【全国百强校】福建省福建师范大学附属中学2019届高三下学期高考模拟(最后一模)数学(理)试题
- 格式:doc
- 大小:2.11 MB
- 文档页数:9
福建省师范大学附属中学2019届高三历史5月模拟(最后一卷)试题(含解析)1.春秋战国时期出现了富民思想,如儒家的“藏富于民”、“恒产论”和“轻徭薄赋”,墨家的“节用利民”思想,道家的“精神足民”论,法家的“以政裕民”思想等。
富民思想的出现表明A. 民本思想成为社会主流思想B. 统治者注重对私产的保护C. 社会阶级矛盾得到有效缓和D. 生产关系领域发生了变革【答案】D【解析】【详解】春秋战国时期随着生产力的提高,私田的开垦越来越多,民众力量的显现产生了富民思想。
故答案为D项。
A项,春秋战国时期并没有特定的社会意识形态主流,民本思想也不可能上升到社会主流思想这个高度,排除;B项,材料不能体现统治者对私产的保护,排除;C项,春秋战国时期,社会动荡,社会矛盾尖锐而不是缓和,排除。
【点睛】表明类选择题是高考中相对稳定的题型,该类选择题主要借用了“表明”一词“相当肯定地显示”的含义。
它要求考生获取材料中比较确切的信息,重点考查学生的概括能力。
试题的题干部分一般描述历史现象,考生需要根据所学知识,用简洁的语言准确、清楚地概括现象背后的历史结论。
一般解题步骤有三步,第一步:概括材料的主体信息;第二步:联系选项,进行“等价转换”。
“等价转换”是指第一步概括出的主体信息要与选项完全匹配;第三步:检验其他选项的正确性。
2.《续封泥考略》所收汉朝齐鲁封泥(古代用泥团制作的印章)中,有“左市”“右市”“南市”“西市”等封泥。
由此可知,汉朝A. 城市建设有明确的设计规划B. 各地城市布局规格完全一致C. 对商业活动进行区域化管理D. 政府限制商业活动的发展空间【答案】C【解析】【详解】由材料‘左市’‘右市’‘南市’‘西市’等封泥可知,政府对商业活动实行区域性的划分,以便更好地进行管理,故选C。
材料仅涉及“市”的划分,这不代表整个城市有明确的规划,排除A;仅从齐鲁封泥无从得知各地城市是否完全一致,排除B;区域化的管理是为了规范商业活动,而不是限制它的发展空间,如果是限制,也就没有必要设置这么多个不同的“市”了,排除D。
2019届福建省福建师大附中高三下学期高考模拟(最后一模)数学(理)试题一、单选题1.设集合{}(5)4,{|}A x x x B x x a =-=≤,若A B B ⋃=,则a 的值可以是( ) A .1 B .2C .3D .4【答案】D【解析】集合(){}|54{x |14},{|}A x x x x B x x a =->=<<=≤,因为A B B ⋃=,所以A B ⊆,故4a ≥,故选D.2.第24届国际数学家大会会标是以我国古代数学家赵爽的弦图为基础进行设计的.如图,会标是由四个全等的直角三角形和一个小正方形拼成的一个大正方形.设直角三角形的一个锐角为θ,且tan 2θ=,若在大正方形内随机取一点,则该点取自小正方形区域的概率为( )A .14 B .15 C .25D .35【答案】B【解析】根据tan 2θ=求出小正方形的边长和直角三角形两直角边的长,进而得到大正方形的边长,然后根据几何概型概率公式求解即可. 【详解】如图,设小正方形的边长为a ,直角三角形较大的锐角为θ、较小的边长为b ,则直角三角形较大的直角边长为+a b ,∵tan 2a bbθ+==, ∴a b =,∴=, 由几何概型概率公式可得,所求概率为215P =. 故选B . 【点睛】解答几何概型概率的关键是分清概率是属于长度型的、面积型的、还是体积型的,然后再根据题意求出表示基本事件的点构成的线段的长度(或区域的面积、空间几何体的体积),最后根据公式计算即可.3.已知i 为虚数单位,a 为实数,复数()()12z i a i =-+在复平面内对应的点为M ,则“2a >”是“点M 在第四象限”的( ) A .充分而不必要条件 B .必要而不充分条件 C .充要条件 D .既不充分也不必要条件 【答案】A【解析】利用复数的乘法法则将复数z 表示为一般形式,由点M 在第四象限可得出关于a 的不等式组,即可求得实数a 的范围,再利用集合的包含关系即可判断. 【详解】()()()()12212z i a i a a i =-+=++-Q ,且点M 在第四象限,20120a a +>⎧∴⎨-<⎩,解得12a >. {}2a a >Q 12a a ⎧⎫>⎨⎬⎩⎭,因此,“2a >”是“点M 在第四象限”的充分而不必要条件.故选:A. 【点睛】本题考查充分不必要条件的判断,同时也考查了利用复数的几何意义求参数,考查运算求解能力与推理能力,属于基础题.4.已知奇函数()f x ,当0x >时单调递增,且(1)0f =,若(1)0f x ->,则x 的取值范围为( ) A .{|012}x x x <或B .{|02}x x x 或C .{|03}x x x 或D .{|11}x x x <->或【答案】A【解析】()f x Q 为奇函数,0x >时,单调递增,0x ∴<时,也单调递增,由()10f =,得()10f -=,()()111,211x x x f x f >⎧∴⇒->>⎨->⎩,()()111,0111x x x f x f <⎧⇒->-<<⎨->-⎩,x \的取值范围为01x <<或2x >,故选A.5.已知Rt ABC ∆,点D 为斜边BC 的中点,63AB =u u u v ,6AC =u u u v ,12AE ED =u u u v u u u v,则AE EB ⋅u u u v u u u v等于( ) A .-14 B .-9C .9D .14【答案】D【解析】利用向量共线及向量的加减法分别表示出()16AE AB AC +=u u u vu u uv u u u v ,5166EB AB AC =-u u u r u u u r u u u r ,再利用0AB AC ⋅=u u u v u u u v即可求得()221536AE EB AB AC ⋅=-u u u u u u r u v u u v u u r u ,问题得解. 【详解】依据题意作出如下图象:因为12AE ED =u u u v u u u v,所以,,A E D 三点共线.()()33111126AE AD AB AC AB AC =⨯=+=+u u u v u u u v u u u v u u u v u u u v u u u v .()51516666EB BE BA AE AB AC AB AC ⎛⎫=-=-+=--+=- ⎪⎝⎭u u u r u u u r u u u r u u u r u u ur u u u r u u u r u u u r又0AB AC ⋅=u u u v u u u v所以()()221156166563AE EB AB AC AB A AC C AB ⎛⎫⋅⋅=- ⎪⎝⎭=+-u u ur u u u r u u u v u u u v u u u v u u u vu u u r u u u r ()()22156361436=⨯-= 故选D 【点睛】本题主要考查了向量的加减法及数乘运算,还考查了向量垂直的数量积关系,考查转化能力及计算能力,属于中档题.6.执行如图所示的程序框图,输出的n 值为A .6B .8C .2D .4【答案】B【解析】程序流程图执行如下:首先初始化数据:0,1,1S a n ===,进入循环体执行循环:第一次循环:2S S a n =++=,不满足10S ≥,执行:1,2222a a n n ====; 第二次循环:142S S a n =++=,不满足10S ≥,执行:1,2424a a n n ====;第三次循环:384S S a n =++=,不满足10S ≥,执行:1,2828a a n n ====;第四次循环:7168S S a n =++=,满足10S ≥,此时跳出循环,输出8n =. 本题选择B 选项.7.学校组织同学参加社会调查,某小组共有5名男同学,4名女同学。
2019-2020学年福建师范大学附属中学高三英语一模试卷及答案解析第一部分阅读(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的A、B、C、D四个选项中选出最佳选项AObesity (肥胖症) is becoming a problem in our busy society, and almost one in three American adults is now considered to be obese. Children obesity is alsoat an all-time high.Obesity means being very overweight. If you are obese, you have too much bodyfat. If you eat more food than your body can use, this will make you put on weight. Food that your body does not need will be stored as fat by your body.The following are the major factors that increase the risk of obesity.What you eat plays a major role in weight gain. Eating a lot of fast food such as hamburgers, sweet drinks, ice creams and other sweet food can increase the risk of becoming obese.If you do not do enough exercise, you will put on weight as the food you eat is not being used to make energy for physical activities.The chances of you being obese are greater if your parents are obese.There are many psychological factors that cause people to eat too much. People who are worried, unhappy or bored will often eat to make themselves feel better. This is known as comfort eating.Age is another factor, as you tend (趋于) to be less active when you get older. When you get older, you need to eat less, and if you do not eat less, you will put on weight. Obesity can cause many health problems such as heart problems, high blood pressure and many other serious medical conditions.1. The underlined sentence in paragraph 1 means that ____.A. obesity does not do harm to health.B. there are more obese children than before.C. all the American children are obese.D. there are less obese children in the USA.2. According to the passage, there are ____ major factors that increase the risk of obesity.A. threeB. fourC. fiveD. six3. What will the writer most probably talk about after the last paragraph?A. How to avoid obesity.B. How to live in the busy USA.C. What illnesses are caused by obesity.D. How doctors treat heart problems.BA year ago I received a full scholarship to attend the University of San Francisco. All of my hard work paid off. My mom had spent a lot on my attending a private high school, so I made sure to push myself: I volunteered, took part in various clubs, and graduated with honors. I was so excited to start a new part of my life.Soon enough, the big day came, but it wasn't like what I had thought. The first two weeks were the most difficult days of my entire life. Every night I would cry myself to sleep. I was missing my family, my home and everything in my hometown so much and I didn’t know how to deal with my broken heart.To distract myself, I threw myself into my studies. I also found a ton of jobs. In any free time, I started forcing myself to go to the gym. I wanted to keep every part of my day busy so I wouldn’t think about how lonely I felt. Soon after, I began to control my eating, considering it another solution to my homesickness. But soon there was something wrong with me.Finally, I went to see a doctor. When the doctor told me I had no choice but to take time away from school, I started to fear. How could I stop? School was what I was best at. “I’m not so bad,” I thought in my head. But the result was that I was taken to hospital again a month later and my mother camewoefully. I had to take a semester off from school, and go to the treatment center near my home.If there are girls who are suffering similarly, I hope you know that there is hope and that you should have a positive attitude towards life. Though you may feel alone, there are so many people who can understand your struggle. That’s why I want to share my story.4. Why did the author push herself during high school?A. She wanted to attend the University of San Francisco.B. It cost too much to study in a private school.C. Her parents controlled much of her life.D. Her family put her under pressure,5. What can we know about the author in the first two weeks?A. She couldn't fall asleep because of pressure.B. She couldn't pay attention to her study.C. She couldn't deal with her homesickness.D. She couldn't catch up with others.6. What does the underlined word “woefully” in paragraph 4 mean?A. Sadly.B. Surprisingly.C. Curiously.D. Happily.7. What is the author's purpose in writing this text?A. To look back on her past life.B. To increase her own confidence.C. To express appreciation to her mother.D. To encourage other girls like her to be positive.CSophie became friends with the gray squirrels during her first week atPennState, after spotting them running around and wondering what they would look like with tiny hats on their heads. Today, everyone at the university knows her as the “Squirrel Girl”.Sophie tried bringing them food, and gradually they began to trust her. She managed to put a hat on a squirrel and take a picture. Thinking that her colleagues could do with something to lift their spirits, she started posting similar photos on Facebook. The response was greatly positive, and before long Sophie and her squirrels became an Internetsensation.Growing up in a neighborhood outside ofState College, Sophie was always fond of birds and animals around her home, but she didn't interact with people very much. She was later diagnosed (诊断) with Asperger's syndrome, but the squirrels changed that. “The squirrels help me break the ice, because I'll be sitting here patting a squirrel and other people will come over and well just start like feeding the squirrels together and chatting about them,” she said, “I am a lot more outgoing.”And in case you're wondering how Sophie is able to get the squirrels to do what she wants for her photos, it has a lot to do with food. For example, whenever she wants them to hold or play with something, she puts peanut butter on the prop (道具), and they'll grab it. In the beginning, she would throw peanuts up the trees on campus and invite the squirrels to come down and get them, but they hesitated to approach her. She had the patience to earn their trust, though.This year, Sophie is graduating with a degree in English and wildlife sciences. She wants to be a science writer and educate people on how to preserve the environment. As for her furry friends, Sophie plans to stay in the area and visit them as often as she can.8. What does the underlined word “sensation” in paragraph 2 mean?A. Event.B. Hit.C. Service.D. Addiction.9. What can we learn about Sophie according to paragraph 3?A. She got lots of friends due to squirrels.B. She used to be a popular girl in her childhood.C. She lived in the far countryside when young.D. She was more outgoing than before.10. How did Mary manage to take photos of squirrels wearing hats?A. By attracting them with food.B. By putting them in cages.C. By playing music to them.D. By dressing like squirrels.11. What do you think of Sophie?A. Tolerant and capable.B. Sociable and aggressive.C. Patient and caring.D. Indifferent and appreciative.DJeff Bezos, Elon Musk and Richard Branson have a combined net worth of 400 billion, roughly the size of the GDP of the entire nation of Ireland. And all three men have decided to put vast sums of their wealth into chasing their space travel dreams, creating a modern space race in which ultra — rich men — rather than countries — shoot for the stars.But why the three billionaires choose the crazy plan? Just for the travel dreams? It’s not that simple.As we all know, the space travel is a mirror of comprehensive national strength, whether it’s the cold war or the present. The first space racespannedmultiple presidents and premiers — Kennedy, Krushchev, Brezhnev, Nixon. It made heroes of astronauts and cosmonauts, and it focused national prides. But obviously, this year’s race between the billionaires features none of that national pride or opposing ideas. It’s tax — averse tycoons (大亨) who want to sell high — priced tickets to rich people interested in experiencing weightlessness. Amazon’s Bezos has said he is funding his portion by selling off large chunks of his Amazon stock (股票) — 1 billion or more a year.What are the odds? If anyone is taking bets on how this will end, and who will have the more profitable space tourism business, consider Amazon’s method of making very little profit in order to eat up competition. So, from this point of view, what’s hiding behind this crazy race is business. As a method of expand influence, it is not a bad idea.Richard Branson has carried out his plan on July 12, and Jeff Bezos announced that he will start his space travel on July 20. Bezos is getting some good press because he’s taking Wally Funk along for the ride. She’s the pilot now in her 80s who was kept out of space in the 1960s because she’s a woman. This may be a bonus for him.As for Musk, he thinks these two above are not real space trips, he wants a larger move.Who will win the race in the end? Let’s wait and see.12. Which of the following people might be the potential client of the space travel?A. Daisy aged 45 with a heavy debt.B. Clarkson who doesn’t enjoy taking risks.C. Billy aged 60 with an ample retirement pension.D. Alexander who knows nothing about space travel.13. What stopped Wally Funk flying to space in the 1960s?A. Age.B. Gender.C. Height.D. Disposition.14. What does the author think of the billionaires race?A. Profit — driven.B. Meaningless.C. Foolish.D. Generous.15. What does the underlined word “spanned” in paragraph 3 probably mean?A. Revolved.B. Turned down.C. Entered.D. Leapt through.第二节(共5小题;每小题2分,满分10分)阅读下面短文,从短文后的选项中选出可以填入空白处的最佳选项。
2019届福建省福建师大附中英语高考模拟试卷20190528满分:150分完卷时间:120分钟第一部分听力(共两节,满分30分)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。
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 woman think of the car journey?A. It’s too long.B. It’s very exciting.C. It’s rather dangerous.2. Why is John late for school?A. He was stuck in traffic.B. He hurt his head.C. He did a good deed.3. What are the speakers talking about?A. A farm.B. Some houses.C. A corn field.4. What does the man say about the movie?A. It’s horrible.B. It’s amusing.C. It’s not good.5. What is the probable relationship between the speakers?A. Acquaintances.B. Classmates.C. A couple.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。
每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项。
听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
听第6段材料,回答第6、7题。
6. What do we know from the conversation?A. The woman will be free tomorrow.B. The man will help the woman tomorrow.C. The woman will have workers to do the job.7. What does the woman probably think of the man?A. Considerate.B. Annoying.C. Stubborn.听第7段材料,回答第8、9题。
福建省福州市福建师范大学附属中学2024届高三下学期校模拟考试数学试时间:120分钟满分:150分命题:高三集备组审核:高三集备组一、单项选择题:本题共8小题,每小题5分,共40分.在每小题给出的选项中,只有一个选项是符合题目要求的.1.设集合,则等于A .B .C .D .或2.已知等差数列满足,则A .B .C .D .3.若函数是奇函数,则的值为A .1B .-1C .D .04.将甲、乙、丙、丁4人分配到3个不同的工作岗位,每人只去一个岗位,每个岗位都要有人去,则甲、乙二人分别去了不同岗位的概率是A.B .C .D .5.没为单位向量,在方向上的投影向量为,则ABCD .6.已知,则A .B .C .D .7.如图,设拋物线的焦点为,不经过焦点的直线上有三个不同的点A ,B ,C ,其中点A ,B 在抛物线上,点在轴上,则与的面积之比是{}{}2210,log 0A x x B x x =->=>∣∣A B ⋂{0}x x >∣{1}x x >∣{1}x x <-∣{1xx <-∣1}x >{}n a 12j 1010a a a a ++++= 11010a a +>11010a a +<3990a a +=5151a =(()ln f x ax =a 1±13122356,a b a b 12b -|2|a b -=311(),(),()552P A P AB P A B === ∣()P B =1525354524y x =F C y BCF ACFA.B .C .D .8.在中,为内一点,,则A .BCD .二.多项选择题:本题共3小题,每小题6分,共18分.在每小题给出的四个选项中,有多个选项符合题目要求,全部选对的得6分,部分选对的得部分分,有选错的得0分.9.已知复数,下列结论正确的是A .若,则B .C .若,则或D .若且,则10.冬末春初,人们容易感冒发热,某公司规定:若任意连续7天,每天不超过5人体温高于,则称没有发生群体性发热.根据下列连续7天体温高于人数的统计量,能判定该公司没有发生群体性发热的为A .中位数是3,众数为2;B.均值小于1,中位数为1;C .均值为3,众数为4;D .均值为2.11.已知,则A .B .C .D .三、填空题:本题共3小题,每小题5分,共15分.12.已知圆台的上、下底面的面积分别为,侧面积为,则该圆台的高为______.||1||1BF AF ++||1||1BF AF --22||1||1BF AF ++22||1||1BF AF --ABC 120,2,ACB BC AC D ︒∠==ABC ,120AD CD BDC ︒⊥∠=tan ACD ∠=12,z z 12z z =2212z z =1212z z z z -=-120z z =10z =20z =10z ≠12z z =2121z z =37.3C ︒37.3C ︒12212log ,log 2baa b ⎛⎫== ⎪⎝⎭22a ba b -+=+22b aa b -+=+121eba+>112eab->4π,36π64π13.的展开式中常数项为______.14.已知函数在区间上单调,其中为正整数,,且.则图象的一个对称中心是______;若的值为______.四、解答题:本题共5小题,共77分.解答应写出文字说明、证明过程或演算步骤.15.(13分)按照《中华人民共和国环境保护法》的规定,每年生态环境部都会同国家发展改革委等部门共同编制《中国生态环境状况公报》,并向社会公开发布.下表是年份2017年2018年2019年2020年2021年年份代码123456.45.55.04.83.8(1)求2017-2021年年份代码与的样本相关系数(精确到0.01);(2)请用样本相关系数说明该组数据中与之间的关系可用一元线性回归模型进行描述,并求出关于的经验回归方程;(3)预测2024年的酸雨区面积占国土面积的百分比.附:①回归直线的斜率和截距的最小二乘法估计公式分别为:,②样本相关系:③参考数据:16.(15分)已知函数.(1)求曲线在点处的切线方程;(2)若恒成立,求的值:17.(15分)如图,在三棱柱中,,E ,F 分别为的中点,且421x x ⎛⎫+- ⎪⎝⎭()sin()f x x ωϕ=+π2π,63⎛⎫⎪⎝⎭ωπ||2ϕ<ππ32f f ⎛⎫⎛⎫=- ⎪ ⎪⎝⎭⎝⎭()y f x =π4f ⎛⎫=⎪⎝⎭ϕix iy i x i y y x y x ()()()121ˆˆ,niii n i i x x yy bay bx x x ==--==--∑∑ nx y r =5521170.6,6ii i i i x yy ====≈∑∑()ln(1)()f x ax x a =--∈R ()y f x =(0,(0))f ()0f x ≥a 111ABC A B C -1AC AB ==11,AC BB平面.(1)求棱BC 的长度;(2)若,且的面积,求二面角的正弦值.18.(17分)设是双曲线的左焦点,经过的直线与相交于M ,N 两点.(1)若M ,N 都在双曲线的左支上,求面积的最小值.(2)是否存在轴上一点,使得为定值?若存在,求出点的坐标;若不存在,说明理由.19.(17分)定义首项为1且公比为正数的等比数列为“—数列”.(1)已知等比数列满足:,求证:数列为“—数列"';(2)已知数列满足:,其中为数列的前项和.①求数列的通项公式;②设为正整数,若存在“—数列”,对任意正整数,当时,都有成立,求的最大值.EF ⊥11AA C C 111BB A B ⊥1A FC 1A FC S =11B A F C --F 221x y Γ-=:F ΓOMN x P PM PN P M {}()*N n a n ∈245321,440a a a a a a =-+={}n a M {}n b 111221,n n n b S b b +==-n S {}n b n {}n b m M {}()*n c n ∈N k k m ≤1k k k c b c +≤≤m2023-2024年高三数学校模拟考参考答案1-8BCCDADBB9.BCD10.BD11.AD12.13.4914.15.【详解】(1)由已知可得,,,(2)由小问1知,与的相关系数接近1,所以与之间具有极强的线性相关关系,可用线性回归模型进行描述.由小问1知,,,所求经验回归方程为.(3)令,则,预测2024年的酸雨区面积占国土面积的百分比为.16.【详解】(1)由,得,因为,所以曲线在点处的切线方程为;(2),①当时,,不符合题意.5π,012⎛⎫⎪⎝⎭π61234535x ++++==522222216.4 5.5 5.0 4.8 3.8 5.112345555ii y x=++++===++++=∑555 5.90.986x x y y x y xyr ----===≈≈y x 0.98,r r ≈-y x ()()()551155222115 5.9ˆ0.59105iii ii i i ii i x x y y x y xybx x xx ====----====---∑∑∑∑ˆˆ 5.1(0.59)3 6.87ay bx =-=--⨯=ˆ0.59 6.87yx =-+8x =ˆ0.598 6.87 2.15y=-⨯+=2.15%()ln(1)f x ax x =--1()(1)1f x a x x'=+<-(0)0,(0)1f f a '==+()y f x =(0,(0))f (1)y a x =+11()(1)11ax a f x a x x x'-++=+=<--0a ≥(1)ln 20f a -=--<②当时,令,解得,当时,在区间上单调递减,当时,在区间上单调递增,所以当时,取得最小值;若恒成立,则,设,则,当时,在区间上单调递增,当时,在区间上单调递减,所以,即的解为.所以;17.【详解】(1)取中点,连接,分别为的中点,则且,又为三棱柱,且分别为的中点,则且,可得且,即四边形DEFB 为平行四边形,故,又平面,则平面,平面,可得,又为AC 的中点,则为等腰三角形,.(2)由(1)可知:,且,即,0a <()0f x '=11x a=+1,1x a ⎛⎫∈-∞+⎪⎝⎭()0,()f x f x '<1,1a ⎛⎫-∞+ ⎪⎝⎭11,1x a ⎛⎫∈+⎪⎝⎭()0,()f x f x '>11,1a ⎛⎫+ ⎪⎝⎭11x a=+()f x 111ln()f a a a ⎛⎫+=++- ⎪⎝⎭()0f x ≥1ln()0a a ++-≥()1ln()(0)x x x x ϕ=++-<11()1x x x xϕ'+=+=(,1)x ∈-∞-()0,()x x ϕϕ'>(,1)-∞-(1,0)x ∈-()0,()x x ϕϕ'<(1,0)-()(1)0x ϕϕ≤-=1ln()0a a ++-≥1a =-1a =-AC D ,ED BD ,D E 1,AC AC 1//DE AA 112DE AA =111ABC A B C - F 1BB 1//BF AA 112BF AA =//DE BF DE BF =//EF DB EF ⊥ 11AA C C DB ⊥11AA C C AC ⊂11AA C C DB AC ⊥D ABC 1BC AB ∴==1BC AB ==AC =222AB BC AC +=,则可得,且,平面平面,则,,由(1)知平面平面,则,又,则又,则,平面,平面,平面,则,且,可得,为直角三角形,则以为坐标原点,向量方向为x 轴,y 轴,z 轴正方向建立空间直角坐标系,则,可得,AB BC ∴⊥EF DB ==1111A B B C⊥EF ⊥ 111,AAC C AC⊂11AA C C 1EF A C ⊥1111122A FCS AC EF AC ∴=⋅== 12A C =DB ⊥111,AAC C AA ⊂11AA C C 1DB AA ⊥11//AA BB 1DB BB ⊥11111,//BB A B AB A B ⊥ 1BB AB ⊥,,AB DB B AB DB ⋂=⊂ABC 1BB ∴⊥ABC AC ⊂ABC 1BB AC ⊥11//AA BB 1AA AC ⊥1AA C ∴ 1AA ==1B 11111,,B C B A B B 1B xyz -111(0,0,0),(0,1,0),(1,0,0),B A C C B F ⎛ ⎝110,,(1,A F A C ⎛=-=- ⎝设平面的一个法向量为,则,令,则,可得,平面的一个法向量为,设二面角的平面角为,可得,,故二面角18.【详解】(1)设直线的方程为.由,由根与系数的关系可知①.此时.原点到直线的距离为,此时.由都在双曲线的左支上知,得,令,则,由于,所以当,即时,此时取最小值,则,当,即时,等号成立.1A FC 1(,,)n x y z = 111100n A F y z n A C x y ⎧=-=⎪⎨⎪=-+=⎩ 1y =1,x z =-=1(n =-11B A F 2(1,0,0)n =11B A F C --(0,π)θ∈121211|cos |212n n n n θ===⨯ sin θ∴==11B A F C --MN ()()1122,,,x my M x y N x y =-221x my x y ⎧=-⎪⎨-=⎪⎩()22110(1)m y m --+=≠±1212211y y y y m +==-()2221||1m MN m +===-O MN d =()222111||221OMNm S d MN m +===- ,M N ()121212220,01x x m y y x x m -+=+-=<=>-11m -<<21(10)m t t -=-≤<2221111144244OMNS t t t ⎛⎫⎛⎫=+=+- ⎪ ⎪⎝⎭⎝⎭ 1(,1]t ∈-∞-11t=-1t =-OMN S ≥ 1t =-0m =(2)假设存在这样的定点.当直线的斜率不为0时,由(1)知②将①代入②可得,此时要想,得.即存在这样的定点满足题意.当直线的斜率为0时,易知,若,则,满足题意.综上,存在满足题意.19.【详解】(1)设等比数列的公比为,所以由,得,解得,因此数列为“—数列”;(2)①因为,所以,(,0)P n ()()()()()()112212121212,,PM PN x n y x n y x n x n y y my n my n y y =--=--+=----+()()2212121))m y y m n y y n =+-++++ 2)PM PN n =++ PM PN 11=-n =12PM PN =- P ⎛⎫⎪ ⎪⎝⎭2(1)(1)1PM PN n n n =+-=- P ⎛⎫ ⎪ ⎪⎝⎭12PM PN =- P ⎛⎫⎪ ⎪⎝⎭{}n a q 10,0a q ≠≠245321440a a a a a a =⎧⎨-+=⎩244112111440a q a q a q a q a ⎧=⎨-+=⎩112a q =⎧⎨=⎩{}n a M 1122n n n S b b +=-0n b ≠由得,则,由,得,当时,由,得,整理得,所以数列是首项和公差均为1的等差数列,因此,数列的通项公式为;②由①知,,因为数列为“—数列”,设公比为,所以,因为,所以,其中,当时,有;当时,有,设,则,则当时,,当时,,故在上单调递增,在上单调递减,因为,所以,取时,,即,令,则,令,则,1111,b S b ==212211b =-22b =1122n n n S b b +=-()112n n n n n b b S b b ++=-2n ≥1n n n b S S -=-()()111122n n n nn n n n n b b b b b b b b b +-+-=---112n n n b b b +-+={}n b {}n b ()*n b n n =∈N *,k b k k =∈N {}n c M q 11,0c q =>1k k k c b c +≤≤1k k q k q -≤≤1,2,3,,k m = 1k =1q ≥2,3,,k m = ln ln ln 1k kq k k ≤≤-ln ()(1)x f x x x =>21ln ()xf x x '-=(1,e)x ∈()0f x '>(e,)x ∈+∞()0f x '<()f x (1,e)(e,)+∞ln 2ln 8ln 9ln 32663=<=max ln 3()(3)3f k f ==q =1,2,3,4,5k =ln ln kq k…k k q ≤ln ()(1)1x g x x x =>-2211(1)ln 1ln ()(1)(1)x x xx x g x x x '----==--1()1ln h x x x =--22111()0xh x x x x'-=-=<故在上单调递减,则,即在上恒成立,即在上单调递减,则,即,因此所求的最大值不小于5,若,分别取,得,且,从而,且,所以不存在,因此所求的最大值小于6,故的最大值为5.()h x (1,)+∞()(1)1100h x h ≤=--=()0g x '<(1,)+∞()g x (1,)+∞min ln 5ln125ln 81ln 3()(5)412123g k g ===<=1ln ln 1k k q q k k -≤≤-m 6m ≥3,6k =33q ≤56q ≤15243q …15216q …q m m。
2019届福建省福建师大附中理科综合高考模拟试卷2019.05.28试卷分值:300分考试时间:150分钟可能用到的相对原子质量:H1 N 14 O 16 Na 23 Mg 24 S 32 Cl 35.5 Mn 55Hg 201 Si 28第I卷一、选择题:本题共13小题,每题6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。
7.化学与生产和生活密切相关。
下列有关说法正确的是A.《本草纲目》中“冬月灶中所烧薪柴之灰,令人以灰淋汁,取碱浣衣”中的碱是KOH B.工艺装饰材料——天然水晶,属硅酸盐产品C.K2FeO4具有强氧化性,被还原后生成的Fe3+水解生成胶状物,可以软化硬水D.秸秆经加工处理成吸水性的材料——植物纤维,可用作食品干燥剂8.下列图示与对应的叙述正确的是A.图甲表示H2O2在有、无催化剂下的分解反应能量变化曲线,b表示有催化剂时的反应B.图乙表示0.1 mol MgCl2·6H2O在空气中充分加热时固体质量随时间的变化C.图丙表示KNO3和NaCl的溶解度曲线,若欲除去NaCl中少量KNO3,可将混合物制成较高温度下的浓溶液再冷却结晶、过滤D.图丁表示碳酸盐的溶解平衡曲线[pM=-lg c(M),p(CO2-3)=-lg c(CO2-3)],则K sp(CaCO3)>K sp(MgCO3)9.下列实验“操作和现象”与“结论”对应关系正确的是A.a图,滴加乙醇,试管中橙色溶液变为绿色,乙醇发生取代反应生成乙酸B.b图,右边试管中产气泡迅速,说明氯化铁的催化效果比二氧化锰好C.c图,根据试管中收集到无色气体,不能验证铜与稀硝酸的反应产物是NOD.d图,试管中先有白色沉淀、后有黑色沉淀生成,能确定K sp(AgCl)>K sp(Ag2S) 10.短周期主族元素W、X、Y、Z的原子序数依次增加。
W、X、Y最外层电子数之和为15,且由这三种元素组成的某盐在净水过程中可以做絮凝剂。
2019 届福建省福建师大附中文科综合高考模拟试卷20190528第 I 卷一、选择题:此题共35 小题,每题 4 分,共140 分。
在每题给出的四个选项中,只有一项是切合题目要求的。
中国进入急速老龄化阶段,估计到2025 年,老龄人口数目将达到3 亿人,以下图为 2010-2025年我国老年人口数目变化展望图(图),据此完1~2 题。
1图 11. 2010-2025年间,老龄化人口均匀每年增添速度最快的时间段是()A. 2010-20l1年B. 2011-2012年C. 2012-2015 年D. 2015-2025年2.对于老龄化迅速发展对社会的影响,以下说法正确的选项是()①教育系统中应当踊跃发展退休前教育和老年教育②创新能力及创新成就在生活中的应用均不会受影响③家产人口由第三家产向第二家产转移④人群生病率增添,医疗保障系统面对较大压力A. ①②B.①③C.②③D.①④以下图为 2018 年 5 月 10 日 2 时亚洲部分地域海平面气压局势图(图2)。
回答3-4题。
图 23.该日,甲地政府部门可能公布()A.台风预警 B .丛林火灾预警 C .寒潮预警 D .滑坡、泥石流预警4.北京市将来两天的天气情况可能是()A.雨过天晴,气温将明显高升B.气压降落,出现连续性降水C.降雨后,可吸入颗粒物减少D.风向转为偏南风,风速降低读“某地 1 月和 7 月均匀气温等值线散布图(图3)”,达成5-6 题。
图 35.等温线与当地等高线的切合程度是指二者的平行程度,平行程度越高则切合度越高,反之则越低。
相对于闭合等温线L2而言, L1与当地等高线切合程度较低。
对图中L1切合度较低影响最小的要素是()A. 冬天风B.太阳辐射C.海陆地点D.地形6.相对于a、 b、 c 山体的南坡而言, d 山体的南坡垂直自然带谱更丰富。
其主要原由最可能是d山体()A. 纬度更低B.距离大海更近C. 光照更充分D.受夏天风影响更明显法国依云小镇是世界特点小镇的模范,以温泉养生功能为主题特点,温泉养生的发展距今已有近200 年历史,目前已逐渐形成了以天然矿泉水制造、温泉休养为主导,商务会展、户外运动、旅行参观、美体保健等为衍生的家产系统。
2019届福建省福建师大附中文科综合高考模拟试卷20190528第I卷一、选择题:本题共35小题,每小题4分,共140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。
中国进入急速老龄化阶段,预计到2025年,老龄人口数量将达到3亿人,下图为2010-2025年我国老年人口数量变化预测图(图1),据此完1~2题。
图11. 2010-2025年间,老龄化人口平均每年增长速度最快的时间段是 ( )A. 2010-20l1年B. 2011-2012年C. 2012-2015年D. 2015-2025年2. 关于老龄化快速发展对社会的影响,以下说法正确的是 ( )①教育体系中应该积极发展退休前教育和老年教育②创新能力及创新成果在生活中的应用均不会受影响③产业人口由第三产业向第二产业转移④人群患病率增加,医疗保障体系面临较大压力A. ①②B. ①③C. ②③D. ①④下图为2018年5月10日2时亚洲部分地区海平面气压形势图(图2)。
回答3-4 题。
图23.该日,甲地政府部门可能发布 ( )A.台风预警 B.森林火灾预警 C.寒潮预警 D.滑坡、泥石流预警4.北京市未来两天的天气状况可能是 ( )A.雨过天晴,气温将显著升高 B.气压下降,出现连续性降水C.降雨后,可吸入颗粒物减少 D.风向转为偏南风,风速降低读“某地1月和7月平均气温等值线分布图(图3)”,完成5-6题。
图35. 等温线与当地等高线的契合程度是指两者的平行程度,平行程度越高则契合度越高,反之则越低。
相对于闭合等温线L2而言,L1与当地等高线契合程度较低。
对图中L1契合度较低影响最小的因素是 ( )A. 冬季风B. 太阳辐射C. 海陆位置D. 地形6. 相对于a、b、c山体的南坡而言,d山体的南坡垂直自然带谱更丰富。
其主要原因最可能是d山体 ( )A. 纬度更低B. 距离海洋更近C. 光照更充足D. 受夏季风影响更显著法国依云小镇是世界特色小镇的典范,以温泉养生功效为主题特色,温泉养生的发展距今已有近200年历史,当前已逐步形成了以天然矿泉水制造、温泉疗养为主导,商务会展、户外运动、旅游观光、美体保健等为衍生的产业体系。
【全国百强校】福建省福建师范大学附属中学2020届高三下学期高考模拟(最后一模)英语试题第一部分(共20小题每,小题1.5分,满分30分)1.The picture looks good ____ the white wall.A.with B.to C.against D.beyond2.They were abroad during the months when we were carrying out the investigation, or they __________to our help.A.would have come B.could comeC.have come D.had come3.The 90’s people seem to have enjoyed the great benefits ________ about by the great level of cultural and economic development.A.brought B.bringingC.to be brought D.having brought4.-- It shouldn’t take long to clear up after the party if we all volunteer to help.--- That’s right._____________.A.Many hands make light work B.Something is better than nothingC.The more the merrier D.The sooner begin, the sooner done5.-----My room gets very cold at night.-----_________________.A.So is mine B.So mine is C.So does mine D.So mine does6.It is reported that the ______ in Libya is hotting up.A.strength B.violence C.power D.activity7.—Amazingly, I’ve managed to start my own shop online!—________ I told you it was easy.A.There you are! B.Believe it or not.C.How come? D.You got me there!8.—People should stop using their cars and start using public transport.—________. The roads are too crowded as it is.A.All right B.ExactlyC.Go ahead D.Fine9.The traffic is heavy these days. I arrive a bit late, so could you save me a place?A.canB.mustC.needD.might10.The laptops made by our company sell best, but nobody could have guessed the place in the market that they ________ 20 years ago.A.had had B.had C.were having D.were to have11.You ________ have booked the tickets in advance; there were plenty left.A.needn’t B.can’tC.shouldn’t D.mustn’t12.The project is far behind schedule. It’s impossible for you to ______ it in a week.A.catch upon B.live up toC.add up to D.hold on to13.In my opinion, parents and teachers are supposed to set an example of good behavior to children _____ they are role models to them.A.although B.even if C.unless D.as14.— I am worn out. — Me too, all work and no play. So it’s time to ________.A.burn the midnight oil B.push the limitsC.go with the flow D.call it a day15.You look beautiful in this dress and there is only one of this kind left here. I wonder if you would buy ________.A.it B.oneC.some D.any16.--- I’ll send you the signed contract personally this time tomorrow.--- Sorry, I ______ a meeting then.A.am attending B.attendedC.have attended D.will be attending17.With a travelling speed of up to 350 kilometres per hour, the railway to be built between Beijing and Shanghai _______ the journey time from 12 hours to 5 hours.A.cuts B.will cut C.is cutting D.has cut18.(2013·天津,9)No one________ be more generous;he has a heart of gold.A.could B.must C.dare D.need19.______almost one hundred jin, the stone was moved by him alone.A.Weighed B.Weighing C.It weighed D.To weigh20.By serving others, a person focuses on someone other than himself or herself, ________ can be very eye-opening and rewarding.A.who B.whichC.what D.that第二部分阅读理解(满分40分)阅读下列短文,从每题所给的A、B、C、D四个选项中,选出最佳选项。
2019届福建省福建师大附中文科综合高考模拟试卷第I卷一、选择题:本题共35小题,每小题4分,共140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。
中国进入急速老龄化阶段,预计到2025年,老龄人口数量将达到3亿人,下图为2010-2025年我国老年人口数量变化预测图(图1),据此完1~2题。
图11. 2010-2025年间,老龄化人口平均每年增长速度最快的时间段是 ( )A. 2010-20l1年B. 2011-2012年C. 2012-2015年D. 2015-2025年2. 关于老龄化快速发展对社会的影响,以下说法正确的是 ( )①教育体系中应该积极发展退休前教育和老年教育②创新能力及创新成果在生活中的应用均不会受影响③产业人口由第三产业向第二产业转移④人群患病率增加,医疗保障体系面临较大压力A. ①②B. ①③C. ②③D. ①④下图为2018年5月10日2时亚洲部分地区海平面气压形势图(图2)。
回答3-4 题。
图23.该日,甲地政府部门可能发布 ( )A.台风预警 B.森林火灾预警 C.寒潮预警 D.滑坡、泥石流预警4.北京市未来两天的天气状况可能是 ( )A.雨过天晴,气温将显著升高 B.气压下降,出现连续性降水C.降雨后,可吸入颗粒物减少 D.风向转为偏南风,风速降低读“某地1月和7月平均气温等值线分布图(图3)”,完成5-6题。
图35. 等温线与当地等高线的契合程度是指两者的平行程度,平行程度越高则契合度越高,反之则越低。
相对于闭合等温线L2而言,L1与当地等高线契合程度较低。
对图中L1契合度较低影响最小的因素是 ( )A. 冬季风B. 太阳辐射C. 海陆位置D. 地形6. 相对于a、b、c山体的南坡而言,d山体的南坡垂直自然带谱更丰富。
其主要原因最可能是d 山体 ( )A. 纬度更低B. 距离海洋更近C. 光照更充足D. 受夏季风影响更显著法国依云小镇是世界特色小镇的典范,以温泉养生功效为主题特色,温泉养生的发展距今已有近200年历史,当前已逐步形成了以天然矿泉水制造、温泉疗养为主导,商务会展、户外运动、旅游观光、美体保健等为衍生的产业体系。
2019届福建省福建师大附中理科数学高考模拟试卷20190528本试卷共4页. 满分150分,考试时间120分钟.注意事项:试卷分第I 卷和第II 卷两部分,将答案填写在答卷纸上,考试结束后只交答案卷.第I 卷 共60分一、选择题:本大题有12小题,每小题5分,在每小题给出的四个选项中,只有一项符合题目要求. 1.设集合,若,则的值可以是( ) A .1 B . 2 C .3 D .42.第24届国际数学大会会标是以我国古代数学家赵爽的弦图为基础进行设计的, 如图,会标是由四个全等的直角三角形和一个小正方形拼成的一个大正方形。
设直角三角形的一个锐角为θ,且tan 2θ=,若在大正方形内随机取一点,则改 点取自小正方形区域的概率为( ) A .14 B .15 C. 25 D .353.已知i 为虚数单位,a 为实数,复数(12)()z i a i =-+在复平面内对应的点为M ,则“2a >”是“点M 在第四象限”的( )A .充分而不必要条件 B.必要而不充分条件 C .充要条件 D.既不充分也不必要条件4.已知奇函数)(x f ,当0>x 时单调递增,且0)1(=f ,若0)1(>-x f ,则x 的取值范围为( ) A .}210|{><<x x x 或 B .}20|{><x x x 或 C. }30|{><x x x 或 D .}11|{>-<x x x 或5、已知Rt ABC ∆,点D 为斜边BC 的中点,63AB =u u u r ,6AC =u u u r,12AE ED =u u u r u u u r ,则AE EB ⋅u u u r u u u r等于( )A. 14-B. 9-C. 9D. 146.执行如图所示的程序框图,输出的n 值为( )A .6B .8C .2D .47.学校组织学生参加社会调查,某小组共有5名男同学,4名女同学.现从该小组中选出3名同学分别到A ,B ,C 三地进行社会调查,若选出的同学中男女均有,则不同的安排方法有( ) A.70种 B.140种 C.840种 D. 420种8. 将函数()2sin(2)(0)f x x ϕϕπ=+<<的图像向左平移6π个单位后得到函数()y g x =的图像,若函数()y g x =为偶函数,则函数()y f x =在0,2π⎡⎤⎢⎥⎣⎦的值域为( ) A .[]1,2- B .[]1,1- C.32⎡⎤⎣⎦,D .3,3⎡⎤-⎣⎦9.在棱长为1的正方体中1111ABCD A B C D -点P 在线段1AD 上运动,则下列命题错误的是( )A .异面直线1C P 和1CB 所成的角为定值 B .直线CD 和平面1BPC 平行C .三棱锥1D BPC -的体积为定值D .直线CP 和平面11ABC D 所成的角为定值10.将函数x y e =(e 为自然对数的底数)的图象绕坐标原点O 顺时针旋转角后第一次与轴相切,则角θ满足的条件是( )A .sin cos e θθ=B .sin cos e θθ=C .sin 1e θ=D .cos 1e θ=11.已知双曲线222210,0)x y a b a b-=>>(的左,右焦点分别为12,F F ,点A 为双曲线右支上一点,线段1AF 交左支于点B .若22AF BF ⊥,且1213BF AF =,则该双曲线的离心率为( )A .2B .655C .355D .312.已知函数46()4sin 2,0,63f x x x ππ⎛⎫⎡⎤=-∈ ⎪⎢⎥⎝⎭⎣⎦,若函数()()3F x f x =-的所有零点依次记为123,,,,n x x x x L ,且123n x x x x <<<<L ,则1231222n n x x x x x -+++++L =( ) A.12763π B.445π C.455π D.14573π第Ⅱ卷 共90分二、填空题:本大题有4小题,每小题5分. 13.25211(1)(1)x x x+++展开式中2x 的系数为 .14.记不等式组22012x y x y +-≥⎧⎪≤⎨⎪≤⎩的解集为D ,若,(1)x y D y a x ∀∈≤+,,则实数的最小值是 .15.已知抛物线2:2(0)E x py p =>的焦点为F ,0(2,)A y 是E 上一点,且AF =2,设点B 是E 上异于点A 的一点,直线AB 与直线3y x =-交于点P ,过点P 作x 轴的垂线交E 于点M 则直线BM 过定点,定点坐标为 .16. 已知在平面四边形ABCD 中,2,2,,AB BC AC CD AC CD ==⊥=.则四边形ABCD 的面积的最大值为______________.三、解答题:解答应写出文字说明、证明过程或演算步骤.17.(本小题满分12分)已知等差数列{}n a 前5项和为50,227=a ,数列{}n b 的前n 项和为n S ,13,111+==+n n S b b . (Ⅰ)求数列{}n a ,{}n b 的通项公式; (Ⅱ)若数列{}n c 满足*+∈=+++N n a b c b c b c n nn ,12211Λ,求122019c c c +++L 的值18.(本小题满分12分)如图1,在四边形ABCD 中,AD BC P ,90BAD ∠=︒,AB =,4BC =,6AD =,E 是AD 上的点,13AE AD =,P 为BE 的中点.将ABE ∆沿BE 折起到1A BE ∆的位置,使得14A C =,如图2.(Ⅰ)求证:平面1A CP ⊥平面1A BE ;(Ⅱ)求二面角1B A P D --的余弦值.19. (本小题满分12分)随着节能减排意识深入人心以及共享单车在饶城的大范围推广,越来越多的市民在出行时喜欢选择骑行共享单车。
为了研究广大市民在共享单车上的使用情况,某公司在我市随机抽取了100名用户进行判断能否在犯错误概率不超过0.05的前提下,认为是否“喜欢骑行共享单车”与性别有关?(2)每周骑行共享单车6次及6次以上的用户称为“骑行达人”,视频率为概率,在我市所有“骑行达人”中,随机抽取4名用户.① 求抽取的4名用户中,既有男生“骑行达人”又有女“骑行达人”的概率;②为了鼓励女性用户使用共享单车,对抽出的女“骑行达人”每人奖励500元,记奖励总金额为X ,求X 的分布列及数学期望.附表及公式:()()()()()22n ad bc Ka b c d a c b d -=++++20.(本小题满分12分)A DB A D如图,已知椭圆221:14x C y +=的左、右顶点为1A ,2A ,上、下顶点为1B ,2B ,记四边形1122A B A B 的内切圆为2C .(1)求圆2C 的标准方程;(2)已知圆2C 的一条不与坐标轴平行的切线l 交椭圆1C 于P ,M 两点. (i )求证:OP OM ⊥; (ii )试探究2211OP OM +是否为定值.21.(本小题满分12分)设函数21()ln ,22x f x a x a R =--∈. (1)若函数()f x 在区间[]1,e ( 2.71828e =L 为自然对数的底数)上有唯一的零点,求实数a 的取值范围;(2)若在[]1,e ( 2.71828e =L 为自然对数的底数)上存在一点0x ,使得()200001122x a f x x x +<---成立,求实数a 的取值范围.请考生在(22)、(23)两题中任选一题作答,如果多做,则按所做的第一题记分. 22.(本小题满分10分)选修4-4:坐标系与参数方程在直角坐标系xOy 中,以坐标原点为极点,x 轴正半轴为极轴建立极坐标系,曲线1C 的极坐标方程为()4cos 0ρθρ=>.M 为曲线1C 上的动点,点P 在射线OM 上,且满足20OM OP ⋅=.(1)求点P 的轨迹2C 的直角坐标方程; (2)设2C 与x 轴交于点D ,过点D 且倾斜角为5π6的直线l 与1C 相交于A ,B 两点,求DA DB ⋅的值.23.(本小题满分10分)选修4-5:不等式选讲 已知函数()1f x x a x =++-.(1)当1a =时,求不等式()4f x x ≥+的解集;(2)若不等式()21f x a ≥-恒成立,求实数a 的取值范围.2019届福建省福建师大附中理科数学高考模拟试卷参考答案20190528一、选择题:本大题有12小题,每小题5分,在每小题给出的四个选项中,只有一项符合题目要求. DBAA D,BDADB,BC三、填空题:本大题有4小题,每小题5分.13. 15 14. 2 15. 16. 310+三、解答题:解答应写出文字说明、证明过程或演算步骤. 17.(1)设等差数列{}n a 的公差为d ,依题意得:11545502622a d a d ⨯⎧+=⎪⎨⎪+=⎩,解得14a =,3d =, 所以1(1)31n a a n d n =+-=+. 当1n =时,21314b b =+=, 当2n ≥时,131n n b S +=+, 131n n b S -=+, 以上两式相减得13n n n b b b +-=,则14n n b b +=,又214b b =,所以14n n b b +=,*n N ∈.所以{}n b 为首项为1,公比为4的等比数列, 所以14n n b -=.(Ⅱ)因为12112n n nc c c a b b b ++++=L ,*n N ∈ 当时,112121n n n c c ca b b b --+++=L ,以上两式相减得13n n n n c a a b +=-=, 所以1334n n n c b -==⨯,2n ≥.当1n =时,121ca b =,所以1217c a b ==,不符合上式,所以2201812201973(444)c c c +++=++++L L 201820194(14)734314-=+⨯=+-.18.(本小题满分12分)解:(Ⅰ)连结CE .在四边形ABCD 中,AD BC P ,90BAD ∠=︒,23AB =,4BC =,6AD =,13AE AD =,∴12A E AE ==,4BE DE ==,……1分 ∴四边形BCDE 为菱形,且BCE ∆为等边三角形.又∵P 为BE 的中点,∴CP BE ⊥.………………2分 ∵1122A P BE ==,23CP =14A C =,满足22211A P CP A C +=,∴1CP A P ⊥,………………3分 又∵1A P BE P =I , ∴CP ⊥平面1A BE .………………4分∵CP ⊂平面1A CP ,∴平面1A CP ⊥平面1A BE .……5分(Ⅱ)以P 为原点,向量,PB PC u u u r u u u r的方向分别为x 轴、y 轴的正方向建立空间直角坐标系P xyz -..6分 则()0,0,0P ,(0,23,0)C ,(4,23,0)D -,(13A -,所以(13PA =-,(4,23,0)PD =-u r ,…7分z yxA1PECB设(),,n x y z =是平面1A PD 的一个法向量,则10,0,n PA n PD ⎧⋅=⎪⎨⋅=⎪⎩u u u r u u u r 即30,4230,x z x y ⎧-+=⎪⎨-+=⎪⎩………8分 取1z =,得(3,2,1)n =.……9分 取平面1A BE 的一个法向量()0,1,0m =.…10分 ∵22cos ,222n m n m n m ===g ,………11分 又二面角1B A P D --的平面角为钝角,所以二面角1B A P D --的余弦值为22-.…………12分19.(1)由图中表格可得22⨯列联表如下:不喜欢骑行共享单车喜欢骑行共享单车合计 男 10 45 55 女 15 30 45 合计2575100将22⨯列联表中的数据代入公式计算得841.303.345557525)10301545(100))()()(()(222<≈⨯⨯⨯⨯-⨯=++++-=d b c a d c b a bc ad n K ,所以在犯错误概率不超过05.0的前提下,不能认为是否“喜欢骑行共享单车”与性别有关.(2)视频率为概率,在我市“骑行达人”中,随机抽取1名用户,该用户为男“骑行达人”的概率为53,女“骑行达人”的概率为52.①抽取的4名用户中,既有男“骑行达人”,又有女“骑行达人”的概率为625528)52()53(144=--=p ;②记抽出的女“骑行达人”人数为Y ,则Y X 500=.由题意得)52,4(~B Y ,ii i C i Y P -==∴4)53()52()(4(4,3,2,1,0=i ),∴Y 的分布列为Y 0 1 2 3 4p62581 62521662521662596 62516∴X 的分布列为X 0500100015002000p62581 62521662521662596 62516所以554)(=⨯=Y E ,所以X 的数学期望800)(500)(==Y E X E 元. 20.(本题满分12分)(1)因为2A ,1B 分别为椭圆221:14x C y +=的右顶点和上顶点,则2A ,1B 坐标分别为(2,0),(0,1),可得直线21A B 的方程为:22x y +=…………………………………………2分则原点O 到直线21A B 的距离为2512d ==+, 则圆2C 的半径5r d ==,故圆2C 的标准方程为2245x y +=.………………………………………………4分(2)(i )可设切线()()1122:(0),,,,l y kx b k P x y M x y =+≠,将直线PM 方程代入椭圆1C 可得22212104k x kbx b ⎛⎫+++-=⎪⎝⎭,由韦达定理得: 1222122214114kbx x k b x x k -⎧+=⎪+⎪⎪⎨-⎪=⎪+⎪⎩则()()()22221212121221414k b y y kx b kx b k x x kb x x b k -+=++=+++=+,……6分 又l 与圆2C 相切,可知原点O 到l 的距离2251d k ==+,整理得22514k b =-,则2122114b y y k -=+,所以12120OP OM x x y y ⋅=+=u u u r u u u u r ,故OP OM ⊥.……………………8分(ii )由OP OM ⊥知1||||2OPM S OP OM =△,①当直线OP 的斜率不存在时,显然||1,||2OP OM ==,此时221154OP OM +=; ②当直线OP 的斜率存在时,设1:OP y k x =代入椭圆方程可得222114x k x +=,则221414x k =+,故()()212222212141114k OP x y k x k+=+=+=+,………………………………………………10分同理()221122211141414114k k OM k k ⎡⎤⎛⎫+-⎢⎥⎪+⎢⎥⎝⎭⎣⎦==+⎛⎫+- ⎪⎝⎭,则()()221122221114411544141k k OP OM k k +++=+=++. 综上可知:221154OP OM +=为定值.…………………………………………………………12分21.【解析】(1)2()a x af x x x x-'=-=,其中[1,e]x ∈.①当1a ≤时,()0f x '≥恒成立,()f x 单调递增,又∵()10f =,.函数()f x 在区间[1,]e 上有唯一的零点,符合题意.②当2a e ≥时,()0f x '≤恒成立,()f x 单调递减,又∵()10f =,.函数()f x 在区间[1,]e 上有唯一的零点,符合题意.…………………………………………………………3分③当21e a <<时,1x <„时,()0f x '<,()f x 单调递减,又∵()10f =,∴(1)0f f <=,∴函数()f x在区间e x <„时,()0f x '>,()f x 单调递增,当()0f e <时符合题意,即21022e a --<, ∴212e a ->时,函数()f x在区间上有唯一的零点;∴a 的取值范围是21|12e a a a ⎧⎫-≤>⎨⎬⎩⎭或.…………………………6分(2)在[1,]e 上存在一点0x ,使得()200001122x a f x x x +<---成立,等价于00001ln 0a x a x x x +-+<在[1,]e 上有解,即函数1()ln ag x x a x x x=+-+在[1,]e 上的最小值小于零.2222211(1)(1)()1a a x ax a x x a g x x x x x x ---+--'=---==,……………………………………8分①当1a e +≥时,即1a e ≥-时,()g x 在[1,]e 上单调递减,所以()g x 的最小值为()g e ,由()10ag e e a e+=+-<可得211e a e +>-,∵2111e e e +>--,∴211e a e +>-; ②当11a +≤时,即0a ≤时,()g x 在[1,]e 上单调递增,所以()g x 的最小值为()1g ,由()1110g a =++<可得2a <-;………………………………………………10分 ③当11a e <+<时,即01a e <<-时,可得()g x 的最小值为()1g a +,∵0ln(1)1a <+<,∴0ln(1)a a a <+<,1(1)1ln(1)2ln(1)211ag a a a a a a a a a +=++-++=+-+>++,所以()10g a +<不成立.综上所述:可得所求a 的取值范围是21(,2),1e e ⎛⎫+-∞-+∞⎪-⎝⎭U .……………………………………12分 请考生在(22)、(23)两题中任选一题作答,如果多做,则按所做的第一题记分.22.(1)设P 的极坐标为()(),0ρθρ>,M 的极坐标为()()110,ρθρ>, 由题设知OP ρ=,14cos OM ρθ==.∴4cos 20ρθ=,即2C 的极坐标方程()cos 50ρθρ=>,∴2C 的直角坐标方程为5x =. (2)交点()5,0D ,∴直线l的参数方程为512x y t ⎧=⎪⎪⎨⎪=⎪⎩(t 为参数),曲线1C 的直角坐标方程()22400x y x x +-=≠,代入得250t -+=,70∆=>, 设方程两根为1t ,2t ,则1t ,2t 分别是A ,B 对应的参数,∴125DA DB t t ⋅==. 23.(1)不等式为114x x x ++-≥+,可以转化为:1114x x x x ≤-⎧⎨---+≥+⎩或11114x x x x -<<⎧⎨+-+≥+⎩或1114x x x x ≥⎧⎨++-≥+⎩, 解得43x ≤-或4x ≥,∴原不等式的解集是443x x x ⎧⎫≤-≥⎨⎬⎩⎭或.(2)()()()min 11f x x a x a =+--=+,∴2211111a a a a a <-⎧+≥-⇔⎨--≥-⎩或2111a a a ≥-⎧⎨+≥-⎩,解得a ∈∅或12a -≤≤. ∴实数a 的取值范围是[]1,2-.。