Question

函數(shù)y=cosx+1(-π≤x≤0)的反函數(shù)是(    )

A.y=-arccos(x-1)(0≤x≤2)                    B.y=π-arccos(x-1)(0≤x≤2)

C.y=arccos(x-1)(0≤x≤2)                     D.y=π+arccos(x-1)(0≤x≤2)

Answer 1

解析：由y=cosx+1,∵x∈［-π,0］,

∴y∈［0,2］,即反函數(shù)的定義域?yàn)椋?,2］,值域?yàn)椋?π,0］.

∴cosx=y-1.

∴x=-arccos(y-1),

即y=-arccos(x-1),x∈［0,2］.

故選A.

答案：A