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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 C1C 2 3 C 3P(X 1) 3 2 33 ,P(X 2) 3 ,P(X 3)
3 1 3 ······················ 4分C5 10 C5 5 C5 10
所以 X 的分布列为：
X 1 2 3
3 3 1
P
10 5 10
所以 E(X ) 1 3 2 3 3 1 9 ································································6分
10 5 10 5
(2) x 1 （2 4 5 1 6 8） 5， y （20 30 60 60 80） 50，······················· 7分
5 5
5
xi yi 2 20 4 30 5 60 6 60 8 80 1460 ·········································· 8分
i 1
5
x 2 5x 2i 4 16 25 36 64 5 25 145 125 20 ····································9分
i 1
b 1460 5 5 50 10.5， a y b x 50 10.5 5 2.5 ······························ 11分
20
所以经验回归方程为 y 10.5x 2.5，··························································· 12分
当 x 10时， y 102.5
答：所求方程为 y 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， an an 1即 an 1 3an ·················································2分3
数学评分标准 第 1页（共 6页）
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因为 a3 2a2 6，所以9a1 6a1 6， a1 2 0 ················································ 3分
a
所以 n 1 3a ，
{an}是首项为 2，公比为3的等比数列，··································· 5分
n
所以 a 2 3n 1n ·························································································7分
（2）因为数列{bn}是以 4为首项， 2为公差的等差数列，
所以bn 2n 2 ··························································································9分
2 3n 1 2
因为 an 2 3
n 1，所以 2m 2 2 3n 1，即m 3n 1 1 ·········· 12分
2
3n 1
所以 cn 3
n 1 1，解得 Sn n ·························································15分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 FD 平面 ABC，CM 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 n×EB = 0 ì
设平面 BED法向量 n = (x, y, z)，则 ，所以 3x + y - 2z = 0í í ，
n×ED = 0 2y + (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，

sinθ | C E ×n | 6 a 6 1= = = ······································· 14分
|CE |×| n | 4 a2 - 2a + 7 4 1 7 2+
a2
-
a
所以，当 a = 7时， sinθ最大值为 7
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）
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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
所以 l : y
x 1
··························································································6分
2
因为 B( 1,0),F (1 ,0), A(1,1),D(0, 1)，故D为 AB中点，
4 2
所以， AB中垂线的方程为 y 2x 1 ， BF 中垂线的方程为 x 3 ，
2 8
因此VABF 的外接圆圆心坐标为 ( 3 5 , )，
8 4
VABF 的外接圆半径等于 ( 3 1)2 (5 5 5 0)2
8 4 8
所以，VABF 外接圆方程为 (x 3 )2 (y 5 125 )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

所以 | P A | 1 ，同理 |Q B | 1 ，
|CP | |CQ |
因为 | PA | |QB | 1，所以1 1 1，所以 1 1 3，
|CP | |CQ |
1 1
设CG mCD(0 m 1)，因为CD CA CB，
2 2

所以CG m CA m CB m CP m CQ
2 2 2 2
因为 P,G,Q三点共线，所以 m m 1，所以m 2 ······································· 14分
2 2 3
2 CG CD，设C(x0 , y0 ), y
2
0 x0 , y0 1,G(x, y)，3
x0
x
因为CG (x x0 , y y0 )，CD ( x0 ,
1
y0 )，所以
3
2 y y0 1
3
数学评分标准 第 3页（共 6页）
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所以 (3y 1)2 3x，···················································································· 16分
因为 A,C 2不重合，所以 y0 1， y ，3
综上，点G的轨迹方程为 (3y 2 1)2 3x(y ) ················································· 17分
3
（解法 2）因为D为 AB中点，
记 |CD | , t |CA | 1 | AP | , t |CB | 1 |QB | 1 2 ，|CG | |CP | |CP | |CQ | |CQ |
则 t1 t2 3因为CD为△ABC 中线，
S△CQP |CQ ||CP | 1 S 且 △CQP S △CGP
S
△CGQ
1 ( 1 1 ) t1 t 2 ，
S△CAB |CB ||CA | t1t2 S△CAB 2S△CAD 2S△CBD 2 t1 t2 2t1t2
故 t1 t2 1 ，所以 3 ，所以G是△ABC 的重心·······································14分
2t1t2 t1t2 2
x0
x
3
设C(x , y ), y 20 0 0 x0 , y0 1,G(x, y)，所以
y y 1 0 ，解得 (3y 1)2 3x ············· 16分
3
y2 x
0 0
因为 A,C 不重合，所以 y0 1， y
2
，
3
综上，点G的轨迹方程为 (3y 1)2 3x(y 2 ) ················································· 17分
3
（解法 3）设G(x, y),P(x1, y1),Q(x , y ),C(y
2
2 2 0 , y0 )(y0 1),G(x, y)，
| PA | 设 ，则 AP PC ，所以 (x1 1, y1 1) (y
2
0 x1, y0 y1)，|CP |
1 y2 1 y
所以 x 0 01 , y ，1 1 1
|QB | 1 y2 y设 ，同理可得： x 02 , y 0 ，|CQ | 1 2 1
2
y 1 y 1 y 0 x 0
直线 PQ的方程为： 1 1 ，
y0 1 y 2 2 0 1 y0 1 y
1
0
1 1 1
又因为 1，化简得：[( ) y (1 )]x [( ) y 20 0 3]y 1 y0 y
2
0 ①，
y 10
CD 1 (2y 1)x y
2
直线 的方程为： y 2 x 0 02 2 ②，·································· 14分y0 2 2y0
数学评分标准 第 4页（共 6页）
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y2
x 0
由①②解得： 31 y ，消去 y0得： (3y 1)
2 3x，······································16分
y 0
3
因为 A,C 不重合，所以 y 20 1， y ，3
综上，点G的轨迹方程为 (3y 1)2 3x(y 2 ) ················································· 17分
3
19．（17分）
sin x 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
sin x
对任意的 x , x [0,1]， |m(x ) -m(x ) |=| 1 - sin x21 2 1 2 |，2
令 f (x) sin x x(0 x 1)， f (x) = cos x -1 0， f (x)在[0,1]上单调递减，
不妨设 0 x1 x2 1，则 f (x1) f (x2 )， sin x1 - x1 sin x2 - x2 ，
所以 x1 - x2 sin x1 - sin x2 0，故 | sin x1 - sin x2 | | x1 - x2 |，
所以 |m(x ) m(x ) | 1 11 - 2 | x2 1
- x2 |，故 λ取 即可．2
所以m(x) P1 ····························································································· 4分
1 1
（2）对任意的 x [0, 1](n N*)， nx [0,1]，m (x) = ( - x), x [0,1]，
n 2n +1 x +1
令m (x) = 0 5 -1 - 5 -1解得： x = 或 x = （舍）··········································· 5分
2 2
所以m(x)在 (0, 5 -1) 5 -1上单调递增，在 ( ，1)上单调递减；
2 2
1 1
又因为m(0) = 0，m(1) = (ln 2 - ) > 0 ，
2n +1 2
所以m(nx) 0．
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
2 2
所以 g(x) g(0) x x= 0， ln(x +1) x， ln(x +1) - x - 1，
2 2
1 1
所以 0 m(nx) ，
2n 1 n
所以 0 1 m(nx) ························································································ 7分
n
1
对任意的 x1， x2 [0, ]，n
数学评分标准 第 5页（共 6页）
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1 2 2 2 2
因为 |m(nx1) -m(nx2) |= | ln(nx1 +1) - ln(nx 1)
n x1 n x
2 + - +
2 | ，
2n +1 2 2
2 2 2 2
又因为 | ln(nx n x1 n x21 +1) - ln(nx2 +1) - + |2 2
2 2 2 2
| ln(nx +1) - ln(nx +1) | | n x+ 1 n x1 2 - 2 |2 2
2 2 2 2
| n x1 n x2 | n其中 - = | nx1 + nx2 || x1 - x | n | x - x | ，2 2 2 2 1 2
令 h(x) = ln x - x(x 1)， h (x) 1 1 0， h(x)在[1,+ )上单调递减，
x
不妨设 0 1 x1 x2 ，1 nxn 1
+1 nx2 +1 2，则 h(nx1 +1) h(nx2 +1)，
所以 ln(nx1 +1) - (nx1 +1) ln(nx2 +1) - (nx2 +1) ，
所以0 ln(nx1 +1) - ln(nx2 +1) nx1 - nx2 ，
所以 | ln(nx1 +1) - ln(nx2 +1) | | nx1 - nx2 |= n | x1 - x2 | ，
所以 |m(nx1) -m(nx2) |
2n
x 2n1 - x2 ，故 λ取[ ,1)内常数即可．2n +1 2n +1
所以m(x) Pn ··························································································· 11分
（3）因为 | an+1 - an |=|m(tan ) -m(tan-1) | λ | an - an-1 | ，
所以 | a a | | a a | n 1n 1 n n n 1 | a2 a1 | ················································· 13分
又因为 | ak+l - ak |=| (ak+l - ak+l-1) + (ak+l-1 - ak+l- 2) + + (ak+1 - ak ) |
| ak+l - ak+l-1 | + | ak+l-1 - ak+l-2 | + + | ak+1 - ak |
λk+l-2 | a2 - a
k+l-3
1 | +λ | a2 - a1 | + + λ
k-1 | a2 - a1 |
= (λk+l-2 + λk+l-3 + + λk-1) | a2 - a1 |
λk-1 - λk+l-1
= | a - a |
1- λ 2 1
λk-1
| a2 - a1 | ···································································17分1- λ
数学评分标准 第 6页（共 6页）
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