四川省合江县马街中学校2026届高三上学期第二次诊断性模拟考试数学试卷（含答案）

高2023级高三第二次教学质量诊断性模拟考试数学试题注意事项：答卷前，考生务必用黑色字迹钢笔或签字笔将自己的姓名、考生号、考场号和座位号填写在答题卡上。2．考生必须保持答题卡的整洁。第=1\*ROMANI卷选择题（58分）一、选择题：本题共8小题，每小题5分，共40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．已知集合，则A． B．C． D．2．下列命题为真命题的是A．若，则 B．若，则C．若，则 D．若，则3．已知，则“”是“”的A．充分不必要条件 B．必要不充分条件C．充要条件 D．既不充分也不必要条件4．记为等差数列的前n项和.若则A． B． C． D．5．已知，且满足，则在上的投影向量为A． B． C． D．6．已知的展开式中，前三项的系数依次成等差数列，则展开式中二项式系数最大的项是A． B． C． D．7．某电视台计划在春节期间某段时间连续播放6个广告，其中3个不同的商业广告和3个不同的公益广告，要求第一个和最后一个播放的必须是公益广告，且商业广告不能3个连续播放，则不同的播放方式有A．144种 B．72种 C．36种 D．24种8．已知的定义域为，且，则A． B． C． D．二、选择题：本题共3小题，每小题6分，共18分。在每小题给出的选项中，有多项符合题目要求。全部选对的得6分，部分选对的得部分分，有选错的得0分。9．若z是非零复数，则下列说法正确的是A．若，则 B．若，则C．若，则 D．若，则10．某校在运动会期间进行了一场“不服来战”对抗赛，由篮球专业的1名体育生组成甲组，3名非体育生的篮球爱好者组成乙组，两组进行对抗比赛.具体规则为甲组的同学连续投球3次，乙组的同学每人各投球1次.若甲组同学和乙组3名同学的命中率依次分别为，则A．乙组同学恰好命中2次的概率为B．甲组同学恰好命中2次的概率小于乙组同学恰好命中2次的概率C．甲组同学命中次数的方差为D．乙组同学命中次数的数学期望为11．在三棱锥中，平面平面，，则A．三棱锥的体积为1B．点到直线AD的距离为C．二面角的正切值为2D．三棱锥外接球的球心到平面的距离为第=2\*ROMANII卷非选择题（92分）三、填空题：本题共3小题，每小题5分，共15分．12．计算：．13．函数（）的最大值是．14．过双曲线：右焦点作直线，且直线与双曲线的一条渐近线垂直，垂足为A，直线与另一条渐近线交于点B．且点A，B位于x轴的异侧，O为坐标原点，若的内切圆的半径为，则双曲线C的离心率为．四、解答题：本题共5小题，共77分．解答应写出文字说明、证明过程或演算步骤．15．（13分）某市统计了2025年4月的空气质量指数（AQI），将其分为，，，的4组，画出频率分布直方图如图所示.若，称当天空气质量达标；若，称当天空气质量不达标.(1)求；(2)从4月的30天中任取2天，求至少有1天空气质量达标的概率；(3)若2025年6月的30天中有8天空气质量达标，请完成下面2×2列联表，根据小概率值的独立性检验，能否认为空气质量是否达标与月份有关联？月份空气质量合计达标不达标4月6月合计附：，0.10.050.012.7063.8416.63516．（15分）如图，直三棱柱的体积为4，的面积为．(1)求A到平面的距离；(2)设D为的中点，，平面平面，求二面角的正弦值．17．（15分）如图，在中，，，，点D在边BC的延长线上.(1)求的面积；(2)若，，求CE的长.18．（17分）已知椭圆（）的离心率为，其上焦点与抛物线的焦点重合．

(1)求椭圆的方程；(2)若过点的直线交椭圆于点，同时交抛物线于点（如图1所示，点在椭圆与抛物线第一象限交点上方），试比较线段与长度的大小，并说明理由；(3)若过点的直线交椭圆于点，过点与直线垂直的直线交抛物线于点（如图2所示），试求四边形面积的最小值．19．（17分）设函数，曲线在点处的切线方程为．(1)求的值；(2)设函数，求的单调区间；(3)求的极值点个数．

高2023级高三第二次教学质量诊断性模拟考试数学试题参考答案一．单选题题号12345678答案CBABDCBB二．多选题题号91011答案BCDBCDACD三．填空题12．13．114．四．解答题15．解：（1）依题意得，，·······································3分解得．··············································································4分（2）由频率分布直方图知，4月份的空气质量达标的天数为：，·································5分则4月份的空气质量不达标的天数为：，···············································6分则任取2天，至少有1天空气质量达标的概率为：．··································8分（3）列联表如下：···········································································10分月份空气质量合计达标不达标4月1218306月82230合计204060零假设：空气质量是否达标与月份无关，则·····································12分所以根据小概率值的独立性检验，没有充分理由推断假设不成立，故不能认为空气质量是否达标与月份有关联．····················································13分16．解：（1）在直三棱柱中，设点A到平面的距离为h，则，································3分解得，·················································································5分所以点A到平面的距离为；····························································6分（2）取的中点E,连接AE,如图，因为，所以,又平面平面，平面平面，·······································7分且平面，所以平面，在直三棱柱中，平面，由平面，平面可得，，·····································8分又平面且相交，所以平面，·········································9分所以两两垂直，以B为原点，建立空间直角坐标系，如图，·····························10分由（1）得，所以，，所以，则,所以的中点，则，,·····················································11分设平面的一个法向量，则，可取，···········································································12分设平面的一个法向量，则，可取，···········································································13分则，···························································14分所以二面角的正弦值为.···············································15分17．解：（1）中，，··························2分因为，，所以，····················································3分所以，因为，·································································5分所以；···································7分（2）方法1：因为,所以,所以，则.················································································9分方法2：在中，由余弦定理得，因为为线段上靠近的三等分点，所以，·············································································11分因为，所以，···········································································12分因为为锐角，所以，························································13分在中，由余弦定理得，，·····························14分所以.············································································15分18．解：（1）由题意得，即：，又，所以，·······························2分由，得，所以椭圆的方程为.···········································3分（2）由题意得过点的直线的斜率存在，设直线方程为，设，，，，联立，消去得：，·············································4分则，，所以.·······································5分抛物线的方程为：，联立，消去得：，则，所以，······················································7分所以，·············································8分即.················································································9分（3）设，，，，当直线的斜率存在且不为零时，设直线方程为，则直线方程为，由（2）的过程可知：，·······················································11分由，以替换，可得，·········································12分所以，·································································14分因为，所以，，；················15分当直线的斜率不存在时，，，所以；·····················································16分综上所述：，所以四边形面积的最小值为.··································17分19．解：（1）因为，所以，··························1分因为在处的切线方程为，·所以，，·······························································2分则，解得，···························································3分所以.（2）由（1）得，则，················4分令，解得，不妨设，，则，易知恒成立，所以令，解得或；令，解得或；所以在，上单调递减，在，上单调递增，即的单调递减区间为和，单调递增区间为和.··········7分（3）由（1）得，，由（2）知在，上单调递减，在，上单调递增，当时，，，即所以在上存在唯一零点，不妨设为，则，此时，当时，，则单调递减；当时，，则单调递增；所以在上有一个极小值点；·························································10分当时，在上单调递减，则，故，所以在上存在唯一零点，不妨设为，则，此时，当时，，则单调递增；当时，，则单调递减；所以在上有一个极大值点；··························································12分当时，在上单调递增，则，故，所以在上存在唯一零点，不妨设为，则，此时，当时，