NCERT Solutions Class 12 Maths (Relation And Functions) Miscellaneous Exercis
We have given the answers of all the questions of NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.
Set 1
Question 1.
Let f : R → R be defined as f(x) = 10x + 7. Find the function
g : R → R such that g o f = f o g = 1R.
Solution:
As, it is mentioned here
f : R → R be defined as
f(x) = 10x + 7
To, prove the function
one-one
Let’s take f(x) = f(y)
10x + 7 = 10y + 7
x = y
Hence f is
one-one.
To, prove the function
onto
y ∈ R, y = 10x+7
So, it means for y ∈ R, there exists 
Hence f is
onto.
As, f is one-one and
onto. This f is invertible function.
Let’s say g : R → R be defined as 
Hence, g : R → R such that g o f = f o g = 1R.
g : R → R is defined as
Question 2.
Let f : W → W be defined as f(n) = n –
1, if n is odd and f(n) = n + 1, if n is even. Show that f is invertible. Find
the inverse of f. Here, W is the set of all whole numbers.
Solution:
The function f is defined as
As, we know f is
invertible, if and only if f is one-one and onto.
ONE-ONE
For the pair of number,
we will deal with three cases:
Case 1: When both numbers p and q are odd numbers.
f(p) = p-1
f(q) = q-1
f(p) = f(q)
p-1 = q-1
p – q = 0
Case 2: When both numbers p and q are even numbers.
f(p) = p+1
f(q) = q+1
f(p) = f(q)
p+1 = q+1
p – q = 0
Case 3: When p is odd and q is even
f(p) = p-1
f(q) = q+1
f(p) = f(q)
p-1 = q+1
p – q = 2
Subtracting an odd
number and even always gives a odd number, not even. Hence, the case 3 result
is impossible.
So, the
function f is one-one, for case 1 and case 2 only.
ONTO
Case 1: When p is odd number
f(p) = p-1
y = p-1
p = y+1
Hence, when p is odd y is
even.
Case 2: When p is even number
f(p) = p+1
y = p+1
p = y-1
Hence, when p is even y
is odd.
So, it means for y ∈ W, there exists p = y+1 and y-1 for odd and even
value of p respectively.
Hence f is
onto.
As, f is one-one and
onto. This f is an invertible function.
Let’s say g : W → W be defined as 
f = g
Hence, The inverse of f is f itself
Question 3.
If f : R → R is defined by f(x) = x2–
3x + 2, find f (f(x)).
Solution:
f(x) = x2– 3x + 2
f(f(x)) = f(x2– 3x + 2)
= (x2– 3x + 2)2 – 3(x2– 3x + 2) + 2
= x4 + 9x2 + 4 -6x3 – 12x + 4x2 – 3x2 + 9x – 6 + 2
f(f(x)) = x4 – 6x3 + 10x2 – 3x
Question 4.
Show that the function f : R → {x ∈ R : – 1 < x
< 1} defined by f(x)= 
,x ∈
R is one one and onto function.
Solution:
As, it is mentioned here
f : R → {x ∈ R : – 1 < x < 1} defined by 
, x ∈ R
As, we know f is
invertible, if and only if f is one-one and onto.
ONE-ONE
For the pair of number,
we will deal with three cases:
Case 1: When both
numbers p and p are positive numbers.
The function f is
defined as
Case 1: When both numbers p and q are positive numbers.
f(p) = f(q)
p(1+q) = q(1+p)
p = q
Case 2: When number p and q are negative numbers.
f(p) = f(q)
p(1-q) = q(1-p)
p = q
Case 3: When p is positive and q is negative
f(p) = f(q)
p(1-q) = q(1+p)
p + q = 2pq
Here, RHS will be
negative and LHS will be positive. Hence, the case 3 result is impossible.
So, the
function f is one-one, for case 1 and case 2.
ONTO
Case 1: When p>0.
Case 2: When p <0
Hence, p is defined for
all the values of y, p∈ R
Hence f is
onto.
As, f is one-one and onto. This f is an invertible
function.
Question 5.
Show that the function f : R → R given by f(x)
= x3 is
injective.
Solution:
As, it is mentioned here
f : R → R defined by
f(x) = x3, x ∈ R
To prove f is injective
(or one-one).
ONE-ONE
The function f is
defined as
f(x) = x3
f(y) = y3
f(x) = f(y)
x3 = y3
x = y
The function f is one-one, so f is injective.
Question 6.
Give examples of two functions f : N → Z and g : Z → Z such that g o f is injective but g is not
injective.
(Hint : Consider f(x) = x and g (x) = | x |).
Solution:
Two functions, f : N → Z
and g : Z → Z
Taking f(x) = x and g(x)
= |x|
Let’s check, whether g
is injective or not
g(5) = |5| = 5
g(-5) = |-5| = 5
As, we can see here that
Taking two integers, 5
and -5
g(5) = g(-5)
but, 5 ≠ -5
So, g is not
an injective function.
Now, g o f: N → Z is
defined as
g o f = g(f(x)) = g(x) =
|x|
Now, as x,y∈ N
g(x) = |x|
g(y) = |y|
g(x) = g(y)
|x| = |y|
x = y (both x and y are
positive)
Hence, g o f is an injective.
Question 7.
Give examples of two functions f : N → N and g : N → N such that g o f is onto but f is not onto.
Solution:
Two functions, f : N → N and g : N → N
Taking f(x) = x+1 and
As, f(x) = x+1
y = x+1
x = y-1
But, when y=1, x = 0.
Which doesn’t satiny this relation f : N → N.
Hence. f is
not an onto function.
Now, g o f: N → N is
defined as
g o f = g(f(x)) = g(x+1)
When x+1=1, we have
g(x+1) = 1 (1∈ N)
And, when x+1>1, we
have
g(x+1) = (x+1)-1 = x
y = x, which also
satisfies x,y∈ N
Hence, g o f is onto.
Question 8.
Given a non empty set X, consider P(X) which is the set of all subsets of X.
Define the relation R in P(X) as follows: For
subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an
equivalence relation on P(X)? Justify your answer.
Solution:
Given, A and B are the
subsets of P(x), A⊂ B
To check the equivalence
relation on P(X), we have to check
- · Reflexive
As, we know that every
set is the subset of itself.
Hence, A⊂ A and B⊂ B
ARA and BRB is reflexive
for all A,B∈ P(X)
- · Symmetric
As, it is given that A⊂ B. But it doesn’t make sure that B⊂ A.
To be symmetric it has
to be A = B
ARB is not symmetric.
- · Transitive
When A⊂ B and B⊂ C
Then of course, A⊂ C
Hence, R is transitive.
So, as R is not
symmetric.
R is not an equivalence relation on P(X).
Question 9.
Given a non-empty set X, consider the
binary operation ∗
: P(X) × P(X) → P(X) given by A ∗ B = A ∩ B ∀ A, B in P(X),
where P(X) is the power set of X. Show that X is the identity element for this
operation and X is the only invertible element in P(X) with respect to the
operation ∗.
Solution:
Given, P(X) × P(X) →
P(X) is defined as A*B = A∩B ∀ A,
B ∈ P(X)
This implies, A⊂ X and B ⊂ X
So, A∩X = A and B∩X = B ∀ A, B ∈ P(X)
⇒ A*X = A and B*X = B
Hence, X is the identity element for intersection of
binary operator.
Question 10.
Find the number of all onto functions
from the set {1, 2, 3, … , n} to itself.
Solution:
Onto function from the
set {1,2,3,…..,n} to itself is just same as the permutations of n.
1×2×3×4×…….×n
Which is n!.
Set 2
Question 11:
Let S = {a, b, c} and T = {1, 2, 3}. Find F–1 of
the following functions F from S to T, if it exists.
(i) F = {(a, 3), (b, 2), (c, 1)}
Solution:
As, F = {(a, 3), (b, 2),
(c, 1)} and S = {a,b,c} and T={1,2,3}
F: S→T is defined as
F(a) = 3, F(b) = 2 and
F(c) = 1
F is one-one and onto.
Taking F-1, so F-1: T→S
a = F-1(3), b = F-1(2) and c = F-1(1)
F-1 =
{(3,a),(2,b),(1,c)}
(ii) F = {(a, 2), (b, 1), (c, 1)}
Solution:
As, F = {(a, 2), (b, 1),
(c, 1)}
F: S→T is defined as
F(a) = 2, F(b) = 1 and
F(c) = 1
Here, F(b) = F(c) but b
≠ c
Hence, F is not one-one.
So, F is not invertible and F-1 doesn’t exists.
Question 12.
Consider the binary operations ∗ : R × R → R and
o : R × R → R defined as a ∗b = |a – b| and a
o b = a, ∀
a, b ∈
R. Show that ∗
is commutative but not associative, o is associative but not commutative.
Further, show that ∀
a, b, c ∈
R, a ∗
(b o c) = (a ∗
b) o (a ∗
c). [If it is so, we say that the operation ∗ distributes over
the operation o]. Does o distribute over ∗? Justify your
answer.
Solution:
Binary operations ∗ : R × R → R defined as a ∗b = |a – b|
a*b = |a-b|
b*a = |b-a| = |-(a-b)| =
|a-b|
a*b = b*a
Hence, ∗ is commutative.
Now, let’s take a=1, b=2
and c=3 for better understanding
a*(b*c) = a*|b-c| =
|a-|b-c|| = |1-|2-3|| = 0
(a*b)*c = |a-b|*c =
||a-b|-c| = ||1-2|-3| = 2
a*(b*c) ≠ (a*b)*c
Hence, ∗ is not associative.
Binary operations o : R
× R → R defined as a o b = a, ∀ a,
b ∈ R
a o b = a
b o a = b
a o b ≠ b o a
Hence, o is not
commutative.
a o (b o c) = a o b = a
(a o b) o c = a o c = a
a o (b o c) ≠ (a o b) o
c
Hence, o is associative.
Let’s check for a ∗ (b o c) = (a ∗ b) o (a ∗ c) a, b, c ∈ R
a ∗ (b o c) = a * b = |a-b|
(a ∗ b) o (a ∗ c) = |a-b| o |a-c| = |a-b|
Hence, a ∗ (b o c) = (a ∗ b) o (a ∗ c)
Now, let’s check for a o
(b * c) = (a o b) * (a o c)
a o (b * c) = a
(a o b) * (a o c) = a *
a = |a-a| = 0
Hence, a o (b * c) ≠ (a
o b) * (a o c)
o does not distribute over ∗
Question 13.
Given a non-empty set X, let ∗ : P(X) × P(X) →
P(X) be defined as A * B = (A – B) ∪ (B – A), ∀ A, B ∈ P(X). Show that
the empty set φ is the identity for the operation ∗ and all the
elements A of P(X) are invertible with A–1 =
A.
(Hint : (A – φ) ∪ (φ – A) = A and
(A – A) ∪
(A – A) = A ∗
A = φ).
Solution:
Set X, such that P(X) ×
P(X) → P(X) be defined as A * B = (A – B) ∪ (B – A), ∀ A, B ∈ P(X)
φ*A = (φ-A) U (A-φ) = φ
U A = A
A*φ = (A-φ) U (φ-A) = A
U φ = A
Hence, φ is the identity
element for the operation * on P(X)
A*A = (A-A) U (A-A) = φ
U φ = φ
⇒ A = A-1
Hence, all the elements A of P(X) are invertible
with A–1 = A.
Question 14.
Define a binary operation ∗ on the set {0, 1, 2, 3, 4, 5} as
Show that zero is the identity for this operation and each element a ≠ 0 of the set is invertible with 6 – a being the inverse of a.
Solution:
Let the set x = {0, 1,
2, 3, 4, 5}
Let’s take i as identity
element, where a*i = a = i*a ∀ a ∈ x
a*0 = a
0*a = a, when (a+0<6)
Hence, zero is the
identity for this operation
An element a ∈ x is invertible if there exists b ∈ x such that a*b = b*a
From above equations, we
have
a = -b or b = 6-a
But, as x = {0, 1, 2, 3,
4, 5} and a,b∈ x. Then a≠-b
Hence, b = 6-a is the
inverse of an element a∈ x
a≠0
a-1 = 6-a
Question 15.
Let A = {– 1, 0, 1, 2}, B = {– 4, – 2, 0, 2} and f, g : A → B be
x ∈ A. Are f and g equal? Justify your answer. (Hint: One may note that two functions f : A → B
and g : A → B such that f(a) = g (a) ∀ a ∈ A, are called
equal functions).
Solution:
Given, f, g : A → B be functions defined by f(x) = x2 – x, x ∈ A and g(x)
x ∈ A
At x = -1
f(0) = (-1)2 – (-1) = 2
g(0) = 
= 2
Here, f(-1) = g(-1) and 2=2
At x = 0
f(0) = 02 – 0 = 0
g(0) = = 0
Here, f(0) = g(0) and
0=0
At x = 1
f(1) = 12 – 1 = 0
g(1) = 
= 0
Here, f(1) = g(1) and 1=1
At x = 2
f(1) = 22 – 2 = 2
g(1) = 
= 2
Here, f(2) = g(2) and 2=2
For, every c∈ A, f(c) = g(c)
Hence, f and g are equal functions.
Question 16.
Let A = {1, 2, 3}. Then number of
relations containing (1, 2) and (1, 3) which are reflexive and symmetric but
not transitive is
(A) 1
(B) 2
(C) 3
(D) 4
Solution:
R =
{(1,1),(1,2),(1,3),(2,1),(2,2),(3,1),(3,3)}
Reflexive : (1,1),
(2,2), (3,3) ∈ R
Symmetric: (1,2), (2,1)∈ R and (1,3), (3,1) ∈ R
R is not Transitive
because, (1,2), (1,3) ∈ R but (3,2) ∉R
So, if we will add (3,2)
and (2,3) or both, then R will become transitive.
New, R =
{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)}
Hence, A is the correct option.
Question 17.
Let A = {1, 2, 3}. Then number of
equivalence relations containing (1, 2) is
(A) 1
(B) 2
(C) 3
(D) 4
Solution:
Smallest equivalence
relations containing (1, 2):
R =
{(1,1),(2,2),(1,2),(2,1),(3,3)}
or R =
{(1,1),(2,2),(1,2),(2,1),(3,2),(2,3)(3,3)}
Hence, B is the correct option.
and g : R → R be the Greatest Integer Function
given by g (x) = [x], where [x] is greatest integer less than or equal to x.
Then, does fog and gof coincide in (0, 1]?
Solution:
Given, f : R → R and g :
R → R
when x ∈ (0,1]
[x] = 1, when x=1
[x] = 0, when
0<x<1
Now, fog(x)=f(g(x)) = f([x])
And, Now gof(x) = g(f(x))
g(1) = [1] = 1
g(0) = [0] = 0
g(-1) = [-1] = -1
When x ∈ (0,1), fog = 0 and gof = 1. fog(1) ≠ gof(1)
Hence, fog and gof do not coincide in (0, 1].
Question 19.
Number of binary operations on the set
{a, b} are
(A) 10
(B) 16
(C) 20
(D ) 8
Solution:
Let A = {a,b}
A x A = {a,b} x {a,b}
R =
{(a,a),(a,b),(b,a),(b,b)}
Number of elements are
4.
Hence, the number of
binary operations on the set will be 24 = 16
Hence, B is the correct option.
