山东省烟台市2021届高三下学期3月高考诊断性测试数学试题答案

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高三数学参考答案（第 1页，共 6页）

2021 年高考诊断性测试

数学参考答案

一、单选题

BDCC BADD

二、多选题

9.ABC 10.BD 11.BC 12.ACD

三、填空题

13.

14.

1.75

15.

[1 3 3, ) 

16.

四、解答题

17.解：若选①:（

）由已知

2 2 1 2b T T  

，

3 3 2 4b T T  

，

所以

 

，通项

2 2 1

22 2 2

n n n

b b q   

   

.·········································· 2 分

故

1 1 1a b 

.···························································································3 分

不妨设

{ }

的公差为

.则

1 2 1 4 14d d   

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

解得

2d

，所以

2 1

a n 

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

（2）由

[lg ]

n n

c a

，则

1 2 3 4 5 0c c c c c    

，

6 7 50 1c c c    

，

51 52 100 2c c c    

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

所以

1 2 3 100

c c c c    

1 45 2 50 145   

.···········································10 分

若选②:（

）由已知

2 2 1 2b T T  

，

3 3 2 4b T T  

，

所以

 

，通项

2 2 1

22 2 2

n n n

b b q   

   

.·········································· 2 分

故

1 1 1a b 

.···························································································3 分

不妨设

{ }

的公差为

，则

4 3

4 1 28



   

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

解得

4d

，所以

4 3

a n 

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

（2）由

[lg ]

n n

c a

，则

1 2 3 0c c c  

，

4 5 25 1c c c    

，

26 27 100 2c c c    

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

所以

1 2 3 100

c c c c    

1 22 2 75 172   

.···········································10 分

若选③:（

）由已知

2 2 1 2b T T  

，

3 3 2 4b T T  

，

所以

 

，通项

2 2 1

22 2 2

n n n

b b q   

   

.·········································· 2 分

故

1 1 1a b 

.···························································································3 分

不妨设

{ }

的公差为

，则

(1 7 ) (1 4 )(1 12 )d d d   

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

因为

0d

，解得

2d

，所以

2 1

a n 

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

高三数学参考答案（第 2页，共 6页）

（2）由

[lg ]

n n

c a

，则

1 2 3 4 5 0c c c c c    

，

6 7 50 1c c c    

，

51 52 100 2c c c    

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

所以

1 2 3 100

c c c c    

1 45 2 50 145   

.···········································10 分

18.解：（1）

( )f x

sin 3 cosx x 

2sin( )



 

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

( )f x

图象向右平移



个单位长度得到

2sin( )

y x



 

的图象，························· 2 分

横坐标缩短为原来的

（纵坐标不变）得到

2sin(2 )

y x



 

图象，

所以

( ) 2sin(2 )

g x x



 

.···········································································3 分

令

2 2 2

2 6 2

k x k

  

 

     

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

解得

3 6

k x k

 

    

所以

( )g x

的单调递增区间为

[ , ]( )

3 6

k k k

 

    Z

；··································5 分

（2）由（1）知，

( ) 2

c g



 

，······································································6 分

因为

sin( )cos( ) cos ( )

3 6 6 4

B B B

  

    

，所以

cos( )

6 2



  

又因为

(0, )B





，所以

( , )

6 6 6

  

 

当

cos( )

6 2



 

时，

6 3

 

 

，





.·················································· 7 分

此时由余弦定理可知，

4 2 2 cos 12

a a



    

解得

3 11a 

.·····················································································8 分

所以

1 3 11

2 ( 3 11) sin

2 6 2

ABC







     

.·········································9 分

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摘要：

高三数学参考答案（第1页，共6页）2021年高考诊断性测试数学参考答案一、单选题BDCCBADD二、多选题9.ABC10.BD11.BC12.ACD三、填空题13.2214.1.7515.[133,)16.56四、解答题17.解：若选①:（1）由已知2212bTT，3324bTT，所以322bqb，通项2212222nnnnbbq.··········································2分故111ab.······························································...

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