四川省泸州市2023届高三下学期第二次教学质量诊断性考试数学(文科）试题及答案

这是一份四川省泸州市2023届高三下学期第二次教学质量诊断性考试数学(文科）试题及答案，文件包含2023二诊文科数学答案doc、四川省泸州市2023届高三下学期第二次教学质量诊断性考试数学文科试题pdf等2份试卷配套教学资源，其中试卷共10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（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AADCCDCBBBCC 二、填空题： 13．；     14．中的任意一个值；      15．；       16．．三、解答题：17．解：（Ⅰ）因为，，成等差数列，所以，·····················································1分即，······················································· 2分解得或（舍去），············································· 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，··············································8分所以是以为首项，公差为2的等差数列，·····························9分所以；····················································12分若选②．因为，···············································8分所以是以为首项，公比为的等比数列，······························9分所以······················································11分．·······················································12分18．解：（Ⅰ）由已知得，，·················································1分，························································2分因为，，···················································3分所以，·····················································4分，························································5分所以所求线性回归方程．········································6分（Ⅱ）当时，；当时，；当时，；当时，；当时，；························8分所以“次数据”有3个，设选取的2个数据恰好是2个“次数据”为事件A，因为从5个数据中选取2个数据共有10种情况，              9分其中从3个“次数据”中取到2个“次数据”有3种情况，····················11分所以．····················································12分19．证明：（Ⅰ）因为，取中点E，连接DE，····················1分所以，····································2分因为平面平面，且交线为，平面，所以平面，···························3分由已知是正三角形，所以，在中，因为CE=1，，所以DE=1,···································4分所以三棱锥的体积为；························6分（Ⅱ）分别延长，，设，连接，······································7分则即为平面与平面的交线，······································8分因为D为棱的中点，，··········································9分所以D为的中点，所以，········································10分由（Ⅰ）知平面，·············································11分所以平面.···················································12分20.解：(Ⅰ)因为，所以，······················································1分（1）当时，，函数在定义域上为增函数；····························2分（2）当时，函数的单调减区间为，·································3分（3）当时，函数的单调减区间为；·································4分(Ⅱ)分别在，上是减函数，在上是增函数，又因为曲线在点处的切线平行，所以，······················································5分不妨设，则，························································6分所以，且，···················································7分所以，······················································8分故，························································9分所以，·····················································10分即，·······················································11分所以．·····················································12分21．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