MCQ
The value of $m$ for which the function $f(x) = \left\{ \begin{array}{l}m{x^2},\,x \le 1\\\,\,\,\,2x,\,x > 1\end{array} \right.$ is differentiable at $x = 1$, is
- A$0$
- B$1$
- C$2$
- ✓Does not exist
$ = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{m\,{{(1 - h)}^2} - m}}{{ - h}}$
$ = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{m\,[1 + {h^2} - 2h - 1]}}{{ - h}}$
$ = \mathop {\lim }\limits_{h \to 0} \,m\,(2 - h) = 2m$ and $R\,f'\,(1) = \mathop {\lim }\limits_{h \to 0} \,\,\frac{{f\,(1 + h) - f(1)}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \,\frac{{2\,(1 + h) - m}}{h}$.
For differentiability, $L\,f'(1) = R\,f'\,(1)$.
But for any value of $m,\,\,R\,\,f'\,(1) = L\,f'(1)$ not possible.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.