试题分析:(1)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240027484021210.png)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748418879.png)
. ……2 分
因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748433574.png)
,以下讨论函数g (x) = –a
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748449304.png)
+ 2ax – a – 1值的情况.
当a = 0时,g (x) =" –1" < 0,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748464549.png)
,所以f (x)在R上是减函数. ……3分
当a > 0时,g (x) = 0的判别式Δ= 4
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748464323.png)
– 4(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748464323.png)
+a) =" –4a" < 0,
所以g(x)<0,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748464549.png)
,所以f(x)在R上是减函数. ……5分
当a < 0时,g (x) = 0有两个根,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748511459.png)
,并且
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748355456.png)
<
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748371434.png)
,
所以,在区间(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748340591.png)
)上,g (x) > 0,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748558537.png)
,f (x)在此区间上 是增函数.
在区间(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748355456.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748371434.png)
)上,g (x) < 0,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748464549.png)
,f (x)在此区间上是减函数.
在区间(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748386585.png)
)上,g (x) > 0,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748558537.png)
,f (x)在此区间上是增函数. ……7分
综上,当a≥0时,f (x)在R上是减函数;
当a < 0时,f (x)在(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748636590.png)
)上单调递增,在(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748355456.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748371434.png)
)上单调递减,在(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748386585.png)
)上单调递增. ……8分
(2)当 – 1 < a < 0时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748683683.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748698754.png)
, ……10分
所以,在区间[1,2]上,函数f (x)单调递减, ……11分
所以,函数f (x)在区间[1,2]上的最小值为f (2) =
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002748402583.png)
. ……12分
点评:在高考解答题中,经常用到分类讨论思想,分类讨论时要准确确定分类标准,分类标准要不重不漏.