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Given; A = {a, b, c, d, e, f}, B = {c, e, g, h} and C = {a, e, m, n} 

(i) A ∪ B = {a, b, c, d, e, f, g, h} 

(ii) B ∪ C = {a, c, e, g, h, m, n} 

(iii) B ∪ C = {a, c, e, g, h, m, n} 

(iv) C ∩ A = {a, e} 

(vi) A ∩ B = {c, e}

Note: In general X ∪ Y = {a: a ϵ X or a ϵ Y} 

X ∩ Y = {a:a ϵ X and a ϵ Y}. 

i. A ∪ B = {x: x ϵ A or x ϵ B} 

= {1, 2, 3, 4, 5, 6, 7, 8} 

ii. A ∪ C = {x: x ϵ A or x ϵ C} 

= {1, 2, 3, 4, 5, 7, 8, 9, 10, 11} 

iii. B ∪ C = {x: x ϵ B or x ϵ C} 

= {4, 5, 6, 7, 8, 9, 10, 11} 

iv. B ∪ D = {x: x ϵ B or x ϵ D} 

= {4, 5, 6, 7, 8, 10, 11, 12, 13, 14} 

v. A ∪ B = {x: x ϵ A or x ϵ B} 

= {1, 2, 3, 4, 5, 6, 7, 8} A ∪ B ∪ C 

= {x: x ϵ A ∪ B or x ϵ C} 

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

vi. A ∪ B = {x: x ϵ A or x ϵ B} 

= {1, 2, 3, 4, 5, 6, 7, 8} A ∪ B ∪ D 

= {x: x ϵ A ∪ B or x ϵ D} 

= {1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14} 

vii. B ∪ C = {x: x ϵ B or x ϵ C} 

= {4, 5, 6, 7, 8, 9, 10, 11} 

B ∪ C ∪ D = {x: x ϵ B ∪ C or x ϵ D} 

= {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14} 

viii. B ∪ C = {x: x ϵ B or x ϵ C} 

= {4, 5, 6, 7, 8, 9, 10, 11} 

A ∩ B ∪ C = {x:x ϵ A and x ϵ B ∪ C}. 

= {4, 5} 

ix. (A ∩ B) = {x:x ϵ A and x ϵ B}. 

= {4, 5} 

(B ∩ C) = {x:x ϵ B and x ϵ C} 

= {7, 8} 

(A ∩ B) ∩ (B ∩ C) = {x:x ϵ (A ∩ B) and x ϵ (B ∩ C)}. 

= ϕ 

x. A ∪ D = {x: x ϵ A or x ϵ D} 

= {1, 2, 3, 4, 5, 10, 11, 12, 13, 14}. 

B ∪ C = {x: x ϵ B or x ϵ C} 

= {4, 5, 6, 7, 8, 9, 10, 11} 

(A ∪ D) ∩ (B ∪ C) 

= {x:x ϵ (A ∪ D) and x ϵ (B ∪ C)}. 

= {4, 5, 10, 11}.

Given; A = {1, 2, 3, 4, 5}, B = {4, 5, 6, 7, 8} and C = {10, 11, 12, 13, 14}

(i) A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8}

(ii) B ∪ C = {4, 5, 6, 7, 8, 10, 11, 12, 13, 14} 

(iii) A ∪ C = {1, 2, 3, 4, 5, 10, 11, 12, 13, 14} 

(iv) B ∪ D 

(v) (A ∪ B) ∪ C = {1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14} 

(vi) (A ∪B) ∩C = Φ or {} 

(vii) (A ∩ B) ∪D = 

(viii) (A ∩ B) ∪ (B ∩ C) = Φ or {} 

(ix) (A ∩ C) ∩ (C ∪ D =

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.

(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

(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

(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

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

( ii ) B = { x : x is an integer; -1/2 <n<9/2}

It can be seen that  -1/2= -0.5 and 9/2= 4.5

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

(iii) C = {x: x is an integer; }

It can be seen that

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

0² = 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}

(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}

B. 7, 4

Given: Two finite sets have m and n elements.

To find: value of m and n

Formula used:

The number of subsets of a set containing x elements is given by 2^x

According to question:

2^m – 2ⁿ = 112

⇒ 2ⁿ (2^m-n – 1) = 16 × 7

⇒ 2ⁿ (2^m-n – 1) = 2⁴ × 7

On comparing:

2ⁿ = 2⁴ and 2^m-n – 1 = 7

⇒ n = 4 and 2^m-n = 8

⇒ 2^m-n = 2³

⇒ m – n = 3

⇒ m – 4 = 3

⇒ m = 7

Hence, value of m and n is 7 and 4 respectively

In this question we cannot find n(A⋂B) as we haven’t required values and information about their union or about universal set. 

So, 

correct answer is D

Here, 

n(A) = 5 

⇒n(P(A) ) = 2⁵ = 32 

So,

A has total 32 subsets but one of them is A itself. 

∴ Proper subsets of A are 32 -1 = 31.

We know x≠x is false for any x. 

∴ the set builder form for null set is {x:x≠x}.

There are two methods of describing a set. 

1. Roster or tabular form: In the roster form, we list all the members of the set within brackets { } and separate them by commas. 

2. Set-Builder form: In the set-builder form, we list the property or properties satisfied by all the elements of the set. 

• N: The set of all-natural numbers. 
• Z: The set of all integers. 
• Q: The set of all rational numbers. 
• M: The set of real numbers. 
• Z⁺: The set of positive integers. 
• Q⁺: Set of positive rational numbers. 
• R⁺: The set of positive real numbers.

A collection of well-defined objects is called a set. The objects in a set are called its members or elements. We denote sets by capital letters A, B, C, X. Y, Z, etc. If ‘a’ is an element of a set A, we write, a ∈ A, which means that a belongs to A or that a is an element of If ‘a ’ does not belong to A, we write, a ∉ A.

Correct option is  B) {1}

Correct option is  C) {4}

Given sets are

A = {1, 2, 3, 4, 5, 6} and B = {2, 4, 6, 8, 10} 

A ∩ B = {1, 2, 3, 4, 5, 6} ∩ {2, 4, 6, 8, 10} = {2, 4, 6}

Sujata:

S = {x | x = 2n- 1, n ∈ N, x < 9}

Hameed:

: f H = {x | x = 2n, n ∈ N, x < 9} 

Mukta: 

M = {x | x = n , n ∈ N, x ≤ 9} 

Nandini:

 N = {x | x ∈ N, x ≤ 9} 

Joseph: 

J = {x | x is a prime number between 1 and 9}

i) A = {51, 52, 53, ……. , 98, 99} 

ii) B = {+2, -2} 

iii) D = {L, O, Y, A}

n(A ∪ B) = n(A) + n(B) – n(A ∩ B) 

= 7 + 13 – 4 

n(A ∪ B) = 16

Correct option is D) 15

i) x ∉ A 

ii) d ∈ B 

iii) 1 ∈ N 

iv) 8 ∉ P

i) B = {1, 2, 3, 4, 5} 

ii) C = {17, 26, 35, 44, 53, 62, 71} 

iii) D = {5, 3} 

iv) E = {B, E, T, R}

Correct option is B) 51

Given sets are 

A = {1, 2, 3, 4} and B = {1, 2, 3, 4, 5, 6, 7, 8} 

A ∪ B = {1, 2, 3, 4} ∪ {1, 2, 3, 4, 5, 6, 7, 8} = {1, 2, 3, 4, 5, 6, 7, 8} = B 

A ∩ B = {1, 2, 3, 4} ∩ {1, 2, 3, 4, 5, 6, 7, 8} = {1, 2, 3, 4} = A 

If A ⊂ B, then A ∪ B = B and A ∩ B = A

i) False 

ii) False 

iii) True 

iv) False

Correct option is  C) 13

Correct option is  A) {1}

Correct option is C) {x / x ∈ μ, x ∈ A}

Correct option is  B) {3, 7, 8}

Correct option is  B) A

Correct option is  D) {2, 4}

Correct option is A) (A ∪ B) ∪ (A ∪ \(\bar{B}\)) = A ∪ Φ

Correct option is B) A – (B ∩ C) = (A – B) ∪ (A – C)

Correct option is A) A – (B ∩ C) = (A – B) ∪ (A – C)

Correct option is A) 1, 3

Correct option is C) 1, 2, 3, 4, 5, 6, 7

Correct option is C) 7

Correct option is C) 4

Correct option is B) 1, 4, 6, 7

Correct option is C) 1

Correct option is B) 3

Correct option is D) 1, 3, 4

False

Since 6 is divisible by both 3 and 2.

Thus, R ∩ S ≠ φ

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

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

If A is a finite set containing n element, then number of subsets of A is 2ⁿ