已知函数f（x）=Asin(ωx+φ)+B(A>0,ω>0）的周期为2π 最大值为f(5π/6,）=3,最小值为-1,

题目

已知函数f（x）=Asin(ωx+φ)+B(A>0,ω>0）的周期为2π 最大值为f(5π/6,）=3,最小值为-1,
（1）求f（x）的解析式
（2）若函数y=f（kx）(k>0)周期为2π/3,当x∈[0,π/3]方程f（kx）=m,恰有两个不同的解,求实数m的取值范围

答案

(1) B=（3-1）/2=1
A=（3+1）/2=2
w=2
f(x)=2sin(2x+θ)+1
x=5π/6
f(5π/6)=2sin(2*5π/6+θ)+1=3
sin(5π/3+θ)=1 ==>5π/3+θ=2kπ+π/2
θ=2kπ-7π/6
θ>0 ==>θ=2π-7π/6=5π/6
f(x)=2sin(2x+5π/6)+1
(2)y=f(kx)=2sin(2kx+5π/6)+1
T=2π/2k=2π/3 ==>k=3/2
y=f(kx)=2sin(3x+5π/6)+1=m
0≤x≤π/3
0≤3x≤π
5π/6≤3x+5π/6≤π+5π/6=11π/6
1/2sin(3x+5π/6)≤1
2≤2sin(3x+5π/6)+1≤3
恰有两个不同的解,2≤m