山东省某重点校2022-2023学年高二上学期期末考试数学评分标准

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高二数学答案第 1页（共 4页）

2022-2023 学年度第一学期期末学业水平检测高二数学评分标准

一、单项选择题：本题共 8小题，每小题 5分，共 40 分。

1--8：B D D A C C A B

二、多项选择题：本题共 4小题，每小题 5分，共 20 分。

9．BC 10．CD 11．AD 12．BCD

三、填空题：本题共 4个小题，每小题 5分，共 20 分。

13．26 ；14．4 5

5；15．2；16．（1）2

2 1

n；（2）1 ( 1)

2 4 2



  ．

四、解答题：本题共 6小题，共 70 分。解答应写出文字说明，证明过程或演算步骤。

17.（10 分）

解：（1）因为 2 2a，所以 2a································································· 1 分

因为渐近线方程为 y x  ，所以 1

a，2b···············································2 分

所以 C的方程为 2 2 2x y  ··········································································· 4 分

（2）由（1）知， C的右焦点坐标为 (2,0) ························································5 分

若直线 l斜率不存在，则直线 l的方程 2x，

此时 (2, 2), (2, 2)A B ，1 2 4 6x x   ，不合题意·········································· 6 分

若直线 l斜率存在，则设直线 l的方程为 ( 2)y k x 

将( 2)y k x  代入 2 2 2x y  得： 2 2 2 2

(1 ) 4 4 2 0k x k x k     ······················ 7 分

所以

1 2 2

2 4 6

x x k



  

···············································································8 分

即24k，解得 2k  ················································································ 9 分

所以，直线 l的方程为

2( 2)y x  

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

18.（12 分）

解：（1）若选择①；

由题知，若数列{ }

a的公比 1q，则 4 1 2 1

4 , 2S a S a  ，

与4 2

15, 3S S  矛盾··················································································· 1 分

数列{ }

a的公比 1q，则

4 2

1 1

4 2

(1 ) (1 )

1 1

a q a q

S S

q q

 

 

  ···································2 分

所以

11 5

Sqq

S q



   

，解得 2q（2q  舍）······································4 分

所以，

(1 2 ) 3

1 2

S

 

，解得 11a···························································· 5 分

所以 1

a

······························································································ 6 分

若选择②；

由题知：数列{ }

a是各项均为正数的等比数列

又因为 1 2 3a a  ，34a，所以 2

1 1

(1 ) 3, 4a q a q   ······································1 分

高二数学答案第 2页（共 4页）

所以

(1 ) 3

a q



，所以

3 4 4 0q q  

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

解得

2q

或

q 

（舍）··········································································· 4 分

所以

3 1 4a a q 

，所以

11a

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

所以

a



······························································································ 6 分

（2）由（1）知：

n n

a



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

所以

0 1 2 2 1

1 2 3 1

...

2 2 2 2 2

nn n

n n

T 



     

1 2 3 1

...

2 2 2 2 2 2

n n

Tn n



     

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

两式相减得：

2 3 2 1

1 1 1 1 1

1 ...

2 2 2 2 2 2 2

n n n

 

       

································· 9 分



 



22n

n

 

所以

T



 

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

19．（12分）

解：（1）因为

(0,1)

在直线

上，所以直线

的方程为：

1y x  

·························· 1 分

因为

CP l

，所以直线

的方程为：

1y x 

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

所以

点的坐标为

( 1,0)

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

设圆

的半径为

，又因为

2 2

| | 2 | | 2 7AB r CP  

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

解得

3r

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

所以，圆

的标准方程为

2 2

( 1) 9x y  

························································ 6 分

（2）若该直线斜率不存在，则其方程为

2x

，显然符合题意······························ 8 分

若该直线斜率存在，设其方程为

( 2) 4y k x  

，设点

到该直线的距离为

因为该直线与圆

相切，所以

| 3 4 | 3



 



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

解得：

k

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

综上，过点

(2,4)Q

与圆

相切的直线的方程为：

2x

或

7 24 82 0x y  

···········12 分

20.（12 分）

解：（1）该校男生支持方案一的概率为

200 1

200+400 3



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

该校女生支持方案一的概率为

300 3

300+100 4



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

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