Question

已知函数f（x）=2cosx*sin(x+π/3)-√3sin^2x+sinx*cosx

(1)求函数f(x)的单调递减区间
（2）将函数f(x)的图像按向量a=（m,0)平移后得到g(x)的图像，求使函数g(x)为偶函数的m的最小正值
速求，要详细答案

Answer 1

1.
f（x）=2cosx*sin(x+π/3)-√3sin^2x+sinx*cosx
= 2cosx*sin(x+π/3)- 2sinx*[(√3/2)sinx-(1/2)cosx]
= 2cosx*sin(x+π/3)- 2sinx*[sin(π/3)sinx-cos(π/3)cosx)]
= 2cosx*sin(x+π/3)+ 2sinxcos(x+π/3)
= 2sin(2x+π/3)
f(x)的单调区间
2kπ - π/2 <= 2x+π/3 < 2kπ + π/2 单调增
(2k+1)π - π/2 <= 2x+π/3 < (2k+1)π + π/2 单调减
k= 0 , +/-1, +/- 2, +/- 3 ......

=>
[2kπ - π/2 - π/3]/2 <= x < [2kπ + π/2 - π/3]/2

[(2k+1)π - π/2 -π/3]/2 <= x < [(2k+1)π + π/2 - π/3]/2

2.
f(x) 平移 a(m,0) 即沿x 轴向正方向移动

g(x)=f(x-m) = 2sin[2(x-m)+π/3]

g(x) 为 偶函数 即 g(x)= g(-x)
=>
2sin[2(x-m)+π/3] = 2sin[2(-x-m)+π/3]
又sin为奇函数
=>
2sin[2(x-m)+π/3] = -2sin{-[2(-x-m)+π/3]}
=>
2sin[2(x-m)+π/3] = 2sin{[2(x+m)-π/3]}
=>
2(x-m) + π/3 + 2kπ = 2(x+m)-π/3
=>
4m= 2π/3 + 2kπ
m=π/6 取最小正值(k=0)