In Exercise 1.7, we shall see more results on operations on sets. Students who are finding difficulty in understanding the concepts can refer to RD Sharma Class 11 Solutions which are prepared by expert tutors at BYJU’S in a simple and understandable language. For further assistance, students can download the RD Sharma Class 11 Solutions, which are available in PDF format for better understanding and to score well in their board exams.
RD Sharma Solutions for Class 11 Maths Exercise 1.7 Chapter 1 – Sets
Access answers to RD Sharma Solutions for Class 11 Maths Exercise 1.7 Chapter 1 – Sets
1. For any two sets A and B, prove that: A‘ – B‘ = B – A
Solution:
To prove, A’ – B’ = B – A
Firstly we need to show
A’ – B’ ⊆ B – A
Let x ∈ A’ – B’
⇒ x ∈ A’ and x ∉ B’
⇒ x ∉ A and x ∈ B (since, A ∩ A’ = ϕ )
⇒ x ∈ B – A
It is true for all x ∈ A’ – B’
∴ A’ – B’ = B – A
Hence Proved.
2. For any two sets A and B, prove the following:
(i) A ∩ (A‘ ∪ B) = A ∩ B
(ii) A – (A – B) = A ∩ B
(iii) A ∩ (A ∪ B’) = ϕ
(iv) A – B = A Δ (A ∩ B)
Solution:
(i) A ∩ (A’ ∪ B) = A ∩ B
Let us consider LHS A ∩ (A’ ∪ B)
Expanding
(A ∩ A’) ∪ (A ∩ B)
We know (A ∩ A’) =ϕ
⇒ ϕ ∪ (A∩ B)
⇒ (A ∩ B)
∴ LHS = RHS
Hence proved.
(ii) A – (A – B) = A ∩ B
For any sets A and B, we have De-Morgan’s law
(A ∪ B)’ = A’ ∩ B’, (A ∩ B) ‘ = A’ ∪ B’
Consider LHS
= A – (A–B)
= A ∩ (A–B)’
= A ∩ (A∩B’)’
= A ∩ (A’ ∪ B’)’) (since (B’)’ = B)
= A ∩ (A’ ∪ B)
= (A ∩ A’) ∪ (A ∩ B)
= ϕ ∪ (A ∩ B) (since A ∩ A’ = ϕ)
= (A ∩ B) (since, ϕ ∪ x = x, for any set)
= RHS
∴ LHS=RHS
Hence proved.
(iii) A ∩ (A ∪ B’) = ϕ
Let us consider LHS A ∩ (A ∪ B’)
= A ∩ (A ∪ B’)
= A ∩ (A’∩ B’) (By De–Morgan’s law)
= (A ∩ A’) ∩ B’ (since A ∩ A’ = ϕ)
= ϕ ∩ B’
= ϕ (since, ϕ ∩ B’ = ϕ)
= RHS
∴ LHS=RHS
Hence proved.
(iv) A – B = A Δ (A ∩ B)
Let us consider RHS A Δ (A ∩ B)
A Δ (A ∩ B) (since E Δ F = (E–F) ∪ (F–E))
= (A – (A ∩ B)) ∪ (A ∩ B –A) (since E – F = E ∩ F’)
= (A ∩ (A ∩ B)’) ∪ (A ∩ B ∩ A’)
= (A ∩ (A’ ∪ B’)) ∪ (A ∩ A’ ∩ B) (by using De-Morgan’s law and associative law)
= (A ∩ A’) ∪ (A ∩ B’) ∪ (ϕ ∩ B) (by using distributive law)
= ϕ ∪ (A ∩ B’) ∪ ϕ
= A ∩ B’ (since, A ∩ B’ = A–B)
= A – B
= LHS
∴ LHS=RHS
Hence Proved
3. If A, B, C are three sets such that A ⊂ B, then prove that C – B ⊂ C – A.
Solution:
Given, ACB
To prove: C – B ⊂ C – A
Let us consider, x ∈ C–B
⇒ x ∈ C and x ∉ B
⇒ x ∈ C and x ∉ A
⇒ x ∈ C – A
Thus, x ∈ C–B ⇒ x ∈ C – A
This is true for all x ∈ C–B
∴ C – B ⊂ C – A
Hence proved.
4. For any two sets A and B, prove that
(i) (A ∪ B) – B = A – B
(ii) A – (A ∩ B) = A – B
(iii) A – (A – B) = A ∩ B
(iv) A ∪ (B – A) = A ∪ B
(v) (A – B) ∪ (A ∩ B) = A
Solution:
(i) (A ∪ B) – B = A – B
Let us consider LHS (A ∪ B) – B
= (A–B) ∪ (B–B)
= (A–B) ∪ ϕ (since B–B = ϕ)
= A–B (since, x ∪ ϕ = x for any set)
= RHS
Hence proved.
(ii) A – (A ∩ B) = A – B
Let us consider LHS A – (A ∩ B)
= (A–A) ∩ (A–B)
= ϕ ∩ (A – B) (since A-A = ϕ)
= A – B
= RHS
Hence proved.
(iii) A – (A – B) = A ∩ B
Let us consider LHS A – (A – B)
Let, x ∈ A – (A–B) = x ∈ A and x ∉ (A–B)
x ∈ A and x ∉ (A ∩ B)
= x ∈ A ∩ (A ∩ B)
= x ∈ (A ∩ B)
= (A ∩ B)
= RHS
Hence proved.
(iv) A ∪ (B – A) = A ∪ B
Let us consider LHS A ∪ (B – A)
Let, x ∈ A ∪ (B –A) ⇒ x ∈ A or x ∈ (B – A)
⇒ x ∈ A or x ∈ B and x ∉ A
⇒ x ∈ B
⇒ x ∈ (A ∪ B) (since, B ⊂ (A ∪ B))
This is true for all x ∈ A ∪ (B–A)
∴ A ∪ (B–A) ⊂ (A ∪ B)…… (1)
Conversely,
Let x ∈ (A ∪ B) ⇒ x ∈ A or x ∈ B
⇒ x ∈ A or x ∈ (B–A) (since, B ⊂ (A ∪ B))
⇒ x ∈ A ∪ (B–A)
∴ (A ∪ B) ⊂ A ∪ (B–A)…… (2)
From 1 and 2, we get,
A ∪ (B – A) = A ∪ B
Hence proved.
(v) (A – B) ∪ (A ∩ B) = A
Let us consider LHS (A – B) ∪ (A ∩ B)
Let, x ∈ A
Then either x ∈ (A–B) or x ∈ (A ∩ B)
⇒ x ∈ (A–B) ∪ (A ∩ B)
∴ A ⊂ (A – B) ∪ (A ∩ B)…. (1)
Conversely,
Let x ∈ (A–B) ∪ (A ∩ B)
⇒ x ∈ (A–B) or x ∈ (A ∩ B)
⇒ x ∈ A and x ∉ B or x ∈ B
⇒ x ∈ A
(A–B) ∪ (A ∩ B) ⊂ A………. (2)
∴ From (1) and (2), We get
(A–B) ∪ (A ∩ B) = A
Hence proved.
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