广东省深圳市福田区2018年一模数学试题.pdf
【福田区三年级数学】2017-2018学年第一学期教学质量检测小学三年级数学试题及答案
福田区2017—2018学年第一学期三年级数学期末考试卷分析一、计算基本功。
（共28分）1.【答案】：1000；0；115；50408；10；26；122.【答案】：24.4；2961；1600；32243.【答案】：9；58；100；180二、精准填空。
（每空1分，共27分）1.【答案】：除；减；732.【答案】：33.【答案】：8【答案】：12；10；大月；315.【答案】：17:30；晚上10:006.【答案】：＞；＞；＝＜；＜；＜7.【答案】：4；80；58.【答案】：2；299.【答案】：0.09；四点一五10.【答案】：165；630；11三、准确判断，在括号内打“√”或“×”。
（每题1分，共6分）【答案】：×2.【答案】：√3.【答案】：√4.【答案】：×5.【答案】：×6.【答案】：×四、动手能力。
（共15分）1.【答案】：【答案】：图形一的周长：12厘米图形一的周长：20厘米（答案不唯一，言之有理即可）3.【答案】：8种五、解决问题1、【答案】：10×6×8=480个2、【答案】：165x2+85=415>400，钱不够。
3、【答案】：122x3+122=488（千克），每天总共需要480千克。
4、【答案】：188+200+110+110+200=808（米），场地周长是808米。
5、【答案】：（1）475+205+220=900（元）；（2）900÷3=300（元）6、【答案】：11时30分。
2018年广东省深圳市数学中考一模试卷2
2018年广东省深圳市数学中考一模试卷扫描二维码，下载客户端，随时随地做题支持iPhone/Android手机1.2018的相反数是（）A. -2018B.C. 2018D.2.下列各图中，可以是一个正方体的平面展开图的是（）A.B.C.D.3.下列计算结果正确的是（）A.B.C.D.4.据报道，我国自行研发的第一艘001A型航空母舰吨位达到6.5万吨，造价30亿美元，用科学记数法表示6.5万吨为（）A. 吨B. 吨C. 吨D. 吨5.下列四个图形分别是四届国际数学家大会的会标，其中属于中心对称图形的有（）A.1个B.2个C.3个D.4个6.如图，一只蚂蚁以均匀的速度沿台爬行，那么蚂蚁爬行的高度h随时间t变化的图象大致是（）A.B.C.D.7.我市某中学九年级（1）班开展“阳光体育运动”，决定自筹资金为班级购买体育器材，全班50名同学筹款情况如下表：则该班同学筹款金额的众数和中位数分别是（）A. 11，20B. 25，11C. 20，25D. 25，208.在中，，以点B为圆心，BC的长为半径作弧，交AB于点D，若点D为AB的中点，则阴影部分的面积是（）A.B.C.D.9.如图所示，在中，，以点B为圆心，BC长为半径做弧，交AB于点D，再以点A为圆心，AD长为半径画弧，交AC于点E，下列结论错误的是（）A.B.C.D.10.下列说法正确的是（）A.真命题的逆命题都是真命题B.在同圆或等圆中，同弦或等弦所对的圆周角相等C.等腰三角形的高线、中线、角平分线互相重合D.对角线相等且互相平分的四边形是矩形11.已知二次函的图象如图所示，它与x轴的两个交点分别，，对于下列命题：①；②；③；④，其中正确的是（）A.3个B.2个C.1个D.0个12.如图，在矩形ABCD中，E是AD的中点，垂足为F，连接DF，下列四个结论；③；④，。
其中正确的是（）A. ①②③B. ②③④C. ①③④D. ①②④13.若一元二次方有两个相等的实数根，则c的值是。
精品解析：广东省深圳市福田区2018届九年级下学期八校第一次联考数学试题(解析版)
2017—2018学年度第二学期初三年级联考数学学科试题一、选择题（本大题共12小题，每小题3分,共36分.每小题给出4个选项,其中只有一个是正确的）1. -3的相反数是A. -3B.C.D.【答案】B【解析】只有符号不同的两个数互为相反数，所以-3的相反数是就3，故选B.2. 分别从正面、左面和上面看下列立体图形，得到的平面图形都一样的是A. B. C. D.【答案】A【解析】球从正面、左面和上面看到的图形都是圆；圆锥从正面和左面看到的图形是等腰三角形，从上面看到的图形是圆和圆心；长方体从正面、左面和上面看到的图形都是矩形，但三个矩形不全等；圆柱从正面和左面看到的图形是矩形，从上面看到的图形是圆，故选A.3. 据统计，我国高新技术产品出口额达40.570亿元将数据40.570亿用科学记数法表示为A. B. C. D.【答案】A【解析】科学记数法的表示形式为a×10n的形式，其中1≤|a|＜10，n为整数．确定n的值时，要看把原数变成a时，小数点移动了多少位，n的绝对值与小数点移动的位数相同．当原数绝对值＞1时，n是正数；当原数的绝对值＜1时，n是负数．所以40.570亿=，故选A.4. 下列平面图形中，既是轴对称图形又是中心对称图形的是A. B. C. D.【答案】B【解析】A.既不是轴对称图形，也不是中心对称图形；B.是轴对称图形，也是中心对称图形；C.是轴对称图形，不是中心对称图形；D.是轴对称图形，不是中心对称图形，故选B.5. 如图，，下列结论：；；；，其中正确的结论有A. B. C. D.【答案】A【解析】因为∠B=∠C，所以AB∥CD，∠A=∠AEC，因为∠A=∠D，所以∠AEC=∠D，所以AE∥DF，∠AMC=∠FNC，因为∠BND=∠FNC，所以∠AMC=∠BND，无法得到AE⊥BC，所以正确的结论有①②④，故选A.6. 关于x的不等式组的解集为，那么m的取值范围为A. B. C. D.【答案】D【解析】解不等式组得，，因为原不等式组的解集为x＜3，所以m≥3，故选D.7. 某商贩同时以120元卖出两双皮鞋，其中一双亏本，另一双盈利，在这次买卖中，该商贩盈亏情况是A. 不亏不盈B. 盈利10元C. 亏本10元D. 无法确定【答案】C【解析】设亏本的皮鞋进价为x，盈利的皮鞋进价为y，则（1-20%）x=120，（1+20%）y=120，解得x=150，y=100，因为120×2-（150+100）=-10，所以亏本10元，故选C.8. 如图，在▱ABCD中，对角线相交于点O，添加下列条件不能判定▱ABCD是菱形的只有A. B. C. D.【答案】C【解析】因为对角线互相垂直的平行四边形是菱形，所以A能够判定▱ABCD是菱形；因为一组邻边相等的平行四边形是菱形，所以B能够判定▱ABCD是菱形；因为对角线相等的平行四边形是矩形，所以C不能够判定▱ABCD是菱形；因为∠1=∠2，OB=OD，所以AB=AD，所以D能够判定▱ABCD是菱形，故选C.9. 下列命题错误的是A. 经过三个点一定可以作圆B. 同圆或等圆中，相等的圆心角所对的弧相等C. 三角形的外心到三角形各顶点的距离相等D. 经过切点且垂直于切线的直线必经过圆心【答案】A10. 在某学校“经典古诗文”诵读比赛中，有21名同学参加某项比赛，预赛成绩各不相同，要取前10名参加决赛，小颖已经知道了自己的成绩，她想知道自己能否进入决赛，只需要再知道这21名同学成绩的A. 平均数B. 中位数C. 众数D. 方差【答案】B学§科§网...学§科§网...学§科§网...学§科§网...学§科§网...11. 如图，将半径为2，圆心角为的扇形OAB绕点A逆时针旋转，点的对应点分别为，连接，则图中阴影部分的面积是A. B. C. D.【答案】C【解析】连接OO′，BO′，由题意得，∠OAO′=60°，所以△OAO′是等边三角形，所以∠AOO′=60°，因为∠AOB=120°，所以∠BOO′=60°，所以△BOO′是等边三角形，所以∠AO′B=120°，所以。
2018-2019初一数学福田区统考试卷
福田区统考考试 答案
一、选择题 题号 答案 1 B 2 A 3 C 4 D 5 A 6 B 7 D 8 A 9 C 10 B 11 C 12 D
二、填空题 题号 答案 13 ③ 14 15 16
15
17
19 6
三、解答题 17． （1） −5 ； （2） −16 ； （3） 12 ． 18．原式 = −3 x + 8 y = 35 ．
9 ； 7
AP = 12 − 4 ( t − 3) = −4t + 24 ， AQ = 12 − ( t − 3) = −t + 15 ，
2 ( −4t + 24 ) = −t + 15 ，解得 t =
当6 t
33 ； 7
39 时， 5
AP = 0 + 4 ( t − 6 ) = 4t − 24 ， AQ = −t + 15 ， 2 ( 4t − 24 ) = −t + 15 ，解得 t = 7 ．
题目整体难度一般， 对大部分知识点都有考察到， 而且比较注重基本概念的理解与理解。 应用题方面考察得较少， 最后的动点题目也考到了分类讨论， 还没有考试的同学值得参考一 下。
考点分析：
试卷难度分析、知识范围、难度情况分析表 题型 题号 1 2 3 4 5 选择题 6 7 8 9 10 11 12 13 填空题 14 15 16 17 18 19 应用题 20 21 22 23 考点 倒数 正方体展开图 科学计数法 调查方式 代数式 同类项 角分线 线段计算 特殊的幂 绝对值和平方非负性 同类项和立体几何 找规律 立体几何 钟面角 打折销售 定义新运算 有理数综合计算 化简求值 列代数式 数据收集与整理 角度计算 一元一次方程应用题 动点问题 难度 ★ ★ ★ ★ ★ ★ ★ ★ ★ ★ ★★ ★★★ ★ ★★ ★★ ★★ ★ ★ ★ ★ ★★ ★★ ★★★ 分值 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 12 5 8 5 6 8 8
2018年广东省深圳市福田区八校中考数学一模试卷带解析答案
11． （3 分）如图，将半径为 2，圆心角为 120°的扇形 OAB 绕点 A 逆时针旋转 60°，点 O，B 的对应点分别为 O′，B′，连接 BB′，则图中阴影部分的 面积是（ ）
第 2 页（共 26 页）
A．
B．2
﹣
C．2
﹣
D．4
﹣
12． （3 分）如图，正方形 ABCD 的边长是 3，BP＝CQ，连接 AQ，DP 交于点 O， 并分别与边 CD，BC 交于点 F，E，连接 AE，下列结论：①AQ⊥DP；②OA2 ＝OE•OP；③S△AOD＝S 四边形 OECF；④当 BP＝1 时，tan∠OAE＝ 结论的个数是（ ） ，其中正确
2． （3 分）分别从正面、左面和上面看下列立体图形，得到的平面图形都一样的 是（ ）
A．
B．
C．
D．
3． （3 分）据统计，我国高新技术产品出口额达 40.570 亿元，将数据 40.570 亿 用科学记数法表示为（ A．4.0570×109 C．40.570×1011 ） B．0.40570×1010 深圳市福田区八校中考一模数学试卷
一、选择题（本大题共 12 小题，每小题 3 分，共 36 分.每小题给出 4 个选项， 其中只有一个是正确的） 1． （3 分）﹣3 的相反数是（ A．﹣3 B．3 ） C． D．
； 再在不等式组
数解中选取一个合适的解作为 a 的取值，代入求值． 19． （7 分）为了了解同学们每月零花钱的数额，校园小记者随机调查了本校部 分同学，根据调查结果，绘制出了如下两个尚不完整的统计图表． 调查结果统计表 组别 A B C D E 分组（单位：元） 0≤x＜30 30≤x＜60 60≤x＜90 90≤x＜120 x≥120 人数 4 16 a b 2
2018初三福田区一模试卷及答案
2018初三福田区一模试卷及答案2018年九年级教学质量检测试卷（4月27日福田区一模）Ⅰ选择填空1.---After seeing Snowflake Boy Manfu’s picture, their class raised money for the poor children.---Yes, they organized a book fair on the playground and sold some books and CDs.A. collectedB. put upC. brought up2.---The boys were crazy when Spiderman came along at the parade in the Universal Studio.---Of course they were. You know, Spiderman was always those boys’ favourite.A. made a speechB. disappearedC. showed up3.---The government will focus on solving the problem of heavy burdens on the students.---That’s good news for us. We have too much work to do after school.A. give a hand toB. give attention toC. take part in4.---As a result of playing computer games too much, Michael is near-sighted and has to wear glasses.---It’s difficult for youngsters to stay away from games.A. Thanks toB. Result inC. Because of5.---What good news! Eric will be a delegate of MUN and will set off for New York next Monday.---Yes, he made it as last. I’m sure his cultural exchange will be educational.A. leave forB. be away fromC. go on a trip to6.---Stephen Hawking, the brilliant British scientist, passed away at the age of 76 last month. ---What a pity! He is regarded as one of the best minds in the world.A. most valuable ideasB. greatest peopleC. wisest thoughts7.---Mother often warned Peter not to play on his phone in the street, but her warning fell on deaf ears.---Unfortunately, Peter was hit by a car and sent to hospital yesterday.A. was heardB. was agreedC. was not noticed8.---Hong Kong’s richest, Li Ka-shing, announced plans to retire before his 90th birthday in July.---It is reported his fortune is up to $37 billion. However, he led a dog’s life during his childhood.A. had a happy lifeB. lived a hard lifeC. lived with a dog9. Your _______ is the way that your face looks at a moment. It shows what you are thinking or feeling.A. expressionB. experienceC. experiment10. If you _______ someone, you invite them to flight or compete with you in some way.A. encourageB. challengeC. introduce11. I’m afraid I can’t go swimming with you, because I have to _______ my little brother at homeA. look forB. make fun ofC. care for12.---Mum, I think I need a bigger room.---You should throw away your old toys. They have _______ too much space.A. taken upB. taken placeC. taken off13.---Jack, you seemed _______ at the party.---You must be kidding. I felt _______ among those successful people.A. confident; out ofB. nervous; out of shapeC. relaxed; out of place14.---What an amazing city Shenzhen is! It is modern and clean with many beautiful parks.---That’s true. We can’t imagine that it _______ be a small fishing village!A. is used forB. used toC. got used to15.---Putin _______ the other candidates and was elected president for his fourth time.---Yes. Putin is very popular. His votes were far _______ the others’ in the election.A. beat; ahead ofB. defeated; better thanC. won; in front ofⅡ完形填空Stephen Hawking, a famous British physicist, died aged 76 on March 14th, 2018. He died peacefully at his home in Cambridge in the early hours of Wednesday, his family said.As one of the well-known _______on space and time in the world, Hawking devoted his whole life to discovering the secrets of the Universe, and he is called the King of the Universe. However, this genius’s life was not smooth and he had been _______ from the disease since he was 21 years old. When he studied maths and science at Oxford University, he became _______ ill, causing his life very harsh: he could neither speak with his mouth nor stand on his legs or move his body. _______, he was given only a few years to live. But he refused to _______ his hope of living and went on to study at Cambridge University after _______ from Oxford University. In 1965, he got a doctor’s degree in philosophy. Then he worked as a professor at Cambridge University. He was known for his work with black holes and relativity, and wrote several popular science books, _______ A Brief History of Time, The Universe in a Nutshell, The Grand Design, etc.Although he lost many most important living conditions ordinary people have, he still felt he was rich, such as a finger that could move, a brain that could think… all these made him _______ to life. Last year, Hawking was invited to China. His self-confidence and humorous conversations _______ us deeply.From Stephen Hawking’s _______ experience, we learn that nobody should lose hope, no matter how bad the situation is. As he once said, “Life is not fair, you just have to do the best you can in your own situation.”16. A. astronauts B. artists C. scientists17. A. separating B. suffering C. supporting18. A. seriously B. probably C. naturally19. A. As a result B. What’s more C. For example20. A. take up B. put up C. give up21. A. graduating B. furthering C. completing22. A. according B. including C. during23. A. thankful B. harmful C. regretful24. A. increased B. improved C. impressed25. A. universal B. unusual C. ordinaryⅢ阅读理解AIn recent years, there have been many breakthroughs in the AI (人工智能) field. Here, let us see four of the most interesting developments in AI in 2017.Magic RadioWant a friend that you can talk to anytime? Then you might like Magic Radio from Microsoft. Magic Radio features a cool cartoon figure. You can talk to it about anything. There is also an app for Magic Radio. With it, you can decorate the cartoon figure and even send it a gift online. Parents can also use the app to talk to their children. Magic Radio also records your life and analyzes(分析) your behavior. This way, it is easy for parents to know what you’re doing when they’re not around.Sophia, RobotSophia is the first robot in the world to get citizenship(公民身份) in Saudi Arabia. In 2015, a US company built Sophia. She looks and acts like a human woman. She can talk to people naturally. She even has her own body language. For example, she may blink her eyes when she is talking about something funny. She also has the ability to learn from people, books and the Internet. Sophia was designed to have her values. “Don’t worry, if you’re nice to me, I’ll be nice to you.”Samsung’s BixbyLike the voice recognition (识别) system, Siri on iPhones, Samsung has a similar system called Bixby. Experts said that it is smarter than any other similar system. If you once asked it to complete a certain task, it will remember that task the next time you talk to it. For example, you might like to listen to certain music while you’re baking a cake. Next time when you tell Bixby “ I want to make a cake”, it will show you how to make a cake and play music for you automatically.AlphaGo ZeroAlphaGo is an AI program that plays the board game GO(). Google Company has created a new version of the AI program called AlphaGo Zero. The new AI program is unique in the way it learned to play Go. AlphaGo Zero mastered Go in just two days without any human knowledge of the game. On the third day, it beat AlphaGo by 100 games to zero, becoming the best Go player on the planet.26. Which of the following developments can help parents learn about their children?A. Magic RadioB. Robot SophiaC. Samsung’s BixbyD. AlphaGo Zero27. What is the most amazing development from AlphaGo to AlphaGo Zero?A. It can record other players’ life and analyzes their behavior.B. It can remember all the tasks it has done and show you again.C. It has the ability to learn from people, books and the Internet.D. It has very strong learning ability and masters Go in two days.28. What can you read about Samsung’s Bixby?A. Samsung’s Bixby is more practical and enjoyable.B. Samsung’s Bixby can talk in the way Sophia does.C. Samsung’s Bixby has a smarter voice recognition system.D. Samsung’s Bixby is better than new version AlphaGo Zero.29. Which of the following can be the best title?A. AlphaGo Zero VS AlphaGoB. Samsung’s Bixby VS Siri On IphoneC. Breakthroughs in the AI FieldD. Welcome to the AI FieldBA young prince had just recently become King. In order to rule his kingdom, he decided to learn all the wisdom of world. Therefore, he gathered all the wise men from his kingdom and abroad and ordered them to look for books for him to read and learn from.Five years passed quickly. The wise men returned with their camels, carrying 5000 books full of wisdom. Seeing so many books, the king was so shocked that he didn’t know how to start. So he ordered the wise men to condense the books and bring them back to him. Another five years passed. The wise men again went to see the king, this time bringing 500 books, but the king still thought there were too many. Five more years passed. The wise men brought back 50 books. At this time, the king was troubled with many problems, but he still felt were too many books.During the next few years, the wise men worked hard to condense the 50 books into one book and present it to the king. The king took no interest in reading this book, nor did he have time to learn from it. More problems broke out in his kingdom---his enemies endlessly attacked and diseases affected his people. He did not have the wisdom to solve these problems. Finally, the king was killed by a ruler in the neighbour country and his country was destroyed.Waiting for wisdom to come to you is just a way of being lazy. If there is no action, there can be no gain. Only if we set out to find wisdom can we make a difference.30. Why did the king gather all the wise man?A. To help him rule the kingdom.B. To teach him some knowledge.C. To look for books for his kingdom.D. To look for books of wisdom for him.31. What does the underlined word “condense” mean?A. To put much information into a small space.B. To make the price of books lower.C. To make the books smaller than ever.D. To make the books easier than ever.32. Why wouldn’t the king read the ONE book made from 50 books?A. Because he didn’t know how to start.B. Because he lost interest and time for it.C. Because he wasn’t sure about that book.D. Because it didn’t have the wisdom he wanted.33. What can we learn from this passage?A. Time and tide wait for no man.B. Wisdom can make a country strong.C. Wisdom won’t come to you by itself.D. The meaning of living is to find wisdom.CFour years ago, Chinese skater Wu Dajing introduced himself to the skating world by winning a silver medal at the Sochi Olympic Winter Games. At this year’s Pyeong Chang Olympics, Wu not only returned, but also made history.On Feb 22th, 2018, Wu won the gold medal in the men’s short-track 500-metre race. He set a new world record with a time of 39.584 seconds. He also became the first Chinese man to take home an Olympic short-track gold medal. BBC said Wu’s win was “flawless”, because he was much faster than all of the other skaters.“I didn’t give them a chance and I kept my speed from the start,” he told the media afterthe match.But China’s short-track teams didn’t do well in general at the Games. Chinese skaters in the women’s 500 metres, 3000-metre relay and men’s 1500 metres all failed to take home the gold. Wu was China’s biggest hope, which put a lot of pressure on him. But he proved himself with his great performance.Wu is now known as a highly talented skater. But things were not always that way. When Wu joined the national team in 2010, he was seen as almost “nothing”compared to gifted skaters like Zhou Yang and Fan Kexin, as his coaches said at the time. Their comments made him quite upset. But as the saying goes, “Winners never quit and quitters never win”. Wu didn’t want to give up and worked as hard as he could. He practiced skating all year round. He even didn’t return to his hometown for the holidays for 10 years. “I believe in myself,” he told the media after his match at the Olympics.34. What does the underlined word “flawless” mean?A. OrdinaryB. ValuableC. PerfectD. Difficult35. According to the passage, which of the following is RIGHT?A. Wu won a gold medal at the Sochi Olympic Winter Games four years ago.B. Wu broke the world record with a time of 39.584 seconds and made history.C. Wu’s coaches placed great hope on him when he joined the national team.D. Chinese skaters in women’s 500 metres, 3000-metre relay took home the gold.36. What does the last paragraph mainly tell us?A. The national team and coaches regarded Wu as a talented skater all the way.B. Wu kept on practicing skating all year round even in the holidays for 10 years.C. Zhou Yang and Fan Kexin were more and more hard-working on skating than Wu Dajing.D. It is the hardwork, confidence and strong will that have made Wu a champion.37. Which one might be the best title of this passage?A. From Zero to Skating HeroB. The Dream of a Skater, WuC. Pyeong Chang Winner OlympicsD. Rising of China’s Bright Skiing StarDCold weather, central heating, and falling leaves---all these things mean that flu season has arrived. And you shouldn’t take flu lightly. Flu season comes every year from October to April. This year has been especially bad, according to the US Centers for Disease Control and Prevention (美国疾病防控中心CDC).The flu is an illness caused by the influenza virus. It is different from the common cold in that it can strike suddenly---and it is much more serious .The flu can cause serious dehydration(脱水) and liver (肝脏)damage ,both of witch can kill people.The flu is especially dangerous for children and elderly people, since they have weaker immune (免疫) system. As of April 20, 142 children in the US had died from the flu during this flu season, according to CDC.These deaths are not limited to the USA. In China, many people die from the flu each year as well. An elderly man in Beijing died from the flu this season, which led to a wide discussion online. The man’s family didn’t know how serious the flu was, just like many others who don’t understand the flu very well.What can you do to keep the flu away? Hygiene is important --- take a shower every dayand wash your hands often. Do not spend time around people who have the flu. And if you can, get a flu vaccine --- this is a type of special medicine that can help prevent the flu. All of these things can keep you safe during flu season. If you are stick with the flu, stay home to prevent spreading it to others.38. What is the main idea of the second paragraph?A. It tells us it’s the flu season now.B. It tells us the flu can kill people.C. It tells us what the cold is.D. It tells us what the flu is.39. Who are easy to catch the flu?A. Kids and the old.B. Adults and children.C. Women and men.D. Doctors and reporters.40. What led to a wide discussion online?A. That the children don’t exercise very often.B. That the flu will cause serious liver damage.C. That an old man in Beijing died during the flu season.D. That the old man’s family don’t know what the flu is.41. What does the last paragraph tell us?A. Get a flu vaccine shot.B. How to prevent the flu.C. Take a shower every dayD. Wash your hands often.ELove it or not, flying is necessary if we want to get to a faraway destination. Taking a flight in China can be a boring experience. This is because Chinese airlines didn’t allow smartphones to be powered on during flights. However, the launch of a new Chinese satellite Shijian 13 has changed the situation. Now, many Chinese airlines allow passengers to use their smartphones freely, that is, we can finally catch up on our favorite books and songs while at 30000 feet in the air.The communication satellite Shijian 13 can transmit (输送) 20 gigabytes (GB千兆) of data per second and cover most parts of China’s land. This makes it the most powerful communication satellite China has ever developed. In fact, Shijian 13 has a higher message capacity (容量) than the total capacity of all China’s communication satellites launched.The powerful satellite has greatly improved the Internet, including that on fast-moving transportation vehicles such as planes and high-speed trains.With shijian 13, people are able to get a download speed of up to 150 MBps on their smartphones. According to Zhou Zhicheng, head of the communications satellite department of the China Academy of Space Technology, the speed means it takes less than one minute to download a 1-GB movie. Watching online TV on planes and high-speed trains has come true.But the satellite’s benefits are far greater that that. China is trying to have 22 communication satellites in the sky by 2025. Shijian 13will be able to cover wider areas, including mountains, oceans and deserts. When people have emergencies while hiking or sailing, the satellite can help them report and ask for help. It will also help people in natural disasters.42. What does the first paragraph tell us?A. Shijian 13 has already enabled us to use smartphones while flying.B. China Airlines allow passengers aboard to use their smartphones.C. Shijian 13 is a kind of rocket which can carry us to a faraway place.D. Shijian 13 can make us catch up with latest books and pop songs.43. What does the underlined word “that” in paragraph 3 refer to?A. Shijian 13B. The InternetC. A high-speed trainD. The powerful satellite44. Which sentence tells Shijian 13 is the most powerful communication satellite?A. With it we can download a 1-GB movie within a second.B. With it we can ask for help when we meet typhoons in sea.C. With it we can do anything we like on high-speed trains.D. With it we can prevent natural disasters from happening.45. What does the story mainly talk about?A. Why China’s satellite technologies are so advanced.B. What the experience has been like on high-speed trains.C. How China develops the new communication satellite.D. What changes Shijian 13 brings to China sooner or later.Ⅳ语法填空Srikanth Bolla, 24 years old, is the first blind CEO in the world, who built a company valued at over $75 million. Today, he considers himself the 46.___________ (lucky) man in the word, not only his 47. ___________ (succeed), but also for having great parents and friends who always stand by him.When Srikanth was born, some of his 47. ___________ (parent) friends and relatives them not to keep him, because it was difficult for a poor family 48. ___________ (raise) a blind baby. But his parents didn’t follow them. They decided to give Srikanth a positive, loving environment.Life was not easy for Srikanth. He had to face many challenges all his life. He 49. ___________ (refuse) by his village school. So he had to go to school for special children. He did very well in studies and also developed many hobbies such as chess and cricket. Later Srikanth found a teacher who was kind enough to turn all his lessons into audio clips, and helped him 50. ___________ his exams.51. ___________ high school was difficult for a blind student, Srikanth didn’t give up. Instead, he worked very hard and tried his best to go to MT(麻省理工), and graduated from 52. ___________ university in 2012. After his graduation, he returned to India, and decided to start to a company 53. ___________ employed disabled people like him.Srikanth said, “Compassion(怜悯) is showing somebody the way to love and giving them the opportunity to grow up. If you do something good, it 54. ___________ (come) back to you.”Ⅴ书面表达假如你是九年级一班的李新，你们的新外教Peter想去北京参观故宫博物院，希望你能给他作一个大概的介绍。
广东省深圳市福田区2018-2019学年第一学期七年级数学教学质量检测卷(扫描版)
2018-2019学年第一学期教学质量检测七年级数学试卷说明：本试卷考试时长90分钟，满分100分.答题必须在答题卷上作答，在试题卷上作答无效•• • •一. 选择题（本题共12小题，每小题3分，共36分.每小题给出4个选项，其中只有一个是正确的）1. 4■的倒数是（）3.中国高速路里程已突破13万公里，居世界第一位，将13万用科学记数法表示为（）4.下列调查中，最适合采用普査方式的是（ ）A. 对全省初中学生每天阅读时间的调查B. 对中秋节期间全国市场上月饼质量悄况的调査C. 对某品牌手机的防水功能的调查D. 对某校七年级2班学生肺活量情况的调査 5.当加=2时，代数式*（加+ 8）的值等于（ ）7.己知射线OC 是AAOB 的平分线，若ZJOC=30° ,则ZAOB 的度数为（&如图所示，C 、D 是线段/13上两点,C 为/D 中点且MB=IOem,贝IJDB=( A. 4cm B. 5cm9. 1.5o = () ■A. 15B. 150第一部分选择题A. 一 6B. 6D. 2.下列图形中不是正方体展开图的是（A. 0.13× IO 5B. 13× IO 4C. 13× IO 5D. I3× IO 4A. 5B. 4C. 3D. 26.下列各组代数式中，属于同类项的是（B. 6m 2 与 一2卅)C. 5pq 2 与-2p'qD. 5a 与5bA. 15°B. 30°C. 45oD. 60°)ACDBC. 6cmD. 7cmC. 90D. 9⅛若 ∕lC=3cm,10. 己知 I -x+ l∣+（^+2）2 = 0,贝 ∣Jx + y=（）A.-3B. -1C. 3D. 111. 下列叙述：① 单项式一誓的系数是一今，次数是3次；② 用一个平面去截一个圆锥，载面的形状可能是一个三角形；③ 在数轴上，点/1、〃分别表示有理数a 、b,若a> b,贝∣L4到原点的距离比B 到原点的距 离大；④ 从八边形的一个顶点出发，最参可以画5条对角线； ⑤ 六棱柱有8个面，18条棱. 其中正确的有（）A. 2个B. 3个C. 4个D. 5个12. 观察下图正方形四个顶点所标的数字规律，可知数字2019应标在（）1415 □ 1316第4个止方形第二部分非选择题二、填空题（本题共4小题，每小题3分，共12分）-13.下列某种儿何体从正面、左面、上面看到的形状图都相同，则这个儿何体是 ▲ （填 写序号）.①三棱锥②圆柱③球14. 当钟面上是6点30分时，时针与分针的夹角是 ▲ 度.15. 某商品进价100元，提价30%后再打九折卖出，则可获利▲元.16. 我们称使f =成立的一对数X ，y 为“甜蜜数对”，记为 a ，Q ，如：当x=y=O 时，等式成立，记为（0, 0）.若5, 3） ,（2, Z?）都是“甜蜜数对”，则加-旳的值为▲•3 __________ 2□ 4 1第1个正方形11 _________ 10□ 12 9 第3个正方形A.第504个正方形的左下角 C.第505个正方形的右上角B.第504个正方形的右下角 D.第505个正方形的左上角第2个正方三、解答题（本题共7小题，其中第17小题12分，第IX小题5分，第19小题X分•第20小题5分,第21小题6分，第22小题8分，第23小题8分，共52分)17.(毎小题4分,共12分)计算:(1)-12-(-9)÷(-2)(2)(-2)3-(-3)1 2 3+ ∣~l I(3)(-36)x(-# + #— )18.(本题5分)先化简，再求值：3(-x + 2/) - 2(3Λ -/) ÷ 6x ,其IpX = -I , y = ~2 .19.(侮小题4分，共8分)解方程:< 1) -3.v - 7 = 2x + 3,r、5 + X 8 —2x I(2) — --------------- 5— = 120.（本题5分）某校最近发布了新的学生午休方案,为了了解学生对方案的了解程度,小明和小颖••起对该学校的学生进行了抽样调査・小明将调査结果整理麻绘制成条形统计图（如图）•昇代农“完金消楚”〃代表“知道一•些" C代表“完全不了解” •1 这次抽样调查共调查了—丄_人：2 小颖想将调爸结果绘制成扇形统计图，那么扇形统计图I Iy部分对应的扇形的閲心角应是多少度？3 若该校一共有IOoO名学生，则根据此次调査，“完全淸楚”的学生大约有参少人？21.（本题6分）如图，ZJO^=180o , ZCOD=Ao O , OD平分ZCOB9 OE平分Z/OC,求ZA OE^ Z EOD的度数.22.（本题8分）一个三位数，十位数字是0,个位数字是百位数字的2倍，如果将这个三位数的个位数字与百位数字交换位置得到一个新的三位数，则这个新的三位数比原三位数的2倍少9,设原三位数的百位数了为x・（I）原三位数可表示为▲,新三位数可表示为_丄_;（2）列方程求解原三位数•23.（本题8分）如下图，∕3=12cm,点C是线段/3上的一点，AC≈3BC.动点P从点/1出发，以4cm∕s的速度向右运动，到达点B后立即返回，以4cm∕s的速度向左运动：动点0从点C 出发，以1 cm/s的速度向右运动，到达点〃后立即返回，以lcm∕s的速度向左运动.设它们同时出发，运动时间为/秒.当点P与点。
广东省深圳市福田区2018届九年级4月教学质量检测数学试卷(扫描版)
2018年九年级数学教学质量检测试卷答案定稿第一部分选择题第二部分非选择题二．填空题：（本大题共4小题，每小题3分，共12分） 13. )(b a b - 14.31 15. -1 16. 16三．解答题：（本题共7小题，其中第17题5分，第18题6分，第19题7分，第20题8分，第21题8分，第22题9分，第23题9分，共52分）17. 解：原式= .........................4分=3 ........................5分18. 解：212(1)211a a a a +÷+-+-＝2112()(1)11a a a a a +-÷+--- .........................2分＝211()(1)1a a a a +-⨯-+.........................4分＝11a - .........................5分 当a 2=时，原式=1.........................6分（注：凡没将a=-1改成a=2的考生，只要化简和计算结果正确，一律给满分）19. （1）8； 0.08 .........................2分 （2） 补全频数分布直方图如图：.........................4分 （3） .........................7分20. 解：∵在Rt △BCD 中，∠DBC =30°，CD =2米 ， ∴BC =4.3323≈=CD 米；.........................3分∵在Rt △ABC 中，∠ABC =67°，∴tan67°=4.3AC,∴AC =3.4×2.4=8.16 .........................6分 ∴AD =8.16-2=6.16≈6米 .........................7分 答：像体AD 的 高度为6米．.........................8分21. 解：（1）根据题意可得：........................3分（2）设每星期的销售利润为元，根据题意可得：.........................5分 ∴w =-30(x -40)(x -70) 当w =0时，x 1=40，x 2=70， ∵a =-30＜0，∴当x =55时，w 有最大值为6750元， ........................7分答：当每件售价定为55元时，每星期的销售利润最大，最大利润为6750元.................8分22.解：（1）当y =0时，-x +6=0，x =6，∴A （6，0）， …………1分 当x =0时，y =6，∴B （0，6）， …………2分 ∵点C 与A 关于y 轴对称，∴C （-6，0） …………3分 （2）连接AF ，由（1）可知：OC =OA ，又∵∠COE =∠AOE =90º，OE =OE ，∴ΔCOE ≌ΔAOE ， …………4分 ∴∠CEO =∠AEO ，∵∠CEO =∠BED ，∴∠BED =∠AEO ，∵四边形ADEF 内接于圆， ∴∠BDE =∠EF A ， ∴ΔBED ∽ΔAEF ， (5)∴EFDE AE BE =∴BE ·EF =DE ·AE . …………6 （3）∵ΔBED ∽ΔAEF ， ∴∠EAF =∠EBD ，∵OA =OB =6，∠AOB =90º， ∴∠ABO =∠OAB =45º，∴∠EAF =45º， …………7分 ∴∠BAE +∠EAO =∠F AO +∠EAO =45º， ∴∠BAE =∠F AO ， ∴tan ∠F AO =tan ∠BAE =31， ∴31=OA OF …………8分 ∵OA =6， ∴OF =2，∴F （0，-2） …………9分23.解：（1）将点P （6，7）代入y =a (x -2)2-9得：a =1, …………1分∴y =(x -2)2-9 …………2分 （2）如图，作CM ⊥l ，BN ⊥l ，垂足分别为M 、N ， 延长CM ，作BG ⊥CM ，垂足为G ， ∴四边形BNMG 为矩形，∴BN =GM ，∴m +n =CM +BN =CM +MG =CG ， ………3分 如图①当直线l 与BC 不垂直时，△BCG 是以BC 为斜边 的直角三角形， ∴CG ﹤BC ，∴m +n ﹤BC如图②当直线l 与BC 垂直时，点M 、N 重合，CM +BN =BC ，∴m +n =BC ，∴m +n 的最大值为线段BC 的长， ………4分 将x=0代入抛物线得：y =-5，∴C （0，-5），将y =0代入抛物线得：(x -2)2-9=0， ∴x 1=-1，x 2=5，∴A （-1，0）、B （5，0），… ∴OB =OC =5， ∴BC =，∴m +n 的最大值为. ………6分（注：如果没有说明m +n 的最大值为线段BC 长的理由扣2分）（3）存在. 当点Q 在点D 上方时，作QG ⊥PD ，由A （-1，0）、P （6，7）可得： 直线AP 为：y =x +1，∴D （0，1），PD =26，∵OE =2，OD =1， ∴tan ∠DEO =21， ∴tan ∠DPQ =21，∴21 PG QG …………6分 设QG =t ，PG =2t ，∵∠QDG =∠ADO =45º， ∴DG =QG =t ，∴t +2t =26，∴t =22， ∴DQ =2DG =4，∴OQ =5，∴Q 1（0，5）， …………7分当点Q 在点D 下方时，作Q 1（0，5）关于直线 P A 的对称点H ，连接DH ， ∵∠QDP =∠HDP =45º， ∴∠QDH =90º，∵DH =DQ =4，OD =1，∴H （4，1） …………8分 由P （6，7）、H （4，1）可得直线PH 为： y =3x -11，当x =0时，y =-11，∴Q 2（0，-11）， 综上所述：Q 1（0，5）、Q 2（0，-11）………9分。
高考最新-2018广东深圳一模 精品
2018年深圳市高三年级第一次调研考试数 学 2018．3本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分．第Ⅰ卷为第1页至第2页，第Ⅱ卷为第3页至第5页．满分150分，考试时间120分钟．第Ⅰ卷 (选择题，共50分)注意事项： 1．答第Ⅰ卷前，考生务必将自己的姓名、考号、考试科目用2B 铅笔涂写在小答题卡上．同时，用黑色钢笔将姓名、考号、座位号填写在模拟答题卡上．2．每小题选出答案后，用2B 铅笔把模拟答题卡上对应题目的答案标号涂黑；最后，用2B 铅笔将模拟答题卡上的答案转涂到小答题卡上，不能答在试题卷上． 3．考试结束后，将模拟答题卡和小答题卡一并交回参考公式：(1)如果事件A 、B 互斥，那么P (A ＋B )＝P (A )＋P (B )； (2)如果事件A 、B 相互独立，那么P (A ·B )＝P (A )·P (B )；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，有且只有一项是符合题目要求的． 1.在复平面内，复数11i+所对应的点位于A ．第一象限 B.第二象限 C ．第三象限 D.第四象限 2．50<<x 是不等式4|4|<-x 成立的A ．充分不必要条件 B.必要不充分条件C ．充要条件 D.既不充分也不必要条件 3. 已知直线l 及三个平面αβγ、、，给出下列命题:①若l //α，l //β，则//αβ ②若,αβαγ⊥⊥，则βγ⊥ ③若,,l l αβ⊥⊥ 则//αβ ④若,//l l ⊂αβ，则//αβ 其中真命题是A. ①B. ②C. ③D. ④4. 已知实数x 、y 满足约束条件⎪⎩⎪⎨⎧≤+≥≥622y x y x ,则y x z 42+=的最大值为A. 24B. 20C. 16D. 125. 已知R 上的奇函数)(x f 在区间（－∞，0）内单调增加，且0)2(=-f ，则不等式()0f x ≤的解集为A. []2,2-B. (][],20,2-∞-⋃C. (][),22,-∞-⋃+∞D. [][)2,02,-⋃+∞6. 某学校要派遣6位教师中的4位去参加一个学术会议，其中甲、乙两位教师不能同时参加，则派遣教师的不同方法数共有 A ．7种 B ．8种 C ．9种 D ．10种7. 按向量)2,6(π=a 平移函数()2sin()3f x x π=-的图象，得到函数()y g x =的图象，则A. ()2cos 2g x x =-+B. ()2cos 2g x x =--C. ()2sin 2g x x =-+D. ()2sin 2g x x =--8. 函数()f x （x ∈R ）由ln ()0x f x -=确定，则导函数()y f x '=图象的大致形状是A. B. C.D.9. 曲线214x y =上的点P 到点(1,A --与到y 轴的距离之和为,d 则d 的最小值是 B.3 C. D.410. 若点A B C 、、是半径为2的球面上三点,且2AB =，则球心到平面ABC 的距离之最大值为A.2第Ⅱ卷(非选择题共100分)注意事项:第Ⅱ卷全部是非选择题，必须在答题卡非选择题答题区域内，用黑色钢笔或签字笔作答，不能答在试卷上，否则答案无效．二． 填空题：本大题共4小题；每小题5分，共20分．11则第3组的频率为 ▲ .12. 14lim14nnn →∞-=+ ▲ . 13. 圆22:2270C x y x y +---=的圆心坐标为 ▲ ，设P 是该圆的过点(3,3)的弦的中点,则动点P 的轨迹方程是 ▲ .14．将给定的25个数排成如右图所示的数表，若 每行5个数按从左至右的顺序构成等差数列，每列 的5个数按从上到下的顺序也构成等差数列，且表 正中间一个数a 33＝1，则表中所有数之和为 ▲ .11121314152122232425313233343541424344455152535455a a a a a a a a a a a a a a a a a a a a a a a a a三．解答题：本大题6小题，共80分．解答应写出文字说明，证明过程或演算步骤． 15．(本小题满分13分)已知向量a ＝)sin ,(cos x x , b ＝)cos ,cos (x x -, c ＝)0,1(-. （Ⅰ）若6π=x ，求向量、的夹角；（Ⅱ）当]89,2[ππ∈x 时，求函数12)(+⋅=b a x f 的最大值.16．(本小题满分13分)已知袋中装有大小相同的2个白球和4个红球.(Ⅰ)从袋中随机地将球逐个取出，每次取后不放回，直到取出两个红球为止，求取球次数ξ的数学期望；(Ⅱ)从袋中随机地取出一个球，放回后再随机地取出一个球，这样连续取4次球，求共取得红球次数η的方差.17． (本小题满分13分)如图，边长为2的等边△PCD 所在的平面垂直于矩形ABCD 所在的平面，BC ＝22，M 为BC 的中点.(Ⅰ)证明：AM ⊥PM ；(Ⅱ)求二面角P －AM －D 的大小； (Ⅲ)求点D 到平面AMP 的距离. 18．（本题满分14分）已知函数()f x x b =+的图象与函数23)(2++=x x x g 的图象相切,记()()()F x f x g x =.（Ⅰ）求实数b 的值及函数()F x 的极值；（Ⅱ）若关于x 的方程k x F =)(恰有三个不等的实数根，求实数k 的取值范围.MPDCＡ19.（本题满分13分）已知椭圆221:36(0)x c y t t+=>的两条准线与双曲线222:536c x y -=的两条准线所围成的四边形之面积为直线l 与双曲线2c 的右支相交于,P Q 两点(其中点P 在第一象限)，线段OP 与椭圆1c 交于点,A O 为坐标原点(如图所示). （I ）求实数t 的值；（II ）若3OP OA =⋅，PAQ ∆的面积26tan S PAQ =-⋅∠求直线l 的方程.20．（本题满分14分）已知数列{}n a 的前n 项和n S 满足：11,S =-121(),n n S S n N *++=-∈数列{}n b 的通项公式为34().n b n n N *=-∈ （I ）求数列{}n a 的通项公式；（II ）试比较n a 与n b 的大小,并加以证明；（III ）是否存在圆心在x 轴上的圆C 及互不相等的正整数n m k 、、，使得三点(,),(,),(,)n n n m m m k k k A b a A b a A b a 落在圆C 上?说明理由.2018年深圳市高三年级第一次调研考试（数学）答案及评分标准说明：一．本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分标准制订相应的评分细则．二．对计算题，当考生的解答在某一步出现错误时，如果后续部分的解答未改变该题的内容和难度，可视影响的程度决定给分，但不得超过该部分正确解答应得分数的一半；如果后续部分的解答有较严重的错误，就不再给分．三．解答右端所注分数，表示考生正确做到这一步应得的累加分数． 四．只给整数分数，选择题和填空题不给中间分数．一．选择题：本大题每小题5分，满分50分．1. D2. A3. C4. B5. B6. C7. A8. C9. B 10. D 二．填空题：本大题每小题5分，满分20分．11. 24.0 12. 1- 13. (1,1)；22(2)(2)2x y -+-= 14. 25 三．解答题：本大题满分80分． 15．(本小题满分13分)已知向量＝)sin ,(cos x x , ＝)cos ,cos (x x -, ＝)0,1(-. （Ⅰ）若6π=x ，求向量、的夹角；（Ⅱ）当]89,2[ππ∈x 时，求函数12)(+⋅=x f 的最大值.解: （Ⅰ）当6π=x 时，2cos ,cos a c a c a c ⋅==⋅ …………………2分 6cos cos π-=-=x ……………………………3分5cos 6π= ……………………………4分∵π≤≤c a,0 ∴65,π=c a…………………………6分（Ⅱ） 1)cos sin cos (212)(2++-=+⋅=x x x x f ……………………8分)1cos 2(cos sin 22--=x x x)42sin(22cos 2sin π-=-=x x x (10)分∵]89,2[ππ∈x∴]2,43[42πππ∈-x ，故]22,1[)42sin(-∈-πx ………………………11分 ∴当4342ππ=-x ,即2π=x 时, 1)(max =x f ………………………13分 16．(本小题满分13分)已知袋中装有大小相同的2个白球和4个红球.(Ⅰ)从袋中随机地将球逐个取出，每次取后不放回，直到取出两个红球为止，求取球次数ξ的数学期望；(Ⅱ)从袋中随机地取出一个球，放回后再随机地取出一个球，这样连续取4次球，求共取得红球次数η的方差.解：(Ⅰ) 依题意，ξ的可能取值为2,3，4 ……………………………1分52)2(2624===A A P ξ； ……………………………3分52)()3(3613221412===A C A C C P ξ； ……………………………5分 51)()4(4613331422===A C A C C P ξ； ……………………………7分 ∴ 514514523522=⨯+⨯+⨯=ξE . 故取球次数ξ的数学期望为14.5…………………………8分(Ⅱ) 依题意，连续摸4次球可视作4次独立重复试验，且每次摸得红球的概率均为32，则η )32,4(B ……………………………10分∴98)321(324=-⨯⨯=ηD . 故共取得红球次数η的方差为8.9……………………………13分17． (本小题满分13分)如图，边长为2的等边△PCD 所在的平面垂直于矩形ABCD 所在的平面，BC ＝22，M 为BC 的中点.(Ⅰ)证明：AM ⊥PM ；(Ⅱ)求二面角P －AM －D 的大小； (Ⅲ)求点D 到平面AMP 的距离.解法1：(Ⅰ) 取CD 的中点E ，连结PE 、EM 、EA ∵△PCD 为正三角形∴PE ⊥CD ，PE=PDsin ∠PDE=2sin60°=3 ∵平面PCD ⊥平面ABCD∴PE ⊥平面ABCD …………………3分 ∵四边形ABCD 是矩形∴△ADE 、△ECM 、△ABM 均为直角三角形 由勾股定理可求得 EM=3，AM=6，AE=3 ∴222AE AMEM =+……………………………5分∴∠AME=90°∴AM ⊥PM ……………………………6分 (Ⅱ)由(Ⅰ)可知EM ⊥AM ，PM ⊥AM∴∠PME 是二面角P －AM －D 的平面角……………………………8分 ∴tan ∠PME=133==EM PE ∴∠PME=45°∴二面角P －AM －D 为45°； ……………………………10分 (Ⅲ)设D 点到平面PAM 的距离为d ，连结DM ，则PAM D ADM P V V --=……………………………11分MPDCBＡEＡBCDPM∴d S PE S PAM ADM ⋅=⋅∆∆3131 而2221=⋅=∆CD AD S ADM在Rt PEM ∆中，由勾股定理可求得PM=6.132PAM S AM PM ∆∴=⋅=, 所以：d ⨯⨯=⨯⨯33132231,∴362=d . 即点D 到平面PAM 的距离为362.……………………………13分 解法2：(Ⅰ) ∵四边形ABCD 是矩形 ∴BC ⊥CD∵平面PCD ⊥平面ABCD∴BC ⊥平面PCD ……………………………2分 而PC ⊂平面PCD ∴BC ⊥PC 同理AD ⊥PD在Rt △PCM 中，PM=62)2(2222=+=+PC MC同理可求PA=32，AM=6 ∴222PA PMAM =+…………………………5分∴∠PMA=90°即PM ⊥AM ……………………6分 (Ⅱ)取CD 的中点E ，连结PE 、EM ∵△PCD 为正三角形∴PE ⊥CD ，PE=PDsin ∠PDE=2sin60°=3 ∵平面PCD ⊥平面ABCD ∴PE ⊥平面ABCD 由(Ⅰ) 可知PM ⊥AM ∴EM ⊥AMEＡBCDPM∴∠PME 是二面角P －AM －D 的平面角……………………………8分 ∴sin ∠PME=2263==PM PE ∴∠PME=45°∴二面角P －AM －D 为45°； ……………………………10分 (Ⅲ)同解法(Ⅰ)解法3：(Ⅰ) 以D 点为原点，分别以直线DA 、DC 为x 轴、y 轴，建立如图所示的空间直角坐标系D xyz -,依题意，可得),0,2,0(),3,1,0(),0,0,0(C P D )0,2,2(),0,0,22(M A ……2分∴)3,1,2()3,1,0()0,2,2(-=-=)0,2,2()0,0,22()0,2,2(-=-=AM …4分∴0)0,2,2()3,1,2(=-⋅-=⋅即AM PM ⊥,∴AM ⊥PM. ……………………………6分 (Ⅱ)设),,(z y x =，且⊥平面PAM ，则⎪⎩⎪⎨⎧=⋅=⋅0即⎪⎩⎪⎨⎧-⋅-⋅)0,2,2(),,()3,1,2(),,(z y x z y x ∴⎪⎩⎪⎨⎧=+-=-+022032y x z y x ⎪⎩⎪⎨⎧==yx yz 23取1=y ，得)3,1,2(=……………………………6分取)1,0,0(=，显然⊥平面ABCD∴2263||||==⋅=p n 结合图形可知，二面角P －AM －D 为45°；……………………………10分(Ⅲ) 设点D 到平面PAM 的距离为d ，由(Ⅱ)可知)3,1,2(=与平面PAM 垂直，则||n d =362)3(1)2(|)3,1,2()0,0,22(|222=++⋅. 即点D 到平面PAM 的距离为362.……………………………13分 18．（本题满分14分）已知函数()f x x b =+的图象与函数23)(2++=x x x g 的图象相切,记 ()()()F x f x g x =.（Ⅰ）求实数b 的值及函数()F x 的极值；（Ⅱ）若关于x 的方程k x F =)(恰有三个不等的实数根，求实数k 的取值范围. 解：（Ⅰ）依题意，令.1,321),()(-=+='='x x x g x f 故得∴函数()f x 的图象与函数()g x 的图象的切点为).0,1(- ……………2分 将切点坐标代入函数()f x x b =+可得 1=b . ……………5分 或:依题意得方程)()(x g x f =，即0222=-++b x x 有唯一实数解………2分故0)2(422=--=∆b ，即1=b …………………5分∴254)23)(1()(232+++=+++=x x x x x x x F ,故)35)(1(3583)(22++=++='x x x x x F , 令0)(='x F ,解得1-=x ,或35-=x . ………………………8分 列表如下 :从上表可知)(x F 在35-=x 处取得极大值274,在1-=x 处取得极小值. ……10分（Ⅱ）由（Ⅰ）可知函数)(x F y =大致图象如下图所示.……………………………12分作函数k y =的图象,当)(x F y =的图象与函数k y =的图象有三个交点时, 关于x 的方程k x F =)(恰有三个不等的实数根.结合图形可知：)274,0(∈k ……………………………14分 19.（本题满分13分）已知椭圆221:36(0)x c y t t+=>的两条准线与双曲线222:536c x y -=的两条准线所围成的四边形之面积为直线l 与双曲线2c 的右支相交于,P Q 两点(其中点P 在第一象限)，线段OP 与椭圆1c 交于点,A O 为坐标原点(如图所示).（I）求实数t的值；（II）若3OP OA=⋅，PAQ∆的面积26S=-⋅求直线l的方程.（I）解：由题意知椭圆221:36(0)xc y tt+=>上，0 1.t∴<<……1分椭圆1c的两条准线的方程为y=y==……3分双曲线222:536c x y-=的两条准线的方程为x=x=，这两条准线相…………4分上述四条准线所围成的四边形是矩形, =1.5t=故实数t的值是15.……………………………5分（II）设(,),A m n由3OP OA=⋅及P在第一象限得(3,3),0,0.P m n m n>>12,,A c P c∈∈∴2222536,54,m n m n+=-=解得2,4,m n==即(2,4),(6,12).A P……………………………8分设(,),Q x y则22536.x y-=①由26tan,S PAQ=-∠得1sin26tan2AP AQ PAQ PAQ⋅⋅∠=-∠，52AP AQ∴⋅=-，即(4,8)(2,4)52,230.x y x y⋅--=-++=②……………………………10分联解① ②得5119319x y ⎧=-⎪⎪⎨⎪=-⎪⎩，或3.3x y =⎧⎨=-⎩因点Q 在双曲线2c 的右支，故点Q 的坐标为(3,3)-. ……………………11分 由(6,12),P (3,3)Q -得直线l 的方程为33,12363y x +-=+-即5180.x y --= ……………………13分 20．（本题满分14分）已知数列{}n a 的前n 和n S 满足：11,S =-121(),n n S S n N *++=-∈数列{}n b 的通项公式为34().n b n n N *=-∈ （I ）求数列{}n a 的通项公式；（II ）试比较n a 与n b 的大小,并加以证明；（III ）是否存在圆心在x 轴上的圆C 及互不相等的正整数n m k 、、，使得三点(,),(,),(,)n n n m m m k k k A b a A b a A b a 落在圆C 上?说明理由.解：（I ）121(),n n S S n N *++=-∈12121,21(),n n n n S S S S n N *+++∴+=-+=-∈两式相减得212120,2().n n n n a a a a n N *+++++==-∈…………………………2分 又111,a S ==-211221231,2.S S a a a a +=+=-=-111,2(),n n a a a n N *+∴=-=-∈即数列{}n a 是首项为1,-公比为2-的等比数列,其通项公式是1(2)().n n a n N -*=--∈ ……………………………4分另解一:111,21(),n n S S S n N *+=-+=-∈111211,2()(),3333n n S S S n N *+∴+=-+=-+∈即数列13n S ⎧⎫+⎨⎬⎩⎭是首项为2,3-公比为2-的等比数列,其通项公式是1(2)().33nn S n N *-+=∈ (2)分当2n ≥时, 111(2)1(2)1(2),3333n n n n n n a S S ---⎡⎤⎡⎤--=-=---=--⎢⎥⎢⎥⎣⎦⎣⎦ 又111,(2)().n n a a n N -*∴=-∴=--∈ ……………………………4分 （II ）(1)1122441,1;2,2;8,8.a b a b a b =-=-====∴当1,2,4n =时,.n n a b = ……………………………6分(2)当21()n k k N *=+∈时, 22121(2)0,610,.k k k n n a b k a b ++=--<=->∴<……………………………7分(3)当2(,3)n k k N k *=∈≥时,252521425012222(11)16()3264,64,k k k k k k a C C k b k ----==⋅+≥+=-=- 2660180,n n a b k ∴-≥-≥>即.n n a b > ……………………………9分（III ）不存在圆心在x 轴上的圆C 及互不相等的正整数n m k 、、，使得三点,,n m k A A A 落在圆C 上. …………10分假设存在圆心在x 轴上的圆C 及互不相等的正整数n m k 、、，使得三点,,n m kA A A 即11(34,(2)),(34,(2)),n n n m A n A m --------1(34,(2))k k A k ----落在圆C 上.不妨设,n m k >>设圆C 的方程为:220x y Dx F +++=. 从而21924164(34)0n n n n D F --+++-+= ①21924164(34)0m m m m D F --+++-+= ②21924164(34)0k k k k D F --+++-+= ③由①-②, ②-③得119()()24()(44)3()0n m n m n m n m n m D --+---+-+-=119()()24()(44)3()0m k m k m k m k m k D --+---+-+-=即11449()2430n m n m D n m---+-++=- ④ 11449()2430m k m k D m k---+-++=- ⑤由④-⑤得111144449()0n m m k n k n m m k-------+-=--整理得14449()()()()()0()()k n k m kn k m k n k n m n m m k n k m k ---⎡⎤-+---+-=⎢⎥----⎣⎦,441,.n k m kn m k n k m k-->>≥∴<-- (12)分作函数4()(1),x f x x x =≥由224ln 444(ln 41)()0(1),x x x x x f x x x x ⋅-⋅-'==>≥ 知函数4()(1)xf x x x=≥是增函数. 441,1,,n k m kn m k n k m k n k m k-->>≥∴->-≥>--产生矛盾. 故不存在圆心在x 轴上的圆C 及互不相等的正整数n m k 、、，使得三点,,n m kA A A 落在圆C 上. ……………………………14分。
2018-2019初二数学福田区统考试卷
C．无限小数都是无理数
D．点 (−2 ，3) 到 y 轴的距离是 2
9．一次函数 y = kx − k （ k 0 ）的图象大致是（ ）
10．小明从 A 地前往 B 地，到达后立刻返回，他与 A 地的距离 y （千
20．（本题 7 分）如图 7，已知点 E 在线段 AD 上，点 P 在直线 CD 上，∠AEF =∠F ，∠BAD =∠CPF ． 求证：∠ABD +∠BDC = 180
21．（本题 7 分）某一天，水果经营户老张用 1600 元从水果批发市场购进猕猴桃和芒果共 50 千克，后
再到水果市场去卖，已知猕猴桃和芒果当天的批发价和零售价如表所示：
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第二部分 非选择题
二、填空题（本题共 4 小题，每小题 3 分，共 12 分）
13．若点 M (a − 3，a +1) 在 y 轴上，则点 M 的坐标为________．
14．如图所示，一次函数
y
=
x
+
2
与
y
=
kx
+
b
的图象相交于点
P
(
m
，4)
，则方程组
y y
= =
x+2 kx + b
以每秒 1 个单位的速度运动到点 P ，再沿线段 PD 以每秒 2 个单位的速度运动到点 D 后停 止，求点 H 在整个运动过程中所用时间最少时点 P 的坐标．
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整体分析
（1） 难度系数：★★，区分度体现在：第16、22、23题（22题最后一问和23题最后一问） （2） 重点考察：二次根式；平面直角坐标系；勾股定理；平行线与三角形；数据与统计；一次函数
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570 A． ( 32 − 2 x )( 20 − x ) =
C． ( 32 − x )( 20 − x ) = 32 × 20 − 570
B． 32 x + 2 × 20 x = 32 × 20 − 570 D． 32 x + 2 × 20 x − 2 x 2 = 570
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11．如图，在 △ ABC 中， ∠ACB = 90° ，按如下步骤操作：①以点 A 为圆心，任意长为半径作弧，分别 交 AC 、 AB 于 D 、 E 两点；②以点 C 为圆心， AD 长为半径作弧，交 AC 的延长线于点 F ；③以 两弧交于点 G ； ④作射线 CG ， 若 ∠FCG = 则 ∠B 为 （ ） 点 F 为圆心，DE 长为半径作弧， 50° ，
B．左视图 D．以上答案都不对
3．2017 年，粤港澳大湾区发展取得显著成效，全年 GDP 将达到 1.4 万亿美元，经济总量有望在未来 几年超越美国纽约湾区，成为全球第二大湾区；1.4 万亿美元用科学记数法表示为（ ） A． 1.4 × 103 亿美元 C． 1.4 × 108 亿美元 4．下列运算正确的是（ A． 2a + 3a = 5a C． ( x − 2 )( x − 3) = x 2 − 6 ） B． ( x − 2 ) =x 2 − 4
6．下列说法中正确的是（ A．8 的立方根是 2 B．函数 y =
）
1 的自变量 x 的取值范围是 x > 1 x −1
C．同位角相等 D．两条对角线互相垂直的四边形是菱形
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7．如图，函数 y = 2 x 和 = y （ ）
2 2 2 ) ，观察图象可知，不等式 < 2 x 的解集为 ( x > 0 ) 的图象相交于点 A ( m， x x
a @ b = a 2 + ab − 1 ． 定义一种运算 “@” 为： 若 x@2 = 0 ， 则 2 x2 + 4 x − 3 = ________． 15． 对于实数 a 、 b，
16．如图，四边形 OABC 中， AB ∥ OC ，边 OA 在 x 轴的正半轴上， OC 在 y 轴的正半轴上，点 B 在第一象限内，点 D 为 AB 的中点， CD 与 OB 相交 于点 E ，若 △BDE 、 △OCE 的面积分别为 1 和 9，反比例函数 y = 图象经过点 B ，则 k = ________．
1 下列结论：① △GCD 和 △FOD 的面积比为 3∶1；② AE 的最大长度为 10 ；③ tan ∠FEO = ；④ 3
当 DA 平分 ∠EAO 时， CG =
3 ，其中正确的结论有（ 2） NhomakorabeaA．①②③ C．②③④
B．②③ D．③④
第二部分 非选择题 二、填空题： （本大题共 4 小题，每小题 3 分，共 12 分） 13．分解因式： ab − b 2 = ________． 14．在一个不透明的空袋子里，放入仅颜色不同的 2 个红球和 1 个白球，从中随机摸出 1 个球后不放 回，再从中随机摸出 1 个球，两次都摸到红球的概率是________．
A． 30° C． 50°
B． 40° D． 60°
12．如图，在平面直角坐标系中，正方形 ABCO 的边长为 3，点 O 为坐标原点，点 A 、 C 分别在 x 轴， 点 B 在第一象限内， 直线 = 线段 BC 交于点 F 、 y 轴上， y kx + 1 分别与 x 轴、y 轴、 D、 G ，AE ⊥ FG ，
2018 年九年级数学质量检测试卷 数学
第一部分 选择题 一、 （本部分共 12 小题，每小题 3 分，共 36 分．每小题给出 4 个选项，其中只有一个是正确的） ） 1．如果“收入 10 元”记作 +10 元，那么支出 20 元记作（ B． −20 元 C． +10 元 D． −10 元 A． +20 元 2．如图所示的圆锥体的三视图中，是中心对称图形的是（ A．主视图 C．俯视图 ）
9．如图，线段 CD 的两个端点的坐标分别为 C (1, 2 ) ， D ( 2,0 ) ，以原点为位似中心，将线段 CD 放大得 到线段 AB ，若点 B 的坐标为（5，0） ，则点 A 坐标为（ ）
A． （2，5） C． （3，5）
B． （3，6） D． （2.5，5）
10．如图，某小区计划在一块长为 32m，宽为 20m 的矩形空地上修建三条同样宽的道路，剩余的空地 ） 上种植草坪，使草坪的面积为 570m2．若设道路的宽为 x m，则下面所列方程正确的是（
2
B． 1.4 × 104 亿美元 D． 1.4 × 1012 亿美元
D． a8 ÷ a 4 = a2
5．我市某小区开展了“节约用水为环保做贡献”的活动，为了解居民用水情况，在小区随机抽查了 10 户家庭的月用水量，结果如右表：则关于这 10 户家庭的月用水量，下列说法错误的是（ ） 月用水量（吨） 户数 A．方差是 4 C．平均数是 9 8 2 9 6 B．极差是 2 D．众数是 9 10 2
k 的 x
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三、解答题： （本题共 7 小题，其中第 17 题 5 分，第 18 题 6 分，第 19 题 7 分，第 20 题 8 分，第 21 题 8 分，第 22 题 9 分，第 23 题 9 分，共 52 分） 1 17． （5 分）计算： − 6 tan 30° + 2 − 2 2
A． x < 0 C． 0 < x < 1
B． x > 1 D． 0 < x < 2
8．如图，已知 AE = CF ， ∠AFD = ∠CEB ，那么添加下列一个条件后，仍无法判定 △ ADF ≌△CBE 的 是（ ）
A． ∠A = ∠C C． BE = DF
B． AD ∥ BC D． AD = CB
−1
(
)
0
+ 12 ．
18． （6 分）先化简，再求值：
a +1 2 ÷ 1 + ，其中 a = −1 ． a − 2a + 1 a − 1
2
19． （7 分）深圳市某校艺术节期间，开展了“好声音”歌唱比赛，在初赛中，学生处对初赛成绩做了 统计分析，绘制成如下频数、频率分布表和频数分布直方图（如图） ，请你根据图表提供的信息， 解答下列问题： 分组