已知函数f(x)=1+sinxcosx,g(x)=cos²(x+π/12)

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  • 已知函数f(x)=1+sinxcosx,g(x)=[cos(x+(π/12))]^2

    f(x)=1+sinxcosx=1+(1/2)sin2x

    g(x)=[cos(x+(π/12)]^2=[cos(2x+(π/6))+1]/2

    (1)设X=Xo是函数y=f(x)图像的一条对称轴,求g(x).

    因为f(x)=1+(1/2)sin2x,对于正弦函数来说,当x=xo为对称轴时函数f(x)取得最大值或者最小值.即:sin2x=1,或者sin2x=-1

    所以,2x=2xo=kπ+(π/2)(k∈Z)

    那么,g(x)=[cos(2x+(π/6))+1]/2=[cos(kπ+(π/2)+(π/6))+1]/2

    =[cos(kπ+(2π/3))+1]/2

    当k为偶数时,g(x)=[cos(2π/3)+1]/2=[(-1/2)+1]/2=1/4

    当k为奇数时,g(x)=[cos(5π/3)+1]/2=[(1/2)+1]/2=3/4

    (2)求h(x)=f(wx/2)+g(wx/2)(w>0)在区间[-2π/3,π/3]上是增函数的w的最大值.

    由前面知,f(x)=1+(1/2)sin2x,g(x)=[cos(2x+(π/6))+1]/2

    所以,f(wx/2)=1+(1/2)sin(2*wx/2)=1+(1/2)sin(wz)

    g(wx/2)=[cos(2*wx/2+(π/6))+1]/2=[cos(wx+(π/6))+1]/2

    所以:f(wx/2)+g(wx/2)=1+(1/2)sin(wx)+(1/2)cos(wx+(π/6))+(1/2)

    =(3/2)+(1/2)[sin(wx)+cos(wx+(π/6))]

    =(3/2)+(1/2)[sin(wx)+cos(wx)*cos(π/6)-sin(wx)*sin(π/6)]

    =(3/2)+(1/2)[sin(wx)+cos(wx)*(√3/2)]-sin(wx)*(1/2)]

    =(3/2)+(1/2)[(1/2)sin(wx)+(√3/2)cos(wx)]

    =(3/2)+(1/2)sin[(wx)+(π/3)]

    =(3/2)+(1/2)sin[w(x+(π/3w))]

    则其周期为T=2π/w

    区间[-2π/3,π/3]的长度为(π/3)-(-2π/3)=π

    要保证其在[-2π/3,π/3]上为增函数,则:

    π/3w≥2π/3,且T/4=π/(2w)≥π

    所以,w≤1/2

    即,w的最大值为1/2