Question

设向量a=(sinx,cosx),b=(sinx,根号3sinx),x属于R,函数f(x)

设向量a=(sinx，cosx)，b=(sinx，根号3sinx)，x属于R，函数f(x)=a(a+2b).(1)求函数f(x)的最大值与单调递增区间;(2)求使不等式f`(x)大余等于2成立的x的取值集合.
速度啊！急

Answer 1

f(x)=a(a+2b)=1+2(sin²x+√3sinxcosx)=1+(1-cos2x)+√3sin2x=2+√3sin2x-cos2x=2+2sin(2x+π/6)
∴由2kπ-π/2≤2x+π/6≤2kπ+π/2得，kπ-π/3≤x≤kπ+π/6
∴单调递增区间为[kπ-π/3，kπ+π/6]
2+2sin(2x+π/6)≥2
得，sin(2x+π/6)≥0
2kπ≤2x+π/6≤2kπ+π，解得kπ-π/12≤x≤kπ+5π/12
∴{x|kπ-π/12≤x≤kπ+5π/12}