NCERT Solutions for Class 11 Maths Chapter 1 Sets

# NCERT Solutions for Class 11 Maths Chapter 1 Sets

Edited By Ramraj Saini | Updated on Sep 21, 2023 08:05 PM IST

## NCERT Sets Class 11 Questions And Answers

NCERT Solutions for Class 11 Maths Chapter 1 Sets are discussd here. 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 a set. In this article, you will get class 11 sets NCERT solutions. This chapter sets class 11 will introduce the concept of sets and different operations on set. Check NCERT solutions that are created by Careers360 Expert team according to latest update on the CBSE syllabus 2023. The knowledge of sets is also required to study geometry, sequences, probability, etc. NCERT solutions for class 11 maths chapter 1 sets will build your basics of data analysis which will be useful in higher education.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics WallahAakash Unacademy

Sets play a crucial role in defining functions and relations, concepts which are introduced in Chapter 1 of the NCERT textbook for Class 11 Maths, as per the CBSE syllabus. This chapter covers fundamental definitions and operations involving sets. Having a sound understanding of sets is imperative for studying sequences, geometry, and probability. Although it is an essential chapter, it is comparatively easier than other chapters in the NCERT Class 11 Maths textbook, making it an opportunity to score maximum marks in the board examination. Here students can find all NCERT solution for Class 11 . The NCERT solutions for class 11 maths chapter 1 set will improve the basics of sets which will be useful in further mathematics courses also.

## Set Class 11 Solutions - Important Formulae

Union of Sets (A∪B): A∪B

Intersection of Sets (A∩B): A∩B

Complement of a Set (A'): A'

De Morgan’s Theorem:

• (A∪B)' = A'∩B'

• (A∩B)' = A'∪B'

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Set Cardinality with Intersection:

• n(A∪B) = n(A) + n(B)

• n(A∪B) = n(A) + n(B) - n(A∩B)

Other Set Formulas:

• A - A = Ø

• B - A = B∩A'

• B - A = B - (A∩B)

• (A - B) = A if A∩B = Ø

• (A - B) ∩ C = (A∩ C) - (B∩C)

• A ΔB = (A-B) ∪ (B-A)

• n(A∪B∪C) = n(A) + n(B) + n(C) - n(B∩C) - n(A∩B) - n(A∩C) + n(A∩B∩C)

• n(A - B) = n(A∪B) - n(B)

• n(A - B) = n(A) - n(A∩B)

• n(A') = n(U) - n(A)

• n(U) = n(A) + n(B) - n(A∩B)

• n((A∪B)') = n(U) + n(A∩B) - n(A) - n(B)

Free download NCERT Solutions for Class 11 Maths Chapter 1 Sets for CBSE Exam.

## Set Class 11 NCERT Solutions (Intext Questions and Exercise)

NCERT sets class 11 questions and answers - 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.

(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

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

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}.

(i) {1, 2, 3, 6} (a) {x : x is a prime number and a divisor of 6}

(ii) {2, 3} (b) {x : x is an odd natural number less than 10}

(iii) {M,A,T,H,E,I,C,S} (c) {x : x is natural number and divisor of 6}

(iv) {1, 3, 5, 7, 9} (d) {x : x is a letter of the word MATHEMATICS}.

(i) 1,2,3,6 all are natural numbers and also divisors of 6.

Hence,(i) matches with (c)

(ii) 2,3 are prime numbers and are divisors of 6.

Hence,(ii) matches with (a).

(iii) M,A,T,H,E,I,C,S are letters of word "MATHEMATICS".

Hence, (iii) matches with (d).

(iv) 1,3,5,7,9 are odd natural numbers less than 10.

Hence,(iv) matches with (b).

NCERT class 11 maths chapter 1 question answer 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(iii) Which of the following are examples of the null set:

{ x : x is a natural numbers, $x< 5$ and $x> 7$ }

No number exists which is less than 5 and more than 7.

Hence, this is 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,

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

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

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.

8 $\in$ 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.

3 $\in$ 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.

NCERT class 11 maths chapter 1 question answer 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

3 $\in$ {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.

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

So, $\phi \notin A$ .

Hence,the above statement is incorrect.

$\phi$ is subset of all sets.

Hence,the above statement is correct.

$\phi$ $\subset$ { 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:

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}

(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 class 11 maths chapter 1 question answer 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.

A $\cup$ 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}

A $\cup$ 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.

A $\cup$ 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}

A $\cup$ C

Here,

A = {1, 2, 3, 4}

C = {5, 6, 7, 8 }

The union can be written as follows

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

B $\cup$ C

Here,

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

C = {5, 6, 7, 8 }

The union of the given sets are

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

B $\cup$ D

Here,

B = {3, 4, 5, 6}

D = { 7, 8, 9, 10 }

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

A $\cup$ 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

A $\cup$ 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}

(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 }

(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.

Class 11 maths chapter 1 NCERT solutions 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 $=$ $\phi$

Class 11 maths chapter 1 NCERT solutions 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.

NCERT class 11 maths ch 1 question answer 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

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 $\phi$

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.

## Class 11 Maths Chapter 1 – Topics

Careers360 offers creative and logical explanations and solutions for chapter 1 of Class 11 Maths, Sets, as well as a downloadable PDF of NCERT Solutions. The chapter includes the following topics and sub-topics:

1.1 Introduction: This section of class 11 maths ch 1 covers the origin, basic definition, and applications of sets.

1.2 Sets and their Representations: Students learn the exact definition of a set and how it can be represented in roster/set-builder notation and denoted using English alphabets.

1.3 The Empty Set: This topic explains when a set is considered an empty set.

1.4 Finite and Infinite Sets: This section provides definitions and examples of finite and infinite sets.

1.5 Equal Sets: Students gain an understanding of equal and unequal sets, along with solved examples.

1.6 Subsets: This section covers the main concepts of subsets, including subsets of a set of real numbers and intervals as subsets of R.

1.7 Power Set: The definition of power set, derived from the concept of subsets, is explained in this section.

1.8 Universal Set: This section uses real-life examples such as population studies to define the universal set and represent it using a letter.

1.9 Venn Diagrams: This section explains the origin and definition of a Venn diagram, as well as its use in illustrating the above topics and sub-topics.

1.10 Operations on Sets: Basic operations on sets are introduced, including union of sets, intersection of sets, and difference of sets.

1.11 Complement of a Set: This section explains the definition and properties of the complement of a set, along with examples and practice problems.

1.12 Practical Problems on Union and Intersection of Two Sets: Practice problems are provided to help students understand the union and intersection of two sets.

The chapter also includes six exercises and a miscellaneous exercise with a total of 64 questions and their solutions.

Exercise 1.1 Solutions 6 Questions

Exercise 1.2 Solutions 6 Questions

Exercise 1.3 Solutions 9 Questions

Exercise 1.4 Solutions 12 Questions

Exercise 1.5 Solutions 7 Questions

Exercise 1.6 Solutions 8 Questions

Miscellaneous Exercise On Chapter 1 Solutions 16 Questions

## Sets Class 11 Maths Solutions - Chapter Wise

Students can find NCERT Solution for class 11th Maths collectively at one place.

 chapter-1 Sets chapter-2 chapter-3 chapter-4 chapter-5 chapter-6 chapter-7 chapter-8 chapter-9 chapter-10 chapter-11 chapter-12 chapter-13 chapter-14 chapter-15 chapter-16

## Key Features of NCERT Solutions for chapter 1 class 11 maths – Sets

To assist students in comprehending the principles of Sets covered in the class 11 chapter 1 maths CBSE Syllabus 2023, this chapter comprises six exercises and one miscellaneous exercise.

Comprehensive Coverage: NCERT Solutions provide a thorough explanation of all the topics covered in the chapter, ensuring students have a complete understanding of sets and related concepts.

Step-by-Step Solutions: The solutions offer step-by-step explanations for each exercise and example in the NCERT textbook. This helps students follow the logical progression of concepts.

Clarity and Simplicity: The solutions are presented in a clear and simple language to make it easy for students to comprehend complex concepts related to sets.

## Sets Class 11 Solutions - Subject Wise

### Benefits of NCERT solutions

• NCERT class 11 maths ch 1 question answer sets is a building block for the set theory which will be useful in higher studies like in computer science: to make an efficient algorithm to solve some problems using programming language and to learn computability theory
• Class 11 sets solutions will help you to strengthen your fundamentals of sets, which are required to understand the concepts of relations and functions in the next chapter also.
• Class 11th maths chapter 1 ncert solutions will develop yous basic concept which will be helpful in further studies of data structures, pattern matching, formal query languages such as relational algebra, relational calculus, statistics, machine learning, etc.
• In computer science, data structure is the efficient implementation of sets operations. NCERT class 11 maths ch 1 question answer sets will build the basic of the operations of sets.

## NCERT Books and NCERT Syllabus

1. What are the topics covered in Chapter 1 of NCERT set class 11 solutions?

Chapter 1 of NCERT Solutions for Class 11 Maths covers the following topics:

1. Introduction
2. Sets and their representations
3. The Empty Set
4. Finite and Infinite Sets
5. Equal Sets
6. Subsets
7. Power Set
8. Universal Set
9. Venn Diagrams
10. Operations on Sets
11. Complement of a Set
12. Practical problems on Union and Intersection of Two Sets
2. Which are the most difficult chapters of NCERT Class 11 Maths syllabus?

Most students consider permutation and combination, trigonometry as the most difficult chapter in the class 11 maths but with consistent practice and the right strategy, they will get command of them also. students can take help from NCERT textbooks, exercises, and solutions. Careers360 provides these solutions freely.

3. Where can I find the complete set class 11 ncert solutions ?

Here you will get the detailed NCERT solutions for class 11 maths by clicking on the link. students can also find class 11 maths chapter 1 NCERT solutions  above in the article that are listed subject-wise and chapter-wise.

4. What applications of Sets are discussed in NCERT Solutions for maths chapter 1 class 11?

ch 1 maths class 11 sets find extensive applications not only in Mathematics but also in real-life scenarios. They are widely used in various branches of Mathematics such as topology, calculus, algebra (including fields, rings, and groups), and geometry. Additionally, sets are also employed in several other fields like Physics, Chemistry, Electrical Engineering, Computer Science, and Biology.

5. Does it important to practice all the problems discussed in NCERT solutions class 11 maths chapter 1?

It is very important to practice sets class 11 solutions as concepts are foundation for higher class topics such as relation and function, probability, permutation and combinations. for ease students can study class 11 maths chapter 1 pdf both offline and online mode.

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