2025届高三年级5月份联合考试数学试卷(pdf版,含详解)

点石联考 5月高三联考 数学
2025 届高三年级 5月份联合考试 数学
说明：
一、本解答给出的解法仅供参考，如果考生的解法与本解答不同，可根据试题的主要考查内
容比照评分标准制订相应的评分细则.
二、对解答题，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；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、单选题（本大题共 8 小题，每小题 5 分，共 40 分。在每小题所给的四个选项中，有且只有一项是符合
题目要求的）
1 2 3 4 5 6 7 8
D B C D C B B B
二、多选题（本大题共 3 小题，每小题 6 分，共 18 分。在每小题所给的四个选项中，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 10 11
AD BCD ABD
三、填空题（本大题共 3 小题，每小题 5 分，共 15 分，第 14 题第 1个空 2 分，第 2 个空 3 分）
12． 4 13． 2, 14．294 ，12
四、解答题（本大题共 5 小题，共 77 分。解答时应写出必要的文字说明、证明过程或演算步骤）
15．（13分）解：
（1）由 3b 2csin B及正弦定理，得 3 sin B 2sinC sin B．················································· 3分
而0 B π，所以 sin B 0,sinC 3 ．··············································································4分
2
π
又VABC
π
为锐角三角形，所以 0 C ，所以C .······························································5分
2 3
a b c 2 3
4
（2）由正弦定理 sin A sin B sinC 3 ，································································· 7分
2
得 a 4sin A,b 4sin B．··································································································8分
因为 sin Asin B
5
，所以 ab 10．··················································································· 10分
8
由余弦定理得 c2 a2 b2 2ab cosC ，·················································································· 11分
12 a2 b2 2 10 1即 ，······························································································ 12分
2
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点石联考 5月高三联考 数学
所以 a2 b2 22 .··········································································································· 13分
16．（15分）解：
（1）证明：由已知可得 ⊥ ,
即 ⊥ , ⊥ ,
又 AE∩PE=E，
故 ⊥ 平面 , ···········································································································2分
又 PA 平面 PAE，所以 ⊥ ,
又由数量关系知 2 + 2 = 2,
即 ⊥ , ·················································································································· 4分
又 AE∩CE=E，所以 ⊥平面 ,
而 平面 ,
所以 ⊥ .················································································································· 6分
（2）解：根据（1）中位置关系，以 为坐标原点， ， ， 的方向分别为 x轴、y轴、z轴正方向，建
立如图所示的空间直角坐标系. ························································································· 7分
由已知, 可知 0,0,0 , 2, 2,0 , 0,1,0 , 2,2,0 , 0,0,2 3 ,
所以 = 2, 2,0 , = 0,0,2 3 , = 2,1,0 , = 2,2, 2 3 . ··································· 9分
设平面 的法向量为 1 = 1, 1, 1 ,
1 = 0则
1
,
= 0
2 1 2 1 = 0所以 = 0 , 取 1 = 1,1,0 ; ················································································ 11分1
设平面 的法向量为 2 = 2, 2, 2 ,
2 = 0则
2
,
= 0
2 2 + 2 = 0
所以 , 取 = 3, 2 3, 1 . ···························································13分
2 2 + 2 2 3
2
2 2 = 0
3 6
因此, cos 1, 2 = = ,2 3+12+1 8
2
所以平面 与平面 6 58夹角的正弦值即为 1 = .················································ 15分
8 8
17.（15分）解：
（1）设事件 ={成品为合格品}， ={成品为不合格品}，
则 ( ) = 0.9 × 0.9 × 0.9 = 0.729，
所以 ( ) = 1 0.729 = 0.271. ······················································································· 4分
（2）设预期收益为 ,分 2种情况：
①成品是合格的,
收益为 1 = 56 4 18 6 = 28，
概率为 ( = 1) = 0.729；······························································································ 6分
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点石联考 5月高三联考 数学
②成品是不合格的,
收益为 2 = 4 18 6 6 = 34，
概率为 ( = 2) = 0.271. ······························································································ 8分
于是预期收益 的数学期望为
( ) = 28 × 0.729 34 × 0.271 = 11.198 (元). ··································································· 9分
（3）设预期收益为 Y，分 5种情况：
①所有检测都是合格的,
收益为 1=56 6 21 9=20；
概率为 ( = 1) = 0.9 × 0.9 × 0.9 = 0.729；····································································· 10分
②零配件 1,2都是合格的, 但是成品不合格,
收益为 2 = 6 21 9 = 36，
概率为 ( = 2) = 0.9 × 0.9 × 0.1 = 0.081；····································································· 11分
③零配件 1不合格, 零配件 2合格, 扔掉零配件 1, 没有成品,
收益为 3 = 6，
概率为 ( = 3) = 0.1 × 0.9 = 0.09； ··············································································12分
④零配件 2不合格,零配件 1合格, 扔掉零配件 2, 没有成品,
收益为 4 = 21，
概率为 ( = 4) = 0.9 × 0.1 = 0.09；·············································································· 13分
⑤零配件 1,2都不合格, 全部扔掉, 没有成品,
收益为 5 = 6 21 = 27，
概率为 ( = 5) = 0.1 × 0.1 = 0.01. ················································································14分
于是预期收益 Y的数学期望为
( ) = 20 × 0.729 36 × 0.081 6 × 0.09 21 × 0.09 27 × 0.01 = 8.964(元). ····················· 15分
18．（17分）解：
（1）设焦距为 2 ,
2
则 = ,2 = 2 2, 且 2 = 2 + 2，
2
解得 = 2, = 1，
2
因此Γ的标准方程为 + 2 = 1. ························································································ 4分
2
（2）在△ , △ 中分别由正弦定理,
= , 得 = ,
sin ∠ sin∠ sin∠ sin ∠
又由 = , 及sin ∠ = sin ∠ ,
得sin ∠ = sin ∠ ,
故∠ = ∠ （互补的情形舍去）. ··············································································7分
又直线 , 的斜率分别为 1, 2,
则两直线的方程分别为 = + 1( = 1,2)，
则 到这两条直线的距离相等,
1+1 = 2+1
2
, +2 +1
2+2 +1
即 平方得 1 12 = 2 2 ,
1+ 2 1+ 2 1+ 1 1+
2
2
1 2
化简得 1 1 2 1 2 = 0, 又 1≠ 2，
则 1 2 = 1. ··············································································································· 10分
（3）证明：设直线 : = + , 1, 1 , 2, 2 ,
= +
与椭圆方程联立得 2 + 2 2 2 = 0,
消去 ，得 2 + 2 + 2 2 = 0,
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点石联考 5月高三联考 数学
即 1 + 2 2 2 + 4 + 2 2 2 = 0, > 0,
4 2 2 2
则 1 + 2 = 2 , 1 2 = 2 , ·············································································· 13分1+2 1+2
= 1 1 2 1 = 1+ 1 2+ 1
2
= 1 2+ 1 1+ 2 + 1
2
因此 1 2 1 2 1 2 1 2
2 2
2 2+ 1 4 2
= 1+2 2 1+2 2
+ 1
= 1,
2 2 2
1+2 2
化简得 m2+2m-3=0，由 ≠ 1, 解得 = 3. ···································································16分
因此直线 过定点 0, 3 . ·························································································· 17分
19. (17分) 解：
1 1
（1）证明：由题意知, =
1+e
+ 1 ,
4
则 ′

= e 1 = 1 1 ≤ 1 1 2 = 0, ··············································2分1+e 4 e +e +2 4 2 e e +2 4
因此 在 上单调递减.
={x|f(x)>f(p)}={x|x ={x|f(x)>f(q)}={x|x又 p
（2）证明：不难发现, + = e 3 1 + + 1 + + + + 1 3 = 2 + 1, ·· 6分1+e 1+e
取 = 2 + 1,下证 与 2 +1 互为对偶集,
∈ , > , 则 2 + 1 > ,
即 < 2 + 1 ,
由定义知 ∈ 2 +1 .··································································································· 8分
同理可知, ∈ 2 +1 , ∈ .
故当 = 2 + 1 时, 与 互为对偶集.············································································10分
（3 1）构造函数 = ,
2
则 1 = 0, + ∞ ，即 > 0 的充分必要条件为 ∈ 0, + ∞ ,
2
若 0 < 0, 又 1 > 0, 由于 的图象在 0,1 上连续不断,
故 0 ∈ 0,1 , 使得 0 = 0,
则与 1 = 0, + ∞ 矛盾,
2
因此 0 = 0, 代入解得 = 0.
3
3 3
< 0, 1 = 1 + 1 = 1若 则 3 3 1 < 0, 不符合题意, 舍去. ······················12分 1 11+e 1+e

若 ≥ 0, 此时 ′ = e 2 + 3 2 > 0在 上恒成立, ······················································· 14分1+e
此时 在 0, + ∞ 上单调递增, 则 g(x)在（0,+∞）上单调递增，
又 g(0)=0，所以 g(x)>0在 0, + ∞ 上恒成立，符合题意. ······················································· 16分
综上， 的取值范围为[0, + ∞ . ······················································································· 17分
— 4—
点石联考 5月高三联考 数学
一、单选题
1. D【解析】已知 (1 + i) = 2i，解得 = 1 + i，计算得 2 = 1+ i 2 = 1 + 2i + i2 = 2i.
2.B【解析】因为 A x x 2 2 x 0 x 4 ，故 A∩Z 1,2,3 .
3.C【解析】因为 19位同学的积分的中位数是第 10名，所以知道中位数即可判断是否在前 10.
2 tan 8 1
4.D【解析】由题意可得 tan 2 ，解得 tan 或 tan 4（舍去）.
1 tan 2 15 4
cos 3sin 1 3tan 1
方法一：故 .
sin 5cos tan 5 21
cos 4sin cos 3sin sin 1方法二：此时 ，故 .
sin 5cos sin 5 4sin 21
5.C【解析】依题意，圆C2的方程等价于 (x a)
2 (y a 2)2 2a2，所以圆C2是以 (a,2 a)为圆心， 2a
为半径的圆.注意到C1是以 (1,1)为圆心， 2 为半径的圆，所以C1，C2的圆心距为
(a 1)2 (a 1)2 2(1 a)，也即圆心距为两半径之和，因此两圆相外切，两相外切的圆有 3条公切线.
6. B【解析】由 = sin + 的最大值为 3，得 = 2，即 = sin + 2. 与 图象的交点
由方程 2 2 + 4 = sin + 2 的根的数量决定，即 2 2 + 2 = sin 在区间 0,2π 上恰有一个根.
令 = 2 2 + 2，其图象是开口向上的抛物线，以直线 x=1 为对称轴，当 x=1 时， ( )min =1，令
= sin ，要使 与 图象只有一个交点，则 的最小值点与 最大值点重合，即 sin k=1，
π
则 = +2nπ，n Z = π∈ . 符合题意.
2 2
1 1
7.B【解析】由题意可得 Sn 2an ，当 n 2时， Sn 1 2an 1 ，两式相减得 an 2a4095 4095 n 1
，
a S 2a 1 a 1 a a 1而 1 1 1 ，解得 1 n 1，因此数列 n 是等比数列， n 2 ，数列 a4095 4095 4095 n 是递增的正项
211 212
数列， a 1,a 1，因此T T T T12 4095 13 4095 1 2 12 13
，所以当Tn取得最小值时， n 12 .
8.B【解析】设正三棱柱的底面边长为 a，侧面高度为 h，则其表面积 S 2 3 a 2 3ah 6 3 ，整理得
4
2 3
h 12 a 3 2 12a a ， 故 正三 棱 柱 的体 积 V a h ， 将 V 看 作 关于 a 的 函 数 V (a) ， 则
2 3a 4 8
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点石联考 5月高三联考 数学
V '(a) 3 3 a 2 ，当 a 2时，V '(a) 0，V (a)单调递增；当 a 2时，V '(a) 0，V (a)单调递减，
2 8
2 3
故当 a 2时，该正三棱柱的体积取得最大值，此时三棱柱上下两底面之间的距离为 .
3
二、多选题
2
9.AD 2 2 2 2【解析】对于A，a b = 1 ( 2) 3 1 1，故A正确；对于B，a 1 3 10, b ( 2) 1 5 ，
2
a b 5 1 4故 ，故 B错误；对于 C， a b ( 1,4)，由于 ，故 C错误；对于 D，
2 1

cos a,b a b 1 2 ，故 D正确.
a b 5 2 10
A(110.BCD 2【解析】将 ,1)的坐标代入C : y 2px(p 0) ，得 p 1，故 A错误；因为 F 是C的焦点，
2
p
所以 F ( ,0)，即 F (1 ,0)，点 A F 2， 横坐标相同，所以 AF y轴，故 B正确；若 y0 1，由于 y0 2x2 2 0
，
1 1 1
所以 2x0 1， x0 ，因为点 A，B均在C上，所以 | AF | 1， | BF | x
1
0 ，则2 2 2 2
| AF | | BF | = 1 x＜0， | AF |＜ | BF |，故 C正确；若 | y | 2x ，则 x 0，所以原条件等价于
2 0 所以 0 0 0
y 20 4x
2 2x 4x 20 ，故 0 0 ，解得 x
1
0 ，由 C知 | AF |＜ | BF |，故 D正确.2
11.ABD【解析】对 A,令 = + 1 + 2，即 = + 1 3 3 + 1 2 + 2 = 3 3 ,则 为奇函数，
对称中心为（0,0），而 的图象可由 的图象向右平移一个单位，向下平移两个单位得到，因此可知
图象的对称中心为（1，-2），A正确;
对 B, 当 a=1 时， = 2 4 + 2 2 4，因为 (2 + ) = 4 + 2 4, g(2 ) = 4 + 2 4,即
(2 + ) = (2 ). 可知 的图象以直线 = 2 为对称轴，B正确;
对 C,当 a=1时, = ( 2)4 + 2 4 , (2) = 4，g(1)=53， 4 = 16, 1所以存在 x1∈( , 2),x2∈(2,4),使2 16 2
g(x1)=g(x2)=0,则 g(x)有两个零点,故 C不正确;
对 D,当 = 0 时，令 = 3 4 2 + 4 = 2 4 + 4 ,令 = 0，解得 x=0或 x=2,即 D正确.
三、填空题
2
2 x
12. 4【解析】因为曲线mx2 y2 1 y 1为双曲线，所以m 0，将双曲线方程化为标准形式为 1 ，所
m
b 1 y a以 a 1， ，所以双曲线的渐近线方程为 x mx，
m b
又因为双曲线的一条渐近线的斜率为 2，所以 m 2，解得m 4 .
— 6—
点石联考 5月高三联考 数学
13. 2, 【解析】因为 a,b 1 1 1 1为正实数，所以 a b b ab 2 2 ab 2 4 ， a ab ab

因此 a
1 1
b 1 1 1 1 的最小值为 4，故存在 ab ，即 ab 1时使得等号成立，此时b ，又因为 m，
b a ab a a b
1 1
所以 a + = m在 0, 上有解，所以由基本不等式可知 a 2,a 1时等号成立，所以m≥ 2，故m的取
a a
值范围是 2, .
14. 294 ，12【解析】当第二轮游戏完成时，丁的罐子里的米粒数为： 7 8 9 10 29 30 36 294 .
将正整数按照以下规律排成数阵：第一行：1,第二行：2,3,第三行：4,5,6，第四行：7,8,9,10，第五行：
n n 1
12,13,14,15, …,设数列 an ： an n，则数阵第 n行的最后一个数为： Sn a1 a2 a3 an .由2
n n 1 S 44 45 1000，解得 n 44，且 n 990 ,所以1000是第 45行的第 10个数，因为 45 4 11 1，2 2
所以当游戏完成时，是进行到第 12轮.
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