试题分析:(1)依题意,知
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003207456447.png)
的定义域为(0,+∞),当
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003207628568.png)
时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003207643934.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240032076591130.png)
……………2分
令
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003207690461.png)
=0,解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003207706323.png)
.(∵
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003207721391.png)
)
当
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003207737436.png)
时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003207830554.png)
,此时
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003207456447.png)
单调递增;当
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003207862359.png)
时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003207877545.png)
,此时
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003207456447.png)
单调递减.
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003207456447.png)
的极大值为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003207565606.png)
,此即为最大值 ……………4分
(2)因为方程
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003207940711.png)
有唯一实数解,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003207955759.png)
有唯一实数解,
设
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003207971902.png)
,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003207986929.png)
.令
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208002542.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208018596.png)
.
因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208049459.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003207721391.png)
, 所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208080884.png)
(舍去),
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208096746.png)
,…… 6分
当
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208111596.png)
时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208142550.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208158442.png)
在(0,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208174334.png)
)上单调递减,
当
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208189649.png)
时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208205564.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208158442.png)
在(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208174334.png)
,+∞)单调递增
当
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208267388.png)
时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208298507.png)
=0,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208158442.png)
取最小值
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208330493.png)
.
则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208345892.png)
既
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240032083611226.png)
……………10分
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208439812.png)
,因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208049459.png)
,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208486678.png)
(*)
设函数
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208501759.png)
,因为当
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003207721391.png)
时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208548484.png)
是增函数,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208564553.png)
至多有一解.
因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208579528.png)
,所以方程(*)的解为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208595384.png)
,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003208626741.png)
,解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824003207581508.png)
………12分
(直接看出x=1时,m=1/2但未证明唯一性的给3分)
点评:典型题,本题属于导数应用中的基本问题,通过研究函数的单调性,明确了极值情况。通过研究函数的单调区间、最值情况,得出方程解的存在情况。涉及对数函数,要特别注意函数的定义域。