高一数学几道函数求值域题!y=(2sinθ-1)/(1-sinθ),y=(3^x)/(1+3^x),y=(2sinθ-1

1个回答

  • 1.

    y=(2sinθ-1)/(1-sinθ)

    =[(2sinθ-2)+1]/(1-sinθ)

    =[-2(1-sinθ)+1]/(1-sinθ)

    =-2+1/[1-sinθ]

    由于:sinθ属于[-1,1]

    则:(1-sinθ)属于[0,2]

    则:1/(1-sinθ)属于[1/2,正无穷)

    则:

    y=-2+1/[1-sinθ]属于[-3/2,正无穷)

    2.

    y=3^x/(1+3^x)

    =[(3^x+1)-1]/(3^x+1)

    =1-1/(3^x+1)

    由于:3^x+1属于(1,正无穷)

    则:1/(3^x+1)属于(0,1)

    则:

    y=1-1/(3^x+1)属于(0,1)

    3.

    y=(2sinθ-1)/(1+cosθ)

    =[4sin(θ/2)cos(θ/2)-1]/[1+2cos^2(θ/2)-1]

    =[4sin(θ/2)cos(θ/2)-sin^2(θ/2)-cos^2(θ/2)]

    /[2cos^2(θ/2)]

    =2tan(θ/2)-(1/2)tan^2(θ/2)-(1/2)

    =-(1/2)[tan(θ/2)-2]^2+3/2

    则:

    Y属于(负无穷,3/2]