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

(ii) B = {1, 2, 3, 4, 5} 

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

(iv) D = {2, 3, 5} (v) E = {T, R, I, G, 0, N, M, E, Y} 

(vi) F = {B, E, T, R} 

Given equation is x² + x – 2 = 0 

⇒ (x – 1)(x + 2) = 0 i.e., = 1,-2 

∴ G = {1,-2} 

(viii) H ={1, 2, 3, 4, 5, 6}

in roster form: 

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

Correct option is C) 16

Given A = {a, b, c, d} 

n(A) = 4 Number of subsets for a set, which is having ‘n’ elements is 2ⁿ. 

So n(A) = 4 

Number of subsets for A is 2⁴ = 16. 

X is a natural number also x = 2n 

∴ Roster form will be {2,4,6,8…..}. 

This an infinite set.

As the collection of all those students of your class whose ages exceed 15 years is known and can be counted, i.e. well – defined. 

∴, this is a set

As the collection of all persons of Kolkata whose assessed annual incomes exceed (say) Rs 20 lakh in the 4 financial years 2016-17 is known and well – defined. 

∴, this is a set.

As the collection of all interesting dramas written by Shakespeare is not well - defined because it depends on person interest. 

∴, this is not a set.

Correct option is A) The collection of all intelligent boys in a class

As the collection of all short boys of your class may vary to person to person. Maybe someone consider short boys of height less than 120 cm and maybe someone consider short boys of height less than 90 cm. 

Here, the set is not well – defined. 

∴, this is not a set

As the collection of all interesting books may vary to person to person. 

∴, this is not a set.

Months of the Year = Jan, Feb, March, April, May, June, July, Aug, Sep, Oct, Nov, Dec 

Months of the year whose names begin with the letter M are: 

• March 

• May 

As, the collection of all the months of the year whose names begin with the letter M is known and can be counted .i.e. well – defined. 

∴, this is a set.

As a collection of 11 best cricket players of India may vary from person to person. 

So, it is not well – defined. 

∴, this is not a set.

Correct option is  C) B ⊆ A

Integers = -3, -2, -1, 0, 1, 2, 3, … 

Integers less than 1 (x < 1) = …-4, -3, -2, -1, 0 

There are infinite integers which are less than 1. 

∴ the given set is infinite.

Given; A = {3, 5, 7, 9, 11}, B = {7, 9, 11, 13} and C = {11, 13, 15}, and D = {15, 17}

(i) A ∩ B = {7, 9, 11} 

(ii) A ∩ C = {11} 

(iii) B ∩ C = {11, 13} 

(iv) B ∩ D = Φ or {} 

(v) B ∩ (C ∪ D) = {11, 13} 

(vi) A ∩ (B ∪ C) = {7, 9, 11}

Integers = -3, -2, -1, 0, 1, 2, 3, … 

The integers lies between -15 and 15 are finite. 

∴ the given set is finite.

Given; A = {x : x ϵ N}, B = {x : x ϵ N and x is even), C = {x : x ϵ N and x is odd} and D = {x : x ϵ N and x is prime} 

(i) A ∩ B = {x : x ϵ N and x is even} 

(ii) A ∩ C = {x : x ϵ N and x is odd} 

(iii) A ∩ D = {x : x ϵ N and x is prime} 

(iv) B ∩ C = Φ or {} 

(v) B ∩ D = {2} [∵ 2 is the only even prime number] 

(vi) C ∩ D = {x : x ϵ N and x is prime and x ≠ 2}

Equal Sets = Two sets A and B are said to be equal if they have exactly the same elements & we write A = B 

We have, 

G = {-1, 1} 

and H = {x : x ϵ Z, x² – 1 = 0} 

Here, x ∈ Z and x² – 1 = 0 

The given equation can be solved as: 

x² – 1 = 0

⇒ x² = 1 

⇒ x = √1 

⇒ x = ± 1 

∴ x = -1 and 1 

∴ H = {-1, 1} 

⇒ G = H because elements of both the sets are equal.

(i) Equivalent Sets can have different or same elements but have the same amount of elements. 

We have,

A = {–1, –2, 0} and B = {1, 2, 3,} 

∴ A and B are equivalent sets because both have 3 elements in their set. 

(ii) Equivalent Sets can have different or same elements but have the same amount of elements.

We have,

C = {x : x ϵ N, x < 3}

Natural numbers = 1, 2, 3, 4, … 

Natural numbers less than 3 (x < 3) = 1, 2 

So, C = {1, 2} 

and D ={x : x ϵ W, x < 3} 

Whole numbers = 0, 1, 2, 3, 4, … 

Whole numbers less than 3 (x < 3) = 0, 1, 2 

So, D = {0, 1, 2} 

∴ C and D are not equivalent sets because their cardinality is not same.

(iii) Equivalent Sets can have different or same elements but have the same amount of elements. 

We have,

E = {a, e, i, o, u} and F = {p, q, r, s, t} 

∴ E and F are equivalent sets because both have 5 elements in their set. 

The given set is the set of all prime numbers and since the set of prime numbers is infinite. Hence, the given set is infinite.