• # question_answer Let $f,g:R\to R$be two functions defined by f(x)=\left\{ \begin{align} & x\sin \left( \frac{1}{x} \right),x\ne 0 \\ & 0,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,x=0 \\ \end{align} \right.,and $g(x)=xf(x)$ Statement I: f is a continuous function at x = 0. Statement II: g is a differentiable function at x = 0.   [JEE Main Online Paper ( Held On 12 Apirl  2014 ) A) Both statement I and II are false. B) Both statement I and II are true. C) Statement I is true, statement II is false. D) Statement I is false, statement II is true.

$f(x)=\left\{ \begin{matrix} x\sin \left( \frac{1}{x} \right),x\ne 0 & x\ne 0 \\ 0, & x=0 \\ \end{matrix} \right.$and $g(x)=xf(x)$ LHL$=\underset{h\to {{0}^{-}}}{\mathop{\lim }}\,\left\{ -h\sin \left( -\frac{1}{h} \right) \right\}$ = 0 × a finite quantity between ?1 and 1 = 0 $RHL=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,h\sin \frac{1}{h}=0$ Also, x = 0 Thus LHL = RHL = f(0) $\therefore$f (x) is continuous at x = 0 $g(x)=\left\{ \begin{matrix} {{x}^{2}}\sin \frac{1}{x}, & x\ne 0 \\ 0, & x=0 \\ \end{matrix} \right.$ For g(x) LHL $=\underset{h\to {{o}^{-}}}{\mathop{\lim }}\,\left\{ -{{h}^{2}}\sin \left( \frac{1}{h} \right) \right\}$ $={{0}^{2}}\times a$a finite quantity between ?1 and 1 = 0 RHL$=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,{{h}^{2}}\sin \left( \frac{1}{h} \right)=0$ Also g(0) = 0 $\therefore$ (x) is continuous at x = 0