已知向量a=(2sinx,2cosx),b=(cosx,sinx)

解
f(x)=a*b
=2sinxcosx+2cos²x
=sin2x+2cos²x-1+1
=sin2x+cos2x+1
=√2sin(2x+π/4)+1
∴最小正周期为：
T=2π/2=π
∵x∈[0,3π/8]
∴2x+π/4∈[π/4,π]
∴sin(2x+π/4)∈[0,1]
∴f(x)=√2sin(2x+π/4)+1∈[1,√2+1]
∴f(x)的值域为：[1,√2+1]