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# Class 11 RD Sharma Solutions – Chapter 1 Sets – Exercise 1.6 | Set 1

• Last Updated : 11 Feb, 2021

### Question 1. Find the smallest set A such that A âˆª {1, 2} = {1, 2, 3, 5, 9}.

Solution:

A âˆª {1, 2} = {1, 2, 3, 5, 9}

The union indicates that the summation of elements of both sets should form RHS.

Elements of A and {1, 2} together give us the resultant set.

Therefore, the smallest set will be,

A = {1, 2, 3, 5, 9} – {1, 2}

Hence, A = {3, 5, 9}

### Question 2. Let A = {1, 2, 4, 5} B = {2, 3, 5, 6} C = {4, 5, 6, 7}. Verify the following identities:

(i) A âˆª (B âˆ© C) = (A âˆª B) âˆ© (A âˆª C)

(iv) A – (B âˆª C) = (A – B) âˆ© (A – C)

(v) A – (B âˆ© C) = (A – B) âˆª (A – C)

Solution:

(i) A âˆª (B âˆ© C) = (A âˆª B) âˆ© (A âˆª C)

Considering LHS,

(B âˆ© C) = {x : x âˆˆ B and x âˆˆ C}

Therefore, we get,

= {5, 6}

A âˆª (B âˆ© C) = {x : x âˆˆ A or x âˆˆ (B âˆ© C)}

= {1, 2, 4, 5, 6}

Considering RHS,

(A âˆª B) = {x : x âˆˆ A or x âˆˆ B}

= {1, 2, 4, 5, 6}.

(A âˆª C) = {x : x âˆˆ A or x âˆˆ C}

= {1, 2, 4, 5, 6, 7}

(A âˆª B) âˆ© (A âˆª C) = {x : x âˆˆ (A âˆª B) and x âˆˆ (A âˆª C)}

= {1, 2, 4, 5, 6}

Hence, LHS = RHS

Considering LHS,

(B âˆª C) = {x: x âˆˆ B or x âˆˆ C}

= {2, 3, 4, 5, 6, 7}

(A âˆ© (B âˆª C)) = {x : x âˆˆ A and x âˆˆ (B âˆª C)}

= {2, 4, 5}

Considering RHS,

(A âˆ© B) = {x : x âˆˆ A and x âˆˆ B}

= {2, 5}

(A âˆ© C) = {x : x âˆˆ A and x âˆˆ C}

= {4, 5}

= {2, 4, 5}

Hence, LHS = RHS

We know,

B – C is defined as {x âˆˆ B : x âˆ‰ C}

Given,

B = {2, 3, 5, 6}

C = {4, 5, 6, 7}

B – C = {2, 3}

Considering LHS,

(A âˆ© (B – C)) = {x : x âˆˆ A and x âˆˆ (B – C)}

= {2}

Considering RHS,

(A âˆ© B) = {x: x âˆˆ A and x âˆˆ B}

= {2, 5}

(A âˆ© C) = {x : x âˆˆ A and x âˆˆ C}

= {4, 5}

= {2}

âˆ´ LHS = RHS

(iv) A – (B âˆª C) = (A – B) âˆ© (A – C)

Considering LHS,

(B âˆª C) = {x : x âˆˆ B or x âˆˆ C}

= {2, 3, 4, 5, 6, 7}

A – (B âˆª C) is defined as {x âˆˆ A : x âˆ‰ (B âˆª C)}

A = {1, 2, 4, 5}

(B âˆª C) = {2, 3, 4, 5, 6, 7}

A – (B âˆª C) = {1}

Considering RHS,

(A – B) = A – B is defined as {x âˆˆ A : x âˆ‰ B}

Given,

A = {1, 2, 4, 5}

B = {2, 3, 5, 6}

A – B = {1, 4}

(A – C)

A – C is defined as {x âˆˆ A : x âˆ‰ C}

A = {1, 2, 4, 5}

C = {4, 5, 6, 7}

A – C = {1, 2}

(A – B) âˆ© (A – C) = {x : x âˆˆ (A – B) and x âˆˆ (A – C)}.

= {1}

Hence, LHS = RHS

(v) A – (B âˆ© C) = (A – B) âˆª (A – C)

Considering LHS,

(B âˆ© C) = {x : x âˆˆ B and x âˆˆ C}

= {5, 6}

A – (B âˆ© C) is defined as {x âˆˆ A : x âˆ‰ (B âˆ© C)}

A = {1, 2, 4, 5}

(B âˆ© C) = {5, 6}

(A – (B âˆ© C)) = {1, 2, 4}

Considering RHS,

(A – B) = A – B is defined as {x âˆˆ A : x âˆ‰ B}

A = {1, 2, 4, 5}

B = {2, 3, 5, 6}

A – B = {1, 4}

(A – C) = A – C is defined as {x âˆˆ A : x âˆ‰ C}

A = {1, 2, 4, 5}

C = {4, 5, 6, 7}

A – C = {1, 2}

(A – B) âˆª (A – C) = {x : x âˆˆ (A – B) OR x âˆˆ (A – C)}.

= {1, 2, 4}

Therefore, LHS = RHS

A = {1, 2, 4, 5} B = {2, 3, 5, 6} C = {4, 5, 6, 7}.

Considering LHS,

B â–³ C = (B – C) âˆª (C – B) = {2, 3} âˆª {4, 7} = {2, 3, 4, 7}

A âˆ© (B â–³ C) = {2, 4}

Considering RHS,

A âˆ© B = {2, 5}

A âˆ© C = {4, 5}

= {2} âˆª {4}

= {2, 4}

Therefore, LHS = RHS

### (ii) (A âˆ© B)â€™ = Aâ€™ âˆª Bâ€™

Solution:

(i) (A âˆª B)â€™ = Aâ€™ âˆ© Bâ€™

Considering LHS,

A âˆª B = {x : x âˆˆ A or x âˆˆ B}

= {2, 3, 5, 7, 9}

(A âˆª B)â€™ means Complement of (A âˆª B) with respect to universal set U.

So, (A âˆª B)â€™ = U – (A âˆª B)â€™

Now,

U – (A âˆª B)â€™ is defined as {x âˆˆ U: x âˆ‰ (A âˆª B)â€™}

U = {2, 3, 5, 7, 9}

(A âˆª B)â€™ = {2, 3, 5, 7, 9}

U – (A âˆª B)â€™ = Ï•

Considering RHS,

We have, Aâ€™ = U – A

(U – A) is defined as {x âˆˆ U : x âˆ‰ A}

U = {2, 3, 5, 7, 9}

A = {3, 7}

Aâ€™ = U – A = {2, 5, 9}

Now, Bâ€™ = U – B

(U – B) is defined as {x âˆˆ U : x âˆ‰ B}

U = {2, 3, 5, 7, 9}

B = {2, 5, 7, 9}

Bâ€™ = U – B = {3}

Aâ€™ âˆ© Bâ€™ = {x : x âˆˆ Aâ€™ and x âˆˆ Câ€™}.

= Ï•

âˆ´ LHS = RHS

(ii) (A âˆ© B)â€™ = Aâ€™ âˆª Bâ€™

Considering LHS,

(A âˆ© B) = {x: x âˆˆ A and x âˆˆ B}.

= {7}

Also,

U – (A âˆ© B) is defined as {x âˆˆ U : x âˆ‰ (A âˆ© B)â€™}

U = {2, 3, 5, 7, 9}

U – (A âˆ© B) = {2, 3, 5, 9}

(A âˆ© B)â€™ = {2, 3, 5, 9}

Considering RHS,

So, Aâ€™ = U – A

(U – A) is defined as {x âˆˆ U : x âˆ‰ A}

U = {2, 3, 5, 7, 9}

A = {3, 7}

Aâ€™ = U – A = {2, 5, 9}

So, Bâ€™ = U – B

(U – B) is defined as {x âˆˆ U : x âˆ‰ B}

U = {2, 3, 5, 7, 9}

B = {2, 5, 7, 9}

Bâ€™ = U – B = {3}

Aâ€™ âˆª Bâ€™ = {x : x âˆˆ A or x âˆˆ B}

= {2, 3, 5, 9}

âˆ´ LHS = RHS

### (iii) A âŠ‚ B â‡’ A âˆ© B = A

Solution:

(i) B âŠ‚ A âˆª B

Considering an element â€˜pâ€™ belonging to B = p âˆˆ B

p âˆˆ B âˆª A

B âŠ‚ A âˆª B

(ii) A âˆ© B âŠ‚ A

Considering an element â€˜pâ€™ belonging to B = p âˆˆ B

p âˆˆ A and p âˆˆ B

(iii) A âŠ‚ B â‡’ A âˆ© B = A

Considering an element â€˜pâ€™ belonging to B = p âˆˆ B

p âˆˆ A âŠ‚ B

Now, x âˆˆ B

Let and p âˆˆ A âˆ© B

x âˆˆ A and x âˆˆ B

x âˆˆ A and x âˆˆ A (since, A âŠ‚ B)

Therefore, (A âˆ© B) = A

### (iv) A âˆ© B = A

Solution:

(i) A âŠ‚ B

We need to prove (i) = (ii), (ii) = (iii), (iii) = (iv), (iv) = (v)

Let us prove, (i) = (ii)

We know, A – B = {x âˆˆ A : x âˆ‰ B} as A âŠ‚ B,

Now, Each element of A is an element of B,

âˆ´ A – B = Ï•

Therefore, (i) = (ii)

(ii) A – B = Ï•

We need to prove that (ii) = (iii)

Let us assume, A – B = Ï•

âˆ´ Every element of A is an element of B

So, A âŠ‚ B and so A âˆª B = B

Hence, (ii) = (iii)

(iii) A âˆª B = B

We need to show that (iii) = (iv)

By assuming A âˆª B = B

âˆ´ AâŠ‚ B and so A âˆ© B = A

Hence, (iii) = (iv)

(iv) A âˆ© B = A

Finally, now we need to show (iv) = (i)

By assuming A âˆ© B = A

Since, A âˆ© B = A, so A âŠ‚ B

Hence, (iv) = (i)

### (ii) A âŠ‚ B â‡’ C – B âŠ‚ C – A

Solution:

(i) A âˆ© B = A âˆ© C need not imply B = C.

Let us assume, A = {1, 2}

B = {2, 3}

C = {2, 4}

Then,

Therefore, A âˆ© B = A âˆ© C, where, B is not equal to C

(ii) A âŠ‚ B â‡’ C – B âŠ‚ C – A

Given: A âŠ‚ B

Let us assume x âˆˆ C – B

â‡’ x âˆˆ C and x âˆ‰ B [by definition C – B]

â‡’ x âˆˆ C and x âˆ‰ A

â‡’ x âˆˆ C – A

Therefore, x âˆˆ C – B â‡’ x âˆˆ C – A for all x âˆˆ C – B.

âˆ´ A âŠ‚ B â‡’ C – B âŠ‚ C – A

### (ii) A âˆ© (A âˆª B) = A

Solution:

(i) A âˆª (A âˆ© B) = A

Since, union is distributive over intersection, we have, A âˆª (A âˆ© B)

(A âˆª A) âˆ© (A âˆª B) [Since, A âˆª A = A]

= A

(ii) A âˆ© (A âˆª B) = A

Since, union is distributive over intersection, we have, (A âˆ© A) âˆª (A âˆ© B)