江苏省扬州市2022-2023学年高一下学期开学考试数学试卷答案

3.0 envi 2024-11-28 4 4 804.48KB 6 页 3知币
侵权投诉
第1页(共6页)
2022—2023 学年第二学期期初考试
高一数学（A）参考答案 2023.2
1.A 2.D 3.C 4.B 5.B 6.C 7.D 8.A
9.AC 10.ABD 11.ACD 12.ACD
13.
2
14.800 15.
2sin 2x
16.
3
1,( ,0)
2
17.解：（1）
11
22
22
1( ) 2 2 2 6x x x
x
   
···························································· 5分
(2) 原式
2
log 3
3 2 lg(4 25) 3 3 2 4   
··························································· 10 分
18.解：（1）选①，
方法一：角
的终边经过点
 
3 , 4 0P m m m
，因为
为第四象限角,故
0m
,
点
P
到原点的距离为
  
22
3 4 5 5m m m m 
, ···················································· 1分
所以
44
sin 55
m
m
 
,
33
cos 55
m
m

···································································· 5分
故
46
sin 2cos 2
55

   
················································································ 6分
方法二：角
的终边经过点
 
3 , 4 0P m m m
，所以
44
tan 33
m
m
 
， ····················· 1分
所以
22
4
sin cos
3
sin cos 1




, 解得
3
cos 5

，
又
为第四象限角，所以
3
cos 5
，
4
sin 5

······················································ 5分
故
46
sin 2cos 2
55

   
············································································ 6分
选②, 由
sin cos 1
sin cos 7


得
4
tan 3

，所以
22
4
sin cos
3
sin cos 1




································· 1分
解得
3
cos 5

，又
为第四象限角，所以
3
cos 5
，
4
sin 5

······························ 5分
故
46
sin 2cos 2
55

   
·········································································· 6分
选③，由
2
3cos 4sin cos 3
 

得
22
4sin cos 3 3cos 3sin
 
 
，
因为
,k k Z


， 所以
sin 0
， 故
4
tan 3

，所以
22
4
sin cos
3
sin cos 1




················ 1分
解得
3
cos 5

，又
为第四象限角，所以
3
cos 5
，
4
sin 5

， ···························· 5分
故
46
sin 2cos 2
55

   
. ············································································· 6分
第2页(共6页)
（2）方法一: 由(1)得
4
tan 3

,
 
 
22
sin
sin cos sin 2 sin 2

   

 


22
sin
sin cos sin cos
 
2
1
sin cos cos
 
·························································································· 9分
22
2
sin cos
sin cos cos

 
2
tan 1
tan 1
25
21

································································· 12 分
方法二:
 
 
22
sin
sin cos sin 2 sin 2

   

 


22
sin
sin cos sin cos
 
2
1
sin cos cos
 
 
····················································································· 9分
由(1)得
4
tan 0
3
 
，所以
为第二或第四象限角
① 若
为第二象限角, 则
3
cos 5

，
4
sin 5
.
所以
式
2
1 25
21
4 3 3
5 5 5
 
   
 
   
   
········································································ 11 分
② 若
为第四象限角, 则
3
cos 5
，
4
sin 5

.
所以
式
2
1 25
21
4 3 3
5 5 5
 
 
 
 
 
········································································· 12 分
说明: 第（2）问中使用方法二，若没有讨论，只有其中一种情况扣 1分
19.解：（1）
2
, 0 ,
3 3 3 3
xx
 
 
 
sin( ) 0,1 ,
3
x
 
 
6sin( ) 0,6 , 6sin( ) 9 9, 3
33
xx

   
 
9, 3B  
． ································· 2分
当
3a
时，
 
2
= +2 15>0 = 3 5 , 5 3 ,
R
A x x x x x x C A x x
  或
···················· 4分
所以
 
 
9,3
RAB
·························································································· 5分
（2）若“
xA
”是“
xB
”的必要不充分条件，则 BA ·············································· 6分
 
  
 
22
= +2 2 0 = + +2 >0A x x x a a x x a x a
 
 

， ··············································· 7分
①当
2aa 
，即
1a
时，
 
, 1 1,A   
，满足 BA； ·························· 8分
②当
2aa 
，即
1a
时，
 
, 2 ,A a a  
，由 BA得
9a-
或
1a
，
又
1a
，所以
11a 
； ·················································································· 10 分
③当
2aa 
，即
1a
时，
 
, 2,A a a  
，由 BA，得
3a
或
7a
，
江苏省扬州市2022-2023学年高一下学期开学考试数学试卷答案.pdf

共6页,预览2页

还剩页未读， 继续阅读

相关推荐

更多
立即下载
作者：envi 分类：分省 价格：3知币 属性：6 页 大小：804.48KB 格式：PDF 时间：2024-11-28

开通VIP享超值会员特权

  • 多端同步记录
  • 高速下载文档
  • 免费文档工具
  • 分享文档赚钱
  • 每日登录抽奖
  • 优质衍生服务

作者详情

相关内容

更多

推荐作者

热门标签

大联考 高考地理 英语真题 听力 语法填空 石家庄 数学竞赛 读后续写 英语阅读 深圳中学
/ 6
评分 收藏
分享
立即下载
客服
关注