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青岛市2025年高三年级部分学生调研检测
数学试题
2025.01
本试卷共4页，19题.全卷满分150分，考试用时120分钟，
注意事项：
1.答卷前，考生务必将自己的姓名、考生号等填写在答题卡和试卷指定位置上，并将准
考证号条形码粘贴在答题卡上的指定位置，
2.回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑，如
需要改动，用橡皮擦干净后，再选涂其它答案标号，回答非选择题时，将答案写在答题卡上.写
在本试卷上无效.
3.考试结束后，将本试卷和答题卡一并交回.
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，
只有一项是符合题目要求的.
1.已知样本数据1,3,5,7,9,11,13,15，则该组数据的中位数是
A.6
B.7
C.8
D.9
名若*
=1+i,其中x,y∈R,则|2y+xi=
2i
A.2
B.5
C.3
√万
D
3己知等差数列{am}的前n项和为Sn,a3=2,则S=
A.7
B.8
C.9
D.10
4.已知圆C:(x-a)2+y2=4与圆C2:x2+(y-b)2-1恰有3条公切线，则ab的最大值是
A.9
B.4
C.2
5引设函数f)=coswx，f(x)-f()归2，-x=),则0可以是
B.1
C.2
D.3
6“椭园”C臣”+学=(a>0)是一种优美的封闭由线
1
如图是当m=2a=h=1时C的图象，点2是C与
y轴正半轴的交点，过原点O的直线交C于点A,B,
则QA·QB的取值范围是
A.g
7
c.[0,g
D.[0.
数学试题第1页共4页
2已知a>0,不等式g≥1n(ax)恒成立，a的最大值是
&.2e
B.e
c.ve
D.1
e
&.实系数一元三次方程ax3+bx2+cx+d=0在复数集内有3个根x,x2,x3,则
古++后+名+出，=-.设，西是方程
b
a
x一2x2+x-1=0的3个根，则7+1+1
2+
。2
A.-4
B.-3
C.3
D.4
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，部分选对得部分分，有选错的得0分.
9.已知随机变量X,Y满足Y=2X+1,X~N(2,1),则
A.P(X≤1)=2
B.E(X)=2
C.E(Y)=5
D.D(Y)=3
1Q,已知点P是圆M:(x+3)2+y=16上一动点，点M(3),线段PN的中垂线1交直线
PM于点Q,若点Q的轨迹为曲线C，则
A曲线C的方程为女y
41
B.线段PN中点的轨迹方程为x2+y2=4
1W的最小值为5
D.直线！与曲线C只有一个公共点
11.曲线f(x)=x2-4在点(xn,f(xn)处的切线与x轴的交点横坐标为x+1,x1=3,则
A.m1=+2
B.数列m名。+3为等差数列
2 Xn
-2
C专-25+
52-1
D.数列{xn-2}的前n项和小于2
数学试题第2页共4页青岛市 2025年高三年级部分学生调研检测
数学参考答案及评分标准
一、单项选择题：本题共 8小题，每小题 5分，共 40分．
1--8：CBDA CDBB
二、多项选择题：本题共 3小题，每小题 6分，共 18分．
9．BC 10．ABD 11．ACD
三、填空题：本题共 3个小题，每小题 5分，共 15分．
3
12．5； 13． ； 14．300．
2
四、解答题：本题共 5小题，共 77分．
15．（13分）
8 8.5 9 9.5 10
解：（1）依题意得： x 9 ·············································· 1分
5
y 11.5 10.5 9 7.5 6.5 9 ···································································2分
5
5
(xi x )(yi y)
b i 1 5
(xi x )2
i 1
(8 9)(11.5 9) (8.5 9)(10.5 9) (9 9)(9 9) (9.5 9)(7 .5 9) (10 9)(6.5 9)

(8 9)2 (8.5 9)2 (9 9)2 (9.5 9)2 (10 9)2
6.5
2.6 ······························································································ 5分
2.5
a y b x 9 2.6 9 32.4，所以 y b x a 2.6x 32.4 ···························7分
（2）由题意，销售利润为 f (x) y (x 7) ······················································· 8分
( 2.6x 32.4)(x 7) 2.6x2 50.6x 226.8 ············································11分
当 x 50.6 9.7时， f (x)取最大值，
5.2
所以预测销售价格是9.7万元/吨时，该产品销售利润最大··································· 13分
16．（15分）
解：（1）由b2 c(a c)得b2 ac c2 ························································· 1分
由余弦定理得b2 a2 c2 2accos B，
得 ac c2 a2 c2 2accos B ····································································· 2分
得 c a 2ccosB ······················································································· 3分
由正弦定理得
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sinC sin A 2sinC cosB sin(B C) 2sinC cosB sin(B C) ·················4分
所以C B C或C B C π，
得 B 2C或 B π（舍掉）···········································································5分
所以 B 2C ································································································6分
（2）由（1）可知 B 2C， A π 3C
2 sin A 2 sin 3C
所以 ·······························································7分
cosC sin B cosC sin 2C
2 sin 2C cosC cos 2C sinC

cosC 2sinC cosC
2 2sinC cos2 C cos 2C sinC

cosC 2sinC cosC
2 4cos2 C 1 3
2cosC ······················································10分
cosC 2cosC 2cosC
0 C π
由题意， 0 2C π ，

0 π 3C π
π
得0 C ，···························································································11分
3
1
所以 cosC 1，··················································································· 12分
2
3
所以 2cosC 3 2 2cosC 2 3，····································· 13分
2cosC 2cosC
3
当且仅当 2cosC，即 cosC 3 时，等号成立，······························ 14分
2cosC 2
2 sin A
所以 的最小值是 2 3 ······························································· 15分
cosC sin B
17．（15分）
解:（1）设 B(x1, y1)，点 B,C 关于 x轴对称，所以 BC x轴....................................... 2分
x2 3y2
所以 1 1 ，解得 x1 6p .............................................................................................4分
y
2
1 2 px1
又因为△OBC 的重心坐标为 (4,0)，所以6p 6， p 1 .............................................. 5分
所以 的方程为： y2 2x ..................................................................................................... 6分
（2）设点 B在点C的左侧，显然 BPD CPD，且 PD x轴，所以直线 PB,PC
的倾斜角 1 , 2互补，斜率互为相反数...................................................................................8分
设直线 PB与圆D的切点为H ，则
数学参考答案 第 2 页 共 6 页
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2 5
sin BPD |DH | 5 5 , tan BPD 1 ，
|DP | 2 5 2
设 B(x1 , y1 ),C(x2 , y2 )，
tan 2 k y 2 y 2 2所以 1 ，即 1 1PB 2，解得 y 1, x
1
....... 11分
x 2 1 11 2 y1 2 y1 2 2
2
同理 tan 2 2 92 ，即 kPC 2，解得 y2 3, x2 ...................................12分y2 2 2
1 9
所以 B( , 1),C( , 3)，直线 BC的方程为： 2x 4y 3 0 .................................. 13分2 2
| BC | 4 16 2 5 ,
点 P BC d | 4 8 3 | 3 5到直线 的距离 .................................................................. 14分
20 2
S 1 | BC | d 1 2 5 3 5 15所以 △PBC .............................................................. 15分2 2 2 2
18．（17分）
解：（1）因为 BD AC1， BD AC ， AC1 AC A，
所以 BD 平面 AA1C1C ················································································3分
又因为 BD 平面 ABCD，所以平面 AA1C1C 平面 ABCD ·······························4分
（2）因为 BD 平面 AA1C1C，所以 AA1 BD，
又因为 AA1 BC ， BC BD B，所以 AA1 平面 ABCD ·····························5分
此时四棱柱 ABCD A1B1C1D1是棱长为 2的为正方体，
（法 1）（ⅰ）以 A为原点， AB, AD, AA1所在直线分别为 x轴， y轴， z轴，建立空间直角坐
标系 A xyz，则 A(0,0,0), A1(0,0, 2),O1(0,1,1),P(1 cos ,1 sin ,0) ，
m A A 0
设平面 A1AP的一个法向量为m (x1, y1, z1)，则
1 ，
m AP 0
z 0
所以 1 ，令 x 1 sin ，
(1 cos )x1 (1 sin ) y1 0
1

得m (1 sin , (1 cos ),0) ····································································· 8分

m AO 0
设平面 APO1的一个法向量为 n (x2 , y2 , z2 )，则
1 ，
m AP 0
数学参考答案 第 3 页 共 6 页
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y2 z 0所以 2 ，令 x2 1 sin ，
(1 cos )x2 (1 sin ) y2 0
得 n (1 sin , (1 cos ),1 cos ) ···························································· 9分
设二面角 A1 AP O1的大小为
2
cos (1 sin ) (1 cos )
2
则
(1 sin )2 (1 cos )2 (1 sin )2 2(1 cos )2
(1 sin )2 (1 cos )2

(1 sin )2 2(1 cos )2
1 (1 cos )
2

(1 sin )2 2(1 cos )2
1 1 2
(1 sin )2
2 2
2
(1 cos )
当且当仅当 sin 1时取，此时点 P为 AB中点·············································13分
π
所以二面角 A1 AP O1的最大值为 ··························································· 14分4
（ⅱ）又因为，点 P,P1,P2均在一个与正方体各棱都相切的球W 上，球W 直径为 2 2
当 P1,P2恰分别为 A1D1,CC1中点时，
则W (1,1,1),P(1,0,0),P1(0,1, 2),P2(2, 2,1) ，显然WP (WP1 WP2)，
所以，W ,P,P1,P2四点共面·········································································· 16分
所以此时△PP1P2外接圆直径的最大，最大值恰为球W 直径 2 2 ························ 17分
（法 2）（ⅰ）以 A为原点， AB, AD, AA1所在直线分别为 x轴， y轴， z轴，建立空间直角坐
标系 A xyz，设直线 AP与直线 BC交点为K，
则 A(0,0,0), A1(0,0, 2),O1(0,1,1),K (2, t,0)(t 0) ，
m A A 0
设平面 A 11AP的一个法向量为m (x1, y1, z1)，则 ，
m AK 0
z1 0所以 ，令 x1 t，

2x1 ty1 0
得m (t, 2,0) ··························································································11分

n AO 0
设平面 APO1的一个法向量为 n (x2 , y2 , z2 )，则
1 ，
n AK 0
y
所以 2
z2 0
，令 x2 t，
2x2 ty2 0
数学参考答案 第 4 页 共 6 页
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得 n (t, 2,2) ···························································································12分
设二面角 A1 AP O1的大小为 ，
2 2
则 cos t 4 t 4 4 2 2 1 2 ，t2 8 t2 4 t 8 t 8 2
等号当仅当 t 0时取，此时点 P为 AB中点···················································· 13分
π
所以二面角 A1 AP O1的最大值为 ··························································· 14分4
（ⅱ）因为点 P,P1,P2均在一个与正方体各棱都相切的球W 上，球W 直径为 2 2 ····15分
当 P1,P2恰分别为 A1D1,CC1中点时，
则W (1,1,1),P(1,0,0),P1(0,1, 2),P2(2,0,1) ，显然WP (WP1 WP2)，
所以，W ,P,P1,P2四点共面·········································································· 16分
所以此时 PP1P2外接圆直径的最大，最大值恰为球W 直径 2 2 ························· 17分
（法 3）（ⅰ）记 A1在平面 APO1上的正投影为H， A1在直线 AP上的投影为G，连结GH ，
则 A1G AP, A1H AP, A1G A1H A1，所以 AP 平面 A1GH ，
所以 AP GH ，即 A1GH为二面角的平面角，记 A1GH ························12分
AH AO 2
显然 sin 1 1 1 ，等号当且仅当H 与O1重合，G 与 A重合时取，A1G A1A 2
此时点 P为 AB中点····················································································13分
π
所以二面角 A1 AP O1的最大值为 ··························································· 14分4
（ⅱ）因为点 P,P1,P2均在一个与正方体各棱都相切的球W 上，球W 直径为 2 2 ····15分
当 P1,P2恰分别为 A1D1,CC1中点时，
以 A为原点，AB, AD, AA1所在直线分别为 x轴，y轴，z轴，建立空间直角坐标系 A xyz，
则W (1,1,1),P(1,0,0),P1(0,1, 2),P2(2,0,1) ，显然WP (WP1 WP2)，
所以，W ,P,P1,P2四点共面·········································································· 16分
所以此时 PP1P2外接圆直径的最大，最大值恰为球W 直径 2 2 ························· 17分
19．（17分）
（1）因为G ai (x) 2 3x x
3 ·······································································1分
3
所以GT a (x) G a (x 1) 2 3(x 1) (x 1)
3 6 6x 3x2 x3 ,
1 i 3 i 3
因为6,6,3是3的倍数，6不是32的倍数，1不是3的倍数，
所以T1 ai 3是3 不可分的············································································· 3分
（2）由(1)知，
T1 ai 3 {6,6,3,1}是3 不可分的，所以T1 ai 3 {6,6,3,1}不是可约的·······················4分
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若 ai 3 2,3,0,1 是可约的，则存在 bi m与 ci k 使得G a (x) G b (x)Gi 3 i m c (x) ······ 6分i k
则GT {a } (x) G a (x 1) G b (x 1)G c (x 1) G1 i 3 i 3 i m i k T1 b (x)GT c (x)，i m 1 i k
此与T1 ai 3 {6,6,3,1}不是可约的矛盾，因此 ai 3 2,3,0,1 不是可约的·················· 9分
（3）对于 ai n 121,11,0, ,0,3 ，若存在 t Z，以及质数 p，使得Tt ai n是 p 不
可分的，由
G ai (x) 121 11x 3x
n ,GT a (x) 121 11(x t) 3(x t)
n
n 1 i n
n 2
121 11t 3t n (3nt n 1 11)x 3 C i t ixn i 3xnn ········································ 11分
i 1
则有3不是 p的倍数；C i int 以及3nt
n 1 11和3t n 11t 121是 p的倍数；
但3t n 11t 121不是 p2的倍数·····································································13分
分两种情况讨论：
1)若 t是 p的倍数，则由3t n 11t 121是 p的倍数，有 p 11，此与3t n 11t 121不
是 p2的倍数矛盾·························································································15分
2)若 t不是 p的倍数，则由C1nt
1是 p的倍数，有 n是 p的倍数；再由3nt n 1 11是 p的
倍数，仍然有 p 11，此与3t n 11t 121是 p的倍数矛盾，
所以原结论成立·························································································· 17分
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