Question 15: Find all real values of parameter m so that the graph of the function y = 2×3-3( m+1) x2+ 6mx has two extreme points A; B such that the line AB is perpendicular to the line y = x + 2.

+ We have derivative y’ = 6x^{2}– 6( m + 1) x + 6m

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

x = 1\\

x = m

\end{array} \right.\)

The condition for the function to have 2 extreme points is: m 1

The coordinates of the two extreme points are A( 1 ; 3m-1) and B ( m ; – m^{3}+ 3m^{2})

+ The slope of line AB is: k = – ( m-1)^{ 2}

+ Line AB is perpendicular to line y = x + 2 if and only if k = -1

Hay – ( m-1) ^{2}= -1( since the two lines are perpendicular, the product of the slopes is -1) \(\left[\begin{array}{l}[\begin{array}{l}

m = 0\\

m = 2

\end{array} \right.\)

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