Question

If $g\left(x\right)=\underset{0}{\overset{x}{\int }}2\left|t\right|dt,$ then which of the following is/are true about $g\left(x\right)$ ?

• A. $g\left(x\right)=x\left|x\right|$
• B. $g\left(x\right)$ is monotonic
• C. $g\left(x\right)$ is differentiable at $x=0$
• D. $g^{\prime }\left(x\right)$ is differentiable at $x=0$

Answer 1

Answer: (A), (B), (C)

$g\left(x\right)$ is differentiable at $x=0$

$g\left(x\right)=\underset{0}{\overset{x}{\int }}2\left|t\right|dt,$
$=\{\begin{array}{cc}\underset{0}{\overset{x}{\int }}-2tdt,& x<0\\ \underset{0}{\overset{x}{\int }}2tdt,& x\ge 0\end{array}=\{\begin{array}{cc}{[-t^2]}_0^x,& x<0\\ {\left[t^2\right]}_0^x,& x\ge 0\end{array}=\{\begin{array}{cc}-x^2,& x<0\\ x^2,& x\ge 0\end{array}=x\left|x\right|$
$g\left(x\right)$ is continuous.
Now,
$g^{\prime }\left(x\right)=\{\begin{array}{cc}-2x,& x<0\\ 2x,& x\ge 0\end{array}$
Clearly, $g\left(x\right)$ is differentiable at $x=0$
Also, $g\left(x\right)$ is monotonic.

$g^{\prime \prime }\left(x\right)=\{\begin{array}{cc}-2,& x<0\\ 2,& x\ge 0\end{array}$
So, $g^{\prime }\left(x\right)$ is non-differentiable at $x=0$