Question

数学高手帮忙解一道函数题目，谢谢

已知函数f(x)=cos^2(x+π/12),g(x)=1+1/2sin2x 1)设x=x0是函数y=f(x)图像的一条对称轴，求g(x0)的值； 2）求函数h(x)=f(x)+g(x)的单调递增区间 麻烦给我很详细的过程，尤其是化简过程，我基础很差，谢谢

Answer 1

①f(x)=cos^2(x+π/12)=(1+
cos(2x+π/6))/2,
三角函数图像的对称轴必定穿过最高点或最低点，所以cos(2x0+π/6)=1或-1。
∴2x0+π/6=
kπ,k∈Z.
x0=
kπ/2-π/12,k∈Z.
g(x0)
=1+1/2sin2x0=1+1/2sin(kπ-π/6)
=1±1/2=3/2或1/2.
②函数h(x)=f(x)+g(x)=
cos^2(x+π/12)+1+1/2sin2x
=(1+
cos(2x+π/6))/2+1+1/2sin2x
=
cos(2x+π/6)/2+1/2sin2x+3/2
=1/2(cos2xcosπ/6-sin2xsinπ/6)
+1/2sin2x+3/2
=√3/4cos2x-1/4sin2x+1/2sin2x+3/2
=√3/4cos2x+1/4sin2x+3/2
=1/2(√3/2cos2x+1/2sin2x)
+3/2
=1/2(cos2xcosπ/6+sin2xsinπ/6)
+3/2
=1/2
cos
(2x-π/6)
+3/2
2kπ-π≤2x-π/6≤2kπ，k∈Z.
kπ-5π/12≤x≤kπ+π/12，k∈Z.
∴函数的单调递增区间是[kπ-5π/12,
kπ+π/12]
，k∈Z.