NCERT Class 11 Maths Chapter 1 Sets Miscellaneous Exercise Solutions

Answer 1.

x² – 8x + 12 = 0

x² – 6x -2x + 12 = 0

x(x-6) – 2(x-6) = 0

(x-2)(x-6) = 0

A ={2,6}

B = {2,4,6}

C = { 2, 4, 6, 8, . . . }

D = { 6 }

A ⊂ B as elements 2 and 6 are contained in set B

A ⊂ C as elements 2 and 6 are contained in set C

B ⊂ C as elements 2,4,6 are contained in set C

D ⊂ A as element 6 is in set A

D ⊂ B as element 6 is in set B

D ⊂ C as element 6 is contained in set C

Answer 2.

(i) False

Example: Let A = {1}

B = {{1}, 2}

Then A ∈ B but 1 ∉ B

(ii) False

Let A = {1}, B = {1,2}, C = {{1,2}, 3} but A ∉ C

(iii) True

Let x ∈ A then x ∈ B, then x ∈ C means x is contained in set C, so, A ⊂ C

(iv) False

Let A = {1,2}, B = {2,3}, C = {1,2,5}

Here A ⊂ C

(v) False

Let A = {1} and B = {2,3} but {1} ⊄ B

(vi) True

Let x ∈ A means x ∈ B but it is given that x ∉ B, so, x ∉ A

Answer 3. Let x ∈ B

We want to show that it must also be in C

And then we’ll do the reverse: take x ∈ C and show it’s in B

Then x ∈ A ∪ B, means, x ∈ A ∪ C as A ∪ B = A ∪ C

Then x ∈ A or x ∈ C

CASE 1 x ∈ A

We started with let x ∈ B and case 1 is x ∈ A

That means as per definition of intersection of 2 sets:

x ∈ A ∩ B

It is given A ∩ B = A ∩ C means x ∈ A ∩ C

Means x ∈ A and x ∈ C

Means B ⊂ C

CASE 2 x ∈ C

In this case also B ⊂ C

Now similarly, if we take x ∈ C, then we can prove that C ⊂ B

B ⊂ C, C ⊂ B

This proves B = C

Answer 4. Let A ⊂ B

To prove: A – B = φ

Let A-B ≠ φ

Let x ∈ A – B means x ∈ A and x ∉ B

But A ⊂ B means every element of A is in B

So, our assumption is wrong, means

A – B = φ

Next, to prove: A ∪ B = B

Given A ⊂ B means every element of A is in B

Let x ∈ A ∪ B means x ∈ A or x ∈ B means x ∈ B or x ∈ B

Means x ∈ A ∪ B ⇒ x ∈ B, means A ∪ B ⊂ B

Also B ⊂ A ∪ B

Means A ∪ B = B

Next, to prove, A ∩ B = A

Given A ⊂ B means every element of A is in B

Let x ∈ A ∩ B means x ∈ A and x ∈ B

So, x ∈ A ∩ B ⇒ x ∈ A

Means A ∩ B ⊂ A

Now, we need to prove A ⊂ A ∩ B

Let x ∈ A and it is given A ⊂ B means x ∈ B

Means x ∈ A ⇒ x ∈ A ∩ B

So, A ⊂ A ∩ B

Therefore, A ∩ B = A

Answer 6. (i)

Step 1: Show that A ⊂ (A ∩ B) ∪ (A – B)

Let x ∈ A.
There are two possibilities:

• Case 1: x ∈ B ⇒ then x ∈ A ∩ B
• Case 2: x ∉ B ⇒ then x ∈ A – B

So in both cases, x ∈ (A ∩ B) ∪ (A – B)

⇒ A ⊂ (A ∩ B) ∪ (A – B)

Step 2: Show that (A ∩ B) ∪ (A – B) ⊂ A

Let x ∈ (A ∩ B) ∪ (A – B)
Then:

• If x ∈ A ∩ B ⇒ clearly x ∈ A
• If x ∈ A – B ⇒ then also x ∈ A

⇒ (A ∩ B) ∪ (A – B) ⊂ A

Conclusion:
Since both directions of inclusion hold,
A = (A ∩ B) ∪ (A – B)

Hence Proved

(ii) A ∪ ( B – A ) = ( A ∪ B )

Step 1: Show that A ∪ (B – A) ⊂ A ∪ B

Let x ∈ A ∪ (B – A)
Then either:

• x ∈ A ⇒ clearly x ∈ A ∪ B
• or x ∈ (B – A) ⇒ so x ∈ B and x ∉ A
• ⇒ therefore x ∈ A ∪ B

⇒ A ∪ (B – A) ⊂ A ∪ B

Step 2: Show that A ∪ B ⊂ A ∪ (B – A)

Let x ∈ A ∪ B
Then either:

• x ∈ A ⇒ clearly x ∈ A ∪ (B – A)
• or x ∈ B

Now two sub-cases:

• If x ∈ A ⇒ already included
• If x ∉ A ⇒ then x ∈ B – A

So x ∈ A ∪ (B – A)

⇒ A ∪ B ⊂ A ∪ (B – A)

Conclusion:
Since both directions of inclusion hold,
A ∪ (B – A) = A ∪ B

Hence Proved