Question bank/JEE Main/Mathematics
JEE MainMathematicsModerateLimits, continuity and derivability
The set of points where x^2 |x| is thrice differentiable is
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Let f(x) = x^2 |x| which can be expressed as
$f(x) = \left\{ \begin{matrix}
- x^{3},x < 0 \\
0,\ \ \ x\ = \ \ 0 \\
x^{3},\ x\ > \ \ 0
\end{matrix} \right.\ $
$f(x) = \left\{ \begin{matrix}
- x^{3}, & x < 0 \\
0, & x = 0 \\
x^{3}, & x > 0
\end{matrix} \right.\ \therefore f^{'}(x) = \left\{ \begin{matrix}
- 3x^{2}, & x < 0 \\
0, & x = 0 \\
3x^{2}, & x > 0
\end{matrix} \right.\ $
So, f '(x) exists for all real x
$f^{''}(x) = \left\{ \begin{matrix}
- 6x, & x < 0 \\
0, & x = 0 \\
6x, & x > 0
\end{matrix} \right.\ $
So, f " (x) exists for all real x.
$f^{'''}(x) = \left\{ \begin{matrix}
- 6,\ x < \ \ 0 \\
0,\ \ \ \ x = \ 0 \\
6,\ \ \ \ x > \ 0
\end{matrix} \right.\ $
$f(x) = \left\{ \begin{matrix}
- x^{3},x < 0 \\
0,\ \ \ x\ = \ \ 0 \\
x^{3},\ x\ > \ \ 0
\end{matrix} \right.\ $
$f(x) = \left\{ \begin{matrix}
- x^{3}, & x < 0 \\
0, & x = 0 \\
x^{3}, & x > 0
\end{matrix} \right.\ \therefore f^{'}(x) = \left\{ \begin{matrix}
- 3x^{2}, & x < 0 \\
0, & x = 0 \\
3x^{2}, & x > 0
\end{matrix} \right.\ $
So, f '(x) exists for all real x
$f^{''}(x) = \left\{ \begin{matrix}
- 6x, & x < 0 \\
0, & x = 0 \\
6x, & x > 0
\end{matrix} \right.\ $
So, f " (x) exists for all real x.
$f^{'''}(x) = \left\{ \begin{matrix}
- 6,\ x < \ \ 0 \\
0,\ \ \ \ x = \ 0 \\
6,\ \ \ \ x > \ 0
\end{matrix} \right.\ $
Source: Grass-G / M-11 Mathematics Paper-2 1Nov.docx
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