Mathematics · Sets and Relations

JEE Main 2024 — 1 February, Shift 2 — Question 14

Consider the relations $R_{1}$ and $R_{2}$ defined as $a R_{1} b$ $\Leftrightarrow a^{2}+b^{2}=1$ for all $a, b, \in R$ and $(a, b) R_{2}(c, d)$ $\Leftrightarrow \mathrm{a}+\mathrm{d}=\mathrm{b}+\mathrm{c}$ for all $(\mathrm{a}, \mathrm{b}),(\mathrm{c}, \mathrm{d}) \in \mathrm{N} \times \mathrm{N}$ . Then

Option A:
Only $R_{1}$ is an equivalence relation
Option B:
Only $R_{2}$ is an equivalence relation
Correct
Option C:
$R_{1}$ and $R_{2}$ both are equivalence relations
Option D:
Neither $R_{1}$ nor $R_{2}$ is an equivalence relation

Answer: B

Step-by-step solution

$\mathrm{aR}_{1} b \Leftrightarrow \mathrm{a}^{2}+\mathrm{b}^{2}=1 ; \mathrm{a}, \mathrm{b} \in \mathrm{R}$

$(\mathrm{a}, \mathrm{b}) \mathrm{R}_{2}(\mathrm{c}, \mathrm{d}) \Leftrightarrow \mathrm{a}+\mathrm{d}=\mathrm{b}+\mathrm{c} ;(\mathrm{a}, \mathrm{b}),(\mathrm{c}, \mathrm{d}) \in \mathrm{N}$ for $\mathrm{R}_{1}$ :

Not reflexive symmetric not transitive for $R_{2}: R_{2}$

is reflexive, symmetric and transitive

Hence only $R_{2}$ is equivalence relation.

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 1 February, Shift 2
Subject: Mathematics
Chapter: Sets and Relations
Topic: Types of Relations