Đáp án A

Ta có \[g'\left( x \right) = 2f'\left( {x + \frac{m}{2}} \right) - 2x - m = 2\left[ {f'\left( {x + \frac{m}{2}} \right) - \left( {x + \frac{m}{2}} \right)} \right]\].

Đặt \[t = x + \frac{m}{2}\] thì \[g'\left( t \right) < 0 \Leftrightarrow f'\left( t \right) < t \Leftrightarrow \left[ \begin{array}{l}t < - 3\\2 < t < 5\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x + \frac{m}{2} < - 3\\2 < x + \frac{m}{2} < 5\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x < - 3 - \frac{m}{2}\\2 - \frac{m}{2} < x < 5 - \frac{m}{2}\end{array} \right..\]

Giả thiết bài toán thỏa mãn khi \[\left[ \begin{array}{l} - 3 - \frac{m}{2} \ge 4\\2 - \frac{m}{2} \le 3 < 4 \le 5 - \frac{m}{2}\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}m \le - 14\\ - 2 \le m \le 12\end{array} \right.\].

Kết hợp điều kiện \[m \in \mathbb{Z},m \in \left[ { - 15;15} \right]\] suy ra \[m = \left\{ { - 14; - 15; - 2; - 1;0;1;2} \right\}\]