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

In Class 11 Maths chapter 1 miscellaneous exercise solutions, you will get a mixture of questions from all the exercises of this chapter. Miscellaneous exercise chapter 1 Class 11 consists of questions related to subsets, power sets, Venn's diagram, operations on sets like intersection and union of the sets, practical problems on the operation on sets, etc. This miscellaneous exercise chapter 1 Class 11 is a bit tougher than the other exercise of this chapter but also very important to get conceptual clarity. You must try to solve the problems of the miscellaneous exercise of this chapter. This NCERT book exercise will check your understanding of this chapter. If you are stuck while solving the NCERT problems, you can go through the NCERT solutions for Class 11 Maths chapter 1 miscellaneous exercise. If you are looking for the NCERT solutions from Classes 6 to 12 for Science and Math, you can click on the NCERT Solutions link.

**Also, see**

- Sets Exercise 1.1
- Sets Exercise 1.2
- Sets Exercise 1.3
- Sets Exercise 1.4
- Sets Exercise 1.5
- Sets Exercise 1.6

**Sets Class 11 Chapter 1-****Miscellaneous Exercise**

## Question:1 Decide, among the following sets, which sets are subsets of one and another:

A = { x : x R and x satisfy – 8x + 12 = 0 }, B = { 2, 4, 6 },

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

Answer:

Solution of this equation are – 8x + 12 = 0

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

X = 2,6

A = { 2,6 }

B = { 2, 4, 6 }

C = { 2, 4, 6, 8, . . . }

D = { 6 }

From the sets given above, we can conclude that A B, A C, D A, D B, D C, B C

Hence, we can say that D A B C

If x A and A B , then x B

Answer:

The given statment is false,

example: Let A = { 2,4 }

B = { 1,{2,4},5}

x be 2.

then, 2 { 2,4 } = x A and { 2,4 } { 1,{2,4},5} = A B

But 2 { 1,{2,4},5} i.e. x B

Question:2(ii) 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.

If A B and B C , then A C

Answer:

The given statement is false,

Let , A = {1}

B = { 1,2,3}

C = {0,{1,2,3},4}

Here, {1} { 1,2,3} = A B and { 1,2,3} {0,{1,2,3},4} = B C

But, {1} {0,{1,2,3},4} = A C

Question:2(iii) 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.

If A B and B C , then A C

Answer:

Let A ⊂ B and B ⊂ C

There be a element x such that

Let, x A

x B ( Because A B )

x C ( Because B C )

Hence,the statement is true that A C

Question:2(iv) 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.

If A B and B C , then A C

Answer:

The given statement is false

Let , A = {1,2}

B = {3,4,5 }

C = { 1,2,6,7,8}

Here, {1,2} {3,4,5 } = A B and {3,4,5 } { 1,2,6,7,8} = B C

But , {1,2} { 1,2,6,7,8} = A C

If x A and A B , then x B

Answer:

The given statement is false,

Let x be 2

A = { 1,2,3}

B = { 4,5,6,7}

Here, 2 { 1,2,3} = x A and { 1,2,3} { 4,5,6,7} = A B

But, 2 { 4,5,6,7} implies x B

Question:2(vi) 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.

If A B and x B , then x A

Answer:

The given statement is true,

Let, A B and x B

Suppose, x A

Then, x B , which is contradiction to x B

Hence, x A.

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.

Answer:

Let A, B, and C be the sets such that A B = A C and A B = A C

To prove : B = C.

A B = A + B - A B = A C = A + C - A C

A + B - A B = A + C - A C

B - A B = C - A C ( since A B = A C )

B = C

Hence proved that B= C.

Question:4 Show that the following four conditions are equivalent :

(i) A B(ii) A – B = (iii) A B = B (iv) A B = A

Answer:

First, we need to show A B A – B =

Let A B

To prove : A – B =

Suppose A – B

this means, x A and x B , which is not possible as A B .

SO, A – B = .

Hence, A B A – B = .

Now, let A – B =

To prove : A B

Suppose, x A

A – B = so x B

Since, x A and x B and A – B = so A B

Hence, A B A – B = .

Let A B

To prove : A B = B

We can say B A B

Suppose, x A B

means x A or x B

If x A

since A B so x B

Hence, A B = B

and If x B then also A B = B.

Now, let A B = B

To prove : A B

Suppose : x A

A A B so x A B

A B = B so x B

Hence,A B

ALSO, A B A B = B

NOW, we need to show A B A B = A

Let A B

To prove : A B = A

Suppose : x A

We know A B A

x A B Also ,A A B

Hence, A B = A

Let A B = A

To prove : A B

Suppose : x A

x A B ( replacing A by A B )

x A and x B

A B

A B A B = A

Question:5 Show that if A B, then C – B C – A.

Answer:

Given , A B

To prove : C – B C – A

Let, x C - B means x C but xB

A B so x C but xA i.e. x C - A

Hence, C – B C – A

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

Answer:

Given, P ( A ) = P ( B )

To prove : A = B

Let, x A

A P ( A ) = P ( B )

For some C P ( B ) , x C

Here, C B

Therefore, x B and A B

Similarly we can say B A.

Hence, A = B

Question:7 Is it true that for any sets A and B, P ( A ) ∪ P ( B ) = P ( A ∪ B )? Justify your answer.

Answer:

No, it is false.

To prove : P ( A ) ∪ P ( B ) P ( A ∪ B )

Let, A = {1,3} and B = {3,4}

A B = {1,3,4}

P(A) = { {},{1},{3},{1,3}}

P(B) = { {},{3},{4},{3,4}}

L.H.S = P(A) P(B) = { {},{1},{3},{1,3},{3,4},{4}}

R.H.S.= P(A B) = { {},{1},{3},{1,3},{3,4},{4},{1,4},{1,3,4}}

Hence, L.H.S. R.H.S

Question:8 Show that for any sets A and B,

A = ( A B ) ( A – B ) and A ( B – A ) = ( A B )

Answer:

A = ( A B ) ( A – B )

L.H.S = A = Red coloured area

R.H.S = ( A B ) ( A – B )

( A B ) = green coloured

( A – B ) = yellow coloured

( A B ) ( A – B ) = coloured part

Hence, L.H.S = R.H.S = Coloured part

A ( B – A ) = ( A B )

A = sky blue coloured

( B – A )=pink coloured

L.H.S = A ( B – A ) = sky blue coloured + pink coloured

R.H.S = ( A B ) = brown coloured part

L.H.S = R.H.S = Coloured part

Question:9(i) Using properties of sets, show that

A ( A B ) = A

Answer:

(i) A ( A B ) = A

We know that A A

and A B A

A ( A B ) A

and also , A A ( A B )

Hence, A ( A B ) = A

Question:9(ii) Using properties of sets, show that

A ( A B ) = A

Answer:

This can be solved as follows

(ii) A ( A B ) = A

A ( A B ) = (A A) ( A B )

A ( A B ) = A ( A B ) { A ( A B ) = A proved in 9(i)}

A ( A B ) = A

Question:10 Show that A B = A C need not imply B = C.

Answer:

Let, A = {0,1,2}

B = {1,2,3}

C = {1,2,3,4,5}

Given, A B = A C

L.H.S : A B = {1,2}

R.H.S : A C = {1,2}

and here {1,2,3} {1,2,3,4,5} = B C.

Hence, A B = A C need not imply B = C.

Question:11 Let A and B be sets. If A X B X and A X B X for some set X, show that A B.

Answer:

Given, A X B X and A X B X

To prove: A = B

A = A (AX) (A X B X)

= A (BX)

= (AB) (AX)

= (AB) (A X )

= (AB)

B = B (BX) (A X B X)

= B (AX)

= (BA) (BX)

= (BA) (B X )

= (BA)

We know that (AB) = (BA) = A = B

Hence, A = B

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

Answer:

Given, A B, B C and A C are non-empty sets

To prove : A B C

Let A = {1,2}

B = {1,3}

C = {3,2}

Here, A B = {1}

B C = {3}

A C = {2}

and A B C

Answer:

n ( taking tea) = 150

n (taking coffee) = 225

n ( taking both ) = 100

n(people taking tea or coffee) = n ( taking tea) + n (taking coffee) - n ( taking both )

= 150 + 225 - 100

=375 - 100

= 275

Total students = 600

n(students taking neither tea nor coffee ) = 600 - 275 = 325

Hence,325 students were taking neither tea nor coffee.

Answer:

n (hindi ) = 100

n (english ) = 50

n(both) = 25

n(students in the group ) = n (hindi ) + n (english ) - n(both)

= 100 + 50 - 25

= 125

Hence,there are 125 students in the group.

(i) the number of people who read at least one of the newspapers.

(ii) the number of people who read exactly one newspaper.

Answer:

n(H) = 25

n(T) = 26

n(I) = 26

n(H I) = 9

n( T I ) = 8

n( H T ) = 11

n(H T I ) = 3

the number of people who read at least one of the newspapers = n(HTI) = n(H) + n(T) + n(I) - n(H I) - n( T I ) - n( H T ) + n(H T I )

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

= 52

Hence, 52 people who read at least one of the newspapers.

(ii) number of people who read exactly one newspaper =

the number of people who read at least one of the newspapers - n(H I) - n( T I ) - n( H T ) + 2 n(H T I )

= 52 - 9- 8 -11 + 6

= 30

Hence, 30 number of people who read exactly one newspaper .

Answer:

n(A) = 21

n(B) = 26

n(C) = 29

n( A B) = 14

n( A C) = 12

n (B C ) = 14

n( A B C) = 8

n(liked product C only) = 29 - 4 -8 - 6 = 11

11 people like only product C.

**More About NCERT Solutions for Class 11 Maths Chapter 1 Miscellaneous Exercise:-**

NCERT syllabus Class 11 Maths chapter 1 miscellaneous solutions are consist of mixed concept questions from all the concepts of this chapter. There are a total of 16 questions in this chapter related to finding the union of sets, the intersection of sets, subsets, power sets, etc. The questions related to the practical application of union and intersection of sets, Venn's diagram are also covered in Class 11 Maths chapter 1 miscellaneous solutions. These questions are not only important for this exercise but also very useful in solving daily life problems using Venn's diagram.

**Also Read| **Sets Class 11th Notes

**Benefits of NCERT Solutions for Class 11 Maths Chapter 1 Miscellaneous Exercise:-**

As the name suggests miscellaneous exercise contain questions from all the topics of this chapter, you can revise all the important concepts of Class 11 Maths chapter 1 miscellaneous exercise solutions.

You may not be able to solve miscellaneous problems at first as these problems are a bit difficult but after going through the Class 11 Maths chapter 1 miscellaneous solutions you will be to solve these problems by yourself.

**Also see-**

**NCERT Solutions of Class 11 Subject Wise**

**Subject Wise NCERT Exampler Solutions**

Happy learning!!!

## Frequently Asked Question (FAQs) - NCERT Solutions for Miscellaneous Exercise Chapter 1 Class 11 - Sets

**Question: **Does miscellaneous exercise is important for the CBSE exam ?

**Answer: **

As most of the questions in the CBSE exam are not asked from the miscellaneous exercise, it is not as important for CBSE exams but it is very important for the competitive exams.

**Question: **How many questions are given in the miscellaneous exercise sets ?

**Answer: **

There is a total of 16 questions and 7 solved examples given in the miscellaneous exercise sets.

**Question: **How many exercises are there in CBSE Class 11 Maths chapter 1 ?

**Answer: **

There are a total of seven exercises including miscellaneous exercise in the CBSE Class 11 Maths chapter 1.

**Question: **What is the use of Venn's diagram ?

**Answer: **

Venn's diagram is useful in solving problems related to the union of sets, the intersection of sets. It is also useful in the chapters like relations and functions, probability.

**Question: **Find all the subsets of the set { –1, 0} ?

**Answer: **

φ, {–1}, {0}, {–1, 0} are the subsets of the given set.

**Question: **Find the power set of set A = {1,0} ?

**Answer: **

Power set of set A = { φ, {1}, {0}, {1, 0} }

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