NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise

NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise is part of NCERT Solutions for Class 11 Maths. Here we have given NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise.

• Sets Class 11 Ex 1.1
• Sets Class 11 Ex 1.2
• Sets Class 11 Ex 1.3
• Sets Class 11 Ex 1.4
• Sets Class 11 Ex 1.5
• Sets Class 11 Ex 1.6

Board	CBSE
Textbook	NCERT
Class	Class 11
Subject	Maths
Chapter	Chapter 1
Chapter Name	Sets
Exercise	Miscellaneous Exercise
Number of Questions Solved	16
Category	NCERT Solutions

NCERT Solutions for Class 11 Maths Chapter 1 Sets Miscellaneous Exercise

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.
Here A = {x : x ∈ R and x satisfies x² – 8x + 12 = 0}
= {x : x ∈ R and (x – 6)(x – 2) = 0} = {2, 6}
B = {2, 4, 6}, C = {2, 4, 6, 8,…….} and D = {6}
Now, A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B and 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.

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.
Solution.
Given that A ∩ B = A ∩C and A ∪ B=A ∪ 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
Solution.
(i) ⇒ (ii)
A – B = {x: x∈ A and x∉B]
Since A⊂B
∴ A – B = φ

(ii) ⇒ (iii)
A – B = φ ⇒ A⊂B ⇒ A∪B = B

(iii) ⇒ (iv)
AuB = B ⇒A⊂B ⇒ A∩B = A

(iv) ⇒ (i)
A∩B = A ⇒ A⊂B
Thus (i) ⇔ (ii) ⇔ (iii) ⇔ (iv).

Question 5.
Show that if A ⊂ B, then C – B ⊂ C – A.
Solution.
Let x ∈ C – B ⇒x ∈ C and x ∉ B
⇒ x ∈ C and x ∉ A [∵ A ⊂ B]
⇒ x ∈ C – A
Hence C – B ⊂C – A

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

Question 7.
Is it true that for any sets A and B, P(A) ∪ P(B) = P(A ∪ B) Justify your answer.
Solution.
No, it is not true.
Take A = {1, 2} and B = {2,3}

Question 8.
Show that for any sets A and B,
A = (A ∩ B) ∪ (A – B) and A ∪ (B – A) = (A ∪ B)
Solution.
(A ∩ B) ∪ (A – B) = (A ∩ B) ∪ (A ∩ B’)
= A ∩ (B ∪ B’) (By distributive law)
= A ∩ U = A
Hence, A = (A ∩ B) ∪ (A – B)
Also A ∪ (B – A) = A ∪ (B ∩ A’)
= (A ∪ B) ∩ (A ∪ A’) (By distributive law)
= (A ∪ B) ∩ U= A ∪ B
Hence, A ∪ (B – A) = A ∪ B.

Question 9.
Using properties of sets, show that
(i) A ∪ (A ∩ B)=A
(ii) A ∩ (A ∪ B) = A.
Solution.
(i) We know that if A ⊂ B, then
A ∪ B = B. Also, A ∩ B ⊂ A
∴ A ∪ (A ∩ B) = A.
(ii) We know that if A ⊂ B,
then A ∩ B = A Also, A ⊂ A ∪ B
∴ A ∩ (A ∪ B) = A.

Question 10.
Show that A ∩ B = A ∩ C need not imply B = C.
Solution.
Let A = {1, 2, 3, 4}, B = {2, 3, 4, 5, 6}, C = {2, 3, 4, 9,10}.
∴ A ∩ B = [1, 2,3,4} ∩ {2,3,4, 5, 6]
= {2, 3, 4}
A ∩ C = {1, 2, 3, 4} ∩ {2, 3, 4, 9, 10} = {2, 3, 4}
Now we have A ∩ B = A ∩ C. But B ≠ C.

Question 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.
Here
A ∪ X = B ∪ X for some set X

Question 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.
Take A = {1, 2}, B = {1, 4} and C = {2, 4}
Now, A ∩ B = {1} ≠ φ, B ∩ C = {4} ≠ φ and
A ∩ C = {2} ≠ φ
But A ∩ B ∩ C = φ.

Question 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 T be the set of students who takes tea and C be the set of students who takes coffee. Here, n(T) = 150, n(C) = 225 and n(C ∩ T) = 100
We know that
n(C ∪ T) = n(C) + n(T) – n(C ∩ T)
= 150 + 225 -100 = 275
∴ Number of students taking either tea or coffee = 275
∴ Number of students taking neither tea nor coffee = 600 – 275 = 325.

Question 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 H be the set of students who know Hindi and E be the set of students who know English.
Here n(H) = 100, n(E) = 50 and n(H ∩ E) = 25
We know that
n(H ∪ E) = n(H) + n(E) – n(H ∩ E)
= 100 + 50 – 25 = 125

Question 15.
In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper H, 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.

Question 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.

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