# NCERT Solutions for Class 11 Maths Chapter 1 Sets

NCERT Solutions for Class 11 Maths Chapter 1 Sets: In our daily life, we have come across situations like arranging books on a shelf with different categories like novels, short stories, science fiction, etc. The well-defined collection of objects is called as set. In this article, you will get NCERT solutions for class 11 maths chapter 1 sets. This chapter will introduce the concept of sets and different operations on set. Solutions of NCERT for class 11 maths chapter 11 sets will help you to learn the concepts of relations and functions which will be taught in the next chapter & class 12th also. The knowledge of sets is also required to study geometry, sequences, probability, etc. CBSE NCERT solutions for class 11 maths chapter 1 sets will build your basics of data analysis which will be useful in higher education. To understand the basics of sets, Let's consider an example of a survey in which 5 data are taken. 1. Gender 2. Age 3. weight. 4. waist diameter 5. Diabatic patient or not. Based on the obtained data, two main sets are formed, namely diabetic patient and non-diabatic patient. Using this data we will be able to predict whether a new person in the given region will have diabetes or not. The new technology of machine learning uses data for predictions and classifications. (The more the data available,the more accurate the prediction). The NCERT solutions for class 11 maths chapter 1 sets will improve the basics of sets which will be useful in further mathematics courses also. Check NCERT solutions provided to help you when you are stuck in the NCERT problems. There are six exercises and a miscellaneous exercise in this chapter which are explained below.

Exercise:1.1

Exercise:1.2

Exercise:1.3

Exercise:1.4

Exercise:1.5

Exercise:1.6

Miscellaneous Exercise

## The main topics of the NCERT for Class 11 Maths Chapter 1 Sets are listed below:

1.1 Introduction

1.2 Sets and their Representations

1.3 The Empty Set

1.4 Finite and Infinite Sets

1.5 Equal Sets

1.6 Subsets

1.7 Power Set

1.8 Universal Set

1.9 Venn Diagrams

1.10 Operations on Sets

1.11 Complement of a Set

1.12 Practical Problem

A brief summary of NCERT solutions for class 11 maths chapter 1 Sets are given below:

• A set is a well-defined collection of objects
• An empty set is a set that does not contain any element. It is represented by $\phi$ or {}. The empty set is a subset of all the sets.
• The power set of A is the collection of all the subsets of A.
• If a set A contains N elements, then the power set of A contains 2n elements

Another important concept of NCERT solutions for class 11 maths chapter 1 is operations on sets. The Venn diagrams for basic operations are shown below

1. U represents universal sets, A and B represents sets A and B respectively

2. Union of A and B is represented by the green colour. Denoted by : $\mathrm{A\cup B}$

3. The intersection of set a and B is represented by green colour. Denoted by $\mathrm{A\cap B}$

4. A-B: A but not B. A-B is represented by red colour

You can solve many problems using operations on sets. An example from the NCERT textbook is given below.

Q)  In a class of 35 students, 24 like to play cricket and 16 like to play football. Also, each student likes to play at least one of the two games. How many students like to play both cricket and football?

Solution: Total number of students in class 35.

Number of students like to play cricket:  n(c)=24

Number of students like to play football: n(f)= 16

35=n(c)+n(f)-Number of students like to play both cricket and football

therefor Number of students like to play both cricket and football=40-35=5

This can be represented by a Venn diagram as follows

## NCERT solutions for class 11 maths chapter 1 sets-Exercise: 1.1

The collection of all the months of a year beginning with the letter J.

The months starting with letter J are:

january

june

july

Hence,this is collection of well defined objects so it is a set.

The collection of ten most talented writers of India.

Ten most talented writers may be different depending on different criteria of determining talented writers.

Hence, this is not well defined so cannot be a set.

A team of eleven best-cricket batsmen of the world.

Eleven most talented cricketers may be different depending on criteria of determining talent of a player.

Hence,this is not well defined so it is not a set.

The collection of all boys in your class.

The collection of boys in a class are well defined and known.

Group of well defined objects is a set.

Hence, it is a set.

The collection of all natural numbers less than 100.

Natural numbers less than 100 has well defined and known collection of numbers.

that is S= {1,2,3............99}

Hence,it is a set.

A collection of novels written by the writer Munshi Prem Chand.

The collection of novels written by Munshi Prem Chand is well defined and known .

Hence,it is a set.

The collection of all even integers.

The collection of even intergers is well defined because we can get even integers till infinite. that is

Hence,it is a set.

The collection of questions in this Chapter.

Collection of questions in a chapter is well defined and known .

Hence ,it is a set.

A collection of most dangerous animals of the world.

A Collection of most dangerous animal is not well defined because the criteria of defining dangerousness of any animal can vary .

Hence,it is not a set.

(i) 5_____A

(ii) 8_____A

(iii) 0_____A

(iv) 4_____A

(v) 2_____A

(vi) 10____A

A = $\left \{ 1,2,3,4,5,6 \right \}$ ,the elements which lie in this set belongs to this set and others do not belong.

(i) 5 $\in$ A

(ii) 8 $\notin$ A

(iii) 0 $\notin$ A

(iv) 4 $\in$ A

(v) 2 $\in$ A

(vi) 10 $\notin$ A

Question:3(i) Write the following sets in roster form

$\, \left ( i \right )\, A= \left \{ x\vdot:x\, is \, an\, integer\, and\, -3\leq x< 7 \right \}$

Elements of this set are:-3,-2,-1,0,1,2,3,4,5,6.

Hence ,this can be written as:

A = $\left \{ -3,-2,-1,0,1,2,3,4,5,6 \right \}$

Question:3(ii) Write the following sets in roster form

$\left ( ii \right )B= \left \{ x:x\, is \, a\, natural\, number\, less\, than\, 6 \right \}$

Natural numbers less than 6 are: 1,2,3,4,5.

This can be written as:

$B = \left \{ 1,2,3,4,5 \right \}$

Question:3(iii) Write the following sets in roster form

C = {x : x is a two-digit natural number such that the sum of its digits is 8}

The two digit numbers having sum equal to 8 are: 17,26,35,44,53,62,71,80.

This can be written as:

$C= \left \{ 17,26,35,44,53,62,71,80 \right \}$

Question:3(iv) Write the following sets in roster form

D = {x : x is a prime number which is divisor of 60}

$60= 2\ast 2\ast 3\ast 5$

Prime numbers which are divisor of 60 are:2,3,5.

This can be written as:

$D= \left \{ 2,3,5 \right \}$

Question:3(v) Write the following sets in roster form

E = The set of all letters in the word TRIGONOMETRY.

Letters of word TRIGONOMETRY are:T,R,I,G,N,O,M,E,Y.

This can be written as :

E = {T,R,I,G,N,O,M,E,Y}

Question:3(vi) Write the following sets in roster form

F = The set of all letters in the word BETTER.

The set of letters of word BETTER are: {B,E,T,R}

This can be written as:

F = {B,E,T,R}

Question:4(i)  Write the following sets in the set builder form

{3, 6, 9, 12}

A = {3,6,9,12}

This can be written as : $\left \{ 3,6,9,12 \right \}= \left \{ x:x= 3n,n\in N\, \, and\,\, 1\leq n\leq 4 \right \}$

Question:4(ii) Write the following sets in the set builder form

{2,4,8,16,32}

$2 = 2^{1}$

$4 = 2^{2}$

$8= 2^{3}$

$16 = 2^{4}$

$32 = 2^{5}$

$\left \{ 2,4,8,16,32 \right \}$ can be written as $\left \{ x:x= 2^{n},n\in N and 1\leq n\leq 5 \right \}.$

Question:4(iii) Write the following sets in the set builder form:

{5, 25, 125, 625}

$5= 5^{1}$

$25= 5^{2}$

$125= 5^{3}$

$625= 5^{4}$

$\left \{ 5,25,125,625 \right \}$ can be written as $\left \{ x:x= 5^{n},n\in N and 1\leq n\leq 4 \right \}$

Question:4(iv) Write the following sets in the set builder form:

{2, 4, 6, . . .}

This is a set of all even natural numbers.

{2,4,6....} can be written as {x : x is an even natural number}

Question:4(v) Write the following sets in the set builder form :

{1,4,9, . . .,100}

$1= 1^{2}$

$4= 2^{2}$

$9= 3^{2}$

.

.

.

.

$100= 10^{2}$

$\left \{ 1,4,9.....100 \right \}$ can be written as $\left \{ x:x= n^{2} ,n\in N and \, 1\leq n\leq 10\right \}$

Question:5(i) List all the elements of the following sets:

A = {x : x is an odd natural number}.

A = { x : x is an odd natural number } = {1,3,5,7,9,11,13.............}

Question:5(ii) List all the elements of the following sets:

B= { x : x is an integer,  $\frac{-1}{2} < x< \frac{9}{2}$ }

Intergers between $-\frac{1}{2}< x< \frac{9}{2}$   are 0,1,2,3,4.

Hence,B = {0,1,2,3,4}

Question:5(iii)  List all the elements of the following sets:

C = {x : x is an integer, $x^{2}\leq 4$}

$\left ( -2 \right )^{2}= 4$

$\left ( -1 \right )^{2}= 1$

$\left ( 0 \right )^{2}= 0$

$\left ( 1\right )^{2}= 1$

$\left ( 2\right )^{2}= 4$

Integers whose square is less than or equal to 4 are : -2,-1,0,1,2.

Hence,it can be written as c = {-2,-1,0,1,2}.

Question:5(iv) List all the elements of the following sets:

D = {x : x is a letter in the word “LOYAL”}

LOYAL has letters L,O,Y,A.

D = {x : x is a letter in the word “LOYAL”} can be written as {L,O,Y,A}.

Question:5(v) List all the elements of the following sets:

E = {x : x is a month of a year not having 31 days}

The months not having 31 days are:

February

April

June

September

November

It can be written as {february,April,June,September,november}

Question:5(vi) List all the elements of the following sets :

F = {x : x is a consonant in the English alphabet which precedes k }.

The consonents in english which precedes K are : B,C,D,F,G,H,J

Hence, F = {B,C,D,F,G,H,J}.

## CBSE NCERT solutions for class 11 maths chapter 1 sets-Exercise: 1.2

Set of odd natural numbers divisible by 2

No odd number is divisible by 2.

Hence, this is a null set.

Question:1(ii) Which of the following are examples of the null set :

Set of even prime numbers.

Even prime number = 2.

Hence, it is not a null set.

Question:1(iv) Which of the following are examples of the null set :

{ y : y is a point common to any two parallel lines}

Parallel lines do not intersect so they do not have any common point.

Hence, it is a null set.

(i) The set of months of a year

(ii) {1, 2, 3, . . .}

(iii) {1, 2, 3, . . .99, 100}

(iv) ) The set of positive integers greater than 100.

(v) The set of prime numbers less than 99

(i) Number of months in a year are 12 and finite.

Hence,this set is finite.

(ii) {1,2,3,4.......} and so on ,this does not have any limit.

Hence, this is infinite set.

(iii) {1,2,3,4,5......100} has finite numbers.

Hence ,this is finite set.

(iv) Positive integers greater than 100 has no limit.

Hence,it is infinite set.

(v) Prime numbers less than 99 are finite ,known numbers.

Hence,it is finite set.

(i) The set of lines which are parallel to the x-axis

(ii) The set of letters in the English alphabet

(iii) The set of numbers which are multiple of 5

(iv) The set of animals living on the earth

(v) The set of circles passing through the origin (0,0)

(i) Lines parallel to the x-axis are infinite.

Hence, it is an infinite set.

(ii) Letters in English alphabets are 26 letters which are finite.

Hence, it is a finite set.

(iii) Numbers which are multiple of 5 has no limit, they are infinite.

Hence, it is an infinite set.

(iv) Animals living on earth are finite though the number is very high.

Hence, it is a finite set.

(v) There is an infinite number of circles which pass through the origin.

Hence, it is an infinite set.

Question:4(i) In the following, state whether A = B or not:

A = { a, b, c, d } B = { d, c, b, a }

Given
A = {a,b,c,d}

B = {d,c,b,a}
Comparing the elements of set A and set B, we conclude that all the elements of  A and all the elements of B are equal.
Hence, A = B.

Question:4(ii) In the following, state whether A = B or not:

A = { 4, 8, 12, 16 } B = { 8, 4, 16, 18}

12 belongs A but 12 does not belong to B

12 $\in$ A but 12 $\notin$ B.

Hence, A $\neq$ B.

Question:4(iii) In the following, state whether A = B or not:

A = {2, 4, 6, 8, 10} B = { x : x is positive even integer and $x \leq 10$}

Positive even integers less than or equal to 10 are : 2,4,6,8,10.

So, B  = { 2,4,6,8,10 } which is equal to A = {2,4,6,8,10}

Hence, A = B.

Question:4(iv) In the following, state whether A = B or not:

A = { x : x is a multiple of 10}, B = { 10, 15, 20, 25, 30, . . . }

Multiples of 10 are : 10,20,30,40,........ till infinite.

SO, A = {10,20,30,40,.........}

B = {10,15,20,25,30........}

Comparing elements of A and B,we conclude that elements of A and B are not equal.

Hence, A$\neq$ B.

Question:5(i) Are the following pair of sets equal ? Give reasons.

A = {2, 3}, B = {x : x is solution of $x^{2}$+ 5x + 6 = 0}

As given,

A = {2,3}

And,

$x^{2}+5x+6= 0$

$x\left ( x+3 \right ) + 2\left ( x+3 \right )= 0$

$\left ( x+2 \right ) ( x+3 \right ))= 0$

x = -2 and -3

B = {-2,-3}

Comparing elements of A and B,we conclude elements of A and B are not equal.

Hence,A$\neq$B.

Question:5(ii) Are the following pair of sets equal ? Give reasons.

A = { x : x is a letter in the word FOLLOW}

B = { y : y is a letter in the word WOLF}

Letters of word FOLLOW are F,OL,W.

SO, A = {F,O,L,W}

Letters of word WOLF are W,O,L,F.

So, B = {W,O,L,F}

Comparing A and B ,we conclude that elements of A are equal to elements of B.

Hence, A=B.

A = { 2, 4, 8, 12},         B = { 1, 2, 3, 4},          C = { 4, 8, 12, 14},             D = { 3, 1, 4, 2}

E = {–1, 1},                   F = { 0, a},                 G = {1, –1},                        H = { 0, 1}

Compare the elements of A,B,C,D,E,F,G,H.

$\in$ A but $8\notin B,8\in C,8\notin D,8\notin E,8\notin F,8\notin G,8\notin H$

Now, 2$\in$ A  but 2$\notin$C.

Hence, A$\neq$B,A$\neq$C,A$\neq$D,A$\neq$E,A$\neq$F,A$\neq$G,A$\neq$H.

$\in$ B,3$\in$D but 3$\notin$C,3$\notin$ E,3$\notin$F,3$\notin$G,3$\notin$H.

Hence,B$\neq$C,B$\neq$E,B$\neq$F,B$\neq$G,B$\neq$H.

Similarly, comparing other elements of all sets, we conclude that elements of B and elements of D are equal also elements of E and G are equal.

Hence, B=D  and E = G.

## Solutions of NCERT for class 11 maths chapter 1 sets-Exercise: 1.3

(i) { 2, 3, 4 } _____ { 1, 2, 3, 4,5 }

(ii) { a, b, c }______ { b, c, d }

(iii) {x : x is a student of Class XI of your school}______{x : x student of your school}

(iv) {x : x is a circle in the plane} ______{x : x is a circle in the same plane with radius 1 unit}

(v) {x : x is a triangle in a plane} ______ {x : x is a rectangle in the plane}

(vi) {x : x is an equilateral triangle in a plane} ______{x : x is a triangle in the same plane}

(vii) {x : x is an even natural number} _____ {x : x is an integer}

A set A is said to be a subset of a set B if every element of A is also an element of B.

(i). All elements {2,3,4} are also elements of  {1,2,3,4,5} .

So, {2,3,4} $\subset$ {1,2,3,4,5}.

(ii).All elements { a, b, c } are not elements of{ b, c, d }.

Hence,{ a, b, c } $\not\subset${ b, c, d }.

(iii) Students of class XI are also students of your school.

Hence,{x : x is a student of Class XI of your school} $\subset$ {x : x student of your school}

(iv). Here, {x : x is a circle in the plane}  $\not\subset$  {x : x is a circle in the same plane with radius 1 unit}   : since a circle in the plane can have any radius

(v). Triangles and rectangles are two different shapes.

Hence,{x : x is a triangle in a plane} $\not\subset$ {x : x is a rectangle in the plane}

(vi). Equilateral triangles are part of all types of triangles.

So,{x : x is an equilateral triangle in a plane}  $\subset$  {x : x is a triangle in the same plane}

(vii).Even natural numbers are part of all integers.

Hence, {x : x is an even natural number} $\subset$ {x : x is an integer}

(i) { a, b } $\not\subset$ { b, c, a }

(ii) { a, e } $\subset$ { x : x is a vowel in the English alphabet}

(iii) { 1, 2, 3 } $\subset$ { 1, 3, 5 }

(iv) { a } $\subset$ { a, b, c }

(v) { a } $\in$ { a, b, c }

(vi) { x : x is an even natural number less than 6}  $\subset$  { x : x is a natural number which divides 36}

(i) All elements of { a, b } lie in  { b, c, a }.So,{ a, b } $\subset${ b, c, a }.

Hence,it is false.

(ii) All elements of { a, e } lie in {a,e,i,o,u}.

Hence,the statements given is true.

(iii) All elements of { 1, 2, 3 } are not present in { 1, 3, 5 }.

Hence,statement given is false.

(iv) Element of { a } lie in { a, b, c }.

Hence,the statement is true.

(v). { a } $\subset$ { a, b, c }

So,the statement is false.

(vi) All elements {2,4,} lies in {1,2,3,4,6,9,12,18,36}.

Hence,the statement is true.

{3, 4} $\subset$ A

$\in$ {3,4} but 3$\notin$ {1,2,{3,4},5}.

SO, {3, 4} $\not \subset$ A

Hence,the statement is incorrect.

{3, 4} $\in$ A

{3, 4} is element of  A.

So, {3, 4} $\in$ A.

Hence,the statement is correct.

{{3, 4}} $\subset$ A

Here,

{ 3, 4 }$\in$ { 1, 2, { 3, 4 }, 5 }

and { 3, 4 } $\in$  {{3, 4}}

So, {{3, 4}} $\subset$ A.

Hence,the statement is correct.

1 $\in$ A

Given, 1 is element of  { 1, 2, { 3, 4 }, 5 }.

So,1 $\in$ A.

Hence,statement is correct.

1 $\subset$ A

Here, 1 is element of set A = { 1, 2, { 3, 4 }, 5 }.So,elements of set A cannot be subset of set A.

1 $\not\subset$ { 1, 2, { 3, 4 }, 5 }.

Hence,the statement given is incorrect.

{1,2,5}$\subset$ A

All elements of {1,2,5} are present in { 1, 2, { 3, 4 }, 5 }.

So, {1,2,5}$\subset$ { 1, 2, { 3, 4 }, 5 }.

Hence,the statement given is correct.

{1,2,5} $\in$ A

Here,{1,2,5} is not an element of  { 1, 2, { 3, 4 }, 5 }.

So,{1,2,5} $\notin$A .

Hence, statement is incorrect.

{1,2,3} $\subset$ A

Here, 3$\in$ {1,2,3}

but  3 $\notin$  { 1, 2, { 3, 4 }, 5 }.

So, {1,2,3} $\not \subset$ A

Hence,the given statement is incorrect.

$\phi \in A$

$\phi$ is not an element of { 1, 2, { 3, 4 }, 5 }.

So,$\phi \notin A$.

Hence,the above statement is incorrect.

$\phi \subset A$

$\phi$ is subset of all sets.

Hence,the above statement is correct.

$\left \{ \phi \right \} \subset A$

$\phi$ $\subset$ { 1, 2, { 3, 4 }, 5 }. but $\phi$ is not an element  of { 1, 2, { 3, 4 }, 5 }.

$\left \{ \phi \right \} \not\subset A$

Hence, the above statement is incorrect.

Question:4(i) Write down all the subsets of the following sets

{a}

Subsets of $\left \{ a \right \} = \phi \, and \left \{ a \right \}$.

Question:4(ii) Write down all the subsets of the following sets:

{a, b}

Subsets of  $\left \{ a,b \right \}\ are \ \phi , \left \{ a \right \},\left \{ b \right \} and \left \{ a,b \right \}$. Thus the given set has 4 subsets

Question:4(iii) Write down all the subsets of the following sets:

{1,2,3}

Subsets of

$\left \{ 1,2,3 \right \} = \left \{ 1 \right \},\left \{ 2 \right \},\left \{ 3 \right \},\phi ,\left \{ 1,2 \right \},\left \{ 2,3 \right \},\left \{ 3,1 \right \},\left \{ 1,2,3 \right \}$

Question:4(iv) Write down all the subsets of the following sets:

$\phi$

Subset  of  $\phi$ is  $\phi$ only.

The subset of a null set is null set itself

Let the elements in set A be m, then n$\left ( A \right ) =$ m

then, the number of elements in power set of A         n$\left ( p\left ( A \right )\right ) =$ $2^{m}$

Here,  A = $\phi$   so    n$\left ( A \right ) =$ 0

n$\left [ P\left ( A \right ) \right ] =$ $2^{0}$ $=$1

Hence,we conclude P(A) has 1 element.

Question:6 Write the following as intervals :

(i) {x : x $\in$R, – 4 $<$ x $\leq$ 6}

(ii) {x : x $\in$ R, – 12 $<$x $<$–10}

(iii) {x : x $\in$ R, 0 $\leq$ x $<$ 7}

(iv) {x : x $\in$ R, 3 $\leq$  x $\leq$  4}

The following can be written in interval as :

(i) {x : x $\in$R, – 4 $<$ x $\leq$ 6} $= \left ( -4 ,6 \right ]$

(ii) {x : x $\in$ R, – 12 $<$x $<$–10} $= \left ( -12,-10 \right )$

(iii) {x : x $\in$ R, 0 $\leq$ x $<$ 7} $= \left \left [ 0,7\right )$

(iv) (iv) {x : x $\in$ R, 3 $\leq$  x $\leq$  4} $=\left [ 3,4 \right ]$

(i) (– 3, 0)

(ii) [6 , 12]

(iii) (6, 12]

(iv) [–23, 5)

The given intervals can be written in set builder form as :

(i) (– 3, 0)  $= \left \{ x:x\in R, -3< x< 0 \right \}$

(ii) [6 , 12]  $= \left \{ x:x\in R, 6\leq x\leq 12\right \}$

(iii) (6, 12] $= \left \{ x:x\in R, 6< x\leq 12\right \}$

(iv) [–23, 5)$= \left \{ x:x\in R, -23 \leq x< 5\right \}$

(i) The set of right triangles.

(ii) The set of isosceles triangles.

(i) Universal set for a set of right angle triangles can be set of polygons or set of all triangles.

(ii)  Universal set for a set of isosceles angle triangles can be set of polygons.

## NCERT solutions for class 11 maths chapter 1 sets-Exercise: 1.4

X = {1, 3, 5}    Y = {1, 2, 3}

Union of X and Y is X $\cup$ Y = {1,2,3,5}

Question:1(ii) Find the union of each of the following pairs of sets :

A = [ a, e, i, o, u}     B = {a, b, c}

Union of A and B is A $\cup$ B = {a,b,c,e,i,o,u}.

Question:1(iii) Find the union of each of the following pairs of sets :

A = {x : x is a natural number and multiple of 3} B = {x : x is a natural number less than 6}

Here ,

A = {3,6,9,12,15,18,............}

B = {1,2,3,4,5,6}

Union of A and B is A $\cup$ B.

$\cup$ B = {1,2,3,4,5,6,9,12,15........}

Question:1(iv) Find the union of each of the following pairs of sets :

A = {x : x is a natural number and 1 $<$ x $\leq$6 }

B = {x : x is a natural number and 6 $<$ x $<$10 }

Here,

A = {2,3,4,5,6}

B = {7,8,9}

$\cup$ B = {2,3,4,5,6,7,8,9}

or it can be written as A $\cup$ B = {x : x is a natural number and 1$<$ x $<$10 }

A = {1, 2, 3}                     B = $\phi$

Here,

A union B is A $\cup$ B.

$\cup$ B = {1,2,3}

Here,

We can see elements of A lie in set B.

Hence, A $\subset$ B.

And, A $\cup$ B = {a,b,c} = B

If A is subset of B then A $\cup$ B will be set B.

A $\cup$ B

Here,

A = {1, 2, 3, 4}

B = {3, 4, 5, 6}

The union of the set can be written as follows

A $\cup$ B = {1,2,3,4,5,6}

$\cup$ C

Here,

A = {1, 2, 3, 4}

C = {5, 6, 7, 8 }

The union can be written as follows

$\cup$ C = {1,2,3,4,5,6,7,8}

B $\cup$

Here,

B = {3, 4, 5, 6},

C = {5, 6, 7, 8 }

The union of the given sets are

$\cup$ C = { 3,4,5,6,7,8}

B $\cup$ D

Here,

B = {3, 4, 5, 6}

D = { 7, 8, 9, 10 }

$\cup$ D = {3,4,5,6,7,8,9,10}

$\cup$ B $\cup$ C

Here,

A = {1, 2, 3, 4},

B = {3, 4, 5, 6},

C = {5, 6, 7, 8 }

The union can be written as

A $\cup$ B $\cup$ C = {1,2,3,4,5,6,7,8}

A $\cup$ B $\cup$ D

Here,

A = {1, 2, 3, 4},

B = {3, 4, 5, 6}

D = { 7, 8, 9, 10 }

The union can be written as

$\cup$ B $\cup$ D = {1,2,3,4,5,6,7,8,9,10}

B $\cup$ C $\cup$ D

Here,B = {3, 4, 5, 6},

C = {5, 6, 7, 8 } and

D = { 7, 8, 9, 10 }

The union can be written as

B $\cup$ C $\cup$ D = {3,4,5,6,7,8,9,10}

(i) X = {1, 3, 5} Y = {1, 2, 3}

(ii) A = [ a, e, i, o, u} B = {a, b, c}

(iii) A = {x : x is a natural number and multiple of 3}     B = {x : x is a natural number less than 6}

(iv) A = {x : x is a natural number and 1 $<$x $\leq$6 }         B = {x : x is a natural number and 6 $<$ x $<$ 10 }

(v) A = {1, 2, 3},       B = $\phi$

(i) X = {1, 3, 5}   Y = {1, 2, 3}

X $\cap$ Y = {1,3}

(ii) A = [ a, e, i, o, u}    B = {a, b, c}

A $\cap$ B = {a}

(iii) A = {x : x is a natural number and multiple of 3}     B = {x : x is a natural number less than 6}

A = {3,6,9,12,15.......}             B = {1,2,3,4,5}

A $\cap$ B = {3}

(iv)A = {x : x is a natural number and 1 $<$x $\leq$6 }         B = {x : x is a natural number and 6 $<$ x $<$ 10 }

A = {2,3,4,5,6}            B = {7,8,9}

A $\cap$ B = $\phi$

(v)   A = {1, 2, 3},       B = $\phi$

A $\cap$ B = $\phi$

(i) A $\cap$ B                     (ii) B $\cap$ C

(iii) A $\cap$C $\cap$ D            (iv) A $\cap$ C

(v) B $\cap$ D                   (vi) A $\cap$ (B $\cup$ C)

(vii) A $\cap$ D                 (viii) A $\cap$ (B $\cup$ D)

(ix) ( A $\cap$ B ) $\cap$ ( B $\cup$ C )  (x) ( A $\cup$ D) $\cap$ ( B $\cup$ C)

Here,  A = { 3, 5, 7, 9, 11 },  B = {7, 9, 11, 13},    C = {11, 13, 15}  and    D = {15, 17}

(i) A $\cap$ B = {7,9,11}                                        (vi) A $\cap$ (B $\cup$ C)  = {7,9,11}

(ii) B $\cap$ C   =  { 11,13}                                    (vii) A $\cap$ D = $\phi$

(iii) A $\cap$C $\cap$ D = $\phi$                                          (viii) A $\cap$ (B $\cup$ D) = {7,9,11}

(iv) A $\cap$ C   = { 11 }                                         (ix) ( A $\cap$ B ) $\cap$ ( B $\cup$ C ) = {7,9,11}

(v) B $\cap$ D  = $\phi$                                                (x) ( A $\cup$ D) $\cap$ ( B $\cup$ C) =  {7,9,11,15}

(i) A $\cap$ B

(ii) A $\cap$ C

(iii) A $\cap$ D

(iv) B $\cap$ C

(v) B $\cap$ D

(vi) C $\cap$ D

Here, A = {1,2,3,4,5,6...........}

B = {2,4,6,8,10...........}

C = {1,3,5,7,9,11,...........}

D = {2,3,5,7,11,13,17,......}

(i) A $\cap$ B = {2,4,6,8,10........} = B

(ii) A $\cap$ C = {1,3,5,7,9.........} = C

(iii) A $\cap$ D  = {2,3,5,7,11,13.............} = D

(iv) B $\cap$ C  = $\phi$

(v) B $\cap$ D  = {2}

(vi) C $\cap$ D = {3,5,7,11,13,..........} = $\left ( x:x \, is\, \, odd\, \, \, prime \, \, number \right )$

Question:8(i) Which of the following pairs of sets are disjoint

{1, 2, 3, 4} and {x : x is a natural number and 4 $\leq$x $\leq$6 }

Here,       {1, 2, 3, 4}     and     {4,5,6}

{1, 2, 3, 4} $\cap$ {4,5,6}  =  {4}

Hence,it is not a disjoint set.

Question:8(ii) Which of the following pairs of sets are disjoint

{ a, e, i, o, u }     and     { c, d, e, f }

Here,   { a, e, i, o, u }     and     { c, d, e, f }

{ a, e, i, o, u } $\cap$ { c, d, e, f } = {e}

Hence,it is not disjoint set.

Question:8(iii) Which of the following pairs of sets are disjoint

{x : x is an even integer }      and      {x : x is an odd integer}

Here, {x : x is an even integer }      and      {x : x is an odd integer}

{2,4,6,8,10,..........}         and     {1,3,5,7,9,11,.....}

{2,4,6,8,10,..........} $\cap$ {1,3,5,7,9,11,.....} = $\phi$

Hence,it is disjoint set.

(i) A – B         (ii) A – C          (iii) A – D        (iv) B – A            (v) C – A           (vi) D – A

(vii) B – C       (viii) B – D       (ix) C – B       (x) D – B            (xi) C – D            (xii) D – C

A = {3, 6, 9, 12, 15, 18, 21}, B = { 4, 8, 12, 16, 20 }, C = { 2, 4, 6, 8, 10, 12, 14, 16 }, D = {5, 10, 15, 20 }

The given operations are done as follows

(i) A – B  = {3,6,9,15,18,21}                                    (vii) B – C = {20}

(ii) A – C = {3,9,15,18,21}                                       (viii) B – D =  {4,8,12,16}

(iii) A – D = {3,6,9,12,18,21}                                    (ix) C – B  = {2,6,10,14}

(iv) B – A = {4,8,16,20}                                            (x) D – B   = {5,10,15}

(v) C – A = {2,4,8,10,14,16}                                     (xi) C – D  = {2,4,6,8,12,14,16}

(vi) D – A = {5,10,20}                                               (xii) D – C = {5,15,20}

(i) X – Y

(ii) Y – X

(iii) X $\cap$ Y

X= { a, b, c, d }                and              Y = { f, b, d, g}

(i) X – Y = {a,c}

(ii) Y – X = {f,g}

(iii) X $\cap$ Y = {b,d}

R = set of real numbers.

Q = set of rational numbers.

R - Q = set of irrational numbers.

{ 2, 3, 4, 5 } and { 3, 6} are disjoint sets.

Here,

{ 2, 3, 4, 5 } and { 3, 6}

{ 2, 3, 4, 5 } $\cap$ { 3, 6} = {3}

Hence,these are not disjoint sets.

So,false.

{ a, e, i, o, u } and { a, b, c, d }are disjoint sets

Here,     { a, e, i, o, u } and { a, b, c, d }

{ a, e, i, o, u } $\cap$ { a, b, c, d } = {a}

Hence,these are not disjoint sets.

So, statement is false.

{ 2, 6, 10, 14 } and { 3, 7, 11, 15} are disjoint sets.

Here,

{ 2, 6, 10, 14 }         and         { 3, 7, 11, 15}

{ 2, 6, 10, 14 }  $\cap$ { 3, 7, 11, 15} = $\phi$

Hence, these are disjoint sets.

So,given statement is true.

{ 2, 6, 10 } and { 3, 7, 11} are disjoint sets.

Here,

{ 2, 6, 10 } and { 3, 7, 11}

{ 2, 6, 10 }  $\cap$ { 3, 7, 11} = $\phi$

Hence,these are disjoint sets.

So,statement is true.

## NCERT solutions for maths class 11 chapter 1 sets-Exercise: 1.5

(i) A′

(ii) B′

(iii) (A $\cup$ C)′

(iv) (A $\cup$ B)′

(v) (A')'

(vi) (B – C)'

U = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = { 1, 2, 3, 4}, B = { 2, 4, 6, 8 } and C = { 3, 4, 5, 6 }

(i) A′ = U - A = {5,6,7,8,9}

(ii) B′ = U - B = {1,3,5,7,9}

(iii) A $\cup$ C = {1,2,3,4,5,6}

(A $\cup$ C)′ = U - (A $\cup$ C) = {7,8,9}

(iv) (A $\cup$ B) = {1,2,3,4,6,8}

(A $\cup$ B)′ = U - (A $\cup$ B) = {5,7,9}

(v) (A')' = A = { 1, 2, 3, 4}

(vi) (B – C) = {2,8}

(B – C)' = U - (B – C) = {1,3,4,5,6,7,9}

(i) A = {a, b, c}

(ii) B = {d, e, f, g}

(iii) C = {a, c, e, g}

(iv) D = { f, g, h, a}

U = { a, b, c, d, e, f, g, h}

(i) A = {a, b, c}

A' = U - A = {d,e,f,g,h}

(ii) B = {d, e, f, g}

B' = U - B = {a,b,c,h}

(iii) C = {a, c, e, g}

C' = U - C = {b,d,f,h}

(iv) D = { f, g, h, a}

D' = U - D = {b,c,d,e}

(i) {x : x is an even natural number}

(ii) { x : x is an odd natural number }

(iii) {x : x is a positive multiple of 3}

(iv) { x : x is a prime number }

(v) {x : x is a natural number divisible by 3 and 5}

Universal set = U = {1,2,3,4,5,6,7....................}

(i) {x : x is an even natural number} = {2,4,6,8,..........}

{x : x is an even natural number}'= U - {x : x is an even natural number}  = {1,3,5,7,9,..........} = {x : x is an odd natural number}

(ii) { x : x is an odd natural number }' = U - { x : x is an odd natural number } = {x : x is an even natural number}

(iii)  {x : x is a positive multiple of 3}' = U -  {x : x is a positive multiple of 3} =  {x : x , x $\in$ N and is not a positive multiple of 3}

(iv) { x : x is a prime number }' = U - { x : x is a prime number } = { x : x is a positive composite number and 1 }

(v) {x : x is a natural number divisible by 3 and 5}' = U - {x : x is a natural number divisible by 3 and 5} = {x : x is a natural number not divisible by 3 or 5}

(vi) { x : x is a perfect square }

(vii) { x : x is a perfect cube}

(viii) { x : x + 5 $=$8 }

(ix) { x : 2x + 5 $=$9}

(x) { x : x $\geq$ 7 }

(xi) { x : x $\in$N and 2x + 1 $>$ 10 }

Universal set = U = {1,2,3,4,5,6,7,8.............}

(vi) { x : x is a perfect square }' = U - { x : x is a perfect square } =  { x : x $\in$ N and x is not a perfect square }

(vii) { x : x is a perfect cube}' = U - { x : x is a perfect cube} =  { x : x $\in$ N and x is not a perfect cube}

(viii) { x : x + 5 $=$8 }' = U - { x : x + 5 $=$8 } = U - {3} = { x : x$\in$ N and x $\neq$ 3 }

(ix) { x : 2x + 5 $=$9}' = U - { x : 2x + 5 $=$9} = U -{2} =  { x : x$\in$ N and x $\neq$ 2}

(x) { x : x $\geq$ 7 }' = U -  { x : x $\geq$ 7 } =  { x : x$\in$ N and x $<$ 7 }

(xi) { x : x $\in$N and 2x + 1 $>$ 10 }' = U - { x : x $\in$N and x  $>$ 9/2 } =  { x : x$\in$ N and x $\leq$  9/2 }

(i) (A $\cup$ B)′ = A′ $\cap$ B′

(ii) (A $\cap$ B)′ = A′ $\cup$ B′

U = {1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = {2, 4, 6, 8} and B = { 2, 3, 5, 7}

(i) (A $\cup$ B)′ = A′ $\cap$ B′

L.H.S = (A $\cup$ B)′ = U - (A $\cup$ B) = {1,9}

R.H.S =  A′ $\cap$ B′ = {1,3,5,7,9} $\cap$ {1,4,6,8,9} = {1,9}

L.H.S = R.H.S

Hence,the statement is true.

(ii) (A $\cap$ B)′ = A′ $\cup$ B′

L.H.S = U - (A $\cap$ B) ={1,3,4,5,6,7,8,9}

R.H.S =  A′ $\cup$ B′ = {1,3,5,7,9} $\cup$ {1,4,6,8,9} = {1,3,4,5,6,7,8,9}

L.H.S = R.H.S

Hence,the statement is true.

(i) (A ∪ B)′

(ii) A′ ∩ B′

(iii) (A ∩ B)′

(iv) A′ ∪ B′

(i) (A ∪ B)′

(A ∪ B) is in yellow colour

(A ∪ B)′ is in green colour

ii) A′ ∩ B′ is represented by the green colour in the below figure

iii) (A ∩ B)′ is represented by green colour in the below diagram and white colour represents (A ∩ B)

iv)  A′ ∪ B′

The green colour represents   A′ ∪ B′

A' is the set of all triangles whose angle is $60\degree$ in other words A' is set of all equilateral triangles.

(i) A $\cup$ A′ $=$ $\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot$

(ii) $\phi '$ $\cap$A $=$ $\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot$

(iii) A $\cap$ A′ $=$$\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot$

(iv) U′ $\cap$ A $=$$\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot$

The following are the answers for the questions

(i) A $\cup$ A′ $=$ U

(ii) $\phi$$\cap$ A  $=$ A

(iii) A $\cap$ A′ $=$ $\phi$

(iv) U′ $\cap$ A $=$ $\phi$

## CBSE NCERT solutions for class 11 maths chapter 1 sets-Exercise: 1.6

n( X ) = 17,       n( Y ) = 23             and  n( X $\cup$ Y ) = 38

n( X $\cup$ Y ) = n( X ) + n( Y ) - n ( X $\cap$ Y )

38  =  17 + 23 - n ( X $\cap$Y )

n ( X $\cap$Y ) =  40 - 38

n ( X $\cap$Y ) = 2.

n( X $\cup$ Y ) = 18

n(X) = 8

n(Y) = 15

n( X $\cup$ Y ) =  n(X) + n(Y) - n(X $\cap$ Y)

n(X $\cap$ Y) = 8 + 15 - 18

n(X $\cap$ Y) = 5

Hindi speaking = 250 = n(A)

English speakind = 200 = n(B)

Group of people = 400 = n(A $\cup$ B)

People speaking both Hindi and English = n(A $\cap$ B)

n(A $\cup$ B) = n(A) + n(B) - n(A $\cap$ B)

n(A $\cap$ B) = 250 + 200 - 400

n(A $\cap$ B) = 450 - 400

n(A $\cap$ B)  = 50

Thus,50 people speak hindi and english both.

n(S) = 21

n(T) = 32

n(S $\cap$  T) = 11

n(S $\cup$ T) = n(S) + n(T) - n(S $\cap$  T)

n(S $\cup$T) = 21 +  32 - 11

n(S $\cup$T) = 53 - 11

n(S $\cup$ T)  = 42

Hence, the set  (S $\cup$ T)  has 42 elements.

n( X $\cup$ Y) = 60

n( X ∩ Y) = 10

n(X) = 40

n( X $\cup$ Y) =  n(X)  +  n(Y)   -  n( X ∩ Y)

n(Y) = 60 - 40 + 10

n(Y) = 30

Hence,set Y has 30 elements.

n(people like coffee) = 37

n(people like tea ) = 52

Total number of people = 70

Total number of people = n(people like coffee)  +   n(people like tea )  - n(people like both tea and coffee)

70  =  37  +   52 - n(people like both tea and coffee)

n(people like both tea and coffee)  =  89 - 70

n(people like both tea and coffee)  =  19

Hence,19 people like both coffee and tea.

Total people = 65

n(like  cricket) = 40

n(like both cricket and tennis) = 10

n(like tennis) = ?

Total people = n(like  cricket)   +   n(like tennis)  -  n(like both cricket and tennis)

n(like tennis) = 65 - 40 + 10

n(like tennis) = 35

Hence,35 people like tennis.

n (people like only tennis) =  n(like tennis) - n(like both cricket and tennis)

n (people like only tennis) = 35 - 10

n (people like only tennis) = 25

Hence,25 people like only tennis.

n(french) = 50

n(spanish) = 20

n(speak both french and spanish) = 10

n(speak at least one of these two languages) = n(french) + n(spanish) - n(speak both french and spanish)

n(speak at least one of these two languages) = 50  + 20 - 10

n(speak at least one of these two languages) = 70 -10

n(speak at least one of these two languages) = 60

Hence,60 people speak at least one of these two languages.

## Solutions of NCERT for class 11 maths chapter 1 sets-Miscellaneous Exercise

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

$\double \rightarrow \arrow$            A + B - A $\cap$ B =  A + C - A $\cap$ C

$\double \rightarrow \arrow$                B - A $\cap$ B =   C - A $\cap$ C              ( since  A $\cap$ B = A $\cap$ C )

$\double \rightarrow \arrow$                      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 $\not = \equal$ $\phi$

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