JEE 2025 — Mathematics PYQ

$f^{\prime }(x)=f(x)$

Multiplying $f(x)$ on both sides, we get
$f^{\prime \prime }(x)f(x)=(f(x){)}^2{f}^{\prime }(x)$

On integration both sides, we get
$\int f^{\prime \prime }(x)f^{\prime }(x)dx=\int (f(x){)}^2{f}^{\prime }(x)dx$

$\Rightarrow \frac{(f^{\prime }(x){)}^2}{2}=\frac{(f(x){)}^2}{2}+C$

C is the constant of integration,

Take $x=0$

$\frac{f^{\prime }(0{)}^2}{2}=\frac{(f(0){)}^2}{2}+C$

$\frac{9}{2}=C$

$(f(x){)}^2=(f(x){)}^2+9$

$f(x)=\sqrt{(f(x){)}^2+9}$ $\therefore f(x)\ge 0$

$\int dx\Rightarrow \ln y+\sqrt{y^2+9}=x+C$ {Let $f(x)=y$}

$\Rightarrow f(0)=0\Rightarrow C=\ln 3$

$\Rightarrow y+\sqrt{y^2+9}=3e^x$

at

$x=\ln 3;y=4$

$\therefore 9(\ln 3)=36$