2021福建省漳州市高三毕业班数学适应性测试题答案

2021届福建省漳州市高三毕业班数学适应性测试题答案

（考试时间：120分钟；满分：150分）

本试卷分第Ⅰ卷和第Ⅱ卷两部分。第Ⅰ卷为选择题，第Ⅱ卷为非选择题。

第Ⅰ卷（选择题60分）

一．单项选择题（本大题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的.）

1．C        2．D        3．D         4．A        5． B       6．C

7．A        8．C

【解析】

1．集合，，不满足，则A错；，则B错；，则C正确；，则D错.故选C.

2．

则，复数在复平面上对应的点为，故复数在复平面上对应的点位于第四象限，故

选D.

3．依题意可知在角的终边上，所以，故选D．

4．点是抛物线内的一点，设点在抛物线准线上的射影为，根据抛物线的定义可知

  ，要求的最小值，即求的最小值. 当，，三点共线时，

  取到最小值. 故选A.

5．由题意得，恰好有6段圆弧或有段圆弧与直线相交时，才恰有个交点，每段圆弧的圆心角都为，且从第1段圆弧到第段圆弧的半径长构成等差数列：，，，

当得到的“螺旋蚊香”与直线恰有个交点时，“螺旋蚊香”的总长度

的最大值为．故选B.

6． 由题意，以所在直线为x轴，的垂直平分线为y轴建立坐标系，

由于，，则，，故

又点是的重心，则

，，，故选C

7．，

即为函数与的图象交点的横坐标

，

即为函数与的图象交点的横坐标

，

即为函数与的图象交点的横坐标

在同一坐标系中画出图象，可得.故选A.

8．如图，点，，，，，分别是边，，，，

，的中点，这两个正四面体公共部分为多面体．

三棱锥是正四面体，其棱长为正四面体棱长的一半，

则

这两个正四面体公共部分的体积为.  

故选C.

二、多项选择题（本大题共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，有选错的得0分，部分选对的得3分。）

9．ACD              10．BCD           11．ABD          12．BD

   【解析】

9． 由于小王选择的是每月还款数额相同的还贷方式，故可知2020年用于房贷方面的支出费用跟2017年

相同，D错；设一年房贷支出费用为，则可知2017年小王的家庭收入为，2020年小王的家庭收入为，，小王一家2020年的家庭收入比2017年增加了50% ，C错；2017、2020年用于饮食的支出费用分别为，A错；2017、2020年用于其他方面的支出费用分别是，B对．故选ACD．

10．由已知，得. 若，则，不满足，故A错；

    由，故B正确；当时，且，则，，

    所以，故C正确；当时，且，则，，所以

    ，所以，则，故D正确. 故选BCD.

11．，A正确；

已知

所以

即，D正确；

若为锐角三角形，

所以 ，若为直角三角

形或钝角三角形时可类似证明，B正确；

，所以，C错．故选ABD.

12．因为的定义域为，，所以是奇函数，

但是，所以不是周期为的函数，故A错误；

当时，，，单调递增，

当时，，，单调递增，

且在连续，故在单调递增，故B正确；

当时，，，

令得，，

当时，，

令得，,

因此，在内有20个极值点，故C错误；

当时，，

设，所以，

令， ，，单调递增，

，所以，在单调递增.

当趋近于时，趋近于， 所以，故D正确. 故选BD.

三．填空题（本大题共4小题，每小题5分，共20分.）

13． 14．          15．  16．，

【解析】

13．展开式的通项为

令，则， 所以的展开式中，的系数为

故答案为.

14．由已知，当时，，不等式等价于

又定义在R上的偶函数，

所以，所以或，解得或

则不等式的解集为

故答案为.

15．因为，，所以，

又因为是等腰直角三角形，，，所以，

因为，又，

所以，所以．又，所以

根据题意可知异面直线与所成角为，

根据余弦定理得，

故答案为.

16．如图所示建立平面直角坐标系，设的中点为，则由双曲线的对称性知，

，

所以，

所以，可得，，；

   的焦距为，所以，. 设，则，

又由，得，

所以，在中，由余弦定理得，

，

即，解得，即．

故答案为，.

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．解：（1）因为，，所以，

因为，所以，······················································2分

又因为，所以．·····················································3分

所以的面积．······················································5分

（2）由（1）可得，

所以

                 ·························································8分

因为，所以，······················································9分

所以当时，有最大值1．··············································10分

18．解：（1）设等比数列的公比为，根据已知条件， ，，依次构成等差数列，

所以，则，························································2分

因为，所以，解得···················································4分

由，即，所以，解得·················································5分

所以····························································6分

（2）选①·························································7分

································································9分

·······························································12分

选②·····························································7分

（*）

································································9分

可得

·······························································11分

所以···························································12分

选③ ····························································9分

·······························································12分

19．解：（1）取的中点为，连接，

则////，且·························································2分

所以四边形为平行四边形， 所以//，又面，面

所以//面.··························································4分

（2）面，则，，

是圆柱底面的直径，是弧的中点，所以，为中点，则

以为原点，，，分别为轴，轴，轴的正方向建立空间直角坐标系，

如图．···························································5分

设，则，，，·····················································6分

，

设平面的一个法向量为，则

，，取，则，，

则······························································8分

，

设平面的一个法向量为，则

，，解得，取，则，

则·····························································10分

所以.···························································11分

所以锐二面角的余弦值为.···········································12分

20．解：（1）

································································2分

································································3分

（2）由题意样本方差，故，所以，·····································4分

由题意，该厂生产的产品为正品的概率

.所以//面.·························································6分

（3）所有可能取值为，，，.··········································7分

            ···························································9分

随机变量的分布列为

·······························································11分

·······························································12分

21．解：（1），，，令，得

当，，在单调递减

当，，在单调递增

所以是的极小值点同时也是最小值点，即································2分

当，即时，在上没有零点；

当，即时，在上只有1个零点；········································3分

因为，所以只有一个零点，

又因为，取，

，得

当，，在单调递增

当，，在单调递减

，所以对，，所以，即

所以，所以内只有一个零点，

所以在上有两个零点.···············································5分

综上所述，当时，在上有两个零点；

当时，函数在上没有零点；

当时，函数在上有一个零点.··········································6分

（2）方法一：

恒成立，

即

································································7分

所以

构造，所以，在上单调递增

只需，即恒成立····················································8分

令， ····························································9分

当时，，所以在单调递减；

当时，，所以在单调递增，

所以，即·························································11分

又，所以. ·······················································12分

方法二：

，有，则当时，····················································7分

令，所以在单调递减，··············································8分

注意到，所以.（必要性） ··············································9分

下面证明（）

令，

当，，所以在上单调递增

当，，所以在上单调递减

所以，

    即对，，即（）得证.

因为，所以，即，即.

当时，··························································11分

.（充分性）························································12分

方法三：

，即恒成立

································································7分

即，恒成立，·····················································8分

注意到在单调递增, ················································9分

当时，，

所以···························································10分

当时，

注意到，存在，使得，矛盾··········································11分

综上，.·························································12分

22．解：（1）由题意可得：，解得 ，，·································2分

则椭圆方程为·····················································3分

（2）设直线的方程为，设，

由，整理得

  ·······························································4分

椭圆的左、右顶点分别为，，

直线方程为：，

又直线与直线交于点，则， ···········································5分

因为,都存在，所以要证，，三点共线，只需证·····························6分

只需证

只需证

只需证

只需证

而

故，，三点共线.···················································8分

（3）由（2）可得

·······························································10分

令（），则

令，函数在区间单调递增，

即当，即时，取到最大值.···········································12分

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