四川省泸州市2023届高三第二次教学质量诊断性考试理科数学试题

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：

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：

，·······················································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