Chapter 1: Sets Exercise 1.2

(i) Set of odd natural numbers divisible by 2 is a null set, as odd numbers are not divisible by 2.

(ii) Set of even prime numbers is not a null set, as 2 is an even prime number.

(iii) {x: x is a natural number, x < 5 and x > 7} is a null set, as a number cannot be both less than 5 and greater than 7.

(iv) {y: y is a point common to any two parallel lines} is a null set, as the parallel lines do not intersect. Therefore, they have no common point.

(i) The set of months of a year is a finite set, as it contains 12 elements.

(ii) {1, 2, 3 …} is an infinite set because it has an infinite number of natural numbers.

(iii) {1, 2, 3 …99, 100} is a finite set, as the numbers from 1 to 100 are finite.

(iv) The set of positive integers greater than 100 is an infinite set, as the positive integers, which are greater than 100, are infinite.

(v) The set of prime numbers less than 99 is a finite set, as the prime numbers which are less than 99 are finite.

(i) The set of lines which are parallel to the x-axis is an infinite set, as the lines which are parallel to the x-axis are infinite.

(ii) The set of letters in the English alphabet is a finite set, as it contains 26 elements.

(iii) The set of numbers which are multiple of 5 is an infinite set, as the multiples of 5 are infinite.

(iv) The set of animals living on the earth is a finite set, as the number of animals living on the earth is finite.

(v) The set of circles passing through the origin (0, 0) is an infinite set, as an infinite number of circles can pass through the origin.

(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.

Therefore, A = B.

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

We know that 12 ∈ A but 12 ∉ B.

Therefore, A ≠ B

(iii) A = {2, 4, 6, 8, 10};

B = {x: x is a positive even integer and x ≤ 10} = {2, 4, 6, 8, 10}

Therefore, A = B

(iv) A = {x: x is a multiple of 10}

B = {10, 15, 20, 25, 30 …}

We know that 15 ∈ B but 15 ∉ A.

Therefore, A ≠ B

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

x² + 5x + 6 = 0 can be written as

x(x + 3) + 2(x + 3) = 0

By further calculation,

(x + 2) (x + 3) = 0

So, we get

x = –2 or x = –3

Here,

A = {2, 3}; B = {–2, –3}

Therefore, 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 which are listed is not significant.

Therefore, A = B.

A = {2, 4, 8, 12}; B = {1, 2, 3, 4}; C = {4, 8, 12, 14}

D = {3, 1, 4, 2}; E = {–1, 1}; F = {0, a}

G = {1, –1}; H = {0, 1}

We know 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

It can be written as

2 ∈ A, 2 ∉ C

Therefore, A ≠ C

3 ∈ B, 3 ∉ C, 3 ∉ E, 3 ∉ F, 3 ∉ G, 3 ∉ H

B ≠ C, B ≠ E, B ≠ F, B ≠ G, B ≠ H

It can be written as

12 ∈ C, 12 ∉ D, 12 ∉ E, 12 ∉ F, 12 ∉ G, 12 ∉ H

Therefore, C ≠ D, C ≠ E, C ≠ F, C ≠ G, C ≠ H

4 ∈ D, 4 ∉ E, 4 ∉ F, 4 ∉ G, 4 ∉ H

Therefore, D ≠ E, D ≠ F, D ≠ G, D ≠ H

Here, 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

Therefore, among the given sets, B = D and E = G.

The other exercises of NCERT Class 11 Maths Solutions for Chapter 1 are as follows.

Exercise 1.1 Solutions 6 Questions

Exercise 1.3 Solutions 9 Questions

Exercise 1.4 Solutions 12 Questions

Exercise 1.5 Solutions 7 Questions

Exercise 1.6 Solutions 8 Questions

Miscellaneous Exercise on Chapter 1 Solutions 16 Questions