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2 Word版含解析

1、“2道”拉分题专练卷(二)1(2013温州八校联考)已知函数f(x)ex2x23x.(1)求曲线yf(x)在点(1，f(x)处的切线方程；(2)当x时，若关于x的不等式f(x)x2(a3)x1恒成立，试求实数a的取值范围解：(1)f(x)ex4x3，则f(1)e1，又f(1)e1，曲线yf(x)在点(1，f(1)处的切线方程为ye1(e1)(x1)，即(e1)xy20.(2)由f(x)x2(a3)x1，得ex2x23xx2(a3)x1，即axexx21.x，a令g(x).则g(x).令(x)ex(x1)x21，则(x)x(ex1)x，(x)0，(x)在上单调递增，(x)0，因此g(x)0，故g

2、(x)在上单调递增，则g(x)g2.a的取值范围是.2经过点f(0,1)且与直线y1相切的动圆的圆心轨迹为m.点a、d在轨迹m上，且关于y轴对称，过线段ad(两端点除外)上的任意一点作直线l，使直线l与轨迹m在点d处的切线平行，设直线l与轨迹m交于点b、c.(1)求轨迹m的方程；(2)证明：badcad；(3)若点d到直线ab的距离等于|ad|，且abc的面积为20，求直线bc的方程解：(1)法一：设动圆圆心为(x，y)，依题意得，|y1|.整理，得x24y.所以轨迹m的方程为x24y.法二：设动圆圆心为p，依题意得点p到定点f(0,1)的距离和点p到定直线y1的距离相等，根据抛物线的定义可知

3、，动点p的轨迹是抛物线且其中定点f(0,1)为焦点，定直线y1为准线所以动圆圆心p的轨迹m的方程为x24y.(2)证明：由(1)得x24y，即yx2，则yx.设点d，由导数的几何意义知，直线l的斜率为kbcx0.由题意知点a，设点c，b，则kbcx0，即x1x22x0.因为kac，kab，所以kackab0，即kackab.所以badcad.(3)法一：由点d到ab的距离等于|ad|，可知bad45.不妨设点c在ad上方，即x2x1，直线ab的方程为yx(xx0)由解得点b的坐标为.所以|ab|(x04)(x0)|2|x02|.由(2)知cadbad45，同理可得|ac|2|x02|.所以ab

4、c的面积s2|x02|2|x02|4|x4|20，解得x03.当x03时，点b的坐标为，kbc，直线bc的方程为y(x1)，即6x4y70.当x03时，点b的坐标为，kbc，直线bc的方程为y(x7)，即6x4y70.法二：由点d到ab的距离等于|ad|，可知bad45.由(2)知cadbad45，所以cab90，即acab.由(2)知kac，kab，所以kackab1.即(x1x0)(x2x0)16.由(2)知，x1x22x0.不妨设点c在ad上方，即x2x1，由解得所以|ab| 2|x02|，同理，|ac|2|x02|.以下同解法一高考资源网版权所有！投稿可联系qq：1084591801c

5、29c91afe4ced1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91afedd1b6c87d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91af95ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9d1b6c8795ad9c91afe4ced1b6c8795a1afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1

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