试题分析:(1)极值点的求法是利用导数知识求解,求出
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857209501.png)
,求得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857209571.png)
的解
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857225324.png)
,然后确定当
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857240382.png)
以及
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857256404.png)
时的
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857209501.png)
的符号,若当
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857240382.png)
时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857303586.png)
,当
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857256404.png)
时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857334578.png)
,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857225324.png)
是极大值点,反之是极小值点;(2)题设中没有其他的已知条件,我们只能设
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857365870.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857365592.png)
,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857381552.png)
的横坐标为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857396542.png)
,利用导数可得出切线的斜率
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857412677.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857427924.png)
,题设要证明的否定性命题,我们用反证法,假设两切线平行,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857443450.png)
,也即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857459999.png)
,下面的变化特别重要,变化的意图是把这个等式与已知函数联系起来,等式两边同乘以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857474422.png)
,得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240428574901055.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240428575051226.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857521714.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857537532.png)
,从而等式变为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240428575521110.png)
,注意到
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857568604.png)
,此等式为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857583741.png)
能否成立?能成立,说明存在平行,不能成立说明不能平行.设
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857599893.png)
,仍然用导数的知识来研究函数的性质,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240428576151222.png)
,即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857630424.png)
是增函数,从而在
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857630356.png)
时,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857661643.png)
,即等式
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857583741.png)
不可能成立,假设不成立,结论得证.
试题解析:(1)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857677797.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240428576931083.png)
2分
令
h’(
x)=0,则4
x2+2
x-1=0,
解出
x1=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857708477.png)
,
x2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857193463.png)
3分
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240428577392258.png)
4分
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240428577552187.png)
5分
所以
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042856944484.png)
的极大值点为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857193463.png)
6分
(2)设
P、
Q的坐标分别是
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857786981.png)
.
则
M、
N的横坐标
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857802592.png)
.
∴
C1在点
M处的切线斜率为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857412677.png)
,
C2在点
N处的切线斜率为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857427924.png)
. 7分
假设
C1在点
M处的切线与
C2在点
N处的切线平行,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857443450.png)
,
即
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857864963.png)
8分
则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240428578642968.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240428578951096.png)
10分
设
t=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857911415.png)
,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857927841.png)
①
令
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240428579421033.png)
则
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408240428579421118.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857958693.png)
∴
r(
t)在[1,+∞)上单调递增,故
r(
t)>
r(1)=0.
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824042857973764.png)
,这与①矛盾,假设不成立,
故
C1在点
M处的切线与
C2在点
N处的切线不平行. 12分