Multiple Choice Questions on Relations and Functions Class 11 OP Malhotra Exe-2I ISC Maths Solutions Ch-2. In this article you would learn to solve MCQs on Relation and Function. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11 Mathematics.

Relations and Functions Class 11 OP Malhotra Multiple Choice Questions ISC Maths Solutions Ch-2

Board	ISC
Publications	S Chand
Subject	Maths
Class	11th
Chapter-2	Relations and Functions
Writer	OP Malhotra
Exe-2(I)	Multiple Choice Questions.

Multiple Choice Questions on Relations and Functions

OP Malhotra ISC Class 11 Maths Solutions

Que-1: If A and B are two non-empty finite sets having 3 and 4 elements respectively, then what would be the possible number of relations of A × B ?

(a) 24
(b) 144
(c) 256
(d) 4096

Sol: (d) 4096
Number of elements in set A = 3
Number of elements in set B = 4
Number of ordered pairs in A × B = 3 × 4 = 12
Number of possible relations = 2ⁿ where n = number of ordered pairs
So, total relations = 2¹² = 4096

Que-2: If there are 2 elements in a set A, then what would be the number of possible relations from the set A to set A ?

(a) 2
(b) 4
(c) 16
(d) 32

Sol:  (c) 16
Let A = {a, b}
⇒ number of elements = 2
Number of ordered pairs in A × A = 2 × 2 = 4
Total number of relations = 2⁴ = 16

Que-3: If f(x) = 5 − 3x and g(x) = 12x + 2, which statement is NOT true?

(a) f(0) > g(0)
(b) g(5) > f(5)
(c) f(1) > g(1)
(d) f(−1) > g(−1)

Sol: (c) f(1) > g(1)
f(x) = 5 − 3x,    g(x) = 12x + 2
(a) f(0) = 5, g(0) = 2 → 5 > 2 ✔ True
(b) f(5) = 5 − 15 = −10, g(5) = 60 + 2 = 62 → 62 > −10 ✔ True
(c) f(1) = 2, g(1) = 14 → 2 > 14 ✖ False
(d) f(−1) = 8, g(−1) = −10 → 8 > −10 ✔ True

Que-4: If A and B be two elements such that A × B consists of 6 elements. If three elements of A × B are (1, 4), (2, 6), and (3, 6), find B × A.

(a) {(1,4),(1,6),(2,4),(2,6),(3,4),(3,6)}
(b) {(4,1),(4,2),(4,3),(6,1),(6,2),(6,3)}
(c) {(4,4),(6,6)}
(d) {(4,1),(6,2),(6,3)}

Sol: (b) {(4,1),(4,2),(4,3),(6,1),(6,2),(6,3)}
Given elements of A × B: (1,4), (2,6), (3,6)
⇒ A = {1, 2, 3}
⇒ B = {4, 6}Now B × A = {(b,a)}
= {(4,1), (4,2), (4,3), (6,1), (6,2), (6,3)}

Que-5: Let n(A) = 5 and n(B) = 7, then the number of relations on A × B is

(a) 2³⁵
(b) 2⁴⁹
(c) 2²⁵
(d) 2⁷⁰

Sol:  (a) 2³⁵
n(A × B) = n(A) × n(B) = 5 × 7 = 35
Total number of relations on a set of 35 elements = 2³⁵
Correct Answer = (a) 2³⁵

Que-6: Let A = {1,2,3,4} and R be the relation on A defined by {(a,b): a,b ∈ A, a×b is an even number}, then find the range of R.

(a) {1,2,3,4}
(b) {2,4}
(c) {2,3,4}
(d) {1,2,4}

Sol: (a) {1,2,3,4}
A product a × b is even if at least one of a or b is even.
Set A = {1,2,3,4}
Even elements in A = {2,4}
So any pair (a,b) is in relation R if
either a is even OR b is even.
Now find all possible second elements (b) that appear:
b = 1 → works when a = 2 or 4
b = 2 → works for all a
b = 3 → works when a = 2 or 4
b = 4 → works for all a
So all elements 1, 2, 3, 4 appear as second elements.
Range = {1,2,3,4}

Que-7: Let A = {x,y,z} and B = {a,b,c,d}, which one of the following is not a relation from A to B?

(a) {(x,a),(x,c)}
(b) {(y,c),(y,d)}
(c) {(z,a),(z,d)}
(d) {(z,b),(y,b),(a,d)}

Sol: (d) {(z,b),(y,b),(a,d)}
A relation from A to B is a subset of A × B.
So, the first element of each ordered pair must belong to set A.
Check each option:
(a) x ∈ A ✔️ valid
(b) y ∈ A ✔️ valid
(c) z ∈ A ✔️ valid
(d) contains (a,d) where a ∉ A ❌ invalid

Que-8: R is a relation on N given by R = {(x,y) | 4x + 3y = 20}. Which of the following belongs to R?

(a) (3,4)
(b) (2,4)
(c) (−4,12)
(d) (5,0)

Sol: (b) (2,4)
Check 4x + 3y = 20 for each option:
(a) (3,4): 4(3) + 3(4) = 12 + 12 = 24 ≠ 20 ❌
(b) (2,4): 4(2) + 3(4) = 8 + 12 = 20 ✔️
(c) (−4,12): 4(−4) + 3(12) = −16 + 36 = 20 ✔️ but −4 ∉ N ❌
(d) (5,0): 4(5) + 3(0) = 20 + 0 = 20 ✔️ but 0 ∉ N ❌

Que-9: Let f : R → R be defined by
f(x) = { 2x,   x > 3
x²,   1 < x ≤ 3
3x,   x ≤ 1 }
then f(−1) + f(2) + f(4) is

(a) 9
(b) 14
(c) 5
(d) 10

Sol: (a) 9
For x = −1 (since x ≤ 1):
f(−1) = 3(−1) = −3
For x = 2 (since 1 < x ≤ 3):
f(2) = 2² = 4
For x = 4 (since x > 3):
f(4) = 2 × 4 = 8
Therefore,
f(−1) + f(2) + f(4) = −3 + 4 + 8 = 9

Que-10: Find whether the function f(x) = x tan³x + x³ cosec x is even or odd or neither.

(a) odd
(b) even
(c) neither
(d) constant

Sol:  (b) even
f(x) = x tan³x + x³ cosec x
Replace x by (−x):
f(−x) = (−x) tan³(−x) + (−x)³ cosec(−x)
Using identities:
tan(−x) = −tan x and cosec(−x) = −cosec x
⇒ f(−x) = (−x)(−tan³x) + (−x³)(−cosec x)
⇒ f(−x) = x tan³x + x³ cosec x
⇒ f(−x) = f(x)
Hence, the function is EVEN.

Que-11: Domain of √(a² − x²) (a > 0) is

(a) (−a, a)
(b) [−a, a]
(c) [0, a]
(d) (−a, 0)

Sol:  (b) [−a, a]
For √(a² − x²) to be real, we need:
a² − x² ≥ 0
⇒ x² ≤ a²
⇒ −a ≤ x ≤ a
Domain = [−a, a]

Que-12: Range of the function f(x) = x² + 2, x ∈ R is

(a) (−2, ∞)   
(b) (2, ∞)   
(c) [2, ∞]   
(d) [0, ∞] 

Sol: (c) [2, ∞]
Since x² ≥ 0 for all real x,
⇒ x² + 2 ≥ 2
Minimum value of f(x) = 2 (when x = 0)
Therefore, Range = [2, ∞)

Que-13: If f : R − {2} → R is a function defined by
f(x) = (x² − 4) / (x − 2), then its range is given by

(a) R   
(b) R − {2}   
(c) R − {4}   
(d) R − {−2, 2} 

Sol: (c) R − {4}
f(x) = (x² − 4)/(x − 2)
Factor numerator:
x² − 4 = (x − 2)(x + 2)
So, for x ≠ 2,
f(x) = (x − 2)(x + 2)/(x − 2) = x + 2
Thus the function becomes:
f(x) = x + 2, where x ≠ 2
Now let y = x + 2 ⇒ x = y − 2
Since x ≠ 2, we get:
y − 2 ≠ 2 ⇒ y ≠ 4
Therefore, the range is all real numbers except 4.

Que-14: If n(A) = 2 and total number of possible relations from set A to set B is 1024, then n(B) is

(a) 20   
(b) 10   
(c) 5   
(d) 512

Sol:  (c) 5
Number of relations from A to B = 2^{n(A) × n(B)}
Given: n(A) = 2 and total relations = 1024
So, 2^{2 × n(B)} = 1024
Since 1024 = 2¹⁰
⇒ 2 × n(B) = 10
⇒ n(B) = 5

Que-15: Let A = {2, 3, 4, 5}, B = {36, 45, 49, 60, 77, 90} and let R be the relation “is factor of” from A to B. Then the range of R is the set

(a) {60}
(b) {36, 45, 49, 60, 90}
(c) {49, 77}
(d) {49, 60, 77}
(e) {36, 45, 49, 60, 77, 90}

Sol: (b) {36, 45, 49, 60, 90}
Check each element of B:
36 → divisible by 2, 3, 4 ✔
45 → divisible by 3, 5 ✔
49 → not divisible by any of A ✖
60 → divisible by 2, 3, 4, 5 ✔
77 → not divisible by any of A ✖
90 → divisible by 2, 3, 5 ✔
Range of R = {36, 45, 60, 90}

Que-16: For which of the following values of x does the function
f(x) = log ( √(25 − x²) / (2 − x) )
have the real values?

(a) −5 < x < 5
(b) −5 < x < 2
(c) x > −2
(d) x < 2

Sol: (b) −5 < x < 2
For log to be real, its argument must be positive.
1. √(25 − x²) must be real ⇒ 25 − x² ≥ 0 ⇒ −5 ≤ x ≤ 5
2. Denominator (2 − x) ≠ 0 ⇒ x ≠ 2
3. The whole expression must be positive:
√(25 − x²) / (2 − x) > 0
Since numerator √(25 − x²) ≥ 0, sign depends on (2 − x)
For expression > 0 ⇒ (2 − x) > 0 ⇒ x < 2
Combine all conditions:
−5 ≤ x < 2

Que-17: If f(x) = logₑ ((1 − x) / (1 + x)), |x| < 1, then
f( 2x / (1 + x²) ) is equal to

(a) 2f(x²)
(b) −2f(x)
(c) (f(x))²
(d) 2f(x)

Sol: (d) 2f(x)
Given: f(x) = ln((1 − x)/(1 + x))
Find: f(2x/(1 + x²))
f(2x/(1 + x²)) = ln[(1 − (2x/(1+x²))) / (1 + (2x/(1+x²)))]
= ln[((1+x² − 2x)/(1+x² + 2x))]
= ln[((1 − x)²/(1 + x)²)]
= ln[(1 − x)/(1 + x)]²
= 2 ln[(1 − x)/(1 + x)]
= 2 f(x)

Que-18: The domain of the function f : R → R defined by
f(x) = √(x² − 7x + 12) is

(a) (−∞, 3] ∪ [4, ∞)
(b) (−∞, 3] ∩ [4, ∞)
(c) (−∞, 3] ∪ [4, ∞)
(d) (3, 4)

Sol:  (a) (−∞, 3] ∪ [4, ∞)
For square root, expression inside must be ≥ 0
x² − 7x + 12 ≥ 0
Factorize :
(x − 3)(x − 4) ≥ 0
This is ≥ 0 when x ≤ 3 or x ≥ 4
Domain = (−∞, 3] ∪ [4, ∞)

Que-19: The domain of the function f(x) = log (3x² − 4x + 7) is

(a) set of real numbers − {3, 2, 7}
(b) set of real numbers
(c) set of positive real numbers
(d) set of non-negative real numbers

Sol:  (b) set of real numbers
For log to be defined:
3x² − 4x + 7 > 0
Check the quadratic:
Discriminant D = b² − 4ac = (−4)² − 4×3×7 = 16 − 84 = −68 < 0
Since D < 0 and leading coefficient (3) is positive,
3x² − 4x + 7 is always positive for all real x
Therefore log is defined for all real x.

Que-20: Which of the following functions is neither even or odd?

(a) f(x) = 5x + sin(4x)
(b) f(x) = 4x³ + 7 tan x
(c) f(x) = 7x⁴ + 8x² − 6x
(d) f(x) = 5x² + cos(6x)

Sol:  (c) f(x) = 7x⁴ + 8x² − 6x
Even function ⇒ f(−x) = f(x)
Odd function ⇒ f(−x) = −f(x)
Short Checking:
(a) 5x (odd) + sin(4x) (odd) ⇒ odd + odd = odd
(b) 4x³ (odd) + tan x (odd) ⇒ odd + odd = odd
(c) 7x⁴ + 8x² (even) − 6x (odd) ⇒ even + odd = neither even nor odd
(d) 5x² (even) + cos(6x) (even) ⇒ even + even = even

Que-21: If S is a set with 10 elements and A = {(x, y) : x, y ∈ S, x ≠ y}, then the number of elements in A is

(a) 100   
(b) 90   
(c) 80   
(d) 150   
(e) 45

Sol: (b) 90
Total ordered pairs (x, y) where x, y ∈ S = 10 × 10 = 100
Pairs where x = y = 10
So, pairs where x ≠ y = 100 − 10 = 90

–: End Relation and Functions Class 11 OP Malhotra Exe-2I ISC Maths Ch-2 Latest editions :–

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