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.

## Relations and Functions Class 11 MCQs Questions with Answers

Relation And Function Class 11 MCQ Question 1.
If f(x) = (a – x)1/n, a > 0 and n ∈ N, then the value of f(f(x)) is
(a) 1/x
(b) x
(c) x²
(d) x1/2

Hint:
Given, f(x) = (a – x)1/n
Now, f(f(x)) = [(a – f(x))n]1/n
⇒ f(f(x)) = [(a – {(a – xn)1/n }n ]1/n
⇒ f(f(x)) = [a – (a – xn)]1/n
⇒ f(f(x)) = [a – a + xn)]1/n
⇒ f(f(x)) = (xn)1/n
⇒ f(f(x)) = x

MCQ On Relation And Function Class 11 Question 2.
The domain of the definition of the real function f(x) = √(log12 x² ) of the real variable x is
(a) x > 0
(b) |x| ≥ 1
(c) |x| > 4
(d) x ≥ 4

Hint:
We have f(x) = √(log12 x²)
Since, loga k ≥ 0 if a > 1, k ≥ 1
or 0 < a < 1 and 0 < k ≤ 1
So, the function f(x) exists if
log12 x² ≥ 0
⇒ x² ≥ 1
⇒ |x| ≥ 1

Relations And Functions Class 11 MCQ Question 3.
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

Hint:
Given, f(x) = ex
and g(x) = log x
fog(x) = f(g(x))
= f (log x)
= elog x
= x
So, fog(1) = 1

Class 11 Maths Chapter 2 MCQ With Answers Question 4.
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

Hint:
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

Relations And Functions Class 11 MCQ Questions Question 5.
A function f(x) is said to be an odd function if
(a) f(-x) = f(x)
(b) f(-x) = -f(x)
(c) f(-x) = k * f(x) where k is a constant
(d) None of these

Hint:
A function f(x) is said to be an odd function if
f(-x) = -f(x) for all x

Class 11 Maths Chapter 2 MCQ Question 6.
If f(x) is an odd differentiable function on R, then df(x)/dx is a/an
(a) Even function
(b) Odd function
(c) Either even or odd function
(d) Neither even nor odd function

Hint:
Given, f(x) is an odd differentiable function on R
⇒ f(-x) = -f(x) for all x ∈ R
differentiate on both side, we get
⇒ -df(-x)/dx = -df(x)/dx for all x ∈ R
⇒ df(-x)/dx = df(x)/dx for all x ∈ R
⇒ df(x)/dx is an even function on R.

MCQ Of Relation And Function Class 11 Question 7.
The function f(x) = sin (‎πx/2) + cos (πx/2) is periodic with period
(a) 4
(b) 6
(c) 12
(d) 24

Hint:
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

Relations And Functions MCQ Class 11 Question 8.
If f(x) = log3 x and A = (3, 27) then f(A) =
(a) (1, 1)
(b) (3, 3)
(c) (1, 3)
(d) (2, 3)

Hint:
Since f(x) = log3 x is an increasing function
So, f(A) = (log3 3, log3 27) = (1, 3)

Relation And Function MCQ Class 11 Question 9.
The domain of tan-1 (2x + 1) is
(a) R
(b) R -{1/2}
(c) R -{-1/2}
(d) None of these

Hint:
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.

Class 11 Maths Ch 2 MCQ Question 10.
the function f(x) = x – [x] has period of
(a) 0
(b) 1
(c) 2
(d) 3

Hint:
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

MCQ Questions On Relations And Functions Class 11 Question 11.
If f(x) =(3x – 2)/(2x – 3) then the value of f(f(x)) is
(a) x
(b) x²
(c) x³
(d) None of these

Hint:
Given, f(x) = (3x – 2)/(2x – 3)
Now, f(f(x)) = f{(3x – 2)/(2x – 3)}
= {(3×(3x – 2)/(2x – 3) – 2)}/{(2(3x – 2)/(2x – 3) – 3)}
= {(9x – 6)/(2x – 3) – 2)}/{((6x – 4)/(2x – 3) – 3)}
= [{(9x – 6) – 2(2x – 3)}/(2x – 3)]/[{(6x – 4) – 3(2x – 3)}/(2x – 3)]
= {(9x – 6) – 2(2x – 3)}/{(6x – 4) – 3(2x – 3)}
= (9x – 6 – 4x + 6)/(6x – 4 – 6x + 9)
= 5x/5
= x
So, f(f(x)) = x

Ch 2 Maths Class 11 MCQ Question 12.
Let R be the set of real numbers. If f(x) = x² and g(x) = 2x + 1, then fog(x) is equal to
(a) 2x + 1
(b) 2x² + 1
(c) (2x + 1)²
(d) None of these

Hint:
Given, f(x) = x² and g(x) = 2x + 1
Now gof(x) = g(f(x)) = f(x²) = 2x² + 1

MCQ Questions For Class 11 Maths Chapter 2 Question 13.
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

Hint:
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}

MCQ Questions For Class 11 Maths With Answers Chapter 2 Question 14.
The number of binary operations on the set {a, b} are
(a) 2
(b) 4
(c) 8
(d) 16

Hint:
Let S is a finite set containing n elements.
Since binary operation on S is a function from S×S to S, therefore total number of
binary operations on S is the
total number of functions from S×S to S = (nn
Given Set = {a, b}
Total number of elements = 2
Total number of binary operations = (2²)² = 24 = 16

Class 11 Relations And Functions MCQ Questions Question 15.
If f is an even function and g is an odd function the fog is a/an
(a) Even function
(b) Odd function
(c) Either even or odd function
(d) Neither even nor odd function

Hint:
Given, f is an even function and g is an odd function.
Now, fog(-x) = f{g(-x)}
= f{-g(x)} {since g is an odd function}
= f{g(x)} for all x {since f is an even function}
So, fog is an even function.

Question 16.
The domain of the function f(x) = 1/(x² – 3x + 2) is
(a) {1, 2}
(b) R
(c) R – {1, 2}
(d) R – {1, -2}

Answer: (c) R – {1, 2}
Hint:
Given, function is f(x) = 1/(x² – 3x + 2)
Clearly, f(x) is not defined when x² – 3x + 2 = 0
⇒ (x – 1)×(x – 1) = 0
⇒ x = 1, 2
So, f(x) is not defined when x = 1, 2
So, domain of function is R – {1, 2}

Question 17.
The domain of the function f(x) = sin-1 (tan x) is
(a) -π/4 ≤ x ≤ π/4
(b) nπ – π/4 ≤ x ≤ nπ + π/4
(c) nπ – π/3 ≤ x ≤ nπ + π/3
(d) -π/3 ≤ x ≤ π/3

Answer: (b) nπ – π/4 ≤ x ≤ nπ + π/4
Hint:
sin-1 (tan x) is defined for -1 ≤ tan x ≤ 1
= -π/4 ≤ x ≤ π/4
The general solution of the above inequality is
nπ -π/4 ≤ x ≤ nπ + π/4

Question 18.
Let A = {-2, -1, 0} and f(x) = 2x – 3 then the range of f is
(a) {7, -5, -3}
(b) {-7, 5, -3}
(c) {-7, -5, 3}
(d) {-7, -5, -3}

Hint:
Given, A = {-2, -1, 0}
and f(x) = 2x – 3
Now, f(-2) = 2 × (-2) – 3 = -4 – 3 = -7
f(-1) = 2 × (-1) – 3 = -2 – 3 = -5
f(0) = 2 × 0 – 3 = -3
So, range of f = {-7, -5, -3}

Question 19.
The range of the function 7-xPx-3 is
(a) {1, 2, 3, 4, 5}
(b) {3, 4, 5}
(c) None of these
(d) {1, 2, 3}

Hint:
The function f(x) = 7-xPx-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}
Now, f(3) = 7-3P3-3 = 4P0 = 1
⇒ f(4) = 7-4P4-3 = 3P1 = 3
⇒ f(5) = 7-5P5-3 = 2P2 = 2
So, the range of the function is {1, 2, 3}

Question 20.
The period of the function f(x) = sin4 3x + cos4 3x is
(a) π/2
(b) π/3
(c) π/4
(d) π/6