Let $f\left(x\right)=x^2\sin \frac{1}{x}forx\ne 0, f\left(0\right)=g\left(0\right)=0$ and $g\left(x\right)\in \{x{\tan }^{-1}\frac{1}{x},x{\cot }^{-1}x,x^2{\tan }^{-1}\frac{1}{x},x{\cot }^{-1}{x}^2\} forx\ne 0$  where g(x) is an even function, then which of the following is/are CORRECT?

Given,  $f\left(x\right)=x^2\sin \frac{1}{x},x\ne 0 as f\left(0\right)=0$

$$\begin{gathered} g\left(x\right)=x{\tan }^{-1}\frac{1}{x},x\ne 0\left(asitisanevenfunction\right) \\ g\left(0\right)=f\left(0\right)=0 \left(given\right) \end{gathered}$$

Now,  $h\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}=\frac{x^2\sin \frac{1}{x}}{x{\tan }^{-1}\frac{1}{x}},x\ne 0$

$\underset{x\to 0}{Lt}h\left(x\right)=\frac{\underset{x\to 0}{Lt}{x}^2\sin \frac{1}{x}}{\underset{x\to 0}{Lt}x{\tan }^{-1}\frac{1}{x}}=\underset{x\text{}\to 0}{Lt}\frac{x^2\sin \frac{1}{x}}{x{\tan }^{-1}\frac{1}{x}}=\frac{0}{nonzero}=0$

Hence  $h\left(0\right)=0forh\left(x\right)tobe continuous$
Again  $\varphi \left(x\right)=\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}forx\ne 0$
For continuity of   $\varphi \left(x\right),\underset{x\text{}\to 0}{Lt}=\varphi \left(x\right)=\underset{x\to 0}{Lt}\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}=\varphi \left(0\right)$
Now  $f\text{'}\left(0^{+}\right)=0andf\text{'}\left(0^{-}\right)=0$
$\Rightarrow f\text{'}\left(0\right)$  exists and it is equal to 0
$\therefore f\text{'}\left(x\right)=[\begin{array}{cc}{x}^2\cos \frac{1}{x}(-\frac{1}{x^2})+\sin \frac{1}{x}.2x& ifx\ne 0\\ 0& ifx=0\end{array}$

Or  $\therefore f\text{'}\left(x\right)=[\begin{array}{cc}2x\sin \frac{1}{x}-\cos \frac{1}{x}& ifx\ne 0\\ 0& ifx=0\end{array}$