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JEE Main 29-Jan-2024 Paper - Shift 2 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 29-Jan-2024 Paper - Shift 24 Marks
Question
Let $f(x)=\sqrt{\lim _{r \rightarrow x}\left\{\frac{2 r^2\left[(f(r))^2-f(x) f(r)\right]}{r^2-x^2}-r^3 e^{\frac{f(r)}{\tau}}\right\}}$ be differentiable in $(-\infty, 0) \cup(0, \infty)$ and $f(1)=1$. Then the value of ea, such that $f(a)=0$, is equal to $........ .$
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Answer

$ (a)\ 2$
$f(1)=1, f(a)=0$
$f^2(x)=\operatorname{Lim}_{r \rightarrow x}\left(\frac{2 r^2\left(f^2(r)-f(x) f(r)\right)}{r^2-x^2}-r^3 e^{\frac{f(r)}{r}}\right)$
$=\operatorname{Lim}_{r \rightarrow x}\left(\frac{2 r^2 f(r)}{r+x} \frac{(f(r)-f(x))}{r-x}-r^3 e^{\frac{f(r)}{r}}\right)$
$f^2(x)=\frac{2 x^2 f(x)}{2 x} f^{\prime}(x)-x^3 e^{\frac{f(x)}{x}}$
$y^2=x y \frac{d y}{d x}-x^3 e^{\frac{y}{x}}$
$\frac{y}{x}=\frac{d y}{d x}-\frac{x^2}{y} e^{\frac{y}{x}}$
Put $y=v x $
$\Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$
$v=v+x \frac{d v}{d x}-\frac{x}{v} e^v$
$\frac{d v}{d x}=\frac{e^v}{v}$
$ \Rightarrow e^{-v} v d v=d x$
Integrating both side
$e^{v}(x+c)+l+v=0$
$f(l)=1 $
$\Rightarrow x=1, y=1$
$\Rightarrow c =-1-\frac{2}{ e }$
$e ^{ v }\left(-1-\frac{2}{ e }+ x \right)+1+ v =0$
$e ^{\frac{ y }{ x }}\left(-1-\frac{2}{ e }+ x \right)+1+\frac{ y }{ x }=0$
$x = a , y =0 $
$\Rightarrow a =\frac{2}{ e }$
$ae =2$

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