山东省青岛市2025年高三年级第三次适应性检测数学答案_2025年5月_250529山东省青岛市2025年高三年级第三次适应性检测（青岛三模）（全科）

2025 年高三年级第三次适应性检测
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：DABC CACA
二、多项选择题：本题共3小题，每小题6分，共18分．
9．BD 10．ACD 11．BCD
三、填空题：本题共3个小题，每小题5分，共15分．
5 7 3
12． ； 13． ； 14．
5100
8 4
四、解答题：本题共5小题，共77分．解答应写出文字说明，证明过程或演算步骤．
15．（13分）
解：(1)销售额不少于60万元的有3天，X 的所有可能取值为1,2,3························1分
C1 3 C1C2 3 C3 1
P(X 1) 3  ,P(X 2) 2 3  ,P(X 3) 3  ······················ 4分
C3 10 C3 5 C3 10
5 5 5
所以X 的分布列为：
X 1 2 3
3 3 1
P
10 5 10
3 3 1 9
所以E(X)1 2 3  ································································6分
10 5 10 5
1 1
(2)x  （24568）5， y  （2030606080）50，·······················7分
5 5
5
x y 2204305606608801460··········································8分
i i
i1
5
x2 5x2 41625366452514512520····································9分
i
i1
14605550
b ˆ 10.5，aˆ  y b ˆ x 5010.552.5······························ 11分
20
所以经验回归方程为yˆ 10.5x2.5，··························································· 12分
当x10时，yˆ 102.5
答：所求方程为yˆ 10.5x2.5，预测销售额为102.5万元································· 13分
16．（15分）
ur 1
解：（1）由直线过原点且方向向量为(3,1)可知l的方程为 y  x·················· 1分
3
1
因此对任意正整数n，a  a 即a 3a ·················································2分
n 3 n1 n1 n
数学评分标准 第1页（共6页）因为a 2a 6，所以9a 6a 6，a 20················································3分
3 2 1 1 1
a
所以 n1 3，{a }是首项为2，公比为3的等比数列，···································5分
a n
n
所以a 23n1 ·························································································7分
n
（2）因为数列{b }是以4为首项，2为公差的等差数列，
n
所以b 2n2··························································································9分
n
23n1 2
因为a  23n1 ，所以2m2 23n1 ，即m 3n1 1·········· 12分
n 2
3n 1
所以c 3n1 1，解得S  n·························································15分
n n 2
17（15分）
解：（1）EA,CD均垂直于平面ABC，
\EA//CD ··································································································1分
取AB中点M ，连接FM,CM ，
FE = FB,AM = BM ，\FM // AE 且FM =1，
又CD// AE且CD =1，故四边形FMCD是平行四边形·····································3分
\FD//CM ，
又FDË平面ABC，CM Ì平面ABC，
\FD//平面ABC·························································································6分
（2）令DC =a(a>0)，取AC中点为O，连接BO，过O作OZ // AE，且交DE于Z，
OZ // AE，AE ^平面ABC，OZ ^平面ABC，
ΔABC 是正三角形，所以AO=CO，
\OB^ AC ，·····························································································7分
  
以O为坐标原点，OB,OC,OZ 方向为x,y,z 轴正方向，建立空间直角坐标系O-xyz，
则B( 3,0,0)，E(0,-1,2)，D(0,1,a)，C(0,1,0)，
 
所以EB=( 3,1,-2)，ED=(0,2,a-2)，··························································9分
 
 ì ïn×EB=0 ì ï 3x+y-2z=0
设平面BED法向量n=(x,y,z)，则í  ，所以í ，
ï în×ED=0 ïî2y+(a-2)z=0

可取n=(a+2, 3(2-a),2 3)，····································································12分

又因为CE =(0,-2,2)，设CE与平面BED所成角为θ ，
 
|CE×n| 6 a 6 1
sinθ =   = ´ = ·······································14分
|CE|×|n| 4 a2 -2a+7 4
1+
7
-
2
a2 a
7
所以，当a=7时，sinθ最大值为
4
7
综上，直线CE与平面EBD所成角正弦值的最大值为 ···································15分
4
数学评分标准 第2页（共6页）18．（17分）
解：（1）因为 A(1,1)在E 上，所以12p，···················································· 2分
所以E 的方程为 y2  x················································································ 3分
（2）若l与x轴平行，不合题意；
若l不与x轴平行，则l与E 相切于 A，
y2  x
设l:x myn，由 ，得 y2 myn 0，
x myn
 m2 4n 0 m 2
所以 ，解得： ，
1mn n1
x1
所以l: y  ··························································································6分
2
1 1
因为B(1,0),F( ,0),A(1,1),D(0, )，故D为AB中点，
4 2
1 3
所以，AB中垂线的方程为y2x ，BF 中垂线的方程为x ，
2 8
3 5
因此VABF 的外接圆圆心坐标为( , )，
8 4
3 5 5 5
VABF 的外接圆半径等于 ( 1)2 ( 0)2 
8 4 8
3 5 125
所以，VABF 外接圆方程为(x )2 (y )2  ········································10分
8 4 64
   
(3)（解法1）设CPCA(01)，CQCB(01)
     
CPCA(CPPA)(01)，故(1)CPPA
 
|PA| 1 |QB| 1
所以   ，同理   ，
|CP|  |CQ| 
|PA| |QB| 1 1 1 1
因为  1，所以  1，所以  3，
|CP| |CQ|    
   1 1
设CG mCD(0m1)，因为CD CA CB，
2 2
 m m m  m 
所以CG  CA CB CP CQ
2 2 2 2
m m 2
因为P,G,Q三点共线，所以  1，所以m ··············14分
2 2 3
 2
CG  CD，设C(x ,y ),y 2  x ,y 1,G(x,y)，
3 0 0 0 0 0
 x
x 0
  1   3
因为CG (xx ,y y )，CD(x ,  y )，所以
0 0 0 2 0  y 1
y 0
 3
数学评分标准 第3页（共6页）所以(3y1)2 3x，····················································································16分
2
因为A,C 不重合，所以y 1，y ，
0 3
2
综上，点G 的轨迹方程为(3y1)2 3x(y )·················································17分
3
（解法2）因为D为AB中点，
|CD| |CA| |AP| |CB| |QB|
记 ,t  1 ,t  1 ，
|CG| 1 |CP| |CP| 2 |CQ| |CQ|
则t t 3因为CD为△ABC 中线，
1 2
S |CQ||CP| 1 S S S 1 1 1 t t
△CQP   且 △CQP  △CGP  △CGQ  (  ) 1 2 ，
S |CB||CA| tt S 2S 2S 2 t t 2tt 
△CAB 1 2 △CAB △CAD △CBD 1 2 1 2
t t 1 3
故 1 2  ，所以 ，所以G是△ABC 的重心·······································14分
2tt  tt 2
1 2 1 2
 x
x 0

3

 y 1
设C(x ,y ),y 2  x ,y 1,G(x,y)，所以y 0 ，解得(3y1)2 3x············· 16分
0 0 0 0 0  3
y2  x
 0 0

2
因为A,C 不重合，所以y 1，y ，
0 3
2
综上，点G 的轨迹方程为(3y1)2 3x(y )·················································17分
3
（解法3）设G(x,y),P(x ,y ),Q(x ,y ),C(y2,y )(y 1),G(x,y)，
1 1 2 2 0 0 0
|PA|  
设 ，则APPC ，所以(x 1,y 1)(y2x ,y y )，
|CP| 1 1 0 1 0 1
1y2 1y
所以x  0 ,y  0 ，
1 1 1 1
|QB| 1y2 y
设 ，同理可得：x  0 ,y  0 ，
|CQ| 2 1 2 1
1y 1y2
y 0 x 0
1 1
直线PQ的方程为：  ，
y 1y 1y2 1y2
0  0 0  0
1 1 1 1
又因为1，化简得：[()y (1)]x[()y23]y1 y y2①，
0 0 0 0
1
y 
直线CD的方程为：y 0 2 x 1  (2y 0 1)x y 0 2 ②，·································· 14分
y2 2 2y2
0 0
数学评分标准 第4页（共6页） y2
x 0
由①②解得： 3 ，消去y 得：(3y1)2 3x，······································16分
1 y 0
y 0
 3
2
因为A,C 不重合，所以 y 1， y ，
0 3
2
综上，点G的轨迹方程为(3y1)2 3x(y )·················································17分
3
19．（17分）
sinx sin1 1
解：（1）m(x)= 在[0,1]上单调递增，且m(0)=0，m(1)= < <1，
2 2 2
所以0m(x)1···························································································1分
sinx -sinx
对任意的x ,x Î[0,1]，|m(x )-m(x )|=| 1 2 |，
1 2 1 2 2
令 f(x)sinxx(0x1)， f ¢ (x)=cosx-1£0 ， f(x)在[0,1]上单调递减，
不妨设0 x  x 1，则 f(x )³ f(x )，sinx -x ³sinx -x ，
1 2 1 2 1 1 2 2
所以x -x £sinx -sinx £0，故|sinx -sinx |£|x -x |，
1 2 1 2 1 2 1 2
1 1
所以|m(x )-m(x )|£ |x -x |，故λ取 即可．
1 2 2 1 2 2
所以m(x)ÎP·····························································································4分
1
（2）对任意的x[0, 1 ](nN*)，nxÎ[0,1]，m ¢ (x)= 1 ( 1 -x),xÎ[0,1]，
n 2n+1 x+1
5-1 - 5-1
令m ¢ (x)=0解得：x= 或x= （舍）···········································5分
2 2
5-1 5-1
所以m(x)在(0, )上单调递增，在( ，1)上单调递减；
2 2
1 1
又因为m(0)=0，m(1)= (ln2- )>0 ，
2n+1 2
所以m(nx)³0．
1
令g(x)ln(x1)x(0x1)，g ¢ (x)= -1£0，所以g(x)在[0,1]上单调递减，
x+1
x2 x2
所以g(x)£g(0)=0，ln(x+1)£ x，ln(x+1)- £ x- £1，
2 2
1 1
所以0m(nx)  ，
2n1 n
1
所以0m(nx) ························································································7分
n
1
对任意的x ，x Î[0, ]，
1 2 n
数学评分标准 第5页（共6页）1 n2x2 n2x 2
因为|m(nx )-m(nx )|= |ln(nx +1)-ln(nx +1)- 1 + 2 | ，
1 2 2n+1 1 2 2 2
n2x2 n2x 2
又因为|ln(nx +1)-ln(nx +1)- 1 + 2 |
1 2 2 2
n2x2 n2x 2
£|ln(nx +1)-ln(nx +1)|+| 1 - 2 |
1 2 2 2
n2x2 n2x2 n
其中| 1 - 2 |= | nx +nx || x - x |£ n| x - x | ，
2 2 2 1 2 1 2 1 2
1
令h(x)=lnx-x(x³1)，h(x) 10，h(x)在[1,+¥)上单调递减，
x
1
不妨设0 x  x  ，1£nx +1£nx +1£2，则h(nx +1)³h(nx +1)，
1 2 n 1 2 1 2
所以ln(nx +1)-(nx +1)³ln(nx +1)-(nx +1) ，
1 1 2 2
所以0³ln(nx +1)-ln(nx +1)³nx -nx ，
1 2 1 2
所以|ln(nx +1)-ln(nx +1)|£|nx -nx |=n|x -x | ，
1 2 1 2 1 2
2n 2n
所以|m(nx )-m(nx )|£ x -x ，故λ取[ ,1)内常数即可．
1 2 2n+1 1 2 2n+1
所以m(x)ÎP ··························································································· 11分
n
（3）因为|a -a |=|m(ta )-m(ta )|£ λ| a - a | ，
n+1 n n n-1 n n-1
所以|a a ||a a |n1|a a | ·················································13分
n1 n n n1 2 1
又因为|a -a |=|(a -a )+(a -a )++(a -a )|
k+l k k+l k+l-1 k+l-1 k+l-2 k+1 k
£|a -a |+|a -a |++|a -a |
k+l k+l-1 k+l-1 k+l-2 k+1 k
£ λk+l-2 |a -a |+λk+l-3 |a - a |++ λk-1| a - a |
2 1 2 1 2 1
=(λk+l-2+λk+l-3++ λk-1)| a - a |
2 1
λk-1-λk+l-1
= |a -a |
1-λ 2 1
λk-1
£ |a -a |···································································17分
1-λ 2 1
数学评分标准 第6页（共6页）