Question

已知函数f(x)=根号3sinwxcoswx+cos^2wx(w>0) 发f(x)最小正周期为π/2

求w及f(x)的最大值
求函数f(x)单调区间

Answer 1

f(x)=(√3/2)sin2wx+(1+cos2wx)/2
=(√3/2)sin2wx+(1/2)cos2wx+1/2
=sin(2wx+π/6)+1/2
最小正周期T=2π/2w=π/2
得：w=2
所以，f(x)=sin(4x+π/6)+1/2
最大值为3/2
递增区间：
-π/2+2kπ<4x+π/6<π/2+2kπ
-2π/3+2kπ<4x<π/3+2kπ
-π/6+kπ/2<x<π/12+kπ/2
所以，递增区间为（-π/6+kπ/2，π/12+kπ/2）k∈Z
递减区间：
π/2+2kπ<4x+π/6<3π/2+2kπ
π/3+2kπ<4x<4π/3+2kπ
π/12+kπ/2<x<π/3+kπ/2
所以，递增区间为（π/12+kπ/2，π/3+kπ/2）k∈Z

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