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2026 年高考适应性考试
数学参考答案和评分标准
一、选择题：本题共8小题，每小题5分，共40分．
1．A 2．C 3．B 4．D 5．A 6．B 7．C 8．D
二、选择题：本大题共3小题，每小题6分，共18分．全部选对的得6分，选对但不
全的得部分分，有选错的得0分．
9．ABD 10．ACD 11．ACD
三、填空题：本题共3个小题，每小题5分，共15分．
12． ； 13．−2； 14．28，
四、解答题：本题共5小题，第15题13分，第16、17小题15分，第18、19小题17
分，共77分．解答应写出文字说明、证明过程或演算步骤．
15．（1）∵ ，
由正弦定理得： ，又 ，·····························1分
∴ ，············································································2分
∴ ，·····························································4分
∴ ，又 ，则 ；···················································6分
（2）方法一：在△BCD中，∵BD=CD，可得 ， ，
又∵ ，·······················································································7分
∴ ， ，····································8分
又在△ACD中，由正弦定理得： ，且CD=2AD，
∴ ，·····································································9分
∴ ，则 ，由（1）知： ，则 ，···························11分
数学答案 第1页（共7页）．·······································································13分
∴
方法二：设AD=x，则BD=CD=2x，·························································7分
△ABC中，由余弦定理得： ，
在
①，······································································8分
∴
又 ，············································································9分
，························································10分
平方得：
∴ ②，································································11分
由①消去 得： ，又 ，所以 ．·····································13分
16．解：（1）∵l过F(−c，0)时，△ABF 的周长为8，则 ， ，···········2分
1 2
设 ，则 ，···············································4分
∴ ，····················································································5分
又 ，则 ，·······················································6分
又 ，可得： ， ，···················································7分
∴E的方程为： ；································································8分
（2）已知直线l的方程为： ，设 ， ，················9分
联立 ，消y整理得： ，························10分
则： ， ，··················································11分
数学答案 第2页（共7页）∴ ··········································12分
································································13分
，····································································14分
∴
k 1，−k
，
k
2成等差数列．······························································15分
数学答案 第3页（共7页）17．解：方法一：（1）设 ， ， ，································1分
∵
，·····························3分
∴
，·················5分
又 ，
∴ ， ，····································································7分
又∵ ，
∴EF⊥平面ABC；···············································································8分
（2）∵EF⊥平面ABC，以E为原点，建立如图所示的空间直角坐标系Exyz，
∴ ， ，
， ，
设 ，由 ，
可得 ，
∴ ，········································································10分
易知平面 的一个法向量 ，········································11分
AEC 的法向量为 ，由 ，
1 1
设平面
得 ，可得一个法向量 ，··13分
∵ ，················································14分
∴平面FAE与平面AEC 的夹角的余弦值为 ．····································15分
1 1 1
方法二：（1）连接AB，AC，易知△AAB≌△AAC，·······························1分
1 1 1 1
AA=1，AB=AC=2，∠BAA=∠CAA=60◦，由余弦定理，
1 1 1
数学答案 第4页（共7页）∴∠AAB=∠AAC=90◦，且AB=AC= ，···············································2分
1 1 1 1
由E为BC中点，则BC⊥AE，
1
延长AF交BC 于点G，则AG⊥BC ，则AG⊥BC，AE∩AG=A ，
1 1 1 1 1 1 1 1 1 1
∴BC⊥平面AGE，EF 平面AGE，
1 1
∴BC⊥EF，EF⊥BC ，········································································4分
1 1
在Rt△ABE中，可得AE= ，····························································5分
1 1
在△AEG中，EG=1，AG= ，则AG2=EG2+AE2，
1 1 1 1
∴AE⊥EG，······················································································6分
1
又F为AG上靠近点G的一个三等分点，FG= ，AF= ，
1 1
可得EG2−GF2=AE2−AF2= ，∴EF⊥AG，··············································7分
1 1 1
又AG∩BC =G，则EF⊥平面ABC ，
1 1 1 1 1 1
∴EF⊥平面ABC；···············································································8分
（2）由（1）知GE=AA=1，AE= ，AG= ，
1 1 1
∴GE2+AE2=AG2，则GE⊥AE，····························································10分
1 1 1
又由（1）知BC⊥AE，BC∩GE=E，BC 平面GEC ，GE 平面GEC ，
1 1 1
∴AE⊥平面GEC ，
1 1
又GE 平面GEC ，···········································································12分
1
∴C E⊥AE，······················································································13分
1 1
∴∠GEC 为平面AEF与平面AGE的夹角，·············································14分
1 1 1
在Rt△EGC 中，cos∠GEC = = = ．········································15分
1 1
18．解：（1）若n=4，k=2时，X=0，1，2，··················································1分
；······································································2分
；································································3分
，·······································································4分
故X的数学分布列为：
X 0 1 2
P
（2） ，·············································5分
∴ ，································································6分
∴ ，···························································7分
∴ ，···································································8分
数学答案 第5页（共7页）又 ，且方程 无正整数根，
∴n=6；·····························································································9分
数学答案 第6页（共7页）（3） ，则 ，
由于两次抽取相互独立，且每个球被抽到的概率均为 ，
∴ ， ，
因此 ，··········································11分
，
可得： ，其中 ， ，
又 ，
∵ ，····································································12分
因此 ，····································································13分
∵ 共有 项， ，
代入得： ，·····························14分
∴ ，·················15分
①当n为偶数时， ，D(X)最大，D(X) ；·····························16分
②当n为奇数时， 或 ，D(X)最大，D(X) ．···17分
19．解：（1）∵ ，
由于 ， ，则 ，································1分
令 ，
要证 ， ，只需证： ，···································2分
，易知 ，
数学答案 第7页（共7页）， ，（其中 为函数 的导函数）
，可得 ，（其中 为函数 的
导函数）
∴ 在 上单调递增， ，
在 上单调递减， ，
∴
在 上单调递增， ，··························3分
∴
当 时， ， ；·············································4分
∴
（2）∵ ，且 ，
，···························5分
∴
（i）∵0为 的极小值点，由于 ，
∴必有 ，即 ， ··················································6分
由于 ，
令 ，则 ，····························································7分
存在 ，使得在 与 上满足 ， 单
∴
调递减；在 上 ， 单调递增．··········································8分
∴存在 ，使得在 与 上有 ， 单调递增；在
与 上有 ， 单调递减．···········································9分
∴ 的极大值点为： ，
由于 ，则 ，
在(1，x)单调递增，则 ．·············································10分
2
由于 ，
由（1）得： ，
∴ ，则 ；·············································11分
（ii）∵ 为 的一个极大值点，
数学答案 第8页（共7页），且 ，由于 ，所以 ，
即 (＊)，···································12分
消去a可得， ，
∴ ，······················································13分
令 ，由于 ，
则T(x)在 单调递增，又T(0)=0，但 ，所以 ，则 ，
将 带入(＊)得到 ，······················································15分
下面检验当 时，代入 得到，
此时 ，
易知 ，
又 ，则 为 的极大值点，····················16分
，则 为 的极大值点，且 ，则符合题意；
∴存在 ，使得 ．··············································17分
数学答案 第9页（共7页）
