令F（x）=2f（x）-x则F′（x）=2f′（x）-1＞0∴F（x）在R上单调递增∵F（1）=2f（1）-1=2-1=1，2f（x）＜x+1∴F（x）=2f（x）-x＜1=F（1）即x＜1故满足2f（x）＜x+1的x的集合为为{x|x＜1}故选B．
