Question

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

Answer 1

Answer: 9

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