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• RD Sharma Class 11 Solutions for Maths

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

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

Solution:

A ∪ {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 ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

(ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

(iii) A ∩ (B – C) = (A ∩ B) -(A ∩ C)

(iv) A – (B ∪ C) = (A – B) ∩ (A – C)

(v) A – (B ∩ C) = (A – B) ∪ (A – C)

(vi) A ∩ (B △ C) = (A ∩ B) △ (A ∩ C)

Solution:

(i) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

Considering LHS,

(B ∩ C) = {x : x ∈ B and x ∈ C}

Therefore, we get,

= {5, 6}

A ∪ (B ∩ C) = {x : x ∈ A or x ∈ (B ∩ C)}

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

Considering RHS,

(A ∪ B) = {x : x ∈ A or x ∈ B}

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

(A ∪ C) = {x : x ∈ A or x ∈ C}

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

(A ∪ B) ∩ (A ∪ C) = {x : x ∈ (A ∪ B) and x ∈ (A ∪ C)}

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

Hence, LHS = RHS

(ii) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

Considering LHS,

(B ∪ C) = {x: x ∈ B or x ∈ C}

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

(A ∩ (B ∪ C)) = {x : x ∈ A and x ∈ (B ∪ C)}

= {2, 4, 5}

Considering RHS,

(A ∩ B) = {x : x ∈ A and x ∈ B}

= {2, 5}

(A ∩ C) = {x : x ∈ A and x ∈ C}

= {4, 5}

(A ∩ B) ∪ (A ∩ C) = {x : x ∈ (A∩B) and x ∈ (A ∩ C)}

= {2, 4, 5}

Hence, LHS = RHS

(iii) A ∩ (B – C) = (A ∩ B) – (A ∩ C)

We know,

B – C is defined as {x ∈ B : x ∉ C}

Given,

B = {2, 3, 5, 6}

C = {4, 5, 6, 7}

B – C = {2, 3}

Considering LHS,

(A ∩ (B – C)) = {x : x ∈ A and x ∈ (B – C)}

= {2}

Considering RHS,

(A ∩ B) = {x: x ∈ A and x ∈ B}

= {2, 5}

(A ∩ C) = {x : x ∈ A and x ∈ C}

= {4, 5}

(A ∩ B) – (A ∩ C) is defined as {x ∈ (A ∩ B) : x ∉ (A ∩ C)}

= {2}

∴ LHS = RHS

(iv) A – (B ∪ C) = (A – B) ∩ (A – C)

Considering LHS,

(B ∪ C) = {x : x ∈ B or x ∈ C}

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

A – (B ∪ C) is defined as {x ∈ A : x ∉ (B ∪ C)}

A = {1, 2, 4, 5}

(B ∪ C) = {2, 3, 4, 5, 6, 7}

A – (B ∪ C) = {1}

Considering RHS,

(A – B) = A – B is defined as {x ∈ A : x ∉ B}

Given,

A = {1, 2, 4, 5}

B = {2, 3, 5, 6}

A – B = {1, 4}

(A – C)

A – C is defined as {x ∈ A : x ∉ C}

A = {1, 2, 4, 5}

C = {4, 5, 6, 7}

A – C = {1, 2}

(A – B) ∩ (A – C) = {x : x ∈ (A – B) and x ∈ (A – C)}.

= {1}

Hence, LHS = RHS

(v) A – (B ∩ C) = (A – B) ∪ (A – C)

Considering LHS,

(B ∩ C) = {x : x ∈ B and x ∈ C}

= {5, 6}

A – (B ∩ C) is defined as {x ∈ A : x ∉ (B ∩ C)}

A = {1, 2, 4, 5}

(B ∩ C) = {5, 6}

(A – (B ∩ C)) = {1, 2, 4}

Considering RHS,

(A – B) = A – B is defined as {x ∈ A : x ∉ B}

A = {1, 2, 4, 5}

B = {2, 3, 5, 6}

A – B = {1, 4}

(A – C) = A – C is defined as {x ∈ A : x ∉ C}

A = {1, 2, 4, 5}

C = {4, 5, 6, 7}

A – C = {1, 2}

(A – B) ∪ (A – C) = {x : x ∈ (A – B) OR x ∈ (A – C)}.

= {1, 2, 4}

Therefore, LHS = RHS

(vi) A ∩ (B △ C) = (A ∩ B) △ (A ∩ C)

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

Considering LHS,

A ∩ (B △ C)

B △ C = (B – C) ∪ (C – B) = {2, 3} ∪ {4, 7} = {2, 3, 4, 7}

A ∩ (B △ C) = {2, 4}

Considering RHS,

A ∩ B = {2, 5}

A ∩ C = {4, 5}

(A ∩ B) △ (A ∩ C) = [(A ∩ B) – (A ∩ C)] ∪ [(A ∩ C) – (A ∩ B)]

= {2} ∪ {4}

= {2, 4}

Therefore, LHS = RHS

### (ii) (A ∩ B)’ = A’ ∪ B’

Solution:

(i) (A ∪ B)’ = A’ ∩ B’

Considering LHS,

A ∪ B = {x : x ∈ A or x ∈ B}

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

(A ∪ B)’ means Complement of (A ∪ B) with respect to universal set U.

So, (A ∪ B)’ = U – (A ∪ B)’

Now,

U – (A ∪ B)’ is defined as {x ∈ U: x ∉ (A ∪ B)’}

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

(A ∪ B)’ = {2, 3, 5, 7, 9}

U – (A ∪ B)’ = ϕ

Considering RHS,

We have, A’ = U – A

(U – A) is defined as {x ∈ U : x ∉ A}

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

A = {3, 7}

A’ = U – A = {2, 5, 9}

Now, B’ = U – B

(U – B) is defined as {x ∈ U : x ∉ B}

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

B = {2, 5, 7, 9}

B’ = U – B = {3}

A’ ∩ B’ = {x : x ∈ A’ and x ∈ C’}.

= ϕ

∴ LHS = RHS

(ii) (A ∩ B)’ = A’ ∪ B’

Considering LHS,

(A ∩ B)’

(A ∩ B) = {x: x ∈ A and x ∈ B}.

= {7}

Also,

(A ∩ B)’ = U – (A ∩ B)

U – (A ∩ B) is defined as {x ∈ U : x ∉ (A ∩ B)’}

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

(A ∩ B) = {7}

U – (A ∩ B) = {2, 3, 5, 9}

(A ∩ B)’ = {2, 3, 5, 9}

Considering RHS,

So, A’ = U – A

(U – A) is defined as {x ∈ U : x ∉ A}

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

A = {3, 7}

A’ = U – A = {2, 5, 9}

So, B’ = U – B

(U – B) is defined as {x ∈ U : x ∉ B}

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

B = {2, 5, 7, 9}

B’ = U – B = {3}

A’ ∪ B’ = {x : x ∈ A or x ∈ B}

= {2, 3, 5, 9}

∴ LHS = RHS

### (iii) A ⊂ B ⇒ A ∩ B = A

Solution:

(i) B ⊂ A ∪ B

Considering an element ‘p’ belonging to B = p ∈ B

p ∈ B ∪ A

B ⊂ A ∪ B

(ii) A ∩ B ⊂ A

Considering an element ‘p’ belonging to B = p ∈ B

p ∈ A ∩ B

p ∈ A and p ∈ B

A ∩ B ⊂ A

(iii) A ⊂ B ⇒ A ∩ B = A

Considering an element ‘p’ belonging to B = p ∈ B

p ∈ A ⊂ B

Now, x ∈ B

Let and p ∈ A ∩ B

x ∈ A and x ∈ B

x ∈ A and x ∈ A (since, A ⊂ B)

Therefore, (A ∩ B) = A

### (iv) A ∩ B = A

Solution:

(i) A ⊂ B

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

Let us prove, (i) = (ii)

We know, A – B = {x ∈ A : x ∉ B} as A ⊂ B,

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

∴ A – B = ϕ

Therefore, (i) = (ii)

(ii) A – B = ϕ

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

Let us assume, A – B = ϕ

∴ Every element of A is an element of B

So, A ⊂ B and so A ∪ B = B

Hence, (ii) = (iii)

(iii) A ∪ B = B

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

By assuming A ∪ B = B

∴ A⊂ B and so A ∩ B = A

Hence, (iii) = (iv)

(iv) A ∩ B = A

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

By assuming A ∩ B = A

Since, A ∩ B = A, so A ⊂ B

Hence, (iv) = (i)

### (ii) A ⊂ B ⇒ C – B ⊂ C – A

Solution:

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

Let us assume, A = {1, 2}

B = {2, 3}

C = {2, 4}

Then,

A ∩ B = {2}

A ∩ C = {2}

Therefore, A ∩ B = A ∩ C, where, B is not equal to C

(ii) A ⊂ B ⇒ C – B ⊂ C – A

Given: A ⊂ B

Let us assume x ∈ C – B

⇒ x ∈ C and x ∉ B [by definition C – B]

⇒ x ∈ C and x ∉ A

⇒ x ∈ C – A

Therefore, x ∈ C – B ⇒ x ∈ C – A for all x ∈ C – B.

∴ A ⊂ B ⇒ C – B ⊂ C – A

### (ii) A ∩ (A ∪ B) = A

Solution:

(i) A ∪ (A ∩ B) = A

Since, union is distributive over intersection, we have, A ∪ (A ∩ B)

(A ∪ A) ∩ (A ∪ B) [Since, A ∪ A = A]

A ∩ (A ∪ B)

= A

(ii) A ∩ (A ∪ B) = A

Since, union is distributive over intersection, we have, (A ∩ A) ∪ (A ∩ B)

A ∪ (A ∩ B) [Since, A ∩ A = A]

= A

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