Practicing the AP Board Solutions Class 11 Maths and Chapter 1 Inter 1st Year Maths Sets Solutions Exercise 1f Pdf Download will help students to clear their doubts quickly.
Intermediate 1st Year Maths Sets Solutions Exercise 1f
Sets Exercise 1f Solutions
Sets Class 11 Exercise 1f Solutions – Sets 1f Exercise Solutions
I.
Question 1.
Decide, among the following sets, which sets are subsets of one another.
A = {x: x ∈ R and x satisfy x2 – 8x + 12 = 0}
B = {2, 4, 6}
C = {2, 4, 6, 8,…}
D = {6}
Solution:
According to the question, we have,
A = {x: x ∈ R and x satisfies x2 – 8x + 12 = 0}
2 and 6 are the 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 ⊂ C, B ⊂ C, D ⊂ A ⊂ 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
Solution:
(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
Question 3.
Show that if A ⊂ B, then (C – B) ⊂ (C – A).
Solution:
Given A ⊂ B
Let us assume that x is any element such that x ∈ C – B
∴ x ∈ C and x ∉ B
So, x ∈ C – A
∴ C – B ⊂ C – A
Hence, proved.
Question 4.
Show that A ∩ B = A ∩ C need not imply B = C.
Solution:
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
II.
Question 1.
Let A, B, and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B = C.
Solution:
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
Question 2.
Show that the following four conditions are equivalent:
(i) A ⊂ B
(ii) A – B = φ
(iii) A ∪ B = B
(iv) A ∩ B = A
Solution:
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
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.
Question 3.
Show that for any sets A and B, A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B).
Solution:
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 x ∈ A ∪ B
x ∈ A or x ∈ B
(x ∈ A or x ∈ B) and (x ∈ A and x ∈ A)
⇒ x ∈ A or (x ∈ B and x ∉ A)
⇒ x ∈ A ∪ (B – A)
Thus, (A ∪ B) ⊂ A ∪ (B – A) ……(iv)
From equations (iii) and (iv), we get
A ∪ (B – A) = A ∪ B
Question 4.
Using properties of sets, show that
(i) A ∪ (A ∩ B) = A
(ii) A ∩ (A ∪ B) = A
Solution:
(i) To show A ∪ (A ∩ B) = A
We know that A ⊂ A
A ∩ B ⊂ A
∴ A ∪ (A ∩ B) ⊂ A
Also, according to the question, we have,
A ⊂ A ∪ (A ∩ B) ⊂ A ……..(i)
Also, according to the question we have,
A ⊂ A ∪ (A ∩ B) ………….(ii)
Hence, from equations (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
Question 5.
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)
Solution:
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 ∪ 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 ……..(ii)
Hence, from equations (i) and (ii),
we obtain A = B.
Question 6.
Find sets A, B, and C such that A ∩ B, B ∩ C, and A ∩ C are non-empty sets and A ∩ B ∩ C = φ.
Solution:
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.