作业帮已知函数F(X)=X^3+AX^2+BX+2与直线4X-Y+5=0相切于X=1处,若X>0时,不等式F(X)≥MX^2-2X+2恒成立,求实数M的取值范围?
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已知函数F(X)=X^3+AX^2+BX+2与直线4X-Y+5=0相切于X=1处,若X>0时,不等式F(X)≥MX^2-2X+2恒成立,求实数M的取值范围?
已知函数F(X)=X^3+AX^2+BX+2与直线4X-Y+5=0相切于X=1处,若X>0时,不等式F(X)≥MX^2-2X+2恒成立,求实数M的取值范围?
已知函数F(X)=X^3+AX^2+BX+2与直线4X-Y+5=0相切于X=1处,若X>0时,不等式F(X)≥MX^2-2X+2恒成立,求实数M的取值范围?
F(X) = X^3+AX^2+BX+2 与直线 4X-Y+5 =0 相切于X=1处
F(1) = 1 + A + B + 2 = 3 + A + B,
4 - Y + 5 = 0,Y = 9,
9 = Y = F(1) = 3 + A + B,
A + B = 6 ...(1)
F'(X) = 3X^2 + 2AX + B,
F'(1) = 3 + 2A + B,
4X - Y + 5 = 0,
Y = 4X + 5.斜率为4,
因此,
4 = F'(1) = 3 + 2A + B,
2A + B = 1 ...(2)
由（1）,（2）解得,
A = -5,
B = 11.
所以,
F(X) = X^3 - 5X^2 + 11X + 2.
令G(X) = F(X) - MX^2 + 2X - 2 = X^3 - 5X^2 + 11X + 2 - MX^2 + 2X -2
= X^3 -(5+M)X^2 + 13X
= X[X^2 -(5+M)X + 13].X>0
则当X>0,G(X) > 0时,总有,
X^2 - (5+M)X + 13 > 0
(5+M)^2 - 4*13 < 0
25 + 10M + M^2 - 52 < 0,
M^2 + 10M - 27 < 0,
Delta = 10^2 + 4*27 = 100 + 108 = 208 = 4^2*13,
[-10 -Delta^(1/2)]/2 < M < [-10 + Delta^(1/2)]/2,
[-10 -4(13)^(1/2)]/2 = -5-2(13)^(1/2) < M < [-10 + 4(13)^(1/2)]/2 = -5 + 2(13)^(1/2),
实数M的取值范围,
-5-2(13)^(1/2) < M < -5 + 2(13)^(1/2).
4x-y+5=0可写为y=4x+5.那么依题意得，F(X)=4x+5。因为x=1时，x同样大于零，所以将x=1代入不等式得F(X)≥M。又因为`F(X)=4x+5，所以4x+5≥M且x大于0，M就大于5 ，可能是错的，见谅！楼上是对的吧