Question

于知函数f(X)=sin²ωx+根号3sinωxsin(ωx+π/2(ω>0)的最小正周期为π。

（1）求ω的值；（2）求函数f(x)在区间[0,2π/3]上的取值范围

Answer 1

f(x)=sin²(ωx)+3^½*{sin(ωx)*[-cos(ωx)]}
=[2sin²(ωx)-1]/2+1/2-3^½*[2sin(ωx)*cos(ωx)]/2
=-cos(2ωx)/2-3^½*sin(2ωx)/2+1/2
=-[sin(π/6)cos(2ωx)+cos(π/6)sin(2ωx)]+1/2
=-sin(2ωx+π/6)+1/2
∵T=(2π)/(2ω)=π
∴ω=1
即 f(x)=-sin(2x+π/6)+1/2
∵0≤x≤2π/3
∴0≤2x≤4π/3
π/6≤2x+π/6≤3π/2
∴sin(2x+π/6)∈[-1，1]
当2x+π/6=π/2，即x=π/6时，有最小值：f(x)(min)=-1+1/2= -1/2
当x=2π/3时，有最大值：f(x)(max)=-sin[2(2π/3)+π/6)+1/2=1+1/2=3/2
所以取值范围为：[-1/2，3/2]

Answer 2

f(x)=sin²ωx+√3sinωxcosωx
=1/2-1/2cos2ωx+√3/2sin2ωx
=1/2+sin2ωxcosπ/6-cos2ωxsinπ/6
=sin(2ωx+π/6)+1/2
根据题意
2π/2ω=π
ω=1
f(x)=sin(2x+π/6)+1/2
0≤x≤2π/3
0≤2x≤4π/3
π/6≤2x+π/6≤3π/2
所以sin（2x+π/6）∈[-1,1]
所以f(x)∈[-1/2,3/2]