河南名校联盟2022-2023年高二下学期期中联考数学参考答案

这是一份河南名校联盟2022-2023年高二下学期期中联考数学参考答案，共7页。试卷主要包含了选择题，填空题，解答题等内容，欢迎下载使用。
2022—2023学年下期期中联考高二数学参考答案一、选择题（本大题共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）题号123456789101112答案BAABBCCCADDB1．【答案】B【解析】，则.故选B.2．【答案】A【解析】买两本，有3种方案；买三本，有1种方案，因此共有方案3＋1＝4(种)．故选A.3．【答案】A【解析】由，得，故选A.4．【答案】B【解析】由题意，所以，所以.故选B.5．【答案】B【解析】由题可知，，则，∴，又，∴函数的图象在点处的切线方程为.故选B.6．【答案】C【解析】，令，得，当，，为减函数，当，，为增函数，则故选C.7．【答案】C【解析】从8个车位里选择5个相邻的车位，共有4种方式，选一种方式将5辆车相邻停放，有种方式，则不同的泊车方案有种，故选C.8．【答案】C【解析】，的展开式的通项公式为，所以展开式中项的系数是，展开式中项的系数是，所以，解得，故选C.9．【答案】A【解析】设，则，所以在上单调递减，又，由，即，所以，所以，故选A.10．【答案】D【解析】在同一坐标系中作出的图像，如图所示：若有两个极值点，则或.故选D.11．【答案】D【解析】由题意，5名实习教师的分配方案为，先将5名实习教师分组，有种，再将分好组的教师分配给3个班级，有种，其中不去甲班则不同的分配方案有种.故选D.12．【答案】B【解析】令，则，则在上单调递增，则，即，即；令，则，则在上单调递减，则，即，即；故，即.故选B.二、填空题（本大题共4小题，每小题5分，共20分）13．【答案】64【解析】因为4科老师布置了作业，在同一时刻每个学生做作业的情况有4种可能，所以3名学生都做作业的可能情况4×4×4=64种．故答案为64.14．【答案】e【解析】由．令，解得．当x变化时，与的变化情况如下表：＋0－单调递增单调递减因此，极大值点为.15．【答案】50【解析】设三位数的回文数为ABA，A有1到9，共9种可能，即1B1、2B2、3B3…其中奇数共5种可能，即1B1，3B3，5B5，7B7，9B9；B有0到9共10种可能，即A0A、A1A、A2A、A3A、…符合题意的有5×10=50个.16．【答案】【解析】由，，所以，①若，，符合题意；②若，令，则在恒成立，∴在单调递增，又，，，∴由，得；若在恒成立，则可化为，令，，在单调递减，单调递增，∴，即有．综上：.三、解答题（本大题共6小题，共70分．解答应写出文字说明、证明过程或演算步骤）17．【解析】（1）先排甲、乙，再排其余6人，共有（种）排法.························5分（2）先排4名男生有种方法，再将4名女生插空，有种方法，故共有（种）排法．······························································10分18．【解析】（1）令，得；令，得，所以.·············································6分（2）对原式两边求导得：令，得；即·································································12分19．【解析】（1）因为所以，···············································································2分由题意可得，，···························································4分解得:，.····················································································6分（2）由(1)可得，所以，且，········································8分易得，当时，，函数单调递减，当时，，函数单调递增，···············································10分又，，且，即最大值为：································································12分20．【解析】（1）根据题意可知，将分别代入两曲线方程得到，．两个函数的导函数分别是，又，，则，解得，，．······································································6分（2）要使抛物线上的点M到直线的距离最短，则抛物线在点M处的切线斜率应该与直线相同，则，解得，又因为点M在抛物线上，解得．所以最短距离即的最小值为点M到直线的距离，代入点到直线的距离公式得．即最短距离为．·············································································12分21．【解析】（1）若在上的单调递减，则在上恒成立，化简得，易知函数在上递增，因此，故·······························································································4分（2）函数，.当时，由得，.·············6分是极小值，所以只要，即.································7分令，则，在内单增.因为，所以当时，；当时，.所以实数的值是.··················································9分当时，.为上的减函数，而，，所以有且只有一个零点.······································································11分综上实数的值是或.·············································································12分22．【解析】（1）的定义域为，·························1分对于二次方程，有．当时，恒成立，在上单调递减．·······························2分当时，方程有两根，若，因为在上，所以在上单调递增；因为在上，所以在上单调递减；·················4分若，因为在与上，所以在与上单调递减，因为在上，所以在上单调递增．·······················································5分（2）证明，即证，因为，所以．····································································6分当时，不等式显然成立．···········································7分当时，因为，所以只需证，即证．···················································································9分令，则，由得；由，得．所以在上为增函数，在上为减函数，所以．，则，易知在上为减函数，在上为增函数，所以，···············································································11分所以恒成立，即．·········································12分