Define f(x)=∫0^x(t-1)(t-2) e^-t dt for x∈mathbbR.

Integrating by parts, $\int(t^2-3t+2)e^{-t}dt=-e^{-t}(t^2-t+1)$, so $f(x)=1-e^{-x}(x^2-x+1)$. (A) $f'(x)=(x-1)(x-2)e^{-x}$ changes sign from $+$ to $-$ at $1$ and from $-$ to $+$ at $2$; correct. (B) As $x\to-\infty$, $e^{-x}(x^2-x+1)\to\infty$, so $f\to-\infty$ and $f$ is unbounded; incorrect. (C) $f(1)=1-e^{-1}\approx0.632$ and $f(2)=1-3e^{-2}\approx0.594$, with $f(0)=0$ and $f\to1$. The level $0.6$ lies strictly between $f(2)$ and $f(1)$, so it is met once on $(0,1)$, once on $(1,2)$ and once on $(2,\infty)$: three roots; correct. (D) $0.5<f(2)$, so $0.5$ is reached only once (on $(0,1)$); incorrect. Answer: A and C.