设A(x1,y1),B(x2,y2)是函数f(x)=12+log2x1-x的图象上满足下面条件的任意两点.若OM=12(OA+OB),则点M的横坐标为12.(1)求证:M点的纵坐标为定植;(2)若Sn

题目简介

设A(x1,y1),B(x2,y2)是函数f(x)=12+log2x1-x的图象上满足下面条件的任意两点.若OM=12(OA+OB),则点M的横坐标为12.(1)求证:M点的纵坐标为定植;(2)若Sn

题目详情

设A(x1,y1),B(x2,y2)是函数f(x)=
1
2
+log2
x
1-x
的图象上满足下面条件的任意两点.若
OM
=
1
2
(
OA
+
OB
)
,则点M的横坐标为
1
2

(1)求证:M点的纵坐标为定植;
(2)若Sn=f(
1
n
)+f(
2
n
)+…+f(
n-1
n
)
,求Sn(n≥2,n∈N*).
(3)已知an=
2
3
(n=1)
1
(Sn+1)(Sn+1+1)
(n≥2)
,(其中n∈N*,又知Tn为数列{an}的前n项和,若Tn<(15)λ(Sn+1+1)对于一切n∈N*.都成立,试求λ的取值范围.
题型:解答题难度:中档来源:不详

答案

(1)∵
OM
=class="stub"1
2
(
OA
+
OB
)
∴M是AB中点,设M为(x,y)
class="stub"1
2
(x1+x2)=x=class="stub"1
2
,得x1+x2=1,∴x1=1-x2或x2=1-x1
y=class="stub"1
2
(y1+y2)

=class="stub"1
2
[f(x1)+f(x2)]

=class="stub"1
2
(class="stub"1
2
+log2class="stub"x
1-x1
+class="stub"1
2
+log2
x2
1-x2
)

=class="stub"1
2
(1+log2
x1
1-x1
+log2
x2
1-x2
)

=class="stub"1
2
(1+log2[
x1
1-x1
x2
1-x2
]

=class="stub"1
2
(1+log2
x1
x2
x2
x1
)=class="stub"1
2

∴M点的纵坐标的定值为class="stub"1
2


(2)由(1)知,x1+x2=1,
则f(x1)+f(x2)=y1+y2=1,
Sn=f(class="stub"1
n
)+f(class="stub"2
n
)++f(class="stub"n-1
n
)
Sn=f(class="stub"n-1
n
)+f(class="stub"n-2
n
)++f(class="stub"1
n
)

上述两式相加,得
2Sn=[f(class="stub"1
n
)+f(class="stub"n-1
n
)]+[f(class="stub"2
n
)+f(class="stub"n-2
n
)]++[f(class="stub"n-1
n
)+f(class="stub"1
n
)]

=1+1++1
Sn=class="stub"n-1
2
(n≥2,n∈N*)


(3)当n=1时,Tn=a1=class="stub"2
3
Sn+1+1=S2+1=class="stub"3
2

由Tn<λ(Sn+1+1),得class="stub"3
2
<class="stub"3
2
λ
,得λ>class="stub"4
9

当n≥2时,an=class="stub"1
(Sn+1)(Sn+1+1)
=class="stub"4
(n+1)(n+2)
=4(class="stub"1
n+1
-class="stub"1
n+2
)

Tn=a1+a2++an=class="stub"2
3
+4(class="stub"1
3
-class="stub"1
n+2
)=class="stub"2n
n+2

由Tn<λ(Sn+1+1),得class="stub"2n
n+2
<λ<class="stub"n+2
n

λ>class="stub"4n
(n+2)2
=class="stub"4n
n2+4n+4
=class="stub"4
n+class="stub"4
n
+4

class="stub"4
n+class="stub"4
n
+4
≤class="stub"4
4+4
=class="stub"1
2
,(当且仅当n=2时,=成立)∴λ>class="stub"1
2

综上所述,若对一切n∈N*.都有Tn<λ(Sn+1+1)成立,由于class="stub"4
9
<class="stub"1
2
,所以λ>class="stub"1
2

更多内容推荐