Question 13: Calculate the sum of the values of the parameter m so that the graph of the function: \(y = – {x^4} + 2m{x^2} – 4m + 1\) has three extreme points. At the same time, those three extreme points together with the origin form a rhombus.

+ Derivative y’ = – 4x^{3}+ 4mx = -4x(x^{2}– m)

Let the function have 3 extreme points when m > 0

+ The coordinates of three extreme points are: A( 0; 1-4m); \(B\left( { – \sqrt m ;\;{m^2} – 4m + 1} \right);\;C\left( {\sqrt m ;\;{m^2} – 4m + 1 } \right)\)

Quadrilateral OBAC already has OB = OC; AB = AC.

So quadrilateral OBAC is a rhombus if and only if :

The sum of the values of m satisfying the title is \(\frac{9}{4}\).

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