已知定义在R上的奇函数f(x),满足f(x-4)=-f(x),且在区间[0,2]上是增函数,则( ) A.f(15)<f(0)<f(-5) B.f(0)<f(15)<f(-5) C.f(-5)<f(15)<f(0) D.f(-5)<f(0
问题描述:
已知定义在R上的奇函数f(x),满足f(x-4)=-f(x),且在区间[0,2]上是增函数,则( )
A. f(15)<f(0)<f(-5)
B. f(0)<f(15)<f(-5)
C. f(-5)<f(15)<f(0)
D. f(-5)<f(0)<f(15)
答
∵f(x)满足f(x-4)=-f(x),∴f(x-8)=f(x),∴函数是以8为周期的周期函数,则f(-5)=f(3)=-f(-1)=f(1),f(15)=f(-1),又∵f(x)在R上是奇函数,f(0)=0,得f(0)=0,又∵f(x)在区间[0,2]上...