Question

已知使函数f(x)＝x3－ax2－1(0≤a≤M0)存在整数零点的实数a恰有3个,则M0的取值范

Answer 1

a=(x^3-1)/x^2=x-1/x^2,

因a>=0,得x>=1,

x=1时,a=0

x=2时,a=2-1/4=7/4

x=3时,a=3-1/9=26/9

x=4时,a=4-1/16=63/16

x>=1增大时,x-1/x^2单调增

因为只有3个a值,故26/9=