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 很高兴为您解答,