Class 11 Maths Chapter 2 Relations and Functions MCQs

Class 11 Maths Chapter 2 Relations and Functions MCQs are given here with the right options and detailed explanations. All these MCQs are given here based on the latest guidelines of the CBSE for Class 11 students. Practising these multiple-choice questions helps the students to score good marks in the examination.

Get MCQs for all the chapters of Class 11 Maths here.

MCQs for Chapter 2 Relations and Functions

MCQs of Class 11 Maths Chapter 2 covers all the concepts of the NCERT curriculum such that they will be helpful for the students of Class 11 who are preparing for the board exam 2022-2023.

Class 11 Maths Chapter 2 Relations and Functions MCQs – Download PDF

Here, you can solve multiple-choice questions on linking pairs of objects from two sets, relations between the two objects in the pair and special relations, which will qualify to be functions and domain and range of functions. Practising MCQs of Class 11 Maths Chapter 2 will help boost your confidence in recognizing the technique to solve any objective type problem in the board exam 2022-23.

Also, check:

• Class 11 Maths Notes Chapter 2 Relations and Functions
• Important Questions for Class 11 Maths Chapter 2

MCQs for Chapter 2 Relations and Functions Class 11 with Answers

1. If f(x) = x³ – (1/x³), then f(x) + f(1/x) is equal to

(a) 2x³

(b) 2/x³

(c) 0

(d) 1

Correct option: (c) 0

Solution:

Given,

f(x) = x³ – (1/x³)

Now,

f(1/x) = (1/x)³ – 1/(1/x)³

= (1/x³) – x³

Thus, f(x) + f(1/x) = x³ – (1/x³) + (1/x³) – x³ = 0

2. Let n (A) = m, and n (B) = n. Then the total number of non-empty relations that can be defined from A to B is

(a) mⁿ

(b) n^m – 1

(c) mn – 1

(d) 2^mn – 1

Correct option: (d) 2^mn – 1

Solution:

Given,

n(A) = m and n(B) = n

We know that,

n(A x B) = n(A). n(B) = mn

Total number of relations from A to B = Number of subsets of A x B = 2^mn

So, the total number of non-empty relations from A to B = 2^mn – 1.

3. If f(x) = x² + 2, x ∈ R, then the range of f(x) is

(a) [2, ∞)

(b) (-∞, 2]

(c) (2, ∞)

(d) (-∞, 2) U (2, ∞)

Correct option: (a) [2, ∞)

Solution:

Given,

f(x) = x² + 2

We know that the square of any number is positive, i.e. greater than or equal to 0.

So, x² ≥ 0

Adding 2 on both sides,

x² + 2 ≥ 0 + 2

f(x) ≥ 2

Therefore, f(x) range is [2, ∞).

4. What will be the domain for which the functions f(x) = 2x² – 1 and g(x) = 1 – 3x are equal?

(a) {-2, 1}

(b) {1/2, -2}

(c) [2, 12]

(d) (-1, 2)

Correct option: (b) {1/2, -2}

Solution:

Given,

f(x) = 2x² – 1

g(x) = 1 – 3x

Now,

f(x) = g(x)

⇒ 2x² – 1 = 1 – 3x

⇒ 2x² + 3x – 2 = 0

⇒ 2x² + 4x – x – 2 = 0

⇒ 2x(x + 2) – 1(x + 2) = 0

⇒ (2x – 1) (x + 2) = 0

Thus the domain for which the function f (x) = g (x) is {1/2, -2}.

5. If [x]² – 5 [x] + 6 = 0, where [ . ] denotes the greatest integer function, then

(a) x ∈ [3, 4]

(b) x ∈ (2, 3]

(c) x ∈ [2, 3]

(d) x ∈ [2, 4)

Correct option: (d) x ∈ [2, 4)

Solution:

Given,

([x] – 2)([x] – 3) = 0

When [x] = 2, 2 ≤ x < 3

When [x] = 3, 3 ≤ x < 4

From the above, x ∈ [2, 4).

6. If f(x) = ax + b, where a and b are integers, f(–1) = – 5 and f(3) = 3, then a and b are equal to

(a) a = – 3, b = –1

(b) a = 2, b = – 3

(c) a = 0, b = 2

(d) a = 2, b = 3

Correct option: (b) a = 2, b = – 3

Solution:

Given,

f(x) = ax + b

And

f(-1) = -5

a(-1) + b = -5

-a + b = -5….(i)

Also, f(3) = 3

a(3) + b = 3

3a + b = 3….(ii)

From (i) and (ii),

a = 2, b = -3

7. The domain of the function f(x) = x/(x² + 3x + 2) is

(a) [-2, -1]

(b) R – {1, 2}

(c) R – {-1, -2}

(d) R – {2}

Correct option: (c) R – {-1, -2}

Solution:

Given f(x) is a rational function of the form g(x)/h(x), where g(x) = x and h(x) = x² + 3x + 2.

Now h(x) ≠ 0

⇒ x² + 3x + 2 ≠ 0

⇒ x² + x + 2x + 2 ≠ 0

⇒ x(x + 2) + 2(x + 1)

⇒ (x + 1) (x + 2) ≠ 0

⇒ x ≠ -1, x ≠ -2

Therefore, the domain of the given function is R – {– 1, – 2}.

8. The range of f(x) = √(25 – x²) is

(a) (0, 5)

(b) [0, 5]

(c) (-5, 5)

(d) [1, 5]

Correct option: (b) [0, 5]

Solution:

Given,

f(x) = √(25 – x²)

The domain of f(x) is [-5, 5] since the given function is defined only when (25 – x²) ≥ 0.

Let y = √(25 − x²)

y² = 25 – x²

or x² = 25 – y²

Since x ∈ [– 5, 5], the range of f(x) is [0, 5].

9. The domain and range of the real function f defined by f(x) = (4 – x)/(x – 4) is given by

(a) Domain = R, Range = {–1, 1}

(b) Domain = R – {1}, Range = R

(c) Domain = R – {4}, Range = {– 1}

(d) Domain = R – {– 4}, Range = {–1, 1}

Correct option: (c) Domain = R – {4}, Range = {– 1}

Solution:

To find the domain, consider the denominator ≠ 0

(x – 4) ≠ 0

x ≠ 4

So, domain = R – {4}

Now,

f(x) = (4 – x)/(x – 4)

= -1(x – 4)/(x – 4)

= -1

Therefore, the range of f(x) = -1.

10. The domain and range of the function f given by f(x) = 2 – |x −5| is

(a) Domain = R+ , Range = ( – ∞, 1]

(b) Domain = R, Range = ( – ∞, 2]

(c) Domain = R, Range = (– ∞, 2)

(d) Domain = R+ , Range = (– ∞, 2]

Correct option: (b) Domain = R, Range = ( – ∞, 2]

Solution:

Given,

f(x) = 2 – |x – 5|

Now x is defined for all real numbers.

Hence the domain of f is R.

To find the range, consider |x – 5| ≥ 0

or

-|x – 5| ≤ 0

Adding 2 on both sides,

2 – |x – 5| ≤ 2

⇒ f(x) ≤ 2

Hence, the range of f(x) is (-∞, 2].