已知函数f(x)=asinxcosx-√3acos²x+√3/2a+b(a＞0） ⑴写出函数的单调递减区间

（1）f(x)=asinxcosx-√3acos²x+√3/2a+b
=(a/2)·sin2x-(√3a/2)cos2x-√3a/2+√3/2a+b
=asin(2x-π/3)+b
由于sinx的单调增区间是[2kπ-π/2,2kπ+π/2],单调减区间是[2kπ+π/2,2kπ+3π/2],又a>0
∴f(x)的单调增区间是[kπ-π/12,kπ+5π/12],单调减区间是[kπ+5π/12,kπ+11π/12]
（2）x∈[0,π/2],2x-π/3∈[-π/3,2π/3]
sin(-π/3)≤sin(2x-π/3)≤sin(π/2)
-√3/2≤sin(2x-π/3)≤1
-√3/2a+b≤f(x)≤a+b
∴-√3/2a+b=-2,a+b=√3
a=2,b=√3-2