The value of m for which the function f(x) = l m x^2 , ,x 1 , , ,… - Vidyadip

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

✓

Answer

Correct option: D.

Does not exist

$Lf'(1) = \mathop {\lim }\limits_{h \to 0} \frac{{f(1 - h) - f(1)}}{{ - h}}$
$ = \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 $Rf'(1) = \mathop {\lim }\limits_{h \to 0} \frac{{f(1 + h) - f(1)}}{h}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{2(1 + h) - m}}{h}$.
For differentiability, $Lf'(1) = Rf'(1)$.
But for any value of $m,Rf'(1) = Lf'(1)$ not possible.

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