Search any question & find its solution
Question:
Answered & Verified by Expert
Let $\mathrm{f}, \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}$ be two functions defined by $f(x)= \begin{cases}x \sin \left(\frac{1}{x}\right) & , x \neq 0 \\ 0, & , x=0\end{cases}$ and $g(x)=x f(x)$
Statement I: $f$ is a continuous function at $\mathrm{x}=0$.
Statement II: $g$ is a differentiable function at $x=0$.
Options:
Statement I: $f$ is a continuous function at $\mathrm{x}=0$.
Statement II: $g$ is a differentiable function at $x=0$.
Solution:
1525 Upvotes
Verified Answer
The correct answer is:
Both statement I and II are true.
Both statement I and II are true.
$$
\begin{aligned}
&f(x)= \begin{cases}x \sin \left(\frac{1}{x}\right), & x \neq 0 \\
0, & x=0\end{cases} \\
&\text { and } g(x)=x f(x) \\
&\text { For } f(x) \\
&\mathrm{LHL}=\lim _{h \rightarrow 0^{-}}\left\{-h \sin \left(-\frac{1}{h}\right)\right\}
\end{aligned}
$$
$=0 \times$ a finite quantity between $-1$ and 1
$$
=0
$$
$$
\mathrm{RHL}=\lim _{h \rightarrow 0^{+}} h \sin \frac{1}{h}=0
$$
Also, $f(0)=0$
Thus $\mathrm{LHL}=\mathrm{RHL}=f(0)$
$\therefore f(x)$ is continuous at $x=0$
$$
g(x)= \begin{cases}x^2 \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0\end{cases}
$$
For $g(x)$
$$
\mathrm{LHL}=\lim _{h \rightarrow 0^{-}}\left\{-h^2 \sin \left(\frac{1}{h}\right)\right\}
$$
$=0^2 \times$ a finite quantity between $-1$ and $1=0$
$$
\text { RHL }=\lim _{h \rightarrow 0^{+}} h^2 \sin \left(\frac{1}{h}\right)=0
$$
Also $g(0)=0$
$\therefore g(x)$ is continuous at $x=0$
\begin{aligned}
&f(x)= \begin{cases}x \sin \left(\frac{1}{x}\right), & x \neq 0 \\
0, & x=0\end{cases} \\
&\text { and } g(x)=x f(x) \\
&\text { For } f(x) \\
&\mathrm{LHL}=\lim _{h \rightarrow 0^{-}}\left\{-h \sin \left(-\frac{1}{h}\right)\right\}
\end{aligned}
$$
$=0 \times$ a finite quantity between $-1$ and 1
$$
=0
$$
$$
\mathrm{RHL}=\lim _{h \rightarrow 0^{+}} h \sin \frac{1}{h}=0
$$
Also, $f(0)=0$
Thus $\mathrm{LHL}=\mathrm{RHL}=f(0)$
$\therefore f(x)$ is continuous at $x=0$
$$
g(x)= \begin{cases}x^2 \sin \frac{1}{x}, & x \neq 0 \\ 0, & x=0\end{cases}
$$
For $g(x)$
$$
\mathrm{LHL}=\lim _{h \rightarrow 0^{-}}\left\{-h^2 \sin \left(\frac{1}{h}\right)\right\}
$$
$=0^2 \times$ a finite quantity between $-1$ and $1=0$
$$
\text { RHL }=\lim _{h \rightarrow 0^{+}} h^2 \sin \left(\frac{1}{h}\right)=0
$$
Also $g(0)=0$
$\therefore g(x)$ is continuous at $x=0$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.