题目

函数y=+1(x≥1)的反函数是(   )A.y=x2-2x+2(x<1)                        B.y=x2-2x+2(x≥1)C.y=x2-2x(x<1)                           D.y=x2-2x(x≥1) 答案：B解析:∵y=+1(x≥1),∴y≥1.又∵y=+1,∴=y-1,x-1=(y-1)2,即x=y2-2y+2.∴所求反函数为y=x2-2x+2(x≥1).

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