东莞市、揭阳市、韶关市2024-2025学年度第一学期期末教学质量检查高三数学参考答案

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东莞市、揭阳市、韶关市 2024-2025 学年度第一学期期末教学质量检查

高三数学参考答案

一、单项选择题

题号

答案

二、多项选择题（全部选对的得 5分，部分选对的得部分分，有选错的得 0分）

题号

答案

ABD

三、填空题

12.

13.

7 2

14.

1 1

[1 ( ) ]

4 3

n

 

（第一空 2分，第二空 3分）

四、解答题

15.解：（1）因为

sin 2 sin

C C=

，即

2sin cos sin

C C C=

，································ 1 分

因为

sin 0C>

，································································································ 2 分

所以

cos 4

，······························································································· 3 分

因为

2 2

sin cos 1C C+ =

，···················································································4 分

即

2 2

sin ( ) 1

C+ =

，·························································································5 分

所以

sin 4

；····························································································6 分

（2）因为

1 15

sin

2 2

ABC

S ab C

D= =

，································································· 7 分

所以

4ab =

，··································································································· 8 分

因为

2 2 2

2 cosa b ab C c  

，··········································································· 9 分

即

2 2

( ) 2 2 cosa b ab ab C c   

，··································································· 10 分

因为

cos 4

，

4ab =

，

2 6

a b c 

，

所以

2 2

2 6

( ) 10

3c c 

，得

6c

，································································ 11 分

所以

2 6 4

a b c  

，·················································································· 12 分

所以

ABC

的周长为

4 6

.··············································································13 分

高三数学参考答案第 2页共 6页

16.解：（1）当

2a

时，

 

ln 2f x x x

  

，所以

 

01f

，······································· 1 分

 

1 2

f



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

所以在

(1, (1))f

切线

 

1 1f 

，··········································································5 分

所以切线方程为

( 1)y x  

，即

1 0x y  

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

（2）因为

 

ln a

f x x a

  

，其中

0x

，则

 

2 2

1a x a

f x x x x



  

，···················· 7 分

①当

0a

时，

 

0f x



恒成立，此时函数

 

f x

在

 

0, 

上单调递增，无极小值，········ 8 分

②当

0a

时，令

 

0f x



，可得

x a

，列表如下：

 

0, a

 

,a

 

f x







 

f x

递减

极小值

递增

所以

   

ln 1f x f a a a   

极小值

，································································· 10 分

由题意可得

ln 1 1a a a   

，即

2ln 0a a a  

.···················································11 分

令

 

2lng a a a a  

(

0a

)，则

 

1 0g

因为

 

1 2 1

2 1 0

a a

g a a a a

 

    



，··························································· 13 分

所以函数

 

g a

在

 

0, 

单调递增，·····································································14 分

所以由

   

1 0g a g 

，得

0 1a 

，所以实数

的取值范围是

 

0,1

.······················· 15 分

17.（1）证明：过点

作

AD PB

，垂足为

．·······················································1 分

因为

PAB PBC平面平面

，

PAB PBC PB平面平面

，

AD PAB平面

，

所以

AD PBC平面

，························································································· 3 分

因为

BC PBC平面

，所以

AD BC

，···································································4 分

又因为

PA ABC平面

，

BC ABC平面

，所以

PA BC

，········································· 5 分

又因为

PA AD A

，

PA PAB平面

，

AD PAB平面

，所以

BC PAB平面

．············· 6 分

（2）解：(方法一)因为

BC PAB平面

，所以

BC AB

，又因为

PA ABC平面

．所以如图，以

点

为坐标原点，分别以

，过

点且平行

的直线，

所在直线为

，

轴建立空间

直角坐标系

A xyz

，···························································································7 分

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