Question

The set of points where the function $f\left(x\right)=x|x|$ is differentiable is

• A. $(-\infty ,\infty )$
• B. $(-\infty ,0)\cup (0,\infty )$
• C. $(0,\infty )$
• D. $[0,\infty ]$

Answer 1

The correct option is A $(-\infty ,\infty )$

We have, $f\left(x\right)=\{\begin{array}{cc}{x}^2,& x\ge 0\\ -x^2& x<0\end{array}$

Clearly, $f\left(x\right)$ is differentiable for all $x>0$ and for all $x<0$. So, we check the differentiability at $x=0$.

Now, $(RHD$ at $x=a)$

$={\left(\frac{d}{dx}\right(x^2\left)\right)}_{x=0}=(2x{)}_{x=0}=0$
$\therefore (LHD$ at $x=0)={\left(\frac{d}{dx}\right(-x^2\left)\right)}_{x=0}=(-2x{)}_{x=0}=0$

$(LHD$ at $x=0)=(RHD$ at $x=0)$

So, $f\left(x\right)$ is differentiable for all $x$ ie, the set of all points where $f\left(x\right)$ is differentiable is $(-\infty ,\infty )$ ie, $R$.