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 âˆ© B and B âˆ© C and A âˆ© C are non-empty sets, and A âˆ© B âˆ© C = Ï•

Solution:

Let us consider the sets,

A = {5, 6, 10}

B = {6, 8, 9}

C = {9, 10, 11}

Now, we have,

A âˆ© B = 6 â‰  Ï•

B âˆ© C = 9 â‰  Ï•

A âˆ© C = 10 â‰  Ï•

And, A âˆ© B âˆ© C = Ï•

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

Question 9. For any two sets A and B, prove that A âˆ© B = Ï• => A âŠ‚ B’

Solution:

Let, a âˆˆ A => a âˆ‰ B

Thus,

A âˆ© B  = Ï•

=> a âˆˆ B’

Thus, a âˆˆ A and a âˆˆ B’ => A âŠ‚ B’

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

Solution:

(i) A – B and A âˆ© B

Let a  âˆˆ A – B => a âˆˆ A and a âˆ‰ B => a âˆ‰ A âˆ© B

Therefore, A – B and A âˆ© B are disjoint sets.

(ii) Let a âˆˆ B – A => a âˆˆ B and a âˆ‰ A => a âˆ‰ A âˆ© B

Hence, B – A and A âˆ© B are disjoint sets.

(iii) A – B and B – A,

A – B = x, x : x âˆˆ A and x âˆ‰ B

A – B and B – A are disjoint sets.

(A âˆª B) âˆ© (A âˆ© B’) = A

Solution:

We have,

LHS = A âˆª B âˆ© A âˆ© B’

Solving this, we get,

= A âˆª B âˆ© A âˆª A âˆª B âˆ© B’

= A âˆª A âˆª B âˆ© B’

Since, B âˆ© B’ = âˆ…

= A âˆª A âˆ© B’

= A

Therefore, LHS = RHS.

B’ âŠ‚ A’ = U => A âŠ‚ B

Solution:

(i) Let a âˆˆ A

= a âˆˆ U

= a âˆˆ A’ âˆª B, because, U = A’ âˆª B

= a âˆˆ B, because a âˆ‰ A’

Hence, A âŠ‚ B

(ii) Let a âˆˆ A

= a âˆ‰ A’

= a âˆ‰ B’, because, B’ is a subset of A’

= a âˆˆ B

Hence, A âŠ‚ B

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

Solution:

Result is False.

Proof:

Let X âˆˆ P(A) âˆª P(B)

= X âˆˆ P(A) or X âˆˆ P(B)

= X âŠ‚ A or X âŠ‚ B

= X âŠ‚ A U B

= X âˆˆ P(A âˆ© B)

Thus, P(A) âˆª P(B) âŠ‚ P(A âˆª B)

Also, Let us assume,

X âˆˆ P(A âˆª B). But, X âˆ‰ P(A) or X âˆ‰ P(B)

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

So, X âˆ‰ P(A) âˆª P(B)

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

A âˆª (B – A) = A âˆª B

Solution:

(i) We have,

RHS = (A âˆ© B) âˆª (A – B)

= (A âˆ© B) âˆª (A âˆ© B)’

= (A âˆ© B) âˆª (A âˆ© A) âˆ© (B âˆª B)’

= A âˆ© (A âˆª B)’ âˆ© (B âˆª B)’

= A âˆ© (A âˆª B)’ âˆ© U

= A âˆ© (A âˆª B)’

= A

Therefore, RHS = LHS

(ii) We have,

LHS = A âˆª (B – A)

= A âˆª (B âˆ© A)’

= (A âˆª B) âˆ© (A âˆª A)’

= (A âˆª B) âˆ© U

LHS = A âˆª B = RHS

Question 15. Each set X, contains 5 elements and each set Y, contains 2 elements and  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

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

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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