Class 11 Maths Chapter 1 Sets Worksheet with Answers, free PDF (CBSE Class 11 and JEE Main; based on NCERT)

A 20-mark worksheet on Sets for about 40 minutes of class or homework: 5 multiple-choice, 3 very short answer and 3 short answer questions, written from the NCERT chapter and checked by a second reviewer. Print the PDF for the class; the answer key carries the marking scheme. Teachers may photocopy it freely.

Question 1

Which of the following is a singleton set?

(a) {x : x ∈ R and x³ = 27}(b) {x : x ∈ Z and x² = 9}(c) {x : x ∈ N and x + 3 = 2}(d) {x : x ∈ N and x divides 9}

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Answer: (a) {x : x ∈ R and x³ = 27}

The only real number whose cube is 27 is 3, so the set is {3}, with one element. The others are {−3, 3}, φ and {1, 3, 9}.

Question 2

The set {x : x ∈ Z and −2 < 2x ≤ 5} in roster form is

(a) {−1, 0, 1, 2}(b) {0, 1, 2, 3}(c) {0, 1, 2}(d) {1, 2}

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Answer: (c) {0, 1, 2}

−2 < 2x ≤ 5 gives −1 < x ≤ 2.5. The integers in this range are 0, 1 and 2. x = −1 fails since 2x = −2 is not greater than −2; x = 3 gives 2x = 6 > 5.

Question 3

Which of the following sets is infinite?

(a) The set of divisors of 1000000(b) The set of points on a line segment 1 cm long(c) The set of districts of India(d) {x : x ∈ N and x < 1000000000}

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Answer: (b) The set of points on a line segment 1 cm long

A line segment, however short, contains infinitely many points. The other three sets are very large but have a definite number of elements, so they are finite.

Question 4

The set {3, 8, 15, 24, 35} in set-builder form is

(a) {x : x = n² − 1, n ∈ N and 1 ≤ n ≤ 5}(b) {x : x = 2n + 1, n ∈ N and 1 ≤ n ≤ 5}(c) {x : x = n² − 1, n ∈ N and 2 ≤ n ≤ 6}(d) {x : x = n² + 2, n ∈ N and 1 ≤ n ≤ 5}

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Answer: (c) {x : x = n² − 1, n ∈ N and 2 ≤ n ≤ 6}

3 = 2² − 1, 8 = 3² − 1, 15 = 4² − 1, 24 = 5² − 1, 35 = 6² − 1, so x = n² − 1 with n from 2 to 6. Option (a) would start at 0 and stop at 24.

Question 5

The number of subsets of the set of prime numbers less than 20 is

(a) 256(b) 255(c) 128(d) 64

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Answer: (a) 256

The primes less than 20 are 2, 3, 5, 7, 11, 13, 17, 19: eight elements. The number of subsets is 2⁸ = 256 (255 is the number of proper subsets).

Question 6 · 2 marks

A set A has 2 more elements than a set B, and A has 48 more subsets than B. Find n(A) and n(B).

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Let n(B) = b, so n(A) = b + 2. Then 2^{b + 2} − 2^b = 4 × 2^b − 2^b = 3 × 2^b = 48, so 2^b = 16 and b = 4. Hence n(B) = 4 and n(A) = 6.

forms 2^{b + 2} − 2^b = 48 (1)
n(B) = 4, n(A) = 6 (1)

Question 7 · 2 marks

Write A = {x : x ∈ R and −3 < x ≤ 5} and B = {x : x ∈ R and 1 ≤ x < 8} as intervals, and write A ∩ B as an interval.

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A = (−3, 5] and B = [1, 8). The real numbers in both satisfy 1 ≤ x ≤ 5, so A ∩ B = [1, 5].

A = (−3, 5] and B = [1, 8) (1)
A ∩ B = [1, 5] (1)

Question 8 · 2 marks

Write in roster form: (i) {x : x ∈ Z and x³ = 4x}, (ii) {x : x ∈ N and x³ = 4x}.

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x³ − 4x = x(x − 2)(x + 2) = 0 gives x = 0, 2, −2. (i) All three are integers: {−2, 0, 2}. (ii) Only 2 is a natural number: {2}.

factorises to get x = 0, 2, −2 (1)
(i) {−2, 0, 2} and (ii) {2} (1)

Question 9 · 3 marks

For subsets A and B of a universal set U, prove that A − B = A ∩ B′.

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Let x ∈ A − B. Then x ∈ A and x ∉ B, so x ∈ A and x ∈ B′, i.e. x ∈ A ∩ B′; hence A − B ⊂ A ∩ B′. Conversely let x ∈ A ∩ B′. Then x ∈ A and x ∈ B′, i.e. x ∈ A and x ∉ B, so x ∈ A − B; hence A ∩ B′ ⊂ A − B. Since each is a subset of the other, A − B = A ∩ B′.

shows A − B ⊂ A ∩ B′ (1)
shows A ∩ B′ ⊂ A − B (1)
concludes equality from the two inclusions (1)

Question 10 · 3 marks

A and B are finite sets with n(A) = 12, n(A ∩ B) = 5 and n(B − A) = 9. Using the fact that A − B, A ∩ B and B − A are mutually disjoint, find n(A − B), n(B) and n(A ∪ B).

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A = (A ∩ B) ∪ (A − B) with the two parts disjoint, so n(A − B) = 12 − 5 = 7. B = (A ∩ B) ∪ (B − A), disjoint, so n(B) = 5 + 9 = 14. A ∪ B is the union of the three disjoint sets, so n(A ∪ B) = 7 + 5 + 9 = 21.

n(A − B) = 7 (1)
n(B) = 14 (1)
n(A ∪ B) = 21 (1)

Question 11 · 3 marks

Using the laws of the algebra of sets, show that A ∩ (A′ ∪ B) = A ∩ B, stating the law used at each step.

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A ∩ (A′ ∪ B) = (A ∩ A′) ∪ (A ∩ B) by the distributive law; = φ ∪ (A ∩ B) since A ∩ A′ = φ (complement law); = A ∩ B since φ is the identity of union.

distributive law applied (1)
A ∩ A′ = φ (1)
φ ∪ (A ∩ B) = A ∩ B (1)

Class 11 Maths Chapter 1: Sets worksheet with answer key