Consider f(x) = l x^2 |x| , ,x 0 , , , , , , ,0, ,x = 0 . - Vidyadip

MCQ

Consider $f(x) = \left\{ \begin{array}{l}\frac{{{x^2}}}{{|x|}},\,x \ne 0\\\,\,\,\,\,\,\,0,\,x = 0\end{array} \right.$

A
$f(x)$ is discontinuous everywhere
✓
$f(x)$ is continuous everywhere
C
$f'(x)$ exists in $( - 1,1)$
D
$f'(x) $ exists in $( - 2,2)$

✓

Answer

Correct option: B.

$f(x)$ is continuous everywhere

b
(b) We have $f(x) = \left\{ {\begin{array}{*{20}{r}}{\frac{{{x^2}}}{{|x|}},}&{x \ne 0}\\{0,}&{x = 0}\end{array} = \left\{ {\begin{array}{*{20}{r}}{\frac{{{x^2}}}{x} = x,}&{x > 0}\\{0,}&{x = 0}\\{\frac{{{x^2}}}{{ - x}} = - x,}&{x < 0}\end{array}} \right.} \right.$

We have $\mathop {\lim }\limits_{x \to 0 - } f(x) = \mathop {\lim }\limits_{x \to 0} \, - x = 0,\,\,\mathop {\lim }\limits_{x \to 0 + } f(x) = \mathop {\lim }\limits_{x \to 0} x = 0$

and $f(0) = 0.$

So $f(x)$ is continuous at $x = 0.$

Also $f(x)$ is continuous for all other values of $x.$

Hence, $f(x)$ is continuous everywhere.

Clearly, $Lf'(0) = - 1$ and $Rf'(0) = 1.$

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

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