试题分析:(Ⅰ)当
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时,
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1分
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.2分
所以曲线
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在点
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处的切线方程
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3分
(Ⅱ)
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4分
当
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时,解
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,得
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,解
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,得
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所以函数
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的递增区间为
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,递减区间为在
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5分
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时,令
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得
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或
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ⅰ)当
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时,
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6分
函数
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的递增区间为
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,
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,递减区间为
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7分
ⅱ)当
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时,
在
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上
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,在
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上
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8分
函数
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的递增区间为
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,递减区间为
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9分
(Ⅲ)由(Ⅱ)知,当
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时,
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在
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上是增函数,在
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上是减函数,
所以
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, 11分
存在
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,使
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即存在
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,使
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,
方法一:只需函数
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在[1,2]上的最大值大于等于
所以有
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即
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解得:
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13分
方法二:将
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整理得
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从而有
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所以
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的取值范围是
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. 13分
点评:中档题,本题属于导数应用中的常见问题,通过研究函数的单调性,明确最值情况。曲线切线的斜率,等于函数在切点处的导函数值。在给定区间,如果函数的导数非负,则函数为增函数,如果函数的导数非正,则函数为减函数。涉及不等式恒成立问题,往往通过构造函数,研究函数的最值,得到确定参数(范围)的目的。对数函数要注意其真数大于0.