题目

对于定义在实数集R上可导函数f(x),满足xf′(x)＜0,则必有A.f(-2)+f(1)＜f(0)                               B.f(-2)+f(1)＞f(0)C.f(-1)+f(1)＜2f(0)                              D.f(-1)+f(1)＞2f(0) 答案：答案：C解析:xf′(x)＜0,当x＞0时,f′(x)＜0,函数单调递减;当x＜0时,f′(x)＞0,函数单调递增.又∵函数在R上可导,∴当x＞0时,f(x)＜f(0),当x＜0时,f(x)＜f(0).∴f(-1)＜f(0),f(1)＜f(0),则有f(-1)+f(1)＜2f(0).数一数,有（　　）个小长方体.