$F(x)=\int_0^x (t-2)f(t)\, dt$ with $f(0)=1$, $f(1)=0$ has an extremum in $(0,3)$? 
$\displaystyle F(x)=\int_0^x (t-2)f(t)\; dt$ with $f(0)=1$, $f(1)=0$ has an extremum in $(0,3)$?

The title explains a lot. Given
$$
\displaystyle F(x)=\int_0^x (t-2)f(t)\,dt
$$
with $f(0)=1$, $f(1)=0$ and $f:\mathbb{R}\to\mathbb{R}$ is a strictly decreasing differentiable function. Then:
1) $F$ is strictly increasing in $[0,3]$
2) $F$ has a unique maximum but no minimum in $(0,3)$
3) $F$ has a unique minimum but no maximum in $(0,3)$
4) $F$ has both maximum and minimum in $(0,3)$
I have done $F''(x)=(x-2)f'(x)+f(x)$ but no idea how to verify the options. Any help is appreciated. 
 A: $$F'(x)=(x-2)f(x)$$
$f$ is strictly decreasing and differentiable (thus continuous). So $f(x) > 0$ on $[0,1)$, $f(1)=0$, and $f(x) <0$ on $(1,3]$.
$F'(x)<0$ on $[0,1)$, $F'(1) =0$, $F'(x) >0$ on $(1,2)$, $F'(2)=0$, and $F'(x)<0$ on $(2,3]$
So $1$ is the minimum point, $2$ is the maximum point. $F(x)$ first decreases, and then reaches local minimum, then increases, then reaches local maximum, and then decreases again. $F''(1) >0$ and $F''(2) < 0$ also confirms this.
A: Since $F'(x) = (x-2)f(x)$, then if $F'(x) = 0$, it implies that either $x = 2$ or $f(x) = 0$. We also know that $f$ is strictly decreasing, so $f$ has at most one root. Given that $f(1) = 0$, then the only solution for $f(x) = 0$ is $x = 1$. Hence $F'(x) = 0$ implies that either $x = 1$ or $x = 2$.
Now, on $(0,3)$, we divide into three cases:


*

*If $0 < x < 1$, then $x-2<0$ and $f(x) > 0$, so $F'(x) < 0$. [Note: $f(x) > 0$ since $f$ is strictly decreasing and $f(0) = 1$ and $f(1) = 0$]

*If $1 < x < 2$, then $x-2<0$ and $f(x) < 0$, so $F'(x) > 0$.

*If $2 < x < 3$, then $x-2>0$ and $f(x) < 0$, so $F'(x) < 0$.


Hence, $F$ cannot be possibly strictly increasing on $(0,3)$. $F$ has a minimum when $x = 1$ and maximum when $x = 2$. The only true statement is (4).
