资源简介

福州三中 2023-2024 学年第二学期期末考
高一数学 参考答案及评分标准
一、选择题：本题共 8 小题，每小题 5 分，共 40 分.在每小题给出的四个选项中，只有一个选项是正
确的．
1. D 2. C 3. B 4. C 5. A 6. C 7. C 8. C
二、选择题：本题共 3 小题，每小题 6 分，共 18 分. 在每小题给出的选项中，有多项符合题目要求.
全部选对得 6 分，部分选对的得部分分，有选错的得 0 分.
11. AC 12. ABD 13. AC
三、填空题：本题共 3小题，每小题 5分，共 15分.
2
14. 12. 25.1 13. ( 2，3)
2
四、解答题：本题共 6 小题，共 77 分. 解答应写出文字说明、证明过程或演算步骤.
15．（本小题满分 10 分）
z 1 m
已知 z 是复数， z +2i 和 均为实数， z1 = z + i（m R）．
1 i m m 1
（1）求复数 z 的共轭复数 z ；
（2）若复数 z1 在复平面内对应的点在第一象限，求m 的取值范围．
解：（1）设 z = a+bi(a,b R) ，············································································ 1 分
则 z + 2i = a+ (b+ 2)i，
因为 z +2i 为实数，所以b + 2 = 0，解得，b = 2， ··················································· 2 分
z a 2i (a 2i)(1+ i) a + 2 a 2
由 = = = + i为实数，
1 i 1 i (1+ i)(1 i) 2 2
a 2
可得， = 0 ，解得， a = 2， ·········································································· 3 分
2
所以 z = 2 2i， ······························································································· 4 分
所以 z = 2+2i； ······························································································· 5 分
1 m 1 m 2m +1 3m 2
（2）由（1）可知， z1 = z + i = 2+ (2+ )i = i， ·················· 7 分
m m 1 m m 1 m m 1
因为复数 z1 在复平面内对应的点在第一象限，
2m +1
0， m
所以 ····························································································· 8 分
3m 2 0，
m 1
1
m 或m 0,
所以 2 ························································································· 9 分
2 m 1,
3
2 2
解得 m 1，故实数m 的取值范围为 ( ,1)． ····················································· 10 分
3 3
16.（本小题满分 12 分）
在△ABC 中，角 A， B ，C 的对边分别为 a，b ， c ， 2bcosC c = 2a．
（1）求 B 的大小；
19
（2）若 a = 3，且 AC 边上的中线长为 ，求△ABC 的面积．
2
解法一：（1）因为 2bcosC c = 2a，
{#{QQABZYyUggCgQJAAAQgCEQWqCAMQkAGACYgORAAMMAAAwBFABAA=}#}
a2 + b2 c2
由余弦定理得 2b c = 2a ， ·································································· 1 分
2ab
化简得 a2 + c2 b2 = ac ， ·················································································· 2 分
a2 + c2 b2 1
所以 cos B = = ， ············································································· 3 分
2ac 2
因为 B (0,π)， ······························································································· 4 分
2π
所以 B = ； ·································································································· 5 分
3
2π
（2）在△ABC 中，由 （1）可得， B =
3
所以由余弦定理可得b2 = a2 + c2 + ac = c2 + 3c + 9①， ················································· 7 分
a2 + b2 c2
又 cosC = ②， ············································ 8 分 A
2ab
取 AC 的中点D，连接 BD， D
b22
BC2 +CD2 2
a + 19
BD
在 CBD中， cosC = = 4 ③， ····· 9 分
2BC CD ab
B C
由②③得 2c2 b2 =1④，
由①④得 c2 3c 10 = 0，解得 c = 5或 c = 2（舍去），
所以 c = 5 ， ·································································································· 11 分
1 1 3 15 3
所以 S ABC = acsin B = 3 5 = ． ························································· 12 分
2 2 2 4
解法二：（1）由正弦定理和 2bcosC c = 2a可得， 2sin BcosC sinC = 2sin A ··················· 1 分
在△ABC 中，由 A+ B +C = π可得， sin A= sin(B+C)，
所以 2sin BcosC sinC = 2sin BcosC + 2sinC cos B ······················································· 2 分
又因为C (0,π)，所以 sinC 0
1
所以 cos B = ， ····························································································· 3 分
2
2
又因为 B (0,π)，所以 B = ； ·········································· 5 分
3 A E
（2）延长 BD使得 BD = BE，连接 AE,CE，则 ABCE 为平行四边形．
···················································································· 7 分
D
CE = AB = c ， BE = 19 ， ·················································· 8 分
2π π
由（1）知 B = ，所以 BCE = ， ···································· 9 分
3 3
B C
在△BCE 中，由余弦定理可得：
π 2
a2 + c2 2accos = ( 19 ) ················································································· 10 分
3
因为 a = 3，所以 c2 3c 10 = 0，解得 c = 5或 c = 2（舍去），
所以 c = 5 ， ·································································································· 11 分
1 1 3 15 3
所以 S ABC = acsin B = 3 5 = ． ························································· 12 分
2 2 2 4
17．（本小题满分 12 分）
小明从一幅扑克牌中挑出 J 和K 共 8 张牌（J 和K 各四个花色：红桃（红色） 方块（红色） 黑
桃（黑色） 梅花（黑色））．现从这 8 张牌中依次取出 2 张，抽到一张红色 J 和一张黑色K 即为游戏
获胜．现有三种游戏方式，如下表：
游戏方式 方式① 方式② 方式③
抽取规则 有放回依次抽取 不放回依次抽取 按颜色等比例分层抽样
{#{QQABZYyUggCgQJAAAQgCEQWqCAMQkAGACYgORAAMMAAAwBFABAA=}#}
获胜概率 p p2 p1 3
（1）分别求出在三种不同游戏方式下获胜的概率；
（2）若三种游戏方式小明各进行一次，第一次采取方式①，后两次采用方式②和方式③，那么
方式②和方式③按照怎样的顺序进行游戏能使得三次游戏中仅连续两次获胜的概率最大？
解：
（1）设方式①的样本空间为Ω1，方式②的样本空间为Ω2 ，方式③的样本空间为Ω3，
则 n (Ω1 ) = 8 8 = 64,n (Ω2 ) = 8 7 = 56,n (Ω3 ) = 4 4+ 4 4 = 32， ······································ 3 分
设事件 A = “抽到一张红色 J 和一张黑色K ”，
A = {（红桃 J，黑桃K ），（红桃 J，梅花K ），（方块 J，黑桃K ），（方块 J，梅花K ），（黑桃K ，红桃
J），（黑桃K ，方块 J），（梅花K ，红桃 J），（梅花K ，方块 J）}
所以 n(A) =8 ···································································································· 4 分
n (A) 8 1 n (A) 8 1 n (A) 8 1
故 p1 = = = , p2 = = = , p3 = = = . ···································· 7 分
n (Ω1 ) 64 8 n (Ω2 ) 56 7 n (Ω3 ) 32 4
（2）按方式①③②抽取概率最大.
设事件“按照方式①，②，③进行游戏获胜”分别为 A1, A2 , A3 ，
B =“小明在三次游戏中仅连续两次获胜”，其中 A1, A2 , A3 相互独立，
1 1 1
P(A1) = p1 = ， P(A2 ) = p2 = ， P(A3) = p3 = ，
8 7 4
若按照方式①②③的顺序进行游戏，
则 B = A1A2 A3 + A1A2 A3 ， A1A2 A3, A1A2A3 互斥， ···························································· 8 分
所以 P(B) = P(A1A2 A3) + P(A1A2A3)
= P(A1)P(A2)P(A3) + P(A1)P(A2)P(A3)
1 1 3 7 1 1 5
= + = ， ···································································· 9 分
8 7 4 8 7 4 112
若按照方式①③②的顺序进行游戏，
则 B = A1A3 A2 + A1A3A2 ， A1A3 A2 , A1A3A2 互斥， ·························································· 10 分
所以 P(B) = P(A1A3 A2) + P(A1A3A2)
= P(A1)P(A3)P(A2) + P(A1)P(A3)P(A2)
1 1 6 7 1 1 13
= + = ， ·································································· 11 分
8 4 7 8 7 4 224
13 5
因为 ，所以按①③②的顺序进行游戏概率最大． ······································· 12 分
224 112
18．（本小题满分 14 分）
已知某工厂一区生产车间与二区生产车间均生产某种型号的零件，这两个生产车间生产的该种
型号的零件尺寸的频率分布直方图如图所示（每组区间均为左开右闭）.
尺寸大于M 的零件用于大型机器制造，尺寸小于或等于M 的零件用于小型机器制造.
（1）若M = 60，试分别估计该工厂一区生产车间生产的 500 个该种型号的零件和二区生产车间
生产的 500 个该种型号的零件中用于大型机器制造的零件个数.
（2）若M (60,70 ， 现有够多的的自一一区生产车间与二区生产车间的零件，分别用于大型机
{#{QQABZYyUggCgQJAAAQgCEQWqCAMQkAGACYgORAAMMAAAwBFABAA=}#}
器 小型机器各 1000 台的制造，每台机器仅使用一个该种型号的零件．现将一区生产车间生产的零件
都用于大型机器制造，其中尺寸小于或等于M 的零件若用于大型机器制造，每台会使得工厂损失 200
元；将二区生产车间生产的零件都用于小型机器制造，其中尺寸大于M 的零件若用于小型机器制造，
每台会使得工厂损失 100 元.求工厂损失费用的估计值H (M )（单位：元）的取值范围．
解：（1）当M = 60时，一区生产车间生产的零件尺寸大于 60 的频率为
(0.020+ 0.024+ 0.020+ 0.020) 10 = 0.84， ································································ 2 分
则该工厂一区生产车间生产的 500 个该种型号的零件用于大型机器中的零件个数为
500 0.84 = 420； ····························································································· 3 分
二区生产车间生产的零件尺寸大于 60 的频率为 (0.024+ 0.016) 10 = 0.4， ······················· 5 分
则该工厂二区生产车间生产的 500 个该种型号的零件用于大型机器中的零件个数为
500 0.40 = 200 . ································································································ 6 分
（2）当M (60,70 时一区生产车间生产的零件尺寸小于或等于M 的频率为
0.004 10+ 0.012 10+ 0.02 (M 60) = 0.02M 1.04 . ··················································· 8 分
二区生产车间生产的零件尺寸大于M 的频率为
0.024 (70 M )+ 0.016 10 =1.84 0.024M . ···························································· 10 分
故 H (M ) = (0.02M 1.04) 200 1000+ (1.84 0.024M ) 100 1000
=1600M 24000 . ················································································ 12 分
因为M (60,70 ，所以H (M )的取值范围是 (72000,88000 （单位：元）. ···················· 14 分
19．（本小题满分 15 分）
如图，四边形 ABCD是边长为 1 的正方形，四边形 ABEF 是等腰梯形， AB / /EF, AF =1，平面
3
ABCD ⊥平面 ABEF ，三棱锥 A BCE 的体积为 .
12
（1）求点 E 到平面 ABCD的距离；
（2）设G 是棱 CD上一点，若二面角G AE B 的正切值是 3，求CG ．
解：（1）设点 E 到平面 ABCD的距离为 h， ···················· 1 分
1 F E
则VA BCE =VE ABC = h S△ABC ······································· 2 分
3
1
因为四边形 ABCD是边长为 1 的正方形，所以 S△ABC = ， 3 分
2
3 A B
又因为三棱锥 A BCE的体积为 ，
12
D C
1 3 3
所以 h = ，解得， h =
6 12 2
3
即点 E 到平面 ABCD的距离为 ； ······································································ 5 分
2
（2）如图，过 E 作 EH ⊥ AB与 AB 的延长线交于H 点， F E
因为平面 ABCD ⊥平面 ABEF ，且
平面 ABCD 平面 ABEF = AB， EH 平面 ABEF ，
所以 OEH ⊥平面 ABCD， ······································· 7 分
3
由（1）知， EH = , A I B H
2
又因为在等腰梯形 ABEF 中， BE =1，
D G C
1
所以由勾股定理可得， BH = ,
2
3
所以 AH = ，
2
{#{QQABZYyUggCgQJAAAQgCEQWqCAMQkAGACYgORAAMMAAAwBFABAA=}#}
在直角三角形 AHE 中，由勾股定理得， AE = 3,
所以 EAB = 30 . ······························································································ 9 分
过G 作GI ⊥ AB ，垂足为 I ，作 IO ⊥ AE ，垂足为O ，
因为平面 ABCD ⊥平面 ABEF ，且
平面 ABCD 平面 ABEF = AB，GI 平面 ABCD，
所以GI ⊥平面 AEB ,
又因为 IO, AE 平面 AEB，所以GI ⊥ IO ，GI ⊥ AE ，
又因为 IO ⊥ AE ，GI IO = I ，所以 AE ⊥平面GIO，
又GO 平面GIO，所以 AE ⊥GO ，
所以 GOI 为二面角二面角G AE B 的平面角， ·················································· 11 分
GI
所以 tan GOI = = 3， ·················································································· 12 分
IO
设CG = x，
因为GI ⊥ AB ，所以GI∥BC ,
所以 BI =CG = x ， AI =1 x ，
又因为 IO ⊥ AE ， EAB = 30 ,
1
所以OI = (1 x) , ··························································································· 13 分
2
1
所以 = 3， ··························································································· 14 分
1
(1 x)
2
1
解得， x = ,
3
1
所以CG = . ·································································································· 15 分
3
20．（本小题满分 14 分）
点 A是直线 PQ外一点，点M 在直线 PQ上（点M 与P,Q 两点均不重合），我们称如下操作为“由A点
AP sin PAM
对 PQ施以视角运算”：若点M 在线段 PQ上，记 (P,Q;M ) = ；若点M 在线段 PQ外，记
AQ sin MAQ
AP sin PAM
(P,Q;M ) = .
AQ sin MAQ
（1）若M 在正方体 ABCD A1B1C1D1 的棱 AB 的延长线上，且 AB = 2BM = 2 ，由 A1对 AB 施以视角运
算，求 (A, B;M )的值；
（2）若M 在正方体 ABCD A1B1C1D1 的棱 AB 上，且 AB = 2，由 A1对 AB 施以视角运算，得到
1 AM
(A, B;M ) = ，求 的值；
2 MB
（3）若M1,M2 ,M3, ,Mn 1是△ABC 边BC的 n(n 2,n N)等分点，由 A对BC施以视角运算，求
(B,C;M k ) (B,C;M n k ) (k =1,2,3, ,n 1)的值.
解：（1）如图 1，
因为 AB = 2BM = 2 ，所以 AM = 3, A1B = 2 2, A1M = 13 .
由正方体的定义可知 AA1 ⊥ AB，则 A1AB = 90 ，
2 2
故 sin AA B = , cos AA B = ， ···················· 1 分 1 1
2 2
3 13 2 13
sin AA M = , cos AA M = . ··················· 2 分 1 1
13 13
{#{QQABZYyUggCgQJAAAQgCEQWqCAMQkAGACYgORAAMMAAAwBFABAA=}#}
因为 BA1M = AA1M AA1B，
26
所以sin BA1M = sin AA1M cos AA1B cos AA M sin AA B = ， ··························· 3 分 1 1
26
3 13
2
A Asin AA M
则 (A, B;M ) = 1 1 = 13 = 3 . ········· 4 分
A1Bsin MA1B 262 2
26
（2）如图 2，设 AM = a (0 a 2)，
a a2 + 4 2 a2 + 4
则 sin AA1M = ,cos AA M = . ··········· 5 分 1
a2 + 4 a2 + 4

因为 BA1M = AA1B AA1M = AA1M ，
4
π (2 a) 2(a
2 + 4)
所以 sin BA1M = sin( AA M ) = ， ················································· 6 分 1
4 2(a2 + 4)
a a2 + 4
2
A Asin AA 21 1M a + 4 a 1(A, B;M ) = = = = 2
则 A Bsin MA B ( ) ( 2 ) 2 a 2 ，解得a = ， ················· 7 分 1 1 2 a 2 a + 4 3
2 2
2(a2 + 4)
AM a 1
故 = = . ···························································································· 8 分
MB 2 a 2
（3）证明：如图 3，
因为M1,M2 ,M n3, ,Mn 1是BC的 等分点，
k n k
所以BM k =CMn k = BC, BM n k =CM k = BC . ······················ 10 分
n n
BM k AB
在 ABMk 中，由正弦定理可得 = ， sin BAM k sin AM k B
则 ABsin BAMk = BMksin AMk B . ······································································ 11 分
在 ACM k 中，同理可得 ACsin CAMk =CMksin AMkC .
因为 AMk B+ AM kC = π，所以sin AMk B = sin AMkC，
ABsin BAM BM sin AM B BM k
则 (B,C;M ) =
k = k kk =
k = . ······································ 12 分
ACsin CAM k CM ksin AM kC CM k n k
BM n k n k
同理可得 (B,C;M n k ) = = . ··································································· 13 分
CM n k k
k n k
故 (B,C;M k ) (B,C;M n k ) = =1(k =1,2,3, ,n 1) ······································· 14 分
n k k
{#{QQABZYyUggCgQJAAAQgCEQWqCAMQkAGACYgORAAMMAAAwBFABAA=}#}福州三中2023-2024学年第二学期期未考试卷
高一数学
注意事项：
1。蓓题前，考生务必将自已的班级、准考证号、姓名填写在答题卡上」
2.第I意每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑；如需改动，用橡皮擦千
净后，寻选涂其他答案标号、第Ⅱ卷必须用0.5毫米黑色签字笔书写作答.若在试题卷上作答，答案
无效
第I卷
一、选择题：本题共8小题，每小题5分，共40分在每小题给出的四个选项中，只有一个选项是正确的。
1.
己知复数z满足（1+i)z=1-i,则z22=
A.1
B.-1
C.-i
D.1
2
已知a,b是不共线的向量，B=a+b,AC=+2b,CD=3a+25.若B,C,D三点共线，则m=
A
C5
2
D
3.
已知4，.I是三条不同的直线，x,B是两个不同的平面，且cx,{c阝，几B=l,
则“{T”是“L‘”的
A.充分不必要条件B.必要不充分条件C.充要条件
D.既不充分也不必要条件
4.
从装有2个红球和2个黑球的口袋内任取2个球，那么互斥而不对立的两个事件是
科至少有一个黑球与都是黑球
后至少有一个黑球与都是红球
C.恰有一个黑球与拾有两个黑球
道.至少有个黑球与至少有一个红球
5.
已知圆锥的表面积为π，它的侧面展开图是·个半圆，则此圆锥的体积为
A.3π
B.9π
6.√3元
D.3
6.
某单位共有AB两部门，1月份进行服务满意度问卷调查，得到两部门服务满意度得分的频率分布
条形图如下.设A,B两部门的服务满意度得分的上四分位数分别为片1，，方差分别为心，s2,则
频率A
扇A部门
0.7
口B部'」
0.6
0.5
0.4
0.3
02
01-
2分3分
4分
5分得分
A.布>，>吃B.>乃，<5
C.<2，
S2D:%s吃
7.
已知通数/儿)=2(@r+pXo>0p=02a在区间（后g）
上单调递减，
则p=
A
B.4
C.liz
18
D.9
8.已知正四棱台ABCD-ABCD,的下底面边长为2√3，侧棱与下底面所成角的大小为45°，则该正四棱
台体积的取值范围是
A.0,v6)
B.(0,35)
C.0,4W6
D.{0,126
二、选择题：本题共3小题，每小题6分，共18分.在每小题给出的选项中，有多项符合题目要求.全部
选对得6分，部分选对的得部分分，有选错的得0分
高一数学第1页（共4页）

展开更多......

收起↑