(Ⅰ)证法一:记
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232204508201041.png)
,
则当x>1时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232204508361011.png)
.
又
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220450945499.png)
有
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220450961530.png)
, 即
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220450992803.png)
证法二:由均值不等式,当x>1时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220451008532.png)
,故
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220451039605.png)
①
令
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220451242664.png)
,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220451273494.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220451366730.png)
.
故
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220451382536.png)
,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220451444454.png)
②
由①②得,当x>1时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220450992803.png)
.
(Ⅱ)(证法一)
记
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232204514911035.png)
,
由(Ⅰ)得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232204515221200.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220451554956.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220451569854.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232204516001082.png)
令
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220451616869.png)
,
则当1<x<3时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232204516321011.png)
因此
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220451663442.png)
在(1,3)内是递减函数,
又由
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220450945499.png)
,得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220450961530.png)
,
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220451725584.png)
因此
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220451741484.png)
在(1,3)内是递减函数,
又由
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220451756532.png)
,得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220451772563.png)
.
于是,当1<x<3时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220450789846.png)
(证法二):
记
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220451819964.png)
则当1<x<3时,由(Ⅰ)得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220451850948.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232204518971250.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232204519441319.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232204519901414.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232204520221024.png)
因此
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220451741484.png)
在(1,3)内单调递减
又
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220451756532.png)
,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220451772563.png)
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823220450789846.png)
.
考点定位:本大题考查导数题目中较为常规的类型题目,考查的切线,单调性,以及最值问题都是课本中要求的重点内容,考查构造函数用求导的方法求最值的能力