List all the elements of the set: C = x : x is an integer , - 1 2 < x < 9 2 .

Given: C = x : x is an integer , - 1 2 < x < 9 2 Integers lying between - 1 2 and 9 2 are 0 , 1 , 2 , 3 , 4 and C is set of integers lying between - 1 2 and 9 2 . ∴ C = 0 , 1 , 2 , 3 , 4

Match each of the sets on the left in the roster form with the same set on the right described in the set-builder form : (i) A , P , L , E (a) x : x + 5 = 5 , x ∈ Z (ii) 5 , – 5 (b) { x : x is a prime natural number and a divisor of 10 } (iii) 0 (c) { x : x is a letter of the word “ R A J A S T H A N ” } (iv) 1 , 2 , 5 , 10 (d) { x : x is a natural number and divisor of 10 } (v) A , H , J , R , S , T , N (e) x : x 2 – 25 = 0 (vi) 2 , 5 (f) { x : x is a letter of the word “ A P P L E ” }

(i) { A , P , L , E } ⇔ { x : x is a letter of word " A P P L E " } The distinct letters of word " A P P L E " are A , P , L , E . (ii) { 5 , - 5 } ⇔ x : x 2 - 25 = 0 The solution set of x 2 - 25 = 0 is x = ± 5 (iii) { 0 } ⇔ { x : x + 5 = 5 , x ∈ z } The solution set of x + 5 = 5 is x = 0 . (iv) { 1 , 2 , 5 , 10 } ⇔ { x : x is a natural and divisor of 10 } The natural numbers which are divisor of 10 are 1 , 2 , 5 , 10 . (v) { A , H , J , R , S , T , N } ⇔ { x : x is a letter of the word " RAJASTHAN " } The distinct letters of word " RAJASTHAN " are A , H , J , R , S , T , N . (vi) { 2 , 5 } ⇔ { x : x is a prime natural number and a divisor of 10 } The prime natural numbers which are divisor of 10 are 2 , 5 .

Write the set 1 2 , 2 5 , 3 10 , 4 17 , 5 26 , 6 37 , 7 50 in the set-builder form.

Set Builder Form: In this form, in order to describe a set, a variable x (say) (to denote each element of the set) is written inside the braces and then after putting a colon the common property P x possessed by each element of the set is written within the braces. Given set is 1 2 , 2 5 , 3 10 , 4 17 , 5 26 , 6 37 , 7 50 Numerator is Natural number from 1 to 7 and denominator is in form of n 2 + 1 can be written as 2 = 1 2 + 1 5 = 2 2 + 1 10 = 3 2 + 1 17 = 4 2 + 1 26 = 5 2 + 1 37 = 6 2 + 1 50 = 7 2 + 1 Therefore, the set-builder form is x : x = n n 2 + 1 , 1 ≤ n ≤ 7 .

E M B I B E

MATHEMATICS for CLASS XI>Chapter 1 - Sets>EXERCISE 1.2

R D Sharma Solutions for Chapter: Sets, Exercise 2: EXERCISE 1.2

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - R D Sharma Solutions for Chapter: Sets, Exercise 2: EXERCISE 1.2

Attempt the practice questions on Chapter 1: Sets, Exercise 2: EXERCISE 1.2 with hints and solutions to strengthen your understanding. MATHEMATICS for CLASS XI solutions are prepared by Experienced Embibe Experts.

Questions from R D Sharma Solutions for Chapter: Sets, Exercise 2: EXERCISE 1.2 with Hints & Solutions

EASY

11th CBSE

IMPORTANT

MATHEMATICS for CLASS XI>Chapter 1 - Sets>EXERCISE 1.2>Q 3

List all the elements of the set:
$C = \{x : x is an integer, - \frac{1}{2} < x < \frac{9}{2}\}$ .

EASY

11th CBSE

IMPORTANT

MATHEMATICS for CLASS XI>Chapter 1 - Sets>EXERCISE 1.2>Q 3

List all the elements of the set:
$D =$ { $x : x$ is a vowel in the word $“ E Q U A T I O N ”$ }.

EASY

11th CBSE

IMPORTANT

MATHEMATICS for CLASS XI>Chapter 1 - Sets>EXERCISE 1.2>Q 3

List all the elements of the set:

$E =$ { $x : x$ is a month of a year not having $31$ days}.

EASY

11th CBSE

IMPORTANT

MATHEMATICS for CLASS XI>Chapter 1 - Sets>EXERCISE 1.2>Q 3

List all the elements of the set:

$F =$ { $x : x$ is a letter of the word $“ M I S S I S S I P P I ”$ }.

EASY

11th CBSE

IMPORTANT

MATHEMATICS for CLASS XI>Chapter 1 - Sets>EXERCISE 1.2>Q 4

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

(i)	$\{A, P, L, E\}$	(a)	$\{x : x + 5 = 5, x \in Z\}$
(ii)	$\{5, - 5\}$	(b)	{ $x : x$ is a prime natural number and a divisor of $10$ }
(iii)	$\{0\}$	(c)	{ $x : x$ is a letter of the word $“ R A J A S T H A N ”$ }
(iv)	$\{1, 2, 5, 10\}$	(d)	{ $x : x$ is a natural number and divisor of $10$ }
(v)	$\{A, H, J, R, S, T, N\}$	(e)	$\{x : x^{2} - 25 = 0\}$
(vi)	$\{2, 5\}$	(f)	{ $x : x$ is a letter of the word $“ A P P L E ”$ }