函数f(x)=根号a*sin(a-1)x+cos(a-1)x的最大值为2,则f(x)的最小正周期是?

题目

函数f(x)=根号a*sin(a-1)x+cos(a-1)x的最大值为2,则f(x)的最小正周期是?

答案

f(x)=根号a*sin(a-1)x+cos(a-1)x
=根号(a+1)[根号a/根号(a+1)*sin(a-1)x+1/根号(a+1)cos(a-1)x]
=根号(a+1)sin[A+(a-1)x]（令cosA=根号a/根号(a+1),0则f(x)max=根号(a+1)=2,则a=3,
此时A=30,f(x)=根号(a+1)sin[A+(a-1)x]=2sin(30+2x),最小正周期为2pai/2=pai