设函数y=f(x)=2x2x+2上两点p1(x1,y1),p2(x2,y2),若op=12(op1+op2),且P点的横坐标为12.(1)求P点的纵坐标;(2)若Sn=f(1n)+f(2n)+…+f(

题目简介

设函数y=f(x)=2x2x+2上两点p1(x1,y1),p2(x2,y2),若op=12(op1+op2),且P点的横坐标为12.(1)求P点的纵坐标;(2)若Sn=f(1n)+f(2n)+…+f(

题目详情

设函数y=f(x)=
2x
2x+
2
上两点p1(x1,y1),p2(x2,y2),若
op
=
1
2
(
op1
+
op2
)
,且P点的横坐标为
1
2

(1)求P点的纵坐标;
(2)若Sn=f(
1
n
)+f(
2
n
)+…+f(
n-1
n
)+f(
n
n
)
,求Sn
(3)记Tn为数列{
1
(Sn+
2
)(Sn+1+
2
)
}
的前n项和,若Tn<a(Sn+2+
2
)
对一切n∈N*都成立,试求a的取值范围.
题型:解答题难度:中档来源:不详

答案

(1)∵
OP
=class="stub"1
2
(
OP1
+
OP2
)
,∴P为P1P2的中点,∴x1+x2=1
∴y1+y2=
2x1
2x1+
2
+
2x2
2x2+
2
=1
∴P的纵坐标为class="stub"1
2

(2)由(1)知,x1+x2=1,y1+y2=1,f(1)=2-
2

Sn=f(class="stub"1
n
)+f(class="stub"2
n
)+…+f(class="stub"n-1
n
)+f(class="stub"n
n
)
Sn=f(class="stub"n
n
)+f(class="stub"n-1
n
)+…+f(class="stub"2
n
)+f(class="stub"1
n
)

2Sn=(n-1)+2(2-
2
)
=n+3-2
2

Sn=
n+3-2
2
2

(3)Sn+
2
=class="stub"n+3
2
Sn+1+
2
=class="stub"n+4
2

class="stub"1
(Sn+
2
)(Sn+1+
2
)
=class="stub"4
(n+3)(n+4)
=4(class="stub"1
n+3
-class="stub"1
n+4

∴Tn=4(class="stub"1
4
-class="stub"1
5
+class="stub"1
5
-class="stub"1
6
+…+class="stub"1
n+3
-class="stub"1
n+4
)=class="stub"n
n+4

Tn<a(Sn+2+
2
)
对一切n∈N*都成立
∴a>
Tn
Sn+2+
2
=class="stub"2
n+class="stub"20
n
+9

设g(n)=n+class="stub"20
n
,则g(n)在[
20
,+∞)上是增函数,在(0,
20
)上是减函数
∴g(n)的最小值为9
class="stub"2
n+class="stub"20
n
+9
<class="stub"1
9

∴a>class="stub"1
9

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