由f（x）= 1 e f′(1)ex?f(0)x+ 1 2 x2，得f′(x)＝ 1 e f′(1)ex?f(0)+x，令x=1，得f′(1)＝ 1 e f′(1)e?f(0)+1，∴f（0）=1．故答案为1．
