试题分析:(Ⅰ)首先考虑定义域.再把
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824031152612346.png)
代入求导.令导函数
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824031152752565.png)
可求得极值点.再通过函数的单调性即可知道函数的极值.
(Ⅱ)由
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824031152643336.png)
.在区间
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824031152659469.png)
上,函数
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824031152628463.png)
的图像在函数
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824031152690718.png)
的图像的下方,可转化为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824031152830902.png)
在区间
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824031152659469.png)
上恒成立的问题.从而令函数F(x)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824031152862807.png)
.通过求导即可求得F(x)函数的最大值.从而可得结论.
试题解析:(Ⅰ)解由于函数f(x)的定义域为(0,+∞), 1分
当a=-1时,f′(x)=x-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824031152877429.png)
2分
令f′(x)=0得x=1或x=-1(舍去), 3分
当x∈(0,1)时,f′(x)<0, 因此函数f(x)在(0,1)上是单调递减的, 4分
当x∈(1,+∞)时,f′(x)>0,因此函数f(x)在(1,+∞)上是单调递增的, 5分
则x=1是f(x)极小值点,
所以f(x)在x=1处取得极小值为f(1)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824031152721338.png)
6分
(Ⅱ)证明 设F(x)=f(x)-g(x)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824031152721338.png)
x
2+ln x-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824031152924382.png)
x
3,
则F′(x)=x+
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824031152940189.png)
-2x
2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824031152940629.png)
, 9分
当x>1时,F′(x)<0, 10分
故f(x)在区间[1,+∞)上是单调递减的, 11分
又F(1)=-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824031152955336.png)
<0, 12分
∴在区间[1,+∞)上,F(x)<0恒成立.即f(x)—g(x)<0恒成立
即f(x)<g(x)恒成立.
因此,
当a=1时,在区间[1,+∞)上,函数f(x)的图像在函数g(x)图像的下方.13分