Let $f:R\to R$ be given by $f\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-5\right).$ Define $F\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt, x>0.$ Then which of the following options is/are correct?

• Correct answer
  A

  $F$ has a local minimum at $x=1$
• Correct answer
  B

  $F$ has a local maximum at $x=2$
• Correct answer
  C

  $F\left(x\right)\ne 0$ for all $x\in \left(0, 5\right)$
• D

  $F$ has two local maxima and one local minimum in $\left(0, \infty \right)$

Solution

$F\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right).dt$
$F^{\text{'}}\left(x\right)=f\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-5\right)$

$\Rightarrow F\left(x\right)$ has maxima at $x=2$
and minima at $x=1$ and $x=5$
now $F\left(2\right)=\underset{0}{\overset{2}{\int }}\left(x^3-8x^2+17x-10\right).dx$
$={\left[\frac{x^4}{4}-\frac{8x^3}{3}+\frac{17x^2}{2}-10x\right]}_0^2$
$F\left(2\right)=\frac{-10}{3}$ which is a maxima in $x\in \left(0, 5\right)$
$F\left(0\right)=0$

Hence $F\left(x\right)<0\forall x\in \left(0, 5\right)$

Hence, the correct answer is options: 'A', 'B', 'C'.