数学.1,若存在过点(1,0)的直线与曲线y=x^3和y=ax^2+15/4x-9都相切,则a等于?

3个回答

  • y=x^3导数为y=3x^2,直线与其切点为(m,m^3)

    则直线过(m,m^3),(1,0)

    求得直线为y=0或者y=27/4*(x-1)

    若y=0.则y=ax^2+15/4x-9顶点在x轴

    得a=-25/64

    若y=27/4*(x-1),斜率为27/4

    y=ax^2+15/4x-9导数为y=2ax+15/4,

    直线与其切点为(n,an^2+15/4n-9)

    2an+15/4=27/4

    n=3/(2a)

    直线过(3/2,27/8),(1,0) (3/(2a),(63-72a)/8a)

    推出a=-1

    所以a=-25/64或者a=-1

    sin^2a+cos^2a=1,得sin^2a=4/5

    f(x)=(1+1/tanx)sin^2-2sin(x+π/4)sin(x-π/4).

    =sinx(cosx+sinx)+2sin(x+π/4)cos(x+π/4)

    =1/2sin2x+sin^2x+sin(2x+π/2)

    =1/2sin2x+1/2-1/2cos2x+cos2x

    =(sin2x+cos2x+1)/2

    tana=sina/cosa=2,所以sin^2a=4cos^2a,即1-cos2a=4*(1+cos2a)得cos2a=-3/5

    sin2a=2sinacosa/(sin^2a+cos^2a)=2tana/(tan^2a+1)=4/5

    所以f(a) =1/2*(4/5-3/5+1)=3/5