试题分析:(1)当x∈(-1, 0)时, - x∈(0, 1).∴由题意可得f(-x)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011805744699.png)
.
又f(x)是奇函数,∴f(x)=" -" f (-x) =
-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011805697517.png)
. 2分
∵f(-0)= -f(0), ∴f(0)=" 0." 3分
又f(x)是最小正周期为2的函数,∴对任意的x有f(x+2)= f(x).
∴f(-1)=" f(-1+2)=" f(1). 另一面f(-1)="-" f (1), ∴- f(1)=" f(1)" . ∴f(1) = f(-1)=0. 5分
∴f(x)在[-1, 1]上的解析式为 f(x)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240118057121822.png)
. 6分
(2)f (x)在(—1, 0)上时的解析式为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011805790532.png)
,∵
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011805806777.png)
,∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240118058221921.png)
,又-1<x<0,∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011805853487.png)
,∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011805868476.png)
,∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240118058841003.png)
,∴f (x)在(—1, 0)上时减函数 10分
(3)不等式f(x)>λ在R上有解的λ的取值范围就是λ小于f(x)在R上的最大值.…12分
由(2)结论可得,当x∈(-1, 0)时,有-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011805915338.png)
< f(x)= -
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011805697517.png)
< -
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011805946370.png)
;
又f(x)是奇函数,当x∈(0, 1)时,有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011805946370.png)
< f(x)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011805697517.png)
<
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011805915338.png)
;
∴f(x)在[-1, 1]上的值域是(-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011805915338.png)
, -
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011805946370.png)
)∪{0}∪(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011805946370.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011805915338.png)
). 14分
由f(x)的周期是2;故f(x)在R上的值域是(-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011805915338.png)
, -
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011805946370.png)
)∪{0}∪(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011805946370.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011805915338.png)
) 15分
∴λ<
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011805728338.png)
时,不等式f(x)>λ在R上有解. 16分
点评:利用奇偶性求函数解析式问题要注意:(1)在哪个区间求解析式,就设在哪个区间里;(2)转化为已知的解析式进行代入;(3)利用
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011806149491.png)
的奇偶性把
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011806227507.png)
写成
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011806243511.png)
或
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011806149491.png)
,从而求出
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824011806149491.png)
.