高一数学题,急有两个函数f(x)=asin(kx+π/3),g(x)=btan(kx-π/3)(k>0),它们的最小正周

2个回答

  • f(x)的最小正周期T1=2π/k

    g(x)的最小正周期T2=π/k

    T1+T2=2π/k+π/k=3π/2 k=2

    f(π/2)=asin(π+π/3)=-asinπ/3=-a*√3/2

    g(π/2)=btan(kx-π/3)=btan(π-π/3)=-b*√3

    f(π/2)=g(π/2),a=2b

    f(π/4)=asin(π/2+π/3)=a/2

    g(π/4)=btan(π/2-π/3)=b*√3/3

    f(π/4)=-根号3×g(π/4)+1

    a/2=-√3*b*√3/3+1

    a=2b

    b=-b+1

    b=1/2 a=1

    1.f(x)=sin(2x+π/3),g(x)=1/2*tan(2x-π/3)

    2.有图像可知 函数f(x)在〔0,π〕上的增区间 【0,π/12】,【7π/12,π)

    3.G(x)=log(1/2) [1/2*tan(2x-π/3)]

    tan(2x-π/3)>0