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2 Word版含解析

1、“2道”拉分題專練卷(二)1(2013溫州八校聯(lián)考)已知函數(shù)f(x)ex2x23x.(1)求曲線yf(x)在點(diǎn)(1，f(x)處的切線方程；(2)當(dāng)x時(shí)，若關(guān)于x的不等式f(x)x2(a3)x1恒成立，試求實(shí)數(shù)a的取值范圍解：(1)f(x)ex4x3，則f(1)e1，又f(1)e1，曲線yf(x)在點(diǎn)(1，f(1)處的切線方程為ye1(e1)(x1)，即(e1)xy20.(2)由f(x)x2(a3)x1，得ex2x23xx2(a3)x1，即axexx21.x，a令g(x).則g(x).令(x)ex(x1)x21，則(x)x(ex1)x，(x)0，(x)在上單調(diào)遞增，(x)0，因此g(x)0，故g

2、(x)在上單調(diào)遞增，則g(x)g2.a的取值范圍是.2經(jīng)過點(diǎn)f(0,1)且與直線y1相切的動(dòng)圓的圓心軌跡為m.點(diǎn)a、d在軌跡m上，且關(guān)于y軸對(duì)稱，過線段ad(兩端點(diǎn)除外)上的任意一點(diǎn)作直線l，使直線l與軌跡m在點(diǎn)d處的切線平行，設(shè)直線l與軌跡m交于點(diǎn)b、c.(1)求軌跡m的方程；(2)證明：badcad；(3)若點(diǎn)d到直線ab的距離等于|ad|，且abc的面積為20，求直線bc的方程解：(1)法一：設(shè)動(dòng)圓圓心為(x，y)，依題意得，|y1|.整理，得x24y.所以軌跡m的方程為x24y.法二：設(shè)動(dòng)圓圓心為p，依題意得點(diǎn)p到定點(diǎn)f(0,1)的距離和點(diǎn)p到定直線y1的距離相等，根據(jù)拋物線的定義可知

3、，動(dòng)點(diǎn)p的軌跡是拋物線且其中定點(diǎn)f(0,1)為焦點(diǎn)，定直線y1為準(zhǔn)線所以動(dòng)圓圓心p的軌跡m的方程為x24y.(2)證明：由(1)得x24y，即yx2，則yx.設(shè)點(diǎn)d，由導(dǎo)數(shù)的幾何意義知，直線l的斜率為kbcx0.由題意知點(diǎn)a，設(shè)點(diǎn)c，b，則kbcx0，即x1x22x0.因?yàn)閗ac，kab，所以kackab0，即kackab.所以badcad.(3)法一：由點(diǎn)d到ab的距離等于|ad|，可知bad45.不妨設(shè)點(diǎn)c在ad上方，即x2x1，直線ab的方程為yx(xx0)由解得點(diǎn)b的坐標(biāo)為.所以|ab|(x04)(x0)|2|x02|.由(2)知cadbad45，同理可得|ac|2|x02|.所以ab

4、c的面積s2|x02|2|x02|4|x4|20，解得x03.當(dāng)x03時(shí)，點(diǎn)b的坐標(biāo)為，kbc，直線bc的方程為y(x1)，即6x4y70.當(dāng)x03時(shí)，點(diǎn)b的坐標(biāo)為，kbc，待添加的隱藏文字內(nèi)容1直線bc的方程為y(x7)，即6x4y70.法二：由點(diǎn)d到ab的距離等于|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.不妨設(shè)點(diǎn)c在ad上方，即x2x1，由解得所以|ab| 2|x02|，同理，|ac|2|x02|.以下同解法一高考資源網(wǎng)版權(quán)所有！投稿可聯(lián)系qq：

5、108459180129c91afe4ced1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91afedd1b6c87d1b6c8795ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9c91fe4ced1b6c8795ad29c91af95ad9c91afe4ced1b6c8795ad91afe4ced1b6c8795ad9d1b6c8795ad9c91afe4ced1b6c8795a1afe4ced1b6c8795ad91afe4ced1b6c8795ad9

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