Exercise Miscellaneous - Sets NCERT Solutions | Mathematics (Maths) Class 11 - Commerce PDF Download

Q1: Decide, among the following sets, which sets are subsets of one and another:
A= {x: x ∈ R and x satisfy x² – 8x + 12 = 0},
B = {2, 4, 6},
C = {2, 4, 6, 8…},
D = {6}.
Ans: According to the question,
We have,
A = {x: x ∈ R and x satisfies x² – 8x + 12 =0}
2 and 6 are the only solutions of x² – 8x + 12 = 0.
Hence, A = {2, 6}
B = {2, 4, 6}, C = {2, 4, 6, 8 …}, D = {6}
Hence, D ⊂ A ⊂ B ⊂ C
Hence, A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C

Q2: In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.
(i) If x ∈ A and A ∈ B, then x ∈ B
(ii) If A ⊂ B and B ∈ C, then A ∈ C
(iii) If A ⊂ B and B ⊂ C, then A ⊂ C
(iv) If A ⊄ B and B ⊄ C, then A ⊄ C
(v) If x ∈ A and A ⊄ B, then x ∈ B
(vi) If A ⊂ B and x ∉ B, then x ∉ A
Ans: (i) False
According to the question,
A = {1, 2} and B = {1, {1, 2}, {3}}
Now, we have,
2 ∈ {1, 2} and {1, 2} ∈ {1, {1, 2}, {3}}
Hence, we get,
A ∈ B
We also know,
{2} ∉ {1, {1, 2}, {3}}

(ii) False
According to the question
Let us assume that,
A {2}
B = {0, 2}
And, C = {1, {0, 2}, 3}
From the question,
A ⊂ B
Hence,
B ∈ C
But, we know,
A ∉ C

(iii) True
According to the question
A ⊂ B and B ⊂ C
Let us assume that,
x ∈ A
Then, we have,
x ∈ B
And,
x ∈ C
Therefore,
A ⊂ C

(iv) False
According to the question
A ⊄ B
Also,
B ⊄ C
Let us assume that,
A = {1, 2}
B = {0, 6, 8}
And,
C = {0, 1, 2, 6, 9}
∴ A ⊂ C

(v) False
According to the question,
x ∈ A
Also,
A ⊄ B
Let us assume that,
A = {3, 5, 7}
Also,
B = {3, 4, 6}
We know that,
A ⊄ B
∴ 5 ∉ B

(vi) True
According to the question,
A ⊂ B
Also,
x ∉ B
Let us assume that,
x ∈ A,
We have,
x ∈ B,
From the question,
We have, x ∉ B
∴ x ∉ A

Q3: Let A, B and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. show that B = C.
Ans: According to the question,
A ∪ B = A ∪ C
And,
A ∩ B = A ∩ C
To show,
B = C
Let us assume,
x ∈ B
So,
x ∈ A ∪ B
x ∈ A ∪ C
Hence,
x ∈ A or x ∈ C
When x ∈ A, then,
x ∈ B
∴ x ∈ A ∩ B
As, A ∩ B = A ∩ C
So, x ∈ A ∩ C
∴ x ∈ A or x ∈ C
x ∈ C
∴ B ⊂ C
Similarly, it can be shown that C ⊂ B
Hence, B = C

Q4: Show that the following four conditions are equivalent:
(i) A ⊂ B (ii) A – B = Φ
(iii) A ∪ B = B (iv) A ∩ B = A
Ans: According to the question,
To prove, (i) ⬌ (ii)
Here, (i) = A ⊂ B and (ii) = A – B ≠ ϕ
Let us assume that A ⊂ B
To prove, A – B ≠ ϕ
Let A – B ≠ ϕ
Hence, there exists X ∈ A, X ≠ B, but since A⊂ B, it is not possible
∴ A – B = ϕ
And A⊂ B ⇒ A – B ≠ ϕ
Let us assume that A – B ≠ ϕ
To prove: A ⊂ B
Let X∈ A
So, X ∈ B (if X ∉ B, then A – B ≠ ϕ)
Hence, A – B = ϕ => A ⊂ B
∴(i) ⬌ (ii)
Let us assume that A ⊂ B
To prove, A ∪ B = B
⇒ B ⊂ A ∪ B
Let us assume that, x ∈ A∪ B
⇒ X ∈ A or X ∈ B
Taking Case I: X ∈ B
A ∪ B = B
Taking Case II: X ∈ A
⇒ X ∈ B (A ⊂ B)
⇒ A ∪ B ⊂ B
Let A ∪ B = B
Let us assume that X ∈ A
⇒ X ∈ A ∪ B (A ⊂ A ∪ B)
⇒ X ∈ B (A ∪ B = B)
∴A⊂ B
Hence, (i) ⬌ (iii)
To prove (i) ⬌ (iv)
Let us assume that A ⊂ B
A ∩ B ⊂ A
Let X ∈ A
To prove, X ∈ A∩ B
Since, A ⊂ B and X ∈ B
Hence, X ∈ A ∩ B
⇒ A ⊂ A ∩ B
⇒ A = A ∩ B
Let us assume that A ∩ B = A
Let X ∈ A
⇒ X ∈ A ∩ B
⇒ X ∈ B and X ∈ A
⇒ A ⊂ B
∴ (i) ⬌ (iv)
∴ (i) ⬌ (ii) ⬌ (iii) ⬌ (iv)
Hence, proved

Q5: Show that if A ⊂ B, then C – B ⊂ C – A.
Ans: To show,
C – B ⊂ C – A
According to the question,
Let us assume that x is any element such that X ∈ C – B
∴ x ∈ C and x ∉ B
Since, A ⊂ B, we have,
∴ x ∈ C and x ∉ A
So, x ∈ C – A
∴ C – B ⊂ C – A
Hence, Proved.

Q6: Show that for any sets A and B,
A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B)
Ans: To Prove,
A = (A ∩ B) ∪ (A – B)
Proof: Let x ∈ A
To show,
X ∈ (A ∩ B) ∪ (A – B)
In Case I,
X ∈ (A ∩ B)⇒ X ∈ (A ∩ B) ⊂ (A ∪ B) ∪ (A – B)
In Case II,
X ∉A ∩ B
⇒ X ∉ B or X ∉ A
⇒ X ∉ B (X ∉ A)
⇒ X ∉ A – B ⊂ (A ∪ B) ∪ (A – B)
∴A ⊂ (A ∩ B) ∪ (A – B) (i)
It can be concluded that, A ∩ B ⊂ A and (A – B) ⊂ A
Thus, (A ∩ B) ∪ (A – B) ⊂ A (ii)
Equating (i) and (ii),
A = (A ∩ B) ∪ (A – B)
We also have to show,
A ∪ (B – A) ⊂ A ∪ B
Let us assume,
X ∈ A ∪ (B – A)
X ∈ A or X ∈ (B – A)
⇒ X ∈ A or (X ∈ B and X ∉A)
⇒ (X ∈ A or X ∈ B) and (X ∈ A and X ∉A)
⇒ X ∈ (B ∪A)
∴ A ∪ (B – A) ⊂ (A ∪ B) (iii)
According to the question,
To prove:
(A ∪ B) ⊂ A ∪ (B – A)
Let y ∈ A∪B
Y ∈ A or y ∈ B
(y ∈ A or y ∈ B) and (X ∈ A and X ∉A)
⇒ y ∈ A or (y ∈ B and y ∉A)
⇒ y ∈ A ∪ (B – A)
Thus, A ∪ B ⊂ A ∪ (B – A) (iv)
∴From equations (iii) and (iv), we get:
A ∪ (B – A) = A ∪ B

Q7: Using properties of sets, show that:
(i) A ∪ (A ∩ B) = A
(ii) A ∩ (A ∪ B) = A.
Ans: (i) To show: A ∪ (A ∩ B) = A
We know that,
A ⊂ A
A ∩ B ⊂ A
∴ A ∪ (A ∩ B) ⊂ A (i)
Also, according to the question,
We have:
A⊂ A ∪ (A ∩ B) (ii)
Hence, from equation (i) and (ii)
We have:
A ∪ (A ∩ B) = A
(ii) To show,
A ∩ (A ∪ B) = A
A ∩ (A ∪ B) = (A ∩ A) ∪ (A ∩ B)
= A ∪ (A ∩ B)
= A

Q8: Show that A ∩ B = A ∩ C need not imply B = C.
Ans: Let us assume,
A = {0, 1}
B = {0, 2, 3}
And, C = {0, 4, 5}
According to the question,
A ∩ B = {0}
And,
A ∩ C = {0}
∴ A ∩ B = A ∩ C = {0}
But,
2 ∈ B and 2 ∉ C
Therefore, B ≠ C

Q9: Let A and B be sets. If A ∩ X = B ∩ X = ϕ and A ∪ X = B ∪ X for some set X, show that A = B.
(Hints A = A ∩ (A ∪ X) , B = B ∩ (B ∪ X) and use Distributive law)
Ans: According to the question,
Let A and B be two sets such that A ∩ X = B ∩ X = ϕ and A ∪ X = B ∪ X for some set X.
To show, A = B
Proof:
A = A ∩ (A ∪ X) = A ∩ (B ∪ X) [A ∪ X = B ∪ X]
= (A ∩ B) ∪ (A ∩ X) [Distributive law]
= (A ∩ B) ∪ Φ [A ∩ X = Φ]
= A ∩ B (i)
Now, B = B ∩ (B ∪ X)
= B ∩ (A ∪ X) [A ∪ X = B ∪ X]
= (B ∩ A) ∪ (B ∩ X) … [Distributive law]
= (B ∩ A) ∪ Φ [B ∩ X = Φ]
= A ∩ B (i)
Hence, from equations (i) and (ii), we obtain A = B.

Q10: Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = Φ.
Ans: Let us assume, A {0, 1}
B = {1, 2}
And, C = {2, 0}
According to the question,
A ∩ B = {1}
B ∩ C = {2}
And,
A ∩ C = {0}
∴ A ∩ B, B ∩ C and A ∩ C are not empty sets
Hence, we get,
A ∩ B ∩ C = Φ