Question

设函数f(x)=sin(π/3x-π/6)-2cos²πx/6, 求(1)f(x)最小正周期及单调递增区间.

设函数f(x)=sin(π/3x-π/6)-2cos�0�5πx/6,
求(1)f(x)最小正周期及单调递增区间.
(2)若函数y=g(x)与y=f(x)的图像关于直线X=2对称,求出X属于[0,1]时,函数y=g(x)的最大值.

Answer 1

∵f(x)=sinxsin(x π/2)-√3cos�0�5(3π x) √3/2
=sinxcosx-√3cos�0�5x √3/2
=(1/2)sin2x-(√3/2)cos2x
=sin(2x-π/3)
(1)∴T=2π/2=π
(2)令2x-π/3=2kπ-π/2，解得x=kπ-π/12，k∈Z
令2x-π/3=2kπ π/2，解得x=kπ 5π/12，k∈Z
∴f(x)的单调递增区间为x∈[kπ-π/12，kπ 5π/12]，k∈Z
(3)令2x-π/3=kπ π/2，解得x=kπ/2 5π/12
∴f(x)的对称轴方程为x=kπ/2 5π/12，k∈Z
令2x-π/3=kπ，解得x=kπ/2 π/6
∴f(x)的对称中心坐标为(kπ/2 π/6，0)，k∈Z