Đáp án: $ - 5 < m < - \dfrac{1}{3}$
Giải thích các bước giải:
$\begin{array}{l}
y = \left( {2 - m} \right).x - \left( {2m + 3} \right).\cos x\\
\Leftrightarrow y' = \left( {2 - m} \right) + \left( {2m + 3} \right).\sin x\\
Khi:y' > 0\forall x\\
\Leftrightarrow \left( {2 - m} \right) + \left( {2m + 3} \right).\sin x > 0\\
\Leftrightarrow \left( {2m + 3} \right).\sin \,x > m - 2\\
+ khi:m = - \dfrac{3}{2} \Leftrightarrow 0 > - \dfrac{7}{2}\left( {tm} \right)\\
+ Khi:m > - \dfrac{3}{2}\\
\Leftrightarrow \sin x > \dfrac{{m - 2}}{{2m + 3}}\\
\Leftrightarrow \dfrac{{m - 2}}{{2m + 3}} < - 1\\
\Leftrightarrow \dfrac{{m - 2 + 2m + 3}}{{2m + 3}} < 0\\
\Leftrightarrow \dfrac{{3m + 1}}{{2m + 3}} < 0\\
\Leftrightarrow \dfrac{{ - 3}}{2} < m < - \dfrac{1}{3}\\
+ Khi:m < - \dfrac{3}{2}\\
\Leftrightarrow \sin \,x < \dfrac{{m - 2}}{{2m + 3}}\\
\Leftrightarrow \dfrac{{m - 2}}{{2m + 3}} > 1\\
\Leftrightarrow \dfrac{{m - 2 - 2m - 3}}{{2m + 3}} > 0\\
\Leftrightarrow \dfrac{{ - m - 5}}{{2m + 3}} > 0\\
\Leftrightarrow \dfrac{{m + 5}}{{2m + 3}} < 0\\
\Leftrightarrow - 5 < m < - \dfrac{3}{2}\\
Vậy\, - 5 < m < - \dfrac{1}{3}
\end{array}$
