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

Relations and Functions Class 11 Maths MCQs Questions with Answers

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 2 Relations and Functions with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have Provided Relations and Functions Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 2 Quiz

Class 11 Maths Chapter 2 MCQ Online Test

You can refer to NCERT Solutions for Class 11 Maths Chapter 2 Relations and Functions to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

Relations and Functions Class 11 Maths MCQ online test

Q1.	The domain of the function ^7-xP_x-3 is
	A.{1, 2, 3}
	B.{3, 4, 5, 6}
	C.{3, 4, 5}
	D.{1, 2, 3, 4, 5}

Ans: {3, 4, 5}
The function f(x) = ^7-xP_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}

Q2.	The domain of tan^-1 (2x + 1) is
	A.R
	B.R – {1/2}
	C.R – {-1/2}
	D.None of these

Ans: 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.

Q3.	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

Ans: 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

Q4.	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

Ans: (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)

Q5.	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

Ans: 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

Q6.	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

Ans: ±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

Q7.	the function f(x) = x – [x] has period of
	A.0
	B.1
	C.2
	D.3

Ans: 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

Q8.	The function f(x) = sin (‎πx/2) + cos (πx/2) is periodic with period
	A.4
	B.6
	C.12
	D.24

Ans: 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

Q9.	The domain of the function f(x) = x/(1 + x²) is
	A.R – {1}
	B.R – {-1}
	C.R
	D.None of these

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

Q10.	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

Ans: x⁴ – 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
= x⁴ – 6x³ + 10x² – 3x

Q11.	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

Ans: 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

Q12.	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

Ans: 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

Q13.	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

Ans: 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

Q14.	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

Ans: (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)

Q15.	The domain of the function ^7-xP_x-3 is
	A.{1, 2, 3}
	B.{3, 4, 5, 6}
	C.{3, 4, 5}
	D.{1, 2, 3, 4, 5}

Ans: {3, 4, 5}
The function f(x) = ^7-xP_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}

Q16.	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

Ans: 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

Q17.	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

Ans: (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}

Q18.	The period of the function f(x) = sin (2πx/3) + cos (πx/3)
	A.3
	B.4
	C.12
	D.None of these

Ans: 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

Q19.	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)

Ans: 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)

Q20.	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

Ans: 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