Miscellaneous Exercise on Chapter 1
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 }.
Solution:
x2 - 8x + 12 = 0
x2 - 6x - 2x + 12 = 0
x(x - 6) -2 (x - 6) = 0
x - 6 = 0, x - 2 = 0;
x = 6 , x = 2
A = {2, 6} , B = { 2, 4, 6 }, C = { 2, 4, 6, 8, . . . }, D = { 6 }.
Every element of A is in B and C
∴ A ⊂ B, A ⊂ C,
Every Element of B is in C
∴ B ⊂ C ,
Every Element of D is in A, B and 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
Solution:
False, Let be A = {1}, and B = {{1}, 2}
then 1 ∈ A and A ∈ B but 1 ∉ B
∴ x ∈ A and A ∈ B but not emply x ∉ B
(ii) If A ⊂ B and B ∈ C , then A ∈ C
Solution: False,
Let A = {1, 2} and B = {1, 2, 3, 4} and C = {{1, 2, 3, 4}, 5, 6}
According to condition,
A ⊂ B and B ∈ C but A ∉ C
A ∈ B but does not emply A ∉ C
(iii) If A ⊂ B and B ⊂ C , then A ⊂ C
Solution: True,
Let A = {a, b} , B = {a, b, c} and C = {a, b, c, d}
Here, A ⊂ B ⇒ x ∈ A and x ∈ B ------------- (i)
Similarily, B ⊂ C ⇒ x ∈ B and x ∈ C ------------------ (ii)
So, A ⊂ B and B ⊂ C ⇒ A ⊂ C
(iv) If A ⊄ B and B ⊄ C , then A ⊄ C
Solution: false;
Let A = {a, b}, B = {b, c} and C = {a, b, d, e}
Here, A ⊄ B and B ⊄ C but A ⊂ C
So, A ⊄ B and B ⊄ C do not emply A ⊄ C
(v) If x ∈ A and A ⊄ B , then x ∈ B
Solution: False;
Let A = {1,2}, B = {2, 3, 5}
1 ∈ A for all x, but 1 ∉ B
Therefore, A ⊄ B which does not imply x ∈ B.
(vi) If A ⊂ B and x ∉ B , then x ∉ A
Solution: True
A ⊂ B given
then x ∈ A and x ∈ B
While 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.
Solution:
A ∪ B = A ∪ C ..................................... (i)
A ∩ B = A ∩ C ................................................. (ii)
taking equation (i)
(A ∪ B) ∩ C = (A ∪ C) ∩ C
Using distributive law;
(A ∩ C) ∪ (B ∩ C) = C [ ∵ (A ∪ C) ∩ C = C ]
(A ∩ B) ∪ (B ∩ C) = C From equ (i) [ ∵ A ∩ B = A ∩ C ]
C = (A ∩ B) ∪ (B ∩ C) ................... (iii)
Now
(A ∪ B) ∩ B = (A ∪ C) ∩ B
using distributive law
(A ∩ B) ∪ (B ∩ B) = (A ∩ B) ∪ (C ∩ B)
(A ∩ B) ∪ B = (A ∩ B) ∪ ( B ∩ C) [∵ (B ∩ B) = B]
B = (A ∩ B) ∪ ( B ∩ C) ....................... (iv) [∵ (A ∩ B) ∪ B = B ]
From equation (iii) and (iv)
B = C proved
Q4. Show that the following four conditions are equivalent :
(i) A ⊂ B
(ii) A – B = φ
(iii) A ∪ B = B
(iv) A ∩ B = A
Solution:
First we try to prove
A ⊂ B ⇔ A - B = φ
Given: A ⊂ B
To prove: A - B = φ
A ⊂ B ⇒ x ∈ A and x ∈ B
then (A ∪ B) = B
(A ∪ B) ∩ B = (B ∩ B)
(A ∩ B ) ∪ (B ∩ B) = (B ∩ B)
(A ∩ B ) ∪ B = B
then (A ∩ B ) = B .................................(i)
Now also, x ∈ A ∩ B
which gives A ∩ B = A .................. (ii) [∵ A ⊂ B ]
L H S = A - B
= A - (A ∩ B ) from eqa .... (i)
= A - A using equa .. (ii)
= φ
∴ A - B = φ proved
(ii) ⇔ (i)
Given : A - B = φ
To prove : A ⊂ B
Solution:
x ∈ A - B ⇒ x ∈ A and x ∉ B
⇒ x ∈ A and x ∈ B'
⇒ A ∩ B' Here x is the element of A and B' both
∴ A ∩ B' = φ [∵ A - B = φ ] given
Again,
Let x ∈ A ⇒ x ∉ B' [∵ A ∩ B' = φ ]
There is no element that is common (A ∩ B') resulting φ.
⇒ x ∈ B
So, x ∈ A ⇒ x ∈ B
∴ A ⊂ B Proved
(i) ⇔ (iii)
Given : A ⊂ B
To prove : A ∪ B = B
Solution:
x ∈ A ⊂ B ⇔ x ∈ A and x ∈ B
clearly, A ∪ B = B All elements of A is in set B. [∵ A ⊂ B ]
(i) ⇔ (iv)
Given : A ⊂ B
To prove : A ∩ B = A
Solution:
x ∈ A ⊂ B ⇔ x ∈ A and x ∈ B
then x ∈ A ∩ B [∵ x is element of set A as well as set B ]
Clearly, which gives A ∩ B = A
Hence Proved
Q5. Show that if A ⊂ B, then C – B ⊂ C – A.
Solution:
Given: A ⊂ B
To show: C – B ⊂ C – A.
x ∈ C – B ⇒ x ∈ C and x ∉ B ................................... (i)
x ∈ C – A ⇒ x ∈ C and x ∉ A ................................... (ii)
while A ⊂ B
then A - B = φ
from equa. (i) and (ii)
x ∈ C – B
x ∈ C – A
∴ C – B ⊂ C – A Proved
Q6. Assume that P ( A ) = P ( B ). Show that A = B
Solution:
Let x is any element of set A and it is an also element of its subset say X
such that x ∈ X
then
X ⊂ A ⇒ X ∈ P(A)
⇒ X ∈ P(B) [∵ P(A) = P(B) ]
⇒ X ⊂ B
⇒ x ∈ B
x ∈ A and x ∈ B
A ⊂ B .............................. (i)
similarily,
y is any elemet of set B and it is an also element of its subset say Y such that y ∈ Y
then
Y ⊂ B ⇒ Y ∈ P(B)
⇒ Y ∈ P(A)
Y ⊂ B ⇒ y ∈ Y and y ∈ A
B ⊂ A .................................................. (ii)
From equation (i) and (ii)
A = B Proved
Q7. Is it true that for any sets A and B, P (A) ∪ P (B) = P (A ∪ B)? Justify your answer.
Solution : False Let A = {0, 1} and B = {1, 2}
∴ 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}}
∴ P(A) ∪ P(B) ≠ P(A ∪ B)
Q8. Show that for any sets A and B, A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B)
Solution :
To show: A = (A ∩ B) ∪ (A – B)
Let x ∈ A
We have to show that x ∈ (A ∩ B) ∪ (A – B)
Case I x ∈ A ∩ B
Then, x ∈ (A ∩ B) ⊂ (A ∪ B) ∪ (A – B)
Case II x ∉ A ∩ B ⇒ x ∉ A or x ∉ B
∴ x ∉ B [x ∉ A] ∴ x ∉ A – B ⊂ (A ∪ B) ∪ (A – B)
∴ A ⊂ (A ∩ B) ∪ (A – B) … (1)
It is clear that A ∩ B ⊂ A and (A – B) ⊂ A
∴ (A ∩ B) ∪ (A – B) ⊂ A … (2)
From (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)
∴ 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)
∴ A ∪ B ⊂ A ∪ (B – A) … (4)
Hence, from (3) and (4), we obtain A ∪ (B – A) = A ∪B.
Q9. 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 … (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)
Q10. 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}
Thus, A ∩ B = {0} and A ∩ C = {0}
Now, A ∩ B = A ∩ C = {0}
However, B ≠ C [2 ∈ B and 2 ∉ C]
Q11. 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:
Let A and B be two sets such that A ∩ X = B ∩ X = f
and A ∪ X = B ∪ X for some set X.
To show: A = B
Here 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) [A ∪ X = B ∪ X]
= (B ∩ A) ∪ (B ∩ X) [Distributive law]
= (B ∩ A) ∪ Φ [B ∩ X = Φ]
= B ∩ A = A ∩ B …… (2)
Hence, from (1) and (2), we have that A = B.
Q12. 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 A = {0, 1}, B = {1, 2}, and C = {2, 0}.
Thus, A ∩ B = {1}, B ∩ C = {2}, and A ∩ C = {0}.
∴ A ∩ B, B ∩ C, and A ∩ C are non-empty.
However, A ∩ B ∩ C = Φ
Q13. In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?
Solution:
Let U be the set of all students who took part in the survey.
Let T be the set of students taking tea.
Let C be the set of students taking coffee.
Then n(U) = 600, n(T) = 150, n(C) = 225, n(T ∩ C) = 100
To find: Number of student taking neither tea nor coffee.
Therefore, we have to find n(T' ∩ C').
n(T' ∩ C') = n(T ∪ C)'
= n(U) – n(T ∪ C)
= n(U) – [n(T) + n(C) – n(T ∩ C)]
= 600 – [150 + 225 – 100]
= 600 – 275
= 325
Hence, 325 students were taking neither tea nor coffee.
Q14. In a group of students 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?
Solution:
Let U be the set of all students in the group.
Let E be the set of all students who know English.
Let H be the set of all students who know Hindi.
∴ H ∪ E = U
Then, n(H) = 100 and n(E) = 50
n( H U E ) = n(H) + n(E) – n(H ∩ E)
= 100 + 50 – 25
= 125
Hence, there are 125 students in the group.
Q15. In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I,11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find:
(i) the number of people who read at least one of the newspapers.
(ii) the number of people who read exactly one newspaper.
Solution:
Let A be the set of people who read newspaper H.
Let B be the set of people who read newspaper T.
Let C be the set of people who read newspaper I.
Then,
n(A) = 25, n(B) = 26, n(C) = 26 , n(A ∩ C) = 9, n(A ∩ B) = 11,
n(B ∩ C) = 8 , n(A ∩ B ∩ C) = 3
Let U be the set of people who took part in the survey.
(i) n(A U B U C) = n(A) + n(B) + n(C) – n(A ∩ B) – n(B ∩ C) – n(C ∩ A) + n(A ∩ B ∩ C)
= 25 + 26 + 26 – 11 – 8 – 9 + 3 = 52
Hence, 52 people read at least one of the newspapers.
(ii) Let a be the number of people who read newspapers H and T only.
Let b denote the number of people who read newspapers I and H only.
Let c denote the number of people who read newspapers T and I only.
Let d denote the number of people who read all three newspapers.
Accordingly, d = n(A ∩ B ∩ C) = 3
Now, n(A ∩ B) = a + d
n(B ∩ C) = c + d
n(C ∩ A) = b + d
∴ a + d + c + d + b + d = 11 + 8 + 9 = 28
Q16. In a survey it was found that 21 people liked product A, 26 liked product B and 29 liked product C. If 14 people liked products A and B, 12 people liked products C and A, 14 people liked products B and C and 8 liked all the three products. Find how many liked product C only.
Solution:
Let A, B, and C be the set of people who like product A, product B, and product C respectively.
Then, n(A) = 21,
n(B) = 26,
n(C) = 29,
n(A ∩ B) = 14,
n(C ∩ A) = 12,
n(B ∩ C) = 14,
n(A ∩ B ∩ C) = 8
The Venn diagram for the given problem-
It can be seen that number of people who like product C only is
{29 – (4 + 8 + 6)} = 11
