高考數(shù)學（理）二輪專題突破演練（浙江專版）第3部分 專題1 第2講“2道”拉分題專練卷1 Word版含解析

1、“2道”拉分題專練卷(一)1已知函數(shù)f(x)ln xax2(2a)x.(1)討論f(x)的單調(diào)性；(2)設a0，證明：當0xf；(3)若函數(shù)yf(x)的圖像與x軸交于a，b兩點，線段ab中點的橫坐標為x0，證明：f(x0)0，所以f(x)在(0，)上單調(diào)遞增；若a0，則由f(x)0得x，且當x時，f(x)0；當x時，f(x)0，所以f(x)在上單調(diào)遞增，在上單調(diào)遞減(2)證明：設函數(shù)g(x)ff，則g(x)ln(1ax)ln(1ax)2ax，g(x)2a.當0x0，而g(0)0，所以g(x)0，故當0xf.(3)證明：由(1)可得，當a0時，函數(shù)yf(x)的圖像與x軸至多有一個交點，故a0，從

2、而f(x)的最大值為f，且f0.不妨設a(x1,0)，b(x2,0)，0x1x2，則0x1f(x1)0.從而x2x1，于是x0.由(1)知，f(x0)b0)的左、右頂點分別為a、b，右焦點為f，過f作垂直于x軸的直線與橢圓相交于r、s兩點，若線段rs的長為.(1)求橢圓c的方程；(2)設q(9，m)是直線x9上的點，直線qa、qb與橢圓c分別交于點m、n，求證：直線mn必過x軸上的一定點，并求出此定點的坐標解：(1)依題意，橢圓過點，故解得所以橢圓c的方程為1.(2)由題意知，直線qa的方程為y(x3)，代入橢圓方程，得(80m2)x26m2x9m27200，設m(x1，y1)，則3x1x1，

3、所以y1(x13)，故點m的坐標為.同理，直線qb的方程為y(x3)，代入橢圓方程，得(20m2)x26m2x9m21800，設n(x2，y2)，則3x2x2，所以y2(x23)，故點n的坐標為.若m240，直線mn的方程為x1，與x軸交于點(1,0)；若m240，直線mn的方程為y，令y0，解得x1.綜上所述，直線mn必過x軸上的定點(1,0)高考資源網(wǎng)版權所有！投稿可聯(lián)系qq：1084591801c29c91afe4ced1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91afedd1b6c8

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