题目内容
设偶函数
上为减函数,且
,则不等式
的解集为( )
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823204101777784.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823204101808544.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823204101840810.png)
A.![]() | B.![]() |
C.![]() | D.![]() |
B
解:∵f(x)是偶函数
∴f(-x)=f(x)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232041019331432.png)
也就是xf(x)>0
①当x>0时,有f(x)>0
∵f(x)在(0,+∞)上为减函数,且f(2)=0
∴f(x)>0即f(x)>f(2),得0<x<2;
②当x<0时,有f(x)<0
∵-x>0,f(x)=f(-x)<f(2),
∴-x>2?x<-2
综上所述,原不等式的解集为:(-∞,-2)∪(0,2)
故选B
![](http://thumb2018.1010pic.com/images/loading.gif)
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