Let f : R → R be given by f x = x - 1 x - 2 x - 5 . Define F x = ∫ 0 x f t d t ,   x >…

1. $F$ has a local minimum at $x=1$
2. $F$ has a local maximum at $x=2$
3. $F\left(x\right)\ne 0$ for all $x\in \left(0, 5\right)$
4. $F$ 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 $F\left(x\right)<0\forall x\in \left(0, 5\right)$

Asked in: JEE Advanced 2019 (Paper 2)

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