ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets Ex 1.5

Utilizing Understanding ISC Mathematics Class 11 Solutions Chapter 1 Sets Ex 1.5 as a study aid can enhance exam preparation.

ML Aggarwal Class 11 Maths Solutions Section A Chapter 1 Sets Ex 1.5

Question 1.
State whether the following statemcnts are true or false :
(i) A ∪ A = A
(ii) A ∩ A = Φ
(iii) A ∪ A’ = ξ
(iv) A ∩ A’ = Φ
(v) A ⊂Φ = A ⇒ A ∪ B = B
(vi) A ⊂ B ⇒ A ∩ B = A
(vii) A ∩ Φ = A
(viii) A ∪ Φ = A
Solution:
(i) A ∪ A = {x : x ∈ A or x ∈ A}
= {x : x ∈ A} = A
∴ Given given statement is true.

(ii) A ∩ A = {x : x ∈ A and x ∈ A}
= {x : x ∈ A) = A
∴ given statement is false.

(iii) A ∪ A’ = {x : x ∈ A or x ∈ A’}
= {x : x ∈ A or x ∉ A}
= {x : x ∈ ξ} = ξ

(iv) A ∩ A’ = {x : x ∈ A and x ∈ A’}
= {x : x ∈ A and x ∉ A} = Φ
∴ given statement is true.

(v) Given A ⊂ B …………………(1)
T.P. A ∪ B = B
∀ x ∈ A ∩ B
⇒ x ∈ A or x ∈ B
⇒ x ∈ B or x ∈ B [∵ by (1) ; A ⊂ B]
⇒ x ∈ B
⇒ A ∪ B ⊂ B ……………..(2)
∀ x ∈ B and A ⊂ B
x ∈ B or x ∈ A
x ∈ A ∪ B
B ⊂ A ∪ B
From (2) and (3) ; we have
A ∪ B = B.
Thus given statement is true.

(vi) Given A ⊂ B …………………(1)
T.P. A ∩ B = A
∀ x ∈ A ∩ B
⇒ x ∈ A and x ∈ B (∵ A ⊂ B)
⇒ x ∈ A [∵ A ⊂ B]
∴ A ∩ B ⊂ A …………………(2)
∀ x ∈ A ⇒ x ∈ A and x ∈ B (∵ A ⊂ B)
⇒ x ∈ A ∩ B
⇒ A ⊂ A ∩ B ………………..(3)
∴ from (2) and (3) ; we have
A ∩ B = A
∴ given statement is true.

(vii) A ∩ Φ = {x : x ∈ A ∩ Φ}
= {x : x ∈ A and x ∈ Φ}
= {x : x ∈ A} = A
∴ given statement is false.

Question 2.
Fill in the blanks to make each of the following a true statement :
(i) Φ’ ∩ A = …………..
(ii) ξ ∩ A = ……………..
Solution:
(i) Now Φ’ = {x : x ∈ Φ’}
= {x : x ∉ Φ}
= {x : x ∈ ξ} = ξ
Also,
Φ’ ∩ A = ξ ∩A
= {x : x ∈ ξ, and x ∈ A}
= {x : x ∈ A) = A

(ii) Now ξ’ = {x : x ∈ ξ’}
= {x : x ∉ ξ}
= {x : x ∈ Φ} = A
∴ ξ’ ∩ A = Φ ∩ A
= {x : x ∈Φ and x ∈ A}
= {x : x ∈ Φ} = Φ

Question 3.
For any two sets A and B, is it truc that P (A) ∪ P (B) = P (A ∪ B) ? Justfy your answer.
Solution:
Let us take A = {1, 2}
and B = {2, 3}
∴ P(A) = {Φ, (1), {2}, {1, 2})
P (B) = {Φ, {2}, {3}, (2, 3))
Thus P (A) ∪ P (B)
= {Φ, {1), {2}, (3), (1,2). {2,3}}
A ∪ B = {1, 2, 3}
P (A ∪ B) = {Φ {1}, {2}, {3}, {1, 2} {1, 3}, {2, 3}, {1, 2, 3}}
Clearly P (A) ∪ P (B) ≠ P (A ∪ B).

Question 4.
Let A, B be two sets. Prove that (A – B) ∪ B = A if and only if B ⊂ A.
Solution:
Let (A – B) ∪ B = A
T.P B ⊂ A
⇒ (A ∩ B’) ∪ B = A
⇒ (A ∪ B) ∩ (B’ ∪ B) = A
⇒ (A ∪ B) ∩ U = A
⇒ A ∪ B = A
⇒ B ⊂ A
Converse :
B ⊂ A
T.P (A – B) ∪ B = A
(A – B) ∪B = (A ∩ B’) ∪ B
= (A ∪ B)∩ (B’ ∪ B)
= (A ∪ B) ∩ U
= A ∪ B = A [∵ B ⊂ A]

Question 5.
For any sets A and B, prove that P(A) ∪ P(B) ⊂ P(A ∪ B).
Solution:
Let X ∈ P(A) ∪ P(B)
⇒ X ∈ P(A) or X ∈ P(B)
⇒ X ⊂ A or X ⊂ B
⇒ X ⊂ A ∪B
⇒ X ∈ P(A ∪ B)
Thus P (A) ∪ P (B) ⊂ P (A ∪ B)

Question 6.
Show that the following four conditions are equivalent:
(i) A ⊂ B
(ii) A – B =
(iii) A ∩ B = A
(iv) A ∪ B = B.
Solution:
Given A ⊂ B
∴ A ∩ B = A
∀ x ∈ A – B
⇔ x ∈ A and x ∉ B
⇔ x ∈ B and x ∉ B
[∵ A ⊂ B]
⇔ x ∈ B and x ∈ B^c
⇔ x ∈ B ∩ B^c = Φ
∴ A – B = Φ

Aliter:
Let A ⊂ B T.P. A – B = Φ
⇒ A ∩ B’ ⊂B ∩ B’ = Φ
⇒ A – B ⊂ Φ
also Φ be a subset of every set.
Φ ⊂ A – B
Thus, A – B = Φ

Conversely:
Let A – B = Φ
T.P. A ⊂ B
Now,
A = A ∩ ξ
= A ∩ (B ∪ B’)
= (A ∩ B) ∪ (A ∩ B’)
[∵ A – B = Φ
⇒ A ∩ B’ = Φ]
A = A ∩ B …………………….(1)
Since A ∩ B ⊂ B = A ⊂ B [using(1)]
∴ A – B = Φ iff A ⊂ B
∴ (i) ↔ (ii)
From (1) ;
If A – B = Φ then A ∩ B = A

Conversely:
When A ∩ B = A T.P. A – B = Φ
Now A – B = A ∩ B’
= A ∩ B ∩ B’
= A ∩ Φ
= Φ [using given]
Thus A – B = Φ iff A ∩ B = A
∴ (ii) ↔ (iii)
Given A ∩ B = A T.P. A ∪ B = B
A ∪ B = (A ∩ B) ∪ B
[∵ A ∩ B = A]
= (A ∪ B) ∩ B = B
[∵ A ∪ B ⊃ B]
Clearly B ⊂ A ∪ B

Conversely :
A ∪ B = B T.P. A ∩ B = A
Clearly A ∩ B ⊂ A ……………….(2)
Now, A ∩ B = A ∩ (A ∪ B)
= (A ∩ A) ∪ (A ∩ B)
= A ∪ (A ∩ B)
[∵ A ∩ B ⊂ A]
Thus, (iii) ↔ (iv)
Given, A ∪ B = B T.P. A ⊂ B
∀ x ∈ A
⇒ x ∈ A or x ∈ B
⇒ x ∈ A ∪B
⇒ x ∈ B
∴ A ⊂ B.

Conversely:
A ⊂ B T.P. A ∪ B = B
⇒ A ∪ B ⊂ B ∪ B
⇒ A ∪ B ⊂ B ………………(3)
and B be a subset of A ∪ B always
∴ B ⊂ A ∪ B ……………………..(4)
From (3) and (4) ; we have
A ∪ B = B
Thus (iv) ↔ (i)
Hence all the four conditions are equivalent.

Long Answer questions (6 to 10) :

Question 7.
For any two sets A and B, prove that
(i) (A – B) ∪ B = A ∪ B
(ii) (A – B) ∩ B = Φ
Solution:
(i) (A – B) ∪ B = (A ∩ B’) ∪ B
= (A ∪ B) ∩ (B’ ∪ B)
= (A ∪ B) ∩ U
[distributive law]
[where U = universal set]
= A ∪ B

(ii) (A – B) ∩ B = (A ∩ B’) ∩ B
= A ∩ (B’ ∩ B) [Associative law]
= A ∩ Φ = Φ

Question 8.
For any two sets A and B, show that
(i) (A ∪ B) – B = A – B
(ii) A – (A ∩ B) = A – B.
Solution:
(i) (A ∪ B) – B = (A ∪ B) ∩ B’
= (A ∩ B’) ∪ (B ∩ B’) [distributive law]
= (A ∩ B’) ∪ Φ
= A ∩ B’
= A – B

(ii) A – (A ∩ B) = A ∩ (A ∩ B)’
= A ∩ (A’ ∪ B’) [using Demorgan’s law]
= (A ∩ A’) ∪ (A ∩ B’)
= Φ ∪ (A ∩ B’)
= A ∩ B’
= A – B

Question 9.
For any sets A, B and C, using properi les of sets, prove that
(i) A – (A – B) = A ∩ B
(ii)(A – B) ∪ (A – C) = A – (B ∩ C)
(iii) (A – B) ∩ (C – B) = (A ∩ C) – B
(iv) A – (B – C) = (A – B) ∪ (A ∩ C).
Solution:
(i) A – (A – B) = A ∩ (A – B)’
= A ∩ (A ∩ B’)’
= A ∩ (A’ ∪ (B’)’) [using demorgan’s law]
= A ∩ (A’ ∪ B)
= (A ∩ A’) ∪ (A ∩ B)
= Φ ∩ (A ∩ B)
= A ∩ B

(ii) (A – B) ∪ (A – C) = (A ∩ B’) ∪ (A ∩ C’)
= A ∩ (B’ ∪ C’) [distributive law]
= A ∩ (B ∩ C)’
= A – (B ∩ C) [using Demorgan’s law]
(A – B) ∩ (C – B) = (A ∩ B’) ∩ (C ∩ B’)
= (B’ ∩ A) ∩ (B’ ∩ C)

(iv) A – (B – C) = A ∩ (B – C)’ = A ∩ (B ∩ C’)’
= A ∩ (B’ ∪ (C’)’) [Demorgan law]
= A ∩ (B’ ∪ C)
= (A ∩ B’) ∪ (A ∩ C)
= (A – B) ∪ (A ∩ C)

Question 10.
For any two sets A and B, prove that (A – B) ∪ (B – A) = (A ∪ B) – (A ∩ B).
Solution:
(A – B)u(B – A) = (A ∩ B’) ∪ (B ∩ A’)
= {(A ∩ B’) ∪ B} ∩ {(A n B’) ∪ A’
= {(A ∪ B) ∩ (B’ ∪ B)} ∩ {(A ∪ A’) ∩ (B’ ∪ A’)}
= {(A ∪ B) ∩ U) ∩ {U ∩(B’ ∪ A’)}
= (A ∪ B) ∩ (B’ ∪ A’)
= (A ∪ B) ∩ (B ∩ A)’ [Dernorgan’s law]
= (A ∪ B) – (B ∩ A)