三角代换
由于(√(x-3))^2+(√(4-x))^2=1
设√(x-3)=cosθ √(4-x)=sinθ θ∈[0,π/2]
y=√(x-3)+√(12-3x)
=cosθ+√3sinθ
=2[(1/2)cosθ+(√3/2)sinθ]
=2sin(θ+π/6)
由0≤θ≤π/2
则π/6≤θ+π/6≤2π/3
1/2≤sin[θ+(π/6)]≤1
所以所求函数的值域为[1,2]