JEE MATHEMATICS MCQ: DIFFERENTIATION

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Analyzing +4 | -1

Let

f(x)= \begin{cases}x e^{a x}, & x \leq 0 \\ x+a x^2-x^3, & x>0\end{cases}

where

a

is a positive constant. The interval in which

f^{\prime \prime}(x) \geq 0

, is

View answer and explanation

Correct option: C

\left[\frac{-2}{a}, \frac{a}{3}\right]

Option C ([

\frac{-2}{a}

\frac{a}{3}

]) is correct because it represents the interval where the second derivative of the given piecewise function is greater than or equal to zero.

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

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x

y

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x y=1

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[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}}