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 第 6 页 共 7 页 {#{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=}#}