Correct option: B.Both statement $I$ and $II$ are true.
b
$f\left( x \right) = \left\{ \begin{array}{l}
x\sin \left( {\frac{1}{x}} \right),\,\,\,x \ne 0\\
0\,,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0
\end{array} \right.$
and $g(x)=xf(x)$
For $f(x)$
$LHL = \mathop {\lim }\limits_{h \to {0^ - }} \left\{ { - h\sin \left( { - \frac{1}{h}} \right)} \right\}$
$ = 0 \times a$ finite quatity between $-1$ and $1$
$=0$
$RHL = \mathop {\lim }\limits_{h \to {0^ + }} h\sin \frac{1}{h} = 0$
Also, $f(0)=0$
Thus $LHL = RHL = f\left( 0 \right)$
$\therefore $ $f(x)$ is continuous at $x=0$
$g\left( x \right) = \left\{ \begin{array}{l}
{x^2}\sin \frac{1}{x},\,\,\,x \ne 0\\
0\,,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 0
\end{array} \right.$
For $g(x)$
$LHL = \mathop {\lim }\limits_{h \to {0^ - }} \left\{ { - {h^2}\sin \left( {\frac{1}{h}} \right)} \right\}$
$ = {0^2} \times a$ a finite quantify between $-1$ and $1$ $=0$
$RHL = \mathop {\lim }\limits_{h \to {0^ + }} {h^2}\sin \left( {\frac{1}{h}} \right) = 0$
Also $g(0)=0$
$\therefore $ $g(x)$ is continuous at $x=0$