Question

If all the real values of $m$ for which the function $f(x)=\frac{x^{3}}{3}-(m-3) \frac{x^{2}}{2}+m x-2018$ is strictly increasing in $\mathrm{x} \in[0, \infty)$ is $[0, \mathrm{k}]$, then find the value of $\mathrm{k}$.

Answer 1

$\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}^{3}}{3}-(\mathrm{m}-3) \frac{\mathrm{x}^{2}}{2}+\mathrm{mx}-2018$
$\therefore \mathrm{f}^{\prime}(\mathrm{x})=\left(\mathrm{x}^{2}-(\mathrm{m}-3) \mathrm{x}+\mathrm{m}\right) \geq 0, \forall \mathrm{x} \in[0, \infty)$
Case-I: When $\mathrm{D} \leq 0 \Rightarrow \mathrm{m} \in[1,9]$
Case-II: When $\mathrm{D} \geq 0 \Rightarrow \mathrm{m} \in[-\infty, 1] \cup[9, \infty)$
$\frac{-\mathrm{b}}{2 \mathrm{a}} \leq 0 \Rightarrow (\mathrm{m}-3) \leq 0 \Rightarrow \mathrm{m} \leq 3$
and $ \mathrm{f}^{\prime}(0) \geq 0 \Rightarrow \mathrm{m} \geq 0$
$\therefore (1) \cap($ ii $) \cap($ iii $)$
$\Rightarrow \mathrm{m} \in[0,1]$
So, finally (I) $\cap$ (II)
$\Rightarrow \mathrm{m} \in[0,9] \equiv[0, \mathrm{k}]$
$\therefore \mathrm{k}=9$