Đáp án:
a)
$\begin{array}{l}
y = \sin 2x - 4\sin x + mx\\
\Rightarrow y' = 2\cos 2x - 4\cos x + m \ge 0\forall x \in \left( {0;\pi } \right)\\
\Rightarrow 2\cos 2x - 4\cos x \ge - m\forall x \in \left( {0;\pi } \right)\\
\Rightarrow g\left( x \right) = 2\cos 2x - 4\cos x \ge - m\forall x \in \left( {0;\pi } \right)\\
g'\left( x \right) = - 4\sin 2x + 4\sin x = 0\\
\Rightarrow {\mathop{\rm sinx}\nolimits} = 0/c{\rm{osx = }}\frac{1}{2} \Rightarrow x = \frac{\pi }{3}\\
\Rightarrow \min {\rm{g}}\left( x \right) = g\left( {\frac{\pi }{3}} \right) = - 3 \ge - m\\
\Rightarrow m \ge 3\\
b){\min _{f\left( x \right)}} \le 2\forall x \in \left[ {1;3} \right]\\
f\left( x \right) = {x^3} - mx + 18\\
\Rightarrow f'\left( x \right) = 3{x^2} - m = 0\\
+ m < 0 \Rightarrow f'\left( x \right) > 0\forall x\\
\Rightarrow {\min _{f\left( x \right)}} = f\left( 1 \right) = 19 - m \le 2\\
\Rightarrow m \ge 17\left( {ktm} \right)\\
+ m > 0 \Rightarrow f\left( x \right) = 0 \Rightarrow x = \pm \sqrt {\frac{m}{3}}
\end{array}$
Thử các TH đều kko có m thỏa mãn
Vậy ko có m thỏa mãn đề bài
