Lời giải chi tiết:

\(\begin{array}{l}y = 2{x^3} + 3{x^2} - 12x + 2 \Rightarrow y' = 6{x^2} + 6x - 12 = 0 \Leftrightarrow \left[ \begin{array}{l}x = 1 \in \left[ { - 1;2} \right]\\x =  - 2 \notin \left[ { - 1;2} \right]\end{array} \right.\\f\left( 1 \right) =  - 5;f\left( { - 1} \right) = 15;f\left( 2 \right) = 6 \Rightarrow \left\{ \begin{array}{l}\mathop {Min}\limits_{\left[ { - 1;2} \right]} y =  - 5 = m\\\mathop {{\rm{Max}}}\limits_{\left[ { - 1;2} \right]}  = 15 = M\end{array} \right. \Rightarrow \frac{M}{m} =  - 3\end{array}\)

Chọn: B