Question

If $f(x)=\cos x \cos 2 x \cos 2^{2} x \cos 2^{3} x \ldots . . \cos 2^{n-1} x$ and $n > 1$, then $f'\left(\frac{\pi}{2}\right)$ is

• A. 1
• B. 0
• C. -1
• D. None of these

Answer 1

Answer: (A)

We have,
$f(x)=\cos x \cos 2 x \cos 2^{2} x \cos 2^{3} x \ldots . . \cos 2^{n-1} x$
$\Rightarrow f(x)=\frac{\sin 2^{n} x}{2^{n} \sin x} $
$\Rightarrow f'(x)=\frac{2^{n} \cos 2^{n} x \sin x-\sin 2^{n} x \cos x}{2^{n} \sin ^{2} x}$
$\Rightarrow f'\left(\frac{\pi}{2}\right)=\frac{2^{n} \cos 2^{n-1} \pi}{2^{n}}=\cos 2^{n-1} \pi=1$