Question

Let $F\left(x\right)=-2+{\int }_0^x(t-1)(t-2{)}^{2}dt$

Answer 1

$F^{\prime }\left(x\right)=(x-1)(x-2{)}^2$.
$F$ is continuous and differentiable.
$F^{\prime }\left(x\right)=0\Rightarrow x=1,x=2$
For $x<1,F^{\prime }\left(x\right)<0$ and for $x>1,F^{\prime }\left(x\right)>0$
At $x=2,F^{\prime }\left(x\right)$ does not change sign, so there is no extremum at $x=2$
$x=1$ is point of minimum
Hence, A - 1
$F\left(x\right)$ increases on $(1,\infty )$
Hence, B - 3

$F\left(x\right)$ decreases on $(-\infty ,-1)$
Hence, C - 2

The minimum value of $F$ is $F\left(1\right)=-2+{\int }_0^1(t-1)(t-2{)}^{2}dt=-2+{\int }_{-1}^{0}t(t-1{)}^{2}dt$
$=\frac{-41}{12}\in (-4,1)$
Hence , D - 4