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The given equation is:

x² + x – 12 = 0 

⇒ x² + 4x – 3x – 12 = 0 

⇒ x(x + 4) – 3(x + 4) = 0 

⇒ (x – 3)(x + 4) = 0 

⇒ x – 3 = 0 or x + 4 = 0 

⇒ x = 3 or x = -4 

therefore, the solution set of the given equation can be written in roaster form as {3, -4}

So, J = {3, -4}

Multiple of 5 are 5, 10, 15, 20, 25, 30, … 

So, 5² = 25 

10² = 100 

15² = 225 

20² = 400 

25² = 625 > 400 

The elements which are multiple of 5 and x² < 400 are 5, 10, 15 

So, K = {5, 10, 15}

Here, 

= {XϵC:X 2 = 1} 

= {1 ,-1} 

B = {xϵC:X 4 = 1} 

= {1,-1,i,- i} 

Now, 

A∖B = A∖(A⋂B) 

= {1,-1}∖{1,-1} 

= ϕ 

And B∖A=B∖(A⋂B) 

= {1,-1,i,- i}∖{1,-1} 

= {i,- i}

Correct option is C) {0, 1, 3, 5, 6, 8}

Correct option is D) (A – B) ∪ (B – A)

Let A = {1, 2} 

B = {2, 3} 

C = {1, 3, 4} 

A∩B = {2} 

A∩C = {1} 

B∩C = {3} 

A∩B∩C = {2}∩{1, 3, 4} = ø 

So the three sets are valid and satisfy the given conditions

Let x be any element of Set A 

And y be any element of Set B 

Now 

x≠y A ∩ B = ϕ 

This means no element of B should be in A. 

Thus, 

x is an element of A and an element of B’ 

As x ϵ B’ 

A B’ 

Hence, Proved.

= A Δ (A ∩ B) [∵ E Δ F =(E–F) ∪ (F–E) ] 

= (A–( A ∩ B)) ∪ (A ∩B –A) [∵ E – F = E ∩ F’] 

= (A ∩ (A ∩ B)’) ∪ (A∩B∩A’) 

= (A ∩ (A’∪B’)) ∪ (A∩A’∩B) 

= ϕ ∪ (A ∩ B’) ∪ ϕ 

= A ∩ B’ [∵A ∩ B’ = A–B] 

= A–B 

= LHS 

∴ LHS = RHS 

Proved

Given as, ACB

To prove the C – B ⊂ C – A

Consider, x ∈ C – B

⇒ x ∈ C and x ∉ B

⇒ x ∈ C and x ∉ A

⇒ x ∈ C – A

Hence, x ∈ C–B ⇒ x ∈ C – A

This is true for all x ∈ C – B

∴ C – B ⊂ C – A

Thus proved.

Given; A = {2, 3, 5, 7, 11} and B = ϕ 

(i) A ∪ B = {2, 3, 5, 7, 11} 

(ii) A ∩ B = ϕ

(i) A ∩ (A’ ∪ B) = A ∩ B

Consider LHS A ∩ (A’ ∪ B)

On expanding

(A ∩ A’) ∪ (A ∩ B)

As we know, (A ∩ A’) =ϕ

⇒ ϕ ∪ (A∩ B)

⇒ (A ∩ B)

∴ LHS = RHS

Thus proved.

(ii) A – (A – B) = A ∩ B

For the any sets A and B we have De-Morgan’s law

(A ∪ B)’ = A’ ∩ B’, (A ∩ B) ‘ = A’ ∪ B’

Considering LHS

= A – (A–B)

= A ∩ (A–B)’

= A ∩ (A∩B’)’

= A ∩ (A’ ∪ B’)’) (Here, (B’)’ = B)

= A ∩ (A’ ∪ B)

= (A ∩ A’) ∪ (A ∩ B)

= ϕ ∪ (A ∩ B) (Here, A ∩ A’ = ϕ)

= (A ∩ B) (Here, ϕ ∪ x = x, for any set)

= RHS

∴ LHS=RHS

Thus proved.

(iii) A ∩ (A ∪ B’) = ϕ

Consider LHS A ∩ (A ∪ B’)

= A ∩ (A ∪ B’)

= A ∩ (A’∩ B’) (By De–Morgan’s law)

= (A ∩ A’) ∩ B’ (here, A ∩ A’ = ϕ)

= ϕ ∩ B’

= ϕ (since, ϕ ∩ B’ = ϕ)

= RHS

∴ LHS=RHS

Thus proved.

(iv) A – B = A Δ (A ∩ B)

Consider RHS A Δ (A ∩ B)

A Δ (A ∩ B) (here, E Δ F = (E–F) ∪ (F–E))

= (A – (A ∩ B)) ∪ (A ∩ B –A) (here, E – F = E ∩ F’)

= (A ∩ (A ∩ B)’) ∪ (A ∩ B ∩ A’)

= (A ∩ (A’ ∪ B’)) ∪ (A ∩ A’ ∩ B) (by using De-Morgan’s law and associative law)

= (A ∩ A’) ∪ (A ∩ B’) ∪ (ϕ ∩ B) (by using distributive law)

= ϕ ∪ (A ∩ B’) ∪ ϕ

= A ∩ B’ (here, A ∩ B’ = A–B)

= A – B

= LHS

∴ LHS=RHS

Thus proved.

Let us prove, A’ – B’ = B – A

Now, firstly we need to show

A’ – B’ ⊆ B – A

Suppose, x ∈ A’ – B’

⇒ x ∈ A’ and x ∉ B’

⇒ x ∉ A and x ∈ B (Here, A ∩ A’ = ϕ )

⇒ x ∈ B – A

Since, it is true for all x ∈ A’ – B’

∴ A’ – B’ = B – A

Thus proved.

(i) A ∪ (A ∩ B) = A

As we know union is distributive over intersection

Therefore, A ∪ (A ∩ B)

(A ∪ A) ∩ (A ∪ B) [Here, A ∪ A = A]

A ∩ (A ∪ B)

A

(ii) A ∩ (A ∪ B) = A

As we know union is distributive over intersection

Therefore, (A ∩ A) ∪ (A ∩ B)

A ∪ (A ∩ B) [Here, A ∩ A = A]

A

(i) A ∩ B = A ∩ C need not imply B = C.

Consider, A = {1, 2}

B = {2, 3}

C = {2, 4}

Now,

A ∩ B = {2}

A ∩ C = {2}

Thus, A ∩ B = A ∩ C, here B is not equal to C

(ii) A ⊂ B ⇒ C – B ⊂ C – A

Given as: A ⊂ B

To prove: C–B ⊂ C–A

Consider x ∈ C– B

⇒ x ∈ C and x ∉ B [by definition C–B]

⇒ x ∈ C and x ∉ A

⇒ x ∈ C–A

Hence x ∈ C–B ⇒ x ∈ C–A. This is true for all x ∈ C–B.

∴ A ⊂ B ⇒ C – B ⊂ C – A

Let 

A = {1, 2} 

B = {2, 3} 

C = {2, 4} 

Then, 

A ∩ B = {2} 

A ∩ C = {2} 

Hence, 

A ∩ B = A ∩ C, but clearly B is not equal to C.

Let 

p ϵ A ⊂ B. 

⇒ x ϵ B

 Let and p ϵ A ∩ B 

⬄ x ϵ A and x ϵ B 

∴ (A ∩ B) = A.

Here, 

A and B are two disjoint sets. 

⇒ A⋂B = ϕ 

⇒ n(A⋂B) = 0 

n(A⋃B) = n(A) + n(B) - n(A⋂B) 

= n(A) + n(B)-0 

= n(A) + n(B)

A⋂(A⋃B)' = A⋂(A'⋂B') 

= (A⋂A' )⋂(A⋂B') 

= ϕ⋂(A⋂B' ) 

= ϕ

Let A be the set of population travels by car(in percentage) and 

B be the set of population travels by bus(in percentage). 

Then A⋂B is the set of population travels by car and bus and 

A⋃B is the set of population travels by car or bus. 

Then,n(A) = 20 , (B) = 50 ,n(A⋂B) =10 

n(A⋃B) = n(A) + n(B) - n(A⋂B) 

= 20 + 50 - 10 

= 60 

∴ The set of population travels by car or bus is 60%

A⋂(A⋂B)^c = A⋂ (A^c⋃B^c ) 

= (A ⋂ A^c )⋃(A ⋂ B^c ) 

= ϕ⋃ (A ⋂ B^c ) 

= A ⋂ B^c

(i) A ∩ B

Since, A ∩ B denotes A intersection B. Common elements of A and B consists in this set.

Given as A ⊂ B, every element of A are already an element of B.

∴ A ∩ B = A

(ii) A ⋃ B

Since, A ∪ B denotes A union B. the elements of either A or B or in both A and B consist in this set.

Given as A ⊂ B, B is having all elements including elements of A.

∴ A ∪ B = B

Let X ∈ P (A ∩ B),then A ⊂ (A ∩ B). 

So, X ⊂ A and X ⊂ B. 

∴ X ∈ P(A) and X ∩ P(B) 

⇒ X ∈ P(A) and P(B) 

Thus, P(A ∩ B) ⊂ P(A) ∩ P(B) ..(i) 

Again, Let Y ∈ P(A) ∩ P(B),then 

Y ∈ P(A) and Y ∈ P(B) 

So, Y ⊂ A and Y ⊂ B. 

∴ Y ⊂ A ∩ B ⇒ Y ∈ P(A ∩ B) 

Then P(A) ∩ P(B) ∈ P(A ∩ B) ..(ii) 

Hence from (i) and (ii), we gets 

P (A ∩ B) = P(A) ∩ P(B)

Let A be the set of percentage of those people who watch a news Channel and B be the set of percentage of those people who watch another channel. 

Given, n(A) = 63, n(B) = 96 and n(A ∩ B) = x 

∴ n(A ∪ B) ≤ 100 

n(A) + n(B) − n(A ∩ B) ≤ 100 

63 + 76 – x ≤ 100 

139 – x ≤ 100 

39 ≤ x 

∴ n(A ∩ B) ≤ n(A) 

x ≤ 63 

∴ 39 ≤ x ≤ 63

Answer is (C) 39 ≤ x ≤ 63 

Answer is False

Answer is True

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)

n(C) = 42 –(7+3+2) 

= 42–12. 

= 30

Let us denote, 

Total number of students by n(P) 

Students studying English n(E) 

Students Studying Hindi n(H) 

Students studying Sanskrit n(S) 

According to the question, 

n(P) =100, n(E–H) = 23, n(E ∩ S) =8, n(E) = 26, n(S) = 48, n(H ∩ S) =8, n(E∪H∪S)’=24 

Number of students studying English only = 18 

Now, 

n(E∪H∪S)’=24 

n(P) – n(E∪H∪S) = 24 

100 – 24 = n(E∪H∪S) 

n(E∪H∪S) = 76 n(E∪H∪S) = n(E)+ n(H)+ n(S)– n(E ∩ S)– n(E ∩ H)– n(H ∩ S)+ (E∩ H ∩ S) 

76 = 26+n(H)+48–3–8–8+3 

n(H) = 76–58 

n(H) = 18 

18 Students are studying Hindi.

From q1 we have n(E ∩ H) = 3 

∴ 3 students study both hindi and English.

Now the equivalent contra positive statement of x ∈ S ⇒ x ∈ P is x ∉ P ⇒ x ∉ S.

Now, we will prove the above contra positive statement by contradiction method 

Let x ∉ P 

⇒ x is a composite number 

Let us now assume that x ∈ S 

⇒ 2^x – 1 = m (where m is a prime number) 

⇒ 2^x = m + 1 

Which is not true for all composite number, say for x = 4 because 

2⁴ = 16 which can not be equal to the sum of any prime number m and 1. 

Thus, we arrive at a contradiction 

⇒ x ∉ S. 

Thus, when x ∉ P, we arrive at x ∉ S 

So S ⊂ P

Answer is True

The correct answer is (B)

Since, n(Xr) = 5, ⋃X_r(for r=1 to 20)=S, we get n(S) = 100

But each element of S belong to exactly 10 of the X_r’s

So, 100/10 = 10 are the number of distinct elements in S.

Also each element of S belong to exactly 4 of the Y_r’ s and each Y_r contain 2 elements. If S has n number of Y_r in it. Then

2n/4 = 10

which gives n = 20