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 $(-\infty,\infty).$
$\Rightarrow\text{f(x)}=\begin{cases}-\text{x}^2, & \text{x}<0\\0 ,& \text{x}= 0\\\text{x}^2,&\text{x}>0\end{cases}$
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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