一、原题展示(2017年高考课标卷Ⅱ文科第21题)设函数f(x)=(1-x~2)e~x.(I)略;(Ⅱ)当x≥0时,f(x)≤ax+1,求a的取值范围.二、试题分析本题考查了函数的单调性和恒成立问题.以含参数不等式问题为载体,既考查学生的分类讨论思想、等价转化思想、数形结合思想和函数方程及不等式思想,又考查学生分析问题和解决问题的能力.三、解法探究解析(I)略;(Ⅱ)(应用分离参数法和洛必达法则) First, the original title display (2017 college entrance examination standard volume II liberal arts twenty-first problem) Let f (x) = (1-x ~ 2) e ~ x. f (x) ≤ax + 1, find a range of values. Second, the test analysis This test examines the monotonicity of the function and the constant establishment of the problem. Transformation thinking, the number of ideas and functional equations and inequality thinking, but also examine the ability of students to analyze and solve problems.Third, the solution to the analysis (I) omitted; (Ⅱ) (the application of the separation of parameter method and Robda’s law)