题目
题型:不详难度:来源:
a |
x |
(Ⅰ)当a=5时,求函数f(x)的单调递增区间;
(Ⅱ)求f(x)的极大值;
(Ⅲ)求证:对于任意a>1,函数f(x)<0在(0,a)上恒成立.
答案
a+1 |
x |
a |
x2 |
(Ⅰ)当a=5时,f′(x)=1-
6 |
x |
5 |
x2 |
x2-6x+5 |
x2 |
(x-1)(x-5) |
x2 |
解得x≥5或x≤1.故函数f(x)在(0,1),(5,+∞)上单调递增. …(2分)
(Ⅱ)令f"(x)≥0,即1-
a+1 |
x |
a |
x2 |
x2-(a+1)x+a |
x2 |
(x-a)(x-1) |
x2 |
当a=1时,上式化为
(x-1)2 |
x2 |
当a>1时,解得x≤1或x≥a.故f(x)在(0,1),(a,+∞)上单调递增,在(1,a)上单调递减.