試題分析:(1)由函數(shù)
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,在點
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處的切線方程為
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.所以對函數(shù)求導,根據(jù)斜率為1以及過點(1,0)兩個條件即可求出結(jié)論.
(2)由函數(shù)
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,對函數(shù)
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求導,并令
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可解得兩個根,由于函數(shù)
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在區(qū)間
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內(nèi)有且僅有一個極值點,
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的根在
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內(nèi)有且僅有一個根.所以通過分類討論即可求
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的取值范圍.
(3)兩曲線在交點
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處的切線分別為
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.若取
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,當直線
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與
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軸圍成等腰三角形時.通過求導求出兩函數(shù)的切線的斜率,即可得到這兩斜率不可能是相等,所以依題意可得到兩切線傾斜角有兩倍的關(guān)系,再通過解方程和函數(shù)的單調(diào)性的判斷即可得到結(jié)論.
(1)
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,∴
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,又
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,
∴
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. 3分
(2)
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;
∴
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由
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得
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,
∴
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或
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. 5分
∵
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,當且僅當
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或
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時,函數(shù)
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在區(qū)間
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內(nèi)有且僅有一個極值點. 6分
若
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,即
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,當
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時
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;當
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時
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,函數(shù)
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有極大值點
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,
若
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,即
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時,當
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時
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;當
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時
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,函數(shù)
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有極大值點
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,
綜上,
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的取值范圍是
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. 8分
(3)當
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時,設(shè)兩切線
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的傾斜角分別為
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,
則
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,
∵
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, ∴
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均為銳角, 9分
當
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,即
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時,若直線
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能與
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軸圍成等腰三角形,則
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;當
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,即
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時,若直線
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能與
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軸圍成等腰三角形,則
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.
由
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得,
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,
得
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,即
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,
此方程有唯一解
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,直線
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能與
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軸圍成一個等腰三角形. 11分
由
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得,
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,
得
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,即
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,
設(shè)
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,
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,
當
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時,
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,∴
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在
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單調(diào)遞增,則
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在
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單調(diào)遞
增,由于
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,且
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,所以
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,則
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,
即方程
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在
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有唯一解,直線
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能與
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軸圍成一個等腰三角形.
因此,當
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時,有兩處符合題意,所以直線
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能與
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軸圍成等腰三角形時,
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值的個數(shù)
有2個. 14分