a=(1,sinθ),b=(3,-1),
|a|=√(1+sin^2θ)
|b|=√(3^2+1^2)=√10
a*b=3-sinθ
|2a-b|=√(2a-b)^2=√(4a^2-4ab+b^2)
=√[4*(1+sin^2θ)-4*(3-sinθ)+10]
=√(4+4sin^2θ-12+4sinθ+10)
=√(4sin^2θ+4sinθ+2)
=√[4(sin^2θ+sinθ)+2]
=√[4(sin^2θ+sinθ+1/4)+2-1]
=√[4(sinθ+1/2)^2+1]
当sinθ=-1/2时得最小值1
当sinθ=1时得最大值√[4(1+1/2)^2+1]=√10
选D