试题分析:解:(
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)因为,
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成立,所以:
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,
由:
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,得
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,
由:
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,得
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解之得:
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从而,函数解析式为:
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(4分)
(2)由于,
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,设:任意两数
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是函数
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图像上两点的横坐标,则这两点处的切线的斜率分别是:
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又因为:
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,所以,
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,得:
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知:
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故,当
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是函数
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图像上任意两点处的切线不可能垂直 (8分)
(3)当
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时,
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且
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此时
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(11分)
当且仅当:
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即
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即,取等号,
所以
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故当
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时,
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有最小值
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(13分)
(或
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)
点评:解决的关键是利用导数的符号确定出函数单调性,以及函数的极值,从而比较极值和端点值的函数值得到最值,属于基础题。