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永州市 2025年上期高一期末质量监测试卷
数学参考答案及评分标准
一、单项选择题
题号 1 2 3 4 5 6 7 8
答案 B A C D B C D C
二、多项选择题
题号 9 10 11
答案 BD ABC ABD
三、填空题

12．15π 13． 14．

部分小题的解析：
8． 解析：在锐角△ABC中, 2S ac sin A c2 sinC ,
所以 ac sin B ac sin A c2 sin c ,即 a sin B a sin A csinC
由正弦定理有 ab a2 c2，又由余弦定理 c2 a2 b2 2abcosC
ab a2 a2 b2 2abcosC， a 2a cosC b
由正弦定理有 sin A 2sin AcosC sin B sin(A C) sin AcosC cos AsinC
sin A sinC cos A cosC sin A sin(C A)
△ABC，所以 A C A，即C 2A

0 A


2
又锐角△ABC 中， 0 B A C 3A

，
2
0 C 2A
2
3
A tan A ,1
6 4 3
tan(C 2 2
7 3
A) tan A 3, ，
tan A tan A 3
故 tan(C 2 A) 的可能取值为 4，答案选 C.
tan A
永州市 2025年上期高一期末质量监测试卷·数学参考答案及评分标准 第 1页（共 7页）
11．解析：
对于 A选项，因为平面PAD 平面ABCD，交线为 AD，
又CD 平面ABCD，CD AD，所以CD 平面PAD ，
又因为 AP 平面PAD ，所以 AP CD，
又因为 AP PD，PD CD D，PD 平面PCD，CD 平面PCD，
所以 AP 平面PCD ,故 A正确；
对于 B选项，取 AD中点为 F ，连接 EF， BF ，又因为 E为 AP中点，
所以 EF // PD所以 BEF为 BE 与 PD所成角（或补角），
又 EF 平面 PCD PCD， PD 平面 PCD，所以 EF //平面 PCD，
又因为 BE //平面 PCD，EF BE E， EF 平面 BEF , BF 平面 BEF ,
所以平面 BEF // 平面 PCD ，又平面 BEF 平面 ABCD BF ，平面 PCD 平面
ABCD CD，所以CD//BF ，又CD 平面 PAD ，所以 BF 平面 PAD ，EF 平面 PAD ，
所以 BF EF， BF AB2 AF 2 2， EF 1，所以 tan BEF
BF
2，所以 B正确
EF
对于 C选项，因为 APC 和 ADC为直角三角形，所以三棱锥 P ACD的外接球球心
3
为 AC的中点，半径为 ，所以三棱锥 P ACD的外接球表面积为 9 ，故 C错误
2
对于 D选项，连接 PF ，则 PF 为三棱锥 P ABCD的高，又 PQ ，故 QF ，
所以动点Q的轨迹是以 F 为圆心 1为半径的半圆，故 D正确；
故正确答案为：ABD
14．解析：
令CM CB，CN CA π， |CM | |CN | ,C
3
所以 CMN 为等边三角形，

| 2CB (1 ) 3 CA | | CM (1 )CN |

记CQ CM (1 )CN，M，N，Q三点共线
当CQ垂直 MN时， |CQ |最小，
则 |CQ |的最小值为 .
永州市 2025年上期高一期末质量监测试卷·数学参考答案及评分标准 第 2页（共 7页）
四、解答题
15. 解析：

（1）由题可知，
a b
a b ····························································································· 2分
a 1,-2 ，b ,1 ············································································3分

a b 1 （-2） 1 0 ········································································5分
解得 2 ···························································································· 6分

（2）由 a (1， 2)， c ( 2，3)，得 a 2c ( 3,4) ··············································· 7分

ka c (k 2， 2k 3) ··········································································· 8分
a c与 ka c平行··············································································9分
a 2c (ka c ) ··············································································· 10分
3 ( 2k 3) 4(k 2) 0 ···································································12分
k ····························································································· 13分

16．解析：
（1）因为 (0.015 a a 0.030 0.003 0.002) 10 1 ··········································1分
所以 a 0.025 ·······················································································2分
参加这次测评学生成绩的众数为 75 分······················································ 3分
由所给频率分布直方图知
100名学生成绩在 ， 的频率为 0.4，在 ， 的频率为 0.65，·················· 4分
所以参加这次问卷调查学生成绩的中位数在 ， 内 ································5分
设中位数为 x，则 0.40 (x 60) 0.025 0.50，·········································· 6分
解得 x 64
所以参加这次问卷调查学生成绩的中位数为 64．·········································7分
（2）在抽查的 100名学生中，成绩在 ， 中的学生有0.003 10 100 3人，·····8分
成绩在 ， 中的学生有0.002 10 100 2人，·····································9分
记 ， 中的 3人为 1，2，3，
记 ， 中的 2人为 a，b ··································································· 10分
所有基本事件有：
永州市 2025年上期高一期末质量监测试卷·数学参考答案及评分标准 第 3页（共 7页）
1, 2 , 1,3 , 1,a , 1,b , 2,3 , 2,a , 2,b , 3,a , 3,b , a,b 共 10种，············· 12分
来自同一组的有: 1, 2 , 1,3 , 2,3 , a,b ，共 4种情况， ·························· 13分
4 2
故恰好来自同一组的概率 p . ···················································15分
10 5
17．解析：
a b c
（1）依题意，由正弦定理 ··················································· 1分
sin A sin B sinC
得 2sin A(sinC cosB sin B cosC) 3 sin A ·················································· 2分
2sin Asin(B C) 3 sin A ·····································································3分
即 2sin Asin A 3 sin A ·········································································· 4分
π
又在锐角 ABC 中，有 A

0

， ，所以 sin A 0，······································5分
2
所以 sinA 3 ，···················································································6分
2
π
所以 A ；·························································································
3 7分
π 2π
（2）结合（1）可得 A ，B C π A ，
3 3
a b c
由 a 3，则根据正弦定理有 2，··································8分
sinA sinB sinC
得b 2sinB，c 2sinC，········································································· 9分
根据余弦定理有a2 b2 c2 2bccos A，得b2 c2 3 bc，··························10分
b2 c2 bc 3 2 2bc 3 8sin B sinC 3 8sin B sin( B) ·························11分
3

3 4 3 sin B cosB 4sin 2 B 3 2 3 sin 2B 2cos2B 5 4sin(2B ) ………12分
6
又 ABC

为锐角三角形，则有 B 0,
π 2π B 0, π π π ，2 3
，得 B ， ，
2 6 2
2B 5 ， ··················································································6 6 6
13分

sin
1
2B

，1 ··············································································14分 6 2
b2 c2 bc 5 4sin 2B 故 7,9 6 ·······················································15分
永州市 2025年上期高一期末质量监测试卷·数学参考答案及评分标准 第 4页（共 7页）
18．解析：
（1）（ⅰ）连接 AC交 BD于点 O，连接 EO，··················································· 1分
因为底面 ABCD为正方形，所以点 O为 AC的中点，·································· 2分
当 PE EC 时，E为 PC中点，所以 PA∥OE ············································· 3分
因为 PA 平面 BDE，OE 平面 BDE
PA∥平面 BDE ··················································································5分
o
（ⅱ）由（ⅰ）知 PA ∥平面 BDE，
所以点 A到平面 BDE的距离即为点 P到平面 BDE的距离····························· 6分
PA 底面 ABCD， PA∥OE
EO 底面 ABCD EO AO，······························································ 7分
因为底面 ABCD为正方形，所以 AO BD，所以 AO 平面 BDE ····················8分
所以 AO即为点 A到平面 BDE的距离，
AO 1 AC 2 ················································································· 9分
2
所以点 P到平面 BDE的距离为 2，························································10分
（2）当 PE 2EC时，点 E为 PC上靠近于 C的三等分点，································· 11分
连接 EO， PC PA AC ，EC ，··································· 12分

又 PB PD，BC DC，PC PC ， PBC PDC，点 E在公共边 PC上···· 13分
BE DE ， EO BD又OC BD
EOC为二面角 P BD C 的平面角···················································· 14分
Rt PAC cos PCA AC 2 2 6在 中， ，
PC 2 3 3
永州市 2025年上期高一期末质量监测试卷·数学参考答案及评分标准 第 5页（共 7页）
在 EOC中，在由余弦定理得：
EO2 OC 2 EC 2 2OC 4 EC cos PCA 2 2 2 2 3 6 2 ··········· 15分
3 3 3 3
2
EO2 OC 2 EC 2 2
4

cos EOC 3 3 3
2EO OC 6 3 ······································16分2 2
3
所以二面角 P BD C 3的余弦值为 ·····················································17分
3
（其它非空间向量方法酌情给分）
19．解析：
2 3
（1）记“接收到的两个数字中有且只有一个正确”为事件A ,由已知 p1 ，p2 ，····1分3 4
事件 A包含两种情况：
0 2 3第一种数字 接收正确数字 1错误，概率为： (1 ) ································· 2分
3 4
第二种数字 0接收错误数字 1正确，概率为： (1 2 3 ) ··································3分
3 4
所以 P(A)
2
(1 3 ) (1 2 ) 3 5 ；·························································· 4分
3 4 3 4 12
（2）（i）由发送的数据为“011”可知，事件 n(X ) 2表示接收到的数据中含两个 0，
包含两种情况：①数字 0接收正确，数字 1有一个正确一个错误，
②数学 0错误，数字 1都错误，·········································5分
所以 P(n(X ) 2) 2p1 p2 (1 p2 ) (1 p1 )(1 p
2
2 ) ·················································6分
事件 n(X ) 3表示接收到的数据中含三个 0，
只有 1种情况：数字 0接收正确数字 1都错误，·········································7分
所以 P(n(X ) 3) p1 (1 p
2
2 ) ······································································ 8分
由 2P n X 2 13P n X 3 得：
2 2p1p2 (1 p2 ) (1 p1 )(1 p
2
2 ) 13p (1 p
2
1 2 )
化简得19p1p2 2 15p1 2p2 0································································· 9分
p 1
1
又因为 1 p2 ， p1 2p ，上式可化为：2 2
永州市 2025年上期高一期末质量监测试卷·数学参考答案及评分标准 第 6页（共 7页）
3
4p 22 23p2 15 0 p2 或 p2 5(舍去）···············································10分4
（ⅱ）当发送的数据为“0101”，事件 n(X ) 2包含以下三种情况：
1
①两个 1传输都正确，且两个 0传输都正确，其概率为 p 2 p 21 2 ··················· 11分4
②有且只有一个 1传输正确，且有且只有一个 0传输正确，
其概率为 2p1(1 p1) 2p2 (1 p2 ) 4p1p2 (1 p1 p2 p1p2 ) 3 2(p1 p2 )，··············· 12分
③两个 1传输都错误，且两个 0传输都错误，其概率为
2
(1 p )2(1 p )2 (1 p p p p 2 3 ，······································ 13分1 2 1 2 1 2)

( p p )
2 1 2
1 3 2
所以 P(n(X ) 2) 3 2( p 1 p2 ) ( p1 p2 ) ，······································· 14分4 2
p p 1 p 1令 x p1 p2，则 1 2 ， 2 ,又0 p1 1且0 p
1
2 1， p1 12 2p1 2
x p p

p p
, ·····························································15分

所以
1 3 2 11
P(n(X ) 2) 3 2( p p ) ( p p ) ( p p 2
4 1 2 2 1 2
1 2 ) 5( p1 p2 )
2
，
2

5 3
(p p )

1 2
2

4
5 2 3 x , 记 f (x) (x ) ， ·····························································16分2 4

由二次函数的性质可知， f (x)在 ， 单调递减，
所以 f (x)
15
得最大值为 f ( 2) 5 2，
2
即 P(n(X ) 2)
15
的最大值为 5 2 ．··························································· 17分
2
永州市 2025年上期高一期末质量监测试卷·数学参考答案及评分标准 第 7页（共 7页）

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