Sets

ICSE Class 8 Mathematics · Selina Solutions Concise Mathematics Class 8 ICSE · 100 solved questions

Exercise

1(i)Exercise 6(A)

{x : x ∈ Z and x² - 4x = 0} is equal to :

(a){4}
(b){0, 4}
(c){0, 2}
(d){2, 4}

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x² - 4x = 0

⇒ x(x - 4) = 0

⇒ x = 0 or (x - 4) = 0

⇒ x = 0 or x = 4

{x : x ∈ Z and x ² - 4x = 0} = {0, 4}

Hence, option 2 is the correct option.

1(ii)Exercise 6(A)

{x : x ∈ Z and x² = 9} is equal to :

(a){3}
(b){0, 3}
(c){0, -3}
(d){-3, 3}

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x² = 9

⇒ x = $\sqrt{9}$

⇒ x = 3 or -3

{x : x ∈ Z and x² = 9} = {-3, 3}

Hence, option 4 is the correct option.

1(iii)Exercise 6(A)

If set A = {x : x ∈ N, x = n - 3, n ∈ N and n < 3}, then set A is :

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

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x = n - 3

When n = 0

x = 0 - 3 = -3

When n = 1

x = 1 - 3 = -2

When n = 2

x = 2 - 3 = -1

As -3, -2, -1 are not natural numbers,

∴ set A = {x : x ∈ N, x = n - 3, n ∈ N and n < 3} = { }

Hence, option 2 is the correct option.

1(iv)Exercise 6(A)

Set {x : x = n² - 1, x ∈ Z (integers) and 2 < n ≤ 5} is equal to :

(a){8, 15, 24}
(b){3, 8, 15}
(c){3, 8, 15, 24}
(d)none of these

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x = n² - 1

When n = 3

⇒ x = 3² - 1

⇒ x = 9 - 1 = 8

When n = 4

⇒ x = 4² - 1

⇒ x = 16 - 1 = 15

When n = 5

x = 5² - 1

⇒ x = 25 - 1 = 24

{x : x = n² - 1, x ∈ Z (integers) and 2 ≤ n ≤ 5} = {8, 15, 24}

Hence, option 1 is the correct option.

1(v)Exercise 6(A)

Set-builder form of set A = $\Big\lbrace \frac{1}{2}, \frac{2}{3}, \frac{3}{4} \Big\rbrace$ is :

(a)$\Big\lbrace x = \frac{n}{n + 1}, n ∈ N \text{ and } n \gt 1 \Big\rbrace$
(b)$\Big\lbrace x = \frac{n - 1}{n - 2}, n ∈ N \text{ and } n \gt 2 \Big\rbrace$
(c)$\Big\lbrace x = \frac{n}{n + 1}, n ∈ N \text{ and } 1 \lt n \lt 2 \Big\rbrace$
(d)$\Big\lbrace x = \frac{n}{n + 1}, n ∈ N \text{ and } 1 \le n \lt 4 \Big\rbrace$

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set A = $\Big\lbrace \frac{1}{2}, \frac{2}{3}, \frac{3}{4} \Big\rbrace$

= $\Big\lbrace \frac{1}{1 + 1}, \frac{2}{2 + 1}, \frac{3}{3 + 1} \Big\rbrace$

= $\Big\lbrace x = \frac{n}{n + 1}, n ∈ N \text{ and } 1 ≤ n \lt 4 \Big\rbrace$

Hence, option 4 is the correct option.

2(i)Exercise 6(A)

Write the following set in roster (Tabular) form :

A₁ = {x : 2x + 3 = 11}

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A₁ = {x : 2x + 3 = 11}

⇒ 2x = 11 - 3

⇒ 2x = 8

⇒ x = $\frac{8}{2}$

⇒ x = 4

A₁ = {4}

2(ii)Exercise 6(A)

Write the following set in roster (Tabular) form :

A₂ = {x : x² - 4x - 5 = 0 }

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A₂ = {x : x² - 4x - 5 = 0 }

⇒ x² - 5x + x - 5 = 0

⇒ x (x - 5) + 1(x - 5) = 0

⇒ (x - 5) (x + 1) = 0

⇒ (x - 5) = 0 or (x + 1) = 0

⇒ x = 5 or x = -1

A₂ = {5, -1}

2(iii)Exercise 6(A)

Write the following set in roster (Tabular) form :

A₃ = {x : x ∈ Z, -3 ≤ x < 4}

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A₃ = {x : x ∈ Z, -3 ≤ x < 4}

= {x : x ∈ Z, -3 ≤ x < 4}

Numbers between -3 and 4 are -3, -2, -1, 0, 1, 2, 3

A₃ = {-3, -2, -1, 0, 1, 2, 3}

2(iv)Exercise 6(A)

Write the following set in roster (Tabular) form :

A₄ = {x : x is a two digit number and sum of the digits of x is 7 }

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A₄ = {x : x is a two digit number and sum of the digits of x is 7}

16 ⇒ 1 + 6 = 7

25 ⇒ 2 + 5 = 7

34 ⇒ 3 + 4 = 7

43 ⇒ 4 + 3 = 7

52 ⇒ 5 + 2 = 7

61 ⇒ 6 + 1 = 7

70 ⇒ 7 + 0 = 7

A₄ = {16, 25, 34, 43, 52, 61, 70}

2(v)Exercise 6(A)

Write the following set in roster (Tabular) form :

A₅ = {x : x = 4n, n ∈ W and n < 4}

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A₅ = {x : x = 4n, n ∈ W and n < 4}

When n = 0

⇒ x = 4 x 0 = 0

When n = 1

⇒ x = 4 x 1 = 4

When n = 2

⇒ x = 4 x 2 = 8

When n = 3

⇒ x = 4 x 3 = 12

A₅ = {0, 4, 8, 12}

2(vi)Exercise 6(A)

Write the following set in roster (Tabular) form :

A₆ = $\Big\lbrace x : x = \frac{n}{n+2}; n ∈ N \text{ and } n \gt 5 \Big\rbrace$

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A₆ = $\Big\lbrace x : x = \frac{n}{n+2}; n ∈ N \text{ and } n \gt 5 \Big\rbrace$

When n = 6

⇒ $\frac{6}{6+2} = \frac{6}{8}$

When n = 7

⇒ $\frac{7}{7+2} = \frac{7}{9}$

When n = 8

⇒ $\frac{8}{8+2} = \frac{8}{10}$

When n = 9

⇒ $\frac{9}{9+2} = \frac{9}{11}$

A₆ = $\Big\lbrace x : x = \frac {6}{8}, \frac {7}{9}, \frac {8}{10}, \frac {9}{11},............... \Big\rbrace$

A₆ = $\Big\lbrace\frac{3}{4}, \frac {7}{9}, \frac {4}{5}, \frac {9}{11},............... \Big\rbrace$

3(i)Exercise 6(A)

Write the following set in set-builder (Rule Method) form :

B₁ = {6, 9, 12, 15, ............... }

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B₁ = {6, 9, 12, 15, ............... }

B₁ = {3 x 1 + 3, 3 x 2 + 3, 3 x 3 + 3, 3 x 4 + 3,...............}

B₁ = {x : x = 3n + 3, n ∈ N}

3(ii)Exercise 6(A)

Write the following set in set-builder (Rule Method) form :

B₂ = {11, 13, 17, 19}

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B₂ = {11, 13, 17, 19}

B₂ = {x : x is a prime number 10 < x < 20}

3(iii)Exercise 6(A)

Write the following set in set-builder (Rule Method) form :

B₃ = $\Big\lbrace \frac{1}{3}, \frac{3}{5}, \frac{5}{7}, \frac{7}{9}, \frac{9}{11}, ............... \Big\rbrace$

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B₃ = $\Big\lbrace \frac{1}{3}, \frac{3}{5}, \frac{5}{7}, \frac{7}{9}, \frac{9}{11}, ............... \Big\rbrace$

B₃ = $\Big\lbrace \frac{1}{1+2}, \frac{3}{3+2}, \frac{5}{5+2}, \frac{7}{7+9}, \frac{9}{9+2},............... \Big\rbrace$

B₃ = $\Big\lbrace \frac{n}{n+2}; \text{where n is an odd natural number}\Big\rbrace$

3(iv)Exercise 6(A)

Write the following set in set-builder (Rule Method) form :

B₄ = {8, 27, 64, 125, 216}

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B₄ = {8, 27, 64, 125, 216}

B₄ = {2³, 3³, 4³, 5³, 6³}

B₄ = {x : x = n³; n ∈ N and 2 ≤ n ≤ 6}

3(v)Exercise 6(A)

Write the following set in set-builder (Rule Method) form :

B₅ = {-5, -4, -3, -2, -1}

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B₅ = {-5, -4, -3, -2, -1}

B₅ = {n; n ∈ Z and -5 ≤ n ≤ -1}

3(vi)Exercise 6(A)

Write the following set in set-builder (Rule Method) form :

B₆ = {............... , -6, -3, 0, 3, 6, ...............}

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B₆ = {............... , -6, -3, 0, 3, 6, ...............}

B₆ = {..............., 3 x (-2) , 3 x (-1), 3 x 0, 3 x 1, 3 x 2, ............... }

B₅ = {x : x = 3n; n ∈ Z}

4(i)Exercise 6(A)

Is {1, 2, 4, 16, 64 } = { x : x is a factor of 32 } ? Give reason.

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No, as 64 is not a factor of 32. So this two given sets are unequal.

Hence, {1, 2, 4, 16, 64 } ≠ { x : x is a factor of 32 }

4(ii)Exercise 6(A)

Is { x : x is a factor of 27 } ≠ { 3, 9, 27, 54 } ? Give reason.

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Yes, as 54 is not a factor of 27. So the two given sets are unequal.

Hence, { x : x is a factor of 27 } ≠ { 3, 9, 27, 54 }

4(iii)Exercise 6(A)

Write the set of even factors of 124.

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Factors of 124 = 1, 2, 4, 31, 62, 124

Even factors of 124 = 2, 4, 62, 124

{2, 4, 62, 124}

4(iv)Exercise 6(A)

Write the set of odd factors of 72.

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Factors of 72 = 1, 2, 3, 4, 6, 8, 9, 12, 18, 36, 72

Odd factors of 72 = 1, 3, 9

{1, 3, 9}

4(v)Exercise 6(A)

Write the set of prime factors of 3234.

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Factors of 3234 = 1, 2, 3, 6, 7, 11, 14, 21, 22, 33, 42, 49, 66, 77, 98, 147, 154, 231, 294, 343, 462, 686, 1029, 1372, 2058, 3234

Prime factors of 72 = 2, 3, 7, 11

{ 2, 3, 7, 11}

4(vi)Exercise 6(A)

Is { x : x² - 7x + 12 = 0 } = { 3, 4 } ?

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x² - 7x + 12 = 0

⇒ x² - 3x - 4x + 12 = 0

⇒ (x² - 3x) - (4x - 12) = 0

⇒ x(x - 3) - 4(x - 3) = 0

⇒ (x - 3)(x - 4) = 0

⇒ (x - 3) = 0 or (x - 4) = 0

⇒ x = 3 or x = 4

Yes, { x : x² - 7x + 12 = 0 } = { 3, 4 }.

4(vii)Exercise 6(A)

Is { x : x² - 5x - 6 = 0 } = {2, 3} ?

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x² - 5x - 6 = 0

⇒ x² - 6x + x - 6 = 0

⇒ (x² - 6x) + (x - 6) = 0

⇒ x(x - 6) + 1(x - 6) = 0

⇒ (x - 6)(x + 1) = 0

⇒ (x - 6) = 0 or (x + 1) = 0

⇒ x = 6 or x = -1

Hence, { x : x² - 5x - 6 = 0 } ≠ { 2, 3 }.

5(i)Exercise 6(A)

Write the following set in Roster form :

The set of letters in the word 'MEERUT'.

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All letters will be written in the set and there should be no repetition. So all letters are - m, e, r, u and t.

Set = {m, e, r, u, t}

5(ii)Exercise 6(A)

Write the following set in Roster form :

The set of letters in the word 'UNIVERSAL'.

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All letters will be written in the set and there should be no repetition. So all letters are - u, n, i, v, e, r, s, a and l.

Set = {u, n, i, v, e, r, s, a, l}

5(iii)Exercise 6(A)

Write the following set in Roster form :

A = {x : x = y + 3, y ∈ N and y > 3}.

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A = {x : x = y + 3, y ∈ N and y > 3}

A = {x : x = 4 + 3, 5 + 3, 6 + 3, ...............}

A = {x : x = 7, 8, 9, ...............}

A = {7, 8, 9, ...............}

5(iv)Exercise 6(A)

Write the following set in Roster form :

B = { p : p ∈ W and p² < 20 }.

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B = { p : p ∈ W and p² < 20 }

B = { p : p² = 0, 1, 4, 9, 16}.

B = { p : p = 0, 1, 2, 3, 4}.

Hence, B = {0, 1, 2, 3, 4}.

5(v)Exercise 6(A)

Write the following set in Roster form :

C = { x : x is a composite number and 5 ≤ x ≤ 21 }

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C = { x : x is a composite number and 5 ≤ x ≤ 21 }

C = { x : x = 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21}

Hence, C = {6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21}.

6(i)Exercise 6(A)

List the elements of the following set :

{ x : x² - 2x - 3 = 0 }

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{ x : x² - 2x - 3 = 0 }

x² - 2x - 3 = 0

x² - 3x + x - 3 = 0

(x² - 3x) + (x - 3) = 0

x (x - 3) + 1 (x - 3) = 0

(x - 3)(x + 1) = 0

x - 3 = 0 or x + 1 = 0

x = 3 or x = -1

Hence, the elements are 3 and -1

6(ii)Exercise 6(A)

List the elements of the following set :

{ x : x = 2y + 5 ; y ∈ N and 2 ≤ y < 6 }

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{ x : x = 2y + 5 ; y ∈ N and 2 ≤ y < 6 }

When y = 2

x = 2 x 2 + 5 = 4 + 5 = 9

When y = 3

x = 2 x 3 + 5 = 6 + 5 = 11

When y = 4

x = 2 x 4 + 5 = 8 + 5 = 13

When y = 5

x = 2 x 5 + 5 = 10 + 5 = 15

Hence, the elements are = 9, 11, 13 and 15.

6(iii)Exercise 6(A)

List the elements of the following set :

{x : x is a factor of 24}

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{x : x is a factor of 24}

The factors of 24 are 1, 2, 3, 4, 6, 8, 12, 24

Hence, the elements are = 1, 2, 3, 4, 6, 8, 12 and 24.

6(iv)Exercise 6(A)

List the elements of the following set :

{x : x ∈ Z and x² ≤ 4}

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{x : x ∈ Z and x² ≤ 4}

When x² = 4

x = $\sqrt{4}$ = 2 or -2

When x² = 1

x = $\sqrt{1}$ = 1 or -1

When x² = 0

x = $\sqrt{0}$ = 0

Hence, the elements are = -2, -1, 0, 1 and 2.

6(v)Exercise 6(A)

List the elements of the following set :

{x : 3x - 2 ≤ 10 and x ∈ N}

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{x : 3x - 2 ≤ 10 and x ∈ N}

3x - 2 ≤ 10

3x ≤ 10 + 2

3x ≤ 12

x ≤ $\frac{12}{3}$

x ≤ 4

Hence, the elements are = 1, 2, 3, and 4.

6(vi)Exercise 6(A)

List the elements of the following set :

{x : 4 - 2x > -6, x ∈ Z}

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{x : 4 - 2x > -6, x ∈ Z}

4 - 2x > -6

⇒ -4 + 4 - 2x > - 6 - 4

⇒ -2x > - 10

⇒ 2x < 10

⇒ x < $\frac{10}{2}$

⇒ x < 5

Hence, the elements are = 4, 3, 2, 1, 0 , -1, -2 ,.......

7Exercise 6(A)

State which of the following sets are finite and which are infinite :

(i) A = {x : x ∈ Z and x < 10}

(ii) B = {x : x ∈ W and 5x - 3 ≤ 20}

(iii) P = {y : y = 3x - 2, x ∈ N and x > 5}

(iv) M = $\Big\lbrace x:x = \frac{3}{n}; n ∈ W \text{ and } 6 \lt n \le 15 \Big\rbrace$

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(i) Infinite.

Reason

The set contains all negative numbers , 0 and natural numbers till 10:

A ={........., -3, -2, -1, 0 , 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

(ii) Finite

Reason

5x - 3 ≤ 20

⇒ 5x ≤ 20 + 3

⇒ 5x ≤ 23

⇒ x ≤ $\frac{23}{5}$

⇒ x ≤ 4.6

B = {0, 1, 2, 3, 4}

(iii) Infinite

Reason

y = 3x - 2

When x = 6

y = 3 x 6 - 2 = 18 - 2 = 16

When x = 7

y = 3 x 7 - 2 = 21 - 2 = 19

When x = 8

y = 3 x 8 - 2 = 24 - 2 = 22

When x = 9

y = 3 x 9 - 2 = 27 - 2 = 25

P = {16, 19, 22, 25,........}

(iv) Finite

Reason

When n = 7

x = $\frac{3}{7}$

When n = 8

x = $\frac{3}{8}$

When n = 9

x = $\frac{3}{9}$

When n = 10

x = $\frac{3}{10}$

When n = 11

x = $\frac{3}{11}$

When n = 12

x = $\frac{3}{12}$

When n = 13

x = $\frac{3}{13}$

When n = 14

x = $\frac{3}{14}$

When n = 15

x = $\frac{3}{15}$

M = $\Big\lbrace \frac{3}{7}, \frac{3}{8}, \frac{3}{9}, \frac{3}{10}, \frac{3}{11}, \frac{3}{12}, \frac{3}{13}, \frac{3}{14}, \frac{3}{15} \Big\rbrace$

Reducing the fractions:

M = $\Big\lbrace \frac{3}{7}, \frac{3}{8}, \frac{1}{3}, \frac{3}{10}, \frac{3}{11}, \frac{1}{4}, \frac{3}{13}, \frac{3}{14}, \frac{1}{5} \Big\rbrace$

8Exercise 6(A)

Find which of the following sets are singleton sets :

(i) The set of points of intersection of two non-parallel straight lines on the same plane.

(ii) A = {x : 7x - 3 = 11}

(iii) B = {y : 2y + 1 < 3 and y ∈ W}

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(i) There will be always one and only point where two non-parallel straight lines meet on the same plane.

Hence, this is a singleton set.

(ii) 7x - 3 = 11

7x = 11 + 3

7x = 14

x = $\frac{14}{7}$

x = 2

A = {2}. Hence, A is a singleton set.

(iii) 2y + 1 < 3

2y < 3 - 1

2y < 2

y < $\frac{2}{2}$

y < 1

B = {0}. Hence, B is a singleton set.

9Exercise 6(A)

Find which of the following sets are empty :

(i) The set of points of intersection of two parallel lines.

(ii) A = {x : x ∈ N and 5 < x ≤ 6}.

(iii) B {x : x² + 4 = 0 and x ∈ N}.

(iv) C = {even numbers between 6 and 10}.

(v) D = {prime numbers between 7 and 11}.

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(i) Parallel lines don't intersect, hence their points of intersection is an empty set.

(ii) A = {6}

A ≠ {}

(iii) x² + 4 = 0

x² = -4

x = $\sqrt{- 4}$

B = {$\sqrt{- 4}$}

As square root of negative number is undefined,

Hence, B = {}

(iv) C = {6, 8, 10}

C ≠ {}

(v) No prime number between 7 and 11.

D = {}

10(i)Exercise 6(A)

Are the sets A = {4, 5, 6} and
B = {x : x² - 5x - 6 = 0} disjoint ?

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A = {4, 5, 6}

B = {x : x² - 5x - 6 = 0}

⇒ B = {x : x² - 6x + x - 6 = 0}

⇒ B = {x : x(x - 6) + 1(x - 6) = 0}

⇒ B = {x : (x - 6)(x + 1) = 0}

⇒ B = {x : (x - 6) = 0 or (x + 1) = 0}

⇒ B = {x : x = 6 or x = -1}

⇒ B = {6 , -1}

As 6 is a common element in A and B,

Hence, A and B are not disjoints set.

10(ii)Exercise 6(A)

Are the sets A = {b, c, d, e} and
B = {x : x is a letter in the word 'MASTER'} joint ?

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A = {b, c, d, e}

B = {x : x is a letter in the word 'MASTER'}

B = {m, a, s, t, e, r}

As e is a common element in A and B,

Hence, A and B are joint sets.

11Exercise 6(A)

State whether the following pairs of sets are equivalent or not :

(i) A = {x : x ∈ N and 11 ≥ 2x - 1} and
B = {y : y ∈ W and 3 ≤ y ≤ 9}.

(ii) Set of integers and set of natural numbers.

(iii) Set of whole numbers and set of multiples of 3.

(iv) P = {5, 6, 7, 8} and M = {x : x ∈ W and x ≤ 4}.

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(i) Not equivalent

Reason

A = {x : x ∈ N and 11 ≥ 2x - 1}

A = {11 ≥ 2x - 1}

A = {11 + 1 ≥ 2x}

A = {12 ≥ 2x}

A = {x ≤ $\frac{12}{2}$}

A = {x ≤ 6}

A = {1, 2, 3, 4, 5, 6}

B = {y : y ∈ W and 3 ≤ y ≤ 9}

B = {3, 4, 5, 6, 7, 8, 9}

There are 6 elements in set A and 7 in set B.

(ii) Not equivalent

Reason

Set of integers contain all negative numbers , 0 and positive numbers.

A = {..............., -2, -1, 0, 1, 2,...............}

Set of natural numbers contain all positive numbers.

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

(iii) Not equivalent

Reason

Set of whole numbers contains all natural numbers and 0.

X = {0, 1, 2, 3, 4,...............}

Set of multiples of 3 contains all the numbers which are multiple of 3.

Y = {3, 6, 9, 12, 15, 18,...............}

Hence, they contain different number of questions.

(iv) Not equivalent

Reason

P = {5, 6, 7, 8}

M = {x : x ∈ W and x ≤ 4}

M = {0, 1, 2, 3, 4}

As P contains 4 elements whereas M contains 5 elements.

12Exercise 6(A)

State whether the following pairs of sets are equal or not :

(i) A = {2, 4, 6, 8} and B = {2n : n ∈ N and n < 5}

(ii) M = {x : x ∈ W and x + 3 < 8} and N = {y : y = 2n - 1, n ∈ N and n < 5}

(iii) E = {x : x² + 8x - 9 = 0} and F = {1, -9}

(iv) A = {x : x ∈ N, x < 3} and
B= {y : y² - 3y + 2 = 0}

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(i) Equal

Reason

A = {2, 4, 6, 8}

B = {2n : n ∈ N and n < 5}

When n = 1

2n = 2 x 1 = 2

When n = 2

2n = 2 x 2 = 4

When n = 3

2n = 2 x 3 = 6

When n = 4

2n = 2 x 4 = 8

B = {2, 4, 6, 8}

Hence, A = B

(ii) Not equal

Reason

M = {x : x ∈ W and x + 3 < 8}

x + 3 < 8

⇒ x < 8 - 3

⇒ x < 5

M = {0, 1, 2, 3, 4}

N = {y : y = 2n - 1, n ∈ N and n < 5}

When n = 1

y = 2 x 1 - 1 = 2 - 1 = 1

When n = 2

y = 2 x 2 - 1 = 4 - 1 = 3

When n = 3

y = 2 x 3 - 1 = 6 - 1 = 5

When n = 4

y = 2 x 4 - 1 = 8 - 1 = 7

N = {1, 3, 5, 7}

Hence, M ≠ N

(iii) Equal

Reason

E = {x : x² + 8x - 9 = 0}

x² + 8x - 9 = 0

⇒ x² + 9x - x - 9 = 0

⇒ x(x + 9) - 1(x + 9) = 0

⇒ (x + 9)(x - 1) = 0

⇒ (x + 9) = 0 or (x - 1) = 0

⇒ x = -9 or 1

E = {1, -9}

F = {1, -9}

Hence, E = F

(iv) Equal

Reason

A = {x : x ∈ N, x < 3}

A = {1, 2}

B = {y : y² - 3y + 2 = 0}

y² - 3y + 2 = 0

⇒ y² - 2y - y + 2 = 0

⇒ y(y - 2) - 1(y - 2) = 0

⇒ (y - 2)(y - 1) = 0

⇒ (y - 2) = 0 or (y - 1) = 0

⇒ y = 2 or y = 1

⇒ y = {1, 2}

Hence, A = B

13Exercise 6(A)

State whether each of the following sets is a finite set or an infinite set :

(i) The set of multiples of 8.

(ii) The set of integers less than 10.

(iii) The set of whole numbers less than 12.

(iv) {x : x = 3n - 2, n ∈ W, n ≤ 8}

(v) {x : x = 3n - 2, n ∈ Z, n ≤ 8}

(vi) $\Big\lbrace x : x = \frac{n - 2}{n + 1}, n ∈ W \Big\rbrace$

Show SolutionHide Solution↓

(i) Infinite

Reason

There are infinite multiple of 8.

A = {8, 16, 24, 32,...............}

(ii) Infinite

Reason

Integers less than 10 includes all negative numbers, 0 and positive number till 9.

B = {..............., -4, -3, -2, -1, 0 , 1, 2, 3, 4, 5, 6, 7, 8, 9}

(iii) Finite

Reason

Whole numbers less than 12 include all the numbers between 0 to 11

C = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

(iv) Finite

Reason

x = 3n - 2

When n = 0

x = 3 x 0 - 2 = 0 - 2 = -2

When n = 1

x = 3 x 1 - 2 = 3 - 2 = 1

When n = 2

x = 3 x 2 - 2 = 6 - 2 = 4

When n = 3

x = 3 x 3 - 2 = 9 - 2 = 7

When n = 4

x = 3 x 4 - 2 = 12 - 2 = 10

When n = 5

x = 3 x 5 - 2 = 15 - 2 = 13

When n = 6

x = 3 x 6 - 2 = 18 - 2 = 16

When n = 7

x = 3 x 7 - 2 = 21 - 2 = 19

When n = 8

x = 3 x 8 - 2 = 24 - 2 = 22

D = {-2, 1, 4, 7, 10, 13, 16, 19, 22}

(v) Infinite

Reason

Integers less than equal to 8 include all negative numbers, 0 and positive numbers till 8.

D = {..............., -8, -5, -2, 1, 4, 7, 10}

(vi) Infinite

Reason

$\Big\lbrace x : x = \frac{n - 2}{n + 1}, n ∈ W\Big\rbrace$

When n = 0

x = $\frac{0 - 2}{0 + 1} = \frac{-2}{1} = -1$

When n = 1

x = $\frac{1 - 2}{1 + 1} = \frac{-1}{2}$

When n = 2

x = $\frac{2 - 2}{2 + 1} = \frac{0}{3} = 0$

When n = 3

x = $\frac{3 - 2}{3 + 1} = \frac{1}{4}$

Similarly n = 4, 5, 6, ................ and hence we get the value of x for every value of n.

14Exercise 6(A)

Answer whether the following statements are true or false. Give reasons.

(i) The set of even natural numbers less than 21 and the set of odd natural numbers less than 21 are equivalent sets.

(ii) If E = {factors of 16} and F = {factors of 20}, then E = F.

(iii) The set A = {integers less than 20} is a finite set.

(iv) If A = {x : x is an even prime number}, then set A is empty.

(v) The set of odd prime numbers is the empty set.

(vi) The set of squares of integers and the set of whole numbers are equal sets.

Show SolutionHide Solution↓

(i) True

Reason

The set of even natural numbers less than 21

A = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}

The set of odd natural numbers less than 21

B = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}

A and B both contains 10 elements. Hence they are equivalent sets.

(ii) False

Reason

E = {factors of 16}

Factors of 16 = 1, 2, 4, 8, 16

E = {1, 2, 4, 8, 16}

F = {factors of 20}

Factors of 20 = 1, 2, 4, 5, 10, 20

F = {1, 2, 4, 5, 10, 20}

Hence, E ≠ F

(iii) False

Reason

A = {integers less than 20}

A contains all negative numbers, 0, positive numbers less than 20.

A = {..............., -5, -4, -3, -2, -1, 0 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19}

Hence, A is infinite set.

(iv) False

Reason

A = {x : x is an even prime number}

A = {2}

Hence, A is not an empty set.

(v) False

Reason

Odd prime numbers = 3, 5, 7, 11, 17, ................

A = {3, 5, 7, 11, 17, ................}

Hence, it is not an empty set.

(vi) False

Reason

The set of squares of integers

M = {1², 2², 3², 4²,................}

M = {1, 4, 9, 16,................}

The set of whole numbers:

N = {0, 1, 2, 3, 4, ................}

M ≠ N

1(i)Exercise 6(B)

A set P has 3 elements. The number of proper subsets of set B is :

(a)3 x 2 = 6
(b)8
(c)9
(d)7

Show SolutionHide Solution↓

If a set has n elements, the number of proper subsets = 2ⁿ - 1

n = 3

Number of proper subsets = 2³ - 1

= 8 - 1

= 7

Hence, option 4 is the correct option.

1(ii)Exercise 6(B)

For sets A and B, where A = {2, 4, 6} and B = {1, 3, 5, 7}, A ∩ B is :

(a)ϕ
(b){0}
(c){1, 2, 3, 4, 5, 6, 7}
(d)none of these

Show SolutionHide Solution↓

A = {2, 4, 6} and B = {1, 3, 5, 7}

No element is common between A and B. Hence,

A ∩ B = {2, 4, 6} ∩ {1, 3, 5, 7}

A ∩ B = {} = ϕ

Hence, option 1 is the correct option.

1(iii)Exercise 6(B)

If set A = {4, 6, 8} and set B = {0}, then A ∪ B is :

(a){4, 6, 8}
(b){4, 6, 8, 0}
(c){ }
(d)none of these

Show SolutionHide Solution↓

set A = {4, 6, 8} and set B = {0}

Elements in both set,

A ∪ B = {4, 6, 8} ∪ {0}

A ∪ B = {4, 6, 8, 0}

Hence, option 2 is the correct option.

1(iv)Exercise 6(B)

If set A = students in class 8 of a particular school and set B = students of this school, then :

(a)A = B
(b)n(A) = n(B)
(c)A ⊂ B
(d)B ⊆ A

Show SolutionHide Solution↓

Set A = students in class 8 of a particular school

Set B = students of this school

All students in class 8 will be the students of the school also.

Hence, A is proper subset of B i.e., A ⊂ B

Hence, option 3 is the correct option.

1(v)Exercise 6(B)

If universal set ξ = {x : x ∈ W, x < 5} and set A = {1, 3}, then complement of set A is equal to :

(a){0, 2, 4, 5}
(b){0, 2, 4}
(c)ϕ
(d){2, 4}

Show SolutionHide Solution↓

Universal set ξ = {x : x ∈ W, x < 5}

Universal set ξ = {0, 1, 2, 3, 4}

Set A = {1, 3}

Complement of A means all the elements of universal set which are not in A.

A' = {0, 2, 4}

Hence, option 2 is the correct option.

1(vi)Exercise 6(B)

If universal set = N, the set of natural numbers, set A = {multiples of 3 less than or equal to 20}
and
Set B = {multiples of 4 less then or equal to 20}, then A - B is equal to :

(a){3, 6, 9, 15, 18}
(b){4, 8, 16, 20}
(c){0}
(d){12}

Show SolutionHide Solution↓

Universal set = N, the set of natural numbers,

= {1, 2, 3, 4, ................}

Set A = {multiples of 3 less than or equal to 20}

= {3, 6, 9, 12, 15, 18}

Set B = {multiples of 4 less then or equal to 20}

= {4, 8, 12, 16}

A - B - all the elements of A except the common elements of A and B

= {3, 6, 9, 15, 18}

Hence, option 1 is the correct option.

2(i)Exercise 6(B)

Find all the subsets of A = {5, 7} .

Show SolutionHide Solution↓

If a set has n elements, the number of subsets = 2ⁿ.

Here the set has 2 elements, the number of subsets = 2² = 2 x 2 = 4.

Subsets of A are ϕ, {5}, {7}, {5, 7}.

2(ii)Exercise 6(B)

Find all the subsets of B = {a, b, c} .

Show SolutionHide Solution↓

If a set has n elements, the number of subsets = 2ⁿ.

Here the set has 3 elements, the number of subsets = 2³ = 2 x 2 x 2 = 8.

Subsets of B are ϕ, {a}, {b}, {c}, {a, b}, {b, c}, {a, c}, {a, b, c}.

2(iii)Exercise 6(B)

Find all the subsets of C = {x : x ∈ W, x ≤ 2} .

Show SolutionHide Solution↓

C = {0, 1, 2}

If a set has n elements, the number of subsets = 2ⁿ.

Here, the set has 3 elements, the number of subsets = 2³ = 2 x 2 x 2 = 8.

Subsets of B are ϕ, {0}, {1}, {2}, {0, 1}, {0, 2}, {1, 2}, {0, 1, 2}.

2(iv)Exercise 6(B)

Find all the subsets of {p : p is a letter in the word 'poor'} .

Show SolutionHide Solution↓

C = {p, o, r}

If a set has n elements, the number of subsets = 2ⁿ.

Here the set has 3 elements, the number of subsets = 2³ = 2 x 2 x 2 = 8.

Subsets of B are ϕ, {p}, {o}, {r}, {p, o}, {o, r}, {r, p}, {p, o, r}.

3Exercise 6(B)

If C is the set of letters in the word 'cooler', find:

(i) set C

(ii) n(C)

(iii) number of its subsets

(iv) number of its proper subsets

Show SolutionHide Solution↓

(i) Set C contains letters of word 'cooler' — c, o, l, e, r.

C = {c, o, l, e, r}

(ii) n(C) is the cardinal number of set C.

n(C) = 5

(iii) Here the set has 5 elements, the number of subsets = 2⁵ = 2 x 2 x 2 x 2 x 2 = 32.

Number of subsets of B are 32.

(iv) Here the set has 5 elements, the number of subsets = 2⁵ - 1 = 2 x 2 x 2 x 2 x 2 - 1 = 32 - 1 = 31.

Number of proper subsets of B are 31.

4Exercise 6(B)

If T = {x : x is a letter in the word 'TEETH'}, find all its subsets.

Show SolutionHide Solution↓

Set T contains all the letters of 'TEETH'.

T = {t, e, h}

If a set has n elements, the number of subsets = 2ⁿ.

Here the set has 3 elements, the number of subsets = 2³ = 2 x 2 x 2 = 8.

Subsets of T are ϕ, {t}, {e}, {h}, {t, e}, {e, h}, {h, t}, {t, e, h}.

5Exercise 6(B)

Given the universal set = {-7, -3, -1, 0, 5, 6, 8, 9}, find :

(i) A = {x : x < 2}

(ii) B = {x : - 4 < x < 6}

Show SolutionHide Solution↓

(i) A contains elements less than 2 i.e, -7, -3, -1 and 0.

A = {-7, -3, -1, 0}

(ii) B contains all the elements between -4 and 6 i.e, -3, -1, 0 and 5.

B = {-3, -1, 0, 5}

6Exercise 6(B)

Given the universal set = { x : x ∈ N and x < 20}, find :

(i) A = {x : x = 3p ; p ∈ N}

(ii) B = {y : y = 2n + 3, n ∈ N}

(iii) C = {x : x is divisible by 4}

Show SolutionHide Solution↓

(i) x = 3p

When p = 1

x = 3p = 3 x 1 = 3

When p = 2

x = 3p = 3 x 2 = 6

When p = 3

x = 3p = 3 x 3 = 9

When p = 4

x = 3p = 3 x 4 = 12

When p = 5

x = 3p = 3 x 5 = 15

When p = 6

x = 3p = 3 x 6 = 18

A = {3, 6, 9, 12, 15, 18}

(ii) y = 2n + 3

When n = 1

y = 2n + 3 = 2 x 1 + 3 = 5

When n = 2

y = 2n + 3 = 2 x 2 + 3 = 7

When n = 3

y = 2n + 3 = 2 x 3 + 3 = 9

When n = 4

y = 2n + 3 = 2 x 4 + 3 = 11

When n = 5

y = 2n + 3 = 2 x 5 + 3 = 13

When n = 6

y = 2n + 3 = 2 x 6 + 3 = 15

When n = 7

y = 2n + 3 = 2 x 7 + 3 = 17

When n = 8

y = 2n + 3 = 2 x 8 + 3 = 19

B = {5, 7, 9, 11, 13, 15, 17, 19}

(iii) x is divisible by 4.

x = 4, 8, 12, 16

C = {4, 8, 12, 16}

7Exercise 6(B)

Find the proper subsets of {x : x² - 9x - 10 = 0}.

Show SolutionHide Solution↓

x² - 9x - 10 = 0

⇒ x² - 10x + x - 10 = 0

⇒ x(x - 10) + 1(x - 10) = 0

⇒ (x - 10)( x + 1) = 0

⇒ (x - 10)= 0 or ( x + 1) = 0

⇒ x = 10 or x = -1

⇒ A = {10, -1}

If a set has n elements, the number of proper subsets = 2ⁿ - 1.

Here the set has 2 elements, the number of proper subsets = 2² - 1 = 2 x 2 - 1 = 3.

Proper subsets of A are ϕ, {10}, {-1}.

8Exercise 6(B)

Given, A = {Triangles}, B = {Isosceles triangles}, C = {Equilateral triangles}. State whether the following are true or false. Give reasons.

(i) A ⊂ B

(ii) B ⊆ A

(iii) C ⊆ B

(iv) B ⊂ A

(v) C ⊂ A

(vi) C ⊆ B ⊆ A.

Show SolutionHide Solution↓

A is a set which contains all triangles.

B is a set which contains all isosceles triangles.

C is a set which contains all equilateral triangles.

(i) False

Reason

As we know each triangle cannot be isosceles triangle.

Hence, A is not a proper subset of B.

(ii) True

Reason

As all isosceles triangles are type of triangles.

Hence, B is a subset of A.

(iii) True

Reason

As all equilateral triangles are isosceles triangles also.

Hence, C is a subset of B.

(iv) True

Reason

As all isosceles triangles are triangle also.

Hence, B is a proper subset of A.

(v) True

Reason

As all equilateral triangles are triangle also.

Hence, C is a proper subset of A.

(vi) True

Reason

As all equilateral triangles are isosceles triangle and all isosceles triangles are triangle also.

Hence, C is a subset of B and B is a subset of A.

9Exercise 6(B)

Given, A = {Quadrilaterals}, B = {Rectangles}, C = {Squares} and D = {Rhombuses}. State, giving reasons, whether the following are true or false.

(i) B ⊂ C

(ii) D ⊂ B

(iii) C ⊆ B ⊆ A

(iv) D ⊂ A

(v) B ⊇ C

(vi) A ⊇ B ⊇ D

Show SolutionHide Solution↓

A is set of all quadrilaterals.

B is set of all rectangles.

C is set of all squares.

D is set of all rhombuses.

(i) False

Reason

As all rectangles are not squares.

Hence, B is a not proper subset of C.

(ii) False

Reason

As all rhombuses are not rectangles.

Hence, D is a not proper subset of B.

(iii) True

Reason

As all squares are rectangles and all rectangles are quadrilateral.

Hence, C is a subset of B and B is a subset of A.

(iv) True

Reason

As all rhombuses are quadrilateral.

Hence, D is a proper subset of A.

(v) True

Reason

As all squares are rectangles.

Hence, C is a subset of B.

(vi) False

Reason

As all rectangles are quadrilateral but all rhombuses are not rectangles.

Hence, B is a subset of A and D is not a subset of B.

10Exercise 6(B)

Given, universal set = {x : x ∈ N, 10 ≤ x ≤ 35}, A = {x ∈ N : x ≤ 16} and B = {x : x > 29}. Find:

(i) A'

(ii) B'

Show SolutionHide Solution↓

Universal set = {x : x ∈ N, 10 ≤ x ≤ 35}

Universal set = {10, 11, 12, 13, 14, 15, 16, ....................., 34, 35}

A = {x ∈ N : x ≤ 16}

B = {x : x > 29}

(i) A = {10, 11, 12, 13, 14, 15, 16}

A' - contains all elements in universal set which are not in set A.

A' = {17, 18, 19, 20, 21,....................., 34, 35}

A' = {x : x ∈ N, 17 ≤ x ≤ 35}

(ii) B = {30, 31, 32, 33, 34, 35}

B' - contains all elements in universal set which are not in set B.

B' = {10, 11, 12, 13, 14, 15, 16,....................., 28, 29}

B' = {x : x ∈ N, 10 ≤ x ≤ 29}

11Exercise 6(B)

Given, universal set = {x ∈ Z : - 6 < x ≤ 6}, N = {n : n is a non-negative number} and P = {x : x is a non-positive number}. Find :

(i) N'

(ii) P'

Show SolutionHide Solution↓

Universal set = {x ∈ Z : - 6 < x ≤ 6}

Universal set = {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6}

N = {n : n is a non-negative number}

P = {x : x is a non-positive number}

(i) N = {0, 1, 2, 3, 4, 5, 6}

N' - contains all elements in universal set which are not in set N.

N' = {-5, -4, -3, -2, -1}

(ii) P = {-5, -4, -3, -2, -1, 0}

P' - contains all elements in universal set which are not in set P.

P' = {1, 2, 3, 4, 5, 6}

12Exercise 6(B)

Let M = {letters of the word REAL} and N = {letters of the word LARE}. Write sets M and N in roster form and then state whether;

(i) M ⊆ N is true.

(ii) N ⊆ M is true.

(iii) M = N is true.

Show SolutionHide Solution↓

M = {letters of the word REAL}

M = {R, E, A, L}

N = {letters of the word LARE}

N = {L, A, R, E}

(i) True

Reason

Every elements in set M is also in set N.

(ii) True

Reason

Every elements in set N is also in set M.

(iii) True

Reason

Both sets contain the same elements.

13Exercise 6(B)

Given A = {x : x ∈ N and 3 < x ≤ 6} and B = {x : x ∈ W and x < 4}. Find :

(i) sets A and B in roster form

(ii) A ∪ B

(iii) A ∩ B

(iv) A - B

(v) B - A

Show SolutionHide Solution↓

A = {x : x ∈ N and 3 < x ≤ 6}

B = {x : x ∈ W and x < 4}

(i) A = {4, 5, 6}

B = {0, 1, 2, 3}

(ii) A ∪ B - contains all the elements in both sets A and B.

A ∪ B = {4, 5, 6} ∪ {0, 1, 2, 3}

A ∪ B = {0, 1, 2, 3, 4, 5, 6}

(iii) A ∩ B - contains all the common elements in both sets A and B.

A ∩ B = {4, 5, 6} ∩ {0, 1, 2, 3}

A ∩ B = ϕ

(iv) A - B - includes all elements in set A but not in set B.

A - B = {4, 5, 6} - {0, 1, 2, 3}

A - B = {4, 5, 6}

(v) B - A - includes all elements in set B but not in set A.

B - A = {0, 1, 2, 3} - {4, 5, 6}

B - A = {0, 1, 2, 3}

14Exercise 6(B)

If P = {x : x ∈ W and 4 ≤ x ≤ 8} and Q = {x : x ∈ N and x < 6}. Find

(i) P ∪ Q and P ∩ Q.

(ii) Is (P ∪ Q) ⊃ (P ∩ Q) ?

Show SolutionHide Solution↓

P = {x : x ∈ W and 4 ≤ x ≤ 8}

P = {4, 5, 6, 7, 8}

Q = {x : x ∈ N and x < 6}

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

(i) P ∪ Q - contains all the elements in both sets P and Q.

P ∪ Q = {4, 5, 6, 7, 8} ∪ {1, 2, 3, 4, 5}

P ∪ Q = {1, 2, 3, 4, 5, 6, 7, 8}

P ∩ Q - contains all the common elements in both sets P and Q.

P ∩ Q = {4, 5, 6, 7, 8} ∩ {1, 2, 3, 4, 5}

P ∩ Q = {4, 5}

(ii) From (i) we can clearly see that every elements in P ∩ Q is also in P ∪ Q.

Hence, (P ∪ Q) ⊃ (P ∩ Q) is correct.

15Exercise 6(B)

If A = {5, 6, 7, 8, 9}, B ={x : 3 < x < 8 and x ∈ W} and C = {x : x ≤ 5 and x ∈ N}. Find :

(i) A ∪ B and (A ∪ B) ∪ C

(ii) B ∪ C and A ∪ (B ∪ C)

(iii) A ∩ B and (A ∩ B) ∩ C

(iv) B ∩ C and A ∩ (B ∩ C)

Is (A ∪ B) u C = A ∪ (B ∪ C) ?

Is (A ∩ B) ∩ C = A ∩ (B ∩ C) ?

Show SolutionHide Solution↓

A = {5, 6, 7, 8, 9}

B ={x : 3 < x < 8 and x ∈ W}

B = {4, 5, 6, 7}

C = {x : x ≤ 5 and x ∈ N}

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

(i) A ∪ B - contains all elements in both set A and B.

A ∪ B = {5, 6, 7, 8, 9} ∪ {4, 5, 6, 7}

A ∪ B = {4, 5, 6, 7, 8, 9}

(A ∪ B) ∪ C - contains all elements in set A, B and C.

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

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

(ii) B ∪ C - contains all elements in both set B and C.

B ∪ C = {4, 5, 6, 7} ∪ {1, 2, 3, 4, 5}

B ∪ C = {1, 2, 3, 4, 5, 6, 7}

A ∪ (B ∪ C) - contains all elements in set A, B and C.

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

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

(iii) A ∩ B - contains common elements in both set A and B.

A ∩ B = {5, 6, 7, 8, 9} ∩ {4, 5, 6, 7}

A ∩ B = {5, 6, 7}

(A ∩ B) ∩ C - contains common elements in set A, B and C.

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

(A ∩ B) ∩ C = {5}

(iv) B ∩ C - contains common elements in both set B and C.

B ∩ C = {4, 5, 6, 7} ∩ {1, 2, 3, 4, 5}

B ∩ C = {4, 5}

A ∩ (B ∩ C) - contains common elements in set A, B and C.

A ∩ (B ∩ C) = {5, 6, 7, 8, 9} ∩ {4, 5}

A ∩ (B ∩ C) = {5}

As we have seen in (i) and (ii), all elements in (A ∪ B) u C and A ∪ (B ∪ C) are same.

(A ∪ B) u C = A ∪ (B ∪ C)

As we have seen in (iii) and (iv), all elements in (A ∩ B) ∩ C and A ∩ (B ∩ C) are same.

(A ∩ B) ∩ C = A ∩ (B ∩ C)

16Exercise 6(B)

Given A = {0, 1, 2, 4, 5}, B = {0, 2, 4, 6, 8} and C = {0, 3, 6, 9}. Show that :

(i) A ∪ (B ∪ C) = (A ∪ B) ∪ C, i.e., the union of sets is associative.

(ii) A ∩ (B ∩ C) = (A ∩ B) ∩ C, i.e., the intersection of sets is associative.

Show SolutionHide Solution↓

A = {0, 1, 2, 4, 5}

B = {0, 2, 4, 6, 8}

C = {0, 3, 6, 9}

(i) B ∪ C - contains all the elements in set B and C.

B ∪ C = {0, 2, 4, 6, 8} ∪ {0, 3, 6, 9}

B ∪ C = {0, 2, 3, 4, 6, 8, 9}

A ∪ (B ∪ C) - contains all the elements in set A, B and C.

A ∪ (B ∪ C) = {0, 1, 2, 4, 5} ∪ {0, 2, 3, 4, 6, 8, 9}

A ∪ (B ∪ C) = {0, 1, 2, 3, 4, 5, 6, 8, 9} ...............(1)

A ∪ B - contains all the elements in set A and B.

A ∪ B = {0, 1, 2, 4, 5} ∪ {0, 2, 4, 6, 8}

A ∪ B = {0, 1, 2, 4, 5, 6, 8}

(A ∪ B) ∪ C - contains all the elements in set A, B and C.

(A ∪ B) ∪ C = {0, 1, 2, 4, 5, 6, 8} ∪ {0, 3, 6, 9}

(A ∪ B) ∪ C = {0, 1, 2, 3, 4, 5, 6, 8, 9} ...............(2)

Hence from (1) and (2), all the elements of A ∪ (B ∪ C) and (A ∪ B) ∪ C are same.

A ∪ (B ∪ C) = (A ∪ B) ∪ C, i.e. the union of sets is associative.

(ii) B ∩ C - contains all the common elements in set B and C.

B ∩ C = {0, 2, 4, 6, 8} ∩ {0, 3, 6, 9}

B ∩ C = {0,6}

A ∩ (B ∩ C) = {0, 1, 2, 4, 5} ∩ {0,6}

A ∩ (B ∩ C) = {0} ...............(3)

A ∩ B - contains all the common elements in set A and B.

A ∩ B = {0, 1, 2, 4, 5} ∩ {0, 2, 4, 6, 8}

A ∩ B = {0, 2, 4}

(A ∩ B) ∩ C = {0, 2, 4} ∩ {0, 3, 6, 9}

(A ∩ B) ∩ C = {0} ...............(4)

Hence from (3) and (4), all the elements of A ∩ (B ∩ C) and (A ∩ B) ∩ C are same.

A ∩ (B ∩ C) = (A ∩ B) ∩ C, i.e., the intersection of sets is associative.

17Exercise 6(B)

If A = {x ∈ W : 5 < x < 10}, B = {3, 4, 5, 6, 7} and C = {x = 2n ; n ∈ N and n ≤ 4}. Find :

(i) A ∩ (B ∪ C)

(ii) (B ∪ A) ∩ (B ∪ C)

(iii) B ∪ (A ∩ C)

(iv) (A ∩ B) ∪ (A ∩ C)

Name the sets which are equal.

Show SolutionHide Solution↓

A = {x ∈ W : 5 < x < 10}

A = {6, 7, 8, 9}

B = {3, 4, 5, 6, 7}

C = {x = 2n ; n ∈ N and n ≤ 4}

C = {2, 4, 6, 8}

(i) B ∪ C - contains all the elements in set B and C.

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

B ∪ C = {2, 3, 4, 5, 6, 7, 8}

A ∩ (B ∪ C) - contains all the common elements in set A and (B ∪ C).

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

A ∩ (B ∪ C) = {6, 7, 8}

(ii) B ∪ A - contains all the elements in set B and A.

B ∪ A = {3, 4, 5, 6, 7} ∪ {6, 7, 8, 9}

B ∪ A = {3, 4, 5, 6, 7, 8, 9}

B ∪ C - contains all the elements in set B and C.

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

B ∪ C = {2, 3, 4, 5, 6, 7, 8}

(B ∪ A) ∩ (B ∪ C) - contains all common elements in set (B ∪ A) and (B ∪ C).

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

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

(iii) A ∩ C - contains all the common elements in set A and C.

A ∩ C = {6, 7, 8, 9} ∩ {2, 4, 6, 8}

A ∩ C = {6, 8}

B ∪ (A ∩ C) - contains all the elements of B and (A ∩ C).

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

B ∪ (A ∩ C) = {3, 4, 5, 6, 7, 8}

(iv) A ∩ B - contains all the common elements in set A and B.

A ∩ B = {6, 7, 8, 9} ∩ {3, 4, 5, 6, 7}

A ∩ B = {6, 7}

A ∩ C - contains all the common elements in set A and C.

A ∩ C = {6, 7, 8, 9} ∩ {2, 4, 6, 8}

A ∩ C = {6, 8}

(A ∩ B) ∪ (A ∩ C) - contains all the elements of (A ∩ B) and (A ∩ C).

(A ∩ B) ∪ (A ∩ C) = {6, 7} ∪ {6, 8}

(A ∩ B) ∪ (A ∩ C) = {6, 7, 8}

As we can see that some sets are having same elements i.e, they are equal.

A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) and B ∪ (A ∩ C) = (B ∪ A) ∩ (B ∪ C)

18Exercise 6(B)

If P = {factors of 36} and Q = {factors of 48} ; find :

(i) P ∪ Q

(ii) P ∩ Q

(iii) Q - P

(iv) P' ∩ Q

Show SolutionHide Solution↓

P = {factors of 36}

P = {1, 2, 3, 4, 6, 9, 12, 18, 36}

Q = {factors of 48}

Q = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

(i) P ∪ Q

P ∪ Q - contains all the elements of P and Q.

P ∪ Q = {1, 2, 3, 4, 6, 9, 12, 18, 36} ∪ {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

P ∪ Q = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48}

(ii) P ∩ Q

P ∩ Q - contains all the common elements in set P and Q.

P ∩ Q = {1, 2, 3, 4, 6, 9, 12, 18, 36} ∩ {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

P ∩ Q = {1, 2, 3, 4, 6, 12}

(iii) Q - P

Q - P - contains all the elements in set Q but not in set P.

Q - P = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48} - {1, 2, 3, 4, 6, 9, 12, 18, 36}

Q - P = {8, 16, 24, 48}

(iv) P' ∩ Q

U is the universal set containing all the elements in set P and Q.

U = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48}

P' - contains all the elements of U which are not in P.

P' = {8, 16, 24, 48}

P' ∩ Q - contains all the common elements of P' and Q

P' ∩ Q = {8, 16, 24, 48} ∩ {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

P' ∩ Q = {8, 16, 24, 48}

19Exercise 6(B)

If A = {6, 7, 8, 9}, B = {4, 6, 8, 10} and C ={x : x ∈ N : 2 < x ≤ 7}; find :

(i) A - B

(ii) B - C

(iii) B - (A - C)

(iv) A - (B ∪ C)

(v) B - (A ∩ C)

(vi) B - B

Show SolutionHide Solution↓

A = {6, 7, 8, 9}

B = {4, 6, 8, 10}

C ={x :x ∈ N: 2 < x ≤ 7}

C = {3, 4, 5, 6, 7}

(i) A - B

A - B - contains all the elements which are in A but not in B.

A - B = {6, 7, 8, 9} - {4, 6, 8, 10}

A - B = {7, 9}

(ii) B - C

B - C - contains all the elements which are in B but not in C.

B - C = {4, 6, 8, 10} - {3, 4, 5, 6, 7}

B - C = {8, 10}

(iii) B - (A - C)

A - C - contains all the elements which are in A but not in C.

A - C = {6, 7, 8, 9} - {3, 4, 5, 6, 7}

A - C = {8, 9}

B - (A - C) - contains all the elements which are in set B but not in (A - C).

B - (A - C) = {4, 6, 8, 10} - {8, 9}

B - (A - C) = {4, 6, 10}

(iv) A - (B ∪ C)

B ∪ C - contains all the elements in set B and C.

B ∪ C = {4, 6, 8, 10} ∪ {3, 4, 5, 6, 7}

B ∪ C = {3, 4, 5, 6, 7, 8, 10}

A - (B ∪ C) - contains all the elements which is in set A but not in (B ∪ C).

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

A - (B ∪ C) = {9}

(v) B - (A ∩ C)

A ∩ C - contains all the common elements in set A and C.

A ∩ C = {6, 7, 8, 9} ∪ {3, 4, 5, 6, 7}

A ∩ C = {6, 7}

B - (A ∩ C) - contains all the elements which is in set B but not in (A ∩ C).

B - (A ∩ C) = {4, 6, 8, 10} - {6, 7}

B - (A ∩ C) = {4, 8, 10}

(vi) B - B

B - B - contains all the elements that are in set B but not in set B.

B - B = {4, 6, 8, 10} - {4, 6, 8, 10}

B - B = ϕ

20Exercise 6(B)

If A = {1, 2, 3, 4, 5}, B = {2, 4, 6, 8} and C = {3, 4, 5, 6} ; verify :

(i) A - (B ∪ C) = (A - B) ∩ (A - C)

(ii) A - (B ∩ C) = (A - B) ∪ (A - C).

Show SolutionHide Solution↓

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

B = {2, 4, 6, 8}

C = {3, 4, 5, 6}

(i) A - (B ∪ C) = (A - B) ∩ (A - C)

Taking LHS : A - (B ∪ C)

B ∪ C - contains all the elements in set B and C.

B ∪ C = {2, 4, 6, 8} ∪ {3, 4, 5, 6}

B ∪ C = {2, 3, 4, 5, 6, 8}

A - (B ∪ C) - contains all the elements which are in set A but not in B ∪ C.

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

A - (B ∪ C) = {1} ...............(1)

Taking RHS : (A - B) ∩ (A - C)

A - B - contains all the elements which are in set A but not in B.

A - B = {1, 2, 3, 4, 5} - {2, 4, 6, 8}

A - B = {1, 3, 5}

A - C - contains all the elements which are in set A but not in C.

A - C = {1, 2, 3, 4, 5} - {3, 4, 5, 6}

A - C = {1, 2}

(A - B) ∩ (A - C) - contains all the common elements in set (A - B) and (A - C).

(A - B) ∩ (A - C) = {1, 3, 5} ∩ {1, 2}

(A - B) ∩ (A - C) = {1} ...............(2)

From (1) and (2), we can see

LHS = RHS

Hence, A - (B ∪ C) = (A - B) ∩ (A - C)

(ii) A - (B ∩ C) = (A - B) ∪ (A - C).

Taking LHS : A - (B ∩ C)

B ∩ C - contains all the common elements in set B and C.

B ∩ C = {2, 4, 6, 8} ∩ {3, 4, 5, 6}

B ∩ C = {4, 6}

A - (B ∩ C) - contains all the elements which are in set A but not in B ∩ C.

A - (B ∩ C) = {1, 2, 3, 4, 5} - {4, 6}

A - (B ∩ C) = {1, 2, 3, 5} ...............(3)

Taking RHS : (A - B) ∪ (A - C)

A - B - contains all the elements which are in set A but not in B.

A - B = {1, 2, 3, 4, 5} - {2, 4, 6, 8}

A - B = {1, 3, 5}

A - C - contains all the elements which are in set A but not in C.

A - C = {1, 2, 3, 4, 5} - {3, 4, 5, 6}

A - C = {1, 2}

(A - B) ∪ (A - C) - contains all the elements in set (A - B) and (A - C).

(A - B) ∪ (A - C) = {1, 3, 5} ∪ {1, 2}

(A - B) ∪ (A - C) = {1, 2, 3, 5} ...............(4)

From (3) and (4), we can see

LHS = RHS

Hence, A - (B ∩ C) = (A - B) ∪ (A - C)

21Exercise 6(B)

Given A = {x ∈ N : x < 6}, B = {3, 6, 9} and C = {x ∈ N : 2x - 5 ≤ 8}. Show that :

(i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

(ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Show SolutionHide Solution↓

A = {x ∈ N : x < 6}

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

B = {3, 6, 9}

C = {x ∈ N : 2x - 5 ≤ 8}

2x - 5 ≤ 8

2x ≤ 8 + 5

2x ≤ 13

x ≤ $\frac{13}{2}$

x ≤ 6.5

C = {1, 2, 3, 4, 5, 6}

(i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

Taking LHS: A ∪ (B ∩ C)

B ∩ C - contains all the common elements in set B and C.

B ∩ C = {3, 6, 9} ∩ {1, 2, 3, 4, 5, 6}

B ∩ C = {3, 6}

A ∪ (B ∩ C) - contains all the elements in set A and B ∩ C.

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

A ∪ (B ∩ C) = {1, 2, 3, 4, 5, 6} .............(1)

Taking RHS : (A ∪ B) ∩ (A ∪ C)

A ∪ B - contains all the elements in set A and B.

A ∪ B = {1, 2, 3, 4, 5} ∪ {3, 6, 9}

A ∪ B = {1, 2, 3, 4, 5, 6, 9}

A ∪ C - contains all the elements in set A and C.

A ∪ C = {1, 2, 3, 4, 5} ∪ {1, 2, 3, 4, 5, 6}

A ∪ C = {1, 2, 3, 4, 5, 6}

(A ∪ B) ∩ (A ∪ C) - contains all the common elements in set (A ∪ B) and (A ∪ C)

(A ∪ B) ∩ (A ∪ C) = {1, 2, 3, 4, 5, 6, 9} ∩ {1, 2, 3, 4, 5, 6}

(A ∪ B) ∩ (A ∪ C) = {1, 2, 3, 4, 5, 6} .............(2)

From (1) and (2), we get

LHS = RHS

Hence, A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

(ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Taking LHS: A ∩ (B ∪ C)

B ∪ C - contains all the elements in set B and C.

B ∪ C = {3, 6, 9} ∪ {1, 2, 3, 4, 5, 6}

B ∪ C = {1, 2, 3, 4, 5, 6, 9}

A ∩ (B ∪ C) - contains all the common elements in set A and B ∪ C.

A ∩ (B ∪ C) = {1, 2, 3, 4, 5} ∪ {1, 2, 3, 4, 5, 6, 9}

A ∩ (B ∪ C) = {1, 2, 3, 4, 5} .............(3)

Taking RHS : (A ∩ B) ∪ (A ∩ C)

A ∩ B - contains all the common elements in set A and B.

A ∩ B = {1, 2, 3, 4, 5} ∩ {3, 6, 9}

A ∩ B = {3}

A ∩ C - contains all the common elements in set A and C.

A ∩ C = {1, 2, 3, 4, 5} ∩ {1, 2, 3, 4, 5, 6}

A ∩ C = {1, 2, 3, 4, 5}

(A ∩ B) ∪ (A ∩ C) - contains all the elements in set (A ∩ B) and (A ∩ C)

(A ∩ B) ∪ (A ∩ C) = {3} ∪ {1, 2, 3, 4, 5}

(A ∩ B) ∪ (A ∩ C) = {1, 2, 3, 4, 5} .............(4)

From (3) and (4), we get

LHS = RHS

Hence, A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

22Exercise 6(B)

If n(A) = 30, n(B) = 20 and n(A ∪ B) = 36, find n(A ∩ B).

Show SolutionHide Solution↓

Given - n(A) = 30

n(B) = 20

n(A ∪ B) = 36

∵ n(A ∩ B) = n(A) + n(B) - n(A ∪ B)

Putting the values, we get

n(A ∩ B) = 30 + 20 - 36

⇒ n(A ∩ B) = 50 - 36

⇒ n(A ∩ B) = 14

∴ n(A ∩ B) = 14

23Exercise 6(B)

If n(A) = 50, n(B) = 30 and n(A ∩ B) = 15, find n(A ∪ B).

Show SolutionHide Solution↓

Given - n(A) = 50

n(B) = 30

n(A ∩ B) = 15

∵ n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

Putting the values, we get

n(A ∪ B) = 50 + 30 - 15

⇒ n(A ∪ B) = 80 - 15

⇒ n(A ∪ B) = 65

∴ n(A ∪ B) = 65

24Exercise 6(B)

If n(A - B) = 30, n(B - A) = 20 and n(A ∩ B) = 10, find :

(i) n(A)

(ii) n(B)

(iii) n(A ∪ B)

Show SolutionHide Solution↓

Given:

n(A - B) = 30

n(B - A) = 20

n(A ∩ B) = 10

(i) n(A - B) = n(A) - n(A ∩ B)

Putting the values, we get

30 = n(A) - 10

n(A) = 30 + 10

n(A) = 40

(ii) n(B - A) = n(B) - n(A ∩ B)

Putting the values, we get

20 = n(B) - 10

n(B) = 20 + 10

n(B) = 30

(iii) n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

Putting the values, we get

n(A ∪ B) = 40 + 30 - 10

⇒ n(A ∪ B) = 70 - 10

⇒ n(A ∪ B) = 60

Hence, n(A ∪ B) = 60

25Exercise 6(B)

If n(A - B) = 30, n(B - A) = 48 and n(A ∩ B) = 15, find n(A ∪ B).

Show SolutionHide Solution↓

n(A - B) = 30

n(B - A) = 48

n(A ∩ B) = 15

∴ n(A - B) = n(A) - n(A ∩ B)

Putting the values, we get

30 = n(A) - 15

n(A) = 30 + 15

n(A) = 45

∴ n(B - A) = n(B) - n(A ∩ B)

Putting the values, we get

48 = n(B) - 15

n(B) = 48 + 15

n(B) = 63

∴ n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

Putting the values, we get

n(A ∪ B) = 45 + 63 - 15

⇒ n(A ∪ B) = 108 - 15

⇒ n(A ∪ B) = 93

∴ n(A ∪ B) = 93

1(i)Test Yourself

If A = {3, 5, 7} and B = {5, 7, 9}, then n(A ∩ B) is :

(a)6
(b)2
(c)4
(d)none of these

Show SolutionHide Solution↓

A = {3, 5, 7}

B = {5, 7, 9}

A ∩ B - contains all the common elements in set A and B.

A ∩ B = {3, 5, 7} ∩ {5, 7, 9}

A ∩ B = {5, 7}

n(A ∩ B) = 2

Hence, option 2 is the correct option.

1(ii)Test Yourself

If n(universal set) = 80 and n(A) = 50, then n(complement of set A) is :

(a)80
(b)50
(c)30
(d)none of these

Show SolutionHide Solution↓

n(universal set) = 80

n(A) = 50

n(complement of set A) = n(universal set) - n(A)

⇒ n(complement of set A) = 80 - 50

⇒ n(complement of set A) = 30

Hence, option 3 is the correct option.

1(iii)Test Yourself

If A = {5, 8, 10} and empty set ϕ. Then ϕ - A is equal to :

(a)A
(b)ϕ
(c){0}
(d)none of these

Show SolutionHide Solution↓

A = {5, 8, 10}

ϕ - A - contains all the elements in set ϕ but not in A.

ϕ - A = ϕ - {5, 8, 10}

ϕ - A = ϕ

Hence, option 2 is the correct option.

1(iv)Test Yourself

If set A = {x : x ∈ W and 0 < x ≤ 4}, then set A is equal to :

(a){0, 1, 2, 3}
(b){0, 1, 2, 3, 4}
(c){1, 2, 3, 4}
(d)none of these

Show SolutionHide Solution↓

Set A = {x : x ∈ W and 0 < x ≤ 4}

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

Hence, option 3 is the correct option.

1(v)Test Yourself

If set P = {factors of 8}, then set P is equal to :

(a){2, 4, 8}
(b){1, 2, 4}
(c){1, 2, 4, 8}
(d)none of these

Show SolutionHide Solution↓

Set P = {factors of 8}

Factors of 8 = 1, 2, 4, 8

Set P = {1, 2, 4, 8}

Hence, option 3 is the correct option.

1(vi)Test Yourself

The elements of the set {x : x ∈ Z and x² ≤ 9} are :

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

Show SolutionHide Solution↓

Set = {x : x ∈ Z and x² ≤ 9}

x² = 0

x = 0

x² = 1

x = 1 or -1

x² = 4

x = 2 or -2

x² = 9

x = 3 or -3

Set = {-3, -2, -1, 0, 1, 2, 3}

Hence, option 2 is the correct option.

1(vii)Test Yourself

If A and B are two equal sets, then A - B is equal to :

(a)A
(b)B
(c){0}
(d){ }

Show SolutionHide Solution↓

A - B - contains all the elements in set A but not in B.

A and B are equal set means all the elements in set A and B are same.

A - B = { }

Hence, option 4 is the correct option.

1(viii)Test Yourself

If n(A) = n(B) then :

(a)A = B
(b)A ≠ B
(c)A - B = {0}
(d)none of these

Show SolutionHide Solution↓

n(A) = n(B)

That means, number of elements in set A and B are same but elements can be different. Hence, none of the option is always correct.

Hence, option 4 is the correct option.

1(ix)Test Yourself

A set has 5 elements, then number of its subsets is :

(a)2⁵
(b)5²
(c)2⁵ - 1
(d)2 x 5

Show SolutionHide Solution↓

If a set has n elements, the number of subsets = 2ⁿ

If a set has 5 elements, the number of subsets = 2⁵

Hence, option 1 is the correct option.

1(x)Test Yourself

Let M = {factors of 12} and N = {factors of 24} then {24} is equal to :

(a)M ∪ N
(b)M ∩ N
(c)M - N
(d)N - M

Show SolutionHide Solution↓

M = {factors of 12}

Factors of 12 = 1, 2, 3, 4, 6, 12

M = {1, 2, 3, 4, 6, 12}

N = {factors of 24}

Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24

N = {1, 2, 3, 4, 6, 8, 12, 24}

M ∪ N = {1, 2, 3, 4, 6, 12} ∪ {1, 2, 3, 4, 6, 8, 12, 24}

= {1, 2, 3, 4, 6, 8, 12, 24}

M ∩ N = {1, 2, 3, 4, 6, 12, 24} ∩ {1, 2, 3, 4, 6, 8, 12, 24}

= {1, 2, 3, 4, 6, 12}

M - N = {1, 2, 3, 4, 6, 12} - {1, 2, 3, 4, 6, 8, 12, 24}

= {}

N - M = {1, 2, 3, 4, 6, 8, 12, 24} - {1, 2, 3, 4, 6, 12}

= {8, 24}

Hence, none of the given options are correct.

1(xi)Test Yourself

Statement 1: The number of subsets of {{1, {0}}, 2} is 8.

Statement 2: A set containing 'n' elements has 2^{n - 1} proper subsets.

Which of the following options is correct?

(a)Both the statements are true.
(b)Both the statements are false.
(c)Statement 1 is true, and statement 2 is false.
(d)Statement 1 is false, and statement 2 is true.

Show SolutionHide Solution↓

Given set : {{1, {0}}, 2}

The set A = {{1, {0}}, 2} has two elements : {1, {0}} and 2.

Thus, n = 2.

The number of subsets of a set with n elements = 2ⁿ = 2² = 4.

Thus, statement 1 is false.

We know that,

The total number of proper subsets of a set with n elements is 2ⁿ - 1.

Thus, statement 2 is true.

Hence, option 4 is the correct option.

1(xii)Test Yourself

Assertion (A) : Let A = {1, {Φ}}, then each of Φ, {1}, {{Φ}} is a proper subset of A.

Reason (R) : The empty set has no proper subset.

(a)Both A and R are correct, and R is the correct explanation for A.
(b)Both A and R are correct, and R is not the correct explanation for A.
(c)A is true, but R is false.
(d)A is false, but R is true.

Show SolutionHide Solution↓

Given set A = {1, {Φ}}

Proper subset of A = Φ, {1}, {{Φ}}

So, assertion (A) is true.

The only subset of empty set is itself, so we can say that empty set does not have a proper subset.

So, reason (R) is true but reason does not explains assertion.

Hence, option 2 is the correct option.

1(xiii)Test Yourself

Assertion (A) : Let A = {factors of 12} and B = {factors of 16}. Then B - A = {8, 16}

Reason (R) : B - A = {x | x ∈ A, but x ∉ B}.

(a)Both A and R are correct, and R is the correct explanation for A.
(b)Both A and R are correct, and R is not the correct explanation for A.
(c)A is true, but R is false.
(d)A is false, but R is true.

Show SolutionHide Solution↓

⇒ A = {factors of 12}

⇒ A = {1, 2, 3, 4, 6, 12}

⇒ B = {factors of 16}

⇒ B = {1, 2, 4, 8, 16}

Set B - A contains all the elements that are in set B but not in set A.

⇒ B - A = {x | x ∈ B, but x ∉ A}

So, reason (R) is false.

⇒ B - A = {8, 16}

So, assertion (A) is true.

Hence, option 3 is the correct option.

1(xiv)Test Yourself

Assertion (A) : Let A = {1, 2, 3, 4, 5, 6} and B = {1, 3, 5, 7, 9} then A ∩ B ⊆ A and A ∩ B ⊆ B, always true for every pair of two sets.

Reason (R) : For any sets A and B, we have A ∩ B ⊆ A and A ∩ B ⊆ B.

(a)Both A and R are correct, and R is the correct explanation for A.
(b)Both A and R are correct, and R is not the correct explanation for A.
(c)A is true, but R is false.
(d)A is false, but R is true.

Show SolutionHide Solution↓

Given : A = {1, 2, 3, 4, 5, 6} and B = {1, 3, 5, 7, 9}

A ∩ B means this set contains elements common to both A and B.

A ∩ B = {1, 3, 5}

A ∩ B ⊆ A means {1, 3, 5} is subset of {1, 2, 3, 4, 5, 6}, which is true.

A ∩ B ⊆ B means {1, 3, 5} is subset of {1, 3, 5, 7, 9}, which is also true.

So, assertion (A) is true.

For any sets A and B, we have A ∩ B ⊆ A and A ∩ B ⊆ B.

This is a fundamental property of set theory :

A ∩ B = {x | x ∈ A and x ∈ B}

So any element of A ∩ B is automatically in both A and B, which implies A ∩ B ⊆ A and A ∩ B ⊆ B are true.

So, reason (R) is true and it clearly explains assertion.

Hence, option 1 is the correct option.

1(xv)Test Yourself

Assertion (A) : Let A = {x | x + 3 = 0, x ∈ N}, B = {x | x ≤ 3, x ∈ W} then A ∩ B = B.

Reason (R) : For any set A, A ∩ Φ = Φ.

(a)Both A and R are correct, and R is the correct explanation for A.
(b)Both A and R are correct, and R is not the correct explanation for A.
(c)A is true, but R is false.
(d)A is false, but R is true.

Show SolutionHide Solution↓

Given : A = {x | x + 3 = 0, x ∈ N}, B = {x | x ≤ 3, x ∈ W}

⇒ x + 3 = 0

⇒ x = -3

Since, x ∈ N. So, A = {}

Whole number = {0, 1, 2, 3, .................} and x ≤ 3

So, B = {0, 1, 2, 3}

A ∩ B means this set contains elements common to both A and B.

⇒ A ∩ B = {}

⇒ A ∩ B ≠ B

So, assertion (A) is false.

For any set A, A ∩ Φ = Φ

The intersection of any set with the empty set is always the empty set.

So, reason (R) is true.

Hence, option 4 is the correct option.

2Test Yourself

If universal set = {all digits in our number system} and set A = {2, 3, 7, 9}. Write the complement of set A.

Show SolutionHide Solution↓

Universal set = {all digits in our number system}

Universal set = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Set A = {2, 3, 7, 9}

Complement of set A = Universal set - Set A

Complement of set A = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 3, 7, 9}

Complement of set A = {0, 1, 4, 5, 6, 8}

3Test Yourself

If A = {factors of 36} and B = {factors of 48}, find :

(i) A ∪ B

(ii) A ∩ B

(iii) A - B

(iv) B - A

Show SolutionHide Solution↓

A = {factors of 36}

Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36

A = {1, 2, 3, 4, 6, 9, 12, 18, 36}

B = {factors of 48}

Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48

B = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

(i) A ∪ B

A ∪ B - contains all the elements in set A and B.

A ∪ B = {1, 2, 3, 4, 6, 9, 12, 18, 36} ∪ {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

A ∪ B = {1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48}

(ii) A ∩ B

A ∩ B - contains all the common elements in set A and B.

A ∩ B = {1, 2, 3, 4, 6, 9, 12, 18, 36} ∩ {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

A ∩ B = {1, 2, 3, 4, 6, 12}

(iii) A - B

A - B - contains all the elements in set A but not in B.

A - B = {1, 2, 3, 4, 6, 9, 12, 18, 36} - {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}

A - B = {9, 18, 36}

(iv) B - A

B - A - contains all the elements in set B but not in A.

B - A = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48} - {1, 2, 3, 4, 6, 9, 12, 18, 36}

B - A = {8, 16, 24, 48}

4Test Yourself

By taking the sets of your own, verify that :

(i) n(A - B) = n(A ∪ B) - n(B)

(ii) n(A ∩ B) + n(A ∪ B) = n(A) + n(B)

Show SolutionHide Solution↓

Lets take the set A = {1, 2, 3, 4, 5, 6}

and set B = {2, 4, 6}

(i) n(A - B) = n(A ∪ B) - n(B)

A - B - contains all the elements in set A but not in B.

A - B = {1, 2, 3, 4, 5, 6} - {2, 4, 6}

A - B = {1, 3, 5}

n(A - B) = 3

A ∪ B - contains all the elements in set A and B.

A ∪ B = {1, 2, 3, 4, 5, 6} ∪ {2, 4, 6}

A ∪ B = {1, 2, 3, 4, 5, 6}

n(A ∪ B) = 6

n(B) = 3

Taking LHS : n(A - B)

n(A - B) = 3

Taking RHS : n(A ∪ B) - n(B)

n(A ∪ B) - n(B) = 6 - 3

n(A ∪ B) - n(B) = 3

∴ LHS = RHS

∴ n(A - B) = n(A ∪ B) - n(B)

(ii) n(A ∩ B) + n(A ∪ B) = n(A) + n(B)

A ∩ B - contains all the common elements in set A and B.

A ∩ B = {1, 2, 3, 4, 5, 6} ∩ {2, 4, 6}

A ∩ B = {2, 4, 6}

n(A ∩ B) = 3

A ∪ B = {1, 2, 3, 4, 5, 6}

n(A ∪ B) = 6

n(A) = 6

n(B) = 3

Taking LHS:

n(A ∩ B) + n(A ∪ B)

n(A ∩ B) + n(A ∪ B) = 3 + 6

n(A ∩ B) + n(A ∪ B) = 9

Taking RHS:

n(A) + n(B)

n(A) + n(B) = 6 + 3

n(A) + n(B) = 9

∴ LHS = RHS

∴ n(A ∩ B) + n(A ∪ B) = n(A) + n(B)

5Test Yourself

If n(A - B) = 24, n(B - A) = 32 and n(A ∩ B) = 10; find n(A ∪ B).

Show SolutionHide Solution↓

n(A - B) = 24

n(B - A) = 32

n(A ∩ B) = 10

∴ n(A - B) = n(A) - n(A ∩ B)

Putting the values, we get

24 = n(A) - 10

n(A) = 24 + 10

n(A) = 34

∴ n(B - A) = n(B) - n(A ∩ B)

Putting the values, we get

32 = n(B) - 10

n(B) = 32 + 10

n(B) = 42

∴ n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

Putting the values, we get

n(A ∪ B) = 34 + 42 - 10

⇒ n(A ∪ B) = 76 - 10

⇒ n(A ∪ B) = 66

∴ n(A ∪ B) = 66

6Test Yourself

If ξ = {x : x ∈ N, x ≤ 10 } , A = {x : x ≥ 5} and B = {x : 3 < x < 6}, then find :

(i) (A ∪ B)'

(ii) A' ∩ B'

Are (A ∪ B)' and A' ∩ B' equal ?

Show SolutionHide Solution↓

ξ = {x : x ∈ N, x ≤ 10 }

ξ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

A = {x : x ≥ 5}

A = {5, 6, 7, 8, 9, 10}

B = {x : 3 < x < 6}

B = {4, 5}

(i) (A ∪ B)'

A ∪ B - contains all the elements in set A and B.

A ∪ B = {5, 6, 7, 8, 9, 10} ∪ {4, 5}

A ∪ B = {4, 5, 6, 7, 8, 9, 10}

(A ∪ B)'- contains all the elements in universal set which are not in A ∪ B.

(A ∪ B)' = ξ - A ∪ B

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

(A ∪ B)' = {1, 2, 3} ...............(1)

(A ∪ B)' = {1, 2, 3}

(ii) A' ∩ B'

A' - contains all the elements in universal set which are not in A.

A' = ξ - A

A' = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {5, 6, 7, 8, 9, 10}

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

B' - contains all the elements in universal set which are not in B.

B' = ξ - B

B' = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {4, 5}

B' = {1, 2, 3, 6, 7, 8, 9, 10}

A' ∩ B' - contains all the common elements in set A and B.

A' ∩ B' = {1, 2, 3, 4} ∩ {1, 2, 3, 6, 7, 8, 9, 10}

A' ∩ B' = {1, 2, 3} ...............(2)

A' ∩ B' = {1, 2, 3}

From (1) and (2) , we can see that

Yes, (A ∪ B)' and A' ∩ B' are equal.

7Test Yourself

Write the elements of the set

{x : x = 3y - 1, y ∈ N and 8 < y ≤ 12}.

Show SolutionHide Solution↓

{x : x = 3y - 1, y ∈ N and 8 < y ≤ 12}

y = 9

x = 3y - 1 = 3 x 9 - 1

x = 27 - 1 = 26

y = 10

x = 3y - 1 = 3 x 10 - 1

x = 30 - 1 = 29

y = 11

x = 3y - 1 = 3 x 11 - 1

x = 33 - 1 = 32

y = 12

x = 3y - 1 = 3 x 12 - 1

x = 36 - 1 = 35

Hence, the elements are 26, 29, 32 and 35

8Test Yourself

If universal set = {x : x ∈ Z, -2 ≤ x < 4},

A = {x : -1 ≤ x < 3}, B = {x : 0 < x < 4} and

C = {x : - 2 ≤ x ≤ 0}; show that :

A - (B ∪ C) = (A - B) ∩ (A - C)

Show SolutionHide Solution↓

Universal set = {x : x ∈ Z, -2 ≤ x < 4}

Universal set = {-2, -1, 0, 1, 2, 3}

A = {x : -1 ≤ x < 3}

A = {-1, 0, 1, 2}

B = {x : 0 < x < 4}

B = {1, 2, 3}

C = {x : - 2 ≤ x ≤ 0}

C = {-2, -1, 0}

To prove:

A - (B ∪ C) = (A - B) ∩ (A - C)

Taking LHS:

A - (B ∪ C)

B ∪ C - contains all the elements in set B and C.

B ∪ C = {1, 2, 3} ∪ {-2, -1, 0}

B ∪ C = {-2, -1, 0, 1, 2, 3}

A - (B ∪ C) - contains all the elements in set A but not in B ∪ C.

A - (B ∪ C) = {-1, 0, 1, 2} - {-2, -1, 0, 1, 2, 3}

A - (B ∪ C) = { }

Taking RHS:

(A - B) ∩ (A - C)

A - B - contains all the elements in set A but not in B.

A - B = {-1, 0, 1, 2} - {1, 2, 3}

A - B = {-1, 0}

A - C - contains all the elements in set A but not in C.

A - C = {-1, 0, 1, 2} - {-2, -1, 0}

A - C = {1, 2}

(A - B) ∩ (A - C) - contains all the common elements in set (A - B) and (A - C).

(A - B) ∩ (A - C) = {-1, 0} ∩ {1, 2}

(A - B) ∩ (A - C) = { }

∴ LHS = RHS

∴ A - (B ∪ C) = (A - B) ∩ (A - C)

9Test Yourself

Let set A = {x : x ∈ Z and x² - 9 = 0} and set B = {x : x ∈ W and x² - 16 < 0}; then find :

(i) A ∪ B

(ii) B ∩ A

Show SolutionHide Solution↓

Set A = {x : x ∈ Z and x² - 9 = 0}

x² - 9 = 0

x² = 9

x = $\sqrt{9}$

x = 3 or -3

Set A = {-3, 3}

Set B = {x : x ∈ W and x² - 16 < 0}

x² - 16 < 0

x² < 16

x < $\sqrt{16}$

x < 4

Set B = {0, 1, 2, 3}

(i) A ∪ B

A ∪ B - contains all the elements of set A and B.

A ∪ B = {-3, 3} ∪ {0, 1, 2, 3}

A ∪ B = {-3, 0, 1, 2, 3}

(ii) B ∩ A

B ∩ A - contains all the common elements in set B and A.

B ∩ A = {0, 1, 2, 3} ∩ {-3, 3}

B ∩ A = {3}

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