# MCQ Questions for Class 11 Maths Chapter 2 Relations and Functions with Answers

Students are advised to solve the Relations and Functions Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Relations and Functions Class 11 with answers will boost your confidence thereby helping you score well in the exam.

Explore numerous MCQ Questions of Relations and Functions Class 11 with answers provided with detailed solutions by looking below.

Question 1.

The domain of the function ^{7-x}P_{x-3} is

(a) {1, 2, 3}

(b) {3, 4, 5, 6}

(c) {3, 4, 5}

(d) {1, 2, 3, 4, 5}

## Answer

Answer: (c) {3, 4, 5}

The function f(x) = ^{7-x}P_{x-3} is defined only if x is an integer satisfying the following inequalities:

1. 7 – x ≥ 0

2. x – 3 ≥ 0

3. 7 – x ≥ x – 3

Now, from 1, we get x ≤ 7 ……… 4

from 2, we get x ≥ 3 ……………. 5

and from 2, we get x ≤ 5 ………. 6

From 4, 5 and 6, we get

3 ≤ x ≤ 5

So, the domain is {3, 4, 5}

Question 2.

The domain of tan^{-1} (2x + 1) is

(a) R

(b) R – {1/2}

(c) R – {-1/2}

(d) None of these

## Answer

Answer: (a) R

Since tan^{-1} x exists if x ∈ (-∞, ∞)

So, tan^{-1} (2x + 1) is defined if

-∞ < 2x + 1 < ∞

⇒ -∞ < x < ∞

⇒ x ∈ (-∞, ∞)

⇒ x ∈ R

So, domain of tan^{-1} (2x + 1) is R.

Question 3.

Two functions f and g are said to be equal if f

(a) the domain of f = the domain of g

(b) the co-domain of f = the co-domain of g

(c) f(x) = g(x) for all x

(d) all of above

## Answer

Answer: (d) all of above

Two functions f and g are said to be equal if f

1. the domain of f = the domain of g

2. the co-domain of f = the co-domain of g

3. f(x) = g(x) for all x

Question 4.

If the function f : R → R be given by f(x) = x² + 2 and g : R → R is given by g(x) = x/(x – 1). The value of gof(x) is

(a) (x² + 2)/(x² + 1)

(b) x²/(x² + 1)

(c) x²/(x² + 2)

(d) none of these

## Answer

Answer: (a) (x² + 2)/(x² + 1)

Given f(x) = x² + 2 and g(x) = x/(x – 1)

Now, gof(x) = g(x² + 2) = (x² + 2)/(x² + 2 – 1) = (x² + 2)/(x² + 1)

Question 5.

Given g(1) = 1 and g(2) = 3. If g(x) is described by the formula g(x) = ax + b, then the value of a and b is

(a) 2, 1

(b) -2, 1

(c) 2, -1

(d) -2, -1

## Answer

Answer: (c) 2, -1

Given, g(x) = ax + b

Again, g(1) = 1

⇒ a × 1 + b = 1

⇒ a + b = 1 ……… 1

and g(2) = 3

⇒ a × 2 + b = 3

⇒ 2a + b = 3 …….. 2

Solve equation 1 and 2, we get

a = 2, b = -1

Question 6.

Let f : R → R be a function given by f(x) = x² + 1 then the value of f^{-1} (26) is

(a) 5

(b) -5

(c) ±5

(d) None of these

## Answer

Answer: (c) ±5

Let y = f(x) = x² + 1

⇒ y = x² + 1

⇒ y – 1 = x²

⇒ x = ±√(y – 1)

⇒ f^{-1} (x) = ±√(x – 1)

Now, f^{-1} (26) = ±√(26 – 1)

⇒ f^{-1} (26) = ±√(25)

⇒ f^{-1} (26) = ±5

Question 7.

the function f(x) = x – [x] has period of

(a) 0

(b) 1

(c) 2

(d) 3

## Answer

Answer: (b) 1

Let T is a positive real number.

Let f(x) is periodic with period T.

Now, f(x + T) = f(x), for all x ∈ R

⇒ x + T – [x + T] = x – [x] , for all x ∈ R

⇒ [x + T] – [x] = T, for all x ∈ R

Thus, there exist T > 0 such that f(x + T) = f(x) for all x ∈ R

Now, the smallest value of T satisfying f(x + T) = f(x) for all x ∈ R is 1

So, f(x) = x – [x] has period 1

Question 8.

The function f(x) = sin (πx/2) + cos (πx/2) is periodic with period

(a) 4

(b) 6

(c) 12

(d) 24

## Answer

Answer: (a) 4

Period of sin (πx/2) = 2π/(π/2) = 4

Period of cos (πx/2) = 2π/(π/2) = 4

So, period of f(x) = LCM (4, 4) = 4

Question 9.

The domain of the function f(x) = x/(1 + x²) is

(a) R – {1}

(b) R – {-1}

(c) R

(d) None of these

## Answer

Answer: (c) R

Given, function f(x) = x/(1 + x²)

Since f(x) is defined for all real values of x.

So, domain(f) = R

Question 10.

If f : R → R is defined by f(x) = x² – 3x + 2, the f(f(y)) is

(a) x^{4} + 6x³ + 10x² + 3x

(b) x^{4} – 6x³ + 10x² + 3x

(c) x^{4} + 6x³ + 10x² – 3x

(d) x^{4} – 6x³ + 10x² – 3x

## Answer

Answer: (d) x^{4} – 6x³ + 10x² – 3x

Given, f(x) = x² – 3x + 2

Now, f(f(y)) = f(x² – 3x + 2)

= (x² – 3x + 2)² – 3(x² – 3x + 2) + 2

= x^{4} – 6x³ + 10x² – 3x

Question 11.

If n is the smallest natural number such that n + 2n + 3n + …. + 99n is a perfect square, then the number of digits in square of n is

(a) 1

(b) 2

(c) 3

(d) 4

## Answer

Answer: (c) 3

Given that

n + 2n + 3n + …. + 99n

= n × (1 + 2 + 3 + …….. + 99)

= (n × 99 × 100)/2

= n × 99 × 50

= n × 9 × 11 × 2 × 25

To make it perfect square we need 2 × 11

So n = 2 × 11 = 22

Now n² = 22 × 22 = 484

So, the number of digit in n² = 3

Question 12.

Let f : R – R be a function defined by f(x) = cos(5x + 2), then f is

(a) injective

(b) surjective

(c) bijective

(d) None of these

## Answer

Answer: (d) None of these

Given, f(x) = cos(2x + 5)

Period of f(x) = 2π/5

Since f(x) is a periodic function with period 2π/5, so it is not injective.

The function f is not surjective also as its range [-1, 1] is a proper subset of its co-domain R

Question 13.

The function f(x) = sin (πx/2) + 2cos (πx/3) – tan (πx/4) is periodic with period

(a) 4

(b) 6

(c) 8

(d) 12

## Answer

Answer: (d) 12

Period of sin (πx/2) = 2π/(π/2) = 4

Period of cos (πx/3) = 2π/(π/3) = 6

Period of tan (πx/4) = π/(π/4) = 4

So, period of f(x) = LCM (4, 6, 4) = 12

Question 14.

If the function f : R → R be given by f(x) = x² + 2 and g : R → R is given by g(x) = x/(x – 1). The value of gof(x) is

(a) (x² + 2)/(x² + 1)

(b) x²/(x² + 1)

(c) x²/(x² + 2)

(d) none of these

## Answer

Answer: (a) (x² + 2)/(x² + 1)

Given f(x) = x² + 2 and g(x) = x/(x – 1)

Now, gof(x) = g(x² + 2) = (x² + 2)/(x² + 2 – 1) = (x² + 2)/(x² + 1)

Question 15.

The domain of the function ^{7-x}P_{x-3} is

(a) {1, 2, 3}

(b) {3, 4, 5, 6}

(c) {3, 4, 5}

(d) {1, 2, 3, 4, 5}

## Answer

Answer: (c) {3, 4, 5}

The function f(x) = ^{7-x}P_{x-3} is defined only if x is an integer satisfying the following inequalities:

1. 7 – x ≥ 0

2. x – 3 ≥ 0

3. 7 – x ≥ x – 3

Now, from 1, we get x ≤ 7 ………4

from 2, we get x ≥ 3 …………….5

and from 2, we get x ≤ 5 ……….6

From 4, 5 and 6, we get

3 ≤ x ≤ 5

So, the domain is {3, 4, 5}

Question 16.

If f(x) = e^{x} and g(x) = log_{e} x then the value of fog(1) is

(a) 0

(b) 1

(c) -1

(d) None of these

## Answer

Answer: (b) 1

Given, f(x) = e^{x}

and g(x) = logx

fog(x) = f(g(x))

= f (logx)

= e^{log x}

= x

So, fog(1) = 1

Question 17.

A relation R is defined from the set of integers to the set of real numbers as (x, y) = R if x² + y² = 16 then the domain of R is

(a) (0, 4, 4)

(b) (0, -4, 4)

(c) (0, -4, -4)

(d) None of these

## Answer

Answer: (b) (0, -4, 4)

Given that:

(x, y) ∈ R ⇔ x² + y² = 16

⇔ y = ±√(16 – x²)

when x = 0 ⇒ y = ±4

(0, 4) ∈ R and (0, -4) ∈ R

when x = ±4 ⇒ y = 0

(4, 0) ∈ R and (-4, 0) ∈ R

Now for other integral values of x, y is not an integer.

Hence R = {(0, 4), (0, -4), (4, 0), (-4, 0)}

So, Domain(R) = {0, -4, 4}

Question 18.

The period of the function f(x) = sin (2πx/3) + cos (πx/3)

(a) 3

(b) 4

(c) 12

(d) None of these

## Answer

Answer: (c) 12

Given, function f(x) = sin (2πx/3) + cos (πx/2)

Now, period of sin (2πx/3) = 2π/{(2π/3)} = (2π × 3)/(2π) = 3

and period of cos (πx/2) = 2π/{(π/2)} = (2π × 2)/(π) = 2 × 2 = 4

Now, period of f(x) = LCM(3, 4) = 12

Hence, period of function f(x) = sin (2πx/3) + cos (πx/2) is 12

Question 19.

If f(x) = ax + b and g(x) = cx + d and f{g(x)} = g{f(x)} then

(a) f(a) = g(c)

(b) f(b) = g(b)

(c) f(d) = g(b)

(d) f(c) = g(a)

## Answer

Answer: (c) f(d) = g(b)

Given, f(x) = ax + b and g(x) = cx + d and

Now, f{g(x)} = g{f(x)}

⇒ f{cx + d} = g{ax + b}

⇒ a(cx + d) + b = c(ax + b) + d

⇒ ad + b = cb + d

⇒ f(d) = g(b)

Question 20.

The domain of the function f (x) = 1/(2 – cos 3x) is

(a) (1/3, 1)

(b) [1/3, 1)

(c) (1/3, 1]

(d) R

## Answer

Answer: (d) R

Given

function is f(x) = 1/(2 – cos 3x)

Since -1 ≤ cos 3x ≤ 1 for all x ∈ R

So, -1 ≤ 2 – cos 3x ≤ 1 for all x ∈ R

⇒ f(x) is defined for all x ∈ R

So, domain of f(x) is R

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