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MHTCET 17 Apr 2026 Shift 1 Paper

Question : 123 of 150

Marks: +1, -0

f(x)=\cos x \cos 2 x \cos 4 x \cos 8 x \cos 16 x

f^{\prime}\left(\frac{\pi}{4}\right)

is equal to

Solution:

Concept:

When one factor becomes zero at the given point, only the derivative of that zero factor contributes to the product rule.

Explanation:

We have

f(x)=\cos x \cos 2x \cos 4x \cos 8x \cos 16x

x=\frac{\pi}{4}

, the factor

\cos 2x = \cos\frac{\pi}{2} = 0

In the product rule, every term except the one differentiating

\cos 2x

contains

\cos 2x

as a multiplier, so those terms vanish.

Hence,

f'\left(\frac{\pi}{4}\right)=\cos\frac{\pi}{4}\cdot \left.\frac{d}{dx}(\cos 2x)\right|_{x=\frac{\pi}{4}}\cdot \cos\pi \cdot \cos 2\pi \cdot \cos 4\pi.

Now compute each value:

\frac{d}{dx}(\cos 2x)=-2\sin 2x, \quad \text{so at } x=\frac{\pi}{4}, \quad -2\sin\frac{\pi}{2}=-2.

Also,

\cos\frac{\pi}{4}=\frac{1}{\sqrt2}, \quad \cos\pi=-1, \quad \cos2\pi=1, \quad \cos4\pi=1.

Substitute these values:

f'\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt2}\cdot(-2)\cdot(-1)\cdot1\cdot1=\frac{2}{\sqrt2}=\sqrt2.

Answer:

f'\left(\frac{\pi}{4}\right)=\sqrt2, \text{ so the correct option is C. }

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