(1)根据
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223505026875.png)
有两个不同的实数根,从而得到b,a的一个不等式,再根据
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223505057487.png)
得到a,b的等式,消去b,可以解出a的取值范围.
(2)直接求其极小值,根据极小值为1,求出a的值即可.
(3)先求出
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223505088987.png)
,然后问题的关键是
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232235048701030.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232235049011726.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232235051661666.png)
下面采用均值不等式进行证明即可.
解:(1)∵
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232235051821071.png)
,∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223505213815.png)
,由题意∴f
/(1)=1+2a-b=1,
∴b=2a. ① ……2分
∵f(x)有极值,∴方程f
/(x)=x
2+2ax-b=0有两个不等实根.
∴△=4a
2+4b>0、 ∴a
2+b>0. ②
由①、②可得,α
2+2a>0.∴a<-2或a>0.故实数a的取值范围是
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223504838728.png)
4分
(2)存在
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223505416500.png)
.……………5分
由(1)可知
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223505587812.png)
,令f
/(x)=0
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223505618902.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232235057128090.jpg)
∴x=x
2时,f(x)取极小值,则f(x
2)=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223505743990.png)
=1,
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232235057591234.png)
……………………………………………………7分
若x
2=0,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223505868641.png)
则a=0(舍).……………………8分
若
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232235058992017.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232235059151794.png)
∴存在实数
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223504854584.png)
,使得函数f(x)的极小值为1 ………9分
(3)∵
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223505962968.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223505977828.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223506008843.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223506196987.png)
…….l0分
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232235062111042.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232235063201757.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232235051661666.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232235049631726.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223504979741.png)
∴其中等号成立的条件为x=1…………………………………………………………13分
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232235049941188.png)
…………………………………………14分