已知向量m=(根号3sinx+cosx.1),n=(cosx,-f(x)),m⊥n

∵m⊥n
∴(√3sinx+cosx)cosx-f(x)=0
∴f(x)=√3sinxcosx+cos²x=√3/2sin2x+1/2+1/2cos2x=sin(2x+π/6)+1/2
由2kπ-π/2≤2x+π/6≤2kπ+π/2得,单调递增区间是[kπ-π/3,kπ+π/6]
由2kπ+π/2≤2x+π/6≤2kπ+3π/2得,单调递减区间是[kπ+π/6,kπ+2π/3]
∵f(A/2)=sin(A+π/6)+1/2=1/2+√3/2
∴A+π/6=π/3或2π/3,解得A=π/6或π/2
∵b>a,∴B>A,∴舍去π/2
由a/sinA=b/sinB得sinB=√2sinA=√2/2,∴B=π/4
∴C=π-π/6-π/4=7π/12
∴S△=1/2absinC=√2/2sin(7π/12)=1/4+√3/4