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Class 11 RD Sharma Solutions – Chapter 1 Sets – Exercise 1.6 | Set 2
  • Last Updated : 11 Feb, 2021

Question 8. Find sets A, B and C, such that A ∩ B and B ∩ C and A ∩ C are non-empty sets, and A ∩ B ∩ C = ϕ

Solution:

Let us consider the sets, 

A = {5, 6, 10}

B = {6, 8, 9}

C = {9, 10, 11}



Now, we have, 

A ∩ B = 6 ≠ ϕ

B ∩ C = 9 ≠ ϕ

A ∩ C = 10 ≠ ϕ

And, A ∩ B ∩ C = ϕ

Now, we have A ∩ B and B ∩ C and A ∩ C as non-empty sets, but A ∩ B ∩ C is empty set. 

Question 9. For any two sets A and B, prove that A ∩ B = ϕ => A ⊂ B’

Solution:

Let, a ∈ A => a ∉ B

Thus, 

A ∩ B  = ϕ

=> a ∈ B’

Thus, a ∈ A and a ∈ B’ => A ⊂ B’

Question 10. Prove the following:

(i) A – B and A ∩ B are disjoint sets

(ii) B – A and A ∩ B are disjoint sets

(iii) A – B and B – A are disjoint sets

Solution:

(i) A – B and A ∩ B 

Let a  ∈ A – B => a ∈ A and a ∉ B => a ∉ A ∩ B

Therefore, A – B and A ∩ B are disjoint sets.

(ii) Let a ∈ B – A => a ∈ B and a ∉ A => a ∉ A ∩ B

Hence, B – A and A ∩ B are disjoint sets.



(iii) A – B and B – A,

A – B = x, x : x ∈ A and x ∉ B

A – B and B – A are disjoint sets. 

Question 11. Using properties of sets, show that for any two sets A and B,

(A ∪ B) ∩ (A ∩ B’) = A

Solution: 

We have, 

LHS = A ∪ B ∩ A ∩ B’

Solving this, we get, 

= A ∪ B ∩ A ∪ A ∪ B ∩ B’

= A ∪ A ∪ B ∩ B’

Since, B ∩ B’ = ∅

= A ∪ A ∩ B’

= A 

Therefore, LHS = RHS.

Question 12.

(i)Show that for any two sets A and B,

A’ U B = U => A ⊂ B 

(ii) Show that for any two sets A and B,

B’ ⊂ A’ = U => A ⊂ B 

Solution:

(i) Let a ∈ A

= a ∈ U

= a ∈ A’ ∪ B, because, U = A’ ∪ B

= a ∈ B, because a ∉ A’

Hence, A ⊂ B

(ii) Let a ∈ A

= a ∉ A’

= a ∉ B’, because, B’ is a subset of A’

= a ∈ B

Hence, A ⊂ B

Question 13. Is it true that for any set A and B, P(A) ∪ P(B) = P(A ∪ B)? Justify your answer.

Solution: 

Result is False.

Proof:

Let X ∈ P(A) ∪ P(B)

= X ∈ P(A) or X ∈ P(B) 

= X ⊂ A or X ⊂ B

= X ⊂ A U B

= X ∈ P(A ∩ B)

Thus, P(A) ∪ P(B) ⊂ P(A ∪ B)

Also, Let us assume, 

X ∈ P(A ∪ B). But, X ∉ P(A) or X ∉ P(B) 

For instance, we have X = 1, 2, 3, 4 and A = 2, 5 and B = 1, 3, 4.

So, X ∉ P(A) ∪ P(B) 

Therefore, P(A ∪ B) doesn’t necessarily have to be a subset of P(A) ∪ P(B).

Question 14. 

(i) Show that For any sets A and B,

A = (A ∩ B) ∩ (A – B)

(ii) Show that For any sets A and B,

A ∪ (B – A) = A ∪ B

Solution:

(i) We have,

RHS = (A ∩ B) ∪ (A – B)

= (A ∩ B) ∪ (A ∩ B)’

= (A ∩ B) ∪ (A ∩ A) ∩ (B ∪ B)’

= A ∩ (A ∪ B)’ ∩ (B ∪ B)’

= A ∩ (A ∪ B)’ ∩ U

= A ∩ (A ∪ B)’

= A

Therefore, RHS = LHS

(ii) We have,\bigcup\limits_{r=1}^{20} X_{r} = S

LHS = A ∪ (B – A)

= A ∪ (B ∩ A)’

= (A ∪ B) ∩ (A ∪ A)’

= (A ∪ B) ∩ U

LHS = A ∪ B = RHS

Question 15. Each set X, contains 5 elements and each set Y, contains 2 elements and  \bigcup\limits_{r=1}^{20} X_{r} = S = \bigcup\limits_{r=1}^{n} Y_{r}  each element of S belongs to exactly 10 of the Xr’s and to exactly 4 of Yr’s, then find the value of n.

Solution:

We have, Each set X contains 5 elements, and \bigcup\limits_{r=1}^{20} X_{r} = S

Therefore, n(S) = 20 x 5 = 100

But, we know, that each of the element of S belong to exactly 10 of the Xr‘s.

Therefore, n(S) = 100/10 = 10        -(1)

Also, Y contains 2 elements and

\bigcup\limits_{r=1}^{n} Y_{r} = S

Therefore, n(S) = n x 2 = 2n 

Each of the element of S belong to exactly 4 of the Yr’s.

n(S) = 2n/4 = n/2         -(2)

From equation (1) and (2)

10 = n/2

n = 20

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