Question

已知函数f(x)=4cosxsin(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(√3/2sinx+1/2cosx)+a
=2√3sinxcosx+2cos²x+a
=2√3sinxcosx+2cos²x-1+1+a
=√3sin2x+cos2x+a+1
=2(sin2x*√3/2+cos2x*1/2)+a+1
=2sin(2x+π/6）+a+1
最大值=2+a+1=2
∴a=-1
f(x)=2sin(2x+π/6）
最小正周期=2π/2=π
(2)
令-π/2+2kπ<=2x+π/6<=π/2+2kπ,k∈Z
-π/3+kπ<=x<=π/6+kπ,k∈Z
∴f(x)增区间是[-π/3+kπ,π/6+kπ],k∈Z
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Answer 2

1
f(x)=4cosxsin(x+π/6)+a
=2根3sinxcosx+2cos~2(x)+a
=2sin(2x+π/6)+1+a
f(x)max=2+1+a=2
所以a=-1
T=2π/2=π
2
单调递增区间为2kπ-π/2<2x+π/6<2kπ+π/2
（k为整数）
所以kπ-π/3<x<kπ+π/6
在区间[0,π]上即：[0,π/6]并[2π/3,π]

Answer 3

f(x)=4cosxsin(x+π/6)+a
=4cosx(√3/2sinx+1/2cosx)+a
=2√3sinxcosx+2cos²x+a
=2√3sinxcosx+2cos²x-1+1+a
=√3sin2x+cos2x+a+1
=2(sin2x*√3/2+cos2x*1/2)+a+1
=2sin(2x+π/6）+a+1
最大值=2+a+1=2
∴a=-1
f(x)=2sin(2x+π/6）
最小正周期=2π/2=π
(2)
令-π/2+2kπ<=2x+π/6<=π/2+2kπ,k∈Z
-π/3+kπ<=x<=π/6+kπ,k∈Z
∴f(x)增区间是[-π/3+kπ,π/6+kπ],k∈Z
如果你认可我的回答，请点击左下角的“采纳为满意答案”，祝学习进步！