Question

If $f(x) = cosx \cdot cos \,2x \cdot cos \,4x \cdot cos \,8x \cdot cos \,16x$, then the value of $f '\left(\frac{\pi}{4}\right)$ is

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

Answer 1

Answer: (B)

We know that,
$cosA\, cos2A \,cos2^2A ... cos2^{n-1}\,A$$=\frac{sin\left(2^{n}\,A\right)}{2^{n}\,sin\,A}$
$\therefore cosx\, cos\, 2x \,cos\, 4x\, cos\, 8x\, cos\, 16x=\frac{sin\,32x}{32\,sin\,x}$
$\Rightarrow f \left(x\right)=\frac{1}{32}\cdot\frac{sin\,32}{sin\,x}$
$\therefore f '\left(x\right)=\frac{1}{32}\times\frac{sin\,x\left(32\,cos\,32x\right)-sin\,32x\cdot cos\,x}{sin^{2}\,x}$
$\therefore f '\left(\frac{\pi}{4}\right)=\sqrt{2}$