How many integer values of \(m\) make the function \(y = \left( {{m^2} – 1} \right){x^3} + \left( {m – 1} \right){x^2} – x + 4\) strictly decreasing over \(\mathbb{R}\)?
A. \(0\) B. \(3\) C. \(2\) D. \(1\)
Solution
Case 1: \({m^2} – 1 = 0 \Leftrightarrow m = \pm 1\)
Case 1.1: \(m = 1\). We have: \(y = – x + 4\), which is the equation of a line with a negative slope \(a\), so the function is strictly decreasing over \(\mathbb{R}\). Therefore, we take \(m = 1\).
Case 1.2: \(m = – 1\). We have: \(y = – 2{x^2} – x + 4\), which is the equation of a parabola, so the function cannot be strictly decreasing over \(\mathbb{R}\). Therefore, we exclude \(m = – 1\).
Case 2: \(m \ne \pm 1\). In this case, the function is strictly decreasing over \(\mathbb{R}\) if and only if \(y’ \le 0\) for all \(x \in \mathbb{R}\), with equality occurring at finitely many points.
$ \Leftrightarrow 3\left( {{m^2} – 1} \right){x^2} + 2\left( {m – 1} \right)x – 1 \le 0$, for all \(x \in \mathbb{R}\).
$ \Leftrightarrow \left\{ \begin{array}{l}a < 0\\\Delta ‘ \le 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{m^2} – 1 < 0\\{\left( {m – 1} \right)^2} + 3\left( {{m^2} – 1} \right) \le 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{m^2} – 1 < 0\\\left( {m – 1} \right)\left( {4m + 2} \right) \le 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} – 1 < m < 1\\ – \frac{1}{2} \le m \le 1\end{array} \right. \Leftrightarrow – \frac{1}{2} \le m < 1$.
Since \(m \in \mathbb{Z}\), we find that \(m = 0\).
So, there are $2$ integer values of \(m\) that make the function strictly decreasing, which are \(m = 0\) or \(m = 1\).
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