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2023一2024学年度第二学期教学质量检查
高一数学
一、单项选择题：本大题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一
项是符合题目要求的.请把正确选项在答题卡中的相应位置涂黑.
1.已知向量a=(2,-1),b=(1,2)(eR),若a6,则1=
A.2
C.-2
2.为了解学生每日参加体育锻炼的情况，学校用比例分配的分层随机抽样方法从高一、高二、
高三年级所有学生中抽取部分学生做抽样调查，已知该学校高一、高二、高三年级学生人数的比
例如图所示，若抽取的样本中高三年级的学生有36人，则抽取的样本容量为
高二
高王
30%
30%
高一
40%
A.90
B.100
C.120
D.160
3.棱长为a的正方体的顶点都在球面上，则球的表面积为
3
A.
B.3m2
C.6m2
D.12m2
4.若(2+i)z=1-i，则zz=
c
8
B.2
D.
5.已知m,n为两条不同的直线，a,B为两个不同的平面，则下列命题正确的是
A.若m/1a,nca,则m//n
B.若mCa,ncB,且m⊥n,则a⊥B
C.若a/1B,nca,则n11B
D.若a⊥B,mca,则m⊥B
6.已知向量O1=a,O丽=五，且同=园=a-万=3，任意点M关于点A的对称点为S,点S关
于点B的对称点为N,则M网=
A.45
B.6
C.25
D.3
7.已知三棱锥P-ABC,PA⊥平面ABC,AB⊥BC,PA=AB=2BC,则异面直线PB与
AC所成角的余弦值为
A.
1-5
.10
②
C.
D.5
5
高一数学第1页（共4页）
器国益
0000000
8.一枚质地均匀的正方体骰子，其六个面分别刻有1,2,3,4,5,6六个数字，投掷这枚骰子
两次，设事件M=“第一次朝上面的数字是奇数”，则下列事件中与M相互独立的是
A.第一次朝上面的数字是偶数
B.第一次朝上面的数字是1
C.两次朝上面的数字之和是8
D.两次朝上面的数字之和是7
二、多项选择题：本大题共3小题，每小题6分，共18分.在每小题给出的四个选项中，有多项
符合题目要求，全部选对的得6分，部分选对的得部分分，有选错的得0分.请把正确选项在答
题卡中的相应位置涂黑.
9.已知某地一周每天的最高温度（单位：℃）分别为：31、27、26、28、27、30、27，则下列关
于这组数据的结论中正确的是
A.众数是27
B.极差是4
C.中位数是28
D.平均数是28
10.已知⊙O的半径为2，△ABC为其内接三角形，则下列结论中正确的是
A.若A=花，则O丽.0C=-2
B.若A=写，则△MBC周长的最大值为65
C.若Aō.B=1,则AB=√2
D.若AO.店=1，则△ABC面积的最大值为3√3
11.如图，在棱长为1的正方体ABCD-ABCD中，点E,F分别为AB,AD的中点，平面C经
过点C,E,F,且与C,D,交于点G,则下列结论正确的是
A.AC,//平面a
B.平面L⊥平面BBF
C.CG:GD=3
D.二面角E-FG-B的正切值为25
三、填空题：本大题共3小题，每小题5分，共15分.请把答案填在答题卡的相应位置上，
12.假设P(0=,P(8)=,且A与B相互独立，则P(A)=
3
13.已知圆台的上底半径为2，下底半径为4，则经过母线中点且与底面平行的平面将圆台分成
上下两部分的体积之比为
14.已知圆0的半径为1，点A是圆O上的动点，BB2…B224为圆0内接正2024边形，则
OA+0B,+0B2++0B2024日
一，B2+瓜2++B242=
一2
高一数学第2页（共4页）
00000002023－2024学年度第二学期教学质量检查
高一数学 参考答案
一、单项选择题
题号 1 2 3 4 5 6 7 8
答案 D C B A C A B D
二、多项选择题（全部选对的得 6分，部分选对的得部分分，有选错的得 0分）
题号 9 10 11
答案 AD ABC BCD
三、填空题（14题第一空 2分，第二空 3分）
1
12. 13. 19：37 14. 1，4048
3
四、解答题
a b c
15. 解：（1）在三角形 ABC中由正弦定理有： t，················· 1分
sin A sinB sinC
则a t sin A,b t sinB,c t sinC，··································································· 2分
代入 acosC 3csin A可得 sin AcosC 3 sinC sin A，········································3分
又因为 A 0, ,所以 sin A 0，······································································· 4分
所以 cosC 3 sinC，······················································································5分
3
所以 tanC ，·····························································································6分
3
又因为C 0, ，所以C .·········································································· 7分
6

【解法 1】因为C ,由余弦定理有 c2 a2 b2 2abcosC 1，···························8分
6
a2 2 3a 3 0，·························································································· 9分
解得 a 3，································································································· 10分
{#{QQABIYSQoggIAJBAAQhCAQUqCkAQkAGAAYgOBEAEMAAAQQFABAA=}#}
S 1所以 AC BC sinC，··············································································· 11分
2
1
所以在三角形 ABC中， S AC BC sinC 1 2 3 sin 3 .····················13分
2 2 6 2
b c
【解法 2】因为C ,由正弦定理有 ，·············································· 8分
6 sin B sinC
解得 sin B 1，································································································· 9分
因为B 0, ，所以 B ，·········································································· 10分
2
所以 a2 b2 c2 3， a 3，········································································11分
ABC S 1 AB BC 3所以在三角形 中， .······················································ 13分
2 2
16.解：（1）根据频率分布直方图中小长方形的面积和等于 1，
得 0.005 a 0.0175 0.0075 20 1，解得： a 0.0125，·········································2分
设第 35百分位数为 x ，则0.005 20 (x 20) 0.0125 35％，·····································4分
解的 x 40 .······································································································· 5分
所以可以允许车辆免费停车 40分钟不收费.······························································ 6分
（2）车辆平均停车时间为
t 0.005 20 10 0.0125 20 30 0.0175 20 50 0.0075 20 70 0.0075 20 90 =50，
······················································································································· 9分
s2 10 50 2 0.1 30 50 2 0.25 50 50 2 0.35 70 50 2 0.15 90 50 2 0.15 560
····················································································································· 12分
∴s 560 4 35 ≈ 24，················································································14分
所以 t0 t s 50 24 74 .··············································································15分
17.解：（1）设 2个白球为w1,w2 ，3个红球为 r1, r2 , r3，则不放回地依次摸出两个球的样本空间为：
{(w1,w2 ), (w1,r1), (w1,r2 ), (w1,r3)，(w2 ,w1), (w2 ,r1), (w2 ,r2 ), (w2 ,r3)，(r1,w1), (r1,w2 ), (r1,r2 ),
(r1,r3), (r2 ,w1), (r2 ,w2 ), (r2 ,r1), (r2 ,r3), (r3,w1), (r3,w2 ), (r3,r1), (r3,r2 )}，其中共有 20个样本点
······················································································································· 3分
{#{QQABIYSQoggIAJBAAQhCAQUqCkAQkAGAAYgOBEAEMAAAQQFABAA=}#}
设一次抽奖中奖为事件 A，则 A {(w1,w2 ), (w2 ,w1), (r1, r2 ), (r1, r3), (r2 , r1), (r2 , r3), (r3 , r1), (r3 , r2 )}，其中
共有 8个样本点，·······························································································5分
因为抽中样本空间中每一个样本点的可能性都相等，所以这是一个古典概型．
因此 P(A) 8 2 ，························································································ 6分
20 5
2
即若某顾客有一次抽奖机会，中奖的概率为 .························································· 7分
5
（2）设在第 i次抽奖时中奖为事件 Ai , i 1,2，
由于每次抽奖的情况相同，由（1）可知 P(Ai )
2
,P(A ) 3i , i 1,2, ······················9分5 5
设两次抽奖至少有一次中奖为事件 B，则 B A1A2 A1A2 A1A2 ，··························10分
其中 A1A2、 A1A2、 A1A2 为互斥事件，
则 P(B) P(A1A2) P(A1A2) P(A1A2) ，····························································11分
因为每次抽奖之间相互独立，
所以 P(A1A2 ) P(A
4
1)P(A2 ) ，····································································· 12分25
P(A1A2 ) P(A1)P(A )
6
2 ，··········································································· 13分25
P(A 61A2 ) P(A1)P(A2 ) ，···········································································14分25
所以 P(B) 4 6 6 16 ，
25 25 25 25
16
即若某顾客有两次抽奖机会，至少有一次中奖的概率为 .·······································15分
25
（备注：利用对立事件求解的相应给分）
18.解： 【解析】（1）在 PDE中，PE 1， PD 2， EPD 60 ，
由余弦定理得DE2 PE2 PD2 2 PE PDcos60 3，
所以 PE2 DE2 PD2，所以 PE DE，································································· 2分
在△PBE中， PE 1，BE 2， PB 5，
所以 PE2 BE2 PB2，所以PE BE，··································································4分
又因为BE DE E， BE、DE 平面BCDE，
所以 PE 平面BCDE，又 PE 平面PDE，····························································6分
所以平面 PDE 平面BCDE .················································································· 7分
（2）在平面BCDE中，过点E作 EF / /CD，交 BC于F ，在平面 PBC 中，过点 F 作 FM / /PC，
交PB于M ，连接ME，····················································································9分
{#{QQABIYSQoggIAJBAAQhCAQUqCkAQkAGAAYgOBEAEMAAAQQFABAA=}#}
如图所示：
因为 EF / /CD，CD 平面 PCD， EF 平面 PCD，所以 EF / /平面 PCD，
同理可得MF / /平面 PCD，·············································································· 11分
又因为 EF MF F， EF ,MF 平面MEF，
所以平面 PCD / /平面MEF，············································································13分
因为ME 平面MEF，所以ME / /平面 PCD，·················································· 14分
即M 即为所求的点，························································································ 15分
CF AE 1
因为 EF / /CD，即 EF / /AC，所以 ，
FB EB 2
PM CF 1 PM 1
又因为 FM / /PC，所以 ，即此时 .···································· 17分
MB FB 2 MB 2

19. 解：（1） a b 1 (1 i) (2 i) ( 2i) 1 i 4i 2 1 5i，·····································1分

a b (2 i, 2 i)，···························································································· 2分
a b (a b) (a b) (2 i) (2 i) (2 i) (2 i) 10，···································· 3分

（2）【解法 1】设 a (x1 y1i, x2 y2i),b (x3 y3i, x4 y 4i),c (x5 y 5i, x 6 y 6i) ，
其中 x j , y j R, j 1,2,3,4,5,6，········································································· 4分

b c (x3 x5 (y3 y5)i, x4 x6 (y4 y6 )i) ······················································ 5分
a (b c) (x1 y1i)(x3 x5 (y3 y5)i) (x2 y2i)(x4 x6 (y4 y6)i)
x1(x3 x5) y1(y3 y5) x1(y3 y5)i (x3 x5)y1i x2(x4 x6) y2(y4 y6) x2(y4 y6)i (x4 x6)y2i
x1x3 x1x5 y1y3 y1y5 x2x4 x2x6 y2y4 y2y6 ( x1y3 x1y5 x3y1 x5y1 x2y4 x2y6 x4y2 x6y2)i
a b (x1 y1i)(x3 y3i) (x2 y2i)(x4 y4i)
x1x3 y1y3 x1y3i x3y1i x2x4 y2 y4 x2 y4i x4 y2i ·······································7分
x1x3 y1y3 x2x4 y2 y4 ( x1y3 x3y1 x2 y4 i x4 y2 )i

同理 a c x1x5 y1y5 x2x6 y2 y6 ( x1y5 x5y1 x2 y6 i x6 y2 )i
{#{QQABIYSQoggIAJBAAQhCAQUqCkAQkAGAAYgOBEAEMAAAQQFABAA=}#}
所以
a

b a c
x1x3 x1x5 y1y3 y1y5 x2x4 x2x6 y2y4 y2y6 ( x1y3 x1y5 x3y1 x5y1 x2y4 x2y6 x4y2 x6y2)i
······················································································································· 8分

所以 a (b c ) a b a c ················································································· 9分

【解法 2】设 a z1, z2 ，b z3 , z4 ， c z5 , z6 ， z1、 z2、 z3、 z4、 z5、 z6 C，
······················································································································· 4分

b c

z3 z5 , z4 z

6 ， a b z1 z3 z2 z4 ，a

c z1 z5 z2 z6
a

(b c) z1(z3 z5) z2 (z4 z6)，································································ 7分
因为 z3 z5 z3 z5 ， z4 z6 z4 z6 ，···························································8分

所以 a

(b c ) z1 z3 z5 z2 z4 z6

z1 z3 z1 z5 z2 z4 z2 z z z

6 1 3 z2 z4 z1 z5 z2 z6 a b a c，结论正确··················· 9分

（3）设 (m ni,m 2 ni)，m,n R则 (2 m (2 n)i, m ni)，······ 11分

[2 m (2 n)i] [2 m (2 n)i] ( m ni) ( m ni) ,························12分
(2 m)2 (2 n)2 ( m)2 n2 2m2 2n2 4m 4n 8
2(m 1)2 2(n 1)2 4 2当且仅当m n 1时，等号成立,·································· 13分

故 的最小值为 2 .····················································································· 14分

此时 (1 i,3 i)则 (1 i, 1 i) ,·························································· 15分

设 (s ti, s 2 ti)， s, t R，则 (1 s (1 t)i,1 s (1 t)i)，············· 16分

则 ( ) ( ) (1 i)[1 s (1 t)i] ( 1 i)[1 s (1 t)i] 0，结论成立.········17分
{#{QQABIYSQoggIAJBAAQhCAQUqCkAQkAGAAYgOBEAEMAAAQQFABAA=}#}

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