1.f(x)=ab+|b|^2-1/2
=5√3cosxsinx+2(cosx)^2+(sinx)^2+4(cosx)^2-1/2
=(5√3/2)sin2x+6(1+cos2x)/2+(1-cos2x)/2-1/2
=(5√3/2)sin2x+(5/2)cos2x+3
=5sin(2x+π/6)+3
-π/6≤x≤π/4,-π/6≤2x+π/6≤2π/3
f(x)在2x+π/6=-π/6即x=-π/6时有最小值:5(-1/2)+3=1/2
在2x+π/6=π/2,即x=π/6时有最大值:5+3=8
所以f(x)的值域:[1/2,8]
2.x=π/6取得最大值,此时a=(15/2,√3/2),b=(1/2,√3)
|2a/t-tb|=√(4a^2/t^2+t^2b^2-4ab)=√[(228/t^2)+(13/4)t^2-21]
≥√[2(√(228/t^2)*(13/4)t^2)-21]
≈5.78