Match each of the set on the left described in the roster form with the same set on the right described in the set-builder form:
$(i) \{ P,R,I,N,C,A,L\} $ $(a) \{ x:x \text{ is a positive integer and is a divisor of } 18\} $ $(ii) \{ 0\} $ $(b) \{ x:x \text{ is an integer and } x^2 - 9 = 0\} $ $(iii) \{ 1,2,3,6,9,18\} $ $(c) \{ x:x \text{ is an integer and } x + 1 = 1\} $ $(iv) \{ 3, -3\} $ $(d) \{ x:x \text{ is a letter of the word } PRINCIPAL\} $
| $(i) \{ P,R,I,N,C,A,L\} $ | $(a) \{ x:x \text{ is a positive integer and is a divisor of } 18\} $ |
| $(ii) \{ 0\} $ | $(b) \{ x:x \text{ is an integer and } x^2 - 9 = 0\} $ |
| $(iii) \{ 1,2,3,6,9,18\} $ | $(c) \{ x:x \text{ is an integer and } x + 1 = 1\} $ |
| $(iv) \{ 3, -3\} $ | $(d) \{ x:x \text{ is a letter of the word } PRINCIPAL\} $ |
(A) In $(d)$,the word $PRINCIPAL$ consists of the letters $P, R, I, N, C, A, L$ (since $P$ and $I$ are repeated,they are written once in the set). Thus,$(i)$ matches $(d)$.
In $(c)$,$x + 1 = 1$ implies $x = 0$. Thus,$(ii)$ matches $(c)$.
In $(a)$,the positive divisors of $18$ are $1, 2, 3, 6, 9, 18$. Thus,$(iii)$ matches $(a)$.
In $(b)$,$x^2 - 9 = 0$ implies $x^2 = 9$,so $x = 3$ or $x = -3$. Thus,$(iv)$ matches $(b)$.
The correct matching is $(i)-(d), (ii)-(c), (iii)-(a), (iv)-(b)$.
In $(c)$,$x + 1 = 1$ implies $x = 0$. Thus,$(ii)$ matches $(c)$.
In $(a)$,the positive divisors of $18$ are $1, 2, 3, 6, 9, 18$. Thus,$(iii)$ matches $(a)$.
In $(b)$,$x^2 - 9 = 0$ implies $x^2 = 9$,so $x = 3$ or $x = -3$. Thus,$(iv)$ matches $(b)$.
The correct matching is $(i)-(d), (ii)-(c), (iii)-(a), (iv)-(b)$.
Explore More
Similar Questions
Make correct statements by filling in the symbols $\subset$ or $\not\subset$ in the blank spaces:
${ x:x \text{ is a circle in the plane} } \dots { x:x \text{ is a circle in the same plane with radius } 1 \text{ unit} }$
${ x:x \text{ is a circle in the plane} } \dots { x:x \text{ is a circle in the same plane with radius } 1 \text{ unit} }$
Easy
View SolutionIf $A, B,$ and $C$ are three sets,then $A \cap (B \cup C) =$
Easy
View SolutionIn the following,state whether $A = B$ or not:
$A = \{2, 4, 6, 8, 10\}$; $B = \{x : x \text{ is a positive even integer and } x \le 10\}$
$A = \{2, 4, 6, 8, 10\}$; $B = \{x : x \text{ is a positive even integer and } x \le 10\}$
Easy
View SolutionThe number of proper subsets of the set $\{1, 2, 3\}$ is
Easy
View SolutionLet $A = \{1, 2, 3, 4, 5, 6\}$. Insert the appropriate symbol $\in$ or $\notin$ in the blank space:
$5 \dots A$
$5 \dots A$
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo