【答案】
分析:(1)根据奇函数的性质f(-x)=f(x),已知条件函数f(x)=ax
3+bx
2+cx(a≠0)在x=1处取得极大值2可以推出f′(1)=0和f(1)=2,代入即可求得函数y=f(x)的解析式;
(2)根据题意对于区间[-2,2]上任意两个自变量的值x
1,x
2都有|f(x
1)-f(x
2)|≤c,将问题转化为)|f(x
1)-f(x
2)|≤|f(x)
max-f(x)
min|,求出f(x)的最大值和最小值即可;
(3)已知关于p的一元二次方程p
2-2mp+4=0两个根均大于1,根据根与系数的关系求出m的范围,利用导数研究函数g(x)的单调性;
解答:解:(1)∵奇函数f(x)=ax
3+bx
2+cx(a≠0)在x=1处取得极大值2,奇函数f(-x)=-f(x),解得b=0,
可得f′(x)=3ax
2+c
由题
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103100332353090686/SYS201311031003323530906019_DA/0.png)
,解得
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103100332353090686/SYS201311031003323530906019_DA/1.png)
,f(x)=-x
3+3x;
(2)|f(x
1)-f(x
2)|≤|f(x)
max-f(x)
min|=4,
根据(1)可得f(x)=-x
3+3x;
求导得f′(x)=-3x
2+3=-3(x
2-1)令f′(x)=0,可得x=1或-1,
当f′(x)>0即-1<x<1,f(x)为增函数,
当f′(x)<0时即x>1或x<-1,f(x)为减函数,
f(x)在x=1处取极大值f(1)=2,在x=-1处取得极小值f(-1)=-,2;
f(-2)=2,f(2)=-2,
∴f(x)
max=2,f(x)
min=-2,
要使对于区间[-2,2]上任意两个自变量的值x
1,x
2都有|f(x
1)-f(x
2)|≤c,
∴|f(x
1)-f(x
2)|≤|f(x)
max-f(x)
min|=4,
故c的最小值为4;
(3)p
2-2mp+4=0两个根均大于1,
则求得
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103100332353090686/SYS201311031003323530906019_DA/2.png)
,g(x)=-x
2+3+mlnx,则x>0.
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103100332353090686/SYS201311031003323530906019_DA/3.png)
.
而
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103100332353090686/SYS201311031003323530906019_DA/4.png)
,则
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103100332353090686/SYS201311031003323530906019_DA/5.png)
时,g'(x)>0,
故
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103100332353090686/SYS201311031003323530906019_DA/6.png)
是g(x)的单调增区间,
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103100332353090686/SYS201311031003323530906019_DA/7.png)
时,g'(x)<0,故
![](http://thumb.1010pic.com/pic6/res/gzsx/web/STSource/20131103100332353090686/SYS201311031003323530906019_DA/8.png)
是g(x)的单调减区间.
点评:此题主要考查利用导数研究函数的单调性,考查的知识点比较全面是一道中档题,这类题是高考的热点问题;