设函数f（x）（x∈R）满足f（x-π）=f（x）+sinx，当0≤x≤π，f（x）=1时，则f(-13π6)=（）A.12B.-12C.32D.-32

Question

设函数f（x）（x∈R）满足f（x-π）=f（x）+sinx，当0≤x≤π，f（x）=1时，则f(-

(13π)/(6))=（　　）

A.

(1)/(2)

B. -

(1)/(2)

C.

(3)/(2)

D. -

(3)/(2)

Answer 1

∵f（x-π）=f（x）+sinx，当0≤x≤π，f（x）=1时，
则f(-(13π)/(6))=f（-(7π)/(6)-π）=f（-(7π)/(6)）+sin（-(7π)/(6)）=f（-(π)/(6)-π）+sin（-(7π)/(6)）
=f（-(π)/(6)）+sin（-(π)/(6)）+sin（-(7π)/(6)）=f（(5π)/(6)-π）+sin（-(π)/(6)）-sin(7π)/(6)
=f（(5π)/(6)）+sin(5π)/(6)+sin（-(π)/(6)）+sin(π)/(6)=1+(1)/(2)-(1)/(2)+(1)/(2)=(3)/(2)，
故选：C．