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Differentiability — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferentiability1 Mark
MCQ
The set of points where the function $f(x) = x |x|$ is differentiable is:
  • ✓
    $(-\infty,\infty)-\infty,\infty$
  • B
    $(-\infty,0)\cup(0,\infty)-\infty,0\cup0,\infty$
  • C
    $(0,\infty)0,\infty$
  • D
    $[0,\infty]0,\infty$
✓

Answer

Correct option: A.
$(-\infty,\infty)-\infty,\infty$
We have,
$\text{f(x)}=\text{x}|\text{x}|$
$\Rightarrow\text{f(x)}=\begin{cases}-\text{x}^2, & \text{x}<0\\0 ,& \text{x}= 0\\\text{x}^2,&\text{x}>0\end{cases}$
When, $x<0$, we have
$f(x)=-x^2$ which being a polynomial function is continuous and differentable in $(-\infty, 0)$
When, $x>0$, we have $f(x)=-x^2$ which being a polynomial function is continuous and differentable in $(0, \infty$,
Thus possible point of non$-$differentiability of $f(x)$ is $x=0$
$\text { Now, LHL (at } x=0)=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}$
$=\lim _{x \rightarrow 0^{-}} \frac{-x^2-0}{x}$
$=\lim _{ h \rightarrow 0} \frac{(-h)^2}{- h }$
$=\lim _{ h \rightarrow 0} h$
$=0$
And $\text{RHL} ($at $x=0)=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}$
$=\lim _{x \rightarrow 0^{+}} \frac{x^2-0}{x}$
$=\lim _{h \rightarrow 0} \frac{h^2}{h}$
$=\lim _{h \rightarrow 0} h$
$=0$
$\therefore \text { LHL (at } x=0 \text { ) = RHL (at } x=0 \text { ) }$
So, $f ( x )$ is also differentiable at $x =0$
i.e. $f(x)$ is differentiable in $(-\infty, \infty)$.

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