Đáp án:
2
Giải thích các bước giải:
$$\eqalign{
& y = {1 \over 3}{x^3} + m{x^2} + \left( {{m^2} - 1} \right)x + 2018 \cr
& y' = {x^2} + 2mx + {m^2} - 1 > 0\,\,\forall x \in \left( {0;1} \right) \cr
& \Delta ' = {m^2} - {m^2} + 1 = 1 \cr
& \Rightarrow \left[ \matrix{
{x_1} = - m + 1 \hfill \cr
{x_2} = - m - 1 \hfill \cr} \right. \cr
& BXD: \cr
& - \infty \,\,\,\,\, + \,\,\,\,\,\,{x_2}\,\,\,\,\,\, - \,\,\,\,\,\,\,{x_1}\,\,\,\,\, + \,\,\,\,\, + \infty \cr
& \Rightarrow De\,\,ham\,\,so\,\,DB/\left( {0;1} \right) \cr
& \Rightarrow {x_2} \le 0 < 1 \le {x_1} \cr
& \Leftrightarrow \left\{ \matrix{
- m - 1 \le 0 \hfill \cr
- m + 1 \ge 1 \hfill \cr} \right. \Leftrightarrow \left\{ \matrix{
m \ge - 1 \hfill \cr
m \le 0 \hfill \cr} \right. \Leftrightarrow - 1 \le m \le 0\,\,\left( {tm\,\,m \in \left[ { - 5;5} \right]} \right) \cr
& m \in Z \Rightarrow m \in \left\{ { - 1;0} \right\} \cr
& Vay\,\,co\,\,2\,\,gia\,\,tri\,m\,\,thoa\,\,man. \cr} $$
