本小题主要考查函数单调性的应用、利用导数研究函数的单调性、导数的几何意义、不等式的解法等基础知识,考查运算求解能力,属于基础题.
(1)先对函数y=f(x)进行求导,然后令导函数大于0(或小于0)求出x的范围,根据f′(x)>0求得的区间是单调增区间,f′(x)<0求得的区间是单调减区间,即可得到答案.
(2)设h(x)=g(x)+x,依题意得出h(x)在(0,2]上是减函数.下面对x分类讨论:①当1≤x≤2时,②当0<x<1时,利用导数研究函数的单调性从及最值,即可求得求a的取值范围.
解:⑴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230059178497.png)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230059194316.png)
-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230059225704.png)
﹥1
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230059241223.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230059178497.png)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230059288935.png)
﹥0
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230059241223.png)
x﹥2或0﹤x﹤
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230059116338.png)
,
所以函数
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230059428491.png)
的单调增区间为(0,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230059116338.png)
)和(2,+∞)……………………………3分
⑵因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230059475823.png)
﹤-1,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232300596931040.png)
﹤0,
所以F
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230059927421.png)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230059943539.png)
在区间(0,2】上是减函数。
① 当1≦x≦2时,F
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230059927421.png)
=ln
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230059990266.png)
+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230100099511.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232301001301053.png)
,
由
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232301001461773.png)
在x∈
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230100161391.png)
上恒成立。
设
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230100177952.png)
,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230100208928.png)
﹥0(1≦x≦2),
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230100224524.png)
在[1,2]上为增函数,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230100302883.png)
②当0﹤x﹤1时,F
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230059927421.png)
=-ln
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230059990266.png)
+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230100099511.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232301003801073.png)
,
由
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230100411824.png)
-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230100426785.png)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230100489551.png)
在x∈(0,1)上恒成立。
令
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230100504472.png)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230100489551.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230100551865.png)
﹥0,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230100504472.png)
在(0,1)上为增函数,所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230100598610.png)
,综上:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230059132283.png)
的取值范围为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230059132283.png)
≧
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823230059163422.png)
…………………12分