Question

已知函数f(x)=acos2ωx+√3asinωxcosωx+b,x∈R(a>0,ω>0)的最小正周期为π,函数f(x)

Answer 1

f(x)=a*(1+cos2wx)/2+(√3/2)*sin2wx+b=a*[(1/2)*cos2wx+(√3/2)*sin2wx]+a/2+b

=a*sin(2wx+π/6)+a/2+b

(3/2)*a+b=7/4

(-1/2)*a+b=3/4

{a=1/2 b=1

2π/2w=π ==>w=1

f(x)=1/2)*sin(2x+π/6)+5/4

由-π/2+2kπ≤2x+π/6≤π/2+2kπ得单调增区间：【-π/3+kπ,π/6+kπ】