Define a relation R on the interval ([0, (π)/(2))) by (x R y) if and only if ( ^2 x- ^2 y=1). Then R is :

Correct answer: B. an equivalence relation aligned & ^2 x- ^2 x=1 ( on replacing y with x) & Reflexive & ^2 x- ^2 y=1 & 1+ ^2 x+1- ^2 y=1 & ^2 y- ^2 x=1 & symmetric & ^2 x- ^2 y=1 & ^2 y- ^2 z=1 aligned Adding both aligned & ^2 x- ^2 y+ ^2 y- ^2 z=1+1 & ^2 x+1- ^2 z=2 & ^2 x- ^2 z=1 & Transitive aligned hence equivalence releation

JEE Main 202529 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 29 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Sets and Relations

JEE Main 2025 · 29 January · Morning Shift · Mathematics · Sets and Relations

Define a relation R on the interval ([0, (π)/(2))) by (x R y) if...

Define a relation R on the interval (\left[0, \frac{\pi}{2}\right)) by (x \mathrm{R} y) if and only if (\sec ^2 x-\tan ^2 y=1). Then R is :

Options

Aboth reflexive and transitive but not symmetric
Ban equivalence relation
Creflexive but neither symmetric not transitive
Dboth reflexive and symmetric but not transitive

Correct answer

B. an equivalence relation

Step-by-step solution

\begin{aligned} & \sec ^2 x-\tan ^2 x=1 \quad(\text{ on replacing } y \text{ with } x) \\ & \Rightarrow \text{ Reflexive } \\ & \sec ^2 x-\tan ^2 y=1 \\ & \Rightarrow 1+\tan ^2 x+1-\sec ^2 y=1 \\ & \Rightarrow \sec ^2 y-\tan ^2 x=1 \\ & \Rightarrow \text{ symmetric } \\ & \sec ^2 x-\tan ^2 y=1 \\ & \sec ^2 y-\tan ^2 z=1 \end{aligned}

Adding both

\begin{aligned} & \Rightarrow \sec ^2 x-\tan ^2 y+\sec ^2 y-\tan ^2 z=1+1 \\ & \sec ^2 x+1-\tan ^2 z=2 \\ & \sec ^2 x-\tan ^2 z=1 \\ & \Rightarrow \text{ Transitive } \end{aligned}

hence equivalence releation

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