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Question

Mathematics Question on Continuity and differentiability

If f(x)=cosxcos2xcos4xcos8xcos16xf(x) = cosx \cdot cos \,2x \cdot cos \,4x \cdot cos \,8x \cdot cos \,16x, then the value of f(π4)f '\left(\frac{\pi}{4}\right) is

A

11

B

2\sqrt{2}

C

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

D

00

Answer

2\sqrt{2}

Explanation

Solution

We know that, cosAcos2Acos22A...cos2n1AcosA\, cos2A \,cos2^2A ... cos2^{n-1}\,A =sin(2nA)2nsinA=\frac{sin\left(2^{n}\,A\right)}{2^{n}\,sin\,A} cosxcos2xcos4xcos8xcos16x=sin32x32sinx\therefore cosx\, cos\, 2x \,cos\, 4x\, cos\, 8x\, cos\, 16x=\frac{sin\,32x}{32\,sin\,x} f(x)=132sin32sinx\Rightarrow f \left(x\right)=\frac{1}{32}\cdot\frac{sin\,32}{sin\,x} f(x)=132×sinx(32cos32x)sin32xcosxsin2x\therefore f '\left(x\right)=\frac{1}{32}\times\frac{sin\,x\left(32\,cos\,32x\right)-sin\,32x\cdot cos\,x}{sin^{2}\,x} f(π4)=2\therefore f '\left(\frac{\pi}{4}\right)=\sqrt{2}