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]

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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 $\mathrm{w}=1$ 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]

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Sol. (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.

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Comments
  • May 14, 2021 at 4:53 pm

    Good

    0
  • January 5, 2021 at 12:34 pm

    Yes adventurino you are right we want more questions

    0
    • January 5, 2021 at 12:36 pm

      You are right

      0
  • September 1, 2020 at 3:17 pm

    Statement 2 is very understandable. Have a great day.

    0
    • January 5, 2021 at 12:32 pm

      What is this only 2 questions we want more

      0