题目
| A.f(-5.5)<f(2)<f(-1) | B.f(-1)<f(-5.5)<f(2) |
| C.f(2)<f(-5.5)<f(-1) | D.f(-1)<f(2)<f(-5.5) |
答案
∴函数f(x)周期为2的偶函数,
∴f(-5.5)=f(0.5)f(2)=f(0)
f(-1)=f(1)
又∵f(x)的区间[0,1]上是增函数,
∴f(0)<f(0.5)<f(1)
即f(2)<f(-5.5)<f(-1)
故选C
| A.f(-5.5)<f(2)<f(-1) | B.f(-1)<f(-5.5)<f(2) |
| C.f(2)<f(-5.5)<f(-1) | D.f(-1)<f(2)<f(-5.5) |