对任意x∈R,函数f(x)满足f(x+1)=f(x)-[f(x)]2+12,设an=[f(n)]2-f(n),数列{an}的前15项的和为-3116,则f(15)=______.-数学

题目简介

对任意x∈R,函数f(x)满足f(x+1)=f(x)-[f(x)]2+12,设an=[f(n)]2-f(n),数列{an}的前15项的和为-3116,则f(15)=______.-数学

题目详情

对任意x∈R,函数f(x)满足f(x+1)=
f(x)-[f(x)]2
+
1
2
,设an=[f(n)]2-f(n),数列{an}的前15项的和为-
31
16
,则f(15)=______.
题型:填空题难度:中档来源:东城区二模

答案

f(x+1)=
f(x)-[f(x)]2
+class="stub"1
2

f(x+1)-class="stub"1
2
=
f(x)-[f(x)]2

两边平方得[f(x+1)-class="stub"1
2
]
2
=f(x)-[f(x)]2

⇒[f(x+1)]2-f(x+1)+class="stub"1
4
=f(x)-[f(x)]2

an+1+an=-class="stub"1
4
,即数列{an}任意相邻两项相加为常数-class="stub"1
4

S15=7×(-class="stub"1
4
)+a15=-class="stub"31
16
a15=-class="stub"3
16

[f(15)]2-f(15)=-class="stub"3
16
⇒f(15)=class="stub"3
4
或f(15)=class="stub"1
4

又由f(x+1)=
f(x)-[f(x)]2
+class="stub"1
2
≥class="stub"1
2

可得f(15)=class="stub"3
4

故答案为:class="stub"3
4

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