题目
| f(x) |
| x |
| A.(0,2) | B.(0,2)∪(2,+∞) | C.(2,+∞) | D.∅ |
答案
即[xf(x)]′=f′(x)•x-f(x)<0,
令g(x)=xf(x),则g(x)在(0,+∞)上为减函数,
又由f(2)=0,则g(2)=2f(2)=0,
即当0<x<2时,有xf(x)<0,
当x>2时,有xf(x)<0,
又由x>0,则
| f(x) |
| x |
即
| f(x) |
| x |
故选C.
| f(x) |
| x |
| A.(0,2) | B.(0,2)∪(2,+∞) | C.(2,+∞) | D.∅ |
| f(x) |
| x |
| f(x) |
| x |