试题分析:(Ⅰ)当
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638264560.png)
时,由
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638108765.png)
有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638295833.png)
,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638311664.png)
3分
f(x)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638327316.png)
的定义域为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638342764.png)
,
令
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638358514.png)
,整理得x
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638358242.png)
+x+1=0,△=-3<0,
因此,不存在x
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638233246.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638342764.png)
使得f(x+1)=f(x)+f(1)成立,所以f(x)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638217612.png)
;…… 6分
(Ⅱ)f(x)=lg
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638420519.png)
的定义域为R,f(1)=lg
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638436401.png)
,a>0, ..7分
若f(x)= lg
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638420519.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638233246.png)
M,则存在x
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638233246.png)
R使得lg
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638498673.png)
=lg
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638420519.png)
+lg
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638436401.png)
,
整理得存在x
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638233246.png)
R使得(a-2)x
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638358242.png)
+2ax+(2a-2)=0. 8分
(1)若a-2=0即a=2时,方程化为8x+4=0,解得x=-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638576338.png)
,满足条件; 10分
(2)若a-2
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638592237.png)
0即a
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638233246.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638623790.png)
时,令△≥0, 12分
解得a
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638233246.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638639922.png)
, 13分
综上,a
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638233246.png)
[3-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638233305.png)
,3+
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824002638233305.png)
]; 14分
点评:综合题,本题以新定义函数为载体,综合考查对数函数的性质,方程解的讨论,对考生数学式子变形能力要求较高。本题较难。