Verified Solution • 2025–26 Session moderate

QJEEM25Ma3c3d: JEE Main 2025 Mathematics - Class 11 JEEM25Ma3c3d Solution: Define a relation R on the interval by R if and only if . Then…

Define a relation R on the interval [0,π2)\left[0, \frac{\pi}{2}\right) by xx R yy if and only if sec⁡2x−tan⁡2y=1\sec^2x - \tan^2y = 1 . Then R is :

Multiple Choice Options

  • A both reflexive and symmetric but not transitive
  • B both reflexive and transitive but not symmetric
  • C reflexive but neither symmetric not transitive
  • D an equivalence relation

Step-by-Step Verified Solution

amp;sec⁡2x−tan⁡2x=1( on replacing y with x)amp;⇒ Reflexive amp;sec⁡2x−tan⁡2y=1amp;⇒1+tan⁡2x+1−sec⁡2y=1amp;⇒sec⁡2y−tan⁡2x=1amp;⇒symmetric⁡amp;sec⁡2x−tan⁡2y=1amp;sec⁡2y−tan⁡2z=1\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 \operatorname{symmetric} \\& \sec ^2 x-\tan ^2 y=1 \\& \sec ^2 y-\tan ^2 z=1\end{aligned}

Adding both

amp;⇒sec⁡2x−tan⁡2y+sec⁡2y−tan⁡2z=1+1amp;sec⁡2x+1−tan⁡2z=2amp;sec⁡2x−tan⁡2z=1amp;⇒ Transitive \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 relation

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