Class 12 Maths Chapter 1 Relations and Functions MCQs

Class 12 Maths Chapter 1 Relations and Functions MCQs are available here with the correct options and explanations. MCQs of Class 12 Maths Chapter 1 covers all the concepts of the NCERT curriculum such that they will be helpful for the students of Class 12 who are preparing for the Term I board exam 2022-23. All these MCQs are given here based on the latest guidelines of the CBSE for Class 12 students. Practising these multiple-choice questions helps the students to score good marks in the examination.

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MCQs for Chapter 1 Relations and Functions

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Chapter 1 of Class 12 Relations and Functions is about the types of relations and functions, the composition of functions, invertible functions and binary operations. Practising MCQs of Chapter 1 Class 12 Maths will help you to boost your confidence in identifying the technique to solve any problem in the board exam 2022-23.

MCQs for Chapter 1 Relations and Functions with Answers

1. A relation R in a set A is called _______, if (a₁, a₂) ∈ R implies (a₂, a₁) ∈ R, for all a₁, a₂ ∈ A.

(a) symmetric

(b) transitive

(c) equivalence

(d) non-symmetric

Correct option: (a) symmetric

Solution:

A relation R in a set A is called symmetric, if (a₁, a₂) ∈ R implies (a₂, a₁) ∈ R, for all a₁, a₂ ∈ A.

2. Let R be a relation on the set N of natural numbers defined by nRm if n divides m. Then R is

(a) Reflexive and symmetric

(b) Transitive and symmetric

(c) Equivalence

(d) Reflexive, transitive but not symmetric

Correct option: (d) Reflexive, transitive but not symmetric

Solution:

Given that n divides m, ∀ n ∈ N, R is reflexive.

Let n = 3 and m = 6

R is not symmetric since for 3, 6 ∈ N, 3 R 6 ≠ 6 R 3.

R is transitive since for n, m, r whenever n/m and m/r ⇒ n/r, i.e., n divides m and m divides r, then n will also divide r.

So, the given relation is reflexive, transitive, but not symmetric.

3. The maximum number of equivalence relations on the set A = {1, 2, 3} are

(a) 1

(b) 2

(c) 3

(d) 5

Correct option: (d) 5

Solution:

Given, set A = {1, 2, 3}

The equivalence relations for the given set are:

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, 2, 3) ⇔ A x A = A²}

Therefore, the maximum number of an equivalence relation is ‘5’.

4. If set A contains 5 elements and the set B contains 6 elements, then the number of one-one and onto mappings from A to B is

(a) 720

(b) 120

(c) 0

(d) none of these

Correct option: (c) 0

Solution:

Given,

n(A) = 5

n(B) = 6

Each element in set B is assigned to only one element in set A for the one-one function.

Here, only ‘5’ elements of set B are assigned to ‘5’ elements of set ‘A’ and one element will be left in set B.

The range of the function must be equal to B. However, for the given sets, it is not possible.

Thus, the range of functions does not contain all ‘6’ elements of set ‘ B’. Therefore, if the function is one-one it cannot be onto.

Hence, the number of one-one and onto mappings from A to B is 0.

5. Let f : [2, ∞) → R be the function defined by f(x) = x² – 4x + 5, then the range of f is

(a) R

(b) [1, ∞)

(c) [4, ∞)

(d) [5, ∞)

Correct option: (b) [1, ∞)

Solution:

Given,

f : [2, ∞) → R

f(x) = x² – 4x + 5

Let f(x) = y

x² – 4x + 5 = y

x² – 4x + 4 + 1 = y

(x – 2)² + 1 = y

(x – 2)² = y – 1

x – 2 = √(y – 1)

x = 2 + √(y – 1)

If the function is a real-valued function, then y – 1 ≥ 0

So, y ≥ 1.

Therefore, the range is [1, ∞).

6. Let f : R → R be defined by f(x) = 1/x ∀ x ∈ R. Then f is

(a) one-one

(b) onto

(c) bijective

(d) f is not defined

Correct option: (d) f is not defined

Solution:

f(x) = 1/x ∀ x ∈ R

Suppose x = 0, then f is not defined.

i.e. f(0) = 1/0 = undefined

So, the function f is not defined.

7. Let A = {1, 2, 3} and consider the relation 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

Correct option: (a) reflexive but not symmetric

Solution:

Given,

A = {1, 2, 3}

R = {1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}.

Let us write the combination of elements to check whether the given relation is reflexive, symmetric, and transitive.

R is reflexive because (1, 1),(2, 2),(3, 3) ∈ R.

R is not symmetric because (1, 2), (2, 3), (1, 3) ∈ R but (2, 1), (3, 2), (3, 1) ∉ R.

R is transitive because (1, 2) ∈ R and (2, 3) ∈ R ⇒ (1, 3) ∈ R

Therefore, R is reflexive, transitive, but not symmetric.

8. If f : R → R be defined by f(x) = 3x² – 5 and g : R → R by g(x) = x/(x² + 1), then g o f is

(a) (3x² – 5)/(9x⁴ – 30x² + 26)

(b) (3x² – 5)/(9x⁴ – 6x² + 26)

(c) 3x²/(x⁴ + 2x² – 4)

(d) 3x²/(9x⁴ + 30x² – 2)

Correct option: (a) (3x² – 5)/(9x⁴ – 30x² + 26)

Solution:

Given,

f(x) = 3x² – 5

g(x) = x/(x² + 1)

g o f = g[f(x)]

= g(3x² – 5)

= (3x² – 5)/ [(3x² – 5)² + 1]

= (3x² – 5)/(9x⁴ – 30x² + 25 + 1)

= (3x² – 5)/(9x⁴ – 30x² + 26)

9. Let f : R → R be given by f (x) = tan x. Then f^–1(1) is

(a) π/4

(b) {n π + π/4 : n ∈ Z}

(c) does not exist

(d) none of these

Correct option: (a) π/4

Solution:

Given,

f(x) = tan x

Let f(x) = y

⇒ y = tan x

⇒ x = tan^-1y

⇒ f^-1(y) = tan^-1y [since f(x) = y ⇒ x = f^-1(y) ]

⇒ f^-1(x) = tan^-1x

Similarly,

f^-1(1) = tan^-1(1)

= tan^-1(tan π/4)

= π/4

10. If f: R → R be given by f(x) = (3 – x³)^1/3, then fof(x) is

(a) x^1/3

(b) x³

(c) x

(d) (3 – x³)

Correct option: (c) x

Solution:

Given,

f(x) = (3 – x³)^1/3, then

fof(x) = f[(3 – x³)^1/3]

= [3 – {(3 – x³)^1/3)}³]^1/3

= [3 – (3 – x³)]^1/3

= [3 – 3 + x³]^1/3

= (x³)^1/3

= x

11. Let f : A → B and g : B → C be the bijective functions. Then (g ∘ f)^–1 is

(a) f^–1 ∘ g^–1

(b) f ∘ g

(c) g^–1 ∘ f^–1

(d) g ∘ f

Correct option: (a) f^–1 ∘ g^–1

Solution:

Given that, f : A → B and g : B → C be the bijective functions.

Let us consider three sets such as A = {1,3,4}, B ={2,5,1} and C = {3,4,2}.

f : A → B is a bijective function.

∴ f = {(1, 2), (3, 5), (4, 1)}

f-1 = {(2,1),(5,3),(1,4)}

g : B → C is a bijective function.

∴ g = {(2, 3), (5, 4), (1, 2)}

g-1 ={(3,2),(4, 5),(2, 1)}

Now,

g∘f (1) = g[f(1)] = g(2) = 3

g∘f (3) = g[f(3)] = g(5) = 4

g∘f (4) = g[f(4)] = g(1) = 2

∴ g∘f = {(1,3),(3,4),(4,2)} ….(1)

Then (g∘f)^-1 = {(3,1),(4, 3),(2, 4)} ….(2)

Now, f^-1∘g^-1(3) = f^-1[g^-1(3)] = f^-1(2) = 1
f^-1∘g^-1(4) = f^-1[g^-1(4)] = f^-1(5) = 3
f^-1∘g^-1(2) = f^-1[g^-1(2)] = f^-1(1) = 4

∴ f^-1∘g^-1 = {(3, 1), (4, 3), (2, 4)} ….(3)

From the (2) and (3), we can conclude that (g ∘ f)^–1 = f^-1∘g^-1.

(a) 9

(b) 14

(c) 5

(d) none of these

Correct option: (a) 9

Solution:

Given,

Let us calculate f(-1), f(2) and f(4).

f(-1) = 3(-1) = -3 [since -1 < 1 and f(x) = 3x for x ≤ 1]

f(2) = 2² = 4 [since 2 < 3 and f(x) = x² for 1 < x ≤ 3]

f(4) = 2(4) = 8 [since 4 > 3 and f(x) = 2x for x > 3]

Therefore, f(– 1) + f(2) + f(4) = -3 + 4 + 8 = 9

13. Consider a binary operation ∗ on N defined as a ∗ b = a³ + b³. Choose the correct answer.

(a) ∗ is both associative and commutative

(b) ∗ is commutative but not associative

(c) ∗ is associative but not commutative

(d) ∗ is neither commutative nor associative

Correct option: (b) ∗ is commutative but not associative

Solution:

Given that the binary operation ∗ on N is defined as a∗b= a³ + b³.

Apply the given binary operation on b∗a.

b∗a= b³ + a³ = a³ + b³

It shows that the value of a∗b is equal to that of b∗a.

So, the operation is commutative.

Consider different values of the variable as a = 1, b = 2 and c = 3.

Apply the given binary operation on (a∗b)∗c.

(a∗b)∗c = ( 1∗2 )∗3 = ( 1³ + 2³ )∗3 = 9³ + 3³ = 729 + 27 = 756

Apply the given binary operation on a∗( b∗c ).

( a∗b )∗c = 1∗( 2∗3 ) =1∗( 2³ + 3³ ) = 1³ + 35³ = 42876

( a∗b )∗c ≠ a∗( b∗c )

So the operation is not associative.

Therefore, the given operation is commutative but not associative.

14. Let f : R → R be 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}

Correct option: (c) {4, –4}, φ

Solution:

Given,

f(x) = x² + 1

Let f ^–1(17) = x

⇒ f(x) = 17 or

x² + 1 = 17

x² = 16

⇒ x = ± 4

or f ^–1(17) = {4, – 4}

Again consider f ^–1(–3) = x

⇒ f(x) = – 3

⇒ x² + 1 = – 3

⇒ x² = – 4

So, f ^–1(– 3) = φ.

15. Set A has 3 elements, and set B has 4 elements. Then the number of injective mappings that can be defined from A to B is

(a) 144

(b) 12

(c) 24

(d) 64

Correct option: (c) 24

Solution:

The total number of injective mappings from the set containing 3 elements into the set containing 4 elements is ⁴P₃ = 4! = 4 × 3 × 2 × 1 = 24.

16. Let f : R → R be defined by f(x) = 3x – 4. Then f^–1 (x) is given by

(a) (x + 4)/3

(b) (x/3) – 4

(c) 3x + 4

(d) None of these

Correct option: (a) (x + 4)/3

Solution:

Given,

f(x) = 3x – 4

Let f(x) = y

3x – 4 = y

3x = y + 4

x = (y + 4)/3

From f(x) = y, we can write as x = f^-1(y).

So, f^-1(y) = (y + 4)/3

Therefore, f^-1(x) = (x + 4)/3

17. The identity element for the binary operation * defined on Q ~ {0} as a * b = ab/2 ∀ a, b ∈ Q ~ {0} is

(a) 1

(b) 0

(c) 2

(d) none of these

Correct option: (c) 2

Solution:

Given,

Given,

Binary operation * defined on Q ~ {0} as a * b = ab/2 ∀ a, b ∈ Q ~ {0}

Let e be the identity element for * such that

a*e = e*a = a….(1)

So, a*e = ae/2

⇒ a = ae/2 [From (1)]

⇒ 2a = ae

⇒ e = 2

18. Number of binary operations on the set {a, b} are

(a) 10

(b) 16

(c) 20

(d ) 8

Correct option: (b) 16

Solution:

Let the given set be A = {a, b}

n(A) = 2

Total number of binary operations = 2^{(2 × Number of elements in the set)}

= 2^{(2 × 2)}

= 2⁴

= 16

Therefore, the number of binary operations on the set {a, b} is 16.

19. Let A = {1, 2, 3}. Then the 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

Correct option: (a) 1

Solution:

Relation R is reflexive as (1, 1), (2, 2), (3, 3) ∈ R.

Relation R is symmetric since (1, 2), (2, 1) ∈ R and (1, 3), (3, 1) ∈ R.

But relation R is not transitive as (3, 1), (1, 2) ∈ R but (3, 2)∉ R.

When we add any one of the two pairs, i.e. (3,2) and (2,3) or both, to relation R, it will become transitive.

Hence, the total number of desired relations is 1.

20. Let f , g : R → R be defined by f(x) = 3x + 1 and g(x) = x² – 2, ∀ x ∈ R, respectively. Then, f o g is

(a) 9x² + 6x – 1

(b) 3x² – 5

(c) 9x² – 6x – 3

(d) 3x²

Correct option: (b) 3x² – 5

Solution:

Given,

f(x) = 3x + 1

g(x) = x² – 2

f o g = f[g(x)]

= f(x² – 2)

= 3(x² – 2) + 1

= 3x² – 6 + 1

= 3x² – 5

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