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QJEEM22Mac2fa: JEE Main 2022 Mathematics - Class 12 JEEM22Mac2fa Solution: The value of at is

The value of log⁡e2ddx(log⁡cos⁡xcosec⁡x)\log _{e} 2 \frac{d}{d x}\left(\log _{\cos x} \operatorname{cosec} x\right) at x=π4x=\frac{\pi}{4} is

Multiple Choice Options

  • A −22-2 \sqrt{2}
  • B 222 \sqrt{2}
  • C −4-4
  • D 4

Step-by-Step Verified Solution

Let f(x)=log⁡cos⁡xcos⁡ec xf(x) = {\log _{\cos x}}\cos ec\,x

=log⁡cos⁡ec xlog⁡cos⁡x= {{\log \cos ec\,x} \over {\log \cos x}}

⇒f′(x)=log⁡cos⁡x . sin⁡x . (−cos⁡ec xcot⁡x−log⁡cos⁡ec x . 1cos⁡x . −sin⁡x)(log⁡cos⁡x)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}}}

at x=π4x = {\pi \over 4}

f′(π4)=−log⁡(12)+log⁡2(log⁡12)2=2log⁡2f'\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 }}

∴\therefore log⁡e2f′(x){\log _e}2f'(x) at x=π4=4x = {\pi \over 4} = 4

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