Question

If $f\left(x\right)=\cos x\cdot \cos 2x\cdot \cos 4x\cdot \cos 8x\cdot \cos 16x$, 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 A $\sqrt{2}$

Given that

$f\left(x\right)=\cos x\cos 2x\cos 4x\cos 8x\cos 16x$

Multiply & divide by $2\sin x$

$f\left(x\right)=\frac{2\sin x\cos x\cos 2x\cos 4x\cos 8x\cos 16x}{2\sin x}$

Since $\sin 2x=2\sin x\cos x$

$f\left(x\right)=\frac{\sin 2x\cos 2x\cos 4x\cos 8x\cos 16x}{2\sin x}$

Multiplying and dividing by $2$, we get

$f\left(x\right)=\frac{2\sin 2x\cos 2x\cos 4x\cos 8x\cos 16x}{2^2\sin x}$

$f\left(x\right)=\frac{\sin 4x\cos 4x\cos 8x\cos 16x}{2^2\sin x}$

Similarly continuing upto $\cos 16x$, we get

$f\left(x\right)=\frac{\sin 2^{5}x}{2^5\sin x}=\frac{\sin 32x}{32\sin x}$

Differentiating w.r. to $x$

$f^{{}^{\prime }}\left(x\right)=\frac{1}{32}\left[\frac{32\sin x\cos 32x-\cos x\sin 32x}{{\left(\sin x\right)}^2}\right]$

$\therefore f^{{}^{\prime }}\left(\frac{\pi }{4}\right)=\frac{1}{32}\left[\frac{32\sin \frac{\pi }{4}\cos 8\pi -\cos \frac{\pi }{4}\sin 8\pi }{{\left(\sin \frac{\pi }{4}\right)}^2}\right]$

$\therefore f^{{}^{\prime }}\left(\frac{\pi }{4}\right)=\frac{2}{\sqrt{2}}=\sqrt{2}$