函数f(x)=2sin(x-π/3）+1,若函数y=f(kx)(k>0)的周期为2π/3,当x∈[0,π/3]时,方程f（kx）=m恰有两个不

题目

函数f(x)=2sin(x-π/3）+1,若函数y=f(kx)(k>0)的周期为2π/3,当x∈[0,π/3]时,方程f（kx）=m恰有两个不
同的解,求实数m的取值范围?

答案

y=f(kx)=2sin(kx-π/3）+1T=2π/k=2π/3k=3y=f(kx)=2sin(3x-π/3）+1x∈[0,π/3]3x-π/3∈[-π/3,2π/3,]sin(3x-π/3）∈[-√3/2,1]y=f(kx)=2sin(3x-π/3）+1）∈[1-√3,3]当x=π/3时,y=√3+1m∈[1+√3,3)...