Search any question & find its solution
Question:
Answered & Verified by Expert
Assertion (A) $: f(x)=x \sin \left(\frac{1}{x}\right)$ is differentiable at $x=0$.
Reason $(\mathbf{R}): \mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$
Options:
Reason $(\mathbf{R}): \mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$
Solution:
1575 Upvotes
Verified Answer
The correct answer is:
A is false but $\mathbf{R}$
Given function : $\mathrm{f}(\mathrm{x})=\mathrm{x} \sin \left(\frac{1}{\mathrm{x}}\right)$
For differentiability at $\mathrm{x}=0 ; \mathrm{LHD}=\mathrm{RHD}$ at $\mathrm{x}=0$
$\begin{aligned} \mathrm{LHD} &=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0-\mathrm{h})-\mathrm{f}(0)}{-\mathrm{h}} \\ &=\lim _{\mathrm{h} \rightarrow 0} \frac{(-\mathrm{h}) \sin \left(-\frac{1}{\mathrm{~h}}\right)}{-\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{h} \sin \left(\frac{1}{\mathrm{~h}}\right)}{-\mathrm{h}} \end{aligned}$
$=\lim _{h \rightarrow 0} \sin \left(\frac{1}{h}\right)=a$ finite value lies between $-1$
and 1 which cannot be qualified exactly.
$\mathrm{RHD}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0+\mathrm{h})-\mathrm{f}(0)}{\mathrm{h}}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{h} \sin \left(\frac{1}{\mathrm{~h}}\right)}{\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \sin \left(\frac{1}{\mathrm{~h}}\right)$
$=$ a finite value lies between $-1$ and 1 which cannot be qualified exactly. $\mathrm{LHD} \neq \mathrm{RHD} \neq$ a definite value.
Hence, $\mathrm{f}(\mathrm{x})$ is not differentiable at $\mathrm{x}=0$. For continuity at $\mathrm{x}=0$ :
$\lim _{x \rightarrow 0}$ LHL $=\lim _{x \rightarrow 0} \mathrm{RHL}=\mathrm{V} . \mathrm{F}$. at $\mathrm{x}=0$
$\mathrm{LHL}=\lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(0-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}-\mathrm{h} \sin \left(-\frac{1}{\mathrm{~h}}\right)$
$=\lim _{h \rightarrow 0} \mathrm{~h} \sin \frac{1}{\mathrm{~h}}=0$
$\mathrm{RHL}=\lim _{\mathrm{x} \rightarrow 0^{\prime}} \mathrm{f}(0+\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \mathrm{~h} \sin \frac{1}{\mathrm{~h}}=0$
$\mathrm{f}(0)=0$
$\Rightarrow \mathrm{LHL}=\mathrm{RHL}=\mathrm{f}(0)$
Hence, $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$ Thus, (A) is false but $(\mathrm{R})$ is true.
For differentiability at $\mathrm{x}=0 ; \mathrm{LHD}=\mathrm{RHD}$ at $\mathrm{x}=0$
$\begin{aligned} \mathrm{LHD} &=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0-\mathrm{h})-\mathrm{f}(0)}{-\mathrm{h}} \\ &=\lim _{\mathrm{h} \rightarrow 0} \frac{(-\mathrm{h}) \sin \left(-\frac{1}{\mathrm{~h}}\right)}{-\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{h} \sin \left(\frac{1}{\mathrm{~h}}\right)}{-\mathrm{h}} \end{aligned}$
$=\lim _{h \rightarrow 0} \sin \left(\frac{1}{h}\right)=a$ finite value lies between $-1$
and 1 which cannot be qualified exactly.
$\mathrm{RHD}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{f}(0+\mathrm{h})-\mathrm{f}(0)}{\mathrm{h}}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{h} \sin \left(\frac{1}{\mathrm{~h}}\right)}{\mathrm{h}}=\lim _{\mathrm{h} \rightarrow 0} \sin \left(\frac{1}{\mathrm{~h}}\right)$
$=$ a finite value lies between $-1$ and 1 which cannot be qualified exactly. $\mathrm{LHD} \neq \mathrm{RHD} \neq$ a definite value.
Hence, $\mathrm{f}(\mathrm{x})$ is not differentiable at $\mathrm{x}=0$. For continuity at $\mathrm{x}=0$ :
$\lim _{x \rightarrow 0}$ LHL $=\lim _{x \rightarrow 0} \mathrm{RHL}=\mathrm{V} . \mathrm{F}$. at $\mathrm{x}=0$
$\mathrm{LHL}=\lim _{\mathrm{x} \rightarrow 0} \mathrm{f}(0-\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0}-\mathrm{h} \sin \left(-\frac{1}{\mathrm{~h}}\right)$
$=\lim _{h \rightarrow 0} \mathrm{~h} \sin \frac{1}{\mathrm{~h}}=0$
$\mathrm{RHL}=\lim _{\mathrm{x} \rightarrow 0^{\prime}} \mathrm{f}(0+\mathrm{h})=\lim _{\mathrm{h} \rightarrow 0} \mathrm{~h} \sin \frac{1}{\mathrm{~h}}=0$
$\mathrm{f}(0)=0$
$\Rightarrow \mathrm{LHL}=\mathrm{RHL}=\mathrm{f}(0)$
Hence, $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$ Thus, (A) is false but $(\mathrm{R})$ is true.
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.