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]

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]

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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