JEE Main 2025 — 4 April, Morning Shift — Question 26

$\because f(x)=1-2 x+\int_{0}^{x} e^{x-t} f(t) d t$

or, $f(x)=1-2 x+e^{x} \int_{0}^{x} e^{-t} f(t) d t$

on differentiating both sides w.r.t. $x$ we get

$f^{\prime}(x)=-2+e^{x} \int_{0}^{x} e^{-t} f(t) d t+e^{x} \cdot e^{-x} f(x)$

$f^{\prime}(x)=-2+f(x)+2 x-1+f(x)$ {from eq. (1)}

$\therefore \quad f^{\prime}(x)-2 f(x)=2 x-3$

I.F. $=e^{\int-2 d x}=e^{-2 x}$

$\therefore \quad e^{-2 x} \cdot f(x)=\int e^{-2 x}(2 x-3) d x$

$e^{-2 x} \cdot f(x)=(2 x-3) \cdot \frac{e^{-2 x}}{-2}-\int 2 \cdot \frac{e^{-2 x}}{-2} d x$

$e^{-2 x} \cdot f(x)=\frac{(2 x-3) e^{-2 x}}{-2}+\frac{e^{-2 x}}{-2}+c$ $f(x)=-x+1+c^{\prime} e^{2 x}$

$\because f(x)=1$ from eq. (1)

$\therefore \quad 1=0+1+c^{\prime} \Rightarrow c^{\prime}=0$

$\therefore f(x)=-x+1$

$\Rightarrow$ Area $=\frac{1}{2}$