广西三新学术联盟2021-2022学年高一上学期1月期末联考试题数学答案(1.10)

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广西三新学术联盟高一 1月期末联考
数学试题 参考答案
一、选择题
题号
1
2
3
4
5
6
7
8
9
10
11
12
答案
B
A
D
C
C
D
B
B
AC
ACD
BCD
ABD
二、填空题
13．
3
14．25 15．
2
5
16．
2 5
,
9 18
 
 
 
 
三、解答题
17．解：（1）∵
{ | 3A x x  
或
1}x
，
∴
 
3 1
RA x x   ð
，·························································································· 1 分
又
1a
即
，················································································ 2 分
∴
 
 
0 1
RA B x x  ð
.·················································································· 4 分
（2）∵
A B A B A  
，··············································································· 5 分
∴又
{ | 3A x x  
或
1}x
当
B 
时，
1 2 3a a 
，解得
4a 
；····································································7 分
当
B 
时，
1 2 3
2 3 3
a a
a
 
 
或
1 2 3
1 1
a a
a
 
 
，解得
4 3a  
或
2a
.····························· 9 分
综上，a的取值范围为
3a 
或
2a
.········································································ 10 分
18．解：
2 2
sin 2 2sin 2 sin cos 2
sin
1 tan 1cos
x x x x sin x
x
x
x
 
··························································2 分
2sin cos (cos sin )
cos sin
x x x x
x x
······················································································· 3 分
2sin cos sin( )
4
cos( )
4
x x x
x
···························································································4 分
3
cos( )
4 5
x
 
，
17 7
12 4
x
 
 
，··········································································· 6 分
4
sin( )
4 5
x
 
，······························································································· 8 分
7
2sin cos 25
x x 
．······························································································ 10 分
2sin cos sin( ) 28
4
75
cos( )
4
x x x
x
 
，
2
sin 2 2sin 28
1 tan 75
x x
x
 
．······················································································ 12 分
19．解：（1）设
 
f n
为前
n
年的总盈利额，单位：万元；············································ 1 分
由题意可得
 
 
 
2 2
95 10 5 90 10 100 90 10 1 9n n nf n n n n n    
，·················3 分
广西三新学术联盟高一 1月期末联考 数学答案 第 2页 共 4页
由
 
0f n
得
1 9n 
，又
*
nN
，···········································································4 分
所以该设备从第
2
年开始实现总盈利；······································································ 5 分
（2）方案二更合理，理由如下：············································································· 6 分
方案一：由（1）知，总盈利额
 
22
100 90 10 5 16010f nn n n    
，···················· 7 分
当
5n
时，
 
f n
取得最大值
160
；此时处理掉设备，则总利润为
160 20 180 
万元；········8 分
方案二：由（1）可得，平均盈利额为
 
2100 90 9 9
10 100 100 2
040
10
nn n
n
f n n
n n n
   
  
 
 
，
························································································································· 10 分
当且仅当
9
nn
，即
3n
时，等号成立；即
3n
时，平均盈利额最大，此时
 
120f n
，
此时处理掉设备，总利润为
120 60 180 
万元；························································· 11 分
综上，两种方案获利都是
180
万元，但方案二仅需要三年即可，故方案二更合适. ··············12 分
20．解：（1）由函数
2
3 1 3 3
( ) 2sin sin cos 3 sin sin cos
2 2 2 2
f x x x x x x x
 
 
 
 
 
························································································································· 1 分
3 1 3
(1 cos 2 ) sin 2
2 2 2
x x 
··········································································· 3 分
sin 2 3
x
 
 
 
 
···································································································· 4 分
令
3
2 2 2 ,
2 3 2
k x k k Z
 
 
   
，解得
5 11
12 12 ,Zx kk k
 
 
，····················· 5 分
所以
 
f x
单调减区间
5 11
, ,
12 12
k k k Z
   
 
 
 
 
．····················································· 6 分
（2）将函数
( )f x
的图象向左平移
3
个单位，得到
sin(2 )
3
y x
 
，···························· 7 分
将函数
 
f x
的图象上每个点的横坐标缩小为原来的
1
2
，得到
( ) sin(4 )
3
y g x x
 
，····· 8 分
因为
04
x
 
，可得
0 4x
 
，则
4
4 +
3 3 3
x
 
，·············································· 9 分
可得
3sin 4 + 1
2 3
x
 
 
 
 
，··············································································· 11 分
所以
 
g x
在
0, 4
 
 
 
上值域为
3,1
2
 
 
 
．··································································· 12 分
21．解：（1）令
0x y 
，得
(0) (0) (0)f f f 
，所以
(0) 0f
，···································1 分
令
y x 
，得
(0) ( ) ( )f f x f x  
，即
0 ( ) ( )f x f x  
，所以
( ) ( )f x f x  
，···················· 2 分
所以函数
 
f x
是
R
上的奇函数. ·············································································· 3 分
（2）任取
1 2
,x x R
,且
1 2
x x
，则
1 2 1 2 1 2
( ) ( ) ( ) ( ) ( )f x f x f x f x f x x  
，···················· 4 分
因为当
0x
时，
( ) 0f x
，而
1 2
x x
，即
1 2 0x x 
，所以
1 2
( ) 0f x x 
，······················ 5 分
所以
1 2
( ) ( )f x f x
，所以
 
f x
在
R
上的单调递减. ······················································ 6 分
（3）由（1）知
 
f x
是
R
上的奇函数，所以
1
( 1) (1) 2
f f  
，所以
1
(1) 2
f 
，
所以
1 1
(2) (1 1) (1) (1) 1
2 2
f f f f    
，··························································· 7 分
所以不等式
 
2 2
( ) 1 1f mx x f x x  
可化为
2 2
( ) ( 1) (2)f mx x f x x f  
，
即
2 2
( ) (2) ( 1)f mx x f f x x  
，所以
2 2
( ) ( 3)f mx x f x x  
，······························· 8 分
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