B
分析:先根據(jù)數(shù)列的周期性,分別計(jì)算a
2010,a
2009,a
2011的值,并利用函數(shù)的對稱性將三個(gè)值化到同一區(qū)間(0,1)上,再利用函數(shù)圖象得函數(shù)f(x)在(0,1)上的單調(diào)性,利用單調(diào)性比較大小即可
解答:∵a
n+3=a
n,∴數(shù)列{a
n}為周期為3的周期數(shù)列,∴a
2010=a
3×670=
,a
2009=
,a
2011=
∴f(a
2011)=f(
),f(a
2009)=f(
)=f(2-
)=f(
),f(a
2010)=f(
)
∵f(x)=f(2-x),∴函數(shù)f(x)的圖象關(guān)于x=1對稱,又∵當(dāng)x≤1時(shí),f(x)=|1-a
x|(a>1),故函數(shù)f(x)的圖象如圖:
函數(shù)f(x)在(0,1)上為增函數(shù),
∵
<
<
,∴f(
)<f(
)<f(
)
即f(a
2011)<f(a
2009)<f(a
2010)
故選 B
點(diǎn)評:本題考查了函數(shù)的對稱性,函數(shù)的單調(diào)性,指數(shù)函數(shù)的圖象和性質(zhì),數(shù)列的周期性,及里用單調(diào)性比較大小的方法