江苏省盐城市2024-2025学年高二下学期期末考试数学试卷参考答案_2025年6月_250625江苏盐城市2024-2025学年高二下学期期末考试（全科）

2024/2025 学年度第二学期高二年级期终考试 数学参考答案 1．B 2．D 3．A 4．C 5．B 6．B 7．D 8．C 27 5 9．AD 10．ABD 11．BCD 12．480 13． 14． e3 6 15. 解：（1）因为a 是等差数列，所以a S 4a 12， n 2 3 2 所以a 3， ··············································································································2分 2 又a 9，所以公差d 2， ···························································································4分 5 于是a a (n2)d 2n1. ·······················································································6分 n 2 1 1 1 1 1 （2）由（1）得   (  )，·················································10分 a a (2n1)(2n1) 2 2n1 2n1 n n1 1 1 1 1 1 1 1 1 1 1 n 所以T  (    L  ) (  ) . ·································13分 n 2 1 3 3 5 2n1 2n1 2 1 2n1 2n1 16. 解：（1）正三棱柱ABC ABC ，点E，G分别为棱AA ，CC 的中点， 1 1 1 1 1 AE CG,AE//CG, 四边形AEGC 为平行四边形,  AC//EG.······································ 2分 又EG 面EBG,AC 面EBG, 1 1 AC//面EBG. 1 同理CF //面EBG.·······································································································4分 1 又ACCF C ,且AC 面ACF ，CF 面ACF , 面AFC//面EBG. ··································································································7分 1 （2）法1：取AC 的中点H ，连接BH ，BH ，AB ，CB ， 1 1 1 正三棱柱ABC ABC 的体积为4 3，且AB 2， 1 1 1 AA  BB CC 4,AA 面ABC , 1 1 1 1 1 1 1 AB 面ABC ,AB  AA,AB  42 22 2 5, 1 1 1 1 1 1 1 1 1 同理CB 2 5, 1 H 为AC 的中点，正三角形 ABC ，B H  AC ,BH  AC， 1 BHB 为二面角B ACB 的平面角.··········································································12分 1 1 57 BH  3,B H  (2 5)2 1 19，在RtBHB 中，cosBHB  ， 1 1 1 19 57 锐二面角B ACB 的余弦值为 .··········································································15分 1 19 法2：取BC的中点O，连接AO，过点O在平面BBCC中作OM //CC , 1 1 1 正三棱柱ABC ABC ， 1 1 1 CC 面ABC，AO面ABC，BC 面ABC， 1 CC  AO,CC  BC，OM  AO,OM  BC， 1 1 O为BC的中点，正三角形ABC，AO  BC， 以{OB,OM,OA}为基底，建立如图所示空间直角坐标系.····················································· 9分 则A(0,0, 3)，C(1,0,0)，B (1,4,0)，C (1,4,0), 1 1 AC (1,0, 3)，AB (1,4, 3)， 1 设面ACB 的法向量n (x,y,z)，n AC 0且n AB 0, 1 1 1 1 1 高二数学答案 第 1 页 共 4 页 {#{QQABSQywwgAQghTACI4LUQUqC0kQsIMhLcokRRAWqAQLSRFABIA=}#} x 3z 0 3 3  ，取x 3,y  ,z 1，n ( 3, ,1)，··························12分  x4y 3z 0 2 1 2 CC 面ABC，面ABC的法向量n CC (0,4,0)， 1 2 1 2 3 57 cosn ,n   1 2 3 19 ，··········································································14分 4 3 1 4 57 锐二面角B ACB 的余弦值为 .··········································································15分 1 19 17.解：（1）记盲盒的外层包装A型为事件A，盲盒的外层包装B型为事件B，盲盒中含限量版商品为事 件C，则P(C) P(C| A)P(A)P(C|B)P(B) ····························································2分 2 4 9 1 1 1      ，所以每个盲盒含限量版商品的概率为 .············································· 4分 5 5 10 5 2 2 1 （2）小王抽中含限量版商品的盲盒数量为随机变量X ，X ~ B(5, )，··································6分 2 则随机变量X 的概率分布为： k 5k 5 1 1 1 P(X k)Ck    Ck  (k 0,1,2,3,4,5) 5 2 2 5 2 X 0 1 2 3 4 5 1 5 5 5 5 1 P 32 32 16 16 32 32 ·································································10分 （3）若单个盲盒含限量版商品，该盲盒外层包装为A型的概率为条件概率 P(C| A)P(A) P(A|C) ··········································································12分 P(C) 2 4  5 5 16   ， ························································································ 14分 1 25 2 16 所以若某个盲盒含限量版商品，则该盲盒外层包装为A型的概率为 . ······················15分 25 c 1 3 1 9 18. 解：（1）椭圆的离心率为e  ，又经过点(1，)，得  1， a 2 2 a2 4b2 x2 y2 又a2 b2 c2，解得a 2,b 3,c1.椭圆方程为  1.···································4分 4 3 3 1 3x 3y 3x 3y （2）①设A(x ,y )，则OB  OA OM ( 1 1, 1)，即B( 1 1, 1), 1 1 4 4 4 4 4 4  x 2 y 2  1  1 1  4 3 1 3 5 又点A，B在椭圆上， 3x 4 3y ，解得x  ， y  ，  ( 1 )2 ( 1)2 1 2 1 4 4 4   1  4 3 高二数学答案 第 2 页 共 4 页 {#{QQABSQywwgAQghTACI4LUQUqC0kQsIMhLcokRRAWqAQLSRFABIA=}#}1 3 5 3 5  A( , )，此时直线l的斜率为 .·····································································8分 2 4 14 ② 设A(x ,y )，B(x ,y )， 1 1 2 2 当直线l斜率为0时，显然B,Q,N 三点共线；····································································10分 x2 y2 当直线l斜率不为0时，设直线l的方程：xmy4，与椭圆E：  1联立， 4 3 整理可得(3m2 4)y2 24my360， 24m 36 y  y  ，y y  ， 144(m2 4)0，·········································12分 1 2 3m2 4 1 2 3m2 4 5 又Q(1,y )，N( ,0)， 1 2 y  y 2y 3y 3y 2y (x 1) 3(y  y )2my y k k  2 1  1  2 1  1 2  1 2 1 2 BQ NQ x 1 3 3(x 1) 3(x 1) 3(x 1) 2 2 2 2 72m 72m   3m2 4 3m2 4 0 ，得k BQ k NQ ，则B,Q,N 三点共线.·············································17分 3(x 1) 2 19. 解：（1）当a1时， f(x) x2 lnx， 1 2x2 1 所以 f(x)2x  (x0)， x x 2 由 f(x)0得x ， ····························································································2分 2 2 2 当0 x 时， f(x)0；当x 时， f(x)0， 2 2 2 2 所以 f(x)在(0, )上单调递减；在( ,)上单调递增. ····················································4分 2 2 1 2ax2 1 （2）由 f(x)2ax  (x0)，因a0， x x 1 所以由 f(x)0得x ， 2a 1 1 与（1）同理可得 f(x)在(0, )上单调递减；在( ,)上单调递增， 2a 2a 1 1 1 1 所以 f(x) f( )  ln ， ···············································7分 2a 2 2 2a 1 1 1 1 1 1 1 1 令g(a)  ln 1  ln   ，只需证g(a)0即可. 2 2 2a 4a 2 2a 4a 2 1 1 1 1 令 t 0，则g(a)(t) lnt t ，只需证(t)0即可. 2a 2 2 2 t1 于是(t) ，类似可得(t)在(0,1)上单调递减；在(1,)上单调递增， 2t 1 所以(t)(1)0，因此 f(x)1 成立. ·····················································10分 4a （3）不等式 f(x)sinx恒成立，即ax2 lnxsinx恒成立， 所以a12 ln1sin1成立，即asin10， 因为a为整数，所以a1. ·································································12分 当a1时，令m(x) f(x)x x2 lnxx， 高二数学答案 第 3 页 共 4 页 {#{QQABSQywwgAQghTACI4LUQUqC0kQsIMhLcokRRAWqAQLSRFABIA=}#}1 2x2 x1 (2x1)(x1) 所以m(x)2x 1  (x0)， x x x 由m(x)0得x1， 类似可得m(x)m(1)0，所以x2 lnx x； ······························································14分 又令h(x) xsinx(x0)， 所以h(x)1cosx0，即h(x)在(0,)上单调递增， 所以h(x)h(0)0，所以xsinx. 因此当a1时， f(x)sinx恒成立， ·········································································16分 所以整数a的最小值1. ························································································17分 高二数学答案 第 4 页 共 4 页 {#{QQABSQywwgAQghTACI4LUQUqC0kQsIMhLcokRRAWqAQLSRFABIA=}#}