2022届四川省泸州市高三二模数学文科试卷及答案

这是一份2022届四川省泸州市高三二模数学文科试卷及答案，共11页。试卷主要包含了 选择题的作答， 填空题和解答题的作答，8．等内容，欢迎下载使用。
泸州市高2019级第二次教学质量诊断性考试数  学（文科）    本试卷分第I卷（选择题）和第II卷（非选择题）两部分．第I卷1至2页，第II卷3至4页．共150分．考试时间120分钟．注意事项：1. 答题前，先将自己的姓名、准考证号填写在试卷和答题卡上，并将准考证号条形码粘贴在答题卡上的指定位置．2. 选择题的作答：每小题选出答案后，用2B铅笔把答题卡上对应题的答案标号涂黑．3. 填空题和解答题的作答：用签字笔直接答在答题卡上对应的答题区域内，作图题可先用铅笔绘出，确认后再用0.5毫米黑色签字笔描清楚，写在试题卷、草稿纸和答题卡上的非答题区域均无效．4．考试结束后，请将本试题卷和答题卡一并上交．第I卷 （选择题  共60分）一、选择题：本题共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．设集合，，则 A．         B．   C．    D．2．复数的虚部为 A．     B．     C．    D． 3. 已知变量，满足，则的最大值为A．3                    B．2                C．1               D．0  4．已知，则A．     B．              C．    D．5．气象意义上从春季进入夏季的标志为“连续5天的日平均温度均不低于”，现有甲、乙、丙三地连续5天的日平均温度的记录数据（记录数据都是正整数，单位： ):①甲地：5个数据的中位数为24，众数为22；②乙地：5个数据的中位数为27，总体均值为24；③丙地：5个数据中有1个数据是32，总体均值为26，总体方差为10.8．其中能够确定进入夏季地区的有 A．①②              B ．②③              C．① ③             D．①②③6．已知曲线在点处的切线方程为，则a的值是A．       B．               C．    D．7．的内角A，B，C的对边分别为a，b，c，已知，，，则的值是A．6     B．8      C．4    D．28．已知双曲线C：的焦点到C的一条渐近线的距离为2，则C的离心率是A．          B．     C．           D．9．已知定义域为R的函数在上为减函数，且函数为偶函数，则  A．       B．   C．       D．10．如图，某几何体的三视图均为边长为2的正方形，则该几何体的体积是A．    B． C．    D． 11．已知等腰直角的顶点都在表面积为的球O的表面上，且球心O到平面的距离为1，则的面积为A．4    B．8      C．    D．12．已知，，且成立，则下列不等式不可能成立的的是A．   B．     C．   D．第II卷 （非选择题　共90分）注意事项：(1)非选择题的答案必须用0.5毫米黑色签字笔直接答在答题卡上，作图题可先用铅笔绘出，确认后再用0.5毫米黑色签字笔描清楚，答在试题卷和草稿纸上无效.(2)本部分共10个小题，共90分.二、填空题（本大题共4小题，每小题5分，共20分．把答案填在答题纸上）.13．已知，则            ．14．写出一个具有下列性质①②③的函数            ．① 定义域为R ； ② 函数是奇函数；③ ． 15．已知向量，，若，则的值为            ．16．已知P为抛物线上一个动点，Q为圆上一个动点，那么点P到点Q的距离与点P到直线的距离之和的最小值是          ．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．（本小题满分12分）设正项数列的前项和为，，且满足______．给出下列三个条件：①，；②；③．请从其中任选一个将题目补充完整，并求解以下问题．（Ⅰ）求数列的通项公式；（Ⅱ）若，且数列的前项和为，求的值．18．（本小题满分12分）某县充分利用自身资源，大力发展优质李子树种植项目．该县农科所为了对比A，B两种不同品种脆红李的产量，各选20块试验田分别种植了A，B两种脆红李，所得的20个亩产数据（单位：100kg）都在内，根据亩产数据得到频率分布直方图如下图：（Ⅰ）从B种脆红李亩产量数据在内任意抽取2个数据，求抽取的2个数据都在内的概率；（Ⅱ）根据频率分布直方图，用平均亩产量判断应选择种植A种还是B种脆红李，并说明理由．19．（本小题满分12分）已知空间几何体ABCDE中，，是全等的正三角形，平面平面，平面平面． （Ⅰ）若，求证：;（Ⅱ）探索，，，四点是否共面？若共面，请给出证明；若不共面，请说明理由．20．（本小题满分12分）已知函数．（Ⅰ）求证：；（Ⅱ）若函数无零点，求a的取值范围．21．（本小题满分12分）已知椭圆的左，右顶点分别为，，且，椭圆过点．（Ⅰ）求椭圆的标准方程；（Ⅱ）斜率不为0的直线与交于M，N两点，若直线的斜率是直线斜率的两倍，证明直线经过定点，并求出定点的坐标．（二）选考题：共10分.请考生在第22、23题中任选一题作答，如果多做，则按所做的第一题计分.22．（本题满分10分）选修4-4：坐标系与参数方程在平面直角坐标系中，曲线的参数方程为（为参数），若曲线上的点的横坐标不变，纵坐标缩短为原来的倍，得到曲线．以坐标原点为极点，轴的非负半轴为极轴建立极坐标系．（Ⅰ）求曲线的极坐标方程；（Ⅱ）已知直线与曲线交于，两点，若，求的值．23．（本题满分10分）选修4-5：不等式选讲已知a，b，c为非负实数，函数．（Ⅰ）当，，时，解不等式；（Ⅱ）若函数的最小值为2，证明：．泸州市高2019级第二次教学质量诊断性考试数  学（文科）参考答案及评分意见评分说明：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参考制订相应的评分细则．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得分数的一半；如果后继部分的解答有较严重的错误，就不再给分．3．解答右侧所注分数，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，选择题和填空题不给中间分．一、选择题：题号123456789101112答案BCBCCDADBCBD 二、填空题：13．1；   14．等；   15．；  16．4．三、解答题：17．解 ：（Ⅰ）选①．由得：，·············································1分所以，因此数列为等比数列，····································2分设数列的公比为，则，由，······································3分解得或（舍去），·············································5分所以；·····················································6分                                                选②．因为，当时，，又，················································1分所以，即，所以，············································2分所以当时，，················································3分两式相减得，················································4分所以，·····················································5分所以数列是，公比为2的等比数列，所以；···························6分 选③．因为，当时，，···················································1分所以，即，··················································2分所以当时，，················································3分所以，·····················································4分即，·······················································5分当时，，上式也成立．； ·······················································6分（Ⅱ）因为·························································7分，························································9分设·························································10分，所以．····················································12分18．解：（Ⅰ）B种脆红李亩产量数据在内的有：，························································1分其中数据在的有：个，记为M，N，································2分数据在的有：个，记为a，b，c，··································3分从B种脆红李亩产量数据在内任意抽取2个数据，基本事件为MN，Ma，Mb，Mc，Na，Nb，Nc，ab，ac，bc，共有10种，·············4分抽取的2个数据都在内包含的基本事件有ab，ac，bc，共有3种，············5分所以抽取的2个数据都在内的概率为；·······························6分（Ⅱ）根据频率分布直方图，A品种脆红李的平均亩产为： 7分， 8分B品种脆红李的平均亩产为： 9分， 10分因为A品种脆红李的平均亩产小于B品种脆红李的平均亩产，所以用平均亩产来判断应选择种植B种脆红李．·······················12分19．解: （Ⅰ）因为，因为，是全等的正三角形，所以，·································1分所以，·····················································2分故，······················································3分因为平面平面，所以BC平面，···············································4分故；······················································5分（Ⅱ），，，四点共面，理由如下：分别取，中点，，连接，，······································6分因为是等边三角形，所以，，····································7分因为平面平面， 所以平面BCD，同理平面BCD，·················8分且，所以，····································9分又，所以，·······························10分所以四边形是平行四边形，·····································11分所以，又，所以，即，，，四点共面．·····································12分 20．证明：（Ⅰ），所以，···················································1分令得：，··················································2分所以在上单调递增，在上单调递减，······························3分当时，取最大值，所以，所以；···················································4分（Ⅱ）因为，所以，···················································5分当时，，在定义域上无零点；··································6分当a > 0时，，所以，令得：，所以在上单调递增，在上单调递减，······························7分当时，取最大值，···········································8分因为无零点，所以，即；······································9分当时，因为，所以，········································10分即，····················································11分所以在定义域上无零点．综上，a的取值范围是．······································12分21．解：（Ⅰ）由题知，，得，·················································1分由椭圆过点，有，得，·········································3分所以椭圆的标准方程为；········································4分（Ⅱ）设，，因为点，点，直线方程为，因为，所以，即，①···········································5分因为，即，把代入 ①式中，得，所以，·····················································6分即，即，   ②··················································7分由 ，消去得，·················································8分所以 ，，    ③··············································9分把③代入②中，得，··········································10分所以（舍去），·············································11分所以直线过定点．············································12分22．解：（Ⅰ）由消去参数可得的普通方程为：，··································1分曲线上的点的横坐标不变，纵坐标缩短为原来的倍，则曲线的直角坐标方程为：，······················2分整理可得：， 即，············································3分因为，，所以，··············································4分所以曲线的极坐标方程为：；····································5分（Ⅱ）设，，则，为方程的两根，········································6分由韦达定理，① ·············································7分，                ②·········································8分由，，    ③················································9分联立①②③可得：，所以，，所以直线的斜率．············································10分23．解：（Ⅰ）当，，时，，·················································1分，························································3分由，解得：，················································4分所以不等式的解集为；·········································5分（Ⅱ），因为的最小值为2，所以．·······································6分证法一：根据柯西不等式可得：··························································7分，························································9分当且仅当：，即，，时取等号．综上，．···················································10分证法二：··························································7分··························································8分， 9分当且仅当，，时取等号，综上，．··················································10分