摘要：8．函数f(x)＝.若f(4x1)+f(4x2)＝1.x1>1.x2>1.则f(x1·x2)的最小值为( ) A. B. C．2 D. 答案:B 解析:依题意得f(x)＝1-. 1-+1-＝1. 由此解得log2x2＝.log2(x2x1)＝log2x2+log2x1＝+log2x1＝+log2x1＝-2++(log2x1+1)≥-2+2＝2.故f(x1x2)＝1-≥1-＝.f(x1·x2)的最小值是.故选B.

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