JEE Main 2024 — Continuity and Differentiability Question with Solution

Given,

$f\left(1\right)=1, f\left(a\right)=0$

And $f\left(x\right)=\sqrt{\underset{r\to x}{\lim }\left\{\frac{2r^2\left[\left(f\right(r){)}^2-f(x\left)f\right(r)\right]}{r^2-x^2}-r^3{e}^{\frac{f\left(r\right)}{r}}\right\}}$

$\Rightarrow f^2\left(x\right)=\underset{r\to x}{\lim }\left(\frac{2r^2\left(f^2\left(r\right)-f\left(x\right)f\left(r\right)\right)}{r^2-x^2}-r^3{e}^{\frac{f\left(r\right)}{r}}\right)$

$\Rightarrow f^2\left(x\right)=\underset{r\to x}{\lim }\left(\frac{2r^{2}f\left(r\right)}{r+x}\frac{\left(f\right(r)-f(x\left)\right)}{r-x}-r^3{e}^{\frac{f\left(r\right)}{r}}\right)$

$\Rightarrow f^2\left(x\right)=\frac{2x^{2}f\left(x\right)}{x+x}\left(\underset{r\to x}{\lim }\frac{\left(f\right(r)-f(x\left)\right)}{r-x}\right)-x^3{e}^{\frac{f\left(x\right)}{x}}$

$\Rightarrow f^2\left(x\right)=\frac{2x^{2}f\left(x\right)}{2x}{f}^{\text{'}}\left(x\right)-x^3{e}^{\frac{f\left(x\right)}{x}}$

Now, taking $f\left(x\right)=y$ we get,

$\Rightarrow y^2=xy\frac{dy}{dx}-x^3{e}^{\frac{y}{x}}$

$\Rightarrow \frac{y}{x}=\frac{dy}{dx}-\frac{x^2}{y}{e}^{\frac{y}{x}}$

Now, let $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$\Rightarrow v=v+x\frac{dv}{dx}-\frac{x}{v}{e}^v$

$\Rightarrow \frac{dv}{dx}=\frac{e^v}{v}$

$\Rightarrow e^{-v}vdv=dx$

Integrating both side

$e^v(x+c)+1+v=0$

Now using $f\left(1\right)=1\Rightarrow x=1,y=1$ we get,

$\Rightarrow c=-1-\frac{2}{e}$

Hence, $e^v\left(-1-\frac{2}{e}+x\right)+1+v=0$

$\Rightarrow e^{\frac{y}{x}}\left(-1-\frac{2}{e}+x\right)+1+\frac{y}{x}=0$

Now taking, $x=a \& y=0$ we get,

$e^0\left(-1-\frac{2}{e}+a\right)+1+0=0$

$\Rightarrow a=\frac{2}{e}$

$\Rightarrow ae=2$