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2026 届高中毕业班适应性练习（四月）
数学参考答案及评分细则
评分说明：
1．本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，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。
2．对计算题，当考生的解答在某一步出现错误时，如果后继部分的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。
3．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。
4．只给整数分数。选择题和填空题不给中间分。
一、单项选择题
题号 1 2 3 4 5 6 7 8
答案 A C D C C B C D
二、多项选择题
题号 9 10 11
答案 BCD AD ACD
三、填空题
12．60° 13． 14．
四、解答题
15.本小题主要考查函数的奇偶性、函数的零点、三角恒等变换、等差数列求和等基础知识，考查运算求
解能力、逻辑推理能力等，考查函数与方程思想、分类与整合思想等，考查逻辑推理、数学运算等
核心素养，体现基础性.满分13分.
解法一：（1）因为 为奇函数，所以 ，··························································1分
即 恒成立.
得 恒成立，···············································································2分
所以 恒成立，···············································3分
所以 恒成立，····························································································4分
所以 ，···········································································································5分
数学参考答案及评分细则 第1页（共 16页）解得 .········································································································6分
（2）因为 ，所以 ，
令 ，则 ，·················································································8分
所以 或 ，·················································10分
解得 或 ，································································11分
令 ， ，则 ，
所以 ，··································12分
所以 ．·······················································13分
解法二：（1）因为 为 上的奇函数，所以 ，························································2分
所以 ，···········································································································3分
解得 ， ·····································································································4分
经检验， 是奇函数，
所以 .········································································································6分
（2）因为 ，所以 ，····································································7分
令 ，则 ，···················································································9分
数学参考答案及评分细则 第2页（共 16页）所以 ，·····························································································10分
所以 或 ，
解得 或 或 ，·····································11分
令 ， ， ，
则 ，
所以 ，
所以 ．··························································13分
解法三：（1）同解法一．···································································································6分
（2）因为 ，所以 ，因为 ，
所以 是 的一个周期，··························································································7分
当 时，令 ，则 ，···························································9分
解得 ，································································································10分
所以 在区间 的零点之和为 ．···············································11分
令 ，
则 是以 为首项， 为公差的等差数列，································································12分
数学参考答案及评分细则 第3页（共 16页）所以
．······································································13分
16.本小题主要考查导数的几何意义、导数的应用等基础知识，考查逻辑推理能力、运算求解能力等，考
查函数与方程思想、化归与转化思想、分类与整合思想等，考查逻辑推理、数学运算等核心素养，
体 现基础性．满分15分．
解法一：（1）函数 的定义域为 ， ．··················································2分
当 时，因为 ，所以 ，····························································3分
又 ，········································································································4分
所以曲线 在点 处的切线方程为 ，
即 ．·································································································7分
（2）(i)当 时， 不符合题意，舍去；························9分
(ii)当 时， 显然成立；···································································11分
(iii)当 时，令 ，得 ，令 ，得 ；
所以 在 单调递减，在 单调递增．····················································13分
所以 ，解得 ．·················································14分
综上所述， 的取值范围为 ．················································································15分
解法二：（1）同解法一．····································································································7分
（2）由已知，得 ．
(i)当 时，可得 ．··············································································8分
因为 ，所以 ，··················································································9分
数学参考答案及评分细则 第4页（共 16页）又因为 时， ，
所以 ；·········································································································10分
(ii)当 时， 恒成立，所以 ；·······················································11分
(iii)当 时，可得 ．
令 ， ，·············································12分
当 时， ， 单调递减；
当 时， ， 单调递增；····································································13分
所以 ，所以 ．·········································································14分
综上所述， 的取值范围为 ．············································································15分
17.本小题主要考查椭圆的定义、直线与椭圆的位置关系、三点共线等基础知识，考查逻辑推理能力、直
观想象能力、运算求解能力等，考查函数与方程思想、数形结合思想、分类与整合思想、转化与化
归思想等，考查逻辑推理、直观想象、数学运算等核心素养，体现基础性与综合性．满分15分．
解法一：（1）当 轴时， ，
所以 ，·································································1分
所以 ，············································································3分
从而 ， ，·······························································································5分
故 的方程为 ．······················································································6分
（2）设 ， ， ，··························································7分
则 ，即 ．·······································································8分
又 ，
数学参考答案及评分细则 第5页（共 16页）所以 ， ， ， . ······························10分
因为 ， ，
所以 ， ，··································12分
两式相加、减，得 ， ，·····························································13分
又因为 ， ，
，··········14分
所以 ，故 三点共线. ···········································································15分
P y
M
Q
F1 O F2 x
解法二：（1）当 轴时， ，
所以 或 ，························································································1分
所以 ①，·······························································································2分
又 ②，···································································································4分
由①②，解得 ， ，··················································································5分
故 的方程为 ．······················································································6分
（2）设 ，则 ，即 ．···········································7分
数学参考答案及评分细则 第6页（共 16页）(i)当直线 ， 斜率均存在时， ， ，
所以直线 ， ，····················································9分
由 得 ，······································································10分
由 得 ， ····································································11分
所以 ， ，
3
6x 2 8(3 x 2)24
3(x 1) 3(x 1) 6x 2 8y 2 24 0 4 0
(x 4)[y 0 ](x 4)[y 0 ] 0 0 0
0 0 y 0 0 y y y
因为 0 0 0 0 ，
所以 ，故 三点共线. ·······································································12分
(ii)当直线 或 斜率不存在时，根据对称性，不妨设 斜率不存在，且 ，
此时点 ， ， ，故直线 ，从而 ， 则 ，
，
所以 三点共线.···························································································14分
综上， 三点共线. ······················································································15分
18.本小题主要考查随机变量的分布列、数学期望、条件概率与全概率公式等基础知识，考查数学建模能
力、运算求解能力等，考查分类与整合思想、概率与统计思想等，考查数学运算、逻辑推理、数据
分 析、数学建模等核心素养等.体现基础性，应用性.满分17分.
解：（1）由题可知， ，··························································1分
化简可得 ， ····················································································2分
当 时， ，
数学参考答案及评分细则 第7页（共 16页）则 ，
即顾客一次性购买文创盲盒数量的平均值为 .···························································4分
（2）（i）设事件 “一次性购买 个文创盲盒”（ ），事件 “顾客为幸运客户”，
·······················································································································5分
则 ， ， ， .
依题意，得 ， ，······································································6分
因为每个盲盒是否为封面款相互独立，
所以 ， ，···········································8分
又由题意知， ，且 两两互斥，···························9分
所以 ，·············11分
由（1）得， ，代入化简可得 ，
所以 ， .······································································12分
（ii）设事件 “一次性购买的文创盲盒全部是封面款”，
依题意，得 ，·········································································13分
且 ， 两两互斥，
所以 ，··············································14分
由（i）得， ，
所以幸运客户中，一次性购买的文创盲盒全部是封面款的概率为
数学参考答案及评分细则 第8页（共 16页），········································································16分
由题意 ，可得 ，解得 ，
又因为 ，所以 .··············································································17分
19.本小题主要考查空间点、线、面位置关系，直线与平面所成角，二面角，平面轨迹方程等基础知识；考查运算求解能力
17分．
解法一：（1） 因为 平面 ， ,所以 ， ．··································1分
不妨设 ，且 ，
因为 ，所以 ， ， ，
所以 ，所以 为△ 的最大内角．···················································2分
由余弦定理，得 ，···········································3分
所以 ，所以△ 是锐角三角形．···························································4分
（2）（i）因为 ， 在 上，且 ，
由对称性知 在同一个轨迹上，且轨迹关于 对称，
故以 为原点， 分别为 轴和 轴的正方向建立如图所示的空间直角坐标系 ．
z
P
y
Q
D
x
A C
γ B
设 ， ，因为 ，所以 ．
因为 是线段 上靠近 的三等分点，
故 ，即 ， ·································································5分
故 ， ，
数学参考答案及评分细则 第9页（共 16页），
依题意得 ，化简得 ，··················································6分
且 ，即 ，故 ，又点 不在直线 上，故 ，
同理， ，且 ，·················································································7分
故在坐标平面 中， 是双曲线 右支上的动点，且 在 轴的两侧，如图．
y
D
A x
B
因为 的两条渐近线分别为 和 ，它们的夹角为 ，
所以 ．·······························································································8分
因为平面 平面 ， ， ，
所以 是二面角 的平面角，所以二面角 为锐角．··························9分
（ii）因为△ 不是任何一个长方体的截面，所以△ 是直角三角形或钝角三角形．·······10分
证明如下：
若△ 为锐角三角形，有 ， ， ，
可令 ， ， ，
则存在以 为共点棱的长方体，△ 为该长方体的截面．
由（1）知，若△ 是长方体的截面，则△ 是锐角三角形，
所以△ 不是任何一个长方体的截面等价于△ 是直角三角形或钝角三角形．···············11分
由（i）知， ，所以 ，又因为 ， ，
所以 ，故 ．····························12分
因为 ，所以 分别是直线 与 所成的角， 即 ，
数学参考答案及评分细则 第10页（共 16页）不妨设 ，则 ，且 ，所以 ， ，····················13分
且 ．
作 于 ，因为平面 ，平面 ， 平面 ，
所以 ，又 ，所以 ．
因为 是线段 上靠近 的三等分点，所以 是线段 上靠近 的三等分点，
所以 ，即直线 过 ，····································································14分
所以 ，所以 ，··············15分
这样，问题等价于在平面直角坐标系 中， 在双曲线 的右支上，直线
过点 ， ， ，求 的最小值．
y
D
M
A x
B
如图，不妨设点 在第四象限，则 ， ．因为 都在双曲线的右支，故
，
即 ，所以 ，又 ， ，
故 解得 即 ，·····································16分
所以 ，
当 ，即 时，等号成立．
数学参考答案及评分细则 第11页（共 16页）故 的最小值为 ．··················································································17分
解法二：（1）因为 平面 ， ，所以 ， ．·······································1分
又因为 ，故可以 为原点， 分别为 轴， 轴和 轴的正方向，建立如图所示的
空间直角坐标系 ．·························································································2分
z
P
Q
y
D
A C
x
γ B
设 ，所以 ，在 中，
，所以 为锐角，
，所以 为锐角，
，所以 为锐角，
所以 是锐角三角形．······················································································4分
（2）（i）同解法一．·····························································································9分
（ii）因为△ 不是任何一个长方体的截面，所以△ 是直角三角形或钝角三角形．·······10分
证明如下：
若△ 为锐角三角形，有 ， ， ，
可令 ， ， ，
则存在以 为共点棱的长方体，△ 为该长方体的截面．
由（1）知，若△ 是长方体的截面，则△ 是锐角三角形，
所以△ 不是任何一个长方体的截面等价于△ 是直角三角形或钝角三角形．···············11分
数学参考答案及评分细则 第12页（共 16页）z
P
y
Q
D
M x
A C
γ B
作 于 ，因为平面 ，平面 ， 平面 ，
所以 ，又 ，所以 ．
因为 是线段 上靠近 的三等分点，所以 是线段 上靠近 的三等分点，
所以 ，即直线 过 ．····································································12分
在平面直角坐标系 中，设直线 的方程为 ，
联立 得 ，
依题意，有 且
因为 ，所以 ．
因为 ，
所以
，·····························································13分
，
同理 ，
不妨设 ，则必有 ．
因为 ，
因为 且 ，所以 ，代入上式得到
数学参考答案及评分细则 第13页（共 16页）····························································14分
，
所以 ，
又因为 ，所以 ．······································································15分
因为 ，所以 分别是直线 与 所成的角，即 ，
因为 ，所以 ，所以 ，所以 ，·······································16分
，
当 ，即 时，等号成立．
故 的最小值为 ．··················································································17分
解法三：（1）因为 平面 ， ，所以 ， ．·································1分
又因为 ，所以在 中，
，所以 为锐角，···········································2分
，所以 为锐角，··········································3分
，所以 为锐角，
所以 是锐角三角形．······················································································4分
（2）（i）同解法一．·····························································································9分
（ii）因为△ 不是任何一个长方体的截面，所以△ 是直角三角形或钝角三角形．·······10分
证明如下：
数学参考答案及评分细则 第14页（共 16页）若△ 为锐角三角形，有 ， ， ，
可令 ， ， ，
则存在以 为共点棱的长方体，△ 为该长方体的截面．
由（1）知，若△ 是长方体的截面，则△ 是锐角三角形，
所以△ 不是任何一个长方体的截面等价于△ 是直角三角形或钝角三角形．···············11分
由（i）知， ，所以 ，又因为 ， ，
所以 ，故 ．····························12分
因为 ，所以 分别是直线 与 所成的角，即 ，
不妨设 ，则 ，且 ，所以 ， ，···················13分
且 ．
z
P
y
Q
D
M x
A C
γ B
作 于 ，因为平面 ，平面 ， 平面 ，
所以 ，又 ，所以 ．
因为 是线段 上靠近 的三等分点，所以 是线段 上靠近 的三等分点，
所以 ，即直线 过 ，····································································14分
所以 ，所以 ．································15分
这样，问题等价于在平面直角坐标系 中， 在双曲线 的右支上，直线
过点 ， ，求 的最小值．
数学参考答案及评分细则 第15页（共 16页）y
D
0
A N M
x
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0
如图，不妨设点 在第四象限，因为 ，所以点 在以 为直径的圆 内（含边界），
记
圆 与双曲线在第四象限的交点为 ，则 ．
因为 在渐近线 的上方，故 ，而 ，故 ，即直线 与双曲
线右支有两个交点 ，符合条件．所以当点 位于点 时， 最大，则 最小．·····16分
联立 ，得 ，解得 或 （舍去），
故当 ，即 时， 的最小值为 ．
故 的最小值为 ．·······················································································17分
数学参考答案及评分细则 第16页（共 16页）
