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Concept:

Selection r things out of n things is denoted by nCr 

nCr = \(\rm \frac {n!}{(n-r)!r!}\)

Calculation:

Set A has 4 elements

Set B has 2 elements

Total number of element in (A × B) = 4 × 2 = 8

Total number of subsets of (A × B) = 2⁸ = 256

Number of subsets having 0 element = ⁸​C₀ ​= 1

Number of subsets having 1 element = ⁸​C₁​ = 8

Number of subsets having 2 elements = ⁸​C₂ ​= (8 × 7)/2 ​= 28

Number of subsets having at least 3 elements

= 256 – 28 – 8 – 1 = 219

Hence, option (3) is correct.

Concept: 

Set theory:

• A ∪ B means set of all the values in the set A and B.
• A ∩ B is the set of common elements of A and B.
• Subset (⊂) is the set such that all the elements of the subset are in the set from which the subset is taken from.
• Number of elements in set A = n(A)
• n(A ∪ B) = n(A) + n(B) - n(A∩B) = n(A) + n(B - A) = n(B) + n(A - B)
• A' = U - A, where U is a universal set given and A is any set
• n(A') = n(U) - n(A)
• A - B = set values of set A not in set B
• n(A - B) = n(A) - n(A ∩ B)

Calculation:

Given: U = {x : x is the natural number less than 20}.

A = {0, 2, 7, 15, 19, 20} and B = {5, 7, 9, 11, 13, 17, 20}

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

A = {0, 2, 7, 15, 19, 20}

B = {5, 7, 9, 11, 13, 17, 20}

(A - B) = {0, 2, 15, 19}

(A - B) ∩ U = {2, 15, 19}

It is very clearly visible (A - B) ∩ U is a subset of U and A

Concept:

(A ⋂ B) are the elements present in both the sets (A and B)

Calculation:

Here, A = {x: x is a natural number and 1 < x ≤ 4},

A = {2, 3, 4}

B = {x: x is a natural number and 4 < x ≤ 7}

B = {5, 6, 7}

Here, not a single element is common in both sets.

So, (A ∩ B) = ϕ 

Hence, option (4) is correct.

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} 

Concept: 

Set theory:

• A ∪ B means set of all the values in the set A and B.
• A ∩ B is the set of common elements of A and B.
• Subset (⊂) is the set such that all the elements of the subset are in the set from which the subset is taken from.
• Number of elements in set A = n(A)
• n(A∪B) = n(A) + n(B) - n(A∩B) = n(A) + n(B - A) = n(B) + n(A - B)
• A' = U - A, where U is a universal set given and A is any set
• n(A') = n(U) - n(A)
• A - B = set values of set A not in set B
• n(A - B) = n(A) - n(A∩B)

Calculation:

Let Set A of people who like apples and Set B of people who like orange and U is set of all people

Given n(A) = 24, n(B) = 18 and n(U) = 50

Also given n(A ∩ B) = 5

n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

⇒ n(A ∪ B) = 24 +18 - 5 

⇒ n(A ∪ B) = 37 

Now, if Set C is of people who likes only grapes

n(C) = n(U) - n(A ∪ B)

n(C) = 50 - 37 = 13

∴ People who likes only grapes n(C) = 13

(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} 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} 

(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} 

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;x² < 4 } 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} } 

(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). 

(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. 

(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 

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 

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. 

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. 

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. 

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  

Concept: 

Set theory:

• A ∪ B means set of all the values in the set A and B.
• A ∩ B is the set of common elements of A and B.
• Subset (⊂) is the set such that all the elements of the subset are in the set from which the subset is taken from.
• Number of elements in set A = n(A)
• n(A ∪ B) = n(A) + n(B) - n(A ∩ B) = n(A) + n(B - A) = n(B) + n(A - B)
• A' = U - A, where U is a universal set given and A is any set
• n(A') = n(U) - n(A)
• A - B = set values of set A not in set B
• n(A - B) = n(A) - n(A ∩ B)

Calculation:

Let Set A of people who like piano and Set B of people who like guitar and U is set of all people

Given n(A') = 15 and n(B - A) = 12

As n(A') = n(U) - n(A)

⇒ 15 = 30 - n(A)

⇒ n(A) = 15 

Now, Set of people who likes at least one A∪B

n(A ∪ B) = n(A) + n(B - A)

n(A ∪ B) = 15 + 12 = 27

∴ People who do not like either of the two N = n(U) - n(A ∪ B) 

N = 30 - 27 = 3

Concept:

Set theory:

• A ∪ B means set of all the values in the set A and B.
• A ∩ B is the set of common elements of A and B.
• Subset (⊂) is the set such that all the elements of the subset are in the set from which the subset is taken from.

Calculation:

Taking any arbitrary example

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

Now, A ∪ B = {1, 3}

Case 1: If x = 3 , then x ∈ A, and A ∉ B, but x ∈ B (option 1 false)

Case 2: If x = 3 , then x ∉ A, and B ⊂ A, then x ∉  B (option 2 true)

Case 3:  A ⊂ C and B ⊂ C, but A ≠ B (option 3 false)

Case 4: A ⊂ C and B ⊂ C, but (A ∪ B) ≠  C (option 4 false)

Concept: 

Set theory:

• A ∪ B means set of all the values in the set A and B.
• A ∩ B is the set of common elements in A and B.
• Subset (⊂) is the set such that all the elements of the subset are in the set from which the subset is taken from.
• Number of elements in set A = n(A)
• n(A ∪ B) = n(A) + n(B) - n(A ∩ B)
• A' = U - A, where U is a universal set given and A is any set
• A - B = set values of set A not in set B

Calculation:

Set A of students who like mathematics, so n(A) = 20

Set B of students who like science, so n(B) = 15

Given n(A ∩ B) = 5

Set of students who like at least one of the 2 given subjects is A∪B 

n(A ∪ B) = n(A) + n(B) - n(A ∩ B)

n(A ∪ B) = 20 + 15 - 5 = 30

Students who do not like either of the subjects = Total students - n(A ∪ B)

N = 50 - 30 = 20

Concept:

The cardinality of a set is a measure of a set's size, meaning the number of elements in the set.

Calculation:

Here, A = {a, b, c, d}, B = {d, f, g, h}

⇒(A ∪ B) = {a, b, c, d, f, g, h}

Cardinality of (A ∪ B) = number of elements in (A ∪ B)

= 7

Hence, option (2) is correct.

Concept:

Let A, and B be any two sets

(A ∩ B) = {x, \(\rm x \in A \;\;and \;\; x \in B\)}

A × B = {(x, y), \(\rm x \in A \;\;and \;\; y \in B\)}

Calculations:

Given, A = {2, 3}, B = {4, 5}, C = {5, 6}

⇒(B ∩ C) = {5}

A × (B ∩ C) = {(2, 5), (3, 5)}

⇒ The number of elements in A × (B ∩ C) = 2

(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 

R: set of real numbers 

Q: set of rational numbers 

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

(i) X – Y = {a, c} 

(ii) Y – X = {f, g} 

(iii) X ∩ Y = {b, d} 

(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} 

(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} 

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. 

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} 

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)] = 20 = 1 

Hence, P(A) has one element. 

(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} 

(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. So, A ∩ B = Φ 

(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] 

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}

(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} 

(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. 

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

(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. 

Here, A = {a, b} and B = {a, b, c} 

Yes, A ⊂ B. 

A ∪ B = {a, b, c} = B 

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]  10 = B ∩ A = A ∩ B …………………………………………………………… (2) Hence, 

from (1) and (2), we obtain A = B. 

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] 

(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)  9 = A 

{from (1)}