Chapter 1 : Sets

 Class 11 Maths Chapter 1 . Sets : Exercise 1.1 , Exercise 1.2 , Exercise 1.3 , Exercise 1.4 , Exercise 1.5 and Miscellaneous Exercise Solutions :

Class 11 Maths Chapter 1 . Sets Exercise 1.1

1. 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.

Solution: (i) The collection of all the months of a year beginning with the letter J are January , June and July . Therefore, the collection is a well-defined . So, it is a sets .
(ii) The collection of ten most talented writers of India. Therefore, the collection is not a well-defined . So, it is not a sets .
(iii) A team of eleven best-cricket batsmen of the world. Therefore, the team is not a well-defined . So, it is not a sets .
(iv) The collection of all boys in your class. Therefore, the collection is a well-defined . So, it is a sets .
(v) The collection of all natural numbers less than 100. Therefore, the collection is a well-defined . So, it is a sets .
(vi) A collection of novels written by the writer Munshi Prem Chand. Therefore, the collection is a well-defined . So, it is a sets .
(vii) The collection of all even integers. Therefore, the collection is a well-defined . So, it is a sets .

(viii) The collection of questions in this Chapter. Therefore, the collection is a well-defined . So, it is a sets .
(ix) A collection of most dangerous animals of the world. Therefore, the collection is not a well-defined . So, it is not a sets .

2. Let A = {1, 2, 3, 4, 5, 6}. Insert the appropriate symbol ∈ or ∉ in the blank spaces:
(i) 5. . .A (ii) 8 . . . A (iii) 0. . .A
(iv) 4. . . A (v) 2. . .A (vi) 10. . .A

Solution: (i) 5 ∈ A

(ii) 8 ∉  A

(iii) 0 ∉ A
(iv) 4 ∈ A   

(v) 2 ∈ A

(vi) 10 ∉ A 

3. Write the following sets in roster form:
(i)   A = {  is an integer and }
(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 .

Solution: (i)   A = {  is an integer and }

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

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

(iv) D = {  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 .

4. 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}

Solution: (i) (3, 6, 9, 12}    

 is a natural number and

(ii)  {2,4,8,16,32}       

 is a natural number and

(iii) {5, 25, 125, 625}

 is a natural number and
(iv) {2, 4, 6, . . .}

 is an even number

(v) {1,4,9, . . .,100 }

 is a natural number and

5. List all the elements of the following sets :
(i) A = {x : x is an odd natural number}
(ii) B = {  is an integer,
(iii) C = {  is an integer,  }
(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 }.

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

(ii) B = {  is an integer,

 
(iii) C = {  is an integer,  }

(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}

 February , April , June , September ,November  
(vi) F = { x : x is a consonant in the English alphabet which precedes k }.

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}.

Solution:

(i) {1, 2, 3, 6}	(c) {x : x is natural number and divisor of 6}
(ii) {2, 3}	(a) {x : x is a prime number and a divisor of 6}
(iii) {M,A,T,H,E,I,C,S}	(d) {x : x is a letter of the word MATHEMATICS}
(iv) {1, 3, 5, 7,9}	(b) {x : x is an odd natural number less than 10}

 Class 11 Maths Chapter 1 . Sets Exercise 1.2

1. 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}

Solution: (i) Set of odd natural numbers divisible by 2 .

Since, the odd natural numbers are does not divisible by 2 . So, it is a null set .
(ii) Set of even prime numbers .

Since, the even prime number is only 2 . So, it is not a null set .
(iii) { x : x is a natural numbers, x < 5 and x > 7 }

There are no natural numbers between x < 5 and x > 7 . So, it is a null set .
(iv) { y : y is a point common to any two parallel lines}

There are no common point in the two parallel lines. So, it is a null set .
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 .

Solution: (i) The set of months of a year .

A = { January , February , March , April , May , June , July , August , September , October , November , December}

So, the sets are finite .
(ii) {1, 2, 3, . . .}

A = {1 , 2 , 3 , ……….}

So, the sets are infinite .
(iii) {1, 2, 3, . . .99, 100}

A = {1 , 2 , 3 , ……….99 , 100}

So, the sets are finite .
(iv) The set of positive integers greater than 100
A ={101 , 102 , 103 , 104 , 105 , 106 , 107 , ……………..}

So, the sets are infinite .

(v) The set of prime numbers less than 99

A = {2, 3 , 5 , 7 , 11 , 13 , …….. 97}

So, the sets are finite .

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)

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

There are infinite many lines drawn parallel to x-axis . So, the sets is infinite .
(ii) The set of letters in the English alphabet .

There are 26 letters in the English alphabet . So, the sets is finite .
(iii) The set of numbers which are multiple of 5 .

A = { 5 , 10 , 15 , 20 , 25 , ………..} . So, the sets is infinite .

(iv) The set of animals living on the earth .
(v) The set of circles passing through the origin (0,0) .

4. 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, . . . }

Solution:   (i) A = B

The order of elements in a set does not matter, so A and B are equal sets.

(ii) A ≠ B

The order of elements in a set matters. Since the order of elements in A and B is different, A is not equal to B.

(iii) A = B

A is the set of positive even integers less than or equal to 10. B is also defined as the set of positive even integers less than or equal to 10. Therefore, A is equal to B.

(iv) A ≠ B

A is the set of multiples of 10, while B is the set of all positive integers that are multiples of 5. These sets are not equal.

5. Are the following pair of sets equal ? Give reasons.
(i) A = {2, 3}, B = {x : x is solution of  }
(ii) A = { x : x is a letter in the word FOLLOW}
B = { y : y is a letter in the word WOLF}

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

Now , 

 

Therefore,

Since  and

Therefore,  A is not equal to B.

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

B = {  is a letter in the word WOLF}

The sets A and B represent the letters in the words FOLLOW and WOLF, respectively.

A = {F, O, L, O, W}  ;   B = {W, O, L, F}

 The order of elements does not matter in sets, and each element is counted only once.

Therefore, A is equal to B.

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}

Solution : Given, 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}

B and D are equal sets .

Since they have the same elements, {1, 2, 3, 4}.

E and G are equal sets

Since they have the same elements, {–1, 1}.

 Class 11 Maths Chapter 1 . Sets Exercise 1.3 :

1. Make correct statements by filling in the symbols ⊂ 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}

Solution : (i) {2, 3, 4} ⊂ {1, 2, 3, 4, 5}

The set {2, 3, 4} is a subset of {1, 2, 3, 4, 5}, as every element in {2, 3, 4} is also in {1, 2, 3, 4, 5}.

(ii) {a, b, c} ⊄ {b, c, d}

The set {a, b, c} is not a subset of {b, c, d}, as 'a' is not an element in {b, c, d}.

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

 The set of students in Class XI is a subset of all students in the school.

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

 The set of all circles in the plane is not a subset of circles with a specific radius (1 unit).

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

The set of all triangles is not a subset of all rectangles in the plane.

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

The set of equilateral triangles is a subset of all triangles in the same plane.

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

The set of even natural numbers is a subset of all integers .

2. Examine whether the following statements are true or false:
(i) { a, b } ⊄ { b, c, a }
(ii) { a, e } ⊂ { x : x is a vowel in the English alphabet}
(iii) { 1, 2, 3 } ⊂ { 1, 3, 5 }
(iv) { a } ⊂ { a, b, c }
(v) { a } ∈ { a, b, c }
(vi) { x : x is an even natural number less than 6} ⊂ { x : x is a natural number which divides 36}

Solution: (i) { a, b } ⊄ { b, c, a }

This statement is false.

The order of elements in a set does not matter, and { a, b } is equal to { b, a }.

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

This statement is true.

  { a, e } is a subset of the set of vowels in the English alphabet, as it only contains the vowels 'a' and 'e.'

(iii) { 1, 2, 3 } ⊂ { 1, 3, 5 }

This statement is false.

 { 1, 2, 3 } is not a subset of { 1, 3, 5 } because it contains the element 2, which is not present in the second set.

(iv) { a } ⊂ { a, b, c }

 This statement is true.

 { a } is a subset of { a, b, c } because it contains only the element 'a,' which is present in the larger set.

(v) { a } ∈ { a, b, c }

This statement is false.

The notation ∈ is used for an element belonging to a set, not for a subset relationship. { a } is an element of { a, b, c }, not a subset.

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

This statement is true.

The set of even natural numbers less than 6 is { 2, 4 }, and these numbers are also natural numbers that divide 36 (2 divides 36 and 4 divides 36).

3. Let A = { 1, 2, { 3, 4 }, 5 }. Which of the following statements are incorrect and why?
(i) {3, 4} ⊂ A     (ii) {3, 4} ∈ A    (iii) {{3, 4}} ⊂ A     (iv) 1 ∈ A    (v) 1 ⊂ A    (vi) {1, 2, 5} ⊂ A     (vii) {1, 2, 5} ∈ A     (viii) {1, 2, 3} ⊂ A     (ix) φ ∈ A    (x) φ ⊂ A     (xi) {φ} ⊂ A

Solution:  (i) {3, 4} ⊂ A

This statement is incorrect.

{3, 4} is not a subset of A because the set A contains the element {3, 4} as a distinct element, not as a separate set.

(ii) {3, 4} ∈ A

This statement is correct .

 {3, 4} is an element of A because A includes the set {3, 4}.

(iii) {{3, 4}} ⊂ A

This statement is correct.

{{3, 4}} is a subset of A because A contains the set {3, 4}.

(iv) 1 ∈ A

This statement is correct .

 The element 1 is in set A.

(v) 1 ⊂ A This statement is incorrect.

The element 1 is not a set, so it cannot be a subset of A.

(vi) {1, 2, 5} ⊂ A This statement is correct.

{1, 2, 5} is a subset of A because all its elements are present in A.

(vii) {1, 2, 5} ∈ A

This statement is incorrect.

 {1, 2, 5} is not an element of A; it is a subset of A.

(viii) {1, 2, 3} ⊂ A

This statement is incorrect.

 {1, 2, 3} is not a subset of A because A contains the set {3, 4}, not the element 3.

(ix) φ ∈ A

This statement is incorrect.

 The empty set φ is an element of A.

(x) φ ⊂ A

This statement is correct.

 The empty set φ is a subset of every set, including A.

(xi) {φ} ⊂ A

This statement is incorrect.

The set containing the empty set {φ} is a subset of A because A contains the set {3, 4}, which is not empty

4. Write down all the subsets of the following sets
(i) {a}     (ii) {a, b}    (iii) {1, 2, 3}     (iv) φ

Solution: (i) {a} Subsets: {φ, {a}}

(ii) {a, b} Subsets: {φ, {a}, {b}, {a, b}}

(iii) {1, 2, 3} Subsets: {φ, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, {1, 2, 3}}

(iv) φ (empty set) Subsets: {φ}

5. Write the following as intervals :
(i) {x : x ∈ R, – 4 < x ≤ 6}

(ii) {x : x ∈ R, – 12 < x < –10}
(iii) {x : x ∈ R, 0 ≤ x < 7}

(iv) {x : x ∈ R, 3 ≤ x ≤ 4}

Solution:  (i) {x : x ∈ R, – 4 < x ≤ 6}

Interval notation: (− 4,6]

(ii) {x : x ∈ R,  – 12 < x < – 10 }

Interval notation: (−12,−10)

(iii) {x : x ∈ R, 0 ≤ x < 7}

Interval notation: [0,7)

(iv) {x : x ∈ R, 3 ≤ x ≤ 4}

Interval notation: [3,4]

6. Write the following intervals in set-builder form :
(i) (– 3, 0)      (ii) [6 , 12]    (iii) (6, 12]      (iv) [–23, 5)

Solution:  (i) (−3,0)

Set-builder form: { x : x ∈ R ,−3 < x < 0}

(ii) [6,12]

Set-builder form: { x : x ∈ R, 6 ≤ x ≤ 12}

(iii) (6,12]

Set-builder form: { x : x ∈ R , 6 < x ≤ 12}

(iv) [−23,5)

Set-builder form: { x : x ∈ R,− 23 ≤ x < 5}

7. What universal set(s) would you propose for each of the following : (i) The set of right triangles. (ii) The set of isosceles triangles.

Solution: (i) The set of right triangles:

Universal set for all triangles: {All triangles in a plane}

Universal set for all polygons: {All polygons in a plane}

(ii) The set of isosceles triangles:

Universal set for all triangles: {All triangles in a plane}

Universal set for all polygons: {All polygons in a plane}

8. 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}

Solution:  Universal sets contains all the elements of A, B, and C.

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

This set does not include 8, which is an element of set C.

 It is not a universal set.

(ii) φ (empty set)

The empty set does not contain the elements of A, B, and C.

It is not a universal set.

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

This set includes all the elements of A, B, and C.

It is a valid universal set.

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

This set does not include 0, which is an element of set C.

 It is not a universal set.

 Class 11 Maths Chapter 1 . Sets Exercise 1.4  :

1. 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 ≤ 6 } , B = {x : x is a natural number and 6 < x < 10 }
(v) A = {1, 2, 3}, B = φ

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

(ii) A={a,e,i,o,u}  , B={a,b,c}

A∪B={a,b,c,e,i,o,u}

(iii) A = {  is a natural number and multiple of 3} = {3,6,9,12,15,…….}

B = {  is natural number less than 6} = {1,2,3,4,5}

(iv) A = {x:x is a natural number and 1 < x ≤ 6} = {2,3,4,5,6}  

B = {x : x is a natural number and 6 < x < 10} = {7,8,9}

(v) A = {1,2,3} , B = φ (empty set)

 

2. Let A = { a, b }, B = {a, b, c}. Is A ⊂ B ? What is A ∪ B ?

Solution:  Yes, A is a subset of B, because every element of A is also an element of B.

A ∪ B = {a , b , c}
3. If A and B are two sets such that A ⊂ B, then what is A ∪ B ?

Solution :  Since  A is a subset of B .

This is because the union of two sets includes all the elements that are in at least one of the sets.

Since A is entirely contained within B , and hence A∪B = B .

4. 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 ∪ B (ii) A ∪ C (iii) B ∪ C (iv) B ∪ D
(v) A ∪ B ∪ C (vi) A ∪ B ∪ D (vii) B ∪ C ∪ D

Solution: Given, A = {1, 2, 3, 4} , B = {3, 4, 5, 6} , C = {5, 6, 7, 8 } and D = { 7, 8, 9, 10 }

(i) A∪B :

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

(ii) A∪C :

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

(iii) B∪C :

B∪C = {3,4,5,6}∪ {5,6,7,8}={3,4,5,6,7,8} 

(iv) B∪D :

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

(v) A∪B∪C :

A∪B∪C = {1, 2, 3, 4} ∪{3, 4, 5, 6} ∪{5, 6, 7, 8 } ={1,2,3,4,5,6,7,8}  

(vi) A∪B∪D :

A∪B∪D= {1, 2, 3, 4} ∪{3,4,5,6}∪{7,8,9,10} = {1,2,3,4,5,6,7,8,9,10} 

(vii) B∪C∪D :

B∪C∪D = {3,4,5,6}∪{5,6,7,8}∪{7,8,9,10} = {3,4,5,6,7,8,9,10}

5. Find the intersection of each pair 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 ≤ 6 } , B = {x : x is a natural number and 6 < x < 10 }
(v) A = {1, 2, 3}, B = φ
Solution:  (i) X = {1,3,5}  , Y = {1,2,3}

 X∩Y= {1,3,5}∩{1,2,3} = {1,3}

(ii) A = {a,e,i,o,u} , B = {a,b,c} 

A∩B = {a} 

(iii) A = { x : x is a natural number and multiple of 3} = {3 , 6 , 9 , 12 , ……}

  B = { x : x is a natural number less than 6) = {1,2,3,4,5}

A ∩B = {3}

(iv) A ={x : x is a natural number and 1<x≤6),

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

 A∩B= {2,3,4,5,6}   

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

 B = φ (empty set)

 A∩B= φ  (empty set, as there are no common elements)

6. If A = { 3, 5, 7, 9, 11 }, B = {7, 9, 11, 13}, C = {11, 13, 15}and D = {15, 17}; find
(i) A ∩ B (ii) B ∩ C (iii) A ∩ C ∩ D
(iv) A ∩ C (v) B ∩ D (vi) A ∩ (B ∪ C)
(vii) A ∩ D (viii) A ∩ (B ∪ D) (ix) ( A ∩ B ) ∩ ( B ∪ C )
(x) ( A ∪ D) ∩ ( B ∪ C)

Solution: (i) A∩B = {7,9,11} 

(ii) B∩C={11,13} 

(iii) A∩C∩D ={11}

(iv) A∩C= {11} 

(v) B∩D= φ (empty set, as there are no common elements)

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

(vii) A∩D=φ  (empty set, as there are no common elements)

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

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

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

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 ∩ B (ii) A ∩ C (iii) A ∩ D
(iv) B ∩ C (v) B ∩ D (vi) C ∩ D

Solution:

A = {x : x is a natural number } = {1,2,3,4,5,………}

 B = {x : x is an even natural number} = {2,4,6,8,……….}
C = {x : x is an odd natural number} = {1,3,5,7,………}

and D = {x : x is a prime number } = {2,3,5,7,11,13, ……….}

(i) A∩B :  

A∩B={1,2,3,4,5,………} ∩ {2,4,6,8,……….} = {2,4,6,8,10,…}  

(ii) A∩C :  

A∩C= {1,2,3,4,5,………} ∩ {1,3,5,7,……….} ={1,3,5,7,9,…}  

(iii) A∩D :

A∩D={1,2,3,4,5,………} ∩ {2,3,5,11,13,……….} ={2,3,5,7,11,13,17,…} 

(iv) B∩C:

 B∩C={2,4,6,8………} ∩ {1,3,5,7,……….}  = φ

(v) B∩D :

 B∩D={2,4,6,8,………} ∩ {2,3,5,7,11,13……….} ={2} 

(vi) C∩D :

 C∩D={1,3,5,7,………} ∩ {2,3,5,7,11,13……….}  ={3,5,7,11,13,17,…}

8. Which of the following pairs of sets are disjoint
(i) {1, 2, 3, 4} and {x : x is a natural number and 4 ≤ x ≤ 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}

Solution: (i) {1, 2, 3, 4} and {x: x is a natural number and 4 ≤ x ≤ 6}

These sets have the element 4 in common, so they are not disjoint.

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

These sets have the element 'e' in common, so they are not disjoint.

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

Every integer is either even or odd, but no integer is both even and odd.

 Therefore, these sets are disjoint.

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

Solution: Given sets:

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}

(i) A−B: Elements that are in A but not in B.

 A−B={3,6,9,15,18,21}

(ii) A−C: Elements that are in A but not in C.

 A−C={3,9,15,18,21}

(iii) A−D: Elements that are in A but not in D.

A−D={3,6,9,12,18,21}

(iv) B−A: Elements that are in B but not in A.

B−A={4,8,16,20}

(v) C−A: Elements that are in C but not in A.

 C−A={2,4,8,10,14,16}

(vi) D−A: Elements that are in D but not in A.

D−A={5,10,20}

(vii) B−C: Elements that are in B but not in C.

B−C={20}

(viii) B−D: Elements that are in B but not in D.

B−D={4,8,12,16}

(ix) C−B: Elements that are in C but not in B.

 C−B={2,6,10,14}

(x) D−B: Elements that are in D but not in B.

D−B={5,10,15}

(xi) C−D: Elements that are in C but not in D.

C−D={2,4,6,8,10,14,16}

(xii) D−C: Elements that are in D but not in C.

D−C={5,15,20}

10. If X= { a, b, c, d } and Y = { f, b, d, g}, find (i) X – Y (ii) Y – X (iii) X ∩ Y

Solution: Given sets:  X= { a, b, c, d } and Y = { f, b, d, g}

(i) X – Y : Elements that are in X but not in Y.

X – Y = {a , c}

(ii) Y –X : Elements that are in Y but not in X.

Y – X = {f , g}

(iii) X ∩ Y : Intersection of X and Y, i.e., elements that are common to both X and Y.

X∩Y ={b ,d }

11. If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q ?
Solution:  The set R – Q  represents the set of real numbers that are not rational, i.e.,  it consists of all irrational numbers.

The set of irrational numbers includes numbers that cannot be expressed as fractions of integers.

So, R – Q represents the set of irrational numbers.
12. 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 .

Solution :   (i) False: Disjoint sets are sets that have no elements in common.

 In this case, both sets {2, 3, 4, 5} and {3, 6} share the element 3, so they are not disjoint.

(ii) False: The sets {a, e, i, o, u} and {a, b, c, d} share the element 'a', so they are not disjoint.

(iii) True: The sets {2, 6, 10, 14} and {3, 7, 11, 15} have no elements in common, so they are disjoint.

(iv) True: The sets {2, 6, 10} and {3, 7, 11} have no elements in common, so they are disjoint.

 Class 11 Maths Chapter 1 . Sets Exercise 1.5  :

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 ∪ C)′ (iv) (A ∪ B)′ (v) (A′)′ (vi) (B – C)′
Solution: 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′ (complement of A):

A′=U−A={5,6,7,8,9}

(ii) B′ (complement of B):

B′=U−B={1,3,5,7,9}

(iii) (A∪C)′ (complement of the union of A and C):

A∪C={1,2,3,4,5,6}

 (A∪C)′=U−(A∪C)={7,8,9}

(iv) (A∪B)′ (complement of the union of A and B):

 A∪B={1,2,3,4,6,8}

 (A∪B)′=U−(A∪B)={5,7,9}

(v) (A′)′ (complement of the complement of A):

 (A′)′=U−A′={1,2,3,4}

(vi) (B−C)′ (complement of the set difference of B and C):

 B – C = {2,4}

(B−C)′ = U − (B – C ) = {1,3,5,6,7,8,9}

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}
Solution: Given, the sets :  U = { a, b, c, d, e, f, g, h}
(i)  Here, A = {a, b, c}

A’= U – A = {d,e,f,g,h}

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

B’ = U – B = {a , b , c , h}
(iii) Here, 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}

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}
(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 ≥ 7 } (xi) { x : x ∈ N and 2x + 1 > 10 }
Solution:  Here, U = {1,2,3,4,5,6,…………..}

 (i) Let A = {x:x is an even natural number } = {2,4,6,8,10,…} 

A’ = U – A = {1,3,7,9,………}

(ii) Let B = {x:x is an odd natural number} = {1,3,5,7,9,…} 

B’ = U – B = {2,4,6,8,…….}

(iii) Let C ={x:x is a positive multiple of 3} = {3,6,9,12,…} 

C’ = U – C = {1,2,4,5,7,10,11,13,……….}

(iv) Let D = {x:x is a prime number} = {2,3,5,7,11,…} 

D’ = U – D = {1,4,6,8,10,13,……….}

(v) Let E = {x:y is a natural number divisible by 3 and 5}  = {15,30,45,60,…} 

E’ = U – E = {1,2,3,…….,16,17,…….29,31,…..}

(vi) Let F = {x:x is a perfect square} = {1,4,9,16,25,…}

F’ = U – F = {2,3,5,6,7,8,10,…….}

(vii) Let G = {x:y is a perfect cube} = {1,8,27,64,…}

G’ = U – G = {2,3,4,5,6,7,9,……..}

(viii) Let H = {x : x+5 = 8} = {3}

H’ = U – H = {1,2,4,5,6,……….}

(ix) Let I = { }={2}

We have 

 

 

I’ = U – I = {1,3,4,5,6,……….}

(x) Let J = {  } = {7,8,9,10,…}

J’ = U – J = {1,2,3,4,5,6}

(xi) Let K = {  and  } = {5,6,7,8,9,…}

We have,

 

 

K’ = U – K = {1,2,3,4}

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 ∪ B)′ = A′ ∩ B′           (ii) (A ∩ B)′ = A′ ∪ B′
Solution: Given, the sets U = {1, 2, 3, 4, 5, 6, 7, 8, 9 }

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

(i)  Here, A∪B={2,3,4,5,6,7,8} ,  A′= U – A = {1,3,5,7,9} , B′= U – B = {1,4,6,8,9}  and A ∩ B = {2}

LHS :   (A∪B)′=U−(A∪B)

  (A∪B)′=U−{2,3,4,5,6,7,8}={1,9}  

RHS : A′∩B′={1,9} 

(A∪B)′=A′∩B′   Verified .

(ii)  LHS :  (A∩B)′=U−{2}

={1,3,4,5,6,7,8,9} 

RHS :  A′∪B′={1,3,5,7,9}∪{1,4,6,8,9}

 ={1,3,4,5,6,7,8,9}  

LHS = RHS Verified .

5. Draw appropriate Venn diagram for each of the following :
(i) (A ∪ B)′, (ii) A′ ∩ B′, (iii) (A ∩ B)′, (iv) A′ ∪ B′
Solution: (i)  (A ∪ B)´ = U – (AUB) = none shaded area .    

(ii) A′ ∩ B′ = (A ∪ B)´ [De Morgan’s law ]

(iii)  (A ∩ B )′ = A′ ∪ B′ = U – (A∩B) = none shaded area

(iv) A′ ∪ B′ = (A ∩ B )′   [De Morgan’s law ]

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′  ?

Solution:  Let U = the universal set of all triangles in a plane

 and  A = be the set of all triangles with at least one angle different from 60°.

The complement of A (A′) would be the set of triangles that have all angles equal to 60°.

So, A′ is the set of all triangles in U that are equilateral (having all angles equal to 60°).Top of Form

7. Fill in the blanks to make each of the following a true statement :
(i) A ∪ A′ = . . . (ii) φ′ ∩ A = . . .
(iii) A ∩ A′ = . . . (iv) U′ ∩ A = . . .

Solution:  (i) A∪A′=U (The union of a set and its complement is the universal set.)

(ii) φ′∩A= A = U (The complement of the empty set is the universal set.)

(iii) A∩A′= φ (The intersection of a set and its complement is the empty set.)

(iv) U′∩A = φ (The complement of the universal set intersected with any set is the empty set.)

 Class 11 Maths Chapter 1 . Sets Miscellaneous Exercise  :

1. Decide, among the following sets, which sets are subsets of one and another:
A = { x : x ∈ R and x satisfy  } , B = { 2, 4, 6 }, C = { 2, 4, 6, 8, . . . }, D = { 6 }.

Solution: Given the sets :  A = { x : x ∈ R and x satisfy  } ,

We have,

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

B = {2, 4, 6} is a subset of A because all elements of B are also in A.

C = {2, 4, 6, 8, ...} is a superset of A because A is a subset of C. A contains only 2 and 6, which are also present in C.

D = {6} is a subset of both A and C because the only element in D, which is 6, is present in both A and C.

2. 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 ∈ A and A ∈ B , then x ∈ B
(ii) If A ⊂ B and B ∈ C , then A ∈ C
(iii) If A ⊂ B and B ⊂ C , then A ⊂ C
(iv) If A ⊄ B and B ⊄ C , then A ⊄ C
(v) If x ∈ A and A ⊄ B , then x ∈ B
(vi) If A ⊂ B and x ∉ B , then x ∉ A

Solution:

(i) If x ∈ A and A ∈ B, then x ∈ B :

False :  Example : Let A = {1} , B = {{1},2} 

1 ∈A and A∈B, but 1∉ B .

So, x ∈ A and A ∈ B need not imply that x ∈ B .

(ii) If A⊂B and B∈C, then A∈C:

False: Example: Let A={1}, B={1,2}, and C={{1,2},3}.

 So, A⊂B and B∈C, but A∈C.

(iii) If A⊂B and B⊂C, then A⊂C:

True: If every element of A is also in B and every element of B is also in C, then every element of A is in C.

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

False: Example: Let A={1,2}, B={2,3}, and C={3,4}.

So, A ⊄ B and B ⊄ C , but A ⊄ C .

(v) If x ∈ A and A ⊄ B , then x ∈ B :

False: Example: Let A={1,2} and B={2,3,4}.

Here,  1 ∈ A and A ⊄ B , but  1 ∉ B .  

So, x ∈ A and A ⊄ B need not imply that x ∈ B . But x∉ B and also x ∉ A .

(vi) If A ⊂ B and x ∈ B, then x ∈ A :

True: If A ⊂ B, then every element of A is also in B. But x ∉ B  and also x ∉ A .

3. Let A, B, and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. Show that B = C.

Solution:  The given A∪B=A∪C  and A∩B=A∩C

Let x∈B . Then x∈A∪B.

Let x∈A∪C .  Then  x∈A or x∈C  .

Since x∈A∩B , we also have x∈A∩C .

Therefore, x∈C  .

Now, let y∈C . Then y∈A∪C  .

Since A∪B=A∪C , we have y∈A∪B .

This implies that y∈A or y∈B.

 Since y∈A∩C , we also have y∈A∩B .

Therefore, y∈B .

Now, we have shown that if x ∈ B , then x∈C , and if y ∈ C , then y∈B .

 This implies B =C  .

Therefore, if A∪B=A∪C and A∩B=A∩C, then B = C .     Proved .

4. Show that the following four conditions are equivalent :
(i) A ⊂ B    (ii) A – B = φ   (iii) A ∪ B = B    (iv) A ∩ B = A

Solution:

(i)⇒(ii) (Assume A⊂ B and prove A –B = φ ):

Let x be an arbitrary element in A−B.

i.e.,  x is in A but not in B.

Since A⊂ B, x cannot be in A and not in B simultaneously.

Therefore, A−B = φ .

(ii)⇒(iii) (Assume A−B= φ and prove A∪B=B):

If A−B= φ ,

i.e.,   every element in A is also in B.

Therefore, A∪B=B, as adding elements from A to B does not change the set B.

(iii)⇒(iv) (Assume A∪B=B and prove A∩B=A):

If A∪B=B,

i.e.,  every element in A or B is in B.

This implies that the intersection of A and B is A, as any element common to A and B must be in A.

(iv)⇒(i) (Assume A∩B=A and prove A⊂ B):

Let x be an arbitrary element in A.

Since A∩B=A, x is also in B (since it is in the intersection).

 Therefore, every element in A is also in B, and A⊂B.

5. Show that if A ⊂ B, then C – B ⊂ C – A.

Solution: Let x be an arbitrary element in C−B.

We have, x is in C but not in B .

Again,  we want to show that x is also in C – A .,

i.e.,  x is not in A .

Since A⊂ B , if x is not in B , it must not be in A as well.

 Therefore, x  is in C – A ,

C – B ⊂ C – A  Proved .

6. Show that for any sets A and B, A = ( A ∩ B ) ∪ ( A – B ) and A ∪ ( B – A ) = ( A ∪ B )

Solution:   (i)  Let x be an arbitrary element in A.

Since x is in A, it is either in A∩B or in A−B (or both).

 Therefore, x is in (A∩B)∪(A−B).

 Let y be an arbitrary element in (A∩B)∪(A−B).

 This means y is either in A∩B or in A−B.

In either case, y is in A.

Therefore , A=(A∩B)∪(A−B).

(ii)  Let x be an arbitrary element in A∪(B−A).

 This means x is either in A or in (B−A).

 If x is in A, it is certainly in A∪B.

 If x is in (B−A), it means x is in B but not in A, so x is in A∪B.

Let y be an arbitrary element in A∪B.

i.e., y is either in A or in B.

 If y is in A, it is certainly in A∪(B−A). If y is in B, it is also in A∪(B−A).

Therefore, A∪(B−A)=A∪B.

7. Using properties of sets, show that  :    (i) A ∪ ( A ∩ B ) = A       (ii) A ∩ ( A ∪ B ) = A.

Solution:  (i) A ∪ ( A ∩ B ) = A

LHS : A ∪ ( A ∩ B ) = (A ∪ A) ∩ (A∪B) = A∩ (A∪B) = A     RHS

(ii) A ∩ ( A ∪ B ) = A.

LHS : A ∩ ( A ∪ B ) = (A ∩ A) ∪ (A∩B) = A∪ (A∩B) = A    RHS 

8. Show that A ∩ B = A ∩ C need not imply B = C.

Solution:  Let A = {1,2}  , B = {1,2,3} , C = {2,3}

A∩B = {1,2} 

A∩C = {2}   

It is clear that A∩B = A∩C , but B ≠ C since B contains the element 3 which is not in C .

Therefore,  A∩B = A∩C does not imply B = C .

9. Let A and B be sets. If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B. (Hints A = A ∩ ( A ∪ X ) , B = B ∩ ( B ∪ X ) and use Distributive law )

Solution:  Given sets A and B, and a set X such that A∩X=B∩X= φ and A∪X=B∪X .

Using the distributive law , we have

A =A∩(A∪X)  and  B=B∩(B∪X) 

A=A∩(A∪X)=(A∩A)∪(A∩X)=A∩X= φ 

Similarly,

B=B∩(B∪X)=(B∩B)∪(B∩X)=B∩X= φ 

Since A = φ and B = φ

 Therefore,  A = B  Proved .

10. Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = φ.

Solution:   Let: A ={1,2}  , B ={2,3}  , C ={1,3}

A∩B={2}  ,  B∩C = {3} , A∩C={1} 

And, A∩B∩C= φ

So,  A∩B, B∩C, and A∩C are non-empty sets, but A∩B∩C is empty .