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JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

If

f(x)=\cos \left(x^{2}-4[x]\right)

for

0 < x < 1

, where

[x]=

G.I.F., then

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

is equal to

View answer and explanation

Correct option: A

-\sqrt{\frac{\pi}{2}}

Option A (

-\sqrt{{\frac{{\pi}}{{2}}}}

) is correct because evaluating the derivative of the function at

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

yields this result.

More Questions from DIFFERENTIATION

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

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

is equal to Let

x

y

be two real numbers such that

x>0

x y=1

. The minimum value of

x+y

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

[x]

denotes the integer part of

x

x=k, k

being an integer, the left derivative of

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}}