ĐỀ BÀI:
9. Cho hàm số \(y = f\left( x \right) = \left| { – \frac{1}{3}{x^3} + \frac{1}{2}\left( {2m + 3} \right){x^2} – \left( {{m^2} + 3m} \right)x + \frac{2}{3}} \right|\). Có bao nhiêu giá trị nguyên của tham số \(m\) thuộc đoạn \(\left[ { – 9\,;\,9} \right]\) để hàm số \(y = f\left( x \right)\) nghịch biến trên khoảng \(\left( {1\,;\,2} \right)\)?
A. \(3\).
B. \(2\).
C. \(16\).
D. \(9\).
Lời giải
Xét \(g\left( x \right) = – \frac{1}{3}{x^3} + \frac{1}{2}\left( {2m + 3} \right){x^2} – \left( {{m^2} + 3m} \right)x + \frac{2}{3}\).
\(g’\left( x \right) = – {x^2} + \left( {2m + 3} \right)x – \left( {{m^2} + 3m} \right)\).
\(g’\left( x \right) = 0\, \Leftrightarrow \,\left[ \begin{array}{l}x = m\\x = m + 3\end{array} \right.\).
Bảng biến thiên:
Hàm số \(\left| {g\left( x \right)} \right|\) nghịch biến trên khoảng \(\left( {1\,;\,2} \right)\) \( \Leftrightarrow \,\left[ \begin{array}{l}\left\{ \begin{array}{l}g’\left( x \right) \ge 0\,,\,\forall \,x \in \left( {1\,;\,2} \right)\\g\left( x \right) \le 0\,,\,\forall \,x \in \left( {1\,;\,2} \right)\end{array} \right.\\\left\{ \begin{array}{l}g’\left( x \right) \le 0\,,\,\forall \,x \in \left( {1\,;\,2} \right)\\g\left( x \right) \ge 0\,,\,\forall \,x \in \left( {1\,;\,2} \right)\end{array} \right.\end{array} \right.\)
\( \Leftrightarrow \,\left[ \begin{array}{l}\left\{ \begin{array}{l}m \le 1 < 2 \le m + 3\\g\left( 2 \right) \le 0\end{array} \right.\\\left\{ \begin{array}{l}m + 3 \le 1\\g\left( 2 \right) \ge 0\end{array} \right.\\\left\{ \begin{array}{l}2 \le m\\g\left( 2 \right) \ge 0\end{array} \right.\end{array} \right.\, \Leftrightarrow \,\left[ \begin{array}{l}\left\{ \begin{array}{l} – 1 \le m \le 1\\ – 2{m^2} – 2m + 4 \le 0\end{array} \right.\\\left\{ \begin{array}{l}m \le – 2\\ – 2{m^2} – 2m + 4 \ge 0\end{array} \right.\\\left\{ \begin{array}{l}m \ge 2\\ – 2{m^2} – 2m + 4 \ge 0\end{array} \right.\end{array} \right.\, \Leftrightarrow \,\left[ \begin{array}{l}\left\{ \begin{array}{l} – 1 \le m \le 1\\m \in \left( { – \infty \,;\, – 2} \right] \cup \left[ {1\,;\, + \infty } \right)\end{array} \right.\\\left\{ \begin{array}{l}m \le – 2\\m \in \left[ { – 2\,;\,1} \right]\end{array} \right.\\\left\{ \begin{array}{l}m \ge 2\\m \in \left[ { – 2\,;\,1} \right]\end{array} \right.\end{array} \right.\)\( \Rightarrow \,\left[ \begin{array}{l}m = – 2\\m = 1\end{array} \right.\).
Vậy \(m \in \left\{ { – 2\,;\,1} \right\}\).
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