Question 9: Calculate the sum of all the real values of the parameter m so that the function y = x3-3mx2+ 3(m2-1) x- m3+ m has the maximum and the distance from the maximum point of the function graph to origin O is equal to \(\sqrt 2 \) times the distance from the minimum point of the function graph to the origin O.

We have y’ = 3x^{2}– 6mx + 3(m^{2}-first).

If a given function has an extreme, the equation y’ = 0 has 2 distinct solutions

$$

there are 2 distinct solutions \( \Leftrightarrow \Delta = 1 > 0,\forall m\)

Then, the maximum point A(m-1; 2-2m) and the minimum point B(m+1; -2-2m)

We have

\(\begin{array}{l}

OA = \sqrt 2 OB \Leftrightarrow {m^2} + 6m + 1 = 0\\

\Leftrightarrow \left[\begin{array}{l}[\begin{array}{l}

m = – 3 + 2\sqrt 2 \\

m = – 3 – 2\sqrt 2

\end{array} \right.

\end{array}\)

The sum of these two values is – 6.

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