f(x)=向量a·向量b-√3/2
=√3cos²x+cosxsinx-√3/2
=(√3/2)(1+cos2x)+(1/2)sin2x-√3/2
=sin(2x+π/3)+(√3/2)-(√3/2)
=sin(2x+π/3)
递增区间是:
2x+π/3∈[-π/2+2kπ,π/2+2kπ]
x∈[-5π/12+kπ,π/12+kπ],k∈Z
此即递增区间
若x∈[0,π/4],则
2x+π/3∈[π/3,5π/6]
∴sin(2x+π/3)∈[1/2,1]
∴f(x)∈[1/2,1]
此即值域