2022宁德高二上学期期末考试数学PDF版含答案

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