2022宁德高二上学期期末考试数学PDF版含答案

这是一份2022宁德高二上学期期末考试数学PDF版含答案，文件包含福建省宁德市2021-2022学年度第一学期期末高二质量检测数学试卷参考答案doc、福建省宁德市2021-2022学年第一学期期末高二质量检测数学试题pdf等2份试卷配套教学资源，其中试卷共15页， 欢迎下载使用。
宁德市2021-2022学年度第一学期期末高二质量检测数学参考答案及评分标准说明:一、本解答指出了每题要考查的主要知识和能力，并给出了一种或几种解法供参考，如果考生的解法与本解法不同，可根据试题的主要考查内容比照评分标准制定相应的评分细则.二、对计算题，当考生的解答在某一部分解答未改变该题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半；如果后继部分的解答有较严重的错误，就不再给分.三、解答右端所注分数，表示考生正确做到这一步应得的累加分数.四、只给整数分数，选择题和填空题不给中间分. 一、单项选择题: 本题共8小题，每小题5分，共40分. 在每小题给出的四个选项中，只有一个选项是符合题目要求的.1. A   2．C    3．B     4．D    5. A     6.C    7．B    ８．C 二、多项选择题：本题共4小题，每小题5分，共20分. 在每小题给出的选项中有多项符合题目要求，全部选对的得5分，部分选对的得2分，有选错的得0分）9． ABC     10. BC11. ABC     12 . BCD 三、填空题:（本大题共4小题，每小题5分，共20分. 把答案填在答题卡的相应位置）13.     14. 30.       15.      16 . （答案为不扣分） 四、解答题：本大题共6小题，共70分. 解答应写出文字说明，证明过程或演算步骤.                17. （本小题满分10分） 解：选①:  ..................................................................2分或（舍去）·························································4分选②: 依题意得························································2分即得································································4分选③: 偶数项的二项式系数之和为·············································2分 ··································································4分（1）展开式的通项公式为 ...................6分令，得····························································7分 展开式中的项的系数为················································8分 （2）展开式中二项式系数最大的项为··································10分  18. （本小题满分12分）解：法一：（1）边上的中线所在的直线的方程为可设···························································1分边上的高所在的直线方程为，其斜率为-2，·····························································3分解得：··························································4分·······························································5分（2）······························································6分······························································8分·····························································10分在以为直径的圆上.···············································12分法二：（1）同法一 ...........................................................................................................5分（2）·····························································7分，·····························································9分的外接圆的圆心为，半径为的外接圆的方程为···············································11分, 即满足上述方程四点共圆.·····················································12分法三：（1）同法一 ...........................................................................................................5分（2）设外接圆的方程为：··········································6分将三点分别代入圆的方程：········································7分解得························································10分的外接圆方程：···············································11分满足上述方程,四点共圆.···················································12分法四：（1）同法一 ...........................................................................................................5分（2）设外接圆的方程为：··············································6分将三点代入圆的方程：············································7分解得..................................................................................................................10分的外接圆方程：················································11分满足上述方程四点共圆.·····················································12分19. （本小题满分12分）解：（1）设的公差为，设的公比为，.....................................................1分 则····························································3分·····························································4分.......................................................................................................................5分····························································6分（2）·····························································7分·····························································8分····························································9分··························································12分(答案是或不扣分）)20.（本小题满分12分）解：（1）设椭圆C的该方程：......................................................１分则························································２分································································４分椭圆C的方程：···············································５分（2）解法一：，设，···············································６分联立························································７分解得：，·····················································９分所以，，··························································１０分··························································１１分··························································１２分解法二：，设，····················································６分···························································７分···························································８分联立························································９分　·····························································１０分...............................................................................................１２分21(本小题满分12分)解：（1），,数列是公比为的等比数列...................................................................              1分................................................................................................... 2分·····························································4分是1为首项，1为公差等差数列·······································5分···························································６分（2）由（1）知，..........................................................................７分所以，所以························································８分即，得...........................................................................................１０分所以·······················································１１分所以·······················································１２分22.（本小题满分12分）解：法一（1）设，动点到直线的距离比到点的距离大1等于到直线的距离··············································１分根据抛物线的定义知，曲线E是以为焦点，直线为准线的抛物线．··········································2分故曲线的方程为．··············································４分（2）显然斜率存在，分别设的直线方程为································５分联立，所以，················································６分同理得：····················································７分因为三点共线，··························································８分，所以，同理得：···································９分若存在满足题设的定点，由抛物线与圆的对称性知定点必在轴上，设定点为由，，三点共线，，························································１０分得，即4即恒成立，············································11分，直线过定点················································12分（备注：能写对定点坐标的给1分）法二：（1）设动点依题意有：····················································2分由几何直观知所以，化简得：................................................................................................... 4分（2）当直线垂直于轴时，由对称性可知也垂直于轴，所以，，的直线方程为，，,同理得：，所以直线的方程为········································5分当直线不垂直于轴时，设的方程为，，，联立·····················6分设的直线方程为，,则，所以，，所以，······················································7分因为，三点共线,·····························································9分····························································10分化简得，····················································11分代入直线的方程为，所以直线恒过综合，得直线恒过··············································12分 法三：（1）设动点依题意有：····················································2分当时，得化简得：当时，得化简得：，舍去所以，·····················································4分（注：若未舍去，扣1分）(2)显然直线不平行于轴，可设的方程为，，，联立················5分设的直线方程为，，所以，·····················································6分，所以，·····················································7分因为，三点共线,····························································9分····························································10分化简得，····················································11分代入直线的方程为，所以直线恒过·································12分