福建省南平市2022届高三5月质量检查数学试题参考答案

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参考答案 1页(共 9页)
南平市 2021-2022 年高中毕业班第三次质量检测
数学参考答案及评分标准
说明:
1.本解答给出了一种或几种解法供参考,如果考生的解法与本解答不同,可根据试题
的主要考查内容比照评分标准制定相应的评分细则.
2.对计算题,当考生的解答在某一步出现错误时,如果后继部分的解答未改变该题的
内容和难度,可视影响的程度决定后继部分的给分,但不得超过该部分正确解答应给分数
的一半;如果后继部分的解答有较严重的错误,就不再给.
3.只给整数分数. 选择题和填空题不给中间.
一、选择题:本题考查基础知识和基本运算,每小题 5,满分 40
1A 2D 3C 4B 5B 6A 7D 8C
二、选择题本大题共 4小题,每小题 5分,满分 20 在每小题给出的四个选项中
多项符合题目要全部选对的5分,部分选对的得 2分,有选错的0
9ABC 10BD 11A B 12A B D
三、填空题:本题考查基础知识和基本运算,每小题 5,满分 20
13
1
2
14
3
3
15
2 ln3
16
3
四、解答题:本大题共 6小题,共 70 ,解答应写出文字说明、证明过程或演算步骤。
17.(本小题满分 10 分)
解:1)选①,因为
( )( ) ( )
sin sin sina b A B c b C+ = −
,由正弦定理,
( )( ) ( )
a b a b c b c+ = −
,所以
··········································· 3
所以
( )
2 2 2 1
cos 0,π
22
b c a
AA
bc
+−
= = 
,故
π
3
A=
. ······································· 5
1)选②,因
2 2 cos 0b c a C− − =
,由正弦定理可
2sin 2sin cos sinB A C C−=
所以,
( )
sin 2sin 2sin cos 2cos sinC A C A C A C= + − =
································· 3
因为
( )
0,πC
,则
sin 0C
,可得
1
cos 2
A=
( )
0,πA
,故
π
3
A=
. ········································································ 5
1)选③,
2 2 2
cos cos sin sin 1 cosB C B C A+ + = +
2 2 2
2 sin sin sin sin 2 sinB C B C A + = −
,即
2 2 2
sin sin sin sin sinB C A B C+ − =
由正弦定理可得
2 2 2
b c a bc+ − =
······························································ 3
由余弦定理可得
2 2 2 1
cos 22
b c a
Abc
+−
==
参考答案 2页(共 9页)
( )
0,πA
,故
π
3
A=
. ·········································································· 5
2)在△ABC 中,由余弦定理得
2 2 2 2 cosBC AB AC AB AC A= +  
因为 AC2
23BC =
π
3
A=
,所以
2π
12 4 4 cos 3
AB AB= + − 
解得 AB4AB=-2(舍) ································································ 7
因为△ACD 与△BCD 积比为 35,所以
3
2
AD =
在三角形 ACD 中,由余弦定理得
2 2 2 2 2
33π13
2 cos 2 ( ) 2 2 cos
2 2 3 4
CD AD AC AD AC A= + −   = + =
13
2
CD =
··················································································· 10
18. (本小题满分 12 分)
解:
1)当
2n
时,由
11
n
n
an
an
++
=
,可得
2
1
2
1
a
a=
3
2
3
2
a
a=
,……,
11
n
n
an
an
=
········································································································· 2
所以
23
11 2 1
n
n
an
n
=  =
···························································· 4
=1n
时,
11a=
适合上式,从
n
an=
················································ 5
2)由已知得
242
2= nb n
,所以当
n
为偶数时,
24= nbn
························ 6
2112222
12 ==
nnb n
,所以当
n
为奇数时,
21= nbn
············· 7
=为偶数
为奇数
nn
nn
bn,24
,21
20 1 2 3 20 1 3 19 2 4 20
( ) ( )S b b b b b b b b b b= + + + + = + + + + + + +
·············· 9
]
2
2)110(10
10)22[(]
2
2)110(10
10)20[(
++
+=
································································································· 11
( 200 90) ( 220 90) 240= − + + − + =
················································· 12
参考答案 3页(共 9页)
19(本小题满分 12 分)
1)由题意得
=55 0.1+65 0.2+75 0.4+85 0.15+95 0.15 75.5
    =
······································································································· 3
75.5
=
,∵
11.5
=
( ) 0.6826
(75.5 87) 0.3413
22
PX
PX
   
−   +
  = = =
········································································································· 5
2)由题意知
( ) ( )
1
2
P X P X

 = =
.
所获赠话费
的可能取值为
10
20
30
40
60
·································· 6
( )
1 3 3
10 2 4 8
P
= =  =
( )
1 3 3 9
20 2 4 4 32
P
= =   =
( )
1 1 1
30 2 4 8
P
= =  =
( )
1 3 1 1 1 3 3
40 2 4 4 2 4 4 16
P
= =   +   =
( )
1 1 1 1
60 2 4 4 32
P
= =   =
.
的分布列为:
10
20
30
40
60
P
3
8
9
32
1
8
3
16
1
32
······································································································· 10
3 9 1 3 1 45
( ) 10 20 30 40 60
8 32 8 16 32 2
E
=  + +  + + =
. ····························· 12
20(本小题满分 12 分)
证明:
1)因为底面
ABCD
是边长为
2
的正方形,所
BDAC
···························· 1
AB BC=
PBCPBA =
PB PB=
可得
PAB PCB△ △
从而
PCPA =
················································································ 3
摘要:

数学参考答案第1页(共9页)南平市2021-2022学年高中毕业班第三次质量检测数学参考答案及评分标准说明:1.本解答给出了一种或几种解法供参考,如果考生的解法与本解答不同,可根据试题的主要考查内容比照评分标准制定相应的评分细则.2.对计算题,当考生的解答在某一步出现错误时,如果后继部分的解答未改变该题的内容和难度,可视影响的程度决定后继部分的给分,但不得超过该部分正确解答应给分数的一半;如果后继部分的解答有较严重的错误,就不再给分.3.只给整数分数.选择题和填空题不给中间分.一、选择题:本题考查基础知识和基本运算,每小题5分,满分40分。1.A2.D3.C4.B5.B6.A7.D8.C二、选...

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