NCERT Solutions for Class 11 Maths Chapter 1 – Sets

ANSWER:

(i) 5 ∈ A

(ii) 8 ∉ A

(iii) 0 ∉ A

(iv) 4 ∈ A

(v) 2 ∈ A

(vi) 10 ∉ A

ANSWER:

(i) A = {x: x is an integer and –3 < x < 7}

The elements of this set are –2, –1, 0, 1, 2, 3, 4, 5, and 6 only.

Therefore, the given set can be written in roster form as

A = {–2, –1, 0, 1, 2, 3, 4, 5, 6}

(ii) B = {x: x is a natural number less than 6}

The elements of this set are 1, 2, 3, 4, and 5 only.

Therefore, the given set can be written in roster form as

B = {1, 2, 3, 4, 5}

(iii) C = {x: x is a two-digit natural number such that the sum of its digits is 8}

The elements of this set are 17, 26, 35, 44, 53, 62, 71, and 80 only.

Therefore, this set can be written in roster form as

C = {17, 26, 35, 44, 53, 62, 71, 80}

(iv) D = {x: x is a prime number which is a divisor of 60}

2	60
2	30
3	15
	5

∴60 = 2 × 2 × 3 × 5

The elements of this set are 2, 3, and 5 only.

Therefore, this set can be written in roster form as D = {2, 3, 5}.

(v) E = The set of all letters in the word TRIGONOMETRY

There are 12 letters in the word TRIGONOMETRY, out of which letters T, R, and O are repeated.

Therefore, this set can be written in roster form as

E = {T, R, I, G, O, N, M, E, Y}

(vi) F = The set of all letters in the word BETTER

There are 6 letters in the word BETTER, out of which letters E and T are repeated.

Therefore, this set can be written in roster form as

F = {B, E, T, R}

ANSWER:

(i) {3, 6, 9, 12} = {x: x = 3n, n∈ N and 1 ≤ n ≤ 4}

(ii) {2, 4, 8, 16, 32}

It can be seen that 2 = 2¹, 4 = 2², 8 = 2³, 16 = 2⁴, and 32 = 2⁵.

∴ {2, 4, 8, 16, 32} = {x: x = 2ⁿ, n∈ N and 1 ≤ n ≤ 5}

(iii) {5, 25, 125, 625}

It can be seen that 5 = 5¹, 25 = 5², 125 = 5³, and 625 = 5⁴.

∴ {5, 25, 125, 625} = {x: x = 5ⁿ, n∈N and 1 ≤ n ≤ 4}

(iv) {2, 4, 6 …}

It is a set of all even natural numbers.

∴ {2, 4, 6 …} = {x: x is an even natural number}

(v) {1, 4, 9 … 100}

It can be seen that 1 = 1², 4 = 2², 9 = 3² …100 = 10².

∴ {1, 4, 9… 100} = {x: x = n², n∈N and 1 ≤ n ≤ 10}

ANSWER:

(i) A = {x: x is an odd natural number} = {1, 3, 5, 7, 9 …}

(ii) B = {x: x is an integer;}

It can be seen that  and 

∴ B 

(iii) C = {x: x is an integer;}

It can be seen that

(–1)² = 1 ≤ 4; (–2)² = 4 ≤ 4; (–3)² = 9 > 4

² = 0 ≤ 4

1² = 1 ≤ 4

2² = 4 ≤ 4

3² = 9 > 4

∴C = {–2, –1, 0, 1, 2}

(iv) D = (x: x is a letter in the word “LOYAL”) = {L, O, Y, A}

(v) E = {x: x is a month of a year not having 31 days}

= {February, April, June, September, November}

(vi) F = {x: x is a consonant in the English alphabet which precedes k}

= {b, c, d, f, g, h, j}

ANSWER:

(i) All the elements of this set are natural numbers as well as the divisors of 6. Therefore, (i) matches with (c).

(ii) It can be seen that 2 and 3 are prime numbers. They are also the divisors of 6.

Therefore, (ii) matches with (a).

(iii) All the elements of this set are letters of the word MATHEMATICS. Therefore, (iii) matches with (d).

(iv) All the elements of this set are odd natural numbers less than 10. Therefore, (iv) matches with (b).

ANSWER:

(i) The set of months of a year is a finite set because it has 12 elements.

(ii) {1, 2, 3 …} is an infinite set as it has infinite number of natural numbers.

(iii) {1, 2, 3 …99, 100} is a finite set because the numbers from 1 to 100 are finite in number.

(iv) The set of positive integers greater than 100 is an infinite set because positive integers greater than 100 are infinite in number.

(v) The set of prime numbers less than 99 is a finite set because prime numbers less than 99 are finite in number.

ANSWER:

(i) The set of lines which are parallel to the x-axis is an infinite set because lines parallel to the x-axis are infinite in number.

(ii) The set of letters in the English alphabet is a finite set because it has 26 elements.

(iii) The set of numbers which are multiple of 5 is an infinite set because multiples of 5 are infinite in number.

(iv) The set of animals living on the earth is a finite set because the number of animals living on the earth is finite (although it is quite a big number).

(v) The set of circles passing through the origin (0, 0) is an infinite set because infinite number of circles can pass through the origin.

ANSWER:

(i) A = {a, b, c, d}; B = {d, c, b, a}

The order in which the elements of a set are listed is not significant.

∴A = B

(ii) A = {4, 8, 12, 16}; B = {8, 4, 16, 18}

It can be seen that 12 ∈ A but 12 ∉ B.

∴A ≠ B

(iii) A = {2, 4, 6, 8, 10}

B = {x: x is a positive even integer and x ≤ 10}

= {2, 4, 6, 8, 10}

∴A = B

(iv) A = {x: x is a multiple of 10}

B = {10, 15, 20, 25, 30 …}

It can be seen that 15 ∈ B but 15 ∉ A.

∴A ≠ B

ANSWER:

(i) A = {2, 3}; B = {x: x is a solution of x² + 5x + 6 = 0}

The equation x² + 5x + 6 = 0 can be solved as:
x(x + 3) + 2(x + 3) = 0
(x + 2)(x + 3) = 0
x = –2 or x = –3
∴A = {2, 3}; B = {–2, –3}
∴A ≠ B

(ii) A = {x: x is a letter in the word FOLLOW} = {F, O, L, W}

B = {y: y is a letter in the word WOLF} = {W, O, L, F}

The order in which the elements of a set are listed is not significant.

∴A = B

ANSWER:

A = {2, 4, 8, 12}; B = {1, 2, 3, 4}; C = {4, 8, 12, 14}

D = {3, 1, 4, 2}; E = {–1, 1}; F = {0, a}

G = {1, –1}; A = {0, 1}

It can be seen that

8 ∈ A, 8 ∉ B, 8 ∉ D, 8 ∉ E, 8 ∉ F, 8 ∉ G, 8 ∉ H

⇒ A ≠ B, A ≠ D, A ≠ E, A ≠ F, A ≠ G, A ≠ H

Also, 2 ∈ A, 2 ∉ C

∴ A ≠ C

3 ∈ B, 3 ∉ C, 3 ∉ E, 3 ∉ F, 3 ∉ G, 3 ∉ H

∴ B ≠ C, B ≠ E, B ≠ F, B ≠ G, B ≠ H

12 ∈ C, 12 ∉ D, 12 ∉ E, 12 ∉ F, 12 ∉ G, 12 ∉ H

∴ C ≠ D, C ≠ E, C ≠ F, C ≠ G, C ≠ H

4 ∈ D, 4 ∉ E, 4 ∉ F, 4 ∉ G, 4 ∉ H

∴ D ≠ E, D ≠ F, D ≠ G, D ≠ H

Similarly, E ≠ F, E ≠ G, E ≠ H

F ≠ G, F ≠ H, G ≠ H

The order in which the elements of a set are listed is not significant.

∴ B = D and E = G

Hence, among the given sets, B = D and E = G.

ANSWER:

(i) False. Each element of {a, b} is also an element of {b, c, a}.

(ii) True. a, e are two vowels of the English alphabet.

(iii) False. 2∈{1, 2, 3}; however, 2∉{1, 3, 5}

(iv) True. Each element of {a} is also an element of {a, b, c}.

(v) False. The elements of {a, b, c} are a, b, c. Therefore, {a}⊂{a, b, c}

(vi) True. {x:x is an even natural number less than 6} = {2, 4}

{x:x is a natural number which divides 36}= {1, 2, 3, 4, 6, 9, 12, 18, 36}

ANSWER:

A = {1, 2, {3, 4}, 5}

(i) The statement {3, 4} ⊂ A is incorrect because 3 ∈ {3, 4}; however, 3∉A.

(ii) The statement {3, 4} ∈A is correct because {3, 4} is an element of A.

(iii) The statement {{3, 4}} ⊂ A is correct because {3, 4} ∈ {{3, 4}} and {3, 4} ∈ A.

(iv) The statement 1∈A is correct because 1 is an element of A.

(v) The statement 1⊂ A is incorrect because an element of a set can never be a subset of itself.

(vi) The statement {1, 2, 5} ⊂ A is correct because each element of {1, 2, 5} is also an element of A.

(vii) The statement {1, 2, 5} ∈ A is incorrect because {1, 2, 5} is not an element of A.

(viii) The statement {1, 2, 3} ⊂ A is incorrect because 3 ∈ {1, 2, 3}; however, 3 ∉ A.

(ix) The statement Φ ∈ A is incorrect because Φ is not an element of A.

(x) The statement Φ ⊂ A is correct because Φ is a subset of every set.

(xi) The statement {Φ} ⊂ A is incorrect because Φ∈ {Φ}; however, Φ ∈ A.

ANSWER:

(i) The subsets of {a} are Φ and {a}.

(ii) The subsets of {a, b} areΦ, {a}, {b}, and {a, b}.

(iii) The subsets of {1, 2, 3} areΦ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, and

{1, 2, 3}

(iv) The only subset of Φ isΦ.

ANSWER:

We know that if A is a set with m elements i.e., n(A) = m, then n[P(A)] = 2^m.

If A = Φ, then n(A) = 0.

∴ n[P(A)] = 2 = 1

Hence, P(A) has one element.

ANSWER:

(i) {x: x ∈ R, –4 < x ≤ 6} = (–4, 6]

(ii) {x: x ∈ R, –12 < x < –10} = (–12, –10)

(iii) {x: x ∈ R, 0 ≤ x < 7} = [0, 7)

(iv) {x: x ∈ R, 3 ≤ x ≤ 4} = [3, 4]

ANSWER:

(i) (–3, 0) = {x: x ∈ R, –3 < x < 0}

(ii) [6, 12] = {x: x ∈ R, 6 ≤ x ≤ 12}

(iii) (6, 12] ={x: x ∈ R, 6 < x ≤ 12}

(iv) [–23, 5) = {x: x ∈ R, –23 ≤ x < 5}

ANSWER:

(i) For the set of right triangles, the universal set can be the set of triangles or the set of polygons.

(ii) For the set of isosceles triangles, the universal set can be the set of triangles or the set of polygons or the set of two-dimensional figures.

ANSWER:

(i) It can be seen that A ⊂ {0, 1, 2, 3, 4, 5, 6}

B ⊂ {0, 1, 2, 3, 4, 5, 6}

However, C ⊄ {0, 1, 2, 3, 4, 5, 6}

Therefore, the set {0, 1, 2, 3, 4, 5, 6} cannot be the universal set for the sets A, B, and C.

(ii) A ⊄ Φ, B ⊄ Φ, C ⊄ Φ

Therefore, Φ cannot be the universal set for the sets A, B, and C.

(iii) A ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

B ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

C ⊂ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

Therefore, the set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} is the universal set for the sets A, B, and C.

(iv) A ⊂ {1, 2, 3, 4, 5, 6, 7, 8}

B ⊂ {1, 2, 3, 4, 5, 6, 7, 8}

However, C ⊄ {1, 2, 3, 4, 5, 6, 7, 8}

Therefore, the set {1, 2, 3, 4, 5, 6, 7, 8} cannot be the universal set for the sets A, B, and C.

ANSWER:

Here, A = {a, b} and B = {a, b, c}

Yes, A ⊂ B.

A∪ B = {a, b, c} = B

ANSWER:

If A and B are two sets such that A ⊂ B, then A ∪ B = B.

ANSWER:

A = {1, 2, 3, 4], B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}

(i) A ∪ B = {1, 2, 3, 4, 5, 6}

(ii) A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}

(iii) B ∪ C = {3, 4, 5, 6, 7, 8}

(iv) B ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}

(v) A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}

(vi) A ∪ B ∪ D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(vii) B ∪ C ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}

ANSWER:

(i) X = {1, 3, 5}, Y = {1, 2, 3}

X ∩ Y = {1, 3}

(ii) A = {a, e, i, o, u}, B = {a, b, c}

A ∩ B = {a}

(iii) A = {x: x is a natural number and multiple of 3} = (3, 6, 9 …}

B = {x: x is a natural number less than 6} = {1, 2, 3, 4, 5}

∴ A ∩ B = {3}

(iv) A = {x: x is a natural number and 1 < x ≤ 6} = {2, 3, 4, 5, 6}

B = {x: x is a natural number and 6 < x < 10} = {7, 8, 9}

A ∩ B = Φ

(v) A = {1, 2, 3}, B = Φ

A ∩ B = Φ

ANSWER:

(i) A ∩ B = {7, 9, 11}

(ii) B ∩ C = {11, 13}

(iii) A ∩ C ∩ D = { A ∩ C} ∩ D = {11} ∩ {15, 17} = Φ

(iv) A ∩ C = {11}

(v) B ∩ D = Φ

(vi) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

= {7, 9, 11} ∪ {11} = {7, 9, 11}

(vii) A ∩ D = Φ

(viii) A ∩ (B ∪ D) = (A ∩ B) ∪ (A ∩ D)

= {7, 9, 11} ∪ Φ = {7, 9, 11}

(ix) (A ∩ B) ∩ (B ∪ C) = {7, 9, 11} ∩ {7, 9, 11, 13, 15} = {7, 9, 11}

(x) (A ∪ D) ∩ (B ∪ C) = {3, 5, 7, 9, 11, 15, 17) ∩ {7, 9, 11, 13, 15}

= {7, 9, 11, 15}

ANSWER:

A = {x: x is a natural number} = {1, 2, 3, 4, 5 …}

B ={x: x is an even natural number} = {2, 4, 6, 8 …}

C = {x: x is an odd natural number} = {1, 3, 5, 7, 9 …}

D = {x: x is a prime number} = {2, 3, 5, 7 …}

(i) A ∩B = {x: x is a even natural number} = B

(ii) A ∩ C = {x: x is an odd natural number} = C

(iii) A ∩ D = {x: x is a prime number} = D

(iv) B ∩ C = Φ

(v) B ∩ D = {2}

(vi) C ∩ D = {x: x is odd prime number}

ANSWER:

(i) {1, 2, 3, 4}

{x: x is a natural number and 4 ≤ x ≤ 6} = {4, 5, 6}

Now, {1, 2, 3, 4} ∩ {4, 5, 6} = {4}

Therefore, this pair of sets is not disjoint.

(ii) {a, e, i, o, u} ∩ (c, d, e, f} = {e}

Therefore, {a, e, i, o, u} and (c, d, e, f} are not disjoint.

(iii) {x: x is an even integer} ∩ {x: x is an odd integer} = Φ

Therefore, this pair of sets is disjoint.

ANSWER:

(i) A – B = {3, 6, 9, 15, 18, 21}

(ii) A – C = {3, 9, 15, 18, 21}

(iii) A – D = {3, 6, 9, 12, 18, 21}

(iv) B – A = {4, 8, 16, 20}

(v) C – A = {2, 4, 8, 10, 14, 16}

(vi) D – A = {5, 10, 20}

(vii) B – C = {20}

(viii) B – D = {4, 8, 12, 16}

(ix) C – B = {2, 6, 10, 14}

(x) D – B = {5, 10, 15}

(xi) C – D = {2, 4, 6, 8, 12, 14, 16}

(xii) D – C = {5, 15, 20}

ANSWER:

(i) X – Y = {a, c}

(ii) Y – X = {f, g}

(iii) X ∩ Y = {b, d}

ANSWER:

R: set of real numbers

Q: set of rational numbers

Therefore, R – Q is a set of irrational numbers.

ANSWER:

(i) False

As 3 ∈ {2, 3, 4, 5}, 3 ∈ {3, 6}

⇒ {2, 3, 4, 5} ∩ {3, 6} = {3}

(ii) False

As a ∈ {a, e, i, o, u}, a ∈ {a, b, c, d}

⇒ {a, e, i, o, u } ∩ {a, b, c, d} = {a}

(iii) True

As {2, 6, 10, 14} ∩ {3, 7, 11, 15} = Φ

(iv) True

As {2, 6, 10} ∩ {3, 7, 11} = Φ

ANSWER:

U = {a, b, c, d, e, f, g, h}

(i) A = {a, b, c}

(ii) B = {d, e, f, g}

(iii) C = {a, c, e, g}

(iv) D = {f, g, h, a}

ANSWER:

U = N: Set of natural numbers

(i) {x: x is an even natural number}´ = {x: x is an odd natural number}

(ii) {x: x is an odd natural number}´ = {x: x is an even natural number}

(iii) {x: x is a positive multiple of 3}´ = {x: x ∈ N and x is not a multiple of 3}

(iv) {x: x is a prime number}´ ={x: x is a positive composite number and x = 1}

(v) {x: x is a natural number divisible by 3 and 5}´ = {x: x is a natural number that is not divisible by 3 or 5}

(vi) {x: x is a perfect square}´ = {x: x ∈ N and x is not a perfect square}

(vii) {x: x is a perfect cube}´ = {x: x ∈ N and x is not a perfect cube}

(viii) {x: x + 5 = 8}´ = {x: x ∈ N and x ≠ 3}

(ix) {x: 2x + 5 = 9}´ = {x: x ∈ N and x ≠ 2}

(x) {x: x ≥ 7}´ = {x: x ∈ N and x < 7}

(xi) {x: x ∈ N and 2x + 1 > 10}´ = {x: x ∈ N and x ≤ 9/2}

ANSWER:

U = {1, 2, 3, 4, 5, 6, 7, 8, 9}

A = {2, 4, 6, 8}, B = {2, 3, 5, 7}

(i)

(A ∪ B)ʹ = {2, 3, 4, 5, 6, 7, 8} = {1, 9}

Aʹ ∩ Bʹ = {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9} = {1, 9}

So, (A ∪ B)ʹ = Aʹ ∩ Bʹ
(ii)

ANSWER:

(i) 

(ii) 

(iii) 

(iv) 

ANSWER:

is the set of all equilateral triangles.

ANSWER:

(i) 

(ii) Φ′ ∩ A = U ∩ A = A

∴ Φ′ ∩ A = A

(iii) A ∩ A′ = Φ

(iv) U′ ∩ A = Φ ∩ A = Φ

∴ U′ ∩ A = Φ

ANSWER:

It is given that:

n(X ∩ Y) = ?

We know that:

ANSWER:

Let H be the set of people who speak Hindi, and

E be the set of people who speak English

∴ n(H ∪ E) = 400, n(H) = 250, n(E) = 200

n(H ∩ E) = ?

We know that:

n(H ∪ E) = n(H) + n(E) – n(H ∩ E)

∴ 400 = 250 + 200 – n(H ∩ E)

⇒ 400 = 450 – n(H ∩ E)

⇒ n(H ∩ E) = 450 – 400

∴ n(H ∩ E) = 50

Thus, 50 people can speak both Hindi and English.

ANSWER:

It is given that:

n(S) = 21, n(T) = 32, n(S ∩ T) = 11

We know that:

n (S ∪ T) = n (S) + n (T) – n (S ∩ T)

∴ n (S ∪ T) = 21 + 32 – 11 = 42

Thus, the set (S ∪ T) has 42 elements.

ANSWER:

It is given that:

n(X) = 40, n(X ∪ Y) = 60, n(X ∩ Y) = 10

We know that:

n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)

∴ 60 = 40 + n(Y) – 10

∴ n(Y) = 60 – (40 – 10) = 30

Thus, the set Y has 30 elements.

ANSWER:

Let C denote the set of people who like coffee, and

T denote the set of people who like tea

n(C ∪ T) = 70, n(C) = 37, n(T) = 52

We know that:

n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

∴ 70 = 37 + 52 – n(C ∩ T)

⇒ 70 = 89 – n(C ∩ T)

⇒ n(C ∩ T) = 89 – 70 = 19

Thus, 19 people like both coffee and tea.

ANSWER:

Let C denote the set of people who like cricket, and

T denote the set of people who like tennis

∴ n(C ∪ T) = 65, n(C) = 40, n(C ∩ T) = 10

We know that:

n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

∴ 65 = 40 + n(T) – 10

⇒ 65 = 30 + n(T)

⇒ n(T) = 65 – 30 = 35

Therefore, 35 people like tennis.

Now,

(T – C) ∪ (T ∩ C) = T

Also,

(T – C) ∩ (T ∩ C) = Φ

∴ n (T) = n (T – C) + n (T ∩ C)

⇒ 35 = n (T – C) + 10

⇒ n (T – C) = 35 – 10 = 25

Thus, 25 people like only tennis.

ANSWER:

Let F be the set of people in the committee who speak French, and

S be the set of people in the committee who speak Spanish

∴ n(F) = 50, n(S) = 20, n(S ∩ F) = 10

We know that:

n(S ∪ F) = n(S) + n(F) – n(S ∩ F)

= 20 + 50 – 10

= 70 – 10 = 60

Thus, 60 people in the committee speak at least one of the two languages.

ANSWER:

(i) False

Let A = {1, 2} and B = {1, {1, 2}, {3}}

Now, 

∴ A ∈ B

However,

(ii) False

Let 

As A ⊂ B

B ∈ C

However,

(iii) True

Let A ⊂ B and B ⊂ C.

Let x ∈ A

∴ A ⊂ C

(iv) False

Let 

Accordingly,and .

However, A ⊂ C

(v) False

Let A = {3, 5, 7} and B = {3, 4, 6}

Now, 5 ∈ A and A ⊄ B

However, 5 ∉ B

(vi) True

Let A ⊂ B and x ∉ B.

To show: x ∉ A

If possible, suppose x ∈ A.

Then, x ∈ B, which is a contradiction as x ∉ B

∴x ∉ A

ANSWER:

Let, A, B and C be the sets such that  and.

To show: B = C

Let x ∈ B

Case I

x ∈ A

Also, x ∈ B

∴ 

∴ x ∈ A and x ∈ C

∴ x ∈ C

∴ B ⊂ C

Similarly, we can show that C ⊂ B.

∴ B = C

ANSWER:

First, we have to show that (i) ⇔ (ii).

Let A ⊂ B

To show: A – B ≠ Φ

If possible, suppose A – B ≠ Φ

This means that there exists x ∈ A, x ≠ B, which is not possible as A ⊂ B.

∴ A – B = Φ

∴ A ⊂ B ⇒ A – B = Φ

Let A – B = Φ

To show: A ⊂ B

Let x ∈ A

Clearly, x ∈ B because if x ∉ B, then A – B ≠ Φ

∴ A – B = Φ ⇒ A ⊂ B

∴ (i) ⇔ (ii)

Let A ⊂ B

To show: 

Clearly, 

Let 

Case I: x ∈ A

∴ 

Case II: x ∈ B

Then, 

Conversely, let 

Let x ∈ A

∴ A ⊂ B

Hence, (i) ⇔ (iii)

Now, we have to show that (i) ⇔ (iv).

Let A ⊂ B

Clearly

Let x ∈ A

We have to show that

As A ⊂ B, x ∈ B

∴ 

∴ 

Hence, A = A ∩ B

Conversely, suppose A ∩ B = A

Let x ∈ A

⇒ 

⇒ x ∈ A and x ∈ B

⇒ x ∈ B

∴ A ⊂ B

Hence, (i) ⇔ (iv).

ANSWER:

Let A ⊂ B

To show: C – B ⊂ C – A

Let x ∈ C – B

⇒ x ∈ C and x ∉ B

⇒ x ∈ C and x ∉ A [A ⊂ B]

⇒ x ∈ C – A

∴ C – B ⊂ C – A

ANSWER:

Let P(A) = P(B)

To show: A = B

Let x ∈ A

A ∈ P(A) = P(B)

∴ x ∈ C, for some C ∈ P(B)

Now, C ⊂ B

∴ x ∈ B

∴ A ⊂ B

Similarly, B ⊂ A

∴ A = B

ANSWER:

False

Let A = {0, 1} and B = {1, 2}

∴ A ∪ B = {0, 1, 2}

P(A) = {Φ, {0}, {1}, {0, 1}}

P(B) = {Φ, {1}, {2}, {1, 2}}

P(A ∪ B) = {Φ, {0}, {1}, {2}, {0, 1}, {1, 2}, {0, 2}, {0, 1, 2}}

P(A) ∪ P(B) = {Φ, {0}, {1}, {0, 1}, {2}, {1, 2}}

∴ P(A) ∪ P(B) ≠ P(A ∪ B)

ANSWER:

To show: A = (A ∩ B) ∪ (A – B)

Let x ∈ A

We have to show that x ∈ (A ∩ B) ∪ (A – B)

Case I

x ∈ A ∩ B

Then, x ∈ (A ∩ B) ⊂ (A ∪ B) ∪ (A – B)

Case II

x ∉ A ∩ B

⇒ x ∉ A or x ∉ B

∴ x ∉ B [x ∉ A]

∴ x ∉ A – B ⊂ (A ∪ B) ∪ (A – B)

∴ A ⊂ (A ∩ B) ∪ (A – B) … (1)

It is clear that

A ∩ B ⊂ A and (A – B) ⊂ A

∴ (A ∩ B) ∪ (A – B) ⊂ A … (2)

From (1) and (2), we obtain

A = (A ∩ B) ∪ (A – B)

To prove: A ∪ (B – A) ⊂ A ∪ B

Let x ∈ A ∪ (B – A)

⇒ x ∈ A or x ∈ (B – A)

⇒ x ∈ A or (x ∈ B and x ∉ A)

⇒ (x ∈ A or x ∈ B) and (x ∈ A or x ∉ A)

⇒ x ∈ (A ∪ B)

∴ A ∪ (B – A) ⊂ (A ∪ B) … (3)

Next, we show that (A ∪ B) ⊂ A ∪ (B – A).

Let y ∈ A ∪ B

⇒ y ∈ A or y ∈ B

⇒ (y ∈ A or y ∈ B) and (y ∈ A or y ∉ A)

⇒ y ∈ A or (y ∈ B and y ∉ A)

⇒ y ∈ A ∪ (B – A)

∴ A ∪ B ⊂ A ∪ (B – A) … (4)

Hence, from (3) and (4), we obtain A ∪ (B – A) = A ∪B.

ANSWER:

(i) To show: A ∪ (A ∩ B) = A

We know that

A ⊂ A

A ∩ B ⊂ A

∴ A ∪ (A ∩ B) ⊂ A … (1)

Also,  A ⊂ A ∪ (A ∩ B) … (2)

∴ From (1) and (2), A ∪ (A ∩ B) = A

(ii) To show: A ∩ (A ∪ B) = A

A ∩ (A ∪ B) = (A ∩ A) ∪ (A ∩ B)

= A ∪ (A ∩ B)

= A {from (1)}

ANSWER:

Let A = {0, 1}, B = {0, 2, 3}, and C = {0, 4, 5}

Accordingly, A ∩ B = {0} and A ∩ C = {0}

Here, A ∩ B = A ∩ C = {0}

However, B ≠ C [2 ∈ B and 2 ∉ C]

ANSWER:

Let A and B be two sets such that A ∩ X = B ∩ X = f and A ∪ X = B ∪ X for some set X.

To show: A = B

It can be seen that

A = A ∩ (A ∪ X) = A ∩ (B ∪ X) [A ∪ X = B ∪ X]

= (A ∩ B) ∪ (A ∩ X) [Distributive law]

= (A ∩ B) ∪ Φ [A ∩ X = Φ]

= A ∩ B … (1)

Now, B = B ∩ (B ∪ X)

= B ∩ (A ∪ X) [A ∪ X = B ∪ X]

= (B ∩ A) ∪ (B ∩ X) [Distributive law]

= (B ∩ A) ∪ Φ [B ∩ X = Φ]

= B ∩ A

= A ∩ B … (2)

Hence, from (1) and (2), we obtain A = B.

ANSWER:

Let A = {0, 1}, B = {1, 2}, and C = {2, 0}.

Accordingly, A ∩ B = {1}, B ∩ C = {2}, and A ∩ C = {0}.

∴ A ∩ B, B ∩ C, and A ∩ C are non-empty.

However, A ∩ B ∩ C = Φ

ANSWER:

Let U be the set of all students who took part in the survey.

Let T be the set of students taking tea.

Let C be the set of students taking coffee.

Accordingly, n(U) = 600, n(T) = 150, n(C) = 225, n(T ∩ C) = 100

To find: Number of student taking neither tea nor coffee i.e., we have to find n(T’ ∩ C’).

n(T’ ∩ C’) = n(T ∪ C)’

= n(U) – n(T ∪ C)

= n(U) – [n(T) + n(C) – n(T ∩ C)]

= 600 – [150 + 225 – 100]

= 600 – 275

= 325

Hence, 325 students were taking neither tea nor coffee.

ANSWER:

Let U be the set of all students in the group.

Let E be the set of all students who know English.

Let H be the set of all students who know Hindi.

∴ H ∪ E = U

Accordingly, n(H) = 100 and n(E) = 50

 = 25

n(U) = n(H) + – n(H ∩ E)

= 100 + 50 – 25

= 125

Hence, there are 125 students in the group.

ANSWER:

Let A be the set of people who read newspaper H.

Let B be the set of people who read newspaper T.

Let C be the set of people who read newspaper I.

Accordingly, n(A) = 25, n(B) = 26, and n(C) = 26

n(A ∩ C) = 9, n(A ∩ B) = 11, and n(B ∩ C) = 8

n(A ∩ B ∩ C) = 3

Let U be the set of people who took part in the survey.

(i) Accordingly,

n(A ∪ B ∪ C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C)

= 25 + 26 + 26 – 11 – 8 – 9 + 3

= 52

Hence, 52 people read at least one of the newspapers.

(ii) Let a be the number of people who read newspapers H and T only.

Let b denote the number of people who read newspapers I and H only.

Let c denote the number of people who read newspapers T and I only.

Let d denote the number of people who read all three newspapers.

Accordingly, d = n(A ∩ B ∩ C) = 3

Now, n(A ∩ B) = a + d

n(B ∩ C) = c + d

n(C ∩ A) = b + d

∴ a + d + c + d + b + d = 11 + 8 + 9 = 28

⇒ a + b + c + d = 28 – 2d = 28 – 6 = 22

Hence, (52 – 22) = 30 people read exactly one newspaper.

ANSWER:

Let A, B, and C be the set of people who like product A, product B, and product C respectively.

Accordingly, n(A) = 21, n(B) = 26, n(C) = 29, n(A ∩ B) = 14, n(C ∩ A) = 12,

n(B ∩ C) = 14, n(A ∩ B ∩ C) = 8

The Venn diagram for the given problem can be drawn as

It can be seen that number of people who like product C only is

{29 – (4 + 8 + 6)} = 11