(1) f(x)=2a·b-1=2√3sinxcosx+2cos²x-1
=√3sin2x+cos2x
=sin(x+π/6)
當x+π/6=π/2+kπ 即 對稱軸：x=π/3+kπ
當x+π/6∈[-π/2+2kπ,π/2+2kπ]時,f(x)單調(diào)遞增,解得：x∈[-2π/3+2kπ,π/3+2kπ]
當x+π/6∈[π/2+2kπ,3/2π+2kπ]時,f(x)單調(diào)遞減,解得：x∈[π/3+2kπ,4/3π+2kπ]
(2)因為 f(x)=1
所以 x+π/6=π/2+2kπ
x=π/3+2kπ
因為x∈[π/6,π/2]
所以x=π/3
(3)因為x∈[0,π/2]
所以 x+π/6∈[π/6,2/3π]
由函數(shù)圖像可知：f(x)max=sin(π/2)=1,f(x)min=sin(2/3π)=-√3/2
所以,值域為：[-√3/2,1]