If f(x)= x 2 x 4 x 8 x 16 x, then f^{}≤ft(π/4)=

Correct Answer is : cosec (π4)≤ft(π/4)(4π)

Correct Answer is : cosec (π4)≤ft(π/4)(4π)

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MHTCET 17 Apr 2026 Shift 2 Paper

Question : 119 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)=

Solution:

Concept:

x = \frac{\pi}{4}

, one factor

\cos 2x

becomes zero, so only one term of the product rule remains.

Explanation:

Let

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

x = \frac{\pi}{4}

, we have:

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

\cos x = \frac{1}{\sqrt{2}}

\cos 4x = \cos \pi = -1

\cos 8x = \cos 2\pi = 1

\cos 16x = \cos 4\pi = 1

When applying the product rule, every term contains

\cos 2x

as a factor except the term where

\cos 2x

is differentiated.

That surviving term is:

\cos x \cdot \frac{d}{dx}(\cos 2x) \cdot \cos 4x \cdot \cos 8x \cdot \cos 16x

Now,

\frac{d}{dx}(\cos 2x) = -2\sin 2x

x = \frac{\pi}{4}

, this becomes

-2\sin \frac{\pi}{2} = -2

Therefore,

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

Also,

\operatorname{cosec}\frac{\pi}{4} = \frac{1}{\sin \frac{\pi}{4}} = \frac{1}{1/\sqrt{2}} = \sqrt{2}

Answer:

Option A:

\operatorname{cosec}\left(\frac{\pi}{4}\right)

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