Đáp án đúng: B

Phương pháp giải:
+) Hàm số \(y = {a^x}\;\left( {0 < a
e 1} \right)\) đồng biến trên \(\mathbb{R}\) khi \(a > 1\) và nghịch biến trên \(\mathbb{R}\) khi \(0 < a < 1.\)
Giải chi tiết:Ta có: \(y = {2^{{x^3} - {x^2} + mx + 1}}\) \( \Rightarrow y' = \left( {3{x^2} - 2x + m} \right){2^{{x^3} - {x^2} + mx + 1}}\)
\( \Rightarrow \) Hàm số đã cho đồng biến trên \(\left( {1;\,\,2} \right) \Leftrightarrow y' \ge 0\,\,\forall x \in \left( {1;\,\,2} \right)\)
\(\begin{array}{l} \Leftrightarrow \left( {3{x^2} - 2x + m} \right){2^{{x^3} - {x^2} + mx + 1}} \ge 0\,\,\,\forall x \in \left( {1;\,\,2} \right)\\ \Leftrightarrow 3{x^2} - 2x + m \ge \,0\,\,\,\forall x \in \left( {1;\,\,2} \right)\\ \Leftrightarrow \left[ \begin{array}{l}\Delta ' \le 0\\\left\{ \begin{array}{l}\Delta ' \ge 0\\\left[ \begin{array}{l}{x_1} < {x_2} \le 1\\2 \le {x_1} < {x_2}\end{array} \right.\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\Delta ' \le 0\\\left\{ \begin{array}{l}\Delta ' \ge 0\\\left[ \begin{array}{l}\left\{ \begin{array}{l}{x_1} + {x_2} < 2\\\left( {{x_1} - 1} \right)\left( {{x_2} - 1} \right) \ge 0\end{array} \right.\\\left\{ \begin{array}{l}{x_1} + {x_2} > 4\\\left( {{x_1} - 2} \right)\left( {{x_2} - 2} \right) \ge 0\end{array} \right.\end{array} \right.\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}\Delta ' \le 0\\\left\{ \begin{array}{l}\Delta ' \ge 0\\\left[ \begin{array}{l}\left\{ \begin{array}{l}{x_1} + {x_2} < 2\\{x_1}{x_2} - \left( {{x_1} + {x_2}} \right) + 1 \ge 0\end{array} \right.\\\left\{ \begin{array}{l}{x_1} + {x_2} > 4\\{x_1}{x_2} - 2\left( {{x_1} + {x_2}} \right) + 4 \ge 0\end{array} \right.\end{array} \right.\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}1 - 3m \le 0\\\left\{ \begin{array}{l}1 - 3m \ge 0\\\left[ \begin{array}{l}\frac{2}{3} < 2\\\frac{m}{3} - \frac{2}{3} + 1 \ge 0\end{array} \right.\\\left[ \begin{array}{l}\frac{x}{3} > 4\,\,\,\left( {ktm} \right)\\\frac{m}{3} - \frac{4}{3} + 4 \ge 0\end{array} \right.\end{array} \right.\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}m \ge \frac{1}{3}\\\left\{ \begin{array}{l}m \le \frac{1}{3}\\\frac{m}{3} \ge  - \frac{1}{3}\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}m \ge \frac{1}{3}\\\left\{ \begin{array}{l}m \le \frac{1}{3}\\m \ge  - 1\end{array} \right.\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}m \ge \frac{1}{3}\\ - 1 \le m \le \frac{1}{3}\end{array} \right. \Leftrightarrow m \ge  - 1.\end{array}\)
Chọn B.