The set of points where the function f (x) = x |x| is differentiable is
(a) $(-\infty, \infty)$
(b) $(-\infty, 0) \cup(0, \infty)$
(c) $(0, \infty)$
(d) $[0, \infty]$
(a) $(-\infty, \infty)$
We have,
$f(x)=x|x|$
$\Rightarrow f(x)=\left\{\begin{array}{cc}-x^{2}, & x<0 \\ 0, & x=0 \\ x^{2}, & x>0\end{array}\right.$
When, $x<0$, we have
$f(x)=-x^{2}$ which being a polynomial function is continuous and differentiable in $(-\infty, 0)$
When, $x>0$, we have
$f(x)=x^{2}$ which being a polynomial function is continuous and differentiable in $(0, \infty)$
Thus possible point of non-differentiability of $f(x)$ is $x=0$
Now, $\operatorname{LHD}($ at $x=0)=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}$
$=\lim _{x \rightarrow 0^{-}} \frac{-x^{2}-0}{x}$
$=\lim _{h \rightarrow 0} \frac{-(-h)^{2}}{-h}$
$=\lim _{h \rightarrow 0} h$
$=0$
And RHD $($ at $x=0)=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}$
$=\lim _{x \rightarrow 0^{+}} \frac{x^{2}-0}{x}$
$=\lim _{h \rightarrow 0} \frac{h^{2}}{h}$
$=\lim _{h \rightarrow 0} h$
$=0$
$\therefore \operatorname{LHD}($ at $x=0)=\mathrm{RHD}($ at $x=0)$
So, $f(x)$ is also differentiable at $x=0$
i. e. $f(x)$ is differentiable in $(-\infty, \infty)$
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