Differentiation Question 339
Question: If $ f(x)=\cos x\cos 2x\cos 4x\cos 8x\cos 16x $ , then $ {f}’( \frac{\pi }{4} ) $ is
[AMU 2005]
Options:
A) $ \sqrt{2} $
B) $ \frac{1}{\sqrt{2}} $
C) 1
D) $ \frac{\sqrt{3}}{2} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ f(x)=\frac{2\sin x.\cos x.\cos 2x.\cos 4x.\cos 8x.\cos 16x}{2\sin x} $
$ =\frac{\sin 32x}{2^{5}\sin x} $
\ $ f’(x)=\frac{1}{32}.\frac{32\cos 32x.\sin x-\cos x.\sin 32x}{{{\sin }^{2}}x} $
$ f’( \frac{\pi }{4} )=\frac{32.\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\times 0}{32{{( \frac{1}{\sqrt{2}} )}^{2}}} $ = $ \sqrt{2} $ .
BETA
