2023泸州高三下学期3月第二次教学质量诊断性考试数学（理）PDF版含答案

这是一份2023泸州高三下学期3月第二次教学质量诊断性考试数学（理）PDF版含答案，文件包含泸州市20级二诊数学答案docx、四川省泸州市高2020级第二次教学质量诊断性考试数学理科2023-03-01pdf等2份试卷配套教学资源，其中试卷共11页， 欢迎下载使用。
泸州市高2020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AADCCDCBBBCC 二、填空题： 13．1；     14．中的任意一个值；   15．；      16．．三、解答题：17．解：（Ⅰ）因为，  ①所以当时，，   ②············································1分由①，②相减得：，··········································· 2分即，······················································· 3分在中令得，，即，············································· 4分所以数列是以为首项，公比为的等比数列，··························· 5分所以；·····················································6分（Ⅱ）若选①．因为··························································7分··························································8分所以······················································10分．·······················································12分若选②．···················································7分··························································8分所以······················································11分．·······················································12分18．解：（Ⅰ）设“该考生报考甲大学恰好通过一门笔试科目”为事件A，“该考生报考乙大学恰好通过一门笔试科目”为事件，根据题意得：，·······················································2分························································3分；·······················································4分（Ⅱ）设该考生报考甲大学通过的科目数为，报考乙大学通过的科目数为，根据题意可知，，所以，，······································5分，························································6分，························································7分，························································8分．························································9分则随机变量的分布列为：0123，·······················································10分若该考生更希望通过乙大学的笔试时，有，··························11分所以，又因为，所以，所以m的取值范围是．···························12分19．证明：（Ⅰ）分别延长B1D，BA，设，连接CE，····················1分则CE即为平面与平面的交线，······················2分因为，取中点F，连接DF，························3分所以，平面，因为平面平面，且交线为，所以平面，·····················································4分因为D为棱的中点，，所以D为的中点，所以，············································5分所以平面；·····················································6分方法一：（Ⅱ）由（Ⅰ）知，因为，，所以， 在平面内过点C作，垂足为G，则平面，·································7分分别以CB，CE，CG所在直线为x，y，z轴，建立如图所示的空间直角坐标系，设，则，，，···················································8分则，，，，·····················································9分设平面的法向量为，则，取，······················································10分设平面的法向量为，则，取，······················································11分所以，即二面角的余弦值为．············································12分方法二：连接BF，因为四边形为菱形，且，所以，·····································7分平面，因为平面平面，且交线为，所以平面，··································8分过点F作，连接，所以，故为二面角的平面角，·············································9分在中，，，，所以，························································10分在中，，所以，·················································11分所以，即二面角的余弦值为．·······································12分20．解：（Ⅰ）因为在C上，所以，·············································1分因为C的左焦点，所以，············································2分所以，，的方程为；··················································4分（Ⅱ）①当直线与x轴重合时，点，，，，，，所以，··················································5分②当直线与x轴不重合时，设直线的方程为，代入消去x得，因为直线与C交于点，，所以，···································6分因为，·····················································7分所以，·····················································8分（1）当m≠0时，同理可得，········································9分，·······················································10分因为，所以的取值范围是，··········································11分（2）当时，，综上知的取值范围是.··········································12分21．解：（Ⅰ），·························································1分因为是函数的一个极值点，所以，得，··················································2分所以，因此在上单减，在上单增，······································3分所以当时，有最小值；·········································4分方法一：（Ⅱ）因为，所以，则在上单增，···········································5分记，当时，，当时，，则，·······················································6分记，当时，；当时，；7分所以存在唯一的，使得，当时，；当时，，所以函数在上单减，在上单增，···································8分若函数有两个零点，只需，即，又，即，···················································9分则，设，则为增函数，，所以当时，，则，即，··················································10分令，，则在上单增，由得，··········································11分所以，所以a的取值范围是．·········································12分方法二：（Ⅱ）若有两个零点，即有两个解，即有两个解，········································5分利用同构式，设函数，·········································6分问题等价于方程有两个解，······································7分恒成立，即单调递增，所以，问题等价于方程有两个解，······································8分即有两个解，设，，即有两个解，令，问题转化为函数有两个零点，·································9分因为，当时，，当时，，则在上递增，在上递减，·······································10分为了使有两个零点，只需，解得，即，解得，············································11分由于，所以在和内各有一个零点．综上知a的取值范围是．········································12分22．解：（Ⅰ）由，得，·····················································1分所以，·····················································2分又，，·····················································3分所以，·····················································4分即的直角坐标方程为；··········································5分（Ⅱ）曲线的普通方程为：，·········································6分直线的参数方程为：为参数），···································7分代入整理得：，··············································8分设A，B两点所对应的参数分别为，，则，因为，所以，即或，·············································9分因为，或，满足，所以或．····················································10分23．解：（Ⅰ）因为，························································1分若对，恒成立，则，···········································2分所以，或，··················································4分所以实数m的取值范围是；······································5分（Ⅱ）由（Ⅰ）知，的最小值为，所以，································6分所以或，因为，所以，即，·······················································7分由柯西不等式得··························································8分··························································9分，所以（当且仅当，，时等号）．··································10分