MCQ Class 11 Mathematics Relations and Functions with Answers

CBSE Class 11 Mathematics Chapter 2 Relations and Functions Multiple Choice Questions with Answers. MCQ Class 11 Mathematics Relations and Functions with Answers was Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 11 Mathematics Relations and Functions MCQs with Answers to know their preparation level.

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MCQ Class 11 Mathematics Relations and Functions with Answers - Set - 5

Question 1: 

The period of the function f(x) = sin (2πx/3) + cos (πx/3)
(a) 3
(b) 4
(c) 12
(d) None of these

Correct Answer – (C)

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 2 : 

If f(x) = ex and g(x) = loge x then the value of fog(1) is
(a) 0
(b) 1
(c) -1
(d) None of these

Correct Answer – (B)

Given, f(x) = ex
and g(x) = logx
fog(x) = f(g(x))
= f (logx)
= elog x
= x
So, fog(1) = 1

Question 3 : 

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

Correct Answer – (D)

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 4 : 

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

Correct Answer – (C)

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 5 : 

The domain of the function f(x) = x/(1 + x²) is
(a) R – {1}
(b) R – {-1}
(c) R
(d) None of these

Correct Answer – (C)

Given, function f(x) = x/(1 + x²)
Since f(x) is defined for all real values of x.
So, domain(f) = R

MCQ Class 11 Mathematics Relations and Functions with Answers

Question 6 : 

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

Correct Answer – (B)

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 7 : 

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

Correct Answer – (A)

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 8 : 

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

Correct Answer – (D)

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 9 : 

If f : R → R is defined by f(x) = x² – 3x + 2, the f(f(y)) is
(a) x4 + 6x³ + 10x² + 3x
(b) x4 – 6x³ + 10x² + 3x
(c) x4 + 6x³ + 10x² – 3x
(d) x4 – 6x³ + 10x² – 3x

Correct Answer – (D)

Given, f(x) = x² – 3x + 2
Now, f(f(y)) = f(x² – 3x + 2)
= (x² – 3x + 2)² – 3(x² – 3x + 2) + 2
= x4 – 6x³ + 10x² – 3x

Question 10 : 

The function f(x) = sin (‎πx/2) + cos (πx/2) is periodic with period
(a) 4
(b) 6
(c) 12
(d) 24

Correct Answer – (A)

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

MCQ Class 11 Mathematics Relations and Functions with Answers

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