河南省鹿邑县第二高级中学
2022年河南省周口市鹿邑县第二高级中学高一数学理测试题含解析
2021-2022学年河南省周口市鹿邑县第二高级中学高一数学理测试题含解析一、选择题：本大题共10小题，每小题5分，共50分。
在每小题给出的四个选项中，只有是一个符合题目要求的1. 在ABC中,若的则ABC是（）A.钝角三角形B.直角三角形C.锐角三角形D.无法确定参考答案：A略2. 巳知全集，集合和的关系的韦恩（Ｖenn）图如右图所示，则阴影部分所示的集合的元素共有（）A．３个Ｂ．２个Ｃ．１个Ｄ．无穷个参考答案：B3. 函数在区间上的图像大致为（A）（B）（C）（D）参考答案：A略4. 的值等于( )A. B. C. D.参考答案：A= ,选A.5. 设集合，,（）A．{0,1} B．{-1,0，1}C．{0,1,2} D．{-1,0,1,2}参考答案：B考点：集合的运算试题解析：所以{-1,0，1}。
故答案为：B6. 若为三角形一个内角，且对任意实数，恒成立，则的取值范围为（）A. B. C. D.参考答案：C试题分析：依题意，方程的，解得或（舍去），又，故有，所以选择C.考点：三角函数与二次函数的综合.7. 已知则的值为（）A. B. C.D.参考答案：B略8. 已知||＝5，与的夹角为60o，在方向上的投影是A． B．3C．－ D．－3参考答案：A9. “x是钝角”是“x是第二象限角”的（）．A．充分非必要条件B．必要非充分条件C．充分必要条件 D．即不充分也不必要条件参考答案：A10. 现有1名女教师和2名男教师参加说题比赛，共有2道备选题目，若每位选手从中有放回地随机选出一道题进行说题，其中恰有一男一女抽到同一道题的概率为（）A．B．C．D．参考答案：C【考点】CB：古典概型及其概率计算公式．【分析】基本事件总数n=23=8，设两道题分别为A，B题，利用列举法求出满足恰有一男一女抽到同一题目的事件个数，由此能求出其中恰有一男一女抽到同一道题的概率．【解答】解：现有1名女教师和2名男教师参加说题比赛，共有2道备选题目，若每位选手从中有放回地随机选出一道题进行说题，基本事件总数n=23=8，设两道题分别为A，B题，所以抽取情况共有：AAA，AAB，ABA，ABB，BAA，BAB，BBA，BBB，其中第1个，第2个分别是两个男教师抽取的题目，第3个表示女教师抽取的题目，一共有8种；其中满足恰有一男一女抽到同一题目的事件有：ABA，ABB，BAA，BAB，共4种，故其中恰有一男一女抽到同一道题的概率为p=．故选：C．【点评】本题考查概率的求法，是基础题，解题时要认真审题，注意列举法的合理运用．二、填空题:本大题共7小题,每小题4分,共28分11. 已知a，b，c是三条不同的直线，α，β，γ是三个不同的平面，那么下列命题中正确的序号为．①若a⊥c，b⊥c，则a∥b；②若α⊥γ，β⊥γ，则α∥β；③若a⊥α，b⊥α，则a∥b；④若a⊥α，α⊥β，则α∥β．参考答案：③④【考点】空间中直线与平面之间的位置关系．【分析】在①中，a与b相交、平行或异面；在②中，α与β相交或平行；在③中，由线面垂直的性质定理得a∥b；在④中，由面面平行的判定定理得α∥β．【解答】解：由a，b，c是三条不同的直线，α，β，γ是三个不同的平面，知：在①中，若a⊥c，b⊥c，则a与b相交、平行或异面，故①错误；在②中，若α⊥γ，β⊥γ，则α与β相交或平行，故②错误；在③中，若a⊥α，b⊥α，则由线面垂直的性质定理得a∥b，故③正确；在④中，若a⊥α，α⊥β，则由面面平行的判定定理得α∥β，故④正确．故答案为：③④．12. 函数在[0,+∞)是增函数，，若，则x的取值范围是．参考答案：由条件知是偶函数，在是增函数，在是增函数，在上减，，则。
2019-2020学年河南省周口市鹿邑县第二高级中学高二数学理期末试卷含解析
2019-2020学年河南省周口市鹿邑县第二高级中学高二数学理期末试卷含解析一、选择题：本大题共10小题，每小题5分，共50分。
在每小题给出的四个选项中，只有是一个符合题目要求的1. 已知椭圆与双曲线的焦点相同，且椭圆上任意一点到两焦点的距离之和为，那么椭圆的离心率等于( )A. B. C. D.参考答案：B2. 已知目标函数z=2x+y且变量x，y满足下列条件，则（）A．z max = 12，z min = 3 B．z max = 12，无最小值C．无最大值，z min = 3 D．无最小值也无最大值参考答案：C3. 若复数，其中i为虚数单位，则下列结论正确的是（）A. 的虚部为B.C. 为纯虚数D. z的共轭复数为参考答案：C【分析】先得到复数的代数形式，然后根据复数的有关概念对给出的四个结论分别进行分析、判断后可得正确的结论．【详解】由题意得．对于A，由得复数的虚部为，所以A不正确．对于B，，所以B不正确．对于C，由于，所以为纯虚数，所以C正确．对于D，的共轭复数为，所以D不正确．故选C．【点睛】本题考查复数的有关概念，解题的关键是得到复数的代数形式和熟悉复数的相关概念，属于基础题．4. 已知抛物线y2=2px（p＞0）的准线和圆x2+y2+6x+8=0相切，则实数p=（）A．p=4 B．p=8 C．p=4或p=8 D．p=2或p=4参考答案：C【考点】抛物线的简单性质．【分析】将圆化成标准方程，得到圆心为C（﹣3，0），半径r=1．再将抛物线化成标准方程，得到抛物线的准线为x=﹣，根据准线与圆相切建立关于p的等式，解之即可得到p的值．【解答】解：圆x2+y2+6x+8=0化成标准方程，得（x+3）2+y2=1，∴圆心为C（﹣3，0），半径r=1，又∵抛物线y2=2px（p＞0），∴抛物线的准线为x=﹣，∵抛物线的准线与圆相切，∴准线到圆心C的距离等于半径，得|3﹣|=1，解之得p=4或p=8．故选C．5. 在中，，，其面积为，则等于( )A．3 B． C． D．参考答案：B略6. 已知抛物线的焦点是椭圆的一个焦点，则椭圆的离心率为（）A． B． C. D．参考答案：B7. 执行如图所示的程序框图，若输入的a值为1，则输出的k值为（）A．1 B．2 C．3 D．4参考答案：B【考点】程序框图．【分析】根据已知的程序框图可得，该程序的功能是利用循环结构计算并输出变量S的值，模拟程序的运行过程，可得答案．【解答】解：输入的a值为1，则b=1，第一次执行循环体后，a=﹣，不满足退出循环的条件，k=1；第二次执行循环体后，a=﹣2，不满足退出循环的条件，k=2；第三次执行循环体后，a=1，满足退出循环的条件，故输出的k值为2，故选：B8. 已知a＞0，b＞0，且ab=1，则函数f（x）=a x与函数g（x）=﹣log b x的图象可能是（）A．B．C．D．参考答案：B【考点】函数的图象；对数函数的图象与性质．【分析】根据对数的运算性质，我们易根据ab=1，进而化简函数g（x）的解析式，然后根据反函数的定义，判断出函数f（x）与g（x）的关系，然后对题目中的四个答案逐一进行比照，即可得到答案．【解答】解：∵ab=1g（x）=﹣log b x=log a x则函数f（x）=a x（a＞0且a≠1）与g（x）=﹣log b x（b＞0且b≠1）互为反函数故函数f（x）=a x（a＞0且a≠1）与g（x）=﹣log b x（b＞0且b≠1）的图象关于直线y=x 对称故选B．【点评】本题考查的知识点是对数函数的图象与性质，指数函数的图象与性质，反函数的图象，其中利用对数运算性质，及反函数的定义，分析出函数f（x）与g（x）的关系，是解答本题的关键．9. 用反证法证明命题: “a, b∈N, 若ab不能被5整除, 则 a与b都不能被5整除”时, 假设的内容应为（）A. a, b都能被5整除B. a, b不都能被5整除C. a, b至少有一个能被5整除D. a, b至多有一个能被5整除参考答案：C10. 设x∈R，则“x＞”是“2x2+x﹣1＞0”的( )A．充分而不必要条件B．必要而不充分条件C．充分必要条件D．既不充分也不必要条件参考答案：A【考点】必要条件、充分条件与充要条件的判断．【专题】简易逻辑．【分析】求出二次不等式的解，然后利用充要条件的判断方法判断选项即可．【解答】解：由2x2+x﹣1＞0，可知x＜﹣1或x＞；所以当“x＞”?“2x2+x﹣1＞0”；但是“2x2+x﹣1＞0”推不出“x＞”．所以“x＞”是“2x2+x﹣1＞0”的充分而不必要条件．故选A．【点评】本题考查必要条件、充分条件与充要条件的判断，二次不等式的解法，考查计算能力．二、填空题:本大题共7小题,每小题4分,共28分11. 若不等式ax2+bx+2>0的解集为{x|-},则a+b=_________.参考答案：-1412. 直线：与曲线：仅有一个公共点，则的取值范围参考答案：13. 已知抛物线y2＝2px(p＞0)上一点M(1，m)，到其焦点的距离为5，双曲线x2－＝1的左顶点为A，若双曲线的一条渐近线与直线AM垂直，则实数a＝__________．参考答案：14. 如图所示，一个空间几何体的主视图和左视图都是边长为1的正方形，俯视图是一个直径为1的圆，那么这个几何体的全面积为__________.参考答案：略15. 如图, 正四棱柱ABCD-A1B1C1D1的高为3cm，对角线A1C的长为cm，则此四棱柱的侧面积为____________.参考答案：24cm216. 为了了解参加运动会的2000名运动员的年龄情况，从中抽取100名运动员；就这个问题，下列说法中正确的有；①2000名运动员是总体；②每个运动员是个体；③所抽取的100名运动员是一个样本；④样本容量为100；⑤这个抽样方法可采用按年龄进行分层抽样；⑥每个运动员被抽到的概率相等．参考答案：④，⑤，⑥【考点】收集数据的方法．【分析】2000名运动员的年龄是总体，每个运动员的年龄是个体，所抽取的100名运动员的年龄是一个样本，样本容量为100，这个抽样方法可采用按年龄进行分层抽样，每个运动员被抽到的概率相等．【解答】解：④，⑤，⑥正确，∵2000名运动员的年龄情况是总体；每个运动员的年龄是个体，所抽取的100名运动员的年龄是一个样本，样本容量为100，这个抽样方法可采用按年龄进行分层抽样，每个运动员被抽到的概率相等．故答案为：④，⑤，⑥．17. ，则a＝________.参考答案：4三、解答题：本大题共5小题，共72分。
河南省周口市鹿邑县第二高级中学2023-2024学年高一下学期同步月考(四)数学试题
河南省周口市鹿邑县第二高级中学2023-2024学年高一下学期同步月考（四）数学试题一、单选题1．已知复数z 满足()34i 5z -=，则z =（ ） A ．43i 55-B ．34i 55-C ．43i 55+D ．34i 55+2．设A ，B 是直线l 上两点，则“A ，B 到平面a 的距离相等”是“//l a ”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3．一组数据：5，1，3，5，2，2，2，3，1，2，则这组数据的85%分位数是（ ） A ．3B ．4C ．4.5D ．54．平面向量a v 与向量b v 满足()3a a b ⋅+=vv v ,且2a =v ，1=v b ,则向量a v 与b v 的夹角为A ．6π B ．3π C ．23π D ．56π 5．若一组数据123,,a a a 的平均数为4，方差为3，那么数据12322,22,22a a a +++的平均数和方差分别是（ ） A ．10，12B ．10，14C ．4，3D ．6，36．用2，3，4这3个数组成没有重复数字的三位数，则事件“这个三位数是偶数”发生的概率为（ ） A ．13B ．12C ．23D ．347．明明同学打靶时连续射击三次，事件“至少有一次中靶”的互斥事件是（ ） A ．三次均未中靶 B ．只有两次中靶 C ．只有一次中靶D ．三次都中靶8．如图，在ABC V 中，AC BC =，D 在边AB 上，3ACB BCD ∠=∠，45AD DB =u u u r u u u r，则c o s A C D ∠=（ ）A ．2425-B ．732-C ．725-D ．1225二、多选题9．下列说法错误的是（ ）A ．若()()1P A PB +=，则事件A 与B 是对立事件 B ．事件发生的可能性越大，它的概率越接近1C ．某种彩票中奖的概率是1%，因此买100张该种彩票一定会中奖D ．事件,A B 同时发生的概率一定比,A B 中恰有一个发生的概率小10．为了研究“同时处理多任务时男女的表现差异”课题，研究组随机抽取男、女志愿者各150名，要求他们同时完成“解题、读地图、接电话”等任务，志愿者完成任务所需时间的分布如图所示，则下列表述正确的是（ ）A ．总体上女性处理多任务平均用时较短B ．处理多任务的能力存在性别差异C ．男性的用时中位数比女性用时中位数大D ．女性处理多任务的用时为正数，男性处理多任务的用时为负数11．在等腰梯形ABCD 中，AB CD ∥，3AB BC ==，1CD =，点1O ，2O 分别为CD ，AB 的中点，以12O O 所在直线为旋转轴，将梯形旋转180︒得到一旋转体，则（ ）A ．该旋转体是一个圆台B ．该旋转体的表面积为6πC ．直线BC 与旋转体的上底面所成角的正切值为D ．该旋转体的外接球的体积为27π2三、填空题12．若事件A 与B 互斥，且()()0.3,0.3P A P B ==，则()P A B =U .13．如图，已知网格小正方形的边长为1，点P 是阴影区域内的一个动点（包括边界），O ，A 在格点上，则OP OA ⋅u u u r u u u r的取值范围是.14．如图，已知在ABC V 中，AD 是BAC ∠的角平分线，与BC 交于点D ，M 是AD 的中点，延长BM 交AC 于点H ，2AC AB =，则AH AC=.四、解答题15．同时掷红、蓝两颗质地均匀的正方体骰子，用(),x y 表示结果，其中x 表示红色骰子向上一面的点数，y 表示蓝色骰子向上一面的点数. (1)写出该试验的样本空间；(2)指出()()()()()(){}1,1,2,2,3,3,4,4,5,5,6,6所表示的事件； (3)写出“点数之和不超过5”这一事件的集合表示.16．如图，四棱柱1111ABCD A B C D -的底面ABCD 是正方形，11A AB A AD ∠=∠．(1)证明：平面1A BD ∥平面11CD B ； (2)证明：平面1A BD ⊥平面11ACC A ．17．在ABC V 中，角A ，B ，C 的对边分别为a ，b ，c ，且πsin 2a C b ⎛⎫-= ⎪⎝⎭．(1)求角A ；(2)若ABC V 的面积为1，求a 的最小值．18．为了估计一批产品的质量状况，现对100个产品的相关数据进行综合评分（满分100分），并制成如图所示的频率分布直方图.记综合评分为80分及以上的产品为一等品.(1)求图中a 的值，并求综合评分的平均数；(2)用样本估计总体，以频率作为概率，按分层随机抽样的思想，先在该条生产线中随机抽取5个产品，再从这5个产品中随机抽取2个产品记录有关数据，求这2个产品中最多有1个一等品的概率；(3)已知落在[)50,60的平均综合评分是54，方差是3，落在[)60,70的平均综合评分为63，方差是3，求落在[)50,70的总平均综合评分z 和总方差2s .19．如图1，在矩形ABCD 中，点E 在边CD 上，22BC DE EC ===，将DAE V 沿AE 进行翻折，翻折后D 点到达P 点位置，且满足平面PAE ⊥平面ABCE ，如图2．(1)若点F 在棱PA 上，PB I 平面CEF G ，求证：//CE FG ； (2)求点E 到平面PAB 的距离．。
考评细则
鹿邑县第二高级中学“发展提升年”活动考评细则考核重点考核内容分值责任处室、年级抓管理每学期制定有详细的工作计划和行事历，能够详细体现阶段性内容，学校每月对教师进行一次业务检查，坚持落实巡课制度，评析及时，制定有学校“发展提升年”活动方案。
4办公室、教导处、教研处校长按要求参加各级各类培训，每学期组织一次班子成员教育法规、理论测试。
3 办公室班主任每学期制定一份班级管理计划，精心营造符合班情、学情的班级文化。
3 政教处、年级班主任每期至少召开两次家长会，两次科任教师联席会，会议有主题，有记录，注重效果。
4 年级班主任每周召开一次主题班会，有记录，班级管理有目标。
班级填写好“班级日志”。
2.5 政教处、年级每学期对所有任课教师开展一次业务能力、执行计划、进度落实、“四清”及工作绩效考核。
有记录及评价资料。
4 教导处、教研处各项制度健全，落实有效。
档案管理规范，入盒入柜。
学校各处室、年级有相应的学期工作计划。
学校对各处室、年级、教师实行目标管理。
4 办公室、目标办贯彻落实《中小学教师职业道德规范》和《中小学教师违反职业道德行为处理办法》，教职工各项业绩档案管理规范，保存完好。
每发现一例违反教师职业道德行为扣1分。
2 教导处、政教处、年级贯彻执行《中小学生守则》和《日常行为规范》，效果好。
3 政教处、团委、年级办学目标明确，观念新颖，三风一训符合校情。
校园文化丰富，符合校情、教情和学情。
4 办公室、团委抓教学学校要制定切实可行的业务培训计划，按要求参加各类教师岗位提高培训。
坚持对教师进行新课程、新技能培训，实施信息技术提升培训。
2教导处、教研处、电教中心每期开展一次师德师风考核。
每学期开展一次师德标兵评比，一次师德报告会，一次师德演讲，一次师德征文。
2.5 教导处、政教处有课改计划、有领导组、有规章制度、班级小组建设有组名、组规、组训等，每学年开展一次“一师一优课，一课一名师”活动。
学校有课改实验班，每期举行一次课改公开擂台赛。
河南省周口市鹿邑县第二高级中学2023-2024学年高二下学期7月期末考试英语试题(含答案)
河南省周口市鹿邑县第二高级中学2023-2024学年度第二学期期末试卷高二下英语试卷及答案时间100分钟总分120分第二部分阅读理解(共两节，满分50分)第一节(共15 小题:每小题 2.5 分，满分37.5 分)阅读下列短文，从每题所给的A、B、C和D四个选项中，选出最佳选项。
ADo you want to see majestic lions roaring, rare rhinos running and colorful birds flying？You'll find this incredible display of nature in the Kenya Lake System in the Great Rift Valley.First stopFly into Nairobi, the capital of Kenya. This great city is known for its fashion markets and beautiful art. And be sure to pick up some rich Kenyan coffee while you're there! Then head for the countryside.Plan your tripThe largest of the three lakes in the system, Lake Nakuru, is less than 175 kilometers fromNairobi. But Lakes Bogoria and Elementaita are equally worth a visit. You can stay at one of several choices of' luxurious cabins in the valley, And if you sign up for a trip, be sure to put flamingos, the white and pink birds, at the top of your list.ParadiseThe Kenya Lake System is shallow and alkaline(碱性的). That means it supports vegetation such as green algae, a food source for many waterfowl. Hundreds of species of birds flourish here, using this area as a major breeding and nesting ground. Water buffalo, leopards.monkeys and the endangered Rothschild giraffe also make this area their home.Other sightsWildlife isn't the only natural wonder in the Great Rift Valley. The valley marks a place where the earth is divided far below the surface. This rock separation below ground leads to noticeable features above ground, Walking around Lake Bogoria especially, you will see steam rising from hot springs.Other activities include golfing, horseback riding and hiking. The fun doesn't stop when the sun goes down, either. The river valley is far from city lights, so many visitors gaze at the stars shining above.If you are ready for a wildlife experience, don't hesitate!21.W hat is Nairobi famous for？A.Coffee markets.B. Beautiful fashion shows.C.Fashion markets.D. Great artists.22. Which does the passage recommend visitors to see first in the Kenya Lake System?A. Majestic lions.B. Rare rhinos.C.Colorful flamingos.D.Luxurious cabins.23.What can visitors do in the Great Rift Valley?A. Horseback riding and hiking.B. Shopping for fashionable clothesC.Admiring beautiful artworks. D, Enjoying Kenyan coffee.BJack Prelutsky is an American poet, He is known for his humorous and fantastic poetry forchildren,which has earned him numerous awards.Jack Prelutsky was born in Brooklyn, New York in 1940. As a child, he did not enjoy poetry, finding it boring and pointless. instead, Prelutsky expressed his artistry through music.taking voice and piano lessons. While Prelutsky claims he got into poetry almost by accident, he also states that he was always a poet. It just took him some time to realize his love for poetry.Prelutsky explored quite a few options before he found his niche in poetry. He was a folksinger and guitar teacher in Greenwich Village in his late teens and early twenties, Prelutsky also held a variety of odd jobs, He worked as a furniture mover, piano mover, taxicab drivercoal shoveler, fruit picker, truck driver and photographer, among other jobs.At the age of23,Prelutsky decided to try illustrating. Just before presenting his illustrations to a children's book editor, he added a few lines of poetry to the pictures. The editor told him that the illustrations were not publishable but the poems showed the talent. Over the next months and years, Prelutsky worked with the editor to create a collection of animal poems inspired by his love of the Bronx Zoo. His first book, entitled A Gopher in the Garden and Other Animal Poems, was published in 1967.In addition to writing, Prelutsky has edited various poetry collections and translated poetry from German and Swedish into English. He currently resides in Seattle, where he writes and performs his work, Prelutsky is considered one of the most popular children's poets. His work pioneered a new era of' children's poetry, In 2006, the Poetry Foundation named Jack Prelutsky the first Children's Poet Laureate, a position designed to promote poetry for children and raise awareness of the power and relevance of poetry for young people.24.Why did Prelutskv dislike poetry in his early years?A. It was too difficult.B. He wasn't good at it.C. He thought it meaningless.D. It took up too much of his time.25, Which of`the following can replace the underlined word “niche" in paragraph 3?A. Mistake.B.Dream.prehension.D.Interest.26. How old was Prelutsky when he published his first book?A.23.B.27.C.40.D.66.27. What is the purpose of the text?A. To introduce a poet.B. To attract poetry lovers.C.To display some excellent poems.D. To encourage readers to write poems.CA popular saying goes, “Sticks and stones may break my bones, but words will never hurt me." However, that's not really true. Words have the power to build us up or tear us down. L doesn't matter if the words come from someone else or ourselves -- the positive and negative effects are just as lasting.We all talk to ourselves sometimes. We're usually too ashamed to admit it, though. In fact we really shouldn’t be because more and more experts believe talking to ourselves out loud is a healthy habit.This “self-talk" helps us motivate ourselves, remember things, solve problems, and calm ourselves down. Be aware, though, that as much as 77% of self-talk tends to be negative. So in order to stay positive, we should only speak words of encouragement to ourselves. We should also be quick to give ourselves a pat on the back. The next time you finish a project, do well in atest, or finally clean your room, join me in saying “Good job!"Often, words come out of our mouths without us thinking about the effect they will have. But we should be aware that our words cause certain responses in others. For example, where returning an item to a store, we might use warm, friendly language during the exchange. And the clerk will probably answer in a similar manner. Or harsh(刻薄的) and critical language will most likely cause the clerk to be defensive.Words have power because of their lasting effect. Many of us regret something we once said. And we remember unkind words said to us! Before speaking, we should always ask ourselves: ls it loving? ls it needed? If what we want to say doesn't pass this test, then it's better left unsaid.Words possess power: both positive and negative. Those around us receive encouragement when we speak positively. We can offer hope, build self-esteem(自尊) and motivate others to do their best. Negative words destroy all those things. Will we use our words to hurt or to heal?The choice is ours.28. The main idea of the first paragraph is that .A.words have a great effect on usB.encouraging words give us confidenceC.not sticks and stones but words' will hurt usD.negative words may let us down29. There is no need for us to feel ashamed when we talk to ourselves because .A.almost everybody has the habit of talking to themselves.B. it does no harm to have “self-talk" when we are alone.C.talking to ourselves always gives us courage.D,talking to ourselves can have a positive effect on us.30. The underlined part in the third paragraph means that we should also timely .A.remind ourselvesB. praise ourselvesC.make ourselves relaxedD. give ourselves happiness31. The author would probably hold the view that .A. it is better to think twice before talking to othersB. negative words may stimulate us to make more progressC. people tend to remember friendly wordsD. encouraging words are sure to lead to kind offersDWHY SHENNONGJA?As we know, in 2016, Shennongjia made it on to the UNESCO World Heritage List. But there are so many important places not vet on this list. So you might ask: why Shennongjia?UNESCO awarded Shennongjia this status because it meets two criteria required by the list It contains a naturally-balanced environment that allows the many and various species to live and prosper. It is also one of the rare locations in the world where scientists can observe in real time the ecological and biological processes that occur as the plants and animals develop and evolve.For the first criterion, Shennongjia is apparently one of the most "complete" natural areasin the world. The region rises from about 400 metres to over 3, 000 metres above sea level giving it the name the "Roof of Central China"For the second criterion, we can see that Shennongjia has incredible biodiversity. Look at this slide. According to official statistics, over 3, 000 plant species have been recorded there This represents more than ten percent of China's total floral richness .It is a challenge to look after so many species. In winter, scientists brave heavy snow and freezing temperatures to supply food to the Golden Snub-nosed Monkey, Thanks to their efforts the monkey's population has doubled since the 1980s.Their number reached over 1. 300 in 201$and continues to grow.But the most impressive aspect of Shennongjia is the local people, who take things from nature without causing damage. I visited a local village which is known for its home-made honey. What is special about the honey is that it is produced by the earliest species of Chinese bee. All of this explains why Shennongjia earned-and deserve suits place on the UNESCO World Heritage List, as well as highlighting how understanding, awareness and hard work have contributed towards protecting a unique and wonderful part of our natural world.32.Why is Shennongjia included in the UNESCO World Heritage List?A.Because its forests stretch on and on like great green seas.B.Because it is the legendary father of Chinese herbal medicine.C.Because it satisfies the two requirements for the status.D. Because it has the mysterious creature called “Yeren”.33. Shennongjia is also called the“.”A. Roof of Central ChinaB. Roof of Northern ChinaC. Roof of Easter ChinaD. Roof of Southern China34. How does the author think of the scientists in Shennongjia?A. Patient.B. Careful.C. Confident.D. Hard-working.35. According to the text, what impressed the author most?A.A local village.B.The local people.C.Some special monkeys.D. The wild bees.第二节(共5小题:每小题2.5分，满分12.5分)阅读下面短文，从选项中选出可以填入空白处的最佳选项“选项中有两项为多杂选项。
河南省周口市鹿邑县第二高级中学2023-2024学年高三上学期12月质量监测考试英语试卷
河南省周口市鹿邑县第二高级中学2023-2024学年高三上学期12月质量监测考试英语试卷学校:___________姓名：___________班级：___________考号：___________一、阅读选择Where to Buy Art OnlineBuying a piece of art is easier than ever. Now, art is at the tip of your fingers thanks to the internet. Here are platforms available for all prices and tastes.ARTFINDERArtFinder focuses on selling original and limited edition artwork from independent artists around the world. With 5,700 five-star review son TrustPilot, ArtFinder is a reliable source for affordable art. Buyers like the simple purchasing process and appreciate the accurate representation of the artwork on the website. If you aren't satisfied with what you receive, the platform has a 14-day return policy.FINE ART AMERICAFine Art America is the world's largest online art marketplace. It is home to hundreds of thousands of artists, photographers and global brands. You can also search thematic collections by high-level brands like Rolling Stone, so it's a good choice if you are looking for a print of a famous photograph. If you aren't satisfied with what you receive, the platform has a 30-day return policy.ARTSYIf you are looking for a high-end gallery experience from home, check out Artsy. This highly respected platform has over 2 million pieces of art available. Artsy aims to make collecting art easy and allows art lovers to buy artwork and even sell pieces from their collections. Once you've purchased or made an offer, all sales are final.SAATCHI ARTIf you want to purchase from emerging and mid-career independent artists, Saatchi Art is a great option. Saatchi has original pieces and art prints ranging from under $500 to over $10,000. With over60,000 international artists on the platform, there is plenty to choose from. Saatchi has a 14-day return policy for original artwork.1．What does ARTFINDER focus on?A．Print of a famous photograph.B．Pieces from personal collections.C．Original pieces and art prints.D．Original and limited edition artwork. 2．Which of the following has no offer of a return policy?A．ARTFINDER.B．FINE ART AMERICA．C．ARTSY.D．SAATCHI ART.3．What is the purpose of the text?A．To educate.B．To introduce.C．To confirm.D．To propose.Everyone has his or her dream and may desire to achieve it anyway. One New Jersey man has proved that to be true by achieving a lifelong goal more than 80 years in the making. Earlier in March, a101-year-old man named Merrill Pittman Cooper received his high school diploma during a graduation ceremony in his honor.Cooper attended Storer College from 1934 to 1938, but life circumstances got in the way and prevented him from finishing school and earning his diploma. Without his father around, it was up to him and his mother to take care of themselves. His mom had taken on work as a live-in housekeep er in order to pay his school tuition and board, but that wasn't enough. And after realizing his mother couldn't afford to make the final tuition payment for his senior year, the teen encouraged her to move them to Philadelphia.In Philadelphia, Cooper took on work to help his mother pay the bills. He was hired as a city trolley car operator in 1945. That job led him to a long and fruitful career in transportation where he became president of the local union and later vice president of the International Transport Workers Union in New York City. But despite such a successful career, he always regretted not getting his high school diploma and struggled to finish all he should learn.But now, after 84 years, Cooper was able to see his lifelong dream come true. His family got the idea to surprise him with the honorary high school diploma. Cooper could hardly hold back his tears when he discovered their plan after arriving at the hotel.The whole thing was a dream come true for Cooper, who now displays his framed diploma proudly. “I can't think of a happier day,” says the inspiring 101-year-old man. “I'm really happy to finally have it.”4．What stopped Cooper from getting his high school diploma when he was young?A．Poverty.B．Illness.C．Laziness.D．War.5．What can we learn from Paragraph 3?A．Cooper changed his work frequently.B．Cooper first worked as a car driver.C．Cooper was a great success in his career.D．Cooper quit studying after getting a job. 6．How did Cooper feel after getting the diploma?A．Confused.B．Regretful.C．Relieved.D．Proud. 7．What can be concluded from the text?A．Practice makes perfect.B．A bird in the hand is worth two in the bush.C．It's never too late to go after your dreams.D．A friend in need is a friend indeed.Have you ever noticed that some songs start with a “one, two, three, four”? You can hear it in songs by artists as different as The Beatles, OutKast, and Taylor Swift. But why are they counting? The counting you hear is a small clue to a big fact—math is behind almost everything in music! Let’s WONDER together!First, let’s start with “counting in.” Musicians will often count to make sure that they all start the song at the same time. Once they start together, everyone has to stay on the same beat. Every song has a steady beat. One of the first ways that young children can start to learn math ideas is by learning how to make a steady beat!A steady beat is one kind of pattern. Other patterns in music relate to math, too. We can see patterns in how far apart notes are from each other. Sometimes we play more than one note at a time.Musicians also need numbers to understand how to play music. If you look at written music, you will see two numbers on the left side. Often, the numbers are 4/4. This is called the time signature. The top number of it tells the musician how many beats are in each part of the line of music. The second number tells us what type of note gets one beat. The “4” on the bottom lets us know that a quarter note gets one beat.It is clear that music is good for our brains, whether or not it makes us better at math. Playing music uses several parts of your brain at the same time. Musicians need to use their brains to see the music, hear the notes, and touch the instrument in just the right ways to make sounds!These are just a few of the ways that music and math are related. As you learn more about music and math, you'll see more connections!8．What is the purpose of the first paragraph?A．To give examples.B．To introduce the main topic.C．To present an academic achievement.D．To list the impact of some famousmusicians.9．What does the second number of the time signature mean?A．What category of note gets one beat.B．What varieties songs should have.C．How many beats are in each measure.D．How the former note is transformed into the next.10．How does the author like learning music?A．Dizzy but powerful B．Complicated but rewarding.C．Time-consuming but relaxing.D．Expensive but pleasant.11．What can be inferred from the text?A．To learn math well, you have no choice but to have a good command of music.B．The performance of one’s music may have an effect on that of one's math.C．The more you learn music, the more broad-minded you will be.D．People’s taste in music varies from person to person.Mahadevan, the Lola England de Valpine Professor of Applied Mathematic s at the Harvard John A．Paulson School of Engineering and Applied Sciences(SEAS), and his team combined physics and machine learning to develop a new 3D-printing technique that can quickly create complex physical patterns including making an exact copy of a Pollock painting by controlling the same natural fluid instability that Pollock used in his work.The issue, as it usually is, is physics. Liquid inks are bound by the rules of fluid dynamics, which means when they fall from a height, they become unstable, folding and coiling(螺旋) in on themselves.Today, most 3D and 4D printing techniques place the print nozzle(喷嘴) near the surface, eliminating the dynamic instability of the liquid stream.But Mahadevan has a motto: use the physics, instead of avoiding it. Pollock composed his famous drip(滴) paintings by placing a canvas on the floor and pouring, dripping paint onto it from above. To the untrained eye, his technique may seem random, but Pollock always claimed he had complete control over the flow of the paint.“If you look at traditional 3D printers, you supply them a path from point A to point B and the nozzle deposits ink along that specified path,” said Chaudhary.“But Pollock’sapproach of throwing paint from a height meant that even if his hand was moving in a specific track, the paint didn't follow that track because of the acceleration gained from gravity.”“With deep learning, the model can learn from its mistakes and get more and more accurate with each trial,” said Chaudhary. The researchers used simple fluid s for this research, but the approach could be expanded to include more complex fluids.As the research continues, there is no telling where Mahadevan may look for inspiration next. “When you're in Maha's lab, nothing is off the table,” said Chaudhary.12．Why is Pollocko's painting mentioned?A．To show his incredible skills.B．To highlight his commitment to art.C．To illustrate the impact of the new technique.D．To compare his painting and the new 3D-printing.13．What can replace the underlined word “eliminating” in paragraph 3?A．Measuring.B．Removing.C．Perceiving.D．Overestimating. 14．What can be concluded from Chaudhary's words in the last paragraph?A．Nothing may be impossible for Mahadevan.B．Nothing can be the inspiration for him.C．He disapproves of Maha's research.D．Mah a pioneers in the research lab. 15．What can be the best title of the text?A．A Development in Complex Painting B．Physics Combined with MachineLearningC．Challenges Faced by Traditional 3D Printers D．A New 3D Printing TechniqueA Rain GardenA garden can be a place for family recreation or well-being of us, but the monsoons (季风) can turn your backyard into a grassy spot. 16 .Don’t drive away slimy pests. The rains also bring with them a flock of visitors, such as snails and ants. Before you lace your garden with insecticide, keep this in mind: Many of these pests are harmless and will do your green space a lot of good. For instance, snails can speed up the process of turning bío-degradable waste into fertilizer. 17 .Make use of rain water. Rainwater is the best kind of water for your plants. It’s as simpleas placing indoor pots on the balcony. You could also put out industrial drums and pots under the rain to load up on water for your garden. 18 . If you have the space, consider creating your own rain garden. Rain gardens collect run-off water from paved surfaces, thus saving you the additional effort of watering your plants.Weed it out. 19 . Besides, they rob your garden of the nutrients required to keep it flourishing. You can get rid of them naturally by pulling them out from the root. For deep-rooted weeds, make use of weeding knives and other tools to remove them.20 . This is the season to plant beauties such as jasmine, bird of paradise and Mountain Roscoe Lily. It’s also the time to cut out unnecessary branches of the plants such as ornamental asparagus... to reduce the burden on the plant. You can pot these shoots to get new plants too.A．Plant and cutB．Protect the new plantsC．Weeds typically crop up after a heavy rainfallD．These tips help you make the most of this seasonE．If you must rid your garden of them, go the chemical-free wayF．Read on for more reasons to have and build an appealing gardenG．Keep them covered with a lid when it’s not pouring, to avoid mosquitoes二、完形填空When you think of the strong actor and politician Arnold Schwarzenegger, what comeswrong 32 that we can do it all alone. In reality, none of us can 33 it.” Schwarzenegger’s message 34 beyond his personal story. He encouraged the audience to reflect on the people who have 35 us in our lives and to pay it forward by offering 36 and assistance to others whenever it is possible.In a society that often 37 individual achievement above all else, Arnold Schwarzenegger’s speech is a 38 reminder that we are interconnected. And our success is often a result of the support we 39 from those around us. It’s a message that urges us to be 40 to those who once helped us and to offer help to those in need. 21．A．impressive B．smooth C．awkward D．accessible 22．A．rare B．tough C．remarkable D．practical 23．A．defeating B．becoming C．delighting D．believing 24．A．favored B．measured C．heard D．delivered 25．A．carelessly B．severely C．randomly D．cautiously 26．A．idea B．altitude C．suggestion D．proof 27．A．gains B．pity C．duty D．efforts 28．A．stressed B．suspected C．denied D．announced 29．A．devoted B．referred C．owed D．turned 30．A．shaking B．helping C．green D．clean 31．A．call B．give C．tell D．leave 32．A．forecast B．argument C．direction D．feeling 33．A．describe B．make C．guess D．deserve 34．A．circles B．wonders C．goes D．traces 35．A．aided B．rejected C．convinced D．distributed 36．A．victory B．function C．treasure D．support 37．A．makes fun of B．thinks much of C．looks forward to D．speaks ill of 38．A．surprising B．comfortable C．powerful D．fearful 39．A．choose B．spread C．upload D．receive 40．A．similar B．mean C．grateful D．familiar三、语法填空阅读下面短文，在空白处填入1个适当的单词或括号内单词的正确形式。
河南省周口市鹿邑县第二高级中学2021-2022学年高三化学月考试卷含解析
河南省周口市鹿邑县第二高级中学2021-2022学年高三化学月考试卷含解析一、单选题（本大题共15个小题，每小题4分。
在每小题给出的四个选项中，只有一项符合题目要求，共60分。
）1. 分类是化学研究的重要方法，下列各组物质的分类正确的是 ( )A．同位素：D2、H2、T2 B．同系物：C．非电解质：NH3、C2H5OH、Cl2 D．含极性键的分子晶体：CO2、CCl4、NH3参考答案：D略2. X、Y、Z、W有如图所示的转化关系，则X、Y可能是①C、CO ②AlCl3、Al(OH)3③N2、NO ④S、SO2A．①② B．②③C．③④ D．①②④．参考答案：A略3. 锑（Sb）在自然界一般以硫化物的形式存在，我国锑的蕴藏量占世界第一。
从硫化物中提取单质锑一般是先在高温下将硫化物转化为氧化物，再用碳还原：2Sb2S3＋3O2＋6Fe＝Sb4O6＋6FeS ……①Sb4O6＋6C＝4Sb＋6CO↑……②关于反应①、反应②的说法正确的是A．反应①②中的氧化剂分别是Sb2S3、Sb4O6B．反应①中每生成3 mol FeS时，共转移6 mol电子C．反应②说明高温下Sb的还原性比C强D．每生成4 mol Sb时，反应①与反应②中还原剂的物质的量之比为4 ：3参考答案：B略4. 下列各组物质的无色溶液，不用其他试剂即可鉴别的是①KOH、Na2SO4、AlCl3②NaHCO3、Ba(OH)2、H2SO4③HCl、NaAlO2、NaHSO4④Ca(OH)2、Na2CO3、BaCl2A．①② B．②③ C．①③④ D．①②④参考答案：A5. 下列说法不正确的是( )A．氯水中加少量的Na2 CO3粉末，可提高溶液中HC1O的浓度B．金属单质都具有一定的还原性，金属阳离子都只有氧化性C．火灾现场自救方法是用湿毛巾捂住口鼻并向地势低的地方撤离D．大气中的N2可作为制硝酸的原料参考答案：BC【考点】氯气的化学性质；金属的通性．【分析】A、氯水中的盐酸可以和碳酸钠之间反应，促使氯气和水的反应：Cl2+H2O HCl+HClO正向移动，据此回答；B、金属单质都具有一定的还原性，金属阳离子大多数均有氧化性，但是还有的具有还原性；C、氯气的密度比空气大，地势低的地方氯气含量高；D、空气中的氮气是制造硝酸和氮肥的重要原料．【解答】解：A、氯水中的盐酸可以和碳酸钠之间反应，促使氯气和水的反应：Cl2+H2O HCl+HClO正向移动，提高溶液中HC1O的浓度，故A正确；B、金属阳离子也可能具有还原性，如Fe2+，故B错误；C、氯气的密度比空气大，地势低的地方氯气含量高，则氯气泄漏，自救方法是用湿毛巾捂住口鼻并向地势高的地方撤离，故C错误；D、空气中的氮气具有广泛的用途，它是制造硝酸和氮肥的重要原料，故D正确；故选BC．【点评】本题涉及氯气性质的考查，注意教材知识的梳理和归纳以及灵活应用是关键，难度中等．6. 某化学合作学习小组讨论辨析以下说法，其中说法正确的是（）①粗盐和酸雨都是混合物；②沼气和水蒸气都是可再生能源；③冰和干冰既是纯净物又是化合物；④不锈钢和目前流通的硬币都是合金；⑤硫酸和食醋既是化合物又是酸；⑥纯碱和熟石灰都是碱。
河南省周口市鹿邑县第二高级中学高三化学下学期期末试卷含解析
河南省周口市鹿邑县第二高级中学高三化学下学期期末试卷含解析一、单选题（本大题共15个小题，每小题4分。
在每小题给出的四个选项中，只有一项符合题目要求，共60分。
）1. 用自来水配制下列溶液，不会变质的是（）A．AgNO3 B．Na2CO3 C．NaCl D．KI参考答案：C解析：A．配制AgNO3溶液时，若用氯水，氯水中含有Cl-离子，则Ag+与Cl-反应生成AgCl沉淀，产生明显的药品变质问题，故A错误； B．配制Na2CO3溶液时，氯水中含有H+，H+与碳酸根离子反应生成水和二氧化碳，则产生明显的药品变质问题，故B错误； C．配制NaCl溶液时，溶液在的离子不会与氯水中的微粒发生反应，则不会产生明显的药品变质问题，故C正确； D．配制KI溶液时，若用氯水，氯水中含有Cl2、HClO、ClO-等微粒，具有氧化性，能氧化碘离子，则产生明显的药品变质问题，故D错误。
2. 右图是制备和收集气体的实验装置，该装置可用于A．用H2O2溶液与MnO2制取O2B．用MnO2和浓盐酸制取Cl2C．用生石灰和浓氨水制取NH3D．用铜片与浓硫酸制取SO2参考答案：A略3. 下列有关热化学方程式的叙述正确的是A．已知2H2O(g) ＝2H2(g)＋O2(g) △H＝+483.6 kJ/mol，则氢气的燃烧热为△H＝－241.8kJ/molB．已知C（石墨，s）＝C（金刚石，s）△H＞0，则金刚石不如石墨稳定C．已知中和热为△H＝－57．4kJ/mol ，则稀醋酸和稀NaOH溶液反应的热化学方程式为：NaOH(aq)＋CH3COOH(aq)＝CH3COONa(aq)＋H2O(l) △H＝－57.4kJ/molD．已知2C(s)＋2O2(g)＝2CO2 (g) △H1；2C(s)＋O2 (g)＝2CO(g) △H2，则△H1＞△H2参考答案：B略4.参考答案：B5. 目前科学家已经开发的便携式固体氧化物燃料电池，它以烷烃气体为燃料，每填充一次燃料，大约可连续几小时甚至几十小时输出50W的电力。
河南省周口市鹿邑县第二高级中学2020-2021学年高三物理联考试题带解析
河南省周口市鹿邑县第二高级中学2020-2021学年高三物理联考试题含解析一、选择题：本题共5小题，每小题3分，共计15分．每小题只有一个选项符合题意1. 如图所示，一吊床用AB、CD两根轻绳悬挂起来，吊床静止时，两轻绳与竖直方向的夹角分别为45°和60°，则AB、CD两绳的拉力之比为A． B．C．1： D．：l参考答案：A2. 一气象探测气球，在充有压强为1.00atm（即76.0cmHg）、温度为27.0℃的氦气时，体积为3.50m3．在上升至海拔6.50km高空的过程中，气球内氦气逐渐减小到此高度上的大气压36.0cmHg，气球内部因启动一持续加热过程而维持其温度不变．此后停止加热，保持高度不变．已知在这一海拔高度气温为﹣48.0℃．求：（1）氦气在停止加热前的体积；（2）氦气在停止加热较长一段时间后的体积．参考答案：解（1）在气球上升至海拔6.50km高空的过程中，气球内氦气经历一等温过程．根据玻意耳﹣马略特定律有：p1V1=p2V2将p1=76.0cmHg，，p2=36.0cmHg带入得：．故氦气在停止加热前的体积为．（2）在停止加热较长一段时间后，氦气的温度逐渐从T1=300K下降到与外界气体温度相同，即T2=225K，这是一等压过程，根据盖﹣吕萨克定律有：将数据带入得：．故氦气在停止加热较长一段时间后的体积为：．【考点】气体的等温变化；气体的等容变化和等压变化．【分析】（1）该上升过程为等温变化，因此根据等温变化的气体状态方程可直接求解．（2）停止加热后高度不变，因此为等压变化，注意经过较长一段时间后，温度与环境温度相等为﹣48.0℃，故根据等压变化可直接求解．3. 如图所示，1、3是圆轨道，2是椭圆轨道.关于人造卫星，下列说法正确的是：卫星在2轨道上的Q点的动能比在3轨道上的P点动能大.卫星在3轨道的P点向心加速度比在2轨道上时P点的加速度大若要卫星从2轨道变到3轨道，应在P点点火加速，这个过程中，引力对卫星做负功若3轨道上原本就有一些卫星，当从3轨道变到2轨道时有可能会发生“撞车”参考答案：A4. （多选）下列说法中正确的是A．β衰变现象说明电子是原子核的组成部分B．目前已建成的核电站的能量来自于重核裂变C．一个氢原子从n=3的激发态跃迁到基态时，能辐射3种不同频率的光子D．卢瑟福依据极少数α粒子发生大角度散射提出了原子核式结构模型E．按照玻尔理论，氢原子核外电子从半径较小的轨道跃迁到半径较大的轨道时，电子的动能减小，原子总能量增大参考答案：BDE5. （单选）某同学用频闪相机拍摄了某中学运动会期间运动员跳远比赛时助跑、起跳、最高点、落地四个位置的照片,简化图如图所示.请你根据体育常识估算运动员正常起跳瞬间消耗的能量最接近A . B.C . D.参考答案：A二、填空题：本题共8小题，每小题2分，共计16分6. 大量氢原子处于不同能量激发态，发生跃迁时放出三种不同能量的光子，其能量值分别是：1.89eV、10.2 eV、12.09 eV。
河南省周口市鹿邑县第二高级中学2024-2025学年高二上学期11月期中考试数学试题
河南省周口市鹿邑县第二高级中学2024-2025学年高二上学期11月期中考试数学试题一、单选题1．已知向量(213)=-，，r a ，(4)x y =,,r b ，且//a b，则x y +=（）A ．4-B ．2-C ．4D ．22．若直线210x ay ++=与直线220x y +-=互相垂直，则实数a 的值是（）A ．1B ．－1C ．4D ．－43．已知椭圆C ：()22104x y m m +=>的离心率为2，则m =（）A ．B ．C ．8或2D ．84．已知向量(1,2,1),(2,0,1)a b =-= ，则向量a 在向量b上的投影向量为（）A ．15br B ．5-C ．5bD ．15b - 5．三棱柱ABC DEF -中，G 为棱AD 的中点，若BA a = ，BC b = ，BD c =，则CG = （）A ．a b c -+-B ．12a b c-++C ．1122a b c-++ D ．1122a b c-+ 6．设点()()1,1,3,1A B --，直线l 过点()1,2P ，且与线段AB 相交，则直线l 的斜率取值范围是（）A ．13,22⎡⎤-⎢⎥⎣⎦B ．31,22⎛⎫- ⎪⎝⎭C ．13,,22⎛⎤⎡⎫-∞-⋃+∞ ⎪⎥⎢⎝⎦⎣⎭D ．13,,22⎛⎫⎛⎫-∞-⋃+∞ ⎪ ⎪⎝⎭⎝⎭7．已知圆C ：()()22349x y -+-=，直线l ：230mx y m +--=．则直线l 被圆C 截得的弦长的最小值为（）A ．B C ．D 8．已知12,F F 是椭圆22:194x yC +=的两焦点，点M 在椭圆C 上，则12MF MF ⋅的最小值是（）A ．5B ．9C ．4D ．3二、多选题9．若直线240mx ny +-=始终平分圆224240x y x y +---=的周长，则mn 的取值可能是()A ．12B ．-12C ．13D ．210．已知直线1:10l x y --=，动直线()()2:10l k x ky k k +++=∈R ，则下列结论错误的有（）A ．不存在k ，使得2l 的倾斜角为90︒B ．存在实数k ，使得1l 与2l 没有公共点C ．对任意的k ，1l 与2l 都不重合D ．对任意的k ，1l 与2l 都不垂直11．已知椭圆2221(03)9x y b b+=<<的左､右焦点分别为12,F F ，过点1F 的直线l 交椭圆于,A B 两点，若AB 的最小值为4，则（）AB ．22AF BF +的最大值为8C ．离心率为2D ．椭圆上不存在点P ，使得1290F PF ∠=三、填空题12．已知空间向量a ，b ，c中两两夹角都是3π，且||4a = ，||6b = ，||2= c ，则||a b c ++=．13．过直线250x y -+=上一点P 向圆22:(1)(2)4C x y -++=作切线，切点为M ，则PM 的最小值为.14．已知F 是椭圆()2222:10x y C a b a b+=>>的右焦点，点B 是短轴的一个端点，线段BF 的延长线交椭圆C 于点D ，且2BF FD =，则椭圆C 的离心率为．四、解答题15．在正四棱柱1111ABCD A B C D -中，1222AA AB BC ===，M 是棱1CC 上的中点．(1)求证：AM BD ⊥；(2)异面直线AM 与BC 所成角的余弦值．16．已知椭圆()2222:10x y C a b a b+=>>的离心率为12e =，上顶点(P ，M 、N 为椭圆上异于点P 且关于原点对称的两点．(1)求椭圆C 的标准方程；(2)求证PM PN k k ⋅为定值．17．已知圆C ：224470x y x y +--+=关于直线10x y -+=的对称圆的圆心为D ，若直线l 过点()1,4.(1)若直线l 与圆C 相切，求直线l 的方程；(2)若直线l 与圆D 交于,A B 两点，AB =l 的方程.18．已知圆E 经过点()()6,0,2,0A B -．且圆心E 在直线y x =-上．（1）求圆E 的一般方程；（2）若圆22:4O x y +=和圆E 相交于点,M N ，求线段MN 的长．19．已知椭圆2222:1(0)x y E a b a b+=>>的左焦点为1F ，右焦点为2F ，焦距12F F 为2，过1F 的直线交椭圆E 于,A B 两点，且2ABF △的周长为(1)求椭圆E 的方程；(2)若直线AB 的斜率为2，求2ABF △的面积．。
2021-2022学年河南省周口市鹿邑县第二高级中学高二物理期末试题含解析
2021-2022学年河南省周口市鹿邑县第二高级中学高二物理期末试题含解析一、选择题：本题共5小题，每小题3分，共计15分．每小题只有一个选项符合题意1. （单选）关于功的概念，下列说法正确的是A．力是矢量，位移是矢量，所以功也是矢量B．功有正、负之分，所以功可能有方向性C．若某一个力对物体不做功，说明该物体一定没有位移D．一个恒力对物体做的功等于这个力的大小、物体位移的大小及力和位移间夹角的余弦三者的乘积参考答案：D2. 火星的质量是地球质量的m倍，它的公转轨道的半径是地球公转轨道半径的n 倍，则太阳对火星的引力是对地球引力的多少倍Ａ．mn倍 B．mn2倍Ｃ．倍Ｄ．倍参考答案：C3. 电流为l的直导线处于磁感应强度为B的匀强磁场中，所受磁场力为F。
关于电流I、磁感应强度B和磁场力F三者之间的方向关系，下列图示中正确的是（）参考答案：B4. 用电高峰期，电灯往往会变暗，其原理可简化为如下物理问题．如图所示，理想变压器副线圈上，通过输电线连接两只相同的灯泡L1和L2，输电线的等效电阻为r，原线圈输入有效值恒定的交流电压，当开关S闭合时，以下说法正确的是（）A．原线圈中电流减小B．r两端的电压增大C．原线圈输入功率不变D．副线圈输出电压减小参考答案：B【考点】变压器的构造和原理．【分析】开关S闭合后改变了副线圈的电流和功率，根据变压器原副线圈电压、电流与匝数比的关系即可求解【解答】解：A、开关S闭合，用户端阻值变小，用户端干路电流增大，则原线圈中电流变大，原线圈输入功率变大，选项A、C均错误；B、用户端干路电流增大则输电线r上电压损失变大，选项B正确；D、由于原线圈输入电压及匝数比不变，故副线圈输出电压不变，选项D错误．故选：B5. 光电效应的实验结论是：对于某种金属A．无论光强多强，只要光的频率小于极限频率就不能产生光电效应B．无论光的频率多低，只要光照时间足够长就能产生光电效应C．超过极限频率的入射光强度越弱，所产生的光电子的最大初动能就越小D．超过极限频率的入射光频率越高，所产生的光电子的最大初动能就越大参考答案：AD二、填空题：本题共8小题，每小题2分，共计16分6. 如图所示,在磁感应强度为的匀强磁场中，导体棒AB的在金属框架上以的速度向右滑动，已知导体棒长为，电阻为，，其他电阻不计，则导体棒端点上的电压大小是V, 电压高的点是______（A或B）参考答案：0.5V B点7. 用升压变压器可以实现高压输电，以减少远距离输电过程中的电能损失。
河南省周口市鹿邑县第二高级中学校2023-2024学年高一下学期月考测试(二)(4月) 数学试题
2023～2024学年度高中同步月考测试卷（二）高一数学测试模块：必修第二册考生注意：1．本试卷分选择题和非选择题两部分.满分150分，考试时间120分钟.2．答题前，考生务必用直径0.5毫米黑色墨水签字笔将密封线内项目填写清楚.3．考生作答时，请将答案答在答题卡上.选择题每小题选出答案后，用2B 铅笔把答题卡上对应题目的答案标号涂黑：非选择题请用直径0.5毫米黑色墨水签字笔在答题卡上各题的答题区域内作答，超出答题区域书写的答案无效，在试题卷、草稿纸上作答无效.4．本试卷主要命题范围：人教A 版必修第二册第六章～第七章.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．在ABC 中，角A ，B ，C 的对边分别为a ，b ，c ，若2a =，30A =︒，45B =︒，则b =（）A ．22B 2C 6D ．622．20232024i i -=（）A ．12i +B ．12iC ．1i--D ．1i-3．已知向量()()1,0,2,1a b == ，若ka b - 与3a b +r r垂直，则k =（）A ．177B ．177-C ．13D ．13-4．在ABC 中，角,,A B C 的对边分别为,,a b c ，若5,4,21a b c ===，则C =（）A ．90B ．45C ．60D ．305．若复数()i R z m m =-∈，zz为实数，则m =（）A ．0B ．1C ．1-D ．2-6．在ABC 中，角,,A B C 的对边分别为,,a b c ，若()sin sin sin b B c A B a A =+-，则ABC 为（）A ．等腰三角形B ．直角三角形C ．锐角三角形D ．钝角三角形7．若向量a ，b 的夹角是π3，a 是单位向量，2b = ，2c a b =+，则向量c 与b 的夹角为（）A ．π6B ．π3C ．2π3D ．3π48．在ABC 中，角A ，B ，C 的对边分别为a ，b ，c ，若2a =，30A =︒，则ABC 面积的最大值为（）A ．B ．C ．3D ．2+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．已知复数2z =，则（）A ．zB ．z 是纯虚数C ．zD ．z 在复平面内对应的点位于第四象限10．在ABC 中，角,,A B C 的对边分别为,,a b c ，则下列对ABC 的个数的判断正确的是（）A ．当4,30a c A === 时，有两解B ．当5,7,60a b A === 时，有一解C ．当4,30a b A === 时，无解D ．当6,4,60a b A === 时，有两解11．在ABC 中，角,,A B C 的对边分别为,,a b c ，已知ABC 的周长为3,60B =︒，则（）A ．若2b a c =+，则ABC 是等边三角形B ．存在非等边ABC 满足2b ac =C ．ABC 内部可以放入的最大圆的半径为6D ．可以完全覆盖ABC 的最小圆的半径为3三、填空题：本题共3小题，每小题5分，共15分.12．已知x 、y ∈R ，若(2)i 1i x y -+=-+，则x y +=．13．在ABC 中，已知向量AB与AC 满足0AB AC BC AB AC⎛⎫+⋅= ⎪⎝⎭，且0AB AC ⋅= ，则角B =.14．若平面有不共线的五点A ，B ，C ，D ，O ，记OA a = ，OB b = ，OC c = ，OD d =-，满足21a b a b ==⋅= .()()2R c a b λλλ=+-∈，4d b += ，则c d + 的最小值为．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．已知复数()()()22i 13i 3i R z m m m =--+-+∈．(1)当m 为何值时，z 为纯虚数?(2)当2m =时，求z z ⋅．16．已知cos 10α=-，()0,πα∈．(1)求sin α和tan α的值；(2)若向量()6,1cos2a α=+ ，()25,sin2cos b αα=-- ，证明：//a b ．17．在ABC 中，角,,A B C 的对边分别为,,a b c，已知2,cos b c C ===(1)求sin B 和a 的值；(2)求ABC 的面积．18．如图、某港口O 要将一件重要物品用小艇送到一艘正在航行的轮船上，在小艇出发时，轮船位于港口O 北偏西30︒方向且与该港口相距30nmile 的A 处，并以20nmile/h 的航行速度沿正东方向匀速行驶，假设该小艇沿直线方向以nmile/h v 的航行速度匀速行驶，经过h t 与轮船相遇.（假设水面平静）(1)要使相遇时小艇的航行距离最短，小艇的航行速度应为多少？(2)假设小艇的速度最快只能达到20nmile/h ，要使小艇最快与轮船相遇，应向哪个方向航行？19．在ABC 中，角,,A B C 的对边分别为,,a b c ，且sin 1sin sin B aA C b c=-++．(1)求角C 的大小；(2)若ABC 为锐角三角形，且4b =，求ABC 周长的取值范围．1．A【分析】由正弦定理解三角形.【详解】ABC 中，由正弦定理sin sin a bA B=，得sin 2sin 45sin sin 30a B b A ︒===︒故选：A.2．C【分析】利用2i 1=-和幂的运算性质计算可得结果【详解】()()1011101220232024202220242210111012i i i i i i i i (1)i (1)i 1-=⋅-=⋅-=-⋅--=--.故选：C 3．A【分析】根据ka b - 与3a b +r r 垂直，由()()30ka b a b -⋅+= 求解.【详解】解：()()()()()()1,02,12,1,31,032,17,3ka b k k a b -=-=--+=+=，ka b -r r Q 与3a b +r r垂直，()()()()32,17,30ka b a b k ∴-⋅+=--⋅=，177k ∴=.故选：A.4．C【分析】根据余弦定理求出答案.【详解】由余弦定理得2221cos 22a b c C ab +-==，因为0180C << ，所以60C = .故选：C.5．A【分析】根据复数代数形式的除法运算化简zz，再根据复数的类型得到2201m m -=+，解得即可.【详解】因为()()()()2222222i i i 2i i 12i i i i i 11m m z m m m m mz m m m m m m ----+--====+++--++，又zz为实数，2201m m -∴=+，0m ∴=.故选：A.6．B【分析】利用诱导公式及正弦定理将角化边即可判断.【详解】因为()sin sin sin b B c A B a A =+-，又()()sin sin πsin A B C C +=-=，即sin sin sin b B c C a A =-，由正弦定理可得222b c a =-，即222+=a b c ，所以ABC 为直角三角形且C ∠为直角.故选：B 7．A【分析】首先根据数量积的定义求出a b ⋅ ，再求出c 、c b ⋅，最后根据夹角公式计算可得.【详解】∵两个向量a ，b 的夹角是π3，a是单位向量，2b = ，∴π1cos 12132a b a b ⋅=⋅=⨯⨯=．∵2c a b =+，∴c == ∴()262224c b a b b a b b ⋅=+⋅=⋅+=+= ．设向量c 与b 的夹角为θ，则cos 2c b c b θ⋅===⋅，∵[]0,πθ∈，∴π6θ=．故选：A ．8．D【分析】结合余弦定理及基本不等式，利用三角形面积公式求解即可.【详解】由余弦定理：222222cos 4a b c bc A b c =+-=+=，因为(22422b c bc bc =+-≥=，当且仅当228b c ==+所以8bc ≤=+，故ABC面积(111sin 82244ABC S bc A bc ==≤⨯+= .即ABC 面积的最大值为2故选：D 9．AC【分析】根据复数的基本概念，以及复数的几何意义，对每个选项进行逐一分析，即可判断和选择.【详解】对A ：由虚部定义知z A 正确；对B ：纯虚数要求实部为0，故B 错误；对C ：z =C 正确；对D ：z 在复平面内对应的点为(，位于第一象限，故D 错误.故选：AC.10．AC【分析】由正弦定理对四个选项一一判断，得到答案.【详解】对于A ，由正弦定理得sin sin a c A C =，即4sin30sin C = ，所以2sin 2C =，又因为0180,C c a <<> ，所以45C = 或135C = ，有两解，故A 正确；对于B ，由正弦定理得7sin 2sin 1510b A B a===>，无解，故B 错误；对于C ，由正弦定理得14sin 2sin 1b A B a ⨯==，无解，故C 正确；对于D ，由正弦定理得4sin 2sin 6b AB a===又b a <，所以B 为锐角，此三角形只有一解，故D 错误.故选：AC.11．ACD【分析】由余弦定理与正弦定理及三角形的面积公式逐项求解即可.【详解】因为ABC 的周长为3，且60B =︒，可得3a b c ++=，由余弦定理得222222cos b a c ac B a c ac =+-=+-．对于A ，因为2b a c =+，所以()2224a c a c ac +=+-，即()20a c -=，则a b c ==，所以ABC 为等边三角形，故A 正确；对于B ，假设2b ac =，则22ac a c ac =+-，即()20a c -=，则a b c ==，此时ABC 为等边三角形，故B 错误；对于C ，由()22223b a c ac a c =+-=--，可得()233ac a c =+-≥，当且仅当a c =时等号成立，解得1ac ≤或9ac ≥（舍去），所以ABC的面积1sin 2S ac B ABC =≤△的内切圆半径为2S a b c ≤++所以ABC内部可以放入的最大圆的半径为6，故C 正确；对于D ，设ABC 外接圆的半径为R ，因为()2232a c a c ac +⎛⎫+-=≤ ⎪⎝⎭，当且仅当a c =时等号成立，所以()()28120a c a c +-++≥，解得2a c +≤或6a c +≥（舍去），由()2231b a c =--≥，可得1b ≥，因为2sin60b R =︒，所以33R ≥，所以可以完全覆盖ABCD 正确．故选：ACD ．12．2【分析】根据相等复数的概念列出方程组，解之即可求解.【详解】由题意，得21111x x y y -=-=⎧⎧⇒⎨⎨==⎩⎩，所以2x y +=.故答案为：2.13．4π##45︒【分析】依题意可得AB AC ⊥，设角A 的平分线交BC 于D ，即可得到AD BC ⊥，从而得到ABC 为等腰直角三角形，即可得解.【详解】设角A 的平分线交BC 于D ，因为0AB AC ⋅= ，故AB AC ⊥，即π2CAB ∠=，又AB AB 表示与AB 同向的单位向量，AC AC表示与AC 同向的单位向量，设AB AE AB =，AC AF AC=（如图所示），AE AF AG +=，因为1AE AF == ，故四边形AEGF 为正方形，所以AG 为角A 的平分线，故G 在AD 上.因为0AB AC BC AB AC⎛⎫⎪+⋅= ⎪⎝⎭，故AD BC ⊥，故AB AC =.综上，ABC 为等腰直角三角形且π2CAB ∠=，所以π4B =.故答案为：π4143223【分析】利用平面向量的几何意义，结合图像即可得到答案.【详解】由21a b a b ==⋅= ,可得1OA OB == ，且1cos 2a b AOB a b ⋅∠==，即π3AOB ∠=.作2,4,OM b OE b MN BA ===，如图所示，则AOB ，ABN ，BMN 均为正三角形，且2OM ME b ==，由()2c a b λλ=+-，得()22OC OA OB OB OB BA OM MN λλλ=-+=+=+ ，化简可得，OC OM MC MN λ-==，所以C 在直线MN 上.由图像可知，4d b OD OE DE +=-+=，所以42DE d b =+= ，可得点D 在以点E 2E 上，所以c d OC OD DC +=-= .如图过E 作MN 垂线垂足为C ，交圆E 于D 点，则显然3EC =此时DC 的最小值为EC ED -= 15．(1)1m =-或32．(2)90.【分析】（1）先化简复数z ，再利用复数的相关概念求解；（2）先求得复数z 和其共轭复数，再利用复数的乘法求解.【详解】（1）解：由已知得()()222331i z m m m m =--+--+，若z 为纯虚数，则22230,310,m m m m ⎧--=⎨--+≠⎩解得1m =-或32．（2）当2m =时，39i z =-，39i z =+，所以()()2239i 39i 381i 98190z z ⋅=-⋅+=-=+=．16．(1)sin α=1tan 3α=-．(2)证明见解析【分析】（1）根据同角三角函数的基本关系计算可得；（2）利用二倍角公式及同角三角函数的基本关系求出2sin2cos 1cos2ααα-+，再根据平面向量共线的坐标表示计算可得.【详解】（1）因为()0,πa ∈，cos α=所以10sin 10α=．所以sin 1tan cos 3ααα==-．（2）证明：因为222sin2cos 2sin cos cos 1cos22cos ααααααα--=+2sin cos 15tan 2cos 26αααα-==-=-，所以()()()26sin2cos 51cos20ααα---⨯+=，所以//a b ．17．(1)sin 33B a ==；3．【分析】（1）根据同角的三角函数关系求出sin C ，结合正、余弦定理计算即可求解；（2）由（1），结合三角形的面积公式计算即可求解.【详解】（1）在ABC 中，由cos 3C =-，可得6sin 3C ==．又由sin sin c b C B =及2b c ==，可得sin 3B =．由余弦定理得2222cos c a b ab C =+-，得2360a +-=，由0a >，解得a =所以36sin 33B a ==（2）由（1）知，,sin 33a C ==，所以ABC 的面积11sin 22333ABC S ab C ==⨯=△．18．(1)(2)航行方向为北偏东30︒【分析】（1）利用余弦定理和二次函数的最值求解；（2）要用时最小，则首先速度最高，然后是距离最短，则由（1）利用余弦定理得到方程解得对应的时间t ，再解得相应角，即可求解.【详解】（1）如图设小艇的速度为v ，时间为t 相遇，相遇点为C ，则由余弦定理得：2222cos OC AC OA AC OA OAC =+-⨯⨯⨯∠，即2222234009001200cos 60400600900400()6754v t t t t t t ︒=+-=-+=-+,当3t 4=时，OC取得最小值，此时速度，此时小艇的航行方向为正北方向，航行速度为.（2）要用时最小，则首先速度最高，即为20nmile/h ，则由（1）可得：2222cos OC AC OA AC OA OAC =+-⨯⨯⨯∠，即()22209004001200cos 60t t t ︒=+-，解得32t =，此时相遇点为B ，此时，在OAB 中，30OA OB AB ===，则30BOD ︒∠=，故可设计航行方案如下:航行方向为北偏东30︒，航行速度为20nmile/h ，小艇能以最短时间与轮船相遇.19．(1)π3C =(2)(6++【分析】（1）利用正弦定理进行角化边，然后根据余弦定理求解出cos C 的值，即可求出角C ；（2）法一：根据正弦定理可得sin sin sin sin a b c b A B C B ++=++，根据三角恒等变换化简可得6tan 2a b c B ++=+，再根据B 的范围求解即可；法二：过点A 作11AB CB ⊥，垂足为1B ，根据直角三角形性质结合图形分析求解.【详解】（1）由正弦定理得1b a a c b c =-++，整理得222a b c ab +-=，所以2221cos 22a b c C ab +-==，又()0,πC ∈，所以π3C =．（2）法一：由（1）知2π3A B +=，即2π3A B =-．因为ABC 为锐角三角形，所以2ππ0,32π0,2B B ⎧<-<⎪⎪⎨⎪<<⎪⎩解得ππ62B <<．由正弦定理sin sin sin a b c A B C ==，得sin sin sin sin a b c b A B C B++=++，则42π43sin sin cos sin sin 3sin 2a b c B B B B B B ⎡⎤⎫⎛⎫++=-+=+⎪⎢⎥ ⎪⎪⎝⎭⎣⎦⎝⎭)22cos1cos2666sin2sin cos tan222BBB BB+=+=+=当ππ62B<<时，ππ1224B<<，则ππtan tan tan1224B<<．又ππtan tanπππ34tan tan2ππ12341tan tan34-⎛⎫=-==-⎪⎝⎭+所以2tan12B<<，所以6tan2B<<+所以6612tan2B+<+<+612a b c+<++<+所以ABC周长的取值范围是(6++．法二：（数形结合）过点A作11AB CB⊥，垂足为1B，在直线1CB上取一点2B，使2B A AC⊥，则1AB CV与2AB C均为直角三角形．ABC为锐角三角形，∴点B在线段12B B上（不含端点）．在1Rt AB C△中，π,43C b==，易得12B C=，1AB=6+在2Rt AB C△中，π,43C b==，易得228,B C AB==12+，所以ABC周长的范围是(6++．。
河南省周口市鹿邑县第二高级中学校2025届高三上学期11月期中考试数学试题
河南省周口市鹿邑县第二高级中学校2025届高三上学期11月期中考试数学试题一、单选题1．已知集合{}205A x x =<<，{}Z 12B x x =∈-<，则A B = （）A ．{}1,0,1,2-B ．{}0,1,2C ．{}1,2D ．{}1,0,1,2,3-2．已知m ，n 是空间两条不同的直线，α，β是两个不重合的平面，有下列命题：p ：若//m α，n ⊂α，则//m n ；q ：若m α⊥，n β⊥，m n ⊥，则αβ⊥.则下列命题是真命题的是（）A ．p q ∧B ．q p ⌝∨C ．q p⌝∧D ．()p q ⌝∨⌝3．若函数2()ln 1f x x ax =+-在区间(1,2)上存在单调递减区间，则实数a 的取值范围是（）A ．1,2⎛⎫-∞- ⎪⎝⎭B ．1,2⎛⎤-∞- ⎥⎝⎦C ．1,8⎛⎫-∞- ⎪⎝⎭D ．1,8⎛⎤-∞- ⎥⎝⎦4．已知函数()()2214,15,1a x a x f x x ax x ⎧-+<=⎨-+≥⎩满足对任意12,x x ，当12x x ≠时都有()()12120f x f x x x ->-成立，则a 的取值范围是（）A ．1,12⎛⎤ ⎥⎝⎦B ．1,22⎛⎫⎪⎝⎭C ．[)2,+∞D ．[]1,25．已知函数()()f x x ∈R 满足()()4f x f x =-，若2y x =-与()y f x =图象的交点为()()()()()()112233445566,,,,,,,,,,,x y x y x y x y x y x y ，则123456x x x x x x +++++=（）A ．4-B ．0C ．8D ．126．已知函数()331f x x x =-+-，下列说法错误的是（）A ．()f x 在1x =-处取得极小值B ．()f x 有3个零点C ．()f x 在区间()2,2-上的值域为()3,1-D ．曲线()y f x =的对称中心为()0,1-7．已知定义在R 上的奇函数()f x 满足()()2f x f x =-．当12x ≤≤时，()()2log 7f x x =+，则()2023f =（）A ．3B ．3-C ．5-D ．58．若对任意的1x ，(]21,3x ∈，当12x x <时，1212ln ln 22a ax x x x ->-恒成立，则实数a 的取值范围是（）A ．[)3,+∞B ．()3,+∞C ．[)6,+∞D ．()6,+∞二、多选题9．对任意两个实数,a b ，定义{},min ,,a a b a b b a b≤⎧=⎨>⎩，若()22f x x =-，()22g x x =-，下列关于函数()()(){}min ,F x f x g x =的说法正确的是（）A ．函数()F x 是偶函数B ．方程()0F x =有两个解C ．方程()F x m =至少有三个根D ．函数()F x 有最大值为0，无最小值10．下列说法正确的是（）A ．命题“21,1x x ∀><”的否定是“21,1x x ∃≤≥”B ．“1a >”是“11a<”的充分不必要条件C ．设,a b ∈R ，则“0a ≠”是“0ab ≠”的必要不充分条件D ．“1x >”是“12x +≥”的既不充分也不必要条件11．设函数()321f x x x ax =-+-，则（）A ．当1a =-时，()f x 有三个零点B ．当13a ≥时，()f x 无极值点C ．a ∀∈R ，曲线()y f x =对称中心的横坐标为定值D ．a ∃∈R ，使()f x 在R 上是减函数三、填空题12．若当1x >时，不等式1211x m x +≥--恒成立，则实数m 的取值范围是.13．已知函数()3sin f x x x x =+-则满足不等式()()221m m f f ≤-成立的实数m 的取值范围是．14．对于三次函数()()320ax bx d a f x cx =+++≠，给出定义：设()f x '是函数()f x 的导数，()f x ''是()f x '的导数，若方程()0f x ''=有实数解0x ，则称点()()00,x f x 为函数()y f x =的“拐点”.经过探究发现：任何一个三次函数都有“拐点”；任何一个三次函数都有对称中心，且“拐点”就是对称中心.设函数()3211533212f x x x x =-+-，则()f x 的拐点为，12320222023202320232023f f f f ⎛⎫⎛⎫⎛⎫⎛⎫++++= ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭.四、解答题15．已知命题:p x ∃∈R ，使2(1)10x a x +-+<；命题:[2,4]q x ∀∈，使2log 0x a -≥.（1）若命题p 为假命题，求实数a 的取值范围；（2）若p q ∨为真命题，p q ∧为假命题，求实数a 的取值范围.16．已知函数()f x 满足()2259f x x x +=++，且()()2g x f x x a =-.(1)求()f x 的解析式；(2)求a 的值，使()g x 在区间[]5,5-上的最小值为1-.17．已知定义在()1,b -上的奇函数()lg a xf x b x-=+．(1)求实数,a b 的值：(2)若()f x 在(),m n 上的值域为()1,-+∞，求实数,m n 的值．18．函数()()241log 2log 2f x x x ⎛⎫=-- ⎪⎝⎭．(1)当[]1,4x ∈时，求该函数的值域；(2)若()4log f x m x >对于[]4,16x ∈恒成立，求m 的取值范围．19．已知函数21()ln ()2f x x ax a R =-∈．（1）若()f x 在点(2，f （2）)处的切线与直线210x y -+=垂直，求实数a 的值；（2）求函数()f x 的单调区间；（3）讨论函数()f x 在区间[1，2]e 上零点的个数．。
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河南省鹿邑县第二高级中学
佚名
【期刊名称】《现代语文（教学研究）》
【年(卷),期】2016(000)0z1
【摘要】学校概况学校创办于1993年6月，是周口市首批公办示范性高中，位于鹿邑县紫气大道东段，东依迎宾大道，西望明道宫，南邻老君台中学，占地70余亩，总建筑面积28200平方米。
学校现有64个教学班，在校学生4600余人。
【总页数】1页(P117-117)
【正文语种】中文
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