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

Answer: (b) x
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

Answer: (b) |x| ≥ 1
Hint:
We have f(x) = √(log₁₂ x²)
Since, log_a 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

Answer: (b) 1
Hint:
Given, f(x) = e^x
and g(x) = log x
fog(x) = f(g(x))
= f (log x)
= e^{log x}
= x
So, fog(1) = 1

Answer: (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

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

Answer: (a) Even 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.

Answer: (a) 4
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

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

Answer: (a) R
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.

Answer: (b) 1
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

Answer: (a) x
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

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

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

Answer: (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)²
Given Set = {a, b}
Total number of elements = 2
Total number of binary operations = (2²)² = 2⁴ = 16

Answer: (a) Even 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.

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}

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

Answer: (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}

Answer: (d) {1, 2, 3}
Hint:
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}
Now, f(3) = ^7-3P_3-3 = ⁴P₀ = 1
⇒ f(4) = ^7-4P_4-3 = ³P₁ = 3
⇒ f(5) = ^7-5P_5-3 = ²P₂ = 2
So, the range of the function is {1, 2, 3}

Answer: (d) π/6
Hint:
Since g(x) = sin⁴ x + cos⁴ x is periodic with period π/2
So, f(x) = sin⁴ 3x + cos⁴ 3x is periodic with period (π/2)/3 = π/6