Practice Questions with Answers - Multiple Choice Questions

MCQ Questions for Class 11 Maths Chapter 1 Sets

MCQ Questions for Class 11 Maths Chapter 1 Sets

Students are advised to solve the Sets Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Sets Class 11 with answers will boost your confidence thereby helping you score well in the exam.

Explore numerous MCQ Questions of Sets Class 11 with answers provided with detailed solutions by looking below.

Question 1.
If f(x) = log [(1 + x)/(1 – x), then f(2x )/(1 + x²) is equal to
(a) 2f(x)
(b) {f(x)}²
(c) {f(x)}³
(d) 3f(x)

Answer

Answer: (a) 2f(x)
Given f(x) = Log [(1 + x)/(1-x)] Now, f{(2x )/(1 + x²)} = Log [{(1 + (2x )/(1 + x²))}/{(1 – (2x )/(1 + x²))}] ⇒ f{(2x )/(1 + x²)} = Log [{(1 + x² + 2x )/(1 + x²))}/{(1 + x² – 2x )/(1 + x²))}] ⇒ f{(2x )/(1 + x²)} = Log [(1 + x² + 2x )/{(1 + x² – 2x )] ⇒ f{(2x )/(1 + x²)} = Log [(1 + x)2 /{(1 – x)2] ⇒ f{(2x )/(1 + x²)} = Log [(1 + x)/{(1 – x)]2
⇒ f{(2x )/(1 + x²)} = 2 × Log [(1 + x)/{(1 – x)] ⇒ f{(2x )/(1 + x²)} = 2 f(x)


Question 2.
The smallest set a such that A ∪ {1, 2} = {1, 2, 3, 5, 9} is
(a) {3, 5, 9}
(b) {2, 3, 5}
(c) {1, 2, 5, 9}
(d) None of these

Answer

Answer: (a) {3, 5, 9}
Given, a set A such that A ∪ {1, 2} = {1, 2, 3, 5, 9}
Now, smallest set A = {3, 5, 9}
So, A ∪ {1, 2} = {1, 2, 3, 5, 9}


Question 3.
Let R= {(x, y) : x, y belong to N, 2x + y = 41}. The range is of the relation R is
(a) {(2n – 1) : n belongs to N, 1 ≤ n ≤ 20}
(b) {(2n + 2) : n belongs to N, 1 < n < 20}
(c) {2n : n belongs to N, 1< n< 20}
(d) {(2n + 1) : n belongs to N , 1 ≤ n ≤ 20}

Answer

Answer: (a) {(2n – 1) : n belongs to N, 1 ≤ n ≤ 20}
Given,
2x + y = 41
⇒ y = 41 – 2x
x : 1 2 3 ………………20
y : 39 37 35 ……………..1
So, range is
{(2n – 1) : n belongs to N, 1 ≤ n ≤ 20}


Question 4.
Empty set is a?
(a) Finite Set
(b) Invalid Set
(c) None of the above
(d) Infinite Set

Answer

Answer: (a) Finite Set
In mathematics, and more specifically set theory, the empty set is the unique set having no elements and its size or cardinality (count of elements in a set) is zero.
So, an empty set is a finite set.


Question 5.
Two finite sets have M and N elements. The total number of subsets of the first set is 56 morethan the total number of subsets of the second set. The values of M and N are respectively.
(a) 6, 3
(b) 8, 5
(c) none of these
(d) 4, 1

Answer

Answer: (a) 6, 3
Let A and B be two sets having m and n numbers of elements respectively
Number of subsets of A = 2m
Number of subsets of B = 2n
Now, according to question
2m – 2n = 56
⇒ 2n( 2m-n – 1) = 2³(2³ – 1)
So, n = 3
and m – n = 3
⇒ m – 3 = 3
⇒ m = 3 + 3
⇒ m = 6


Question 6.
If the number of elements in a set S are 5. Then the number of elements of the power set P(S) are?
(a) 5
(b) 6
(c) 16
(d) 32

Answer

Answer: (d) 32
Given, the number of elements in a set S are 5
Then the number of elements of the power set P(S) = 25 = 32


Question 7.
Every set is a ___________ of itself
(a) None of the above
(b) Improper subset
(c) Compliment
(d) Proper subset

Answer

Answer: (b) Improper subset
An improper subset is a subset containing every element of the original set.
A proper subset contains some but not all of the elements of the original set.
Ex: Let a set {1, 2, 3, 4, 5, 6}. Then {1, 2, 4} and {1} are the proper subset while {1, 2, 3, 4, 5} is an improper subset.
So, every set is an improper subset of itself.


Question 8.
If x ≠ 1, and f(x) = x + 1 / x – 1 is a real function, then f(f(f(2))) is
(a) 2
(b) 1
(c) 4
(d) 3

Answer

Answer: (d) 3
Given f(x) = (x + 1)/(x – 1)
Now, f(2) = (2 + 1)/(2 – 1) = 3
Now since f(2) is independent of x
So, f(f(f(2))) = 3


Question 9.
In 3rd Quadrant?
(a) X < 0, Y < 0 (b) X > 0, Y < 0
(c) X < 0, Y > 0
(d) X < 0, Y > 0

Answer

Answer: (a) X < 0, Y < 0
MCQ Questions for Class 11 Maths Chapter 1 Sets with Answers 1
In the 3rd quadrant,
X < 0, Y < 0


Question 10.
IF A ∪ B = A ∪ C and A ∩ B = A ∩ C, THEN
(a) none of these
(b) B = C only when A I C
(c) B = C only when A ? B
(d) B = C

Answer

Answer: (d) B = C
If A ∪ B = A ∪ C and A ∩ B = A ∩ C
Then B = C