Question:

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

Updated On: Jul 6, 2022
  • $1$
  • $\sqrt{2}$
  • $\frac{1}{\sqrt{2}}$
  • $0$
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The Correct Option is B

Solution and Explanation

We know that, $cosA\, cos2A \,cos2^2A ... cos2^{n-1}\,A$$=\frac{sin\left(2^{n}\,A\right)}{2^{n}\,sin\,A}$ $\therefore cosx\, cos\, 2x \,cos\, 4x\, cos\, 8x\, cos\, 16x=\frac{sin\,32x}{32\,sin\,x}$ $\Rightarrow f \left(x\right)=\frac{1}{32}\cdot\frac{sin\,32}{sin\,x}$ $\therefore f '\left(x\right)=\frac{1}{32}\times\frac{sin\,x\left(32\,cos\,32x\right)-sin\,32x\cdot cos\,x}{sin^{2}\,x}$ $\therefore f '\left(\frac{\pi}{4}\right)=\sqrt{2}$
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