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# Relation - JEE Main Previous Year Question with Solutions

JEE Main Previous Year Question of Math with Solutions are available at eSaral. Practicing JEE Main Previous Year Papers Questions of mathematics will help the JEE aspirants in realizing the question pattern as well as help in analyzing weak & strong areas. eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th both for all subjects. Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advance, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more. Download eSaral app for free study material and video tutorials.
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]
Ans. (3) For $\mathrm{R}, \mathrm{xRy} \Rightarrow \mathrm{x}=\mathrm{wy}$ For reflexive $\mathrm{xRx} \Rightarrow \mathrm{x}=\mathrm{wx}$ Which is true then $\mathrm{w}=1$ For symmetric consider $\mathrm{x}=0, \mathrm{y} \neq 0$ $\mathrm{xRy} \Rightarrow$ oRy $\Rightarrow 0=\mathrm{wy}$ which is true when w=0 Now y 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 relation Now for $S$ For reflexive $\frac{m}{n} S \frac{m}{n} \Rightarrow m n=n m$ which is true For symmetric Let $\frac{m}{n} S \frac{m}{n} \Rightarrow q m=n p$ $\frac{p}{q} S \frac{m}{n} \Rightarrow p n=m q$ which is true Relation is symmetric For transitive Let $\frac{\mathrm{m}}{\mathrm{n}} \mathrm{S} \frac{\mathrm{p}}{\mathrm{q}} \Rightarrow \mathrm{qm}=\mathrm{pn} \quad \ldots(1)$ $\frac{\mathrm{p}}{\mathrm{q}} \mathrm{S} \frac{\mathrm{r}}{\mathrm{s}} \Rightarrow \mathrm{ps}=\mathrm{rq} \quad \quad \ldots$ (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]
Ans. (1) $y-x=$ integer $\Rightarrow x-y=$ integer reflex $x-x=0=$ integer symmetric $y-x=l,$ and $z-y=12$ $z-x=I_{1}+I_{2} \Rightarrow x R z$ transitive A is equivalence $x=\alpha y$ is equivalence only for $\alpha=1$ not for other values.