题目

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?

提问时间：2021-09-09

答案

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,...