Question

已知函数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

Answer 1

f(kx）=2sin(kx-π/3）+1首先周期t=2π/3 因为 t=2π/k=2π/3所以 k=3 因为x∈[0,π/3] 所以 kx∈[0,π]

记kx=n 故f(n）=2sin(n-π/3）+1 记u=n-π/3∈[-π/3,2π/3] 有两个不同解就是y=m与之有两个交点 根据函数图像可知 m∈[根号3+1,3]