高考數(shù)學(xué)（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(shè)a0，證明：當(dāng)0xf；(3)若函數(shù)yf(x)的圖像與x軸交于a，b兩點(diǎn)，線段ab中點(diǎn)的橫坐標(biāo)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(dāng)x時(shí)，f(x)0；當(dāng)x時(shí)，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(shè)函數(shù)g(x)ff，則g(x)ln(1ax)ln(1ax)2ax，g(x)2a.當(dāng)0x0，而g(0)0，所以g(x)0，故當(dāng)0xf.(3)證明：由(1)可得，當(dāng)a0時(shí)，函數(shù)yf(x)的圖像與x軸至多有一個(gè)交點(diǎn)，故a0，從

2、而f(x)的最大值為f，且f0.不妨設(shè)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(diǎn)分別為a、b，右焦點(diǎn)為f，過f作垂直于x軸的直線與橢圓相交于r、s兩點(diǎn)，若線段rs的長為.(1)求橢圓c的方程；(2)設(shè)q(9，m)是直線x9上的點(diǎn)，直線qa、qb與橢圓c分別交于點(diǎn)m、n，求證：直線mn必過x軸上的一定點(diǎn)，并求出此定點(diǎn)的坐標(biāo)解：(1)依題意，橢圓過點(diǎn)，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(shè)m(x1，y1)，則3x1x1，

3、所以y1(x13)，故點(diǎn)m的坐標(biāo)為.同理，直線qb的方程為y(x3)，代入橢圓方程，得(20m2)x26m2x9m21800，設(shè)n(x2，y2)，則3x2x2，所以y2(x23)，故點(diǎn)n的坐標(biāo)為.若m240，直線mn的方程為x1，與x軸交于點(diǎn)(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(diǎn)(1,0)高考資源網(wǎng)版權(quán)所有！投稿可聯(lián)系qq：1084591801c29c91afe4ced1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91afedd1b6c8

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