JEE Main 2023 — Continuity and Differentiability Question with Solution

Given:

$f\left(x\right)=\left\{\begin{array}{l}{x}^2\sin \left(\frac{1}{x}\right); x\ne 0\\ 0; x=0\end{array}\right.$

$\underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\left[x^2\sin \left(\frac{1}{x}\right)\right]$

$\Rightarrow \underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }{x}^2\times \left[\mathrm{number} \mathrm{between} -1 \mathrm{to} 1\right]$

$\Rightarrow \underset{x\to 0}{\lim }f\left(x\right)=0=f\left(0\right)$

Hence, $f\left(x\right)$ is continuous at $x=0$.

Now,

$\mathrm{RHD}=f^{\text{'}}\left(0^{+}\right)=\underset{h\to 0}{\lim }\left[\frac{f\left(0+h\right)-f\left(0\right)}{h}\right]$

$\Rightarrow \mathrm{RHD}=f^{\text{'}}\left(0^{+}\right)=\underset{h\to 0}{\lim }\left[\frac{h^2\sin \left({\displaystyle \frac{1}{h}}\right)-0}{h}\right]$

$\Rightarrow \mathrm{RHD}=f^{\text{'}}\left(0^{+}\right)=\underset{h\to 0}{\lim }\left[h\sin \left(\frac{1}{h}\right)\right]=0$

And,

$\mathrm{LHD}=f^{\text{'}}\left(0^{-}\right)=\underset{h\to 0}{\lim }\left[\frac{f\left(0-h\right)-f\left(0\right)}{-h}\right]$

$\Rightarrow \mathrm{LHD}=f^{\text{'}}\left(0^{-}\right)=\underset{h\to 0}{\lim }\left[\frac{-h^2\sin \left({\displaystyle \frac{1}{h}}\right)-0}{-h}\right]=0$

Since, $\mathrm{LHD}=\mathrm{RHD}=\mathrm{finite}$

So, $f\left(x\right)$ is differentiable at $x=0$.

Now,

$f^{\text{'}}\left(x\right)=\left\{\begin{array}{l}2x\sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right); x\ne 0\\ 0; x=0\end{array}\right.$

Now,

$\underset{x\to 0}{\lim }{f}^{\text{'}}\left(x\right)=\underset{x\to 0}{\lim }\left[2x\sin \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right)\right]$

$\Rightarrow \underset{x\to 0}{\lim }{f}^{\text{'}}\left(x\right)=\underset{x\to 0}{\lim }\left[2x\left\{\frac{1}{x}-\frac{1}{3!}\frac{1}{x^3}+\frac{1}{5!}\frac{1}{x^5}-....\right\}-\left\{1-\frac{1}{2!}\frac{1}{x^2}+\frac{1}{4!}\frac{1}{x^4}-....\right\}\right]$

$\Rightarrow \underset{x\to 0}{\lim }{f}^{\text{'}}\left(x\right)=\underset{x\to 0}{\lim }\left[\left\{1-\frac{2}{3!}\frac{1}{x^2}+\frac{2}{5!}\frac{1}{x^4}-....\right\}-\left\{-\frac{1}{2!}\frac{1}{x^2}+\frac{1}{4!}\frac{1}{x^4}-....\right\}\right]$

This limit does not exists finitely, hence $f^{\text{'}}\left(x\right)$ is discontinuous at $x=0$.