题目内容
(本小题满分14分)
已
知二次函数
的图象经过点
、
与点
,设函数
在
和
处取到极值,其中
,
。
(1)求
的二次项系数
的值;
(2)比较
的大小(要求按从小到大排列);
(3)若
,且过原点存在两条互相垂直的直线与曲线
均相切,求
。
已
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(1)求
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
(2)比较

(3)若
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解:(1)由题意可设
,
又函数图象经过点
,则
,得
.……… 2分
(2)由(1)可得
。
所以
,
, ………… 4分
函数
在
和
处取到极值,
故
, ………… 5分

,

………… 7分

又
,故
。 …… 8分
(3)设切点
,则切线的斜率
又
,所以切线的方程是
…… 9分
又切线过原点,故
所以
,解得
,或
。 ………… 10分
两条切线的斜率为
,
,
由
,得
,
,

,
………………………… 12分
所以
,
又两条切线垂直,故
,所以上式等号成立,有
,且
。
所以
。 ………… 14分

又函数图象经过点
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(2)由(1)可得
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所以
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函数
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故
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
又
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(3)设切点
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又
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又切线过原点,故

所以
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两条切线的斜率为
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
由
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所以

又两条切线垂直,故
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所以
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略
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