Question

If $f\left(x\right)=\cos x\cos 2x\cos 4x\cos 8x\cos 16x$, then $f\text{'}\left(\frac{\mathrm{\pi }}{4}\right)$ is equal to

• A. $\sqrt{2}$
• B. $\frac{1}{\sqrt{2}}$
• C. $0$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is C

$0$

Explanation for the correct option.

Step 1. Simplify the given function.

The value of the function $f\left(x\right)=\cos x\cos 2x\cos 4x\cos 8x\cos 16x$ for $x=\frac{\mathrm{\pi }}{4}$ is given as:

$\begin{array}{rcl}f\left(\frac{\mathrm{\pi }}{4}\right)& =& \cos \left(\frac{\mathrm{\pi }}{4}\right)\cos \left(2\times \frac{\mathrm{\pi }}{4}\right)\cos \left(4\times \frac{\mathrm{\pi }}{4}\right)\cos \left(8\times \frac{\mathrm{\pi }}{4}\right)\cos \left(16\times \frac{\mathrm{\pi }}{4}\right)\\ & =& \cos \left(\frac{\mathrm{\pi }}{4}\right)\cos \left(\frac{\mathrm{\pi }}{2}\right)\cos \left(\mathrm{\pi }\right)\cos \left(2\mathrm{\pi }\right)\cos \left(4\mathrm{\pi }\right)\\ & =& \frac{1}{\sqrt{2}}\times 0\times \left(-1\right)\times 1\times 1\\ & =& 0\end{array}$

Now, in the function $f\left(x\right)=\cos x\cos 2x\cos 4x\cos 8x\cos 16x$ take logarithm of both sides.

$$\begin{gathered} \log \left(f\left(x\right)\right)=\log \left(\cos x\cos 2x\cos 4x\cos 8x\cos 16x\right) \\ \Rightarrow \log \left(f\left(x\right)\right)=\log \left(\cos x\right)+\log \left(\cos 2x\right)+\log \left(\cos 4x\right)+\log \left(\cos 8x\right)+\log \left(\cos 16x\right) \end{gathered}$$

Step 2. Find the value of $f\text{'}\left(\frac{\mathrm{\pi }}{4}\right)$.

Differentiate both sides of the function $\log \left(f\left(x\right)\right)=\log \left(\cos x\right)+\log \left(\cos 2x\right)+\log \left(\cos 4x\right)+\log \left(\cos 8x\right)+\log \left(\cos 16x\right)$.

$$\begin{gathered} \frac{d}{dx}\log \left(f\left(x\right)\right)=\frac{d}{dx}\left(\log \left(\cos x\right)+\log \left(\cos 2x\right)+\log \left(\cos 4x\right)+\log \left(\cos 8x\right)+\log \left(\cos 16x\right)\right) \\ \Rightarrow \frac{1}{f\left(x\right)}f\text{'}\left(x\right)=\frac{1}{\cos x}\left(-\sin x\right)+\frac{1}{\cos 2x}\left(-\sin 2x\right)+\frac{1}{\cos 4x}\left(-\sin 4x\right)+\frac{1}{\cos 8x}\left(-\sin 8x\right)+\frac{1}{\cos 16x}\left(-\sin 16x\right) \\ \Rightarrow f\text{'}\left(x\right)=f\left(x\right)\left[-\tan x-\tan 2x-\tan 4x-\tan 8x-\tan 16x\right] \end{gathered}$$

Now, the value of $f\text{'}\left(\frac{\mathrm{\pi }}{4}\right)$ is given as:

$\begin{array}{rcl}f\text{'}\left(\frac{\mathrm{\pi }}{4}\right)& =& f\left(\frac{\mathrm{\pi }}{4}\right)\left[-\tan \frac{\mathrm{\pi }}{4}-\tan \left(2\times \frac{\mathrm{\pi }}{4}\right)-\tan \left(4\times \frac{\mathrm{\pi }}{4}\right)-\tan \left(8\times \frac{\mathrm{\pi }}{4}\right)-\tan \left(16\times \frac{\mathrm{\pi }}{4}\right)\right]\\ & =& 0\times \left[-\tan \frac{\mathrm{\pi }}{4}-\tan \left(2\times \frac{\mathrm{\pi }}{4}\right)-\tan \left(4\times \frac{\mathrm{\pi }}{4}\right)-\tan \left(8\times \frac{\mathrm{\pi }}{4}\right)-\tan \left(16\times \frac{\mathrm{\pi }}{4}\right)\right]\left[f\left(\frac{\pi }{4}\right)=0\right]\\ & =& 0\end{array}$

Hence, the correct option is C.