河南省洛阳市孟津县第二高级中学2020-2021学年高二数学9月周练试题【含答案】

河南省洛阳市孟津县第二高级中学2021-2022高二数学9月周练试题注意事项：本试卷分第I 卷（选择题）和第Ⅱ卷（非选择题）两部分。

考试时间120分钟，满分150分。

考生应首先阅读答题卡上的文字信息，然后在答题卡上作答，在试题卷上作答无效。

交卷时只交答题卡。

第Ⅰ卷一、选择题：（本大题12个小题，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．） 1．等差数列{n a }中，3a ＝2，5a ＝7，则7a ＝A ．10B ．20C ．16D ．12 2．设集合A ＝{x ｜0<x<1}，B ＝{x ｜0<x<3}，那么“m ∈A ”是“m ∈B ”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．已知△ABC 的周长为20，且顶点B （0，－4），C （0，4），则顶点A 的轨迹方程是A ．213620x 2y ＋＝（x ≠0）B ．212036x 2y ＋＝（x ≠0） C ．21620x 2y ＋＝（x ≠0） D ．21206x 2y ＋＝（x ≠0） 4．不等式312x x－－≥0的解集是 A ．123x x ⎧⎫≤≤⎨⎬⎩⎭｜B ．123x x ⎧⎫≤⎨⎬⎩⎭｜＜ C ．123x x x ⎧⎫≤⎨⎬⎩⎭｜或＞ D ．13x x ⎧⎫⎨⎬⎩⎭｜＞ 5．在等比数列{n a }中，若357911243a a a a a ＝，则2911a a 的值为A ．9B ．1C ．2D ．36．三角形两条边长分别为2和3，其夹角的余弦值是方程22x －3x ＋1＝0的根，则此三角形周长为AB ． 7C ．5D ．5＋7．若实数x 、y 满足2,,y ,x x ≤⎧⎪≤⎨⎪≥⎩y 3＋1则S ＝2x ＋y －1的最大值为A ．6B ．4C ．3D ．2 8．在△ABC 中，a 、b 、c 分别是∠A 、∠B 、∠C 所对应的边，∠C ＝90°，则a bc＋的取值范围是A ．（1，2）B ．（1）C ．（1]D ．[1] 9．在△ABC ，若cos cos A B ＝b a ＝43，则△ABC 是 A ．直角三角形 B ．等腰三角形 C ．等腰或直角三角形 D ．钝角三角形 10．设定点M （3，103）与抛物线2y ＝2x 上的点P 的距离为1d ，P 到抛物线准线l 的距为2d ，则1d ＋2d 取最小值时，P 点的坐标为A ．（0，0）B ．（1）C ．（2，2）D ．（18，－12） 11．已知1F 、2F 是椭圆2221x a b2y ＋＝（a>b>0）的两个焦点，以线段1F 2F 为边作正三角形 M 1F 2F ，若边M 1F 的中点在椭圆上，则椭圆的离心率是A ．12 B 1 C ．12D 1 12．空间直角坐标系中，O 为坐标原点，已知两点坐标为A （3，1，0），B （－1，3，0），若点C 满足OC ＝αOA ＋βOB ，其中α，β∈R ，α＋β＝1，则点C 的轨迹为 A ．平面 B ．直线 C ．圆 D ．线段第Ⅱ卷二、填空题：（本大题共4个小题，每小题5分，共20分．将答案填在答题卡上相应位置） 13．数列{n a }的通项公式为n a ＝2n －9，n ∈N ﹡，当前n 项和n S 达到最小时，n 等于_________________.14．若双曲线214x m 2y －＝的右焦点与抛物线2y ＝12x 的焦点重合，则m ＝______________．15．在△ABC 中,∠A ＝60°，b ＝1，ABC S ∆cos aA＝_______________． 16．在下列命题中，①0x ∃∈R ，20x ＋20x ＋2≤0的否定；②若m>0，则方程2x ＋x －m ＝0有实根的逆命题;③渐近线方程为y＝34x的双曲线的离心率为54；其中真命题的序号是__________________．三、解答题：（本大题共6个小题，共70分）17．（本小题满分10分）给定两个命题，p：对任意实数x都有2x＋ax＋1>0恒成立；q：函数y＝logax（a>0且a≠1）为增函数，若p假q真，求实数a的取值范围．18．（本小题满分12分）在△ABC中，BC，AC＝3，sinC＝2sinA．（Ⅰ）求边长AB的值；（Ⅱ）求△ABC的面积．19．（本小题满分12分）某公园计划建造一个室内面积为800m2的矩形花卉温室．在温室内，沿左、右两侧与后侧内墙各保留1m宽的通道。

2020年河南省洛阳市孟津二高高二数学理下学期期末试卷含解析一、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. 在右图的正方体中，棱BC与平面ABC1D1所成的角为（）A．30° B．45° C．90° D． 6 0°参考答案：B2. 已知函数f（x）的导函数为f′（x），且满足f（x）=2xf′（1）+lnx，则f′（1）=（）A．﹣e B．﹣1 C．1 D．e参考答案：B【考点】导数的乘法与除法法则；导数的加法与减法法则．【分析】已知函数f（x）的导函数为f′（x），利用求导公式对f（x）进行求导，再把x=1代入，即可求解；【解答】解：∵函数f（x）的导函数为f′（x），且满足f（x）=2xf′（1）+ln x，（x ＞0）∴f′（x）=2f′（1）+，把x=1代入f′（x）可得f′（1）=2f′（1）+1，解得f′（1）=﹣1，故选B；【点评】此题主要考查导数的加法与减法的法则，解决此题的关键是对f（x）进行正确求导，把f′（1）看成一个常数，就比较简单了；3. 若椭圆的焦点在x轴上，且离心率e=，则m的值为（）A．B．2 C．﹣D．±参考答案：B【考点】椭圆的简单性质．【分析】通过椭圆的焦点在x轴上，利用离心率，求出m的值．【解答】解：因为椭圆的焦点在x轴上，且离心率e=，所以，解得m=2．故选B．【点评】本题考查椭圆的简单性质的应用，离心率的求法，考查计算能力．4. 用反证法证明命题“同一平面内，不重合的两条直线，都和直线垂直，则与平行”时，否定结论的假设应为（）A. 与垂直B. 与是异面直线C. 与不垂直D.与相交参考答案：D5. 已知函数，若，则函数在定义域内( )A．有最小值，但无最大值． B．有最大值，但无最小值．C．既有最大值，又有最小值． D．既无最大值，又无最小值．参考答案：A6. 右图是2011年在某大学自主招生面试环节中，七位评委为某考生打出的分数的茎叶统计图，去掉一个最高分和一个最低分后，所剩数据的平均数和方差分别为（）A、84，4.84B、84，1.6C、85，1.6D、85，1.5参考答案：C7. 已知集合，，则A∪B=A. {1}B. {1,2}C. {0,1,2,3}D. {－1,0,1,2,3}参考答案：C试题分析：集合，而，所以，故选C.【名师点睛】集合的交、并、补运算问题，应先把集合化简再计算，常常借助数轴或韦恩图进行处理.8. 在正方体ABCD﹣A1B1C1D1中，直线AB1与平面ABC1D1所成的角的正弦值为（）A．B．C．D．参考答案：D【考点】直线与平面所成的角．【分析】如图所示，建立空间直角坐标系．不妨时AB=1，取平面ABC1D1的法向量==（1，0，1），则直线AB1与平面ABC1D1所成的角的正弦值=|cos＜，＞|=，即可得出．【解答】解：如图所示，建立空间直角坐标系．不妨时AB=1，则D（0，0，0），A（1，0，0），B1（1，1，1），A1（1，0，1）．则=（0，1，1），取平面ABC1D1的法向量==（1，0，1），则直线AB1与平面ABC1D1所成的角的正弦值=|cos＜，＞|===．故选：D．【点评】本题考查了空间位置关系、法向量的应用、线面角、向量夹角公式，考查了推理能力与计算能力，属于中档题．9. 在中，面积，则A、 B、75 C、55 D、49参考答案：C10. 已知函数，则“”是“为偶函数”的（）A. 必要不充分条件B. 充分不必要条件C. 充要条件D. 既不充分也不必要条件参考答案：B【分析】根据充分条件与必要条件的定义，结合函数奇偶性的定义和性质，进行判断即可.【详解】若，则为偶函数；当，时，为偶函数，但不成立；所以“”是“为偶函数”的充分不必要条件.故选B【点睛】本题主要考查充分条件与必要条件的判断，熟记定义即可，属于基础题型. 二、填空题:本大题共7小题,每小题4分,共28分11. 若关于x的方程x+b=恰有一个解，则实数b的取值范围为．参考答案：[﹣2，0）∪{﹣1}考点：根的存在性及根的个数判断．专题：计算题；函数的性质及应用．分析：方程x+b=解的个数即函数y=x+b与y=的交点的个数，作图求解．解答：解：方程x+b=解的个数即函数y=x+b与y=的交点的个数，作函数y=x+b与y=的图象如下，由图可知，直线在y=x的右侧或直线与半圆相切，故实数b的取值范围为[﹣2，0）∪{﹣1}．故答案为：[﹣2，0）∪{﹣1}．点评：本题考查了方程的根与函数的图象的关系，属于基础题12. ①若椭圆+=1的左右焦点分别为F1、F2，动点P满足|PF1|+|PF2|＞10，则动点P 不一定在该椭圆外部；②椭圆+=1（a＞b＞0）的离心率e=，则b=c（c为半焦距）；③双曲线﹣=1与椭圆+y2=1有相同的焦点；④抛物线y2=4x上动点P到其焦点的距离的最小值为1．其中真命题的序号为．参考答案：②③④【考点】命题的真假判断与应用．【分析】根据点与椭圆的位置关系，可判断①；根据离心率，求出b，c关系，可判断②；求出椭圆和双曲线的焦点，可判断③；求出抛物线上点到焦点的最小距离，可判断④【解答】解：①若椭圆+=1的左右焦点分别为F1、F2，动点P满足|PF1|+|PF2|＞10，则动点P一定在该椭圆外部，故错误；②椭圆+=1（a＞b＞0）的离心率e=，则b=c=a（c为半焦距），正确；③双曲线﹣=1与椭圆+y2=1有相同的焦点（，0），正确；④抛物线y2=4x上动点P到其焦点的距离的最小值为=1，正确．故答案为：②③④．13. 已知都是正实数, 函数的图象过点,则的最小值是_ __.参考答案：14. 若以原点为圆心，椭圆的焦半径c为半径的圆与该椭圆有四个交点，则该椭圆的离心率的取值范围为：．参考答案：（，1）【考点】椭圆的简单性质．【专题】分析法；不等式的解法及应用；圆锥曲线的定义、性质与方程．【分析】设椭圆的方程为+=1（a＞b＞0），与圆方程为x2+y2=c2，联立方程组，解得x，y，由题意可得c＞b，再由离心率公式，计算即可得到所求范围．【解答】解：设椭圆的方程为+=1（a＞b＞0），以原点为圆心，椭圆的焦半径c为半径的圆方程为x2+y2=c2，联立两方程，可得y2=，x2=，由题意可得x2＞0，y2＞0，结合a＞b＞0，a＞c＞0，可得c2＞b2，即有c2＞a2﹣c2，即为a＜c，则离心率e=＞，由0＜e＜1，可得＜e＜1．故答案为：（，1）．【点评】本题考查椭圆的离心率的范围，注意运用圆与椭圆方程联立，通过方程组有解，考查运算能力，属于中档题．15. 某地区对两所高中学校进行学生体质状况抽测，甲校有学生600人，乙校有学生700人，现用分层抽样的方法在这1300名学生中抽取一个样本．已知在甲校抽取了42人，则在乙校应抽取学生人数为．参考答案：49【考点】B3：分层抽样方法．【分析】根据分层抽样原理，列方程计算乙校应抽取学生人数即可．【解答】解：甲校有学生600人，乙校有学生700人，设乙校应抽取学生人数为x，则x：42=700：600，解得x=49，故在乙校应抽取学生人数为49．故答案为：49．16. .参考答案：略17. 如图，在正方体ABCD﹣A′B′C′D′中，异面直线AC与BC′所成的角为．参考答案：60°【考点】异面直线及其所成的角．【专题】计算题；转化思想；综合法；空间角．【分析】连结A′B、A′C′，由AC∥A′C′，得∠A′C′B是异面直线AC与BC′所成的角，由此能求出异面直线AC与BC′所成的角．【解答】解：在正方体ABCD﹣A′B′C′D′中，连结A′B、A′C′，∵AC∥A′C′，∴∠A′C′B是异面直线AC与BC′所成的角，∵A′B=BC′=A′C′，∴∠A′C′B=60°，∴异面直线AC与BC′所成的角为60°．故答案为：60°．【点评】本题考查异面直线所成角的求法，是基础题，解题时要认真审题，注意空间思维能力的培养．三、解答题：本大题共5小题，共72分。

英语试卷第二部分阅读理解（共两节，满分40分）AWhether they are already household names or a hidden figure deserving of more recognition, the following ladies changed the world with their enormous contributions.Ali StrokerAli Stroker took the theater world — and, indeed, the very Internet — by storm when, on June 9, 2019, she became the first performer in a wheelchair to take home a Tony Award. After becoming the first actor in a wheelchair in Broadway history in 2015, she won the award for her powerhouse performance in the revival of Oklahoma \Junko TabeiTwenty-two years after the first-ever successful mission to the top of Mount Everest, Japanese mountaineer Junko Tabei became the first woman to reach the peak. She led a team of 15 women, accompanied by six Sherpas （夏尔巴人），and reached the summit with one of the Sherpas on May 16, 1975.Gertrude EderleThe Queen of Waves, who also happened to be deaf, was the first woman to swim across the English Channel. Fighting through cold temperatures and strong tides that change direction every six hours for 22 miles, she clocked a time of 14 hours and 34 minutes. Virginia Apgar Generations of parents owe this American doctor a huge thank you, as she developed the Apgar Score, the first standardized system of tests to assess if newborn babies were healthy once they made their way from womb to world. Apgar, who was a gifted cellist and violinist in her spare time, also happens to hold the title of the first woman to be hired as a full professor at the medical school at Columbia University.21. Whose story may inspire the disabled?A. Stroker and Tabei.B. Stroker and Ederle.C. Ederle and Apgar.D. Ederle and Tabei.22. Why should Apgar be appreciated by parents?A. She took home a big award.B. She saved many babies' lives.C. She developed the Apgar Score.D. She became the first full professor.23. Who won the title of the Queen of Waves?A. Ali Stroker.B. Junko Tabei.C. Virginia Apgar.D.Gertrude Ederle.BMy students were taking midterms when my phone erupted with urgent messages. " A student is having a panic attack," texted a teaching assistant. I ran out of my office, down a flight of stairs and found the student —a pupil in my 350-person organic chemistry class —lying motionless on the ground outside the exam hall. " Did my exam really trigger a panic attack?" I asked myself. "Why am I not prepared to deal with a situation like this?”It was my first time teaching the course. But I knew that the subject was challenging for my students. This was a source of stress for premedical students in particular, who feared that a low grade in organic chemistry would keep them from getting into medical school.The following day, I was scheduled to lecture to the same class. I knew that I had to address what had happened during the midterm. So, I started by saying: "I want to take some time today to talk about something important. How many of you think that this is a weed-out course?" Half of my students raised their hands carefully. "I'm sorry to hear that,” I continued. "I want you all to know that I do not consider any of you to be weeds; you all deserve to be here. ”I flashed a slide of flowers in various shapes. I smiled at my students and said: "I think of you as flowers —different flowers with different needs. You may not bloom at the same time, but you will bloom! You may not do well in the midterm exam, but you will learn from your mistakes and do better in the final exam. I believe this. I believe in you."From that point on, my office hours were packed. Some asked about lecture topics and study strategies; others opened up about personal issues.I was amazed that a simple, frank discussion in lecture could make such a difference.24. What made the pupil have a panic attack?A. Hiding personal issues.B. The stress for high grades.C. Lacking study strategies.D. Failing to handle the situation.25. What does the underlined word "trigger" in Paragraph 1 most probably mean?A. Cure.B. Prevent.C. Frighten.D. Cause.26. Why did the author go to the same class the next day?A. To give the lesson according to the arrangement.B. To apologize and explain to the panicked student.C. To give a speech on what happened in the test.D. To persuade all the students to stay in the class.27. Which paragraph mainly shows the author's encouragement to students?A. Paragraph 2.B. Paragraph 3.C. Paragraph 4.D. Paragraph5.CThe cognitive health and development of boys may be affected by their mothers ' body mass index ( BMI) (体重指数)while pregnant with them, according to research from Columbia University and the University of Texas at Austin.The study, which was published in the journal BMC Pediatrics on Friday, observed 368 subjects from low-income African American and Dominican women during the second half of their pregnancies, and then evaluated their children three and seven years later. Researchers found that the sons of women whose BMIs indicated that they were overweight or obese when they became pregnant were more likely to show less developed athletic skills as 3-year-olds and lower intelligence as 7-year-olds compared to boys whose mothers were at "normal" weights during pregnancy.Among boys, the study found, mothers' overweight and obesity connected with IQ scores between 4.6 and almost 9 points lower thanthose of boys whose mothers' weights were in the "normal" range before pregnancy. Researchers did not observe the same phenomenon among daughters whose mothers had been obese."These findings aren't meant to shame or scare anyone, "Elizabeth Widen, assistant professor of nutritional sciences at UT Austin and one of the study's co-authors, said in a press release. " We are just beginning to understand some of these interactions between mothers' weight and the health of their babies."Why mothers' obesity appeared to affect childhood IQ was unclear, but earlier research has suggested that there is a relationship between a mother's diet and her child's later IQ, according to Columbia University. Researchers did not control for what the mothers ate, the press release noted.The study's authors wrote that because childhood IQ has been shown to be an indicator of later success in life, studying how a mother's obesity could affect the IQ of her child is worthwhile.28. How did researchers carry out the study?A. By measuring mothers' body mass index.B. By watching mothers and babies for years.C. By comparing 3-year-old babies with 7-year-olds.D. By evaluating the health of mothers and their babies.29. What's the main purpose of the study?A. To show links between mothers' weight and babies' IQ.B. To make those overweight mothers shameful and scared.C. To warn some fat mothers to keep a balanced diet.D. To persuade more obese mothers to lose weight.30. What do the researchers think of the study?A. Doubtful.B. Worrying.C. Significant.D. Interesting.31. In which section of a newspaper may the text appear?A. Entertainment.B. Novel.C. Education.D. Health.DMore than half of the birds in Washington are at risk of extinction because of climate change. That's according to a new national report fromthe Audubon Society, which gives detailed analysis of climate effects on about 600 species of North American birds.It's based on more than 140 million observations of birds across the US, Mexico and Canada. Audubon scientists looked at the likely effects of sea-level rise, urbanization, drought, extreme spring heat, increased fires, heavy rain and other factors.But it doesn't just spell out a doomsday scenario （世界末日）.Instead, it offers a range of effects and warming, depending on how much carbon humans add to the atmosphere."It is truly an existential threat （威胁），not only to birds but to people,” said Doug Santoni, board chair of Audubon Washington, who looked into the report as soon as it came out.Santoni says he was struck to see the vulnerability （脆弱）of a common “ backyard bird" , the dark-eyed junco. It's one that many first-time birders become familiar with as they learn how to identify species based on their markings and other traits. Currently in Washington, you can count on juncos to show up at your feeder, year round. Extreme spring heat, increased fires and heavy rain are the kinds of changes that will force birds like these north, or kill them off if they fail to adapt.Trina Baya rd, director of bird conservation at Audubon’s Washington chapter, says, "It's certainly a very serious warning report," but adds that there’s still hope. “If we can stabilize current temperatures and decrease our emissions （排放），we can really reduce the effects to these birds --- that's very motivating. ”32. What can we know about the new report?A. It analyses the species of birds in detail.B. It's issued by watching 600 bird species.C. It shows the end of North American birds.D. It reports the threat some birds are facing.33. What may Santoni probably agree with?A. Climate change is a threat only to birds.B. It's too late to take action to save the birds.C. The current situation of the birds is worrying.D. It's common that birds are affected by climate change.34. Which of the following can help these birds according to Trina?A. Lowering present temperatures.B. Reducing our daily emissions.C. Making them adapt to climate change.D. Encouraging people to protect them.35. What can be the best title for the text?A. Climate change threatens many Washington bird speciesB. A new report about 600 species of North American birdsC. Different attitudes towards the situation of bird speciesD. Climate change makes different kinds of species at risk第二节（共5小题；每小题2分，满分10分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。

化学试题班级_______ 考号______ 姓名_______第I卷选择题(共50分)一、选择题（每小题2分，共20分）1．对于可逆反应3H2＋N22NH3，下列措施能使反应物中活化分子百分数、化学反应速率和化学平衡常数都变化的是( )A．增大压强B．充入更多N2C．使用高效催化剂D．降低温度2．一定条件下，反应2A(g)＋2B(g)3C(g)＋D(g)在容积不变的密闭容器中进行，达到化学平衡状态的标志是( )A．单位时间内生成2n mol B，同时消耗3n mol C B．容器内压强不随时间变化C．混合气体的密度不随时间变化 D．单位时间内生成2n mol A，同时生成n mol D3．一定条件下的密闭容器中，反应3H2(g)＋3CO(g)CH3OCH3(g)＋CO2(g) ΔH<0达到平衡，要提高CO的转化率，可以采取的措施是( )A．升高温度B．加入催化剂C．减小CO2的浓度D．增加CO的浓度4．下列有关说法中正确的是( )A．2CaCO3(s)＋2SO2(g)＋O2(g)===2CaSO4(s)＋2CO2(g)在低温下能自发进行，则该反应的ΔH<0B．NH4Cl(s)===NH3(g)＋HCl(g)在室温下不能自发进行，则该反应的ΔH<0C．若ΔH>0，ΔS<0，化学反应在任何温度下都能自发进行D．加入合适的催化剂能降低反应活化能，从而改变反应的焓变5.以下对影响反应方向因素的判断不正确的是()A．有时焓变对反应的方向起决定性作用B．有时熵变对反应的方向起决定性作用C．焓变和熵变是判断反应方向的两个主要因素D．任何情况下，温度都不可能对反应的方向起决定性作用6． 298 K时，在2 L固定体积的密闭容器中，发生可逆反应：2NO2(g)N2O4(g) ΔH＝－a kJ/mol(a>0)。

N2O4的物质的量浓度随时间变化如图。

达平衡时，N2O4的浓度为NO2的2倍，若反应在398 K进行，某时刻测得n(NO2)＝0.6 mol，n(N2O4)＝1.2 mol，则此时下列关系正确的是( )A．v(正)>v(逆) B．v(正)<v(逆)C．v(正)＝v(逆) D．v(正)、v(逆)大小关系不确定7.已知反应I2(g)＋H2(g)2HI(g)ΔH<0，下列说法正确的是()A．降低温度，正向反应速率减小倍数大于逆向反应速率减小倍数B．升高温度将缩短达到平衡的时间C．达到平衡后，保持温度和容积不变，充入氩气，正逆反应速率同等倍数增大D．达到平衡后，保持温度和压强不变，充入氩气，HI的质量将减小8．反应2SO2+O22SO3经一段时间后,SO3的浓度增加了0.8 mol·L-1,在这段时间内用O2表示的反应速率为0.04 mol·L-1·s-1,则这段时间为()。

河南省洛阳市轴第二中学2020-2021学年高二数学理模拟试卷含解析一、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. 过所在平面外一点，作,垂足为,连接。

若则点（）A.垂心B. 外心C.内心D. 重心参考答案：B2. 圆的切线方程中有一个是（）A．x－y＝0 B．x＋y＝0 C．x＝0 D．y＝0参考答案：C略3. 在等差数列{a n}中，若a4＋a6＋a8＋a10＋a12＝120，则2a10－a12＝（）A．24 B．22 C．20D．18参考答案：A4. 已知函数，其导函数的图象如图所示，则()A．在上为减函数B．在处取极小值C．在上为减函数D．在处取极大值参考答案：C略5. 设满足约束条件，则的取值范围是（）A． B． C． D．参考答案：D略6. 在中，，则最短边的边长等于（）第1页共4页第2页共4页共A. B. C. D.参考答案：B7. 如图，长方形的四个顶点为，曲线经过点．现将一质点随机投入长方形中，则质点落在图中阴影区域的概率是（）A． B． C． D．参考答案：D8. 在中，，则（）A． B． C． D．参考答案：A9. 焦距为，离心率，焦点在轴上的椭圆标准方程是（）参考答案：D10. 某公司在甲、乙、丙、丁四个地区分别有150个、120个、180个、150个销售点，公司为了调查产品销售的情况，需从这600个销售点中抽取一个容量为100的样本，记这项调查为(1)；在丙地区中有20个特大型销售点，要从中抽取7个调查其销售收入和售后服务情况，记这项调查为(2)。

则完成(1)、(2)这两项调查宜采用的抽样方法依次是 ( )A、分层抽样法，系统抽样法B、分层抽样法，简单随机抽样法C、系统抽样法，分层抽样法D、简单随机抽样法，分层抽样法参考答案：B略二、填空题:本大题共7小题,每小题4分,共28分11. 对取某给定的值,用秦九韶算法设计求多项式的值时,应先将此多项式变形为参考答案：略12. 在等差数列中已知，a7=8，则a1=_______________参考答案：D略13. 如图，在梯形中，，点在的内部（含边界）运动，则的取值范围是．参考答案：14. 已知抛物线的焦点为，抛物线的准线与轴的交点为，点A在抛物线上且，则的面积是．参考答案：8略15. 除以的余数是＿＿＿＿.参考答案：116. 利用数学归纳法证明不等式1+++…+＜f（n）（n≥2，n∈N*）的过程中，由n=k变到n=k+1时，左边增加的项是．参考答案：【考点】RG：数学归纳法．【分析】依题意，由n=k+1时，不等式左边为1+++…++，与n=k时不等式的左边比较即可得到答案．【解答】解：用数学归纳法证明等式1+++…+＜f（n）（n≥2，n∈N*）的过程中，假设n=k时不等式成立，左边=1+++…+，则当n=k+1时，左边=1+++…++，∴由n=k递推到n=k+1时不等式左边增加了：，故答案为：．17. 已知复数z=x+yi（x，y∈R，x≠0）且|z﹣2|=，则的范围为．参考答案：考点：复数求模．专题：计算题．分析：利用复数的运算法则和模的计算公式、直线与圆有公共点的充要条件即可得出．解答：解：∵|z﹣2|=|x﹣2+yi|，，∴．∴（x﹣2）2+y2=3．设，则y=kx．联立，化为（1+k2）x2﹣4x+1=0．∵直线y=kx与圆有公共点，∴△=16﹣4（1+k2）≥0，解得．∴则的范围为．故答案为．点评：熟练掌握复数的运算法则和模的计算公式、直线与圆有公共点的充要条件是解题的关键．三、解答题：本大题共5小题，共72分。

绝密★启用前数学试题注意事项：1、答题前填写好自己的姓名、班级、考号等信息 2、请将答案正确填写在答题卡上一、选择题（每小题5分，共60分）。

1．在△ABC 中，若a = 2 ,b =,030A = , 则B 等于（ ） A ．60 B ．60或 120 C ．30 D ．30或150 2．在数列55,34,21,,8,5,3,2,1,1x 中，x 等于（ ）A ．11B ．12C ．13D ．143．等差数列{}n a 的前n 项和为S n ，若a 3+a 17=10，则S 19= （ ） A ．55 B ．95 C ．100 D ．不能确定 4．若∆ABC 中，sin A :sin B :sin C =2:3:4，那么cos C =（ ） A. 14-B. 14C.23-D.235.设n S 为等差数列n a ｛｝的前n 项和，132s 4a a 2==-，，则9a =（ ）A ．6/5 B.16/5 C.32/5 D.26．等差数列{}n a 的前m 项和为30，前2m 项和为100，则它的前 3m 项和是( )A.130B.170C.210D.2607. 数列n a ｛｝是首项为2，公差为3的等差数列，数列{}n b 是首项为-2，公差为4的等差数列。

若n n a b =,则n 的值为（ ） A.4 B.5 C.6 D.78.在△ABC 中，060,2||,1||===B BC AB ，则||AC 的值是( ) A. 2 B. 3 C.325- D.39．等差数列{}n a 中，a 1＞0，d ≠0，S 3=S 11，则S n 中的最大值是 ( ) A ．S 7 B ．S 7或S 8 C ．S 14 D ．S 8 10．等差数列{a n }和{b n }的前n 项和分别为S n 和T n ，且132+=n nT S nn，则55b a =（ ） A. 32 B.149 C. 3120 D. 9711．已知A 、B 、C 是△ABC 的三个内角，且sin 2cos sin A B C =，则（ ） A . B =C B . B ＞C C . B ＜C D . B ,C 的大小与A 的值有关 12．给出下列三个命题（1）若tan A tan B >1，则△ABC 一定是钝角三角形； （2）若sin 2A ＋sin 2B ＝sin 2C ，则△ABC 一定是直角三角形； （3）若cos(A －B )cos(B －C )cos(C －A )＝1，则△ABC 一定是等边三角形.以上正确命题的个数有（ ）A ．0B ．1C ．2D ．3 二．填空题。

河南省洛阳市孟津县第二高级中学2020-2021学年高二英语9月周练试题本试卷共12页，全卷满分150分，考试用时120分钟﹡祝考试顺利﹡注意事项：1.答题前，先将自己的姓名、准考证号填写在试题和答题卡上，并将准考证号条形码粘贴在答题卡上的指定位置。

用2B铅笔将答题卡上试卷类型A后的方框涂黑。

2.选择题作答：每小题选出答案后，用2B铅笔将答题卡上对应题目的答案标号涂黑。

写在试题卷、演草纸和答题卡上的非答题区域均无效。

3.非选择题的作答：用签字笔直接答在答题卡上对应的答题区域内。

写在试题卷、演草纸和答题卡上的非答题区域均无效。

4.考试结束后，请将本试卷和答题卡一并上交。

第一部分听力（共两节，满分20分）做题时，先将答案标在试卷上。

录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节（共5小题；每小题1.5分，满分7.5分）听下面5段对话。

每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. What does the man advice woman to do?A. Continue looking for the dogB. Give up looking for the dogC. Adopt another dog2. What do you think the speakers will buy for Steve most probably?A. A grey suitB. A pair of green shoesC. A black suit3. Why does the man think Susan might have taken the book?A. He saw her take itB. She likes reading booksC. He’s watching TV4. Who made the Mother’s Day card?A. The manB. The sellerC. The woman5. Where did the man use to work?A. In a schoolB. At a restaurantC. At a store第二节（共15小题；每小题1.5分，满分22.5分）听下面5段对话或独白。

洛阳市孟津县第二高级中学2020-2021学年高二9月周练物理试卷一．选择题（本大题共12小题，每小题4分，共48分，全对的给4分，部分对的给2分，不答的或答错的给零分）1 、(2019·上海市金山中学高二上学期期中)某电解池中，如果在1 s 内共有5×1018个二价正离子和1×1019个一价负离子通过面积为0.1 m 2的某截面，那么通过这个截面的电流是( D )A ．0B ．0.8 AC ．1.6 AD ．3.2 A2、如图所示为一质量分布均匀的长方体金属导体，在导体的左右两端加一恒定的电压，使导体中产生一恒定电流，其电流的大小为I 。

已知金属导体的横截面积是S ，导体单位长度的自由电子数为n ，金属内的自由电子的电荷量为e ，自由电子做无规则热运动的速率为v 0，则下列说法中正确的是( D )A ．自由电子定向移动的速率为v 0B ．自由电子定向移动的速率为v ＝I neSC ．自由电子定向移动的速率为真空中的光速cD ．自由电子定向移动的速率为v ＝I ne3、有关电压和电动势的说法中错误的是( A )A ．电压和电动势的单位都是伏特，故电动势与电压是同一物理量B ．电动势公式E ＝W q 中的W 与电压U ＝W q 中的W 是不同的，前者为非静电力做功，后者为静电力做功C ．电动势是反映电源把其他形式的能转化为电能的本领的物理量D ．断路时的路端电压等于电源的电动势的大小4、两电阻R 1、R 2的电流I 和电压U 的关系如图所示，可知两电阻R 1∶R 2等于( A )A ．1∶3B ．3∶1C ．1∶ 3D ．3∶15、对于欧姆定律的理解，下列说法中错误的是( C )A ．由I ＝U R 知通过电阻的电流强度跟它两端的电压成正比，跟它的电阻成反比B ．由U ＝IR 知对一定的导体，通过它的电流强度越大，它两端的电压也越大C ．由R ＝U I 知导体的电阻跟它两端的电压成正比，跟通过它的电流强度成反比D ．对一定的导体，它两端的电压与通过它的电流强度的比值保持不变6、(2019·浙江省绍兴市诸暨中学高一下学期检测)如图是某导体的伏安特性曲线，下列说法正确的是( B )A ．当通过导体的电流是5×10－3 A 时，导体的电阻是1 000 ΩB ．当通过导体的电流是5×10－3 A 时，导体的电阻是2 000 ΩC ．随着电流的增大，导体的电阻增大D ．当通过导体的电流是0.001 A 时，导体两端的电压是1 V7、(2019·北京市中关村中学高二上学期期中)如图所示，甲、乙两个电路都是由一个灵敏电流计G 和一个变阻器R 组成，它们中一个是测电压的电压表，另一个是测电流的电流表，那么以下结论中正确的是( D )A ．甲表是电流表，R 增大时量程增大B ．甲表是电压表，R 增大时量程减小C ．乙表是电流表，R 增大时量程减小D ．乙表是电压表，R 增大时量程增大 8、(2019·广东省惠州市高二上学期期末)额定电压都是110 V ，额定功率P A ＝100 W ，P B ＝40 W 的电灯两盏，若接在电压是220 V 的电路上，使两盏电灯均能正常发光，且电路中消耗功率最小的电路是下图中的哪一个( C )9、(多选)(2019·广东省台山市华八侨中学高二上学期期中)横截面积为S 的导线中通有电流I ，已知导线每单位体积中有n 个自由电子，每个自由电子的电荷量是e ，自由电子定向移动的速率是v ，则在时间Δt 内通过导线横截面的电子数是( AC )A ．nSv ΔtB ．nv ΔtC ．I Δt eD ．I Δt Se10、(多选)以下说法中正确的是( ABD )A ．电源的作用是维持导体两端的电压，使电路中有持续的电流B ．在电源内部正电荷能从负极到正极是因为电源内部存在非静电力和静电力C ．静电力与非静电力都可以使电荷移动，所以本质上都是使电荷的电势能减少D ．静电力移动电荷做功电荷电势能减少，非静电力移动电荷做功电荷电势能增加11、(多选)现代家庭电器化程度越来越高，用电安全是一个十分重要的问题。

河南省洛阳市孟津县第二高级中学2020-2021学年高二生物9月周练试题一选择题1．真核生物进行有性生殖时，通过减数分裂和随机受精使后代(）A．增加发生基因自由组合的概率B．继承双亲全部的遗传性状C．从双亲各获得一半的DNAD．产生不同于双亲的基因组合2．有丝分裂与减数分裂过程中均要发生的现象是（）①DNA复制和有关蛋白质的合成②纺锤体的形成③同源染色体配对和分离④着丝点的分裂⑤非同源染色体的自由组合⑥同源染色体间的交叉互换A．①②③ B．①②④ C．①③⑤ D．①②④⑥3．某动物精原细胞在减数分裂过程中形成了四个四分体，则其减数第二次分裂后期的次级精母细胞中染色体数、染色单体数和ＤＮＡ分子数依次为A．４、８、４B．4、0、4 C．８、１６、１６D．８、０、８4．在形成卵细胞的减数分裂过程中，细胞质中的遗传物质的分配特点是①有规律地分配②随机分配③均等分配④不均等分配⑤交换分配A．①③ B．②③ C．③⑤ D．②④5．现有某动物减数分裂过程中产生的一个极体,染色体数为M，DNA分子数为N个，且M不等于N，则该动物的一个初级卵母细胞中的染色体数和一个卵原细胞中的DNA分子数分别为A．M和N B．M和2N C．2M和N D．2M 和2N6．下面四幅图是来自于同一生物体内的、处于四个不同状态的细胞分裂图。

下列有关叙述中，正确的是（）A．该生物的正常体细胞中含有16条染色体B．图①与图③所示细胞中DNA含量比例为1：2C．图②与图④所示过程仅发生在某些器官中D．由图④可知，该生物一定是雄性个体7．在减数分裂中，家兔的初级卵母细胞有22个四分体，则其卵细胞中染色体数为：（）A．11个B．11对22个C．不成对的22个D．44个8．在生物传种接代过程中，能够使染色体保持一定的稳定性和连续性的重要生理过程是：( ）①有丝分裂②无丝分裂③减数分裂④受精作用A．①②③ B．①③④ C．③④ D．①②③④9．下图表示一个细胞进行分裂的过程中，细胞核内染色体数目以及染色体上DNA分子数目的变化情况。

20XX年高中测试
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河南省2021年上学期洛阳市孟津县第二高级中学高二生物9月周练试题答案
1--10 DBDDC CCCDC 11--20 DDBBD DBDDA
21--30 BAABB ABDBB
31.(1) 动物减数第一次分裂后期次级精母细胞 2 8 同源染色体 2 4
（2）有丝分裂后期 4 0 体
（3）减数第二次分裂后期0 4 4 着丝点分裂，姐妹染色单体分开
精细胞
32.（1）有丝分裂
（2）5、2、1、4
（3）a. b. c
（4）4或8 8
33.（1）自花授粉、闭花授粉; 有易于区分的性状
（2）二黄色
（3）1/3
（4）Y:y=1:1 去雄相同
（5）3/5
34.（1）多只无添加剂的食物喂养的黑体雌性
（2）无添加剂的食物
（3）①全灰②全黑③有黑有灰
35.（1）去雄套袋
（2）二或三有芒
（3）Bb×Bb
（4）213
（5）1。

河南省洛阳市孟津县第二高级中学2020-2021学年高二历史9月周练试题一、选择题(每小题1.5分，共48分)1．战国时期诸子百家的思想观点实质上反映了( )A．统治阶级的利益 B．不同阶级阶层的利益C．知识分子的不同认识 D．新兴地主阶级的利益2．针对当时“天下无道……礼乐征伐自诸侯出”的社会现状，孔子主张( )A．克己复礼B．为政以德 C．仁者爱人D．有教无类3．孔子的教育思想中与其“贵贱有序”的政治主张相矛盾的是( )A．因材施教B．有教无类 C．学而不思则罔D．知之为知之4．有学者赞扬孟子“提倡民权，为孔子所未及焉”。

该学者赞扬孟子的主要依据是( ) A．提出“民贵君轻”和“仁政” B．被尊称为“亚圣”C．在伦理观上主张“性本善” D．以孔子的继承人自居5．“所谓不知《春秋》，不能涉世；不精《老》《庄》，不能忘世；不参禅，不能出世。

”从中可以看出儒家思想的特点是( )A．注重以人为本的理念 B．注重研究社会现实C．着重研究人与自然的关系 D．重视研究人的前世来生6．某思想家说：“只有‘有’是发挥不了大用处的，唯有‘有’与‘无’配合才能产生大作用。

”据此判断该思想家应是( )A．孔子B．孟子 C．荀子D．老子7．“文王行仁义而王天下，偃王行仁义而丧其国，是仁义用于古而不用于今也。

故曰：‘世异则事异’。

”这段话反映了( )A．孟子的“仁政”学说B．墨子的“兼爱”思想C．韩非的变法革新主张D．庄子的“齐物”观点8．“欲求兴天下之利，除天下之害，当若繁为攻伐，此实天下之巨害也。

”这一观点出自先秦( )A．儒家B．法家 C．墨家D．道家9．战国后期，诸子已开始尝试以自己的学说统一思想。

《吕氏春秋·不二》篇宣称“听众人议以治国，国危无日矣”。

“故一则治，异则乱。

一则安，异则危”。

思想大一统被提到了十分醒目的位置。

这一现象产生的原因是( )①建立地主阶级统治的需要②封建经济发展的需要③适应国家统一的需要④适应兼并战争的需要A．①② B．①②③ C．②③④D．①②③④10．史载汉初曹参为相时，任官选用不善辞令的忠厚长者，而对追逐功名利禄之徒一概不用；省事节用，量刑从宽；对下属不苛察细过，虚怀自持。

河南省洛阳市孟津县第二高级中学2020-2021学年高二英语9月月考试题第二部分阅读理解（共两节，满分40分）AWhether they are already household names or a hidden figure deserving of more recognition, the following ladies changed the world with their enormous contributions.Ali StrokerAli Stroker took the theater world —and, indeed, the very Internet —by storm when, on June 9, 2019, she became the first performer in a wheelchair to take home a Tony Award. After becoming the first actor in a wheelchair in Broadway history in 2015, she won the award for her powerhouse performance in the revival of Oklahoma \Junko TabeiTwenty-two years after the first-ever successful mission to the top of Mount Everest, Japanese mountaineer Junko Tabei became the first woman to reach the peak. She led a team of 15 women, accompanied by six Sherpas （夏尔巴人），and reached the summit with one of the Sherpas on May 16, 1975.Gertrude EderleThe Queen of Waves, who also happened to be deaf, was the first woman to swim across the English Channel. Fighting through cold temperatures and strong tides that change direction every six hours for 22 miles, she clocked a time of 14 hours and 34 minutes. Virginia ApgarGenerations of parents owe this American doctor a huge thank you, as she developed the Apgar Score, the first standardized system of tests to assess if newborn babies were healthy once they made their way from womb to world. Apgar, who was a gifted cellist and violinist in her spare time, also happens to hold the title of the first woman to be hired as a full professor at the medical school at Columbia University.21. Whose story may inspire the disabled?A. Stroker and Tabei.B. Stroker and Ederle.C. Ederle and Apgar.D. Ederle and Tabei.22. Why should Apgar be appreciated by parents?A. She took home a big award.B. She saved many babies' lives.C. She developed the Apgar Score.D. She became the first full professor.23. Who won the title of the Queen of Waves?A. Ali Stroker.B. Junko Tabei.C. Virginia Apgar.D. Gertrude Ederle.BMy students were taking midterms when my phone erupted with urgent messages. "A student is having a panic attack," texted a teaching assistant. I ran out of my office, down a flight of stairs and found the student — a pupil in my 350-person organic chemistry class — lying motionless on the ground outside the exam hall. " Did my exam really trigger a panic attack?" I asked myself. "Why am I not prepared to deal with a situation like this?”It was my first time teaching the course. But I knew that the subject was challenging for my students. This was a source of stress for premedical students in particular, who feared that a low grade in organic chemistry would keep them from getting into medical school.The following day, I was scheduled to lecture to the same class. I knew that I had to address what had happened during the midterm. So, I started by saying: "I want to take some time today to talk about something important. How many of you think that this is a weed-out course?" Half of my students raised their hands carefully. "I'm sorry to hear that,” I continued. "I want you all to know that I do not consider any of you to be weeds; you all deserve to be here. ”I flashed a slide of flowers in various shapes. I smiled at my students and said: "I think of you as flowers — different flowers with different needs. You may not bloom at the same time, but you will bloom! You may not do well in the midterm exam, but you will learn from your mistakes and do better in the final exam. I believe this.I believe in you."From that point on, my office hours were packed. Some asked about lecture topics and study strategies; others opened up about personal issues. I was amazed that a simple, frank discussion in lecture could make such a difference.24. What made the pupil have a panic attack?A. Hiding personal issues.B. The stress for high grades.C. Lacking study strategies.D. Failing to handle the situation.25. What does the underlined word "trigger" in Paragraph 1 most probably mean?A. Cure.B. Prevent.C. Frighten.D. Cause.26. Why did the author go to the same class the next day?A. To give the lesson according to the arrangement.B. To apologize and explain to the panicked student.C. To give a speech on what happened in the test.D. To persuade all the students to stay in the class.27. Which paragraph mainly shows the author's encouragement to students?A. Paragraph 2.B. Paragraph 3.C. Paragraph 4.D. Paragraph 5.CThe cognitive health and development of boys may be affected by their mothers ' body mass index ( BMI) (体重指数)while pregnant with them, according to research from Columbia University and the University of Texas at Austin.The study, which was published in the journal BMC Pediatrics on Friday, observed 368 subjects from low-income African American and Dominican women during the second half of their pregnancies, and then evaluated their children three and seven years later. Researchers found that the sons of women whose BMIs indicated that they were overweight or obese when they became pregnant were more likely to show less developed athletic skills as 3-year-olds and lower intelligence as 7-year-olds compared to boys whose mothers were at "normal" weights during pregnancy.Among boys, the study found, mothers' overweight and obesity connected with IQ scores between 4.6 and almost 9 points lower than those of boys whose mothers' weights were in the "normal" range before pregnancy. Researchers did not observe the same phenomenon among daughters whose mothers had been obese."These findings aren't meant to shame or scare anyone, "Elizabeth Widen, assistant professor of nutritional sciences at UT Austin and one of the study's co-authors, said in a press release. " We are just beginning to understand some of these interactions between mothers' weight and the health of their babies."Why mothers' obesity appeared to affect childhood IQ was unclear, but earlier research has suggested that there is a relationship between a mother's diet and her child's later IQ, according to Columbia University. Researchers did not control for what the mothers ate, the press release noted.The study's authors wrote that because childhood IQ has been shown to be an indicator of later success in life, studying how a mother's obesity could affect the IQ of her child is worthwhile.28. How did researchers carry out the study?A. By measuring mothers' body mass index.B. By watching mothers and babies for years.C. By comparing 3-year-old babies with 7-year-olds.D. By evaluating the health of mothers and their babies.29. What's the main purpose of the study?A. To show links between mothers' weight and babies' IQ.B. To make those overweight mothers shameful and scared.C. To warn some fat mothers to keep a balanced diet.D. To persuade more obese mothers to lose weight.30. What do the researchers think of the study?A. Doubtful.B. Worrying.C. Significant.D. Interesting.31. In which section of a newspaper may the text appear?A. Entertainment.B. Novel.C. Education.D. Health.DMore than half of the birds in Washington are at risk of extinction because of climate change. That's according to a new national report from the Audubon Society, which gives detailed analysis of climate effects on about 600 species of North American birds.It's based on more than 140 million observations of birds across the US, Mexico and Canada. Audubon scientists looked at the likely effects of sea-level rise, urbanization, drought, extreme spring heat, increased fires, heavy rain and other factors.But it doesn't just spell out a doomsday scenario （世界末日）.Instead, it offers a range of effects and warming, depending on how much carbon humans add to the atmosphere."It is truly an existential threat （威胁），not only to birds but to people,” said Doug Santoni, board chair of Audubon Washington, who looked into the report as soon as it came out.Santoni says he was struck to see the vulnerability （脆弱）of a common “ backyard bird" , the dark-eyed junco. It's one that many first-time birders become familiar with as they learn how to identify species based on their markings and other traits. Currently in Washington, you can count on juncos to show up at your feeder, year round.Extreme spring heat, increased fires and heavy rain are the kinds of changes that will force birds like these north, or kill them off if they fail to adapt.Trina Bayard, director of bird conservation at Audubon’s Washington chapter, says, "It's certainly a very serious warning report," but adds that there’s still hope. “If we can stabilize current temperatures and decrease our emissions （排放），we can really reduce the effects to these birds --- that's very motivating. ”32. What can we know about the new report?A. It analyses the species of birds in detail.B. It's issued by watching 600 bird species.C. It shows the end of North American birds.D. It reports the threat some birds are facing.33. What may Santoni probably agree with?A. Climate change is a threat only to birds.B. It's too late to take action to save the birds.C. The current situation of the birds is worrying.D. It's common that birds are affected by climate change.34. Which of the following can help these birds according to Trina?A. Lowering present temperatures.B. Reducing our daily emissions.C. Making them adapt to climate change.D. Encouraging people to protect them.35. What can be the best title for the text?A. Climate change threatens many Washington bird speciesB. A new report about 600 species of North American birdsC. Different attitudes towards the situation of bird speciesD. Climate change makes different kinds of species at risk第二节（共5小题；每小题2分，满分10分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。

河南省洛阳市孟津县第二高级中学2021-2022高二化学9月周练试题（无答案）一．选择题（每题只有一个正确选项，每题2分，共20分）1.下列反应中不属于可逆反应的是( )A.Cl2溶于水B.NH3溶于水C.SO2溶于水D.电解水生成H2和O2，点燃H2和O2的混合物生成水2、反应4A(g)+3B(g)=2C(g)+5D(g)在四种不同条件下的反应速率分别为( )①v(A)=0.02mol·L-1·s-1; ②v(B)=0.6mol·L-1·min-1;③v(C)=0.3mol·L-1·min-1; ④v(D)=1.2mol·L-1·min-1其中表示该反应速率最慢的是( )A.①B.②C.③D.④3.合成氨是工业上的重要反应：N2(g)+3H2(g)2NH3(g)，下列说法不正确的是( )A.反应达平衡状态后，各物质浓度不变，反应未停止B.反应达平衡状态后，单位时间内生成1 mol N2的同时消耗3 mol H2C.若改用新型高效催化剂，N2有可能100%转化为NH3D.使用催化剂是为了加快反应速率，提高生产效率4.反应N2(g)+3H2(g)2NH3(g)经过一段时间后,NH3的浓度增加了0.6 mol·L-1,在此段时间内用H2表示的平均反应速率为0.45 mol·L-1·s-1,则此段时间是( )A.1 sB.2 sC.44 sD.1.33 s5.一定条件下，对于可逆反应：X(g)+3Y(g)2Z(g)，若X、Y、Z的起始浓度分别为c1、c2、c3(均不为零)，达到平衡时，X、Y、Z的浓度分别为0.1 mol·L-1、0.3 mol·L-1、0.08 mol·L-1，则下列判断正确的是( )A.c1∶c2=3∶1B.平衡时，Y和Z的生成速率之比为2∶3C.X、Y的转化率不相等D.c1的取值范围为0<c1<0.14 mol·L-16.反应N2(g)+3H2(g)2NH3(g) ΔH<0，若在恒压绝热容器中发生，下列选项表明反应一定已达平衡状态的是( )A.容器内的温度不再变化B.容器内的压强不再变化C.相同时间内，断开H—H键的数目和生成N—H键的数目相等D.容器内气体的浓度c(N2)∶c(H2)∶c(NH3)=1∶3∶27.在一定温度下，向2 L固定容积的密闭容器中通入1 mol CO2、3 mol H2，发生反应CO2(g)+3H2(g)CH3OH(g)+H2O(g) ΔH<0。

河南省洛阳市孟津县老城高级中学2020-2021学年高二数学文联考试题含解析一、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. 在某种信息传输过程中，用4个数字的一个排列（数字允许重复）表示一个信息，不同排列表示不同信息，若所用数字只有0和1，则与信息0110至多有两个对应位置上的数字相同的信息个数为( )A.10B.11C.12D.15参考答案：B略2. 等比数列中，，，则等于（ )A. B. C. D.参考答案：A3. 已知定义在R上的偶函数f(x)满足，当时，.函数，则f(x)与g(x)的图象所有交点的横坐标之和为（）A. 3B. 4C. 5D. 6参考答案：A【分析】根据题意，分析可得与的图象都关于直线对称，作出两个函数的图象，分析其交点的情况即可得答案．【详解】根据题意，函数满足，则的图象关于直线对称，函数的图象也关于直线对称，函数的图象与函数的图象的位置关系如图所示，可知两个图象有3个交点，一个在直线上，另外2个关于直线对称，则两个函数图象所有交点的横坐标之和为3；故选：A．【点睛】一般地，如果函数满足，那么的图像关于对称，如果函数满足，那么的图像关于点对称.刻画函数图像时，注意利用上述性质.4. 已知下列等式：，，，，…，，则推测（）A．109 B．1033 C．199 D．29参考答案：A根据题意，分析所给的等式，可归纳出等式，（且是正整数），将代入，可得，从而可以求得，于是，故选A.5. 一个网站针对“是否同意恢复五一长假”进行了随机调查,在参加调查的2 600名男性公民中有1 600 名持反对意见,在2 400名女性公民中有1 300人持反对意见,在运用这些数据分析说明“是否同意恢复五一长假”与性别有无关系时,比较适合的方法是( ).Ａ.平均数与方差Ｂ.独立性检验Ｃ.回归分析Ｄ.条件概率参考答案：B略6. 过点，且圆心在直线上的圆的标准方程为A. B.C. D.参考答案：B7. 在一个列联表中，由其数据计算得，则其两个变量间有关系的可能性为（）A．99% B．95% C．90% D．无关系参考答案：A8. 如图，在梯形ABCD中，，，P是BC中点，则（）A. B.C. D.参考答案：D【分析】由平面向量基本定理及线性运算可得：，得解.【详解】因为是中点，所以. 故选D. 【点睛】本题考查了平面向量基本定理，属基础题.9. 把一枚硬币掷三次，三次都出现正面的概率为（）（Ａ）（Ｂ）（Ｃ）（Ｄ）参考答案：A略10. 在极坐标系中,直线与曲线相交于两点, 为极点,则的大小为（）参考答案：C二、填空题:本大题共7小题,每小题4分,共28分11. 若P是以F1F2为焦点的椭圆＋＝1上一点，则 PF1F2的周长等于__________。

河南省洛阳市孟津县第二高级中学2021-2022高二英语9月周练试题本试卷共12页，全卷满分150分，考试用时120分钟﹡祝考试顺利﹡注意事项：1.答题前，先将自己的姓名、准考证号填写在试题和答题卡上，并将准考证号条形码粘贴在答题卡上的指定位置。

用2B铅笔将答题卡上试卷类型A后的方框涂黑。

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4.考试结束后，请将本试卷和答题卡一并上交。

第一部分听力（共两节，满分20分）做题时，先将答案标在试卷上。

录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节（共5小题；每小题1.5分，满分7.5分）听下面5段对话。

每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

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1. What does the man advice woman to do?A. Continue looking for the dogB. Give up looking for the dogC. Adopt another dog2. What do you think the speakers will buy for Steve most probably?A. A grey suitB. A pair of green shoesC. A black suit3. Why does the man think Susan might have taken the book?A. He saw her take itB. She likes reading booksC. He’s watching TV4. Who made the Mother’s Day card?A. The manB. The sellerC. The woman5. Where did the man use to work?A. In a schoolB. At a restaurantC. At a store第二节（共15小题；每小题1.5分，满分22.5分）听下面5段对话或独白。

2020-2021学年河南省洛阳市偃师第二高级中学高二数学理模拟试卷含解析一、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. 已知抛物线y=ax2（a＞0）的焦点到准线距离为1，则a=（）A．4 B．2 C．D．参考答案：D【考点】K8：抛物线的简单性质．【分析】抛物线y=ax2（a＞0）化为，可得．再利用抛物线y=ax2（a＞0）的焦点到准线的距离为1，即可得出结论．【解答】解：抛物线方程化为，∴，∴焦点到准线距离为，∴，故选D．2. 在中，已知，则的形状是（）等腰三角形直角三角形等腰直角三角形等腰三角形或直角三角形参考答案：D3. 下列四个命题中，正确的是.已知函数，则；.设回归直线方程为，当变量增加一个单位时，平均增加个单位；.已知服从正态分布,，且，则.对于命题：,使得，则：，均有参考答案：A4. 下列四个结论：⑴两条直线都和同一个平面平行，则这两条直线平行。

⑵两条直线没有公共点，则这两条直线平行。

⑶两条直线都和第三条直线垂直，则这两条直线平行。

⑷一条直线和一个平面内无数条直线没有公共点，则这条直线和这个平面平行。

其中正确的个数为（）A．B．C．D．参考答案：A 解析：⑴两条直线都和同一个平面平行，这两条直线三种位置关系都有可能⑵两条直线没有公共点，则这两条直线平行或异面⑶两条直线都和第三条直线垂直，则这两条直线三种位置关系都有可能⑷一条直线和一个平面内无数条直线没有公共点，则这条直线也可在这个平面内5. 过双曲线(a＞0，b＞0)的右焦点F，作渐近线y=x的垂线与双曲线左右两支都相交，则双曲线离心率e的取值范围为( )A．(1,2) B．(1,) C．(,+∞)D．(2,+∞)参考答案：C6. 已知数列，则其前是A． B．C． D．B略7. 抛物线的焦点到准线的距离是（）A． B． C． D．参考答案：B 解析：，而焦点到准线的距离是8. 已知双曲线：（）的离心率为，则的渐近线方程为....参考答案：C略9. 执行如图所示的程序框图，若输入A的值为2，则输出的P值为()A．2 B．3C．4 D．5参考答案：C 10. 在一次独立性检验中，得出列联表如下：且最后发现，两个分类变量和没有任何关系，则的可能值是（）A. B. C.D.参考答案：B略二、填空题:本大题共7小题,每小题4分,共28分11. 如图，用6种不同的颜色给图中的4个格子涂色，每个格子涂一种颜色．要求最多使用3种颜色且相邻的两个格子颜色不同，则不同的涂色方法共有种（用数字作答）．参考答案：390【考点】D5：组合及组合数公式．【分析】由题意选出的颜色只能是2种或3种，然后分别求出涂色方法数即可．【解答】解：用2色涂格子有C62×2=30种方法，用3色涂格子，第一步选色有C63，第二步涂色，从左至右，第一空3种，第二空2种，第三空分两张情况，一是与第一空相同，一是不相同，共有3×2（1×1+1×2）=18种，所以涂色方法18×C63=360种方法，故总共有390种方法．故答案为：39012. 在的二项展开式中，的系数为_____参考答案：－84先求出展开式的通项公式为，再令的幂指数等于3求出的值，即可求得的系数．【详解】二项式的展开式的通项公式为．令，解得，展开式中的系数为，故答案为：-84【点睛】本题主要考查二项式定理的应用，二项式展开式的通项公式，求展开式中某项的系数，属于 中档题．13. 双曲线﹣y 2=1的离心率等于．参考答案：【考点】双曲线的简单性质．【分析】根据双曲线的方程，求出a ，b ，c ，即可求出双曲线的离心率． 【解答】解：由双曲线的方程可知a 2=4，b 2=1， 则c 2=a 2+b 2=4+1=5， 则a=2，c=，即双曲线的离心率e==，故答案为：14. 方程x 2+y 2﹣x+y+m=0表示一个圆，则m 的取值范围是 ．参考答案：（﹣∞，）【考点】二元二次方程表示圆的条件． 【分析】根据圆的一般方程即可得到结论． 【解答】解：若方程x 2+y 2﹣x+y+m=0表示一个圆， 则满足1+1﹣4m ＞0，即m ＜，故答案为：（﹣∞，）．15. 过圆外一点P （5，－2）作圆x 2+y 2－4x －4y=1的切线，则切线方程为__________。