东北三省精准教学2025届高三理想树联考-数学答案_2025年5月_250529东北三省精准教学联盟2025届高三5月联考（理想树）（全科）

高三数学 x2
所以 fx ，··············································································································2分
x2
参考答案 当x2时， fx0，当0x2时， fx0，···································································· 4分
所以 f x在0,2上单调递减，在2,+上单调递增，·······························································5分
1 2 3 4 5 6 7 8 9 10 11
所以 f x的最小值为 f 21ln 2.························································································ 6分
D B C C A C B B ABD ABC BCD
(备注： f x求导正确给2分；判断出 f x的单调性给3分；求出 f x的最小值给1分）
12. 13 （5分） （2）由题意知，函数 f x的定义域为(0,)，求导得 fx ax2x2 ，·································7分
2 x2
因为x,x （x x ）是 f x的两个极值点，
13.y 2e（5分） 1 2 1 2
14.10（5分）
所以x
1
,x
2
（x
1
x
2
）是方程ax2x20的两个正根，
p
15.【答案】(1)88000枚（7分） （2） （6分） 
p 1 p q Δ18a 0，
( )

【解析】(1) 设事件 A：芯片合格，记 X 为该生产批次合格芯片的数量，则每个芯片通过测试的概率为 则有  x x  1 0， ·············································································································9分
   1 2 a
P A  p(1 p)q 0.8(10.8)0.40.88，····································································3分
 2
xx  0，
 1 2 a
于是X ~ B(105，0.88)，··········································································································5分
1
则 1050.88=88000，·································································································6分 解得0a .······················································································································10分
8
所以  预  估=合格芯片的数量为88000枚.······················································································7分 1 2
x x 2 x22xx x2 x x  a   1
  因为 1 2  1 1 2 2  1 2 2   ，
（备注：推导出P A 给3分；推导出E(X)给3分；下结论给1分） xx xx x x 2 2a
1 2 1 2
2 1
a
（2）记事件 ：芯片合格，事件 ：通过测试I,事件C：通过测试II.·············································8分
x x 1
由题意得P  A    P  B  P  B  P  C  |B   p(1 p)q，································································10分 所以 2  1  2，·······································································································12分
x x 2a
1 2
     
P AB  P B P A|B  p1 p，·························································································12分 x x x 1 5
x 2  1  2  ≥ 1 5
又 2≥2，易知 x x x x 2，即 2≥ ，
  x 1 2 1 2 2a 2
  P AB p p 1 x
则P B|A   ，故所求概率为 .·················································13分 1
 
P A p 1 p q p 1 p q
( ) ( )
1
     
所以0a ≤ ，················································································································· 14分
（备注：设事件给1分；推导出P A 给2分；推导出P AB 给2分；推导出P B|A 给1分,过程酌情给分) 9
1
1 故a  .·························································································································15分
16.【答案】(1)1ln2（6分）（2） （9分） max 9
9
x x 1 1
2 （备注：f x求导正确给1分；列出根与系数的关系给2分；化简出 2  1  2给3分；推导出0a ≤
【解析】（1）当a0时， f xln x ，定义域为(0,)，······················································1分 x x 2a 9
1 2
x
给2分；下结论给1分，过程酌情给分）
第 1 页 共 4 页17.【答案】（1）EF  5 2 （4分）（2）① 3 （5分）② 6 （6分） 易知平面ABC的一个法向量为m (0,0,1).···············································································13分
2 5 9
设二面角EACB的平面角的大小为，由题图可得为钝角，
【解析】（1）在正方体ABCD ABCD中，
1 1 1 1
mn 2 6
则 cos  cosm,n = =  ，
因为平面ABCD//平面A 1 B 1 C 1 D 1 ，又平面ACE平面ABCD AC，平面ACE平面A 1 B 1 C 1 D 1  EF， m  n 13 6 9
所以EF//AC，····················································································································2分 6 6
则cos ，所以二面角E ACB的余弦值为 .·······················································15分
9 9
连接AC （图略），又因为AC// AC ，所以EF // AC .
1 1 1 1 1 1 3
（备注：①判断出四边形ACEF 为等腰梯形给1分；写出EF，CE给2分；求出 给2分.
1 5
又 ，即E为DC 的中点，所以EF 为△D AC 的中位线，
2 1 1 1 1 1 ②建系并写出点坐标给 1 分；求出平面 ACE的法向量给 2 分；写出平面 ABC 的法向量给 1 分；求出二面角
E ACB的余弦值给2分，过程酌情给分）
5 2
所以F为D A 的中点，则EF  .······················································································4分
1 1 18.【答案】(1)y2 4x（3分）(2)①32（7分） ②直线PQ过定点，定点坐标为(3,0)（7分）
2
(备注：推导出EF//AC给2分；得出EF  5 2 给2分） 【解析】（1）根据题意，圆心坐标为 .·············································································1分
2  −4
2 ,0
 x4  2 x4
又因为该圆经过点 和 ,所以  0   0 y 2  4 ，····························· 2分
（2）①由（1）知，EF//AC，D E  D F ，所以C E  AF ，所以CE  AF ，所以四边形ACEF 为等腰梯形，  2  2
1 1 1 1
−4,0  0, 
········································································································································ 5分 化简得 y2 4x，所以点M(x,y)的轨迹E的方程为 y2 4x.························································3分
由D 1 E D 1 C 1 ，得D 1 E  D 1 F = 5，则 EF 5 2，C 1 E  A 1 F 5(1)， (备注：列出关系式   x4 0  2   0 y 2  x4 4 给2分；推导出点M的轨迹方程给1分）
 2  2
所以CE  AF = 25(1)225=5 222 ，········································································· 7分
1
（2）①因为直线l ,l 的斜率一定存在且不为0，故设l：y k(x1)，l ：y  (x1)，G(x ,y )，H(x ,y )，
1 2 1 2 k 1 1 2 2
所以等腰梯形ACEF 的周长为5 2  1  10 2 22 =8 22 29，
J(x ,y )，K(x ,y ).········································································································· 4分
3 3 4 4
3
又01，所以 .······································································································9分
5 y k(x1)，
②以D为原点，DA,DC,DD
1
所在直线分别为x轴、y轴、z轴，建立空间直角坐标系（图略），
联立方程


y2 4x，
消x得ky2 4y4k 0，····································································5分
4
则E(0,3,5)，A(5,0,0)，C(0,5,0)，·······················································································10分 则
Δ
16(k2 1)0，y
1
 y
2

k
，y
1
y
2
4.·········································································6分
则AC (5,5,0)，AE (5,3,5).
1 1 1 4
2
 1 
所以 GH  1  y  y  1   y  y 2 4y y  1    44 41 ，··7分
k2 1 2 k2 1 2 1 2 k2 k  k2 
  nAC 5x5y 0，
设平面ACE的法向量为n(x,y,z)，则 令x5，则 y 5，z 2，所以平面ACE 同理 JK 4(k2 1)，··········································································································8分
 nAE 5x3y5z 0，
1  1   1 
的一个法向量为n(5,5,2)，································································································ 12分 所以S
四边形GJHK

2
GH  JK 8

k2 
k2
2

≥8
 
2 k2
k2
2
 
32，
第 2 页 共 4 页当且仅当k 1时，四边形GJHK 的面积最小，最小值为32.·························································10分 所以b (n1)(b b )（n≥3），····················································································6分
n n2 n1
②易知当直线PQ斜率不存在时，直线l，l 关于x轴对称，此时①中k2 1，得直线PQ：x3； ········11分
1 2
 
所以b nb b (n1)b （n≥3），·········································································7分
当直线PQ斜率存在时，设直线l ：y mxn， n n1 n1 n2
PQ
又b 2b 1，·················································································································· 8分
y mxn， kmkn 2 1
联立方程 得 y  ，···············································································13分
y k(x1)， P km 所以{b (n1)b }是以1为首项，1为公比的等比数列.·························································10分
n1 n
y  y 2  
又 y
P
 1
2
2 
k
，得(mn)k2 2k2m0，···································································14分 （备注：推导出b
n
(n1)(b
n2
b
n1
)（n≥3）给4分；推导出b
n
nb
n1
b
n1
(n1)b
n2
（n≥3）给1
分；求出b 2b 1给1分；推导出{b (n1)b }是等比数列给2分，过程酌情给分）
2 1 n1 n
2
同理可得(mn)   1   2   1   2m0，···········································································15分 （3）由（2）知b n1 (n1)b n (1)n1，···············································································11分
 k   k 
b b (1)n1 (1)n1
等式两边同除(n1)!得 n1  n  = ，···························································12分
(n1)！ n! (n1)！ (n1)！
1
所以k, 是方程(mn)x2 2x2m0的两根，···································································16分
k
b b n  b b  n (1)i1
由累加法得 n1  1   i1  i   ，·························································14分
(n1) ! 1 ! (i1) ! i ! (i1) ！
2m i1 i1
所以 1，即3mn0，则l ：y m(x3),所以直线PQ过定点(3,0).
mn PQ
b n (1)i1
则 n1  ，······································································································15分
(n1) ! (i1) ！
综上，直线PQ过定点(3,0).··································································································17分 i1
n (1)i1
（备注：①设出直线l 1 ,l 2 的方程给1分；求出 GH 给3分；求出 JK 给1分；求出四边形GJHK 面积的最小值
即b
n1
=(n1)!  
(i1) ！
，
i1
给2分.
n1 (1)i1
②写出直线 PQ 斜率不存在的情况给 1 分；推导出点 P 的纵坐标给 2 分； 推导出 k, 1 是方程
则b
n
=n !  
i1 (i1) ！
（n≥2）.······························································································17分
k
(mn)x2 2x2m0的两根给3分；求出定点坐标给1分，过程酌情给分） （备注：写出 b (n1)b (1)n1给 1 分；由累加法推导出 b n1 =  n (1)i1 给 4 分；推导出
n1 n (n1) ! (i1) ！
i1
19.【答案】（1）b 0，b =1 （2分）（2）证明见解析 （8分） （3）证明见解析（7分）
1 2
n1 (1)i1
【解析】（1）n=1时，显然b 1 0；n2时，“2元全错位数列”只能是2,1，所以b 2 =1.···················2分 b n =n !   (i1) ！ （n≥2）给2分）
i1
（备注：求出b 0给1分；求出b =1 给1分）
1 2
（2）当n≥3时，为了得到“n元全错位数列”，我们分两步来完成正整数1,2,3,,n的排列：
①将正整数n放到第m（mn）个位置，有(n1)种排法；·······················································3分
②考虑正整数m，有两种放法.
若放到第n个位置，则余下(n2)个正整数放到余下(n2)个位置，有b 种排法；
n2
若不放到第n个位置，这时对于这(n1)个正整数，共有b 种排法.·············································5分
n1
第 3 页 共 4 页题号 题型 分值 考查的主要内容及知识点 难度
1 单选 5 集合的交集运算 易
2 单选 5 复数的概念与运算 易
3 单选 5 等差数列基本量 易
4 单选 5 三角恒等变换 易
5 单选 5 圆台的侧面积 易
6 单选 5 回归分析、最小二乘估计 中
7 单选 5 三角函数的图象与性质 中
8 单选 5 新定义 中难
9 多选 6 解三角形 易
10 多选 6 椭圆、双曲线的第一定义与几何性质 中
11 多选 6 利用导数研究函数的性质 中难
12 填空 5 三棱锥的外接球 易
13 填空 5 导数的几何意义 中
14 填空 5 等比数列前n项和 中难
15 解答 13 二项分布的期望、条件概率 易
16 解答 15 函数与导数 易
17 解答 15 证明线线平行、求参数、求二面角的余弦值 中
18 解答 17 求轨迹方程、求四边形面积最小值、研究定点问题 中
19 解答 17 排列组合与数列综合 中难
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