Question

已知函数fx=Acos(wx+ψ),x∈R(其中A>0,w>0,-π/2

Answer 1

（Ⅰ）由最低点为M(2π/3,-2)得A=2由T=π
得ω=2π/T=2π/π=2
由点M(2π/3,-2)在图象上得
2sin(4π/3+φ)=-2即sin(4π/3+φ)=-1
所以4π/3+φ=2kπ-π/2
故φ=2kπ-11π/6 (k∈Z)
又φ∈(0,π/2),所以φ=π/6
所以f(x)=2sin(2x+π/6)
（Ⅱ）因为x∈[0,π/12],可得2x+π/6∈[π/6,π/3]
所以当2x+π/6=π/6时,即x=0时,f（x）取得最小值1；
当2x+π/6=π/3,即x=π/12时,f（x）取得最大值√3 .