已知函数f(x)=2acos^2x+bsinxcosx(a>0 ,b>0) f(x)最大值为1+a,最小为-1/2

f(x)=2acos^2x+bsinxcosx=a(1+cos2x)+(b/2)sin2x
=√[a^2+(b/2)^2]sin(2x+φ)+a
按照题意有:
√[a^2+(b/2)^2]+a=1+a .(1)
-√[a^2+(b/2)^2]+a=-1/2 .(2)
联立(1)(2)解得 a=1/2,b=√3
所以,
f(x)=(1/2)(1+cos2x)+(√3/2)sin2x
=sin(2x+π/6)+1/2
1.f(x)的最小正周期是:2π/2=π.
2.2kπ-π/2≤2x+π/6≤2kπ+π/2 (k∈Z)
==> kπ-π/3≤x≤kπ+π/6 (k∈Z)
所以f(x)的单调递增区间是[kπ-π/3,kπ+π/6](k∈Z)