高考数学（理）二轮专题突破演练（浙江专版）第3部分 专题1 第2讲“2道”拉分题专练卷1 Word版含解析

1、“2道”拉分题专练卷(一)1已知函数f(x)ln xax2(2a)x.(1)讨论f(x)的单调性；(2)设a0，证明：当0xf；(3)若函数yf(x)的图像与x轴交于a，b两点，线段ab中点的横坐标为x0，证明：f(x0)0，所以f(x)在(0，)上单调递增；若a0，则由f(x)0得x，且当x时，f(x)0；当x时，f(x)0，所以f(x)在上单调递增，在上单调递减(2)证明：设函数g(x)ff，则g(x)ln(1ax)ln(1ax)2ax，g(x)2a.当0x0，而g(0)0，所以g(x)0，故当0xf.(3)证明：由(1)可得，当a0时，函数yf(x)的图像与x轴至多有一个交点，故a0，从

2、而f(x)的最大值为f，且f0.不妨设a(x1,0)，b(x2,0)，0x1x2，则0x1f(x1)0.从而x2x1，于是x0.由(1)知，f(x0)b0)的左、右顶点分别为a、b，右焦点为f，过f作垂直于x轴的直线与椭圆相交于r、s两点，若线段rs的长为.(1)求椭圆c的方程；(2)设q(9，m)是直线x9上的点，直线qa、qb与椭圆c分别交于点m、n，求证：直线mn必过x轴上的一定点，并求出此定点的坐标解：(1)依题意，椭圆过点，故解得所以椭圆c的方程为1.(2)由题意知，直线qa的方程为y(x3)，代入椭圆方程，得(80m2)x26m2x9m27200，设m(x1，y1)，则3x1x1，

3、所以y1(x13)，故点m的坐标为.同理，直线qb的方程为y(x3)，代入椭圆方程，得(20m2)x26m2x9m21800，设n(x2，y2)，则3x2x2，所以y2(x23)，故点n的坐标为.若m240，直线mn的方程为x1，与x轴交于点(1,0)；若m240，直线mn的方程为y，令y0，解得x1.综上所述，直线mn必过x轴上的定点(1,0)高考资源网版权所有！投稿可联系qq：1084591801c29c91afe4ced1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91afedd1b6c8

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