Question

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

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

Answer 1

Answer: (A)

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