题目

已知定义在实数R上的函数y=f(x)不恒为零,同时满足f(x+y)=f(x)f(y),且当x＞0时,f(x)＞1,那么当x＜0时,一定有(    )A.f(x)＜-1                         B.-1＜f(x)＜0C.f(x)＞1                          D.0＜f(x)＜1 答案：解析：令x=0,y＞0,得f(0+y)=f(0)f(y),因为f(y)＞1,则f(0)=1.设x＜0,则-x＞0,有f(0)=f(x+(-x))=f(x)f(-x).故f(x)=.因为-x＞0,所以f(-x)＞1,故f(x)∈(0,1).答案：D

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