Question 17: Given the function y = x4- 2( 1 - m2) x2+ m+1. There exists a value of m so that the function has a maximum, a minimum and the extreme points of the graph of the function form a triangle with the largest area. Which statement is correct then?

Question 17: Given the function y = x4- 2( 1 – m2) x2+ m+1. There exists a value of m so that the function has a maximum, a minimum and the extreme points of the graph of the function form a triangle with the largest area. Which statement is correct then?

A. m is a positive integer

B.m is not an integer

C. m = 1

D. All wrong

We have the derivative y’ = 4x³– 4(1 – m .)²) x

$y’ = 0 \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
{x^2} = 1 – {m^2}
\end{array} \right.$

Hàm số có cực đại, cực tiểu khi và chỉ khi -1 < m <1

Tọa độ điểm cực trị

$A (0; m + 1); B (\sqrt{1 - m^{2}}; - m^{4} + 2 m^{2} + m); C (- \sqrt{1 - m^{2}}; - m^{4} + 2 m^{2} + m); \vec{B C} = (- 2 (\sqrt{1 - m^{2}}; 0)$

Phương trình đường thẳng BC: y+ m4- 2m2- m=0

d( A: BC) = m⁴-2m²+ 1,

$B C = 2 \sqrt{1 - m^{2}} \Rightarrow S_{∆ A B C} = \frac{1}{2} B C . d [A, B C] = \sqrt{1 - m^{2}} (m^{4} - 2 m^{2} + 1)=\sqrt{{(1 - m^{2})}^{5}}\leq ;1$

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