2022届四川省泸州市高三二模数学理科试卷及答案

这是一份2022届四川省泸州市高三二模数学理科试卷及答案，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（理科）   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，，全集为，则 A．   B．   C．    D．2．已知，则复数的虚部为  A．    B．     C．    D． 3．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．其中能够确定进入夏季地区的有 A．①②              B ．① ③             C．②③            D．①②③4. 已知变量，满足，则的最大值为A．0               B．1                C．2              D．3  5．已知命题p:，．命题q: 某物理量的测量结果服从正态分布，则该物理量在一次测量中落在与落在的概率相等．下列命题中的假命题是 A．    B．            C．      D． 6．设双曲线C：的左，右焦点分别是，，点M是C上的点，若是等腰直角三角形，则C的离心率是A．            B．2    C．          D．7．已知，则  A．     B．    C．     D．8．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． 9．如图，航空测量的飞机航线和山顶在同一铅直平面内，已知飞机飞行的海拔高度为，速度为 ．某一时刻飞机看山顶的俯角为，经过420 s 后看山顶的俯角为，则山顶的海拔高度大约为（，） A．7350m    B．2650 m    C．3650 m        D．4650 m10．2022年北京冬奥会速度滑冰、花样滑冰、冰球三个项目竞赛中，甲，乙，丙，丁，戊五名同学各自选择一个项目开展志自愿者服务，则甲和乙均选择同一个项目，且三个项目都有人参加的不同方案总数是A．     B．     C．    D．11．已知中，，，其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2，则的面积为A．2     B．4      C．8    D．1012．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．的展开式中的常数项为          （用数字作答）． 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. 等边三角形ABC的边长为1，，，，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，是数列的前项和，求证：．18．（本小题满分12分）某县种植的脆红李在2021年获得大丰收，依据扶贫政策，所有脆红李由经销商统一收购．为了更好的实现效益，质监部门从今年收获的脆红李中随机选取100千克，进行质量检测，根据检测结果制成如图所示的频率分布直方图．下表是脆红李的分级标准，其中一级品、二级品统称为优质品．等级四级品三级品二级品一级品脆红李横径/mm经销商与某农户签订了脆红李收购协议，规定如下：从一箱脆红李中任取4个进行检测，若4个均为优质品，则该箱脆红李定为类；若4个中仅有3个优质品，则再从该箱中任意取出1个，若这一个为优质品，则该箱脆红李也定为类；若4个中至多有一个优质品，则该箱脆红李定为类；其他情况均定为类．已知每箱脆红李重量为10千克，类、类、类的脆红李价格分别为每千克10元、8元、6元．现有两种装箱方案：方案一：将脆红李采用随机混装的方式装箱；方案二：将脆红李按一、二、三、四等级分别装箱，每箱的分拣成本为1元．以频率代替概率解决下面的问题．（Ⅰ）如果该农户采用方案一装箱，求一箱脆红李被定为类的概率；（Ⅱ）根据统计学知识判断，该农户采用哪种方案装箱收入更多，并说明理由．19．（本小题满分12分）已知空间几何体中，，是全等的正三角形，平面平面，平面平面． （Ⅰ）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；（Ⅱ）若，求二面角的余弦值．20．（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，探究直线是否过定点？若过定点，求出定点的坐标；若不过定点，请说明理由．21．（本小题满分12分）已知函数．（Ⅰ）求证：；（Ⅱ）若函数有两个零点，求a的取值范围．（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（理科）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．；   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分则·························································10分，所以．····················································12分18．解：（Ⅰ）从脆红李中任意取出一个，由脆红李频率分布直方图知，则该脆红李为优质品的概率是，              2分记“该农户采用方案一装箱，一箱脆红李被定为类”为事件，则························································5分；·······················································6分（Ⅱ）采用方案一装箱：记“该农户采用方案一装箱，一箱脆红李被定为类”为事件，“该农户采用方案一装箱，一箱脆红李被定为类”为事件，则，·······················································7分，·······················································8分所以该农户每箱脆红李收入的数学期望为：，·······················································9分采用方案二装箱：由题意可知，一箱脆红李被定为类的概率为，被定为类的概率为，······················10分所以该农户每箱脆红李收入的数学期望为，························11分 所以，该农户采用方案二装箱收入更多．··························12分19．解: （Ⅰ） 、、、四点共面，理由如下：分别取，中点，，连接，，·····································1分因为是等边三角形，所以，，···································2分因为平面平面， 所以平面，同理平面，且，所以，又，所以，··························3分所以四边形是平行四边形，···················4分所以，又，·······························5分所以，、、、四点共面；·····················6分（Ⅱ）取中点，连接，则，因为，，即，··············································7分，，又由（Ⅰ）可知平面，········································8分以M为坐标原点，以，，所在直线为轴，轴，轴建立如图所示坐标系，则：，，，，，······················9分所以，，，，设平面的法向量，则：，所以，令，得，·················································10分同理，设平面的法向量，则：，可得：，令，得，·················································11分，因为二面角为锐二面角，所以二面角的余弦值为．···················12分法二：（Ⅱ）由（Ⅰ）可知：四边形是等腰梯形，过点作，·························7分则，在中，，················································8分取中点，取中点，连接，，因为四边形是等腰梯形，，因为，所以，所以就是二面角的平面角，·····················9分，因为，所以，所以是以为直角的直角三角形，又为的中点，所以，····················································10分又在中，，·················································11分在中，由余弦定理可得：，所以二面角的余弦值为．·······································12分20．解：（Ⅰ）由题知，，得，·················································1分由椭圆过点，有，解得，········································3分所以椭圆的标准方程为；········································4分（Ⅱ）直线过定点，证明如下：设，，因为点，点，直线方程为，因为，所以，即，    ①········································5分因为，即，··················································6分把代入①式中，得，所以，·····················································7分即，即，   ②··················································8分由 ，消去得，·················································9分所以 ，，        ③··········································10分把③代入②中，得，··········································11分所以，或n=-2（舍去），所以直线过定点．··························12分21．证明：（Ⅰ）因为，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分因为，，所以，令得：，所以在上单调递增，在上单调递减，当时，取最大值，···········································6分因为有两个零点，所以，即，下面证明在和上分别有一个零点，因为，由（Ⅰ）知，，即，·····································7分所以， 因为在上单调递增，所以在必有一个零点；·························8分当时，，令，其中，则，，···················································9分所以在上单调递增，，所以在上单调递增，， 因此当时，，··············································10分因为，所以，即，所以，由得，，即，所以，···················································11分因为，在上必有一个零点，综上，a的取值范围是．······································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