福建省龙岩市第一中学2023届高三上学期第二次月考数学答案

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龙岩一中 2023 届高三上学期第二次月考数学答案
1-8：DDCC BABD 9.ABD 10.AB 11.AC 12.BC
13. 14.
(
−∞,0
)
和
(
0,1
)
15. 16. 13
17．（1）因为 ，由余弦定理可得 .
故······································································································································ 4
（2）因为 ，所以 .由余弦定理得 ，
则.······································································································································ 6
因为 的周长为 ，所以 ，解得 .··························8
所以 的面积为 ········································································10
18.（1）因为函数
························································································································· 3
令 ，解得 ，即对称中心 ·································5
4
3
2
2 2 2
5
8
b c a bc 
2 2 2
- 5
cos 2 16
b c a
Abc
 
5
cos 16
A
sin 2sinC B
2c b
2 2 2 2
15
2 cos 4
a b c bc A b  
15
2
a b
ABC
6 15
15
3 6 15
2
b b  
2b
ABC
2
1 5 231
2 1
2 16 4
b b  
 
 
 
 
2
π π
2sin sin 3 sin cos cos
4 4
f x x x x x x
 
 
 
 
2
2 2 2 2 3
2 cos sin cos sin sin 2 cos
2 2 2 2 2
x x x x x x
  
  
  
  
  
 
2 2 3 1 cos 2
cos sin sin 2
2 2
x
x x x
 
π 1
3 sin 2 3 2
x
 
 
 
 
π
2 π,
3
x k k  Z
ππ
6 2
k
x 
π 1
π,
6 2 2
kk
 
 
 
Z，
当 时，则 ，再结合三角函数图像可得
所以，函数对称中心： ， ，值域： .··········································7
（2）因为函数 的图像与函数 的图像关于 y轴对称，
则 ，·······················································································9
令 ， ，解得 ······························11
当 时，即为 所以当 时， 的单调递增区间： .·················12
19.（1）由题表知，随着时间 x的增大，y的值随 的增大，先减小后增大，而所给的函数
， 和 在 上显然都是单调函数，
不满足题意，故选择 ．·····················································································3
（2）把 ， ， 分别代入 ，得
解得 ， ，
∴ ， ．·································································5
∴当 时，y有最小值，且 ．
故当该纪念章上市 10 天时，市场价最低，最低市场价为每枚 70 元．·······································7
（3）令 ，·······························································8
π
0, 2
x 
 
 
π π 4π
2 ,
3 3 3
x 
  
 
 
1
1, 3 2
f x  
 
 
π 1
π,
6 2 2
k
 
 
 
 
kZ
1
1, 3 2
 
 
 
 
g x
 
f x
 
π 1
3 sin 2 3 2
g x f x x
 
 
 
 
π π 3π
2 π 2 2 π
2 3 2
k x k  
kZ
7π π
π π,
12 12
k x k k     Z
1k
5π 11π
,
12 12
 
 
 
 
0, πx
 
g x
5π 11π
,
12 12
 
 
 
x
( 0)y ax b a 
 
log 0, 0, 1
b
y a x a b b   
( 0)
a
y b a
x
 
(0, )
 
2
0y ax bx c a  
 
2,102
 
6, 78
 
20,120
2
y ax bx c  
4 2 102,
36 6 78,
400 20 120,
a b c
a b c
a b c
 
 
 
1
2
a
10b 
120c
 
2
2
1 1
10 120 10 70
2 2
y x x x   ,( )0x 
10x
min
70y
     
1 70
10
10 2 10
f x
g x x
x x
 
 
(10, )x
 
因为存在 ，使得不等式 成立，
则 ．································································································································ 9
又
∴ 当 时， 取得最小值，且最小值为 ，
∴ ．··································································································································· 12
20.解：（1）由 ，可得 ．
因为 ， ，
所以切点坐标为 ，切线方程为： ，
因为切线经过 ，所以 ，解得 ．··································································4
（2）解：由题可知 的定义域为 ， ，
令 ，则 ，解得 或 ，·······················································6
因为 所以 ，所以 ，
令 ，即 ，解得： ，
令 ，即 ，解得： 或 ，··················································8
又 的定义域为 ，所以， 增区间为 ，减区间为 ．
因为 ，所以函数 在区间 的最大值为 ，·································9
函数 在 上单调递增，故在区间 上 ，·······················10
 
10,x 
 
0g x k 
 
min
k g x
   
1 70 1 70
10 2 10 2 35
2 10 2 10
g x x x
x x
 
 
10 2 35x 
 
g x
2 35
2 35k
2
1
( ) 2ln (2 1) ( 0)
2
f x x ax a x a  
2
( ) 2 1f x ax a
x
 
(1) 2 2 1 1f a a a
 
1 3
(1) 2 1 1
2 2
f a a a 
3
(1, 1)
2
a
 
31 1 ( 1)
2
a
y a x
 
 
 
 
(0,0)
31 1
2
aa 
4a
( )f x
(0, )
2
1
( ) [ (2 1) 2]f x ax a x
x
 
( ) 0f x
2
(2 1) 2 0ax a x  
1
xa
 
2x
0,a
10
a
 
12
a
 
( ) 0f x
2
(2 1) 2 0ax a x  
12x
a
 
( ) 0f x
2
(2 1) 2 0ax a x  
1
xa
 
2x
( )f x
(0, )
( )f x
(0, 2)
(2, )
 
2
2
( ) 2 1 1g t t t t  
( )g t
(0,2]
0
( )f s
(0, 2)
(0, 2]
max
( ) (2) 2ln 2 2 2f s f a  
摘要：

龙岩一中2023届高三上学期第二次月考数学答案1-8：DDCCBABD9.ABD10.AB11.AC12.BC13.14.(−∞,0)和(0,1)15.16.1317．（1）因为，由余弦定理可得.故······································································································································4（2）因为，所以.  由余弦定理得，则.··············································...

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