Question

高中数学题，帮忙！

有些符号不会打，用汉字写出来！ →已知向量a=(2,cosx),向量b=(sin(x+圆周率/6),-2),函数f(x)=a*b(x属于R)。(1)求函数f(x)的单调递增区间;(2)当x属于(0，圆周率)时，求函数f(x)的最大值

Answer 1

(1)f(x)=2sin(x+π/6)-2cosx=2sinxcosπ/6+2cosxsin π/6-2cosx
=2sin(x-π/6)
增区间2kπ-π/2<x-π/6<2kπ+π/2
2kπ+π/3<x<2kπ+2π/3 k整数
（2）-π/6<x-π/6<5π/6
x=2π/3 f(x)最大=2

Answer 2

1.f(x)=2sin(x+π/6)-2cosx=2sinxcosπ/6+2cosxsin π/6-2cosx
=2sin(x-π/6)
增区间2kπ-π/2<x-π/6<2kπ+π/2
2kπ+π/3<x<2kπ+2π/3 （k属于Z）
2.-π/6<x-π/6<5π/6
x=2π/3 f(x)最大=2