题目内容
设y=(log2x)2+(t-2)log2x-t+1,若t在[-2,2]上变化时,y恒取正值,求x的取值范围.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824035410409665.png)
解:设y=f(t)=(log2x-1)t+(log2x)2-2log2x+1,
则f(t)是一次函数,当t∈[-2,2]时,
f(t)>0恒成立,则有![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240354104241017.png)
即![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240354104401321.png)
解得log2x<-1或log2x>3.
∴0<x<
或x>8,
∴x的取值范围是
∪(8,+∞).
则f(t)是一次函数,当t∈[-2,2]时,
f(t)>0恒成立,则有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240354104241017.png)
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240354104401321.png)
解得log2x<-1或log2x>3.
∴0<x<
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824035410456339.png)
∴x的取值范围是
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824035410409665.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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