Referring to the AP Inter 1st Year Maths Study Material Chapter 1 Sets Exercise 1f Solutions makes it easier to understand complex problems.
AP Inter 1st Year Maths Sets Solutions Exercise 1f
I.
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}.
Solution:
A = {x : x ∈ R and x satisfies x2 – 8x + 12}
2 and 6 are the only solutions of x2 – 8x + 12 = 0. ∴ A = {2, 6)
B = {2, 4, 6}, C = {2, 4, 6, 8, ……..}, D = {6}. ∴ 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.
Solution:
i) False
Let A = {1, 2} and B = {1, {1, 2), {3}}.
Now 2 ∈ {1, 2} and {1, 2} ∈ {{3}, 1, {1, 2}}.
∴ A ∈ B. However, 2 ∉ {{3}, 1, {1, 2}}
ii) False
Let A = {2}, B = (0, 2} and C = {1, {0, 2}, 3}. As A ⊂ B.
∴ B ∈ C. However, A ∉ C.
iii) True
Let A ⊂ B and B ⊂ C.
Let x ∈ A ⇒ x ∈ B [∵ A ⊂ B]
⇒ x ∈ C [∵ B ⊂ C]
∴ A ⊂ C.
iv) False
Let A = {1, 2, 9} and B = {0, 6, 8} and C = {0, 1, 2, 6, 9}
Accordingly, A ⊄ B and B ⊄ C. However, A ⊂ C
v) False
Let A = {1, 2} and B = {1, 3}. Here x = 2 ⊄ A and A ct B but x ∉ B.
vi) True
A ⊂ B ⇒ all elements of A are in B.
∴ If x ∈ B; it means x is not an element of B.
∵ all elements of A are also in B.
If x is not in B, it cannot be in A.
Question 3.
Show that if A ⊂ B, then (C – B) ⊂ (C – A).
Solution:
Let A c B. To show that C – B ⊂ C – A.
Let x ∈ C – B
⇒ x ∈ C and x ∉ B
⇒ x ∈ C and x ∉ A [A ⊂ B]
⇒ x ∈ C – A
∴ C – B ⊂ C – A.
Question 4.
Show that A ∩ B = A ∩ C need not imply B = C.
Solution:
Let A = {0, 1}, B = {0, 2, 3} and C = {0, 4, 5}
Accordingly, A ∩ B = {0} and A ∩ C = {0}
Here, A ∩ B = A ∩ C = {0}
However, B ≠ C [2 ∈ B and 2 ∉ 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:
Given that A ∪ B = A ∪ C and A ∩ B = A ∩ C
Let x ∈ B
⇒ x ∈ A ∪ B [∵ B ⊂ A ∪ B]
⇒ x ∈ A ∪ C [∵ A ∪ B = A ∪ C]
⇒ x ∈ A or x ∈ C
⇒ If x ∈ A then x ∈ A ∩ B [∵ x ∈ B]
⇒ x ∈ A ∩ C [∵ A ∩ B = A]
⇒ x ∈ C
∴ B ⊂ C
Similarly, we can show 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:
To prove (i) is equivalent to (ii)
From (i) A ⊂ B.
It means that all elements of A are in B. i.e., there is no element of A that is not in B. i.e., A – B = Φ (which is (ii)).
Thus (i) ⇔ (ii).
To prove (i) is equivalent to (iii).
From (i) A ⊂ B.
It means that all elements of A are in B.
Then A ∪ B = set of all elements of A and B = B (which is (iii)).
Thus, (i) ⇔ (iii).
To prove (i) is equivalent to (iv) :
From (i) A ⊂ B.
It means that all elements of A are in B.
Then A ∩ B = Set of all elements common to A and B.
= A (which is (iv)).
Thus, (i) ⇔ (iv)
We have proved that (i) ⇔ (ii), (i) ⇔ (iii) and (i) ⇔ (iv).
Thus, (i) ⇔ (ii) ⇔ (iii) ⇔ (iv).
Question 3.
Show that for any sets A and B.
Solution:
A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B).
Solution:
i) To show that A = (A ∩ B) ∪(A-B).
Consider R.H.S. = (A ∩ B) ∪ (A – B).
= (A ∩ B) ∪ (A ∩ B’) (∵ By Def.of difference of sets A – B = A ∩ B’).
= A ∩ (B ∪ B’) [∵ By distributive law],
= A ∩ U [∵ A ∪ A’ – U].
= A = L.H.S
∴ A = (A ∩ B) ∪ (A – B).
ii) To show that A ∪ (B – A) = (A ∪ B)
Consider A ∪ (B – A)
= A ∪ (B ∩ A’) (∵ By def. of difference of sets, A – B = A ∩ B’).
= (A ∪ B) ∩ (A ∪ A’) [∵ By the distributive law],
= (A ∪ B) ∩ U (∵ A ∪ A’ = U).
= A ∪ B
∴ 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 that, A ∪ (A ∩ B) = A
We know that,
A ⊂ A
A ∩ B ⊂ A
∴ A ∪ (A ∩ B) ⊂ A …………(1)
Also, A ⊂ A ∪ (A ∩ B) ………….(2)
From (1) and (2), A ∪ (A ∩ B) = A
ii) To show : A ∩ (A ∪ B) = A
A ∩ (A ∪ B) = (A ∩ A) ∩ (A ∪ B) = A ∪ (A ∩ B) = A {From (i)}.
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:
Given A and B be two sets such that A ∩ X = B ∩ X = Φ and A ∪ X = B ∪ X for some set X.
To show that A = B
It can be seen that
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) ……….(1)
Now, B = B ∩ (B ∪ X)
= B ∩ (A ∪ X)
= (B ∩ A) ∪ (B ∩ X)
= (B ∩ A) ∪ Φ
= (B ∩ A) …………(2)
Hence, from (1) 8s (2), we obtain A = B.
Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and
A ∩ B ∩ C = Φ.
Let A = {0, 1}, B = {1, 2} and C = {2, 0}
Accordingly, A ∩ B = {1}, B ∩ C = {2} & A ∩ C = {0}
A ∩ B, B ∩ C and A ∩ C are non-empty.
However, A ∩ B ∩ C = Φ