题目(2014年高考陕西卷理科第21题)设函数f(x)=ln(1+x),g(x)=xf′(x),x≥0,其中f′(x)是f(x)的导函数.(1)g1(x)=g(x),gn+1(x)=g(gn(x)),n∈N+,求gn(x)的表达式;(2)若f(x)≥ag(x)恒成立,求实数a的取值范围;(3)设n∈N+,比较g(1)+g(2)+…+g(n)与n-f(n)的大小,并加以证明.原题提供的答案第(2)问是利用分类讨论的思想,通过对参数的缜密讨论,确定参数的取值范围.分类讨论向来是同学 (F) = ln (1 + x), g (x) = xf ’(x), x ≧ 0, where f’ (x) is f (x) (1) g1 (x) = g (x), gn + 1 (x) = g (gn (x)), n∈N + (1) + g (2) + ... + g (n) and nf (n) are satisfied if f (x) ≥ag (x) (2) Q is the use of the idea of ​​classification, through careful discussion of the parameters to determine the range of parameters. Category discussion has always been the students