Question

Let f : ℝ $/to$ ℝ be given by f(x) = (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 $x\in \left(0,5\right)$

Answer 1

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

$f(x)=(x-1)(x-2)(x-5)$
$f(x)=$ $\underset{0}{\overset{x}{\int }}$ $f(t)dt,x>0$
$F^{\prime }\left(x\right)=f\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-5\right),x>0$
clearly F(x) has local minimum at x = 1,5
$F(x)$ has local maximum at $x=2$
$f\left(x\right)=x^3-8x^2+17x-10$
$\Rightarrow F\left(x\right)=$$\underset{0}{\overset{x}{\int }}$$\left(t^3-8t^2+17t-10\right)dt$
$F\left(x\right)=\frac{x^4}{4}-\frac{8x^3}{3}+\frac{17x^2}{2}-10x$
from the graph of $y=F(x)$, clearly $F\left(x\right)\ne 0\forall x\in \left(0.5\right)$