题目内容
若函数f(x)=|x|+
-
(a>0)没有零点,则实数a的取值范围为________.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824052720476499.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824052720492344.png)
(0,1)∪(2,+∞)
在平面直角坐标系中画出函数y=
(a>0)的图象(其图象是以原点为圆心、
为半径的圆,且不在x轴下方的部分)与y=
-|x|的图象.观察图形可知,要使这两个函数的图象没有公共点,则原点到直线y=
-x的距离大于
,或
>
.又原点到直线y=
-x的距离等于1,所以有0<
<1,或
>
,由此解得0<a<1或a>2.所以,实数a的取值范围是(0,1)∪(2,+∞).
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240527208665790.jpg)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824052720476499.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824052720492344.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824052720492344.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824052720492344.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824052720663339.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824052720663339.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824052720492344.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824052720492344.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824052720663339.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824052720663339.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824052720492344.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140824/201408240527208665790.jpg)
![](http://thumb.zyjl.cn/images/loading.gif)
练习册系列答案
相关题目