已知函数y=f(x+12)为奇函数,设g(x)=f(x)+1,则g(12011)+g(22011)+g(32011)+g(42011)+…+g(20102011)=()A.1005B.2010C.20

题目简介

已知函数y=f(x+12)为奇函数,设g(x)=f(x)+1,则g(12011)+g(22011)+g(32011)+g(42011)+…+g(20102011)=()A.1005B.2010C.20

题目详情

已知函数y=f(x+
1
2
)
为奇函数,设g(x)=f(x)+1,则g(
1
2011
)+g(
2
2011
)+g(
3
2011
)+g(
4
2011
)+…+g(
2010
2011
)
=(  )
A.1005B.2010C.2011D.4020
题型:单选题难度:中档来源:蓝山县模拟

答案

∵函数y=f(x+class="stub"1
2
)
为奇函数
∴f(-x+class="stub"1
2
)=-f(x+class="stub"1
2
),即f(x)+f(1-x)=0
则f(class="stub"1
2011
)+f(class="stub"2010
2011
)=0,f(class="stub"2
2011
)+f(class="stub"2009
2011
)=0,
根据g(x)=f(x)+1可得
g(class="stub"1
2011
)+g(class="stub"2
2011
)+g(class="stub"3
2011
)+g(class="stub"4
2011
)+…+g(class="stub"2010
2011
)
=2010,
故选B.

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