Do you need some help in preparing for your upcoming Class 11 Maths exams? We’ve compiled a list of MCQ on Sets Class 11 MCQs Questions with Answers to get you started with the subject. You can download NCERT MCQ Questions for Class 11 Maths Chapter 1 Sets with Answers Pdf free download, and learn how smart students prepare well ahead with MCQ Questions for Class 11 Maths with Answers.

## Sets Class 11 MCQs Questions with Answers

Solving the MCQ Questions of Sets Class 11 with answers can help you understand the concepts better.

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 = 2^{m}

Number of subsets of B = 2^{n}

Now, according to question

2^{m} – 2^{n} = 56

⇒ 2^{n}( 2^{m-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) = 2^{5} = 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

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

Question 11.

A set is known by its _______.

(a) Elements

(b) Values

(c) Members

(d) Letters

## Answer

Answer: (a) Elements

A set is known by its elements.

Question 12.

If the set has P elements, B has q elememts then number of elements in A × B is

(a) pq

(b) p + q

(c) p + q + 1

(d) p²

## Answer

Answer: (a) pq

Given the set A has p elements, B has q elements.

then number of elements in A × B = pq

Question 13.

Which from the following set has closure property w.r.t multiplication?

(a) {0, -1}

(b) {1, -1}

(c) {-1}

(d) {-1,-1}

## Answer

Answer: (b) {1, -1}

The set {1, -1} has closure property w.r.t multiplication.

This is because -1 × 1 = -1 which is an element in the given set.

Question 14.

Consider the set A of all determinants of order 3 with entries 0 or 1 only. Let B be the subset of containing all determinants with value 1. Let C be the subset of containing all determinants with value -1 then

(a) B has many elements as C

(b) A = B ∪ C

(c) B has twice as many elements as C.

(d) C is empty

## Answer

Answer: (a) B has many elements as C

The matrix C is not empty because

|1 0 0|

|1 0 1| = -1

|1 1 0|

Let Δ ∈ B

So, Δ = 1

Again let Δ_{1} be the determinant obtained by interchanging any two rows and columns of Δ

So, Δ_{1} = -1 ⇒ n(B) ≥ n(C)

Similarly, we can show that n(C) ≥ n(B)

So, n(B) = n(C)

Question 15.

If A and B are two sets containing respectively M and N distinct elements. How many different relations can be defined for A and B?

(a) 2^{m + n}

(b) 2^{m / n}

(c) 2^{m – n}

(d) 2^{mn}

## Answer

Answer: (d) 2^{mn}

Given A and B are two sets containing respectively m and n distinct elements.

Then number of different relations can be defined for A and B = 2^{mn}

Question 16.

Let R be a relation N define by x + 2y = 8. The domain of R is

(a) {2, 4, 6, 8}

(b) {1, 2, 3, 4}

(c) {2, 4, 8}

(d) {2, 4, 6}

## Answer

Answer: (d) {2, 4, 6}

Given R be a relation N define by x + 2y = 8

⇒ 2y = 8 – x

⇒ y = 8/4 – x/2

⇒ y = 4 – x/2

Now, the pair of x any satisfying the above equation is:

(2, 3), (4, 2),(6, 1)

So R = {(2, 3), (4, 2),(6, 1)}

Now, Dom(R) = {2, 4, 6}

Question 17.

In 2nd quadrant?

(a) X < 0, Y < 0

(b) X < 0, Y > 0

(c) X > 0, Y > 0

(d) X > 0, Y < 0

## Answer

Answer: (b) X < 0, Y > 0

In the second quadrant,

X < 0, Y > 0

Question 18.

A survey shows that 63% of the americans like cheese whereas 76% like apples. If X% of the americans like both cheese and apples, then we have

(a) 39 ≤ x ≤ 63

(b) x ≤ 63

(c) x ≤ 39

(d) none of these.

## Answer

Answer: (a) 39 ≤ x ≤ 63

Given,

Number of americans who like cheese n(C) = 63

Number of americans who like apple n(A) = 76

Total number of person = 100 (since 100%)

Number of americans who like both n(A ∩ B) = x

Now, n(A ∪ B) = n(A) + n(B) – n(A ∩ B)

⇒ 100 = 63 + 76 – x

⇒ 100 = 139 – x

⇒ x = 139 – 100

⇒ x = 39

This is the minimum value of x

Now, let us look for the highest value of x, the intersection or the common portion

between A and C, it would be larger when one set takes in more of the other.

Thus, when the smaller set gets completely absorbed into the larger and in that situation then x = 63

So 39 ≤ x ≤ 63

Question 19.

Which from the following set has closure property w.r.t addition?

(a) {0}

(b) {1}

(c) {1, 1}

(d) {1, -1}

## Answer

Answer: (a) {0}

A set is closed under addition if, when we add any two elements, we always get another element in the set.

(a) Closed under addition. The only possible way to add two numbers in the set is 0 + 0 = 0, which is in the set.

(b) Not closed under addition. For example, 1 + 1 = 2, which is not in the set.

(c) Not closed under addition. For example, 1 + 1 = 2, which is not in the set.

(d) Not closed under addition. For example, 1 + (-1) = 0, which is not in the set.

Question 20.

A’ will contain how many elements from the orginal set A

(a) 0

(b) All elements in A

(c) 1

(d) Infinite

## Answer

Answer: (a) 0

A’ will contain zero many elements from the original set A.

Let A = {1, 2, 3, 4}, then |A| = 4

Now A’ = {}, then |A| = 0

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