Question

If $f\left(x\right)=\cos x.\cos 2x.\cos 4x.\cos 8x.\cos 16x$, then $f^{{}^{\prime }}\left(\frac{\pi }{4}\right)$ is

• A. $\sqrt{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $1$
• D. none of these

Answer 1

The correct option is D none of these

Given, $f\left(x\right)=\cos x.\cos 2x.\cos 4x.\cos 8x.\cos 16x\Rightarrow f\left(\frac{\pi }{4}\right)=0$
Apply $\ln$ on both sides, we get
$\ln \left(f\right(x\left)\right)=\ln \left(\cos x\right)+\ln \left(\cos 2x\right)+\ln \left(\cos 4x\right)+\ln \left(\cos 8x\right)+\ln \left(\cos 16x\right)\frac{1}{f\left(x\right)}{f}^{\prime }\left(x\right)=\frac{1}{\cos x}(-\sin x)+\frac{1}{\cos 2x}(-2\sin 2x)+.......+\frac{1}{\cos 16x}(-\sin 16x)f^{\prime }\left(x\right)=-f\left(x\right)(\tan x+\tan 2x+\tan 4x+\tan 8x+\tan 16x)$

$\Rightarrow f^{\prime }\left(\frac{\pi }{4}\right)=0$