JEE 2024 — Mathematics PYQ

\begin{aligned}
 & f(x)=\sqrt{\operatorname*{lim}_{r\to x}\frac{2r^{2}[(f(x))^{2}-f(x).f(x)]}{r^{2}-x^{2}}-r^{3}e^{\frac{f(z)}{r}}} \\
 & =\sqrt{\lim_{r\to x}\frac{2r^{2}f(x)[f(x)-f(x)]}{(r-x)-(r+x)}-\lim_{x\to\infty}r^{3}e^{\frac{r(x)}{r}}} \\
 & =\sqrt{\lim_{x\to\infty}\frac{2r^{2}-f(x)}{r+x}.\left(\frac{f(r)-f(x)}{r-x}\right)-x^{3}e^{\frac{f(x)}{x}}} \\
 & =\sqrt{\frac{2x^{2}.f(x)}{2x}.f^{1}(x)-x^{3}e\frac{f(x)}{x}} \\
\text{Squaring both sides} \\
f^{2}(x) & =x.f(x).f^{1}(x)-x^{3}e^{\frac{f(x)}{x}} \\
y^{2} & =xy\frac{dy}{dx}-x^{3}e^{\frac{y}{x}} \\
x^{3}e^{x}dx & =xydy-y^{2}dx \\
x^{3}dx & =y\frac{(xdy-ydx)}{e^{\frac{y}{x}}} \\
x^{3}dx & =y.(xdy-ydx).e^{-\frac{y}{x}} \\
\frac{x}{y}dx & =\frac{(xdy-ydx)}{xz}.e-\frac{y}{x}....(\mathbf{i}) \\
e^{\frac{y}{x}} & =t
\end{aligned}

$\Rightarrow -\frac{y}{x}=lnt$

$e^{\frac{y}{x}}(-1)\left(\frac{xdy-ydx}{x^2}\right)=dt$

$e^{\frac{y}{x}}\left(\frac{xdy-ydx}{x^2}\right)=-dt$

Now eqn (i) becomes

$\frac{dx}{lnt}=dt$

$dx=lnt\cdot dt$

$\int dx=\int lntdt$

$x=[lnt-1]+C$

\begin{aligned}
 & \mathrm{,} & & c=e^{-\frac{y}{x}}\left(-\frac{y}{x}-1\right)+\mathrm{C} \\
 & x+\left(1+{\frac{y}{x}}\right)e^{-{\frac{y}{x}}} & & \mathrm{=C} \\
 & f(1) & & =1 \\
 & x & & =1 \\
 & \mathrm{y} & & =1 \\
 & 
\begin{array}
{l}{\mathrm{isfies~eqn~(i)}} \\
{1+(1+1)\mathrm{e}^{-1}}
\end{array} & & \mathrm{=C} \\
 & \text{一} \\
 & \mathrm{Now}\quad f(a) & & =0 \\
 & \mathrm{x} & & =a \\
 & \mathrm{y} & & =0 \\
 & a+(1+0)\mathrm{e}^{-0} & & \mathrm{=C} \\
 & a+1 & & =\mathbb{C} \\
 & & & z=C-1 \\
 & & & a=1+2\mathrm{e}^{-1}-1 \\
 & & & a=2\mathrm{e}^{-1} \\
 & & & a\mathbf{e}=2
\end{aligned}