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.

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This Story also Contains
  1. NCERT Sets Class 11 Questions And Answers
  2. Sets Class 11 Questions And Answers PDF Free Download
  3. Set Class 11 Solutions - Important Formulae
  4. Set Class 11 NCERT Solutions (Intext Questions and Exercise)
  5. Class 11 Maths Chapter 1 – Topics
  6. Sets Class 11 Maths Solutions - Chapter Wise
  7. Key Features of NCERT Solutions for chapter 1 class 11 maths – Sets
  8. Sets Class 11 Solutions - Subject Wise
  9. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 11 Maths Chapter 1 Sets
NCERT Solutions for Class 11 Maths Chapter 1 Sets

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.

Also read:

Sets Class 11 Questions And Answers PDF Free Download

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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

Question:1(i) Which of the following are sets ? justify your answer

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

Answer:

The months starting with letter J are:

january

june

july

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

Question:1(ii) Which of the following are sets ? justify your answer

The collection of ten most talented writers of India.

Answer:

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.

Question:1(iii) Which of the following are sets ? justify your answer

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

Answer:

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.

Question:1(iv) Which of the following are sets ? justify your answer

The collection of all boys in your class.

Answer:

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.

Question:1(v) Which of the following are sets ? justify your answer

The collection of all natural numbers less than 100.

Answer:

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.

Question:1(vi) Which of the following are sets? Justify your answer

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

Answer:

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

Hence,it is a set.

Question:1(vii) Which of the following are sets? Justify your answer

The collection of all even integers.

Answer:

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

Hence,it is a set.

Question:1(viii) Which of the following are sets? Justify your answer

The collection of questions in this Chapter.

Answer:

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

Hence ,it is a set.

Question:1(ix) Which of the following are sets? Justify your answer

A collection of most dangerous animals of the world.

Answer:

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.

Question:2 \, Let A = \left \{ 1,2,3,4,5,6 \right \}.Insert\, the\, appropriate\, symbol \in or\notin in\, the\, blank\, space :

(i) 5_____A

(ii) 8_____A

(iii) 0_____A

(iv) 4_____A

(v) 2_____A

(vi) 10____A

Answer:

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

1646200832601

Answer:

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 \}

Answer:

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}

Answer:

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}

Answer:

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.

Answer:

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.

Answer:

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}

Answer:

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}

Answer:

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}

Answer:

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, . . .}

Answer:

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}

Answer:

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}.

Answer:

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} }

Answer:

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 }

Answer:

\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”}

Answer:

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}

Answer:

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 }.

Answer:

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

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

Question:6 Match each of the set on the left in the roster form with the same set on the right described in set-builder form:

(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}.

Answer:

(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

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

Set of odd natural numbers divisible by 2

Answer:

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.

Answer:

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 }

Answer:

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}

Answer:

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

Hence, it is a null set.

Question:2 Which of the following sets are finite or infinite:

(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

Answer:

(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.

Question:3 State whether each of the following set is finite or infinite:

(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)

Answer:

(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 }

Answer:

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}

Answer:

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 }

Answer:

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, . . . }

Answer:

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}

Answer:

As given,

A = {2,3}

And,

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

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

(x+2)(x+3)=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}

Answer:

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.

Question:6 From the sets given below, select equal sets :

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}

Answer:

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

Question:1 Make correct statements by filling in the symbols \subset or \not\subset in the blank spaces :

(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}

Answer:

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}

Question:2 Examine whether the following statements are true or false:

(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}

Answer:

(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.

Question:3(i) Let A = { 1, 2, { 3, 4 }, 5 }. Which of the following statements are incorrect and why?

{3, 4} \subset A

Answer:

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

SO, {3, 4} \not \subset A

Hence,the statement is incorrect.

Question:3(ii) Let A = { 1, 2, { 3, 4 }, 5 }. Which of the following statements are incorrect and why?

{3, 4} \in A

Answer:

{3, 4} is element of A.

So, {3, 4} \in A.

Hence,the statement is correct.

Question:3(iii) Let A = { 1, 2, { 3, 4 }, 5 }. Which of the following statements are incorrect and why?

{{3, 4}} \subset A

Answer:

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.

Question:3(iv) Let A = { 1, 2, { 3, 4 }, 5 }. Which of the following statements are incorrect and why?

1 \in A

Answer:

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

So,1 \in A.

Hence,statement is correct.

Question:3(v) Let A = { 1, 2, { 3, 4 }, 5 }. Which of the following statements are incorrect and why?

1 \subset A

Answer:

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.

Question:3(vi) Let A = { 1, 2, { 3, 4 }, 5 }. Which of the following statements are incorrect and why?

{1,2,5} \subset A

Answer:

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.

Question:3(vii) Let A = { 1, 2, { 3, 4 }, 5 }. Which of the following statements are incorrect and why?

{1,2,5} \in A

Answer:

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

So,{1,2,5} \notin A .

Hence, statement is incorrect.

Question:3(viii) Let A = { 1, 2, { 3, 4 }, 5 }. Which of the following statements are incorrect and why?

{1,2,3} \subset A

Answer:

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.

Question:3(ix) Let A = { 1, 2, { 3, 4 }, 5 }. Which of the following statements are incorrect and why?

\phi \in A

Answer:

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

So, \phi \notin A .

Hence,the above statement is incorrect.

Question:3(x) Let A = { 1, 2, { 3, 4 }, 5 }. Which of the following statements are incorrect and why?

\phi \subset A

Answer:

\phi is subset of all sets.

Hence,the above statement is correct.

Question:3(xi) Let A = { 1, 2, { 3, 4 }, 5 }. Which of the following statements are incorrect and why?

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

Answer:

\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}

Answer:

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

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

{a, b}

Answer:

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}

Answer:

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

Answer:

Subset of \phi is \phi only.

The subset of a null set is null set itself

Question:5 How many elements has P(A), if A = \phi ?

Answer:

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}

Answer:

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} =[0,7)

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

Question:7 Write the following intervals in set-builder form :

(i) (– 3, 0)

(ii) [6 , 12]

(iii) (6, 12]

(iv) [–23, 5)

Answer:

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 \}

Question:8 What universal set(s) would you propose for each of the following :

(i) The set of right triangles.

(ii) The set of isosceles triangles.

Answer:

(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

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

X = {1, 3, 5} Y = {1, 2, 3}

Answer:

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}

Answer:

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}

Answer:

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 }

Answer:

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 }

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

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

Answer:

Here,

A union B is A \cup B.

A \cup B = {1,2,3}

Question:2 Let A = { a, b }, B = {a, b, c}. Is A \subset B ? What is A \cup B ?

Answer:

Here,

We can see elements of A lie in set B.

Hence, A \subset B.

And, A \cup B = {a,b,c} = B

Question:3 If A and B are two sets such that A \subset B, then what is A \cup B ?

Answer:

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

Question:4(i) If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8 }and D = { 7, 8, 9, 10 }; find

A \cup B

Answer:

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}

Question:4(ii) If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8 }and D = { 7, 8, 9, 10 }; find

A \cup C

Answer:

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}

Question:4(iii) If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8 }and D = { 7, 8, 9, 10 }; find

B \cup C

Answer:

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}

Question:4(iv) If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8 }and D = { 7, 8, 9, 10 }; find

B \cup D

Answer:

Here,

B = {3, 4, 5, 6}

D = { 7, 8, 9, 10 }

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

Question: 4(v) If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8 }and D = { 7, 8, 9, 10 }; find

A \cup B \cup C

Answer:

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}

Question:4(vi) If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8 }and D = { 7, 8, 9, 10 }; find

A \cup B \cup D

Answer:

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}

Question:4(vii) If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8 }and D = { 7, 8, 9, 10 }; find

B \cup C \cup D

Answer:

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}

Question:5 Find the intersection of each pair of sets of question 1 above

(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

Answer:

(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

Question:6 If A = { 3, 5, 7, 9, 11 }, B = {7, 9, 11, 13}, C = {11, 13, 15}and D = {15, 17}; find

(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)

Answer:

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}

Question:7 If A = {x : x is a natural number }, B = {x : x is an even natural number} C = {x : x is an odd natural number} and D = {x : x is a prime number }, find

(i) A \cap B

(ii) A \cap C

(iii) A \cap D

(iv) B \cap C

(v) B \cap D

(vi) C \cap D

Answer:

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 }

Answer:

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 }

Answer:

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}

Answer:

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.

Question:9 If 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 }; find

(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

Answer:

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}

Question:10 If X= { a, b, c, d } and Y = { f, b, d, g}, find

(i) X – Y

(ii) Y – X

(iii) X \cap Y

Answer:

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}

Question:11 If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q ?

Answer:

R = set of real numbers.

Q = set of rational numbers.

R - Q = set of irrational numbers.

Question:12(i) State whether each of the following statement is true or false. Justify your answer.

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

Answer:

Here,

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

{ 2, 3, 4, 5 } \cap { 3, 6} = {3}

Hence,these are not disjoint sets.

So,false.

Question:12(ii) State whether each of the following statement is true or false. Justify your answer.

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

Answer:

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.

Question:12(iii) State whether each of the following statement is true or false. Justify your answer.

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

Answer:

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.

Question:12(iv) State whether each of the following statement is true or false. Justify your answer.

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

Answer:

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

Question:1 Let 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 }. Find

(i) A′

(ii) B′

(iii) (A \cup C)′

(iv) (A \cup B)′

(v) (A')'

(vi) (B – C)'

Answer:

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}

Question:2 If U = { a, b, c, d, e, f, g, h}, find the complements of the following sets :

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

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

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

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

Answer:

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}

Question:3 Taking the set of natural numbers as the universal set, write down the complements of the following sets:

(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}

Answer:

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}

Question:3 Taking the set of natural numbers as the universal set, write down the complements of the following sets:

(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 }

Answer:

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 }

Question:4 If U = {1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = {2, 4, 6, 8} and B = { 2, 3, 5, 7}. Verify that

(i) (A \cup B)′ = A′ \cap B′

(ii) (A \cap B)′ = A′ \cup B′

Answer:

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.

Question:5 Draw appropriate Venn diagram for each of the following :

(i) (A ∪ B)′

(ii) A′ ∩ B′

(iii) (A ∩ B)′

(iv) A′ ∪ B′

Answer:

(i) (A ∪ B)′

(A ∪ B) is in yellow colour

(A ∪ B)′ is in green colour

1646201280340

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

1646201310933

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

1646201329212

iv) A′ ∪ B′

1646201343847

The green colour represents A′ ∪ B′

Question:6 Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from 60\degree , what is A′?

Answer:

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

Question:7 Fill in the blanks to make each of the following a true statement :

(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

Answer:

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

Question:1 If X and Y are two sets such that n ( X ) = 17, n ( Y ) = 23 and n ( X \cup Y ) = 38, find n ( X \cap Y ).

Answer:

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.

Question:2 If X and Y are two sets such that X \cup Y has 18 elements, X has 8 elements and Y has 15 elements ; how many elements does X \cap Y have?

Answer:

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

Question:3 In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English?

Answer:

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.

Answer:

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.

Question:5 If X and Y are two sets such that X has 40 elements, X \cup Y has 60 elements and X \cap Y has 10 elements, how many elements does Y have?

Answer:

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.

Question:6 In a group of 70 people, 37 like coffee, 52 like tea and each person likes at least one of the two drinks. How many people like both coffee and tea?

Answer:

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.

Question:7 In a group of 65 people, 40 like cricket, 10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?

Answer:

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.

Question:8 In a committee, 50 people speak French, 20 speak Spanish and 10 speak both Spanish and French. How many speak at least one of these two languages?

Answer:

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

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 }.

Answer:

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

Question:2(i) In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.

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

Answer:

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

Question:2(ii) In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.

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

Answer:

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

Question:2(iii) In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.

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

Answer:

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

Question:2(iv) In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.

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

Answer:

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

Question:2(v) In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.

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

Answer:

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

Question:2(vi) In each of the following, determine whether the statement is true or false. If it is true, prove it. If it is false, give an example.

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

Answer:

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.

Question:3 Let A, B, and C be the sets such that A \cup B = A \cup C and A \cap B = A \cap C. Show that B = C.

Answer:

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.

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

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

Answer:

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

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

Answer:

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

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

Answer:

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

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

Answer:

No, it is false.

To prove : P ( A ) ∪ P ( B ) \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 )

Answer:

A = ( A \cap B ) \cup ( A – B )

L.H.S = A = Red coloured area1646201732372 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

1646201753850 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

1646201771347

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

1646201783603

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

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

A \cup ( A \cap B ) = A

Answer:

(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

Answer:

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

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

Answer:

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.

Question:11 Let A and B be sets. If A \cap X = B \cap X = \phi and A \cup X = B \cup X for some set X, show that A = B.

Answer:

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

Question:12 Find sets A, B and C such that A \cap B, B \cap C and A \cap C are non-empty sets and A \cap B \cap C = \phi

Answer:

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

Question:13 In a survey of 600 students in a school, 150 students were found to be taking tea and 225 taking coffee, 100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?

Answer:

n ( taking tea) = 150

n (taking coffee) = 225

n ( taking both ) = 100

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

= 150 + 225 - 100

=375 - 100

= 275

Total students = 600

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

Hence,325 students were taking neither tea nor coffee.

Question:14 In a group of students, 100 students know Hindi, 50 know English and 25 know both. Each of the students knows either Hindi or English. How many students are there in the group?

Answer:

n (hindi ) = 100

n (english ) = 50

n(both) = 25

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

= 100 + 50 - 25

= 125

Hence,there are 125 students in the group.

Question:15 In a survey of 60 people, it was found that 25 people read newspaper H, 26 read newspaper T, 26 read newspaper I, 9 read both H and I, 11 read both H and T, 8 read both T and I, 3 read all three newspapers. Find:

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

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

Answer:

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 .

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.

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

Happy Reading !!!

Frequently Asked Questions (FAQs)

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. 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.

4. 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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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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