Relation – JEE Main Previous Year Question with Solutions
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Q. Consider the following relations:-$\mathrm{R}=\{(\mathrm{x}, \mathrm{y}) | \mathrm{x}, \text { y are real numbers and } \mathrm{x}=\mathrm{~ w y ~ f o r ~ s o m e ~ r a t i o n a l ~ n u m b e r ~ w \} ~}$$\mathrm{S}=\left\{\left(\frac{\mathrm{m}}{\mathrm{n}}, \frac{\mathrm{p}}{\mathrm{q}}\right) | \mathrm{m}, \mathrm{n}, \mathrm{p} \text { and } \mathrm{q} \text { are integers such that } \mathrm{n}, \mathrm{q} \neq 0 \text { and } \mathrm{qm}=\mathrm{pn}\right\}Then :(1) R is an equivalence relation but S is not an equivalence relation(2) Neither R nor S is an equivalence relation(3) S is an equivalence relation but R is not an equivalence relation(4) R and S both are equivalence relations. [AIEEE – 2010] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (3)For \mathrm{R}, \mathrm{xRy} \Rightarrow \mathrm{x}=\mathrm{wy}For reflexive\mathrm{xRx} \Rightarrow \mathrm{x}=\mathrm{wx}Which is true then \mathrm{w}=1For symmetricconsider \mathrm{x}=0, \mathrm{y} \neq 0$$\mathrm{xRy} \Rightarrow$ oRy $\Rightarrow 0=\mathrm{wy}$which is true when w=0Nowy Rx $\Rightarrow$ yR $0 \Rightarrow y=w \times 0$There is no rational value of $w$for which $y=w \times 0$Hence relation is not symmetric and hence not an equivalence relationNow for $S$For reflexive$\frac{m}{n} S \frac{m}{n} \Rightarrow m n=n m$which is trueFor symmetricLet $\frac{m}{n} S \frac{m}{n} \Rightarrow q m=n p$$\frac{p}{q} S \frac{m}{n} \Rightarrow p n=m qwhich is trueRelation is symmetricFor transitiveLet \frac{\mathrm{m}}{\mathrm{n}} \mathrm{S} \frac{\mathrm{p}}{\mathrm{q}} \Rightarrow \mathrm{qm}=\mathrm{pn} \quad \ldots(1)$$\frac{\mathrm{p}}{\mathrm{q}} \mathrm{S} \frac{\mathrm{r}}{\mathrm{s}} \Rightarrow \mathrm{ps}=\mathrm{rq} \quad \quad \ldots$ (2)From Equation ( 1) and equation ( 2)$\Rightarrow \mathrm{ms}=\mathrm{nr}$$\therefore \frac{\mathrm{m}}{\mathrm{n}} \mathrm{S} \frac{\mathrm{r}}{\mathrm{s}}S is transitive\therefore S is equivalence. Q. Let R be the set of real numbers.Statement-1: \mathrm{A}=\{(\mathrm{x}, \mathrm{y}) \in \mathrm{R} \times \mathrm{R}: \mathrm{y}-\mathrm{x} \text { is an integer’ is an equivalence relation on } \mathrm{R} .Statement- 2: \mathrm{B}=\{(\mathrm{x}, \mathrm{y}) \in \mathrm{R} \times \mathrm{R}: \mathrm{x}=\alpha \mathrm{y} \text { for some rational number } \alpha\} is an equivalence relation on \mathrm{R} .(1) Statement-1 is true, Statement-2 is false.(2) Statement-1 is false, Statement-2 is true(3) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1(4) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1. [AIEEE – 2011] Download eSaral App for Video Lectures, Complete Revision, Study Material and much more... Sol. (1)y-x= integer \Rightarrow x-y= integer reflexx-x=0= integer symmetricy-x=l, and z-y=12$$z-x=I_{1}+I_{2} \Rightarrow x R z$ transitiveA is equivalence$x=\alpha y$ is equivalence only for $\alpha=1$ not for other values.

• May 14, 2021 at 4:53 pm

Good

• January 5, 2021 at 12:34 pm

Yes adventurino you are right we want more questions

• January 5, 2021 at 12:36 pm

You are right

• September 1, 2020 at 3:17 pm

Statement 2 is very understandable. Have a great day.

• January 5, 2021 at 12:32 pm

What is this only 2 questions we want more