福建省福清市2023-2024学年高二下学期期末质量检测数学试题

2023-2024学年第二学期高二年期末质量检测数学学科试卷注意事项：1.答题前，考生务必在试题卷、答题卡规定的地方填写自己的准考证号、姓名，考生要认真核对答题卡上粘贴的条形码的“准考证号、姓名”与考生本人准考证号、姓名是否一致.2.第Ⅰ卷每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑.如需改动，用橡皮擦干净后、再选涂其它答案标号.第Ⅱ卷用0.5毫米黑色签字笔在答题卡上书写作答.在试题卷上作答，答案无效；3.考试结束，考生必须将答题卡交回.第Ⅰ卷一、单项选择题：本大题共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．已知数列满足，，则A．7 B．8 C．10 D．112．福厦高铁全线共设8个客运站：福州南、福清西、莆田、泉港、泉州东、泉州南、厦门北、漳州，则铁路部门应为福厦高铁线上的这8个站间准备不同的火车票的种数为（）A．28 B．56 C．64 D．1123．已知函数，则（）A．0 B．2 C．3 D．44．将4个不同的小球全部投入3个不同的盘子（每个盒子容纳的小球的个数不限），则所有的投放方法数为（）A． B． C． D．5．已知函数，则“”是“在上单调递增”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6．地图涂色是一类经典的数学问题.如图，用4种不同的颜色涂所给图形中的4个区域，要求相邻区域的颜色不能相同，则不同的涂色方法有（）A．24种 B．48种 C．72种 D．84种7．在等差数列中，，则（）A．7 B．11 C．14 D．168．已知函数，则（）A． B．C． D．二、多项选择题：本题共3小题，每小题6分，共18分。在每小题给出的四个选项中，有多项是符合题目要求的。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．下列求导运算正确的是（）A．若，则 B．若，则C．若，则 D．若，则10．已知函数，则（）A．的极大值点为 B．的极大值为C．有两个零点 D．直线是曲线的一条切线11．如图，满足，，以的斜边为第2个直角三角形的直角边，且，再以的斜边为第3个直角三角形的直角边，且，依此方法一直继续下去，记第个直角三角形为，则（）A． B． C． D．第Ⅱ卷三、填空题：本大题共3小题，每小题5分，共15分.12．在等比数列中，，，则__________.13．甲、乙、丙、丁4人到三所学校去应聘，若每人恰被一所学校录用，每所学校至少录用其中1人，则所有不同的录用情况种数为__________（用数字作答）.14．已知函数有且仅有一个零点，则实数的取值范围为__________.四、解答题：本大题共5小题，共7分。解答应写出文字说明、证明过程或演算步骤.15．（本小题满分13分）按要求列出式子，再计算结果，用数字作答.（1）在5件产品中，有3件正品，2件次品，从这5件产品中任意抽取3件.（ⅰ）抽出的3件中恰有1件正品的抽法有多少种？（ⅱ）抽出的3件中至少有1件次品的抽法有多少种？（2）现有，，等5人排成一排照相，按下列要求各有多少种不同的排法.（ⅰ）若，之间恰有一人，有多少种不同的排法？（ⅱ）不站左端，且不站右端，有多少种不同的排法？16．（本小题满分15分）设为数列的前项和，已知，.（1）求证：是等差数列；（2）求数列的前项和.17．（本小题满分15分）设是函数的导函数，是函数的导数，若方程有实数解，则称点为的“拐点”.经过探究发现：任何一个三次函数都有“拐点”且“拐点”就是三次函数图象的对称中心.已知三次函数的对称中心为.（1）求实数，的值；（2）求的极值.18．（本小题满分17分）已知数列与等差数列，若，，.（1）求，的通项公式；（2）求数列的前项和.19．（本小题满分17分）已知函数.（1）若，求曲线在点处的切线的斜率；（2）若，讨论的单调性；（3）若，且时，恒成立，求实数的取值范围.2023—2024学年第二学期高二年期末质量检测数学参考答案及评分细则评分说明：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标准制定相应的评分细则.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给分数的一半；如果后继部分的解答有较严重的错误，就不再给分.3．解答右端所注分数，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、单项选择题：本题共8小题，每小题5分，满分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．A2．B3．C4．A5．A6．D7．C8．D二、多项选择题：本题共3小题，每小题6分，满分18分．在每小题给出的四个选项中，有多项是符合题目要求的．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．BC10．ABD11．ABD二、填空题：每小题5分，满分15分．12．413．3614．三、解答题：本大题共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．（1）（ⅰ）抽出的3件中恰有1件次品是指1件正品，2件次品，则有种不同的抽法；······················································3分（ⅱ）解法一：抽出的3件中至少有1件次品的抽法有两种情况：只有1件次品的抽法和2件次品的抽法，由（ⅰ）得有2件次品的抽法为种不同的抽法，只1件次品的抽法为种不同的抽法，共有种不同的抽法.······················································6分解法二：抽出的3件中至少有1件次品的抽法数，是在5件产品中任意抽出3件的抽法数，减去抽出的3件产品全是正品的抽法数，所以共有种不同的抽法.··············6分（2）（ⅰ）将A、某人、B看作一个整体，进行捆绑，再将另外两人一起排列，所以一共有36种排法．······················································9分（ⅱ）解法一：因为5个人全排列有排法，······················································10分且A站左端有种排法，B站右端有种排法，A站左端且B站右端有种排法，···············11分所以A不站左端，且B不站右端有种排法.···································13分解法二：依题意可得：整件事可分为B站左端，和B不站左端．若B站左端，则其他4人全排列，有种排法；····························································10分若B不站左端，则其他3人中选出1人站在左端，有种选法，又由于B不站左端，也不站右端，有种排法，剩下3人有有种排法，所以B不站左端有排法；·······························11分所以A不站左端，且B不站右端有排法.·····································13分（注：如若没给解释说明，但是式子正确，也给满分.）16．（1）当时，，则.···············································1分因为①.所以时，②···························································2分由①－②得时，，即.···························4分因为，所以，即.············································6分故是以1为首项，1为公差的等差数列.··················································7分（2）由（1），得.························································9分所以，·············································································10分························································12分··········································································14分.·················································································15分17．（1）因为，所以，··································································1分所以，························································2分又因为函数的对称中心为，所以，············································6分即，解得.···························································8分（2）由（1）知，，····················································9分所以，····························10分由，得或，·································································11分当变化时，，的变化情况如下表所示：12002················································································································14分因此，的极大值为，极小值为.·····································15分18．（1）因为，所以，······································1分又，得.················································································2分所以数列是以2为首项，2为公比的等比数列.·······································3分所以，故.·······································4分则.····························································································5分设等差数列的公差为，则，解得.····························6分所以.···································································7分（2）由（1）知，，，所以，···································································8分所以············································9分···································10分两式相减，得·····································12分····················································13分··············································································14分故.········································································15分19．（1）因为，，所以，所以，···································································1分又因为函数的图象过点，所以，····················································································2分即，解得，························································3分所以，····················································································4分即曲线在点处的切线的斜率为；······································5分（