答案为D
找个特殊函数代进去就可以了
比如 g(x) = e^x
g'(x) = g(x)
只要令 f(x) = g(x)+1 = e^x + 1
就满足 f(x)>f'(x)
f(0) = 2
f(2013) = e^2013+1
e^2013 * f(-2013) = e^2013 * (e^-2013 + 1) = 1+ e^2013
显然
e^2013 * f(-2013) > f(0)
f(2013) < e^2013 * f(0)
所以答案是D