本试题主要是考查了函数与方程的思想的综合运用。
(1)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231109124857.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231109155598.png)
+3即
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231109171570.png)
,对于定义域分段讨论得到解的情况。
(2)因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231108905454.png)
是定义域(0,2)上的单调函数,结合函数与图像的关系式得到结论。
(3)关于x的方程
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231108937516.png)
在(0,2)上有两个不同的解
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231108968422.png)
,那么借助于图像得到结论。
解(1)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231109124857.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231109155598.png)
+3即
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231109171570.png)
当
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231109311447.png)
时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231109342647.png)
,此时该方程无解. ……1分
当
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231109405449.png)
时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231109717623.png)
,原方程等价于:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231109919439.png)
此时该方程的解为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231108983336.png)
.
综上可知:方程
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231108890579.png)
+3在(0,2)上的解为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231108983336.png)
.……3分
(2)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231110029222.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231110044821.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232311100751444.png)
………4分
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231110091639.png)
,…………5分
可得:若
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231108905454.png)
是单调递增函数,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231110138683.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231110153551.png)
…6分
若
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231108905454.png)
是单调递减函数,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231110200722.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231110231599.png)
,………7分
综上可知:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231108905454.png)
是单调函数时
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231108921313.png)
的取值范围为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231109015778.png)
.…8分
(2)[解法一]:当
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231110481444.png)
时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231110497425.png)
,①
当
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231110512448.png)
时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231110528647.png)
,②
若k=0则①无解,②的解为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231110699813.png)
故
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231110731396.png)
不合题意。…………9分
若
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231110746421.png)
则①的解为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231110887466.png)
,
(Ⅰ)当
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231110902622.png)
时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231110918406.png)
时,方程②中
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231110933703.png)
故方程②中一根在(1,2)内另一根不在(1,2)内,…………10分
设
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231110965757.png)
,而
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231110980661.png)
则
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232311110111419.png)
又
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231110918406.png)
,故
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231109108614.png)
,………11分
(Ⅱ)当
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231111121649.png)
时,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231111152463.png)
或
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231111167362.png)
0时,方程②在(1,2)须有两个不同解,12分
而
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231110980661.png)
,知方程②必有负根,不合题意。……13分
综上所述,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231109108614.png)
………14分
[略解法二]
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231111323878.png)
,………9分
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232311113551258.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232311113861109.png)
………10分
分析函数的单调性及其取值情况易得解(用图象法做,必须画出草图,再用必要文字说明)……………13分
利用该分段函数的图象得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823231111401782.png)
……………………14分