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    【题目】【2017高考课标2 文21】设函数.

    （1）讨论的单调性；

    （2）当时，，求的取值范围.

    【答案】（Ⅰ）在 和单调递减，在单调递增（Ⅱ）

    【解析】

    试题分析：（1）先求函数导数，再求导函数零点，列表分析导函数符号确定单调区间（2）对分类讨论，当a≥1时，，满足条件；当时，取，当0＜a＜1时，取，.

    试题解析：（1）

    令得

    当时，；当时，；当时，

    所以在 和单调递减，在单调递增

    (2)

    当a≥1时，设函数h(x)=（1-x）ex，h’(x)= -xex＜0（x＞0），因此h(x)在[0，+∞)单调递减，而h(0)=1，

    故h(x)≤1，所以

    f(x)=（x+1）h(x)≤x+1≤ax+1

    当0＜a＜1时，设函数g（x）=ex-x-1，g’（x）=ex-1＞0（x＞0），所以g（x）在在[0，+∞)单调递增，而g(0)=0，故ex≥x+1

    当0＜x＜1，，，取

    则

    当时，取

    综上，a的取值范围[1，+∞）

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