Đáp án D

HD: Ta có \(y' = 2\cos x - \sin x + {x^2} - 2\left( {m + 2} \right)x + {m^2} + 4m + 7 \ge 0,\forall x \in R\)
\( \Leftrightarrow - m \le 4 + \cos x + \dfrac{1}{2}\left( {2{{\cos }^2}x - 1} \right) + \dfrac{1}{3}\left( {4{{\cos }^3}x - 3\cos x} \right)\)
\( \Rightarrow - m \le \dfrac{4}{3}{\cos ^3}x + {\cos ^2}x + \dfrac{7}{2} = \dfrac{4}{3}{t^3} + {t^2} + \dfrac{7}{2} = f\left( t \right);{\rm{ }}t = \cos x \in \left[ { - 1;1} \right]\)
\(f'\left( t \right) = 4{t^2} + 2t = 0 \Rightarrow \left[ \begin{array}{l}
t = 0\\
t = - \dfrac{1}{2}
\end{array} \right.\)
\(f\left( { - 1} \right) = \dfrac{{19}}{6};f\left( 1 \right) = \dfrac{{35}}{6};f\left( 0 \right) = \dfrac{7}{2};f\left( { - \dfrac{1}{2}} \right) = \dfrac{{43}}{{12}} \Rightarrow - m \le \dfrac{{19}}{6} \Leftrightarrow m \ge - \dfrac{{19}}{6} \Rightarrow m \in \left\{ { - 3; - 2; - 1} \right\}.\) Chọn D