Let f(x) = * 20 c 0, & x < 0 x^2 , & x 0 . , then for all values of x

Let f(x) = * 20 c 0, & x < 0 x^2 , & x 0 . , then for all values …

MCQ

Let $f(x) = \left\{ {\begin{array}{*{20}{c}}{0,}&{x < 0}\\{{x^2},}&{x \ge 0}\end{array}} \right.$ , then for all values of $x$

A
$f$ is continuous but not differentiable
B
$f$ is differentiable but not continuous
✓
$f'$ is continuous but not differentiable
D
$f'$ is continuous and differentiable

✓

Answer

Correct option: C.

$f'$ is continuous but not differentiable

c
(c) $f(x) = \left\{ {\begin{array}{*{20}{c}}{\;\,0,}&{x < 0}\\{{x^2},}&{x \ge 0}\end{array}} \right.$;

$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{h \to 0} f(0 - h) = 0$

and $\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{h \to 0} f(0 + h) = \mathop {\lim }\limits_{h \to 0} {(0 + h)^2} = 0$

==> $\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} f(x) = f(0)$

Hence $f(x)$ is continuous function at $x = 0$.

$L\,f'(x) = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{f(x) - f(0)}}{{x - 0}} = \mathop {\lim }\limits_{h \to 0} \frac{{f(0 - h) - 0}}{{ - h}} = \mathop {\lim }\limits_{h \to 0} \frac{{0 - 0}}{{ - h}} = 0$

$Rf'(x) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{f(x) - f(0)}}{{x - 0}}$

$ = \mathop {\lim }\limits_{h \to 0} \frac{{f(0 + h) - f(0)}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{{{(0 + h)}^2} - 0}}{h} = 0$

==> $Lf'(x) = Rf'(x)$

Hence $f(x)$ is differentiable at $x = 0$.

Now $f'(x) = \left\{ {\begin{array}{*{20}{c}}{0\,\,\,\,,}&{x < 0}\\{2x\,\,,}&{x \ge 0}\end{array}} \right.$;

$\mathop {\lim }\limits_{x \to {0^ - }} f'(x) = \mathop {\lim }\limits_{h \to 0} f'(0 - h) = 0$

and $\mathop {\lim }\limits_{x \to {0^ + }} f'(x) = \mathop {\lim }\limits_{h \to 0} f'(0 + h) = \mathop {\lim }\limits_{h \to 0} 2(0 + h) = 0$

==> $\mathop {\lim }\limits_{x \to {0^ - }} f'(x) = \mathop {\lim }\limits_{x \to {0^ + }} f'(x) = 0$

Hence $f'(x)$ is continuous function at $x = 0$.

Now $L\,f''(x) = \mathop {\lim }\limits_{x \to {0^ - }} \frac{{f(x) - f(0)}}{{x - 0}}$

$ = \mathop {\lim }\limits_{h \to 0} \frac{{f(0 - h) - f(0)}}{{ - h}}$

$ = \mathop {\lim }\limits_{h \to 0} \frac{{0 - 0}}{{ - h}} = 0$

$R\,f''(x) = \mathop {\lim }\limits_{x \to {0^ + }} \frac{{f(x) - f(0)}}{{x - 0}}$

$ = \mathop {\lim }\limits_{h \to 0} \frac{{f(0 + h) - f(0)}}{h}$

$ = \mathop {\lim }\limits_{h \to 0} \frac{{2(0 + h) - 0}}{h}$

$ = \mathop {\lim }\limits_{h \to 0} \frac{{2h}}{h} = 2$==> $Lf''(x) \ne Rf''(x)$

Hence $f'(x)$ is not differentiable at $x = 0$.

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