对于定义在R上的奇函数f(x),满足f(x+3)=f(x),若f(-1)=1,则f(1)+.f(10)=

题目

对于定义在R上的奇函数f(x),满足f(x+3)=f(x),若f(-1)=1,则f(1)+.f(10)=

答案

解,因为f(x)是奇函数,所以f(-x) = -f(x)所以f(1) = -f(-1) = -1f(2) = f((-1) + 3) = f(-1) = 1因为f(0) = f(3) = -f(-3) = -f(-3+3) = -f(0)所以f(0) = 0而f(10) = f(7) = f(4) = f(1)f(9) = f(6) = f(3) = f(0)f(8...