广西桂林市第十八中学2019-2020学年高二数学下学期开学考试试题 文（PDF）

桂林十八中2019-2020学年度18级高二下学期开学考试卷数学（文科）命题：曾祥富审题：易斌注意：1、本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，满分150分。考试时间：120分钟。答卷前，考生务必将自己的姓名和考号填写或填涂在答题卷指定的位置，将条形码张贴在指定位置2、选择题答案用2B铅笔把答题卡上对应题目的答案标号涂黑；如需改动，用橡皮擦干净后，再选涂其他答案；不能答在试题卷上。3、主观题必须用黑色字迹的钢笔或签字笔在答题卷上作答，答案必须写在答题卷各题目指定区域内的相应位置上，超出指定区域的答案无效；如需改动，先划掉原来的答案，然后再写上新的答案。第I卷（选择题，共60分）一、选择题（本题包括12小题。每小题只有一个选项符合题意。每小题5分，共60分）1.{|}{|}.(2,1).(2,3).(1,1).(1–213,3)1AxxBxxxAABCDB若集合，或，则∩00002.:,cos1,.,cos1.,cos1.,cos1.,cos1pxRxpAxRxBxRxCxRDxRx已知命题则是3.(1)2,||.1.2.3.2zziizABCD若复数满足则2224.,cos1213....2222ABCabcabCABCD在中，已知则5.402810.3.4.5.7ABCD某校田径队共有运动员人，其中男运动员人，现用分层抽样方法抽取容量为的样本，则应该从女运动员中抽取的人数为21536.{},53.1.3.3.5naaayxxaABCD已知数列是等比数列，是函数的两个零点，则7.,ˆˆˆˆ.1.52.1.52ˆˆˆˆ.1.52.1.52xyAyxByxCyxDyx已知变量的散点图如图所示，则其回归直线方程可能是8.3456kABCD若执行如图所示的程序框图，则输出的值为．．．．9..4.5.6.7ABCD某个几何体的三视图如图所示，则该几何体的外接球表面积为222110.044.3.22.2.3xymyyxmABCD已知圆与抛物线准线相切，则实数22122222111.:1(0,0)..3.2.5.6xyCabFFFCabMOMOFaCABCD已知双曲线的左右焦点分别为、，过作的一条渐近线的垂线，垂足为，为坐标原点若的面积为，则的离心率为212.()(0,),()()()+()0,()().(0,1).(2,).(1,2).(1,)fxfxfxfxxfxfxxfxABCD已知的定义域为是的导函数，且满足则不等式的解集是第7题图第8题图第9题图第II卷（非选择题，共90分）二、填空题（本题包括4题。共20分）713.cos.62,33__________.014.,xyxyxyxyy已知实数，满足条件则的最大值为1115.ln2(,).22yxx曲线在处切线方程为2212122216.:1(02)4,||||5.xyCbFFFlbCABAFBFb已知，左右焦点分别为、，过的直线交椭圆于两点，若的最大值为，则椭圆三、计算题（本题包括6题，共70分）1323117.({}3627.(1){}1(2){}l{.10)og}nnnnnnnnaaaaabbanTbb已知为等比数列，且，求数列的通项公式；若数列满足，求数列的前项本小题和满分分218.,,,,sincos()cos().(1)cos(2)902,.abcABCABCABCBCacAAaABCo已知分别是的内角的对边，若，求的值；设，且求的面积19.(12).800100.本小题满分分眼保健操是一种眼睛保健体操，主要是通过按摩眼部穴位，调整眼及头部的血液循环，调节肌肉，改善眼的疲劳，达到预防近视等眼部疾病的目的某学校为了调查推广眼保健操对改善学生视力的效果，在应届高三的全体名学生中随机抽取了名学生进行视力检查，并得到如图的频率分布直方图(1)5.0(2)0.005若直方图中后三组的频数成等差数列，试估计全年级视力在以上的人数；为了研究学生的视力与眼保健操是否有关系，对年级不做眼保健操和坚持做眼保健操的学生进行了调查，得到上面的列联表，根据表中的数据，能否在犯错的概率不超过的前提下认为视力与做眼保健操有关系？22()()()()()nadbcKabcdacbd附：20.(,//,,2,1.(1122))//().BCABDEABDEAEABFBCABCBAEDEDFACEDACEd如图，面为的中点，求证：平面本小题满分；求点到平面的离分距2212)1||321.((1)1(2)(1,.||)3.MQlMABQAPxyDPyDDMMDPDPBl求点的轨迹方程；过点作直线与点的轨迹相交于本小题满分分如图所、两点，点示，点在圆上运动，轴，为垂足，点恰好为弦中点，求直线在延长上，且的方程线2222.(()1.(1)()(2)(201)).xfxexxRefxxxfxkxxk已知函数，，为自然对数的底数证明：；若对任意的恒成立，本小题满求实数的取分值范围分2()PKk0.100.050.0250.0100.005k2.7063.8415.0246.6357.879桂林十八中2019-2020学年度18级高二下学期开学考---参考答案数学（文科）一、选择题1-5：ADBAA6-10：BCBCD11-12：CD二、填空题13．3214.415.y=3x-116.3三、解答题232111117.(1){}1627627,333351111(2)(1),(1)nnnnnnaqaaaqaqaqabnbbnnnLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL解：设等比数列的公比为，（分）由得：（分）将代入，解得，（4分）所以；（分）由得1223171111111111822311191101nnnnTbbbbbbnnnnnLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL（分）（分）（分）（分）22222218.(1)sincos()cos()sin2sinsin223241cos.624ABCBCABCabcacacbbcaAbcQLLLLLLLLLQLLLLLLLLL解：，（分）由正弦定理得：，（分）又，（分）由余弦定理得（分）20222(2)(1)29082111.1222abcAbcaabcABCSbcQLLLLLLLLLQLLLLLLLLLLL由知，，，（分），解得，（10分）的面积（分）19.(1)3727242100-(3727)63()324,21,1855.00.186LLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL解：由图知，第一组有人，第二组有人，第三组有人，第四组有人，（分）因为后三组频数成等差数列，且共有人，（分）所以后三组频数依次为，（分）故全年级视力在以上的频率为，（分）所以全年级5.08000.18144().7LLLLLLLL视力在以上的人数约为人（分）22100(4418326)(2)7.8957.87911505076204.00125kLLLLLLLLLLLLLLLLLLLLL能在犯错的概率不超过的前提下认为视力与做眼保健操有关，（分）因系此（.分）20.(1),11,212,23ACGEGFGFGABFGABDEABDEABDEFGDEFGEDFGDFEGLLLLLLLLLLLLLLLLLLQLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL证明：设中点为，连接，（分）则∥且，∥，，∥，（分）四边形为平行四边形，（分）∥，45//6DFACEEGACEDFACELLQLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL（分）平面，平面，（分）平面，（分）(2)7,,822122112//AECAFCABCDACEFABCABDEBCAEAEABABBCBAEABCAEACACSACAESSDFACEVVQLLLLLLLLLLLLLLLLLLLLLLLLLLLQLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLQ面，，（分），平面，（分）由已知得，，，（9分）平面，101111332122CEEACFAECAFCVSdSAEdLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL，（分），（分），（分）00000002222000021.:(1)(,),(,)(0,)13||3,02||0,33(,)11,MxyPxyDyxxDMDMyxyyDPxxxyyPxyxyxyLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLQLLLLLLL解设，则，（分）由轴，得且，（分）所以且①（分）在圆上，所以②222241,591(0).69xyxMyxLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL（分）由①②得（分）的轨迹方程为：（分）2222221122121(1,)31((2)1(1)731(1)31911(19)18()9()9083318(,),(,)1,)03ykxykxxykxkkxkkAxyBllxyxxLLLLLLLLLLLLLLLLLLLQ由题意知直线的斜率存在，且过点，设的直线方程为在椭圆方法一：，（分）联立，得：，内部（分）点设，所以，则：恒成立，212226919210186219111311(1)1320.123(1,)33kkABxxkkkklyxxyLLLLLLQLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL，（分）点中点，，（分），解得，（分）因此直线的方程为，即（为分）1122121222112222121212121(2)(,),(,),(1,)322,8319,919()()()()0109AxyBxyQABxxyyxyABMxyxxxxyyyylLLLLLLLLLLLLLLLLLLLLLQLLLLLLLLLLLLLLLLLLLLLLLL方法二：设由为中点得：（分）都在的轨迹上，（分）从而有（分）由题意得直线斜率存121212121119()311(1)320.1233yyxxxxyylyxxyLLLLLLLLLLLLLLLLLLLLLLL在，故：（分）因此直线的方程为，即（分）222.(1)()10,1()1,()1,2()00,(-,0)()0,()(0,)()0,()4()g(0)0,xxxfxxxexgxexgxegxxxgxgxxgxgxgxLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL