黑龙江省哈尔滨市2024-2025学年高二下学期7月期末学业质量检测数学试卷答案_2025年7月_250717黑龙江省哈尔滨市2024-2025学年高二下学期期末质量检测（全科）

参考答案 题号 1 2 3 4 5 6 7 8 答案 C D D B D B A A 题号 9 10 11 答案 ABD BC ACD 12. 5 13.12 14.2022 4.【详解】奇函数图像关于原点对称，所以在关于原点对称区域内单调性相同，函数y f(x)是奇函数且 在区间5,1上单调递增，则函数y f(x)在区间1,5上单调递增，由于区间为半开半闭区间，所以存 在最大值.故选B. 1 2r 2r S 3 2 2 5.【详解】设圆锥底面半径为r，则高也为r,    6,所以r  3,母线长为 V 1 r r2r 3 2r  6.故选D 6.【详解】由题可先将1号球分给甲，再将情况进行分类： 第一类，甲只有一个球，则另外三个球分给乙和丙两位同学，有C2A2 =6种方法， 3 2 第二类，甲有两个球，则需要将3个球分发给3位同学的方法数有A3 6种， 3 所以共有6+6=12种分法.故选B. 7.【详解】随机变量X 的密度曲线关于x2对称， PX 0PX 3  PX  4PX 3  PX  4PX  4P3 X  4 13 1 1P3 X 4 ，所以P3 X 4 .故选A 12 12 8.【详解】设a ，a ，a ，a 平均数为a，因为s2  1  4 a2 a 2 ，得a 2 25，由a0，故a5.则 1 2 3 4 4 i i1 样本数据4a 1，4a 1，4a 1，4a 1的平均数为4a121.故选A 1 2 3 4 2 P(B A ) 11.【详解】由题意可知P(A)P(A )P(A ) 1 ,P(B A)  4 ,P(C A)  1 , 2 5 ,A正确，B 1 2 3 3 1 5 1 5 11 错误；P(B A ) 1，P(B)P(A)P(B A)P(A )P(B A )P(A )P(B A ) ,C正确； 3 1 1 2 2 3 3 15 1 4  P(AB) 3 5 4 P(A B) 1   ,D正确；故选ACD. 1 P(B) 11 11 15 14.【详解】 f x21为奇函数，所以 f x关于点2,1对称， f 2x1为偶函数，所以 f x关于直线 x1对称，故 f x的周期为4， f 2 f41， f 1 f32， f 2025 f12 2025 所以 f(i)506(f(1) f(2) f(3) f(4)) f(2025)506422022 ，故答案为2022. i1 15.(本小题满分13分) （1）由S 25可得a 5，·····························································································2分 5 3 由a a a a 12可得d 2，····················································································4分 2 5 3 4 故a 2n1.·················································································································6分 n （2） 由（1）得S n2,································································································8分 n 高二数学 第 1 页 共 4 页 {#{QQABTQYg4gi4wBbACY4LUQU4CwoQsJMjLcokwQASKAQKCRFABKA=}#}1 1 1 1 1 1 故b     (  )··············································11分 n 4S 1 4n21 (2n1)(2n1) 2 2n1 2n1 n 1 1 1 1 1 1 n 则T  (1     ) .··························································13分 n 2 3 3 5 2n1 2n1 2n1 16.(本小题满分15分) a 解：(1) fx2x2a1 , f11a 3，所以a 2·············································2分 x f x x2 3x2lnx ， f 14，切点为1,4 ··································································4分 切线方程为 y43x1，即3x y10········································································6分  lnx (2)x2 xlnxmx ，x0，m  x1  ··························································8分  x  min lnx x2 1lnx 设gx x1 ,x0,，gx ·························································9分 x x2 2x2 1 设hx x2 1lnx,x0,，hx  0，hx在0,上单调递增············11分 x h10，x0,1时hx0即gx0，gx在0,1上单调递减 x1,时hx0即gx0，gx在1,上单调递增············································13分 gx  g1 2，m2························································································15分 min 17.(本小题满分15分) 答：（1）由频率分布直方图可知，0.004a0.0180.0220.0220.028101，解得a0.006，···2分 该校学生满意度打分不低于70分的人数为：10000.280.220.18680.··································4分 （2）平均数为：x450.04550.06650.22750.28850.22950.1876.2；················6分 所以0.040.060.220.280.6,0.040.060.220.280.220.82， 0.750.6 所以75%的分位数为80 1086.82 .······································································8分 0.820.6 （3）由频率分布直方图可知，打分在60,80和80,100内的频率分别为0.5和0.4， 所以打分在60,80和80,100内的频率之比为5:4，·································································9分 5 所以在打分60,80中抽取的人数为 95人； 9 4 在打分80,100中抽取的人数为 94人.···········································································11分 9 设“至少有一份答卷成绩为优秀”为事件A，···········································································12分 C1C1C2C0 13 13 则P(A) 4 5 4 5  .至少有一份答卷成绩为优秀 .······················································15分 C2 18 18 9 18.(本小题满分17分) 【解】（1）因为 f xex，则 fxex，fxex，·····························································1分 f0 1 2 K    所以  1f0 2 3 2  112 3 2 4 .················································································3分   （2）因为 f xex，则 fxex， fxex ， fx ex K   所以 3 3 ，····················································································4分  1fx 2 2  1e2x 2   高二数学 第 2 页 共 4 页 {#{QQABTQYg4gi4wBbACY4LUQU4CwoQsJMjLcokwQASKAQKCRFABKA=}#}t g(t) 令t ex(t 0)， 3  1t2 2 3 3 1  1t2 2t  1t2 22t 则gt 2  1t23t2  12t2 ，·························································6分  1t23  1t2 5 2  1t2 5 2 当0t 2 时，g't0，gt单调递增， 2 当t 2 时，g't0，gt单调递减，···············································································8分 2 所以gt  g   2  2 3 ，所以曲率K最大值为 2 3 .··························································9分 max   2   9 9 1 32m 1 1 （3）h(x)(x2)ex x3( )x2 x2lnx，h(x)(x1)ex x2(1m)xxlnx 6 4 2 2 h(x)xexxlnxm·····································································································10分 因为hx不存在曲率为0的点，所以hx0在(0,)无实数解，令t(x)xexxlnx 1 xex x2ex x1 (xex 1)(x1) t(x)exxex1   ，··························································12分 x x x 令(x)xex1,x0 ，求导得(x)ex(x1)0，函数(x)在(0,)上单调递增， 1 1 1 1 而( ) e10,(1)e10，则存在x ( ,1)，使(x )0，即t(x )0，此时ex0  ， 2 2 0 2 0 0 x 0 当x(0,x )时，t(x)0；当x(x ,) 时，t(x)0， 0 0 函数t(x)在(0,x )上单调递减，在(x ,)上单调递增，····························································15分 0 0 1 因此t(x) t(x ) x ex0 x lnx ，由ex0  ，得x lnx ， min 0 0 0 0 x 0 0 0 则t(x )1x x 1，m1，所以m的取值范围是(,1).·····················································17分 0 0 0 19.(本小题满分17分) （1）记选择方案二学生的得分为X P(X 100)(p p )2······································································································1分 0 1 (p  p )2 1 1 p p  0 1  ，当且仅当 p  p  时取等，························································2分 0 1 4 4 0 1 2 1 此时满分概率为 ···········································································································3分 16 （2）在方案一中，记A类题得分为Y ,B类题得分为Y ,选择方案一学生得分为Z 1 2 1 1 1 1 7 P(Y 50)   ,P(Y 0)1  1 2 4 8 1 8 8 1 1 1 1 1 1 1 1 1 P(Y 0)   ,P(Y 25)C1   ,P(Y 50)   2 2 2 4 2 2 2 2 2 2 2 2 4 由题可知Z 可能的取值为0、25、50、75、100 7 1 7 7 1 7 7 1 1 1 1 P(Z 0)   ,P(Z 25)   ,P(Z 50)     ， 8 4 32 8 2 16 8 4 8 4 4 1 1 1 1 1 1 P(Z 75)   ,P(Z 100)   ·································································7分 8 2 16 8 4 32 高二数学 第 3 页 共 4 页 {#{QQABTQYg4gi4wBbACY4LUQU4CwoQsJMjLcokwQASKAQKCRFABKA=}#}Z 的分布列为 Z 0 25 50 75 100 ···················································································8分 7 7 1 1 1 P 32 16 4 16 32 （3）方法一： 记A、B类题得分分别为事件A、B,当A类题只需答对一道得50分为事件A' 1 由2p q 1(p q )可得0 p  ··············································································9分 0 0 0 0 0 3 E(Z)E(A2B)E(A)2E(B),E(X)E(A)E(A') 故E(Z)E(X)2E(B)E(A')······················································································11分 E(B)25q 0(1q )25q 0 0 0 E(A')501(1 p )(1 p )0(1 p )(1 p )5012(1 p )p q  ·································13分 0 1 0 1 0 0 0 所以E(Z)E(X)2E(B)E(A')50q 50100(1 p )p q 0 0 0 0 50(12p )50100(1 p )p (12p )100(1 p )p (12p )100p 0 0 0 0 0 0 0 0 100p (1 p )(12p )1100p 2(2p 3)0 0 0 0 0 0 所以E(Z)E(X)应选择方案二·······················································································17分 方法二： 1 由2p q 1(p q )可得0 p  ··············································································9分 0 0 0 0 0 3 若选择方案一，学生得分Z 可能的取值分别是0、25、50、75、100 P(Z 0)(1 p p )(1q )2,P(Z 25)(1 p p )C1q (1q ) , 0 1 0 0 1 2 0 0 P(Z 50) p p (1q )2(1 p p )q 2,P(Z 75) p pC1q (1q ) 0 1 0 0 1 0 0 1 2 0 0 P(Z 100) p pq 2,所以E(Z)50q 50p p ································································11分 0 1 0 0 0 1 若选择方案二，学生得分X 可能的取值分别是0、50、100 P(X 50) p p (1 p )(1 p )(1 p p )1(1 p )(1 p ) ， 0 1 0 1 0 1 0 1 P(X 100) p p 1(1 p )(1 p ) ， 0 1 0 1 所以E(X)5050p p 50(1 p )(1 p ) ······································································13分 0 1 0 1 所以E(Z)E(X)50q 5050(1 p )(1 p )50q 50100(1 p )p q 0 0 1 0 0 0 0 50(12p )50100(1 p )p (12p )100(1 p )p (12p )100p 0 0 0 0 0 0 0 0 100p (1 p )(12p )1100p 2(2p 3)0 0 0 0 0 0 所以E(Z)E(X)应选择方案二·······················································································17分 高二数学 第 4 页 共 4 页 {#{QQABTQYg4gi4wBbACY4LUQU4CwoQsJMjLcokwQASKAQKCRFABKA=}#}