MCQ
જો $f(x) = \left\{ \begin{array}{l}\,\,\,\,\,\,\,{e^x};\,\,\,\,x \le 0\\|1 - x|;\,\,x > 0\end{array} \right.$, તો
- A$f(x)$ એ $x = 1$ આગળ સતત છે.
- B$f(x)$ એ $x = 0$ આગળ સતત છે.
- ✓$(a)$ અને $(b)$ બંને
- Dઆપેલ પૈકી એક પણ નહીં
$Rf'(0) = \mathop {\lim }\limits_{h \to 0} \frac{{f(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{1 - h - 1}}{h} = - 1$
$Lf'(0) = \mathop {\lim }\limits_{h \to 0} \frac{{f(0 - h) - f(0)}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{ - h}} - 1}}{{ - h}} = 1$
So, it is not differentiable at $x = 0$.
Similarly, it is not differentiable at $x = 1$.
But it is continous at $x = 0$, $1$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.