Question

Let $f,g:R\to R$ be two functions defined by $f\left(x\right)=\{\begin{array}{cc}xsin\left(\frac{1}{x}\right)& x\ne 0\\ 0& x=0\end{array}$, and $g\left(x\right)=xf\left(x\right)$
Statement I : $f$ is a continuous function at $x=0$
Statement II : $g$ is a differentiable function at $x=0$

• A. Both statements I and II are false.
• B. Both statements I and II are true.
• C. Statement I is true, statement II is false.
• D. Statement I is false, statement II is true.

Answer 1

The correct option is B Both statements I and II are true.

RHL$=\underset{h\to 0^{+}}{lim}h\sin \left(\frac{1}{h}\right)=0\times$ finite number $=0$
LHL$=\underset{h\to 0^{-}}{lim}(-h)\sin (-\frac{1}{h})=\underset{h\to 0^{-}}{lim}h\sin \left(\frac{1}{h}\right)=0\times$ finite number $=0$
$f\left(0\right)=0$
Hence,$f$ is continuous at $x=0$
$g\left(x\right)=xf\left(x\right)=x^2\sin \frac{1}{x}$ .
Clearly, $g\left(0\right)=0$
$g^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{g(x+h)-g\left(x\right)}{h}$
$g^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{{\left(h\right)}^{2}sin\left(\frac{1}{h}\right)-0}{h}$
$=0$ (finite )
Hence $g\left(x\right)$ is differentiable at $x=0$