JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

f(x)=\cos x \cos 2 x \cos 2^2 x \cos 2^3 x \ldots \ldots

\cos 2^{n-1} x

and

n>1

, then

f^{\prime}\left(\frac{\pi}{2}\right)

View answer and explanation

Correct option: A

Option A (1) is correct because by using the identity

\sin(2x) = 2\sin(x)\cos(x)

repeatedly, we can simplify

f(x)

\frac{{\sin(2^n x)}}{{2^n \sin(x)}}

, then differentiate and evaluate at

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

More Questions from DIFFERENTIATION

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f(x)=|x|^{|\tan x|}

f^{\prime}\left(\frac{-\pi}{6}\right)

x

y

x>0

x y=1

x+y

f(x)=[x] \sin (\pi x)

[x]

x

x=k, k

f(x)

y = {\sec ^{ - 1}}\left[ {\mathrm{cosec}x} \right] + \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sec x} \right] + {\sin ^{ - 1}}[\cos x] + {\cos ^{ - 1}}[\sin x],

\frac{{dy}}{{dx}}

\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right)

z = y + f\left( v \right),

v = \left( {\frac{x}{y}} \right),

v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}