Verified Solution • 2025–26 Session moderate QJEEM22Mac2fa: JEE Main 2022 Mathematics - Class 12 JEEM22Mac2fa Solution: The value of at is The value of loge2ddx(logcosxcosecx)\log _{e} 2 \frac{d}{d x}\left(\log _{\cos x} \operatorname{cosec} x\right)loge2dxd(logcosxcosecx) at x=π4x=\frac{\pi}{4}x=4π is Multiple Choice Options A −22-2 \sqrt{2}−22 B 222 \sqrt{2}22 C −4-4−4 D 4 Step-by-Step Verified Solution Let f(x)=logcosxcosec xf(x) = {\log _{\cos x}}\cos ec\,xf(x)=logcosxcosecx =logcosec xlogcosx= {{\log \cos ec\,x} \over {\log \cos x}}=logcosxlogcosecx ⇒f′(x)=logcosx . sinx . (−cosec xcotx−logcosec x . 1cosx . −sinx)(logcosx)2\Rightarrow f'(x) = {{\log \cos x\,.\,\sin x\,.\,\left( { - \cos ec\,x\cot x - \log \cos ec\,x\,.\,{1 \over {\cos x}}\,.\, - \sin x} \right)} \over {{{(\log \cos x)}^2}}}⇒f′(x)=(logcosx)2logcosx.sinx.(−cosecxcotx−logcosecx.cosx1.−sinx) at x=π4x = {\pi \over 4}x=4π f′(π4)=−log(12)+log2(log12)2=2log2f'\left( {{\pi \over 4}} \right) = {{ - \log \left( {{1 \over {\sqrt 2 }}} \right) + \log \sqrt 2 } \over {{{\left( {\log {1 \over {\sqrt 2 }}} \right)}^2}}} = {2 \over {\log \sqrt 2 }}f′(4π)=(log21)2−log(21)+log2=log22 ∴\therefore∴ loge2f′(x){\log _e}2f'(x)loge2f′(x) at x=π4=4x = {\pi \over 4} = 4x=4π=4 Related Learning Resources & Exam Preparation JEE Main PYQ Question Bank Browse year-wise authentic previous years' questions with solutions Authentic CBT Mock Tests Simulate official exam environment with live timer and percentile scoring