Access to comprehensive Class 12 ISC Maths Solutions Chapter 1 Relations and Functions MCQs encourages independent learning.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 1 Relations and Functions MCQs

Choose the correct answer from the given four options in questions (1 to 45) :

Question 1.
If R is a relation on the set of all straight lines drawn in a plane defined by l1 and l2 iff l1 ⊥ l2, then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) an equivalence relation
Solution:
(b) symmetric
Given R is a relation on set of all straight lines drawn in a plane defined by l1 R l2 iff l1 ⊥ l2
Since every line is not ⊥ to itself
∴ (l, l) ∉ R
∴ R is not reflexive on L.
Now l1 R l2
⇒ l1 ⊥ l2
⇒ l2 ⊥ l1
⇒ (l2, l1) ∈ R
Thus R is symmetric on L.
Now (l1, l2 ), (l2, l3 ) ∈ R
⇒ l1 ⊥ l2 and l2 ⊥ l3
∴ l1 || l3
∴ (l1, l3) ∉ R.

Question 2.
If R is a relation on N (set of all natural numbers) defined by n R m iff n divides m, then R is
(a) reflexive and symmetric
(b) transitive and symmetric
(c) reflexive and transitive
(d) equivalence relation
Solution:
(c) reflexive and transitive

Given R is a relation on N defined by n R m iff n divides m.
Reflexive :
since n | n
⇒ (n, n) ∈ R V n ∈ N
⇒ R is reflexive on N.

Symmetric : (n, m) ∈ R
⇒ n | m ⇏ m | n
⇒ (m, n) ∉ R
Since 2 is a factor of 4 but 4 is not a factor of 2.
∴ R is not symmetric on N.

Transitive :
(n, m) ∈ R and (m, l) ∈ R ∀ n, m, l ∈ N
⇒ n | m
⇒ m = nk where k ∈ N
and (m, l) ∈ R
⇒ m | l
⇒ l = mk’, k’ ∈ N
⇒ l = nkk’ = np where p = kk’ ∈ N
⇒ n | l
⇒ (n, l) ∈ R
Thus R is transitive on N.

Question 3.
If R is a relation on Z (set of all integers) defined by xRy if f | x-y | < 1, then R is
(a) reflexive and symmetric
(b) reflexive and transitive
(c) symmetric and transitive
(d) an equivalence relation
Solution:
(a) reflexive and symmetric

Given relation R on Z defined by x ∈ R iff | x – y | ≤ 1
Reflexive :
since | x – x | = 0 ≤ 1
∴ xRx ⇒ R is reflexive on Z.

Symmetric :
(x, y) ∈ R ∀ x, y ∈ Z
⇒ |x – y | ≤ 1
⇒ |- (y – x) | ≤ 1
⇒ | y – x | ≤ 1
⇒ (y, x) ∈ R
∴ R is symmetric on Z.

Transitive :
Since (2, 3) ∈ R
as | 2 – 3| = | – 1 | = 1 ≤ 1 (3, 4) ∈ R as | 3 – 4 | = 1 ≤ 1 but (2, 4) ∉ R as | 2 – 4 | = | – 2 | = 2 > 1
R is transitive on Z.
Thus R is reflexive and symmetric on Z.

Question 4.
If R is a relation on R (set of all real numbers) defined by a R b iff a ≥ b, then R is
(a) an equivalence relation
(b) reflexive, transitive but not symmetric
(c) symmetric, transitive but not reflexive
(d) neither reflexive nor transitive but symmetric
Solution:
(b) reflexive, transitive but not symmetric

Given relation R on set of real numbers defined by aRb iff a ≥ b

Reflexive :
since a ≥ a
⇒ (a, a) ∈ R
⇒ R is reflexive

Symmetric :
Now (a, b) ∈ R
⇒ a ≥ b ⇏ b ≥ a
R is not symmetric.

Transitive :
Now (a, b), (b, c) ∈ R
⇒ a ≥ b and b ≥ c
⇒ a ≥ b ≥ c
⇒ a ≥ c
⇒ (a, c) ∈ R
∴ R is transitive.

Question 5.
Which, of the following is not an equivalence relation on I (set of all integers) ?
(a) a R b iff a + b is even
(b) a R b iff a – b is even
(c) a R b iff a < b
(d) a R b iff a = b
Solution:
(c) a R b iff a < b

Given a relation R on I defined by (a, b) ∈ R iff a < b
Clearly 1, 2 ∈ I s.t (1, 2) ∈ R as 1 < 2
but 2 ≮ 1
∴ (2, 1) ∉ R
∴ R is not symmetric on I.
∴ R is not an equivalence relation on I.

Question 6.
If R is a relation on the set T of all triangles drawn in a plane defined by a R b iff a is congruent to b for all a, b e T, then R is
(a) reflexive but not transitive
(b) reflexive but not symmetric
(c) symmetric but not transitive
(d) equivalence relation
Solution:
(d) equivalence relation

Given R is relation on T defined by a R b iff a is congruent to b ∀ a, b ∈ T

Reflexive :
Since every triangle is congruent to itself
∴ (a, a) ∈ R
⇒ R is reflexive on T.

Symmetric :
(a, b) ∈ R ∀ a, b ∈ T
⇒ a is congruent to b
⇒ b is congruent to a
⇒ (b, a) ∈ R
⇒ R is symmetric on T.

Transitive :
Let (a, b) ∈ R and (b, c) ∈ R ∀ a, b, c ∈ T
∴ a is congruent to b and b is congruent to c
⇒ a is congruent to c
⇒ (a, c) ∈ R
∴ R is transitive on T.
Therefore, R forms an equivalence relation on T.

Question 7.
If R is a relation on R (set of all real numbers) defined by x Ry iff x – y + √2 is an irrational number, then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) none of these
Solution:
(a) reflexive

Since x – x + √2 = √2,
which is an irrational number
⇒ x R x
∴ R is reflexive.

Question 8.
If R is a relation on N × N defined by (a, b) R (c, d) iff a + d = b + c, then
(a) reflexive
(b) symmetric
(c) transitive
(d) all of these
Solution:
(d) all of these

Given R = {(1, 1), (2, 2), (3, 3)} be a relation on set A = {1, 2, 3}
Clearly R is reflexive, symmetric and transitive.

Question 9.
If R is a relation on the set A = {1, 2, 3} defined by R = {(1, 2)}, then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) all of these
Solution:
(c) transitive

Since 1 ∈ A but (1, 1) ∉ R
∴ R is not reflexive on A.
Since (1, 2) ∈ R but (2, 1) ∉ R
∴ R is not symmetric on A.
∀ (a, b) ∈ R there is no (b, c) ∈ R s.t (a, c) ∈ R
∴ R is transitive on A.

Question 10.
If R is a relation on the set A = {1, 2, 3} given by R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}, then R is
(a) reflexive but not symmetric
(b) reflexive but not transitive
(c) symmetric and transitive
(d) neither symmetric nor transitive
Solution:
(a) reflexive but not symmetric

Given A = {1, 2, 3} since (1, 1), (2, 2) and
(3, 3) ∈ R
∴ R is reflexive on A.
Now (1, 2) ∈ R but (2, 1) ∉ R
∴ R is not symmetric on A.
∴ R is reflexive but not symmetric.

Question 11.
If R is a relation on the set A = {1, 2, 3} given by R = {(1, 1), (2, 2), (1, 3)}, then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) all of these
Solution:
Now 3 ∈ A but (3, 3) ∉ R
∴ R is not reflexive on A.
Since (1, 3) ∈ R but (3, 1) ∉ R
∴ R is not symmetric on A.
Also, (a, b), (b, c) ∈ R
⇒ (a, c) ∈ R
∴ R is transitive on A.

Question 12.
If R is a relation on the set A = {1, 2, 3} given by R = {(1, 1), (1, 2), (2, 1)}, then R is
(a) reflexive
(b) symmetric
(c) transitive
(d) all of these
Solution:
(b) symmetric

Clearly 2, 3 ∈ A but (2, 2), (3, 3) ∉ R
∴ R is not reflexive on A.
Now (1, 2) ∈ R
⇒ (2, 1) ∈ R
∴ R is symmetric on A.
Now (2, 1), (1, 2) ∈ R but (2, 2) ∉ R
∴ R is not transitive on A.

Question 13.
If R is a relation on the set A = {1, 2, 3} given by R = {(1,1), (2,2), (3,3)}, then R is
(a) reflexive only
(b) symmetric only
(c) transitive only
(d) equivalence relation
Solution:
(d) equivalence relation

Clearly (1, 1), (2, 2), (3, 3) ∈ R
∴ R is reflexive on A.
Clearly (a, b) ∈ R
⇒ (b, a) ∈ R
∴ R is symmetric on A.
∀ (a, b) ∈ R there is no (b, c) ∈ R s.t (a, c) ∈ R
∴ R is transitive on A.
Thus, R forms on equivalence relation on A.

Question 14.
If A = {1, 2, 3}, then which of the following relations are equivalence relation on A ?
(a) {(1, 1), (2, 2), (3, 3)}
(b) {(1,1), (2, 2), (3, 3), (1, 2), (2,1)}
(c) {(1,1), (2, 2), (3, 3), (2, 3), (3, 2)}
(d) all of these
Solution:
(d) all of these

Clearly (1, 1), (2, 2), (3, 3) ∈ R
∴ R is reflexive on A.
Clearly (a, b) ∈ R
⇒ {b, a) ∈ R
∴ R is symmetric on A.
∀ (a, b) ∈ R there is no (b, c) ∈ R s.t (a, c) ∈ R
∴ R is transitive on A.
Thus, R forms on equivalence relation on A.
Clearly (1, 1), (2, 2), (3, 3) ∈ R
∴ R is reflexive on A.
Clearly (1, 2), (2, 1) ∈ R
∴ R is symmetric on A.
Further, (a, b), (b, c) ∈ R
⇒ (a, c) ∈ R ∀ a, b, c ∈ A
∴ R is transitive on A.
Clearly option (b) forms an equivalence relation on A.
Similarly in option (c);
(1, 1), (2, 2), (3, 3) ∈ R
∴ R is reflexive on A.
Now (2, 3), (3, 2) ∈ R
⇒ R is symmetric on A.
Further, (a, b) ∈ R and (b, c) ∈ R
⇒ (a, c) ∈ R
∴ R forms an equivalence relation on A.

Question 15.
If A = {1, 3, 5}, then the number of equi-valence relations on A containing (1, 3) is
(a) 1
(b) 2
(c) 4
(d) 5
Solution:
(b) 2

Since we know that, R forms an equivalence relation on A iff R is reflexive, symmetric and transitive.
Given A = {1, 3, 5}
For reflexivity : (1, 1), (3, 3), (5, 5) ∈ R
For symmetry : (1, 3), (3, 1) ∈ R
Thus equivalence relations on A are :
R1 = {(1, 1), (3, 3), (5, 5), (1,3), (3,1)}
R2 = {(1, 1), (3, 3), (5, 5), (1,3), (3, 1), (3, 5), (5, 3), (1, 5), (5, 1)}.

Question 16.
If A = {1, 2, 3}, then the maximum number of equivalence relations on A is
(a) 2
(b) 3
(c) 4
(d) 5
Solution:
(d) 5

Given A = {1, 2, 3}
If R is reflexive, symmetric and transitive on A.
Then R forms equivalence relation on A
∴ R1 = {(1, 1), (2, 2), (3, 3)}
R2 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}
R3 = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)}
R4 = {(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)}
R5 = {(1, 1), (2, 2), (3, 3), (1,2), (2, 1), (2, 3), (3, 2), (1,3), (3, 1)}

Question 17.
The relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1), (3, 3)} is
(a) symmetric and transitive, but not reflexive
(b) reflexive and symmetric, but not transitive
(c) symmetric, but neither reflexive nor transitive
(d) an equivalence relation
Solution:
(c) symmetric, but neither reflexive nor transitive

Clearly (1, 1), (2, 2) ∉ R
∴ R is not reflexive on A.
Clearly (1, 2), (2, 1) ∈ R
∴ R is symmetric on A.
Now (2, 1), (1, 2) ∈ R but (2, 2) ∉ R
∴ R is not transitive on A.

Question 18.
Let R be the relation in the set N, given by R = {(x, y) : x = y + 3, y > 5}. Choose the correct answer from the following :
(a) (7, 4) ∈ R
(b) (9, 6) ∈ R
(c) (4, 7) ∈ R
(d) (8, 5) ∈ R
Solution:
(b) (9, 6) ∈ R

Clearly 9 = 6 + 3
⇒ (9, 6) ∈ R

Question 19.
If A= {1,2, 3} and B = {1, 3, 4, 7} and R is a relation from A to B defined by ‘x is greater than then the range of R is
(a) {1, 3, 4, 7}
(b) {3, 4, 7}
(c) {1}
(d) none of these
Solution:
(c) {1}

Relation R from A to B defined by x > y
Clearly 3 > 1
∴ (3, 1) ∈ R
∴ range of R = {1}.

Question 20.
If A = {1, 2, 3} and B = {a, b}, then the number of functions from A to B is
(a) 3
(b) 6
(c) 8
(d) 12
Solution:
(c) 8

We know that,
No. of functions from A to B = [O (B)]O (A)
∴ No. of functions from A to B = [O (B)]O (A)
= 23 = 8
[Here O (A) = 3 ; O (B) = 2]

Question 21.

(a) f is a function from A to B
(b) f is a one-one function from A to B
(c) f is an onto function from A to B
(d) f is not a function from A to B
Solution:
(d) f is not a function from A to B

Since element a in A have two images 1 and 3 in B.
Thus every element in A does not have unique image in B.
∴ f is not a function from A to B.

Question 22.
If a function f : R → R is defined by f(x) = $$\left\{\begin{array}{cc} 2 x, & x>3 \\ x^2, & 1 \leq x 3 \\ 3 x, & x \leq 1 \end{array}\right.$$
Then the value of f (- 1) + f (2) + f (4) is
(a) 9
(b) 14
(c) 5
(d) none of these
Solution:
(a) 9

When x = – 1 < 1
∴ f (- 1) = 3 × (- 1) = – 3
When x = 2
∴ 1 ≤ x ≤ 3
∴ f (2) = 22 = 4
When x = 4 > 3
∴ f (4) = 2 × 4 = 8
∴ f (- 1) + f (2) + f (4)
= – 3 + 4 + 8 = 9.

Question 23.
If a function f : R → R is defined by f(x) = x2 + 1, then pre-images of 17 and – 3 respectively, are
(a) Φ {4, – 4}
(b) {3, – 3}, Φ
(c) {4, – 4}, Φ
(d) {4, – 4}, {2, – 2}
Solution:
(c) {4, – 4}, Φ

Given function f : R → R defined by f (x) = x2 + 1
Now f (x) = 17
⇒ x2 + 1 = 17
⇒ x2 = 16
⇒ x = ± 4
∴ pre-image of 17 are {- 4, 4}
Now f (x) = – 3
⇒ x2 + 1 = – 3
⇒ x2 = – 4
⇒ x = ± 2i but x ∈ R
Thus pre-image of – 3 be Φ.

Question 24.
If a function f : C → C is defined by f (x) = 3x2 – 1, where C is the set of complex numbers, then the pre-images of – 28 are
(a) 3, – 3
(b) 3i, – 3i
(c) 3i only
(d) – 3i only
Solution:
(b) 3i, – 3i

Given a function f : C → C defined by f(x) = 3x2 – 1
Now f(x) = – 28
⇒ 3x2 – 1 = – 28
⇒ x2 = – 9
⇒ x = ± 3i
Thus, pre-images of – 28 are 3i and – 3i.

Question 25.
If a function f : [2, ∞) → R is defined by f(x) = x2 – 4x + 5, then the range of f is
(a) R
(b) [1, ∞)
(c) [4, ∞)
(d) [5, ∞)
Solution:
(b) [1, ∞)

Given f : [2, ∞) R is defineed by
f(x) = x2 – 4x + 5
Now f(x) = (x – 2)2 + 1
since x ∈ [2, ∞)
⇒ x ≥ 2
⇒ x – 2 ≥ 0
⇒ (x – 2)2 ≥ 0
⇒ (x – 2)2 + 1 ≥ 1
⇒ f(x) ≥ 1
∴ Rf = [1, ∞)

Question 26.
If a function f : Q → R is defined by f(x) = $$\frac{2 x-1}{2}$$ and function g : Q → R is defined by g (x) = x + 2, then (gof) ($$\frac{3}{2}$$) is
(a) 1
(b) 2
(c) $$\frac{7}{2}$$
(d) none of these
Solution:
(d) none of these

Given f : Q → R defined by f(x) = $$\frac{2 x-1}{2}$$
and g : Q → R defined by g (x) = x + 2
∴ gof exists if Rf ⊂ Dg but R ⊄ Q.

Question 27.
If function f : R → R is defined by f (x) = sin x and function g : R → R is defined by g (x) = x2, then (fog) (x) is
(a) x2 sin x
(b) (sin x)2
(c) sin x2
(d) $$\frac{\sin x}{x^2}$$
Solution:
(c) sin x2

Here Df = Dg = Rf = Rg = R
∴ fog exists.
(fog) (x) = f (g (x)) = f (x2) = sin x2

Question 28.
If f : R → R is defined by f (x) = 3x2 – 5 and g : R → R is defined by g (x) = $$\frac{x}{x^2+1}$$, then (gof) (x) is
(a) $$\frac{3 x^2-5}{9 x^4-30 x^2+26}$$
(b) $$\frac{3 x^2-5}{9 x^4-6 x^2+26}$$
(c) $$\frac{3 x^2}{x^4+2 x^2-4}$$
(d) $$\frac{3 x^2}{9 x^4+20 x^2-2}$$
Solution:
gof exists as Rf ⊂ Dg
∴ (gof) (x) = g (f (x)) = g(3x2 – 5)
= $$\frac{3 x^2-5}{\left(3 x^2-5\right)^2+1}$$
= $$\frac{3 x^2-5}{9 x^4-30 x^2+26}$$

Question 29.
If function f : N → N is defined by f(x) = 2x + 3, for all x ∈ N, then f is
(a) surjective
(b) injective
(c) bijective
(d) none of these
Solution:
(b) injective

Given f : N → N defined by f (x) = 2x + 3 ∀ x ∈ N
one-one :
∀ x, y ∈ N s.t f(x) = f (y)
⇒ 2x + 3 = 2y + 3
⇒ x = y
∴ f is one-one.

onto :
Now 1 ∈ N (codomain of f)
Let x ∈ N (domain of f) s.t. f (x) = 1
⇒ 2x + 3 = 1
⇒ x = – 1 ∉ N.
Thus element 1 has no pre-image in N (Df)
∴ f is not onto.

Question 30.
If function f : Z → Z ¡s defined by f(x) = $$\begin{cases}\frac{x}{2}, & \text { if } x \text { is even } \\ 0, & \text { if } x \text { is odd }\end{cases}$$, then f is
(a) one-one but not onto
(b) onto but not one-one
(c neither one-one nor onto
(d) a bijection
Solution:

Given f : Z → Z defined by f(x) = $$\begin{cases}\frac{x}{2}, & \text { if } x \text { is even } \\ 0, & \text { if } x \text { is odd }\end{cases}$$
Clearly 1, 3 ∈ Z (Df) have same image 0 i.e. all odd integers have same image 0.
Thus different elements in Z have same image 0 (codomain of f)
∴ f is not one-one.
Consider x ∈ Z (codomain of f) be any arbitrary element.
∃ 2x ∈ Z, 2x is even.
s.t f(2x) = $$\frac{2 x}{2}$$ = x
∴ f is onto.

Question 31.
If a function f : R → R is defined by f(x) = $$\frac{x^2-5}{x^2+4}$$ then f is
(a) one-one but not onto
(b) onto but not one-one
(c) neither one-one nor onto
(d) a bijection
Solution:
(c) neither one-one nor onto

Given f : R → R defined by f (x) = $$\frac{x^2-5}{x^2+4}$$
Clearly f (1) = $$\frac{1^2-5}{1^2+4}=-\frac{4}{5}$$
f (- 1) = $$\frac{1-5}{1+4}=-\frac{4}{5}$$
Thus elements 1 and – 1 have same image $$-\frac{4}{5}$$.
∴ f is not one – one.

Onto:
Now I ∈ R (codomain of f)
Let x ∈ R (codomain of f) s.t. f(x) = 1
⇒ $$\frac{x^2-5}{x^2+4}$$ = 1
⇒ x2 – 5 = x2 + 4
⇒ – 5 = 4 which is false.
Thus there is no pre-image in R. (domain of f)
s.t. f(x) = 1
∴ f is not onto.

Question 32.
If f : [0, 1] → [0, 1] is defined by f(x) = $$\left\{\begin{array}{cc} x, & \text { if } x \text { is rational } \\ 1-x, & \text { if } x \text { is irrational } \end{array}\right.$$, then (fof) (x) is
(a) x
(b) 1 + x
(c) constant
(d) none of these
Solution:
(a) x

f : [0, 1] → [0, 1] defined by f(x) = $$\left\{\begin{array}{cc} x, & \text { if } x \text { is rational } \\ 1-x, & \text { if } x \text { is irrational } \end{array}\right.$$
∴ (fof) (x) = f (f (x)) = f (x) = x
when x is rational
(fof) (x) = f (f (x)) = f (x) = x
when x is irrational
(fof) (x) = f (f (x)) = f ( 1 – x)
= 1 – (1 – x) = x
From both cases ;
(fof) (x) = x.

Question 33.
If A = (1, 2, 3, ………….., n}, n ≥ 2 and B = {a, b}, then the number of surjections from A to B is
(a) nP2
(b) 2n – 2
(c) 2n – 1
(d) none of these
Solution:
(b) 2n – 2

We know that, No. of onto surjections from set A containing n elements to a set B containing 2 elements = 2n – 2
Since No. of functions from set A containing n elements to set B containing 2 elements = [n (B)]n (A) = 2n

Thus out of 2n functions, 2 functions are not onto.

Question 34.
If A = {a, b, c} and B = {- 3, – 1, 0, 1, 3}, then the number of injections that can be defined from A to B is
(a) 125
(b) 243
(c) 60
(d) 120
Solution:
(c) 60

We know that,
No. of one-one functions from set A containing m elements to a set B containing n elements = $$\left\{\begin{array}{cc} { }^n \mathrm{P}_m ; & n \geq m \\ 0 ; & n<m \end{array}\right.$$
Given O (A) = 3 = m ;
O (B) = 5 = n
∴ No. of one-one functions = nPm
= 5P3
= $$\frac{5 !}{2 !}$$
= 5 × 4 × 3 = 60.

Question 35.
If A and B are two sets such that O (A) = 5 and O (B) = 6, then the number of one- one and onto mapping from A to B is
(A) 120
(b) 720
(c) 0
(d) none of these
Solution:
(c) 0

We know that,
No. of bijections from set A containing m elements to set B containing n elements.
= $$\left\{\begin{array}{cl} m ; & m=n \\ 0 ; & m \neq n \end{array}\right.$$
Since O (A) ≠ O (B)
∴ No. of bijections from A to B = 0.

Question 36.
If function f : A → B is a bijection, then f-1 of is
(a) the identity function on A
(b) the identity function on B
(c) equal to fof-1
(d) none of these
Solution:
(a) the identity function on A

Since f : A → B is bijective f is one-one and onto.
Then f (x) = y
⇔ x = f-1 (y)
∴ (f-1of) (x) = f-1 (f (x)) = f-1 (y)
= x ∀ x ∈ A
∴ f-1of = IA

Question 37.
If function f : R → R is defined by f(x) = 3x – 4, then f-1 (x) is given by
(a) 3x + 4
(b) $$\frac{x}{3}$$ – 4
(c) $$\frac{x+4}{3}$$
(d) none of these
Solution:
(c) $$\frac{x+4}{3}$$

one-one :
∀ x, y ∈ R s.t f (x) = f (y)
⇒ 3x – 4 = 4y – 4
⇒ x = y
f is one-one.

onto :
Let y ∈ R (codomain of f) be any arbitrary.
Let f (x) = y
⇒ 3x – 4 = y
⇒ x = $$\frac{y+4}{3}$$
as y ∈ R
⇒ $$\frac{y+4}{3}$$ ∈ R
s.t. f(x) = $$f\left(\frac{y+4}{3}\right)=3\left(\frac{y+4}{3}\right)$$ – 4 = y
Thus ∀ y ∈ R 3 x ∈ R s.t f (x) = y
∴ f is onto.
Therefore f is bijective and hence inversible.

To find f-1 :
Let y = f (x) = 3x – 4
⇒ x = $$\frac{y+4}{3}$$
Now y = f (x) and f is invertible
⇒ x = f-1 (y)
⇒ f-1 (y) = $$\frac{y+4}{3}$$
⇒ f-1 (x) = $$\frac{x+4}{3}$$

Question 38.
Which of the following functions from Z → Z is a bijection ?
(a) f (x) = x3
(b) f (x) = x + 2
(c) f (x) = 2x + 1
(d) f (x) = x2 + 1
Solution:
(b) f (x) = x + 2

Given f : Z → Z defined by f (x) = x3
Clearly – 2 ∈ Z (codomain of f) s.t f (x) = – 2
x3 = – 2
⇒ x = (- 2)1/3 ∉ Z
So element – 2 has no pre-image in Z
∴ f is not onto.

Clearly 4 ∈ Z (codomain of f) s.t f x) = 4
⇒ 2x + 1 = 4
⇒ x =$$\frac{3}{2}$$ (domain of f)
Thus f : Z → Z defined by f (x) = 2x + 1 is not onto.
Clearly f : Z → Z defined by f (x) = x2 + 1
Now f (1) = 12 + 1 = 2 and
f (- 1) = (- 1)2 + 1 = 2
So element 1 and – 1 hs same image 2.
∴ f is not one-one.

Given f : Z → Z defined by f (x) = x + 2 ∀ x, y ∈ Z s.t f (x) = f (y)
⇒ x + 2 = y + 2
⇒ x = y
∴ f is one-one

onto :
∀ y ∈ Z (codomain of f) be any arbitrary element.
Let f (x) = y
⇒x + 2 = y
⇒ x = y – 2
since y ∈ Z
⇒ y – 2 ∈ Z
⇒ x ∈ Z s.t f(x) = f(y – 2) = y – 2 + 2 = y
Thus, ∀ y ∈ Z ∃ x ∈ Z s.t f (x) = y
∴ f is onto
Hence f is bijective.

Question 39.
If f : R → R is a function defined by f (x) = x3 + 5, then f-1 (x) is
(a) (x + 5)$$\frac{1}{3}$$
(b) (x – 5)$$\frac{1}{3}$$
(c) (5 – x)$$\frac{1}{3}$$
(d) 5 – x
Solution:
(b) (x – 5)$$\frac{1}{3}$$

Given f : R → R defined by f(x) = x3 + 5
one-one :
∀ x, y ∈ R s.t f (x) = f (y)
⇒ x3 + 4= y3 + 5
⇒ x3 = y3
⇒ (x – y) (x2 + xy + y2) = 0
⇒ x – y = 0
⇒ x = y
∴ f is one-one.

onto :
Clearly Rf = codomain of f = R
∴ f is onto
Thus f is one-one and onto and hence inversible.
∴ f-1 exists.

To find f-1 :
Let y = f (x) = x3 + 5
⇒ x3 = y – 5
⇒ x = (y – 5)1/3
Now f (x) = y and/is invertible
x = f-1 (y)
= (y – 5)1/3
⇒ f-1 (x) = (x – 5)1/3

Question 40.
If f : R → R is defined by f (x) = ax + b, a ≠ 0, then f-1 (x)
(a) is given by $$\frac{x-b}{a}$$
(b) is given by $$\frac{1}{a x+b}$$
(c) does not exist as f is not onto
(d) does not exist as f is not one-one
Solution:
(a) is given by $$\frac{x-b}{a}$$

Given f : R → R defined by
f(x) = ax + b, a ≠ 0
one-one :
∀ x, y ∈ R s.t f (x) = f (y)
⇒ ax + b = ax + b
⇒ x = y [∵ a ≠ 0]
∴ f is one-one.

onto :
Let y ∈ R (codomain of f) be any arbitraiy element
Let y = f (x) = ax + b
⇒ x = $$\frac{y-b}{a}$$
since y ∈ R, a ≠ 0
∴ $$\frac{y-b}{a}$$ ∈ R
⇒ x ∈ R
Thus, ∀ y ∈ R ∃ 3x ∈ R
s.t f(x) = $$f\left(\frac{y-b}{a}\right)$$
= a $$\left(\frac{y-b}{a}\right)$$ + b = y
∴ f is onto.

To find f-1:
Let f(x) = y
⇒ ax + b = y
⇒ x = $$\frac{y-b}{a}$$
Now f(x) = y and f is invertible
∴ x = f-1 (y)
⇒ x = f-1 (y) = $$\frac{y-b}{a}$$

Question 41.
If A = R – {I} and function f : A → A is defined by f(x) = $$\frac{x+1}{x-1}$$, then f-1 (x) is given by
(a) $$\frac{1}{x-1}$$
(b) $$\frac{1}{1-x}$$
(c) $$\frac{x+1}{1-x}$$
(d) $$\frac{x+1}{x-1}$$
Solution:
(d) $$\frac{x+1}{x-1}$$

Given f : A → A defined by f(x) = $$\frac{x+1}{x-1}$$ ∀ x ∈ A

one-one :
∀ x, y ∈ A s.t f(x) = f(y)
⇒ 
⇒ (x + 1) (y – 1) = (x – 1) (y + 1)
⇒ xy – x + y – 1 = xy + x – y – 1
⇒ 2x = 2y
⇒ x = y
∴ f is one-one.

onto:
Let y ∈ A (codomain of f) be any arbitrary element.
Let y = f(x) = $$\frac{x+1}{x-1}$$
⇒ x + 1 = y (x – 1)
⇒ x + 1 = xy – y
⇒ x (1 – y) = – y – 1
⇒ x = $$\frac{y+1}{y-1}$$
Now y ∈ A
⇒ $$\frac{y+1}{y-1}$$ ∈ A [∵ y ≠ 1]
[if $$\frac{y+1}{y-1}$$ = 1
⇒ y + 1 = y – 1
⇒ 1 = – 1, which is false]
Thus, ∀ y ∈ A ∃ x = $$\frac{y+1}{y-1}$$ ∈ A
s.t. f (x) = f ($$\frac{y+1}{y-1}$$)
= $$\frac{\frac{y+1}{y-1}+1}{\frac{y+1}{y-1}-1}$$
= $$\frac{2 y}{2}$$ = y
Thus f is one-one and onto and hrnce invertible.

To find f-1:
Let f(x) = y
⇒ $$\frac{x+1}{x-1}$$ = y
⇒ x = $$\frac{y+1}{y-1}$$
Now f(x) = y and f-1 exists
∴ x = f-1 (y)
= $$\frac{y+1}{y-1}$$
⇒ f-1 (x) = $$\frac{x+1}{x-1}$$

Question 42.
If f : R – {- $$\frac{1}{2}$$} → R – {$$\frac{1}{2}$$} is defined by f(x) = $$\frac{x-3}{2 x+1}$$, then f-1 (x) is
(a) $$\frac{x-3}{2 x-1}$$
(b) $$\frac{x+3}{2 x-1}$$
(c) $$\frac{x+3}{1-2 x}$$
(d) $$\frac{x-3}{1-2 x}$$
Solution:
(c) $$\frac{x+3}{1-2 x}$$

one-one:
∀ x, y ∈ R – {$$\frac{1}{2}$$} s.tf (x) = f(y)
⇒ $$\frac{x-3}{2 x+1}=\frac{y-3}{2 y+1}$$
⇒ (x – 3) (2y + 1) = (y – 3) (2x + 1)
⇒ 2xy + x – 6y – 3 = 2xy + y – 6x – 3
⇒ 7x = 7y
⇒ x = y
∴ f is 1 – 1.

onto:
Let y ∈ R – {$$\frac{1}{2}$$} be any arbitrary element.
Let y = f(x) = $$\frac{x-3}{2 x+1}$$
⇒ y (2x+ 1) = x – 3
⇒ x (2y – 1) = – 3 – y
⇒ x = $$\frac{y+3}{1-2 y}$$
since y ∈ R – {$$\frac{1}{2}$$}; y ≠ $$\frac{1}{2}$$}
∴ x = $$\frac{y+3}{1-2 y}$$ ∈ R
Further, if $$\frac{y+3}{1-2 y}$$ = – $$\frac{1}{2}$$
⇒ 2y + 6 = – 1 + 2y
⇒ 6 = – 1 which is impossible
Thus x = $$\frac{y+3}{1-2 y}$$ ∈ R – {- $$\frac{1}{2}$$}
Hence ∀ y ∈ R – {$$\frac{1}{2}$$} ∃ x ∈ R – {- $$\frac{1}{2}$$}
s.t f(x) = f ($$\frac{y+3}{-2 y}$$)
= $$\frac{\frac{y+3}{-2 y+1}-3}{2\left(\frac{y+3}{1-2 y}\right)+1}$$
= $$\frac{7 y}{7}$$ = y
∴ f is onto.
Thus ∀ y ∈ R – {$$\frac{1}{2}$$} ∃ x ∈ R – {- $$\frac{1}{2}$$}
s.t f(x) = y
∴ f is onto
Hence, f is bijective and hence inversible.

To find f-1:
Let f(x) = y = $$\frac{x-3}{2 x+1}$$
⇒ x = $$\frac{y+3}{1-2 y}$$
Now f(x) = y and f-1 exists
∴ x = f-1 (y)
⇒ x = f-1 (y) = $$\frac{y+3}{1-2 y}$$
⇒ f-1 (x) = $$\frac{x+3}{1-2 x}$$

Question 43.
If A = R – {b} and B = R – {1} and function f : A → B is defined by f (x)= $$\frac{x-a}{x-b}$$, a ≠ b,then f is
(a) one-one but not onto
(b) onto but not one-one
(c) both one-one and onto
(d) neither one-one nor onto
Solution:
(c) both one-one and onto

Given f : A → B defined by f (x) = $$\frac{x-a}{x-b}$$, a ≠ b
∀ x, y ∈ A = R – {b} s.t f(x) = f(y)
⇒ $$\frac{x-a}{x-b}=\frac{y-a}{y-b}$$
⇒ (x – a) (y – b) = (y – a) (x – b)
⇒ xy – bx – ay + ab = xy – by – ax + ab
⇒ (a – b) x = (a – b) y
⇒ x = y [∵ a ≠ b]
∴ f is one-one.
Let y ∈ B = R – {1}
∴ y ∈ R, y ≠ 1 be any arbitrary element.
Let y = f (x) = $$\frac{x-a}{x-b}$$
⇒ y (x – b) = x – a
⇒ x (y – 1) = by – a
⇒ x = $$\frac{b y-a}{y-1}$$ y ∈ R, y ≠ 1
∴ $$\frac{b y-a}{y-1}$$ ∈ R
Furthrt if $$\frac{b y-a}{y-1}$$ = b
⇒ by – a = by – b
⇒ a = b which is false
∴ x = $$\frac{b y-a}{y-1}$$ ∈ A = R – {b}
and f(x) = f $$\left(\frac{b y-a}{y-1}\right)$$
= $$\frac{\frac{b y-a}{y-1}-a}{\frac{b y-a}{y-1}-b}$$
= $$\frac{(b-a) y}{b-a}$$ = y
Thus ∀ y ∈ B ∃ x ∈ A s.t y = f (x)
∴ f is onto.
Thus, f is one-one and onto.

Question 44.
If A = R – {b} and B = R – {1} and function f : A → B is defined by f (x) = $$\frac{x-a}{x-b}$$ a ≠ b, then f-1 (x) is
(a) $$\frac{b x-a}{x-1}$$
(b) $$\frac{a-b x}{x-1}$$
(c) $$\frac{a x-b}{x-1}$$
(d) none of these
Solution:
(a) $$\frac{b x-a}{x-1}$$

Let f (x) = y = $$\frac{x-a}{x-b}$$
⇒ y (x – b) = x – a
⇒ x = $$\frac{b y-a}{y-1}$$
Now f (x) = y and f is invertible
∴ x = f-1 (y) = $$\frac{b y-a}{y-1}$$
⇒ f-1 (x) = $$\frac{b x-a}{x-1}$$

Question 45.
If f : A → B and g : B → C are both bijective functions, then (gof)-1 is bijective functions, then (gof)-1 is
(a) g-1 of-1
(b) f-1 og-1
(c) fog
(d) gof
Solution:
Given f : A → B and g : B → C both are bijective functions and gof exists and one- one, onto.
∴ (gof)-1 exists
Now (f-1 og-1) o (gof) = f-1 o (g-1 o g of)
= f-1 o (g-1 og) of
[∵ Composition of functions is associative]
= f-1 o IB of
= (f-1 o IB) of
= f-1 of
= IA
(gof)-1 = f-1 og-1.