湖北省武汉市七彩联盟2024-2025学年度下学期高一年级6月联考数学答案_2024-2025高一（7-7月题库）_2025年7月_250702湖北省武汉市七彩联盟2024-2025学年度下学期高一年级6月联考

七彩联盟高一期末考试数学答案 2025.06.25
1 2 3 4 5 6 7 8
单选
A C C B B D A B
9 10 11
多选
BCD ACD AC
3 
填空 12.【答案】100 13.【答案】 13.【答案】2 2
2 2
1
15.【答案】（1）a ,b1； （2）a4,b3
2
【详解】（1）解法一：由已知有，方程az2bz10的两根为1i，··························2分
 b
 1i1i2

 a
由根与系数的关系得 ····························································5分
 1 1i1i2
 a
1
a ,b1··································································································6分
2
解法二：由已知有a1i2 b1i10，····························································2分
 b10
b12abi0由复数的相等得 ，···················································5分
2ab0
1
a ,b1··································································································6分
2
（2）设zabia,bR，················································································7分
3a a2 b2 7
3abi a2 b2 79i，由复数的相等有 ，································9分
 3b9
解得a 4,b3.······························································································13分
5
（注：a 是增根，未舍去的扣2分）
4
16.【答案】（1）证明见解析； （2）2 2
5
【详解】（1）证明：在ABC中，∵AB2AC2 BC2 ，∴AC  AB，·························2分
在直三棱柱ABCABC 中，AA 平面ABC，∴AA AC，
1 1 1 1 1
∵AA ,AB是平面AABB内两条相交直线，
1 1 1
∴AC 平面AABB，∴AC  AB. ····································································5分
1 1 1
又知四边形AABB是正方形，∴AB AB.
1 1 1 1
∵AC,AB 是平面ABC内两条相交直线，∴AB 平面ABC. 得证.·····························8分
1 1 1 1（2）在直三棱柱ABCABC 中，BC //BC，
1 1 1 1 1
∴BC 与平面ABC所成的角等于BC平面ABC所成的角，设大小为，·······················10分
1 1 1 1
设AB ABO，连接OC.
1 1
由（1）知，BO平面ABC，∴BC平面ABC所成的角为BCO，·····························13分
1 1
BO 2 2
在RtABC中，sinsinBCO   . ·························································15分
BC 5
17.【答案】（1）a0.012； （2）93；（3）792
【详解】（1）解法一：设成绩在90,150的频率为p，则成绩在30,90的频率为1 p，
根据题目x 65，x 115，平均成绩 x97 ，
1 2
有p1151 p6597，解得p0.64；···························································· 2分
则根据频率分布直方图有0.014a0.006200.64，
解得a0.012. ································································································4分
解法二：根据频率分布直方图以及x 115，得
2
0.014 a 0.006
x  100 120 140115，···············2分
2 0.014a0.006 0.014a0.006 0.014a0.006
解得a0.012. ································································································4分
（2）设获得优胜奖的成绩为Y 分，
易计算得频率分布直方图成绩在90,110,110,130,130,150的频率分别为0.28、0.24、0.12；
则优胜奖成绩Y 位于90,110中，·········································································6分
Y 90 0.640.60 0.640.60 20
由此有  ，解得Y 90 90 93，
11090 0.01420 0.014 7
故以样本估计总体，估计获得优胜奖的成绩为93分. ···············································9分
（3）样本方差s2
10.64
x x
2
s2

0.64

x x
2
s2

，·······························12分
 1 1  2 2
代入有s2 0.36 65972200 0.64 115972225 351.36440.64792，
   
则样本的方差s2 792. ·····················································································15分
（注：为降低难度，减小运算量，本题增加了条件）
  
18.【答案】（1）A ,C ； （2） AD 2 31
3 4
 1
S  bcsinA3 3
【详解】（1）由已知得 ABC 2 ，················································2分
 

 ABAC bccosA6两式相除得tanA 3，

又0 A，∴A ；···················································································· 4分
3
cosA cosB cosC
又已知3  2 ，（下用两种方法求C，任一种都算对）
a b c
解法一：根据余弦定理有
b2c2a2 a2c2b2 a2b2c2
3  2 ，化简得2a2 3c2，∴ 2a  3c················· 7分
2abc 2abc 2abc
2 2
由正弦定理得，sinC  sinA ，
3 2
2 
又∵0C  ，∴C . ················································································9分
3 4
cosA cosB cosC
解法二：由正弦定理得，3  2 ()，
sinA sinB sinC

∵A ，
3
3tanC
∴tanB tan

AC

tan AC  ，代入()，
1 3tanC
1 3tanC 1
得 3 2 ，化简得tan2C1.·······················································7分
3tanC tanC
2 
又0C  ，∴C . ···················································································9分
3 4
  5
（2）∵A ，C  ，∴BAB ，
3 4 12
sinB 31
∴b c c；
sinC 2
 
又由（1）得bc12，∴c2 12 31 ，······························································· 11分
2
∴b2    31 c   ，化简得4b2   31 2 c2 24  31  ，即b2 6  31  ；·············13分
 
 2 
    
∵BD 2DC，∴3AD 2AC AB，·································································15分
两边同时平方有
      
2  2 2 2
9 AD  2ACAB 4 AC  4ABAC AB
   
4b2 46c2  24 31 2412 31 ，
 
36 31

化简得， AD 2 31.················································································· 17分  m     1 
19.【详解】（1）证明：∵OP mPM  OM ，由NRtRO知OR ON .
1m 1t
   m  1
又P,Q,R三点共线，∴OQOP1OR OM  ON①，
1m 1t
   1  n 
由MQ nQN 知OQ OM  ON ②，
1n 1n
 m 1 1 m
 
    1m 1n 1nm
∵OM,ON 不共线，由①②及平面向量基本定理有 ，
1 n 1tn
 1

1t 1n 1n
1m 1tn
∴1  ，化简得mnt 1. 得证. ·····················································5分
1nm 1n
（2）①证明：∵AC//，平面ACB经过AC且与平面相交于EG，
∴AC//EG.
又E为AB的中点，
∴G为BC中点.······························································································ 7分
∵F 为CD中点，∴FG//BD.
∵GF ,BD
∴BD//. 得证.····························································································· 9分
②由已知有AC,EG,TF三线相交于点K.
AE BG CK CK AT DF
由（1）有   1，   1,
EB GC KA KA TD FC
∵E,F分别为AB，CD中点，
BG AT
∴  .···································································································11分
GC TD
CG DT
记  ，四面体ABCD的体积为V ，多面体ETBGFD的体积为V'.
CB DA
V BE S 1 S  1 
连接EF,ED，则有 EBGFD   四边形BGFD  1 CGF  1 ，···························14分
V AB S 2 S  2 2
BCD BCD
V AE S 1 DTDF 
EFDT   FDT    ，··································································· 16分
V AB S 2 DADC 4
DAC
V' V V 1
∴  EBGFD EFDT  .
V V 2
即平面平分四面体ABCD的体积. 得证. ·····························································17分