JEE Main 2023 — Mathematics Calculus

Given:

f(x)={\begin{matrix}{x}^{2}\mathrm{sin}(\frac{1}{x});x\neq 0 \\ 0;x=0\end{matrix}

$$\underset{x\rightarrow 0}{\mathrm{lim}}f(x)=\underset{x\rightarrow 0}{\mathrm{lim}}[{x}^{2}\mathrm{sin}(\frac{1}{x})]$$x→0lim​f(x)=x→0lim​[x2sin(x1​)]

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}f(x)=\underset{x\rightarrow 0}{\mathrm{lim}}{x}^{2}\times [\mathrm{number}\mathrm{between}-1\mathrm{to}1]$$⇒x→0lim​f(x)=x→0lim​x2×[numberbetween−1to1]

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}f(x)=0=f(0)$$⇒x→0lim​f(x)=0=f(0)

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

Now,

$$\mathrm{RHD}={f}^{'}({0}^{+})=\underset{h\rightarrow 0}{\mathrm{lim}}[\frac{f(0+h)-f(0)}{h}]$$RHD=f′(0+)=h→0lim​[hf(0+h)−f(0)​]

$$\Rightarrow \mathrm{RHD}={f}^{'}({0}^{+})=\underset{h\rightarrow 0}{\mathrm{lim}}[\frac{{h}^{2}\mathrm{sin}(\frac{1}{h})-0}{h}]$$⇒RHD=f′(0+)=h→0lim​[hh2sin(h1​)−0​]

$$\Rightarrow \mathrm{RHD}={f}^{'}({0}^{+})=\underset{h\rightarrow 0}{\mathrm{lim}}[h\mathrm{sin}(\frac{1}{h})]=0$$⇒RHD=f′(0+)=h→0lim​[hsin(h1​)]=0

And,

$$\mathrm{LHD}={f}^{'}({0}^{-})=\underset{h\rightarrow 0}{\mathrm{lim}}[\frac{f(0-h)-f(0)}{-h}]$$LHD=f′(0−)=h→0lim​[−hf(0−h)−f(0)​]

$$\Rightarrow \mathrm{LHD}={f}^{'}({0}^{-})=\underset{h\rightarrow 0}{\mathrm{lim}}[\frac{-{h}^{2}\mathrm{sin}(\frac{1}{h})-0}{-h}]=0$$⇒LHD=f′(0−)=h→0lim​[−h−h2sin(h1​)−0​]=0

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

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

Now,

{f}^{'}(x)={\begin{matrix}2x\mathrm{sin}(\frac{1}{x})-\mathrm{cos}(\frac{1}{x});x\neq 0 \\ 0;x=0\end{matrix}

Now,

$$\underset{x\rightarrow 0}{\mathrm{lim}}{f}^{'}(x)=\underset{x\rightarrow 0}{\mathrm{lim}}[2x\mathrm{sin}(\frac{1}{x})-\mathrm{cos}(\frac{1}{x})]$$x→0lim​f′(x)=x→0lim​[2xsin(x1​)−cos(x1​)]

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}{f}^{'}(x)=\underset{x\rightarrow 0}{\mathrm{lim}}[2x{\frac{1}{x}-\frac{1}{3!}\frac{1}{{x}^{3}}+\frac{1}{5!}\frac{1}{{x}^{5}}-....}-{1-\frac{1}{2!}\frac{1}{{x}^{2}}+\frac{1}{4!}\frac{1}{{x}^{4}}-....}]$$⇒x→0lim​f′(x)=x→0lim​[2xx1​−3!1​x31​+5!1​x51​−....−1−2!1​x21​+4!1​x41​−....]

$$\Rightarrow \underset{x\rightarrow 0}{\mathrm{lim}}{f}^{'}(x)=\underset{x\rightarrow 0}{\mathrm{lim}}[{1-\frac{2}{3!}\frac{1}{{x}^{2}}+\frac{2}{5!}\frac{1}{{x}^{4}}-....}-{-\frac{1}{2!}\frac{1}{{x}^{2}}+\frac{1}{4!}\frac{1}{{x}^{4}}-....}]$$⇒x→0lim​f′(x)=x→0lim​[1−3!2​x21​+5!2​x41​−....−−2!1​x21​+4!1​x41​−....]

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