A

- Hướng dẫn giải

Với \(u = {x^3} - 3{x^2} + m\) có \(u'=3x^2-6x=0\) khi x=0; x=2

Do đó \(\left\{ \begin{array}{l} \mathop {\min }\limits_{\left[ {1;3} \right]} u = \min \left\{ {u\left( 1 \right);u\left( 3 \right);u\left( 2 \right)} \right\} = \min \left\{ {m - 2;m;m - 4} \right\} = m - 4\\ \mathop {\max }\limits_{\left[ {1;3} \right]} u = \min \left\{ {u\left( 1 \right);u\left( 3 \right);u\left( 2 \right)} \right\} = \min \left\{ {m - 2;m;m - 4} \right\} = m \end{array} \right.\)

Nếu

 \(\begin{array}{l} m - 4 \ge 0 \Leftrightarrow m \ge 4\\ \Rightarrow \mathop {\min }\limits_{\left[ {1;3} \right]} f(x) = m - 4 \le 2020\\ \Leftrightarrow m \le 2024 \Rightarrow m \in \left\{ {4,...,2020} \right\} \end{array}\)

Nếu

\(\begin{array}{l} m \le 0 \Leftrightarrow m \ge 4\\ \Rightarrow \mathop {\min }\limits_{\left[ {1;3} \right]} f(x) = - m \le 2020\\ \Leftrightarrow - 2020 \le m \Rightarrow m \in \left\{ { - 2020,...,0} \right\} \end{array}\)

Nếu

\(\begin{array}{l} 0 < m < 4\\ \Rightarrow \mathop {\min }\limits_{\left[ {1;3} \right]} u < 0;\,\,\mathop {\max }\limits_{\left[ {1;3} \right]} u > 0\\ \Rightarrow \mathop {\min }\limits_{\left[ {1;3} \right]} f(x) = 0\,\,(tm) \end{array}\)

Vậy \(m \in \left\{ { - 2020;...;2024} \right\}\) có tất cả 4045 số nguyên thỏa mãn,