x = 9
or 3 x = 27 or x2 − 3 = − 1
⇒ x =2 ⇒ x 2 = 4 or x2 = 2
or x=3 Hence, x=±2
∴Solution pair is {2, 3}. or x=± 2 245
5 98. The given equation can be written as
5
3
3x
5
x
+ =2
3
⇒
⇒
x2 + 1 − x
x
= (integer) k
x 2 − (k + 1) x + 1 = 0
[say]
Objective Mathematics Vol. 1
x Since, x is real, so (k + 1)2 − 4 ≥ 0
5
On putting = t, the equation becomes ⇒ k 2 + 2 k − 3 ≥ 0 ⇒ (k + 3)(k − 1) ≥ 0
3
t3 + t −2 = 0 ⇒ k ≤ −3 or k ≥ 1
Therefore, number of solutions is infinite.
⇒ t − 1 + ( t − 1) = 0
3
⇒ ( t − 1) ( t + t + 1) + ( t − 1) = 0
2 103. The given equation can be written as
3/ x 1/ x
⇒ ( t − 1) ( t + t + 2) = 0
2 3 3
+ =2
2 2
⇒ t = 1 or t + t + 2 = 0
2
1/ x
3
But t + t + 2 = 0 does not have real solutions.
2
Put = t, the equation becomes t 3 + t − 2 = 0
x
2
5
Therefore, t = 1 ⇒ = 1 ⇒ x=0 ⇒ (t − 1) (t 2 + t + 2 ) = 0
3
But t 2 + t + 2 = 0 has no real root. [Q t = 1]
99. Since, α and β are the roots of the equation 1/ x
3
x 2 − px + q = 0 ⇒ =1
2
∴ α+β=p 1
and αβ = q ⇒ =0
x
Now, (α 1/ 4 + β1/ 4 )4 = [(α 1/ 4 + β1/ 4 )2 ]2 which is not possible for any value of x.
= [α 1/ 2 + β1/ 2 + 2(αβ)1/ 4 ]2 104. Here,α + β = − b < 0 [Qb > 0]
= [ α + β + 2 αβ + 2(αβ ) 1/ 4 2
] αβ = c < 0 [Qc < 0]
Since, αβ < 0 and α < β ⇒ α is a negative root.
= [ p + 2 q + 2(q ) 1/ 4 2
]
Ta rg e t E x e rc is e s
α + β < 0 and α < 0 ⇒ β <|α |
= p + 6 q + 4q 1/ 4
p+2 q Hence, α < 0 < β < |α |
∴ α 1/ 4 + β1/ 4 = [ p + 6 q + 4 q1/ 4 p + 2 q ]1/ 4 105. Given, x 2 − 2 x + sin 2 α = 0
⇒ sin 2 α = 2 x − x 2
100. The function y = x ( x − 3) is defined for x ( x − 3) ≥ 0
Since, 0 ≤ sin 2 α ≤ 1
i.e. x ≤ 0 or x≥3 ...(i)
⇒ 0 ≤ 2 x − x2 ≤ 1
The given equation can be written as
9| x|2 − 19| x| + 2 = 0 ⇒ x − 2 x ≤ 0 and
2
x2 − 2 x + 1 ≥ 0
⇒ (9| x| − 1) (| x| − 2 ) = 0 ⇒ x ∈[0, 2 ] and x ∈ R
1 ⇒ x ∈[0, 2 ]
⇒ | x| = 2 or | x| =
9 106. (m + 1)2 − 4 (m + 4) ≥ 0
1
∴Solutions of the given equation are ± 2, ± . ⇒ (m − 1)2 ≥ 42
9 ⇒ | m − 1| ≥ 4
1 ⇒ m − 1 ≤ − 4 or m − 1 ≥ 4
In the domain (i), the required solutions are −2, − .
9 ⇒ m ≤ − 3 or m ≥ 5
101. We have, 2 x + 2 ⋅ 33 x /( x − 1) = 9 = 32 ∴Their product > 0, i.e. m + 4 > 0
3x ⇒ m>−4
⇒ ( x + 2 ) log 2 + log 3 = 2 log 3 Hence, −4 < m ≤ − 3
x −1
107. If x − a < 0,| x − a| = − ( x − a)
3x
⇒ ( x + 2 ) log 2 + − 2 log 3 = 0 ∴Equation becomes x 2 + 2 a( x − a) − 3 a2 = 0
x −1
⇒ x 2 + 2 ax − 5 a2 = 0
1
⇒ ( x + 2 ) log 2 + log 3 = 0 ⇒ x = − (1 + 6 ) a, (− 1 + 6) a
x −1 Q x<a ≤0
log 3 ∴ x = (− 1 + 6 ) a
⇒ x = − 2 or x = 1−
log 2 If x − a ≥ 0,| x − a| = x − a
∴The equation becomes x 2 − 2 a ( x − a) − 3 a2 = 0
102. We have, f ( x ) + f = 1
1
x ⇒ x 2 − 2 ax − a2 = 0
1 1 ⇒ x = (1 + 2 ) a, (1 − 2 ) a
⇒ x − [x] +
− =1
x x ⇒ x = (1 + 2 ) a, (1 − 2 ) a
1 1 Q x ≥ a and a ≤ 0
⇒ x + − 1 = [x] +
246 x x ∴ x = (1 − 2 ) a
108. For x ≥ a, the given equation becomes
x 2 − 2 a ( x − a) − 3 a2 = 0
115. sin (e x ) = 5 x + 5 − x, let 5 x = t , then
sin (e x ) = t +
1 1
= t − +2
t t
2
[Q 5 x > 0 ]
5
Inequalities and Quadratic Equation
⇒ x − 2 ax − a = 0
2 2
⇒ x = a (1 ± 2 ) ⇒ sin (e x ) ≥ 2, which is not possible for any real x.
As a ≥ 0 and x ≥ a, the possible root is a (1 + 2 ). 116. D = a2 + b2 + c 2 = a2 + (a + 1)2 + a2 (a + 1)2
For x < a, the given equation becomes = a2 {1 + (a + 1)2 } + (a + 1)2
x + 2 a( x − a) − 3 a = 0
2 2
= a2 (a2 + 2 a + 2 ) + (a2 + 2 a + 1)
⇒ x + 2 ax − 5 a = 0
2 2
= a4 + 2 a3 + 3 a2 + 2 a + 1
⇒ x = a (− 1 ± 6) = (a2 + a + 1)2
As a≥0
∴ D = a(a + 1) + 1 = an odd integer
and x<a
n 2 1 1 1
∴Possible root is a (− 1 − 6 ). 117. ∑ x − + x+
k − 1 k
k(k − 1)
x x k=2
109. + 2 + 1 = 1, which is of the form
2 1
5+ 2 2 5+ 2 2 where, constant term is given by
k ⋅ (k − 1)
∞
cos x α + sin x α = 1 1 1 1 1
∴ ∑ = + + +…
k(k − 1) 1 ⋅ 2 2 ⋅ 3 3 ⋅ 4
∴ x =2 k=2
110. [ x ] − [ x ] − 2 = 0
2 1 1 1 1 1
= 1 − + − + − + ... ∞
2 2 3 3 4
[ x ] = 2, − 1
=1
⇒ x ∈ [− 1, 0 ) ∪ [2, 3) 2
+ 7 x + 12
118. y x =1
111. Let e sin x = t
⇒ x 2 + 7 x + 12 = 0
Targ e t E x e rc is e s
1
t − − 4= 0 ⇒ x = − 3, − 4
t
⇒ t =2 ± 5 ⇒ y = 9, 10 [when y ≠ 1]
⇒ e sin x = 2 + 5, 2 − 5 Again, when y = 1, x = 5
But e sin x cannot be negative. ∴ Solutions are (− 3, 9) (− 4, 10 ) and (5, 1.)
∴ e sin x = 2 + 5 119. As (λ + 1) x 2 + 2 = λx + 3 has only one solution.
⇒ e sin x > e D=0
⇒ λ2 − 4 (λ + 1) (−1) = 0
⇒ sin x > 1, which is not possible.
⇒ λ2 + 4λ + 4 = 0
So, the given equation has no real solution.
⇒ (λ + 2 )2 = 0
112. Given, x x
= xx
∴ λ = −2
⇒ x x
= ( x x )1/ 2 , x > 0 120. Given quadratic expression is
⇒ x x
= x x/ 2, x > 0 x 2 + 2 (a + b + c ) x + 3 (bc + ca + ab) .
⇒ Either x = 1 This quadratic expression will be a perfect square, if the
x discriminant of its corresponding equation is zero.
or x = ,x>0
2 Hence,
Hence, x = 1and 4 4 (a + b + c )2 − 4 × 3 (bc + ca + ab) = 0
6 ± 36 − 32 6 ± 2 1 1 ⇒ (a + b + c )2 − 3 (bc + ca + ab) = 0
113. sec θ = = = ,
16 16 2 4 ⇒ a2 + b2 + c 2 + 2 ab + 2 bc + 2ca
But sec θ ≤ − 1or sec θ ≥ 1 − 3 (bc + ca + ab) = 0
Hence, the given equation has no roots. ⇒ a2 + b2 + c 2 − ab − bc − ca = 0
1
114. Given, log 2 x +
1
= 3+
1 ⇒ [2 a2 + 2 b2 + 2c 2 − 2 ab − 2 bc − 2ca] = 0
log 2 x 3 2
1
1 ⇒ [(a2 + b2 − 2 ab) + (b2 + c 2 − 2 bc )
= log 2 y + 2
log 2 y + (c 2 + a2 − 2ca)] = 0
1 1
⇒ log 2 x = 3 and log 2 y = [Q x ≠ y ] ⇒ [(a − b)2 + (b − c )2 + (c − a)2 ] = 0
3 2
⇒ x = 2 3 and y = 21/ 3 which is possible only when (a − b)2 = 0 , (b − c )2 = 0
Hence, x + y 3 = 8 + 2 = 10 and (c − a)2 = 0 i.e. a = b = c
247
5 121. ( x 2 + x − 2 )( x 2 + x − 3) = 12
Put x + x = y, so that Eq.(i) becomes
2
( y − 2 )( y − 3) = 12
…(i) Now, for equation x 2 − 4 qx + 2 q 2 − r = 0, product of
roots is
2q 2 − r = 2 (αβ)2 − (α 4 + β 4 )
Objective Mathematics Vol. 1
⇒ y 2 − 5y − 6 = 0 = − (α 2 − β 2 )2 < 0
⇒ ( y − 6) ( y + 1) = 0 ⇒ y = 6, − 1
As product of roots is negative, so the roots must be
When y = 6, we get real.
x2 + x − 6 = 0
128. A = a(b − c ) (a + b + c )
⇒ ( x + 3) ( x − 2 ) = 0 ⇒ x = − 3, 2
B = b(c − a) (a + b + c )
When y = − 1, we get
C = c ( a − b) ( a + b + c )
x2 + x + 1 = 0
Now, Ax 2 + Bx + C = 0
which has non-real roots and sum of roots is − 1.
⇒ (a + b + c ) [a (b − c ) x 2 + b (c − a) x + c (a − b)] = 0
122. Here, x = 0 is not a root. Divide both the numerator and
denominator by x and put x + 3/ x = y to obtain Given that roots are equal.
4 5 3 ∴ D=0
+ = ⇒ y = − 5, 3
y + 1 y−5 2 ⇒ b2 (c − a)2 − 4ac (b − c ) (a − b) = 0
x + 3/ x = − 5 has two irrational roots and x + 3/ x = 3 ⇒ b2c 2 − 2 ab2c + b2 a2 − 4a2 bc + 4acb2
has imaginary roots. + 4a2c 2 − 4abc 2 = 0
123. Given α and β are roots of equation ⇒ b c + b a + 4a c + 2 ab c − 4a2 bc − 4abc 2 = 0
2 2 2 2 2 2 2
x − 2x + 3 = 0
2
⇒ (bc + ab − 2 ac )2 = 0
⇒ α 2 − 2α + 3 = 0 …(i) ⇒ bc + ab = 2 ac
and β 2 − 2β + 3 = 0 …(ii) 1 1 2
⇒ + =
⇒ α 2 = 2α − 3 a c b
⇒ α 3 = 2α 2 − 3 α Hence, a, b and c are in HP.
⇒ P = (2α 2 − 3 α ) − 3 α 2 + 5 α − 2 129. Q ax 2 − bx + c = 0
Ta rg e t E x e rc is e s
= − α + 2α − 2 = 3 − 2 = 1
2
b
∴ α+β=
[using Eq. (i)] a
Similarly, we have Q = 2 c
and αβ =
Now, sum of roots is 3 and product of roots is 2. a
Hence, the required equation is x 2 − 3 x + 2 = 0. Also, (a + cy )2 = b2 y
124. Since, α and β are the roots of the equation ⇒ c 2 y 2 − (b2 − 2 ac ) y + a2 = 0
2 x − 35 x + 2 = 0.
2
c
2 b 2 c
⇒ y 2 − − 2 y + 1 = 0
∴ 2α 2 − 35 α = − 2 a a a
2
⇒ 2α − 35 = − ⇒ (αβ)2 y 2 − (α 2 + β 2 )y + 1 = 0
α
and 2 β 2 − 35 β = − 2 ⇒ y 2 − (α −2 + β −2 )y + α −2β −2 = 0
−2
⇒ 2β − 35 = ⇒ ( y − α −2 ) ( y − β −2 ) = 0
β
3 3 Hence, the roots are α −2 and β −2 .
− 2 − 2
Now, (2α − 35)3 (2β − 35)3 =
α β 130. Here, α 4 + β 4 = (α 2 + β 2 )2 − 2α 2β 2
8 × 8 64
= 3 3 = = 64 [Qα β = 1] = [(α + β )2 − 2αβ ]2 − 2 (αβ )2
α β 1
2
2
−3 2
−3 1 1
125. (31 + 8 15 )x + 1 = (32 + 8 15 )x = p2 + 2 −
2 2 2 p 2 p4
−3 −3 −3
⇒ (31 + 8 15 )x + 1x = (32 + 8 15 )x
1
⇒ x − 3=1
2
or x = ± 2 [Q a + b = (a + b)n ]
n n = p4 + +2
2 p4
c
126. aα 2 + c = − bα, aα + b = − 1
2
α = p2 − + 2 + 2 ≥2 + 2
Hence, the given expression is 2 p2
b 2 b (b2 − 2 ac ) 131. x1( x − x2 )2 + x2 ( x − x1 )2 = 0
(α + β 2 ) =
c a2c
⇒ x 2 ( x1 + x2 ) − 4 x x1 x2 + x1 x2 ( x1 + x2 ) = 0
127. Since, α and β are roots of x 2 + px + q = 0.
D = 16 ( x1 x2 )2 − 4 x1 x2 ( x1 + x2 )2 > 0 [Q x1 x2 < 0 ]
∴ α + β = − p and αβ = q
Now, α and β 4 are roots of x 2 − rx + q = 0.
4 The product of roots is x1 x2 < 0.
248 ∴ α +β =r
4 4
and α β = q4 4 Thus, the roots are real and of opposite signs.
132. ( x + a)( x + 1991) + 1 = 0
⇒
⇒
( x + a)( x + 1991) = − 1
( x + a) = 1
141. α ± iβ are complex roots of equation x 3 + qx + r = 0 and
γ is its real root.
⇒ ( x − γ )( x 2 − 2αx + α 2 + β 2 ) = x 3 + qx + r
5
Inequalities and Quadratic Equation
and x + 1991 = − 1 On comparing coefficients, we get
⇒ a = 1993 − γ − 2α = 0
⇒ γ = − 2α
or x + a = −1
Since, γ 3 + qγ + r = 0
and x + 1991 = 1
⇒ (− 2α )3 + q(− 2α ) + r = 0
⇒ a = 1989
i.e. 2α is root of cubic equation x 3 + qx − r = 0.
133. Let f ( x ) = − 3 + x − x 2 . Then, f ( x ) < 0 for all x, because
coefficient of x 2 is less than 0 and D < 0. Thus, LHS of 142. Since, α and β are the roots of x 2 + px + q = 0.
the given equation is always positive, whereas the RHS ∴ α + β = − p, αβ = q …(i)
is always less than zero. Hence, there is no solution. α 2 n + pnα n + q n = 0
134. x = 3 cos θ and y = 3 sin θ β 2 n + pn . β n + q n = 0
On subtraction,
z = 2 cos φ, t = 2 sin φ
α n + β n = − pn
∴ 6 cos θ sin φ − 6 sin θ cos φ = 6 α
⇒ sin (φ − θ ) = 1 Now, is a root of x n + 1 + ( x + 1)n = 0.
β
⇒ φ = 90 ° + θ n
αn α
⇒ P = xz = − 6 sin θ cos θ = − 3 sin 2θ ∴ + 1 + + 1 = 0
βn β
⇒ Pmax = 3
α n + β n (α + β )n
⇒ + =0
135. D > 0 ⇒ (a − 3)2 + 4 (a + 2 ) > 0 βn βn
⇒ a2 − 6 a + 9 + 4 a + 8 > 0 ⇒ − pn + (− p)n = 0
⇒ a2 − 2 a + 17 > 0 This is true only, if n is an even integer.
Targ e t E x e rc is e s
⇒ a ∈R 143. Let α and β be the roots of x 2 + bx + c = 0.
a2 + 1 1 1
∴ = 1− 2 ≥ Then, α + β = − b, αβ = c
a +2
2
a +2 2
Roots of the required equation are α 3 and β 3 . So, the
equation is x 2 − (α 3 + β 3 )x + α 3β 3 = 0
136. x = 2 + 3
⇒ ( x − 2 )2 = 3 ⇒ x 2 − {(α + β)3 − 3 αβ (α + β)} x + (αβ)3 = 0
⇒ x − 4x + 1 = 0
2
...(i) ⇒ x 2 − (− b3 + 3 cb)x + c 3 = 0
⇒ ( x − 2 )4 = 9 ⇒ x 2 + b(b2 − 3 c )x + c 3 = 0
⇒ x 4 − 8 x 3 + 24 x 2 − 32 x + 16 = 9 144. Since,α and β are roots of the equation ax 2 + bx + c = 0.
⇒ x 4 − 8 x 3 + 18 x 2 − 8 x + 2 + 6 ( x 2 − 4 x + 1) − 1 = 0 b
∴ α+β=−
From Eq. (i), we get a
x 4 − 8 x 3 + 18 x 2 − 8 x + 2 = 1 c
and αβ =
n n a
137. x 3 + (− x )3 = 0, if n ≥ 0 and n is an integer.
The required equation is
138. At x = 2, 1 1 1 1
x2 − + x+ ⋅ =0
x 3 − 2(1 + α )x 2 + (4α + α 2 + β 2 )x − 2(α 2 + β 2 ) aα + b aβ + b aα + b aβ + b
= 8 − 8 (1 + α ) + 8 α + 2 α 2 + 2β 2 − 2α 2 − 2β 2 = 0 ⇒ { a2αβ + ab (α + β) + b2 } x 2
∴ x = 2, z0 and z0 are three roots.
− { a (α + β ) + 2 b} x + 1 = 0
139. Let α = − 4, β = − 4 ω and γ = − 4ω 2 ⇒ (ca − b2 + b2 )x 2 − (2 b − b)x + 1 = 0
2 2
α 1 α 1 ⇒ cax 2 − bx + 1 = 0
∴ = 2 = ω and = 4 =ω
2
β ω γ ω
145. x = 2 + 21/ 3 + 2 2 / 3
∴Required equation is
⇒ x − 2 = 21/ 3 + 2 2 / 3
x 2 − (ω + ω 2 )x + ω 3 = 0 ⇒ x2 + x + 1 = 0
⇒ ( x − 2 )3 = (21/ 3 + 2 2 / 3 )3
140. Let f ( x ) = x 4 + ax 3 + bx 2 + cx − 1
⇒ x 3 − 8 − 6x ( x − 2)
Since, ( x − 1)3 is a factor.
= (21/ 3 )3 + (2 2 / 3 )3 + 3 (21/ 3 )⋅ (2 2 / 3 ) (21/ 3 + 2 2 / 3 )
∴ f(1) = 0, f′ (1) = 0, f ″ (1) = 0
⇒ a + b + c = 0, 3 a + 2 b + c = − 4 On putting 21/ 3 + 2 2 / 3 = x − 2 in above equation, we get
and 3a + b = − 6
x 3 − 6 x 2 + 12 x − 8 = 6 + 6 ( x − 2 )
⇒ a = − 2, b = 0, c = 2
So, other factor is ( x + 1.) ⇒ x 3 − 6x 2 + 6x = 2 249
5 146. 1, a1, a2 ,..., an − 1 are roots of x n − 1 = 0
⇒
x −1n
x −1
= ( x − a1 ) ( x − a2 )...( x − an − 1 )
On putting x = i and then x = − i , we get
1 − 4i + 6 + 7 i − 9 = (i − α ) (i − β )(i − γ )(i − σ )
and 1 + 4i + 6 − 7 i − 9 = (− i − α )(− i − β )(− i − γ )(− i − σ )
Objective Mathematics Vol. 1
On multiplying these two equations, we get
⇒ x n − 1 + x n − 2 + x n − 3 + ... + x + 1
(−2 + 3i ) (−2 − 3i ) = (1 + α 2 )(1 + β 2 ) (1 + γ 2 ) (1 + σ 2 )
= ( x − a1 ) ( x − a2 )... ( x − an − 1 )
⇒ 13 = (1 + α 2 )(1 + β 2 ) (1 + γ 2 ) (1 + σ 2 )
Put x = 1on the both sides, we get
1 + 1 + 1 + ... + 1 = (1 − a1 ) (1 − a2 ) ... (1 − an − 1 ) 152. tan θ1 + tan θ 2 + tan θ 3 = (a + 1)
⇒ (1 − a1 ) (1 − a2 ) ... (1 − an − 1 ) = n Σ tan θ1 tan θ 2 = (b − a)
tan θ1 tan θ 2 tan θ 3 = b
2 cos θ ± 4 cos θ − 4 2
Σ tan θ1 − ∏ tan θ1
147. x = = cos θ ± i sin θ ∴ tan(θ1 + θ 2 + θ 3 ) =
2 1 − Σ tan θ1 tan θ 2
Take α = cos θ + i sin θ, then ( a + 1 − b)
β = cos θ − i sin θ = =1
1 − (b − a)
⇒ α n = (cos θ + i sin θ)n π
⇒ θ1 + θ 2 + θ 3 =
= cos nθ + i sin nθ 4
Also, β n = (cos θ − i sin θ)n
153. Σα = 1, Σαβ = 0, αβγ = 1
= cos nθ − i sin nθ 2
1+ α −α + 1 − 2
Now, α n + β n = 2 cos nθ and α nβ n = 1 Σ = −Σ =Σ − 1
1− α 1− α 1− α
∴Required equation is x 2 − (α n + β n )x + α nβ n = 0 1
= 2Σ −3
⇒ x 2 − (2 cos nθ) x + 1 = 0 1− α
xn − 1 1 1 1 3x 2 − 2 x
148. = ( x − a1 ) ( x − a2 )...( x − an − 1 ) Now, + + = 3
x −1 ( x − α ) ( x − β) ( x − γ ) x − x2 − 1
⇒ x n − 1 + x n − 2 + ... + x + 1 1 1 1 3−2
⇒ + + = = −1
Ta rg e t E x e rc is e s
= ( x − a1 ) ( x − a2 ) ...( x − an − 1 ) 1− α 1− β 1− γ 1−1−1
Put x = − 1, we get 1+ α
⇒ = −5
(−1)n − 1 (1 + a1 ) (1 + a2 )...(1 + an − 1 ) 1− α
= 1 − 1 + 1 − ... + 1 154. Here, x → x + 1
0, if n is even ⇒ ax 2 + (b + 2 a)x + a + b + c = 0
=
1, if n is odd But equation given is 2 x 2 + 8x + 2 = 0
n −1
Thus, (1 + ω) (1 + ω )...(1 + ω
2
)= n ⇒ a = 2, b + 2 a = 8, a + b + c = 2
149. Let y = − 5 + 4i ⇒ y + 10 y + 41 = 0
2 ⇒ b = 4, c = − 4
∴ f ( x ) = x 4 + 9 x 3 + 35 x 2 − x + 4 ⇒ b= −c
= x 2 ( x 2 + 10 x + 41) − x ( x 2 + 10 x + 41) 155. x 2 + x + 1 = 0 has roots ω, ω 2 .
+4 ( x 2 + 10 x + 41) − 160 ∴ ax 2 + bx + c = 0
= − 160 ∴Both roots common.
∴f (− 5 + 4i ) = − 160 a b c
⇒ = =
150. Let x = cos θ, we get 4 cos 3 θ − 3 cos θ = p 1 1 1
⇒ p = cos 3 θ 156. We have, x 3 + 3 x 2 + 3 x + 2 = 0
1 −1
⇒ θ = cos ( p) and x = cos θ ⇒ ( x + 1)3 + 1 = 0
3
1 ⇒ ( x + 1 + 1) {( x + 1)2 − ( x + 1) + 1} = 0
⇒ x = cos cos − 1 ( p) ...(i)
3 ⇒ ( x + 2 ) ( x 2 + x + 1) = 0
Since, − 1 ≤ p ≤ 1 ⇒ 0 ≤ cos − 1 p ≤ π −1 ± 3 i
⇒ x = − 2,
1 π 2
or 0 ≤ cos −1 p ≤ , as we know cos x is decreasing.
3 3 ⇒ x = − 2, ω, ω 2
1 π Since, a, b, c ∈ R, ax 2 + bx + c = 0 cannot have one
∴ cos 0 ≥ cos cos − 1 p ≥ cos …(ii)
3 3 real and one imaginary root. Therefore, two common
From Eqs. (i) and (ii), we get roots of ax 2 + bx + c = 0 and x 3 + 3 x 2 + 3 x + 2 = 0 are
1 1 ω and ω 2 .
≤ x ≤ 1 ⇒ x ∈ ,1
2 b
2 Thus, − = ω + ω2 = − 1
a
151. Since, α, β, γ and σ are the roots of the given equation, c
⇒ a = b and = ω ⋅ ω2 = 1 ⇒ c = a
therefore a
250 x 4 + 4 x 3 − 6 x 2 + 7 x − 9 = ( x − α ) ( x − β) ( x − γ ) ( x − σ ) ⇒ a= b=c
157. As the coefficients are in reverse order, the roots of
ax + bx + c = 0 are α and β, while the roots of
2
1 1
163. x 2 + 2 bx + c > 0
So, the given equation has no roots.
⇒ D < 0 ⇒ 4b2 − 4 c < 0
5
Inequalities and Quadratic Equation
cx 2 + bx + a = 0 are and .
α β ⇒ b2 < c
One negative root is common. 164. f (0 ) = 10 > 0, as f ( x ) = 0 does not have distinct real
1
⇒ α= <0 roots.
α
⇒ f (5) ≥ 0
So, α = −1 ⇒ a− b+ c = 0
⇒ 25 a + 5 b + 10 ≥ 0 ⇒ 5 a + b ≥ − 2
158. Let α be the common root. 165. f (2 ) > 0 and f (4) > 0
Then, α 2 + iα + a = 0,
⇒ a2 < 4 and 13 a2 < 8
α 2 − 2α + ia = 0
−2 2 2 2
α2 α 1 ⇒ − 2 < a < 2 and <a<
⇒ = = 13 13
a a(1 − i ) − 2 − i
⇒ a2 (1 − i )2 = a (− 2 − i ) 2 2 2 2
⇒ − <a<
1 13 13
⇒ a= −i
2 166. Since, the coefficient of x 2 is 1 which is positive.
159. α + 2α + 3λ = 0 and 2α + 3 α + 5λ = 0
2 2 ∴The given expression is positive for all real values of x,
α 2
α 1 if D < 0.
⇒ = =
λ λ −1 ⇒ (− a)2 − 4 (1 − 2 a2 ) < 0
⇒ α = 1 and α = − λ ⇒ 9 a2 − 4 < 0
⇒ λ = −1
⇒ (3 a + 2 ) (3 a − 2 ) < 0
160. For the equation x 2 + 2 x + 3 = 0, 2 2
Discriminant = (2 )2 − 4 ⋅ 1 ⋅ 3 < 0 ⇒ − <a<
3 3
Targ e t E x e rc is e s
∴ Roots of x + 2 x + 3 = 0 are imaginary. Since, the
2
167. x 2 − 6 x + 5 ≤ 0 and x 2 − 2 x > 0
equations x 2 + 2 x + 3 = 0 and ax 2 + bx + c = 0 are
given to have a common root, therefore both roots will ⇒ 1≤ x ≤ 5
be common. Hence, both the equations are identical. and x < 0 or x > 2
⇒ a : b : c = 1: 2 : 3 ⇒ 2<x≤5
1 ⇒ x ∈{ 3, 4, 5}
161. Let α and β be the roots of x 2 − px + q = 0 and α and Hence, there are three solutions.
β
be the roots of x 2 − ax + b = 0. 168. Denominator = ( x − 1)2 + 5 > 0, ∀x
Then, α + β = p and αβ = q ⇒ 2 x 2 + 3 x − 27 = (2 x + 9)( x − 3) ≤ 0
1 α 9
Also, α + = a and =b which gives − ≤ x ≤ 3 leading to 0 ≤ 4 x 2 ≤ 81
β β 2
2 2
α 1 169. Given,|2 x − 3| < 1 ⇒ − 1 < 2 x − 3 < 1
Now, (q − b)2 = αβ − = α 2 β −
β β ⇒ 1< x < 2
2
α 1 ⇒ x ∈(1, 2 )
= ⋅ βα (α + β) − α + β
β 170. | x − 1| − 1 ≤ 1
= bq ( p − a)2
⇒ 0 ≤ | x − 1| ≤ 2
162. For x > 0, f ( x ) > 0, so there is no positive real root of ⇒ | x − 1| ≤ 2
f( x) = 0
⇒ − 1≤ x ≤ 3
Now, f(0 ) = 1 ≠ 0, so x = 0 is not a root x x
f ( x ) = 1 + 2 x + 3 x 2 + K + (n + 1)x n 171. + ≥ 1
…(i) 5 12
13 13
Multiplying Eq. (i) by x, we get
xf ( x ) = x + 2 x 2 + K + nx n + (n + 1)x n + 1 ∴ cos x α + sin x α ≥ 1
5
⇒ (1 − x )f ( x ) = (1 + x + x 2 + K + x n ) − (n + 1)x n + 1 where, cos α =
13
1 − xn + 1
= − (n + 1)x n + 1 and sin α =
12
1− x 13
1 − x n + 1 (n + 1)x n + 1
⇒ f( x) = − Equality holds for x = 2.
(1 − x )2 (1 − x )
If x < 2, both cos α and sin α increase (being positive
1 − (n + 2 )x n + 1 + (n + 1)x n + 2
= fractions).
(1 − x )2
So, cos x α + sin x α > 1
So, the equation f ( x ) > 0, ∀x.
Hence, no solution. If x < 2. Thus, x ≤ 2. 251
5 172. Here, xy ≥ 100
x+ y
2
≥ xy
179. Let the roots be α and β.
∴
Given,
α + β = − 2a and αβ = b
|α − β| ≤ 2m
Objective Mathematics Vol. 1
⇒ x + y ≥ 2 ⋅ 10 ⇒ |α − β|2 ≤ (2m)2
⇒ x + y ≥ 20
⇒ (α + β ) − 4ab ≤ 4m2
2
x2 − x + 1
173. Let 2 =m ⇒ 4a2 − 4b ≤ 4m2
x + x+1
⇒ a2 − m2 ≤ b
Discriminant ≥ 0 and discriminant D>0 or 4a2 − 4b > 0
⇒ (1 + m)2 − 4(1 − m)2 ≥ 0
⇒ a − m ≤ b and
2 2
b < a2
1
⇒ (m − 3) m − ≤ 0 Hence, b ∈ [a2 − m2 , a2 ]
3
1 180. Given, a < b < c < d . Let
⇒ m∈ ,3
3 f ( x ) = ( x − a)( x − c ) + 2( x − b)( x − d )
tan x tan x (1 − 3 tan 2 x ) ⇒ f (b) = (b − a)(b − c ) < 0
174. y = = and f (d ) = (d − a)(d − c ) > 0
tan 3 x 3 tan x − tan 3 x
1 Hence, f ( x ) = 0 has one root in (b, d ). Also, f (a)f (c ) < 0.
⇒ y< or y > 3 So, the other root lies in (a, c ). Hence, roots of the
3 equation are real and distinct.
1
⇒ −∞<y< or 3 < y < ∞
3 (a − 4) − (a − 4)2 − 4(4 − 2 a)
181. tan x =
175. The given inequation is valid only when 2
a− 4− a
x≥0 ...(i) = = a − 2, − 2
2
The given inequation can be written in the form
∴ tan x = a − 2 [Q tan x ≠ − 2]
372 − x − x > 1
π
Ta rg e t E x e rc is e s
⇒ 72 − x − x > 0 [Q 3 > 1] x ∈ 0,
4
Q
⇒ x + x − 72 < 0 ∴ 0 ≤ a−2 ≤1
⇒ ( x + 9) ( x − 8) < 0 ⇒ 2≤a≤3
But x + 9 > 0, ∀x ≥ 0
∴ x − 8< 0 ⇒ x <8 182. We know that, ax 2 + bx + c ≥ 0, ∀ x ∈ R
∴ 0 ≤ x < 64 If a > 0 and b2 − 4ac ≤ 0.
176. f(1) > 0, f(2 ) < 0, f(3) > 0 and D > 0 1
∴ mx − 1 + ≥ 0
x
⇒ − 12 + λ > 0,
mx 2 − x + 1
λ < 16, ⇒ ≥0
λ > 12 and 16 (16 − λ ) > 0 x
⇒ 12 < λ < 16 ⇒ mx 2 − x + 1 ≥ 0 as x > 0
∴ λ ∈{13, 14, 15} Now, mx 2 − x + 1 ≥ 0, if m > 0 and 1 − 4m ≤ 0
Hence, three integral solutions. 1
⇒ m > 0 and m ≥
4
177. k − 2 > 0 and 1
64 − 4 (k − 2 ) (k + 4) < 0 Thus, the minimum value of m is .
4
k > 2 and 16 − (k 2 + 2 k − 8) < 0
183. The equation on simplifying gives
k > 2 and k 2 + 2 k − 24 > 0
x ( x − b) ( x − c ) + x ( x − c ) ( x − a) + x ( x − a) ( x − b)
k > 2 and (k < − 6 or k > 4) − ( x − a) ( x − b) ( x − c ) = 0 ...(i)
⇒ k>4 Let f ( x ) = x ( x − b) ( x − c ) + x ( x − c ) ( x − a)
∴ Least integral value of k = 5 + x ( x − a) ( x − b) − ( x − a) ( x − b) ( x − c )
178. (e a − e 2 a + e a − 1) (4 e a − 2 e 2 a + e a − 1) < 0 We can assume without loss of generality that
a < b < c . Now,
⇒ (e 2 a − 2e a + 1) (2e 2 a − 5 e a + 1) < 0
f (a) = a (a − b) (a − c ) > 0
Let x = ea
f (b) = b (b − c ) (b − a) < 0
⇒ ( x − 1)2 (2 x 2 − 5 x + 1) < 0 f (c ) = c (c − a) (c − b) > 0
5 − 17 5 + 17 So, one root of Eq. (i) lies in (a, b) and one root in (b, c ).
⇒ ( x − 1)2 x − x − <0
4 4 Obviously the third root must also be real.
5 − 17 5 + 17 184. As a < b < c < d are in AP.
⇒ <x<
4 4 ∴f ( x ) = ( x − a)( x − c ) + 2( x − b)( x − d )
5 − 17 5 + 17 ⇒ f (a) = +ve, f (b) = − ve, f (c ) = − ve and f (d ) = + ve
⇒ log < a < log
252 4 4 ⇒ One root lies in (a, b) and other in (c , d ).
185. cos 4 x − cos 2 x + 1 − p = 0; 0 ≤ cos 2 x ≤ 1,
∴
The roots of y 2 − y + 1 − p = 0 lie between 0 and 1.
α ≥ 0, β ≥ 0
191. Since, f (1) = − f (2 ),
i.e. f (1)⋅ f (2 ) < 0,
Hence, the equation must have a root x ∈(1, 2 ).
5
Inequalities and Quadratic Equation
α − 1≤ 0 ∴Two distinct real roots.
β − 1≤ 0 192. 6 x 2 − xy − y 2 − 6 x + 8 y − 12 = 6 ( x + λy − 2 )( x − µy + 1)
and D≥0 ⇒ λ=
1
,µ =
1
⇒ α+β≥0 3 2
αβ ≥ 0, α + β − 2 ≤ 0, 193. The given equation can be written as
αβ − (α + β ) + 1 ≥ 0 y 2 + y(2 x + m) + (2 x − 3) = 0
and D≥0 Here, D = (2 x + m)2 − 4(2 x − 3)
∴ 1 ≥ 0, 1 − p ≥ 0, = 4 x + 4 xm − 8 x + m + 12
2 2
1 − 2 ≤ 0, 1 − p − 1 + 1 ≥ 0 = 4 x 2 + 4 x(m − 2 ) + (m2 + 12 )
and 1 − 4 (1 − p) ≥ 0 = [4 x 2 + 4 x(m − 2 ) + (m − 2 )2 ] + (m2 + 12 ) − (m − 2 )2
⇒
3
≤ p≤1 = [2 x + (m − 2 )]2 + [12 + 4m − 4]
4
For rational factors, D should be a perfect square.
186. Since, both roots of f ( x ) = 0 exceed a. Therefore, ∴ 8 + 4m = 0 ⇒ m = − 2
Discriminant > 0 …(i) Aliter
f (a) > 0 …(ii) Use abc + 2 fgh − af 2 − bg 2 − ch 2 = 0
1
− >a …(iii) ⇒ m= −2
2
On solving Eqs. (i), (ii) and (iii), we get 194. The given expression
a< −2 x 2 y
2
y x
= z 2 a + b + c + 2 a + 2 b
187. f (0 )⋅ f (1) > 0 z z z z
Targ e t E x e rc is e s
(− 1) (1 − a2 + 2 a − 1) > 0 x y
_ + 2c
⇒ a2 − 2 a > 0 + + z z
0 2
⇒ a(a − 2 ) > 0 = z 2 [aX 2 + bY 2 + c + 2 aY + 2 bX + 2cXY ] …(i)
⇒ a<0 x y
or a>2 where, = X and =Y
z z
188. Since, a lies between the roots of the given equation. The given expression can be resolved into rational
∴ 2 f (a) < 0 factors when the expression under the brackets given
⇒ f (a) < 0 in Eq. (i) can be resolved into rational factors.
Here, f ( x ) = 2 x 2 − 2(2 a + 1) x + a(a − 1) The condition for this is
abc + 2 ⋅ a ⋅ b ⋅ c − a ⋅ a2 − b ⋅ b2 − c ⋅ c 2 = 0
∴ f (a) = 2 a2 − 2(2 a + 1) a + a(a − 1) < 0
[Q here, f = a, g = b, h = c ]
⇒ −a − 3 a < 0
2
⇒ a3 + b3 + c 3 = 3abc
⇒ a (a + 3) > 0
⇒ a ∈ (− ∞, − 3) ∪ (0, ∞ ) 195. Given equation is
x 2 + 9y 2 − 4x + 3 = 0
189. Since, α < − 1and β > 1
or x 2 − 4x + 9y 2 + 3 = 0 ...(i)
∴ α + λ = − 1and β = 1 + µ, where λ , µ > 0
c b Since, x is real.
Now, 1 + + = 1 + αβ + |α + β | ∴ (− 4)2 − 4(9 y 2 + 3) ≥ 0
a a
= 1 + (− 1 − λ ) (1 + µ) + |− 1 − λ + 1 + µ| ⇒ 16 − 4 (9 y 2 + 3) ≥ 0
= 1 − 1 − µ − λ − λµ + |µ − λ| ⇒ 4 − 9y 2 − 3 ≥ 0
− µ − λ − λµ + µ − λ , if µ > λ ⇒ 9y 2 − 1 ≤ 0
=
− µ − λ − λµ + λ − µ, if λ > µ ⇒ (3 y − 1) (3 y + 1) ≤ 0
c b −1 1
∴ 1+ + = − 2 λ − λµ or − 2µ − λµ ⇒ ≤y≤
a a 3 3
c b Eq. (i) can also be written as
In both cases, 1 + + <0 [Q λ , µ > 0]
a a 9y 2 + 0 y + x 2 − 4x + 3 = 0
ax 3 bx 2 Since, y is real.
190. Let f ( x ) = + + cx + d
∴ 0 2 − 4 × 9 ( x 2 − 4 x + 3) ≥ 0
3 2
where, f (0 ) = f (1) = d ⇒ x 2 − 4x + 3 ≤ 0
∴f ′ ( x ) has atleast one root [0, 1] ⇒ ( x − 1) ( x − 3) ≤ 0
or ax 2 + bx + c has atleast one root [0, 1]. ⇒ 1≤ x ≤ 3 253
5 1
2
−15
Case I y = a + b ⇒ (a + b )x =a+ b
196. If mr , satisfy the given equation
m r ⇒ x − 15 = 1 ⇒ x = ± 4
2
x 2 + y 2 + 2 gx + 2 fy + c = 0, then
Objective Mathematics Vol. 1
1
Case II y = a − b ⇒ y =
1 2f a+ b
mr2 + 2 + 2 gmr + +c =0
mr mr ⇒ y = (a + b )−1
⇒ mr4 + 2 gmr3 + cmr2 + 2 fmr + 1 = 0 ⇒ (a + b )x
2
−15
= (a + b )− 1
Now, roots of given equation are m1, m2 , m3 and m4 . ⇒ x − 15 = − 1
2
The product of roots,
Constant term 1 ⇒ x = ± 14
m1m2 m3 m4 = = =1
4
Coefficient of mr 1 202. The roots of ax + bx + c = 0 are imaginary.
2
197. Let f ( x ) = 0, where f ( x ) = x 2 − 3 x + k has two roots α and ∴ b2 − 4ac < 0
β in (0, 1) for some value of k, then by Rolle’s theorem −b± b2 − 4ac b 4ac − b2
f ′ ( x ) = 2 x − 3 = 0 has a root in (0, 1)[Q α , β ∈ (0, 1]) but Roots are =− ± i
2a 2a 2a
3
f ′ ( x ) = 0 gives x = ∉ (0, 1), which is contradiction. b 4ac − b2
2 Let α=− + i
2a 2a
Hence, f ( x ) = 0 has no root in (0, 1) for any value of k.
b 4ac − b2
198. and β=− − i
Y y = ex 2a 2a
y=x ∴ | α | = |β| [Q α = β ]
b2 4ac − b2 c
X′ X Also, |α| =2
+ 2
= >1
O 4a 4a a
[Q 0 < a < b < c ]
|α| = | β | > 1
Y′
Ta rg e t E x e rc is e s
∴Option (c) is false.
∴ No solution.
203. We have, tan α + tan β = − p and tan α tan β = q
199.
y = |x|
Y tan α + tan β −p p
∴ tan (α + β ) = = =
1 − tan α tan β 1 − q q − 1
X′ X ∴Option (b) is true.
O
It can be easily checked that options (c) and (d) are
false. LHS of expression in option (a)
Y′ y = log0.5 x = cos 2 (α + β ) [tan 2 (α + β ) + p tan (α + β ) + q ]
∴Only one solution.
1 p 2 p
200. We have,(log 5 x )2 + log 5 x < 2 = + p + q
q − 1 q − 1
2
log 5 x = a, then a2 + a < 2 p
Put 1+
q − 1
⇒ a2 + a − 2 < 0
⇒ (a + 2 ) (a − 1) < 0 1
= [ p2 + p2 (q − 1)2 + q (q − 1)2 ] = q
⇒ − 2 < a < 1 or − 2 < log 5 x < 1 (q − 1)2 + p2
1 204. We have, (ax 2 + c ) y + (a′ x 2 + c ′ ) = 0
∴ 5− 2 < x < 5 i.e. <x<5
25 ∴ (ay + a′ ) x 2 + (cy + c ′ ) = 0
( a − b) ( a + b) a −b
2
1
201. a − b = = = Since, x is a rational function, the discriminant must be a
a+ b a+ b a+ b perfect square.
∴ Given equation reduces to Discriminant = 0 − 4 (ay + a′ ) (cy + c ′ )
2 1
(a + b )x −15 + 2
= 2a = − 4 [acy 2 + (ac ′ + a′ c )y + a′ c ′ ]
(a + b )x − 15 ∴ Roots of acy 2 + (ac ′ + a′ c )y + a′ c ′ = 0 must be
1
⇒ y + = 2a equal.
y
2
−15
∴ Discriminant = 0, i.e.(ac ′ + a′ c )2 − 4ac ⋅ a′ c ′ = 0
where, y = (a + b )x
⇒ (ac ′ − a′ c )2 = 0
⇒ y 2 − 2 ay + 1 = 0
⇒ ac ′ − a′ c = 0
2a ± 4a2 − 4 ⇒ ac ′ = a′ c
⇒ y= =a± a −1
2
2 a c a a′
⇒ = ⇒ =
⇒ y =a± b [Q a2 − b = 1] a′ c ′ c c′
254
From Eq. (ii), putting a = 1, we get
5
2 2
x x
205. We have, + = a (a − 1) 1/r + r + 1 = q …(v)
x + 1 x − 1
From Eqs. (iv) and (v), we have
Inequalities and Quadratic Equation
2
x x x x − p=q ⇒ p+ q = 0
⇒ + −2 = a (a − 1)
x + 1 x − 1 x + 1 x − 1 Hence, option (a) is correct.
2 Now, as r > 1
2 x2 2 x2 a/ r = 1 / r < 1
⇒ 2 − 2 = a (a − 1)
x − 1 x −1 and ar = r > 1
Hence, option (d) is correct.
2 x2
⇒ z 2 − z − a (a − 1) = 0, where z = 209. Since α, β, γ and δ are in HP. So, 1 / α, 1 / β, 1 / γ and 1 / δ
x2 − 1
⇒ z=a or a−1 are in AP and they may be taken as
2 x2 a − 3 d , a − d , a + d , a + 3 d . Replacing x by 1/ x, we get
z=a ⇒ =a the equation whose roots are 1 / α , 1 / β, 1 / γ, 1 / δ.
x2 − 1
Therefore, equation x 2 − 4 x + A = 0 has roots
⇒ 2 x 2 = ax 2 − a a − 3 d , a + d and equation x 2 − 6 x + B = 0 has roots
a a − d , a + 3 d . Sum of the roots is
⇒ x=±
a−2 2(a − d ) = 4 and 2(a + d ) = 6
2 x2 ∴ a = 5 / 2, d = 1 / 2
z = 1− a ⇒ 2 = 1− a
x −1 Product of the roots is
⇒ 2 x 2 = (1 − a) x 2 − 1 + a (a − 3 d ) (a + d ) = A = 3
a−1
⇒ x=± (a − d ) (a + 3 d ) = B = 8
a+1
210. (1 + k ) tan 2 x − 4 tan x − 1 + k = 0 ...(i)
a a−1
∴ x=± ,± Since, roots are real, we have
a−2 a+1
Targ e t E x e rc is e s
(− 4)2 − 4 (1 + k ) (− 1 + k ) ≥ 0
⇒ a < −1 ⇒ All roots are real ⇒ 16 − 4 (k 2 − 1) ≥ 0
a a−1
1< a < 2 ⇒ x = ± i, ± ⇒ k2 ≤ 5
2−a a+1
−4 4
⇒ Only two roots are real. We have, tan x1 + tan x2 = − =
1 + k 1 + k
⇒ a > 2 ⇒ All roots are real. − 1+ k
and tan x1 ⋅ tan x2 =
206. Symmetric functions are those which do not change by 1+ k
interchanging α and β. 4
207. sec 2 θ + cosec 2θ = sec 2 θ ⋅ cosec 2θ 1+ k 4
∴ tan ( x1 + x2 ) = = =2
Sum of the roots is equal to their product and the roots − 1 + k 2
1−
are real. 1+ k
b c
∴ − = ⇒ b+c =0 For k = 2, from Eq. (i),
a a
3 tan 2 x − 4 tan x + 1 = 0
Also, b2 − 4ac ≥ 0
1
⇒ c 2 − 4ac ≥ 0 ⇒ tan x = 1,
3
⇒ c (c − 4a) ≥ 0 π 1
∴ x1 =, x2 = tan − 1
⇒ c − 4a ≥ 0 [Qc > 0] 4 3
Further b2 + 4ab ≥ 0 For k = 1, from Eq. (i), 2 tan 2 x − 4 tan x = 0
⇒ b + 4a ≤ 0 [Q b < 0] ⇒ tan x = 0, 2 ⇒ x1 = 0, x2 = tan −1 2
208. Let the roots be a/r, a, ar, where a > 0, r > 1. 211. We have, u + v = − p and uv = q
Now, a/ r + a + ar = − p ...(i) 1 1 u+v −p 1 1 1 1
(a) + = = and ⋅ = =
a(a/ r ) + a(ar )(ar )(a/ r ) = q ...(ii) u v uv q u v uv q
(a/ r ) (a) (ar ) = 1 ...(iii) ∴ The required equation is
⇒ a3 = 1 ⇒ a = 1 p 1
x2 − − x + = 0
Hence, option (c) is correct. q q
From Eq. (i), putting a = 1, we get ⇒ qx 2 + px + 1 = 0
1 (b) (u + v ) + uv = − p + q and (u + v ) uv = − pq
− p− 3> 0 Qr + >2
r ∴ The required equation is
⇒ p< −3 x 2 − (− p + q ) x + (− pq ) = 0
Hence, option (b) is not correct. ⇒ x 2 + ( p − q ) x − pq = 0
Also, 1/ r + 1 + r = − p ...(iv) 255
⇒ ( x + p) ( x − q ) = 0
5 (c) u 2 + v 2 = (u + v )2 − 2uv = p2 − 2q
and u v = (uv ) = q
2 2
∴ The required equation is
2 2
216. cos x − y 2 − y − x 2 − 1 ≥ 0
Now, y − x 2 − 1 is defined when y − x 2 − 1 ≥ 0 or
...(i)
Objective Mathematics Vol. 1
x 2 − ( p2 − 2q ) x + q 2 = 0 y ≥ x + 1. So, minimum value of y is 1.
2
From Eq. (i),
u v u 2 + v 2 p2 − 2q u v
(d) + = = and ⋅ =1 cos x − y 2 ≥ y − x2 − 1
v u uv q v u
∴ The required equation is where, cos x − y 2 ≤ 0 [as when cos x is maximum (= 1)
p2 − 2q and y 2 is minimum (= 1,) so cos x − y 2 is maximum].
x2 − x + 1= 0
q Also, y − x2 − 1 ≥ 0
⇒ qx 2 − ( p2 − 2q ) x + q = 0 Hence, cos x − y 2 = y − x2 − 1 = 0
( x − a) ( x − b) ⇒ y = 1 and cos x = 1, y = x 2 + 1
212. =k
x −c ⇒ x = 0, y = 1
⇒ ( x − a) ( x − b) − k ( x − c ) = 0 217. Y
⇒ x 2 − (a + b + k ) x + ab + kc = 0
⇒ { − (a + b + k )} 2 − 4 ⋅ 1⋅ (ab + kc ) ≥ 0 [Q x is real]
⇒ k 2 + 2 (a + b − 2c ) + (a − b)2 ≥ 0
X
This is true for all real k.
∴ Discriminant = 4 (a + b − 2c )2 − 4 ⋅ 1 (a − b)2 ≤ 0
From the graph,
⇒ (a + b − 2c + a − b) (a + b − 2c − a + b) ≤ 0 f (0 ) = c > 0 ...(i)
⇒ 2 (a − c ) 2 (b − c ) ≤ 0 Also, the graph is concave downward.
⇒ (a − c ) (b − c ) ≤ 0 ∴ a<0 ...(ii)
b
Further, abscissa of the vertex =
Ta rg e t E x e rc is e s
⇒ (c − a) (c − b) ≤ 0 ...(iii)
2a
⇒ a ≤ c ≤ b or b≤c ≤ a From Eqs. (i), (ii) and (iii), we get
213. Since, P( x ) divides both of them. So, P( x ) also divides ac < 0, ab < 0 and bc > 0
(3 x 4 + 4 x 2 + 28 x + 5) − 3 ( x 4 + 6 x 2 + 25) 218. For ax 2 + bx + c = 0 and a1 x 2 + b1 x + c1 = 0 have a
= − 14 x 2 + 28 x − 70 common root
(a1c − ac1 )2 = 4 (ab1 − ba1) (bc1 − b 1c ) ...(i)
= − 14( x 2 − 2 x + 5)
a b c
which is a quadratic. If , , are in AP.
a1 b 1 c 1
Hence, P( x ) = x 2 − 2 x + 5
ba1 − ab 1 cb 1 − c 1 b
⇒ P(1) = 4 = =k
a1b 1 b 1c 1
214. Roots of 4 x 2 − x − 1 = 0 are irrational. So, one root cb 1 − ac1
common implies both roots are common. and = 2k
a1c1
4 −1 −1 −3
∴ = = ⇒ λ= ,µ = 0 On putting these values in Eq. (i), we have
3 λ + µ λ −µ 4
c1a1 = b 12
215. 2 x 2 + 6 xy + 5 y 2 = 1 ...(i)
219. Statement I Given equation is x 2 − bx + c = 0
Eq. (i) can be rewritten as
Let α and β be two roots such that
2 x 2 + (6 y )x + 5 y 2 − 1 = 0 lα − βl = 1
Since, x is real. ⇒ (α + β)2 − 4αβ = 1
36 y 2 − 8 (5 y 2 − 1) ≥ 0 ⇒ b2 − 4 c = 1
⇒ y2 ≤ 2 Statement II Given equation is
⇒ − 2≤y≤ 2 4abc x 2 + (b2 − 4ac ) x − b = 0.
Eq. (i) can also be rewritten as ∴ D = (b2 − 4ac )2 + 16 ab2c
5 y + (6 x )y + 2 x − 1 = 0
2 2
= (b2 + 4ac )2 > 0
Since, y is real. Hence, roots are real and unequal.
∴ 36 x 2 − 20 (2 x 2 − 1) ≥ 0
Solutions (Q.Nos. 220-222)
⇒ 36 x 2 − 40 x 2 + 20 ≥ 0 Let the other roots of f { f ( x )} = x be λ and δ.
⇒ − 4 x 2 ≥ − 20 ⇒ f { f (λ )} = λ
⇒ x2 ≤ 5 Let f(λ ) = γ
256 ⇒ − 5≤x≤ 5 ⇒ f (γ ) = γ
Other roots γ and δ lie on the line
y=−x+c
There must be two points C and D on the parabola
229. If a = 2, then
b−c =1 5
Inequalities and Quadratic Equation
y = ax 2 + bx + c which are image of each other in the 230. Investigating the nature of the cubic equation of a
line y = x let f (a) = a3 − 4ab + 8 c
Y
B f ′ (a) = 3a2 − 4b
(β, β) If b < 0, then f ′ (a) > 0
A
(α, α) ∴The equation a3 − 4ab + 8 c = 0 has only one real root.
X
O 231. On substituting c = 1 − b in Eq. (i), we get
(a + 2 ) [(a − 1)2 + 3 − 4b] = 0
⇒ If α and β are real so are λ and δ. ⇒ 4b − 3 > 0
If α + β = λ + δ ⇒ middle points of AB and CD become 3
⇒ b>
same. 4
This is not possible ⇒ α, β and λ , δ cannot be real. ⇒ c<
1
Also, if α and β are equal to then λ , δ cannot be real. 4
223. f ′′ (c 2 ) f ′′ (c1 ) < 0 and f ′ (c1 ) = f ′ (c 2 ) = 0 232. A. Roots of the given equation will be opposite sign, if
⇒ f ′′ (c1 ) − f ′′(c 2 ) > 0 D ≥ 0 and product of roots < 0.
⇒ f ′′ (c1 ) > 0 (a3 + 8a − 1)2 − 8 (a2 − 4a) ≥ 0
and f ′′ (c 2 ) > 0 a2 − 4a
and <0
⇒ c 2 is local maximum and c1 is local minimum for f ( x ) 2
⇒ a2 − 4a < 0
⇒ f ′ ( x ) = 0 has atleast four times in [c1 − 1, c 2 + 1].
Q a2 − 4a < 0
224. Here, c1 is local maximum and c 2 is local minimum.
⇒ (a + 8 a − 1) − 8 (a − 4a) ≥ 0
3 2 2
⇒ f ′ ( x ) = 0 has atleast two roots in [c1 − 1, c 2 + 1].
Targ e t E x e rc is e s
⇒ 0<a<4
225. As c 2 is local maximum and c1 is local minimum.
B. Equation has negative roots of discriminant ≥ 0,
⇒ f ( x ) = 0 has atleast two solutions. α < 0, β < 0
Solutions (Q. Nos. 226-228) i.e. α + β < 0 and f(0 ) > 0
(β − α) = ((β + δ) − (α + δ)) ∴ 4(a + 1)2 − 36 a + 20 > 0
(β + α )2 − 4αβ = [(β + δ ) + (α + δ)]2 − 4 (β + δ) (α + δ)
⇒ a2 − 7 a + 6 ≥ 0
(− b1 ) − 4 c1 = ( − b2 ) − 4 c 2
2 2
⇒ a ≤ 1 or a≥6 ...(i)
D1 = D2 α+β<0
Least value of f ( x ) is ⇒ a + 1> 0
D 1 ⇒ a> −1
− 1 =− ⇒ D1 = 1 ⇒ D2 = 1 ...(ii)
4 4 and f(0 ) > 0
∴Least value of g( x ) is 9a − 5 > 0
D 1
− 2 =− ⇒ a>
5
...(iii)
4 4 9
Least value of g( x ) accurs at
b 7 From Eqs. (i), (ii) and (iii), we get
x=− 2 = ⇒ b2 = − 7 a≥6
2 2
b22 − 4 c 2 = D2 ⇒ 49 − 4 c 2 = 1 C. Given, x 2 − 2(4a − 1)x + 15 a2 − 2 a − 7 > 0
48 ∴ Discriminant < 0 [Q coefficient of x 2 > 0]
⇒ = c 2 ⇒ c 2 = 12
4 ⇒ 4 (4a − 1) − 4 (15 a − 2 a − 7 ) < 0
2 2
x 2 − 7 x + 12 = 0 ⇒ x = 3, 4 ⇒ a2 − 6 a + 8 < 0
Solutions (Q. Nos. 229-231) ⇒ 2<a<4
Let x + ax + bx + cx + d
4 3 2
x + 2x + a
2
D. Let y =
= ( x − x1 ) ( x − x2 ) ( x − x3 ) ( x − x4 ) x 2 + 4 x + 3a
Let ( x − x1 ) ( x − x2 ) = x 2 + px + q ⇒ x 2 ( y − 1) + 2 (2 y − 1)x + a (3 y − 1) = 0
and ( x − x3 ) ( x − x4 ) = x 2 + px + r
As x ∈ R, D ≥ 0
∴ q = x1 x2 and r = x3 x4 ∴ 4(2 y − 1)2 − 4( y − 1) a(3 y − 1) ≥ 0
x 4 + ax 3 + bx 2 + cx + d 4
= x 4 + 2 px 3 + ( p2 + q + r )x 2 + p(q + r )x + qr ⇒ (4 − 3 a) y 2 − 4(1 − a) y + 1 − a ≥ 0 ⇒ a <
3
∴ a = 2 p; b = p2 + q + r ; c = p(q + r ); d = qr and a (a − 1) ≤ 0
Clearly, a3 − 4ab + 8 c = 0 ...(i) ∴ 0 ≤ a≤1 257
5 233. A. Since, f ( x ) = x 2 + 2(k + 1)x + (9k − 5)
negative roots.
has
(i) Discriminant ≥ 0 (ii) α < 0, β < 0 (iii) f(0 ) > 0
both ⇒
and
1
8 k2 + k + ≥ 0
2
−k −k<0
2
Objective Mathematics Vol. 1
i.e. α+β<0 This is always true and k > 0 or k < − 1.
∴Discriminant ≥ 0 E. y = − x 2 + 3 x − 2
⇒ 4(k + 1)2 − 36k + 20 > 0 9 9
= − x 2 − 3x + − 2
⇒ k 2 − 7k + 6 ≥ 0 4 4
⇒ (k − 1) (k − 6) ≥ 0 1 x − 3 1
y= − ≤
⇒ k ≤ 1 or k ≥ 6 …(i) 4 2 4
α+β<0 234. Let f ( x ) = x 3 − x 2 + βx + γ ...(i)
⇒ − 2 (k + 1) < 0
⇒ k + 1> 0 f ( x ) = 0 has three positive real roots in GP.
⇒ k > −1 ...(ii) ⇒ f ′ ( x ) = 0 will have two distinct real roots.
and f(0 ) > 0 ⇒ 3 x 2 − 2 x + β = 0 has two distinct roots.
⇒ 9k − 5 > 0
5 ∴ D>0
⇒ k> …(iii) 1
9 4 − 12β > 0 ⇒ β < ...(ii)
From Eqs. (i), (ii) and (iii), we get 3
k≥6 Also, from Eq. (i),
B. f ( x ) > 0 x1 x2 + x2 x3 + x3 x1 = β
⇒ Discriminant < 0 x1 + x2 + x3 = 1
⇒ 4(4k − 1)2 − 4(15k 2 − 2 k − 7 ) < 0 x1 x2 x3 = − γ
⇒ k 2 − 6k + 8 < 0 ⇒ x23 = − r > 0 [Q x22 = x1 x3 ]
⇒ 2<k<4 ⇒ r >0
x2 ( x1 + x3 ) + x22 = β
Ta rg e t E x e rc is e s
C. f ( x ) = x 2 − 2 (k − 1) + (2 k + 1) has both roots Also,
positive, if ⇒ x2 (1 − x2 ) + x22 = β
(i) Discriminant > 0 ⇒ x2 = β > 0
(ii) Sum of the roots > 0 1
From Eqs. (ii) and (iv), we get 0 < β < and γ < 0
(iii) f(0 ) > 0 3
[ β ]max = 0, [γ ]max = − 1
⇒ [ β ] + [γ ] + 2 = [0 ] + [− 1] + 2 = 1
235. x1 + x2 + x1 x2 = a ...(i)
X
α β
1 2 x1 x2 + x12 x2 + x1 x22 = b ...(ii)
x12 x22 = c ...(iii)
Now, discriminant ≥ 0
From Eq. (ii), we get
⇒ 4 (k − 1)2 − 4 (2 k + 1) ≥ 0
x1 x2 (1 + x1 + x2 ) = b
⇒ k 2 − 4k ≥ 0
x1 x2 (1 + a − x1 x2 ) = b
⇒ k ≤ 0 and k ≥ 4 ...(i)
x1 x2 (1 + a) − c = b
Sum of roots > 0 b+c
⇒ 2(k − 1) > 0 x1 x2 = , which is clearly a rational number.
a+1
⇒ k >1 …(ii)
Q b + c = 2 (a + 1)
and f(0 ) > 0
∴ x1 x2 = 2
⇒ (2 k + 1) > 0
1 α + β = − p
⇒ k>− …(iii) 236. Here, ⇒α + β = γ + δ
2 γ + δ = − p
From Eqs. (i), (ii) and (iii), we get Now, (α − γ ) (α − δ ) = α 2 − α (γ + δ ) + γδ
k≥4
= α 2 − α (α + β ) + r
D. Since, k lies between the roots of
f ( x ) = 2 x 2 − 2(2 k + 1)x + k(k + 1) = − αβ + r
= − (− q ) + r = q + r
⇒ Discriminant > 0 and f (k ) < 0 By symmetry of the results
⇒ 4 (2 k + 1)2 − 8k(k + 1) > 0 (β − γ )(β − δ ) = q + r
and 2 k 2 − 2 k(2 k + 1) + k(k + 1) < 0 Hence, the required ratio is 1.
258
Entrances Gallery
1. Let y = x 5 − 5 x and g( x ) = − a. Then, it is clear from the 1 1 1
5
Inequalities and Quadratic Equation
graph that f ( x )has only real roots, if a > 4 or a < − 4 and
6. Let 4 − 4− 4− ... = y
3 2 3 2 3 2
f ( x ) has three real roots, if − 4 < a < 4. 1
(–1,4) So, 4− y = y2 [Q y > 0 ]
g(x) = 4 3 2
1 8
⇒ y2 + y − 4= 0 ⇒ y =
–1 1 3 2 3 2
1 8
(1,– 4) g(x) = – 4 So, the required value is 6 + log 3 / 2 ×
3 2 3 2
4
2. P( x ) = ax 2 + b with a, b of same sign. = 6 + log 3 / 2 = 6 − 2 = 4
9
P(P( x )) = a (ax 2 + b)2 + b
If x ∈ R or ix ∈ R
7. (2 x )In 2 = (3 y )In 3 …(i)
⇒ x ∈R
2 3 In x = 2 In y …(ii)
⇒ P( x ) ∈ R ⇒ (log x ) (log 3) = (log y ) log 2
⇒ P { P( x )} ≠ 0 (log x ) (log 3)
⇒ log y = …(iii)
log 2
Hence, real or purely imaginary number cannot satisfy
P(P( x )) = 0. In Eq. (i), taking log on both sides, we get
(log 2 ) {log 2 + log x} = log 3 {log 3 + log y}
3. Let f ( x ) = x − x sin x − cos x ⇒ f ′ ( x ) = 2 x − x cos x
2
(log 3)2 (log x )
lim f ( x ) → ∞ ⇒ (log 2 )2 + (log 2 ) (log x ) = (log 3)2 +
x→ ∞ log 2
lim f ( x ) → ∞ [from Eq. (iii)]
x → −∞ (log 3)2 − (log 2 )2
f(0 ) = − 1 ⇒ (log 2 )2 − (log 3)2 = (log x )
Targ e t E x e rc is e s
log 2
⇒ − log 2 = log x
1 1
⇒ x= ⇒ x0 =
(0,–1) 2 2
8. an = α n − β n
Hence, 2 solutions.
α 2 − 6α − 2 = 0
4. log 2 3 x = ( x − 1) log 2 4 = 2( x − 1)
Multiply with α 8 on both sides, we get
⇒ x log 2 3 = 2 x − 2
2 α 10 − 6 α 9 − 2α 8 = 0 …(i)
⇒ x=
2 − log 2 3 Similarly, β10 − 6 β 9 − 2β 8 = 0 ...(ii)
After rearranging, we get From Eqs. (i) and (ii), we get
x=
2 α 10 − β10 − 6(α 9 − β 9 ) = 2 (α 8 − β 8 )
1 a − 2 a8
2− ⇒ a10 − 6 a9 = 2 a8 ⇒ 10 =3
log 3 2 2 a9
2 log 3 2
= 9. x 2 + bx − 1 = 0
2 log 3 2 − 1
Rearranging, again and x2 + x + b = 0 …(i)
1 Common root is
b+1
x=
log 3 4
=
log 4 3
=
1 (b − 1) x − 1 − b = 0 ⇒ x=
log 3 4 − 1 1 1 − log b−1
−1 4 3
log 4 3 This value of x satisfies Eq. (i).
5. Let 1 + a = y (b + 1)2 b + 1
∴ + + b=0
(b − 1)2 b−1
⇒ ( y1/ 3 − 1) x 2 + ( y1/ 2 − 1) x + y1/ 6 − 1 = 0
⇒ b = 3 i, − 3 i, 0
y1/ 3 − 1 2 y1/ 2 − 1 y1/ 6 − 1
⇒ x + x+ =0
10. Since,f − ⋅ f − < 0, so lies in − , − .
1 3 3 1
y −1 y −1 y −1
2 4 4 2
Taking lim on both the sides, we get
y →1 11. α 3 + β 3 = q
1 2 1 1
x + x+ =0 ⇒ (α + β)3 − 3αβ (α + β) = q
3 2 6
⇒ − p3 + 3 pαβ = q
⇒ 2 x 2 + 3x + 1 = 0
q + p3
1 ⇒ αβ =
∴ x = − 1, − 3p
2 259
5 ⇒
⇒
α β
x2 − + x + ⋅ = 0
β α
x −
(α 2 + β 2 )
α β
β α
x + 1= 0
14. Given equations are
and
x2 + 2 x + 3 = 0
ax 2 + bx + c = 0
…(i)
…(ii)
Objective Mathematics Vol. 1
2
αβ Since, Eq. (i) has imaginary roots.
(α + β)2 − 2αβ So, Eq. (ii) will also have both roots same as Eq. (i).
⇒ x2 − x + 1= 0
αβ Thus,
a b c
= =
2 p3 + q 1 2 3
p −2 Hence, a : b : c is 1 : 2 : 3
3p
⇒ x2 − x + 1= 0 15. Given equation is
p3 + q
e sin x − e − sin x = 4
3p 1
⇒ e sin x − sin x = 4
⇒ ( p + q ) x − (3 p − 2 p − 2q ) x + ( p3 + q ) = 0
3 2 3 3
e
⇒ ( p3 + q ) x 2 − ( p3 − 2q )x + ( p3 + q ) = 0 Now, let y = e sin x
1
12. Given, α and β are the roots of the equation ∴ y− =4
y
x − 6 x − 2 = 0.
2
⇒ y 2 − 4y − 1 = 0
Q an = α n − β n for n ≥ 1
∴ a10 = α 10 − β10 4 ± 16 + 4
⇒ y=
a8 = α − β 8 ⇒ a9 = α 9 − β 9
8 2
⇒ y =2 ± 5
Now, consider
a10 − 2 a8 α 10 − β10 − 2(α 8 − β 8 ) On substituting the value of y, we get
= e sin x = 2 ± 5
2 a9 2(α 9 − β 9 )
α 8 (α 2 − 2 ) − β 8 (β 2 − 2 ) Now, sine is a bounded function,
= i.e. − 1 ≤ sin x ≤ 1.
2(α 9 − β 9 )
Ta rg e t E x e rc is e s
∴ e − 1 ≤ e sin x ≤ e
Q α and β are the roots of
2 1
⇒ e sin x ∈ , e
x − 6x − 2 = 0 e
So, x 2 = 6x + 2
α 8 ⋅ 6 α − β8 ⋅ 6 β 16. Let z = x + iy, given Re( z ) = 1
= ⇒ α = 6α + 2
2
∴ x =1
2(α − β )
9 9
⇒ α − 2 = 6α ⇒ z = 1 + iy
2
and β2 = 6 β = 2 Since, the complex roots are conjugate of each other.
∴ z = 1 + iy and 1 − iy are two roots of z 2 + α z + β = 0
⇒ β 2 − 2 = 6 β
Product of roots = β
6 α 9 − 6 β9 6 ⇒ (1 + iy ) (1 − iy ) = β
= = =3
2(α 9 − β 9 ) 2 ∴ β = 1 + y2 ≥ 1
Aliter ⇒ β ∈ (1, ∞ )
Since, α and β are the roots of the equation
17. Let the quadratic equation be
x 2 − 6x − 2 = 0
ax 2 + bx + c = 0
⇒ x = 6x + 2 ⇒ α 2 = 6α + 2
2
⇒ α 10 = 6 α 9 + 2α 8 …(i) Sachin made a mistake in writing down constant terms.
∴Sum of roots is correct.
Similarly, β10 = 6 β 9 + 2 β 8 …(ii)
i.e. α + β =7
On subtracting Eq. (ii) from Eq. (i), we get
Rahul made mistake in writing down coefficient of x.
α 10 − β10 = 6(α 9 − β 9 ) + 2(α 8 − β 8 ) ∴Product of roots is correct.
⇒ a10 = 6 a9 + 2 a8 [Q an = α n − β n ] i.e. αβ = 6
a − 2 a8 ⇒ Correct quadratic equation is
⇒ a10 − 2 a8 = 6a9 ⇒ 10 =3
2 a9 x 2 − (α + β) x + αβ = 0
13. Q a2 = 3 t 2 − 2 t ⇒ x 2 − 7 x + 6 = 0 having roots 1 and 6.
18. Since, α and β are roots of the equation x 2 − x + 1 = 0.
1 ⇒ α + β = 1, αβ = 1
1 ± 3i
(0,0) (2/3,0) ⇒ x=
2
For non-integral solution, 1 + 3i 1 − 3i
0 < a2 < 1 ⇒ a ∈ (− 1, 0 ) ∪ (0, 1) ⇒ x= or
2 2
260 ⇒ x = − ω or − ω 2
Ø It is assumed that a real solution of given equation exists.
Thus, α = − ω 2 , then β = − ω
or
Hence, α
α = − ω,
2009
+β 2009
then β = − ω
= (− ω ) 2009
2
[where, ω = 1]
+ (− ω )
2 2009
3
23. Since, both roots of equation x 2 − 2 mx + m2 − 1 = 0 are
∴
greater than − 2 but less than 4.
D ≥ 0, − 2 < −
b
< 4,
5
Inequalities and Quadratic Equation
2a
= − [(ω 3 )669 ⋅ ω 2 + (ω 3 )1339 ⋅ ω ]
f(4) > 0 and f(− 2 ) > 0
= − [ω 2 + ω] = − (− 1) = 1 Now, D ≥ 0 ; 4m2 − 4m2 + 4 ≥ 0
19. Given bx 2 + cx + a = 0 has imaginary roots ⇒ 4 > 0, ∀ m ∈ R …(i)
⇒ c 2 − 4ab < 0 ⇒ c 2 < 4ab b
−2 < − <4
⇒ − c 2 > − 4ab …(i) 2a
2 m
Let f ( x ) = 3b2 x 2 + 6bcx + 2c 2 ⇒ −2 < <4
2 ⋅ 1
Here, 3b > 02
⇒ −2 < m< 4 …(ii)
So, the given expression has a minimum value. f(4) > 0
−D ⇒ 16 − 8m + m2 − 1 > 0
∴Minimum value =
4a
⇒ m2 − 8m + 15 > 0
4ac − b2 4 (3b2 ) (2c 2 ) − 36b2c 2
= = ⇒ (m − 3) (m − 5) > 0
4a 4(3b2 )
2 2
⇒ − ∞ < m < 3 and 5 < m < ∞ …(iii)
12 b c f(− 2 ) > 0
=− = − c 2 > − 4 ab [from Eq. (i)] and
12 b2 ⇒ 4 + 4m + m2 − 1 > 0
20. Let the roots of x 2 − 6 x + a = 0 be α, 4 β and that of ⇒ m2 + 4m + 3 > 0
x 2 − cx + 6 = 0 be α, 3 β. ⇒ (m + 3) (m + 1) > 0
∴ α + 4β = 6 and 4α β = a ⇒ − ∞ < m < − 3 and − 1 < m < ∞ …(iv)
and α + 3β = c and 3 αβ = 6 From Eqs. (i), (ii), (iii) and (iv), we get m lies between − 1
a 4 and 3.
⇒ = ⇒ a=8
Targ e t E x e rc is e s
6 3 24. Let α and β be the roots of equation
∴ x − 6x +
2
8= 0
x 2 − (a − 2 )x − a − 1 = 0
⇒ ( x − 4) ( x − 2 ) = 0
Then, α + β = a − 2 and αβ = − a − 1
⇒ x = 2, 4
Now, α 2 + β 2 = (α + β)2 − 2αβ
and x 2 − cx + 6 = 0
⇒ α 2 + β 2 = (a − 2 )2 + 2(a + 1)
If x = 2, then 2 2 − 2c + 6 = 0 ⇒ c = 5
∴ x 2 − 5x + 6 = 0 ⇒ α 2 + β 2 = a2 − 2 a + 6
⇒ x = 2, 3 ⇒ α 2 + β 2 = (a − 1)2 + 5
Hence, the common root is 2. The value of α 2 + β 2 will be least, if a − 1 = 0
21. Letα andβ be the roots of equation x 2 + ax + 1 = 0, then ⇒ a=1
α + β = − a and αβ = 1. Aliter
Now, α − β = (α + β)2 − 4αβ Since, α + β = (a − 2 ) and αβ = − a − 1
Let f (a) = α 2 + β 2
⇒ α − β = a2 − 4
= (α + β)2 − 2αβ
According to given condition,
= (a − 2 )2 + 2(a + 1)
a2 − 4 < 5
= a2 − 2 a + 6
⇒ a2 − 4 < 5
⇒ f ′ (a) = 2 a − 2
⇒ a2 < 9 For maxima or minima, put f ′ (a) = 0.
⇒ a <3 ∴ 2a − 2 = 0 ⇒ a = 1
⇒ a ∈ (− 3, 3) Now, f ″ (a) = 2 ⇒ f ″ (1) = 2 > 0
22. Since, tan 30° and tan 15° are the roots of equation ∴ f (a) is minimum at a = 1.
x 2 + px + q = 0. 25. Let n and (n + 1) be the two consecutive roots of
∴ tan 30 ° + tan 15° = − p x 2 − bx + c = 0. Then, n + (n + 1) = b and n (n + 1) = c
and tan 30 ° tan 15° = q ∴ b2 − 4 c = (2 n + 1)2 − 4 n (n + 1)
Now, 2 + q − p = 2 + tan 30 ° tan 15° = 4 n2 + 4 n + 1 − 4 n2 − 4 n = 1
+ (tan 30 ° + tan 15° )
26. Let f ( x ) = x 2 − 2 kx + k 2 + k − 5
= 2 + tan 30 ° tan 15° + 1 − tan 30 ° tan 15°
Since, both roots are less than 5.
tan 30 ° + tan 15° b
Q tan 45° = 1 − tan 30 ° tan 15° Then, D ≥ 0, − <5
2a
⇒ 2+q − p=3 and f(5) > 0
261
5 Now,
⇒
D = 4k 2 − 4(k 2 + k − 5)
= − 4k + 20 ≥ 0
k≤5 …(i)
Then,
and
z1 + z2 = − a
z1 z2 = b
On putting these values in Eq. (i), we get
Objective Mathematics Vol. 1
−
b
<5 ⇒ k<5 …(ii) (− a)2 = 3 b
2a ⇒ a2 = 3b
and f(5) > 0
32. Given equation is ax 2 + bx + c = 0
⇒ 25 − 10 k + k 2 + k − 5 > 0
Let α and β be the roots of the equation.
⇒ k 2 − 9k + 20 > 0 b c
⇒ (k − 5) (k − 4) > 0 Then, α+β=− and αβ =
a a
⇒ k < 4 and k > 5 …(iii) 1 1 α 2 + β2
Also given, α + β = 2 + 2 =
From Eqs. (i), (ii) and (iii), we get α β α 2β 2
k<4 2
α + β 2
27. Let f ( x ) = an x + an − 1 x
n n −1
+ ... + a1 x = 0 ... (i) ⇒ α+β= −
αβ αβ
f (0 ) = 0 b − b / a
2
2
and f (α) = 0 ⇒ − = −
a c/a c/a
According to the Rolle’s theorem, 2
b b 2a
f′ ( x) = 0 ⇒ = −
−
a c c
has atleast one root between (0, α ).
2a b b c
⇒ f ′ ( x ) = 0 has a positive root less than α. ⇒ = +
c c c a
28. Since, (1 − p) is a root of quadratic equation 2a b c
x 2 + px + (1 − p) = 0 …(i) ⇒ = +
b c a
So, (1 − p) satisfied the above equation. ⇒
c a
, and
b
are in AP.
∴ (1 − p)2 + p (1 − p) + (1 − p) = 0 a b c
Ta rg e t E x e rc is e s
⇒ (1 − p) (1 − p + p + 1) = 0 a b c
⇒ , and are in HP.
⇒ (1 − p) 2 = 0 c a b
⇒ p=1 33. Given equation is x 2 − 3 x + 2 = 0
On putting the value of p in Eq. (i), we get x 2 + x = 0 Case I When x > 0, then x = x
⇒ x = 0, − 1 ∴ x 2 − 3x + 2 = 0
29. Since, one of the roots of equation x 2 + px + 12 = 0 is 4. ⇒ ( x − 1) ( x − 2 ) = 0
∴ 16 + 4 p + 12 = 0 ⇒ x = 1, 2
⇒ 4 p = − 28 Case II When x < 0, then x = − x
⇒ p= −7 ∴ x 2 + 3x + 2 = 0
So, the other equation is x 2 − px + q = 0 whose roots ⇒ ( x + 1) ( x + 2 ) = 0
are equal. Let the roots be α and α. ⇒ x = − 1, − 2
7 Hence, four solutions are possible.
∴ Sum of roots = α + α =
1 34. Since, one root of the quadratic equation
7 (a2 − 5a + 3)x 2 + (3a − 1) x + 2 = 0 is twice as large as
⇒ α=
2 the other, then let their roots be α and 2α.
and product of roots = α ⋅ α = q (3 a − 1)
2 ∴ α + 2α = − 2
7 49 (a − 5 a + 3)
⇒ =q ⇒ q =
2 4 (3 a − 1)
⇒ 3α = − 2
30. Let α and β be two numbers whose arithmetic mean is 9 (a − 5 a + 3)
and geometric mean is 4. 2
and α ⋅ 2α = 2
∴ α + β = 18 and αβ = 16 (a − 5 a + 3)
∴Required equation is 2
⇒ 2α = 2
2
x 2 − (α + β) x + (αβ) = 0 (a − 5 a + 3)
⇒ x 2 − 18 x + 16 = 0 (3 a − 1)2 1
⇒ = 2
31. Since, origin z1 and z2 are the vertices of an equilateral 9 (a − 5 a + 3)
2 2
(a − 5 a + 3)
triangle, then ⇒ (3 a − 1)2 = 9 (a2 − 5 a + 3)
z12 + z22 = z1 z2 ⇒ 9 a2 − 6 a + 1 = 9 a2 − 45 a + 27
⇒ ( z1 + z2 )2 = 3 z1 z2 …(i) ⇒ 45 a − 6 a = 27 − 1
26 2
Again, z1 and z2 are the roots of the equation ⇒ a= =
262 z 2 + az + b = 0 39 3
35. Since, α 2 = 5 α − 3
⇒
⇒
α 2 − 5 α + 3 = 0 and β 2 = 5 β − 3
β − 5β + 3 = 0
2
39. Let α and β be the roots of the equation
x 2 − (a − 2 ) x − a + 1 = 0
∴ α + β = a−2
5
Inequalities and Quadratic Equation
and αβ = − (a − 1)
These two equations show that α and β are the roots of
the equation ⇒ s = α 2 + β 2 = (α + β )2 − 2αβ
x 2 − 5x + 3 = 0 = (a − 2 )2 + 2 (a − 1)
∴ α + β = 5 and αβ = 3 = a2 − 4a + 4 + 2 a − 2
α β α 2 + β2 = a2 − 2 a + 2
Now, + =
β α αβ ds
Now, = 2a − 2
(α + β)2 − 2αβ da
=
αβ For maximum and minimum, put ds / da = 0
25 − 6 19 ⇒ 2a − 2 = 0 ⇒ a = 1
= = d 2s
3 3 Also, =2 > 0
α β da2
and ⋅ =1
β α Hence, at a = 1, s will have minimum value.
α β
Thus, the equation having roots and is given by 40. Let two consecutive odd integers be(α + 1)and(α + 3).
β α
∴Sum of roots, α + 1 + α + 3 = + a
α β α β a−4
x2 − + x + ⋅ = 0 ⇒ 2α = a − 4 ⇒ α =
β α β α 2
19
⇒ x −
2
x + 1= 0 and product of roots, (α + 1) (α + 3) = b
3 ⇒ α 2 + 4α + 3 = b
⇒ 3 x 2 − 19 x + 3 = 0 2
a − 4 a − 4
2 x2 − 7x + 7 ⇒ + 4 + 3= b
36. Given, 3 =3 2
2 2
Targ e t E x e rc is e s
⇒ 2 x2 − 7 x + 7 = 2 a2 + 16 − 8 a 4 a − 16
⇒ 2 x2 − 7 x + 5 = 0 ⇒ + + 3= b
4 2
Now, D = b2 − 4ac ⇒ a + 16 − 8 a + 8 a − 32 + 12 = 4b
2
= (− 7 )2 − 4 × 2 × 5 ∴ a2 − 4b = 4
= 49 − 40 = 9 > 0
41. Since, α and β are the roots of x 2 − ax + b2 = 0.
Hence, it has two real roots.
Then, α+β=a
37. Let y = 20 301 and αβ = b2
Number of digits = Integral part of (301 log10 20 ) + 1 Now, α + β 2 = (α + β )2 − 2αβ
2
= Integral part of [301 (log10 10 + log10 2 )] + 1 = (a)2 − 2 (b)2 = a2 − 2 b2
= Integral part of [301 (1 + 0.3010 )] + 1
= Integral part of [301 × 13010
. ]+ 1 42. Since, α and β are the roots of the equation
= Integral part of [391601
. ]+ 1 x 2 + 3 x − 4 = 0.
= 391 + 1 = 392 Then, α + β = − 3 and αβ = − 4
1 1 β+α −3 3
38. Given, log101 log 7 ( x + 7 + x)= 0 Now, + = = =
α β αβ −4 4
∴ log 7 ( x + 7 + x ) = (101) 0
43. Since, α and β are the roots of x 2 + 2 bx + c = 0.
⇒ log 7 ( x + 7 + x)= 1
Then, α + β = − 2b
⇒ ( x+7 + x ) = 71
and αβ = c
⇒ x + 7 + x =7 α + β
2
On squaring both sides, we get
Now, b2 − c = − αβ
−2
( x + 7 ) + x + 2 x 2 + 7 x = 49 α 2 + β 2 + 2αβ − 4αβ
=
⇒ 2 x − 42 = − 2 x + 7 x2
4
α 2 + β 2 − 2αβ (α − β )2
⇒ x − 21 = − x2 + 7 x = =
4 4
Again, squaring on both sides, we get 44. Since, a, b and c are in GP.
x 2 + 441 − 42 x = x 2 + 7 x ∴ b2 = ac
⇒ 49 x = 441 Given equation is
441 (loge a) x 2 − (2 loge b) x + (loge c ) = 0
⇒ x=
49 Put x = 1, we get
∴ x=9 loge a − 2 loge b + loge c = 0
263
5 ⇒
⇒
2 loge b = loge a + loge c
loge b2 = loge ac
⇒ b2 = ac , which is true. ⇒ Sn + 1 =
= −
b
a
c
Sn − Sn − 1
a
− bS n − cS n − 1
Objective Mathematics Vol. 1
a
Hence, one of the root of given equation is 1.
∴ aS n + 1 + bS n + cS n − 1 = 0
Let another root be α.
2 loge b loge b2 48. Since,sin α andcos α are the roots of ax 2 + bx + c = 0.
∴ Sum of roots, 1 + α = = b c
loge a loge a ∴ sin α + cos α = − and sin α cos α =
loge ac a a
⇒ α= −1 b
2
loge a ⇒ (sin α + cos α ) = −
2
a
(loge a + loge c )
= −1 b2
loge a ⇒ sin 2 α + cos 2 α + 2 sin α cos α =
loge c a2
= = log a c c b2
loge a ⇒ 1+ 2⋅ =
a a2
Hence, roots are 1 and log a c.
⇒ a (a + 2c ) = b2
45. Given α , β are the roots of ax 2 + bx + c = 0 and
⇒ b2 − a2 = 2 ac
α + h, β + h are the roots of px 2 + qx + r = 0.
b c 49. Given, 32 x − 2 (3 x + 2 ) + 81 = 0
∴ α + β = − , αβ =
a a ∴ (3 x )2 − 2 (3 x ) 32 + 81 = 0
q r Let 3x = y
and α + h + β + h = − , (α + h ) (β + h ) =
p p ∴ y − 18 y + 81 = 0
2
Now, (α + h ) − (β + h ) = α − β
⇒ ( y − 9)2 = 0
⇒ [(α + h ) − (β + h )]2 = (α − β )2
⇒ y = 9 ⇒ 3 x = 32 [Q 3 x = y]
⇒ [(α + h ) + (β + h )]2 − 4 (α + h ) ( β + h )
∴ x =2
Ta rg e t E x e rc is e s
= (α + β )2 − 4αβ
50. Let the roots of the equation 2 x 2 + 3 x + 1 = 0 beα andβ.
q 2 4r b2 4 c −3
⇒ − = − Then, α+β= …(i)
p2 p a2 a 2
q − 4 pr p − 4ac
2 2 1
⇒ = and αβ = …(ii)
p2 a2 2
∴ α + β 2 = (α + β )2 − 2αβ
2
b − 4ac a
2 2
∴ = 2 − 3
2
1 9 5
q 2 − 4 pr p = − 2 = − 1=
2 2 4 4
Hence, the ratio of the square of their discriminants is 2
1 1
a2 : p2 . and α 2β 2 = =
2 4
46. Given, f ( x ) = 2 x 2 + 5 x + 1 …(i) ∴ Required equation is
Also, f ( x ) = a ( x + 1) ( x − 2 ) + b ( x − 2 ) ( x − 1) x 2 − (α 2 + β 2 ) x + α 2β 2 = 0
+ c ( x − 1) ( x + 1)
5 1
= a ( x 2 − x − 2 ) + b ( x 2 − 3 x + 2 ) + c ( x 2 − 1) ⇒ x2 − x + = 0
4 4
⇒ f ( x ) = (a + b + c ) x 2 + (− a − 3b) x
⇒ 4x 2 − 5x + 1 = 0
+ (− 2 a + 2 b − c ) …(ii)
On equating the coefficients of x 2 , x and constant 51. Given equation can be rewritten as
r2
terms in Eqs. (i) and (ii), we get x2 − x + 1= 0
a + b + c = 2, − a − 3b = 5 and− 2 a + 2 b − c = 1 pq
On solving above equations, we get r2
∴ tan A + tan B = and tan A tan B = 1
a = − 4, b = − 1/ 3 and c = 19 / 3 pq
Hence, exactly one choice for each of a, b and c. We know that, A + B + C = 180 °
47. Given, α and β are the roots of equation ⇒ ( A + B) = 180 ° − C
⇒ tan ( A + B) = tan (180 ° − C )
ax 2 + bx + c = 0.
tan A + tan B
b c ⇒ = − tan C
∴ α+β=− and αβ = 1 − tan A tan B
a a
Now, S n + 1 = α n + 1 + β n + 1 r 2 / pq
⇒ = − tan C
1− 1
= α n + 1 + β n + 1 + α nβ + β nα − α nβ − β nα
⇒ ∞ = − tan C
= α n (α + β ) + β n (α + β ) − αβ (α n − 1 + β n − 1 ) ∴ C = 90 °
= (α + β ) (α n + β n ) − αβ (α n − 1 + β n − 1 ) Hence, ∆ABC is a right angled triangle.
264
52. Q Sum of roots, α + β = − p and αβ = q
∴ (α 3 + β 3 ) = (α + β )3 − 3 αβ (α + β )
= (− p) − 3 q (− p) = − p + 3 pq
3 3
Then, sum of the roots,
S = α + β = a + a2 + a3 + a4 + a5 + a6
a (1 − a6 ) a − a7 a − 1
5
Inequalities and Quadratic Equation
⇒ S = = = = −1 [Q a7 = 1]
and α 4 + α 2β 2 + β 4 = (α 4 + β 4 ) + (αβ )4 1− a 1− a 1− a
= (α 2 + β 2 )2 − (αβ )2 Product of the roots,
= [(α + β ) − 2αβ ] − (αβ )
2 2 2 P = αβ = (a + a2 + a4 ) (a3 + a5 + a6 )
= [(− p)2 − 2q ]2 − 3 (q )2 = a4 + a5 + 1 + a6 + 1 + a2 + 1 + a + a3 [Q a7 = 1]
= [ p2 − 2q ]2 − 3 q 2 = 3 + (a + a + a + a + a + a ) = 3 − 1 = 2
2 3 4 5 6
= p4 − 4 p2q + 4 q 2 − q 2 Hence, the required quadratic equation is
x 2 + x + 2 = 0.
= p4 − 4 p2q + 3 q 2
= p4 − 3 p2q − p2q + 3 q 2 57. Since, α is a root of x 2 + 3 p2 x + 5 q 2 = 0, then
= p2 ( p2 − 3 q ) − q ( p2 − 3 q ) α 2 + 3 p2α + 5 q 2 = 0 …(i)
= ( p − q ) ( p − 3q )
2 2 and β is a root of x + 9 p x + 15 q = 0, then
2 2 2
∴ β 2 + 9 p2β + 15 q 2 = 0 …(ii)
53. Given equation is x + 1 − x − 1 = 4 x − 1 ...(i)
Let f ( x ) = x 2 + 6 p2 x + 10 q 2
On squaring both sides, we get Then, f (α ) = α 2 + 6 p2α + 10 q 2
( x + 1) + ( x − 1) − 2 x 2 − 1 = 4 x − 1 = (α 2 + 3 p2α + 5 q 2 ) + 3 p2α + 5 q 2
⇒ 2 x − 2 x − 1 = 4x − 1
2
= 0 + 3 p2α + 5 q 2 [from Eq. (i)]
⇒ − 2 x − 1 = 2x − 1
2
⇒ f (α ) > 0 and f ( β ) = β 2 + 6 p2β + 10 q 2
Again, squaring on both sides, we get = ( β 2 + 9 p2 β + 15 q 2 ) − (3 p2β + 5 q 2 )
4 ( x 2 − 1) = 4 x 2 + 1 − 4 x = 0 − (3 p2β + 5 q 2 )
⇒ − 4 = + 1 − 4x ⇒ f ( β) < 0
Targ e t E x e rc is e s
⇒ 4x = 5 ⇒ x = 5 / 4 Thus, f ( x ) is a polynomial such that f (α ) > 0 and
5 f ( β ) < 0.
But when we put x = in Eq. (i), we get
4 Therefore, there exists γ satisfying α < γ < β such that
5 5 5 f (γ ) = 0.
+ 1− −1= 4× −1
4 4 4 58. Given equation is λ2 + 8λ + µ 2 + 6 µ = 0.
9 1
⇒ − = 5−1 Since, roots are real.
4 4 ∴ b2 − 4ac ≥ 0
3 1
⇒ − = 2 ⇒ 1 = 2, which is not true. ⇒ (8)2 − 4 ( µ 2 + 6 µ ) ≥ 0
2 2
⇒ 64 − 4 ( µ 2 + 6 µ ) ≥ 0
Hence, no value of x satisfy the given equation.
⇒ 16 − ( µ 2 + 6 µ ) ≥ 0
54. Given, 9 x = 12 + 147
⇒ µ 2 + 6 µ − 16 ≤ 0
On squaring both sides, we get
⇒ µ 2 + 8 µ − 2 µ − 16 ≤ 0
81x = 12 + 147 + 2 12 147
⇒ 81x = 159 + 84 ⇒ µ ( µ + 8) − 2 ( µ + 8) ≤ 0
⇒ 81x = 243 ⇒ ( µ + 8) ( µ − 2 ) ≤ 0
∴ x=3 ∴ − 8≤µ ≤2
Aliter 59. Given,| x − 2| + | x + 2| < 4
9 x = 12 + 147
Case I When x < − 2
⇒ 9 x =2 3 + 7 3 ⇒ − ( x − 2) − ( x + 2) < 4
⇒ 9 x =9 3 ⇒ − 2x < 4
⇒ x = 3 ⇒ − x < 2 ⇒ x > − 2, which is not true.
∴ x=3 Hence, no value of x exist.
55. x 2 − 3 x + 1 = 0 Case II When − 2 < x < 2
3± 3− 4 3±i ⇒ − ( x − 2) + x + 2 < 4
⇒ x= = ⇒ 4 < 4, which is not true.
2 2
Hence, no value of x exist.
2π 2π
56. Given, a = cos + i sin Case III When x > 2
7 7
∴ a7 = cos 2 π + i sin 2 π [Q e iθ
= cos θ + i sin θ ] ⇒ x −2 + x + 2 < 4
⇒ 2 x < 4 ⇒ x < 2, which is not true.
=1
So, no value of x exist.
Also, α = a + a2 + a4 and β = a3 + a5 + a6 Hence, the number of solutions of the inequation is 0. 265
5 60. 2 x − 5 ≤
⇒
4x − 7
3
6 x − 15 ≤ 4 x − 7
Now, on substituting the value of k in Eq. (iii), we get
2+
c −a
2a
+
2 a
1
+1
=−
b
a
Objective Mathematics Vol. 1
⇒ 2x ≤ 8 ⇒ x ≤ 4 c −a
61. 3 x + 2 > − 16 c −a c −a b
⇒ 2+ + =−
⇒ x>−6 2a a+ c a
and 2 x − 3 ≤ 11 2 (2 a) (a + c ) + (c − a) (c + a) + 2 a (c − a) b
⇒ =−
⇒ x ≤7 2 a (a + c ) a
∴ x ∈ (− 6, 7 ] ⇒ a2 + c 2 + 6ac = − 2 ab − 2 bc
1
62. We can write, 5 x + 5− x = 5 x + x ⇒ a2 + b2 + c 2 + 2 ab + 2 bc + 2ca = b2 − 4ac
5
1 ∴ (a + b + c )2 = b2 − 4ac
= 5x + x − 2 + 2
5 66. Since, the second equation has imaginary roots.
x 1
2
2 a − 3b 4 c
= 5 − x + 2 ∴ = = =k
5 3 −4 5
3k 4k 5k
This means 5 x + 5− x > 1 ⇒ a= ,b= ,c =
2 3 4
Thus, sin (e ) > 1, which is not possible.
x
3k 4k
+
a+ b 3 = 34
63. By hypothesis, a2α 2 + bα + c = 0 …(i) ∴ = 2
b + c 4k + 5k 31
and a β − bβ − c = 0
2 2
…(ii)
3 4
where, 0 <α <β
9x − 1 + 7
and α ≠ 0. 67. log 2 = 2 log 2 2
Let f ( x ) = a2 x 2 + 2 bx + 2c 3 x − 1 + 1
(32 )x − 1 + 7 (3 x − 1 )2 + 7
Ta rg e t E x e rc is e s
Clearly, it is continuous on [α , β ]. ⇒ =4 ⇒ =4
(3 x − 1 + 1) 3x − 1 + 1
∴ f (α ) = a2α 2 + 2 bα + 2c = − α 2 a2 < 0
Let 3x − 1 = y
and f (β ) = a2β 2 + 2 bβ + 2c = 3 a2β 2 > 0 y +7
2
∴ =4
Therefore, by the intermediate value theorem of y+1
continuous function, f ( x ) = 0 has a root γ between α ⇒ y 2 + 7 = 4y + 4
and β. ⇒ y − 4y + 3 = 0
2
64. Let α and β be the roots, then ⇒ y = 1, 3
α + β = 16 ∴ 3 x − 1 = 30 and 3 x − 1 = 31
and αβ = 25 ⇒ x = 1, 2
∴ x 2 − x (α + β ) + αβ = 0 68. Since, a, b and c are in AP.
⇒ x 2 − 16 x + 25 = 0 ∴ 2b = a + c
65. We have, Let α and β be the roots of the given equation.
k+1 k+2 b 2b
+ =− ...(i) ∴ α+β=
k k+1 a a
k+1 k+2 c a+c c
and ⋅ = ...(ii) = = 1+ ...(i)
k k+1 a a a
c
From Eq. (i), we have and αβ =
1 1 b a
1+ + 1+ =− Now, (α − β )2 = (α + β )2 − 4 αβ
k k+1 a 2 2
1 1 b c c c
⇒ 2+ + =− …(iii) = 1 + − 4 ⋅ = 1 −
k k+1 a a a a
c
From Eq. (ii), we have ⇒ α − β = 1− ...(ii)
k+2 c a
= On adding Eqs. (i) and (ii), we get
k a
2 c 2α = 2 ⇒ α = 1
⇒ 1+ =
k a On subtracting Eq. (ii) from Eq. (i), we get
2 c 2c
⇒ = −1 2β =
k a a
2a c
∴ k= ∴ β=
266 c −a a
69. Let α and β be the roots of the given equation.
∴ α+β=
4+ 5
5+ 2
73. Here, α + β = 2 and αβ = 4
Now, (α − β ) = (α + β )2 − 4αβ 5
Inequalities and Quadratic Equation
= 4 − 16 = 2 3i
8+ 2 5 1 3
and αβ = ∴ α =2 + i
5+ 2 2 2
Now, harmonic mean of α and β =
2αβ π π
= 2 cos + i sin
α+β 3 3
16 + 4 5 2 − 2 3i
and β=
5+ 2 2
=
4+ 5 π π
= 2 cos − i sin
5+ 2 3 3
16 + 4 5 (4 − 5 ) π π
n
= × ∴ α n + β n = 2 cos + i sin
4+ 5 (4 − 5 ) 3 3
64 − 16 5 + 16 5 − 4 × 5 n
= π π
16 − 5 + 2 cos − i sin
3 3
44
= =4 nπ
11 = 2 n 2 cos
3
70. α 3 + β 3 + γ 3 − 3 αβγ
nπ
= (α + β + γ ) (α 2 + β 2 + γ 2 − αβ − βγ − γα ) = 2 n + 1 cos
3
⇒ α 3 + β 3 + γ 3 − 3 αβγ = 0
74. Since, α, β and γ are the roots of x 3 − 2 x + 1 = 0.
⇒ α 3 + β 3 + γ 3 = 3 αβγ
We have, Σα = 0 ...(i)
= 3 × (− 2 ) = − 6
Σαβ = − 2
Targ e t E x e rc is e s
...(ii)
71. (α + β ) + (α − β ) = − p [given] αβγ = − 1 ...(iii)
⇒ 2α = − p 1 1
p Now, Σ = Σ
⇒ α=− α + β − γ − γ − γ
2
and [(α + β ) (α − β )] = q [given] − 1 − 1 1
= Σ = Σ
⇒ α2 − β = q 2γ 2 γ
⇒ β = α2 − q − 1 1 1 1
= + +
2 α β γ
2
p
= − − q
2
− 1 Σαβ − 1 − 2
p2 = = = −1
= −q 2 αβγ 2 − 1
4
⇒ p2 − 4 q = 4 β 75. (i) b must be even integer.
(ii) Its discriminant is a perfect square of an even number.
The given equation is
So, D = b2 + 4 ⋅ 16
( p2 − 4 q ) ( p2 x 2 + 4 px ) − 16 q = 0
= b2 + 64
⇒ 4 β ( p2 x 2 + 4 px ) − 16 (α 2 − β ) = 0 [Qα 2 − β = q]
⇒ β (4 α 2 x 2 − 8 αx ) − 4 (α 2 − β ) = 0 [Q p = − 2α] Here, 0 and 6 only the even integers value of b, on which
the discriminant of given quadratic equation have
⇒ α βx − 2αβx + β = α
2 2 2
perfect square of an even number.
⇒ (αx β − β )2 = α 2
So, the number of integral values of b is 2.
⇒ αx β − β = ± α
1 1 76. On taking option (c), it is clear that,
∴ x= ± 2
α β 1 1
a + b + c = 0
2 2
72. Given, x 2 − 2 2 kx + 2e 2 log k − 1 = 0
⇒ a + 2b + 4c = 0
∴ Product of roots = 2e 2 log k − 1 = 31 [given]
77. 2(3)2 − c (3) + 3 = 0
⇒ 2e 2 log k = 32
2 ⇒ c =7
⇒ e log k = 16 and 2 x2 − 7 x + d = 0
⇒ k 2 = 16 Also, B 2 = 4 AC
∴ k=±4
⇒ 49 = 4 × 2 × d
But k = − 4, log k is not defined. 49
Hence, required value of k is 4. ∴ d =
8 267
5 78.
( 3 + 1+
( 3 + 1) − (2
3 − 1)2 ( 3 − 2 )
3 − 1)2
84. ax 2 + bx + c = 0
Replacing x by
1 − bx
ax
, we get the required equation
[Qc ≠ 0]
Objective Mathematics Vol. 1
( 3 + 1 + 3 − 1 + 2 2)( 3 − 2)
= 2
( 3 + 1) − ( 3 − 1) 1 − bx 1 − bx
a + b +c =0
ax ax
2( 3 + 2 ) ( 3 − 2 )
= =1
2 ⇒ a(1 + b2 x 2 − 2 bx ) + ax(b − b2 x ) + ca2 x 2 = 0
79. We have,|2 x − 3| < | x + 5| ⇒ a + ab2 x 2 − 2 abx + abx − ab2 x 2 + a2cx 2 = 0
Case I If − ∞ < x < − 5 ⇒ acx 2 − bx + 1 = 0
–5
85. Since, one root of the equation x 2 + px + q = 0 is
3/2
⇒ − 2x + 3 < − x − 5 2 + 3, then the other root will be 2 − 3.
⇒ −x<−8 ∴ Sum of roots, 2 + 3 + 2 − 3=− p
∴ x>8 …(i) ⇒ p=−4
So, no solution exist. and product of roots, (2 + 3 ) (2 − 3) = q
Case II If − 5 < x < 3 / 2 ∴ q =1
− 2x + 3 < x + 5 86. α + β = − p, αβ = q
−2
⇒ − 3x < 2 ⇒ x>
3 ∴ α 2 + β 2 = (α + β )2 − 2αβ
2 3 = p2 − 2q
∴ − <x< …(ii)
3 2
3 ⇒ (α − β )2 = α 2 + β 2 − 2αβ
Case III If <x<∞
2 = ( p2 − 2q ) − 2q = p2 − 4 q
⇒ 2x − 3 < x + 5
87. a + b = − a ...(i)
⇒ x<8
Ta rg e t E x e rc is e s
3 ab = b ...(ii)
∴ <x<8 ...(iii) From Eq. (ii), a=1 [Q b ≠ 0]
2
From Eq. (i), b= −2
From Eqs. (i), (ii) and (iii), we get
−2 88. Let α and β be the roots, then
x ∈ , 8
3 α + β = b, αβ = c
Given, |α − β| = 1
k k α + β − b / a
80. We have, + = k = k ⇒ (α + β )2 − 4 αβ = 1
α β αβ c/a
⇒ b2 − 4 c = 1
− kb k k k 2 k 2a
= and ⋅ = = 89. Qα = ω, β = ω 2 will satisfy the given equation.
c α β αβ c
− kb
2
k a Now, α 19 = ω19 = ω, β 7 = ω14 = ω 2
∴ x2 − x+ =0
c c ∴Required equation is
⇒ cx 2 + kbx + k 2 a = 0 x 2 − (ω + ω 2 )x + ω 3 = 0
81. 3 ≤ 3 t − 18 ≤ 18 ⇒ 21 ≤ 3 t ≤ 36 ⇒ x2 + x + 1 = 0
⇒ 7 ≤ t ≤ 12 ⇒ 8 ≤ t + 1 ≤ 13 90. Q x 2 + 15| x| + 14 = | x|2 + 15| x| + 14 > 0
x + 11
82. >0 for all real x.
x−3
So, given equation has no solution.
⇒ ( x − 3) ( x + 11) > 0
⇒ x < − 11, x > 3 91. Since, roots are equal.
⇒ x ∈ (− ∞, − 11) ∪ (3, ∞ ) ∴ (2 6 )2 = 4 ⋅ 2 ⋅ a
logc + b a + logc − b a ⇒ 24 = 8 a ⇒ a = 3
83. 3 7 3 7
2 logc + b a ⋅ logc − b a 92. Let α = + i and β = − i
2 2 2 2
log a log a 9 49 29
+ ∴ α + β = 3, αβ = + =
log(c + b) log(c − b) 4 4 2
=
log a log a 6 b 29
2⋅ ⋅ ⇒ = 3, =
log(c + b) log(c − b) a a 2
log a{log(c − b) + log(c + b)} ⇒ a = 2, b = 29
=
2(log a)2 ∴ a + b = 31
log(c 2 − b2 ) log a2 93. Discriminant (D ) = (− 2 3 )2 + 88
= = [Q a2 + b2 = c 2 ]
2 log a log a2 = 100 = 10 2
268 So, roots are real, irrational and unequal.
=1
6
Permutation and
Combination
Fundamental Principles of Counting (FPC)
Fundamental Principle of Multiplication Chapter Snapshot
If there are two jobs such that one of them can be completed in m ways and when it has ● Fundamental Principles of
been completed in anyone of these m ways, second job can be completed in n ways, Counting (FPC)
then the two jobs in succession can be completed in m × n ways. ● Factorial
X Example 1. In a class, there are 15 boys and 10 girls. The teacher wants to ● Exponent of Prime p in
Factorial n
select a boy and a girl to represent the class in a function, then the total number of
selection by teacher is ● Representation of Symbols
n
(a)15 (b)10 (c) 25 (d)150 Pr and n Cr
● Some Basic Arrangements
Sol. (d) Here, the teacher is to perform two jobs:
and Selections
(i) Selecting a boy among 15 boys.
(ii) Selecting a girl among 10 girls. ● Summation of Numbers
(3 different ways)
The first job can be performed in 15 ways and the second in 10 ways. Therefore, by
fundamental principle of multiplication, the total number of selection = 15 × 10 = 150 ways. ● Permutations under Certain
Conditions
Fundamental Principle of Addition ● Circular Permutations
If there are two jobs such that they can be performed independently in m and n ways ● Geometrical Applications of
respectively, then either of the two jobs can be performed in ( m + n) ways. n
Cr
● Selection of One or More
X Example 2. There are 20 students in a class, in which 12 boys and 8 girls. The Objects
class teacher select either a boy or a girl for monitor of the class. In how many
● Number of Divisors and the
ways, the teacher can make this selection? Sum of the Divisors of a Given
(a)120 (b)12 (c) 96 (d) 20 Natural Number
Sol. (d) By fundamental principle of addition, the number of ways in which either a boy or a girl ● Division of Objects into
can be chosen as a monitor = 12 + 8 = 20 ways. Groups
● Dearrangements
X Example 3. Three dices are rolled. The number of possible outcomes in which
● Number of Integral Solutions
atleast one dice shows 5, is
of Linear Equations and
(a) 36 (b)18 (c) 90 (d) 91 Inequations
6 Sol. (d) When a die is rolled, there are six possible
outcomes. So, by FPC, the total number of outcomes
when three dice is rolled = 6 × 6 × 6 = 216.
Sol. (a) x1
9
x2
9
x3
9
... xn
9
Objective Mathematics Vol. 1
Now, the number of possible outcomes in which atleast one
dice shows 5 = Total number of possible outcomes − The digit x1 can be selected in 9 ways as ‘0’ cannot be
Number of possible outcomes in which 5 does not appear selected. The digit x2 can be selected in 9 ways as 0 can
on any dice be selected but digit in position x1 cannot be selected.
= 63 − 53 = 216 − 125 = 91 Similarly, each of the remaining digit can also be selected
in 9 ways. Thus, the total number of such numbers is 9n .
X Example 4. How many four-digit numbers can X Example 6. The number of polynomials of the
be formed using the digits 0, 1, 2, 3, 4, 5, if repetition
form x 3 + ax 2 + bx + c that are divisible by x 2 + 1
of digits is not allowed?
where a, b, c ∈{1, 2, 3, ...,9, 10}, are
(a)120 (b)1080 (c) 300 (d) 600
(a) 7 (b) 8
Sol. (c) In a four-digit numbers, 0 cannot appear in the (c) 9 (d)10
thousand’s place. So, thousand’s place can be filled in
5 ways (i.e. 1, 2, 3, 4, 5). Since, repetition of digits is not x+ a
allowed and 0 can be used at hundred’s place. So, Sol. (d) x + 1 x + ax2 + bx + c
2 3
hundred’s place can be filled in 5 ways. Now, anyone of x3 + x
the remaining 4 digits can be used to fill up ten’s place. − −
So, ten’s place can be filled in 4 ways. One’s place can be ax2 + (b − 1) x + c
filled from the remaining 3 digits in 3 ways. Hence, by
fundamental principle of multiplication, the required ax2 + a
− −
number of numbers = 5 × 5 × 4 × 3 = 300.
(b − 1)x + c − a
X Example 5. The total number of n-digit Now, remainder (b − 1)x + c − a must be zero for any x,
numbers ( n >1) having the property that no two then
consecutive digits are same, is b − 1 = 0 and c − a = 0
⇒ b = 1 and c=a
(a) 9 n (b)10 n
Now, c or a can be selected in 10 ways. Hence, number
(c)10 − 9
n n
(d) None of these of polynomials are 10.
Work Book Exercise 6.1
1 Twenty buses run between New Delhi to Meerut. 5 A letter lock consists of three rings, each marked
If a man goes from Meerut to New Delhi by a bus with 10 different letters, the number of ways in
and comes back to Meerut by another bus. The which it is possible to make an unsuccessful
total possible ways is attempt to open the lock is
20 40 39 380 999 899
479 None of these
2 Ten different letters of an alphabet are given. 6 How many two-digit even numbers can be
Words with five letters are formed from these formed from the digits 1, 2, 3, 4, 5, if the digits can
given letters. Then, the total number of words be repeated?
which have atleast one letter repeated, is 120 20 25 10
30240 69760
99748 None of these 7 A five-digit numbers divisible by 3 is to be formed
using the digits 0, 1, 2, 3, 4 and 5 without
3 An examination paper contains 8 questions of repetition. The total number of ways in which this
which 4 have 3 possible answers, 3 have 2 can be done, is
possible answers and the remaining one 216 600 240 3125
question has 5 possible answers. The total 8 If p, q ∈(1, 2, 3, 4), then the number of equations of
number of possible answers to all the question is the form px 2 + qx + 1 = 0 having real roots, is
2880 178 4 5 6 7
94 3240
9 Find the number ordered pairs ( x, y ), if
4 There are stalls for 10 animals in a ship. The x, y ∈ { 0, 1, 2, 3,...,10} and if| x − y| > 5.
number of ways in which ship load can be made, 30 20
if there are cows, calves and horses to be 15 None of these
transported, animals of each kind being not less
than 10, is 10 Find the total number of ways in which two small
410 squares can be selected on the normal chessboard,
1310 if they are not in same row or same column.
130 64 1296
270 None of the above 1568 None of these
Factorial
For any natural number n, we define factorial of Sol. (a) We have,
6
Permutation and Combination
(2n)! 1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 ⋅ 6 ⋅ 7 ⋅ 8... (2 n − 2 ) (2 n − 1) (2 n)
n ( n! or n ) as follows: =
n! n!
n! = n ( n − 1) ( n − 2)... 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 {1 ⋅ 3 ⋅ 5 ⋅ 7...(2 n − 1)} ⋅ {2 ⋅ 4 ⋅ 6 ⋅ 8...(2 n − 2 ) (2 n)}
=
∴ 1! =1, 2! = 2 × 1 = 2, 3! = 3 × 2 × 1 = 6 and so on. n!
{1 ⋅ 3 ⋅ 5 ⋅ 7...(2 n − 1)} ⋅ 2 n ⋅ n!
Ø We can extend the definition of the factorial to zero and negative =
integers as follows: n!
(i) 0 ! = 1 = {1 ⋅ 3 ⋅ 5 ⋅ 7...(2 n − 1)} ⋅ 2 n
(ii) Factorial of negative integers and fractions are not defined.
(iii) n ! = n (n − 1)! = n(n − 1)(n − 2)!
X Example 11. The one’s and ten’s digits in
n! 1! + 2! + 3! + ... + 49! are respectively
(iv) = n (n − 1)(n − 2)..., n > r
r! (a) 4 and 1 (b) 3 and 1 (c) 4 and 2 (d) 5 and 1
Sol. (b) We know that n! is divisible by 100 for n ≥ 10 and
X Example 7. If (n + 1)! = 12 × (n − 1)!, then n is
1! + 2 ! + 3! + 4! + 5! + 6! + 7 ! + 8! + 9! = 409113.
equal to
Therefore,
(a) 4 (b) 5
1! + 2 ! + 3! + ... + 49! = 409113 + a multiple of 100
(c) 3 (d) 6
Hence, one’s and ten’s digits in 1! + 2 ! + ... + 49! are 3
Sol. (c) We have, (n + 1)! = 12 × (n − 1)! and 1, respectively.
⇒ (n + 1) (n) (n − 1)! = 12 × (n − 1)!
⇒ n2 + n = 12 X Example 12. If
⇒ n + n − 12 = 0
2
S = (1) (1!) + (2) (2!) + (3) (3!) + ... + n( n!), then
⇒ (n + 4) (n − 3) = 0 S+1
∴ n = 3, n = − 4 (a) ∈ integer
n!
1 1 x S+1
X Example 8. If + = , then x is equal to (b) ∉ integer
8! 9! 10! n!
(a) 100 (b)10 S +1
(c) cannot be discussed
(c) 90 (d) None of these n!
Sol. (a) We have, (d) None of the above
1 1 x
+ = Sol. (a) We have,
8! 9 × 8! 10 × 9 × 8! n n
⇒
1
1+ =
x
⇒
10
=
x S= ∑ k (k !) = ∑ {(k + 1) − 1}(k !)
9 10 × 9 9 10 × 9 k =1 k =1
n
∴ x = 100 = ∑ {(k + 1)! − k !}
k =1
X Example 9. Find the remainder when
= (n + 1)! − 1
1! + 2! + 3! + ...+ n! is divided by 15, if n ≥5.
⇒ S + 1 = (n + 1)!
(a) 13 (b) 12 (c) 3 (d) 5
S+1
Thus, ∈ Integer
Sol. (c) Let N = 1! + 2 ! + 3! + ...+ n! n!
N 1! + 2 ! + 3! + 4! + 5! + 6! + ...+ n!
⇒ =
15 15 X Example 13. S n = 1! + 2! + 3! + 4! + K + n! cannot
N 1! + 2 ! + 3! + 4! 5! + 6! + 7 ! + L + n! be the square of natural number except for n, is
⇒ = +
15 15 15 (a) 1, 4 (b) 2, 3 (c) 1, 3 (d) 2, 4
N 1 + 2 + 6 + 24
⇒ = + 8k Sol. (c) For n = 1, we have
15 15
S1 = 1! = 1, which is a perfect square.
N 33
⇒ = [as 5!, 6!, 7 !, 8! are divisible by 15] For n = 2, we have
15 15
S 2 = 1! + 2 ! + = 1 + 2 = 3, which is not a perfect square.
Hence, remainder is 3.
For n = 3, we have
(2n)! S 3 = 1! + 2 ! + 3! = 1 + 2 + 6 = 9
X Example 10. The value of is which is a perfect square.
n!
For n = 4, we have
(a) {1 ⋅ 3 ⋅ 5... (2n − 1)}2 n (b) {1 ⋅ 3 ⋅ 5... ( n − 1)}2 n S 4 = 1! + 2 ! + 4! = 1 + 2 + 6 + 24 = 33
(c) {1 ⋅ 3 ⋅ 5... (2n + 1)}2 n (d) None of these which is not a perfect square. 271
6 For n ≥ 5, we find that the digits at unit’s place in n! is 0
and S 4 = 1! + 2 ! + 3! + 4! has 3 as the digit at unit’s
place. Therefore, for n ≥ 5, S n has 3 at unit’s place. But
X Example 16. The exponent of 12 in 50! is
(a) 22 (b) 23 (c) 45 (d) 48
Objective Mathematics Vol. 1
the square of any natural number does not have 3 at
unit’s place.
Sol. (a) We have,
12 = 2 2 × 3 = 2 × 2 × 3 and 50! = 2 a × 3 b × 5 c × ...
Therefore, S n is not a perfect square for n ≥ 5.
Now, a = E2 (50!) = + 2 + 3 + 4 + 5
Hence, S n = 1! + 2 ! + 3! + K + n! is not a perfect square 50 50 50 50 50
of a natural number except for n = 1, 3. 2 2 2 2 2
= 25 + 12 + 6 + 3 + 1 = 47
X Example 14. Let p be a prime number such that and b = E3 (50!) = + 2 + 3
50 50 50
3 3 3
p ≥11. Let n = p! + 1. The number of primes in the
list n +1, n + 2, n + 3, ... n + p −1, is = 16 + 5 + 1 = 22
(a) p −1 (b) 2 ∴ 50! = 2 47 × 322 × 5 c × ...
(c) 1 (d) None of these ⇒ 50! = (2 2 × 3)22 × 2 3 × 5 c × K
Sol. (d) For 1 ≤ i ≤ p − 1, p! is divisible by (i + 1.) ⇒ 50! = (12 )22 × 2 3 × 5 c × ...
Thus, n + i = p! + (i + 1) is divisible by (i + 1)
Hence, E12 (50!) = 22
for 1≤ i ≤ p − 1
∴None of n + 1, n + 2,..., n + p − 1 is prime. X Example 17. Let
1 1 1 2 1 3
Exponent of Prime p in Factorial n E = + + + + + +…+ 50 terms,
3 50 3 50 3 50
Let p be a prime number and n be a positive then the exponent of 2 in E ! is
integer. (a) 17 (b) 25 (c) 15 (d) 20
Let E p n! denotes the exponent of the prime p in the 50
1 + x
positive integer n. Then, Sol. (c) Let E = ∑ 50
x =1 3
n n n n
E p ( n!) = + 2 + 3 + ... + s For 1 ≤ x ≤ 33,
1 1
< +
x
<1
p p p p 3 3 50
∴ 1 + x = 0, for 1 ≤ x ≤ 33
where, s is the largest positive integer such that 3 50
ps ≤ n < ps + 1. 1 x 4
For 34 ≤ x ≤ 50, 1 < + <
3 50 3
X Example 15. The exponent of 3 in 100! is ⇒ 1 + x = 1, for 34 ≤ x ≤ 50
3 50
(a) 33 (b) 48 (c) 45 (d) 50
Thus, E = 17 and p = 2
Sol. (b) We have, n = 100!
Q 2 4 < 17 < 2 5
100 100 100 100
E3 (100!) = + + + ∴ a=4
3 32 33 34
Exponent of 2 in 17 ! = + + +
17 17 17 17
= 33 + 11 + 3 + 1 = 48 2 4 8 16
Hence, the exponent of 3 in 100! = 48.
= 8 + 4 + 2 + 1 = 15
Work Book Exercise 6.2
1 The exponent of 7 in 100! is 4 The number of zero at the end of 70! is
14 15 16
16 None of these 5
7
1 1 x
2 If + = , then x is equal to 70
6! 7 ! 8! n
49
64
36
None of these
5 Sum of the series ∑ (r 2 + 1) r ! is
r =1
3 If n ! , 3 × n ! and (n + 1)! are in GP, then n !, 5 × n ! ( n + 1)!
( n + 2 )! − 1
and (n + 1)! are in
n ( n + 1)!
AP GP
None of the above
HP None of these
272
Representation of Symbols n Pr 6
and nCr
Permutation and Combination
Symbol nPr or P ( n, r ) If n is a natural number and
viii. n
C r = n C s implies r = s or r + s = n
r is an integer satisfying 0 ≤ r ≤ n, then the natural
n! n n −1 n ( n − 1) n − 2
number is denoted by the symbol n Pr or P ( n, r ). ix. n
Cr = Cr − 1 = ⋅ C r − 2 =…
( n − r )! r r ( r − 1)
n
n! Cr n − ( r − 1)
i.e. n
Pr = P ( n, r ) = x. =
( n − r )! n
C r −1 r
4!
e.g. 4
P2 = P ( 4, 2) = r = n
( 4 − 2)! , if n is even
xi. n
C r is greatest = 2
4! 4 × 3 × 2 ! n ±1
= = = 12 r = , if n is odd
2! 2! 2
7! 7!
7
P3 = P ( 7, 3) = = xii. n
C 0 + n C1 + n C 2 + ... + n C n = 2 n
( 7 − 3)! 4!
7 × 6 × 5 × 4! xiii. n
C 0 + n C 2 + ... = n C1 + n C 3 + ... = 2 n − 1
= = 210
4!
2n + 1 2n + 1
xiv. C0 + C1 + ... + 2n + 1 C n = 2 2n
Symbol nCr or C ( n, r ) If n is a natural number and r
is an integer satisfying 0 ≤ r ≤ n, then the natural xv. r
Cr + r +1
C r + r + 2 C r + ... + n C r = n +1
Cr + 1
n!
number is denoted by the symbol n C r or
( n − r )! r ! xvi. The number of combinations of n distinct
C ( n, r ). objects taken r ( ≤ n) at a time, when
n! k (0 ≤ k ≤ r ) particular objects always occur, is
i.e. n
C r = C ( n, r ) = n−k
Cr − k .
( n − r )! r !
8! 8 × 7 × 6! xvii.
e.g. 8
C 2 = C (8, 2) = = = 28 The number of combinations of n distinct
2! (8 − 2)! 2 × 1 × 6! objects taken r at a time, when k (1 ≤ k ≤ r )
10! objects never occur, is n − k C r .
10
C 7 = C (10, 7) =
7! (10 − 7)!
xviii. The total number of selections of one or more
10! 10 × 9 × 8 × 7!
= = =120 objects from n different objects
7! 3! 7! × 3 × 2 × 1
= 2 n − 1 = ( C1 + C 2 + C 3 +... + n C n )
Some Properties Related to n Pr and n Cr X Example 18. If 56
Pr + 6 : 54 Pr + 3 = 30800 :1,
then r is equal to
i. n
Pn = n! (a) 39 (b) 40
n −1 n −1 (c) 41 (d) None of these
ii. n
Pr = Pr + r Pr − 1
56! (51 − r )!
n Sol. (c) We have, ⋅ = 30800
(50 − r )!
iii. ∑ r r Pr n +1 54!
= Pn + 1 − 1
r =1 ⇒ 56 × 55 (51 − r ) = 30800
30800 30800
⇒ 51 − r = = = 10
iv. n
C r is a natural number and exists only if 280 × 11 3080
n ≥ r ; n, r ( ≥ 0) ∈ integers. ∴ r = 51 − 10 = 41
4n
v. n
C0 =1 = nCn X Example 19. C 2n : 2n C n is equal to
{1 ⋅ 3 ⋅ 5... ( 4n − 1)} {1 ⋅ 3 ⋅ 5... ( 4n − 1)}2
vi. n
Cr = nCn − r , 0 ≤ r ≤ n (a) (b)
{1 ⋅ 3 ⋅ 5... (2n − 1)}2 {1 ⋅ 3 ⋅ 5... (2n − 1)}
n +1
vii. n
C r −1 + n C r = Cr , 1 ≤ r ≤ n {1 ⋅ 3 ⋅ 5... (2n − 1)}2 273
(c) (d) None of these
{1 ⋅ 3 ⋅ 5... ( 4n − 1)}
6 ⇒ (n − 2 ) (n − 3) (n − 11) < 0 [ Q n + 2 > 0 for n ∈ N ]
4n
C2n 4 n! n! n!
Sol. (a) We have, = ×
2n
Cn 2 n! 2 n! 2 n! ⇒ n ∈ (−∞, 2 ) ∪ (3, 11)
{1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 ⋅ 6 ... (4 n − 1) 4 n!} n! n! – + – +
=
Objective Mathematics Vol. 1
(1 ⋅ 2 ⋅ 3 ⋅ 4 ⋅ 5 ⋅ 6 ... 2 n)2 2 n! 2 3 11
{1 ⋅ 3 ⋅ 5 ... (4n − 1)} {2 ⋅ 4 ⋅ 6 ... 4n} n! n! ⇒ n ∈ (0, 2 ) ∪ (3, 11) ⇒ n = 1, 4, 5, 6, 7, 8, 9,10
=
n −1
{1 ⋅ 3 ⋅ 5 ... (2 n − 1)}2 {2 ⋅ 4 ⋅ 6 ... 2 n}2 2 n! But C 4 and n − 2 P2 both are meaningful for n ≥ 5.
{1 ⋅ 3 ⋅ 5 ... (4n − 1)} 2 2 n 2 n! n! n! Hence, n = 5, 6, 7, 8, 9, 10.
=
{1 ⋅ 3 ⋅ 5 ... (2 n − 1)}2 2 2 n {1 ⋅ 2 ⋅ 3 ... n}2 2 n! 20
Cn
=
{1 ⋅ 3 ⋅ 5 ... (4 n − 1)} 2 2 n 2 n! n! n! X Example 23. The value of when both 25
{1 ⋅ 3 ⋅ 5 ... (2 n − 1)}2 2 2 n (n!)2 2n! Cm
{1 ⋅ 3 ⋅ 5 ... (4 n − 1)} numerator and denominator have their greatest
= values, is
{1 ⋅ 3 ⋅ 5 ... (2 n − 1)}2
140 140 143 140
(a) (b) (c) (d)
X Example 20. The number of positive terms in 403 4025 4025 2045
n+3
195 P3 Sol. (c) We know that,
the sequence x n = − n +1
, n ∈ N are n C , if n is even
4 ⋅ Pn
n
Pn + 1 Greatest value of n C r = n n / 2
C( n + 1)/ 2 , if n is odd
(a) 2 (b) 3 20
∴ 20
C n is greatest for n = = 10
(c) 4 (d) None of these 2
n+ 3 25 + 1
P3 25
C m is greatest for m = = 13
Sol. (c) We have, xn = 195 − n+1
and
2
n
4 ⋅ Pn Pn + 1
For n = 10 and m = 13, we have
195 (n + 3) (n + 2 ) (n + 1) 195 (n + 3) (n + 2 ) 20
Cn 20
C 20! 12 ! × 13! 143
= − = − = 25 10 = × =
4 ⋅ n! (n + 1)! 4 ⋅ n! n! 25
Cm C13 10! × 10! 25! 4025
195 − 4 n2 − 20 n − 24 171 − 4 n2 − 20 n
= =
4 ⋅ n! 4 ⋅ n! X Example 24. (n!)! is divisible by
Q xn is positive. (a) 28(2n −1) n − 1 (b) (3n)!
171 − 4 n2 − 20 n
∴ > 0 ⇒ 4 n2 + 20 n − 171 < 0 (c) ( n!) n!
(d) ( n!) ( n − 1)!
4 ⋅ n!
which is true for n = 1, 2 , 3, 4. Sol. (d) Clearly, (n!)! is the product of natural numbers from 1
Hence, the given sequence has 4 positive terms. to n!.
∴ (n!)! = {1 × 2 × 3 × K × n} × {(n + 1) (n + 2 ) K(2 n)}
X Example 21. The value of × {(2 n + 1) (2 n + 2 )...(3 n)}
M M M
( 7 C 0 + 7C1 ) + ( 7 C1 + 7C 2 ) + ... + ( 7 C 6 + 7C 7 ) is × {(n! − n + 1)(n! − n + 2 ) (n! − n + 3)... n!} …(i)
(a) 2 8 − 2 (b) 2 8 − 1 (c) 2 8 + 1 (d) 2 8 We observe that,
Last term of first bracket on RHS of Eq. (i) is n.
Sol. (a) ( 7C 0 + 7C1 ) + ( 7C1 + 7C 2 ) + ... + ( 7C 6 + 7C 7 ) Last term of second bracket on RHS of Eq. (i) is 2 ⋅ n.
= 8C1 + 8C 2 + ... + 8C 7 Last term of third bracket on RHS of Eq. (i) is 3 ⋅ n and so on.
= 8C 0 + 8C1 + 8C 2 + ... + 8C 7 + 8C 8 − ( 8 C 0 + 8C 8 ) Last term of the last bracket on RHS of Eq. (i) is n! i.e.
n(n − 1)!. It is clear from above pattern that there are(n − 1)!
= 2 8 − (1 + 1) = 2 8 − 2
brackets on the RHS of Eq.(i) and in each bracket, there
is product of n consecutive natural numbers.
X Example 22. The value of n for which From Eq. (i), we have
n −1 5
C 4 − n − 1C 3 − ⋅ n − 2 P2 < 0, where n ∈ N , is ( n − 1)
4 (n!)! = ∏ [{r − 1}n + 1]{(r − 1) n + 2}
r =1
(a) {5, 6, 7, 8, 9, 10} (b) {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} {(r − 1) n + 3}…[(r − 1) n + n]
(c) (0, 2) ∪ (3, 11) (d) ( − ∞, 2) ∪ (3, 11) We know that, the product of n consecutive natural
5 n−2 numbers is divisible by n!.
Sol. (a) We have, n − 1C 4 − n −1
⋅C3 − P2 < 0
4 So, let
(n − 1)(n − 2 )) (n − 3)(n − 4) (n − 1) (n − 2 ) (n − 3) {(r − 1)n + 1}{(r − 1)n + 2}…{(r − 1)n + n} = n! Ir
⇒ −
4! 3! where, r = 1, 2,…,(n − 1)!
− 5/ 4 (n − 2 )(n − 3) < 0 ( n − 1)! ( n − 1)!
⇒
(n − 2 ) (n − 3)
{(n − 1) (n − 4) − 4 (n − 1) − 30} < 0
∴ (n!)! = ∏ n! Ir = (n!)n − 1! ∏ Ir
r =1 r =1
24
⇒ (n − 2 ) (n − 3) (n2 − 9n − 22 ) < 0 = (n!)( n − 1)! × A natural number
274 ⇒ (n − 2 ) (n − 3) (n − 11) (n + 2 ) < 0 Thus, (n!)! is divisible by (n!)( n − 1)! .
Work Book Exercise 6.3 6
Permutation and Combination
n −1
1 If 15 Pr = 14 P8 + 8 ⋅ 14 P7, then r is equal to 5 If Cr = (k 2 − 3) nCr + 1, then k belongs to
a 5 b 6 c 7 d 8 a [− 3, 3 ] b (−∞, −2)
c (2, ∞) d ( 3, 2 )
2 Which of the following is incorrect?
a n
C r = nC n − r b Cr =
n n +1
C r + nC n − r 6 If n is even and n C0 < nC1 < nC2 < ... < nCr
c Cr =
n n−1
Cr + n−1
C r − 1 d r ! nC r = nPr > nCr + 1 > nCr + 2 > ... > nCn , then r is equal to
n n−1 n−2 n+2
n+ 2
3 If C8 : n − 2 P4 = 57 : 16, then the value of n is a
2
b
2
c
2
d
2
a 20 b 19 c 18 d 17
7 The number of positive integers satisfying the
n+1
4 The exponent of 7 in 100
C50 is inequality Cn − 2 − n + 1Cn − 1 ≤ 100, are
a 0 b 2 a 9 b 8
c 4 d None of these c 5 d None of these
Some Basic Arrangements and Selections
Combinations
Each of the different selections made by taking We find that the sub-job J1 can be completed in
n
some or all of a number of distinct objects or items, C r ways and sub-job J 2 can be completed in
irrespective of their arrangements or order in which r! ways. Therefore, number of ways of arranging
they are placed is called a combination. n distinct items, taking r at a time
= n C r × r!
Permutations
Each of the different arrangements which can be Clearly, it is consistent with the result stated in
made by taking some or all of a number of point (1), because
distinct objects is called a permutation. n
C r × r ! = n ( n − 1) ( n − 2)... ( n − r − 1) = n Pr
1. Let r and n be positive integers such that 1 ≤ r ≤
n. Then, the number of all permutations of n X Example 25. Seven atheletes are participating
distinct items or objects taken r at a time is in a race. In how many ways can the first three
n ( n − 1) ( n − 2) ( n − 3) ... ( n − r − 1) prizes be won?
(a) 420 (b) 210
e.g. We have, n ( n − 1) ( n − 2) K ( n − r − 1 ) (c) 35 (d) 105
n ( n − 1) ( n − 2)... ( n − r − 1) ( n − r )!
= Sol. (b) The total number of ways in which first three prizes
( n − r )! can be won is equivalent to the number of arrangements
n! of seven different things taken 3 at a time
= = n Pr = 7P3 =
7!
=
7!
= 210
( n − r )! (7 − 3)! 4!
So, the total number of arrangements
(permutations) of n distinct items, taking r at X Example 26. How many numbers between 400
time is n Pr or P ( n , r ). and 1000 can be made with the digits 2, 3, 4, 5, 6
and 0?
2. The number of all permutations (arrangements)
(a) 120
of n distinct objects taken all at a time is n!.
(b) 720
3. The number of ways of selecting r items or objects
(c) 60
from a group of n distinct items or objects is
n! (d) None of the above
= nCr
( n − r )! r ! Sol. (c) Any number between 400 and 1000 will be three
digits.
e.g. Let there be n distinct objects which are to
Since, the number should be greater than 400, therefore
be arranged in a row by taking r at a time. hundred’s place can be filled up by any one of the
The job of arranging n items by taking r at a three digits 4, 5 and 6 in 3 ways. Remaining two places
time can be divided into two ordered can be filled up by remaining five digits in 5 P2 ways.
sub-jobs J1 and J 2 given by
∴ Required number = 3 × 5 P2
J1 : Selecting r items from n distinct items. 5!
J 2 : Arranging r selected items. =3× = 60 275
3!
(n − 2 )(n − 3)
6 X Example 27. The number of numbers lying
between 40000 and 70000 that can be made from
⇒
⇒
1⋅2
= 78
n2 − 5n + 6 = 156
Objective Mathematics Vol. 1
the digits 0, 1, 2, 4, 5, 7, if digits can be repeated
⇒ n2 − 5n − 150 = 0
in the same number, is
⇒ (n − 15) (n + 10) = 0
(a) 2591 (b) 932 ∴ n = 15 [Q n ≠ − 10 ]
(c) 766 (d) None of these
Sol. (a) The numbers are of five digits having 4 or 5 in ten
X Example 31. The maximum number of points
thousand’s place and the remaining digits are any of the of intersection of 8 circles are
given six digits. (a) 56 (b) 28
∴Number of ways to fill ten thousand’s place = 2 P1 = 2 (c) 24 (d) 16
Number of ways to fill other four places = 64 Sol. (a) Maximum number of points
∴Required number of numbers = 2 × 64 − 1 = 2591 = 8 P2 = 8 × 7 = 56
X Example 28. In a network of railways, a small X Example 32. On a railway route, there are
island has 15 stations. The number of different 15 stations. The number of tickets required in order
types of tickets to be printed for each class, if every that it may be possible to book a passenger from
station must have tickets for other station, are every station to every other, is
(a) 230 15! 15!
(a) (b)
(b) 210 13!2! 13!
(c) 340 15!
(d) None of the above (c)15! (d)
2!
Sol. (b) For each pair of stations, two different types of
tickets are required. Now, the number of selections of Sol. (a) Required number of tickets = 15C 2 = 15!
2 !13!
2 stations from 15 stations = 15 C 2 .
∴Required number of types of tickets X Example 33. If a denotes the number of
15!
= 2 ⋅ 15C 2 = 2 ⋅ = 15 × 14 = 210 permutations of ( x + 2) things taken all at a
2 !13!
time,b denotes the number of permutations of x
X Example 29. In a certain test, a i students gave things taken 11 at a time and c denotes the number
wrong answers to atleast i questions, where of permutations of ( x −11) things taken all at a
i =1, 2, 3,...,k. No student gave more than k wrong time such that a =182 bc, then the value of x is
answers. The total number of wrong answers given (a) 15 (b) 12
is (c) 10 (d) 19
(a) a1 + a 2 + ... + a k x+ 2
Sol. (b) We have, Px + = a ⇒ a = ( x + 2 )!
(b) a1 + a 2 + ... + a k − 1 2
(c) a1 + a 2 + K + a k + 1 P11 = b
x
Also,
x!
(d) None of the above ⇒ b=
( x − 11)!
Sol. (a) Total number of wrong answers and x − 11
Px − 11 = c
= 1 ⋅ (a1 − a2 ) + 2 ⋅ (a2 − a3 ) + K + (k −1) (ak − 1 − ak ) + kak
= a1 + a2 + a3 + K + ak ⇒ c = ( x − 11)!
Q a = 182 bc
Example 30. In a class tournament, the x!
X ∴ ( x + 2 )! = 182 ⋅ ⋅ ( x − 11)!
participants were to play one game with another, ( x − 11)!
two class players fell ill, having played 3 games ⇒ ( x + 2 ) ( x + 1) = 182 = 14 × 13
⇒ x + 1 = 13
each. If the total number of games played is 84, and x + 2 = 14
then the number of participants at the beginning, ∴ x = 12
was
(a) 22 (b) 15 X Example 34. In a certain test, there are n
(c) 17 (d) None of these questions. In this test, 2 n − i students gave wrong
answers to atleast i questions, where i =1, 2, 3, ..., n.
Sol. (b) Suppose, the two players did not play at all, so that
the remaining (n − 2 ) players played n−2
C 2 matches. If the total number of wrong answers given is
Since, these two players played 3 matches each, hence 2047, then n is equal to
the total number of matches (a) 10 (b) 11
276 = n − 2C 2 + 3 + 3 = 84 [given] (c) 12 (d) 13
Sol. (b) The number of students answering exactly
i (1 ≤ i ≤ n − 1) questions wrongly is 2 n − i − 2 n − i −1. The
number of students answering all n questions wrongly is 2 0 .
Method 3
Applicable only if the digits used are such that they 6
Permutation and Combination
Hence, the total number of wrong answer is have the same common difference. (Valid even, if the
n −1 digits are repeating)
∑ i (2
n−i
− 2 n − i −1 ) + n(2 0 ) = 2047
i =1 Writing all the numbers in ascending order of
⇒ 2 n −1 + 2 n − 2 + 2 n − 3 + K + 2 0 = 2047 magnitude,
⇒ 2 n − 1 = 2047 S = (13579 + 13597) + ... + 97513 + 97531
⇒ 2 = 2048 ⇒ 2 n = 211
n
S = (13579 + 99531) + (13597 + 97513) + ...
∴ n = 11 = (111110) 60 times = 6666600
n
X Example 35. The number of numbers of 9 S = ( l + L), where n = Number of numbers,
2
different non-zero digits such that all the digits in
the first four places are less than the digit in the l = Smallest, L = Largest
middle and all the digits in the last four places are X Example 36. The sum of all the numbers that
greater than that in the middle, is can be formed with the digits 2, 3, 4, 5 taken all at
(a) 2 (4!) (b) ( 4!) 2 a time, is equal to
(c) 8! (d) None of these (a) 93324
(b) 93328
Sol. (b) The required number of numbers = 4 P4 × 4 P4 = (4!)2 (c) 92324
(d) None of the above
Summation of Numbers Sol. (a) The total number of numbers formed with the digits
2, 3, 4, 5 taken all at a time = Number of arrangements of
(3 different ways) 4 digits taken all at a time = 4 P4 = 4! = 24
Sum of all the numbers greater than 10000 formed To find the sum of these 24 numbers, we will find the
by the digits 1, 3, 5, 7, 9, no digit being repeated. sum of digits at unit’s, ten’s, hundred’s and thousand’s
places in all these numbers.
Method 1 Consider the digits in the unit’s place in all these
numbers. Each of the digits 2, 3, 4, 5 occur in 3! i.e.
All possible numbers = 5! = 120
6 times in the unit’s place.
If one occupies the unit’s place, then total number So, total for the digits in the unit’s place in all the
= 24 numbers = (2 + 3 + 4 + 5) × 3! = 84.
Hence,1 enjoys unit’s place 24 times. Since, each of the digits 2, 3, 4, 5 occurs 3! times in
anyone of the remaining places. So, the sum of the
1
digits in the ten’s, hundred’s and thousand’s places in
all the numbers = (2 + 3 + 4 + 5) × 3! = 84
Similarly, 1 enjoys each place 24 times. Hence, the sum of all the numbers
Sum due to 1 = 1 × 24 (1 + 10 + 10 2 + 10 3 + 10 4 ) = 84 (100 + 101 + 102 + 103 ) = 93324
Similarly, sum due to the digit
Ø Sum of all the numbers that can be formed with n (non-zero)
3 = 3 × 24 (1 + 10 + 10 2 + 10 3 + 10 4 ) digits, when repetition is not allowed and each digit being used
M M M M M M M M (10 n − 1)
once = (n − 1)! (sum of digits)
Required total sum = 24 (1 + 10 + 10 + 10 3 + 10 4 )
2
(10 − 1)
(1 + 3 + 5 + 7 + 9) = n − 1! (sum of digits) × (111 … n times)
Method 2 X Example 37. The sum of all the numbers
greater than 10000 formed by using digits 0, 2, 4,
In 1st column, there are twenty four 1’s, twenty 6, 8, no digit repeated in any number, is equal to
four 3’s and so on and their sum = 24 × 25 = 600
(a) 5199960
Hence, add in vertical column normally, we get (b) 5209960
Sum = 6666600 (c) 5199980
5th 2nd 1st (d) 5299960
X X X X X
X X X X X Sol. (a) The total number of numbers formed by arranging
120 Number the digits 0, 2, 4, 6, 8 taken all at a time is 5! = 120. Some
of these numbers are less than 10000 and some are
X X X X X greater than 10000. Infact all those numbers having 0 at
666 6 6 0 0 = 6666600 ten thousand’s place are four-digit numbers. 277
6 ∴ Required sum = Sum of the numbers formed by using
digits 0, 2, 4, 6, 8 − Sum of the numbers
formed by using digits 2, 4, 6, 8
105 − 1
X Example 38. There are 5 boys and 3 girls. In
how many ways can they be seated in a row, so
that all the three girls do not sit together?
Objective Mathematics Vol. 1
= (0 + 2 + 4 + 6 + 8)(5 − 1)! (a) 8! (b) 8!5!
10 − 1
(c) 8! − 5! (d) 50 ⋅ 6!
104 − 1
− (2 + 4 + 6 + 8) (4 − 1)!
10 − 1
Sol. (d) Total number of persons = 5 + 3 = 8
When there is no restriction, then they can be seated in
105 − 1 104 − 1
= 480 − 120 a row in 8! ways.
10 − 1 10 − 1 But when all the three girls sit together, consider the
= 5333280 − 133320 = 5199960 three girls as one person, we have only (5 + 1) = 6
persons. These 6 persons can be arranged in a row in
6! ways. But the three girls can be arranged among
Permutations under Certain themselves in 3! ways.
∴Number of ways when three girls are together = 6! 3!
Conditions ∴Required number of ways in which all the three girls
The number of all permutations (arrangements) of n do not sit together
different objects taken r at a time, = 8! − 6! 3! = 6! (56 − 6) = 6! ⋅ 50 = 50 ⋅ 6!
1. when a particular object is to be always included X Example 39. In how many ways can 10
in each arrangement, is n − 1 C r − 1 × r !. examination paper be arranged, so that the best
2. when a particular object is never taken in each and the worst papers never come together?
arrangement, is n − 1 C r × r !. (a) 9 ⋅ 8! (b) 8 ⋅ 9!
3. when two specified objects always occur together (c)10 ⋅ 9! (d) 9 ⋅ 10!
is n − 2 C r − 2 × ( r − 1)! × 2!. Sol. (b) The number of permutation of 10 papers
when there is no restriction = 10!
Proof When the best and the worst papers come together,
1. In order to arrange n distinct items by taking r at consider the two as one paper, we have only 9 papers.
a time when a particular object is to be always These 9 papers can be arranged in 9! ways.
included in each arrangement, But these two papers can be arranged among
themselves in 2! ways.
M M M … M ∴Number of arrangement when the best and the worst
paper do not come together = 10! − 9! 2 ! = 8 ⋅ 9!
1st 2nd 3rd nth
Let us first put aside the specified item and select X Example 40. The number of ways in which a
( r −1) items from the remaining ( n −1) items. This committee of 5 can be chosen from 10 candidates
can be done in n − 1 C r − 1 ways. Now, we have r so as to exclude the youngest, if it includes the
items, namely, one specified item and ( r −1) oldest, is
selected items. These r items can be arranged in (a) 196
r! ways. Hence, the required number of
(b) 178
arrangements is n − 1 C r − 1 × r !.
(c) 202
2. If a particular objects is never taken in each (d) None of the above
arrangement, then we first select r objects from
the remaining ( n −1) objects and then they are Sol. (a) There are two different ways of forming the
committee:
arranged. So, required number of arrangements is
n −1 (i) oldest may be included
C r × r !.
(ii) oldest may be excluded
3. In order to arrange n distinct items by taking r at (i) If oldest is included, then youngest has to be
a time when two specified objects always occur excluded and we are to select 4 candidates out
together, we first select ( r − 2) objects from the of 8.
This can be done in
remaining ( n − 2) objects. This can be done in 8! 8×7 × 6× 5
n−2 8
C4 = = = 70 ways
C r − 2 ways. Now, we consider two specified 4! 4! 4× 3×2
objects temporarily as a single object and mix it (ii) If oldest is excluded, then we are to select 5
with ( r − 2) selected objects. Now, these ( r −1) candidates from 9 which can be done in
objects can be arranged in ( r −1)! ways. But two 9
C5 =
9!
=
9× 8×7 × 6
= 126 ways
specified objects can be put together in 2! ways. 5! 4! 4 × 3 × 2 × 1
Hence, required number of arrangements is Hence, the total number of ways in which
n−2 committee can be formed
278 C r − 2 × ( r − 1)! × 2! = 126 + 70 = 196
X Example 41. The number of ways in which the
letters of the word VOWEL can be arranged so that
Sol. (b) There are 8 chairs on each side of the table. Let the
sides be represented by A and B. Let four persons sit on
side A, then number of ways of arranging 4 persons on 8
6
Permutation and Combination
the letters O, E occupy only even places, is chairs on side A = 8 P4 and then two persons sit on side B.
(a) 12 (b) 18 The number of ways of arranging 2 persons on 8 chairs on
(c) 24 (d) None of these side B = 8 P2 and the remaining 10 persons can be
arranged in remaining 10 chairs in 10! ways.
Sol. (a) V O W E L
Hence, the total number of ways in which the person can
1 2 3 4 5
be arranged
× × 8! 8!10!
= 8 P4 × 8 P2 × 10! =
QO, E occupy only even places. 4! 6!
∴O and E can be arranged in two × marked places in
2
P2 ways = 2 ! = 2. X Example 45. The number of permutations of n
Also, the remaining 3 letters can be arranged in different things taken r at a time, in which
themselves in 3! ways = 3 × 2 × 1 = 6 p ( r ≤ n − p) particular things will never occur, is
∴Required number of words = 3! = 2 × 6 = 12 (a) P ( n − p, r )
X Example 42. The number of ways in which the (b) P ( n, r ) − P ( n, p)
letters of word FRACTION be arranged so that no (c) P ( n, r ) × P ( n, p)
two vowels are together, is (d) P ( n − p, r ) × P ( n, n − p)
(a) 17330 (b) 14400 Sol. (a) As, p things are excluded.
(c) 16440 (d) None of these
∴Number of things left = (n − p), out of which r are to be
Sol. (b) The word ‘FRACTION’ consists of 8 different letters arranged.
out of which A, I, O are 3 vowels and F, R, C, T, N are Now, the number of permutations = P(n − p, r )
5 consonants.
First of all, we arrange the consonants among X Example 46. The number of permutations of n
themselves.
different objects taken r ( ≥ 3) at a time which
This can be done in P (5, 5) = 5! = 120 ways.
include 3 particular objects, is
Let one such way be × F × R × C × T × N ×
Now, no two vowels are together, if they are put at
(a) n Pr ⋅ 3! (b) n Pr − 3 ⋅ 3!
places marked (× ). The 3 vowels can fill up these n−3 n−3
6 places in P (6, 3) ways.
(c) Pr − 3 ⋅ 3! (d) r P3 ⋅ Pr − 3
Hence, the total number of words Sol. (d) First, we arrange 3 particular things in r places.
= 120 × P(6, 3) = 120 × 6 × 5 × 4 = 120 × 120 = 14400 This can be done in r P3 ways. Then, remaining n − 3
things can be arranged taken r − 3 at a time in n − 3 Pr − 3
X Example 43. A boat is to be managed by eight
men of whom 2 can only row on bow side and 3 ways.
can only row on stroke side, the number of ways in ∴Total number of ways = r P3 ⋅n − 3 Pr −3
which the crew can be arranged, is X Example 47. We are required to form different
(a) 4360 (b) 5760
words with the help of the letters of the word
(c) 5930 (d) None of these INTEGER. Let m1 be the number of words in which
Sol. (b) First, we have to select 2 men for bow side and 3 I and N are never together and m2 be the number
for stroke side.
of words which begin with I and end with R, then
∴ The number of selections of the crew for two sides
m1
= 5C 2 × 3C 3 is given by
For each selection, there are 4 persons, each on both
m2
sides who can be arranged in 4! × 4! ways. 1
∴Required number of arrangements
(a) 42 (b) 30 (c) 6 (d)
30
5×4
= 5C 2 × 3C 3 × 4! × 4! = × 1 × 24 × 24 = 5760
2 Sol. (b) The number of words in which I and N are together
6!
X Example 44. A teaparty is arranged for =
× 2 ! = 6! = 720
2!
16 people along two sides of a large table with
∴The number of words in which I and N are never together,
8 chairs on each side. Four men want to sit on one 7!
particular side and two on the other side. The m1 = − 720 = 1800
2!
number of ways in which they can be seated, is The number of words which begin with I and end with R,
6!8!10! 8!8!10! 5!
(a) (b) m2 = = 60
4! 6! 4! 6! 2!
8! 8! 6! m1 1800
∴ = = 30
(c) (d) None of these m2 60
6! 4! 279
6 X Example 48. The number of six-digit numbers
that can be formed from the digits 1, 2, 3, 4, 5, 6
and 7, so that digits do not repeat and the terminal
X Example 50. The number of ways in which two
10 paise, two 20 paise, three 25 paise and one
50 paise coins can be distributed among 8 children,
Objective Mathematics Vol. 1
digits are even, is so that each child gets only one coin, is
(a) 144 (b) 72 (c) 288 (d) 720 (a) 1720 (b) 1680
(c) 1570 (d) None of these
Sol. (d) Terminal digits are the first and last digits.
Since, terminal digits are even. Sol. (b) Total number of coins = 2 + 2 + 3 + 1 = 8
∴1st place can be filled in 3 ways and last place can Q2 coins are of 10 paise, 2 coins are of 20 paise, 3 coins
be filled in 2 ways and remaining places can be filled in are of 25 paise and 1 coin is of 50 paise.
5
P4 = 120 ways. ∴Required number of ways
Hence, the number of six digit numbers, so that the 8! 8×7 × 6× 5× 4× 3×2 ×1
= =
terminal digits are even = 3 × 120 × 2 = 720. 2 ! × 2 ! × 3! × 1! 2 × 1 × 2 × 1 × 3 × 2 × 1 × 1
= 8 × 7 × 6 × 5 = 1680
Permutations of Objects not all Distinct
X Example 51. A boy has 3 library tickets and 9
1. The number of mutually distinguishable
books of his interest in the library. Out of these 8,
permutations of n things taken all at a time, of he does not want to borrow Chemistry part II,
which p are alike of one kind, q alike of second unless Chemistry part I is also borrowed. The
n! number of ways in which he can choose the three
kind such that p + q = n, is .
p! q ! books to be borrowed, is
2. The number of permutations of n things, of which (a) 41 (b) 32
p1 are alike of one kind, p2 are alike of second (c) 51 (d) None of these
kind, p3 are alike of third kind,..., pr are alike of Sol. (a) The following are the different possibilities in which
rth kind such that three books can be borrowed.
n! (i) When Chemistry part II is selected, then Chemistry
p1 + p2 + ... + pr = n, is . part I is also borrowed and the third book is
p1 ! p2 ! p3 !... pr ! selected from the remaining 6 books = C(6, 1) = 6.
3. The number of permutations of n things, of which (ii) When Chemistry part II is not selected, in this case,
he has to select the three books from the remaining
p are alike of one kind, q are alike of second kind 7 books. First choice can be made in
n! 7 ×6×5
and remaining all are distinct, is . C(7, 3) = = 35 ways.
p! q ! 1× 2 × 3
4. Suppose there are r things to be arranged, ∴ Total number of ways in which he can choose the
three books to be borrowed = 6 + 35 = 41.
allowing repetitions. Let further p1 , p2 , ..., pr be
the integers such that the first object occurs X Example 52. There are three copies, each of 4
exactly p1 times, the second occurs exactly p2 different books. The number of ways in which they
can be arranged on a shelf, is
times, etc. Then, the total number of permutations
12! 11!
of these r objects to the above condition is (a) 4
(b)
( p1 + p2 + ... + pr )! (3!) (3!) 2
p1 ! p2 ! p3 !... pr ! 9!
(c) (d) None of these
(3!) 2
X Example 49. The number of arrangements
which can be made out of the letters of the word Sol. (a) Total number of books = 3 × 4 = 12 in which each
ALGEBRA, without changing the relative order of 4 different books is repeated 3 times.
(positions) of vowels and consonants, is Hence, the required number of arrangements
(a) 72 (b) 54 12 ! 12 !
= =
(c) 36 (d) 62 3! × 3! × 3! × 3! (3!)4
Sol. (a) We have to arrange the 3 vowels in their places, i.e. X Example 53. A library has a copies of one
in the 1st, 4th, 7th places and the 4 consonants in their book, b copies of two books, c copies of each of
places. The number of ways to arrange 3 vowels in 1st, 4th three books and single copy of d books. The total
3! number of ways in which these books can be
and 7th places = [Qthere are 2A’s]
2! distributed, is
The number of ways to arrange 4 consonants in their ( a + 2b + 3c + d )! ( a + b + c + d )!
places = 4!
(a) (b)
a ! b! c! a ! b! c!
∴Required number of arrangements ( a + 2b + 3c + d )!
=
3!
× 4! = 3 × 24 = 72 (c) (d) None of these
280 2! a ! ( b!) 2 ( c!) 3
a6
Sol. (c) Total number of books = a + 2 b + 3c + d
Since, there are b copies each of two books, c copies
each of three books and single copy of d books.
a5
6
Permutation and Combination
∴Total number of arrangements an–2 a4
(a + 2 b + 3c + d )!
=
a! (b !)2 (c !)3 an–1 a3
an a2
X Example 54. How many different a1
arrangements can be made out of the letters in the Clearly, this circular permutation provides n linear
expansion A 2 B 3C 4 , when written in full? permutations as given below:
9! 9! a1 , a 2 , a 3 , ..., a n − 1 , a n
(a) (b)
2! + 3! + 4! 2 ! 3 ! 4! a 2 , a 3 , a 4 , ..., a n , a1
(c) 2! + 3! + 4!(2!3! 4!) (d) 2!3! − 4 a 3 , a 4 , a 5 , ..., a n , a1 , a 2
a 4 , a 5 , a 6 , ..., a n , a1 , a 2 , a 3
Sol. (b) A 2 B3C 4 written in full is AA BBB CCCC. … … … …
9!
∴Required number of ways = … … … …
2 ! 3! 4! a n , a1 , a 2 , a 3 , ..., a n − 1
X Example 55. The number of ways in which the Thus, each circular permutation gives n linear
six faces of a cube is painted with six different permutations. But there are x circular permutations, so
colours, is that number of linear permutations is xn. But the number
(a) 6 of linear permutations of n distinct objects is n!.
(b) 6! n!
∴ xn = n! ⇒ x = = ( n − 1)!
(c) 6 C 2 n
(d) None of the above Ø In the above theorem, anti-clockwise and clockwise orders of
Sol. (d) Number of ways = 6! = 1 [all faces are alike] arrangements are considered as distinct permutations.
6!
X Example 56. The number of numbers greater
Difference between Clockwise and
than a million that can be formed with the digits Anti-clockwise Arrangements
2, 3, 0, 3, 4, 2 and 3, is Consider the following circular permutations:
(a) 360 (b) 340 We observe that in both the order of the circular
(c) 370 (d) None of these arrangements is a1 , a 2 , a 3 a 4 . In Fig. (i), the order is
Sol. (a) Any number greater than a million must be of 7 or anti-clockwise whereas in Fig. (ii), the order is
more than 7 digits. Here, number of given digits is seven, clockwise.
therefore we have to form numbers of seven digits only. a3 a3
Now, there are seven digits of which 3 occurs thrice and
two occurs twice.
7! a4 a2 a2 a4
∴Number of numbers formed = = 420
2 ! 3!
But this also includes those numbers of seven digits,
whose first digit is zero and so infact, they are only six a1 a1
digit numbers. (i) (ii)
Number of numbers of seven digits having zero in the
6! Thus, the number of circular permutation of n
first place = 1 × = 60 things in which clockwise and anti-clockwise
3!2 !
Hence, required number = 420 − 60 = 360
arrangements give rise to different permutations is
( n −1)!. e.g. The number of permutations of 5 persons
seated around a table is (5 − 1)! = 4!, because with
Circular Permutations respect to the table, the clockwise and anti-clockwise
The number of circular permutations of n distinct arrangements are distinct.
objects is ( n −1)!. If anti-clockwise and clockwise orders of
Proof Let a1 , a 2 , a 3 ,…,a n − 1 , a n be n distinct arrangements are not distincts, then the number of
1
objects. circular permutations of n distinct items is {( n − 1)!}.
2
Let the total number of circular permutations be x.
e.g. Arrangements of beads in a necklace, arrangements
Consider one of these x permutations as shown in figure.
of flowers in a garland. 281
6 X Example 57. The Chief Minister of 11 states
of India meet to discuss the language problem. The
Sol. (c) The two particular persons can be arranged among
themselves in 2 P2 ways i.e. 2! ways.
Taking them as one person and keeping him fixed, we
Objective Mathematics Vol. 1
number of ways they can seat themselves at a
can arrange the remaining 5 persons among
round table so that Punjab and Chennai Chief themselves in 5! ways.
Ministers sit together, is Hence, the required number of ways of which
(a) 9! × 2! 2 particular persons come together
(b) 9! = 5! × 2 ! = 120 × 2 = 240
(c)10!
(d) None of the above
X Example 60. Three boys and three girls are to
be seated around a table, in a circle. Among them,
Sol. (a) Treat the Punjab and Chennai Chief Ministers as the boy X does not want any girl neighbour and
one (P, C) + 9 others.
the girl Y does not want any boy neighbour. The
So, we have to arrange 10 persons around round table.
number of such arrangements possible is
This can be done in (10 − 1)! = 9! ways. Corresponding to
each of these 9! ways, Punjab and Chennai Chief
(a) 4
Ministers can interchange their places in 2! ways. (b) 6
∴ Associating the two operations, total number of ways (c) 8
= 9! × 2 !. (d) None of the above
X Example 58. The number of ways in which Sol. (a) As shown in figure, 1, 2 and X are the three boys
n things of which r are alike, can be arranged in a and 3, 4 and Y are three girls. Boy X will have neighbours
circular order, is as boys 1 and 2 and the girl Y will have neighbours as girls
( n −1)! 3 and 4.
(a) X
r!
n!
(b)
r! 2
1
n!
(c)
( r −1)!
(d) None of the above 3 4
Sol. (a) Let the required number of arrangements = x
Y
Consider anyone of these x arrangements.
If in this arrangements, r alike things are replaced by 1 and 2 can be arranged in
r different things, then these r things can be arranged P(2, 2 ) = 2 ! = 2 × 1 = 2 ways
among themselves in r! ways.
Also, 3 and 4 can be arranged in
Since, one arrangement gives rise to r! arrangements.
P(2, 2 ) = 2 ! = 2 × 1 = 2 ways
∴x arrangements will give rise to xr! arrangements.
But if all the n things are different, then the number of Hence, required number of permutations
circular arrangements = 2 ×2 = 4
= (n − 1)!
Thus, x ⋅ r ! = (n − 1)! X Example 61. There are n seats round a table
(n − 1)! numbered 1, 2, 3,..., n. The number of ways in
Hence, x=
r! which m ( ≤ n) persons can take seats, is
(a) n Pm (b) n C m ⋅ ( m − 1)!
X Example 59. The number of ways in which
1
7 people can be arranged at a round table, so that (c) ⋅ n Pm (d) n − 1 Pm
2 particular persons may be together, is 2
(a) 132 Sol. (a) Since, the seats are numbered.
(b) 148 So, the arrangement is not circular.
(c) 240 Hence, the required number of ways = The number of
(d) None of the above arrangements of n things taken m at a time = n Pm
282
Work Book Exercise 6.4 6
Permutation and Combination
1 If the letters of the word ‘VARUN’ are written in all 10 In a unique hockey series between India and
possible ways and then are arranged as in a Pakistan, they decide to play on till a team wins 5
dictionary, then the rank of the word ‘VARUN’ is matches. The number of ways in which the series
98 99 100 101 can be won by India, if no match ends in a draw, is
126 252
2 5 Indian and 5 American couples meet at a party 225 None of these
and shake hands. If no wife shakes hands with
her own husband and no Indian wife shakes 11 The number of ways in which n things of which r
hands with a male, then the number of hand alike and the rest different can be arranged in a
shakes that takes place in the party, is circle distinguishing between clockwise and
95 110 135 150 anti-clockwise arrangement, is
( n + r − 1)! ( n − 1) !
3 If m denotes the number of 5-digit numbers, if r! r
each successive digits are in their descending ( n − 1) ! ( n − 1) !
order of magnitude and n is the corresponding ( r − 1) ! r!
figure, when the digits are in their ascending
order of magnitude, then (m − n ) has the value 12 A gentleman invites a party of m + n (m ≠ n )
10
C4 9
C5 10
C3 9
C3 friends to dinner and places m at one table T1 and
n at another table T2 , the table being round. If not
4 The number of ways in which 9 different prizes be all people shall have the same neighbour in any
given to 5 students, if one particular boy receives two arrangement, then the number of ways in
4 prizes and the rest of the students can get any which he can arrange the guests, is
numbers of prizes, is ( m + n) ! 1 ( m + n) !
⋅
9
C 4 ⋅ 2 10 9
C 5 ⋅ 54 4mn 2 mn
( m + n) !
4 ⋅ 45 None of these 2⋅ None of these
mn
5 The number of ways in which 8 non-identical
apples can be distributed among 3 boys such 13 There are 12 guests at a dinner party. Supposing
that every boy should get atleast 1 apple and that the master and mistress of the house have
atmost 4 apples is k ⋅ 7P3 , where k has the value fixed seats opposite one another and that there
are two specified guests who must always, be
equal to
placed next to one another, the number of ways
88 66 44 22
in which the company can be placed, is
6 A rack has 5 different pairs of shoes. The number 20 ⋅ 10 !
of ways in which 4 shoes can be chosen from it 22 ⋅ 10 !
so that there will be no complete pair, is 44 ⋅ 10 !
None of the above
1920 200
110 80 14 The number of signals that can be generated by
7 The number of ways in which 7 people can using 6 differently coloured flags, when any
occupy six seats, 3 seats on each side in a first number of them may be hoisted at a time, is
class railway compartment, if two specified 1956 1957
1958 1959
persons are to be always included and occupy
adjacent seats on the same side, is (5 !)⋅ k, then k 15 The sum of all five-digit numbers that can be
has the value equal to formed using the digits 1, 2, 3, 4, 5 when
2 4 repetition of digits is not allowed, is
8 None of these 366000 660000
360000 3999960
8 The number of different ways in which five
‘dashes’ and eight ‘dots’ can be arranged, using 16 By using the digits 0, 1, 2, 3, 4 and 5 (repetitions
only seven of these 13 ‘dashes’ and ‘dots’, is not allowed), any number of digits being used,
1287 119 the number of non-zero numbers that can be
120 1235520 formed, is
9 There are n identical red balls and m identical 1030 1630 1200 1530
green balls. The number of different linear 17 The total number of numbers of not more than
arrangements consisting of n red balls but not 20 digits that are formed by using the digits 0, 1,
necessarily all the green balls is x Cy , then 2, 3 and 4, is
x = m+ n, y = m 5 20
x = m+ n + 1, y = m 5 20 − 1
x = m+ n + 1, y = m + 1 5 20 + 1
None of the above 283
x = m+ n, y = n
6 18 Six boys and six girls sit along a line alternately in
x ways and along a circle (again alternatively in y
ways), then
24 The number of ways in which we can select four
numbers from 1 to 30, so as to exclude every
selection of four consecutive numbers, is
Objective Mathematics Vol. 1
x= y y = 12 x x = 10 y x = 12 y 27378 27405
27399 None of these
19 The number of times the digit 3 will be written
when listing the integers from 1 to 1000, is 25 The number of ways in which 10 candidates
269 300 271 302 A1, A 2 , …, A 10 can be ranked, so that A 1 is
always above A 2 , is
20 The number of ways in which a mixed double
10!
game can be arranged from amongst 9 married 10 !
couples, if no husband and wife play in the same 2
game, is 9!
756 1512 None of the above
3024 None of these
26 There are n white and n black balls marked
21 The number of different seven digit-numbers that 1, 2, 3, …, n. The number of ways in which we
can be written using only the three digits 1, 2 and can arrange these balls in a row, so that
3 with the condition that the digit 2 occurs twice neighbouring balls are of different colours, is
in each number, is n! (2 n)!
2
P2 ⋅ 2 5 7
C2 ⋅25 2
2( n!)
(2 n)!
7
C 2 ⋅ 52 None of these ( n!)2
22 The number of ways one can put 5 different balls 27 A man invites in a party of (m + n ) friends to
in 5 different boxes such that atmost three boxes dinner and places m at one round table and n at
is empty, is another. The number of ways of arranging the
3000 3010 2990 3120 guests is
( m + n)!
23 Given that n is odd, the number of ways in which m! n !
three numbers in AP can be selected from ( m + n)!
1, 2, 3, …, n, is ( m − 1)!( n − 1)!
( n − 1)2 ( n + 1)2 ( n + 1)2 ( n − 1)2 ( m − n)!( n − 1)!
2 4 2 4 None of the above
Geometrical Applications of n Cr
Some basic Geometrical Applications on n C r are as 8. Number of parallelogram formed by two system
follow: of parallel lines (when 1st set contains m parallel
1. Out of n non-concurrent and non-parallel straight lines and 2nd set contains n parallel lines)
lines, points of intersection are n C 2 . = n C 2 × mC 2
2. Out of n points, the number of straight lines 9. Number of squares formed by two system of
(when three points are not collinear) are n C 2 . parallel lines in which 1st set is perpendicular
3. If out of n points, m are collinear, then 2nd sets of lines (when 1st set contains m parallel
lines and 2nd set contains n parallel lines)
Number of straight lines = n C 2 − mC 2 + 1 m−1
4. In a polygon, total number of diagonals out of = ∑ (m − r ) (n − r ); m < n
n points (when three points are not collinear) r =1
n ( n − 3)
= X Example 62. The number of parallelograms
2
that can be formed from a set of four parallel lines
5. Number of triangles formed from n points (when
intersecting another set of three parallel lines, is
three points are not collinear) are n C 3 . (a) 6 (b) 18
6. Number of triangles out of n points in which m (c) 12 (d) 9
are collinear, are n C 3 − mC 3 . Sol. (b) Required number of parallelograms
4! 3!
7. Number of triangles that can be formed out of = 4C 2 × 3C 2 = ×
2 ! 2 ! 2 ! 1!
n points (when none of the side is common to the 4×3 3
= × = 18
284 sides of polygon) are n C 3 − n C1 − n C1 ⋅ n − 4C1 . 2 ×1 1
X Example 63. The number of diagonals that
can be drawn by joining the vertices of an octagon,
Sol. (b) 4 lines intersect each other in 4 C 2 = 6 points and
4 circles intersect in 4 P2 = 12 points.
Each line cuts 4 circles into 8 points.
6
Permutation and Combination
is
∴ 4 lines cut 4 circles into 32 points.
(a) 28 (b) 48
∴ Required number = 6 + 12 + 32
(c) 20 (d) None of these = 50
Sol. (c) The total number of lines obtained by joining
8 vertices of octagon X Example 67. The straight lines I 1 , I 2 , I 3 are
= 8C 2 = 28 ways parallel and lie in the same plane. A total number of
Out of these 28 lines, 8 are sides and remaining m points are taken on I 1 , n points on I 2 , k points on
diagonals. I 3 . The maximum number of triangles formed with
Hence, the number of diagonals vertices at these points, are
= 28 − 8 = 20
(a) m + n + k C 3
m+ n + k
X Example 64. There are n concurrent lines and (b) C 3 − mC 3 − n C 3 − k C 3
another line parallel to one of them. The number of (c) m C 3 + n C 3 + k C 3
different triangles that will be formed by the ( n +1)
lines, is (d) None of the above
( n −1) n ( n − 1)( n − 2) Sol. (b) Total number of points are m + n + k. The triangles
(a) (b) formed by these points = m+ n+ k
C3 .
2 2
n( n +1) ( n + 1)( n + 2) Joining 3 points on the same line gives no triangle,
(c) (d) such triangles are m C 3 + nC 3 + kC 3 .
2 2
∴ Required number of triangles
Sol. (b) The number of triangles = Number of selections of 2 = m + n + kC 3 − mC 3 − nC 3 − kC 3
lines from the (n − 1) concurrent lines which are cut by the
another line
(n − 1)! X Example 68. Two straight lines intersect at a
= n − 1C 2 =
2 ! (n − 3)! point O. Points A1 , A2 , …, An are taken on one
(n − 1) (n − 2 ) line and points B1 , B 2 , …, B n on the other. If the
=
2 point O is not to be used, then the number of
triangles that can be drawn using these points as
X Example 65. In a plane, there are two sets of
vertices, is
parallel lines, one of m lines and the other of
n lines. If the lines of one set cut those of the other, (a) n ( n −1)
then the number of different parallelograms that (b) n ( n −1) 2
can be formed is (c) n 2 ( n − 1)
mn ( m − 1) ( n − 1) (d) n 2 ( n − 1) 2
(a)
2
m ( m − 1)( n − 1) Sol. (c) Number of triangles = 2n
C 3 − nC 3 − nC 3
(b) 2 n(2 n − 1)(2 n − 2 ) 2 n(n − 1)(n − 2 )
6 = −
6 6
nm ( m − 1) ( n − 1) 1
(c) = n(n − 1)(3n) = n (n − 1)
2
4 3
(d) None of the above
Sol. (c) The number of parallelograms = Number of
X Example 69. There are 10 points in a plane, of
selections of 4 lines, two from each set which no 3 points are collinear and 4 points are
Q parallelogram has two concyclic. The number of different circles that can
sets of parallel sides
be drawn through atleast 3 of these points, is
n ( n − 1) m (m − 1) (a) 116
= nC 2 × mC 2 = ×
2 2 (b) 120
mn (m − 1)(n − 1)
= (c) 117
4
(d) None of the above
X Example 66. The maximum number of points Sol. (c) The required number of circles
into which 4 circles and 4 straight lines intersect, is = (10 C 3 − 4C 3 ) + 1
(a) 26 (b) 50 (c) 56 (d) 72 = 117
285
6 Work Book Exercise 6.5
m+ n−2 m+ n−2
Objective Mathematics Vol. 1
1 The interior angles of a regular polygon measure
150° each. The number of diagonals of the m+ n m+ n−1
polygon is m+ n−2 m ( n − 1)
35 44 m+ n+2 ( m + 1) ( n + 1)
54 78
4 Let there be 9 fixed points on the circumference
2 18 points are indicated on the perimeter of a of a circle. Each of these points is joined to every
∆ ABC (see figure). How many triangles are there one of the remaining 8 points by a straight line
with vertices at these points? and the points are so positioned on the
circumference that atmost 2 straight lines meet in
A
any interior point of the circle. The number of
such interior intersection points is
126
351
756
None of the above
5 The greatest possible number of points of
B C
intersection of 9 different straight lines and
331 408 710 711 9 different circles in a plane, is
117
3 There are m points on a straight line AB and 153
n points on the line AC, none of them being the 270
point A. Triangles are formed with these points as None of the above
vertices, when
6 The number of triangles whose vertices are at the
(i) A is excluded. vertices of an octagon but none of whose sides
(ii) A is included.
happen to come from the sides of the octagon, is
Then, the ratio of number of triangles in the two 24 52
cases is 48 16
Selection of One or More Objects
Selection from Different Items X Example 71. A man has 6 friends. In how many
The number of ways of selecting one or more items ways he can invite one or more of them at dinner?
from a group of n distinct items is 2 n − 1. (a) 64 (b) 65 (c) 63 (d) 61
Out of n items, 1 item can be selected in n C1 ways, Sol. (c) The man has to select some or all of his 6 friends.
2 items can be selected in n C 2 ways, 3 items can be So, required number of ways = 2 6 − 1 = 63
selected in n C 3 ways and so on.
Ø ● The number of ways of selecting r items out of n identical
Hence, the required number of ways items is 1.
= n C1 + n C 2 + n C 3 + K + n C n ● The total number of ways of selecting zero or more i.e. items
from a group of n identical items is (n + 1) .
= ( n C 0 + n C1 + n C 2 + K + n C n ) − n C 0 = 2 n − 1 ● The total number of selections of one or more out of
p + q + r items, where p are alike of one kind, q are alike of
X Example 70. Nishi has 5 coins, each of the second kind and rest are alike of third kind is
different denomination. The number of different [(p + 1)(q + 1)(r + 1)] − 1.
sums of money, she can form, is
(a) 32 Selection from Identical and Distinct
(b) 25 Objects
(c) 31 The total number of ways of selecting one or more
(d) None of the above items from p identical items of one kind, q identical
Sol. (c) Required number of ways items of second kind, r identical items of third kind and
= 5C1 + 5C 2 + 5C 3 + 5C 4 + 5C 5 n different items is
= 2 5 − 1 = 31 ( p + 1) ( q + 1) ( r + 1) 2 n − 1
286
Number of Divisors and the Sum
of the Divisors of a Given
Sol. (b) We have,
2160 = 2 4 × 33 × 51
The total number of divisors is same as the number of
6
Permutation and Combination
ways selecting same or all out of four 2’s, three 3’s and
Natural Number one 5’s.
The number of ways
n n n n
Let n = p1 1 ⋅ p2 2 ⋅ p3 3 K pk k …(i) = (4 + 1)(3 + 1)(1 + 1) − 1 = 39
where, p1 , p2 , …, pk are distinct prime numbers X Example 73. The total number of proper
and n1 , n2 , …, nk are positive integers.
factors of 7875 is
Clearly, any divisor of n is of the form
m m m m (a) 23
d = p1 1 ⋅ p2 2 ⋅ p3 3 … pk k …(ii) (b) 24
where, m1 , m2 , …, mk are natural numbers such (c) 22
that (d) 21
0 ≤ mi ≤ ni , i =1, 2, 3, …, k Sol. (c) We have,
Therefore, the total number of divisors of Eq. (i) 7875 = 32 × 53 × 71
will be equal to the number of ways of selecting atleast The total number of factor
one from n1 identical primes p1 , n2 identical primes p2 = (2 + 1)(3 + 1)(1 + 1) − 1 = 23
and so on, finally nk identical primes pk . The number But this includes the given number itself.
of such ways is ∴ Number of proper factors = 23 − 1 = 22
( n1 + 1) ( n2 + 1) K ( nk + 1)
Hence, the total number of divisors of
X Example 74. The number of factors (excluding
n n n n 1 and the expression itself) of the product a 7 b 4 c 3
n = p1 1 ⋅ p2 2 ⋅ p3 3 K pk k is
def , where a, b, c, d, e and f are all prime
( n1 + 1) ( n2 + 1) ( n3 + 1) K ( nk + 1) numbers, is
This includes 1 and n as divisors. Therefore, (a)1279
number of divisors other than 1 and n is (b)1278
( n1 + 1) ( n2 + 1) ( n3 + 1) K ( nk + 1) − 2 (c)1280
The sum of all divisors of Eq.(i) is given by (d)1277
n1 n2 n3 nk Sol. (a) The total number of factors of the product a7b 4c 3def
∑ ∑ ∑ K ∑ r
p11 ⋅
r
p22 ⋅
r
p33 K
r
pkk is equal to number of ways of selecting atleast one from
r1 = 0 r2 = 0 r3 = 0 rk = 0 seven a’s, four b’s, three c’s, d, e, f are different.
p n1 + 1 − 1 p2n2 + 1 − 1 p3n3 + 1 − 1 ∴ The number of ways
= 1 = (7 + 1)(4 + 1)(3 + 1)2 3 − 1 = 1279
p1 − 1 p2 − 1 p3 − 1
X Example 75. The number of even divisors of
pknk + 1 − 1
K 10800 is
pk − 1 (a) 12 (b) 24 (c) 36 (d) 48
The number of ways of expressing n as a product of two Sol. (d) We have,
natural numbers 10800 = 2 4 × 33 × 52
1 For a divisor to be even, we must have atleast one
(n1 + 1)(n2 + 1)…(nk + 1), if nis not perfect square 2 out of four 2’s any number of 3’s and 5’s.
= 2 ∴ Total number of even divisors
1
[(n1 + 1)(n2 + 1)…(nk + 1) + 1], if nis perfect square = 4(3 + 1)(2 + 1) = 48
2
e.g. To find number of ways in which the number 94864 can X Example 76. The sum of divisors of 2520 is
be resolved as a product of two factors. (a) 9360 (b) 5040
94864 = 24 × 7 2 × 112 (c) 9600 (d) None of these
So, 94864 is a perfect square.
1 Sol. (a) We have,
Hence, the number of ways = [(4 + 1)(2 + 1)(2 + 1) + 1] = 23 2520 = 2 3 × 32 × 5 × 7
2
The sum of divisors of 2520
X Example 72. The total number of factors = (1 + 2 + 2 2 + 2 3 )(1 + 3 + 32 )(1 + 5)(1 + 7 )
(excluding 1) of 2160 is = 15 × 13 × 6 × 8
(a) 40 (b) 39 (c) 41 (d) 38 = 9360 287
6 Division of Objects into Groups The number of ways in which mn different items
can be divided equally into m groups, each containing n
objects and the order of groups is important, is
Division of Items into Groups of Unequal
Objective Mathematics Vol. 1
( mn)! 1 ( mn)!
Size × m! =
( n!) m m! ( n!) m
Number of ways in which ( m + n) distinct items
can be divided into two unequal groups containing m
( m + n)! X Example 78. The number of ways in which a
and n items is . pack of 52 cards can be divided equally into four
m! n!
groups, is
Proof 52! 52!
(a) (b)
The number of ways in which ( m + n) items are (13!) 4
(13!) 4 × 4!
divided into two groups containing m and n items is 52!
same as the number of combinations of ( m + n) things (c)
(13!) × 4!
(d) None of these
taken m at a time. Thus, the required number
( m + n)! Sol. (b) Here, 52 cards are to be divided in four equal
= m + nCm = . groups and order of the group is not important.
m! n! 52 !
So, required number of ways =
(13!)4 × 4!
Ø ● The number of ways in which (m + n + p) items can be
divided into unequal groups containing m, n, p items, is X Example 79. The number of ways 12 different
m+ n + p (m + n + p) !
C m⋅ n + pC n = books can be distributed equally among 4 persons,
m !n ! p ! is
● The number of ways to distribute (m + n + p) items among 12! 12! 12! 12!
3 persons in the groups containing m, n and p items (a) (b) (c) (d)
( 4!) 3
(3!) × 3
4
(3!) 4
( 4!) 4
= (Number of ways to divide) × (Number of groups)!
(m + n + p) ! Sol. (c) Here, 12 different books can be divided equally
= ×3! among 4 persons that each person will get 3 books.
m !n ! p !
Hence, required number of distribution
12 !
X Example 77. The number of ways to 16 = 12C 3 ⋅ 9C 3 ⋅ 6C 3 ⋅ 3C 3 =
(3!)4
different things to three persons A, B, C so that B
gets one more than A and C gets two more than Division of Identical Objects into Groups
B, is
(a)
16 !
(b)
16! i. The total number of ways of dividing n
4 !5 ! 7 ! 3 !5 !8 ! identical items among r persons, each one of
(c) 4 !5 ! 7 ! (d) None of these whom, can receive 0, 1, 2 or more items ( ≤ n)
is n + r − 1 C r − 1 .
Sol. (a) Suppose, A gets x things, then B gets ( x + 1) and C
gets ( x + 3) things. Or
∴ x + x + 1 + x + 3 = 16 The total number of ways of dividing n
⇒ x=4 identical objects into r groups, if blank groups
Thus, we have to distribute 16 things to A, B and C in are allowed, is n + r − 1 C r − 1 .
such that A gets 4, B gets 5 and C gets 7 things.
(4 + 5 + 7 )!
∴Required number of ways = X Example 80. The total number of ways can
4! 5! 7 !
16! 15 identical Mathematics books be distributed
=
4! 5! 7 ! among six students, is
15! 20!
(a) (b)
Division of Objects into Groups of 6! 9! 5!15!
Equal Size (c) 15P6 (d) None of these
The number of ways in which mn different objects Sol. (b) We know that the total number of ways in which
can be divided equally into m groups, each containing n things all alike can be distributed into r different boxes is
n + r −1
C r − 1, when blank box is allowed. Hence, number of
n objects and the order of groups is not important, is
ways in which 15 identical books be distributed among six
( mn)! 1 students
20!
( n!) m m! = 15 + 6 − 1C 6 − 1 = 20C 5 =
288 5!15!
ii. The total number of ways of dividing n
identical items among r persons, each one of
If r (0 ≤ r ≤ n) objects occupy the places assigned to
them i.e. their original places and none of the remaining
( n − r ) objects occupies its original places, then the
6
Permutation and Combination
whom, receives atleast one item is n − 1 C r − 1 .
number of such ways is D (n − r ) = n C r ⋅ D ( n − r )
Or 1 1 1 1
= nC r ⋅(n − r )! 1 − + − + K + (−1)n − r
The number of ways in which n identical items 1! 2 ! 3! (n − r )!
can be divided into r groups such that blank
groups are not allowed is n − 1 C r − 1 .
X Example 83. There are four balls of different
colours and four boxes of colours, same as those of
the balls. The number of ways in which the balls,
X Example 81. The number of ways in which
one each in a box, could be placed such that a ball
10 identical balls can be distributed in 4 distinct
does not go to a box of its own colour is
boxes, so that none of the boxes remain empty, is
(a) 9
(a) 10 C 4 (b) 10 P4
(b) 13
(c) 9 P3 (d) None of these (c) 8
Sol. (d) The required number of ways of distributing (d) None of the above
10 identical items to three persons such that each person Sol. (a) The required number of ways
receives atleast one item.
= 4! 1 − + +
1 1 1 1
−
So, required number of ways 1! 2 ! 3! 4!
9!
= 10 − 1C 4 − 1 = 9C 3 = = 84 = 12 − 4 + 1 = 9
3! 6!
iii. The number of ways in which n identical X Example 84. The number of ways can 10 letters
items can be divided into r groups so that no be placed in 10 marked envelopes, so that no letter
group contains less than m items and more than is in the right envelope, are
k ( m < k ) items is coefficient of x n in the 1 1 1 1
(a)10! 1 − + − + K +
expansion of ( x m + x m + 1 + K + x k ) r . 1! 2! 3! 10!
1 1 1 1
X Example 82. Five distinct letters are to be (b)10! 1 + − + − K −
1! 2 ! 3 ! 10!
transmitted through a communication channel. A
1 1 1 1
total number of 15 blanks is to be inserted between (c) 1 + − + − K −
the letters with atleast three between every two. 1! 2! 3! 10!
The number of ways in which this can be done, is 1 1 1 1
(d) 9! 1 + − + − K −
(a) 1200 (b) 1800 1! 2! 3! 10!
(c) 2400 (d) 3000
Sol. (a) The required number of ways is equal to the
Sol. (c) For 1 ≤ i ≤ 4, let xi ( ≥ 3) be the number of blanks number of dearrangements of 10 objects.
between ith and (i + 1) th letters. Then, Hence, required number of ways
x1 + x2 + x3 + x4 = 15 …(i)
= 10! 1 − +
1 1 1 1
− + K+
The number of solutions of Eq. (i) 1! 2 ! 3! 10!
= Coefficient of t 15 in (t 3 + t 4 + K )4
= Coefficient of t 3 in (1 − t )−4 X Example 85. Ajay writes letters to his five
= Coefficient of t in (1 + C1 + C 2 t + C 3 t + K )
3 4 5 2 6 3
friends and addresses the corresponding. The
But 5 letters can be permuted in 5! = 120 ways. number of ways can the letters be placed in the
Thus, the required number of arrangements envelopes, so that atleast two of them are in the
= (120) (20) = 2400 wrong envelopes, are
(a)120
Dearrangements (b)125
(c)119
If n distinct objects are arranged in a row, then the (d) None of the above
number of ways in which they can be dearranged, so
that none of them occupies its original place is Sol. (c) Required number of ways
5
1 1 1 1 1 = ∑
5
C 5 − r D(r )
n!1 − + − + − K + (−1) n r =2
1! 2! 3! 4! n! 5
5! 1 1 1 (−1)r
and it is denoted by D ( n).
= ∑ r !(5 − r )! r ! 1 − 1! + 2 ! − 3! + K + r!
289
r =2
6 1 (−1)r
5 3
5! 1 1 − xn + 1 1 − x2 n + 1
= ∑ (5 − r )! 2 ! − 3! + K +r!
= Coefficient of x3 n in
r =2 1− x 1− x
= + − + − +
5! 1 5! 1 1 5! 1 1 1
= Coefficient of x3 n in (1 − xn + 1 )3 (1 − x2 n + 1 ) (1 − x)−4
Objective Mathematics Vol. 1
3! 2 ! 2 ! 2 ! 3! 1! 2 ! 3! 4!
= Coefficient of x3 n in
+ − −
5! 1 1 1 1
+
0! 2 ! 3! 4! 5! [(1 − 3 xn + 1 + 3 x2 n + 2 − x3 n + 3 ) (1 − x2 n + 1 ) (1 − x)−4 ]
= 10 + 20 + (60 − 20 + 5) + (60 − 20 + 5 − 1) = Coefficient of x3 n in
= 10 + 20 + 45 + 44 = 119 [(1 − 3 xn + 1 − x2 n + 1 + 3 x2 n + 2
+ … ) (1 − x)−4 ]
= Coefficient of x3 n in (1 − x)−4
Number of Integral Solutions of − 3 Coefficient of x2 n − 1 in (1 − x)−4
Linear Equations and Inequations − Coefficient of xn − 1 in (1 − x)−4
+ 3 Coefficient of xn − 2 in (1 − x)−4
Consider the equation = 3n + 4 − 1
C4 − 1 − 3 × 2n − 1 + 4 − 1
C4 − 1 − n −1+ 4 −1
C4 − 1
x1 + x 2 + x 3 + x 4 + K + x r = n …(i) + 3× n − 2 + 4 −1
C4 − 1
where, x1 , x 2 , x 3 , x 4 , …, x r and n are non-negative = 3n + 3
C 3 − 3 × 2 n + 2C 3 − n + 2C 3 + 3 × n + 1C 3
integers. (3n + 3)(3n + 2 )(3n + 1) (2 n + 2 )(2 n + 1)(2 n)
= − 3⋅
This equation may be interpreted as that n identical 6 6
(n + 2 )(n + 1)(n) (n + 1)(n)(n − 1)
objects are to be divided into r groups, where a group − + 3⋅
6 6
may contain any number of objects. Therefore, 1
= (n + 1)(5n + 10n + 6)
2
Total number of solutions of Eq. (i) 6
= Coefficient of x n in ( x 0 + x 1 + K + x n ) r
X Example 87. The number of ways of selecting
= n + r − 1C n or n + r − 1 C r − 1
n objects out of 3n objects, of which n are alike and
Ø ● The total number of non-negative integral solutions of the rest are different, is
equation x1 + x 2 + K + x r = n is n + r − 1C r − 1 and the total (2n − 1)! (2n − 1)!
(a) 2 2n − 1 + (b) 2 2n − 1 −
number of solutions of the same equation in the set N of n!( n − 1)! n!( n − 1)!
natural numbers is n − 1C r − 1. ( 2 n + 1)!
(c) 2 2n + 1 + (d) None of these
● If the upper limit of a variable in solving an equation of the n!( n + 1)!
form x1 + x 2 + K + x m = n subject to the conditions
ai ≤ x ≤ b i ; i = 1, 2 , K , m is more than or equal to the sum Sol. (a) The required number of ways
required and lower limit of all the variables are non-negative,
then upper limit of that variable can be taken as infinite. = Coefficient of xn in ( x0 + x1 + x2 + K + xn )( x0 + x)2 n
● In order to solve inequations of the form 1 − x n + 1
= Coefficient of x n in (1 + x)
2n
x1 + x 2 + K + x m≤ n 1− x
We introduce a dummy (artificial) variable x m+ 1 such that = Coefficient of x n in (1 − x)−1 (1 + x)2 n
x1 + x 2 + K + x m + x m+ 1 = n, where x m+ 1 ≥ 0. ∞
= Coefficient of x n in ∑ xr (1 + x)2 n
The number of solutions of this equation are same as the r = 0
number of solutions of Eq. (i).
n
= Coefficient of x n in ∑ xr (1 + x)2 n
X Example 86. In an examination, the maximum
r = 0
marks for each of three papers is n and that for n
= Coefficient of x n in ∑x (1 + x)2 n
r
fourth paper is 2n. Then, the number of ways in
r =0
which a candidate can get 3n marks, is n
1
∑
n−r
(a) ( n − 1) (5n 2 + 10n + 6) = Coefficient of x in (1 + x)2 n
6 r =0
1 n
(b) ( n + 1) (5n 2 + 10n + 6) = ∑
2n
Cn − r
6 r =0
1 = Cn + Cn + 1 + K + C1 +
(c) ( n + 1) (5n 2 + n + 6)
2n 2n 2n 2n
C0
6 1 2n + 1
(d) None of the above = [2 n C n + {2 n C 0 + C1 + K +
2n 2n
Cn + Cn + 1
2
Sol. (b) The total number of ways of getting 3n marks + K+ 2n
C 2 n }] [ Q nC r = nC n − r ]
= Coefficient of x3 n in 1 2n 1 2 n! (2 n − 1)!
= [ Cn + 2 2n ] = + 2 2n = 2 2n − 1 +
( x0 + x1 + x2 + K + xn )3 × ( x0 + x1 + K + x2 n ) 2 2 n! n! n! (n − 1)!
290
6
4
X Example 88. The number of ways of selecting 1 − x5
= Coefficient of x4 in
4 letters from the word EXAMINATION, is 1− x
−4
= Coefficient of x in [(1 − x ) (1 − x) ]
Permutation and Combination
4 5 4
(a)136 (b)140 (c)126 (d)102
4⋅ 5⋅ 6⋅7
Sol. (a) There are 11 letters in the word EXAMINATION viz. = Coefficient of x4 in (1 − x)−4 = 7C 3 = = 35
4!
2 A’s, 2 I’s, 2 N’s, E, X, M, T, O.
Aliter
Therefore, x1 + x2 + x3 + x4 = 16, where x1, x2 , x3 , x4 ≥ 3
Number of ways of selecting 4 letters from these letters or x1 − 3 = a1, x2 − 3 = a2 , x3 − 3 = a3 , x4 − 3 = a4
= Coefficient of x4 in ( x0 + x1 + x2 )3 (1 + x)5 ∴ a1, a2 , a3 , a4 ≥ 0 or a1 + a2 + a3 + a4 = 8
3
1 − x3 ⇒ Number of solutions = 4 + 4 − 1C 4 − 1 = 7C 3
= Coefficient of x4 in (1 + x)
5
1− x 7 ×6×5
= = 35
−3 3×2
= Coefficient of x in (1 − x ) (1 − x)
4 3 3
(1 + x) 5
= Coefficient of x in (1 − 3 x + 3 x − x9 )(1 + x)5 (1 − x)−3
4 3 6
X Example 90. If N is the number of positive
= Coefficient of x4 in (1 − 3 x3 ) (1 + 5C1 x + 5C 2 x2 integral solution of x1 x 2 x 3 x 4 = 770. Then,
+ 5C 3 x3 + 5C 4 x4 ) (1 − x)−3 (a) N is 250 (b) N is 252
= Coefficient of x in (1 + 5 x + 10 x2 + 7 x3 − 10 x4 )(1 − x)−3
4
(c) N is 254 (d) N is 256
4 + 3 −1 3 + 3 −1 2 + 3 −1
= C3 − 1 + 5 × C 3 − 1 + 10 × C3 − 1 Sol. (d) We have, 770 = 2 ⋅ 5 ⋅ 7 ⋅ 11
1+ 3 −1
+7× C 3 − 1 − 10 We can assign 2 to x1 or x2 or x3 or x4 i.e. 2 can be
= 6C 2 + 5 × 5C 2 + 10 × 4C 2 + 7 × 3C 2 − 10 assigned in 4 ways.
= 15 + 50 + 60 + 21 − 10 = 136 Similarly, each of 5, 7 or 11 can be assigned in 4 ways.
Thus, the number of positive integral solution of
X Example 89. The number of ways in which 16 x1 x2 x3 x4 = 770 is 44 = 2 8 = 256
identical things can be distributed among 4
persons, if each person gets atleast 3 things, is Aliter
x1 x2 x3 x4 = 21 ⋅ 51 ⋅ 71 ⋅ 111
(a) 33 (b) 35
∴ a1 + a2 + a3 + a4 = 1
(c) 38 (d) None of these
i.e. 21 can be assigned to a1, a2 , a3 , a4 .
Sol. (b) Required number ⇒ 1+ 4 −1
C 4 − 1 = 4C 3 = 4
= Coefficient of x 16
in ( x + x + K + x )
3 4 7 4
Similarly, for 51, 71, 111
= Coefficient of x4 in (1 + x + K + x4 )4 ∴ Total number of ways = 4 ⋅ 4 ⋅ 4 ⋅ 4 = 256
Work Book Exercise 6.6
1 The total number of ways of dividing 15 different 4 The number of ways, 3 persons A, B and C
things into groups of 8, 4 and 3 respectively, is having 6 one rupee coins, 7 one rupee coins and
15 ! 15 ! 8 one rupee coins respectively can donate 10
7 ! 3 !2 ! 9! 5! 4! one rupee coins collectively, is
15 ! a 7
C3 8
C3
None of these
8! 4! 3! 10
c C3 None of these
2 The number of ways of distributing 50 identical
things among 8 persons in such a way that three
5 In an examination, the maximum marks for each
of them get 8 things each, two of them get 7 of the three papers are 50. Maximum marks for
things each and remaining 3 get 4 things each, is the fourth paper is 100. The number of ways in
8! 8! which a candidate can score 60% marks on the
( 3 !)2 (2 !) (2 !)2 3 ! whole, is
8! a 100550 101551
None of these
( 4 !)2 3 ! c 110551 d None of these
3 In how many ways can we put 12 different balls in 6 The number of ways of dividing 20 persons into
three different boxes such that the first box 10 couples, is
contains exactly 5 balls? 20 ! 20
a C10
12
C 3 ⋅ 57 12
C5 ⋅27 2 10
20 !
c 12
C 5 ⋅ 52 None of these c None of these
(2 !)9
291
6 7 The number of integer solutions for the equation
a
x + y + z + t = 20, where x, y, z, t ≥ − 1, is
20
C4 23
C3
18 The number of non-negative integral solutions of
x + y + z ≤ n, where n ∈ N is
a n +3
C3 b n +4
C4
Objective Mathematics Vol. 1
27 27 n +5
c C4 C3 c C5 d None of these
8 The number of ways in which mn students can be 19 If n objects are arranged in a row, then the
distributed equally among m sections, is number of ways of selecting three of these
( mn)! ( m n)! objects, so that no two of them are next to each
a
n! ( n!)m other, is
( mn)! n− 2 n− 3
c ( mn)m a C3 b C2
m! n ! n− 3
c C3 d None of these
9 The number of integers which lie between 1 and
10 6 and which have the sum of the digits equal to 20 There are three piles of identical red, blue and
12, is green balls and each pile contains atleast
10 balls. The number of ways of selecting
a 8550 5382 c 6062 8055
10 balls, if twice as many red balls as green balls
10 The number of integral solutions of are to be selected, is
x + y + z = 10, with x ≥ − 5, y ≥ − 5, z ≥ − 5, is a 3 b 4
a 135 136 c 455 105 c 6 d 8
11 The number of ways in which a score of 11 can 21 Two classrooms A and B having capacity of 25
be made from a throw by three persons, each and (n − 25) seats, respectively. Let A n denotes
throwing a single die once, is the number of possible sitting arrangements of
a 45 b 18 room A, when n students are to be seated in
c 27 d None of these these rooms, starting from room A which is to be
filled up full to its capacity. If
12 If x, y, z, … are (m + 1) distinct prime numbers,
A n − A n − 1 = 25!(49 C25 ), then n equals
then then number of factors of x n y z K is
a 49 b 48 c 50 d 51
a m( n + 1) b 2 nm
c ( n + 1)2 m
d n⋅2 m 22 Between two junction stations A and B, there are
12 intermediate stations. The number of ways in
13 There are m copies of each of n different books in which a train can be made to stop at 4 of these
a university library. The number of ways in which stations, so that no two of these halting stations
one or more than one book can be selected, is are consecutive, is
a mn + 1 b ( m + 1)n − 1 a 8
C4 b 9
C4
c ( n + 1) − m n
d m c 12
C4 − 4 d None of these
14 The number of ways in which one or more balls 23 The sum of the divisors of 2 5 ⋅ 34 ⋅ 52 is
can be selected out of 10 white, 9 green and
a 3 2 ⋅ 7 1 ⋅ 112
7 blue balls, is
b 63 ⋅ 121 ⋅ 31
a 892 b 881
c 3 ⋅ 7 ⋅ 11 ⋅ 31
c 891 d 879
d None of the above
15 The number of all three element subsets of the 24 The number of ways arranging m positive and
set { a1, a2 , a3 , K , an } which contain a3 , is
n < m + 1negative signs in a row, so that no two
n n−1
a C3 b C3 negative signs are together, is
n−1
c C2 d None of these a m +1
Pn
n+1
16 If one quarter of all three element subsets of the b Pm
m +1
set A = { a1, a2 , a3 , K , an } contains the element a3 , c Cn
then n is equal to n +1
d Cm
a 10
b 12
25 If ai, i = 1, 2, 3, 4 is four real numbers of the same
ai
c 14 sign, then the minimum value of Σ ,
d None of the above aj
i, j ∈{1, 2, 3, 4}, i ≠ j , is
17 The number of positive integral solutions of
a 8
abc = 30 is
b 6
a 30 b 27 c 12
292 c 8 None of these d None of the above
WorkedOut Examples
Type 1. Only One Correct Option
Ex 1. Number of positive integer n less than 17, for Ex 4. ∑ ∑∑∑ is equal to
which n! + ( n + 1)! + ( n + 2)! is an integral 0≤i < j <k <l ≤n
multiple of 49, is (a) n + 1 C 4 (b) n n + 1C 4
(a) 0 (b) 3
(c) n + 1 C 3 (d) n ( n + 1)
(c) 5 (d) 2
Sol. Here, n! + (n + 1)! + (n + 2)! Sol. As ∑ ∑ ∑ ∑n = n∑ ∑ ∑ ∑ represents
= n![1 + (n + 1) + (n + 2)(n + 1)] = n!(n + 2)2 0≤ i < j < k< l ≤ n 0≤ i < j < k< l ≤ n
Either 7 divides (n + 2) or 49 divides n !. selection of four terms from {0, 1, 2, ..., n} i.e. (n + 1)
i.e. n = 5 ,12, 14, 15, 16 [Qn < 17] terms.
Thus, ∑ ∑ ∑ ∑ n = n ⋅n + 1 C 4
Thus, number of solutions is five.
0≤ i < j < k< l ≤ n
Hence, (c) is the correct answer.
Hence, (b) is the correct answer.
2 2 1 2a
Ex 2. If + + = , where a, b ∈ N , then Ex 5. Let A denotes the property that two elements of
9! 3! 7! 5!5! b!
the ordered pair ( a, b) is A = {1, 5, 9, 13, ..., 1093} add upto 1094. Then,
maximum number of elements in A can be
(a) (9, 10)
(a) 273 (b) 136
(b) (10, 9) (c) 137 (d) 138
(c) (7, 10)
(d) (10, 7) Sol. Since, A = {1, 5, 9, 13, ..., 1093} is an arithmetic
progression.
2
+
2
+
1 1093 − 1
Sol. Here, ∴ Number of terms = +1
9! 3!7! 5!5! 4
=
1
+
1
+
1
+
1
+
1 = 274
1!9! 3!7! 5!5! 3!7! 9!1! Since, sum of equidistant terms in AP is equal to sum of
1 10! 10! 10! 10! 10! first and last terms = 1 + 1093 = 1094.
= + + + +
10! 9!1! 7!3! 5!5! 3!7! 9! 1 ! ∴ Maximum number of elements in A that add upto 1094
1 10 274
= { C 1 + 10C 3 + 10C 5 + 10C 7 + 10C 9 } = = 137
10! 2
1 29 2a Hence, (c) is the correct answer.
= ⋅ (210 − 1 ) = , but it is .
(10)! 10! b!
Ex 6. Let n1 = x1 x 2 x 3 and n2 = y1 y2 y3 be two 3-digit
⇒ a = 9 and b = 10 numbers, then the pairs of n1 and n2 can be
Hence, (a) is the correct answer. formed, so that n1 can be subtracted from n2
without borrowing, is
Ex 3. Number of ordered triplets (x, y, z ) such that
(a) 55⋅ 54 (b) 45⋅ 55
x, y, z are primes and x y + 1 = z, is
(a) 0 (c) 45⋅ ( 55) 2 (d) 55⋅ ( 45) 2
(b) 1 Sol. Here, n1 = x1x2x3 and n 2 = y1 y2 y3
(c) 3 ⇒ n1 can be subtracted from n2 without borrowing, if
(d) None of the above yi ≥ xi for i = 1, 2, 3.
r = 0, 1, 2, ..., 9, for x2 and x3
Sol. Here, x y + 1 = z, where x , y, z are prime. ∴ Let xi = r ⇒
Thus, y cannot be odd, if y is prime. r = 1, 2, 3, ..., 9, for x1
⇒ x y + 1is divisible by ( x + 1). ∴ yi = r, r + 1, ...,9
Now, z must be odd. Thus, for y1, y2 and y3, we have (10 − r) choices each.
Now, total number of ways for choosing yi and xi
⇒ x must be even. [Qx y = z − 1]
9 9 9
Thus, only even value (i.e. prime) is x = 2. = ∑ (10 − r) ∑ (10 − r) ∑ (10 − r)
⇒ x = 2, y = 2, z = 5 r = 1 r = 0 r = 0
So, there is only one such triplet (2, 2, 5). = 45 ⋅ 55 ⋅ 55 = 45 ⋅ (55)2
Hence, (b) is the correct answer. Hence, (c) is the correct answer. 293
6 Ex 7. Let
E = (2n + 1)(2n + 3)(2n + 5)K ( 4n − 3)( 4n − 1) ;
n >1, then 2 n E is divisible by
Ex 10. A seven-digit numbers divisible by 9 is to be
formed by using 7 out of numbers {1, 2, 3, 4, 5,
Objective Mathematics Vol. 1
6, 7, 8, 9}. The number of ways in which this
(a) n C n/ 2 (b) 2n
Cn (c) 3n
Cn (d) 4n
C 2n can be done, is
Sol. Here, E = (2n + 1)(2n + 3)(2n + 5) K (4 n − 3)(4 n − 1) (a) 7! (b) 2⋅ ( 7)!
(2n + 1)(2n + 2)(2n + 3)(2n + 4 )K (4 n − 1)(4 n) (c) 3⋅ ( 7)! (d) 4 ⋅ 7!
or E =
(2n + 2)(2n + 4 )K(4n)
Sol. Sum of 7 digits = a multiple of 9
(2n)!
⋅ We know, sum of numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 is 45.
(2n)!
So, two left numbers should also have sum as 9. The pairs
(4 n)!
= to be left are (1, 8), (2, 7), (3, 6), (4, 5) with each pair left
(2n)! ⋅ 2n ⋅ (n + 1)(n + 2)...(2n) number of 7 digit numbers is 7!.
(4 n)! ⋅ n! So, with all 4 pairs = 4 × 7!
E=
(2n)!2n ⋅ (2n)! Hence, (d) is the correct answer.
(4 n)!
⇒ 2n E = ⋅ n! = 4nC 2n ⋅ n! Ex 11. Among 9! permutations of the digits
(2n)!(2n)!
1, 2, 3,..., 9. Consider those arrangements
∴2n E is divisible by 4n
C 2n. which have the property that if we take any
Hence, (d) is the correct answer. five consecutive positions, the product of the
digits in those positions is divisible by 7, the
∑ ∑
10
Ex 8. C j j C i is equal to number of such arrangements is
0 ≤ i ≤ j ≤ 10
(a) 7! (b) 3⋅ 7! (c) 8! (d) 4 ⋅ 7!
(a) 2 10
−1 (b) 210
Sol. Since, amongst 9! arrangements of digits 1, 2, 3, 4, 5, 6,
(c) 310 − 1 (d) None of these 7, 8, 9 i.e. the number formed is of the form
x1 x2 x3 x4 x5 x6 x7 x8 x9 . Taking any five consecutive
∑ ∑
10
Sol. We have, C j jC i positions product is divisible by 7 only, if x5 = 7.
0 ≤ i ≤ j ≤ 10 So, the required arrangements = 8!
= 10C 0 ⋅ 0C 0 + C 1 (1C 0 + 1C 1 ) +
10
C 2 (2C 0 + 2C 1 + 2C 2 )
10 Hence, (c) is the correct answer.
+ 10
C 3 (3C 0 + 3C 1 + 3C 2 + 3C 3 ) + K + 10
C 10 n
S
( C0 +
10
C1 +
10 10
C2 + K + 10
C 10 )
Ex 12. Let S n = ∑ r! ;
( n > 6), then S n − 7 n
7
r =1
= 10C 0⋅ 20 + 10C 1⋅ 21 + 10
C 2⋅ 22 + 10
C 3 ⋅ 23 + K+ 10C 10⋅ 210 (where, [ ] denotes the greatest integer
function) is equal to
= (1 + 2)10 = 310
n n
Hence, (d) is the correct answer. (a) (b) n ! − 7
7 7
∑ ∑ ( i + 1)( j + 1) (c) 5 (d) 3
1≤ i ≤ j ≤ n
Ex 9. lim is equal to Sol. All the number r! 1; (r ≥ 7) will be multiple of 7.
n→ ∞ n4 So, for remainder, consider
1 1 S6 = 1 + 2 + 6 + 24 + 120 + 720 = 873
(a) (b)
4 6 which leaves the remainder 5, when divided by 7.
1 Hence, (c) is the correct answer.
(c) (d) None of these
8
Ex 13. The number of ways of arranging m numbers
∑ ∑ (i + 1)( j + 1) out of 1, 2, 3, ..., n, so that maximum is ( n − 2)
1 ≤ i≤ j ≤ n
Sol. Here, lim and minimum is 2 (repetitions of numbers is
n→ ∞ n4
2 allowed) such that maximum and minimum
n
both occur exactly once ( n > 5, m > 3), is
n
∑ (i + 1) − ∑ (i + 1)2
n−3
i = 1 i=1 (a) Cm − 2 (b) m
C 2 ( n − 3) m − 2
= lim
n→ ∞ 2n4 (c) m( m − 1)( n − 5) m − 2 (d) n C 2 ⋅ n C m
2
(n + 2)(n + 1) (n + 1)(n + 2)(2n + 3)
− 1 − − 1 Sol. First, we take one number as 2 and one as (n − 2) and put
= lim
2 6
them in m(m − 1) ways. Now, remaining (m − 2),
n→ ∞ 2n4
numbers can be anyone from, 3, 4, ..., (n − 4 ), (n − 3).
1
n4 + n3 K and decreasing powers of n Which we can do in (n − 5)(m − 2) .
4 1
= lim = ∴ Total number of ways = m(m − 1)(n − 5)m − 2
n→ ∞ 2n4 8
294 Hence, (c) is the correct answer.
Hence, (c) is the correct answer.
Ex 14. Total number of integer n such that 2 ≤ n ≤ 2000
and HCF of n and 36 is one, is equal to
Sol. The number of ways of choosing first couple is
(15C 1 )(15C 1 ) = 152, the number of ways of choosing 2nd
couple is (14C 1 )(14C 1 ) = 14 2 and so on. Thus, the
6
Permutation and Combination
(a) 666
number of ways of choosing the couple is
(b) 667
152 + 14 2 + 132 + K + 22 + 12
(c) 665 15 × (15 + 1)[ 2(15) + 1]
(d) None of the above = = 1240
6
Sol. Q 36 = 22 ⋅ 32 Hence, (a) is the correct answer.
Total number of integers ∈[ 2, 2000 ] that are divisible by Ex 18. The number of ordered pairs (m, n),
2 is
m, n ∈ {1, 2, ..., 50} such that 6 n + 9 m is a
2000
= 1000
2
multiple of 5, is
(a) 2500 (b) 1250 (c) 625 (d) 500
The integer ∈[ 2, 2000 ] that are divisible by 3 is
2000 Sol. All the numbers of the form 6 will end with 6 and 9m
n
= 666
3 will end with 9, if m is odd and will end with 1, if m is
Integers that are divisible by 2 and 3 and contained in even, so 6n + 9m will end with 5, if nis any number and m
2000 is odd.
[ 2, 2000 ] = = 333
6 So, ordered pairs will be 50 × 25 = 1250.
Hence, (b) is the correct answer.
Thus, total number of integers that are neither divisible
by 2 nor by 3 Ex 19. n is selected from the set {1, 2, 3, ..., 100} and
= 1999 − (1000 + 666 − 333) = 666 the number 2 n + 3 n + 5 n is formed. Total
Hence, (a) is the correct answer. number of ways of selecting n, so that the
1 2 3 4 30 31 formed number is divisible by 4, is equal to
Ex 15. If E = ⋅ ⋅ ⋅ K ⋅ = 8 x , then value of (a) 50 (b) 49
4 6 8 10 62 64
x is (c) 48 (d) None of these
(a) − 7 (b) − 9 Sol. If n is odd, 3n = 4 λ 1 − 1, 5n = 4 λ 2 + 1
(c) − 10 (d) − 12 ⇒ 2n + 3n + 5n is not divisible by 4, as 2n + 3n + 5n will
Sol. We have, be in the form of 4 λ + 2.
31! 1 1 Thus, total number of ways of selecting n is 49.
E= = = Hence, (b) is the correct answer.
231 (32)! 231 (32) 236
= 2−36 = (23 )−12 = 8−12, but it is 8x. Ex 20. The number of ordered pairs (m, n),
Thus, x = − 12 m, n ∈{1, 2, ..., 100} such that 7 m + 7 n is
Hence, (d) is the correct answer.
divisible by 5, is
Ex 16. The number of rational numbers lying in the (a) 1250 (b) 2000 (c) 2500 (d) 5000
interval (2002, 2003) all whose digits after the Sol. Note that 7 (r ∈ N ) ends in 7, 9, 3 or 1 (corresponding to
r
decimal points are non-zero and are in r = 1, 2, 3 and 4, respectively).
decreasing order, is
9 10 Thus, 7m + 7n cannot end in 5 for any values of m, n ∈ N .
(a) ∑ 9
Pi (b) ∑ 9
Pi In other words, for 7m + 7n to be divisible by 5, it should
i=1 i=1 end in 0.
(c) 29 − 1 (d) 210 − 1 For 7m + 7n to end in 0, the forms of m and n should be as
follows:
Sol. A rational number of the desired category is of the form
2002. x1x2K xk, where 1 ≤ k ≤ 9 and 9 ≥ x1 > x2 > S.No. m n
K > xk ≥ 1. We can choose k digits out 9 in 9 C k ways and 1. 4r 4s + 2
arrange them in decreasing order in just one way. Thus, 2. 4r + 1 4s + 3
the desired number of rational numbers is 3. 4r + 2 4s
9
C 1 + 9C 2 + ... + 9C 9 = 29 − 1 4. 4r + 3 4s + 1
Hence, (c) is the correct answer. Thus, for a given value of m, there are just 25 values of n
for which 7m + 7n ends in 0.
Ex 17. The number of ways of dividing 15 men and
[for instance, if m = 4 r, then n = 2, 6, 10, ..., 98]
15 women into 15 couples, each consisting of a
man and a woman, is ∴ There are 100 × 25 = 2500 ordered pairs (m, n) for
(a) 1240 (b) 1840 (c) 1820 (d) 2005 which 7m + 7n is divisible by 5.
295
Hence, (c) is the correct answer.
6 Ex 21. If the number of ways of selecting K coupons
out of an unlimited number of coupons bearing
the letters A, T, M, so that they cannot be used
Sol. Let the boys gets a, b and c toys, respectively.
⇒ a + b + c = 14, a, b, c ≥ 1and a, b and care distinct.
Let a < b < c and x1 = a, x2 = b − a, x3 = c − b
Objective Mathematics Vol. 1
to spell the word MAT is 93, then K is equal to So, 3x1 + 2x2 + x3 = 14; x1 , x2 , x3 ≥ 1
∴ The number of solutions
(a) 3 (b) 5
(c) 7 (d) None of these = Coefficient of t 14 in {(t 3 + t6 + t 9 + ...) (t 2 + t 4 + ...)
(t + t 2 + ...)}
Sol. Here, A, T, M bears unlimited number of coupons.
Thus, selecting K in which word MAT is not spelled. = Coefficient of t 8 in {(1 + t 3 + t6 + K )
If atleast one letter is not selected. i.e. 3C 1. ...(i) (1 + t 2 + t 4 + ...) (1 + t + t 2 + ...)}
From the remaining two we have to select K i.e. for each = Coefficient of t 8 in {(1 + t 2 + t 3 + t 4 + t 5 + 2t6
selection of K coupons we have 2 ways.
+ t 7 + 2t 8 ) (1 + t + t 2 + K + t 8 )}
For K coupons, we have 2K ways but to select atleast one,
we have = 1 + 1 + 1 + 1 + 1 + 2 + 1 + 2 = 10
⇒ 2K − 1 …(ii) Since, three distinct numbers can be assigned to three
Words cannot be used to spell MAT boys in 3! ways.
⇒ 3(2K − 1) = 93 ⇒ K = 5 So, total number of ways = 10 × 3! = 60
Hence, (c) is the correct answer.
Hence, (b) is the correct answer.
Ex 25. Total number of positive integral solutions of
Ex 22. In a shooting competition, a man can score 5, 4,
15 < x1 + x 2 + x 3 ≤ 20 is equal to
3, 2 or 0 points for each shot. Then, the number
of different ways in which he can score 30 in (a) 1125 (b) 1150
seven shots, is (c) 1245 (d) 685
(a) 419 (b) 418 (c) 420 (d) 421 Sol. 15 < x1 + x2 + x3 ≤ 20
Sol. Number of ways of making 30 in 7 shots ⇒ x1 + x2 + x3 = 16 + r; r = 0, 1, 2, 3, 4
= Coefficient of x in (x + x + x + x + x )
30 0 2 3 4 5 7 Now, number of positive integral solutions of
= Coefficient of x 30 in {(x 0 + x 2 + x 3 ) + x 4 (x + 1)}7 x1 + x2 + x3 = 16 + r is 13 + r + 3 − 1C 13 + r
15 + r 15 + r
= Coefficient of x 30 in i.e. C 13 + r = C2
{x 28 (x + 1)7 + 7C 1x 24 (x + 1)6 (1 + x 2 + x 3 ) Thus, total solutions
4
+ C 2x (x + 1) (x + x + 1) + K} ∑ 15 + r
7 20 5 3 2
= C 2 = 15C 2 + C2 +
16 17
C2 + C2 +
18 19
C2
= C 5 + C 1 ( C 3 + C 2 + C 0 ) + C 2 ( C 1 + 2)
7 7 6 6 6 7 5 r=0
= 21 + 252 + 147 = 420 = 20C 3 − 15C 3 = 685
Hence, (c) is the correct answer. Hence, (d) is the correct answer.
Ex 23. The number of ways in which the sum of upper Ex 26. Ten persons numbered 1, 2, ..., 10 play a chess
faces of four distinct dices can be six, is tournament, each player playing against every
(a) 10 (b) 4 other player exactly one game. It is known that
(c) 6 (d) 7 no game ends in a draw. Let w1 , w2 , ..., w10 be
Sol. The number of required ways will be equal to the the number of games won by players 1, 2, 3, ...,
number of solutions of 10 respectively and l1 , l2 , ..., l10 be the number
x1 + x2 + x3 + x4 = 6 of games lost by the players 1, 2, ..., 10,
where, x1 , x2 , x3 , x4 ≥ 1 respectively. Then,
Let, x1 − 1 = a1, x2 − 1 = a2, x3 − 1 = a3, x4 − 1 = a4 ≥ 0
(a) Σw i2 = 81 − Σli2 (b) Σw i2 = 81 + Σli2
⇒ (a1 + 1) + (a2 + 1) + (a3 + 1) + (a4 + 1) = 6
where, a1 , a2 , a3 , a4 ≥ 0 (c) Σw i2 = Σli2 (d) None of these
⇒ a1 + a2 + a3 + a4 = 2
Sol. Clearly, each player will play 9 games and total number
∴ Number of solutions = 2 + 4 − 1C 4 − 1 = 5C 3 = 10
of games = 10C 2 = 45
Hence, (a) is the correct answer.
Clearly, wi + li = 9
Ex 24. The number of ways can 14 identical toys be and Σwi = Σli = 45
distributed among three boys, so that each one ⇒ wi = 9 − li
gets atleast one toy and no two boys get equal ⇒ wi2 = 81 + li2 − 18li
number of toys, is ⇒ Σwi2 = 81⋅ 10 + Σli2 − 18Σli
(a) 45 (b) 48 = 810 + Σli2 − 18 ⋅ 45 = Σli2
(c) 60 (d) None of these Hence, (c) is the correct answer.
296
Ex 27. The number of subsets of the set
A = {a1 , a 2 , ..., a n } which contain even
number of elements, is
Sol. There are exactly (n − 1) subsets of S containing two
elements having 1 as least element, exactly (n − 2)
subsets of S having 2 as least element and so on.
6
Permutation and Combination
Thus,
(a) 2n − 1 (b) 2n − 1 (c) 2n − 2 (d) 2n
∑ min ( A ) = 1(n − 1) + 2(n − 2) + K + (n − 1)(1)
Sol. For each of the first (n − 1) elements a1 , a2 , K , an − 1, we A ∈X
n− 1 n− 1 n− 1
have two choices : either ai (1 ≤ i ≤ n − 1) lies in the
subset or ai does not lie in the subset. For the last element = ∑ r(n − r) = n ∑ r − ∑ r2
r=1 r=1 r=1
we have just one choice. If even number elements have
already been taken, we do not include an, in the subset, 1 1
= n n(n − 1) − n(n − 1)(2n − 1)
otherwise (when odd number of elements have been 2 6
added), we include it in the subset. 1
= n(n − 1){3n − 2n + 1}
Thus, the number of subsets of A = {a1 , a2 , ..., an} 6
which contain even number of elements is equal to 2n − 1. 1
= (n + 1)n(n − 1) = n + 1C 3
Hence, (a) is the correct answer. 6
Hence, (a) is the correct answer.
Ex 28. A is a set containing n elements. A subset P of
A is chosen. The set A is reconstructed by Ex 31. The sum of factors of 8! which are odd and are
replacing the elements of P. A subset Q of A is of the form 3m + 2, where m is a natural
again chosen. The number of ways of choosing number, is
P and Q, so that P ∩ Q = φ, is (a) 40
(a) 22n − 2n
Cn (b) 2n (b) 8
(c) 45
(c) 2n − 1 (d) 3n
(d) 35
Sol. Let A = {a1 , a2 , ..., an}. For ai ∈ A, we have the Sol. Here, 8! = 27 ⋅ 32 ⋅ 51 ⋅ 71
following choices:
(i) ai ∈ P and ai ∈ Q Obviously, the factors are not multiple of either 2 or 3.
(ii) ai ∈ P and ai ∉ Q So, the factors may be 1, 5, 7, 35 of which 5 and 35 are of
(iii) ai ∉ P and ai ∈ Q the form 3m + 2.
(iv) ai ∉ P and ai ∉ Q So, the sum = 40
Out of these, only (ii), (iii) and (iv), ai ∉ P ∩ Q. Hence, (a) is the correct answer.
Therefore, the number of required subsets is 3n.
Ex 32. The number of positive integral solutions of
Hence, (d) is the correct answer.
the equation x1 x 2 x 3 x 4 x 5 = 420 is
Ex 29. The number of integral points that lie exactly (a) 1875
in the interior of the triangle with vertices (b) 1600
(0, 0), (0, 21), (21, 0), is (c) 1250
(a) 133 (b) 190 (d) None of the above
(c) 233 (d) 105
Sol. Since, x1 x2 x3 x4 x5 = 420
Sol. The integral points lying on the line x = 1which also lie ∴ x 1 x 2 x 3 x 4 x 5 = 22 × 3 × 5 × 7
in the interior of ∆OAB are
(1, 1), (1, 2), ..., (1, 19). Each of 3, 5 or 7 can take 5 places and 22 can be
∴ There are 19 points lying on the line x = 1. disposed in 15 ways.
So, required number of solutions
Similarly, the number of integral points x = 2is 18 and on
= 53 (5C 1 + 5C 2 ) = 53 × 15 = 1875
x = 3 is 17 and so on. The number of integral points on
x = 19 is just 1. Hence, (a) is the correct answer.
Aliter
Thus, number of integral points lying in the interior of the
triangle is x 1 x 2 x 3 x 4 x 5 = 22 × 3 × 5 × 7
1 Here, 22 can be assigned to a1 , a2 , a3 , a4 , a5
19 + 18 + K + 1 = (19)(19 + 1) = 190
2 ⇒ a1 + a2 + a3 + a4 + a5 = 2.
2+ 5− 1
Hence, (b) is the correct answer. ⇒ C 5 − 1 = 6C 4 = 15.
Ex 30. Let S = {1, 2, 3, ..., n}. If X denotes the set of all Also, 31 can be assigned to b1 , b2 , b3 , b4 , b5
subsets of S containing exactly two elements, ⇒ b1 + b2 + b3 + b4 + b5 = 1
then the value of ∑ (min A ) is ⇒ 1+ 5− 1
C 5 − 1 = 5C 4 = 5
A ∈X
n+1 Similarly, for 5 and 7.
(a) C3 (b) n C 3
∴Total number of positive integral solutions
(c) n C 2 (d) None of these = 15 × 5 × 5 × 5 = 1875
297
6 Ex 33. The number of numbers lying in (2, 3), where
all the digits after decimal are non-zero and in
increasing order, is
Ex 37. Total number of positive integral solutions of
x1 ⋅ x 2 ⋅ x 3 = 60 is equal to
Objective Mathematics Vol. 1
(a) 27 (b) 54
9 10
(c) 64 (d) None of these
(a) ∑ 9
Pi (b) 511 (c) ∑ 10
Pi (d) 1023
i=1 i=1 Sol. Q x1 ⋅ x2 ⋅ x3 = 22 ⋅ 3 ⋅ 5
Sol. Since, the digits required is non-zero. Total number of positive integral solutions
∴ To select amongst {1, 2, 3, 4, 5, 6, 7, 8, 9} = 3 ⋅ 3 ⋅ (3C 1 + 3C 2 ) = 54
Thus, any number is of the form Hence, (b) is the correct answer.
2x1 x2 K xk 3, where 1 ≤ x1 < x2 < x3 < K < x9 ≤ 9 Aliter
i.e. number of numbers lying between (2, 3) x 1 ⋅ x 2 ⋅ x 3 = 22 ⋅ 3 ⋅ 5
= 9C 1 + 9C 2 + 9C 3 + K + 9C 9
Here, 22 can be assigned to a1 , a2 , a3.
= 29 − 1 = 1023
⇒ a1 + a2 + a3 = 2
Hence, (d) is the correct answer. ⇒ 2+ 3− 1
C 3 − 1 = 4C 2 = 6
Ex 34. The number of divisors of n = 2 7 ⋅ 3 5 ⋅ 5 3 , and 31 can be assigned to b1 , b2 , b3.
which are of the form 4t + 1( t ∈ N ), is ⇒ b1 + b2 + b3 = 1 ⇒ 1+ 3− 1
C3 − 1 = 3
(a) 12 (b) 10 ∴ x1 ⋅ x2 ⋅ x3 = 2 ⋅ 3 ⋅ 5
2 1 1
(c) 11 (d) 14
Total number of positive integral solutions
Sol. Any number will be of the form 4 t + 1, if it does not = 6 × 3 × 3 = 54
contain any power of 2 but contain even number of odds
of the form 4 k + 3 and any number of odds of the form Ex 38. The number of positive integral solutions of
4 k + 1, so total factors = 1 × 3 × 4 = 12 the equation x1 x 2 x 3 x 4 x 5 = 1050, is
But one will be 1, also so required number = 12 − 1 = 11 (a) 1800 (b) 1600
Hence, (c) is the correct answer. (c) 1400 (d) None of these
Aliter Sol. Using prime factorisation of 1050, we can write the
Odd numbers of the type 4 t + 1 or 4 t + 3 and they are given equation as
symmetrical. x1x2x3x4x5 = 2 × 3 × 52 × 7
Number of odd numbers = (1 + 5)(1 + 3) = 24
We can assign 2, 3 or 7 to any of variables. We can assign
[any number of 5’s and 3’s and no 2’s]
entire 52 to just one variable in 5 ways or can assign
Required number = 12 but 1 is included.
52 = 5 × 5 to two variables in 5C 2 ways. Thus, 52 can be
∴Required number of numbers = 12 − 1 = 11
assigned in
Ex 35. Total number of divisors of n = 2 5 ⋅ 3 4 ⋅ 510 ⋅ 7 6
5
C 1 + 5C 2 = 5 + 10 = 15 ways
that are of the form 4λ + 2, λ ≥ 1, is equal to Hence, required number of solutions
(a) 385 (b) 55 (c) 384 (d) 54 = 5 × 5 × 5 × 15 = 1875
Sol. Q 4 λ + 2 = 2 (2λ + 1) = odd multiple of 2 Hence, (d) is the correct answer.
Thus, total divisors = 1⋅ 5 ⋅ 11⋅ 7 − 1 = 384 Aliter
[one is subtracted because there will be case x 1 ⋅ x 2 ⋅ x 3 ⋅ x 4 ⋅ x 5 = 2 × 3 × 52 × 7
when selected powers of three, five and seven 5+ 2− 1
For 5 , the number of ways =
2
C 5 − 1 = 6C 4 = 15
are zero each and this will make λ = 0]
1+ 5− 1
Hence, (c) is the correct answer. For 21, the number of ways = C 5 − 1 = 5C 4 = 5
∴Positive integral solutions for x1 ⋅ x2 ⋅ x3 ⋅ x4 ⋅ x5
Ex 36. Total number of divisors of n = 3 5 ⋅ 5 7 ⋅ 7 9 that
= 2 × 3 × 52 × 7 is 15 × 5 × 5 × 5 = 1875
are of the form 4λ + 1, λ ≥ 0, is equal to
(a) 240 (b) 30 (c) 120 (d) 15 Ex 39. Let a be a factor of 120, then the number of
Sol. Q 3 n1 n1
= (4 − 1) = 4 λ 1 + (− 1) , n1
positive integral solutions of x1 x 2 x 3 = a is
5n2 = (4 + 1)n2 = 4 λ 2 + 1 (a) 160 (b) 320 (c) 480 (d) 960
and 7 n3
= (8 − 1) n3
= 4 λ 3 + (− 1) n3 120
Sol. Let x4 be such that x4 = , then the number of positive
a
Hence, any positive integer power of 5 will be in the form
integral solutions of x1x2x3 = a is same as that of number
of 4 λ 2 + 1. Even powers of 3 and 7 will be in the form of
of positive integral solutions of x1x2x3x4
4 λ + 1 and odd powers of 3 and 7 will be in the form of
4 λ − 1. Thus, required number of divisors = 120 = 23 × 3 × 5
= 8 ⋅ (3 ⋅ 5 + 3 ⋅ 5) = 240 We can assign 3 and 5 to unknown quantities in 4 × 4
298 Hence, (a) is the correct answer. ways.
We can assign all 2 to one unknown in 4C 1 ways, to two
unknown in 4C 2 and to three unknown in 4C 3 ways.
Ex 43. The function f :{1, 2, 3, 4, 5} → {1, 2, 3, 4, 5} that
are onto and f ( i) ≠ i, is equal to 6
Permutation and Combination
Hence, the number of required solutions (a) 9
= 4 × 4 × [ 4C 1 + (4C 2 )(2) + 4C 3 ] = 4 × 4 × 20 = 320 (b) 44
Hence, (b) is the correct answer. (c) 16
Aliter (d) None of the above
x1 ⋅ x2 ⋅ x3 ⋅ x4 = 120 Sol. Total number of required functions
⇒ x 1 ⋅ x 2 ⋅ x 3 ⋅ x 4 = 23 × 3 × 5 = Number of dearrangement of 5 objects
For 2 , the number of ways =
3 3+ 4− 1
C 4 − 1 = C 3 = 20
6 1 1 1 1
= 5! − + − = 44
1+ 4− 1
2! 3! 4 ! 5!
For 3 , the number of ways =
1
C 4 − 1 = 4C 3 = 4
Hence, (b) is the correct answer.
∴Total number of integral solutions = 20 × 4 × 4 = 320
Ex 44. The total number of functions f from the set
Ex 40. The number of ways of choosing triplets {1, 2, 3} into the set {1, 2, 3, 4, 5} such that
( x, y, z ) such that z ≥ max {x, y} f ( i) ≤ f ( j), ∀ i < j, is equal to
and x, y, z ∈{1, 2, ..., n, n + 1}, is (a) 35 (b) 30
n+1 n+2
(a) C3 + C3 (b) n( n + 1)( 2n + 1) (c) 50 (d) 60
(c) 11 + 22 +K+ ( n − 1) 2 (d) None of these Sol. Let 1 is associated with r, r ∈{1, 2, 3, 4 , 5}, then 2 can be
associated with r, r + 1, ..., 5.
Sol. Triplets with x = y < z, x < y < z, y < z < x can be
n+ 1 n+ 1 n+ 1 Let 2 is associated with j, then 3 can be associated with j,
chosen in C2, C3, C 3 ways.
∴There n + 1C 2 + 2(n + 1C 3 ) = n + 2C 3 + n+ 1
C 3 ways j + 1, ..., 5. Thus, required number of functions
5 5 5
(6 − r)(7 − r)
= ∑ ∑ (6 − j ) = ∑
Hence, (a) is the correct answer.
r = 1 j = r r=1 2
Ex 41. The number of maps (functions) from the set
A = {1, 2, 3} into the set B = {1, 2, 3, 4, 5, 6, 7} 1 5
such that f ( i) ≤ f ( j) whenever i < j, is
=
∑
2 r = 1
(42 − 13r + r2 )
(a) 84 (b) 90
1 6 ⋅ 5 5 ⋅ 6 ⋅ 11
(c) 88 (d) None of these = 42 ⋅ 5 − 13 ⋅ +
2 2 6
Sol. If the function is one-one, then select any three from the = 35
set B in 7 C 3 ways i.e. 35 ways . Hence, (a) is the correct answer.
If the function is many-one, then there are two
Ex 45. The number of functions f from the set
possibilities.
A = {0, 1, 2} into the set B = {0, 1, 2, 3, 4, 5, 6, 7}
All three corresponds to same element number of such
such that f ( i) ≤ f ( j) for i < j and i, j ∈ A, is
functions = 7C 1 = 7 ways.
(a) 8 C 3
Two corresponds to same element. Select any two from
(b) 8 C 3 + 2( 8 C 2 )
the set B. The larger one corresponds to the larger and the
smaller one corresponds to the smaller the third may (c) 10 C 3
corresponds to any two. (d) None of the above
Number of such functions = 7C 2 × 2 = 42
Sol. A function f : A → B such that f (0) ≤ f (1) ≤ f (2) falls
So, the required number of maps = 35 + 7 + 42 = 84
in one of the following four categories.
Hence, (a) is the correct answer.
Case I f (0) < f (1) < f (2)
Ex 42. If f :{1, 2, 3, 4, 5} → {x, y, z, t}, then the total There are 8 C 3 functions in this category.
number of onto functions is equal to Case II f (0) = f (1) < f (2)
(a) 242 (b) 245 (c) 102 (d) 240
There are 8 C 2 functions in this category.
Sol. Total functions = 4 5
Case III f (0) < f (1) = f (2)
Total functions when any one element is the left out
There are again 8 C 2 functions in this category.
= 4C 1 ⋅ 35
Total functions when any two elements are left out Case IV f (0) = f (1) = f (2)
= 4C 2 ⋅ 25 There are 8 C 1 functions in this category.
∴ Total number of onto functions Thus, the number of desired functions is
= 4 5 − (4C 1 ⋅ 35 − 4C 2 ⋅ 25 + 4C 3 ) = 240 8
C 3 + 8C 2 + 8C 2 + 8C 1 = 9C 3 + 9C 2 = 10C 3
Hence, (d) is the correct answer. Hence, (c) is the correct answer. 299
6 Ex 46. The integers from 1 to 1000 are written in order
around a circle. Starting at 1, every fifteenth
number is marked (i.e. 1, 16, 31 etc.) This
Ex 48. The number of ways in which we can choose
3 squares on a chessboard such that one of the
squares has its two sides common to other two
Objective Mathematics Vol. 1
process is continued until a number is reached squares, is
which has already been marked, then (a) 290 (b) 292
unmarked numbers are (c) 294 (d) 296
(a) 200 (b) 400 (c) 600 (d) 800
Sol. Either we have to choose or
Sol. In first round all the integers, which leaves the remainder
1 when divided by 15, will be marked.
(i) Every square of (2 by 2) will contribute four
Last number of this category is 991.
Next number to be marked is (991 + 15 − 1000) = 6 shaped figures by removing anyone square
Again, second round of integers which leaves the
remainder 6 when divided by 15, will be marked. out of it.
Last number of this category is 996. Number of ways to choose these squares
= 7 × 7 = 49
Next number to be marked is (996 + 15 − 1000) = 11
Thus, third round of integers which leaves the remainder So, total shaped figures = 49 × 4 = 196
11 when divided by 15, will be marked last number of this
(ii) In every line it is possible to have 6 shaped
category is 986.
Next number to be marked is 986 + 15 − 1000 = 1, figures.
which is already been marked. So, total number of such figures = 6 × 8 × 2 = 96
Hence, the numbers of the form 15λ + 1, 15λ + 6, ∴Total number of cases = 196 + 96 = 292
15λ + 11are marked. All these numbers can also be put in Hence, (b) is the correct answer.
the form of 5λ 1 + 1(λ 1 ≥ 0).
Hence, total number of marked numbers Ex 49. A man on the verge of dying. He has unlimited
999 amount of money and intends to distributed
=1+ = 200
5 some of it among his n relatives, so that total
∴Unmarked numbers = 800 money that is to be distributed is a positive
Hence, (d) is the correct answer. multiple of four and no relative gets more than
` ( 4n − 1), then the number of ways in which
Ex 47. Let f (n) denotes the number of different ways the rich man can write his will, is
the positive integer n can be expressed as the (a) 4 n − 1 − 1
sum of 1’s and 2’s. For example f ( 4) = 5, since (b) n n − 1
4 = 1 + 1 + 1 + 1, 1 + 1 + 2, 1 + 2 + 1, 2 + 1 + 1, 2 + 2
(c) 4 n − 1 ⋅ n n
(Note Order of 1’s and 2’s is important). Then,
f { f (6)} is (d) 4 n − 1 ⋅ n n − 1
(a) f ( 6) (b) f (12) Sol. Let us assume the total money that is to be distributed is
(c) f (13) (d) None of these 4K; (K ∈Natural numbers) and the relative xi receives
`pi .
Sol. As f (4 ) = 5 [given] n
∴ f (6) can be written using 1’s and 2’s as ⇒ ∑ pi = 4 K and 0 ≤ pi ≤ 4 n − 1
i=1
Number of
Number of 1’s Number of 2’s Thus, to obtain the number of integral solutions of
arrangements
p1 + p2 + p3 + ... + pn = 4 K , K = 1, 2, K is
0 3 3!
=1 Coefficient of x 4K in (1 + x + x 2 + K + x 4n − 1 )n ...(i)
3!
4n2 − n
2 2 4!
2 !2 !
=6 Let (1 + x + x 2 + K + x 4n − 1 )n = ∑ ar x r ...(ii)
r=0
4 1 5! Now, required coefficients = a4 + a8 + a12 + K
=5
4! Putting x = 1, − 1, i and − i in Eq. (ii) and adding, we get
6 0 6! 4 (a0 + a4 + a8 + ...) = (4 n)n + 0 + 0 + 0
=1
6! ⇒ a0 + a4 + a8 + K = 4 n − 1 ⋅ nn ...(iii)
Total 13 From Eqs. (i) and (iii), we get
Coefficient of x 4K in (1 + x + x 2 + K + x 4n − 1 )n is
∴ f (6) = 13
∴ f { f (6)} = f (13) (4 n − 1 ⋅ nn − 1).
Hence, (c) is the correct answer. Hence, (d) is the correct answer.
300
Ex 50. A seven-digit number made up of all distinct
digits 8, 7, 6, 4, 2, x and y is divisible by 3.
Sol. It is possible in two mutually exclusive cases.
Case I 2 children get none, one child gets three and all
remaining children get one each.
6
Permutation and Combination
Then, possible number of order pair ( x, y) is
Case II 2 children get none, 2 children get 2 each and all
(a) 4 remaining children get one each.
(b) 8 10!
In first case, number of ways = ⋅ 10!
(c) 2 3!2! 7!
(d) None of the above 10!
In second case, number of ways = ⋅ 10!
Sol. We have, 8, 7, 6, 4, 2, x and y (2!)4 ⋅ 6!
Any number is divisible by 3, if sum of digits is divisible Thus, total number of ways
by 3. 1 1
= (10!)2 +
i.e. x + y + 27 is divisible by 3. 3!2! 7! (2!)4 6!
x and y can take values from 0, 1, 3, 5, 9. Hence, (b) is the correct answer.
∴ Possible pairs are (5, 1), (3, 0), (9, 0), (9, 3), (1, 5),
Ex 52. A delegation of five students is to be formed
(0, 3), (0, 9) and (3, 9). from a group of 10 students. If three particular
Hence, (b) is the correct answer. students want to remain together whereas two
particular students do not want to remain
Ex 51. 10 different toys are to be distributed among together, then the number of selection is
10 children. Total number of ways of (a) 10 (b) 20
distributing these toys so that exactly (c) 30 (d) None of these
2 children do not get any toy, is equal to Sol. Let A , B and C want to remain together and D, E do not
1 1 want to remain together.
(a) (10!) 2 +
3! 2! 7! ( 2!) 5 6! In Out Number of selections
A, B, C , D E 5
C1 = 5
1 1
(b) (10!) 2
+ A, B, C , E D 5
C1 = 5
3! 2! 7! ( 2!) 4 6!
D A, B, C , E C4 = 5
5
1 1
(c) (10!) 2
+ E A, B, C , D C4 = 5
5
3! 7! ( 2!) 5 6!
∴ Required number of selections
1 1 = 5C 1 + 5C 1 + 5C 4 + 5C 4 = 20
(d) (10!) 2
+
3! 7! ( 2!) 4 6! Hence, (b) is the correct answer.
Type 2. More than One Correct Option
Ex 53. If m and n are positive integers more than or Ex 54. Let n be a positive integer with
equal to 2, m > n, then ( mn)! is divisible by f ( n) = 1! + 2! + 3! + K + n! and P ( x ), Q ( x ) be
(a) ( m!) n polynomials in x such that f ( n + 2) = P ( n)
f ( n +1) + Q ( n) f ( n) for all n ≥1. Then,
(b) ( n !) m
(a) P ( x ) = x + 3
(c) ( m + n )! (b) Q ( x ) = − x − 2
(d) ( m − n )! (c) P ( x ) = − x − 2
Sol.
(mn)!
is the number of ways of distribution mn distinct (d) Q ( x ) = x + 3
(m!)n
Sol. Q f (n) = 1! + 2 ! + 3! + K + n!
objects in n persons equally.
f (n + 1) = 1! + 2! + 3! + K + (n + 1)!
(mn)!
Hence, is an integer ⇒ (m!)n /(mn)! f (n + 2) = 1! + 2! + 3! + K + (n + 2)!
(m!)n
∴ f (n + 2) − f (n + 1) = (n + 2)! = (n + 2)(n + 1)!
Similarly, (n!)m /(mn)!
= (n + 2)[ f (n + 1) − f (n)]
Further m + n < 2m ≤ mn
⇒ (m + n)!/(mn)! ⇒ f (n + 2) = (n + 3) f (n + 1) − (n + 2) f (n)
and m − n < m < mn ⇒ P (x ) = x + 3, Q (x ) = − x − 2
Hence, (a), (b), (c) and (d) are the correct answers. Hence, (a) and (b) are the correct answers.
301
6 Ex 55. The number of ways in which we can choose 2
distinct integers from 1 to 100, such that the
difference between them is atmost 10, is
Ex 56. n locks and n corresponding keys are available.
But the actual combination is not known. The
Objective Mathematics Vol. 1
maximum numbers of trials that are needed to
(a) 100
C2 − 90
C2 assign the keys to their corresponding locks,
(b) 100
C 98 − 90
C 88 are
(c) 100
C2 − 90
C 88 (a) n C 2
n
(d) None of the above (b) ∑ ( k − 1)
Sol. Let the chosen integers be x1 and x2. k =2
Let a be an integer before x1 , b be integer between x1 and (c) n !
x2 and c be integer after x2. (d) n + 1 C 2
∴ a + b + c = 98, where a ≥ 0, b ≥ 10, c ≥ 0.
Now, if we consider the choices, where difference is Sol. First key will be tried for atmost (n − 1) locks. Second
atleast 11, then the number of solutions is key will be tried for atmost (n − 2) locks and so on. Thus,
88 + 3 − 1
C 3 − 1 = 90C 2 the maximum number of trials needed
∴ The number of ways in which b is less than 10 is = (n − 1) + (n − 2) + K + 1
n(n − 1) n
100
C 2 − 90C 2 which is equal to options (a), (b) and (c). = = C2
2
Hence, (a), (b) and (c) are the correct answers. Hence, (a) and (b) are the correct answers.
Type 3. Assertion and Reason
Directions (Ex. Nos. 57-61) In the following Ex 58. Statement I If there are six letters,
examples, each example contains Statement I (Assertion) L1 , L2 , L3 , L4 , L5 , L6 and their corresponding
and Statement II (Reason). Each example has 4 choices six envelopes are E1 , E 2 , E 3 , E 4 , E 5 , E 6 .
(a), (b), (c) and (d) out of which only one is correct. The Letters having odd value can be put into odd
choices are value envelopes and even value letters can be
(a) Statement I is true, Statement II is true; Statement II put into even value envelopes, so that no letter
is a correct explanation for Statement I go into the right envelopes, the number of
(b) Statement I is true, Statement II is true; Statement II arrangements will be equal to 4.
is not a correct explanation for Statement I
Statement II If Pn is the number of ways in
(c) Statement I is true, Statement II is false
which n letter can be put in n corresponding
(d) Statement I is false, Statement II is true envelopes, such that no letters goes to correct
envelopes, then
Ex 57. Statement I The maximum number of points
of intersection of 8 circles of unequal radii 1 1 ( −1) n
is 56. Pn = n! 1 − + − K +
1! 2! n!
Statement II The maximum number of
Sol. L1 L3 L5 L2 L4 L6
points into which 4 circles of unequal radii and E1 E3 E5 E2 E4 E6
4 non-coincident straight lines intersect is 50. ∴ Number of ways
Sol. Statement I Two circles intersect in 2 points. 1 1 1 1 1 1
= 3! 1 − + − ⋅ 3! 1 − + − =4
∴Maximum number of points of intersection 1! 2! 3! 1! 2! 3!
= 2 × Number of selection of two circles from 8 circles Hence, (a) is the correct answer.
= 2 × 8C 2 = 2 × 28 = 56
Ex 59. Statement I The maximum value of k, such
Statement II 4 lines intersect each other in that (50) k divides 100! is 2.
4
C 2 = 6 points and 4 circles intersect each other in
Statement II If p is any prime number, then
2 × 4C 2 = 12 points.
power of p in n! is equal to
n n n
Further, one line and one circle intersect in two points.
So, 4 lines will intersect four circles in 32 points.
p + 2 + 3 + ... , where [ ] represents
∴Maximum number of points = 6 + 12 + 32 = 50 p p
Hence, (b) is the correct answer. the greatest integer function.
302
n n
Sol. If power of p is l in n!, then l = + 2 +...
p p
Statement II 3 m + 7 n has last digit zero,
when m is of 4k − 2 type and n is of 4l type
where k , l ∈W.
6
Permutation and Combination
Power of 2 in 100!
100 100 100 100 100 100
= + + + + + Sol. 3m gives 1 at unit place, if m = 4 k
2 22 23 24 25 26
and these are 5 in numbers.
= 50 + 25 + 12 + 6 + 3 + 1 = 97
So, power of 5 in 100! is 24. 7n gives 9 at unit place, if n = 4 l + 2
∴ Exponent of 25 in (100)! = 12 and these are 5 in numbers.
⇒ Maximum value of k is 12. ∴ Number of ways = 25
Hence, (d) is the correct answer. 3m gives 3 at unit place, if m = 4 k + 1
and these are 5 in numbers.
Ex 60. Statement I A five-digit number divisible by
3 is to be formed using the digits 0, 1, 2, 3, 4 7n gives 7 at unit place, if n = 4 l + 1
and 5 with repetition.The total number formed and these are 5 in numbers.
are 216. ∴ Number of ways = 25
Statement II If sum of digits of any number 3m gives 7 at unit place, if m = 4 k + 3
is divisible by 3, then the number must be and these are 5 in numbers.
divisible by 3. 7n gives 3 at unit place, if n = 4 l + 3
Sol. Number of numbers form by using 1, 2, 3, 4, 5 = 5! = 120 and these are 5 in numbers.
Number of numbers formed by using 0, 1, 2, 4 , 5
∴ Number of ways = 25
4 4 3 2 1 = 4 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 96
3m gives 9 at unit place, if m = 3k + 2
Statement I is incorrect and Statement II is correct.
Hence, (d) is the correct answer. and these are 5 in numbers.
7n gives 1 at unit place, if n = 4 l
Ex 61. Statement I The number of ordered pairs
(m, n); m, n ∈{1, 2, 3, ..., 20}, such that and these are 5 in numbers.
3 m + 7 n is a multiple of 10, is equal to 100. Hence, (c) is the correct answer.
Type 4. Linked Comprehension Based Questions
Passage I (Ex. Nos. 62-64) One of the most 100 100 100 100 100 100
= + + − − −
important techniques of counting is the principle of 3 5 7 15 21 35
exclusion and inclusion. Let A1, A2 , ..., A n be m sets and 100
+ = 55
n( A i ) represents the cardinality of the set A i (the number 105
of elements in the set A i ), then according to the principle ∴ Desired number = 100 − 55 = 45
of exclusion and inclusion Hence, (c) is the correct answer.
m
n( A1 ∪ A2 ∪ K ∪ A n ) = ∑ n( Ai ) − ∑ n( Ai ∩ Aj )
Ex 63. A six letters word is formed using the letters of
i =1 i ≠j
the word ‘ALMIGHTY’ with or without
+ ∑ n( Ai ∩ Aj ∩ Ak )
repetition. The number of words that contain
i < j <k
− K + ( − 1)m + 1n( A 1 ∩ A 2 ∩ K ∩ A m )
exactly three different letters, is
(a) 15600
In particular, if A , B, C are three sets, then
n( A ∪ B ∪ C ) = n( A) + n(B ) + n(C ) − n( A ∩ B ) (b) 30240
− n(B ∩ C ) − n(C ∩ A) + n( A ∩ B ∩ C ) (c) 8 P6 − 8 P3
Principle of exclusion and inclusion must be applied (d) None of the above
whenever there is a chance of repeated counting of Sol. We can select three different letters in 8 C 3 ways.
some of the samples. Suppose we select A, L, M.
Ex 62. The number of numbers from 1 to 100, which Let X = Set of words in which A is absent,
are neither divisible by 3 nor by 5 nor by 7, is Y = Set of words in which L is absent and
(a) 67 (b) 55 (c) 45 (d) 33 Z = Set of words in which M is absent.
Then, n ( X ∪ Y ∪ Z )= (26 + 26 + 26 ) − (16 + 16 + 16 ) + 0
Sol. A : Set of numbers, which are divisible by 3.
B : Set of numbers, which are divisible by 5. = 189
C : Set of numbers, which are divisible by 7. So, the number of six letter words formed by using A, L
Number of numbers, which are divisible by atleast one of and M = 36 − 189 = 540
3 or 5 or 7 ∴The desired number of words = 8C 3 × 540
= n( A ∪ B ∪ C ) = n( A ) + n( B ) + n(C ) − n( A ∩ B ) = 30240
− n( B ∩ C ) − n(C ∩ A ) + n( A ∩ B ∩ C ) 303
Hence, (b) is the correct answer.
6 Ex 64. The number of natural numbers less than or
equal to 2985984, which are neither perfect
squares nor perfect cubes, is
Sol. Consonants can be placed in
7!
2!2!
ways.
Then, there are 8 places and 4 vowels.
Objective Mathematics Vol. 1
7! 8 4!
(a) 2984124 ∴Number of ways = ⋅ C4 ⋅
2!2! 2!
(b) 2984244 Hence, (a) is the correct answer.
(c) 2959595
Passage III (Ex. Nos. 68-70) Let set S = {1, 2, 3, ...,
(d) None of the above n} be a set of first n natural and A ⊆ S . Suppose, n( A)
Sol. Desired number represents cardinal number and min (A) represents least
= 2985984 − { 2985984 + 3 2985984 − 6 2985984 } number among the elements of set A.
= 2985984 − 1860 = 2984124
[Q 2985984 = 212 ⋅ 36] Ex 68. The greatest value of min (A), where A ⊆ S and
Hence, (a) is the correct answer. n( A ) = r, 1 ≤ r ≤ n, is
(a) r
Passage II (Ex. Nos. 65-67) Consider the letters of
the word ‘MATHEMATICS’. There are eleven letters (b) ( n − r )
some of them are identical. Letters are classified as (c) n − r + 1
repeating and non-repeating letters. Set of repeating (d) r + 1
letters = {M , A , T } and set of non-repeating letters
= {H, E, I, C, S }. Sol. Consider, A = {n − r + 1, n − r + 2, ..., n} ⊂ S
Clearly, n( A ) = r and min ( A ) = n − r + 1
Ex 65. Possible number of words taking all letters at a ∴ Greatest value of min ( A ) = n − r + 1
time such that atleast one repeating letter is at Hence, (c) is the correct answer.
odd position in each word, is
9! 11! Ex 69. The number of subsets A of S for which
(a) (b)
2! 2! 2! 2! 2! 2! n( A ) = r and min ( A ) = k , is
11! 9! 9! n−k
(c) − (d) (a) Cr − 1
2! 2! 2! 2! 2! 2! 2! n
(b) C r − 1
Sol. Since, there are 5 even places and 3 pairs of repeated n− k +1
letters, therefore atleast one of these must be at an odd (c) Cr − 1
place. n− k −1
11! (d) Cr − 1
∴The number of ways =
2!2!2! Sol. k is the least member of A and rest of the r − 1members
Hence, (b) is the correct answer. of A are among the numbers k + 1, k + 2, k + 3, ..., n.
Ex 66. Possible number of words taking all letters at a ∴Number of ways = n − kC r − 1
time such that in each word both M’s are Hence, (a) is the correct answer.
together and both T’s are together but both A’s
are not together, is
11! 10!
Ex 70. The value of ∑ {min( A)} = k is
n( A) = r
8
(a) 7!⋅ C 2 (b) −
2! 2! 2! 2! 2! (a) n ⋅ n − k C r − 1
6! 4 ! 9!
(c)
2! 2!
(d)
2! 2! 2! (b) ( n + 1) ⋅ n − k C r − 1 − r ⋅ n − k + 1C r
Sol. Make a bundle of both M’s and another boundle of T’s. (c) k ⋅ n − k C r − 1 + n ⋅ n − k + 1C r
Then, except A’s we have 5 letters remaining, so M’s, T’s
and the letters except A’s can be arranged in 7! ways. (d) n C r
∴Total number of arrangements = 7! × 8C 2 n− r + 1
Hence, (a) is the correct answer. Sol. Now, ∑ [min ( A )] = ∑ (k ⋅ n − kC r − 1 )
A⊆ S k=1
|A | = r
Ex 67. Possible number of words in which no two n− k n− k
vowels are together, is Now, k C r − 1 = [(n + 1) − (n − k + 1)]⋅ Cr − 1
n− k n− k
(a)
7! 8
⋅ C4 ⋅
4! = (n + 1) ⋅ C r − 1 − (n − k + 1) ⋅ Cr − 1
2! 2! 2! (n − k + 1) ⋅ (n − k )!
7! 8 4! Also, (n − k + 1) ⋅ n − kC r − 1 =
(b) ⋅ C 4 ⋅ (r − 1)![ n − k − r + 1]!
2! 2! r ⋅ (n − k + 1)!
4! = = r ⋅ n − k + 1C r
(c) 7!⋅ C 4 ⋅
8 r!(n − k + 1 − r)!
2! ∴ k ⋅ n − kC r − 1 = (n + 1) ⋅ n − kC r − 1 − r ⋅ n − k + 1C r
7! 8 4!
(d) ⋅ C4 ⋅ Hence, (b) is the correct answer.
304 2! 2! 2! 2!
Type 5. Match the Columns
Ex 71. Match the statements of Column I with values Sol. A. Required number of ways
6
Permutation and Combination
of Column II. = (2 + 1)(3 + 1)(4 + 1) − 1
= 59
Column I Column II B. The number of ways of selecting 3 points out of
A. The number of five-digit numbers having p. 77 12 points is 7 C 3 ways.
the product of digits 20 is
Hence, the number of triangles formed is
B. A man took 5 space plays out of an engine q. 31 12
C 3 − 7C 3 = 185
to clean them. The number of ways in which
he can place atleast two plays in the engine C. Required number of ways
from where they came out, is = Coefficient of x 10 in (1 + x + x 2 + K )4
C. The number of integers between 1 and r. 50 = Coefficient of x 10 in (1 − x )−4
1000 inclusive in which atleast two 10 + 4 − 1
consecutive digits are equal, is = C4 − 1
1 = C 3 = 286
13
D. The value of ∑ ∑ i ⋅ j is
15 1 ≤ i ≤ j ≤ 9
s. 181
D. Factorising the given number, we have
38808 = 23 ⋅ 32 ⋅ 72 ⋅ 11
Sol. A. There are two cases:
5! The total number of divisors of this number is same
Case I 5, 4, 1, 1, 1, = 20 as the number of ways of selecting some or all of
3!
two 2’s, two 3’s, two 7’s and one 11. Therefore, the
5!
Case II 5, 2, 2, 1, 1 = 30 total number of divisors
2!2! = (3 + 1)(2 + 1)(2 + 1)(1 + 1)
∴ Total = 20 + 30 = 50 = 72
B. 5! − D5 − 5 ⋅ D4 Hence, the required number of divisors
[D5 stands for dearrangements of 5 things] = 72 − 2
= 120 − 44 − 5 × 9 = 31 = 70
C. 1000 − 93 − 92 − 9 = 181 A → p; B → p, q, r; C → p, q, r, s; D → p
1 1
D. ∑ ∑
15 1 ≤ i ≤ j ≤ 9
i⋅ j = ∑ ∑ i⋅ j
15 1 ≤ i ≤ j ≤ 9
Ex 73. Match the statements of Column I with values of
Column II. Consider the word ‘HONOLULU’.
+ (12 + 22 + K + 92 ) Column I Column II
(1 + 2 + K + 9)2 − (12 + 22 + .. + 92 ) A. Number of words that can be formed using p. 26
the letters of given word in which
1 + (12 + 22 + K + 92 ) consonants and vowels are alternate, is
=
15 2 B. Number of words that can be formed q. 144
without changing the order of vowels, is
1 9 × 10
2
9 × 10 × 19 C. Number of ways in which 4 letters can be r. 840
= + = 77 selected from the letters of the given word,
30 2 6 is
∴ A → r; B → q; C → s, D → p
D. Number of words in which two O’s are s. 900
together but U’s are separated, is
Ex 72. Match the statements of Column I with values
of Column II. 4! 4!
Sol. A. Required number of ways = 2!⋅ ⋅ = 144
Column I Column II 2! 2!2!
4!
A. The total number of selections of p. greater than 50 B. Required number of words = 8C 4 ⋅ = 840
fruits which can be made from, 3 2!
bananas, 4 apples and 2 oranges, C. Required number of ways of
is
Case I 2 alike + 2 alike = 3C 2
B. If 7 points out of 12 are in the the q. greater than 100
same straight line, then the Case II 2 alike + 2 different = 3C 14C 2
number of triangles formed is Case III All different = 5C 4 = 3 + 18 + 5
C. The number of ways of selecting r. greater than 150 = 26
10 balls from unlimited number of 5! 6
red, black, white and green balls, is D. Required number of words = ⋅ C 2 = 900
2!
D. The total number of proper divisors s. greater than 200
A → q; B → r; C → p; D → s
of 38808 is
305
6 Type 6. Single Integer Answer Type Questions
Ex 74. The possible number of ordered triads Coefficient of x 12 in
Objective Mathematics Vol. 1
( m, n, p), where m, n, p ∈ N is (6250) k, such (1 + x 3 + x6 + ...)(x 2 + x 4 + x6 + ...)(x + x 2 + x 3 + ...)
that 1 ≤ m ≤ 100, 1 ≤ n ≤ 50, 1 ≤ p ≤ 25 and = 12
2 m + 2 n + 2 p is divisible by 3, then k is _____. So, total ways = 1 + 2 + 4 + 12 = 19
Hence, λ = 1
Sol. (5) Here, 2m + 2n + 2p = (3 − 1)m + (3 − 1)n + (3 − 1)p
= 3k + (− 1)m + (− 1)n + (− 1)p (k ∈ I ) Ex 76. If the number of ordered triplet (a, b, c), such
So that 2m + 2n + 2p is divisible by 3, if m, n, p all are that LCM ( a, b) =1000, LCM ( b, c) = 2000 and
odd or all are even. LCM ( c, a ) = 2000 is 10 k, then k is ________.
∴Number of possible ordered triads
= 50 × 25 × 12 + 50 × 25 × 13 = 31250 = 6250 × 5 Sol. (7)Q 1000 = 23 ⋅ 53 , 2000 = 24 ⋅ 53
But it is 6250k. So, a = 2A ⋅ 5R , b = 2B ⋅ 5S , c = 2C ⋅ 5T
∴ k =5 max( A , B ) = 3, max( A , C ) = max ( B , C ) = 4
max(R , S ) = max(R , T ) = max (S , T ) = 3
Ex 75. If the total number of garlands that can be There are 10 ways of choosing R , S , T .
formed using 3 flowers of 1st kind and
1 with all 3, 3 with one 2, 3 with one 1, 3 with one 0.
12 flowers of the 2nd kind is 19λ, then λ is
_______. C must be 4. There are 7 ways of choosing A , B.
Sol. (1) This is a case equivalent to distributing 12 identical 1 with both 3, 3 with A = 3, B not, 3 with B = 3 and A not
flowers into 3 identical boxes is same as x + y + z = 12 thus, 7 ways of choosing A , B , C , hence 7 ⋅ 10 = 70 ways
taking only different combinations of x , y and z. ⇒ k =7
Case I x = y = z = 4 only 1 way Ex 77. N is the number of ways in which a person can
Case II x = y ≠ z walk up a stairway which has 7 steps. If he can
2x + z = 12 take 1 or 2 steps up the stairs at a time, then the
(a) x > z, x − z = a, a ≥ 1 value of N /3 is _________ .
⇒ x=z+ a
So, 2x + z = 12 Sol. (7) If x denotes the number of times, he can take unit step
⇒ 3z + 2a = 12 , a ≥ 1, z ≥ 0 and y denotes the number of times, he can take 2 steps,
i.e. coefficient of x 12 in then x + 2 y = 7.
We must have x = 1, 3, 5.
(1 + x 3 + x6 + K )(x 2 + x 4 + K ) = 2
If x = 1, then steps will be 1, 2, 2, 2.
(b) x < z, z − x = a 4!
⇒ Number of ways = =4
⇒ z=x+ a 3!
So, 3x + a = 12, a ≥ 1, x ≥ 0 If x = 3, then steps will be 1, 1, 1, 2, 2.
i.e. coefficient of x 12 in ⇒ Number of ways =
5!
= 10
(1 + x 3 + x6 + x 9 + ...)(x + x 2 + ...) = 4 3! 2!
Case III x + y + z = 12, x ≠ y ≠ z If x = 5, then steps will be 1, 1, 1, 1, 1, 2.
Taking only different combinations of x , y and z. ⇒ Number of ways = 6C 1 = 6
Let x > y > z If x = 7, then steps will be 1, 1, 1, 1, 1, 1, 1.
⇒ y−z=a ⇒ Number of ways = 7C 0 = 1
⇒ y = z + a, a ≥ 1 Hence, total number of ways (N ) = 4 + 10 + 6 + 1 = 21
x − y=b
N 21
⇒ x = z + a + b, b ≥ 1 ⇒ =
3 3
x + y + z = 12
⇒ 3z + 2a + b = 12 ; a, b ≥ 1, z > 0 =7
306
Target Exercises
Type 1. Only One Correct Option
1. If α = mC 2 , then α
C 2 is equal to 11. Let A be a set of n (≥ 3) distinct elements. The
(a) m+ 1
C4 (b) m−1
C4 (c) 3 m+ 2
C4 (d) 3 m+ 1
C4
number of triplets (x, y, z) of the elements of A in
which atleast two coordinates same, is
n+1
2. If n C 3 + nC 4 > C 3 , then (a) n P3 (b) n3 − nP3 (c) 3n2 − 2n (d) 3n2 (n − 1)
(a) n > 6 (b) n > 7
(c) n < 6 (d) None of these
12. The total number of six-digit numbers that can be
formed, having the property that every succeeding
n−1
3. The value of ∑ nCr / ( nCr + nC r + 1 ) equals digit is greater than the preceding digit, is
(a) 9 C 3 (b) 10C 3 (c) 9 P3 (d) 10
P3
r=0
(a) n + 1 (b) n / 2 13. The number of four-digit numbers that can be made
(c) n + 2 (d) None of these
with the digits 1, 2, 3, 4 and 5 in which atleast two
n n
Pr
∑
4. The value of is digits are identical, is
r =1 r! (a) 4 5 − 5! (b) 505
(c) 600 (d) None of these
(a) 2n (b) 2n − 1 (c) 2n − 1 (d) 2n + 1
Targ e t E x e rc is e s
2n + 1 2n − 1 14. The total number of five-digit numbers of different
5. If Pn − 1 : Pn = 3 : 5, then n is equal to
digits in which the digit in the middle is the largest, is
(a) 4 (b) 6 (c) 3 (d) 8 9
(a) ∑ n P4 (b) 33(3!)
6. The number less than 1000 that can be formed using n= 4
the digits 0, 1, 2, 3, 4, 5 when repetition is not (c) 30(3!) (d) None of these
allowed is equal to
15. The number of nine non-zero digit numbers such that
(a) 130 (b) 131
(c) 156 (d) 155 all the digits in the first four places are less than the
digit in the middle and all the digits in the last four
7. The numbers greater than 1000 but not greater than places are greater than that in the middle, is
4000, which can be formed with the digits 0, 1, 2, 3, 4 (a) 2(4 !) (b) 3(7!) / 2 (c) 2(7!) (d) 4 P4 × 4P4
(repetition of digits is allowed), are
(a) 350 (b) 375 16. The total number of three-digit numbers, the sum of
(c) 450 (d) 576 whose digits is even, is equal to
8. A variable name in certain computer language must (a) 450 (b) 350 (c) 250 (d) 325
be either an alphabet or an alphabet followed by a 17. The permutations of n objects taken
decimal digit. The total number of different variable (i) atleast r objects at a time
names that can exist in that language is equal to (ii) atmost r objects at a time
(a) 280 (b) 290
(if repetition of the objects is allowed)
(c) 286 (d) 296
are respectively
9. The number of five-digit numbers that contain 7 nn − r + 1 nr + 1 − 1
(a) and
exactly once is n−1 n−1
(a) (41) (93 ) (b) (37) (93 ) nn− r + 1 nr (nr + 1 − 1)
(b) and
(c) (7) (94 ) (d) (41) (94 ) n −1 n−1
n− r
r
n (n − 1) nr + 1 − 1
10. The total number of flags with three horizontal strips (c) and
n−1 n−1
in order, which can be formed by using 2 identical red,
(d) None of these
2 identical green and 2 identical white strips, is equal
to 18. How many ten-digit numbers can be written by using
(a) 4! the digits 1 and 2?
(b) 3 × (4 !)
(a) 10
C 1 + 9C 2 (b) 210
(c) 2 × (4 !)
(d) None of these (c) 10
C2 (d) 10! 307
6 19. A five-digit number divisible by 3 is to be formed
using the numerals 0, 1, 2, 3, 4 and 5 without
repetition. The total number of ways this can be done
30. The number of 5-digit telephone numbers having
atleast one of their digits repeated is
(a) 90000 (b) 100000 (c) 30240 (d) 69760
Objective Mathematics Vol. 1
is
(a) 216 (b) 240 (c) 600 (d) 3125 31. If letters of the word ‘KUBER’ are written in all
possible orders and arranged as in a dictionary, then
20. The total number of 9-digit numbers which have all rank of the word ‘KUBER’ will be
different digits is (a) 67 (b) 68 (c) 65 (d) 69
(a) 10! (b) 9! (c) 9 ⋅ 9! (d) 10 ⋅ 10!
32. The total number of 4-digit numbers that are greater
21. Eight chairs are numbered 1 to 8. Two women and than 3000, that can be formed using the digits 1, 2, 3,
three men wish to occupy one chair each. First the 4, 5, 6 (no digit is being repeated in any number) is
women choose the chairs from amongst the chairs equal to
marked 1 to 4 and then the men select the chairs from (a) 120 (b) 240 (c) 480 (d) 80
amongst the remaining. The number of possible
arrangements is 33. In a country, no two persons have identical set of
(a) C 3 × C 2
4 4
(b) C 2 × P3
4 4 teeth and there is no person without a tooth, also no
(c) 4 P2 × 4P3 (d) None of these person has more than 32 teeth. If shape and size of
tooth is disregarded and only the position of tooth is
22. From 4 Officers and 8 Jawans, a committee of 6 is to considered, then maximum population of that
be chosen to include exactly one officer. The number country can be
of such committee is (a) 232 (b) 232 − 1
(a) 160 (b) 200 (c) 224 (d) 300 (c) Cannot be determined (d) None of these
23. Seven different lecturers are to deliver lectures in 34. n different toys have to be distributed among
seven periods of a class on a particular day. A, B and n children. Total number of ways in which these toys
Ta rg e t E x e rc is e s
C are three of the lecturers. The number of ways in can be distributed, so that exactly one child gets no
which a routine for the day can be made such that A toy, is equal to
delivers his lecture before B and B before C, is (a) n! (b) n !⋅n C 2
(a) 420 (b) 120 (c) (n − 1) ! ⋅ nC 2 (d) n !⋅ n − 1C 2
(c) 210 (d) None of these
24. The number of arrangements of letters of the word 35. The total number of permutations of k different
‘BANANA’ in which the two N’s do not appear things, in a row, taken not more than r at a time (each
together is thing may be repeated any number of times) is equal
(a) 40 (b) 60 (c) 80 (d) 100
to
kr − 1 k (k r − 1)
(a) k r − 1 (b) k r (c) (d)
25. How many different nine-digit numbers can be k −1 (k − 1)
formed from the number 223355888 by rearranging
its digits, so that odd digits occupy even positions? 36. A person predicts the outcome of 20 cricket matches
(a) 16 (b) 36 (c) 60 (d) 180 of his home team. Each match can result either in a
win, loss or tie for the home team. Total number of
26. Let A = {x | x is a prime number and x < 30}. The
number of different rational numbers whose ways in which he can make the predictions, so that
numerator and denominator belong to A, is exactly 10 predictions are correct, is equal to
(a) 90 (b) 180 (a) 20
C 10 ⋅ 210 (b) 20
C 10 ⋅ 320
(c) 91 (d) None of these (c) 20
C 10 ⋅ 310 (d) 20
C 10 ⋅ 220
27. Let S be the set of all functions from the set A to the 37. A team of four students is to be selected from a total
set A. If n ( A ) = k, then n ( S ) is of 12 students. Total number of ways in which team
(a) k ! (b) k k (c) 2k − 1 (d) 2k can be selected such that two particular students
refuse to be together and other two particular
28. Let A be the set of 4-digit numbers a1 a 2 a 3 a 4 , where
students wish to be together only, is equal to
a1 > a 2 > a 3 > a 4 , then n ( A ) is equal to
(a) 220 (b) 182
(a) 126 (b) 84
(c) 226 (d) None of these
(c) 210 (d) None of these
38. The total number of numbers that are less than 3⋅ 108
29. An n-digit number is a positive number with exactly and can be formed using the digits 1, 2, 3, is equal to
n-digits. Nine hundred distinct n-digit numbers are to 1 9 1 9
be formed using only the three digits 2, 5 and 7. The (a) (3 + 4 ⋅ 38 ) (b) (3 − 3)
2 2
smallest value of n for which this is possible is 1 1
308 (c) (7 ⋅ 38 − 3) (d) (39 − 3 + 38 )
(a) 6 (b) 7 (c) 8 (d) 9 2 2
39. Four couples (husband and wife) decide to form a
committee of four members. The number of different
committees that can be formed in which no couple
48. The number of different pairs of words
(****, ***) that can be made with the letters of
the word ‘STATICS’ is
6
Permutation and Combination
finds a place, is (a) 828 (b) 1260
(a) 10 (b) 12 (c) 14 (d) 16 (c) 396 (d) None of these
40. In an examination of 9 papers, a candidate has to pass 49. A shopkeeper sells three varieties of perfumes and he
in more papers than the number of papers in which he has a large number of bottles of the same size of each
fails in order to be successful. The number of ways in variety in his stock. There are 5 places in a row in his
which he can be unsuccessful, is showcase. The number of different ways of
(a) 255 (b) 256 (c) 193 (d) 319 displaying the three varieties of perfumes in the
showcase is
41. The number of 5-digit numbers that can be made
(a) 6 (b) 50
using the digits 1 and 2 and in which atleast one digit (c) 150 (d) None of these
is different, is
(a) 30 (b) 31 50. The number of ways in which 3 boys and 3 girls (all
(c) 32 (d) None of these are of different heights) can be arranged in a line, so
that boys as well as girls among themselves are in
42. In a club election, the number of contestants is one decreasing order of height from left to right, is
more than the number of maximum candidates for (a) 1 (b) 6!
which a voter can vote. If the total number of ways in (c) 20 (d) None of these
which a voter can vote be 62, then the number of
candidates is 51. The number of words of four letters containing equal
(a) 7 (b) 5 number of vowels and consonants, repetition
(c) 6 (d) None of these allowed, is
(a) 1052 (b) 210 × 243
43. Two players P1 and P2 play a series of ‘2n’ games.
Targ e t E x e rc is e s
(c) 105 × 243 (d) None of these
Each game can result in either a win or loss for P1 .
Total number of ways in which P1 can win the series 52. There are three teams each of a chairman, a
of these games, is equal to supervisor and a worker three different companies.
1 2n 2n 1 2n There are nine bonus of different denominations to
(a) (2 − C n ) (b) (2 − 2 ⋅ 2nC n ) be paid to these nine persons in all. In how many
2 2
1 1 ways can this be done with due respect to superiority
(c) (2n − 2nC n ) (d) (2n − 2 ⋅ 2nC n ) is given in every team?
2 2
(a) 865 (b) 129
44. In the decimal system of numeration, the number of (c) 1680 (d) None of these
6-digit numbers in which the sum of digits is
53. If two numbers are selected from numbers 1 to 25,
divisible by 5, is then the number of ways that their difference does
(a) 180000 (b) 540000 not exceed 10 is
(c) 5 × 105 (d) None of these
(a) 105 (b) 195
(c) 15C 2 (d) None of these
45. The number of possible outcomes in a throw of n
ordinary dice in which atleast one of the dice shows 54. The number of ways in which a mixed double game
an odd number, is can be arranged amongst 9 married couples, if no
(a) 6n − 1 (b) 3n − 1 husband and wife play in the same game, is
(c) 6n − 3n (d) None of these (a) 756 (b) 1512
(c) 3024 (d) None of these
46. The number of different matrices that can be formed
55. In a certain test, there are n questions. In this test
with elements 0, 1, 2 or 3, each matrix having 4
2n − i students gave wrong answers to atleast i
elements, is
question, where i = 1, 2, ... , n. If the total number of
(a) 3 × 24 (b) 2 × 4 4
wrong answers given is 2047, then n is equal to
(c) 3 × 4 4 (d) None of these (a) 10 (b) 11 (c) 12 (d) 13
47. There are 20 questions in a question paper. If no two 56. The number of natural numbers which are less than
students solve the same combination of questions but 2 ⋅ 108 and which can be written by means of the digit
solve equal number of questions, then the maximum 1 and 2, is
number of students who appeared in the (a) 772 (b) 870 (c) 900 (d) 766
examination, is 57. The number of times digit 3 will be written when
20
(a) C9 (b) 20C 11 listing the integer from 1 to 1000, is
20 309
(c) C 10 (d) None of these (a) 269 (b) 300 (c) 271 (d) 302
6 58. Five balls of different colours are to be placed in
three boxes of different sizes. Each box can hold all
five balls. The number of ways in which we can place
66. The value of 10 C 5 / 11 C 6 , when numerator and
denominator takes their greatest value, is
Objective Mathematics Vol. 1
6 5 10 10
(a) (b) (c) (d)
the balls in the boxes (order is not considered in the 11 11 6 5
box) so that no box remains empty is
(a) 150 (b) 300 67. The number of subsets {1, 2, 3, ..., n} having least
(c) 200 (d) None of these element m and greatest element k, 1≤ m < k ≤ n, is
(a) 2n − (k − m) (b) 2k − m − 2 (c) 2k − m− 1 (d) 2k − m + 1
59. The number of permutations of the letters of the word
‘HINDUSTAN’ such that neither the pattern ‘HIN’ 68. A class has 21 students. The class teacher has been
nor ‘DUS’ nor ‘TAN’ appear, are asked to make n groups of r students each and go to
(a) 166674 (b) 168474 (c) 166680 (d) 181434 zoo taking one group at a time. The size of group zoo
i.e. the value of r) for which the teacher goes to the
60. Two teams are to play a series of 5 matches between maximum number of times is (no group can go to the
them. A match ends in a win or loss or draw for a zoo twice)
team. A number of people forecast the result of each (a) 9 or 10 (b) 10 or 11 (c) 11 or 12 (d) 12 or 13
match and no two people make the same forecast for
the series of matches. The smallest group of people 69. A class has n students. We have to form a team of the
in which one person forecasts correctly for all the students including atleast two students and also
matches will contain n people, where n is excluding atleast two students. The number of ways
(a) 81 (b) 243 of forming the team is
(c) 486 (d) None of these (a) 2n − 2n (b) 2n − 2n − 2
(c) 2n − 2n − 4 (d) None of these
61. In a class of 20 students, every student had a hand
shake with every other student. The total number of 70. From 4 gentlemen and 6 ladies, a committee of five is
hand shakes were to be selected. The number of ways, in which the
Ta rg e t E x e rc is e s
(a) 180 (b) 190 (c) 200 (d) 210 committee can be formed so that gentlemen are in
majority, is
62. There are 10 persons among whom two are brother.
(a) 66 (b) 156
The total number of ways in which these persons can (c) 60 (d) None of these
be seated around a round table so that exactly one
person sit between the brothers, is equal to 71. If the total number of m elements subsets of the set
(a) (2!) (7!) (b) (2!) (8!) A = {a1 , a 2 , a 3 , ... , a n } is k times the number of
(c) (3!) (7!) (d) (3!) (8!) m elements subsets containing a 4 , then n is
(a) (m − 1) k (b) mk
63. The number of ways in which a couple can sit around (c) (m + 1) k (d) None of these
a table with 6 guests, if the couple take consecutive
seats, is 72. An urn contains 3 red pens, 4 green pens and 6
(a) 1440 (b) 720 yellow pens. The number of ways of drawing 4 pens
(c) 5040 (d) None of these from the urn, if atleast one red pen is to be included
in the draw, is (all the pens are different from each
64. The number of different garlands, that can be formed
other)
using 3 flowers of one kind and 3 flowers of other
(a) 500 (b) 505
kind, is (c) 510 (d) None of these
(a) 60 (b) 20
(c) 4 (d) 5 73. Two numbers are chosen from 1, 3, 5, 7, ... , 147, 149
and 151 and multiplied together in all possible ways.
65. In the next world cup of cricket, there will be 12
The number of ways which will give us the product a
teams, divided equally in two groups. Teams of each
multiple of 5, is
group will play a match against each other. From
(a) 1710 (b) 2900
each group, 3 top teams will qualify for the next (c) 1700 (d) None of these
round. In this round each team will play against
others once. Four top teams of this round will qualify 74. Let Tn denotes the number of triangles which can be
formed using the vertices of a regular polygon of n
for the semifinal round, where each team will play
sides. If Tn + 1 − Tn = 21, then n equals
against the other three. Two top teams of this round
(a) 5 (b) 7 (c) 6 (d) 4
will go to the final round, where they will play the
best of three matches. The minimum number of 75. If a polygon has 44 diagonals, then the number of its
matches in the next world cup will be sides are
(a) 54 (b) 53 (a) 11 (b) 7
310 (c) 38 (d) None of these (c) 8 (d) None of these
76. The sides AB, BC, CA of a ∆ABC have 3, 4, 5 interior
points, respectively on them. Total number of
triangles that can be formed using these points as
85. The sides AB, BC and CA of a ∆ABC have a, b and c
interior points on them respectively, then the number
of triangles that can be constructed using these
6
Permutation and Combination
vertices, is equal to interior points as vertices, is
ab + bc + ca
(a) 135 (b) 145 (c) 178 (d) 205 (a) (b) Σ ab (a + b − 2)
2
77. ABCD is a convex quadrilateral 3, 4, 5 and 6 points (c) Σ ab (a + b) (d) None of these
are marked on the sides AB, BC, CD and DA,
respectively. The number of triangles with vertices 86. The number of points in the cartesian plane with
on different sides is integral coordinates satisfying the inequalities
(a) 270 (b) 220 x ≤ k , y ≤ k , x − y ≤ k, is
(c) 282 (d) None of these (a) (k + 1)3 − k 3 (b) (k + 2)3 − (k + 1)3
(c) (k 2 + 1) (d) None of these
78. In a polygon, no three diagonals are concurrent. If
the total number of points of intersection of 87. Given 5 different green dyes, 4 different blue dyes
diagonals interior to the polygon is 70, then the and 3 different red dyes. The number of
number of diagonals of the polygon is combinations of dyes, which can be chosen taking
(a) 20 (b) 28 atleast one green and one blue dye, is
(c) 8 (d) None of these
(a) 3600 (b) 3720
79. n lines are drawn in a plane such that no two of them (c) 3800 (d) None of these
are parallel and no three of them are concurrent. The 88. Six X ′s have to be placed in the squares of the figure
number of different points at which these lines will given below such that each row contains atleast
cut, is one X ′. The number of ways in which this can be
n
(a) ∑k (b) n (n − 1) done is
Targ e t E x e rc is e s
k=1
2
(c) n (d) None of these
80. If n objects are arranged in a row, then the number of
ways of selecting three objects, so that no two of
them are next to each other, is
(n − 2) (n − 3) (n − 4 ) n− 2
(a) (b) C3
3 (a) 26 (b) 27
(c) n − 2C 3 + n − 3C 2 (d) None of these (c) 22 (d) None of these
81. There are three coplanar parallel lines. If any p 89. The total number of six-digit numbers x1 x 2 x 3 x 4 x 5 x 6
points are taken on each of the lines, the maximum having the property x1 < x 2 ≤ x 3 < x 4 < x 5 ≤ x 6 , is
number of triangles with vertices at these points is (a) 10
C6 (b) 12C6
(a) 3 p2 ( p − 1) + 1 (b) 3 p2 ( p − 1) (c) 11
C6 (d) None of these
(c) p2 (4 p − 3) (d) None of these
90. A class contains three girls and four boys. Every
82. In a plane, there are two families of lines y = x + r, Saturday five students go on a picnic, a different
y = − x + r, where r ∈{0, 1, 2, 3, 4}. The number of group of students is being sent each week. During the
squares of diagonals of length 2 formed by the lines, picnic, each girl in the group is given doll by the
is accompanying teacher. All possible groups of five
(a) 9 (b) 16 have gone once, the total number of dolls the girls
(c) 25 (d) None of these have got, is
83. Line L1 contains l1 point and line L2 contains l2 point. (a) 21 (b) 45 (c) 27 (d) 24
If the points on L1 are joined to the points on L2 , then 91. The total number of ways of selecting two numbers
number of points of intersection of new lines, is from the set {1, 2, 3, 4, ...., 3n}, so that their sum of
(a) l1 C 2 × l2 C 2 divisible by 3, is
(b) 4 ⋅ l1C 2 × l2 C 2 2n2 − n 3n2 − n
(a) (b) (c) 2n2 − n (d) 3n2 − n
(c) 2 ⋅ l1C 2 × l2 C 2 2 2
(d) None of the above
92. If r > p > q, the number of different selections of
84. The number of triangles whose vertices are at the p + q things taking r at a time, where p things are
vertices of an octagon but none of whose sides identical and q other things are identical, is
happen to come from the octagon, is (a) p + q − r (b) p + q − r + 1
(a) 16 (b) 28 (c) 56 (d) 70 (c) r − p − q + 1 (d) None of these 311
6 93. The number of proper divisors of 2 p ⋅ 6q ⋅ 15r is
(a) ( p + q + 1) (q + r + 1) (r + 1)
(b) ( p + q + 1) (q + r + 1) (r + 1) − 2
102. A person writes letters to 6 friends and addresses the
corresponding envelopes. The number of ways in
which all 5 letters can be placed in wrong envelopes,
Objective Mathematics Vol. 1
(c) ( p + q) (q + r) r − 2 is
(d) None of the above (a) 264 (b) 210
(c) 206 (d) None of these
94. The number of even proper divisors of 1008 is
(a) 23 (b) 24 103. The number of ways in which m + n ( n ≤ m + 1)
(c) 22 (d) None of these different things can be arranged in a row such that no
95. The number of ways to fill each of the four cells of two of the n things may be together, is
(m + n)! m! (m + 1)!
the table with a distinct natural number such that the (a) (b)
m ! n! (m + n)!
sum of the numbers is 10 and the sum of the numbers
m ! n!
places diagonally are equal, is (c) (d) None of
(m − n + 1)!
these
104. Number of divisors of the form 4n + 2 ( n ≥ 0) of the
integer 240, is
(a) 2! × 2! (b) 4! (a) 4 (b) 8
(c) 2(4 !) (d) None of these (c) 10 (d) 3
96. In the figure, two 4-digit numbers are to be formed 105. Total number of non-negative integral solutions of
by filling the places with digits. The number of x1 + x 2 + x 3 = 10 is equal to
12
different ways in which the places can be filled by (a) C3 (b) 10C 3
12
digits so that the sum of the numbers formed is also a (c) C2 (d) 10C 2
4-digit number and in no place the addition is with
Ta rg e t E x e rc is e s
carrying, is 106. Total number of positive integral solutions of
Th H T U x1 + x 3 + 2x 3 = 15 is equal to
(a) 42 (b) 70
(c) 32 (d) None of these
+ 107. Total number of ways in which 15 identical blankets
can be distributed among 4 persons, so that each of
(a) 554 (b) 36 ⋅ (55)3 them gets atleast two blankets, is
(c) 454 (d) None of these (a) 10
C3 (b) 9 C 3
11
97. A bag contains 2 apples, 3 oranges and 4 bananas. (c) C3 (d) None of these
The number of ways in which 3 fruits can be
108. Total number of 3 letter words that can be formed
selected, if atleast one banana is always in the
from the letter of the word ‘AAAHNPRRS’ is equal to
combination (assume fruit of same species to be
(a) 210 (b) 237 (c) 247 (d) 227
alike) is
(a) 6 (b) 10 (c) 29 (d) 7 109. 15 identical balls have to be put in 5 different boxes.
Each box can contain any number of balls. Total
98. The number of divisors of the number 38808
number of ways of putting the balls into box, so that
(excluding 1 and the number itself), is each box contains atleast 2 balls, is equal to
(a) 70 (b) 72
(a) 9 C 5 (b) 10
C5
(c) 71 (d) None of these
(c) 6C 5 (d) 10
C6
99. The number of integral solutions of x1 + x 2 + x 3 = 0,
with x i ≥ − 5, is 110. The minimum marks required for clearing a certain
(a) 15C 2 (b) 16
C2 (c) 17
C2 (d) 18
C2
screening paper is 210 out of 300. The screening
paper consists of 3 sections each of Physics,
100. If 33! is divisible by 2n , then the maximum value of n Chemistry and Mathematics. Each section has 100 as
is equal to maximum marks. Assuming, there is no negative
(a) 30 (b) 31 (c) 32 (d) 33 marking and marks obtained in each section are
integers, the number of ways in which a student can
101. Let p be a prime number such that p ≥ 3. Let n = p ! + 1. qualify the examination, assuming no cut-off limit, is
The number of primes in the list (a) 210C 3 − 90C 3
n + 1, n + 2, n + 3, ...., n + p − 1is (b) 93C 3
(a) p − 1 (b) 2 (c) 213C 3
312 (c) 1 (d) None of these
(d) (210)3
111. Number of ways of distributing 10 identical objects
among 8 persons (one or many persons may not be
getting any object), is
115. The number of non-negative integral solutions of
a + b + c + d = n, n ∈ N , is
n− 1
6
Permutation and Combination
(a) P2
(a) 810 (b) 108 (c) 17
C7 (d) 10
C8 (n + 1) (n + 2) (n + 3)
(b)
6
112. A bag contains 3 black, 4 white and 2 red balls, all (c) n − 1C n − 4
the balls being different. The number of selections of
(d) None of these
atmost 6 balls containing balls of all the colours, is
(a) 42 (4 !) 116. The number of points ( x, y, z ) in space, whose each
(b) 26 × 4 ! coordinate is a negative integer such that
(c) (26 − 1) (4 !) x + y + z + 12 = 0, is
(d) None of the above (a) 385 (b) 55
(c) 110 (d) None of these
113. The number of ways to give 16 different things to
three persons, so that B gets 1 more than A and C 117. If a, b, c are three natural numbers in AP and
gets 2 more than B, is a + b + c = 21, then the possible number of values of
16!
(a) the ordered triplet (a, b, c) is
4 ! 5! 7! (a) 15
(b) 4 ! 5! 7! (b) 14
16! (c) 13
(c)
3! 5! 8! (d) None of the above
(d) None of the above
118. If a, b, c, d are odd natural numbers such that
114. The number of positive integral solutions of a + b + c + d = 20, then the number of values of the
x + y + z = n, n ∈ N , n ≥ 3, is ordered 4-tuple (a, b, c, d) is
(a) n − 1C 2 (a) 165
Targ e t E x e rc is e s
(b) n − 1 P2 (b) 455
(c) 310
(c) n (n − 1)
(d) None of the above
(d) None of the above
Type 2. More than One Correct Option
119. Number of ways in which three numbers in AP can 122. If x is the number of 5-digit numbers sum of whose
be selected from 1, 2, 3,..., n, is digits is even and y is the number of 5-digit numbers
2
n − 1 sum of whose digits is odd, then
(a) , if n is even0
2 (a) x = y
n (n − 2) (b) x + y = 90000
(b) , if n is even (c) x = 45000
4 (d) x < y
(n − 1)2
(c) , if n is odd
4 123. If n C α = nCβ , then it may be true that
(d) None of the above
(a) α = β
120. Kanchan has 10 friends among whom two are (b) α + β = 2n
(c) α + β = n
married to each other. She wishes to invite five of (d) All of the above
them for a party. If the married couples refuse to
attend separately, then the number of different ways 124. There are n married couples at a party. Each person
in which she can invite five friends, is shakes hand with every person other than her or his
(a) 8 C 5 (b) 2 × 8C 3 spouse. The total number of hand shakes must be
(c) 10C 5 − 2 × 8C 4 (d) None of these (a) 2nC 2 − n
(b) 2nC 2 − (n − 1)
121. A forecast is to be made of the results of five cricket (c) 2n (n − 1)
matches, each of which can be a win or a draw or a (d) 2nC 2
loss for Indian team. Let
125. There are n lines in a plane, no two of which are
p = number of forecasts with exactly 1 error parallel and no three of which are concurrent. If
q = number of forecasts with exactly 3 errors and plane is divided in u n parts, then
r = number of forecasts with all five errors (a) u4 = 11
Then, the correct statement(s) is/are (b) u3 = 7
(a) 2q = 5r (b) 8p = q (c) u2 = 3
(c) 8 p = 5r (d) 2( p + r) > q (d) un = un − 1 + 4 313
6 Type 3. Assertion and Reason
Directions (Q. Nos. 126-133) In the following 129. Statement I The number of ways of writing 1400
Objective Mathematics Vol. 1
questions, each question contains Statement I as a product of two positive integers is 12.
(Assertion) and Statement II (Reason). Each question Statement II 1400 is divisible by exactly three
has 4 choices (a), (b), (c) and (d) out of which only one is prime factors.
correct. The choices are
130. Statement I Let E = {1, 2, 3, 4} and F = {a, b}.
(a) Statement I is true, Statement II is true; Statement II
is a correct explanation for Statement I
Then, the number of onto functions from E to F is 14.
(b) Statement I is true, Statement II is true; Statement II Statement II Number of ways in which four
is not a correct explanation for Statement I distinct objects can be distributed into two different
boxes, if no box remains empty is 14.
(c) Statement I is true, Statement II is false
(d) Statement I is false, Statement II is true 131. Statement I Number of ways in which India can
win the series of 11 matches, if no match is drawn
126. Statement I The number of positive integral
is 210 .
solutions of abc = 30 is 27.
Statement II Number of ways in which three Statement II For each match, there are two
prizes can be distributed among three persons is 33 . possibilities, either India wins or loses.
127. Statement I Number of ways in which 10 identical 132. Statement I Number of ways in which Indian team
toys can be distributed among three students, if each (11 players) can bat, if Yuvraj wants to bat before
receives at least two toys, is 9 C 2 . Dhoni and Pathan wants to bat after Dhoni is 11!/3!.
Statement II Number of positive integral solutions Statement II Yuvraj, Dhoni and Pathan can be
of x + y + z + w = 7 is 6 C 3 . arranged in batting order in 3! ways.
Ta rg e t E x e rc is e s
128. Statement I ( n 2 )! / ( n !) n is a natural number, 133. Statement I When number of ways of arranging
21 objects of which r objects are identical of one type
∀ n ∈ N.
and remaining are identical of second type is
Statement II Number of ways in which n 2 objects maximum, then maximum value of 13 C r is 78.
can be distributed among n persons equally is
2n + 1
( n 2 )!/( n !) n . Statement II C r is maximum, when r = n.
Type 4. Linked Comprehension Based Questions
Passage I (Q. Nos. 134-136) We have to choose 11 Passage II (Q. Nos. 137-139) 12 persons are to be
players for cricket team from eight batsmen, six bowlers, arranged around two round tables such that one table
four all rounders and two wicket keepers in the following can accommodate seven persons and another five
questions. persons only.
134. The number of selections, when atmost one all 137. Number of ways in which these 12 persons can be
rounder and one wicket keeper will play, is arranged is
(a) 4C 1 × 14C 10 + 2C 1 × 14C 10 + 4C 1 × 2C 1 × 14C 9 + 14
C 11 (a) 12
C 5 6! 2! (b) 6!4 !
(b) 4C 1 × 15C 11 + 15C 11 (c) 12
C 5 6! 4 ! (d) None of these
(c) 4C 1 × 15C 10 + 15C 11
(d) None of the above 138. Number of ways of arrangements, if two particular
persons A and B do not want to be on the same
135. Number of selections when two particular batsmen table, is
do not want to play when a particular bowler will 10
(a) C 4 6! 4 ! (b) 210C6 6! 4 !
play, is 11
(a) 17
C 10 + 19
C 11 (b) 17
C 10 + 19
C 11 + 17
C 11 (c) C6 6! 4 ! (d) None of these
(c) 17
C 10 + 20
C 11 (d) 19
C 10 + 19
C 11 139. Number of ways of arrangement, if two particular
persons A and B want to be together and consecutive,
136. Number of selections when a particular batsman and a
is
particular wicket keeper do not want to play together,
(a) 10C 7 6! 3!2! + 10C 5 4 ! 5! 2!
is
(b) 10C 5 6! 3! + 10C 7 4 ! 5!
(a) 218 C 10 (b) 19 C 11 + 18C 10
(c) 10C 7 6! 2! + 10C 5 5! 2!
(c) 19 C 10 + 19
C 11 (d) None of these
314 (d) None of the above
Passage III (Q. Nos. 140-142) Number of ways of
distributing n different things into r different groups is r n
when blank groups are taken into account and is
141. 8 different balls can be distributed among 3 children
so that every child receives atleast 1 ball, is
(a) 3 8 8
(b) C 3
6
Permutation and Combination
r n − r C1 (r − 1)n + r C 2 (r − 2) 2 − .. + ( − 1)r − 1 r Cr − 1 when (c) 83 (d) None of these
blank groups are not permitted.
142. 5 letters can be posted into 3 letter boxes in
140. 4 candidates are competing for two managerial posts.
(a) 35 ways
In how many ways can the candidates be selected? (b) 53 ways
(a) 4 2 (b) 4C 2 (c) 5C 3 ways
(c) 24 (d) None of these (d) None of the above
Type 5. Match the Columns
143. Match the following. 145. Match the statements of Column I with values of
Column I Column II
Column II.
A. Number of straight lines joining any two p. 56
of 20 points of which four points are Column I Column II
collinear, is
B. Maximum number of points of q. 60
A. The value of p. ( n + 2 ) 2 n − 1 − ( n + 1)
intersection of 20 straight lines in the
1 ⋅ 1! + 2 ⋅ 2 ! + 3 ⋅ 3 ! + ...
plane, is
+ n ⋅ n! is
C. Maximum number of points of r. 185
intersection of 8 circles in the plane, is
2n
D. Maximum number of points of s. 190 B. The value of q. Cn
intersection of six parabolas, is n ⋅ n C1 + ( n − 1) ⋅ n C 2 + ( n − 1)
⋅ nC 3 + ... + 1 ⋅ nC n is
Targ e t E x e rc is e s
144. A function is defined as
f :{a1 , a 2 , a 3 , a 4 , a 5 , a 6 } → {b1 , b2 , b3 }.
C. The value of r. ( n + 1)! − 1
Then, match the following. 2
C 2 + 3C 2 + 4C 2 + ... + nC 2 is
Column I Column II
A. Number of surjective functions is p. divisible by 6
n n +1
s. C3
∑( C r )
n 2
B. Number of functions in which q. divisible by 2 D. The value of is
f( ai ) ≠ b i , is r=0
C. Number of invertible functions is r. divisible by 4
D. Number of many-one functions is s. divisible by 3 t. Cn − 1 − 1
2n
t. not possible
Type 6. Single Integer Answer Type Questions
146. A class has three teachers, Mr. P, Ms. Q and Mrs. R 149. There are n distinct white and n distinct black balls. If
and six students A, B, C, D, E, F. Number of ways in the number of ways of arranging them in a row, so
which they can be seated in a line of 9 chairs, if that neighbouring balls are of different colours, is
between any two teachers, there are exactly two 1152, then the value of n is _________ .
students, is k ! (18), then the value of k is 150. There are 4 oranges, 5 apples and 6 mangoes in a
____________ . fruit basket. If (21k + 20) ways can a person make a
147. Consider, the five points comprising the vertices of a selection of fruits from among the fruits in the
square and the intersection point of its diagonals. basket, when all fruits of the same type, are identical,
How many triangles can be formed using these then k refers to _________ .
points? 151. The number of ways of filling three boxes (named A,
B and C) by 12 or less number of identical balls, if no
148. If number of selections of 6 different letters that can
box is empty, box B has atleast 3 balls and box C has
be made from the words ‘SUMAN’ and ‘DIVYA’, so atmost 5 balls, is 55 λ, where λ r refers to _______ .
that each selection contains 3 letters from each word
is N 2 , then the value of N is _________ . 152. The number of non-negative integral solutions
2x + y + z = 21is 22( k ), where k refers to _______ .
315
Entrances Gallery
JEE Advanced/IIT JEE
1. Let n1 < n 2 < n 3 < n 4 < n 5 be positive integers such 3. Six cards and six envelopes are numbered 1, 2, 3, 4,
that n1 + n 2 + n 3 + n 4 + n 5 = 20. Then, the number 5, 6 and cards are to be placed in envelopes so that
of such distinct arrangements (n1 , n 2 , n 3 , n 4 , n 5 ) each envelope contains exactly one card and no card
is placed in the envelope bearing the same number
is _________ . [2014] and moreover the card numbered 1 is always placed
in envelope numbered 2. Then, the number of ways it
2. Let n ≥ 2 be an integer. Take n distinct points on a
can be done is [2014]
circle and join each pair of points by a line segment.
(a) 264 (b) 265 (c) 53 (d) 67
Colour the line segment joining every pair of
adjacent points by blue and the rest by red. If the 4. Consider the set of eight vectors
V = {a$i + b$j + ck$ ; a, b, c ∈ {−1, 1}}. Three
number of red and blue line segments are equal, then
non-coplanar vectors can be chosen from V in 2 p
the value of n is _________ . [2014]
ways. Then, p is _________ . [2013]
JEE Main/AIEEE
5. The number of integers greater than 6000 that can be (a) Statement I is true, Statement II is true; Statement II is
formed, using the digits 3, 5, 6, 7 and 8 without not a correct explanation for Statement I
repetition, is [2015] (b) Statement I is true, Statement II is false
(c) Statement I is false, Statement I is true
(a) 216 (b) 192
Ta rg e t E x e rc is e s
(d) Statement I is true, Statement II is true; Statement II is
(c) 120 (d) 72 a correct explanation for Statement I
6. Let A and B be two sets containing 2 elements and 11. There are 10 points in a plane, out of these 6 are
4 elements, respectively. The number of subsets of collinear. If N is the number of triangles formed by
A × B having 3 or more elements, is [2013] joining these points, then [2011]
(a) 256 (b) 220 (a) N > 190 (b) N ≤ 100
(c) 219 (d) 211
(c) 100 < N ≤ 140 (d) 140 < N ≤ 190
7. The total number of ways in which 5 balls of
12. There are two urns. Urn A has 3 distinct red balls and
different colours can be distributed among 3 persons
urn B has 9 distinct blue balls. From each urn, two
so that each person gets atleast one ball, is [2012]
balls are taken out at random and then transferred to
(a) 75 (b) 150 (c) 210 (d) 243 the other. The number of ways in which this can be
Directions (Q. Nos. 8-9) Let an denotes the number done is [2010]
of all n-digit positive integers formed by the digits 0, 1 or (a) 3 (b) 36
both, such that no consecutive digits in them are 0. Let (c) 66 (d) 108
bn be the number of such n digit integers ending with
13. From 6 different novels and 3 different dictionaries,
digit 1 and cn be the number of such n digit integers
4 novels and 1 dictionary are to be selected and
ending with digit 0. [2012]
arranged in a row on the shelf so that the dictionary is
8. Which of the following is correct? always in the middle. Then, the number of such
(a) a17 = a16 + a15 arrangements is [2009]
(b) c17 ≠ c16 + c15 (a) atleast 500 but less than 750
(c) b17 ≠ b16 + c16 (b) atleast 750 but less than 1000
(d) a17 = c17 + b16 (c) atleast 1000
(d) less than 500
9. The value of b6 is
14. In a shop, there are five types of ice-creams
(a) 7 (b) 8 (c) 9 (d) 11 available. A child buys six ice-creams.
10. Statement I The number of ways distributing Statement I The number of different ways the
10 identical balls in 4 distinct boxes such that no box child can buy the six ice-creams is 10 C 5 .
is empty is 9 C 3 . Statement II The number of different ways the
Statement II The number of ways of choosing any child can buy the six ice-creams is equal to the
number of different ways of arranging 6A’s and 4B’s
316 3 places from 9 different places is 9 C 3 . [2011] in a row. [2008]
(a) Statement I is true, Statement II is true; Statement II is
a correct explanation for Statement I
(b) Statement I is true, Statement II is true; Statement II is
18. If the letters of the word ‘SACHIN’ are arranged in
all possible ways and these words are written out as
in dictionary, then the word ‘SACHIN’ appears at
6
Permutation and Combination
not a correct explanation for Statement I
(c) Statement I is true, Statement II is false
serial number is [2005]
(d) Statement I is false, Statement II is true (a) 602 (b) 603 (c) 600 (d) 601
7−x
15. How many different words can be formed by 19. The range of the function f ( x ) = Px − 3 is [2004]
jumbling the letters in the word ‘MISSISSIPPI’ in
(a) {1, 2, 3} (b) {1, 2, 3, 4, 5, 6}
which no two S are adjacent? [2008] (c) {1, 2, 3, 4} (d) {1, 2, 3, 4, 5}
(a) 8 ⋅ 6C 4 ⋅ 7C 4
(b) 6 ⋅ 7 ⋅ 8C 4
20. How many ways are there to arrange the letters in the
word ‘GARDEN’ with the vowels in alphabetical
(c) 6 ⋅ 8 ⋅ 7C 4
order? [2004]
(d) 7 ⋅ 6C 4 ⋅ 8C 4
(a) 120 (b) 240 (c) 360 (d) 480
16. The set S :{1, 2, 3,... , 12} is to be partitioned into three
21. The number of ways of distributing 8 identical balls
sets A, B, C of equal size.
in 3 distinct boxes, so that none of the boxes is
Thus, A ∪ B ∪ C = S , A ∩ B = B ∩ C = A ∩ C = φ. empty, is [2004]
The number of ways to partition S is [2007] (a) 5 (b) 21 (c) 38 (d) 8 C 3
3 4
(a) 12!/ 3! (4 !) (b) 12!/ 3! (3!)
(c) 12!/ (4 !)3 (d) 12!/ (3!)4
22. The number of ways in which 6 men and 5 women
can dine at a round table, if no two women are to sit
17. At an election, a voter may vote for any number of together, is given by [2003]
candidates not greater than the number to be elected. (a) 6! × 5! (b) 30 (c) 5! × 4 ! (d) 7! × 5!
There are 10 candidates and 4 are to be elected. If a n
23. If C r denotes the number of combinations of n
Targ e t E x e rc is e s
voter votes for atleast one candidate, then the number things taken r at a time, then the expression
of ways in which he can vote, is [2006] n
C r + 1 + nC r − 1 + 2 × nC r equals [2003]
(a) 6210 (b) 385 (c) 1110 (d) 5040 n+ 2 n+ 2 n+ 1 n+ 1
(a) Cr (b) C r + 1 (c) Cr (d) Cr + 1
Other Engineering Entrances
24. How many 5-digit telephone numbers can be 28. 10 different toys are to be distributed among
constructed using the digits 0 to 9, if each number 10 children. Total number of ways of distributing
starts with 67 and no digit appears more than once? these toys, so that exactly two children do not get any
[Karnataka CET 2014] toy, is [Manipal 2014]
(a) 335 (b) 336 (c) 338 (d) 337 10! 10!
(a) (b)
2! 3! 7! (2!)4 × 6!
25. A student is allowed to select atmost n books from a
1 1 10! × 10! 25
collection of ( 2n + 1) books. If the number of ways in (c) (10!)2 + (d)
(2!) × 6! 2! × 3!
4
(2!)2 × 6! 84
which he can do this, is 64, then the value of n is
[BITSAT 2014] 29. Out of 7 consonants and 4 vowels, the number of
(a) 6 (b) n words (not necessarily meaningful) that can be made,
(c) 3 (d) None of these each consisting of 3 consonants and 2 vowels, is
26. S = {1, 2, 3, ... , 20} is to be partitioned into four sets A, [WB JEE 2014]
(a) 24800 (b) 25100 (c) 25200 (d) 25400
B , C and D of equal size. The number of ways it can
n
∑ k C r is equal to
be done, is [Manipal 2014]
30. m
Cr + 1 + [AMU 2014]
20! 20!
(a) (b) k =m
4 ! × 5! 45 n+ 1
20! 20! (a) nC r + 1 (b) Cr + 1
(c) (d) n
(5!)4 (4 !)5 (c) C r (d) None of these
27. In how many ways can a student choose a program of C1 C C Cn
31. + 2 2 + 3 3 + ... + n is equal to
5 courses, if 9 courses are available and 2 specific C0 C1 C2 Cn − 1 [AMU 2014]
courses are compulsory for every student? n (n − 1) n (n + 1)
[Manipal 2014] (a) (b)
2 2
(a) 34 (b) 36 (n + 1) (n + 2)
(c) (d) None of these
(c) 35 (d) 37 2 317
6 32. If n is an integer with 0 ≤ n ≤ 11, then the minimum
value of n ! (11 − n )! is attained, when a value of n is
42. A teacher takes 3 children from her class to the zoo at
a time as often as she can, but she does not take the
same three children to the zoo more than once. She
Objective Mathematics Vol. 1
[EAMCET 2014]
(a) 11 (b) 5 finds that she goes to the zoo 84 times more that a
(c) 7 (d) 9 particular child goes to the zoo. The number of
33. Find n C 21 , if n C10 = nC11 . children in her class is [GGSIPU 2011]
[J&K CET 2014]
(a) 12
(a) 1 (b) 0 (b) 10
(c) 11 (d) 10 (c) 60
(d) None of the above
34. Determine n, if 2n
C 2 : nC 2 = 9 : 2. [J&K CET 2014]
(a) 5 (b) 4 43. A student is allowed to select atmost n books from a
(c) 3 (d) 2 collection of ( 2n + 1) books. If the total number of
ways in which he can select atleast one book is 225,
35. Out of thirty points in a plane, eight of them are
then the value of n is [J&K CET 2011]
collinear. The number of straight lines that can be
(a) 6 (b) 5
formed by joining these points, is [EAMCET 2014] (c) 4 (d) 3
(a) 296 (b) 540
(c) 408 (d) 348 44. In how many ways can 5 prizes be distributed among
four students, when every student can take one or
36. The number of positive integers which can be formed
more prizes? [BITSAT 2011]
by using any number of digits from 0, 1, 2, 3, 4, 5 but
(a) 1024 (b) 625
using each digit not more than once in each number (c) 120 (d) 600
is [Karnataka CET 2013]
(a) 1200 (b) 1500 45. The number of ways in which the digits 1, 2, 3, 4, 3,
(c) 1600 (d) 1630 2, 1 can be arranged so that the odd digits always
Ta rg e t E x e rc is e s
37. The total number of ways in which five ‘+’ and three occupy the odd places, is [MP PET 2011]
‘−’ signs can be arranged in a line such that no two (a) 6 (b) 12
(c) 18 (d) 24
‘−’ sign occur together, is [AMU 2013]
(a) 10 46. The number of integers greater than 6000 that can be
(b) 20 formed with 3, 5, 6, 7 and 8, where no digit is
(c) 15
(d) None of the above
repeated, is [Kerala CEE 2011]
(a) 120 (b) 192
38. A student is to answer 10 out of 13 questions in an (c) 216 (d) 72
examination such that he must choose atleast 4 from (e) 202
the first five questions. The number of choices 47. If 56
Pr + 6 : 54
Pr + 3 = 30800 : 1, then the value of r is
available to him is [Manipal 2013] [Kerala CEE 2010]
(a) 140 (b) 196 (a) 40 (b) 51
(c) 280 (d) 346 (c) 41 (d) 510
(e) 101
39. A box contains 2 white balls, 3 black balls and 4 red
balls. In how many ways can 3 balls be drawn from 48. The number of permutations by taking all letters and
the box, if atleast one black ball is to be included in keeping the vowels of the word ‘COMBINE’ in the
the draw? [AMU 2013] odd places, is [WB JEE 2010]
(a) 64 (b) 24 (c) 3 (d) 12 (a) 96
(b) 144
40. Four speakers will address a meeting, where speaker (c) 512
Q will always speak after speaker P. Then, the (d) 576
number of ways in which the order of speakers can
49. From 12 books, the difference between number of
be prepared is [WB JEE 2012]
ways of selection of 5 books when one specified
(a) 256 book is always excluded and one specified book is
(b) 128
(c) 24 always included, is [Kerala CEE 2010]
(d) 12 (a) 64 (b) 118
(c) 132 (d) 330
41. Sum of digits in the unit’s place formed by the digits (e) 462
1, 2, 3 and 4 taken all at a time is
50. If n − 1 C 3 + n−1
[OJEE 2012]
C 4 > nC 3 , then n is just greater than
(a) 30
(b) 60 integer [WB JEE 2010]
(c) 59 (a) 5 (b) 6
318 (d) 61 (c) 4 (d) 7
Answers
Work Book Exercise 6.1
1. (d) 2. (b) 3. (d) 4. (b) 5. (a) 6. (d) 7. (a) 8. (d) 9. (a) 10. (c)
Work Book Exercise 6.2
1. (c) 2. (c) 3. (a) 4. (a) 5. (c)
Work Book Exercise 6.3
1. (d) 2. (b) 3. (b) 4. (a) 5. (d) 6. (a) 7. (b)
Work Book Exercise 6.4
1. (c) 2. (c) 3. (b) 4. (a) 5. (d) 6. (d) 7. (c) 8. (c) 9. (b) 10. (a)
11. (d) 12. (a) 13. (a) 14. (a) 15. (d) 16. (b) 17. (a) 18. (d) 19. (b) 20. (b)
21. (b) 22. (d) 23. (d) 24. (a) 25. (b) 26. (c) 27. (d)
Work Book Exercise 6.5
1. (c) 2. (d) 3. (a) 4. (a) 5. (c) 6. (d)
Work Book Exercise 6.6
1. (c) 2. (a) 3. (b) 4. (a) 5. (c) 6. (d) 7. (d) 8. (b) 9. (c) 10. (b)
11. (c) 12. (c) 13. (b) 14. (d) 15. (c) 16. (b) 17. (b) 18. (a) 19. (a) 20. (b)
21. (c) 22. (b) 23. (b) 24. (c) 25. (c)
Targ e t E x e rc is e s
Target Exercises
1. (d) 2. (a) 3. (b) 4. (b) 5. (a) 6. (b) 7. (b) 8. (c) 9. (a) 10. (a)
11. (b) 12. (a) 13. (b) 14. (d) 15. (d) 16. (a) 17. (d) 18. (b) 19. (a) 20. (c)
21. (d) 22. (c) 23. (d) 24. (a) 25. (c) 26. (c) 27. (b) 28. (c) 29. (b) 30. (d)
31. (a) 32. (b) 33. (b) 34. (b) 35. (d) 36. (a) 37. (c) 38. (c) 39. (d) 40. (b)
41. (a) 42. (c) 43. (a) 44. (a) 45. (c) 46. (c) 47. (c) 48. (b) 49. (c) 50. (c)
51. (d) 52. (d) 53. (d) 54. (b) 55. (b) 56. (d) 57. (b) 58. (a) 59. (b) 60. (b)
61. (b) 62. (b) 63. (a) 64. (d) 65. (b) 66. (a) 67. (d) 68. (b) 69. (b) 70. (a)
71. (b) 72. (b) 73. (d) 74. (b) 75. (a) 76. (d) 77. (d) 78. (a) 79. (d) 80. (b)
81. (c) 82. (a) 83. (c) 84. (a) 85. (d) 86. (a) 87. (b) 88. (a) 89. (c) 90. (b)
91. (b) 92. (b) 93. (b) 94. (a) 95. (a) 96. (b) 97. (a) 98. (a) 99. (c) 100. (b)
101. (d) 102. (a) 103. (d) 104. (a) 105. (c) 106. (a) 107. (a) 108. (c) 109. (a) 110. (b)
111. (c) 112. (a) 113. (a) 114. (a) 115. (b) 116. (b) 117. (c) 118. (a) 119. (b,c) 120. (b,c)
121. (a,b,d) 122. (a,b,c) 123. (a,c) 124. (a,c) 125. (a,b) 126. (a) 127. (d) 128. (a) 129. (b) 130. (a)
131. (b) 132. (a) 133. (d) 134. (a) 135. (b) 136. (b) 137. (c) 138. (b) 139. (a) 140. (b)
141. (d) 142. (a) 143. (*) 144. (**) 145. (***) 146. (6) 147. (8) 148. (8) 149. (4) 150. (9)
151. (2) 152. (6)
* A → r; B → s; C → p; D → q
** A → p,q,r,s; B → p,q,r,s; C → p,q,r,s,t; D → s
*** A → r; B → p; C → s; D → q
Entrances Gallery
1. (7) 2. (5) 3. (c) 4. (5) 5. (b) 6. (c) 7. (b) 8. (a) 9. (b) 10. (a)
11. (b) 12. (d) 13. (c) 14. (d) 15. (d) 16. (c) 17. (b) 18. (d) 19. (a) 20. (c)
21. (b) 22. (a) 23. (b) 24. (b) 25. (c) 26. (c) 27. (c) 28. (d) 29. (c) 30. (b)
31. (b) 32. (b) 33. (a) 34. (a) 35. (c) 36. (d) 37. (b) 38. (b) 39. (a) 40. (d)
41. (b) 42. (d) 43. (c) 44. (a) 45. (c) 46. (b) 47. (c) 48. (d) 49. (c) 50. (d)
319
Explanations
Target Exercises 8. Total number of variables, if only alphabet is used is 26.
m(m − 1) Total number of variables, if alphabets and digits both
1. We have, α = C2 ⇒ α =
m
are used is 26 × 10. Hence, the total number of variables
2
α (α − 1) 1 m(m − 1) m(m − 1) is 26 (1 + 10 ) = 286.
α
∴ C2 = = ⋅ − 1 9. If 7 is used at first place, the number of numbers is 94
2 2 2 2
1 and for any other four places, it is 8 × 93 .
= m(m − 1) (m − 2 ) (m + 1) Now, required number of numbers = 94 + 4 × 8 × 93
8
1 = 93 (9 + 32 ) = 41 (93 )
= (m + 1) m (m − 1) (m − 2 ) = 3 m + 1C4
8
10. If all strips are of different colours, then number of flags
n+1
2. Given, n C3 + nC4 > C3 is 3 ! = 6 . When two strips are of same colour, then
n+1 n+1 n+1 number of flags is 3 C1 × (3 !/ 2 ) × 2C1 = 18.
⇒ C4 > C3 [Q Cr + Cr + 1 =
n n
Cr + 1]
n+1
∴Total number of flags = 6 + 18 = 24 = 4!
C4 n −2
⇒ n+1
>1 ⇒ >1 ⇒ n > 6 11. Total number of triplets without restriction is n × n × n.
C3 4 The number of triplets with all different coordinates is
n −1 n n −1 n
P3 .
Cr 1
3. We have, ∑ n
= ∑
Cr + nCr + 1 r = 0 n
Cr + 1
Therefore, the required number of triplets is n 3 − n P3 .
r =0
1+ n 12. x1 < x2 < x3 < x4 < x5 < x6 , when the number is
Cr
n −1 x1 x2 x3 x4 x5 x6 . Clearly, no digit can be zero. Also, all the
1
= ∑ n−r
digits are distinct. So, let us first select six digits from the
r = 0 1+ list of digits 1, 2, 3, 4, 5, 6, 7, 8, 9 which can be done in
r +1 9
C6 ways.
n −1 n −1
r +1
Ta rg e t E x e rc is e s
1 After selecting these digits, they can be put only in one
= ∑
n + 1 n + 1 r∑
= (r + 1)
order.
r =0 =0
1 n Thus, total number of such numbers = 9C6 × 1 = 9C3
= [1 + 2 + K + n ] =
(n + 1) 2 13. The number of numbers when repetition is allowed is 54 .
n The number of numbers when digits cannot be repeated
Pr n is 5 P4 .
4. We know that, = Cr
r! Therefore, the required number of numbers is
n
54 − 5 ! = 505.
∴ ∑ nCr = 2n − 1 [1 corresponding to n C0 ]
r =1
14. Number of
(2 n + 1)! (n − 1)! 3 Digits available ways of
5. Given, = Middle
for remaining four Pattern filling
(n + 2 )! (2 n − 1)! 5 digit
places remaining
(2 n + 1) (2 n ) 3
= four places
(n + 2 ) (n + 1) n 5
4 0, 1, 2, 3 4 3 × 3P3
⇒ 10 (2 n + 1) = 3 (n 2 + 3 n + 2 )
⇒ 3 n − 11 n − 4 = 0
2
⇒ (n − 4) (3 n + 1) = 0 5 0, 1,...,4 5 4 × 4 P3
∴ n=4
6. The number of one-digit numbers = 0 6 0, 1,...,5 … 5 × 5P3
The number of two-digit numbers = 5 × 5 = 25.
The number of three-digit numbers = 5 × 5 × 4 = 100. 7 0, 1,...,6 … 6 × 6P3
Hence, the total numbers are 131. 8 0, 1,...,7 … 7 × 7P3
7. Numbers greater than 1000 and less than or equal to
9 0, 1,...,8 … 8 × 8P3
4000 will be of 4-digits and will have either 1 or 2 or 3 in
the 1st place. After fixing 1 at first place, then 2nd place Thus, number of required numbers = 5292
can be filled by any of the 5-digits. Similarly, the 3rd
15. According to the given conditions, numbers can be
place can be filled up in 5 ways and 4th place can be formed by the following format:
filled up in 5 ways. Thus, there will be 5 × 5 × 5 = 125
ways in which 1 will be in first place but this also includes
1000. Hence, there will be 124 numbers having 1 in the
first place. Similarly, 125 for each 2 or 3. One number will
be there in which 4 will be in the first place i.e. 4000.
Hence, the required number of numbers Filled with 1, 2, 3, 4 Filled with 6, 7, 8, 9
= 124 + 125 + 125 + 1
320 ∴ Number of required numbers = 4 P4 × 4 P4
= 375
16. Total number of digits (sum of digits is even) = 450
17. (i) Required number of permutations = Number of
permutations of n objects (taken r at a time + taken
26. A = {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}. A rational number
is made by taking any two in any order. 6
Permutation and Combination
So, the required number of rational numbers
(r + 1) at a time + taken (r + 2 ) at a time ) = 10 P2 + 1 = 9 [including 1]
+ .... + taken n at a time]
27. Each element of the set A (domain) can be give the
= n r + n r + 1 + n r + 2 + ....+ n n image in the set A (codomain) in k ways. So, the
n−r +1 required number of functions.
r
n (n − 1)
=
n −1 i.e. n(S ) = k × k × ....(k times) = k k
(ii) Required number of permutations = Number of
28. Any selection of four digits from the ten digits 0, 1, 2, 3,
...,9 gives one such number. So, the required number of
permutations of n objects (taken 0 at a time
numbers = 10C4 = 210
+ taken 1 at a time + K + taken r at a time)
1 × (n r + 1 − 1) 29. Number of distinct n digit numbers using 2, 5, 7 = 3n .
= n 0 + n1 + n 2 + K + n r =
n −1 Now consider, 3n > 900 ⇒ n = 7 [for minimum]
18. Total number of required numbers 30. Number of numbers with repetition
= 2 × 2 × ... 10 times = 210 = 10 × 10 × 10 × 10 × 10 = 100000
Number of numbers without repetition
19. Number of numbers using 1, 2, 3, 4, 5 = 5 ! = 120 = 10 × 9 × 8 × 7 × 6 = 30240
Number of numbers using 0, 1, 2, 4, 5
∴Number of numbers with atleast one digit repeated
= 5 ! − 4 ! = 120 − 24 = 96
= 100000 − 30240 = 69760
[Q there are 4! numbers beginning with ‘0’]
∴Total number of required numbers = 120 + 96 = 216 31. Total words starting with B = 4 ! = 24
Total words starting with E = 4 ! = 24
20. Number of ways of filling first place = 9
Total words starting with KB = 3 ! = 6
Number of ways of filling remaining 8 places Total words starting with KE = 3 ! = 6
Targ e t E x e rc is e s
= 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 = 9! Total words starting with KR = 3 ! = 6
∴Total number of numbers = 9 × 9! The first word starting with KU will be KUBER.
21. Number of ways for women = 4 P2 So, rank of the word KUBER
Number of ways for men = 6 P3 = 24 + 24 + 18 + 1 = 67
∴ Total number of ways = 4 P2 × 6 P3 32. Total number of such numbers
= 4 ⋅ 5 ⋅ 4 ⋅ 3 = 240
22. Number of committees = 4C1 × 8C5 = 4 × 56 = 224 33. Maximum population = 2 32 − 1[as each position of tooth
23. As the order of A, B and C is not to change, therefore has two options]
number of ways of assigning three periods to them is 34. Total number of ways
7 n!
C3 . Now, we left with 4 periods for 4 lecturers that can = nC2 × nC1 × (n − 1) ! ⋅ ⋅ n ! = n !⋅ nC2
assign in 4! ways. 2 ! (n − 2 )!
∴Required number of ways = 7C3 × 4 ! 35. Total number of permutations = 0 + k + k 2 + k 3 +K+ k r
=
7!
× 4! =
7!
= 840 k (k r − 1)
=
3! × 4! 3! (k − 1)
24. Total number of arrangements of the letters of the given 36. Matches whose predictions are correct can be selected
word in 20 C10 ways. Now, each wrong prediction can be made
6!
= = 60 in 2 ways. Thus, total number of ways = 20
C10 ⋅ 210 .
3! 2 !
Number of arrangements in which two N’s are together 37. LetS1 andS 2 refuse to be together andS 3 andS 4 want to
5! be together only. Total number of ways, when S 3 and S 4
= = 20
3! are selected
∴ Number of arrangements in which two N’s are not = (8 C2 + 2C1 ⋅ 8C1 ) = 44
together Total number of ways when S 3 and S 4 are not selected
= 60 − 20 = 40 = (8 C2 + 2C1 ⋅ 8C3 ) = 182
25. There are 5 odd places and that can be filled by using Thus, total number of ways = 44 + 182 = 226
5! 38. Formed number can be atmost of nine digits.
2, 2, 8, 8, 8 in = 10 ways and remaining 4 even
2 ! 3! Total number of such numbers
places can be filled by using 3, 3, 5, 5 in
4!
= 6 ways = 3 + 32 + 33 + K + 38 + 2 ⋅ 38
2! × 2! 3 (38 − 1) 39 − 3 + 4 ⋅ 38 7 ⋅ 38 − 3
= + 2 ⋅ 38 = =
∴Total number of numbers = 10 × 6 = 60 3−1 2 2 321
6 39. The number of committees of 4 gentlemen = 4C4 = 1
The number of committees of 3 gentlemen, 1 wife
= 4C3 × 1C1
after selecting 3 gentlemen, only 1
49. Possibilities
One triplet, two different
Selections
3
C1 × 2C 2 3
Arrangements
C1 × 2C 2 ×
5!
3!
= 60
Objective Mathematics Vol. 1
wife is left, which can be included
Two pairs, one different 3
C 2 × 1C1 C 2 × 1C1 ×
3 5!
= 90
2 !2 !
The number of committees of 2 gentlemen, 2 wives
= 4C2 × 2C2 ∴Required number of ways = 150
The number of committees of 1 gentleman, 3 wives 6!
= 4C1 × 3C3
50. Total number of ways = = 20
3! × 3!
The number of committees of 4 wives = 1
51. The number of selections of 1 pair of vowels and 1 pair of
∴The required number of committees
consonants = 5C1 × 21C1
= 1 + 4 + 6 + 4 + 1 = 16
40. The candidate is unsuccessful, if he fails in 9 or 8 or 7 or ∴Required number of words
4!
6 or 5 papers. = 5C1 × 21C1 × + 5C2 × 21C2 × 4 !
2!2!
∴The number of ways to be unsuccessful
= 9C9 + 9C8 + 9C7 + 9C6 + 9C5 52. Required number of ways = 3 ! ⋅ 3 ! ⋅ 3 ! = 216
= 1 + 9 + 36 + 84 + 126 = 256 53. If x = 1, y ∈{1, 2,...,11}
41. Total number of numbers of without restriction = 2 5 x = 2, y ∈ {1, 2,...,12}
Two numbers have all the digits equal. So, the required ... ... ... ...
number of numbers = 2 5 − 2 x = 11, y ∈ {1, 2,..., 21}
= 32 − 2 = 30
∴Number of ways = 11 + 12 + K + 21
42. 62 = nC1 + nC2 + nC3 + K + nCn − 1
54. We can choose two men out of 9 in 9 C2 ways. Since, no
= 2 − C0 − Cn = 2 − 2
n n n n
husband and wife are to play in the same game, two
⇒ 62 + 2 = 2 n women out of the remaining 7 can be chosen in
∴ 2 n = 64 = 2 6 ⇒ n = 6
7
Ta rg e t E x e rc is e s
C2 ways. If M1, M2 , W1 and W2 are chosen, then a team
may consist of M1 and W1 or M1 and W2 . Thus, the number
43. Total number of ways
n
of ways of arranging the game is
9× 8 7 × 6
= ∑ Cn + r =
2n 2n
Cn + 1 + K + 2n
C2 n 9
C2 ⋅7 C2 = × × 2 = 1512
r =1 2 2
1 2n
= (2 − 2n
Cn ) 55. The number of students answering exactly i (1 ≤ i ≤ n − 1)
2
× × × × × × questions wrongly is 2 n − i − 2 n − i − 1. The number of
44. Ways 9 10 10 10 10 2 students answering all n questions wrongly is 2 0 .
0 cannot be placed in the first place. In the other four
places, any digit can go. After filling the first five places, Thus, the total number of wrong answers is
the last place can be filled by 0 or 5. 1 (2 n − 1 − 2 n − 2 ) + 2 (2 n − 2 − 2 n − 3 )
∴Required number of numbers = 9 × 10 4 × 2 = 18000 + K + (n − 1) (21 − 2 0 ) + n(2 0 )
n −1 n−2
45. Total number of outcomes = 6 × 6 × ... to n times = 6n =2 +2 + 2n − 3 + K + 20 = 2n − 1
Total number of outcomes which show only even Now, 2 n − 1 = 2047
numbers ⇒ 2 n = 2048 = 211
= 3 × 3 × 3 ...to n times = 3n
∴ n = 11
46. The matrix will be of the order 4 × 1 or 1 × 4 or 2 × 2. 56. The required numbers are
For each order, the number of different matrices
= The number of ways to fill four places by 0, 1, 2, 3 1, 2, 11, 12, 21, 22, ...,122222222
= 4× 4× 4× 4 Let us calculate how many numbers are these.
47. If r questions are solved by each student, then the There are 2 one-digit such numbers. There are 2 2
number of possible selections of questions is 20 Cr . two-digit such numbers and so on.
∴The number of students = 20Cr
There are 2 8 eight-digit such numbers. Also, the 9-digit
Q each student has solved different numbers beginning with 1 and written by means of 1
combinations of questions
and 2 are smaller than 2 ⋅ 10 8 . Thus, there are 2 8 such
∴The maximum number of students nine-digit numbers.
= Maximum value of 20 Cr = 20C10 Hence, the required number of numbers
because 20 C10 is the largest among 20 C0 , 20
C1,...,20 C20 = 2 + 2 2 + 2 3 + ....+ 2 8 + 2 8
being the middle one.
2 (2 8 − 1)
48. From every arrangement of 7 letters, we get a pair by = + 28
2 −1
putting a sign (,).
7! = 29 − 2 + 28
∴The required number of pairs = = 1260
322 2 !2 ! = 766
57. Since, 3 does not occur in 1000, we have to count the
number of times 3 occurs when we list the integers from
1 to 999. Any number between 1 and 999 is of the form
65. The number of matches in the first round = 6C2 + 6C2
The number of matches in the next round = 6C2
The number of matches in the semifinal round = 4C2
6
Permutation and Combination
xyz, where 0 ≤ x, y, z ≤ 9. Let us first count the numbers
in which 3 occurs exactly once. Since, 3 can occur at So, the required number of matches
one place in 3 C1 ways, there are 3 C1 (9 × 9) = 3 × 92 such = 6C2 + 6C2 + 6C2 + 4C2 + 2
numbers. Next, 3 can occur exactly two places in = 53
10
(3 C2 ) (9) = 3 × 9 such numbers. C5 6
66. 11
=
Lastly, 3 can occur in all three-digits in one number only. C6 11
Hence, the number of times 3 occurs is 67. Required set is { m, m + 1, ..., k}
1 × (3 × 92 ) + 2 × (3 × 9) + 3 × 1 = 300
⇒ Number of subsets = 2 k − ( m − 1)
58. One possible arrangements is
2 2 3 68. Number of groups each having r students = 21Cr
Three such arrangements are possible. Therefore, the Let G = Number of times the teacher can go to the zoo
number of ways is (5 C2 ) (3 C2 ) (1C1 ) (3) = 90 Z = Number of groups she can form = 21Cr
∴ G is maximum, when Z is maximum.
The other possible arrangements is 21 − 1 21 + 1
1 1 3 i.e. r= or
2 2
Three such arrangement are possible. In this case, the
i.e. ] r = 10 or 11
number of ways is (5 C1 ) (4 C1 ) (3 C3 ) (3) = 60
69. Required number of ways
Hence, the total number of ways = 90 + 60 = 150 = nC2 + nC3 + ... + nCn − 3 + nCn − 2
9! = 2 n − (n C0 + nC1 + nCn − 1 + nCn )
59. Total number of permutations =
2!
= 2 n − (1 + n + n + 1)
Number of permutation of those containing ‘HIN’ = 7 !
7! = 2 n − 2(n + 1)
Number of permutation of those containing ‘DUS’ =
Targ e t E x e rc is e s
2! 70. In a committee of 5 persons, the gentlemen will be in
Number of permutation of those containing ‘TAN’ = 7 ! majority, if the number of gentlemen ≥ 3.
Number of permutation of those containing ‘HIN’ and
‘DUS’ = 5! Total number of ways of forming committee
Number of permutation of those containing ‘HIN’ and = 4C3 × 6C2 + 4C4 × 6C1
‘TAN’ = 5! = 4 × 15 + 1 × 6
Number of permutation of those containing ‘TAN’ and = 66
‘DUS’ = 5!
71. Total number of m elements subsets of A = nCm
Number of permutation of those containing ‘HIN’,
‘DUS’ and ‘TAN’ = 3! Number of m elements subsets of A each containing the
9! 7 ! element a4 = Number of ways of choosing remaining
∴Required number = − 7 ! + 7 ! + − 3 × 5 ! − 3 !
2! 2 (m − 1) elements of the subset, out of remaining (n − 1)
= 168474 elements of A
60. The smallest number of people = n − 1Cm − 1
n −1
= Total number of possible forecast Given that, n
Cm = k ×Cm − 1
= Total number of possible results n! (n − 1)!
= 3 × 3 × 3 × 3 × 3 = 243 =k⋅
m ! (n − m)! (m − 1)! (n − m)!
61. Since, every pair of the students gives us a hand shake ⇒ n = km
∴Total number of hand shakes 72. Required number of ways
= Total number of pairs of students = Total number of ways of choosing 4 pens
= Number of ways of choosing two students out of − Number of ways of choosing 4 non-red pens
20 ! 3+ 4+ 6 4+ 6
= 20C2 = = C4 − C4 = 13C4 − 10C4
2 !⋅ 18 !
20 × 19 × 18 ! 13 ⋅ 12 ⋅ 11 ⋅ 10 10 ⋅ 9 ⋅ 8 ⋅ 7
= = 190 = −
2 × 18 ! 4⋅ 3⋅2 ⋅1 4⋅ 3⋅2 ⋅1
62. Total number of ways = 8 ! 2 ! = 715 − 210 = 505
63. A couple and 6 guests can be arranged in (7 − 1)! ways. 73. Out of the given numbers, the numbers which are
multiple of 5, are 5, 10, 15, 20,...,145.
But the two people forming the couple can be arranged
among themselves in 2 ! ways. This is an AP whose first term a = 5,
Common difference = 5
∴ Required number of ways = 6 ! × 2 ! = 1440
nth term, t n = 145
5! ∴ 145 = 5 + (n − 1) × 5
64. Required number of garlands = =5
2 ⋅ 2 !⋅ 3 ! ⇒ n = 29 323
6 Now, if total number of numbers is m, then
⇒
151 = 1 + (m − 1) × 2
m − 1=
151 − 1
⇒ m = 76
81. The number of triangles with vertices on different lines
= pC1 × pC1 × pC1 = p3
The number of triangles with two vertices on one line
Objective Mathematics Vol. 1
2
and the third vertex on any one of the other two lines
So, number of ways in which product is a multiple of 5 p ( p − 1)
= 3C1 { p C2 × 2 pC1} = 6 p ⋅
= Both two numbers from 5, 10, 15, 25,...,145 2
+ One number from 5, 10, 15, 25,...,145 and one from So, the required number of triangles
remaining numbers
= p3 + 3 p2 ( p − 1) = p2 (4 p − 3)
= 29C2 + 29 C1 × 76C1
= 29 × 14 + 29 × 76= 29 × 90 = 2610 82. There are two sets of five parallel lines at equal
distances.
74. Q n + 1C3 − nC3 = 21
m5
(n + 1) (n − 1) n n(n − 1) (n − 2 )
∴ − = 21 4 m4
6 6 m3
n(n − 1) m2
⇒ [(n + 1) − (n − 2 )] = 21 3
6 m1
⇒ n(n − 1) = 42
2
⇒ n 2 − n − 42 = 0 l5
⇒ n = − 6, 7 l4
1
∴ n =7 l3
n(n − 1) l2
75. Q n C2 − n = 44 ⇒ − n = 44
2 O l1
∴ n 2 − 3 n − 88 = 0
⇒ n = − 8 , 11 Clearly, lines like l1, l 3 , m1, m3 form a squares whose
∴ n = 11 diagonals length is 2. So, the number of required
squares = 3 × 3 = 9.
76. Total number of triangles =12 C3 − 3C3 − 4C3 − 5C3 = 205
Ta rg e t E x e rc is e s
Since, choices are (l1, l 3 ),(l 2 , l 4 ),(l 3 , l 5 ) for one set.
77. The number of triangles with vertices on sides
AB, BC, CD = C1 × C1 × C1
3 4 5 83. 1 2 3 4 5 l1
L1
Similarly, for other cases, the total number of triangles
= 3C1 × 4C1 × 5C1 + 3C1 × 4C1 × 6C1 + 3C1 × 5C1 × 6C1
+ 4C1 × 5C1 × 6C1 L2
= 342 1 2 3 4 5 l2
78. A selection of four vertices of the polygon gives an Required number of points on intersection of new lines
interior intersection.
= 2 × Number of choosing 1 point on L1 and 2 point on L2
∴The number of sides = n
⇒ n
C4 = 70 = 2 ⋅ l1C1 × l 2C2
⇒ n (n − 1) (n − 2 ) (n − 3) = 24 × 70 84. Required number of triangles
= 8×7 × 6× 5 = Total number of triangles
∴ n=8
− Number of triangles having two sides common
∴ The number of diagonals = 8 C2 − 8 = 20
− Number of triangles having one side common
79. n C2 is point of intersection. = 8C3 − 8 − 8 × 4 = 16
80. Let x0 be the number of objects to left of the first object 85. A
chosen, x1 be the number of objects between the first c
and the second, x2 be the number of objects between 1
the second and the third and x3 be the number of
objects to the right of the third object. We have, 2
x0 , x3 ≥ 0, x1, x2 ≥ 1 3
3 2
and x0 + x1 + x2 + x3 = n − 3 ...(i)
1
The number of solutions of Eq. (i) a
= Coefficient of y n − 3 in (1 + y + y 2 + ... ) (1 + y + y 2 + ... ) B C
1 2 3 b
( y + y 2 + y 3 + ... ) ( y + y 2 + y 3 + ... )
n−3
= Coefficient of y in y 2 (1 + y + y 2 + y 3 + ... )4 Required number of triangles
n−5
= Coefficient of y in (1 − y )−4 = Total number of ways of choosing 3 points
= Coefficient of y n − 5 in (1 + 4C1 y + 5C2 y 2 + 6C3 y 3 + ... ) − Number of ways of choosing all the 3 points
(n − 2 ) (n − 3) (n − 4) from AB or BC or CA
= n − 5 + 3Cn − 5 = n − 2C3 = = a+ b+c
C3 − (a C3 + bC3 + c C3 )
324 6
86. | x| ≤ k ⇒ − k ≤ x ≤ k
and | y| ≤ k ⇒ − k ≤ y ≤ k
| x − y| ≤ k ⇒ | y − x| ≤ k
...(i)
...(ii)
94. 1008 = 2 4 × 32 × 7
∴The required number of even proper divisors
= Total number of selections of atleast one 2 and any
6
Permutation and Combination
and
number of 3’s or 7’s
⇒ −k≤y−x≤k = 4 × (2 + 1) × (1 + 1) − 1 = 23
⇒ x−k≤y≤x+k ...(iii)
95. The natural numbers are 1, 2, 3 and 4.
Y
Clearly, in one diagonal we have to place 1, 4 and in the
other 2, 3.
k The number of ways = 2 ! × 2 ! = 4
3 96. If 0 is placed in the unit’s place of the upper number,
2
then the unit’s place of the lower number can be filled in
1
10 ways (filling by any one of 0, 1, 2, ..., 9).
X¢ X
–k –3 –2 –1 1 2 3 k If 1 is placed in the units place of the upper number, then
–1
–2 the unit’s place of the lower number can be filled in 9
–3 ways (filling by any one of 0, 1, 2, ..., 8) etc.
∴The unit’s column can be filled in 10 + 9 + 8 + .... + 1,
–k i.e. 55 ways. Similarly, for the second and the third
(2 –k)
(k
(1
,–
,–
,
Y¢ columns. The number of ways for the fourth column
k)
k)
= 8 + 7 + ... + 1 = 36
∴Number of points having integral coordinates ∴The required number of ways
= (2 k + 1)2 − 2[k + (k − 1) + ...+ 2 + 1] = 55 × 55 × 55 × 36
= (3 k 2 + 3 k + 1)
97. Coefficient of x 3 in
87. Total number of ways = 31 × 15 × 8 = 3720 ( x 0 + x1 + x 2 ) ( x 0 + x1 + x 2 + x 3 )( x1 + x 2 + x 3 + x 4 )
88. Total number of squares = 8. We are to choose 6 =6
Targ e t E x e rc is e s
squares for six X’s and this can be done is 8 C6 = 28 98. 38808 = 2 3 × 32 × 7 2 × 111
ways. This also include the possibility that either the top ∴ Divisors = 4 × 3 × 3 × 2 − 2 = 70
horizontal row does not have any X or the bottom 15 + 3 − 1
99. C3 − 1 = 17C2
horizontal row have no X. These possibilities are not
100. + + + +
required. 33 33 33 33 33
∴Total number of ways = 28 − 2 = 26 2 4 8 16 32
= 16 + 8 + 4 + 2 + 1 = 31
89. Total number of such numbers
101. For 1 ≤ k ≤ p − 1, n + k = p ! + k + 1, is clearly divisible
= 9C6 + 9C5 + 9C5 + 9C4 = 10C6 + 10
C5 = 11 C6
by k + 1. Therefore, there is no prime number in the
90. The number of times, a particular girl goes on picnic = 6C4 given list.
∴Total number of dolls given to the girls 102. The number of ways in which all 5 letters be placed in
6!
= 6 C4 ⋅ 3 = ⋅ 3 = 45 wrong envelopes
4! 2 ! 1 1 1 1 1
= 6 C5 . 5 ! 1 − + − + −
1! 2 ! 3 ! 4 ! 5 !
91. Case I 1 4 7 ... 3 n − 2
Case II 2 5 8 ... 3 n − 1 1 1 1 1
= 720 − + − = 264
2 6 24 120
Case III 3 6 9 ... 3 n
That means we must take 2 numbers from last row or 103. There is (m + 1) space between them (including the
one number each from first and second rows.
beginning and the end) and n things can be placed in all
Total number of ways = nC2 + nC1 ⋅ nC1z
possible spaces in m + 1 Pn ways.
n (n − 1) 3 n2 − n
= + n2 = ∴Required number of ways
2 2 m !⋅ (m + 1)!
= m ! ⋅ m + 1Pn =
92. The number of selections of p things from q identical (m + 1 − n )!
things are (r − p) things from q identical things = 1 × 1 104. We have, 240 = 2 4 × 31 × 51
p→ p p − 1... r − q
∴ There are 4 divisors of 240 which of the form
q → r − p r − p + 1... q 4n + 2, n ≥ 0.
Similarly, in all other cases, total number of ways
= p − (r − q ) + 1or q − (r − p) + 1 = p + q − r + 1 105. Number of non-negative integral solutions
= 12C10 = 12C2
93. 2 p ⋅ 6 q ⋅ 15r = 2 p + q ⋅ 3 q + r ⋅ 5r
∴The number of proper divisors 106. Q x1 + x2 + 2 x3 = 15 ⇒ x1 + x2 = 15 − 2 x3
= {Total number of selections from ( p + q ) twos, (q + r ) where, x3 can be 1, 2, ..., 7.
threes and r fives} − 2 Let x3 = r
= ( p + q + 1)(q + r + 1)(r + 1) − 2 ⇒ x1 + x2 = 15 − 2 r 325
6 Number of positive integral solutions of this equation
= (14 − 2 r )
∴Total number of solutions
or p+q + r + s=8
The required number of solutions
= The number of non-negative integral solutions of
Objective Mathematics Vol. 1
7 ⋅ 8
7
= ∑ (14 − 2 r ) = 14 ⋅ 7 − 2 ⋅ = 42 ( p + q + r + s = 8)
2 (8 + 3) (8 + 2 ) (8 + 1)
r =1
= = 165
6
107. x1 + x2 + x3 + x4 = 15 and xi ≥ 2
⇒ ( x1 − 2 ) + ( x2 − 2 ) + ( x3 − 2 ) + ( x4 − 2 ) = 7 119. If a, b and c are in AP, then a and c both are odd or both
⇒ y1 + y2 + y3 + y4 = 7, where yi = xi − 2 ≥ 0 are even.
⇒ 10 C7 = 10C3 are number of non-negative. Case I When n is even.
108. Coefficient of x in 3 The number of ways of selection of two even numbers, a
and c is n / 2 C2 . Number of ways of selection of two odd
{( x 0 + x1 + x 2 + x 3 ) ( x 0 + x1 + x 2 ) ⋅ ( x 0 + x1 )4 } numbers is n / 2 C2 .
= 247 Hence, the number of AP’s
109. Let the balls put in the box be x1, x2 , x3 , x4 and x5 . n n
− 1
2 2 n (n − 2 )
We have, 2 ⋅ C2 = 2 ⋅
n/ 2
=
x1 + x2 + x3 + x4 + x5 = 15, xi ≥ 2 2 4
⇒ ( x1 − 2 ) + ( x2 − 2 ) + ( x3 − 2 ) + ( x4 − 2 ) + ( x5 − 2 ) = 5 Case II When n is odd.
⇒ y1 + y2 + y3 + y4 + y5 = 5, yi = xi − 2 ≥ 0
The number of ways of selection of two odd numbers a
= 5 + 5 − 1C5 − 1 = 9C4 and c is ( n + 1)/ 2 C2 . The number of ways of selection of
110. Coefficient of x 210 in{( x 0 + x1 + x 2 + x 3 + ... + x100 )3 } two even numbers a and c is ( n − 1)/ 2 C2 .
= 93
C3 Hence, the number of AP’s = ( n + 1)/ 2C2 + ( n − 1)/ 2C2
n + 1 n + 1 n − 1 n − 1
111. Required number of ways = Number of non-negative − 1 − 1
Ta rg e t E x e rc is e s
integral solution of x1 + x2 + x3 + x4 + ... + x8 = 10 2 2 2 2
= +
2 2
= 10 + 8 − 1C8 − 1 = 17C7
1 (n − 1)2
= (n − 1)[(n + 1) + (n − 3)] =
112. The required number of selections 8 4
= 3 C1 × 4C1 × 2C1 (6 C3 + 6C2 + 6C1 + 6C0 ) 120. The number of ways of inviting, when the couple not
113. x + y + z = 16, y = x + 1 and z = y + 2 included is 8 C5 . The number of ways of inviting with the
∴ x = 4, y = 5, z = 7 couple included is 8 C3 . Therefore, the required number
Now, number of ways to give 16 different things of ways = 8C5 + 8C3 = 8C3 + 8C3 [Q 8C5 = 8C3 ]
=16 C4 × 12C5 × 7C7 10 ! 8!
Also, 10 C5 − 2 × 8C4 = −2 ×
114. x + y + z = n 5! 5! 4! 4!
10 × 9 × 8 × 7 × 6 8×7 × 6× 5
x, y, z ≥ 1or x − 1= a, y − 1 = b, z − 1 = c ≥ 0 = −2 ×
⇒ a+ b+c =n−3 120 24
8!
The required number of solutions = 9 × 4 × 7 − 140 = 112 = 2 ×
3! 5!
= n − 3 + 3 − 1C3 − 1 = n − 1C2
121. Q p = 5C4 × 2C1 = 10
115. The number of non-negative integral solutions of q = 5C2 (2 C1 )3 = 80
a+ b+c +d =n
n+ 3 (n + 3)! (n + 3) (n + 2 ) (n + 1) and r = 5C0 (2 C1 )5 = 32
⇒ Cn = =
n ! 3! 6 ∴ 2q = 5r , 8 p = q
and 2( p + r ) > q
116. The required number of solutions
= The number of positive integral solution of 122. By principle of homogenuity, x = y is evident and since
(a + b + c = 12 ) sum of digits is either or odd.
(12 − 1) (12 − 2 ) ⇒ x + y = Total 5-digit numbers
= = 55 = 9 × 10 × 10 × 10 × 10
2
⇒ 2 x = 90000
117. a + a + d + a + 2d = 21or a + d = 7 ∴ x = 45000
∴ a + c = 14
123. Since, n Cr = nCn − r , hence options (a) and (c) follow.
and b=7
The number of positive integral solutions of (a + c = 14) 124. Total hand shakes possible = 2n
C2 . This will include n
is 13. hand shakes in which a person shakes hand with her or
118. Let a = 2 p + 1, b = 2q + 1, c = 2 r + 1, d = 2 s + 1, where p his spouse.
q, r and s are non-negative integers. ∴Required number of hand shakes
326 ∴ 2 p + 1 + 2q + 1 + 2 r + 1 + 2 s + 1 = 20 = 2 n C2 − n = 2 n (n − 1)
125. The nth line will rise to n additional parts, when u n − 1
parts are already there.
⇒ un = un − 1 + n
131. India must win atleast 6 matches out of 11 matches.
Then, number of ways in which India can win the series
= 11C6 + 11C7 + ... + 11C11 = 210
6
Permutation and Combination
⇒ Option (d) is incorrect.
Again, u 0 = 1, u1 = 2 Thus, both the statements are true, but Statement II is
We easily get u 2 = 4, u 3 = 7, u 4 = 11 not a correct explanation of Statement I.
Hence, options (a) and (b) are correct and option (c) is 132. The batting order of 11 players can be decided in 11!
false. ways. Now, Yuvraj, Dhoni and Pathan can be arranged
126. We have, 30 = 2 × 3 × 5. So, 2 can be assigned to either in 3! ways. But the order of these three players is fixed
a or b or c i.e. 2 can be assigned in 3 ways. Similarly, i.e. Yuvraj-Dhoni-Pathan. Now, 11! answer is 3! times
each of 3 and 5 can be assigned in 3 ways. Thus, the more, hence the required answer is 11!/3!.
number of solutions is 3 × 3 × 3 = 27. 133. Number of ways of arranging 21 identical objects, when
Also, the number of ways in which three prizes can be r is identical of one type and remaining are identical of
distributed among three persons 21!
second type, is = 21Cr , which is maximum
= 3 × 3 × 3 = 33 r ! (21 − r )!
127. As each student receives atleast two toys. So, let us first when r = 10 or 11.
give each student one toy. Now, we are left with 7 toys, Therefore, 13 Cr = 13C10 or 13 C11, hence maximum value
which can be distributed among three students such of 13 Cr is 13 C10 = 286.
that each receives atleast one toy, which is equivalent to
number of positive integral solutions of the equation Hence, Statement I is false. Obviously, Statement II is
x + y + z + w = 7, which is given by 7−1C4 − 1 = 6C3 . true.
Hence, Statement I is false and Statement II is correct. 134. When one all rounder and ten players from bowlers and
128. Number of ways of dividing n 2 objects into n groups of batsman play, number of ways is 4 C1 14 C10 .
2
(n ) !
same size is . When one wicketkeeper and 10 players from bowlers
(n !)n n !
Targ e t E x e rc is e s
and batsman play, number of ways is 2 C114C10 .
Now, number of ways of distributing these n groups
When one all rounder, one wicketkeeper and nine from
among n persons is
batsman and bowlers play, number of ways is
(n 2 )! (n 2 )!
n ! = , which is always an integer.
4
C1 2C1 14C9 .
n
(n !) n ! (n !)n
When all eleven players play from bowlers and batsmen
Also, we know that product of r is divisible by r ! . then number of ways is 14 C11.
Now, (n 2 ) ! = 1 × 2 × 3 × 4 × ... × n 2
Total number of selections
= 1 × 2 × × 3.... n × (n + 1) (n + 2 )... 2 n
× (2 n + 1) (2 n + 2 )... 3 n × { n 2 − (n 2 − 1)} = 4C1 14C10 + 2C1 14C10 + 4C1 2C1 14C9 + 14
C11
{ n 2 − (n 2 − 1)} ... n 2 135. If the particular bowler plays, then two batsmen will not
2 play. So, rest of 10 players can be selected from 17
Thus, in n ! , there are n rows each consisting product of
n integers. Each row is divisible by n ! . other players. Number of such selections is 17C10 .
(n 2 )! If the particular bowler does not play, then number of
Hence, (n 2 )! is divisible by (n !)n or is a natural
(n !)n selections = 19C11
number. If all the three players do not play, then number of
Hence, both statements are correct and Statement II is selections = 17C11
the correct explanation of Statement I.
∴Total number of selections = 17C10 + 19
C11 + 17C11
129. Q1400 = 2 3 527
The number of ways in which 1400 can be expressed as 136. If the particular batsman is selected, then rest of
a product of two positive integers is 10 players can be selected in 18 C10 ways.
(3 + 1) (2 + 1) (1 + 1) If particular wicketkeeper is selected, then rest of 10
= 12
2 players can be selected in 18 C10 ways.
Statement II is correct but does not explain Statement I
as it just gives the information of prime factor about If both are not selected, then number of ways is 18 C11.
which 1400 is divisible. Hence, total number of ways
= 2 ⋅18 C10 + 18C11 = 19C11 + 18
C10
130. In onto functions, each image must be assigned to
12
atleast one pre-image. Now, if we consider the images a 137. Seven persons can be selected for first table in C7
and bas two different boxes, then four distinct objects 1, ways. Now, these seven persons can be arranged in
2, 3 and 4 (pre-images) can be distributed, such that no 6! ways. The remaining five persons can be arranged on
box remains empty, in 2 4 − 2C1 (2 − 1)4 = 14 ways. the second table in 4! ways.
Hence, both statements are correct and Statement II is Hence, total number of ways = 12C5 6 ! 4 ! 327
the correct explanation of Statement I.
6 138. Here, A can sit on first table and B on the second or A on
second table and B on the first table.
If A is on the first table, then remaining six for first table
Let C = (n − 1)⋅ nC1 + (n − 2 )⋅ nC2 + ....+ 1⋅ nCn − 1 ...(i)
We use n Cr = nCn − r and rewrite the terms in the
Objective Mathematics Vol. 1
can be selected in 10 C6 ways. Now, these seven reverse order, to obtain
persons can be arranged in 6! ways. Remaining five can C = 1⋅ nC1 + 2 ⋅ nC2 + K+ (n − 1)⋅ nCn − 1 ...(ii)
be arranged on the other table in 4! ways. On adding Eqs. (i) and (ii), we get
Hence, total number of ways is 2 10
C6 6 ! 4 ! . 2C = n (n C1 + nC2 + ....+ n Cn − 1 )
139. If A, B are on the first table, then remaining five can be = n (2 n − nC0 − nCn ) = n(2 n − 2 )
selected in 10
C5 ways. ⇒ C = n (2 n − 1 − 1)
Now, seven persons including A and B can be arranged Thus, S = n(2 n − 1 − 1) + (2 n − 1)= (n + 2 ) 2 n − 1 − (n + 1)
on the first table in which A and B are together in 2 ! 5 ! C. We have,
ways. Remaining five can be arranged on the second n+1
2
C2 + 3C2 + 4C2 + K + nC2 = C3
table in 4! ways. Total number of ways is 10 C5 4!5!2!.
n n
If A and B are on the second table, then remaining three D. We have, ∑ (nCr )2 = ∑ (nCr ) (nCn − r )
can be selected in 10 C3 ways. r =0 r =0
Now, five persons including A and B can be arranged on ∴ The number of ways of selecting n persons out on
the second table in which A and B are together in 2!3! n men and n women = 2nCn
ways. Remaining seven can be arranged on the first
146. Let T and S denote teacher and student, respectively.
table in 6! ways. Total number of ways for first table is
10 Then, we have following possible patterns according to
C7 6!3!2!.
questions
∴Total number of ways = 10C7 6 ! 3 ! 2 ! + 10
C5 4 ! 5 ! 2 ! (i) T S S T S S T S S
140. 2 candidates can be selected out of 4 in 4C2 ways. (ii) S T S S T S S T S
Ta rg e t E x e rc is e s
(iii) S S T S S T S S T
141. Required number of ways = 38 − 3C1 ⋅ 2 8 + 3C2
Hence, total number of arrangements are
142. Each letter has 3 choices of letter boxes in 35 ways. 3 ⋅ (3 !)6 ! = 18 × 6 ! ⇒ k = 6
143. A. 20 C2 − 4C2 + 1 = 190 − 6 + 1 = 185 147. To form a triangle, 3 points out of 5 can be chosen in
B. 1 × 20
C2 = 190 5
C3 = 10 ways.
C. 2 × C2 = 56
8
But of these, the three points lying on the 2 diagonals
D. 6 C2 × 4 = 60 will be collinear.
So, 10 − 2 = 8 triangles can be formed.
144. A. Number of surjective functions is
148. Here, A is common letter in words SUMAN and DIVYA.
36 − 3C1 (3 − 1)6 + 3C2 (3 − 2 )6 = 729 − 192 + 3 = 540
Now, for selecting six different letters, we must select A
B. If f (ai ) ≠ bi, then pre-image a1, a2 , a3 cannot be either from word SUMAN or from word DIVYA.
assigned to images b1, b2 , b3 , respectively. Hence, for possible selections, we have
Hence, each of a1, a2 , a3 can be assigned to images A excluded from SUMAN + A included in SUMAN
in 2 ways. a4 , a5 , a6 can be assigned to images in 3 = 4C3 ⋅ 5C3 + 4C2 ⋅ 4C3
ways each.
= 40 + 24 = 64
Hence, number of functions is 2 3 33 = 216 Hence, N 2 = 64 ⇒ N = 8
C. One-one functions are not possible, as pre-images 149. Number of arrangements are 2 n ! n !.
are more than images.
Given that, 2 n ! n ! = 1152
D. Number of many one functions ⇒ (n !)2 = 576
= Total number of functions
− Number of one-one functions ∴ n ! = 24 ⇒ n = 4
= 36 − 0 = 729 150. Zero or more oranges can be selected out of 4 identical
oranges in 4 + 1 = 5 ways.
145. A. For k ≥ 1, we have
Zero or more apples can be selected out of 5 identical
k ⋅ k ! = [(k + 1) − 1] k ! = (k + 1)! − k ! apples in 5 + 1 = 6 ways.
Thus, 1⋅ 1! + 2 ⋅ 2 ! + 3 ⋅ 3 ! + K + n ⋅ n ! Zero or more mangoes can be selected out of 6 identical
= (2 ! − 1) + (3 ! − 2 !) + (4 ! − 3 !) + K+ {(n + 1)! − n !} mangoes in 6 + 1 = 7 ways.
= (n + 1) ! − 1 ∴Total number of selections when all the three types of
B. Let S = n ⋅ nC1 + (n − 1)⋅ nC2 + (n − 2 )⋅ nC3 + K+ 1⋅ nCn fruits are selected = 5 × 6 × 7 = 210
= (n − 1) ⋅n C1 + (n − 2 )⋅ nC2 + K + 1 ⋅ nCn − 1 But in one of these selections number of each type of
fruit is zero and hence this selection must be excluded.
328 + (n C1 + nC2 + K + nCn )
∴ Required number of ways = 210 − 1 = 209
151. Suppose box Ahas x1 balls, box B has x2 balls and boxC
has x3 balls. Then,
x1 + x2 + x3 ≤ 12, x1 ≥ 1, x2 ≥ 3, 1 ≤ x3 ≤ 5
= Coefficient of x 7 in
= C7 − C2 = 110
10 5
(1 − x 5 ) (1 + 4C1 x + 5C2 x 2 + 6C3 x 3 + . .. ) 6
Permutation and Combination
Let x4 = 12 − ( x1 + x2 + x3 ). 152. Clearly, x = 0, 1, 2, 3, ..., 10. Let x = k, then 0 ≤ k ≤ 10
Then, x1 + x2 + x3 + x4 = 12 When x = k, y + z = 21 − 2 k
[1 ≤ x1 ≤ 8, 3 ≤ x2 ≤ 10, 1 ≤ x3 ≤ 5 and 0 ≤ x 4 ≤ 7 ] The number of non-negative integral solutions in above
∴Required number of ways = Coefficient of x12 in equation = Number of ways to distribute (21 − 2 k )
identical things (each thing is number 1) among 2
( x1 + x 2 + ... + x 8 ) ( x 3 + x 4 + ... + x10 )
persons
( x1 + x 2 + ... + x 5 ) ( x 0 + x1 + ... + x 7 ) = 21 − 2 k + 2 − 1
C2 − 1 = 22 − 2 k
C1 = 22 − 2 k
= Coefficient of x in (1 + x + x + ...+ x )
7 2 7 10
(1 + x + x 2 + ... + x 7 ) (1 + x + x 2 + x 3 + x 4 )
∴Required number of solutions = ∑ (22 − 2 k )
k=0
(1 + x + x 2 + ... + x 7 ) = 22 + 20 + 18 + ... + 2
= Coefficient of x 7 in (1 − x )− 4 (1 − x 5 )(1 − x 8 ) = 2 (1 + 2 + 3 + ... + 11)
11 × 12
= Coefficient of x 7 in (1 − x )−4 (1 − x 5 ) =2 × = 132
2
Entrances Gallery
1. When n5 takes value from 10 to 6 the carry forward Aliter
8× 6× 4
moves from 0 to 4 which can be arranged in Required value =
4 3!
C1 4 C2 4
C3 4
C4
4
C0 + + + + =7 ∴ p=5
4 3 2 1
Aliter 5. The integer greater than 6000 may be of 4 digit or 5 digit.
Targ e t E x e rc is e s
Possible solutions are So, here two cases arise.
1, 2, 3, 4, 10 Case I When number is of 4 digit.
1, 2, 3, 5, 9 Four-digit number can starts from 6, 7 or 8.
1, 2, 3, 6, 8 6,7 or 8
1, 2, 4, 5, 8
1, 2, 4, 6, 7
1, 3, 4, 5, 7
2, 3, 4, 5, 6 3 4 3 2
Hence, there are 7 solutions. Thus, total number of 4-digit number, which are greater
2. Number of red lines = nC2 − n than 600 = 3 × 4 × 3 × 2 = 72
Case II When number is of 5 digit.
Number of blue lines = n
Total number of five-digit number which are greater
Hence, n
C2 − n = n than 6000 = 5 ! = 120
⇒ n
C2 = 2 n ∴Total number of integers = 72 + 120 = 192
n (n − 1) 6. Given, n( A) = 2, n(B) = 4
⇒ = 2n
2 ∴ n ( A × B) = 8
⇒ n − 1= 4 Now, the number of subsets of A × B having 3 or more
∴ n=5 elements
3. Required number of ways = 8C3 + 8C4 + ... + 8C8
= 5 ! − { 4 ⋅ 4 ! − 4C2 ⋅ 3 ! + 4C3 ⋅ 2 ! − 1} = 53 = (8 C0 + 8C1 + ... + 8C8 ) − (8 C0 + 8C1 + 8C2 )
4. Let (1, 1, 1), (− 1, 1, 1), (1, − 1, 1), (−1, − 1, 1) be vectors = 2 8 − 8C0 − 8C1 − 8C2 [Q n C0 + nC1 + ... + nCn = 2 n ]
a , b, c, d and rest of the vectors be − a , − b, − c, − d and = 256 − 1 − 8 − 28 = 219
let us find the number of ways of selecting coplanar 7. Number of ways = 35 − 3C1 ⋅ 2 5 + 3C2 ⋅ 15
vectors.
= 243 − 96 + 3 = 150
Observe that out of any three coplanar vectors, two will
be collinear (anti-parallel). 8. As, an = bn + c n
Number of ways of selecting the anti-parallel pair = 4 ⇒ an = 1 ... (1 or 0)
Number of ways of selecting the third vector = 6 ⇒ an = 1_ _ _ _ _ _ _1 = an − 1 + 1 _ _ _ _ _ _ _ 1 0 = an − 2
Total = 24 14
4244
3
( n − 1) places
14
4244
3
( n − 2 ) places
∴Number of non-coplanar selections
an = an −1 + an − 2
= 8C3 − 24 = 32 = 2 5 , p = 5 ∴ a17 = a16 + a15
329
6 9. b6 = Six-digit numbers ending with 1.
⇒ 1 – – – – 1
∴Required number of words = 8C4 ×
= 8C4 ×
7!
4 !2 !
7 × 6!
Objective Mathematics Vol. 1
Now, the four places are to be filled. 4 !2 !
Case I 1 ................... 1 1 = 7 ⋅ 8C4 ⋅ 6C4
For 3 places, all 1’s are used in 1 way. 16. Required number of ways = 12C4 × 8C4 × 4C4
One zero is used = 3C1 = 3 12 ! 8! 12 !
= × ×1 =
Two zeros are used = 1 way 0 1 0 1... 8! × 4! 4! × 4! (4 !)3
Total = 5 ways
17. Total number of ways
Case II 1 ................. 0 1 =10 C1 + 10C2 + 10C3 + 10C4
For 3 places, = 10 + 45 + 120 + 210
All 1’s are used = 1 way = 385
One zero is used = 2C1 = 2 ways
∴ Total ways = 3 ways
18. In the word SACHIN, order of alphabets is A, C, H, I, N,
Thus, b6 = 5 + 3 = 8 S. Number of words start with A = 5! so with C, H, I, N.
Now, words start with S and after that ACHIN are in
10. The number of ways of distributing n identical objects ascending orders of position, so 5 ⋅ 5 ! = 600 words are in
among r persons such that each person gets atleast
dictionary before words with start S. Thus, position of
one object is same as the number of ways of selecting
given word is 601.
(r − 1)places out of(n − 1)different places, i.e. is n −1Cr −1.
7− x
19. Given that, f ( x ) = Px − 3
11. If out of n points, m are collinear.
The above function is defined, if
Number of triangles = nC3 − mC3 7 − x ≥ 0 and x − 3 ≥ 0
∴ Number of triangles = 10 C3 − 6C3 = 120 − 20 and 7−x≥x−3
⇒ N = 100 ⇒ x ≤ 7 and x≥3
Ta rg e t E x e rc is e s
12. and x≤5
∴ Df = { 3, 4, 5}
Now, f (3) = 4 P0 = 1
3 distinct 9 distinct f (4) = 3 P1 = 3
red balls blue balls
and f (5) = 2 P2 = 2
∴ Rf = {1, 2, 3}
Urn A Urn B
20. Total number of ways in which all letters can be
The number of ways in which two balls from urn A and arranged = 6!
two balls from urn B can be selected There are two vowels (A, E ) in the word GARDEN. Total
= 3C2 × 9C2 = 3 × 36 = 108 number of ways in which these two vowels can be
13. The number of ways in which 4 novels can be selected arranged = 2 !
= C4 = 15
6 ∴ Total number of required ways
6!
The number of ways in which 1 dictionary can be = = 360
2!
selected
= 3C1 = 3 8 −1 7!
21. Required number of ways = C3 −1 = 7C2 =
4 novels can be arranged in 4! ways. 2 ! 5!
7⋅6
[since, dictionary is fixed in the middle] = = 21
2 ⋅1
∴ Total number of ways = 15 × 3 × 4!
= 15 × 3 × 24 = 1080 22. First, we fix the position of men, the number of ways in
which men can sit = 5!.
14. The number of different ways the child can buy the six
Now, the number of ways in which women can sit = 6 P5
ice-creams = Total number of non-negative integral M
solution of the equation x1 + x2 + x3 + x4 + x5 = 6
= 6 + 5 − 1C5 −1 = 10C4 M
M
and number of ways to arrange 6A’s and 4B’s in a row
10 ! 10
= = C4 M
6! 4! M
∴Statement I is false and Statement II is true.
M
15. Given word is MISSISSIPPI.
Here, I = 4 times, S = 4 times, P = 2 times, M = 1time ∴ Total number of ways = 5 ! × 6 P5
330 _M_I_I_I_I_P_P_ = 5! × 6!
6
n
23. Now, n Cr + 1 + nCr − 1 + 2 nCr
= Cr + 1 + Cr + Cr −1 + Cr
n n n n
30. We have, mCr + 1 + ∑ k Cr = (mCr + 1 + mCr ) + m+1
Cr
k=m
= n +1Cr + 1 + n + 1Cr = n + 2Cr + 1 + m + 2Cr + ... + nCr
Permutation and Combination
m+1 m+1 m+ 2
24. Since, telephone number start with 67, so two digits is =( Cr + 1 + Cr ) + Cr + ... + nCr
already fixed. Now, we have to arrange three digits from m+ 2 m+ 2
= Cr + 1 + Cr + ... + nCr
remaining eight digits i.e. 0, 1, 2, 3, 4, 5, 8, 9 in 8 P3 ways M M M
8! n+1
= = 8 × 7 × 6 = 336 ways = Cr + 1
5!
C1 C C C
Thus, required number of telephone numbers = 336 31. We have, + 2 2 + 3 3 + ... + n n
C0 C1 C2 Cn − 1
25. Number of ways n n n n
= 2n + 1
C0 + 2n + 1
C1 + 2n + 1
C2 + ... + 2n + 1 C1 C C C
Cn = n
+ 2 n 2 + 3 n 3 + ... + n n n
1 2n + 1 C0 C1 C2 Cn −1
=
(2 ) = 22n n (n − 1) (n − 2 )
2 n (n − 1)
Thus, 2 2n = 64 ⇒ 2 2 n = 2 6 n 2 3×2 1
= +2× + 3× + ... + n ×
On comparing, 2 n = 6 1 n n ( n − 1) n
∴ n=3 2
n (n + 1)
26. Set S = {1, 2, 3, ..., 20} is to be partitioned into four sets = n + (n − 1) + (n − 2 ) + ... + 1 = Σ n =
2
of equal size i.e. having 5-5 numbers. The number of
ways in which it can be done 32. We know, 11Cn is maximum when n = 5.
11!
= C5 × 15C5 × 10C5 × 5C5
20 ∴ 11
Cn =
n ! (11 − n )!
20 ! 15 ! 10 !
= × × ×1 ∴ n ! (11 − n )! is minimum, when n = 5.
15 ! × 5 ! 10 ! × 5 ! 5 ! × 5 !
(20 !) 33. Given, n C10 = nC11
=
Targ e t E x e rc is e s
(5 !)4 ⇒
n!
=
n!
10 ! (n − 10 )! 11! (n − 11)!
27. Total number of available courses = 9
Out of these, 5 courses can be chosen. But it is given ⇒ 11 × 10 ! (n − 11)! = 10 ! (n − 10 ) (n − 11)!
that 2 courses are compulsory for every student i.e. ⇒ 11 = n − 10 ⇒ n = 11 + 10 = 21
you have to choose only 3 courses instead of 5, out of Now, n
C21 = 21C21 = 1
7 instead of 9.
2n
7 × 6× 5 C2 9
It can be done in 7C3 ways = 34. Given, 2n
C2 : nC2 = 9 : 2 ⇒ n
=
6 C2 2
= 35 ways (2 n )! 2 ! (n − 2 )! 9
⇒ × =
28. There may be two cases. 2 ! (2 n − 2 )! n! 2
(2 n ) (2 n − 1) (2 n − 2 )! 2(n − 2 )! 9
Case I Two children get none and one get three and ⇒ × =
others get one each. Then, total number of ways 2(2 n − 2 )! n (n − 1) (n − 2 )! 2
10 ! 2(2 n − 1) 9
= × 10 ! ⇒ = ⇒ 4 (2 n − 1) = 9 (n − 1)
2 ! × 3! × 7 ! n −1 2
⇒ 8n − 4 = 9n − 9 ⇒ 9n − 8n = − 4 + 9
Case II Two get none and two get 2 each and the
∴ n=5
others get one each. Then, total number of ways
10 ! 35. Total number of points in a plane is 30.
= × 10 !
(2 !)4 × 6 ! Out of them, 8 points are collinear.
∴ Total number of straight lines formed
Hence, total number of ways 30 × 29 8 × 7
(10 !)2 (10 !)2 = 30C2 − 8C2 + 1 = − +1
= + 2 2 ×1
2 ! × 3 ! × 7 ! (2 !)4 × 6 ! = 435 − 28 + 1= 408
(10 !)2 × 25 36. We have, 0, 1, 2, 3, 4, 5 i.e. six numbers, one zero and 5
=
(2 !)2 × 6 ! × 84 positive numbers.
29. 3 consonants can be selected from 7 consonants One-digit number = 5 P1 = 5
in 7C3 ways. Two-digit numbers = 6 P2 − 5 P1 = 25
2 vowels can be selected from 4 vowels in 4C2 ways. Three-digit numbers = 6 P3 − 5 P2 = 100
∴ Required number of words = 7C3 × 4C2 × 5 ! Four-digit numbers = 6 P4 − 5 P3 = 300
Five-digit numbers = 6 P5 − 5 P4 = 600
selected 5! letters can be arranged
in 5! ways, to get different words Six-digit numbers = 6 P6 − 5 P5 = 600
∴ Total possible numbers
= 35 × 6 × 120 = 25200 = 5 + 25 + 100 + 300 + 600 + 600 = 1630 331
6 37. First, we fix the 5 plus ‘+’ in five alternate positions.
... + ... + ... + ... + ... + ...
Now, select the three positions for three ‘minus’ out of 6
⇒
∴
n
3
= 84
n = 252
Objective Mathematics Vol. 1
blank positions. 43. Since, the student is allowed to select atmost n books
∴ Required number of ways out of (2 n + 1) books. Therefore, in order to select atleast
6
P3 6× 5× 4 one book, he has the choice to select one, two, three, ...,
= 1× = 1 × 6C3 = 1 × = 1 × 5 × 4 = 20
3! 3×2 ×1 n books. Thus, if T is the total number of ways of
selecting atleast one book, then
38. Case I When he chooses 4 questions from first five
2n + 1
questions. T= C1 + 2 n + 1C2 + ... + 2 n + 1Cn = 255 ...(i)
2n + 1
Now, the number of choices = 5C4 × 8C6 Now, C0 + 2 n + 1C1 + ... + 2 n + 1Cn + 2 n + 1Cn + 1
8×7 + 2n + 1
Cn + 2 + ... + 2n + 1
C2 n + 1
= 5× = 140
2 ×1 2n + 1 2n + 1
= (1 + 1) =2
Case II When he chooses 5 questions from first five
2n + 1 2n + 1 2n + 1
questions. ∴ C0 + 2( C1 + ... + Cn )
2n + 1
Now, the number of choices + C2 n + 1 = 22n + 1
8×7 × 6
= 5C5 × 8C5 = 1 × 8C3 = = 56
3×2 ×1 ⇒ 1 + 2T + 1 = 2 2 n + 1
Now, the total number of choices available ⇒ 1 + 2 (255) + 1 = 2 2 n + 1 [from Eq. (i)]
= 140 + 56 = 196
⇒ 255 + 1 = 2 2n
⇒ 256 = 2 2n
⇒ 4 = 4n4
39. Case I When 3 drawn balls contain 1 black ball
∴ n=4
= 3C1 ⋅ 2C2 ⋅ 4C0 + 3C1 ⋅ 2C1 ⋅ 4C1 + 3C1 ⋅ 4C2 ⋅ 2C0
44. Clearly, first prize can be distributed among 4 students
= 3 ⋅ 1 + 3 ⋅ 2 ⋅ 4 + 3 ⋅ 6 = 3 + 24 + 18 = 45 in 4 C1 = 4 ways
Case II When 3 drawn balls contains 2 black balls
and second prize can be distributed among 4 students
= 3C2 ⋅ 2C1 ⋅ 4C0 + 3C2 ⋅ 4C1 ⋅ 2C0
in 4 C1 = 4 ways.
Ta rg e t E x e rc is e s
= 3 ⋅ 2 ⋅ 1 + 3 ⋅ 4 ⋅ 1 = 6 + 12 = 18
Similarly, 3rd, 4th and 5th prizes can be distributed in
Case III When 3 drawn balls contain all black balls 4 ways each.
= 3C3 ⋅ 2C0 ⋅ 4C0 = 1 Hence, by fundamental principle of counting, the total
On adding all three cases, we get required ways number of ways = 4 × 4 × 4 × 4 × 4 = 45 = 1024
= 45 + 18 + 1 = 64
4! 3!
Aliter 45. Required number of ways = × = 18
2 !2 ! 2 !
Case I When 3 drawn balls contains 1 black ball
6× 5 46. The number of integers greater than 6000
= 6C2 × 3C1 = × 3 = 45
2 ×1 = 5 ! + (3 × 4 × 3 × 2 ) = 192
Case II When 3 drawn balls contain 2 black balls 56
Pr + 6
3× 2 47. Q 54 = 30800
= 6C1 × 3C2 = 6 × = 18 Pr + 3
2 ×1
56 ! (51 − r )!
Case III When 3 drawn balls contain all black balls ∴ × = 30800
= 3C3 = 1 54 ! (50 − r )!
On adding all three cases, we get ⇒ 56 × 55 × (51 − r ) = 56 × 55 × 10
Required ways = 45 + 18 + 1 = 64 ⇒ 51 − r = 10
∴ r = 41
40. Four speakers will address the meeting in 4! ways = 24
different ways in which half number of cases will be such 48. The vowels in the word COMBINE are O, I, E which can
that P speaks before Q and half number of cases will be be arranged at 4 places in 4 P3 ways and other words
such that P speaks after Q. can be arranged in 4! ways.
24 Hence, total number of ways
∴Required number of ways = = 12
2 = 4 P3 × 4 ! = 4 ! × 4 ! = 576
41. Required sum = 3 ! (1 + 2 + 3 + 4) = 6 (10 ) = 60 49. Required number of ways
11! 11!
42. Number of ways of selecting 3 children from n children = 11C5 − 11C4 = − = 132
= C3
n 5 ! 6 ! 4 !7 !
n −1 n −1
Since, the number of times she will go to the garden 50. C3 + C4 > nC3
= 84 × (Number of ways that a particular child goes to n+1
the zoo) ⇒ n
C4 > nC3 [Q n Cr + nCr + 1 = Cr + 1]
n −1 (n − 1)(n − 2 ) ⇒
n!
>
n!
⇒ C3 = 84 × (1 ×
n
C2 ) ⇒ C3 = 84 ×
n
2 (n − 4)! 4 ! (n − 3)! 3 !
n(n − 1)(n − 2 ) (n − 1)(n − 2 ) ⇒ (n − 3) (n − 4)! > (n − 4)! 4
⇒ = 84 ×
332 3×2 ×1 2 ∴ n >7
7
Mathematical
Induction
Introduction
There are two basic processes of reasoning. Chapter Snapshot
Deduction (result is deduced from general to particular) and induction (result is
generalised from a particular result). ● Introduction
Deduction is the application of a general case to a particular case i.e. given a statement ● Statement
to be proven, often called a conjecture or a theorem in Mathematics, valid deductive ● Principle of Mathematical
steps are derived and a proof may or may not be established. Induction
In contrast to deduction, induction depends on working with each case and developing a ● Algorithm for Mathematical
conjecture by observing incidences till each and every case is observed. Induction
Induction is a key aspect of scientific reasoning, where collecting and analysing data is ● Types of Problems
the norm. Thus, induction means the generalisation from particular cases or facts.
In algebra or in other disciplines of Mathematics, there are certain results or statements
that are formulated in terms of n (where, n is a positive integer).
To prove such statements, the well suited principles that is used based on the specific
technique, is the principle of mathematical induction.
Statement
A sentence or description which can be judged to be true or false, is called a statement.
e.g. (i) 3 divides 9.
(ii) Jaipur is the capital of Rajasthan.
A statement involving mathematical relations is known as the mathematical
statement. e.g. 3 divides 9.
Principle of Mathematical Induction
Suppose there is a given statement P ( n) involving the natural number n, such that
(i) statement is true for n =1 i.e. P (1) is true and
(ii) if the statement is true for n = k (where, k is some positive integer), then
statement is also true for n = k +1 i.e. truth of P ( k ) implies the truth of P ( k +1).
7 Then, P ( n) is true for all natural numbers n.
i. It is simply a statement of fact. There may be
Now, show that the result is true, for n = k + 1.
P(k + 1): 1 + 2 + 3 + ... + k + (k + 1)
=
(k + 1) (k + 2 )
Objective Mathematics Vol. 1
a situation when a statement is true for all 2
Then, LHS = 1 + 2 + 3 + ... + k + (k + 1)
n ≥ 4. In this case, (i) will start from n = 4 and k(k + 1)
we shall verify the result for n = 4 i.e. P ( 4). = + (k + 1) [from Eq. (i)]
2
k + 1
ii. It is a conditional property, it does not assert = (k + 2 ) = RHS
2
that the given statement is true for n = k but
Therefore, P(k + 1) is true, whenever P(k ) is true.
only that, if it is true for n = k , then it is also
true for n = k +1. So, to prove that property Hence, from the principle of mathematical induction, the
statement is true for all natural numbers.
holds, only prove that conditional proposition.
Ø The step ‘if the statement is true for n = k, then it is also true for X Example 2. For a natural number n, which
n = k + 1 ’ is sometimes referred as the inductive step and the one is the correct statement?
assumption in this step that statement is true for n = k is called (a)13 + 2 3 + 3 3 +… + n 3 = (1 + 2 + 3 +… + n) 2
the inductive hypothesis.
(b)13 + 2 3 + 3 3 +… + n 3 > (1 + 2 + 3 +… + n) 2
Algorithm for Mathematical (c)13 + 2 3 + 3 3 +… + n 3 < (1 + 2 + 3 +… + n) 2
(d)13 + 2 3 + 3 3 +… + n 3 ≠ (1 + 2 + 3 +… + n) 2
Induction 2
The steps used to prove that a statement is true for Sol. (a) We have, 13 + 2 3 + 33 + … + n3 = n(n + 1)
2
all natural numbers by using principle of mathematical = (1 + 2 + 3 + … + n)2
induction are given below:
Aliter
Step I Consider the given statement as P ( n). Let P(n) : 13 + 2 3 + 33 + … + n3 = (1 + 2 + 3 + … + n)2
Step II Put n =1, then verify that it is a true statement. For n = 1, we have
Step III Suppose that the statement is true, for n = k P(1) : 13 = 12 ⇒ 1 = 1, which is true and for n = 2, we have
(any arbitrary number). P(2 ) : 13 + 2 3 = (1 + 2 )2 ⇒ 9 = 9, which is true.
Step IV Show that the statement is true for n = k +1. Similarly, we can prove that P(n) is true for all n ∈ N.
Combining the results of steps II and IV
conclude by principle of mathematical Divisibility Type Problems
induction that P ( n) is true for all n ∈ N .
In this type of problems, we use principle of
Ø This algorithm for principle of mathematical induction is used to mathematical induction to show that given statement
solve problems which can be categorised into the following types. P ( n) is divisible by given number or P ( n) is a multiple
of given number.
Types of Problems This process can also be explained with the
following examples:
Identity Type Problems
In this type of problems, we use principle of X Example 3. The greatest positive integer,
mathematical induction to show LHS is equal to RHS. which divides ( n + 2) ( n + 3) ( n + 4) ( n + 5) ( n + 6)
for all n ∈ N , is
X Example 1. 1 + 2 + 3 + ... + n is equal to (a) 4 (b) 120
n( n −1) n( n +1) n( n +1) n( n −1 ) (c) 240 (d) 24
(a) (b) (c) (d)
2 2 3 3 Sol. (b) Let P(n) : (n + 2 ) (n + 3) (n + 4) (n + 5) (n + 6)
Sol. (b) By the help of AP, we have Put n = 1, P(1) = (1 + 2 )(1 + 3)(1 + 4)(1 + 5)(1 + 6)
n(n + 1) = 3 × 4 × 5 × 6 × 7 = 120 × 21
P(n) : 1 + 2 + 3 + ... + n =
2 Put n = 2, P(2 ) = (2 + 2 )(2 + 3)(2 + 4)(2 + 5)(2 + 6)
Put n = 1, LHS = 1 = 4 × 5 × 6 × 7 × 8 = 120 × 56 and so
( + 1)
11 on.
and RHS = = 1 = LHS
2 Hence, P(n) is always divisible by 120.
Therefore, P(1) is true. Aliter
Suppose that the statement is true, for n = k, k ∈ N. Qn consecutive positive integer is divided by n!.
k(k + 1) ∴5 consecutive positive integer is divided by 5!
∴ P(k ) : 1 + 2 + 3 + ... + k = …(i)
334 2 = 5 × 4 × 3 × 2 × 1= 120
X Example 4. If n ∈ N , then 11n + 2 + 12 2n + 1 is
divisible by
(a) 113 (b) 123
X Example 7. If un + 1 = 3un − 2un − 1 and u0 = 2,
u1 = 3, then un is equal to
7
Mathematical Induction
(c) 133 (d) None of these (a)1 − 2 n (b) 2 n + 1 (c) 2 n − 1 (d) 2 n + 2
Sol. (c) Putting n = 1in 11n + 2 + 12 2 n + 1, we get Sol. (b) We have, u n + 1 = 3u n − 2u n − 1 …(i)
113 + 12 3 = 3059, which is divisible by 133. Step I Given, u1 + 1 = 3 = 2 + 1 = 2 + 11
Inequality Type Problems which is true, for n = 1.
Put n = 1 in Eq. (i), then u1 + 1 = 3u1 − 2u1 − 1
In this type of problems, we have inequality
⇒ u 2 = 3u1 − 2u 0
symbols like >, <, ≥ or ≤ and use the principle of
= 3⋅ 3 − 2 ⋅2 = 5 = 22 + 1
mathematical induction to show that LHS and RHS
satisfies the given inequality. which is true, for n = 2.
Therefore, the result are true, for n = 1and n = 2.
X Example 5. For each n ∈ N , the correct
Step II Assume it is true for n = k, then it is also true for
statement is n = k − 1.
(a) 2 n < n Then, uk = 2 k + 1 ...(ii)
(b) n 2 > 2n and u k − 1 = 2 k −1 + 1 …(iii)
(c) n 4 < 10 n Step III On putting n = k in Eq. (i), we get
u k + 1 = 3 u k − 2u k − 1
(d) 2 3n > 7n + 1
= 3(2 k + 1) − 2(2 k − 1 + 1) [from Eqs. (ii) and (iii)]
Sol. (c) Let n = 1, then options (a), (b) and (d) do not satisfy. = 3⋅2k + 3 − 2 ⋅2k − 1 − 2
Only option (c) satisfied. = 3 ⋅ 2 k + 3 − 2 k − 2 = (3 − 1)2 k + 1
= 2 ⋅2k + 1 = 2k + 1 + 1
X Example 6. For any natural number n,
This shows that the result is true for n = k + 1, hence by the
2 n ( n − 1)! < n n , if principle of mathematical induction, the result is true for all
(a) n < 2 (b) n > 2 (c) n ≥ 2 (d) Never n ∈ N.
Sol. (b) Check through option, the condition 2 n (n − 1)! < nn X Example 8. If a1 = 2, a n = 5 a n − 1 , for all
is satisfied for n > 2. natural numbers n ≥ 2, then the value of a 4 is equal
to
Recursion Type Problems (a) 150 (b) 50
In this type of problems, we get a sequence in (c) 200 (d) None of these
which later terms are deduced from earlier ones by
Sol. (d) We have, a1 = 2
using the principle of mathematical induction.
Then, a2 = 5 a2 − 1 = 5 a1 = 10
This process can be explained by following a3 = 5 a3 − 1 = 5 a2 = 5 × 10 = 50
examples: ∴ a4 = 5 a4 − 1 = 5 a3 = 5 × 50 = 250
Work Book Exercise
1 Let P(n ) be a statement and let P(n ) → P(n + 1) is 4 For every positive integral value of n, 3 n > n 3 when
true for all natural numbers n, then P(n ) is true a n> 2 b n≥ 3 c n≥ 4 d n< 4
a for all n
b for all n > 1 5 For every positive integer n, 2 n < n ! when
c for all n > m, m being a fixed integer a n< 4 b n≥ 4
d Nothing can be said c n< 3 d None of these
2 The value of the natural numbers n such that the 6 For every natural number n, n(n + 1) is always
inequality 2 n > 2 n + 1 is valid, is a even b odd
a for n ≥ 3 b for n < 3 c multiple of 3 d multiple of 4
c for n ∈ N d for any n n
n + 1
3n − 3 n −1 7 If n is a natural number, then > n ! is true,
3 If n ∈ N, then 7 2n
+2 ⋅3 is always divisible 2
by when
a 25 b 35 a n< 1 b n≥ 1
c 45 d None of these c n≥ 2 d n< 3
335
WorkedOut Examples
Type 1. Only One Correct Option
Ex 1. The value of sum in the nth bracket of n (n + 1) (n2 + n + 2) 2(4 )
Sol. If n = 1, then = =1
(1) + (2 + 3) + ( 4 + 5 + 6 + 7) + (8 + 9 + 10 8 8
+… 15) +… , is Hence, (a) is the correct answer.
(a) 2n ( 2n + 2n − 1 − 1)
Ex 4. Sum of nth bracket of
(b) 2n − 1 ( 2n + 2n − 1 − 1)
(c) 2n − 2 ( 2n + 2n − 1 − 1) 1 1 1 1 1
(1) + + 2 + 3 + 4 + 5 +… is
(d) None of the above 3 3 3 3 3
( 3 − 1)
n 3
( 3n − 1)
Sol. If n = 1, then 2n (2n + 2n − 1 − 1) (a) (b)
= 2 (2 + 1 − 1) = 2(2) = 4, 2⋅ 4 n − 1 2⋅ 3(n − 1)(n + 2)/ 2
2n − 1 (2n + 2n − 1 − 1) = 20 (2 + 20 − 1) = (2 + 1 − 1) = 2 ( 3 + 1)
n
(c) (d) None of these
and 2n − 2 (2n + 2n − 1 − 1) = 2− 1 (21 + 20 − 1) 3⋅ 7n − 1
= 2− 1 (2) = 1 (3n − 1)3 (2)3
Sol. If n = 1, then = = 4;
∴ Sum of nth bracket = 2n − 2 (2n + 2n − 1 − 1) 2⋅ 4n − 1 2
(3n − 1) 2 3n + 1 4
Hence, (c) is the correct answer. = = 1 and =
2 ⋅ 3(n − 1) (n + 2) / 2 2 ⋅ 30 3 ⋅ 7n − 1 3
Ex 2. If (1) + (2 + 3 + 4) + (5 + 6 + 7 + 8 + 9)+… , then (3n − 1)
∴ Sum of nth bracket = (n − 1) (n + 2) / 2
the sum of terms in the nth bracket is 2⋅ 3
(a) ( n − 1) 3 + n 3 Aliter
nth bracket
(b) ( n + 1) 3 + 8n 2 1 − 1/ 3n
1 1 1
( n + 1) ( n + 2) = n(n − 1) / 2 + … + ( n + 1) n/ 2 − 1 = n(n − 1) / 2
(c) 3 3 3 1 − 1/ 3
6n
3(3n − 1) 3n − 1
(d) ( n + 1) 3 + n 3 = n n (n − 1) / 2 = (n − 1) (n + 2) / 2
2⋅ 3 3 2⋅ 3
Sol. If n = 1, then (n − 1)3 + n3 = 0 + 1 = 1, Hence, (b) is the correct answer.
(n + 1)3 + 8n2 = 8 + 8 = 16, Ex 5. The sum of (12 ) + (12 + 2 2 ) + (12 + 2 2 + 3 2 )
(n + 1) (n + 2) 6 +… n brackets is equal to
= = 1,
6n 6 n ( n + 1) ( n + 2) n( n + 1) 2 ( n + 2)
and (n + 1) + n = 8 + 1 = 9
3 3 (a) (b)
6 12
If n = 2, then (n − 1)3 + n3 = 1 + 8 = 9, n( n + 1) ( 2n + 1)
(c) n ( n + 1) ( n + 2) (d)
(n + 1) (n + 2) 12 6
= =1
6n 12
Sol. Q tn = 12 + 22 + … + n2
∴ Sum of nth bracket = (n − 1) + n 3 3
n (n + 1) (2n + 1) n(2n2 + 3n + 1)
Hence, (a) is the correct answer. = =
6 6
n3 n2 n
Ex 3. (1) + (2 + 3) + (4 + 5 + 6) + ... n brackets is equal = + +
3 2 6
to n3 n2 n
n( n + 1) ( n 2 + n + 2) ∴S n = ∑ + +
(a) 3 2 6
8
n2 (n + 1)2 n(n + 1)(2n + 1) n(n + 1)
n ( n + 1) ( n 2 − n + 2) = + +
(b) 12 12 12
8 n(n + 1)
n( n − 1) ( n 2 + n + 2) = [ n(n + 1) + 2n + 1 + 1]
(c) 12
8 n(n + 1) (n2 + 3n + 2) n(n + 1)2 (n + 2)
n( n − 1) ( n 2 − n + 2) = =
(d) 12 12
336 8 Hence, (b) is the correct answer.
Ex 6. (1) + (1 + 3) + (1 + 3 + 5) + ... n brackets is equal
to
n( n + 1) ( n + 2)
Ex 9. 2 ⋅ 12 + 3 ⋅ 2 2 + 4 ⋅ 3 2 +… + (n + 1)n 2 is equal to
(a)
n( n + 1) ( n + 2) ( 3n + 5) 7
Mathematical Induction
(a) 12
6 n( n + 1) ( n + 2) ( n + 3)
(b)
n( n + 1) ( 3n 2 + 23n + 46) 4
(b)
12 (c) 2n( n + 1) ( n + 2) ( n + 3)
n( n + 1) ( n + 2) ( 3n + 1)
n( 27n 3 + 90n 2 + 45n − 50) (d)
(c) 12
4
n( n + 1) ( 2n + 1) Sol. Q tn = (n + 1)n2 = n3 + n2
(d)
6 n2 (n + 1)2 n(n + 1) (2n + 1)
= +
4 6
Sol. n th bracket = 1 + 3 + 5 + ... + (2n − 1) = n2 n(n + 1)
n(n + 1) (2n + 1) = [ 3n(n + 1) + 2(2n + 1)]
∴ S n = Σ n2 = 12
6 n(n + 1) (3n2 + 7n + 2)
Hence, (d) is the correct answer. =
12
n(n + 1) (n + 2) (3n + 1)
Ex 7. (1) + (1 + 2) + (1 + 2 + 3) +… n brackets is equal =
12
to Hence, (d) is the correct answer.
n( n + 1) ( n + 2)
(a) Ex 10. 1 ⋅ 2 2 + 2 ⋅ 3 2 + 3 ⋅ 4 2 +… + n(n + 1) 2 is equal to
6
n( n + 1) ( 3n 2 + 23n + 46) n( n + 1) ( n + 2) ( 3n + 5)
(b) (a)
12 12
n( n + 1) ( n + 2) ( n + 3)
n( 27n 3 + 90n 2 + 45n − 50) (b)
(c) 4
4
n( n + 1) ( 2n + 1) (c) 2n( n + 1) ( n + 2) ( n + 3)
(d) n ( n + 1) ( n + 2) ( 3n + 1)
16 (d)
12
Sol. If n = 1, then the sum of the terms = (1) = 1,
Sol. Q tn = n (n + 1)2 = n(n2 + 2n + 1) = n3 + 2n2 + n
n(n + 1) (n + 2) 1 × 2 × 3
∴ = ∴ S n = Σn3 + 2Σn2 + Σn
6 6
n2 (n + 1)2 2n(n + 1)(2n + 1) n(n + 1)
=1 = + +
4 6 2
Hence, (a) is the correct answer. n(n + 1)
= [ 3n(n + 1) + 4 (2n + 1) + 6 ]
12
Ex 8. 1 ⋅ 3 ⋅ 4 + 2 ⋅ 4 ⋅ 5 + 3 ⋅ 5 ⋅ 6 +… upto n terms is
n(n + 1) (3n2 + 11n + 10)
equal to =
12
n( n + 1) ( n + 2) n(n + 1) (n + 2) (3n + 5)
(a) =
6 12
n( n + 1) ( 3n 2 + 23n + 46) Hence, (a) is the correct answer.
(b)
12
12 12 + 2 2 12 + 2 2 + 3 2
n( 27n + 90n 2 + 45n − 50)
3
Ex 11. + + +… upto n terms is
(c) 3 5 7
4 equal to
n( n + 1) ( 2n + 1) n( n + 1) ( n + 2) n( n + 1) ( n + 2)
(d) (a) (b)
6 18 6
Sol. Given expression n( n + 1) ( n + 2) 2n ( n + 1) ( n + 2)
(c) (d)
Σn(n + 2) (n + 3) = Σ (n3 + 5n2 + 6n) 3 3
n2 (n + 1)2 5n(n + 1) (2n + 1) 6n(n + 1) 12 + 22 + … + n2 n(n + 1)
= + + Sol. Q tn = =
4 6 2 2n + 1 6
n(n + 1) (n2 + n) 1 n(n + 1)(2n + 1) n(n + 1)
= [ 3n (n + 1) + 10(2n + 1) + 36 ] ∴ Sn = Σ = +
12 6 6 6 2
n(n + 1) (3n2 + 23n + 46) n(n + 1) (n + 2)
= =
12 18
Hence, (b) is the correct answer. Hence, (a) is the correct answer.
337
7 Ex 12. 1 ⋅ 2 + 2 ⋅ 3 + 3 ⋅ 4 +… upto n terms is equal to
(a)
n( n + 1) ( n + 5)
(b)
n( n + 1) ( n + 2)
Sol.
⇒
S n = 1 + 4 + 13 + 40 + … + tn
Sn = 1 + 4 + 13 + … + tn + 1 + tn
0 = 1 + 3 + 9 + 27 + … − tn
Objective Mathematics Vol. 1
3 3 (3n − 1) (3n − 1)
n( 4n 2 + 6n − 1) ⇒ tn = =
(c) (d) n( n + 1) ( n + 2) (3 − 1) 2
3 (3n − 1)
∴ Sn = Σ
Sol. If n = 1, then the sum of the terms = 1⋅ 2 = 2 2
n(n + 1) (n + 2) 1 × 2 × 3 3(3n − 1) n
= =2 = −
3 3 2(3 − 1) 2
Hence, (b) is the correct answer. (3n + 1 − 3 − 2n)
=
4
Ex 13. 1 ⋅ 3 + 2 ⋅ 3 2 + 3 ⋅ 3 3 +… + n ⋅ 3 n is equal to
Hence, (b) is the correct answer.
( 2n − 1)3n + 1 + 3 ( 2n + 1)3n + 1 + 3
(a) (b) (3n 2 − 3n + 2)
4 4 Ex 17. 1 + 4 + 10 + 19 +… + is equal to
( 2n + 1)3n + 1 − 3 ( 2n − 1)3n + 1 − 3 2
(c) (d) n( n + 1) ( 2n + 1)
4 4 (a)
6
Sol. If n = 1, then n 2 ( n + 1) 2
(b)
(2n − 1)3n + 1 + 3 (2n + 1)3n + 1 + 3 15 4
= 3, = ,
4 4 2 n ( n + 1) ( n + 2)
(c)
(2n + 1)3n + 1 − 3 (2n − 1)3n + 1 − 3 3 6
= 6, =
4 4 2 n( n 2 + 1)
(2n − 1)3n + 1 + 3 (d)
∴ Sum = 2
4
Hence, (a) is the correct answer. (3n2 − 3n + 2)
Sol. Sum = Σ
2
Ex 14. 1 ⋅ 1! + 2 ⋅ 2! + 3 ⋅ 3! +… + n ⋅ n! is equal to 3 2 3
= Σn − Σn + Σ1
(a) ( n + 1)! − 1 2 2
3n (n + 1) (2n + 1) 3n (n + 1)
(b) ( n − 1)! + 1 = − +n
12 4
(c) ( n + 1)! + 1 n
(d) ( n − 1)! − 1 = [ 2n2 + 3n + 1 − 3n − 3 + 4 ]
4
Sol. If n = 1, then (n + 1)! − 1 = 2 − 1 = 1, (n − 1)! + 1 = 2, n n(n2 + 1)
= [ 2n2 + 2 ] =
(n + 1)! + 1 = 2 + 1 = 3, (n − 1)! − 1 = 1 − 1 = 0 4 2
∴ Sum = (n + 1)! − 1 Hence, (d) is the correct answer.
Hence, (a) is the correct answer.
1 1 1 1
Ex 18. 1 − 1 − 1 − … 1 −
2 3 4 n + 1
is equal to
Ex 15. 1 ⋅ 3 + 3 ⋅ 5 + 5 ⋅ 7 +… upto n terms is equal to
n( n + 1) ( n + 5) 1
(a) (a)
3 n+1
n( n + 1) ( n + 2) n
(b) (b)
3 n+1
n( 4n 2 + 6n − 1) n
(c) (c)
3 2n + 1
(d) n( n + 1) ( n + 2) n
(d)
n(n + 1) (n + 5) n(n + 1) (n + 2) 3n + 1
Sol. If n = 1, then = 4, = 2,
3 3
1 1 n 1
n(4 n2 + 6n − 1) Sol. If n = 1, then = , = ,
= 3, n(n + 1) (n + 2) = 6 n+1 2 n+1 2
3
n 1 n 1
Hence, (c) is the correct answer. = , =
2n + 1 3 3n + 1 4
Ex 16. 1 + 4 + 13 + 40 +… upto n terms is equal to If n = 2, then
1 1 n
= , =
2
3n + 1 − 2n ( 3n + 1 − 2n − 3) n+1 3 n+1 3
(a) (b) n
2n 4 ∴ Sum =
n+1
3n − 1 + 3n ( 3n − 1 + 2n 2 )
338 (c) (d) Hence, (b) is the correct answer.
9 8
Target Exercises
Type 1. Only One Correct Option
3 15 63 x x( x + a1 )
1. + + +K n terms is equal to 9. 1 + + +…+
4 16 64 a1 a1 a 2
(a) n −
1 n 1
4 − (b) n +
1 −n
4 −
1 x( x + a1 ) ( x + a 2 )… ( x + a n − 1 )
3 3 3 3 is equal to
1 n 1 1 1 a1 a 2… a n
(c) n + 4 − (d) n − 4 − n +
3 3 3 3 (x + a1 )(x + a2 )… (x + an )
(a)
1 2 2 3 3 4 a1a2… an
⋅ ⋅ ⋅ (x − a1 )(x − a2 )… (x − an )
(b)
2. 2 32 + 32 2 3 + 3 2 32 3 + ... n terms is equal to a1a2… an
1 1 +2 1 +2 +3 (c) (x + a1 )(x + a2 ) … (x + an )
n2 n3 n 1 (d) (x − a1 )(x − a2 ) … (x − an )
(a) (b) (c) (d)
(n + 1)2 (n + 1)3 n+1 n+1
10. Using mathematical induction, the numbers a n ’s are
1 + 2 + 3 + ... + n
2 2 2 2
defined by a 0 = 1, a n + 1 = 3n 2 + n + a n , ( n ≥ 0).
3. Σ is equal to
1+ 2 + 3 + K + n Then, a n is equal to
(n2 + 2n) n2 − 2n (a) n3 + n2 + 1 (b) n3 − n2 + 1 (c) n3 − n2 (d) n3 + n2
(a) (b)
3 6
Targ e t E x e rc is e s
1 5 1 3 7
n2 + 11 11. For n ∈ N , n + n + n is
(c) (d) None of these 5 3 15
12n
(a) an integer (b) a natural number
1⋅ 22 + 2⋅ 32 + 3⋅ 4 2 + … + n( n + 1) 2 (c) a positive fraction (d) None of these
4. is equal to
12 ⋅ 2 + 22 ⋅ 3 + 32 ⋅ 4 + … + n 2 ( n + 1) 12. If n ∈ N , then x n − y n is divisible by
3n + 5 3n + 1 (a) x − 1 (b) x − y (c) x + y (d) y − 1
(a) (b)
3n + 1 3n + 5
(c) (3n + 1)(3n + 5) (d) None of these 13. x + a is divisible by x + a for n is any
n n
(a) positive integer (b) integer
14 24 34 n4
5. + + +K+ is equal to (c) odd positive integer (d) None of these
1⋅ 3 3⋅ 5 5⋅ 7 ( 2n + 1)( 2n − 1)
14. x n + 1 + ( x + 1) 2n − 1 is divisible by
n(n + 1)(n + 2)
(a) (a) x (b) x + 1 (c) x 2 + x + 1 (d) x 2 − x + 1
6n
n(n + 1)(n2 + n + 1)
(b) 15. If n ∈ N , then 2n
6(2n + 1)
(c) n(n + 2)(n + 3)2 (a) = n (b) > n
(c) < n (d) None of these
(d) None of the above
1 1 1
1 1 1 16. If n ∈ N , then 1+ + +…+
6. + + + K n terms is equal to 2 3 n
1⋅ 2 2⋅ 3 3⋅ 4
1 n n n (a) = n (b) ≥ n
(a) (b) (c) (d) (c) ≤ n (d) None of these
n+1 n+1 2n + 1 3n + 1
( 2n )!
7.
1
+
1
+
1
+ K n terms is equal to 17. ≤
1⋅ 3 3⋅ 5 5⋅ 7 2 ( n !) 2
2n
1 n n n 1 1 1 1
(a) (b) (c) (d) (a) (b) (c) (d)
n+1 n+1 2n + 1 3n + 1 3n + 1 3n + 2 3n + 4 3n + 5
8. cos θ cos 2θ cos 4θ …cos 2n − 1 θ is equal to 18. If n > 1, then
sin 2 θn
sin 2 θn (2n)! 4n (2n)! 4n
(a) (b) (a) = (b) <
2n sin θ sin θ (n!) 2
2n + 1 (n!) 2
2n + 1
cos 2nθ cos 2nθ (2n)! 4n
(c) n (d) n (c) > (d) None of these
2 cosθ 2 sin θ (n!)2 2n + 1 339
7 Type 2. More than One Correct Option
19. If a1 = 1 and a n = na n−1 + 5 for all positive integer (a) n ≥ 1(n ∈ Z ) (b) n > 2
Objective Mathematics Vol. 1
n ≥ 2 , then a 5 is equal to (c) all n ∈ N (d) None of these
(a) 550 21. If p is prime number, then n p − n is divisible by p,
(b) 120 when n is
(c) 5a4 + 5 (a) natural number greater than 1
(d) 24 (b) irrational number
(c) complex number
20. x( x n−1 − na n−1 ) + a n ( n − 1) is divisible by ( x − a ) 2 (d) positive integer greater than or equal to 2
for
Type 3. Assertion and Reason
Directions (Q. Nos. 22-24) In the following 22. Statement I Sum of 20th group of {1}, {1, 2},
questions, each question contains Statement I {1, 2, 3},... is 110.
(Assertion) and Statement II (Reason). Each question n( n + 1)
Statement II Sum of n natural numbers is .
has 4 choices (a), (b), (c) and (d) out of which only one is 2
correct. The choices are
23. Statement I If n is a positive integer, then 32n + 7 is
(a) Statement I is true, Statement II is true; Statement II divisible by 8.
is a correct explanation for Statement I
Statement II HCF of 16 and 88 is 8.
(b) Statement I is true, Statement II is true; Statement II
is not a correct explanation for Statement I
24. Statement I If n ∈ N , product of n( n + 1)( n + 2) is
divisible by 6.
(c) Statement I is true, Statement II is false
Statement II Product of 3 consecutive positive
Ta rg e t E x e rc is e s
(d) Statement I is false, Statement II is true integers is divisible by 3!.
Type 4. Linked Comprehension Based Questions
Passage I (Q. Nos. 25-26) If an = 7 + 7 + 7 + … 27. S ( k ) can be expressed as
having n radical signs, then by methods of mathematical (a) k 2
induction, answer the following questions. (b) 2k 2
(c) 2 + k 2
25. Which of the following is true? (d) None of the above
(a) an > 7, ∀ n ≥ 1
(b) an > 3, ∀ n ≥ 1 28. If S ( k ) expressed as 3 + k 2 , then
(c) an < 4 , ∀ n ≥ 1 (a) S (1) is true (b) S (k ) ⇒
/ S (k + 1)
(d) an < 3, ∀ n ≥ 1 (c) S (k ) ⇒ S (k + 1) (d) None of these
26. The value of a1 + a 2 + a 3 is
n
Passage III (Q. Nos. 29-30) Consider an = 2 2 + 1, if
(a) < 6 n > 1.
(b) < 4
(c) < 12 29. The last digit of a 2 is
(d) None of the above (a) 1 (b) 2
Passage II (Q. Nos. 27-28) Consider S(k) is the sum (c) 3 (d) None of these
of first k odd natural numbers expressed as 30. The last digit of a n is
S (k ): 1 + 3 + 5 + … + (2k − 1) (a) 2 (b) 3 (c) 7 (d) 4
340
Type 5. Match the Column 7
Mathematical Induction
31. Match the following:
Column I Column II
n +1
A. The smallest positive integer for which the statement 3 < 4 holds is
n
p. 1
B. If x − 1is divisible by x − k. Then, the least positive integral value of k is
n
q. 4
C. If 49 + 16n + k is divisible by 64 for n ∈ N. Then, the numerically least
n
r. 2
negative integral value of k is
D. For all n ∈ N, 2 4 n − 15n − 1 is divisible by 15 k . Then, the maximum s. 3
integer value of k is
Type 6. Single Integer Answer Type Questions
( n + 2)!
32. If divisible by n, n ∈ N and 1≤ n ≤ 9, then n is___________ .
6( n − 1)!
33. If n ∈ N and 1< n ≤ 9, then n 3 + 2n is divisible by___________ .
34. The minimum value of n for which 10n + 3⋅ 4 n+ 2 + 5 is an integral multiple of 97, is___________ .
Entrances Gallery
Targ e t E x e rc is e s
JEE Main/AIEEE
1. 23n − 7n − 1 is divisible by [2014] Statement II For every natural number n ≥ 2
(a) 64 (b) 36 n( n + 1) < n + 1. [2008]
(c) 49 (d) 25
(a) Statement I is correct, Statement II is correct;
2. Statement I The sum of the series Statement II is a correct explanation for Statement I
1 + (1 + 2 + 4 ) + ( 4 + 6 + 9) + ( 9 + 12 + 16) (b) Statement I is correct, Statement II is correct;
+ .... + ( 361 + 380 + 400) is 8000. Statement II is not a correct explanation for Statement I
n (c) Statement I is correct, Statement II is incorrect
Statement II ∑ [ k 3 − ( k − 1)3 ] = n 3 , for any (d) Statement I is incorrect, Statement II is correct
k =1
natural number n. [2012] 1 0 1 0
5. If A = and I = , then which one of the
(a) Statement I is true, Statement II is true; Statement II is
not a correct explanation for Statement I
1 1 0 1
(b) Statement I is true, Statement II is false following holds, ∀ n ≥ 1 by the principle of
(c) Statement I false, Statement II is true mathematical induction? [2005]
(d) Statement I is true, Statement II is true; Statement II is a
(a) A n = 2n − 1 A + (n − 1)I (b) A n = nA + (n − 1)I
correct explanation for Statement I
(c) A n = 2n − 1 A − (n − 1)I (d) A n = nA − (n − 1)I
3. Statement I For each natural number n,
( n + 1) 7 − n 7 − 1is divisible by 7. 6. Let S ( k ) = 1 + 3 + 5 + ... + ( 2k − 1) = 3 + k 2 .
Statement II For each natural number n, n 7 − n is Then, which of the following is true? [2004]
divisible by 7. [2011] (a) S (1) is correct
(a) Statement I is incorrect, Statement II is correct (b) Principle of mathematical induction can be used to
(b) Statement I is correct, Statement II is correct; prove the formula
Statement II is correct explanation for Statement I (c) S (k ) ⇒
/ S (k + 1)
(c) Statement I is correct, Statement II is correct; (d) S (k ) ⇒ S (k + 1)
Statement II is not a correct explanation for Statement I
(d) Statement I is correct, Statement II is incorrect 7. 1⋅ 2⋅ 3 + 2⋅ 3⋅ 4 + 3⋅ 4 ⋅ 5 + ... n terms is equal to [2002]
n(n + 1)(n + 2)(3n + 5) n(n + 1)(n + 2)(n + 3)
4. Statement I For every natural number n ≥ 2 (a) (b)
12 4
1 1 1 n(n + 1)(n + 2)(3n + 1)
+ +K+ > n. (c) 2n(n + 1)(n + 2)(n + 3) (d)
1 2 n 12
341
7 Other Engineering Entrances
8. If n is a positive integer, then n 3 + 2n is divisible by 12. If a, b and n are natural numbers, then a 2n − 1 + b 2n − 1
Objective Mathematics Vol. 1
[BITSAT 2014] is divisible by [EAMCET 2011]
(a) 2 (b) 6 (c) 15 (d) 3 (a) a + b
(b) a − b
9. 10n × 3( 4 n + 2 ) + 5 is divisible by ( n ∈ N )
(c) a3 + b3
[BITSAT 2014]
(a) 7 (b) 5 (c) 9 (d) 17 (d) a2 + b2
10. Let n ≥ 2 be an integer, 13. Which of the following result is valid? [AMU 2011]
(a) (1 + x ) > (1 + nx ), for all natural numbers n
n
2π 2π
cos sin 0 (b) (1 + x )n ≥ (1 + nx ), for all natural numbers n, where x > − 1
n n
(c) (1 + x )n ≤ (1 + nx ), for all natural numbers n
2π 2π
A = − sin cos 0
(d) (1 + x )n < (1 + nx ), for all natural numbers n
n n
0 0 1 14. If n is a natural number, then [AMU 2011,13]
and I is the identity matrix of order 3. Then, n3
[WB JEE 2014] (a) 1 + 2 + ... + n <
2 2 2
3
(a) A n = I and A n − 1 ≠ I n3
(b) A m ≠ I for any positive integer m (b) 1 + 2 + ...+ n =
2 2 2
3
(c) A is not invertible (c) 12 + 22 + ...+ n2 > n3
(d) A m = 0 for a positive integer m n3
(d) 12 + 22 + ...+ n2 >
1 1 1 kn 3
11. If + + + ... n terms = , then
2× 4 4 × 6 6× 8 n+1 15. If a1 = 1 and a n = na n − 1 for all positive integers n ≥ 2,
Ta rg e t E x e rc is e s
k is equal to [EAMCET 2012]
1 1 then a 5 is equal to [Kerala CEE 2010]
(a) (b) (a) 125 (b) 120
4 2
1 (c) 100 (d) 115
(c) 1 (d) (e) None of these
8
Answers
Work Book Exercise
1. (d) 2. (a) 3. (a) 4. (c) 5. (b) 6. (a) 7. (c)
Target Exercises
1. (b) 2. (c) 3. (a) 4. (a) 5. (b) 6. (b) 7. (c) 8. (a) 9. (a) 10. (b)
11. (b) 12. (b) 13. (c) 14. (c) 15. (b) 16. (b) 17. (a) 18. (c) 19. (a,c) 20. (a,c)
21. (a,d) 22. (d) 23. (a) 24. (a) 25. (c) 26. (c) 27. (a) 28. (c) 29. (d) 30. (c)
31. (*) 32. (1) 33. (3) 34. (2)
* A → q; B → p; C → p; D → r
Entrances Gallery
1. (c) 2. (d) 3. (b) 4. (b) 5. (d) 6. (d) 7. (b) 8. (d) 9. (c) 10. (a)
11. (a) 12. (a) 13. (b) 14. (d) 15. (b)
342
Explanations
Target Exercises
1. If n = 1, then n(n + 1)(n + 2 )(3 n + 1)
=
1 1 4 1 −2 12
n − 4n − = 1 − − = , n (n + 1)(n + 2 )(3 n + 5)
3 3 3 3 3
Σn (n + 1)2 12
1 1
n + 4− n − = 1 +
1 1 9 3
− = = , ∴ =
3 3 12 3 12 4 Σn 2 (n + 1) n(n + 1)(n + 2 )(3 n + 1)
1 1 4 1 12
n + 4n − = 1 + − = 2, 3n + 5
3 3 3 3 =
1 1 1 1 15 5 3n + 1
and n − 4− n + = 1 − + = =
3 3 12 3 12 4 n4
1 −1 5. S n = Σ
∴ Sum = n + 4−n (2 n + 1)(2 n − 1)
3 3
n4
2. If n = 1, then =Σ 2
4n − 1
n2 1
= , n2 1 1
(n + 1)2
4 =Σ + +
4 16 16 (2 n + 1)(2 n − 1)
n3 1
= , n(n + 1)(2 n + 1) n (2 n + 1) − (2 n − 1)
(n + 1)3 8 = + +Σ
24 16 32(2 n + 1)(2 n − 1)
n 1
= , n(n + 1)(2 n + 1) n 1 1
n+1 2 = + + 1 −
24 16 32 2 n + 1
1 1
and = n(n + 1)(2 n + 1) n n
n+1 2 = + +
Targ e t E x e rc is e s
24 16 16 (2 n + 1)
If n = 2, then
n(n + 1)(2 n + 1) n(2 n + 2 )
n
=
2 = +
n+1 3 24 16 (2 n + 1)
1 1 n(n + 1) (2 n + 1) + 3
2
= =
n+1 3 24 2 n + 1
∴ Sum =
n n(n + 1)(n 2 + n + 1)
=
n+1 6 (2 n + 1)
12 + 2 2 + 32 + K + n 2 1
3. We have, 6. t n =
1+ 2 + 3 + K + n n(n + 1)
n(n + 1)(2 n + 1)2 (2 n + 1) 2 −1 3−2 4− 3 (n + 1) − n
= = ∴ Sn = + + + ... +
n(n + 1) 6 3 1⋅ 2 2⋅3 3⋅ 4 n(n + 1)
(Σn ) n n(n + 1) n 1 1 1 1 1
Sn = 2 + = + = 1 − + − + ... + −
3 3 3 3 2 2 3 n n+1
(n + 2 n )
2
1 n
= = 1− =
3 n+1 n+1
4. Σ n(n + 1)2 = Σ (n 3 + 2 n 2 + n ) 7. If n = 1, then
n 2 (n + 1)2 2 n(n + 1)(2 n + 1) n(n + 1) 1 1
= + + = ,
4 6 2 n+1 2
n(n + 1) n 1
= [3 n(n + 1) + 4(2 n + 1) + 6] = ,
12
n+1 2
n(n + 1)(3 n 2 + 11 n + 10 ) n 1
= = ,
12 2n + 1 3
n(n + 1)(n + 2 )(3 n + 5) n 1
= and =
12 3n + 1 4
Σn 2 (n + 1) = Σ(n 3 + n 2 ) n
∴ Sum =
n (n + 1)
2 2
n(n + 1)(2 n + 1) 2n + 1
= +
4 6 8. If n = 1, then
n(n + 1) sin 2 n θ sin 2θ
= [3 n(n + 1) + 2(2 n + 1)]
12 =
2 n sin θ 2 sin θ
n(n + 1)(3 n 2 + 7 n + 2 ) 2 sin θ cos θ
= = = cos θ 343
12 2 sin θ
7 9. If n = 1, then given sum = 1 +
10. Given, a0 = 1, an + 1 = 3 n 2 + n + an
x x + a1
a1
=
a1
∴
and
1
3n + 5
(2 n )!
=
≤
1
8
1
Objective Mathematics Vol. 1
2n
2 (n !) 2
3n + 1
⇒ a1 = 3 (0 ) + 0 + a0 = 1
18. If n = 2, then
⇒ a2 = 3 (1)2 + 1 + a1
(2 n )! 24
= 3 + 1+ 1= 5 = =6
(n !)2 4
From option (b), 4n 8
Let P(n ) = n 3 − n 2 + 1 and =
2n + 1 3
∴ P(0 ) = 0 − 0 + 1 = 1 = a0 (2 n )! 4n
⇒ >
P(1) = 13 − 12 + 1 = 1 = a1 (n !)2 2 n + 1
and P(2 ) = (2 )3 − (2 )2 + 1 = 5 = a2
19. Given, an = nan − 1 + 5
∴ an = n 3 − n 2 + 1
For n = 2, then
11. If n = 1, then a2 = 2 a1 + 5 = 7 [Q a1 = 1]
1 5 1 3 7 a3 = 3 a2 + 5 = 21 + 5 = 26
n + n + n
5 3 15 a4 = 4a3 + 5 = 104 + 5 = 109
1 1 7
= + + ∴ a5 = 5a4 + 5 = 545 + 5 = 550
5 3 15
3+ 5+ 7 20. Check through options, the condition is satisfied,
= ∀ n ∈ N.
15
=1 21. n p − n is divisible by p for any natural number greater
12. If n = 1, then than 1.
x − y = x − y, divisible by x − .y.
n n It is Fermet’s theorem.
Aliter
Ta rg e t E x e rc is e s
13. If n = 1, then Let n = 4 and p = 2
x n + an = x + a Then, (4)2 − 4 = 16 − 4 = 12
If n = 3, then It is divisible by 2. So, it is true for any natural number
greater than 1.
x n + an = x 3 + a3 = ( x + a)( x 2 + a2 + ax )
20 ⋅ 21
∴ x n + an is divisible for odd positive integer n. 22. Sum of 20th group = = 210
2
n(n + 1)
14. If n = 1, then Sum of n natural numbers =
2
x n + 1 + ( x + 1)2 n − 1 1⋅ 2
It is true, for n = 1, =1
= x 2 + x + 1, divisible by x 2 + x + 1. 2
15. If n = 1, then Assume, it is true, for n = k.
k(k + 1)
2 n = 2, > 1 = n So, 1 + 2 + 3 + ... + k =
2
⇒ 2n > n Now, for k + 1, 1 + 2 + 3 + ... + k + k + 1
k(k + 1) (k + 1)(k + 2 )
16. If n = 2, then given sum = 1 +
1 = +k+1=
2 2 2
2 +1 2+ 2 So, Statement II is true, for all n ∈ N.
= =
2 2 23. Let statement P(n ) = 32 n + 7
=
3.414 For P(n = 1) = 16
2 For P(n = 2 ) = 88
= 1.707 ∴HCF of 16 and 88 is 8.
⇒ 2 = 1.414 24. Let P(n ) ≡ n(n + 1)(n + 2 )
∴ Sum ≥ n
P(n = 1) ≡ 1⋅ 2 ⋅ 3 = 3 !
17. If n = 1, then P(n = 2 ) ≡ 2 ⋅ 3 ⋅ 4 = 2 ⋅ (1⋅ 2 ⋅ 3) = 2 ⋅ 3 !
(2 n )!
= =
2 1 P(n = 3) ≡ 3 ⋅ 4 ⋅ 5 = 2 ⋅ 5 ⋅ (1⋅ 2 ⋅ 3) = 10 ⋅ 3 !
2n 2
2 (n !) 4 2 So, Statement I is true, Statement II is also true and
1 1 1 Statement II is a correct explanation of Statement I.
Now, = = ,
3n + 1 4 2 25. We have, a1 = 7 < 4. Assume ak < 4 for some natural
1 1 number k.
= ,
3n + 2 5 We have,
1 1 ak + 1 = 7 + ak < 7 + 4 < 4
= ,
344 3n + 4 7 ∴ an < 4 , ∀ n ≥ 1
7
k
26. Q a1 < 4, a2 < 4, a3 < 4 30. Assume that ak = 2 2 + 1 = 10 m + 7
∴ a1 + a2 + a3 < 4 + 4 + 4 = 12 where, k > 1and m is some positive integer.
k +1 k
k Now, ak + 1 = 2 2 + 1 = (2 2 )2 + 1
Mathematical Induction
27. Consider 1 + 3 + 5 + ... + (2 k − 1) = [1 + (2 k − 1)]
2 = (10 m + 6)2 + 1
Q if a and l are the first term and last
= 10(10 m2 + 12 m+ 3)+ 7
n
term of an AP, then S n = (a + l ) Thus, last digit of an is 7, ∀ n > 1.
2
⇒ 1 + 3 + 5 + ... + (2 k − 1) = k 2 31. A. If n = 1, then 32 </ 4
If n = 2, then 27 </ 16
Thus, S (k ) can be expressed as k 2 . If n = 3, then 81 </ 64
i.e. S (k ): 1 + 3 + 5 + ...+ (2 k − 1) = k 2 If n = 4, then 243 < 256
B. x n − 1is divisible by x − 1, so k = 1
28. Consider S (k ): 1 + 3 + 5 + ... + (2 k − 1) = 3 + k 2
C. Value of k = − 1, so numerically 1 is smallest.
Then, S(1) : 1 = 3 + 12 , which is not true.
D. If n = 1⇒ 2 4n − 15 n − 1 = 0; n = 2
Suppose S (k ) is true, then
1 + 3 + 5 + ... + (2 k − 1) = 3 + k 2 ⇒ 256 − 30 − 1 = 225 = 152
So, maximum value of k is 2.
On adding (2 k + 1) both sides, we get
(n + 2 )! 3! 6
1 + 3 + 5 +…+ (2 k − 1) + (2 k + 1) = 3 + k 2 + (2 k + 1) 32. If n = 1, then = = = 1, divisible by 1.
6 (n − 1)! 6 × 0 ! 6
= 3 + (k + 1)2
which is S (k + 1.) 33. If n = 1, then n 3 + 2 n = 1 + 2 = 3, divisible by 3.
Thus, S (k ) ⇒ S (k + 1)
2
34. For n = 1, we have given expression = 207 = 9 × 23
29. For n = 2, a2 = 2 2 + 1 = 2 4 + 1 and for n = 2, we have given expression = 873 = 9 × 97
= 16 + 1 = 17 Thus, for n = 2 given expression is an integral multiple
Thus, the last digit of a2 is 7. of 97.
Targ e t E x e rc is e s
Entrances Gallery
1. Let P(n ) = 2 3 n − 7 n − 1 which is divisible by 7.
⇒ P(1) = 0, P(2 ) = 49 For n = k, P2 (k ) = k 7 − k
P(1) and P(2 ) are divisible by 49. Let P2 (k ) be divisible by 7.
Let P(k ) = 2 3 k − 7 k − 1 = 49I ∴ k 7 − k = 7λ …(i)
P(k + 1) = 2 3 k + 3 − 7 k − 8 For n = k + 1, P2 (k + 1) = (k + 1)7 − (k + 1)
= 8 (49 I + 7 k + 1) − 7 k − 8 = (k 7 − k ) + 7(k 6 + 3k 5 + ... + k )
= 49 (8I ) + 49k = 49λ = 7 λ + 7(k 6 + 3k 5 + ....+ k ) [from Eq. (i)]
[where λ = 8I + k, which is an integer] ⇒ Divisible by 7.
n
∑ [k
1 1 1
2. We have, 3
− (k − 1) ]
3
4. Let P(n ) = + + ... +
k =1 1 2 n
= (13 − 0 3 ) + (2 3 − 13 ) + (33 − 2 3 ) + ...+ [n 3 − (n − 1)3 ] = n 3 ∴ P(2 ) =
1
+
1
∴1 + (1 + 2 + 4) + (4 + 6 + 9) + (9 + 12 + 16) 1 2
+ ... + (361 + 380 + 400 ) = 1707
. > 2
= (13 − 0 3 ) + (2 3 − 13 ) + (33 − 2 3 ) + (43 − 33 ) Let us assume that
+ ... + (20 3 − 193 ) 1 1 1
P(k ) = + + ... + > k is true, for n = k + 1
= (20 ) = 8000
3
1 2 k
3. Let P1(n ) = (n + 1)7 − n 7 − 1 LHS =
1
+
1
+ ... +
1
+
1
= (7C0 n 7 + 7C1n 6 + ... + 7 C6 n + 1) − n 7 − 1 1 2 k k+1
= (n 7 + 7C1n 6 + ... + 7 C6 n + 1) − n 7 − 1 1 k(k + 1) + 1
> k + =
= 7C1n 6 + ....+ 7C6 n k+1 (k + 1)
7(7 − 1) 5 k+1
= 7n6 + n + ... + 7 n > [Q k(k + 1) > k, ∀ k ≥ 0]
2 k+1
which is divisible by 7. = (k + 1)
Now, let P2 (n ) = (n )7 − n
∴ P(k + 1) > k + 1
for n = 1, P2 (1) = 0
345
7 By mathematical induction, Statement I is true, ∀ n ≥ 2.
Now, let
∴
α(n ) = n(n + 1)
α(2 ) = 2(2 + 1)
9. For n = 1, 10 n + 3 ⋅ 4n + 2 + 5 = 10 + 3 ⋅ 43 + 5
= 207, which is divisible by 9.
So, by induction, the result is divisible by 9.
Objective Mathematics Vol. 1
= 6<3
2π 2π
cos n sin n 0
Let us assume that
α(k ) = k(k + 1) < (k + 1) is true.
2π 2π
For n = k + 1 10. Given, A = − sin cos 0
n n
LHS = (k + 1) (k + 2 ) < (k + 2 ) [Q(k + 1) < (k + 2 )] 0 0 1
∴ α(k + 1) < (k + 2 )
By mathematical induction, Statement II is true but 2π 2π
Statement II is not a correct explanation for Statement I. cos n sin n 0
1 0 1 0 1 0 2π 2π
5. A2 = = Now, A × A = − sin cos 0
n n
1 1 1 1 2 1
0 0 1
1 0 1 0 1 0
A3 = =
2 1 1 1 3 1
2π 2π
cos n sin n
M M M 0
1 0
An = can be verified by induction. 2π 2π
n 1 × − sin cos 0
n n
Now, taking options 0 0 1
n 0 n − 1 0
(b) nA + (n − 1) I = +
n n 0 n − 1
Arrange properly in a line
2 n − 1 0
= ≠ An 2 2π 2 2π
− 1 cos n − sin n
Ta rg e t E x e rc is e s
n 2 n
n 0 n − 1 0 = 2π 2π 2π 2π
(d) nA − (n − 1) I = −
−
sin cos − sin cos
n n 0 n − 1 n n n n
1 0 0
= = A
n
n 1 2π 2π 2π 2π
cos sin + sin cos 0
n n n n
6. 1 + 3 + 5 + K + (2 k + 1)
2π 2π
= [1 + 3 + 5 + K + (2 k − 1)] + (2 k + 1) − sin 2 + cos 2 0
n n
= 3 + k 2 + 2k + 1 1
[given] 0
= 3 + (k + 1)2 2π 2π 2π
cos 2 × 2 sin cos 0
∴ S (k ) ⇒ S (k + 1) n n n
2π 2π 2π
7. t n = n(n + 1)(n + 2 ) = −2 sin cos cos 2 × 0
n n n
= n(n 2 + 3 n + 2 ) 0 0 1
= n3 + 3 n2 + 2n
∴S n = Σ n 3 + 3 Σ n 2 + 2 Σ n 2π 2π
cos 2 × n sin 2 × n 0
n 2 (n + 1)2 3n(n + 1)(2 n + 1) 2 n(n + 1)
= + + 2π 2π
4 6 2 = − sin 2 × cos 2 × 0
n n
n(n + 1) 1
= [n(n + 1) + 4 n + 2 + 4] 0 0
4
n(n + 1)(n 2 + 5n + 6)
= Similarly,
4 n − 1 2π n − 1 2π
n(n + 1)(n + 2 )(n + 3) cos 2 × sin 2 × 0
= n n
4
n − 1 2π n − 1 2π
A = − sin 2 × × 0
n
cos 2
8. Let P(n ) = n 3 + 2 n n n
Now, P(1) = 1 + 2 = 3 0 0 1
P(2 ) = 8 + 4 = 12
P(3) = 27 + 6 = 33 1 0 0
Clearly, we see that all these numbers are divisible = 0 1 0 = I
by 3. 0 0 1
346
n − 2 2π
cos 2
×
and An − 1 = − sin 2 n − 2 ×
n
2π
sin 2 n − 2 ×
cos 2 n − 2 ×
2π
n
2π
0
0 ≠ I
For n = k, let P(k ) : (1 + x )k ≥ (1 + kx ) is true.
For n = k + 1, P(k + 1) : (1 + x )k + 1 ≥ {1 + (k + 1)x} is also
true.
7
Mathematical Induction
n n We will show P(k + 1) is true.
0 0 1
Consider (1 + x )k + 1 = (1 + x )k ⋅ (1 + x ) ≥ (1 + kx )(1 + x )
[if x > − 1]
kn 1 1 1 = 1 + x + kx + kx 2 ≥ 1 + x + kx [Q k > 0 and x > − 1]
11. = + + + ... n terms
n + 1 2 ⋅ 4 4⋅ 6 6⋅ 8 = 1 + (k + 1)x
1 4 − 2 6 − 4 8 − 6 2n + 2 − 2n Thus, (1 + x )k + 1 ≥ 1 + (k + 1)x, if x > − 1
= + + + ... +
2 2⋅4 4⋅ 6 6⋅ 8 2 n(2 n + 2 )
14. By taking option (d),
1 1 1 1 1 1 1 1 1 1
= − + − + − + ... + − When n = 1, then 1 > [true]
2 2 4 4 6 6 8 2 n 2 n + 2 3
8
1 1 1 When n = 2, then 5 > , [true]
= −
2 2 2 (n + 1)
3
When n = 3, then 14 > 9, [true]
n 1
= ⇒ k= 64
4(n + 1) 4 When n = 4, then 30 > = 2133
. , [true]
3
12. 2 n − 1 is an odd positive integer. 15. Given, an = nan − 1
⇒ a2 n − 1 + b2 n − 1 is divisible by a + b.
⇒ a2 = 2 a1 = 2 [Q a1 = 1]
13. Let P(n ) : (1 + x )n ≥ (1 + nx ) a3 = 3 a2 = 3 (2 ) = 6
For n = 1, (1 + x )1 = 1 + x = 1 + 1⋅ x ≥ 1 + 1⋅ x a4 = 4(a3 ) = 4 (6) = 24
∴ (1 + x )1 ≥ 1 + 1⋅ x ∴ a5 = 5 (a4 ) = 5 (24) = 120
Targ e t E x e rc is e s
347
8
Binomial
Theorem
Binomial Theorem for Positive Integral Index
If x and a are real numbers, then for all n Î N , Chapter Snapshot
( x + a ) n = n C 0 x n a 0 + n C1 x n -1 a 1 + n C 2 x n - 2 a 2
● Binomial Theorem for Positive
+K + n C r x n - r a r + K + n C n -1 x 1 a n -1 + n C n x 0 a n Integral Index
n
i.e. (x + a ) n = å n C r x n - r a r . ● Multinomial Theorem
r =0 ● Greatest Coefficient
X Example 1. x 10 + 10x 8 a + 40x 6 a 2 + 80x 4 a 3 + 80x 2 a 4 + 32a 5 is equal to
● Greatest Term
(a) ( x 2 + a ) 5 (b) ( x + a )10 (c) ( x 2 + 2a ) 5 (d) ( x 2 - 2a ) 5 ● R-f Factor Relation
● Divisibility Problems
Sol. (c) Using binomial theorem, we have
( x2 + 2 a)5 = 5C 0 ( x2 )5 (2 a)0 + 5C1( x2 )4 (2 a)1 + 5C 2 ( x2 )3 (2 a)2 + 5C 3 ( x2 )2 (2 a)3 ● Properties of Binomial
5 2 4 5 2 0 5 Coefficients
+ C 4 ( x )(2 a) + C 5 ( x ) (2 a)
= x 10
+ 5( x )(2 a) + 10( x )(4a ) + 10( x )(8a ) + 5( x2 )(16a4 ) + 32 a5
8 6 2 4 3 ● Binomial Theorem for any
10 8 6 2 4 3 2 4 5 Index
= x + 10 x a + 40 x a + 80 x a + 80 x a + 32 a
● Approximation
Properties of Binomial Expansion ● Exponential Series
i. The above expansion is also valid when x and a are complex numbers. ● Logarithmic Series
n
ii. We have, (x + a ) n = å n C r x n - r a r . Since, r can have values from 0 to n, the
r =0
total number of terms in the expansion is ( n +1).
iii. The sum of the indices of x and a in each term is n.
9
e.g. (2x + 3 y) 9 = å 9 C r (2x ) 9- r (3 y) r and it has (9 + 1) i.e. 10 terms.
r =0
Also, the sum of indices of 2x and 3y in each term is 9.
iv. Replacing a by -a, in the expansion of ( x + a ) n , we get
( x - a ) n = n C 0 x n a 0 - n C1 x n -1 a 1 + n C 2 x n - 2 a 2 - n C 3 x n - 3 a 3
+ K + ( -1) r n
C r x n - r a r + K + ( -1) n n
Cn x 0a n
n
i.e. ( x - a ) n = å ( -1) r n
C r x n -r a r
r =0
v. Putting x =1 and a = x in the expansion of
( x + a ) n , we get
X Example 3. The number of terms in the
expansion of (1 + 5 2x ) 9 + (1 - 5 2x ) 9 are 8
Binomial Theorem
(1 + x ) n = n C 0 + n C1 x + n C 2 x 2 (a) 3 (b) 6
+K + nCr x r +K + nCn x n (c) 5 (d) None of these
n Sol. (c) Since, here n = 9 is odd, therefore the expansion of
i.e. (1 + x ) n = å n C r x r (1 + 5 2 x)9 + (1 - 5 2 x)9 has æç
9 + 1ö
÷ = 5 terms.
r =0 è 2 ø
known as the expansion of (1 + x ) n in X Example 4. Using binomial theorem, expand
ascending powers of x.
{( x + y) 5 + ( x - y) 5 } and hence find the value of
vi. Putting a =1 in the expansion of ( x + a ) n , we
{( 2 + 1) 5 + ( 2 - 1) 5 }.
get
( x + 1) n = n C 0 x n + n C1 x n -1 + n C 2 x n - 2 + K Sol. We have,
( x + y)5 + ( x - y)5 = 2 [5 C 0 x5 + 5C 2 x3 y2 + 5C 4 x1 y4 ]
+ n C r x n - r + K + n C n -1 x + n C n
n = 2( x5 + 10 x3 y2 + 5 xy4 )
n n n -r
i.e. ( x + 1) = å C r x Putting x = 2 and y = 1, we get
r =0 ( 2 + 1)5 + ( 2 - 1)5 = 2 {( 2 )5 + 10( 2 )3 + 5 2 }
known as the expansion of ( x +1) n in = 2(4 2 + 20 2 + 5 2 ) = 58 2
descending powers of x.
x. We have,
vii. Putting x =1 and a = - x in the expansion of
( x + a ) n = n C 0 x n a 0 + n C1 x n -1 a 1 + n C 2 x n - 2 a 2
( x + a ) n , we get
+ K+ n C r x n - r a r + K + n C n x 0 a n
(1 - x ) n = n C 0 - n C1 x + n C 2 x 2 - n C 3 x 3 + K
Here, ( r +1)th term is given by n C r x n - r a r .
+ ( -1) r n C r x r + K + ( -1) n n C n x n
n If Tr +1 denotes the ( r +1)th term, then
i.e. (1 - x ) n = å ( -1) r n
Cr x r Tr +1 = n C r x n - r a r
r =0
This is called the general term, because by
X Example 2. Expand (2x - 3 y) 4 by binomial giving different values to r we can determine
theorem. all terms of the expansion.
Sol. Using binomial theorem, we have Also, the general term can be expressed as
4
(2 x - 3 y)4 = {2 x + (- 3 y)}4 = å(- 1)
r 4
C r (2 x)4 - r (3 y)r n! s r
x a , where r + s = n.
r =0
r ! s!
= 4C 0 (2 x)4 (3 y)0 - 4C1(2 x)3 (3 y) + C 2 (2 x)2 (3 y)2
1 4
xi. In the binomial expansion of ( x - a ) n , the
- 4C 3 (2 x)(3 y)3 + 4C 4 (2 x)0 (3 y)4
general term is given by
= 16 x4 - 4(8 x3 )(3 y) + 6(4 x2 )(9 y2 ) - 4(2 x)(27 y3 ) + 81y4
= 16 x4 - 96 x3 y + 216 x2 y2 - 216 xy3 + 81y4
Tr +1 = ( -1) r n C r x n - r a r
xii. In the binomial expansion of (1 + x ) n , we have
viii. We have,
(x + a ) n + (x - a ) n = 2 [ n C 0 x n a 0 Tr +1 = n C r x r
+ n C 2 x n - 2 a 2 + K] xiii. In the binomial expansion of (1 - x ) n , we have
and ( x + a ) n - ( x - a ) n = 2 [ n C1 x n -1 a 1
Tr +1 = ( -1) r n
Cr x r
+ n C 3 x n - 3 a 3 + K]
Ø To find the terms free from radicals or rational indicies in the
ix. If n is odd, then {( x + a ) n + ( x - a ) n } and expansion of (a1/ p + b1/ q )n , where a and b are prime numbers, first
{( x + a ) n - ( x - a ) n } both have the same N- r r
N 1/ p N - r 1/ q r N p q
æ n + 1ö find the general termTr + 1 = C r(a ) (b ) = C r a b , then
number of terms equal to ç ÷ whereas, if n
è 2 ø put the values of r, where 0 £ r £ N, for which indicies of a and b
æn ö become integers.
is even, then {( x + a ) n + ( x - a ) n } has ç + 1÷
è2 ø X Example 5. If the rth term in the expansion of
terms and 10
æx 2 ö 4
æ nö ç - 2 ÷ contains x , then r is equal to
{( x + a ) n - ( x - a ) n } has ç ÷ terms. è3 x ø
è 2ø
(a) 2 (b) 1 (c) 3 (d) 5 349
8 Sol. (c) Clearly, the general term in the binomial expansion
x 2
of æç - 2 ö÷
è3 x ø
10
is given by
Sol. (a) Let the general term i.e. (r + 1) th contains x11.
We have, Tr +1 =
-2
C r ( x3 )12 - r æç 2 ö÷
12
èx ø
r
Objective Mathematics Vol. 1
10 - r r
x æ- 2ö = 12C r x36 - 3 r - 2 r (- 1)r 2 r
Tr + 1 = 10
C r æç ö÷ ç 2 ÷
è 3ø èx ø
= 12
C r (- 1)r 2 r x36 - 5 r …(i)
10 - r
1
C r æç ö÷
10 r 10 - r - 2 r 11
= (- 2 ) x to find the coefficient of x , put 36 - 5r = 11
è 3ø
10 - r
Þ 5r = 25 Þ r = 5
1
= C r æç ö÷
10
(- 2 )r x10 - 3 r On substituting the value of r in Eq. (i), we get
è 3ø T6 = 12C 5 (- 1)5 2 5 x11
Now, replace r by r - 1, we get the rth term i.e. Thus, the coefficient of x11 is
11 - r
1 - 12 ´ 11 ´ 10 ´ 9 ´ 8
Tr = 10C r -1 æç ö÷
12
(- 2 )r -1 x13 - 3 r C 5 (- 1)5 2 5 = ´ 32
è 3ø 5´ 4´ 3´2
Since, the r th term contains the x4 , therefore we have = - 25344
13 - 3 r = 4
Þ 3r = 9 Þ r = 3 X Example 8. The coefficient of x p and x q
(p and q are positive integers) in the expansion of
X Example 6. The number of integral terms in (1 + x ) p + q are
642
æ 1 1ö
(a) equal
the expansion of ç 5 2 + 7 6 ÷ is (b) equal with opposite signs
ç ÷
è ø (c) reciprocal of each other
(a) 106 (b) 108 (d) None of the above
(c) 103 (d) 109 Sol. (a) Coefficient of xp and xq in the expansion of
Sol. (b) Clearly, the general term in binomial expansion of (1 + x)p +q
are p +q
C P and p +q
Cq
642 p +q p +q ( p + q )!
æ 1 1ö and Cp = Cq =
ç52 + 76 ÷ is given by p! q !
ç ÷
è ø
æ 1ö
642 - r
æ 1ö
r xvii. Independent term or constant term of a
Tr + 1 = 642
Cr ç52 ÷ ç7 6 ÷ binomial expansion is the term in which
ç ÷ ç ÷ exponent of the variable is zero. i.e. the
è ø è ø
Obviously, r should be a multiple of 6. coefficient of x 0 .
642
Total numbers = = 107, but first term for r = 0 is
6 X Example 9. The ratio of the coefficient of x 15
also integral. Hence, total terms are 107 + 1 = 108 15
æ 2ö
to the term independent of x in ç x 2 + ÷ is
xiv. The coefficient of x k in the expansion of è xø
( x + a ) n can be obtained by equating the (a) 12 : 32 (b) 1 : 32
power of x in the general term to k. This will (c) 32 : 12 (d) 32 : 1
give value of r and on substituting this value in Sol. (b) Clearly, the general term in the expansion of
general term we can find the coefficient of x k . æ x2 + 2 ö
15
ç ÷ is given by
è xø
Ø If r is not a positive integer, then there will be no term containing
r
2
x k and hence coefficient of x k will be zero. Tr + 1 = 15
C r ( x2 )15 - r æç ö÷
è xø
xv. The coefficient of ( r +1)th term in the = 15
C r (2 )r x30 - 3 r …(i)
expansion of (1 + x ) n is n C r . Now, for the coefficient of term containing x15 ,
30 - 3r = 15, i.e. r = 5
Therefore, 15 C 5 (2 )5 is the coefficient of x15 [from Eq.(i)]
xvi. The coefficient of x r in the expansion of
Now, to find the term independent of x, put 30 - 3 r = 0
(1 + x ) n is n C r .
Thus, 15 C10 210 is the term independent of x [from Eq.(i)]
15
X Example 7. The coefficient of x 11 in the C52 5 1 1
Now, the ratio is 15
= =
12 C10 210 25 32
æ 3 2 ö
expansion of ç x - ÷ is
è x2 ø xviii. In the binomial expansion of ( x + a ) n , the rth
(a) - 25344 (b) 25344 term from the end is ( n - r + 2)th term from
350 (c) 23544 (d) - 23544 the beginning.
X Example 10. The 4th term from the end in the
æ x3
expansion of ç
2 ö
9
- ÷ is ii.
Some another form of binomial expansion
(x + y + z ) n = å
n!
x r ys z t
8
Binomial Theorem
r ! s! t !
è 2 x2 ø r + s+ t = n
n + 3-1
670 671 The above expansion has C 3-1
(a) (b) n +2
3
x x3 = C 2 terms
672 n!
(c) (d) None of these iii. ( x + y + z + u) n = x p yq z r us
x3 å p! q ! r ! s!
p + q + r + s= n
Sol. (c) Clearly, the 4th term from the end is 9 - 4 + 2 i.e. 7th n + 4-1
term from the beginning, which is given by
There are C 4-1 = n + 3C 3
terms in the
æ x3 ö
3 6 9 above expansion.
T7 = 9C 6 çç ÷÷ æ -2 ö = 9C x × 64
ç 2÷ 3 Above result can be generalized in the
è2 ø èx ø 8 x12
9´ 8´7 64 672 following form (known as multinomial
= ´ 3 = 3 theorem)
3´2 ´1 x x
( x1 + x 2 + K + x k ) n
xix. Middle term in a binomial expansion If n is n! r r r
an even natural number, then in the binomial = å r ! r !K r ! x11 x 22 K x kk
r1 + r2 +K+ rk = n 1 2 k
æn ö
expansion of ( x + a ) n , ç + 1÷ th term is the
è2 ø The general term in the above expansion is
middle term. n!
x1r1 x 2r2 x 3r3 K x krk
r1 ! r2 ! r3 !K rk !
xx. If n is odd natural number, then
æ n + 1ö æ n + 3ö The number of terms in the above expansion is
ç ÷ th and ç ÷ th are the middle terms in equal to the number of non-negative integral
è 2 ø è 2 ø
solution of the equation r1 + r2 + K + rk = n,
the binomial expansion of ( x + a ) n . because each solution of this equation gives a
term in the above expansion. The number of
Ø When there are two middle terms in the expansion, then their
binomial coefficients are equal. such solutions is n + k -1 C k -1 .
X Example 11. The middle term (terms) in the Ø n
The coefficient of x1 1 × x 2n2 K x nmm in the expansion of
9
æ p xö n!
expansion of ç + ÷ is/are (x1 + x 2 + K + x m )n is
è x pø n1 ! n2 ! n3 ! K nm!
126 p 126x 125 p 125x X Example 12. The number of distinct terms in
(a) (b) (c) (d)
x p x p the expansion of ( x + y - z )16 is
Sol. (a, b) Since, the power of binomial is odd, therefore we (a) 136 (b) 153 (c) 16 (d) 17
have two middle terms which are 5th and 6th terms. These
are given by
Sol. (b) ( x + y - z)16 = 16C 0 x16 + 16C1 x15 ( y - z) + …
p æ xö
5 4
p 126 p + 16
C r x16 - r ( y - z)r + .... + 16
C16 ( y - z)16
T5 = 9C 4 æç ö÷ ç ÷ = 9C 4 =
è x ø è pø x x Clearly, all the terms are distinct.
4 5 \ The number of distinct terms
p æ xö x 126 x
and T6 = 9C 5 æç ö÷ ç ÷ = 9C 5 = = 1 + 2 + 3 + K + 17
è x ø è pø p p 17 ´ 18
= = 153
2
Multinomial Theorem Aliter
Number of terms in ( x1 + x2 + K + xm )n
= n + m -1C m -1
i. We know,
\ Number of terms in ( x + y - z)16
n
n n n -r r 18 ´ 17
(x + a ) = å C r x a , nÎ N = 16 + 3 -1C 3 -1 = 18C 2 = = 153
r =0 2
n
n! Example 13. The coefficient of a 8 b 6 c 4 in the
=å x n -r a r X
r =0 ( n - r )! r ! expansion of ( a + b + c)18 is
n! s r (a) 18 C14 × 14C 8 (b) 18 C14
= å x a , where s = n - r
r + s= n r ! s! (c) 12
C8 (d) None of these 351
8 Sol. (a) In (a + b + c )18 , general term is given by
(18)!
(r1 )!(r2 )!(r3 )!
(a)r1 (b )r 2 (c )r 3 …(i)
Sol. (d) (1 - x + x2 )5 = {1 + x( x - 1)}5
= 5C 0 + 5C1 x( x - 1) + 5C 2 x2 ( x - 1)2 + 5C 3 x3 ( x - 1)3 + K
\ The coefficient of x3 = - 2 ×5 C 2 - 5C 3 = - 30
Objective Mathematics Vol. 1
\ Coefficient of a8 b 6c 4 is obtained by putting Aliter
r1 = 8, r2 = 6, r3 = 4 in Eq.(i) General term of (1 - x + x2 )5 is
(18)! 8 6 4 5!
i.e. a bc (1)r1 (- x)r 2 ( x2 )r 3 …(i)
8! × 6!× 4! (r1 )! (r2 )!(r3 )!
(18)! 14! 18 \ For coefficient of x3 , r2 + 2 r3 = 3
Thus, required coefficient = ´ = C14 ´14 C 8
8! × 6! × 4! 14!
where r1 + r2 + r3 = 5
X Example 14. The coefficient of x 3 y 4 z 2 in the Let r3 = 0, then r2 = 3 and r1 = 2
expansion of (2x - 3 y + 4 z ) 9 is r3 = 1, then r2 = 1 and r1 = 3
(a) 12063600 (b) 13063680 If r3 = 2, 3, 4, 5, then r2 is not possible as negative.
\ Only two terms contains x5 i.e.
(c) 11063600 (d) None of these
(r1 = 2, r2 = 3, r3 = 0) and (r1 = 3, r2 = 1, r3 = 1)
Sol. (b) The general term in the expansion of Thus, from Eq.(i),
(2 x - 3 y + 4 z)9 =
5!
(- 1)3 +
5!
(- 1)1 = - 10 - 20 = - 30
9! 2 ! 3! 0! 3!1!1!
= × (2 x)a1 (- 3 y)a 2 (4 z)a 3
a1 ! a2 ! a3 !
9!
X Example 16. The total number of terms in the
= × 2 a1 (- 3)a 2 4a 3 × xa1 ya 2 za 3
a1 ! a2 ! a3 ! expansion of ( a + b + c + d ) n , n Î N is
\ Coefficient of x3 y4 z2 n( n + 1)( n + 2) n( n + 1)( n + 2) ( n + 3)
(a) (b)
9! 6 6
= × 2 3 (- 3)4 × 42 [taking a1 = 3, a2 = 4, a3 = 2]
3! 4!2 ! ( n + 1)( n + 2) ( n + 3)
= 1260 ´ 10368 = 13063680 (c) (d) None of these
6
X Example 15. The coefficient of x 3 in the Sol. (c) The number of terms in the expansion of
expansion of (1 - x + x 2 ) 5 is (a + b + c + d )n
n + 4 -1 n+ 3 (n + 1) (n + 2 ) (n + 3)
(a) 10 (b) 8 (c) - 50 (d) - 30 = C4 - 1 = C3 =
6
Work Book Exercise 8.1
1 If the second, third and fourth terms in the 7 The coefficient of x 5 in the expansion of
expansion of ( x + y )n are 135, 30 and 10 / 3 (1 + x )21 + (1 + x )22 + ¼ + (1 + x )30 is
respectively, then 51 9
a C5 b C5
a n=7 b n=5 31 21 30 20
c C6 - C6 d C5 + C5
c n=6 d None of these
2 The total number of dissimilar terms in the 8 If the nineth term in the expansion of
expansion of ( x1 + x2 + K + xn )3 is log 25 x - 1 + 7 x -1
3 2
[3 3 + 3-1/ 8 log 3( 5 + 1) 10
]
n + 3n
a n3 b is equal to 180 and x > 1, then x equals
4
n( n + 1) ( n + 2 ) n2( n + 1)2 a log10 15 b log 5 15
c d c loge 15 d None of these
6 4
3 If the coefficients of rth and (r + 1) th term in the 9 The expression
expansion of (3 + 7 x )29 are equal, then r equals ìæ 1 + 4 x + 1 ö 7 æ1 - 4 x + 1 ö 7 ü
1 ï ï
íç ÷÷ - çç ÷÷ ý
a 15 b 21 c 14 d 25 4 x + 1 ï çè 2 ø è 2 ø ï
î þ
4 If (1 + 2 x + 3 x ) = a0 + a1 x + a2 x + ¼ + a20 x 20 ,
2 10 2
is a polynomial in x of degree
then a1 equals
a 7 b 5
a 10 b 20 c 210 d 110
c 4 d 3
5 The number of terms which are free from radical ì æ 1 + 4x + 1ö n æ 1 - 4x + 1ö nü
signs in the expansion of ( y1/ 5 + x1/ 10 )55 is 1 ï ï
10 If íç ÷ -ç ÷ ý
a 5 b 6 c 7 d 4 4x + 1 ï è 2 ø è 2 ø ï
î þ
6
6 The coefficient of x in the expansion of 5
= a0 + a1 x + ¼ + a5 x , then n is equal to
(1 + x 2 - x 3 )8 is a 11 b 9
a 80 b 84 c 88 d 92 c 10 d 8
352
Greatest Coefficient
In the binomial expansion of ( x + a ) n , the
Now, Tr + 1 > = < Tr Þ
Tr + 1
> = <1
8
Binomial Theorem
coefficient n C 0 , n C1 , n C 2 , …, n C n are known as the Tr
binomial coefficients. n - r +1 a
For given value of n, Þ × > = <1
r x
(a) If n is even, then greatest coefficient = n C n / 2
ìæ n + 1ö ü a
(b) If n is odd, then greatest coefficient are Þ íç ÷ - 1ý > = < 1
n n
îè r ø þ x
Cn - 1 Cn + 1 n +1 x n +1 xö
and . æ
Þ -1 > = < Þ > = < ç1 + ÷
2 2 r a r è aø
Ø ● Binomial coefficient of middle term is the greatest binomial n +1 n + 1ö
coefficient. Þ > = < r Þ r > = <æ
x ç x÷
● The greatest coefficient in the expansion of 1+ ç1 + ÷
n! a è aø
(x1 + x 2 + ¼ + x m )n is , where q and r are
m- r
(q !) [(q + 1)!]r Thus, Tr + 1 > = < Tr according as
the quotient and remainder respectively when n is divided æn + 1 ö
by m. r>=< ç ÷ …(i)
ç x ÷
1ç + ÷
X Example 17. Find the value of r for which è aø
200
C r is the greatest. Now, two cases arise.
Sol. Coefficient of terms in the expansion of ( x + y)200 are n +1
200 200 200
Case I When is an integer.
C0, C1, C 2 , …, 200 C 200 × x
1+
Since, middle term has greatest coefficient. a
\ Greatest coefficient = Coefficient of middle term n +1
= Coefficient of 101st term = 200C100 Let = m. Then, from Eq. (i), we have
x
Hence, 200
C r is the greatest, when r = 100 1+
a
X Example 18. The greatest coefficient in the mth and ( m +1)th terms are greatest terms.
expansion of ( x + y + z + w)15 is Case II When
n +1
is not an integer.
15! 15! x
(a) (b) 1+
3
3!( 4!) (3!) 3 4! a
n +1
15! Let m be the integral part of . Then, from
(c) (d) None of these x
2!( 4!) 2 1+
a
Sol. (a) The greatest coefficient is Eq. (i), we have ( m +1)th term is the greatest term.
n!
= [here, n = 15, q = 3, r = 3, k = 4] Above discussion suggests the following algorithm
(q !)k - r [(q + 1)!]r
to find the greatest term in a binomial expansion.
15! 15!
= =
(3!)4 - 3 [(3 + 1)!]3 3!(4!)3
Algorithm
Step I Write Tr + 1 and Tr from the given expansion.
Greatest Term Step II Find
Tr + 1
Let Tr + 1 and Tr be ( r +1)th and rth terms, Tr
respectively in the expansion of ( x + a ) n . Then, Tr + 1
Step III Put >1
Tr
Tr + 1 = n C r x n - r a r and Tr = n C r - 1 x n - r + 1 a r - 1
Step IV Solve the inequality in step III for r to get an
Tr + 1 n
Cr x n - r a r
\ = inequality of the form r < m or r > m.
Tr n
Cr - 1x n - r + 1a r - 1 If m is an integer, then mth and ( m +1)th terms
n! ( r - 1)!( n - r + 1)! a are equal in magnitude and these two are the
= ´ ×
( n - r )! r ! n! x greatest terms. If m is not an integer, then
n - r +1 a obtain the integral part of m, say k. In this
= × case, ( k +1) th term is the greatest term. 353
r x
8 Shortcut Method
To find the greatest term (numerically) in the
X Example 20. The greatest term (numerically)
3
in the expansion of (2 + 3x ) 9 , when x = , is
2
Objective Mathematics Vol. 1
expansion of (1 + x ) n .
| x | ( n + 1) 5 ´ 311 5 ´ 313
(a) Calculate . (a) (b)
| x | +1 2 2
13
7´3
(b) If m is integer, then Tm and Tm + 1 are equal and (c) (d) None of these
both are greatest term. 2
Sol. (c) We have,
(c) If m is not integer, there T[ m] + 1 is the greatest 9 9
3xö 9ö éQ x = 3 ù
term, where [ ] denotes the greatest integral part. (2 + 3 x)9 = 2 9 æç1 + 9æ
÷ = 2 ç1 + ÷ êë
è 2 ø è 4ø 2 úû
n
n n næ yö ½ x(n + 1) ½ éQ - 1 < 0ù
Ø To find greatest term in (x + y) , express (x + y) = x ç1 + ÷ . \ r =½ ½ êë úû
è xø ½ ( x + 1) ½ 3
n
æ yö ½ æ 9 ö(9 + 1) ½
Then, find the greatest term in ç1 + ÷ . ½ çè ÷ø ½ 90 12
è xø =½ 4 ½= =6 ¹ Integer
æ 9
½ ç ÷+1 ½ö 13 13
X Example 19. The greatest term (numerically) ½ è 4ø ½
1 The greatest term in the expansion is T[r ] + 1 = T6 + 1 = T7
in the expansion of (3 - 5x )15 , when x = is Hence, the greatest term = 2 9 × T7
5 6 6
9 9
15
(a) C 3 ´ 3 10 15
(b) C 3 ´ 3 11 = 2 9 × T6 + 1 = 2 9 ×9 C 6 æç ö÷ = 2 9 ×9 C 3 æç ö÷
è 4ø è 4ø
(c) 15
C 3 ´ 312 (d) None of these 9 × 8 × 7 312 7 ´ 313
= 29 × × =
1× 2 × 3 212 2
Sol. (c) Let Tr + 1 and Tr denote the (r + 1th
) and rth terms
respectively. Then,
Tr + 1 = 15C r 315 - r (-5 x)r and Tr = 15
Cr - 13
15 - r + 1
(-5 x)r - 1
R-f Factor Relation
Tr 15
C r 315 - r (-5 x)r
To find the integral and fractional parts of an
+1
Þ = 16 - r irrational number of the form ( a + b c ) n , where a, b, c
Tr 15
Cr - 13 (-5 x)r - 1
Tr 15 - r + 1 æ -5 x ö
and n are natural number, follow the following
+1
Þ = ç ÷ algorithm.
Tr r è 3 ø
Tr 16 - r æ 5 1 ö 1
Þ
Tr
+1
=
r
´ ç - ´ ÷, when x =
è 3 3ø 5
Algorithm
Tr 16 - r 1
Step I Write the given expression equal to I + f ,
+1
Þ = ´ , numerically where I is its integral part and f is the
Tr r 3
fractional part.
[neglecting minus sign]
Now, Step II Define G by replacing ‘+’ sign in the given
Tr +1 expression by ‘-’. Note that G always lies
> 1 (numerically)
Tr between 0 and 1.
16 - r 1 Step III Either add G to the expression in step I or
Þ ´ >1
r 3 subtract G from the expression in step I so that
Þ 16 > 4r RHS is an integer.
Þ r<4 Step IV If G is added to the expression in step I, then
Since, 4 is an integer. Therefore, 4th and 5th terms are
numerically greatest terms.
G + f will always come out to be equal to 1
Now, T4 = T3 + 1 = 15C 3 315 - 3 (-5 x)3 i.e. G = 1 - f . If G is subtracted from the
3
expression in step I, then G will always come
1 1
Þ T4 = C 3 ´ 312 æç -5 ´ ö÷ , when x =
15 out to be equal to f .
è 5ø 5
Step V Obtain the value of the desired expression
Þ T4 = 15C 3 ´ 312 (numerically)
after getting f in terms of G.
and T5 = T4 + 1 = 15
C 4 315 - 4 (-5 x)4
X Example 21. Integral part of (2 + 5 ) 2n + 1 is
4
1 1
= C 4 311 æç -5 ´ ö÷ , when x =
15 (n Î N )
è 5ø 5 (a) an even number
= 15C 4 ´ 311 (numerically) (b) an odd number
15
= C 3 ´ 312 (c) an even or an odd number depending upon the
From above, we see that the values of both greatest value of n
354 terms are equal. (d) None of the above
Sol. (a) Here, " n Î N, (2 + 5 )2 n + 1 Ï N
\ We denote (2 + 5 ) 2n + 1
by I + f
where I is an integer and f Î R such that 0 < f < 1
e.g. (i) (1 + 7) 51 - 1 = 8 51 - 1 is M(7) i.e.
multiple of 7 8
Binomial Theorem
Q 0 < ( 5 - 2) < 1 (ii) (1 + 7) 51 - 1-51 ´ 7 = 8 51 - 358 is M ( 7 2 )
\ We denote ( 5 - 2 )2 n + 1 by G i.e. multiple of 49.
51 ´ 50
where G Î R such that 0 < G < 1 (iii) (1 + 7) 51 - 1 - 51 ´ 7 - ´ 72
Now, I + f = ( 5 + 2 )2 n + 1 2
= ( 5 )2 n + 1 + 2 n + 1C1( 5 )2 n × 2 = 8 51 - 358 - 62475
2n + 1
+ C 2 ( 5 )2 n - 1 × 2 2 + ¼ …(i) = 8 51 - 62833 is M ( 7 3 ) i.e. multiple of 343.
2n + 1
G = ( 5 - 2) = ( 5 )2 n + 1 - 2n + 1
C1( 5 )2 n × 2
2n + 1
+ C 2 ( 5 )2 n - 1 × 2 2 - ¼ …(ii) X Example 23. Which of the following
On subtracting Eq. (ii) from Eq. (i), we get expression is divisible by 1225?
I + f - G = 2 [2 n + 1C1(5)n × 2 + 2 n + 1C 3 (5)n - 1 × 2 3 + ¼ ] (a) 6 2n - 35n - 1 (b) 6 2n - 35n + 1
= 2k, where k is an integer.
(c) 6 2n - 35n (d) 6 2n - 35n + 2
QI is an integer.
\ f - G = 2 k - I is an integer. …(iii) Sol. (a) Consider expression = 62n - 35n - 1
0< f <1
Now, 0 < f < 1ü = (62 )n - 35n - 1 = 36n - 35n - 1
ý Þ -1 < -G < 0 …(iv)
and 0 < G < 1þ -1 < f - g < 1 So, 62n - 35n - 1 = (1 + 35)n - 35n - 1
From Eqs. (iii) and (iv), f - G = 0 = 1 + 35n + nC 2 × 352 + ¼ + 35n - 35n - 1
Now, from Eq. (iii), I = 2 k - (f - G ) = 2 k - 0 = 2 k = 352 [n C 2 + nC 3 × 35 + ¼ + 35n - 2 ]
= 1225 ´ a positive integer, if n ³ 2
X Example 22. The integral part of ( 2 + 1) 6 is
If n = 1, then 62n - 35n - 1 = 0, which is divisible by 1225.
(a) 198 (b) 196
(c) 197 (d) 199 So, options (b), (c) and (d) are not divisible by 1225.
Sol. (c) Let ( 2 + 1)6 = I + f, where I Î N and 0 < f < 1. X Example 24. If 7103 is divided by 25, then the
Clearly, I is the greatest integer less than or equal to remainder is
( 2 + 1)6 . (a) 20 (b) 16 (c) 18 (d) 15
Let G = ( 2 - 1)6 . Then,
Sol. (c) We have, 7103 = 7(49)51 = 7(50 - 1)51
I + f + G = ( 2 + 1)6 + ( 2 - 1)6
= 7{5051 - 51
C1 5050 + 51
C 2 5049 - ¼ - 1}
Þ I + f + G = 2{6 C 0 ( 2 )6 + 6C 2 ( 2 )6 - 2 + ¼}
Þ I + f + G = An even integer = 7{(5051 - 51
C1 5050 + 51
C 2 5049 - ¼ ) - (7 + 18 - 18)}
Þ f + G =1 = k + 18 (say) [Qk is divisible by 25]
Again, I + f + G
\ Remainder is 18.
= 2{6 C 0 ( 2 )6 + 6C 2 ( 2 )6 - 2 + 6C 4 ( 2 )6 - 4 + 6 C 6 ( 2 )6 - 6 }
Þ I + 1 = 2(8 + 15 ´ 4 + 15 ´ 2 + 1) [Q f + G = 1] 2 4n
X Example 25. The fractional part of is
Þ I = 197 15
1 2
(a) (b)
Divisibility Problems 15 15
4
From the expansion, (c) (d) None of these
15
(1 + a ) n = 1 + n C1a + n C 2a 2 +¼ + n C n a n
4n
16n (1 + 15)n
We can conclude that Sol. (a) 2 = =
15 15 15
i. (1 + a ) n - 1 = n C1a + n C 2a 2 +¼ + n C n a n is 1 + nC115 + nC 2 152 + ¼ + nC n 15n
=
divisible by a i.e. it is a multiple of a. 15
1 + 15k 1
= , where k Î N = + k
In short, we can write that (1 + a ) n - 1 = M (a ) 15 15
ì2 4n ü ì 1 ü 1
ii. (1 + a ) n - 1 - na = n C 2a 2 + n C 3a 3 \ í ý = í + ký =
î 15 þ î 15 þ 15
+¼ + n C n a n = M (a 2 )
X Example 26. Larger of 99 50 + 100 50 and 10150
n n( n - 1) 2
iii. (1 + a ) - 1 - na - a is
2
(a)10150 (b) 99 50 + 100 50
= n C 3a 3 + n C 4a 4 +¼ + n C n a n = M (a 3 ) 355
(c) both are equal (d) None of these
8 Sol. (a) We have, 10150 = (100 + 1)50
= 10050 + 50 × 10049 +
50 × 49
1× 2
× 10048 + ¼ …(i)
X Example 28. The greatest integer which
divides the number 101100 - 1 is
Objective Mathematics Vol. 1
and 9950 = (100 - 1)50
(a) 100
50 × 49
= 10050 - 50 × 10049 + × 10048 - ¼ …(ii) (b) 1000
1× 2
On subtracting Eq. (ii) from Eq. (i), we get (c) 10000
50 × 49 × 48 (d) 100000
10150 - 9950 = 2 éê 50 × 10049 + ´ 10047 + ¼ùú
ë 1× 2 × 3 û
50 × 49 × 48
Sol. (c) By binomial theorem,
50
= 100 + 2 × × 100 + ¼> 10050
47
n(n - 1) 2
1× 2 × 3 (1 + x)n = éê1 + nx + × x + ¼ + xn ùú
ë 2 û
Þ 10150 > 9950 + 10050
n( n - 1)
Þ (1 + x)n - 1 = nx + x2 + ¼ + xn
X Example 27. If 9 7 + 7 9 is divisible by 2
n(n - 1) 2
(a) 6 (b) 24 (c) 64 (d) 72 If x = n, (1 + n)n - 1 = n2 + n + ¼ + nn
2
Sol. (c) We have, 97 + 7 9 = (1 + 8)7 - (1 - 8)9 n(n - 1)
(1 + n)n - 1 = n2 éê1 + + ¼ + nn - 2 ùú
= (1 + 7C1 × 81 + 7C 2 × 82 + ¼ + 7C 7 × 87 ) ë 2 û
- (1 - 9C1 × 81 + 9C 2 × 82 - ¼ - 9C 9 × 89 ) Put n = 100,
= 16 ´ 8 + 64[( 7C 2 + ¼ + 7C 7 × 85 ) 100(100 - 1)
9 9 7
(1 + 100)100 -1 = (100)2 éê1 + +¼ + 10098 ùú
- ( C 2 - ¼ - C 9 × 8 )] ë 2 û
= 64 (an integer) Clearly, (101)100 - 1 is divisible by (100)2 = 10000
Work Book Exercise 8.2
1 If n is even, then the greatest coefficient in the 6 If R = (6 6 + 14)2 n + 1 and f = R - [R] , where [×]
n
expansion of ( x + a) is denotes the greatest integer function, then Rf
n n
a Cn b Cn equals
+1 -1
2 2 a 20 n b 20 2n
n
c Cn d None of these c 20 2n + 1 d None of these
2
2 If n is even positive integer, then the condition 7 If R = ( 2 + 1)2 n + 1 and f = R - [R], where [ ]
that the greatest term in the expansion of (1 + x )n denotes the greatest integer function, then [R]
equals
may have the greatest coefficient also is
1
n n+2 a f +
a < x< f
n+2 n 1
n+1 n b f -
b < x< f
n n+1 1
c -f
n n+ 4 f
c < x<
n+ 4 4 d None of the above
d None of the above
8 If (5 + 2 6 )n = I + f, where I Î N, n Î N and
n
3 If n > 1, then (1 + x ) - nx - 1 is divisible by 0 £ f < 1, then I equals
a x5 b x2 c x3 1 1
a -f b -f
d x4 f 1+ f
1 1
4 The positive integer just greater than c -f d + f
1- f 1- f
(1 + 0.0001)10000 is
a 3 9 If R = (7 + 4 3 )2 n = I + f , where I Î N and
b 4 0 < f < 1, then R(1 - f ) equals
c 5
1
d None of the above a (7 - 4 3 )2n b
(7 + 4 3 )2n
5 If [ x ] denotes the greatest integer less than or c 1 d None of these
equal to x, then [(1 + 0.0001)10000 ] equals
10 The digit at unit’s place in the number
a 3
b 2
171995 + 111995 - 71995 is
c 0 a 0 b 1
d None of the above c 2 d 3
356
Properties of the Binomial Coefficients
X Example 30. If the sum of the coefficients in
8
Binomial Theorem
i. In the expansion of (1 + x ) n the coefficients of
the expansion of ( a 2 x 2 - 2ax + 1) 51 vanishes, then
terms equidistant from the beginning and the the value of a is
end are equal. (a) 2 (b) -1 (c) 1 (d) -2
Proof
Sol. (c) The sum of the coefficients of the polynomial
We have, (a2 x2 - 2 ax + 1)51 is obtained by putting x = 1in
(1 + x ) n = C 0 + C1 x + C 2 x 2 +¼ + C r x r (a2 x2 - 2 ax + 1)51.
n -1 n
+¼ + C n - 1 x + Cn x By hypothesis, (a2 - 2 a + 1)51 = 0
In this expansion the coefficient of ( r +1)th Þ a=1
term from the beginning is n C r . The ( r +1)th iii. The sum of the coefficients of the odd terms in
term from the end is ( n - r +1)th term from the the expansion of (1 + x ) n is equal to the sum of
beginning. Therefore, its coefficients is the coefficients of the even terms and each is
n
C n - r . But, we have n C r = n C n - r . equal to 2 n - 1 .
Hence, the coefficients of terms equidistant i.e.
from the beginning and the end are equal. C 0 + C 2 + C 4 +¼ = C1 + C 3 + C 5¼ = 2 n - 1
ii. The sum of the binomial coefficients in the Proof
expansion of (1 + x ) n is 2 n . We have, (1 + x ) n = C 0 + C1 x + C 2 x 2 + C 3 x 3
i.e. C 0 + C1 + C 2 +¼ + C n = 2 n +¼ + C n - 1 x n - 1 + C n x n
n Putting x =1 and -1 respectively in the above
or å n Cr = 2n expansion, we get
r=0 n
2 = C 0 + C1 + C 2 + C 3 +¼ + C n - 1 + C n
Proof
We have, (1 + x ) n = C 0 + C1 x + C 2 x 2 + C 3 x 3 and 0 = C 0 - C1 + C 2 - C 3 +¼ + ( -1) n - 1 C n - 1
+¼ + C n - 1 x n - 1 + C n x n …(i) + ( -1) n C n
Putting x =1 in this expansion, we get On adding and subtracting these two
expressions, we get
2 n = C 0 + C1 + C 2 + C 3 +¼ + C n - 1 + C n
2 n = 2(C 0 + C 2 + C 4 +¼ )
Ø ● It follows from the above result that and 2 n = 2(C1 + C 3 + C 5 +¼ )
n n
●
å n Cr = 2n , å n - 1Cr - 1 = 2n - 1 etc. Þ C 0 + C 2 + C 4 +¼ = 2 n -1 = C1 + C 3 + C 5 +¼
r= 0 r=1
n n n -1 n - 2
● By putting x = - 1in Eq. (i), we get iv. n
Cr = ×n - 1 Cr - 1 = × × C r - 2 and
n
n
r r r -1
å(-1) Cr = 0 i.e. C0 - C1 + C2 - C3 + ¼ = 0 so on.
r= 0
● Sum of coefficients in the expansion of (a + bx + cx 2)n is Proof
(a + b + c)n . n n!
We have, Cr =
X Example 29. The sum of the last ten ( n - r )! r !
coefficients in the expansion of (1 + x )19 when n( n - 1)!
expanded in ascending powers of x, is =
r ( r - 1)!{( n - 1) - ( r - 1)}!
(a) 218 (b) 219
n
(c) 218 - 19C10 (d) None of these = ×n - 1 Cr - 1
r
Sol. (a) The required sum n -1 n - 2
19 19 19 Similarly, we have n - 1 C r - 1 = × Cr - 2
= C10 + C11 + K + C19 ...(i) r -1
19 19 19 19
= C0 + C1 + C2 + K + C9 ...(ii)
n
[Q nC r = nC n - r ] \ n C r = × n - 1C r - 1
On adding Eqs. (i) and (ii), we get
r
2 ´ (Required sum) = 19C 0 + 19C1 + K + 19
C19 = 219 n n -1 n - 2
= × × Cr - 2
\ Required sum = 2 18 r r -1 357
8 X Example 31.
1
+
1
+
1
n! 2!( n - 2)! 4!( n - 4)!
+K is X Example 33. If
(1 + x ) n = C 0 + C1 x + C 2 x 2 + K + C n x n , then
Objective Mathematics Vol. 1
equal to
2n - 1 2n å< å (C i + C j ) 2 is equal to
(a) (b) 0£ i j £ n
n! ( n + 1)!
(a) ( n - 1) × 2n C n + 2 2n
n n-2
2 2
(c) (d) (b) n × 2n C n + 2 2n
n! ( n - 1)!
(c) ( n + 1) × 2n C n + 2 2n
Sol. (a) Given expression (d) None of the above
1 n 2n - 1
= [ C 0 + nC 2 + nC 4 + ...] = Sol. (a) 2
n! n! å å(C i + Cj)
0 £ i < j£ n
X Example 32. The sum = n(C 02 + C12 + K + C n2 ) + 2 å åC i C j
0 £ i < j£ n
1 × 20C1 - 2 × 20C 2 + 3 × 20C 3 - K - 20 × 20C 20 is = n × 2 nC n + [(C 0 + C1 + K + C n )2 - (C 02 + C12 + K + C n2 )]
equal to = n × 2 nC n + (2 n )2 - 2n
Cn
19
(a) 2 (b) 0 = (n - 1) × 2n
Cn + 2 2n
(c) 2 20 - 1 (d) None of these
2n
vii. C 0C1 + C1C 2 + C 2C 3 +¼+ C n - 1C n = Cn + 1
Sol. (b) Using r × nC r = n × n - 1C r - 1
20 20 Proof On putting r =1 in Eq. (iii), we get
r -1 r -1
\ å(- 1) r ×20 C r = å(- 1) × 20 × 19C r -1 2n
r =1 r =1
C 0C1 + C1C 2 + C 2C 3 +¼+ C n - 1C n = Cn + 1
19 19 19 19
= 20{ C 0 - C1 + C2 - K - C19 } (2n)!
= 20 ´ 0 = 0
=
( n + 1)!( n - 1)!
2n
v. C 0C r + C1C r + 1 +¼+ C n - r C n = Cn + r
viii. C 0C 2 + C1C 3 + C 2C 4 +¼+ C n - 2C n = 2n
Cn + 2
(2n)!
= Proof On putting r = 2 in Eq. (iii), we get
( n + r )!( n - r )! 2n
C 0C 2 + C1C 3 + C 2C 4 +¼+ C n - 2C n = Cn + 2
1
Proof Replacing x by in Eq. (i), we get (2n)!
x =
n
C1 C 2 C ( n + 2)!( n - 2)!
æ 1ö
ç1 + ÷ = C 0 + + +¼ + n …(ii)
è xø x x 2
xn ix. C 0 + 2C1 + 3C 2 +¼ + ( n + 1)C n = 2 n - 1 ( n + 2)
On multiplying Eqs. (i) and (ii), we get Proof Consider,
(1 + x ) 2n (1 + x ) n = C 0 + C1 x + C 2 x 2 +¼ + C n x n
= (C 0 + C1 x + C 2 x 2 +¼)
n
x Now, on multiplying both sides by x, we get
æ C1 C 2 ö
çC 0 + +
2
+¼÷ x (1 + x ) n = C 0 x + C1 x 2 + C 2 x 3 +¼ + C n x n + 1
è x x ø
On differentiating both sides w. r. t. x, we get
Now, comparing coefficient of x r on both
(1 + x ) n + nx (1 + x ) n - 1 = C 0 + 2C1 x + 3C 2 x 2
sides, we get
C 0C r + C1C r + 1 +¼ + C n - r C n = 2n C n + r +¼ + ( n + 1)C n x n
(2n)! Now, putting x =1, we get
= …(iii)
( n + r )!( n - r )! 2 n + n(2) n - 1 = C 0 + 2C1 + 3C 2 +¼+ ( n + 1)C n
vi. Sum of squares of coefficients is 2n
C n i.e. Þ C 0 + 2C1 + 3C 2 +¼+ ( n + 1)C n
= 2 n - 1 ( n + 2)
2 2 2 2 (2n)!
C 0 + C1 + C 2 +¼ + C n =
n! n! C1 C 2 C 2n + 1 - 1
x. (i) C 0 + + +¼+ n =
Proof On putting r = 0 in Eq. (iii), we get 2 3 n +1 n +1
(2n)! C1 C 2 C 3 ( -1) n C n 1
C 02 + C12 + C 22 +¼ + C n2 = (ii) C 0 - + - +¼+ =
n! n! 2 3 4 n +1 n +1
358
Proof (i) Integrating expansion of (1 + x ) n
w.r.t. x, between the limits 0 to 1, we get
X Example 35. If n is a positive integer and
n
n
3 æ Ck ö
C k = C k , then the value of å k çç
2 8
Binomial Theorem
1 ÷÷ is
é (1 + x ) n + 1 ù é x2 x3
ê ú = C
ê 0 x + C 1 + C 2 k =1 è Ck - 1 ø
ë n +1 û 0 ë 2 3
n( n + 2)( n + 1) 2 n( n + 1)( n + 2) 2
1 (a) (b)
xn +1 ù 12 12
+¼+ C n ú n( n + 1)( n + 2)
n +1û 0 (c) (d) None of these
n +1 12
2 -1 C C C
Þ = C 0 + 1 + 2 +¼ + n æ n ö n
2
2
n +1 2 3 n +1 Sol. (a) å k ç C k ÷ = å k 3 æç n - k + 1ö÷
3
çC ÷
k =1 è k -1ø k =1 è k ø
(ii) Integrating expansion of (1 + x ) n between
é n
C n - k + 1ù
limits -1 to 0, we get êQ n k = ú
0 êë C k - 1 k úû
é (1 + x ) n + 1 ù é x2 x3 n n
ê ú = êC 0 x + C1 + C2 = å k(n - k + 1)
2
= å k [(n + 1)
2
- 2 k(n + 1) + k 2 ]
ë n + 1 û -1 ë 2 3 k =1 k =1
n n n
0 2 2 3
C xn +1 ù = (n + 1) å k - 2(n + 1) å k + åk
+¼+ n ú k =1 k =1 k =1
n + 1 úû n(n + 1) n(n + 1)(2 n + 1) n2 (n + 1)2
-1 = (n + 1)2 × - 2(n + 1) × +
2 6 4
1 C C C ( -1) n C n n(n + 1)2
Þ = C 0 - 1 + 2 - 3 +¼+ = [6(n + 1) - 4(2 n + 1) + 3n]
n +1 2 3 4 n +1 12
2
n(n + 1) n(n + 2 )(n + 1)2
= × (n + 2 ) =
12 12
Some Other Important Properties
X Example 36. If
xi. n C r = n C s , then either r = s or r + s = n
n
(1 + x ) n = C 0 + C1 x + C 2 x 2 + K + C n x n , then
Cr n - r +1
xii. n
= (C 0 + C1 )(C1 + C 2 ) K (C n - 1 + C n )
Cr - 1 r = k × C1C 2C 3 K C n , where k is equal to
n +1
( n +1) n nn
xiii. n
Cr + nCr - 1 = Cr (a) (b)
n! ( n -1)!
n
xiv. C 02 - C12 + C 22 - C 32 +... (c)
( n + 1)
(d) None of these
ì0, if n is odd ( n - 1)!
=í n/2 n n+1 n+1
î(-1) C n / 2 , if n is even Sol. (a) We have, C 0 + C1 = C1,C1 + C 2 = C 2 , ... ,
n+1
Cn - 1 + Cn = Cn
X Example 34. The value of \ Given expression = n + 1C1n + 1C 2 K n + 1C n ...(i)
3 5
(n + 1)! (n + 1)n!
47
C4 + å 50 - j C 3 + å 56 - k C 53 - k is Now, n + 1C r = =
r !(n - r + 1)! r !(n - r + 1)(n - r )!
j=0 k =0
n+1 n+1 n
57 57 57 57 or Cr = Cr ...(ii)
(a) C4 (b) C3 (c) C5 (d) C6 n-r+1
3 5 Putting n = 1, 2, 3, ..., n in Eq. (ii), we get
50 - j 56 - k
Sol. (a) We have, 47C 4 + å C3 + å C 53 - k (C 0 + C1 )(C1 + C 2 ) K (C n - 1 + C n )
j=0 k =0
n+1 n+1 n+1 (n + 1)n
= 47
C 4 + ( 50 C 3 + 49
C3 + 48
C3 + 47
C3 ) + = × C1 × C2 K Cn = C1C 2C 3 K C n
n n-1 1 n!
( 56 C 53 + 55
C 52 + 54
C 51 + 53
C 50 + 52
C 49 + 51
C 48 )
= 47 47
C4 + ( C3 + 48
C3 + 49
C3 + 50
C3 )
X Example 37. The value of
15
+ ( 56 C 3 + 55
C3 + 54
C3 + 53
C3 + 52
C3 + 51
C3 ) C 02
- 15
C12 + C 22 -K - 15C15
15 2
is
[Q n C r = nC n - r ] (a) 15 (b) -15 (c) 0 (d) 51
47 47 48 49 50
= ( C4 + C3 ) + ( C3 + C3 + C3 )
Sol. (c) As we know that,
+ ( 51C 3 + 52
C3 + ¼ + 56
C3 ) C 02 - C12 + C 22 - C 32 + K + (- 1)n C n2 = 0 (if n is odd)
n n n+1
[Q C r + C r -1 = Cr ] Here, n = 15 (odd)
= ( C4 +48 48
C3 ) + 49
C3 + 50
C3 + ¼ + 56
C3 = 57
C4 \ C 02 - C12 + C 22 - K - C15
2
=0 359
8 X Example 38. If
(1 + x - 2x 2 ) 6 = 1 + a1 x + a 2 x 2 + a 3 x 3 +¼,
X Example 40. If
A = 2n C 0 × 2n C1 + 2n C1 × 2n -1 C1 + 2n
C 2 × 2n - 2 C1
Objective Mathematics Vol. 1
then the value of a 2 + a 4 + a 6 + K + a12 will be +K , then A is
n 2n
(a) 32 (b) 31 (c) 64 (d) 1024 (a) 0 (b) 2 (c) n × 2 (d) 1
2 6 2
Sol. (b) We have, (1 + x - 2 x ) = 1 + a1 x + a2 x Sol. (c) A = Coefficient of x in
Putting x = 1, we get [2 n C 0 (1 + x)2 n + 2n
C1(1 + x)2 n -1 + ...]
0 = 1 + a1 + a2 + a3 + .... + a12 …(i) = Coefficient of x in ([1 + (1 + x)]2 n
Putting x = - 1, we get
= Coefficient of x in (2 + x)2 n
64 = 1 - a1 + a2 - a3 + .... + a12 ...(ii)
2n
x
On adding Eqs. (i) and (ii), we get = Coefficient of x in 2 2 n æç1 + ö÷ = n × 2 2n
64 = 2 + 2 a2 + 2 a4 + 2 a6 + .... + 2 a12 è 2ø
Þ a2 + a4 + a6 + ....+ a12 = 31 n
X Example 39. The sum
X Example 41. å n C r sin rx cos (n - r ) x is equal
r=0
20
C 0 + 20C1 + 20C 2 + ....+ 20C10 is equal to to
(a) 2 20 +
20! (a) 2 n - 1 sin ( n - 1) x (b) 2 n sin nx
(10!) 2 (c) 2 n -1 sin nx (d) None of these
1 20!
(b) 219 - × Sol. (c) We have,
n
n
2 (10!) 2 å C r sin r x cos (n - r )x
r =0
(c) 219 + 20C10 1 n
= [( C 0 sin 0 xcos nx + nC n sin nxcos 0 x)
(d) None of the above 2
20 20 20 20 + ( nC1 sin x cos (n - 1)x + n C n -1 sin (n - 1)x × cos x)
Sol. (d) Let S = C0 + C1 + C2 + K + C10 ...(i)
20 20 20 20 +¼+ ( n C n sin nx cos 0 x + nC 0 sin 0 x cos nx)]
S= C 20 + C19 + C18 + K + C10
1 n n
= [ C 0 sin nx + C1 sin nx +¼+ nC n sin nx]
[Q nC r = nC n - r ]…(ii) 2
On adding Eqs. (i) and (ii), we get
1 n n 2 n sin nx
2S = ( 20 C 0 + 20C1 + K + 20C 20 ) + 20
C10 = [ C0 + C1 +¼+ nC n ] sin nx =
2 2
= 2 20 + 20
C10 n
n
20
C10 \ å C r sin rx cos (n - r ) x = 2 n -1 sin nx
19
\ S =2 + r =0
2
Work Book Exercise 8.3
n
1 If the sum of the coefficients in the expansion of (-1)r
n
(a + b) is 4096, then the greatest coefficient in
4 If n is an odd natural number, then å n
equals
r = 0 Cr
the expansion is
1
a 924 b 792 a 0 b
n
c 1594 d None of these n
c d None of these
2 If the sum of the coefficients in the expansion of 2n
n
(1 + 2 x )n is 6561, the greatest term in the (-1)r
1
5 If n is an even natural number, then å n
expansion for x = , is r = 0 Cr
2 equals
a 4th b 5th 1
c 6th d None of these a 0 b
n
( -1)n/ 2
3. If the sum of the coefficients in the expansion of c n
d None of these
C n/ 2
n
(1 + 2 x ) is 6561, then the greatest coefficient in
n n
the expansion is 1 r
6 If an = å nC , then å nC equals
a 896 r =0 r r =0 r
b 3594
a ( n - 1) an b n an
c 1792 n
d None of the above c an d None of these
2
360
7 The value of 1× C1 + 3 × C3 + 5 × C5 + 7 × C7 + ¼,
where C0 , C1, C2 , ¼, Cn are the binomial
coefficients in the expansion of (1 + x )n , is
13 P is a set containing n elements. A subset A of P
is chosen and the set P is reconstructed by
replacing the elements of A. A subset B of P is
8
Binomial Theorem
a n×2n - 1 b n×2n - 2 chosen again. The number of ways of choosing A
c n( n - 1) 2 n - 1 d None of these and B such that A = B, is
a 2n b 3n
8 If Cr is the coefficient of x r in (1 + x )n , then the c 2nC n d None of these
n
value of å (r + 1)2Cr is 14 P is a set containing n elements. A subset A of P
r =0
is chosen and the set P is reconstructed by
a ( n + 1)( n + 4)2 n - 2 b ( n + 1)( n + 4)2 n - 1
replacing the elements of A. A subset B of P is
c ( n + 1)22 n - 2 d None of these chosen again. The number of ways of choosing A
and B such that B is a subset of A, is
9 If Cr stands for n Cr , then the sum of the series
a 2n b 3n
æ nö æ nö c 2 nC n d None of these
2ç ÷!ç ÷!
è2ø è2ø
[C02 - 2C12 + ¼ + (-1)n (n + 1) Cn2 ] ,
n! 15 P is a set containing n elements. A subset A of P
is chosen and the set P is reconstructed by
where n is an even positive integer, is equal to
replacing the elements of A. A subset B of P is
a 0 b ( -1)n/ 2( n + 1)
chosen again. The number of ways of choosing A
c ( -1)n/ 2( n + 2 ) d ( -1)n n and B such that B contains just one element
more than A, is
10 In the expansion of (1 + x )n (1 + y )n (1 + z )n , the 2n
a Cn - 1 b 3n c (2 n )2 d 2n
Cn
sum of the coefficients of the terms of degree r is
a ( nC r )3 b 3 × nC r 16 P is a set containing n elements. A subset A of P
c 3 nC r d nC 3 r is chosen and the set P is reconstructed by
replacing the elements of A. A subset B of P is
11 There are two bags each of which contains n chosen again. The number of ways of choosing A
balls. A man has to select an equal number of and B such that A and B have equal number of
balls from both the bags. The number of ways in elements, is
which a man can choose atleast one ball from
a 2n b 3n c (2 n )2 d 2n
Cn
each bag is
a 2n
Cn b ( nC n )2 17 If n > 3, then
2n
c C1 d 2nC n - 1 xyC0 - ( x - 1)( y - 1) C1 + ( x - 2 )( y - 2 ) C2
- ( x - 3)( y - 3) C3 +¼+ (-1)n ( x - n )( y - n ) Cn
12 P is a set containing n elements. A subset A of P
is chosen and the set P is reconstructed by equals
replacing the elements of A. A subset B of P is a xy ´ 2 n b n xy
chosen again. The number of ways of choosing A c xy d None of these
and B such that A and B have no common 18 If n > 3, then xyzC0 - ( x - 1)( y - 1)( z - 1) C1
elements is
+ ( x - 2 )( y - 2 )( z - 2 ) C2 - ( x - 3)( y - 3)( z - 3) C3
a 2n
b 3n + ¼ + (-1)n ( x - n )( y - n )( z - n ) Cn equals
c 4n a xyz b 0
d None of the above c - xyz d None of these
Binomial Theorem for any Index
Statement (Binomial theorem for any index) ● If first term is not 1, then make first term unity in the
following ways,
Let n be a rational number and x be a real number such n
æ xö x
that | x | <1, then (x + a)n = an ç1 + ÷ , if½ ½< 1.
è a ø ½ a½
n( n - 1) 2 n( n - 1)( n - 2) 3
(1 + x ) n = 1 + nx + x + x ● Expansion of (x + a)n for any index
2! 3! a
n( n - 1)( n - 2)¼ ( n - r + 1) r Case I When | x | >| a| i.e. <1
+¼ + x +¼ x
r! n n
ì æ a öü æ aö
Ø ● The condition | x | < 1is unnecessary, if n is a whole number In this case, (x + a)n = í x ç1 + ÷ý = x n ç1 + ÷
while the same condition is essential if n is a rational number î è x øþ è xø
other than a whole number. ì
ï
2 3
a n(n - 1) æ a ö n(n - 1)(n - 2) æ a ö üï üï
= x n í1 + n × + ç ÷ + ç ÷ ý + ¼ý
● There are infinite number of terms in the expansion of x 2! è x ø 3! èxø ï
(1 + x)n , when n is a negative integer or a fraction. îï þ ïþ
361
8 Case II When | x | <| a| i.e.
a
x
< 1..
n n
iv. Putting n =1, 2, 3 in the expansion of (1 + x ) - n ,
we get
Objective Mathematics Vol. 1
ì æ x öü æ xö
In this case, (x + a)n = í a ç1 + ÷ý = an ç1 + ÷ (1+ x ) -1 = 1 - x + x 2 - x 3 + x 4 -¼+ ( -1) r x r +¼
î è ø
a þ è aø
(1 + x ) -2 = 1 - 2x + 3x 2 - 4x 3 + 5x 4 -¼
2 3
ïì x n(n - 1) æ x ö n(n - 1)(n - 2) æ x ö ïü
+ ( -1) r ( r + 1) x r +¼
= an í1 + n × + ç ÷ + ç ÷ ¼ý
ïî a 2! è aø 3! è aø ï
þ (1 + x ) -3 = 1 - 3x + 6x 2 - 10x 3
● If n is a positive integer, the expansion of (1 + x)n contains ( r + 1)( r + 2) r
(n + 1) terms and coincides with +¼ + ( -1) r x +¼
2
(1 + x)n = nC 0 + nC 1x + nC 2x 2 + ¼ + nC n x n ,
Replacing x by -x in the above expansions,
because n C 0 = 1 , n C 1 = n , we get
n n(n - 1)
C2 = , (1 - x ) -1 = 1 + x + x 2 + x 3 +¼ + x r +¼
2!
n(n - 1)(n - 2) (1 - x ) -2 = 1 + 2x + 3x 2 +¼ + ( r + 1) x r +¼
n
C3 = ,…
3! ( r +1)( r +2) r
(1- x ) -3 = 1 + 3x + 6x 2 +¼ + x +¼
● The general term in the expansion of (1 + x)n is given by 2
n(n - 1)(n - 2)¼{n - (r - 1)} r In all these expansion, it is assumed that | x | <1.
Tr + 1 = x
r! Students are advised to learn these expansions.
1
Some Important Deductions X Example 42. The expansion of ,
( 4 - 3x ) 1/ 2
i. Replacing n by - n in the expansion for binomial theorem will be valid, if
(1 + x ) n , we get (a) x <1 (b) | x| <1
n( n+1) 2 n( n + 1)( n + 2) 3 2 2
(1+ x ) - n = 1 - nx + x - x (c) - <x< (d) None of these
2! 3! 3 3
n( n + 1)( n + 2)¼ ( n + r - 1) r
+¼+( -1) r x +¼ Sol. (d) The given expression can be written as
r! - 1/ 2
3
The general term in this expansion is 4-1/ 2 æç1 - xö÷ and it is valid only when
è 4 ø
n( n + 1)( n + 2)¼ ( n + r - 1) r 3 4 4
Tr + 1 = ( -1) r x x <1 Þ - < x<
r! 4 3 3
ii. Replacing x by -x and n by -n in the expansion 1
X Example 43. Evaluate .
of (1 + x ) n , we get 3 6 - 3x
n( n+1) 2 n( n+1)( n + 2) 3 é x 2x 2 ù é x 2x 2 ù
(1- x ) - n = 1+ nx + x + x (a) 61/ 3 ê1 + + + Kú (b) 6 -1/ 3 ê1 + + + Kú
2! 3! 2 2
ë 6 6 û ë 6 6 û
n(n + 1)( n + 2)¼ ( n + r - 1) r
+¼ + x +¼ é x 2x 2 ù é x 2x 2 ù
r! (c) 61/ 3 ê1 - + - Kú (d) 6 -1/ 3 ê1 - + - Kú
2 2
The general term in this expansion is ë 6 6 û ë 6 6 û
n( n + 1)( n + 2)¼ ( n + r - 1) r 1 x
- 1/ 3
Tr + 1 = x Sol. (b) = (6 - 3 x)-1/ 3 = 6-1/ 3 æç1 - ö÷
r! ( 6 - 3 x)1/ 3 è 2ø
é 1
æ- ö æ- ö 4 ù
iii. Replacing x by -x in the expansion of (1 + x ) n , -1/ 3
ê æ
ç ÷ç
1ö æ x ö è 3ø è 3ø æ x ö
÷ 2 ú
=6 ê1 + ç - ÷ ç - ÷ + ç - ÷ + Kú
we get ê è 3ø è 2 ø 2 ×1 è 2ø ú
n( n - 1) 2 n( n - 1)( n - 2) 3 ëê úû
(1 - x ) n = 1 - nx + x - x é x 2 x 2 ù
2! 3! = 6-1/ 3 ê1 + + 2 + Kú
n( n-1)( n-2)¼ ( n - r + 1) r ë 6 6 û
+¼ +( -1) r x +¼
r! 1
X Example 44. If = 1 + a1 x + a 2 x 2 +K,
The general term is 1 - 2x + x 2
n( n - 1)( n - 2)¼ ( n - r + 1) r then the value of a r is
Tr + 1 = ( -1) r x
362 r! (a) 2r (b) r +1 (c) r (d) r -1
Sol. (b) 1
1 - 2 x + x2
= (1 - x)-2 = 1 + 2 x + 3 x2 + K
= 1 + a1 x + a2 x2 + K Þ ar = r + 1
Thus, if x is so small that its squares and higher
powers may be neglected, then (1 + x ) n = 1 + nx,
approximately. This is an approximate value of
8
Binomial Theorem
(1 + x ) n .
X Example 45. The coefficient of x n in the
1 X Example 47. If x is so small such that its
expansion of is
(1 - x )(1 - 2x )(1 - 3x ) square and higher powers may be neglected, then
1 1 (8 + 3x ) 2/ 3
(a) (2 n + 2 - 3 n + 3 + 1) (b) (3 n + 2 - 2 n + 3 + 1) the value of is
2 2 (2 + 3x )( 4 - 5x )1/ 2
1 n+3 n+2 3 5
(c) (2 -3 + 1) (d) None of these (a)1 - x (b)1 + x
2 2 8
1 5
Sol. (b) We have, (c)1 - x (d) None of these
(1 - x)(1 - 2 x)(1 - 3 x) 8
1 4 9
= - + (8 + 3 x)2 / 3
2(1 - x) 1 - 2 x 2(1 - 3 x) Sol. (c) We have,
(2 + 3 x) (4 - 5 x)
[by resolving into partial fractions] 2/ 3
3xö
1 9
= (1 - x) - 4(1 - 2 x)-1 + (1 - 3 x)-1
-1 82 / 3 æç1 + ÷
è 8ø
2 2 =
3 5xö
1
= [1 + x + x2 + K + xn + ...] 2 æç1 + xö÷ 2 æç1 - ÷
2 è 2 ø è 4ø
- 4[1 + 2 x + (2 x)2 + K + (2 x)n + ...] 3xö
2/ 3
æ1 + 3 xö æ1 - 5 xö
-1 -1/ 2
= æç1 + ÷ ç ÷ ç ÷
9 è 8ø è 2 ø è 4 ø
+ [1 + (3 x) + (3 x)2 + K + (3 x)n + ...]
2 2 3 3 5
1 = æç1 + × x + Kö÷ æç1 - x + Kö÷ æç1 + x + Kö÷
\ Coefficient of xn = [1 - 8 × 2 n + 9 × 3n ] è 3 8 øè 2 øè 8 ø
2 1 3 5 5
1 = 1 + æç - + ö÷ x = 1 - x
= [1 - 2 n + 3 + 3n + 2 ] è 4 2 8ø 8
2
(1 - 3x )1/ 2 + (1 - x ) 5/ 3
X Example 46. If x is nearly equal to one, then X Example 48. If is
mx m - nx n 4-x
the approximate value of is approximately equal to a + bx for all small values
m-n
of x, then
(a) x m (b) x n 35 35
(c) x m+ n (d) None of these (a) a = 1, b = - (b) a = - , b = 1
24 24
Sol. (c) Since, x is very nearly equal to one, let x = 1 + h, 35
(c) a = 1, b = (d) None of these
where h is very nearly equal to zero, so that h2 may be 24
neglected.
Sol. (a) Since, the given expression is equal to a + bx, so we
mxm - nxn m(1 + h)m - n(1 + h)n
Now, = have to neglect the terms x2 and higher powers of x.
m- n m- n
(1 - 3 x)1/ 2 + (1 - x)5 / 3
m(1 + mh) - n(1 + nh) \
= 4- x
m- n
= [(1 - 3 x)1/ 2 + (1 - x)5 / 3 ][4 - x]-1/ 2
[neglecting h2 and higher powers]
-1/ 2
(m - n) [1 + (m + n) h] 1 5 x
= = éê1 + (- 3 x) + 1 + (- x)ùú(4)- 1/ 2 æç1 - ö÷
m- n ë 2 3 û è 4ø
= 1 + (m + n)h = (1 + h)m + n = xm + n 1 3 5xö é æ 1 ö æ x öù
= æç2 - x - ÷ 1 + ç- ÷ ç- ÷ú
2è 2 3 ø êë è 2 ø è 4 øû
1æ 19 ö æ xö
Approximation = ç2 -
2è
x÷ ç1 + ÷
6 øè 8ø
n( n - 1) 2 n( n - 1)( n - 2) 3 1æ x 19 x ö 1 æ 35 x ö 35 x
= ç2 + - ÷ = ç2 - ÷ = 1-
(1 + x ) n = 1 + nx + x + x +¼ 2è 4 6 ø 2è 12 ø 24
1× 2 1× 2 × 3 35 x 35
\ 1- = a + bx Þ a = 1 and b = -
If x <1, the terms of the above expansion go on 24 24
decreasing and if x is very small, a stage may be
reached when we may neglect the terms
X Example 49. Cube root of 217 is
containing higher powers of x in the expansion. (a) 6.01 (b) 6.04
363
(c) 6.02 (d) None of these
8
1/ 3
Sol. (a) Consider, (217 )1/ 3 = (63 + 1)1/ 3 = 6 æç1 + 13 ö÷ 22 24 26
è 6 ø + + 1+
+K
X Example 52. The value of 2 ! 3! 4!
On expanding by binomial theorem, we get
1 2 22
Objective Mathematics Vol. 1
æ 2 ö
(217 )1/ 3 = 6 ç1 +
1 1´ 2 æ 1 ö 1+ + + +K
ç
- ç ÷ + K÷÷ = 6.01 2! 3! 4!
è 3 ´ 216 3 ´ 3 ´ 2 è 216 ø ø
is
X Example 50. The approximate value of (a) e 2 (b) e 2 + 1 (c) e 2 - 1 (d) e +1
( 7.995)1/ 3 correct to four decimal places is 2 4 6
Sol. (c) Numerator = 1 + 2 + 2 + 2 + K
(a) 1.9995 (b) 1.9996 (c) 1.9990 (d) 1.9991 2! 3! 4!
1/ 3
Sol. (a) (7.995) 1/ 3 1/ 3 1/ 3 é 0.005 ù 1 ì 24 26 28 ü
= (8 - 0.005) = (8) êë1 - 8 úû = 2 í2 2 + + + +Ký
2 î 2 ! 3! 4 ! þ
é 1 æ1 ö ù ì 2 2 2 2 3 2 4 ü
ê ç - 1÷ 2 ú 1 (2 ) (2 )
= 2í + +
(2 )
+
(2 )
+ Ký
1 0.005 3 è 3 ø æ 0.005 ö
= 2 ê1 - ´ + ç ÷ - ...ú 2 î 1! 2! 3! 4! þ
ê 3 8 2! è 8 ø ú
êë úû 1 22
= 2 {e - 1}
é 1 1 ù 2
´ 2
ê 0.005 3 3 æ 0.005 ö ú e4 - 1
= 2 ê1 - - ´ç ÷ - K ú = ...(i)
24 1 è 8 ø 4
ê ú
ë û
and denominator
= 2(1 - 0.000208) = 2 ´ 0.999792 = 1.9995
1 2 22 1 ì 22 23 24 ü
= 1+ + + + K = 2 í2 2 + + + + Ký
2 ! 3! 4! 2 î 2! 3! 4!
Exponential Series 1 ïì æ 22 23 24 ö ïü
þ
= í1 + çç1 + 2 + + + + K÷÷ ý
x x2 x3
(a) e x = 1 + + + +¼ , where x may be any 4 ïî è 2! 3! 4! ø ïþ
1! 2! 3! 1
n = {1 + e 2 } ...(ii)
æ 1ö 4
real or complex and e = lim ç1 + ÷
n ®¥ è nø From Eqs. (i) and (ii), we get
22 24 26 1 4
x x2 2 x3 3 1+ + + +K (e - 1)
(e 2 + 1)(e 2 - 1)
(b) a x = 1 + ln a + ln a + ln a +¼ where 2! 3! 4! = 4 =
1! 2! 3! 1 2 2 2 1 2
(e + 1) (e 2 + 1)
1+ + + +K
a >0 2! 3! 4! 4
Ø 1 1 1 = e2 - 1
● e = 1+ + + +¼
1! 2 ! 3 !
● e is an irrational number lying between 2.7 and 2.8. Its value
correct upto 10 places of decimal is 2.7182818284. Logarithmic Series
æ 1 1 1 ö x2 x3 x4
● e + e -1 = 2 ç1 + + + + ¼÷ (a) ln (1 + x ) = x - + - +¼
è 2! 4! 6! ø
2 3 4
-1 æ 1 1 1 ö where, -1 < x £ 1
● e - e = 2 ç1 + + + + ¼ ÷
è 3! 5! 7! ø
x2 x3 x4
● Logarithms to the base e are known as the Napierian system, (b) ln (1 - x ) = - x - - - -¼
so named after Napier, their inventor. They are also called 2 3 4
natural logarithm. where, -1 £ x < 1
X Example 51. If x <1, then find the coefficient (1 + x ) æ x3 x5 ö
(c) ln = 2 çx + + +¼÷ , | x | < 1
ex (1 - x ) è 3 5 ø
of x n in the expansion of .
1- x
x
Remember
Sol. We have, e = e x (1 - x)-1 1 1 1
1- x (a) 1 - + - +¼= ln 2 (b) e ln x = x
2 3 4
æ x x2 xn - 2 xn - 1 xn ö
= çç1 + + + K+ + + +K÷ (c) ln 2 = 0693
. (d) ln 10 = 2.303
è 1! 2 ! (n - 2 )! (n - 1)! n! ø
¥ ¥
(1 + x + x2 + K + xn - 2 + xn - 1 + xn + ...) n n2
(e) å = e (f) å = 2e
ex n = 1 n! n = 1 n!
\ Coefficient of xn in
(1 - x) ¥ ¥
n3 n4
364
= 1+ +
1
1! 2 !
1
+ .. +
1
+
(n - 1)! n!
1 (g) å n! = 5 e (h) å n! = 15e
n =1 n =1
X Example 53. The value of
æ 1 1ö 1 æ 1 1ö 1 æ 1 1 ö 1
=-
1
2
3
loge æç ö÷ + loge (3) = loge 3 - loge
è 4ø
æ 2 ö
= loge 3 + loge ç ÷ = loge ç 3 ×æ 2 ö
3
4
÷ = loge (2 3 )
8
Binomial Theorem
1+ ç + ÷ + ç + ÷ + ç + ÷ +K is è 3ø è 3ø
è 2 3ø 4 è 4 5ø 42 è 6 7ø 43
= loge ( 12 )
1
(a) log e 12 (b) log e 12
3 X Example 54. In the expansion of
1 2 log e x - log e ( x + 1) - log e ( x - 1), the coefficient of
(c) log e 4 (d) None of these
4 x -4 is
Sol. (a) Consider, 1
(a)
1 1 1 1 1 1 1 1 1 2
1 + æç + ö÷ + æç + ö÷ 2 + æç + ö÷ 3 + K
è2 3 ø 4 è 4 5 ø 4 è6 7ø 4 (b) -1
1 1 1 1 1 1
= æç × + × 2 + × 3 + Kö÷
(c) 1
è2 4 4 4 6 4 ø (d) None of the above
1 1 1 1 1 1
æ
+ ç1 + × + × 2 + × 3 + ...ö÷
è 3 4 5 4 7 4 ø Sol. (a) 2 loge x - loge ìíæç1 + 1 ö÷ xüý - loge ìíæç1 - 1 ö÷ xüý
2 3 î è x þ ø î x þ è ø
1 é 1 1 æ 1ö 1 æ 1ö ù
= ê + ç ÷ + ç ÷ + ...ú ì æ 1ö ü
2 êë 4 2 è 4 ø 3 è 4ø úû = 2 loge x - íloge ç1 + ÷ + logexý
î è xø þ
é 1 1 æ 1ö3 1 æ 1ö5 ù ì 1 ü
+ 2 ê + ç ÷ + ç ÷ + Kú - íloge æç1 - ö÷ + loge xý
êë 2 3 2è ø è
5 2 ø úû î è xø þ
æ1 + 1 ö ì æ 1ö æ 1 öü ì 1 1 ü
= - íloge ç1 + ÷ + loge ç1 - ÷ ý = 2 í 2 + + Ký
1ì 1 ü ç ÷ î è xø è x øþ î2 x 4 x4 þ
= í- loge æç1 - ö÷ ý + loge ç 2÷
2î è ø
4 þ ç1 - ÷ 1 -4 1 1
Thus, coefficient of x = 2 × =
è 2ø 4 2
Work Book Exercise 8.4
1 5 If the binomial expansion of (a + bx )-2 is
1 4/ 3
can be expanded by binomial
æ 2 1ö 1
çx + ÷ - 3 x + ¼, then the values of a and b are
è xø 4
theorem, if a 2, 12 b 2, 10
c 1, 12 d None of these
a x<1 b | x| < 1
c x>1 d | x| > 1 6 In the expansion of (e x - 1)(e - x + 1,) the
2
æ 2 x ö n(n + 1) æ 2 x ö coefficient of x 3 is
2 If| x| < 1, then 1 + n ç ÷+ ç ÷ +¼ 1
è1+ x ø 2! è1+ x ø a 0 b
3
is equal to c
2
d
1
n n
æ1 - x ö æ1 + x ö 3 6
a ç ÷ b ç ÷
è1 + x ø è1 - x ø e 7x + e 3 x
n n 7 In the expansion of , the constant term
1 + xö æ 2x ö e5x
c æç ÷ d ç ÷
è 2x ø è1 + x ø is
2 a 0 b 1
(1 + 3 x )
3 The coefficient of x 3 in the expansion of c 2 d None of these
1- 2x
1 1 1 1 1
will be 8 The value of + × + × + ¼ is
a 8 b 32 1× 3 2 3 × 5 3 5 × 7
c 50 d None of these a 2 loge 2 - 1 b loge 2 - 1
r c loge 2 d None of these
4 The coefficient of x in the expansion of
(1 - 4 x )-1/ 2 is 9 The coefficient of n -r in the expansion of
(2 r )! æ n ö
a
( r !)2
log10 ç ÷ is
è n - 1ø
b 2r C r + 1
1 -1
1 × 3 × 5¼ (2 r - 1) a b
c r loge 10 r loge 10
2 r × r! -1
c d None of these
d None of the above r ! loge 10 365
WorkedOut Examples
Type 1. Only One Correct Option
1
Ex 1. The coefficient of x 50 in the expansion ⇒ a0 + nd =
2n + 1
(1 + x ) 1000 + 2x (1 + x ) 999 + 3x 2 (1 + x ) 998
1
+...+1001x 1000 is ⇒ an =
2n + 1
1002 1002 1005 1005
(a) C 50 (b) C 51 (c) C 50 (d) C 48 [Q an = (n + 1) th term = a0 + (n + 1 − 1)d ]
Sol. Let S = (1 + x ) 1000
+ 2x (1 + x ) 999
+ 3x (1 + x )
2 998 Hence, (a) is the correct answer.
+ ...+ 1000(1 + x )x 999 + 1001x 1000 ...(i)
Ex 3. The coefficient of x r {0 ≤ r ≤ (n − 1)} in the
xS
⇒ = x (1 + x )999 + 2x 2 (1 + x )998 expansion of
1+ x
( x + 2) n − 1 + ( x + 2) n − 2 ( x + 1) + ( x + 2) n − 3
x 1001
+ ...+ 1000x 1000 + 1001 ...(ii) ( x + 1) 2 +...+ ( x + 1) n − 1 is equal to
1+ x
On subtracting Eq.(ii) from Eq.(i), we get (a) ( 2n + r + 1) n C r
S−
xS
= {(1 + x )1000 + x (1 + x )999 + x 2 (1+ x )998 (b) ( 2n − r + 1) n C r
1+ x
(c) ( 2n − r − 1) n C r
x 1001
+ ... + x 1000} − 1001 (d) ( 2n + r − 1) n C r
1+ x
x 1001 Sol. We have,
1 − (x + 2)n − 1 + (x + 2)n − 2 (x + 1) + (x + 2)n − 3 (x + 1)2
S 1 + x x 1001
⇒ = (1 + x )1000 − 1001 + ...+ (x + 1)n − 1
1+ x 1−
x
1+ x
1+ x x + 1 n
1 −
n − 1 x + 2
= (x + 2)
1001
x+1
S x
⇒ = (1 + x )1001 − x 1001 − 1001
1+ x 1+ x 1−
x+2
⇒ S = (1 + x )1002 − x 1001 (1 + x ) − 1001x 1001
⇒ S = (1 + x )1002 − 1002x 1001 − x 1002 = (x + 2)n − (x + 1)n
∴Coefficient of x 50 in S = Coefficient of x 50 in ∴Coefficient of x r in the given expression
(1 + x )1002 = Coefficient of x r in (x + 2)n
= 1002
C 50 − Coefficient of x r in (x + 1)n
n− r
Hence, (a) is the correct answer. = Cr 2
n
− Cr n
n− r
Ex 2. If (1 − x + x ) = a 0 + a 1 x + a 2 x +...+a 2n x ,
2 n 2 2n = (2 − 1)nC r
where a 0 , a 1 , a 2 , ..., a 2n are in AP, then a n is Hence, (c) is the correct answer.
equal to
1 1 2 1 Ex 4. If k and n are positive integer and
m
sk = 1k + 2 k +.... + n k , then ∑ m + 1C r sr is equal to
(a) (b) (c) (d)
2n + 1 2n − 1 2n − 1 n+1
r=1
Sol. We have, (a) ( n + 1) m + 1 − ( n + 1)
(1 − x + x 2 )n = a0 + a1x + a2 x 2 + ...+ a2n x 2n
(b) ( n + 1) m + 1 + ( n + 1)
Putting x = 1on both sides, we get
(c) ( n − 1) m + 1 − ( n − 1)
a0 + a1 + a2 + ...+ an + ...+ a2n = 1
2n + 1 (d) None of the above
⇒ [ a0 + a2n ] = 1 m
2
[Qa0 , a1 ,..., a2n are in AP]
Sol. We have, ∑ m + 1C r sr
r=1
2 m
⇒ a0 + a2n =
2n + 1 = ∑ m + 1C r (1r + 2r +...+ nr )
r=1
2
⇒ a0 + [ a0 + (2n + 1 − 1)d ] = n m
2n + 1
366
= ∑ ∑ m + 1C r k r
where, d denotes the common difference of the AP. k = 1r = 1
=
n m + 1
k = 1 r = 0
∑ ∑ m + 1C r k r −
m +1
C0 − m +1
C m + 1k m + 1
If the general term in the above expansion contains
x 3 y4z5, then
r + t = 3, r + s = 4 and s + t = 5
8
Binomial Theorem
n n Also, r+ s+t=6
= ∑ {(1 + k )m + 1 − 1 − k m + 1} Q ∑ nC r x r = (1 + x )n On solving these equations, we get
k=1 r=0
n n
r = 1, s = 3, t = 2
= ∑ {(1 + k )m + 1 − k m + 1} − ∑ 1 ∴Coefficient of x y z =
3 4 5 6!
=
6!
= 60
k=1 k=1 1!3!2! 2!3!
n
Hence, (b) is the correct answer.
= ∑ {(1 + k )m + 1 − k m + 1} − n
k=1
Ex 7. If (1 + x ) n = C 0 + C 1 x + C 2 x 2 +...+ C n x n , n ∈ N ,
= [(2m + 1 − 1m + 1 ) + (3m + 1 − 2m + 1 )
then for 2 ≤ m ≤ n,
+ … + {(n + 1)m + 1 − nm+ 1}] − n
m+ 1
C 0 − C 1 + C 2 − C 3 + K + ( −1) m − 1 C m − 1 is
= {(n + 1) − 1} − n = (n + 1)m + 1 − (n + 1)
equal to
Hence, (a) is the correct answer.
(a) ( −1) m − 1 ⋅ n − 1 C m − 1 (b) n− 1
Cm − 1
Ex 5. If C 0 , C 1 , C 2 , K, C n denote the binomial (c) ( −1) ⋅ m n− 1
Cm − 1 (d) None of these
coefficients in the expansion of (1 + x ) n , then
Sol. We have, (1 − x )n (1 − x )−1
n
1 + r log e 10
∑
r=0
( −1) r n C r
(1 + log e 10 n ) r
is equal to = {C 0 − C 1x + C 2x 2 − C 3x 3 + ...+ (−1)m − 1C m − 1x m − 1
+ ...+ (−1)nC nx n}
(a) 0 × {1 + x + x 2 + x 3 + K + x m − 1 + x m + K}
(b) 1
Equating coefficients of x m − 1 on both sides, we get
(c) 2 (−1)m − 1 ⋅ n − 1C m − 1 = C 0 − C 1 + C 2 − C 3
(d) Cannot be discussed
+ K + (−1)m − 1 C m − 1
Sol. Let loge 10 = x. Then, ⇒ C 0 − C 1 + C 2 − C 3 + ... + (−1)m − 1 C m − 1
n
1 + r log 10 n
1 + rx
∑ (−1)r nC r (1 + log 10e n )r = ∑ (−1)r nC r (1 + nx )r = (−1)m − 1
(n − 1)!
r=0 e r=0 (n − m)! (m − 1)!
r
n
1 n
n rx = (−1)m − 1 ⋅n − 1 C m − 1
= ∑ (−1)r nC r 1 + nx + ∑ (−1)r r
n− 1
Cr − 1
(1 + nx )r
r=0 r=0 Hence, (a) is the correct answer.
r
n
1
= ∑ (−1)r n
Cr Ex 8. For any positive integers m, n with n ≥ m, let
1 + nx
r=0 n n
1
r−1 = C m . Then,
m
n
nx
− ⋅ ∑ (−1)r − 1 n− 1
C r − 1
1 + nx r = 1 1 + nx n n − 1 n − 2 m
n n− 1 + + + K + is equal to
= 1 −
1 nx 1 m m m m
− 1 −
1 + nx 1 + nx 1 + nx (a) n C m (b) n C m + 1
n n
nx nx n+ 1
= − =0 (c) Cm + 1 (d) None of these
1 + nx 1 + nx
Hence, (a) is the correct answer. Sol. We know that,
C r = Coefficient of x r in (1 + x )n
n
n n − 1 n − 2
3 4 5
Ex 6. The coefficient of x y z in the expansion of m
∴ + + +K+
( xy + yz + zx )6 is m m m m
(a) 70 = Coefficient of x m in (1 + x )n + Coefficient of x m in
(b) 60 (1 + x )n − 1 + K + Coefficient of x m in (1 + x )m
(c) 50 = Coefficient of x m in
(d) None of the above [(1 + x )n + (1 + x )n − 1 + K + (1 + x )m ]
(1 + x )n − m + 1 − 1
Sol. We have, = Coefficient of x m in (1 + x )m
6! (1 + x ) − 1
(xy + yz + zx )6 = ∑ r ! s ! t !
(xy)r ( yz)s (zx )t
= Coefficient of x m + 1 in {(1 + x )n + 1 − (1 + x )m}
r + s + t =6
6! r + t r + s s + t = Coefficient of x m + 1 in (1 + x )n + 1 = n+ 1
Cm + 1
= ∑ r ! s !t !
x y z
367
r + s + t =6 Hence, (c) is the correct answer.
8 Ex 9. Let n be an even integer and k =
k
3n
2
. Then, the
On adding and subtracting these two, we get
C 0 + C 2 + C 4 + K = 2n − 1
C 1 + C 3 + C 5 + .... = 2n − 1
value of ∑ ( −3) r − 1
...(ii)
Objective Mathematics Vol. 1
3n
C 2r − 1 is
r=1 Putting x = i in Eq. (i), we get
(a) 0 (b) 1 (c) 2 (d) 3 (C 0 −C 2 + C 4 − C6 + ...) + i (C 1 − C 3 + C 5 −...) = (1 + i )n
⇒ (C 0 − C 2 + C 4 − C6 + ...)
Sol. Since, n is an even integer. Therefore, n = 2m, m ∈ N
3n + i (C 1 − C 3 + C 5 − C 7 + ...)
Also, k= ⇒ k = 3m n/ 2 nπ nπ
2 = 2 cos + i sin
k 4 4
∴ ∑ (−3)r − 1 3nC 2r − 1 On equating imaginary parts on both sides, we get
r=1 nπ
3m C 1 − C 3 + C 5 − C 7 + K = 2n/ 2 sin ...(iii)
∑ (−3)r − 1 6 mC 2r − 1
= 4
r=1 On subtracting Eq. (iii) from Eq. (ii), we get
nπ
= 6m
C1 − 6m
C 3 (3) + 6m
C 5 (3)2 − 6m
C 7 (3)3 2 (C 3 + C 7 + C 11 + ...) = 2n − 1 − 2n/ 2 sin
4
+ K + (−3)3m − 1 6m
C6m − 1
1 n− 1 nπ
⇒ C 3 + C 7 + C 11 + K = 2 − 2n/ 2 sin
2 4
6m
Now, (1 + i 3 )6m = ∑ 6mC r (i 3 )r
r=0 Hence, (a) is the correct answer.
6m
π π
⇒ 26m cos + i sin = 6mC 0 + 6m
C 1 (i 3 ) Ex 11. The integer just greater than ( 3 + 1) 2m
3 3
contains
+ 6mC 2 (i 3 )2 + 6m
C 3 (i 3 )3 (a) 2m + 2 as a factor (b) 2m + 1 as a factor
(c) 2m + 3 as a factor (d) None of these
+K+ 6m
C6m (i 3 )6m
⇒ 26m (cos 2mπ + i sin 2mπ ) Sol. Let ( 3 + 1)2m = I + F , where I ∈ N and 0 < F < 1
Let G = ( 3 − 1)2m. Then,
= {6mC 0 − 6mC 2 3 + C 4 32 − ...}
6m
I + F + G = ( 3 + 1)2m + ( 3 − 1)2m
+ i 3 {6mC 1 −6m C 3 (3) + 6mC 5 (3)4 −...}
= 2m (2 + 3 )m + 2m (2 − 3 )m
[using De-Moivre’s theorem on LHS]
= 2m + 1 × an integer ...(i)
Equating imaginary parts on both sides, we get
⇒ I + F + G = an even integer
26m sin 2mπ = 3 {6mC 1 − 6mC 3 (3) + 6mC 5 (3)4 − ...}
⇒ F + G = an even integer − I
⇒ 6m
C 1 − 6mC 3 (3) + 6m
C 5 (3)4 − ... = 0 [Q sin 2mπ = 0 ] ⇒ F + G = an integer
3m ⇒ F + G =1 [Q 0 < F < 1, 0 < G < 1]
⇒ ∑ (−3)r − 1 6mC 2r − 1 = 0 Putting F + G = 1in Eq. (i), we get
r=1
k
I + 1 = 2m + 1 × an integer
⇒ ∑ (−3)r − 1 3nC 2r − 1 = 0, ⇒ 2m + 1 is a factor of the integer just greater than
r=1 ( 3 + 1)2m.
3n
where n = 2m and k= Hence, (b) is the correct answer.
2
2n 2n
Hence, (a) is the correct answer. Ex 12. If ∑
r=0
a r ( x − 2) r = ∑ br ( x − 3) r
r=0
and a k =1 for
Ex 10. The value of C 3 + C 7 + C 11 +K is
n n n
1 n− 1 nπ
all k ≥ n, then bn is equal to
(a) 2 − 2n/ 2 sin (a) n+ 1
C 2n + 1 (b) 2n + 1
Cn + 1
2 4
2n + 1
1 n− 1 nπ (c) Cn + 2 (d) None of these
(b) 2 + 2n/ 2 sin
2 4
Sol. Clearly, bn is the coefficient of (x − 3)n in the expression
1 n+ 1 n π
(c) 2 − 2n/ 2 sin 2n
4 4 ∑ br (x − 3)r . Therefore,
r=0
(d) None of the above 2n
Sol. We have, (1 + x )n = C 0 + C 1x + C 2x 2 + K + C nx n ...(i) bn = Coefficient of (x − 3)n in ∑ ar (x − 2)r ...(i)
r = 0
Putting x = 1and − 1respectively in Eq. (i), we get
= Coefficient of (x − 3)n in
C 0 + C 1 + C 2 + K + C n = 2n
n − 1 2n
C 0 − C 1 + C 2 − C 3 + K + (−1)n C n = 0 ∑ ar (x − 2) + ∑ ar (x − 2)r
r
368 r = 0 r=n
= Coefficient of (x − 3)n in
n− 1 2n
∑ ar (x − 2) + ∑ (x − 2)
r
r
Ex 14. The greatest term in the expansion of (1 + x ) 10 ,
2
when x = , is
8
Binomial Theorem
r = 0 r=n 3
4 4
[Q ak = 1 for all k ≥ n ] 2 2
(a) 210 (b)
2n 3 3
= Coefficient of (x − 3)n in ∑ (x − 2)r 1
4
2
6
r=0
(c) 210 (d) 210
(x − 2)n + 1 − 1
3 3
= Coefficient of (x − 3)n in (x − 2)n
(x − 2) − 1 Sol. Let Tr and Tr + 1 denote the rth and (r + 1)th terms in the
expansion of (1 + x )10. Then,
(x − 2)2n + 1 − (x − 2)n
= Coefficient of (x − 3)n in Tr = 10C r − 1x r − 1 and Tr + 1 = 10C r x r
x−3 10
Tr + 1 C r xr
= Coefficient of (x − 3)n + 1 in {(x − 2)2n + 1 − (x − 2)n} ∴ =
Tr 10
C r − 1x r − 1
= Coefficient of (x − 3)n + 1 in (x − 2)2n + 1 10
Tr + 1 Cr
= Coefficient of (x − 3)n + 1 in [(x − 3) + 1] 2n + 1 ⇒ = 10
⋅x
Tr Cr − 1
2n + 1
= Coefficient of (x − 3) n+ 1
in ∑ 2n + 1C r (x − 3)r Tr + 1 11 − r
⇒ = ⋅x
r = 0 Tr r
2n + 1
= Cn + 1 Tr + 1 11 − r 2 2
⇒ = × Qx =
Tr r 3 3
Hence, (b) is the correct answer.
Tr + 1
Now, >1
Ex 13. Let f (x ) = a 0 + a 1 x + a 2 x 2 + K + a 2n x 2x and Tr
g (x ) = b0 + b1 x + b2 x 2 + K + bn − 1 x n − 1 11 − r 2
⇒ × >1
r 3
+ x n + x n + 1 + K + x 2n
⇒ 22 > 5r
If f ( x ) = g ( x + 1), then a n in terms of n is equal to 2
⇒ r<4
(a) 2n + 1 C n + 1 (b) 2n − 1 C n − 1 5
(c) 2n − 1
Cn + 1 (d) n+ 1
Cn + 1 ∴ (4 + 1) th i.e. 5th term is the greatest term.
Putting r = 4 in Tr + 1, we get
Sol. We have, f (x ) = g (x + 1) T5 = 10C 4 x 4
2n n− 1 2n 4
2 2
⇒ ∑ a1x r = ∑ br (x + 1)r + ∑ (x + 1)r ⇒ T5 = 10C 4
3
Q x=
3
r=0 r=0 r=n
n− 1 4
2n 2
⇒ T5 = 210
⇒ ∑ ar x r = ∑ br (x + 1)r + (x + 1)n + (x + 1)n + 1 3
r=0 r=0
Hence, (a) is the correct answer.
+ K + (x + 1)2n
n− 1
2n (x + 1)n + 1 − 1 Ex 15. The greatest term in the expansion of (1 + x ) 2n
⇒ ∑ ar x r = ∑ br (x + 1)r + (x + 1)n
r=0 r=0 x + 1−1 has also the greatest coefficient, then
2n n− 1
(x + 1)2n + 1 − (x + 1)n n n + 1 n n − 1
(a) x ∈ (b) x ∈
⇒ ∑ ar x r = ∑ br (x + 1)r + x
,
n − 1 n
,
n − 1 n
r=0 r=0
2n n n + 1 n n + 1
Now, an = Coefficient of x n in ∑ ar x r (c) x ∈ , (d) x ∈ ,
r=0
n + 1 n n − 1 n
= Coefficient of x in n
Sol. Let (r + 1)th term be the greatest term in the expansion
(x + 1)2n + 1 − (x + 1)n
n− 1
of (1 + x )2n. Then,
∑ br (x + 1) +
r
r = 0 x Tr + 1 = 2nC r x 2n − r
(x + 1) 2n + 1
− (x + 1)n The coefficient of Tr + 1 is 2nC r . Clearly, 2nC r will be
= Coefficient of x n in 2n
x greatest, if r = =n
2
= Coefficient of x n + 1 in {(x + 1)2n + 1 − (x + 1)n}
Thus, Tn + 1 = 2nC nx n has the greatest coefficient.
= Coefficient of x n + 1 in (x + 1)2n + 1
2n + 1
Now, Tn + 1 will be the greatest term, if
= Cn + 1 Tn < Tn + 1 > Tn + 2
Hence, (a) is the correct answer. ⇒ Tn + 1 > Tn and Tn + 2 < Tn + 1 369
8 ⇒
Tn + 1
Tn
>1 and
Tn + 2
Tn + 1
<1
C n + 1x n + 1
n
r=0
Ex 17. The value of ∑ (−2) r
Cr
n
r+ 2
is
Cr
Objective Mathematics Vol. 1
2n 2n
C nx n 1 1
⇒ > 1 and <1 (a) , n is odd (b) , n is even
2n
C n − 1x n − 1 2n
C nx n n+1 n+2
n + 1 n 1
⇒ x > 1 and x <1 (c) , n is even (d) None of these
n n + 1 n+1
n n+1
⇒ x> and x< Cr n
n+1 n Sol. For r ≥ 0, we have r+ 2
Cr
n n+1
⇒ <x< n! r ! 2!
n+1 n = ×
(n − r)! r! (r + 2)!
n n + 1 n! × 2!
⇒ x ∈ , =
n+ 1 n (n − r)! (r + 2)!
Hence, (c) is the correct answer. 2 (n + 1) (n + 2) n!
= ×
n n (n + 1) (n + 2) {(n + 2) − (r + 2)}! (r + 2)!
Cr
Ex 16. ∑ (−1) r r+ 3
is equal to =
2
× n + 2C r + 2
r=0 Cr (n + 1) (n + 2)
3! n n
C
(a)
2 ( n − 3)
∴ ∑ (−2)r r + 2Cr
r=0 r
3! 2 n
(b)
2 ( n + 3) = ∑
(n + 1) (n + 2) r = 0
n+ 2
C r + 2 (−2)r
3! n
(c) 1
( n + 3) = ∑ n + 2C r + 2 (−2)r + 2
2(n + 1)(n + 2) r = 0
(d) None of the above n+ 2
1
C n
Sol. We have, ∑ (−1)r r + 3 r
n = ∑ n + 2C s (−2)s
2 (n + 1) (n + 2) s = 2
r=0 Cr
[putting r + 2 = s]
n
n ! ⋅ 3! 1
= ∑ (−1) r
=
r=0
(n − r )! (r + 3)! 2 (n + 1) (n + 2)
n n + 2
= 3! ∑ (−1)
n!r × ∑ n + 2C s (−2)s − n + 2C 0 (−2)0 − n + 2C 1 (−2)1
r=0
(n − r )! (r + 3)! s = 0
1
(−1)r (n + 3)! [(1 − 2)n + 2 − 1 + 2 (n + 2)]
n
3! =
= ∑
(n + 1) (n + 2) (n + 3) r = 0{(n + 3) − (r + 3)}! (r + 3)!
2 (n + 1) (n + 2)
1
(2n + 4 ), if n is even
3! n
2 (n + 1) (n + 2)
= ∑
(n + 1) (n + 2) (n + 3) r = 0
(−1)r n+ 3
Cr + 3 =
1
(2n + 3 − 1), if n is odd
3!
n+ 3 2 (n + 1) (n + 2)
(n + 1) (n + 2) (n + 3) s∑
= (−1)s − 3 n+ 3
Cs
1
=3 , if n is even
n + 1
n+ 3 =
3! (−1)3
(n + 1) (n + 2) (n + 3) s∑
n+ 3
= (−1)s Cs 1 , if n is odd
=3 n + 2
− 3! n + 3 Hence, (c) is the correct answer.
(n + 1) (n + 2) (n + 3) s∑
= (−1)s ⋅ n+ 3
C s
n
n
Cr
=0 Ex 18. The sum of ∑ (−1) r r+ 2
is
n+ 3 n+ 3 n+ 3 r=0 Cr
− C0 + C1 − C2
2 1
(a) (b)
=
3! n+1 n+2
2 (n + 3) 2 2
(c) (d)
Hence, (b) is the correct answer. n−2 n+2
370
Sol. We have,
n
∑ (−1)r r + 2Cr
r=0
n
C
r
Sol. We have,
an = (log 3)n ∑
r=1
n
r2
r! (n − r)!
8
Binomial Theorem
n
n! 2! r !
= ∑ (−1)r (n − r)! r! × (r + 2)! (log 3)n n n!
r=0 = ∑
n! r = 1(n − r)! r!
⋅ r2
n
n!
= 2 ∑ (−1)r
(n − r)! (r + 2)! (log 3)n n n
n! r∑
r=0 = C r ⋅ r2
2 n
(n + 2)! =1
= ∑
(n + 1) (n + 2) r = 0
(−1)r
{(n + 2) − (r + 2)}! (r + 2)! (log 3)n
= [ n (n − 1) 2n − 2 + n ⋅ 2n − 1 ]
n n!
2
= ∑
(n + 1) (n + 2) r = 0
(−1)r + 2 n+ 2
Cr + 2 n
Q ∑ r ⋅ C r = n (n − 1) 2
n− 2
2 n
+ n ⋅ 2n − 1
n+ 2 r = 1
2
(n + 1) (n + 2) s∑
n+ 2
= (−1)s Cs (log 3) n
=2 = [ n (n − 1) + 2n ] 2n − 2
n!
2
= =
1
(log 3)n 2n − 2 +
2
(log 3)n ⋅ 2n − 2
(n + 1) (n + 2) (n − 2)! (n − 1)!
n + 2 (2 log 3)n − 2 (2 log 3)n − 1
∑ (−1)s n+ 2
C s − (n + 2C 0 − n+ 2
C 1 ) = (log 3)2 + (log 3)
s = 0
(n − 2)! (n − 1)!
2 ∴Required sum
= ∞
n+ 2 = ∑ an
Hence, (d) is the correct answer. n= 1
∞
(log 9)n − 2 ∞
(log 9)n − 1
Ex 19. If n C 0 , nC 1 , nC 2 , ..., nC n denote the binomial = (log 3)2 ∑ + (log 3) ∑
n= 2
(n − 2)! n= 1
(n − 1)!
coefficients in the expansion of (1 + x ) n and
n = (log 3)2 elog 9
+ (log 3)elog 9
p + q =1, then ∑ r 2 n
Cr p q r n− r
is ∞ xn − 1 ∞
xn − 2
r=0 Q ∑ = ∑ = ex
(a) np (b) npq n = 1(n − 1)! n = 2(n − 2)!
(c) n 2 p 2 + npq (d) None of these = (1 + log 3) (9 log 3)
n
Hence, (a) is the correct answer.
Sol. We have, ∑ r2 nC r pr qn − r n
C0 2 C1 3
r=0 Ex 21. If (1 + x ) n = ∑ n C r x n , then 2 + 2
n r=0 1⋅ 2 2 ⋅3
= ∑ [ r (r − 1) + r ] nC r pr qn − r C
+ 2 24 + K +
Cn
2 n + 2 is equal to
r=0
n n
3⋅4 ( n + 1) ( n + 2)
= ∑ r (r − 1) nC r pr qn − r + ∑ r ⋅ nC r pr qn − r (a)
3n + 2 + 2n − 5
r=0 r=0
( n + 1) ( n + 2)
n
n n−1
= ∑ r (r − 1) r . r − 1 n − 2C r − 2 pr qn − r 3n + 2 − 2n + 5
r=2 (b)
n ( n + 1) ( n + 2)
n
+ ∑r. r n− 1
C r − 1 pr qn − r
3n + 2 − 2n − 5
r=1 (c)
= n (n − 1) p2 ( p + q)n − 2 + np ( p + q)n − 1 ( n + 1) ( n + 2)
= n (n − 1) p2 + np [Q p + q = 1] (d) None of the above
= n2 p2 − np2 + np Sol. We have,
C0 2 C1 3 C2 4 Cn
= n p + npq
2 2
[Q p + q = 1] 2 + 2 + 2 +K+ 2n + 2
1⋅ 2 2⋅ 3 3⋅ 4 (n + 1) (n + 2)
Hence, (c) is the correct answer. n
1
r n 2 = ∑ (r + 1) (r + 2) nC r ⋅ 2r + 2
Ex 20. Let a n = (log 3) n ∑ , n ∈ N . Then, r=0
r = 1 r ! ( n − r )! 1 n
n+ 2 n+1 n
the sum of the series a 1 + a 2 + a 3 + a 4 +K is
= ∑ ⋅
(n + 1) (n + 2) r = 0 r + 2 r + 1
C r 2r + 2
(a) (1 + log 3) ( 9 log 3) (b) (1 + log 3) 1 n
(c) (1 − log 3) ( 9 log 3) (d) (1 + log 3) (log 3) = ∑
(n + 1) (n + 2) r = 0
n+ 2
C r + 2 2n + 2
371
8 =
1 n + 2
∑
(n + 1) (n + 2) s = 2
n+ 2
C s 2s
Ex 24. If (1 + x ) n = C 0 + C 1 x + K + C n x n , then the
value of ∑
n n
∑ (C r + C s ) is equal to
Objective Mathematics Vol. 1
1 r=0 s=0
=
(n + 1) (n + 2)
(a) ( n + 1) 2n + 1
n + 2
∑ n + 2C s 2s − {n + 2C 0 20 + n + 2C 1 21} (b) ( n − 1) 2n + 1
s = 0 (c) ( n + 1) 2n
1
= [(1 + 2)n + 2 − {1 + 2 (n + 2)}] (d) None of the above
(n + 1) (n + 2) n n
=
3n + 2 − 2n − 5 Sol. ∑ ∑ (C r + C s )
r=0 s=0
(n + 1) (n + 2)
n n n n
Hence, (c) is the correct answer. = ∑ ∑Cr + ∑ ∑Cs
r=0 s=0 r=0 s=0
Ex 22. The value of n
n n n
n− 1 n− 1 n− 1 = ∑ ∑ C r + ∑ ∑ C s
C0 ⋅ C1 + n
C1 ⋅ C2 + C2 ⋅ C3 n n
s = 0 r = 0 r = 0 s = 0
n− 1
+ .... + C n − 1 ⋅ C n is equal to
n
n n
(a) 2n
Cn
= ∑ 2n + ∑ 2n
s=0 r=0
2n − 1
(b) Cn − 1 = (n + 1) 2n + (n + 1) 2n = (n + 1) 2n + 1
2n
(c) Cn + 1 Hence, (a) is the correct answer.
(d) None of the above Ex 25. If (1 + x ) n = C 0 + C 1 x + K + C n x n , then the
n n
Sol. We have,
n− 1 n− 2 value of ∑ ∑ C r C s is equal to
(C 0 x + C 1 x
n
+ C2 x + K + C n − 1x + C n )
r=0 s=0
n− 1 n− 1 n− 1 n− 1
×( C0 + C 1x + C 2 x2 + K + C n − 1x n − 1 ) (a) 2n (b) 22n (c) 2n + 1 (d) 23n
= (1 + x )n (1 + x )n − 1 n n n n
⇒ ( C 0 x + C 1x
n n n n− 1
+ C2 x
n n− 2
+ K + C n − 1x + C n )
n n Sol. We have, ∑ ∑ C rC s = ∑ C r ∑ C s
r=0 s=0 r = 0 s=0
n− 1 n− 1 n− 1 n− 1
×( C0 + C 1x + C2 x + K +
2
C n − 1x n − 1 ) n n
= (1 + x )2n − 1 = ∑ 2n ⋅ C r = 2n ∑ C r
r=0 r = 0
On equating the coefficient of x n − 1 both sides, we get
n− 1 = 2 ⋅ 2 = (2 ) = 22n
n n n 2
C 0nC 1 + n − 1C 1nC 2 + K+ n − 1C n − 1nC n = 2n − 1C n − 1
Aliter
Hence, (b) is the correct answer. n n n n
Ex 23. The value of ∑ ∑ C rC s = ∑ C r ∑ C s = 2n ⋅ 2n = 22n
r=0 s=0 r = 0 s = 0
n− 1
C 0 ⋅ nC 2 + n − 1 C 1 ⋅ nC 3 + n− 1
C2⋅ n C4 Hence, (b) is the correct answer.
+ K + n − 1C n − 2 ⋅ nC n is
(a) 2n
Cn − 2 Ex 26. If (1 + x ) n = C 0 + C 1 x + K + C n x n , then the
(b) 2n − 1
Cn value of ∑ ∑ C r C s is equal to
0≤ r< s ≤ n
2n − 1
(c) Cn − 2
1 1 2n 2n
(d) None of the above (a) [ 22n − 2n
Cn ] (b) [2 − C n ]
2 4
Sol. We have, 1 1
(C 0 x n + C 1 x n − 1 + C 2 x n − 2 + K + C n − 1 x + C n ) (c) [ 22n + 2n
Cn ] (d) [ 2n − 2nC n ]
2 2
n− 1 n− 1 n− 1 n− 1 n− 1
×( C0 + C1 x + C2 x + K+
2
Cn − 1 x )
n− 1
Sol. We have,
= (1 + x ) (1 + x ) n
n n n
⇒ (nC 0 x n + nC 1 x n − 1 + nC 2 x n − 2 + K + nC n − 1x + nC n ) ∑ ∑ C rC s = ∑ C r2 + 2 ∑ ∑ C rC s
r=0 s=0 r = 0 0≤ r < s ≤ n
× (n − 1C 0 + n− 1
C1 x + n− 1
C 2 x2 + K + n− 1
C n − 1 xn − 1)
⇒ 2 = 2n 2n
Cn + 2 ∑ ∑ C rC s
= (1 + x )2n − 1 0≤ r < s ≤ n
On equating the coefficient of x n − 2 both sides, we get 1
n− 1
⇒ ∑ ∑ C rC s = 2 [ 22n − 2nC n ]
C 0nC 2 + n − 1C 1nC 3 + K + n − 1C n − 2nC n = 2n − 1C n − 2 0≤ r < s ≤ n
372 Hence, (a) is the correct answer.
Hence, (c) is the correct answer.
Ex 27. If (1 + x ) n = C 0 + C 1 x + C 2 x 2 + K + C n x n , then
the value of ∑ ∑ ( r ⋅ s) C r C s is
Ex 30. If (1 + x ) n = C 0 + C 1 x + C 2 x 2 + K + C n x n , then
n n
∑ ∑ (r + s) C r C s is equal to
8
Binomial Theorem
0≤ r < s ≤ n r=0 s=0
1 (a) 22n (b) n ⋅ 22n
(a) n 2 22n − 3 + 2n − 2C n − 1
2 (c) n ⋅ 2n (d) None of these
1
(b) 22n − 3 − 2n − 2C n − 1
n n
2
Sol. We have, ∑ ∑ (r + s) C rC s
r=0 s=0
1
(c) n 2 22n − 3 − 2n − 2C n − 1 n n
2 = ∑ ∑ (r ⋅ nC r ⋅ nC s + s ⋅ nC r ⋅ nC s )
r=0 s=0
(d) None of the above n n n n
= ∑ ∑ r ⋅ nC r C s + ∑ ∑ s nC r C s
n n
Sol. We have, ∑ ∑ (r ⋅ s) C rC s = ∑ ∑ (r ⋅ nC r ) (s ⋅ nC s ) r=0 s=0 r=0 s=0
0≤ r < s ≤ n 0≤ r < s ≤ n
n n n n n n
= ∑∑ r⋅
r
n− 1
C r − 1
s⋅
s
n− 1
C s − 1
= ∑ r ⋅ nC r ∑ nC s + ∑ s ⋅ nC s ∑ nC r
0≤ r < s ≤ n r=0 s = 0 s=0 r = 0
1
= n ⋅ [ 22 (n − 1) − 2 (n − 1)C n − 1 ]
2 n n
2 = ∑ r ⋅ nC r ⋅ 2n + ∑ s ⋅ nC s ⋅ 2n
2 2n − 3 1 r=0 s=0
=n 2 − 2n − 2C n − 1
2 n n
= 2n ∑ r ⋅ nC r + 2n ∑ s ⋅ nC s
Hence, (c) is the correct answer.
r = 0 s = 0
Ex 28. If (1 + x ) n = C 0 + C 1 x + C 2 x 2 + K + C n x n , then n n n n
= 2n ∑ r ⋅ n− 1
C r − 1 + 2n ∑ s ⋅ n − 1C s − 1
the value of ∑ ∑ (C r ± C s ) 2 is
r = 0 r
s = 0 s
0 ≤ r< s ≤ n
(a) ( n + 1) 2nC n (b) ( n m 1) 2nC n ± 2n = 2n (n ⋅ 2n − 1 ) + 2n (n ⋅ 2n − 1 )
(c) ( n m 1) 2nC n (d) ( n m 1) 2nC n ± 22n = n ⋅ 22n − 1 + n ⋅ 22n − 1 = n ⋅ 22n
Hence, (b) is the correct answer.
Sol. We have, ∑ ∑ (C r ± C s )2
0≤ r < s ≤ n
Ex 31. If (1 + x ) n = C 0 + C 1 x + C 2 x 2 + K + C n x n , then
= (C 0 ± C 1 )2 + (C 0 ± C 2 )2 + K + K + (C 0 ± C n )2
+ (C 1 ± C 2 )2 + (C 1 ± C 3 )2 + K + (C 1 ± C n )2 ∑ ∑ (r + s) C r C s is equal to
0≤ r < s ≤ n
+ (C 2 ± C 3 )2 + K + (C 2 ± C n )2
(a) n [ 22n − 1 − 2n − 1
Cn − 1 ]
..........................................+........................................
2n − 1 2n − 1
..........................................+........................................ (b) n [ 2 + Cn − 1 ]
+ (C n − 2 + C n − 1 )2 + (C n − 2 ± C n )2 + (C n − 1 ± C n )2 (c) 2n [ 2 2n − 1
− 2n − 1
Cn − 1 ]
=n (C 02 + C 12 +K+ C n2 ) ± 2∑ ∑C rS s (d) None of the above
0≤ r < s ≤ n
n n
= n⋅ Cn ± [2 − Cn ] =
2n 2n 2n
(n m 1) 2nC n ± 22n Sol. We have, ∑ ∑ (r + s) C rC s
Hence, (d) is the correct answer. r=0 s=0
n n n
Ex 29. The value of ∑ ∑ n s
C s C r is = ∑ 2r C r2 + 2 ∑ ∑ (r + s) C rC s
r=0 s=1 r=0 0 ≤ r <s ≤ n
r≤s n n n
(a) 3n − 1 (b) 3n + 1 ⇒ ∑ ∑ (r + s) C rC s = 2 ∑ r ⋅ C r2
r=0 s=0 r=0
(c) 3n (d) None of these
+ 2∑ ∑ (r + s) C rC s
n n
0≤ r < s ≤ n
∑ ∑ n s
Sol. We have, Cs Cr
n
r=0 s=1 ⇒ n ⋅ 22n = 2 ⋅ 2nC n + 2 ∑ ∑ (r + s) C rC s
r ≤s 2
0≤ r < s ≤ n
n
= ∑ nC s (s C 0 + C 1 + sC 2 + K + sC s )
s
[Qr ≤ s] ⇒ ∑ ∑
1
(r + s) C rC s = [ n ⋅ 22n − n ⋅ 2nC n ]
s=1 0≤ r <s ≤ n
2
n n
2n 2n 2n − 1
= ∑ n
C s (2)s = ∑ nC s 2n − nC 0 20 =
n
2 − Cn − 1
s=1 s=0 2 n
2n − 1 2n − 1
= (1 + 2)n − 1 = 3 − 1 n
= n [2 − Cn − 1 ]
Hence, (a) is the correct answer. Hence, (a) is the correct answer. 373
8 Ex 32. If (1 + x ) n = C 0 + C 1 x + C 2 x 2 + K + C n x n , then
the value of ∑ ∑ ( r + s) (C r + C s ) is
0≤ r < s ≤ n
Sol. We have, (1 + x + x 2 )n = a0 + a1x + a2x 2 + K + a2nx 2n
Putting x = 1on both sides, we get
a0 + a1 + a2 + K + a2n = 3n
Objective Mathematics Vol. 1
Hence, (a) is the correct answer.
(a) n ⋅ 2 2 n
(b) n ⋅ 2n
(c) n 2 ⋅ 22n (d) None of these Ex 35. Let n be a positive integer such that
(1 + x + x 2 ) n = a 0 + a 1 x + a 2 x 2 + K + a 2n x 2n ,
n n
Sol. We have, ∑ ∑ (r + s) (C r + C s )
r=0 s=0 then a r is
(a) a 2n , 0 ≤ r ≤ 2n (b) a 2n − r ,0 ≤ r ≤ 2n
n n
= ∑ ∑ (r C r + r C s + s C r + s C s ) (c) a n − r , 0 ≤ r ≤ 2n (d) None of these
r=0 s=0
n n n n n Sol. We have,
= ∑ ∑ r Cr + r ∑Cs + Cr ∑ s + ∑ s Cs 2n
r=0
s = 0 s=0 s=0 s=0 (1 + x + x 2 )n = ∑ ar x r ...(i)
r=0
n
n (n + 1)
= ∑
(n + 1) r ⋅ C r + r2n + C r + n ⋅ 2n − 1
1 1 2n
a
n
1
r=0
2 ⇒ 1 + + 2 = ∑ rr replacing x by
x x x
n (n + 1) n (n + 1) n r=0x
= (n + 1) (n ⋅ 2n − 1 ) + 2n + 2 2n
2 2
+ n 2n − 1 (n + 1)
⇒ (1 + x + x 2 )n = ∑ ar x 2n − r ...(ii)
r=0
= n (n + 1) 2n + n (n + 1) 2n = 2n (n + 1) 2n …(i) [multiplying both sides by x 2n]
n n
∑ ∑ (r + s) (C r + C s )
2n 2n
Also,
r=0 s=0
⇒ ∑ ar x r = ∑ ar x 2n − r [from Eqs. (i) and (ii)]
r=0 r=0
n
= ∑ 4 r C r + 2 ∑ ∑ (r + s) (C r + C s ) On equating the coefficient of x 2n − r both sides, we get
r=0 0≤r< s ≤ n a2n − r = ar , for 0 ≤ r ≤ 2n
∴ 2n (n + 1) 2n = 4 n ⋅ 2n − 1 + 2 ∑ ∑ (r + s) (C r + C s ) Hence, (b) is the correct answer.
0≤ r < s ≤ n
⇒ ∑ ∑ (r + s) (C r + C s ) = n 2
⋅2 n
Ex 36. Let n be a positive integer such that
0≤ r < s ≤ n (1 + x + x 2 ) n = a 0 + a 1 x + a 2 x 2 + K + a 2n x 2n ,
Hence, (a) is the correct answer.
then a 0 + a 1 + a 2 + K + a n − 1 is
Ex 33. If (1 + x ) n = C 0 + C 1 x + C 2 x 2 + K + C n x n , then 1 n 1 3n
(a) (3 + an ) (b) (3 − an )
the value of ∑ ∑ ( r + s) (C r + C s + C r C s ) is 2 2
1
0≤ r < s ≤ n (c) ( 3n − a n ) (d) None of these
n 2
(a) n 2 ⋅ 2n − [ 22n − 2nC n ]
2 Sol. We have, ar = a2n − r for 0 ≤ r ≤ 2n
n n− 1 n− 1
(b) n ⋅ 2 + [ 22n − 2nC n ]
2 n
2 ⇒ ∑ ar = ∑ a2n − r
n r=0 r=0
(c) n 2 ⋅ 2n + [ 2n + 2nC n ]
2 ⇒ a0 + a1 + K + an − 1 = a2n + a2n − 1 + K + an + 1
(d) None of the above 2n
⇒ 2 (a0 + a1 + K + an − 1 ) + an = 3n Q ∑ ar = 3n
Sol. We have, ∑ ∑ (r + s) (C r + C s + C rC s ) r = 0
0≤ r < s ≤ n 1 n
⇒ a0 + a1 + K + an − 1 = (3 − an )
=∑ ∑ (r + s) (C r + C s ) + ∑ ∑ (r + s) C rC s 2
0≤ r < s ≤ n 0≤ r < s ≤ n Hence, (c) is the correct answer.
n
= n ⋅ 2 + [ 22n − 2nC n ]
2 n
2 Ex 37. Let n be a positive integer such that
Hence, (b) is the correct answer. (1 + x + x 2 ) n = a 0 + a 1 x + a 2 x 2 + K + a 2n x 2n ,
Ex 34. Let n be a positive integer such that then ( n − r ) a r + (2n − r + 1) a r − 1 , 0 < r < 2n is
(1 + x + x 2 ) n = a 0 + a 1 x + a 2 x 2 + K + a 2n x 2n , (a) ( r + 1) a r + 1 (b) ( r − 1) a r + 1
(c) ( 2r + 1) a r + 1
2n
then ∑ a r is
(d) None of these
r=0 2n
(a) 3 n Sol. We have, (1 + x + x 2 )n = ∑ ar x r
r=0
(b) 3n − 1
Differentiating both sides w.r.t. x, we get
3n 2n
(c)
374
2 n (1 + x + x 2 )n − 1 (1 + 2x ) = ∑ r ar x r − 1
(d) None of the above r=0
8
2n
2n
⇒ n (1 + 2x ) (1 + x + x ) = (1 + x + x ) ∑ r ar x r − 1
2 n 2 Ex 40. If (1 + x + x 2 ) n = ∑ a r x r , then
r = 0 r=0
n
∑ (−1) r a r nC r = nC n/ 3 , where n is
Binomial Theorem
[multiplying both sides by (1 + x + x )]
2
r=0
2n 2n
⇒ n (1 + 2x ) ∑ ar x r = (1 + x + x 2 ) ∑ r ar x r − 1 (a) 3k + 1 (b) 3k + 2
r = 0 r = 0 (c) 3k (d) None of these
Equating the coefficients of x r (0 < r < 2n) on both Sol. We have,
sides, we get (1 + x + x 2 )n = a0 + a1x + a2x 2 + K + a2nx 2n ...(i)
nar + 2nar − 1 = (r + 1) ar + 1 + rar + (r − 1) ar − 1 and (x − 1)n = nC 0x n − nC 1x n − 1 + nC 2x n − 2
⇒ (r + 1) ar + 1 = (n − r) ar + (2n − r + 1) ar − 1 − K + (−1)n nC n ...(ii)
Hence, (a) is the correct answer. On multiplying Eqs. (i) and (ii), we get
2n − 2 (1 + x + x 2 )n (x − 1)n
Ex 38. The value of C 0 ⋅ C n − C 1 ⋅
n 2n
Cn n
2n − 4 = (a0 + a1x + a2x 2 + K + a2nx 2n )
+ C2 ⋅
n
C n − Kis
× {nC 0x n − nC 1x n − 1 + nC 2x n − 2 − K + (−1)n nC n}
(a) 3n ⇒ (x 3 − 1)n = (a0 + a1x + a2x 2 + K + a2nx 2n )
(b) 4 n × {nC 0x n − nC 1x n − 1 + K + (−1)n nC n} ...(iii)
(c) 2n
Now, coefficient of x n on RHS of Eq. (iii)
(d) None of the above
= a0 nC 0 − a1 nC 1 + a2 nC 2 − K + (−1)n annC n
Sol. We have,
2n − 2 2n − 4
LHS of Eq. (iii)
C 0 ⋅ 2nC n − nC 1 ⋅
n
C n + nC 2 ⋅ Cn n
2n − 2n = (x 3 − 1)n = (−1)n (1 − x 3 )n = (−1)n ∑ nC r (− x 3 )r
− K+ (−1) n n
Cn Cn r=0
2n − 2
= Coefficient of x in [ C 0 (1 + x ) − C 1 (1 + x )
n` n 2n n n
= (−1) n
∑ (−1) r n
C rx 3r
...(iv)
+ nC 2 (1 + x )2n − 4 − K + (−1)n nC n (1 + x )2n − 2n ] r=0
= Coefficient of x n in [(1 + x )2 − 1]n Clearly, if n is not a multiple of 3, then x n does not
occur in Eq. (iv).
= Coefficient of x n in (2n + x 2 )n
∴ Coefficient of x n in LHS = 0
= Coefficient of x 0 in (2 + x )n = 2n When n is not a multiple of 3.
Hence, (c) is the correct answer. If n is a multiple of 3 i.e. if n = 3m, then
LHS = (−1)3m (−1)m 3mC m
Ex 39. The value of n C 0 ⋅ 2nC r − nC 1 ⋅ 2n − 2C r
= 3m
C m = nC n/ 3 [Qn = 3m]
+ nC 2 ⋅ 2n − 4C r + ... + ( −1) n nC n 2n − 2nC n is n
Thus, equating the coefficients of x on both sides, we
22n − r nC r − n , if r > n get
(a)
if r < n a0nC 0 − a1nC 1 + a2nC 2 − a3nC 3 + K + (−1)n annC n
0,
32n − r nC r − n , if r > n 0, if n is not a multiple of 3
(b) = n
if r < n C n/ 3 , if n is multiple of 3
0,
Hence, (c) is the correct answer.
2 2n + r n
C r + n , if r > n
(c) np
0, if r < n Ex 41. If n ∈ N and (1 + x + x 2 + K + x p ) n = ∑ a r x r ,
r=0
(d) None of the above
if 0 ≤ r ≤ np and r is not a multiple of p +1, then
2n − 2 2n − 4
Sol. We have, n
C 0 ⋅ 2nC r − nC 1 ⋅ C r + nC 2 ⋅ Cr n
C 0 a r − nC 1 a r − 1 + nC 2 a r − 2 − K + ( −1) r a 0 nC r
2n − 2n
+ K+ (−1) n n
Cn Cn is
= Coefficient of x in [ C 0 (1 + x ) − C 1 (1 + x )
r n 2n n 2n − 2 (a) 3 n (b) 2n
(c) 0 (d) None of these
+ nC 2 (1 + x )2n − 4 − K + (−1)n nC n (1 + x )2n − 2n ]
Sol. We have,
= Coefficient of x r in [(1 + x )2 − 1]n (1 − x )n = nC 0 − nC 1x + nC 2x 2 − nC 3x 3 + K + (−1)n nC nx n
= Coefficient of x in (2x + x )
r 2 n
and (1 + x + x 2 + K + x p )n = (a0 + a1x + a2x 2
= Coefficient of x r − n in (2 + x )n + K + anp x np )
22n − r nC r − n , if r > n ∴ (1 − x )n (1 + x + x 2 + K + x p )n
= = (nC 0 − nC 1x + nC 2x 2 − nC 3x 3 + K + (−1)n nC nx n )
0, if r < n
Hence, (a) is the correct answer. × (a0 + a1x + a2x 2 + K + anp x np ) ...(i) 375
8 Now, coefficient of x r on RHS of Eq. (i)
= ar nC 0 − ar − 1nC 1 + ar − 2nC 2 − K + a0 (−1)r nC r
= (25 )3m + 1 = 215m + 5
= 23 (5m + 1) ⋅ 22= (23 )5m + 1 ⋅ 22 = (7 + 1)5m + 1 × 4
= {5m + 1C 0 75m + 1 + 5m + 1C 1 75m
Objective Mathematics Vol. 1
LHS of Eq. (i)
= (1 − x )n (1 + x + x 2 + K + x p ) +K+ 5m + 1
C 5m 7 + 5m + 1
C 5m + 1 70} × 4
1 − x p + 1 n
= (7n + 1) × 4 ,
= (1 − x )n
1− x where, n = 5m + 1C 0 75m + 1 + K + 5m + 1
C 5m 7
= (1 − x p + 1 )n = 28n + 4
( 32 )
Thus, when 32(32) is divided by 7, the remainder
Since, r is not a multiple of ( p + 1). Therefore, the
is 4.
expansion of (1 − x p + 1 )n does not contain x r in any
Hence, (c) is the correct answer.
term.
∴ Coefficient of x r on LHS of Eq. (i) = 0 Ex 43. The coefficient of x n y n in the expansion of
Hence, {(1 + x ) (1 + y) ( x + y)}n is
n
C 0ar − nC 1ar − 1 + nC 2ar − 2 − K + (−1)r nC r a0 = 0, n n
(a) ∑ C r2 (b) ∑ C r3
if r is not a multiple of ( p + 1). r =0 r =0
Hence, (c) is the correct answer. n
Ex 42. When 32 (32)
( 32)
is divided by 7, then the
(c) ∑ n
C r C s2
n
(d) None of these
r + s =0
remainder is
Sol. We have, {(1 + x ) (1 + y) (x + y)}n
(a) 2
= (C 0 + C 1x + C 2 x 2 + ....+ C n x n )
(b) 8
(c) 4 × (C 0 yn + C 1 yn − 1 + .... + C n − 1 y + C n )
(d) None of the above × (C 0 x n + C 1 x n − 1 y + ... + C n − 1xyn − 1C n yn )
Sol. We have, 3232 = (25 )32 = 2160 = (3 − 1)160 Clearly, coefficient of x n yn on RHS
n
= C 0 3160 − C 1 3159 + K − C 159 3 + = C 03 + C 13 + C 23 + K + C n3 = ∑ C r3
160 160 160 160
C 160 30
( 32 ) r =0
= 3m + 1, where m ∈ N 32(32) = (32)3m + 1
Hence, (b) is the correct answer.
Type 2. More than One Correct Option
(n − 1) (n − 2) K (n − m + 1)
Ex 44. The largest term in the expansion of (3 + 2x ) 50 , Sol.
(m − 1)!
1
where x = is (n − 1) (n − 2) K (n − m + 1) (n − m) K 2 ⋅ 1
5 =
(n − m)! (m − 1)!
(a) 5th (b) 51st (c) 7th (d) 6th n− 1
= Cm − 1
50
2x
= Coefficient of x m − 1 in (1 + x )n − 1
Sol. Consider, (3 + 2x )50 = 350 1 +
3
= Coefficient of x m − 1 in (1 + x )n (1 + x )−1
Let Tr + 1 be the largest term in the expansion.
Tr + 1 Now, (1 + x )n = C 0 + C 1x + C 2x 2
∴ ≥ 1 ⇒ 102 − 2r ≥ 15r ⇒ r ≤ 6 + K + C m − 1x m − 1 + K + C nx n ...(i)
Tr
(1 + x )−1 = 1 − x + x 2 − x 3 + K + (−1)m − 1 x m − 1 +…
Hence, (a) and (d) are the correct answers.
…(ii)
Ex 45. If (1 + x ) n = C 0 + C 1 x + C 2 x 2 +K+ C n x n , n ∈ N , Collecting the coefficients of x m − 1 in the product of
Eqs. (i) and (ii), we get
then C 0 − C 1 + C 2 −… + ( −1) m − 1 C m − 1
(−1)m − 1C 0 + (−1)m − 2C 1 + K + C m − 1
is equal to ( m < n)
( n − 1) ( n − 2) K ( n − m + 1) = Coefficient of x m − 1 in (1 + x )n − 1 = n − 1C m − 1
(a) ( −1) m − 1
( m − 1)! ∴ C 0 − C 1 + C 2 − K + (−1)m − 1C m − 1
n− 1
(b) C m − 1 ( −1) m − 1 = n− 1
C m − 1 (−1)m − 1
( n − 1) ( n − 2) K ( n − m ) (n − 1) (n − 2) K (n − m + 1)
(c) ( −1) m − 1 = (−1)m − 1
( m − 1)! (m − 1)!
n− 1
(d) C n − m ( −1) m − 1 Hence, (a), (b) and (d) are the correct answers.
376
Ex 46. For which of the following values of x, 5th
term is the numerically greatest term in the
expansion of (1 + x / 3)10 ?
Ex 48. The term independent of x in the expansion of
1
n
(1 + x ) n⋅ 1 − is
x
8
Binomial Theorem
(a) −2 (b) 1.8
(c) 2 (d) −1.9 (a) 0, if n is odd
n− 1
Sol. Let T5 be numerically the greatest term in the expansion (b) ( −1) 2 ⋅ nC n − 1 , if n is odd
of (1 + x / 3) .
10
2
Then,
T5
≥1 (c) ( −1) n/ 2 n
, C n/ 2 , if n is even
T4 (d) None of the above
T6
and ≤1 (1 − x 2 )n
T5 Sol. Expression = (−1)n ⋅
xn
Tr + 1 10 − r + 1 x
Now, = ⋅ ∴ The term independent of x
Tr r 3 = The coefficient of x n in (−1)n ⋅ (1 − x 2 )n
7 x
⇒ × ≥1 = The coefficient of x n in
4 3
6 x (−1)n {nC 0 + nC 1 (− x 2 ) + nC 2 (− x 2 )2 + K + nC n (− x 2 )n},
and ⋅ ≤1 and the expansion contains only even powers of x.
5 3
12 Hence, (a) and (c) are the correct answers.
⇒ |x| ≥
7
5 Ex 49. Let R = (8 + 3 7 ) 20 and [R ] = The greatest
and |x| ≤ ...(i) integer less than or equal to R. Then,
2
12 5 (a) [ R ] is even
⇒ ≤ |x| ≤
7 2 (b) [ R ] is odd
5 12 12 5 1
⇒ x∈ − ,− ∪ , (c) R − [ R ] = 1 −
2 7 7 2 ( 8 + 3 7 ) 20
Hence, (a), (b), (c) and (d) are the correct answers. (d) None of the above
Ex 47. The coefficient of a 8 b6 c 4 in the expansion of Sol. R = [ R ] + g = (8 + 3 7 )20
( a + b + c) 18 is = 20C 0 820 + C 1 819 (3 7 ) + ..., where 0 < g < 1
20
(a) 18
C 14 ⋅ C 8
14
(b) 18
C 10 ⋅ C610
f = (8 − 3 7 )20 = 20C 0 820 − 20C 1 819 (3 7 ) + ...,
(c) 18
C6 ⋅ C 8
12
(d) 18
C 4 ⋅ C614
where 0 < f < 1
∴ [ R ] + g + f = 2 [ 20C 0 820 + 20
C 2 818 (3 7 )2 + K ]
Sol. Given expression
= an even integer
= {a + (b + c)}18 = C 0a18 +
18
C 1a17 (b + c)
18
∴ g + f = an integer = 1as 0 < g < 1, 0 < f < 1
+K+ 18
C 10a8 (b + c)10 + K ∴ [ R ] = an even integer − 1 = an odd integer
18! Also, R − [ R ] = g = 1 − f
∴ The coefficient of a8 b6c4 = C 10 ⋅ 10C6 =
18
1
8! 6 ! 4 ! = (8 − 3 7 )20 = 1 −
(8 + 3 7 )20
Hence, (a), (b), (c) and (d) are the correct answers.
Hence, (b) and (c) are the correct answers.
Type 3. Assertion and Reason
Directions (Ex. Nos. 50-54) In the following Ex 50. Statement I Coefficient of a 2 b 3 c 4 in the
examples, each example contains Statement I 8!
expansion of ( a + b + c) 8 is .
(Assertion) and Statement II (Reason). Each example 2! 3 ! 4 !
has 4 choices (a), (b), (c) and (d) out of which only one is
Statement II Coefficient of a α b β c γ , where
correct. The choices are α + β + γ = n, in the expansion of ( a + b + c ) n is
(a) Statement I is true, Statement II is true; Statement II n!
is a correct explanation for Statement I .
α !β ! γ !
(b) Statement I is true, Statement II is true; Statement II
n!
is not a correct explanation for Statement I Sol. Q (a + b + c)n = ∑ ⋅ a p bq cr , p + q + r = n
p!q!r!
(c) Statement I is true, Statement II is false
In Statement I, p + q + r exceeds n.
(d) Statement I is false, Statement II is true Hence, (d) is the correct answer. 377
8 Ex 51. Statement I If q =
∑
15
1
3
1
and p + q =1, then
r 15C r p r q 15 − r = 15 × = 5
Ex 53. Statement I The
40
greatest value of
C 0 ⋅ 60C r + 40C 1 ⋅ 60C r − 1K 40C 40 ⋅ 60C r − 40 is
Objective Mathematics Vol. 1
100
r=0 3 C 50 .
Statement II If p + q = 1, 0 < p < 1, then Statement II The greatest value of 2n
C r occurs
at r = n.
n
∑ r nC r p r q n − r = np
r=0 Sol. The number of ways of selecting committee of r persons
n among 40 women and 60 men = 100C r . This will assume
Sol. ∑ r nC r pr qn − r greatest value at r = 50 .
r=0 Hence, (a) is the correct answer.
n
= np ∑ n − 1C r − 1 pr − 1qn − r = np (q + p)n − 1 = np Ex 54. Statement I If x = nC n − 1 + n + 1C n − 1
r=1
x +1
Hence, (d) is the correct answer. + K + 2nC n − 1 , then is an integer.
2n + 1
n+ 1
Ex 52. Statement I Coefficient of x 51 in the Statement II n
C r + nC r − 1 = Cr
expansion of ( x − 1) ( x 2 − 2) ( x 3 − 3) K n
and C r is divisible by n, if n and r are coprime.
( x 10 − 10) is −1. Sol. 1 + x = nC n + nC n − 1 + n+ 1
Cn − 1 + K + 2n
Cn − 1
n (n + 1)
−4
Statement II Coefficient of x 2 , n ≥ 4, in = 2n + 1C n
the expansion of Since, 2n + 1 and n are coprime for every natural
number n .
( x − 1) ( x 2 − 2) ( x 3 − 3) K ( x n − n ) is ∴ 2n + 1C n is divisible by 2n + 1 .
− 4 + ( −1) ( −3) = − 1 x+1
∴ is an integer.
2n + 1
Sol. Statement II is obviously true and it explains Statement I.
Hence, (a) is the correct answer.
Hence, (a) is the correct answer.
Type 4. Linked Comprehension Based Questions
1
6m
Passage I (Ex. Nos. 55-57) Consider the identity 3m
1 1
6m Sol. ∑ (−1)r 6mC 2r = 2 ( 2 )6m
2
+i
2
(1 + x ) 6m = ∑ 6m Cr ⋅ x r . By putting different values of x on r=0
r =0 6m
1 1
both sides, we can get summation of several series + ( 2 )6m −i
2 2
involving binomial coefficient. For example, By putting
6m 3mπ 0, if m is odd
x = , we get ∑ 6m Cr ⋅ r = .
6m
1 1 3 = 23m cos =
(−1) 2 , if m is even
m/ 2 3m
2 2
2 r =0 2
Hence, (b) is the correct answer.
6m
Ex 55. The value of ∑ 6m C r 2 r / 2 is equal to 3m
r=0
Ex 57. The value of ∑ (−3) r − 1 6m
C 2r − 1 is
6m r=1
3
(a) (a) 0
2
(b) 1
(b) (1 + 2 ) 3m
(c) depends on m
(c) ( 3 + 2 2 ) 3m
(d) None of the above
(d) None of the above
3m
Sol. ∑
6m
6m
Cr 2 r/ 2
put x = 2 = (1 + 2)6m
= (3 + 2 2 ) 3m Sol. ∑ (−3)r − 1 6mC 2r − 1
r=1
r=0
1
Hence, (c) is the correct answer. = { 3i 6m
C 1 + ( 3i )3 6m
C 3 + ( 3i )5 6mC 5
3i
3m
Ex 56. The value of ∑ (−1) r + K + ( 3i )6m − 1
6m 6m
C 2r is C6m − 1}
r=0
(1 + 3i )6m
= C0 +
6m
3i 6m
C 1 + ( 3i ) 2 6m
C2
(a) 23m + ( 3i )3 6m
C3 + K
(b) 0, if m is odd
(1 − 3i ) 6m
= C 0 − 3i
6m
C 1 + ( 3i )
6m 2 6m
C2
(c) −23m , if m is even
378 − ( 3i )3 6m
C3 + K
(d) None of the above
∴ (1 + 3i )6m − (1 − 3i )6m
∴Given expression
= 2 [ 3i 6mC 1 + ( 3i )3 6m
C3 + K ]
Ex 59. If n is a multiple of 3, then C 0 + C 3 + C6 +K is
equal to
2 +2 2 −2
8
Binomial Theorem
n n
1 (a) (b)
= [(1 + 3i )6m − (1 − 3i )6m ] 3 3
2 3i
26m
2n + 2 ( −1) n 2n − 2 ( −1) n
= (c) (d)
2 3i 3 3
(cos 2mπ + i sin 2mπ − cos 2mπ + i sin 2mπ )
Ex 60. Sum of values of x, which we should substitute
=0
in Eq. (i) to give the sum of the series
Hence, (a) is the correct answer.
C 0 + C 4 + C 8 + C 12 +K is
Passage II (Ex. Nos. 58-60) (a) 2
If (1 + x )n = C 0 + C1x + C 2x 2 + K + Cn x n …(i) (b) 2 (1 + i )
Then, sum of the series C 0 + Ck + C 2k + K can be (c) 2 (1 − i )
obtained by putting all the roots of the equation x k − 1 = 0 (d) 0
in Eq. (i) and then adding vertically. Sol. (Ex. Nos. 58-60) x 3 − 1
For example, Sum of the series C 0 + C 2 + C 4 + K can be x = 1, ω , ω 2 or x = ω , ω 2 , ω 3
obtained by putting roots of the equation x 2 − 1 = 0 x = 1, C 0 + C 1 + C 2 + C 3 + C 4 + K = 2n
⇒ x = ± 1 in Eq. (i) x = ω ,C 0 + C 1 ω + C 2 ω 2 + C 3 ω 3 + C 4ω 4 + K = (1 + ω )n
x = 1, C 0 + C1 + C 2 + C 3 + ... = 2n x = ω 2 , C 0 + C 1 ω 2 + C 2 ω 4 + C 3 ω6 + K = (1 + ω 2 )n
x = − 1, C 0 − C1 + C 2 − C 3 + K = 0 Now, 3 (C 0 + 0 + 0 + C 3 + 0 + 0 + C6 + ...)
2 (C 0 + C 2 + C 4 + ...) = 2n = 2n + (− ω 2 )n + (− ω )n = 2n + (−1)n + (−1)n
C 0 + C 2 + C 4 + K = 2n − 1
2n + 2 (−1)n
∴ C 0 + C 3 + C6 + K =
Ex 58. The values of x, we should substitute in Eq. (i) 3
to get the sum of the series x4 − 1 = 0
C 0 + C 3 + C6 + C 9 +K, are ⇒ x = ± 1, ± i
(a) 1, − 1 , ω (b) ω , ω 2 , ω 3 ∴ Sum of values of x = 1 + (−1) + i + (− i ) = 0
(c) ω , ω 2 , − 1 (d) None of these 58. (b) 59. (c) 60. (d)
Type 5. Match the Columns
Ex 61. Match the statement of Column I with values = Coefficient of x m in (mC 0 − [ mC 0 − mC 1 (1 + x )n
of Column II. + mC 2 (1 + x )2n − mC 3 (1 + x )3m
Column I Column II + K + (−1)m mC m (1 + x )mn ])
n 2n = Coefficient of x m in (1 − [1 − (1 + x )n ]m )
A. m
C1 C m − m
C2 Cm p. the coefficient of x m in
+ m 3n
C 3 C m... is the expansion of = Coefficient of x m in (1 − (1 + x )n )m
((1 + x )n − 1)m n− 1 n− 2
B. n
Cm + Cm + C m + K + mC m is the
n n −1 n−2 m
B. Cm + Cm + Cm q. the coefficient of x in coefficient of x m
in the series
+ K + m C m is (1 + x )n + 1 (1 + x )n + (1 + x )n − 1 (1 + x )n − 2 + K + (1 + x )m
x
= (1 + x )m [1 + (1 + x ) + (1 + x )2
C. C 0C n + C1C n − 1 r. the coefficient of x inn
+ K+ (1 + x )n − m ]
+ K + C nC 0 is (1 + x )2 n
1 − (1 + x )n − m + 1
k n k −1 n n −1 k = (1 + x )m
D. 2 C0 − 2 C1 C k − 1 s. the coefficient of x in 1 − (1 + x )
n−k the expansion of
K ( −1)k nC k C 0 is
(1 + x )n (1 + x )n + 1 − (1 + x )m
=
x
Sol. A. (mC 1nC m − mC 22nC m + mC 33nC m (1 + x )m
m−1 m
Q does not have x m,
− K (−1) Cm mn
Cm) x
= Coefficient of x in [ C 1 (1 + x ) − C 2 (1 + x )2n
m m n m ∴ nC m + n − 1C m + n − 2C m + K + mC m is the
+ mC 3 (1 + x )3n − K + (−1)m − 1 mC m ⋅ (1 + x )mn ] (1 + x )n + 1
coefficient of x m in the series ⋅
x
379
8 C. (1 + x )n = nC 0 + nC 1x + nC 2 x 2 + nC 3x 3
+ K + C nx ...(i)
(1 + x )n = nC 0 + nC 1x + nC 2x 2 + nC 3x 3
n n
C.
Column I
If the second term in the expansion
1
n
r.
Column II
divisible by 10
Objective Mathematics Vol. 1
a13 + a is14a 5 / 2 , then the value
+ K + n C nx n ...(ii)
a −1
On multiplying Eqs. (i) and (ii) and equate of n is
coefficient of x n both sides, we get D. The coefficient of x 4 in the expression s. a prime number
Coefficient of x n in (1 + x )n (1 + x )n = nC 0nC n (1 + 2 x + 3 x 2 + 4 x 3 + K ∞ )1/ 2 is
c, c ∈ N, then c + 1 (where| x | < 1) is
+ nC 1nC n− 1 + nC 2nC n − 2 + K + nC nnC 0
7 7 7 7 r
∴Coefficient of x n in (1 + x )2n − 1 − 2 K − r + 1 x
2 2 2 2
= C 0C n + C 1C n − 1 + K + C nC 0 Sol. A. Tr + 1 =
r!
D. 2 k n
C k = Coefficient of x k in (1 + 2x )n 7 9
First negative term, if − r + 1 < 0 i.e. r >
2k − 1 n− 1
C k − 1 = Coefficient of x k − 1 in 2 2
(1 + 2x )n − 1 Hence, r = 5
r
1
= Coefficient of x k in x (1 + 2x )n − 1 B. Tr + 1 = 5C r ( y2 )5 − r = 5C r y10 − 3r
y
Q nC 0 (1 + 2x )n − n C 1x (1 + 2x )n − 1
∴ 10 − 3r = 1
+ nC 2 x 2 (1 + 2x )n − 2 − .... = (1 + 2x − x )n ⇒ r=3
Hence, given expression is coefficient of x k in So, coefficient of y = 5C 3 = 10
(1 + x )n. C. T2 = 14 a5/ 2 ⇒ nC 1 (a1/ 13 )n − 1 (a3/ 2 )1 = 14 a5/ 2
A → p; B → q; C → r; D → s n −1
+
3
13 2
⇒ naa = 14 a5/ 2
Ex 62. Match the statement of Column I with values
⇒ n = 14
of Column II.
D. (1 + 2x + 3x + 4 x + K )1/ 2
2 3
Column I Column II
= [(1 − x )−2 ]1/ 2 = (1 − x )−1
A. If ( r + 1) th term is the first negative term p. divisible by 2
= 1 + x + x2 + K + xn + K
in the expansion of (1 + x )7/ 2 , then the
value of r (where| x | < 1) is Hence, coefficient of x 4 = 1
B. The coefficient of y in the expansion of q. divisible by 5 ∴ c=1
( y 2 + 1 / y)5 is Hence, c + 1 = 2
A → q, s; B → p, q, r; C → p; D → p, s
Type 6. Single Integer Answer Type Questions
Ex 63. If 6 83 + 8 83 is divided by 49, then the sum of the digits of remainder is ________ .
Sol. (8) (1 + 7)83 + (7 − 1)83 = (1 + 7)83 − (1 − 7)83
= 2 [ 83C 1 ⋅ 7 + 83
C 3 ⋅ 73 + ... + C 83 ⋅ 783 ]
83
= (2 ⋅ 7 ⋅ 83) + 49I, where I is an integer.
∴ 14 × 83 = 1162
1162 35
⇒ = 23
49 49
∴ Remainder is 35.
⇒ Sum of digits = 8
Ex 64. The remainder, when (15 23 + 23 23 ) is divided by 19, is ________ .
Sol. (0) E = (19 − 4 )23 + (19 + 4 )23
= 2 [1923 + C 2 ⋅ 1921 ⋅ 4 2 + K +
23
C 22 ⋅ 19 ⋅ 4 22 ]
23
= 2 ⋅ 19 [1922 + 23
C 2 ⋅ 1920 ⋅ 4 2 + K + C 22 ⋅ 4 22 ]
23
⇒ E is divisible by 19.
⇒ Remainder = 0
380
Ex 65. The term independent of x in the expansion of 9x −
1
18
, x > 0, is α times the corresponding binomial
3 x
8
Binomial Theorem
coefficient. Then, α is _________ .
(x )−r / 2
Sol. (1) Tr + 1 = 18C r ⋅ (9x )18 − r ⋅ (−1)r ⋅
3r
3r
(−1)r 18 −
Tr + 1 = 18
C r ⋅ 918 − r ⋅ ⋅x 2
3r
3r
∴ 18 − =0
2
⇒ r = 120
6
9
18
C 12 ⋅ 12 = α ⋅ 18C 12
3
∴ α =1
Ex 66. (1 + x ) (1 + x + x 2 ) (1 + x + x 2 + x 3 )...(1 + x + x 2 + K + x 100 ) when written in the ascending power of x, then
the highest exponent of x is λ (1010), where λ equals ___ .
Sol. (5) Highest exponent in the product of first two is
3 =1+ 2
Highest exponent in the product of first three is
6 =1+ 2 + 3
Similarly, highest exponent in the product of first hundred
= 1 + 2 + K + 100 = 5050 = 5 (1010)
⇒ λ=5
Ex 67. Given (1 − 2x + 5x 2 − 10x 3 ) (1 + x ) n = 1 + a 1 x + a 2 x 2 + K and that a 12 = 2a 2 , then the value of n is ________ .
Sol. (6) (1 − 2x + 5x 2 − 10x 3 ) [C 0 + C 1x + C 2x 2 + K ]
= 1 + a1x + a2x 2 + K
⇒ a1 = n − 2
n (n − 1)
and a2 = − 2n + 5
2
Now, a12 = 2a2. Therefore, we have
(n − 2)2 = n (n − 1) − 4 n + 10
⇒ n − 4 n + 4 = n2 − 5n + 10
2
∴ n=6
Ex 68. If (1 + x − 3x 2 ) 2145 = a 0 + a 1 x + a 2 x 2 + K , then a 0 − a 1 + a 2 − a 3 +K ends with _______ .
Sol. (3) Put x = − 1, (−3)2145 = a0 − a1 + a2 − a3 + ...;
− (3)2145 = − (34 )536 ⋅ 3
which ends with 3.
381
Target Exercises
Type 1. Only One Correct Option
20
9. The coefficient of the ( r + 1)th term of x +
1. The expansion 1
[ x + ( x 3 − 1)1/ 2 ] 5 + [ x − ( x 3 − 1)1/ 2 ] 5
is a x
polynomial of degree when expanded in the descending powers of x, is
(a) 5 (b) 6 equal to the coefficient of the 6th term of
(c) 7 (d) 8 10
2 1
2.The value of x, for which the 6th term in the expansion x + 2 + 2 when expanded in ascending
x
7
(9 x − 1 + 7) 1 powers of x. The value of r is
of 2log 2 + x− 1 is 84, is
2(1/ 5) log 2 (3 + 1) (a) 5
(b) 6
equal to (c) 14
(a) 4 (b) 3 (d) None of the above
(c) −2 (d) 1
10. Let n ∈ N . If (1 + x ) n = a 0 + a1 x + a 2 x 2 + ... + a n x n
3. The 10th term in the expansion of ( x − 1) 11
and a n − 3 , a n − 2 , a n − 1 are in AP, then
(in decreasing powers of x) is (a) a1 , a2 , a3 are in AP
(a) −x (b) −11x (b) a1 , a2 , a3 are in HP
Ta rg e t E x e rc is e s
(c) −x 2 (d) − 11C 9 x 2 (c) n = 17
(d) n = 14
4. If the third term in the expansion of [ x + x log 10 x ] 5 is 11. The number of terms with integral coefficients in the
106 , then x must be expansion of ( 71/ 3 + 51/ 2 ⋅ x ) 600 is
(a) 1 (b) 10 (a) 100 (b) 50
(c) 10 (d) 10−3/ 5 (c) 101 (d) None of these
5. If the 4th term in the expansion of ( px + x −1 ) m is 2.5 12. The sum of the rational terms in the expansion of
for all x ∈ R , then ( 2 + 5 10 )10 is
5
(a) p = , m = 3 (a) 32 (b) 9
2 (c) 41 (d) None of these
1
(b) p = , m = 6 6
2 3
1 13. The coefficient of x in x 5 +
3
is
(c) p = − , m = 6 x3
2
(d) None of the above (a) 0 (b) 120 (c) 420 (d) 540
n
6. If the coefficient of 4th term in the expansion of x
n 14. If the coefficients of x 7 and x 8 in 2 + are equal,
α 3
x + is 20, then the respective values of α and n
2x then n is equal to
are (a) 56 (b) 55
(a) 2, 7 (b) 5, 8 (c) 45 (d) 15
(c) 3, 6 (d) 2, 6 2n
7. If coefficients of 2nd, 3rd and 4th terms in the
15. If (1 + 2x + x 2 ) n = ∑ a r x r , then a r is equal to
r=0
expansion of (1 + x ) 2n are in AP, then
(a) (nC r )2 (b) nC r ⋅ nC r + 1
(a) 2n2 + 9n + 7 = 0 (b) 2n2 − 9n + 7 = 0 (c) 2nC r (d) 2nC r + 1
(c) 2n2 − 9n − 7 = 0 (d) 2n2 + 9n − 7 = 0
16. If in the expansion of (1 + x ) m (1 − x ) n , the
8. If the numerical coefficient of the pth term in the coefficients of x and x 2 are 3 and − 6 respectively,
expansion of ( 2x + 3) 6 is 4860, then the value(s ) of p then m is
is/are (a) 6 (b) 9
(a) 2 (b) 3 (c) 12 (d) 24
382 (c) 4 (d) 5
17. The coefficient of x 53 in
100
∑ 100 C r ( x − 3)100 − r 2r is
r=0
26. If f ( x ) = 1 − x + x 2 − x 3 + K − x15 + x16 − x17 .
Then, the coefficient of x 2 in f ( x − 1) is
(a) 826 (b) 816
8
Binomial Theorem
(a) 100
C 51 (b) 100
C 52 (c) − 100C 53 (d) 100
C 54 (c) 822 (d) None of these
18. If the coefficients of x 2 and x 3 in the expansion of 30 30 30 30
27. The value −
of
( 3 + kx ) 9 are equal, then the value of k is 0 10 1 11
(a) −9/7 (b) 9/7 30 30 30 30
(c) 7/9 (d) None of these + + K + is equal to
2 12 20 30
19. The coefficient of x 6 in (a) 60C 20 (b) 30
C 10
{(1 + x ) 6 + (1 + x ) 7 + ... + (1 + x )15 } is (c) 60C 30 (d) 40
C 30
(a) 16C 9
28. If the coefficient of x n in (1 + x )101 (1 − x + x 2 )100 is
(b) 16C 5 − 6C 5
non-zero, then n cannot be of the form
(c) 16C6 − 1
(a) 3r + 1 (b) 3r
(d) None of the above (c) 3r + 2 (d) None of these
20. The coefficient of x13 in the expansion of n
29. If the expansion of x 2 + for positive n has a
2
(1 − x ) 5 (1 + x + x 2 + x 3 ) 4 is x
(a) 4 (b) − 4 term independent of x, then n is
(c) 0 (d) None of these
(a) 23 (b) 18
21. The coefficient of x 6 ⋅ y −2 in the expansion of (c) 16 (d) 0
10
x2 y
12 x 3
− is 30. The term independent of x in + is
y x 3 2x 2
Targ e t E x e rc is e s
(a) 1
(a) 12C6 (b) − 12C 5
(b) 10C 1
(c) 0 (d) None of these
(c) 5/12
6 (d) None of these
22. Let f ( x ) = x 2 + 1 + x 2 − 1
31. The term independent of x in the expansion of
6
2 [( t −1 − 1) x + ( t −1 + 1) −1 x − 1 ] 8 is
+ . Then,
x 2 + 1 + x 2 − 1
3 3
1 − t 1 + t
(a) 56 (b) 56
1 + t 1 − t
(a) f (x ) is a polynomial of the fourth degree in x 4 4
(b) f (x ) has exactly two terms 1 − t 1 + t
(c) 70 (d) 70
(c) f (x ) is not a polynomial in x 1 + t 1 − t
(d) coefficient of x6 is 46
32. The term that is independent of x, in the expansion of
23. The coefficient of x 4 in the expansion of 3 2 1
9
(1 + x + x 2 + x 3 ) n is x − is
2 3x
(a) nC n (b) nC 4 + nC 2 5 4
3 1
(c) nC 4 + nC 1 + nC 4 − nC 2 (d) nC 4 + nC 2 + nC 1 ⋅ nC 2 (a) 9 C6 −
2 3
3
24. The coefficient of λ n µ n in the expansion of 1
(b) 9 C6
6
(1 + λ ) n (1 + µ ) n ( λ + µ ) n is 4 5
3 1
n n
(c) 9 C 4 −
(a) ∑ n
( Cr ) 2
(b) ∑ ( C r −2)
n 2
2 3
r=0 r=0 6 6
3 1
n n (d) 9 C6 −
(c) ∑ (nC r + 3 )2 (d) ∑ (nC r )3 2 3
r=0 r=0
33. Total number of terms, that are dependent on the
25. If in the expansion of (1 + x ) n , a, b and c are three n
1
consecutive coefficients, then n is equal to value of x, in the expansion of x 2 − 2 + 2 is
x
ac + ab + bc 2ac + ab + bc
(a) (b)
b2 + ac b2 − ac equal to
ab + ac (a) 2n + 1 (b) 2n
(c) 2 (d) None of these (c) n (d) n + 1
b − ac 383
8 34. The range of the values of the term independent of x
in the expansion of x sin −1 α +
cos −1 α
10
,
41. In the expansion of 3 4 +
1
4
6
20
Objective Mathematics Vol. 1
x (a) the number of rational terms = 4
α ∈ [ −1, 1] , is (b) the number of irrational terms = 18
(c) the middle term is irrational
10C 5π 10 10
C 5π 10 10C 5π 2 10
C 5π 2 (d) the number of irrational terms = 17
(a) − , (b) − ,
25 220 220 25
42. If n is an integer between 0and 21, then the minimum
(c) [1, 2 ] (d) (1, 2) value of n ! ( 21 − n ) ! is attained for n, is equal to
35. The term independent of x in the expansion of (a) 1 (b) 10
(c) 12 (d) 20
10
1
(1 − x ) 2 x + n
43. If the middle term in the expansion of x 2 + is
is 1
x
11
x
(a) C5 (b) 10C 5
10 924 x 6 , then n is equal to
(c) C4 (d) None of these
(a) 10 (b) 12
36. The term independent of x in the expansion of (c) 14 (d) None of these
1
n
44. Middle term in the expansion of (1 + 3x + 3x 2 + x 3 ) 6
(1 + x ) n ⋅ 1 − is
x is
(a) 0, if n is odd (a) 4th (b) 3rd
n− 1 (c) 10th (d) None of these
(b) (−1) 2 ⋅ nC n − 1, if n is odd
45. The coefficient of a8b4c9d 9 in
2
(c) (−1) . C n
n/ 2 n
, if n is even
( abc + abd + acd + bcd )10 is
−1 10!
Ta rg e t E x e rc is e s
2 (a) 10! (b)
(d) None of the above 8! 4 ! 9! 9!
(c) 2520 (d) None of these
37. The term independent of x in the expansion of
n 46. The coefficient of x 2 y 3 in the expansion of
1
(1 + x ) 1 + is
n
(1 − x + y ) 20 is
x
20! 20!
(a) C 02 + 2 C 12 + 3C 22 + ... + (n + 1) C n2 (a) (b) −
2! 3 ! 2! 3!
(b) (C 0 + C 1 + ...+ C n )2 20!
(c) C 02 + C 12 + ... + C n2 (c) (d) None of these
5! 2! 3 !
(d) None of the above
n 47. The coefficient of x 2 in the expansion of
1 t 7 from the 1st 1 (1 + x + x 2 )10 is
38. 6 3 2 + , = , n is equal to
3
3 t 7 from the last 6 (a) 10
(b) 10C 1
C2
(a) 9 (b) 10 (c) 11
C2 (d) None of these
(c) 8 (d) 7
39. If sum of middle terms is S in the expansion of 48. The coefficient of x 4 in the expansion of
9 (1 + x − 2x 2 ) 7 is
a2
2a − , then the value(s) of S is/are (a) − 81 (b) − 91
4 (c) 81 (d) 91
63 63 49. The total number of different terms in the expansion
(a) a14 (8 + a) (b) a13 (8 + a)
32 32 of ( x1 + x 2 + x 3 + ... + x n ) m is equal to
63 63 n+m n + m −1
(c) a14 (8 − a) (d) a13 (8 − a) (a) Cm (b) C n− 1
32 32
n + m+ 1 n + m+ 1
(c) Cm + 1 (d) Cn + 1
40. If the rth term is the middle term in the expansion of
20 50. The coefficient of a 3 b 4 c in the expansion of
2 1
x − , then the ( r + 3)th term is (1 + a − b + c ) 9 is equal to
2x
9! 9!
1 1 (a) (b)
(a) 20
C 14 ⋅ ⋅x (b) C 12 ⋅
20
⋅ x2 3! 6! 4 ! 5!
214 212
1 20 9! 9!
(c) − ⋅ C 13 ⋅ x (d) None of these (c) (d)
3!5! 3!4 !
213
384
51. The number of distinct terms in the expansion of
( x + 2 y − 3z + 5w − 7u ) n is
(a) n + 1 n− 4
62. The value of
32003
28
, where {}denotes the fractional 8
Binomial Theorem
(b) C4 part, is equal to
(c) n + 4C n + 2 (d) None of these 15 5 19 9
(a) (b) (c) (d)
28 28 28 28
52. If n is an even integer and a, b and c are distinct
numbers, then the number of distinct terms in the 63. p is prime number and n < p < 2n, if N = 2n
C n , then
expansion of ( a + b + c ) n + ( a + b − c ) n is (a) p divides N 2
(b) p divides N
2
n + 2 (c) p cannot divide N (d) None of these
(a) (b) n + 2
2 64. Let f ( n ) = 10n + 3⋅ 4 n + 2 + 5, n ∈ N . The greatest
n+ 4
(c) (d) None of these value of the integer which divides f ( n ) for all n is
2
(a) 27 (b) 9
53. In the expansion of ( x + y + z ) 25 (c) 3 (d) None of these
(a) every term is of the form 25C r⋅ rC k⋅ x 25− r⋅ yr − k⋅ zk 65. 260 when divided by 7 leave the remainder
(b) the coefficient of x 8 y9 z9 is 25C 8 8C 17 (a) 1 (b) 6 (c) 5 (d) 2
(c) the number of terms is 325
66. If sum of coefficients in the expression of
(d) None of the above
(αx 2 − 2x + 1) 35 is equal to sum of coefficients in the
54. The greatest coefficient in the expansion of (1 + x ) 2n expansion of ( x − αy ) 35 , then α is equal to
is (a) 0 (b) 1
1⋅ 3 ⋅ 5 ...(2n − 1) n
(a) ⋅2 (b) 2nC n − 1 (c) any real number (d) None of these
n!
2n
(c) C n + 1 (d) None of these 67. The sum of the coefficient of the binomial expansion
n
of + 2x is equal to 6561. The constant term in
1
Targ e t E x e rc is e s
55. The largest term in the expansion of ( 3 + 2x ) 50 , x
where x = 1/ 5, is the expansion is
(a) 15th (b) 51st (c) 7th (d) 6th (a) 8 C 4 (b) 16 ⋅ 8C 4
2n + 1 (c) 6C 4 ⋅ 24
56. If n is positive integer and ( 3 3 + 5) = α + β, (d) None of these
where α is an integer and 0 < β < 1 , then
68. The sum of the numerical coefficients in the
(a) α is an even integer 12
(b) (α + β )2 is divisible by 22n + 1 x 2 y
expansion of 1 + + is
(c) α is an odd integer 3 3
(d) None of the above
(a) 1 (b) 2
57. If x = ( 2 + 3 ) n , then the value of x − x 2 + x [ x] , (c) 212 (d) None of these
where [ ] denotes the greatest integer function, is 69. The sum of the coefficients x 2r , r = 1, 2, 3, ... , in the
equal to
expansion of (1 + x ) n is
(a) 1 (b) 2
(c) 22n (d) 2n (a) 2n (b) 2n − 1 − 1
(c) 2n − 1 (d) 2n − 1 + 1
58. The last digit of ( 2137) 754 is
(a) 2 (b) 3 70. The sum of the coefficients in the polynomial
(c) 7 (d) 9 expansion of (1 + x − 3x 2 ) 2163 is
(a) 1 (b) −1
59. The remainder when 337 is divided by 80 is
(c) 0 (d) None of these
(a) 78 (b) 3
(c) 2 (d) 35 71. If (1 + x − 2x 2 ) 8 = a 0 + a1 x + a 2 x 2 + ... + a16 x16 ,
60. The remainder when 4101 is divided by 101 is then the sum a1 + a 3 + a 5 + ...+ a15 is equal to
(a) 4 (b) 64 (a) −27 (b) 27
(c) 84 (d) 36 (c) 28 (d) None of these
61. The remainder obtained when 1! + 2 ! + 3! + ... + 95! 72. If (1 + x )10 = a 0 + a1 x + a 2 x 2 + ....+ a10 x10 , then
is divided by 15, is ( a 0 − a 2 + a 4 − a 6 + a 8 − a10 ) 2
(a) 3 + (a1 − a3 + a5 − a7 + a9 )2 is equal to
(b) 14
(c) 1 (a) 310 (b) 210
(d) None of the above (c) 29 (d) None of these
385
8 73. In the expansion of (1 + x ) 50 , let S be the sum of
coefficients of odd power of x, then S is
(b) 249 (c) 250 (d) 251
82. The sum of the series
n
∑ ( − 1)r − 1 . nC r
r =1
( a − r ) is
Objective Mathematics Vol. 1
(a) 0 equal to
74. If (1 + x − 2x 2 ) 20 = a 0 + a1 x + a 2 x 2 + a 3 x 3 (a) n ⋅ 2n − 1 + a (b) 0
(c) a (d) None of these
+ ... + a 40 x 40 , then a1 + a 3 + a 5 + ... + a 39 is equal
n
to 83. The sum ∑ r ⋅ 2n
C r is equal to
(a) − 2 19
(b) − 2 20
(c) − 2 21
(d) − 2 18
r =1
75. If n ∈ N and n is not a multiple of three and (a) n ⋅ 22n − 1 (b) 22n − 1
2n (c) 2n − 1 + 1 (d) None of these
(1+ x + x 2 ) n = ∑ a r x r , then the value of
r =0 84. If (1 − x + x 2 ) n = a 0 + a1 x + a 2 x 2 + ... + a 2n x 2n ,
n then a 0 + a 2 + a 4 + ... + a 2n is equal to
∑ ( −1)r ⋅ a r ⋅ n C r is equal to 3n + 1 3n − 1
r =0 (a) (b)
2 2
(a) 2 (b) −2 (c) 1 (d) None of these 1 1
(c) 3 −
n
(d) 3 +
n
76. If (1 + x ) 2n = a 0 + a1 x + a 2 x 2 + ... + a 2n x 2n , then 2 2
(a) a0 + a2 + a4 + .... = 2n 85. aC 0 + ( a + b ) C1 + ( a + 2b ) C 2 + ... + ( a + nb ) C n is
(b) an + 1 > an equal to
(c) an − 3 = an + 3
(a) (2a + nb) 2n (b) (2a + nb) 2n − 1
(d) None of the above
(c) (na + 2b) 2n (d) (na + 2b)2n − 1
n
77. The value of ∑ n C r sin rx ⋅ cos ( n − r ) x is equal to 86. If C1 , C 2 , ... , C n are coefficients in the expansion of
r=0
(1 + x ) n = 1 + C1 x + C 2 x 2 + ... + C n x n , then the value
Ta rg e t E x e rc is e s
(a) 2n − 1 ⋅ sin (nx ) (b) 2n + 1 ⋅ sin (nx )
(c) 2n ⋅ sin (nx ) (d) None of these of C12 − 2C 22 + 3 C 32 − ... − 2nC 22n is
n
(a) n2 (b) (− 1)n − 1 n
∑ nCr (c) 2 (− 1)n − 1 n ⋅ 2n − 1
78. The value of (sin rx ) is equal to Cn (d) − n2
r=0
x nx x nx
87. ( 10 C 0 ) 2 − ( 10 C1 ) 2 +... − ( 10 C 9 ) 2 + ( 10 C10 ) 2 equals
(a) 2 ⋅ cos
n
⋅ sin
n
(b) 2 ⋅ sin
n n
⋅ cos
2 2 2 2 (a) 10
C5 (b) − 10C 5 (c) (10C 5 )2 (d) (n!)2
x nx x nx
(c) 2n + 1 ⋅ cosn ⋅ sin (d) 2n + 1 ⋅ sin n ⋅ cos 88. The sum to ( n + 1) terms of the series
2 2 2 2
C 0 C1 C 2 C 3
n
r ⋅ nC r − + − + ... is
79. The value of ∑ nC is equal to 2 3 4 5
r =1 r −1 1 1
(a) (b)
n (n + 1) n (n − 1) n+1 n+ 2
(a) (b) 1
2 2 (c) (d) None of these
(c) n (n + 1) (d) n (n − 1) n (n + 1)
2n
1
80. The value of ∑ r ⋅ ( 2 n C r ) ⋅ ( r + 2) is equal to 89. If C 0 , C1 , C 2 , ... , C n denote the coefficients in the
r=0 expansion of (1 + x ) n , then the value of
22n + 1 (2n2 + 2n − 1) C 1 + 2C 2 + 3C 3 + ... + nC n is
(a)
(2n + 1) (2n + 2) (a) n 2n − 1 (b) (n + 2) 2n
2n + 1
2 (2n2 − n + 1) − 2 (c) (n + 1)2n − 1 (d) (n + 2) 2n − 1
(b)
(2n + 1) (2n + 2)
22n + 1 (2n2 + 2n + 1)
90. The value of
(c)
(2n + 1) (2n + 2)
C 0 + 3C1 + 5C 2 + 7C 3 + ... + ( 2n + 1) C n is equal to
(d) None of the above (a) 2n (b) 2n + n 2n − 1
(c) 2n (n + 1) (d) None of these
81. In ∆ABC, the value of
n 50
∑ nCr a r ⋅ b n − r ⋅ cos {rB − ( n − r ) A} is equal to 91. The value of ∑ (100 C r ⋅ 200C150 + r ) is equal to
r=0 r=0
(a) c n
(b) zero (a) 300
C 50 (b) 100C 50 ⋅ 200C 150
(c) an (d) bn (c) 100
C 50 (d) None of these
386
92. The value of
2n
∑ ( −1 )r ⋅ ( 2n C r )2 is equal to
r=0
102. ∑ ∑ n C i is equal to
0 ≤i < j ≤n
n− 1
(b) (n + 1) 2 n− 1
8
Binomial Theorem
(a) n 2
(a) 4nC 2n (b) 2nC n
(c) (n + 1) 2n (d) n 2n
(c) (− 1)n ⋅ 2nC n (d) None of these
2n 103. The value of ∑ ∑ ∑ ∑ 2 is equal to
93. The value of ∑ r ⋅ ( 2n C r )2 is equal to 0 ≤i < j < k < l ≤ n
r=0
(a) 2 (n + 1)3 (b) 2 n + 1C 4
(a) n ⋅ C 2n
4n
(b) 4 n ⋅ C 2n4n
(c) 2 (n + 1)4 (d) None of these
(c) 2n ⋅ 4nC 2n (d) None of these
104. The value of ∑ ∑ ( i ⋅ j )n C i ⋅ nC j is equal to
n− r
94. The value of ∑ n− i
C r , r ∈ [1, n] is equal to 1 ≤i < j ≤n − 1
2
i=0 n
(a) (22n − 2 − 2n − 2C n − 1 )
n+ 1 n n n+ 1 2
(a) Cr (b) C r (c) C r + 1 (d) Cr + 1
n2 2n − 2 2n − 2
(b) (2 + Cn − 1)
95. If C 0 , C1 , C 2 ,... , C n are the coefficients of the 2
n (c) n2 (22n − 2 + 2n − 2C n − 1 )
Ck
expansion of (1 + x ) n , then the value of ∑ is (d) n2 (22n − 2 − 2n − 2
Cn − 1)
k =0 k +1
2n − 1 n C 0 + nC1 + nC 2 + ... + n C n
∞
(a) 0 (b)
n
105. If S = ∑ n
Pn
, then S
n=1
2n + 1 − 1
(c) (d) None of these is equal to
n+1
(a) 2e (b) 2e − 1
(c) 2e + 1 (d) None of these
96. If (1 + x ) 2n
= a 0 + a1 x + a 2 x + ... + a 2n x , then
2 2n
Targ e t E x e rc is e s
n
a1 a 2 a 3 a 2n 106. The value of ∑ r ( n − r ) ⋅ ( n C r )2 is equal to
1 + 1 + 1 + ... 1 + is equal to
a0 a1 a 2 a 2n − 1 r=0
2n − 1
(a) n2 ⋅ Cn (b) n2 ⋅ 2nC n − 1
nn (2n + 1)2n
(a) (b) 2n − 1 2n − 2
(2n)! (2n)! (c) n2 ⋅ Cn− 1 (d) n2 ⋅ Cn
nn + 1 n+1 n−1
(c) (d) None of these 107. If C r + 1 : nC r : C r − 1 = 11: 6 : 3, then nr is
(2n + 1) !
equal to
97. The value of ∑ ∑ i ⋅ nC j is equal to (a) 20 (b) 30 (c) 40 (d) 50
0≤ i < j ≤ n
f 1 (1) f 2 (1)
(a) n (n + 1) ⋅ 2n − 3 (b) n2 ⋅ 2n − 3 108. If f ( x ) = x n , then the value of f (1) + +
(c) n (n − 1) ⋅ 2n − 3 (d) None of these 1! 2!
f n (1)
98. The value of ∑ +K+ , where f r ( x ) denotes the r th order
∑ j⋅ n
C j is equal to n!
0 ≤i ≤ j ≤n
derivative of f ( x ) with respect to x, is
(a) n ⋅ 2n − 3 (b) n (n + 3) ⋅ 2n − 3 (a) n (b) 2n
(c) (n + 3) ⋅ 2n − 3 (d) None of these (c) 2n − 1 (d) None of these
n
99. The value of ∑ ( − 1)r ⋅ r 2 ⋅ nC r is equal to 109. The value of ∑ ∑
∞ k
1
( k C r ) is
r=0 k
k =1 r = 0 3
(a) n (b) − n
(c) (n + 1) (d) None of these 2 4
(a) (b) (c) 2 (d) 1
3 3
100. If (1 + x ) = C 0 + C1 x + C 2 x + ... + C n x ,
n 2 n
then
1 1× 4 2 1× 4 × 7 3
110. 1 + x+ x + x + ... is equal to
∑ ∑ (C i − C j ) 2
3 3× 6 3× 6× 9
0 ≤i < j ≤n
(a) x (b) (1 + x )1/ 3 (c) (1 − x )1/ 3 (d) (1 − x )−1/ 3
(a) (n + 1) 2nC n − 22n (b) (n + 1)2 × 2nC n − 2n
(c) (n + 1) × 2nC n + 2n (d) None of these 5 5⋅ 7 5⋅ 7⋅ 9
111. If α = + 2
+ 3
+ ... , then α 2 + 4α is
2! 3 3! 3 4!3
101. If A = 2nC 0 2nC1 + 2n
C1 2 n − 1 C1 + 2n
C 2 2n − 2C1 + ... ,
then A is equal to
(a) 0 (b) 2n (c) n ⋅ 22n (d) 1 (a) 21 (b) 23 (c) 25 (d) 27 387
8 Type 2. More than One Correct Option
112. For the expansion ( x sin p + x −1 cos p )10 , ( p ∈ R )
n
118. If f ( n ) = ∑ [ r ( n n − 1C r − 1 − r nC r −1 ) + ( 2r + 1)
Objective Mathematics Vol. 1
(a) the greatest value of the term independent of x is r =1
10!/ 25 (5!)2 n
C r ], then
(b) the least value of sum of coefficient is zero
(c) the greatest value of sum of coefficient is 32 (a) f (n) = n2 − 1
(d) the least value of the term independent of x occurs
π (b) f (n) = (n + 1)2 − 1
when p = (2n + 1) , n ∈ Z 10
4
a
n (c) ∑f (n) = 495
113. In the expansion of x + 2 ( a ≠ 0), if term n= 1
x 10
independent of x does not exist, then n must be (d) ∑f (n) = 374
n= 1
(a) 20 (b) 16 (c) 15 (d) 10
n+1 n+2 n+k
n n 119. The value of n
C0 + C1 + C 2 + ... + Ck
α β
114. If the expansion of x + and x + 2 in is equal to
x x (a) n+ k+ 1
Ck (b) n+ k+ 1
Cn + 1
powers of n have one term independent of x, then n is n+ k
(c) Cn + 1 (d) None of these
divisible by
(a) 2 (b) 3 120. Number of values of r satisfying the equation
(c) 1 (d) None of these 69
C 3r − 1 − 69C r 2 = 69C r 2 − 1 − 69C 3r is
115. If n is a positive integer, then in the trinomial (a) 1 (b) 2 (c) 3 (d) 7
expansion of ( x 2 + 2x + 2) n coefficient of
121. Suppose x1 , x 2 ,... , x n ( n > 2) are real numbers such
(a) x is 2n ⋅ n (b) x 2 is n2 ⋅ 2n − 1 that x i = − x n − i + 1 for 1≤ i ≤ n. Consider the sum
Ta rg e t E x e rc is e s
(c) x 3 is 2n ⋅ n + 1C 3 (d) None of these S = ΣΣΣx i x j x k (1≤ i, j, k ≤ n ) ( i, j, k distinct), then
116. Which of the following will not be true? which of the following is not true?
(a) S = n! x1 , x2 , ..., xn
(a) The last two digits of 3100 will be 73
(b) S = (n − 3) (n − 4 )
(b) The last two digits of 350 will be 51 (c) S = (n − 3) (n − 4 ) (n − 5)
(c) The last two digits of 350 will be 49 (d) S = 0 for all n
(d) The last three digits of 350 will be 249
m
30 20 p
117. If ac > b 2 , then the sum of the coefficients in the 122. If f ( m) = ∑ 30 − i m − i , where = pC q ,
q
i=0
expansion of ( aα 2 x 2 + 2bαx + c ) n , where
then
a, b, c, α ∈ R and n ∈ N , (a) maximum value of f (m) is 50C 25
(a) positive, if a > 0 (b) f (0) + f (1) + K + f (50) = 250
(b) positive, if c > 0
(c) negative, if a < 0 and n is odd (c) f (m) is always divisible by 50 (1 ≤ m ≤ 49)
50
(d) positive, if c < 0 and n is even (d) The value of ∑ ( f (m))2 = 100C 50
m=0
Type 3. Assertion and Reason
Directions (Q. Nos.123-128) In the following 123. Statement I Greatest term in the expansion of
questions, each question contains Statement I (1 + x )12 , when x = 11/ 10 is 7th.
(Assertion) and Statement II (Reason). Each question
has 4 choices (a),(b),(c) and (d) out of which only one is Statement II 7th term in the expansion of (1 + x )12
correct. The choices are has the factor 12 C 6 which is greatest value of 12 C r .
(a) Statement I is true, Statement II is true; Statement II
is a correct explanation for Statement I 124. Statement I If n is an odd prime, then integral part
(b) Statement I is true, Statement II is true; Statement II of ( 5 + 2) n − 2n + 1 is divisible by 20n.
is not a correct explanation for Statement I
Statement II If n is prime, then
(c) Statement I is true, Statement II is false n
C1 , nC 2 , nC 3 , K , nC n − 1 must be divisible by n.
(d) Statement I is false, Statement II is true
388
125. Statement I Any positive integral power of
( 2 − 1) can be expressed as N − N − 1 for some
127. Statement I
( − 1)
If S n = nC1 −
n n−1
C2
2
+
nn
C3
3 8
Binomial Theorem
Cn 1 1 1
natural number N > 1. − ... + and Tn = 1 + + + ... + ,
n 2 3 n
Statement II Any positive integral power of 2 − 1 then S n = Tn for all n.
can be expressed as A + B 2, where A and B are
Statement II S n + 1 − S n = Tn + 1 − Tn
integers.
126. Two integers n and m are such that m = n 2 − n, where i j
128. Statement I ∑ ∑
nC
+ n is equal to
C j
n ≥ 3. 0≤i < j ≤n i
Statement I The number m ( m − 2) is divisible by n2 n
1
a, where a = ∑ n
24. 2 r = 0 Cr
Statement II Product of r successive integers are n
r n
n−r
divisible by r!.
Statement II ∑ nC = ∑ n
Cr
r=0 r r=0
Type 4. Linked Comprehension Based Questions
Passage I (Q. Nos.129-131) Consider following two 134. C 0 − C 2 + C 4 − C 6 + .... is equal to
infinite series in real r and θ (a) 2n/ 2 (b) 2n/ 2 (1 + i )
r 2 cos 2θ r 3 cos 3θ (c) 2n/ 2 cos π / 4 (d) None of these
C = 1 + r cos θ + + + ...
2! 3! n n
1
S = r sin θ +
r 2 sin 2θ r 3 sin 3θ
+ + ...
135. If (1 + x ) n = ∑ nCr xr and ∑ nC = a, then the
2! 3! r=0 r=0 r
Targ e t E x e rc is e s
If θ remains constant and r varies, then i j
value of ∑ ∑
nC
+ n
is equal to
dC dS 0 ≤i ≤n 0 ≤ j ≤n C j
129.The expression C +S is equal to i
dr dr n2
(a) C + S 2 2
(b) (C + S ) cosθ
2 2 (a) n2a (b)
2a
(c) (C 2 + S 2 ) sin 2 θ (d) 1 n
(c) a (d) None of these
2
dS dC n
−S
130.The expression C
dr dr
is equal to Passage III (Q. Nos. 136-138) If (1 + x )n = ∑ n Cr x r ,
r =0
(a) C 2 + S 2 (b) (C 2 + S 2 ) cosθ
then the sum of the binomial coefficients can be
(c) (C 2 + S 2 ) sin θ (d) CS obtained by substituting x = 1. But in case we have to
2 2 find the sum of coefficients in some particular order, we
dC dS can substitute x by ± ix or ± ω x or ± ω 2 x depending
131. + is equal to
dr dr upon certain requirements.
(a) C 2 + S 2 (b) (C 2 + S 2 ) cos2 θ 136. The sum of binomial coefficients
(c) (C 2 + S 2 ) sin 2 θ (d) 1 n
C 0 + nC 4 + nC 8 + ... must be
Passage II (Q. Nos. 132-135) π nπ
(a) 2n/ 2 cos (b) 2n/ 2 sin
n 8 8
If ( x + y )n = ∑ Cr x n − r y r , where Cr = nCr .
n
nπ −1 nπ
r =0
(c) 2n + 2n/ 2 cos (d) 2n − 2 + 2 2 cos
4 4
C1 C 3 C 5 C 7
132. + + + + ... is equal to 137. The sum of the binomial coefficients
2 4 6 8 n
C 0 + nC 3 + nC 6 + ... must be
2n − 1 2n + 1 2n
(a) (b) (a)
n+1 n+1 3
2n + 1 − 1 1 n nπ
(c) (d) None of these (b) 2 + cos
n+1 3 3
1 nπ
n
r + 2 28 − 1 (c) 2n + 2 cos
133. If ∑ C r = , then n is equal to 3 3
r = 0 r + 1 6 1 n nπ
(d) 2 + 2 sin
(a) 8 (b) 6 (c) 5 (d) 4 3 3
389
8 138. If n is an even positive integer and k =
n
3n
2
, then the 140. The value of
s n
∑ ∑ n C s ⋅ sC r (when r ≤ s) is
r = 0 s=1
∑ ( − 3) r − 1
Objective Mathematics Vol. 1
3n
value of C 2r − 1 is (a) 3n
r =1 (b) n 3n − 1
(a) 3 n
(b) 3n − 1 (c) 6n (d) zero (c) n 3n − 1 − 1
(d) 3n − 1
Passage IV (Q. Nos. 139-141) Let n be positive
integer such that n n n n
(1 + x )n = C 0 + C1x + C 2 x 2 + K + Cn x n . 141. The value of ∑ ∑ ∑ ∑ (1) is
r =0 s=0 t =0 u=0
n n
139. The value of ∑ ∑ (C r + C s ) is (a) nC 4
(b) n + 1C 4
r =0 s=0
(a) (n + 1) 2n + 1 (b) n 2n (c) 4n + 1C 4
(c) (n + 1) 2n (d) n 2n − 1 (d) (n + 1)4
Type 5. Match the Columns
142. Consider an expression f ( x ) = x n + x n + 1 , n ∈ N . 144. Match the following.
f ( x ) is differentiated successively an arbitrary Column I Column II
number of times, then multiplied by ( x + 1) and again A. The last non-zero digit in the sum p. 2
differentiated successively till it attains the form of 2 ∑ ∑ i ⋅ j 10C i 10C k is
Ax + B. It is found that A − B is always divisible by a 0≤i ≤ j ≤10
proper integer λ which depends on n. Now, in B. The largest real value for x such that q. 4
4
3 4 − k x k 32
Ta rg e t E x e rc is e s
Column I different values of n are given and in ∑ (4 − k ) ! k ! = 3
Column II different values of λ are given, match the k=0
corresponding values of n and λ . C. If the expansion of (1 + x + x 2 )n be written r. 1
as a0 + a1 x + a2 x 2 + K + a2n x 2n,
Column I Column II then the value of
a0 + a1 + a3 + a4 + a6 + a7 + ...
A. 5 p. 15 is
a2 + a5 + a8 + ...
B. 7 q. 81
D. The value of s. 3
47C 5 56 − k C
53 − k
C. 9 r. 49 3 50 − j
C3
57 4 + ∑ 57 + ∑ 57 is
D. 13 s. 91 C 4 j=0 C 53 j=0 C4
143. Match the following. 145. Match the statements of Column I with values of
Column I Column II Column II.
A. If x, y, z ∈ N, then number of p. 19 Column I Column II
ordered triplet ( x, y, z) satisfying
xyz = 243 is A. 32
C 02 − 32
C12 + 32
C 22 −K+ 32 2
C 32 = p. 63
C 32
B. The number of terms in the q. 28 B. 32
C 02 + 32
C12 + 32
C 22 + K + 32 2
C 32 = q. 32
C16
expansion of ( x + y + z)6 is
1 r. 0
C. (1 × 32C12 + 2 × 32C 22
C. If x ∈ N, then number of solutions of r. 21 32
x 2 + x − 400 ≤ 0 is + K + 32 × 32C 32
2
)=
D. If x, y, z ∈ N, then number of s. 36
D. 31
C 02 − C12 +
31
C 22 − K −
31 31 2
C 31 = s. 64
C 32
solutions of x + y + z = 10 is
390
Type 6 Single Integer Answer Type Questions
1
151. If the middle term in the expansion of + 2
x
8
8
Binomial Theorem
146. Let a = 3 223 + 1, for all n ≥ 3 and let is
n−1
2
f (x ) = C 0 a
n
− nC1 a n − 2 + nC 2 a n − 3 − ...
1120, then the sum of possible real values of x is ___.
+ ( −1) n − 1 nC n − 1 a 0 .
If the value of f ( 2007) + f ( 2008) = 3k , where k ∈ N , 152. The number formed by last two digits of the number
then the value of k is _______ . (17) 256 is ab, then ( a + b ) is _______ .
147. The sum of roots of x for which the sixth term of
m
153. The remainder, if 1 + 2 + 22 + 23 + ... + 21999 is
log (10 − 3x ) (x − 2) log 3
2 + 5 2 is equal to 21 and divided by 5, is _______ .
binomial coefficients of second, third and fourth 154. The largest real value for x = 2 2 − λ such that
terms are the first, third and fifth terms of an 4
54 − k x k 8
arithmetic progression, is _______ . ∑ = , then λ is _______ .
(4 − k )! k ! 3
k=0
148. The fractional part of the sum of all the rational terms
in the expansion of ( 31/ 4 + 41/ 3 )12 , is _______ . 155. The value of
149. The coefficient of x in (1 + x + x + x + x )
83 2 3 4 n (1 + x ) (1 + 2x )
C 0 − C1 + C2 ⋅ −
( x − 1) n + 3 , is − nC 2λ , then λ is _______ . (1 + nx ) (1 + nx ) 2
150. The coefficient of x 4 in the expansion of (1 + 3x )
(1 + x + x 2 + x 3 )11 is λ. Then, the number of divisors C3 ⋅ + K = 1, then λ is ….
(1 + nx ) 3
of λ of the form 9 k, is _______ .
Targ e t E x e rc is e s
Entrances Gallery
JEE Advanced/IIT JEE
1. Coefficient of x11 in the expansion of 3. For r = 0, 1, ... , 10, let A r , B r and C r denote,
(1 + x 2 ) 4 (1 + x 3 ) 7 (1 + x 4 )12 is [2014] respectively the coefficient of x r in the expansions of
(a) 1051 (b) 1106 (1 + x )10 , (1 + x ) 20 and (1 + x 30 ). Then,
(c) 1113 (d) 1120 10
∑ A r ( B10 B r − C10 A r ) is equal to [2010]
2. The coefficients of three consecutive terms of r =1
(1 + x ) n+ 5 are in the ratio 5 : 10 : 14. Then, n is…. (a) B10 − C 10 2
(b) A10 ( B10 − C 10 A10 )
[2013] (c) 0 (d) C 10 − B10
JEE Main/AIEEE
4. The sum of coefficients of integral powers of x in the 6. If X = {4 n − 3n − 1: n ∈ N } and Y = {9 ( n − 1) : n ∈ N },
binomial expansion of (1 − 2 x ) 50 is [2015] where N is the set of natural numbers, then
(a)
1 50
(3 + 1)
1
(b) (350 ) X ∪ Y is equal to [2014]
2 2 (a) N
1
(c) (350 − 1)
1
(d) (250 + 1) (b) Y − X
2 2 (c) X
(d) Y
5. If the coefficients of x 3 and x 4 in the expansion of
(1 + ax + bx 2 ) (1 − 2x )18 in powers of x are both zero, 7. The term independent of x in expansion of
10
then ( a, b ) is equal to [2014] x+1 x−1
2/ 3 − is [2014]
251 251 x − x1 / 3 + 1 x − x1 / 2
(a) 16, (b) 14 ,
3 3
(a) 4 (b) 120
272 272
(c) 14 , (d) 16, (c) 210 (d) 310
3 3
391
8 8. If n is a positive integer, then ( 3 + 1) 2n − ( 3 − 1) 2n
is [2012]
15. If the expansion in powers of x of the function
1
(1 − ax ) (1 − bx )
is a 0 + a1 x + a 2 x 2 + a 3 x 3 + ... , then
Objective Mathematics Vol. 1
(a) an irrational number
(b) an odd positive integer a n is [2006]
(c) an even positive integer an − bn an + 1 − bn + 1
(d) a rational number other than positive integers (a) (b)
b−a b−a
9. The coefficient of x 7 in the expansion of bn + 1 − a n + 1 bn − an
(c) (d)
(1 − x − x 2 + x 3 ) 6 is [2011] b−a b−a
(a) −132 (b) −144 16. For natural numbers m, n, if
(c) 132 (d) 144
10 10 (1 − y ) m (1 + y ) n = 1 + a1 y + a 2 y 2 + K and
10. Let S 1 = ∑ j( j −1) 10
C j ,S 2 = ∑ j 10
Cj a1 = a 2 = 10, then ( m, n ) is [2006]
j =1 j =1 (a) (35, 20)
10 (b) (45, 35)
and S 3 = ∑ j 2 10C j (c) (35, 45)
j =1 (d) (20, 45)
Statement I S 3 = 55 × 2 9 6
Statement II S 1 = 90 × 28 and S 2 = 10 × 29 [2010]
17. The value of 50
C4 + ∑ 56 − r C 3 is [2005]
r =1
(a) Statement I is false, Statement II is true (a) 56
C4 (b) 56
C3
(b) Statement I is true, Statement II is true; Statement II is 55 55
(c) C3 (d) C4
a correct explanation for Statement I
(c) Statement I is true, Statement II is true; Statement II is 18. If the coefficients of r th, ( r + 1) th and ( r + 2) th terms
not a correct explanation for Statement I
(d) Statement I is true, Statement II is false in the binomial expansion of (1 + y ) m are in AP, then
Ta rg e t E x e rc is e s
m and r satisfy the equation [2005]
11. The remainder left out when 82n − ( 62) 2n + 1 is
(a) m2 − m (4 r − 1) + 4 r2 + 2 = 0
divided by 9, is [2009]
(a) 0 (b) 2 (b) m2 − m (4 r + 1) + 4 r2 − 2 = 0
(c) 7 (d) 8 (c) m2 − m (4 r + 1) + 4 r2 + 2 = 0
n (d) m2 − m (4 r − 1) + 4 r2 − 2 = 0
12. Statement I ∑ ( r + 1) nC r = ( n + 2) 2n − 1 11
r=0 1
19. If the coefficient of x 7 in ax 2 + equals the
n bx
Statement II ∑ ( r + 1) nC r x r = (1 + x ) n
1
11
r=0 coefficient of x −7 in ax − 2 , then a and b
bx
+ nx (1 + x ) n − 1 [2008]
satisfy the relation [2005]
(a) Statement I is false, Statement II is true a
(b) Statement I is true, Statement II is true; Statement II is (a) ab = 1 (b) = 1
a correct explanation for Statement I b
(c) Statement I is true, Statement II is true; Statement II is (c) a + b = 1 (d) a − b = 1
not a correct explanation for Statement I
20. If x is so small that x 3 and higher powers of x may be
(d) Statement I is true, Statement II is false 3
1
13. In the binomial expansion of ( a − b ) n , n ≥ 5, the sum (1 + x ) 3/ 2 − 1 + x
2
a neglected, then may be
of 5th and 6th terms is zero, then equals [2007] (1 − x )1/ 2
b
5 6
approximated as [2005]
(a) (b) x 3 2 3 2
n−4 n−5 (a) − x (b) − x
2 8 8
n−5 n−4 3 3
(c) (d) (c) 3x + x 2 (d) 1 − x 2
6 5 8 8
14. The sum of the series 21. The coefficient of the middle term in the binomial
20
C 0 − 20C 1 + C 2 − 20C 3 + K +
20 20
C 10 is [2007] expansion in powers of x of (1 + αx ) 4 and of (1 − αx ) 6
(a) − 20C 10 is the same, if α equals [2004]
1 5 10
(b) 20C 10 (a) − (b)
2 3 3
(c) 0 3 3
(c) − (d)
(d) 20C 10 10 5
392
22. The coefficient of x n in expansion of (1 + x ) (1 − x ) n
is
(a) (n − 1)
[2004]
(b) (−1) (1 − n)
n
25. If x is positive, the first negative term in the
expansion of (1 + x ) 27/ 5 is [2003] 8
Binomial Theorem
(a) 7th term (b) 5th term
(c) (−1)n − 1 (n − 1)2 (d) (−1)n − 1 n (c) 8th term (d) 6th term
n
1 n
r tn 26. The coefficient of x 5 in (1 + 2x + 3x 2 + K ) −3/ 2 is
23. If sn = ∑ nC and t n = ∑ nC , then
sn
is equal [2002]
r=0 r r=0 r (a) 21
to [2004] (b) 25
n n 2n − 1 (c) 26
(a) (b) − 1 (c) n − 1 (d) (d) None of the above
2 2 2
27. If | x| < 1, then the coefficient of x n in expansion of
24. The number of integral terms in the expansion of (1 + x + x 2 + x 3 + K ) 2 is [2002]
( 3 + 8 5 ) 256 is [2003] (a) n (b) n − 1 (c) n + 2 (d) n + 1
(a) 32 (b) 33 (c) 34 (d) 35
Other Engineering Entrances
28. The sum of the coefficients in the expansion of 35. The term independent of x in the expansion of
( 5x − 4 y ) n , where n is a positive integer, is 2
18
x− is [EAMCET 2014]
[BITSAT 2014] x
(a) 0 (b) n (c) 1 (d) −1
(a) − 18 C 9 29 (b) 18
C 9 212
29. If 21st and 22nd terms in the expansion of (1 + x ) 44 (c) 18 C6 26 (d) 18
C 9 28
are equal, then x is equal to [Karnataka CET 2014]
(a) 8/7 (b) 21/22 (c) 7/8 (d) 23/24 36. The value of the sum ( n C1 ) 2 + ( n C 2 ) 2 + ( n C 3 ) 2
Targ e t E x e rc is e s
13
+ K + ( n C n ) 2 is
30. If the coefficient of x 8 in ax 2 + is equal to the
1 [WB JEE 2014]
bx (a) (2nC n )2 (b) 2nC n
13
1 (c) 2nC n + 1 (d) 2nC n − 1
coefficient of x −8 in ax − 2 , then a and b will
bx 2 n 22 23 2n + 1 n
satisfy the relation [WB JEE 2014] 37. Let S = C0 + n
C1 + n
C 2 +K+ Cn.
(a) ab + 1 = 0 (b) ab = 1 1 2 3 n+1
(c) a = 1 − b (d) a + b = − 1
Then, S equals [WB JEE 2014]
31. The coefficient of x 3 in the infinite series expansion 2n + 1 − 1 3n + 1 − 1
(a) (b)
2 n+1 n+1
of , for | x| < 1is [WB JEE 2014]
(1 − x ) ( 2 − x ) 3n − 1 2n − 1
(c) (d)
1 15 1 15 n n
(a) − (b) (c) − (d)
16 8 8 16 n
32. If ( a + bx ) −3 =
1 1
+ x + ... , then the ordered pair
38. m
Cr + 1 + ∑ k C r is equal to [AMU 2014]
k =m
27 3
( a, b ) equals [EAMCET 2014] (a) nC r + 1
n+ 1
1 (b) Cr + 1
(a) (3, − 27) (b) 1, (c) (3, 9) (d) (3, − 9)
3 n
(c) C r
(d) None of the above
33. If the term free from x in the expansion of
10 C1 C C C
k 39. + 2 2 + 3 3 + K + n n is equal to
x − is 405, then the value of k is
x2 [Manipal 2014]
C0 C1 C2 Cn − 1 [AMU 2014]
(a) ± 1 (b) ± 3 n (n − 1) n (n + 1)
(a) (b)
(c) ± 4 (d) ± 2 2 2
n (n + 1) (n + 2)
(c) (d) None of these
1
34. Middle term in the expansion of x 2 + 2 + 2 is 2
x
40. If n is an integer with 0 ≤ n ≤ 11, then the minimum
[Manipal 2014]
n! (2n)! value of n ! (11 − n )! is attained when a value of n
(a) (b) equals [EAMCET 2014]
(n!)2 (n!)2
(2n − 1)! (2n)! (a) 11 (b) 5
(c) (d) (c) 7 (d) 9
n! n! 393
8 41. If the 6th term in the expansion of the binomial
x
[ 2log (10 − 3 ) + 5 2(x − 2) log 3 ] m is equal to 21 and it is
49. The 13th term in the expansion of x 2 + is
2
x
n
Objective Mathematics Vol. 1
known that the binomial coefficient of 2nd, 3rd and independent of x, then the sum of the divisors of n is
4th terms in the expansion represent, respectively the [Karnataka CET 2012]
(a) 36 (b) 37 (c) 38 (d) 39
first, third and fifth terms of an AP, then x is equal to
10 7
∑ C r x r and (1 + x )7 = ∑ dr xr .
[BITSAT 2013]
50. Let (1 + x )10 = If
(a) 0 (b) 1 (c) − 2 (d) 3
r=0 r=0
n
42. If the 4th term in the expansion of ax + is
1 5 3
P
x
P= ∑ C 2r and Q = ∑ d 2r + 1 , then Q is equal to
r=0 r=0 [WB JEE 2012]
5
, ∀ x ∈ R, then the values of a and n are [AMU 2013] (a) 4 (b) 8 (c) 16 (d) 32
2
1 1 51. If the binomial coefficients of ( 3r ) th and ( r + 2) th
(a)
2
,6 (b) 6,
2 terms in the binomial expansion of (1 + x ) 2 n are
(c) 2, 6 (d) None of these equal, then [AMU 2012]
(a) n = r (b) n = r + 1
43. ∑ ( n C r + nC s ) is equal to [Manipal 2013] (c) n = 2r (d) n = 2r − 1
0≤r <s≤n
52. If A and B are coefficients of x n in the expansions of
(a) 2n + 1 (b) 2n + 1 − 1
(1 + x ) 2n and (1 + x ) 2n − 1 respectively, then A / B is
(c) n2n (d) (n + 1) 2n
equal to [WB JEE 2011; AMU 2011]
C C C Cn (a) 4 (b) 2
44. 1 + 1 1 + 2 1 + 3 K 1 + is equal
C 0 C1 C 2 C n − 1
(c) 9 (d) 6
n
1
to [Manipal 2013] 53. In the expansion of x 3 − 2 , n ∈ N , if the sum of
n+1 (n + 1)n x
Ta rg e t E x e rc is e s
(a) (b)
n! (n − 1)!
5
the coefficients of x and x10 is 0, then n is
[GGSIPU 2011]
(n + 1)n (n − 1)n
(c) (d) (a) 25 (b) 20
n! n! (c) 15 (d) None of these
ex
45. If = B 0 + B1 x + B 2 x 2 + K + B n x n + K, then 54. 5th term from the end in the expansion of
1− x 12
x3 2
B n − B n − 1 equals [Manipal 2012] − 2 is [GGSIPU 2011]
1 1 2 x
(a) (b)
n! (n − 1) ! (a) − 7920x −4 (b) 7920x 4
1
(c) −
1
(d) 1 (c) 7920x −4 (d) −7920x 4
n! (n − 1) !
55. The coefficient of p n q n in the expansion of
10
46. The coefficient of x in the expansion of [(1 + p ) (1 + q ) ( p + q )] n is [J&K CET 2011]
1 + (1 + x ) + K + (1 + x ) 20 is [WB JEE 2012] n n
(a) 19
C9 (b) 20
C 10
(a) ∑ [C (n, k )]2 (b) ∑ [C (n, k + 2)] 2
21 22 k=0 k=0
(c) C 11 (d) C 12 n n
47. If r th and ( r + 1) th terms in the expansion of
(c) ∑ [C (n, k + 3)] 2
(d) ∑ [C (n, k )] 3
k=0 k=0
( n + 1) q
( p + q ) n are equal, then is 56. In the expansion of (1 − 3x + 3x 2 − x 3 ) 2n , the middle
r ( p + q)
term is [MP PET 2011]
[Karnataka CET 2012]
(a) (n + 1) th term (b) (2n + 1) th term
(a) 0 (b) 1 (c) (3n + 1) th term (d) None of the above
1 1
(c) (d)
4 2 57. If n C 4 , nC 5 and n C 6 are in AP, then n is
27 2 [WB JEE 2011]
48. If (1 + ax ) n = 1 + 6x + x + K + a n x n , then the (a) 7 or 14 (b) 7
2 (c) 14 (d) 14 or 21
values of a and n are, respectively [Kerala CEE 2012]
(a) 2, 3 (b) 3, 2
3
(c) , 4 58. 15
C3 + 15
C5 + K + 15
C15 will be equal to
2 [WB JEE 2011]
3
(d) 1, 6 (e) , 6 (a) 2 14
(b) 2 14
− 15 (c) 2 14
+ 15 (d) 214 − 1
2
394
59. Let C 0 , C1 , ... , C n denote the binomial coefficients in
the expansion of (1 + x ) n . The value of
C1 − 2C 2 + 3C 3 − 4C 4 + K (upto n terms) is
62. The coefficient of the middle term in the expansion
of ( x + 2 y ) 6 is [Kerala CEE 2010]
8
Binomial Theorem
(a) 6C 3
[UP SEE 2011] (b) 8 6C 3
(a) 2n (b) 2−n (c) 8 6C 5
(c) 0 (d) 1
(d) 6C 4
60. The expression n
C 0 + 2⋅ nC1 + 3⋅ n C1 + ... (e) 8 6C 4
+ ( n + 1) ⋅ nC n is equal to [AMU 2011]
63. Let (1 + x ) n = 1 + a1 x + a 2 x 2 + K + a n x n . If a1 , a 2
(a) (n + 1) 2n (b) 2n (n + 2)
and a 3 are in AP, then the value of n is [BITSAT 2010]
(c) (n + 2) 2n − 1 (d) (n + 2) 2n + 1
(a) 4
61. If in the expansion of ( a − 2b ) n , the sum of the 5th (b) 5
(c) 6
a
and 6th terms is zero, then the value of is (d) 7
b
[WB JEE 2010] 64. If the sum of the coefficients in the expansion of
n−4 2 (n − 4 ) ( a 2 x 2 − 6ax + 11)10 , where a is constant is 1024, then
(a) (b)
5 5 the value of a is [Kerala CEE 2010]
5 5 (a) 5 (b) 1 (c) 2
(c) (d)
n−4 2 (n − 4 ) (d) 3 (e) 4
Targ e t E x e rc is e s
395
Answers
Work Book Exercise 8.1
1. (b) 2. (c) 3. (b) 4. (b) 5. (b) 6. (b) 7. (c) 8. (b) 9. (d) 10. (a)
Work Book Exercise 8.2
1. (c) 2. (a) 3. (b) 4. (a) 5. (b) 6. (c) 7. (c) 8. (c) 9. (c) 10. (b)
Work Book Exercise 8.3
1. (a) 2. (b) 3. (c) 4. (a) 5. (d) 6. (c) 7. (b) 8. (a) 9. (c) 10. (c)
11. (d) 12. (b) 13. (a) 14. (b) 15. (a) 16. (d) 17. (d) 18. (b)
Work Book Exercise 8.4
1. (d) 2. (a) 3. (c) 4. (a) 5. (a) 6. (b) 7. (c) 8. (a) 9. (a)
Target Exercises
1. (c) 2. (d) 3. (d) 4. (c) 5. (b) 6. (d) 7. (b) 8. (d) 9. (a) 10. (a)
11. (c) 12. (c) 13. (d) 14. (b) 15. (c) 16. (c) 17. (c) 18. (b) 19. (a) 20. (a)
21. (c) 22. (b) 23. (d) 24. (d) 25. (b) 26. (b) 27. (b) 28. (c) 29. (b) 30. (d)
31. (c) 32. (b) 33. (c) 34. (a) 35. (a) 36. (a) 37. (c) 38. (b) 39. (d) 40. (c)
41. (b) 42. (b) 43. (b) 44. (c) 45. (c) 46. (d) 47. (c) 48. (b) 49. (b) 50. (d)
51. (d) 52. (d) 53. (a) 54. (a) 55. (c) 56. (a) 57. (a) 58. (d) 59. (b) 60. (a)
61. (a) 62. (c) 63. (a) 64. (b) 65. (a) 66. (b) 67. (b) 68. (c) 69. (b) 70. (b)
Ta rg e t E x e rc is e s
71. (a) 72. (b) 73. (b) 74. (a) 75. (d) 76. (c) 77. (a) 78. (a) 79. (a) 80. (b)
81. (a) 82. (c) 83. (a) 84. (a) 85. (b) 86. (c) 87. (b) 88. (d) 89. (a) 90. (c)
91. (a) 92. (c) 93. (a) 94. (d) 95. (c) 96. (b) 97. (c) 98. (b) 99. (d) 100. (a)
101. (c) 102. (a) 103. (b) 104. (a) 105. (d) 106. (d) 107. (d) 108. (b) 109. (c) 110. (d)
111. (b) 112. (a,b,c) 113. (a,b,d) 114. (a,b) 115. (a,b,c) 116. (a,b) 117. (a,b) 118. (b, c) 119. (a,b) 120. (c,d)
121. (a,b,c) 122. (a,b,d) 123. (b) 124. (a) 125. (b) 126. (a) 127. (a) 128. (a) 129. (b) 130. (c)
131. (a) 132. (a) 133. (c) 134. (d) 135. (d) 136. (d) 137. (c) 138. (d) 139. (a) 140. (d)
141. (d) 142. (*) 143. (**) 144. (***) 145. (****) 146. (9) 147. (2) 148. (0) 149. (8) 150. (8)
151. (0) 152. (9) 153. (0) 154. (5) 155. (0)
* A → p; B → p,r; C → p, q; D → p, q, s
** A → r; B → q; C → p; D → s
*** A → q; B → r; C → p; D → p
**** A → q; B → s; C → p; D → r
Entrances Gallery
1. (c) 2. (6) 3. (d) 4. (a) 5. (d) 6. (d) 7. (c) 8. (a) 9. (b) 10. (b)
11. (b) 12. (b) 13. (d) 14. (b) 15. (c) 16. (c) 17. (a) 18. (b) 19. (a) 20. (b)
21. (c) 22. (b) 23. (a) 24. (b) 25. (c) 26. (d) 27. (d) 28. (c) 29. (c) 30. (a)
31. (b) 32. (d) 33. (b) 34. (b) 35. (a) 36. (d) 37. (b) 38. (b) 39. (b) 40. (b)
41. (a) 42. (a) 43. (c) 44. (c) 45. (a) 46. (c) 47. (b) 48. (c) 49. (d) 50. (b)
51. (c) 52. (b) 53. (c) 54. (c) 55. (d) 56. (c) 57. (a) 58. (b) 59. (c) 60. (c)
61. (b) 62. (b) 63. (d) 64. (d)
396
Explanations
Target Exercises
5 5 2n 2n
1. x + x 3 − 1 + x − x 3 − 1 7. Since, C1, C2 , 2 n C3 are in AP.
∴ 2 2 n C2 = 2 nC1 + 2 nC3
2 4
= 2 5 C0 x 5 + 5C2 x 3 x 3 − 1 + 5C4 x x 3 − 1 ⇒ 2n − 9 n + 7 = 0
2
8. Tp = T( p − 1) + 1 = 6Cp − 1(2 x )6 − ( p − 1) 3 p − 1
= 2 [ x 5 + 10 x 3 ( x 3 − 1) + 5 x( x 3 − 1)2 ]
This is a polynomial of degree 7. = 6 Cp − 1 2 7 − p 3 p − 1 x 7 − p
7 ∴ Coefficient of Tp = 6Cp − 1 2 7 − p 3 p − 1 = 4860 [given]
1
2. Given expression = 9 x − 1 + 7 + Clearly, p = 5 is the solution.
(3 x − 1 + 1)1/ 5
r
9. Tr + 1 = x 20 ⋅20 Cr
7− 5
5 1
1 in descending powers of x.
∴ T6 = T5 + 1 = C5 9 x − 1 + 7
7
x −1 x2
(3 + 1)
1/ 5
1
1 T6 = ⋅ 20
C5 ( x 2 )15 in ascending powers of x
= 21(9 x − 1 + 7 )⋅ x − 1 x 20
3 +1 ⇒ 20
Cr = 20C5
2x − 2
21(3 + 7) ∴ r =5
= = 84 [given]
3x − 1 + 1
10. In an expansion, binomial coefficients equidistance
⇒ 32 x − 2 + 7 = 4 (3 x − 1 + 1) from the beginning and the end are equal.
⇒ y 2 + 7 − 4 y − 4 = 0 , where y = 3 x − 1 ∴ an − 1 = a1, an − 2 = a2, an − 3 = a3
∴ y = 1 ⇒ 3x − 1 = 1 Thus, a1, a2 , a3 are in AP.
⇒ x − 1= 0 ⇒ x = 1 ⇒ 2 ⋅n C2 = nC1 + nC3
Targ e t E x e rc is e s
and y = 3 ⇒ 3x − 1 = 3 n
C n
C
⇒ 2= n 1 + n 3
⇒ x − 1= 1 ⇒ x = 2 C2 C2
3. T10 = T9 + 1 = 11C9 ( x )11 − 9 (− 1)9 = − 11C9 x 2 2 n −2
⇒ 2= +
n −1 3
4. T3 = T2 + 1 = 5C2 x 5 − 2 ( xlog10 x )2 ⇒ n − 9 n + 14 = 0
2
= 10 x 3 x 2 log10 x = 10 x( 3 + 2 log10 x ) = 10 6 [given] ∴ n = 2 or 7, neglecting n = 2 ;we get n = 7
3 + 2 log10 x = 10 5
⇒ x 600 − r r
⇒ (3 + 2 log10 x )log10 x = 5 log10 10 = 5 11. Tr + 1 = 600
Cr ⋅ 7 3 ⋅ 52 x r
5
⇒ log10 x = − ⇒ x = 10 − 5 / 2 r r
2 Here, 0 < r < 600 and , 200 − are integers.
2 3
and log10 x = 1 ⇒ x = 101 = 10
∴ r should be a multiple of 6.
∴ x = 10 − 5 / 2 , 10 ∴ r = 0, 6, 12, ..., 600 i.e. 101 terms
m 10 − r r
5. In px + , t 4 = 2.5
1 r r
12. Tr + 1 = 10Cr ⋅ 2 2 ⋅ 35 . This is rational, if and 5 − are
x 5 2
1
3
integers. Also, 0 < r < 10.
∴ m
C3 ( px )m − 3 = 2.5
x Now, r must be a multiple of 10 (LCM of 2 and 5).
So, r = 0, 10
⇒ mC3 pm − 3 x m − 6 = 2.5, which is independent of x.
∴The rational terms are
∴ m− 6= 0 ⇒ m=6
5 5
10
C0 ⋅ 2 5 ⋅ 30, 10
C10 ⋅ 2 0 ⋅ 32
∴ 6
C3 p3 = ⇒ 20 p3 =
2 2 ∴Sum of rational terms = 32 + 9 = 41
1 1
p3 = ⇒ p= r
13. Tr + 1 = 6Cr ( x 5 / 2 )6 − r
or 3
3/ 2
8 2
1 x
Thus, p= and m=6 5r 3r
2 15 − −
3 3
= 6Cr 3 r x 2 2
α α
6. T4 = nC3 x n − 3
r 15 − 4 r
⇒ n
C3 x n − 6 = 20 = Cr 3 x
6
2 x 2
∴ 15 − 4 r = 3
Since, independent of x. ⇒ r =3
⇒ n − 6 = 0 or n = 6 ∴ Tr + 1 = T3 + 1 = 6C3 (3)3 x15 − 4( 3 )
3
α
α3 = 20 × 27 × x 3
∴ 6
C3 = 20 ⇒ 20 ⋅ 3 = 20
2 2 = 540 x 3
⇒ α =2 ∴Coefficient of x = 540
3 397
8
12 − r
2n − r r
n r
14. Tr + 1 in 2 + = nCr 2 n − r = nCr x2
x x r
y
x 21. Tr + 1 = 12Cr −
3 3
r
3 y x
n−r
2 = 12Cr x 24 − 3 r ⋅ y 2 r − 12 (− 1)r
Objective Mathematics Vol. 1
∴ Coefficient of x r = nCr r
3 ∴ Coefficient of x 6 y − 2 is obtained,
2n − 7 n 2n − 8 when 24 − 3 r = 6 and 2 r − 12 = − 2
∴ n
C7 ⋅ 7 = C8 ⋅ 8
3 3 ⇒ r = 6 and r = 5, which is not possible.
2 1
⇒ = 22. f ( x ) = ( x 2 + 1 + x 2 − 1)6 + ( x 2 + 1 − x 2 − 1)6
n −7 8× 3
⇒ n − 7 = 48 After expansion and simplification
⇒ n = 55 f ( x ) = 64 x 6 − 48 x 2
2n
23. (1 + x + x 2 + x 3 ) n = [(1 + x )(1 + x 2 )]n = (1 + x )n (1 + x 2 ) n
15. We have, (1 + 2 x + x 2 )n = ∑ ar x r = (n C0 + nC1 x + nC2 x 2 + nC3 x 3 + nC4 x 4 + ... )
r =0
2n × (n C0 + nC1 x 2 + nC2 x 4 + ... )
⇒ (1 + x )2 n = ∑ ar xr ∴ Coefficient of x = C0 ⋅n C2 + nC2 ⋅n C1 + nC4 ⋅n C0
4 n
r =0
= nC2 + nC1 ⋅n C2 + nC4
Comparing the coefficients of x r , we get
2n
Cr = ar 24. Coefficient of λnµ n in (1 + λ )n (1 + µ)n (λ + µ)n
⇒ Coefficient of λnµ n in
m(m − 1) 2
16. (1 + x ) (1 − x ) = 1 + mx +
m n
x + K
{(n C0 + nC1λ + nC2 λ2 +...+ n Cn λn )
2
− (n C0µ n + nC1µ n − 1 +...+ nCn )
n( n 1) 2
× 1 − nx + x + K (n C0 λn + nC0 λn − 1 µ + ... + nCnµ n )}
2
∴ Coefficient of x = m − n = 3 [given] Clearly, coefficient of λnµ n is (C03 + C13 + C23 + ... + Cn3 )
n(n − 1) m(m − 1) 25. Here, a = nCr , b = nCr + 1 and c = nCr + 2
Coefficient of x =2
− mn + =−6
Ta rg e t E x e rc is e s
2 2
Put n = 2, r = 0, then option (b) holds the condition
[given] 2 ac + ab + bc
⇒ n 2 − n − 2 mn + m2 − m = − 12 i.e. n=
b2 − ac
⇒ (m − n )2 − (m + n ) = − 12
⇒ (3)2 − (m + n ) = − 12 26. f ( x ) = 1 − x + x 2 − x 3 + ... − x15 + x16 − x17
1 − x18
⇒ m + n = 21 =
∴ Solving m − n = 3, m + n = 21, we get 1+ x
m = 12 1 − ( x − 1)18
⇒ f ( x − 1) =
n=9 x
100 Therefore, required coefficient of x 2 is equal to
17. ∑ 100Cr ( x − 3)100 − r 2 r coefficient of x 3 in 1 − ( x − 1)18 , which is given by
r =0 18
C3 = 816.
= {( x − 3) + 2}100 = ( x − 1)100 = (1 − x )100
100
27. (1 − x )30 = 30C0 x 0 − 30C1 x1 + 30
C2 x 2
∴ Coefficient of x 53 = (−1)53 C53 = – 100C53 30
− ... + (− 1)30 C30 x 30 …(i)
18. Coefficient of x 2 = Coefficient of x 3 ( x + 1)30 = 30
C0 x 30 + C1 x + 30C2 x 28
30 29
9−2 9 −3
∴ 9
C2 3 k = C3 3
2 9
k 3
+ ... + 30C10 x 20 + ... + 30C30 x 0 …(ii)
⇒ 36 = 28k On multiplying Eqs. (i) and (ii) and equating the
9
⇒ k= coefficient of x 20 on both sides, we get required sum is
7
equal to coefficient of x 20 in (1 + x 2 )30 , which is given
(1 + x )6 {1 − (1 + x )10 }
19. Expression = by 30
C10 .
1 − (1 + x )
[sum of 10 terms of GP] 28. We have,
(1 + x )101 (1 − x + x 2 )100 = (1 + x ) {(1 + x ) (1 − x + x 2 )}100
(1 + x )16 − (1 + x )6
= = (1 + x ) (1 + x 3 )100 = (1 + x ) {C0 + C1 x 3 + C2 x 6
x
∴Required coefficient + ... + C100 x 300}
n n n
= Coefficient of x 7 in {(1 + x )16 − (1 + x )6 } = (1 + x ) ∑ nCr x 3r = ∑ nCr x 3r + ∑ nCr x 3r + 1
= 16C7 = 16C9 r =0 r =0 r =0
20. (1 − x ) (1 + x ) (1 + x )
5 4 2 4 Hence, there will be no term containing 3 r + 2.
= (1 − x 2 )4 (1 + x 2 )4 (1 – x ) 29. n Cr 2 r ⋅ x 2 n − 3 r ⇒ 2 n − 3 r = 0
= (1 − x 4 )4 (1 − x ) ⇒ 2n = 3 r
398 n is multiple of 3/2.
∴Coefficient of x13 in {(1 − x )(1 − 4 x 4 + 6 x 8 − 4 x12 )} = 4
8
10 − r r 5
x 3 9π 2 − π 2 − 10C5 π 10
30. Tr + 1 = 10Cr ⇒ Minimum value is − 10C5 =
3 2 x2 16 25
− π 10
Binomial Theorem
1
10 − r r
3 5 − 2
r
−r
10
C5 π 10
= 10Cr x and maximum value is − 10C5 =
3 2 165 2 20
1
Let Tr + 1 be the term independent of x. 35. Expression = (1 − 2 x + x 2 )(1 + x 2 )10
r 3r x10
∴ 5− − r = 0 ⇒ =5 1 2 1
2 2 = 10 (1 + x 2 )10 − 9 (1 + x 2 )10 + 8 (1 + x 2 )10
10 x x x
⇒ r=
3 ∴The term independent of x
∴ r is not a whole number. = 10C5 − 2 × 0 + 10C4 = 11C5
∴ Given expression does not have term independent
(1 − x 2 )n
of x. 36. Expression = (− 1)n ⋅
xn
31. [(t − 1 − 1)x + (t − 1 + 1)− 1 x − 1 ]8
8
∴Term independent of x
1 −1
1 1 = Coefficient of x n in (− 1)n ⋅ (1 − x 2 )n
= − 1 x + + 1
t = Coefficient of x n in
t x
8−r r (− 1)n { n C0 + nC1(− x )2 + nC2 (− x 2 )2 + K + nCn (− x 2 )n }
1 1 − 1 1
Tr + 1 = Cr − 1 x
8
+ 1 Since, the expansion contains only even powers of x.
t
t x Thus, if n is odd ⇒ Coefficient is zero.
8−r −r n
1 1
37. (1 + x )n 1 + = (C0 + C1 x + C2 x 2 + ... + Cn x n )
1
= 8Cr − 1 + 1 x 8 − 2r
t t x
Let Tr + 1 be the term independent of x. C C C
× C0 + 1 + 22 + ... + nn
∴ 8 − 2r = 0 x x x
i.e. r =4 Term independent of x on the RHS
Targ e t E x e rc is e s
8−4 −4
1 1 8 − 2( 4 ) = C02 + C12 + C22 + ... + Cn2
∴ Tr + 1 = T4 + 1 = 8C4 − 1 + 1 x
t t n 6
1 1
4 4 38. Tn in 6 3 2 + = nC6 (21/ 2 31/ 6 )n − 6 1/ 3
1− t t 1− t
4
3 3
= 70 = 70 3
t 1+ t 1+ t n
1
7th term from the end in 6 3 2 + 3
9 9 9−r r 3
32. x 2 − = ∑ 9 Cr x 2 1
3 1 3
− n n−6
2 3x 2 3x 1 1
r =0 = Tn in + 3 2 = nC6 1/ 3
6
(31/ 621/ 2 )6
3 3 3
For the term that is independent of x, we must have 6
18 − 2 r − r = 0 1
n
C6 [21/ 2 31/ 6 ]n − 6 1/ 3
⇒ r =6 3 1
3 6 ∴ n−6
=
3 1 1 6
Thus, required term = 9C6 x 2 − n
C6 1/ 3 [21/ 2 31/ 6 ]6
2 3x 3
3 6 3
3 1 1 n−6 n−6
i.e. 9
C6 = 9C6 2 2 3 6 × [3 ]
−2
2 3 6
1
n ⇒ =
33. x 2 − 2 + 2 = 2 n ( x 4 − 2 x 2 + 1)n
1 1 n − 6 6
−
x x 3 3 2 3 ⋅ 31
n−6 n−6
( x 2 − 1)2 n
= 2 2 3 6 1 1
x 2n ⇒ 6−n
⋅ =
32 6
Total number of terms that are dependent on 3 3 23 ⋅ 31
x = number of terms in the expansion of ( x 2 − 1)2 n n−6
that have degree of x different from 2n. 2 2 1 1 1
⇒ ⋅ ⋅ =
= (n + 1) − 1= n 6−n n − 6
− 2 3 ⋅ 33 6
3 3 6
34. Term independent of x = 10C5 (sin − 1 α )5 (cos − 1 α )5 3
n−6
5
π 2 2 2 3 ⋅ 33
= 10C5 (sin − 1 α )5 − sin − 1 α ⇒ =
2 6−n
6
5 3 2
π
= 10C5 sin −1 α − (sin − 1 α)2 n−6 n−6
2 ⇒ 2 2 ⋅ 3 2 = 2 2 × 32
5
π
2
π2 Comparing the coefficients of 2 and 3, we get
− C5 sin − 1 α − −
10
n−6
4 16 = 2 ⇒ n = 10
2
399
8 39. Middle terms are T9 + 1 = T5 and T9 + 3 = T6 10
(1 − x )(1 + x + x 2 )
47. (1 + x + x 2 )10 =
2
4
2
5 1− x
a2 a2
= (1 − x 3 )10 (1 − x )− 10
Objective Mathematics Vol. 1
∴ S = 9C4 (2 a)9 − 4 − + 9C5 (2 a)9 − 5 −
4 4 Thus, coefficient of x 2 in (1 + x + x 2 )10 = coefficient of
a8 a10 x 2 in (1 − x )− 10 i.e. 10 + 2 − 1C2 = 11C2
= 126 × 32 a5 × − 126 × 16 a4 ×
256 1024 7
=
63 13
a −
63 14 63
a = (8 − a) a13 48. (1 + x − 2 x 2 )7 = ∑ 7Cr (1 + x )7 − r (− 2 x 2 )r
4 32 32 r =0
40. As the index = 20 , the 11th term is the middle term. So, Thus, required coefficient
we have to find the 14th term. = 7C0 ⋅7 C4 + 7C1(− 2 )⋅6 C2 + 7C2 ⋅ (− 2 )2 ⋅ 5C0
13 = 1⋅ 35 − 7 ⋅ 2 ⋅ 15 + 21⋅ 4 ⋅ 1 = − 91
1
T14 = C13 ( x 2 )20 − 13 −
20
49. Total number of different terms in the expansion of
2 x
1 ( x1 + x2 + K + xn )m is same as the number of
= − C13 x ⋅ 13 13
20 14
non-negative integral solutions of y1 + y2 +…+ yn = m
2 x n + m −1
1 20 i.e. Cn − 1
= − 13 C13 x
2 50. (1 + a − b + c )9
20 − r r 40 − 2 r r r
− − − 9!
41. Tr + 1 = 20Cr ⋅4 3 ⋅6 4 = 20
Cr ⋅2 3 ⋅2 4 ⋅3 4 =∑ ⋅ (1)x1 (a)x2 (− b)x3 (c )x4
x1 ! x2 ! x3 ! x4 !
160 − 11r r
− 9! 9!
=
Cr ⋅ 2 12 ⋅ 3 4
20
⇒ Coefficient of a3 b4c = =
1! 3 ! 4 !1! 3 ! 4 !
r 160 − 11r
For rational terms, and must be integers 51. Number of terms = n + 5 −1
C5 − 1 = n+ 4
C4
4 12
and 0 < r < 20.
r 52. (a + b + c ) + (a + b − c ) = 2{ C0 c (a + b)n
n n n 0
So, = integer ⇒ r = 0, 4, 8, 12, 16, 20 + nC2 c 2 (a + b)n − 2 + K + nCnc n (a + b)0 }
Ta rg e t E x e rc is e s
4
160 − 11r Number of terms = (n + 1) + (n − 1) + (n − 3) + K + 1
Clearly, for r = 8, 16 and 20, is also an
12 n + 2 (n + 2 )2
= [1 + (n + 1)] =
integer. 2 2
∴ Only three terms are rational. So, 21 − 3, i.e. 18
terms are irrational. 53. ( x + y + z )25 = { x + ( y + z )} 25
n ! (21 − n )! 21! = 25C0 x 25 + 25C1 x 24 ( y + z ) + K+ 25Cr x 25 − r ( y + z )r + K
42. n ! (21 − n )! = 21! = 21 , which is minimum
21! Cn = K + 25Cr x 25 − r (K + r Ck y r − k z k + K ) + K
when Cn is maximum, which occurs when n = 10.
21 8 + 9 + 9 ≠ 25. So, there is no term like x 8 y 9 z 9 .
The number of terms
43. Since, n is even, therefore + 1 th term is middle term.
n
2 = 1 + 2 + 3 + K + 26
26 × 27
n/ 2 =
1 2
Hence, n Cn / 2 ( x 2 )n / 2 = 924 x 6
x = 351
⇒ x n/ 2 = x 6 ⇒ n = 12 54. The greatest coefficient
44. (1 + 3 x + 3 x + x ) = {(1 + x ) } = (1 + x )
2 3 6 3 6 18 = Coefficient of the middle term
(2n )!
= 2nCn =
Hence, the middle term is 10th. n!n!
10 (2 n )(2 n − 1)(2 n − 2 )(2 n − 3)(2 n − 4) K 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1
45. a10 b10c 10d 10 +
1 1 1 1 =
+ + n!n!
a b c d
2 n{ n ⋅ ( n − 1) K 2 ⋅ 1}{(2 n − 1)(2 n − 3)(2 n − 5) K 5 ⋅ 3 ⋅ 1}
Therefore, the required coefficient is equal to the =
10 n! n!
1 1 1 1
coefficient of a− 2 b− 6c − 1d − 1 in + + + , 2 n ⋅ n !⋅ {(2 n − 1)(2 n − 3)(2 n − 5) K 5 ⋅ 3 ⋅ 1}
a b c d =
n!n!
10 ! 10 × 9 × 8 × 7
which is given by = = 2520 2 n {1⋅ 3 ⋅ 5 ⋅ K ⋅ (2 n − 1)}
2 ! 6 !1!1! 2 =
n!
46. The general term in the expansion of (1 − x + y )20 is n−r +1 y
55. For largest term ⋅ >1
20 ! r r x
1 (− x )s ( y ) t , where r + s + t = 20
r ! s !t ! 51 − r 2 x 51 − r 2 1
⇒ ⋅ ≥ 1; if ⋅ ≥1
For x 2 y 3 , we have the term r 3 r 3 5
20 ! 15
1 (− x )2 ( y )3 if 102 − 2 r ≥ 15 r; if 17 r ≤ 102
15 !2 ! 3 ! if r ≤ 6
Hence, the coefficient of x 2 y 3 is
20 !
. ∴ We take r = 6
400
15 !2 ! 3 ! ∴ Largest term = Tr + 1 = T6 + 1 = T7
56. α + β = (3 3 + 5)2 n + 1
β1 = (3 3 − 5)
⇒
2n + 1
, where β1 < 1
α + β + β1 = Even integer, β − β1 = 0
65. 2 60 = (1 + 7 )20
= 20C0 + 20C1 ⋅ 7 + 20C2 ⋅ 7 2 + ... +
∴ The remainder = 20C0 = 1
20
C20 ⋅ 7 20 8
Binomial Theorem
⇒ α = Even integer
66. (α − 2 + 1)35 = (1 − α )35
57. x − x 2 + x[ x ] = x − x( x − [ x ]), = x − x{ x} ⇒ (α − 1)35 = − (α − 1)35 ⇒ α =1
= x (1 − { x})
x = (2 + 3 )n = nC0 2 n + nC12 n − 1 3 67. The sum of all the coefficients in an expansion is
obtained by putting x = 1in the expression.
+ nC2 2 n − 2 ( 3 )2 + K n
1
Let x1 = (2 − 3 )n ∴ + 2 ⋅ 1 = 6561
1
= nC0 2 n − nC12 n − 1 3 + nC2 2 n − 2 ( 3 )2 + K
∴ 3n = 38 ⇒ n = 8
x + x1 = 2(n C0 2 n + nC2 2 n − 2 ( 3 )2 + K ) 8 8−r
1 1
= Even integer In + 2 x , Tr + 1 = 8Cr ⋅ ⋅ (2 x )r
x x
` Clearly, x1 ∈ (0, 1), ∀ n ∈ N This is a constant, if 8 − 2 r = 0 i.e. r = 4
⇒ [ x ] + { x} + x1 = Even integer ∴The constant term = T5 = 8C4 ⋅ 2 4
⇒ { x} + x1 = Integer
{ x} ∈ (0, 1,) x1 ∈(0, 1) 68. The sum of all the numerical coefficients in the
⇒ { x} + x1 ∈ (0, 2 ) expansion is obtained by putting x = 1, y = 1
⇒ { x} + x1 = 1 1 2
12
1 + + = (2 )
12
⇒ x1 = 1 − { x} i.e.
3 3
⇒ x(1 − { x}) = x ⋅ x1 = (2 + 3 )n (2 − 3 )n = 1
69. Sum of coefficient of x 2 , x 4 , x 6 , x 8 , …, is
58. End digit of 7 4 n + 1 is 7 of 7 4 n + 2 is 9, n
C2 + nC4 + nC6 + K = 2 n − 1 − 1
4n + 3
of 7 = 1 and 7 4n
= 7n ∈ N
70. The sum of coefficient is obtained by putting variable
Now,754 = 4 ⋅ 188 + 2. Thus, end digit of(2137 )754 will be term as 1.
Targ e t E x e rc is e s
4n + 2 ∴ (1 + x − 3 x 2 )2163 = (1 + 1 − 3)2163 = − 1,
same as that of 7 i.e. 9.
4 ⋅9
as sum of coefficient.
59. 3 = 3
37
⋅ 3 = 3 ⋅ (81) = 3 (80 + 1)9
9
71. Putting x = 1, x = − 1 and adding, we get
= 3 ( C0 80 9 + 9C1 80 8 + K + 9C9 )
9
1
Thus, required remainder is equal to 3. Sum = {(a0 + a1 + a2 + ... + a16 )
2
60. 4101 = 4100 ⋅ 4, clearly 101 ÷ 4100 − (a0 − a1 + a2 − ... + a16 )}
∴ a1 + a3 + a5 + .... + a15
Thus, required remainder is 4. 1 −1 8
= {(1 + 1 − 2 )8 − (1 − 1 − 2 )8 } = (2 ) = − 2 7
61. 1! + 2 ! + 3 ! + 4 ! = 1 + 2 + 6 + 24 = 33 2 2
⇒ Remainder = 3 72. Putting x = i , − i and multiplying both the results, we get
62. 32003 = 32001 ⋅ 32 = 9 (27 )667 = 9 (28 − 1)667 the value of the required expression as,
= 9 [ C0 28 + C1(28) + K + C667(−1)
667 667 667 666 667 667
] (a0 − a2 + a4 − a6 + a8 − a10 )2 + (a1 − a3 + a5 − a7 + a9 )2
When 32003 is divided by 28, remainder is 19. = (1 + i )10 (1 − i )10 = (1 + 1)10 = 210
32003 19 73. Sum of coefficients of odd powers of x.
Thus, =
28 28
50
C1 + C3 + ...+
50
C49 = 2 50 − 1 = 2 49
50
(2 n )! (n + 1)(n + 2 ) K (n + n ) 74. (1 + x − 2 x 2 )20 = a0 + a1 x + a2 x 2 + ... + a40 x 40
63. N = 2n
Cn = =
(n !)2 (n !) a0 + a1 + a2 + a3 + K + a40 = 0
⇒ (n !) N = (n + 1)(n + 2 ) K (n + n ) a0 − a1 + a2 − a3 + K + a40 = 2 20
∴ a1 + a3 + K + a39 = − 219
Since, n < p < 2 n
2n
⇒ p /(n + 1)(n + 2 ) K (n + n )
⇒ p/ N
75. (1 + x + x 2 )n = ∑ ar x r …(i)
r =0 n
64. Expression = 10 n − 1 + 3 ⋅ 4n + 2 + 6 We know that, (1 − x )n = ∑ (− 1)r ⋅ nCr x r
2n + 3
= (10 − 1) + 6 (2
n
+ 1) r =0
n
= (10 − 1){10 n − 1 + 10 n − 2 + K + 10 + 1} = ∑ (− 1)n − r ⋅n Cr x n − r …(ii)
+ 6 (2 + 1){2 2 n + 2 − 2 2 n + 1 + K − 2 + 1} r =0
10 n − 1 + 10 n − 2 + K + 10 + 1 is an odd number which is On multiplying Eqs. (i) and (ii), we get
n
not divisible by any odd number of all n.
2n + 2 2n + 1
∑ (− 1)− r ⋅n Cr ar = Coefficient of x n in (1 − x 3 )n
Again, 2 −2 + K − 2 + 1 is also an odd r =0
number which is not divisible by any odd number for Since, n ≠ M(3)
n n
all n.
∴ The expression is divisible by 9.
∴ ∑ (− 1)− r ⋅ ar ⋅n Cr =0 ⇒ ∑ (− 1) r ⋅ ar ⋅n Cr =0
401
r =0 r =0
8 76. a0 + a1 + a2 + ... = 2 2 n and a0 + a2 + a4 + ... = 2 2 n − 1 n n
82. Sum = a ∑ (−1 )r − 1 ⋅ nCr − ∑ (−1 )r − 1 ⋅ r ⋅ nCr
an = Cn = The greatest coefficient, being the middle
2n
r =1 r =1
coefficients.
Objective Mathematics Vol. 1
= a { C1 − C 2 + C 3 − K } − n{ n − 1C 0 −
n n n n−1
C1 + K }
an − 3 = 2n
Cn − 3 = 2n
C2 n − ( n − 3 ) = 2n
Cn + 3 = an + 3 n −1
[Q r ⋅ Cr = n ⋅
n
Cr − 1 and C0 − C1 + C2 − … = 0]
n n n
n
= a ⋅ nC0 = a
77. I = ∑ nCr sin r x ⋅ cos(n − r )x
r =0
n
83. r ⋅ 2 nCr = 2 n ⋅ 2 n − 1Cr − 1
= ∑ nCn − r sin(n − r )x ⋅ cos r x n
r =0 ∴ ∑ r ⋅ 2r Cr = 2 n { 2 n − 1C0 + 2n − 1
C1 + K + 2n − 1
Cn − 1}
n r =1
⇒ 2I = ∑ nCr ⋅ sin nx = sin nx ⋅ 2 n = 2 n ⋅ 2( 2 n − 1) − 1 = n ⋅ 2 2 n − 1
r =0
⇒ I =2 n −1
⋅ sin nx 84. Putting x = 1and − 1, we get
n n 1 = a0 + a1 + a2 + a3 + K + a2 n
78. ∑ n
Cr sin r x = Im ∑ n Cr e irx
and 3n = a0 − a1 + a2 − a3 + K + a2 n
r =0 r = 0 Adding, we get 1 + 3n = 2 (a0 + a2 + a4 + K + a2 n )
n 3n + 1
= Im ∑ n Cr (e ix )r = Im((1 + e ix )n ) ∴ a0 + a2 + a4 + K + a2 n =
r = 0 2
n n
= Im(1 + cos x + i sin x )n n
n
85. ∑ (a + rb) nCr = a (2 n ) + b ∑ r ⋅ ⋅ n − 1Cr − 1
r
x x x r =0 r =0
= Im 2 cos 2 + 2 i sin ⋅ cos
2 2 2 = a ⋅ 2 n + nb ⋅ 2 n − 1 = (2 a + nb) 2 n − 1
n
x x x 86. (1 + x )2 n = C0 + C1 x + C2 x 2 + K + C2 n x 2 n
= Im 2 cos cos + i sin
2 2 2 On differentiating w.r.t. x, we get
Ta rg e t E x e rc is e s
x nx
= 2 n ⋅ cos n ⋅ sin 2 n(1 + x )2 n − 1 = C1 + 2C2 x + K + 2 nC2 n x 2 n − 1 …(i)
2 2
n
r ⋅n Cr n
n−r +1 Also, (1 − x )2 n = C0 − C1 x + C2 x 2 − K + C2 n x 2 n
79. ∑ n
= ∑r⋅
Cr − 1 r = 1 r 1
r =1 Replacing x by , we get
n n x
= (n + 1)∑ 1 − ∑ r 2n
1 C1 C2 C
r =1 r =1 ∴ 1 − = C0 − + 2 − K + 22 nn …(ii)
n(n + 1) n(n + 1) x x x x
= n(n + 1) − =
2 2 On multiplying Eqs. (i) and (ii), we get
2n 2n 2n 2n
1 C 1
80. ∑ r. 2nCr ⋅ r + 2 = ∑ (r + 2 − 2 )⋅ r + 2r 2 n(1 + x )2 n − 1 1 −
r =0 r =0 x
2n 2n 2n
Cr = (C1 + 2C2 x + K + 2 nC2 n x 2 n − 1 )
= ∑ 2nCr − 2⋅∑
+
r =0 r =0
r 2 C C C
× C0 − 1 + 22 − K + 22 nn
1 x x x
=2 2n
− 2⋅∫ x(1 + x ) dx2n
0 1
Coefficient of on RHS
(1 + x )2 n + 1 1 (1 + x )2 n + 2 1
x
= 2 2 n − 2 x ⋅ − (2 n + 1)(2 n + 2 )
2 n + 1 = − C12 + 2 C22 − 3 C32 + K + 2 nC22n
0 0
2n + 1
= − (C12 − 2 C22 + 3 C32 − K − 2 nC22n )
2 (2 n − n + 1) − 2
2
= 1
(2 n + 1)(2 n + 2 ) Also, coefficient of on LHS
x
n
81. ∑ nCr ar ⋅ bn − r ⋅ cos{ rB − (n − r )A} 1 2 n(1 + x )2 n − 1( x − 1)2 n
= Coefficient of in
r =0 x x 2n
n
= Re ∑ n Cr ar ⋅ bn − r ⋅ e i{ rB − ( n − r )A} = Coefficient of x 2 n − 1 in 2 n(1 − x 2 )2 n − 1(1 − x )
r = 0 = 2n 2 n −1
Cn (−1)n −1
n 2n − 1
= Re ∑ n Cr (a ⋅ e iB )r ⋅ (b ⋅ e − iA )n − r ∴ C12 − 2 C22 + 3 C32 − K − 2 nC22n = 2 n Cn (−1)n − 1
r = 0 87. (1 + x )10 = 10C0 + 10C1 x + 10C2 x 2 + K + 10C10 x10 …(i)
= Re (ae iB + be − iA )n
Also,
= Re (a cos B + ia sin B + b cos A − ib sin A)n
(1 − x )10 = 10C0 − 10C1 x + 10
C2 x 2 − K + 10
C10 x10 …(ii)
402 = (a cos B + b cos A)n = c n
8
n−r
On multiplying, we get
(1 − x ) = ( C0 +
2 10 10 10
C1 x + 10
C2 x + K +
2 10 10
C10 x )
94. ∑ n −i
Cr = nCr + n −1
Cr + n−2
Cr + ... + r + 1 Cr + r Cr
i=0
n+1
=
Binomial Theorem
× ( C0 − C1 x +
10 10 10
C2 x − K +
2 10 10
C10 x ) Cr + 1
Equating the coefficients of x10 , we get Cr 1 nn
95. Here, Tr + 1 = = ⋅ Cr
10
C5 (−1) = C0 C10 − C1 C9 +
5 10 10 10 10 10 10
C2 C8 r +1 r +1
1 n+1
− K + 10C10 10C0 = Cr + 1
n+1
⇒ − C5 = ( C0 ) − ( C1 ) + ( C2 ) − K + (10 C10 )2
10 10 2 10 2 10 2
Putting r = 0, 1, 2, …, n and adding, we get
88. (1 − x )n = C0 − C1 x + C2 x 2 − C3 x 3 + K n terms n
C n
1
∴ x(1 − x )n = C0 x − C1 x 2 + C2 x 3 − C3 x 4 + K ∑ k +k 1 = ∑ n + 1 n+ 1Ck + 1
k=0 k=0
1
∴ ∫0 x(1 − x )n dx =
1 n+1
[ C1 + n + 1C2 + n + 1C3 + K + n+1
Cn + 1 ]
1 n+1
= ∫ (C0 x − C1 x 2 + C2 x 3 − C3 x 4 + ... )dx 1 2n + 1 − 1
0
= [2 n + 1 − n + 1C0 ] =
0 n+1 n+1
⇒ − ∫ (1 − t )t ndt
1
1 96. Clearly, ar = 2n
Cr
x2 x3 x4 x5
= C0 − C1 + C2 − C3 + ... ar 2n
Cr (2 n − r + 1)
2 3 4 5 ⇒ = =
0 ar − 1 2n
Cr − 1 r
(where t = 1 − x) ar 2n + 1
tn +1 tn + 2
0
C C C C ⇒ 1+ =
⇒ − − = 0 − 1 + 2 − 3 + .... ar − 1 r
n + 1 n + 2 2 3 4 5
1 2n a 2 n (2 n + 1) (2 n + 1)2 n
C C C C
∴ 0 − 1 + 2 − 3 +K
⇒ ∏ 1 + a r = ∏ =
r =1 r − 1 r =1 r (2 n )!
2 3 4 5
Targ e t E x e rc is e s
1 1 n
= −
n+1 n+2 97. ∑ ∑ i ⋅ nC j = ∑ n
Cr (0 + 1 + 2 + ...+ r − 1)
0 ≤ i< j ≤ n r =1
1
= 1
(n + 1)(n + 2 ) = { n(n − 1)⋅ 2 n − 2 + n ⋅ 2 n − 1 − n ⋅ 2 n − 1}
2
89. In (1 + x )n = C0 + C1 x + C2 x 2 + ... + Cn x n = n(n − 1)⋅ 2 n − 3
On differentiating both sides w.r.t. x, we get n −1
n(1 + x )n − 1 = C1 + 2C2 x + K + nCn x n − 1 98. ∑ ∑ j ⋅ nCi = ∑ n
Cr [(r + 1) + (r + 2 ) + ... + (n )]
0 ≤ i< j ≤ n r =0
Put x =1
n −1
n ⋅ 2 n − 1 = C1 + 2C2 + .... + nCn (n − r )
= ∑ n
Cr
2
(r + 1 + n )
90. In (1 + x 2 )n x = C0 x + C1 x 3 + C2 x 5 + .... + Cn x 2 n + 1 r =0
n −1
On differentiating both sides w.r.t. x and putting x = 1 (n + 1) r ( n − r )
(n + 1)⋅ 2 n = C0 + 3 C1 + 5 C2 + .... + (2 n + 1) Cn
= ∑ n
Cr
2
(n − r ) +
2
r =0
100
Cr ⋅ 200
C150 + r = 100
Cr ⋅ 200 n+1 n n n 1 n
91. C50 − r = ∑
2 r=0
( n − r ) ⋅ nC r − ∑ r ⋅ nC r + ∑ r 2 ⋅ nC r
2r=0 2r=0
50
Thus, ∑ 100Cr ⋅200 C50 − r = 300
C50 n+1 n n n 1 n
r =0 = ∑
2 r =0
r ⋅ nCr − ∑ r ⋅ nCr + ∑ r 2 ⋅ nCr
2r =0 2 r =0
2n 2n
92. ∑ (−1)r ⋅ (2nCr )2 = ∑ (−1)r ⋅2n Cr ⋅2n Cr 1 n n
2 r∑ ∑
r =0 r =0 = r ⋅ nCr + r 2 ⋅ nCr = n(n + 3)⋅ 2 n − 3
2n =0 r =0
= ∑ (−1)r ⋅2n Cr ⋅2n C2n − r n
r =0 99. ∑ (−1)r ⋅ r 2 ⋅ nCr
= Coefficient of x 2n
in (1 − x ) ⋅ (1 + x )
2n 2n
r =0
n
= Coefficient of x 2 n in (1 − x 2 )2 n = (−1)n ⋅2 n Cn
2n 2n
Since, (1 − x )n = ∑ (−1)r ⋅ nCr xr
2n r =0
93. I = ∑ r ⋅ (2nCr )2 = ∑ r ⋅ r
2n − 1
Cr − 1 ⋅2 n Cr On differentiating both sides w.r.t. x, we get
r =0 r =0 n
= 2n ∑
2n
2n − 1
Cr − 1 ⋅ 2n
Cr
− n(1 − x )n − 1 = ∑ (−1)r ⋅ nCr ⋅ r ⋅ x r − 1
r =0
r =0 n
∑ (−1)r ⋅ nCr ⋅ r ⋅ x r
4n − 1
= 2n ⋅ C2 n ⇒ − nx(1 − x )n − 1 =
⇒ I = n ⋅ C2 n 4n r =0 403
8 On differentiating both sides again w.r.t. x, we get
n
∑ (−1)r ⋅ nCr ⋅ r 2 ⋅ x r −1 = − n(1 − x )n −1 + n(n − 1)x(1 − x )n − 2
r =0
= n 2 ⋅ Coefficient of x n − 2 in (1 + x )n − 1 ⋅ (1 + x )n − 1
= n 2 ⋅ 2 n − 2Cn − 2
Objective Mathematics Vol. 1
2n − 2
Now, putting x = 1on both sides, we get = n2 ⋅ Cn
n
∑ (−1)r ⋅ nCr ⋅ r 2 = 0 107. Given,
n+1 n
r =0
n+1 × Cr
Cr + 1 11 r +1 11
n n = ⇒ =
∑ ∑ (Ci − C j )2 − 0 Cr n
6 n
Cr 6
i=0j=0
100. ∑ ∑ (Ci − C j ) 2
=
2
⇒ 6 n + 6 = 11 r + 11 ⇒ 6 n − 11 r = 5
Also,
…(i)
0≤i< j≤n
n
=
∑∑ Ci2 + ∑∑ C j2 − 2 ∑ ∑ Ci C j
n −1
Cr
=
6
2 Cr − 1 3
= (n + 1) 2 nCn − 2 2 n n n −1
× Cr − 1
r 6
101. Coefficient of x in ⇒ n −1
=
Cr − 1 3
{ 2nC 0(1 + x )2n + 2n
C1(1 + x )2n − 1 + C 2(1 + x )2n − 2 + ...}
2n
⇒ n = 2r …(ii)
⇒ Coefficient of x in (2 + x ) 2n
= n ⋅ 22n From Eqs. (i) and (ii),
n −1 r = 5 and n = 10
102. ∑ ∑ n
Ci = ∑ nCr (1
1 + 1 + 1 + .... + 1)
4442444 3 ∴ nr = 50
0≤i< j≤n r =0 ( n − r ) times
n −1
108. We have, f ( x ) = x n . So,
f 1( x ) = nx n − 1 ⇒ f 1(1) = n
= ∑ (n − r )⋅n Cr
r =0 f 2 ( x ) = n(n − 1 )x n − 2 ⇒ f 2 (1) = n(n − 1)
n n
f ( x ) = n(n − 1) (n − 2 )x n − 3 ⇒ f 3 (1) = n(n − 1) (n − 2 )
3
= ∑ (n − r )⋅ nCr = ∑ { n − (n − r )} ⋅ nCn − r
Ta rg e t E x e rc is e s
r =0 r =0
KKKK
n f n ( x ) = n(n − 1)(n − 2 ) K 1
= ∑ r ⋅n Cr = n ⋅ 2n − 1 ⇒ f n (1) = n(n − 1) (n − 2 ) K 1
r =0
f 1(1) f 2 (1) f n (1)
⇒ f (1) + + +K+
103. ∑ ∑ ∑ ∑ 2 = 2 n + 1C4 1 2! n!
0 ≤ i< j < k < l ≤ n n n(n − 1) n(n − 1)(n − 2 )
= 1+ + +
104. ∑ ∑ (i ⋅n Ci ) ( j ⋅n C j ) 1 2! 3!
n(n − 1) (n − 2 ) K 1
1≤ i < j ≤ n −1 +K+
n!
= n2 ∑ ∑ n − 1Ci − 1 ⋅n − 1 C j − 1 = nC0 + nC1 + nC2 + K + nC = 2 n
1≤ i < j ≤ n −1
∞ k
2 2( n − 1) − 2( n − 1)Cn − 1 1
= n 2 ⋅
109. ∑ ∑ 3k (k Cr )
2 k =1r = 0
∞ 1 k
105. Q n C0 + nC1 + nC2 + ... + nCn = 2 n = ∑ 3k ∑ k Cr
and n
Pn = n ! k = 1 r = 0
∞
∞
C0 + C1 + C2 + K + Cn
n n n n 2k
= ∑ k
∴ S = ∑ n
Pn
k = 1 3
n =1
2
∞
2 n 21 2 2 2 3 2 2
= ∑ = + + + ... + +K
=
3 3
n = 1
n ! 1! 2 ! 3 !
2/ 3
= =2
x x2 x3 2
Q e = 1 + 1! + 2 ! + 3 ! + K
x
1−
= e2 − 1 3
2 22 23 1 1× 4 2 1× 4 × 7 3
so e = 1 + 1! + 2 ! + 3 ! + K
2
110. Let (1 + y )n = 1 + x + x + x +K
3 3× 6 3× 6× 9
n n n(n − 1) 2
= 1 + ny + y +K
106. ∑ r(n − r )(nCr )2 = ∑ r ⋅ nCr (n − r )⋅ nCn − r 2!
r =0 r =0
Comparing the terms, we get
n
1 n(n − 1) 2 1 × 4 2
=∑ n⋅ n −1
Cr − 1 ⋅ n ⋅ n −1
Cn − r − 1 ny = x, y = x
r =0
3 2! 3× 6
n Solving, n = − 1 / 3, y = − x
= n2 ∑ n −1
Cr − 1 ⋅ n −1
Cn − r − 1 Hence, the given series is (1 − x )− 1/ 3 .
404 r =0
8
n
113. The general term of x +
111. Given that, a
5 5⋅ 7 5⋅ 7 ⋅ 9 is
α= + + +K …(i) x2
2 ! 3 3 ! 32 4 ! 33 r
a
Binomial Theorem
Tr + 1 = nCr 2 x n − r = nCr ar x n − 3 r
We know that, x
nx n(n − 1) 2
(1 + x )n = 1 + + x If the term independent of x does not exist, then n − 3 r
1! 2! must not be zero for positive integers n and r.
n(n − 1) (n − 2 ) 3
+ x + K …(ii) n
⇒ r = must not be an integer. Now, the integers
3! 3
On comparing Eqs. (i) and (ii), with respect to factorial given in options (a), (b) and (d) are not divisible by 3.
5
n(n − 1)x 2 = …(iii) Options (a), (b) and (d) are correct options.
3 Since, 15 is divisible by 3, so option (c) is not true.
5⋅ 7
n(n − 1)(n − 2 ) x 3 = 2 …(iv) α
n
3 114. In x + ,
5⋅ 7 ⋅ 9 x
and n(n − 1)(n − 2 ) (n − 3)x 4 = …(v) r
33 α
Tr + 1 = nCr x n − r
On dividing Eq. (iv) by Eq. (iii) and Eq. (v) by Eq. (iv), x
7
we get (n − 2 )x = ... (vi) = nCr x n − 2 r α r
3
Tr + 1 is independent of x, if n − 2 r = 0
and (n − 3)x = 3 …(vii)
⇒ 2r ⇒ n is divisible by 2.
Again, dividing Eq. (vi) by Eq.(vii), we get n
n −2 7 β
= Again, in x + 2
x
n−3 9
r
⇒ 9 n − 18 = 7 n − 21 β
Tr + 1 = nCr x n − r 2 = nCr x n − 3 r − β r
3 x
⇒ 2n = − 3 ⇒ n = −
2 Tr + 1 is independent of x, if n − 3 r = 0 i.e. n = 3 r
On putting the value of n in Eq. (vi), we get n is divisible by 3.
Targ e t E x e rc is e s
3 7 115. ( x 2 + 2 x + 2 )n = {2 + x(2 + x )} n
− − 2 x =
2 3
= 2 n + nC1 2 n − 1 x(2 + x ) + nC2 2 n − 2 x 2 (2 + x )2
2
⇒ x=− + nC3 2 n − 3 x 3 (2 + x )3 + ...
3
∴ From Eq. (ii), we get It is easy to observe that coefficient of x = n ⋅2 n and the
− 3/ 2
5⋅ 7 coefficient of x 2
2 5
1 − = 1+ 1+ + +K n(n − 1) 2 n − 2
3 2 ! 3 3 ! 32 = n ⋅ 2n − 1 + = n2 ⋅ 2n − 1
5 5⋅ 7 2
⇒ 33 / 2 − 2 = + +K Coefficient of x 3 = nC2 ⋅ 2 n − 2 ⋅ 4 + nC3 ⋅ 2 n − 3 ⋅ 2 3
2 ! 3 3 ! 32
⇒ α = 33 / 2 − 2 [from Eq. (i)] = 2 n ⋅ n + 1C3
Now, α + 4 α = (33 / 2 − 2 )2 + 4(33 / 2 − 2 )
2
116. (a) 3100 = 950 = (10 − 1)50
= 27 + 4 − 4 ⋅ 33 / 2 + 4 ⋅ 33 / 2 − 8 = 10 50 − 50C1 10 49 + K − 50 C49 101 + 50C50
= 23 ⇒ A multiple of 100 + 1.
112. ( x sin p + x − 1 cos p)10 ⇒ Last two digits of 3100 are 01.
⇒ Choice (a) is wrong.
The general term in the expansion is
Again, 350 = 925 = (10 − 1)25
Tr + 1 = 10Cr ( x sin p)10 − r ( x − 1 cos p)r
= 10 25 − 25C1 10 24 + K + 25C24 10 − 25C25
For the term independent of x, we have 10 − 2 r = 0 = A multiple of 100 + 249
or r = 5.
⇒ The remainder when 350 divided by 100 is 49
Hence, the independent term is
⇒ Last two digits of 350 are 49.
sin 5 2 p
10
C5 sin 5 p cos 5 p = 10C5 ⇒ Option (b) is false and option (c) becomes true.
32 (d) From the same expansion, we get
which is the greatest when sin 2 p = 1 350 = A multiple of 1000 + 249
sin 5 2 p 10 ! ⇒ The remainder when 350 is divided by 1000 is
The least value of 10 C5 is − 5 ,
32 2 (5 !)2 249.
π ∴The last three digits are 249.
when sin 2 p = − 1or p = (4n − 1) , n ∈ Z.
4 117. Sum of coefficients in (aα 2 x 2 + 2 bαx + c )n is
Sum of coefficients is (sin p + cos p)10 , when x = 1 (aα 2 + 2 bα + c )
or (1 + sin 2 p) , which is least, when sin 2 p = − 1
5
Let f (α ) = aα 2 + 2 bα + c
Hence, least sum of coefficients is zero. Greatest sum Such that sum = { f (α )} n
of coefficients occurs, when sin 2 p = 1. Now, f (α ) > 0, ∀ n and α ∈ R, then a > 0
Hence, the greatest sum is 2 5 = 32. 405
Also, f (α ) < 0, ∀ n ∈odd and α ∈ R, if a < 0
8
n
124. Let( 5 + 2 )n = N + f, where N is an integer and 0 < f < 1
118. f (n ) = ∑ {(r + 1)2 Cr − r 2 nCr − 1}
n
Let ( 5 − 2 )n = f ′, then 0 < f ′ < 1
r =1
= (n + 1)2 nCn − 1 = (n + 1)2 − 1 Let ( 5 + 2 )n − ( 5 − 2 )n = integer [Q n is odd]
Objective Mathematics Vol. 1
10 where N = ( 5 + 2 )n − ( 5 − 2 )n [Q f = f ′]
Now, ∑ f (n ) = (2 2 − 1) + (32 − 1) + (42 − 1) + (52 − 1) n − 1 n − 3
n =1 = 2 nC1 ⋅ 2 ⋅ 5 2 + nC3 ⋅ 2 3 ⋅ 5 2 + ...
+ (62 − 1) + (7 2 − 1) + (82 − 1) + (92 − 1)
+ (10 2 − 1) + (112 − 1) ⇒ N is divisible by 20n on using Statement II.
11⋅ (11 + 1)(22 + 1) [if n is prime and r < n, then there is no factor
= − 11 = 495
6 which will cut n, so nCr will be divisible by n]
119. The given expression is the coefficient of x n in 125. Statement I is true, since
(1 + x )n + (1 + x )n + 1 + K + (1 + x )n + k ( 2 − 1)2 = 3 − 2 2 = 9 − 8
(1 + x )k + 1 − 1 ( 2 − 1)3 = 5 2 − 7 = 50 − 49
⇒ Coefficient of x n in (1 + x )n
1+ x − 1 [this could be proved by induction]
(1 + x )n (1 + x )k + 1 − (1 + x )n Statement II is also true, since any integral power of
= ( 2 − 1) will have rational and irrational parts. The
x
⇒ Coefficient of x n + 1 in [(1 + x )n + k + 1 − (1 + x )n ] irrational part will have only one surd 2. Thus,
n+ k+1 ( 2 − 1)n = A + B 2 , where A and B are integers.
i.e. Cn + 1 or n + k + 1Ck
But, Statement II does not explain the Statement I.
120. Given, 69
C3 r − 1 + 69
C3 r = Cr 2 − 1 +
69 69
Cr 2 , 70
C3 r = 70
Cr 2 126. X = (n 2 − n )(n 2 − n − 2 ) = n(n − 1)(n − 2 )(n + 1)
Either 3 r = r or r + 3 r = 70
2 2 = (n − 2 )(n − 1)n(n + 1)
X n+1
i.e. either r = 0, 3 or r = 7, −10 ⇒ = C4 ∈ I
24
But the given equation is not defined for
Hence, m(m − 2 ) is divisible by 24.
r = 0, −10
Ta rg e t E x e rc is e s
Hence, r = 3 or 7 127. Since, S1 = 1C1 and T1 = 1 ⇒ S1 = T1
121. Since, i , j and k are distinct, n − i + 1, n − j + 1, n − k + 1 Thus, if Statement II is true and if S n = Tn for some n,
then S n + 1 = Tn + 1
are also distinct and they all lie from 1 to n. ⇒ S n = Tn for all n.
Now, S = Σ Σ Σ (− xn − i + 1 )(− xn − j + 1 )(− xn − k + 1 )
Thus, Statement I is true, if Statement II is true. Let us
= − Σ Σ Σ xi x j xk = − S ⇒ S = 0 for all n prove Statement II.
Options (a), (b) and (c) are correct since the 1
Now, Tn + 1 − Tn =
statements given in options (a), (b) and (c) are not n+1
correct. n+1 n+1
C2 C3
and S n + 1 − S n = n + 1C1 − + −K
122. We have, 2 3
n n+1
m
30 20 m
30 20 (−1)n −1 n + 1Cn (−1) Cn + 1
…+ +
f ( m) = ∑ 30 − i m − i ∑ i m − i =
= 50
Cm
n n+1
i=0 i=0
n
C2 n
C3 (−1)n − 1 nCn
Now, f (m) is the greatest, when m = 25. − n C1 − + −K+
Also, f (0 ) + f (1) + ... + f (50 ) 2 3 n
n+1 1 n+1 1 n+1
= 50C0 + 50C1 + K + 50C50 = 2 50 =( C1 − C1 ) − (
n
C2 − C1 ) + (
n
C3 − C3 )
n
50 2 3
Also, Cm is not divisible by 50 for any m as 50 is not a n −1
(−1) (−1)n
prime number. −… + ( n + 1Cn − nCn ) +
50 n n+1
∑ (f (m))2 = (50C0 )2 + (50C1)2 + K + (50C50 )2 = 100C50 −
n n n n
C C ( 1) C
= nC0 − 1
+ 2
−K+ n
m=0 2 2 n+1
12 − r + 1 11
1 1
123.
Tr + 1
= ⋅ = ∫ (1 − x )n dx =
Tr r 10
0 n+1
Let Tr + 1 ≥ Tr ⇒ 13 − r ≥ 11.x Thus, Statement II is true.
⇒ 13 ≥ 2.1r ⇒ r ≤ 6.19 i j n−i n−j
128. S = ∑ ∑ n + n = ∑ ∑ n + n
Hence, the greatest term occurs for r = 6. Hence, 7th 0 ≤ i < j ≤ n Ci C j 0 ≤ i < j ≤ n Cn − i Cn − j
term is the greatest term. Also, the binomial coefficient 1 1
of 7th term is 12 C6 , which is the greatest binomial = n ∑ ∑ n + n − S
coefficient. 0 ≤ i < j ≤ n Ci Cj
But this is not the reason for which T7 is the greatest. n 1 1
⇒ S = ∑ ∑ n + n
Here, it is coincident that the greatest term has the 2 0 ≤ i < j ≤ n Ci Cj
greatest binomial coefficient.
n r n n n n2
n − 1
n−r n
Hence, Statement I is true, Statement II is true; but = ∑ n + ∑ n = ∑ n = a
406 Statement II is not a correct explanation of Statement I. 2 r = 0 Cr r = 1 Cr
2
r = 0 Cr 2
r + 1 + 1 n
8
n n n
Solutions (Q. Nos. 129-131) 1
We have,
133. Now, ∑ r + 1
Cr = ∑ Cr +
n
( n + 1 )
∑ n+1
Cr + 1
r =0 r =0 r =0
r 2e 2 iθ r 3e 3 iθ
iθ iθ 1
Binomial Theorem
C + iS = 1 + re + + + K = e re …(i) = 2n + (2 n + 1 − 1)
2! 3! (n + 1)
r 2e −2 iθ r 3e −3 iθ − iθ n
r + 2 n 28 − 1
and C − iS = 1 + re −iθ + + + K = e re Since, ∑
2! 3! Cr = [given]
r =0
r + 1 6
…(ii)
Clearly, C 2 + S 2 2 n +1 − 1 2 8 − 1
∴ 2n + =
iθ − iθ iθ − iθ n+1 6
= e re ⋅ e re = e r (e + e ) = e 2 r cos θ …(iii)
⇒ n=5
On differentiating Eqs. (i) and (ii) w.r.t. r, we get
dC dS r 2e 3 iθ iθ
134. Using, (1 + x )n = C0 + C1 x + C2 x 2 + C3 x 3 + C4 x 4 + K
+i = e iθ + re 2 iθ + + K = e iθe re …(iv) On putting x = i , we get
dr dr 2!
(1 + i )n = (C0 − C2 + C4 − C6 + ... )
dC i dS r 2e −3 iθ
− = e −iθ + re −2 iθ + +… + i (C1 − C3 + C5 − C7 + ... )
dr dr 2!
n nπ nπ
r 2e −2 iθ ( 2 ) cos + i sin = (C0 − C2 + C4 − C6 + ... )
= e − iθ 1 + r ⋅ e − iθ + + … 4 4
2! + i (C1 − C3 + C5 − C7 + ... )
− iθ
= e − iθ ⋅ e re …(v) On equating the real part, we get
n
On multiplying Eqs. (iv) and (v), we get nπ
C0 − C2 + C4 − C6 + K = 2 2 ⋅ cos
dC
2
dS
2 4
− iθ re − iθ iθ
+ = e ⋅e ⋅ e iθ ⋅ e re n
dr dr 1
( − iθ + iθ ) − iθ
+e iθ
135. Given, a = ∑ nC
=e ⋅e r (e )
r =0 r
= e 2 r cos θ = C 2 + S 2 i j
Let y= ∑
∑ n
+ n
C j
Targ e t E x e rc is e s
Now, on multiplying Eqs. (ii) and (iv), we get 0≤i≤n0≤ j≤n Ci
dC dS iθ r ( e iθ + e − iθ ) n−i n−j
+i (C − iS ) = e e
dr dr = ∑ ∑ + n
0≤i≤n0≤ j≤n
n
Cn − i Cn − j
= e i θe 2 r cos θ
1 1
= (cos θ + i sin θ )e 2 r cos θ …(vi) =n ∑ ∑ + −y
n Cn − j
n
0 ≤ i ≤ n 0 ≤ j ≤ n Cn − i
On equating real and imaginary parts from Eq. (vi)
dC dS ⇒ 2 y = n{ a ⋅ (n + 1) + a ⋅ (n + 1)} ⇒ y = na(n + 1)
C +S = (C 2 + S 2 )cos θ
dr dr 136. Since, (1 + x )n = C0 + C1 x + C2 x 2 + K + Cn x n
dS dC
and C −S = (C 2 + S 2 )sin θ Replace x by ix, we get
dr dr
(1 + ix )n = C0 + iC1 x − C2 x 2 − iC3 x 3 + C4 x 4 + iC5 x 5 − K
n
⇒ (1 + ix )n = (C0 − C2 x 2 + C4 x 4 + K )
132. Given, ( x + y )n = ∑ n
Cr x n − r y r , where Cr = nCr
r =0 + i (C1 x − C3 x 3 + C5 x 5 + K ) …(i)
n
Similarly, (1 − ix ) = (C0 − C2 x 2 + C4 x 4 + K )
n
Using (1 + x )n = ∑ n
Cr x r …(i)
− i (C1 x − C3 x 3 + C5 x 5 + K ) …(ii)
r =0
On integrating between 0 to 1, we get On adding Eqs. (i) and (ii), we get
(1 + x )n + 1
1 (1 + ix )n + (1 − ix )n = 2(C0 − C2 x 2 + C4 x 4 − K )
C1 C2 Cn
n + 1 = C0 + 2 + 3 + K + n + 1 On putting x = 1, we get
0 C0 − C2 + C4 − C6 + C8 − K
C1 C2 C3 C4 Cn 2n + 1 − 1 nπ
∴ C0 + + + + +K+ = = (1 + i )n + (1 − i )n = 2 n / 2 cos …(iii)
2 3 4 5 n+1 n+1 4
…(ii) Also, C0 + C2 + C4 + C6 + C8 + K = 2 n − 1 …(iv)
Also, integrating Eq. (i) between 0 to −1, we get On adding Eqs. (iii) and (iv), we get
−1
(1 + x )n + 1 C1 C2 C3 n
−1 nπ
n + 1 = − C0 + 2 − 3 + 4 − ... C0 + C4 + C8 + C12 + K = 2 n − 2 + 2 2 cos
0 4
C1 C2 C3 C4 1 1 n
∴ − C0 + − + − + ... = − …(iii) 137. n
C0 + nC3 + nC6 + K = [2 + (−1)n ω 2 n + (−1)n ω n ]
2 3 4 5 n+1 3
1 n
On adding Eqs. (ii) and (iii), we get = [2 + (−1)n (ω n + ω 2 n )]
n+1
−2 3
C C C 2
2 1 + 3 + 5 + K = 1
= [2 n + (−1)n (ω n + ω n )] [Qω 2 = ω]
2 4 6 n+1 3
C1 C3 C5 2n − 1 1 nπ
∴ + + +K= = 2 n + 2 cos
2 4 6 n+1 3 3 407
8
n 3
1− t 8
138. ∑ (−3)r − 1 ⋅3n C2r − 1 = Coefficient of t 10 in t 3
r =1 1− t
n n
Objective Mathematics Vol. 1
1 = Coefficient of t 7 in (1 + 3C1 t + 3C2 t 2 + K )
= ∑( 3 i )2 r − 2 ⋅3 n C2 r − 1 =
3i
∑( 3 i )2 r − 1 ⋅3 n C2 r − 1
r =1 r =1 = 9C7 = 36
1
=
2i 3
Imaginary part of {(1 + i 3 )2 n − (1 − i 3 )2 n } 144. A. Let S = ∑ ∑ i ⋅ j ⋅ nCi ⋅ nC j
0 ≤ i ≤ j ≤ 10
=0 n n n
= ∑ ∑ i ⋅ j ⋅ nCi ⋅ C j − S + ∑ i 2 (n Ci )2
n n n n n n n
139. ∑ ∑ (Cr + Cs ) = ∑ ∑ Cr + ∑ ∑ Cs i=0j=0 i= 0
r =0 s=0 r =0 s=0 r =0 s=0 n n n
n n n
n ⇒ 2S = ∑ ∑ i ⋅ j ⋅ nCi ⋅ nCk − ∑ i 2(nCi )2
= ∑ ∑ Cr + ∑ ∑ C s i=0j=0 i=0
s = 0 r = 0 r =0 s=0 n
= (n + 1)2 n + (n + 1)2 n ⇒ 2S = (n2 n − 1 )⋅ (n2 n − 1 ) − ∑ i 2(nCi )2
i=0
= (n + 1)2 n + 1 n
s n where, ∑ i 2(nCi )2 is the coefficient of x 0 in the
∑ ∑ nC s Cr
s
140. i=0
r = 0 s =1 n −1
1
n n product of n (1 + x )n − 1 n 1 + .
x
= ∑ nC s ( sC0 + C1 + ... + sC s ) =
s
∑ nC s (2 s )
s =1 s =1 n 2 (1 + x )2 n − 2
n Coefficient of x 0 in the product of
xn − 1
= ∑ nC s2 0 − 5C02 0 = (1 + 2 )n − 1 = Coefficient of x n −1
in the product of n 2 (1 + x )2 n − 1
s=0
2n − 2
= 3n − 1 = n2 Cn − 1
∴ 2S = n {2 2 n − 2 −
2 2n − 2
Ta rg e t E x e rc is e s
n n n n n n n n Cn − 1}
141. ∑ ∑ ∑ ∑ (1) = ∑ (1) ∑ (1) ∑ (1) ∑ (1) For n = 10, we have
r =0 s=0 t =0 u =0 r =0 s=0 t =0 u =0
2S = 100{218 − 18C9 } × 100
= (n + 1) (n + 1) (n + 1) (n + 1)
= (n + 1)4 Last digit in 218 is 4 and that the last digit in 18
C9 is
n+1 zero (0).
142. Given, f ( x ) = x + x n
Hence, last non-zero digit in 218 − 18C9 is 4.
n! (n + 1)!
f k( x) = xn − k + xn + 1 − k 4
34 − k x k
(n − k )! (n + 1 − k )! B. ∑ (4 − k )! k !
⇒ (1 + x )f k ( x ) k=0
n! (n + 1)! 4
34 − k x k 4 !
= xn − k + x n + 1 − k (1 + x ) = g( x ) = ∑ (4 − k )! k ! 4 !
(n − k )! (n + 1 − k )! k=0
[say] 4 4
Ck ⋅ 34 − k ⋅ x k (3 + x )4
g ( n + 1 − k) (n + 1)!
( x) =
(n + 2 − k )! x
= ∑ 4!
=
4!
k=0
(n + 1 − k )!
According to the question,
n! (n + 1)!
+ + (n + 1 − k )! (3 + x )4 32
(n − k )! (n + 1 − k )! =
4! 3
(n + 1)! ⇒ (3 + x )4 = 256
∴ A−B= (n + 2 − k )!
(n + 1 − k )! ⇒ x + 3= 4
n! (n + 1)! ⇒ x =1
− + (n + 1 − k )!
(n − k )! (n + 1 − k )! C. Since, (1 + x + x 2 )n
= (n !) n(n + 1 − k ) = a0 + a1 x + a2 x 2 + K + a2 n x 2 n …(i)
143. A. Since, xyz = 3 5
On substituting x = ω, ω 2 and 1 and then adding them
Number of solutions of (x, y, z) = 3 + 5 − 1C5 = 21 together
B. Number of terms = 6 + 3 − 1C3 − 1 = 28 a0 + a3 + a6 + K = 3n − 1
C. Since, x 2 + x − 400 ≤ 0 On multiplying Eq. (i) by x 2 and then repeating the
⇒ x( x + 1) ≤ 400 same process again, we get
Solutions are 1, 2, 3, …, 19. a1 + a4 + a7 + K = 3n − 1
Number of solutions = 19 ⇒ a0 + a3 + a6 + K = a1 + a4 + a7 + K
D. Since,x + y + z = 10 = a2 + a5 + a8 + K
Number of solutions 2 ⋅ 3n − 1
∴The required ratio is n − 1 = 2
408 = Coefficient of t 10 in ( t + t 2 + K + t 8 )3 3
D. Let E =
C4 + ∑
47
57
3
1
C4
j=0
50 − j
5
j=0
C3 + ∑ 56 − k C53 − k
147. The sixth term of the given binomial expansion is
m
x
C5 2log(10 − 3 )
m−5
⋅( 5 2 x − 2 log 3 )5 = 21 …(i) 8
Binomial Theorem
1
⇒ E = 57 47C4 + (47C4 + 48C4 + 49C4 + 50C4 ) The other given condition is mC1 + mC3 = 2 ⋅ mC2
C4
Q m ≠ 0, so m2 − 9 m + 14 = 0 ⇒ (m − 7 )(m − 2 ) = 0
5
Q m = 2 is not possible.
+ ∑ 56 − k C53 − k Putting m = 7 in Eq. (i), we get x = 0, 2
j=0
1 Hence, the sum of roots = 0 + 2 = 2
⇒ E = 57 [47C4 + (47C4 + 48
C4 + C4 +
49 50
C4 ) 12 − r
C4 1
148. Tr + 1 = Cr 34
12
⋅ (4)r / 3
+ (51C3 + 52
C3 + 53
C3 + 54
C3 + 55
C3 + 56
C3 )]
= 12Cr ⋅ 3 ( 3 − r / 4 ) ⋅ (4)r / 3
57
C
∴ E = 57 4 = 1
C4 For rational term, r = 0 or 12
⇒ 2E = 2 ∴ Sum = T1 + T13 = 12C0 ⋅ 33 + 12C12 ⋅ 44
145. We know that n C02 + nC12 + ... + nCn2 = 2n
Cn = 33 + 44 = 27 + 256 = 283
⇒ {283} = 0
and n
C02 − n
C12 + ...+ (− 1)n nCn2
0 , if n is odd 149. We have, (1 + x + x 2 + x 3 + x 4 ) n ( x − 1)n + 3
= n 1 − x5
n
Cn / 2 (−1) , if n is even
n
= (1 − x )n + 3 = (1 − x 5 )n (1 − x )3
1− x
From this,
n
C02 − 31C12 + − ... − 31C31 =0 = (− x 3 + 3 x 2 − 3 x + 1) ∑ n Cr (−1)r x 5 r
31 31 2 2
C2
r =0
32
C02 − 32
C12 + 32
C22 − ... + 32 2
C32 = 32
C16 n n
32
C02 + 32
C12 + 32
C22 + ... + 32 2
C32 = 64
C32 =− ∑ nCr (−1)r x 5r + 3 + 3 ∑ nCr (−1)r x 5r + 2
r =0 r =0
Targ e t E x e rc is e s
Also, (1 / 32 ) (1 × 32
C12 +2× 32
C22 + ... + 32 × 32 2
C32 ) n n
32 32
− 3 ∑ n Cr (−1)r x 5 r + 1 + 3 ∑ n Cr (−1)r x 5 r
1 1
=
32 ∑ r (32Cr )2 = 32 ∑ r 32Cr32C32 − r r =0
For term containing x 83 , we have
r =0
r =1 r =1
1 32 5 r + 3 = 83
=
32 ∑ 32 31
Cr − 132C32 − r ⇒ r = 16
r =1 whereas 5 r + 2 = 83, 5 r + 1 = 83 and 5 r = 83 give no
= 63
C31 = 63
C32 integral value of r.
n −1 Hence, their is only one term containing x 83 whose
146. f (n ) = C0 a
n
− C1 an − 2 + n C2 an − 3
n
coefficient
− K+ (− 1)n − 1 nCn − 1a0 = − nC16 = − nC2 λ
1 n ⇒ 2 λ = 16
= ( C0 an − nC1 an − 1 + nC2 an − 2
a ∴ λ=8
− K+ (− 1)n − 1 nCn − 1 a)
150. We have, coefficient of x 4 in (1 + x + x 2 + x 3 )11
1
1
1
= {(a − 1)n − (− 1)n nCn } = (3223 )n − (− 1)n = Coefficient of x 4 in (1 + x 2 )11 (1 + x )11
a a = Coefficient of x 4 in (1 + x )11 + Coefficient of x 2 in
n
11⋅ (1 + x )11 + Constant term in 11C2 ⋅ (1 + x )11
3223 − (− 1)n = C4 + 11⋅ 11C2 + 11C2
11
∴ f (n ) = 1
3223 + 1 = 990 = 2 × 32 × 5 × 11
2007 2008 ∴ Number of divisors of the form 9k is
3 223 + 1 3 223 − 1 2 × 2 × 2 × 1= 8
⇒ f (2007 ) + f (2008) = +
151. Middle term is + 1 th
n
1 1
3223 +1 3223 + 1 2
2007 2008 1
9+
3 223 + 3 223 39 + 3 223 i.e. (4 + 1)th i.e. T5
= = x
4
1 1
∴ T5 = 8C4 ⋅ 2 4 = 1120
3223 +1 3223 +1 2
1 8⋅ 7 ⋅ 6⋅ 5 4
⇒ ⋅ x = 1120
39 1 + 3223 4⋅ 3⋅ 2 ⋅ 1
⇒ x4 =
1120
= 16
= = 39
1 70
1 + 3223 ⇒ ( x + 4) ( x − 4) = 0
2 2
∴ x = ± 2 only as x ∈ R
⇒ 39 = 3k ⇒ k=9 409
Thus, the sum of possible real values of x is 0.
8 152. (17 )256 = (289)128
= (300 − 11) 128
= 128C0 (−11)128 + 100 m, for some integer m
155. C0 − C1
(1 + x )
(1 + nx )
= C0 −
C1
+ C2
+
(1 + 2 x )
(1 + nx )
C2
2
− C3
−
(1 + 3 x )
(1 + nx )3
C3
+ K
+K
Objective Mathematics Vol. 1
(1 + nx )
144444444244444444 + 2
+ 3
3
= 11128 + 100 m (1 nx ) (1 nx )
= (10 + 1)128 + 100 m I1
x 2C2 3 C3
= 128C0 1128 + 128C110 + 100 m1 + 100 m + − C1 + 1 + nx − + K …(i)
1 + nx 144444 (1 + nx )2
for some integer m1 42444444 3
= 1 + 1280 + 100 k, m + m1 = k I2
= 1281 + 100 k Q (1 − y )n = C0 − C1 y + C2 y 2 − C3 y 3 + ...+ (−1)n y n …(ii)
Hence, the required number is 81 = ab On differentiating w.r.t. y, we get
⇒ a+ b=9 − (1 − y )n − 1 = − C1 + 2C2 y − 3 C3 y 2 + K …(iii)
1(2 2000 − 1) 1 nx
n n
153. = 2 2000 − 1 Now, I1 = 1 − = …(iv)
1 1 + nx 1 + nx
= (5 − 1)1000 − 1 = (1 − 5)1000 − 1
1 1
= 1 − 1000C1 ⋅ 5 + 1000C2 ⋅ 52 + K + 1000
C1000 ⋅ 51000 − 1 and I2 = − C1 + 2C2 ⋅ − 3 C3 ⋅ +K
1 + nx (1 + nx )2
which is divisible by 5. n −1 n −1
1 nx
= (− n )1 − = −n …(v)
4
54 − k x k 4
54 − k x k 4 ! 1 + nx 1 + nx
154. ∑ = ∑ Putting I1 and I2 from Eqs. (iii) and (v) in Eq. (i), we get
k = 0
(4 − k )! k ! k = 0 (4 − k )! k ! 4 !
(1 + x ) (1 + x ) (1 + 3 x )
4 4
Ck ⋅ 54 − k ⋅ x k (5 + x )4 C0 − C1 + C2 − C3
(1 + nx )
= ∑ 4!
=
4!
(1 + nx )2 (1 + nx )3
1 + nx
k=0
+ K + (−1)n Cn
(5 + x )4 8 (1 + nx )n
Now, = n n −1
4! 3 nx x nx
Ta rg e t E x e rc is e s
⇒ (5 + x ) = 64
4 = − ⋅n⋅
1 + nx 1 + nx 1 + nx
⇒ x + 5 = 23/ 2 = 2 2 n n
⇒ x =2 2 − 5 nx nx
= − =0 ⇒ λ=0
∴ λ=5 1 + nx 1 + nx
Entrances Gallery
1. 2 x1 + 3 x2 + 4 x3 = 11 10
Possibilities are (0, 1, 2); (1, 3, 0); (2, 1, 1); (4, 1, 0).
3. Let y = ∑ A r (B10 Br − C10 A r )
r =1
∴ Required coefficients 10
= (4 C0 × 7C1 × 12C2 ) + (4 C1 × 7C3 × 12C0 ) ∑ A r B r = Coefficient of x 20 in {(1 + x )10 ( x + 1)20 } − 1
r =1
+ (4 C2 × 7C1 × 12C1 ) + (4 C4 × 7C1 × 12C0 ) 10
= (1 × 7 × 66) + (4 × 35 × 1) + (6 × 7 × 12 ) + (1 × 7 ) = C20 − 1 = C10 − 1 and ∑ ( A r )2 = Coefficient of x10 in
r =1
= 462 + 140 + 504 + 7
{(1 + x )10 ( x + 1)10 } − 1 = B10 − 1
= 1113
⇒ y = B10 (C10 − 1) − C10 (B10 − 1) = C10 − B10
2. Let Tr − 1, Tr , Tr + 1 be three consecutive terms of
4. Let Tr +1 be the general term in the expansion of
(1 + x )n + 5 .
(1 − 2 x )50 .
Tr − 1 = n + 5Cr − 2 ( x )r − 2 , Tr = n+ 5
Cr − 1 x r − 1,
n+ 5
∴ Tr + 1 = 50Cr (1)50 − r (−2 x1/ 2 )r = 50Cr 2 r x r / 2 (−1)r
Tr + 1 = Cr x r
For the integral power of x, r should be even integer.
n+ 5 n+ 5 n+ 5
where, Cr − 2 : Cr − 1 : Cr = 5 : 10 : 14 25
n+ 5 n+ 5
∴Sum of coefficients = ∑ 50
C2 r (2 )2 r
Cr − 2 Cr − 1 n+ 5
Cr r =0
∴ = =
5 10 14 1
n+ 5 n+ 5
= [(1 + 2 )50 + (1 − 2 )50 ]
Cr − 2 Cr − 1 2
Now, = 1
5 10 = [30 50 + 1]
2
⇒ n − 3r = − 9 …(i)
n+ 5 Aliter
Cr − 1 n + 5 Cr
and = We have, (1 − 2 x )50 = C0 − C12 x + C2 (2 x )2
10 14 − ... + C50 (2 x )50 …(i)
⇒ 5 n − 12 r = − 30 …(ii)
and (1 + 2 x )50 = C0 + C12 x + C2 (2 x )2
From Eqs. (i) and (ii), n = 6 + ... + C50 (2 x )50 …(ii)
410
On adding Eqs. (i) and (ii), we get
(1 − 2 x )50 + (1 + 2 x )50 = 2[C0 + C2 (2 x )2
+ ... + C50 (2 x )50 ]
For independent of x, put
10 − r r
3
− = 0 ⇒ 20 − 2 r − 3 r = 0
2
8
Binomial Theorem
⇒ 20 = 5 r ⇒ r = 4
(1 − 2 x ) + (1 + 2 x )
50 50
⇒ = C0 + C2 (2 x )2 10 × 9 × 8 × 7
2 ∴ T5 = 10C4 = = 210
+ ... + C50 (2 x )50 4× 3×2 ×1
On putting x = 1, we get 8. ( 3 + 1)2 n = 2n
C0 ( 3 )2 n + 2n
C1 ( 3 )2 n − 1
(1 − 2 1)50 + (1 + 2 1)50 + C2 ( 3 )2 n − 2 + ... +
2n 2n
C2 n ( 3 )2 n − 2 n
= C0 + C2 (2 )2 + ... + C50 (2 )50
2
⇒ ( 3 − 1) 2n
= 2n
C0 ( 3 ) 2n
(− 1) + 0 2n
C1 ( 3 )2 n − 1 (− 1)1
(−1)50 + (3)50
⇒ = C0 + C2 (2 )2 + ... + C50 (2 )50 + 2n
C2 ( 3 )2 n − 2 (− 1)2 + ... + C2 n ( 3 )2 n − 2 n (− 1)2 n
2n
2
1 + 350 On subtracting both the binomial expansions, we get
⇒ = C0 + C2 (2 )2 + ... + C50 (2 )50
2 ( 3 + 1)2 n − ( 3 − 1)2 n
5. Consider, (1 + ax + bx 2 ) (1 − 2 x )18 = 2 [2 n C1 ( 3 )2 n − 1 + C3 ( 3 )2 n − 3
2n
= 11
( − 2 x )18 + ax(1 − 2 x )18 + bx 2 (1 − 2 x )18 + 2n
C5 ( 3 )2 n − 5 + ... + 2n
C2 n − 1 ( 3 )2 n − ( 2 n − 1) ]
Coefficient of x 3 which is most certainly an irrational number because of
odd powers of 3 in each of the terms.
(− 2 )3 18C3 + a(− 2 )2 18C2 + b(− 2 )18 C1 = 0
4 × (17 × 16) 17 9. Here, (1 − x − x 2 + x 3 )6 = {(1 − x ) − x 2 (1 − x )} 6
− 2a ⋅ + b=0 …(i)
(3 × 2 ) 2 = {(1 − x ) (1 − x 2 )} 6 = (1 − x )6 ⋅ (1 − x 2 )6
Coefficient of x 4 6 6
= ∑ (−1)r 6Cr ⋅ x r ∑ (−1)s 6C s ⋅ x 2 s
(− 2 )4 18C4 + a(− 2 )3 C3 + b(− 2 )2 18C2 = 0
18
r = 0 s = 0
16 6 6
(4 × 20 ) − 2 a ⋅ + b = 0 …(ii)
3 = ∑ ∑ (−1)r + s ⋅ 6Cr ⋅ 6C s ⋅ x r + 2 s
Targ e t E x e rc is e s
From Eqs. (i) and (ii), we get r =0 s=0
17 × 8 16 17 Now, consider r + 2 s = 7
4 − 20 + 2 a − = 0
3 3 2 i.e. (s = 1, r = 5)
17 × 8 − 60 2 a (−19) or (s = 2, r = 3) or (s = 3, r = 1)
⇒ 4 + =0 ∴Coefficient of x 7 is
3 6
4 × 76 × 6 {(− 1)5 + 1 ⋅ 6C5 ⋅ 6C1} + {(− 1)3 + 2 ⋅ 6C3 ⋅ 6C2 }
⇒ a= ⇒ a = 16 + {(− 1)1 + 3 ⋅ 6C1 ⋅ 6C3 }
3 × 2 × 19
2 × 16 × 16 = (36) − (20 )(15) + 6(20 )
⇒ b= − 80
3 = 36 − 300 + 120 = − 144
272 10
= 10. QS1 = ∑ j ( j − 1)
10 !
3 j ( j − 1) ( j − 2 )! (10 − j )!
j =1
6. Set X contains elements of the form 10
8!
4n − 3 n − 1 = (1 + 3)n − 3 n − 1 = 90 ∑ ( j − 2 )! { 8 − ( j − 2 )} ! = 90 ⋅ 2 8
= 3n + nCn − 1 3n − 1 + K + nC2 32 j=2
10
n−2 n−3 10 !
= 9 (3 + Cn − 1 3
n
+ K + C2 )
n
and S2 = ∑j j ( j − 1)! { 9 − ( j − 1)} !
Set X has natural numbers which are multiples of 9 j =1
(not all) 10
9!
Set Y has all multiples of 9 = 10 ∑ ( j − 1)! { 9 − ( j − 1)} ! = 10 ⋅ 2 9
j =1
X∪Y=Y
10
10 !
∑ [ j ( j − 1) +
10
x+1 x −1 Also, S 3 = j]
7. Consider, 2 / 3 − j =1
j ! (10 − j )!
x − x1/ 3 + 1 x − x1/ 2
10 10
∑ j ( j − 1) 10C j = ∑ j10C j
10
( x1/ 3 )3 + 13 {( x )2 − 1} =
= 2/ 3 − j =1 j =1
x −x +1
1/ 3
x ( x − 1)
10 = 90 ⋅ 2 8 + 10 ⋅ 2 9
( x + 1)
= ( x1/ 3 + 1) − = 90 ⋅ 2 8 + 20 ⋅ 2 8 = 110 ⋅ 2 8 = 55 ⋅ 2 9
x
− 1/ 2 10 11. 82 n − (62 )2 n + 1 = (1 + 63)n − (63 − 1)2 n + 1
= (x − x
1/ 3
)
= (1 + 63)n + (1 − 63)2 n + 1
∴ The general term is
= [1 + nC1 63 + nC2 (63)2 + ... + (63)n ]
Tr + 1 = 10Cr ( x1/ 3 )10 − r (− x − 1/ 2 )r
10 − r r + [1 − ( 2 n + 1)C1 63 + ( 2 n + 1)
C2 (63)2
−
= 10Cr (− 1)r x 3 2
+ ... + (− 1) (63)( 2 n + 1) ] 411
8 = 2 + 63 [n C1 + nC2 (63) + ... + (63)n − 1 − ( 2 n + 1)C1
∴ Remainder is 2.
+ ( 2 n + 1)C2 (63) − ... − (63)( 2 n ) ]
16. (1 − y )m (1 + y )n = 1 + a1 y + a2 y 2 + a3 y 3 + ...
On differentiating w.r.t. y, we get
− m (1 − y )m − 1 (1 + y )n + (1 − y )m n(1 + y )n − 1
Objective Mathematics Vol. 1
n
= a1 + 2 a2 y + 3a3 y 2 + ... …(i)
12. Since, ∑ nCr ⋅ x r = (1 + x )n
r =0 On putting y = 0 in Eq. (i), we get
On multiplying by x, we get − m + n = a1 = 10 …(ii)
n [Q a1 = 10, given ]
∑ nCr ⋅ x r + 1 = x(1 + x )n On again differentiating Eq. (i), we get
r =0
− m [− (m − 1) (1 − y )m − 2 (1 + y )n
On differentiating w.r.t. x, we get
n + (1 − y )m − 1 n(1 + y )n − 1 ] + n[− m(1 − y )m − 1(1 + y )n − 1
∑ (r + 1)⋅ n
Cr ⋅ x = (1 + x ) + nx(1 + x )
r n n −1
+ (1 − y )m(n − 1) (1 + y )n − 2 ]
r =0
= 2 a2 + 6 a3 y + ... …(iii)
Statement II is true. On putting y = 0 in Eq. (iii), we get
If x = 1, then − m [− (m − 1) + n ] + n[− m + (n − 1)] = 2 a2 = 20
n
∑ (r + 1)⋅ nCr = 2 n + n(2 )n − 1 = (n + 2 )2 n − 1 ⇒ m(m − 1) − mn − mn + n(n − 1) = 20
r =0 ⇒ m2 + n 2 − m − n − 2 mn = 20
∴ Statement I is true and Statement II is a correct ⇒ (m − n )2 − (m + n ) = 20
explanation of Statement I. ⇒ 100 − (m + n ) = 20
13. Since, in a binomial expansion of (a − b)n , n ≥ 5, the sum ⇒ m + n = 80 …(iv)
of 5th and 6th terms is equal to zero. On solving Eqs. (ii) and (iv), we get
m = 35 and n = 45
∴ n
C4 an − 4 (− b)4 + nC5 an − 5 (− b)5 = 0
n! n! 17. Now, C4 + 55C3 + 54C3 + 53C3 + 52C3 + 51C3 + 50C3
50
⇒ an − 4 ⋅ b4 − an − 5 ⋅ b5 = 0
(n − 4)! 4 ! (n − 5)! 5 ! = C3 + 50C4 + 51C3 + 52C3 + 53C3 + 54C3 + 55C3
50
a b = 51C4 + 51C3 + 52C3 + 53C3 + 54C3 + 55C3
Ta rg e t E x e rc is e s
n!
⇒ an − 5 ⋅ b4 − =0
(n − 5)! 4 ! n − 4 5 [Q nCr + nCr − 1 = n + 1Cr ]
a n−4 = 52
C4 + 52
C3 + 53
C3 + 54
C3 + 55
C3
⇒ =
b 5 = 53
C4 + 53
C3 + 54
C3 + 55
C3
14. We know that, = 54
C4 + 54
C3 + 55
C3 = 55
C4 + C3 =
55 56
C4
(1 + x )20 = C0 +
20 20
C1 x + ... + 20
C10 x10
18. Since, the coefficient of given terms are
+ ... + 20C20 x 20 m
Cr − 1, mCr , mCr + 1 respectively and they also in AP.
On putting x = − 1in the above expansion, we get
0 = 20C0 − 20C1 + ... − 20C9 ∴ m
Cr − 1 + mCr + 1 = 2 mCr
+ 20C10 − 20C11 + ... + 20C20 m! m!
⇒ +
⇒ 0 = C0 − C1 + ... − 20C9 + 20C10
20 20 (r − 1)! (m − r + 1)! (r + 1)! (m − r − 1)!
− 20C9 + ... + 20C10 m!
= 2⋅
⇒ 0 = 2 (20 C0 − 20C1 + ... − 20C9 ) + 20 C10 r ! (m − r )!
⇒ 20
C10 = 2 (20 C0 − 20C1 + ... + 20C10 ) 1 1 2
⇒ + =
1 (m − r + 1) (m − r ) (r + 1) r r (m − r )
⇒ 20
C0 − 20C1 + ... + 20C10 = 20C10 r (r + 1) + (m − r + 1) (m − r ) 2
2 ⇒ =
r (r + 1) (m − r + 1) (m − r ) r (m − r )
15. Now, (1 − ax )− 1 (1 − bx )− 1
= (1 + ax + a2 x 2 + ... ) (1 + bx + b2 x 2 + ... ) ⇒ r 2 + r + m2 + r 2 − 2 mr + m − r
Hence, an = Coefficient of x n in (1 − ax )− 1 (1 − bx )− 1 = 2(mr − r 2 + r + m − r + 1)
= a0 bn + abn − 1 + ... + an b0 ⇒ 4 r 2 − 4 mr − m − 2 + m2 = 0
a a
2 ⇒ m2 − m(4 r + 1) + 4 r 2 − 2 = 0
= a0 bn 1 + + + ...
b b 19. Let x 7 is contained in (r + 1) th term in the expansion of
a n + 1 2 1
11
− 1 ax + .
b bx
= a0 bn
a
a11 − r 22 − 3 r
r
−1 1
∴ Tr + 1 = 11Cr (ax 2 )11 − r = 11Cr ⋅x
b bx br
bn (an + 1 − bn + 1 ) b For coefficient of x 7, put
= ⋅ n+1
a−b b 22 − 3 r = 7 ⇒ r = 5
an + 1 − bn + 1 a6
= ∴ T6 = 11C5 5 ⋅ x 7
412 a−b b
8
n
1
11
n−r n
r
∴ Coefficient of x 7 in the expansion of ax 2 +
is
bx
⇒ nsn = ∑ nC + ∑ nC
r =0 n−r r =0 r
a6 n 1
Binomial Theorem
11 n −1 n
r
C5 5 .
b ⇒ nsn = n
Cn
+ n
Cn − 1
+ ...+ n +
C1
∑ nC
r =0 r
Similarly, coefficient of x − 7 in the expansion of
11 n
r
1 a5
ax − 2 is C6 6 .
11 ⇒ nsn = t n + t n Q t n = ∑ n , given
bx b r = 0 Cr
According to the given condition, tn n
⇒ nsn = 2 t n ⇒ =
a6 a5 sn 2
11
C5 5 = 11C6 6
b b 24. The general term of ( 3 + 8 5 )256 is
a6 a5
⇒ = ⇒ ab = 1 Tr + 1 = Cr (3)( 256 − r )/ 2 (5)r / 8
256
b5 b6
256 − r r
3 For integral terms, and are both positive
1 2 8
(1 + x ) − 1 + x
3/ 2
2 integers.
20.
(1 − x )1/ 2 i.e. r = 0,8, 16, 24, 32, ..., 256
3 1 Hence, total number of terms are 33.
⋅
2 x 2 − 1 + 3 x + 3 ⋅ 2 ⋅ x
2
3
1 + x + 2 25. Since, (r + 1)th term in the expansion of (1 + x )27/ 5
2 2 2 2 4 27 27 27
− 1 ...
− r + 1
=
5 5 5 r
(1 − x )1/ 2 = x
r!
[neglecting higher powers of x] Now, this term will be negative, if the last factor in
3x 2 − 1/ 2 numerator is the only one negative factor.
=− (1 − x ) 27
8 ⇒ − r + 1< 0
5
Targ e t E x e rc is e s
1 3
⋅
3x 2 1 2 2
3x 2 ⇒
32
<r
=− 1 + x + ⋅x =−
2
5
8 2 2 8
⇒ 6.4 < r ⇒ least value of r is 7.
[Q higher powers of x 2 can be neglected] Thus, first negative term will be 8th.
21. The coefficient of the middle term in powers of x of 26. (1 + 2 x + 3 x 2 + ... )− 3 / 2 = [(1 − x )− 2 ]− 3 / 2
(1 + αx )4 = 4C2α 2 = (1 − x )3
The coefficient of the middle term in powers of x of ∴Coefficient of x in (1 + 2 x + 3 x 2 + ... )− 3 / 2
5
(1 − αx )6 = 6C3 (− α )3 = Coefficient of x 5 in (1 − x )3 = 0
According to the given condition,
4
C2α 2 = 6C3 (− α)3
27. (1 + x + x 2 + x 3 + ... )2 = [(1 − x )− 1 ]2
4! 2 6! 3 = (1 − x )− 2
⇒ α =− α ⇒ 6 α 2 = − 20α 3
2 !2 ! 3! 3! Coefficient of x in (1 + x + x 2 + ... )2
n
⇒ α=−
6
⇒ α=−
3 = Coefficient of x n in (1 − x )− 2
20 10 = n + 2 − 1C2 − 1 = n + 1C1 = n + 1
22. The coefficient of x n in the expansion of (1 + x ) (1 − x )n
28. Using binomial theorem,
= Coefficient of x n in (1 − x )n (5 x − 4 y )n = nC0 (5 x )n + nC1(5 x )n − 1(− 4 y )
+ Coefficient of x n − 1 in (1 − x )n + nC2 (5 x )n − 2 (− 4 y )2 + ... + nCn (− 4 y )n
= (− 1)n nCn + (− 1)n − 1 nCn − 1 Sum of coefficients
n! = nC0 5n + nC1 5n − 1(− 4) + nC2 5n − 2 ⋅ (− 4)2
= (− 1)n 1 − = (− 1) (1 − n )
n
1! (n − 1)! + ... + nCn (− 4)n
n = (5 − 4) = 1 = 1
n n
1
23. Given that, sn = ∑ nC 29. Given, in the expansion of (1 + x )44 , 21st and 22nd terms
r =0 r
n
are equal.
1 T21 = T22
sn = ∑ n [Q Cr = Cn − r ]
n n i.e.
44
r = 0 Cn − r C20
44
C20 x 20
= 44
C21 x 21 ⇒ x = 44
n
n C21
⇒ nsn = ∑ nC 44 !
r =0 n−r
20 ! × 24 ! 21! × 23 ! 21 7
n−r ⇒ x= = = =
n
r 44 ! 20 ! × 24 ! 24 8
⇒ nsn = ∑
nC
n−r
+ n
Cn − r 21! × 23 !
r =0
413
8 ⇒ a=3
13
30. The general term in ax 2 +
1
is 1 1
bx and − 3C1 4 × b =
r 3 3
1
Objective Mathematics Vol. 1
Tr + 1 = 13Cr (ax 2 )13 − r 1 1
bx ⇒ − 3× 4 b=
13 − r −r 26 − 3 r
3 3
= Cr a
13
× b ( x) ⇒ − b = 32 ⇒ b = − 9
For coefficient of x , put 26 − 3 r = 8
8
∴ (a, b) = (3, − 9)
⇒ 3 r = 18 ⇒ r = 6 10
33. General term in the expansion of x −
k
∴ T7 = 13C6 a13 − 6 b− 6 x 8 = 13C6 a7b− 6 x 8 is
x2
13
1 − k
r
Now, the general term in ax − 2 is Tr + 1 = 10Cr ( x )10 − r ⋅ 2
bx x
r
1 10 − r
Tr′ + 1 = 13Cr (ax )13 − r − 2
bx = 10Cr x 2 ⋅ (− k )r ⋅ x − 2 r
13 − r 10 − 5 r
= Cr a
13
× b− r × ( x )13 − 3 r (− 1)r
2
= Cr (− k ) x
10 r
For coefficient of x − 8 , put 13 − 3 r = − 8
The term is free from x
⇒ 3 r = 21 ⇒ r = 7 10 − 5 r
Put = 0 ⇒ r =2
∴ T8 ′ = (− 1)7 13C7 a13 − 7b− 7 x − 8 2
= (− 1)7 13C7 a6 b− 7 x − 8 Now, 10
C2 (− k )2 = 405
According to the given condition, 10 × 9 2 405
⇒ ⋅ k = 405 ⇒ k 2 = =9
1
13 1× 2 45
Coefficient of x 8 in ax 2 +
bx ∴ k=±3
13
1 34. The given expression is
= Coefficient of x − 8 in ax − 2 n
bx 2 1
n 1
2
1
2n
x + 2 + 2 = x + = x +
Ta rg e t E x e rc is e s
∴ 13
C6 a7b− 6 = − 13C7 a6 b− 7 x x x
a7 a6
⇒ 13
C7 6 = − 13C7 7 The binomial contains (2 n + 1) terms in its expansion.
b b The middle term is Tn + 1.
a7 b6 1 n
⇒ =− 7 ⇒ a=− ⇒ ab + 1 = 0 1
a6 b b ∴ Tn + 1 = 2 nCn ( x )2 n − n
x
31. Given expression can be rewritten as (2 n )! (2 n )!
−1 = Cn =
2n
=
x (2 n − n )! n ! (n !)2
2 (1 − x )− 1 × 2 − 1 1 −
2 18
2
= [1 + x + x 2 + x 3 + ... ] 35. General term in the expansion of x − is
x
x x
2
x
3 r
1 + + + + ... 2
Tr + 1 = 18Cr ( x )18 − r −
2 2 2
x
∴Coefficient of x in the above expression
3
18 − r
−
r
1 3 1 2 1 = 18Cr ( x ) 2 (− 2 )r ( x ) 2 = 18Cr (− 2 )r x 9 − r
= + + + 1
2 2 2 For independent of x, put
1 1 1 1 + 2 + 4 + 8 15 9− r = 0 ⇒ r = 9
= + + + 1 = = ∴ T10 = 18C9 (− 2 )9 = − 18C9 2 9
8 4 2 8 8
−3 36. We know that,
32. Now, (a + bx )− 3 = a− 3 1 +
bx
(1 + x )n = nC0 + nC1 x + nC2 x 2 + ... + nCn x n …(i)
a
1 bx and ( x − 1)n = nC0 x n + nC1 x n − 1 + nC2 x n − 2
=1 − C1 + ...
3
+ ... + n Cn …(ii)
a3 a
1 3 1 bx On multiplying Eqs. (i) and (ii), we get
= 3 − C1 3 + ... (1 + x )2 n = (n C0 + nC1 x + nC2 x 2 + ... + nCn x n )
a a a
1 1 × (n C0 x n + nC1 x n − 1 + nC2 x n − 2 + ... + nCn )
= + x + ... [given]
27 3 Coefficient of x n in RHS
On equating constant terms and coefficients of x both = (n C0 )2 + (n C1 )2 + ... + (n Cn )2
sides, we get and coefficient of x n in LHS = 2n Cn
1 1
= 2n !
a3 27 ∴ (n C0 )2 + (n C1 )2 + ... + (n Cn )2 =
1 b 1 n!n!
and − 3C1 3 = (2 n )!
a a 3 ⇒ (n C1 )2 + ... + (n Cn )2 = − 1 = 2 nCn − 1
414 n!n!
37. We know that,
(1 − x ) = C0 + x C1 + x C2 + ... + x Cn
n n n
On integrating both sides from 0 to 2, we get
2 n n n
⇒
Now, put
⇒
32 x − 3 x ⋅ 10 + 9 = 0
t = 3x
t − 10 t + 9 = 0
2
8
Binomial Theorem
2
(1 + x )n + 1 ⇒ ( t − 1)( t − 9) = 0
n+1 ∴ t = 1 or 9
0
2 ⇒ 3 x = 1 or 3 x = 9
x2 n x3 n xn + 1 n ⇒ x = 0 or x = 2
= x nC0 + C1 + C2 + ... + Cn
2 3 n+1 0 n
42. Given expression is ax + .
1
(3)n + 1 1 2 2
2 3
x
= − + 2 n C0 = n
C1 + n
C2
n+1 n+1 2 3 3
1 5
2n + 1 n Here, T4 = T( 3 + 1) = nC3 (ax )n − 3 = [given]
+ ... + Cn − 0 x 2
n+1 5
⇒ n
C3 ⋅ an − 3 ⋅ x n − 6 = …(i)
2 22 n 23 n 2n + 1 n 2
= nC0 + C1 + C2 + ... + Cn
1 2 3 n+1 Since, x ∈ R
∴ n − 6= 0 ⇒ n = 6
3n + 1 − 1
= Now, for a, on substituting the value of n in Eq. (i),
n+1 we get
n 5 5 3×2 ×1
6
C3 a3 = ⇒ a3 = ×
38. mCr + 1 + ∑ k Cr = (mCr + 1 + mCr ) + m+1
Cr 2 2 6× 5× 4
k=m
3
+ m+ 2
Cr + ... + nCr 1 1 1
⇒ a3 = = ⇒ a=
=( m+1
Cr + 1 + m+1
Cr ) + m+ 2
Cr + ... + Cr n 8 2 2
= (m + 2 Cr + 1 + m+ 2
Cr ) + ... + nCr 43. We have,
n n n
M
n+1
M M ∑ ∑ (Cr + Cs ) = ∑ (Cr + Cr ) + 2 ∑ ∑ (Cr + Cs )
=
Targ e t E x e rc is e s
Cr + 1 r =0 s=0 r =0 0 ≤ r< s≤ n
C1 C C C n
39. + 2 2 + 3 3 + ... + n n ⇒ (n + 1)⋅ 2 n + 1 = 2 ∑ Cr + 2 ∑ ∑ (Cr + C s )
C0 C1 C2 Cn − 1 r = 0 0 ≤ r< s≤ n
(n + 1)⋅ 2 n + 1 = 2 ⋅ 2 n + 2 ∑ ∑ (Cr
n n n n
=
C1 C C
+ 2 n 2 + 3 n 3 + ... + n n n
C + Cs )
n 0 ≤ r< s≤ n
C0 C1 C2 Cn − 1
n(n − 1) n(n − 1) (n − 2 ) ⇒ ∑ ∑ (Cr + C s ) = n2 n
0 ≤ r< s≤ n
n 2 3×2 1
= +2× + 3× + ... + n ×
1 n n( n − 1 ) n C1 C C Cn
44. 1 + 1 + 2 1 + 3 L 1 +
2 C0 C1 C2 Cn − 1
n(n + 1)
= n + (n − 1) + (n − 2 ) + ... + 1 = Σn = n n(n − 1) 1
2 = 1 + 1 + L 1 +
1 2n n
40. We know that, 11Cn is maximum when n = 5
1 + n 1 + n 1 + n (1 + n )
n
11! = ... =
Q 11
Cn = 1 2 n n!
n ! (11 − n )!
∴ n ! (11 − n )! is minimum, when n = 5. ex
45. We have, = B0 + B1 x + B2 x 2 + ... + Bn x n + ...
1− x
41. By given condition, 2 ⋅ mC2 = mC1 + mC3
m (m − 1) (m − 2 ) ⇒ e x (1 − x )− 1 = B0 + B1 x + B2 x 2 + ... + Bn x n + ...
⇒ m (m − 1) = m +
1⋅ 2 ⋅ 3 x x2
⇒ 1 + + + ... (1 + x + x 2 + ... )
⇒ m2 − 9 m + 14 = 0 1 ! 2 !
⇒ (m − 2 ) (m − 7 ) = 0 = B0 + B1 x + B2 x 2 + ...
⇒ m = 2, 7 x x2
Since, 6th term is 21, m = 2 is ruled out, so we have ⇒ (1 + x + x 2 + ... ) + + + ...
1! 1!
x
T6 = 21 = 7C5 [ 2log (10 − 3 ) ]7 − 5 × [5 2( x − 2 ) log 3 ]5 x2 x3
7 × 6 log (10 − 3 x ) ( x − 2 ) log 3 + + + ... + ... = B0 + B1 x + B2 x 2 + ...
⇒ 21 = 2 ⋅2 2! 2!
2 ×1 1 1 1
⇒ 1 + 1 + x + 1 + + x 2 + ...
⇒ 2 log (10 − 3
x
) + ( x − 2 ) log 3
= 1 = 20 1! 1! 2 !
⇒ log (10 − 3 x ) = (2 − x ) log 3 = B0 + B1 x + B2 x 2 + ...
On comparing, we get
⇒ log (10 − 3 x ) = log 32 − x 1 1 1
B0 = 1, B1 = 1 + , B2 = 1 + + + ...
⇒ 10 − 3 x = 32 − x ⇒ 3 x ⋅ 10 − 32 x = 9 1! 1! 2 ! 415
8 1 1 1 5
∴ Bn = 1 ++ + ... + 210
1! 2 ! n! 50. P = ∑ C2r = 10C0 + 10
C2 + ... + 10
C10 =
2
= 29
r =0
1 1
Now, Bn − Bn − 1 = 1 + + ... + 3
Objective Mathematics Vol. 1
n !
1! Q= ∑ d 2 r + 1 = d1 + d 3 + d 5 + d 7
1 1 1 r =0
− 1 + + ... + = 27
1! (n − 1)! n ! = C1 + 7C3 + 7C5 + 7C7 =
7
= 26
2
46. The given series is in GP. P 29
Hence, its sum ∴ = = 23 = 8
Q 26
1 {(1 + x )20 + 1 − 1}
S = [Q a = 1, r = 1 + x] 51. Given expansion is (1 + x )2 n .
(1 + x ) − 1
According to the given condition,
(1 + x )21 − 1
=
2n
C3 r −1 = 2 nCr + 1
x
⇒ (2 n − 3 r + 1) = (r + 1)
Therefore, the required coefficient of x10 in the
⇒ 2 n = 4r ⇒ n = 2 r
(1 + x )21 − 1
expansion of = Coefficient of x11 in the 52. Here, A = 2n
Cn and B = 2 n − 1Cn
x
2n
expansion of (1 + x )21 − 1 = 21C11 A C 2n
∴ = 2n − 1 n = =2
B Cn n
47. We have, Tr = Tr + 1 r
− 1
⇒ n
Cr − 1 pn − r + 1 ⋅ q r − 1 = nCr pn − r q r 53. Tr + 1 = nCr ( x 3 )n − r
x2
⇒ n
Cr − 1 pn − r ⋅ pq r ⋅ q −1 = nCr pn − r q r
= nCr x 3 n − 3 r (− 1)r x − 2 r = nCr x 3 n − 5 r (− 1)r
n
Cr p
⇒ n
= For coefficient of x 5 and x10 , substitute 3 n − 5 r = 5 and
Cr − 1 q
10 respectively, we get coefficient of
(r − 1)! (n − r + 1)! p 3n − 5
⇒ =
Ta rg e t E x e rc is e s
r ! (n − r )! q x 5 = nC3 n − 5 (− 1) 5 and coefficient of
n−r +1 p 5
⇒ = 3 n − 10
r q
x10 = nC3 n − 10 (− 1) 5 .
⇒ nq + q − qr = pr 5
⇒ (n + 1) q = r ( p + q ) Coefficient of x 5 + Coefficient of x10 = 0
(n + 1) q 3n − 5 3 n − 10
∴ =1
r (p + q) ⇒ n
C3 n − 5 (− 1) 5 = − nC3 n − 10 (− 1) 5
n(n − 1) 5 5
48. Here, (1 + ax )n = 1 + nax + (ax )2 + ... 3n − 5 3n − 5
2! ⇒ n
C3 n − 5 (− 1) 5 = C3 n − 10 (− 1)
n 5
On comparing the coefficients of like powers in the 5 5
given equation, we get
27 n(n − 1) 2 ⇒ n
C3 n − 5 = nC3 n − 10
na = 6 and = ⋅a 5 5
2 2 3 n − 5 3 n − 10
9 ⇒ + =n
⇒ (n − 1) a = [Q na = 6] 5 5
2 6 n − 15
(n − 1)6 9 ⇒ =n
⇒ = 5
n 2
3 ⇒ 6 n − 15 = 5 n
∴ n = 4 and a = ∴ n = 15
2
n 54. 5th term from end = (12 − 5 + 2 ) th term from beginning
49. 13th term in the expansion of x 2 + is given by
2
x = 9th term
12 − 8
2
12
212 x3 − 2
8
T13 = nC12 ( x 2 )n − 12 = nC12 x 2 n − 24 12 = 12C8 −4
2 = C8 x (2 )
12 4
x x 2 x
= nC12 x 2 n − 24 −12 (212 ) 12 × 11 × 10 × 9
= × 2 4 × x − 4 = 7920 x − 4
= nC12 x 2 n − 36 (212 ) 4× 3×2 ×1
If the 13th term will be independent of x, then x 2 n − 36 55. For n = 1, we have
must be 1.
= [(1 + p) (1 + q ) ( p + q )] n
i.e. 2 n − 36 = 0
⇒ 2 n = 36 = (1 + q + p + pq ) ( p + q )
⇒ n = 18 = p + q + pq + q 2 + p2 + pq + p2q + pq 2
The divisors of n = 18 are 1, 2, 3, 6, 9, 18 and their sum = p + q + 2 pq + q 2 + p2 + p2q + pq 2
416 is 1 + 2 + 3 + 6 + 9 + 18 = 39. ∴ Coefficient of p′ q′ = 2 = 1 + 1 = [C (1, 0 )]3 + [C (1, 1)]3
For n = 2, we have
[(1 + p) (1 + q ) ( p + q )]2 = (1 + p)2 (1 + q )2 ( p + q )2
= (1 + p2 + 2 p) (1 + q 2 + 2q ) ( p2 + q 2 + 2 pq )
60. We know, (1 + x )n = nC0 + nC1 x + nC2 x 2 + ... + nCn x n
⇒ x(1 + x )n = nC0 x + nC1 x 2 + nC2 x 3 + ... + nCn x n + 1 8
Binomial Theorem
On differentiating w.r.t. x, we get
= (1 + q 2 + 2q + q 2 + p2q 2 + 2 p2q
(1 + x )n + nx(1 + x )n − 1 = nC0 + 2 n C1 x +... + (n + 1)n Cn x n
+ 2 p + 2 pq 2 + 4 pq ) ( p2 + q 2 + 2 pq )
Put x = 1, we get
∴Coefficient of p2q 2 = 1 + 1 + 8 = 10
2 n + n 2 n − 1 = nC0 + 2 n C1 + ... + (n + 1)n Cn
= [C (2, 0 )]3 [C (2, 2 )]3 + [C (2, 1)]3
2 ⇒ 2 n − 1 (n + 2 ) = nC0 + 2 ⋅ nC1 + ... + (n + 1) nCn
= ∑ [C (2, k )]3 61. Here, T5 + T6 = 0
k=0
⇒ n C4 an − 4 (− 2 b)4 + nC5 an − 5 (− 2 b)5 = 0
Thus, the coefficient of pnq n in [(1 + p) (1 + q ) ( p + q )]n is
n ⇒ 16 ⋅ nC4 an − 4 b4 = 32 nC5 an − 5 b5
∑ [(n, k )] . 3
⇒
n
C5 an − 5 b5 1
⋅ n−4 4 = ⇒
b 1 n C4
= ⋅
k=0 n
C4 a b 2 a 2 n C5
56. Given expansion is (1 − 3 x + 3 x 2 − x 3 )2 n = [(1 − x )3 ]2 n n!
2⋅
= (1 − x )6 n a 5 ! (n − 5)! 4 ! (n − 4)! 2(n − 4)
(6 n + 2 )th ⇒ = = ×2 =
∴ Middle term = term = (3 n + 1)th term b n! 5 ! (n − 5)! 5
2 4 ! (n − 4)!
[Q here, 6 n is even]
62. In the expansion of ( x + 2 y )6 , + 1 th term is the
6
57. Since, n C4 , nC5 and n C6 are in AP. 2
∴ 2 n C5 = nC4 + nC6
middle term.
2 1 1
⇒ = + So, T4 = T3 + 1 = 6C3 x 6 − 3 (2 y )3 = 8 (6 C3 ) ( xy )3
5 (n − 5) (n − 4) (n − 5) 30
2 30 + n 2 − 9 n + 20 ∴ Coefficient of middle term = 8 (6 C3 )
⇒ =
Targ e t E x e rc is e s
5 (n − 5) 30 (n − 4) (n − 5) 63. a1 = nC1, a2 = nC2 , a3 = nC3
⇒ 12 (n − 4) = 30 + n 2 − 9 n + 20 Q a1, a2 and a3 are in AP.
⇒ 12 n − 48 = n 2 − 9 n + 50 ∴ 2 a2 = a1 + a3
⇒ n 2 − 21 n + 98 = 0 ⇒ 2 ⋅ nC2 = nC1 + nC3
⇒ (n − 14) (n − 7 ) = 0 n(n − 1) n(n − 1) (n − 2 )
⇒ n = 14 or 7 ⇒ 2⋅ =n+
2! 3!
58. Consider, 15 C3 + 15C5 + ... + 15C15 ⇒ n 2 − 9 n + 14 = 0
= (15 C1 + ... + 15C15 ) − 15C1 ⇒ (n − 2 ) (n − 7 ) = 0
215 ∴ n = 7 [Q n = 2 is not possible]
= − 15 = 214 − 15
2 64. (a2 − 6 a + 11)10 = 1024
59. Since, (1 + x )n = C0 + C1 x + C2 x 2 + ... + Cn x n ⇒ (a2 − 6 a + 11)10 = 210
On differentiating w.r.t. x, we get ⇒ a2 − 6 a + 11 = 2
n(1 + x )n − 1 = C1 + 2C2 x + ... + nCn x n − 1 ⇒ a2 − 6 a + 9 = 0
On putting x = − 1, we get ⇒ (a − 3)2 = 0
0 = C1 − 2C2 + ... + n(− 1)n − 1Cn ∴ a=3
417
9
Trigonometric
Functions
and Equations
Chapter Snapshot
Introduction
● Introduction
Trigonometry is a branch of Mathematics in which we study triangles and relationships
between the lengths of their sides and the angles between the sides. ● Measure of Angles
The word ‘Trigonometry’ is derived from two Greek words ‘trigonon’ and ‘metron’. ● Systems of Measurement of
The word ‘trigonon’ means a triangle and the word ‘metron’ means a measure. i.e. Angles
trigonometry means measuring the angles and sides of a triangle. Trigonometry defines ● Trigonometric Ratios
the trigonometric functions, which describe the relationships and have applicability to ● Trigonometric Function
physical phenomena such as waves.
● Graph of Trigonometric
Functions
Measure of Angles ● Trigonometrical Identities
The measure of angle is the amount of rotation from the direction of one ray of the ● Trigonometric Ratios of Allied
angle to the other. The initial and final positions of the revolving rays are respectively Angles
called the initial and terminal sides. If initial and final positions of the revolving ray are
Trigonometrical Ratios of
OP and OQ, then the angle formed will be ∠POQ.
●
Compound Angles
Q ● Trigonometric Ratios of
Multiples of an Angle
● Maximum and Minimum
Values of Trigonometrical
Expressions
O P ● Trigonometric Equations
● General Solution of
If the rotation is in clockwise direction, then the angle measured is negative and if the Trigonometric Equations
rotation is in anti-clockwise direction, then the angle measured is positive. ● Solution of Trigonometric
Inequality
Systems of Measurement of
Angles
Ø ● When an angle is expressed in radians, the word radian is
generally omitted.
e.g. Ic = 1, 22 c = 22
9
Trigonometric Functions and Equations
● We have,
There are three systems for measuring angles: π radian = 180 °
(i) Sexagesimal system In this system, a right angle 180 ° 180 °
∴ 1 radian = = × 7 = 57 °16 ′ 45 ′′(approx)
is divided into 90 equal parts, called degrees. The π 22
symbol 1° is used to denote one degree. Thus, one ● We have, 180° = π radian
1 π
degree is one-ninetieth th part of a right ⇒ 1° = radian
90 180
angle. Each degree is divided into 60 equal parts, 22
= radian = 0.01746 radian
called minutes and one minute is divided into 60 7 × 180
equal parts, called seconds. The symbols 1′ and 1′′
are used to denote one minute and one second Relation among Three Systems of
respectively. Measurement of an Angle
Thus, 1 right angle = 90 degree ( = 90° ) Let D be the number of degree, R be the number of
1° = 60 minutes ( = 60′ ) radian and G be the number of grade in an angle θ.
1′ = 60 seconds ( = 60 ′ ′ ) Now, 90° = 1 right angle
(ii) Centesimal system In this system, a right angle 1
is divided into 100 equal parts, called grades. ⇒ 1° = right angle
90
Each grade is subdivided into 100 minutes and
D
each minute is subdivided into 100 seconds. ⇒ D° = right angle
90
The symbols 1g , 1′ and 1′′ are used to denote a
D
grade, a minute and a second, respectively. ⇒ θ= right angle …(i)
Thus, 1 right angle =100 grades ( =100 g ) 90
1 grade =100 minutes ( = 100′ ) Again, π radian = 2 right angles
1 minute =100 seconds ( = 100 ′′ ) ⇒
2
1 radian = right angles
(iii) Circular system If the angle subtended by an π
arc of length l to the centre of circle of radius r is 2R
⇒ R radian = right angles …(ii)
l
θ, then θ = . π
r and 100 grades =1 right angle
1
l ⇒ 1 grade = right angle
100
θ G
O r ⇒ G grades = right angles
100
G
⇒ θ= right angles …(iii)
100
If the length of arc is equal to the radius of the From Eqs. (i), (ii) and (iii), we get
circle, then the angle subtended at the centre of D G 2R
the circle will be one radian. One radian is = =
90 100 π
denoted by 1c .
This is the required relation between the three
The ratio of the circumference of the circle to the
systems of measurement of an angle.
diameter of the circle is denoted by a Greek
letter π and it is a constant quantity. X Example 1. If the three angles of a
Circumference of circle 5π
∴ =π quadrilateral are 60°, 60 g and radian. Then,
Diameter of circle 6
This constant quantity π is an irrational quantity the fourth angle is
22 (a) 60°
and generally its approximate value is and its
7 (b) 96°
general value upto six places of decimal is (c) 96 g
3.142857. (d) None of the above 419
9 Sol. (b) First angle = 60°
Except these six ratios, there are two other terms
which are known as ‘versed sine’ and ‘coversed sine’,
respectively.
Objective Mathematics Vol. 1
90
Second angle = 60g = 60 × = 54°
100 i.e. ver sin θ = 1 − cos θ
5π
Third angle = radian and cover sin θ = 1 − sin θ
6
5 × 180° Ø Trigonometric ratios of any angle is always constant.
= = 150°
6
Fourth angle = 360° − (60° + 54° + 150° ) X Example 3. In a right angled ∆ABC, if base is
= 360° − 264° 6 and perpendicular height is 8, then the
= 96° trigonometric ratios sin A, cos A and tan A,
respectively are
X Example 2. The angle between the hour hand 4 3 4 2 3 4
and the minute hand of a clock at half past three, (a) , , (b) , ,
is 5 5 3 5 5 5
4 1 2
(a) 75° (c) , , (d) None of these
(b) 60° 5 5 3
(c) 30° Sol. (a) By Pythagoras theorem,
(d) 175° C
Sol. (a) The angle traces by the hour hand in 12 h is of 360°.
∴ The angle traced by the hour hand in 3 h 30 min. 8
360 7 °
h =
7
i.e. × = 105°
2 12 2
The angle traced by the minute hand in 60 min is 360°. A 6 B
∴ The angle traced by the minute hand in 30 min AC = AB + BC
2 2 2
°
= × 30 = 180°
360
60 = 62 + 82 = 36 + 64 = 100
Hence, the required angle between two hands ⇒ AC = 10
= 180° − 105° = 75° ∴ sin A =
Perpendicular BC
= =
8
=
4
Hypotenuse AC 10 5
Trigonometric Ratios cos A =
Base
=
AB
Hypotenuse AC 10 5
=
6
=
3
Let us take a right angled ∆ABC right angled at B. tan A =
Perpendicular BC 8 4
= = =
C Base AB 6 3
Trigonometric Functions
Let X ′ OX and YOY ′ be the coordinate axes.
θ Taking O as the centre and a unit radius, draw a circle,
A B cutting the coordinate axes at A, B, A′ and B ′, as shown
in the figure.
Let ∠CAB = θ, then
Y
Perpendicular BC
sin θ = =
Hypotenuse AC B
P(x, y)
Base AB
cos θ = = 1 y
Hypotenuse AC X'
q A
X
A' O x M
Perpendicular BC
tan θ = =
Base AB
B'
1 Base AB
cot θ = = =
tan θ Perpendicular BC Y'
1 Hypotenuse AC
sec θ = = = Let ∠AOP = θ
cos θ Base AB
arc AP θ l
Q ∠AOP = radius OP = 1 = θ , using θ = r
1 Hypotenuse AC c
420 cosec θ = = =
sin θ Perpendicular BC
Now, the six trigonometric functions may be
defined as under:
OM
Facts related to sin x
(a) Period = 2π 9
Trigonometric Functions and Equations
(i) cos θ = =x (b) Graph of sin x is continuous for all real
OP values of x.
PM (ii) Graph of cos x
(ii) sin θ = =y
OP Y
OP 1
(iii) sec θ = = , x ≠0 (– 2p, 1) (0, 1) (2p, 1)
y=1
OM x D
OP 1 0 2p
X' X
(iv) cosecθ = = , y≠0 –2p – 3p –p – p p p 3p
PM y 2 2 2 2 y = –1
(– p, –1) (p, –1)
PM y
(v) tan θ = = , x ≠0
OM x Y'
OM x
(vi) cot θ = = , y≠0 Facts related to cos x
PM y
(a) Period = 2π
Domain and Range of Trigonometrical (b) Graph of cos x is continuous for all real
values of x.
Functions
(iii) Graph of tan x
The domain of a trigonometric ratio is the set of all
values of angle θ for which it is meaningful and the Y
range is the set of all values of the trigonometric ratio
for different values of θ for which it is meaningful. 1
The following are domain and ranges of
X' –p X
trigonometric ratios defined in the above section: – 3p –p –p 0 p p p 3p
2 2 4 4 2 2
Trigonometric ratio Domain Range –1
sinθ R [−1, 1]
cosθ R [−1, 1] Y'
π
tan θ R − (2 n + 1) , n ∈ I R
2 Facts related to tan x
cosec θ R − { nπ, n ∈ I} R − ( −1, 1) (a) Period = π
sec θ π
R − (2 n + 1) , n ∈ I R − ( −1, 1) mπ
(b) Graph of tan x is discontinuous at x = ,
2 2
cot θ R − { nπ, n ∈ I} R where m is an odd integer.
(iv) Graph of cot x
Ø | sin θ | ≤ 1 , | cos θ | ≤ 1 , | sec θ | ≥ 1 , | cosec θ | ≥ 1 for all values of θ
for which these functions are defined. Y
Graph of Trigonometric Functions
(i) Graph of sin x X' X
–2π – 3π –π –π O π π 3π 2π
Asymptotes
Y 2 2 2 2
( )
– 3p , 1
2 ( p2,1)
y=1
B D Y'
h1 h1
X' 0 X
A p C Facts related to cot x
–2p – 3p –p –p p 3p
2p
2 2 2
2
y=–1 (a) Period = π
(– 2p , –1) ( )
3p , –1 (b) Graph of cot x is discontinuous at x = mπ,
2
Y' where m is an integer.
421
9 (v) Graph of sec x
Y
(iv) tan θ =
sin θ
cos θ
cos θ
Objective Mathematics Vol. 1
(–2p, 1) (2p, 1) (v) cot θ =
sin θ
(vi) cos 2 θ + sin 2 θ = 1 or 1 − cos 2 θ = sin 2 θ
y=1
1
X' X or 1 − sin 2 θ = cos 2 θ
–2p – 3p –p – 0p
p p 3p 2p
2 2 2 2
O y = –1
(vii) 1 + tan 2 θ = sec 2 θ
–1
(viii) 1 + cot 2 θ = cosec 2 θ
(–p, –1) (p, –1)
Y'
X Example 4. (sin θ + cosec θ) 2 + (cos θ + sec θ ) 2
Facts related to sec x
is
(a) Period = 2π (a) ≥ 9
mπ (b) ≤ 9
(b) Graph of sec x is discontinuous at x = ,
2 (c) = 9
where m is an odd integer. (d) None of the above
(vi) Graph of cosec x
Sol. (a) (sin θ + cosec θ)2 + (cos θ + sec θ)2
Y
= sin2 θ + cosec 2θ + 2 + cos 2 θ + sec 2 θ + 2
= 7 + tan2 θ + cot 2 θ ...(i)
tan θ + cot θ
2 2
Now, ≥ tan2 θ ⋅ cot 2 θ [Q AM ≥ GM]
p ,1 2
( )
2 ⇒ tan2 θ + cot 2 θ ≥ 2 …(ii)
y=1
1 From Eq. (i), we get
X' –2p –p 0 p 3p X (sin θ + cosec θ)2 + (cos θ + sec θ)2 ≥ 7 + 2
– 3p –p/2 p 2p
2 2 2 [from Eq. (ii)]
–1 y = –1
⇒ (sin θ + cosec θ)2 + (cos θ + sec θ)2 ≥ 9
– p,–1
( 2 ) ( )
3p,–1
2
X Example 5. Prove that
1 + sin θ π π
= sec θ + tan θ, − < θ < .
Y' 1 − sin θ 2 2
Facts related to cosec x Sol. LHS = 1 + sin θ
1 − sin θ
(a) Period = 2π
(b) Graph of cosec x is discontinuous at x = mπ, 1 + sin θ 1 + sin θ
= ⋅
where m is an integer. 1 − sin θ 1 + sin θ
(1 + sin θ)2
=
Trigonometrical Identities 1 − sin2 θ
(1 + sin θ)2
An equation involving trigonometric functions =
cos 2 θ
which is true for all those angles for which the functions
1 + sin θ
are defined, is called a trigonometrical identity. =
cos θ
Following are some fundamental trigonometrical 1 sin θ
= +
identities: cos θ cos θ
1 1 = sec θ + tan θ = RHS
(i) sin θ = or cosec θ = Hence proved.
cosec θ sin θ
1 1 Signs of Trigonometrical Ratios
(ii) cos θ = or sec θ = We have introduced six trigonometric ratios. Signs
sec θ cos θ
of these ratios depend on the quadrant in which the
1 1
(iii) cot θ = or tan θ = terminal side of the angle lies. We always take the
422 tan θ cot θ length OP = r to be positive.
Thus, sin θ =
y
r
x
has the sign of y, cos θ = has the
r
sign of x. The sign of tan θ depends on the signs of x
cos θ =
x
r
y
< 0,
9
Trigonometric Functions and Equations
and y and similarly the signs of other trigonometric tan θ = > 0,
x
ratios are decided by the signs of x and/or y. Thus, we r
have the following: cosec θ = < 0,
y
In first quadrant
r
We have, x > 0, y > 0 sec θ = < 0
x
y
∴ sin θ = > 0, x
r and cot θ = > 0
y
x
cos θ = > 0, Thus, in the third quadrant, all trigonometric
r
functions are negative except tangent and cotangent.
y
tan θ = > 0, In fourth quadrant
x
r We have, x > 0, y < 0
cosec θ = > 0, y
y ∴ sin θ = < 0,
r r
sec θ = > 0 x
x cos θ = > 0,
x r
and cot θ = > 0 y
y tan θ = < 0,
x
Thus, in the first quadrant, all trigonometric r
functions are positive. cosec θ = < 0,
y
In second quadrant
r
We have, x < 0, y > 0 sec θ = > 0
x
y
∴ sin θ = > 0, x
r and cot θ = < 0
y
x
cos θ = < 0, Thus, in the fourth quadrant, all trigonometric
r
functions are negative except cosine and secant.
y
tan θ = < 0,
x AID to Memory
r
cosec θ = > 0, The signs of trigonometrical ratios in different
y quadrants is the four word phrases ‘AFTER SCHOOL
r TO COLLEGE’. The first letter of the first word in this
sec θ = < 0 phrase is ‘A’. This may be taken to indicate that all
x
x trigonometric ratios are positive in the first quadrant.
and cot θ = < 0 The first letter of the second word is ‘S’. This indicates
y
that sine and its reciprocal are positive in the second
Thus, in the second quadrant, sine and cosecant quadrant. The first letter of third word is ‘T’. This may
functions are positive and all others are negative. be taken as to indicate that tangent and its reciprocal
In third quadrant are positive in the third quadrant. The first letter of the
We have, x < 0, y < 0 fourth word in the phrase is ‘C’ which may be taken as
to indicate that only cosine and its reciprocal are
y
∴ sin θ = < 0, positive in the fourth quadrant.
r
423
Y
9 II Quadrant I Quadrant
Objective Mathematics Vol. 1
sin, cosec are All positive
positive and the rest
are negative
X¢ X
O
III Quadrant IV Quadrant
tan, cot are cos, sec are
positive and the rest positive and the rest
are negative are negative
Y¢
Variation of Values of Trigonometrical Ratios in Different Quadrants
Y
II I
sine decreases from 1 to 0 sine increases from 0 to 1
cosine decreases from 0 to −1 cosine decreases from 1 to 0
tangent increases from −∞ to 0 tangent increases from 0 to ∞
cotangent decreases from 0 to −∞ cotangent decreases from ∞ to 0
secant increases from −∞ to −1 secant increases from 1 to ∞
cosecant increases from 1 to ∞ cosecant decreases from ∞ to 1
X′ X
III O IV
sine decreases from 0 to –1 sine increases from −1 to 0
cosine increases from −1 to 0 cosine increases from 0 to 1
tangent increases from 0 to ∞ tangent increases from −∞ to 0
cotangent decreases from ∞ to 0 cotangent decreases from 0 to −∞
secant decreases from −1 to −∞ secant decreases from ∞ to 1
cosecant increases from −∞ to −1 cosecant decreases from −1 to −∞
Y′
Trigonometric Ratios of π
sin − θ = cos θ
π
cos − θ = sin θ
2 2
Allied Angles
π π
Two angles are said to be allied when their sum or tan − θ = cot θ cot − θ = tan θ
difference is either zero or a multiple of 90°. 2 2
The angles −θ, 90° ± θ, 180° ± θ, 360° ± θ, etc., are π π
sec − θ = cosec θ cosec − θ = sec θ
angles allied to the angle θ, if θ is measured in degree. 2 2
However, if θ is measured in radian, then the angles
π π π
sin + θ = cos θ cos + θ = − sin θ
allied to θ are − θ, ± θ, π ± θ, 2π ± θ, etc. 2 2
2
Using trigonometric ratios of allied angles, we can π π
tan + θ = − cot θ cot + θ = − tan θ
find the trigonometric ratios of angles of any 2 2
magnitude.
π π
Following are the trigonometrical ratios of allied sec + θ = − cosec θ cosec + θ = sec θ
2 2
angles:
sin ( − θ ) = − sin θ cos ( − θ ) = cos θ sin ( π − θ ) = sin θ cos ( π − θ ) = − cos θ
tan ( − θ ) = − tan θ cot ( − θ ) = − cot θ tan ( π − θ ) = − tan θ cot ( π − θ ) = − cot θ
cosec ( − θ ) = − cosec θ sec ( − θ ) = sec θ sec ( π − θ ) = − sec θ cosec ( π − θ ) = cosec θ
424
sin ( π + θ ) = − sin θ
tan ( π + θ ) = tan θ
sec ( π + θ ) = − sec θ
cos ( π + θ ) = − cos θ
cot ( π + θ ) = cot θ
cosec ( π + θ ) = − cosec θ
X Example 6. The value of cosec (−1410° ) is
(a) 3 (b) 2 (c) −3 (d) −1
9
Trigonometric Functions and Equations
Sol. (b) cosec (−1410° )
3π 3π = − cosec (1410° ) [Qcosec (−θ) = − cosec θ]
sin − θ = − cos θ cos − θ = − sin θ
2 2 = − cosec (360 × 4 − 30)°
= − (−cosec 30° ) [Qcosec (2nπ − θ) = − cosec θ]
3π 3π
tan − θ = cot θ cot − θ = tan θ = cosec 30° = 2
2 2
3π 3π X Example 7. The value of
sec − θ = − cosec θ cosec − θ = − sec θ tan 1° tan 2° tan 3° K tan 89° is
2 2
(a) 0 (b)1 (c) 2 (d) 3
3π 3π
sin + θ = − cos θ cos + θ = sin θ Sol. (b) tan1° tan 2 ° tan 3° K tan 89°
2 2
= (tan1° tan 89° ) (tan 2 ° tan 88° )
3π 3π K(tan 44° tan 46° )tan 45°
tan + θ = − cot θ cot + θ = − tan θ
2 2 = (tan 1° cot 1° ) (tan 2 ° cot 2 ° ) K tan 45°
= 1⋅ 1 K 1 = 1
3π 3π
sec + θ = cosec θ cosec + θ = − sec θ
2 2 X Example 8. The value of
3π π π
sin (2π − θ ) = − sin θ cos (2π − θ ) = cos θ 2 sin 2 + 2 cos 2 + 2 sec 2 is
tan (2π − θ ) = − tan θ cot (2π − θ ) = − cot θ 4 4 3
(a)10 (b) −10 (c)12 (d)14
sec (2π − θ ) = sec θ cosec (2π − θ ) = − cosec θ
sin (2π + θ ) = sin θ cos (2π + θ ) = cos θ Sol. (a) 2 sin2 3 π + 2 cos 2 π + 2 sec 2 π
4 4 3
tan (2π + θ ) = tan θ cot (2π + θ ) = cot θ 2 π 2 π π
= 2 sin π − + 2 cos + 2 sec 2
sec (2π + θ ) = sec θ cosec (2π + θ ) = cosec θ 4 4 3
2
π
+ 2 ×
1
Ø sin {nπ + (−1) θ} = sin θ , n ∈I
n = 2 sin2 + 2(2 )
2
[Qsin ( π − θ) = sin θ]
2
●
4
● cos (2nπ ± θ) = cos θ , n ∈I 2
= 2 ×
1 1
● tan (nπ + θ) = tan θ, n ∈I + 2 × + 2 × 4 = 1 + 1 + 8 = 10
2 2
Work Book Exercise 9.1
1 The value of cos 10 ° − sin 10 ° is 5 In a right angled triangle, if the hypotenuse is four
a positive b negative times as long as the perpendicular drawn to it
c 0 d 1 from the opposite vertex. One of the acute angle
is
2 A circular wire of radius 7 cm is cut and bend a 15° b 30°
again into an arc of a circle of radius 12 cm. The c 45° d None of these
angle subtended by the arc at the centre is
a 50° b 210° 6 If the interior angles of a polygon are in AP with
c 100° d 60° common difference 5° and the smallest angle
120°, then the number of sides of the polygon is
3 Which of the following is correct? a 9 or 16 b 9
a sin1° > sin1 b sin1° < sin1 c 13 d 16
π
c sin1° = sin1 d sin1° = sin1
180 7 The value of x for which f ( x ) = sin x − cos x
sec 8 A − 1 defined, is
4 is equal to π π 5π
sec 4 A − 1 a 0, b ,
4 4 4
tan 8 A
a π 5π
tan 2 A c , d None of these
4 4
cot 8 A
b
cot 2 A 8 If A, B, C, D are angles of a cyclic quadrilateral,
tan 2 A
c
tan 8 A then cos A + cos B + cos C + cos D equals
a 0 b 1
d None of the above
c −1 d None of these
425
9 9 The value of
log tan 1° + log tan 2 ° + ... + log tan 89° is
a 0 b −1
11 The value of
e log10 tan 1° + log10 tan 2 ° + log10 tan 3 ° + ... + log10 tan 89 ° is
a 0 b e
Objective Mathematics Vol. 1
c 1 d ∞ c
1
d None of these
e
10 If 1 + sin x + sin 2 x + ... = 4 + 2 3, 0 < x < π, then
x is equal to 12 If (sec A − tan A) (sec B − tan B) (sec C − tan C )
a
π
b
π = (sec A + tan A) (sec B + tan B) (sec C + tan C ),
6 4 then each side is equal to
π π π 2π a 0 b 1
c or d or
3 6 3 3 c −1 d ±1
Trigonometrical Ratios of ⇒ cos 2 ( A − B ) − 2 cos ( A − B ) + 1 + sin 2 ( A − B )
Compound Angles = cos 2 B + cos 2 A − 2 cos A cos B + sin 2 B
The algebraic sum of two or more angles are + sin 2 A − 2 sin A sin B
generally called compound angles and the angles are ⇒ 2 − 2 cos ( A − B ) = 2 − 2 cos A cos B − 2 sin A sin B
known as the constituent angles. ⇒ cos ( A − B ) = cos A cos B + sin A sin B
Cosine of the Difference and Sum of Two Hence, cos ( A − B ) = cos A cos B + sin A sin B
Angles Ø Above formula is true for all values of angles A and B whether
positive, zero or negative.
(i) cos ( A − B ) = cos A cos B + sin A sin B
(ii) cos ( A + B ) = cos A cos B − sin A sin B for all (ii) We have,
angles A and B. cos ( A + B ) = cos [ A − ( −B )]
= cos A cos ( −B ) + sin A sin ( −B )
Proof (i) Let X ′ OX and YOY ′ be the coordinate axes. [from part (i)]
Consider a unit circle with O as the centre. = cos A cos B − sin A sin B
Let P1 , P2 and P3 be three points on the circles such Hence, cos ( A + B ) = cos A cos B − sin A sin B
that ∠XOP1 = A, ∠XOP2 = B and ∠XOP3 = A − B .
Using above formula, a list of formula is formed:
The terminal side of any angle intersects the circle
(i) sin ( A + B ) = sin A cos B + cos A sin B
with centre O and unit radius at a point whose coordinates
are respectively the cosine and sine of the angle. (ii) sin ( A − B ) = sin A cos B − cos A sin B
Therefore, coordinates of P1 , P2 and P3 are (cos A, sin A ), (iii) cos ( A + B ) = cos A cos B − sin A sin B
(cos B , sin B ) and [cos ( A − B ), sin ( A − B )], respectively. (iv) cos ( A − B ) = cos A cos B + sin A sin B
We know that equal chords of a circle make equal tan A + tan B
tan ( A + B ) =
angles at its centre. Since, chords P0 P3 and P1 P2 1 − tan A tan B
subtend equal angles at O. Therefore, (v)
tan A − tan B
Chord P0 P3 = Chord P1 P2 tan ( A − B ) =
Y
1 + tan A tan B
P1(cos A, sin A)
π π
P3[cos (A – B), sin (A – B)] where, A ≠ nπ + , B ≠ nπ +
2 2
P2 (cos B, sin B) π
A and A ± B ≠ mπ +
B 2
X' X
O P0 (1, 0) cot A cot B − 1
cot ( A + B ) =
cot A + cot B
(vi)
cot A cot B + 1
cot ( A − B ) =
Y'
cot B − cot A
where, A ≠ nπ , B ≠ nπ and A ± B ≠ nπ
⇒ {cos ( A − B ) − 1}2 + {sin ( A − B ) − 0}2
(vii) sin ( A + B )sin ( A − B ) = sin 2 A − sin 2 B
= (cos B − cos A ) 2 + (sin B − sin A ) 2 = cos 2 B − cos 2 A
⇒ {cos ( A − B ) − 1}2 + sin 2 ( A − B )
(viii) cos ( A + B )cos ( A − B ) = cos 2 A − sin 2 B
= (cos B − cos A ) + (sin B − sin A )
2 2
= cos 2 B − sin 2 A
426
(ix) sin ( A + B + C ) = sin A cos B cos C
+ cos A sin B cos C + cos A cos B sin C
X Example 10. If cos θ + sin φ = m and
1
sin θ + cos φ = n, then ( m 2 + n 2 − 2) is equal to
9
Trigonometric Functions and Equations
− sin A sin B sin C 2
Also, sin ( A + B + C ) = cos A cos B cos C (a) sin (θ − φ ) (b) sin (θ + φ )
(tan A + tan B + tan C − tan A tan B tan C )
(c) cos (θ − φ ) (d) cos (θ + φ )
(x) cos ( A + B + C )
= cos A cos B cos C − sin A sin B cos C Sol. (b) Given, m = cos θ + sin φ …(i)
and n = sin θ + cos φ …(ii)
− sin A cos B sin C − cos A sin B sin C On squaring and adding Eqs. (i) and (ii), we get
Also, cos ( A + B + C ) m2 + n2 = cos 2 θ + sin2 φ + 2 sin φcos θ
= cos A cos B cos C (1 − tan A tan B + sin2 θ + cos 2 φ + 2 sin θ cos φ
− tan B tan C − tan C tan A ) ⇒ m2 + n2 = 2 + 2 (sin θ cos φ + cos θsin φ)
(xi) tan ( A + B + C ) ⇒ sin (θ + φ) =
m2 + n2 − 2
tan A + tan B + tan C − tan A tan B tan C 2
=
1 − tan A tan B − tan B tan C − tan C tan A X Example 11. If tan A − tan B = x and
(xii) sin ( A1 + A2 +…+ An ) cot B − cot A = y, then cot ( A − B ) is equal to
1 1 1 1
= cos A1 cos A2… cos An ( S 1 − S 3 + S 5 − S 7 +… ) (a) − (b) −
y x x y
(xiii) cos ( A1 + A2 +… + An )
1 1
= cos A1 cos A2… cos An (1 − S 2 + S 4 − S 6 +… ) (c) + (d) None of these
x y
S − S 3 + S 5 − S 7 +…
(xiv) tan( A1 + A2 +… + An ) = 1 Sol. (c) We have, tan A − tan B = x and cot B − cot A = y
1 − S 2 + S 4 − S 6 +…
Now, cot B − cot A = y
where, S 1 = tan A1 + tan A2 +… + tan An tan A − tan B x
⇒ = y ⇒ = y
tan A tan B tan A tan B
= sum of the tangents of the separate x
angles ⇒ tan A tan B =
y
S 2 = A1 tan A2 + tan A1 tan A3 +… Now, cot( A − B) =
1
=
1 + tan A tan B
= sum of the tangents taken two at a tan( A − B) tan A − tan B
x
time 1+
= y = x+ y= 1+ 1
S 3 = tan A1 tan A2 tan A3 x xy x y
+ tan A2 tan A3 tan A4 +… X Example 12. If sin α sin β − cos α cos β + 1 = 0,
= sum of the tangents taken three at a then prove that 1 + cot α tan β = 0.
time and so on. Sol. Given, sin α sin β − cos α cos β + 1 = 0
If A1 = A2 =… = An = A, we have ⇒ cos α cos β − sinα sin β = 1
⇒ cos (α + β ) = 1
S 1 = n tan A,S 2 = C 2 tan A,
n 2
cos α sin β
Now, 1 + cot α tan β = 1 + ×
S 3 = n C 3 tan 3 A, … sin α cos β
sin α cos β + cos α sin β
=
X Example 9. The value of sin α cos β
sin(α + β ) 0
cos ( 45° − A )cos ( 45° − B ) − sin ( 45° − A ) = =
sin α cos β sin α cos β
sin ( 45° − B ) is
=0 Hence proved.
(a) sin ( A + B )
(b) cos ( A + B )
(c) sin ( A − B ) Transformation Formulae
(d) cos ( A − B ) In the previous section, we have learnt about the
Sol. (a) We have, following formulae:
cos (45° − A)cos (45° − B) − sin (45° − A) sin (45° − B) sin A cos B + cos A sin B = sin ( A + B ) …(i)
= cos {(45° − A) + (45° − B)} sin A cos B − cos A sin B = sin ( A − B ) …(ii)
= cos {90° − ( A + B)}
cos A cos B − sin A sin B = cos ( A + B ) …(iii) 427
= sin ( A + B)
9 cos A cos B + sin A sin B = cos ( A − B )
On adding Eqs. (i) and (ii), we obtain
2sin A cos B = sin ( A + B ) + sin ( A − B )
…(iv) X Example 14. Prove that
cos 18° − sin 18° = 2 sin 27°.
Objective Mathematics Vol. 1
Sol. Consider,
On subtracting Eq. (ii) from Eq. (i), we get cos 18° − sin18° = cos 18° − sin (90° − 72 ° )
2cos A sin B = sin ( A + B ) − sin ( A − B ) = cos 18° − cos 72 °
18° + 72 ° 72 ° − 18°
On adding Eqs. (iii) and (iv), we get = 2 sin sin
2 2
2cos A cos B = cos ( A + B ) + cos ( A − B ) = 2 sin 45° sin 27 °
On subtracting Eq. (iii) from Eq. (iv), we get =2⋅
1
⋅ sin 27 °
2sin A sin B = cos ( A − B ) − cos ( A + B ) 2
= 2 sin 27 ° Hence proved.
Thus, we obtain the following formulae:
2sin A cos B = sin ( A + B ) + sin ( A − B ) π
X Example 15. If α, β, γ ∈ 0, , then
2
2cos A sin B = sin ( A + B ) − sin ( A − B )
sin α + sin β + sin γ
2cos A cos B = cos ( A + B ) + cos ( A − B )
(a) > sin (α + β + γ ) (b) < sin (α + β + γ )
2sin A sin B = cos ( A − B ) − cos ( A + B ) (c) = sin (α + β + γ ) (d) ≤ sin (α + β + γ )
These four formulae convert the product of two
Sol. (a) We have,
sines or two cosines or one sine and one cosine into the
sin α + sin β + sin γ − sin (α + β + γ )
sum or difference of two sines or two cosines. α+β α −β α+β α + β + 2γ
= 2 sin cos − 2 sin cos
In the above formulae, if we substitute A + B = C 2 2 2 2
and A − B = D, we obtain the following formulae to α + β α −β α + β + 2γ
= 2 sin cos − cos
convert the sum into the product. 2 2 2
α + β α+ γ β + γ
C + D C − D = 2 sin 2 sin sin
sin C + sin D = 2 sin cos 2 2 2
2 2 α+β β+ γ γ+α
= 4sin sin sin
C − D C + D 2 2 2
sin C − sin D = 2 sin cos We have, α, β, γ ∈ 0, π
2 2 2
C + D C − D α + β β + γ γ + α π
⇒ , , ∈ 0,
cos C + cos D = 2 cos cos 2 2 2 2
2 2
α + β β + γ , sin γ + α > 0
⇒ sin , sin
C + D C − D 2 2 2
cos C − cos D = − 2 sin sin α + β
2 2 ⇒ 4sin β + γ sin γ + α > 0
sin
2 2 2
or ⇒ sin α + sin β + sin γ − sin (α + β + γ ) > 0
C + D D −C ∴ sin α + sin β + sin γ > sin (α + β + γ )
cos C − cos D = 2 sin sin
2 2
Trigonometric Ratios of
X Example 13. The value of
π
2 cos cos
9π
+ cos
3π
+ cos
5π
is
Multiples of an Angle
13 13 13 13 2 tan θ
(a) 2 (b) 0 (i) sin 2θ = 2 sin θ cos θ =
1 + tan 2 θ
(c)1 (d) 3
(ii) cos 2θ = cos 2 θ − sin 2 θ = 1 − 2 sin 2 θ
Sol. (b) We have,
π 9π
+ cos
3π
+ cos
5π 1 − tan 2 θ
2 cos cos = 2 cos 2 θ − 1 =
13 13 13 13 1 + tan 2 θ
9π π 9π π 3π 5π
= cos + + cos − + cos + cos
13 13 13 13 1 1
13 13
(iii) cos 2 θ = (1 + cos 2θ ), sin 2 θ = (1 − cos 2θ )
10 π 8π 3π 5π 2 2
= cos + cos + cos + cos
13 13 13 13 2 tan θ
= cos π −
3π 5π 3π 5π (iv) tan 2θ =
+ cos π − + cos + cos
1 − tan 2 θ
13 13 13 13
3π 5π 3π 5π cot 2 θ − 1
= − cos − cos + cos + cos
13 13 13 13 (v) cot 2θ =
=0
2 cot θ
428
(vi) sin 3θ = 3 sin θ − 4 sin 3 θ
(vii) cos 3θ = 4 cos θ − 3 cos θ
3
X Example 16. Prove that
(i)
1 + sin 2θ + cos 2θ
= cot θ.
9
Trigonometric Functions and Equations
3 tan θ − tan 3 θ 1 + sin 2θ − cos 2θ
(viii) tan 3θ =
1 − 3 tan 2 θ 1 + sin θ − cos θ θ
(ii) = tan .
1 + sin θ + cos θ 2
cot 3 θ − 3 cot θ 3 cot θ − cot 3 θ
(ix) cot 3θ = = cos θ π θ
3 cot 2 θ − 1 1 − 3 cot 2 θ (iii) = tan − .
1 + sin θ 4 2
Trigonometric Ratios of Submultiple Angles Sol. (i) LHS = 1 + sin 2θ + cos 2θ = (1 + cos 2θ) + sin 2θ
1 1 + sin 2θ − cos 2θ (1 − cos 2θ) + sin 2θ
2 tan θ
1 1 2 2 cos θ + 2 sin θ cos θ
2
2 cos θ (cos θ + sin θ)
(i) sin θ = 2 sin θ cos θ = = =
2 2 1 2 sin2 θ + 2 sin θcos θ 2sin θ (cos θ + sin θ)
1 + tan 2 θ
2 = cot θ = RHS Hence proved.
1 1 1 1 + sin θ − cos θ (1 − cos θ) + sin θ
(ii) cos θ = cos 2 θ − sin 2 θ = 1 − 2 sin 2 θ (ii) LHS = =
2 2 2 1 + sin θ + cos θ (1 + cos θ) + sin θ
1 θ θ θ
= 2 cos 2 θ − 1 2 sin2 + 2 sin cos
2 = 2 2 2
θ θ θ
1 2 cos 2 + 2 sin cos
1 − tan 2 θ 2 2 2
θ θ θ
= 2 2 sin sin + cos
1 2 2 2 θ
1 + tan 2 θ = = tan = RHS
θ θ θ 2
2 2 cos sin + cos
2 2 2
1
2 tan θ Hence proved.
(iii) tan θ = 2 cos θ
1 (iii) LHS =
1 − tan 2 θ 1 + sin θ
2 π
sin − θ
1 2
cot 2 θ − 1 =
π
(iv) cot θ = 2 1 + cos − θ
1 2
2 cot θ π θ π θ
2 2 sin − cos −
4 2 4 2
A A =
(v) sin A = 3 sin − 4 sin 3 π θ
3 3 2 cos 2 −
4 2
A A π θ
(vi) cos A = 4 cos 3 − 3 cos = tan − = RHS Hence proved.
3 3 4 2
A A
3 tan − tan 3 X Example 17. The value of
3 3
(vii) tan A = π
A
2 2 + 2 + 2 + 2 cos 8θ , where 0 < θ < , is
1 − 3 tan 8
3
(a) 2cos θ (b) cos θ
Ø ● sin(α) + sin (α + β) + sin (α + 2β) +… + sin [α + (n − 1)β] (c) 2sin θ (d) −2cos θ
Sol. (a) We have,
β nβ
sin α + (n − 1) sin
2 2 2+ 2+ 2 + 2 cos 8θ
=
β Q1 + cos 8θ = 2 cos 2 8θ
sin = 2+ 2+ (4cos 2 4θ)
2 2
● cos(α) + cos(α + β) + cos (α + 2β)+… + cos [α + (n − 1)β] = 2+ 2 + 2 cos 4θ = 2 + 2(1 + cos 4θ)
β nβ = 2+ 2(2 cos 2 2 θ) [Q1 + cos 4θ = 2 cos 2 2 θ]
cos α + (n − 1) sin
= 2 2
= 2 + 2 cos 2 θ = 2(1 + cos 2 θ)
β
sin = 2(2 cos 2 θ) = 2 cos θ
2 429
9 X Example 18. The value of
tan 2θ tan 8θ
sec 8θ − 1
sec 4θ − 1
tan 8θ
is
tan 6θ
Sol. (d) We have, cos2 θ
=
1 − tan2 θ
1 + tan2 θ
Objective Mathematics Vol. 1
(a) (b) (c) (d) 1 − (2 tan2 φ + 1)
tan 8θ tan 4θ tan 2θ tan 2θ = [Qtan2 θ = 2 tan2 φ + 1 ]
1 + 2 tan2 φ + 1
Sol. (c) We have, sec 8θ − 1 =
−2 tan2 φ
=
− tan2 φ
= − sin2 φ
sec 4θ − 1
2 + 2 tan φ
2
sec 2 φ
1
−1 ∴ cos 2θ + sin2 φ = 0
cos 8θ
=
1
−1 X Example 21. The smallest positive value of x
cos 4θ
1 − cos 8θ cos 4θ (in degree) for which
= ⋅
cos 8θ 1 − cos 4θ tan ( x + 100° ) = tan ( x + 50° ) tan x tan ( x − 50° ), is
2 sin2 4θ cos 4θ (a) 60° (b) 90°
= ⋅
cos 8θ 1 − cos 4θ (c) 45° (d) 30°
(2 sin 4θcos 4θ) sin 4θ Sol. (d) We have,
= ×
cos 8θ 2 sin2 2 θ tan ( x + 100° ) = tan ( x + 50° )tan x tan ( x − 50° )
2 sin 4θcos 4θ 2 sin 2 θ cos 2 θ tan ( x + 100° )
= ⇒ = tan ( x + 50° )tan x
× tan ( x − 50° )
cos 8θ 2 sin2 2 θ
sin ( x + 100° ) cos ( x − 50° ) sin ( x + 50° ) sin x
sin 2(4θ) cos 2 θ sin 8θ cos 2 θ ⇒ =
= × = × cos ( x + 100° ) sin ( x − 50° ) cos ( x + 50° ) cos x
cos 8θ sin 2 θ cos 8θ sin 2 θ sin (2 x + 50° ) cos 50°
tan 8θ ⇒ =
= tan 8θ ⋅ cot 2 θ = sin150° − cos (2 x + 50° )
tan 2 θ
[applying componendo and dividendo]
p ⇒ sin (2 x + 50° )cos(2 x + 50° ) = − sin150° cos 50°
X Example 19. If tan α = , where α = 6β and α ⇒ 2 sin (2 x + 50° )cos(2 x + 50° ) = − cos 50°
q
⇒ sin (4 x + 100° ) = sin (270° − 50° )
1 ⇒ sin (4 x + 100° ) = sin 220°
being an acute angle, then [ p cosec 2β − q sec 2β]
2 ⇒ 4 x + 100° = 220° ⇒ x = 30°
is equal to
(a) p 2 + q 2 (b) p 2 − q 2
X Example 22. The value of
θ
(c) 2 p 2 + q 2 ∞ tan n
(d) None of these 2
∑ θ
is
Sol. (a) We have, tan α = p n =1 n −1
2 cos n − 1
q 2
p q
∴ sin α = and cos α = 2 1 2 1
p2 + q 2 p2 + q 2 (a) − (b) +
sin 2θ θ sin 2θ θ
1
Now, [ p cosec 2β − q sec 2β ] 1 1 2 1
2 (c) − (d) −
sin 2θ θ sin θ θ
p2 + q 2 p q
Sol. (a) We have,
= cosec 2β − sec 2β
θ
p2 + q 2 p +q tan r
2 2 2
n
2
p +q
2 2
sin α cos α ∑ θ
= sin2 β − cos 2 β r = 1 2 r −1 cos
r −1
2 2
θ
= p2 + q 2
sin(α − 2 β ) sin r
2 sin2 β cos 2 β n
2
= ∑
r = 1 2 r − 1 cos
θ θ
sin (6β − 2β ) r cos r −1
= p2 + q 2 [Qα = 6β] 2 2
sin 4β 2 θ
2 sin r
= p2 + q 2
n
2
= ∑
r =1 2 r − 1 θ θ θ
2 sin r cos r cos r − 1
X Example 20. If tan 2 θ = 2 tan 2 φ + 1, then 2 2 2
θ
cos 2θ + sin 2 φ is equal to 1 − cos r − 1
n
2
= ∑
(a)1 (b) 2 θ θ
430 r =1 2 r − 1
sin r − 1 cos r − 1
(c) −1 (d) None of these 2 2
9
1
= [2 sin2 n −1 A cos 2 n −1 A]
n 1 1 2 n sin A
= ∑ −
r = 1 2 r − 2 sin
θ θ 1
r −2 2
r −1
sin r − 1 = n ⋅ sin(2 ⋅ 2 n −1 A)
Trigonometric Functions and Equations
2 2 2 sin A
1 sin2 n A
= n ⋅ sin2 n A = n
2 1 1 2 sin A 2 sin A
=
1
− + −
sin2θ sinθ sinθ 2 sin θ Remark
2 Students are advised to learn above result as standard
formula.
1
+ −
1
π 3π 5π
θ θ X Example 24. sin + sin + sin + ... to n
2 sin
2 2 sin 2 n n n
2 2
terms is equal to
1 1 (a)1 (b) 2
+K + −
2 n − 2 sin θ 2 n −1 sin θ (c) 3 (d) 0
2n − 2 2 n − 1
2 1
Sol. (d) sin π + sin 3 π + sin 5 π + ... to n terms
= − n n n
sin 2θ 2 n −1 sin θ n⋅2 π π 2π
n −1
2 sin
2 n 2 ⋅ n + (n − 1) n
∞ = ⋅ sin
tan(θ / 2 )
n n
tan(θ /2 r ) 2π
∴ ∑ n −1 θ = nlim ∑ θ
sin
2
r = 1 2 r −1 cos
→∞ 2n
n=1 2 cos n −1 r −1
2 2 sin π π + π − π 2
π
⋅ sin =
2 2 n 2 sin
= =0
π 2 n π
sin sin
2 1 n n
= lim −
π 3π 5π
n→ ∞ sin 2θ θ
2 n −1
sin ∴ sin + sin + sin + K to n terms = 0
n − 1 n n n
2
2 θ / 2 n −1 2 1 Computation of Trigonometrical Ratios of
= − lim = −
sin 2θ n→ ∞ θ ⋅ sin θ sin 2θ θ
n −1
Some Useful Angles
2
By using the formulae introduced in the previous
sections, we can now find the trigonometrical ratios of
X Example 23. The value of
some important angles.
cos A cos 2 A cos 2 2 A cos 2 3 A ... cos 2 n −1 A is
Value of sin 18°
cos 2 n A sin 2 n A Let θ = 18°. Then,
(a) n (b) n
2 cos A 2 sin A 5θ = 90°
n
sin 2 A cos 2 n A ⇒ 2θ + 3θ = 90°
(c) n (d) n
2 cos A 2 sin A ⇒ 2θ = 90° − 3θ
Sol. (b) We have, ⇒ sin 2θ = sin (90° − 3θ)
cos A cos 2 A cos 2 2 A cos 2 3 A K cos 2 n − 1 A ⇒ sin 2θ = cos 3θ
=
1
[(2 sin A cos A)cos 2 A cos 2 2 A cos 2 3 AK
⇒ 2 sin θ cos θ = 4 cos 3 θ − 3 cos θ
2 sin A
cos 2 n −1 A]
⇒ cos θ (2 sin θ − 4 cos 2 θ + 3) = 0
=
1
[(sin2 A cos 2 A) cos 2 2 A cos 2 3 A ...cos 2 n −1 A] ⇒ 2 sin θ − 4 cos 2 θ + 3 = 0 [Qcos θ = cos 18° ≠ 0]
2 sin A
1 ⇒ 2 sin θ − 4(1 − sin θ ) + 3 = 0
2
= [(2 sin2 A cos 2 A) cos 2 2 A cos 2 3 AK cos 2 n −1 A]
2 2 sin A ⇒ 4 sin 2 θ + 2 sin θ − 1 = 0
1
= [sin2(2 A) ⋅ cos 2 2 A cos 2 3 A ...cos 2 n −1 A]
2 2 sin A −2 ± 4 + 16
1 ⇒ sin θ =
= [(2 sin2 2 A cos 2 2 A)cos 2 3 A ...cos 2 n −1 A] 8
2 3 sin A
−1 ± 5
=
1
[sin(2 ⋅ 2 2 A)cos 2 3 A ...cos 2 n −1 A] ⇒ sin θ =
2 3 sin A 4
=
1
[(sin2 3 A cos 2 3 A cos 2 4 A ...cos 2 n −1 A] −1 ± 5 5 −1
2 3 sin A ⇒ sin θ = = [Qsin18° > 0]
4 4
M M M
1 n −1 n −1 5 −1
= [sin2 A cos 2 A] Hence, sin18° =
2 n −1 sin A 4 431
9 Value of cos 36°
Since, cos 2θ = 1 − 2 sin 2 θ
Value of sin 36°
Putting θ = 36° in sin θ = 1 − cos 2 θ , we obtain
Objective Mathematics Vol. 1
∴ cos 36° = 1 − 2 sin 2 18°[putting θ = 18°] sin 36° = 1 − cos 2 36°
2 2
5 − 1 5 + 1
⇒ cos 36° = 1 − 2 = 1−
4 4
6 −2 5 3 − 5 16 − (6 + 2 5 )
=1 − 2 =1 − =
16 4 16
5 +1 10 − 2 5
= =
4 4
5 +1 10 − 2 5
Hence, cos 36° = Hence, sin 36° =
4 4
Trigonometrical Ratios of Some More Angles between 0° and 90°
π π π π π π 3π π 3π 2π 5π π
Angle 0°/0 15°/ 18°/ 22.5°/ 30°/ 36°/ 45°/ 54°/ 60°/ 67.5°/ 72°/ 75°/ 90°/
12 10 8 6 5 4 10 3 8 5 12 2
3 −1 5 −1 2 − 2 1 10 − 2 5 1 5+1 3 2 + 2 10 + 2 5 3+1
sin θ 0 1
2 2 4 2 2 4 2 4 2 2 4 2 2
3+1 10 + 2 5 2 + 2 3 5+1 1 10 − 2 5 1 2 − 2 5 −1 3 −1
cos θ 1 0
2 2 4 2 2 4 2 4 2 2 4 2 2
3 −1 5 −1 1 10 − 2 5 10 − 2 5 10 + 2 5 3+1
tan θ 0 2 −1 1 3 2 +1 ∞
3+1 10 + 2 5 3 5+1 5+1 5 −1 3 −1
3+1 10 + 2 5 5+1 5+1 1 5 −1 3 −1
cot θ ∞ 2 +1 3 1 2 −1 0
3 −1 5 −1 10 − 2 5 10 − 2 5 3 10 + 2 5 3+1
2 2 4 4−2 2 2 4 2 2
sec θ 1 5 −1 2 2 4+2 2 5+1 ∞
3+1 10 + 2 5 3 10 − 2 5 3 −1
2 2 4+2 2 4 2 4 2 2 1
cosec θ ∞ 5+1 2 2 5 −1 4−2 2
3 −1 10 − 2 5 3 10 + 2 5 3+1
X Example 25. The value of X Example 26. Prove that
cos 12° + cos 84° + cos 156° + cos 132° is 4 sin 27° = 5 + 5 − 3 − 5 .
1
(a)1 (b) Sol. Consider, (cos 27 ° + sin 27 ° )2 = sin2 27 ° + cos 2 27 °
2
−1 + 2 sin 27 ° cos 27 °
(c) −1 (d) = 1 + sin 54° = 1 +
5+1 1
= (5 + 5 )
2 4 4
Sol. (d) Consider, 1
∴ cos 27 ° + sin27 ° = 5+ 5 …(i)
cos 12 ° + cos 84° + cos 156° + cos 132 ° 2
[Qsin 27 ° > 0, cos27 ° > 0]
= (cos 12 ° + cos 132 ° ) + (cos 84° + cos 156° ) Similarly, we have
12 ° + 132 ° 132 ° − 12 °
= 2 cos cos (cos 27 ° − sin 27 ° )2 = 1 − sin 54°
2 2
5+1 3− 5
° + ° = 1− =
+ 2 cos
84 156 156° − 84°
cos 4 4
2 2 1
∴ cos 27 ° − sin 27 ° = 3− 5 …(ii)
= 2 cos 72 ° cos 60° + 2 cos 120° cos 36° 2
Q 0 < θ < π , cos θ > sin θ
= 2 sin18° cos 60° + 2 cos 120° cos 36° 4
5 − 1 1 −1 5 + 1 On subtracting Eq. (ii) from Eq. (i), we get
=2 + 2
4 2 2 4 2 sin 27 ° =
1
5+ 5 −
1
3− 5
2 2
−1
= ⇒ 4 sin 27 ° = 5+ 5 − 3− 5 Hence proved.
2
432
X Example 27. If θ =
θ
− 4π
3
, justify whether the X Example 29. Prove that
tan 6° tan 42° tan 66° tan 78° = 1.
Sol. tan 6° tan 42 ° tan 66° tan78°
9
Trigonometric Functions and Equations
relation 2 sin = − 1 + sin θ + 1 − sin θ is true.
2 sin 6° sin 42 ° sin 66° sin78°
=
cos 6° cos 42 ° cos 66° cos 78°
Sol. θ = − 4 π = − 240° ⇒ θ = − 120° (2 sin 66° sin 6° )(2 sin78° sin 42 ° )
3 2 =
θ (2 cos 66° cos 6° )(2 cos 78° cos 42 ° )
∴ sin = sin (−120° ) = − sin120° = − sin (180° − 60° )
2 (cos 60° − cos 72 ° )(cos 36° − cos 120° )
=
= − sin60° = −
3 (cos 60° + cos 72 ° )(cos 120° + cos 36° )
2 1 − sin18° cos 36° + 1
θ
cos = cos (−120° ) = cos 120° = cos (180° − 60° ) 2
2 = 2
1 + sin18° − 1 + cos 36°
1
= − cos 60° = − 2 2
2
1 5 − 1 5 + 1 1
∴
θ θ
cos + sin = − −
1 3
=−
3+1
<0 − +
2 4 4 2
2 2 2 2 2 =
θ θ 3 −1 1 5 − 1 5 + 1 1
and cos − sin = − +
1 3
= >0 + −
2 2 2 2 2 2 4 4 2
θ θ (3 − 5 ) (3 + 5 )
Thus, cos + sin = − 1 + sin θ [Q 1 + sin θ > 0] …(i) =
2 2 (1 + 5 ) ( 5 − 1)
θ θ
and cos − sin = 1 − sin θ ...(ii) 9−5 4
2 2 = = =1 Hence proved.
5−1 4
On subtracting Eq. (ii) from Eq. (i), we get
θ
2 sin = − 1 + sinθ − 1 − sinθ, X Example 30. Prove that
2
which shows that the given relation is not true. π
2 cos = 2 + 2 + 2 + 2 .
32
π
2 = 2 1 +
X Example 28. Show that 1
Sol. Consider, 2+ = 2 1 + cos
π 2 4
tan = 6 − 3 + 2 − 2. 2
π π
24 = 2 cos
= 2 ⋅ 2 cos 2
8 8
Sol. Consider, sin15° = sin(45° − 30° ) π π
∴ 2 cos = 2 + 2 Qcos > 0
= sin 45° cos 30° − cos 45° sin 30° 8 8
1 3 1 1 3 −1 π π
= ⋅ − ⋅ = Again, 2+ 2 + 2 = 2 + 2 cos = 2 1 + cos
2 2 2 2 2 2 8 8
cos 15° = cos(45° − 30° ) π π
2
= 2 × 2 cos 2 = 2 cos
= cos 45° cos 30° + sin 45° sin 30° 16 16
1 3 1 1 π
= ⋅ + ⋅ ∴ 2+ 2+ 2 = 2 cos
2 2 2 2 16
3+1 π
= Now, 2 + 2+ 2+ 2 = 2 + 2 cos
2 2 16
π
sin
π
2 sin2
π
1 − cos
π = 2 1 + cos
π 16
∴ tan = 24 = 24 = 12
24 cos π 2 sin π cos π π 2
π π
sin = 2 ⋅ 2 cos 2 = 2 cos
24 24 24 12 32 32
3+1
1− π Qcos π > 0
1 − cos 15° 2 2 ∴ 2 cos = 2+ 2+ 2+ 2
= = 32 32
sin15° 3 −1
Hence proved.
2 2
2 2 − 3 −1
=
3 −1 Maximum and Minimum Values
=
( 3 + 1) (2 2 − 3 − 1)
( 3 + 1) ( 3 − 1)
of Trigonometrical Expressions
2 6−3− 3+2 2 − 3 −1 In this section, we shall discuss problems on
= finding the maximum and minimum values of various
2
2( 6 − 3 + 2 − 2 ) trigonometrical expressions.
=
2 Consider the expression a cos θ ± b sin θ, where θ is
= 6 − 3 + 2 −2 Hence proved. a variable. 433
9 Let y = a cos θ ± b sin θ
Further, let a = r cos α and b = r sin α. Then,
r = a 2 + b 2 and tan α =
b
⇒ −7 + 3 ≤
13
2
cos θ −
3 3
2
sinθ + 3 ≤ 7 + 3, ∀ θ
π
⇒ −4 ≤ 5 cos θ + 3 cos θ + + 3 ≤ 10, ∀ θ [from Eq.(i)]
3
Objective Mathematics Vol. 1
a Hence proved.
∴ y = r cos α cos θ ± r sin α sin θ
⇒ y = r cos (θ m α ) X Example 33. Find a and b such that the
We know that, inequality holds good for all θ,
π
−1 ≤ cos (θ m α) ≤ 1, ∀ θ a ≤ 3 cos θ + 5 sin θ − ≤ b.
6
⇒ −r ≤ r cos (θ m α ) ≤ r, ∀ θ
⇒ − a 2 + b2 ≤ y ≤ a 2 + b2 , ∀ θ Sol. We have, 3cos θ + 5sin θ − π
6
⇒ − a + b ≤ a cos θ ± b cos θ π π
= 3cos θ + 5 sin θcos − cos θsin
2 2
6 6
≤ a 2 + b 2 for all θ. 3 1
= 3 cos θ + 5 sin θ − cos θ
If follows from the above discussion that 2 2
1 5 3
− a 2 + b 2 and a 2 + b 2 are minimum and maximum = cos θ + sin θ
2 2
values of a cos θ ± b sin θ for varying values of θ. 2 2
1 + 5 3 ≤ 1 cosθ + 5 3
We have,
2 2
Ø The maximum and minimum values of a cos θ ± b sin θ + c are 2 2
c + a2 + b 2 and c − a2 + b 2 respectively i . e. 2 2
sinθ ≤ 1 + 5 3 , ∀ θ
2 2
c − a2 + b 2 ≤ a cos θ ± b sin θ + c ≤ c + a2 + b 2 .
π
⇒ − 19 ≤ 3cos θ + 5sin θ − ≤ 19, ∀ θ
6
X Example 31. The maximum and minimum
values of 7 cos θ + 24 sin θ are Hence, a = − 19 and b = 19.
(a) 25 and − 25 (b) 24 and – 24 X Example 34. Show that for all real value θ,
(c) 5 and – 5 (d) None of these the expression a sin 2 θ + b sin θ cos θ + c cos 2 θ lies
Sol. (a) We know that the maximum and minimum values of 1
acos θ + b sin θ are a2 + b 2 and − a2 + b 2 respectively.
between {( a + c) − b 2 + ( a − c) 2 } and
2
So, the maximum and minimum values of 1
{( a + c) + b 2 + ( a − c) 2 }.
7 cos θ + 24 sin θ are 7 2 + 242 = 25 2
and − 7 2 + 242 = − 25 respectively. Sol. Let y = asin2 θ + b sin θcos θ + c cos 2 θ. Then,
1 − cos 2θ b 1 + cos 2θ
y = a + sin2θ + c
X Example 32. Prove that 2 2 2
π a+c 1
5 cos θ + 3 cos θ + + 3 lies between −4 and 10. ⇒ y= + {(c − a)cos 2θ + b sin2θ}
3 2 2
Now, − (c − a)2 + b 2 ≤ (c − a)cos 2θ + b sin2θ
Sol. We have, 5cos θ + 3cos θ + π + 3 ≤ (c − a)2 + b 2 , ∀ θ ∈ R
3
1 1
π π ⇒ − (c − a)2 + b 2 ≤ {(c − a)cos 2θ + b sin2θ}
= 5cos θ + 3 cos θcos − sin θsin + 3 2 2
3 3
1
3 3 3 ≤ (c − a)2 + b 2 , ∀ θ ∈ R
= 5cos θ + cos θ − sin θ + 3 2
2 2 a+c 1 a+c
⇒ − (c − a)2 + b 2 ≤
13 3 3
= cos θ − sin θ + 3 …(i) 2 2 2
2 2 1
+ {(c − a)cos 2θ + b sin2θ}
2 2 2
− 13 + 3 3 ≤ 13 cos θ − 3 3 sin θ a+c 1
Now,
2 2 2 2 ≤ + (c − a)2 + b 2 , ∀ θ ∈ R
2 2
2 2 1
≤ 13 + 3 3 ⇒ {(a + c ) − b 2 + (c − a)2 }
2 2 2
≤ asin2 θ + b sin θ cos θ + c cos 2 θ
13 3 3
⇒ −7 ≤ cos θ − sinθ ≤ 7, ∀ θ 1
2 2 ≤ [(a + c ) + b 2 + (c − a)2 ], ∀ θ ∈ R Hence proved.
2
434
X Example 35. Show that for varying θ and fixed
α, the expression cos θ (sin θ + sin θ + sin α ) 2 2
X Example 36. If in a ∆ABC,
A B C
tan , tan and tan are in HP, then find the
9
Trigonometric Functions and Equations
2 2 2
always lies between − 1 + sin 2 α and 1 + sin 2 α. B
minimum value of cot .
2
Sol. Let y = cos θ (sin θ + sin2 θ + sin2 α ).
Sol. In ∆ABC, we have
Then, y = sin θcos θ + cos θ sin2 θ + sin2 α cot
A B C A
+ cot + cot = cot cot cot
B C
…(i)
2 2 2 2 2 2
⇒ ( y − sinθcos θ)2 = cos 2 θ (sin2 θ + sin2 α ) A B C
It is given that, tan , tan and tan are in HP.
⇒ y2 − 2 ysin θcos θ = cos 2 θ sin2 α 2 2 2
A B C
⇒ y2 − 2 ysin θ cos θ + cos 2 θ = cos 2 θ + cos 2 θ sin2 α ⇒ cot , cot and cot are in AP.
2 2 2
⇒ y2 sec 2 θ − 2 y tan θ + 1 = 1 + sin2 α ⇒
B A
2 cot = cot + cot
C
…(ii)
[dividing throughout by cos 2 θ] 2 2 2
From Eqs. (i) and (ii), we have
⇒ y2 tan2 θ − 2 y tan θ + 1 = (1 + sin2 α ) − y2
A B C B A C
cot cot cot = 3cot ⇒ cot cot = 3
⇒ ( y tan θ − 1)2 = (1 + sin2 α ) − y2 2 2 2 2 2 2
⇒ (1 + sin2 α ) − y2 ≥ 0 Since, AM ≥ GM
A C
cot + cot
⇒ y2 − ( 1 + sin2 α )2 ≤ 0 2 2 ≥ cot A cot C
⇒
2 2 2
⇒ − 1 + sin2 α ≤ y ≤ 1 + sin2 α
cot ≥ 3 Qcot + cot = 2 cot
B A C B
⇒
Hence, − 1 + sin2 α ≤ cos θ(sin θ + sin2 θ + sin2 α ) 2 2 2 2
B
≤ 1 + sin2 α, ∀ θ Hence proved. Hence, the minimum value of cot is 3.
2
Work Book Exercise 9.2
1 1 6 The value of 3 cosec20° − sec 20 ° is
1 If sin A = and sin B = , where A, B are
10 5 a 2 b 1
acute angles, then A + B equals c 4 d −4
π π
a b x2 + y2
4 2 7 The equation sin 2 θ = is possible, if
π 2 xy
c d π
3 a x= y b x=−y
c 2x = y d None of these
2 If the angle α is in the third quadrant and
tan α = 2, then sin α is equal to sin ( x + y ) a + b tan x
8 If = , then is equal to
a
2
b
2 sin ( x − y ) a − b tan y
5 3 b a
a b
5 a b
c d None of these
2 c ab d None of these
3 If θ lies in the first quadrant, then which of the 9 If tan 2 θ = 2 tan 2 φ + 1, then cos 2θ + sin 2 φ equals
following is not true?
a −1
θ θ θ θ
a < tan b < sin b 0
2 2 2 2
θ θ θ c 1
c θ cos 2 < sinθ d θ sin < 2 sin d None of the above
2 2 2
4 cos 2θ + 2 cos θ is always 10 The value of
π 3π 5π 7π 9π 11π 13π
3 3 sin sin sin sin sin sin sin is
a >− b ≥ 14 14 14 14 14 14 14
2 2
3 1
c ≤− d None of these a
2 16
1
b
5 If A = tan 6° tan 42 ° and B = cot 66° cot 78°, then 64
1 1
a A = 2B b A= B c
3 128
c A=B d 3A = 2B d None of the above
435
9 11 If x cos α + y sin α = 2 a, x cos β + y sin β = 2 a and
2 sin
α
2
β
sin = 1, then
2 then k 2 is equal to
1
2
1
16 If a sin 2 θ + bsinθ cos θ + c cos 2 θ − (a + c ) ≤ k,
2
Objective Mathematics Vol. 1
2 ax a b 2 + ( a − c )2
a cos α + cos β =
x + y
2 2
b a 2 + ( b − c )2
2 a 2 − y2
b cos α cos β = c c 2 + ( a − b )2
x 2 + y2
d None of the above
c y = 4a(α − x )
2
d cos α + cos β = 2 cos α cos β 17 If α, β, γ and δ are the smallest positive angles in
12 If sin θ − cos θ < 0, then θ lies between ascending order of magnitude which have their
3π π sines equal to the positive quantity λ, then the
a nπ − and nπ + , n ∈ I α β γ δ
4 4 value of 4 sin + 3 sin + 2 sin + sin is
π π 2 2 2 2
b nπ − and nπ + , n ∈ I
4 4 a 1+ λ
3π π
c 2 nπ − and 2 nπ − , n ∈ I b 2 1+ λ
4 4
3π π c 1− λ
d 2 nπ − and 2 nπ + , n ∈ I
4 4 d 2 1− λ
θ φ
18 If tan = and tan = , then the value of
A A 5 3
13 If 2 cos = 1 + sin A + 1 − sin A, then lies
2 2 2 2 2 4
between (n ∈ I ) cos(θ + φ ) is
π 3π 364
a 2 nπ + and 2 nπ + a −
4 4 725
π π
b 2 nπ − and 2 nπ + 627
4 4 b −
3π π 725
c 2 nπ − and 2 nπ − 240
4 4 c −
d − ∞ and + ∞ 339
d None of the above
14 The angle θ whose cosine equals to its tangent,
π
is given by 19 If α, β, γ ∈ 0, , then the value of
2
a cos θ = 2 cos 18° b cos θ = 2 sin18°
c sinθ = 2 sin18° d sinθ = 2 cos 18° sin(α + β + γ )
is
sin α + sin β + sin γ
15 The value of tan 5θ is
a <1 b >1
5 tanθ − 10 tan3 θ + tan5 θ
a c =1 d None of these
1 − 10 tan2 θ + 5 tan4 θ
5 tanθ + 10 tan3 θ − tan5 θ 20 If cos x = tan y, cos y = tan z,cos z = tan x, then
b the value of sin x is
1 + 10 tan2 θ − 5 tan4 θ
a 2 cos 18°
5 tan5 θ − 10 tan3 θ + tanθ
c b cos18°
1 − 10 tan2 θ + 5 tan4 θ c sin18°
d None of the above d 2 sin18°
Trigonometric Equations
An equation involving one or more trigonometrical Solutions or Roots of a Trigonometric
ratios of unknown angle is called a trigonometric Equation
equation.
A value of the unknown angle which satisfies the
e.g. sin 2 x − 4 sin x = 1
given equation, is called a solution or root of the
It is to be noted that a trigonometrical identity is
satisfied for every value of the unknown angle, whereas equation.
trigonometric equation is satisfied only for some values The trigonometric equation may have infinite
(finite or infinite) of unknown angle.
number of solutions and can be classified as
e.g. sin 2 x + cos 2 x = 1 is a trigonometrical identity
(i) Principal solution (ii) General solution
as it is satisfied for every value of x ∈ R .
436
(i) Principal solution The value of unknown angle
which satisfies the given equation and lies in the
interval [0, 2π ) , is called a principal solution of
X Example 37. If the equation sin 2 θ − cos θ = ,
1
4
then the values of θ lying in the interval 0 ≤ θ ≤ 2π
9
Trigonometric Functions and Equations
trigonometric equation. are
(ii) General solution We know that trigonometric π 5π π 2π 4π 5π 3π π
(a) , (b) , (c) , (d) ,
functions are periodic and solutions of 3 3 3 3 3 3 5 5
trigonometric equations can be generalised with
Sol. (a) The given equation can be written as
the help of the periodicity of the trigonometric 1
functions. 1 − cos 2 θ − cos θ =
4
The solution consisting of all possible solutions 3
⇒ cos 2 θ + cos θ − =0
of a trigonometric equation is called its general 4
solution. ⇒ 4cos 2 θ + 4cos θ − 3 = 0
−4 ± 16 + 48 1 3
Ø A function f (x) is said to be a periodic function, if a least positive ⇒ cosθ = = ,−
8 2 2
real numberTis such that f (x +T ) = f (x), thenTis known as period 3
of function f (x) . Since, cosθ = − is not possible.
2
1 π
∴ cos θ = = cos
General Solution of 2
π
3
⇒ θ = 2 nπ ±
Trigonometric Equations 3
For the given interval, n = 0 and n = 1
(i) sin θ = 0 θ = nπ, n ∈ I π 5π
π ⇒ θ= ,
(ii) cos θ = 0 θ = (2 n + 1) , n ∈ I 3 3
2
(iii) tanθ = 0 θ = nπ, n ∈ I X Example 38. The solution of
(iv) sinθ = sinα π π
θ = nπ + ( −1) α, α ∈ − , , n ∈ I
n tan x + tan 2x + tan 3x = tan x tan 2x tan 3x is
2 2 nπ
(a) x = nπ , n ∈ I (b) x = , n ∈ I
(v) cos θ = cos α θ = 2 nπ ± α,α ∈ [0, π ], n ∈ I 2
(vi) tanθ = tanα π π
θ = nπ + α, α ∈ − , , n ∈ I
nπ
2 2 (c) x = , n ∈ I (d) None of these
3
(vii) sin2 θ = sin2 α θ = nπ ± α, n ∈ I
Sol. (c) We know that,
cos 2 θ = cos 2 α tan x + tan2 x
2
tan θ = tan α
2 tan 3 x = tan ( x + 2 x) =
1 − tan x tan2 x
(viii) sinθ = 1 π ⇒ tan 3 x − tan x tan2 x tan 3 x = tan x + tan2 x
θ = ( 4n + 1) , n ∈ I
2 ⇒ tan x tan2 x tan 3 x = tan 3 x − tan x − tan2 x …(i)
(ix) cosθ = 1 θ = 2 nπ, n ∈ I Now, tan x + tan2 x + tan 3 x = tan x tan2 x tan 3 x
⇒ tan x + tan2 x + tan 3 x = tan 3 x − tan x − tan2 x
(x) cosθ = − 1 θ = (2 n + 1) π, n ∈ I
[from Eq. (i)]
(xi) sin x = sinα and x = 2 nπ + α ⇒ 2 tan x = − 2 tan2 x
cos x = cos α
⇒ tan2 x = tan(− x)
2 x = nπ + (− x) ⇒ 3x = nπ
π π nπ
Ø ● If θ is an odd multiple of i . e . when θ = (2n + 1) , n ∈I, then ∴ x= , where n ∈ I
2 2 3
secθ and tanθ are not defined.
π
● If θ is an even multiple of i . e . when θ = nπ, n ∈I, then cosec θ X Example 39. The number of distinct solution
2
and cotθ are not defined. π
of sin 5θ cos 3θ = sin 9θ cos 7θ in 0, , is
● The general solution of equations containing cosec θ, secθ and 2
cotθ is equivalent to that of the equation involving sinθ, cosθ
and tanθ. (a) 4 (b) 5
● When cosθ = 0, then sinθ = 1or −1 (c) 8 (d) 9
θ = n + π
1
∴ Sol. (d) sin 8θ + sin2 θ = sin16θ + sin2 θ or sin16θ = sin 8θ
2
∴ 16θ = nπ + (−1)n 8θ
If sinθ = 1, then n is even and if sinθ = − 1, then n is odd. ⇒ 8 θ = 2 mπ, when n is even, n = 2 m
Similarly, when sinθ = 0, then cosθ = 1or −1 24θ = (2 m + 1)π, when n is odd, n = (2 m + 1)
∴ θ = nπ mπ (2 m + 1)π
If cosθ = 1, then n is even and if cosθ = − 1, then n is odd. ∴ θ= , , when m ∈ I
4 24
The general solution should be given unless the solution is π π π π 5 π 9 π 7 π 11π
= 0, , and , , , , , 437
required in a specified interval or range. 2 4 24 8 24 24 24 24
9 X Example 40. If 0 ≤ x ≤
sin 2 x cos2 x
+ 81 = 30, then x is equal to
π
2
and ∴ sin α =
a + b2
2
b
Objective Mathematics Vol. 1
81 a
π π π and cos α =
(a) (b) (c) (d) 0 a 2 + b2
6 2 4
2 2 From Eq. (i), we get
Sol. (a) Given, 81 sin x
+ 81cos x
= 30
c
⇒ 81sin
2
+ 811−sin x = 30
x 2
cos θ cos α + sin θ sin α =
2 81 a 2 + b2
⇒ 81sin x + 2
= 30
81sin x c
sin 2 x ⇒ cos (θ − α ) =
Let 81 = y
a 2 + b2
81
∴ y+ = 30
y If | c | > a 2 + b 2
⇒ y − 30 y + 81 = 0
2
⇒ ( y − 27 ) ( y − 3) = 0 Then, equation a cos θ + b sin θ = c has no solution
⇒ y = 27 or y = 3 and if
2 2
⇒ 81sin x
= 27 or 81sin x
=3 | c | ≤ a 2 + b 2 , then let
2 2
⇒ 34 sin = 33 or 34 sin x = 31
x
| c|
⇒ sin2 x =
3
or 4sin2 x = 1
cos φ =
4 a + b2
2
1 π
Q x ∈ 0,
3
⇒ sin x = or sin x = ⇒ cos(θ − α ) = cos φ
2 2 2
π π ⇒ θ − α = 2nπ ± φ
⇒ x= or x =
3 6 ⇒ θ = 2nπ ± φ + α, n ∈ I
tan 3θ − 1
X Example 41. If = 3, then the Ø ● While solving a trigonometric equation, squaring the
tan 3θ + 1 equation at any step should be avoided as far as possible.
● If squaring is necessary, check the solution for extraneous
general value of θ is values.
nπ π 7π nπ 7π π Never cancel terms containing unknown terms on the two
− (b) nπ + + (d) nπ +
●
(a) (c) sides, which are in product. It may cause loss of genuine
3 12 12 3 36 12 solution.
Sol. (c) Given, tan 3θ − 1 = 3 ● The answer should not contain such values of angles which
tan 3θ + 1 make any of the terms undefined or infinite.
● Domain should not change. If it changes, necessary
⇒ tan 3θ − 1 = 3 tan 3θ + 3 corrections must be made.
⇒ tan 3θ − 1 − 3 tan 3θ − 3 = 0 ● Check that denominator is not zero at any stage while
⇒ tan 3θ (1 − 3 ) − (1 + 3 ) = 0 solving equations.
1+ 3
⇒ tan3θ =
1− 3 X Example 42. If sin x + 3 cos x = 2, then x is
7π
⇒ tan 3θ = tan105° = tan equal to
12 π π 5π π
∴ 3θ = nπ +
7π
⇒ θ=
nπ 7 π
+ (a) 2nπ + , 2nπ − (b) 2nπ + , 2 nπ −
12 3 36 12 3 12 12
π π
(c) 2nπ − , 2nπ − (d) None of the above
Solution of Trigonometric Equation of the 12 12
Form a cos θ + b sin θ = c Sol. (b) Given, 3 cos x + sin x = 2
Let the equation be ⇒
3 1
cos x + sin x =
2
a cos θ + b sin θ = c 2 2 2
π π
⇒ cos x − = cos
On dividing both sides by a 2 + b 2 , we get 6 4
π π
a b c ⇒ x− = 2 nπ ±
cos θ + sin θ = …(i) 6 4
a 2 + b2 a 2 + b2 a 2 + b2 ⇒
π
x = 2 nπ ± +
π
4 6
b 5π π
Let tan α = ⇒ x = 2 nπ + , 2 nπ − , where n ∈ I
a 12 12
438
X Example 43. General solution of
sin x + cos x = min {1, a 2 − 4a + 6} is
a ∈R
X Example 45. If the equation tan θ + sec θ = 3,
then the value of θ ∈(0, 2π ) is
9
Trigonometric Functions and Equations
π 3π
nπ π π (a) (b)
(a) + ( −1) n (b) 2nπ + ( −1) n 4 2
2 4 4
π
n +1 π n π π (c) (d) None of these
(c) nπ + ( −1) (d) nπ + ( −1) − 6
4 4 4
Sol. (d) Given, Sol. (c) Given, sin θ + 1
= 3 [Qcosθ ≠ 0]
cos θ cos θ
sin x + cos x = min {1, a − 4a + 6}
2
a ∈R ⇒ 3 cos θ − sin θ = 1
Now, a2 − 4a + 6 = (a − 2 )2 + 2 On dividing both sides by 2, we have
∴ min {1, a2 − 4a + 6} = min{1, 2} = 1 3 1 1
a ∈R cos θ − sin θ =
Now, sin x + cos x = 1 2 2 2
π π π
sin x + = cos θ + = cos
1
⇒ ⇒
4 2 6 3
π π π π
⇒ x+ = nπ + (−1)n ⋅ ⇒ θ+ = 2 nπ ±
4 4 6 3
n π π
⇒ x = nπ + (−1) − Taking positive sign,
4 4 π π
θ+ = 2 nπ +
6 3
X Example 44. The equation 3 sin x + cos x = 4 ⇒ θ = 2 nπ +
π
…(i)
has 6
(a) only one solution Taking negative sign,
π π
(b) two solutions θ+ = 2 nπ −
6 3
(c) infinitely many solutions π
⇒ θ = 2 nπ −
(d) no solution 2
π
Sol. (d) We know that, ⇒ θ = (4n − 1) …(ii)
2
acos x + b sin x = c
From Eqs. (i) and (ii), values of θ between 0 and 2 π are
has solutions, if| c | ≤ a2 + b 2 π 3π
, .
So, given equation is 6 2
3 sin x + cos x = 4 3π
But when θ = , then tan θ and sec θ does not exist, so
∴ a2 + b 2 = 12 + ( 3 )2 = 4 =2 2
it is rejected.
and | c| = 4 ⇒ | c| > a + b
2 2
π
Hence, given equation has no solution. Hence, θ=
6
Work Book Exercise 9.3
1 In a ∆ABC, ∠A is greater than ∠B. If the values of 4 General solution of the equation
the angles A and B satisfy the equation ( 3 − 1)sin θ + ( 3 + 1)cos θ = 2 is
3 sin x − 4 sin 3 x − k = 0, 0 < k < 1, then the π π π π
a 2 nπ ± + b nπ + ( −1)n +
measure of ∠C is 4 12 4 12
π π π π π π
a b c 2 nπ ± − d nπ + ( −1)n −
3 2 4 12 4 12
2π 5π
c d 5 In a right angled triangle, the hypotenuse is 2 2
3 6
times the length of perpendicular drawn from the
2 The equation (cos p − 1)x 2 + (cos p)x + sin p = 0, opposite vertex on the hypotenuse, then the
where x is a variable, has real roots. Then, p lies
other two angles are
in π π
π π
c − , d (0, π)
a ,
a ( 0, 2 π) b ( − π, 0) 3 6
2 2 π π
b ,
4 4
3 The smallest positive root of the equation π 3π
c ,
tan x − x = 0, lies in 8 8
π π 3π 3π π 5π
a 0, b , π c π, d , 2 π d , 439
2 2 2 2 12 12
9 6 The solution set of (2 cos x − 1) (3 + 2 cos x ) = 0 in
the interval 0 ≤ x ≤ 2 π, is
π
8 General solution of 4 cot 2θ = cot 2 θ − tan 2 θ is
a nπ, n ∈ I b nπ + ( −1)n
π
4
, n∈I
Objective Mathematics Vol. 1
a π
3 c nπ ± , n∈I d None of these
π , 5π 4
b
3 3 9 The number of solutions of 2 sin 2 x + sin 2 2 x = 2,
π 5π 3
c , , cos −1 − x ∈[0, 2π ] is
3 3 2
a 4 b 5
d None of the above c 7 d 6
7 The most general value of θ which satisfies both sin
x x
+ cos − i tan x
1 2 2
the equations tan θ = − 1and cos θ = , will be 10 If the expression is real,
2 x
1 + 2 i sin
7π 2
a nπ +
4 then x is equal to
7π
b nπ + ( −1)n a 2 nπ + 2 tan−1 k, k ∈ R, n ∈ Z
4
7π b 2 nπ + 2 tan−1 k, where k ∈ ( 0, 1), n ∈ Z
c 2 nπ +
4 c 2 nπ + 2 tan−1 k, where k ∈ (1, 2 ), n ∈ Z
d None of the above d 2 nπ + 2 tan−1 k, k ∈ (2, 3), n ∈ Z
Solving Simultaneous Equations X Example 47. If r > 0, −π ≤ θ ≤ π and r, θ satisfy
Here, we will discuss problems related to the r sin θ = 3 and r = 4 (1 + sin θ ), then the number of
solution of two equation satisfied simultaneously. We possible solutions of the pair ( r, θ ) is
may divide the problems in two categories : (a) 2 (b) 4 (c) 0 (d) infinite
(i) Two equations in one ‘unknown’ satisfied
Sol. (a) On eliminating θ from two given equations,
simultaneously.
we get
(ii) Two equations in two ‘unknowns’ satisfied
r = 4 1 +
3
simultaneously. r
1 ⇒ (r − 6) (r + 2 ) = 0 ⇒ r = 6
X Example 46. If the equations sin θ = − and
2 Neglecting r = − 2, as r > 0
1 1 π 5π
tan θ = , then most common general value of θ is ∴ sin θ = ⇒ θ= ,
2 6 6
3
π 5π
7π So, (r, θ) = 6, , 6,
6
(a) 2nπ ± 6
6
7π X Example 48. If 3 sin 2 θ + 2 sin 2 φ = 1 and
(b) 2nπ −
6 π π
7π 3 sin 2θ = 2 sin 2φ, 0 < θ < and 0 < φ < , then the
(c) 2nπ + 2 2
6 value of θ + 2φ is
(d) None of the above π π
(a) (b)
Sol. (c) First find the values of θ lying between 0 and 2 π and 2 4
satisfying the two given equations separately. Select the value (c) 0 (d) None of these
of θ which satisfies both the equations, then generalise it.
1
Sol. (a) Here, 3sin2 θ = cos 2 φ and 3sinθ cos θ = sin2 φ
sin θ = − On squaring and adding, we get
2
7π 11π 9 sin2 θ (sin2 θ + cos 2 θ) = 1
⇒ θ= or
6 6 1 2 2
1 i.e. sin θ =
and cosθ =
tan θ = 3 3
3
1 1 2 2
π 7π ∴ cos2 φ = 3 ⋅ = and sin2 φ =
⇒ θ= or 9 3 3
6 6
[values between 0 and 2 π] Now, cos(θ + 2 φ) = cos θcos 2 φ − sin θsin2 φ
7π =
2 2 1 1 2 2
⋅ − ⋅ =0
Common value of θ =
6 3 3 3 3
The required solution is 3π
and θ + 2φ <
7π 2
θ = 2 nπ +
6 π
440 ∴ θ + 2φ =
2
Solution of Trigonometric Inequality
To solve trigonometric inequality, we must use
X Example 51. The solution set of inequality
1
sin x > is
9
Trigonometric Functions and Equations
2
graph of the trigonometric functions which would be π 5π
clear from the following examples: (a) U 2nπ + , nπ +
n ∈I 6 6
X Example 49. If 2 sin x + 1 ≥ 0 and x ∈[0, 2π ], π 5π
then the solution set for x is (b) U 2nπ − , 2nπ +
n ∈I 6 6
7π 7π 11π
(a) 0, (b) 0, ∪ , 2π π 5π
6 6 6 (c) U nπ + , nπ +
n ∈I 6 6
11π
(c) , 2π (d) None of these π 5π
6 (d) U 2nπ + , 2nπ +
n ∈I 6 6
Sol. (b) 2 sin x + 1 ≥ 0 ⇒ sin x ≥ − 1
2 Sol. (d) From the graph of y = sin x, it obvious that between
The value scheme for this is shown below: 0 and 2 π,
1 Y
1
sin x ≥ − 1 –π π
y = 1/2
2 X' –3π –2π 2π 3π 4π
π 5π X
0 0 6 6
–1
–1
–1 2 Y'
2
1 π 5π
–1 sin x > for < x <
2 6 6
From the figure, 1 π 5π
11π 7π Hence, sin x > ⇒ 2 nπ + < x < 2 nπ +
≤ x ≤ 2 π or 0 ≤ x ≤ 2 6 6
π 5π
The required solution set is U 2 nπ + , 2 nπ +
6 6
7 π 11π .
∴
x ∈ 0, ∪ ,2π n ∈I 6 6
6 6
Problems on Extremum Values
X Example 50. If cos x − sin x ≥1 and 0 ≤ x ≤ 2π,
then the solution set for x is
X Example 52. If equation sin 4 x = 1 + tan 8 x,
π 7π 3π 7π then x is equal to
(a) 0, ∪ , 2π (b) , ∪ {0} π π π
4 4 2 4 (a) φ (b) (c) (d)
2 6 4
3π
(c) , 2π ∪ {0} (d) None of these Sol. (a) sin4 x = 1 + tan8 x only when
2 sin4 x = 1 and 1 + tan8 x = 1
Sol. (c) cos x + π ≥ 1 ⇒ sin2 x = 1 and tan8 x = 0
4 2 which is never possible, since sin x and tan x vanish
The value scheme for this is shown below: simultaneously. Therefore, the given equation has no
0 solution.
1
√2 X Example 53. If sin 2 x + cos 2 y = 2 sec 2 z, then
π
–1 1
(a) x = (2m + 1) , y = nπ and z = tπ
2
π
1 (b) x = ( m + 1) , y = 2nπ and z = tπ
√2 2
π
0 (c) x = (2m − 1) , y = nπ and z = 2tπ
π π π 2
From the figure, − ≤ x+ ≤
4 4 4 (d) None of the above
π π π Sol. (a) sin2 x + cos 2 y = 2 sec 2 z only when
and in general, 2 nπ − ≤ x + ≤ 2 nπ +
4 4 4 sin2 x = 1, cos 2 y = 1, sec 2 z = 1
π
∴ 2 nπ − ≤ x ≤ 2 nπ ⇒ cos 2 x = 0, sin2 y = 0, cos 2 z = 1
2
π 3π ⇒ cos x = 0, sin y = 0, sin z = 0
For n = 0, − ≤ x ≤ 0; for n = 1, ≤ x≤ 2π π
2 2 ∴ x = (2 m + 1) , y = nπ and z = tπ
2
where, m, n and t are integers. 441
9 X Example 54. The number of solutions of the
equation x 3 + x 2 + 4x + 2 sin x = 0 in 0 ≤ x ≤ 2π, is
Sol. (a) LHS = 2 cos 2 x sin2 x
2
= (1 + cos x)sin2 x < 2 [Q1 + cos x < 2, sin2 x ≤ 1]
Objective Mathematics Vol. 1
(a) 0 (b) 1 RHS = x2 + x−2 = x2 + 2 ≥ 2
1
(c) 2 (d) 4 x
Hence, the given equation has no solution.
Sol. (b) Here, x3 + ( x + 2 )2 + 2 sin x = 4
Clearly, x = 0 satisfies the equation. X Example 57. If sin 6 x = 1 + cos 4 3x, then x is
If 0 < x ≤ π, x3 + ( x + 2 )2 + 2 sin x > 4
equal to
If π < x ≤ 2 π, x3 + ( x + 2 )2 + 2 sin x > 27 + 25 − 2
π 3π
So, x = 0 is the only solution. (a) (2n − 1) , n ∈ I (b) (2n + 1) , n ∈ I
4 4
X Example 55. The number of real solutions of π
(c) (2n + 1) , n ∈ I (d) None of these
sin e x ⋅ cos e x = 2 x − 2 + 2 − x − 2 is 2
(a) 0 (b) 1 Sol. (c) sin6 x = 1 + cos 4 3 x ≤ 1
(c) 2 (d) infinite sin6 x = 1 + cos 4 3 x ≤ 1 is possible only when
Sol. (a) Here, 1 sin (2e x ) = 1 (2 x + 2 − x ) ≥ 1 ⋅ 2 = 1 sin6 x = 1 + cos 4 3 x = 1
2 4 4 2 ⇒ sin6 x = 1 and 1 + cos 4 3 x = 1
So, sin (2e x ) ≥ 1 ⇒ sin2 x = 1 and cos 4 3 x = 0
But sin (2e x ) >/ 1
⇒ cos 2 x = 0 and cos 3 x = 0
∴ sin (2e x ) = 1
But equality can hold when 2 x = 2 − x = 1 ⇒ cos x = 0 and cos 3 x = 0
π
i.e. x=0 ⇒ x = (2 m + 1)
2
sin (2 ⋅ e 0 ) = 1
π
which is not true. and 3 x = (2 n + 1) , where m, n ∈ I
2
π
x ⇒ x = (2 m + 1)
X Example 56. If 2 cos 2 sin 2 x = x 2 + x −2 , 2
2 π
and x = (2 n + 1)
π
0 < x ≤ , then x is equal to 6
π
2 Common value of x is (2 n + 1) , where n ∈ I.
π π π π 2
π
(a) φ (b) (c) (d) , The required solution is x = (2 n + 1) , n ∈ I.
2 3 4 3 2
Work Book Exercise 9.4
1 If A and B are acute positive angles satisfying the 1
4 If cos(α − β) = 1and cos(α + β) = , then the
equations 3 sin A + 2 sin B = 1 and
2 2
e
3 sin 2 A − 2 sin 2 B = 0, then A + 2 B is equal to number of ordered pairs (α , β) ∈ [− π, π ] is
π a 0 b 1
a 0 b c 2 d 4
2
π π
c d 5 The number of all possible ordered pairs
4 3
( x, y ), x, y ∈ R satisfying the system of equations
2 If sin A = sin B, cos A = cos B, then the value of A 2π 3
x+ y= , cos x + cos y = is
in terms of B is 3 2
a nπ + B a 0
b nπ + ( −1)n B b 1
c 2 nπ + B c infinite
d 2 nπ − B d None of the above
3 The number of pairs (x, y) satisfying the 6 The number of distinct roots of the equation
equations sin x + sin y = sin( x + y ) and A sin 3 x + B cos 3 x + C = 0 no two of which differ
| x | + | y | = 1is by 2π, is
a 2 a 3
b 4 b 4
c 6 c infinite
d infinite d 6
442
WorkedOut Examples
Type 1. Only One Correct Option
= 2cos
Ex 1. The maximum value of x x
Sol. Q 1 + cos x = 2 cos2
x x 2 2
4 sin 2 x + 3 cos 2 x + sin + cos is
and 1 − cos x = 2 sin 2 = 2 sin
2 2 x x
(a) 4 + 2 (b) 3 + 2 2 2
(c) 9 (d) 4 1 + cos x + 1 − cos x
∴
1 + cos x − 1 − cos x
Sol. Maximum value of 4 sin 2 x + 3 cos2 x i.e. sin 2 x + 3 is 4
cos x + sin x x x
x x
+ cos is
1
+
1
= 2, both − cos + sin
and that of sin 2 2 2 2
2 2 2 2 = =
π cos x − sin x − cos x − sin x
attained at x = . So, the given function has maximum 2 2 2 2
4
value 4 + 2. x x
cos − sin
= 2 2
Hence, (a) is the correct answer. x x
cos + sin
Ex 2. If in a ∆ABC, ∠C = 90°, then the maximum 2 2
x
value of sin A sin B is 1 − tan
1 = 2 = tan π − x
(a) (b) 1 x 4 2
2 1 + tan
2
(c) 2 (d) None of these π π x π x
= cot − + = cot +
1 2 4 2 4 2
Sol. sin A sin B = × 2 sin A sin B
2 π
1 ∴ a=
= [cos ( A − B ) − cos ( A + B )] 4
2 Hence, (a) is the correct answer.
1
= [cos ( A − B ) − cos 90° ] π
2 Ex 5. Let 0 < A, B < satisfying the equation
1 1 2
= cos ( A − B ) ≤
2 2 3sin 2 A + 2 sin 2 B = 1 and 3 sin 2 A −2 sin 2B = 0,
∴ Maximum value of sin A sin B =
1 then A + 2B is equal to
2 π
Hence, (a) is the correct answer. (a) π (b)
2
Ex 3. The maximum value of sin (cos x ) is π
(c) (d) 2π
(a) sin 1 (b) 1 4
1 3 Sol. From the second equation, we have
(c) sin (d) sin 3
2 2 sin 2 B = sin 2 A …(i)
2
Sol. cos x ∈ [ −1, 1], ∀ x ∈ R and sin x is increasing in and from the first equality,
π π 3 sin 2 A = 1 − 2 sin 2 B = cos 2 B …(ii)
− , .
2 2 Now, cos ( A + 2 B ) = cos A cos 2 B − sin A sin 2 B
∴ Maximum value of sin (cos x ) = sin 1 3
= 3 cos A sin 2 A − ⋅ sin A sin 2 A
Hence, (a) is the correct answer. 2
= 3 cos A sin 2 A − 3 sin 2 A cos A = 0
Ex 4. If x ∈(π, 2π ) and π 3π
⇒ A + 2B = or
1 + cos x + 1 − cos x x 2 2
= cot a + , then a π π
1 + cos x − 1 − cos x 2 Given that, 0 < A < and 0 < B <
2 2
is equal to π
⇒ 0 < A + 2B < + π
π π 2
(a) (b) π 3π
4 2 So, A + 2 B = , neglecting
π 2 2
(c) (d) None of these Hence, (b) is the correct answer. 443
3
9 Ex 6. If α, β, γ and δ are the solutions of the equation
π
tan θ + = 3 tan 3θ, no two of which have
⇒
⇒
π
2 2p
p≥
π
π
= cos x ± ≤ 1
4
4
Objective Mathematics Vol. 1
2 2
equal tangents, then the value of π
Now, the smallest value of p =
tan α + tan β + tan γ + tan δ is 2 2
(a) 1 (b) −1 Hence, (d) is the correct answer.
(c) 2 (d) 0
Ex 9. If α and β are solutions of
Sol. We have,
π sin x + a sin x + b = 0 as well that of
2
tan θ + = 3 tan 3θ
4 cos 2 x + c cos x + d = 0, then (α + β) is equal to
1 + tan θ 3 tan θ − tan 3 θ 2bd a2 + c2
⇒ = 3⋅ (a) (b)
1 − tan θ 1 − 3 tan 2 θ b2 + d 2 2ac
1+ t 3t − t 3
b2 + d 2 2ac
⇒ = 3 [putting t = tanθ] (c) (d)
1− t 1 − 3t 2 2bd a + c2
2
⇒ 3t 4 − 6t 2 + 8t − 1 = 0
0 Sol. According to the given condition,
So, t1 + t2 + t3 + t4 = = 0 sin α + sin β = − a and cosα + cosβ = − c
3
α+β α −β
∴ tan α + tan β + tan γ + tan δ = 0 ⇒ 2 sin cos = −a
Hence, (d) is the correct answer. 2 2
α+β α −β
= −c
π and 2 cos cos
Ex 7. For all θ in 0, , cos (sin θ ) is 2 2
2 α+β a
⇒ tan =
(a) < sin (cos θ ) (b) > sin (cos θ ) 2 c
α+β
(c) = sin (cos θ ) (d) None of these 2 tan
2ac
⇒ sin (α + β ) = 2 = 2
Sol. We have, α + β a + c2
1 + tan 2
1 1 2
cosθ + sin θ = 2 cos θ + sin θ
2 2 Hence, (d) is the correct answer.
π π π 2π ( n − 1) π
= 2 cos cos θ + sin sin θ Ex 10. If S = cos 2 + cos 2 +… + cos 2
4 4
n n n
,
π
= 2 cos θ − then S equals
4 n 1
π (a) ( n + 1) (b) ( n − 1)
⇒ cos θ + sin θ ≤ 2 < 2 2
2 1 n
π (c) ( n − 2) (d)
⇒ cos θ < − sin θ 2 2
2
π π 2π π
⇒ sin (cosθ ) < sin − sin θ Sol. S = cos2 + cos2 + … + cos2 (n − 1)
2 n n n
⇒ sin (cos θ ) < cos (sin θ ) 1 2π 4π 6π
= 1 + cos + 1 + cos + 1 + cos
⇒ cos (sin θ ) > sin (cos θ ) 2 n n n
Hence, (b) is the correct answer. π
+ … + 1 + cos 2(n − 1)
n
Ex 8. The smallest positive number p for which the
1 2k π
n− 1
equation cos ( p sin x ) = sin ( p cos x ) has a =n − 1 + ∑ cos
2 n
solution x ∈[0, 2π ], is equal to k=1
π π π 1 1
(a) π (b) (c) (d) = [ n − 1 − 1] = (n − 2)
2 2 2 2 2 2
Hence, (c) is the correct answer.
Sol. We have, cos ( p sin x ) = sin ( p cos x )
⇒
π
sin ± p sin x = sin ( p cos x )
Ex 11. If sin θ = 3 sin (θ + 2α ), then the value of
2 tan (θ + α ) + 2 tan α is
π (a) 3 (b) 2 (c) 1 (d) 0
⇒ ± p sin x = p cos x
2
Sol. Given, sin θ = 3 sin (θ + 2α )
π
⇒ = p cos x ± p sin x ⇒ sin (θ + α − α ) = 3 sin (θ + α + α )
2 ⇒ sin (θ + α )cosα − cos (θ + α )sin α
π 1 1
⇒ = cos x ± sin x = 2 cos x ± sin x = 3 sin (θ + α )cosα + 3 cos (θ + α )sin α
444 2p 2 2 ⇒ −2 sin (θ + α )cosα = 4 cos (θ + α )sin α
sin (θ + α ) 2 sin α 2 cos y − 1
⇒ − =
cos (θ + α ) cosα
⇒ tan (θ + α ) + 2 tan α = 0
Ex 15. If cos x =
x y
2 − cos y
, where x, y ∈ (0, π ), then
9
Trigonometric Functions and Equations
Hence, (d) is the correct answer. tan cot is equal to
2 2
Ex 12. The numerical value of tan 20° tan 80° cot 50° is (a) 2 (b) 3
1 1 1
(a) 3 (b) (c) (d)
3 2 3
1 2 cos y − 1
(c) 2 3 (d) Sol. Given, cos x =
2 3 2 − cos y
Sol. tan 20° tan 80° cot 50° x 2(1 − tan 2 y/ 2)
1 − tan 2 −1
1
2 = 1 + tan y/ 2
2
= tan (50° − 30° )tan (50° + 30° ) ⇒
tan 50° x 1 − tan 2 y/ 2
1 + tan 2 2−
tan 50° − 1/ 3 tan 50° + 1/ 3 1 2 1 + tan 2 y/ 2
= ⋅
1 + tan 50°/ 3 1 − tan 50°/ 3 tan 50° y x
⇒ 6 tan 2 = 2 tan 2
3 tan 2 50° − 1 2 2
= = − cot (150° ) = cot 30° = 3
3 tan 50° − tan 3 50° x y
⇒ tan ⋅ cot = 3
Hence, (a) is the correct answer. 2 2
Hence, (b) is the correct answer.
π
Ex 13. If A + B = , where A, B ∈ R + , then the Ex 16. If sin x + sin 2 x = 1, then the value of
3
minimum value of sec A + sec B is cos 12 x + 3 cos 10 x + 3 cos 8 x + cos 6 x is
2 4 (a) 1 (b) −1
(a) (b)
3 3 (c) 2 (d) −2
(c) 2 3 (d) None of these Sol. Let I = cos12 x + 3 cos10 x + 3 cos8 x + cos6 x
Sol. For y = sec x , x ∈ [ 0, π / 2), tangent drawn to it any = cos12 x + 3 ⋅ cos8 x (cos2 x + 1) + cos6 x
point lies completely below the graph of y = sec x, thus = (cos4 x )3 + 3 cos6 x (cos4 x + cos2 x ) + (cos2 x )3
sec A + sec B A + B = (cos4 x + cos2 x )3 = [cos2 x (cos2 x + 1)]3
≥ sec
2 2 We have, sin x = 1 − sin 2 x = cos2 x
π 4 −1 + 5
⇒ sec A + sec B ≥ 2 ⋅ sec = ⇒ sin x = cos2 x =
6 3 2
Hence, (b) is the correct answer. 3
5 − 1 5 + 1
π ⇒ I = =1
Ex 14. If A + B = , where A, B ∈ R + , then the 2 2
2 Hence, (a) is the correct answer.
maximum value of sin A + sin B is
1 + sin 2x
(a) 1 Ex 17. If = cot 2 ( a + x ),
1 − sin 2x
(b) 3
π
(c) 2 ∀ x ∈ R ~ nπ + , n ∈ N , then a is equal to
4
(d) None of the above
π
Sol. For y = sin x; x ∈ [ 0, π / 2 ] (a)
4
R
Q π
P
(b)
S 2
3π
(c)
4
(d) None of the above
O 2
A A + B B π/2 1 + sin 2x (sin x + cos x )2 1 + cot x
2 Sol. = =
1 − sin 2x (sin x − cos x )2 1 − cot x
2 2
From the above curve, 1 + tan x π 2 π
sin A + sin B A + B π ⇒ = tan + x = tan + x
≤ sin = sin 1 − tan x 4 4
2 2 4
π π 3π
⇒ sin A + sin B ≤ 2 = cot 2 + + x = cot 2 + x
2 4 4
Hence, (c) is the correct answer. Hence, (c) is the correct answer. 445
9 Ex 18. If sin α sin β − cos α cos β + 1 = 0, then the value
of cot α tan β is
(a) −1
Ex 21. If a cos 2θ + b sin 2θ = c has α and β as its
solutions, then tan α + tan β and tan α tan β
Objective Mathematics Vol. 1
(b) 0 are, respectively
(c) 1 (d) None of these 2b c − a c+a c−a
(a) , (b) ,
Sol. Given, sin α sin β − cosα cosβ + 1 = 0 c+a c+a 2b c + a
⇒ cos (α + β ) = 1 ⇒ α + β = 2nπ 2b c − a 2b c + a
cosα sin β (c) , (d) ,
1 + cot α tan β = 1 + ⋅ c−a c+a c+a c−a
sin α cosβ
sin α cosβ + cosα sin β sin (α + β ) Sol. We have, a cos 2θ + b sin 2θ = c
= = ⇒ a (cos2 θ − sin 2 θ ) + 2b sin θ cos θ = c
sin α cosβ sin α cosβ
sin (2nπ ) ⇒ a (1 − tan 2 θ ) + 2b tan θ = c sec2 θ = c (1 + tan 2 θ )
= =0
sin α cosβ ⇒ tan 2 θ (c + a) − 2b tan θ + c − a = 0
⇒ cot α tan β = − 1 This equation has tanα and tanβ as its roots.
Hence, (a) is the correct answer. 2b c − a
⇒ tan α + tan β = and tan α tan β =
Ex 19. If tan x = n tan y, n ∈ R + , then maximum value (c + a) c + a
Hence, (a) is the correct answer.
of sec 2 ( x − y) is
( n + 1) 2 ( n + 1) 2 ( n + 1) 2 ( n + 1) 2 tan 3 A sin 3A
(a) (b) (c) (d) Ex 22. If = k , then is equal to
2n n 2 4n tan A sin A
2k 2k
Sol. Given, tan x = n tan y (a) (b)
We know that, cos (x − y) = cos x cos y + sin x sin y k −1 k +1
⇒ cos (x − y) = cos x cos y (1 + tan x tan y) 3k
(c) (d) None of these
= cos x cos y (1 + n tan 2 y) k +1
sec2x sec2 y
⇒ sec2 (x − y) = tan 3 A 3 tan A − tan 3 A 1
(1 + n tan 2 y)2 Sol. = ⋅ =k
tan A 1 − 3 tan 2 A tan A
(1 + tan 2 x )(1 + tan 2 y)
= 3 − tan 2 A
(1 + n tan 2 y)2 ⇒ =k
1 − 3 tan 2 A
(1 + n2 tan 2 y)(1 + tan 2 y)
= ⇒ 3 − tan 2 A = k (1 − 3 tan 2 A )
(1 + n tan 2 y)2
⇒ (3k − 1)tan 2 A = k − 3
(n − 1)2 tan 2 y
=1+ k −3
(1 + n tan 2 y)2 ∴ tan 2 A =
2
3k − 1
1 + n tan y 2
sin 3 A 3 sin A − 4 sin 3 A
Now, ≥ n tan 2 y Now, =
2 sin A sin A
tan 2 y 4
⇒ ≤
1 = 3 − 4 sin A = 3 −
2
1
(1 + n tan y)
2 2
4n 1+
tan 2 A
(n − 1)2 (n + 1)2 4 (k − 3)
⇒ sec2 (x − y) ≤ 1 + = =3−
4
=3−
4n 4n 3k − 1 4 (k − 1)
Hence, (d) is the correct answer. 1+
k −3
Ex 20. If f (x ) = cos x (sin x + sin 2 x + sin 2 θ ), where 3k − 3 − k + 3 2k
= =
θ is a given constant, then maximum value of k −1 k −1
Hence, (a) is the correct answer.
f ( x ) is
π
(a) 1 + cos 2 θ (b) 1 + sin 2 θ Ex 23. If 1 + cot θ ≤ , for 0 < θ < π, then θ is equal to
2
(c) |cos θ| (d) |sin θ| π π π π
(a) (b) (c) (d)
Sol. f (x ) = cos x (sin x + sin x + sin θ )
2 2
6 4 2 8
⇒ ( f (x ) ⋅ sec x − sin x )2 = sin 2 x + sin 2 θ θ
2 tan
⇒ f (x ) ⋅ sec x − 2 f (x ) ⋅ tan x = sin θ
2 2 2
Sol. Q tan θ = 2
θ
⇒ f (x ) ⋅ tan x − 2 f (x ) ⋅ tan x + f (x ) − sin θ = 0
2 2 2 2 1 − tan 2
2
⇒ 4 f 2 (x ) ≥ 4 f 2 (x )[ f 2 (x ) − sin 2 θ ] θ
−1
cot 2
⇒ cot θ = 2
⇒ f 2 (x ) ≤ 1 + sin 2 θ ⇒ | f (x ) | ≤ 1 + sin 2 θ θ
2 cot
446 Hence, (b) is the correct answer. 2
θ
θ
Now, 1 + cot θ − cot = 1 +
2
cot 2
2
2 cot
−1
θ
− cot
θ
2
π
8
3π
Ex 25. The value of 1 + cos 1 + cos
8 9
Trigonometric Functions and Equations
2 5π 7π
θ θ θ 1 + cos 1 + cos is
2 cot + cot 2 − 1 − 2 cot 2 8 8
= 2 2 2
θ 1 1 1 1
2 cot (a) (b) (c) (d)
2 4 6 8 2
θ
2
5π π π π
− cot − 1 Sol. Using cos = cos + = − sin
2 8 2 8 8
= ≤ 0, for 0 < θ < π
θ 3π π π π
2 cot ⇒ cos = cos − = sin
2 8 2 8 8
θ 7π π π
⇒ 1 + cot θ − cot ≤ 0 ⇒ cos = cos π − = − cos
2 8 8 8
θ π π π π
Equality holds when cot − 1 = 0
2 ∴ 1 + cos 1 − cos 1 + sin 1 − sin
8 8 8 8
π
∴ θ= π π π π
2 = 1 − cos2 1 − sin 2 = sin 2 cos2
8 8 8 8
Hence, (c) is the correct answer. 2
1 π 1 1 1
2π 4π 8π 16π = sin = ⋅ =
Ex 24. The value of cos cos cos cos is 4 4 4 2 8
15 15 15 15 Hence, (c) is the correct answer.
1 1
(a) (b) 3
16 32 Ex 26. Let cos A + cos B + cos C = in a ∆ABC. Then,
1 1 2
(c) (d) the triangle is
64 8
(a) isosceles (b) right angled
2π 4π 8π 16π
Sol. cos cos cos cos (c) equilateral (d) None of these
15 15 15 15
1 2π 2π 4π 8π 16π Sol. In a ∆ABC , A + B + C = π
= ⋅ 2 sin cos cos cos cos A + B
2π 15 15 15 15 15 ⇒ cos A + cos B + cos C = 2 cos
2 sin 2
15
1 4π 4π 8π 16π A − B 3
= ⋅ sin cos cos cos cos + cosC =
2π 15 15 15 15 2 2
2 sin
15 π C A − B 2C 3
⇒ 2 cos − cos + 1 − 2 sin =
1 8π 8π 16π 2 2 2 2 2
= ⋅ sin cos cos
2π 15 15 15
4 sin C C A − B
15 ⇒ 4 sin 2 − 4 sin cos + 1= 0 …(i)
2 2 2
1 16π 16π
= ⋅ sin cos C
2π 15 15 Now, sin is real.
8 sin 2
15
A − B
1 32π ⇒ 16 cos2 − 16 > 0
= ⋅ sin 2
2π 15
16 sin A − B
15 ⇒ cos2 −1≥ 0
2π 2
1
= ⋅ sin 2π +
2π 15 A − B
16 sin ⇒ cos2 = 1 = cos 0
15 2
1 2π 1 ∴ A=B …(ii)
= ⋅ sin =
2π 15 16 Similarly, it can be shown that B = C , C = A. So, the
16 sin
15 triangle is equilateral.
Aliter Hence, (c) is the correct answer.
Use cos A cos 2 A cos 22 A… cos 2n − 1 A =
sin 2n A π 2π 3π
Ex 27. tan 2 + tan 2 + tan 2
2n sin A 7 7 7
π
sin 24 ⋅ 2π 2π 3π
2π 4π 8π 16π 1
∴cos cos cos cos = 15 = cot + cot 2 + cot 2 is equal to
15 15 15 15 4 π 16 7 7 7
2 sin
15 (a) −105 (b) 105
Hence, (a) is the correct answer. (c) ±105 (d) None of these 447
9 Sol. Letθ =
⇒
nπ
7
7θ = nπ
Now consider,
2 sin
2π
7
sin
4π
7
+ 2 sin
4π
7
sin
8π
7
+ 2 sin
8π
7
sin
2π
7
Objective Mathematics Vol. 1
⇒ 4θ + 3θ = nπ 2π 6π 4π 12π
= cos − cos + cos − cos
⇒ tan 4θ = tan (nπ − 3θ ) 7 7 7 7
⇒ tan 4θ = − tan 3θ 6π 10π
+ cos − cos
4 tan θ − 4 tan 3 θ 3 tan θ − tan 3 θ 7 7
⇒ =−
1 − 6 tan θ + tan θ
2 4
1 − 3 tan 2 θ 2π 4π 2π
= cos + cos − cos 2π −
4 z − 4 z3 3z − z3 7 7 7
⇒ = −
1 − 6z2 + z4 1 − 3z2 4π
− cos 2π −
[where, tanθ = z (say)]
7
⇒ (4 − 4 z2 )(1 − 3z2 ) = − (3 − z2 )(1 − 6z2 + z4 ) 2π 4π 2π 4π
= cos + cos − cos − cos =0
7 7 7 7
⇒ z6 − 21z4 + 35z2 − 7 = 0 …(i)
2π 4π 8π
This is a cubic equation in z i.e. in tan θ. Therefore,
2 2 ⇒ x 2 = sin 2 + sin 2 + sin 2
π 2π 7 7 7
the roots of this equation are tan 2 , tan 2 and 4π 8π 16π
7 7 1 − cos 1 − cos 1 − cos
3π = 7 + 7 + 7
tan 2 . 2 2 2
7 1 4π 8π 16π
− (−21) = 3 − cos + cos + cos
From Eq. (i), sum of the roots = = 21 2 7 7 7
1
π 2π 3π
⇒ tan 2 + tan 2 + tan 2 = 21 …(ii) 1 1 π 2π π 4π
7 7 7 = 3 − 2 sin cos + 2 sin cos
π
1 2
2 sin 7 7 7 7
On putting in place of z in Eq. (i), we get
y 7
π 6π
−7 y6 + 35 y4 − 21 y2 + 1 = 0 + 2 sin cos
7 7
⇒ 7 y6 − 35 y4 − 21 y2 + 1 = 0 …(iii)
This is a cubic equation in y2 i.e. in cot 2 θ. Therefore, 1 1 3π π
= 3 − sin − sin
π 2π 2 π 7 7
the roots of this equation are, cot 2 , cot 2 and 2 sin
7 7 7
3π 5π 3π 7π 5π
cot 2 . + sin − sin + sin − sin
7 7 7 7 7
35
From Eq. (iii), sum of the roots = 1 1 7 7
7 = 3 + = ⇒ x=
2 2 4
2π 2 2π 2 3π
2
⇒ cot + cot + cot =5 …(iv)
7 7 7 Hence, (a) is the correct answer.
By multiplying Eqs. (ii) and (iv), we get
Ex 29. If 3 sin θ + 5 cos θ = 5, then the value of
2π 2π 3π
tan + tan 2 + tan 2 5 sin θ − 3 cos θ is
7 7 7
(a) 5 (b) 3
2π 2π 3π
cot + cot 2 + cot 2 = 21 × 5 = 105 (c) 4 (d) None of these
7 7 7
Hence, (b) is the correct answer. θ θ 3
Sol. 3 sin θ = 5(1 − cosθ ) = 5 × 2 sin 2 ⇒ tan =
2π 4π 8π 2 2 5
Ex 28. sin + sin + sin is equal to θ 2 θ
2 tan 1 − tan
7 7 7 2 2
∴ 5 sin θ − 3 cosθ = 5 × −3
7 2θ 2θ
(a) 1 + tan 1 + tan
2 2 2
3 9
(b) 7 2× 3 × 1 −
5 − 25
(c) 2 =5× =3
9 9
7 1+ 1+
(d) 25 25
4 Hence, (b) is the correct answer.
2π 4π 8π
Sol. Let x = sin + sin + sin Ex 30. In a ∆ABC, the maximum value of
7 7 7
A B C
On squaring both sides, we get a cos 2 + b cos 2 + c cos 2
2π 4π 8π 2π 4π 2 2 2 is
x 2 = sin 2 + sin 2 + sin 2 + 2 sin sin a +b+c
7 7 7 7 7
4π 8π 8π 2π 1 3
448 + 2 sin sin + 2 sin sin (a) (b) 2 (c) (d) 1
7 7 7 7 4 4
cos ( B + C ) 1 − p
Sol.
A
a cos2
2
a+ b+ c
B
+ b cos2 + c cos2
2
C
2
⇒
⇒
1+ p
=
cos ( B − C ) 1 + p
= cos ( B − C ) …(i)
9
Trigonometric Functions and Equations
a (1 + cos A ) + b(1 + cos B ) + c(1 + cosC ) 2 ( p − 1)
=
2(a + b + c) 3π
Since, B or C can vary from 0 to
a cos A + b cos B + c cosC 1 4
= + 3π
2(a + b + c) 2 0≤ B −C <
R [sin 2 A + sin 2 B + sin 2C ] 1 4
= + 1
2(a + b + c) 2 ⇒ − < cos ( B − C ) ≤ 1
2
4 R sin A sin B sin C 1
= + Eq. (i) will now lead to
2 ⋅ 2s 2 1 p+1
abc 1 r 1 1 1 3 − < ≤1
= + = + ≤ + = as R ≥ 2r 2 2 ( p − 1)
8sR 2 2 2R 2 4 2 4
2p
A B
a cos2 + b cos2 + c cos2
C ⇒ >0
2 2 2 ≤ 3 p−1
⇒ ⇒ p < 0 or p > 1
a+ b+ c 4 …(ii)
p + 1 − 2 ( p − 1)
Thus, the maximum value is .
3 Also, ≤0
4 2 ( p − 1)
Hence, (c) is the correct answer. {p − ( 2 + 1)2}
⇒ ≥0
( p − 1)
Ex 31. If A + B + C = π, then the value of
⇒ p < 1 or p ≥ ( 2 + 1)2 …(iii)
cos 2 A + cos 2 B + cos 2 C is
Combining Eqs. (ii) and (iii), we get p ≥ ( 2 + 1)2
(a) 1 − cos A cos B cos C Hence, (b) is the correct answer.
(b) 1 − 2sin A sin B sin C
(c) 1 − sin A sin B sin C Ex 33. If A, B and C are angles of anyone of the
(d) 1 − 2cos A cos B cos C triangle such that tan A + tan B + tan C =100.
Then,
Sol. We have, cos2 A + cos2 B + cos2 C (a) there exist exactly three non-similar
= cos2 A + (1 − sin 2 B ) + cos2 C isosceles triangles
= (cos2 A − sin 2 B ) + cos2 C + 1 (b) there exist exactly two non-similar isosceles
triangles
= cos ( A + B ) cos ( A − B ) + cos2 C + 1
(c) there exist exactly three similar isosceles
= cos (π − C ) cos ( A − B ) + cos2 C + 1 triangles
= − cosC cos ( A − B ) + cos2 C + 1 (d) None of the above
= − cosC [cos ( A − B ) − cosC ] + 1
Sol. Let A = B, then 2 A + C = 180°
= − cosC [cos ( A − B ) − cos {π − ( A + B )}] + 1
and 2 tan A + tan C = 100
= − cosC [cos ( A − B ) + cos ( A + B )] + 1
Now, 2 A + C = 180°
= − 2 cosC cos A cos B + 1
⇒ tan 2A = − tan C …(i)
Hence, (d) is the correct answer.
Also, 2 tan A + tan C = 100
π ⇒ 2 tan A − 100 = − tan C …(ii)
Ex 32. Let A, B and C be three angles such that A = From Eqs. (i) and (ii), we have
4
2 tan A
and tan B tan C = p. Then, all possible values 2 tan A − 100 =
1 − tan 2 A
of p such that A, B and C are the angles of a
2x
triangle, is Let tan A = x, then = 2x − 100
1 − x2
(a) ≤ ( 2 + 1) 2 (b) ≥ ( 2 + 1) 2
⇒ x 3 − 50x 2 + 50 = 0
(c) > ( 2 + 1) 2
(d) < ( 2 + 1) 2 Let f (x ) = x 3 − 50x 2 + 50 = 0
Sol. Q A+ B+C =π Then, f ′ (x ) = 3x 2 − 100x
3π 100
⇒ B+C = So, f ′ (x ) = 0 has roots 0, .
4 3
3π 100
⇒ 0 < B, C < Also, f (0) ⋅ f <0
3
4
Thus, f (x ) = 0 has exactly three distinct real roots.
Also, tan B tan C = p
Therefore, tan A and hence A has three distinct
sin B sin C p
⇒ = values.
cos B cosC 1
So, there exist exactly three non-similar isosceles
cos B cosC − sin B sin C 1 − p
⇒ = triangles.
cos B cosC + sin B sin C 1 + p
Hence, (a) is the correct answer. 449
9
2 4 6
Ex 34. The sum of the series Ex 36. If exp[(sin x + sin x + sin x + …) ln 2] satisfies the
sin θ sec3θ + sin 3θ sec3 2 θ + sin 3 2 θ sec 3 3 θ equation y 2 − 9 y + 8 = 0, then the value of
+… n terms, is π
Objective Mathematics Vol. 1
cos x
1 , 0 < x < , is
(a) [tan 3n θ − tan 3n−1 θ] cos x + sin x 2
2
3−1
(b) [tan 3n θ − tan θ] (a) 3 + 1 (b)
1 2
(c) [tan 3n θ − tan θ] (c) 3 − 1 (d) None of these
2
(d) None of the above sin 2 x
Sol. sin 2 x + sin 4 x + sin6 x + …∞ = = tan 2 x
Sol. Here, sin θ sec 3θ + sin 3θ sec 32θ + sin 32θ sec 33θ 1 − sin 2 x
tan 2 x
+ … n terms
2
x + sin 4 x + …) ln 2] 2
⇒ exp[(sin = etan x ln 2
= eln 2
n n
sin 3r − 1θ Given equation, y2 − 9 y + 8 = 0
= ∑ sin 3r − 1θ ⋅ sec 3r θ = ∑
r = 1 cos 3 θ
r
r=1 ⇒ ( y − 1)( y − 8) = 0
r−1 r−1 2
n
2 cos 3 θ ⋅ sin 3 θ Either y = 1, then 2tan = 1 = 20
x
= ∑ 2 cos 3r − 1θ ⋅ cos 3r θ
r=1 π
⇒ tan 2 x = 0, but x ∈ 0,
2
1 n
sin(2 ⋅ 3r − 1θ )
` =
2 ∑ cos 3r − 1θ ⋅ cos 3r θ ∴ Neglecting x = 0
r=1
or y = 23 ⇒ tan 2 x = 3
1 n
sin(3r θ − 3r − 1θ )
=
2 ∑ cos 3r − 1θ ⋅ cos 3r θ ⇒ tan x = ± 3
r=1 π π
⇒ x = , as 0 < x <
1 n
sin 3r θ ⋅ cos 3r − 1θ − cos 3r θ ⋅ sin 3r − 1θ
=
2 ∑ cos 3r − 1θ ⋅ cos 3r θ
3
cos x
2
1/ 2 1 3 −1
r=1
⇒ = = =
1 n cos x + sin x 1/ 2 + 3 / 2 3+1 2
=
2 ∑ (tan 3r θ − tan 3r − 1θ ) Hence, (b) is the correct answer.
r=1
1
= [(tan 3θ − tan θ ) + (tan 32θ − tan 3θ ) Ex 37. If tan (π cos θ ) = cot (π sin θ ), then the value(s)
2
π
+ … + (tan 3nθ − tan 3n − 1θ )] of cos θ − is/are
1
4
= [tan 3nθ − tan θ ] 1 1
2 (a) (b)
Hence, (c) is the correct answer. 2 2
1
Ex 35. The set of values of x for which (c) ± (d) None of these
2 2
tan 3x − tan 2x
= 1, is Sol. We have, tan (π cos θ ) = cot (π sin θ )
1 + tan 3x tan 2x
π
(a) φ ⇒ tan (π cos θ ) = tan − π sin θ
2
π
(b) π
4 ⇒ π cosθ = − π sin θ + nπ , n ∈ Z
2
π 2n + 1
(c) nπ + , n = 1, 2, 3, … ⇒
1
cosθ +
1
sin θ = , n ∈Z
4 2 2 2 2
π π 2n + 1
(d) 2nπ + , n = 1, 2, 3, … ⇒ cos θ − = , n ∈Z
4 4 2 2
π 1
tan 3x − tan 2x ⇒ cos θ − = ±
Sol. We have, =1 4 2 2
1 + tan 3x tan 2x
[for n = 0 and n = − 1]
⇒ tan(3x − 2x ) = 1 ⇒ tan x = 1
Hence, (c) is the correct answer.
π
⇒ tan x = tan
4 Ex 38. The set of all x in (−π, π ) satisfying
π
⇒ x = nπ + | 4 sin x − 1| < 5 is given by
4
π π 3π π 3π
But for this value of x, tan 2x = tan 2nπ + = ∞ (a) − , (b) ,
2 10 10 10 10
which does not satisfy the given equation as it π 3π
(c) , − (d) None of these
450 reduces to indeterminate form. 10 10
Hence, (a) is the correct answer.
Sol. We have, | 4 sin x − 1 | < 5
⇒ − 5 < 4 sin x − 1 < 5
5 − 1 5 + 1
Ex 41. The equation p cos x − q sin x = r admits of a
solution for x only, if
(a) r < max{ p, q}
9
Trigonometric Functions and Equations
⇒ − < sin x <
4 4
(b) − p 2 + q 2 < r < p2 + q2
π π
⇒ − sin < sin x < cos (c) r 2 = p 2 + q 2
10 5
π π π (d) None of the above
⇒ sin − < sin x < sin −
10 2 5
Sol. Let p = a cos θ and −q = a sin θ
π 3π ∴ a(cos θ cos x + sin θ sin x ) = r
⇒ sin − < sin x < sin
10 10 ⇒ a cos (x − θ ) = r
π 3π Thus, the condition for solution is
⇒ x ∈− ,
10 10 −a ≤ r ≤ a
Hence, (a) is the correct answer. where, a = p 2 + q2
Ex 39. Total number of solutions of Thus, equation p cos x − q sin x = r admits solution
for
cos x = 1 − sin 2x in [0, 2π ], is
− p 2 + q2 ≤ r ≤ p 2 + q2
(a) 2 (b) 3
Hence, (b) is the correct answer.
(c) 5 (d) None of these
Sol. cos x = 1 − sin 2x = | sin x − cos x | Ex 42. If 4 cos 2 x sin x − 2 sin 2 x = 3 sin x, then x is
Case I sin x ≤ cos x equal to
⇒ cos x = cos x − sin x 3π 3π
(a) nπ + (b) nπ + ( −1) n + 1
⇒ sin x = 0 10 10
π 5π 3π
where, x ∈ 0, ∪ , 2π (c) nπ −
4 4 (d) None of these
10
∴ sin x = 0
⇒ x = 0, 2π Sol. We have, [ 4 cos2 x − 2 sin x − 3 ]sin x = 0
Case II sin x > cos x Either sin x = 0 ⇒ x = nπ
⇒ tan x = 2 or 4 (1 − sin 2 x ) − 2 sin x − 3 = 0
π 5π ⇒ 4 sin 2 x + 2 sin x − 1 = 0
where, x ∈ ,
4 4
−2 ± 4 + 16
∴ tan x = 2 ⇒ sin x =
8
⇒ x = tan −1 (2)
−1 ± 5 5 −1 1+ 5
Thus, the given equation has three solutions. = = ,−
4 4 4
Hence, (b) is the correct answer. π
⇒ sin x = sin
Ex 40. Total number of solutions of 10
π
1 ⇒ x = nπ + (−1)n
| cot x | = cot x + , x ∈[0, 3π ] , is 10
sin x π π
or sin x = − cos = sin −
(a) 1 (b) 2 5 10
(c) 3 (d) 0 3π
⇒ x = nπ + (−1)n + 1
1 10
Sol. | cot x | = cot x +
sin x Hence, (b) is the correct answer.
1
Let cot x > 0 ⇒ cot x = cot x + =0 3x x
sin x Ex 43. If cos 3x + cos 2x = sin + sin , 0 ≤ x ≤ 2π,
1 2 2
⇒ = 0, which is not possible. then the number of values of x is
sin x
1 (a) 6 (b) 7
Let cot x ≤ 0 ⇒ − cot x = cot x +
sin x (c) 4 (d) 5
1 5x x x
⇒ −2 cot x = Sol. We have, 2 cos cos = 2 sin x cos
sin x 2 2 2
1 x
⇒ cos x = − Either cos =0
2 2
2π 8π x π
⇒ x= , ⇒ = (2n + 1) ⇒ x = (2n + 1)π
3 3 2 2
So, the number of solutions is 2. 5x π
or cos = sin x = cos − x 451
Hence, (b) is the correct answer. 2 2
9 ⇒
⇒
5x
7x
2
= 2nπ +
π
= 2nπ ± − x
2
π
⇒ x=
4 nπ π
+
Ex 46. The general solution of the equation
x
cos − 2 sin x sin x
Objective Mathematics Vol. 1
2 2 7 7 4
3x π 4 nπ π x
or = 2nπ − ⇒ x= − + 1 + sin − 2 cos x cos x = 0 is
2 2 3 3 4
π 5π 9π 13π
For 0 ≤ x ≤ 2π, x = , , , ,π (a) 2π + 8mπ (b) 2π + mπ
7 7 7 7
Hence, (d) is the correct answer. (c) 2π + 4mπ (d) None of these
Sol. The given equation can be simplified to
Ex 44. The number of points of intersection of the 5x
π π sin + cos x = 2
curves y = cos x, y = sin 3x, if − ≤ x ≤ , is 4
2 2 5x
(a) 3 (b) 4 Since, the greatest value of sin and cos x is 1,
4
(c) 5 (d) 6 5x
their sum is equal to 2 only, if sin = 1 and
Sol. The point of intersection is given by 4
π 5x π
sin 3x = cos x = sin − x cos x = 1 simultaneously i.e. = 2nπ + and
2 4 2
π x = 2kπ ; n, k ∈ I .
⇒ 3x = nπ + (−1)n − x
2 Since, we have to choose those values of x which
satisfy both of these equations, we have
π 8nπ 2π 4n + 1
Case I Let n be even i.e. n = 2 m ⇒ 3x = 2mπ + − x
2 2kπ = + ⇒ k=
5 5 5
mπ π
⇒ x= + where, k and n are integers.
2 8 n−1
We write k =n−
Case II Let n be odd i.e. n = 2m + 1 5
π π n−1
⇒ 3x = (2m + 1)π − − x = 2mπ + + x For = m, we have k = 1 + 4 m (m ∈ I )
2 2 5
π Therefore, x = 2π + 8mπ (m ∈ I )
⇒ x = mπ + Hence, (a) is the correct answer.
4
π π
Now, − ≤x≤ Ex 47. The values of x between 0 and 2π which
2 2
π π −3π satisfy the equation sin x 8 cos 2 x = 1 are in
⇒ x= , ,
8 4 8 AP with common difference
Points of intersection are π π 3π 5π
(a) (b) (c) (d)
π π π π 3π 3π 4 8 8 8
, cos , cos , − , cos .
8 8 4 4 8 8
Hence, (a) is the correct answer.
Sol. We have, sin x 8 cos2 x = 1
1
⇒ sin x | cos x | =
Ex 45. The number of values of θ in the interval 2 2
π π Case I When cos x > 0
− , , satisfying the equation
2 2 In this case, sin x cos x =
1
2 2 2
(1 − tan θ )(1 + tan θ ) sec 2 θ + 2 tan θ = 0 is 1
⇒ sin 2x =
(a) 2 (b) 3 2
(c) 4 (d) None of these π 3π 9π 11π
⇒ 2x = , , ,
2 4 4 4 4
θ
Sol. (1 − tan θ )(1 + tan θ ) sec2 θ + 2tan =0 π 3π 9π 11π
⇒ x= , , ,
tan 2 θ 8 8 8 8
⇒ (1 − tan θ )(1 + tan θ ) + 2
2 2
=0
π 3π
tan 2 θ As x lies between 0 and 2π and cos x > 0, x = , .
⇒ 1+ 2 = tan 4 θ 8 8
By observation, we have tan 2 θ = 3 Case II When cos x < 0
1
π In this case, sin x | cos x | =
⇒ θ = nπ ± 2 2
3 1 1
Moreover there will be values of θ, satisfying, ⇒ sin x cos x = − or sin 2x = −
2 2 2
3 < tan 2 θ < 4 and satisfying the given equation as if 5π 7π 13π 15π
f (x ) = x 2 − 2x − 1, then f (3+ ) f (4 − ) < 0. ⇒ x= , , ,
8 8 8 8
So, the number of values of θ is 4. 5π 7π
⇒ x= ,
Hence, (c) is the correct answer. 8 8
452
Thus, the values of x satisfying the given equation
which lie between 0 and 2π are ,
π 3π 5π 7π
,
8 8 8 8
, .
Ex 50. The equation
2 2
3 sin 2x + 2 cos x + 31 − sin 2x + 2 sin x = 28
9
Trigonometric Functions and Equations
π
These are in AP with common difference . is satisfied for the values of x given by
4
(a) cos x = 0, tan x = − 1 (b) tan x = 0
Hence, (a) is the correct answer.
(c) tan x = 1 (d) None of these
Ex 48. The values of θ, for which the system Sol. The given equation is
(sin 3θ ) x − y + z = 0, (cos 2θ ) x + 4 y + 3 z = 0 3sin 2x + 2 cos
2
x
+ 31 − sin 2x + 2 sin
2
x
= 28
and 2x + 7 y + 7 z = 0, has a non-trivial 2 2
sin 2x + 2 cos x 3 − (sin 2x + 2 cos x)
solution, are ⇒ 3 +3 = 28
π π 27
(a) nπ , nπ + ( −1) n
(b) nπ , nπ − (1)n
⇒ y+ = 28
6 6 y
n π π 2
(c) 2nπ , nπ + ( −1) (d) nπ , nπ + ( −1) n where, y = 3sin 2x + 2 cos x ⇒ y2 − 28 y + 27 = 0
6 3 ⇒ y = 27or 1
2
Sol. The system has non-trivial solutions, if If y = 27, then 3sin 2x + 2 cos x
= 33
sin 3θ −1 1 ⇒ sin 2x + 2 cos x = 3 2
cos 2θ 4 3 = 0 ⇒ sin 2x + cos 2x = 2
2 7 7 ⇒ sin 2x = cos 2x = 1
On expanding, we get which is not possible for any value of x and so y ≠ 27.
7 sin 3θ + 14 cos 2θ − 14 = 0 Also, we have y = 1
⇒ sin 3θ + 2 cos 2θ − 2 = 0
2
⇒ 3sin 2x + 2 cos x
= 1 = 30
⇒ sin θ (4 sin 2 θ + 4 sin θ − 3) = 0 ⇒ sin 2x + 2 cos2 x = 0
⇒ sinθ = 0 ⇒ θ = nπ ⇒ 2 cos x (sin x + cos x ) = 0
or 4 sin 2 θ + 4 sin θ − 3 = 0 Either cos x = 0 or tan x = − 1
⇒ (2 sin θ − 1)(2 sin θ + 3) = 0 Hence, (a) is the correct answer.
⇒ 2 sinθ − 1 = 0
⇒
1 π
sin θ = = sin
Ex 51. If cos 4 x + a cos 2 x + 1 = 0 has atleast one
2 6 solution, then
or 4 sin θ + 4 sin θ − 3 = 0 , which is not possible.
2
(a) a ∈ [ 2, ∞ ) (b) a ∈ [ −2, 2]
π (c) a ∈ ( −∞ , − 2] (d) a ∈ R ~ ( −2, 2)
⇒ θ = nπ + (−1)n
6
π Sol. cos4 x + a cos2 x + 1 = 0
⇒ θ = nπ or nπ + (−1)n
6 Let t = cos2 x, then t ∈[ 0, 1]
Hence, (a) is the correct answer. Thus, we have t 2 + at + 1 = 0
1 This quadratic must have atleast one root in [0, 1].
Ex 49. A ∆ABC is such that sin (2 A + B ) = . ⇒ a2 − 4 ≥ 0, 1 + a + 1 ≤ 0
2
If A, B and C are in AP, then the values of ⇒ a ∈ (−∞ , − 2 ] ∪ [ 2, ∞ ), a ≤ −2
A, B and C are ⇒ a ∈ (−∞ , − 2 ]
π π 5π π π π Hence, (c) is the correct answer.
(a) , , (b) , ,
4 3 12 2 3 6 Ex 52. Total number of solutions of
π π π 1
(c) , , (d) None of these cos x cos 2x cos 3x = in [0, π ] is
2 4 4 4
1 π (a) 4 ` (b) 6
Sol. sin(2 A + B ) = = sin
2 6 (c) 8 (d) None of these
π 1
⇒ 2 A + B = nπ + (−1)n …(i) Sol. Given, cos x cos 2x cos 3x =
6 4
Also, A + B + C = π and 2B = A + C ⇒ 4 cos 3x cos x cos 2x = 1
π
⇒ 3B = π ⇒ B = ⇒ 2(cos 4 x + cos 2x )cos 2x = 1
3 ⇒ 2(2 cos2 2x − 1 + cos 2x )cos 2x = 1
5π
From Eq. (i), for n = 1, 2 A + B = ⇒ 4 cos3 2x + 2 cos2 2x − 2 cos 2x − 1 = 0
6
π ⇒ 2 cos2 2x (2 cos 2x + 1) − (2 cos 2x + 1) = 0
⇒ A= ⇒ (2 cos 2x + 1)(2 cos2 2x − 1) = 0
4
5π ⇒ cos 4 x (2 cos 2x + 1) = 0
and C = π
12 ⇒ cos 4 x = 0 ⇒ 4 x = (2n1 + 1) 453
Hence, (a) is the correct answer. 2
π π 3π 5π 7π
9 ⇒
or
x = (2n1 + 1)
cos 2x = −
1
= ,
8 8 8 8 8
⇒ 2x = 2n2π ±
,
2π
, Sol. We have, x + 2 tan x =
Y
π
2
π x
⇒ tan x = −
4 2
Objective Mathematics Vol. 1
y = tanx
2 3
π π 2π π/4
⇒ x = n2π ± ⇒ x= ,
3 3 3
Thus, there are six solutions. X
Hence, (b) is the correct answer. 0 π 2π
Ex 53. Total number of solutions of
sin 4 x + cos 4 x = sin x cos x in [0, 2π ] is π– x
y=
(a) 2 (b) 4 (c) 6 (d) 8 4 2
π x
Now, the graph of the curve y = tan x and y = − ,
Sol. sin 4 x + cos4 x = sin x cos x 4 2
⇒ (sin 2 x + cos2 x )2 − 2 sin 2 x cos2 x = sin x cos x in the interval [ 0, 2π ] intersect at three points. The
abscissa of these three points are the roots of the
sin 2 2x sin 2x
⇒ 1− = ⇒ sin 2 2x + sin 2x − 2 = 0 equation.
2 2 Hence, (c) is the correct answer.
⇒ (sin 2x + 2)(sin 2x − 1) = 0 ⇒ sin 2x = 1
π π |x|
⇒ 2x = (4 n1 + 1) ⇒ x = (4 n1 + 1) Ex 57. Total number of solutions of sin x = is
2 4 10
π 5π (a) 4 (b) 6
⇒ x= ,
4 4 (c) 7 (d) None of these
Thus, there are two solutions. |x |
Hence, (a) is the correct answer. Sol. Graph of y = sin x and y = meet exactly six times.
10
Ex 54. Total number of integral values of n so that So, there are six solutions.
sin x (sin x + cos x ) = n has atleast one solution, Y
is y = |x| y = sin x
(a) 2 (b) 1 10
(c) 3 (d) 0 – 4π –2π 4π
X′
–π 0 π 2π 3π 5π X
Sol. sin x (sin x + cos x ) = n –3π
⇒ sin x + sin x cos x = n
2
Y′
1 − cos 2x sin 2x
⇒ + =n Hence, (b) is the correct answer.
2 2
⇒ sin 2x − cos 2x = 2n − 1 Ex 58. Total number of solutions of sin {x} = cos {x},
⇒ − 2 ≤ 2n − 1 ≤ 2
where { } denotes the fractional part, in [0, 2π ] is
1− 2 1+ 2 (a) 5 (b) 7
⇒ ≤n≤
2 2 (c) 8 (d) None of these
⇒ n = 0, 1
Hence, (a) is the correct answer. Sol. sin {x} = cos {x}
The graph of y = sin {x}and y = cos {x}meet exactly
Ex 55. The equation tan 4 x − 2 sec 2 x + a 2 = 0 will 7 times in [ 0, 2π ] .
have atleast one solution, if Y
(a) | a |≤ 4 (b) | a |≤ 2
(c) | a |≤ 3 (d) None of these
X
Sol. tan 4 x − 2 sec2x + a2 = 0 0 1 2 3 4 5 6 2π
⇒ tan 4 x − 2(1 + tan 2 x ) + a2 = 0 Hence, (b) is the correct answer.
⇒ tan x − 2 tan x + 1 = 3 − a
4 2 2
Ex 59. The number of all possible 5-tuples
⇒ (tan 2 x − 1)2 = 3 − a2 ⇒ 3 − a2 ≥ 0 ( a1 , a 2 , a 3 , a 4 , a 5 ) such that
⇒ a2 ≤ 3 ⇒ | a | ≤ 3 a1 + a 2 sin x + a 3 cos x + a 4 sin 2x + a 5 cos 2x = 0
Hence, (c) is the correct answer. holds for all x, is
(a) zero (b) 1
Ex 56. The number of roots of the equation (c) 2 (d) infinite
π
x + 2 tan x = in the interval [0, 2π ] is Sol. Since, the equation
2 a1 + a2 sin x + a3 cos x + a4 sin 2x + a5 cos 2x = 0 holds
(a) 1 (b) 2 for all values of x.
454 (c) 3 (d) infinite a1 + a3 + a5 = 0 [on putting x = 0]
π
⇒
a1 − a3 + a5 = 0
a3 = 0 and a1 + a5 = 0
π
Putting x = and
3π
, we get
[on putting x = π]
…(i) Ex 61. If 0 ≤ x ≤
is equal to
2
2 2
and 81sin x + 81cos x = 30, then x
9
Trigonometric Functions and Equations
2 2 π π π 2π
a1 + a2 − a5 = 0 and a1 − a2 + a5 = 0 (a) , (b) ,
⇒ a1 = 0 and a2 − a5 = 0 …(ii) 6 3 3 3
From Eqs. (i) and (ii), 5π π
(c) , (d) None of these
a1 = a2 = a3 = a5 = 0 6 6
The given equation reduces to a4 sin 2x = 0. This is 2
true for all values of x, therefore a4 = 0. Sol. Let 81sin x
= y
So, a1 = a2 = a3 = a4 = a5 = 0 Then, 81cos
2
x
= 811 − sin
2
x
= 81 y− 1
Thus, the number of 5-tuples is one.
Hence, (b) is the correct answer. So that, the given equation can be written as
y2 − 30 y + 81 = 0
Ex 60. The equation e sin x − e − sin x − 4 = 0 has ⇒ y=3 or y = 27
(a) no solution (b) two solutions ⇒ 81 sin 2 x
= 3 or 27
(c) three solutions (d) None of these 4 sin 2 x
⇒ 3 = 3 or 33 1
Sol. The given equation can be written as
1 3
e2 sin x
− 4 esin x − 1 = 0 ⇒ sin 2 x = or
4 4
4 ± 16 + 4
⇒ esin x = =2+ 5 1 3
2 ⇒ sin x = ± or ±
2 2
⇒ sin x = ln (2 + 5 )
1 3 1 3 π
[Q ln (2 − 5 ) is not defined as 2 − 5 is − ve] ⇒ sin x = , neglecting − , − as 0 ≤ x ≤
2 2 2 2 2
Now, 2 + 5 > e ⇒ ln (2 + 5 ) > 1 ⇒ sin x > 1
π π
which is not possible. ⇒ x= or
6 3
∴No real solution exists.
Hence, (a) is the correct answer.
Hence, (a) is the correct answer.
Type 2. More than One Correct Option
Ex 62. If x = (sec φ − tan φ ) and y = cosec φ + cot φ, 2b
Ex 63. If tan x = ( a ≠ c),
then a−c
y +1 y−1 y = a cos 2 x + 2b sin x cos x + c sin 2 x and
(a) x = (b) x =
y−1 y+1 z = a sin x − 2b sin x cos x + c cos x, then
2 2
1+ x
(c) y = (d) xy + x − y + 1 = 0 (a) y = z
1− x (b) y + z = a + c
1 − sin φ 1 + cos φ (c) y − z = a − c
Sol. We have, x = , y=
cos φ sin φ (d) y − z = ( a − c ) 2 + 4b 2
On multiplying, we get
(1 − sin φ )(1 + cos φ ) Sol. Adding the expression for y and z, we get
xy = y + z = a(cos2 x + sin 2 x ) + c(sin 2 x + cos2 x ) = a + c
cos φ sin φ
1 − sin φ + cos φ − sin φ cos φ + sin φ cos φ and subtracting them ,
⇒ xy + 1 =
cos φ sin φ y − z = a(cos2 x − sin 2 x ) + 4 b sin x cos x
1 − sin φ + cos φ − c(cos2 x − sin 2 x )
=
cos φ sin φ = a cos 2x + 2b sin 2x − c cos 2x
(1 − sin φ )sin φ − cos φ (1 + cos φ ) = (a − c) cos 2x + 2b sin 2x
and x − y = 1 − tan 2 x (a − c)2 − 4 b2
cos φ sin φ Now, cos 2x = =
sin φ − sin 2 φ − cos φ − cos2 φ 1 + tan 2 x (a − c)2 + 4 b2
= 2 tan x 4 b(a − c)
cos φ sin φ and sin 2x = =
sin φ − cos φ − 1 1 + tan x (a − c)2 + 4 b2
2
= = − (xy + 1)
cos φ sin φ (a − c)[(a − c)2 − 4 b2 ] + 8b2 (a − c)
⇒ y−z=
Thus, xy + x − y + 1 = 0 (a − c)2 + 4 b2
y−1
⇒ x= (a − c)[(a − c)2 + 4 b2 ]
y+1 = =a−c
1+ x (a − c)2 + 4 b2
and y= As a ≠ c, we get y ≠ z
1− x
Hence, (b), (c) and (d) are the correct answers. Hence, (b) and (c) are the correct answers. 455
9 Ex 64. If x = sin (α − β)sin (γ − δ ),
y = sin (β − γ )sin (α − δ ) and
z = sin ( γ − α )sin (β − δ ), then
⇒
⇒
−13 ≤
1
a
< 0 and
1
0<
0 < 5 cos x + 12 sin x ≤ 13
≤ 13
Objective Mathematics Vol. 1
a
(a) x + y + z = 0 (b) x + y − z = 0 1 1
∴ a≤− and a ≥
(c) y + z − x = 0 (d) x 3 + y 3 + z 3 = 3xyz 13 13
Hence, (a) and (b) are the correct answers.
Sol. 2x = cos (α − β − γ + δ ) − cos (α − β + γ − δ )
2 y = cos (β − γ − α + δ ) − cos (β − γ + α − δ ) Ex 67. Let [x ] be the greatest integer less than or
and similarly for 2z,
equal to x. The equation
On adding, we get 2x + 2 y + 2z = 0
sin x = [1 + sin x ] + [1 − cos x ] has
⇒ x + y+ z=0
π π
Hence, (a) and (d) are the correct answers. (a) no solution in − ,
2 2
15π 7π 3π
Ex 65. sin sin sin is equal to π
32 16 8 (b) no solution in , π
1 1 2
(a)
15π
(b)
π 3π
(c) no solution in π ,
8 2 cos 8sin 2
32 32
1 π 1 π (d) no solution for x ∈ R
(c) cosec (d) cosec
4 2 16 8 2 32 π 3π
Sol. At x = − , ; [1 + sin x ] = 0, [1 − cos x ] = 1
2 2
15π 7π 3π
Sol. sin sin sin ∴ sin x = 0 + 1 ⇒ −1 = 1 [absurd]
32 16 8
At x = 0, [1 + sin x ] = 1, [1 − cos x ] = 0
π 15π π 7π π 3π
= cos − cos − cos − ∴ sin x = 1 + 0 ⇒ 0 = 1 [absurd]
2 32 2 16 2 8
π
π π π At x = , [1 + sin x ] = 2, [1 − cos x ] = 1
= cos cos cos 2
32 16 8
π π π π ∴ sin x = 2 + 1 = 3 [absurd]
2 sin cos cos cos At x = π , [1 + sin x ] = 1, [1 − cos x ] = 2
32 32 16 8
= ∴ sin x = 1 + 2 = 3 [absurd]
π
2 sin π
32 In − , 0 , [1 + sin x ] = 0, [1 − cos x ] = 0
2
π π π π π
2 sin cos cos 2 sin cos ∴ sin x = 0 + 0 = 0 [absurd]
= 16 16 8 = 8 8
π π π
4 sin 8 sin In 0, , [1 + sin x ] = 1, [1 − cos x ] = 0
32 32 2
π ∴ sin x = 1 + 0 = 1 [absurd]
sin
= 4 = 1
=
1 π
π π 15π In , π , [1 + sin x ] = 1, [1 − cos x ] = 1
8 sin 8 2 sin 8 2 cos 2
32 32 32 ∴ sin x = 1 + 1 = 2 [absurd]
Hence, (a) and (d) are the correct answers. 3π
In π , , [1 + sin x ] = 0, [1 − cos x ] = 1
1 2
Ex 66. If a = , then for all real x
5 cos x + 12 sin x ∴ sin x = 0 + 1 = 1 [absurd]
1 Hence, (a), (b), (c) and (d) are the correct answers.
(a) the least positive value of a is
13 7π
1 Ex 68. If sin θ = a for exactly one value of θ ∈ 0, ,
(b) the greatest negative value of a is − 3
13 then the value of a is
1
(c) a ≤ 3
13 (a) (b) 1
1 1 2
(d) − ≤ a ≤ (c) 0 (d) −1
13 13
Sol. Clearly, −1 ≤ a ≤ 1
Sol. We know that, For any value of a other than 1, − 1 we know sinθ has
− 52 + 122 ≤ 5 cos x + 12 sin x ≤ 52 + 122 two values (in quadrants I, II or III, IV).
∴ −13 ≤ 5 cos x + 12 sin x < 0 Hence, (b) and (d) are the correct answers.
456
Type 3. Assertion and Reason
Directions (Ex. Nos. 69-77) In the following Statement II
9
Trigonometric Functions and Equations
examples, each example contains Statement I tan α tan β + tan β tan γ + tan γ tan α =1
(Assertion) and Statement II (Reason). Each example
π
has 4 choices (a), (b), (c) and (d) out of which only one is Sol. Statement II α + β = −γ
correct. The choices are 2
tan α − tan β 1
(a) Statement I is true, Statement II is true; Statement II is =
1 − tan α tan β tan γ
a correct explanation for Statement I
⇒ Σ tan α tan β = 1
(b) Statement I is true, Statement II is true; Statement II is ∴ Statement II is true.
not a correct explanation for Statement I a! b!
Statement I tan α tan β = , tan β tan γ =
(c) Statement I is true, Statement II is false 6 2
c!
(d) Statement I is false, Statement II is true and tan α tan γ =
3
Ex 69. Statement I The number of integral values of a! b! c!
+ + =1
λ, for which the equation 6 2 3
7 cos x + 5 sin x = 2λ + 1 has a solution, is 8. ⇒ a! = 1, b! = 1, c! = 1
Statement II a cos θ + b sin θ = c has atleast ⇒ tan α tan β , tan γ tan α and tan β tan γ are in AP.
∴Statement I is false.
one solution, if | c | > a 2 + b 2 Hence, (d) is the correct answer.
Sol. 7 cos x + 5 sin x = 2λ + 1 Ex 72. The angles of a triangle are given by the three
|2λ + 1 | ≤ 49 + 25 values of x obtained from the equation
⇒ |2λ + 1 | ≤ 74
tan 3 x − 3 3 tan 2 x + 3 3 tan x − 3 = 0.
− 74 ≤ 2λ + 1 ≤ 74
−9.6 ≤ 2λ ≤ 7.6, − 4.8 ≤ λ ≤ 3.8 Statement I The triangle so obtained is an
λ = − 4 , − 3, − 2, − 1, 0, 1, 2, 3 equilateral triangle.
Also, a cosθ + b sin θ = c Statement II If roots of the equation are tan A,
has no solution, if | c | > a2 + b2 tan B and tan C, then
Hence, (c) is the correct answer. tan A + tan B + tan C = 3 3
Ex 70. Statement I sin 2 > sin 3 Sol. tan A + tan B + tan C = 3 3 and tan A tan B tan C = 3
π ∴ tan A + tan B + tan C ≠ tan A tan B tan C
Statement II If x, y ∈ , π , x < y, then
2 So, triangle does not exists.
Hence, (b) is the correct answer.
sin x > sin y.
π
Sol. Y Ex 73. Statement I x = nπ + , n ∈ I satisfies the
3
trigonometric equation
tan x sin x + 3 = 3 sin x + tan x.
Statement II tan x sin x + 3 = 3 sin x + tan x
sin 2
sin 3 ⇒ (tan x − 3 )(sin x − 1) = 0
X ∴ General solution is given by
2 2 3
π π
x = nπ + , n ∈ I ∪ x = rπ + , r ∈ I .
Hence, (a) is the correct answer. 6 2
π Sol. tan x sin x + 3 = 3 sin x + tan x
Ex 71. Let α, β, γ > 0 and α + β + γ = …(i)
2 ⇒ tan x (sin x − 1) + 3 (1 − sin x ) = 0
Statement I If ⇒ (sin x − 1)(tan x − 3 ) = 0
tan α tan β − a !
tan β tan γ − b!
+ ∴ sin x = 1, tan x = 3, but sin x = 1is not possible as
π
6 2 tan x is not defined at x = (2n + 1) .
2
c! π
+ tan γ tan α − ≤ 0, where n! = 1 ⋅ 2… n, i.e. domain of Eq. (i) is R − (2n + 1) .
3 2
then tan α tan β, tan β tan γ , tan γ tan α are in π
∴ tan x = 3 = tan
AP. 3 457
π
9 ⇒ x = nπ +
3
, n ∈I
∴ Statement I is true and Statement II is false.
Also,
⇒
⇒
logsin x (sec x + 8) > 0
sec x + 8 < 1
sec x < −7 …(ii)
Objective Mathematics Vol. 1
Hence, (c) is the correct answer. π
Clearly, x = + nπ satisfy Eq. (ii).
kπ 4
Ex 74. Statement I x = , k ∈ I does not represent ∴ Statement I is true and Statement II is false.
13
the general solution of trigonometric equation Hence, (c) is the correct answer.
sin 13x − sin 13x cos 2x = 0. Ex 76. Statement I If f (x ) = cos x + 2 cos kx, k ∈θ
kπ k ≠ 0, then f ( x ) = 3 has infinite solutions.
Statement II Both x = rπ, r ∈ I and x = ,
13 Statement II The function is periodic.
k ∈ I satisfies the trigonometric equation
Sol. If f (x ) = cos x + 2 cos kx , k ∈ θ , k ≠ 0, then f (x ) = 3
sin 13x − sin 13x cos 2x = 0. has only one solution when x = 0. It has infinite
Sol. sin 13x (1 − cos 2x ) = 0 solutions, if k ∈I but it is given as k ∈θ.
⇒ sin13x = 0 or cos 2x = 1 ∴ Statement I is false.
∴ 13x = kπ or cos 2x = 1 Hence, (d) is the correct answer.
kπ Ex 77. Statement I The maximum value of
x = , k ∈I or 2x = 2nπ
13 sin 2x + sin ax cannot be 2 (where, a is a
⇒ x = nπ, n ∈I
So, Statement I is false and Statement II is true. positive rational number).
Hence, (d) is the correct answer. 2
Statement II is irrational.
a
Ex 75. Statement I Common value(s) of x satisfying
the equations Sol. f (x ) = sin 2x + sin ax
log sin x (sec x + 8) > 0 and 2π
The period of sin 2x =
log sin x cos x + log cos x sin x = 2 in (0, 4π ) does 2
not exist. 2π
and period of sin ax =
Statement II On solving above trigonometric a
equations we have to take intersection of f (x ) has period as
trigonometric chains given by sec x >1 and
2π 2π
π LCM of , does not exist.
x = nπ + , n ∈ I . 2 a
4
As 2 is irrational and a is rational.
Sol. logsin xcos x + logcos xsin x = 2 only when
∴ Maximum value of f (x ) cannot be 2 as both cannot
sin x = cos x
π be 1.
⇒ x = nπ + …(i)
4 Hence, (a) is the correct answer.
Type 4. Linked Comprehension Based Questions
Passage I (Ex. Nos. 78-80) If θ is an angle, one Ex 79. The wheel of a train is 1 m in diameter and it
measured in radian and θ ∈[0, 2 π ], then r θ is length of makes 5 revolutions per second. Then, the
arc AB, of circle of radius r, subtending angle θ at the speed of the train is (approximately)
1 (a) 57 km/h (b) 66 km/h
centre O, of the circle. Area of sector OAB is r 2θ.
2 (c) 68 km/h (d) 42.6 km/h
1
Ex 78. The angle between minute hand and hour hand Sol. Distance covered in 1s = 5 2π ⋅ = 5π m
2
of a clock at ‘half past 4’ equals 5π
(a) 42° Distance covered in 1h = × 60 × 60 = 56.52 km
1000
(b) 43° So, speed of the train is 57km /h.
(c) 44° Hence, (a) is the correct answer.
(d) None of the above Ex 80. Two lines drawn through a point on the
Sol. Angle subtended by two consecutive marks at centre circumference of a circle divided the circle
= 30° into three regions of equal area. Then, the
angle θ between the lines is given by
So, at ‘half past 4’, the angle is 45°.
(a) 3θ + 3sin θ = π (b) 6θ + 3sin θ = π
458 Hence, (d) is the correct answer. (c) 2θ + sin θ = π (d) θ + sin θ = π / 2
π 2π 3π 11π
Sol. Area of region ABC =
1
Area of OAB = r2 2θ = r2θ
πr 2
3
Ex 83. cos
11
cos cos … cos
1
11 11
1
11
1
is equal to
1
9
Trigonometric Functions and Equations
2 (a) − (b) (c) (d) −
1
32 512 1024 2048
Area of ∆OAC = r2 sinθ = Area of ∆OBC π 2π 3π 11π
2 Sol. cos cos cos … cos
1 2 1 2 πr 2 11 11 11 11
⇒ r sin θ + r sin θ + r2θ = 2
2 2 3 π 2π 3π 4π 5π
= cos cos cos cos cos
C 11 11 11 11 11
2
θ π 2π 4π 8π 5π
= cos cos cos cos cos
11 11 11 11 11
2 2
16π 5π 5π
sin 5π 2 sin cos
O = 11 ⋅ cos = 11 11 = 1
16 sin π 11 π 1024
32 sin
11 11
A B
Hence, (c) is the correct answer.
⇒ 3 sinθ + 3θ = π Passage III (Ex. Nos. 84-88) y = ax 2 + bx + c = 0 is a
Hence, (a) is the correct answer. quadratic equation which has real roots if and only if
b 2 − 4ac ≥ 0. If ( x , y ) = 0 is a second degree equation,
Passage II (Ex. Nos. 81-83) Given, then using above fact we can get the range of x and y
sin 2n + 1θ by treating it as quadratic equation in y or x . Similarly,
cos 2m θ cos 2m + 1θ …cos 2n θ = , where
2n − m + 1 sin 2m θ ax 2 + bx + c ≤ 0, ∀ x ∈ R, if a > 0 and b 2 − 4ac ≤ 0.
2m θ ≠ kπ, n, m, k ∈ l .
Ex 84. If 0 < α, β < 2π , then the number of ordered
9π 11π 13π
Ex 81. sin sin sin is equal to pairs (α, β) satisfying
14 14 14 sin 2 (α + β) − 2 sin α sin (α + β)
1 1
(a) (b) − + sin 2 α + cos 2 β = 0, is
64 64
1 1 (a) 2 (b) 0
(c) (d) − (c) 4 (d) None of these
8 8
9π 11π 13π 5π 3π π Sol. On solving it, we get
Sol. sin sin sin = sin sin sin
14 14 14 14 14 14 sin (α + β ) = sin α ± − cos2 β
π 2π 3π π 3π
= cos cos cos ⇒ cosβ = 0 ⇒ β = ,
7 7 7 2 2
π 2π 4π π π 5π
= − cos cos cos If β = , then tan α = 1, α ∈ ,
7 7 7 2 4 4
8π 3π 3π 7π
sin If β = , then tan α = − 1, α ∈ ,
=− 7 =1 2 4 4
π 8 Hence, (c) is the correct answer.
8 sin
7
Hence, (c) is the correct answer. Ex 85. If A + B + C = π , then the maximum value of
π π π π cos A + cos B + k cos C (where, k >1/ 2) is
Ex 82. cos 2 3 cos 2 4 cos 2 5 … cos 210 is 1 k 2k 2 + 1
10 10 10 10 (a) + (b)
equal to k 2 3
1 1 k2 + 2 1
(a) (b) (c) (d) +k
128 256 3 2k
Sol. Let cos A + cos B + k cos C = y
1 π 5 − 1 3π
(c) sin (d) sin C A − B C
512 10 512 10 ⇒ 2k sin 2 − 2 cos sin + y − k = 0
2 2 2
π π π π
Sol. cos 23 cos 24 ⋅ cos 25 … cos 210 As D≥0
10 10 10 10 1 A − B 1
π ⇒ k ( y − k ) ≤ cos2 ≤
sin 211 2 2 2
10 1
= =
3 π
1
256 sin 2 256 ⇒ y≤ +k
10 2k
Hence, (b) is the correct answer. Hence, (d) is the correct answer. 459
9 Ex 86. A circle with radius | a | and centre on Y-axis
slides along it and a variable line through
( a, 0) cuts the circle at points P and Q. The
Passage IV (Ex. Nos. 89-93) In the figure, it is given
that ∠C = 90° , AD = DB, ED is perpendicular to AB,
AB = 20 and AC = 12.
Objective Mathematics Vol. 1
region in which the point of intersection of C
tangents to the circle at points P and Q lies is E
δ
represented by θ
(a) y 2 ≥ 4( ax − a 2 ) α
β γ
(b) y 2 ≤ 4( ax − a 2 ) A
D
B
(c) y ≥ 4( ax − a ) 2
Ex 89. Area of ∆AEC is
(d) y ≤ ( ax − a 2 )
(a) 24 sq units (b) 21 sq units
Sol. Let the centre be (0, α) and equation of circle be 21
(c) 42 sq units (d) sq units
x 2 + ( y − α )2 = | a |2 2
Y Sol. Given, AD = BD
P(h, k)
Now, BC 2 = AB 2 − AC 2 = 162
C i.e. BC = 16
∆AED ≅ ∆BED [by SAS property]
AC 12 3
tan γ = = = …(i)
BC 16 4
γ =β …(ii)
X BD 10 25
O A(a, 0) cos γ = ⇒ BE = = …(iii)
BE 4 /5 2
Equation of chord of contact for P (h, k ) is DE 3 15
xh + yk − α ( y + k ) + α 2 − a2 = 0 tan γ = ⇒ DE = 10 = …(iv)
BD 4 2
It passes through (a, 0). 25 7
∴ α 2 − αk + ah − a2 = 0 CE = BC − BE = 16 − =
2 2
As a is real ⇒ k 2 − 4 (ah − a2 ) ≥ 0 1 1
Area of ∆AEC = ( AC )(CE ) = × 12 × = 21
7
Thus, required equation is 2 2 2
y2 ≥ 4 (ax − a2 ) Hence, (b) is the correct answer.
Hence, (a) is the correct answer. Ex 90. The value of tan γ tan (60° − γ ) tan (60° + γ ) is
117 17 3 5
Ex 87. In above question let Aa represents the region (a) − (b) (c) (d)
44 4 4 4
asked in the problem (obviously | a | is radius
of circle), then the region represented by Sol. tan γ tan (60° − γ )tan (60° + γ )
3 − tan γ 3 + tan γ
∩ Aa will be = tan γ
a ∈R 1 + 3 tan γ 1 − 3 tan γ
(a) h 2 − k 2 ≤ 0 (b) h 2 − k 2 ≥ 0 3
Put tan γ = , we get
(c) h + 2k < 0 (d) h + 2k > 0 4
117
Sol. Intersection of all region is nothing but the region which tan γ tan (60° − γ )tan (60° + γ ) = −
44
satisfy k 2 − 4 ah + 4 a2 ≥ 0 for all values of a ∈ R. Hence, (a) is the correct answer.
⇒ 16h2 − 16k 2 ≤ 0
⇒ h2 − k 2 ≤ 0 Ex 91. The value of cos (α + θ ) is
4 3 117 44
Hence, (a) is the correct answer. (a) (b) (c) (d)
5 5 125 125
Ex 88. If 3 2x + 2 + (a 2 − 4a − 2)3 x + 1 > 0, ∀ x ∈ R , Sol. We have, cos (α + θ ) = cos α cos θ − sin α sin θ
CE 7 7
then ⇒ tanα =
= =
AC 2 × 12 24
(a) a ∈ R (b) a ∈ R + 24 7
(c) a ∈ [1, ∞ ) (d) a ∈ R − {2} ⇒ cos α = ; sin α =
25 25
1 10 × 2 4
Sol. 3x 9 ⋅ 3x + + (a2 − 4 a − 2) > 0 ⇒ tanθ = =
3x 15 3
1
2 ⇒
3
cos θ = ; sin θ =
4
⇒ 3x 3 ⋅ 3x/ 2 − x/ 2 (a − 2)2 > 0
3 5 5
24 3 7 4 72 − 28 44
⇒ a≠2 ⇒ cos (α + θ ) = ⋅ − ⋅ = =
25 5 25 5 125 125
460 Hence, (d) is the correct answer. Hence, (d) is the correct answer.
Ex 92. The value of
α +θ + γ +δ
sin
α +θ − γ −δ
sin is
= − cos (α + θ ) = −
44
125
Hence, (d) is the correct answer.
9
Trigonometric Functions and Equations
2 2
Ex 93. Which of the following is true?
4 3
(a) (b) (a) γ + δ = θ + α (b) θ + δ > γ + α
5 5 (c) α + θ > γ + δ (d) α + β > γ + δ
117 44
(c) (d) − Sol. Since, AC > CE
125 125 δ >α …(i)
α + θ + γ + δ α + θ − γ − δ Since, AD > DE
Sol. sin sin θ >β
2 2
Since, β = γ
1
= [cos (γ + δ ) − cos (α + θ )] θ>γ …(ii)
2 On adding Eqs. (i) and (ii), we get
1
= [ − cos (α + θ ) − cos (α + θ )] δ + θ >α + γ
2
[Q γ + δ = π − (α + θ )] Hence, (b) is the correct answer.
Type 5. Match the Columns
Ex 94. Match the statements of Column I with values C. 4 cos 36° − 4 cos 72° + 4 sin 18° cos 36°
of Column II. 5 + 1 5 − 1 5 − 1 5 + 1
= 4 − 4 + 4
4 4 4 4
Column I Column II
= 5 + 1− 5 + 1+ 1= 3
A. The number of real roots of the equation p. 1 D. cosec x = 1 + cot x
cos 7 x + sin4 x = 1 in ( − π, π ) is 1 sin x + cos x
⇒ =
sin x sin x
B. The value of 3 cosec 20° − sec 20° is q. 4 ⇒ sin x + cos x = 1and sin x ≠ 0
π 1
cos x − =
C. 4 cos 36° − 4 cos 72 °+ 4 sin18° cos 36° r. 3 4 2
equals π π π
⇒ x − = − 2π + ,
4 4 4
D. The number of values of x, where s. 2
x ∈ [−2 π, 2 π ],
π π π
Q x − 4 ∈ −2π − 4 , 2π − 4
which satisfy
cosec x = 1 + cot x, is
3π π
⇒ x=− ,
Sol. A. cos7 x + sin 4 x = 1 2 2
cos7 x = (1 + sin 2 x ) cos2 x Hence, A → r; B → q; C → r; D → s
⇒ cos x = 0 Ex 95. Match the statements of Column I with values
or cos5 x = 1 + sin 2 x of Column II.
cos x = 0
π π Column I Column II
⇒ x=− , A. The number of solutions of the equation p. No solution
2 2 1
or cos5 x = 1 + sin 2 x |cot x| = cot x + ( 0 < x < π ) is
sin x
⇒ x=0 [QLHS ≤ 1 and RHS ≥ 1] 1 1
B. Ifsinθ + sin φ = andcos θ + cos φ = 2, then q.
π π
∴ x = − , 0, 2 3
θ + φ
2 2 cot is equal to
2
B. 3 cosec 20° − sec 20°
π π r. 1
C. sin α + sin − α sin + α is equal to
2
3 1 3 3
= −
sin 20° cos 20°
D. If tanθ = 3 tan φ, then maximum value of s. 3
3 cos 20° − sin 20° tan2(θ − φ) is
= 4
sin 20° cos 20°
3 1
1 Sol. A. |cot x | = cot x +
2 cos 20° − sin 20° sin x
2 2 π
= If 0 < x < , then cot x > 0
sin 20° cos 20° 2
4 cos 50° 1
= =4 So, cot x = cot x +
sin 40° sin x
461
9 If
⇒
π
1
sin x
= 0, no solution
< cot x < π , then − cot x = cot x +
1
C. 3 sin x + 4 cos x
Maximum value = 5
D. tan 15° + tan 30° + tan 15° tan 30°
Objective Mathematics Vol. 1
2 sin x
2 cos x
+
1
=0 3 −1 1 3 −1 1
sin x sin x = + + ×
3+1 3 3+1 3
2π
⇒ 1 + 2 cos x = 0 and sin x ≠ 0 ⇒ x =
3 3 − 1 1 1
= + 1 +
1
B. Since, sin φ + sin θ = and cos θ + cos φ = 2 has 3 + 1 3 3
2
no solution. 3 −1 3+1 1
= × + =1
π π 3+1 3 3
C. sin 2 α + sin − α sin + α
3 3
A → r; B → p; C → s; D → q
π 3
= sin 2 α + sin 2 − sin 2 α =
3 4 Ex 97. Match the statements of Column I with values
D. tan θ = 3 tan φ of Column II.
tan θ − tan φ 2 tan φ
tan (θ − φ ) = = Column I Column II
1 + tan θ tan φ 1 + 3 tan 2 φ
2
= [maximum, if tanθ > 0] A. Maximum value of − x 2 − 2 x + 2 is p. 3
cot φ + 3 tan φ
cot φ + 3 tan φ
≥ 3 [using AM ≥ GM] B. Number of real solutions of q. 2
2 cos (e x ) = 2 x + 2 − x is
1
⇒ (cot φ + 3 tan φ )2 ≥ 12 ⇒ tan 2 (θ − φ ) ≤
3 C. Number of solutions of equation r. 1
A → r; B → p; C → s; D → q | x | = cos x is
Ex 96. Match the statements of Column I with values D. Minimum value of x 2 − 2 x + 4, ∀ x ∈ R, is s. 0
of Column II.
D
Column I Column II Sol. A. Maximum value = − =3
4a
A. The value of p. 4 Y
cos 1° cos 2 ° cos 3°… cos 170° is
B. The value of 3 cosec 20° − sec 20° is q. 1
X' X
O
C. Maximum value of 3 sin x + 4 cos x is r. 0
D. tan15° + tan 30° + tan15° tan 30° is s. 5 Y'
Sol. A. cos1° cos 2° cos 3°… cos 170° = 0 B. LHS ≤ 1, whereas RHS ≥ 2
So, no solution.
B. 3 cosec 20° − sec 20°
C. Number of solutions = 2
3 1 3 cos 20° − sin 20°
= − = D 12
sin 20° cos 20° cos 20° sin 20° D. Minimum value = − =− − =3
4a 4
4[sin 60° cos 20°− cos 60° sin 20° ] 4 sin 40°
= = =4 A → p; B → s; C → q; D → p
2 cos 20° sin 20° sin 40°
Type 6. Single Integer Answer Type Questions
A B C A+ B A−B
Ex 98. In a ∆ABC, 2 sin + sin + sin has the Since, sin is real, 4 cos2 − 8(k − 1) ≥ 0
2 2 2 4 4
A−B
maximum value λ, then λ is equal to ______. ⇒ 2(k − 1) ≤ cos2 ≤1
4
A B C
Sol. (3) Let sin + sin + sin = k ⇒ k≤
3
2 2 2 2
A+ B A−B A+ B A−B A+ B A+ B
or 2 sin cos + cos =k So, 2 cos sin − 2 sin 2 + 1≤
3
4 4 2 4 4 4 2
A+ B A−B A+ B
⇒ 2 sin 2 − 2 cos sin + k −1= 0 Hence, λ = 3
4 4 4
462
θ φ
Ex 99. If a + b tan
2
= a − b tan and
2
( b − a sec θ ) ( b + a cos φ ) has maximum value
Ex 102. If x ∈[0, 2π ] for which
2 cos x ≤ | 1 + sin 2x − 1 − sin 2x | ≤ 2
9
Trigonometric Functions and Equations
λ, then λ is equal to _______. λπ µπ
has solution set x ∈ , , then µ − λ is
Sol. (0) Given that, 4 4
θ φ
a + b tan = a − b tan ⇒ a≥b …(i) equal to _______ .
2 2
θ Sol. (6) Let y = | 1 + sin 2x − 1 − sin 2x |
1 − tan 2
⇒ cos θ = 2 = a cos φ + b
2θ a + b cos φ ⇒ y2 = 2 − 2 |cos 2x |
1 + tan
2 π 3π 5π 7π
If x ∈ 0, or , or , 2π
⇒ (b − a sec θ ) (b + a cos φ ) = b2 − a2 4 4 4 4
⇒ (b − a sec θ ) (b + a cos φ ) ≤ 0 [from Eq. (i)] cos 2x is non-negative.
So, the maximum value is 0. So, y2 = 2 − 2 cos 2x − 4 sin 2 x
∴ λ=0 y = 2 |sin x |
Ex 100. If log | sin x | |cos x | + log | cos x | |sin x | = 2, then ⇒ cos x ≤ |sin x |
π 7π
Except for x in 0, and , 2π
| tan x | is equal to _______. 4 4
1 3π 5π
Sol. (1) Here, log|sin x | |cos x | + =2 so that leaves ,
log|sin x | |cos x | 4 4
1 1
⇒ y+ = 2, where y = log|sin x | |cos x | In which we certainly have, sin x ≤
y 2
⇒ y=1 π 3π 5π 7π
If x ∈ , or , , then
⇒ log|sin x | |cos x | = 1 4 4 4 4
⇒ |sin x | = |cos x | cos 2x is negative, so y2 = 2 + 2 cos 2x = 4 cos2 x
⇒ |tan x | = 1 ⇒ y = 2 | cos x |
π 3π
Ex 101. If 16 cos θ − cos cos θ − cos So the first inequality certainly holds, the second also
8 8 holds.
5π 7π π 7π
Thus, solution set is x ∈ , .
cos θ − cos cos θ − cos = λ cos 4θ, 4 4
8 8
⇒ µ−λ=6
then the value of λ is ________.
π 3π Ex 103. If 2 7 cos 3 θ sin 5 θ = a sin 8θ − b sin 6θ
Sol. (2) LHS = 16 cos θ − cos cos θ − cos + c sin 4θ + d sin 2θ
8 8
5π 7π and θ is real, then the value of a + b + c + d
cos θ − cos cosθ − cos must be ________.
8 8
π 3π Sol. (7) Let z = eiθ
= 16 cos θ − cos cos θ − cos
8 8 1 1
∴ 2 cos θ = z + and 2i sinθ = z −
3π π z z
cos θ + cos cos θ + cos 3 5
8 8 1 1
Now, (2 cos θ ) ⋅ (2i sin θ ) = z + z −
3 5
2 π 3 π z z
= 16 cos θ − cos2 cos2 θ − cos2
8 8 8 1 6 1 4 1 2 1
= z − 8 − 2 z − 6 − 2 z − 4 + 6 z − 2
π π z z z z
= 16 cos4 θ − cos2 θ + sin 2 cos2
8 8 = 2i sin 8θ − 4 i sin 6θ − 4 i sin 4θ + 12i sin 2θ
1 π ∴ 27 cos3 θ sin 5 θ = sin 8θ − 2 sin 6θ
= 16 cos4 θ − cos2 θ + sin 2
4 4 − 2 sin 4θ + 6 sin 2θ
1 On comparing, a = 1, b = 2, c = − 2, d = 6
= 16 cos4 θ − cos2 θ +
8 ⇒ a+ b+ c+ d =7
1 − sin 2 2θ 1
= 16 − cos2 θ sin 2 θ + = 16 + Ex 104. If sin x + sin y ≥ cos α cos x, ∀ x ∈ R , then
8 4 8
sin y + cos α is equal to _______.
1 − 2 sin 2 2θ cos2 2θ − sin 2 2θ
= 16 = 16 Sol. (1) sin x + sin y ≥ cos α cos x , ∀ x ∈ R
8 8 π
16 cos 4θ Let x=−
= = 2 cos 4θ 2
8 ∴ −1 + sin y ≥ 0
Hence, the value of λ is 2. ⇒ sin y ≥ 1 ⇒ sin y = 1
463
9 If sin y = 1, then
sin x + 1 ≥ cos α cos x
⇒ cos α cos x − sin x ≤ 1 …(i)
Sol. (1) As 1 ≤ |sin x | + |cos x | ≤ 2
∴ [|sin x | + |cos x | ] = 1
Objective Mathematics Vol. 1
⇒ log|x − 1 | | x 2 − 1 | = 1
Using a sin θ + b sin θ ≤ a2 + b2 , ∀ θ ∈ R
cos α cos x − sin x ≤ 1 + cos2 α ⇒ |x2 − 1 | = |x − 1 |
⇒ x = 1, 0, − 2, but x ≠ 0, 1
∴ 1 + cos2 α ≤ 1
∴ Number of solutions is one.
⇒ cos α = 0
⇒ sin y + cos α = 1 π π
Ex 108. For a given sector OABC, if θ ∈ , , then
6 3
Ex 105. If 0 ≤ A, B , C ≤ π and A + B + C = π, then the
minimum area of segment ABCA is
minimum value of sin 3 A + sin 3B + sin 3C is
______. π 3
r2 − , then λ is _______.
Sol. (−2) Let y = sin 3 A + sin 3 B + sin 3C for sin 3A to be 6 λ
non-positive. O
We have, 2π ≤ 3 A ≤ 3π
2π 2θ
⇒ <A<π
3
Since, A+ B+C =π
All of sin 3 A , sin 3 B , sin 3C can not be negative.
Let us take sin 3 A = − 1 A C
π π
⇒ A= ⇒ sin 3 B = − 1 ⇒ B = B
2 2
∴ C =0 Sol. (4) Area of segment ABCA
i.e. sin 3 A = − 1, sin 3 B = − 1, sin 3C = 0 = Area of sector AOC − Area of ∆AOC
2θ r2
Hence, minimum value of = × πr2 − sin 2θ
sin 3 A + sin 3 B + sin 3C = − 2 360° 2
2 1 π π
= r θ − sin 2θ , θ ∈ , …(i)
Ex 106. If θ 1 , θ 2 and θ 3 are three values lying in [0, 2π ] 2 6 3
for which tan θ = λ, then 1
Let f (θ ) = θ − sin 2θ
2
θ1 θ θ θ θ θ ⇒ f ′ (θ ) = 1 − cos 2θ > 0
tan tan 2 + tan 2 tan 3 + tan 1 tan 3 i [increasing]
3 3 3 3 3 3 2π 1 2π
Hence, area of segment ABCA = r − sin
s equal to _______. 6 2 3
π 3
2 π 3
θ θ = r2 − =r − ⇒ λ=4
3 tan − tan 3 λ
3 3 6 4 6
Sol. (3) tan θ =
θ
1 − 3 tan 2 Ex 109. The number of solutions for
3
θ θ θ 2 sin 2 x + sin 2 2x = 2 when −π < x < π, is ____.
⇒ tan 3 − 3λ tan − 3 tan + λ = 0
3 3 3
θ1 θ2 θ2 θ3 θ3 θ Sol. (6) We have, 2 sin 2 x + sin 2 2x = 2
∴ tan tan + tan tan + tan tan 1 = − 3 ⇒ 1 − cos 2x + 1 − cos2 2x = 2
3 3 3 3 3 3
θ1 θ2 θ2 θ3 θ1 θ 3 ⇒ cos 2x + cos2 2x = 0
⇒ tan tan + tan tan + tan tan =3
3 3 3 3 3 3 ⇒ cos 2x (1 + cos 2x ) = 0
⇒ cos 2x = 0, cos 2x = − 1
Ex 107. The number of solutions of the equation π 3π
⇒ 2x = ± , ± , 2x = ± π
log | x − 1 | | x 2 − 1 | = [|sin x | + |cos x | ] 2 2
π 3π π
where, [ ] denotes greatest integer function, ⇒ x=± ,± ,±
4 4 2
is ______. Hence, the number of solutions is six.
464
Target Exercises
Type 1. Only One Correct Option
1. The numerical value of 11. The value of
p 2p 4p 8p 6 (sin 6 q + cos 6 q ) - 9 (sin 4 q + cos 4 q ) + 4 is
tan + 2 tan + 4 tan + 8 tan is
3 3 3 3 (a) - 3 (b) 0 (c) 1 (d) 3
5
(a) - 5 3 (b) - 2sin a 1 - cos a - sin a
3 12. If = x, then is equal
5 1 + cos a + sin a cos a
(c) 5 3 (d)
3 to
1
2. A quadratic equation whose roots are cosec 2 q and (a)
x
(b) x
sec 2 q, can be (c) 1 - x (d) None of these
(a) x 2 - 2x + 2 = 0 (b) x 2 - 3x + 3 = 0
13. Which of the following is not correct?
(c) x 2 - 5x + 5 = 0 (d) None of these 1
(a) sin q = - (b) cos q = 1
3p 4p 5
3. cos 2 + cos 2 is equal to 1
5 5 (c) secq = (d) tan q = 20
2
4 5 5 3
(a) (b) (c) (d)
5 2 4 4 14. log (sin 1° ) log (sin 2° ) log (sin 3° ) ¼ log (sin 179° )
Targ e t E x e rc is e s
4. The value of sin 600° cos 330°+ cos 120° sin 150° is is equal to
(a) 1 (b) 0 (c) 2 (d) - 1
1 3
(a) -1 (b) 1 (c) (d)
2 2 é æ 3p ö ù
15. The value of 3 êsin 4 ç - a ÷ + sin 4 ( 3p + a )ú
1 - 4 sin 10° sin 70° ë è 2 ø û
5. The value of expression is
2sin 10° é æp ö ù
1 -2 êsin 6 ç + a ÷ + sin 6 ( 5p - a )ú is
(a) (b) 1 ë è 2 ø û
2
(c) 2 (d) None of these (a) 0 (b) 1
(c) 3 (d) sin 4 a + sin 6 a
6. The value of the expression ( 3 sin 75° - cos 75° ) is
16. If tan x + cot x = 2 , then sin 2n x + cos 2n x is equal to
(a) 2 sin 15° (b) 1 + 3 (c) 2 sin 105° (d) 2 1
(a) 2n (b) -
1 3 2
7. If sin a = and sin b = , then b - a lies in the 1
5 5 (c) (d) None of these
2
interval
æ p ö æ 3p ö é p 3p ù p 3p 15 12
(a) ç 0, ÷ È ç , p÷ (b) ê , 17. If < a < p , p < b < ; sin a = and tan b = ,
è 4ø è 4 ø ë 2 4 úû 2 2 17 5
æp pö æ 5p ù then the value of sin (b - a ) is
(c) ç , ÷ (d) ç p,
è 4 2ø è 4 úû 171 21 21 171
(a) - (b) - (c) (d)
221 221 221 221
8. The numerical value of 3 cosec 20° - sec 20° is
(a) 2 (b) 4 18. If cos (q - a ) = a and sin (q - b ) = b, then
(c) 6 (d) None of these cos 2 (a - b ) + 2ab sin (a - b ) is equal to
(a) 4 a2b2 (b) a2 - b2
9. Which of the following numbers is rational?
(c) a2 + b2 (d) - a2b2
(a) sin 15° (b) cos 15°
(c) sin 15° cos 15° (d) sin 15° cos 75° cos 6x + 6cos 4x + 15cos 2x + 10
19. The expression is
p 4p 5p cos 5x + 5cos 3x + 10cos x
10. The value of cos cos cos is
7 7 7 equal to
1 1 1 1 (a) cos 2x (b) 2cos x
(a) (b) (c) - (d)
2 4 8 8 (c) cos2 x (d) 1 + cos x 465
9 20. sin 47° + sin 61° - sin 11° - sin 25° is equal to
(a) sin 36°
(c) sin 7°
(b) cos 36°
(d) cos 7°
p
29. If A + B = , where A , B Î R + , then the minimum
4
value of (1 + tan A ) (1 + tan B ) is
Objective Mathematics Vol. 1
(a) 2 (b) 4
æp ö æp ö (c) 1 (d) None of these
21. The value of cos y cos ç - x÷ - cos ç - y÷ cos x
è2 ø è2 ø
sin x 1 cos x 3 æ pö
æp ö æp ö 30. If = , = , where x, y Î ç 0, ÷, then the
+ sin y cos ç - x÷ + cos x sin ç - y÷ is zero, if sin y 2 cos y 2 è 2ø
è2 ø è2 ø
value of tan ( x + y ) is
(a) x = 0 (b) y = 0
p (a) 13 (b) 14 (c) 17 (d) 15
(c) x = y (d) x = np - + y
4 2p
31. In DABC , if ÐC = , then the value of
11 3
22. If cosec A + cot A = , then tan A is equal to
2 cos 2 A + cos 2 B - cos A cos B is
21 15 44 117 3 3 1 1
(a) (b) (c) (d) (a) (b) (c) (d)
22 16 117 43 4 2 2 4
sin 3 x cos 3 x 10
pr
23. The value of the expression +
1 + cos x 1 - sin x
is 32. The value of å cos 3 3
is
r=0
ép ù ép ù 9 7 9 1
(a) 2 cos ê - x ú (b) 2 cos ê + x ú (a) - (b) - (c) - (d) -
ë4 û ë4 û 2 2 8 8
ép ù
(c) 2 sin ê - x ú (d) None of these 33. If in any DABC, the value of ÐA is obtained from the
ë4 û
equation 3cos A + 2 = 0, then the equation whose
24. In D ABC, the value of the expression roots are sin A and tan A, is
cosec A (sin B cos C + cos B sin C ) is (a) 6x 2 + 5x - 5 = 0 (b) 6x 2 + 5x - 5 = 0
Ta rg e t E x e rc is e s
c (c) x 2 + 5x - 5 = 0 (d) None of these
(a) 1 (b)
a
(c)
a
(d) None of these 34. If cot 2 x = cot ( x - y )cot ( x - z ), then cot 2x is equal
c to ( where, x ¹ ± p / 4 )
3 1 1
25. If cos ( A - B ) = and tan A tan B = 2 , then (a) (tan y + tan z) (b) (cot y + cot z)
5 2 2
1 2 1
(a) cos A cos B = (b) sin A sin B = - (c) (sin y + sin z) (d) None of these
5 5 2
1 p
(c) cos ( A + B ) = (d) None of these 35. If a + b = and b + g = a , then tan a is equal to
5 2
æ 2p ö æ 4p ö (a) 2(tan b + tan g ) (b) tan b + tan g
26. f (q ) = sin 2 q + sin 2 çq + ÷ + sin 2 çq + ÷, then (c) tan b + 2 tan g (d) 2tan b + tan g
è 3ø è 3ø
æpö 36. If sin q = n sin (q + 2a ), then tan (q + a ) is equal to
f ç ÷ is equal to
è 15 ø æ 2 + nö æ 1 + nö
2 3
(a) ç ÷ tan a (b) ç ÷ tan a
(a) (b) è 2 - nø è1 - nø
3 2 æ1 - nö
1 (c) tan a (d) ç ÷ tan a
(c) (d) None of these è 1 + nø
3
x y x y
27. If cos ( x - y ) , cos x and cos ( x + y ) are in HP, then 37. If cos a + sin a = 1, cos b + sin b = 1 and
a b a b
æ yö
cos x × sec ç ÷ is equal to cos a cos b sin a sin b
è 2ø + = 0, then
a2 b2
(a) ±1 (b) ± 2
b2 (x 2 + a2 )
(c) ± 3 (d) None of these (a) tan a tan b = (b) x 2 + y2 = a2 + b2
a2 ( y2 + b2 )
n
sin 3 x sin 3x = a2
28. If å a r cos ( rx ), " x Î R, where (c) tan a tan b = 2
b
(d) None of these
r =0
a 0 , a1 , ¼ , a n are constants and a n ¹ 0, then 38. Ifx = sin (a - b )sin ( g - d ), y = sin (b - g ) sin (a - d )
(a) n = 5 and z = sin ( g - a ) sin (b - d ) , then
(b) n = 6
(a) x + y + z = 0 (b) x + y - z = 0
466 (c) n = 7
(c) y + z - x = 0 (d) None of these
(d) None of the above
æp
è4
ö
ø
æp
è4
ö
ø
æp
è4
ö
ø
æp
è4
ö
39. cos ç - x÷ cos ç - y÷ - sin ç - x÷ sin ç - y÷ is
ø
48. If sin x + cos x =
2
7 é pù x
, where x Î ê 0, ú, then tan is
ë 4û 2 9
Trigonometric Functions and Equations
equal to equal to
(a) sin (x - y) (b) sin (x + y) 3- 7 7-2
(c) cos (x + y) (d) cos (x - y) (a) (b)
3 3
40. If x Î ( 0, p ) and sin x cos 3 x > cos x sin 3 x , then (c)
7- 7
(d) None of these
complete set of values of x is 4
æ p ö æ p 3p ö 1 tan ( A + B )
(a) x Î ç 0, ÷ È ç , ÷ 49. If sin B = sin ( 2A + B ) , then is equal
è 4ø è 2 4 ø
5 tan A
æ p p ö æ 3p ö
(b) x Î ç , ÷ È ç , p÷ to
è 4 2ø è 4 ø
5 2 3 3
æ p p ö (a) (b) (c) (d)
(c) x Î ç , ÷ 3 3 2 5
è 4 2ø
(d) None of the above 1 1 p
50. If tan a = and sin b = , where 0 < a, b < , then
¥ ¥ 7 10 2
p
41. For 0 < f < , if x = å cos 2n f, y = å sin 2n f and 2b is equal to
2 n=0 n=0 p 3p p 3p a
¥
(a) -a (b) -a (c) -a (d) -
4 4 8 8 2
z= å cos 2n f sin 2n f, then 3cos q + cos 3q
n=0
51. is equal to
(a) xyz = xz + y (b) xyz = xy + y 3sin q - sin 3q
(c) xyz = x + y + z (d) xyz = yz + x (a) 1 + cot 2 q (b) cot 4 q (c) cot 3 q (d) 2cot q
x y z
42. If = = , then a +b
cos q æ 2p ö æ 2p ö tan
cos çq - ÷ cos çq + ÷
Targ e t E x e rc is e s
52. If 3sin a = 5sin b, then 2 is equal to
è 3 ø è 3 ø
a -b
x + y + z is equal to tan
2
(a) 1 (b) 0
(c) -1 (d) None of these (a) 1 (b) 2 (c) 3 (d) 4
1æ 1ö 1æ 2 1ö cos ( A + C )
43. If cos q = çx + ÷, then ç x + 2 ÷ is equal to 53. If cos 2B = , then
2è x ø 2 è x ø cos ( A - C )
(a) sin 2q (b) cos 2q (a) tan A , tan B and t an C are in AP
(c) tan 2q (d) None of these (b) tan A , tan B and tan C are in GP
(c) tan A , tan B and tan C are in HP
44. If - p £ x £ p, - p £ y £ p and cos x + cos y = 2 , then (d) None of the above
the value of cos ( x - y ) is b
(a) -1 (b) 0 54. If tan x = , then the value of a cos 2x + b sin 2x is
a
(c) 1 (d) None of these
(a) a (b) a - b
45. If cos x + cos y + cos a = 0 and (c) a + b (d) b
æ x + yö 55. The graph of the function
sin x + sin y + sin a = 0, then cot ç ÷ is equal to 2
è 2 ø cos x cos ( x + 2) - cos ( x + 1) is
(a) sin a (b) cos a (a) a straight line passing through (0, - sin 2 q) with slope 2
æx + yö
(b) a straight line passing through (0, 0)
(c) cot a (d) sin ç ÷
è 2 ø (c) a parabola with vertex (1, - sin 2 1)
æp ö
(d) a straight line passing through the point ç , - sin 2 1÷
46. If a + b - g = p, then sin 2 a + sin 2 b - sin 2 g is è2 ø
and parallel to the X-axis
equal to
(a) 2sin a sin b cos g (b) 2cos a cosb cos g 56. If a cos 2 3q + b cos 4 q = 16cos 6 q + 9cos 2 q is an
(c) 2sin a sin b sin g (d) None of these identity, then
47. If a + b + g = 2p , then (a) a = 1, b = 18 (b) a = 1, b = 24
(c) a = 3, b = 24 (d) a = 4, b = 2
a b g a b g
(a) tan + tan + tan = tan tan tan n
2 2 2 2 2 2
a b b g g a
57. Let n be an odd integer. If sin n q = å br sin r q for
(b) tan tan + tan tan + tan tan = 1 r=0
2 2 2 2 2 2
a b g a b g every value of q, then
(c) tan + tan + tan = - tan tan tan (a) b0 = 1, b1 = 3 (b) b0 = 0, b1 = n
2 2 2 2 2 2 467
(d) None of the above (c) b0 = - 1, b1 = n (d) b0 = 0, b1 = n2 - n + 3
9 58. If 0£q £
p
2
y = X sin q - Y cos q
and x = X cos q + Y sin q,
such that
67. If
2
x=a , b satisfies both
2
cos x + a cos x + b = 0 and sin x + p sin x + q = 0,
the equation
Objective Mathematics Vol. 1
then the relation between a, b, p and q is
x 2 + 2xy + y 2 = aX 2 + bY 2 , where a and b are
(a) 1 + b + a2 = p2 - q (b) a2 + b2 = p2 + q2
constants, then (c) b + q = a2 + p2 - 2 (d) None of these
p
(a) a = - 1, b = - 3 (b) q =
2 68. If f (q ) = | sin q | + |cos q | , q Î R, then
p
(c) a = 3, b = - 1 (d) q = (a) f (q) Î [ 0, 2 ] (b) f (q) Î [ 0, 2 ]
3
(c) f (q) Î [ 0, 1] (d)
59. If q 1 and q 2 are two values lying in [0, 2p] for which f (q) Î [1, 2 ]
q q
tan q = l , then tan 1 tan 2 is equal to 69. If f (q ) = (sin q + cosec q ) 2 + (cos q + sec q ) 2 , then
2 2
minimum value of f (q ) is
(a) 0 (b) -1 (c) 2 (d) 1
(a) 7 (b) 8
60. If a Î ( 0, p / 2), then the expression (c) 9 (d) None of these
tan 2 a 70. If f (q ) = sin 4 q + cos 4 q + 1, then range of f (q ) is
x2 + x + is always greater than or equal to
x2 + x é3 ù
(a) ê , 2ú
é 3ù
(b) ê1, ú
(a) 2tan a (b) 2cos a ë2 û ë 2û
(c) 1 (d) sec2 a (c) [1, 2 ] (d) None of these
A B C 71. If f ( x ) = sin 6 x + cos 6 x + 1, then range of f (q ) is
61. In DABC, tan , tan and tan are in HP, then the
2 2 2 é1 ù é 1 3ù
(a) ê , 1ú (b) ê , ú
A C ë4 û ë4 4û
value of cot cot is
é3 ù
Ta rg e t E x e rc is e s
2 2 (c) ê , 1ú (d) None of these
(a) 1 (b) 2 ë4 û
(c) 3 (d) 4
æ pö æ pö
72. The maximum value of sin ç x + ÷ + cos ç x + ÷ in
62. The expression è 6ø è 6ø
cos 2 f + cos 2 ( a + f ) - 2cos a cos f cos ( a + f ) is æ pö
the interval ç 0, ÷ is attained at
independent of è 2ø
(a) f (b) a p p p p
(c) both a and f (d) None of these (a) (b) (c) (d)
12 6 3 2
x
63. If cos x - sin a cot b sin x = cos a, then tan is equal 73. The number of integral value of k for which the
2
equation 7cos x + 5 sin x = 2 k + 1 has a solution, is
to
a b b a (a) 4 (b) 8 (c) 10 (d) 12
(a) cot tan (b) cot tan
2 2 2 2 sin 2 q cos 2 q x
a b 2
(c) tan tan
2 2
(d) None of these 74. If f ( x ) = cos q x sin 2 q , q Î ( 0, p / 2), then
x sin 2 q cos 2 q
p
64. If A + B = and cos A + cos B = 1, then
3 roots of f ( x ) = 0 are
1 2 (a) 1/ 2, - 1, 0 (b) 1/ 2, - 1
(a) cos ( A - B ) = (b) |cos A - cos B | = (c) -1/ 2, 1, - 1 (d) -1/ 2, - 1, 1
3 3
2 1
(c) cos ( A - B ) = (d) None of these 75. If f (q ) = , then range of f (q ) is
3 2 - 3cos q
65. Let I ( n ) = 2cos nx, " n Î N , then I (1) × I ( n + 1) - I ( n ) æ 1ù é 1ù
(a) ç - ¥ , ú È [1, ¥ ) (b) ê -1, ú
è 5û ë 5û
is equal to
(a) I (n + 3) (b) I (n + 2) æ1 ù
(c) (- ¥ , - 1] È ç , ¥ ú (d) None of these
(c) I (n + 1) × I (2) (d) I (n + 2) × I (2) è5 û
66. The equation (cos p - 1) x 2 + (cos p ) x + sin p = 0 in æ pö
76. If f (q ) = 5cos q + 3cos çq + ÷ + 3, then range of
x has real roots. Then, the set of values of p is è 3ø
(a) [ 0, 2p ] (b) [ - p , 0 ] f (q ) is
é p pù (a) [ - 5, 11] (b) [ - 3, 9 ]
(c) ê - , ú (d) [ 0, p ] (c) [ - 2, 10 ] (d) [ - 4 , 10 ]
468 ë 2 2û
77. If a sin x + b cos ( x + q ) + b cos ( x - q ) = d], then the
minimum value of |cos q | is
87. If sin q =
x2 + y2
x2 - y2
, then
9
Trigonometric Functions and Equations
1 1
(a) d 2 - a2 (b) d 2 - a2 (a) x ¹ y (b) x ¹ y and x , y Î R
2| b | 2| a | (c) x = 0 or y = 0 (d) x ¹ y and x ¹ 0, y ¹ 0
1
(c) d 2 - a2 (d) None of these 88. The most general values of q satisfying the equation
2| d |
(1 + 2sin q ) 2 + ( 3 tan q - 1) 2 = 0 is/are given by
78. The minimum value of the expression p 7p
(a) np ± (b) np + (-1)n
sin a + sin b + sin g , where a , b and g are real 6 6
numbers satisfying a + b + g = p, is 7p 11p
(c) 2np + (d) 2np +
(a) positive (b) zero 6 6
(c) negative (d) None of these
89. The number of solutions of the equation
p
79. If A ³ 0, B ³ 0, A + B = and y = tan A tan B, then tan 2 x - sec10 x + 1 = 0 in ( 0, 10), is
3
(a) 3 (b) 8
æ 1ù 1
(a) y Î ç -¥ , ú È [3, ¥) (b) 0 £ y £ (c) 10 (d) None of these
è 3û 3
1 90. Total number of solutions of tan x + cot x = 2 cosec x
(c) y > (d) None of these
3 in [ - 2p , 2p ] is
(a) 2 (b) 4 (c) 6 (d) 8
80. If ( a sin 4 q + b ) ( a + b ) = ab cos 4q , then
(a) | a | ³ | b | 91. Total number of solutions of the equation
5p
(b) | b | = | a | 3x + 2 tan x = in x Î[ 0, 2p ], is
(c) | a | £ | b | 2
(a) 1 (b) 2 (c) 3 (d) 4
(d) None of the above
Targ e t E x e rc is e s
a 92. If a, b Î [ 0, p ] and the equation
81. If sin q = x + , " x Î R ~ {0}, then x 2 + 4 + 3 sin ( ax + b ) - 2x = 0 has atleast one
x
1 1 solution, then the value of ( a + b ) can be
(a) a ³ (b) a ³ 7p 3p
4 2 (a) (b)
1 1 2 2
(c) a £ (d) a £ 9p
4 2 (c) (d) None of these
2
2
82. If sin x + a sin x + 1 = 0 has no real number solution, 1
then 93. If sin q, cos q and tan q are in GP, then the general
6
(a) | a | ³ 2 (b) | a | ³ 1 value(s) of q is/are
(c) | a | < 2 (d) None of these p p
(a) 2np ± , n ÎI (b) 2np ± , n ÎI
83. If sin q 1 - sin q 2 = a and cos q 1 + cos q 2 = b, then 3 6
p p
(a) a2 + b2 ³ 4 (b) a2 + b2 £ 4 (c) 2np + (-1)n , n Î I (d) np + , n Î I
3 3
(c) a2 + b2 ³ 3 (d) a2 + b2 £ 2
1
94. The general values of x for which cos 2x, and sin 2x
84. If 0 £ a £ 3, 0£ b £ 3 and the equation 2
x 2 + 4 + 3cos ( ax + b ) = 2x has atleast one solution, are in AP, are given by
p p
then the value of a + b is (a) np, np + (b) np, np +
2 4
p p
(a) 0 (b) (c) np + (d) np
2 4
(c) p (d) None of these
95. The most general values of q satisfying
85. If tan 2 x + sec x - a = 0 has atleast one solution, then
æ 3p ö
the complete set of values of a is tan q + tan ç + q ÷ = 2 are
è 4 ø
(a) (- ¥ , 1] (b) [1, 1]
(c) [ - 1, 1] (d) [ -1, ¥ ) p
(a) np ± , n ÎI
3
4xy p
86. sec 2 q = is true if and only if (b) 2np ± , n Î I
(x + y )2 6
p
(a) x + y ¹ 0 (c) 2np ± , n Î I
(b) x = y, x ¹ 0 3
(c) x = y p
(d) 2np + (-1)n , n Î I 469
(d) x ¹ 0, y ¹ 0 3
9 96. The most general solutions of the equation
sec x - 1 = ( 2 - 1) tan x are given by
p p
105. If x, y Î [ 0, 2p ] , then total number of ordered pairs
( x, y ) satisfying sin x cos y = 1is
Objective Mathematics Vol. 1
(a) 1 (b) 3 (c) 5 (d) 7
(a) np + , np (b) 2np , 2np +
8 4
(c) 2np, np (d) None of these 106. The number of values of x for which
sin 2x + cos 4x = 2, is
97. If n tan y = tan x, n Î R + , then the maximum value of (a) 0 (b) 1 (c) 2 (d) infinite
sec 2 ( x - y ) is equal to
107. If |tan x | £ 1and x Î [ - p , p ] , then the solution set for
(n + 1)2 (n + 1)2
(a) (b) x is
2n n é 3p ù é p p ù é 3p ù
(n + 1)2 (n + 1)2 (a) ê - p , - È - , È , pú
(c) (d) ë 4 úû êë 4 4 úû êë 4 û
2 4n é p p ù é 3p ù
(b) ê - , ú È ê , pú
98. Let a , b be any two positive values of x for which ë 4 4û ë 4 û
2 cos x, |cos x | and 1 - 3cos 2 x are in GP. The é p pù
(c) ê - , ú
minimum value of |a - b | is ë 4 4û
p p (d) None of the above
(a) (b)
3 4
p 108. If 4 sin 2 x - 8sin x + 3 £ 0, 0 £ x £ 2p, then the
(c) (d) None of these
2 solution set for x is
é pù é 5p ù é 5p ù é p 5p ù
99. The value(s) of x Î [ -2p , 2p ] such that (a) ê 0, ú (b) ê 0, ú (c) ê , 2p ú (d) ê ,
ë 6û ë 6û ë 6 û ë 6 6 úû
sin x + i cos x
, i = -1 is purely imaginary, is/are
1+ i 109. The set of values of x for which
given by sin x cos 3 x > cos x sin 3 x, 0 £ x £ 2p, is
Ta rg e t E x e rc is e s
p p æ pö
(a) np - (b) np + (a) (0, p) (b) ç 0, ÷
4 4 è 4ø
(c) np (d) None of these æp ö
(c) ç , p ÷ (d) None of these
100. The number of solutions of the equation è4 ø
x
1 + sin x sin 2 = 0 in [ - p , p ] is 110. If x1 and x 2 are two positive values of x for which
2
2cos x, | cos x | and 3sin 2 x - 2 are in GP. The
(a) 0 (b) 1
(c) 2 (d) 3 minimum value of | x1 - x 2 | is
4p p
(a) (b)
101. The number of solutions of cos x = | 1 + sin x | , 3 3
0 £ x £ 3p is æ 2ö æ 2ö
(c) 2 cos-1 ç ÷ (d) cos-1 ç ÷
(a) 3 (b) 2 è 3ø è 3ø
(c) 4 (d) None of these
111. If 3sin x + 4 cos ax = 7 has atleast one solution, then a
102. The most general solutions of has to be necessarily a/an
2
x + |cos x |3 + ¼ ¥ (a) odd integer
21 + |cos x | + cos = 4 are given by (b) even integer
p p (c) rational number
(a) np ± ,n Î I (b) 2np ± ,n Î I
3 3 (d) irrational number
2p
(c) 2np ± ,n Î I (d) None of these 112. If the equation a1 + a 2 cos 2x + a 3 sin 2 x = 1 is
3 satisfied by every real value of x, then the number of
np 2 possible values of the triplet ( a1 , a 2 , a 3 ) is
103. If x ¹ and (cos x ) sin x - 3 sin x + 2 = 1, then all (a) 0 (b) 1 (c) 3 (d) infinite
2
solutions of x is/are given by 2
113. If sin q - 2 sin q - 1 = 0 is to be satisfied for exactly
p p
(a) 2np + (b) (2n + 1) p - 4 distinct values of q Î [ 0, np ], n Î N, then the least
2 2 value of n is
p
(c) 2np + (-1)n (d) None of these (a) 2 (b) 6 (c) 4 (d) 15
2
114. If 2 tan 2 x - 5sec x is equal to 1 for exactly 7 distinct
104. If 0 £ x £ 3p, 0 £ y £ 3p and cos x sin y = 1, then the
é np ù
possible number of values of the ordered pair ( x, y ) is values of x Î ê 0, ú, n Î N , then the greatest value
ë 2û
of n is
470 (a) 6 (b) 12
(c) 8 (d) 15 (a) 6 (b) 12 (c) 13 (d) 15
115. If sin a, 1 and cos 2a are in GP, then a is equal to
p
(a) np + (-1)n , n Î I
2
(b) np + (-1)n - 1
p
2
, n ÎI
123. If sin 2 (q - a ) cos a = cos 2 (q - a )sin a
= m sin a cos a , then 9
Trigonometric Functions and Equations
1
(c) 2np , n Î I (d) None of these (a) | m | £
2
116. If max{5sin q + 3sin (q - a )} = 7, then set of 1
q ÎR
(b) | m | ³
2
possible values of a is (c) | m | ³ 1
ì p ü (d) | m | £ 1
(a) í x | x = 2np ± ; n Î I ý
î 3 þ cot b sin x 3 x
ì 2p ü 124. If cos x - = , then the value of tan is
(b) í x | x = 2np ± ; n Î Iý 2 2 2
î 3 þ
b
é p 2p ù (a) tan tan 15°
(c) ê , 2
ë 3 3 úû b
(d) None of the above (b) tan
2
117. The number of solutions of |cos x | = sin x, (c) tan15°
(d) None of the above
0 £ x £ 4 p is
(a) 8 (b) 4 125. If the expression n sin 2 q + 2n cos (q + a ) sin a sin q
(c) 2 (d) None of these
+ cos 2(a + q ) is independent of q, then the value of n
118. The values of a for which the equation is
cos 4 x - ( a + 2)cos 2 x - ( a + 3) = 0 possesses a (a) 1
(b) 2
solution, lies in (c) 3
(a) [ - 3, - 2 ] (b) (- 3, - 2) (c) (0, 2) (d) (0, 3) (d) 4
119. The least positive non-integral solution of 126. For any real value of q ¹ p, the value of the
Targ e t E x e rc is e s
sin p ( x 2 + x ) - sin px 2 = 0 is cos 2 q - 1
(a) rational expression y = is
(b) irrational of the form p
cos 2 q + cos q
p -1 (a) 1 £ y £ 2
(c) irrational of the form , where p is an odd integer (b) y < 0 and y > 2
4 (c) - 1 £ y £ 1
p+1 (d) y ³ 1
(d) irrational of the form , where pis an even integer
4
127. Solution of equation [sin x] = [1 + sin x] + [1 - cos x],
120. If q Î[ 0, 5p ] r Î R such that
and 0 £ x £ 2p is
4 2
2sin q = r - 2r + 3, then the maximum number of 3p
(a) x =
values of the pair ( r, q ) is 2
(a) 8 3p
(b) x =
(b) 10 4
(c) 6 (c) no real solution
(d) None of the above (d) None of the above
121. Total number of solutions of [sin x] = cos x, where [ ] 128. The general solution of the equation
denotes the greatest integer function in [ 0, 3p ] , is sin 50 x - cos 50 x = 1 is
(a) 2 p
(b) 6 (a) 2np +
2
(c) 5 p
(d) None of the above (b) 2np +
3
3 p p
(c) np +
122. If 1 - sin x = x - + a has no solution, where 2
2 2 p
(d) np +
a Î R + , then 3
(a) a Î R +
129. Number of ordered pairs ( a, x ) satisfying the
3 p
(b) a > - equation sec 2 ( a + 2) x + a 2 - 1 = 0 , - p < x < p is
2 3
(a) 2
æ 3 p ö
(c) a Î ç 0, + ÷ (b) 1
è 2 3ø (c) 3
(d) None of the above (d) infinite
471
9 Type 2. More than One Correct Option
130. If f n (q ) = tan
q
(1 + sec q ) (1 + sec 2q )(1 + sec 4q ) ¼ 138. If tan a and tan b are the roots of the equation
Objective Mathematics Vol. 1
2 x 2 + px + q = 0 with p ¹ 0, then
(1 + sec 2n q ) , then
(a) sin 2 (a + b ) + p sin (a + b )cos (a + b ) + q cos2 (a + b ) = q
æpö æpö p
(a) f2 ç ÷ = 1 (b) f3 ç ÷ = 1 (b) tan (a + b ) =
è 16 ø è 32 ø q-1
æ p ö æ p ö (c) cos (a + b ) = 1 - q
(c) f4 ç ÷ = 1 (d) f5 ç ÷ =1
è 64 ø è 128 ø (d) sin (a + b ) = - p
p 139. If y = sin 2 x + cos 4 x , then for all real x
131. Let 0 £ q £ and x = X cos q + Y sin q ,
2 (a) the maximum value of y is 2
y = X sin q - Y cos q such that 3
(b) the minimum value of y is
x 2 + 4xy + y 2 = aX 2 + bY 2 , where a and b are 4
(c) y £ 1
constants. Then, 1
(d) y ³
p 4
(a) a = - 1, b = 3 (b) q =
4
p 140. If k1 = sin x cos 3 x and k 2 = cos x sin 3 x, then
(c) a = 3, b = - 1 (d) q =
3 æ pö
(a) k 1 - k 2 > 0, x Î ç 0, ÷
2 2 2 è 4ø
sin x + a - 2 a
132. = has a solution, if æ pö
cos 2x 1 - tan 2 x (b) k 1 - k 2 < 0, x Î ç 0, ÷
è 4ø
(a) a £ - 1 (b) a ³ 1 æ pö
(c) a = 1/ 2 (d) a is any real number (c) k 1 + k 2 > 0, x Î ç 0, ÷
è 2ø
Ta rg e t E x e rc is e s
n æ pö
133. If cos 3 x sin 2x = (d) k 1 + k 2 < 0, x Î ç 0, ÷
å a m sin mx is identity in x. Then, è 4ø
m=1
3 1 141. If a Î [ - 2p , 2p ] and
(a) a3 = , a2 = 0 (b) n = 6, a1=
8 2 a a
1 3 cos + sin = 2 (cos 36° - sin 18° ) , then a value
(c) n = 5, a1 = (d) San = 2 2
4 4 of a is
7p p
134. If 0 £ x, y £ 180° and sin ( x - y ) = cos ( x + y ) = 1/ 2, (a) (b)
6 6
then x and y are given by 5p p
(c) (d) -
(a) x = 45° , y = 15° (b) x = 45° , y = 135° 6 6
(c) x = 165° , y = 15° (d) x = 165° , y = 135°
142. If cos x = 1- sin 2x , where 0 £ x £ p, then a value of
x x
135. If tan = cosec x - sin x, then tan 2 is equal to x is
2 2 (a) p
(a) 2 - 5 (b) 5 - 2 (b) 0
(c) (9 - 4 5 ) (2 + 5) (d) (9 + 4 5 ) (2 - 5 ) (c) tan -1 2
(d) None of the above
1 - sin 4 A + 1 143. If sin q + 3 cos q = 6x - x 2 - 11, 0 £ q £ 4p, x Î R,
136. If y = , then one of the values of y is
1 + sin 4 A - 1 then holds for
(a) - tan A (b) cot A (a) no value of x and q
æp ö æp ö (b) one value of x and two values of q
(c) tan ç + A ÷ (d) - cot ç + A ÷ (c) two values of x and two values of q
è4 ø è4 ø
(d) two pairs of values of (x , q)
137. If x = a 2 cos 2 a + b 2 sin 2 a + a 2 sin 2 a + b 2 cos 2 a 144. If 0 £ x £ 2p and | cos x | £ sin x, then
2 2 2 2 2 2 ép pù
, then x = a + b + 2 p ( a + b ) - p , where p (a) the set of values of x is ê , ú
is equal to ë 4 2û
p
(a) a2 cos2 a + b2 sin 2 a (b) the number of solutions that are integral multiples of
4
(b) a2 sin 2 a + b2 cos2 a is three
1 3p
(c) [ a2 + b2 + (a2 - b2 ) cos 2a ] (c) the sum of the largest and the smallest solution is
2 4
1 é p p ù é p 3p ù
(d) [ a2 + b2 - (a2 - b2 )cos 2a ] (d) x Î ê , ú È ê ,
472 2 ë 4 2 û ë 2 4 úû
Type 3. Assertion and Reason
Directions (Q. Nos. 145-151) In the following 147. Statement I The numbers sin 18° and - sin 54°
9
Trigonometric Functions and Equations
questions, each question contains Statement I are equations of such quadratic equations with
(Assertion) and Statement II ( Reason). Each question integer coefficients.
has 4 choices (a), (b), (c) and (d ) out of which only one is Statement II If x = 18°, then 5x = 90° and if
correct. The choices are
y = - 54°, then 5 y = - 270°.
(a) Statement I is true, Statement II is true; Statement II
is a correct explanation for Statement I 148. Statement I cos 36° > tan 36°
(b) Statement I is true, Statement II is true; Statement II Statement II cos 36° > sin 36°
is not a correct explanation for Statement I
149. Statement I The minimum value of the expression
(c) Statement I is true, Statement II is false sin a + sin b + sin g , where a , b and g are real
(d) Statement I is false, Statement II is true numbers such that a + b + g = p, is negative.
145. Statement I If A , B and C are the angles of a Statement II a , b and g are angles of a triangle.
triangle such that ÐA is obtuse, then tan B tan C > 1.
150. Statement I The equation sin x + cosec x = 2
tan B + tan C é pù é 5p ù
Statement II In any triangle, tan A = possesses unique solution, if x Î ê 0, ú È ê 2p , ú.
tan B tan C - 1 ë 2û ë 2û
q 1
146. Statement I If = 1 + sin q + 1 - sin q ,
2sin Statement II x + ³ 2, x > 0 equality holds, only
2 x
q p 3p when x = 1.
then lies between 2np + and 2np + .
2 4 4
151. Statement I If (sin x ) 2 + (tan x ) 2 = 0, then x = np,
q p 3p q
Targ e t E x e rc is e s
Statement II If runs from to , then sin > 0. Statement II If a 2 + b 2 = 0, then a, b = 0.
2 4 4 2
Type 4. Linked Comprehension Based Questions
Passage I (Q. Nos. 152-154) For each positive real 154. Let the particle starts from the point A (2, 0) and
number k, let Ck denotes the circle with centre at origin p
moves units on circle C 2 in the counter-clockwise
and radius k units. On a circle Ck , particle a moves 2
k units in the counter-clockwise direction. After direction, then moves on circle C 3 along the
completing its motion on Ck , the particle moves onto the tangential path. Let this straight line (tangential path
circle Ck + g in some well defined manner, where g > 0. traced by particle) intersect the circle C 3 at points A
The motion of the particle continues in this manner. and B and tangents drawn at A and B intersect at
152. Let k Î I + and g = 1, particle moves in the radical (a) ( 2 , 2 ) (b) (9, 9)
direction from circle C k to C k + 1 . If particle starts æ 9 9 ö
(c) ç , ÷ (d) (9 2 , 9 2 )
from the point P ( - 1, 0), then è2 2 2 2ø
(a) it will cross the X-axis again at (3, 0)
Passage II (Q. Nos. 155-157) Let f (q, a ) = 2 sin2 q
(b) it will cross the X-axis again at (4, 0)
+ 4 cos(q + a ) sin q sin a + cos 2(q + a )
(c) it will cross the positive Y-axis at (0, 4) p
(d) it will cross the positive Y-axis at (0, 6) General solution of cos x = 0 is x = (2n + 1) : n ÎI
2
General solution of sin x = 0 is x = np : n Î I .
153. Let k Î I + and g Î R, particle moves tangentially æp pö
155. The value of f ç , ÷ is
from the circle C k to C k + y , such that length of è 3 4ø
tangent is equal to k unit itself. If particle starts from (a) 0 (b) 1 (c) 2 (d) 3
the point Q (1, 0), then
(a) the particle will cross positive X-axis again for æp ö
156. The general solution of f ç , a ÷ = 0 is
2 2<x<4 è3 ø
(b) the particle will cross positive X-axis again at x = 2 2 p p
(a) a = (2n + 1) , n Î I (b) a = (2n + 1) , n Î I
(c) the particle will cross X-axis again at x = 4 2 4
(d) the particle will cross X-axis again at x = 3 p
(c) a = (2n + 1)p , n Î I (d) a = (2n + 1) , n Î I
3
473
9 æ pö 1
157. If f çq , ÷ = , then
è 6ø 2
Therefore, sin x + sin y = 2 if and only if sin x = 1 and
sin y = 1 Û x = 2np +
2
p
and y = 2mp +
p
2
Objective Mathematics Vol. 1
(a) q = np : n Î I
np which is the required solution of equation. To solve the
(b) q = : n ÎI Eq. (i), we have used the boundedness of sin x rather
2
(c) q = 2np : n Î I than using conventional methods of solving equations.
(d) q is any real In general, we employ one or more of the following
Passage III (Q. Nos. 158-160) If A, B and C are angles extreme value conditions
of a triangle such that tan A + tan B + tan C = k , then find (i) - 1 £ sin x £ 1 Þ |sin x | £ 1 and sin2 x £ 1
the interval in which k should lie. (ii) - 1 £ cos x £ 1 Þ |cos x | £ 1 and cos 2 x £ 1
158. There exists exactly one isosceles DABC, if (iii) - a 2 + b 2 £ a sin x + b cos x £ a 2 + b 2
(a) k < 0 or k = 3 3
Þ | a sin x + b cos x | £ a 2 + b 2
(b) k > 0
(c) - 1 < k < 0
(d) 0 < k < 3 161. The minimum value of 27cos 2x × 81sin 2x is
1 1 1
159. There exist exactly two isosceles DABC, if (a) 1 (b) (c) (d)
9 81 243
(a) k < 3
(b) k > 3 3 162. Number of roots of the equation cos 7 x + sin 4 x = 1in
(c) k < 3 3 the interval [ 0, 2p ] is
(d) 0 < k < 3 3 (a) 0 (b) 1 (c) 2 (d) 4
160. There exist three or more non-similar isosceles 163. The smallest positive number p for which the
triangles for equation cos ( p sin x ) = sin ( p cos x ) has a solution in
(a) no value of k [ a, 2p ] , is
Ta rg e t E x e rc is e s
(b) 0 < k < 3 3 p
(a)
(c) k > 3 4
(d) k > 0 p
(b)
3
Passage IV (Q. Nos. 161-163) Use boundedness for p
solving trigonometric equations. Consider an equation (c)
4 2
sin x + sin y = 2 …(i) p
(d)
We know that, sin x £ 1, sin y £ 1 and sin x + sin y £ 2, " x 2 2
and y .
Type 5. Match the Columns
164. Match the items of Column I with that of Column II. 165. Match the items of Column I with that of Column II.
Column I Column II Column I Column II
A. If in a D ABC, p. right angled
A. The number of pairs ( x, y) satisfying p. 3
sin2 A + sin2 B = sin ( A + B), then
the equation sin x + sin y = sin( x + y)
| x | + | y | = 1, is the triangle must be
B. The number of values of x for which q. 8 B. If in a D ABC, q. equilateral
bc
f( x ) is valid, where = b 2 + c 2 - 2 bc cos A, then
1 - | x |ö 2 cos A
f( x ) = sec -1 æç ÷ , is the triangle must be
è 2 ø
C. If in a D ABC, r. isosceles
C. If x, y Î [0, 2 p ], then total number of r. ¥ A B C
tan + tan + tan = 3, then the
ordered pairs ( x, y) satisfying 2 2 2
sin x cos y = 1, is triangle must be
D. f( x ) = sin x - cos x - kx + b s. 6 D. If in a triangle, the sides and the s. obtuse angles
decreases for all values of real values altitudes are in AP, then the triangle
of x when 4 2 k is always greater than must be
474
166. Match the items of Column I with that of Column II.
Column I Column II
168. Match the items of Column I with that of Column II.
Column I Column II
9
Trigonometric Functions and Equations
A. The ratio of greatest value of p. 0
A. The number of solutions of p. 1 k
x sin x p 2 - cos x + sin2 x to its least value is ,
+ = in [-p, p ], is 4
2 cos x 4 then k equals
B. The number of solutions of equation q. 0 B. The equation cos 2 x + p sin x = 2 p - 7 q. 12
sin-1 (| x 2 - 1|) + cos -1 possesses a solution, then number of
p integral values of p satisfying equation is
(|2 x 2 - 5 |) = is
2 C. Ifcos a = 3, then value of16 sin a sin 3a is r. 13
4 2 2
C. The number of solutions of r. 3
p D. If a is a real constant and A, B, C are s. 5
x 4 - 2 x 2 sin2 x + 1 = 0 is variable angles and
2
a 2 - 4 tan A + a tan B
D. The number of solutions of s. 2
+ a 2 - 4 tanC = 6a, then least value of
x 2 + 2 x + 2 sec 2 px + tan2 px = 0 is
tan2 A + tan2 B + tan2 C is
167. Match the items of Column I with that of Column II. 169. Match the items of Column I with that of Column II.
Column I Column II
Column I Column II
A. sin x cos 3 x > cos x p. é - p,- 3p ù
êë 4 úû 3 p. 0
sin3 x, 0 £ x £ 2 p, is A. If cos A = , then the value of
p p 4
È é- , ù A 5A
êë 4 4 úû sin sin is
3p 2 2
È é , pù
Targ e t E x e rc is e s
êë 4 úû
p 2p 3p q. 11
B. The value of sin + sin + sin is
B. 2
4 sin x - 8 sin x + 3 £ 0, q. é 3p , 2 p ù È { 0} 7 7 7 32
0 £ x £ 2 p, is êë 2 úû
C. If tan2 q = 2 tan2 f + 1, then r. 1 p
cot
C. |tan x | £ 1and x Î [- p, p ], is r. æ 0, p ö
ç ÷ cos 2 q + sin2 f equals 2 14
è 4ø
1 tan( A + B) s. 3
D. cos x - sin x ³ 1 and s. é p , 5p ù D. If sin B = sin(2 A + B), then
êë 6 6 úû 5 tan A 2
0 £ x £ 2 p, is
is equal to
170. Match the items of Column I with that of Column II.
Column I Column II
A. The number of solutions of the equation|tan2 x | = sin x, x Î[0, p ] is p. 1
p p p p q. 4
B. The value of 4 tan - 4 tan3 + 6 tan2 - tan4 - 1 is
16 16 16 16
p 4
C. If the equation tan ( p cot x ) = cot ( p tan x ) has a solution in ( 0, p ) - ìí üý, then pmax is equal to r. 3
î2 þ p
2x 2 2 2 2
s. 2
D. The value of in [0, 2 p ], if 5cos 2x + 2sin x + 5 2cos x + sin 2x = 126 has a solution, is
p
Type 6. Single Integer Answer Type Questions
171. The number of real solutions of p
3 - tan 2
[ x] 2 - 5[ x] + 6 - sin x = 0, where [ ] denotes the 173. If 7 = l cos p , then l is ________ .
p 7
greatest integer function, is . 1 - tan 2
7
2sin 2q
172. If f ( nq ) = and
cos 2q - cos 4nq 174. The number of integral values of a such that
sin x cos 3x - a cos x sin 3x = 0 does not have any real
sin lq p
f (q ) + f ( 2q ) + f ( 3q ) + ¼ + f ( nq ) = , root other than ( 2n + 1) , n Î I for any real value of
sin q sin mq 2
then the value of m - l is . x, is ______. 475
9 175. The expression n sin 2 q + 2n cos (q + a ) sin a sin q
+ cos 2 (a + q ) is independent of q, the value of n is
182. If the values of x and y satisfy the equation
tan 2 ( x + y ) + cot 2 ( x + y ) = 1 - 2x - x 2 , then the
value of x 2 - 3x + 2 is ______.
Objective Mathematics Vol. 1
________.
176. For any real value of q ¹ p, the value of the 183. If the smallest positive values of x and y satisfying
2 p lp mp
cos q - 1 x - y = and cot x + cot y = 2are x = and y = ,
expression y = , y Î R - [ a, b] , then 4 12 6
cos 2 q + cos q then l + m is equal to ______.
the value of ( b - a ) is _______.
184. The system of simultaneous equations for x and y,
177. The number of solutions of equation 1
æ pö
[sin x] = [1 + sin x] + [1 - cos x], 0 £ x £ 2p, is _____. where x, y Î ç 0, ÷ , are 4 sin x + 3 cos y = 11 and
è 2ø
p 1
178. In a DABC, ÐB = and sin A sin C = l, then the set
3 5× 16sin x - 2× 3 cos y = 2 .
of all possible values of l Î[ a, b], then ( b - a ) is p
_______. If x + y = , then l is equal to _______.
l
A B C
179. In a DABC, cot cot cot ³ l 3, then the value 185. The number of solutions for
2 2 2
æ pö æ 3p ö ü
of l is _____. sin ç x - ÷ - cos ç x + ÷ = 1ï
è 4ø è 4 ø ï in (0, 2p ) is _____ .
A B C ý
180. If in a DABC, tan , tan , tan are in HP and 2cos 7x cos 2x
2 2 2 >2 ï
B cos 3 + sin 3 ïþ
cot ³ l , then the value of l is _______.
2 186. The integral value of p for which
Ta rg e t E x e rc is e s
181. If a and b satisfy the equation 12sin a + 5cos a p cos x - 2sin x = 2 + 2 - p
= 2 b 2 - 8 b + 21, then the value of b 2 - 2 b is
has a solution, is _______.
_______.
Entrances Gallery
JEE Advanced/IIT JEE
1. For x Î ( 0, p ), the equation sin x + 2sin 2x - sin 3x = 3 4. Let f : ( -1, 1) ® IR be such that
has [2014] 2 æ pö æp pö
f (cos 4q ) = for q Î ç 0, ÷ È ç , ÷.
(a) infinitely many solutions 2 - sec q2 è 4 ø è 4 2ø
(b) three solutions
(c) one solution æ 1ö
Then, the value(s) of f ç ÷ is/are [2012]
(d) no solution è 3ø
3 3 2 2
2. The number of points in ( - ¥ , ¥ ) , for which (a) 1 - (b) 1 + (c) 1 - (d) 1 +
2 2 3 3
x 2 - x sin x - cos x = 0, is [2013]
5. The positive integer value of n > 3 satisfying the
(a) 6 (b) 4
(c) 2 (d) 0 1 1 1
equation = + is ______.
æpö æ 2p ö æ 3p ö
3. Let q, f Î[ 0, 2p ] be such that sin ç ÷ sin ç ÷ sin ç ÷
ènø è nø è nø [2011]
2 æ q qö
2cos q (1 - sin f ) = sin q ç tan + cot ÷ cos f - 1,
è 2 2ø 6. Let f ( x ) = x 2 and g ( x ) = sin x, for all x Î R. Then, the
3 set of all x satisfying ( fogogof) ( x ) = (gogof) ( x ),
tan ( 2p - q ) > 0 and - 1 < sin q < - . Then, f where ( fog) ( x ) = f [ g ( x )] , is [2011]
2
(a) ± np , n Î {0, 1, 2, ¼}
cannot satisfy [2012]
p p 4p (b) ± np , n Î { 1, 2, ¼}
(a) 0 < f < (b) < f < p
2 2 3 (c) + 2np , n Î {..., - 2, - 1, 0, 1, 2, ¼}
4p 3p 3p 2
(c) <f< (d) < f < 2p (d) 2np , n Î {..., - 2, - 1, 0, 1, 2, ¼}
3 2 2
476
7. The number of all possible values of q, where
0 < q < p, for which the system of equations
æ p pö
8. The number of values of q in the interval ç - , ÷
np
è 2 2ø 9
Trigonometric Functions and Equations
( y + z ) cos 3q = ( xyz ) sin 3q such that q ¹ for n = 0 , ±1, ± 2 and tan q = cot 5q
5
2cos 3q 2sin 3q
x sin 3q = + as well as sin 2q = cos 4q , is . [2010]
y z
9. The maximum value of the expression
( xyz )sin 3q = ( y + 2z )cos 3q + y sin 3q 1
is . [2010]
have a solution ( x 0 , y 0 , z 0 ) with y 0 z 0 ¹ 0, is . sin 2 q + 3sin q cos q + 5cos 2 q
[2010]
JEE Main/AIEEE
1 17. The number of values of x in the interval [ 0, 3p ]
10. Let f k ( x ) = (sin k x + cos k x ) , where, x Î R and
k satisfying the equation 2sin 2 x + 5sin x - 3 = 0 is
k ³ 1. Then, f 4 ( x ) - f 6 ( x ) equals [2014]
[2006]
1 1 1 1
(a) (b) (c) (d) (a) 6 (b) 1 (c) 2 (d) 4
6 3 4 12
tan A cot A 18. A triangular park is enclosed on two sides by a fence
11. The expression + can be written and on the third side by a straight river bank. The
1 - cot A 1 - tan A
two sides having fence are of same length x. The
as [2013]
maximum area enclosed by the park is [2006]
(a) sin A cos A + 1 (b) sec A cosec A + 1
(c) tan A + cot A (d) sec A + cosec A x3 1 3
(a) (b) x 2 (c) px 2 (d) x 2
8 2 2
12. In a DPQR, if 3sin P + 4 cos Q = 6 and
Targ e t E x e rc is e s
4 sin Q + 3cos P = 1, then ÐR is equal to [2012]
1
19. If 0 < x < p and cos x + sin x = , then tan x is equal to
5p p p 3p 2
(a) (b) (c) (d)
6 6 4 4 [2006]
(4 - 7 ) (4 + 7 )
13. If A = sin 2 x + cos 4 x, then for all real x [2011] (a)
3
(b) -
3
13 (1 + 7 ) (1 - 7 )
(a) £ A £1 (b) 1 £ A £ 2 (c) (d)
16 4 4
3 13 3
(c) £ A £ (d) £ A £1
4 16 4 20. Let a and b be such that p < a - b < 3p. If
21 27
14. The possible values of q Î ( 0, p ) such that sin a + sin b = - and cos a + cos b = - , then
65 65
sin q + sin 4q + sin 7q = 0 are [2011]
æa - b ö
2p p 4 p p 3p 8p the value of cos ç ÷ is [2004]
(a) , , , , , è 2 ø
9 4 9 2 4 9
p 5p p 2p 3p 8p 3 3
(b) , , , , , (a) - (b)
4 12 2 3 4 9 130 130
2p p p 2p 3p 35p 6 6
(c) , , , , , (c) (d) -
9 4 2 3 4 36 65 65
2p p p 2p 3p 8p
(d) , , , , ,
9 4 2 3 4 9 21. If f : R ® S , defined by f ( x ) = sin x - 3 cos x + 1, is
4 5 onto, then the interval of S is [2004]
15. Let cos (a + b ) = and sin (a - b ) = , where (a) [ 0, 3 ] (b) [ -1, 1]
5 13 (c) [ 0, 1] (d) [ -1, 3 ]
p
0 £ a ,b £ . Then, tan 2a is equal to [2010] 4xy
25
4
56 19 20
22. sin 2 q = is true if and only if [2002]
(a) (b) (c) (d) (x + y )2
16 33 12 7
(a) x - y ¹ 0
16. Let A and B denote the statements (b) x = - y
A :cos a + cos b + cos g = 0 (c) x + y ¹ 0
B :sin a + sin b + sin g = 0 (d) x ¹ 0, y ¹ 0
3
If cos (b - g ) + cos ( g - a ) + cos (a - b ) = - , then 1 - tan 2 15°
2 [2009] 23. The value of is [2002]
1 + tan 2 15°
(a) A is true and B is false (b) A is false and B is true
(c) both A and B are true (d) both A and B are false (a) 1 (b) 3 (c) 3 / 2 (d) 2 477
9 4
4
24. If tan q = - , then sin q is
3
4
[2002] 26. If y = sin 2 q + cosec 2 q, q ¹ 0, then
(a) y = 0
(c) y ³ - 2
(b) y £ 2
(d) y ³ 2
[2002]
Objective Mathematics Vol. 1
(a) - but not
5 5
4 4 27. The equation a sin x + b cos x = c, where |c | > a 2 + b 2
(b) - or
5 5 has [2002]
4 4
(c) but not - (a) a unique solution
5 5 (b) infinite number of solutions
(d) None of the above (c) no solution
1 (d) None of the above
25. If sin (a + b ) = 1 and sin (a - b ) = , then
2 28. If a is a root of 25cos 2 q + 5cos q - 12 = 0,
tan (a + 2b ) tan ( 2a + b ) is equal to [2002] p
(a) 1 < a < p , then sin 2a is equal to [2002]
2
(b) -1
(c) 0 24 24 13 13
(a) (b) - (c) (d) -
(d) None of the above 25 25 18 18
Other Engineering Entrances
¥
æn !pö 34. tan 81° - tan 63° - tan 27° + tan 9° equals
29. The sum of the series å sin çè 720 ÷ø is [WB JEE 2014]
[EAMCET 2014]
n=1
æ p ö æ p ö æ p ö (a) 6 (b) 0 (c) 2 (d) 4
(a) sin ç ÷ + sin ç ÷ + sin ç ÷
è 180 ø è 360 ø è 540 ø 35. If sin A + cos A = 2, then the value of cos 2 A is
æpö æpö æ p ö æ p ö
(b) sin ç ÷ + sin ç ÷ + sin ç ÷ + sin ç ÷ [RPET 2014]
Ta rg e t E x e rc is e s
è 6ø è 30 ø è 120 ø è 360 ø
(a) 2 (b) 1/2 (c) 4 (d) -1
æpö æpö æ p ö
(c) sin ç ÷ + sin ç ÷ + sin ç ÷
è 6ø è 30 ø è 120 ø a b
36. If tan and tan are the roots of 8x 2 - 26x + 15 = 0 ,
æ p ö æ p ö 2 2
+ sin ç ÷ + sin ç ÷
è 360 ø è 720 ø then cos (a + b ) is equal to [Manipal 2014]
æ p ö æ p ö 627 627
(d) sin ç ÷ + sin ç ÷ (a) (b) -
è 180 ø è 360 ø 726 725
(c) -1 (d) None of these
30. If cos x = tan y, cot y = tan z and cot z = tan x, then
3 3p x
sin x is equal to [EAMCET 2014] 37. If tan x = , p < x < , then the value of cos is
4 2 2
5+1 5 -1 5+1 5 -1
(a) (b) (c) (d) [Karnataka CET 2014]
4 4 2 2 1 3 1 3
(a) - (b) (c) (d) -
31. The value of 10 10 10 10
2cot 2 ( p / 6) + 4 tan 2 ( p / 6) - 3cosec ( p / 6) is 2p 4p 6p
38. cos + cos + cos [WB JEE 2014]
[J&K CET 2014] 7 7 7
(a) 2 (b) 4 (c) 4/3 (d) 3/4 (a) is equal to zero (b) lies between 0 and 3
(c) is a negative number (d) lies between 3 and 6
32. Find the value of cos (29 p / 3). [J&K CET 2014]
p 2p 4p
(a) 1 (b) 0 (c) 3 / 2 (d) 1 / 2 39. The value of tan + 2 tan + 4 cot is
5 5 5
33. If sin q = sin a , then [Karnataka CET 2014] [WB JEE 2014]
q+ a p q-a p 2p 4p 3p
(a) is any odd multiple of and is any (a) cot (b) cot (c) cot (d) cot
2 2 2 5 5 5 5
multiple of p 3
q+ a q-a 40. If x and y are acute angles such that cos x + cos y =
(b) is any odd multiple of p and is any 2
2 2
multiple of p
3
and sin x + sin y = , then sin ( x + y ) equals
q+ a p q-a 4
(c) is any multiple of and is any even
2 2 2 [Karnataka CET 2014]
multiple of p 2 3
q+ a p q-a (a) (b)
(d) is any even multiple of and is any odd 5 4
2 2 2 3 4
multiple of p (c) (d)
478 5 5
41. Find the value of cos ( x / 2), if tan x = 5 / 12 and x lies
in third quadrant. [J&K CET 2014]
1
52. If cos a + 2cos b + 3cos g = 0, sin a + 2sin b
+ 3sin g = 0 and a + b + g = p, then 9
Trigonometric Functions and Equations
(a) 5 13 (b) 5 / 26 (c) 5 / 13 (d) sin 3 a + 8sin 3 b + 27 sin 3g is equal to
26 [Karnataka CET 2012]
(a) -18 (b) 0 (c) 3 (d) 9
42. Let n be a positive integer such that
p p n æ A + Bö æB - Aö
sin + cos = , then [AMU 2014] 53. If cos A = mcos B and cot ç ÷ = l/tan ç ÷,
è 2 ø è 2 ø
2n 2n 2
(a) 6 £ n £ 8 (b) 4 < n £ 8 then l is equal to [Manipal 2012]
(c) 4 £ n < 8 (d) 4 < n < 8 m m+1
(a) (b)
m -1 m
43. The sum of the solutions in ( 0, 2p ) of the equation m+1
(c) (d) None of these
æp ö æp ö 1 m -1
cos x cos ç - x÷ cos ç + x÷ = is [EAMCET 2014]
è3 ø è3 ø 4
(a) 4p (b) p (c) 2p (d) 3p 54. The least value of 3sin 2 q + 4 cos 2 q is [OJEE 2012]
(a) 2 (b) 3 (c) 0 (d) 1
44. Which one of the following is not correct?
[AMU 2013] 55. If sin 2x = 4 cos x, then x is equal to
(a) |sin x | £ 1 [Karnataka CET 2012]
(b) - 1 £ cos x £ 1 np p
(a) ± , n ÎZ (b) no value
(c) |sec x | < 1 2 4
(d) cosec x ³ 1 or cosec x £ - 1 p p
(c) np + (-1)n , n Î Z (d) 2np ± , n ÎZ
p 3 4 4 2
45. If 0 < b < a < , cos (a + b ) = and cos (a - b ) = ,
4 5 5 56. Number of solutions of the equation
then sin 2a is equal to [Karnataka CET 2013] tan x + sec x = 2cos x, x Î[ 0, p ] is [WB JEE 2012]
Targ e t E x e rc is e s
(a) 0 (b) 1 (a) 0 (b) 1
(c) 2 (d) None of these (c) 2 (d) 3
46. The most general solutions of the equation 57. The value of q satisfying cos q + 3 sin q = 2 is
sec 2 x = 2 (1 - tan 2 x ) are given by [BITSAT 2013] [OJEE 2012]
p p (a) 30° (b) 60°
(a) np ± (b) 2np +
4 4 (c) 45° (d) 90°
p
(c) np ± (d) None of these 2t
8 58. If sin q = and q lies in second quadrant, then
1+ t 2
47. The number of solutions of the equation (1 - cos 2x ) cos q is equal to [WB JEE 2011]
(cos 2x + cos 2 x ) = 0, 0 £ x £ 2p is [Manipal 2013] 1 - t2 t2 - 1
(a) (b)
(a) 3 (b) 2 (c) 1 (d) 0 1 + t2 1 + t2
48. If log cos x tan x + log sin x cot x = 0, then the most - |1 - t 2 | 1 + t2
(c) (d)
general solution of x is [UP SEE 2013] 1 + t2 |1 - t 2 |
3p p
(a) 2np - , n ÎZ (b) 2np + , n ÎZ 59. If sin q = 3sin(q + 2a ), then the value of
4 4
p tan(q + a ) + 2 tan a is [Kerala CEE 2011]
(c) np - , n Î Z (d) None of these
4 (a) 3 (b) 2 (c) -1
(d) 0 (e) 1
49. The number of real solutions of the equation
x 3 + x 2 + 4x + 2sin x = 0 in 0 £ x £ 2p is [OJEE 2013] 60. If a , b , g Î [ 0, p ] and a , b , g are in AP, then
(a) 4 (b) 2 (c) 1 (d) 0 sin a - sin g
is equal to [Kerala CEE 2011]
cos g - cos a
50. The general solution of sin 3x + sin x - 3sin 2x
(a) sinb (b) cosb (c) cotb
= cos 3x + cos x - 3cos 2x is [AMU 2013]
(d) cosec b (e) 2cosb
np p np p
(a) + for n integer (b) - for n integer
2 8 2 8 61. If A , B and C are the angles of a triangle such that
p p
(c) np + for n integer (d) np - for n integer sec ( A - B ), sec A and sec( A + B ) are in AP, then
6 6 [UP SEE 2011]
2 B 2 B
51. If sin A + sin B + sin C = 3, then (a) cosec A = 2cosec (b) 2sec A = sec2
2
2 2
cos A + cos B + cos C is equal to [MP PET 2012] B A
(a) 3 (b) 2 (c) 1 (d) 0 (c) 2cosec2 A = cosec2 (d) 2sec2 B = sec2 479
2 2
9 62. If
1
sin q =
2
and tan q =
general value of q is
1
3
, " n Î I, then most
[BITSAT 2011]
71. The value of
1
sin 55° - cos 55°
sin 10°
is [VITEEE 2010]
Objective Mathematics Vol. 1
(a) (b) 2
p p 2
(a) 2np + ," n Î I (b) 2np + ," n Î I
6 4 (c) 1 (d) 2
p p
(c) 2np + ," n Î I (d) 2np + ," n Î I æ pö 4 12
3 5 72. If a , b Î ç 0, ÷, sin a = and cos (a + b ) = - ,
è 2ø 5 13
63. A value of q satisfying sin 5q - sin 3q + sin q = 0
then sin b is equal to [Kerala CEE 2010]
p 63 61
such that 0 < q < is [Karnataka CET 2011] (a) (b)
2 65 65
p p p p 3 5
(a) (b) (c) (d) (c) (d)
12 6 4 2 5 13
8
64. The number of solutions of 2sin x + cos x = 3 is (e)
65
[WB JEE 2011] b p
(a) 1 (b) 2 73. If tan a = , a > b > 0 and 0<a < , then
(c) infinite (d) no solution a 4
a+b a-b
65. If sin q + cos q = 0 and 0 < q < p, then q is equal to - is equal to [Kerala CEE 2010]
a-b a+b
[WB JEE 2011] 2 sin a 2 cos a
p p 3p (a) (b)
(a) 0 (b) (c) (d) cos 2a cos 2a
4 2 4
2 sin a 2 cos a
(c) (d)
66. The solution of trigonometric equation sin 2a sin 2a
cos 4 x + sin 4 x = 2cos ( 2x + p )cos ( 2x - p ) is 2 tan a
Ta rg e t E x e rc is e s
(e)
[UP SEE 2011] cos 2a
np æ 1ö cot x - tan x
(a) x = ± sin -1 ç ÷ 74. The value of is [WB JEE 2010]
2 è 5ø
cot 2x
np (-1)n æ ±2 2ö (c) -1
(b) x = + sin -1 ç ÷ (a) 1 (b) 2 (d) 4
4 4 è 3 ø
np -1 1
75. The number of points of intersection of 2 y = 1 and
(c) x = ± cos
2 5 y = sin x in - 2p £ x £ 2p is [VITEEE 2010]
np (-1)n æ 1ö (a) 1 (b) 2 (c) 3 (d) 4
(d) x = - cos-1 ç ÷
4 4 è 5ø
æ pö
76. The value of x in ç 0 , ÷ satisfying the equation
sin x cos x cos x è 2ø
67. If cos x sin x cos x = 0 , then the number of 1
sin x cos x = is [Kerala CEE 2010]
cos x cos x sin x 4
p p p p
distinct real roots of this equation in the interval (a) (b) (c) (d)
6 3 8 4
- p / 2 < x < p / 2 is [AMU 2011] p
(e)
(a) 1 (b) 3 (c) 2 (d) 4 12
cos A cos B 1 p p 77. The number of solutions of cos 2q = sin q in ( 0, 2p ) is
68. If = = - < A < 0 and - < B < 0,
3 4 5 2 2 [Kerala CEE 2010]
then the value of 2sin A + 4 sin B is [BITSAT 2010] (a) 1 (b) 2 (c) 3 (d) 4
(a) 4 (b) -2 (c) -4 (d) 0 (e) 0
cot 54° tan 20° 78. If sin 6q + sin 4q + sin 2q = 0, then the general values
69. The value of + is [VITEEE 2010]
tan 36° cot 70° of q are [WB JEE 2010]
(a) 0 (b) 2 np p
(c) 3 (d) 1 (a) , np ±
4 3
np p
70. The value of tan 40° + tan 20° + 3 tan 20° tan 40° (b) , np ±
4 6
is [BITSAT 2010] np p
1 (c) , 2np ±
(a) 12 (b) 4 3
3 np p
(d) , 2np ±
(c) 2 (d) 3 4 6
480
Answers
Work Book Exercise 9.1
1. (a) 2. (b) 3. (b) 4. (a) 5. (a) 6. (b) 7. (b) 8. (a) 9. (a) 10. (d)
11. (d) 12. (d)
Work Book Exercise 9.2
1. (a) 2. (d) 3. (b) 4. (a) 5. (c) 6. (c) 7. (a) 8. (b) 9. (b) 10. (b)
11. (c) 12. (d) 13. (a) 14. (c) 15. (a) 16. (a) 17. (b) 18. (b) 19. (a) 20. (d)
Work Book Exercise 9.3
1. (c) 2. (d) 3. (c) 4. (a) 5. (c) 6. (b) 7. (c) 8. (c) 9. (d) 10. (c)
Work Book Exercise 9.4
1. (b) 2. (c) 3. (c) 4. (d) 5. (a) 6. (c)
Target Exercises
1. (a) 2. (c) 3. (d) 4. (a) 5. (b) 6. (d) 7. (a) 8. (b) 9. (c) 10. (d)
11. (c) 12. (d) 13. (c) 14. (b) 15. (b) 16. (d) 17. (d) 18. (c) 19. (b) 20. (d)
21. (d) 22. (c) 23. (a) 24. (a) 25. (a) 26. (b) 27. (b) 28. (b) 29. (a) 30. (d)
Targ e t E x e rc is e s
31. (a) 32. (d) 33. (a) 34. (b) 35. (c) 36. (b) 37. (b) 38. (c) 39. (b) 40. (a)
41. (c) 42. (b) 43. (b) 44. (c) 45. (c) 46. (a) 47. (a) 48. (b) 49. (c) 50. (a)
51. (c) 52. (d) 53. (b) 54. (a) 55. (d) 56. (b) 57. (b) 58. (c) 59. (b) 60. (a)
61. (c) 62. (a) 63. (c) 64. (b) 65. (b) 66. (d) 67. (c) 68. (d) 69. (c) 70. (a)
71. (a) 72. (a) 73. (b) 74. (b) 75. (c) 76. (d) 77. (a) 78. (c) 79. (a) 80. (a)
81. (c) 82. (c) 83. (b) 84. (c) 85. (d) 86. (b) 87. (c) 88. (c) 89. (a) 90. (c)
91. (c) 92. (b) 93. (a) 94. (b) 95. (a) 96. (b) 97. (d) 98. (d) 99. (a) 100. (a)
101. (a) 102. (a) 103. (d) 104. (a) 105. (b) 106. (a) 107. (a) 108. (d) 109. (b) 110. (c)
111. (c) 112. (d) 113. (c) 114. (d) 115. (b) 116. (a) 117. (b) 118. (a) 119. (c) 120. (c)
121. (d) 122. (b) 123. (b) 124. (a) 125. (b) 126. (b) 127. (c) 128. (c) 129. (c) 130. (all)
131. (b,c) 132. (a,b) 133. (a,c,d) 134. (a,c,d) 135. (b,c) 136. (all) 137. (all) 138. (a,b) 139. (b,c) 140. (a,c)
141. (a,d) 142. (b,c) 143. (b,d) 144. (b,d) 145. (d) 146. (b) 147. (a) 148. (b) 149. (c) 150. (d)
151. (c) 152. (b) 153. (a) 154. (c) 155. (a) 156. (b) 157. (d) 158. (a) 159. (b) 160. (a)
161. (d) 162. (d) 163. (d) 164. (*) 165. (*) 166. (*) 167. (*) 168. (*) 169. (*) 170. (*)
171. (1) 172. (1) 173. (4) 174. (3) 175. (2) 176. (2) 177. (0) 178. (1) 179. (3) 180. (3)
181. (0) 182. (6) 183. (6) 184. (2) 185. (1) 186. (2)
* 164. A ® s; B ® r; C ® p; D ® q 165. A ® p; B ® r; C ® q; D ® q 166. A ® r; B ® s; C ® s; D ® q
167. A ® r; B ® s; C ® p; D ® q 168. A ® r; B ® s; C ® s; D ® q 169. A ® q; B ® r; C ® p; D ® s
170. A ® q; B ® s; C ® p; D ® p,r
Entrances Gallery
1. (d) 2. (c) 3. (a,c,d) 4. (a,b) 5. (7) 6. (a) 7. (3) 8. (3) 9. (2) 10. (d)
11. (b) 12. (b) 13. (d) 14. (a) 15. (b) 16. (c) 17. (d) 18. (b) 19. (b) 20. (a)
21. (d) 22. (c) 23. (c) 24. (b) 25. (a) 26. (d) 27. (c) 28. (b) 29. (c) 30. (d)
31. (c) 32. (d) 33. (a) 34. (d) 35. (b) 36. (b) 37. (a) 38. (c) 39. (a) 40. (d)
41. (d) 42. (d) 43. (c) 44. (c) 45. (b) 46. (c) 47. (d) 48. (b) 49. (c) 50. (a)
51. (d) 52. (b) 53. (c) 54. (b) 55. (d) 56. (c) 57. (b) 58. (c) 59. (d) 60. (c)
61. (b) 62. (a) 63. (b) 64. (d) 65. (d) 66. (b) 67. (c) 68. (c) 69. (b) 70. (d)
71. (d) 72. (a) 73. (a) 74. (b) 75. (d) 76. (e) 77. (c) 78. (a)
481
Explanations
Target Exercises
π 2π 4π 8π π 3π
1. tan + 2 tan + 4 tan + 8 tan ⇒ 0 < (β − α) < ⇒ < (β − α) < π
3 3 3 3 4 4
π π π π 3π
= tan + 2 tan π − + 4 tan π + ⇒ ( β − α ) ∈ 0, ⇒ ( β − α) ∈ , π
3 3 3 4 4
π 3 cos 20 ° − sin 20 °
+ 8 tan 3π − 8. 3 cosec 20 ° − sec 20 ° =
3 sin 20 ° cos 20 °
π π π π 4 ( 3 / 2 cos 20 ° − 1 / 2 sin 20 ° ) 4 ⋅ sin 40 °
= tan − 2 tan + 4 tan − 8 tan = = =4
3 3 3 3 2 ⋅ sin 20 ° cos 20 ° sin 40 °
π
= − 5 tan = − 5 3 1
3 9. We have, sin 15° = (1 − cos 30 ° )
1 1 4 2
2. sec 2 θ + cosec 2 θ = + = ≥4
cos θ sin θ sin 2 2θ
2 2 1 3 1
= 1 − = 2− 3
4 2 2 2
Also, sec 2 θ cosec 2 θ = ≥4
sin 2 2θ which is an irrational number.
Thus, required quadratic equation will be 1
cos 15° = 2 + 3 is also an irrational number.
x 2 − t x + t = 0, where t ≥ 4 2
Hence, x 2 − 5 x + 5 = 0 can be the required equation. 1
sin 15° cos 15° = sin 30 ° =
1
3π 4π 2 4
3. cos 2 + cos 2 = cos 2 108° + cos 2 144°. which is a rational number.
5 5
Further, sin 15° cos 75° = sin 15° cos(90 ° − 15° )
Ta rg e t E x e rc is e s
2 2
5 − 1 5 + 1 1
= sin 2 18° + cos 2 36° = + = sin 15° sin 15° = (2 − 3 )
4 4 4
2(5 + 1) 3 which is an irrational number.
= =
16 4 10. We have,
4. sin (2 × 360 ° − 120 ° )cos (360 ° − 30 ° ) π 4π 5π π 2π 4π
cos cos cos = − cos cos cos
+ cos 120 ° sin (180 ° − 30 ° ) 7 7 7 7 7 7
5π 2π 2π
= − sin 120 ° cos 30 ° + cos 120 ° sin 30 ° As cos = cos π − = − cos
7 7 7
= − sin (90 ° + 30 ° )cos 30 ° + cos (90 ° + 30 ° )sin 30 °
= − cos 2 30 ° − sin 2 30 ° 1 π π 2π 4π
=− 2 sin cos cos cos
π 7
= −1 2 sin 7 7 7
7
5. Given expression 1 4π 4π 1 8π
1 =− sin cos =− sin
1 − 2 − cos 80 ° π π 7
1 − 2(cos 60 ° − cos 80 ° ) 2
2
2 sin 7 7 8 sin
= = 7 7
2 sin 10 ° 2 sin 10 ° 1 π 1 π 1
2 cos 80 ° cos(90 ° − 10 ° ) sin 10 ° =− sin π + = − − sin =
= = = =1 π 7 π 7 8
2 sin 10 ° sin 10 ° sin 10 ° 8 sin 8 sin
7 7
6. Given expression 11. 6 (sin 6 θ + cos 6 θ ) − 9 (sin 4 θ + cos 4 θ ) + 4
3 1
= 2 sin 75° − cos 75° = 6 [(sin 2 θ + cos 2 θ )3 − 3 sin 2 θ cos 2 θ (sin 2 θ + cos 2 θ )]
2 2
− 9 [(sin 2 θ + cos 2 θ )2 − 2 sin 2 θ cos 2 θ ] + 4
= 2(cos 30 ° sin 75° − sin 30 ° cos 75° )
1 = 6 [1 − 3 sin 2 θ cos 2 θ ] − 9 [1 − 2 sin 2 θ cos 2 θ ] + 4
= 2 sin (75° − 30 ° ) = 2 sin 45° = 2 × = 2 = 6− 9+ 4=1
2
2 sin α
7. Given, sin α =
1
and sin β =
3 12. Given, =x
5 5 1 + cos α + sin α
2 4 2 sin α (1 − cos α − sin α )
⇒ cos α = and cos β = ⇒ =x
5 5 (1 + cos α + sin α )(1 − cos α − sin α )
∴ sin(β − α ) = sin β cos α − cos β sin α 2 sin α (1 − cos α − sin α )
3 2 4 1 2 ⇒ =x
= − = 1 − sin 2 α − cos 2 α − 2 sin α cos α
5 5 5 5 5 5
1 1 − cos α − sin α
Clearly, 0 < sin( β − α ) < ⇒ =−x
482 2 cos α
13. Since, sec θ ∉ (− 1, 1)
∴ sec θ ≠
1
2
2
π
21. cos y cos π
2
π
− x − cos − y cos x
π
+ sin y cos − x + cos x sin − y = 0
9
Trigonometric Functions and Equations
2 2
14. Since,log(sin 90° )is a factor in the given expression and
the value of which is log(1) = 0. Hence, the value of the ⇒ cos y sin x − cos x sin y
given expression is 0. + sin y sin x + cos x cos y = 0
⇒ sin( x − y ) + cos( x − y ) = 0
3π
15. We have, 3 sin 4
− α + sin 4 (3π + α ) ⇒ tan( x − y ) = − 1
2 π
⇒ x = nπ − + y
6 π 4
−2 sin + α + sin 6 (5π − α )
2 11
22. cosec A + cot A =
= 3 [(− cos α )4 + (− sin α )4 ] − 2[cos 6 α + sin 6 α ] 2
= 3 [(cos 2 α + sin 2 α )2 − 2 sin 2 α cos 2 α ] ⇒ cosec 2 A − cot 2 A = 1
− 2[(cos 2 α + sin 2 α )3 − 3 cos 2 α sin 2 α 2
⇒ cosec A − cot A =
(cos 2 α + sin 2 α )] 11
117 44
= 3 − 6 sin α cos α − 2 + 6 sin α cos 2 α = 3 − 2 = 1
2 2 2 ⇒ 2 cot A = ⇒ tan A =
22 117
1 sin 3 x cos 3 x
16. Since, tan x + =2 23. Let + = A, then
tan x 1 + cos x 1 − sin x
π
⇒ tan x = 1 ⇒ x= (sin 3 x + cos 3 x ) + (cos 4 x − sin 4 x )
4 A=
1 1 (1 + cos x )(1 − sin x )
∴ sin x = and cos x =
2 2 {(sin 3 x + cos 3 x )}{(cos x + sin x )
1 1 1 (cos x − sin x )(cos 2 x + sin 2 x )}
Hence, sin x + cos x = n + n = n − 1
2n 2n
⇒ A=
2 2 2 (1 + cos x )(1 − sin x )
Targ e t E x e rc is e s
15 12 (sin x + cos x ){(1 − sin x cos x )
17. Given, sin α = , tan β =
17 5 + (cos x − sin x )}
π 3π ⇒ A=
Since, < α < π , π < β < 1 + cos x − sin x − sin x cos x
2 2
8 12 ⇒ A = sin x + cos x
∴ cos α = − , sin β = − 1 1
17 13 ⇒ A= 2 sin x + cos x ...(i)
2 2
5
and cos β = − π π
13 ⇒ A = 2 cos sin x + sin cos x
4 4
Now, sin(β − α ) = sin β cos α − cos β sin α
12 −8 −5 15 π
=− = 2 sin + x
− 4
13 17 13 17
96 75 171 Again, by Eq. (i), we get
= + = π π π
221 221 221 A = 2 sin sin x + cos cos x = 2 cos −x
4 4
4
18. cos 2 (α − β) + 2 ab sin(α − β)
= cos 2 (α − β ) + 2 cos(θ − α )sin(θ − β )sin(α − β ) 24. Given expression = cosec A sin(B + C )
= cos (α − β ) + [sin{2θ − (α + β )}
2 = cosec A ⋅ sin(π − A) [Q A + B + C = π]
= cosec A ⋅ sin A = 1
+ sin (α − β )]sin(α − β )
3 4
= 1 + sin{2θ − (α + β )} sin(α − β ) 25. cos( A − B) = ⇒ tan( A − B) =
= 1 + sin{(θ − α ) + (θ − β )} sin{(θ − β ) − (θ − α )} 5 3
tan A − tan B 4
= 1 + sin 2 (θ − β ) − sin 2 (θ − α ) ⇒ =
1 + tan A tan B 3
= cos 2 (θ − α ) + sin 2 (θ − β ) = a2 + b2
⇒ tan A − tan B = 4
19. Put x = 60 °, then expression reduces to 1, therefore (b) Now, (tan A + tan B)2 = (tan A − tan B)2
is the appropriate choice. + 4 tan A tan B = 24
20. sin 47 ° + sin 61° − sin 11° − sin 25° ⇒ tan A + tan B = 2 6
∴ tan A = 2 + 6 and tan B = 6 − 2
= 2 sin 54° cos 7 ° − 2 sin 18° cos 7 ° 1
cos A =
= 2 cos 7 ° (sin 54° − sin 18° ) 11 + 4 6
= 2 cos 7 ° (cos 36° − sin 18° ) 1
5 + 1 5 − 1 cos B =
= 2 cos 7 ° − 11 − 4 6
4 4 1 1
∴ cos A cos B = =
= cos 7 ° 121 − 96 5 483
9 sin A sin B =
(2 + 6 )( 6 − 2 ) 2
5
∴ cos( A + B) = cos A cos B − sin A sin B
=
5
31. Given, ∠C =
2π
3
⇒ A+ B=
π
3
Now, cos 2 A + cos 2 B − cos A cos B
Objective Mathematics Vol. 1
1 2 1 1
= − =− = [2 + cos 2 A + cos 2 B − cos( A + B) − cos( A − B)]
5 5 5 2
2π 4π 1
26. f(θ ) = sin 2 θ + sin 2 θ + + sin 2 θ +
1
= 2 + 2 cos( A + B)cos( A − B) − − cos( A − B)
3 3 2 2
1 4π 8π 1 3 3
= 3 − cos 2θ − cos + 2θ − cos + 2θ = + cos( A − B) − cos( A − B) =
2 3 3 2 2 4
1
2 π 10
1 πr πr
= 3 − cos 2θ − 2 cos(2 π + 2θ )cos
2 3 32. Let I = ∑ 4 cos 3 3
+ 3 cos
3
r =0
1 3 10
= (3 − cos 2θ + cos 2θ ) = 1 πr 1
2 2 = ∑ 4 cos πr + 3 cos = (I1 + I2 )
3 4
π 3 r =0
⇒ f = 10
15 2
2 1 1
∴ I1 = ∑ cos πr = 1 − 1 + 1 − 1 + K − 1 + 1 = 1
r =0
27. We have, = +
cos x cos( x − y ) cos( x + y ) 10 π 11 π
10 cos ⋅ ⋅ sin
πr 2 3 6 1
=
2 cos x cos y and I2 = ∑ cos = =−
3 π 2
cos 2 x − sin 2 y r =0 sin
6
⇒ 2 cos 2 x − 2 sin 2 y = 2 cos 2 x cos y 1 3 1
⇒ 2 cos 2 x(1 − cos y ) = 2 sin 2 y ∴ I = 1 − = −
4 2 8
y y y
⇒ 2 cos 2 x 2 sin 2 = 8 sin 2 cos 2 33. The given equation is
2 2 2 −2
y 3 cos A + 2 = 0 ⇒ cos A =
Ta rg e t E x e rc is e s
⇒ cos 2 x = 2 cos 2 3
2
y Since, A is angle of triangle.
⇒ cos 2 x sec 2 = 2
2 ∴ A < π ⇒ sin A is positive.
y 5
⇒ cos x sec = ± 2 ∴ sin A = (1 − cos 2 A) =
2 3
1 sin A − 5
28. sin 3 x sin 3 x = sin 2 x(cos 2 x − cos 4 x ) ⇒ tan A ==
2 cos A 2
1
= (1 − cos 2 x )(cos 2 x − cos 4 x ) − 5
4 sin A + tan A =
1 6
= (cos 2 x − cos 4 x − cos 2 2 x + cos 2 x cos 4 x ) −5
4 sin A tan A =
1 1 + cos 4 x 1 6
= cos 2 x − cos 4 x − + (cos 6 x + cos 2 x ) Hence, the equation whose roots are sin A and tan A, is
4 2 2
x 2 − (sin A + tan A)x + sin A tan A = 0
13 3 1 1
= cos 2 x − cos 4 x + cos 6 x − 5 −5
4 2 2 2 2 ⇒ x2 + x+ =0
1 3 3 1 6 6
= − + cos 2 x − cos 4 x + cos 6 x ⇒ n = 6
8 8 8 8 ⇒ 6x 2 + 5x − 5 = 0
π 34. cot 2 x = cot( x − y )⋅ cot( x − z )
29. A + B = ⇒ tan( A + B) = 1
4 cot x cot y + 1 cot x cot z + 1
⇒ tan A + tan B = 1 − tan A tan B ⇒ cot 2 x =
cot y − cot x cot z − cot x
⇒ tan A + tan B + tan A tan B + 1 = 2
⇒ cot 2 x cot y cot z − cot 3 x cot y − cot 3 x cot z
⇒ (1 + tan A)(1 + tan B) = 2
+ cot 4 x = 0
sin x 1 cos x 3 tan x 1
30. = , = ⇒ = ⇒ cot x cot y cot z + cot x cot y + cot x cot z + 1 = 0
2
sin y 2 cos y 2 tan y 3
tan x + tan y 4 tan x ⇒ cot x cot y(1 + cot 2 x ) + cot x cot z(1 + cot 2 x )
tan( x + y ) = =
1 − tan x tan y 1 − 3 tan 2 x + 1 − cot 4 x = 0
2 ⇒ cot x(cot y + cot z )(1 + cot x ) + (1 − cot 2 x )
2
Also, sin y = 2 sin x and cos y = cos x
3 (1 + cot 2 x ) = 0
4 cos 2 x ⇒ cot x(cot y + cot z ) + (1 − cot 2 x ) = 0
⇒ sin y + cos y = 4 sin x +
2 2 2
=1
9 cot 2 x − 1 1
⇒ = (cot y + cot z )
5 2 cot x 2
⇒ 27 tan 2 x = 5 ⇒ tan x =
3 3 1
484 ⇒ (cot y + cot z ) = cot 2 x
⇒ tan( x + y ) = 15 2
π
35. Since, α + β = , then
2
π
tan β = tan − α = cot α
Now,
⇒
x 2 + y 2 = a2 + b2
x −a
x 2 − a2 = − ( y 2 − b2 )
2 2
9
Trigonometric Functions and Equations
2 ⇒ = −1
y 2 − b2
and β + γ =α ⇒ γ =α −β
tan α − tan β b2 ( x 2 − a2 ) b2
∴ tan γ = tan(α − β ) = ⇒ = − = tan α tan β [from Eq. (iii)]
1 + tan α tan β a2 ( y 2 − b2 ) a2
tan α − tan β tan α − tan β 38. 2 x = cos(α − β − γ + δ ) − cos(α − β + γ − δ )
= =
1 + tan α cot α 2 2 y = cos(β − γ − α + δ ) − cos(β − γ + α − δ )
Hence, tan α = tan β + 2 tan γ and similarly for 2 z,
36. We have, sin θ = n sin(θ + 2α ) On adding, we get 2 x + 2 y + 2 z = 0
sin (θ + 2α ) 1 ⇒ x+ y+ z=0
⇒ =
π π π π
sin θ n
39. cos − x cos − y − sin − x sin − y
sin (θ + 2α ) + sin θ 1 + n 4 4 4 4
⇒ =
sin (θ + 2α ) − sin θ 1 − n π π
Let − x = A and −y=B
2 sin (θ + α ) cos α 1 + n 4 4
⇒ =
2 sin α cos (θ + α ) 1 − n Then, cos A cos B − sin A sin B = cos( A + B)
π π
1 + n = cos − x + − y
⇒ tan (θ + α ) = tan α 4 4
1 − n
π π
= cos − x − y = cos − ( x + y ) = sin( x + y )
x
37. Given, cos α +
y
sin α = 1 ...(i) 2 2
a b
x y 40. sin x cos 3 x > cos x sin 3 x
cos β + sin β = 1 ...(ii)
a b ⇒ sin x cos x(cos 2 x − sin 2 x ) > 0
cos α cos β sin α sin β ⇒ sin x cos x ⋅ cos 2 x > 0 ⇒ cos x cos 2 x > 0
+ =0
Targ e t E x e rc is e s
and ...(iii)
a2 b2 + – + –
From Eqs. (i) and (ii), we get
sin α − sin β cos β − cos α 0 π
— 3π
— π
— π
x=a and y = b 4 2 4
sin(α − β ) sin(α − β )
a2 (sin α − sin β )2 + b2 (cos β − cos α )2 π π 3π
Now, x 2 + y 2 = ∴ x ∈ 0, ∪ ,
4 2 4
sin 2 (α − β )
a2 (sin 2 α + sin 2 β ) + b2 (cos 2 α + cos 2 β ) 41. We have,
∞
=
− 2(a2 sin α sin β + b2 cos α cos β ) x= ∑ cos 2n φ = 1 + cos 2 φ + cos 4 φ + ...
sin 2 (α − β ) n=0
a (sin α + sin β ) + b2 (cos 2 α + cos 2 β )
2 2 2 1 1
= = =
sin 2 (α − β ) 1 − cos 2 φ sin 2 φ
1 1
[from Eq. (iii)] Similarly, y = =
1 − sin φ cos 2 φ
2
a2 {sin 2 α + sin 2 β − sin 2 (α − β )}
1
and z=
+ b {cos α + cos β − sin (α − β )}
2 2 2 2
1 − sin 2 φ cos 2 φ
=a +b +
2 2
sin 2 (α − β ) =
1
=
xy
⇒ xyz = xy + z
1 1 1 xy − 1
= a2 + b2 + 1− ⋅
sin 2 (α − β ) x y
1 1 1
[a2 (sin 2 α + sin 2 β − sin 2 α cos 2 β − cos 2 α sin 2 β Also, x + y = + = = xy
sin 2 φ cos 2 φ sin 2 φ cos 2 φ
+ 2 sin α sin β cos α cos β) + b2 {cos 2 α(1 − sin 2 β)
Thus, xyz = xy + z = x + y + z
+ cos 2 β(1 − sin 2 α ) + 2 sin α sin β sin α sin β cos α x y z
cos β} ] 42. We have, = =
cos θ 2π 2π
2 a2 sin α sin β cos(α − β ) + 2 b2 cos θ − cos θ +
3 3
cos α cos β cos(α − β )
= a2 + b2 + Therefore, each ratio
sin 2 (α − β ) x+ y+ z
=
2 a b cos(α − β ) sin α sin β cos α cos β
2 2 2π 2π
= a2 + b2 + + cos θ + cos θ − + cos θ +
sin 2 (α − β ) b2 a2 3 3
x+ y+ z x+ y+ z
[from Eq. (iii)] = = =0
= a2 + b2 + 0 = a2 + b2 2π 0
cos θ + 2 cos θ cos
3
Thus, x 2 + y 2 = a2 + b2 485
9 43. Given that, cos θ = x +
⇒ x+
1
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