2022年5月福州市高中毕业班质量检测数学试卷【答案】
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高二数学参考答案（第1页 共15页） 2022年5月福州市高中毕业班质量检测
数学参考答案及评分细则
评分说明：
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．解答右端所注分数，表示考生正确做到这一步应得的累加分数。
4．只给整数分数。
一、单项选择题：本题共8小题，每小题5分，共40分．
1．B 2．A 3．A 4．D
5．C 6．D 7．A 8．C
二、多项选择题：本题共4小题，每小题5分，共20分．
9．ABD 10．AC 11．AC 12．BCD
三、填空题：本大题共4小题，每小题5分，共20分．
13
．31 14
．5 15．0.52 16．π
2
四、解答题：本大题共6小题，共70分．
17. 【命题意图】本题主要考查等差数列、等比数列的概念、通项公式，数列求和等基础
知识．考查运算求解能力，考查化归与转化思想，涉及的核心素养有数学抽象、数学
运算、逻辑推理等，体现基础性，综合性．满分10分．
【解答】解法一：选①②作条件证明③．
设等差数列
ln
na
的公差是d
，则
21lnlndaa
， ····································· 1
分
因为
212aa
， 所以2
1lnln2a
d
a
，
所以
1lnlnln2
nnaa
，2n≥， ····························································· 3
分 高二数学参考答案（第2页 共15页） 所以
12n
na
a
，2n≥， ·········································································· 4
分
所以
na
是首项为
1a
，公比为2的等比数列， ············································ 5
分 所以1(12)
12n
na
S
， ············································································ 6
分
所以
112n
nSaa，即
112n
nSaa． ···················································· 7
分
设
1nnbSa，则
12n
nb
b
，2n≥， ························································· 8
分
又
1120ba
， ··················································································· 9
分
所以
1nSa
是首项是
12a
，公比为2的等比数列． ··································· 10
分
解法二：选①③作条件证明②．
设等比数列
1nSa
的公比是q
（0q
）， 所以21
11Sa
q
Sa
， ················································································ 1
分 所以12
12
2aa
q
a
，
因为
212aa
，所以2q
， ···································································· 3
分
又因为
1112Saa
，
所以数列
1nSa
的通项公式为1
111222nn
nSaaa
， ·························· 4
分
所以
nS
112n
aa． ··········································································· 5
分
当2n≥时，11
1111222nnn
nnnaSSaaa
， ······································· 6
分
又当1n时，11
112aa
，符合上式， ······················································ 7
分
所以1
12,n
naan
N． ········································································· 8
分
所以1
111lnlnln(2)ln(2)ln2nn
nnaaaa
， ··········································· 9
分
所以
ln
na
是等差数列． ····································································· 10
分
解法三：选②③作条件证明①．
因为数列
ln
na
是等差数列，则
1lnln
nnaa
为常数，2n≥， ······················· 1
分 高二数学参考答案（第3页 共15页） 所以
1lnn
na
a
为常数，2n≥， 即
1n
na
a
为常数，2n≥， ········································································· 3
分 令2
1(0)a
a
，
所以
na
为首项为
1a
，公比为q
的等比数列， ············································ 4
分
此时1
1n
naaq
． ················································································· 5
分
因为数列
1nSa
是等比数列，
所以2
211131()()()SaSaSa， ··························································· 6
分
故22
111[(2)]2[(2)]aqaaqq，························································· 7
分
即22
(2)2(2)qqq， ····································································· 8
分
化简得2
20qq，
因为0q
，解得2q
， ········································································ 9
分 所以2
12a
a
，即
212aa
． ··································································· 10
分
18. 【命题意图】本小题主要考查独立性检验、独立事件、随机变量的数学期望、二项分
布等基础知识，考查数据处理能力、运算求解能力、应用意识，考查统计与概率思想，
涉及的核心素养有数学抽象、逻辑推理、数学建模、数学运算、数据分析等，体现综
合性、应用性．满分12分．
【解答】（1
）设男性患者有x
人，则女性患者有2x
人，22列联表如下：
A型病 B型病 合计
男 5
6x
6x
女 2
3x
4
3x
2x
合计 3
2x
3
2x
3x
x
