Question

$f\left(x\right)=\frac{x^2}{|x|}$ , $x\ne 0;f\left(0\right)=0$ then:

• A. $f\left(x\right)$ is discontinuous everwhere
• B. $f\left(x\right)$ is continuous everywhere
• C. $f^{\prime }\left(x\right)$ exists in $(-1,1)$
• D. $f^{\prime }\left(x\right)$ exists in $(-2,2)$

Answer 1

The correct option is B $f\left(x\right)$ is continuous everywhere

$f\left(x\right)=\frac{x^2}{|x|}$
$RHL={lim}_{x\to 0^{+}}f\left(x\right)$
$={lim}_{h\to 0}\frac{{0+h}^2}{|0+h|}$
$={lim}_{h\to 0}\frac{h^2}{h}=0$
$LHL={lim}_{x\to 0^{-}}f\left(x\right)$
$={lim}_{h\to 0}\frac{{0-h}^2}{|0-h|}$
$={lim}_{h\to 0}\frac{h^2}{h}=0$
Given $f\left(0\right)=0$
Here, $LHL=RHL=f\left(0\right)$
Hence, $f\left(x\right)$ is continuous at $x=0$
So, it is continuous everywhere. (Rational functions are continuous in its domain.)