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JEE Main 202226 JulyEvening Shift
Exam
JEE Main
Year
2022
Date
26 July
Shift
Evening Shift
Subject
Mathematics
Chapter
Differentiation

JEE Main 2022 · 26 July · Evening Shift · Mathematics · Differentiation

The value of _ e 2(d)/(dx)( _ x cosec x) at x = (π)/(4) is

The value of log⁡e2ddx(log⁡cos⁡xcosecx)\log_{e}2\dfrac{d}{dx}\left(\log_{\cos x} cosec x\right) at x=π4x = \dfrac{\pi}{4} is

Options

  1. A−22- 2\sqrt{2}
  2. B222\sqrt{2}
  3. C−4- 4
  4. D44

Correct answer

D. 44

Step-by-step solution

For log⁡e2ddx(log⁡cos⁡xcosecx)\log_{e}2\dfrac{d}{dx}\left(\log_{\cos x} cosec x\right)

let, y=log⁡cos⁡xcosecxy = \log_{\cos x} cosec x

i.e. y=−ln⁡(sin⁡x)ln⁡(cos⁡x)y = - \dfrac{\ln \left(\sin x\right)}{\ln \left(\cos x\right)}

⇒dydx=−[cot⁡x⋅ln⁡(cos⁡x)+tan⁡x⋅ln⁡(sin⁡x)](ln⁡(cos⁡x))2⇒ \dfrac{dy}{dx} = - \dfrac{\left[\cot x \cdot \ln \left(\cos x\right) + \tan x \cdot \ln \left(\sin x\right)\right]}{{\left(\ln \left(\cos x\right)\right)}^{2}}

Now (dydx)x=π4=−[cotπ4⋅ln⁡(cosπ4)+tanπ4⋅ln⁡(sinπ4)](ln⁡(cosπ4))2{\left(\dfrac{dy}{dx}\right)}_{x = \dfrac{\pi}{4}} = - \dfrac{\left[cot\dfrac{\pi}{4} \cdot \ln \left(cos\dfrac{\pi}{4}\right) + tan\dfrac{\pi}{4} \cdot \ln \left(sin\dfrac{\pi}{4}\right)\right]}{{\left(\ln \left(cos\dfrac{\pi}{4}\right)\right)}^{2}}

=4ln⁡2= \dfrac{4}{\ln 2}

⇒log⁡e2⋅4ln⁡2=4⇒ \log_{e}2 \cdot \dfrac{4}{\ln 2} = 4

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