JEE 2025 — Mathematics PYQ

Let $f(x)=\frac{1}{e^{x-1}}=e^{1-x}$
If $f(x)=1\Rightarrow 1-x=0\Rightarrow x=1$
If $f(x)=2\Rightarrow e^{1-x}=2\Rightarrow 1-x=\ln 2\Rightarrow x=1-\ln 2$
$f(0)=e^1=2.71$
$f(e^3)=e^{1-e^3}\in (0,1)$
$I={\int }_0^{1-\ln 2}[e^{1-x}]dx+{\int }_{1-\ln 2}^1[e^{1-x}]dx+{\int }_1^{e^3}[e^{1-x}]dx$

$I={\int }_0^{e^3}\left[\frac{1}{e^{x-1}}\right]dx=\alpha -{\log }_{e}2$
Let $f(x)=e^{1-x}$
$I={\int }_0^{1-\ln 2}2dx+{\int }_{1-\ln 2}^{1}1dx+{\int }_1^{e^3}0dx$
$I=2(1-\ln 2)+1(\ln 2)=2-\ln 2$
$\alpha -\ln 2=2-\ln 2\Rightarrow \alpha =2$
${\alpha }^3=8$