河南省南阳市2023年02月高一期末考试数学试卷参考答案11118

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2022 年秋期高中一年级期终质量评估

数学参考答案

一、选择题：1—4：ABBC 5—8：BCCD.

二、选择题：9AC 10BCD 11BD 12ABD

三、填空题：13. 58 14. 8 15. 0.0615 3.1547 16.（

6,

）

四、解答题

17.解：（1）选择①

1a

作为已知条件，

则

 



  





1 2 0 1 2 ,A x x x x x       

························································································2 分





1 4 ,B t t  又

···································································································································4 分





1 2 .A B x x   

·······························································································································5 分

选择②

2a

作为已知条件，

则



  





4 0 0 4A x x x x x     

，

··································································································2 分





1 4 ,B t t  又





1 4 .A B x x   

（2）因为“

x A

”是“

x B

”的必要不充分条件，所以

B A





，

易知





1 4 ,B t t  

方程

  

2 2 0x a x a   

的根

1 2

2, 2 ,x a x a  

·······································7 分

分三种情况讨论：

①当

2 2a a 

，即

2a 

时：





,不满足题设，舍去；

②当

2 2a a 

，即

2a 

时：





2 2A x a x a   

,须满足

2 1,

4 2

 







解得：

2 3a 

；

③当

2 2a a 

，即

2a 

时：





2 2A x a x a   

,须满足

2 1 ,

4 2





 



无解；····························9 分

综上：

 

2,3 .a

·············································································································································10 分

18. 解：（1）即

4 2 3 0

x x

a

   

对任意

x R

恒成立，

4 3 3

2 2

x x

a

   

恒成立，················································································································3 分

又

3 3

2 2 2 2 3,

2 2

x x

   

当且仅当



时“=”成立，························································5 分

故所求



, 3a

  

.·····································································································································6 分

（2）令

 

t x 

，则

 

t x

在





0, 

单调递增且

1,t

高一数学参考答案第 2页共（4）页

又

22 3y t at  

图象开口向上，对称轴为

t a

，

···············································································8 分



函数

 

f x

单调增区间是





0, ,

22 3y t at    单调增区间是





1, , 1,t a   

·············································································11 分

1.a故

························································································································································12 分

19. 解:（1）设事件

表示“

2 2

a b

”,

是从

0,1, 2,3

四个数中任取的一个数，

是从

0,1, 2

三个数中任取的一个数,

所以样本点一共有 12 个：（0,0），（0,1），（0,2），（1,0），（1,1），

（1,2），（2,0），（2,1），（2,2），（3,0），（3,1），（3,2），···············································3 分

其中第一个数表示

的取值，第二个数表示

的取值,符合古典概型模型，

事件

包含其中 3 个样本点：（0,1），（0,2），（1,2）.···································································4 分

故事件

发生的概率为：

 

3 1

12 4

P A  

.···········································································································································6 分

（2）若方程

2 2

2 0x ax b  

有实数根，则需

2 2

4 4 0a b   

，即

2 2 ,a b

······································8 分

记事件“方程

2 2

2 0x ax b  

有实数根”为事件

，

由（1）知，

,B A

·····································································································································10 分

故所求

 

P B P A P A   

.·····································································································12 分

20. 解：（1）函数

 

22 ( 0)f x ax bx a a   

是“倒戈函数”，理由如下：·········································1分

由

   

0 0

f x f x  

得：

   

 

0 0 0 0

2 2a x b x a ax bx a       

化简得：

 

2 2 0,a x  

解得：

02,x 

·······························································································3分

所以存在实数

满足

   

0 0 ,f x f x  

故函数

 

22 ( 0)f x ax bx a a   

是“倒戈函数”.··············································································5 分

（2）

   

log 2 2 1f x x t   

是定义在

2, 2

 



 

上的“倒戈函数”，



关于

的方程

   

f x f x  

即

   

2 2

log 2 2 1 log 2 2 1x t x t        

有解，

·················7 分

等价于

    

 

2 2

2 2 1 log 2 2 log 4 , 2, 2t x x x x  

        

 

   

有解，

·······························9 分

又因为

2 4 4x  

，所以

 

1 2 2 1 log 4 2,t x    

······························································11 分

解得：

3,1

t 

 

 

.········································································································································12 分

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