\(y=\left|-2x^3+3mx-2\right|=\left|2x^3-3mx+2\right|\)
Xét hàm \(f\left(x\right)=2x^3-3mx+2\)
\(y=\left|f\left(x\right)\right|\) đồng biến trên khoảng đã cho khi::
TH1: \(\left\{{}\begin{matrix}f\left(x\right)\ge0;\forall x>1\\f'\left(x\right)\ge0;\forall x>1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}f'\left(x\right)\ge0\\f\left(1\right)\ge0\end{matrix}\right.\) ;\(\forall x>1\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x^2-3m\ge0\\2-3m+2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\le\min\limits_{x>1}\left(2x^2\right)\\m\le\dfrac{4}{3}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m\le2\\m\le\dfrac{4}{3}\end{matrix}\right.\)
\(\Rightarrow m\le\dfrac{4}{3}\)
TH2: \(\left\{{}\begin{matrix}f\left(x\right)\le0\\f'\left(x\right)\le0\end{matrix}\right.\) ;\(\forall x>1\)
\(\Leftrightarrow\left\{{}\begin{matrix}f'\left(x\right)\le0\\f\left(1\right)\le0\end{matrix}\right.\) ;\(\forall x>1\)
\(\Leftrightarrow\left\{{}\begin{matrix}6x^2-3m\le0\\4-3m\le0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\ge\max\limits_{x>1}\left(2x^2\right)\\m\ge\dfrac{4}{3}\end{matrix}\right.\) \(\Rightarrow\) không tồn tại m thỏa mãn
Vậy \(m\le\dfrac{4}{3}\)
