发布时间 : 星期六 文章福建省福州市2019—2020学年度第一学期高三期末质量检测理科数学答案更新完毕开始阅读bc03d80aa100a6c30c22590102020740be1ecdd3
2019－2020学年度第一学期福州市高三期末质量检测
数学（理科）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。除第16题外，选择题和填空题不给中间分。
一、选择题：本大题共12小题，每小题5分，共60分．在每小题给出的四个选项中，只
有一项是符合题目要求的． 1．B 7．C
2．D 8．C
3．C 9．A
4．A 10．B
5．D 11．B
6．B 12．A
二、填空题：本大题共4小题，每小题5分，共20分．把答案填在题中的横线上．
13．e2?2
14．3
15．
83
16．286?； 6729三、解答题：本大题共6小题，共70分．解答应写出文字说明、证明过程或演算步骤．第
17～21题为必考题，每个试题考生都必须作答．第22、23题为选考题，考生根据要求作答．
17．【命题意图】本题主要考查利用正弦定理和余弦定理解三角形，任意三角形的面积，考查学生的逻辑推理能力与数学运算能力,考查的核心素养是逻辑推理、直观想象、数学运
理科数学参考答案及评分细则（第1页 共11页）
算.
【解析】解法一：（1）在△ABC中，由正弦定理及题设得
ACBC17，故， ································································ 3分 ??sinBsin150?sinBsinA解得sinB?127， ····················································································· 4分
3327321. ······················································ 6分14
又0?＜B＜30?，所以cosB??（2）设AD?CD?x，则BD?2x． 在△ABC中，由余弦定理得， BC2`?AB2?AC2?2AB?ACcosA，
即7?9x2?1?6xcosA，① ··········································································· 7分
1AC12?在等腰△ACD中，有cosA?，② ···················································· 8分
AD2x联立①②，解得x?1或x??1（舍去）． ························································· 9分 所以△ACD为等边三角形，所以A?60?， ···················································· 11分 1133所以S△ABC??AB?ACsinA??3?1?sin60??． ··································· 12分
224解法二：（1）同解法一． ············································································· 6分
（2）设AD?x，则CD?x,BD?2x, 因为?ADC????BDC，
所以cos?ADC??cos?BDC， ·········································································· 7分
由余弦定理得，
4x2?x2?72x2?1得， ·········································································· 8分 ??4x22x2理科数学参考答案及评分细则（第2页 共11页）
2．······················································ 9分 所以x?1，解得x?1或x??1（舍去）
所以△ACD为等边三角形，所以A?60?， ···················································· 11分 1133所以S△ABC??AB?ACsinA??3?1?sin60??． ··································· 12分
22418.【命题意图】本题考查等差数列和等比数列的通项公式、性质，错位相减法求和,考查学生的逻辑推理能力,化归与转化能力及综合运用数学知识解决问题的能力.考查的核心素养是逻辑推理与数学运算. 【解答】(1)依题意得: b32?b2b4，
所以(a1?6)2?(a1?2)(a1?14) ， ·································································· 1分
22所以a1?12a1?36?a1?16a1?28,
解得a1?2. ································································································ 2分 ································································································ 3分 ?an?2n. ·
设等比数列?bn?的公比为q，所以q?b3a48???2, ········································ 4分 b2a24又b2?a2?4,?bn?4?2n?2?2n. ····································································· 5分 (2)由（1）知，an?2n,bn?2n. 因为
ccc1c2???????n?1?n?2n?1 ① a1a2an?1ancc1c2··························································· 6分 ??????n?1?2n ② ·
a1a2an?1当n≥2时,
由①?②得，
cnn?1······························································· 7分 ?2n,即cn?n?2, ·
an3又当n?1时，c1?a1b2?2不满足上式,
?8,n?1,?cn?? ···················································································· 8分 n?1n?2,n?2.?342021数列?cn?的前2020项的和S2020?8?2?2?3?2?????2020?2
理科数学参考答案及评分细则（第3页 共11页）
?4?1?22?2?23?3?24?????2020?22021 ·············· 9分
设T2020?1?22?2?23?3?24?????2019?22020?2020?22021 ③， 则2T2020?1?23?2?24?3?25?????2019?22021?2020?22022 ④，
23420212022由③?④得：?T2020?2?2?2?????2?2020?2 ··································· 10分
22(1?22020)??2020?22022
1?2································································ 11分 ??4?2019?22022 ·
所以T2020?2019?22022?4,
2022所以S2020? T2020?4?2019?2?8. ··························································· 12分
19.【命题意图】本题考查空间直线和直线、直线和平面、平面和平面的垂直的证明，二面角等基础知识,考查学生的逻辑推理能力,化归与转化能力和空间想象能力.考查的核心素养是直观想象、逻辑推理与数学运算.
【解析】解法一：（1）因为PA?底面ABCD，BC?平面ABCD，
所以PA?BC. ··························································································· 1分 因为ABCD为正方形，所以AB?BC， 又因为PA···················································· 2分 AB?A，所以BC?平面PAB. ·
因为AE?平面PAB，
所以AE?BC. ··························································································· 3分 因为PA?AB，E为线段PB的中点，
所以AE?PB， ························································································· 4分 又因为PBBC?B，
所以AE?平面PBC. ··················································································· 5分 又因为AE?平面AEF，
理科数学参考答案及评分细则（第4页 共11页）