Question

若函数f(x)= sin²ax-sinaxcosax(a＞0)的图象与直线y=m(m为常数)相切，并且切点的横坐标依次成公差为的等差数列.

（1）求m的值；

（2）若点A（x₀,y₀）是y=f(x)图象的对称中心，且x₀∈［0,］,求点A的坐标.

Answer 1

解:（1）f(x)=［(1-cos2ax)-sin2ax］?

=-(sin2ax+cos2ax)+?

=-sin(2ax+)+.                                                                                           ?

∵y=f(x)的图象与y=M相切,?

∴M为f(x)的最大值或最小值,?

即M=1+或M=1-.                                                                                   ?

（2）又∵切点横坐标依次成公差为的等差数列,?

∴f(x)最小正周期为.?

又T==,a＞0,∴a=2,                                                                                      ?

即f(x)=-sin(4x+)+.                                                                                   ?

令sin(4x+)=0,则4x₀+=kπ(k∈Z),?

x₀=-.                                                                                                           ?

由0≤-≤π及k∈Z.得k=1,2,3.?

因此对称中心为(π, )、(π,)、(π,).