JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Differential Equations

Write $y=f(x)$. Inside the limit, as $r\to x$ the fraction $\frac{2r^2\left[f(r{)}^2-f(x)f(r)\right]}{r^2-x^2}$ is a $\frac{0}{0}$ form (numerator vanishes at $r=x$). Differentiate numerator and denominator with respect to $r$ and let $r\to x$:

Numerator derivative at $r=x$: $4xf(x{)}^2+4x^{2}f(x)f^{\prime }(x)-4xf(x{)}^2-2x^{2}f(x)f^{\prime }(x)=2x^{2}f(x)f^{\prime }(x)$.

Denominator derivative: $2x$.

So the limit of that fraction is $xf(x)f^{\prime }(x)$, and

$f(x{)}^2=xf(x)f^{\prime }(x)-x^3{e}^{f(x)/x}$.

Divide through by $x^{2}f(x)$ after rearranging: $xf(x)f^{\prime }(x)=f(x{)}^2+x^3{e}^{f(x)/x}\Rightarrow f^{\prime }(x)=\frac{f(x)}{x}+\frac{x^2{e}^{f(x)/x}}{f(x)}$.

Put $v=\frac{f(x)}{x}$, so $f(x)=vx$, $f^{\prime }(x)=v+x\frac{dv}{dx}$:

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

Integrating by parts, $\int v{e}^{-v}dv=-e^{-v}(v+1)$, so

$-e^{-v}(v+1)=x+k$ for a constant $k$.

At $x=1$, $f(1)=1\Rightarrow v=1$: $-e^{-1}(2)=1+k\Rightarrow k=-1-\frac{2}{e}$.

We need $f(a)=0$, i.e. $v=0$ at $x=a$: $-e^0(1)=a+k\Rightarrow -1=a-1-\frac{2}{e}\Rightarrow a=\frac{2}{e}$.

So $ea=2$.