Phương pháp giải:

Lời giải chi tiết:

\(\begin{array}{l}g\left( x \right) = {\left[ {f\left( x \right)} \right]^3} - 3{\left[ {f\left( x \right)} \right]^2} + 2\\g'\left( x \right) = 3{\left[ {f\left( x \right)} \right]^2}.f'\left( x \right) - 6f\left( x \right).f'\left( x \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\, = 3f\left( x \right).f'\left( x \right)\left[ {f\left( x \right) - 2} \right]\\\,\,\,\,\,\,\,\,\,\,\,\,\, = 3\left( {\dfrac{1}{3}{x^3} - 2{x^2} + mx + m - 2} \right)\left( {{x^2} - 4x + m} \right)\left( {\dfrac{1}{3}{x^3} - 2{x^2} + mx + m - 4} \right) \ge 0\,\,\forall x < 0\end{array}\)

TH1: \(\left\{ \begin{array}{l}{x^2} - 4x + m \le 0\\\left( {\dfrac{1}{3}{x^3} - 2{x^2} + mx + m - 2} \right)\left( {\dfrac{1}{3}{x^3} - 2{x^2} + mx + m - 4} \right) \le 0\end{array} \right.\,\,\forall x < 0\)

Trường hợp này vô nghiệm vì \(\mathop {\lim }\limits_{x \to - \infty } \left( {{x^2} - 4x + m} \right) = + \infty \) nên không thể đúng với mọi \(x < 0\).

TH2: \(\left\{ \begin{array}{l}{x^2} - 4x + m \ge 0\\\left( {\dfrac{1}{3}{x^3} - 2{x^2} + mx + m - 2} \right)\left( {\dfrac{1}{3}{x^3} - 2{x^2} + mx + m - 4} \right) \ge 0\end{array} \right.\,\,\forall x < 0\)

\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l}{x^2} - 4x + m \ge 0\\\left[ \begin{array}{l}\dfrac{1}{3}{x^3} - 2{x^2} + mx + m - 2 \le 0\\\dfrac{1}{3}{x^3} - 2{x^2} + mx + m - 2 \ge 2\end{array} \right.\end{array} \right.\,\,\,\forall x < 0\\ \Leftrightarrow \left\{ \begin{array}{l}{x^2} - 4x + m \ge 0\,\,\,\,\left( 1 \right)\\\dfrac{1}{3}{x^3} - 2{x^2} + mx + m - 2 \le 0\,\,\,\left( 2 \right)\end{array} \right.\,\,\,\forall x < 0\end{array}\)

Ta có: \(f\left( x \right) = \dfrac{1}{3}{x^3} - 2{x^2} + mx + m - 2\) \( \Rightarrow f'\left( x \right) = {x^2} - 4x + m \ge 0\,\,\forall x < 0\)

\( \Rightarrow \left( 2 \right) \Leftrightarrow f\left( 0 \right) \le 0 \Leftrightarrow m - 2 \le 0 \Leftrightarrow m \le 2\).

Và

\(\begin{array}{l}\left( 1 \right) \Leftrightarrow m \ge g\left( x \right) = - {x^2} + 4x\,\,\forall x < 0\\ \Leftrightarrow m \ge g\left( 0 \right) = 0\end{array}\)

Vậy \(0 \le m \le 2\), mà \(m \in \mathbb{Z}\) nên \(m \in \left\{ {0;1;2} \right\}\).

Chọn B.