Let $f:[0,\infty) \to \mathbb{R}$ be a continuous function such that $f(x) = 1 - 2x + \int_0^x e^{x-t} f(t)\,dt$ for all $x \in [0,\infty)$. Then, which of the following statement(s) is(are) TRUE?

Let f:[0,∞) → R be a continuous function such that f(x) = 1 - 2x + ∫₀ˣ eˣ⁻ᵗ f(t) dt for all x ∈ [0,∞). Then, which of the following statement(s) is(are) TRUE? $\text{Given: } f(x) = 1 - 2x + \displaystyle\int_0^x e^{x-t} f(t)\,dt,\quad f \text{ continuous on } [0,\infty)$ $\text{To find: which of (1)--(4) are TRUE}$ $1.\ \text{differentiate to get an ODE}$ $2.\ \text{solve for } f(x)$ $3.\ \text{check points, then the area}$ $f(x) = 1 - 2x + e^x \displaystyle\int_0^x e^{-t} f(t)\,dt$ $f'(x) = -2 + e^x\displaystyle\int_0^x e^{-t}f(t)\,dt + e^x\cdot e^{-x}f(x)$ $f'(x) = -2 + (f(x)-1+2x) + f(x) = 2f(x) + 2x - 3$ $f(0) = 1$ $f' - 2f = 2x - 3,\qquad \text{IF} = e^{-2x}$ $\dfrac{d}{dx}\left(f e^{-2x}\right) = (2x-3)e^{-2x}$ $f e^{-2x} = e^{-2x}(1-x) + C$ $x=0:\ 1 = 1 + C \Rightarrow C = 0$ $f(x) = 1 - x$ $f(1) = 0 \neq 2 \Rightarrow \text{(1) FALSE}$ $f(2) = -1 \Rightarrow \text{(2) TRUE}$ $A = \displaystyle\int_0^1 \left(\sqrt{1-x^2} - (1-x)\right)dx$ $A = \dfrac{\pi}{4} - \dfrac12 = \dfrac{\pi-2}{4} \Rightarrow \text{(3) TRUE},\ \text{(4) FALSE}$ $\boxed{\text{Answer: (2) and (3)}}$