福建省泉州市2020届高三上学期单科质量检查数学（文）（答题全析——解答题部分）.pdf

《福建省泉州市2020届高三上学期单科质量检查数学（文）（答题全析——解答题部分）.pdf》由会员上传分享，免费在线阅读，更多相关内容在教育资源-天天文库。

1、准考证号________________姓名________________（在此卷上答题无效）保密★启用前泉州市2020届高中毕业班单科质量检查文科数学2020.1三、解答题：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．（一）必考题：共60分．17．（12分）记数列a的前n项和为S．若2S3a3.nnnn（1）证明：a为等比数列；n1（2）设bnlog9an，求数列的前n项和Tn．bbnn1【命题意图】本题主要考察等比数列的定义及通项公式、数列求和的裂项相消法

2、等基础知识，考查运算求解能力，考查化归与转化思想，体现基础性，导向对发展数学抽象、数学运算及数学建模等核心素养的关注．【试题简析】解：（1）由已知，得2S3a3，……①nn当n≥2时，2S3a3，……②·········································································2分n1n1①—②，得2S2S(3a3)(3a3)，即2a3a3a，nn-1nn-1nnn-1an整理，得3n≥2，···············································

3、··········································5分an1又由2S3a3，得a=30，所以a是以3为首项，3为公比的等比数列.·················6分111nnnn（2）由（1）得a=3，所以blog3，····························································8分nn921411所以=4()，·······································································10分bnbn1n

4、n1nn1市单科质检数学（文科）试题第1页（共11页）11111114n故T=4()+4()++4()4(1).····································12分n1223nn1n1n118．（12分）△ABC的内角A,B,C所对的边分别为a,b,c.已知a(3cosC)ccosA．b（1）求；a（2）求cosA的最小值．【命题意图】本小题主要考查正弦定理，余弦定理等解三角形的基础知识，结合考察了基本不等式的基础知识，考查推理论证能力与运算求解能力等，考查数形结合思想、化归与转化思想、函数与方程思想等，体现基础性、综合性与应

5、用性，导向对发展直观想象、逻辑推理、数学运算及数学建模等核心素养的关注．【试题简析】abc解法1:（1）在△ABC中，由正弦定理，得，··········································1分sinAsinBsinC从而由a(3cosC)ccosA，可得sinA(3cosC)sinCcosA，······························2分整理，得3sinAsinAcosCcosAsinC，即3sinAsin(AC)，····························4分又因为ABCπ，所以sinB3si

6、nA，································································5分b所以3．············································································································6分a（2）由（1）不妨设b3a3，则31c31，··············································7分222bca在△ABC中，由余弦定理，得cosA，·····

7、··················································8分2bc2222(3)c12c126所以cosA(c)≥，·············································11分23c23c23c326当c即c2时，等号成立，故cosA取到最小值为．··········································12分c3解法2：（1）同解