A) \[\sqrt{2}\]
B) \[\frac{1}{\sqrt{2}}\]
C) 1
D) \[\frac{\sqrt{3}}{2}\]
Correct Answer: A
Solution :
\[f(x)=\frac{2\sin x.\cos x.\cos 2x.\cos 4x.\cos 8x.\cos 16x}{2\sin x}\] \[=\frac{\sin 32x}{{{2}^{5}}\sin x}\] \ \[f'(x)=\frac{1}{32}.\frac{32\cos 32x.\sin x-\cos x.\sin 32x}{{{\sin }^{2}}x}\] \[f'\left( \frac{\pi }{4} \right)=\frac{32.\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\times 0}{32{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}}\] =\[\sqrt{2}\].
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