Question

Let $f,g:R\to R$ be two functions defined by
$f(x)=\{\begin{array}{llc}x& sin\left(\frac{1}{x}\right),x\ne 0,& \\ 0,& x=0& \end{array}$ 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
$f(x)=\{\begin{array}{ll}x\sin \left(\frac{1}{x}\right)& x\ne 0\\ 0& x=0\end{array}$
$\underset{x\to 0}{\lim }x\sin \left(\frac{1}{x}\right)=0=f(0)$
Hence $f(x)$ is continuous function
$g(0)=0=\underset{x\to 0}{\lim }g(x)$
$\underline{\text{ LHD : }}$
$\underset{x\to 0}{\lim }\frac{g(0-h)-g(0)}{-h}$
$\underset{x\to 0}{\lim }\frac{h^2\sin \left(-\frac{1}{h}\right)-0}{-h}$
$LHD=0$
$\underline{\text{ RHD : }}$
$\underset{h\to 0}{\lim }\frac{g(0+h)-g(0)}{h}$
$\underset{h\to 0}{\lim }\frac{h^2\sin \left(\frac{1}{n}\right)-0}{h}$
$RHD=0$
$LHD=RHD$
Hence $g(x)$ at $x=0$ is diff. function.