Đáp án đúng:
Giải chi tiết:Ta có \(f\left( x \right) = \dfrac{{m{x^3}}}{3} - 3{x^2} + mx - 5 \Rightarrow f'\left( x \right) = m{x^2} - 6x + m\)
a) \(f'\left( x \right) > 0\,\,\forall x \in \mathbb{R} \Leftrightarrow m{x^2} - 6x + m > 0\,\,\forall x \in \mathbb{R}\).
TH1: \(m = 0 \Rightarrow - 6x > 0\,\,\forall x \in \mathbb{R} \Leftrightarrow x < 0\,\,\forall x \in \mathbb{R}\) (không thỏa mãn).
TH2: \(m \ne 0\), khi đó \(m{x^2} - 6x + m > 0\,\,\forall x \in \mathbb{R}\) \( \Leftrightarrow \left\{ \begin{array}{l}m > 0\\\Delta ' = 9 - {m^2} < 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m > 0\\\left[ \begin{array}{l}m > 3\\m < - 3\end{array} \right.\end{array} \right. \Leftrightarrow m > 3\).
Vậy \(m > 3\).
b) \(f'\left( x \right) < 0\,\,\forall x \in \mathbb{R} \Rightarrow m{x^2} - 6x + m < 0\,\,\forall x \in \mathbb{R}\)
TH1: \(m = 0 \Rightarrow - 6x < 0\,\,\forall x \in \mathbb{R} \Leftrightarrow x > 0\,\,\forall x \in \mathbb{R}\) (không thỏa mãn).
TH2: \(m \ne 0\), khi đó \(m{x^2} - 6x + m < 0\,\,\forall x \in \mathbb{R}\) \( \Leftrightarrow \left\{ \begin{array}{l}m < 0\\\Delta ' = 9 - {m^2} < 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}m < 0\\\left[ \begin{array}{l}m > 3\\m < - 3\end{array} \right.\end{array} \right. \Leftrightarrow m < - 3\).
Vậy \(m < - 3\).
