Question

已知函数f(x)=.

(1)解不等式f(x)＞1;

(2)设函数F(x)=f(x)-aln(x+1)在(0,+∞)上单调递增，求实数a的取值范围.

Answer 1

解：（1）∵＞1,∴＞0.?

∴(x²-4x+1)(x+1)＞0.                                                                                               ?

令(x²-4x+1)(x+1)=0,则x₁=2+,x₂=2-,x₃=-1，?

∴不等式的解集为{x|-1＜x＜2-,或x＞2+}.                                            ?

（2）F(x)=-aln(x+1),在(0,+∞)上单调递增，则?

F′(x)= -.                                                                                   ?

当F′(x)＞0时，x²+2x-5-a(x+1)＞0在（0，+∞）上恒成立.?

∴a＜.                                                                                                 ?

令g(x)=,g′(x)=＞0,?

∴g(x)在(0,+∞)上为增函数.又g(x)在x=0处连续，?

g(x)＞g(0)=-5,∴a≤-5.                                                                                       ?

当a＞-5时，经检验F(x)在(0,+∞)上不是单调增函数.?

综上,a的取值范围是a≤-5.