Let $f(x)=x-1$ and $g(x)=e^x$ for $x \in \mathbb{R}$. If $\frac{d y}{d x}=\left(e^{-2 \sqrt{x}}…

Let $f(x)=x-1$ and $g(x)=e^x$ for $x \in \mathbb{R}$. If $\frac{d y}{d x}=\left(e^{-2 \sqrt{x}} g(f(f(x)))-\frac{y}{\sqrt{x}}\right), y(0)=0$, then $y(1)$ is :-
  1. $\frac{1-e^2}{e^4}$
  2. $\frac{2 e-1}{e^3}$
  3. $\frac{e-1}{e^4}$
  4. $\frac{1-e^3}{e^4}$

Solution

$\begin{aligned}
& f(x)=x-1 \\ & f(f(x))=f(x)-1=x-1-1=x-2 \\ & g(f(f(x)))=e^{x-2} \\ & \therefore \frac{d y}{d x}=e^{-2 \sqrt{x}} \times e^{x-2}-\frac{1}{\sqrt{x}} y \\ & \frac{d y}{d x}+\frac{1}{\sqrt{x}} y=e^{x-2 \sqrt{x}-2} \text { which is L.D.E } \\ & \text { I.F. }=e^{\int \frac{d y}{\sqrt{x}}}=e^{2 \sqrt{x}}
\end{aligned}$
Its solution is
$\begin{aligned}
& y \times e^{2 \sqrt{x}}=\int e^{2 \sqrt{x}} \times e^{x-2 \sqrt{x}-2} d x+c \\ & y \times e^{2 \sqrt{x}}=\int e^{x-2} d x+c \\ & y \times e^{2 \sqrt{x}}=e^{x-2}+c
\end{aligned}$
$\text { Given } \mathrm{x}=0, \mathrm{y}=0 \Rightarrow 0=\mathrm{e}^{-2}+\mathrm{c} \quad ; \mathrm{c}=-\mathrm{e}^{-2}$
$\therefore \mathrm{y} \times \mathrm{e}^{2 \sqrt{\mathrm{x}}}=\mathrm{e}^{\mathrm{x}-2}-\mathrm{e}^{-2}$
$\text { when } \mathrm{x}=1, \mathrm{y} \times \mathrm{e}^2=\mathrm{e}^{-1}-\mathrm{e}^{-2}$
$y=\frac{e^{-1}-e^{-2}}{e^2}=\frac{\frac{1}{e}-\frac{1}{e^2}}{e^2}=\frac{e^2-e}{e^5}=\frac{e-1}{e^4}$
Option (1) is correct

Asked in: JEE Main 2025 (08 Apr Shift 2)

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