Class 11 Maths NCERT Solutions Chapter 1 SETS: Exercise PDFs with Solution, Topics, Weightage & Formulae

This chapter has multiple topics listed below:

NCERT Maths 11th Sets Relations and Functions Trigonometric Functions Principle of Mathematical Induc...Complex Numbers and Quadratic E...Linear Inequalities Permutations and Combinations Binomial Theorem Sequences and Series Straight Lines Conic Sections Introduction to Three Dimension...Limits and Derivatives Mathematical Reasoning Statistics Probability

Updated on Feb 5, 2025 16:47 IST

By nitesh singh, Senior Executive

Class 11 Chapter 1 Sets NCERT Solutions: Students can access complete NCERT solutions for Class 11 Chapter 1 Sets prepared by our experts in Shiksha. Class 11 Sets have a good weightage in annual exams and combined with Relation and Function it's also important in competitive exams such as JEE Mains and others. Class 11 Math Sets delve into topics such as Fundamentals, Types, Finites & Infinite Sets, Venn Diagrams, and operations on set.

Class 11 Math Sets NCERT Solutions will be very useful for both board and entrance exam students and We have designed the solutions in such a way that concepts are easy to understand. Shiksha has prepared complete class 11 Sets NCERT Solutions by our experts to help students. Students can access the complete Exercise-wise NCERT Class 11 Chapter 11 Sets Solution PDF through the link below.

Students can use the NCERT Solutions of Sets Class 11 Math textbook to strengthen their concepts for CBSE Boards as well as other Entrance exams such as JEE Main, CUSAT CAT, etc. Students must check Class 11 Mathematics topics such as Relation and Functions, and Permutation and combination for more inclusive understanding. The German mathematician Georg Cantor developed the theory of sets. A set is a well-defined collection of objects. A set that does not have any element is defined as an empty set. A set that consists of a definite number of elements is called a finite set, otherwise, the set is called an infinite set

Students also check the Class 11 Sets NCERT solutions, which can help to get a better understanding of several other chapters in mathematics. Students can access the complete Class 11 Math chapter-wise Solutions and Class 12 Chapter-wise Math solutions on Shiksha. For more information check below;

Class 11 Math-Related Chapter Solutions
Class 11 Chapter 2 Relation and Function	Class 11 Chapter 3 Trigonometric Functions	Class 11 Chapter 5 Linear Inequalities

Table of content

Class 11 Chapter 1 Sets: Key Topics, Weightage and Important formulae
Class 11 Chapter 1 Sets NCERT Solution PDF: Free PDF Download
Class 11 Chapter 1 Sets Exercise-wise NCERT Solutions
Class 11 Chapter 1 Sets Exercise 1.1 Solutions
Class 11 Chapter 1 Sets Exercise 1.2 Solutions
Class 11 Chapter 1 Sets Exercise 1.3 Solutions
Class 11 Chapter 1 Sets Exercise 1.4 Solutions
Class 11 Chapter 1 Sets Exercise 1.5 Solutions
Class 11 Chapter 1 Sets Exercise 1.6 Solutions
Class 11 Chapter 1 Sets Miscellaneous Exercise NCERT Solutions
Mathematics Prep Tips and Recommended Books for Boards and Competitive Exams

Class 11 Chapter 1 Sets: Key Topics, Weightage and Important formulae

Class 11th Math Sets chapter carries around 6-8 marks weightage in the annual exams. Class 11 Sets chapter is also important to develop a basic understanding of Relation and Function, their formation, domain, and ranges. Sets combined with Relation and Function are also important for JEE Mains. Several key issues are discussed in the exercises;

Class 11 Math Sets Key Topics

Basic Concepts of Sets: Definition and Representation of Sets (Roster & Set Builder Form)
Types of Sets: Finite, Infinite, Equal, Null, Singleton, Universal, etc.
Operations on Sets: Union, Intersection, Difference, Complement of a Set
Venn Diagrams: Representation, and Solving Problems using Venn Diagrams
Subsets and Power Sets
Algebra of Sets: Idempotent Law, Domination Law, Commutative Law, Associative Law, Distributive Law, De Morgan’s Laws, Cartesian Product of Sets
Ordered Pairs: Cartesian Product, Number of Elements in A × B (|A × B| = |A| × |B|)

Check out the Dropped Topics for Sets – 1.7 Power Set, Exercise 1.3 Ques. 5, 1.12 Practical Problems on Union and Intersection of Two Sets(Exercise 1.6), Examples 31–34 and Ques. 6–7 & Ques. 13–16 (Miscellaneous Exercise)

Class 11th Important Formulae

Number of Elements Formula
- $|A \cup B| = |A| + |B| - |A \cap B|$
- $|A - B| = |A| - |A \cap B|$
De Morgan’s Laws
- $(A \cup B)' = A' \cap B'$
- $(A \cap B)' = A' \cup B'$
Subset Formula
- Total subsets of a set with ‘n’ elements = $2^n$
- Total proper subsets = $2^n - 1$
Properties of Union and Intersection
- $A \cup B = B \cup A$ (Commutative Property)
- $A \cap B = B \cap A$ (Commutative Property)
- $(A \cup B) \cup C = A \cup (B \cup C)$ (Associative Property)
- $(A \cap B) \cap C = A \cap (B \cap C)$ (Associative Property)

Class 11 Chapter 1 Sets NCERT Solution PDF: Free PDF Download

Students can find the solutions to all the exercises of class 11 Math Sets in one place. This Class 11 Chapter 1 Sets NCERT solution PDF is a very useful document for all categories of students whether on boards or competitive exams such as JEE Mains, BITSAT and others. access the Sets NCERT solution PDF below;

Class 11 Math Chapter 1 Sets NCERT Solution: Free PDF Download

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Class 11 Chapter 1 Sets Exercise-wise NCERT Solutions

Class 11 Sets chapter deals with many important topics, such as the definition and representation of Sets, and types of Sets; Finite, Infinite, Singleton, Universal, etc., Venn Diagrams, operations on sets. . Students will encounter problems related to different concepts and properties of Set Theory in these exercises. Candidates can check the exercise-wise Chapter 1 Class 11 math solutions below;

Class 11 Chapter 1 Sets Exercise 1.1 Solutions

Class 11 Chapter 1 Sets Exercise 1.1 focuses on building the basic conceptual understanding of Set, relation and function. Sets Exercise 1.1 deals with fundamental concepts such as the representation of sets (roster form and set-builder form), types of sets (finite, infinite, empty, equal, and equivalent sets), and Venn diagrams. the Class 11 Set Exercise 1.1 also focuses on operations on sets like union, intersection, and complement of sets. Exercise 1.1 Solutions includes a total of 6 Questions all descriptive. Students can check the complete solution of Exercise 1.1 Below:

Class 11 Chapter 1 Set Exercise 1.1 NCERT Solutions

Q1. Which of the following are sets? Justify your answer.

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

(ii)The collection of ten most talented writers of India.

(iii)A team of eleven best-cricket batsmen of the world.

(iv)The collection of all boys in your class.

(v)The collection of all natural numbers less than 100.

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

(vii)The collection of all even integers.

(viii)The collection of questions in this Chapter.

(ix)A collection of most dangerous animals of the world.

A.1. (i)The collection of all months of a year with J as initial are January, June and July. Hence, it is a well-defined and is therefore a set $\{J a n u a r y, J u n e, J u l y\}$ .

(ii)The collection of ten most talented writers of India is not well-defined as it may vary from one person to another. Hence, it is not a set.

(iii)The team of 11 best-cricket batsmen of the world is not well-defined as it may vary from one parson to another as they may vary from one person to another. Hence, it is not a set.

(iv)The collection of all boys in your class is well-defined as your-class is fixed. Hence, it is a set.

(v)The collection of all natural numbersless than 100 will be from 1 to 99. Hence, it is well-defined and is therefore a set.

(vi)Novels written by writer Munshi Prem cham is a well-defined collection as he is no more so. Hence it is a set.

(vii)The collection of all even integers is well-defined and hence is a set.

(viii)The collection of questions in this chapter is well-defined and hence is a set.

(ix)The collection of most dangerous animals of the world is not well-defined as it may vary from one person to another.

Q2. Let A = {1, 2, 3, 4, 5, 6}. Insert the appropriate symbol $\in$ or $\notin$ in the blank spaces:

(i)5. . .A(ii)8 . . . A (iii)0. . .A

(iv) 4. . . A (v) 2. . .A (vi) 10. . .A

A.2. (i)5 $\in$ A(ii)8 $\notin$ A(iii)0 $\notin$ A

(iv)4 $\in$ A(v)2 $\in$ A(vi)10 $\notin$ A

Q3. Write the following sets in roster form:

(i)A = {x : x is an integer and –3< x < 7}

(ii) B = {x : x is a natural number less than 6}

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

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

(v) E = The set of all letters in the word TRIGONOMETRY

(vi) F = The set of all letters in the word BETTER

A.3. (i)A = {– 3, –2, –1, 0, 1, 2, 3, 4, 5, 6}

(ii)B = {1, 2, 3, 4, 5}

(iii)C = {17, 26, 35, 44, 53, 62, 71, 80}

(iv)x = Prime number which are divisor of 60

Factors of 60 are 1,2,3,4,5,6,10,12,15,20,30,60

Hence, x = 2, 3, 5

⸫D = {2, 3, 5}

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

(vi) F = {B, E, T, R}

Q4. Write the following sets in the set-builder form:

(i)(3, 6, 9, 12} (ii) {2,4,8,16,32} (iii) {5, 25, 125, 625}

(iv) {2, 4, 6, . . .} (v) {1,4,9, . . .,100}

A.4. (i) {3,6,9,12}={3 × 1, 3 × 2, 3 × 3, 3 × 4}

= {x : x = 3n, n is natural number and 1≤ n ≤ 4}

(ii){2,4,8,16,32}={2¹, 2², 2³, 2⁴, 2⁵}

={x : x = 2ⁿ, n is natural number and 1 ≤ n ≤ 5}

(iii){5,25,125,625}={5¹, 5², 5³, 5⁴}

={x : x = 5ⁿ, n is natural number and 1 ≤ n ≤ 4}

(iv){2,4,6,}={2 × 1, 2 × 2, 2 × 3, …}

={x : x = 2n, n is a natural number}

(v){1,4,9, …, 100}={1², 2², 3², …, 10²}

= {x : x = n2, x is a natural number and 1 ≤ n ≤ 10}

Q5. List all the elements of the following sets :

(i)A = {x : x is an odd natural number}

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

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

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

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

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

A.5. (i)A = {1, 3, 5, 7, …}

(ii)B ={0,1,2,3,4} (as $\frac{9}{2}$ =4.5 and $- \frac{1}{2}$ = –0.5)

(iii)x² ≤ 4

x² ≤ 2²

x ≤ ± 2

So, C ={–2, –1,0,1,2}

(vi)D = {L, O, Y, A}

(v)E = {February, April, June, September, November}

(vi)F = {b, c, d, f, g, h, j}

Q6. 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}.

A.6. (i)(i) (c)

(ii) (a)

(iii) (d)

(iv) (b)

Class 11 Chapter 1 Sets Exercise 1.2 Solutions

Class 11 Chapter 1 Sets Exercise 1.2 introduces important operations and properties of Set Theory. This Exercise 1.2 focuses on subsets, power sets, universal sets, and Venn diagrams, which are key to understanding Set Theory. Exercise 1.2 of class 11 Sets also explores operations on sets, including union, intersection, difference, and complement of set. Class 11 Set Exercise 1.2 Solutions consists of answers to 6 Questions. Students can check the complete solution of Exercise 1.2 Below;

Class 11 Chapter 1 Set Ex 1.2 NCERT Solutions

Q1. Which of the following are examples of the null set

(i) Set of odd natural numbers divisible by 2

(ii) Set of even prime numbers

(iii) {x : x is a natural numbers, x < 5 and x > 7}

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

A.l. (i)There is no odd number that can be divided by 2.

⸫The given set is a null set.

(ii)An even prime number is 2. Hence, the set will have 2 as element. ⸫ The given set is not a null set.

(iii)As there is no natural number which is both less than 5 and greater than 7. ⸫ The given set is a null set.

(iv)Two parallel lines never meets and hence no common point. ⸫ The given set is a null set.

Q2. 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

A.2. (i)The set of months of a year has 12 elements. Here, the set is finite.

(ii)The given set has the natural number as its elements. Hence, the set is infinite.

(iii)The given set has 100 elements i.e., from 1 to 100. Hence, the set is finite.

(iv)There are infinite numbers of positive integers greater than 100. Hence the set is infinite.

(v)The numbers of prime number less than 99 is finite. Hence, the set is finite.

Q3. 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)

A.3. (i)We can draw infinite number of lines parallel to x-axis.

⸫The set is infinite.

(ii)There are 26 letters in the English alphabet.

⸫The set is finite.

(iii)There are infinite number which are a multiple of 5.

⸫The set is infinite

(iv)The numbers of animals living on earth are finite.

⸫The set is finite.

(v)There can be infinite number of circular passing through origin (0,0).

⸫The set is infinite.

Q4. In the following, state whether A = B or not:

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

(ii) A = { 4, 8, 12, 16 } B = { 8, 4, 16, 18}

(iii) A = {2, 4, 6, 8, 10} B = { x : x is positive even integer and x < 10}

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

A.4. (i)As A and B has a, b, c and d as elements and are exactly the same, Hence, A = B.

(ii)Hence, 12 $\in$ A but 12 $\notin$ B

Similarly 18 $\in$ B but 18 $\notin$ A.

So, A ≠ B.

(iii)A = {2,4,6,8,10} and B = {2,4,6,8,10}

So, A=B.

(v)A={10,20,30,40, …} and B={10,15,20,25,30, …}

So, A ≠ B.

Q5. Are the following pair of sets equal? Give reasons.

(i) A = {2, 3}, B = {x : x is solution of x²+ 5x + 6 = 0}

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

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

A.5. (i)A = {2, 3)

B = {x : x is solution of x² + 5x + 6 = 0}

So, x² + 5x + 6 = 0

x² + 2x + 3x + 6 = 0

x(x+2) + 3(x+2) = 0

(x+2)(x+3) = 0

x = –2, –3

So, B = {–2, –3}

So, A ≠ B.

(ii)A = {x : x is a letter in word FOLLOW}

A = {F, O, L, W}

B = {x : x is a letter in word WOLF}

B = {W, O, L, F}

So, A = B as the elements are all same.

Q6.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}

A.6. B = D = {1, 2, 3, 4} = {3, 1, 4, 2}

E = G = {–1, 1} = {1, –1}.

Class 11 Chapter 1 Sets Exercise 1.3 Solutions

Class 11 Math Chapter 1 Sets Exercise 1.3 focuses on the algebra of sets, introducing important properties and identities that govern set operations. This section explores laws of union and intersection, such as commutative, associative, and distributive laws, which help simplify complex set expressions. Additionally, it covers De Morgan’s Laws, which provide a systematic way to find the complement of set operations.

Exercise 1.3 Solutions 9 Questions

Students can check the complete solution of Exercise 1.3 Below

Class 11 Set Exercise 1.3 Solutions

Q1. Make correct statements by filling in the symbols $\subset$ or ₡ 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}

A.1. (i){2,3,4} $\subset$ {1,2,3,4,5}

(ii){a, b, c} ₡ {b, c, d}

(iii){x : x is a student of class XI of yours school} $\subset$ {x : x is a student of your school}

(iv)As any circle in the plane can have radius more or less than 1 unit.

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

(v)As a triangle can never be a rectangle.

{x : x is a triangle in a plane} ₡ {x : x is a triangle in the same plane}

(vi)Any triangle in a plane can be scaler, isosceles, equilateral.

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

(vii)As all even natural number is also an integer.

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

Q2. Examine whether the following statements are true or false:

(i) { a, b } ₡ { 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}

A.2. (i)False as every element of set {a, b} is also an element of {b, c, a} hence {a, b } $\subset$ { b, c, a }.

(ii)True as every element in {a, e} is also a vowel in English alphabet.

(iii)False as 2 $\in$ {1, 2, 3} but 2 $\notin$ {1, 3, 5}.

(iv)True as a $\in$ {a} is also a $\in$ {a, b, c}

(v)False as a $\in$ {a, b, c} but {a} $\in$ {a, b, c}

(vi){x : x is an even natural number less than 6} = {2, 4}

{x : x is a natural number which divides 36} = {1, 2, 3, 4, 6, 9, 12, 18, 36}

As {2, 4} $\subset$ {1, 2, 3, 4, 6, 9, 12, 18, 36}

It is true.

Q.3. Let A = { 1, 2, { 3, 4 }, 5 }. Which of the following statements are incorrect and why?

(i) {3, 4} $\subset$ A (ii) {3, 4} $\in$ A (iii) {{3, 4}} $\subset$ A

(iv) 1 $\in$ A (v) 1 $\subset$ A (vi) {1, 2, 5} $\subset$ A

(vii) {1, 2, 5} $\in$ A (viii) {1, 2, 3} $\subset$ A (ix) $ϕ \in$ A

(x) $ϕ \subset$ A (xi) { $ϕ$ } $\subset$ A

A.3. (i)False as 3 $\notin$ A and 4 $\notin$ 4. So, {3, 4} $\subset$ A.

(ii)True as {3, 4} $\in$ A. i.e, {3, 4} is an element of A.

(iii)True as {3, 4} $\in$ A so, {3, 4} $\subset$ A.

(iv)True as 1 is an element of A.

(v)False as 1 is not a set so it cannot be a subset of A.

(vi)True as 1 $\in$ A, 2 $\in$ A and 5 $\in$ A. so, {1, 2, 5} $\subset$ A.

(vii)False as {1, 2, 5} is not an element of A.

(viii)False of 3 $\notin$ A.

(ix)False as $ϕ$ is not an element of A.

(x)True, $ϕ$ ⸦ A as $ϕ$ is a subset of every set.

(xi)False, as $ϕ$ is not an element of A.

Q4. Write down all the subsets of the following sets

(i) {a}(ii) {a, b}(iii) {1, 2, 3} (iv) $ϕ$

A.4. (i)Subset of {a} = {a}, $ϕ$ .

(ii)Subset of {a, b} = $ϕ$ ,{a},{b},{a, b}

(iii)Subset of {1,2,3} = $ϕ$ ,{1},{2},{3}, {1,2}, {1,3}, {2,3}, {1,2,3}.

(iv)Subset of $ϕ$ = $ϕ$ .

Q5. How many elements has P(A), if A = $ϕ$ ?

A.5. As A= $ϕ$ , no elements

n(A) = 0 = m

So, n[P(A)] = 2^m= 2⁰ = 1.

Q6. Write the following as intervals:

(i) {x : x $\in$ R, – 4 <x < 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}

A.6. (i)As x does not include –4 while 6 is included.

(–4, 6].

(ii)As x does not include both –12 and –10.

(–12, –10)

(iii)As x includes 0 but does not includes 7.

[0, 7)

(iv)As includes both 3 and 4

[3, 4]

Q7. Write the following intervals in set-builder form :

(i) (– 3, 0) (ii) [6 , 12] (iii) (6, 12] (iv) [–23, 5)

A.7. (i)[–3, 0] = {x : x such that x $\in$ R, –3 < x <0}

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

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

(iv)[–23, 5) = {x : x $\in$ R, –23 ≤ x < 5}

Q8. What universal set(s) would you propose for each of the following:

(i)The set of right triangles. (ii) The set of isosceles triangles.

A.8. (i)The set of all triangles. U= {x : x is a triangle in a plane}

(ii) The set of all triangles. U= {x : x is a triangle in a plane}

Q9. Given the sets A = {1, 3, 5}, B = {2, 4, 6} and C = {0, 2, 4, 6, 8}, which of the following may be considered as universal set (s) for all the three sets A, B and C

(i) {0, 1, 2, 3, 4, 5, 6}

(ii) $ϕ$

(iii) {0,1,2,3,4,5,6,7,8,9,10}

(iv) {1,2,3,4,5,6,7,8}

A.9. Universal set of A, B and C must includes all elements of A, B and C, i.e. 0,1,2,3,4,5,6,8.

So, the universal set of A, B and C is (iii) {0,1,2,3,4,5,6,7,8,9,10}.

Class 11 Chapter 1 Sets Exercise 1.4 Solutions

Class 11 Chapter 1 Sets Exercise 1.4 focuses on the practical applications of set theory, particularly in solving problems using the principle of inclusion and exclusion. Class 11 Math Exercise 1.4 is very important to understand counting elements in unions and intersections of multiple sets efficiently. Many important concepts such as finite and infinite sets, cardinality of sets, and methods to determine the number of elements in given set operations are discussed in Exercise 1.4 of Class 11 Sets. Class 11 Sets Exercise 1.4 consists of 12 questions. Students can check solutions to all 12 questions in Exercise 1.4 Below.

Sets Ex 1.4 Class 11 Math Solutions

Q1. Find the union of each of the following pairs of sets:

(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 = $ϕ$

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

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

(iii)A = {3, 6, 9, 12 …}

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

So, A∪ B = {3, 6, 9, 12, …}∪ {1, 2, 3, 4, 5}

= {1, 2, 3, 4, 5, 6, 9, 12 …}

(iv)A = {2, 3, 4, 5, 6}

B = {7, 8, 9}

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

(v)A ∪B = {1, 2, 3} ∪ $ϕ$ = {1, 2, 3}.

Q2. Let A = { a, b }, B = { a, b, c}. Is A $\subset$ B ? What is A $\cup$ B ?

A.2. Given,A = {a, b}

B = {a, b, c}

Yes A $\subset$ B as a, b $\in$ A and a, b $\in$ B.

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

Q3.If A and B are two sets such that A $\subset$ B, then what is A $\cup$ B ?

A.3.If A ⸦ B then let a $\in$ A and also a $\in$ B.

but b $\in$ B and b $\notin$ A i.e, A = {a} and B = {a, b}

So, A $\cup$ B = {x : x $\in$ A or x $\in$ B}

= {a, b}

= B

Q4. If A = {1, 2, 3, 4}, B = {3, 4, 5, 6}, C = {5, 6, 7, 8 } and D = { 7, 8, 9, 10 }; find

(i) A $\cup$ B (ii) A $\cup$ C (iii) B $\cup$ C (iv) B $\cup$ D

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

A.4. (i)A $\cup$ B = {1, 2, 3, 4} $\cup$ {3, 4, 5, 6}

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

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

= {1, 2, 3, 4, 5, 6, 7, 8}

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

= {3, 4, 5, 6, 7, 8}

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

= {3, 4, 5, 6, 7, 8, 9, 10}

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

= {1, 2, 3, 4, 5, 6, 7, 8}

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

= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

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

= {3, 4, 5, 6, 7, 8, 9, 10}

Q5. Find the intersection of each pair of sets of question 1 above.

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

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

(iii)A ∩ B = {3,6,9,12 …} ∩ {1,2,3,4,5}

= {3}

(iv)A ∩ B = {2,3,4,5,6} ∩ {7,8,9} = $ϕ$

(v)A ∩ B = {1,2,3} ∩ $ϕ$ = $ϕ$

Q6. 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)

A.6. (i)A ∩ B = {3,5,7,9,11} ∩ {7,9,11,13}

= {7, 9, 11}

(ii)B ∩ C = {7,9,11,13} ∩ {11,13,15}

= {11,13}

(iii)A ∩ C ∩ D = (A ∩ C) ∩ D

= [{3,5,7,9,11} ∩ {11,13,15}] ∩ {15,17}.

= {11} ∩ {15,17} = $ϕ$ .

(iv)A ∩ C = {3,5,7,9,11} ∩ {11,13,15}.

= {11}

(v)B ∩ D = {7,9,11,13} ∩ {15,17}= $ϕ$

(vi)A ∩ (B∪ C) = {3,5,7,9,11} ∩ [{7,9,11,13}∪ {11,13,15}]

= {3,5,7,9,11} ∩ {7,9,11,13,15}.

= {7,9,11}

(vii)A ∩ D = {3,5,7,9,11} ∩ {15,17} = $ϕ$ .

(viii)A ∩ (B ∪ D) = {3,5,7,9,11} ∩ [{7,9,11,13} ∪ {15,17}]

= {3,5,7,9,11} ∩ {7,9,11,13,15,17}

= {7,9,11}

(ix)(A ∩ B) ∩ (B ∪ C) = [{3,5,7,9,11} ∩{7,9,11,13}] ∩ [{7,9,11, 13} ∪ {11,13, 15}]

= {7,9,11} ∩ {7,9,11,13,15}

= {7,9,11}.

(x)(A ∪ D) ∩ (B ∪ C) = [{3,5,7,9,11} ∪ {15,17}] ∩ [{7, 9, 11, 13} ∪ {11, 13, 15}

={3,5,7,9,11,15,17} ∩ {7,9,11,13,15}

={7,9,11,15}.

Q7. 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

A $\cap$ B (ii) A $\cap$ C (iii) A $\cap$ D

(iv) B $\cap$ C (v) B $\cap$ D (vi) C $\cap$ D

A.7. A = {1,2,3,4,5, 6, …}

B = {2,4,6, …}

C = {1,3,5, …}

D = {2,3,5, …,}

(i)A ∩ B = {1,2,3,4 …} ∩ {2,4,6, …} = {2,4,6 …} = B.

(ii)A ∩ C = {1,2,3,4, …} ∩ {1,3,5 …} = {1,3,5, …} = C.

(iii)B ∩ C = {2,4,6, …} ∩ {1,3,5, …} = $ϕ$ .

(iv)B ∩ D = {2,4,6 …} ∩ {2,3,5 …} = {2}

(v)C ∩ D = {1,3,5, …} ∩ {2,3,5 …} = {3,5,7 …} = {x : x is odd prime number}

Q.8. Which of the following pairs of sets are disjoint

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

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

(iii) {x : x is an even integer} and {x : x is an odd integer}

A.8. (i){1,2,3,4} ∩ {x : x is a natural number and 4 ≤ x ≤ 6}

{1, 2, 3, 4} ∩ {4, 5, 6}

{4} ≠∅

Hence, the given pair of set is not disjoint.

(ii){a, e, i, o, u} ∩ {c, d, e, f}

{e} ≠∅

Hence, the given pair of set is not disjoint.

(iii){x: x is an even integer} ∩ {x: x is are odd integer}.

=∅

As there is no integer which is both even and odd at the same time.

⸫ Given pair of set are disjoint.

Q9. 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

A.9. (i)A – B = {3,6,9,12,15,18,21} – {4,8,12,16,20}

= {3,6,9,15,18,21}

(ii)A – C = {3,6,9,12,15,18,21} – {2,4,6,8,10,12,14,16}

= {3,9,15,18,21}

(iii)A – D = {3,6,9,12,15,18,21} – {5,10,15,20}

= {3,6,9,12,18,21}

(iv)B – A = {4,8,12,16,20} – {3,6,9,12,15,18,21}

= {4,8,16,20}

(v)C – A = {2,4,6,8,10,12,14,16} – {3,6,9,12,15,18,21}

= {2,4,8,10,14,16}

(vi)D – A = {5,10,15,20} – {3,6,9,12,15,18,21}

={5,10,20}

(vii)B – C={4,8,12,16,20} – {2,4,6,8,10,12,14,16}

={20}

(viii)B – D ={4,8,12,16,20} – {5,10,15,20}

= {4,8,12,16}

(ix)C – B = {2,4,6,8,10,12,14,16} – {4,8,12,16,20}

= {2,6,10,14}

(x)D – B = {5,10,15,20} – {4,8,12,16,20}

= {5,10,15}

(xi)C – D = {2,4,6,8,10,12,14,16} – {5,10,15,20}

= {2,4,6,8,12,14,16}

(xii)D – C = {5,10,15,20} – {2,4,6,8,10,12,14,16}

= {5,15,20}

Q10. If X= { a, b, c, d } and Y = { f, b, d, g}, find

(i)X – Y (ii)Y – X (iii) X ∩Y

A.10. (i)X – Y = {a, b, c, d} – (f, b, d, g}

= {a, c}

(ii) Y – X = {f, b, d, g} – {a, b, c, d}

= {f, g}

(iii)X ∩ Y = {a, b, c, d} ∩ {f, b, d, g}

= {b, d}.

Q11. If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?

A.11. R – Q = {x: x is a real number but not rational number}

= {x: x is an irrational number}

Since real number = rational number + irrational number

Q12. State whether each of the following statement is true or false. Justify your answer.

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

(ii) {a, e, i, o, u} and {a, b, c, d}are disjoint sets.

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

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

A.12. (i)False, as {2,3,4,5} ∩ {3,6} = {3} ≠ $ϕ$ .Hence sets are not disjoint.

(ii)False as {a, e, i, o, u} ∩ {a, b, c, d} = {a} ≠ $ϕ$ Hence sets are not disjoint.

(iii)True as {2,6,10,14} ∩ {3,7,11,15} = $ϕ$ . Hence sets are disjoint.

(iv)True as {2,6,10} ∩ {3,7,11}= $ϕ$ . Hence sets are disjoint.

Class 11 Chapter 1 Sets Exercise 1.5 Solutions

Class 11 Math Chapter 1 Sets Exercise 1.5 deals with the concept of Cartesian products of sets. The Class 11 Math Exercise 1.5 contains problems based on the idea of ordered pairs and how they form new sets through Cartesian products. Through Cartesian product and ordered pairs exercise 1.5 of class 11 Maths. Class 11 Math Exercise 1.5 will consist of 7 Questions and students will find the Exercise 1.5 solution PDF here also. Students can check the complete solution of Exercise 1.5 Below

Exercise 1.5 Class 11 Maths Solutions

Q1. Let U = { 1, 2, 3, 4, 5, 6, 7, 8, 9 }, A = { 1, 2, 3, 4}, B = { 2, 4, 6, 8 } andC = { 3, 4, 5, 6 }.

Find (i) A $'$ (ii)B $'$ (iii)(A $\cup$ C) $'$ (iv)(A $\cup$ B) $'$

(v) (A $'$ ) $'$ (vi) (B – C) $'$

A.1. (i)A $'$ = U – A = {1,2,3,4,5,6,7,8,9} – {1,2,3,4}

= {5,6,7,8,9}

(ii)B $'$ = U – B = {1,2,3,4,5,6,7,8,9} – {2,4,6,8}

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

(iii)(A $\cup$ C) $'$ = A $'$ ∩ C $'$

= {5,6,7,8,9} ∩ [U – C] [⸪ (i)]

= {5,6,7,8,9} ∩ [{1,2,3,4,5,6,1,8,9} – {3,4,5,6}]

= {5,6,7,8,9} ∩ {1,2,7,8,9}

= {7,8,9}

(iv)(A $\cup$ B) $'$ = A $'$ ∩ B $'$ [By demorgan’s law]

= {5,6,7,8,9} ∩ {1,3,5,7,9} [⸪ (i) and (ii)]

= {5,7,9}.

(v)(A $'$ ) $'$ = U – A $'$ = {1,2,3,4,5,6,7,8,9} – {5,6,7,8,9} [⸪ (1)]

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

(A $'$ ) $'$ = A.

(vi)(B – C) $'$ = U – (B – C) = {1,2,3,4,5,6,7,8,9} – [{2, 4, 6, 8} – {3, 4, 5, 6}]

= {1,2,3,4,5,6,7,8,9} – {2,8}

= {1,3,4, 5, 6,7,9}.

Q2. 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}

A.2. (i)A $'$ = U – A = {a, b, c, d, e, f, g, h} – {a, b, c}

= {d, e, f, g, h}

(ii)B $'$ = U – B = {a, b, c, d, e, f, g, h} – {d, e, f, g}

= {a, b, c, h}.

(iii)C $'$ = U – C = {a, b, c, d, e, f, g, h} – {a, c, e, g}.

= {b, d, f, h}

(iv)D $'$ = U – D = {a, b, c, d, e, f, g, h} – {f, g, h, a}

= {b, c, d, e}

Q3. 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}

(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 }

A.3. (i){x : x is an odd natural number}

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

(iii){x : x is not a multiple of 3}

(iv){x : x is a positive composite number and x = 1}

(v){x : x is a natural number not divisible by 3 and 5}.

(vi){x : x is not a perfect square}

(vii){x : x is not a perfect cube}

(viii)We have, x + 5 = 8.

x = 8 – 5 = 3

x = 3

⸫ {x : x ≠ 3, x ϵ N}

(ix)We have,

2x + 5 = 9

2x = 9 – 5

2x = 4

x = 2

⸫{x : x ϵ N and x ≠ 2}

(x){x : x < 7} = {1,2,3,4,5,6}

(xi)We have,

2x + 1 >10

2x >10 – 1

x > $\frac{9}{2}$

⸫ ${x : x \in x < \frac{9}{2}}$ ={1,2,3,4}

Q4. 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 $'$ ∩ B $'$ (ii) A ∩ B) $'$ = A $'$ $\cup$ B $'$

A.4. (i)L.H.S = (A $\cup$ B) $'$ = U – (A $\cup$ B)

= {1,2,3,4,5,6,7,8,9} – [{2,4,6,8) $\cup$ {2,3,5,7}]

= {1,2,3,4,5,6,7,8,9} – {2,3,4,5,6,7,8}

= {1,9}

R.H.S. = A $'$ ∩ B $'$ = [U – A] ∩ [U $\cup$ B]

= $[{1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 4, 6, 8}]$ ∩ $[{1, 2, 3, 4, 5, 6, 7, 8, 9} - {2, 3, 5, 7}]$

= {1,3,5,7,9} ∩ {1,4,6,8,9}

= {1,9}

⸫ L.H.S. = R.H.S.

(A B) = A ∩ B.

(ii)L.H.S. = (A ∩ B) $'$ = U – (A ∩ B)

= {1,2,3,4,5,6,7,8,9} – [{2,4,6,8} ∩ {2,3,5,7}]

= {1,2,3,4,5,6,7,8,9} – {2}

= {1,3,4,5,6,7,8,9}

R.H.S. = A $'$ $\cup$ B $'$

= [U – A] $\cup$ [U – B]

= [{1,2,3,4,5,6,7,8,9} – {2,4,6,8}] [{1,2,3,4,5,6,7,8,9} – {2,3,5,7}]

= {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.

(A ∩ B) $'$ = A $'$ $\cup$ B $'$ .

Q5. Draw appropriate Venn diagram for each of the following :

(i) (A $\cup$ B) $'$ , (ii) A $'$ $\cup$ B $'$ , (iii) (A $\cup$ B) $'$ , (iv) A $'$ $\cup$ B $'$

A.5. (i)(A $\cup$ B) $'$ = U – (A $\cup$ B)

(ii)A $'$ ∩ B $'$ = (AU B) $'$ = U – (AU B)

(iii)(A ∩ B) $'$ = U – (A ∩ B)

(iv)A $'$ U B $'$ = (A ∩ B) $'$ = U – (A ∩ B) $'$ .

Q.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°, , what is A $'$ ?

A.6. A $'$ = U – A

= Set of all triangle in a plane - Set of all triangle with at least the angle different from 60°.

= Set of all triangle with each angle 60°.

A´ = set of all equilateral triangle.

Q7. Fill in the blanks to make each of the following a true statement :

(i) A $\cup$ A $'$ = . . . (ii) $ϕ' \cap$ A = . . .

(iii) A $\cap$ A $'$ = . . . (iv)U $'$ $\cap$ A = . . .

A.7. (i)A $\cup$ A $'$ = U

(ii) $ϕ$ ∩ A = U ∩ A = A.

(iii)A ∩ A $'$ = $ϕ$ .

(iv)U $'$ ∩ A = $ϕ$ ∩ A = $ϕ$ .

Class 11 Chapter 1 Sets Exercise 1.6 Solutions

Class 11 Math Exercise 1.6 focuses on the concept of relations through the extension of ordered pairs and cartesian products. In Exercise 1.6 Students will learn how to represent relations using roster form, set-builder form, and Cartesian products. Class 11 Sets 1.6 exercise also covers types of relations, such as reflexive, symmetric, and transitive relations. Exercise 1.6 Solutions PDF is also available on this page for all 8 Questions. Students can check the complete solution of Exercise 1.6 Below

Class 11 Sets Exercise 1.6 Solutions

Q1. 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).

A.1. Given, n (X)= 17

n (Y) = 23

n(X $\cup$ Y)= 38.

So, n(X $\cup$ Y) = n(X) + n(Y) n(X $\cap$ Y)

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

= 17 + 23 38.

= 2

Q2. 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?

A.2. Given, n(X $\cup$ Y) = 18.

n(X) = 8.n(Y) = 15.n(X $\cap$ Y) = ?

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

= 8 + 15 18.

= 5.

Q3. In a group of 400 people, 250 can speak Hindi and 200 can speak English. How many people can speak both Hindi and English?

A.3. Let H and E be set if people who speak Hindi and English respectively. there,

n(H) = 250, people speak Hindi n(E) = 200, people speak English

n(H $\cup$ E) = 400, people speaks either Hindi or English

So, n(H $\cap$ E) = n(H) + n(E) n(H $\cap$ E)

= 250 + 200 400

= 50

So, 50 people can speak both Hindi and English.

Q4. If S and T are two sets such that S has 21 elements, T has 32 elements, and S ∩T has 11 elements, how many elements does S $\cup$ T have?

A.4. Given,n(S) = 21

n(T) = 32

n(S $\cap$ T) = 11.

Using, n(S $\cup$ T) = n(S) + n(T) n(S $\cap$ T)

= 21 + 32 11

= 42.

Q5. 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?

A5. Given, n(X) = 40

n(Y $\cup$ Y) = 60.

n(X $\cap$ Y) = 10.

n(Y) = ?

Using, n(X $\cup$ Y) = n(X) + n(Y) n(X $\cap$ Y)

60 = 40 + n(y) 10.

n(Y) = 60 40 + 10

n(Y) = 30.

Q6. 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?

A.6. Let A and B but the set of people who likes coffee and tea.

Then, n(A) = 37. no. of people who like coffee

n(B) = 52, no. of people who like tea.

As each person likes at least one of the two drink,

n(A $\cup$ B) = 70.

So using, n(A $\cup$ B) = n(A) + n(B) n (A $\cap$ B)

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

= 37 + 52 70

= 89 70

= 19

So, 19 people likes both coffee and tea.

Q7. 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?

A.7. Let C and T be the set of people who likes cricket and tennis respectively. Then,

n(C) = 40, people who likes cricket

n(C $\cup$ T) = 65, people who likes either cricket or tennis

and n(C $\cap$ T) = 10, people who likes both

So, n(C $\cup$ T) = n(C) + n(T) n(C $\cap$ T)

65 = 40 + n(T) 10

n(T) = 35

So, 35 people likes tennis

And number of people who likes tennis only and not cricket

(TC) = n(T) n(C $\cap$ T) = 35 10 = 25.

Q8. 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?

A.8. Let A and B be set of people who speaks Frenchand Spanish respectively. Then,

n(A) = 50, people speak French

n(B) = 20, people speak Spanish.

n(A $\cap$ B) = 10, speaks both French& Spanish.

So, number of people who speaks at least one of these two languages

= n(A $\cup$ B)

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

= 50 + 20 10

= 60

Class 11 Chapter 1 Sets Miscellaneous Exercise NCERT Solutions

Class 11 Chapter 1 Sets Miscellaneous Exercise serves as a comprehensive review of all key concepts related to sets and their operations. This section includes a variety of problems that test students' understanding of set representation, types of sets, subsets, power sets, and universal sets. Additionally, it involves questions on set operations such as union, intersection, difference, and complement, along with applications of De Morgan’s Laws and Venn diagrams in problem-solving. this exercise includes 16 questions and their solutions are given below;

Miscellaneous Exercise Chapter 1 Solutions

Q1. Decide, among the following sets, which sets are subsets of one and another:

A = { x: x $\in$ R and x satisfy x²– 8x + 12 = 0 },

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

A.1. A = {x: xR and x satisfy x² - 8x + 12 = 0}

So, x² - 8x + 12 = 0

x² - 6x 2x + 12 = 0

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

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

x = 6, 2.

So, A = {2, 6)B = {2, 4, 6}C = {2, 4, 6, 8, ….}

D = {6}

D⊂ A ⊂B ⊂C

i e,D ⊂A, D⊂ B,D⊂ C, A⊂ B, A ⊂C and B⊂ C.

Q2. 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.

(i) If x $\in$ A and A $\in$ B, then x $\in$ B

(ii) If A $\subset$ B and B $\in$ C, then A $\in$ C

(iii) If A $\subset$ B and B $\subset$ C , then A $\subset$ C

(iv) If A $⊄$ B and B $⊄$ C, then A $⊄$ C

(v) If x $\in$ A and A $⊄$ B , then x $\in$ B

(vi) If A $\subset$ B and x $\in$ B , then x $\notin$ A

A.2. (i) False Let A = {a}, a $\in$ A then B = {{a}, b} i e, $a \inB.$

(ii) False. Let A. = {a}, if A $\subset$ B, B = {a, b} and B $\in$ C i e, C = {{a, b}, c} i e, A = {a}. $\notin C .$

(iii) True. Let x $\in$ A, if A $\subset$ B then x $\in$ B and if B $\subset$ C, x $\in$ C i e, elements of A are also elements of C.

A $\subset$ C.

(iv) False. Let A = {a} and B = {b} then A $⊄$ $B$ . Let C = {a, c} then B $⊄$ $\subset$ but a $\in$ A and a c, i.e., A $\subset$ C.

(v) False. Let A = {a} and B = {b} so, A $⊄$ $B$ i.e., A $\notin$ $B$ .

(vi) True. Let A $\subset$ B such that y $\in$ B i.e., y A But x $\in$ $B$ and suppose x $\in$ A.

Then by above definition,

A $\subset$ B i.e., x $\in$ B and x $\in$ A which is not the case

Q3. Let A, B, and C be the sets such that A $\cup$ B = A $\cup$ C and A $\cap$ B = A $\cap$ C. Showthat B = C.

A.3. Let x $\in$ b

As B $\subset$ A $\cup$ B we can write

Let x $\in$ A $\cup$ B.

as A $\cup$ B = A $\cup$ C.

x $\in$ A $\cup$ C.

i.e., x $\in$ A or x $\in$ C

when x $\in$ A, and x $\in$ B,

x $\in$ A $\cap$ B

But A $\cap$ B = A $\cap$ C

So, x $\in$ A $\cap$ C

i.e., x $\in$ A and x $\in$ C

x $\in$ C

when, x $\in$ C

as x $\in$ B and x $\in$ C

So, B $\subset$ C

Similarly, C $\subset$ B

So, B = C

Q4. Show that the following four conditions are equivalent :

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

A.4. (i) Let A- B≠ $ϕ$ , our assumption.

i.e., x $\in$ A But x≠ B where x is an element.

But as A B, the above condition of assumption is wrong if A $\subset$ B then A - B = $ϕ$

(ii) Let x $\in$ A.

As A - B = $ϕ$ we can say that x $\in$ B because if $x \in B,$ A - B ≠ $ϕ$

if A - B = $ϕ$ then A $\subset$ B.

(iii) We know that,

B $\subset$ A $\cup$ B always true

Let x $\in$ A $\cup$ B i.e., x $\in$ A or x $\in$ B.

As A $\subset$ B,

If x $\in$ A then x $\in$ B, all elements of A are among the elements of B

So, (A $\cup$ B) = B

(iv) We know that,

(A $\cap$ B) $\subset$ A as A $\subset$ B.

Let x $\in$ A then x $\in$ B

So, x $\in$ (A $\cap$ B)

i.e., A $\subset$ (A $\cap$ B)

So, A = (A $\cap$ B)

Hence, A $\subset$ B

A - B = $ϕ$ .

A $\cup$ B = B

A $\cap$ B = A.

i.e., the 4 conditions are equivalent.

Q5. Show that if A $\subset$ B, then C – B $\subset$ C – A.

A.5. Given, A $\subset$ B.

Let x $\in$ C - B then x $\in$ C but X $\notin$ $B$ .

However, A $\subset$ B, elements of B should have elements of A

i.e., X $\notin$ $B$ $x \notin A$

So, x $\in$ C A i.e., x $\in$ C but X $\notin$ $A$

C - B $\subset$ C- A

Q6. Assume that P (A) = P (B). Show that A = B.

A.6. Given, P(A) = P(B) where P is power set

Let x $\in$ A.

Then, {x} $\subset$ P(A) $\subset$ P(B)

i.e., x $\in$ B

A $\subset$ B

Similarly, B $\subset$ A

A = B

Q7. Is it true that for any sets A and B, P (A) $\cup$ P(B) = P (A $\cup$ B) ? Justify your answer.

A.7. Let A = {a}, B = {b}.

So, P(A) = ${ϕ, {a}} B (A) = {ϕ, (b}}$

So, P(A) $\cup$ P {B} = ${ϕ, {a}, {b}}$ ______(1)

Now, A $\cup$ B = {a, b}.

P(A $\cup$ B) = ${ϕ, {a}, {b}, {a, b}}$ ____(2)

So. From (1) and (2) we see that,

P(A) $\cup$ P(B) ≠P(A $\cup$ B)

Q8. Show that for any sets A and B,

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

A.8. Here,

(A $\cap$ B) $\cup$ (A - B) = (A $\cap$ B) $\cup$ (A $\cap$ B’) as (A - B) = A $\cap$ B’

= A $\cap$ (B $\cup$ B’)[ by converse of distributive law]

= A $\cap$ U[ B $\cup$ B’ = U, sample space set or universal set]

= A

And (A $\cup$ (B - A)) = A $\cup$ (B $\cap$ A’)[as B - A = B $\cap$ A’]

= (A $\cup$ B) $\cap$ (A $\cup$ A’)

= (A $\cup$ B) $\cap$ U[ A $\cup$ A’ = U, universal set]

= A $\cup$ B.

Q9. Using properties of sets, show that

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

A.9. (i) We know that,

A $\subset$ A

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

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

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

and also

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

So, A $\cup$ (A $\cap$ B) = A.

(ii) A $\cap$ (A $\cup$ B) = (A $\cap$ A) $\cup$ (A $\cap$ B)[By distributive law]

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

= A as (A $\cap$ B) $\subset$ A

Q10. Show that A $\cap$ B = A $\cap$ C need not imply B = C.

A.10. Let A = {a}, B = {a, b}, C = {a, c}

So, A $\cap$ B = {a} $\cap$ {a, b} = {a}

A $\cap$ C = {a} $\cap$ {a,c} = {a}

i.e., A $\cap$ B = A $\cap$ C = {a}

But B ≠C. as b $\in$ B but $b \neq C$ vice-versa

Q11. Let A and B be sets. If A $\cap$ X = B $\cap$ X = $ϕ$ and A $\cup$ X = B $\cup$ X for some set X, show that A = B.

(Hints A = A $\cap$ (A $\cup$ X) , B = B $\cap$ (B $\cup$ X ) and use Distributive law )

A.11. Let A, B and x be sets such that,

A $\cap$ x = B $\cap$ x = $ϕ$ and A $\cup$ x = B $\cup$ x.

We know that,

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

= A $\cap$ (B $\cup$ x)[∵ A $\cup$ x = $\cup$ Bx]

= (A $\cap$ B) $\cap$ (A $\cap$ x)[by distributive law]

= (A $\cap$ B) ∪ ⌽[∵A∩x = ⌽]

=> A = A∩ B [∵A ∪ ⌽ = A]

And B = $\cap$ (B $\cup$ x)

= B $\cap$ (A $\cup$ x)[∵ B $\cup$ x = A $\cup$ x]

= (B $\cap$ A) $\cup$ (B $\cap$ x)[By distributive law]

= (B $\cap$ A) ∪ $ϕ$ [∵ B $\cap$ x = ⌽]

B = B $\cap$ A[∵ A $\cup$ $ϕ$ = A]

So, A = B = A $\cap$ B.

Q12. 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 = $ϕ$

A.12. Let A = {x, y}

B = {y, z}

C = {x, z}

So, A $\cap$ B = {x, y} $\cap$ {y, z} = {y}≠ $ϕ$

B $\cap$ C = {y, z} $\cap$ {x, z} = {z}≠ $ϕ$

A $\cap$ C = {x, y} $\cap$ {x, z} = {x}≠ $ϕ$

But A $\cap$ B $\cap$ C = (A $\cap$ B) $\cap$ C

= {y} $\cap$ (x, z}

= $ϕ$

Q13. In a survey of 600 students in a school, 150 students were found to be taking teaand 225 taking coffee, 100 were taking both tea and coffee. Find how manystudents were taking neither tea nor coffee?

A.13. Let T and C be sets of students taking tea and coffee.

Then,n(T) = 150, number of students taking tea

n(C) = 225, number of students taking coffee

n(T $\cap$ C) = 100, number of students taking both tea and coffee.

So, Number of students taking either tea or coffee is.

n(T $\cup$ C) = n(T) + n(C) n(T $\cap$ C)

= 150 + 225 100

= 275

Number of students taking neither tea coffee

= Total number of students No of students taking either tea or coffee

= 600 275

= 325.

Q14. In a group of students, 100 students know Hindi, 50 know English and 25 knowboth. Each of the students knows either Hindi or English. How many studentsare there in the group?

A.14. Let H and E be set of students who known Hindi and English respectively.

Then, number of students who know Hindi = n(H) = 100

Number of students who know English = n(E) = 50

Number of students who know both English & Hindi = 25 = n(H $\cap$ E)

As each of students knows either Hindi or English,

Total number of students in the group,

n(H $\cup$ E) = n(H) + n(E) - n(H $\cap$ E)

= 100 + 25 25

= 125,

Q15. In a survey of 60 people, it was found that 25 people read newspaper H, 26 readnewspaper 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.

A.15. Let H,T and I be of people who reads newspaper H,T and I respectively.

Then,

number of people who reads newspaper H, n(H) = 25.

number of people who T, n(T) = 26.

number of people who I , n(I) = 26

number of people who both H and T, n(H $\cap$ I) = 9

number of people who both H and T, n(H $\cap$ T) = 11

number of people who both T and I, n(T $\cap$ I) = 8

number of people who reads all newspaper, n(H $\cap$ T $\cap$ I) = 3.

Total no. of people surveyed = 60

The given sets can be represented by venn diagram

(i) The number of people who reads at least one of the newspaper.

in (H∪T $\cup$ I) = n(H) + n(T) + n(I) n(H $\cap$ T) n(H $\cap$ I) n(T $\cap$ I) + n(H $\cap$ T $\cap$ I)

= 25 + 26 + 26 11 9 8 + 3

= 80 28

= 52

(ii) From venn diagram,

a = number of people who reads newspapers H and T only.

b = number of people who reads newspapers H and I only.

c = number of people who reads newspapers T and I only.

d = number of people who reads all newspaper.

So, n(H $\cap$ T) = a + d.

n(H $\cap$ I) = b + d

n(T $\cap$ I) = c + d

So, a + d + c + d + b + d = n(H $\cap$ T) + n(H $\cap$ I) + n(T $\cap$ I)

a + d + c + b + 2d = 11 + 9 + 8

a + b + c +d = 28 - 2d

= 28 2 3 [d = n(H $\cap$ T $\cap$ I) = 3]

= 28 - 6

= 22

Number of people who reads exactly one newspaper

= Total no. of people - No. of people who reads more than one newspaper

= 52 - (a + b +c + d)

= 52 - 22

= 30

Q16. In a survey it was found that 21 people liked product A, 26 liked product B and29 liked product C. If 14 people liked products A and B, 12 people liked productsC and A, 14 people liked products B and C and 8 liked all the three products.Find how many liked product C only.

A.16. Let A, B and C be the set of people who like product A, B and C respectively.

Then,

Number of people who like product A, n(A) = 21

Number of people who like product B, n(B) = 26

Number of people who like product C, n(C) = 29.

Number of people likes both product A and B, n(A $\cap$ B) = 14

Number of people likes both product A and C, n(A $\cap$ C) = 12

Number of people likes both product B and C, n(B $\cap$ C) = 14.

No. of people who likes all product, n(A $\cap$ B $\cap$ C) = 8

a→n(A $\cap$ B)

b→n(A $\cap$ C)

d→n(B $\cap$ C)

c→n(A $\cap$ B $\cap$ C)

From the above venn diagram we can see that,

Number of people who likes product C only

= n(C) - b - d + c

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

= 29 - 12 - 14 + 8

= 11

Mathematics Prep Tips and Recommended Books for Boards and Competitive Exams

We bring a list of the most recommended books for competitive exam math preparation. Shiksha experts understand that choosing the right study materials and a proper preparation plan is important for success in both board exams and competitive exams like JEE, COMEDK, and more. Below are some useful preparation tips and book recommendations for Maths prep.

Important Maths Reference Books For Preparation

RD Sharma Mathematics for Class 12 ( ( Vol.I and II)
Objective Mathematics by R.D. Sharma
Cengage Mathematics Book Series by G. Tewani
Problems in Calculus of One Variable by I.A. Maron
Trigonometry and Co-ordinate Geometry by SL Loney

Preparation Tips for Class 12 Mathematics

Regular Practice: Mathematics requires regular practice. Students must focus on solving problems from the NCERT books and other reference materials consistently to enhance their problem-solving skills and speed.
Focus on Important Chapters: Students should prioritize important chapters like Calculus (Differentiation, Integration), Vectors and 3D Geometry, and Linear Programming are crucial for both board exams and competitive exams.
Time Management: Practice solving problems within a set time limit to improve your speed and accuracy. This is especially important for competitive exams like JEE or NEET.
Revision: Students should focus on revision. Regular revision of formulas, theorems, and important concepts is very useful for students. Students should make concise notes for last-minute revision..
Solve PYPS and Mock Test: Practice solving past 5–10 years' board exam question papers. This will help you understand the exam pattern and frequently asked questions.
Work on Weak Areas: Identify your weak areas and devote extra time to improving them. Seek help from teachers, classmates, or online resources if needed.