已知函数f(x)=1+sinxcos,g(x)=cos^2 (x+π/12)
2个回答

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

g(x)=cos^2 (x+π/12)=[1+cos(2x+π/6)]/2

(1)x=x1是函数y=f(x)的图像的一条对称轴,所以x1=kπ+π/2(k∈Z)

所以,g(x1)=[1+cos(2kπ+π+π/6)]/2=(2-√3)/4.

(2)h(x)=f(ωx/2)+g(ωx/2)

=1+(1/2)sinωx+[1+cos(ωx+π/6)]

=...

=(1/2)sin(ωx+π/3)+3/2

-π/2≤ωx+π/3≤π/2

==> -5π/(6ω)≤x≤π/(6ω)

要使h(x)在[-3π/2,π/3]上是增函数,须

-5π/(6ω)≤3π/2且π/3≤π/(6ω)

解得ω≤1/2

所以ω的最大值是1/2