福建省三明市2025年普通高中高三毕业班质量检测数学答案_2025年5月_250510福建省三明市2025年普通高中高三毕业班质量检测（全科）

三明市 2025 年普通高中高三毕业班质量检测
数学参考答案及评分细则
评分说明：
1．本解答给出了一种或几种解法供参考，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定相应的评分细则．
2．对计算题，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，但不得超过该部分正确解答应给分数的一半；如果后继部分的
解答有较严重的错误，就不再给分．
3．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．
4．只给整数分数．选择题和填空题不给中间分．
一、选择题：本大题考查基础知识和基本运算．每小题5分，满分40分．
1．C 2．D 3．B 4．A 5．B 6．A 7．A 8．D
二、选择题：本大题考查基础知识和基本运算．每小题6分，满分18分．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．AC 10．ACD 11．BCD
三、填空题：本大题考查基础知识和基本运算．每小题5分，满分15分．
3  1
12．5 13． 14． 0,

3  4
四、解答题：本大题共5小题，共77分．解答应写出文字说明、证明过程或演算步骤．
5 2
15.（1）证明：因为2na  ( )n 2n1a ，
n1 3 3 n
5
所以a 2a  ，············································································1分
n1 n 3n1
1 6 1
a  2a  （2 a  ），·························································· 3分
n1 3n1 n 3n1 n 3n
1 1 1
因为a  ，所以a  0,a  0,······················································4分
1 3 1 3 n 3
1
a 
n1 3n1
所以 2··················································································5分
1
a 
n 3n
1 1
所以数列{a  }为等比数列是首项为a  ，公比为2的等比数列.···············6分
n 3n 1 3
1
（2）由（1）可得数列{a  }为首项为2，公比为2的等比数列,
n 3n
1 1
所以a  2n,即a  2n  ································································7分
n 3n n 3n
1 1
（ ）n1
所以S  22n1  3 3  2n1  1 ( 1 )n  3 ，（注：S  2n1  3 ( 1 )n1  3 应得分）
n 12 1 2 3 2 n 2 3 2
1
3
··········································································································9分
因为数列{S }为单调递增数列，······························································10分
n
第 1 页 共 7 页
{#{QQABBQYswggwwhSACB5KQ0GyCgsQkIGiJYoEgRAUOAxigJNIBKA=}#}1 1 3
当n 9时，S  210  ( )9  1024  2025，
9 2 3 2
1 1 3
当n 10时，S 211 ( )10  2025，··············································12分
10 2 3 2
所以，使不等式S 2025成立的n的最小值为10.······································13分
n
16．解：（1）设A“一次回答问题，AI软件答对问题”，
B“选出语文问题让AI回答”，（注：若设事件合理则得分）························1分
2 3
依题意,P(B) ,P(B) ,·····································································2分
5 5
3 1
P(A|B) ,P(A|B) ,·······································································3分
4 2
2 3 3 1 3
所以P(A) P(B)P(A|B)P(B)P(A|B)     ,······························5分
5 4 5 2 5
3
依题意，X 的所有可能取值为0，1，2，3，4，X ~ B(4, )，·····················7分
5
3 12
所以E(X)4  .··········································································· 8分
5 5
（2）设C “共回答5道题后停止，其中最后2道题AI软件均答对”，
D “共回答5道题后停止，其中最后3道题AI软件均答错”,······················9分
那么P(Y 5) P(C）P(D)，·································································10分
3 2 216
所以P(C）C1( )3 ( )2  ，·······················································12分
2 5 5 3125
3 2 48
所以P(D） ( )4  ，·································································14分
5 5 3125
264
所以P(Y 5) .············································································15分
3125
17．（15分）证明：（1）ABC中，AC 2 3,AB 3,CB  3 ,

由余弦定理得ACB  ，······································································1分
3

又因为直线AD 与直线AC 所成角为 ，AD// AD ,
1 1 3 1 1
所以DAC为直线AD 与直线AC 所成角或其补角，
1 1
 
又因为CD  AD,所以DAC（0， ），则DAC  ,·································· 3分
2 3
所以BC//FA ,又BC//BC ,所以FA //BC ,·················································4分
1 1 1 1
又BC 平面E FA ,FA 平面E FA,所以BC //平面E FA .····························6分
1 1 1 1 1 1 1
第 2 页 共 7 页
{#{QQABBQYswggwwhSACB5KQ0GyCgsQkIGiJYoEgRAUOAxigJNIBKA=}#}π π
（2）（法一）因为AC 2 3,AB 3,CB  3 , 由勾股定理得ABC  ,且DAB  ,
2 2
··········································································································7分
因为ABCD ABC D 为直四棱柱,故以A为原点，AD,AB,AA 所在直线分别为x轴、y轴、
1 1 1 1 1
z轴，建立如图所示的空间直角坐标系，
则A(0,0,0),B (0,3,3),F( 3,0,0),H(0,0,t),t[0,3].·····································8分
1
11 21
HB (0,3,3t)，AC ( 3,3,3),因为HB 在AC 投影向量的模为 ，
1 1 1 1 14
 
HB 1 AC 1 11 21 183t 11 21 1 23
所以   即  ,得t  或t  （舍去）,············ 10分
AC 14 21 14 2 2
1
 1 5
所以FH ( 3,0, ),HB (0,3, ),
2 1 2
因为B 、E 、F 、H 四点共面,
1 1

所以设平面BE FH 的法向量为n(x,y,z)，
1 1
 1
   3x z 0
 nHB 0   2
由  1 得 ,
nFH 0 
3y
5
z 0
 2

令x 3，得n( 3,5,6),···································································12分
E ( 3cos,3 3sin,3)，FE ( 3cos 3,3 3sin,3)，······················13分
1 1
 
由nFE 0得 （3 3cos- 3）5(3 3sin)180 ，·······························14分
1
3
化简得3cos5 3sin0，所以tan .··········································· 15分
5
π π
(法二)因为AC 2 3,AB 3,CB  3 , 由勾股定理得ABC  ,且DAB  ,
2 2
··········································································································7分
因为ABCD ABC D 为直四棱柱,故以B为原点，BC,AB,BB 所在直线分别为x轴、y轴、
1 1 1 1 1
z轴，建立如图所示的空间直角坐标系，则A(0,3,0)，B (0,0,3)，F( 3,3,0)，
1
H(0,3,t),t[0,3]．················································································8分
第 3 页 共 7 页
{#{QQABBQYswggwwhSACB5KQ0GyCgsQkIGiJYoEgRAUOAxigJNIBKA=}#}FB ( 3,3,3),HB (0,3,3t),AC ( 3,3,3),
1 1 1
11 21
因为HB 在AC 投影向量的模为 ，
1 1 14
 
HB 1 AC 1 11 21 11 21 183t 1 23
所以   ，  ，得t  或t  （舍去）,···········10分
AC 14 14 21 2 2
1
 1 5
所以FH ( 3,0, ),HB  (0,3, )，
2 1 2
因为B 、E 、F 、H 四点共面,
1 1

所以设平面BE FH 的法向量为n(x,y,z)，
1 1
 5
  3y z 0
 nHB 0   2
由  1 得 ,
nFH 0 
 3x
1
z 0
 2

令x 3,得n( 3,5,6),····································································12分
E ( 3cos, 3sin,3)，FE ( 3cos 3, 3sin3,3)····························13分
1 1
 
由nFE 0得 （3 3cos- 3）5( 3sin3)180 ，·······························14分
1
3
化简得3cos5 3sin0，所以tan .··········································· 15分
5
18. 解：（1）设Mx,y，直线 y 3 x 与直线 y 3 x 的夹角为 π ，即M OM = π ,··
3 3 3 1 2 3
··········································································································1分
π 2π
又因为OM M OM M  ，所以M MM  ,········································ 2分
1 2 2 1 2 3
 x 3y  x 3y
又因为 , ,······················································4分
MM  MM 
1 2 2 2
所以   x 3y x 3y 1 3,化简得 x23y2 3,·······················6分
MM MM    ( )
1 2 2 2 2 8
由于M 位于第一象限，M 位于第四象限，
1 2
所以M 的轨迹方程C: x2 y2 1  x 3 .······················································8分
3
第 4 页 共 7 页
{#{QQABBQYswggwwhSACB5KQ0GyCgsQkIGiJYoEgRAUOAxigJNIBKA=}#}（2）由题可知直线斜率不为0,故设直线BD方程为xmy2(m0)，············9分
B(x ,y ),D(x ,y ),A(x ,y ),G(x ,0)，
1 1 2 2 2 2 0
xmy2

联立直线BD与曲线C，可得x2 且m0,

 y2 1
 3
化简得（m2 3)y2 4my10，
4m 1
m2 30， 0，y  y  ，y y  0,·······························11分
1 2 m2 3 1 2 m2 3
y  y
设直线AB方程为y  1 2 (xx ) y ，
x x 1 1
1 2
 y (x x ) my (y  y )
令y 0，得x  1 1 2 x  1 1 2 my 2，··························12分
0 y  y 1 y  y 1
1 2 1 2
2my y 3
所以x  1 2 2 ，······································································13分
0 y  y 2
1 2
1 1 1 1 3(m2 1)
所以 S  GH  y  y   | y  y |  ，························14分
BGD 2 1 2 2 2 1 2 2 3m2
3 4t 3 2 1 3
令t 3m2（0，3），(S )2    (  )2  ,····························15分
BGD 4 t2 4 t 4 64
2 2 1 3
所以 （ ，），(S )2  ,S  ,
t 3 BGD 12 BGD 6
3
综上，BGD面积的取值范围为（ ，）.·············································· 17分
6
19.解:（1）令 f xx24 x ，其中xR，则 f xx2 4 x  x2 4 x  f x，
所以，函数 f xx24 x 为偶函数，··························································2分
函数 f(x)x24 x 的一条2阶临界直线方程为y4.····································· 3分
1
（2）若 f(x)exlnx， f(x)e（x lnx )，··············································· 4分
x
1 2 1
令g(x)ex(lnx )，g(x)ex(lnx  ),············································ 5分
x x x2
2 1 x2 2x2
令h(x)lnx  ，h(x) 0,··············································6分
x x2 x3
第 5 页 共 7 页
{#{QQABBQYswggwwhSACB5KQ0GyCgsQkIGiJYoEgRAUOAxigJNIBKA=}#}所以x[1,)时g(x)单调递增，
所以g(x)  g(1)e0，x[1,)时， f(x)单调递增，···························7分
min
所以不存在x ,x [1,)，使得 f(x ) f(x )，
1 2 1 2
综上可知函数y f x不存在2阶临界直线.·················································8分
（3）当x[0,1)，f(x)2(x2sinx)，令g(x) x2sinx，g(x)12cosx12cos10，
所以 f(x)在x[0,1)单调递减,
因为 f(0)0, f(1)24sin1,所以x[0,1)， f(x)（24sin1,0],··················9分
2
因为4cos134 30，
2
    
4cos13（24sin1)4(sin1cos1)54 2sin( 1)5， 1（ ，  ）,
4 4 2 4 3
 2
所以4 2sin( 1)54 2 (1 3)50,
4 4
则存在x [0,1)，使 f(x )4cos13，····················································11分
0 0
当x[n,n1),nN*, f(x) f(x1)4cos13,所以 f(x) f(x1),
所以 f(x ) f(x 1) f(x n),················································12分
0 0 0
因为函数 f(x)在(x , f(x ))处的切线方程为y  f(x )(xx ) f(x )，
0 0 0 0 0
函数 f(x)在(x n, f(x n))处的切线方程y  f(x n)(xx n) f(x n),
0 0 0 0 0
又因为 f(x n) f(x )(4cos13)n,······················································13分
0 0
所以 f(x n)(xx n) f(x n) f(x )(xx )nf(x ) f(x n)
0 0 0 0 0 0 0
 f(x )(xx )(4cos13)n f(x )(4cos13)n  f(x )(xx ) f(x ),
0 0 0 0 0 0
所以 f(x n)(xx n) f(x n) f(x )(xx ) f(x )，··························15分
0 0 0 0 0 0
所以直线y  f(x n)(xx n) f(x n)与直线y  f(x )(xx ) f(x )重合，
0 0 0 0 0 0
则x ,x ,,x ,为方程 f(x) f(x )(xx ) f(x )的解，
1 2 n 0 0 0
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{#{QQABBQYswggwwhSACB5KQ0GyCgsQkIGiJYoEgRAUOAxigJNIBKA=}#}且k  f(x ) f(x 1) f(x n),··············································16分
0 0 0
又因为n可趋近于无穷大，所以存在直线y  f(x )(xx ) f(x )为函数y f x的无限阶
0 0 0
临界直线.···························································································17分
第 7 页 共 7 页
{#{QQABBQYswggwwhSACB5KQ0GyCgsQkIGiJYoEgRAUOAxigJNIBKA=}#}