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angelthas881. SETS AΔB = (A B) − (A ∩ B).
Set containing no elements is called an empty set A set X is said to be a finite set if it has k elements
or a void set denoted by or { }. for some k W. We say the finite set X is of
Every element of set A is an element of set B. cardinality k and is denoted by n(X).
A set is an infinite set if it is not finite. For an
A B. A is a subset of B and B is a super set of A.
infinite set A, the cardinality is infinity.
For any two sets A and B,
If n(A) = 1, then it is called a singleton set.
if A B and B A, then the two sets are equal.
n() = 0 and n({}) = 1.
For any set A,
PROPERTIES OF SET OPERATIONS
Empty set and set A are always subsets of A.
Commutative i) A B = B A ii) A∩B= B∩ A.
These two subsets are called trivial subsets.
Associative i) (A B) C = A (B C)
A is a proper subset of B. If A is a subset of B and A
ii) (A ∩ B) ∩ C = A ∩ (B ∩ C).
B. B contains all elements of A
Distributive i) A (B ∩ C) = (A B)∩(A C)
At least one element which is not in A.
ii) A ∩ (B C) = (A ∩ B)(A ∩ C).
Any set is a subset of itself A A
Identity i) A = A ii) A ∩ U = A.
This subset is called an improper subset.
Two sets A and B are disjoint if they do not have
Idempotent i) A A = A ii) A ∩ A = A.
any common element. Absorption i) A(A∩B) = A ii)A∩ (A B) = A.
A and B are disjoint if A ∩ B = De Morgan Law i) (A B)’ = A’ ∩ B’
If A is a set, set of all subsets of A is called power ii) (A ∩ B)’ = A’ B’
set of A denoted as P(A).
iii) A − (B C) = (A − B) ∩ (A − C)
That is, P(A) = {B : B A}.
iv) A − (B ∩ C) = (A − B) (A − C).
No. of elements in P(A) = 2n Symmetric Difference
where n = No. of elements in A i) AΔB = BΔA
All sets under consideration are assumed to be ii) (AΔB)ΔC = AΔ(BΔC)
subsets of some fixed set. This basic set is called
iii) A ∩ (BΔC) = (A ∩ B)Δ(A ∩ C).
universal set. denoted by U
On Empty Set and Universal Set
One of the sets N,W, Z,Q,R may be taken as a
universal set for set of prime numbers i) ’ = U ii) U’ =
A be a subset of universal set U. iii) A A’ = U iv) A ∩ A’ =
Complement of A w.r.t U is denoted as A’ or Ac v) A U = U vi) A ∩ U = A.
Defined as A’ = {x : x U and x A} On Cardinality
i) For any two finite sets A and B,
Set difference denoted by A − B or A \ B
n(A B) = n(A) + n(B) − n(A ∩ B).
Defined as A − B = {a : a A and a B}.
ii) If A and B are disjoint finite sets, then
i) U − A = A’ ii) A − A = iii) − A =
n(A B) = n(A) + n(B).
iv) A − = A v) A − U = .
iii) For any three finite sets A,B and C,
Symmetric difference between two sets A and B is
denoted by AΔB n(AB C) = n(A)+n(B)+n(C)−n(A∩B)−n(A∩ C)
Defined as AΔB = (A − B) (B − A). −n(B ∩ C)+n(A∩B ∩ C).
1. Write the following in roster form. 5. Prove that
i) {x N : x2 < 121 and x is a prime}. ((AB’C)∩(A∩B’∩C’))((ABC’)∩(B’∩C’))=B’∩C
{2, 3, 5, 7}
A ∩ B’ ∩ C’ A A B’ C
ii) set of all positive roots of (x−1)(x+1)(x2 − 1) = 0.
(A B’ C) ∩ (A ∩ B’ ∩ C’) = A ∩ B’ ∩ C’.
{1}
B’ ∩ C’ C’ A B C’
iii) {x N : 4x + 9 < 52}.
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10} (A B C’) ∩ (B’ ∩ C’) = B’ ∩ C’.
x4 A ∩ B’ ∩ C’ B’ ∩ C’,
iv) {x : 3 , x R − {−2}}.
x2
((AB’C)∩(A∩B’∩C’))((ABC’)∩(B’∩C’))=B’∩C
{−5}
6. If X = {1, 2, 3, . . . 10} and A = {1, 2, 3, 4, 5}, find
2. Write the set {−1, 1} in set builder form.
the number of sets B X such that A − B = {4}
{x R: x2 = 1}
For every subset C of {6, 7, 8, 9, 10},
3. State whether given sets are finite or infinite.
(i) {x N : x is an even prime number}. let B = C {1, 2, 3, 5}.
finite A − B = {4}.
ii) {x N : x is an odd prime number}. No of sets B X such that A − B = {4} and No. of
infinite subsets of {6, 7, 8, 9, 10} are the same.
iii) {x Z : x is even and less than 10}. No. of sets B X such that A − B = {4}
infinite =25 = 32.
iv) {x R : x is a rational number}. 7. Find the number of subsets of A if
infinite A = {x : x = 4n + 1, 2 ≤ n ≤ 5, n N}.
v) {x N : x is a rational number}.
A { x : x 4n 1, n 2, 3, 4, 5}
infinite {9, 13, 17, 21}.
4. In a survey of 5000 persons in a town, 45% of the n(A) 4.
persons know Language A, 25% know Language n(P(A)) 2 4
B, 10% know Language C, 5% know Languages 16.
A and B, 4% know Languages B and C, and 4%
8. If n(P(A)) = 1024, n(A B) = 15 and n(P(B)) = 32,
know Languages A and C. If 3% of the persons
then find n(A ∩ B).
know all the three Languages, find the number
of persons who knows only Language A. 2 n 210 2 n 25
n(A) = 45% of 5000 n( A) 210 n( B) 5
= 2250 Using n(AUB) n(A) n(B) - n(A B)
n(B) = 1250, n(C) = 500, (A B) 0
n(A B) = 250, n(B C) = 200
9. IfP(A) denotes power set of A, find n(P(P(P()))).
n(C A) = 200
n(A B C) = 150. () contains 1 element
No. of persons who . P(P()) contains 21 elements
knows only Language A
P(P(P())) contains 22 elements.
n(A B' C' ) = n{A (B C)' }
= n(A) - n{A (B C)}. That is, 4 elements.
= n(A) - n(A B ) - n(A C) + n(A B C) 10. If A and B are two sets so that
= 2250 - 250 - 200 + 150 n(B − A) = 2n(A − B) = 4n(A ∩ B) and if
No of persons = 1950
n(A B) = 14, then find n(P(A)).
Let n(A B) = k. CARTESIAN PRODUCT
n(A - B) = 2k
Cartesian product of 2 sets is set of ordered pairs
n(B - A) = 4k.
n(A B) = n(A - B) + n(B - A) + n(A B) Cartesian product of 3 sets is set of ordered triplets.
7k. Let A,B and C be three non-empty sets.
n(A B) = 14. (Given) Cartesian product of A with B
7k = 14
A B = {(a, b) : a A, b B}.
k = 2.
n(A - B) = 4 Cartesian product A B C
n(B - A) = 8. A B C = {(a, b, c) : a A, b B, c C}.
n(A) = n(A - B) + n(A B),
A A = {(a, b) : a, b A}.
n(A) = 6
Elements of the Cartesian product are ordered and hence,
n(P(A)) = 2 6 = 64.
for non-empty sets,
11. Two sets have m and k elements. If total number
A B B A, unless A = B.
of subsets of the first set is 112 more than that of
the second set, find the values of m and k. A B = B A only if A = B.
A , B two sets R R = {(x, y) : x, y R}.
n(A) m and n(B) k.
R R R = {(x, y, z) : x, y, z R}.
m k (A has more elements than B)
R R can be represented as R2
2 m 2 k 112.
2 k (2 m k 1) 2 4 7 R R R as R3.
2 k 2 4 , 2 m k 1 7 R R is a set of ordered pairs
k 4 m k 3
2 2 , 2 2 R R R is a set of ordered triplets.
Equating powers :
n(A B) = n(A)n(B),
k 4, m k 3
k 4, m 7 n(A B C) = n(A)n(B)n(C),
1. If A = {1, 2, 3} and B = {2, 4, 6} then Find A B
12. If n(A) = 10 and n(A∩ B) = 3, find n((A ∩ B)’ ∩ A). AB = {(1, 2),(1,4),(1,6),(2,2),(2,4),(2,6),(3,2),(3,4), (3, 6)}.
(A ∩ B)’ ∩ A = (A’ B’) ∩ A 2. Let A = {1,2,3}, B = {3,4} and C = {4,5,6}. Find
= (A’ ∩ A) (B’ ∩ A) i) A × (B C) ii) (A × B) (A × C)
= (B’ ∩ A) iii) A × (B C) iv) (A × B) (A × C)
= (B’ ∩ A) i) B C = {4}.
= A − B. A×(B C) = {(1,4), (2,4), (3,4)}.
n((A ∩ B)’ ∩ A) = n(A − B) ii) A × B = {(1,3), (1,4), (2,3), (2,4), (3,3), (3,4)}
= n(A) − n(A ∩ B) A×C = {(1,4)(1,5)(1,6)(2,4)(2,5)(2,6), (3,4) (3,5)(3,6)}
n((A ∩ B)’ ∩ A) = 7. (A × B) (A × C) = {(1, 4), (2, 4), (3, 4)}.
13. If n(A ∩ B) = 3 and n(A B) = 10, then find iii) (B C) = {3, 4, 5, 6}
n(P(AΔB)).
A × (B C) = {(1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5),
n(AB) n(A B) (A B))
(2,6), (3,3), (3,4), (3,5), (3,6)}.
7
iv) (A × B) (A × C) = {(1,3), (1,4), (1,5), (1,6), (2,3),
n( P(AB)) 2 7 128
(2,4), (2,5), (2,6), (3,3), (3,4), (3,5), (3,6)}
3. For a set A, A A contains 16 elements,two of its If a relation contains single element, it is transitive.
elements are (1, 3) and (0, 2). Find elements of A. Inverse of relation
A A = {(1, 3) (0, 2) ….(2,3)…(0,3)…(3,1).. If R is a relation from A to B, then inverse of
A= {0, 1, 2,3} relation R−1 defined from B to A by
4. If A A has 16 elements, S = {(a, b) A A : a < R−1 = {(b, a) : (a, b) R}
b} ; (−1, 2) and (0, 1) are two elements of S, then Discuss the relations for reflexivity,
find the remaining elements of S. symmetricity and transitivity
{(−1, 0), (−1, 1), (0, 2), (1, 2)}
1. Let X = {1, 2, 3, 4}
5. Let A and B be two sets such that n(A) = 3 and
R = {(1, 1), (2, 1), (2, 2), (3, 3),(1, 3),(4,4),(1,2),(3, 1)}.
n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A B, find A
and B, where x, y, z are distinct elements. As (1,1),(2,2), (3, 3),(4, 4) are all in R, it is reflexive.
A B = (x, 1), (y, 2), (z, 1) For each pair (a, b) R pair (b, a) is also in R.
A = set of first elements = {(x, , y, , z } So R is symmetric.
B = set of second elements = {1,2 }. (2, 1), (1, 3) R and (2, 3) R, R is not transitive.
6. If A = {1, 2, 3, 4} and B = {3, 4, 5, 6}, find Thus R is not an equivalence relation.
n((A B) (A ∩ B) (AΔB)). 2. P denote set of all straight lines in a plane R be
relation defined on P as lRm if l is parallel to m
n(A B) = 6, n(A ∩ B) = 2
This relation is reflexive, symmetric,transitive.
n(AΔB) = 4.
Thus it is an equivalence relation.
n((AB)(A∩B)(AΔB)) = n(AB)n(A∩B)n(AΔB)
3. P be set of all triangles in a plane and R be the
= 6 2 4 = 48. relation defined on P as aRb if a is similar to b.
RELATION This relation is reflexive, symmetric,transitive.
Let A and B be any two non-empty sets. Thus it is an equivalence relation.
A relation R from A to B is defined as a subset of 4. P denote set of all straight lines in a plane. R
cartesian product of A and B, R A B. defined by lRm if l is perpendicular to m
Basic relations. Not reflexive; symmetric; not transitive
Let S be any non-empty set. R be a relation on S. 5. Let A be the set consisting of children and
elders of a family. Let R be relation defined by
R is reflexive if a is related to a for all a S.
aRb if a is a sister of b.
(a, a) R for all a S
A woman is not a sister of herself. So it is not
R is said to be symmetric if a is related to b implies reflexive. It is not symmetric. It is not transitive.
that b is related to a. So it is not an equivalence relation.
(a, b) R (b, a) R 6. A be set consisting of all members of a family R
R is said to be transitive if “a is related to b and b is defined by “aRb if a is not a sister of b”.
related to c” implies that a is related to c. Reflexive; not symmetric ;not transitive
(a, b), (b, c) R (a, c) R 2. A be set consisting of all female members of a
family. Relation R defined by “aRb if a is not a
A relation on S is said to be an equivalence relation
sister of b”.
if it is reflexive, symmetric and transitive.
A woman is not a sister of herself. So it is not
Empty relation is symmetric, transitive reflexive. It is not symmetric. It is not transitive.
Universal relation is always an equivalence relation. So it is not an equivalence relation.
3. Prove that relation friendship is not equivalence 10. Check R = {(1,1),(2,2),(3,3), . . . , (n, n)} defined on
relation on set of people in Chennai. S = {1, 2, 3,..., n} for three basic relation
A friend is not a friend of him/herself. So it is not As (a, a) R for all a S, R is reflexive.
reflexive. It is not symmetric. It is not transitive.
For every pair (a, b) R, (b, a) is also in R. Thus R
So it is not an equivalence relation.
is symmetric.
4. On the set of natural numbers R be the relation
We cannot find two pairs (a, b) and (b, c) in R,
defined by xRy if x + 2y = 21.
such that (a, c) R. Hence R is transitive.
R= (1, 10), (3, 9),(5, 8), (7, 7), (9, 6), (11, 5), (13, 4),
Since R is reflexive, symmetric and transitive, this
(15, 3), (17, 2), (19, 1). relation is an equivalence relation.
As (1, 1) R it is not reflexive 11. Let S = {1, 2, 3}, ρ={(1, 1), (1, 2), (2, 2), (1, 3), (3, 1)}.
As (1, 10) R and (10, 1) R it is not symmetric. i) Is ρ reflexive? If not, state the reason and write the
As (3, 9) R, (9, 6) R but (3, 6) R, the relation minimum set of ordered pairs to be included to ρ so as
to make it reflexive.
is not transitive.
5. Relation R defined on the set of all positive ρ is not reflexive because (3, 3) is not in ρ. (1, 1) (2,
integers by “mRn if m divides n”. 2) are in ρ, including (3, 3) makes it reflexive.
reflexive; not symmetric; transitive (ii) Is ρ symmetric? If not, state the reason, write
minimum number of ordered pairs to be included to ρ
6. On the set of natural numbers the relation R
and write minimum number of ordered pairs to be
defined by “xRy if x + 2y = 1”.
deleted from ρ so as to make it symmetric.
R is an empty set
ρ is not symmetric because (1, 2) is in ρ, but (2, 1)
Not reflexive; symmetric; transitive is not in ρ. Including (2, 1) makes it symmetric.
7. On the set of natural numbers let R be the Removing (1, 2) from ρ to make it symmetric
relation defined by aRb if 2a + 3b = 30.
iii) Is ρ transitive? If not, state the reason, write minimum
a) Write down the relation by listing all the pairs. number of ordered pairs to be included and write
{(3, 8), (6, 6), (9, 4), (12, 2)}; minimum number of ordered pairs to be deleted from ρ
so as to make it transitive.
b) Check if it is reflexive, symmetric, transitive
ρ is not transitive because (3, 1) (1, 3) are in ρ, but
Not reflexive, not symmetric, transitive
(3,3) is not in ρ. To make it transitive include (3, 3)
Not an equivalence relation
(3, 1) and (1, 2) are in ρ, but (3, 2) is not in ρ. To
8. On the set of natural numbers let R be the make it transitive we have to include (3, 2) in ρ.
relation defined by aRb if a + b ≤ 6.
If we remove (3, 1) from ρ, it becomes transitive.
a) Write down the relation by listing all the pairs.
iv) Is ρ an equivalence relation? If not, write the minimum
R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), ordered pairs to be included to ρ so as to make it an
(2, 3), (2, 4),(3, 1),(3, 2), (3, 3), (4, 1), (4, 2), (5, 1)} equivalence relation.
b) Check if it is reflexive, symmetric, transitive to make ρ reflexive, we have to include (3, 3)
Not reflexive; symmetric; not transitive to make ρ symmetric, we have to include (2, 1)
Not equivalence to make ρ transitive, to include (3, 3) , (3, 2).
9. Let X = {1, 2, 3, 4} and R = , empty set. Relation becomes
As (1, 1) R it is not reflexive. {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1), (1, 3), (3, 1), (3, 2)}
We cannot find a pair (x, y) in R such that (y, x) But this relation is not symmetric because (3, 2) is
R, it is symmetric, transitive. in the relation and (2, 3) is not in the relation.
So we have to include (2, 3) For a relation on {0, 1, 2, 3} to be reflexive, it
New relation becomes must have pairs (0, 0), (1, 1), (2, 2), (3, 3).
{(1,1),(2,2),(3,3),(1,2),(2, 1), (1, 3), (3, 1), (3, 2), (2, 3)} insert (1, 2) and (2, 3).
This relation is reflexive, symmetric and {(0, 0), (1, 1), (2, 2), (3, 3), (1, 2), (2, 3)}
transitive, and hence it is an equivalencerelation. is reflexive; it is not symmetric not transitive.
12. Let X = {a, b, c, d} and R = {(a, a), (b, b), (a, c)}. vi) reflexive, not symmetric, transitive.
Write down the minimum number of ordered
{(0, 0), (1, 1), (2, 2), (3, 3), (1, 2)}
pairs to be included to R to make it
vii) reflexive, symmetric, not transitive.
i) reflexive (c, c) and (d, d)
{(0,0),(1,1), (2,2), (3, 3), (1, 2), (2, 3), (2, 1), (3, 2)}
ii) symmetric (c, a)
viii) reflexive, symmetric, transitive.
iii) transitive nothing to include
{(0, 0), (1, 1), (2, 2), (3, 3)}
iv) equivalence (c, c), (d, d), (c, a)
15. In the set Z of integers, define mRn if m− n is a
13. Let A = {a, b, c} and R = {(a, a), (b, b), (a, c)}.
multiple of 12. Prove that R is an equivalence
Write down the minimum number of ordered
relation.
pairs to be included to R to make it
m−m=0 and 0 = 0 12
i) reflexive (c, c)
Hence mRm proving that R is reflexive.
ii) symmetric (c, a)
For mRn, m − n = 12k for some integer k;
iii) transitive nothing
n − m = 12(−k) and hence nRm.
iv) equivalence (c, c) and (c, a)
This shows that R is symmetric.
14. Let A = {0, 1, 2, 3}.
For mRn and nRp;
Construct relations on A of the following types:
m − n = 12k and n − p = 12l for integers k and l.
i) not reflexive, not symmetric, not transitive.
m − p = 12(k +l) hence mRp.
Let us use the pair (1, 2)
This shows that R is transitive.
Relation {(1, 2)}. is transitive.
Thus R is an equivalence relation.
If we include (2, 3) and not include (1, 3), then
the relation is not transitive. 16. In set Z of integers,define mRn if m − n is
divisible by 7. Prove R is equivalence relation
{(1, 2), (2, 3)} is not reflexive, not symmetric
and not transitive. 17. Let A = {a, b, c}.
ii) not reflexive, not symmetric, transitive. a) What is the equivalence relation of smallest
cardinality on A?
{(1, 2)}
{(a, a), (b, b), (c, c)}
iii) not reflexive, symmetric, not transitive.
b) What is the equivalence relation of largest
start with the pair (1, 2). include (2, 1). cardinality on A?
(1, 1), (2, 2) are not here, not transitive. AA
{(1, 2), (2, 1)} is not reflexive; If R = {(1, a), (2, b), (2, c), (3, a)}, Find R – 1
it is symmetric; and it is not transitive. R−1 = {(a, 1), (b, 2), (c, 2), (a, 3)}.
iv) not reflexive, symmetric, transitive. Domain of R becomes the range of R−1
{(1, 2), (2, 1), (1, 1), (2, 2)} Range of R becomes the domain of R−1
v) reflexive, not symmetric, not transitive.
CONSTANTS AND VARIABLES ii) Set of all real numbers greater than 5 and less
than 7.
A quantity that remains unaltered throughout a
mathematical process is called a constant. finite interval, open
A quantity that varies in a mathematical process iii) Set of all real numbers x such that 1 ≤ x ≤ 3.
is called a variable. finite interval, closed
A variable is an independent variable when it takes iv) Set of all real numbers x such that 1 < x ≤ 2.
any arbitrary value not depending on any other
finite interval, neither open nor closed
variables,
Type of Intervals
If its value depends on other variables, then it is
called a dependent variable
Example
Area of a triangle A = ½ bh.
Here ½ is a constant and A, b, h are variables.
b and h are independent variables
A is a dependent variable
Intervals and Neighbourhoods
Real numbers are represented by points on a line
and a point on the line can be related to a unique
real number.
[a, b] contains both a and b and in between real
Any real number can be identified as a point on numbers.
line called as the real line.
(a, b) does not contain a and b but contains all in
A subset I of R is said to be an interval if between the numbers.
i) I contains at least two elements (a, b] does not contain a but contains b and all in
ii) a, b I and a < c < b then c I. between numbers.
Geometrically, intervals correspond to rays and −∞ and ∞ are used to indicate ends of real line.
line segments on the real line. Intervals (a, b) and [a, b] are subsets of R.
unfilled circle “◦”indicates point is removed
intervals correspond to line segments are finite Filled circle “•”indicates point is included
intervals whereas the intervals that correspond to It is not possible to draw line if a point is
rays and the entire real line are infinite intervals. removed
A finite interval is said to be closed if it contains Neighbourhood
both of its end points and open if it contains Neighbourhood of a point ‘a’ is any open interval
neither of its end points. containing ‘a’.
( ) parentheses indicate open interval and [ ] If is a very small positive number, -neigh
square brackets indicate closed interval
bourhood of ‘a’ is the open interval (a − , a +).
State weather type of Interval
The set (a −, a+) − {a} is called deleted
i) Set of all real numbers greater than 0. neighbourhood of ‘a’ denoted as 0 < |x − a| <
infinite interval, open
FUNCTION A student cannot get two different marks.
Let A and B be two sets. For every a A, there is definitely only one b
B such that (a, b) R.
A relation f from A to B, a subset of A B, is called
a function from A to B if it satisfies 2. Suppose that 120 students are studying in 4
sections of eleventh standard in a school. Define
i) for all a A, there is an element b B such
a relation from A to B as “x related to y if the
that (a, b) f.
student x belongs to the section y”.
ii) if (a, b) f and (a, c) f then b = c.
i) Is it a function
A function is a relation in which each element in
domain is mapped to exactly one element in Let A denote the set of students and B denote
range. the set of the sections.
A is called the domain of f and B is called the co- Student x is related to mark y if x belongs to y
domain of f. Every student got a class
If (a, b) is in f, then we write f(a) = b; For every x A, there is an element y B
Element b is called the image of a and element a is such that (x, y) R.
called a pre-image of b and f(a) is value of f at a.
For every x A, there is definitely only one y
Set {b : (a, b) f for some a A} is called the range B such that (x, y) R.
of the function.
A student cannot get two different classes
If B is a subset of R, it is a real-valued function
Hence the relation is a function
Functions f and g are equal functions if domains
ii) What can you say about the inverse relation?
are same f(a) = g(a) for all a in the domain.
For every y B, there is definitely not only
If f is a function with domain A and co-domain B,
one x A such that (x, y) R.
f:A→B
Thus inverse is not a function
f is from A to B or f is a function from A to B or f
3. Let f = {(a, 1), (b, 2), (c, 2), (d, 4)}. Is f a function?
maps A into B.
This is a function from set {a, b, c, d} to {1, 2, 4}.
If f(a) = b, f maps a to b or a is mapped onto b by f
This is not a function from {a, b, c, d, e} to {1,2, 3, 4}
Range of function is the collection of all elements
in the co-domain which have pre-images. because e has no image.
Range of a function is a subset of the co-domain. This is not a function from {a, b, c, d} to {1, 2, 3, 5}
because the image of d is not in the co-domain;
By first condition, domain of relation R from a set
A to a set B as the set of all elements of A having f is not a subset of {a, b, c, d} {1, 2, 3, 5}.
images and not as A. 4. Which of the following are functions
By second condition, an element in the domain (i) y = 2x (ii) y = x2 (iii) y2 = x.
cannot have two or more images.
1. A test is written by number of students.
Define a relation from A to B as “a related to b
if student a belongs to mark b. if
Let A= set of students for an examination
B = {0, 1, 2, 3, . . . 100} set of possible marks.
Student a is related to mark b if a got b marks
i) and ii) are functions
Every student got a mark
iii) not a function, if domain & co-domain are R
For every a A, there is an element b B such
that (a, b) R. In (iii) for the same x, we have two y values
if we split into y = √x and y = −√x both become x2−5x+6 is positive.
functions with same domain and co-domains
ii) Take any point in (2, 3), say x = 2.5.
[0,∞) and (−∞, 0]
x2−5x+6 is negative.
iii) Take any point in (3,∞), say x = 4.
x2−5x+6 is positive.
For all x, in the intervals (−∞, 2) and (3,∞),
x2−5x+6 is positive.
REPRESENTATION OF A FUNCTION
At x = 2, 3 value of x2−5x+6 is zero.
Analytical Representation of a Function Thus, x 2 5x 6 is defined for all x in
Series of symbols denoting mathematical (−∞, 2] [3,∞).
operations in a definite sequence on numbers, Hence domain is (−∞, 2] [3,∞).
letters designate constants or variable quantities
7. Find the domain of
Domain and range of certain functions
1 1
i) f(x) = ii) f(x) =
1 2 cos x 1 2 sin x
i) Function is defined for all x R
except 1 − 2 cos x = 0.
1
except cos x = 2
except x = 2nπ 3 , n Z
Domain R − {2nπ 3 }, n Z
ii) Function is defined for all x R
except 1 − 2 sin x = 0.
1
5. Find domain except sin x = 2
i) y = x3 + 3 is (−∞,∞) except x = nπ (−1)n
,nZ
6
ii) y = x4 − 2 is (−∞,∞) Domain R − { nπ (−1)n
}, n Z
6
x 1
iii) y = is R − {−1} 8. Find the range of
x 1
1 1
iv) y = 4 x2 is −2 ≤ x ≤ 2. i) f(x) = . ii) f(x) = .
1 3 cos x 2 cos x 1
6. Find the largest possible domain for the real i) −1 ≤ cos x ≤1
2
valued function f defined by f(x) x 5x 6. 3 ≥ −3 cos x ≥ −3
For square root, −3 ≤ −3 cos x ≤3
x2−5x+6 ≥ 0 for all x in the domain 1 − 3 ≤ 1 − 3 cos x ≤ 1 + 3
Solving x2−5x+6 = 0, −2 ≤ 1 − 3 cos x, 1 − 3 cos x ≤ 4.
x = 2 , 3. Taking reciprocals
1 1
≤ − 12 , ≥ 14
1 3 cos x 1 3 cos x
Intervals. (−∞, 2), (2, 3) and (3,∞)
Range of f : (−∞,− 12 ] [ 14 ,∞).
i) Take any point in (−∞, 2), say x = 1.
ii) −1 ≤ cos x ≤1 x must lie on (−∞,−3) (3,∞).
−2 ≤ 2 cos x ≤2 Combining these two conditions,
− 2−1 ≤ 2 cos x−1 ≤ 2 −1 [−2, 2] ∩ ((−∞,−3) (3,∞))
−3 ≤2 cos x−1, 2 cos x−1 ≤ 1. domain
Taking reciprocals Piece-wise defined functions
1 1 Depending upon value of x, select the formula
≤ − 13 , ≥1
2 cos x 1 2 cos x 1 to find value of f at any point x.
Range of f : (−∞,− 13 ] [1,∞). To find value of f at any real number, first find
to which interval x belongs to
9. a) Show that the relation xy = −2 is a function Using corresponding formula find value of f
for a suitable domain. at that point.
xy 2 11. Write the values of f at f(6) , f(−1), f(−5)
2
y 0 if - x 2
x
f(x) 2x if -2 x 3
i) for all x A, there is an element y B such 2
that (x, y) f. x if 3 x
ii) if (x, y) f and (x, z) f then y = z. To find f(6) : x belongs to interval 3 ≤ 6 ≤ ∞
b) Find domain and the range of the function. f(x) = x2
R − {0},R − {0} f(6) = 36.
10. Find the largest possible domain for the real f(−1) = −2,
valued function given by f(−5) = 0
2 2
9x 4x 12. Write the values of f at −4, 1,−2, 7, 0 if
i) f(x) = ii) f(x)
2
x 1 x 92
x 4 if x 3
x 4 if 3 x 2
i) If x < −3 or x > 3, x2> 9, 9 − x2 ≤ 0
2
f(x) x x if 2 x 1
It has no square root in R. 2
x x if 1 x 7
x must lie on the interval [−3, 3]. 0 otherwise
If x ≥ −1 or x ≤ 1, x2 − 1 ≤ 0 f(−4) = 8 f(1) = 0
It has no square root in R. f(−2) = 6 f(7) = 0
f(0) = 0
x must lie outside [−1, 1].
13. Write the values of f at −3, 5, 2,−1, 0 if
x must lie on (−∞,−1) (1,∞).
x 2 x 5 if x ( , 0 )
2
Combining these two conditions, x 3x 2 if x ( 3,)
f(x) 2
[−3, 3] ∩ ((−∞,−1) (1,∞)) x if x ( 0 , 2 )
2
domain [−3,−1) (1, 3]. x 3 otherwise
ii) If x < −2 or x > 2, 4 − x2 ≤ 0 f(−3) = 1 f(5) = 38
f(2) = 1 f(−1) = −5
It has no square root in R.
f(0) = −3
x must lie on the interval [−2, 2].
Graphical Representation of a Function
If x ≥ −3 or x ≤ 3, x2 − 9 ≤ 0
If the function is defined from R or a subset of R
It has no square root in R. we can draw the graph of the function.
x must lie outside [−3, 3].
14. Represent graphically and find domain, range
i) 2x − y = 0
Finding pre image, domain
Analytical relation between x and y Let x be a point in the domain.
y = 2x. Draw a vertical line through the point x.
Values of y depends on values of x. Let it meet the curve at P.
y = f(x). Point at which horizontal line drawn through
Set of all points that lie on the straight line P meets the y-axis is f(x).
{(x, 2x) : x R}. Using horizontal lines through a point y in the
co-domain, find pre-images of y.
This is a subset of R R.
Domain and range are R and R 16. Plot the curve and find domain and pre image of
ii) x2 − y = 0. f(x) = x2 + 4, x ≥ 0.
Relation between x and y is y = x2. 17. If f : R → R is defined as f(x) = 2x2 − 1, find pre-
images of 17, 4 and −2.
Set of all points on curve is {(x, x2) : x R}
Solving 2x 2 1 17.
This is a subset of Cartesian product R R.
Domain is R and range is [0,∞) x 3 , -3
preimages of 17 under f : 3 and3
iii) x − y2 = 0
2x 2 1 4
5 5
preimages of 4 under f : . ,
2 2
2x 2 1 2.
1
x2
2
y2 = x or y = √x, x ≥ 0. It has no solution in R
y = +√x and y = −√x. Because square of number cannot be negative
Set of all points on the curve is the union of sets Hence −2 has no pre-image under f.
{(x,√x)} and {(x,−√x)}, Vertical Line Test
Subsets of Cartesian product R R. If the vertical line through a point x in domain
Domain is [0,∞) and the range is R meets the curve at more than one point, we
will get more than one value for f(x) for one x.
15. if f : [0, 4] → R is defined by This is not allowed in a function.
x If the vertical line through a point x in the
f(x) = + 1, Draw the function graphically
2 domain does not meet the curve, there will be
x no image for x; this is not possible in function.
Plot the points (x, + 1) for all x [0, 4].
2 if the vertical line through a point x in the domain
meets the curve at more than one point or does not
A straight line joining (0, 1) and (4, 3) is got meet the curve, curve will not represent a function
18. Test if the given represents a function or not Function f : R → R defined by f(x) = |x|,
where |x| is modulus or absolute value of x,
|x| is defined as
-x if x 0
|x| 0 if x 0
x if x 0
-x if x 0 -x if x 0
or |x| or |x|
x if x 0 x if x 0
i) Curve does not represent a function from [0,
4] to R because a vertical line meets the curve
at more than one point
ii) Curve does not represent a function from [0, iv) Signum function
4] to R because a vertical line drawn through Function f : R → R defined by
x = 2.5 in [0, 4] does not meet the curve
x if x 0
TYPES OF FUNCTIONS
f(x) |x|
0 if x 0
I. Elementary Functions
i) Identity function denoted by sgn.
Let X be any non-empty set.
Function f : X → X defined by f(x) = x for all x
X is identity function on X denoted by IX or I.
v) greatest integer function
Function f : R → R defined by f(x) is greatest
integer less than or equal to x denoted by x
ii) constant function
Let X and Y be two sets. Let c be a fixed
element of Y ,function f : X → Y defined by
f(x) = c for all x X
vi) smallest integer function
f : R → Rdefined by f(x) is the smallest integer
greater than or equal to x denoted by x
Value of constant function same for all values of x
If X is any set, constant function defined by
f(x) = 0 for all x is called zero function, a
particular case of constant function. II. Types of Functions
iii) Modulus function or absolute value function i) A function f : A → B is one-to-one if
x, y A, x y f(x) f(y) or f(x) = f(y) x = y 7) Let X be a finite set with k elements. Then, we
Another names : injective function; have a bijection from X to {1, 2, . . . k}.
2. Check following functions one-to-one and onto.
ii) A function f : A → B is onto, if for each b B
there exists at least one element a A such i) f : N → N defined by f(n) = n + 2.
that f(a) = b. That is, the range of f is B. If f(n) = f(m)
Another names : surjective function. n+2 =m+2
iii) A function f : A → B is said to be bijective if it is m = n.
both one-to-one and onto.
Thus f is one-to-one.
iv) A function which is not onto is into function.
1 has no pre-image, this function is not onto.
Range of function is proper subset of codomain
ii) f : N {−1, 0} → N defined by f(n) = n + 2.
Simple results.
As above, this function is one-to-one.
Let A and B be two sets with m and n elements.
If m is in co-domain, m − 2 is in domain
i) No one-to-one function from A to B ifm > n.
f(m − 2) = (m − 2) + 2 = m
ii) If one-to-one function from A to B, then m ≤ n.
iii) There is no onto function from A to B ifm < n. m has pre-image and this function is onto.
iv) If onto function from A to B, then m ≥ n. 3. Check for one-to-oneness and ontoness.
v) Bijection from A to B, if and only if, m = n. i) f : N → N defined by f(n) = n2.
vi) No bijection from A to B if and only if, m n. f(m) = f(n) m2 = n2 m = n since m, n N.
1. Write the type of functions
Thus f is one-to-one. But, non-perfect
1) X = {1, 2, 3, 4}, Y = {a, b, c, d, e}
Square elements in the co-domain do not have
f = {(1, a), (2, c), (3, e), (4, b)}. pre-images and hence not onto.
one-to-one but not onto. ii) f : R → R defined by f(n) = n2.
2) X = {1, 2, 3, 4}, Y = {a, b} Two different elements in the domain have
f = {(1, a), (2, a), (3, a), (4, a)}. same images and hence f is not one-to-one.
one-to-one; it is not onto. Range of f is a proper subset of R.
3) X = {1, 2, 3, 4}, Y = {a} Thus it is not onto.
f = {(1, a), (2, a), (3, a), (4, a)}. 4. Check whether one-to-oneness and ontoness.
1
Not one-to-one but it is onto. i) f : R → R defined by f(x) =
x
co-domain decides whether function is onto or not.
This is not at all a function
4) X = {1, 2, 3, 4}, Y = {a, b, c, d, e}
Because f(x) is not defined for x = 0.
f = {(1, a), (2, c), (3, b), (4, b)}.
1
ii) f : R − {0} → R defined by f(x) =
Neither one-to-one nor onto. x
5) X = {1, 2, 3, 4}, Y = {a, b, c, d} This function is one-to-one but not onto
f = {(1, a), (2, c), (3, d), (4, b)}. Because 0 has no pre-image.
Both one-to-one and onto. 5. Check for one-tooneness and ontoness. If it is
6) X = {1, 2, 3, 4}, Y = {a, b, c, d, e} and not a function, state why?
f = {(1, a), (2, c), (3, e)}. i) A = {a,b, c} and f = {(a, c),(b, c),(c, b)}; (f : A→ A)
Not at all a function, only a relation. function
not one–to–one and not onto 8. If f : [−2, 2] → B is given by f(x) = 2x3, find B so
ii) X = {x, y, z} , f = {(x, y), (x, z), (z, x)}; (f : X → X). that f is onto.
Not a function Minimum value f(−2) = 2(-2)3 = - 16
x Maximum value f(2) = 2(2)3 = 16
6. If f : R − {−1, 1} → R is defined by f(x) = 2 ,
x 1 B is [−16, 16].
verify whether f is one-to-one or not.
9. Give a function from A → B for the following
Assume f(x) f(y).
Let A = {1, 2, 3, 4} and B = {a, b, c, d}.
x y
2
2 i) neither one-to-one nor onto.
x 1 y 1
. x(y 2 - 1) y(x 2 - 1) {(1, a), (2, a), (3, a), (4, a)}
2 2
. xy - x - yx y 0 ii) not one-to-one but onto.
. (y - x)(xy 1) 0
not possible
This implies that x y or xy -1.
Select x and y so that xy -1 then iii) one-to-one but not onto.
f(x) f(y) not possible
possible pairs :
iv) one-to-one and onto.
( 2 , -21 ), ( 7 , -71 ), (-2 , 1
2 )
{(1, a), (2, b), (3, c), (4, d)}
2
f( 2 ) f( -21 )
3 Horizontal Test
f(x) f(y) does not imply x y.
i) If the horizontal line through a point y in the
Hence it is not one - to - one. co-domain does not meet the curve, there will
7. Check whether the function f(x) = x|x| defined be no pre-image for y and function is not onto.
on [−2, 2] is one-to-one or not. If it is one-to-one,
find a suitable co-domain so that the function
becomes a bijection.
Let x, y [-2 , 2 ] such that
f(x) = f(y).
If y = 0 , then x = 0. ii) If the horizontal line through atleast one point
Let y 0 and x 0. meets the curve at more than one point, then
function is not one-to-one.
x x = y|y| .
x |y|
= .
y |x|
y x
Since > 0, > 0
x y
x and y are both positive
or both negative
iii) If for all y in the range the horizontal line
hence x = y 2 .
2
through y meets the curve at only one point,
So if f(x) = f(y), we must have then the function is one-to-one.
x 2 = y 2.
This is possible if x = y.
Thus f is one - to - one.
When x < 0 , f(x) = -x 2
when x 0 , f(x) = x 2 .
So the range : [-4 , 4 ].
f is bijection : from [-2 , 2 ] to [-4 , 4 ]. 10. Identify type of function using Horizontal Test
i) f : R → R defined by f(x) = 2x We cannot assign any value to f−1(y).
one-to-one and onto function. Working Rule to Inverse of Functions from R to R
ii) f : R → R defined by f(x) = x2 Let f : R → R be the given function.
neither one-to-one nor onto. i. write y = f(x);
iii) f : [0,∞) → R defined by f(x) = +√x ii. write x in terms of y;
one-to-one but not onto function. iii. write f−1(y) = the expression in y.
iv) f : [0,∞) → [0,∞) defined by f(x) = +√x iv. replace y as x.
one-to-one and onto function. Algebra of Functions
v) f : [0,∞) → R defined by f(x) = −√x A function whose co-domain is R or a subset of R
one-to-one but not onto function. is called a real valued function.
vi) f : [0,∞) → (−∞, 0] defined by f(x) = −√x Let X be any set.
one-to-one and onto function. Let f and g be real valued functions defined on X.
Define, for all x X
OPERATIONS OF FUNCTIONS
Composition (f + g)(x) = f(x) + g(x).
Let f : X → Y , g : Y → Z be two functions. (f − g)(x) = f(x) − g(x).
function h :X→Z defined as h(x) = g(f(x)) for every (fg)(x) = f(x)g(x).
xX is composition of f with g denoted by g ◦ f
f f (x)
g ◦ f is defined if and only if the range of f is (x) = where g(x) 0.
g g( x )
contained in the domain of g.
Inverse of Function (cf)(x) = cf(x), where c is a real constant.
Let f : X → Y be a bijection. (−f)(x) = −f(x).
Function g : Y → X defined by g(y) = x if Properties.
f(x) = y, is called inverse of f and denoted by f−1. (f + g) + h = f + (g + h)
Let f : X → Y be bijection g : Y → X inverse. f+g = g+f
g◦f = IX and f ◦g = IY 0+f = f + 0, where 0 is zero function
where IX , IY = identity functions on X and Y f + (−f) = (−f) + f = 0
Both f and g are bijections and they are f(g + h) = fg + fh
inverses to each other; (c1 + c2)f = c1f + c2f , c1 ,c2 real constants.
That is f−1 = g and g−1 = f. 1. Find g ◦ f and f ◦ g if
−1
If f is a bijection, then f (y) is pre-image of y f = {(1, 2), (3, 4), (2, 2)}, g = {(2, 1), (3, 1), (4, 2)}.
under f.
Domain of f = {1, 2, 3} Range of f = {2, 4}
Inverses are defined only for bijections.
Domain of g = {2, 3, 4} Range of g = {1, 2}
i) If f is not one-to-one , then there exists a
Range of f is contained in the domain of g
and b such that a b and f(a) = f(b).
Thus g ◦ f is defined
We cannot define f−1(y) because both a and
Image of 1 under f is 2
b are pre-images of y under f, as f−1 cannot
its image under g is 1.
assume two different values for y.
(g ◦ f)(1) = g(f(1))
ii) If f is not onto, then there will be a y in Y
= g(2) = 1.
without a pre-image.
(g ◦ f)(2) = 1 5. If f, g : R → R are defined by f(x) = |x| + x and
g(x) = |x| − x, find g ◦ f and f ◦ g.
(g ◦ f)(3) = 2.
g◦f = {(1, 1), (2, 1), (3, 2)}. x x if x 0
f(x)
x x if x 0
f◦g = {(2, 2), (3, 2), (4, 2)}.
0 if x 0
Thus f(x)
2. Let f = {(1, 4), (2, 5), (3, 5)} and g = {(4, 1), (5, 2), (6, 2x if x 0
4)}. Find g ◦ f. Can you find f ◦ g? x (-x) if x 0
Also g(x)
g ◦ f = {(1, 1), (2, 2), (3, 2)}. x x if x 0
f ◦ g is not defined as range of g = {1, 2, 4} is 2x if x 0
Thus g(x)
not contained in the domain of f = {1, 2, 3}. 0 if x 0
Let x 0. Then
3. Let f and g be two functions from R to R defined (g . f)(x) g(f(x))
by f(x) = 3x − 4, g(x) = x2 + 3. Find g ◦ f and f ◦ g.
g( 0 )
(g ◦ f)(x) = g(f(x)) 0.
= g(3x − 4) Let x 0. Then
= (3x − 4)2 +3 (g . f)(x) g(f(x))
= 9x2 − 24x + 19. g(x)
0.
(f ◦ g)(x) = f(g(x))
Thus (g . f)(x) 0 for all x
= f(x2 + 3)
= 3(x2 + 3) − 4 6. f : A → B and g : B → C be two functions. If f and
g are one-to-one, then g ◦ f is one-to-one - Prove
= 3x2 + 5.
f ◦g g ◦f. Let x y in A.
Composition of functions is not commutative. Since f is one-to-one, f(x) f(y).
4. Let f, g : R → R be defined as f(x) = 2x − |x| and Since g is one-to-one, g(f(x)) g(f(y)).
g(x) = 2x + |x|. Find f ◦ g.
That is, x y (g ◦ f)(x) (f ◦ g)(y).
-x if x 0
x Hence g ◦ f is one-to-one.
x if x 0
2x - (-x) if x 0 7. Show that the statement, “if f and g ◦ f are one-
So f(x) to-one, then g is one-to-one” is not true.
2x - x if x 0
3x if x 0
Thus f(x)
x if x 0
2x (-x) if x 0
Also g(x)
2x x if x 0
x if x 0
Thus g(x)
3x if x 0
Let x 0. Then f and g ◦ f are one-to-one.
(g . f)(x) g(f(x))
But g is not one-to-one.
g( 3x)
3x. Thus the statement is not true.
Let x 0. Then 8. Function for exchanging American dollars for
(g . f)(x) g(f(x)) Singapore Dollar is f(x) = 1.23x, where x is
g(x) number of American dollars. Function for
3x. exchanging Singapore Dollar to Indian Rupee is
Thus (g . f)(x) 3x for all x. g(y) = 50.50y, where y is number of Singapore
dollars. Write a function to give exchange rate One-to-one : Let f(x) = f(y).
of American dollars in terms of Indian rupee.
3x − 5 = 3y − 5
f(x) = 1.23x
x = y.
g(y) = 50.50y
Thus f is one-to-one.
(g ◦ f)(x) = g(f(x))
Onto : Let y R.
= 50.50(1.23x)
y5
Let x =
= 62.115x 3
9. Find invers of f if y5
f(x) = 3( )−5
3
A = {1, 2, 3, 4}
f(x) = y.
f = {(1, 2), (2, 4), (3, 1), (4, 3)}.
Thus f is onto.
Range of f = {1, 2, 3, 4}
Inverse Let y = 3x − 5.
inverse of f = {(1, 3), (2, 1), (3, 4), (4, 2)}.
y + 5 = 3x
10. If f : R → R is defined by f(x) = 2x − 3 prove that
f is a bijection and find its inverse. y5
x =
3
One-to-one : Let f(x) = f(y).
y5
Then 2x−3 = 2y−3 f−1(y) =
3
this implies that x = y. Byreplacing y as x,
That is, f(x) = f (y) x5
f−1(x) =
implies that x = y. 3
Thus f is one-to-one. 12. A simple cipher uses the function f(x) = 3x−4.
Find the inverse of this function, determine
Onto : Let y R.
whether the inverse is also a function
y3
Let x = Working out as Problem No: 10
2
x4
y3 f−1(x) =
f(x) = 2( )−3 3
2
13. Formula for converting from Fahrenheit to
f(x) = y. 5x 160
Celsius temperatures is y Find
Thus f is onto. 9 9
inverse of this function and determine whether
Inverse Let y = 2x − 3. the inverse is also a function.
y + 3 = 2x One-to-one : Let f(x) = f(y).
y3 5 5
x = ( x 32) = ( y 32)
2 9 9
x = y.
−1 y3
f (y) = Thus f is one-to-one.
2
Byreplacing y as x, Onto : Let y R.
9
−1 x3 Let x = y 32
f (x) = 5
2
5 9
11. Iff : R → R is defined by f(x) = 3x − 5, prove that f(x) = ( y 32 32)
9 5
f is a bijection and find its inverse. f(x) = y.
Thus f is onto. 17. Owner of a restaurant can prepare a meal at a
5 cost of Rupees 100. If the menu price of the
Inverse Let y = ( x 32)
9 meal is x rupees, number of customers who will
9 order that meal at that price in an evening is
x = y 32
5 given by the function D(x) = 200−x. Express his
9 day revenue, total cost and profit on this meal as
f−1(y) = y 32
5 functions of x.
Byreplacing y as x, day revenue = (200 − x)x
9
f−1(x) = x 32 total cost = 100(200 − x)
5
14. If X is a set of students of a class, f and g total profit = (200 − x)x − 100(200 − x)
functions representing the marks obtained by 18. Weight of muscles of a man is a function of his
the students in two tests, represent the total body weight x expressed as W(x) = 0.35x.
marks of the students in two tests. Determine the domain of this function.
By addition property of functions x>0
(f + g)(x) = f(x) + g(x). 19. If f and g are real-valued functions, prove that
15. Total cost of airfare is comprised of base cost C f(g + h) = fg + fh.
and the fuel surcharge S in rupee. Both C and S Let X be any set
are functions of the mileage m; C(m) = 0.4m + 50
f and g be real-valued functions defined on X.
and S(m) = 0.03m. Determine a function for the
total cost of a ticket in terms of the mileage and Let x X.
find the airfare for flying 1600 miles. (f(g + h))(x) = f(x)(g + h)(x) (product)
C(m) = 0.4m + 50 = f(x)[g(x) + h(x)] (addition)
S(m) = 0.03m = f(x)g(x) + f(x)h(x) (distributivity)
Total cost (C+S)m = C(m) + S(m) = (fg)(x) + (fh)(x) (product)
= 0.43m + 50, = (fg + fh)(x) (addition)
Airfare (C+S)1600 = 0.43(1600) + 50 (f(g + h))(x) = (fg + fh)(x) for all x X
= 738 f(g + h) = fg + fh.
16. A salesperson whose annual earnings can be Special Functions
represented by function A(x) = 30, 000+0.04x,
where x is the rupee value of the merchandise i) polynomial function
he sells. His son is also in sales and his earnings Function f : R → R defined by
are represented by the function S(x) = 25, 000 + f(x) = a0xn + a1xn−1 + a2xn−2 + . . . + an−1x + an
0.05x. Find (A + S)(x) and determine the total where ai are constants,
family income if they each sell Rupees 1, 50, 00,
ii) linear function
000 worth merchandise.
Function f : R → R defined by
A(x) = 30, 000+0.04x
f(x) = ax + b where a 0, b are constants
S(x) = 25, 000 + 0.05x.
A function which is not linear is called a non-
(A + S)(x) = A (x) + S (x)
linear function.
= 55,000 + 0.09x
iii) Exponential function
total family income = (A+S)(150 00 000)
Let a be a non-negative constant.
=14, 05, 000
Consider the function f : R → R defined by
f(x) = ax. f(−x) = f(x) for all x R.
If a = 0, x 0, f(x) = 1 zero function Results.
if a = 1, f(x) = 1 constant function i) Sum of two odd functions is an odd function.
If a > 1, f(x) = ax exponential function ii) Sum of two even functions is an even function
iii) Product of two odd functions is an even
function.
iv) Product of two even functions is an even
function.
v) Product of an odd function and an even
function is an odd function.
iv) logarithmic function vi) The only function which is both odd and even
function is the zero function.
Let a > 1 be a constant.
vii) The product of a positive constant and an
function f : (0,∞) → R defined by
even function is an even function.
f(x) = loga x is logarithmic function
viii) The product of a negative constant and an
Inverse of exponential function f(x) = a is x
even function is also an even function.
logarithmic function.
ix) The product of a constant and an odd function
is an odd function.
x) There are functions which are neither odd nor
even.
1. Classify the functions defined by
f(x) = x, f(x) = 2x , f(x) = x3 + 2x , f(x) = x2, f(x) = 3,
f(x) = x4 + x2 , f(x) = |x| , f(x) = x + x2 ,
v) rational function
odd functions.
Real valued function f defined by
f(x) = x f(x) = 2x f(x) = x3 + 2x
p(x)
f(x) = on a suitable domain even functions
q(x)
f(x) = x2 f(x) = 3 f(x) = x4 + x2 f(x) = |x|
where p(x) and q(x) are polynomials, q(x) 0,
neither even nor odd.
vi) reciprocal function
f(x) = x + x2
If f is a real valued function such that f(x) 0,
the real valued function g defined by 2. Product of an odd function and an even function
is an odd function.
1
g(x) = on a suitable domain
f(x) Let f be an odd function, g be an even function.
1 Let h = fg.
Largest domain of f(x) = is R − {1}.
f(x) 1
h(−x) = (fg)(−x)
vii) Odd and Even function = f(−x)g(−x)
A function f : R → R is odd function if = −f(x)g(x)
f(−x) = −f(x) for all R. as f is odd and g is even
It is said to be an even function if h(−x) = −h(x) Thus h is an odd function.
Graphing Functions using Transformations
REFLECTION
Reflection of the graph of a function with respect
to a line l is the graph that is symmetric to it w.rt l
A reflection is the mirror image of the graph
where line l is the mirror of the reflection.
For f(x) =√x, f(−x) =√−x
Reflections of y = f(x)
f(−x)=√−x is the reflection of f(x) =√x about y-axis.
i) y = −f(x) is the reflection of of f about x-axis.
3. Plot the graphs of i) y = ex ii) y = loge x.
ii) y = f(−x) is the reflection of f about y-axis.
y = ex is the inverse function of y = loge x
iii) y = f−1(x) is the reflection of f in y = x.
y = ex is the reflection of y = loge x about y = x
TRANSLATION
A translation of a graph is a vertical or horizontal
shift of graph that produces congruent graphs.
Translation of y = f(x)
y = f(x + c), c > 0 causes the shift to the left.
y = f(x − c), c > 0 causes the shift to the right.
4. Plot the the functions:
y = f(x) + d, d > 0 causes the shift to the upward.
i) f(x) =|x| ii) f(x) =|x − 1|
y = f(x) − d, d > 0 causes shift to the downward.
iii) f(x) = |x + 1|
DILATION
f(x) = |x − 1|
Multiplying a function by a positive constant
causes the graph of the function f(x) = |x|
vertically stretches or compresses its graph
shifts to the right for one unit.
Dilation of y = f(x)
f(x) = |x + 1|
If the positive constant is greater than one,
causes the graph of the function f(x) = |x|
the graph moves away from the x-axis.
shifts to the left for one unit.
If the positive constant is less than one, the
graph moves towards the x-axis.
1. Plot the graphs of i) y = x2
ii) y = −x2.
5. Plot the functions:
i) f(x) = |x| ii) f(x) = |x| − 1
iii) f(x) = |x| + 1 iv) y = |x − 1| + 1
v) y = |x + 1| − 1 vi) y = |x + 2| − 3.
f(x) = |x| − 1
f(x) = x2, −f(x) = −x2.
causes the graph of the functionf(x) = |x|
y = −x2 is the reflection of y = x2 about x-axis. shifts to the downward for one unit.
2. Plot the graphs of y2 = x , y2 = −x. f(x) = |x| + 1
causes the graph of the function f(x) = |x| f(x) = x2 + 1 causes the graph of f(x) = x2 shifts to
shifts to the upward for one unit. the upward for one unit.
f(x) = (x + 1)2 causes the graph of f(x) = x2 shifts to
the left for one unit.
8. Compare and contrast y = x2 − 1, y = 4(x2 − 1)
and y = (4x)2 − 1
6. Plot the graphs of
y = x2 − 1, y = 4(x2 − 1) graphs
1 2 look identical until we
i) f(x) = x2 ii) f(x) = x iii) f(x) = 2x2
2
compare scales on the y-axis.
1 2
f(x) = x causes the graph of f(x) = x2 Scale is four times large,
2
reflecting the multiplication of
stretches towards the x-axis since multiplying
the original function by 4
1
factor is less than 1.
2 Effect looks different when
f(x) = 2x2 causes the graph of f(x)= x2 functions plotted on same
compresses towards moves away from x−axis scale
since the multiplying factor 2 is greater than 1 In the graph of y = (4x)2 − 1 x-
scale has now changed, by
the same factor of 4
When plotted on the same set
of axes parabola y = (4x)2 −1
looks thinner. Here, x-
intercepts are different, but y-
intercepts are same
9. Draw the graphs of
7. Plot the graphs of i) f(x) = x2
i) y = sinx ii) y = sin2x
ii) f(x) = x2 + 1
iii) y = sin(−x) iv) y = −sin(−x)
iii) f(x) = (x + 1) .2
vi) y = sin|x| (Hint: sin(−x) = −sin x.)
iv) y = 3(x − 1)2 + 5 x intercepts for y = sinx are nπ
x intercepts for y = sin2x are ½ nπ, n Z.
Graph the functions f(x) = x3 and g(x) = 3√x on the
same coordinate plane. Find f ◦ g and graph it on
the plane as well. explain your results.
v) y = sin + x iv) y = sin −x which are cos x
2 2
From the curve y = x, draw
10. Draw the graph of y = 2 sin(x − 1) + 3.
i) y = −x ii) y = 2x iii) y = x + 1 iv) y =1/2 x + 1
First we draw y = sinx.
v) 2x + y+3 = 0.
Draw the curve y = sin(x − 1)Draw y = 2 sin(x − 1)
1
For the curve y = x 3 , draw
Draw y = 2 sin(x − 1) + 3.
11. For the curve y = x3, draw
i) y = −x3 ii) y = x3 + 1 iii) y = x3 − 1 iv) y = (x + 1)3
1 1 1 1
i) y = − x 3 ii) y = x 3 +1 iii) y = x 3 − 1 iv) y = ( x 1)3
2.1. REAL NUMBER SYSTEM Decimal representation of an irrational number
will neither be terminating nor infinite periodic.
Natural Numbers
2.2.4 Properties of Real Numbers
N = {1, 2, 3, . . . }
i) For any a, b R, a + b R and ab R.
Integers
[Sum or Multiplication of two real numbers is
Z = {. . ,−4,−3,−2,−1, 0, 1, 2, . .}
again a real number]
Whole numbers
ii) For any a, b, c R,
W = {0, 1, 2, 3, ….. }
(a + b) + c = a + (b + c) and a(bc) = (ab)c.
Rational numbers
[we can add (or multiply)real numbers by
Q= { mn | m, n Z, n 0} grouping them in any order.]
Examples of Rational numbers iii) For all a R, a + 0 = a and a(1) = a.
7 22 iv) For every a R, a + (−a) = 0
−5, , 0, , 7, 12.
3 7
1
Terminating or infinite periodic decimals. for every b R − {0}, b = 1.
b
25 v) For any a, b R, a + b = b + a and ab = ba.
−5.0, −2.333 ...., = 0.252525 ....,
99
vi) For a, b, c R, a(b + c) = ab + ac.
2
= 0.66666 ...., 7.14527836231231231 ....
3 vii) For a, b R, a < b if and only if b −a > 0.
Number Line viii) For any a R, a2 ≥ 0.
It is a horizontal line with origin, to represent 0, ix) For any a, b R, a = b or a < b or a > b.
another point marked to right of 0 to represent 1.
x) If a, b R and a < b, then a +c< b+cfor all c R
Distance from 0 to 1 defines one unit of length.
xi) If a, b R and a < b, then ax < bx for all x > 0.
For any x, y Q if x < y, then x is to the left of y;
xii) If a, b R and a < b, then ay > by for all y < 0.
x y
x< <y 1. Classify each element of as a member of N, Q, R
2
1 22
Between any two distinct rational numbers there is − Q or Z. {√7, , 0, 3.14, 4, }
4 7
another rational number between them.
1
√7 R−Q Q,
Irrational Numbers 4
There are points on the number line that are not 0 Z,Q 3.14 Q
represented by rational numbers. 22
4 N, Z,Q Q
Numbers on number line that do not correspond 7
to rational numbers are irrational numbers. 2. Proof that 2 is not a rational number.
Set of all irrational numbers is denoted by Q’ Suppose that √2 is a rational number.
m
N W Z Q R. √2 = n where m, n positive integers with
Every real number is either a rational number or no common factors greater than 1.
an irrational number, but not both. m2 = 2n2,
implies that m2 is even and hence m is even.
The set R of real numbers can be visualized as the
set of points on the number line such that if x < y, m = 2k.
then x lies to left of y. 2n2 = 4k2
R = Q Q’ and Q ∩ Q’ = n2 = 2k2.
Thus, n is also even. 2.2 EXPONENTS AND RADICALS
m,n even numbers having a common factor 2. Exponents
Thus, we arrived at a contradiction. Let n N, a R. an = a . a ....a (n times).
Hence,√2 is an irrational number. For m, n Z and a 0, a m a n a m n .
3. Prove that √3 is an irrational number. am
For m, n Z and m >n, n
a m-n.
Suppose that √3 is a rational number. a
1
√3 = m
where m, n positive integers with If m is a negative integer am .
n
a -m
no common factors greater than 1.
For any a 0, a 0 1.
m2 = 3n2,
For all x, y R, (a x ) y a xy
m2 is even and hence m is even.
For all x R, (ab) x a x b x
m = 3k.
Radicals
3n2 = 9k2
i) For n N, n even/odd, and b > 0, there is a
n = 9k .
2 2 Thus, n is also even. unique a > 0 such that an = b.
m,n even numbers having a common factor 3. ii) For n N, n odd, b R, there is a unique a
R such that an = b.
Hence,√3 is an irrational number.
a is called nth root of b or radical denoted by
4. Are there two distinct irrational numbers such 1
that their difference is a rational number?. a bn or a n b
Let 4 +√3, 2 +√3 be distinct irrational numbers iii) If n = 2, then nth root is called square root
Difference = 4 +√3 – (2 +√3) if n = 3, then it is called cube root.
= 4 +√3 – 2 –√3 ii) x2 = a2, has two solutions x = a, x = −a;
=2 is a rational number but a 2 |a|
5. Find two irrational numbers such that their sum 1
| a| if n is even ,
is a rational number. iv) For n N and a 0 ( a n ) n .
a if n is odd
Assume 2 +√3, 2 −√3 are two irrational numbers Conjugates.
Sum = 2 +√3+2 −√3 Let u, v, b be rational numbers , b is positive.
= 4 is a rational number u + v√b, u − v√b are called conjugates.
6. Find two irrational numbers whose product is a (u + v√b)(u − v√b) = u2 − bv2 is rational.
rational number.
If u + v √b appears in the denominator
Assume 2 +√3, 2 −√3 are two irrational numbers
multiply both Nr. and Dr by its conjugate
Product = (2 +√3) (2 −√3) u−v√b, to get rational number in denominator.
= 22 – √32 Exponential Function
= 1 is a rational number For any a > 0 and x R, If a = 1, define 1x = 1.
1 ax, x R for 0 < a 1 is called exponential
7. Find a positive number smaller than 1000 .
2 function with base a.
1
1000 .
2 1000 . ax may not be defined if a < 0 and x = 1
2
2 1001 . Special Exponential Function
1 f(x) = ex, x R
10001
2
1. Exponential funcion is a one-to-one and onto 1
3 3
1
iii) (-1000) 3 (-10 )
f(x) a x , x R Let a 2
-10.
f(x) 2 x , x R. . 3 3
1
Suppose f(u) f(v) for some u, v R. iv) 49 2 49 2
2u 2 v,
3
7
2u
1, 343
2v
3
2 u-v 1.
v) (-49) 2 :
2 u-v 2 0 There is no real number x as
u-v 0
x 2 - 49 is meaningless
Hence u v. 2 2
Thus f is a one - to - one function. vi) (125 ) 3 ( 5 3 ) 3
From the graph values of 25
3 3
f(x) = 2x increase as x
vii) 16 4 (24 ) 4
values increase and the
range of f is (0, ∞). 2 3
As 20 = 1, 2x > 1 for all x > 0 1
and 2x < 1 for all x < 0. 8
2 2
Thus f : R → (0,∞) is onto. viii) (-1000 ) 3 (-10 3 ) 3
2. Value of e by Compound Interest formula ( 10) 2
r nt 1
A P 1 ......(1)
100
n
1
A total amount ix) ( 3 -6 ) 3 3 - 2
P principal 1
n No. of compounding periods 9
2 2
t number of years
27 3 (3 3 ) 3
When P 1, r 1, t 1 x) 1
1
1 n 27 3 (3 3 ) 3
A P 1 ....(2)
n 32
When, n 10,100,1000,.... 3 1
A 2.593,2.704, 2.718,.... 3 21
(1) becomes, A Pe rt
1
e irrational number 3
1 3 3
e 2.718
xi) ( 256 ) -1 / 2 4 ( 2 8 ) 8
1 1
3. Simplify : i) 4 (-2) 4 ii) 343 3 iii) (-1000) 3 8
1
3 3 2 3
iv) 49 2 v) (-49) 2 vi) (125) 3 vii) 16 4 4. Simplify: (x 8 y 4 ) 4 ; where x, y ≥ 0
1 1 1
2 4 2
2 1
27 3 1
3 (x 8 y 4 ) 4 (x 2 ) 4 (y 2 ) 4
vii) (-1000 ) 3 ix) ( 3 -6 ) 3 x) xi)
( 256 ) -1 / 2 4
1
x 2 | y|
27 3
5. Simplify: (x1 / 2 y 3 )1 / 2 ; where x, y ≥ 0.
1
i) 4 (-2) 4
24 4 (x1 / 2 y 3 )1 / 2 x 1 / 2
1/ 2
( y 3 )1 / 2
2 1
1 1 x4
ii) 343 3 7 3 3 3
y2
7
1
6. Simplify by rationalising the denominator.
x 2 10x 25 ( x 5) 2
5 7 6
i) ii) |x 5|
6 2 3 2
1 1 1 1 1 9
iii) 10. If, (x 1 / 2 x -1 / 2 ) 2 find (x 1 / 2 x -1 / 2 ) for x > 1.
3 8 8 7 7 6 6 5 52 2
1 2 9
i)
5
5
6 2 (x1 / 2 1/2 )
x 2
6 2 6 2 6 2
1 9
5( 6 2 ) x 2
x 2
( 6 )2 ( 2) 2
1 5
30 10 x
x 2
4 2
2( x 1) 5x
7 6 7 6 3 2
ii) 2 x 2 5x 2 0
3 2 3 2 3 2
1
21 7 2 3 6 12 x 2, x
2
92
x 2, as x 1
21 7 2 3 62 3
1 1
7 Now , x 1 / 2 1 / 2 2
1 1 x 2
iii) 3 8 8 7 1
3 8 8 7
1 1 2
7 6 6 5 3 2 n 9 2 3 -n
7 6 6 5 11. Find the value of n if 27 .
1 33 n
52 2
52 3 2 n 3 2 3 -n 3 3n 33
Adding 3 8 8 7 7 6 3 2 n 4 n 3n 3 3
.
6 5 52 3 4 2 n 33
5 4 2n 3
x2 1
7. If x 2 3 Find 1
x2 2 n
2
2
x 2 2 3 12. Find the square root of 7 − 4√3.
232 6 5 2 6 74 3 a b 3
2
x 1 5 2 6 1 Squaring on both sides
x2 2 5 2 6 2 7 4 3 a 2 3b 2 2ab 3.
62 6 Comparing both sides
3 2 6 2ab 4 a - 2/b.
6 2 6 32 6 2 2
a 3b 7
3 2 6 3 2 6
(-2 /b) 2 3b 2 7
18 6 6 12
4
9 24 3b 2 7
22 6 b2
3b 4 -7b 2 4 0.
5
8. Find the radius of the spherical tank whose ( 3b 2 1) ( 4b 2 3) 0
volume is 32 / 3 units.
2
Radius r b 1 or b
3
π Since b is rational, b 1 , a 2.
Volume 32
3
Since 7 - 4 3 0,
4 3 π
r 32
3 3 7 - 4 3 2 - 3.
3
r 8
r 2
9. Simplify: x 2 10x 25
2.3. LOGARITHM iv) Power Rule
With a base 0 < a 1, the exponential function For any x > 0 and r R, loga xr = r loga x.
f(x) = ax is defined on R having range (0, ∞). Let log a x u.
As f(x) is a bijection, it has an inverse. au x
Its inverse function is called as logarithmic x r (au )r
function denoted by log a(.). x r a ru.
Taking logwith base a
If f(x) takes x to y = ax, then loga (.) takes y to x.
log a x log a a ru
r
For 0 < a 1, y = ax is equivalent to ru
r
loga y = x. log a x r log a x
Properties of Logarithm v) Change of base
i) a loga x = x for all x (0, ∞) For any x > 0, with a and b as bases,
log a a y y for all y R. logb x =loga x loga b .
log a 1 0 Let log b x v.
b v x.
ii) Product Rule
Taking log with base a, log a b v log a x.
For any x, y > 0, loga(xy) = loga x + loga y.
by Power rule, v log a b log a b
For x, y 0 Exponential form log a x
u
v
let log a x u, a x log a b
log a y v, av y log a x
log b x , b 0
log a b
log a (xy) w. a w xy.
a w xy 1
log ab x log a x
b
au a v
1
au v log a x
log x a
w uv log a x. log b y. log x b 1
log a(xy) log a x log a y log y x. log z y. log a z log a x
iii) Quotient Rule vi) common logarithm: log10 x
x
For any x, y > 0, loga = loga x − loga y vii) natural logarithm : loge x denoted by ln x.
y
When we write log x we mean loge x.
For x, y 0 Exponential form
viii) binary logarithm log2 x used in computer
let log a x u, au x
Graph of logarithmic and exponential functions
log a y v, av y
x
log a w.
y
x
aw .
y
au
av
a uv
w uv
x
log a log a x log a y
y
50 7
1. Evaluate : i) log a 35 ii) log a log 2 x log 4 x log 16 x
3 2
iii) log a 22 x iv) log 5 50 1
change of base, log a x
log x a
log a 35 log a ( 7 5 )
1 1 1 7
log a 7 log a 5 ( Product Rule)
log x 2 log x 2 2 log x 2 4 2
50
log a log a 50 - log a 3 (Quotient Rule) 1 1 1 7
3
log x 2 2 log x 2 4 log x 2 2
log a 22 x x log a 22 (Power Rule)
Let a log x 2.
log10 10
log 5 50 (Change of base) 1 1 1 7
log10 5
a 2a 4a 2
2. If the logarithm of 324 to base a is 4, then find a. 7 7
log 2 3 1728 6 .
4a 2
log a 324 4 1
a
a 4 324 2
1
34 ( 2 )4 log x 2.
2
a 4 (3 2 ) 4 1
x2 2
a .3 2
x4
75 5 32
3. Prove log 2 log log log2 . 7. Find the logarithm of 1728 to the base 2√3.
16 9 243
Let log 2 3 1728 x.
75 9 2 32 x
log 2 3 1728
16 5 243
2 6 33
52 3 34 25
log 4 2 5
2 5 3 2 6 ( 3) 6
x
log 2 2 3 (2 3)6 .
a2 b2 c2 x 6
4. Prove log log log = 0.
bc ca ab .
logx logy log(xy) 8. Solve x log 3 x 9
2 2 2 2 2 2
a b c a b c Let log 3 x y.
log log log log . .
bc ca ab bc ca ab x 3y
log 1 But x log 3 x 9 (given)
0
(3 y ) y 9
5. Prove that 2
16 25 81 3y 32 .
log 2 16 log 12 log 7 log 1
15 24 80 y2 2
using product and power rule
y 2
16 16 25 12 81 7
log 2 Thus x 3 2 , 3 2
15 24 80
9. Compute log 3 5 log 25 27
2 4 16 5 2 12 34 7
log 2 log 3 5 log 25 27 log 3 5 log 25 33 .
3 5 2 3 3 2 4 5 .
log 3 5 3 log 25 3
64 24 28
2 5 3 By Changing Base
log 21 16 16 36 12 28 7
3 5 2 3 2 5 3 log 25 5
65 24 28
2 5 3 3 log 5 5
log 28 23
64
3 5 2 log 5 25
log 10 3
1 2 log 5 5
7 3
6. If log 2 x log 4 x log 16 x find the value of x. .
2 2
10. Given that log102 = 0.30103, log103 = 0.47712 1 1 1
LHS log a a . log b b. log c c
(approximately), find number of digits in 2 8 312 . 2 2 2
Let N 10 n b where 1 b 10. 1 1 1
.
Taking log to base 10 2 2 2
1
log N log (10 n b)
8
n log 10 log b n(n 1)
15. log a log a 2 log a 3 .... log a n log a
n log b. 2
LHS log a 2 log a 3 log a .... n log a
log N log 2 8 312
(1 2 3 ..... n ) log a
8 log 2 12 log 3
n(n 1)
8 0.30103 12 0.47712 log a.
2
8.13368. log x log y log z
16. If then prove that xyz = 1
n log b 8.13368. y-z z-x x-y
Since 1 b 10 log x log y log z
k
Number of digits is 9. y-z z-x x-y
11. Compute log 9 27 log 27 9 Let log a x k( y z )
log 9 27 log 27 9 log 3 2 27 log 33 9 x a k ( y z )
1 Simillarly y a k ( z x )
using log a b x log a x, log a a 1
b
z a k(xy )
1 1
log 9 27 log 27 9 log 3 33 log 3 32 xyz a k ( y z z x x y )
2 3
3 2 1
log 3 3 log 3 3
2 3 17. Solve log 2 x 3 log 1 x 6
2
3 2
log 2 x 3 log 1 x 6
2 3
2
5
log 2 x 3 log 2 1/ x 6 log 2 2
6
12. Solve log8 x + log4 x + log2 x = 11. 1
log 2 x log 2 3
log 2 2 6
log 8 x log 4 x log 2 x 11 x
log 23 x log 22 x log 2 x 11 1
log 2 ( x 3 ) log 2 2 6
x
1 1
log 2 x log 2 x log 2 x 11 log 2 x 4 log 2 64
3 2
1 1 x 4 64
log 2 x 1 11
2 3 x 4 64 0
6
log 2 x 11 ( x 2 8)( x 2 8) 0
11
6
x 2 64 x2 2
a b 1 18. Solve log 5 -x (x 2 -6x 65 ) 2.
13. If a2 + b2 = 7ab, S.T log (log a log b ) .
3 2 log 5 -x (x 2 -6x 65 ) 2
a 2 b 2 7 ab x 2 -6x 65 (5 x ) 2
.
( a b ) 2 7 ab 2ab x 2 -6x 65 25 x 2 10x 0
a b 3 ab x 10
taking log on bothsides 19. Solve log 4 2 8x 2 log 2 8 .
1 log 4 2 8x 2 log 2 8.
log( a b ) log 3 (log a log b )
2 3
1 log 2 2 2 8x 2 log 2 2
log( a b ) log 3 (log a log b )
2 1
8x log 2 2 2 3 log 2 2
a b 1 2
log (log a log b )
3 2 4x 8
1 x2
14. log a2 a log b 2 b log c 2 c
8
2.4 POLYNOMIAL FUNCTIONS f(x) = q(x)g(x).
I. Polynomial If g(x) = x−a, remainder r(x) should have degree
zero and hence r(x) is a constant.
anxn + an−1xn−1 + ....+ a0, where ai R, i = 0, 1, 2,
...., n, is a polynomial in variable x. To determine the constant
f(x) = (x − a)q(x) + c.
When an 0, polynomial has degree n.
Substituting x=a
a0, a1, . . . , an R are coefficients of polynomial.
c = f(a).
a0 is constant term
VI. Remainder Theorem
an is leading coefficient
If a f(x) is divided by x−a, remainder is f(a).
P(x) = anxn +an−1xn−1 +....+a0 is a polynomial function
defined from R to R. Remainder c = f(a) = 0
II. Polynomials Based on Degeree if and only if x − a is a factor for f(x).
Polynomial with degree 1 is linear polynomial. A real number a is said to be zero of polynomial f(x)
Polynomial with degree 2 is quadratic polynomial. if f(a) = 0.
A cubic polynomial has degree three. If x = a is a zero of f(x), then x − a is a factor for f(x).
Degree 4 polynomials are quartic VII. Conjugate of a Zero
Degree 5 quintic polynomials Let P(x) is a polynomial with rational coefficients.
III. Like and Equal Polynomials Let zero of P(x) = a +b√p
f(x) = anxn + an−1xn−1 + ....+ a0, an 0 where a, b Q, p is a prime,
g(x) = bmxm + bm−1xm−1 + ....+b0, bm 0 Its conjugate a − b√p is also a zero.
i) f(x) andg(x) are like if and only if n = m VIII. Multiplicity
ii) f(x) = g(x) for all x R. A polynomial function of degree n can have at
most n distinct real zeros. polynomial P(x) = x2 +
if and only if , n = m ak =bk,k=0,1,2, ..n.
1 has no real zeros
IV. Addition, Subtraction, Multiplication, Division
f(x) is expressed as f(x) = (x − a)k.g(x) , g(a) 0,
i) Degree of P(x)+Q(x) is at most the
Value of k depends on a, It cannot exceed
maximum of degrees of P(x) and Q(x).
degree of f(x).
ii) Degree of P(x)Q(x) is the sum of the
k is called the multiplicity of the zero a.
degrees of P(x) and Q(x),
Root which has multiplicity 1, is a simple root.
p(x)
iii) is a rational function, for x R, g(x) 0.
q(x) IX. Important Identities
V. Division Algorithm 1. (x + a)3 = (x + a)2(x + a)
= x3 + 3x2a + 3xa2 + a3
For polynomials f(x) and g(x), where g(x) 0, there
exist two polynomials q(x) and r(x) such that = x3 + 3xa(x + a) + a3
2. (x − b)3 = x3 − 3x2b + 3xb2 − b3
f(x) = q(x)g(x)+r(x)
= x3 − 3xb(x − b) + b3
where degree of r(x) < degree of g(x).
3. x3 + a3 = (x + a)(x2 − xa + a2)
q(x) is quotient polynomial
4. x3 − b3 = (x − b)(x2 + xb + b2)
r(x) is remainder polynomial. 5. xn− an = (x − a)(xn−1 + xn−2a + ....+ xn−k−1ak +....+an−1)
r(x) is zero polynomial, q(x), g(x) are factors of f(x) 6. xn+bn=(x+b)(xn−1−xn−2b+...+xn−k−1(−b)k +...+(−b)n−1)
1. Write the Degrees of given Polynomials f(x) 4x 2 25.
i) 100x7 − πx5 + 20√2x2 + 7x + 1.22 ( 2x 5)( 2x 5)
2 2
Degree 7. x ,
5 5
ii) (17x − 3)(x + 3)(2x −√π)(x + 2.3)
6. Find quadratic polynomial which has one of its
Degree 4. zeros 1 +√5 and it satisfies p(1) = 2.
iii) (x2 + x + 1)(x3 + 2x + 2)(x5 − 5x +√3) One of the zero 1 5 , other Zero 1 5
p( x ) d( x - )( x )
Degree 10.
d( x 1 5 )( x 1 5 )
2. If P(x) = 2x3 + 7x2 − 5, Q(x) = x4 − 2x3 + x2 x+ 1,
d[( x 1) 2 ( 5 ) 2 ]
Find their sum and product
d( x 2 2x 4)
P(x)+Q(x) = x4 + 8x2 + x−6
Also p(1) 2 (given)
P(x)Q(x) = 2x7 +3x6 − 12x5 +4x4 +19x3 +2x2 − 5x − 5 d[12 2(1) 4] 2
3. If f(x) = x3−3px+2q is divisible by g(x) = 2
d
x2+2ax+a2, prove that ap+q = 0 5
2 2
Degree of f ( x ) 3 Rqd polynomial : ( x 2 x 4)
5
leading coefficient 1. 7. Find a quadratic polynomial f(x) such that,
f ( x ) ( x b)g( x ), for some b x f(0) = 1, f(−2) = 0 and f(1) = 0.
3 2 2
x 3px 2q ( x b)( x 2 ax a ).
f( 0 ) 1, f( 2 ) 0 , f(1) 0.
Equating like coefficients
i.e Zeros are x 2 , 1
2a b 0
f ( x ) d( x - )( x )
2
a 2ab 3p d( x 1 )( x 2)
2q ba 2 . Also f (0) 1 (given)
Thus, b 2a, d( 0 1 )( 0 2) 1
2
p a 1
d
q a . 3 2
1 2
Now, q a 3 Rqd polynomial : ( x x 2)
2
a( a 2 )
8. Construct a cubic polynomial function having
q - ap
2
ap q 0. zeros at x = ,1 +√3 such that f(0) = −8.
5
4. If x2 + x + 1 is a factor of the polynomial 3x3 + 8x2 One of the zero 1 3 , other Zero 1 3
+ 8x + a, then find the value of a. p( x ) d( x - )( x )( x )
f ( x ) 3x 3 8x 2 8x a 2
d( x 1 3 )( x 1 3 )( x )
5
g( x ) x 2 x 1
2
Degree of f ( x ) 3 d[( x 1) 2 ( 3 ) 2 ]( x )
5
leading coefficient 3. 2
d( x 2 2x 2)( x )
f ( x ) (3x b ) g( x ), for b x 5
3x 8x 8x a (3x b )( x 2 x 1)
3 2 Also f (0) 8 (given)
2
Equating x 2 , 3x 2 bx 2 8x 2 d( 0 0 2)(0 ) 8
5
b 38 b5 d 10
Equating constant, a b 2
Rqd polynomial : 10( x )( x 2 2x 2)
Thus, a 5 5
5. Find the zeros of polynomial f(x) = 4x2 − 25. 10x 24x 2 12x 8
3
9. Find the roots of the polynomial equation Now, S( n + 1) - S(n) = n + 1
(x−1)3(x+1)2(x+ 5) = 0. State their multiplicity. a+b( n+1) + c(n+1) 2 [ a+bn+cn 2 ] = n + 1
b + 2cn + c = n + 1
Let f(x) = (x−1)3(x+1)2(x+ 5)
Equating like coefficien ts :
x = 1,−1,−5. 1
Thus, 2c = 1 c
Roots are : 1 with multiplicity 3, 2
1
−1 with multiplicity 2 b+c=1 b=
2
−5 with multiplicity 1. S(1) = 1
a+b+c=1
10. If x = −2 is one root of x3 − x2 − 17x = 22, then
a=0
find the other roots of equation.
1 1
S(n) = n + n 2
f(x) x 3 x 2 17x 22 2 2
Let g(x) ( x 2) [ x - 2 is a root of p(x)] n(n + 1)
= , n N
Degree of f ( x ) 3 2
leading coefficien t 1 2.5. QUADRATIC FUNCTIONS
f(x) g( x )( h( x ) I. Quadratic function.
x x 17 x 22 ( x 2)( x 2 bx c )
3 2
P(x) = ax2 + bx + c,
Comp. coeff. 2c 22 c 11
where a, b, c R are constants and a 0
cx 2bx 17 x
If P(t) = 0 for some t R, t is a zero of P(x).
11 2b 17 b 3
II. Formation of Quadratic Equation
x 3 x 2 17 x 22 ( x 2)( x 2 3x 11)
Using quadratic Formula x 2 Sx P 0
x 53 S Sum of roots
Other Roots x P Product of Roots
2
If roots/Zeros are ,
11. Find the real roots of x4 = 16.
2
x ( )x 0
x 4 16
III. Relation between roots and Coefficients
x 4 16 0
If and are roots of ax 2 bx c 0,
x 2 2 42 0
b
( x 2 4)( x 2 4) 0 sum of roots
a
( x 2 4)( x 2)( x 2) 0 c
Product of roots
a
x 2 4 R
IV. Quadratic Formula
Thus x 2
If P(x) = ax2 + bx + c, its solution
12. Use the method of undetermined coefficients to
b b 2 4ac
find the sum of x
2a
Let S(n) = n + (n - 1) + (n - 2) + .... + 2 + 1 V. Nature of Roots
= n+(n-1)+(n-2 )+ ....+[n -(n-2 )]+[n-(n-1)] if b 2 4ac > 0, real and distinct roots
n-1 n-2 n-(n-2 ) n-(n-1)
= n 1+
2
+ +....+ if b 4ac = 0, roots are real and equal
n n n n
if b 2 4ac < 0, no real root
n-1 n-2
since < 1, < 1, ....
n n VI. Condition for graphs of f(x) to intersect x-axis
S(n) = n [1 + 1 + .... + 1] if b2 − 4ac > 0 intersects x-axis in two places
Thus, S(n) = n 2 . if b2 − 4ac = 0 touches x-axis at only one point
2
Let S(n) = a + bn + cn , where a, b, c, R. if b2 − 4ac < 0 does not intersect x-axis
1. Solve (2x + 1)2 − (3x + 2)2 = 0.
( 2x 1) 2 ( 3x 2 ) 2 0 6. Factorize: x4 + 1.
4x 2 4x 1 9x 2 12x 4 0 x 4 1. [ x 2 ]2 12
5x 2 8x 3 0 ( x 2 1 2x 2 ) 2 x 2
(5x 3)( x 1) 0 ( x 2 1) 2 ( 2x ) 2
3
x , or 1 ( x 2 1 2x )( x 2 1 2x )
5
( x 2 2x 1)( x 2 2x 1)
2
2. Solve the equation 6 4x x = x + 4.
7. Construct quadratic equation with roots 7, −3
(x + 4) ≥ 0 and 6 − 4x − x2 = (x + 4)2 .
x 2 ( )x 0
x ≥ −4 and x2 + 6x+5 = 0. x 2 4x 21 0
Thus, x = −1,−5. 8. Discuss the nature of roots of
x = −1 satisfies both the conditions. i) −x2 + 3x+1 = 0
Hence, x = −1. Discriminant b 2 4ac
94
3. Solve: x = √x + 20 for x R.
13
√x + 20 is defined only if x + 20 ≥ 0. 0 real and distinct
√x + 20 ≥ 0 is positive. ii) 4x2 − x − 2 = 0
So, x is positive.
real and distinct
squaring x2 = x + 20.
iii) 9x2 + 5x = 0.
x2 − x − 20 = 0
real and distinct
(x − 5)(x + 4) = 0 9. Find the complete set of values of a for which
x = 5, x = −4 the quadratic x2 − ax + a + 2 = 0 has equal roots.
Since, x is positive, solution x = 5. x 2 ax a 2 0 has equal roots.
4. Solve by completing square : ax2 + bx + c=0 So 0.
2
ax 2 bx c 0 b 4ac 0
2
b c a 4( a 2) 0.
x2 x 0
a a a 2 4a 8 0
b b 2 c b 2
x 2 2x using quadatic formula
2a 2a a 2a
4 16 32
2 2
x b b 4 ac a
2
2a 4a 2 2 12., 2 12
b b 2 4 ac
x 10. Find whether graphs of given functions will
2a 2a
intersect x-axis and if so in how many points.
b b 2 4ac
x i) y = x2 + x + 2
2a 2a
b b 2 4 ac
not intersect
x
2a ii) y = x2 − 3x − 7
5. Write f(x) = x2 + 5x + 4 in completed square form. intersect at two points
2
f ( x ) x 5x 4 iii) y = x2 + 6x + 9.
5 5 2 5 2 touch at only one point
x 2 2x. 4
2 2 2
iv) y = x2 − 4x + 5.
2
5 25
x 4 Since the graph does not intersect the x−axis,
2 4 x2 −4x+5 = 0 has no real roots.
5 2 3 2
x
2 2
11. If a and b are the roots of the equation x2 − px + q k(x 1) 2 5x 7
1 1
= 0, find the value of kx 2 2kx k 5x 7 0
a b
kx 2 ( 2k 5)x ( k 7 ) 0
Sum of roots a b p
2 (Given)
Product of roots ab q
2k 5
1 1 ab sum of roots 2
k
a b ab
2k 5
p
3k
q
2k 5 2
12. If α and β are the roots of the quadratic equation Taking square, 2 ......(1)
3k
x2 +√2x+3 = 0, form a quadratic polynomial with k7
1 1 product of roots ( 2)
zeroes , k
k 7
2 .....(2)
For given Equation : 2k
Sum of roots 2 2k 5 2 k 7
Sub (1) in (2),
Product of roots 3 3k 2k
For Required equation : 4 k 2 20 k 25 k 7
1 1 9k 2 2k
Sum of roots 8k 2 40k 50 9k 2 63k
α β
k 2 23k 50 0
( k 25)( k 2) 0
2 k 25, k 2
3 15. If the difference of roots of 2x2 −(a+1)x + a −1 = 0
1 1 is equal to product, prove that a = 2.
Product of roots
α β 2x 2 ( a 1)x ( a 1) 0
1 a 1
2
1 a 1
3 2
Form of Required equation : a 1
[given]
x 2 ( Sum of roots)x Product of roots 0 2
squaring both sides
2 1
x2 x 0 2
3 3 2 2 2 4 a 2 1
2 1
x2 x 0 a 1 2
3 3 ( ) 2 4
13. Find values of p for which difference between 2
2 2
the roots of the equation x2 + px+8 = 0 is 2. a 1 a 1 4 a 1
2 2 2
Let α and β be the roots of x2 + px+8 = 0.
a 2 2a 1 ( a 2 2a 1) 8( a 1)
α+β = −p 4 4
2 2
a 2 a 1 a 2 a 1 8a 8
αβ = 8
a2
|α − β| = 2. [Given] 16. Find the condition that one of the roots of ax2 +
(α + β)2 − 4αβ = (α − β)2, bx + c may be
i) negative of the other,
p2 − 32 = 4. b
a
p = 6.
b
[Given :
14. If one root of k(x − 1)2 = 5x − 7 is double the a
other root, show that k = 2 or −25. b0
ii) thrice the other, 18. The equations x2 −6x+a = 0 and x2 −bx+6 = 0 have
3 one root in common. The other root of the first
c and the second equations are integers in the
a ratio 4 : 3. Find the common root.
c
3. Let be the common root.
a
c
2 .....(1) Let , 3β be the roots of x2 − bx+6 = 0.
3a
b 3β = 6
3
a
β = 2
b
4
a Let , 4β be the roots of x2 − 6x + a = 0.
b2
By Squaring 2 ...(2) 4β = a
16 a
c b2 a = 8.
From (1) and (2)
3a 16 a
Roots of x2 −6x+8 = 0 are 2, 4.
3b 2 16 ac
iii) reciprocal of the other. If = 2, β =1
c
If = 4 β = ½ which is not an integer.
a
1 c 1 Hence, common root is 2.
a [given ]
2.6. LINEAR INEQUALITIES
c
1
a f(x) = ax + b, a, b R are constants, is a linear
ac function, because its graph is a straight line.
17. If x2 −ax+b = 0 and x2 −ex+f = 0 have one root in When x a , x ( a , ) a R
common and if the second equation has equal
roots, then prove that ae = 2(b + f). x a , x ( , a )
x 2 ex f 0 has equal roots x a, x [ a, )
x a , x ( , a ]
e2 4 f 0
For i ) x a and x b
e2 4 f
x (b , ) if a b
e0
Also x [b 2 4ac 0] x ( a , ) if b a
2
e e ii ) x a or x b
x ,
2 2 x ( a , ) if a b
e x (b , ) if b a
Let Common root .....(1)
2
Refer to Constants and Variable on Page : 7
Consider x 2 ax b 0
a 1. Represent the inequalities in interval notation:
a (i) x ≥ −1 and x<4
e [−1, 4)
from (1), a ....(2)
2 (ii) x ≤ 5 and x ≥ −3
b
[−3, 5]
b
(iii) x < −1 or x < 3
e 2b (−∞, 3)
from (1) and (2) a
2 e (iv) −2x > 0 or 3x − 4 < 11.
2b e
a (−∞, 5)
e 2
2ae 4b e 2 2. Solve 23x < 100 when
4b 4 f (i) x is a natural number,
ae 2(b f ) 1, 2, 3,4
(ii) x is an integer. 9. A girl A is reading a book having 446 pages and
. . . ,−3,−2,−1, 0, 1, 2, 3, 4 she has already finished reading 271 pages.
What is the minimum number of pages she
3. Solve −2x ≥ 9 when should read per day to complete reading the
(i) x is a real number, [−∞,−9/2] book within a week?
(ii) x is an integer, . . . ,−7,−6,−5 x = number of pages should read per day.
(iii) x is a natural number. no solution 7x + 271 ≥ 446.
4. Solve 3x − 5 ≤ x + 1 for x. x ≥ 25 must read at least 25 pages/ day
3x 5 x 1 10. To secure A grade one must obtain an average of
2x 6. 90 marks or more in 5 subjects each of maximum
100 marks. If one scored 84, 87, 95, 91 in first
x 3 four subjects, what is the minimum mark scored
solution set : ( , 3] in the fifth subject to get A grade in the course?
3(x 2 ) 5( 2 x) minimum mark to be scored in 5th subject x
5. Solve:
5 3 84 87 95 91 x
90
9(x 2 ) 25( 2 x) 5
9x 18 50 25x 357 x 450
34x 68 x 93
x2 11. A manufacturer has 600 litres of a 12 percent
(−∞, 2] solution of acid. How many litres of a 30 percent
acid solution must be added to it so that the acid
5x x content in the resulting mixture will be more
6. Solve: 4.
3 2 than 15 percent but less than 18 percent?
5x x8 30% acid solution to be added x liters
3 2 Then Total mixture (x 600) litres
10 2x 3x 24
30% x 12% of 600 15% of ( x 600)
5x 34
30% x 12% of 600 18% of ( x 600)
34
( , ) 30 12 15
5 or x (600) ( x 600)
100 100 100
7. Solve 3x − 9 ≥ 0, 4x − 10 ≤ 6. 30 12 18
and x (600) ( x 600)
3x − 9 ≥ 0 4x − 10 ≤ 6 100 100 100
30 x 7200 15x 9000
3x ≥ 9, 4x ≤ 16
and 30x 7200 18x 10800
x ≥ 3. x ≤ 4.
or 15x 1800 and 12 x 3600
Solution set [3,∞) (−∞, 4].
or x 120 and x 300,
intersection of these intervals [3, 4]. i.e. 120 x 300
8. A monthly electricity bill contains a basic between120 and 300
charge, which does not change with number of 12. Find all pairs of consecutive odd natural
units used, charge that depends on how many numbers both of which are larger than 10 and
units we use. Electricity Board charges Rs.110 as their sum is less than 40.
basic charge Rs. 4 for each unit we use. To keep
electricity bill below Rs.250, what should be smaller odd number x
electricity usage? other consecutive one x 2.
x Number of units used, x 0. x 10 ... (1)
Electricity bill Rs. 110 4x. x ( x 2) 40 ... (2)
Bill has to be below Rs.250, Solving (2), 2 x 2 40
110 4x 250. i.e., x 19 ... (3)
4x 140 From (1) and (3), 10 x 19
0 x 35. x is an odd number, x 11, 13, 15, and 17.
Usage shold be below 35 units Possible pairs :
(11, 13), (13, 15), (15, 17), (17, 19) vii) If the inequality is satisfied, then the required
13. A model rocket is launched from the ground. half plane is the one that contains P; otherwise
The height h reached by the rocket after t the required half plane is the one that does not
seconds from lift off is given by h(t) = −5t2+100t, contain P.
0 ≤ t ≤ 20. At what time the rocket is 495 feet
viii) When c 0, take P to be the origin.
above the ground?
1. Shade the region given by the inequality x ≥ 2.
h(t) 495 feet
i) Consider x = 2.
5t 2 100t 495
It is parallel to y axis at 2 units from it.
t 2 20t 99 0
(t 11)(t 9) 0 ii) It divides the cartesian plane into two parts.
t 9seconds, 11seconds iii) Substituting (0, 0) , 0 ≥ 2 which is false.
14. A plumber will be paid rupees 500 plus rupees iv) Region which does not contain the origin is
70 per hour in the first scheme, and in the represented by the inequality x ≥ 2.
second scheme he will paid rupees 120 per hour. v) Since x ≥ 2, points on x = 2 are also solutions.
If he works x hours, then for what value of x
does the first scheme give better wages?
Let x Working Hours of Plumber
500 70x 120x
500 50x
x 10
less than10 hours
15. A and B are working on similar jobs but their
salaries differ by more than Rs 6000. If B earns 2. Shade the region given by x + 2y > 3.
rupees 27000 per month, then what are the i) x + 2y = 3 divides cartesian plane into two half
possibilities of A’s salary per month? planes.
A' s salary per month x ii) Substitute (0, 0), 0 > 3 which is false.
B' s salary per month 27000
iii) Region given by x + 2y > 3 is the half plane
Salary Differnce 6000 which does not contain the origin.
x - 27000 6000
6000 x - 27000 6000
21000 x 33000
less than Rs.21, 000 or greater than Rs.33000
2.7 GRAPHICAL REPRESENTATION OF
LINEAR INEQUALITIES
i) A straight line ax + by = c divides the
Cartesian plane into two parts.
ii) Each part is an half plane.
iii) Vertical line x = c will divide the plane in left
and right half planes
3. Solve the linear inequalities and exhibit the
iv) Horizontal line y = k will divide the plane into solution set graphically:
upper and lower half planes.
x + y ≥ 3, 2x − y ≤ 5, −x + 2y ≤ 3.
v) A point in the cartesian plane which is not on
ax + by = c will lie in exactly one of the two i) (3, 0) and (0, 3) can be easily identified as two
half planes determined by line and satisfies points on the straight line x + y = 3.
the inequalities ax + by < c or ax + by > c. ii) Draw the three straight lines x + y = 3, 2x − y =
vi) To identify half plane represented by ax + by < 5 and −x + 2y = 3.
c, choose a point P in any one half planes and iii) (0, 0) does not satisfy x + y ≥ 3.
substitute coordinates of P in the inequality.
iv) Half plane bounded by x+y= 3, not containing 1
iii) Label −2 and 3 on Number line.
the origin, is the solution set of x + y ≥ 3.
v) Half-plane bounded by 2x − y ≤ 5 containing
origin is the solution set of the 2x−y ≤ 5.
vi) The region represented by −x + 2y ≤ 3 is the
half space bounded by the straight line the
line −x + 2y = 3 that contains the origin.
vii) Region common to three half planes is the
solution set of the given linear inequalities.
iv) Check the sign of (x + 2)(x − 1/3 ) by picking
arbitrary point anywhere in the interval.
Sign of x+2 1 3x2 +5x−2
Interval Sign of x − 3
(−∞,−2) - - +
(−2, 1/3) + - -
(1/3,∞) + + +
v) Inequality is satisfied in [−2, 13 ].
2. Solve 2x2 + x − 15 ≤ 0.
Proceeding as above
4. Solve 3x − 5 ≤ x + 1 graphically.
2x2 + x − 15 ≤ 0.
Let f(x) = 3x − 5 and g(x) = x + 1
2x2 +6x − 5x - 15 ≤ 0
(2x – 5)(x+3) ≤0
5
critical points −3 and 2
5
Inequality is satisfied in [−3, 2 ].
3. Solve −x2 + 3x − 2 ≥ 0.
x2 − 3x + 2 ≤0
(x −2) (x −1) ≤ 0
Find all the x-values for which f is below g. x = 2, 1
Solution [1, 2]
2.8.QUADRATIC INEQUALITIES 4. Solve x 14 < x+ 2.
i) First solve ax2 + bx + c = 0. x 14 is defined for x + 14 ≥ 0.
ii) If there are no real solutions, then one of the x ≥ −14, x + 2 > 0 implies x ≥ −2.
above inequality holds for all x R
(x + 14) < (x + 2)2
iii) If there are real solutions called critical points,
x2 + 3x − 10 > 0.
then label those points on the number line.
(x + 5)(x − 2) > 0.
iv) Critical points divide number line into disjoint
intervals. critical points x = −5 and x = 2.
v) Choose one number from each interval. Substituting a reference point in the sub-interval
vi) Substitute that numbers in the inequality. solution set to be x < −5 and x > 2.
vii) Identify intervals where inequality is satisfied. Since x ≥ −2, solution is x > 2.
1. Solve 3x2 + 5x − 2 ≤ 0. 2.9 RATIONAL INEQUALITIES
3x2 + 5x − 2 ≤ 0. Rational expression of x is the ratio of two
i) 3(x + 2)(x − 13 ) ≤ 0. polynomials in x, P(x) and Q(x) where Q(x) 0.
1 If degree of Nr. P(x) Dr. Q(x), then
ii) critical points −2 and 3
P(x) = f(x)Q(x)+r(x) 2.10. ABSOLUTE VALUE INEQUALITIES
where r(x) = 0 or degree of r(x) < Q(x) There is an order preserving one-to-one
Dividing both sides by Q(x) correspondence between elements of R and
points on the number line.
P(x) r(x)
f(x)
Q(x) Q(x) For each x R, x and −x are equal distance
from the origin.
x 1
Solve 3 Distance of number a R from 0 on the
x 3
number line is called the absolute value of that
x 1
3 . number a denoted by |a|.
x 3
x 1 x if x 0 ,
3 0 For any x R, |x| = .
x 3 x if x 0
x 1 3( x 3)
0 |.| is absolute value function from R onto [0, ∞)
x 3
2x 8 i) For any x R, |x| = | − x|
0
x3 thus, |x| = |y| if and only if x = y or x = −y
x 4
0 ii) |x − a| = r if and only if r ≥ 0
x 3
x−a = r or x − a = −r.
x + 4 and x + 3 are both positive or both negative.
Results For Absolute Value
To find out the signs of x + 3 and x + 4
i) If x, y R, |y + x| = |x − y|, then xy = 0.
x x+3 x+4 x 4
x 3 ii) For any x, y R, |xy| = |x||y|.
x < −4 − − + iii) |x||y| , for all x, y R and y _= 0.
−4 < x < −3 − + − iv) For any x, y R, |x + y| ≤ |x| + |y|.
x > −3 + + + Inequalities Involving Absolute Value
x = −4 − 0 0 i) |x| < r if and only if −r < x < r.
Solution set (−∞,−4) (−3, ∞). 1) If x ≥ 0, |x| = x, |x| < r implies x < r.
x 3(x 1) 2) If x < 0, |x| = −x, |x| < r implies x > −r
Find all values of x for which 0.
(x 2 ) Thus |x|< r if and only if −r < x < r,
x 3(x 1)(x 2 ) ii) |x| > r if and only if x < −r or x > r.
0.
(x 2 ) 2 1) If x ≥ 0, then |x| = x > r.
x 3(x 1)(x 2 ) 0 2) If x < 0, then |x| = −x > r, that is, x < −r.
Solution : (0, 1) (2, ) So |x| > r, if and only if x < −r or x > r,
Find all values of x that satisfies the inequality iii) For any a R,
2x - 3 1) |x − a| ≤ r if and only if −r ≤ x − a ≤ r if
0
(x - 2 )(x - 4 ) and only if x [a − r, a + r].
( 2x - 3)(x - 2 )(x - 4 ) 2) |x − a| ≥ r is equivalent to x − a ≤ −r or x −
0
(x - 2 ) 2(x - 4 ) 2 a ≥ r if and only if x (−∞,a−r][a + r, ∞).
( 2x - 3)(x - 2 )(x - 4 ) 0 1. Solve |2x − 17| = 3 for x.
, 3 (2, 4) 2x 17 3.
2 2x 17 3
x2 - 4 x 10 or x 7.
Solve 0.
x 2 - 2x - 15 2. Solve |2x − 3| = |x − 5|.
( x - 2)( x 2)( x 5)( x 3) 0. |x| = |y| if and only if x = y or x = −y.
( 3,2] [ 2,5) . |2x − 3| = |x − 5|
2x −3 = x − 5 or 2x − 3 = 5 − x.
8 −4 < x < 10
x = −2 and x =
3 10. Solve |4x − 5| ≥ −2.
8
Solutions. : −2, 4x − 5 ≥−2
3
for |x| > r,If r < 0, every x R satisfies inequality
3. Solve 3|x − 2|+7 = 19 for x.
11. Solve −3|x| + 5 ≤ −2 and graph the solution set
3 x 2 7 19.
in a number line.
19 - 7
x2 3 x 5 2
3
4. 3 x 7
x 2 4 or x 2 4. 7
x
Solutions are x 2 and x 6. 3
7 7
x
4. Solve |x − 9| < 2 for x. 3 3
−2 < x− 9 < 2 7 7
x (- , ) ( , )
3 3
7 < x < 11
12. Solve 2|x + 1| − 6 ≤ 7 and graph the solution set
2 in a number line.
5. Solve 1, x 4 .
x4
2x 1 6 7
2 > |x − 4|. 2 x 1 13
−2 < x− 4 < 2 x 4. 13
x 1
2<x<6 x 4. 2
13 13
solution set : (2, 4) (4, 6). x 1
2 2
3 1 15 11
6. Solve 3 - x x
4 4 2 2
1 3 1 1
3 - x 13. Solve 10x 2 1 .
4 4 4 5
1 3 1
3 3 - x 3 1
4 4 4 10x 2 1
13 3 11 5
x 10x 2 5
4 4 4
11 13 5 10x 2 5
x
3 3 3 10 x 7
7. Solve |x| − 10 < −3. 3 7
x 7 x
10 10
7x7 14. Solve |5x − 12| < −2.
1
8. Solve 6 and express the solution No solution
2x 1
using the interval notation. 15. Find the number of solutions of x2 + |x − 1| = 1.
1 i) For x 1, |x 1| x 1.
2x 1
6 2
1 1 x x 2 0.
2x 1 ( x 2)( x 1) 0,
6 6
7 5 x 2 or 1.
2x
6 6 x 1 , as x 1
7 5 ii) For x 1, |x 1| 1 x
x
12 12
5 7 x2 1 x 1
x (- , ) ( , )
12 12 x( x 1) 0
9. Solve |3 − x| < 7 x 0 or x 1,
x 0, as x 1
-7 < 3 − x < 7
Solution set : {0,1} i.e two solutions
-10 < -x <4
2.11. PARTIAL FRACTIONS x
4.
( x 1) 3
f(x)
A rational expression is called a proper x A B C
g(x) 3
2
( x 1) ( x 1) ( x 1) ( x 1) 3
fraction if the degree of f(x) < degree of g(x),
where g(x) can be factored into linear factors and x A( x 1) 3 B( x 1) C
quadratic factors without real zeros. Put x 1, x 0, x 1
f(x) C 1, B 1, A 0
can be expressed in simpler terms, as a sum
g(x) x 1 1
3
2
of expressions of the form ( x 1) ( x 1) ( x 1) 3
Result is called partial fraction decomposition. x 12
5.
Type 1: Linear factors, none of which is repeated (x 1) 2(x 2 )
x 12 A B C
x 2
2
1. (x 1) (x 2 ) x 1 ( x 1) x2
(x 3)(x 4 )
x A B x 12 A( x 2)( x 1) B( x 2) C ( x 1) 2
(x 3)(x 4 ) (x 3) (x 4 ) Put x 2, x 1, x 0
x A(x 4 ) (x 3) 14 11 14
A ,B , C
(x 3)(x 4 ) (x 3)(x 4 ) 9 3 9
x A(x 4 ) (x 3)B x1
6. 2
When x 4, x ( x 1)
4 A( 4 4 ) ( 4 3)B x1 A B C
2
2
x ( x 1) x x x 1
4
B
7 x 1 Ax(x - 1) B(x - 1) Cx 2
When x - 3, When x 0, B - 1
4 A( 3 4 ) ( 3 3 )B when x 1, C 2.
3 When x - 1,
A
7 2A - 2B C 0
x 3 4
Thus A - 2.
(x 3)(x 4 ) 7(x 3 ) 7(x 4 )
1 x1 2 1 2
2. Thus 2
2
x2 a2 x ( x 1) x x x 1
1 A B Type 3: Quadratic factors, none of which is repeated
2 2
x a x a x a
2x
1 A( x a ) B( x a ) 7. 2
(x 1)(x 1)
1
When x a , B 2x A Bx C
2a 2
2
1 (x 1)(x 1) ( x 1) (x 1)
When x a , A
2a 2x ( Ax B)( x 1) C ( x 2 1)
1 1 1 1 When x 1,
Thus
x a2 2 2 a ( x a x a A 1.
3x 1 When x 0,
3.
( x 2)( x 1)
A-C0
3x 1 A B
C 1.
( x 2)( x 1) x 2 x 1
When x - 1
3x 1 B( x 2) A( x 1)
2A - 2(C - B) - 2,
7 2
Sub. value of x , A , B B - 1.
3 3
3x 1 7 2 2x 1 1 x
2
2
( x 2)( x 1) 3( x 2) 3( x 1) (x 1)(x 1) ( x 1) (x 1)
Type 2: Linear factors, some of which are repeated x
8.
( x 2 1)( x 1)( x 2)
x A B Cx D 2x 2 5x 11 x5
2
2 2 2
( x 1)( x 1)( x 2) x 1 x 2 x 1 2
x 2x 3 x 2x 3
x A( x 2 1)( x 2) ( x 2 1)( x 1) x5 A B
(Cx D)( x 1)( x 2) ( x 3)( x 1) x 3 x 1
Sub. x 1, x 2, x 1 x 5 A( x 1) B( x 3)
x 1 2 3x 1 Put x 1, x 3
B 1, A 2
( x 1)( x 1)( x 2) 6( x 1) 15( x 2) 10( x 2 1)
2
2
1 2x 5x 11 2 1
9. 2
4 2
x 2x 3
x 3 x 1
x 1
2
1 Ax B C D x x 1
2 2 2 2
2 13. 2
( x ) (1 ) x 1 x 1 x 1 x 5x 6
1 ( Ax b )( x 2 1) C ( x 1)( x 2 1) x2 x 1 6x 5
2
1
x 5x 6 ( x 3)( x 2)
D( x 1)( x 2 1)
6x 5 A B
Sub x 1, x 1, x 0
( x 3)( x 2) x 3 x 2
1 1 1
B , A 0, C , D 6x 5 A( x 2) B( x 3)
2 4 4
1 1 1 1 Put x 3, x 2
4
2
A 7 , B 13
x 1 4( x 1) 4( x 1) 4( x 1)
2
( x 1) 2 x x 1 13 7
10. 2
1
3 x 5x 6 x 3 x 2
x x 3
( x 1) 2 x 2x 1
A Bx C 14. 2
2
x 5x 6
x( x 1) x x2 1
x 3 2x 1 21x 31
( x 1) 2 A( x 2 1) ( Bx C )x 2
( x 5)
x 5x 6 ( x 3)( x 2)
Put x 1, x 0
21x 31 A B
1
A 1, B 0, C ( x 3)( x 2) x 3 x 2
2 21x 31 A( x 2) B( x 3)
( x 1) 2 1 1
2
2
Put x 2, x 3
x( x 1) x 2( x 1)
A 32, B 11
7x
11. 3
x 2x 1 32 11
(1 x)(1 x 2 ) 2
( x 5)
x 5x 6 x3 x2
7x A Bx C
2
2
6x x 1
(1 x)(1 x ) (1 x) (1 x)(1 x 2 ) 15. 3
x x2 x 1
7 x A(1 x 2 ) ( Bx c )(1 x)
6x 2 x 1 6x 2 x 1
Put x -1 A 3
x 3 x 2 x 1 ( x 1)( x 2 1)
Put x 0
6x 2 x 1 A Bx C
A C 7 C 4 2
2
( x 1)( x 1) x 1 ( x 1)
Put x 1
6x 2 x 1 A( x 2 1) ( Bx C )( x 1)
A B C 4 B 3
Put x 1, x 0 , x 1
7x 3 4 3x
2
A 4, C 3, B 2
(1 x)(1 x ) 1 x (1 x 2 )
2
Type 4: improper fraction. 6x x 1 4 2x 3
3 2
2
x x x 1 x 1 x 1
Degree of numerator Degree of denominator,
2x 2 5x 11
12.
x 2 2x 3
Divide Nr by Dr
3.1 BASIC RESULTS OF TRIGONOMETRY One minute 1 corresponds to 1/60 of a degree
I. Angles A second (1’) corresponds to 1/60 of a minute (or)
AOB is formed by two rays OA and OB sharing 1/3600 of a degree.
common point O called vertex of angle.
IV. Classification of a pair of Angles
i) congruent angles
Two angles that have the exact same measure
ii) complementary angles
Two angles having their measures adding to 900
iii) supplementary angles.
If we rotate ray OA about its vertex O and takes
position OB, OA and OB are called initial side Two angles that have measures adding to 1800
and terminal side of the angle produced. iv) conjugate Angles
An anticlockwise rotation generates a positive Two angles between 0 and 360 are conjugate if
angle (positive sign), clockwise rotation generates their sum equals 360.
negative angle (negative sign).
V. Angles in Standard Position
One full anticlockwise or clockwise rotation of
An angle is in standard position if its vertex is at
OA back to itself is one complete rotation
origin and initial side is along the positive x-axis.
II. Different Systems of measurement of angle
An angle is in first quadrant, if in standard
i) Sexagesimal system position, terminal side falls in the first quadrant.
Right angle is divided into 90 equal Degrees. Angles in standard position having terminal sides
Degree is divided into 60 equal Minutes along x-axis or y-axis are quadrantal angles.
Each minute into 60 equal Seconds. 0, 90, 180, 270 and 360 are quadrantal angles.
1, 1’ and 1” denote a degree, a minute, a second Degree of quadrantal angle is multiple of 90.
ii) Centesimal system VI. Coterminal angles
Right angle is divided into 100 equal Grades One complete rotation of a ray in anticlockwise
direction results in an angle measuring of 360.
Each grade is subdivided into 100 Minutes
Each minute is subdivided into 100 Seconds.
1g denotes a grade.
iii) Circular system
Radian measure of an angle is introduced using
arc lengths in a circle of radius r. By continuing the anticlockwise rotation, angles
1c denotes 1 radian measure. larger than 360 can be produced.
III. Degree Measure If we rotate in clockwise direction, negative
angles are produced.
Unit of angle measurement represented by degree
Angles in standard position that have the same
One complete rotation is split into 360 equal parts terminal sides are coterminal angles .
and each part is one degree, denoted by 1. If and β are coterminal angles, β = + k(360), k
1 is 1/360 of one complete rotation. is an integer.
To measure a fraction of angle minutes and Coterminal angles differ by integral multiple of
seconds are introduced. 360
1. Express 59.0854into degrees, minutes, seconds I. Radian measure of an angle
59.0854 = 59 + 0.0854 Ratio of arc length it subtends, to radius of the circle in
60' which it is the central angle.
= 59 +.0854 o
1
Consider a circle of radius r
= 59 + 5.124’
Its arc length s
= 59 +5’ + 0.124’
60" arc length
= 59 +5’ +0.124’ Angle subtended at centre.
1' radius
s
59.0854 = 595’7.44” θ radian
r
2. Express Complementary of 3618’47”into Hence, s r
degrees, minutes, seconds
II. Properties of Radian measure
Complementary = 90 − 3618’47”
= 5341’13” i) All circles are similar.
3. Express Supplementary of 34.8597into degrees, For a given central angle in any circle, the ratio of the
minutes, seconds intercepted arc length to the radius is always constant.
34.8597 = 3451’35” ii) When s = r, θ = 1 radian.
Supplementary = 180 –3451’35” 1 radian is the angle made at the centre of a circle by
= 1458’25” an arc with length equal to radius of the circle.
4. Identify the quadrant in which given angle lies iii) s and r have same unit
(i) 25 (i) first quadrant
θ is unitless and no notation to denote radians.
(ii) 825 (ii) second quadrant
iv) θ = k radian measure, if s = kr.
(iii) −55 (iii) fourth quadrant
θ = 1 radian measure, if s = r
(iv) 328 (iv) fourth quadrant
(v) −230 (v) second quadrant θ = 2 radian measure, if s = 2r
5. Check whether 417 and −303 are coterminal. If Radian measure tells us how many radius lengths,
so find another coterminal angle need to sweep out along circle to subtend the angle θ.
If , β coterminal angles, β - = k(360) v) Radian angle measurement can be related to the
417− (−303) = 720 edge of the unit circle.
= 2 (360) we measure an angle by measuring the distance
417 and −303 are coterminal travelled along the edge of the unit circle to where the
terminal side of the angle intercepts the unit circle .
57, 417 and −303 have same initial side and
terminal side but with different amount of III. Difference b/w Degree and Radian Measures
rotations, so they are coterminal angles.
In measuring temperature, Celsius unit is better
6. Write coterminal angles pairs for 30, 280, -85
than Fahrenheit as Celsius was defined using 0
30, 390 280, 1000
and 100 for freezing and boiling points of water.
−85, 275
Radian measure is better for conversion and
7. Find a coterminal angle with measure of θ such
calculations.
that 0 ≤ θ < 360
(i) 395 35
Radian measure is convenient for analysis;
Degree measure is convenient to communicate
(ii) 525 165
the concept between people.
(iii) 1150 70
Scale used in radians is much smaller than the
(iv) −270 90
scale in degrees. Smaller scale makes graphs of
(v) −450 270 trigonometric functions more visible and usable.
IV. To convert radians into degrees or degrees into Solved Problems
radians 1. Express following angles in radian measure:
Example : i) 18 ii) −108
In unit circle, a full rotation corresponds to 360
i)1o radians. ii) 1o radians.
or 2π radians, the circumference of the unit circle. 180 180
18 108 o
2 π radians 360 o 18 o radians 108 o radians
180 180
radians 180 o 3
radians radians
180 o 180x o 10 5
1 radian x radian 3
i) 30 radians ii) 135 radians
x 6 4
1o radians. xo radians 41 5
180 180 iii) −205 radians iv) 150 radians
36 6
V. Properties of π 11
v) 330 radians
6
i) The ratio of the circumference of any circle to its 2. Find degree measure of given radian measures
diameter is always a constant. radians 180 o
o
This constant is denoted by irrational number π. 180
1 radians
22
ii) Values of π and correct to four decimal places o
7 180
i) 6 radians 6
are 3.1416 and 3.1429
o
22 180
π and are approximately equal correct upto 22 6
7 7
two decimal places. 7 o
343
22 11
Hence, π ≈ 180 o
7 ii) radians 36o iii) radians 30o
5 5 6
iii) 1 radian ≈ 57O 17’45” 2
iv) radians 20o v) radians 72o
1o ≈ 0.017453 radian 9 5
7 10
vi) radians 420o vii) radians 200 o
1’ ≈ 3 9
180 60 3. What must be the radius of a circular running
path, around which an athlete must run 5 times
≈ 0.000291 radian.
in order to describe 1 km?
Radius of circular path r
1” ≈
180 60 60 5 Circumference 1km
≈ 0.000005 radian. 5 2r 1000m
1000 7
VI. Radian measures and corresponding degree r
22 5 2
measures for some known angles 31.82 meters
4. In a circle of diameter 40 cm, a chord is of length
20 cm. Find the length of the minor arc of chord.
Ends of chord make equilateral traiangle
VII. Area of the sector Radius of circle 20cm
Central Angle 60 o
Area of sector in degree r 2
360 Length of minor arc r
Area of sector in radian r 2
2 60 o 20
2 180 0
r
20
2
3
Perimeter 2 r 20.95 cm
360
5. Find the length of an arc of a circle of radius 5 10. The perimeter of a sector of a circle is equal to
cm subtending a central angle measuring 15 the length of the arc of a semi-circle having the
radiusof circle r 5 cm same radius. Express the angle of the sector in
degrees, minutes and seconds.
central angle 15 o
r Radius sector and semicircle
15 Central angle of sector
180
Length of the arc s r Central angle ofsemicircle
Perimeter Length of arc of semicircle
5 15
180 r 2r r
2
cm
12 3.1416 - 2
6. What is the length of the arc intercepted by
1.1416 radian
central angle 41 in a circle of radius 10 ft?
radiusof circle r 10 ft Angle of sector 65 o 27'16"
central angle 41o 11. An airplane propeller rotates 1000 times per
minute. Find the number of degrees that a point
Length of the arc s r
on the edge of propeller will rotate in 1 second.
1 rotation 360 o deg
10 41
180 o
1000 rotation 360 o 1000 deg/ min
22 1
41 ft 360 o 1000
7 18 deg/ sec
7.16feet 60
7. Find the degree measure of the angle subtended 6000 0 / sec
at the centre of circle of radius 100 cm by an arc
of length 22 cm. 12. A train is moving on a circular track of 1500 m
radiusof circle r 100 cm radius at the rate of 66 km/hr. What angle will it
central angle turn in 20 seconds?
Length of the arc s 22 cm Radius r 1500m
speed 66 km/hr
100 22
180 66 1000
Distance covered in 1s
9 7 60 60
22
5 22 Distance covered in 20s 18.33 20
(12.6) o s r radian
12 o 0.6 60' 18.33 20
radian
1500
12 o 36'
8. If arcs of same lengths in two circles subtend 14 o
central angles 30, 80, find ratio of their radii. 13. Circular metallic plate of radius 8 cm thickness
r1 ,r2 radii of the two given circles 6 mm is melted and molded into a pie of radius
l1 , l 2 length of the arc. 16 cm thickness 4 mm. Find angle of sector
circular metallic plate with thickness is a cylinder
1 , 2 central anglestwo given circles
Sector : Cylinder :
l1 l 2 (Given)
Radius r 16cm Radius R 8cm
r1 1 r2 2
height h 0.4cm height H 0.6cm
r1 30 r2 80 Volume of sector Volume of Cylinder
180 180
r1 80 Area height Area height
r2 30 r 2 h R 2 H
r1 : r2 8 : 3
360
9. If in two circles, arcs of the same length subtend 8 8 0.6 360
angles 60,75 at centre, find ratio of their radii. 16 16 0.4
r1 60 o r2 75 o 135 o
r1 : r2 5 : 4 3
4
BASIC TRIGONOMETRIC RATIOS & IDENTITIES 5. If sin θ + cos θ = m, show that
tan sec - 1 1 sin 4 - 3 (m 2 - 1) 2
1. Prove that cos 6 sin 6 where m2 ≤ 2.
tan - sec 1 cos 4
tan sec (sec tan 2 )
2 Let, sin θ cos θ m
LHS
tan sec 1 ( sin θ cos θ) 2 m 2
(tan sec ) (sec tan )(sec tan ) sin 2 θ cos 2 θ 2 sin θ cos θ m 2
tan sec 1
2 sin θ cos θ m 2 1
(tan sec )[1 (sec tan )]
(m 2 1) 2
tan sec 1 sin 2 θ cos 2 θ .....(1)
(tan sec )[tan sec 1] 4
Now, cos 6 θ sin 6 θ ( cos 2 θ) 3 ( sin 2 θ)3
tan sec 1
tan sec ( cos 2 θ sin 2 θ)
1 sin 3 sin 2 θ cos 2 θ( sin 2 θ cos 2 θ)
cos
(m 2 1) 2
2. Prove that 1 3 using (1)
4
(secA−cosecA)(1+tanA+cotA)=tanAsecA−cotAcosecA
( sec A cos ecA)(1 tan A cot A) 4 - 3 (m 2 - 1) 2
4
sec A cos ecA sec tan A cos ecA tan A
2 sin 1 cos sin
sec A cot A cos ecA cot A 6. If y prove y
1 cos sin 1 sin
sec A sec A 2 sin 1 cos sin
sec A cos ecA sec cos ecA y
cos ecA cos ecA 1 cos sin 1 cos sin
cos ecA cos ecA multiply by (1 cos sin )
sec A cos ecA
sec A sec A 2 sin (1 cos sin )
sec A
sec A cos ecA sec sec A (1 cos sin )(1 cos sin )
cos ecA
2 sin (1 cos sin )
cos ecA
cos ecA cos ecA (1 sin cos )(1 sin cos )
sec A
sec tan A cos ecA cot A 2 sin (1 cos sin )
3. Eliminate θ from a cos θ = b and c sin θ = d, (1 sin ) 2 cos 2
a cos b c sin θ d 2 sin (1 cos sin )
b2 d2 1 sin 2 2 sin (1 sin 2 )
cos 2 ....(1) sin 2 ...(2)
a2 c2 2 sin (1 cos sin )
d2 b2 2 sin 2 sin 2
(1) ( 2), cos 2 sin 2 2 2
c a 1 cos sin
a d b 2c 2
2 2
1 sin
1
a 2c 2 cos 4 sin 4
7. If 1 prove that
a2c 2 a2d 2 b 2c 2 cos 2 sin 2
4. If a cos θ − b sin θ = c, show that cos 4 sin 4
i) sin 2 sin 2 2 sin 2 sin 2 ii)
a sin b cos a 2 b 2 c 2 . cos 2 sin 2
( a cos b sin ) 2 c 2 cos 4 sin 4
Let cos 2 sin 2
a 2 cos 2 b 2 sin 2 c 2 2ab sin cos ....(1) cos 2 sin 2
( a sin b cos ) 2 b 2 cos 2 a 2 sin 2 cos 4 2 2 sin 4
cos sin
2ab sin cos cos 2 sin 2
Using (1) cos 2 2 2 2 2 sin 2
(cos cos ) ( sin sin )
( a sin b cos ) 2 b 2 cos 2 a 2 sin 2 cos 2 sin 2
a 2 cos 2 b 2 sin 2 c 2 cos 2 sin 2
2
(cos 2 cos 2 ) (1 cos 2 1 cos 2 )
b 2 (cos 2 sin 2 ) cos sin 2
a 2 (cos 2 sin 2 ) c 2 cos 2 2 2 2 2 sin 2
(cos cos ) ( cos cos )
( a sin b cos ) 2 a 2 b 2 c 2 cos 2 sin 2
( a sin b cos ) a 2 b 2 c 2 .
cos 2 sin 2 cos 2 sin 2 1
(cos 2 cos 2 ) 2
2
0
cos sin cos sin 1 sin cos 2
2 2 2
cos 2 cos 2 0 1 sin 2 cos 2 cos 2 sin 2
cos 2 cos 2 ........(1) sin 2 cos 2 (1 sin 2 cos 2 )
cos 2 sin 2 1
xyz
cos 2 sin 2 sin cos (1 sin 2 cos 2 )
2 2
sin 2 cos 2 sin 2 cos 2 10. If sec θ + tanθ = p, obtain the values of sec θ, tan
θ and sin θ in terms of p.
sin 2 (1 sin 2 ) sin 2 (1 sin 2 ) 1
sec tan p sec tan
sin 2 sin 2 sin 2 sin 2 sin 2 sin 2 p
sin 2 sin 2 .....( 2) 1
By Adding 2 sec p
4 4 2
i) sin sin sin sin sin sin 2 2 2 p
sin 2 sin 2 sin 2 sin 2 p2 1
sec
2p
2 sin 2 sin 2
1
cos 4 sin 4 cos 4 sin 4 sin 1
ii) sec 2
cos 2 sin 2 cos 2 sin 2
4p 2
cos 4 sin 4 1
1 ( p 2 1) 2
cos 2 sin 2
8. Show that p 4 2p 2 1 4p 2
3 / 2.
sec tan 3 cos ec 2 k 2 if tan 2 1 k 2 ( p 2 1) 2
tan 2 1 k 2 ( p 2 1) 2
1 tan 2 1 1 k 2 ( p 2 1) 2
sec 2 2 k 2 p2 1
sec 2 k 2 p2 1
2 2
sin 3 1 sin p 1 p 1
3
sec tan cos ec sec . tan 2
cos p 1 2p
cos 3 sin
sec tan 2 sec p2 1
sec (1 tan 2 ) 2p
11. If cot θ (1 + sinθ) = 4m and cot θ (1 − sin θ) = 4n,
sec (sec 2 ) then prove that
(sec ) 3 cot (1 sin ) 4m ......(1)
3 / 2. cot cos 4m
2 k 2
Squaring both sides :
9. If 0 and x cos 2 n θ,y sin 2 n θ
2 n 0 n 0
cot 2 cos 2 2 cot cos 16m 2 ...(2)
2n 2n
z n 0
cos θ sin θ Prove that xyz x y z cot (1- sin ) 4n ..(3)
cot - cos 4n
x n0
cos 2 n θ
cot 2 cos 2 2 cot cos 16n 2 ...(4)
x 1 cos cos 2 ... is an infinite G.P
2 2
( 2) ( 4),4 cot cos 16(m n )
1
1 x x 2 x3 . . . cot cos
1 x (m 2 n 2 )
4
1 1
x x cot 2 cos 2
1 cos 2
sin 2 (m 2 - n 2 ) 2 ...(5)
16
1 1
y 2
z Now, (1) (3) Gives :
cos 1 cos sin 2
2
cot (1 sin ) cot (1 sin ) 16mn
1 1 1
xyz cot 2 (1 sin 2 )
cos 2 sin 2 1 sin 2 cos 2
mn
16
(m 2 - n 2 ) 2 mn
12. If cos ec sin a 3 and sec cos b 3 , TRIGONOMETRIC FUNCTIONS
2 2 2 2
then prove that a b ( a b ) 1 I. Trig. Functions in Cartesian coordinates
3
cos ec sin a
The trigonometric ratios to any angle in terms of radian
1 3
sin a measure are called trigonometric functions.
sin
1 sin 2 Let P(x, y) be a point other than the origin on the
a3
sin
terminal side of an angle θ in standard position .
cos 2 3
a
sin Let OP r.
2
2 cos 2 3 2 r x2 y2
Raising power to ,
3 sin
a3 3
y x
sin cos
2 2 r r
cos 2 32 2 sin 2 3 y x
sin a , Thus cos b tan cot
x y
2 2 2
r r
2 2 2 cos 2 sin 2 3 cos 2 3
2 sin 2 3 cos ec sec
a b ( a b )
sin cos sin cos y x
2 2
i) Since |x| ≤ r, |y| ≤ r,
2
cos 2 3 sin 2 3
(cos sin ) 3 |sin θ| ≤ 1 and |cos θ| ≤ 1.
sin cos
2
2
2
2
ii) Trigonometric functions have positive or negative
cos 2 3 3
sin 2 3 3
sin (cos sin ) cos (cos sin ) values depending on the quadrant in which the
2 2 point P(x, y) on the terminal side of θ lies.
2 2
cos 3 sin 2 3
(cos sin ) 3
sin cos iv) Trigonometric functions is independent of the
2 2 2 2 points on the terminal side of the angle.
cos
2 3 (cos ) 3
cos 2
3 (sin ) 3
2 2 II. Trigonometric ratios of Quadrantal angles
cos sin
1 Angle in its standard position for which terminal side
13. Eliminate θ from the equations a secθ− ctan θ= b
coincides with one of the axes, is a Quadrantal angle.
and b sec θ + d tan θ = c.
a sec θ - c tan b Consider the unit circle x2+y2 = 1.
b sec d tan c Let P(x, y) be a point on unit circle where terminal
Using cross multiplication
side of θ intersects the unit circle.
a sec θ - c tan b 0
b sec d tan c 0 x
cos x ( x - coordinate of P)
1
sec θ tan 1
2
2
y
c bd b ac ad bc sin y ( y - coordinate of P)
1
c 2 bd b 2 ac
sec , tan
ad bc ad bc Coordinates of any point P(x, y)
Squaring and subtractinging
P(cos θ, sin θ).
2 2
c 2 bd b 2 ac
sec 2 tan 2 III. Values of Trig. functions of quadrantal angles.
ad bc ad bc
2 2
c 2 bd b 2 ac
1
ad bc ad bc
( ad bc ) 2 ( ac b 2 ) 2 ( ac b 2 ) 2
(c 2 bd ) 2 ( ad bc ) 2 ( ac b 2 ) 2
i) x and y coordinates of all points on the unit circle The value of a trigonometric function of a real
lie between −1 and 1. number t is its value at the angle t radians.
Hence, −1 ≤ cos θ ≤ 1, −1 ≤ sin θ ≤ 1, no matter s = rθ
whatever be the value of θ. Arc length t = θ.
ii) When θ = 360, terminal side coincides with sin t = sinθ
positive x- axis.
cos t = cos θ.
Hence, sine has equal values at 0 and at 360.
sin t = sinθ = y
Other trigonometric functions follow it.
cos t = cos θ = x.
iii) If two angles differ by an integral multiple of 360
or 2π, then each trigonometric function will have B(x, y) = B(cos t, sin t) is a point on the unit circle.
equal values at both angles. −1 ≤ cos t ≤ 1 , −1 ≤ sin t ≤ 1 for any real number t.
IV. Geometrical generalization VI. Signs of Trigonometric functions
Consider a unit circle with centre at the origin.
Let θ be in standard position.
P(x, y) be point on unit circle corresponding to θ.
values of x and y are positive or negative depending
on the quadrant in which P lies.
tan θ is not defined when cos θ = 0
So,tan θ is not defined when θ = (2n +1)π/2,nZ cos θ = x,
V. Trigonometric Functions of real numbers sin θ = y
tan θ = y/x
Consider a unit circle with centre at the origin.
First quadrant:
Let the angle zero (in radian) be associated with
the point A(1, 0) on the unit circle. cos θ = x > 0 (positive)
Draw a tangent to unit circle at the point A(1, 0). sinθ = y > 0 (positive)
Let t be a real number such that t is y- coordinate cos θ and sin θ and all
of a point on the tangent line. trigonometric
functions are positive
For each real number t, identify a point B(x, y) on
unit circle such that the arc length AB is equal to t. Second quadrant:
If t is positive, choose B(x, y) in anti clockwise cos θ = x < 0 (negative)
direction, or choose in clockwise direction. sinθ = y > 0 (positive)
Let θ be the angle subtended by arc AB at centre. sin θ and cosec θ are positive and others are negative.
A function w(t) associating a real number t to a Third quadrant:
point on the unit circle.
cos θ = x < 0 (negative)
Such a function is called a wrapping function .
sinθ = y < 0 (positive)
Using sin t and cos t, other trigonometric
tan θ and cot θ are positive and others are negative
functions is defined as functions of real numbers.
Fourth quadrant:
Wrapping function w(t) is analogous to wrapping
cos θ = x > 0 (negative)
a line around a circle.
sinθ = y < 0 (positive) IX. Trigonometric ratios of (90 − θ), 0 < θ <π2
cos θ and sec θ are positive and others are negative sin(90 − θ) = cosθ,
If sin θ and cos θ are known, then the reciprocal identities cos(90 − θ) = sinθ,
and quotient identities can be used to find the other four
tan(90 − θ) = cot θ
trigonometric values.
cosec(90 − θ) = sec θ,
Pythagorean identities are used to find trigonometric
values when one trigonometric value and quadrant known. sec(90 − θ) = cosec θ,
VII. Allied Angles cot(90 − θ) = tanθ.
Two angles are said to be allied if their sum or X. Trig.ratios of of the form (90 + θ), 0 < θ <π/2
difference is a multiple of π/2radians. Let AOL = θ and AOR = (90 + θ).
Any two angles of θ such as, Let P(a, b) be a point on OL
3 Choose a point P’ on OR
, , , ....
2 2
such that OP = OP’.
are all allied angles .
Draw rs PM and P’N
VIII. Trigonometric ratios of −θ in terms of θ
from P and P’ on Ox and
Let AOL = θ and AOM = −θ.
Ox’
Let P(a, b) be a point on OL.
AOP’ = 90 + θ.
Choose a point P’ on OM
OPM and P’ON are congruent.
such that OP = OP’.
ON = MP nd NP’ = OM
Draw PN rOA intersecting
Cordinates of P and P’ are P(a, b) and ’(−b, a),
OM at P’.
ycoordinate xcoordinate
AOP=AOP’ sin(90 o ) cos(90 o )
OP' OP'
PON= P’ON and PON , P’ON are congruent. a b
OP OP
Thus, PN = P’N and point P’ is P’(a,−b) cos sin
By definition of trig. functions
tan(90 + θ) = −cot θ cosec (90+θ) = sec θ
b a
sin cos sec(90 + θ) = −cosec θ cot(90 + θ) = −tan θ.
OP OP
b a XI. Trig.function of other allied angles
sin( ) cos( )
OP' OP'
b a
OP' OP'
sin( ) sin cos( ) cos
Thus, tan(-) - tan , cosec(-) - cosec ,
sec(-) sec , cot(-) - cot
−θ is same as θ but it is on other side of x- axis.
Flipping (x, y) to other side of x- axis makes it into (x,−y)
So y-coordinate is negated and sine is negated
x-coordinate is same and cosine is unchanged.
Thus, sin(−θ) = −sin θ, and cos(−θ) = cosθ.
XII. Summary of Trig.function of allied angles XIV. Graph and Important Datas of T- Ratios
i) Allied angles of the form 2n θ, n Z
2
That is, −θ, π θ, 2πθ
Form of trigonometric ratio is unaltered
i.e., sine remains sine, cosine remains cosine etc.
ii) Allied angles of the form (2n+1) θ, n Z
2
θ θ,3 θθ
2 2
Form of trig. ratio altered to its complementary
Add “co” if it is absent
Remove “co” if it is already present
i.e., sine becomes cosine, cosine become sine etc.
iii) For determining sign,
Find out quadrant
Attach appropriate sign(+or−) according to ASTC
iv) Characteristics of Trig. Functions
i) Sine and cosine functions are complementary
sin (90 − θ) = cos θ and cos (90 − θ) = sinθ.
ii) cos θ and sin θ are satisfy inequalities
−1 ≤ cos θ ≤ 1 and −1 ≤ sin θ ≤ 1.
cos θ, sin θ [−1, 1]
XIII. Periodicity of Trigonometric Functions
A function f is said to be a periodic function with
period p, if there exists a smallest positive number
p such that f(x + p) = f(x) for all x in the domain.
sin (x 2 ) sin(x 4) sin(x 6) . . .
sinx
i.e sin (x 2n) sinx, n Z.
sin x , cos x,cosecx,sec x are periodic functions
with period 2π.
tan x,cot x are periodic functions with period π.
1. The terminal side of an angle θ in standard
position passes through the point (3,−4).Find the
six trigonometric function values at an angle θ.
Let B(x, y) = B(3,−4), OA be the initial side
OB = terminal side of θ in standard position.
AOB is the angle θ
θ lies in the IV quadrant.
r x2 y2
32 ( 4) 2
5
y 4 x 3
sin , cos ,
r 5 r 5
y 4
tan and so on
x 3
5 2 6
2. , is a point on the terminal side of an
7 7
angle θ in standard position. Determine the
trigonometric function values of angle θ.
5 2 6
Let B( x , y ) = B , , OA be the initial side
7 7
lies in the I quadrant.
2 2
5 2 6
r
7 7
25 24
1
XV. Odd and Even trigonometric functions 49 49
2 6 5
sin , cos ,
Real valued function f(x) is even function if it 7 7
satisfies f(−x) = f(x) for all real number x 2 6 7 2 6
tan and so on
7 5 5
A real valued function f(x)odd function if it
3
satisfies f(−x) = −f(x) for all real number x. 3. If sin θ and angle θ is in second quadrant,
5
As, cos(−x) = cosx for all x, cos x is even function then find other trigonometric functions.
Also sec x is even function sin 2 θ cos 2 θ 1
sin(−x) = −sin x for all x , sin x is odd function. cosθ 1 sin 2 θ
9
tan x, cosec x and cot x are odd functions. 1
25
f(t) = t − cos t is neither even nor odd function 4
5
A function is an even function if its graph is cos θ negative in 2 nd quadrant
unchanged under reflection about the y-axis. 4 3 5
Thus, cos θ - , tan , cos ec etc
A function is odd if its graph is symmetric about 5 5 3
the origin 4. Find values of other trigonometric functions
1
i) cos - , θ lies in the III quadrant.
2
sin 2 θ cos 2 θ 1 5. Find the values of
sinθ 1 cos 2 θ i) sin(−45) ii) cos(−45) iii) cot(−45)
1 sin(-45) - sin(45)
1
4 1
3 2
2 cos(-45) cos(45)
sin θ negative in 3 nd quadrant 1
3 2
Thus, sin θ - , tan 3 , and so on cot(-45) cot(45)
2
ii) cos θ =2/3, θ lies in the I quadrant. -1
6. Find the value of
4
sinθ 1
9 i) sin 150 ii) cos 135 iii) tan 120.
5
Using (90 θ) Using (180 θ)
3
sin θ ve in I st quadrant sin 150 o sin(90 o 60 o ) sin 150 o sin (180 o - 30 o )
5 3 cos(60 o ) sin(30 o )
Thus, sin θ , tan , and so on 1 1
3 2
2 2
iii) sin θ = −2/3, θ lies in the IV quadrant.
cos 135 o cos(90 o 45 o ) cos 135 o cos (180 o - 45 o )
4
cosθ 1 - sin(45o ) - cos(45o )
9
1 1
5 -
2 2
3
tan 120 o tan (90 o 30 o ) tan 120 o tan (180 o - 60 o )
cos θ ve in IV th quadrant
cot 30 o - tan(60 o )
5 3
Thus, cos θ , tan , and so on 3 - 3
3 2
iv) tan θ = −2, θ lies in the II quadrant. 7. Find the value of:
2 2
sec θ tan θ 1 iii) cos(300) i) sin(480) v) cot(660)
2
sec θ 1 tan θ i) sin 765 iv) tan(1050) ii)sin(−1110)
1 4 5 ii) cosec (−1410)
1
cos 15 19 11
5 iii) cot - vi) tan vii) sin
4 3 3
cos θ negative in 2 nd quadrant
1 cos 300 = cos(360 -60)
Thus, cos θ - ,
5
= cos60
sin tan cos
1 2 1
- 2 and so on =
2
5 5
v) sec θ =13/5, θ lies in the IV quadrant. sin 480 = sin(360 +120)
5 = sin120
cos
13
= sin(180- 60)
25
sin θ 1
169 3
12 =
2
13
sin θ ve in IV th quadrant sin 765 = sin(2 360 + 45)
- 12 12 = sin45
Thus, sin θ , tan , and so on
13 5
1 cot(180 ) sin(90 ) cos(-)
= cos 2 cot
2 sin(270 ) tan(-)cosec(360 )
cot 660 = cot (2360 -60) cot(180 ) sin(90 ) cos(-)
sin(270 ) tan(-)cosec(360 )
= -cot 60
cot cos cos
1
=- (-cos ) (-tan)cosec
3
cot cos 2
tan (1050) = tan (3360 - 30) cot tan
= -tan 30 cos 2 cot
1 10. Find all the angles between 0 and 360 which
=-
3 satisfy the equation sin2 θ =34
sin (−1110) = −sin (1110) 3
sin 2
= −sin (3360 + 30) 4
3
= - sin 30 sin
2
1 3
= when sin , 60 o
2 2
cosec(−1410) = −cosec (1410) 3
when sin , 120 o
2
= −cosec (4360 − 30)
7 4
= cosec 30 11. Prove sin 2 sin 2 sin 2 sin 2 2
18 9 18 9
=2
15 sin 2 sin 2 sin 2 sin 2
cot - = - cot 4 - 18 2 18 9 2 9
4 4
sin 2 cos 2 sin 2 cos 2
18 18 9 9
= cot
4 1 1 2
19
tan = tan 6 12. Determine whether the functions are even, odd
3 3
or neither i) sin2 x − 2 cos2 x − cos x ii) sin (cos(x))
iii) cos (sin(x)) iv) sin x + cos x
= tan
3
i) Let f(x) = sin2 x − 2 cos2 x − cos x
= 3
sin(−x) = −sin x and cos(−x) = cos x
11
sin = sin 4 f(−x) = f(x)
3 3
Thus, f(x) is even.
= sin
3 ii) Let f(x) = sin (cos(x))
3 f(−x) = f(x)
=
2 f(x) is an even function.
8. Prove: tan315cot(−405)+cot 495tan(−585)= 2 iii) f(x) = cos (sin(x)) ,
= tan(360−45)[−cot(360+45)]+cot(360+135) f(−x) = f(x),
[−tan (360 + 225)] Thus, f(x) is an even function.
= [−tan 45] [−cot 45] + [−tan 45] [−tan 45] iv) f(x) = sinx + cos x
= (−1)(−1) + (−1)(−1) f(−x) f(x) and f(−x) ) − f(x)
= 2. Thus, f(x) is neither even nor odd.
9. Prove that
Trigonometric Identities i) If = β, cos2 + sin2 = 1.
Compound Angles formulas ii) If = 0.β = x, cos(−x) = cos x
Compound angles are algebraic sum of two or more angles. 3. Prove: sin( + β) = sin cos β + cos sin β
Trigonometric functions do not satisfy the functional sin(+β) = cos[ 2 − ( + β) ]
relations like f(x + y) = f(x) + f(y) and f(kx) = kf(x), k is = cos[( 2 −)− β) ]
a real number. = cos( 2 −)cosβ+sin( 2 −) sinβ
cos(+β) cos +cos β sin( + β) = sin cos β + cos sin β
sin(2) 2sin If +β = 2 , cos2 + sin2 = 1
tan 3 3tan, . . .. 4. Prove: sin( − β) = sin cos β − cos sin β
1. Prove that cos( + β) = cos cos β − sin sin β sin(−β) = sin[ + (−β)]
Consider the unit circle with centre at O. = sin cos (−β) + cos sin (−β)
Let P = P(1, 0). sin( − β) = sin cos β − cos sin β
Let Q,R and S be points on the unit circle tanα tanβ
5. Prove tan(α β)
such that POQ = , 1 tanαanαt
POR = +β POS = −β sin( β )
tan ( β)
cos( β )
angles , + β and − β are in standard positions.
Points Q,R and S are given by sin cos cos sin
cos cos
Q(cos , sin )
cos cos sin sin
R (cos( + β), sin( + β)) cos cos
S (cos(−β), sin(−β) sin cos cos sin
cos cos cos cos
cos cos sin sin
cos cos cos cos
tan tan β
tan ( β)
1 tan tan β
tanα tanβ
6. Prove tan(α β)
1 tanαanαt
tan tan( β )
tan ( β)
Since ’POR and ’SOQ are congruent. 1 tan tan( β )
PR SQ tan tan β
1 tan tan β
PR 2 SQ 2
cotAcotB - 1
[cos( ) - 1] 2 sin 2 ( ) [cos - cos(-)]2 [sin - sin(-)]2 7. Prove cot(A B)
cotA cotB
- 2 cos( ) 2 2 - 2 cos cos 2sin sin cot ( β) tan 2 ( β)
cos( ) cos cos - sin sin
tan 2 β)
Arc lengths PR and SQ, subtends angles +β, +(−β) tan 2 tan β
Thus, PR = SQ. 1 tan 2 tan β
Distance b/w (cos , sin ) and (cos(−β), sin(−β)) is 1
cot
same as distance b/w (cos(+β), sin( +β)) and (1, 0). cot
1
2. cos( − β) = cos cos β + sin sin β 1 cot
cot
cos( + β) = cos cos β − sin sin β cot cot 1
cos( − β) = cos[ + (−β)] cot cot
8. Expand sin(A B C)
= cos cos(−β) − sin sin(−β)
cos( − β) = cos cos β + sin sin β.
sin(A B C) sin[A (B C)] 13. Prove that
sinAcos(B C) cosAsin(B C) cos(A B) cos(A - B) cos 2 A - sin 2 B cos 2 B - sin 2 A
sinAcosB cosC cosAsinB cosC (cosA cosB - sinA sinB) (cosA cosB sinA sinB)
cosAcosB sinC - sinAsinB sinC cos 2 A cos 2 B - sin 2 A sin 2 B
9. Expand : cos(A B C) cos 2 A (1 - sin 2 B) - (1 - cos 2 A) sin 2 B
cos 2 A - sin 2 B
cos(A B C) cos[ A ( B C )]
cosAcos(B C) sinAsin(B C) 14. Prove that sin 2 (A B) - sin 2 (A - B) sin2Asin2B
cosAcosB cosC cosAsinB sinC sin 2 (A B) - sin 2 (A - B)
sin AsinB cosC - sinAcosB sinC sin(A B A - B) sin(A B A B)
sin2Asin2B
if, A B C
2
15 12
cos(A B C) 0 15. If sin x , cos y , 0 < x <π/2, 0 < y <π/2,
17 13
0 cosAcosB cosC cosAsinB sinC
Find i) sin(x + y) ii) cos(x − y) iii) tan(x + y).
sin AsinB cosC - sinAcosB sinC
x,y are in In 1st quadrant
cosAcosB cosC sinAsinB cosC sinB sinC cosA
In 1st quadrant, sin y, cos x is always positive
sinC sinAcos B,
15 12
. sin x cos y
17 13
10. Expand : tan(A B C) 15 2 12 2
cos x 1 sin y 1
tan( A B C ) tan[A (B C)] 17 13
tan A tan(B C ) 8 5
1 - tanAtan(B C) 17 13
sin(x y) sin x cos y cos x sin y
tanB tanC
tan A 15 12 8 5 220
1 - tanB tanC
tanB tanC 17 13 17 13 221
1 - tanA cos(x y) cos x cos y sin x sin y
1 - tanB tanC
tan A tan B tan C- tan A tan B tan C 8 12 15 5 171
1- tan A tan B- tan B tan C- tan C tan A 17 13 17 13 221
cos(x y) cos x cos y sin x sin y
i) If A+B+C = 0 or π,
8 12 15 5 21
tan(A B C) 0
17 13 17 13 221
Hence tanA tanB tanC tanAtanB tan C. sin(x y )
tan(x y )
ii) in case of oblique triangles cos( x y )
220 221 220
tan(x − y) + tan(y − z) + tan(z − x)
221 21 21
= tan(x − y) tan(y − z) tan(z − x 16. If sinA = 3 and cosB = 9, 0 < A <π2, 0 < B <π/2
find the value of i) sin(A + B) ii) cos(A − B).
iii) If A + B + C = π/2 In 1st quadrant, sinB, cosA is always positive
tanAtanB + tanB tanC + tanC tanA = 1 3 9
sin A cos B
11. Evaluate : cos(A + x) − cos(A − x) 5 41
9 81
cos(A x ) - cos(A - x ) cos A cos x sinA sin x cos A 1 sin B 1
25 1681
- cos A cos x sinA sin x 4 40
- 2 sinAsin x 5 41
12. Prove that sin(A B) sin(A - B) sin 2 A - sin 2 B sin(A B) sin A cos B cos A sin B
3 9 4 40 187
(sin A cosB cosA sinB) (sinA cosB - cosA sinB)
5 41 5 41 205
sin 2 A cos 2 B - cos 2 A sin 2 B cos(A B) cos A cos B sin A sin B
sin 2 A (1 - sin 2 B) - (1 - sin 2 A) sin 2 B 4 9 3 40 156
2 2
sin A - sin B 5 41 205
5 41
17. If sin x = 4/5( in I quadrant ), cos y =−12/13( in II tan 165o tan(120 o 45o )
quadrant), find (i) sin(x − y), (ii) cos(x − y).
tan 120 o tan 45o
In 1st quadrant, cos x is always positive
1 - tan 120 o tan 45o
In 2nd quadrant, siny is always positive tan 120 o tan(90 o 30 o )
4 12 - cot 30 o - 3
sin x cos y -
5 13
tan 45o 1
16 144
cos x 1 sin y 1 1 3
25 169 Thus, tan 165o
3 5 1 3
cos 105 cos(45o 60 o )
o
5 13
sin(x y ) sin x cos y cos x sin y cos45o cos 60 o sin45o sin 60 o
4 - 12 3 5 63
1 1 1 3
. .
5 13 5 13 65 2 2 2 2
cos(x y ) cos x cos y sin x sin y 1 3
3 - 12 4 5 16 2 2
5 13 5 13 65 sin 105 o sin(45o 60 o )
18. Find cos(x − y), if cos x = −4/5 with π < x <3π/ 2
sin45o cos 60 o cos45 o sin 60 o
and sin y = −24/25with π < y <3π/2
In 3rd quadrant, cosy, sin x is always negative 1 1 1 3
. .
2 2 2 2
4 24
cos x - sin y - 3 1
5 25
2 2
16 576 7
sin x 1 cos y 1 tan tan
25 625
12 3 4
3 7
5 25 tan tan
3 4
cos(x y ) cos x cos y sin x sin y
1 - tan tan
4 -7 3 24 4 3 4
5 25 5 25 5 3 1
19. Find sin(x − y), given that sin x = 8/17with 0 < x
1 3
<π/2 and cos y = −24/25with π < y < 3π/ 2 1 3 1 3
In 1st quadrant, cos x is always positive
1 3 1 3
In 3rd quadrant, sin y is always negative ( 2 3)
8 24 21. Prove that
sin x cos y - 3 cos x- sin x
17 25 i) cos( 30 o x)
64 576 2
cos x 1 sin y 1 cos( 30 x) cos 30 cos x sin 30 o sin x
o o
289 625
15 7 3 1
cos x sin x
17 25 2 2
sin(x y ) sin x cos y cos x sin y 3 cos x- sin x
8 - 24 15 7 87 2
ii) cos(π + θ) = −cos θ
17 25 17 25 425
20. Find the value of i) cos 15◦ ii) tan 165◦. cos( ) cos cos sin sin
iii) cos 105◦ iv) sin 105◦ iv) tan 7π/12 cos cos(90 90) sin sin(90 90)
cos 15o cos(45o - 30 o ) sin 90 1 cos 90 0
cos45o cos 30 o sin45o sin 30 o cos( ) ( 1) cos (0) sin
1 3 1 1 cos
. . iii) sin(π + θ) = −sin θ.
2 2 2 2
sin ( ) sin cos cos sin
3 1
(0) cos ( 1) sin
2 2
sin
22. Find quadratic equation with roots sin15,cos15 25. Prove that sin 105◦ + cos 105◦ = cos 45◦.
let sin15o sin(45o - 30 o ) 1 3 1 3
cos 105 o , sin 105 o
sin45o cos30o cos45o sin 30 o 2 2 2 2
[ See Problem No : 20 (iii), (iv)]
1 3 1 1 3 1
. . 1 3 1 3
2 2 2 2 2 2 sin 105 o cos 105
3 1 2 2 2 2
cos15o [see problem :20(i) 1
2 2
2
3 1 3 1 2 3
cos 45 o
2 2 2 2 2
26. Prove that sin 75◦ − sin 15◦ = cos 105◦ + cos 15◦.
3 2 6
3 1
2 2 2 sin 15 o [ See problem No : 22]
2 2
3 1 3 1
sin 75 o sin(90 - 15)
2 2 2 2
1 cos 15 o
4 3 1
Required QE : 2 2
x 2 ( )x 0 3 1 3 1
sin 75 o sin 15 o
6 1 2 2 2 2
x2 x 0 1
2 4
2
x 2 6x 1 0 2
27. Show that tan 75◦ + cot 75◦ = 4.
23. Prove that
tan 75 o tan (45o 30 o )
i) sin(45◦ + θ) − sin(45◦ − θ) =√2 sinθ.
sin(45o ) sin 45 o cos cos 45 o sin 3 1
[ See problem No : 20]
cos sin 3 1
cot 75 o cot (45o 30 o )
2
sin(45 ) sin 45 cos cos 45o sin
o o cot 45 o cot 30 o - 1
cos sin cot 45 o cot 30 o
..
2 3 1
cos sin cos sin
sin(45o ) - sin(45o ) 3 1
2 2 3 1 3 1
tan 75 o cot 75 o
2 sin 3 1 3 1
ii) sin(30◦ + θ) + cos(60◦ + θ) = cos θ ( 3 1) ( 3 1) 2
2
sin(30 o ) sin 30 o cos cos 30 o sin
( 3) 2 12
cos 3 sin 4
...(1)
2 28. Prove that
sin(60 o ) sin 60 o cos cos 60 o sin cos(A+B) cosC − cos(B +C)cosA=sinB sin(C −A)
3 cos sin cos(A B)cosC ( cos A cos B sin A sin B)cosC
....(2) cos A cos BcosC sin A sin BcosC
2
cos 3 sin cos 3 sin cos(B C) cosA cosA cos B cos C cosA sin B sin C
(1) ( 2),
2 2 Subtract, sinB sin C cos A cosCsinB sin C
cos sinB sin(C - A)
24. If a cos(x + y) = b cos(x − y), show that
29. Prove that
(a + b) tanx = (a − b) cot y.
sin(n+1)θ sin(n−1)θ +cos(n+1)θ cos(n−1)θ = cos2θ
a cos(x y) b cos(x - y)
a cos x cos y a sin x sin y b cos x cos y b sin x sin y = cos(n+1)θ cos(n−1)θ + sin(n+1)θ sin(n−1)θ
( a b ) cos x cos y ( a b ) sin x sin y = cos[(n+1)θ – (n−1)θ]
cos y
( a b) ( a b ) sin x = cos2 θ
sin y cos x
(a b) tan x (a - b) cot y. 30. If x cos θ = y cos_θ +2π3_= z cos_θ +4π3_ , find
the value of xy + yz + zx.
2 4 3
x cos y cos z cos 33. Prove that tan tan 1.
3 3 4 4
2
tan 4 tan
x cos y cos tan
3 4 1 tan 4 tan
x 1 tan
cos cos cos 23 sin sin 23
y 1 tan
x
y
1
cos cos
2 2
3
sin
3
4
tan tan
4
x 1
3
tan ....(1)
tan
4
y 2 2
tan4
4
x cos z cos 3 1 tan 1 tan
3 tan tan
4 4 1 tan 1 tan
x
cos cos cos 43 sin sin 43 1
z 34. Prove that cos 8θ cos 2θ = cos2 5θ − sin23θ
x 1 3
cos cos sin cos 8 cos 2 cos(5 3) cos(5 3)
y 2 2
x 1 3 cos(A B)cos(A B) cos 2 A cos 2 B
tan .....(2)
z 2 2 cos 8 cos 2 cos 2 5 sin 2 3
Adding (1) and (2)
35. Find the values of tan(α+β), given that
x x
1
y z cot 1 , sec - 5 , ( 0, ), ( , )
2 3 2 2
xy yz zx 1
cot ,
xy yz zx 0 2
31. Show that tan 2 [ is in Quadrant I]
i) tan(45 o A)
1 tanA sec - 5
1 - tanA 3
tan 45 o tan A tan sec 2 1
tan(45o A)
1 - tan 45 o tan A 25
1
1 tanA 9
1 - tanA 4
[ is in Quadrant II]
1 tanA 3
ii) tan(45o A)
1 tanA tan tan
tan( )
o tan 45 o tan A 1 tan tan
tan(45 A)
1 tan 45 o tan A 2
4
1 tanA 3 2
8 11
1 tanA 1
n 1 3
32. If tan x , tan y find tan(x + y) 36. If cos(α − β) + cos(β − γ) + cos(γ− α) = −3/2, prove
n 1 2n 1
n 1 that cos α + cos β + cos γ = sinα + sinβ + sinγ = 0
tan x , tan y
n 1 2n 1
2cos(α − β) + 2cos(β − γ) + 2cos(γ− α) +3=0
tan x tan y
tan (x y). 2 cos cos 2 sin sin 2 cos cos 2 sin sin
1 tan x tan y
2 cos cos 2 sin sin sin 2 cos 2
n 1
n 1 2n 1 sin 2 cos 2 sin 2 cos 2 0
n 1
1 sin 2 sin 2 sin 2 2 sin sin 2 sin sin
n 1 2n 1
2 sin sin
2n 2 n n 1
cos 2 cos 2 cos 2 2 cos cos 2 cos cos
[ 2n 2 2n n 1] n
2 cos cos 0
2n 2 2n 1
(cos α + cos β + cos γ) + (sinα + sinβ + sinγ)3 = 0
3
2n 2 2n 1
1 cos α + cos β + cos γ = sinα + sinβ + sinγ = 0
37. If θ + φ = α and tan θ = k tan φ, then prove that 3.5. MULTIPLE -SUBMULTIPLE ANGLES
k -1 I. Double-Angle Identities
sin ( ) sin
k 1
i) sin 2 A sin (A A)
sin( ) sin sin A cos A sin A cos A
tan k tan sin 2A 2sinAcos A .
tan ii) cos 2 A cos (A A)
k
tan cos A cos A - sin A sin A
tan tan cos2A cos 2 A - sin 2 A
k 1
tan
cos 2 A - (1 - cos 2 A )
tan tan
k 1 cos 2A 2cos 2
A-1
tan
k - 1 tan tan
( 1- sin 2 A) - sin 2 A
k 1 tan tan
cos 2A 1 - 2 sin 2 A .
sin cos cos sin
iii)tan 2 A tan (A A)
sin cos cos sin
tan A tan A
sin( )
1 - tan A tan A
sin( )
2 tanA
sin( ) tan2A
1 - tan 2 A
sin
2 sin A cos A
k -1 sin( ) iv) sin 2 A
sin sin sin 2 A cos 2 A
k 1 sin
sin ( ) 2 sin A cos A
38. Show that cos 2 A
cos 2 A
cos2A+cos2 B−2cosAcosB cos(A+B) = sin2(A+B) 2 tanA
sin2A
sin 2 (A B) 1 - tan 2 A
[ sin(A B)]2 cos 2 A - sin 2 A
v) cos 2 A
(sin A cos B cos A sin B) 2 sin 2 A cos 2 A
cos 2 A - sin 2 A
sin 2 A cos 2 B cos 2 A sin 2 B
cos 2 A
2 sin A cos B cos A sin B
cos 2 A sin 2 A
(1 cos 2 A) cos 2 B cos 2 A(1 cos 2 B) cos 2 A
2 sin A cos B cos A sin B 1 - tan 2 A
cosA .
cos 2 B cos 2 A cos 2 B cos 2 A cos 2 A cos 2 B 1 tan 2 A
2 sin A cos B cos A sin B II. Power reducing or Reduction identities
cos 2 A cos 2 B 2 cos 2 A cos 2 B 2 cos 2 A - 1 cos 2 A
2 sin A cos B cos A sin B 1 cos 2 A
2 2
cos 2 A
cos A cos B 2 cos A cos B 2
( cos A cos B sin A sin B) 1- 2 sin 2 A. cos 2 A
cos 2 A cos 2 B 2 cos A cos B cos( A B) 1 cos 2 A
sin 2 A
2
1 cos 2 A
sin 2 A 2
2
cos A 1 cos 2A
2
1 cos 2A
tan 2 A
1 cos 2 A
III. Triple-Angle Identities
i) sin 3 A sin ( 2 A A) x x x
1. Prove that sin x 210 sin 10
cos cos ... cos 10
sin 2 A cos A cos 2 A sin A 2 2 2
sin A cos A cos A (1 2 sin 2 A) sin A x x
sin x 2 sin cos
2 2
2 sin A cos 2 A sin A - 2 sin 2 A Again apply half angle,
2 sin A(1 sin 2 A) sin A - 2 sin 2 A x x x
3
sin 2sin 2 cos 2
3 sin A - 4 sin A 2 2 2
ii) cos 3 A cos ( 2 A A) x x x
Thus, sin x 2 2 sin 2 cos 2 cos
cos 2 A cos A - sin 2 A sin A 2 2 2
Again apply half angle repeatedly
( 2 cos 2 A - 1) cos A - 2 sin A cos A sin A
x x x
2 cos 3 A - cos A -2 cos A(1- cos 2 A) sin x 210 sin 10 cos cos ... cos 10
2 2
2
4 cos 3 A - 3 cos A. sin θ sin 2θ
2. tan θ
iii) tan 3A tan( 2 A A) 1 cos θ cos 2θ
sin θ 2 sin θ cos θ
tan 2 A tan A
1 cos θ 2 cos 2 θ 1
1 tan 2 A tan A
2 tan A sin θ(1 2 cos θ )
tan A
2 cos θ(1 2 cos θ )
1 tan A
2 tan A tan θ
1 tan A 1 sin 3 x cos 3 x
1 tan 2 A 3. 1 sin 2x
2 sin x cos x
3 tan A tan 3 A
sin x cos 3 x
3
1 3 tan 2 A RHS
sin x cos x
IV. Half-Angle Identities
2A A A sin x cos x (sin 2 x cos 2 x sin x cos x)
i) sin 2 sin cos .
2 2 2 sin x cos x
A A 1 sin x cos x
sin A 2 sin cos .
2 2 2 sin x cos x
1
2 A A 2
ii) cos A cos - sin 2
2 2 1
1 sin 2x
2 A 2
cos A 2 cos -1
2 4. Find x such that −π ≤ x ≤ π and cos 2x = sinx
A cos 2x sin x
cos A 1- 2 sin 2
2 1 2 sin 2 x sin x
A
2 tan 2 sin 2 x sin x 1 0
iii) tan 2 A 2
2 A
Using quadratic formula :
1 - tan
2 1 3
sinx
A 4
2 tan
iv) sin A 2 1
1 or
2 A 2
1 - tan
2
if sin x is - 1, x
2 A 2
1 - tan
v) cos 2 A 2 . 1 5
A if sin x is , x ,
1 tan 2 2 6 6
2 5
V. Prove that sin 4 A 4 sin A cos3 A- 4 cos A sin 3 A Thus x , ,
2 6 6
sin 4 A 4 sin A cos 3 A - 4 cos A sin 3 A 5. Find the values of
RHS 4 sin A cos A(cos 2 A - sin 2 A) i) sin 18◦ ii) cos 18◦ iii) sin 72◦iv) cos 36◦ v) sin 54◦
or
4 sin A cos A cos 2 A
Verify that
2( 2 sin A cos A) cos 2 A
2( sin 2 A) cos 2 A sin 18◦ = cos 72◦, cos 18◦ = sin72◦ and cos 36◦ = sin54◦
sin 4 A.
18o 1 a cos a
7. tan tan , cos
5 90 o 2 1 a 2 1 a cos
1 a
3 2 90 o tan tan ,
2 1 a 2
2 90 o - 3 2
1 tan 2
cos
sin 2 sin ( 90 o - 3) 1 tan 2 2
sin 2 cos 3 1 a
1 2
tan 2
3
2 sin cos 4 cos - 3 cos . 1 a
1 a 2
2 sin 4 cos 2 - 3 1 tan 2
1 a
2 sin 4(1 - sin 2 ) - 3 2
1 a sin 2
4 sin 2 2 sin - 1 0 1
1 a cos 2 2
- 2 4 - 4(4)(-1) 2
1 a sin 2
sin
2( 4) 1
1 a cos 2 2
5 1
or sin 18o (1 a ) cos 2 2 (1 a ) sin 2 2
4
(1 a) cos 2 2 (1 a ) sin 2
2
cos 18 o 1 - sin 2 18o
2
(cos 2 2 sin 2 2 ) a(sin 2 2 cos 2 2 )
5 1
1 (cos 2 2 sin 2 2 ) a(sin 2 2 cos 2 2 )
4 cos a
10 2 5 1 a cos
4 sin 2 n A
8. cos A cos 2 A cos 2 2 A cos 2 3 A... cos 2 n-1 A
Using cos 2 A 1 2 sin 2 A 2 n sin A
cos 36 o 1 2 sin 2 18 o 1
2 sin A cos A cos 2 A cos 2 2 A cos 2 3 A... cos 2 n-1 A
2 2 sin A
5 1 1
1 sin 2 A cos 2 A cos 2 2 A cos 2 3 A... cos 2 n-1 A
4 2 sin A
5 1 1
cos 36 o 2 sin 4 A cos 2 2 A cos 2 3 A... cos 2 n-1 A
4 2 sin A
sin 72 sin(90 o - 18 o )
o
Continuing the process, we get
cos18o sin 2 n A
10 2 5 2 n sin A
sin 72 o
4 o
1
9. Find the value of sin 22
sin 54 sin(90 o - 36 o ) cos 36 o
o
2
o 5 1 1 2 sin 2 2 cos
sin 54
4
o o 1 cos
6. 3 cos ec 20 sec 20 sin , take 45 o
1 1 2 2
3 o
45 o 1 cos 45 o
sin 20 cos 20 o sin
3 cos 20 o sin 20 o 2 2
1 2 1
sin 20 o cos 20 o
3 1 2
cos 20 o sin 20 o o
1 2 2
4 2 2 sin 22
2 sin 20 o cos 20 o 2 2
12
o o 0 o
10. Find sin 2 θ if sin
sin 60 cos 20 cos 60 sin 20 13
4 5
2 sin 20 o cos 20 o cos 1 144
169
13
sin(60 o 60 o ) sin 2 2 sin cos
4 o
sin(2 20 ) 12 5 120
4 2
13 13 169
11. If A + B = 45°, show that (1 + tanA) (1 + tanB) = 2 3713
o
A B 45 4225
tan (A B) tan45 15 4
i) cosA ii) sinA
17 5
tanA tanB
1 cos 2 A 2 cos A 1 cos 2 A 1 2 sin 2 A
2
1 - tanA . tanB
. tanA tan B 1 - tanA . tanB 161 7
Add 1 on both sides 289 25
o
1
. 1 tanA tanB 2 - tanA tanB 16. Prove that cot 7 2 3 4 6
1 tanA tanB tanA tanB 2 2
o
(1 tan A) (1 tanB) 2 7 12 2 15 o
12. (1 tan 1o )(1 tan 2 o )(1 tan 3 o )...(1 tan 44 o ) is a 1 cos 2
cot
multiple of 4 – Prove sin 2
3 1
Let A B 45 o 1 cos15 o 1 2 2
3 1
tan (A B) tan45 sin 15 o 2 2
1 tanA tanB tanA tanB 2 2 2 3 1
(1 tan A) (1 tanB) 2 3 1
(1 tan 1o )(1 tan 44 o )][(1 tan 2 o )(1 tan 43 o )]... 2 2 3 1 3 1
3 1 3 1
(1 tan 22 o )(1 tan 23 o )] o
1
2 2 2....upto 22 terms cot 7 2 3 4 6
2
2 22 1 1 1 3 1
17. Prove if cos a , cos 3 a 3
411 2 a 2 a
1 - tan 2 1
13. Prove sin 4 4 tan a 2 cos
2 2 a
(1 tan ) 3
sin 4 2 sin cos a 1 2 cos 3
a
2 tan 1 - tan 2
2 1 1
1 tan 2 1 tan 2 a 2 3 8 cos 3 3 a
a a
1 - tan 2 1
4 tan 2
2 a 2 3 8 cos 3 3 2 cos
(1 tan ) a
1 3 1
14. Prove that tan tan 2 tan 2 3
a 3 4 cos 3 cos
3
4 4 2 a
tan tan 1 3 1
a 3 cos 3
4 4 2 a
tan 4 tan tan 4 tan
18. (1 sec 2θ)(1 sec 4θ)...(1 sec 2 n θ) tan 2 n θ cot θ
1 tan 4 tan 1 tan 4 tan 1 cos 2θ 1 cos 2 2 θ 1 cos 2 n θ
...
1 tan 1 tan cos 2θ cos 2 2 θ cos 2 n θ
1 tan 1 tan 2 cos 2 θ 2 cos 2 2θ 2 cos 2 2 2 θ cos 2 2 n1 θ
(1 tan ) 2 (1 tan ) 2 cos 2θ cos 2 2 θ cos 2 3 θ... cos 2 n θ
1 tan 2 cos ( 2 cos )( 2 cos 2)(2 cos 2 2 )..
2 cos 2 2 n θ
1 tan 2 2 tan 1 tan 2 2 tan cos 2 n
cos 2 cos 2 2 n θ
1 tan 2 ( 2 cos sin )(2 cos 2)( 2 cos 2 2 )..
sin cos 2 n
2 tan
cos 2 n
2 cos 2 θ
1 tan 2 ( 2 sin 2 cos 2)( 2 cos 2 2 )..
sin cos 2 n
2 tan 2 Proceeding like this
15. Find cos 2A, A lies in the first quadrant, when
cos 1
15 4 16 ( 2 sin 2 n 1 cos 2 n1 )
i)cosA ii) sinA iii) tanA sin cos 2 n
17 5 63
sin 2 n
16 1 tan 2 A cot
iii) tanA cos 2 A cos 2 n
63 1 tan 2 A cot θ tan 2 n θ
PRODUCT TO SUM AND SUM TO PRODUCT
19. Prove 32( 3) sin cos cos cos cos 3
48 48 24 12 6 sin(A B) sinA cosB cosA sinB
sin(A - B) sinA cosB - cosA sinB
16( 3 ) 2 sin cos cos cos cos
48 48 24 12 6
sin(A B) sin (A - B) 2 sinA cosB. ..I
8( 3)2 sin cos cos cos sin(A B) - sin(A - B) 2 cosA sinB ..II
24 24 12 6
cos(A B) cosA cosB - sinA sinB
4( 3)2 sin cos cos cos(A - B) cosA cosB sinA sinB
12
12 6
cos(A B) cos(A - B) 2 cosA cosB III
2 3)2 sin cos
6 6 cos(A B) - cos(A - B) - 2sinA sinB......IV
Let A B C .....(1),
2 3 sin
6 A - B D ....(2)
3
CD
20. Prove that cos 5θ 16 cos 5 θ - 20 cos 3 θ 5 cos θ (1) (2) gives , A
2
cos(3 cos 2)
CD
cos 3 cos 2 sin 3 sin 2 (1) (2) gives , B
2
( 4 cos 3 3 cos )( 2 cos 2 1) Substistute in I, II, III, IV
(3 sin 4 sin 3 )2 sin cos C D C - D
sin C sin D 2 sin cos
2 2
8 cos 5 6 cos 3 4 cos 3 3 cos
C D C - D
(6 sin 2 cos 8 sin 4 cos ) sinC - sinD 2 cos sin
2 2
8 cos 5 6 cos 3 4 cos 3 3 cos C D C D
cos C cosD 2 cos cos
(6(1 cos 2 ) cos 8(1 cos 4 ) cos ) 2 2
C D C D
8 cos 5 6 cos 3 4 cos 3 3 cos cosC - cosD - 2sin sin
2 2
6 cos 6 cos 3 8 cos 8 cos 3 1. Express as a sum or difference
16 cos 5 θ - 20 cos 3 θ 5 cos θ x 3x
21. If θ is an acute angle, then find
i) sin40 o cos30 o ii) cos110 o sin55 o iii) sin cos
2 2
1 8 o o
i) sin , sin ii) cos , sin iv)sin35 cos 28 v) sin4x cos2x vi) sin10θ cos2θ
4 2 25 4 2 9
vii)cos 5θ cos 2θ viii)sin 5θ sin 4θ.
sin sin cos sin cos 2 sinA cosB sin(A B) sin(A - B)
4 2 4 2 2 4
1
1 i) sin40 o cos30 o [sin(40 o 30 o ) sin(40 o 30 o )]
sin cos sin 2
4 2 2 2 2
1
squaring both sides [sin70 o sin10 o ]
2
1 x 3x 1 x 3x x 3x
sin 2 cos 2 sin 2 2 cos sin iii) sin cos sin sin
4 2 2 2 2 2 2 2 2 2 2 2 2 2
1 1
1 2 cos sin [ sin 2x - sin x]
2 2 2 2
1 1
sin 2 1 sin iv)sin35 o cos 28 o [sin63 o sin7 o ]
4 2 2 2
1 1 1
1 v) sin 4x cos 2x [sin 6x sin2x]
2
2 25
1
2 3 vi) sin10θ cos2θ [sin 12 sin8]
sin 2
4 2 5
2 cosAsinB sin(A B) - sin(A - B)
proceeding as above
ii) 2cos110 o sin55o sin(110 o 55 o ) - sin(110 o 55 o )
1
cos cos sin 1
4 2 2 2 2 [sin165 o sin55 o ]
2
1
cos 2 cosAcosB cos(A B) cos(A - B)
4 2 3 2
1
vii)cos 5θ cos 2θ [cos 7 cos3]
2
2 sinAsinB cos(A B) cos(A B) 5. tan(60 o - A) tanAtan(60 o A) tan3A
1
viii)sin 5θ sin 4θ. [cos cos9] sin(60 o - A) sin(60 o A) sinA
2
cosA[ cos(60 o - A)cos(60 o A)]
2. Express as Product
sin 3 A
o o 3x 9x 4
i) sin50 sin20 ii) cos 6θ cos2θ iii) cos - cos cos 3 A
2 2 4
iv)sin75 - sin 35 v)cos65 cos15 vi)sin 50 sin40 o
o o o o
tan3A
1
vii) cos 35 o - cos 75 o viii)si n 34 o cos 64 o - cos 4 o 6. cos 36 o cos 72 o cos 108 o cos 144 o
16
CD C-D
sinC sinD 2sin cos cos 36 o cos 72 o cos 108 o cos 144 o
2 2
50 o
20 o
50 o 20 o cos36 o cos(90 o - 18 o )cos(90 o 18 o )cos(180 o - 36 o
i) sin50 o sin20 o 2sin cos
2 2 cos 36 o sin 18 o sin 18 o cos 36 o
o o
2 sin 35 cos15 cos 2 36 o sin 2 18 o
vi) sin 50 o sin40 o 2sin45 o cos 5 o 2 2
5 1 5 1
CD C-D
sinC sinD 2cos
sin 4 4
2 2
o o o o
1
iv) sin75 - sin 35 2cos55 sin 20
16
C D C D 1
cos C cosD 2 cos cos 7. sin 12 o sin 48o sin 54 o
2 2 8
ii) cos 6θ cos2θ 2 cos 4 cos 2 1
[cos 36 o cos 60 o ] sin 54 o
v) cos65 o cos15 2 cos 40 o cos 25 o 2
1
DC D C [ cos 60 o sin 54 o cos 36 o sin 54 o ]
cosD - cosC 2sin sin 2
2 2
1 1 1
3x 9x 9x
3x 9x
3x - sin 54 o (sin 90 o sin 18 o )
iii) cos - cos 2 sin 2 2 sin 2 2 2 2 2
2 2 2 2
3x 14 [1 - (sin 54 o sin 18 o )]
2 sin 3x sin 2
14 [1 2 sin 18 o cos 36 o ]
vii) cos 35 o - cos 75 o 2sin55 o sin 20 o
sin 34 o cos 64 o - cos 4 o sin 34 o [ cos 64 o - cos 4 o ] 1 2 sin 18 o cos 18 o cos 36 o
[1 ]
4 cos18 o
sin 34 o 2sin 34 o sin 30 o
1 2 sin 36 o cos 36 o
0 [1 ]
Prove that
4 2cos18 o
o 1o
1 sin 72 o
3. cos(60 - A) cosAcos(60 A) cos 3A [1 ]
4 4 2cos[90 - 72 o )
cosA[ cos(60 o - A)cos(60 o A)] 1 sin 72 o
[1 ]
cos A·(cos 2 60 - sin 2 A) 4 2sin72 o
cos A14 - (1 - cos 2 A) 1
8
cos A 34 cos 2 A 3
3 3 8. cos 10 o cos 30 o cos 50 o cos 70 o
4 cos A cos A 16
o o o o
1 1 cos 30 [cos 10 cos 50 cos 70 ]
4
cos 3 A - 3cosA cos 3A
4 cos 30 o [cos 10 o cos (60 o - 10 o ) cos(60 o 10 o ]
1
4. sin(60 o - A) sinAsin(60 o A) sin 3A 3 1
4 cos 3(10 o )
o
2 4
sin(60 - A) sin(60. A) sinA
3
1 cos 30 0
[cos 2A - cos 120 o ] sinA 8
2
3
1 1
cos 2AsinA sinA 16
2 2
sin 75 o - sin 15 o 1
1 1 1 9.
sin 3A sin 3A o
cos 75 cos 15 o
3
2 2 4
75 o 15 o 75 o 15 o (1 cos 2x ) (cos 4x cos 6x )
2sin sin
2 2 2 cos 2 x 2 cos 5x cos x
o o
75 15 75 15 o
o
2 cos x(cos x cos 5x )
2cos cos
2 2
o o
4 cos x cos 2x cos 3x.
cos 45 sin 30 1 θ 7θ 3θ 11θ
o o
tan 30 o 16. sin sin sin sin sin 2 sin
cos 45 cos 30 3 2 2 2 2
2 3 4 5 6 7 1 1
10. cos cos cos cos cos cos cos cos 3 cos cos 4 cos 7
15 15 15 15 15 15 15 128 2
2 4 1 8 3 6 1
cos cos cos cos( ) cos cos cos 3 cos 7
15 15 15 2 15 15 15 2
1 2 4 8 3 6 sin 2 sin
cos cos cos cos ) cos cos
2 15 15 15 15 15 15 17. cos ( 30 o -A) cos ( 30 o A) cos ( 45 o -A) cos ( 45 o A) cos 2 A 1
4
4 2 3
1 sin 2 15 sin 2 15 cos 2 30 o - sin 2 A cos 2 45 o sin 2 A
2 4 2 3 3 1
2 sin 15 2 sin 15 2 sin 2 A
4 2
16 12 1
1 sin 15 sin 15 1 2 sin 2 A
4
128 3 1
sin 15 sin 15 cos 2 A
4
sin sin 3 sin x sin 3x sin 5x sin 7x
1
15
15 18.
cos x cos 3x cos 5x cos 7x
tan 4x.
128 3 sin x sin 3x sin 5x sin 7 x
sin sin
15 15 cos x cos 3x cos 5x cos 7x
3 (sin 3x sin x ) (sin 7 x sin 5x )
1 sin 15 sin 15 1
(cos 3x cos x ) (cos 7 x cos 5x )
128 3 128
sin 15 sin 15 2 sin 4x cos 2x 2 sin 12x cos 2x
sin 8x cos x - sin 6x cos 3x 2 cos 4x cos 2x 2 cos 12x cos 2x
11. tan2x. 2 cos 2x( sin 4x sin 12x)
cos 2x cos x - sin 3x sin 4x
1
[sin 9x cos 7 x - sin 9x cos 3x ] 2 cos 2x(cos 4x cos 12x )
21 sin 4x sin 12x
2 [cos 3x cos x cos 7 x cos x ]
cos 7 x cos 3x cos 4x cos 12x
2 sin 4x cos 2x
cos 3x cos 7 x
2 sin 10x sin 2x 2 cos 4x cos 2x
tan 4x.
2 cos10x cos 2x
sin (4A - 2B) sin(4B - 2A)
sin 2x 19. tan(A B).
cos (4A - 2B) cos (4B - 2A)
cos 2x
(4A - 2B 4B - 2A) (4A - 2B 4B 2A)
tan2x. 2sin 2
cos 2
(cos θ - cos 3θ ) (sin8θ sin2θ ) (4A - 2B 4B - 2A) (4A - 2B 4B 2A)
12. 1. 2cos 2
cos 2
(sin 5θ - sinθ ) (cos 4θ - cos 6θ )
sin( A B)
(2sin 2θ sin θ ) (2sin5θ cos3θ ) tan(A B).
cos( A B)
(2cos 3θ sin2θ ) (2sin θ sin5θ )
4 cos 2A
1. 20. cot (A 15 o ) - tan (A - 15 o )
1 2sin2A
13. sin x sin 2x sin 3x sin 2x (1 2 cos x ).
cos (A 15 o )cos (A - 15 o ) - sin (A - 15 o ) sin (A 15 o )
(sin x sin 3x ) sin 2x
cos (A - 15 o ) sin (A 15 o )
2 sin 2x. cos x sin 2x
cos (A 15 o A 15 o )
sin 2x (1 2 cos x ). 1
[sin 2 A sin 30]
sin 4x sin 2x 2
14. tan 3x 2 cos 2 A
cos 4x cos 2x
2 sin 3x cos x sin 2 A 12
2 cos 3x cos x 4 cos 2A
tan 3x 1 2sin2A
15. 1 cos 2x cos 4x cos 6x 4 cos x cos 2x cos 3x.
CONDITIONAL TRIGONOMETRIC IDENTITIES cos( 2 A2 ) cos( 2 B2 ) cos( 2 C2 )
If A B C 180 o prove that 2 cos( 2 A B B A
1 2 sin 2 ( 4 C4 )
4 ) cos( 4 )
1. sin 2A sin2B sin2C 4sinAsinB sinC
1 2 sin( 4 C4 ) cos( B4A ) 2 sin 2 ( 4 C4 )
(sin2A sin2B) sin2C
1 2 sin( 4 C4 )[cos( B4A ) sin( 4 C4 )]
2sin(A B) cos(A - B) sin2C
1 2 sin( 4 C4 )[cos( B4A ) cos( 2 A B
4 )]
2sin( - C) cos(A - B) sin2C
B A
2sinC cos(A - B) 2sinC cosC 2 A4 B B A
2 AB
4
1 2 sin( 4 C4 ) 2 sin 4 sin 4
2 2
2sinC {cos(A - B) cosC}
1 sin 4A sin 4B sin 4C
2 sinC {cos(A - B) cos(180 - A B)}
2 sinC {cos(A - B) - cos(A B)} 7. sin 2 A sin 2 B sin 2 C 2 2cosAcosB cosC
1
2sinC {2 sinA sinB} [1 2cos 2A 1 cos2B 1 cos2C ]
2
4 sinA sinB sinC 32 12 [cos 2A cos2B cos2C ]
A B C
2. cosA cosB - cosC - 1 4cos
cos sin 32 12 [ 2 cos (A B) cos (A - B) 2cos 2 C - 1]
2 2 2
AB AB 2 C
2 cos 2 cos 2 2sin 2 1 32 12 [ 2 cos ( - C) cos (A - B) 2cos 2 C - 1]
2 cos ( 2 - C2 ) cos A 2 B 2sin 2 C 1 32 12 [ - 2 cos C cos (A - B) 2cos2C - 1]
2 sin C2 cos A 2 B 2sin 2 C
1 32 12 [ - 2 cos C [cos (A - B) - cos C - 1]
2
2 sin C2 [cos A 2 B sin C2 ] 1 32 12 [-2cos C [cos (A - B) - cos { - (A B)}] - 1]
2sin C2 [cos A 2 B sin { 2 - A 2 B }] 1 32 12 [-2cos C [cos (A - B) cos (A B)] - 1]
2sin C2 [cos A 2 B cos A 2 B ] 1 2 2cosAcosB cosC
8. sin 2 A sin 2 B - sin 2 C 2sinAsinB cosC
- 1 4cos A2 cos B2 sin C2
sin 2 A sin 2 B - sin 2 C
A B C
3. cosA cosB cosC - 1 4sinsin sin sin 2 A sin( B C)sin(B - C)
2 2 2
AB AB 2C
2 cos 2 cos 2 2sin 2 1 sin 2 A sin A sin (B - C)
2 cos ( 2 - C2 ) cos A 2 B 2sin 2 C 1 sin A[sinA - sin (B - C)]
2 sin C2 cos A 2 B 2sin 2 C2 1 sin A[sin(B C) - sin (B - C)]
2 sin C2 [cos A 2 B - sin C2 ] 1 2 sin A sin B cos C
A B B C C A
2sin C2 [cos A 2 B - sin { 2 - A 2 B }] 1 9. tan tan tan tan tan tan 1
2 2 2 2 2 2
2sin C2 [cos A 2 B cos A 2 B ] 1 A B C
2
2
1 4sin A2 sin B2 sin C2 tan A B2 C tan 2
A B C 1 tan A2 tan B2 tan C2 - tan A2 tan B2 tan C2
4. sin sin sin 1
2 2 2 8 A B B C C B
1 tan 2 tan 2 tan 2 tan 2 tan 2 tan 2
sin A2 sin B2 sin C2 u
1 tan A2 tan B2 tan B2 tan C2 tan C2 tan B2
- 12 [cos A 2 B cos A 2B ]cos A 2 B u 0
tan A2 tan B2 tan C2 - tan A2 tan B2 tan C2
cos 2 AB
2
cos A 2 B cos A 2 B 2u 0
1 tan A2 tan B2 tan B2 tan C2 tan C2 tan B2 0
2
b 4ac 0
tan A2 tan B2 tan B2 tan C2 tan C2 tan B2 1
cos 2 A 2 B 8u 0
10. sinA sinB sinC 4 cos A2 cos 2B cos C2
u 18 cos 2 A 2 B 18
2sin A 2 B cos A 2 B 2sin C2 cos C2
3
5. 1 cosA cosB cosC 2sin( 2 - C2 ) cos A 2 B 2 sin C2 cos C2
2
cosA cosB cosC 1 2 cos C2 cos A 2 B 2 sin C2 cos C2
2 cos C2 {cos A 2 B sin C2 }
cosA cosB cosC 1 4 1
8 2 cos C2 {cos A 2 B sin( 2 - A2 B )}
1 cosA cosB cosC 3 2 cos C2 {cos A 2 B cos A2 B}
2
A B C -A -B -C 2 cos C2 {2 cos A2 cos B2 }
6. sin sin sin 1 sin sin sin
2 2 2 4 4 4 4 cos A2 cos B2 cos C2
11. cos 2 A cos 2 B cos 2 C 1 - 2 cosAcosB cos C. 2 cosC {cos(A - B) sin( 2 - ( A B)}
1 2 cos C {cos(A - B) cos(A B)}
[1 2cos 2A 1 cos2B 1 cos2C ]
2 2cosC {2 cosA cosB}
32 12 [cos 2A cos2B cos2C ]
4cosAcosB cosC
32 12 [ 2 cos (A B) cos (A - B) 2cos 2 C - 1] 16. cos 2A cos2B cos2C 1 4sinAsinB sinC
32 12 [ 2 cos ( - C) cos (A - B) 2cos 2 C - 1] 2 cos (A B) cos (A - B) 2sin 2 C 1
32 12 [ - 2 cos C cos (A - B) 2cos2C - 1] 2 cos ( 2 - C) cos (A - B) 2sin 2 C 1
32 12 [ - 2 cos C [cos (A - B) - cos C - 1] 2 sin C cos (A - B) 2sin 2 C 1
32 12 [-2cos C [cos (A - B) - cos { - (A B)}] - 1] 2 sin C [cos (A - B) - sin C] 1
32 12 [-2cos C [cos (A - B) cos (A B)] - 1] 2sin C [cos (A - B) - sin { 2 - (A B)} 1
1 2cosAcosB cosC 2sin C [cos (A - B) cos (A B)] 1
12. sin(B C - A) sin(C A - B) sin(A B - C) 4sinAsinB sinC 1 4sinAsinB sin C
sin(B C - A) sin(C A - B) sin(A B - C)
If ABC is a right triangle and if A , then
sin( - A - A) sin( - B - B) sin ( - C - C) 2
sin( - 2A) sin( - 2B) sin ( - 2C) prove that
sin 2A sin 2B sin 2C 17. cos 2 B cos 2 C 1
4 sin A sin B sin C 90 o B C
13. If A + B + C = 2s, then prove that B C 0 o
sin(s − A) sin(s − B) + sins sin(s − C) = sinAsinB. B 0 o C
sin(s - A)sin(s - B) sins sin(s - C)
cos B cos(0 o C )
1
[cos s - A - s B cos s - A s B cos B sin C
2 2 2
cos s - s C cos s s - C ] cos 2 B cos 2 C sin 2 C cos 2 C
2 2
1
1 2s - (A B)
[cos B - A cos cos C cos 2s - C ] 18. sin 2 B sin 2 C 1
2 2 2 2 2
Proceeding as above
1
[cos B - A cos C cos C cos B A ] sin B sin C cos 2 C sin 2 C
2 2
2 2 2 2 2
1 1
[cos B - A cos B A ]
2 2 2 B C
19. cosB - cosC - 1 2 2cos sin
sinAsinB 2 2
14. If x + y + z = xyz, then prove that cosnsider, CosA cosB - cosC
2x 2y 2z 2x 2y 2z
. 2 cos A B cos A B 2sin 2 C 1
1 - x2 1 - y2 1 - z2 1 - x2 1 - y2 1 - z2 2 2 2
Let x tan A, y tan B, z tan C C
2 cos ( - ) cos A B 2
2sin C 1
2 2 2
Now x y z xyz
2 sin C cos A B 2sin 2 C 1
tan A tan B tan C tan A tan B tan C 2 2 2
A B C n 2 sin C [cos A B sin C ] 1
2 2 2
2 A 2B 2C 2n
2sin C [cos A B sin { - A B }] 1
tan 2 A tan 2B tan 2C tan 2 A tan 2B tan 2C 2 2 2 2
2x 2y 2z 2x 2 y 2 z C
2sin [cos A B cos A B ] 1
2
2
2
. 2 2 2
1-x 1-y 1-z 1 - x2 1 - y2 1 - z2
- 1 4cos A cos B sin C
2 2 2
If A + B + C = , prove the following
2 B C
Cos cosB - cosC - 1 4cos cos sin
15. sin 2A sin2B sin2C 4cosAcosB cosC 2 4 2 2
(sin2A sin2B) sin2C B C
cosB - cosC - 1 2 2cos sin
2sin(A B) cos(A - B) sin2C 2 2
2sin( 2 - C) cos(A - B) sin2C
2cosC cos(A - B) 2sinC cosC
2cos C {cos(A - B) sinC}
TRIGONOMETRIC EQUATIONS tan tan
sin sin
Principal Solution
cos cos
Smallest numerical value of unknown angle sin cos cos sin 0
satisfying the equation in the interval [−π, π] is
sin( ) 0
taken for defining principal solution.
n , n Z
Principal values
n , n Z
Function interval Quadrant
4. Solve an equation of the form a cos θ+b sin θ = c
Sine [ 2 , 2 ] I or IV
Take a r cos , b r sin
Cosine [0, π] I or II b
r a 2 b 2 , tan
tangent [ 2 , 2 ] I or IV a
a cos b sin c
General solution of trigonometric equations.
r cos cos r sin sin c
c
cos ( -)
r
c
cos ( -)
a b2
2
cos ( -) cos (say)
- 2n , n z
2n , n z
5. Find the principal solution of
1 1
1. Solve the equation sin θ = k (−1 ≤ k ≤ 1) : i) sin ii)cosec - 2, iii)cos
2 2
Let sin = k, numerically smallest angle 1
i) sin 0 principal value of sin lies in[ -2 , 2 ]
sin sin 2
sin - sin 0 principal value lies in I quadrant
2 cos sin 0 sin sin
2 2 6
Either cos 0 ( 2n 1) , n Z
2 2 2 6
( 2n 1) , n Z...(1) 1
ii) - 2,
sin
or sin 0 n , n Z 1
2 2 sin - 0
2
2n , n Z...(2)
principal value lies in IV quadrant
Combining , n (-1)n , n Z.
sin sin
2. Solve equation of the form cos θ = k (−1 ≤ k ≤ 1) 6
cos cos
6
cos cos 0
1
iii)cos 0, principal value of cos lies in[0, 2 ]
2 sin sin 0 2
2 2 principal value lies in I quadrant
Either sin 0 n , n Z
2 2 cos cos
3
2n , n Z...(1)
or sin 0 n , n Z 3
2 2
2n , n Z...(2) 6. Find the general solution of
Combining , 2n , n Z. 3
i) sin - ii) sec θ = −2 iii) tan θ =√3
2
3. Solve equation of the form tan θ = k (−∞ < k < ∞)
8. Solve the following equations for which
sin sin -
3 solutions lies in the interval 0◦ ≤ θ < 360◦
general solution : i) sin4 x = sin2 x
nπ (-1) n , n Z sin 4 x sin 2 x 0
nπ (-1) n - , n Z sin 2 x(sin 2 x 1) 0
3
Either, sin 2 x 0
or nπ (-1) n1 , n Z x
3
ii) sec θ = −2 or sin 2 x 1 0
1
cos - sin 2 x sin 2
2 2
cos cos - x
3 2
2
cos cos x 0, , ,
3 2 2
general solution 2nπ , n Z ii) 2 cos2 x + 1 = −3 cos x
2
2nπ , n Z 2 cos 2 x 3 cos x 1 0
3
(cos x 1)( 2 cos x 1) 0
iii) tan θ =√3
Either, (cos x 1) 0
tan tan cos x cos x
3
general solution : Or 2 cos x 1 0
nπ , n Z 1 2
cos x x
2 3
nπ , n Z 2 4
3 x , ,
3 3
7. Find the principal solution and general solution
iii) 2 sin2x+1 = 3sinx
1 1
i) sin ii) tan iii) cot 3
2 3 2 sin 2 x 3 sin x 1 0
1 (sinx 1)( 2sin 1) 0
i) sin
2 4 Either, (sinx 1) 0
general solution :
sinx sin 2 , x
2
nπ (-1) n , n Z
Or 2 sin x 1 0
nπ (-1) - , n Z
n
1
3 sinx x
2 6
or nπ (-1) n1 , n Z 5
x , ,
3 2 6 6
1
ii) tan 0 iv) cos 2x = 1− 3 sinx.
3
1 2 sin 2 x 1- 3 sin x
tan tan
6 6 2 sin 2 x 3 sin x 0
general solution nπ , n Z sinx ( 2sinx 3) 0
nπ , n Z Either, sinx 0
3 x 0,
iii) cot 3
Or 2 sin x 3 0
1
tan 0 3
3 sinx Not possible
2
tan tan x 0,
6 6
Solve the following equations:
general solution nπ , n Z
6 1. 3 cos2 θ = sin2 θ
3 cos 2 1- cos 2 tan 2x - cot( x 3 )
1 tan 2x tan( 2 x 3 )
cos 2
4 tan 2x tan( 56 x )
1 cos 2 1
2x n 56 x , n z
2 4
1 x n 56 , n z
cos 2
2 6. sin x − 3 sin2x + sin3x = cos x − 3 cos 2x + cos3x
cos 2 cos( 3 ) sin x - 3 sin 2x sin 3x cos x - 3 cos 2x cos 3x
2 2n 23 , n z sin 3x sin x - 3 sin 2x cos 3x cos x -3 cos 2x
n 3 , n z 2 in2x cos x - 3 sin 2x 2 cos 2x cos x - 3 cos 2x
( sin 2x- cos 2x)( 2 cos x-3) 0
2. sin x + sin5x = sin3x
sin x sin5x sin3x . sin 2x - cos 2x 0 sin ce 2 cos x -3 0
2 sin3x cos 2x sin3x sin 2x cos 2x
tan 2x 1
sin 3x (2 cos 2x - 1) 0
n
1 x , n Z.
Either sin 3x 0 (or) cos 2x 2 8
2
If sin 3x 0, then 3x n . 7. 3 tan 2 ( 3 - 1)tan - 1 0
n 3 tan 2 ( 3 - 1)tan - 1 0
x , nZ .
3 3 tan 2 3 tan - tan - 1 0
1
If cos 2x , cos 2x cos ( 3 tan - 1)(tan 1) 0
2 3
Either ( 3 tan - 1 0
2x 2n x n , n Z 1
3 6 If( 3 tan - 1 0, tan
n 3
general solution x (or) x n , n Z
3 6
tan
3. cos x + sinx = cos2x + sin2x 6
cos x sin x cos 2x sin 2x
n , n z
cos x - cos 2x sin 2x - sin x 6
2 sin 2 x2 x sin 2 x2x 2 cos 2 x2 x sin 2 x2x or tan 1 0
If tan 1 0 then tan -1
2 sin 32x sin x2 2 cos 32x sin 2x
sin x2 [sin 32x cos 32x ] 0 n - n z
4
Either sin x2 0 8. sin x + cos x = 1 + sinx cos x
x Let sin x cos x t
n x 2n, n z
2 t2 1
Or [sin 32x cos 32x ] 0 1 2 sin x cos x t 2 sin x cos x
2
sin 32x cos 32x t2 1
sub in given eqn, t 1
tan 32x 1 32x n 4 , 2
2n t 2t 1 0 (t 1) 2 0
2
x ,n z
3 6 sin x cos x 1
4. sin 9θ = sinθ 1 1
sin 9 sin 2 sin x cos x 1
2 2
sin 9 - sin 0 1
cos sin x cos cos x
2 cos 5 sin 4 0 4 4 2
Either cos 5 0
cos x cos
4 4
5 ( 2n 1) ( 2n 1) n z
2 10 x 2n
(or) sin 4 0 4 4
n
4 n ,n z x 2n , x 2n , n Z
4 2
9. √ 3 sinθ − cos θ =√2
5. tan 2x = −cot ( x 3 )
a -1; b 3 c 2 ; r a 2 b 2 2. 14. 2 cos2 x − 7 cos x+3 = 0
2 cos 2 x-7 cos x 3 0
3 1 2
sin cos (cos x 3)( 2 cos x 1) 0
2 2 2
1 Either, cos x 3 Not possible
sin cos cos sin
6 6 2 Or, 2 cos x 1 0
sin sin cos x
1
x 2n
6 4 2 3
15. 2 cos2 θ + 3sinθ − 3 = 0
n (-1) n
6 4 2 cos 2 3 sin 3 0
n (-1) n n Z 2 - sin 2 3 sin 3 0
6 4
10. sin θ + cos θ =√2 sin 2 3 sin 1 0
1 1 2 ( sin 1)( 2sin 1) 0
sin cos
2 2 2 Either, ( sin 1) 0
sin 1
sin sin cos cos 1
4 4
sin sin 2 , n ( 1) n
2
cos cos 0 Or 2 sin 1 0
4
1
2n 0 sin n ( 1) n
4 2 6
16. 2 sin2 x + sin2 2x = 2
(8n 1) , n Z 2 sin 2 x sin 2 2x 2
4
11. sin θ +√3 cos θ = 1 2 sin 2 x ( 2 sin x cos x) 2 2
1 3 1
sin cos cos 2 x[sin 2 x - 1] 0
2 2 2
1 Either cos x 0
sin sin cos cos
6 6 2 x ( 2n 1) , n Z
2
cos cos 1
6 3 or sin 2 x
2
2n
6 3 sin 2 x sin 2
4
2n , n Z
3 6 x n , n Z
12. cot θ + cosecθ = √3 4
cos 1 17. sin 5x − sin x = cos3x
3 sin 5x sinx cos 3x .
sin sin
cos 3 sin 1 2 sin2x cos 3x cos3x
cos 3x (2sin 2x - 1) 0
1 3 1
cos sin Either cos 3x 0
2 2 2
1
sin sin cos cos 3x (2n 1) , x (2n 1) , n Z
3 3 2 2 6
1
cos cos (or) sin 2x
3 3 2
2
2n sin 2x sin
3 3 6
2
2n ,n Z 2x 2n x n , n Z
3 3 6 12
5 1
13. cos 2θ = solution x (2n 1) or x n , n Z
4 6 12
18. cos θ + cos3θ = 2cos2θ
cos 2 cos 36 o [ see page : 62, Problem : 5]
[cos cos 3]- 2 cos 2 0
2 2n 20
2 cos 2 cos - 2 cos 2 0
n 10
2 cos 2( cos -1) 0
cos 2 0 PROPERTIES OF TRIANGLE
2 ( 2n 1) , n Z 1. Law of Sines
2
In any triangle, lengths of the sides are proportional to the
( 2n 1) , n Z sines of the opposite angles
4
or, cos -1 0 cos 1 a b c
2R
2n , n Z sin A sin B sin C
19. sin θ + sin3θ + sin5θ = 0
sin sin5 - sin3
2 sin3 cos 2 sin3
sin 3 (2 cos 2 1) 0
Either sin 3 0 (or) cos 2 12
If sin 3 0, then 3 n . Case I : A is acute.
n Produce BO to meet the circle at D.
, nZ .
3 BDC BAC
1 2
If cos 2 , cos 2 cos BDC A [angles in same segment]
2 3
2 BCD 90 o .[angle in a semicircle]
2 2n n , n Z
3 3 BC
sin BDC
n BD
general solution x (or) n , n Z
3 3 a
or sin A
20. sin 2θ − cos 2θ − sin θ + cos θ = 0 2R
(sin 2 sin ) ( cos2 cos) 0 a
2R
2cos 32 sin 2 2 sin 32 sin 2 0 sin A
Case II : A is right angle.
2 sin 2 (cos 32 sin 32 ) 0
O is on the side BC of ABC.
Either sin 2 0
a BC
n 2n , n Z sin A sin 90 o
2
2R
or cos 32 sin 32 0
1
tan 32 1 a
2R
3 n sin A
2
4 Case III : A is obtuse .
23 n 6 n Z Pr oduce BO to meet the circle at D.
21. tan θ + tan (
3) + tan ( 23 ) =√3 BDC BAC 180 o
tan tan( 3 ) tan( 23 ) 3 BDC 180 o - A
tan tan 3 tan tan 23 BC
tan 3 sin BDC
1 tan tan 3 1 tan tan 23 BD
tan 3 tan 3 a
tan 3 or sin (180 o - A)
1 3 tan 1 3 tan 2R
a
8 tan sin A
tan 3 2R
1 3 tan 2
a
tan 3 tan 3 8 tan 2R
3 sin A
1 3 tan 2 b c
simillarly 2R
3 tan tan 3 1 sin B sin C
1 3 tan 2 3 2. Napier’s Formula
AB ab C BC b c A
tan 3 tan cot , tan cot
6 2 ab 2 2 bc 2
3 n 6 CA ca B
n
tan cot
3 18 2 ca 2
a b Area of triangle 12 ab sin C 12 bc sin A
sine Law 2R 2R
sin A sin B 12 ac sin B
a 2R sin A b 2R sin B
In ABC, draw AD BC
ab C 2 R sin A 2 R sin B C
cot cot AD
ab 2 2R sin A 2R sin B 2 In ADC, sin C AD b sin C
sin A sin B C AC
cot Thus , 12 base height
sin A sin B 2
2 cos A2 B sin A2B C 12 ab sin C.
cot
A B
2 sin 2 cos 2 AB 2 6. Area of the segment of a circle
C Area of segment ABD Sector Area - Area ofOAB
cot A B tan A B cot
2 2 2 12 r 2 θ - 12 r 2 sin θ
2 2
cot C tan A B cot
2
C
2
12 r 2 (θ(θ sin θθ
7. Half-Angle formula
C
tan C tan A B cot In ABC if s semi - perimeter ofΔfABC
2 2 2
ab C abc
cot tan A B s
2
ab 2 2
3. Law of Cosines A s(s - a) A (s - b)(s - c)
i) cos ii) tan
b 2 c 2 -a 2 c 2 a 2 -b 2 2 bc 2 s(s - a)
cos A , cos B
2bc 2ca A (s - b)(s - c)
iii) sin
2
b a -c2 2 2 bc
cos C
2ab A A
sin sin 2
Draw AD BC in ABC 2 2
In ΔABD , AB 2 AD 2 BD 2 1 cos A
2
c2 AD 2 BD 2
AD 1 b 2 c 2 a2
1
sin C 2 2bc
AC
AD b sin C 2bc b 2 c 2 a 2
BD BC - DC 4bc
a - b cos C a 2 (b c ) 2
c 2 (b sin C)2 (a - b cos C) 2 4bc
( a b c )( a b c )
b 2 sin 2 C a 2 b 2 cos 2 C - 2ab cos C
4bc
a 2 b 2 (sin 2 C cos 2 C )- 2ab cos C
( a b c 2c )( a b c 2b )
a 2 b 2 - 2 ab cos C
4bc
c 2 a 2 b 2 - 2ab cos C ( 2s - 2b)( 2s - 2c)
b 2 c 2 -a 2 4bc
cos A (s - b)(s - c)
2bc
4. Projection Formula bc
i) a b cos C c cos B, A A
sin A 2 sin cos
ii) b c cos A a cos C 2 2
(s - b)(s - c) s(s - a)
iii) c a cos B b cos A 2
bc bc
In ABC,a BC
2 s(s - a)(s - b)(s - c)
Draw AD BC
bc
a BC 8. Heron’s formula
BD DC 12 ab sin C.
BD DC 21 ab 2 sin C2 cos C2
AB AC
AB AC (s - b)(s - a) s(s - c)
( cos B)c ( cos C)b ab
ab ab
a b cos C c cos B s(s - a)(s - b)(s - c)
5. Area of the Triangle
In a ABC, prove that 5. The angles of a triangle ABC, are in Arithmetic
1. b2 sin 2C+c2 sin 2B = 2bc sinA Progression and if b : c = √3 :√2, find A.
Sine Law, A , B, C are in A.P
a b c 2B A C
2R
sin A sin B sin C 2B B A C B
a 2R sin A; b 2R sin B; c 2 R sin C
3B 180 o
2 2 2 2
4R sin B sin2C 4R sin C sin2B
B 60 o
2 2 2
4R 2sin B sinCcosC 2sin CsinBcosB
b:c 3: 2
8R 2 sinB sinC (sinB cosC sinC cosB) b c
Sin Law
8R 2 sinB sinC sin(B C) sin B sin C
8R2 sinB sinC sin ( - A) b sin B
c sin C
8R 2 sinB sinC sinA
3 sin 60 o
b c C 45 o
8R 2 sinA 2 sin C
2R 2R A (B C )
2bc sin A.
B-C b-c A A 75 o
2. sin cos a 2 b 2 1 cos (A - B) cos C
2 a 2 6.
Sine Law : a 2 c 2 1 cos (A - C) cos B
a b c a b c
2R Sine Law 2R
sin A sin B sin C sin A sin B sin C
a 2 R sin A; b 2R sin B; c 2R sin C a 2 R sin A; b 2 R sin B; c 2 R sin C
b-c A 2RsinB - 2RsinC A a 2 b 2 ( 2R sin A) 2 ( 2 R sin B) 2
cos cos
a 2 2RsinA 2 a 2 c 2 ( 2R sin A) 2 ( 2 R sin C) 2
B-C BC sin 2 A sin 2 B
2sin cos
2 2 cos A
A A sin 2 A sin 2 C
2
2 sin cos 1 - cos 2 A sin 2 B
2 2
B-C A 1 - cos 2 A sin 2 C
sin cos
2 2 2 1 - (cos 2 A - sin 2 B)
A 1 - (cos 2 A - sin 2 C)
sin
2 1 - cos(A B) cos(A - B)
B-C
sin 1 - cos(A C) cos(A - C)
2
3. If the three angles in a triangle are in the ratio 1 1 cos (A - B) cos C
: 2 : 3, then prove that the corresponding sides 1 cos (A - C) cos B
are in the ratio 1 :√3 : 2. sin B c - a cos B
7.
Let the angles , 2, 3 sin C b - a cos C
2 3 180. c - a cos B ( a cos B b cos A) - a cos B
b - a cosC ( a cos C c cos A) - a cos C
30 o
b cos A
angles 30 o , 60 o , 90 o
c cos A
Sine Law : 2R sin B
a b c 2R sin C
o
o
sin B
sin 30 sin 60 sin 90 o
sin C
a : b : c sin 30 o : sin 60 o sin 90 o
8. a cosA + b cosB + c cosC = 2a sinB sinC.
1 3 a cos A b cos B c cos C
: :2
2 2 2R sin A cos A 2R sin B cos B 2R sin C cos C
a : b : c 1: 3 :1
R(sin 2 A sin 2B sin 2C )
4. (b+c) cosA+(c+ a) cosB + (a + b) cosC = a + b + c
R( 4 sin A sin B sin C )
= b cosA+c cosA+c cosB+a cosB + a cosC+b cosC
= b cosC +c cosB+c cosA+a cosC+b cosA+a cosB 4R( 2aR sin B sin C )
= a + b + c [by projection formula] 2a sin B sin C
2 2 2
a 2 -c 2 sin (A - C) b 2 c 2 - a 2 (2RsinB) (2RsinC) - (2RsinA)
9.
b2 sin (A C) 2bc 2 (2RsinB) (2RsinC)
a b c sin 2 B sin(C A) sin(C - A)
sine Law : 2R
sin A sin B sin C 2 sinB sinC
a 2R sin A; b 2 R sin B; c 2 R sin C sinB [sinB sin(C - A)]
a 2 -c 2 4R 2 sin 2 A 4 R 2 sin 2 C 2 sinB sinC
b2 4R 2 sin 2 B sin(C A) sin(C - A)
sin 2 A sin 2 C 2 sinC
cosA
sin 2 B
sin( A C ) sin( A C ) 13. Derive Projection formula from Law of sines,
a
sin 2 B Sine Law , 2R
sin A
sin( A C ) sin( A C )
a 2R sin A
sin( [ A C ])
2R sin( [ B C ])
sin( A C ) sin (A - C)
2R sin[B C ]
sin 2 (A C)
2R sin B cos C 2 R sin C cos B
sin (A - C)
a b cos C c cos B
sin (A C)
14. Derive Projection formula from Law of cosines.
a sin (B - C) b sin (C - A) c sin (A -B)
10. a 2 c 2 -b 2 b 2 a 2 -c 2
b2- c 2 c 2 - a2 a 2- b 2 cosine law, cos B , cos C
a sin (B - C) 2 R sin A sin (B - C) 2 ac 2ab
2 2
a c -b b a 2 -c 2
2 2 2 2
b -c 4R 2 sin 2 B- 4R 2 sin 2 C c cos B b cos C c b
1 sin A sin (B - C) 2 ac 2ab
2R sin(B C ) sin(B C ) a 2 c 2 -b 2 b 2 a 2 -c 2
1 sin A 2a
2R sin( A) 2a 2
1 2a
2R c cos B b cos C a
b sin (C - A) 1 sin B sin (C - A) 15. Using Heron’s formula, show that equilateral
c 2- a2 2R sin(C A) sin(C A) triangle has maximum area for fixed perimeter.
1 Let ABC be a triangle with constant perimeter 2s.
s(s - a)(s - b)(s - c)
2R
c sin (A -B) 1 is maximum, when (s−a)(s−b))(s−c) is maximum
a2 - b2 2R G.M A.M.
a sin (B - C) b sin (C - A) c sin (A -B) 1 (s - a) (s - b) (s - c) 3 s 3
(s- a)(s-b)(s - c)
b2- c 2 c 2- a2 a2 - b2 2R 3 27
ab A B A B
tan
3
11. cot s
ab 2 2 (s - a)(s - b)(s - c)
27
a b 2R sin A 2R sin B Equality occurs when s − a = s − b = s − c.
a b 2 R sin A 2R sin B s3
sin A sin B when a = b = c, maximum of(s − a)(s − b))(s − c)=
27
sin A sin B For fixed perimeter 2s, area is maximum if a = b = c
A B A B
2 sin cos ss 3 s2
2 2 Maximum area sq.units.
27 3 3
A B A B
2 cos sin sinA sin(A - B)
2 2 16. In a ABC, if , prove that a2, b2,
sinC sin(B - C)
A B A B
tan cot c2 are in Arithmetic Progression.
2 2 sin( - [B C]) sin(A - B)
12. Derive cosine formula using the law of sines
sin( - [A B]) sin(B - C)
Sine Law :
sin(B C) sin(A - B)
a b c
2R sin(A B) sin(B - C)
sin A sin B sin C
a 2R sin A; b 2R sin B; c 2R sin C sin(B C)sin(B - C) sin(A - B)sin(A B)
sin 2 B sin 2 C sin 2 A sin 2 B 21. (a2 − b2 + c2)tanB = (a2 + b2 − c2) tanC
a b c a 2 b 2 2ca cos B c 2
sine law, sinA , sinB , sinC
2R 2R 2R sin B
(a 2 b 2 c 2 ) tan B ( 2ca cos B c 2 c 2 )
b 2 c 2 a2 b2 cos B
2ca sin B
2b 2 a 2 c 2
a , b , c are in Arithmetic Progression.
2 2 2 b sin C
2ca
sinA c
17. If cosC , show that triangle is isosceles. 2 ab sin C
2 sinB
sine law : sin C
(a 2 b 2 c 2 ) tan C ( a 2 2ab cos C a 2 )
a b c cos C
sin A , sin B , sin C 2 ab sin C
2R 2R 2R
cosine law : 22. An Engineer has to develop a triangular shaped
park with a perimeter 120 m in a village. The park
b 2 a 2 -c 2
cos C to be developed must be of maximum area. Find
2ab out the dimensions of the park.
sinA
Now cosC For fixed perimeter, and maximum area the
2 sinB triangle should be equilateral
2 2 2
b a -c a 2R Perimeter 120
2 ab 2 R 2b a b c 120
b 2 a 2 -c 2 a 2 since a b c
2 2
b c 3 a 120
ca a 40
Hence isoceles 23. A rope of length 12 m is given. Find the largest
B - C area of the triangle formed by this rope and find
18. Prove that b c 2a cos if A = 60o
2 the dimensions of the triangle so formed.
b c a cos C c cos A a cos B b cos A Perimeter 12
( b c ) cos A a(cos B cos C ) 3 a 12
BC B-C a4
( b c ) cos 60 o a( 2 cos cos )
2 2 s2
bc A B-C Max Area
2a cos cos 3 3
2 2 2 62
bc 60 o
B-C
2 a cos cos 3 3
2 2 2
4 3 sq.m
B - C
2 a cos 60 o cos 24. Government plans a circular zoological park of
2
diameter 8 km. Find separate area in the form of a
B - C
b c 2 a cos
segment formed by a chord of length 4 km is to be
2
allotted for a veterinary hospital in the park.
A
19. a( cos B cos C) 2(b c) sin 2 Let AB chord
2
a( cos B cos C) a cos B a cos C O centre of circular park.
c b cos A b c cos A .AOB , r 4 m
( b c )(1 cos A) Area of segment Area of sector - Area of OAB.
A 1 1
( b c )(1 1 sin 2 ) r 2 r 2 sin
2 2 2
A 8[ sin ]
2(b c) sin 2
2 42 42 - 42 1
cos
A A 2(4)(4) 2
20. a sin B (b c) sin
2 2
A
(b c) sin 3
2 3
A Area of segment 8
a sin B 3 2
2
43 [ 2 3 3]m 2
3.11.APPLTICATION OF TRIGNOMETRY 4. In a ABC, if a = 2√2, b = 2√3 and C = 75, find
Working Rule: the other side and the angles
In a right triangle, two sides determine third side cos C cos 75 o
via Pythagorean theorem and one acute angle a2 b2- c 2 3 1
determine other using the fact that acute angles in
2ab 2 2
a right triangle are complementary. 2
8 12 - c 3 1
c 2( 3 1)
If all sides of triangle are given, use either cosine 8 6 2 2
formula or half-angle formula to calculate all the 12 8 4 3 8
angles of the triangle. cos A
4 3 2( 3 1)
If two angles and one of sides opposite to given 4 3( 3 1)
angles given, use sine formula to find other sides. 4 3 2( 3 1)
If two sides and included angle given, use law of 1
cos A A 45 o
cosines to calculate other side and other angles of 2
triangle. In this case we have a unique triangle. B 180 ( A C ) B 60 o
Solving an oblique triangle require that length of
atleast one side must be provided. 5. In a ABC, if a =√3 − 1, b =√3 + 1 and C = 60,
find the other side and other two angles.
cos C cos 60 o
( 3 1) 2 ( 3 1) 2 - c 2 1
c 6
2( 3 1)( 3 1) 2
SOLVED PROBLEMS
( 3 1) 2 6 ( 3 1) 2
1. In a ABC, a = 3, b = 5 and c = 7. Find the values cos A
of cos A, cosB and cos C 2 3 2( 3 1)
By Cosine formula : 2 33
3 2( 3 1)
b2 c2 - a2
cosA 3( 3 2)
2bc
52 72 - 32 3 2( 3 1)
.
2(5)(7) 3 1
cos A A 15 o
13 2 2
14 B 180 75 0 B 105 o
11 1
cosB , cosC - 6. Find area of triangle : sides 13 cm, 14 cm,15 cm.
14 2
2. In ABC, A = 30,B = 60,c = 10, Find a and b. abc
s
2
A 30 o , B 60 o , C 180 o - 90 o 90 o 13 14 15
a b c 21 cm.
sine law 2
sin A sin B sin C Area of triangle s(s - a)(s - b)(s - c)
a b 10
o
o
21(21 - 13)(21 - 14)(21 - 15)
sin 30 sin 60 sin 90 o
a 10 84 sq.cm.
a5
sin 30 o
sin 90 o 7. In a ABC, if a = 12 cm, b = 8 cm and C = 30,
b 10 then show that its area is 24 sq.cm.
o
b5 3
sin 60 sin 90 o 1
ab sin C
3. If the sides of a ABC are a = 4, b = 6 and c = 8, 2
then show that 4 cosB + 3cosC = 2. 1
12 8 sin 30 o
2
Projection Formula :
24sq.cm
b cos C c cos B a
8. In a ABC, if a = 18 cm, b = 24 cm and c = 30 cm,
6 cos C 8 cos B 4
then show that its area is 216 sq.cm.
3 cos C 4 cos B 2
abc x distance of phone’ s position from highway
s
2 x
18 24 30 sin60 o
36 cm. 5
2 . x 5sin60 o
Area of triangle s(s - a)(s - b)(s - c)
5 3
36 18 12 6 km
2
216 sq.cm. 13. Two radar stations located 100km apart, detect a
82 fighter aircraft between them. Angle of
9. InABC prove that a cos A b cos B c cos C elevation measured by first station is 30. Angle
abc
Refer to problem 8, page : 76 of elevation measured by the second station is
a cos A b cos B c cos C 2 a sin B sin C 45. Find altitude of the aircraft at that instant.
2 2 A 180 o - (30 o 45 o ) 105 o
2 a
ac ab a 100
o
82 sin 45 sin 105 o
a cos A b cos B c cos C 2a 100 2 2
abc
1 3 1
b2 c 2- a2
10. In any ABC, prove that the area = a 100( 3 1)
4 cot A
x
b 2 c 2- a2 sin 30 o
cosine law, cos A a
2bc
x 50( 3 1)
b c 2- a2
2
bc 14. Two soldiers A and B in two underground
2 cos A
1 bunkers spot an intruder atthe top of a hill. The
Area of Triangle bc sin A
2 angle of elevation of the intruder from A and B
1 b2 c 2- a2 to the ground level in the eastern direction are
sin A
2 2 cos A 30 and 45. If A and B stand 5km apart, find the
b 2 c 2- a2 distance of the intruder from B.
4 cot A C position of intruder during detection
11. A boat travels 10 km from the port towards east A 180 o - (30 o 45 o ) 105 o
and then turns 60 to its left. If the boat travels BC 5
further 8 km, how far from the port is the boat? o
sin 30 sin 105 o
BP required distance. 2BC 5 2 2
Using cosine formula, 1 3 1
BP 10 2 8 2 - 2 10 8 cos 120 o
2
5 2
BC
244km . 3 1
BP 2 61km 15. A researcher wants to determine the width of a
pond from east to west. From a point P, he finds
12. Two cell phone towers located at 6 km apart,
the distance to eastern-most point of the pond to
east to west. A cell phone is north of highway.
be 8 km, distance to the western most point from
The signal is 5 km from the first tower and √31
P to be 6 km. If the angle between the two lines
km from the second tower. Determine the
of sight is 60, find the width of pond.
position of the cell phone north and east of the
first tower and how far it is from the highway. Using cosine formula,
2
31 5 2 6 2 - 2 5 6 cos a b 2 c 2 - 2bc cos A
2
31 25 36 - 60 cos a 2 8 2 6 2 - 2 6 8 cos 60 o
1 52km .
cos
2 a 2 13km
60 o 16. Two Navy helicopters A and B are flying over
the Bay of Bengal at same altitude to search a
missing boat apart 10 km from each other. If the
distance of boat from A is 6 km and if the line Using cosine formula,
segment AB subtends 60 at the boat, find the a 2 b 2 c 2 - 2bc cos A
distance of the boat from B.
a 2 1 2 2 2 - 2 1 2 cos 45 o
Using cosine formula,
5 - 2 2 km .
a b 2 c 2 - 2 bc cos A
2
a 5 - 2 2 km
a 2 6 2 10 2 - 2 6 10 cos 60 o
21. A man starts his morning walk at a point A
76km . reaches two points B and C and finally back to A
a 2 19km such that A = 60 and B = 45 , AC = 4km in
17. A surveyor observes two extremities A and B of the ABC. Find the total distance he covered
the tunnel to be built from point P in front of during his morning walk.
mountain. If AP = 3km, BP = 5 km APB = 120, C 180 105 75 o
find the length of the tunnel to be built. BC 4
Using cosine formula, o
sin 60 sin 45 o
a 2 b 2 c 2 - 2bc cos A 3
BC 4 2
a 2 3 2 5 2 - 2 6 8 cos 120 o 2
49km . 2 6
a 7km AB 4
o
18. A farmer purchase a triangular shaped land sin 75 sin 45 o
with sides 120feet and 60feet and angle included 3 1
AB 4 2
between these two sides is 60. If the land costs 2 2
Rs.500/sq.ft, find amount he needed to purchase 2( 3 1)
the land. Also find the perimeter of the land. Total Distance 4 2 6 2( 3 1)
Using cosine formula, 22. Two vehicles leave the same place P along two
2
a 14400 3600 - 7200 roads. One vehicle moves at an average speed of
10800 60km/hr and the other vehicle moves at an
average speed of 80 km/hr. After half an hour
a 60 3
vehicle reach the destinations A and B. If AB
120 60 60 3 subtends 60 at the initial point P, then find AB.
s
2
Using cosine formula,
90 30 3
a b 2 c 2 - 2bc cos A
2
s(s a )( s b )( s c )
a 2 80 2 60 2 - 4800
(90 30 3 )(90 30 3)(30 3 30)(30 3 30)
5200km .
30 30 (3 3)(3 3 )( 3 1)( 3 1)
a 20 13km
900 6 2
23. A satellite in space, an earth station and the
1800 3 centre of earth all lie in the same plane. Let r be
1 ft 2 Rs.500, Cost 1800 3 500 1558800 radius of earth and R be distance from the
centre of earth to satellite. Let d be distance from
19. A fighter jet has to hit a small target by flying a
horizontal distance. When the target is sighted, the earth station to the satellite. Let 30 be angle
of elevation from the earth station to satellite. If
angle of depression is 30. If after 100 km, target
the line segment connecting earth station and
has an angle of depression of 45, how far is the
satellite subtends angle α at the centre of earth,
target from the fighter jet at that instant?
Refer to Problem 13 , page 79 then prove that d R 1 ( Rr ) 2 2 Rr cos
20. A plane is 1 km from one landmark and 2 km d 2 R 2 r 2 2rR cos
from another. From the planes point of view the r2 r
land between them subtends an angle of 45. R 2 (1 2
2 cos
R R
How far apart are the landmarks?
d R 1 ( Rr ) 2 2 Rr cos
FUNDAMENTAL PRINCIPLES OF COUNTING Typically in a tree the branches represent the
1. The Sum Rule various possibilities.
5. Pigeonhole Principle:
Consider two tasks which need to be
completed. Suppose a flock of pigeons fly into a set of
First task is completed in M different ways pigeonholes. If there are more pigeons than
pigeonholes then there must be at least one
Second in N different ways,
pigeonhole with at least two pigeons in it.
if these cannot be performed simultaneously,
In any group of 27 English words, there must
then ways of doing either task are
be at least 2words begin with same letter
M+N (There are only 26 letters in English alphabet).
If there are n non-simultaneous tasks T1, T2, If six meetings are held on weekdays only,
T3, …., Tn performed in m1,m2, …,mn ways, then there must be at least two meetings held
Number of ways of doing one of these tasks is on the same day.
m1 + m2 + ….. + mn. Generalised form
2. The Product Rule If k + 1 or more objects are placed in k boxes,
Suppose a task comprises of two procedures. then there is at least one box containing two or
more of the objects.
If the first procedure can be completed in M
6. Notion of a ’string’.
different ways
Second procedure can be done in N different A string is formed by writing given letters one
ways after the first procedure is done, by one in a sequence.
Total No. of ways of completing task is Strings of length three formed out of letters
a,b,c & d are
MN
aaa, abb, bda, dca, cdd ….
If a task comprises of n procedures P1, P2, P3,
…, Pn performed in m1,m2,…….,mn ways Problems
Procedure Pi can be done after procedures P1, 1. Suppose one girl or one boy has to be selected
P2, P3, ….., Pi−1 are done, for a competition from a class comprising 17
boys and 29 girls. In how many different ways
Number of ways of completing the task is
can this selection be made?
m1 m2 ….mn.
First task : selecting a girl done in 29 ways.
3. The Inclusion-Exclusion Principle
Second task : selecting a boy done in 17 ways.
Suppose two tasks A and B can be performed By sum rule,
simultaneously.
No.of ways of making selection= 17+29
Let n(A), n(B) be No. of ways of perform ing
= 46
tasks A and B independent of each other.
2. In the menu card of a restaurant, the person saw
n(A ∩ B) be No. of ways of performing two
10 Indian and 7 Chinese food items. In how
tasks simultaneously.
many ways the person can select either an
Using the notation of set theory Indian or a Chinese food?
n(A B) = n(A) + n(B) − n(A ∩ B). the person can select Indian food in 10 ways
4. Tree Diagrams: Chinese food in 7 ways
Tree diagrams are helpful in representing the Either Indian or Chinese food = 10+7 ways
possibilities in a counting problem.
= 17 ways route A_1. Find the number of routes of
3. A School library has 75 books on Mathematics,
commuting from place A to place C via place B
35 books on Physics. A student can choose only without using similar mode of transportation.
one book. In how many ways a student can Bus route connecting Ato B are B1 and B2.
choose a book on Mathematics or Physics? Bus route connecting Bto C are B’1
A student can choose a Mathematics book in Train route connecting Ato B are T1 and T2.
“75” different ways.
Train route connecting Bto C are T’1 and T’2.
A student can choose a Physics book in “35”
Air route connecting Ato B are A1
different ways.
Air route connecting Bto C are A’1
By Sum Rule
i) If the person chooses B1 to travel from Ato C,
No. of ways a student can choose a book is
possible choices are
75 + 35 = 110.
(B1, T’1), (B1, T’2), (B1, A’1).
4. Consider 3 cities Chennai, Trichy and
Tirunelveli. In order to reach Tirunelveli from ii) If the person chooses B2 to travel from Ato C,
Chennai, one has to pass through Trichy. There possible choices are
are 2 roads connecting Chennai with Trichy and (B2, T’1), (B2, T’2), (B2, A’1).
there are 3 roads connecting Trichy with
iii) If the person chooses T1 to travel from Ato C,
Tirunelveli. What are the total number of ways
possible choices are
of travelling from Chennai to Tirunelveli?
(T1, B’1), (T1, A’1).
i) Roads connecting Chennai to Trichy are
iv) If the person chooses T2 to travel from Ato C,
R1 and R2.
possible choices are
Roads connecting Trichy to Tirunelveli are
(T2, B’1), (T2, A’1).
S1, S2 and S3.
v) If the person chooses A1 to travel from Ato C,
possible choices are
(A1, B’1), (A1, T’1), (A1, T’2)
By sum rule, No.of ways of travelling from
ii) Suppose a person chooses R1 to travel from Chennai to Tirunelveli = 3+ 3+2+2+3
Chennai to Trichy and further choose any of 3 = 13
roads S1, S2 or S3 to travel from Trichy to
6. Find the number of positive integers divisible
Tirunelveli. Possible road choices are
by 2 or 7 (but not both), upto 1000.
(R1, S1), (R1, S2), (R1, S3).
Let n(A) = No. of integers divisible by 2
If the person chooses R2 to travel from
n(B) = No. of integers divisible by 7
Chennai to Trichy, possible road choices are
n(A ∩ B) = No. of integers divisible by 2 and 7.
(R2, S1), (R2, S2), (R2, S3).
No. of positive integers divisible by 2 or 7 is
iii) By sum rule, No.of ways of travelling from
n(A B) = n(A) + n(B) − n(A ∩ B)
Chennai to Tirunelveli = 3 + 3 = 6
= 500 + 142 − 71
5. To travel from a place A to place B, there are two
different bus routes B1,B2, two different train = 571.
routes T1, T2 and one air route A1. From place B 7. How many numbers are there between 1 and
to place C there is one bus route say B1, two 1000 (both inclusive) which are divisible neither
different train routes say T_1, T_2 and one air by 2 nor by 5?
No. of integers divisible by 2, n(A) = 500 There are 110 smaller capacity transformers
No. of integers divisible by 5 , n(B) = 200 attached to a larger capacity transformer.
Divisible by 2 and 5. n(A ∩ B) = 100 As each smaller capacity transformer can be
linked with only 100 consumers, for 109
i.e divisible by 10
transformers, there will be
No. of positive integers divisible by 2 or 5 is
109 100 = 10900 links.
n(A B) = n(A) + n(B) − n(A ∩ B)
For 110th transformer there are only 29
= 500 + 200 − 100 consumers linked.
= 600. Total No. of consumers linked to the 238th
No. of positive integers not divisible by 2 or 5 larger capacity transformer is
= 1000 – 600 10900 + 29 = 10929
= 400 10. A person wants to buy a car. There are two
brands of car available in the market and each
8. In a village, out of the total number of people,
brand has 3 variant models and each model
80 percentage of people own Coconut groves 65
comes in five different colours In how many
percent of the people own Paddy fields. What is
ways she can choose a car to buy?
the minimum percentage of people own both?
% of people who own Coconut groves n(C) = 80
% of people who own Paddy fields and n(P) = 65.
By the rule of inclusion – exclusion
n(C P) = n(C) + n(P) −n(C ∩ P)
n(C ∩ P) = 80 + 65 − 100
= 45.
A brand can be chosen in 2 ways
Minimum % of people who own both is 45.
A model can be chosen in 3 ways
9. If an electricity consumer has the consumer
number say 238:110: 29, then describe the A colour can be chosen in 5 ways.
linking and count the number of house By the rule of product
connections upto the 29th consumer connection
A person can buy a car = 2 3 5 ways
linked to the larger capacity transformer
number 238 subject to the condition that each = 30 ways
smaller capacity transformer can have a 11. A Woman wants to select one silk saree and one
maximal consumer link of say 100. sungudi saree from a textile shop located at
Electricity distribution network Kancheepuram. In that shop, there are 20
different varieties of silk sarees and 8 different
varieties of sungudi sarees. In how many ways
she can select her sarees?
She can select a silk saree in 20 ways
Sungudi saree in 8 ways.
By the rule of product,
No. of ways of selecting these 2 sarees is
20 8 = 160.
12. Find the number of strings of length 4, formed
using the letters of the word BIRD, without
repetition of the letters. No. of strings of length ending with R is
The first place can be filled in 4 different ways 5 4 3 2 1 = 120
by any one of the letters B,I,R,D.
c) If a string starts with F and ends with R, other
Second place can be filled in by remaining 3 4 positions are to be filled
letters in 3 different ways,
Third place can be filled in 2 different ways,
Fourth place can be filled in 1 way. No. of strings of length of 6 starting with F and
By the rule of product ending with R is
No. of ways in which 4 places can be filled, 4 3 2 1 = 24.
4 3 2 1 = 24. d) By the principle of inclusion - exclusion,
Hence, the required No. of strings is 24. No. of strings of length 6, either starting with
F or ending with R is
13. How many strings of length 5 can be formed out
of the letters of the word PRIME taking all the 120 + 120 − 24 = 216.
letters at a time without repetition. ii) neither starts with F nor ends with R?
i) First place can be filled in 5 ways by any one All possible strings of length 6 formed out of the
of five letters P,R,I,M,E letters, F,L,O,W,E,R, where no letter is repeated.
ii) 4 letters are left to be placed in the second Total No. of letter strings
place, 3 letters are left for the third place, 2
6 5 4 3 2 1 = 720
letters are left to be put in the fourth place,
remaining 1 letter can be placed in fifth place. Neither start with F Total either start with F
-
nor ending with R strings or end with R
iii) No. of ways filling up five places is
720 - 216
5 4 3 2 1 = 120. 504.
14. How many strings of length 6 can be formed 15. How many strings can be formed using the
using letters of the word FLOWER if letters of the word LOTUS if the word (a) either
i) either starts with F or ends with R? starts with L or ends with S? (b) neither starts
with L nor ends with S?
ii) neither starts with F nor ends with R?
i) either starts with L or ends with S
i) either starts with F or ends with R?
a) If a string starts with L, other five positions
a) If a string starts with F, other five positions are
are to be filled with the letters O, T, U,S.
to be filled with the letters L, O, W, E, R.
L _ _ _ _
1way 4 3 2 1
2nd , 3rd , 4th , 5th, 6th places can be filled up in 2nd , 3rd , 4th , 5th places can be filled up in
5, 4, 3, 2 and 1 ways. 4, 3, 2 and 1 ways.
By rule of product, No. of strings starting with L = 1 4 3 2 1
No. of strings of length 6 starting with F is = 24.
5 4 3 2 1 = 120. b) _ _ _ _ S
b) If a string ends with R, other five positions are 4way 3 2 1 1
to be filled with the letters F,L,O,W,E. No. of strings ending with R = 4 3211
= 24 iv) 100th place filled in 8 ways by remaining
c) If a string starts with L and ends with S, numbers
L _ _ _ S v) 10th place can be filled in 7 ways.
1 3 2 1 1 Required No. of numbers is
No. of strings startwith L and endwith R 1 8 7 2 = 112.
18. How many 4 - digit even numbers can be
1 3 2 11 = 6.
formed using the digits 0, 1, 2, 3 and 4, if
d) By the principle of inclusion - exclusion, repetition of digits are not permitted?
No. of strings either starting with F or ending 1000th place cannot be 0.
with R 24 + 24 − 6 = 42.
1000th place can be filled in 4 ways
ii) neither starts with F nor ends with R?
100th place can be filled in 3 ways
Possible strings formed out of L,O,T,U,S where
10th place in 2 ways.
no letter is repeated
unit place can be either 0, 2 or 4 (even number)
Total No. of letter strings =5 4 3 2 1
i) If unit place is filled by 0,
= 120
Neither start with L Total either start with L
-
nor ending with S strings or end with S
120 - 42 78 No.of 4-digit numbers having 0 at unit place is
16. How many licence plates may be made using = 4 3 2 1 = 24.
either two distinct letters followed by four
ii) If unit place is filled with 2 or 4,
digits or two digits followed by 4 distinct letters
where all digits and letters are distinct? No. of ways is 2
i) No. of licence plates having two letters
followed by four digits is
26 25 10 9 8 7 = 32, 76, 000. No.ofways of filling the 1000th place is 3 ways
ii) No. of licence plates having two digits (excluding ’0’), 100th place in 3 ways and 10th
followed by four letters is place in 2 ways.
10 9 26 25 24 23 = 3, 22, 92, 000. 4-digit numbers without 0 at unit place
iii) Total No. of licence plates is = 3 3 2 2 = 36.
(26251098 7) + (1092625 24 23) iii) By the rule of sum,
= 3, 55, 68, 000. No. of 4 digit even numbers = 24+36 = 60.
17. Count the number of positive integers greater 19. Find the total number of outcomes when 5 coins
than 7000 and less than 8000 which are divisible are tossed once.
by 5, provided that no digits are repeated. When a coin is tossed, outcomes are {Head, Tail}.
i) 7000<4-digit number < 8000. By the rule of product rule,
ii) 1000th place will be the digit 7. No. of outcomes when 5 coins are tossed is
iii) unit place will be either 0 or 5. [divisible by 5 ] 2 2 2 2 2 = 25
= 32.
If n coins are tossed, No. of outcomes is 2n.
20. In how many ways (a) 5 different balls be 23. How many two-digit numbers can be formed
distributed among 3 boxes? (b) 3 different balls using 1,2,3,4,5 without repetition of digits?
be distributed among 5 boxes? 1st digit can be formed in 5 ways
i) Each ball can be placed into any one of the 2st digit can be formed in 4 ways
three boxes in 3 different ways.
Two digit number formed = 5 4 ways
By rule of product,
= 20 ways
No. of ways of distributing 5 balls among 3 boxes
24. Three persons enter in to a conference hall in
= 33333 which there are 10 seats. In how many ways they
= 35 can take their seats?
= 243. 1st person can takes seats in 10 ways
ii) Each ball can be placed into any one of the 2nd person can takes seats in 9 ways
five boxes in 5 different ways. 3rd person can takes seats in 8 ways
By rule of product, No. of wayss they can take seats = 10 98 ways
No. of ways of distributing 3 balls among 5 boxes = 720 ways
= 555 25. In how many ways 5 persons can be seated in a
= 53 row?
= 125. 1st person can be seated in a row in 5way
if n different objects are to be placed in m places, 2nd person can be seated in a row in 4way
No. of ways of placing is mn. ……… ….
21. There are 10 bulbs in a room. Each one of them 5th person can be seated in a row in 1way
can be operated independently. Find number of 5 person can be seated in row in = 54321way
ways in which the room can be illuminated.
= 120 ways
Each bulb can be operated in off mode or on
26. A mobile phone has a passcode of 6 distinct
mode that is in 2 ways.
digits. What is the maximum number of
Total No. of doing ‘on’ to illuminate = 210 attempts one makes to retrieve the passcode?
Total No. of doing ‘off’ to illuminate = 20 Passscode digits : 0,1,2,3…..,9
[Keeping all “off”, room cannot be illuminated] 1st digit can beretrieve in 10way
Total No. of ways of illuminating 2st digit can beretrieve in 9way
= 210 − 20 …..
= 1024 − 1 6st digit can beretrieve in 5way
= 1023. Max. No. of attempts to retrieve passcode
22. There are 3 types of toy car and 2 types of toy
= 1098765
train available in a shop. Find the number of
ways a baby can buy a toy car and a toy train? = 151200
27. Given 4flags of different colours, how many
A baby can buy a toy car in 3 ways
different signals can be generated if each signal
A baby can buy a toy train in 2 ways requires use of 3 flags, one below the other?
Atoy car and a toy train = 3 2 ways
= 6 ways
30. How many three-digit numbers are there with 3
in the unit place? (a) with repetition (b) without
repetition.
1st vacant place occupied in 4 ways a) with repetition __3
2n vacant place occupied in 3 ways Hundreds place filled in 9 ways (except 0)
32 vacant place occupied in 2 ways Tens place filled in 10 ways (include 0)
Total No.of signals = No. of ways of filling 3 Unit place filled in 1 ways (3)
vacant places Ways to form 3 digit number = 9101 ways
= 432 = 64 ways
= 24 b) without repetition. _ _ 3
28. Four children are running a race (a)In how many Hundreds place filled in 8 ways (except 0, 3)
ways can first two places be filled? (b)In how Tens place filled in 8 ways (include 0)
many different ways could they finish the race?
Unit place filled in 1 ways (3)
a) 1st place filled in 4ways
Ways to form 3 digit number = 881 ways
2nd place filled in 3ways
= 64 ways
No.of ways first two places be filled = 43
31. How many numbers are there between 100 and
= 12 500 with the digits 0, 1, 2, 3, 4, 5 ? if (a) repetition
b) 1st children can finish the race in 4ways of digits is not allowed (b) repetition of digits
allowed
2nd children can finish the race in 3ways
a) repetition of digits is not allowed _ _ _
…., …….
Hundreds place filled in 4 ways (except 0,5)
4 children can finish race in = 4321ways
Tens place filled in 5 ways (include 0)
= 24 ways
Unit place filled in 4 ways
29. Count the number of three-digit numbers which
can be formed from the digits 2,4,6,8 if (a) Ways to form 3 digit number = 454 ways
repetitions of digits is allowed (b) repetitions of = 80 ways
digits is not allowed
b) repetition of digits allowed
a) repetitions of digits is allowed _ _ _
Hundreds place filled in 4 ways (except 0,5)
Hundreds place filled in 4 ways
Tens place filled in 6 ways (include 0)
Tens place filled in 4 ways
Unit place filled in 6 ways
Unit place filled in 4 ways
Ways to form 3 digit number = 466 ways
Ways to form 3 digit number= 444 ways
= 144 ways
= 64 ways
32. How many three-digit odd numbers can be
a) repetitions of digits is not allowed _ _ _ formed by using the digits 0, 1, 2, 3, 4, 5 ? if (a)
Hundreds place filled in 4 ways the repetition of digits is not allowed (b) the
repetition of digits is allowed.
Tens place filled in 3 ways
a) the repetition of digits is not allowed
Unit place filled in 2 ways
Hundreds place filled in 4 ways (except 0)
Ways to form 3 digit number= 432 ways
Tens place filled in 4 ways (include 0)
= 24 ways
Unit place filled in 3 ways (1,3,5) unit place can be either 0, 5 (Divisible by 5)
Ways to form 3 digit number = 443 ways i) If unit place is filled by 0,
= 48 ways No.of 3-digit numbers having 0 at unit place is
b) the repetition of digits is allowed. = 5 4 1 = 20.
Hundreds place filled in 5 ways (except 0) ii) If unit place is filled with 5
Tens place filled in 6 ways (include 0) 0 _ 0/5
Unit place filled in 3 ways (1,3,5) No.ofways of filling the 100th place is 4 ways
Ways to form 3 digit number = 563 ways (excluding ’0’), 10th place in 4 ways and unit
place in 1 ways.
= 90 ways
3 digit numbers without 0 at unit place
33. Count the numbers between 999 and 10000
subject to the condition that there are (a) no = 4 4 1= 16.
restriction. (b) no digit is repeated. (c) at least
iii) By the rule of sum,
one of the digits is repeated.
No. of 3 digit even numbers = 20+16 = 36.
a) no restriction. _ _ _ _
b) repetition of digits are allowed
1000 place filled in 9 ways (except 0)
0 _ 0/5
Hundreds place filled in 10 ways (include 0)
100th place is fillef in 5 ways (excluding ’0’),
Tens place filled in 10 ways
10th place in 6 ways
Unit place filled in 10 ways
unit place in 2 ways (0/5)
Ways to form 3 digit number = 9101010
3 digit numbers = 5 6 2= 60.
= 9000 ways
35. Count the total number of ways of answering 6
b) no digit is repeated.
objective type questions, each question having 4
1000 place filled in 9 ways (except 0) choices.
Hundreds place filled in 9 ways (include 0) Each question can be answred in 4 ways
Tens place filled in 8 ways 6 question can be answred in 46 ways
Unit place filled in 7 ways 36. In how many ways 10 pigeons can be placed in 3
Ways to form 3 digit number = 9987 different pigeon holes ?
= 4356 ways Each pigeon holes can be occupied in 10 ways
c) at least one of the digits is repeated. 3pigeon holes can be occupied in 310 ways
Numbers if at least formed with formed with 37. Find the number of ways of distributing 12
- distinct prizes to 10 students?
1 digits is repeated out repetition repetition
Ways to form 3 digit number= 9000-4356 Each student can get prize in 12 ways
= 4464 ways 10 student can get prize in 1012 ways
34. How many three-digit numbers, which are
divisible by 5, can be formed using the digits 0,
1, 2, 3, 4, 5 if (a) repetition of digits are not
allowed? (b) repetition of digits are allowed?
a) repetition of digits are not allowed
0 _ 0/5
2. FACTORIALS 6!
3. If 6 then find the value of n.
n!
Factorial of a natural number n is the product of 6!
the first n natural numbers. It is denoted by n!. 6
n!
Read as “n factorial” or “factorial of n” 1.2.3.4.5.6.
6
n! = 1 2 3……..n. 1.2.3...n
n! 5!
For a positive integer n
n5
n! = n (n − 1) (n − 2)……3 2 1 4. If n! + (n − 1)! = 30, then find the value of n.
= n(n − 1)! for n > 1 n! (n - 1)! 30
n( n 1)! (n 1)! 30
= n(n − 1)(n − 2)! for n > 2
(n 1)(n-1)! 6 5
= n(n − 1)(n − 2)(n − 3)! for n > 3
equating (n - 1)! 6
(n + 1)!= (n + 1) n!
(n - 1)! 3! ,
sub. N = 0 in the equation as n4
1! = (0 + 1) 0! 5. Find the value of n if (n + 1)! = 20(n − 1)!
(n 1)! 20(n - 1)!
1!
0! = (n 1)n( n 1)! 20(n - 1)!
1
(n 1) n 5 4
0! = 1
equating n 4
1. Find the value of
6. What is unit digit of the sum 2!+3!+4!+.....+ 22!
i) 5! ii) 6! iii) 6! - 5! iv)4! 5! v) 3!-2!
7! 8! 12! From 5! onwards for all n! the unit digit is zero
vi) 3! 4! vii) viii) ix)
2! 5! 2! 9! 3! Contribution to unit digit is through
i) 5! 5 4 3 2 1 ii) 6! 720
2! + 3! + 4! = 2 + 6 + 24
120.
iii) 6! - 5! 6 5! - 5! v) 3!-2! 4 = 32 .
(6 - 1) 5! vi) 3!4! 3 2 4 3 2 Required unit digit is 2.
5 120 144 n!
600. 7. Evaluate when i) n = 7, r = 5
r!(n - r)!
7! 7 6 5 4 3 2! ii) n = 50, r = 47 iii) n = 6, r = 2 iv) n = 10, r = 3
iv) 4! 5! 4! 5 4! vii)
2! 2!
v) For any n with r = 3 vi) For any n with r = 2.
4! (1 5) 2520
6 4321 i) When n 7, r 5 ii) When n 50, r 47
144 n! 7! n! 50!
8! 8 7 6 5! 12! 12 11 10 9! r!(n - r)! 5!( 7 5)! r! (n - r)! 47! (50 47 )!
viii) ix)
5! 2! 5! 2! 9! 3! 9! 3 2 7 6 5! 50 49 48 47!
87 6 12 11 10
5!2! 47!3!
2 6 21 19600
168. 110 iii) n 6, r 2 iv) n 10, r 3
2. Let N denote the No. of days. If the value of N! 6! 6 5 4! 10! 10 9 8 7!
is equal to the total No. of hours in N days then
2!(6 2)! 2!4! 3!(10 3)! 3!7!
find the value of N?
15 120
N! 24 N.
v) When n n, r 3
For N 1, 2, 3, 4, N! 24 N.
n! n!
For N 5, N! 5!
r!(n - r)! 3!( n 3)!
4! 5
n(n 1)( n 3)!
24N.
3! (n 3)!
ForN 5, N! 5! N 24 N.
n(n 1)( n 2)( n 3)!
Hence N 5.
6
vi) When n n, r 2 PERMUTATIONS
n! n! 1. Meaning & Definition
r!(n - r)! 2! (n 2)!
A permutation is an ordering.
n( n 1)( n 3)!
A permutation of n distinct objects taken r at a
2!( n 2)!
time is formed by filling r positions, in a row
n( n 1)( n 2)!
with objects from given n distinct objects.
4
1 1 A If 3 objects have to be arranged in a row there
8. Find the value of A if
7! 8! 9! are 321 = 3! possible permutations.
A 1 1
If n objects have to be arranged in a line there
9! 7! 8!
A 1 1 are n (n − 1) (n − 2) ….3 2 1 = n!
possible arrangements or permutations.
9 8 7! 7! 8 7!
1 A 1 1 In terms of function on finite set S = {x1, x2,
1
7! 9 8 7! 8 ...xn}, a permutation can be defined as a
A 1 bijective mapping on the set S onto itself.
1
98 8
Number of permutation on set S is same as
A 81
1 1 n total number of bijective mappings on set S.
9. Find the value of A if
8! 9! 10! 2. No. of permutations of n distinct objects taken r
1 1 n at a time is n(n−1)(n −2)…(n−r+1)
8! 9! 10!
n 1 1 If n, r are positive integers and r ≤ n,
10 9 8! 8! 9 8!
1 n 1 1
1
8! 10 9 8! 9
n 1 n objects can be filled in 1st position.
1
10 9 9
Remaining n − 1 objects in second position
n 100
(2n)! n − 2 objects for third position.
10. Prove that 2n(1.3.5· ·· (2n - 1)).
n! Finally (n − (r − 1)) possible objects is placed
(2n)! 1.2.3.4··· (2n - 2).(2n - 1)2n
in rth position.
n! n!
(1.3.5.· ·· (2n - 1)) (2.4.6.· ·· (2n - 2).2n) By the rule of product
n! nPr = n (n − 1) (n − 2) ……..(n − r + 1).
(Grouping odd & even numbers separately)
3. Number of distinct permutations of r objects
(1.3.5.· ·· (2n - 1)) 2n (1.2.3.··· (n - 1).n)
n!
n! made from n distinct objects n Pr
(n - r)!
(taking out the 2’ s)
n
(1.3.5.· ·· (2n - 1)) 2n n! Pr n(n - 1)(n - 2 )· · ·(n - r 1)
n! multiply and divide by (n - r)(n-r-1)..2.1
2n (1.3.5.· ·· (2n - 1)) n(n-1)(n-2 )···(n-r 1)(n - r)(n-r-1)..2.1
Double Factorial of n: (n - r)(n-r-1)..2.1
n!
(n - r)!
4. No. of permutations of n different objects taken
r at a time where repetition is allowed, is nr.
If n, r are positive integers and r ≤ n,
We can fill the first position with n objects.
For 2nd position (we can use object used in No. of permutations of n different objects, taken all at a
first position), there are n objects time, when m specified objects are always together
On rth position can be filled with n objects. • Consider a string of m specified objects as a
single unit
By rule of product,
• Permute (n− m+1) objects in (n − m + 1)! ways.
No. of permutations of n different objects
taken r at a time when repetition allowed is • Permute m specified objects between
themselves in m! ways.
n n n ….n (r times) = nr.
• Finally, the answer is m! (n − m + 1)!
5. Properties of Permutation
n 7. No two things are together (Gap method)
i) Pn n Pn-1.
n n! No. of permutations of n different objects when no two
Pn-1.
(n - (n - 1))! of k given objects occur together and there are no
n! restrictions on remaining m = n − k objects
1! • Arrange m objects on which there is no
n! restriction in a row.
n!
These m objects permuted in mPm = m! ways.
(n - n)!
n
Pn • Count No. of gaps between every two of m
n objects on which there is no restriction
ii) Pr n n-1 Pr-1
including end positions.
(n - 1)!
n n-1 Pr-1 n
((n - 1) - (r - 1))! No.of gaps will be 1 more than m i.e (m+1).
n! In m + 1 gaps, permute k objects in m+1Pk ways.
(n - r)!
n Pr . • Then the required No. of ways are m! (m+1) Pk.
Continuing this process, 8. Rank of a word in the dictionary.
n n-1
Pr n Pr-1 It is the place at which given word comes when
n (n - 1) n- 2 Pr- 2 writing all strings formed by letters of given
n(n - 1)(n - 2 ) n- 3 Pr-3 ... (n - (r - 1)) n-r P0 word in dictionary or lexicographic order.
n (n - 1) (n - 2 ) ·· · (n - r 1).
9. Permutations of not all distinct objects
n
Pr n (n - 1) (n - 2 ) ·· · (n - r 1).
n n-1
i) No. of permutations of n objects, where p
iii) Pr Pr r n-1 Pr-1
objects are same kind and rest are all different
n-1 (n - 1)! (n - 1)!
Pr r n-1 Pr-1 r n!
((n - 1) - r)! (n - r)!
p!
(n - 1)! (n - 1)!
r
(n - 1 - r)! (n - r)! ii) No. of permutations of n objects, where p1
(n - 1)! (n - r) (n - 1)! objects are one kind, p2 objects are second
r
(n - 1 - r)! (n - r) (n - r)! kind, … pk, are of kth kind and rest of it are
(n - 1)! (n - r) (n - 1)! different kind is
r
(n - r)! (n - r)! n!
(n - 1)! ((n - r) r) p1! p2! ·· · pk!
(n - r)!
(n - 1)!n iii) Sum of all r-digit numbers that can be formed
using the given n non zero digits is
(n - r)!
n! (n−1)
P(r−1) (sum of digits) 111 …….1(r times)
(n - r)! iv) If 0 is one digit among given n digits, then we
n Pr get that sum of r-digits numbers that can be
6. Objects always together (String method) formed using given n digits (including 0) is
{(n−1)P(r−1)(sum of digits)111...1(r times)} −
{(n−2)P (sum digits) 111 ….1((r-1) times)}
(r−2)
(n-1 )
Solved Problems P3 1
n
1. Three friends A, B,C have to stand in line for a P4 10
photograph. In how many order can they stand? (n - 1)(n 2)(n 3) 1
n(n - 1)(n 2)(n 3) 10
n 10
10 7
6. If Pr Pr 2 find r.
10 7
Pr Pr 2
10! 7!
(10 - r)! ( 5 - r)!
Some possible arrangements (from left to right) 10 9 8 7! 7!
A, B, C: A, C, B: B, A, C (10 - r)(9 - r)(8 - r)( 7 - r)( 6 - r)( 5 - r)! ( 5 - r)!
B, C, A: C, B, A: C, A, B (10 - r)( 9 - r)(8 - r)( 7 - r)( 6 - r) 10 9 8
(10 - r)( 9 - r)( 8 - r)( 7 - r)( 6 - r) 6 5 4 3 2
Thus there are six possible ways
10 - r 6
2. From 7 letters A,B,C,D,E,F,G a 4 letter string is
r 4.
made. Express number of permutations of 4 10 6
letters chosen from 7 letters in generalised form 7. If Pr-1 2 Pr , find r .
10
7 choices for the 1st letter. Pr-1 2 6 Pr
10! 6!
6 choices for the second letter. 2
(10 r 1)! (6 r )!
Proceeding this way,
10 9 8 7 6! 6!
4 choices for the 4th letter. 2
(11 r )(10 - r)( 9 - r)(8 - r)( 7 - r)( 6 - r)! ( 6 - r)!
No. of permutations of 4 letters chosen from 7 (11 r )(10 - r)( 9 - r)( 8 - r)( 7 - r) 2 10 9 8 7
7654 (11 r )(10 - r)( 9 - r)( 8 - r)( 7 - r) 7 6 5 4 3
7 6 5 4 3 2 1 11 - r 7
321 r 4.
7! 8. Suppose 8 people enter an event in a swimming
3! meet. In how many ways could the gold, silver
7! and bronze prizes be awarded?
(7 - 4)! 8 people can be awarded 3 prizes in
8!
3. Evaluate i) 4 P4 ii) 5 P3 iii) 8 P4 iv) 6 P5 . 8P3 ways =
(8 3)!
i) 4 P4 4 3 2 1 iv) 6 P5 6 5 4 3 2
= 336
4! 6! 9. Three men have 4 coats, 5 waist coats and 6 caps
24. 720. In how many ways can they wear them?
5 8
ii) P3 5 4 3 iii) P4 8 7 6 5 4coatscan be worn by 3men in 4p3 ways
60. 1680. 5 waist coatscan be worn by 3men in 5p3 ways
(n 2 ) n 6capscan be worn by 3men in 6p3 ways
4. If P4 42 P2 , find n
(n 2 )
Total No. of ways = 4p3 5p3 6p3
P4 42 n P2 = 172800
(n 2 )
P4 10. Determine the No. of permutations of the letters
42
P2 n of the word SIMPLE if all are taken at a time?
(n 2 )(n 1)(n)(n - 1) 6 letters of SIMPLE permutated in 6P6 ways
42
n(n - 1) = 6!
. (n 2 )(n 1) 7 6 = 720
n 2 7 11. A test consists of 10 multiple choice questions.
In how many ways can the test be answered if
n 5
(n-1 ) n No. of permutations of n different objects taken r
5. If P3 : P4 1:10 , find n
at a time when repetition allowed is
n n n ….n (r times) = nr.
i) Each question has four choices? No. of permutations of different sequences
No. of permutations of 4 different objects 8!
containing six heads and two tails =
taken 10 is 410 6! 2!
ii) The first four questions have three choices = 28
and the remaining have five choices? 14. How many ‘letter strings’ together can be
formed with the letters of the word “VOWELS”
No. of permutations of 3 different objects
so that i) the strings begin with E
taken 4 is 34
ii) strings begin with E and end with W.
No. of permutations of 5 different objects
taken 6 is 56 Given strings contains 6 letters (V,O,W,E,L,S).
By product rule i) Since all strings must begin with E, remaining
5 letters can be arranged in 5P5 = 5! Ways
Total No. of permutation = 3456
iii)Question number n has n + 1 choices?
Question n can be answered in (n+1) ! ways
Total No. of strings with E as starting letter is
For n =10
5! = 120.
A question can be answered in (10+1)! ways
ii) Since all strings must begin with E, and end
12. A student appears in an objective test which with W, we need to fix E and W.
contain 5 multiple choice questions. Each
question has four choices out of which one Remaining 4 letters arranged in 4P4 = 4! ways.
correct answer.
i) What is the maximum No. of different
answers can the students give?
Total No. of strings with E as the starting
No. of permutations of 4 different objects letter and W as the final letter is
taken 5 is 45
4! = 24.
ii) How will the answer change if each question
may have more than one correct answers? 15. A No. of four different digits is formed with the
use of the digits 1,2,3,4 and 5 in all possible
No.of options if a question has 1 correct ways. Find the following
answer i.e 1,2,3,4 is 4 i) How many such numbers can be formed?
No.of options if a question has 2 correct ii) How many of these are even?
answer that is (1,2,(1,3),(1,4) (2,3)(2,4)(3,4) is 6
iii) How many of these are exactly divisible by 4
No.of options if a it has 3 correct answer is 4
i) No. of permutations taking 4 digits out of 5
No.of options if a it has 4 correct answer is 1 digits
Total ways of arrangring 1 question 15
Total ways of arrangring 5 question 155
5P4 = 5432
13. A coin is tossed 8 times,
= 120
i) How many different sequences of heads and
ii) For even number last digits must be 2 or 4
tails are possible?
No. of permutations of 2 sequence taken 8 at a
time is 28
ii) How many different sequences containing six Filled in 2P1 ways and remaining 3 places
heads and two tails are possible? filled from remaining 4 digits in 4P3 ways.
Required number of ways is
2P1 4P3 = 2 24 iv) No two vowels are together.
= 48. ii) There are 12 letters in INTERMEDIATE
iii) Since the number divisible by 4, thenlast two 6 vowels 3E, A,2I, 6 consonants D,M,N,R,2T
digit must be divisible by 4. Considering 6 vowels as one letter, 7 letters
Last two digits become 12,24,32,52 ( 4 ways). 7!
arranged in ways.
2!
6!
Each 3E, A,2I can be put in = ways.
3!2!
The remaining first two places filled from 7! 6!
No. of words = = 151200
remaining 3 digits in 3P2 ways. 2! 3!2!
Required No. of numbers of divisible by 4 iii) Total No. of strings formed by all 12 letters of
12!
= 4P1 3P2 INTERMEDIATE =
2!3!2!
= 46 =24.
= 19958400.
16. In how many ways 4 mathematics books, 3
No. of strings vowels Total No. of strings
physics books, 2 chemistry books and 1 biology -
book can be arranged on a shelf so that all never together strings vowels together
books of the same subjects are together. = 19958400 −151200
Total no of books = 10 = 19807200
Considering 4 math book as 1 unit, and so on 19. How many different strings can be formed
together using the letters of the word
Books of 4 subjects can be arranged in 4! ways “EQUATION” so that i) vowels always come
4 maths books can be arranged in 4! together? ii) the vowels never come together?
3 physics books can be arranged in 3! i) There are 8 letters in the word “EQUATION”
5 vowels E,U,A,I,O, 3 consonants Q,T,N
2 chemistry books can be arranged in 2!
1 bioology book can be arranged in 1! Considering 5 vowels as one letter, 4 letters
arranged in 4P4 = 4! ways.
Total no. of arrangements
Each E,U,A,I,O can be put in 5P5 = 5! ways.
4!4!3!2!1! = 6912
17. In how many ways can the letters of the word No. of words = 4! 5!
SUCCESS be arranged so that all Ss are = 24 120 = 2880.
together?
ii) Total No. of strings formed by all 8 letters of
There are 7 letters in which there are 3S,2C, 1E,1U EQUATION = 8P8
Considering 3SSS as a letter there are 5 letters
= 8! = 40320.
5!
Which can be arranged in
2! No. of strings vowels Total No. of strings
-
3SSS can be arranged in 1 way never together strings vowels together
= 40320 − 2880
5!
Total ways = 1 = 60
2! = 37440.
18. How many strings are there using the letters of 20. There are 15 candidates for an examination. 7
the word INTERMEDIATE, if candidates are appearing for mathematics
examination remaining 8 are appearing for
i) The vowels and consonants are alternative
different subjects. In how many ways can they
ii) All the vowels are together be seated in a row so that no two mathematics
candidates are together?
iii) Vowels are never together
8-non-mathematics candidates in 8P8 = 8! ways.
Each of these arrangements create 9 gaps. 9!
8! 9 P7 8!
2!
7 mathematics candidates can be placed in 9 gaps
8! 9!
in 9P7 ways.
2!
By the rule of product, the required No. of 23. 4 boys and 4 girls form a line with the boys and
arrangements is girls alternating. Find the number of ways of
making this line.
4 boys can be arranged in 4P4 = 4! ways.
9!
8! 9 P7 8! By keeping boys as first 4 gaps are created.
2!
8! 9! In 4 gaps, 4 girls arranged in 4P4 = 4! ways.
2!
Keeping boys as first, total No. of arrangements
21. In how many ways 5 boys and 4 girls can be
seated in a row so that no two girls are together. = 4!4!.
5 boys can be seated in the row in 5P5 = 5! ways. Keeping girls first, total No. of arrangements are
In each of these arrangements 6 gaps are created. = 4!4!.
Since no two girls are to sit together, we may By the rule of sum, keeping either a boy or a girl
arrange 4 girls in this 6 gaps. first, the total No. of arrangements are
This can be done in 6P4 ways. (4! 4!) + (4! 4!) = 2(4!)2
Hence, the total No. of seating arrangements = 1152.
5! 6P4 = 120 360 24. A van has 8 seats. It has two seats in the front
with two rows of three seats behind.. How
= 43200. many ways can seven members, F,M, S1, S2, S3,
22. 8 women and 6 men are standing in a line. D1,D2 sit in the van if i) There are no
i) How many arrangements are possible if any restriction? ii) Either F or M drives the van? iii)
D1,D2 sits next to window, F is driving?
individual can stand in any position?
i) As there 8 seats to be occupied out of which one
Total individuals = 8+6 = 14
seat is for driver. Since there are no restrictions
No of arrangement 14P14= 14!
No. of ways of occupying driver seat 7P1 = 7 ways
ii) In how many arrangements will all 6 men be
No. of ways of occupying the remaining 7 seats
standing next to one another?
by the remaining 6 people is 7P6 = 5040.
Considering 6 men as one unit, 9 individuals
Hence the total No. of ways the family can be
arranged in 9P9 = 9! ways.
seated in the car is 7 5040 = 35280.
Each men can be put in 6P6 = 6! ways.
ii) As the driver seat can be occupied by only F or M,
No.Men = 9! 6! there are only two ways it can be occupied.
iii) In how many arrangements will no two men
Hence the total No. of ways the family can be
be standing next to one another?
seated in the car is 2 5040 = 10080.
6 men arranged in in 6P6 =6! ways.
iii) only 5 window seats available for D1&D2
Each of these arrangements create 7 gaps.
No. of ways of seated near windows by the two
6 men can be placed in 7 gaps in 9P7 ways. family members is 5P2 = 20.
By the rule of product, the required No. of Driver seat is occupied by F, remaining 4 people
arrangements is can be seated in the available 5 seats in 5P4 = 120.
Hence the total No. of ways the family can be
seated in the car is 20 1 120 = 2400.
25. If the letters of the word TABLE are permuted = 360 + 19= 379.
in all possible ways and the words thus formed
iI) rank of the word DANGER
are arranged in dictionary order find the ranks
of the words (i) TABLE, (ii) BLEAT A−−−−− = 5! = 120ways
Dictionary order of the letters is A, B, E, L, T. D AE−−− = 3! = 6 ways
If we fill the first place with A, remaining 4 letters DAG −−− = 3! = 6 ways
(B, E, L, T) can be arranged in 4! ways.
DANE−−= 2!=2 way
i) The rank of the word TABLE
DANGER= 0!=1 way
A−−−− = 4! = 24 ways
rank of the word DANGER =135
B −−−− = 4! = 24 ways 27. Find the No. of strings that can be made using
E −−−− = 4! = 24 ways all letters of the word THING. If these words
are written as in a dictionary, what will be the
L−−−− = 4! = 24 ways
85th string?
TABEL = 1 way i) There are 5 lettters in the word THING
TABLE = 1 way 5 letters can be arranged in 5! Ways
Rank of TABLE = 4 4! + 1 + 1 = 98. No of string = 120
ii) The rank of the word BLEAT ii) Dictionary order of the letters is G, H, I, N, T.
A−−−− = 4! = 24 ways If we fill the first place with G, remaining 4 letters
BA−−− = 3! = 6 ways (H, I, N, T) can be arranged in 4! ways.
BE −−− = 3! = 6 ways i) Rank of the word THING
BLA−− = 2! = 2 ways G−−−− = 4! = 24 ways
H −−−− = 4! = 24 ways
BLEAT = 1 way
I −−−− = 4! = 24 ways
Rank of the word BLEAT =24 + 6 + 6 + 2 + 1 = 39.
NG−−− = 3! = 6 ways
26. If the letters of the word GARDEN are
NH−−− = 3! = 6ways
permuted in all possible ways and the strings
thus formed are arranged in the dictionary NIGHT = 1! = 1 ways
order, then find the ranks of the words Rank of the word NIGHT
Dictionary order of the letters is A, D, E, G, N, R = 3 4! + 6 + 6+1 = 85.
If we fill the first place with A, remaining 5 letters 28. If the letters of the word FUNNY are permuted
(D, E, G, N, R) can be arranged in 5! ways. in all possible ways and the strings thus
formed are arranged in the dictionary order,
i) The rank of the word GARDEN find the rank of the word FUNNY.
A−−−−− = 5! = 120ways Dictionary order of the letters is F,N,N,U,Y
D −−−−− = 5! = 120 ways i) Rank of the word FUNNY
E −−−−− = 5! = 120 ways FN−−− = 3! = 6 ways
GAD−−−= 3!=6 way FUNNY = 1! = 1 ways
GAE−−−= 3!=6 way Rank of the word FUNNY = 7 ways
GAN−−−= 3!=6 way = 4 4! + 1 + 1 = 98.
GARDEN = 0!= 1 way 29. Find the No. of ways of arranging the letters of
Rank of the word GARDEN the word BANANA.
This word has 6 letters
there are 3 A’S, 2 N’s and one B. There are 7 places in that there are 3 even places
6! we have A,E,I as vowels numbers.
No. of ways of arrangements = 60.
3! 2! No. of ways of permuting A,E,I in 3 even places
30. Find the No. of ways of arranging the letters of in = 3! = 6 ways.
word RAMANUJAN so that relative positions
Remaining R, T, C, L can be permuted in the
of vowels and consonants are not changed.
remaining 4 places in 4 != 24 ways.
There are 4 vowels (A,A,U,A)in that 3 A’s, 1 U
Required No. of ways is 24 6 = 144
5 consonants (R,M,N,J,N) in that two N’s and rest
34. How many paths are there from start to end on
are distinct.
a 64 grid How Many Paths?
4!
4 vowels (A,A,A,U) arranged in = 4 ways. Such path comprises 6 horizontal unit lengths
3!
and 4 vertical unit lengths.
5!
5consonants (R,M,N,J,N) arranged in = 60 ways
2! This each path consists of 10 unit lengths where 6
4! 5! are of one kind (horizontal) and 4 are of another
No. of required arrangements are
3!
2! kind (vertical).
= 4 60 = 240. Thus the total No. of paths is 10! 4! 6! = 210.
31. Three twins pose for a photograph standing in 35. If the different permutations of all letters of
a line. How many arrangements are there (i). the word BHASKARA are listed as in a
when there are no restrictions. (ii). when each dictionary, how many strings are there in this
person is standing next to his or her twin? list before the first word starting with B?
(i) 6 persons without any restriction arranged in Required No. of strings is total No. of strings
6P6 = 6! = 720 ways. starting with A and using letters A,A,B,H,K,R,S is
ii) 3 twins as T1, T2, T3. 7!2!= 2520.
Each twin is considered as a single unit and these 36. If the letters of the word IITJEE are permuted
three can be permuted in 3! ways. in all possible ways and the strings thus
formed are arranged in the lexicographic order,
Again each twin can be permuted between find the rank of the word IITJEE
themselves in 2! ways.
The lexicographic order of the letters of given
Hence, the total No. of arrangements is word is E, E, I, I, J, T. In the lexicographic order,
3! 2! 2! 2! = 48 ways. the strings which begin with E come first. If we
fill the first place with E, remaining 5 letters (E, I,
32. How many numbers can be formed using the
digits 1,2,3,4,2,1 such that, even digits occupies I, J, T) can be arranged in 5!2!ways. On
even place? proceeding like this we get,
There are 6 places in that there are 3 even places E −−−−− =5!2!= 60 ways
we have 2,4,2 as even numbers. IIE −−− = 3! = 6 ways
The No. of ways of permuting 2,4,2 in the 3 even IIJ −−− =3!2!= 3 ways
places in 3! 2! = 3 ways.
IITE −− = 2! = 2 ways
The remaining numbers 1,3,1 can be permuted in
IITJEE = 1 way
the remaining 3 places in 3! 2! = 3 ways.
Rank of the word IITJEE is 60 + 6 + 3 + 2 + 1 = 72
Hence, the required No. of numbers is 3 3 = 9.
37. Find the sum of all 4-digit numbers that can be
33. How many strings can be formed from the
formed using the digits 1, 2, 4, 6, 8.
letters of the word ARTICLE, so that vowels
occupy the even places? i) No. of 4-digit numbers using given 5 digits
= 5P4 = 120.
ii) Sum of digits in unit place =586608-14652
By filling 1 in unit place, remaining 3 places =571956
can be filled with remaining 4 digits in
40. Find the distinct permutations of the letters of
= 4P3 = 24 ways. the word MISSISSIPPI?
iii) Each of 2, 4, 6, 8 appear 24 times in units place There are 11 letters in which there are 4 S, 2P, 4I
iv) Sum of all unit digits of all 120 numbers. 11!
Total No. of Permutation =
=(4P3 1)+(4P3 2)+(4P3 4) + (4P3 6)+(4P3 8) 4!2!4!
= 4P3 (1 + 2 + 4 + 6 + 8) = 34650
= 4P3 (sum of the digits) 41. How many ways can the product a2b3c4 be
= 4P3 21. expressed without exponents?
v) Sum of the digits in 10th place as a2b3c4 = aa bbb cccc , Total = 9
= 4P3 21. product a2b3c4 be expressed without exponents as
Since it is in 10th place, its value is 9! 9.8.7.6.5
= 4P3 21 10. 2!3!4! 3.2.2
1260
vi) Sum of digits in 100th place
42. Each of the digits 1, 1, 2, 3, 3 and 4 is written on
=4P3 21100 a separate card. The six cards are then laid out
vii) Sum of digits in 1000thplace in a row to form a 6-digit number.
= 4P3 211000 i) How many distinct 6digit numbers are there?
viii)Sum of all 4 digit numbers by digits 1, 2,4,6, 8 There are 6 digits in which 2 are 1’s, 2are 3’s
(4P321)+(4P32110)+(4P321100)+(4P3211000)
Distinct 6 digit numbers can be formed in
= 4P3 (21 1111)
6!
= 24 21 1111 180 ways
2!2!
= 559944.
ii) How many of these 6-digit numbers are even?
38. Find the sum of all 4-digit numbers that can be
_ _ _ _ _ 2/4 (2ways)
formed using digits 1, 2, 3, 4, and 5 repetitions
not allowed? Unit place can be filled in 2 ways
Sum of all r-digit numbers that can be formed Remaining places can be filled with any of 5
using the given n non zero digits is digits of which two are 1’s and two are 3’s
(n−1)P(r−1) (sum of digits) 111 …….1(r times) 5P1
2 P1
Sum of all 4-digit numbers that can be formed 2!2!
using the given 5 non zero digits is iii) How many of these 6-digit numbers are
= (5−1) P(4−1) (1+2+3+4+5) 1111 divisible by 4?
=2415 1111 Numbers divisible by 4 from 1,1,3,3,4,2 are
= 399960 12, 32, 24
39. Find the sum of all 4-digit numbers that can be 4!
_ _ _ _ 1 2 No.of digits formed =
formed using digits 0, 2, 5, 7, 8 without 2!
repetition? 4!
_ _ _ _ 3 2 No.of digits formed =
2!
Sum of r-digits numbers that can be formed using
given n digits including 0 4!
_ _ _ _ 2 4 No.of digits formed =
2!2!
={(n−1)P(r−1)(sum of digits)111...1(r times)} −
{(n−2)P
4!
(r−2) (sum digits) 111 ….1((r-1) times)}
6 digit nos. divisible by4 = [1+1+ ½]
2!
=[(5−1)P(4−1)221111]− [(5−2)P(4−2) 22 111] = 30
=[24221111]− [622 111]
4.5. COMBINATIONS 10! 15!
(i) 10 C 3 (ii) 15 C 13
No. of combinations of n different objects taken r 7! 3! 2! 13!
at a time = nCr. 10 9 8 7! 15 14 13!
1. Relation bbetween permutation & combination 7! 3 2 1 2 1 13!
nPr =n Cr r!.
120 105
100 99! 50!
(iii) 100 C 99 (iv) 50 C 50
99! 50! 0!
100 1
2. Find the value of 5C2 and 7C3 using the property 5
n n
C r n-1C r-1.
r
5 7
5
C 2 5 -1C 2 -1 7
C 3 6C 2
2 3
5 4 7 6
5C 2 1
2 1 3 2
10 35
3. If nC4 = 495, What is n?.
n
C 4 495.
n (n - 1) (n - 2 ) (n - 3 )
495
4 3 2 1
n (n - 1) (n - 2 ) (n - 3) 495 4 3 2 1
n (n-1) (n-2 ) (n-3 ) 12 11 10 9
n 12
2. No. of combinations of n distinct objects taken r 4. If nC12 =n C9 find 21Cn.
at a time If nC x n C y then x y n
n
n Pr 12 9 n
Cr
r! n 21
n!
Thus, nC r , 0 r n 21
C 21 1 , since n C n 1
r!(n - r)!
3. Properties of Combinations 5. If 15C2r−1 =15 C2r+4, find r.
n! If nC x n C y , then x y n
i) nC 0 1.
0!(n - 0 )! 2r 1 2r 4 15
n! n! r 3
ii) n C n
n!(n - n)! n!0! 6. If nPr = 11880 and nCr = 495, Find n and r.
1 n Pr
r!
n(n - 1)(n - 2 ) · · · (n - (r - 1)) nCr
iii) nC r 11880
r! r!
n(n - 1)(n - 2 ) · · · (n - r 1) 495
r! 24
r!
n! r! 4!, r 4.
iv) nC n-r
(n - r)!(n - (n - r))! n
C 4 495
n!
n Cr n (n - 1) (n - 2 ) (n - 3)
495
(n - r)!r
4 3 2 1
v)If nC x n C y , either x y or x y n. n (n - 1) (n - 2 ) (n - 3) 495 4 3 2 1
n
C y n C n-y n (n-1) (n-2 ) (n-3) 12 11 10 9
n
C x n C n-y n 12
7. If nPr = 720, and nCr = 120, find n, r.
x y or x n-y
n Pr 720
x y or x y n r! r!
nCr 120
SOLVED PROBLEMS r! 6
1. Evaluate the following: r! 3!, r 3.
(i) 10 C 3 (ii) 15 C 13 (iii) 100 C 99 (iv) 50 C 50 .
n = 26C4+26 C3 +27 C3 +28 C3
C 3 120.
n (n - 1) (n - 2 ) = 27C4 +27 C3 +28 C3
120
3 2 1 = 28C4 +28 C3
n (n - 1) (n - 2 ) 120 3 2 1 = 29C4
n (n-1) (n-2 ) 10 9 8 4
35 (39- r)
12. Prove that C5 C 4 40 C 5
n 10 r 0
(n 2 ) (n-1)
8. If C 7 : P4 13 : 24 find n.
= 35C4+39 C3 +38 C3 +37 C3 +36 C3 +35 C3
(n 2 )
C7 13 = 35C4+35 C3+36 C3 +37 C3 +38 C3 +39 C3
(n-1)
P4 24
= 40C5
(n 2 )! (n - 5 )! 13
13. Prove that 10C2 + 2 10 C3 +10 C4 =12 C4
(n - 5 )! 7! (n - 1)! 24
= 10C2 + (10C3 +10 C3) +10 C4
(n 2 )(n 1)n(n - 1)! 13
= (10C2 +10 C3) + (10C3 +10 C4)
(n - 1)! 7! 24
13 = 11C3 +11 C4
(n 2 )(n 1)(n) 7!
24 = 12C4
(n 2 )(n 1)(n) 13 14 15 14. Prove that 15C3 + 2 15 C4 +15 C5 =17 C5.
n 2 15 =15C3 + 15 C4 +15 C4 +15 C5
n 13 =15C3 + 15 C4 +15 C4 +15 C5
9. If (n+1)C8 :(n−3) P4 = 57 : 16, find the value of n. = 16C4 +16 C5
(n 1 )
C8 57
(n- 3 )
=17 C5.
P4 16
n
( n 1)! (n 7)! 57 15. Prove that n C r (n - 1) C (r - 1)
r
8!(n 7 )! ( n 3)! 16 n (n-1) n (n 1)!
C (r-1 )
(n 1).n.(n 1)(n 2)(n 3)! 57 r r r(r - 1)! ((n - 1) - (r - 1))!
8!(n 3)! 16 n(n - 1)!
57 r(r - 1)!(n - r)!
( n 1).n.( n 1)( n 2) 8!
16 n Cr
( n 1).n.( n 1)( n 2) 21 20 19 18 2n 1 3 ·· · (2n - 1)
n 20 16. Prove that 2n C n
n!
n n n1
10. Prove that C r C r - 1 Cr . 2n ( 2n)!
Cn
n n! n! n! n!
C r n C r-1
r!.(n - r)! (r - 1)!.(n - (r - 1))! 1.2.3.4 · · · ( 2n - 2 ).( 2n - 1)2n
n! n! n! n!
r! (n - r)! (r - 1)! (n - r 1) (1.3.5. · · · ( 2n - 1)) ( 2.4.6. · · · ( 2n - 2 ).2n)
n! n!
n! n!
r(r -1)!(n - r)! (r - 1)!(n-r)!(n-r 1)
(1.3.5..( 2n - 1)) 2 n (1.2.3.....(n - 1).n)
n! 1 1
n! n!
(r - 1)! (n - r)! r (n - r 1)
(1.3.5. · · · ( 2n - 1)) 2 n n!
n! (n - r 1 r)
n! n!
(r - 1)! (n - r)! r(n - r 1)
n! (n 1) 2 n 1 3 ·· · ( 2n - 1)
(r - 1)! (n - r)! r(n - r 1) n!
(n 1)! 17. Prove that if 1 ≤ r ≤n then n (n−1)Cr−1=(n−r+ 1)nCr−1
r! (n 1 - r)! n(n 1)!
n (n-1) C (r-1 )
n n
C r C r-1 n1
Cr. (n -r)!(r - 1)!
4 (n r 1)n!
24 (28 - r)
C 3 29 C 4
11. Prove that C4
r 0
( n r 1)( n r )!(r - 1)!
= 24C4+28 C3 +27 C3 +26 C3 +25 C3 +24 C3 (n r 1)n!
( n r 1)!(r - 1)!
= 24C4+24 C3 +25 C3 +26 C3 +27 C3 +28 C3
= 25C4 +25 C3 +26 C3 +27 C3 +28 C3 ( nr 1) C r 1
18. Find the total No. of subsets of a set with There are 8 girls and 7 boys
nC0 +n C1 +n C2 + …..+n Cn = 2n
No. of ways of selecting 3 girls and 3 others
i) 4 elements = 24
ii) 5 elements = 25 = 8C3 7 C3
iii) n elements. = 2n = 56 35
19. There are four persons A,B,C and D and we have = 1960.
to select three of them to be a part of a
25. Find No. of ways of forming a committee of 5
committee. In how many ways can we make this
members out of 7 Indians and 5 Americans, so
selection?
that always Indians will be the majority in the
Here the order of selection is immaterial.
committee.
A,B,C is same as B, A,C or C, A,B as long as the
same three persons are selected. i) 4 Indians from 7 1Americans from 5
Possible distinct choices or selections are = 7C4 5C1 ways = 175
A,B,C; A,B,D; A,C,D and B,C,D. ii) 3 Indians from 7 2Americans from 5
There are 4 ways of selecting 3 people out of 4. = 7C3 5C2 ways = 350
4 C3 = 4 ii) 5 Indians from 7 0 Americans from 5
20. A salad at a certain restaurant consists of 4 of the = 7C5 5C0 ways = 21
following fruits: apple, banana, guava,
Total Selection = 546
pomegranate, grapes, papaya and pineapple.
Find the total possible No. of fruit salads. 26. How many ways can a team of 3 boys,2 girls and
1 transgender be selected from 5 boys, 4 girls
To select 4 fruits from 7fruits for fruit salad.
and 2 transgenders?
Total No. of possible ways of making fruit salad
3 boys can be selected from 5 in 5C3 ways
7 C4 = 7 C3
2 girls can be selected from 4 in 4C2 ways
= 35
1 trangender can be selected from 2 in 2C1 ways
21. Kabaddi coach has 14 players ready to play.
Total ways = 5C3 4C2 2C1
How many different teams of 7 players could
the coach put on the court? = 120 ways
14C7=3432 27. In rating 20 brands of cars, a car magazine picks
a first, second, third, fourth and fifth best brand
22. There are 15 persons in a party and if each 2 of
and then 7 more as acceptable. In how many
them shakes hands with each other, how many
ways can it be done?
handshakes happen in the party?
15C2= 15C13
Picking of 5 brands for first, second, third, fourth
and fifth best brand from 20 brands in 20P5 ways.
= 105
Select 7 acceptable in 15C7 ways from remaining 15
23. In a parking lot one hundred , one year old cars,
By the rule of product it can be done in
are parked. Out of them five are to be chosen at
random for to check its pollution devices. How = 20 P5 15 C7 ways.
many different set of five cars can be chosen? 28. From a class of 25 students, 10 students are to be
5 cars from 100 cars can be selected in 100C5 ways chosen for an excursion party. There are 4
students who decide that either all of them will
24. A Mathematics club has 15 members. In that 8
join or none of them will join. In how many
are girls. 6 members are selected for competition
ways can the excursion party be chosen?
and half of them should be girls. How many
ways of these selections are possible? i) All 4 students will got to the excursion party
To select 6 students out of 21 students can be = 24 + 24 + 4 = 52.
21!
done in 21 C 6 ways. 31. A box contains two white balls, three black balls
6! 15!
and four red balls. In how many ways can three
ii) All 4 students will not go to excursion party balls be drawn from the box, if at least one
To select 10 students out of 21 students can be black ball is to be included in the draw?
21! B = Black ball, W = White ball, R = Red Ball
done in 21 C 10
10! 11!
Possible Selections :
Hence, the total No. of ways
1B, 1W, 1R = 3c12C14C1 = 24
21 21! 21!
C 6 21 C 10 1B, 0W, 2R = 3c12C04C2 = 18
6! 15! 10! 11!
29. A box of one dozen apple contains a rotten 1B, 2W, 0R = 3c12C24C0 = 12
apple. If we are choosing 3 apples 2B, 1W, 0R = 3c22C14C0 = 6
simultaneously, in how many ways, one can get
only good apples. 2B, 0W, 1R = 3c22C04C1 = 3
No. of ways of selecting 3 apples from 12 apples 3B, 0W, 0R = 3c22C04C0 = 1
= 12C3 Total Selection = 24
= 220. 32. Out of 7 consonants and 4 vowels, how many
strings of 3 consonants 2 vowels can be formed?
No. of ways of getting a rotten apple
No. of ways of selecting (3 consonants out of 7)
when selecting 3 apples from 12 apples =
and (2 vowels out of 4) = 7C3 4 C2
Selecting 1 rotten apple and remaining
2 apples can be selected from 11 apples Each string contains 5 letters.
= 11C2 = 55. No.of ways of arranging 5 letters among themself
Total No. of ways of getting only good apples is = 5! = 120
=12C3 − 11C2 Required No. of ways = 7C3 4 C2 5!
= 220 − 55 = 165 = 35 6 120
30. An exam paper contains 8 questions, 4 in Part A = 25200
and 4 in Part B. Examiners are required to 33. Find No. of strings of 5 letters that can be
answer 5 questions. In how many ways can this formed with letters of the word PROPOSITION.
be done if
There are 11 letters with respect to No. of
i) There are no restrictions. repetitions of letters there are
There are 8 questions in both Part A and Part B. 4 distinct letters (R, S, T, N)
No. of ways of attempting 5 from 8 questions is 2 sets of two alike letters (PP,II)
8 C5 = 8C3 = 56. 1 set of three alike letters (OOO).
ii) At least 2 questions from Part A to be answered
Part A Part B Number of selections
2 3 4C2 4 C3
3 2 4C3 4 C2
4 1 4C4 4 C1
Required No. of ways of answering
=4 C2 4 C3 +4 C3 4 C2 +4 C4 4 C1
Total No. of strings = 455 ways
= 2520 + 20 + 300 + 450 + 3600 38. How many triangles can be formed by 15 points,
in which 7 of them lie on one line and the
= 6890.
remaining 8 on another parallel line?
34. Find No. of strings of 4 letters that can be
formed with the letters of EXAMINATION?. No. of triangls obtainedby joining 15 points taken
3 at a time 15C3 ways = 455
EXAMINATION = 11 letters
No. of triangls obtainedby joining 7 points taken
5 distinct letters E, X,M,T,O
3 at a time 7C3 ways = 35
3 set of 2 alike letters NN,II,AA
No. of triangls obtainedby joining 8 points taken
Letter option Selection Arrangements 3 at a time 8C3 ways = 56
4 distinct 8 C4 8C3 4! = 1680 Collinear points can’t form a triangle
E,X,M,T,O,N,I,A Rqd No.of triangles = 455 – 35 – 56
2 sets of 2 alike 3 C2 4! = 364
3C 2 18
2!2!
NN,II,AA 39. There are 11 points in a plane. No three of these
lies in the same straight line except 4 points,
1set of 2 alike
which are collinear. Find,
(NN,II,AA) 3 C1 7C2 4!
3C 1 7C 2 i) No. of straight lines that can be obtained
2!
2 distinct (E, from the pairs of these points?
756
X,M,T,O) and
No. of straight lines obtainedby joining 11
remaining 2 in 2
points taken 2 at a time
alikes
11C2 ways = 475
Total No. of strings = 1680+18+756
No. of straight lines obtainedby joining 4
35. If a set of m parallel lines intersect another set of
points 4C2 ways = 6
n parallel lines (not parallel to first set), find No.
of parallelograms formed in this lattice Collinear points when joined pairwise give
structure. only one line
Selecting 2 lines from first set of m lines and 2 ii) No. of triangles that can be formed for which
lines from second set of n lines, one parallelogram the points are their vertices?
is formed. No. of triangls obtainedby joining 11 points
No. of parallelograms formed = mC2 n C2. taken 3 at a time
11C3 ways = 165
36. How many chords can be drawn through 20
points on a circle? No. of triangls obtainedby joining 4 points
4C3 ways = 4
A chord can be drawn using 2 points
No.of chord with 2 points from 20 points Collinear points can’t form a triangle
20C2 = 20C18 Rqd No.of triangles = 165 – 4
= 190 = 161
37. How many triangles can be formed by joining 40. How many diagonals are there in a polygon
15 points on the plane, in which no line joining with n sides?
any three points? A polygon of n sides has n vertices.
Using 3 points can be drawn 1 traingle By joining any two vertices of a polygon, either a
Using 15 points can be drawn in 15C3 traingle side or a diagonal of the polygon is obtained
No.of line segments obtained by joining vertices Rqd5 books selected from Remaining 10 in
of a n sided polygon taken two at a time is 10C5 ways = 252
n n(n - 1)
C2 45. There are 5 teachers and 20 students. Out of
2
them a committee of 2 teachers and 3 students is
Out of these lines, there are n sides of polygon. to be formed.
n(n - 1) i) Find No. of ways in which this can be done.
No. of diagonals of polygon n
2
n(n - 3 ) 2 teachers from 5 and 3 students from 20
2 = 5C2 20C3 ways = 11400
41. A polygon has 90 diagonals. Find No. of sides? ii) a particular teacher is included?
Proceeding as above, If 1 teacher is included other 1 teacher selected
No. of diagonals of polygon 90 from remaining 4 teachers in 4C1 ways
n(n - 3 ) 3 students selected from 20 in 20C3 ways
90
2
n 2 3n 180 0 Total selection = 4C1 20C3 ways = 4560
(n 15)( n 12) 0 iii) a particular student is excluded?
n 15 , since n 12 If 1 student excluded rqd 3 students selected
42. A trust has 25 members. from remaining 19 students in 19C3 ways
i) How many ways 3 officers can be selected? 2 teachers selected from 5 in 5C2 ways
3 officers out of 25 members can be selected in Total selection = 5C2 19C3 ways = 9690
25C3 ways 46. In an examination a student has to answer 5
questions, out of 9 questions in which 2 are
ii) In how many ways can a President, Vice
compulsory. In how many ways a student can
President and a Secretary be selected?
answer the questions?
3 persons out of 25 members can be selected in
As 2 questions out 9 are compulsory, the student
25C3 ways can answer other 3 questions from remaining 7
43. How many ways a committee of six persons qusestions in 7C3 ways = 35
from 10 persons can be chosen along with a 47. Determine No. of 5 card combinations out of a
chair person and a secretary? deck of 52 cards if there is exactly three aces in
chair person and secretary chosen in 10P2 ways each combination.
Remaining 4 persons chosen from 8 in 8C4 ways 3 aces out 4 from 52 cards selected in 4C3 ways
Total no.of ways = 10P2 8C4 = 3150 Other 2 cards selected from remaining 48 cards in
48C2 ways
44. How many different selections of 5 books can
be made from 12 different books if, Total Selection = 4C3 48C2 ways = 4512
Total books = 12 , Rqd books = 5 48. A committee of 7 peoples has to be formed from
i) Two particular books are always selected? 8 men and 4 women. In how many ways can this
be done when the committee consists of
2 books always selected
i) exactly 3 women?
Remaining 3 books selected from 10 in
10C3
3 women from 4 and 4 men from 8
ways = 120
= 4C3 8C4
ii) Two particular books are never selected?
= 280
2 books never selected
ii) at least 3 women? MATHEMATICAL INDUCTION
a) 3 women and 4 men = 4C3 8C4 Step 1: Verify that the statement is true for n = 1,
b) 4 women and 3 men = 4C4 8C3 That is, verify that P(1) is true.
Total ways = 4C3 8C4 +4C4 8C3 Step 2: Assume it is true for n = k,
where k is a positive integer.
= 336
Step 3: Verify that the statement is true for n = k + 1
iii) at most 3 women?
To prove that P(k + 1) is true whenever P(k) is true.
a) 3 women and 4 men = 4C3 8C4
Step 4: P(n) is true for all positive integers n
b) 2 women and 5 men = 4C2 8C5
If steps 1 and 2 have been established
c) 1 women and 6 men = 4C1 8C6 By principle of mathematical induction, prove that
d) 0 women and 7 men = 4C0 8C7 n(n 1)
1. 1 2 3 · ·· n
2
Total ways = 4C3 8C4 + 4C2 8C5+
n(n 1)
4C1 8C6+ 4C0 8C7
P(n) 1 2 3 ··· n .
2
i) To prove : P(n)is true for n 1
= 735
1(1 1)
49. 7 relatives of a man comprises 4 ladies and 3 P(1) 1, P(1) istrue.
1
gentlemen, his wife also has 7 relatives; 3 of ii) Assume : P(n) is true for n k.
them are ladies and 4 gentlemen. In how many k(k 1)
P(k) 1 2 3 ... k
ways can they invite a dinner party of 3 ladies 2
and 3 gentlemen so that there are 3 of man’s iii) To prove : P(n)is true forP(k 1)
relative and 3 of the wife’ s relatives? P(k 1) 1 2 3 .. k (k 1)
k(k 1)
Man Relatives Wife relatives Arrangement (k 1).
2
L G L G k(k 1) 2(k 1)
P(k 1)
3 0 0 3 4C3 4C3= 16 2
(k 1)(k 2)
2 1 1 2 (4C2 3C1)2=324
2
1 2 2 1 (4C1 3C2)2=144 P(k 1) is true.
n(n 1)
0 3 3 0 3C3 3C3= 1 iv) 1 2 3 ... n
2
2. Sum of first n positive odd numbers is n2
Let P(n) : 1 3 5 ··· (2n - 1) n 2
Total No. of ways of Inviting = 16+324+144+1
To prove : P(n ) is true for n 1
= 485 P(1) 1 2 Hence P(n) true for n 1
Assume : P(n ) is true for n k
P(k) 1 3 5 ...(2k - 1)
k2
To prove : P(n ) is true for n k 1
P(k 1) 1 3 5 ...(2(k 1) - 1)
1 3 5 7 4 ... (2k - 15) 2k 1
P(k) 2k 1
k 2 2k 1
(k 1) 2
This implies, P(k 1) is true
3. Sum of first n non zero even numbers n 2 n
P(n) : 2 4 6 ...... 2n n 2 n To Prove : P(n ) is true for n k 1
To prove : P(n) is true for n 1 P( k 1) 13 2 3 33 ..... k 3 ( k 1)3
P(1) : 2(1) 1 2 1 k(k 1) 2
( k 1) 3
2 2, Hence P(n) is true for n 1 2
Assume : P(n) is true for n k k 2 4k 4
( k 1) 2
P(n) : 2 4 6 ...... 2 k k 2 k 4
To prove : P(n) is true for n k 1 ( k 1) 2 ( k 2) 2
P(k 1) 2 4 6 ...... 2 k 2( k 1) 4
n( 2n 1)( 2n 1)
k 2 k 2k 2 6. 12 3 2 5 2 ..... ( 2n 1) 2
3
k( k 1) 2( k 1) n( 2 n 1)( 2n 1)
P(n) 12 3 2 5 2 ..... ( 2n 1) 2
( k 1)( k 2) 3
n(n 1)(2n 1) To prove : P(n)is true for n 1
4. 1 2 2 2 3 2 · ·· n 2
6 1(1) ( 3)
P(1) 3 , Hence, P(1) istrue.
n(n 1)(2n 1) 3
P(n) 1 2 2 2 ·· · n 2
6 Assume : P(n) is true for n k.
i) To Prove : P( n) is true for n 1 k( 2 k 1)( 2 k 1)
P( k ) 12 3 2 ... ( 2 k 1) 2
1(1 1)(2(1) 1) 3
P(1) 1,
6 To prove : P(n)is true for n k 1
Hence, P(1) is true. P(k 1) 12 3 2 ... ( 2k 1) 2 ( 2k 1) 2
ii) Assume : P(n) is true for n k. k( 2 k 1)( 2k 1)
k(k 1)(2k 1) ( 2k 1) 2
2 2
P(k) 1 2 · ·· k 2 3
6 k( 2k 1) 3( 2k 1)
iii)To Prove : P( n) is true for n k 1 ( 2 k 1)
3
P(k 1) 1 2 2 2 3 2 · ·· k 2 (k 1) 2 (2k 1)(2k 2 5k 3)
P(k) (k 1) 2 3
k(k 1)(2k 1) (k 1)(2k 1)(2k 3)
(k 1) 2
6 3
n(n 1)(n 2 )
k(k 1)(2k 1) 6(k 1) 2 7. 1.2 2.3 3.4 .... n.(n 1)
3
6 n(n 1)(n 2 )
(k 1) (k(2k 1) 6(k 1)) P(n) : 1.2 2.3 .. n(n 1)
3
6 To prove : P( n) is true for n 1
2
(k 1) (2k 7k 6)
1(1 1) (1 2)
6 1(1 1)
3
(k 1) [(k 2)(2k 3)]
2 2, Hence P(1) is true.
6
Assume : P(n ) is true for n k.
(k 1) [((k 1) 1)(2(k 1) 1)]
P(k) 1.2 2.3 3.4 k(k 1)
6
(k 1)((k 1) 1) (2(k 1) 1) k(k 1) (k 2)
P(k 1)
6 3
n(n 1) 2 To prove : P( n) is true for n k 1
5. 13 2 3 33 ..... n 3
2 P(k 1) 1.2 2.3 .. k(k 1) (k 1)(k 2)
2
3 3 3 3 n(n 1) [1.2 2.3 .. k(k 1)] (k 1)(k 2)
P( n) : 1 2 3 ..... n
2 k(k 1) (k 2)
(k 1) (k 2)
To Prove : P(n ) is true for n 1 3
1 2 k(k 1) (k 2) 3(k 1) (k 2)
P(1) : 13 1 Hence, P(1) is true
3
2
Assume : P(n ) is true for n k (k 1) (k 2) (k 3)
3
k(k 1) 2
P( k ) : 13 2 3 33 ..... k 3
8. For any natural number n 2,
2
1 1 1 1 1 1 ....1 1 n 1 Assume : P(n) is truefor n k.
2 2 3 2 4 2 n 2 2n 1 1 1 1 k
P(k) ...
1 1 1 n 1 1.2 2.3 3.4 k( k 1) k 1
P( n ) : 1 2 1 2 ....1 2
2 3 n 2n To prove : P(n ) is true for n k 1
To prove : P(n) is true for n 2 1 1 1 1
P(k 1) ...
1 1 2 1 3 3 , P(1) is true 1.2 2.3 k( k 1) ( k 1)( k 2)
2 2 2( 2) 4 4 1
P( k )
Assume : P(n) is true for n k ( k 1)( k 2)
1 1 1 k 1 k 1
P( k ) : 1 2 1 2 ....1 2
2 3 k 2 k k( k 1) ( k 1)( k 2)
To prove : P(n) is true for n k 1 1 k 1
1 1 1 1 k 1 1 k 2
P( k 1) 1 2 1 2 ....1 2 1
2 3 k ( k 1) 2 1 k 2 2 k 1
k 1 k 2
k 1 k 2 2 k 1 1
2k ( k 1) 2
2
1 ( k 1)
k2 k 1 k 2
2( k 1) k 1
1 1 1 1 n 1 k2
9. .... n( n 3)
1 2 1 2 3 1 2 3 4 1 2 3 ... n n 1 11. 1 1 1 ... 1
1 1 1 n 1 1 .2 . 3 2 . 3 .4 3.4. 5 n( n 1)( n 2 ) 4( n 1)(n 2)
P( n) : ....
1 2 1 2 3 1 2 ... n n 1 1 1 1 n(n 3)
P (n ) : ...
1 1 2 n 1 1.2.3 2.3.4 n(n 1)( n 2) 4(n 1)( n 2)
P( n) : ....
1 2 1 2 3 n( n 1) n 1 To prove : P(n) is true for n 1
To prove : P(n) is true for n 1 1 1(1 3)
P(1) :
2 2 1 1(1 1)(1 2) 4(1 1)(1 2)
P(1) :
2( 2 1) 2 1 1 1
, P(1) is true
1 1 6 6
P(1) : , Hence P(1) is true Assume : P(n) is true for n k
3 3
Assume : P(n) is true for n k 1 1 1 k( k 3)
P( k ) : ...
1 1 2 k 1 1.2.3 2.3.4 k( k 1)( k 2) 4( k 1)( k 2)
P( k ) : ....
1 2 1 2 3 k( k 1) k 1 To prove : P(n) is true for n k 1
To prove : P(n) is true for n k 1 1 1 1
P( k 1) ...
1.2.3 2.3.4 3.4.5
1 1 2 2
P( k 1) .... 1 1
1 2 1 2 3 k( k 1) ( k 1)( k 2)
k( k 1)( k 2) ( k 1)( k 2)( k 3)
k 1 2
k( k 3) 1
k 1 ( k 1)( k 2)
4( k 1)( k 2) ( k 1)( k 2)( k 3)
1 k 1 2
1 k 3 3k 1
k 1 1 k 2
( k 1)( k 2) 4 ( k 3)
1 k 2 k 2 2
1 k 2 6k 2 9k 4
k 1 k2
( k 1)( k 2) 4( k 3)
k
1 ( k 1) 2 ( k 4)
k2
1 1 1 1 n ( k 1)( k 2) 4( k 3)
10. ...
1.2 2.3 3.4 n( n 1) n 1 ( k 1) 2 ( k 4)
1 1 1 1 n 4( k 1)( k 2)( k 3)
P(n) ...
1.2 2.3 3.4 n( n 1) n 1 P( k 1) is true
To prove : P(n) is true for n 1 1 1 1 1 n
12. ....
1 1 2.5 5.8 8.11 (3n 1)(3n 2) 6n 4
P(1) , Hence, P(1) is true
1.2 2
1 1 1 1 n n3
P(n) : .... 15. 1 2 2 2 3 2 · ·· n 2
2.5 5.8 8.11 (3n 1)(3n 2) 6n 4 3
To prove : P(n) is true for n 1 n3
P(n) :12 2 2 ... n 2 , n N
1 1 3
P(1) : , 1 1, P(1) is true. 1
2.5 6 4 since 1 2 , P(n) is true for n 1
Assume : P(n) is true for n k. 3
Assume : P(k) is true
1 1 1 k
P(k) ....
2.5 5.8 (3k 1)(3k 2) 6k 4 k3
P(k): 12 2 2 ... k 2
, n N
To prove : P(n) is true for n k 1 3
To prove : P(k 1) is true whenever P(k) is true
1 1 1 1
P(k 1) .. P(k 1) 12 2 2 ... k 2 ( k 1) 2
2.5 5.8 (3k 1)(3k 2) (3k 2)(3k 5)
k 1 k3
( k 1) 2
6k 4 (3k 2)(3k 5) 3
1 k 1 1
[ k 3 3k 2 6k 3]
3k 2 2 3k 5 3
1 ( k 1) 3
1 3k 2 5k 2 [( k 1) 3 3k 2]
3 3
3k 2 2(3k 5)
16. an − bn is divisible by a − b, where a > b
1 ( k 1)(3k 2) P(n) a n - b n , is divisible by a - b.
3k 2 2(3k 5)
To prove : P(n) is true for n 1
k 1
P(1) a - b, divisible by a - b, P(1) is true.
6 k 10
13. 1! (2 2! ) (3 3! ) ... (n n! ) (n 1)! - 1 Assume : the statement is true for n k
P(n) : 1! (2 2! ) (3 3! ) ... (n n! ) (n 1)! - 1 P(k) a k - b k , is divisible by a - b.
Prove : P(n) is true for n 1 P(k) a k - b k (a - b)
P(n) : (1 1! ) (n 1)!-1, To prove : P(n) is true for n k 1
1 1, Hence P(1) is true P(k 1) a k 1 - b k 1
Assume : P(n) is true for n k a k 1 - ab k ab k - b k 1
P(k) : 1! (2 2! ) (3 3! ) ... (k k! ) (k 1)! - 1
a(a k - b k ) b k (a - b)
Prove : P(n) is true for n k 1
a( (a - b)) b k (a - b)
P(k 1) 1! (2 2! ) ... (k k! ) (k 1)(k 1)!
(a - b)(a bk)
(k 1)! - 1 (k 1)(k 1)!
(a - b) 1 , 1 a b k , 1 N
(k 1)![(k 1) 1] - 1
17. x 2n - y 2n is divisible by x y
(k 1)! (k 2) - 1 (k 2)!-1
P(n ) x 2n - y 2n is divisible by x y
14. For any integer n ≥ 2, 3n2 > (n + 1)2
P(n) : 3n 2 (n 1) 2 with n 2. To prove : P(n) is true for n 1
To prove : P(n) is true for n 2 P(1) x 2 y 2 ( x y )( x y ) divisible by x y
P(2) 3 2 2 12 and 3 2 9. Assume : P(n) is true for n k
As 12 9, P(2) is true. P( k ) x 2 k - y 2 k is divisible by x y
Assume : P(n) is true for n k. x 2k - y 2 k ( x y )
P(k) : 3k 2 (k 1) 2 To prove : P(n) is true for n k 1
To prove : P(n) is true for n k 1 P( k 1) x 2 k 2 - y 2 k 2
P(k 1) 3(k 1) 2 x 2 k 2 -x 2 k y 2 x 2 k y 2 y 2 k 2
3k 2 6k 3 x 2 k x 2 --x 2 k y 2 x 2 k y 2 y 2 k y 2
(k 1) 2 6k 3 x 2 k (x 2 y 2 ) y 2 ( x 2 k y 2 k )
k 2 8k 4 x 2 k ( x 2 y 2 ) y 2 ( x y )
k 2 4k 4 4k ( x y )[ x 2 k ( x y ) y 2 ]
(k 2) 2 4k (k 2) 2 since k 0. 18. n3 − 7n + 3, is divisible by 3, for all n.
P(n) : n 3 7n 3 is divisible by 3 21. 5n+1+46n when divided by 20 leaves a
To prove : P(n) is true for n 1 remainder 9, for all natural numbers
P(1) 13 7(1) 3 is divisible by 3 P(n) 5 n1 4 6 n when divided by 20
Hence P(1) is true reminder is 9
Assume : P(n) is true for n k To Prove : P(n) is true for n 1
P(k) : k 3 7 k 3 3 P(1) 511 4 6 1
To prove : P(n) is true for n k 1 49 40 9 P(1) is true.
Assume : P(n ) is true for n k.
P( k 1) ( k 1)3 7( k 1) 3
P(k) 5 n 1 4 6 n 20 9, k 1 N
k 3 3k( k 1) 1 7( k 1) 3
20 5 n1 4 6 n 9
k 3 7 k 3 3k 2 3k 6
To Prove : P(n) is true for n k 1
3 3( k 2 k 2)
2 P(k 1) 5 k 2 4 6 k 1 9
3[ k k 2]
5 k 1 5 4 6 k 6 9
3 , , N
19. 32 n+2 − 8n − 9 is divisible by 8 for all n ≥ 1. 5 k 1 5 4 6 k (5 1) 45 45 9
P(n) 3 2n 2 - 8n - 9, is divisible by 8. 5( 5 k 1 4 6 k 9) 4 6 k 36
To Prove : P(n) is true for n 1 5( 20 ) 4 6 k 36
P(1) 3 2 2 - 8(1) - 9 64, divisible by 8. divisible by 20
Hence, P(1) is true. Let 4(9 6 k ) be divisible by 20
Assume : P(n ) is true for n k. for k 1, 60 divisible by 20
P(k) 3 2k 2 - 8k - 9, is divisible by 8.
for k 1, 4(9 6 k 1 )
2k 2
3 - 8k - 9 8k1, k1 N 4[9 6 k (5 1)]
3 2k 2 8k 1 8k 9. 4[9 6 k ] 20 6 k
To Prove : P(n) is true for n k 1
divisible by 20 divisible by 20
2(k 1) 2
P(k 1) 3 - 8(k 1) - 9
22. For natural number n, with assumption i2 = −1,
2 2k 2
3 3 - 8k - 8 - 9 (r(cos θ + i sin θ))n = rn (cos nθ + i sin nθ) ,
3 2 (8k 1 8k 9) - 8k - 17 (r(cos i sin )) n r n (cos n i sin n)
72k 1 64k 64
P(n) (r(cos i sin) n
8(9k 1 8k 1)
r n (cos n i sin n))
8k 2 , k 2 9k 1 8k 1 N
Sub. n 1, in the statement
20. 10 + 3 4n+2 + 5, is divisible by 9, for all n.
n
P(n) 10 n 3 4 n 2 5, is divisible by 9. P(1) (r(cos i sin))1
To Prove : P(n) is true for n 1 r (cos i sin)
Hence, P(1) is true.
P(1) 10 3 4 3 5
Assume : the statement is true for n k
207 , divisible by 9 P(1) is true.
Assume : P(n) is true for n k. r (cos i sin)k r k (cos k i sink) ,
k
To Prove : P(k 1) is true.
P(k) 10 k 3 4 k 2 5 9k1, k1 N
P(k 1) (r(cos i sin ))k 1
10 k 9k1 3 4 k 2 5
To Prove : P(n) is true for n k 1 r k (cos i sin )k r(cos i sin)
P(k 1) 10 k 1 3 4 k 3 5 r k 1 (cosk i sink)(cos i sin)
10 k10 3 4 k 2 4 5 r k 1 (cos k cos i 2 sin k sin)
i(sin k cos cos k sin )
[9k1 3 4 k 2 5 ]10 12 4 k 2 5
r k 1 ((cos k cos - sin k sin)
90k1 30 4 k 2 50 12 4 k 2 4 5
i(sin k cos cos k sin))
9[ k1 2 4 k 2 5]
9k 2 N r k 1 (cos(k 1) i sin(k 1))
cos cos ( ) cos ( 2) · · · To prove : P(n) is true for n 1
sin sin n12
23. n
( n1) sin 2
cos ( (n - 1) ) cos 2
sin 2
sin
sin 12
P(n) : cos cos ( ) cos ( 2 ) ... Hence P(n) is true for n 1
cos ( (n-1) )
cos ( n21) sin n2 Assume : P(n) is true for n k
sin 2 P( k ) : sin sin ( 6 ) sin ( 2
6 ) .....
To prove : P(n) is true for n 1
sin (
( k 1)
)
sin ( k12
1)
sin 12
k
sin
6
sin12
2
cos cos To prove : P(n) is true for n k 1
sin
2
k
Hence P(n) is true for n 1 sin sin ( 6 ) sin ( 2
6 ) .... sin ( 6 )
Assume : P(n) is true for n k
sin ( k12
1)
sin k
12
sin ( k
)
P( k ) : cos cos ( ) cos ( 2 ) ...
sin 12 6
cos ( k 21) sin k2
sin
( k 1)
12 sin
sin12 sin ( k6 )
k
12
cos ( (k-1)) sin12
sin 2
To prove : P(n) is true for n k 1 sin( 12
k
) 12 sin12
k
sin12 sin ( k6 )
sin12
cos cos ( ) cos ( 2 ) ... cos ( (k-1) )
cos ( (n - 1)) cos ( k)
k
[sin( 12 ) cos 12 cos( 12 k
) sin 12 ]sin12
k
sin12 sin ( k6 )
sin12
cos
( k 1)
2 sin cos ( k)
k
2 k k k k
) sin 12 sin 12
sin ( k6 )
sin 12 sin( 12 ) cos 12 cos( 12 sin 12
sin 2
sin 12
cos
( k 1)
2 sin cos ( k) sin
k
2
2 k
sin 12 sin( 12k
) cos 12 sin 12 [sin ( k6 ) cos( 12 k k
) sin 12 ]
sin 2
sin12
cos( k2 )
cos ( k) sin 2
k
2 sin 2 k
sin 12 k
sin( 12
sin
) cos 12 212 [ 2 sin ( k6 ) sin sin k6 ]
sin 2 sin12
k k k
[cos( 2 ) cos 2 sin( 2 ) sin 2 ] sin 2 cos ( k) sin 2 sin12
k
sin( 12k ) cos 12 sin212 [sin sin k6 ]
sin 2
sin12
sin 2
sin( 12k ) cos 12 sin212 [ 2 sin 12k cos 12k ]
k k
{ 2 sin( k2 ) sin k2 2 cos ( k)
cos( 2 ) cos 2 sin 2 2 sin12
k
sin 2 sin12
)[sin12 cos 12 sin12 cos 12k ]
k k
k k sin 2 sin( 12
cos( 2 ) cos 2 sin 2 2 [cos cos( k) 2 cos ( k)
sin12
sin 2
k
sin( 12 )[sin12
k
12 ]
cos( k k
sin
2 ) cos 2 sin 2 2 2 {cos cos ( k )} sin12
sin 2 k
sin( 12 )[sin
( k 1)
]
12
cos( k k
2 ) cos 2 sin 2 sin 2
2 { 2 cos( k
}
k
2 ) cos 2
sin
12
sin
2
cos( k
2 ){cos 2 sin 2
k
sin 2 {cos( k2 )}
sin 2
( k 1)
cos( k2 ) sin 2
sin 2
( n 1)
sin sin ( 6 ) sin ( 2
6 ) ..... sin ( 6 )
24.
sin ( n 1)
12 sin n
12
sin n12
P( n ) : sin sin ( 6 ) sin ( 26 ) .....
sin (
( n1)
)
sin 12 sinn12
( n1)
6
sinn12
5.1.BINOMIAL THEOREM when x = 1, 0 = nC 0 − n C1 + n C 2 − .. + ( −1)n nC n
Binomial theorem for any positive integer n,
n n n 0 n n−1 1 n 0 n General Terms in the expansion (a + b)n
(a + b) = C 0 a b + C 1 a b + ... + C n a b
P(n) : (a + b) = C 0 a b + C1 a
n n n 0 n n −1 1
b + ... + C n a b
n 0 n First term = nC0an, second term = nC1an–1b,
Using Mathematical Induction, third term = nC2an–2b2,…. (r + 1)th term = nCran–rbr.
To prove : P(n) is true for n = 1 (r + 1)th term is also called general term
1
C 0 = 1 and C1 = 1 1
Thus Tr+1 = nCr an–rbr
( a + b) = a b + a b
1 1 0 0 1
Middle term in the expansion (a + b)n
a + b = a + b , Hence P(1) is true.
Assume : P(n) is true for n = k i) If n is even,No. of terms = n+1. So n + 1 is odd
k −1 1
P(k) : (a + b) = C 0 a b + C1 a
k k k 0 n
b + ... + C k a b
k 0 k
n + 1 + 1 th th
Middle term = term i.e n + 1 term
To prove : P(n) is true for n = k + 1 2 2
k +1
(a + b) = (a + b)(a + b)
k
ii) If n is odd, n + 1 is even
k −1 1
= (a + b)[ C 0 a b + C1 a
k k 0 n
b + ... + C k a b k 0 k
]
n −1 n +1
th th
k
C0 a
k +1 0
b + C1 a b + ... + C k a b
k k 1 k 1 k Middle term = + 1 and + 1 term
=
2 2
k −1 2 0 k +1
C 0 a b + C1 a b + ... + C k a b
k k 1 k k
+
Greatest coefficient in the expansion (a + b)n
n +1
C r = C r + C r-1
n n
nC
k +1 k +1 0
(a + b)
k
b + [ C1 + C 0 ]a b + ..
k k k 1 n
= C0 a
i) If n is even, Greatest coef. = 2
[ k C k + k C k −1 ]a 1b k + k C k a 0 b k +1
nC and nC n +1
k +1 k +1 0 k +1 k +1) − r 1 k +1 0 k +1 n −1
= C0 a b + C 1 a( b + ... C k +1 ab ii) If n is odd, Greatest coeff. = 2 2
Observations
n n
(a + b) = C 0 a b + C1 a
n 0 n n −1 1
b + ... + C k a
n n−k k
b
Term independent of x :
1. n n Term independent of x contains no x. Hence find
= ∑ nk= 0 C k a n − k b k or ∑ nk= 0 C k b n − k a k
the value of r for which exponent of x is zero.
2. The coefficients nCr occuring in the binomial
theorem are known as binomial coefficients. Equidistant Coefficients
3. There are (n+1) terms in the expansion of (a+b)n, In the expansion of (a+b)n, coefficients at equi
i.e., one more than the index. distant from beginning and from end are equal
4. In the successive terms of the expansion the index i.e nCr = nCn−r .
of a goes on decreasing by unity. Index of b
SOLVED PROBLEMS
increases by unity 1. Expand :(2x + 3)5
5. In the expansion of (a+b)n, sum of the indices of a ( 2x)5 + 5( 2x)4 3 + 10( 2x)3 32 + 10( 2x)2 33 + 5( 2x)34 + 35
and b is n in every term of the expansion. 32x 5 + 240x 4 + 720x 3 + 1080x 2 + 810x + 243
Particular cases of Binomial Theorem 2. Evaluate 984.
98 4 = (100 - 2) 4
i) Replace b by (−b), in the expansion of (a + b)n
= 4 C 0 100 4 - 4 C 1 100 3 2 + 4 C 2 100 2 2 2
(a - b) = C0 a b - C1 a b + .. + (-1)n Cn a b
n n n 0 n n-1 1 n 0 n
- 4 C 3 100 1 2 3 + 4 C 4 2 4
‘+’ and ‘−’ appear alternately in the expansion of = 100000000 - 8000000 + 240000 - 3200 + 16
(a − b)n = 92236816.
ii) Replace a by 1, b by x, in the xpansion of (a + b)n 3. Evaluate 994.
99 4 = (100 - 1) 4
(1 + x)n = nC 0 + n C1 x + n C 2 x 2 + .. + n C r x r .. + n C n x n
= 4 C 0 100 4 - 4 C 1 100 3 1 + 4 C 2 100 2 12
when x = 1, 2 = C 0 + C1 + C 2 + .. Cn
n n n n n
- 4 C 3 100 1 13 + 4 C 4 1 4
iii) Replace a by 1, b by (- x) in the xpansion of (a + b)n = 10000000 - 4000000 + 60000 - 400 + 1
= 96059601.
(1 − x)n = nC 0 − n C1 x + n C 2 x 2 + ... + ( −1)n nC n x n
4. Evaluate 1024.
102 4 = (100 + 2) 4 10. Which is larger (1.01)1000000 or 10000.
= 4 C 0 100 4 + 4 C 1 100 3 2 + 4 C 2 100 2 2 2 (1.01)1000000 = (1 + 0.01)1000000
+ 4 C 3 100 1 2 3 + 4 C 4 2 4 = 1000000 C 0 + 1000000 C 1 (0.01) + positive terms
= 10000000 + 8000000 + 240000 + 3200 + 16 = 1 + 1000000 × (0.01) + other positive terms
= 108243216 = 1 + 10000 (0.01) + other positive terms
5. Evaluate 97. (1.01)1000000 > 10000
9 7 = (10 - 1)7 11. Find middle term in the expansion of (x + y)6.
= 7 C 0 10 7 - 7 C 1 10 6 1 + 7 C 2 10 5 1 2 - 7 C 3 10 4 1 3 + n = 6; which is even.
th
C 4 10 3 1 4 − 7 C 5 10 2 1 5 + 7 C 6 10 1 16 − 7 C 7 17
7
middle term = + 1 term
n
= 4782969 . 2
6. Expand ( 2x − 21x )4 = 4 th term i.e (3 + 1 ) th term
= 4 C 0 ( 2x)4 + 4C1 ( 2x)3 (− 21x ) + 4 C 2( 2x)2 ( − 21x )2 Tr +1 = nC r a n−r b r
1
+ 4 C 3( 2x)1( − 21x )3 + 4 C 4 ( 2x)0 ( − 21x )4 T3+1 = 6C 3 x 3 y 3 = 20x 3 y 3
12. Find middle term in the expansion of (x + y)7
= ( 2x)4 - 4( 2x)3 ( 21x ) + 6( 2x)2 ( 21x )2 - 4( 2x)( 21x )3 + ( 21x )4
n = 7 (odd)
= 16x 4 - 16x 2 + 6 - 1
+ 1
Here a = x , b = y ,
x2 16 x 4
7. Expand ( 2x − 2 3 3
x) Mmiddle term = Tn −1 +1 and Tn+1 +1 terms
2 2
= C 0( 2 x ) −
3 2 3 3
C1( 2x 2 ) 2 x3 ( )+ 3
C 2( 2 x 2
)( 3x )2 − 3
C 3 ( x3 )3 r = 3 and 4
= 8x 6 - 36x 3 + 54 - 27
x3 = 7C3 x 4 y 3 and7C3 x3 y 4
8. Expand ( x 2 + 1 − x 2 )5 + ( x 2 − 1 − x 2 )5 = 35x 4 y 3 and 35x 3 y 4
( x 2 + 1 − x 2 ) 5 = 5 C 0 ( x 2 ) 5 + 5 C 1 ( x 2 ) 4 ( 1 − x 2 )1 + 13. Find coeff. of x6 in the expansion of (3 + 2x)10
Tr +1 = nC r a n−r b r
5
C 2 x 6 ( 1 − x 2 ) 2 + 5 C 3 (x 2 ) 2 ( 1 − x 2 )3
n = 10, a = 3, b = 2x
5
C 4 ( x 2 )1 ( 1 − x 2 ) 4 + 5 C 5 ( 1 − x 2 ) 5
For coeff. of x 6 = take r = 6
= x + 5x
10 8
1 − x + 10x (1-x ) + 10x (1-x )
2 6 2 4 2
10
C6 = 10
C4
1 − x + 5x (1-x ) + (1-x )
2 2 2 2 2 2
1− x 2
coefficient of x = 6 10
C 4 34 ( 2x ) 6
( x 2 − 1 − x 2 ) 5 = x10 − 5x 8 1 − x 2 + 10x 6(1-x 2 ) − 10x 4(1-x 2 ) = 210 × 34 × 2 6 x 6
1 − x 2 + 5x 2(1-x 2 ) 2 − (1-x 2 ) 2 1 − x 2 coefficient of x 6 = 210 × 3 4 2 6
Adding two resulats in 14. Find coeff.of x3 in the expansion of (2 − 3x)7.
Tr +1 = nC r a n−r b r
= 2[x 10 + 10x 6(1-x 2 ) + 5x 2(1-x 2 ) 2 ]
n = 7 , a = 2 , b = − 3x
= 2[x 10 + 10x 6 -10x 8 + 5x 2 (1-2x 2 + x 4 ]
For coeff. of x 3 = take r = 3
= 2[x -10x + 15x − 10x + 5x ]
10 8 6 4 2
coefficient of x 3 = 7 C 4 2 4 ( −3x )3
9. Expand ( 2x 2 + 3 1 − x 2 ) 4 + ( x 2 − 3 1 − x 2 ) 4
= 35 × 16 × ( −27)x 3
( 2 x 2 + 3 1 − x 2 )4 + ( x 2 − 3 1 − x 2 )4
coefficient of x 3 = - 15120
( 2x + 3 1 − x ) = C 0 ( 2x ) + C1 ( 2x ) (3 1 − x ) +
2 2 4 4 2 4 4 2 3 2 1
1 10
15. Find coeff. of x15 in the expansion of x 2 + 3
4
C 2 ( 2 x 2 ) 2 (3 1 − x 2 ) 2 + 4 C 3 ( 2 x 2 ) x
n = 10, a = x , b = x3
2 1
(3 1− x 2 )3 + 4 C 4 (3 1 − x 2 ) 4
1 r
= 16x 8 + 96x 6 1-x 2 + 216x 4(1-x 2 ) Tr +1 = 10 C r ( x 2 )10 − r ( )
x3
+ 54x 2(1-x 2 ) 1-x 2 + 81(1-x 2 ) 2 =10 C r x 20 −5r
( 2x 2 − 3 1 − x 2 ) 4 = 16x 8 − 96x 6 1-x 2 + 216x 4(1-x 2 ) For coefficient of x 15 :
− 54x 2(1-x 2 ) 1-x 2 + 81(1-x 2 ) 2 20 − 5r = 15 ⇒ r =1
Adding two results in T1+1 =10 C 1 x 20 −5(1)
= 2[16x 8 + 216x 4(1-x 2 ) + 81(1-x 2 ) 2 ] = 10x 15
1 6 21. a and b are distinct integers, prove that a − b is a
16. Find coeff. of x2 in the expansion of x 2 − 3 factor of an − bn, where n is a positive integer.
x
n = 6, a = x 2 , b = − x13 a n = ( a − b + b )n
1
2 6− r
= n C 0 ( a − b )n + n C1 ( a − b ) n−1 b + ..... + n C n b n
Tr +1 = C r ( x ) (− 3 )r
6
x = ( a − b )[( a − b )n−1 + n C1 ( a − b ) n−1 b + .....] + b n
12 − 5r
= Cr x
6
( −1) r
a n − b n = ( a − b )( k )
For coeff.of x 2 : ( a - b) is a factor of a n − b n
12 − 5r = 2 ⇒ r=2 22. Find the last two digits of the number 7 400 .
T2 +1 = 6 C 2 x12 −5( 2 ) ( −1) 2 7 400 = (7 2 ) 200
= 15x 2 = (50 - 1) 200
17. Find coeff. of x4 in (1 + x 3 )50 ( x 2 + 1x )6 = 200
C 0 50 200 - 200
C 1 50 199 +···
(1+ x ) ( x +
3 50 2 1 5
x) = [1 + C1 x + C 2 x + C 3 x ...]
50 3 50 6 50 9
+ 200
C 199 50(-1) 199 + 200
C 200 (-1) 200
[ 5 C 0 x10 + 5 C1 xx + 5 C 2 xx 2 + .....]
8 6
= 50 2 ( 200 C 0 50 198 - 200 C 1 50 197 +···
= [1+ 50x 3 + 1225x 6 + ..][x10 + 5x 7 .. + 200
C 198 (-1) 198 ) - 200 × 50 + 1.
Coeff. of x 5 = 1× 10 + 1225 × 5 + 50 × 10 + 19600 As 502 and 200 are divisible by 100,
= 26235 last two digits: 0 1.
1 5 23. Find the last two digits of the number 3600
18. Find constant termof 2x 3 − 2 3 600 = (3 2 ) 300
3x
n = 5, a = 2x , b = − 3 x1 2
3
= (10 - 1) 300
1 r = 300
C 0 10 300 - 300
C 1 10 299 +·· ·
Tr +1 = 5C r ( 2x 3 ) 5− r ( − )
3x 2 + 300
C 299 10(-1) 299 + 300
C 300 (-1) 300
= 5 C r x15 −5 r ( −31 ) r ( 2) 5−r = 10 2 ( 300 C 0 10 298 - 300 C 1 10 297 +· ··
For constant term :
+ 300
C 298 (-1) 298 ) - 300 × 10 + 1.
15 − 5r = 0 ⇒ r =3
As 102 and 300 are divisible by 100,
T3+1 = 5
C 3 x15 −15 ( −31 ) 3 ( 2) 5−3 last two digits: 0 1.
40 24. If n is an odd positive integer, prove that
=− coefficients of middle terms in the expansion of
27
19. Prove that 6 −5n always leaves remainder 1
n (x + y)n are equal.
when divided by 25 for all positive integer n. Middle Terms = Tn−1 and Tn+1 [n is odd]
+1 +1
2 2
Let 6n -5n = 25k + 1 for some integer k. Consider
coefficient of Tn−1 = n C n −1
(1 + x) = C 0 + C1x + C 2 x .... C n-1x
n n n n 2 n n-1
+ Cn x
n n
2
+1
2
Taking x = 5 =
n!
( n2−1 )!( n − [ n2−1 ])!
(1 + 5 )n = nC 0 + n C1 5 + n C 2 52....n C n-1 5n-1 + n C n 5n
n!
6n = 1 + 5n + 25( n C 2 + 5n C 3 ... + n C n 5n
-2
=
( n2−1 )!( n2+1 )!
-2
6n − 5n = 1 + 25( n C 2 + 5n C 3 ... + n C n 5n coefficient of Tn+1 = n C n +1
+1
= 1 + 25k 2 2
20. If n is a positive integer, show that, 9n+1 − 8n − 9 n!
= n +1
is always divisible by 64. ( 2 )!(n - ( n2+1 ))!
9 n+1 = (1 + 8)n +1 n!
=
= n+1
C 0 + n+1 C1 8 + n+1 C 2 8 2...n+1C n 8n ( n2−1 )!( n2+1 )!
Tn −1 = Tn+1
+ n+1 C n +1 8n+1 +1 +1
2 2
2 n+1 n+1 n− 2
= 1 + 8(n + 1) + 8 ( C 2 + ... Cn 8 25. If n is a positive integer and r is a nonnegative
+ n+1 C n +1 8n−1 ) integer, prove that the coefficients of xr and xn− r
in the expansion of (1 + x)n are equal.
= 1 + 8 + 8n + 64( k )
n+1
9 − 8n − 9 = 64k
n! 29. In the binomial coefficients of (1 + x)n, the
coefficient of x r , nC r =
r!(n − r )! coefficients of the 5th, 6th and 7th terms are in AP.
n! Find all values of n.
coefficient of x n−r , nC n−r = coefficien t of 5 th term = nC 4
(n - r)!(n - (n - r))!
n! coefficien t of 6 th term = nC 5
=
(n - r)!r coefficien t of 6 th term = nC 6
C n−r = C r
n n
They are in A.P
26. In the binomial expansion of (a + b)n the
coefficients of the 4th and 13th terms are equal to 2. C 5 = n C 4 + n C 6
n
n
each other, find n. C5 n C 4 n C6
2n = +
coefficient of 4 th term = nC 3 C5 n C 5 n C5
coefficient of 13 th term = nC12 5 n−5
2= +
n−4 6
coefficient of 4 th term = coefficient of 13 th term
n − 21n + 98 = 0
2
n
C 3 = n C12
n = 7 or 14
If nC x = n C y then x + y = n 2n!
30. Prove that C 02 + C12 + C 22 + ... + C n2 =
Thus n = 15 ( n! ) 2
nd rd th
27. 2 , 3 and 4 terms in the binomial expansion
(1 + x ) n = C 0 + C1 x + C1x 2 + ....C n x n
of (x + a)n are 240, 720 and 1080. Find x, a and n.
T2 = 240 ⇒ nC1 x n-1 a = 240 (1) ( x + 1) n = C 0 x n + C1 x n−1 + ..... + C n
Multiplying both
T3 = 720 ⇒ nC 2 x n- 2 a 2 = 720 ( 2 )
(1 + x ) 2n
= (C 0 + C1 x + C1 x 2 + ....C n x n )
T4 = 1080 ⇒ nC 3 x n-3 a 3 = 1080 ( 3)
Dividing (2) by (1) (C 0 x n + C1 x n−1 + ..... + C n )
n
C 2 x n- 2 a 2 720 a 6 Equating Coeff of x n ,
= ⇒ = ( 4)
n n-1
C1 x a 240 x n −1 C 02 + C12 + C 22 + ... + C n2 = 2 nC n
Dividing (3) by (2) 2n!
=
n
C 3 x n- 3 a 3 1080 a 9 ( n! ) 2
= ⇒ = ( 5)
n n- 2 2
C2 x a 720 x 2(n − 2)
sub (4) in (5)
5.4 FINITE SEQUENCES
9 6
= § A list of elements with a particular order.
2(n − 2) n − 1
§ A function whose domain is either the set of first
9n − 9 = 12n − 24 ⇒ n = 5, x = 2, a = 3
n natural numbers or N.
28. If the binomial coefficients of three consecutive 1. Basic definitions of sequences and Progressions.
terms in the expansion of (a + x)n are in the ratio
§ If X is any set and n∈ N, any function f : {1, 2, 3, .
1 : 7 : 42, then find n.
. . , n} X is a finite sequence on X and any
3 consecutive terms : (r–1)th , r th ,(r + 1)th terms
function g : N X is an infinite sequence on X.
coefficient(r–1)th term = nC r– 2
§ The value f(n) of the function f at n is denoted by
coefficient of r th term = nC r–1
an and the sequence itself is denoted by (an).
coefficient ( r + 1) th term = nC r
§ If X is a set of real numbers, sequence is a
n
C r– 2 : nC r–1 :n C r = 1 : 7 : 42 numerical sequence or sequence of real numbers.
n
C r– 2 1
n
= § Every sequence is a function
C r–1 7
§ Terms in a sequence may be repeated. A sequence
n – 8r + 9 = 0 ... (1)
n
in which all terms are same is constant sequence.
C r–1 7
n
=: § Progressions are Special cases of sequences where
Cr 42
terms are in increasing or decreasing form.
n – 7r + 1 = 0 ... (2)
Solving (1) and (2) n = 55. 2. Arithmetic Progression (AP)
§ A sequence of the form 1 1 1
§ General H.P : , , ,.... where, a+kd ≠ 0
a a + d a + 2d
a, a + d, a + 2d, a + 3d, . . . , a + (n − 1)d, a + nd, . . .
§ HP are not reciprocals of AP as denominator of a
is an arithmetic progression or arithmetic sequence.
fraction cannot be 0
§ Each term (other than first term) is obtained by
§ If a, b, c are in HP, then b = 2ac
adding a constant to its previous term a+c
§ d is common difference and a is initial or first term. § In triangle, if altitudes are in AP, sides are in HP
§ nth term of A.P, Tn = a + (n − 1)d. 6. Arithmetic Mean
§ Tn = an + b where a and b are prime, form an AP § Given two numbers a and b. We can insert a
§ It contains infinitely many prime numbers along number A between them so that a, A, b is an A.P.
with infinitely many composite numbers. A – a = b – A,
3. Geometric Progression (GP)
A= a+b
2
§ A sequence of the form
§ Such a number A is called the arithmetic mean
a, ar, ar2, ar3 , …, arn−1, arn, .. (A.M.) of the numbers a and b.
is a geometric progression or a geometric sequence. § Given any two numbers a and b, we can insert
§ Each term other than first term is obtained by many numbers between them such that resulting
multiplying its previous term by a constant; sequence is an A.P.
§ r is common ratio and a is initial term or first term. § Let A1, A2, A3, …, An be n numbers between a and
§ nth term of a G.P, Tn = arn−1. b such that a, A 1, A2, A3, …,An, b is an A.P.
§ Taking logarithm of each term in a G.P with § Here, b is (n + 2) th term,
yields an arithmetic progression. i.e., b = a + [(n + 2) – 1]d
§ If a, ar, ar2, . . . is a GP with r > 0, log a, log(ar), = a + (n + 1) d
log(ar2),…is an AP with common difference log r.
d = b−a
n +1
§ Constant sequence c, c, c, . . . is an arithmetic
sequence and also a geometric sequence if c ≠ 0. n numbers between a and b are
4. Arithmetico-Geometric Progression (AGP) A 1, A2, A3, …, An
a + d, a + 2d, a +3 d, a + nd
§ A sequence of the form
a,(a+d)r,(a+2d)r2,(a+3d)r3….(a+(n−1)d)rn−1,(a+nd)rn a + b − a , a + 2 b − a , a +3 b − a , a + n b − a
n +1 n +1 n +1 n +1
is arithmetico-geometric progression or sequence. Geometric Mean (G .M.)
§ Given two numbers a and b. We can insert a
§ nth term of an AGP
number G between them so that a, G, b is an G.P
Tn = (a + (n − 1)d)rn−1
a G
= ⇒ G = ab
§ a is the initial term, d is the common difference and r G b
is the common ratio of the AGP.
G is called Geometric mean (G.M.) of a and b.
§ All APs and all GPs are AGPs. § We can insert many numbers between a and b
For r = 1, AGP becomes AP such that resulting sequence is an G.P.
For d = 0, AGP becomes GP. § Let G1, G2, G3, …, Gn be n numbers between a and
5. Harmonic Progression (HP) b such that a, G1, G2, G3, …, Gn ,b is G.P.
§ Here, b is (n + 2) th term,
§ A sequence is in harmonic progression if its
1
reciprocals are in AP b = ar n +1 ⇒ r = ( ba ) n +1
§ A sequence h1, h2, h3, . . . is said to a harmonic § n numbers between a and b are
1 1 1
sequence or progression if , , … is an A.P G 1, G2, G3, …, Gn
h1 h2 h3
1 2 3 n
a ( ba ) n +1 , a ( ba ) n+1 , a ( ba ) n+1 ,.. a ( ba ) n+1 1. Write the first 6 terms of the sequences whose
nth terms are given. Classify them as AP, GP,
Harmonic mean AGP, HPand none of them.
H is called the harmonic mean between a and b if a, H, b 1
1 1 1 i) n+1
are in H.P. If a, H, b are in H.P then , , are in A.P 2
a H b
1 1 1 1 1 1 1
1 1 n +1
= 0 +1
, 1+1
, 2 +1
, 3+1
, 4 +1
, 5+1
+ 2 2 2 2 2 2 2
1 a b 2 ab
= ⇒H= 1 1 1 1 1 1
H 2 a+b = , , , , ,
2 4 8 16 32 64
H1, H2, Hn are called n harmonic means between a G.P
and b if a, H1, H2, Hn, b are in H.P. ( n + 1)(n + 2)
ii)
Relation between A.M., G.M. and H.M. (n + 3)( n + 4)
i) A.M > G.M > H.M ( n + 1)(n + 2) 2 6 12
= , ,
For any set of n non-negative numbers, the arithmetic I (n + 3)(n + 4) 12 20 30
mean is greater than or equal to the geometric mean None
a+b 1
AM − GM = − ab iii) 4( )n
2 2
a + b − 2 ab 1 1 1
4( ) n = 4 ,2,1, ,
= 2 2 4
2
=
( a− b )2 GP
2 ( −1)n
AM − GM ≥ 0 iv)
n
AM ≥ GM
( −1) n 1 1 −1
For any set of n positive numbers, the geometric mean = , ( −1), , ,...
n 0 2 3
is greater than or equal to the harmonic mean”. None
GM − HM = ab −
2 ab 2n + 3
a+b v)
3n + 4
ab[ a + b − 2 ab ] 2n + 3 3 5 7
= = , , ,...
a+b 3n + 4 4 7 10
=
ab ( a− b )
2
None
2 vi) 2018
GM − HM ≥ 0
3n − 2
GM ≥ HM vii)
3n−1
ii) A.M., G.M., H.M. are in G.P.
3n − 2 4 7
n−1
= −6 ,1, , ....
GM ab 3 3 9
=
AM a + b AGP
2
2. Write the first 6 terms of the sequences whose
2 ab
= nth term an is given below
a+b
2ab n + 1 if n is odd
HM a + b an =
= n if n is even
GM ab
2 ab = (1+1), 2,(3+1), 4,(5+1),6
=
a+b = 2, 2, 4, 4, 6, 6, . . .
GM HM
= 1 if n = 1
AM GM
GM 2 = AM.HM a n = 2 if n = 2
a + if n > 2
n -1 an - 2
= 1, 2, (1+2), 2+(1+2), ….
= 1, 2, 3, 5, 8, 13, . . 6. Find seven numbers 4 numbers G1,G2,G3,G4 so
that the sequence 12,G1,G2,G3,G4, 3 8 is in GP
n if n is 1, 2 or 3
an =
a = 12 , b = 3
,n = 6
a n -1 + a n -2 + a n -3 if n > 3 8
12r = 83
5
⇒ r = 5 1
32 .
=1, 2, 3, 6, 11, 20, . . .
1
3. Write the nth term of the following sequences r=
2
i) 2, 2, 4, 4, 6, 6, G1 = ar G 2 = ar 2 G 3 = ar 3
n + 1 if n is odd = 12 × 12 = 12( 12 ) 2 = 12( 12 ) 3
an = =6 =3
n if n is even = 1 12
G 4 = 34
1 2 3 4 5
ii) , , , , ,.....
2 3 4 5 6 3 5 7
7. Write nth term of sequence 2 2
, 2 2 , 2 2 ,... as
n 1 2 2 3 3 4
an = , ∀n ∈ N
n + 1 a difference of two terms.
1 3 5 7 9 ( 2n + 1)
iii) , , , , ,... n th Term t n =
2 4 6 8 10 n 2 ( n + 1) 2
2n - 1 Using partial fraction,
an = , ∀n ∈ N
2n ( 2n + 1) A B
= 2-
iv) 6, 10, 4, 12, 2, 14, 0, 16, −2, . . . n (n + 1)
2 2
n (n + 1) 2
A = 1, B = −1
7 - n if n is odd
an = 1 1
8 + n if n is even Hence, t n = -
n (n + 1) 2
2
4. If the 5th and 9th terms of a harmonic progression
are 191 and 351 , find the 12th term of the equence. 8. If the product of the 4th, 5th and 6th terms of a GP
is 4096 and if the product of the 5th, 6th and 7th-
1
n th term of HP, h n = terms of it is 32768, find the sum of first 8 terms
a + (n − 1)
of the geometric progression.
n th term of AP, a n = a + (n − 1)
1 n th term of GP t n = a r n −1
h5 = ⇒ a 5 = 19
19 product of 4 th , 5 th , 6 th terms = 4096
a + 4d = 19 .....(1) ( ar 3 )( ar 4 )( ar 5 ) = 4096
1
h9 = ⇒ a 9 = 35 a 3 r 12 = 4096____(1)
35
a + 8d = 35 .....(2) product of 5 th , 6 th , 7 th terms = 32768
Solving (1) & (2), a = 3, d = 4. ( ar 4 )( ar 5 )( ar 6 ) = 32768
a 12 = a + 11d = 47 a 3 r 15 = 32768___(2)
1
12 th term of HP, h 12 = (2) a 3 r 15 32768
gives, 3 12 =
47 (1) 4096
a r
5. Find seven numbers A1, A2, . . . , A7 so that the r = 2, a = 1
sequence 4, A1, A2, . . . , A7, 7 is in A.P
a(r n − 1)
sum of first n terms , S n =
a = 4 , b = 7 , n = 9. r −1
4 + ( 9 –1) d = 7 1( 2 8 − 1)
sum of first 8 terms , S 8 =
d = 3
. 2 −1
8
= 255
A1 = a + d A2 = a + 2d A3 = a + 3d
=4+ 3
=4+ 2( 83 ) = 4 + 3 38 9. The product of three increasing numbers in GP
8
is 5832. If we add 6 to the second number and 9
= 4 83 = 4 68 = 5 18
to the third number, then resulting numbers
A 4 = 5 84 A5 = 5 78 A6 = 6 82 A7 = 6 58 form an AP. Find the numbers in GP.
a
3 incresing Numbers = , a , ar AM = GM + 10 .....(1)
r AM = HM +16 .....(2)
Their product = 5832
Sub(1)in( 2)
a
× a × ar = 5832 GM + 10 = HM + 16
r
HM = GM − 6
a 3 = 5832
a = 18 But GM 2 = AM × HM
a = (GM + 10)(GM − 6)
, a + 6, ar + 9 are in AP
r GM = 15
18 ab = 15
(18 + 6) − = ( 18r + 9) − (18 + 6)
r
ab = 225 ....(3)
18r − 39r + 18 = 0
2
AM = GM + 10
( 2r − 9)(3r − 2) = 0
a + b = 50 ....(4)
3 2
r= or Solving (3) and (4)
2 3
For incresing GP, r = 32 a = 45 or 5, b = 5 or 45
18 13. If the roots of the equation (q − r)x2 + (r − p)x + p
Numbers = 3
,18,18( 32 )
2 − q = 0 are equal, show that p, q and r are in AP.
= 12 ,18, 27
(q - r)x 2 + (r - p)x + p - q
10. If tk is the kth term of a GP, then show that tn− k,
A = q − r , B = r − p,C = p − q
tn, tn+k also form a GP for any positive integer k.
B2 − 4 AC = 0
t k = ar k −1
( r − p )2 = 4(q − r )( p − q )
t n− k = ar n− k −1
r 2 + p 2 − 2rp = 4 pq − 4q 2 − 4 pr + 4qr
t n = ar n−1
r 2 + p 2 + 4q 2 + 2rp − 4 pq − 4qr = 0
t n+ k = ar n+ k −1
( p + r − 2q ) 2 = 0
(t n −k )(t n+ k ) = ( ar n−k −1 )( ar n+ k −1 )
p + r − 2q = 0
= a 2 r n− k−1+n+ k −1
2q = p + r
= a 2 r 2( n−1)
14. If a, b, c are pth, qth and rth terms of GP, show that
= (t n ) 2
(q − r) log a + (r − p) log b + (p − q) log c = 0.
t n− k , t n , t n+ k are in GP
1
a = AR p−1 ⇒ log a = log A + ( p − 1) log R
1 1
11. If a, b, c are in GP, and if a = b = c , then x y z
b = AR q −1 ⇒ log b = log A + ( q − 1) log R
prove that x, y, z are in arithmetic progression.
c = AR r −1 ⇒ log c = log A + (r − 1) log R
1
LHS = (q - r) log a + (r - p) log b + (p - q) log c
1 1
a x = b y = c z = k ( say)
1 = (q - r)[log A + ( p − 1) log R ]
Let a = k
x
+ (r - p)[log A + ( q − 1) log R ]
a = k x , and b = k y , c = k y
+ (p - q)[log A + (r − 1) log R ]
a , b , c , are in GP,
= (q - r) log A + (q - r)( p − 1) log R
b c
= + (r - p) log A + (r - p)( q − 1) log R
a b
b 2 = ac + (p - q) log A + (p - q)(r − 1) log R
= log A[ p − q + q − r + r − p ]
( k y ) 2 = ( k x )( k y )
+ log R[(q - r)( p − 1) + (r - p)(q − 1)(p - q)( r − 1)]
Equating powers
2y = x + z = log A[ 0] + log R[0]
=0
Thus, x, y, z are in arithmetic progression
12. AM of two numbers exceeds their GM by 10
and HM by 16. Find the numbers.
n
∑ k( k1+ 1)
FINITE SERIES
3. Find
Sum of Arithmetic and Geometric Progressions k =1
§ A series is said to be an arithmetic series if the
t k = k th term of given series
terms of the series form an arithmetic sequence.
1
§ Sum Sn of first n terms of the arithmetic sequence tk =
n k( k + 1)
S n = [ 2 a + ( n − 1)d ]
2 By using partial fraction
§ A series is said to be a geometric series if the terms 1 1 1
= −
of the series form a geometric sequence. k( k + 1) k ( k + 1)
§ Sum Sn of first n terms of the geometric sequence
t1 + t 2 + .... + t n = − + − + ... + −
1 1 1 1 1 1
a[r n − 1]
Sn = if r > 1 1 2 2 3 n n + 1
r −1 1
=1−
S n = na if r = 1 n +1
§ A series is said to be arithmetico-Geometric series if 4. Find sum of the first 20-terms of the arithmetic
the terms form arithmetico-Geometric sequence. progression having the sum of first 10 terms as
§ Sum of first n terms of AGP 52 and the sum of the first 15 terms as 77.
n
a - (a + (n - 1)d)r n 1 - r n-1 Sn = [ 2 a + (n − 1)d ]
Sn = + dr
2
2
1-r (1 - r) S10 = 52 S15 = 77
Telescopic Summation for Finite Series 10a + 45d = 52 30a + 210d =154
§ Telescopic summation is used for summing a
Solving,
series either for finite or infinite terms.
2 133
§ This technique expresses sum of n terms of a d=− ,a = ,
given series just in as first and last term, by 75 25
S 20 = 2
133 2 304
making intermediate terms cancel each other. + 19 − =
§ After canceling intermediate terms, we bring the 25 75 3
last term very close to the first term 5. Find the sum up to the 17th term of the series
Solved Problems 13 13 + 2 3 13 + 2 3 + 33
+ + ....
6 11 16 1 1+ 3 1+ 3 + 5
1. Find the sum up to n terms 1 + + + + .....
7 49 343 t n = n th term of the series
It is an AGP
13 + 2 3 + 3 3 + ..... + n 3
1 tn =
Here a = 1, d = 5, r = 1 + 3 + 5 + .... + ( 2n − 1)
7
n( n + 1)
2
a - (a + (n - 1)d)r n 1 - r n-1 2
Sn = + dr
2
1-r (1 - r) =
n
[(1 + ( 2n − 1)]
=
1 - (1 + 5(n -
1 − 71
1))( 71 ) n
+5 1
7
() 1 − ( 71 ) n−1
(1 − 71 ) 2
2
(n + 1) 2 1 2
tn = = ( n + 2n + 1)
7n - 5n + 4 5(7 - 1)
n -1 4 4
= +
1
n n n
7 n -1 × 6 7 n -2 × 36
2. Find the sum of the first n terms of the series 1 4 1 ∑
S n = n 2 + 2 n + 1
1
∑
1
∑
1 1 1
+ + + ... 1 n(n + 1)( 2n + 1) 2n(n + 1)
1+ 2 2+ 3 3+ 4 = + + n
4 6 2
k th term of given series n
= ( 2n 2 + 9n + 13)
1 24
tk = .
k + k +1 S17 = 527
1 k − k +1 Compute the sum of first n terms of
= ×
k + k +1 k − k +1 S n = 8 + 88 + 888 + ....... to n terms
k − k +1 S n = 8(1 + 11 + 111 + ....... to n terms)
=
k − k −1 =
8
(9 + 99 + 999 + ....... to n terms)
= k +1 − k 6. 9
8
t1 + t 2 + ... + tn = 2 − 1 + 3 − 2 + ... + n + 1 − n = (10 - 1) + (100 - 1) + (1000 - 1)..... ton terms]
9
= n +1 −1 8
= [(10 + 10 2 + 10 3 + ...... n terms) − n ]
9
It is a G.P with a = 10, r = 10 11. Find the value of n, if the sum to n terms of the
series 3 + 75 + 243 + ...is 435 3
8 10(10 − 1)
n
Sn = − n 3 + 5 3 + 9 3 + ... = 435 3
9 9
It is A.P, Here a = 3 , d = 4 3
7. Compute the sum of first n terms of
S n = 6 + 66 + 666 + ....... to n terms S n = 435 3
S n = 6(1 + 11 + 111 + ....... to n terms) n
[ 2( 3) + (n − 1)4 3 = 435 3
6 2
= (9 + 99 + 999 + ....... to n terms) 2n 2 − n − 435 = 0
9
2
= [(10 - 1) + (100 - 1) + (1000 - 1)..... ton terms] 1 ± 3481
3 n=
4
2
= [(10 + 10 2 + 10 3 + ...... n terms) − n ] n = 15
3
12. In a race, 20 balls are placed in a line at intervals
It is a G.P with a = 10, r = 10 of 4 meters, with the first ball 24 meters away
2 10(10 n − 1) from the starting point. A contestant is required
Sn = − n to bring the balls back to the starting place one
3 9 at a time. How far would the contestant run to
8. Find the general term and sum to n terms of the bring back all balls?
4 7 10 Distance to pick 1st ball = 24 × 2
sequence : 1, , , ,....
3 9 27 2 nd ball = 2( 24 + 4)
It is an AGP,
3 rd ball = 2( 24 + 2 × 4)
a = 1, d = 3, r = 3
48,56,64,.....20 terms
a + ( n − 1)d 3n − 2
Tn = n−1
⇒ Tn = n−1 it is an AP
r 3
1 − r n−1 20
a − [ a + (n − 1)d ]r n
Sn = [ 2( 48) + ( 20 − 1)8]
Sn = + dr 2
2
1− r (1 − r ) = 2480
3n − 3n − 2 3 n −1 − 1 13. The number of bacteria in a certain culture
= + doubles every hour. If there were 30 bacteria
2.3 n−1 8.3 n−3
9. Show that the sum of (m + n)th and (m − n)th term present in the culture originally, how many
of an AP. is equal to twice the mth term. bacteria will be present at the end of 2nd hour,
4th hour and nth hour?
a m = 2a + ( m − 1)d
No. of bacteria present initially = 30
a m +n = a + (m + n − 1)d
No. of bacteria@ end of1st hour = 2(30)
a m−n = a + (m − n − 1)d
a m+ n + a m−n = a + (m + n − 1)d + a + (m − n − 1)d No. of bacteria@ end of 2 nd hour = 2(2(30))
= 2a + d( m + n − 1 + m − n − 1) = 30(2 2 )
= 2 a + 2 d( m − 2 ) = 120
= 2[ a + (m − 1)d ] No. of bacteria@end of 4 th hour t 4 = 30 × 2 4
= 2a m = 480
10. A man repays an amount of Rs.3250 by paying No. of bacteria @ end of every hour : G.P.
Rs.20 in the first month and then increases the common ratio r = 2.
payment by Rs.15 per month. How long will it
No. of bacteria after n hours = t n
take him to clear the amount?
20 + ( 20 + 15) + ( 20 + 2 × 15) + ...... = 3250 General term of G.P. t n = 30 (2 n )
It' s an AP. Hence, S n = 3250 14. What will Rs.500 amounts to in 10 years after its
deposit in a bank which pays annual interest
n
[ 2a + ( n − 1)d ] = 3250 rate of 10% compounded annually?
2
n[ 2 × 20 + (n − 1)15] = 6500 10
interest for one year = 500 = 50
15n 2 + 25n − 6500 = 0 100
3n 2 + 5n − 1300 = 0 principal for 2nd year = principal for 1st year + Interest
= 500 + 500
(n − 20)(3n + 65) = 0 10
100
n = 20
= 5001 +
10 nd INFINITE SEQUENCES AND SERIES
interest for 2 year
100 Fibonacci Sequence
10 10 § A number other than first two terms, is found by
= 5001 +
100 100 adding up two numbers before it.
principal for 3rd year = 5001 + 10 10 10 1, 1, 2, 3, 5, 8, 13,..
+ 5001 +
100 100 100
§ Rule: xn = xn−1 + xn−2, n ≥ 3 with x0 = 1, x1 = 1
2
10
= 5001 + § Every nth number is a multiple of nth term.
100
Infinite Series
n −1
10 § If (an) is infinite sequence of numbers,
principal for n–1year = 5001 +
100 ∞
th
Principal for n year = 5001 +
10
n −1
+ 5001 +
10
n −1
10
a1+a2+….. is called infinite series denoted as ∑a
k =1
k
100 100 100
∞
∑a
n
= 500
11 § For n , sn = a1 + a2 + a3 +….. + an
10 n=1
∑a
10
11 Sequence (sn) is partial sum sequence of
Amount after 10 year = 500 n
10 n=1
15. In a certain town, a viral disease spread in § If (sn) converges and if the series is a convergent
Geometric Progression. Amount of infectious series and s is called the sum of the series.
virus particle gets doubled each day, 5 particles ∞
on first day. Find the day when the infectious § ∑ ( −1)
n=1
n+1
does not converge because partial sum
virus particles just grow over 1,50,000 units?
5,10, 20,40 ,.......n terms ≤ 150000 sequence 1, 0, 1, 0, 1, 0, . . . does not converge.
∞
It is a GP, a = 5, r = 2
S n ≤ 150000
§ ∑x
n=1
n
converges if x = ½ .Not converge if x = 2.
5(2 n − 1) Infinite Geometric Series
≤ 150000
2 −1
§ Σxn is a geometric series .
2 n − 1 ≤ 30000
1-x n
2 ≤2n 15 § If sn = x0 + x1 + x2 + …+ xn, then s n = . As xn
1-x
n ≤ 15 1
tends to 0 if |x| < 1, sn tends to if |x| < 1.
16. Compute the sum of first n terms of 1− x
1 + (1 + 4) + (1 + 4 + 42) + (1 + 4 + 42 +43) + … ∞
§ ∑x
n=1
n
converges for all real number x satisfying
1
|x| < 1, = 1+x + x2 + x3 + …..
1− x
∞
§ ∑ ( −1)
n=1
n
xn converges for real number x
1
satisfying |x| < 1, = 1− x + x2 − x3 + …..
1+ x
∞
§ ∑ ( 2x)
n=1
n
converges for real number x satisfying
1
|x|< 1 = 1 + 2x + 4x2 + 8x3 +……
1 − 2x
∞
§ ∑ ( −1)
n=1
n
x converges only for x = 0.
Infinite Arithmetico-Geometric Series
a dr
S = lim Sn = +
n→∞ 1 - r (1 - r)2
Binomial Theorem for Rational Exponent 1. Expand in ascending powers of x.Find condition
on x for which binomial expansion is valid.
For any rational number n,
1
i)
n(n-1) 2 n(n-1)(n- 2 ) 3
(1 + x) n = 1 + nx + x + x + .. (1 + 3x ) 2
2! 3! 1
1. By taking - x in the place of x = (1 + 3x ) −2
(1 + 3x ) 2
n(n-1) 2 n(n-1)(n-2 ) 3 2( 2 + 1) 2( 2 + 1)( 2 + 2 )
(1 − x) n = 1 − nx + x − x + .. = 1- 2( 3x) + ( 3x) 2 - ( 3x) 3 ..
2! 3! 2! 3!
2. By taking - n in the place of n ,
= 1- 6x + 27 x 2 -108x 3 + 405x 4 -..|x|< 13
n(n + 1) 2 n(n + 1)(n + 2 ) 3
(1 + x) −n = 1 − nx + x − x +. ii)
1
2! 3! 5 + x
3. by taking - x and - n in the places of x and n 1 1
=
n(n + 1) 2 n(n + 1)(n + 2 ) 3 5 + x [5(1 + x )]1
(1 − x) −n = 1 + nx +x + x +.
2! 3! 1
p = [1 + x ] −1
Binomial Theorem For any rational number q , 5
1 x x2 x3
p
p p(p - q) 2 p(p - q)(p - 2q) 3 = [1 − + − + ....] |x| < 5
(1 + x ) q = 1 + x + 2 x + x + ... 5 5 25 125
q q 2! q 3 3! 1
iii)
p
p p(p - q) 2 p(p - q)(p - 2q) 3 (3 + 2 x ) 2
(1 − x ) q = 1 − x + 2 x − x + ...
q q 2! q 3 3! 1 1
=
Binomial Theorem Special Cases (3 + 2 x ) [3(1 + 23 x )]2
2
(1 + x)−1 = 1− x + x2 − x3 + …. 1
= [1 + 23 x ]-2
9
(1 − x)−1 = 1+x + x2 + x3 + …. 1
= [1 − 2( 2 x ) + 3( 2 x ) 2 − 4( 2 x )3 + 5( 2 x ) 4 + ..]
(1 − x)−2 = 1 + 2x + 3x2 + 4x3 + 5x4 + 6x5 + …. 9 3 3 3 3
1 4
(1 + x)−2 = 1− 2x + 3x2 − 4x3 + 5x4 − 6x5 + …. = − x + 4 x 2 − 32 x 3 + 80 x 4 ...
9 27 27 243 729
∞ n 2
Exponential Series : ∑ xn!
n=1
iv)
(3 + 4 x ) 2
2 2 3
e =1 +
1 1 1
+ + + ... = [1 + 43 x ] - 2 |x|<
1! 2! 3! (3 + 4x ) 2 9 4
1
1 1 1
e −1 = 1 − + − + ... = [1 − 2( 4 x ) + 3( 4 x ) 2 − 4( 4 x ) 3 + 5( 4 x ) 4 + ..]
1! 2! 3! 9 3 3 3 3
e + e −1 e − e −1 1 1 1
2
1 1 1
= 1 + + + ,.... = + + .. iii) (1 + x ) 3
2 2! 4! 6! 2 1! 3! 5! − 1) 2 23 ( 23 − 1)( 23 − 2) 3
2 2
3(3
2
x x 2
x 3 (1 + x ) 3 = 1 + 23 x + x + x + ..
ex =1+ + + + ... 2! 2!
1! 2! 3!
= 1 + 2 x − 1 x 2 + 4 x3
x x2 x3 3 9 81
e −x = 1 − + − + ... 2
1! 2! 3! iv) (5 + x 2 ) 3
e x + e −x x2 x4 e x − e −x x x 3 2 2
=1+ + + ..., = + + .... (5 + x 2 ) 3 = [5(1 + x5 ) 3 ]
2
2 2! 4! 2 1! 3!
∞
xn
2
(1 − 23 ) x 2 2
= 5 3 [1 + 2 ( x5 ) − 3
2
∑ ( 5 ) + ..]
2
Logarithmic Series : ( −1) n +1 3 2!
n=1
n
1 4
= 5 3 [1 + 2 x 2 −
2
x + ..] x 2 < 5
x2 x3 x4 15 225
log (1 + x) = x - + − + ... − 23
2 3 4 v) ( x + 2)
2
x x x4
3
( x + 2)
− 23
= [ 2(1 + 2x )]
− 23
log (1 − x) = − x - − − − ...
2 3 4
3 (1 + 3 ) x 2 (1+ 32 )( 2 + 32 ) x 3
2 2 2
−2
1+ x 3 5
= 2 3 [1− 32 ( 2x ) + −3
log
x x (2) (2)
= 2 x + + + ... 2! 3!
1 − x 3 5 − 23
= 2 [1 − + x 5 2
36 x − 815 x 3 + ....
4x 8x 16x 4
2 3 3
log(1 + 2x ) = 2x - + − + ... 3
2 3 4 2. Find 3
1001 approximately (2decimal places).
1− x
1
3
1001 = (1000 + 1) 3 x2
6. Prove that =1− x + if x is very small.
1 1+ x 2
= 1000 1 +
1 3
1
3
1− x
1 1
−
1000 = (1 − x ) 2 (1 + x ) 2
1+ x
1 1 ( − 1 1 2
1 1
= 101 + × +3 3 × + ..
1
1 ( − 1) 2
1 1
3 1000 2! 1000 (1 − x ) 2 = 1 − x + 2 2 x + ..
2 2!
≈ 10 + 0.0033
1 1 2
= 1 − x − x + ..
3
10001 = 10.0033 2 8
3. Find 3 65 approximately(2decimal places) −
1
1 − 1 ( − 1 − 1) 2
1 (1 + x ) 2 = 1 − x + 2 2 x + ..
3
65 = (64 + 1) 3 2 2!
1
1 3
= 64 3 1 +
1
1 3 = 1 − x + x 2 + ..
64 2 8
1 1
1 1 1 (1 − 1 1 2
− x x2 x 3x 2
(1 − x ) 2 (1 + x ) 2 =1−
− + ..][1 − + + ..]
= 41 + × + 3 3 × + .. 2 8 2 8
3 64 2! 64 x x x 2 x 2 3x 2
1 1 =1− − + − +
=4+ − ≈ 4 + 0.02 2 2 4 8 8
48 36864 + ...
2x − x + 3x
2 2 2
3
65 = 4.02 =1− x +
8
1
4. Prove that 3 x 3 + 7 − 3 x 3 + 4 ≅ 2 if x is large. x 2
x =1− x +
1
2
3
x 3 + 7 = [ x 3 (1 + x73 )]3 7. Write the first 6 terms of the exponential series
2
− 1) 7 2 5x ( 5x ) ( 5x ) 3
e 5x = 1 + + + + ...
1 1
3(3
= x(1 + 13 x73 + [ x 3 ] + .....) 1! 2! 3!
2! i)
25x 2 125x 3 625x 4
=x+
7
−
49
+ .... e 5 x = 1 + 5x + + + + ...
3x 2 9x 6 2 6 24
2
1
2x ( 2x ) ( 2x ) 3
x 3 + 4 = [ x 3 (1 + x43 )]3 e −2 x = 1 − + − + ...
3
ii) 1! 2! 3!
1 1
3(3 − 1) 4 2 4x 3 2 x 4
= x(1 + 13 x43 + [ x 3 ] + .....) e −2 x = 1 − 2x + 2x 2 − + + ...
2! 3 3
4 16
=x+ − + .... 1
1
x ( 1 x )2 ( 12 x )3
e2 = 1 + 2 + 2 + + ...
x
3x 2 9x 6 1! 2! 3!
iii)
3 x 3 + 7 − 3 x 3 + 4 = ( x + 7 − 49 + ..) − ( x + 4 − 16 + .)
1 1 x3 x 4
3x 2 9x 6 3x 2 9x 6 e −2 x =1 + x + x 2 + + + ...
2 8 48 384
1
≅ 2 8. Write the first 4 terms of the logarithmic series
x Find intervals on which expansions are valid.
1
5. Prove that 3 x 3 + 6 − 3 x 3 + 3 ≅ 2 if x is large. ( 4x ) 2 ( 4x ) 3 ( 4x ) 4
x log (1 + 4x) = 4x - + − + ...
1 i) 2 3 4
3
x 3 + 6 = [ x 3 (1 + x63 )]3 64 64
= 4x − 8x 2 + x 3 − x 4 |x|< 14
1 1
3(3 − 1) 6 2 3 3
= x(1 + 13 x63 + [ x 3 ] + .....) ( 2 x ) 2 ( 2x ) 3 ( 2 x ) 4
2! log (1 − 2x) = − 2x - − − − ..
2 4 2 3 4
=x+ − + .... ii)
x 2 x6 4x 2 8 3 16x 4
= −2x − − x − |x|< 12
3
x 3 + 3 = [ x 3 (1 + x33 )]3
1
2 3 4
1 + 3x (3x ) 3 (3x ) 5
1 1
3(3 − 1) 3 2 log = 23x + + + ...
= x(1 + 13 x33 + [ x 3 ] + .....) 1 − 3x 3 5
2! iii)
3 5
1 1 27 x 243x
=x+
− + .... = 2[3x + + + ..], |x|< 13
x 2 x6 3 5
2 4 1 1 1 − 2x ( 2x ) 3 ( 2 x ) 5
3 x 3 + 6 − 3 x 3 + 3 = ( x + 2 − 6 + .) − ( x + 2 − 6 + .) log = − 2 2 x + + + ...
x x x x 1 + 2x 3 5
iv)
1
≅ 2 8x 3
32x 5
x = −2[3x + + + ..] |x|< 12
3 5
x 2 x3 x 4
9. If y = x + + + + .... then show that
2 3 4
y 2 y3 y 4
x=y− + − + ....
2! 3! 4!
10. If p − q is small compared to either p or q, show
p (n + 1)p + (n-1)q 15
that n ≈ Hence find 8
q (n-1)p + (n + 1)q . 16
3- 4 x + x 2
11. Find coeff.of x4 in the expansion of
e 2x
∞
1 1 1
12. Find the value of
n=1
∑2n − 1
n −1 + 2n−1
9 9
6.1. LOCUS OF A POINT (ii) three units from the y-axis.
Locus P ( h, k ) = point on the locus.
Path traced out by a moving point under certain k = 2 Þ y = 2,
conditions is called the locus of that point. h =3 Þ x=3
When a point moves in accordance with a 3. Find path traced out by the point (ct , ct ) , t is
geometrical law, its path is called locus.
parameter and c is a constant
The plural of locus is loci.
Let P (h, k) = a point on locus.
Important loci in mathematics
c
A moving point P Graph Name of the path h = ct , k =
c
under given c
condition ( h ) ( k ) = (ct )
c
A point P Perpendicular hk = c 2
moves such that bisector of the line Locus is xy = c 2
it is equidistant segment AB
from two fixed 4. Find the locus of a point P moves such that its
points A and B distances from two fixed points A(1, 0) and B (5,
0) , are always equal.
A point P Angle bisector of Given Points : A(1, 0) and B (5, 0)
moves such that the angle∠xoy Point on required path = P (h, k)
it is equidistant
AP 2 = BP 2
from two fixed
lines ox and oy ( h - 1) 2 + ( k - 0) 2 = ( h - 5) 2 + ( k - 0) 2 .
h=3
A point P Circle Locus of P, x = 3, a straight
moves equi
5. Find the equation of the locus of a point such
distant from a
that the sum of the squares of the distance from
fixed point O
the points (3, 5), (1,−1) is equal to 20
Procedure to find equation of locus of a point
Let A(3, 5), B(1,-1), P (h, k) be 3 points :
i) For locus of point P, assign coordinates, (h, k) to P
AP 2 + BP 2 = 20
ii) Express given conditions as equations in terms of
( h - 3) 2 + ( k - 5) 2 + ( h - 1) 2 + ( k + 1) 2 = 20 .
the known quantities and unknown parameters.
h 2 + y 2 - 4h- 4k + 8 = 0
iii) Eliminate the parameters, so that the resulting
equation contains only h, k and known quantities. x 2 + y 2 - 4x - 4 y + 8 = 0
iv) Replace h by x, and k by y, in resulting equation. 6. Find equation of the locus of the point P such
Solved Problems that the line segment AB, joining the points
A(1,−6) and B(4,−2), subtends a right angle at P.
1. Find the locus of a point which moves such that
Let A be(1,-6), B(4,-2) P (h, k) :
its distance from the x-axis is equal to the
distance from the y-axis. AP 2 + BP 2 = AB 2
( h - 1) 2 + ( k + 6) 2 + ( h - 4) 2 + ( k + 2) 2 = ( 4 - 1) + (-2 + 6) 2
2
P ( h, k ) = point on the locus.
A, B = foot of ^ rs drawn fromP on x, y - axes h 2 + y 2 - 5h + 8k + 16 = 0
P = (OA, OB) = (BP, AP) x 2 + y 2 - 5x + 8y + 16 = 0
= ( h, k ) 7. The sum of the distance of a moving point from
AP = BP (Given) the points (4, 0) and (−4, 0) is always 10 units.
k = h Find equation of the locus of the moving point.
Sub. h = x and k = y Let P be(h, k), A(4,0) ; B(-4,0)
Locus y = x , line thro' origin AP + BP = 10 (given)
AP = 10 – BP
2. Find locus of a point P that moves at a distant of
(i) two units from the x-axis AP 2 = (10 – BP) 2
AP 2 = 100 – 20BP + BP 2 11. If θ is a parameter, find the equation of the
locus of a moving point, whose coordinates are x
(h – 4 ) 2 + k 2 = 100– 20BP + (h + 4 ) 2 + k 2
= a cos3 θ, y = a sin3 θ.
h 2 – 8h + 16 + k 2 = 100 – 20BP + h 2 + 8h + 16 + k 2
Let P (h, k) = any point on rqd path.
5BP = 25 + 4h
2 2
h = a cos 3 q, k = a sin 3 q
25BP = ( 25 + 4h) 1 1
2
25 (h + 4 ) + k 2
= 625 + 200h + 16h 2 æ h ö 3 = cos q, æ k ö 3 = sin q
ç ÷ ç ÷
èaø èaø
25h 2 + 8h + 16 + k 2 = 625 + 200h + 16h 2
squaring and adding,
9x 2 + 25y 2 = 225 2 2
2 æ h ö 3 + æ k ö 3 = sin 2 q + cos 2 q
x y
2 ç ÷ ç ÷
+ =1 è aø è aø
25 9 2 2
8. If θ is a parameter, find eqn of locus of a æ h ö3 + æ k ö 3 = 1
ç ÷ ç ÷
è aø è aø
moving point, (a sec θ, b tan θ).
2 2 2
Let P (h, k) = any point on rqd path. x3 + y 3 = a3
h = a sec θ, k = b tan θ 12. Coordinates of a moving point P are [ 2a (cosecθ
h k sinθ) , b
(cosecθ − sin θ) ] where θ is a variable
= secq , = tan q 2
a b
squaring and subtracting, parameter. Show that the equation of the locus
2 2 P is b2x2 − a2y2 = a2b2 .
æ h ö - æ k ö = sec 2 q - tan 2 q
ç ÷ ç ÷ Let P be (h, k)
èaø èbø
2 2 a b
æhö -ækö =1 (cos ecq + sin q) = h (cos ecq + sin q) = k
ç ÷ ç ÷ 2 2
èaø èbø 2 2
2h 2k
x2 y
2
(cos ecq + sin q) 2 = æç ö÷ (cos ecq + sin q) 2 = æç ö÷
- =1 è a ø è b ø
a2 b2
4h 2 4k 2
9. Find the locus of P, if for all values of α, the co- cos ec 2 q + sin 2 q + 2 = 2 cos ec 2 q + sin 2 q - 2 = 2
a b
ordinates of a moving point P is (9cosα , 9 sinα) On Subtracting,
Let P (h, k) = any point on rqd path. 4h 2 4k 2
- 2 =4
h = 9 cos a, k = 9 sin a a2 b
2
h k 2
x - y =1
= cos a, = sin a
9 9 a2 b2
squaring and adding,
2 2
13. If θ is a parameter, find the equation of the
æç h ö÷ + æç k ö÷ = sin 2 a + cos 2 a locus of a moving point, whose coordinates are
è9ø è9ø (a(θ − sin θ), a(1 − cos θ)).
2 2
æç h ö÷ + æç k ö÷ = 1 P (h, k) be any point on the requiredpath.
è9ø è9ø
k = a(1 - cos q) ....(1)
x + y2 = 9
2
a-k
cosq =
10. θ is a parameter, find eqn of locus of moving a
point, whose coordinates are (9cosα , 6 sinα). a-k ö
q = cos -1 æç ÷
Let P (h, k) = any point on rqd path. è a ø
h = a( q - sin q) ....(2)
h = 9 cos a, k = 6 sin a
h k é a-k ö 2 ak-k 2 ù
= cos a, = sin a h = aêcos -1 æç ÷- ú ....(1)
9 6 êë è a ø a úû
2 2
æ h ö + æ k ö = sin 2 a + cos 2 a The locus of (h, k)
ç ÷ ç ÷
è9ø è6ø
æ a-y ö
2 2 x = a cos -1 ç ÷ - 2 ay-y
2
æç h ö÷ + æç k ö÷ = 1 è a ø
è9ø è6ø
2 14. A straight rod of the length 6 units, slides with
x2 y its ends A and B always on the x and y axes
+ =1
81 36
respectively. If O is the origin, then find the Let R be (a, b), O(0,0), P(h, k) :
locus of the centroid of ΔOAB midpoint of line segment OR = P(h, k)
coordinates of O, A, B = (0, 0) , (a, 0), (0, b) æç a + 0 , b + 0 ö÷ = P(h, k)
Centroid of DOAB. = (h, k) è 2 2 ø
æç 0 + a + 0 , 0 + 0 + b ö÷ = (h, k). a = 2h, b = 2k
è 3 3 ø R is a point on y 2 = 4x
a = 3h, b = 3k b 2 = 4a
2 2 2
In rt DOAB, OA + OB = AB 4 k 2 = 8h
( 3h) 2 + ( 3k)2 = ( 6 )2 i.e y 2 = 2x
2 2
h + k = 4
18. If P(2,−7) is a given point and Q is a point on 2x2
Locus of (h, k) , x 2 + y 2 = 4, a circle + 9y2 = 18, then find the equations of the locus of
15. A straight rod of length 8 units slides with its the mid-point of PQ.
ends A and B always on x and y axes. Find the Let P be (2,-7), Q(a, b) :
locus of the mid point of the line segmentAB mid - point of PQ = M (h, k)
æ 2 + a , - 7 + b ö = (h , k )
ç ÷
è 2 2 ø
a = 2h - 2 , b = 2k + 7
Q is point on 2x 2 + 9y 2 = 18
2 a 2 + 9b 2 = 18
2( 2h - 2) 2 + 9( 2 k + 7) 2 = 18
Let A be (a,0), B(b,0) P(h, k) 8h 2 - 16h + 8 + 36 k 2 + 72k + 441 = 18
Midpoint of AB = P(h, k) 8x 2 + 36 y 2 - 16y + 72y + 431 = 0
æ a + 0 , b + 0 ö = ( h, k) 19. If R is any point on x-axis and Q is any point on
ç ÷
è 2 2 ø
the y-axis and P is a variable point on RQ with
a = 2h , b = 2 k
RP = b, PQ = a. then find equation of locus of P.
OA 2 + OB 2 = AB 2
( a - 0) 2 + ( b - 0) 2 = 8 2
a 2 + b 2 = 64
( 2h ) 2 + ( 2k ) 2 = 64
Or x 2 + y 2 = 16
16. Find the value of k and b, if the points P(−3, 1) P(h , k ) any point on locus
and Q(2,b) lie on the locus of x2 − 5x + ky = 0. AP BP
In DPAR, sinq = In DPAR, cos q =
2 PR QP
(–3,1) lies on locus x – 5x + ky = 0
k h
(-3)2 - 5(-3) + k(1) = 0 sin q = ....(1) cos q = ....( 2)
b a
9 + 15 + k = 0 squaring and adding
k = - 24 k2 h2 x2 y
2
sin 2 q + cos 2 q = + Þ + =1
Equation of locus, x 2 –5x - 24 y = 0 b 2 a2 a2 b 2
Q(2, b) l ies on locus x 2 –5x - 24 y = 0 20. Find the points on the locus of points that are 3
2
2 – 5(2) – 24b = 0 units from x-axis and 5 units from point (5, 1).
- 24b = 6
b = -1
4
17. If O is origin and R is a variable point on y2 = 4x,
then find the equation of the locus of the mid-
point of the line segment OR.
Let Q be (h, k) , P be(5,1) 6.2. Straight Lines
If Q is above line, Q is (h,3) Equations of straight line
2 § Linear equations can be rewritten using the laws
PQ = 25
(5 - h ) + (1 - 3) 2 = 25
2 of elementary algebra into several different forms
referred to as “equations of straight line.”
(5 - h ) = ±21
§ General form the linear equation
h = 5 ± 21
ax + by + c = 0
Rqd points ( 5 + 21,5 - 21)
§ Set of solutions forms a straight line in the plane
If Q is below line, Q is (h,-3)
Angle of inclination and slope
PQ 2 = 25
(5 - h )2 + (1 + 3)2 = 25
(5 - h) = ± 9
h = 2 or h = 8
Rqd points (2,-3) or (8,-3)
21. If the points P(6, 2) and Q(−2, 1) and R are the § Angle of inclination of a straight line is the angle,
vertices of a ΔPQR and R is the point on the θ, made by tline with the x-axis measured in the
locus y = x2 − 3x + 4, then find the equation of counter clockwise (positive) direction.
the locus of centroid of ΔPQR § Slope or gradient of a straight line is a number
Vertices of Dle = P(6, 2) , Q(-2, 1)R(a, b) that measures its “direction and steepness”.
Centroid of Dle = (h, k) § Slope of a line in the plane containing the x and y
æç 6 - 2 + a , 2 + 1 + b ö÷ = ( h , k ) axes represented by the letter m.
è 3 3 ø Measurement of Slope
a = 3h - 4, b = 3k - 3
( a , b )lieson y = x 2 -3x + 4
b = a 2 -3a + 4
3k - 3 = (3h - 4) 2 - 3(3h - 4) + 4
9h 2 + 33h - 3k + 35 = 0
Replacing h , k by x , y
9x 2 + 33x - 3y + 35 = 0 § If θ is the angle of inclination of the line with x-
22. If Q is a point on the locus of x2 + y2 + 4x − 3y + 7 axis in counter clockwise direction
= 0, then find the equation of locus of P which slope m = tan q
divides segment OQ externally in the ratio § If (x1, y1) and (x2, y2) are two points on line
3:4,where O is origin.
vertical change
Let Q be (a, b), O (0,0) , P (h, k) : Slope =
horizontal change
Pdivides OQ externally in 3 : 4, y 2 - y1
m=
æç 3a - 0 , 3b - 0 ö÷ = P(h , k ) x 2 - x1
è 3- 4 3- 4 ø § For general form of linear equation ax + by+c = 0
-h -k a
a= ,b= m=-
3 3 b
2 2
Q lies on x + y + 4 x-3y + 7 = 0 Types of Slope
a 2 + b 2 + 4 a-3b + 7 = 0
2 2
æ - h ö÷ + æç - k ö÷ + 4æç - h ö÷-3æç - k ö÷ + 7 = 0
ç
è 3 ø è 3 ø è 3 ø è 3 ø
Replacing by (x, y) :
x 2 + y 2 - 12x + 9y + 63 = 0
Collinearity
§ In a plane three or more points are said to be 1. Find the slope of the straight line passing
collinear if they lie on a same straight line. through the points (5,7) and (7,5). Also find the
angle of inclination of the line with the x-axis.
§ Let A,Band C be any three points on a plane.
( x1 , y1 ), ( x 2 , y 2 ) = (5,7) and (7,5)
Slope of AB = Slope of BC (or AC), 5-7
Slope m =
Intercepts of a Line 7-5
=-1
tan q = - 1
q = tan -1 (-1)
3p
= or 135.
4
§ The intercept of a line is the point at which the 2. Find equation of a straight line cutting an
line crosses either the x-axis or the y-axis. intercept of 5 from negative direction of the y-
§ x-intercept is a point where y =0, axis and is inclined at an angle 150◦ to the x-axis
Negative y intercept = 5
y-intercept is a point where the x = 0.
i.e., b = - 5 and q = 150.
§ Intercepts of a line are points where line
intersects, or crosses, horizontal and vertical axes. slope m = tan 150 o
§ Equation of the y-axis is x = 0. = tan(18o. - 30 o )
Equation of the x-axis is y = 0. = - tan 30 o
Different Forms of an equation of a straight line 1
=-
Information Equation of Graph 3
given straight lines Slope - intercept form :
Slope(m) y- y = mx + b y = mx + b.
intercept b 1
y=- +5
3
x + 3y + 5 3 = 0
Slope (m) y - y1 = m(x - x1 ) 3. Show the points (0 , -23 ) , (1,−1), ( 2, -21 ) are collinear.
point (x1, y1)
Let A (0 , -23 ) , B(1,−1), C ( 2, -21 )
- 1 + 32
slope of AB =
points(x1, y1) y - y1 x - x1 1-0
=
(x2, y2) y 2 - y 1 x 2 - x1 1
=
2
- 12 + 1
slope of BC =
2-1
1
=
2
x-intercept a x + y =1 slope of AB = slope of BC
y-intercept b a b \A, B and C are collinear
4. Show that points (1, 3), (2, 1) and ( 12 , 4) are
collinear, by using
Normal x cos a + y sin a = p Let A(1, 3), B(2, 1) C ( 12 ,4) be the points
length p , (i) concept of slope
angle a 1-3
slope of AB =
2-1
= -2
4 -1
Parametric y - y1 x - x1 slope of BC = 1
parameter-r
= =r 2 - 2
sin q cos q
= -2
slope of AB = slope of BC
\A, B and C are collinear
ii) using straight line
Equation of A(1, 3), B(2, 1) : ntercepts cut off from axes = a, b
x - x1 y - y1 a+b=1
=
y 2 - y1 y 2 - y1 a =1 - b
x -1 y - 3 Intercept form of eqn :
=
1 -2 x y
- 2x + 2 = y - 3 + =1
1- b b
2x + y - 5 = 0 it passes through (8,3) :
1 8 3
C( , 4) must satisfy AB : + =1
2 1- b b
1 8b + 3 - 3b = b - b 2
= 2æç ö÷ + 4 - 5
è 2ø
b 2 + 4b + 3 = 0
=0
b = - 1, - 3
Hence they are collinear
a = 2, 4
iii) any other method
Area of Dle = 0, if 3 points collinear Equation of the line :
ì x1 x 2 x 3 x1 ü 3x - 4 y = 12 , x - 2y = 2
í ý=0 8. Find equation of line through (1, 5) and also
îy1 y 2 y 3 y 1 þ
divides the co-ordinate axes in the ratio 3:10.
5. Find equations of the straight lines, making the Intercepts = a, b
y- intercept of 7 and angle between the line and
the y-axis is 30◦ a 3
= Þ a = 103 b
Two straight lines making 30◦ with the y-axis. b 10
two lines make 60◦ and 120◦ with the x-axis Intercept form of eqn :
b=7 x y
3
+ =1
Slope m1 = tan 60 o 10 b b
it passes through (1,5) :
= 3
1 5
Eqn of line, y = m1 x + b 3
+ =1
10 b b
y = 3x + 7 10b + 15b = 3b 2
o
Slope m2 = tan 120 3b 2 - 25b = 0
= tan(180 o - 60 o ) b = 0 , 253 Þ a = 0, 52
= - tan 60. Equation of line10x + 3y = 25
=- 3 9. Find equation of line, if perpendicular from
Eqn of line, y = m2 x + b origin makes 30◦ with x-axis and its length is 12.
y = - 3x + 7 Here, p = 12 and a = 30 o
6. Find equation of straight line passing through (− x cos a + y sin a = p
1, 1) and cutting off equal intercepts, but x cos 30 o + y sin 30 o = 12
opposite in signs with two coordinate axes
3 1
ntercepts cut off from axes = a, - a x + y = 12
2 2
Intercept form of eqn : 3x + y = 24
x y 10. Length of perpendicular drawn from the origin
+ =1
a -a to a line is 12 and makes an angle 150◦ with
x-y = a positive direction of x-axis. Find the equation of
it passes through (-1, 1) : the line Area of the triangle formed by a line
. (-1) - (1) = a with the coordinate axes, is 36 square units.
Here, p = 12 and a = 150.
a = - 2.
x cos a + y sin a = p
Equation of the line :
x - y + 2 = 0. x cos 150 o + y sin 150 o = 12
7. Find the equation of straight lines passing x cos 30 o - y sin 30 o = 12
through (8, 3) having intercepts whose sum is 1 3x - y + 24 = 0
11. Straight line L with negative slope passes length of ^ r from origin = p
through the point (9, 4) cuts the positive
a = 45 o
coordinate axes at the points P and Q. As L
varies, find the minimum value of Equation of line :
|OP|+|OQ|, where O is the origin. x cos a + y sin a = p
Slope of the line L = m x cos 45 + y sin 45 = p
It passes through (9, 4) x + y = 2p
Equation of line L, y - 4 = m(x - 9).
It cuts coordinate axes at : A( 2p, 0) B(0, 2p )
4
Points P and Q are : (9 - , 0) and (0, 4 - 9m) . Area of the ÄOAB = 36
m
4 1
OP + OQ = 9 - +|4 - 9m| ´ 2p ´ 2p = 36
m 2
m < 0, take m = - k, k > 0 p = 6 ( p is positive)
4 Rqd line x + y = 6 2
= 9 + +|4 + 9k|
k 14. Find equation of line
4
= (9 + ) + (4 + 9k) i) passing through (1,1) with y-intercept (−4)
k x y
4 Intercept Form, + = 1
= (4 + 9) + ( + 9K) a b
k y - intercept , b = -4
4 Passes thro : (1,1)
³ 13 + 2 K
K
1 1 4
Arithmetic mean ³ Geometric mean + =1 Þ a =
a -4 5
Min. absolute value of|OP|+|OQ|= 25. x y
Rqd Eqn, + =1
12. Find equation of lines make 60◦ with positive x- 4 /5 - 4
axis and at a distance 5 √ 2 units measured from 5x - y - 4 = 0
the point (4, 7), along the line x − y+3 = 0. ii) passing through point (1,1) with slope 3
Slope of x - y + 3 = 0 Slope point Form, ( y - y1 ) = m( x - x1 )
m=1 Slope , m = 3
q = tan -1 (1) Passes thro : (1,1)
Angle of inclination = 45., ( y - 1) = 3( x - 1)
Point on the line : (4, 7). Rqd Eqn, 3x - y - 2 = 0
Using parametric form : iii) passing through point (1,1) and (-2, 3)
x - x1 y - y 1 y - y1 x - x1
= =r 2 point Form, =
cosq sin q y 2 - y1 x 2 - x1
x- 4 y- 7 Passes thro : (1,1) & (-2,3)
1
= 1 =± 5 2
2 2
y -1 x -1
=
x - 4 = y - 7 = ± 5. 3 -1 - 2 -1
Rqd Eqn, 2x + 3y + 1 = 0
Points at a distance 5 2 ü
ý = (4 + 5,7 + 5), (4 - 5, 7 - 5) iv) through point (1,1) perpendicular from the
unitseither side of (4, 7) þ origin makes an angle 60◦ with x- axis.
Points on lines are : (9, 12) and (-1, 2) Slope point Form, ( y - y1 ) = m( x - x1 )
given slope = tan 60. Slope = tan 60 o Þm= 3
m = 3. Passes thro : (1,1)
Rqd equations : Eqn of perpen.line ( y - 1) = 3( x - 1)
3x - y + (12 - 9 3 ) = 0 3x - y - ( 3 - 1) = 0
3x - y + (2 + 3) = 0 Rqd line form, x + 3y + k = 0
13. Area of the triangle formed by a line with the Passes thro : (1,1)
coordinate axes, is 36 square units. Find
equation of line if the perpendicular drawn 1+ 3 + k = 0 Þ k = -(1 + 3)
from the origin to the line makes an angle of 45◦ Eqn of Rqd line , x + 3y = 1 + 3
with positive the x-axis.
15. If P(r, c) is mid point of a line segment between i) Slope and Intercept form
x y 3x - y + 4 = 0
the axes, then show that + = 2
r c
y = 3x + 4
Mid point = (r , c )
Comp. with y = mx + b
æ a + 0 , b + 0 ö = (r , c )
ç ÷ Slope = 3 and y - intercept = 4
è 2 2 ø
a = 2r , b = 2c ii) Intercept form
Eqn in intercept form : 3x - y + 4 = 0
x y x y 3x - y = - 4
+ =1Þ + = 2
2r 2c r c x y
16. If p is length of perpendicular from origin to the + =1
4 4
line whose intercepts on the axes are a and b, -
3
1 1 1
then show that 2 = 2 + 2 x y
p a b Comp. with + = 1
a b
x y 4
Eqn of line, + = 1 x - intercept = - , y - intercept = 4
a b 3
x y iii) Normal form
+ -1= 0
a b 3x - y + 4 = 0
Distance b/w point ü ax1 + by1 + c
ý p= (- 3 )x + y = 4
& line, þ a2 + b 2 Comp. Ax + By + C = 0.
1 1
Sub, ( x1 , y 1 ) = ( 0 ,0), a = , b = , c = -1 A = - 3 , and B = 1,
a b
Thus, p = 1 12 1 A 2 + B2 = 2
a2
+ b2 3x y
Dividing by 2, - + = 2
1 1 1 2 2
2
= 2 + 2
p a b Comp. x cos a + y sin a = p
17. A straight line is passing through the point A(1, - 3 1
cos a = , sin a =
2) with slope 125 . Find points on the line which 2 2
are 13 units away from A. 5p
a = 150 0 = ,p= 2
6
5p 5p
x cos + y sin = 2
6 6
19. Rewrite √3x + y+ 4 = 0 in to normal form
3x + y + 4 = 0
p is always positive :
Passes thro : (1,2), Slope , m = 5
12 Thus, - 3x - y = 4
5
Slope point Form, ( y - 2) = 12 ( x - 1) Comp. Ax + By + C = 0.
2
( y - 2) 2 = 5 ( x - 1) .....(1) A 2 + B2 = 2
12
Distance b/w P( x , y ) and A(1,2) = 13 3x y
Dividing by 2, - - = 2
2 2
(x - 1) 2 + ( y - 2) 2 = 13 2 .....( 2) Comp. x cos a + y sin a = p
25
From (1), ( x - 1) 2 + ( x - 1) 2 = 169 - 3 -1
144 cos a = , sin a =
2 2
25 ù
( x - 1) 2 é1 + = 169 0 7p
êë 144 úû a = 210 = ,p= 2
6
( x - 1) 2 = 144 7p 7p
x cos + y sin = 2
x - 1 = ±12 6 6
x = 13,-11 20. Pamban Sea Bridge is of length 2065 m
Rameswaram to Mandapam,. Bridge is restricted
Sub in (2),y = 7 , -3
to a uniform speed of only 12.5 m/s. If a train of
Rqd points : (13,7) & (-11,-3) length 560m starts at the entry point of the
18. Express the equation √3x −y+4 = 0 in bridge from Mandapam, find
i) equation of the motion of the train. iii) relation between Slope-common difference
x and y axis = time and distance Slope of line = - 3
negative y - intercept, b = 560m = common difference
slope of motion of train m = 12.5 m/s 23. Quantity demanded of a certain type of
Eqn of line y = mx + b Compact Disk is 22,000 units when a unit price
is Rs 8. Customer will not buy disk, at a unit
y = 12 5x - 56 0
price of Rs 30 or higher. Manufacturer will not
ii) When does the engine touch island market disk if the price is Rs 6 or lower. If price
y = 2065 and b = 0 Rs14 manufacturer can supply 24,000 units. Find
2065 = 12.5x i) demand equation
x = 165.2 seconds. x & y axes represent = No. of units & price
iii) when does the last coach cross the entry let ( x1 , y1 ) ( x 2 , y 2 ) = (22,8)and (0,30)
point of the bridge y - 8 x - 22
When y = 0, x = ? Eqn. of line =
30 - 8 0 - 22
0 = 12.5x - 560 demand function y D = -x + 30
x = 44.8 seconds. ii) supply equation
iv) what is time taken by train to cross bridge ( x1 , y1 ) ( x 2 , y 2 ) = (0,6) and (24,14)
When y = 2065, x = ?
y-6 x-0
2065 = 12.5x - 560 Eqn of line =
14 - 6 24 - 0
x = 210 seconds. 1
21. A 150m long train is moving with constant supply function y S = x + 6
3
velocity of 12.5 m/s. Find iii) market equilibrium quantity and price.
i) the equation of the motion of the train, At market equilibrium :
x and y axis = time and distance demand = supply
negative y - intercept, b = 150m yD = yS
slope of motion of train m = 12.5 m/s 1
-x + 30 = x + 6
Eqn of line y = mx + b 3
y = 12 5x - 150 x = 18 and y = 12.
ii) time taken to cross a pole. Market equilibrium price = Rs12
When y = 150, b = 0, x = ? No. of quantity = 18,000 units.
150 = 12.5x - 0 iv) Demand and supply when price is Rs 10.
x = 44.8 seconds. when y = 10 , x = ?
iii) time taken to cross the bridge of length 850m By demand function - x + 30 = 10
When y = 850, x = ? x = 20.
850 = 12.5x - 150 Demand = 20,000 units.
x = 80 seconds. 1
22. The seventh term of an arithmetic progression is By supply function, x + 6 = 10
3
30 and tenth term is 21. Find
x = 12.
i) first three terms of an A.P.
x & y - axes denote = No.of term & value of term Supply = 12,000 units.
(x 1 , y 1 ) and (x 2 , y 2 ) = (7,30) and (10,21) 24. A spring was hung in the ceiling. Different
weights were attached to the spring to make it
y - y1 x - x1
Eqn of line = stretch, and the total length measured each time
y 2 - y 1 x 2 - x1 Weight, (kg) 2 4 5 8
y - 30 x - 7 Length, (cm) 3 4 4.5 6
=
21 - 30 10 - 7 i) Draw a graph showing the results.
y = - 3x + 51
Sub x = 1, 2 and 3
first 3 terms of AP = 48, 45, 42
ii) Which term of the A.P. is zero
- 3x + 51 = 0
x = 17.
17th term of A.P. = 0
ii) Find the equation relating the length of the iii) Slopes of escalator at the turning points.
spring to the weight on it. 9
Eqn of motion y = x + 0
Eqn of line Passes thro : (2,3) & (4,4) 40
y-3 x -2 Compare with y = mx + c
=
4-3 4-2 9
slope m =
Rqd Eqn, x - 2 y + 4 = 0 40
27. Normal boiling point of water is 100◦C or 212◦F
x = weight, y = Length
and freezing point of water is 0◦C or 32◦F. Find
iii) What is the actual length of the spring.
i) linear relationship between C and F
If x = 0 , y = ?
Denote C and F as points (100, 212) & (0 ,32)
(0) - 2y + 4 = 0 Þ y = 2 cm
F - 212 C - 100
iv) If the spring has to stretch to 9 cm long, how Eqn of line =
32 - 232 0 - 100
much weight should be added? 9
If y = 9, x = ? F = C + 32
5
x - 2( 9) + 4 = 0 Þ x = 14kg ii) value of C for 98.6◦F
v) How long will the spring be when 6 kg of F = 98.6 , C = ?
weight on it? 9
C + 32 = 98.6
If x = 6, y = ? 5
6 - 2y + 4 = 0 Þ y = 5 cm C = 37
iii) the value of F for 38◦C
25. A family is using LPG of weight 14.2 kg for
C = 38, F = ?
consumption.. If it is use with constant rate then
it lasts for 24 days. 9
F = (38) + 32 Þ F = 100.4 o
i) Equation relating quantity of gas to the days. 5
Take 2 points as (0,14.2) and (24,0) 28. An object was launched from a place P. At the
15th second it was 1400m away from target and at
y - 14.2 x - 0
Eqn of line, = the 18th second 800m away. Find VB
0 - 14.2 24 - 0 i) distance between the place and the target
7. 1 Denote t and D as points (15 ,1400) & (18,800)
y= - x + 14.2
12 D - 1400 t - 15
ii) Draw the graph for first 96days. Eqn of line =
800 - 1400 18 - 15
D = - 200t + 4400
When t = 0, D = - 200(0) + 4400
= 4400m
ii) distance covered by it in 15 seconds.
When t = 15, D = - 200(15) + 4400
26. In a shopping mall a hall of cuboid shape with = 1400m
dimension 800´800´720 units, needs to be added iii) time taken to hit the target.
the facility of an escalator in the path. Find When D = 0, t = ?
i) minimum total length of the escalator. - 200t + 4400 = 0 Þ t = 22 sec
In rt.Dle OAB, OA 2 = OB 2 + AB 2 29. Population of a city in the years 2005 and 2010
= 3200 2 + 720 2 are 1,35,000 and 1,45,000 respectively. Find the
= 10758400 approximate population in the year 2015.
Let x denote year, y denote population
Total length, OA = 3280 m
( 2005 ,135000 ) ( 2010 , 145000 ) ( 2015 ,y)
ii) heights at which escalator
changes direction. 1 Eqn of line thro' (2005,1,35,000) (2010 , 1,45,000)
Taking 2 points as (0,0)(3200,720) x - 2005 y - 135000
=
y-0 x-0 2010 - 2005 145000 - 135000
Eqn of st line, = y = 2000x - 3875000
720 - 0 3200 - 0
9 (2015, y) lies on the line
y= x
40 y = 2000( 2015) - 3875000
When x = 800, y = 180 y = 155000
When x = 1600, y = 360 i.e In 2015 population will be » 155000
When x = 2400, y = 540
6.3.ANGLE BETWEEN 2 STRAIGHT LINES If c > 0, P (x1, y1) is on the origin side or non orign
Let eqns of 2 straight lines be side of the line ax + by + c = 0, according as ax1 +
y = m1 x + c1 , y = m 2 x + c 2 by1 + c is positive or negative.
q 1 , q 2 = Angles of two lines with x - axis i) a1a2 + b1b2 < 0, angle between them is acute
m 1 = tanq 1 , m 2 = tanq 2 ii) a1a2 + b1b2 > 0, angle between them is obtuse.
f = Angle between two straight lines
Distance Formulas
f = q2 - q1 Distance between two points (x1, y1) and (x2, y2)
tan f = tan(q 2 - q 1 )
d = ( x1 - x 2 ) 2 + ( y1 - y 2 ) 2
m2 - m1
tan f = Distance from point P (x1, y1) to line ax + by + c = 0
1 + m2m1 Eqn of AB :
æ m - m1 ö ax + by + c = 0
f = tan - 1 ç 2 ÷
è 1 + m2m1 ø Normal Eqn of AB :
m 2 - m1
§ > 0 , f is the acute angle x cos a + y sin a = p
1 + m2m1
coef. proportional :
m2 - m1
§ < 0 , f is the otuse angle cosa sin a p
1 + m2m1 = =
a b -c
Condition for Parallel Lines
If two lines are parallel ü cosa sin a p cos 2 a + sin 2 a
= = =
ý = 0 or p a b -c ± a2 + b 2
angle b/w lines þ
±a
tan f = 0 cos a =
a2 + b2
m2 - m1
=0 ±b ±c
1 + m 2m 1 simillarly, sin a = ,p =
2 2
a +b a + b2
2
m2 - m1 = 0
Normal Eqn of CD :
m2 = m1
x cos a + y sin a = p'
§ For a1x + b1y +c1 = 0, a2x + b2y+c2 = 0, to be parallel
passes through P ( x1 , y 1 )
m1 = m2
x1 cos a + y 1 sin a = p'
a1 a 2
= Þ a1b 2 = a 2 b 2 Required distance = p - p'
b1 b 2
c ax1 by1
=± ± ±
§ lines parallel to ax + by + c = 0 is ax + by = k. 2
a +b 2 2
a +b 2
a2 + b2
Condition for perpendicular Lines ax1 + by1 + c
=
If two lines are perpendi ü p
ý= 2 a2 + b 2
cular angle b/w lines þ Distance between parallel lines a1x + b1y + c1 = 0
m2 - m1 p and a1x + b1y + c2 = 0
= tan
1 + m2m1 2 c1 - c 2
D=
1 + m2m1 p a2 + b 2
= cot
m2 - m1 2 Bisector of angle b/w a1x+ b1y +c1 = 0, a2x+ b2y+c2 = 0
1 + m2m1 = 0 a1x + b1 y + c1 a x + b2 y + c 2
=± 2
m 2 m 1 = -1 a12 + b12 a22 + b 22
§ For a1x + b1y + c1 = 0, a2x+b2y +c2 = 0, to be ^lar Coordinates of nearest point & Image
m1 m 2 = -1 Coordinates of nearest point (foot of perpendicular)
on ax + by + c = 0 from (x1, y1)
a1 a 2
´ = -1 Þ a1 a 2 + b1 b 2 = 0 x - x1 y - y1 ax1 + by1 + c
b1 b 2 = =
a b a2 + b 2
§ lines ^lar to ax + by + c = 0 is bx – ay = k.
Coordinates of image of (x1, y1) w.rt. ax+by+c = 0
Position of a point with respect to a straight line x - x1 y - y 1 2( ax1 + by1 + c )
= =-
A point P (x1, y1) is on origin side or non-origin a b a2 + b2
side of ax + by + c = 0 according as ax1 + by1 + c
and c are of same sign or opposite sign.
1. Show that the lines are 3x + 2y+9 = 0 and 12x + 5. Find equation of the line through the
8y − 15 = 0 are parallel lines. intersection of the lines 3x+2y+5 = 0 and 3x −
Slope of 3x + 2 y + 9 = 0 4y+6 = 0 and the point (1,1).
3 Eqns of lines thro' point of intersection :
m1 = -
2 (a1 x + b1 y + c1 ) + l(a 2 x + b 2 y + c 2 ) = 0
Slope of 12x + 8y - 15 = 0 ( 3x + 2y + 5 ) + l( 3x - 4 y + 6 ) = 0
12 3 Rqd eqn passes thro' (1,1) :
m2 = - = -
8 2
{3 + 2(1) + 5} + ë {3(1) - 4(1) + 6} = 0
m1 = m2 , parallel lines.
l=- 2
2. Find equations of a parallel line and a
Rqd equation as 3x - 10 y + 7 = 0
perpendicular line passing through the point (1,
2) to the line 3x+ 4y = 7. 6. Find eqn of lines passing through point of inter
section of 4x − y + 3 = 0 and 5x+2y+7 = 0, and
Parallel line to 3x + 4 y = 7 is
3x + 4y + k = 0 4x - y + 3 = 0
This passes thro (1, 2) : 5x + 2 y + 7 = 0
3(1) + 4(2) + k = 0 Solving, x = -1, y = -1
k = -11 Pt. of intersection : ( -1, -1)
Rqd line 3x + 4 y = 11 (i) through the point (−1, 2)
Perpendicular line to 3x + 4 y = 7 is y - y1 x - x1
=
4x - 3y + k = 0 y 2 - y1 x 2 - x1
This passes thro (1, 2) : y +1 x +1
=
4(1) - 3( 2 ) + k = 0 2 +1 -1+1
x +1= 0
k=2
(ii) Parallel to x−y+5 = 0
4x - 3y = -2
Any line parallet to x - y + 5 = 0 is
3. Find the equation of the straight line parallel to
x-y+k = 0
5x − 4y+3 = 0 and having x-intercept 3.
This passes thro : ( -1,-1)
Parallel line to 5x - 4y + 3 = 0 is
-1+1 + k = 0 Þ k =0
5x - 4y + k = 0
Rqd Eqn, x - y = 0
This passes thro (3, 0) :
(iii) Perpendicular to x − 2y+1 = 0
5(3) - 4(0) + k = 0
Line perpendicular to x - 2 y + 1 = 0 is
k = -15
2x + y + k = 0
Rqd line 5x - 4 y - 15 = 0
This passes thro : ( -1,-1)
4. Write equation of lines through the point (1,−1)
- 2 -1 + k = 0 Þ k =3
i) parallel to x + 3y − 4 = 0
Rqd Eqn, 2 x + y + 3 = 0
Parallel line to x + 3y - 4 = 0 is
7. Find the equation of a straight line parallel to 2x
x + 3y + k = 0
+ 3y = 10 and sum of intercepts on the axes is 15.
This passes thro (1, - 1) :
Any line parallel to 2x + 3y = 10 is
( 1) + 3( -1) + k = 0
2x + 3y + k = 0
k=2
2x + 3y = - k
Rqd line x + 3y + 2 = 0
x y
Intercept form + =1
ii) perpendicular to 3x + 4y = 6 - k / 2 - k /3
Perpendicular line to 3x + 4y = 6 is -k -k
a= ,b =
4x - 3y + k = 0 2 3
sum of intercepts = 15
Passes thro (1,-1) :
-k -k
4(1) - 3( - 1) + k = 0 Þ k = -7 + = 15 Þ k = -18
2 3
4x - 3y -7 = 0 Rqd Eqn, 2x + 3y - 18 = 0
8. Find the distance x - x1 y - y 1 ax + by + c
= = - 1 2 12
i) between two points (5, 4) and (2, 0) a b a +b
x-0 y - 0 [ 2( 0) + 1(0) - 5]
D = (x 2 -x1 ) 2 + (y 2 - y1 ) 2 = =-
2 1 2 2 + 12
= 32 + 42 x y
= =1
D=5 2 1
9. Find the distance between x = 2, y = 1 Þ ( 2,1)
i) line 5x + 12y − 3 = 0 a point (1, 2) 12. Find image of (−2, 3) about x + 2y − 9 = 0
Distance b/w ( x1 ,y1 ) and ax + by + c = 0
x - x1 y - y 1 2( ax1 + by1 + c )
ax1 + by1 + c = =-
D= a b a2 + b 2
a2 + b2 x+2 y-3 2[(-2) + 2(3) - 9]
= =-
.Distance b/w (1, 2) and 5x + 12 y-3 = 0 is 1 2 12 + 2 2
5(1) + 12(2) - 3 x+2 y-3
D= = =2
1 2
5 2 + 12 2 x + 2 = 2, y - 3 = 4
D=2
x = 0, y = 7 Þ ( 0 ,7 )
ii) line 4x + 3y+4 = 0, and a point (−2, 4)
.Distance b/w (-2, 4) and 4x + 3y + 4 = 0 is 13. Find equation of the bisector of the acute angle
between 3x+4y+2 = 0 and 5x + 12y − 5 = 0.
4(-2) + 3( 4) + 4
D= Bisector of angle b/w a1x+ b1y +c1 = 0, a2x+ b2y+c2 = 0
4 2 + 32 a1 x + b1 y + c1 a x + b2 y + c 2
8 =± 2
D= a12 + b12 a22 + b 22
5
iii) line 4x + 3y+4 = 0 (7,−3) 3x + 4 y + 2 -5x - 12y + 5
=±
Distance b/w (7,-3) and 4x + 3y + 4 = 0 is 2
3 +4 2
52 + 12 2
4(7) + 3( -3) + 4 a1 a 2 + b1b2 = -15 - 48 < 0 ,
D=
42 + 32 3x + 4 y + 2 -5x - 12y + 5
=+
23 5 13
D= 64x + 112 y + 1 = 0
5
10. Find the distance between the parallel lines 14. Find the points on the line x + y = 5, that lie at a
i) 3x + 4y = 12 , and 6x + 8y+1 = 0 distance 2 units from the line 4x + 3y − 12 = 0
Distance b/w two parallel lines
Any point on x + y = 5 is
c1 - c 2 x = t, y = 5- t
D=
a2 + b 2 Distance from(t,5 - t) to 4x + 3y - 12 = 0 is 2 units
- 12 - 12
D= 4(t) + 3( 5 - t) - 12
=2
3 2 + 42 4 2 + 32
D = 52 units t + 3 = ± 10
ii) 12x + 5y = 7 and 12x + 5y+7 = 0 t = - 13, t = 7
Distance b/w 12x + 5y - 7 = 0 ,12x + 5y + 7 = 0
x = - 13,7
7 - ( -7 ) y =18, - 2
D=
12 2 + 5 2
15. A straight line passes through a fixed point (6,
14
D= units 8). Find locus of the foot of the perpendicular
13
drawn to it from the origin O.
iii) 3x − 4y+5 = 0 and 6x − 8y − 15 = 0.
Distance b/w 3x-4 y + 5 = 0 ,3x - 4 y - 15
= 0 Eqn of lines thro' (6 , 8) :
2
5 - ( - 152 ) y - y1 = m(x - x1 )
D= y -8 = m(x - 6 )....(1)
32 + 4 2
point on rqd locus = P(h, k)
5
D= units k-0
2 Slope of OP =
11. Find the nearest point on 2x + y = 5 from origin. h-0
Since (1) and OP are ^ r, 19. Find the length of the perpendicular and the co-
k -h ordinates of the foot of the perpendicular from
m ´ = -1 Þ m = (−10,−2) to the line x + y − 2 = 0
h k
P(h, k) lies on (1) :
h
k -8 = - (h - 6 )
k
. k (k - 8 ) = -h(h - 6 )
h 2 + k 2 - 6h - 8k = 0
Line perpendicular to x + y - 2 = 0
Locus of P(h, k) :
x -y+ k =0
x 2 + y 2 - 6 x - 8y = 0
This passes thro : P( -10, -2)
16. If (−4, 7) is one vertex of a rhombus and if the - 10 + 2 + k = 0 Þ k = 8
equation of one diagonal is 5x − y +7 = 0, then Eqn of perpendicular x - y + 8 = 0
find the equation of another diagonal.
x +y -2=0
one diagonal 5x- y + 7 = 0 Solving, x = -3, y = 5
Diagonals of a rhombus ^ r to each other foot of perpendicular : Q(-3,5)
Other diagonal x + 5y + k = 0 Length of perpendicular = PQ
This passes thro : (-4,7)
(-3 + 10) 2 + (5 + 2) 2
(-4) + 5(7) + k = 0
k = - 31 =7 2
Eqn of other diagonal x + 5y -31 = 0 20. If p1 and p2 are the lengths of the perpendiculars
from origin to the lines x sec θ + y cosec θ = 2a
17. Find the equations of two straight lines which and x cos θ − y sin θ = a cos 2θ, then prove that
are parallel to the line 12x + 5y +2 = 0 and at a p 21 + p 22 = a 2
unit distance from the point (1, − 1).
x sec q + y cos ec q = 2a
Line parallel to 12x + 5y + 2 = 0 is
length of perpen :
12x + 5y + k = 0
Distance from (1,-1) = 1 units (0) sec q + (0) cos ec q - 2a
p1 =
ax1 + by1 + c sec 2 q + cos ec 2 q
=1 4 a2
a2 + b 2 p12 =
sec q + cos ec 2 q
2
12(1) + 5( -1) + k = 4a 2 sin 2 q cos 2 q
=1
12 2 + 5 2 x cos q - y sin q = a cos 2q
7 + k = 13 length of perpen :
k = 6 or - 20
(0) cos q + ( 0) sin q - a cos 2q
Rqd eqn 12x + 5 y + 6 = 0 , p2 =
cos 2 q + sin 2 q
12x + 5 y - 20 = 0
p 22 = a 2 cos 2 2q
18. Find the equations of straight lines which are
= a 2 (cos 2 q - sin 2 q)2
perpendicular to the line 3x+ 4y −6 = 0 and are at
a distance of 4 units from (2, 1). a 2 (cos 4 q + sin 4 q - 2 sin 2 q cos 2 q)
Line perpendicular to3x + 4y - 6 = 0 is p21 + p22 = a 2 (cos 4 q + sin 4 q + 2 sin 2 q cos 2 q)
4 x - 3y + k = 0 = a 2 (cos 2 q + sin 2 q)2
Distance from (2, 1) = 4 units = a2
4( 2) - 3(1) + k 21. Find the family of straight lines
=4
4 2 + 32 i) Perpendicular to ii) Parallel to 3x + 4y − 12 = 0.
5 + k = 20 Lline Perpendicular to 3x + 4y- 12 = 0 is
k = 15 or - 25 4 x - 3y + k = 0 , k Î R
Rqd eqn 4x - 3y + 15 = 0 , Anyline Parallel to 3x + 4y - 12 = 0 is
4x - 3y - 25 = 0 3x + 4y + k = 0 , k Î R
22. Find equations of the straight lines in the family Given points : A(2,0) and B(3,1)
of the lines y = mx + 2, for which m and the x- y 2 - y1
coordinate of the point of intersection of the Slope of PQ = = 1.
x2 - x1
lines with 2x+3y = 10 are integers
Inclination of line PQ = tan - 1 (1)
y = mx + 2
Sub in 2x + 3y = 10 = 45 o
2x + 3( mx + 2) = 10 Slope in new line m = tan(45o + 15o )
4 Slope = tan 60 o
x=
3m + 2 m= 3
3m + 2 is a divisor of 4 ( ±1, ± 2 and ± 4)
line passes thro' (2, 0) :
3m + 2 = ±1, 3m + 2 = ±2 , 3m + 2 = ±4
y - 0 = 3( x - 2 )
m = {-2,-1, 0}
equations y = -2 x + 2 , y = -x + 2 , y = 2 3x - y - 2 3 = 0
23. Find atleast two equations of the straight lines 26. If a line joining two points (3, 0) and (5, 2) is
rotated about the point (3, 0) in counter
in the family of the lines y = 5x + b, for which
clockwise direction through an angle 15◦ , then
and the x-coordinate of the point of intersection
find the equation of the line in the new position
of the lines with 3x − 4y = 6 are integers.
Given points : P (3, 0) , Q (5, 2)
y = 5x + b
y 2 - y1
Sub in 3x -4 y = 6 Slope of PQ = =1.
x 2 - x1
3x - 4(5x + b ) = 6 Inclination of line PQ = tan - 1 (1)
- ( 6 + 4b ) p
x= =
17 4
x will be 17 or multiple of 17 = 45
x = ±17 , ± 34,...
Slope in new line m = tan(45 o + 15 o )
- 6 + 4b = ± 17 Þ b = - 23/4 or - 11/4
Slope = tan 60 o
Since b is integer,
m= 3
- 6 + 4b = ± 34 Þ b = 10 or - 7
line passes thro' (3, 0) :
Eqns 5x - y + 10 = 0 , 5x - y - 7 = 0
y - 0 = 3( x - 3)
24. Find all the equations of the straight lines in the
family of the lines y = mx − 3, for which m and 3x - y - 3 3 = 0
the x-coordinate of the point of intersection of 27. A ray of light coming from the point (1,2) is
the lines with x − y = 6 are integers. reflected at a point A on the x-axis and it passes
through the point (5,3). Find co-ordinates of A.
y = mx - 3
Sub in x - y = 6
3
x -mx + 3 = 6 Þ x=
m -1
m - 1 is a divisor of 3 ( ±1, ± 3 )
m - 1 = ±1, m - 1 = ±3
m = {2,1, 4}
Coordinate of A : ( x ,0)
equations y = 2x - 3, y = -3, y = 4x - 3
Reflection of A : A' (1, -2)
25. If the line joining two points A(2,0) and B(3,1) is
This passes thro (5,3)
rotated about A in anticlockwise direction
x -1 y + 2
through an angle of 15o, then find the equation Eqn of St. line =
of the line in new position. 1- 5 - 2 - 3
5x - 4y - 13 = 0
A ( x ,0) lies on the line
5x - 4( 0) - 13 = 0 Þ x = 13 / 5
13
Coordinate of A : ( , 0)
5
28. A line is drawn perpendicular to 5x = y+7. Find 30. Govt has decided to erect a new Electrical Power
equation of line if the area of triangle formed by Transmission Substation to provide power
this line with co-ordinate axes is 10 sq. units. supply to two villages A and B. The substation
has to to be on the line l. The distances A and B
from the foot of the perpendiculars P and Q on
line l are 3 km and 5 km. Distance between P
and Q is 6 km. (i) What is the smallest length of
cable required to connect the two villages. (ii)
Find the equations of the cable lines that
connect the power station to two villages.
Vertices of Dle :O(0,0), A( x ,0), B(0, y )
Line perpen. to 5x - y - 7 = 0 is
x + 5y + k = 0
This passes thro' : A( x ,0)
x + 5(0 ) + k = 0 Þ x = -k
This passes also thro' : B( 0, y ) Take PQas x - axis, PAas y - axis, P as origin
0 + 5y + k = 0 Þ y = - k /5 Coordinates are P(0, 0), A(0, 3) and B(6, 5)
Area of Dle = 10 sq. unit Image of A about x - axis A' (0,-3).
1 point of intersection of AB andx - axis = R
´ base ´ height = 10
2
AR and BR are path of the cable (road)
-k
-k´ = 20
5 Shortest length of cable = AR + BR
k = ± 10 = BR + RA'
Eqn of AB , x + 5 y = ± 10 = BA'
29. A photocopy store charges Rs. 1.50 per copy for = (6 - 0) 2 + (5 + 3) 2
the first 10 copies and Rs.1.00 per copy after the = 10km
10th copy. Let x be the number of copies, and let y - (-3) (x - 0)
y be the total cost of photocopying. Equation of AB =
5 - (-3) 6 - 0
i) Draw cost graph as x goes from 0 to 50 copies 4x - 3y = 9
No. of copies = x When y = 0, R (9/4, 0)
total cost of photocopying = y Substation is at = 2.25 km from P.
y = 1.5x for 0 £ x £ 10 Equation of AR is 4x + 3y = 9
After 10 copies = Rs.1/copy Eqns of cable lines RA, RB :
y = 1.5(10 + 1( x - 10) 4x - 3y = 9 and 4x + 3y = 9
y = x + 5 for x > 10 31. A car rental firm has charges Rs.25 with 1.8 free
kilometers, Rs.12 for every additional kilometer.
Find the equation relating the cost y to the
number of kilometers x. Also find the cost to
travel 15 kilometers
Up to 1.8 kilometers fixed rent = Rs 25.
y = 25 , 0 £ x £ 1.8
Every additional kilometer after 1.8 km = Rs.12
y = 25 + 12(x - 1.8), x > 1.8
Combined equation :
ii) Find the cost of making 40 copies
ì25, 0 £ x £ 1.8
Cost for 40 copies : y=í
î25 + 12(x - 1.8), x > 1.8
y=x+5 When x = 15, y = 25 + 12(15 - 1.8)
= 40 + 5 cost to travel 15km = Rs 183.40
y = 45
6.4. PAIR OF STRAIGHT LINES § Equation of bisectors is the locus of points from
Pair of Lines Passing through the Origin which perpendicular drawn to two straight lines
are equal.
y - m1 x = 0
y - m2 x = 0 § Let P(p, q) be any point on the locus of bisectors.
Combined equation : § perpendiculars from P(p, q) to y − m1x = 0 is equal
( y - m1 x ) ( y - m2 x ) = 0 to perpendicular from P(p, q) to y − m2x = 0
y - (m1 + m 2 ) xy + m1 m 2 x 2 = 0
2
q - m1 p q - m2 p
± =±
§ General equation of a pair of straight lines 1+ m12 1 + m22
passing through origin with slopes m1 and m2, ( q - m1 p) 2 ( 1 + m 22 ) = (q - m 2 p )2 (1 + m 12 )
ax2 + 2hxy + by2 = 0 æ 1 - m2 m1 ö
p 2 - q 2 = 2 pqç ÷
§ It is a homogenous equation of degree two, è m1 + m 2 ø
æ1 - a ö
§ Implying that the degree of each term is 2. p 2 - q 2 = 2 pqçç -2 hb ÷÷
è b ø
Angle between Pair of Straight Lines 2 2
p -q pq
=
ax 2 + 2hxy + by 2 = 0 a-b h
2 2
slopes of these two lines = m1 , m 2 x -y xy
locus of P(p, q) is =
y a-b h
Divide byx 2 , substitute =m Combined equation of the pair of straight lines
x
bm 2 + 2hm + a = 0 Individual equations of any two straight lines
Let Quadrati roots = m1 , m 2 l1x + m1y + n1 = 0 and l2x + m2y + n2 = 0
- 2h
m1 + m 2 = Combined equation
b
a (l1x + m1y + n1) (l2x + m2y + n2) = 0
m1m 2 =
b l1l2x2+(l1m2+l2m1)xy+m1m2y2+(l1n2+l2n1)x+(m1n2+m2n1)y+n1n2=0
Angle between two lines = q
General form of Pair of Straight Lines
m - m1
tan q = 2 ax2 + 2hxy + by2 + 2gx + 2fy + c = 0,
1 + m1 m 2
where a, b, c, f, g, h are constants.
( m1 + m 2 ) 2 - 4m1 m2
= Condition for a general second degree equation in x
1 + m1 m 2 and y to represent a pair of straight lines
( - b2 h ) 2 - 4( ab ) Treating General equation as a quadratic in x,
=
1 + ba ax 2 + 2(hy + g)x + by 2 + 2 fy + c = 0
2 h 2 - ab -(hy + g) ± (hy + g) 2 - a (by 2 + 2 fy + c)
tan q = x =
a+b a
Conclusion ax + hy + g = ± (hy + g) 2 - a (by 2 + 2 fy + c)
1. Lines are real and distinct, if m1 and m2 are real ax + hy + g = ± (h 2 - ab) y 2 + 2(gh - af) y + g 2 - ac
and distinct, that is if h2 − ab > 0
Since each of above equations represents a straight
2. The lines are real and coincident, if m1 and m2 are line, they must be of the first degree in x and y.
real and equal, that is if h2 − ab = 0 Expression under the radical sign should be a perfect
3. The lines are not real (imaginary), if m1 and m2 are square and the condition for this is
not real, that is if h2 − ab < 0 4(gh-af)2 -4h 2 -abg 2 - ac = 0
Simpliyf and divide by a :
4. Lines are parallel since both pass through origin,
abc + 2 fgh-af 2 - bg 2 - ch 2 = 0
Coincident lines if tan θ = 0, that is h2 − ab = 0,
a h g
perpendicular if cot θ = 0 that is a + b = 0
h b f =0
Equation of angle bisectors b/w ax2 + 2hxy + by2 = 0
g f c
1. Separate the equations 5x2 + 6xy + y2 = 0 Sub (3) & (4) in (2) :
5x 2 + 6 xy + y 2 = 0 a 2 2h
y - xy + x 2 = 0
5x 2 + 5xy + xy + y 2 = 0 b b
5x (x + y) + y (x + y) = 0 Rqd eqn ay 2 - 2hxy + bx 2 = 0
( 5x + y) (x + y) = 0 7. Show that the straight lines x2− 4xy +y2 = 0 and x
+ y = 3 form an equilateral triangle
5x + y = 0 , x + y = 0
Let x + y = 3 intersects pair of line
2. Separating the equations 2x2 + 2xy + y2 = 0.
x 2 -4xy + y 2 = 0 at A and B.
Values of are not real , therefore no line will exist
Angle betweenlines x 2 -4xy + y 2 = 0
with joint equation 2x2 + 2xy + y2 = 0
2 h 2 - ab
This equation represents imaginary lines. tan q =
a+b
3. Find separate equation of 3x2 + 2xy − y2 = 0 2 4 -1
3x 2 + 2xy - y 2 = 0 =
2
x(3x - y ) + y(3x - y ) = 0 q = tan -1 3
( 3x - y )(x + y ) = 0 q = 60 o
4. Find the separate equation of
Angle bisectors of ÐAOB :
2x 2 - xy - 3y 2 - 6x + 19y - 20 = 0
x 2 -y 2 xy
Consider 2 x 2 - xy - 3y 2 = ( 2x - 3y )( x + y ) =
a-b h
Seperate Eqns : ( 2x - 3y + l )( x + y + m) 2 2
x -y = 0
2x 2 - xy - 3y 2 - 6 x + 19y - 20 = ( 2x - 3y + l )( x + y + m ) ( x + y )(x - y ) = 0
Equating coeff. of x , y : x − y = 0 is perpendicular to given line thro’AB,
2m + l = -6 x + y = 3 Þ ΔOAB is isosceles.
- 3m + l = 19 Þ ÐABO = ÐBAO = 60o
Solving, m = - 5, l = 4 8. If the pair of lines x2 − 2cxy − y2 = 0 and x2 − 2dxy
− y2 = 0 be such that each pair bisects the angle
Seperate Eqns : ( 2x - 3y + 4)(x + y - 5)
between the other pair, prove that cd = −1.
5. Find separate equation of
angle bisectors of x 2 -2cxy-y 2 = 0 is
6(x - 1) 2 + 5(x - 1)(y - 2 ) - 4(y - 2 ) 2 = 0
x 2 -y 2 xy
6(x - 1) 2 - 3(x - 1)(y - 2 ) + 8(x - 1)(y - 2 ) - 4(y - 2 ) 2 = 0 =
2 -c
3(x -1)[ 2(x - 1) - (y - 2 )] + 4(y - 2 )[ 2(x - 1) - (y - 2 )] = 0 2 2
cx + 2xy - cy = 0 .....(1)
3(x - 1)[ 2x - y)] + 4(y - 2 )[ 2x - y ] = 0
Given that angle bisector :
[ 2x - y ](3x + 4y - 11) = 0
x 2 -2dxy-y 2 = 0 ....( 2)
6. Find equation of the pair of lines through origin
perpendicular to pair of lines ax2 + 2hxy + by2 = 0 (1)& (2)represent same equation :
slopes of these two lines = m1 ,m 2 1 - 2d - 1
Comparing like terms, = =
y - m1 x = 0 c 2 -c
cd = 1
y - m 2 x = 0 ....(1)
9. If the equation λx2− 10xy + 12y2+ 5x− 16y− 3 = 0
perpendicular Lines : represents a pair of straight lines, find
m1 y + x = 0 i) value of λ and separate equations of the lines
m2 y + x = 0 General equation :
Combined equation : ax 2 + 2hxy + by 2 + 2 gx + 2 fy + c = 0
(m1 y + x ) (m 2 y + x ) = 0 Comp, lx 2 -10xy + 12 y 2 + 5x - 16y - 3 = 0
m1 m 2 y 2 + (m1 + m 2 )xy + x 2 = 0 ....( 2) Condition abc + 2 fgh - a f 2 - bg 2 - ch 2 = 0
Given, ax 2 + 2hxy + by 2 = 0 5
a = l, b = 12 , c = -3 , h = -5 , g = , f = -8
-2 h 2
m1 + m 2 = ...(3) 2
b 5 5
l (12)(-3) + 2(8)æç ö÷5 - l (8)2 - 12 æç ö÷ + 3(-5)2 = 0
a 2
è ø è2ø
m1m 2 = ...( 4)
b . - 36l + 200 - 64l - 75 + 75 = 0,
l=2 12. Find the combined equation of straight lines
ii) point of intersection of the lines whose separate equations x−2y−3= 0, x+y+5= 0
2x 2 -10xy + 12 y 2 + 5x -16y-3 = 0.
( x - 2 y - 3)( x + y + 5) = 0
2x 2 - 10xy + 12y 2 = ( x - 2 y )( 2x - 6y )
x - xy - 2y 2 + 2x - 13y - 15 = 0
2
2 2
2x -10xy + 12 y + 5x -16 y-3 = ( x- 2y + l )(2x - 6 y + m)
2l + m = 5 13. Show that 4x2 + 4xy + y2 − 6x − 3y − 4 = 0
3l + m = 8 represents a pair of parallel lines.
Solving , l = 3, m = - 1. 4x 2 + 4xy + y 2 - 6x - 3y - 4 = 0
Separate equations : 3
a = 4, h = 2 , b = 1, g = - 3, f = - , c = -4
x - 2y + 3 = 0 2
2x - 6y - 1 = 0 Condition to be parallel : bg 2 = af 2
2
-7 3
Solving, Pt. of intersection : (-10, ) 1( -3) 2 = 4æç ö÷
2 è 2ø
iii) angle between the lines Hence pair of st lines are parallel
2 25 - 24
tan q = 14. Show that 2x2 + 3xy − 2y2 + 3x + y+1 = 0
2 + 12
represents a pair of perpendicular lines.
1
q = tan -1 æç ö÷
è 7ø 2x 2 + 3xy - 2y 2 + 3x + y + 1 = 0
10. If one of the straight lines of ax2+2hxy+by2 = 0 is a = 2, b = -2
perpen. px+qy = 0, show that ap2 + 2hpq + bq2 = 0. a+b=0
- 2h a They are ^ r lines
m1 + m 2 =, m1 m 2 =
b b
-p 15. Show that the equation 2x2 −xy−3y2 −6x+19y−20 =
Slope of of px + qy , m = 0 represents a pair of intersecting lines. Show
q
Since one of them ^ r, further that the angle between them is tan−1(5).
mm1 = -1 or mm2 = -1 2x 2 - xy - 3y 2 - 6x + 19 y - 20 = 0
mm1 + 1 = 0 or mm 2 + 1 = 0 a = 2, b = -3, h = -1 / 2, g = -3, f = 19 / 2 , c = -20
. (mm1 + 1)(mm2 + 1) = 0 Condition for intersection :
2
. (m1m 2 )m + (m1 + m 2 )m + 1 = 0 abc + 2 fgh - af 2 - bg 2 - ch 2 = 0
2
After Sub. rqd values & simplify as
æç a ö÷æç - p ö÷ + æç -2h ö÷æç - p ö÷ = 0
è b øè q ø è b øè q ø 57 361
= 120 + - + 27 + 5
2 2
ap 2 + 2hpq + bq 2 = 0
=0
11. Show that the straight lines joining the origin to
2 h 2 - ab
the points of intersection of 3x − 2y+2 = 0 and q = tan -1
a+b
3x2 + 5xy − 2y2 + 4x + 5y = 0 are at right angles
= tan -1 5
3x - 2 y + 2 = 0
16. Find equation of pair of straight lines passing
3x - 2 y
= 1....(1) through the point (1, 3) and perpendicular to the
-2
lines 2x − 3y+1 = 0 and 5x + y − 3 = 0
3x 2 + 5xy - 2 y 2 + ( 4x + 5y ) (1) = 0
æ 3x - 2 y ö Line perpendicular to 2x - 3y + 1 = 0 is
3x 2 + 5xy - 2 y 2 + ( 4x + 5y )ç ÷ = 0
è -2 ø 3x + 2y + a = 0
( -2)(3x 2 + 5xy - 2 y 2 ) + ( 4x + 5y )(3x - 2y ) = 0 It passes thro' (1,3) 3(1) + 2( 3) + a = 0 Þ a = -9
Line perpendicular to 5x + y - 3 = 0 is
On simplification, 2x 2 + xy - 2y 2 = 0. .
x - 5y + b = 0
a = 2 , b = -2
It passes thro' (1,3) 1 - 5( 3) + b = 0 Þ b = 14
a + b = 0
Comb. Eqn, 3x + 2 y - 9) ( x - 5y + 14) = 0
Hence lines are at right angles :
3x 2 - 13xy - 10 y 2 + 33x + 72y - 126 = 0
17. Slope of one of the straight lines ax2 + 2hxy + by2 6x 2 + 5xy - py 2 + 7 x + qy - 5 = 0
= 0 is twice that of the other, show that 8h2 = 9ab. a = 6 , b = - p , h = 5 / 2, g = 7 / 2, f = q / 2 , c = -5
-2 h Condition for perpendicular :
m1 + m2 = ...(1)
b a + b =0
a
m1 m2 = ...( 2) 6- p=0 Þp =6
b
Slope of Line 1 = 2 ´ Slope of Line 2 Condition for intersection :
m 1 = 2m 2 ...(3) abc + 2 fgh - af 2 - bg 2 - ch 2 = 0
-2 h After Sub. p = 6 & simplify as
Sub in (1), 2m 2 + m2 =
b 1139 + 35q - 6q 2 = 0
- 2h
m2 = ...( 4) ( 6q + 67 )(q - 17 ) = 0
3b
- 2h öæ -2h ö a 67
Sub in (2), 2æç = q = - , 17
÷ç ÷ 6
è 3b øè 3b ø b
8h 2 = 9 ab 22. For what value of k does the equation 12x2+2kxy
+2y2+11x−5y+2 = 0 represent two straight lines.
18. Slope of one of the straight lines ax2 + 2hxy + by2
= 0 is three times the other, show that 3h2 = 4ab. 12x 2 + 2 kxy + 2 y 2 + 11x - 5y + 2 = 0
- 2h a a = 12 , h = k , b = 2, g = -11 / 2, f = - 5 / 2 , c = 2
m1 + m2 = , m1 m 2 =
b b Condition for pair of stlines :
m1 = 3m 2 ...(3) abc + 2 fgh - af 2 - bg 2 - ch 2 = 0
-h
Sub in (1), m 2 = ...( 4) After Sub. & simplify as
2b
- 2 h öæ - 2 h ö a 4 k 2 + 55k + 175 = 0
Sub in (2), 2æç ÷ç ÷=
è 3b øè 3b ø b ( k + 5)(4 k + 35) = 0
3h 2 = 4ab k =-
35
,- 5
4
19. Prove that the straight lines joining the origin to
the points of intersection of 3x2 + 5xy − 3y2 + 2x + 23. Show that 9x2−24xy+16y2−12x+16y−12 = 0
3y = 0 and 3x − 2y − 1 = 0 are at right angles. represents a pair of parallel lines. Find the
3x - 2y = 1 distance between them.
3x 2 + 5xy - 3y 2 + ( 2x + 3y ) (1) = 0 9x 2 - 24xy + 16 y 2 - 12x + 16 y - 12 = 0
3x 2 + 5xy - 3y 2 + ( 2x + 3y ) (3x - 2 y ) = 0 a = 9, h = - 12, b = 16 , g = - 6 , f = 8, c = -12
3x 2 + 5xy - 3y 2 + 6x 2 + 5xy - 6y 2 = 0 Condition to be parallel : bg 2 = af 2
9x 2 + 10xy - 9y 2 = 0 576 = 576
a = 9 , b = -9 pair of st lines are parallel
a+b=0 g 2 - ac
Hence they are at 900 Distance b/w Parallel lines = 2
a( a + b
20. Find value of k, if 12x2 + 7xy − 12y2 − x+ 7y + k = 8
0 represents a pair of straight lines. Find D=
5
whether these lines are parallel or intersecting,
24. Show that the equation 4x2 +4xy + y2 − 6x − 3y −4
12x 2 + 7xy - 12y 2 - x + 7 y + k = 0
= 0 represents a pair of parallel lines. Find the
12x 2 + 7xy - 12 y 2 = ( 4x - 3y )(3x + 4y ) distance between them.
12x 2 + 7xy - 12y 2 - x + 7 y + k = ( 4x - 3y + l )(3x + 4y + m)
4x 2 + 4xy + y 2 - 6x - 3y - 4 = 0
4m + 3l = - 1
3
- 3m + 4l = 7 a = 4, h = 2 , b = 1, g = - 3, f = , c = -4
2
By solving, l = 1, m = -1 Condition for pair of st lines to be parallel :
k = lm Þ k = -1 9 =9 pair of st lines are parallel
21. Find p and q, if 6x2 + 5xy − py2 + 7x + qy − 5 = 0 f 2 - bc
Distance = 2 = 5
represents a pair of perpendicular lines b( a + b)
25. Prove that one of the straight lines given by ax2 P(1 + 3
,1 - 3 3 3
3 3 ) and Q(1 - 3 ,1 + 3 )
+ 2hxy + by2 = 0 will bisect the angle between
æ 1 + 33 + 1 - 3
1+ 3
+1- 3 ö
the co-ordinate axes if (a + b)2 = 4h2 Midpoint of PQ = ç 3
, 3 3 ÷
ç 2 2 ÷
è ø
(a + b) 2 = 4h 2
= M (1,1)
a + b = 2h ...(1)
Eqn of median O(0,0)M(1,1) :
ax 2 + 2hxy + by 2 = 0 y-0 x-0
=
From (1), ax 2 + ( a + b ) xy + by 2 = 0 1- 0 1-0
( ax + by )(x + y ) = 0 y=x
by = - ax , y = - x 28. Prove that the equation to straight lines through
the origin, each of which makes an angle a with
Slope of y = -x straight line y = x is x2 − 2xy sec 2α + y2 = 0
m = -1
tan q = -1
q = 135
26. If the pair of straight lines x2 − 2kxy − y2 = 0
bisect the angle between the pair of straight
lines x2 − 2lxy − y2 = 0, Show that the later pair
also bisects the angle between the former.
Angle b/w x axis and AB, q = 45 - a
x 2 - 2kxy - y 2 = 0
m1 = tan(45 - a )
a = 1, h = - k , b = -1
tan 45 - tan a
=
x 2 - y 2 xy 1 + tan 45 tan a
Eqn. of Ð bisector, =
a-b h 1 - tan a
m1 =
x 2 - y 2 xy 1 + tan a
= Angle b/w x axis and CD, q = 45 + a
2 -k
2
kx + 2xy - ky = 02 m 2 = tan(45 + a )
Ð bisector, x 2 - 2lxy - y 2 = 0 (given) 1 + tan a
m2 =
1 - tan a
k 2 -k Eqn of stline :
comp. coeff., = =
1 - 2l - 1
kl = -1 ( m1 x - y )( m 2 x - y ) = 0
Thus later pair also bisects angle between former m1m 2 x 2 - xy(m1 + m 2 ) + y 2 = 0
27. A ΔOPQ is formed by the pair of straight lines 1 + tan a 1 - tan a
m1 + m 2 = +
1 - tan a 1 + tan a
x2 −4xy +y2 = 0 and the line PQ. Equation of PQ
= 2 sec 2a
is x + y − 2 = 0. Find the equation of the median
29. A student when walks from his house, at an
of the triangle ΔOPQ drawn from the origin O.
average speed of 6 kmph, reaches his school ten
x 2 - 4xy + y 2 = 0 ....(1) minutes before the school starts. When his
x+ y-2=0 average speed is 4 kmph, he reaches his school
y = 2 - x ...( 2) five minutes late. If he starts to school every day
Sub (2) in (1), at 8.00 A.M, then find (i) the distance between
x 2 - 4 x( 2 - x ) + ( 2 - x ) 2 = 0 his house and the school (ii) the minimum
3x 2 - 6x + 2 = 0 average speed to reach the school on time and
time taken to reach the school (iii) the time
3± 3
x= school gate closes (iv) pair of straight lines of
3
his path of walk.
3 3
If x = 1 + , y =1-
3 3
3 3
If x = 1 - , y =1+
3 3
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