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(i) True 

Explanation: We have A ⊂ B since A is a subset of B then all elements of A should be in B. 

Let A = {1,2} and B = {1,2,3} 

Let x=4∉B 

Also we observe that 4∉A. 

Hence, If A ⊂ B and x∉B than x ∉A. 

(ii) True

Explanation: We have, A ⊆ ϕ

Now, A is a subset of null set , this implies A is also an empty set.

⇒ A =ϕ 

(iii) False 

Explanation: Let A = {a}, B = {{a}, b}

here , A ϵ B

Now, let C = {{a}, b, c}. 

Since, {a},b is in B and also in C thus, B ⊂ C. 

But, A ={a} and {a} is an element of C, since the element of a set cannot be a subset of a set. 

Hence,A ⊄ C. 

(iv) False

Explanation: Let A = {a},B = {a, b} and C = {{a, b}, c}. 

Then,A⊂ B and B ϵC. But, A ∉ C since {a} is not an element of C. 

(v) False. 

Explanation: Let A = {a}, B = {b, c} and C = {a, c}. 

Since a ∈ A and a ∉ B.Then, A ⊄ B 

Now, b ∈ B and b ∉ C ⇒ B ⊄C. 

But, A ⊂ C since, a ∈ A and a ∈ C. 

(vi) False. 

Explanation: Let A = {x}, B = {{x}, y} 

Now, x ϵ A and {x} is an element of B ⇒ A ϵ B 

But, x is not an element of B. Thus, x∉B.

1) {11, 2, 3, 4, 5}

2) {105, 110, 115, 120}

3) {1³, 2³, 3³, 4³, 5³} 

{1, 8, 27, 64, 125}

4) Ramanujan’s number is 1729 

{1, 2, 7, 9}

(i) False

Since, 1 is not an element of A.

(ii) True

Hence, {1,2,3} ∈ A this is correct form.

(iii) True.

Hence, {6, 7, 8} ∈ A.

(iv) True

Since, {{4, 5}} is a subset of A.

(v) False

Since, Φ is a subset of A, not an element of A.

(vi) True

Hence, Φ is a subset of every set, therefore it is a subset of A.

(i) In this a isn’t subset of given set but belongs to the given set. 

∴ The correct form would be 

a ϵ {a,b,c} 

(ii) In this {a} is subset of {a,b,c} 

∴ The correct form would be 

{a} ⊂ {a,b,c} 

(iii) In this a is not the element of the set. 

∴ The correct form would be 

{a} ϵ {{a},b} 

(iv) In this {a} is not athe subset of given set 

∴ The correct form would be 

{a} ϵ {{a},b} 

(v) {b,c} is not a subset of given set. But it belongs to the given set. 

∴ The correct form would be 

{b,c} ϵ {a,{b,c}} 

(vi) {a,b} is not a subset of given set 

∴ The correct form would be 

{a,b} {a,{b,c}} 

(vii) ϕ does not belong to given set but it is subset 

∴ The correct form would be 

ϕ ⊂ {a,b} 

(viii) True ϕ is subset of every set 

(ix) X + 3 = 3 

X = 0 

{0} 

It is not ϕ 

∴ The correct form would be 

{x:x + 3 = 3} = {0}

(i) True 

1 belongs to the given set as it is present in it. 

(ii) False 

{a} ⊂ {b,c,a} 

This is the write form to write it or a ϵ {b,c,a} 

(iii) False 

{a} ⊂ {b,c,a} 

This is the write form to write it or a ϵ {b,c,a} 

(iv) True 

We do not repeat same elements in a given set. 

∴ The given set can be written as {a,b} 

(v) False 

X+8 = 8 

i.e. x = 0 

{0} It is not a null 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.

A ⊂ B 

Explanation: we have, A = set of real numbers and B = set of complex numbers, a combination of the real and imaginary number in the form of a+ib, where a and b are real, and i is imaginary. 

Since, any real number can be expressed as complex number, A ⊂ B.

A = {….-2, -1, 0, 1, 2, 3….} 

B = {-∞, ……..0, ……∞ } 

A ⊂ B as integers are contained in rational numbers.

The all the possible subsets of A are: 

ϕ, {5}, {6}, {5,6}

Number of elements in power set of A = 2ⁿ 

= 2³ = 8 

here, n = 3 

Or 

The number of elements in power set of A = {a,b,c} [n(A) = 3]

Correct option is B) {Rectangles}

Here, A = {D, E, H, I, L} and B = {D, L, O}

(i)  A ∪ B = {D, E, H, I, L} and B ⋃ {D, L, O} 

= {D, E, H, I, L, O}

(ii)  A ∩ B {D, E, H, I, L} ∩ B ⋃ {D, L, O} 

= {D, L}

(iii)  A – B = {D, E, H, I, L} − {D, L, O} 

= {E, H, I}

Correct option is C) A ∩ Φ = A

A ∪ B = {x: x ϵ A or x ϵ B} 

= {2, 3, 4, 5, 6, 7, 8} 

(A∪B)’ means Complement of (A∪B) with respect to universal set U. 

So, 

(A∪B)’ = U – (A∪B)’ 

U – ( A∪B)’ is defined as {x ϵ U : x ∉ (A∪B)’} 

U = {1, 2, 3, 4, 5, 6, 7, 8, 9} 

(A∪B)’ = {2, 3, 4, 5, 6, 7, 8} 

U – ( A∪B)’ = {1, 9} 

Now 

A’ means Complement of A with respect to universal set U. 

So, 

A’ = U – A 

U – A is defined as {x ϵ U : x ∉ A} 

U = {1, 2, 3, 4, 5, 6, 7, 8, 9} 

A = {2, 4, 6, 8} 

A’ = {1, 3, 5, 7, 9} 

B’ means Complement of B with respect to universal set U. 

So, 

B’ = U – B 

U – B is defined as {x ϵ U : x ∉ B} 

U = {1, 2, 3, 4, 5, 6, 7, 8, 9} 

B = {2, 3, 5, 7}. 

B’ = {1, 4, 6, 8, 9} 

A’ ∩ B’ = = {x:x ϵ A’ and x ϵ C’}. 

= {1, 9} 

Hence verified.

LHS = A ∩ (A∪B)’ 

Using De-Morgan’s law (A∪B)’ = (A’ ∩ B’) 

⇒ LHS = A ∩ (A’ ∩ B’) 

⇒ LHS = (A ∩ A’) ∩ (A ∩ B’) 

We know that A ∩ A’ = ϕ 

⇒ LHS = ϕ ∩ (A ∩ B’) 

We know that intersection of null set with any set is null set only

Let (A ∩ B’) be any set X hence 

⇒ LHS = ϕ ∩ X 

⇒ LHS = ϕ 

⇒ LHS = RHS 

Hence proved

Let , 

Total number of People n(P) = 500. 

People who watch Basketball n(B) =115. 

People who watch Football n(F) = 285. 

People who watch Hockey n(H) = 195. 

People who watch Basketball and Hockey n(B ∩ H) = 50 

People who watch Football and Hockey n(H ∩ F) = 70 

People who watch Basketball and Football n(B ∩ F) = 45 

People who do not watch any games. n(H∪B∪F)= 50 

Now, 

n(H∪B∪F)’ = n(P) – n(H∪B∪F) 

50 = 500–( n(H)+n(B)+n(F) – n (H ∩ B)– n (H ∩ F)– n (B ∩ F)+ n (H ∩B ∩ F)) 

50 = 500–(285+195+115–70–50–45 +n (H ∩B ∩ F)) 

50 = 500–430 + n (H ∩B ∩ F)) 

n (H ∩B ∩ F) = 70–50 n 

(H ∩B ∩ F)) = 20 

∴ 20 People watch all three games. 

Number of people who only watch football 

= 285–(50+20+25) 

= 285–95 

= 190. 

Number of people who only watch Hockey 

= 195–(50+20+30) 

= 195–100 

= 95. 

Number of people who only watch Basketball 

= 115–(25+20+30) 

= 115–75 

= 40. 

Number of people who watch exactly one of the three games 

As the sets are pairwise disjoint we can write 

= number of people who watch either football only or hockey only or Basketball only 

=190+95+40 

=325 

∴ 325 people watch exactly one of the three games.

Correct option is C) (B ∩ C) – A

Correct option is A) element

Correct option is D) {6, 7, 8, 9, 10, 11}

Correct option is B) Unique

Perfect squares are: 

0² = 0 

1² = 1 

2² = 4 

3² = 9 

4² = 16 

5² = 25 

6² = 36 

7² = 49 

8² = 64 > 50 

The elements which are perfect square and x < 50 are 0, 1, 2, 3, 4, 5, 6, 7

Prime number = Those number which is divisible by 1 and the number itself.

Prime numbers are 2, 3, 5, 7, 11, 13,… 

The numbers 80 < x < 100 are 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100 

The elements which are prime and lies between 80 and 100 are 83, 89, 97 

So, G = {83, 89, 97}

Correct option is D) 5

Correct option is C) 4 : 3

Correct option is B) 27

There are 11 letters in the word MATHEMATICS, out of which M, A and T are repeated. 

So, F = {M, A, T, H, E, I, C, S} Q. 3 G.

Integers = …, -4, -3, -2, -1, 0, 1, 2, 3, 4, … 

x = -4, x² = (-4)² = 16 > 9 

x = -3, x² = (-3)² = 9 

x = -2, x² = (-2)² = 4 

x = -1, x² = (-1)² = 1 x = 0, 

x² = (0)² = 0

x = 1, x² = (1)² = 1 

x = 2, x² = (2)² = 4 

x = 3, x² = (3)² = 9 

x = 4, x² = (4)² = 16 

The elements of this set are -3, -2, -1, 0, 1, 2, 3 

So, D = {-3, -2, -1, 0, 1, 2, 3}

Prime number = Those number which is divisible by 1 and the number itself. 

Prime numbers are 2, 3, 5, 7, 11, 13, … 

Divisor of 42: 

42 = 1 × 42 

42 = 2 × 21 

42 = 3 × 14 

42 = 6 × 7 

So, divisors of 42 are 1, 2, 3, 6, 7, 14, 21, 42 

The elements which are prime and divisor of 42 are 2, 3, 7 

So, E = {2, 3, 7}

Correct option is (D) 16

Correct option is (C) {2, 3}

Correct option is B) {x : x ∈ A or x ∈ B}

Correct option is B) 2ⁿ – 1