Question

Let $f:R\to R$be given by $f\left(x\right)=(x-1)(x-2)(x-5)$. Define $F\left(x\right)={\int }_0^{x}f\left(t\right)dt,x>0$. Then, which of the following options is/are correct?

• A. $F$ has a local minimum at $x=1$
• B. $F$ has a local maximum at $x=2$
• C. $F\left(x\right)\ne 0\forall x\in (0,5)$
• D. $F$ has two local maxima and one local minimum in $(0,\infty )$

Answer 1

Answer: (A), (B), (C)

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

Determine the correct option

Given that:

$f\left(x\right)=(x-1)(x-2)(x-5)$

$$\begin{gathered} F\left(x\right)={\int }_0^{x}f\left(t\right)dt,x>0 \\ \Rightarrow F\left(x\right)={\int }_0^x(t-1)(t-2)(t-5)dt \\ \Rightarrow F\left(x\right)={\int }_0^x\left(t^3-3t^2+17t-10\right)dt \\ \Rightarrow F\left(x\right)=\frac{t^4}{4}-t^3+17\frac{t^2}{2}-10t \\ \Rightarrow F\left(x\right)=\frac{x^4}{4}-x^3+17\frac{x^2}{2}-10x \\ whenx\in \left[0,5\right], \\ F\left(x\right)\ne 0 \end{gathered}$$

Therefore, Option (C) is correct.

Now, finding the local maxima and minima:

At $x=1,F\left(x\right)$will be negative.

After that alternate sign inversion will follow.

Therefore, $F\left(x\right)$ has two minima at $x=1$and $x=5$

$F\left(x\right)$ has a maxima at $x=2$.

Hence, Option (A), (B) and (C) are correct.