Question 3: Find all real values of parameter m so that the graph of the function \(y = {x^4} – 2m{x^2}\) has three extreme points forming a triangle with area area less than 1.

+ The condition for the function to have 3 extremes is m > 0

\(\begin{array}{l}

y’ = 4{x^3} – 4mx\\

y’ = 0 \Rightarrow \left[\begin{array}{l}[\begin{array}{l}

{x_1} = 0\\

{x_2} = – \sqrt m \\

{x_3} = \sqrt m

\end{array} \right. \Rightarrow \left[\begin{array}{l}[\begin{array}{l}

{y_1} = 0\\

{y_2} = – {m^2}\\

{y_3} = – {m^2}

\end{array} \right.

\end{array}\)

+ The extreme points form an isosceles triangle with base equal to \(2\sqrt m \), height equal to m^{2}. (as picture below)

We get \({S_{ABC}} = \frac{1}{2}AC.BD = \sqrt m . {m^2}.\)

+ Let the triangle have area less than 1 then

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