Question

Let $f:R\to R$ be given by$f\left(x\right)=(x-1)(x-2)(x-5).$ 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?

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

Answer 1

Answer: (A), (B), (D)

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

$f\left(x\right)=(x-1)(x-2)(x-5)$
$F\left(x\right)=\underset{0}{\overset{x}{\int }}f\left(t\right)dt,x>0$
$F^{\prime }\left(x\right)=f\left(x\right)=x^3-8x^2+17x-10$
$F^{\prime \prime }\left(x\right)=3x^2-16x+17$
For $x=1,F^{\prime \prime }\left(x\right)>0$
Therefore $F$ has a local minimum at $x=1.$
For $x=2,F^{\prime \prime }\left(x\right)<0$
Therefore $F$ has a local maximum at $x=2.$
For $x=5,F^{\prime \prime }\left(x\right)>0$
Therefore $F$ has a local minimum at $x=5.$

$F\left(x\right)=\frac{x^4}{4}-\frac{8x^3}{3}+\frac{17x^2}{2}-10x$
From the graph of $y=F\left(x\right),$
we can easily say $F\left(x\right)\ne 0\forall x\in (0,5).$