题目内容
已知函数
.
(1)求
的单调区间;
(2)设
,若对任意
,均存在
,使得
,求
的取值范围.
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(1)求
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(2)设
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(1) 函数
的单调递增区间为
,单调递减区间为
(2)
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(2)
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本试题主要是考查了导数在研究函数中的运用。
(1)函数
,求解定义域和导数,然后利用导数的正负号判定单调性。
(2)由已知,转化为
.,然后分别求解最值得到参数的范围。
解:(1)
, ………………2分
①当
时,由于
,故
,
………………3分
所以,
的单调递增区间为
. ………………4分
②当
时,由
,得
. ………………5分
在区间
上,
,在区间
上
,
所以,函数
的单调递增区间为
,单调递减区间为
.…………7分
(2)由已知,转化为
. ………………8分
………………9分
由(1)知,当
时,
在
上单调递增,值域为
,故不符合题意.
(或者举出反例:存在
,故不符合题意.) ………………11分
当
时,
在
上单调递增,在
上单调递减,
故
的极大值即为最大值,
, ………14分
所以
,解得
. ………15分
(1)函数
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(2)由已知,转化为
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解:(1)
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①当
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所以,
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②当
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在区间
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所以,函数
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(2)由已知,转化为
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由(1)知,当
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(或者举出反例:存在
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当
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故
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所以
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