Let f ( x ) = { x 2 sin ( 1 x ) , x 0 0 , x = 0 Then at x = 0 [2023]

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Solution

Q.
Let $f (x) =$ ${\begin{matrix} x^{2} \sin (\frac{1}{x}), & x \neq 0 \\ 0, & x = 0 \end{matrix}$ Then at $x$ = 0 [2023]

1 $f$ is continuous but not differentiable

2 $f$ is continuous but $f'$ is not continuous

3 $f'$ is continuous but not differentiable

4 $f$ and $f'$ both are continuous

Ans.
(2)

$f (x) =$ ${\begin{matrix} x^{2} \sin (\frac{1}{x}), & x \neq 0 \\ 0, & x = 0 \end{matrix}$

LHD = $\lim_{h \to 0} \frac{f (0 - h) - (0)}{- h} = \lim_{h \to 0^{-}} - \frac{h^{2} \sin (1 / h)}{- h} = 0$

RHD = $\lim_{h \to 0} \frac{f (0 + h) - f (0)}{h} = \lim_{h \to 0^{+}} \frac{h^{2} \sin (1 / h)}{h} = 0$

$\Rightarrow f (x) is continuous and differentiable at x = 0 .$

Now, $f' (x) = {\begin{matrix} 2 x \sin (\frac{1}{x}) - \cos (\frac{1}{x}), & x \neq 0 \\ 0, & x = 0 \end{matrix}$

$\lim_{x \to 0} f' (x) = \lim_{x \to 0} [2 x \sin (\frac{1}{x}) - \cos (\frac{1}{x})] = 0 - [- 1, 1] \neq 0$

$f' (0) = 0$

$\Rightarrow f' (x) is discontinuous at x = 0 .$

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