Ta có:\[y = {2^{{x^3} - {x^2} + mx + 1}} \Rightarrow y' = \left( {3{x^2} - 2x + m} \right){2^{{x^3} - {x^2} + mx + 1}}\]

⇒ Hàm số đã cho đồng biến trên\[\left( {1;\,\,2} \right) \Leftrightarrow y' \ge 0\,\,\forall x \in \left( {1;\,\,2} \right)\]

\[ \Leftrightarrow (3{x^2} - 2x + m){2^{{x^3} - x2 + mx + 1}} \ge 0\forall x \in (1;2)\]

\[ \Leftrightarrow 3{x^2} - 2x + m \ge 0\forall x \in (1;2)\]

\( \Leftrightarrow {\left[ {\begin{array}{*{20}{c}}{\Delta \prime \le 0}\\{\left\{ {\begin{array}{*{20}{c}}{\Delta \prime \ge 0}\\{\left[ {\begin{array}{*{20}{c}}{{x_1} < {x_2} \le 1}\\{2 \le {x_1} < x2}\end{array}} \right.}\end{array}} \right.}\end{array}} \right._{}} \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\Delta \prime \le 0}\\{\left\{ {\begin{array}{*{20}{c}}{\Delta \prime \ge 0}\\{\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{{x_1} + {x_2} < 2}\\{({x_1} - 1)({x_2} - 1) \ge 0}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{{x_1} + {x_2} > 4}\\{({x_1} - 1)({x_2} - 1) \ge 0}\end{array}} \right.}\end{array}} \right.}\end{array}} \right.}\end{array}} \right.\)</>

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{\Delta \prime \le 0}\\{\left\{ {\begin{array}{*{20}{c}}{\Delta \prime \ge 0}\\{\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{{x_1} + {x_2} < 2}\\{{x_1}{x_2} - ({x_1} + {x_2}) + 1 \ge 0}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{{x_1} + {x_2} > 4}\\{{x_1}{x_2} - ({x_1} + {x_2}) + 1 \ge 0}\end{array}} \right.}\end{array}} \right.}\end{array}} \right.}\end{array}} \right.\)</>

\( \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{1 - 3m \le 0}\\{\left\{ {\begin{array}{*{20}{c}}{1 - 3m \ge 0}\\{\left[ {\begin{array}{*{20}{c}}{\frac{2}{3} < 2}\\{\frac{m}{3} - \frac{2}{3} + 1 \ge 0}\end{array}} \right.}\\{\left[ {\begin{array}{*{20}{c}}{\frac{x}{3} > 4(ktm)}\\{\frac{m}{3} - \frac{4}{3} + 4 \ge 0}\end{array}} \right.}\end{array}} \right.}\end{array}} \right.\)</>

\( \Rightarrow \left[ {\begin{array}{*{20}{c}}{m \ge \frac{1}{3}}\\{\left\{ {\begin{array}{*{20}{c}}{m \le \frac{1}{3}}\\{\frac{m}{3} \ge - \frac{1}{3}}\end{array}} \right.}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{m \ge \frac{1}{3}}\\{\left\{ {\begin{array}{*{20}{c}}{m \le \frac{1}{3}}\\{m \ge - 1}\end{array}} \right.}\end{array}} \right. \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{m \ge \frac{1}{3}}\\{ - 1 \le m \le \frac{1}{3}}\end{array} \Leftrightarrow m \ge - 1.} \right.\)

Đáp án cần chọn là: B

Hàm số y = 2^ ln x + x^2 có đạo hàm là