题目内容
设A(x1,y1),B(x2,y2)是函数f(x)=| 1 |
| 2 |
| x |
| 1-x |
| OM |
| 1 |
| 2 |
| OA |
| OB |
| 1 |
| 2 |
(1)求点M的纵坐标;
(2)若Sn=f(
| 1 |
| n |
| 2 |
| n |
| n-1 |
| n |
①求Sn;
②已知an=
|
分析:(1)由题设条件知M是AB的中点,由中点坐标公式可以求出M点的给坐标.
(2)①Sn=
f(
)=f(
)+f(
)++f(
),即 Sn=f(
)+f(
)++f(
)以上两式相加后两边再同时除以2就得到Sn.②当n≥2时,根据题设条件,由Tn<λ(Sn+1+1)得
<λ•
,∴λ>
=
=
,再由均值不等式求出λ的取值范围.
(2)①Sn=
| n-1 |
| i=1 |
| i |
| n |
| 1 |
| n |
| 2 |
| n |
| n-1 |
| n |
| n-1 |
| n |
| n-2 |
| n |
| 1 |
| n |
| 2n |
| n+2 |
| n+2 |
| 2 |
| 4n |
| (n+2)2 |
| 4n |
| n2+4n+4 |
| 4 | ||
n+
|
解答:解:(1)依题意由
=
(
+
)知M为线段AB的中点.
又∵M的横坐标为
,A(x1,y1),B(x2,y2)即
=
?x1+x2=1
∴y1+y2=1+log2(
•
)=1+log21=1?
=
即M点的纵坐标为定值
.
(2)①由(Ⅰ)可知f(x)+f(1-x)=1,
又∵n≥2时Sn=f(
)+f(
)+…+f(
)
∴Sn=f(
)+f(
)+••+f(
)
两式想加得,2Sn=n-1
Sn=
②当n≥2时,an=
=
=4(
-
)
又n=1时,a1=
也适合.
∴an=4(
-
)
∴Tn=
+
++
=4(
-
+
-
++
-
)=4(
-
)=
(n∈N*)
由
≤λ(
+1)恒成立(n∈N*)?λ≥
而
=
≤
=
(当且仅当n=2取等号)
∴λ≥
,∴λ的最小正整数为1.
| OM |
| 1 |
| 2 |
| OA |
| OB |
又∵M的横坐标为
| 1 |
| 2 |
| x1+x2 |
| 2 |
| 1 |
| 2 |
∴y1+y2=1+log2(
| x1 |
| 1-x1 |
| x2 |
| 1-x2 |
| y1+y2 |
| 2 |
| 1 |
| 2 |
即M点的纵坐标为定值
| 1 |
| 2 |
(2)①由(Ⅰ)可知f(x)+f(1-x)=1,
又∵n≥2时Sn=f(
| 1 |
| n |
| 2 |
| n |
| n-1 |
| n |
∴Sn=f(
| n-1 |
| n |
| n-2 |
| n |
| 1 |
| n |
两式想加得,2Sn=n-1
Sn=
| n-1 |
| 2 |
②当n≥2时,an=
| 1 |
| (Sn+1)(Sn+1+1) |
| 4 |
| (n+1)(n+2) |
| 1 |
| n+1 |
| 1 |
| n+2 |
又n=1时,a1=
| 2 |
| 3 |
∴an=4(
| 1 |
| n+1 |
| 1 |
| n+2 |
∴Tn=
| 4 |
| 2×3 |
| 4 |
| 3×4 |
| 4 |
| (n+1)(n+2) |
| 1 |
| 2 |
| 1 |
| 3 |
| 1 |
| 3 |
| 1 |
| 4 |
| 1 |
| n+1 |
| 1 |
| n+2 |
| 1 |
| 2 |
| 1 |
| n+2 |
| 2n |
| n+2 |
由
| 2n |
| n+2 |
| n |
| 2 |
| 4n |
| n2+4n+4 |
而
| 4n |
| n2+4n+4 |
| 4 | ||
n+
|
| 4 |
| 4+4 |
| 1 |
| 2 |
∴λ≥
| 1 |
| 2 |
点评:本题考查了数列与函数、函数的图象、不等式等综合内容,函数图象成中心对称的有关知识,考查相关方法,考查了数列中常用的思想方法,如倒序相加法,裂项相消法求数列前n项的和,利用函数与方程的思想,转化与化归思想解答热点问题--有关恒成立问题.
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