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# NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 1 Sets

NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 1 Sets are the most important for the preparation of class 11 maths for examination purposes because NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 1 Sets contains the solutions of all type of questions available in Miscellaneous Exercise Chapter 1 Sets .Miscellaneous Exercise Chapter 1 Sets of class 11 maths is the extract of all the exercises from 1.1 to 1.6 hence study of this exercise enable you to clear all the concepts on sets that is necessary for proper understanding of the questions when you would face your maths exam.

## NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 1 Sets

Q1. Decide ,among the following sets, which sets are subsets of one and another:

A = {x : x ∈ R and x satify x²-8x + 12 =0}

B ={2,4,6}

C ={2,4,6,8}

D ={6}

Ans. A = {x : x ∈ R and x satify x²-8x + 12 =0}

Solving the equation

x²-8x + 12 =0

x² -6x -2x +12 =0

x(x -6) -2(x -6) = 0

( x-6)(x -2) =0

x =6,2

∴ A = {2,6}

B ={2,4,6},C ={2,4,6,8},D ={6}

It is clear that

A⊂ B,A⊂C, B ⊂ 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, 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

Ans.

(i) We are given that if  x ∈A and A ∈B Then x∈ B

Let  A = {1,2} and B={{1,2},3,4}

1∉ B or 2 ∉ B,so it is  false

(ii) If A⊂ B and B ∈C then A∈C

Let  A ={1,2},B={1,2,3},C={5,6,{1,2,3},7}

Here A⊂B and B ∈C but A ∉C,so  it is false

(iii)If A⊂B and B⊂ C then A⊂C

Let A  ={1,2} ,B ={1,2,,3}  ,C ={1,2,,3,4}

Here A⊂C,so  it  is  true

(iv)If A ⊄ B and B ⊄ C then A ⊄C

Let A={1,2},B={3,4},C={1,2,5,6}

Here A⊂C,so  it is  false

(v) If x ∈A and A ⊄ B then x∈B

Let  A={1,2,3}, B={1,2,4,5}

Here 3 ∉B,so it is false

(vi) If A ⊂B and x∉B then x ∉ A

Let  A={1,2,3}, B={1,2,3,4,5}

Here 6 ∉B  6 ∉A, so it is true

NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 1 Sets

Q3. Let A, B, and C be the sets such that A B = A C and A ∩ B = A ∩ C. show that B = C.

Ans. The given sets A, B, and C are such that A ∪ B = A ∪ C and A ∩ B = A ∩ C

In order to prove B=C, let’s prove B⊂C and C⊂B

Let x ∈B

x∈ A ∪ B (Every element belongs to B also the elements of A ∪ B)

x ∈A ∪ C(Since A ∪ B = A ∪ C)

⇒ x∈A or x ∈C

In the first case x ∈ A

Also x∈B

Therefore x∈A ∩ B ( x is common in the set A and set B)

Since A ∩ B = A ∩ C

∴ x∈A ∩ C

⇒x ∈A and x∈C

∴ If x ∈ B then x∈C

Therefore B⊂ C (Every element of B belongs to the set C)….(i)

Similarly we can solve C⊂B….(ii)

From (i) and (ii) equation

B = C,Hence proved

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Q4. Show that the following four conditions are equivalent:

(i) A B (ii) A – B = Φ

(iii) A B = B (iv) A ∩ B = A

Ans. Showing the condition (i) A ⊂ B  and (ii) A – B = Φ are equivalent or not

A ⊂ B shows that every elements of A is available in B

Therefore A – B = Φ

Hence both conditions are equivalent

Showing the condition (ii) A – B = Φ and A ∪ B = B are equivalent

A ∪ B = B shows that all the elements of A are in B

Therefore A -B=Φ

NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 1 Sets

Q5.Show that if A  B, then C – B  C – A.

Ans. The condition given to us is

A⊂B (all the elements of A are in the set B)

Let x ∈ C – B…..(i)

⇒x ∈ C and x∉B

Since A⊂B

∴ x∈C and x∉A

Therefore x∈ C -A….(ii)

From (i) and (ii)

C – B ⊂ C -A,Hence proved

Q6.Assume that P (A) = P (B). Show that A = B

Ans. We are given that

P (A) = P (B)

In order to prove A = B,let’s prove A⊂B and B⊂A

P(A) is the power set of the set A implies all subsets of A belongs to P(A)

A is one of the subset of A

∴ A ∈ P(A) [ e.g Let A ={1,2} then P(A) ={{1},{2},{1,2},Φ}⇒{1,2}∈P(A) ]

Since P (A) = P (B)

∴ A ∈ P (B)

If set A is in the power set of B then all the elements of set A is in set B

∴A⊂ B…..(i)

Similarly we can show B⊂A….(ii)

From (i) and (ii)

A = B

NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 1 Sets

Q7. Is it true that for any sets A and B, P (A)  P (B) = P (A  B)? Justify your answer.

Ans. For justifying P (A) ∪ P (B) = P (A ∪ B),let us take an examples

Let A = {1,2} and B ={2,3}

∴ A∪B ={1,2,3}….(i)

P(A) = {{1},{2},{1,2},Φ} and P(B) = {{2},{3},{2,3},Φ}

P (A) ∪ P (B) = {{1},{2},{1,2},{3},{2,3},Φ}….(ii)

From (i) equation

P (A ∪ B) = {{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3},Φ}…(iii)

From equation (ii) and the equation (iii),we have

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)

Ans. For showing

A = (A ∩ B) ∪ (A – B)

Taking RHS

(A ∩ B) ∪ (A – B)

=(A ∩ B) ∪ (A∩ B’)[ Since A -B =(A∩ B’)]

Applying the distributive property of the sets (X ∩ Y) ∪ (X∩ Z)=X ∩ (Y∪Z)

= A∩(B∪B’)

= A∩U ( Since U =B∪B’)

= A =RHS

For showing A ∪ (B – A) = (A ∪ B)

Taking LHS

A ∪ (B – A)

=A∪(B∩A’) [ Since A -B =(A∩ B’)]

=(A∪B)∩ (A∪A’)[ Distributive property of the sets]

=(A∪B)∩U[Since A∪A’= U]

=A∪B= RHS,Hence proved

Q9. Using properties of sets, show that:
(i) A
(A ∩ B) = A
(ii) A ∩ (A
B) = A

Ans. (i) We have to show A ∪ (A ∩ B) = A

Taking LHS A ∪ (A ∩ B)

Applying distributive property

= (A∪A) ∩(A∪B)

= A ∩(A∪B) (since A∪A =A)

= A=RHS (since all elements of A are in A∪B)

(ii) We have to show A ∩ (A ∪ B) = A

Applying distributive property

= (A∩A)∪(A∩B)

= A ∪(A∩B)

= A=RHS [ from  (i) ]

NCERT Solutions for Class 11 Maths Miscellaneous Exercise Chapter 1 Sets

Q10.Show that A ∩ B = A ∩ C need not imply B = C.

Ans.Let A = {1,2,3}, B = (2,3,4},C={2,3,5}

Here A ∩ B ={2,3} and A ∩ C ={2,3}

Hence A ∩ B = A ∩ C but B≠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)

Ans.We are given three sets A,B and X such that

If A ∩ X = B ∩ X = ϕ and A ∪ X = B ∪ X

We have to prove

A = B

Taking LHS

A = A ∩ (A ∪ X) [ Using the property of the sets]

= (A∩A) ∪ (A∩X) (Using the distributive property of the sets]

=A∪(A∩X)

=A∪(B∩X) (it is given to us A∩X =B∩X)

=(A∪B)∩(A∪X)[Using the distributive property of the sets]

=(B∪A)∩(B∪X)[Since A∪B = B∪A and it is given A ∪ X = B ∪ X]

=B∩(A∪X)

=B∩(B∪X)[it is given that A∪X =B∪X]

=B( property of the sets B= B∩(B∪X)

Hence 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.

Ans.We are given A ∩ B, B ∩ C and A ∩ C are non-empty sets

Let A = {1,2}, B = (1,3},C = {2,3}

According to the question

A ∩ B = {1}

B ∩ C ={3}

A ∩ C ={2}

Here A ∩ B, B ∩ C and A ∩ C are non-empty sets

Therefore we get A ∩ B ∩ C = Φ

See the video for the solutions of the questions from 8 to 12

See the video for NCERT Solutions of the Chapter 1 Exercise 1.6

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

Ans. The survey was taken of 600 students ,n(U) = 600

The number of students who took tea = n(T) = 150

The number of students who took coffee = n(C) = 225

The number of students who took both tea and coffee = n(T∩C) = 100

For getting students who  were taking neither tea nor coffee,let’s determine the students who were taking either tea or coffee, i.e n(T∪C)

We have

n(T∪C) = n(T) + n(C) – n(T∩C)

n(T∪C)= 150 + 225 – 100 = 275

The number of persons who were taking neither tea nor coffee =n(U) –n(T∪C) = 600 -275 = 325

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?

Ans.The  number of students who know Hindi,n(H) =100

The  number of students who know English,n(E) =50

The  number of students who know both Hindi and English  n(H∩E) =25

The number of students in the group are n(H∪E) =?

We have

n(H∪E) = n(H) + n(E)  – n(H∩E)

=100  +50 -25 =125

Hence the number of students in the group are 125

Q15.In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read the 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.

Ans. Total number of people surveyed are 60,so n(U) = 60

The number of the people who read news paper H,n(H) = 25

The number of the people who read news paper T,n(T) = 26

The number of the people who read news paper I,n(I) = 26

The number of the people who read both news paper H and I,n(H∩I) = 9

The number of the people who read both news paper H and T,n(H∩T) = 11

The number of the people who read both news paper T and I,n(T∩I) = 8

The number of the people who read all three news paper n(H∩T∩I) = 3

(i) We have to determine the number of people who read at least one of the newspapers,n(H∪T∪I) =?

n(H∪T∪I) = n(H) +n(T)+n(I)-n(H∩T)-n(T∩I) -n(H∩I)+n(H∩T∩I)

= 25 +26 +26 -11-8-9 + 3

= 80 -28 = 52

Hence the number of people who read at least one of the newspaper is 52

(ii) Now, we have to determine the number of people who read exactly one newspaper.

Drawing the venn diagram

It is shown in the venn diagram, we have to find out the colour portions

Here  n(H∩T)=a+d=11, n(H∩I)=b+d=9 and n(T∩I)=d+c=8,d =n(H∩T∩I)=3

We are required to find out

n(H∩T) + n(H∩I)+ n(T∩I) =11+9+8 =28

a +d + b+d +d+c =28

But as it is seen in the venn diagram 28 is the sum of a,b,c and d

a +b+c+d =28 -2d

It is given to us,d =n(H∩T∩I)=3

a + b+c +d = 28 -2×3 =28 -6 =22

The number of people who read exactly one newspaper = 52 -22 = 30

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.

Ans.The number of people who liked product A ,n(A) = 21

The number of people who liked product B ,n(B) = 26

The number of people who liked product C,n(C) = 29

The number of people who liked product A and B ,n(A∩B) = 14

The number of people who liked product B and C ,n(B∩C) = 14

The number of people who liked product C and A ,n(C∩A) = 12

The number of people who liked all three products A and B and C ,n(A∩B∩C) = 8

Drawing the venn diagram

n(A∩B) = 14⇒b +d = 14

d = n(A∩B∩C) = 8

b= 14 – 8 = 6

n(B∩C) = 14 ⇒d +c = 14

8 +c =14

c = 14 – 8 = 6

n(C∩A) = 12⇒a +d =12

a + 8 =12

a = 12 -8 = 4

The people who liked product C only = n(C) – (a +d +c) =29 -(4 +8+6) =29-18=11

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