Question

设x₁、x₂是函数f(x)=x³+x²-a²x(a＞0)的两个极值点,且｜x₁｜+｜x₂｜=2.

(1)求a的取值范围;

(2)求证:｜b｜≤.

Answer 1

(1)解:易得f′(x)=ax²+bx-a²,                                                 

∵x₁、x₂是f(x)的两个极值点,

∴x₁、x₂是f′(x)=0的两个实根.又a＞0,

∴x₁x₂=-a＜0,x₁+x₂=.                                                     

∴｜x₁｜+｜x₂｜=｜x₁-x₂｜=.

∵｜x₁｜+｜x₂｜=2,

∴+4a=4,即b²=4a²-4a³=4a²(1-a).

∵b²≥0,∴0＜a≤1.                                                          

(2)证明:设b²=g(a)=4a²-4a³,则

g′(a)=8a-12a²=4a(2-3a).

由g′(a)＞0,得0＜a＜;由g′(a)＜0,得＜a≤1.

∴g(a)在(0,)上单调递增,在(,1)上单调递减.                                 

∴［g(a)］_max=g()=.

∴｜b｜≤.