Question

对于实数集R上可导函数f(x),满足xf′(x)＜0,则必有

A.f(-1)+f(1)＞2f(0)                               B.f(-1)+f(1)＜2f(0)

C.f(-2)+f(1)＜f(0)                                D.f(-2)+f(1)＞f(0)

Answer 1

B

解析:x＞0时,f′(x)＜0,∴f(x)在(0,+∞)上为减函数;

x＜0时,f′(x)＞0,∴f(x)在(-∞,0)上为增函数.

又x∈R,且f(x)在x=0处可导,

∴f(x)在x=0处连续.

∴f(0)为f(x)的极大值.

∴f(-1)＜f(0),f(1)＜f(0).∴f(-1)+f(1)＜2f(0).