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Let A be the set with n elements. 

Each member of A has two possibilities either present or absent. 

⇒ Total possible subsets of A are 2 × 2 × 2 ×…n times = 2ⁿ

Here,

A = {(x,y):y = e^x,xϵR} and

B = {(x,y)y = e^-x,xϵR} 

A⋂B = {(x,y):y = e^x ,xϵR} and {(x,y):y=e^-x ,xϵR}

= {(x,y):y = e^x and y = e^-x xϵR} 

= {(x,y): e^x = e^-x xϵR}

= {(x,y):e^x = \(\frac{1}{e^x}\),xϵR}

= {(x,y):e^2x = 1,xϵR} 

= {(0,1)} (∵e⁰ = 1  )

Here, 

n(A) =115 , n(B) = 326 , n(A∖B) = 47 

n(A∖B) = n(A) - n(A⋂B) 

∴ n(A⋂B) = n(A) - n(A∖B) 

= 115 - 47 

= 68 

Now, 

n(A⋃B) = n(A) + n(B) - n(A⋂B) 

= 115 + 326 - 68 

= 373

A is not correct answer because 4 is not in A. 

B is not correct answer because {4} is not in A. 

C is not correct answer because 2 is in B but not in A. 

So, 

D is correct answer.

P = (P : P is a letter in the word “PERMANENT”} or P = {p, e, r, m, a, n, t) n (P) = 7

We first show that A ∪ (B ∩ C) ⊂ (A ∪ B) ∩ (A ∪ C) 

Let x ∈ A ∪ (B ∩ C). Then 

x ∈ A or x ∈ B ∩ C 

⇒ x ∈ A or (x ∈ B and x ∈ C) 

⇒ (x ∈ A or x ∈ B) and (x ∈ A or x ∈ C) 

⇒ (x ∈ A ∪ B) and (x ∈ A ∪ C) 

⇒ x ∈ (A ∪ B) ∩ (A ∪ C) 

Thus, A ∪ (B ∩ C) ⊂ (A ∪ B) ∩ (A ∪ C) ... (1) 

Now we will show that (A ∪ B) ∩ (A ∪ C) ⊂ (A ∪ C) 

Let x ∈ (A ∪ B) ∩ (A ∪ C) 

⇒ x ∈ A ∪ B and x ∈ A ∪ C 

⇒ (x ∈ A or x ∈ B) and (x ∈ A or x ∈ C) 

⇒ x ∈ A or (x ∈ B and x ∈ C) 

⇒ x ∈ A or (x ∈ B ∩ C) 

⇒ x ∈ A ∪ (B ∩ C) 

Thus, (A ∪ B) ∩ (A ∪ C) ⊂ A ∪ (B ∩ C) ... (2) 

So, from (1) and (2), we have 

A ∩ (B ∪ C) = (A ∪ B) ∩ (A ∪ C)

False

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

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

Every element of M and B is same

Hence, M = B and B ⊂ M

So, the given statement is false.

True

Every set is subset of itself.

Hence, A ⊂ A is true

(i) A∩B will contain the common elements of A and B 

A∩B = {c, e} 

(ii) AU(B∩C) 

B∩C = {c, d, g} 

AU(B∩C) = {a, b, c, d, e, f, g} 

(iii) A - B implies the set of all elements in A that are not in B 

A - B = {a, b, f} 

(iv) B - A implies the set of all elements in B that are not in A 

B - A = {d, g} 

(v) A - (B∩C) denotes elements of A that are not in B∩C 

A - (B∩C) = {a, b, e, f} 

(vi) (B - C)U(C - B) implies the union of sets B - C and C – B 

B - C = {d, e} 

C - B = {b, f} 

(B - C)U(C - B) = {b, d, e, f}

Total number of People n(P) = 100. 

People who Read Magazine A n(A) =28. 

People who Read Magazine B n(B) = 30. 

People who Read Magazine C n(C) = 42. 

People who Read Magazine A and B n(A ∩ B) = 8 

People who Read Magazine B and C n(B ∩ C) = 10 

People who v Read Magazine A and C n(A ∩ C) = 5 

n(A∩B∩C)= 3 

Now, 

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

= 28+30+42–8–10–5+3 

= 100–20 

= 80 

People who read none of the three magazines 

= n(A∪B∪C)’

 = n(P) – n(A∪B∪C) 

= 100 –80 

= 20.

 ∴ People who read no magazine = 20

(i) False

(ii) False

(iii) True

(iv) True (each set is the empty set)

(v) True

(vi) False (6 is an even natural number which is divisible by 3)

(vii) True (no positive number can be less than 0)

(viii) False

(i) Finite

(ii) Infinite

(iii) Finite

(iv) Finite

(v) Infinite

(vi) Finite

Here, 

Question is (A')' = ?

Now, 

A'=U∖A 

⇒(A' )' = (U∖A)' = U'∖A' 

⇒(A' )' = U'∖(U∖A) 

⇒(A' )' = ϕ∖U+A 

⇒(A' )' = A

A∖B is the set of elements which are in A but not in B 

⇒A∖B is the set of elements which are in A and in B' 

⇒A∖B is the set of elements of A⋂B'

∴ A/B = A⋂B'

(i) (-4, 6]

(ii) (-12, -10)

(iii) (0, 7)

(iv) [3, 4]

(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) (–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) A = {x : x ϵR, –2 < x < 3} 

(ii) B = {x : x ϵ R, 4 ≤ x ≤ 10} 

(iii) C= {x : x ϵ R, –1 ≤ x < 8} 

(iv) D = {x : x ϵ R, 4 < x ≤ 9} 

(v) E = {x : x ϵ R, –10 ≤ x < 0} 

(vi) F = {x : x ϵ R, 0 < x ≤ 5}

(i) True 

Explanation: we have, A = {3, {4, 5}, 6} 

Let {4,5} = x 

Now, A = {3, x, 6} 

4,5 is not in A, {4,5} is an element of A and element cannot be subset of set,thus {4, 5} ⊄A. 

(ii) True 

Explanation: we have, A = {3, {4, 5}, 6} 

Let {4,5} = x 

Now, A = {3, x, 6} 

Now, x is in A. 

So, x ∈ A. 

Thus, {4, 5} ϵ A 

(iii) True

Explanation: {4,5} is an element of set {{4,5}}. 

Let {4,5} = x 

{{4,5}} = {x} 

we have, A = {3, {4, 5}, 6} 

Now, A = {3, x, 6} 

So, x is in {x} and x is also in A. 

So , {x} is a subset of A. 

Hence, {{4, 5}} ⊆A 

(iv) False 

Explanation: 4 is not an element of A.

(v) True 

Explanation: 3 is in {3} and also 3 is in A. 

(vi) False 

Explanation: ϕ is an element in { ϕ} but not in A. 

Thus, {ϕ} ⊄ A 

(vii) True 

Explanation: ϕ is a subset of every set. 

(viii) False 

Explanation: we have, A = {3, {4, 5}, 6} 

Let {4,5} = x 

Now, A = {3, x, 6} 

4,5 is in {3,4,5} but not in A, thus {3,4, 5} ⊄A. 

(ix) True 

Explanation: 3,6 is in {3,6} and also in A, thus {3, 6} ⊆A.

The collection of all subsets of a set A is called the power set of A. It is denoted by P(A). 

Now, We know that ϕ is a subset of every set. So, ϕ is a subset of {a, b, c}. 

Also, {a},{b},{c},{a,b},{b,c},{a,c} are also subsets of {a, b, c} 

We know that every set is a subset of itself. So, {a, b, c} is a subset of {a, b, c}. 

Thus, the set {a, b, c} has, in all eight subsets, viz. ϕ , {a},{b},{c},{a,b},{b,c},{a,c},{a, b, c}. 

∴ P(A) = { ϕ , {a},{b},{c},{a,b},{b,c},{a,c},{a, b, c}} 

Now, n{P(A)} = 2^m , where m = n(A) = 3

⇒ n{P(A)} = 2³ = 8

Let {2,3} = x 

Now, A = {1,x} 

Subsets of A are ϕ , {1} , {x} , {1,x} 

⇒ Subsets of A are ϕ , {1} , {2,3} , {1, {2,3}} 

Now, n{P(A)} = 2^m , where m = n(A) = 2 

⇒ n{P(A)} = 2² = 4

We have, A = ϕ , i.e. A is a null set. 

Then, n(A)= 0 

∴ n{P(A)} = 2^m , where m = n(A) 

⇒ n{P(A)} = 2⁰ = 1. 

Thus, P(A) has one element.

Elements of A+B+C = {1,3,5,2,4,6,0,8} 

Thus, the universal set for A,B and C = {0,1,2,3,4,5,6,8}

We have A ⊆B , B⊆ C and C⊆ A 

Now , A is a subset of B and B is a subset of C, So A is a subset of C. 

Given that C⊆ A.

Hence, A = C.

Let A ⊆ ϕ

A is a subset of the null set , then A is also an empty set.

⇒ A =ϕ

Now, let A =ϕ 

⇒ A is an empty set. 

Since, every set is a subset of itself.

⇒ A ⊆ ϕ 

Hence, for any set A, A⊆ϕ ⇔ A =ϕ

Let A⊆ ϕ, 

If A is a subset of an empty set, then A is the empty set. 

∴ A = ϕ 

Now let A = ϕ, 

This means A is an empty set. 

As we know that every set is a subset of itself. 

∴ A ⊆ ϕ 

Thus, we have, 

A⊆ ϕ ⇔ A=ϕ 

Hence, Proved.

We have A ⊂ B, B ⊂ C and C ⊂ A

∴ A ⊂ B ⊂ C ⊂ A. 

Now, A is a subset of B and B is a subset of C, 

so 

A is a subset of C, i.e., A ⊂ C 

Also, C ⊂ A 

Hence C = A.

Empty set has zero element. 

∴ Power set of ϕ has 2⁰ = 1 element

B. 290

Given:

Total number of persons are 840

Persons who read Hindi and English are 450 and 300 respectively

Persons who read both are 200

To find: number of persons who read neither

Let U be the total number of persons, H and E be the number of persons who read Hindi and English respectively

n(U) = 840, n(H) = 450, n(E) = 300, n(H ∩ E) = 200

Number of persons who read either of them = n(H ∪ E)

= n(H) + n(E) – n(H ∩ E)

= 450 + 300 – 200

= 550

Number of persons who read neither

= Total – n(H ∪ E)

= 840 – 550

= 290

Hence, there are 290 persons who read neither Hindi nor English.

Answer is 

(i) →(c), (ii) → (a), (iii) → (d), (iv) → (b).

Correct option is C) Set of even prime numbers