吉林省吉林市2023-2024学年高三上学期第二次模拟考试数学答案

（Ⅱ）由题意，样本空间中共有个样本点，设表示两小区室内温度，用表吉林地区普通高中2023—2024学年度高三年级第二次模拟考试20x1,x2A,B(x1,x2)示可能的结果.数学试题参考答案C{(21,19),(22,19),(24,19),(21,20),(23,20),(24,20)}，n(C)6，一、单项选择题：本大题共8小题，每小题5分，共40分．n(C)63所以，事件C的概率P(C).··················································6分n()201012345678（Ⅲ）（选择A）从供热状况角度选择生活地区居住，应建议选择A小区，理由如下：CDBDABCD①在20天的数据中，A小区室温大于B小区室温的有14天，B小区室温大于A小区室温的7二、多项选择题：本大题共4小题，共20分．全部选对的得5分，部分选对的得2分，有选错的得0有5天，由此可以估计，每天A小区室温大于B小区室温的概率为P，B小区室温大110分．1于A小区室温的概率为P，P远远小于P；24219101112②随机抽取的20天中，A小区室温平均数为TA22.05C，B小区室温平均数为BDACABDACDTB20.7C，TATB；三、填空题：本大题共4小题，每小题5分，共20分．其中第15题的第一个空填对得2分，第二③在随机抽取的20天中，B小区供热等级达到“舒适”的天数为9天，远小于A小区供热等级个空填对得3分．达到“舒适”的天数；9小区室温中位数为，小区室温中位数为，分④AZA22.5CBZB20CZAZB1013．1014．4（选择B）从供热状况角度选择生活地区居住，应建议选择B小区，理由如下：28222在天的数据中，小区中存在供热不达标的情况，而小区供热等级全部达标①20AB.15．3；4016．eea1或1aee②随机抽取的天中，小区室温平均数为，小区室温平均数为222220ATA22.05CB（注：16题或写成{a|eea1或1aee}，或写成(ee,1)(1,ee)）TB20.7C，在TA,TB全部达标的情况下，A小区室温方差大于B小区室温方差，B小四、解答题2区室温波动较小，说明B小区供热更加稳定.（A小区室温方差为sA7.84，B小区室温方17．【解析】差为s24.01，以上数值仅作参考，不要求计算方差具体值）.·····························10分（Ⅰ）A小区当年随机抽取的20天数据中，供热等级达到“舒适”的有15天，所以可以估计A小B153赋分说明：区一天中供热等级达到“舒适”的概率为，··················································2分204①只做判断没能说明理由的不给分；那么，在当年的供热期内，3②给出一个正确理由的给3分，给出两个及以上正确理由的给4分；A小区供热等级达到“舒适”的天数约为172129天········································3分4③除以上理由外，其它符合统计概率知识的判断依据都可酌情给分.高三年级第二次模拟考试数学试题参考答案第1页（共5页）18．【解析】nABx3y0,x3y,则nAB,nAC，（Ⅰ）证明：nACx3z0,x3z,取BC中点G，连接AG,EG，取z1，则x3,y1，n(3,1,1)是平面ABC的一个法向量，CC11为中点，，，EB1CGE//BB1GEBB1332AEn13(1)16GEcosAE,n22，在三棱柱中，，|AE||n|105ABCA1B1C1AA1//BB1,AA1BB152AFA1为中点，，FAA1GE//AF,GEAFAAB6B1设直线AE与平面ABC所成角为，则sin|cosAE,n|，四边形AGEF为平行四边形，EF//GA，5又平面，平面平面分GAABCEFABCEF//ABC.··································56即直线AE与平面ABC所成角的正弦值为.··················································12分（Ⅱ）解：5在平行四边形ABBA中，ABAA,ABB,19．【解析】11113（Ⅰ）sin2Bsin2CsinBsinCsin2A平行四边形ABB1A1为菱形，zCC1由正弦定理可得b2c2bca2b2c2a2bc连接AB1，则ΔAA1B1为正三角形，E222为中点，，bcabc1FAA1B1FAA1由余弦定理得cosA2bc2bc2同理可证，xAFA1CFAA1A(0,)A······················································································5分BB13又，，B1FACACAA1Ay3设ΔABC外接圆半径为R，则R3，由正弦定理得a2RsinA233B1F平面AA1C1C2····················································································································6分以F为原点，FA,FB,FC所在直线分别为x轴，y轴，z轴，建立如图所示空间直角坐1注意：求角未写范围扣1分.标系Fxyz，（Ⅱ）由（Ⅰ）知a3,A，由余弦定理a2b2c22bccosA333F(0,0,0),A(1,0,0),B(2,3,0),B(0,3,0),C(0,0,3),E(0,,)，得9b2c22bccos9(bc)23bc12233(bc)2333bc(bc)29AB(1,3,0),AC(1,0,3),AE(1,,)，··································8分422(bc)236bca3bc6.设n(x,y,z)是平面ABC的法向量，当且仅当bc3时取等号··············································································8分高三年级第二次模拟考试数学试题参考答案第2页（共5页）3mbc令Tm2024，得m4048······································································8分112又由等面积法可知bcsinA(abc)rr222abc②当n4k1,(kN*)时，3(bc)292(bc)93Tn103(0507)(09011)0(n2)0nbc，r23(bc3)····························10分3bc36n122223320bc33,0(bc3)n3个6243m1r的取值范围为(0，]···············································································12分令T2024，得m4047································································9分2m220．【解析】③当n4k2，(kN*)时，（Ⅰ）a24··········································································································1分Tn10(3050)(7090)(n2)0(n1)0a39··········································································································2分n2nnn（Ⅱ）由，1(2)(2)(2)1an12ancos2sin2222n2个nn4可得acos2a2sinn12n2m（n1）nn令，得舍去分即（）（*）分Tm2024m4048··························································10an1sin2an2sin2ansin,nN·····················42222又因为asin20④当n4k3，(kN*)时，12n所以{asin}是首项为2，公比为2的等比数列············································5分Tn1(0305)(0709)(0(n2)0n)n2nn所以n，即n，*·········································6分n1n1ansin2an2sinnN1(2)(2)(2)12222n1nn*个（Ⅲ）n(a2)nsin，(nN)·····································································7分4n2①当n4k，(kN*)时，m1令T2024，得m4049舍去······················································11分m2T(1030)(5070)(n3)0(n1)0n综上：m4048或4047··············································································12分21．【解析】n22221n（Ⅰ）由题可知S2cbbc2····························································1分个ΔPF1F242高三年级第二次模拟考试数学试题参考答案第3页（共5页）36k2t216t2216在椭圆上cosFPFcos2OPF2cosOPF1cosOPFMC2222112223233(23k)2(23k)b6即24t2(23k2)6(23k2)223k204t223k2在RtOPF中，cosOPF·····························································2分22a34t223k2t2，符合0a2b2c2································································································3分6kt3k4t1x，y·································