NCERT Solutions for Class 11 Maths
Chapter 1- Sets
Miscellaneous Exercise on Chapter 1 (Page No-24)
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 =0}
We will solve the equation x2 -8x+12=0
We can write this equation as
X2-6x-2x+12=0
X(x-6)-2(x-6)=0
(x-6)(x-2)=0
Therefore x=2 and 6
Hence, A = {2, 6}
B = {2, 4, 6}, C = {2, 4, 6, 8 …}, D = {6}
Therefore, D ⊂ A ⊂ B ⊂ C
Hence, A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C
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
For satisfying the given condition,
Let x=2 and
A = {2, 3} and B = {1, {2, 3}, 4}
Here,
2 ∈ {2, 3} and {2, 3} ∈ {1, {2, 3}, 4}
Hence, we get,
A ∈ B
We also know,
2 ∉ {1, {2, 3}, 4}
Hence it is False
(ii) False
For satisfying the given condition,
Let,
A = {1,2}
B = {1, 2, 3}
And, C = {{1, 2, 3}, 4}
Here,,
A ⊂ B
B ∈ C
But,
A ∉ C
Hence, It is false.
(iii) True
For satisfying the given condition,
Let
A= {1,2}
B={1,2,3}
C={1,2,3,4}
Here
A ⊂ B
B ⊂ C
And A ⊂ C
Hence it is true
(iv) False
For satisfying the given condition,
Let
A = {1, 2}
B = {4,5}
C = {1,2,6,7}
Here,
A ⊄ B
B ⊄ C
But,
A ⊂ C
Hence, it is False
(v) False
For satisfying the given condition,
Let
X=1
A= {1,2}
B={2,3}
Here,
x ∈ A
A ⊄ B
But
x ∉ B
Hence, it is False
(vi) True
For satisfying the given condition,
Let
A= {1,2}
B={1,2,3}
And
x=4
Here
A ⊂ B
x ∉ B
and
x ∉ B
Hence, it is True
3. Let A, B and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. show that B = C.
Solution:
As per question,
A ∪ B = A ∪ C
And,
A ∩ B = A ∩ C
We have to show,
B = C
Let,
x ∈ B
So,
x ∈ A ∪ B
x ∈ A ∪ C
Hence,
x ∈ A or x ∈ C
When x ∈ A, then,
And we have assumed that x ∈ B
∴ x ∈ A ∩ B
As, A ∩ B = A ∩ C
So, x ∈ A ∩ C
∴ x ∈ A and x ∈ C
∴ x ∈ C
∴ B ⊂ C
Similarly, we can show that C ⊂ B
Hence, B = C
4. Show that the following four conditions are equivalent:
(i) A ⊂ B (ii) A – B = Φ
(iii) A ∪ B = B (iv) A ∩ B = A
Solution:
As per the question,
We have to prove, (i) ⬌ (ii)
Here, (i) = A ⊂ B and (ii) = A – B = ϕ
Let A ⊂ B
and
Let A – B ≠ ϕ
Hence, there exists X ∈ A, X ≠ B,
But it is given that A⊂ B,
So it is not possible
∴ A – B = ϕ
And A⊂ B ⇒ A – B = ϕ
Let A – B = ϕ
We have to prove A ⊂ B
Let X∈ A
So, X ∈ B (if X ∉ B, then A – B ≠ ϕ)
Therefore A ⊂ B
Hence, A – B = ϕ => A ⊂ B
∴ (i) ⬌ (ii)
Let A ⊂ B
∴ B ⊂ A ∪ B ————-(a)
We have to prove, A ∪ B = B
Let, x ∈ A∪ B
∴ x ∈ A or x ∈ B
Case I: x ∈ B
∴ A ∪ B ⊂ B ——————-(b)
Now take Case II: X ∈ A
∴ X ∈ B
Because A ⊂ B
∴ A ∪ B ⊂ B —————–(c)
From (a), (b), (c)
AUB=B
Now to prove AUB=B ⇒ A ⊂ B
Let A ∪ B = B
Let X ∈ A
X ∈ A ∪ B
Because A ⊂ A ∪ B
⇒ X ∈ B
Because A ∪ B = B
∴ A⊂ B
Hence, (i) ⬌ (iii)
Now we have to prove (i) ⬌ (iv)
Let A ⊂ B
∴ A ∩ B ⊂ A —————–(d)
Let X ∈ A
Since, A ⊂ B
Therefore X ∈ B
Hence, X ∈ A ∩ B
∴ A ⊂ A ∩ B ————–(e)
From (d) and (e)
A = A ∩ B
Let A ∩ B = A
Let X ∈ A
⇒ X ∈ A ∩ B (as A ∩ B = A)
⇒ X ∈ B and X ∈ A
⇒ A ⊂ B
∴ (i) ⬌ (iv)
∴ (i) ⬌ (ii) ⬌ (iii) ⬌ (iv)
Hence, proved.
5. Show that if A ⊂ B, then C – B ⊂ C – A.
Solution:
Let 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.
6. Assume that P (A) = P (B). Show that A = B
Solution:
Let x ∈ A
Since, P (A) is the power set of set A, So it has all the subsets of set A.
A ∈ P (A)
A ∈ P (B) Because P(A)=P(B)
x ∈ B
∴ A ⊂ B
Similarly, we have:
Let x ∈ B
Since, P (B) is the power set of set B, So it has all the subsets of set B.
B ∈ P (B)
B ∈ P (A) Because P(A)=P(B)
x ∈ A
∴ B ⊂ A
SO,
If A ⊂ B and B ⊂ A
∴ A = B
7. Is it true that for any sets A and B, P (A) ∪ P (B) = P (A ∪ B)? Justify your answer.
Solution:
Let,
A = {1, 2}
And, B = {2, 3}
∴ A ∪ B = {1, 2, 3}
As per the question,
We have,
P (A) = {ϕ, {1}, {2}, {1, 2}}
P (B) = {ϕ, {2}, {3}, {2, 3}}
∴ P (A ∪ B) = {ϕ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, {1, 2, 3}}
Also,
P (A) ∪ P (B) = {ϕ, {1}, {2}, {3}, {1, 2}, {2, 3}}
∴ P (A) ∪ P (B ≠ P (A ∪ B)
Hence, the given statement is false.
8. 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)
Let X ∈ A
There may be two cases X ∈ (A ∩ B) or X ∉A ∩ B
In Case I,
X ∈ (A ∩ B)
⇒ X ∈ (A ∩ B) ∪ (A – B)
Because (A ∩ B) ⊂ (A ∩ B) ∪ (A – B)
∴ A ⊂ A ∩ B) ∪ (A – B)
In Case II,
X ∉A ∩ B
⇒ X ∉ B and X ∉ A
This is not possible as we have already assumed X ∈ A
Therefore only case I will prevail.
i.e. A ⊂ (A ∩ B) ∪ (A – B) ———————(i)
let X ∈ A ∩ B
therefore
X ∈ (A ∩ B) ∪ (A – B)
That means
(X ∈ A and X ∈ B) or (X ∈ A and X ∉ B)
Therefore
X ∈ A
Hence
(A ∩ B) ∪ (A – B) ⊂ A ——————————-(ii)
From (i) and (ii),
(A ∩ B) ∪ (A – B) = A
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 and X ∉A)
⇒ X ∈ (A ∪ B )
∴ A ∪ (B – A) ⊂ (A ∪ B) ————–(iii)
According to the question,
Let Y ∈ A∪B
Y ∈ A or Y ∈ B
(Y ∈ A or Y ∈ B) and (Y ∈ A and Y∉A)
⇒ Y ∈ A or (Y ∈ B and Y ∉A)
⇒ Y ∈ A ∪ (B – A)
∴ A ∪ B ⊂ A ∪ (B – A) —————-(iv)
From equations (iii) and (iv), we get:
A ∪ (B – A) = A ∪ B
9. Using properties of sets, show that:
(i) A ∪ (A ∩ B) = A
(ii) A ∩ (A ∪ B) = A.
Solution:
- A ∪ (A ∩ B) = ,(AUA) ∩ (A U B)
= A ∩ (A U B)
= A
Hence
A ∪ (A ∩ B) = A
(ii) A ∩ (A ∪ B) = (A ∩ A) ∪ (A ∩ B)
= A ∪ (A ∩ B)
= A
10. Show that A ∩ B = A ∩ C need not imply B = C.
Solution:
Let,
A = {1, 2}
B = {1, 3, 4}
And, C = {1, 5, 6}
According to the question,
A ∩ B = {1}
And,
A ∩ C = {1}
∴ A ∩ B = A ∩ C = {1}
But,
3 ∈ B and 3 ∉ C
Therefore, B ≠ C
11. 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,
A = A ∩ (A ∪ X) = A ∩ (B ∪ X) [Given A ∪ X = B ∪ X]
= (A ∩ B) ∪ (A ∩ X) [Using Distributive law]
= (A ∩ B) ∪ Φ [Given A ∩ X = Φ]
= A ∩ B ———–(i)
Now, B = B ∩ (B ∪ X)
= B ∩ (A ∪ X) [Given A ∪ X = B ∪ X]
= (B ∩ A) ∪ (B ∩ X) [Using Distributive law]
= (B ∩ A) ∪ Φ [Given B ∩ X = Φ]
= A ∩ B—————- (i)
Hence, from equations (i) and (ii), A = B.
12. 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= {1, 2}
B = {2, 3}
And, C = {3, 1}
According to the question,
A ∩ B = {2}
B ∩ C = {3}
And,
A ∩ C = {1}
∴ A ∩ B, B ∩ C and A ∩ C are not empty sets
and
A ∩ B ∩ C = Φ
13. 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 = set of all students who took part in the survey
T = set of students taking tea
C = set of the students taking coffee
Total number of students in a school who took part in the survey, n (U) = 600
Number of students taking tea, n (T) = 150
Number of students taking coffee, n (C) = 225
As 100 students were taking both tea and coffee
n (T ∩ C) = 100
number of students taking coffee or tea will be n (T U C)
n (T U C) = n (T) + n(C) – n (T ∩ C)
= 150 + 225 – 100
= 275
∴ Number of students taking neither coffee nor tea = n (U) – n (T ∩ C) = 600-225=375
14. 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 = the set of all students in the group
E = the set of students who know English
H = the set of the students who know Hindi
∴ H ∪ E = U
∴ n(H ∪ E) = n(U)
It is given that,
Number of students who know Hindi n (H) = 100
Number of students who knew English, n (E) = 50
Number of students who know both, n (H ∩ E) = 25
n (U) = n(H ∪ E) = n(H) + n(E) – n(H ∩ E)
= 100 + 50 – 25
= 125
∴ Total number of students in the group = n(U) = 125
15. 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:
(i) Let,
h = the set of people who read newspaper H
t = the set of people who read newspaper T
i = the set of people who read newspaper I
According to the question,
Number of people who read newspaper H, n (h) = 25
Number of people who read newspaper T, n (t) = 26
Number of people who read the newspaper I, n (i) = 26
Number of people who read both newspaper H and T, n (h ∩ t) = 11
Number of people who read both newspaper T and I, n (t ∩ i) = 8
Number of people who read both newspaper H and I, n (h ∩ i) = 9
And, Number of people who read all three newspaper H, T and I, n (h ∩ t ∩ i) = 3
Now, we have to find the number of people who read at least one of the newspaper
∴, we get.
n(h U t U i ) = n (h) + n (t) + n (i) – n (h ∩ t) – n (t ∩ i) – n (t ∩ i) + n (h ∩ t ∩ i)
= 25 + 26 + 26 – 11 – 8 – 9 + 3
= 80 – 28
= 52
∴ There are 52 students who read at least one newspaper.
(ii) Let,
a = the number of people who read newspapers H and T only
b = the number of people who read newspapers T and I only
c = the number of people who read H and I newspapers only
According to the question,
d = the number of people who read H, T and I newspapers = n (h ∩ t ∩ i) = 3
Now, we have:
n(h ∩ t) = a + d = 11
n(t ∩ i) = b + d = 8
n(h ∩ i) = c + d = 9
As per venn diagram,
Number of people read exactly one newspaper = n(h U t U i ) – (a+b+c+d)
∴ a + d + b +d + c + d = 11 + 8 + 9
a + b + c + d = 28 – 2d
= 28 – 6
= 22
∴ Number of people read exactly one newspaper = n(h U t U i ) – (a+b+c+d)
= 52 – 22
= 30
16. 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 = the set of people who like product A
B = the set of people who like product B
C = the set of people who like product C
So,
Number of students who like product A, n (A) = 21
Number of students who like product B, n (B) = 26
Number of students who like product C, n (C) = 29
a = the Number of students who like product A and B only
b = the Number of students who like product B and C only
c = the Number of students who like product C and A only
d = Number of students who like all three product = 8
Therefore,
Number of students who like both products A and B, n (A ∩ B) = 14= a+d
Number of students who like both products A and C, n(C ∩ A) = 12 = c+d
Number of students who like both product C and B, n (B ∩ C) = 14 = b+d
From the Venn diagram, we get,
c+d+b+d=n(C ∩ A)+ n (B ∩ C)=12+14
(b+c+d)+d=26
b+c+d=26-d= 26-8 =18
Number of students who only like product C = n (C) – [b+c+d]
= 29-18
= 11