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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability1 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 $x = 0$
Now, $\text{LHL}\ ($at $x = 0) =\lim\limits_{\text{x}\rightarrow0^{-}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim\limits_{\text{x}\rightarrow0^{-}}\frac{-\text{x}^2-0}{\text{x}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{-(-\text{h})^2}{-\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}$
$=0$
And $\text{RHL}\ ($at $x = 0) =\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{f(x)}-\text{f}(0)}{\text{x}-0}$
$=\lim\limits_{\text{x}\rightarrow0^{+}}\frac{\text{x}^2-0}{\text{x}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{\text{h}^2}{\text{h}}$
$=\lim\limits_{\text{h}\rightarrow0}\text{h}$
$=0$
$\therefore \text{LHL}\ ($at $x = 0) = \text{RHL}\ ($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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