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

这是一份福建省福清市2023-2024学年高二下学期期末质量检测数学试题，共8页。试卷主要包含了考试结束，考生必须将答题卡交回等内容，欢迎下载使用。
注意事项：
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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．7B．8C．10D．11
2．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．28B．56C．64D．112
3．已知函数，则（ ）
A．0B．2C．3D．4
4．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．7B．11C．14D．16
8．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．A 2．B 3．C 4．A
5．A 6．D 7．C 8．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．BC 10．ABD 11．ABD
二、填空题：每小题5分，满分15分．
12．4 13．36 14．
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：
················································································································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
（2）因为，所以，所以，························6分
当时，，在区间上单调递增；······················7分
当时，令，解得，······································8分
当时，；当时，；
所以函数在单调递减，在单调递增；·····················10分
综上：当时，在区间上单调递增；
当时，在上单调递减，在上单调递增. ············11分
（3）因为，所以，所以，
设，则，
当时，，所以在上单调递增，且；······12分
①当时，，即，
所以函数在上单调递增，····························································13分
所以当时，，所以符合题意，.························14分
②当时，又在上单调递增，且，
当时，，
，使得，·································································15分
，，即，所以在上单调递减，
，，即，所以在上单调递增，
所以，所以不合题意. ········································16分
综上，实数的取值范围为.···································································17分1
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