JEE MATHEMATICS MCQ: DIFFERENTIATION

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Analyzing +4 | -1

Let

f(x)=\int_0^{g(x)} \frac{d t}{\sqrt{1+t^2}}

where

g(x)=\int_0^{\cos x}\left(1+\sin t^2\right) d t

. Also,

h(x)=e^{-|x|}

and

f(x)=x^2 \sin \frac{1}{x}

, if

x \neq 0

and

f(0)=0

, then

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

is equal to

View answer and explanation

Correct option: C

h^{\prime}\left(0^*\right)

Option C (

h^{\prime}(0^+)

) is correct because the derivative of f(x) at

\frac{\pi}{2}

is equal to -1, which is also the right-hand derivative of h(x) at 0.

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}}