Question

Let $f,g : R\rightarrow R$ be two functions defined by
$f(x) = \begin{cases} x \,&sin\left(\frac{1}{x}\right), x\ne0, & \text{} \\[2ex] 0, & x=0& \text{} \end{cases}$ and $g(x) = xf(x)$
Statement I : $f$ is a continuous function at $x = 0$.
Statement II : $g$ is a differentiable function at $x = 0$.

• 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

Answer: (B)

Statement-I
$f(x)=\begin{cases}x \sin \left(\frac{1}{x}\right) & x \neq 0 \\ 0 & x=0\end{cases}$
$\displaystyle\lim _{x \rightarrow 0} x \sin \left(\frac{1}{x}\right)=0=f(0)$
Hence $f ( x )$ is continuous function
$g(0)=0=\displaystyle\lim _{x \rightarrow 0} g(x)$
$\underline{\text { LHD : }}$
$\displaystyle\lim _{x \rightarrow 0} \frac{g(0-h)-g(0)}{-h}$
$\displaystyle\lim _{x \rightarrow 0} \frac{h^{2} \sin \left(-\frac{1}{h}\right)-0}{-h}$
$LHD =0$
$\underline{\text { RHD : }}$
$\displaystyle\lim _{h \rightarrow 0} \frac{g(0+h)-g(0)}{h}$
$\displaystyle\lim _{h \rightarrow 0} \frac{h^{2} \sin \left(\frac{1}{n}\right)-0}{h}$
$RHD =0$
$LHD = RHD$
Hence $g(x)$ at $x=0$ is diff. function.