已知函數(shù)f(x)對(duì)任意x∈R都有f(x+4)-f(x)=2f(2),若y=f(x-1)的圖象關(guān)于直線x=1對(duì)稱,且f(1)=2,則f(1003)=( )
A.2
B.3
C.4
D.6
【答案】分析:由函數(shù)f(x-1)的圖象關(guān)于直線x=1對(duì)稱且由y=f(x-1)向左平移1個(gè)單位可得y=f(x)的圖象可知函數(shù)y=f(x)的圖象關(guān)于x=0對(duì)稱即函數(shù)y=f(x)為偶函數(shù),在已知條件中令x=-2可求f(2)=0,從而求得函數(shù)的周期,利用所求周期以及偶函數(shù)的性質(zhì)即可求解.
解答:解::∵函數(shù)f(x-1)的圖象關(guān)于直線x=1對(duì)稱且把y=f(x-1)向左平移1個(gè)單位可得y=f(x)的圖象,
∴函數(shù)y=f(x)的圖象關(guān)于x=0對(duì)稱,即函數(shù)y=f(x)為偶函數(shù).
∵f(x+4)-f(x)=2f(2),
令x=-2可得f(2)-f(-2)=2f(2),則f(2)=0.
從而可得f(x+4)=f(x)即函數(shù)是以4為周期的周期函數(shù),
∴f(1003)=f(250×16+3)=f(3)=f(-1)=f(1)=2,
故選A.
點(diǎn)評(píng):本題主要考出了函數(shù)的圖象的平移及函數(shù)圖象的對(duì)稱性的應(yīng)用,利用賦值求解抽象函數(shù)的函數(shù)值,函數(shù)周期的求解是解答本題的關(guān)鍵所在,屬于中檔題.