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2025年普通高等学校招生全国统一考试模拟测试（二）
数学 参考答案
一、单项选择题
题号 1 2 3 4 5 6 7 8
答案 D C A B C B A D
二、多项选择题
题号 9 10 11
答案 BC BCD ACD
三、填空题
12. 13. 14. ，
四、解答题
15.解：（1）由余弦定理得 ，···················································2分
所以 ，····································································4分
（没变形但答案正确不扣分）
所以 .········································································································6分
（2）由正弦定理得 ，······················7分
整理得 ，··································8分
（正弦定理与正弦二倍角公式各1分）
因为 ，所以 ，···9分
所以 ，··············································································10分
（化简最后结果正确即可得分）
高三数学 第1页（共9页）因为 ，所以 ，解得 ，···············································11分
所以 ，·····························································································12分
（ 与 都没有扣1分）
所以 .································································13分
（答案正确，面积公式没写不扣分）
16. 解：（1）由已知得 ，································································1分
设点 ，则 ，且 ，所以 ，
·······················································································································2分
因为 ，所以 ，············································3分
因为 ，所以 ，所以 .··········································5分
另解：因为 ，所以 ，·································3分
因为 ，所以 ，所以 . ··································5分
（x范围写错，但题目过程思路正确，给3分）
（2）解法一：
假设四边形 是以 为对称轴的，
则 ，且 的中点 在 上,···································································6分
设 ， ，则 ， ，
，···················································································7分
高三数学 第2页（共9页）（说明M是AC中点，且写出中点坐标关系即可得分）
所以 ，····················································································8分
因为点 在第一象限，且 ，所以 ，····················································9分
所以 ，·········································································10分
所以 ，············································································11分
因为 三点共线，所以 ，·························································12分
所以 ，···········································································13分
因为 ，与 矛盾，故假设不成立，···················································14分
所以四边形 不能是以 为对称轴································································15分
解法二：
假设四边形 是以 为对称轴的，
则 ，且 的中点 在 上，································································6分
因为点 在第一象限，所以直线 斜率存在且直线 不过原点，·····························7分
设直线 的方程为 ，···························································8分
联立得 ，消去 并整理得 ，·························5分
（注：设直线AC方程时，没有说明ｋ存在，扣１分）
设 ，则 ········································10分
所以 的中点为 ，································································11分
高三数学 第3页（共9页）因为 三点共线，所以 ，·······················································12分
所以 ，············································································13分
因为 ，与 矛盾，故假设不成立，···················································14分
所以四边形 不能是以 为对称轴································································15分
另解：
因为 ，所以 ，则直线OB方程为： ································12分
因为 的中点 在 上，所以 代入直线OB方程得：
，解得ｍ＝０，即直线AC过原点O·······································14分
所以A、O、C三点共线，与OABC为四边形矛盾，
所以四边形 不能是以 为对称轴································································15分
17.解：（1）如图，连接 ， ， ， 和 ················································1分
因为 ， ， 为 中点，所以 ， ··························2分
又因为 ， 平面 ，所以 平面 ·······························3分
同理 平面 ····························································································4分
因为平面 与平面 有公共点 ，且垂直于同一条直线 ································5分
所以 ， ， ， 四点共面···············································································6分
注：①第5分，原因有道理即可给分，若没有写，这1分不给
（2）法一：如图，以 ， 分别为 轴， 轴，以过点 且垂直于平面 的直线 轴，建
立如图所示空间直角坐标系，则 ， ， ，························8分
由（1）得 为二面角 的平面角
高三数学 第4页（共9页）设 ，则点 ···································································9分
故 ， ， ，
设平面 的法向量为 ，
则 ，即 ，解得 ,
取 ,得
······················································································································11分
设直线 与平面 所成角为 ，
则
······················································································································12分
其中 ， ，·········································································13分
当 时， 取得最大值 ，·······················································14分
所以直线 与平面 所成角的正弦值的最大值为 .·········································15分
注：②第8分，建系正确就给1分，建系正确且D点坐标正确就给2分
③以下是x轴和y轴正方向不同时，对应的A，C，D的坐标及平面ACD的法向量，供参考
高三数学 第5页（共9页）（2）法二：如图，以 ， 分别为 轴， 轴，以过点 且垂直于平面 的直线 轴，建
立如图所示空间直角坐标系，则 ， ， ，························8分
设点 ， ··········································································9分
故 ， ， ，
设平面 的法向量为 ，
则 ，即 ，解得 ,
取 ,得
······················································································································11分
设直线 与平面 所成角为 ，
则 ·····································12分
求 的最大值，即求 的最大值
令 ，即 表示一条直线， 表示圆，所以当直线与圆相切
时， 取得最大值·····························································································13分
则圆心到直线的距离等于圆的半径3，即 ，所以 ·······························14分
所以直线 与平面 所成角的正弦值的最大值为 .···········································15分
高三数学 第6页（共9页）18. 解：（1）质点运动3次后停在原点右侧的情况有4种，分别是：0次不动、3次向右、0次
向左；1次不动、2次向右、0次向左；0次不动、2次向右、1次向左；2次不动、1次向右、0
次向左.·············································································································2分
所以质点运动3次后停在原点右侧的概率
.·····································4分
说明：每1种情况分析、列式正确得1分.
（分类不完整，每少一种情况扣2分；分类完整，概率计算错误扣1分）
（2）①质点在运动过程中出现在原点左侧就停止运动且运动5次后停在原点右侧的情况有4
种：5次向右；第1次向右、后4次有3次向右1次向左；前2次向右、后3次有1次向右2次
向左；第1次向右、第2次向左、第3次向右、后2次有1次向右1次向左.····················6分
所以质点在运动过程中出现在原点左侧就停止运动且运动5次后停在原点右侧的概率
.································································································8分
说明：每1种情况分析、列式正确得1分.
（分类不完整，每少一种情况扣2分；分类完整，概率计算错误扣1分）
②第一轮游戏结束进入第二轮游戏的情况有2种，分别是3次向右；2次向右，1次向左.
其概率为 ；····································································9分
设两轮游戏最终得分的随机变量为 ，则 的所有可能取值为0，1，3，
易知 的期望 仅与 1，3的概率只有关，因此
，
，·············································································10分
所以最终得分 的期望
高三数学 第7页（共9页），···········11分
（计算每个概率时未乘 ，给1分）
因为 ，所以 ，即 ，
所以当 时， ；当 时， ；·······································12分
记 ，
求导得
记 ，··············································································13分
解法一、①当 时， ，
因为 ， ， ，
所以由零点存在定理，存在 ，使得 ；存在 ，使得 ，
当 时， ；当 时， ，
所以 在 上单调递增，在 上单调递减，
所以 是 的极大值点，
所以 .····································································································15分
②当 时， ，
因为 ， ， ，
所以由零点存在定理，存在 ，使得 ；存在 ，使得 ，
若要使得 在 上存在极大值点，
则 ，
高三数学 第8页（共9页）解得 或 ，
因为 ，所以 .
综上所述 .······················································································17分
解法二、令 ，得 ，
令 ， ，
①当 时， ， ， ， ，易得 在 上
单调递增， ，
（ⅰ）当 时，当 时， ，由极大值点的定义判断，函数 在 上
存在唯一极大值点；
（ⅱ）当 时， ，
，当 时， 恒成立，即
，由极大值点的定义判断，函数 在 上存在唯一极大值点；
所以 .····································································································15分
②当 时， ， ， ， ，易得 在 上
单调递增，
当 时， ，由极大值点的定义判断，只需 即可，即
高三数学 第9页（共9页），解得解得 或 ，因为 ，所以 .
综上所述 .······················································································17分
（①②两种情况中，未区分当 时， ；当 时， 两种情况，对
进行分析，酌情给1-2分）
19.解：（1）由题意得 , ，·······························2分
（ 与 每算对一个给1分）
所以 ， .····························································································4分
（ 与 每算对一个给1分）
（2）由题得 ，
所以 .···················5分
因此 中的数对 必由 中的数对 经运算 得到， 中的数对
必由 中的 或数对 经运算 得到，··························································6分
（能说明“要产生 ，必需 ，要产生 ，则需要 或 给2分）
因为 是 数组，其中有一半的项为0，即 个0，经过两次运算 能在 中产生
个数对 ，·································································································7分
因为 中数对 的个数为 个，经过两次运算 能在 中产生 个数对 ，
高三数学 第10页（共9页）·······················································································································8分
所以 ，即 ，··································································9分
所以 ，（少此步扣1分）
所以数列 是以2为首项，2为公比的等比数列.··········································10分
（3）当 为奇数时， ，累加得
因为 ，所以 （ 为奇数）.···························································11分
当 为偶数时， ，累加得
因为 ，所以 （ 为偶数）.···········································12分
所以 ，
故 . ·······································································13分
（
高三数学 第11页（共9页）或 均得1分）
因 为 当 且 为 偶 数 时 ，
.
······················································································································14分
①当 时， ；···············································································15分
②当 且 为奇数时，
；····················16分
③当 且 为偶数时，因为对任意的 都有 ，所以
;
综上所述，对任意的 都有 .·······························17分
高三数学 第12页（共9页）
