QJEEM24Ma4c61: JEE Main 2024 Mathematics - Class 12 JEEM24Ma4c61 Solution: Let f ( x ) = lim r → x { 2 r 2 [ ( f ( r ) ) 2 − f ( x ) f ( r …

$\begin{aligned}& f(1)=1, f(a)=0 \\& {f^2}(x) = \mathop {\lim }\limits_{r \to x} \left( {{{2{r^2}({f^2}(r) - f(x)f(r))} \over {{r^2} - {x^2}}} - {r^3}{e^{{{f(r)} \over r}}}} \right) \\& = \mathop {\lim }\limits_{r \to x} \left( {{{2{r^2}f(r)} \over {r + x}}{{(f(r) - f(x))} \over {r - x}} - {r^3}{e^{{{f(r)} \over r}}}} \right) \\& f^2(x)=\frac{2 x^2 f(x)}{2 x} f^{\prime}(x)-x^3 e^{\frac{f(x)}{x}} \\& y^2=x y \frac{d y}{d x}-x^3 e^{\frac{y}{x}} \\& \frac{y}{x}=\frac{d y}{d x}-\frac{x^2}{y} e^{\frac{y}{x}}\end{aligned}$

Put $y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$

$\begin{aligned}& v=v+x \frac{d v}{d x}-\frac{x}{v} e^v \\& \frac{d v}{d x}=\frac{e^v}{v} \Rightarrow e^{-v} v d v=d x\end{aligned}$

Integrating both side

$\begin{aligned}& \mathrm{e}^{\mathrm{v}}(\mathrm{x}+\mathrm{c})+1+\mathrm{v}=0 \\& \mathrm{f}(1)=1 \Rightarrow \mathrm{x}=1, \mathrm{y}=1\end{aligned}$

$\begin{aligned}& \Rightarrow c=-1-\frac{2}{e} \\& e^v\left(-1-\frac{2}{e}+x\right)+1+v=0 \\& e^{\frac{y}{x}}\left(-1-\frac{2}{e}+x\right)+1+\frac{y}{x}=0 \\& x=a, y=0 \Rightarrow a=\frac{2}{e} \\& a e=2\end{aligned}$