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宁德市2023-2024学年度第二学期期末高一质量检测
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. C 2．A 3．D 4．D 5. B 6.C 7．C 8．B
第8题解析：
由 可得， ，
所以 .
因为 为钝角，所以 ，且 不共线，
所以 ，即 ，且 .
当 时，有 且 ，所以 可取2，4，5，6；
当 时，有 ， 可取5，6；
当 ， ， ， 时， ，此时无解.
综上所述，满足条件的 有6种可能.
又先后抛掷两次，得到的样本点数共36种，
所以 为钝角的概率
二、多项选择题：本题共3小题，每小题6分， 共18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6分，部分选对的得部分分，有选错的得0分）
9． AD 10. ACD 11. AC
第11题解析：
对于A ．该“刍童”的表面积为 ，所以正确
对于B ．由轴截面的等腰梯形 可知，其高 ，如图所示：
高二数学参考答案 第1 页 共8页
学学学科科科网网网（（（北北北京京京）））股股股份份份有有有限限限公公公司司司能够被完整放入该“刍童”内的圆台的最大的体积为 ,所以不正确
对于C ．该“刍童”的外接球的球心到平面 的距离为 ，而平面 的外接圆
的圆心恰为线段 的中点，故该“刍童”的外接球的球心到平面 的距离为 ，所
以正确．
对于D ，若正四面体在此容器内部可以任意转动，则正四面体的外接球可以放进容器，棱
长为 的正四面体的外接球直径为 ，由轴截面的等腰梯形 可知，其高
，如下图所示：
可知此棱台可放入的最大球的直径为 ，小于正四面体的外接球直径，
故不可以在此空心棱台容器内部任意转动，所以D不正确．
故选：AC
三、填空题:（本大题共3小题，每小题5分，共15分. 把答案填在答题卡的相应位置）
12. 2800 13. 14.
第14题解析：
如图示，可知： 是边长为2的菱形，且 ，
，所以
四、解答题：本大题共5小小题，共77分. 解答应写出文字说明，证明过程或演算步骤
.
15. （本小题满分13分）
高二数学参考答案 第2 页 共8页
学学学科科科网网网（（（北北北京京京）））股股股份份份有有有限限限公公公司司司（1）因为 ，································································2分
所以 ·····························································································4分
所以 ······························································6分
（2）因为 ···························································7分
·················································································9分
所以 ······················································································11分
所以 ························································································13分
16. （本小题满分15分）
解：（1）证明：由于底面 为矩形，所以 ··································1分
又有 平面 ， 平面 ，所以 ···································4分
因为 ，所以 平面 ·······················································6分
因为 平面 ，所以平面 平面 ·······································7分
（2）解：由于 即 为异面直线 和 所成角所成的角．·············8分
因为异面直线 和 所成角的正切值为 ，
由于 平面 所以
所以在 中， ，所以 ····················································9分
由于 是边长为4的正三角形，取 的中点 ，连接 ，
所以 ···········································································10分
因为平面 平面 ， 平面 ，所以 平面 ．···········11分
取 的中点 ，连接 ， ， ， ，所以
所以 为二面角 的平面角，··················································13分
在 中， ，所以 ． ·································14分
所以 为二面角 的平面角 ·················································15分
高二数学参考答案 第3 页 共8页
学学学科科科网网网（（（北北北京京京）））股股股份份份有有有限限限公公公司司司17.（本小题满分15分）
解：（1）根据正弦定理可得： ， ················2分
··············································3分
因为 ， ··········································································4分
所以 ··························································································5分
即 ···································································································7分
法二:根据余弦定理可得 ················2分
即 ，························································································3分
所以 ··························································································4分
因为 ，······················································································5分
所以 .······························································································7分
(2)解：因为 ，
所以 ····························································8分
解得 ·····························································································9分
因为 平分
在 中根据正弦定理得：
在 中根据正弦定理得：
所以 ，······················································································11分
高二数学参考答案 第4 页 共8页
学学学科科科网网网（（（北北北京京京）））股股股份份份有有有限限限公公公司司司所以 ，·························································································12分
所以 ···················13分
所以 ，··············································14分
解得 ，即 ；·································································15分
法二:因为 ，
所以 ····························································8分
解得 ·····························································································9分
因为 平分
所以 ·······························12分
整理得： ················································································14分
即 ····························································································15分
18. （本小题满分17分）
（1）根据频率分布直方图有，男生成绩样本数据的平均数
························4分
所以男生成绩样本数据的平均数为71.（列式正确，计算错误扣1分）
（2）在区间 和 内的男生成绩样本数据分别有4个和2个，···········5分
分别用 和 表示，则在这6个数据中随机抽取两个的样本空间 包含的样本点
有 ，
，
个数为 ，····················································································7分
高二数学参考答案 第5 页 共8页
学学学科科科网网网（（（北北北京京京）））股股股份份份有有有限限限公公公司司司记事件 “这两个样本来自同一区间”，
则事件 包含的样本点有
个数为 ，······················································································9分
所以 ；·············································································10分
（3）设男生成绩样本数据为 ， ，…， ，其平均数为71，方差为
女生成绩样本数据为 ， ，…， ，其平均数为 ，
方差为 ；总样本的平均数为 ，方差为 .··········································11分
由按比例分配分层随机抽样总样本平均数与各层样本平均数的关系，
得 .·············································································13分
················································14分
·············································15分
.··································································································17分
所以总样本的平均数和方差分别为 和 .
19.（本小题满分17分）
（1）解：由于 是正方体，
两直线 ， 与面 所成的角相等
即 ，由于 ·········································1分
法1： ，即 ····································3分
法2：所以 ，又有 是 的中点， ········3分
即 ，
依题意平面内点 到两定点 距离之比为2,故点 的轨迹是圆，
而点 是正方体表面 上一动点（包括边界），
即点 的轨迹是一段阿波罗尼斯圆的弧.··················································4分
高二数学参考答案 第6 页 共8页
学学学科科科网网网（（（北北北京京京）））股股股份份份有有有限限限公公公司司司画出上图弧线即给分，不要求精确·························································5分
（2）依题意可知：圆心 在 所在的直线上，············································6分
法一：作圆 与 交于点 ，与 的延长线上交于点 ,显然 恰为圆 的直
径，故依 , 恰好为 线段的三分之一分点， ,
, , , ·················································7分
法二：易知此圆 与 的交点为 ，与 的交点为 ,
则满足： ,故在 , ,
在 中， ,
故 为正三角形，故 , ， ··············7分
法一：设 与 所成的角为 ，可知 ·····························8分
·················································9分
·········································································10分
法二：（建系）以 为平面直角的坐标原点，分别以 ,过点 垂直于 的直线
为 ， 轴，建立平面直角坐标系，故可设 , , ,
·····················8分
其中 ················9分
高二数学参考答案 第7 页 共8页
学学学科科科网网网（（（北北北京京京）））股股股份份份有有有限限限公公公司司司故 ······································································10分
（3）由（2)可知，当线段 的长最短时，即点 在直线 上，
故延长 交 于点 ,过点 做 ,交 于点 ,交 于点 ,交 于
点 ,连接 交 于点 ,所求的截面即为五边形 .
以下证明 平面 ,
由于 , 平面 , 平
所以 平面 ,·······································································11分
故有 , ,
在 中， 12分
············································
在 中， 13分
·········································
在 中， ·············································14分
在 中， 15分
············································
在 中， ···········································16分
所以所求的截面五边形 的周长
············································································17分
高二数学参考答案 第8 页 共8页
学学学科科科网网网（（（北北北京京京）））股股股份份份有有有限限限公公公司司司
