Cho hàm số \(f\left( x \rig...

B

- Hướng dẫn giải

Có \(f'\left( x \right) = 3{x^2} + 6x = 3x\left( {x + 2} \right),f'\left( x \right) = 0 \Leftrightarrow \left[ {\begin{array}{*{20}{c}}{x = 0}\\{x = - 2}\end{array}} \right. \Rightarrow f'\left( x \right) > 0,\forall x \in \left[ {1;3} \right]\).

Vậy trên [1;3] hàm số luôn đồng biến.

Có \(f\left( 1 \right) = 5 - 2m;\,f\left( 3 \right) = 55 - 2m\).

- TH1: \(\left( {5 - 2m} \right)\left( {55 - 2m} \right) \le 0 \Leftrightarrow \frac{5}{2} \le m \le \frac{{55}}{2}\)

Khi đó \(\mathop {\min }\limits_{\left[ {1;3} \right]} f\left( x \right) = 0\) và \(\left[ \begin{array}{l}\mathop {\max }\limits_{\left[ {1;3} \right]} \left| {f\left( x \right)} \right| = \left| {5 - 2m} \right| = 2m - 5\\\mathop {\max }\limits_{\left[ {1;3} \right]} \left| {f\left( x \right)} \right| = \left| {55 - 2m} \right| = 55 - 2m\end{array} \right.\)

Ta có \(2m - 5 > 55 - 2m \Leftrightarrow m > 15\).

Với \(15 < m \le \frac{{55}}{2}\) thì \(\mathop {\max }\limits_{\left[ {1;3} \right]} \left| {f\left( x \right)} \right| = 2m - 5\)

\(\mathop {\max }\limits_{\left[ {1;3} \right]} \left| {f\left( x \right)} \right| + \mathop {\min }\limits_{\left[ {1;3} \right]} \left| {f\left( x \right)} \right| \ge 10 \Leftrightarrow 2m - 5 + 0 \ge 10 \Leftrightarrow m \ge \frac{{15}}{2}\).

Do đó \(15 < m \le \frac{{55}}{2}\).

Với \(\frac{5}{2} \le m \le 15\) thì \(\mathop {\max }\limits_{\left[ {1;3} \right]} \left| {f\left( x \right)} \right| = 55 - 2m\)

\(\mathop {\max }\limits_{\left[ {1;3} \right]} \left| {f\left( x \right)} \right| + \mathop {\min }\limits_{\left[ {1;3} \right]} \left| {f\left( x \right)} \right| \ge 10 \Leftrightarrow 55 - 2m + 0 \ge 10 \Leftrightarrow m \le \frac{{45}}{2}\).

Do đó \(\frac{5}{2} \le m \le 15\).

Vậy \(\frac{5}{2} \le m \le \frac{{55}}{2}\).

-TH2: \(5 - 2m > 0 \Leftrightarrow m < \frac{5}{2}\).

Thì \(\mathop {\max }\limits_{\left[ {1;3} \right]} \left| {f\left( x \right)} \right| + \mathop {\min }\limits_{\left[ {1;3} \right]} \left| {f\left( x \right)} \right| \ge 10 \Leftrightarrow 55 - 2m + 5 - 2m \ge 10 \Leftrightarrow m \le \frac{{25}}{2}\). Vậy \(m < \frac{5}{2}\).

- TH3: \(55 - 2m < 0 \Leftrightarrow m > \frac{{55}}{2}\).

Thì \(\mathop {\max }\limits_{\left[ {1;3} \right]} \left| {f\left( x \right)} \right| + \mathop {\min }\limits_{\left[ {1;3} \right]} \left| {f\left( x \right)} \right| \ge 10 \Leftrightarrow - 5 + 2m - 55 + 2m \ge 10 \Leftrightarrow m \ge \frac{{35}}{2}\). Vậy \(m > \frac{{55}}{2}\).

Tóm lại S = R. Vậy trong [-30;30], S có 61 giá trị nguyên.