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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

又0?＜B＜30?,所以cosB?3327?321. ························································ 6分14

理科数学参考答案及评分细则（第1页 共10页）

（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 ??4x22x22························································ 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. ·

理科数学参考答案及评分细则（第2页 共10页）

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?1an当n≥2时,

cc1c2··························································· 6分 ??????n?1?2n ② ·

a1a2an?1cnn?1································································· 7分 ?2n,即cn?n?2, ·

an由①?②得,

又当n?1时,c1?a1b2?23不满足上式,

?8,n?1,?cn?? ···················································································· 8分 n?1n?2,n?2.?数列?cn?的前2020项的和S2020?8?2?23?3?24?????2020?22021

?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,

由③?④得：?T2020?22?23?24?????22021?2020?22022 ··································· 10分

22(1?22020)??2020?22022

1?2································································ 11分 ??4?2019?22022 ·

所以T2020?2019?22022?4,

所以S2020? T2020?4?2019?22022?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,

理科数学参考答案及评分细则（第3页 共10页）