问题 解答题
设A(x1,y1),B(x2,y2)是函数f(x)=
1
2
+log2
x
1-x
图象上任意两点,且
OM
=
1
2
(
OA
+
OB
)
,已知点M的横坐标为
1
2

(1)求点M的纵坐标;
(2)若Sn=f(
1
n
)+f(
2
n
)+…+f(
n-1
n
)
,其中n∈N*且n≥2,
①求Sn
②已知
1
12
,其中n∈N*,Tn为数列{an}的前n项和,若Tn≤λ(Sn+1+1)对一切n∈N*都成立,试求λ的最小正整数值.
答案

(1)依题意由

OM
=
1
2
(
OA
+
OB
)知M为线段AB的中点.

又∵M的横坐标为

1
2
,A(x1,y1),B(x2,y2)即
x1+x2
2
=
1
2
x1+x2=1

y1+y2=1+log2(

x1
1-x1
x2
1-x2
)=1+log21=1⇒
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
)+f(
2
n
)+…+f(
n-1
n
)

Sn=f(

n-1
n
)+f(
n-2
n
)+••+f(
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)
=4(
1
n+1
-
1
n+2

又n=1时,a1=

2
3
也适合.

∴an=4(

1
n+1
-
1
n+2
)                                                                                     

Tn=

4
2×3
+
4
3×4
++
4
(n+1)(n+2)
=4(
1
2
-
1
3
+
1
3
-
1
4
++
1
n+1
-
1
n+2
)=4(
1
2
-
1
n+2
)=
2n
n+2
(n∈N*)

2n
n+2
≤λ(
n
2
+1)恒成立(n∈N*)⇒λ≥
4n
n2+4n+4

4n
n2+4n+4
=
4
n+
4
n
+4
4
4+4
=
1
2
(当且仅当n=2取等号)

λ≥

1
2
,∴λ的最小正整数为1.

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