试题分析:解:(Ⅰ)
f (
x)的定义域为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002756077559.png)
.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002756077826.png)
=-
a(
x-1)[
x-(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002755984327.png)
-1)]. ……2分
当0<
a<
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002755937338.png)
时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002755984327.png)
-1>1,
∴
f (
x)在(0,1),(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002755984327.png)
-1,+¥)递减;在(1,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002755984327.png)
-1)递增; ……4分
(Ⅱ)
f (
x)在区间
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002756155529.png)
上不具有单调性等价于
f (
x)在区间
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002756155529.png)
内至少有一个极值点. ……5分
①当
a=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002755937338.png)
时,
f ¢(
x)=-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002755937338.png)
(
x-1)
2≤0Þ
f (
x)在
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002756218566.png)
上递减,不合题意; …7分
②当
a≥1时,
f ¢(
x)=0的两根为
x1=1,
x2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002755984327.png)
-1,∵
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002756233774.png)
,故不合题意;③当
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002756249438.png)
,且
a≠
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002755937338.png)
时,
f (
x)在区间
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002756155529.png)
上不具有单调性等价于:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002756296452.png)
或
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002756311642.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240027563271006.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002756327509.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002756342465.png)
,且
a≠
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002755937338.png)
. ……11分
综上可知,所求
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002756374283.png)
的取值范围是(0,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002755937338.png)
)∪(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002755937338.png)
,1). ……12分
点评:这类问题的解决一般主要涉及两类题型,求解单调区间,同时证明不等式恒成立问题。前者经常要对于参数分类讨论,注意对于一元二次不等式的熟练运用,是解决这个题型的关键,后者主要是求解函数的最值来证明不等式。如果递增,则说明函数在给定区间上导数恒大于等于零,反之,则恒小于等于零。来分离参数的思想求解参数的范围。