For a differentiable function $f$ on $\mathbf{R}$ there is an $x_0$ with $f^{\prime}\left(x_0\right)=\frac{1}{2}\left[f^{\prime}(0)+f^{\prime}(1)\right]$

JEE MATHEMATICS MCQ: DIFFERENTIATION

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For a differentiable function

f

\mathbf{R}

there is an

x_0

with

f^{\prime}\left(x_0\right)=\frac{1}{2}\left[f^{\prime}(0)+f^{\prime}(1)\right]

View answer and explanation

Correct option: D

f

is continuously differentiable

Option D (if

f

is continuously differentiable) is correct because the Intermediate Value Theorem can be applied to the derivative of a continuously differentiable function.

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

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and

x y=1

. The minimum value of

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is Let

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

where

[x]

denotes the integer part of

x

. Then at

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being an integer, the left derivative of

f(x)

is If

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],

then

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

is equal to

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

is equal to If

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

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

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