试题分析:(1)先求导函数
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,进而根据题中条件得出
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,从可即可求解出
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的值,注意,根据函数在某点取得极值去求参数的值时,往往必须进行检验,也就是将所求得的
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的值代回原函数,看看是否真的在该点处取得极值,如果不是必须舍去,如果是则保留;(2)先将
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对任意
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恒成立等价转化为
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在
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恒成立,进而求出导函数并进行因式分解得到
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,进而分
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、
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两类分别确定
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的单调性,随之确定
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,然后分别求解不等式
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,解出
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的取值范围,最后取这两种情况下的
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的取值范围的并集即可.
(1)
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,依题意有:
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,即
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解得:
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检验:当
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时,
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此时:函数
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在
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上单调递减,在
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上单调递增,满足在
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时取得极值
综上:
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5分
(2)依题意:
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对任意
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恒成立等价转化为
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在
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恒成立 6分
因为
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令
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得:
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8分
当
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即
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时,函数
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在
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恒成立,则
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在
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单调递增,于是
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,解得:
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,此时:
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10分
②当
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即
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时,函数
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在
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单调递减,在
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单调递增,于是
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,不合题意,此时:
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综上所述:实数
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的取值范围是
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12分.
说明:本题采用参数分离法或者先用必要条件
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缩小参数范围也可以.