Lời giải:Xét \(g\left( x \right)={{x}^{3}}-3{{x}^{2}}+4x+a\) trên \(\left[ 0;2 \right]\).
Ta có \(g'\left( x \right)=3{{x}^{2}}-6x+4>0,\forall x\in \left[ 0;2 \right]\). Suy ra \(g\left( x \right)\in \left[ a;a+4 \right],\forall x\in \left[ 0;2 \right]\).
TH1: Nếu a>0 thì \(M=a+4;\,\,m=a\).
Từ gt: \(M\le 2m \Leftrightarrow a+4\le 2a\Leftrightarrow a\ge 4\). Vì \(\left\{ \begin{align} & a\in \mathbb{Z} \\ & a\in \left[ -10;10 \right] \\ \end{align} \right.\) nên \(a\in \left\{ 4;5;6;7;8;9;10 \right\}\).
TH2: Nếu a<-4 thì \(M=-a;\,\,m=-a-4\).
Từ gt \(M\le 2m \Leftrightarrow -a\le 2\left( -a-4 \right)\Leftrightarrow a\le -8\). Vì \(\left\{ \begin{align} & a\in \mathbb{Z} \\ & a\in \left[ -10;10 \right] \\ \end{align} \right.\) nên \(a\in \left\{ -10;-9;-8 \right\}\).
TH3: Nếu \(-4\le a\le 0\) thì \(M=m\text{ax}\left\{ \left| a \right|;\left| a+4 \right| \right\};\,\,m=0\).
\(M=m\text{ax}\left\{ \left| a \right|;\left| a+4 \right| \right\}=m\text{ax}\left\{ \left| -a \right|;\left| a+4 \right| \right\}\,\ge \frac{\left| -a \right|+\left| a+4 \right|}{2}\ge \frac{-a+a+4}{2}=2>m=0.\)
Vậy \(a\in \left\{ -10;-9;-8;4;5;6;7;8;9;10 \right\}\).
