Question 46: Find all real values ​​of parameter m so that the graph of the function (y = m{x^3} – 3m{x^2} + 3m – 3) has two extreme points A, B such that (2A{B^2} – left( {O{A^2} + O{B^2}} right) = 20) (where O is the origin).

We have: \(y’ = m\left( {3{x^2} – 6x} \right)\)

For all m 0, we have y′ = 0

\( \Leftrightarrow \left[\begin{array}{l}[\begin{array}{l}
x = 0 \Rightarrow y = 3m – 3\\
x = 2 \Rightarrow y = – m – 3
\end{array} \right.\) .

So the function always has two extreme points.

Suppose A(0;3m−3); B(2;−m−3)

We have

\(\begin{array}{l}
2A{B^2} – \left( {O{A^2} + O{B^2}} \right) = 20\\
\Leftrightarrow 11m^2 + 6m – 17 = 0\\
\Leftrightarrow \left[\begin{array}{l}[\begin{array}{l}
m = 1\\
m = – \frac{{17}}{{11}}
\end{array} \right.\left( {tm} \right)
\end{array}\)

So the value of m to look for is: m = 1 or \(m = – \frac{{17}}{{11}}\)