题目内容
(本小题满分12分)设函数
的图象上两点P1(x1,y1)、P2(x2,y2),若
,且点P的横坐标为
.
(1),求证:P点的纵坐标为定值,并求出这个定值;
(2),求![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232221341051281.png)
(3),记Tn为数列
的前n项和,若
对一切n∈N*都成立,试求a的取值范围。
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222134058799.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222134074799.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222134089331.png)
(1),求证:P点的纵坐标为定值,并求出这个定值;
(2),求
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232221341051281.png)
(3),记Tn为数列
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222134136990.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222134152819.png)
(1)见解析;(2)
;(3)
.
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222134183800.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222134199467.png)
本试题主要考查了函数,与向量,以及数列的知识的综合运用。以函数为模型,确定点的坐标关系式,进一步结合向量得到结论,并利用倒序相加法求解和,同时利用裂项求和得到不等式的证明。
(1)由于点在函数图像上,同时满足
,那么利用坐标化简得到结论。
(2)根据f (x1)+f (x2)=y1+y2=1,f (1)=2-
,结合倒序相加法求解得到结论。
(3)根据已知的和式得到
,裂项求和的数学思想得到证明。
(1)证:∵
,∴P是P1P2的的中点Þx1+x2=1------(2分)
∴![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232221342922436.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232221343081377.png)
∴
.-----------------------------(4分)
(2)解:由(1)知x1+x2=1,f (x1)+f (x2)=y1+y2=1,f (1)=2-
,
,![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232221344011226.png)
相加得![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232221344172729.png)
(n-1个1)
∴
.------------(8分)
(3)解:![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232221345112566.png)
--------------------(10分)
Û
∵
≥8,当且仅当n=4时,取“=” ∴
,因此,
-------------------(12分)
(1)由于点在函数图像上,同时满足
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222134074799.png)
(2)根据f (x1)+f (x2)=y1+y2=1,f (1)=2-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222134230317.png)
(3)根据已知的和式得到
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232221342612508.png)
(1)证:∵
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222134074799.png)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232221342922436.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232221343081377.png)
∴
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222134339827.png)
(2)解:由(1)知x1+x2=1,f (x1)+f (x2)=y1+y2=1,f (1)=2-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222134230317.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232221343701251.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232221344011226.png)
相加得
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232221344172729.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222134448626.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222134464543.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222134183800.png)
(3)解:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232221345112566.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232221345261677.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222134152819.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232221345731549.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222134589490.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222134604971.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823222134199467.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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