NCERT Solutions for Miscellaneous Exercise Chapter 1 Class 11 - Sets

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

Edited By Vishal kumar | Updated on Nov 15, 2023 09:42 AM IST

## NCERT Solutions for Class 11 Maths Chapter 1: Sets Miscellaneous Exercise- Download Free PDF

NCERT Solutions for Class 11 Maths Chapter 1 : Sets Miscellaneous Exercise- 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 class 11 chapter 1 maths miscellaneous solutions 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.

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

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

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

Solution of this equation are $x^{2}$ – 8x + 12 = 0

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

X = 2,6

$\therefore$ A = { 2,6 }

B = { 2, 4, 6 }

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

D = { 6 }

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

Hence, we can say that D $\subset$ A $\subset$ B $\subset$ C

If x $\in$ A and A $\in$ B , then x $\in$ B

The given statment is false,

example: Let A = { 2,4 }

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

x be 2.

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

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

If A $\subset$ B and B $\in$ C , then A $\in$ C

The given statement is false,

Let , A = {1}

B = { 1,2,3}

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

Here, {1} $\subset$ { 1,2,3} = A $\subset$ B and { 1,2,3} $\in$ {0,{1,2,3},4} = B $\in$ C

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

If A $\subset$ B and B $\subset$ C , then A $\subset$C

Let A ⊂ B and B ⊂ C

There be a element x such that

Let, x $\in$ A

$\Rightarrow$ x $\in$ B ( Because A $\subset$ B )

$\Rightarrow$ x $\in$ C ( Because B $\subset$ C )

Hence,the statement is true that A $\subset$ C

If A $\not\subset$ B and B $\not\subset$ C , then A $\not\subset$ C

The given statement is false

Let , A = {1,2}

B = {3,4,5 }

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

Here, {1,2} $\not\subset$ {3,4,5 } = A $\not\subset$ B and {3,4,5 } $\not\subset$ { 1,2,6,7,8} = B $\not\subset$ C

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

If x $\in$ A and A $\not\subset$ B , then x $\in$ B

The given statement is false,

Let x be 2

A = { 1,2,3}

B = { 4,5,6,7}

Here, 2 $\in$ { 1,2,3} = x $\in$ A and { 1,2,3} $\not \subset$ { 4,5,6,7} = A $\not \subset$B

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

If A $\subset$ B and x $\notin$ B , then x $\notin$ A

The given statement is true,

Let, A $\subset$ B and x $\notin$ B

Suppose, x $\in$ A

Then, x $\in$ B , which is contradiction to x $\notin$ B

Hence, x $\notin$ A.

Let A, B, and C be the sets such that A $\cup$ B = A $\cup$ C and A $\cap$ B = A $\cap$ C

To prove : B = C.

A $\cup$ B = A + B - A $\cap$ B = A $\cup$ C = A + C - A $\cap$ C

A + B - A $\cap$ B = A + C - A $\cap$ C

B - A $\cap$ B = C - A $\cap$ C ( since A $\cap$ B = A $\cap$ C )

B = C

Hence proved that B= C.

(i) A $\subset$ B(ii) A – B = $\phi$ (iii) A $\cup$ B = B (iv) A $\cap$ B = A

First, we need to show A$\subset$ B $\Leftrightarrow$ A – B = $\phi$

Let A $\subset$ B

To prove : A – B = $\phi$

Suppose A – B $\neq$ $\phi$

this means, x $\in$ A and x $\not =$ B , which is not possible as A $\subset$ B .

SO, A – B = $\phi$.

Hence, A $\subset$ B $\implies$ A – B = $\phi$.

Now, let A – B = $\phi$

To prove : A $\subset$ B

Suppose, x $\in$ A

A – B = $\phi$ so x $\in$ B

Since, x $\in$ A and x $\in$ B and A – B = $\phi$ so A $\subset$ B

Hence, A$\subset$ B $\Leftrightarrow$ A – B = $\phi$.

Let A$\subset$ B

To prove : A $\cup$ B = B

We can say B $\subset$ A $\cup$ B

Suppose, x $\in$ A $\cup$ B

means x $\in$ A or x $\in$ B

If x $\in$ A

since A$\subset$ B so x $\in$ B

Hence, A $\cup$ B = B

and If x $\in$ B then also A $\cup$ B = B.

Now, let A $\cup$ B = B
To prove : A$\subset$ B

Suppose : x $\in$A

A $\subset$ A $\cup$ B so x $\in$ A $\cup$ B

A $\cup$ B = B so x $\in$ B

Hence,A$\subset$ B

ALSO, A$\subset$ B $\Leftrightarrow$ A $\cup$ B = B

NOW, we need to show A $\subset$ B $\Leftrightarrow$ A $\cap$ B = A

Let A $\subset$ B

To prove : A $\cap$ B = A

Suppose : x $\in$ A

We know A $\cap$ B $\subset$ A

x $\in$ A $\cap$ B Also ,A $\subset$ A $\cap$ B

Hence, A $\cap$ B = A

Let A $\cap$ B = A

To prove : A $\subset$ B

Suppose : x $\in$ A

x $\in$ A $\cap$ B ( replacing A by A $\cap$ B )

x $\in$ A and x $\in$ B

$\therefore$ A $\subset$ B

A $\subset$ B $\Leftrightarrow$ A $\cap$ B = A

Given , A $\subset$ B

To prove : C – B $\subset$ C – A

Let, x $\in$ C - B means x$\in$ C but x$\notin$B

A $\subset$ B so x$\in$ C but x$\notin$A i.e. x $\in$ C - A

Hence, C – B $\subset$ C – A

Given, P ( A ) = P ( B )

To prove : A = B

Let, x$\in$ A

A $\in$ P ( A ) = P ( B )

For some C $\in$ P ( B ) , x $\in$ C

Here, C $\subset$ B

Therefore, x $\in$ B and A $\subset$ B

Similarly we can say B $\subset$ A.

Hence, A = B

No, it is false.

To prove : P ( A ) ∪ P ( B ) $\neq$ P ( A ∪ B )

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

A $\cup$ B = {1,3,4}

P(A) = { {$\phi$},{1},{3},{1,3}}

P(B) = { {$\phi$},{3},{4},{3,4}}

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

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

Hence, L.H.S.$\neq$ R.H.S

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

A = ( A $\cap$ B ) $\cup$ ( A – B ) and A $\cup$ ( B – A ) = ( A $\cup$ B )

A = ( A $\cap$ B ) $\cup$ ( A – B )

L.H.S = A = Red coloured area

R.H.S = ( A $\cap$ B ) $\cup$ ( A – B )

( A $\cap$ B ) = green coloured

( A – B ) = yellow coloured

( A $\cap$ B ) $\cup$ ( A – B ) = coloured part

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

A $\cup$ ( B – A ) = ( A $\cup$ B )

A = sky blue coloured

( B – A )=pink coloured

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

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

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

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

A $\cup$ ( A $\cap$ B ) = A

(i) A $\cup$ ( A $\cap$ B ) = A

We know that A $\subset$ A

and A $\cap$ B $\subset$ A

A $\cup$ ( A $\cap$ B ) $\subset$ A

and also , A $\subset$ A $\cup$ ( A $\cap$ B )

Hence, A $\cup$ ( A $\cap$ B ) = A

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

A $\cap$ ( A $\cup$ B ) = A

This can be solved as follows

(ii) A $\cap$ ( A $\cup$ B ) = A

A $\cap$ ( A $\cup$ B ) = (A $\cap$ A) $\cup$ ( A $\cap$ B )

A $\cap$ ( A $\cup$ B ) = A $\cup$ ( A $\cap$ B ) { A $\cup$ ( A $\cap$ B ) = A proved in 9(i)}

A $\cap$ ( A $\cup$ B ) = A

Let, A = {0,1,2}

B = {1,2,3}

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

Given, A $\cap$ B = A $\cap$ C

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

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

and here {1,2,3} $\not =$ {1,2,3,4,5} = B $\not =$ C.

Hence, A $\cap$ B = A $\cap$ C need not imply B = C.

Given, A $\cap$ X $=$B $\cap$ X $=$ $\phi$ and A $\cup$ X $=$ B $\cup$ X

To prove: A = B

A = A $\cap$(A$\cup$X) (A $\cap$ X $=$B $\cap$ X)

= A $\cap$(B$\cup$X)

= (A$\cap$B) $\cup$ (A$\cap$X)

= (A$\cap$B) $\cup$ $\phi$ (A $\cap$ X $=$ $\phi$)

= (A$\cap$B)

B = B $\cap$(B$\cup$X) (A $\cap$ X $=$B $\cap$ X)

= B $\cap$(A$\cup$X)

= (B$\cap$A) $\cup$ (B$\cap$X)

= (B$\cap$A) $\cup$ $\phi$ (B $\cap$ X $=$ $\phi$)

= (B$\cap$A)

We know that (A$\cap$B) = (B$\cap$A) = A = B

Hence, A = B

Given, A $\cap$ B, B $\cap$ C and A $\cap$ C are non-empty sets

To prove : A $\cap$ B $\cap$ C $=$ $\phi$

Let A = {1,2}

B = {1,3}

C = {3,2}

Here, A $\cap$ B = {1}

B $\cap$ C = {3}

A $\cap$ C = {2}

and A $\cap$ B $\cap$ C $=$ $\phi$

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.

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.

n(H) = 25

n(T) = 26

n(I) = 26

n(H $\cap$ I) = 9

n( T $\cap$ I ) = 8

n( H $\cap$ T ) = 11

n(H $\cap$ T $\cap$ I ) = 3

the number of people who read at least one of the newspapers = n(H$\cup$T$\cup$I) = n(H) + n(T) + n(I) - n(H $\cap$ I) - n( T $\cap$ I ) - n( H $\cap$ T ) + n(H $\cap$ T $\cap$ 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 $\cap$ I) - n( T $\cap$ I ) - n( H $\cap$ T ) + 2 n(H $\cap$ T $\cap$ I )

= 52 - 9- 8 -11 + 6

= 30

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

n(A) = 21

n(B) = 26

n(C) = 29

n( A $\cap$ B) = 14

n( A $\cap$ C) = 12

n (B $\cap$ C ) = 14

n( A $\cap$ B $\cap$ 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

## Topics Covered in NCERT class 11 Chapter 1 Maths Miscellaneous Solutions

1. Sets and Their Representations: Understanding the concept of sets and how they are represented.

2. The Empty Set, Finite and Infinite Sets, Equal Sets: Exploring the empty set, finite and infinite sets, and the notion of equal sets.

3. Subsets, Power Set, Universal Set: Discussing subsets, power sets, and the concept of a universal set.

4. Venn Diagrams: Utilizing Venn diagrams as a visual tool for set representation.

5. Operations on Sets: Learning about various operations that can be performed on sets.

6. The Complement of a Set: Understanding the complement of a set and its significance.

7. Practical Problems on Union and Intersection of Two Sets: Solving real-world problems related to the union and intersection of sets.

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

• As the name suggests miscellaneous exercise class 11 chapter 1 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 class 11 maths miscellaneous exercise chapter 1 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.

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## Key Features of Class 11 Maths ch 1 Miscellaneous Exercise Solutions Free PDF Access

1. Comprehensive Solutions: Careers360 provides in-depth and comprehensive solutions for the miscellaneous exercise class 11 chapter 1. This ensures a thorough understanding of the various mathematical concepts covered.

2. Expert-Crafted Solutions: The miscellaneous exercise chapter 1 class 11 solutions are meticulously crafted by subject matter experts, ensuring accuracy, clarity, and adherence to the prescribed syllabus. This helps students grasp complex mathematical concepts more effectively.

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## Subject Wise NCERT Exampler Solutions

Happy learning!!!

1. Does miscellaneous exercise is important for the CBSE exam ?

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.

2. How many questions are given in the miscellaneous exercise sets ?

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

3. How many exercises are there in CBSE Class 11 Maths chapter 1 ?

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

4. What is the use of Venn's diagram ?

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.

5. Find all the subsets of the set { –1, 0} ?

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

6. Find the power set of set A = {1,0} ?

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

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