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NCERT Solutions Class 11 Maths Chapter 1 - Sets

EXERCISE - 1.1

Q1: Which of the following are sets? Justify our answer.
(i) The collection of all 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 Premchand.
(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.
Ans:
(i) The collection of all months of a year beginning with the letter J is a well-defined collection of objects because one can definitely identify a month that belongs to this collection. Hence, this collection is a set.

(ii) The collection of ten most talented writers of India is not a well-defined collection because the criteria for determining a writer’s talent may vary from person to person.

Hence, this collection is not a set.

(iii) A team of eleven best cricket batsmen of the world is not a well-defined collection because the criteria for determining a batsman’s talent may vary from person to person.

Hence, this collection is not a set.

(iv) The collection of all boys in your class is a well-defined collection because you can definitely identify a boy who belongs to this collection.

Hence, this collection is a set.

(v) The collection of all natural numbers less than 100 is a well-defined collection because one can definitely identify a number that belongs to this collection.

Hence, this collection is a set.

(vi) A collection of novels written by the writer Munshi Prem Chand is a well-defined collection because one can definitely identify a book that belongs to this collection.

Hence, this collection is a set.

(vii) The collection of all even integers is a well-defined collection because one can definitely identify an even integer that belongs to this collection.

Hence, this collection is a set.

(viii) The collection of questions in this chapter is a well-defined collection because one can definitely identify a question that belongs to this chapter.

Hence, this collection is a set.

(ix) The collection of most dangerous animals of the world is not a well-defined collection because the criteria for determining the dangerousness of an animal can vary from person to person.
Hence, this collection is not a set.

Q2: 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

Ans:
(i) 5 ∈ A
(ii) 8 ∉ A
(iii) 0 ∉ A
(iv) 4 ∈ A
(v) 2 ∈ A
(vi) 10 ∉ A

Q3: Write the following sets in roster form:
(i) A = {: x is an integer and –3 < < 7}.
(ii) B = {: x is a natural number less than 6}.
(iii) C = {: x is a two-digit natural number such that the sum of its digits is 8}
(iv) D = {: 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.
Ans:
(i) A = {: x is an integer and –3 < x < 7}
The elements of this set are –2, –1, 0, 1, 2, 3, 4, 5, and 6 only.
Therefore, the given set can be written in roster form as
A = {–2, –1, 0, 1, 2, 3, 4, 5, 6}

(ii) B = {: x is a natural number less than 6}
The elements of this set are 1, 2, 3, 4, and 5 only.
Therefore, the given set can be written in roster form as
B = {1, 2, 3, 4, 5}

(iii) C = {: x is a two-digit natural number such that the sum of its digits is 8}
The elements of this set are 17, 26, 35, 44, 53, 62, 71, and 80 only.
Therefore, this set can be written in roster form as
C = {17, 26, 35, 44, 53, 62, 71, 80}

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

NCERT Solutions Class 11 Maths Chapter 1 - Sets
∴ 60 = 2 × 2 × 3 × 5
The elements of this set are 2, 3, and 5 only.
Therefore, this set can be written in roster form as D = {2, 3, 5}.

(v) E = The set of all letters in the word TRIGONOMETRY
There are 12 letters in the word TRIGONOMETRY, out of which letters T, R, and O are repeated.
Therefore, this set can be written in roster form as
E = {T, R, I, G, O, N, M, E, Y}

(vi) F = The set of all letters in the word BETTER
There are 6 letters in the word BETTER, out of which letters E and T are repeated.
Therefore, this set can be written in roster form as
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}
Ans:
(i) {3, 6, 9, 12} = {: x = 3n, n∈ N and 1 ≤ n ≤ 4}
(ii) {2, 4, 8, 16, 32}
It can be seen that 2 = 21, 4 = 22, 8 = 23, 16 = 24, and 32 = 25.
∴ {2, 4, 8, 16, 32} = {: x = 2n, n∈ N and 1 ≤ n ≤ 5}
(iii) {5, 25, 125, 625}
It can be seen that 5 = 51, 25 = 52, 125 = 53, and 625 = 54.
∴ {5, 25, 125, 625} = {: x = 5n, n∈N and 1 ≤ n ≤ 4}
(iv) {2, 4, 6 …}
It is a set of all even natural numbers.
∴ {2, 4, 6 …} = {: x is an even natural number}
(v) {1, 4, 9 … 100}
It can be seen that 1 = 12, 4 = 22, 9 = 32 …100 = 102.
∴ {1, 4, 9… 100} = {: x = n2, n∈N and 1 ≤ n ≤ 10}

Q5: List all the elements of the following sets:
(i) A = {: x is an odd natural number}
(ii) B = {x:x is an integer, - 1/2<x<9/2}
(iii) C = {: x is an integer, x2 ≤ 4}
(iv) D = {: x is a letter in the word “LOYAL”}
(v) E = {: x is a month of a year not having 31 days}
(vi) F = {: x is a consonant in the English alphabet which proceeds k}.
Ans:
(i) A = {: x is an odd natural number} = {1, 3, 5, 7, 9 …

(ii) B = {x:x is an integer, - 1/2<n<9/2}
It can be seen that -1/2 = -0.5 and 9/2 = 4.5
∴ B = {0, 1, 2, 3, 4}

(iii) C = {: x is an integer; x2 ≤ 4}
It can be seen that
(–1)2 = 1 ≤ 4; (–2)2 = 4 ≤ 4; (–3)2 = 9 > 4
02 = 0 ≤ 4
12 = 1 ≤ 4
22 = 4 ≤ 4
32 = 9 > 4
∴ C = {–2, –1, 0, 1, 2}

(iv) D = (: x is a letter in the word “LOYAL”) = {L, O, Y, A}

(v) E = {: x is a month of a year not having 31 days}
= {February, April, June, September, November}

(vi) F = {: x is a consonant in the English alphabet which precedes k}
= {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:
NCERT Solutions Class 11 Maths Chapter 1 - Sets

Ans:
(i) All the elements of this set are natural numbers as well as the divisors of 6. Therefore, (i) matches with (c).
(ii) It can be seen that 2 and 3 are prime numbers. They are also the divisors of 6.
Therefore, (ii) matches with (a).
(iii) All the elements of this set are letters of the word MATHEMATICS. Therefore, (iii) matches with (d).
(iv) All the elements of this set are odd natural numbers less than 10. Therefore,
(iv) matches with (b)

EXERCISE - 1.2

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) {: is a natural numbers, < 5 and > 7}
(iv) {: is a point common to any two parallel lines}
Ans:
(i) A set of odd natural numbers divisible by 2 is a null set because no odd number is divisible by 2.
(ii) A set of even prime numbers is not a null set because 2 is an even prime number.
(iii) {x : x is a natural number, x < 5 and x > 7} is a null set because a number cannot be simultaneously less than 5 and greater than 7.
(iv) {y : y is a point common to any two parallel lines} is a null set because parallel lines do not intersect. Hence, they have no common point.

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
Ans:
(i) The set of months of a year is a finite set because it has 12 elements.
(ii) {1, 2, 3 …} is an infinite set as it has infinite number of natural numbers.
(iii) {1, 2, 3 …99, 100} is a finite set because the numbers from 1 to 100 are finite in number.
(iv) The set of positive integers greater than 100 is an infinite set because positive integers greater than 100 are infinite in number.
(v) The set of prime numbers less than 99 is a finite set because prime numbers less than 99 are finite in number.

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)
Ans:
(i) The set of lines which are parallel to the x-axis is an infinite set because lines parallel to the x-axis are infinite in number.
(ii) The set of letters in the English alphabet is a finite set because it has 26 elements.
(iii) The set of numbers which are multiple of 5 is an infinite set because multiples of 5 are infinite in number.
(iv) The set of animals living on the earth is a finite set because the number of animals living on the earth is finite (although it is quite a big number).
(v) The set of circles passing through the origin (0, 0) is an infinite set because infinite number of circles can pass through the origin.

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 = {: is positive even integer and ≤ 10}
(iv) A = {: is a multiple of 10}; B = {10, 15, 20, 25, 30 ...}
Ans:
(i) A = {a, b, c, d}; B = {d, c, b, a}
The order in which the elements of a set are listed is not significant.
∴ A = B
(ii) A = {4, 8, 12, 16}; B = {8, 4, 16, 18}
It can be seen that 12 ∈ A but 12 ∉ B.
∴ A ≠ B
(iii) A = {2, 4, 6, 8, 10}
B = {: x is a positive even integer and x ≤ 10}
= {2, 4, 6, 8, 10}
∴ A = B
(iv) A = {: x is a multiple of 10}
B = {10, 15, 20, 25, 30 …}
It can be seen that 15 ∈ B but 15 ∉ A.
∴ A ≠ B

Q5: Are the following pair of sets equal? Give reasons.
(i) A = {2, 3}; B = {: is solution of x2 +5x +6 = 0}
(ii) A = {: is a letter in the word FOLLOW}; B = {: is a letter in the word WOLF}
Ans:
(i) A = {2, 3}; B = {: x is a solution of x2 + 5x + 6 = 0}
The equation x2 + 5x + 6 = 0 can be solved as:
x(x + 3) 2(x + 3) = 0
(x + 2)(x + 3) = 0
x = –2 or x = –3
∴ A = {2, 3}; B = {–2, –3}
∴ A ≠ B
(ii) A = {x : x is a letter in the word FOLLOW} = {F, O, L, W}
B = {y : y is a letter in the word WOLF} = {W, O, L, F}
The order in which the elements of a set are listed is not significant.
∴ A = B

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}
Ans: 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}; A = {0, 1}
It can be seen that
8 ∈ A, 8 ∉ B, 8 ∉ D, 8 ∉ E, 8 ∉ F, 8 ∉ G, 8 ∉ H
⇒ A ≠ B, A ≠ D, A ≠ E, A ≠ F, A ≠ G, A ≠ H
Also, 2 ∈ A, 2 ∉ C
∴ A ≠ C
3 ∈ B, 3 ∉ C, 3 ∉ E, 3 ∉ F, 3 ∉ G, 3 ∉ H
∴ B ≠ C, B ≠ E, B ≠ F, B ≠ G, B ≠ H
12 ∈ C, 12 ∉ D, 12 ∉ E, 12 ∉ F, 12 ∉ G, 12 ∉ H
∴ C ≠ D, C ≠ E, C ≠ F, C ≠ G, C ≠ H
4 ∈ D, 4 ∉ E, 4 ∉ F, 4 ∉ G, 4 ∉ H
∴ D ≠ E, D ≠ F, D ≠ G, D ≠ H
Similarly, E ≠ F, E ≠ G, E ≠ H
F ≠ G, F ≠ H, G ≠ H
The order in which the elements of a set are listed is not significant.
∴ B = D and E = G
Hence, among the given sets, B = D and E = G

EXERCISE - 1.3

Q1: 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}
Ans:
(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 is 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 in a triangle in the same plane}
(vii) {x : x is an even natural number} ⊂ {x : x is an integer}

Q2: 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}
Ans: (i) False. Each element of {a, b} is also an element of {b, c, a}.
(ii) True. a, e are two vowels of the English alphabet.
(iii) False. 2∈ {1, 2, 3}; however, 2∉ {1, 3, 5}
(iv) True. Each element of {a} is also an element of {a, b, c}.
(v) False. The elements of {a, b, c} are a, b, c. Therefore, {a} ⊂ {a, b, c}
(vi) True. {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}

Q3: 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
Ans: A = {1, 2, {3, 4}, 5}
(i) The statement {3, 4} ⊂ A is incorrect because 3 ∈ {3, 4}; however, 3∉A.
(ii) The statement {3, 4} ∈A is correct because {3, 4} is an element of A.
(iii) The statement {{3, 4}} ⊂ A is correct because {3, 4} ∈ {{3, 4}} and {3, 4} ∈ A.
(iv) The statement  1∈A  is correct because 1 is an element of A.
(v) The statement 1⊂ A is incorrect because an element of a set can never be a subset of itself.
(vi) The statement {1, 2, 5} ⊂ A is correct because each element of {1, 2, 5} is also an element of A.
(vii) The statement {1, 2, 5} ∈ A is incorrect because {1, 2, 5} is not an element of A.
(viii) The statement {1, 2, 3} ⊂ A is incorrect because 3 ∈ {1, 2, 3}; however, 3 ∉ A.
(ix) The statement Φ ∈ A is incorrect because Φ is not an element of A.
(x) The statement Φ ⊂ A is correct because Φ is a subset of every set.
(xi) The statement {Φ} ⊂ A is incorrect because Φ∈ {Φ}; however, Φ ∈ A.

Q4: Write down all the subsets of the following sets:
(i) {a}
(ii) {a, b}
(iii) {1, 2, 3}
(iv) Φ
Ans: (i) The subsets of {a} are Φ and {a}.
(ii) The subsets of {a, b} are Φ, {a}, {b}, and {a, b}.
(iii) The subsets of {1, 2, 3} are Φ, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3}, and{1, 2, 3}
(iv) The only subset of Φ is Φ.

Q5: How many elements has P(A), if A = Φ?
Ans: e know that if A is a set with m elements i.e., n(A) = m, then n[P(A)] = 2m.
If A = Φ, then n(A) = 0.
∴ n[P(A)] = 20 = 1
Hence, P(A) has one element.

Q6: 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}
Ans: (i) {x : x ∈ R, –4 < x ≤ 6} = (–4, 6]
(ii) {x : x ∈ R, –12 < x < –10} = (–12, –10)
(iii) {x : x ∈ R, 0 ≤ x < 7} = [0, 7)
(iv) {x : x ∈ R, 3 ≤ x ≤ 4} = [3, 4]

Q7: Write the following intervals in set-builder form:
(i) (–3, 0)
(ii) [6, 12]
(iii) (6, 12]
(iv) [–23, 5)
Ans: (i) (–3, 0) = {x : x ∈ R, –3 < x < 0}
(ii) [6, 12] = {x : x ∈ R, 6 ≤ x ≤ 12}
(iii) (6, 12] = {x : x ∈ R, 6 < x ≤ 12}
(iv) [–23, 5) = {x : x ∈ 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
Ans: (i) For the set of right triangles, the universal set can be the set of triangles or the set of polygons.
(ii) For the set of isosceles triangles, the universal set can be the set of triangles or the set of polygons or the set of two-dimensional figures.

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 universals 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}
Ans: (i) It can be seen that A ⊂ {0, 1, 2, 3, 4, 5, 6}
B ⊂ {0, 1, 2, 3, 4, 5, 6}
However, C ⊄ {0, 1, 2, 3, 4, 5, 6}
Therefore, the set {0, 1, 2, 3, 4, 5, 6} cannot be the universal set for the sets A, B, and C.

(ii) A ⊄ Φ, B ⊄ Φ, C ⊄ Φ
Therefore, Φ cannot be the universal set for the sets A, B, and C.

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

(iv) A ⊂ {1, 2, 3, 4, 5, 6, 7, 8}
B ⊂ {1, 2, 3, 4, 5, 6, 7, 8}
However, C ⊄ {1, 2, 3, 4, 5, 6, 7, 8}
Therefore, the set {1, 2, 3, 4, 5, 6, 7, 8} cannot be the universal set for the sets A, B, and C.

EXERCISE - 1.4

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 ≤ 6}
B = {x : x is a natural number and 6 < x < 10}
(v) A = {1, 2, 3}, B = Φ
Ans:
(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 = {x : x is a natural number and multiple of 3} = {3, 6, 9 …}
As B = {x : x is a natural number less than 6} = {1, 2, 3, 4, 5, 6}
A ∪ B = {1, 2, 4, 5, 3, 6, 9, 12 …}
∴ A ∪ B = {x : x = 1, 2, 4, 5 or a multiple of 3}
(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}
A∪ B = {2, 3, 4, 5, 6, 7, 8, 9}
∴ A∪ B = {x : x ∈ N and 1 < x < 10}
(v) A = {1, 2, 3}, B = Φ
A∪ B = {1, 2, 3}

Q2: Let A = {a, b}, B = {a, b, c}. Is A ⊂ B? What is A ∪ B?
Ans: Here, A = {a, b} and B = {a, b, c}
Yes, A ⊂ B.
A∪ B = {a, b, c} = B

Q3: If A and B are two sets such that A ⊂ B, then what is A ∪ B?
Ans: If A and B are two sets such that A ⊂ B, then 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 ∪ B  
(ii) A ∪ C
(iii) B ∪ C
(iv) B ∪ D
(v) A ∪ B ∪ C
(vi) A ∪ B ∪ D
(vii) B ∪ C ∪ D
Ans: A = {1, 2, 3, 4], B = {3, 4, 5, 6}, C = {5, 6, 7, 8} and D = {7, 8, 9, 10}
(i) A ∪ B = {1, 2, 3, 4, 5, 6}
(ii) A ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}
(iii) B ∪ C = {3, 4, 5, 6, 7, 8}
(iv) B ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}
(v) A ∪ B ∪ C = {1, 2, 3, 4, 5, 6, 7, 8}
(vi) A ∪ B ∪ D = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(vii) B ∪ C ∪ D = {3, 4, 5, 6, 7, 8, 9, 10}

Q5: 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
Ans: (i) X = {1, 3, 5}, Y = {1, 2, 3}
X ∩ Y = {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 …}
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} = {2, 3, 4, 5, 6}
B = {x : x is a natural number and 6 < x < 10} = {7, 8, 9}
A ∩ B = Φ
(v) A = {1, 2, 3}, B = Φ
A ∩ B = 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 ∩ 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)
Ans:
(i) A ∩ B = {7, 9, 11}
(ii) B ∩ C = {11, 13}
(iii) A ∩ C ∩ D = { A ∩ C} ∩ D = {11} ∩ {15, 17} = Φ
(iv) A ∩ C = {11}
(v) B ∩ D = Φ
(vi) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
= {7, 9, 11} ∪ {11} = {7, 9, 11}
(vii) A ∩ D = Φ
(viii) A ∩ (B ∪ D) = (A ∩ B) ∪ (A ∩ D)
= {7, 9, 11} ∪ Φ = {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}

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
(i) A ∩ B
(ii) A ∩ C
(iii) A ∩ D
(iv) B ∩ C
(v) B ∩ D
(vi) C ∩ D
Ans: 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, 9 …}
D = {x : x is a prime number} = {2, 3, 5, 7 …}
(i) A ∩B = {x : x is a even natural number} = B
(ii) A ∩ C = {x : x is an odd natural number} = C
(iii) A ∩ D = {x : x is a prime number} = D
(iv) B ∩ C = Φ
(v) B ∩ D = {2}
(vi) C ∩ D = {x : x is odd prime number}

Q8: 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}
Ans: (i) {1, 2, 3, 4}
{x : x is a natural number and 4 ≤ x ≤ 6} = {4, 5, 6}
Now, {1, 2, 3, 4} ∩ {4, 5, 6} = {4}
Therefore, this pair of sets is not disjoint.
(ii) {a, e, i, o, u} ∩ (c, d, e, f} = {e}
Therefore, {a, e, i, o, u} and (c, d, e, f} are not disjoint.
(iii) {x : x is an even integer} ∩ {x : x is an odd integer} = Φ
Therefore, this pair of sets is 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
Ans:
(i) A – B = {3, 6, 9, 15, 18, 21}
(ii) A – C = {3, 9, 15, 18, 21}
(iii) A – D = {3, 6, 9, 12, 18, 21}
(iv) B – A = {4, 8, 16, 20}
(v) C – A = {2, 4, 8, 10, 14, 16}
(vi) D – A = {5, 10, 20}
(vii) B – C = {20}
(viii) B – D = {4, 8, 12, 16}
(ix) C – B = {2, 6, 10, 14}
(x) D – B = {5, 10, 15}
(xi) C – D = {2, 4, 6, 8, 12, 14, 16}
(xii) D – C = {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
Ans:
(i) X – Y = {a, c}
(ii) Y – X = {f, g}
(iii) X ∩ Y = {b, d}

Q11: If R is the set of real numbers and Q is the set of rational numbers, then what is R – Q?
Ans: R: set of real numbers
Q: set of rational numbers
Therefore, R – Q is a set of irrational numbers.

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.
Ans: (i) False
As 3 ∈ {2, 3, 4, 5}, 3 ∈ {3, 6}
⇒ {2, 3, 4, 5} ∩ {3, 6} = {3}
(ii) False
As a ∈ {a, e, i, o, u}, a ∈ {a, b, c, d}
⇒ {a, e, i, o, u } ∩ {a, b, c, d} = {a}
(iii) True
As {2, 6, 10, 14} ∩ {3, 7, 11, 15} = Φ
(iv) True
As {2, 6, 10} ∩ {3, 7, 11} = Φ

EXERCISE - 1.5

Q1: 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)'
Ans:
U = {1, 2, 3, 4, 5, 6, 7, 8, 9}
A = {1, 2, 3, 4}
B = {2, 4, 6, 8}
C = {3, 4, 5, 6}
(i) A' = {5, 6, 7, 8 ,9}
(ii) B' = {1, 3, 5, 7, 9}
(iii) A ∪ C = {1, 2, 3, 4, 5, 6} ∴ (A ∪ C)' = {7, 8, 9}
(iv) A ∪ B = {1, 2, 3, 4, 6, 8}  (A ∪ B)' = {5, 7, 9}
(v) (A')' = A = {1, 2, 3, 4}
(vi) B - C = {2, 8}   ∴ (B - C)' = {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}
Ans: U = {a, b, c, d, e, f, g, h}
(i) A = {a, b, c} A' = {d, e, f, g, h}      
(ii) B = {d, e, f, g}  ∴ B' = {a, b, c, h}
(iii) C = {a, c, e, g}   ∴ C' = {b, d, f, h}  
(iv) D = {f, g, h, a}  ∴ D' = {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 perfect cube}
(viii) {x : x + 5 = 8}
(ix) {x : 2x + 5 = 9}
(x) {x : x ≥ 7}
(xi) {x : x ∈ N and 2x + 1 > 10}
Ans: U = N: Set of natural numbers
(i) {x : x is an even natural number}´ = {x : x is an odd natural number}
(ii) {x : x is an odd natural number}´ = {x : x is an even natural number}
(iii) {x : x is a positive multiple of 3}´ = {x : x ∈ N and x is not a multiple of 3}
(iv) {x : x is a prime number}´ = {x : x is a positive composite number and x = 1}
(v) {x : x is a natural number divisible by 3 and 5}´ = {x : x is a natural number that is not divisible by 3 or 5}
(vi) {x : x is a perfect square}´ = {x : x ∈ N and x is not a perfect square}
(vii) {x : x is a perfect cube}´ = {x : x ∈ N and x is not a perfect cube}
(viii) {x : x + 5 = 8}´ = {x : x ∈ N and x ≠ 3}
(ix) {x : 2x + 5 = 9}´ = {x : x ∈ N and x ≠ 2}
(x) {x : x ≥ 7}´ = {x : x ∈ N and x < 7}
(xi) {x : x ∈ N and 2x + 1 > 10}´ = {x : x ∈ N and x ≤ 9/2}

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 U B)' = A' ∩ B'
(ii) (A ∩ B)' = A' U B'

Ans: It is given that
U = {1, 2, 3, 4, 5,6,7,8, 9}
A = {2, 4, 6, 8}
B = {2, 3, 5, 7}
(i) (A U B)' = {2, 3, 4, 5, 6, 7, 8}' = {1, 9}
A' ∩ B' = {1, 3, 5, 7, 9} ∩ {1, 4, 6, 8, 9} = {1, 9}
Therefore, (A U B)' = A' ∩ B'.
(ii) (A ∩ B)' = {2}' = {1, 3, 4, 5, 6, 7, 8, 9}
A' U B' = {1, 3, 5, 7, 9} U {1, 4, 6, 8, 9} = {1, 3, 4, 5, 6, 7, 8, 9}
Therefore, (A ∩ B)' = A' U B'.

Q5: Draw appropriate Venn diagrams for each of the following:
(i) (A ∪ B)'
(ii) A' ∩ B'
(iii) (A ∩ B)'
(iv) A' ∪ B'
Ans:

(i) In the diagrams, shaded portion represents (A ∪ B)'

NCERT Solutions Class 11 Maths Chapter 1 - Sets

(ii) In the diagrams, shaded portion represents A' ∩ B'

NCERT Solutions Class 11 Maths Chapter 1 - Sets

(iii) In the diagrams, shaded portion represents (A ∩ B)'

NCERT Solutions Class 11 Maths Chapter 1 - Sets

(iv) In the diagrams, shaded portion represents  A' ∪ B'

NCERT Solutions Class 11 Maths Chapter 1 - Sets

Q6: 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 60º what is A’?
Ans: Given: U = {x : x is a triangle}
A = {x : x is a triangle and has at least one angle different from 60º}
∴ A' = U – A = {x : x is a triangle and has all angles equal to 60º}
= Set of all equilateral triangles

Q7: 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' = ____
Ans:

(i) A'∪ A' = U
(ii) ϕ'∩ A = U ∩ A = A
(iii) A'∩ A' = ϕ
(iv) U'∩ A' = ϕ∩ A = ϕ


Old NCERT Questions

EXERCISE - 1.5

Q1: If X and Y are two sets such that n(X) = 17, n(Y) = 23 and n(X ∪ Y) = 38, find n(X ∩ Y).
Ans: 
Given

n (X) = 17

n (Y) = 23

n (X U Y) = 38

We can write it as

n (X U Y) = n (X) + n (Y) – n (X ∩ Y)

Substituting the values

38 = 17 + 23 – n (X ∩ Y)

By further calculation

n (X ∩ Y) = 40 – 38 = 2

So we get

n (X ∩ Y) = 2

Q2: If X and Y are two sets such that X ∪Y has 18 elements, X has 8 elements and Y has 15 elements; how many elements does X ∩ Y have?

Ans: Given

n (X U Y) = 18

n (X) = 8

n (Y) = 15

We can write it as

n (X U Y) = n (X) + n (Y) – n (X ∩ Y)

Substituting the values

18 = 8 + 15 – n (X ∩ Y)

By further calculation

n (X ∩ Y) = 23 – 18 = 5

So we get

n (X ∩ Y) = 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?
Ans: 
Consider H as the set of people who speak Hindi

E as the set of people who speak English

We know that

n(H ∪ E) = 400

n(H) = 250

n(E) = 200

It can be written as

n(H ∪ E) = n(H) + n(E) – n(H ∩ E)

By substituting the values

400 = 250 + 200 – n(H ∩ E)

By further calculation

400 = 450 – n(H ∩ E)

So we get

n(H ∩ E) = 450 – 400

n(H ∩ E) = 50

Therefore, 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 ∪ T have?

Ans: We know that

n(S) = 21

n(T) = 32

n(S ∩ T) = 11

It can be written as

n (S ∪ T) = n (S) + n (T) – n (S ∩ T)

Substituting the values

n (S ∪ T) = 21 + 32 – 11

So we get

n (S ∪ T)= 42

Therefore, the set (S ∪ T) has 42 elements.

Q5: If X and Y are two sets such that X has 40 elements, X ∪Y has 60 elements and X ∩Y has 10 elements, how many elements does Y have?

Ans: We know that

n(X) = 40

n(X ∪ Y) = 60

n(X ∩ Y) = 10

It can be written as

n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)

By substituting the values

60 = 40 + n(Y) – 10

On further calculation

n(Y) = 60 – (40 – 10) = 30

Therefore, the set Y has 30 elements.

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?
Ans: 
Consider C as the set of people who like coffee

T as the set of people who like tea

n(C ∪ T) = 70

n(C) = 37

n(T) = 52

It is given that

n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

Substituting the values

70 = 37 + 52 – n(C ∩ T)

By further calculation

70 = 89 – n(C ∩ T)

So we get

n(C ∩ T) = 89 – 70 = 19

Therefore, 19 people like 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?

Ans: Consider C as the set of people who like cricket

T as the set of people who like tennis

n(C ∪ T) = 65

n(C) = 40

n(C ∩ T) = 10

It can be written as

n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

Substituting the values

65 = 40 + n(T) – 10

By further calculation

65 = 30 + n(T)

So we get

n(T) = 65 – 30 = 35

Hence, 35 people like tennis.

We know that,

(T – C) ∪ (T ∩ C) = T

So we get,

(T – C) ∩ (T ∩ C) = Φ

Here

n (T) = n (T – C) + n (T ∩ C)

Substituting the values

35 = n (T – C) + 10

By further calculation

n (T – C) = 35 – 10 = 25

Therefore, 25 people like only tennis.

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?
Ans: 
Consider F as the set of people in the committee who speak French

S as the set of people in the committee who speak Spanish

n(F) = 50

n(S) = 20

n(S ∩ F) = 10

It can be written as

n(S ∪ F) = n(S) + n(F) – n(S ∩ F)

By substituting the values

n(S ∪ F) = 20 + 50 – 10

By further calculation

n(S ∪ F) = 70 – 10

n(S ∪ F) = 60

Therefore, 60 people in the committee speak at least one of the two languages

EXERCISE - 1.6

Q1: If X and Y are two sets such that n(X) = 17, n(Y) = 23 and n(X ∪ Y) = 38, find n(X ∩ Y).
Ans: 
Given

n (X) = 17

n (Y) = 23

n (X U Y) = 38

We can write it as

n (X U Y) = n (X) + n (Y) – n (X ∩ Y)

Substituting the values

38 = 17 + 23 – n (X ∩ Y)

By further calculation

n (X ∩ Y) = 40 – 38 = 2

So we get

n (X ∩ Y) = 2

Q2: If X and Y are two sets such that X ∪Y has 18 elements, X has 8 elements and Y has 15 elements; how many elements does X ∩ Y have?
Ans: 
Given

n (X U Y) = 18

n (X) = 8

n (Y) = 15

We can write it as

n (X U Y) = n (X) + n (Y) – n (X ∩ Y)

Substituting the values

18 = 8 + 15 – n (X ∩ Y)

By further calculation

n (X ∩ Y) = 23 – 18 = 5

So we get

n (X ∩ Y) = 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?
Ans: 
Consider H as the set of people who speak Hindi

E as the set of people who speak English

We know that

n(H ∪ E) = 400

n(H) = 250

n(E) = 200

It can be written as

n(H ∪ E) = n(H) + n(E) – n(H ∩ E)

By substituting the values

400 = 250 + 200 – n(H ∩ E)

By further calculation

400 = 450 – n(H ∩ E)

So we get

n(H ∩ E) = 450 – 400

n(H ∩ E) = 50

Therefore, 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 ∪ T have?

Ans: We know that

n(S) = 21

n(T) = 32

n(S ∩ T) = 11

It can be written as

n (S ∪ T) = n (S) + n (T) – n (S ∩ T)

Substituting the values

n (S ∪ T) = 21 + 32 – 11

So we get

n (S ∪ T)= 42

Therefore, the set (S ∪ T) has 42 elements.

Q5: If X and Y are two sets such that X has 40 elements, X ∪Y has 60 elements and X ∩Y has 10 elements, how many elements does Y have?
Ans: 
We know that

n(X) = 40

n(X ∪ Y) = 60

n(X ∩ Y) = 10

It can be written as

n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)

By substituting the values

60 = 40 + n(Y) – 10

On further calculation

n(Y) = 60 – (40 – 10) = 30

Therefore, the set Y has 30 elements.

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?

Ans: Consider C as the set of people who like coffee

T as the set of people who like tea

n(C ∪ T) = 70

n(C) = 37

n(T) = 52

It is given that

n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

Substituting the values

70 = 37 + 52 – n(C ∩ T)

By further calculation

70 = 89 – n(C ∩ T)

So we get

n(C ∩ T) = 89 – 70 = 19

Therefore, 19 people like 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?
Ans: 
Consider C as the set of people who like cricket

T as the set of people who like tennis

n(C ∪ T) = 65

n(C) = 40

n(C ∩ T) = 10

It can be written as

n(C ∪ T) = n(C) + n(T) – n(C ∩ T)

Substituting the values

65 = 40 + n(T) – 10

By further calculation

65 = 30 + n(T)

So we get

n(T) = 65 – 30 = 35

Hence, 35 people like tennis.

We know that,

(T – C) ∪ (T ∩ C) = T

So we get,

(T – C) ∩ (T ∩ C) = Φ

Here

n (T) = n (T – C) + n (T ∩ C)

Substituting the values

35 = n (T – C) + 10

By further calculation

n (T – C) = 35 – 10 = 25

Therefore, 25 people like only tennis.

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?
Ans: 
Consider F as the set of people in the committee who speak French

S as the set of people in the committee who speak Spanish

n(F) = 50

n(S) = 20

n(S ∩ F) = 10

It can be written as

n(S ∪ F) = n(S) + n(F) – n(S ∩ F)

By substituting the values

n(S ∪ F) = 20 + 50 – 10

By further calculation

n(S ∪ F) = 70 – 10

n(S ∪ F) = 60

Therefore, 60 people in the committee speak at least one of the two languages.

The document NCERT Solutions Class 11 Maths Chapter 1 - Sets is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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FAQs on NCERT Solutions Class 11 Maths Chapter 1 - Sets

1. What are sets in mathematics?
Ans. Sets in mathematics are collections of distinct objects, considered as an object in their own right. These objects can be anything from numbers to letters to even other sets.
2. How are sets represented in mathematical notation?
Ans. Sets are typically represented using curly braces { } to enclose the elements of the set, separated by commas. For example, the set of even numbers less than 10 can be written as {2, 4, 6, 8}.
3. What is the cardinality of a set?
Ans. The cardinality of a set is the number of elements in the set. It is denoted by |S|, where S is the set. For example, the cardinality of the set {1, 2, 3} is 3.
4. How are sets compared in terms of equality and subsets?
Ans. Two sets are equal if they have the same elements, regardless of the order. A set A is a subset of set B if every element of A is also an element of B. If A is a subset of B but not equal to B, it is called a proper subset.
5. How are sets used in solving problems in JEE exams?
Ans. Sets are an important concept in JEE exams, particularly in topics like probability and permutations. Understanding how to work with sets, including operations like union, intersection, and complement, is crucial for success in these exams.
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