NCERT Solutions Class 11 Mathematics Sets Exercise 1-Misc

Question 1:
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}.

Answer:
A = {x: x ∈ R and x satisfies x²– 8x + 12 = 0}
Now, x² – 8x + 12 = 0
⇒ x² – 2x – 6x + 12 = 0
⇒ x(x – 2) – 6(x – 2) = 0
⇒ (x – 2)(x – 6) = 0
⇒ x = 2, 6
So, 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}
So, D ⊂ A ⊂ B ⊂ C
Hence, A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C

Question 2:
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

Answer:
(i) False.
Let A = {1, 2} and B = {1, {1, 2}, {3}}
Now, 2 ∈ {1, 2} and {1, 2} ∈ {1, {1, 2}, {3}}
Hence, A = B
However, 2 ∉ {1, {1, 2}, {3}}
(ii) False.
Let A = {2}, B = {0, 2} and C = {1, {0, 2}, {3}}
As A ⊂ B and B ⊂ C
However, A ∉ C
(iii) True.
Let A ⊂ B and B ⊂ C.
Let x ∈ A
⇒ x ∈ B           [Since A ⊂ B]
⇒ x ∈ C           [Since B ⊂ C]
So, A ⊂ C
(iv) False.
A = {1, 2}, B = {0, 6, 8} and C = {0, 1, 2, 6, 9}
Let A ⊄ B and B ⊄ C
Accordingly,
However, A ⊂ C
(v) False.
Let A = {3, 5, 7} and B = {3, 4, 6}
Now, 5 ∈ A and A ⊄ B
However, 5 ∉ B
(vi) True.
Let A ⊂ B and x ∉ B.
To show: x ∉ A
If possible, suppose x ∈ A.
Then, x ∈ B, which is a contradiction as x ∉ B
So, x ∉ A

Question 3:
Let A, B, and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B = C.

Answer:
Let, A, B and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C.
To show: B = C
Let x ∈ B
⇒ x ∈ A U B         [B ⊂ A U B]
⇒ x ∈ A U C         [A U B = A U C]
⇒ x ∈ A or x ∈ C
Case 1: x ∈ A
Also, x ∈ B
So, x ∈ A ∩ B
⇒ x ∈ A ∩ C              [Since A ∩ B = A ∩ C]
⇒ x ∈ A and x ∈ C
So, x ∈ C
Hence, B ⊂ C
Similarly, we can show that C ⊂ B.
So, B = C

Question 4:
Show that the following four conditions are equivalent:
(i) A ⊂ B                (ii) A – B = φ                   (iii) A ∪ B = B                       (iv) A ∩ B = A

Answer:
First, we have to show that (i) ⇔ (ii).
Let A ⊂ B
To show: A – B ≠ φ
If possible, suppose A – B ≠ φ
This means that there exists x ∈ A, x ≠ B, which is not possible as A ⊂ B.
∴ A – B = φ
∴ A ⊂ B ⇒ A – B = φ
Let A – B = φ
To show: A ⊂ B
Let x ∈ A
Clearly, x ∈ B because if x ∉ B, then A – B ≠φ
∴ A – B = φ ⇒ A ⊂ B
So, (i) ⇔ (ii)
Let A ⊂ B
To show: A U B = B
Clearly, B ⊂ A U B
Let x ∈ A U B
Case 1: x ∈ A
⇒ x ∈ B            [Since A ⊂ B]
So, A U B ∈ B
Case 2: x ∈ B
Then, A U B = B
Conversely, let A U B = B
Let x ∈ A
⇒ x ∈ A U B          [Since A ⊂ A U B]
⇒ x ∈ B                 [Since A U B = B]   
So, A ⊂ B
Hence, (i) ⇔ (iii)
Now, we have to show that (i) ⇔ (iv).
Let A ⊂ B
Clearly A ∩ B ⊂ A
Let x ∈ A
We have to show that x ∈ A ∩ B
As A ⊂ B, x ∈ B
So, x ∈ A ∩ B
⇒ A ⊂ A ∩ B
Hence, A = A ∩ B
Conversely, suppose A ∩ B = A
Let x ∈ A
⇒ x ∈ A ∩ B
⇒ x ∈ A and x ∈ B
⇒ x ∈ B
So, A ⊂ B
Hence, (i) ⇔ (iv).

Question 5:
Show that if A ⊂ B, then C – B ⊂ C – A.

Answer:
Let A ⊂ B
To show: C – B ⊂ C – A
Let x ∈ C – B
⇒ x ∈ C and x ∉ B
⇒ x ∈ C and x ∉ A        [Since A ⊂ B]
⇒ x ∈ C – A
Hence, C – B ⊂ C – A

Question 6:
Assume that P(A) = P(B). Show that A = B

Answer:
Let P(A) = P(B)
To show: A = B
Let x ∈ A
A ∈ P(A) = P(B)
Hence, x ∈ C, for some C ∈ P(B)
Now, C ⊂ B
So, x ∈ B
Hence, A ⊂ B
Similarly, B ⊂ A
So, A = B

Question 7:
Is it true that for any sets A and B, P(A) ∪ P(B) = P(A ∪ B)? Justify your answer.

Answer:
It is false.
Let A = {0, 1} and B = {1, 2}
Now, A ∪ B = {0, 1, 2}
P(A) = { φ, {0}, {1}, {0, 1}}
P(B) = { φ, {1}, {2}, {1, 2}}
P(A ∪ B) = { φ, {0}, {1}, {2}, {0, 1}, {1, 2}, {0, 2}, {0, 1, 2}}
P(A) ∪ P(B) = { φ, {0}, {1}, {0, 1}, {2}, {1, 2}}
So, P(A) ∪ P(B) ≠ P(A ∪ B)

Question 8:
Show that for any sets A and B,
A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B)
 

Answer:
To show: A = (A ∩ B) ∪ (A – B)
Let x ∈ A
We have to show that x ∈ (A ∩ B) ∪ (A – B)
Case 1: x ∈ A ∩ B
Then, x ∈ (A ∩ B) ⊂ (A ∪ B) ∪ (A – B)
Case 2: x ∉ A ∩ B
⇒ x ∉ A or x ∉ B
⇒ x ∉ B [x ∉ A]
⇒ x ∉ A – B ⊂ (A ∪ B) ∪ (A – B)
So, A ⊂ (A ∩ B) ∪ (A – B)  ………. (1)
It is clear that
A ∩ B ⊂ A and (A – B) ⊂ A
So, (A ∩ B) ∪ (A – B) ⊂ A   …….. (2)
From equation (1) and (2), we obtain
A = (A ∩ B) ∪ (A – B)
To prove: A ∪ (B – A) ⊂ A ∪ B
Let 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 or x ∉ A)
⇒ x ∈ (A ∪ B)
So, A ∪ (B – A) ⊂ (A ∪ B)  ………. (3)
Next, we show that (A ∪ B) ⊂ A ∪ (B – A).
Let y ∈ A ∪ B
⇒ y ∈ A or y ∈ B
⇒ (y ∈ A or y ∈ B) and (y ∈ A or y ∉ A)
⇒ y ∈ A or (y ∈ B and y ∉ A)
⇒ y ∈ A ∪ (B – A)
So, A ∪ B ⊂ A ∪ (B – A)   ………. (4)
Hence, from equation (3) and (4), we get
A ∪ (B – A) = A ∪ B.

Question 9:
Using properties of sets, show that
(i) A ∪ (A ∩ B) = A                                (ii) A ∩ (A ∪ B) = A.

Answer:
(i) To show: A ∪ (A ∩ B) = A
We know that
A ⊂ A
A ∩ B ⊂ A
So, A ∪ (A ∩ B) ⊂ A   …….. (1)
Also, A ⊂ A ∪ (A ∩ B)   ……. (2)
Now, from equation (1) and (2), we get
A ∪ (A ∩ B) = A
(ii) To show: A ∩ (A ∪ B) = A
A ∩ (A ∪ B) = (A ∩ A) ∪ (A ∩ B)  = A ∪ (A ∩ B)    = A        [from equation (1)]