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NCERT

Set Relations and Functions

MATHEMATICS
CLASS 11 & 12

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1 Set Relations and Functions

QUICK LOOK
Power set: If S is any set, then the family of all the subsets
of S is called the power set of S. The power set of S is
SET THEORY denoted by P(S). Symbolically, P(S) = {T : T ⊆ S}.
A set is well defined class or collection of objects. A set is
Obviously φ and S are both elements of P(S).
often described in the following two ways.

Subsets (Set inclusion): Let A and B be two sets. If every

Roster method or Listing method: In this method a set is
element of A is an element of B, then A is called a subset of B.
described by listing elements, separated by commas, within
If A is subset of B, we write A ⊆ B, which is read as “A is a
braces {}. The set of vowels of English alphabet may be
subset of B” or “A is contained in B”. Thus, A ⊆ B ⇒ a ∈ A
described as {a, e, i, o, u}.

Set-builder method or Rule method: In this method, a set ⇒ a ∈ B.
is described by a characterizing property P(x) of its
Proper and improper subsets: If A is a subset of B and
elements x. In such a case the set is described by {x : P(x) A ≠ B, then A is a proper subset of B. We write this
holds} or {x | P(x) holds}, which is read as ‘the set of all x as A ⊂ B. The null set φ is subset of every set and every set
such that P(x) holds’. The symbol ‘|’ or ‘:’ is read as ‘such is subset of itself, i.e., φ ⊂ A and A⊂ A for every set A.
that’. They are called improper subsets of A. Thus every non-
The set A = {0,1, 4,9,16,...} can be written as empty set has two improper subsets. It should be noted that
A = { x 2 | x ∈ Z }. φ has only one subset φ which is improper. All other
U+
subsets of A are called its proper subsets. Thus,
Types of sets if A ⊂ B , A ≠ B , A ≠ φ , then A is said to be proper subset of

Null set or Empty set: The set which contains no element at
ED

B.
all is called the null set. This set is sometimes also called the
‘empty set’ or the ‘void set’. It is denoted by the symbol φ or Venn-Euler diagrams
{}. The combination of rectangles and circles are called Venn-

Singleton set: A set consisting of a single element is called Euler diagrams or simply Venn-diagrams. If A and B are not
a singleton set. The set {5} is a singleton set. equal but they have some common elements, then to represent

Finite set : A set is called a finite set if it is either void set A and B we draw two intersecting circles. Two disjoints sets are
or its elements can be listed (counted, labelled) by natural represented by two non - intersecting circles.
number 1, 2, 3, … and the process of listing terminates at a U
certain natural number n (say).
A

Cardinal number of a finite set: The number n in the
above definition is called the cardinal number or order of a Figure 1.1: Venn-Euler diagrams
finite set A and is denoted by n(A) or O(A).

Infinite set: A set whose elements cannot be listed by the Operations on sets
natural numbers 1, 2, 3, ..., n, for any natural number n is
Union of sets: Let A and B be two sets. The union of A and
called an infinite set. B is the set of all elements which are in set A or in B. We

Equivalent set: Two finite sets A and B are equivalent if denote the union of A and B by A ∪ B , which is usually read
their cardinal numbers are same i.e. n(A) = n(B). as “A union B”.

Equal set: Two sets A and B are said to be equal iff every Symbolically, A ∪ B = {x : x ∈ A or x ∈ B}.
element of A is an element of B and also every element of B U
is an element of A. Symbolically, A = B if x ∈ A ⇔ x ∈ B. A∪B

Universal set: A set that contains all sets in a given context
is called the universal set. It should be noted that universal A B
set is not unique. It may differ in problem to problem. Figure 1.2: Operations on sets

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2 Quick Revision NCERT- MATHEMATICS

Intersection of sets: Let A and B be two sets. The
Total number of relations: Let A and B be two non-empty
intersection of A and B is the set of all those elements that finite sets consisting of m and n elements respectively. Then
belong to both A and B. A × B consists of mn ordered pairs. So, total number of
U subset of A × B is 2mn. Since each subset of A × B defines
relation from A to B, so total number of relations from A to
A∩B
B is 2mn. Among these 2mn relations the void relation φ and
A B the universal relation A × B are trivial relations from A to B.
Figure 1.3: Intersection of sets
Domain and range of a relation: Let R be a relation from a
The intersection of A and B is denoted by A ∩ B (read as “A set A to a set B. Then the set of all first components or
intersection B”). coordinates of the ordered pairs belonging to R is called the
domain of R, while the set of all second components or
Thus, A ∩ B = {x: x ∈ A and x ∈ B}.
coordinates of the ordered pairs in R is called the range of R.

Disjoint sets: Two sets A and B are said to be disjoint, if
Thus, Dom (R) = {a: (a, b) ∈ R} and Range (R) = {b : (a, b)
A ∩ B = φ . If A ∩ B ≠ φ , then A and B are said to be non-
∈ R}.
intersecting or non-overlapping sets.

Difference of sets: Let A and B be two sets. The difference of Inverse relation
A and B written as A – B, is the set of all those elements of A Let A, B be two sets and let R be a relation from a set A to a set
which do not belong to B. B. Then the inverse of R, denoted by R–1, is a relation from B to
Thus, A – B = {x : x ∈ A and x ∉ B} A and is defined by R −1 = {(b, a) : (a, b) ∈ R}
Similarly, the difference B − A is the set of all those elements of Clearly (a,b)∈R⇔(b,a) ∈R–1. Also, Dom (R) = Range ( R −1 ) and
B that do not belong to A i.e., B − A = {x ∈ B : x ∉ A} . Range (R) = Dom (R −1 )
U+
U U

A–B B–A
Types of relations

Reflexive relation: A relation R on a set A is said to be
ED

A B A B reflexive if every element of A is related to itself.

Figure 1.4: Difference of sets
Thus, R is reflexive ⇔ (a, a) ∈ R for all a ∈ A.

Symmetric difference of two sets: Let A and B be two sets. Example: Let A = {1, 2, 3} and R = {(1, 1); (1, 3)}
The symmetric difference of sets A and B is the set Then R is not reflexive since 3 ∈ A but (3, 3) ∉ R
( A − B ) ∪ (B − A ) and is denoted by A∆B . Thus, A∆B = A reflexive relation on A is not necessarily the identity
( A − B) ∪ (B − A) = { x : x ∉ A ∩ B} . relation on A.

Complement of a set: Let U be the universal set and let A The universal relation on a non-void set A is reflexive.
be a set such that A ⊂ U . Then, the complement of A with
Symmetric relation: A relation R on a set A is said to be a
respect to U is denoted by A′ or Ac or C(A) or U – A and is symmetric relation iff (a, b) ∈ R ⇒ (b, a) ∈ R for all a, b ∈
defined the set of all those elements of U which are not in A. A i.e., aRb ⇒ bRa for all a, b ∈ A.
Thus, A′ = {x ∈ U : x ∉ A}. it should be noted that R is symmetric iff R −1 = R
Clearly, x ∈ A′ ⇔ x ∉ A The identity and the universal relations on a non-void set
U are symmetric relations.
A’
A A reflexive relation on a set A is not necessarily symmetric.

Anti-symmetric relation: Let A be any set. A relation R on
set A is said to be an anti-symmetric relation iff (a, b) ∈ R
Figure 1.5: Complement of a set
and (b, a) ∈ R ⇒ a = b for all a, b ∈ A.
RELATIONS
Thus, if a ≠ b then a may be related to b or b may be related
Let A and B be two non-empty sets, then every subset of A × B
to a, but never both.
defines a relation from A to B and every relation from A to B is

Transitive relation: Let A be any set. A relation R on set A
a subset of A × B. Let R ⊆ A × B and (a, b) ∈ R. Then we say
is said to be a transitive relation iff
that a is related to b by the relation R and write it as a R b . If
(a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for all a, b, c ∈ A i.e.,
(a, b) ∈ R , we write it as a R b . aRb and bRc

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Set Relations and Functions 3
⇒ aRc for all a, b, c ∈ A. FUNCTION
Transitivity fails only when there exists a, b, c such that a R If f : A → B where A ⊆ R, B ⊆ R is a real function then f is a rule
b, b R c but a R/ c . according to which, corresponding to each x ∈ A there is a

Identity relation: Let A be a set. Then the relation IA = {(a, unique real f ( x) ∈ B. f (a) is the value of the function at
a) : a ∈ A} on A is called the identity relation on A. x = a, i.e., x = a ∈ A corresponds to f (a) ∈ B. y = f ( x), i.e.,
In other words, a relation IA on A is called the identity relation
x ∈ A corresponds to y ∈ B then x is the independent variable
if every element of A is related to itself only. Every identity
and y is the dependent variable.
relation will be reflexive, symmetric and transitive.

Example: On the set = {1, 2, 3}, R = {(1, 1), (2, 2), (3, 3)} Domain and range of a real-valued function
is the identity relation on A . It is interesting to note that If y = f ( x ) be a real function then domain of f = the set of real
every identity relation is reflexive but every reflexive x for which f ( x ) is real range of f = the set of real values
relation need not be an identity relation.
of f ( x ) for x ∈ domain f = [min f ( x), max f ( x)]

Equivalence relation : A relation R on a set A is said to be
an equivalence relation on A iff Note
It is reflexive i.e. (a, a) ∈ R for all a ∈ A The above result on the range of f is true for continuous
It is symmetric i.e. (a, b) ∈ R ⇒ (b, a) ∈ R, for all a, b ∈ A functions. If the domain of φ ( x) be D1 and the domain of
It is transitive i.e. (a, b) ∈ R and (b, c) ∈ R ⇒ (a, c) ∈ R for ψ ( x ) be D2 then the domain of f ( x) = φ ( x) ±ψ ( x) is
all a, b, c ∈ A. D1 ∩ D2 the domain of f ( x) = φ ( x) ×ψ ( x) is D1 ∩ D2 the
φ ( x)
domain of f ( x) = is D1 ∩ D2 − E , where E = the set of
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Congruence modulo (m): Let m be an arbitrary but fixed
ψ ( x)
integer. Two integers a and b are said to be congruence modulo
zeros of ψ ( x).
m if a − b is divisible by m and we write a ≡ b (mod m).

Equality of functions
ED

Thus a ≡ b (mod m) ⇔ a − b is divisible by m. For

Two functions f (x) and φ ( x) are equal if
example, 18 ≡ 3 (mod 5) because 18–3=15 which is divisible by
(a) Domain of f = domain of φ and
5. Similarly, 3 ≡ 13 (mod 2) because 3 – 13 = –10 which is
(b) f ( x) = φ ( x) for all x ∈ the common domain
divisible by 2. But 25 ≠ 2 (mod 4) because 4 is not a divisor of
25 – 3 = 22. The relation “Congruence modulo m” is an Example: f ( x) = log x 2 , x > 0 and φ ( x) = 2log x, x > 0 are
equivalence relation.
equal function because they have the same
domain (0, + ∝) and for each x ∈ (0, + ∝) we have
Equivalence classes of an equivalence relation
Let R be equivalence relation in A(≠ φ ). Let a ∈ A. Then the f ( x) = φ ( x) as log x 2 = 2log x. But f ( x) = log x 2 , x ∈ and φ/
equivalence class of a, denoted by [a] or {a } is defined as the (x) = 2log x , x > 0 are not equal because they do not
have the same domain.
set of all those points of A which are related to a under the
A function may be defined in any one of the following ways:
relation R. Thus [a] = {x ∈ A: x R a}.
(a) Uniform definition
It is easy to see that
(b) Piecewise definition

b ∈ [a] ⇒ a ∈ [b]
(c) General definition given by a property of the function.

b ∈ [a] ⇒ [a] = [b]

Two equivalence classes are either disjoint or identical. Example Let
(a) f ( x) = x 2 + 1
Composition of relations: Let R and S be two relations from (b) g ( x) = 2 x − 1, x < 0 x + 3, x ≥ 0
sets A to B and B to C respectively. Then we can define a
(c) h( x + y ) = h( x).h( y ) for all x, y ∈ R.
relation So R from A to C such that (a, c) ∈ So R ⇔ ∃ b ∈ B
In (a), the definition is uniform. For every
such that (a, b) ∈ R and (b, c) ∈ S. This relation is called the
x ∈ R, f ( x) = x + 1.
2
composition of R and S.

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4 Quick Revision NCERT- MATHEMATICS
In (b), the definition is piecewise. For negative values of ⇒ 3 x1 − 1 = 3 x2 − 1
x, g ( x) = 2 x − 1 is to be used while for non-negative values ⇒ x1 = x2 and g ( x1 ) = g ( x2 )
of x, g ( x) = x + 3 is to be used. In (c), the definition is ⇒ x12 + 1 = x22 + 1
general. The function is described by no rules but by a
⇒ x12 = x22
property of h( x). Clearly h( x) = e x obeys the property, but
⇒ x1 = x2 , − x2
there may be other functions satisfying the same property.

Into and onto (subjective) functions: A function

Some special piecewise functions:- Modulus function
f : A → B is an into function if for at least one β ∈ B there
f ( x) =| x | i.e., f ( x) = x, x > 00, x = 0 − x, x < 0 Sign
is no α ∈ A such that f (α ) = β . If there is no such
function f ( x) = 1, x > 0, 0, x = 0 −1, x < 0 Step function or
β ∈ B then f is an onto (surjective) function
greatest integer function f ( x) = [ x], where x = greatest
Example: y = f ( x) = 2 x − 1 is an onto function from R to R
integer less than or equal to x, i.e., f ( x) = n, n ≤ x < n + 1.
but y = g ( x) = x 2 + 1 is not onto, i.e., into function from R
Dirichlet function f ( x) = 1, x is rational 0, x is irrational
f ( x) + 1 y + 1

Even, odd and periodic functions to R because, x = = and so for every y ∈ R we
A function f (x) is even if 2 2
f (− x) = f ( x) for all x ∈domain  y +1 
have the real x =   , and x = g ( x) − 1 = y − 1 and so
 2 
A function f ( x) is odd if
for y < 1 we have no real x = ( y − 1). In order to check
f (− x) = − f ( x) for all x ∈ domain
whether y = f ( x) from the set A to B is onto or not, write x
A function f ( x) is periodic if f ( x + k ) = f ( x) for all x ∈
in terms of y and see if for every y ∈ B, x ∈ A. If so, it is
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domain and k is a positive constant.
If k is the least possible positive constant then k is the period onto. Otherwise, it is into. A function f which is one-one and
of the function. onto (i.e., injective as well as subjective) is called bijective.
ED

Example: cos x,| x |, x 2 − 1 are even functions sin x, x3 are Inverse function: Let y = f ( x) be a function whose domain is
odd functions x + 2, e are neither even nor odd functions
x
A and whose range is B. If for every y ∈ B there exists a single
sin x, tan x, x − [ x] are periodic functions. Some standard values of x such that f ( x) = y then an inverse function of
periodic functions with their periods: f ( x) from B to A is defined, given by x = f −1 ( y ).
f
Function Period Function Period X Y
sin x,cos x 2p sin ax,cos ax 2π / a
a a1
sin x,cosecx 2p sec ax,cos exax 2π / a
b b1
tan x,cot x π tan ax, cot ax π /a
x − [ x] 1 constant c c1
f −1
intermediate

If f ( x) and g ( x ) are two trigonometrical functions of Figure 1.6: Inverse function

periods 1 and µ then af ( x) + bg ( x) is periodic and has the Note

period = 1 c.m. of {λ , µ} For the existence of inverse function, it should be one-one and
onto.

One-one (injective and many-one functions:-A function
Inverse of a bijection is also a bijection function.
f ( x) is one-one (injective) if f ( x1 ) = f ( x2 ) ⇒ x1 = x2 ;
Inverse of a bijection is unique.(f–1)–1= f
otherwise it is many-one.
If f and g are two bijections such that (gof) exists then
Example: f ( x) = 3x − 1 is one-one but g ( x) = x + 1 is 2
(gof)–1=f–1og–1.
many-one because f ( x1 ) = f ( x2 )
If f : A→ B is a bijection then f–1: B → A is an inverse

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Set Relations and Functions 5
function of f. f –1of = IA and fof–1=IB. Here IA, is an x is an integer ⇒ x = [ x] ⇒ { x} = 0 ⇒ {[ x ]} = 0
identity function on set A, and IB, is an identity function on If x ∈ [ 0, 1) , then [ x] = 0 ⇒ {x} = x
set B.
. x ∈ [1, 2), then [ x] = 1 ⇒ {x} = x − 1
Composite function: If f ( x) be a function whose domain is A Y
and range is B, and g ( y ) be a function whose domain is B and
the range is C then ( g o f ) ( x) is a composite function whose y = {x}
(0, 1)
domain is A and range is C such that ( g o f ) ( x) = z ∈ C where 0 0 0 0

f ( x) = y and g ( y ) = z.
X
x1 –2 –1 1 2 3
y1 z1
Figure 1.8: Fractional Part
x2 y2 z2
x3 y3 z3 Domain → R; Range → [0,1);
f g
Period → 1; Nature → neither even nor odd
( gof )( x )
Figure 1.7: Composite function
Periodic Function:- A function f : X → Y is said to be a
Properties of composition of Function periodic function provided there exists a positive real

f is even, g is even ⇒ fog even function. number T such that f (x + T) = f (x), for all x ∈ X . The least

f is odd, g is odd ⇒ fog is odd function. of all such positive numbers T is called the principal period

f is even, g is odd ⇒ fog is even function. or fundamental period or simply period of f.

To check the periodicity of a function put f (T + x) = f ( x)
U+

f is odd, g is even ⇒ fog is even function.

Composite of functions is not commutative i.e., fog ≠ gof and solve this equation to find the positive values of t

Composite of functions is associative i.e., (fog)oh = fo( goh) independent of x. If positive values of T independent of x
ED

If f : A → B is bijection and g : B → A is inverse of f. Then are obtained, then f(x) is a periodic function and the least
positive value of T is the period of the function f(x). If no
fog = I B and gof = I A . where, IA and IB are identity
positive value of T independent of x is obtained then f(x) is
functions on the sets A and B respectively. non-periodic function.

If f : A → B and g : B → C are two bijections, then
A constant function is periodic but does not have a well-
gof : A → C is bijection and A ∩ B = A defined period.

fog ≠ gof but if fog = gof then either f −1 = g or g −1 = f
If f(x) is periodic with period p, then f(ax + b) where
also, ( fog )( x) = ( gof )( x) = ( x). a, b ∈ R(a ≠ 0) is also period with period p / | a | .

If f ( x) is periodic with period p, then a f ( x) + b where
Greatest integer and fractional part a, b ∈ R( a ≠ 0) is also periodic with period p.

Greatest integer: Any real number x can always think of

If f ( x) is periodic with period p, then f (ax + b) where
lying between two consecutive integers say P and P +1. i.e.
P ≤ x < ( P + 1). That means, there always exist an integer, a, b ∈ R ( a ≠ 0) is also period with period p / | a | .
say ‘P’ which is just less than or equal to x. This unique ‘P’ Let f ( x) has period p = m / n ( m, n ∈ N and co-prime) and
is called the greatest integral value of x and is symbolically g ( x) has period q = r / s ( r , s ∈ N and co-prime) and let t be
denoted as [x] i.e. [x] stands for the greatest integer that is
LCM of (m, r )
less than or equal to x. the LCM of p and q i.e. t = , then t shall
HCM of ( r , s)
be the period of f + g provided there does not exist a

Fractional Part
Fractional Part of any real number is defined as the positive number) k (< t ) for which f (k + x) + g ( k + x)
difference between the number ‘x’ and it’s integral = f ( x) + g ( x), else k will be the period. The same rule is c
value‘[x]’and is symbolically denoted as {x}.Thus, applicable for any other algebraic combination of f(x) and
{x} = x − [ x], e.g. if x = 5.68, then [x] = 5 and {x} = 0.68. If f(x).

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6 Quick Revision NCERT- MATHEMATICS
MULTIPLE CHOICE QUESTIONS 10. If X = {4 – 3n – 1 : n ∈ N} and Y ={9(n – 1) : n ∈ N}
n

Set and Types of Sets then X ∪ Y is equal to:

a. X b. Y
1. The set of intelligent students in a class is: c. N d. None of these
a. A null set
b. A singleton set Relations of Number of Elements in Sets
c. A finite set
d. Not a well defined collection 11. In a town of 10,000 families it was found that 40% family
buy newspaper A, 20% buy newspaper B and 10%
2. Which of the following is the empty set? families buy newspaper C, 5% families buy A and B, 3%
a. { x : x is a real number and x 2 − 1 = 0} buy B and C and 4% buy A and C. If 2% families buy all
b. {x : x is a real number and x 2 + 1 = 0} the three newspapers, then number of families which buy:
A only is
c. {x : x is a real number and x 2 − 9 = 0}
a. 3100 b. 3300
d. {x : x is a real number and x2 = x + 2} c. 2900 d. 1400

3. If X = {8n – 7 – 1: n ∈ N): n∈N} and Y = {49(n–1): n∈N} 12. In a city 20 percent of the population travels by car, 50
then: percent travels by bus and 10 percent travels by both car
a. X ⊆ Y b. Y ⊆ X and bus. Then persons travelling by car or bus is:
c. X = Y d. None of these a. 80 percent b. 40 percent
c. 60 percent d. 70 percent
Venn-Euler Diagrams 13. Suppose A1 , A 2 , A 3 ,........, A 30 are thirty sets each having 5
U+
4. Given the sets A = {1, 2, 3}, B = {3, 4} , C = {4, 5, 6}, then: elements and B1 , B 2 , ......., Bn are n sets each with 3
A ∪ ( B ∩ C ) is 30 n
elements. Let ∪ Ai = ∪ B j = S and each elements of S
a. {3} b. {1, 2, 3, 4} i =1 j =1
ED

c. {1, 2, 4, 5} d. {1, 2, 3, 4, 5, 6} belongs to exactly 10 of the Ai' s and exactly 9 of the B 'j s .
5. If A ⊆ B , then A ∪ B is equal to: Then n is equal to:
a. A b. B ∩ A a. 15 b. 3
c. B d. None of these c. 45 d. None of these

6. If A and B are any two sets, then A ∪ ( A ∩ B ) is equal to: 14. In a class of 55 students, the number of students studying
a. A b. B c. A c d. B c different subjects are 23 in mathematics, 24 in physics, 19
in chemistry, 12 in mathematics and physics, 9 in
7. If the sets A and B are defined as A = {( x, y ) : y
mathematics and chemistry, 7 in physics and chemistry
1
= ,0 ≠ x ∈ R} B = {( x , y ) : y = − x , x ∈ R } , then: and 4 in all the three subjects. The number of students
x who have taken exactly one subject is:
a. A ∩ B = A b. A ∩ B = B a. 6 b. 9
c. A ∩ B = φ d. None of these c. 7 d. 5
8. Let A = [x : x ∈ R, |x| < 1] B = [x : x ∈ R, |x – 1| ≥ 1] and
A ∪ B = R − D, then the set D is: Laws of Algebra of Sets
a. [ x : 1 < x ≤ 2] b. [ x : 1 ≤ x < 2] 15. If A,B and C are any three sets, then A×(B∩C) is equal to:
c. [ x : 1 ≤ x ≤ 2] d. None of these a. (A × B) ∪ (A × C) b. (A × B) ∩ (A × C)
c. (A ∪ B) × (A ∪ C) d. (A ∩ B) × (A ∩ C)
9. If the sets A and B are defined as:
16. If A,B and C are non-empty sets, then (A–B) ∪ (B – A)
A = {( x, y ) : y = e x , x ∈ R}
equals?
B = {( x, y ) : y = x, x ∈ R}, then a. (A ∪ B) – B b. A – (A ∩ B)
a. B ⊆ A b. A ⊆ B c. A ∩ B = φ d. A ∪ B = A c. (A ∪ B) – (A ∩ B) d. (A ∩ B) ∪ (A ∪ B)

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Set Relations and Functions 7
Cartesian Product of Sets 25. With reference to a universal set, the inclusion of a subset
in another, is relation, which is?
17. If A = {2, 4 , 5}, B = {7, 8 , 9}, then n( A × B ) is equal to:
a. Symmetric only
a. 6 b. 9
b. Equivalence relation
c. 3 d. 0
c. Reflexive only
18. If the set A has p elements, B has q elements, then the d. None of these
number of elements in A × B is: 26. Let P = {( x, y ) | x 2 + y 2 = 1, x, y ∈ R} .Then P is:
a. p + q b. p + q + 1
a. Reflexive b. Symmetric
c. pq d. p2
c. Transitive d. Anti-symmetric
19. If P, Q and R are subsets of a set A, then R×(Pc∪Qc)c =? 27. Let R be a relation on the set N of natural numbers defined by
a. (R × P) ∩ (R × Q) b. (R × Q ) ∩ (R × P ) nRm ⇔ n is a factor of m (i.e., n|m). Then R is:
c. (R × P ) ∪ (R × Q ) d. None of these a. Reflexive and symmetric
b. Transitive and symmetric
Relation and Inverse Relation c. Equivalence
d. Reflexive, transitive but not symmetric
20. Let A = {1, 2, 3}. The total number of distinct relations
that can be defined over A is: 28. Let X be a family of sets and R be a relation on X defined
by ‘A is disjoint from B’. Then R is:
a. 29 b. 6
a. Reflexive b. Symmetric
c. 8 d. 5
c. Anti-symmetric d. Transitive
21. Let X = {1, 2, 3, 4, 5} and Y = {1, 3, 5, 7, 9} . Which of the 29. Let R and S be two non-void relations on a set A. Which
U+
following is/are relations from X to Y? of the following statements is false:
a. R1 = {( x, y ) | y = 2 + x, x ∈ X , y ∈ Y } a. R and S are transitive ⇒ R ∪ S is transitive
b. R2 = {(1,1), (2,1), (3, 3), (4, 3), (5, 5)} b. R and S are transitive ⇒ R ∩ S is transitive
ED

c. R3 = {(1,1), (1, 3)(3, 5), (3, 7), (5, 7)} c. R and S are symmetric ⇒ R ∪ S is symmetric
d. R and S are reflexive ⇒ R ∩ S is reflexive
d. R4 = {(1, 3), (2, 5), (2, 4), (7, 9)}
30. The solution set of 8 x ≡ 6(mod 14 ), x ∈ Z , are:
22. Given two finite sets A and B such that n(A) = 2, n(B) = 3.
a. [8] ∪ [6] b. [8] ∪ [14]
Then total number of relations from A to B is?
c. [6] ∪ [13] d. [8] ∪ [6] ∪ [13]
a. 4 b.8
c. 64 d. 9 31. If R be a relation < from A = {1,2, 3, 4} to B = {1, 3, 5}
i.e., (a, b) ∈ R ⇔ a < b, then RoR −1
is:
23. Let A = {1, 2, 3}, B = {1, 3, 5}. A relation R : A → B is
a. {(1, 3), (1, 5), (2, 3), (2, 5), (3, 5), (4, 5)}
defined by R = {(1, 3), (1, 5), (2, 1)}. Then R −1 is defined
b. {(3, 1) (5, 1), (3, 2), (5, 2), (5, 3), (5, 4)}
by:
c. {(3, 3), (3, 5), (5, 3), (5, 5)}
a. {(1,2), (3,1), (1,3), (1,5)}
d. {(3, 3) (3, 4), (4, 5)}
b. {(1, 2), (3, 1), (2, 1)}
c. {(1, 2), (5, 1), (3, 1) Characteristics of Function
d. None of these
32. If A contains 10 elements then total number of functions
Types of Relations defined from A to A is:
a. 10 b. 210
24. The relation R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)} 10
c. 10 d. 210 –
on set A = {1, 2, 3} is:
a. Reflexive but not symmetric 1
33. If f ( y ) = log y, then f ( y ) + f   is equal to:
b. Reflexive but not transitive  y
c. Symmetric and Transitive a. 2 b. 1
d. Neither symmetric nor transitive c. 0 d. –1

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8 Quick Revision NCERT- MATHEMATICS
1 1 41. Domain of definition of the function
34. If f ( x) = + for x > 2, then
x + 2 2x − 4 x − 2 2x − 4 f ( x ) = sin −1 (2 x ) +
π
, for real value x , is:
6
f (11) = ?
 1 1  1 1
7 5 a. − ,  b.  − , 
a. b.  4 2  2 2
6 6
 1 1  1 1
c.  − ,  d. − , 
6 5  2 9  4 4
c. d.
7 7
x2 + x + 2
42. Range of the function f ( x) = ; x ∈ R is:
Domain, Co-domain and Range of Function x2 + x + 1
a. (1, ∞) b. (1,11 / 7)
1
35. Domain of the function is: c. (1, 7 / 3] d. (1, 7 / 5]
x2 −1
a. (−∞, − 1) ∪ (1, ∞) b. (−∞, − 1] ∪ (1, ∞) Kinds of Function
c. (−∞ , − 1) ∪ [1, ∞ ) d. None of these
43. Function f : N → N , f ( x) = 2 x + 3 is:
1 a. One-one onto b. One-one into
36. The domain of the function f ( x ) = is:
| x | −x c. 0 Many-one onto d. Many –one into

a. R + b. R − 44. The function f:R→R defined by f ( x) =

c. R 0 d. R ( x − 1)( x − 2)( x − 3) is:
U+
a. One-one but not onto b. Onto but not one-one
37. The domain of the function f ( x) = x − x 2 c. Both one-one and onto d. Neither one-one nor onto
+ 4 + x + 4 − x is:
ED

45. The function f : R → R defined by f (x ) = e x is:

a. [−4, ∞) b. [ − 4, 4] a. Onto b. Many-one
c. [0, 4] d. [0,1] c. One-one and into d. Many one and onto

38. The domain of the function log( x 2 − 6 x + 6) is: Even and Odd Function
46. Which of the following is an even function?
a. (−∞, ∞) b. (−∞, 3 − 3) ∪ (3 + 3, ∞)
 ax −1 
c. ( −∞,1] ∪ [5, ∞) d. (−∞,1] ∪ [3, ∞ ) a. x  x  b. tan x
 a +1
39. The domain of the derivative of the function a x − a− x ax + 1
c. d.
 tan x −1
, | x | ≤1 2 ax − 1

f ( x) =  1 is: 47. Which of the following is an even function?
 (| x | −1) , | x | > 1
2  ax − 1 
ax + 1
a. f ( x ) = b. f (x ) = x  x 

a. R − {0} b. R − {1} ax − 1  a +1
c. R − {−1} d. R − {−1, 1} a x − a− x
c. f (x ) = d. f ( x ) = sin x
a x + a− x
40. The domain of the function f ( x) = log 3+ x ( x − 1) is: 2

a. (−3, − 1) ∪ (1, ∞) Periodic Function

b. [−3, − 1) ∪ [1, ∞) 1
48. The period of the function f ( x) = 2 cos ( x − π ) is:
3
c. (−3,−2) ∪ (−2, − 1) ∪ (1, ∞)
a. 6π b. 4π
d. [−3, − 2) ∪ (−2, − 1) ∪ [1, ∞]
c. 2π d. π

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Set Relations and Functions 9
πx πx πx 58. Suppose that g ( x) = 1 + x and f ( g ( x )) = 3 + 2 x + x,
49. The function f ( x) = sin + 2cos − tan is periodic
2 3 4
then f(x) is:
with period:
a. 6 b. 3 a. 1 + 2 x 2 b. 2 + x 2
c. 4 d. 12 c. 1 + x d. 2 + x

50. The period of | sin 2 x | is:

Inverse Function
π π
a. b. 59. If f : R → R is given by f ( x) = 3 x − 5, then f −1 (x ) ?
4 2
c. π d. 2π
1
51. The period of the function f (x ) = sin 2 x is: a. Is given by
3x − 5
π
a. b. π x+5
2 b. Is given by
c. 2π d. 3π 3
c. Does not exist because f is not one-one
52. The period of f (x ) = x − [x ], if it is periodic, is:
d. Does not exist because f is not onto
a. f (x ) is not periodic b. 1/2
c. 1 d. 2 60: Let f : R → R be defined by f ( x) = 3 x − 4, then f −1 (x ) is:
1
53. The period of f ( x ) = sin 
 πx   πx 
 + cos  , n ∈ Z , n > 2 is:
a. 3 x + 4 b. x−4
− 3
 n 1   n 
a. 2πn (n − 1) b. 4 n (n − 1) 1 1
c. ( x + 4) d. (x − 4 )
U+
3 3
c. 2n(n − 1) d. None of these

Composite Function NCERT EXEMPLAR PROBLEMS

ED

54. If f : R → R, f ( x ) = 2 x − 1 and g : R → R, g( x ) = x 2
then More than One Answer
( gof ) ( x ) equals? x+2
61. If y = f ( x) = , then:
a. 2 x − 1 2
b. (2 x − 1) 2
x −1
c. 4 x − 2 x + 1
2
d. x 2 + 2 x − 1 a. x = f(y)
b. f(1) = 3
 π
55. f ( x) = sin x + sin  x +  + cos x cos
2 2
c. y increases will x for x < 1
 3
d. f is a rational function of x
 π 5
 x +  and g   = 1, then (gof )( x ) is equal to: 2x −1
 3  4 62. If S is the set of all real x such that is
2 x + 3x 2 + x
3

a. 1 b. –1
positive, then S contains:
c. 2 d. – 2
 3  3 1
a.  −∞, −  b.  − , − 
1  2  2 4
56. If g(x)x2 +x – 2 and ( gof ) ( x) = 2x2 – 5x + 2, then f (x ) is
2
 1 1 1 
equal to: c.  − ,  d.  ,3 
 4 2 2 
a. 2 x − 3 b. 2 x + 3
c. 2 x 2 + 3 x + 1 d. 2 x 2 − 3 x − 1 63. Let g(x) be a function defined on [–1,1]. If the area of the
equilateral triangle with two of its vertices at (0, 0) and
2x − 3
57. If f ( x ) = , then [ f { f (x )}] equals:
x−2 [ x, g ( x)] is 3 / 4, then the function g ( x ) is:
a. x b. –x
a. g ( x ) = ± 1 − x 2 b. g ( x ) = 1 − x 2
x 1
c. d. −
2 x c. g ( x) = − 1 − x 2 d. g ( x) = 1 + x 2

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10 Quick Revision NCERT- MATHEMATICS
64. If f ( x) = cos[π ]x + cos[−π ]x, where [ x] stands for the
2 2 Assertion and Reason
greatest integer function, then:
Note: Read the Assertion (A) and Reason (R) carefully to mark
a. f (π / 2) = −1 b. f (π ) = 1
the correct option out of the options given below:
c. f ( −π ) = 0 d. f (π / 4) = 1
a. If both assertion and reason are true and the reason is the
b−x correct explanation of the assertion.
65. Let f : (0,1) → R be defined by f ( x) = , where b is
1 − bx b. If both assertion and reason are true but reason is not the
a constant such that 0 < b < 1. Then: correct explanation of the assertion.
a. f is not invertible on (0, 1) c. If assertion is true but reason is false.
−1 1
b. f ≠ f one (0, 1) and f '(b) = d. If the assertion and reason both are false.
f '(0)
e. If assertion is false but reason is true.
1
c. f = f −1 on (0, 1) and f '(b) = 70. Let F(x) be an indefinite integral of sin2 x:
f '(0)
Assertion: The function F(x) satisfies F(x + π) = F (x) for
d. f −1 is differentiable on (0, 1)
all real x.
66. For every integer n, let an and bn be real numbers. Let Reason: sin2 (x + π) = sin2 x for all real x.
function f : R → R be given by
71. Let f(x) = 2 + cos x for all real x:
f ( x) = {ab ++ cos
n
n
sin π x, for x ∈ [2n, 2n + 1]
π x, for x ∈ (2n − 1, 2n) ,
for all integers n.
Assertion: For each real t, there exists a point c in [t,t +π]
If f is continuous, then which of the following hold(s) for such that f’(c) = 0.
U+
all n? Reason: f’(t) = f (t + 2π) for each real t.
a. an −1 − bn −1 = 0 b. an − bn = 1
c. an − bn +1 = 1 d. an −1 − bn = −1 x2
ED

72. Assertion: The curve y = − + x + 1 is symmetric with

2
67. For every pair of continuous function f , g : [0,1] → R such respect to the line x = 1.
that max { f ( x) : x ∈ [0,1]} = max{g ( x) : x ∈ [0,1]}. Reason: A parabola is symmetric about its axis.
The correct statement: (s) is (are)
a. [ f (c)]2 + 3 f (c) = [ g (c)]2 + 3 g (c) for some c ∈ [0,1] 73. Consider the following relations. R= {(x,y)| x,y}are real
numbers and x=xy for some rational number w}
b. [ f (c )]2 + f (c) = [ g (c )]2 + 3 g (c ) for some c ∈ [0,1]
c. [ f (c )]2 + 3 f (c) = [ g (c )]2 + g (c ) for some c ∈ [0,1]  m p  
S =  ,   m,n,p,q are integer such that n.q ≠ 0 and
 n q  
d. [ f (c)]2 = [ g (c)]2 for some c ∈ [0,1]
qm = pn}
 π π
68. Let f :  − ,  → R be given byf(x)=[log(sec x+tan x)]3. Assertion: S is an equivalence relation but R is not an
 2 2
equivalence relation.
Then:
a. f(x) is an odd function Reason: R and S both are symmetric.
b. f(x) is a one-one function
74. Let R be a relation on the set N of natural numbers defined
c. f(x) is an onto function
d. f(x) is an even function by n Rm ⇔ n is a factor of m (i.e., n | m) :
Assertion: R is not an equivalence relation
69. Let L be the set of all straight lines in the Euclidean plane.
Reason: R is not symmetric
Two lines l1 and l2 are said to be related by the relation R
if l1 is parallel to l2. Then the relation R is: 75. Let A = {1,2,3}and B = {3,8}?
a. Reflexive c. Symmetric Assertion: (A∪B) × (A∩B) = {(1,3), (2,3), (3,3)(8,3)}
b. Transitive d. Equivalence
Reason: (A×B) ∩ (B×A) = {(3,3)}

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Set Relations and Functions 11
76. Assertion: f:R→R is a function defined by f ( x) = 5 x + 3. 1995  r  
83. Let f(x) = g9(x), then the value of  ∑    is: (where
x −3  r =1  1996  
If g = f −1 , then g ( x) = .
5 [.] denotes the greatest integer function)
Reason: If f : A → B is a bijection and g : B → A is the a. 995 b. 996
inverse of f, then fog is the identity function on A. c. 997 d. 998

77. Let X and Y be two sets: 1996

 r 
Assertion: X ∩ (Y ∩ X ) ' = φ
84. Let f ( x) = g 4 ( x), then ∑ f  1997  is:
r =1

Reason: If X∪Y has m elements and X∩Y has n elements a. zero b. even
then symmetric difference X ∆Y has m − n elements c. odd d. none of these

78. Let f be a function defined by f ( x) = ( x − 1) 2 + 1,( x ≥ 1) 85. The value of g5 (x) + g5 (1–x) is:
−1
Assertion: The set {x : f ( x) = f ( x)} = {1, 2} a. 1 b. 5
c. 10 d. none of these
Reason: f is a bijection and f −1 ( x) = 1 + x − 1, x ≥ 1
2 n −1
 r 
79. Consider the following relation R on the set of real square 86. The value of ∑ 2 f  2n  is:
r =1

matrices of order 3. R = {( A, B ) : A = P −1 BP for some a. 0 b. 2n–1

invertible matrix P} c. 2n d. none of these
Assertion: R is an equivalence relation.
87. If the value of ∑ f 
2n
r  1
U+
Reason: For any two invertible 3 × 3 matrices M and N = + 987, then the value
r =0  2 n + 1  1+ a
sin x of n is:
80. Let f ( x) = sin x + cos x, g ( x) = ?.
1 − cos x
ED

a. 493 b. 494
Assertion: f is neither an odd function nor an even c. 987 d. 988
function.
Reason: g is an odd function.
Paragraph-II
81. Assertion: A function f:R→R satisfied the equation f (x) – f(y) f ( x)
Let F ( x) = f ( x ) + g ( x), G ( x) = f ( x) − g ( x ) and H ( x) = ,
= x – y ∀ x, y ∈ R and f (3) = 2, then f ( xy ) = xy − 1 g ( x)

Reason: f ( x) = f (1/ x)∀x ∈ R, x ≠ 0, and f (2) = 7 / 3 if where f ( x) = 1 − 2sin 2 x and g ( x) = cos 2 x, ∀f : R → [ −1,1] and

x2 + x + 1 g : R → [−1,1].
f ( x) =
x2 − x + 1
88. Domain and range of H (x) are respectively:
82. Assertion: Let A{2, 3, 7, 9}and B = {4, 9, 49, 81} f:A → a. R and {1}
2
B is a function defined as f(x) = x . Then is a bijection b. R and {0, 1}
from A to B. π
c. R ∼ {(2n + 1) }, and{1}, n ∈ I
Reason: A function f from a set A to a set B is a bijection 4
if f(A) = B and f(x1) ≠ f(x2) if x1 ≠ x2 for all x1, x2 ∈A and π
d. R ∼ (2 n + 1)  , and{0,1}, n ∈ I
n(A) = n(B).  2

89. If F: R → [–2, 2], then:

Comprehension Based
a. F (x) is one – one function
Paragraph-I b. F (x) is onto function
ax c. F (x) is into function
Let f be a function satisfying f ( x) = = g a ( x)(a > 0)
ax + a d. none of the above

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12 Quick Revision NCERT- MATHEMATICS
90. Which statement is correct? 95. Let f1 : R → R, f 2 :[0, ∞] → R, f3 : R → R and f 4 : R → [0, ∞) be
a. period of f(x), g(x) and F(x) makes are AP with
| x |, if x < 0
common difference π/3 defined by f1 ( x) =  x ; f 2 ( x) = x 2 ;
 e , if x ≥ 0
b. period of f(x), g(x) and F(x) are same and is equal to 2π
sin x,if x < 0
c. sum of periods of f(x), g(x) and F(x) is 3π f 3 ( x) =  and
d. sum of periods of f(x), g(x) and F(x) is 6π  x, if x ≥ 0
 f [ f ( x )], if x < 0
91. Which statement is correct? f 4 ( x) =  2 1
f [
 2 1f ( x )] − 1, if x≥0
a. the domain of G(x) and H(x) are same
Column I Column II
b. the rang of G(x) and H(x) are same
c. the union of domain of G(x) and H(x) are all real (A) f4 is 1. onto but not one-one

d. the union of domain of G(x) and H(x) are rational (B) f3 is 2. neither continuous nor one-one
numbers (C) f2 o f1 is 3. differentiable but not one-one
(D) f2 is 4. continuous and one-one
92. If the solutions of F (x) – G (x) = 0 are x1,x2,x3,…xn where
a. A→3; B→1; C→4; D→2
x∈[0, 5π], then:
b.A→1; B→3; C→4; D→ 2
a. x1,x2,x3,…xn are in AP with common difference π/4
c. A→3; B→1; C→2; D→4
b. the number of solution of F (x) – G (x) = 0 is 10, ∀x∈[0, 5π].
d. A→1; B→3; C→2; D→4
c. the sum of all solutions of F ( x) − G ( x) = 0, ∀x ∈ [0,5π ]
is 25π Integer
U+
d. (b) and (c) are correct
96. If maximum and minimum values of
16 − x 20 −3 x
Match the Column f ( x) = C2 x −1 + C4 x −5 are λ and µ respectively, then
ED

93. Let the functions defined in Column I have domain the value of λ + µ must be:

(–π/2, π/2) and range (–∞,∞)?

97. The least period of the function
Column I Column II
 π [ x]  πx  π [ x ]  is λ, then the value of
(A) 1 + 2x 1. onto but not one-one sin   + cos   + tan  
 12   4   3 
(B) tan x 2. one-one but not onto
201λ must be (where [⋅] denotes the greatest integer
3. one-one and onto
function:
4. neither one-one nor onto
a. A→ 2; B→ 3 b. A→ 2; B→ 4  y2 y2 
98. If f  2 x 2 + , 2 x 2 −  = xy, then the value of f(60, 48)
c. A→ 1; B→ 3 d. A→ 4; B→ 1  8 8 

x2 − 6 x + 5 + f(80, 48) + f(13, 5) must be:

94. Let f ( x) = .?
x2 − 5x + 6
99. Let f be a function form the set of positive integers to the
Column I Column II
set of real numbers i.e., f: N→R, such that (i) f (1) = 1 ,
(A) If –1< x <1, then f(x) satisfies 1. 0 < f (x) < 1
(B) If 1 < x < 2, then f(x) satisfies 2. f (x) < 0 (ii) f (1) + 2 f (2) + 3 f (3) + … + nf (n) = n(n + 1) f (n) from
(C) If 3 < x < 5, then f(x) satisfies 3. f (x) > 0 1
n ≥ 2, then the value of must be:
(D) If x > 5, then f(x) satisfies 4. f (x) < 1 f (1004)
a. A→3; B→1; C→2; D→2
(61x + 80 x 2 )
1
b.A→1; B→2; C→2; D→1 100.If f ( x) = x + ∫ ( xy 2 + x 2 y ) f ( y ) dy and x + , the
c. A→1; B→3; C→2; D→4 0
λ
d.A→4; B→1; C→3; D→2 n the value of λ must be:

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Set Relations and Functions 13
ANSWER 6. (a) A ∩ B ⊆ A .
1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Hence, A ∪ ( A ∩ B ) = A .
d b a b c a c b c b
11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 1 1
7. (c) Since y = , y = − x meet when − x =
b c c d b c b c a, b a x x
21. 22. 23. 24. 25. 26. 27. 28. 29. 30.
⇒ x = −1 , which does not give any real value of x
2
a,b,c c c a d b d b a c
31. 32. 33. 34. 35. 36. 37. 38. 39. 40.
Hence, A ∩ B = φ .
c c c c a b d c c c
8. (b) A = [ x : x ∈ R, − 1 < x < 1]
41. 42. 43. 44. 45. 46. 47. 48. 49. 50.
a c b b c a b a d b B = [ x : x ∈ R : x − 1 ≤ −1
51. 52. 53. 54. 55. 56. 57. 58. 59. 60. Or x − 1 ≥ 1] = [ x : x ∈ R : x ≤ 0 or x ≥ 2]
b c c b a a a b b c
∴ A∪B = R−D
61. 62. 63. 64. 65. 66. 67. 68. 69. 70.
Where, D = [ x : x ∈ R ,1 ≤ x < 2]
a, d a, d b, c a, c a b, d a, d a,b,c All d
71. 72. 73. 74. 75. 76. 77. 78. 79. 80.
9. (c) Since, y = e x and y = x do not meet for any x ∈ R
b a c a b c b a b b
81. 82. 83. 84. 85. 86. 87. 88. 89. 90.
∴ A∩B =φ .
b a c b a b c c d c
10. (b) Since, 4n − 3n − 1 = (3 + 1)n − 3n − 1
91. 92. 93. 94. 95. 96. 97. 98. 99. 100.
c d a b d 2345 4824 112 2008 119 = 3n + n C1 3n −1 + n C2 3n − 2 + ..... + n Cn −1 3 + n Cn − 3n − 1
= n C2 32 + n C3 .33 + ..... + n Cn 3n
SOLUTION
( n C0 = n Cn , n C1 = n Cn −1 etc.)
U+
Multiple Choice Questions
1. (d) Since, intelligence is not defined for students in a class = 9[ n C2 + n C3 (3) + ..... + n Cn 3n −1 ]
i.e., Not a well defined collection. ∴ 4 n − 3 n − 1 is a multiple of 9 for n ≥ 2 .
ED

For n = 1, 4 n − 3 n − 1 = 4 − 3 − 1 = 0 ,
2. (b) Since x 2 + 1 = 0, gives x 2 = −1
⇒
For n = 2, 4 n − 3 n − 1
x = ±i
∴ x is not real but x is real (given) = 16 − 6 − 1 = 9
∴ No value of x is possible. ∴ 4 n − 3 n − 1 is a multiple of 9 for all n ∈ N
∴ X contains elements which are multiples of 9 and clearly Y
3. (a) Since 8n − 7n − 1 = (7 + 1) n − 7n − 1 contains all multiples of 9.
= 7 n + n C1 7 n −1 + n C 2 7 n − 2 + ..... + n C n−1 7 + n C n − 7 n − 1 ∴ X ⊆ Y , ∴ X ∪Y = Y .

= n C 2 7 2 + n C 3 7 3 + ..... + n C n 7 n ( n C 0 = n C n , n C1 = n C n −1 etc.) 11. (b) n(A) = 40% of 10,000 = 4,000

= 49 [ C 2 + C 3 (7) + ...... + C n 7
n n n n−2
] n(B) = 20% of 10,000 = 2,000
n(C) = 10% of 10,000 = 1,000
∴ 8 − 7 n − 1 is a multiple of 49 for n ≥ 2 .
n
n (A ∩ B) = 5% of 10,000
For n = 1 , 8 n − 7 n − 1 = 8 − 7 − 1 = 0 ;
= 500, n (B ∩ C) = 3% of 10,000 = 300
For n = 2, 8 n − 7 n − 1 = 64 − 14 − 1 = 49 n(C ∩ A) = 4% of 10,000 = 400, n(A ∩ B ∩ C)
∴ 8n − 7 n − 1 is a multiple of 49 for all n ∈ N . = 2% of 10,000 = 200
∴ X contains elements which are multiples of 49 and clearly We want to find n(A ∩ Bc ∩ Cc) = n[A ∩ (B ∪ C)c]
Y contains all multiplies of 49. = n(A) – n[A ∩ (B ∪ C)] = n(A) – n[(A ∩ B) ∪ (A ∩ C)]
∴ X ⊆Y . = n(A) – [n(A ∩ B) + n(A ∩ C) – n(A ∩ B ∩ C)]
4. (b) B ∩ C = {4 } , = 4000 – [500 + 400 – 200] = 4000 – 700 = 3300.

∴ A ∪ ( B ∩ C ) = {1, 2, 3, 4}. 12. (c) n(C) = 20, n(B) = 50, n(C ∩ B) = 10

Now, n(C ∪ B) = n(C) + n(B) – n(C ∩ B)
5. (c) Since A ⊆ B
= 20 + 50 – 10 = 60.
⇒ A∪B =B. Hence, required number of persons = 60%.

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14 Quick Revision NCERT- MATHEMATICS
 30  1 22. (c) Here n( A × B) = 2 × 3 = 6
13. (c)O(S) = O  ∪ Ai  = (5 × 30 ) = 15
 i =1  10 Since, every subset of A × B defines a relation from A to
Since, element in the union s belongs to 10 of ai' s B, number of relation from A to B is equal to number of
 n  3n n subsets of A × B = 26 = 64, which is given in (c).
Also, O(S) = O ∪ B j  = =
 j =1  9 3
23. (c) ( x, y ) ∈ R ⇔ ( y, x) ∈ R −1 ,
n
∴ = 15 ⇒ n = 45 . ∴ R −1 = {(3,1), (5,1), (1, 2)} .
3

14. (d) n(M) = 23, n(P) = 24, n(C)= 19 24. (a) Since (1, 1); (2, 2); (3, 3) ∈ R therefore R is reflexive.
n(M∩P) = 12, n(M∩C)= 9, n(P ∩ C)=7 n(M ∩ P∩ C) = 4 (1, 2) ∈ R but (2, 1) ∉ R, therefore R is not symmetric. It
We have to find n(M∩P′∩C′), n(P ∩M ′∩C′ ), n (C∩M ′∩ P ′) can be easily seen that R is transitive.
Now n (M ∩ P′ ∩ C′) = n[M ∩ (P ∪ C)′] 25. (d) Since A ⊆ A
= n(M)– n(M ∩ (P ∪ C)) ∴ Relation ' ⊆' is reflexive.
= n ( M ) − n[( M ∩ P ) ∪ ( M ∩ C )]
Since, A ⊆ B , B ⊆ C ⇒ A ⊆ C
= n(M) – n(M ∩ P)– n(M ∩ C) + n(M ∩ P ∩ C)
∴ Relation ' ⊆' is transitive.
= 23 –12 – 9 + 4 = 27 –21 = 6
But, A ⊆ B, ⇒ B ⊆ A ,
n(P ∩ M′ ∩ C′) = n[P ∩ (M ∪ C)′]
= n(P)– n[P ∩ (M ∪ C)] = n( P ) − n[( P ∩ M ) ∪ ( P ∩ C )] ∴ Relation is not symmetric.
= n(P) – n(P ∩ M) – n(P ∩ C) + n(P ∩ M ∩ C) 26. (b) Obviously, the relation is not reflexive and transitive
= 24 – 12 – 7 + 4 = 9 but it is symmetric, because x 2 + y 2 = 1 ⇒ y 2 + x 2 = 1 .
n(C ∩ M′ ∩ P′)
U+
27. (d) Since n | n for all n ∈ N , therefore R is reflexive. Since
= n(C) – n(C ∩ P) – n(C ∩ M)+ n(C ∩ P ∩ M)
= 19 – 7 – 9 + 4 = 23 – 16 = 7 2 | 6 but 6 | 2 , therefore R is not symmetric.
Let n R m and m R p ⇒ n|m
15. (b) A × (B ∩ C) = (A × B) ∩ (A × C). It is distributive law.
ED

and m|p ⇒ n|p ⇒ nRp. So R is transitive.

16. (c) (A – B) ∪ (B – A) = (A ∪ B) – (A ∩ B).
28. (b) Clearly, the relation is symmetric but it is neither
reflexive nor transitive.
A B
29. (a) Let A = {1, 2, 3} and R = {(1, 1), (1, 2)}, S
= {(2, 2) (2, 3)} be transitive relations on A.
A∩B
A–B
B–A Then R ∪ S = {(1, 1); (1, 2); (2, 2); (2, 3)}
Obviously, R ∪ S is not transitive.
17. (b) A × B = {(2, 7), (2, 8), (2, 9), (4, 7), (4, 8), (4, 9), (5,7),
Since (1, 2) ∈ R ∪ S and (2, 3) ∈ R ∪ S but (1, 3) ∉ R ∪ S .
(5, 8), (5, 9)}
n(A × B) = n(A) . n(B) = 3 × 3 = 9. 30. (c) 8 x − 6 = 14 P (P ∈ Z )
18. (c) n( A × B) = pq . 1
⇒ x = [14 P + 6 ] , x ∈ Z
8
19. (a, b) R × ( P ∪ Q ) = R × [( P ) ∩ (Q ) ]
c c c c c c c
1
⇒ x = (7 P + 3 )
= R × ( P ∩ Q) = ( R × P) ∩ ( R × Q) 4
= ( R × Q) ∩ ( R × P) ⇒ x = 6, 13, 20, 27, 34, 41, 48,.....
∴ Solution set = {6, 20, 34, 48,..} ∪ {13, 27, 41, ...}
20. (a) n( A × A) = n( A).n( A) = 3 2 = 9 = [6] ∪ [13].
So, the total number of subsets of A × A is 2 9 and a subset of Where [6],[13] are equivalence classes of 6 and 13 respectively.
A × A is a relation over the set A.
31. (c) We have, R={(1,3); (1,5); (2,3); (2,5); (3,5);(4,5)}
21. (a, b, c) R4 is not a relation from X to Y, R −1 = {(3, 1), (5, 1), (3, 2), (5, 2); (5, 3); (5, 4)}
because (7, 9) ∈ R 4 but (7, 9) ∉ X × Y . Hence RoR −1 = {(3, 3); (3, 5); (5, 3); (5, 5)}

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Set Relations and Functions 15
n
32. (c) According to formula, total number of functions = n 1 1 1
f ′(1 − 0) = = ; f ′(1 + 0) =
Here, n = 10 So, total number of functions = 1010. 1 + (1 − 0) 2
2 2
33. (c) Given f (y ) = log y ∴ f ′(−1) does not exist.
⇒ f (1/ y ) = log (1/ y ) ∴ domain of f ′( x ) = R − {−1} .
1
then f (y ) + f   = log y + log(1 / y ) = log 1 = 0. 40. (c) f (x ) is to be defined when x 2 − 1 > 0
  y
⇒ x 2 > 1, ⇒ x < −1 or x > 1 and 3 + x > 0
1 1 ∴ x > −3 and x ≠ −2
34. (c) f ( x) = +
x + 2 2x − 4 x − 2 2x − 4 ∴ Dr = (−3, − 2) ∪ (−2, − 1) ∪ (1, ∞) .
1 1
f (11) = + π π 1
41. (a) − ≤ sin −1 (2 x ) ≤ ⇒− ≤ 2x ≤ 1
11 + 2 18 11 − 2 18 6 2 2
1 1 3− 2 3+ 2 6  1 1
= + = + = . ⇒ x ∈ − ,  .
3+ 2 3− 2 7 7 7  4 2
1
35. (a) For domain, x 2 − 1 > 0 ⇒ (x − 1)( x + 1) > 0 42. (c) f ( x ) = 1 + 2
⇒ Range = (1, 7 / 3 ] .
 1 3
⇒ x < −1 or x > 1 ⇒ x ∈ (−∞,−1) ∪ (1, ∞) . x+  +
 2 4

36. (b) For domain, | x | − x > 0 ⇒ | x | > x . 43. (b) f is one-one because f ( x1 ) = f ( x2 )
−
This is possible, only when x ∈ R . ⇒ 2 x1 + 3 = 2 x2 + 3 ⇒ x1 = x2
x−3
U+
37. (d) f ( x) = x − x 2 + 4 + x + 4 − x clearly f ( x) is Further f −1 ( x) = ∉ N (domain) when x = 1, 2, 3 etc.
2
defined if 4 + x ≥ 0 ⇒ x ≥ −4 ∴ f is into which shows that f is one-one into.
4 −x ≥0 ⇒ x ≤ 4
44. (b) We have f ( x) = ( x − 1)( x − 2)( x − 3)
ED

x (1 − x ) ≥ 0 ⇒ x ≥ 0 and x ≤ 1
∴ Domain of f = (−∞ , 4 ] ∩ [−4 , ∞ ) ∩ [0 , 1] = [0 , 1] . ⇒ f (1) = f (2) = f (3) = 0 ⇒ f (x ) is not one-one
For each y ∈ R, there exists x ∈ R such that f (x ) = y .
38. (c) The function f ( x) = log( x − 6 x + 6) is defined
2
Therefore f is onto.
when log( x 2 − 6 x + 6) ≥ 0 Hence, f : R → R is onto but not one-one.

⇒ x 2 − 6 x + 6 ≥ 1 ⇒ ( x − 5)( x − 1) ≥ 0 45. (c) Function f : R → R is defined by f (x ) = e x . Let

This inequality hold if x ≤ 1 or x ≥ 5 . Hence, the domain x 1 , x 2 ∈ R and f ( x 1 ) = f ( x 2 ) or e 1 = e 2
x x
or x 1 = x 2 .
of the function will be (−∞,1] ∪ [5, ∞) . Therefore f is one-one. Let f (x ) = e x = y . Taking log on
both sides, we get x = log y . We know that negative real
1
 2 ( − x − 1), x < −1 numbers have no pre-image or the function is not onto and

39. (c) f ( x ) =  tan −1 x, − 1 ≤ x ≤ 1 zero is not the image of any real number. Therefore
1 function f is into.
 ( x + 1), x > 1
 2  ax − 1 
46. (a) We have: f (x ) = x  x 

 1  a +1 
− 2 , x < −1
 1 
  a− x − 1   −1
 1 f (− x ) = − x  − x  = −x ax 
⇒ f ′( x) =  , −1 < x < 1 a +1  1 
1 + x  
2
 + 1 
1  ax 
 , x >1
2  1 − ax   ax − 1 
= − x  =
 x  x  = f (x )

1+ a  a +1 
x
1 1 1 
f ′(−1 − 0) = − ; f ′(−1 + 0) = =
2 1 + ( −1 + 0) 2 2 So, f (x ) is an even function.

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16 Quick Revision NCERT- MATHEMATICS
47. (b) In option 2π
⇒ sin 2 x has period = =π
−x
a +1 1+ a a +1 x x 2
(a) f (− x) = = =− x = − f ( x)
a− x −1 1 − a x a −1 Now, if f (x ) has period p then | f ( x ) | has period
p
So, It is an odd function. 2
π
In option ⇒ | sin 2 x | has period = .
2
a− x −1 (1 − a x ) (a x − 1)
(b) f (− x) = (− x) − x = −x = x = f ( x)
a +1 1+ ax (a x + 1) 1 − cos 2 x 2π
51. (b) sin 2 x = ⇒ Period = = π.
So, It is an even function. 2 2
In option 52. (c) Let f (x ) be periodic with period T.
a− x − a x f ( x + T ) = f ( x) for all x∈R
(c) f (− x ) = − x = − f (x ) Then,
a + ax
⇒ x + T − [x + T ] = x − [x ] for all x ∈ R
So, It is an odd function.
In option ⇒ x + T − x = [x + T ] − [x ]
(d) f (− x ) = sin(− x ) = − sin x = − f (x ) ⇒ [x + T ] − [x ] = T for all x ∈ R ⇒ T = 1, 2, 3, 4 ,........
So, It is an odd function. The smallest value of T satisfying,
f (x + T ) = f (x ) for all x ∈ R is 1.
1 x π
48. (a) f ( x ) = 2 cos (x − π ) = 2 cos  −  Hence f (x ) = x − [ x ] has period 1.
3 3 3
Now, since cos x has period 2π  πx  π x 
53. (c) f ( x) = sin   + cos  
x π 2π  n −1   n 
⇒ cos  −  has period = 6π
 3 3  1
U+
 πx  2π
3 Period of sin = = 2(n − 1 )
n −1   π 
x π  
⇒ 2 cos  −  has period = 6 π . n −1
3 3
 π x  2π
ED

49. (d) sin x has period = 2π

Period of cos  = = 2n
 n  π 
πx 2π  
⇒ sin has period = =4 n
2 π
Hence period of f (x ) is LCM of 2 n and 2(n − 1) ⇒ 2n(n − 1) .
2
∵ cos x has period = 2π 54. (b) gof (x ) = g{ f (x )} = g (2 x − 1) = (2 x − 1) 2 .
πx 2π
⇒ cos has period = =6
3 π 55. (a) f ( x ) = sin 2 x + sin 2 (x + π / 3) + cos x cos( x + π / 3)
3 1 − cos 2 x 1 − cos(2 x + 2π /3)
πx = +
⇒ 2 cos has period = 6 2 2
3
1
∵ tan x has period = π + {2 cos x cos( x + π / 3)}
2
πx π
⇒ tan has period = =4. 1
4 π = [1 − cos 2 x + 1 − cos(2 x + 2π /3)
4 2
L.C.M. of 4, 6 and 4 =12, period of f (x ) = 12. + cos(2 x + π / 3) + cos π / 3]
1 5  2π   π 
=  − {cos 2 x + cos  2 x + } + cos  2 x +  
(1 − cos 4 x) 2 2  3   3 
50. (b) Here | sin 2 x |= sin 2 2 x =
2 1 5  π π  π 
= − 2 cos  2 x +  cos + cos  2 x +   = 5 / 4
π 2  2  3 3  3 
Period of cos 4 x is .
2 for all x.
π ∴ gof ( x ) = g( f ( x )) = g(5 / 4 ) = 1 [∵ g(5/4) =1 (given)]
Hence, period of | sin 2 x | will be
2
Hence, gof (x ) = 1, for all x.
∵ sin x has period = 2π

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Set Relations and Functions 17
56. (a) g(x ) = x + x − 2
2 NCERT Exemplar Problems

⇒ ( gof ) ( x) = g[ f ( x)] = [ f ( x)]2 + f ( x) − 2 More than One Answer

x+2
Given,
1
( gof ) ( x) = 2 x 2 − 5 x + 2 61. (a, d) Given, y = f ( x) =
2 x −1
⇒ yx − y = x + 2 ⇒ x( y − 1) = y + 2
1 1
∴ [ f ( x)]2 + f ( x) − 1 = 2 x 2 − 5 x + 2
2 2 y+2
⇒ y= ⇒ x = f ( y ).
⇒ [ f ( x)]2 + f ( x) = 4 x 2 − 10 x + 6 y −1
Here, f (1) does not exist, so domain ∈ R − {1}
⇒ f ( x)[ f ( x) + 1] = (2 x − 3)[(2 x − 3) + 1]
f :R→R. dy ( x − 1) ⋅1 − ( x + 2) ⋅1 3
⇒ = =−
dx ( x − 1) 2 ( x − 1) 2
 2x − 3  ⇒ f ( x) is decreasing for all x ∈ R − {1}.
2 −3
x−2 
57. (a) f [ f ( x)] =  =x Also, f is rational function of x.
 2x − 3 
 −2 + – + – + ∞
 x−2  62. (a, d) −∞
–1 –1/2 0 1/2
2x −1
58. (b) g ( x) = 1 + x Since, >0
2 x3 + 3x 2 + x
and f ( g ( x)) = 3 + 2 x + x . . . (i) (2 x − 1)
⇒ >0
⇒ f (1 + x ) = 3 + 2 x + x x(2 x 2 + 3x + 1)

Put 1 + x = y (2 x − 1)
⇒ >0
U+
x (2 x + 1)( x + 1)
⇒ x = ( y −1)2
Hence, the solution set is,
then, f ( y) = 3 + 2( y −1) + ( y −1)
2
x ∈ (−∞, −1) ∪ (−1/ 2, 0) ∪ (1/ 2, ∞)
ED

= 2 + y2 63. (b, c) A

therefore, f ( x) = 2 + x .
2

59. (b) Clearly, f : R → R is a one-one onto function. So, it is

invertible.
B C
Let f (x ) = y. (0,0) (x, g(x))
y+5 3
then, 3 x − 5 = y ⇒ x = Since, area of equilateral triangle = ( BC ) 2
3 4
y +5 3 3 2
⇒ f −1 ( y ) = . ⇒ = ⋅ [ x + g 2 ( x )] ⇒ g 2 ( x ) = 1 − x 2 c
3 4 4
x+5 ⇒ g ( x ) = 1 − x 2 or − 1 − x
2
Hence, f −1 ( x) = .
3
64. (a, c) Since, f ( x) = cos[π 2 ]x + cos[−π 2 ]x
60. (c) f (x ) = 3 x − 4 = y ⇒ f ( x) = cos(9) x + cos(−10) x,
⇒ y = 3x − 4
(using [π 2 ] = 9 and [−π 2 ] = −10 )
y+4
⇒ x= π  9π
3 ∴ f   = cos + cos 5π = −1
2 2
y+4 f (π ) = cos 9π + cos10π = −1 + 1 = 0
⇒ f −1 ( y ) =
3 f (−π ) = cos 9π + cos10π = −1 + 1 = 0
x+4 π  9π 10π 1 1
⇒ f −1 ( x) = . f   = cos + cos = +0 =
3 2 2 4 2 2

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18 Quick Revision NCERT- MATHEMATICS
b−x 68. (a, b, c) (a) If f '( x) > 0, ∀x ∈ (a, b), then f ( x) is an
65. (a) Here, f ( x) = , where 0 < b < 1, 0 < x < 1
1 − bx increasing function in (a,b) and thus f ( x) is one-one
For function to be invertible it should be one-one onto.
function in (a, b).
∴ Check Range:
(b) If range of f ( x) = codomain of f ( x), then f ( x) is an
b−x
Let f ( x) = y ⇒ y = onto function.
1 − bx
(c) A function f ( x) is said to be odd function, if
⇒ y − bxy = b − x ⇒ x(1 − by ) = b − y
f (− x) = − f ( x), ∀x ∈ R i.e.,
b− y
⇒ x= , where 0 < x < 1 f (− x) + f ( x) = 0, ∀x ∈ R
1 − by
b− y b− y b− y f ( x) = [ln(sec x + tan x)]3
∴ 0< < 1⇒ > 0 and <1
1 − by 1 − by 1 − by 3[ln(sec x + tan x )]2 (sec x tan x + sec 2 x )
f '( x) =
1 (sec x + tan x)
⇒ y < b or y > . . .(i)
b  −π π 
f '( x ) = 3sec x[ln (sec x + tan x )]2 > 0, ∀x ∈  , 
(b − 1)( y + 1) 1  2 2
< 0 −1 < y < . . .(ii)
1 − by b f ( x) is an increasing function.
From Equation (i) and (ii), we get ∴ f ( x) is an one-one function.
 1 π x 
y ∈  −1,  ⊂ codomain ; Thus, f ( x) is not invertible. (sec x + tan x ) = tan  +  ,
 b  4 2
66. (b, d) f (2n) = an , f (2n + ) = an f (2n − ) = bn + 1  π π π x 
x ∈  − ,  , then 0 < tan  +  < ∞
U+
As
⇒ an − bn = 1 f (2n + 1) = an  2 2   4 2
⇒ 0 < sec x + tan x < ∞
f ((2n + 1) − ) = an f ((2 n + 1) + ) = bn +1 − 1
⇒ −∞ < ln (sec x + tan x ) < ∞
ED

⇒ an = bn +1 − 1 or an − bn +1 = −1
−∞ < [ln (sec x + tan x )]3 < ∞
Or an −1 − bn = −1
⇒ −∞ < f ( x) < ∞
67. (a, d) Plan if a continuous function has value of opposite Range of f ( x) is R and thus f ( x) is an onto function.
sign inside an interval, f , g : [0,1] → R 3
  1 
We take two cases. Let f and g attain their common f (− x) = [ln (sec x − tan x )]3 =  ln  
maximum value at p.   sec x + tan x  
⇒ f ( p ) = g ( p ), where p ∈ [0,1] f (− x) = −[ln (sec x + tan x)]3
Let f and g attain their common maximum value at f ( x) + f (− x) = 0
different points. ⇒ f ( x) is an odd function.
⇒ f (a ) = M and g (b ) = M
69. (a, b, c, d) Here l1 Rl2
⇒ f (a ) − g (a) > 0 and f (b) − g (b) < 0 l1 is parallel l2 and also l2 is parallel to l1 , so it is symmetric.
⇒ f (c) − g (c) = 0 for some c ∈ [0,1] as f and g are Clearly, it is also reflexive and transitive. Hence it is
continuous functions. equivalence relation.
⇒ f (c) − g (c) = 0 for some c ∈ [0,1] for all cases. . . . (i)
Assertion and Reason
Option (a) ⇒ f 2 (c) − g 2 (c) + 3[ f (c) − g (c)] = 0 which is
1 − cos 2 x
true from Equation (i). 70. (d) Given, F ( x ) = ∫ sin 2 x dx = ∫ dx
2
Option (d) ⇒ f 2 (c) − g 2 (c) = 0 which is true from 1
⇒ F ( x) = (2 x − sin 2 x ) + c
Equation (i) 4
Now, if we take f ( x) = land g ( x) = 1, ∀x ∈ [0,1] Since, F ( x + π ) ≠ F ( x) Hence, Assertion is false.
Option (b) and (c) does not hold. But reason is true as sin 2 x is periodic with period π .

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Set Relations and Functions 19
71. (b) Given, f ( x) = 2 + cos x, ∀ x ∈ R y −3
76. (c) Let y = 5 x + 3 ⇒ x =
Assertion: 5
y −3
There exists a point c ∈ [t , t + π ], where f '(c) = 0 ⇒ g ( y) =
5
Hence, Assertion is true. x−3
Or g ( x) = , so Assertion is true.
Reason: f (t ) = f (t + 2π ) is true. But reason not a correct 5
explanation for assertion. Reason is false because g : B → A
And f : A → B
2
x 3 1 ⇒ for: B→B and g = f –1 →fog is an identity function on B.
72. (a) y = − + x + 1 ⇒ y − = − ( x − 1) 2
2 2 2
⇒ It is symmetric about x = 1. 77. (b) X ∩ (Y ∪ X )' = X ∩ (Y '∩ X ') = X ∩ X '∩ Y ' = φ .
⇒ Assertion is true. X∆Y = (X ~ Y) ∪(Y ~ X) = (X ∪Y) ~(X ∩Y)
73. (c) Since (0,1) ∈ R but (1,0) ∉ R, R is not symmetric and ⇒ Number of elements in X ∆Y = m − n
hence is not an equivalence relation so Reason is false. ⇒ Reason is true but does explain Assertion.
Next, For the relation S, qm = pm
78. (a) Let y = f ( x) = ( x − 1) 2 + 1
m p
⇒ = ⇒ y − 1 = ( x − 1)2 ⇒ x = 1 + y − 1, y ≥ 1
n q

m p m p Thus f −1 ( x) = 1 + x − 1, x ≥ 1.
Thus  ,  ∈ S ⇒ = which shows that S is
n q n q So, Reason is true.
Now f ( x) = f −1 ( x)
U+
reflexive and symmetric
m p ⇒ ( x − 1) 2 = x − 1
Again,  ,  ∈ S
n q ⇒ x − 1[( x − 1)3/ 2 − 1] = 0
ED

 p r ⇒ x = 1, 2.
And  ,  ∈ S
 q s So, Assertion is true and Reason is a correct explanation for
Assertion.
m p r m r 
⇒ = = ⇒  , ∈S
n q s  n s  79. (b) Reason in true
Thus S is transitive and hence S is an equivalence relation. In Assertion, A = I −1 A I
So Assertion is true. For all real square matrices A of order 3.
⇒ is ( A, A) ∈ R ⇒ R reflexive, Next let ( A, B) ∈ R
74. (a) Reason is true as 2 | 6 ⇒ 2r 6 but 6 χ 2 (6 does not
⇒ ∃ a invertible matrix P of order 3.
divide 2) so R is not symmetric ⇒ 6 is not an equivalence Such that A = P −1 B P
relation and the Assertion is also true.
⇒ B = P A P −1 = ( p −1 ) −1 A( p −1 )
75. (b) A ∪ B = {1, 2,3,8}, A ∩ B = {3} ⇒ R in symmetric

⇒ ( A ∪ B) × ( A ∩ B) = {(1,3),(2,3),(3,3),(8,3)} If now ( A, B) ∈ R
And ( B, C ) ∈ R
⇒ Assertion is true. ( x, y ) ∈ ( A × B ) ∩ ( B × A)
Then ∃ invertible matrices P and Q or order 3 such that
⇒ ( x, y ) ∈ A × B and ( x, y ) ∈ B × A
A = P −1 B P and B = Q −1 C Q.
⇒ x ∈ A ∩ B, y ∈ A ∩ B
⇒ A = P −1Q −1C Q P = (QP ) −1 C QP (From Reason)
⇒ {(3,3) = ( A × B) ∩ ( B × A)}
⇒ ( A, C ) ∈ R. and thus R in transitive Hence R is an
⇒ Reason is also true but is not a correct explanation for equivalence relation and the Assertion in also true but
Assertion. Reason is not a correct explanation for it.

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20 Quick Revision NCERT- MATHEMATICS
80. (b) f (− x) = − sin x + cos x ≠ f ( x)  r 
1996
 1   2   3 
84. (b) ∑ f  = f  + f  + f   +… +
Or − f ( x), ⇒ f is neither odd nor even. So Assertion is true r =1  1997   1997   1997   1997 

− sin x  1994   1995   1996 

g (− x) = = −( g ( x)) ⇒ g is an odd function f + f  + f  
1 − cos x  1997   1997   1997 
⇒ Reason is also True but does not lead to Assertion.   1   1996     2   1995  
=f  + f  +  f  + f  
  1997   1997     1997   1997  
81. (b) Taking y = 3, f ( x) − f (3) = x − 3
  3   1994  
⇒ f ( x) = x − 3 + f (3) = x − 3 + 2 = x − 1 + f  + f   + …
  1996   1996  
⇒ f ( xy ) = xy − 1 ⇒ = 1 + 1 + 1 + … 998 times
x2 + x + 1 = 998 (even)
Assertion is true f ( x ) =
x2 − x + 1 85. (a) g5 ( x) + g 5 (1 − x) = f ( x) + f (1 − x) = 1
1 + x + x2
⇒ f (1/ x) = = f ( x)
1 − x + x2 2 n −1   1   2   3   2 −3
86. (b) ∑ 2 f   = 2  f   +
r f  + f   +… + f +
And f (2) = 7 / 3
⇒ r =1  2n    2n   2n   2n   2n 
Reason is also true, but does not lead  to Assertion.
1   2   3   2n − 3   2n − 2   2n − 1  
=  f   + f   + f   +… + f  + f  + f  
  2 n 
82. (a) Reason is true by definition of a bijective mapping,  2 n   2 n   2 n   2n   2n  
using which Assertion is also true.   1   2n − 1     2   2n − 2   
=  f   + f   +  f  + f   
∵ f ( x) + f (1 − x) g a ( x) + g a (1 − x)   2   2n     2n   2n   
U+
ax a1− x   3   2n − 3  
= + + f  + f   + …
1− x
a + a
x
a + a   2 n   2 n  

  1 
x
a a
ED

= + =1 . . .(i) = 2 1
+ 1
+ 1
+…  +1+ f  
a + a
x
a+ a x
 ( n −1)times  2 
 1
Comprehension Based = 2  ( n − 1) +  = 2 n − 2 + 1 = (2n − 1)
 2
1995
 r   1   2 
83. (c) ∑ f  1996  = f + f 
 1996 

 1996 
r =1 2n
 r  2n
 r 
87. (c) ∑ f  2n + 1  = f (0) + ∑ f  
 2n + 1 
 3   1993   1994   1995  r =0 r =1
+f   +… + f  f + f 
 1996   1996   1996   1996  1 1
= +n = + 987 (given)
  1  1+ a 1+ a
 1995     2   1994  
=f  + f   +  f  +  ∴ n = 987
  19996   1996     1996   1996  
  3   1993   f ( x) 1 − 2sin 2 x cos 2 x
=f  + f   88. (c) H ( x) = = = =1
  1996   1996   g ( x) cos 2 x cos 2 x
  997   999    998  but cos 2 x ≠ 0
+… +  f  + f   + f  
  1996   1996    1996  π
⇒ cos 2 x ≠ 0 ⇒ 2 x ≠ nπ + ,n∈ I
1 2
= 1
+ 1 + 1
… +1 +1+ f  
2  π 
997 times
∴ x ∈ R ∼ (2 n + 1) , n ∈ I  And Range ={1}
 4 
1
= 997 + [from Equation (i)]
2 89. (d) F ( x) = f ( x) + g ( x) = 1 − 2sin 2 x + cos 2 x
= 997.5 = 2 cos 2 x −1 ≤ cos 2 x ≤ 1 ⇒ −2 ≤ 2 cos 2 x ≤ 2
1995
 r  Range of F ( x) co-domain of F ( x)
∴  ∑ f  1996   = 1997
 r =1   ⇒ F ( x) is onto function.

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Set Relations and Functions 21
90. (c) Periods of f (x), g(x) and F (x) are π. 95. (d) Plan (a) For such questions, we need to properly
∴ sum of periods = π + π + π = 3 π define the functions and then we draw their graphs.
(n) From the graphs, we can examine the function for
91. (c) The domain of G(x) is R and domain of H(x)is
continuity, differentiability, one-one and onto.
R ∼ {x : g ( x) = 0}
∴ DG ∪ DH = R

92. (d) F (x) – G(x) = 0⇒2g(x) = 0

⇒ g(x) = 0
⇒ cos 2x = 0  − x, x<0
f1 ( x ) =  x
e , x≥0
π
⇒ 2 x = nπ + f 2 ( x) = x 2 , x ≥ 0
2

Or x = (2n + 1)
π
4
,n∈ I f 3 ( x) = {
sin x,
x,
x<0
x≥0
∵ x ∈ [0,5π ]  f ( f ( x )), x<0
f 4 ( x) =  2 1
π 3π 5π 7π 9π 11π 13π 15π 17π 19π  f 2 ( f1 ( x )) − 1, x≥0
∴ Solutions are , , , , , , , , ,
4 4 4 4 4 4 4 4 4 4  x2 , x < 0  x2 , x<0
∴ Number of solutions = 10 Now, f 2 ( f1 ( x)) =  2 x ⇒ f4 =  2 x
 e , x ≥ 0  e − 1, x≥0
10  π 19π 
Sum of solutions =  +  = 25π  2 x, x < 0
As f 4 ( x) is continuous f '4 ( x ) =  2 x
2 4 4 
 2e , x > 0
Match the Column f4' (0) is not defined. Its range is [0,∞).
U+
93. (a) y = 1 + 2x is linear function therefore, it is one-one and Thus, range = co-domain = [0,∞) thus, f4 is onto.
its range is (–π + 1, π + 1). Therefore, (1+2x) is one-one Also, horizontal line (drawn parallel to x-axis) meets the
but not onto so (A→2). curve more than once thus function is not one-one.
ED

Again, see the figure. It is clear from the graph that y – tan Integer
x is one-one and onto, therefore (B→3). 96. (2345) For f (x) to be defined
y 16 – x > 0 ⇒ x < 16
y =1+2x 1
2x −1 ≥ 0 ⇒x≥
2
x' x 17
−π o π 16 − x ≥ 2 x − 1 ⇒x≤
2 2 3
20
20 − 3x > 0 ⇒x<
y'
3
5
( x − 1)( x − 5) 4x − 5 ≥ 0 ⇒x≥
94. (b) Given, f ( x) = 4
( x − 2)( x − 3) 25
20 − 3x ≥ 4 x − 5 ⇒x≤
y 7
5 25
Combining all we get ≤ x ≤
y =1 4 7
x' x ∵ x∈I
0 1 2 3 5
∴ x = 2, 3
y' ∴ Domain of f (x) = {2, 3}
The graph of f(x) is shown
∴ Range of f (x) = {f (2), f (3)}
= {14 C 3 + 14 C 3, 13
C 5 + 11C 7 }
(A) If –1 < x < 1 ⇒ 0 < f (x) < 1
(B) If 1 < x < 2 ⇒ f (x) < 0 = {2 ⋅ 14 C3 , 13C5 + 11C 4 } = {728, 1617}
(C) If 3 < x < 5 ⇒ f (x) < 0 ∴ λ = 1617, µ = 728,
(D) If x > 5 ⇒ 0 < f (x) < 1 Then λ + µ = 2345

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22 Quick Revision NCERT- MATHEMATICS
 π [ x]  2π Substituting the value of 2 f (2),3 f (3),… in terms of
97. (4824) The period of sin   is = 24 as
 12  π /12 nf (n) in equation (i), we have
 π [ x + 24]   π ([ x] + 24)  f (1) + (n − 1)n f (n) = n(n + 1) f (n)
sin   = sin  
 12   12  ⇒ f (1) = 2nf (n)
 π [ x]   π [ x] 
= sin  2π +  = sin   1 1 2n
 12   12  ⇒ = =
f ( n) f (1) 1
 π [ x]  1
Similarly the period of tan   is 3 and the period of ∴ = 2 × 1004 = 2008
 3  f (1004)
 π x  2π
cos   is =8 1
 4  π /4 100. (119) Given f ( x) = x + ∫ ( xy 2 + x 2 y ) f ( y ) dy
Hence, the period of the given function 0
1 1
λ = LCM of 24, 8, 3 = 24
Or f ( x) = x + x ∫ y 2 f ( y ) dy + x 2 ∫ yf ( y ) dy
∴ 201 λ = 201 × 24 = 4824 0 0
1 1
 y 2
y  2
Let A = ∫ y 2 f ( y ) dy and B = ∫ yf ( y ) dy
98. (112) f  2 x 2 + , 2 x 2 −  = xy
 8 8  0 0

2 2 Then f ( x) = x + Ax + Bx 2
. . .(i)
 y2   y2 
=  2 x2 +  −  2x2 −  1 1
 8   8  ∴ A = ∫ y 2 f ( y ) dy = ∫ y 2 ( y + Ay + By 2 ) dy
∴ f (60, 48) + f (80, 48) + f (13, 5) 0 0

1 A B
U+
⇒ A= + +
= (60)2 − (48)2 + (80)2 − (48)2 + (13)2 − (5)2 4 4 5
=36 + 64 +12 =112 3A B 1
Or − + . . .(ii)
4 3 4
99. (2008) Given f (1) + 2 f (2) + 3 f (3) + … + nf ( n)
ED

1 1

= n (n + 1) f (n) . . .(i) and B = ∫ yf ( y ) dy = ∫ y ( y + Ay + By 2 ) dy

0 0
Replacing n by n +1, then
1 A B
f (1) + 2 f (2) + 3 f (3) + … + nf ( n) + ( n + 1) f (n + 1) = + +
3 3 4
= (n + 1)(n + 2) f (n + 1) . . .(ii) A 3B 1
Or − + = . . . (ii)
Subtracting equation (i) form from equation (ii), then we 3 4 3
get 61 80
Solve equations (ii) and (iii), we get A = ,B=
(n + 1)(n + 1) f ( n + 1) = (n + 1)(n + 2) f (n + 1) − n(n + 1) f (n) 119 119

nf (n) = (n + 1) f ( n + 1) From which we concluded that 61x + 80 x 2

⇒ Now from equation (i), f ( x) = x + = 119 .
119
2 f (2) = 3 f (3) = 4 f (4) = … = nf (n)

***

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