(1)f(x)=asinxcosx-(√3)acos²x+[(√3)a/2]+b
=asin2x/2-(√3)acos2x/2+b
=asin(2x-π/6)+b,a>0
當(dāng)π/2+2kπ≤2x-π/6≤3π/2+2kπ即π/3+kπ≤x≤5π/6+kπ,k∈Z時(shí)單調(diào)遞減
當(dāng)-π/2+2kπ≤2x-π/6≤π/2+2kπ即-π/6+kπ≤x≤π/3+kπ,k∈Z時(shí)單調(diào)遞減
(2)當(dāng)k=0時(shí),f(x)的單調(diào)遞減區(qū)間為[π/3,5π/6],遞增區(qū)間為[-π/6,π/3]
所以f(x)在[0,π/3]遞增,[π/3,π/2]遞減
即當(dāng)x=π/3,f(x)最大,有f(π/3)=a+b=√3
因f(0)=b-0.5a,f(π/2)=0.5a+b
因?yàn)閍>0
所以當(dāng)x=0時(shí),f(x)最小,有b-0.5a=-2
解方程組,得a=(4+2√3)/3,b=(√3-4)/3