Question

已知函数f（x）=4cosxxsin(x+π/6 )+a的最大值为2

已知函数f（x）=4cosxxsin(x+π/6 )+a的最大值为2
1.求a的值及f(x)的最小正周期
2.求f(x)的单调递增区间

Answer 1

f（x）=4cosxsin(x+π/6 )+a
=4cosx(sinxcosπ/6+cosxsinπ/6)+a
=4cosx(sinx√3/2+cosx/2)+a
=2√3sinxcosx+2cosxcosx+a
=√3sin2x+cos2x+1+a
=2sin(2x+π/6)+a+1
1、因为f（x）=4cosxxsin(x+π/6 )+a的最大值为2
即2sin(2x+π/6)+a+1的最大值为2
sin(2x+π/6)最大为1
2sin(2x+π/6)+a+1的最大值=2*1+a+1=2
所以a= -1
f（x）=2sin(2x+π/6)的最小正周期为2π/2=π

2、f(x)=2sin(2x+π/6)
递增区间为
-π/2+2πn≤2x+π/6≤π/2+2πn
解得 -π/3+πn≤x≤π/6+πn
所以递增区间为 【-π/3+πn，π/6+πn】