Question

If $f\left(x\right)={\int }_{-1}^x\left|t\right|dt,x\ge -1$, then

• A. $f$ and $f^{{}^{\prime }}$ are continuous for $x+1>0$
• B. $f$ is continuous but $f^{{}^{\prime }}$ is not continuous for $x+1>0$
• C. $f$ and $f^{{}^{\prime }}$ are not continuous at $x=0$
• D. $f$ is continuous at $x=0$ but $f^{{}^{\prime }}$ is not so

Answer 1

The correct option is A $f$ and $f^{{}^{\prime }}$ are continuous for $x+1>0$

$f\left(x\right)={\int }_{-1}^x\left|t\right|dt,x\ge -1$

Given function can be defined as,
$f\left(x\right)=\{\begin{array}{c}\frac{-1}{2}(x^2-1),-1\le x<0\\ \frac{1}{2}(x^2+1),x\ge 0\end{array}$

$\Rightarrow f^{\prime }\left(x\right)=\{\begin{array}{c}-x,-1<x<0\\ 0,x=0\\ x,x>0\end{array}$

At $x=0$ $f\left(0^{-}\right)=\frac{1}{2},f\left(0^{+}\right)=\frac{1}{2},f\left(0\right)=\frac{1}{2}$
Therefore, $f$ is continuous at $x=0$.
For $f^{\prime },$

$f^{\prime }\left(0^{-}\right)=f^{\prime }\left(0\right)=f^{\prime }\left(0^{+}\right)=0$
Therefore, $f^{\prime }$ is also continuous at $x=0$.
Thus, both $f$ and $f^{\prime }$ are continuous at $x=0$.
Hence, both are continuous for $x>-1$.