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 l_{1} and l_{2} iff l_{1} ⊥ l_{2}, 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 l_{1} R l_{2} iff l_{1} ⊥ l_{2}

Since every line is not ⊥ to itself

∴ (l, l) ∉ R

∴ R is not reflexive on L.

Now l_{1} R l_{2}

⇒ l_{1} ⊥ l_{2}

⇒ l_{2} ⊥ l_{1}

⇒ (l_{2}, l_{1}) ∈ R

Thus R is symmetric on L.

Now (l_{1}, l_{2} ), (l_{2}, l_{3} ) ∈ R

⇒ l_{1} ⊥ l_{2} and l_{2} ⊥ l_{3}

∴ l_{1} || l_{3}

∴ (l_{1}, l_{3}) ∉ 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 :

R_{1} = {(1, 1), (3, 3), (5, 5), (1,3), (3,1)}

R_{2} = {(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

∴ R_{1} = {(1, 1), (2, 2), (3, 3)}

R_{2} = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}

R_{3} = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 1)}

R_{4} = {(1, 1), (2, 2), (3, 3), (2, 3), (3, 2)}

R_{5} = {(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)}

= 2^{3} = 8

[Here O (A) = 3 ; O (B) = 2]

Question 21.

The adjoining diagram shows that

(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) = 2^{2} = 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) = x^{2} + 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) = x^{2} + 1

Now f (x) = 17

⇒ x^{2} + 1 = 17

⇒ x^{2} = 16

⇒ x = ± 4

∴ pre-image of 17 are {- 4, 4}

Now f (x) = – 3

⇒ x^{2} + 1 = – 3

⇒ x^{2} = – 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) = 3x^{2} – 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) = 3x^{2} – 1

Now f(x) = – 28

⇒ 3x^{2} – 1 = – 28

⇒ x^{2} = – 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) = x^{2} – 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) = x^{2} – 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

∴ R_{f} = [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 R_{f} ⊂ D_{g} 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) = x^{2}, then (fog) (x) is

(a) x^{2} sin x

(b) (sin x)^{2}

(c) sin x^{2}

(d) \(\frac{\sin x}{x^2}\)

Solution:

(c) sin x^{2}

Here D_{f} = D_{g} = R_{f} = R_{g} = R

∴ fog exists.

(fog) (x) = f (g (x)) = f (x^{2}) = sin x^{2}

Question 28.

If f : R → R is defined by f (x) = 3x^{2} – 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 R_{f} ⊂ D_{g}

∴ (gof) (x) = g (f (x)) = g(3x^{2} – 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 (D_{f})

∴ 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 (D_{f}) 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

⇒ x^{2} – 5 = x^{2} + 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) ^{n}P_{2}

(b) 2^{n} – 2

(c) 2^{n} – 1

(d) none of these

Solution:

(b) 2^{n} – 2

We know that, No. of onto surjections from set A containing n elements to a set B containing 2 elements = 2^{n} – 2

Since No. of functions from set A containing n elements to set B containing 2 elements = [n (B)]^{n (A)} = 2^{n}

Thus out of 2^{n} 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 = ^{n}P_{m}

= ^{5}P_{3}

= \(\frac{5 !}{2 !}\)

= 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^{-1}of) (x) = f^{-1} (f (x)) = f^{-1} (y)

= x ∀ x ∈ A

∴ f^{-1}of = I_{A}

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) = x^{3}

(b) f (x) = x + 2

(c) f (x) = 2x + 1

(d) f (x) = x^{2} + 1

Solution:

(b) f (x) = x + 2

Given f : Z → Z defined by f (x) = x^{3}

Clearly – 2 ∈ Z (codomain of f) s.t f (x) = – 2

x^{3} = – 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) = x^{2} + 1

Now f (1) = 1^{2} + 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) = x^{3} + 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) = x^{3} + 5

one-one :

∀ x, y ∈ R s.t f (x) = f (y)

⇒ x^{3} + 4= y^{3} + 5

⇒ x^{3} = y^{3}

⇒ (x – y) (x^{2} + xy + y^{2}) = 0

⇒ x – y = 0

⇒ x = y

∴ f is one-one.

onto :

Clearly R_{f} = 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) = x^{3} + 5

⇒ x^{3} = 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 I_{B} of

= (f^{-1} o I_{B}) of

= f^{-1} of

= I_{A}

(gof)^{-1} = f^{-1} og^{-1}.