试题分析:(I)求导函数,对参数a进行讨论,利用导数的正负,确定函数的单调区间;
(II)确定f(x)的极大值为f(0)=a+b,f(x)的极小值为f(a)=a+b-a
3,要使f(x)有三个不同的零点,则f(0)>0,f(a)<0,从而得证;
(III)先确定|x
1-x
2|=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731553540.png)
,并求得其最小值,假设存在实数m满足条件,则m
2+tm+1≤(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731553540.png)
)
min,即m
2+tm+1≤4,即m
2+tm-3≤0在t∈[-1,1]上恒成立,从而可求m的范围.
解:(I)∵
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731569892.png)
,
当a=0时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731584582.png)
≥0,于是
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731600448.png)
在R上单调递增;
当a>0时,x∈(0,a),
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731615547.png)
,得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731600448.png)
在(0,a)上单调递减;
x∈(-∞,0)∪(a,+∞),
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731647537.png)
,得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731600448.png)
在(-∞,0),(a,+∞)上单调递增;
当a<0时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731693565.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731615547.png)
,得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731600448.png)
在(0,a)上单调递减;
x∈(-∞,a)∪(0,+∞),
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731740552.png)
得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731600448.png)
在(-∞,a),(0,+∞)上单调递增.
综上所述:当a=0时,f(x)的增区间为(-∞,+∞);
当a>0时,f(x)的增区间为(-∞,0),(a,+∞);f(x)的减区间为(0,a);
当a<0时,f(x)的增区间为(-∞,a),(0,+∞);f(x)的减区间为(a,0).……3分
(II)当a>0时,由(I)得f(x)在(-∞,0),(a,+∞)上是增函数,f(x)在(0,a)上是减函数;则f(x)的极大值为f(0)=a+b,f(x)的极小值为f(a)=a+b-a
3.
要使f(x)有三个不同的零点,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731756857.png)
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731787889.png)
可得-a<b<a
3-a.…8分
(III)由2x
3-3ax
2+a+b=x
3-2ax
2+3x+a+b,得x
3-ax
2-3x=0即x(x
2-ax-3)=0,
由题意得x
2-ax-3=0有两非零实数根x
1,x
2,则x
1+x
2=a,x
1x
2=-3,
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240007318031566.png)
.∵ f (x)在[1,2]上是减函数,
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731818936.png)
≤0在[1,2]上恒成立,
其中x-a≤0即x≤a在[1,2]上恒成立,∴ a≥2.∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731834518.png)
≥4.
假设存在实数m满足条件,则m
2+tm+1≤(
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731849510.png)
)
min,即m
2+tm+1≤4,即m
2+tm-3≤0在t∈[-1,1]上恒成立,
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731865880.png)
解得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731896842.png)
.
∴ 存在实数m满足条件,此时m∈[
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731522516.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824000731537515.png)
]. …………………14分
点评:解决该试题的关键是利用导数的正负对于函数单调性的影响得到函数单调区间,进而分析极值问题,以及构造函数的思想求证函数的最值,解决恒成立问题的运用。