【错解分析】∵不等式f(x)≤1,∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002204000457.png)
≤1 + ax.两边平方,得x
2 + 1≤(1 + ax)
2 ,
即x·[(a
2 - 1)x + 2a]≥0.∵a > 0,∴当a > 1时,x ≥ 0,或x ≤-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002204016527.png)
;
当0 < a < 1时,0 ≤ x ≤
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002203985531.png)
.
【正解】不等式f(x)≤1,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002204000457.png)
≤1 + ax.
由此得1≤1 + ax,即ax≥O,其中a > 0.
∴原不等式等价于不等式组
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240022040631033.png)
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240022040781168.png)
∴当0 < a <1时,原不等式的解集为{x|0≤x≤
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002203985531.png)
};
当a≥1时,原不等式的解集为{x|x≥O}.
【点评】准确挖掘题干中的隐含条件往往能使很多问题的求解变得非常简单,甚至关系到求解正确与否。错解中就是未能从已知条件中挖掘出隐含条件:“1 + ax ≥ 1”,即“ax≥0”,
进而由a > 0可得x≥0,导致求解错误的。