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Question: If f(x) = cos x . cos 2x . cos 4x .cos 8x . cos 16x then \(f'\left( \frac{\pi}{4} \right)\) is –...

If f(x) = cos x . cos 2x . cos 4x .cos 8x . cos 16x

then f′(π4)f'\left( \frac{\pi}{4} \right) is –

A

2\sqrt{2}

B

12\frac{1}{\sqrt{2}}

C

1

D

None of these

Answer

2\sqrt{2}

Explanation

Solution

f(x) = 2sin⁡x.cos⁡x.cos⁡2x.cos⁡4x.cos⁡8x.cos⁡16x2sin⁡x\frac{2\sin x.\cos x.\cos 2x.\cos 4x.\cos 8x.\cos 16x}{2\sin x}= sin⁡32x25sin⁡x\frac{\sin 32x}{2^{5}\sin x}

∴ f ′(x) = 132.32cos⁡32x.sin⁡x−cos⁡x.sin⁡32xsin⁡2x\frac{1}{32}.\frac{32\cos 32x.\sin x - \cos x.\sin 32x}{\sin^{2}x}

∴ f′(π4)f'\left( \frac{\pi}{4} \right) = 32.12−12.032.(12)2\frac{32.\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}.0}{32.\left( \frac{1}{\sqrt{2}} \right)^{2}} = 2\sqrt{2}