Question

已知f(x)=√3sinωx(√3sinωx+cosωx)+t,在函数的图像中,两条对称轴间的最小距离为π/2,

且当x∈[0,π/3]时f(x)的最小值为-1,求函数解析式

Answer 1

f(x)=√3sinωx(√3sinωx+cosωx)+t
=3sin²ωx + √3sinωxcosωx + t
=(3/2)(1-cos2ωx) + (√3/2)sin2ωx + t
=(√3/2)sin2ωx -(3/2)cos2ωx+ 3/2 + t
=√3[(1/2)sin2ωx -(√3/2)cos2ωx] +3/2 + t
=√3sin(2ωx-π/3) +3/2 + t
∵两条对称轴间的最小距离为π/2
∴T/2=π/2，T=π，即2π/(2ω)=π
∴ω=1
∴f(x)=√3sin(2x-π/3) +3/2 + t
0≤x≤π/3
0≤2x≤2π/3
-π/3≤2x-π/3≤π/3
-√3/2≤sin(2x-π/3)≤√3/2
-3/2≤√3sin(2x-π/3)≤3/2
t≤√3sin(2x-π/3)+3/2 + t≤3+t
∴t=-1
∴f(x)=√3sin(2x-π/3) +1/2