Question

If $y = \left(\log_{\cos x} \sin x \right)\left(\log_{nx } \cos x\right) + \sin^{-1} \frac{2x}{1+x^{2}} ,$ then $ \frac{dy}{dx}$ at $ x = \frac{\pi}{2} $ is equal to

• A. $\frac{8}{(4 + \pi^2)}$
• B. 0
• C. $ - \frac{8}{(4 + \pi^2)}$
• D. 1

Answer 1

Answer: (A)

$y = \left(\log_{\cos x} \sin x \right)\left(\log_{nx } \cos x\right) + \sin^{-1} \frac{2x}{1+x^{2}} , $
$ \Rightarrow y = \frac{\log\sin x}{\log \cos x} . \frac{\log\cos x}{\log nx} + \sin^{-1} \frac{2x}{1+x^{2}}$
$ \Rightarrow y = \frac{\log\sin x}{\log nx} + \sin^{-1} \frac{2x}{1+x^{2}}$
$ \left( \text{Let} \, u = \frac{\log\sin x}{\log nx} , v =\sin^{-1} \frac{2x}{1+x^{2}}\right) \Rightarrow y =u+v $
$ \left[\frac{dy}{dx}\right]_{\pi/2} = \left[\frac{du}{dx}\right]_{\pi/2} + \left[\frac{dv}{dx}\right]_{\pi/2} + ....\left(1\right)$
$ u = \frac{\log \sin x}{\log nx} $
$ \Rightarrow \frac{du}{dx} = \frac{\left(\log nx\right)\cot x -\frac{.\log\sin x}{x}}{\left(\log nx\right)^{2}}$
$ \therefore \left[\frac{du}{dx}\right]_{x=\pi/2} = 0 $
$v = \sin^{-1} \left(\frac{2x}{1+x^{2}} \right); $ Let $x =\tan\theta$
$ \therefore v =\sin^{-1} \left(\sin 2\theta\right) = 2\theta=2 \tan^{-1} x$
$ \therefore \frac{dv}{dx} = \frac{2}{1+x^{2}}$
$ \therefore \left[\frac{dv}{dx}\right]_{x =\left(\pi/2\right)} = \frac{2}{1+ \left(\pi^{2}/4\right)} = \frac{8}{\pi^{2}/4} $
$ \therefore \frac{dy}{dx} $ at $\left(x = \frac{\pi}{2}\right) = \frac{8}{\pi^{2} + 4}$