Đáp án đúng: D

Phương pháp giải:

- Đặt \(g\left( x \right) = {x^3} - 3{x^2} + m\), tìm \(\mathop {\min }\limits_{\left[ {1;3} \right]} g\left( x \right);\,\,\mathop {\max }\limits_{\left[ {1;3} \right]} g\left( x \right)\).- Suy ra \(\mathop {\min }\limits_{\left[ {1;3} \right]} f\left( x \right) = \left\{ {\left| {\mathop {\min }\limits_{\left[ {1;3} \right]} g\left( x \right)} \right|;\,\,\left| {\mathop {\max }\limits_{\left[ {1;3} \right]} g\left( x \right)} \right|} \right\}\).- Xét các TH:+ TH1: \(\mathop {\min }\limits_{\left[ {1;3} \right]} f\left( x \right) = \left| {\mathop {\min }\limits_{\left[ {1;3} \right]} g\left( x \right)} \right| \Leftrightarrow \left\{ \begin{array}{l}\left| {\mathop {\min }\limits_{\left[ {1;3} \right]} g\left( x \right)} \right| \le 3\\\left| {\mathop {\min }\limits_{\left[ {1;3} \right]} g\left( x \right)} \right| \le \left| {\mathop {\max }\limits_{\left[ {1;3} \right]} g\left( x \right)} \right|\end{array} \right.\).+ TH2: \(\mathop {\min }\limits_{\left[ {1;3} \right]} f\left( x \right) = \left| {\mathop {\max }\limits_{\left[ {1;3} \right]} g\left( x \right)} \right| \Leftrightarrow \left\{ \begin{array}{l}\left| {\mathop {\max }\limits_{\left[ {1;3} \right]} g\left( x \right)} \right| \le 3\\\left| {\mathop {\max }\limits_{\left[ {1;3} \right]} g\left( x \right)} \right| \le \left| {\mathop {\min }\limits_{\left[ {1;3} \right]} g\left( x \right)} \right|\end{array} \right.\).Giải chi tiết:Đặt \(g\left( x \right) = {x^3} - 3{x^2} + m\) ta có \(g'\left( x \right) = 3{x^2} - 6x = 0 \Leftrightarrow \left[ \begin{array}{l}x = 0 \notin \left[ {1;3} \right]\\x = 2 \in \left[ {1;3} \right]\end{array} \right.\).Ta có: \(g\left( 2 \right) = m - 4;\,\,g\left( 1 \right) = m - 2,\,\,g\left( 3 \right) = m\) \( \Rightarrow \left\{ \begin{array}{l}\mathop {\min }\limits_{\left[ {1;3} \right]} g\left( x \right) = m - 4\\\mathop {\max }\limits_{\left[ {1;3} \right]} g\left( x \right) = m\end{array} \right.\).\( \Rightarrow \mathop {\min }\limits_{\left[ {1;3} \right]} f\left( x \right) = \left\{ {\left| {m - 4} \right|;\left| m \right|} \right\}\).TH1: \(\mathop {\min }\limits_{\left[ {1;3} \right]} f\left( x \right) = \left| {m - 4} \right| \Leftrightarrow \left\{ \begin{array}{l}\left| {m - 4} \right| \le 3\\\left| {m - 4} \right| \le \left| m \right|\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} - 3 \le m - 4 \le 3\\{\left( {m - 4} \right)^2} \le {m^2}\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l}1 \le m \le 7\\ - 8m + 16 \le 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}1 \le m \le 7\\m \ge 2\end{array} \right. \Leftrightarrow 2 \le m \le 7\).TH2: \(\mathop {\min }\limits_{\left[ {1;3} \right]} f\left( x \right) = \left| m \right| \Leftrightarrow \left\{ \begin{array}{l}\left| m \right| \le 3\\\left| m \right| \le \left| {m - 4} \right|\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} - 3 \le m \le 3\\{m^2} \le {\left( {m - 4} \right)^2}\end{array} \right.\)\( \Leftrightarrow \left\{ \begin{array}{l} - 3 \le m \le 3\\ - 8m + 16 \ge 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} - 3 \le m \le 3\\m \le 2\end{array} \right. \Leftrightarrow  - 3 \le m \le 2\).Vậy \( - 3 \le m \le 7\). Vậy có 11 giá trị nguyên của \(m\) thỏa mãn yêu cầu bài toán.Chọn D