Question 48: Find the set of all real values of the parameter m so that the graph of the function y = x⁴ – 2(m + 1) x² + m has three extreme points A, B, C such that OA = BC; in where O is the origin, A is the extreme point on the vertical axis, and B and C are the other two extreme points.

TXĐ: \(D=\mathbb{R}\)

\(\begin{array}{l} y’ = 4{x^3} – 4\left( {m + 1} \right)x = 4x\left( {{x^2} – m – 1} \ right) = 0\\ \Leftrightarrow \left[\begin{array}{l}x=0\\{x^2}=m+1\\\\\\(1)\end{array}\right\end{array}\)

The graph of the function has 3 extreme points when entering only if (1) has two distinct solutions other than 0\(\Leftrightarrow m>-1\,\,\,\,\,(2)\).

Then:

\(\begin{array}{l} A(0;m),B\left( { – \sqrt {m + 1} ; – {m^2} – m – 1} \right),C\left( {\sqrt {m + 1} ; – {m^2} – m – 1} \right)\\ O{A^2} = {m^2},B{C^2} = 4(m + 1 )\\ OA = BC \Leftrightarrow O{A^2} = B{C^2}\\ \Leftrightarrow {m^2} = 4\left( {m + 1} \right)\\ \Leftrightarrow \left[\begin{array}{l}m=2-2\sqrt2\\\\rm{deal\(2)}\\m=2+2\sqrt2\\\\rm{save\(2)}\end{array}\right\end{array}\)

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