山西省范亭中学复数经典试题(含答案) 百度文库
2025届山西省范亭中学高三六校第一次联考数学试卷含解析
2025届山西省范亭中学高三六校第一次联考数学试卷注意事项1．考生要认真填写考场号和座位序号。
2．试题所有答案必须填涂或书写在答题卡上，在试卷上作答无效。
第一部分必须用2B 铅笔作答；第二部分必须用黑色字迹的签字笔作答。
3．考试结束后，考生须将试卷和答题卡放在桌面上，待监考员收回。
一、选择题：本题共12小题，每小题5分，共60分。
在每小题给出的四个选项中，只有一项是符合题目要求的。
1． “幻方”最早记载于我国公元前500年的春秋时期《大戴礼》中．“n 阶幻方()*3,n n ≥∈N ”是由前2n 个正整数组成的—个n 阶方阵，其各行各列及两条对角线所含的n 个数之和（简称幻和）相等，例如“3阶幻方”的幻和为15（如图所示）．则“5阶幻方”的幻和为（ ）A ．75B ．65C ．55D ．452．已知集合2{|1}M x x ==．N 为自然数集，则下列表示不正确的是（ ） A ．1M ∈B ．{1,1}M =-C ．M ∅⊆D ．M N ⊆3．已知双曲线2222:1x y C a b-=（0a >，0b >），以点P （,0b ）为圆心，a 为半径作圆P ，圆P 与双曲线C 的一条渐近线交于M ，N 两点，若90MPN ∠=︒，则C 的离心率为（ ） A 2B 3C 5D 7 4．已知(1)2i ai bi -=+（i 为虚数单位，,a b ∈R ），则ab 等于（ ） A ．2B ．-2C ．12D ．12-5．已知()()()[)3log 1,1,84,8,6x x f x x x ⎧+∈-⎪=⎨∈+∞⎪-⎩ 若()()120f m f x ⎡⎤--≤⎣⎦在定义域上恒成立，则m 的取值范围是（ ）A ．()0,∞+B ．[)1,2C ．[)1,+∞D ．()0,16．已知x ，y R ∈，则“x y <”是“1xy<”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7．函数()231f x x x =-+在[]2,1-上的最大值和最小值分别为（ ） A ．23，-2 B ．23-，-9 C ．-2，-9 D ．2，-28．已知奇函数()f x 是R 上的减函数，若,m n 满足不等式组()(2)0(1)0()0f m f n f m n f m +-≥⎧⎪--≥⎨⎪≤⎩，则2m n -的最小值为（ ）A ．-4B ．-2C ．0D ．49．已知集合M ＝{y ｜y ＝，x ＞0}，N ＝{x ｜y ＝lg （2x －）}，则M∩N 为（ ） A ．（1，＋∞）B ．（1，2）C ．[2，＋∞）D ．[1，＋∞）10．已知圆锥的高为3，底面半径为3，若该圆锥的顶点与底面的圆周都在同一个球面上，则这个球的体积与圆锥的体积的比值为( ) A ．53B ．329C ．43D ．25911．如图，2AB =是圆O 的一条直径，,C D 为半圆弧的两个三等分点，则()AB AC AD ⋅+=（ ）A ．52B ．4C ．2D ．13+12．已知双曲线的中心在原点且一个焦点为7,0)F ，直线1y x =-与其相交于M ，N 两点，若MN 中点的横坐标为23-，则此双曲线的方程是 A ．22134x y -= B ．22143x y -= C ．22152x y -=D ．22125x y -=二、填空题：本题共4小题，每小题5分，共20分。
复数经典试题(含答案)百度文库
，.
故选：B.
解析：B
【分析】
利用复数的除法法则可化简 ，即可得解.
【详解】
， .
故选：B.
3．D
【分析】
由复数乘法化复数为代数形式，然后根据复数的分类求解．
【详解】
，它为纯虚数，
则，解得．
故选：D．
解析：D
【分析】
由复数乘法化复数为代数形式，然后根据复数的分类求解．
【详解】
，它为纯虚数，
则 ，解得 ．
【详解】
因为 ，所以其虚部是 .
故选：A.
6．C
【分析】
由复数除法法则计算出后可得其虚部．
【详解】
因为，
所以复数z的虚部是．
故选：C．
解析：C
【分析】
由复数除法法则计算出 后可得其虚部．
【详解】
因为 ，
利用复数的除法法则化简，再求的共轭复数，即可得出结果.
一、复数选择题
1．已知复数 ，则 （）
A．2B． C．4D．5
2．已知复数 ，若 为虚数单位，则 （）
A． B． C． D．
3．若复数 是虚数单位 为纯虚数，则实数 的值为（）
A． B．
C． D．
4． 是虚数单位，复数 （）
A． B． C． D．
5．已知i为虚数单位，则复数 的虚部是（）
A． B． C． D．
则，
故选：A.
解析：A
【分析】
将 代入 ，利用复数的除法运算化简，再利用复数的求模公式求解.
【详解】
由 ，得 ，
则 ，
故选：A.
10．C
【分析】
对复数进行化简根据实部为零，虚部不为零建立等量关系和不等关系即可得解.
山西省原平市范亭中学2024_2025学年高二英语4月月考试题
山西省原平市范亭中学2024-2025学年高二英语4月月考试题本试卷分为第I卷（选择题）和第II卷（非选择题）两部分, 共150分。
考试时间120分钟。
第I卷（共100分）第一部分阅读理解（共两节，满分60分）第一节（共15小题，每小题3分，满分45分）AA new app aims to help parents interpret what their baby wants based on the sound of their cry. The free app ChatterBaby, which was released last month, analyzes the acoustic (声学的) features of a baby’s cry, to help parents understand whether their child might be hungry, fussy or in pain. While critics say caregivers should not rely too much on their smartphone, others say it’s a helpful tool for new or tired parents.Ariana Anderson, a mother of four, developed the app. She originally designed the technology to help deaf parents better understand why their baby was upset, but soon realized it could be a helpful tool for all new parents.To build a database, Anderson and her team uploaded 2,000 audio samples of infant(婴儿) cries. She used cries recorded during ear piercings and vaccinations to distinguish pain cries. And to create a baseline for the other two categories, a group of moms had to agree on whether the cry was either hungry or fussy.Anderson’s team continues to collect data and hopes to make the app more accurate by asking parents to get specific about what certain sounds mean.Pediatrician Eric Ball pointed out that evaluating cries can never be an exact science. “I think that all of the apps and technology that new parents are using now can be helpful but need to be taken seriously,” Ball said ,“ I do worry that some parents will get stuck in big data and turn their parenting into basically a spreadsheet(电子表格) which I think will take away the love and caring that parents are supposed to be providing for the children. ”But Anderson said the aim of the app is to have parents interpret the results, not to provide a yes o r no answer. The Bells, a couple using this app, say it’s a win-win. They believe they are not only helping their baby now but potentially othersin the future.1．How does the app judge what babies want?A．By collecting data. B．By recording all the sounds.C．By analyzing the sound of their cries. D．By asking parents about specific messages.2．What was the app designed for in the beginning?A．All new parents. B．Deaf parents.C．Ariana Anderson. D．Crying babies.3．What i s Ball’s opinion about the app?A．Parents should use the app wisely.B．The app can create an accurate result.C．Parents and babies are addicted to the app.D．The app makes babies lose love and caring.4．What is the text mainly about?A．Parents should not rely too much on their smartphones.B．A new app helps parents figure out why their babies are crying.C．Parents can deal with babies’ hunger with the help of a new app.D．A new app called ChatterBaby can prevent babies from crying.BMany people spend more than four hours per day on We Chat, and it is redefining the word “friend.” Does friending someone on social media make him or her your friend in real life?Robin Dunbar, a professor at Oxford University, found that only 15, of the 150 Facebook friends the average user has, could be counted as actual friends and only five as close friends. We Chat may show a similar pattern.Those with whom you attended a course together, applied for the same part-time job, went to a party and intended to cooperate but failed take up most of your WeChat friends. In chat records, the only message may be a system notice, “You have accepted somebody’s friend request”. Sometimes when seeing some photos shared on “Moments”, you even need several minutes to think about when you became friends. Also, you maybe disturbed by mass messages (群发信息) sent from your unfamiliar “friends”, including requests for voting for their children or friends, links from Pinduoduo (a Chinese e-commerce platform that allows users to buy items at lower prices if they purchase in groups) and cookie-cutter (一模一样的) blessings in holidays.You would have thought about deleting this type of “friends” and sort out your connections. But actually you did not do that as you were taught that social networkingis valuable to one’s success. Besides, it would be really awkward if they found thatyou have unfriended them already. Then, you keep increasing your “friends” in social media and click “like” on some pictures that you are not really interested. Butthe fact is that deep emotional connections do not come with the increasing numberof your friends in social media.If the number of your friends reaches 150, maintaining these relationships canbe tough to you, and sometimes even will make you anxious. According to Robin Dunbar,150 is the limit of the number of people with whom one can maintain stable social relationships.5．What can we learn from Robin Dunbar's finding in Paragraph 2?A．A Facebook user has 250 friends on average.B．Most of the social media friends can be actual friends.C．Among our social media friends, only a few people matter.D．Only 15 people of a person’s Facebook friends can be close friends.6．What does the third paragraph tell us about most of your WeChat friends?A．You have deep communication with them.B．You benefit a lot from their mass messages.C．You just have a nodding acquaintance with them.D．You become friends with them in important occasions.7．What does the underlined word “that” in Paragraph 4 refer to?A．Removing unfamiliar friends in WeChat.B．Strengthening ties with your We Chat friends.C．Keeping increasing your friends in social media.D．Clicking “like” on pictures posted by your friends.8．What can we infer from the last paragraph?A．We will be anxious if we make friends online.B．We should avoid making any friends in social media.C．We should make as many friends as possible in social media.D．We have difficulty managing relationships with over 150 people.CLast week, Vodafone started a test of the UK's first full 5G service, available for use by businesses in Salford. It is part of its plan to trial the technology in seven UK cities. But what can we expect from the next generation of mobile technology?One thing we will see in the preparation for the test is lots of tricks with the new tech. Earlier this year, operators paid almost £ 1.4 billion for the 5G wavelengths, and to compensate for that cash, they will need to catch the eye of consumers. In September, Vodafone used its bit of the range to display the UK's first hologram (全息) call. The Manchester City captain Steph Houghton appeared as a hologram in Newbury. It isn't all holograms, however: 5G will offer faster internet access, with Ofcom (英国通讯管理局) suggesting that video that takes a minute to download on 4G will be available in just a second.The wider application is to support connected equipment on the "internet of things" -not just the internet-enabled fridge that can reorder your milk for you, but the network that will enable driverless cars and delivery drones (无人机) to communicate with each other.Prof William Webb has warned that the technology could be a case of the emperor's new clothes. Much of the speed increase, he claims, could have been achieved by putting more money in the 4G network, rather than a new technology. Other different voices have suggested that a focus on rolling out wider rural broadband access and addressing current network coverage would be more beneficial to the UK as a whole.Obviously, 5G will also bring a cost to consumers. It requires a handset for both 5G and 4G, and the first 5G-enabled smart phones are expected in the coming year. With the slow pace of network rollout so far, it is likely that consumers will end up upgrading to a new 5 G phone well before 5 G becomes widely available in the nextcouple of years.9．Why does Prof William Webb say "the technology could be a case of the emperor's new clothes" ?A．He is in favor of the application of the new technology.B．5G will bring a cost to consumers in their daily life.C．5G helps people communicate better with each other.D．He prefers more money to be spent on 4G networks.10．The underlined word "addressing" in the fourth paragraph has the closest meaning to________A．making a speech to B．trying to solveC．managing to decrease D．responding to11．The last paragraph indicates thatA．it'll take several years .to make 5G accessible to the public in the UK B．5G service shows huge development potential and a broad marketC．customers are eager to use 5G smart phones instead of 4G onesD．it's probable that 5G network rollout is speeding up in BritainDZebra crossings (斑马线) — the alternating dark and light stripes on the road surface — are meant to remind drivers that pedestrians may be trying to get across. Unfortunately, they are not very effective. A 1998 study done by the Department of Traffic Planning and Engineering at Sweden’s Lund University showed that three out of four drivers kept the same speed or even speeded up as they were approaching a crossing. Even worse? Only 5% stopped even when they saw someone trying to get across.Now a mother-daughter team in Ahmedabad, India has come up with a clever way to get drivers to pay more attention — a 3D zebra crossing with an optical illusion (视错觉). Artists Saumya Pandya Thakkar and Shakuntala Pandya were asked to paint the crosswalks by IL&FS, an Indian company that manages the highways in Ahmedabad. The corporation was looking for a creative solution to help the city’s residents to cross the busy accident-prone (易出事故的) roads safely. Thakkar and Pandya, who had previously seen images of 3D zebra crossings that gave drivers the illusion oflogs of wood on the streets in Taizhou， China, decided to test if a similar way would work in India.Sure enough, in the six months when the 3D crosswalks have been painted across four of the city’s most dangerous highways, there have been no accidents reported! The artists say that while it may appear that the zebra crossing could cause the drivers to brake suddenly and endanger the vehicles behind, such is not the case. Because of the way the human eye works, the illusion is only visible from a distance. As they get closer, the painting looks just like any other ordinary zebra crossing. The creators hope that their smart design will become increasingly common throughout India and perhaps even the world. So let’s look forward to it.12．What can we learn from the first paragraph?A．Most drivers will slow down at zebra crossings.B．Common zebra crossings don’t function well.C．Drivers have to stop when approaching zebra crossings.D．About 95% of the drivers choose to speed up when approaching zebra crossings. 13．Why do drivers seeing the 3D zebra crossings slow down according to Para. 2?A．Because the drivers consider the safety of pedestrians.B．Because the drivers mistake them for logs of wood on the streets.C．Because the drivers are afraid of being fined for breaking the traffic rules.D．Because the drivers don’t want to brake suddenly and endanger the vehicles behind.14．The last paragraph is mainly about ________.A．the theory of the 3D zebra crossingsB．the popularity of the 3D zebra crossingsC．the shortcoming of the 3D zebra crossingsD．the positive effect of the 3D zebra crossings15．What is the author’s attitude towards the 3D zebra cross ings?A．Cautious. B．Doubtful. C．Approving. D．Disapproving.其次节（共5小题；每小题3分，满分15分）依据短文内容，从短文后的选项中选出能填入空白处的最佳选项。
山西省原平市范亭中学2018-2019学年高二4月月考数学(文)试题 Word版含解析
2018—2019学年度第二学期月考试题高二文科数学一．选择题1.若复数3z i =-，则z 在复平面内对应的点位于( ) A. 第一象限 B. 第二象限 C. 第三象限 D. 第四象限【答案】D 【解析】【详解】复数在复平面内对应的点是()3,1-，在第四象限，故选D.2.用演绎法证明函数3y x =是增函数时的小前提是( ) A. 增函数的定义B. 函数3y x =满足增函数的定义 C. 若12x x <，则12()()f x f x < D. 若12x x >，则12()()f x f x >【答案】B 【解析】解：因为用演绎法证明函数3y x =是增函数，可以根据函数3y x =满足增函数的定义，得到结论。
3.计算1i1i-+的结果是 （ ） A. i B. i -C. 2D. 2-【答案】B【解析】()()()21121112i i i i i i i ---===-++-,故选B.4.有下列关系：①人的年龄与他（她）拥有的财富之间的关系； ②曲线上的点与该点的坐标之间的关系； ③苹果的产量与气候之间的关系；④森林中的同一种树木，其横断面直径与高度之间的关系，其中有相关关系的是（）A. ①②③B. ①②C. ②③D. ①③④【答案】D【解析】试题分析：本题考查学生对变量间相关关系的理解，所谓相关关系，指的是介于确定性的函数关系与无必然关系之间的一类非确定性关系.根据变量间相关关系的含义可知，①③④中两个变量具有相关关系；而②，曲线上的点定了，该点的坐标也就随之确定，这两个变量间的关系为确定性关系，故选D.考点：变量间的相关性.=++中，下列说法正确的是( )．5.在线性回归模型y bx a eA. 是一次函数；B. 因变量y是由自变量x唯一确定的；C. 因变量y除了受自变量x的影响外，可能还受到其它因素的影响；这些因素会导致随机误差e的产生；D. 随机误差e是由于计算不准确造成的，可通过精确计算避免随机误差e的产生。
复数考试题目大全及答案
复数考试题目大全及答案一、选择题1. 下列哪个选项是复数的共轭？A. 2 + 3iB. 2 - 3iC. 3 + 2iD. 3 - 2i答案：B2. 复数 \( z = 3 + 4i \) 的模是：A. 5B. 7C. 8D. 9答案：A3. 复数 \( z_1 = 2 + i \) 和 \( z_2 = 1 - 2i \) 的和是：A. 3 - iB. 3 + iC. 1 + 3iD. 1 - 3i答案：A二、填空题1. 复数 \( z = a + bi \) 中，\( a \) 称为复数的______，\( b \) 称为复数的______。
答案：实部，虚部2. 复数 \( z = -4 + 3i \) 的共轭复数是______。
答案：-4 - 3i3. 若复数 \( z \) 的模为 10，且 \( z \) 的虚部为 6，则 \( z \) 的实部为______。
答案：±8三、简答题1. 解释什么是复数的模，并给出计算公式。
答案：复数的模是复数在复平面上到原点的距离，计算公式为\( |z| = \sqrt{a^2 + b^2} \)，其中 \( z = a + bi \)。
2. 描述如何计算两个复数的乘积。
答案：两个复数 \( z_1 = a + bi \) 和 \( z_2 = c + di \) 的乘积计算公式为 \( z_1 \cdot z_2 = (a + bi)(c + di) = ac - bd+ (ad + bc)i \)。
四、计算题1. 计算复数 \( z = 1 + 2i \) 的模和共轭复数。
答案：复数 \( z \) 的模为 \( |z| = \sqrt{1^2 + 2^2} =\sqrt{5} \)，共轭复数为 \( 1 - 2i \)。
2. 求复数 \( z_1 = 3 - 4i \) 和 \( z_2 = 1 + i \) 的乘积。
答案：\( z_1 \cdot z_2 = (3 - 4i)(1 + i) = 3 + 3i - 4i -4i^2 = 3 - i + 4 = 7 - i \)。
复数考试题及答案
复数考试题及答案一、选择题（每题4分，共20分）1. 复数z=1+i的模长为（）。
A. 1B. √2C. 2D. √3答案：B2. 下列复数中，实部为1的复数是（）。
A. 2+3iB. 1-iC. 3-2iD. 4+i答案：B3. 复数z=a+bi（a，b∈R）是纯虚数的充要条件是（）。
A. a=0，b≠0B. a≠0，b=0C. a=0，b=0D. a≠0，b≠0答案：A4. 复数z=3+4i的共轭复数为（）。
A. 3-4iB. -3+4iC. -3-4iD. 3+4i答案：A5. 复数z=1+i的平方为（）。
A. 2iB. -2iC. 0D. 2答案：D二、填空题（每题4分，共20分）1. 复数z=2+3i的模长为______。
答案：√132. 复数z=1-2i的共轭复数为______。
答案：1+2i3. 复数z=a+bi（a，b∈R）是实数的充要条件是______。
答案：b=04. 复数z=3+4i与复数z'=1-2i的和为______。
答案：4+2i5. 复数z=2+i的平方为______。
答案：3+4i三、计算题（每题10分，共30分）1. 计算复数z1=3+4i与z2=1-2i的乘积。
答案：z1 * z2 = (3+4i)(1-2i) = 3 - 6i + 4i - 8i^2 = 3 - 2i + 8 = 11 - 2i 2. 计算复数z=2+3i的模长和辐角。
模长：|z| = √(2^2 + 3^2) = √(4 + 9) = √13辐角：arg(z) = arctan(3/2)3. 计算复数z=1+i的立方。
答案：z^3 = (1+i)^3 = (1+i)(1+i)(1+i) = (1+2i+i^2)(1+i) = (1+2i-1)(1+i) = 2i(1+i) = 2i + 2i^2 = 2i - 2 = -2 + 2i四、解答题（每题15分，共30分）1. 已知复数z1=3+4i和z2=1-2i，求它们的和与差，并计算结果的模长。
山西省范亭中学高三英语上学期第二次月考试题
山西省范亭中学2018-2019学年高三英语上学期第二次月考试题本试卷分为第I卷（选择题）和第II卷（非选择题）两部分, 共150分。
考试时间120分钟。
第I卷第一部分听力（略）第二部分阅读理解（共两节，满分60分）第一节（共15小题，每小题3分，满分45分）ATears ran down from my eyes as I saw the essay my high school English teacher had just handed back. A big F was written on top. I was hopeless. I was stupid! My face burned with shame when my classmates called me stupid.“I’m the only one who doesn’t know my ABCs!”I sobbed to Mom.I’ll help,”she promised.Every day I sat with her, but to me, cat looked like cta and red was reb. Frustrated, I would return to my bedroom and draw, filling the paper with houses, restaurants and offices.“When I grow up, I want my own store,”I told Mom, pointing to my drawings.“That’s great !”she said,“ but first you have to learn to read.”Later, I was diagnosed with dyslexia (失读症). So Mom took me to a learning centre, where I was given reading exercises. But I still had a hard time. Finally, I graduated, but I was afraid of my reading skills.“I’ll never get a job!”I cried to Mom later again.“Don’t focus on what you can’t do,” she comforted, “Concentrate on what you really can.”But what can I do? I wondered. Suddenly, I thought of the drawings I’d made as a child and my dream of h aving my own store. I enjoyed sales so much that over the next few years, I tried my hand at other businesses. Today, I watch over seven branches. We have 187 employees and $15 million in sales.While I’ll never be what my teachers might have wanted, I am a success--on my own terms. The other day a student sent me a card, reading: You gave me so much confidence. I hope to be like you when I am big. Tears of joy filled my eyes. This was my A, and I smiled.21.Why did the author feel so ashamed at school?A.His teacher ignored him in class.B.His teacher didn’t grade his essay.C.He failed to finish his essay in time.D.His classmates looked down upon him.22.Which of the following best describes the author’s feeling to his mother?A.Grateful.B.Guilty.C.Doubtful.D.Regretful.23.Which of the following can match the text?A.No pains, no gains.B.Never too old to learn.C.Every man has his value.D.Two heads are better than one.BWhen other nine-year-old kids were playing games, she was working at a petrol station. When other teens were studying or going out, she fought to find a place to sleep on the street. But she beat these terrible setbacks to win a highly competitive scholarship and gain entry into Harvard University. And her amazing story has inspired a movie, “Homeless to Harvard: The Liz Murray Story”.Liz Murray, a 22-year-old American girl, has been writing a real-life story of willpower and determination. Liz grew up with two drug-addicted parents. There was never enough food or warm clothes in the house. Liz was the only member of the family who had a job. Her mother had AIDS and died when Liz was just l5 years old. The effect of that loss became a turning point in her life. Connecting the environmentin which she had grown up with how her mother had died. She decided to do something about it.Liz went back to school. She threw herself into her studies, never telling her teachers that she was homeless. At night, she lived on the streets. “What drove me to live on had something to do with understanding, and by understanding that there was a whole o ther way of being. I had only experienced a small part of the society,” she wrote in her book Breaking Night.She admitted that she used envy to drive herself on. She used the benefits that come easily to others, such as a safe living environment, to encourage herself that “next to nothing could hold me down”. She finished high school in just two years and won a full scholarship to study at Harvard University. But Liz decided to leave her top university a couple of months earlier this year in order to take care of her father, who has also developed AIDS. “I love my parents so much. They are drug addicts. But I never forget that they love me all the time. ”Liz wants moviegoers to come away with the idea that changing your life is “as simple as making a decision”.24.In which order did the following things happen to Liz?a. Her mother died of AIDS.b. She got admitted into Harvard.c. She worked at a petrol station.d. The movie about her life was put on.e. She had trouble finding a place to sleep.A.c, a, e, b, dB.a, b, c, e, dC.c, d, b, a, eD.b, e, a, d, c25.What actually made her go towards her goal?A.Envy and encouragement.B.Willpower and determination.C.Decisions and understanding.D.Love and respect for her parents.26.What does Liz mean by saying “What drove me to live on...I had only experienceda small part of the society”?A.She had little experience of social life.B.She could hardly understand the society.C.She would do something for her own life.D.She needed to travel more around the world.27.What does the passage mainly tell us?A.Why Liz loved her parents so much.B.How Liz made efforts to change her life.C.What a hard time Liz had in her childhood.D.How Liz managed to enter Harvard University.CA heated debate is currently going on in our town. Should we allow the cinema to be constructed in the Havenswood Shopping Center? There is just one large lot left to build on, and the theater would use up all of that space. Some people are excited at the idea of finally having our own movie theater. Others would rather travel ten miles to the nearest theater to keep our quiet town the way it is. They say it is enough to have Marvin’s Movie Video Rentals. After all, Marvin’s store keeps thousands of the latest videos.There are certainly benefits to renting videos. For one thing, you can plan your own schedule when it is convenient for you. You can relax on your sofa, and take a break whenever you need one.You can also talk to others without bothering any strangers seated nearby. In addition, it is a less expensive way to view a movie compared to going to a theater.On the other hand, seeing a movie in a theater is an experience all its own. First, you can see the movie on wide screen as the filmmaker intended. To be viewed on a television screen, a film must be changed in some way to make it smaller. One is the “pan-and-scan” method, which involves removing some of the details in the picture. The other way, called “letterboxing”, keeps the image the way it is onthe big screen, with one annoying exception; because the big-screen version is wide, the same picture on a television screen must be long and narrow.Another problem is sound. The sound from a television cannot compare to the sound system in a theater. Your experience of a movie improves when you can clearly hear all of the sounds. Furthermore, at home, viewing companions often talk during a movie, which makes you miss out on what’s happening in the film.Besides, having a movie theater will not mean that you can’t still go to Marvin’s! You will just have a choice that you didn’t have before. Isn’t it time for Havenswood residents to enjoy a little progress?28.The underlined wor d “lot” in Paragraph 1 means “__________”.A.a great numberB.a complete groupC.an area of landD.a result of chance29.How does the author mainly state his point?A.By using examples.B.By making a comparison.C.By carrying out a surveyD.By presenting d ifferent people’s view.30.Why do some people think it is enough to have Marvin’s Movie Video Rentals?A.Marvin’s store stocks lots of recent videos.B.It’ very convenient to go to Marvin’s store.C.the owner of the store is friendly and helpful.D.It is more comfortable to see films at home than in the theater.31.What is the author’s attitude towards building a new cinema?A.Cautious.B.Approving.C.Doubtful.D.Worried.DStaying positive through the cold season could be your best defense against getting ill, a new American study suggests.In an experiment that exposed healthy volunteers to a cold or flu virus, researchers found that people with a generally sunny character were less likely to fall ill. The findings, published in the journal Psychosomatic Medicine, build on evidence that a “positive emotional style” can help ward off the common cold and other illness.Researchers believe the reasons may be both objective―as in happiness improving immune function(免疫功能)―and subjective―as in happy people being less troubled by a sore throat or runny nose.“People with a positive emotional style may have different immune responses to the virus,” explained lead ing study author Dr Sheldon Cohen of Carnegie Mell on University in Pittsburgh.“And when they do get a cold, they may interpret their illness as being less severe.”Cohen and his colleagues have found in a previous study that happier people seemed less likely to catch a cold, but some questions remained as to whether the emotional quality itself had the effect.For the new study, the researchers had 193 healthy adults complete standard measures of personality qualities, physical health, and emotional “style”. Those who tended to be happy, energetic and easy-going were judged as having a positive emotional style, while those who were often unhappy, tense, and hostile had a negative style.Afterwards, the researchers gave them nose drops containing either a cold virus or a particular flu virus. Over the next six days, the volunteers reported on any aches, pains, sneezing they had, while the researchers collected objective data. Cohen and his colleagues found that happy people were less likely to develop a cold.What’s more, when happy folks did develop a cold, their symptoms were less severe than expected based on objective measures. On the contrary, people with negative characters were not at increased risk of developing a cold based on objective measures, though they did tend to get down about their symptoms.“We find that it’s really positive emotions that have the big effect,” Cohen said,“not the negative ones.”So can a bad-tempered person fight a cold by deciding to be happy?32.The purpose of Cohen’s new study was to _____.A.find effective ways to fight illnessesB.test people’s different immune responses to cold virusC.tell differences between happy people and unhappy peopleD.examine whether health was related to emotional styles33.The underlined phrase “ward-off” in Paragraph 2 can be replaced by “_____”.A.get close toB.keep away fromC.get used toD.go on with34.How did Cohen reach his conclusion?A.By comparing the experimental results of different groups.B.By asking the volunteers to complete a form.C.By collecting data among people with a cold.D.By observing the volunteers’ symptoms.35.Cohen’s new study showed that_____.A.an emotional style is difficult to changeB.happy people are immune to cold virusC.people attitudes towards illnesses are differentD.happiness itself helps protect people from cold第二节、七选五根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。
山西省范亭中学2025届高三一诊考试数学试卷含解析
山西省范亭中学2025届高三一诊考试数学试卷注意事项:1．答题前，考生先将自己的姓名、准考证号码填写清楚，将条形码准确粘贴在条形码区域内。
2．答题时请按要求用笔。
3．请按照题号顺序在答题卡各题目的答题区域内作答，超出答题区域书写的答案无效；在草稿纸、试卷上答题无效。
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5．保持卡面清洁，不要折暴、不要弄破、弄皱，不准使用涂改液、修正带、刮纸刀。
一、选择题：本题共12小题，每小题5分，共60分。
在每小题给出的四个选项中，只有一项是符合题目要求的。
1．秦九韶是我国南宋时期的数学家，普州（现四川省安岳县）人，他在所著的《数书九章》中提出的多项式求值的秦九韶算法，至今仍是比较先进的算法．如图的程序框图给出了利用秦九韶算法求某多项式值的一个实例，若输入x 的值为2，则输出的v 值为( )A ．10922⨯-B ．10922⨯+C ．11922⨯+D ．11922⨯-2．在平面直角坐标系xOy 中，将点()1,2A 绕原点O 逆时针旋转90︒到点B ，设直线OB 与x 轴正半轴所成的最小正角为α，则cos α等于( ) A ．255-B ．55-C ．55D ．25-3．直线经过椭圆的左焦点，交椭圆于两点，交轴于点，若，则该椭圆的离心率是（） A ．B ．C ．D ．4．设函数()2ln x e f x t x x x x ⎛⎫=-++ ⎪⎝⎭恰有两个极值点，则实数t 的取值范围是（ ） A ．1,2⎛⎤-∞ ⎥⎝⎦B ．1,2⎛⎫+∞⎪⎝⎭ C ．1,,233e e ⎛⎫⎛⎫+∞⎪ ⎪⎝⎭⎝⎭D ．1,,23e ⎛⎤⎛⎫-∞+∞ ⎪⎥⎝⎦⎝⎭5．过抛物线()220y px p =>的焦点F 作直线与抛物线在第一象限交于点A ，与准线在第三象限交于点B ，过点A 作准线的垂线，垂足为H .若tan 2AFH ∠=，则AF BF=（ ）A ．54B ．43 C ．32D ．26．已知ABC ∆的内角,,A B C 的对边分别是,,,a b c 且444222222a b c a b ca b +++=+，若c 为最大边，则a b c +的取值范围是（ ）A ．2313⎛⎫⎪ ⎪⎝⎭,B ．()1,3C ．2313⎛⎤⎥ ⎝⎦,D ．(1,3]7．集合}{220A x x x =--≤，{}10B x x =-<，则A B =( )A ．}{1x x < B ．}{11x x -≤< C ．{}2x x ≤ D ．{}21x x -≤<8．函数的图象可能是下列哪一个？（ ）A ．B ．C ．D ．9．已知12,F F 是双曲线2222:1(0,0)x y C a b a b -=>>的左、右焦点，,A B 是C 的左、右顶点，点P 在过1F 且斜率为34的直线上，PAB △为等腰三角形，120ABP ∠=︒，则C 的渐近线方程为（ ） A ．12y x =±B ．2y x =±C ．33y x =±D ．3y x =±10．已知向量a ，b ，b =(1，3)，且a 在b 方向上的投影为12，则a b ⋅等于（ ） A ．2B ．1C ．12D ．011．下列不等式成立的是（ ）A ．11sin cos 22>B ．11231122⎛⎫⎛⎫> ⎪ ⎪⎝⎭⎝⎭C ．112311log log 32< D ．11331123⎛⎫⎛⎫> ⎪ ⎪⎝⎭⎝⎭12．设O 为坐标原点，P 是以F 为焦点的抛物线24y x =上任意一点，M 是线段PF 上的点，且PM MF =，则直线OM 的斜率的最大值为（ ） A ．1B ．12C ．22D ．52二、填空题：本题共4小题，每小题5分，共20分。
山西省忻州市范亭中学高二数学文月考试题含解析
山西省忻州市范亭中学高二数学文月考试题含解析一、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. 如图，一个空间几何体的主视图和左视图都是边长相等的正方形，俯视图是一个圆，那么这个几何体是A．棱柱 B．圆柱 C．圆台 D．圆锥参考答案：B略2. 设实数x，y满足，则xy的最大值为（）A. 14B.C.D.参考答案：D【分析】先由约束条件作出可行域，再利用基本不等式进行求解，即可得出结果.【详解】由约束条件作出可行域如下：由图像可得，则，当且仅当，时，取等号；经检验，在可行域内，所以的最大值为.故选D【点睛】本题主要考查简单的线性规划，熟记基本不等式即可求解，属于常考题型.3. 已知是等比数列，前项和为，，则A. B. C. D.参考答案：B4. 已知椭圆与双曲线有相同的焦点，则a的值为A．B．C．4 D．10参考答案：C5. 如图，在△ABC中，D是边AC上的点，且AB=AD，2AB=BD，BC=2BD，则sinC的值为（）A．B．C．D．参考答案：D考点：三角形中的几何计算．专题：解三角形．分析：根据题中条件，在△ABD中先由余弦定理求出cosA，利用同角关系可求sinA，利用正弦定理可求sin∠BDC，然后在△BDC中利用正弦定理求解sinC即可解答：解：设AB=x，由题意可得AD=x，BD=△ABD中，由余弦定理可得∴sinA=△ABD中，由正弦定理可得?sin∠ADB=∴△BDC中，由正弦定理可得故选：D．点评：本题主要考查了在三角形中，综合运用正弦定理、余弦定理、同角基本关系式等知识解三角形的问题，反复运用正弦定理、余弦定理，要求考生熟练掌握基本知识，并能灵活选择基本工具解决问题6. 一个直棱柱被一个平面截去一部分后所剩几何体的三视图如图所示，则该几何体的体积为A.9B.10C.11D.参考答案：C7. 点P(2,3)到直线：的距离为最大时，与的值依次为（）A．3,-3 B．5,1 C．5,2D．7,1参考答案：B8. 已知i为虚数单位，则复数（）A. B. C. D.参考答案：C9. 定义在R上的函数及其导函数的图象都是连续不断的曲线，且对于实数，有．现给出如下结论：①；②；③；④.其中结论正确的个数是A． 1 B． 2 C． 3 D． 4参考答案：B10. 要使与轴的两个交点分别位于原点的两侧,则有（）A． B．C．D．参考答案：D略二、填空题:本大题共7小题,每小题4分,共28分11. 设是定义在R上的奇函数，且当时，，若对任意的，不等式恒成立，则实数t的取值范围是___________．参考答案：【分析】根据奇函数的定义求出函数的解析式，可得，可将对任意的均成立转化为对任意的恒成立，即可求解.【详解】由题意得：当时，，所以是上的增函数且为奇函数，的解析式为.由题意得成立，从而原不等式等价于对任意的均成立，即对任意的恒成立∴对恒成立∴.【点睛】本题主要考查利用奇函数求解析式的方法.解答本题的关键是利用转化思想，将对任意的均成立转化为对任意的恒成立. 12. 设点P是边长为2的正三角形ABC的三边上的动点，则?（+）的取值范围为．参考答案：[﹣，2]【考点】平面向量数量积的运算．【分析】以AB 中点为坐标原点，建立如图所示的直角坐标系，可得A （﹣1，0），B （1，0），C （0，），讨论P在AB，BC，CA上，分别设P的坐标，可得向量PA，PB，PC的坐标，由向量的坐标表示，化为二次函数在闭区间上的最值问题，即可得到所求取值范围．【解答】解：以AB中点为坐标原点，建立如图所示的直角坐标系，可得A（﹣1，0），B（1，0），C（0，），当P在线段AB上，设P（t，0），（﹣1≤t≤1），=（﹣1﹣t，0），=（1﹣t，0），=（﹣t，），即有?（+）=（﹣1﹣t，0）?（1﹣2t，）=（﹣1﹣t）（1﹣2t）+0×=2t2+t﹣1=2（t﹣）2﹣，由﹣1≤t≤1可得t=取得最小值﹣，t=﹣1时，取得最大值0；当P在线段CB上，设P（m，（1﹣m）），（0≤m≤1），=（﹣1﹣m，（m﹣1）），=（1﹣m，（m﹣1）），=（﹣m，m），即有?（+）=（﹣1﹣m，（m﹣1））?（1﹣2m，（2m﹣1））=（﹣1﹣m ）（1﹣2m ）+（m ﹣1）×（2m ﹣1）=2（2m ﹣1）2，由0≤m≤1可得m=取得最小值0，m=0或1时，取得最大值2； 当P 在线段AC 上，设P （n ，（1+n ）），（﹣1≤n≤0）， =（﹣1﹣n ，﹣（1+n ）），=（1﹣n ，﹣（1+n ）），=（﹣n ，﹣ n ），即有?（+）=（﹣1﹣n ，﹣（1+n ））?（1﹣2n ，﹣（1+2n ））=（﹣1﹣n ）（1﹣2n ）+（1+n ）×（1+2n ）=8n 2+10n+2=8（n+）2﹣，由﹣1≤n≤0可得n=﹣取得最小值﹣，n=0时，取得最大值2； 综上可得?（+）的取值范围是[﹣，2]．故答案为：[﹣，2]．【点评】本题考查向量数量积的坐标表示，考查坐标法的运用，同时考查分类讨论和转化思想，转化为二次函数在闭区间上的最值问题是解题的关键，属于中档题．13. 函数的图像的相邻两条对称轴之间的距离等于参考答案：略14. 用等值算法求294和84的最大公约数时，需要做次减法. 参考答案：415. 若，则函数的最小值为__________．参考答案：4设，∵，∴，函数可化为，由于对称轴为，∴时，函数有最小值4，故答案为4.16. 已知命题则是 .参考答案：17. 已知的外接圆的圆心为，则．参考答案：略三、 解答题：本大题共5小题，共72分。
2016-2017学年山西省原平市范亭中学高二下学期期末考试数学(文)试题(解析版)
2016-2017学年山西省原平市范亭中学高二下学期期末考试数学（文）试题一、单选题1．已知复数2211612x y +=，则z 的共轭复数等于 ( ) A. i- B. i C. 2i - D. 2i 【答案】D【解析】复数2i- ，共轭复数为2i ；根据复数的共轭的概念； 2．{}{}2|13,|4P x x Q x x =≤≤=≥ R P C Q ⋃ =（ ）A. [2,3]B. [)1,2C. (]2,3-D. ][(),21,-∞-⋃+∞ 【答案】C【解析】()--2][2+Q =∞⋃∞，，， R C Q =-22（，） ， R P C Q ⋃ (]=2,3-， 故选C ；3．将函数5sin 26y x π⎛⎫=+ ⎪⎝⎭的图像向左平移3π 个单位长度，所得函数图像对应解析式为（ ） A. 5,sin 26A y x π⎛⎫=+⎪⎝⎭B. cos2y x =-C. cos2y x =D. sin 26y x π⎛⎫=- ⎪⎝⎭【答案】A【解析】函数5s i n 26y xπ⎛⎫=+ ⎪⎝⎭的图像向左平移3π 个单位长度，得到3sin 2=-cos22y x x π⎛⎫=+⎪⎝⎭， 故选A ；4．命题*,,x R n N ∀∈∃∈ 使得2n x ≥ 的否定形式是 （ ） A. *,,x R n N ∀∈∃∈使得2n x < B. *,,x R n N ∀∈∀∈使得2n x < C. 使得2n x < *,,x R n N ∃∈∃∈ D. *,,x R n N ∃∈∀∈使得2n x < 【答案】D【解析】特称命题的否定是，换量词，否结论，不改变条件， *,,x R n N ∃∈∀∈ 使得2n x <。
故选D ；5．已知定义在R 上的奇函数f （x ）满足f （x+2）＝－f （x ），且在区间[0，2]上是增函数，则（ ）A ．f （－25）<f （10）<f （80）B ．f （80）<f （10）<f （－25）C ．f （10）<f （80）<f （－25）D ．f （－25）<f （80）<f （10） 【答案】D【解析】试题分析：4)2()()2(=⇒+=-=-T x f x f x f ， 所以0)0()80(),2()10(),1()25(===-=-f f f f f f又)(x f 在区间[-2，2]上是增函数，所以)10()80()25(f f f <<-，选D 【考点】函数性质综合应用6．PA ⊥的展开式中含ABCD 项的系数是 （ ） A. 240 B. 240- C. 192 D. 192- 【答案】D【解析】二项式6⎛⎝的展开式的通项公式为(()()6366*1*kkk k k kC C x --⎛=- ⎝， ∴当k=1时x 2项的系数是﹣192，故选：D ．7．“9k <”是“方程221259x y k k +=--表示双曲线”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 【答案】A【解析】试题分析：∵方程221259x y k k +=--为双曲线，∴(25)(9)0k k --<， ∴9k <或25k >，∴“9k <”是“方程221259x y k k +=--为双曲线”的充分不必要条件， 故选A.【考点】双曲线的标准方程、充分必要条件.8．已知某几何体的三视图如图所示,则该几何体的体积是 ( )（正视图与侧视图的形状一样，都是边长为2的正方形，竖线为中线）A. 4π+B. 24π+C. 2π+D. 22π+ 【答案】A【解析】该几何体是半个圆柱（底面半径为1，高为2）和一个三棱柱（底面积为1，高为2）的组合体，组合体的体积为两者之和，半圆柱体积为*22ππ= ，一个三棱柱体积为2 ，故总和为,2π+； 故选A.9．已知n S 为数列{}n a 的前n 项和且22n n S a =-，则54S S -的值为（ ） A. 8 B. 10 C. 16 D. 32 【答案】D【解析】 22n n S a =-1122n n S a ++∴=-两式相减得： 12n n a a +=12a ={}n a ∴是首项为2，公比为2的等比数列 5432S S ∴-=故选D.10．若实数,x y 满足约束条件1{2 22x y x y ≤≤+≥，则22z x y =+的最小值是（ ）A.5B. 45C. 1D. 4【答案】B【解析】由约束条件约束条件1{222x y x y ≤≤+≥，作出可行域如图，由图可知, 222z x y =+的最小值为原点O(0,0)到直线220x y +-=的距离的平方，等于245= ,故选：B.11．若椭圆2231x ky += 的一个焦点的坐标是()0,1,则其离心率等于( )A. 2B. . 12C. .D. 【答案】D【解析】依题意可知，b=13 ，a=1k =1，∴=3 ∴e=c a= 故选B ．点睛：根据题意可知a 和b ，进而根据c ，进而根据e=ca求得e ． 12．已知函数()*93,xxf x m =- 若存在非零实数0x ,使得()()00f x f x -= 成立,则实数m 的取值范围是( ) A. . 12m ≥B. 2m ≥C. 102m << D. 02m <<【答案】C【解析】由题意存在非零实数x 0，使得f （﹣x 0）=f （x 0）成立，可得m•9x ﹣3x =m•9﹣x ﹣3﹣x 有解，即m （9x ﹣9﹣x ）=（3x ﹣3﹣x ）有解． 可得1m =3x +3﹣x ≥2 ①，求得0＜m≤12． 再由x 0为非零实数，可得①中等号不成立，故0＜m ＜12， 故选：B ．点睛：由题意可得m•9x ﹣3x =m•9﹣x ﹣3﹣x有解，可得1m=3x +3﹣x，利用基本不等式求得m 的范围．主要考查指数函数的综合应用，基本不等式的应用，注意检验等号成立条件是否具备．二、填空题13．已知向量()()2,1,3,a b λ== ， 若(2-)a b b ⊥则λ=_______________【答案】-1或3【解析】因为()2-1,2a b λ=- ， (2-)a b b ⊥ ，则(2-)*a b b = ，()320,1,3.λλλ∴+-==-故结果为-1或3. 14．设函数()()()22211{ log 1(1)x x f x x x -+≥-<，若f （a ）=﹣1，则a=____________【答案】1或2 【解析】函数f （x ）=()()22211{ log 1(1)x x x x -+≥-<，当a ≥1时，f （a ）=﹣1，可得﹣2a 2+1=﹣1，解得a=1；当a ＜1时，f （a ）=﹣1，可得log 2（1﹣a ）=﹣1，解得a=12； 故答案为：5；1或12． 点睛：直接利用分段函数，由里及外求解函数值，通过方程求出方程的根即可， 考查函数的值的求法，方程的根的求解，分段函数的应用，考查计算能力．15．离心率12e = 的椭圆，它的焦点与双曲线2213x y -= 的焦点重合，则此椭圆的方程是_______【答案】2211612x y += 【解析】由题意知此椭圆的焦点坐标是F 1（﹣2，0），F 2（2，0）， ∵离心率12e =，∴a=4，b 2=12， ∴此椭圆的方程为】2211612x y +=． 点睛：由题意知此椭圆的焦点坐标是F 1（﹣2，0），F 2（2，0），再由离心率12e =，知此椭圆的方程为2211612x y +=.根据数据求得回归直线方程为A xyz -当产量为80吨时，预计需要生产消耗为____________________吨. 【答案】59【解析】由题意，=45，=36.25，代入y ∧=0.65x+a ，可得a=7，∴当产量为80吨时，预计需要生成能耗为0.65×80+7=59， 故答案为：59．点睛：求出x ，y 的平均数，代入y 关于x 的线性回归方程，求出a ，把x=80代入，能求出当产量为80吨时，预计需要生成的能耗．三、解答题17．已知,,A B C 为ABC 的三内角，且其对边分别为(),,x y z =、AEC 、m ，3,37A c a π==． （1）求sinC 的值。
山西省原平市范亭中学2020-2021学年高二4月月考数学(文)试题
山西省原平市范亭中学2020-2021学年高二4月月考数学（文）试题学校:___________姓名：___________班级：___________考号：___________一、单选题1．若复数3z i =-，则z 在复平面内对应的点位于( ) A ．第一象限B ．第二象限C ．第三象限D ．第四象限2．用演绎法证明函数3y x =是增函数时的小前提是（ ） A ．函数3y x =满足增函数的定义 B ．增函数的定义C ．若12x x <，则12()()f x f x <D ．若12x x >，则12()()f x f x >3．计算1i1i-+的结果是 （ ） A ．iB ．i -C ．2D ．2-4．有下列关系：①人的年龄与他（她）拥有的财富之间的关系； ②曲线上的点与该点的坐标之间的关系； ③苹果的产量与气候之间的关系；④森林中的同一种树木，其横断面直径与高度之间的关系， 其中有相关关系的是（ ） A ．①②③B ．①②C ．②③D ．①③④5．在线性回归模型y bx a e =++中，下列说法正确的是( )． A ．是一次函数；B ．因变量y 是由自变量x 唯一确定的；C ．因变量y 除了受自变量x 的影响外，可能还受到其它因素的影响；这些因素会导致随机误差e 的产生；D ．随机误差e 是由于计算不准确造成的，可通过精确计算避免随机误差e 的产生． 6．对相关系数r ，下列说法正确的是（ ） A ．r 越大，线性相关程度越大 B ．r 越小，线性相关程度越大C ．r 越大，线性相关程度越小，r 越接近0，线性相关程度越大D ．1r ≤且r 越接近1，线性相关程度越大，r 越接近0，线性相关程度越小 7．用反证法证明命题：“一个三角形中不能有两个直角”的过程归纳为以下三个步骤：(1)9090180A B C C ++=︒+︒+>︒，这与三角形内角和为180︒相矛盾，90A B ==︒不成立（2）所以一个三角形中不能有两个直角（3）假设三角形的三个内角A .B .C 中有两个直角，不妨设90A B ==︒，正确顺序的序号为（ ） A ．(1)(2)(3)B ．(3)(1)(2)C ．(1)(3)(2)D ．(2)(3)(1)8．已知,,a b c 均大于1，且1log log 4c c a b ⋅=，则下列不等式一定成立的是（ ） A ．ac b ≥B ．bc a≥C ．ab c ≥D ．ab c ≤9．ABC ∆中，角,,A B C 对应边,,a b c ，若,,a b c 成等差数列，则角B 的取值范围是（ ） A ．0,4π⎛⎤ ⎥⎝⎦B ．,32ππ⎡⎫⎪⎢⎣⎭C ．0,3π⎛⎤ ⎥⎝⎦D ．,2ππ⎛⎫⎪⎝⎭10．在独立性检验中，统计量2K 有两个临界值：3.841和6.635；当2 3.841K >时，有95%的把握说明两个事件有关，当2 6.635K >时，有99%的把握说明两个事件有关，当2 3.841K ≤时，无把握认为两个事件有关．在一项打鼾与患心脏病的调查中，共调查乐2000人，经计算的220.87K =，根据这一数据分析，认为打鼾与患心脏病之间（ ） A ．约有95%的把握认为两者有关 B ．约有95%的打鼾者患心脏病 C ．约有99%的把握认为两者有关 D ．约有99%的打鼾者患心脏病11．若定义运算：()()a a b a b b a b ≥⎧⊗=⎨<⎩，例如233⊗=，则下列等式可能不成立的是（ ）A ．a b b a ⊗=⊗B ．()()a b c a b c ⊗⊗=⊗⊗C ．222()a b a b ⊗=⊗D ．()()()c a b c a c b ⋅⊗=⋅⊗⋅(0)c >12．指数曲线bx y ae =进行线性变换后得到的回归方程为10.6u x =-，则函数2y x bx a =++的单调递增区间为（ ）A ．()0,∞+B ．3,10⎛⎫+∞⎪⎝⎭C ．1(,)2+∞D ．()1,+∞二、填空题 13．若11abi i=--，其中,a b 都是实数，i 是虚数单位，则a bi +=__________．14．在等差数列{}n a 中，若100a =，则有：121219n n a a a a a a -++⋅⋅⋅+=++⋅⋅⋅+（19n <，且*n N ∈）成立.类比上述性质，在等比数列{}n b 中，若91b =，则有______. 15．已知,,1a b R a b ∈+<，求证20x ax b ++=的两根的绝对值都小于1，用反证法证明可假设__________16．一个车间为了规定工时定额，需要确定加工零件所花费的时间，为此进行了8次试验，收集数据如下：设回归直线方程为y bx a =+，若23b =，则点(,)a b 在直线45200x y --=的________方三、解答题17 18．设z C ∈，满足1z R z +∈，且14z -是纯虚数，求z 19．S 为ABC ∆所在平面外的一点，SA ⊥平面ABC ，平面SAB ⊥平面SBC ，求证：AB BC ⊥20．假定小麦基本苗数x 与成熟期有效穗y 之间存在相关关系，今测得5组数据如下：（1）以x 为解释变量，y 为预报变量，画出散点图 （2）求y 与x 之间的回归方程（3）当基本苗数为56.7时预报有效穗（注：1221ni ii ni i x y nxyb x nx ==-=-∑∑， a y bx =-）5522115101.56,9511.43iii i xy ====∑∑，22189.25,921.7296y x ==，516746.76i i i x y ==∑21．在平面几何中，研究三角形内任意一点与三边的关系时，有真命题：边长为a 的正a 。
山西省范亭中学2018-2019学年高三上学期第二次月考英语试题(含答案)
范亭中学高三月考试题英语本试卷分为第I卷（选择题）和第II卷（非选择题）两部分, 共150分。
考试时间120分钟。
第I卷第一部分听力（略）第二部分阅读理解（共两节，满分60分）第一节（共15小题，每小题3分，满分45分）ATears ran down from my eyes as I saw the essay my high school English teacher had just handed back. A big F was written on top. I was hopeless. I was stupid! My face burned with shame when my classmates called me stupid.“I’m the only one who doesn’t know my ABCs!”I sobbed to Mom.I’ll help,”she promised.Every day I sat with her, but to me, cat looked like cta and red was reb. Frustrated, I would return to my bedroom and draw, filling the paper with houses, restaurants and offices.“When I grow up, I want my own store,”I told Mom, pointing to my drawings.“That’s great !”she said,“ but first you have to learn to read.”Later, I was diagnosed with dyslexia (失读症). So Mom took me to a learning centre, where I was given reading exercises. But I still had a hard time. Finally, I graduated, but I was afraid of my reading skills.“I’ll never get a job!”I cried to Mom later again.“Don’t focus on what you can’t do,” she comforted, “Concentrate on what you really can.”But what can I do? I wondered. Suddenly, I thought of the drawings I’d made as a child and my dream of h aving my own store. I enjoyed sales so much that over the next few years, I tried my hand at other businesses. Today, I watch over seven branches. We have 187 employees and $15 million in sales.While I’ll never be what my teachers might have wanted, I am a success--on my own terms. The other day a student sent me a card, reading: You gave me so much confidence. I hope to be like you when I am big. Tears of joy filled my eyes. This was my A, and I smiled.21.Why did the author feel so ashamed at school?A.His teacher ignored him in class.B.His teacher didn’t grade his essay.C.He failed to finish his essay in time.D.His classmates looked down upon him.22.Which of the following best describes the author’s feeling to his mother?A.Grateful.B.Guilty.C.Doubtful.D.Regretful.23.Which of the following can match the text?A.No pains, no gains.B.Never too old to learn.C.Every man has his value.D.Two heads are better than one.BWhen other nine-year-old kids were playing games, she was working at a petrol station. When other teens were studying or going out, she fought to find a place to sleep on the street. But she beat these terrible setbacks to win a highly competitive scholarship and gain entry into Harvard University. And her amazing story has inspired a movie, “Homeless to Harvard: The Liz Murray Story”.Liz Murray, a 22-year-old American girl, has been writing a real-life story of willpower and determination. Liz grew up with two drug-addicted parents. There was never enough food or warm clothes in the house. Liz was the only member of the family who had a job. Her mother had AIDS and died when Liz was just l5 years old. The effect of that loss became a turning point in her life. Connecting the environment in which she had grown up with how her mother had died. She decided to do something about it.Liz went back to school. She threw herself into her studies, never telling her teachers that she was homeless. At nigh t, she lived on the streets. “What drove me to live on had something to do with understanding, and by understanding that there was a whole o ther way of being. I had only experienced a small part of the society,” she wrote in her book Breaking Night.She admitted that she used envy to drive herself on. She used the benefits that come easily to others, such as a safe living environment, to encourage herself that “next to nothing could hold me down”. She finished high school in just two years and won a fu ll scholarship to study at Harvard University. But Liz decided to leave her top university a couple of months earlier this year in order to take care of her father, who has also developed AIDS. “I love my parents so much. They are drug addicts. But I never forget that they love me all the time. ”Liz wants moviegoers to come away with the idea that changing your life is “as simple as making a decision”.24.In which order did the following things happen to Liz?a. Her mother died of AIDS.b. She got admitted into Harvard.c. She worked at a petrol station.d. The movie about her life was put on.e. She had trouble finding a place to sleep.A.c, a, e, b, dB.a, b, c, e, dC.c, d, b, a, eD.b, e, a, d, c25.What actually made her go towards her goal?A.Envy and encouragement.B.Willpower and determination.C.Decisions and understanding.D.Love and respect for her parents.26.What does Liz mean by saying “What drove me to live on...I had only experienced a smallpart of the society”?A.She had little experience of social life.B.She could hardly understand the society.C.She would do something for her own life.D.She needed to travel more around the world.27.What does the passage mainly tell us?A.Why Liz loved her parents so much.B.How Liz made efforts to change her life.C.What a hard time Liz had in her childhood.D.How Liz managed to enter Harvard University.CA heated debate is currently going on in our town. Should we allow the cinema to be constructed in the Havenswood Shopping Center? There is just one large lot left to build on, and the theater would use up all of that space. Some people are excited at the idea of finally having our own movie theater. Others would rather travel ten miles to the nearest theater to keep our quiet town the way it is. They say it is enough to have Marvin’s Movie Video Rentals. After all, Marvin’s store keeps thousands of the latest videos.There are certainly benefits to renting videos. For one thing, you can plan your own schedule when it is convenient for you. You can relax on your sofa, and take a break whenever you need one.You can also talk to others without bothering any strangers seated nearby. In addition, it is a less expensive way to view a movie compared to going to a theater.On the other hand, seeing a movie in a theater is an experience all its own. First, you can see the movie on wide screen as the filmmaker intended. To be viewed on a television screen, a film must be changed in some way to make it smaller. One is the “pan-and-scan” method, which involves removing some of the details in the picture. The other way, called “letterboxing”, keeps the image th e way it is on the big screen, with one annoying exception; because the big-screen version is wide, the same picture on a television screen must be long and narrow.Another problem is sound. The sound from a television cannot compare to the sound system in a theater. Your experience of a movie improves when you can clearly hear all ofthe sounds. Furthermore, at home, viewing companions often talk during a movie, which makes you miss out on what’s happening in the film.Besides, having a movie theat er will not mean that you can’t still go to Marvin’s! You will just have a choice that you didn’t have before. Isn’t it time for Havenswood residents to enjoy a little progress?28.The underlined word “lot” in Paragraph 1 means “__________”.A.a great numberB.a complete groupC.an area of landD.a result of chance29.How does the author mainly state his point?A.By using examples.B.By making a comparison.C.By carrying out a surveyD.By presenting different people’s view.30.Why do some people think it is enough to have Marvin’s Movie Video Rentals?A.Marvin’s store stocks lots of recent videos.B.It’ very convenient to go to Marvin’s store.C.the owner of the store is friendly and helpful.D.It is more comfortable to see films at home than in the theater.31.What is the author’s attitude towards building a new cinema?A.Cautious.B.Approving.C.Doubtful.D.Worried.DStaying positive through the cold season could be your best defense against getting ill, a new American study suggests.In an experiment that exposed healthy volunteers to a cold or flu virus, researchers found that people with a generally sunny character were less likely to fall ill. The findings, published in the journal Psychosomatic Medici ne, build on evidence that a “positive emotional style” can help ward off the common cold and other illness.Researchers believe the reasons may be both objective―as in happiness improving immune function(免疫功能)―and subjective―as in happy people being l ess troubled by a sore throat or runny nose.“People with a positive emotional style may have different immune responses to the virus,” explained lead ing study author Dr Sheldon Cohen of Carnegie Mellon University in Pittsburgh.“And when they do get a cold, they may interpret their illness as being less severe.”Cohen and his colleagues have found in a previous study that happier people seemed less likely to catch a cold, but some questions remained as to whether the emotional quality itself had the effect.For the new study, the researchers had 193 healthy adults complete standard measures of personality qualities, physical health, and emotional “style”. Those who tended to be happy, energetic and easy-going were judged as having a positive emotional style, while those who were often unhappy, tense, and hostile had a negative style.Afterwards, the researchers gave them nose drops containing either a cold virus or a particular flu virus. Over the next six days, the volunteers reported on any aches, pains, sneezing they had, while the researchers collected objective data. Cohen and his colleagues found that happy people were less likely to develop a cold.What’s more, when happy folks did develop a cold, their symptoms were less severe than expected based on objective measures. On the contrary, people with negative characters were not at increased risk of developing a cold based on objective measures, though they did tend to get down about their symptoms.“We find that it’s really positive emotions that have the big effect,” Cohen said,“not the negative ones.”So can a bad-tempered person fight a cold by deciding to be happy?32.The purpose of Cohen’s new study was to _____.A.find effective ways to fight illnessesB.test people’s di fferent immune responses to cold virusC.tell differences between happy people and unhappy peopleD.examine whether health was related to emotional styles33.The underlined phrase “ward-off” in Paragraph 2 can be replaced by “_____”.A.get close toB.keep away fromC.get used toD.go on with34.How did Cohen reach his conclusion?A.By comparing the experimental results of different groups.B.By asking the volunteers to complete a form.C.By collecting data among people with a cold.D.By obser ving the volunteers’ symptoms.35.Cohen’s new study showed that_____.A.an emotional style is difficult to changeB.happy people are immune to cold virusC.people attitudes towards illnesses are differentD.happiness itself helps protect people from cold第二节、七选五根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。
山西省忻州市范亭中学高一英语联考试卷含解析
山西省忻州市范亭中学高一英语联考试卷含解析一、选择题1. In the past five years China ________ great _________.A. made, progressB. has made, progressesC. made, progressesD. has made, progress参考答案：B略2. No potatoes for me--I’m _______ a diet.A. atB. inC. onD. with参考答案：C3. –Has Ms Lin arrived already?--Yes. _____ I bring her in now or just let her wait outside, sir?A. WouldB. WillC. MustD. Shall参考答案：D句意：---林女士已经到了吗？----是的。
我是现在带她进来还是让她在外面等呢，先生？shall可用在问句中表示征求对方意见，主要用于第一、三人称，有“…好吗？”，“要不要”等意思。
故选D。
4. ________ is often the case, we have worked out the production plan.A. WhichB. WhenC. WhatD. As参考答案：d略5. Evidence shows that the _____ of a parent adds to the probability that a youngster will commit criminal activities.A. disappearanceB. absenceC. attentionD. concern 参考答案：B6. 单词辨音（共5小题，每小题1分，满分5分）1. July A diary B energy C reply D daily2. medicine A twice B medical C perfect D clinic3. reach A breathe B really C pleasure D heaven3. united A use B ugly C upstairs D put5. thirty A theatre B thus C although D father参考答案：1—5 CAAAA略7. By the end of next September, the university ______ 200,000 graduates.A. have turned outB. will turn outC. will have turned outD. had turned out参考答案：C8. We are considering ______ to Spain for our holidays.A. goingB. to goC. wentD. gone参考答案：A9. John gave Mary many new books.A. 状语B. 定语C. 宾语D. 宾补参考答案：B考查形容词。
山西省忻州市范亭中学2022年高二数学文月考试题含解析
山西省忻州市范亭中学2022年高二数学文月考试题含解析一、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. 已知，，，则（）（A）（B）（C）（D）参考答案：D略2. 直线x＋y-1=0到直线x sinα＋y cosα－1=0 (<α<)的角是（）A.α－B.－αC.α－D.－α参考答案：D3. 在空间直角坐标系O﹣xyz中，平面OAB的法向量为，O为坐标原点．已知P（﹣1，﹣3，8），则P到平面OAB的距离等于（）A．4 B．2 C．3 D．1参考答案：A【考点】点、线、面间的距离计算．【分析】直接利用空间点到平面的距离公式d=求解即可．【解答】解：平面OAB的一个法向量为=（2，﹣2，1），已知点P（﹣1，﹣3，8），则点P到平面OAB的距离d====4．故选：A．4. 在中，若，则的形状一定是（）A．锐角三角形 B．钝角三角形 C．直角三角形 D．等腰三角形参考答案：D5. 按流程图的程序计算，若开始输入的值为，则输出的的值是（）A. 95,57 B．47, 37 C．59,47D．47，47参考答案：A略6. 已知复数z满足，则z = （）A、-5B、5C、-3D、3参考答案：B7. 椭圆2x2+y2=6的焦点坐标是（）A．（±，0）B．（0，±）C．（±3，0）D．（0，±3）参考答案：B【考点】K4：椭圆的简单性质．【分析】根据题意，将椭圆的方程变形为标准方程，分析可得其焦点位置以及c的值，由焦点坐标公式即可得答案．【解答】解：根据题意，椭圆2x2+y2=6的标准方程为+=1，其焦点在y轴上，且c==，则其焦点坐标为（0，±），故选：B．8. （理）已知点P（x，y）在不等式组表示的平面区域上运动，则z=x-y的取值范围是（）A.［-2，-1］B.［-2，1］C.［-1，2］D.［1，2］参考答案：A略9. 算法的有穷性是指（）A．算法必须包含输出 B．算法中每个操作步骤都是可执行的C．算法的步骤必须有限 D．以上说法均不正确参考答案：C10. 中，，，，则的值是（）A． B． C． D．或参考答案：B略二、填空题:本大题共7小题,每小题4分,共28分11. 在平面直角坐标系中，曲线在处的切线方程是___________．参考答案：【分析】根据导数几何意义得切线斜率，再根据点斜式得结果. 【详解】因为，所以，因此在x＝0处的切线斜率为，因为x＝0时，所以切线方程是【点睛】本题考查导数几何意义，考查基本求解能力.属基础题.12. 若函数在[1，＋∞)上的最大值为，则a的值为________．参考答案：当x>时，f′(x)<0，f(x)单调递减；当－<x<时，f′(x)>0，f(x)单调递增；当x＝时，不合题意，∴13. 以下属于基本算法语句的是。
山西省忻州市范亭中学高三数学理联考试卷含解析
山西省忻州市范亭中学高三数学理联考试卷含解析一、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. 如图，网格纸上小正方形的边长为1，粗线画出的是某几何体的三视图，则此几何体的体积为A．B．C．D．3参考答案：D2. 已知函数是定义在上的奇函数，且当时，（其中是的导函数），若，，，则，，的大小关系是（）A．B．C．D．参考答案：C3. 已知、为双曲线：的左、右焦点，点为双曲线右支上一点，直线与圆相切，且，则双曲线的离心率为（）A．B．C．D．2参考答案：C4. 已知，平面上任意向量都可以唯一地表示为，则实数的取值范围是（）A． B． C． D．参考答案：C5. 若函数是R上的增函数，则实数a的取值范围为A.(0,3)B. (1,3)C.D. (1,+∞)参考答案：C6. 若，则直线与轴、轴围成的三角形的面积小于的概率是A. B. C. D.参考答案：C略7. 设集合，，则（）A．B．C．D．参考答案：【知识点】交集的运算.A1A解析：因为集合，化简为，所以，故选A.【思路点拨】先化简集合M，再求其交集即可。
8. 已知实数a,b满足则的零点所在的区间是( )A.(－2,－1)B. (－1,0)C.(0,1)D.(1,2)参考答案：B9. 已知函数f（x）=，若函数g（x）=f（x）+2x﹣a有三个零点，则实数a的取值范围是（）A．（0，+∞）B．（﹣∞，﹣1）C．（﹣∞，﹣3）D．（0，﹣3）参考答案：C【考点】根的存在性及根的个数判断．【分析】由题意可得需使指数函数部分与x轴有一个交点，抛物线部分与x轴有两个交点，判断x≤0，与x＞0交点的情况，列出关于a的不等式，解之可得答案．【解答】解：g（x）=f（x）+2x﹣a=，函数g（x）=f（x）+2x﹣a有三个零点，可知：函数图象的左半部分为单调递增指数函数的部分，函数图象的右半部分为开口向上的抛物线，对称轴为x=﹣a﹣1，最多两个零点，如上图，要满足题意，函数y=2x+2x是增函数，x≤0一定与x相交，过（0，1），g（x）=2x+2x﹣a，与x轴相交，1﹣a≥0，可得a≤1．还需保证x＞0时，抛物线与x轴由两个交点，可得：﹣a﹣1＞0，△=4（a+1）2﹣4（1﹣a）＞0，解得a＜﹣3，综合可得a＜﹣3，故选：C．10. 已知函数，定义函数给出下列命题：①；②函数是奇函数；③当时，若，，总有成立，其中所有正确命题的序号是A．②B．①③C．②③D．①②参考答案：C二、填空题:本大题共7小题,每小题4分,共28分11. 设是一元二次方程的两个虚根.若，则实数．参考答案：12. 在△ABC中，B(10，0)，直线BC与圆Γ：x2＋(y－5)2＝25相切，切点为线段BC的中点．若△ABC 的重心恰好为圆Γ的圆心，则点A的坐标为．参考答案：【答案解析】(0，15) 或 (－8，－1)解析：由已知得过点B与圆相切的切线长为10，则以B为圆心，切线长为半径的圆的方程为与已知圆的方程联立解得切点坐标为（0,0）或（4,8），所以C点坐标为（－10,0）或（－2,16），又已知圆心坐标为（0,5）设A点坐标为(x,y)，利用三角形重心坐标公式得A点坐标为(0，15) 或 (－8，－1).【思路点拨】本题的关键是先求切点坐标，可转化为两圆的交点问题，联立方程求切点坐标.13. 已知函数,则函数的图象在点处的切线方程是 .参考答案：4x-y-8=0略14. 已知角的终边上一点的坐标为，则角的最小正值为 .参考答案：因为点的坐标为，所以，即，所以当时，得角的最小正值为。
山西省原平市范亭中学2018-2019学年高二下学期期末考试数学(文)试题 含解析
高二第二学期数学（文科）期末试题第Ⅰ卷（共60分）一、选择题（每小题5分，共60分）1.已知集合{}|1 2 A x x =-<<， {}2|20 B x x x =+≤，则A B =I （ ）A. {}|0 2 x x <<B. {}|0 2 x x ≤<C. {}|10 x x -<<D.{}|10 x x -<≤【答案】D 【解析】 【分析】先求出集合A ， B ，然后根据交集的定义求出A B ⋂【详解】{}|12A x x =-<<Q ， {}{}2|20|20B x x x x x =+≤=-≤≤{}|10A B x x ∴⋂=-<≤故选D 【点睛】本题主要考查了集合的交集运算，属于基础题2.若201824(1)2i z i i =+-+，则复数z 在复平面内对应的点位于（ ） A. 第一象限 B. 第二象限C. 第三象限D. 第四象限【答案】B 【解析】 【分析】利用复数的四则运算可得 【详解】201824(1)242112221i i i i z i i ii =+-=-=-=+-+--，故复数z 在复平面内对应的点位于第二象限，故选B. 【点睛】本题考查复数的运算，属于基础题.3.已知双曲线2221x y -=的一个焦点为F ,则焦点F 到其中一条渐近线的距离为（ ）A. 2B. 1 D.12【答案】C 【解析】 【分析】求得双曲线的a ，b ，c ，焦点F 的坐标和一条渐近线方程，由点到直线的距离公式计算即可得到所求．【详解】双曲线2221x y -=的a=1，b=2，，右焦点F ，0），一条渐近线方程为x 0=，则F 到渐近线的距离为=2．故选：C ．【点睛】本题考查双曲线的方程和性质，主要考查渐近线方程的运用，点到直线的距离公式，属于基础题．4.设函数()(1)xf x x e =+，则'(1)f =（ ）A. 1B. 2C. 3+eD. 3e【答案】D 【解析】 【分析】对函数()()1xf x x e =+求导，然后把1x =代入即可.详解】()()()()112,xxxxf x x e e x e x e =+=++=+'Q()()1'1123,f e e ∴=+=故选C.【点睛】本题考查函数在某一点出的导数，属基础题.5.已知4cos 45πα⎛⎫-= ⎪⎝⎭，则 sin 2α=（ ） A.725B.2425C. 725±D. 2425±【答案】A 【解析】 【分析】利用两角差的余弦可得sin cos αα+的值，平方后得到sin 2α的值.【详解】因为4cos 45πα⎛⎫-= ⎪⎝⎭，故()4cos sin 25αα+=即cos sin 5αα+=， 故2232sin 2sin cos cos 25αααα++=即7sin 225α=，故选A. 【点睛】三角函数的中的化简求值问题，我们往往从次数的差异、函数名的差异、结构的差异和角的差异去分析，处理次数差异的方法是升幂降幂法，解决函数名差异的方法是弦切互化，而结构上差异的处理则是已知公式的逆用等，最后角的差异的处理则往往是用已知的角去表示未知的角.6.若,a b 表示直线，α表示平面，且b α⊂，则“a b ∥”是“a P α” （ ）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【答案】D 【解析】 【分析】依据充分条件和必要条件的定义去判断.【详解】“a b ∥”推不出“a P α”，因为a α⊂可能成立，“a P α”也推不出“a b ∥”，,a b 可能异面，故“a b ∥”是“a P α”的既不充分也不必要条件，故选D.【点睛】充分性与必要性的判断，可以依据命题的真假来判断，若“若p 则q ”是真命题，“若q 则p ”是假命题，则p 是q 的充分不必要条件；若“若p 则q ”是真命题，“若q 则p ”是真命题，则p 是q 的充分必要条件；若“若p 则q ”是假命题，“若q 则p ”是真命题，则p 是q 的必要不充分条件；若“若p 则q ”是假命题，“若q 则p ”是假命题，则p 是q 的既不充分也不必要条件.7.某雷达测速区规定：凡车速大于或等于70/km h 的汽车视为“超速”，并将受到处罚，如图是某路段的一个检测点对200辆汽车的车速进行检测所得结果的频率分布直方图，则从图中可以看得出将被处罚的汽车大约有 ( )A. 80辆B. 60辆C. 40辆D. 20辆【答案】C 【解析】 【分析】根据车速大于或等于70/km h 的汽车的频率可得将被处罚的汽车数量. 【详解】车速大于或等于70/km h 的汽车的频率为0.02100.2⨯=， 故将被处罚的汽车数量为2000.240⨯=（辆），故选C. 【点睛】本题考查频率分布直方图的应用，属于基础题.8.已知{}n a 是正项等比数列，若134a a =，2416a a =，则10S 的值是（ ） A. 1024 B. 1023C. 512D. 511【答案】B 【解析】 【分析】根据题设条件算出基本量公比q 及1a ，利用公式可求10S . 【详解】设{}n a 的公比为()0q q >，则211311416a a q a q a q ⎧⨯=⎨⨯=⎩，故121q a =⎧⎨=⎩，所以()1010112102312S ⨯-==-， 故选B.【点睛】等差数列或等比数列的处理有两类基本方法：（1）利用基本量即把数学问题转化为关于基本量的方程或方程组，再运用基本量解决与数列相关的问题；（2）利用数列的性质求解即通过观察下标的特征和数列和式的特征选择合适的数列性质处理数学问题．9.过点(3,1)A -且在两坐标轴上截距相等的直线有（ ） A. 1条 B. 2条C. 3条D. 4条【答案】B 【解析】当截距相等均为0时，直线方程为13y x =-； 当截距相等不为0时，设方程为()10x ya a a+=≠，代入点()3,1-得2a =，直线方程为2x y +=，所以共有2条，故选择B.10.设sin5a π=，b =，231=4c ⎛⎫ ⎪⎝⎭，则（ ）A ． a c b <<. B. b a c << C. c ＜a ＜b D. c ＜b ＜a 【答案】C 【解析】 【分析】利用三角函数、对数函数、指数函数的单调性直接求解．【详解】∵24331111 11? 265422sin a sin b c ＜＜，，（）（）＜，ππ====== ∴c＜a ＜b ． 故选：C ．【点睛】本题考查三个数的大小的求法，是基础题，解题时要认真审题，注意函数性质的合理运用．11.如图，,E F 分别是三棱锥P ABC -的棱AP BC 、的中点，10PC =，6AB =，7EF =，则异面直线AB 与PC 所成的角为（ ）A. 120oB. 60oC. 45oD. 30o【答案】B 【解析】 【分析】取PB 的中点H ，连接,EH FH ，利用余弦定理可求EHF ∠的余弦值，从而得到异面直线AB 与PC 所成的角．【详解】取PB 的中点H ，连接,EH FH ，因为,F H 为中点，故152FH PC ==且FH PC P 同理3EH =，EH AB P ， 故EHF ∠或其补角为异面直线所成的角．在FEH ∆中，925491cos 2352EHF +-∠==-⨯⨯，因为0180EHF ︒<∠<︒，所以120EHF ∠=︒，故异面直线AB 与PC 所成的角为60o ， 故选B ．【点睛】空间中的角的计算，可以建立空间直角坐标系把角的计算归结为向量的夹角的计算，也可以构建空间角，把角的计算归结平面图形中的角的计算．12.已知圆()()22:1C x a y b -+-=，设平面区域70,{30,0x y x y y +-≤Ω=-+≥≥，若圆心C ∈Ω,且圆C 与x 轴相切，则22a b +的最大值为 （ ） A. 5 B. 29C. 37D. 49【答案】C 【解析】试题分析：作出可行域如图，圆C ：（x －a ）2＋（y －b ）2＝1的圆心为，半径的圆，因为圆心C∈Ω，且圆C 与x 轴相切，可得，所以所以要使a 2＋b 2取得的最大值，只需取得最大值，由图像可知当圆心C 位于B 点时，取得最大值，B 点的坐标为，即时是最大值.考点：线性规划综合问题.第II 卷（共90分）二、填空题（每小题5分,共20分）13.某校对高三年级1 600名男女学生的视力状况进行调查，现用分层抽样的方法抽取一个容量是200的样本，已知样本中女生比男生少10人，则该校高三年级的女生人数是________． 【答案】760 【解析】设样本中女生有x 人，则男生有(10)x +人，则(10)200x x ++=，即95x =， 设该校高三年级的女生有y 人，则由分层抽样的特点（等比例抽样），得951600200y =，解得760y =，即该校高三年级的女生人数是760.14.不等式242133x x x+-+⎛⎫> ⎪⎝⎭的解集为__________．【答案】（-1，4） 【解析】分析：利用指数函数的单调性，转化为二次不等式问题. 详解：由242133x xx+-+⎛⎫> ⎪⎝⎭可得：22433xxx -+-->∴224x x x -+>--，即2340x x --< ∴不等式242133x x x+-+⎛⎫> ⎪⎝⎭解集为（-1，4）故答案为：（-1，4）点睛：本题考查指数型不等式的解法，解题关键是利用指数函数的单调性转化为一元二次不等式问题即可.15.已知函数()213,1log ,1x x x f x x x ⎧-+≤⎪=⎨>⎪⎩ ，若对任意的x ∈R ，不等式()f x m ≤恒成立，则实数m 的取值范围为_______． 【答案】14m ≥ 【解析】 【分析】求()f x 的最大值后可得实数m 的取值范围．【详解】当1x ≤时，()22111244f x x x x ⎛⎫=-+=--+≤ ⎪⎝⎭，当12x =时等号成立， 当1x >时，()13log 0f x x =<，故()max 14f x =，故14m ≥，填14m ≥． 【点睛】本题考查分段函数的最值，注意不等式的恒成立问题可以归结为函数的最值问题进行讨论．16.已知函数()()323321f x x ax a x =++++恰有三个单调区间，则实数a 的取值范围是__________．【答案】1a <-或2a > 【解析】分析：求出函数的导函数，利用导数有两个不同的零点，说明函数恰好有三个单调区间，从而求出a 的取值范围．详解：∵函数()()323321f x x ax a x =++++，∴f′（x ）=3x 2+6ax+()32a +，由函数f （x ）恰好有三个单调区间，得f′（x ）有两个不相等的零点， ∴3x 2+6ax+()32a +=0满足：△=236a ﹣()362a +＞0，解得1a <-或2a >， 故答案为：1a <-或2a >．点睛：本题考查了单调性与极值点的关系，解题关键利用图象分析出恰有三个单调区间等价于函数()f x 有两个极值点.三、解答题（解答应写出必要的文字说明、证明过程或演算步骤）17.已知，圆C ：228120x y y +-+=，直线l ：20ax y a ++=.（1）当a 为何值时，直线l 与圆C 相切；（2）当直线l 与圆C 相交于A 、B两点，且AB =l 的方程． 【答案】（1）34a =-（2）7140x y -+=或20x y -+=． 【解析】 【分析】（1）直线与圆相切的等价条件为圆心到直线距离等于半径，根据该等价条件建立关于a 的方程即可求出.（2）利用关系2222AB d r ⎛⎫+= ⎪⎝⎭，求出圆心到直线距离d，再由d =即可求出a ，从而求出直线l 的方程.【详解】（1）根据题意，圆C ：x 2+y 2-8x+12=0，则圆C 的方程为22(4)4x y -+=，其圆心为（4，0），半径r=2；若直线l 与圆C=2，解可得a =-34； （2）设圆心C 到直线l 的距离为d ，则有（AB 2）2+d 2=r 2，即2+d 2=4，解可得则有，解可得a =-1或-7；则直线l 的方程为x-y-2=0或x-7y-14=0．【点睛】主要考查了直线方程的求解，以及直线与圆的位置关系，属于基础题.18.某学生对其亲属30人的饮食习惯进行一次调查，并用如图所示的茎叶图表示30人的饮食指数（说明：图中饮食指数低于70的人，饮食以蔬菜为主；饮食指数高于70的人，饮食以肉类为主）（1）根据以上数据完成下列22⨯列联表.（2）能否有99%的把握认为其亲属的饮食习惯与年龄有关？并写出简要分析.主食蔬菜主食肉食总计50岁以下50岁以上总计()2P K k>0.150.100.050.0250.0100.0050.001 k 2.072 2.706 3.841 5.024 6.6357.87910.828参考公式：22()()()()()n ad bcKa b c d a c b d-=++++，其中n a b c d=+++【答案】（1）列联表见解析；（2）有.【解析】【分析】（1）根据茎叶图可得列联表.（2）利用列联表可计算2κ的值，利用临界值表可知有99%把握认为亲属的饮食习惯与年龄有关.【详解】（1）主食蔬菜主食肉食总计（2）2230(8128)10 6.63512182010κ-==>⨯⨯⨯ 有99%把握认为亲属的饮食习惯与年龄有关.【点睛】本题考查独立性检验，属于容易题.19.已知函数()f x 是定义在R 上的偶函数，()00f =，当0x >时，()()13log 1f x x =+.（1）求函数()f x 的解析式； （2）解不等式()212f x ->-.【答案】（1）1313log (1),0(){0,0log (1),0x x f x x x x +>==-<（2）33x -<<【解析】分析：（1）设x ＜0，可得﹣x ＞0，则f （﹣x ）=()13log 1x -+，再由函数f （x ）是偶函数求出x ＜0时的解析式；（2）由()82f =-，f （x ）是偶函数，不等式f （x 2﹣1）＞﹣2可化为f （|x 2﹣1|）＞()8f ．利用函数f （x ）在（0，+∞）上是减函数，可得218x -<，求解绝对值的不等式可得原不等式的解集．详解：（1）当0x <时()()()130,log 1x f x f x x ->=-=-+()()()1313log 1,00,0log 1,0x x f x x x x ⎧+>⎪⎪∴==⎨⎪-<⎪⎩,（2）()()()()2282118f f x fxf =-∴-=->Q又()f x 在()0,+∞单调递减 218x ∴-<2818x ∴-<-< 2933x x ∴<∴-<<点睛：本题考查函数解析式的求法，考查了利用函数的单调性求解不等式，体现了数学转化思想方法，属于中档题．20.已知：三棱锥A BCD -中，等边ABC ∆边长为2，2,2BD DC AD ===.（1）求证：AD BC ⊥；（2） 求证：平面ABC ⊥平面BCD .【答案】（1）证明见解析；（2）证明见解析. 【解析】 【分析】（1）取BC 中点E ，连接,AE DE ，可以证明BC ⊥平面ADE ，从而可证AD BC ⊥. （2）可证AE ⊥平面BCD ，从而得到平面ABC ⊥平面BCD .【详解】（1）取BC 中点E ，连接,AE DE ，则AE BC ⊥，DE BC ⊥，AE DE E =I ，所以BC ⊥平面ADE ，因为AD ⊂平面ADE ，AD BC ⊥；（2）21AD AE DE ===Q ,，所以222AD AE DE =+，故AE DE ⊥，又AE BC ⊥，DE BC E ⋂=，所以AE ⊥平面BCD ， 又 AE ⊂平面ABC , ∴ 平面ABC ⊥平面BCD .【点睛】线线垂直的判定可由线面垂直得到，也可以由两条线所成的角为2π得到，而线面垂直又可以由面面垂直得到，解题中注意三种垂直关系的转化. 面面垂直的判定可由线面垂直或两个平面构成的二面角为直二面角得到.21.已知函数3()31f x x x =--，其定义域是[]3,2-.（1）求()f x 在其定义域内的极大值和极小值；（2）若对于区间[]3,2-上的任意12,x x ，都有12()()f x f x t -≤，求t 的最小值. 【答案】（1）()f x 极大值为()11f -=,极小值为(1)3f =-；（2）20. 【解析】 【分析】（1）求出()f x '，讨论其符号后可得函数的极值.（2）求出()f x 在[]3,2-上的最大值和最小值后可得t 的最小值.【详解】（2）求导得2()33f x x '=- 令()0f x '=得1x =±, ∴1x =±为极值点，令()0f x '>得31x -≤<-或12x <≤，令()0f x '<得11x -<<， 列表讨论如下：所以()f x 极大值为()11f -=,极小值为(1)3f =- （2）需max min ()()f x f x t -≤即可， 由(1)可知max min ()1,()19f x f x ==-,max min ()()1(19)20t f x f x ≥-=--=,即20t ≥，所以t 的最小值为20 .【点睛】函数的极值刻画了函数局部性质，它可以理解为函数图像具有“局部最低”的特性，用数学语言描述则是：“在0x 的附近的任意x ，有()()0f x f x >（()()0f x f x <）” ．另外如果()f x 在0x 附近可导且0x 的左右两侧导数的符号发生变化，则0x x =必为函数的极值点．函数不等式的证明，可归结为函数的最值来处理.22.已知曲线122cos :12sin x t C y t=-+⎧⎨=+⎩（t 为参数），曲线2:4cos sin 10C ρθρθ--=.（设直角坐标系x 正半轴与极坐系极轴重合）（1）求曲线1C 普通方程与直线2C 的直角坐标方程； （2）若点P 在曲线1C 上，Q 在直线2C 上，求PQ 的最小值.【答案】（1）22(2)(1)4x y ++-=, 410x y --=；（22-. 【解析】 【分析】（1）消去参数t 后可得曲线1C 的普通方程，利用cos sin x y ρθρθ=⎧⎨=⎩可得曲线2C 的直角方程.（2）因为1C 为圆，2C 为直线，所以最小值为圆心到直线的距离减去圆的半径. 【详解】（1）对于1C ，消去参数t 可得()2222(2)(1)4cos sin 4x y t t ++-=+=，因为cos sin x y ρθρθ=⎧⎨=⎩，故410x y --=.（2）圆心1C 到直线410x y --=的距离为d ==.故PQ 的最小值为217-，填217-. 【点睛】极坐标方程与直角方程的互化，关键是cos sin x y ρθρθ=⎧⎨=⎩，必要时需在给定方程中构造cos ,sin ρθρθ．与圆有关的最值问题，可以转为圆心到几何对象的距离最值问题．已知曲线的参数方程，求其普通方程时，应利用平方、反解等方法消去参数.。
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一、复数选择题1．若复数(1)()(i a i i -+是虚数单位)为纯虚数，则实数a 的值为（ ） A ．2 B ．1 C ．0D ．1-2．设复数(,)z a bi a R b R =+∈∈，它在复平面内对应的点位于虚轴的正半轴上，且有1z =，则a b +=（ ）A ．-1B ．0C ．1D ．23．i =（ ）A ．i -B ．iC i -D i4．212ii+=-（ ） A ．1B ．−1C ．i -D ．i5．若复数z 满足()13i z i +=+（其中i 是虚数单位），复数z 的共轭复数为z ，则（ ） A ．z 的实部是1 B ．z 的虚部是1C ．z =D ．复数z 在复平面内对应的点在第四象限6．欧拉是瑞士著名数学家，他首先发现：e cos isin i θθθ=+（e 为自然对数的底数，i 为虚数单位），此结论被称为“欧拉公式”，它将指数函数的定义域扩大到复数集，建立了三角函数和指数函数的关系．根据欧拉公式可知，i e π＝（ ） A ．1B ．0C ．－1D ．1＋i7．若复数(2)z i i =+（其中i 为虚数单位），则复数z 的模为（ ）A ．5B C ．D ．5i8．已知复数()123z i i +=- （其中i 是虚数单位），则z 在复平面内对应点在（ ） A ．第一象限 B ．第二象限 C ．第三象限 D ．第四象限 9．已知a 为正实数，复数1ai +（i 为虚数单位）的模为2，则a 的值为（ ）A B ．1C ．2D ．310．若复数()()24z i i =--，则z =（ ） A ．76i -- B ．76-+iC ．76i -D ．76i +11．若(1)2z i i -=，则在复平面内z 对应的点位于（ ）A ．第一象限B ．第二象限C ．第三象限D ．第四象限12．复数12iz i=+（i 为虚数单位）在复平面内对应的点位于（ ）A ．第一象限B ．第二象限C ．第三象限D ．第四象限13．复数z 对应的向量OZ 与(3,4)a =共线，对应的点在第三象限，且10z =，则z =（ ） A ．68i +B ．68i -C ．68i --D ．68i -+14．已知i 是虚数单位，设复数22ia bi i-+=+，其中,a b ∈R ，则+a b 的值为（ ） A ．75B ．75-C ．15D ．15-15．题目文件丢失！二、多选题16．已知复数202011i z i+=-(i 为虚数单位)，则下列说法错误的是（ ）A ．z 的实部为2B ．z 的虚部为1C ．z i =D ．||z =17．若复数351iz i-=-，则（ ）A ．z =B ．z 的实部与虚部之差为3C ．4z i =+D ．z 在复平面内对应的点位于第四象限 18．下面是关于复数21iz =-+的四个命题，其中真命题是（ ）A ．||z =B ．22z i =C ．z 的共轭复数为1i -+D ．z 的虚部为1-19．已知复数12z =-，则下列结论正确的有（ ）A ．1z z ⋅=B ．2z z =C ．31z =-D ．2020122z =-+20．已知复数122z =-+（其中i 为虚数单位，，则以下结论正确的是（ ）． A ．20zB ．2z z =C ．31z =D ．1z =21．设复数z 满足1z iz+=，则下列说法错误的是（ ） A ．z 为纯虚数B ．z 的虚部为12i -C ．在复平面内，z 对应的点位于第三象限D ．2z =22．若复数z 满足()234z i i +=+（i 为虚数单位），则下列结论正确的有（ ）A ．z 的虚部为3B ．z =C ．z 的共轭复数为23i +D ．z 是第三象限的点23．已知复数1cos 2sin 222z i ππθθθ⎛⎫=++-<< ⎪⎝⎭（其中i 为虚数单位），则（ ）A ．复数z 在复平面上对应的点可能落在第二象限B ．z 可能为实数C ．2cos z θ=D ．1z 的实部为12- 24．下列关于复数的说法，其中正确的是（ ） A ．复数(),z a bi a b R =+∈是实数的充要条件是0b = B ．复数(),z a bi a b R =+∈是纯虚数的充要条件是0b ≠ C ．若1z ，2z 互为共轭复数，则12z z 是实数D ．若1z ，2z 互为共轭复数，则在复平面内它们所对应的点关于y 轴对称 25．已知1z ，2z 为复数，下列命题不正确的是（ ） A ．若12z z =，则12=z z B ．若12=z z ，则12z z =C ．若12z z >则12z z >D ．若12z z >，则12z z >26．已知复数z 满足(2i)i z -=(i 为虚数单位)，复数z 的共轭复数为z ，则（ ）A ．3||5z =B ．12i5z +=-C ．复数z 的实部为1-D ．复数z 对应复平面上的点在第二象限 27．以下为真命题的是（ ） A ．纯虚数z 的共轭复数等于z -B ．若120z z +=，则12z z =C ．若12z z +∈R ，则1z 与2z 互为共轭复数D ．若120z z -=，则1z 与2z 互为共轭复数 28．已知i 为虚数单位，下列说法正确的是( ) A ．若,x y R ∈，且1x yi i +=+，则1x y == B ．任意两个虚数都不能比较大小C ．若复数1z ，2z 满足22120z z +=，则120z z == D ．i -的平方等于129．已知复数z 满足23z z iz ai ⋅+=+，a R ∈，则实数a 的值可能是（ ） A ．1B ．4-C ．0D ．530．对任意1z ，2z ，z C ∈，下列结论成立的是（ ） A ．当m ，*n N ∈时，有m n m n z z z +=B ．当1z ，2zC ∈时，若22120z z +=，则10z =且20z = C ．互为共轭复数的两个复数的模相等，且22||||z z z z ==⋅D ．12z z =的充要条件是12=z z【参考答案】***试卷处理标记，请不要删除一、复数选择题 1．D 【分析】由复数乘法化复数为代数形式，然后根据复数的分类求解． 【详解】 ，它为纯虚数， 则，解得． 故选：D ． 解析：D 【分析】由复数乘法化复数为代数形式，然后根据复数的分类求解． 【详解】2(1)()1(1)i a i a i ai i a a i -+=+--=++-，它为纯虚数，则1010a a +=⎧⎨-≠⎩，解得1a =-． 故选：D ．2．C 【分析】根据复数的几何意义得． 【详解】∵它在复平面内对应的点位于虚轴的正半轴上，∴，又，∴， ∴． 故选：C ．解析：C 【分析】根据复数的几何意义得,a b ． 【详解】∵z 它在复平面内对应的点位于虚轴的正半轴上，∴0a =，又1z =，∴1b =， ∴1a b +=． 故选：C ．【分析】由复数除法运算直接计算即可. 【详解】.故选：B.解析：B【分析】由复数除法运算直接计算即可.【详解】()211iii i++==--.故选：B.4．D【分析】利用复数的除法运算即可求解.【详解】，故选：D解析：D【分析】利用复数的除法运算即可求解.【详解】()()()()2221222255 121212145i ii i i iii i i i+++++====--+-，故选：D5．C【分析】利用复数的除法运算求出，即可判断各选项. 【详解】，，则的实部为2，故A错误；的虚部是，故B错误；，故C正；对应的点为在第一象限，故D错误.故选：C.【分析】利用复数的除法运算求出z ，即可判断各选项. 【详解】()13i z i +=+，()()()()3132111i i i z i i i i +-+∴===-++-， 则z 的实部为2，故A 错误；z 的虚部是1-，故B 错误；z ==，故C 正；2z i =+对应的点为()2,1在第一象限，故D 错误.故选：C.6．C 【分析】利用复数和三角函数的性质，直接代入运算即可 【详解】 由题意可知＝， 故选C解析：C 【分析】利用复数和三角函数的性质，直接代入运算即可 【详解】由题意可知i e π＝cos sin 101i ππ+=-+=-， 故选C7．B 【分析】由已知等式，利用复数的运算法则化简复数，即可求其模. 【详解】 ，所以， 故选：B解析：B 【分析】由已知等式，利用复数的运算法则化简复数，即可求其模. 【详解】(2)21z i i i =+=-，所以|z |=故选：B8．D先由复数的运算化简复数z，再运用复数的几何表示可得选项. 【详解】由已知得，所以复数z在复平面上所对应的点为，在第四象限，故选：D.解析：D【分析】先由复数的运算化简复数z，再运用复数的几何表示可得选项.【详解】由已知得()()()()312317171+21+212555i ii iz ii i i----====--，所以复数z在复平面上所对应的点为17,55⎛⎫-⎪⎝⎭，在第四象限，故选：D.9．A【分析】利用复数的模长公式结合可求得的值.【详解】，由已知条件可得，解得.故选：A.解析：A【分析】利用复数的模长公式结合0a>可求得a的值.【详解】a>，由已知条件可得12ai+==，解得a=故选：A.10．D【分析】由复数乘法运算求得，根据共轭复数定义可求得结果. 【详解】，.故选：.解析：D【分析】由复数乘法运算求得z，根据共轭复数定义可求得结果.【详解】()()2248676z i i i i i =--=-+=-，76z i ∴=+.故选：D .11．B 【分析】先求解出复数，然后根据复数的几何意义判断. 【详解】 因为，所以，故对应的点位于复平面内第二象限. 故选：B. 【点睛】本题考查复数的除法运算及复数的几何意义，属于基础题. 化简计解析：B 【分析】先求解出复数z ，然后根据复数的几何意义判断. 【详解】因为(1)2z i i -=，所以()212112i i i z i i +===-+-， 故z 对应的点位于复平面内第二象限. 故选：B. 【点睛】本题考查复数的除法运算及复数的几何意义，属于基础题. 化简计算复数的除法时，注意分子分母同乘以分母的共轭复数.12．A 【分析】对复数进行分母实数化，根据复数的几何意义可得结果. 【详解】 由，知在复平面内对应的点位于第一象限， 故选：A. 【点睛】本题主要考查了复数除法的运算以及复数的几何意义，属于基础题解析：A 【分析】对复数z 进行分母实数化，根据复数的几何意义可得结果. 【详解】 由()()()122112121255i i i z i i i i -===+++-，知在复平面内对应的点21,55⎛⎫⎪⎝⎭位于第一象限， 故选：A. 【点睛】本题主要考查了复数除法的运算以及复数的几何意义，属于基础题.13．D 【分析】设，根据复数对应的向量与共线，得到，再结合求解. 【详解】 设，则复数对应的向量， 因为向量与共线， 所以， 又， 所以， 解得或，因为复数对应的点在第三象限， 所以， 所以，，解析：D 【分析】设(,)z a bi a R b R =+∈∈，根据复数z 对应的向量OZ 与(3,4)a =共线，得到43a b =，再结合10z =求解.【详解】设(,)z a bi a R b R =+∈∈， 则复数z 对应的向量(),OZ a b =， 因为向量OZ 与(3,4)a =共线， 所以43a b =， 又10z =， 所以22100+=a b ， 解得68a b =-⎧⎨=-⎩或68a b =⎧⎨=⎩， 因为复数z 对应的点在第三象限，所以68a b =-⎧⎨=-⎩，所以68z i =--，68z i =-+， 故选：D14．D 【分析】先化简，求出的值即得解. 【详解】 ， 所以. 故选：D解析：D 【分析】 先化简345ia bi -+=，求出,ab 的值即得解. 【详解】22(2)342(2)(2)5i i ia bi i i i ---+===++-，所以341,,555a b a b ==-∴+=-. 故选：D15．无二、多选题 16．AC 【分析】根据复数的运算及复数的概念即可求解. 【详解】 因为复数， 所以z 的虚部为1，， 故AC 错误，BD 正确. 故选：AC解析：AC 【分析】根据复数的运算及复数的概念即可求解. 【详解】因为复数2020450511()22(1)11112i i i z i i i i +++=====+---，所以z 的虚部为1，||z =故AC 错误，BD 正确.故选：AC17．AD【分析】根据复数的运算先求出复数z ，再根据定义、模、几何意义即可求出.【详解】解：，，z 的实部为4，虚部为，则相差5，z 对应的坐标为，故z 在复平面内对应的点位于第四象限，所以AD 正解析：AD【分析】根据复数的运算先求出复数z ，再根据定义、模、几何意义即可求出.【详解】 解：()()()()351358241112i i i i z i i i i -+--====---+，z ∴==z 的实部为4，虚部为1-，则相差5，z 对应的坐标为()41-,，故z 在复平面内对应的点位于第四象限，所以AD 正确， 故选：AD.18．ABCD【分析】先根据复数的除法运算计算出，再依次判断各选项.【详解】，，故A 正确；，故B 正确；的共轭复数为，故C 正确；的虚部为，故D 正确； 故选：ABCD.【点睛】本题考查复数的除法解析：ABCD【分析】先根据复数的除法运算计算出z ，再依次判断各选项.【详解】()()()2121111i z i i i i --===---+-+--，z ∴==，故A 正确；()2212z i i =--=，故B 正确；z 的共轭复数为1i -+，故C 正确；z 的虚部为1-，故D 正确；故选：ABCD.【点睛】本题考查复数的除法运算，以及对复数概念的理解，属于基础题.19．ACD【分析】分别计算各选项的值，然后判断是否正确，计算D 选项的时候注意利用复数乘方的性质.【详解】因为，所以A 正确；因为，，所以，所以B 错误；因为，所以C 正确；因为，所以，所以D 正确解析：ACD【分析】 分别计算各选项的值，然后判断是否正确，计算D 选项的时候注意利用复数乘方的性质.【详解】因为111312244z z ⎛⎫⎛⎫=+= ⎪⎪ ⎪⎪⎝⎭⎭=⎝⋅，所以A 正确；因为22112222z ⎛⎫-=-- ⎪ ⎪⎝⎭=，12z =，所以2z z ≠，所以B 错误；因为3211122z z z ⎛⎫⎛⎫=⋅=-=- ⎪⎪ ⎪⎪⎝⎭⎝⎭，所以C 正确；因为6331z z z =⋅=，所以()202063364431112222z z z z z ⨯+⎛⎫===⋅=-⋅-=-+ ⎪ ⎪⎝⎭，所以D 正确，故选：ACD.【点睛】本题考查复数乘法与乘方的计算，其中还涉及到了共轭复数的计算，难度较易.20．BCD【分析】计算出，即可进行判断.【详解】，，故B 正确，由于复数不能比较大小，故A 错误；，故C 正确；，故D 正确.故选：BCD.【点睛】本题考查复数的相关计算，属于基础题.解析：BCD【分析】 计算出23,,,z z z z ，即可进行判断.【详解】122z =-+， 221313i i=2222z z ，故B 正确，由于复数不能比较大小，故A 错误； 33131313i i i 1222222z ，故C 正确； 2213122z，故D 正确.故选：BCD.【点睛】 本题考查复数的相关计算，属于基础题.21．AB【分析】先由复数除法运算可得，再逐一分析选项，即可得答案.【详解】 由题意得：，即，所以z 不是纯虚数，故A 错误；复数z 的虚部为，故B 错误；在复平面内，对应的点为，在第三象限，故C 正确解析：AB【分析】先由复数除法运算可得1122z i =--，再逐一分析选项，即可得答案. 【详解】由题意得：1z zi +=，即111122z i i -==---， 所以z 不是纯虚数，故A 错误； 复数z 的虚部为12-，故B 错误； 在复平面内，z 对应的点为11(,)22--，在第三象限，故C 正确；2z ==，故D 正确. 故选：AB【点睛】本题考查复数的除法运算，纯虚数、虚部的概念，复平面内点所在象限、复数求模的运算等知识，考查计算求值的能力，属基础题.22．BC【分析】利用复数的除法求出复数，利用复数的概念与几何意义可判断各选项的正误.【详解】，，所以，复数的虚部为，，共轭复数为，复数在复平面对应的点在第四象限. 故选：BD.【点睛】本题考解析：BC【分析】利用复数的除法求出复数z ，利用复数的概念与几何意义可判断各选项的正误.【详解】()234z i i +=+，34232i z i i+∴=-=-+，所以，复数z 的虚部为3-，z =共轭复数为23i +，复数z 在复平面对应的点在第四象限.故选：BD.【点睛】 本题考查复数的四则运算、虚部、模、共轭复数以及几何意义，考查计算能力，属于基础题.23．BC【分析】由可得，得，可判断A 选项，当虚部，时，可判断B 选项，由复数的模计算和余弦的二倍角公式可判断C 选项，由复数的运算得，的实部是，可判断D 选项.【详解】因为，所以，所以，所以，所以A 选解析：BC【分析】 由22ππθ-<<可得2πθπ-<<，得01cos22θ<+≤，可判断A 选项，当虚部sin 20θ=，,22ππθ⎛⎫∈- ⎪⎝⎭时，可判断B 选项，由复数的模计算和余弦的二倍角公式可判断C 选项，由复数的运算得11cos 2sin 212cos 2i z θθθ+-=+，1z 的实部是1cos 2122cos 22θθ+=+，可判断D 选项.【详解】 因为22ππθ-<<，所以2πθπ-<<，所以1cos21θ-<≤，所以01cos22θ<+≤，所以A 选项错误；当sin 20θ=，,22ππθ⎛⎫∈- ⎪⎝⎭时，复数z 是实数，故B 选项正确；2cos z θ===，故C 选项正确：()()111cos 2sin 21cos 2sin 21cos 2sin 21cos 2sin 21cos 2sin 212cos 2i i z i i i θθθθθθθθθθθ+-+-===+++++-+，1z 的实部是1cos 2122cos 22θθ+=+，故D 不正确. 故选：BC【点睛】本题主要考查复数的概念，复数模的计算，复数的运算，以及三角恒等变换的应用，属于中档题.24．AC【分析】根据复数的有关概念和充分条件和必要条件的定义进行判断即可．【详解】解：对于：复数是实数的充要条件是，显然成立，故正确；对于：若复数是纯虚数则且，故错误；对于：若，互为共轭复数解析：AC【分析】根据复数的有关概念和充分条件和必要条件的定义进行判断即可．【详解】解：对于A ：复数(),z a bi a b R =+∈是实数的充要条件是0b =，显然成立，故A 正确；对于B ：若复数(),z a bi a b R =+∈是纯虚数则0a =且0b ≠，故B 错误；对于C ：若1z ，2z 互为共轭复数，设()1,z a bi a b R =+∈，则()2,z a bi a b R =-∈，所以()()2122222z a bi a bi a b b z i a =+-=-=+是实数，故C 正确； 对于D ：若1z ，2z 互为共轭复数，设()1,z a bi a b R =+∈，则()2,z a bi a b R =-∈，所对应的坐标分别为(),a b ，(),a b -，这两点关于x 轴对称，故D 错误；故选：AC【点睛】本题主要考查复数的有关概念的判断，利用充分条件和必要条件的定义是解决本题的关键，属于基础题．25．BCD【分析】根据两个复数之间不能比较大小，得到C 、D 两项是错误的，根据复数的定义和复数模的概念，可以断定A 项正确，B 项错误，从而得到答案.【详解】因为两个复数之间只有等与不等，不能比较大小解析：BCD【分析】根据两个复数之间不能比较大小，得到C 、D 两项是错误的，根据复数的定义和复数模的概念，可以断定A 项正确，B 项错误，从而得到答案.【详解】因为两个复数之间只有等与不等，不能比较大小，所以C 、D 两项都不正确；当两个复数的模相等时，复数不一定相等， 比如11i i -=+，但是11i i -≠+，所以B 项是错误的；因为当两个复数相等时，模一定相等，所以A 项正确；故选：BCD.【点睛】该题考查的是有关复数的问题，涉及到的知识点有两个复数之间的关系，复数模的概念，属于基础题目.26．BD【分析】因为复数满足，利用复数的除法运算化简为，再逐项验证判断.【详解】因为复数满足，所以所以，故A 错误；，故B 正确；复数的实部为 ，故C 错误；复数对应复平面上的点在第二象限解析：BD【分析】因为复数z 满足(2i)i z -=，利用复数的除法运算化简为1255z i =-+，再逐项验证判断. 【详解】因为复数z 满足(2i)i z -=， 所以()(2)1222(2)55i i i z i i i i +===-+--+所以5z ==，故A 错误； 1255z i =--，故B 正确； 复数z 的实部为15- ，故C 错误； 复数z 对应复平面上的点12,55⎛⎫- ⎪⎝⎭在第二象限，故D 正确. 故选：BD【点睛】本题主要考查复数的概念，代数运算以及几何意义，还考查分析运算求解的能力，属于基础题. 27．AD【分析】根据纯虚数的概念即可判断A 选项；根据实数、复数的运算、以及共轭复数的定义即可判断BCD 选项.【详解】解：对于A ，若为纯虚数，可设，则，即纯虚数的共轭复数等于，故A 正确；对于B解析：AD【分析】根据纯虚数的概念即可判断A 选项；根据实数、复数的运算、以及共轭复数的定义即可判断BCD 选项.【详解】解：对于A ，若z 为纯虚数，可设()0z bi b =≠，则z bi z =-=-，即纯虚数z 的共轭复数等于z -，故A 正确；对于B ，由120z z +=，得出12z z =-，可设11z i =+，则21z i =--， 则21z i =-+，此时12z z ≠，故B 错误；对于C ，设12,z a bi z c di =+=+，则()()12a c b d i R z z =++++∈，则0b d +=， 但,a c 不一定相等，所以1z 与2z 不一定互为共轭复数，故C 错误；对于D ，120z z -=，则12z z =，则1z 与2z 互为共轭复数，故D 正确.故选：AD.【点睛】本题考查与复数有关的命题的真假性，考查复数的基本概念和运算，涉及实数、纯虚数和共轭复数的定义，属于基础题. 28．AB【分析】利用复数相等可选A ，利用虚数不能比较大小可选B ，利用特值法可判断C 错误，利用复数的运算性质可判断D 错误．【详解】对于选项A ，∵，且，根据复数相等的性质，则，故正确；对于选项B ，解析：AB【分析】利用复数相等可选A ，利用虚数不能比较大小可选B ，利用特值法可判断C 错误，利用复数的运算性质可判断D 错误．【详解】对于选项A ，∵,x y R ∈，且1x yi i +=+，根据复数相等的性质，则1x y ==，故正确；对于选项B ，∵虚数不能比较大小，故正确；对于选项C ，∵若复数1=z i ，2=1z 满足22120z z +=，则120z z ≠≠，故不正确； 对于选项D ，∵复数()2=1i --，故不正确；故选：AB ．【点睛】本题考查复数的相关概念，涉及复数的概念、复数相等、复数计算等知识，属于基础题. 29．ABC【分析】设，从而有，利用消元法得到关于的一元二次方程，利用判别式大于等于0，从而求得a 的范围，即可得答案.【详解】设，∴，∴，∴，解得：，∴实数的值可能是.故选：ABC.【点解析：ABC【分析】设z x yi =+，从而有222()3x y i x yi ai ++-=+，利用消元法得到关于y 的一元二次方程，利用判别式大于等于0，从而求得a 的范围，即可得答案.【详解】设z x yi =+，∴222()3x y i x yi ai ++-=+， ∴222223,23042,x y y a y y x a ⎧++=⇒++-=⎨=⎩， ∴244(3)04a ∆=--≥，解得：44a -≤≤， ∴实数a 的值可能是1,4,0-.故选：ABC.【点睛】本题考查复数的四则运算、模的运算，考查函数与方程思想，考查逻辑推理能力和运算求解能力.30．AC【分析】根据复数乘法的运算律和复数的模及共轭复数的概念可判断出答案A 和C 正确；C 中可取，进行判断；D 中的必要不充分条件是.【详解】解：由复数乘法的运算律知，A 正确；取，；，满足，但且不解析：AC【分析】根据复数乘法的运算律和复数的模及共轭复数的概念可判断出答案A 和C 正确；C 中可取11z =，2z i =进行判断；D 中12z z =的必要不充分条件是12=z z .【详解】解：由复数乘法的运算律知，A 正确；取11z =，；2z i =，满足22120z z +=，但10z =且20z =不成立，B 错误； 由复数的模及共轭复数的概念知结论成立，C 正确；由12z z =能推出12=z z ，但12||||z z =推不出12z z =，因此12z z =的必要不充分条件是12=z z ，D 错误.故选：AC【点睛】 本题主要考查复数乘法的运算律和复数的基本知识以及共轭复数的概念，属于基础题.。