试题分析:(1)先将原式化成求解导数f‘(x),再利用导数的正负与函数单调性的关系,即可求得函数f(x)的最小值;
(2)原题等价于x
2+2x+a>0对x∈[1,+∞)恒成立,再结合二次函数的单调性只须g(1)>0,从而求得实数a的取值范围;
解(Ⅰ)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000752582453.png)
时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240007527691522.png)
(因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000752785358.png)
)
所以,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000752613447.png)
在
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000752816474.png)
上单调递增,故
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000752675323.png)
时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000752613447.png)
取得最小值
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000752707377.png)
.
(Ⅱ) 因为对任意
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000752629588.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000752644538.png)
恒成立,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000752925648.png)
恒成立,只需
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000752941559.png)
恒成立,只需
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000752956772.png)
,因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000752972791.png)
,
所以,实数
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000752660283.png)
的取值范围是
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000752722560.png)
.
点评:解决该试题的关键是是对于同一个问题的不同的处理角度,可以运用均值不等式得到最值,也可以结合导数的工具得到最值,对于恒成立问题一般都是转换为求解函数的 最值即可得到。