MCQ
જો $y = {\cot ^{ - 1}}\left( {\frac{{\log \left( {\frac{e}{{{x^2}}}} \right)}}{{\log \left( {e{x^2}} \right)}}} \right) + {\cot ^{ - 1}}\left( {\frac{{\log \left( {e{x^4}} \right)}}{{\log \left( {\frac{{{e^2}}}{{{x^2}}}} \right)}}} \right),$ તો $\frac{{dy}}{{dx}} = ......0 < \log x < \frac{1}{2}$
  • A
    $ - 1$
  • $0$
  • C
    $1$
  • D
    $2$

Answer

Correct option: B.
$0$
$y = \cot ^{-1} \left(\frac{\log \left(\frac{e}{x^2}\right)}{\log (ex^2)}\right) + \cot^{-1} \left(\frac{\log (ex^4)}{\log \left(\frac{e^2}{x^2}\right)}\right)$
$= \cot ^{-1} \left(\frac{\log e^e - \log x^2}{\log e^e + \log x^2} \right)+ \cot^{-1} \left(\frac{\log e^e + \log x^4}{\log e^2 - \log x^2}\right)$
$= \cot ^{-1} \left(\frac{1 - \log x^2}{1 + \log x^2} \right)+ \cot^{-1} \left(\frac{1+2 \log x^2}{2 - \log x^2}\right)$
$= \tan ^{-1} \left(\frac{1 + \log x^2}{1 - \log x^2} \right)+ \tan^{-1} \left(\frac{2 -\log x^2}{1+2 \log x^2}\right)$
$= \tan ^{-1} 1 + \tan^{-1} (\log x^2) + \tan^{-1} 2 - \tan^{-1} (\log x^2)$
$= \tan ^{-1} 1 + \tan^{-1}2$
$\frac{dy}{dx} = 0$

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