Đáp án:
\[1 < m < 10\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
y = \left( {m - 1} \right){x^3} - \left( {m - 1} \right){x^2} + 3x + 2011\\
\Rightarrow y' = 3\left( {m - 1} \right){x^2} - 2\left( {m - 1} \right)x + 3\\
y' > 0,\,\,\,\forall x \in R\\
\Leftrightarrow 3\left( {m - 1} \right){x^2} - 2\left( {m - 1} \right)x + 3 > 0,\,\,\,\forall x \in R\\
\Leftrightarrow \left\{ \begin{array}{l}
a > 0\\
\Delta ' < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
3\left( {m - 1} \right) > 0\\
{\left( {m - 1} \right)^2} - 3.\left( {m - 1} \right).3 < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 1\\
\left( {m - 1} \right)\left[ {\left( {m - 1} \right) - 9} \right] < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 1\\
\left( {m - 1} \right)\left( {m - 10} \right) < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
m > 1\\
1 < m < 10
\end{array} \right.\\
\Leftrightarrow 1 < m < 10
\end{array}\)
Vậy \(1 < m < 10\)
