英语卷·2019届江苏省泰州中学高一下学期期中考试
江苏省泰州2019-2020年高一下学期期末考试英语试题
高一下学期期末模拟英语试卷（考试时间：120分钟；总分：120分）第I卷（选择题三部分共75分）第一部分听力（共两节，满分20分）做题时，先将答案标在试卷上。
录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题纸上。
第一节（共5小题；每小题1分，满分5分）听下面5段对话。
每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。
听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. What kind of pet does the woman suggest?A. A dog.B. A fish.C. A cat.2. Which place is the woman looking for?A. A grocery store.B. A movie theater.C. The railway station.3. What did the man buy for the woman's birthday?A. A fruit cake.B. Some apple pies.C. A bunch of flowers.4. What is the relationship between the speakers?A. Classmates..B. Parent and child.C. Teacher and student.5. Where is the woman?A. In a car.B. In a lift.C. In a bookstore.第二节（共15小题；每小题1分，满分15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，并标在试卷的相应位置。
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。
2019-2020学年泰州二中高三英语期中考试试题及答案解析
2019-2020学年泰州二中高三英语期中考试试题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AThe COVID -19 pandemic has affected all aspects of life, including the way we travel. But for those who are looking to expand their horizons while still staying safe, the following three travel trends in 2021 may provide inspirations. Let’s take a look.StaycationWith many travel restrictions during the pandemic, people preferred traveling to nearby places in 2020. This trend continues in 2021. According to search data, 62 percent of people are interested in taking a vacation within driving distance of home. People who live in large cities want to get back in touch with nature. Travelers are looking for places different from their everyday accommodations, for example, farm stays, villas and cottages.Pod travelWhile 2020 saw a rise in solo travel and isolated adventures, 2021 shows that people want to be more connected. “Pod travel”, or gathering in isolated spaces with loved ones, is growing in popularity. 85 percent of survey respondents favor traveling with family or friends, and over half of the trips searched include three or more people. Pod travel is here to stay for those who want to safely be together while reducing risks associated with socializing with others.Remote working and travelingMany people worked and learned from home in 2020 because of the pandemic. Remote working blurs the line between working and traveling. There was a 128 percent increase in the mention of phrases such as “relocation”, “relocate”, “remote work” and “trying a new neighborhood”. People are actively booking longer stays (e. g. two plus week trips) in small to mid—size cities with access to immersive natural surroundings and wide—open spaces.1.What can we learn about Staycation?A.Travelling to the countryside.B.Taking an isolated adventure.CHaving holidays in nearby places. D.Staying indoors all by oneself.2.What’s special about Pod travel?A.Traveling alone.B.Traveling far away.C.Traveling while working.D.Traveling with loved ones.3.Where might we find the text in a magazine?A.Medicine.cation.C.TourismD.Career.BUnderstanding the link between a clean environment and human life is not a new concept. In fact, it was noticed as early as ancient Rome. Today we see how green living has infiluenced our everyday lives. There is a growing community of people who embrace a zero waste lifestyle and make changes to the way they live to reduce their carbon footprint.Living a zero waste lifestyle means doing one’s best to achieve the aim of not sending anything to a landfill. People who adopt this lifestyle ultimately cut down on their waste by reducing what they need and want. They reuse what they own, sending few things to be recycled.Many people who adopt the zero waste lifestyle claim to be frustrated by the many harmful chemical substances found in beauty and cleaning products. They also find the uses of disposable items and excessive packaging. For example, how many times have we had to peel away layers of plastic wrap and cardboard before finally taking out the item which we had bought? Instead of buying pre-packed food and goods, those who identify with the zero waste philosophy tend to shop in stores that allow them to make purchases and bring their own cloth bags and glass jars to store their purchases.Many people may have the misconception that it is easier to live a zero waste lifestyle in the West. Nevertheless, Malaysian environmental journalist, Ms. Aurora Tin, has proven that a zero waste lifestyle is possible even in the Asian context. Instead of going to the supermarket to buy pre-packaged foods, Ms. Tin now visits the wet market and brings her own bags for vegetables. She has even stopped using store-bought toothpaste and makes her own toothpaste from coconut oil and baking soda. This lifestyle may be too big a change for the average person, but we could follow her suit to make gradual changes to our own lives.4. Which of the following is a zero waste lifestyle?A. Bringing a resuable container to take away food.B. Choosing appliances that cost less money.C. Turning off a device to stop using power.D. Classifying the garbage before throwing it away.5. What may disappoint a person who adopts a zero waste lifestyle?A. Recycable carboard.B. Excessive packaging.C. Glass jars to store purchases.D. Natural substances in cleaning products.6. What is the main idea of the last paragraph?A. How do people live a zero waste lifestyle.B. Why Ms. Tin chooses to live a zero waste lifestyle..C. We can also practice a zero waste lifestyle in Asia.D. It is easy to live a zero waste lifestyle in the West.7. What is the best title of the passage?A. Living a zero waste lifestyle.B. Going green ismore than a fashion.C. A zero waste lifetyle is easy to achieve.D. Making environmentally-conscious decisions.CWhen 36-year-old J Andy Duran decided to return to his favorite high school hobby—skateboarding, the only trouble he expected to have was his own ability to get back on the skateboard after such a long time. However, the 340-pound skateboarder soon realizedthatwas the least of his problems.Duran's problems began before he even stepped on a skateboard. He couldn't find anything for fat skaters. What Duran did find was a belief that plus-size people should not be skating. Determined to do something to change the image(形象) of plus-size people in sports, Duran set up Chub Rollz—a skating and skateboarding community for overweight skaters. He knew that not only did he need to get back into it to prove people wrong, but he needed to create a safe space where others can haverepresentation as well.To encourage plus-size people to take part in the fun sport, Duran created a list of recommended products for fat skaters. He also hosted roller skating and skateboarding classes to teach beginners.After an article about his thoughts in the San Francisco Chronicle, Duran received lots of messages from strangers thanking him for giving them the courage to take up skateboarding. He has also been contacted by some skateboarding brands offering to create larger clothing sizes and beenoffered free equipment by skating organizations like "Skate Like a Giri ".Though encouraged, Duran believes a lot more needs to be done to remove body image stereotypes(刻板印象).“I want to see more changes in communities. Maybe skate shops create a more welcoming environment for all types of skaters. Or boards are made in a variety of strengths and sizes—everyone is making thinner, lighter products, but sometimes we need those heavy-duty choices to stay available," he explains.For those hesitant of taking up their desired activity due to their body size, Duran has this to say:“Be kind to yourself. Just because you don't see it doesn't mean you can't be it.”8. What does the underlined word "that" in paragraph 1 refer to?A. The skateboard.B. His hobby.C. His ability.D. The time.9. Why did Duran found Chub Rollz?A. To realize his childhood dream.B. To help plus-size people lose weight.C. To fight for the equal right for fat people.D. To change people's impression of the fat in sports.10. Which word can best describe Duran?A. Generous.B. Inspiring.C. Adaptable.D. Talented.11. Which is most likely Duran's belief?A. Everyone has a gift for sports.B.No one is too fat to enjoy sports.C. Fat people do deserve social concern.D. Skateboarding is most suitable for fat people.DA nurse has fulfilled (实现) a promise she made to her patient four years ago to one day attend her daughter's graduation from nursing school.Edina Habibovic, 22, graduated from Chamberlain University's College of Nursing in 2020. Her mother, Sevala Habibovic, 46, died in2017 after a two year fight with breast cancer.“I thought the medical field wasn't for me. Then, my mom got sick and I had all the experience going in and out of the hospital, ” Edina toldGood MorningAmerica. “When my mom passed away, I thought, ‘I want to dothis.’”she said.Sanja Josipovic, who at the time worked as a home health nurse with Northwestern Medicine in Winfield, Illinois, cared for Sevala inside her home. They often chatted and shared the latest news with each other over six months of care.“She was most worried about Edina because she was young and hadn't finished school yet, ” Sanja said. “We are like sisters; we care about and trust each other. She was a powerful and strong minded woman. She wasn't scared to die; she was just worried about her kids and husband.”Edina said her mother lived for being with her family and taking care of people. “When Sanja was working,my mom would still try to make her something to eat, no matter how sick she was, ” Edina added. When Sevala's life was coming to an end, she asked Sanja to take her place at her youngest daughter's nursing school graduation. “That was the only thing she was going to miss. Edina's graduation, ” said Sanja, who is a mother of three herself. She agreed.Due to COVID -19, there was no graduation or pinning ceremony. Edina's manager at Marianjoy Rehabilitation Hospital decided to host a pinning ceremony for her and have Sanja present the pin. “Sanja has fulfilled her promise, ” Edina said.Edina and Sanja are now caring for patients alongside one another as colleagues at Marianjoy.12. What does the underlined word “this” in paragraph 3 refer to?A. Leaving the hospitalB. Working as a nurseC. Facing death positivelyD. Caring for Edina's mother13. What can be learned about Sanja and Sevala?A. They enjoyed volunteeringB. They were cancer survivorsC. They had unhappy marriagesD. They developed a close bond14. What would be Sevala's regret?A. The loss of the chance to study medicineB. Her absence from Edina's school graduationC. Failing to keep the promise made to SanjaD. Never cooking a good meal for her husband15. How did Sanja fulfill her promise?A. By taking care of Edina and her familyB. By helping Edina enter her dream hospitalC. By attending a special ceremony for EdinaD. By managing to become Edina's colleague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。
江苏省泰州中学2019-2020学年高一下学期期中数学试题(解析版)
江苏省泰州中学2019-2020学年度第二学期期中考试高一数学试题一、单项选择题（本题共10小题，每小题5分，共50分）1.在ABC ∆中，若30B°?，AB =2AC =，则满足条件的三角形有（ ）个 A.B. 1C. 2D. 不确定 【答案】C【解析】【分析】直接利用sin AB AC AB B >>来判断三角形解得情况.【详解】在ABC ∆中，30B°?，AB =2AC =，则sin AB AC AB B >>，所以，ABC ∆有两解.故选：C.【点睛】本题考查的知识要点：三角形解的情况的应用，属于基础题．2.正方体被平面所截得的图形不可能是（ ）A. 正三角形B. 正方形C. 正五边形D. 正六边形 【答案】C【解析】【分析】平面与正方形相交与不同的位置，可以出现正三角形，正方形，正六边形，不可能出现正五边形【详解】如图所示，平面与正方形相交与不同的位置，可以出现正三角形，正方形，正六边形，不可能出现正五边形，故选C 项【点睛】本题考查正方形的截面图形，空间想象能力，属于基础题.3.10y +-=的倾斜角为（ ）A. 30°B. 60°C. 120°D. 150°【答案】C【解析】【分析】由直线的一般式方程得到直线的斜率k ，再由tan θk =求解倾斜角.10y +-=的斜率=k -tan [0,180)o o k θθ∴==∈，∴120θ︒=.故选：C【点睛】本题考查了直线的一般式方程、直线的斜率和直线的倾斜角的关系，考查了学生转化，运算的能力，属于基础题.4.以()3,1A -，()2,2B -为直径的圆的方程是A. 2280x y x y +---=B. 2290x y x y +---=C. 2280x y x y +++-=D. 2290x y x y +++-= 【答案】A【解析】【分析】设圆的标准方程，利用待定系数法一一求出,,a b r ，从而求出圆的方程.【详解】设圆的标准方程为222()()x a y b r -+-=，由题意得圆心(,)O a b 为A ，B 的中点， 根据中点坐标公式可得32122a -==，12122b -+==，又||2AB r ===，所以圆的标准方程为： 221117()()222x y -+-=，化简整理得2280x y x y +---=， 所以本题答案为A.【点睛】本题考查待定系数法求圆的方程，解题的关键是假设圆的标准方程，建立方程组，属于基础题.5.过两直线1l ：310x y -+=，2l ：260x y ++=的交点且与310x y +-=平行的直线方程为（ ）A. 310x y -+=B. 370x y ++=C. 3110x y --=D. 3130x y ++=【答案】D【解析】【分析】求出两直线1l 、2l 的交点坐标，再设与310x y +-=平行的直线方程为30x y m ++=，代入交点坐标求出m 的值，即可写出方程. 【详解】解：两直线1l ：310x y -+=，2l ：260x y ++=的交点为310260x y x y -+=⎧⎨++=⎩解得41x y =-⎧⎨=-⎩，即()4,1--； 设与310x y +-=平行的直线方程为30x y m ++=则3(4)(1)0m ⨯-+-+=解得13m =所求的直线方程为3130x y ++=.故选：D【点睛】本题考查了直线方程的应用问题，是基础题.6.将棱长为1的正方体木块切削成一个体积最大的球，则该球的体积为（ ）A. 2B.C. 43πD. 6π 【答案】D【解析】【分析】依题意最大的球为与正方体各个面相切，直径为正方体的棱长，即可求解.【详解】将棱长为1的正方体木块切削成一个体积最大的球， 该球为正方体的内切球，其半径为12，所以球体积为341()326ππ⨯=.故选：D .【点睛】本题考查多面体与球的“接”“切”问题，属于基础题.7.在ABC V 中，2cos22B a c c +=（a ，b ，c 分别为角A ，B ，C 的对边），则ABC V 的形状为（ ） A. 等边三角形 B. 直角三角形 C. 等腰三角形或直角三角形D. 等腰直角三角形 【答案】B【解析】【分析】由二倍角公式和余弦定理化角为边后变形可得． 【详解】∵2cos 22B a c c +=，∴22cos 2B a c c +=，1cos a c B c ++=，22212a c b a c ac c+-++=，整理得222+=a b c ，∴三角形为直角三角形．故选：B ．【点睛】本题考查三角形形状的判断，考查二倍角公式和余弦定理，用余弦定理化角为边是解题关键． 8.一竖立在水平面上的圆锥物体的母线长为2m ，一只蚂蚁从圆锥的底面圆周上的点P 出发，绕圆锥表面爬行一周后回到P点，蚂蚁爬行的最短路径为，则圆锥的底面圆半径为（ ） A. 1m B. 2m 3 C. 43m D. 3m 2【答案】B【解析】【分析】将圆锥展开后的扇形画出，结合母线及最短距离，即可确定圆心角大小；进而求得弧长，即为底面圆的周长，由周长公式即可求得底面圆的半径.【详解】将圆锥侧面展开得半径为2m 的一扇形，蚂蚁从P 爬行一周后回到P （记作1P ），作1OM PP ⊥，如下图所示：的由最短路径为，即12PP OP ==， 由圆的性质可得13POM POM π∠=∠=，即扇形所对的圆心角为23π， 则圆锥底面圆的周长为24233l ππ=⨯=， 则底面圆的半径为423223l r πππ===， 故选：B.【点睛】本题考查了了圆锥侧面展开图、扇形弧长公式的简单应用，属于基础题.9.在ABC V 中，内角A ，B ，C 所对的边分别为a ，b ，c ，且cos sin a B b A c +=.若2a =，ABC V 的面积为1)，则b c +=（ ）A. 5B. C. 4 D. 16 【答案】C【解析】【分析】 根据正弦定理边化角以及三角函数公式可得4A π=,再根据面积公式可求得6(2bc =-,再代入余弦定理求解即可.【详解】ABC V 中,cos sin a B b A c +=,由正弦定理得sin cos sin sin sin A B B A C +=,又sin sin()sin cos cos sin C A B A B A B =+=+,∴sin sin cos sin B A A B =,又sin 0B ≠,∴sin A cos A =,∴tan 1A =,又(0,)A π∈, ∴4A π=.∵1sin 1)24ABC S bc A ===-V , ∴bc=6(2,∵2a =,∴由余弦定理可得22()22cos a b c bc bc A =+--,∴2()4(2b c bc +=++4(26(216=++⨯-=,可得4b c +=.故选：C 【点睛】本题主要考查了解三角形中正余弦定理与面积公式的运用,属于中档题.10.在平面直角坐标系xOy 中，圆1C ：224x y +=，圆2C ：226x y +=，点(1,0)M ，动点A ，B 分别在圆1C 和圆2C 上，且MA MB ⊥，N 为线段AB 的中点，则MN 的最小值为A. 1B. 2C. 3D. 4【答案】A【解析】【分析】由MA MB ⊥得0MA MB ⋅=u u u r u u u r ，根据向量的运算和两点间的距离公式，求得点N 的轨迹方程，再利用点与圆的位置关系，即可求解MN 的最小值，得到答案．【详解】设11(,)A x y ，22(,)B x y ，00(,)N x y ，由MA MB ⊥得0MA MB ⋅=u u u r u u u r ，即1212121x x y y x x +=+-，由题意可知，MN 为Rt △AMB 斜边上的中线，所以12MN AB =,则2222222121211221122()()22AB x x y y x x x x y y y y =-+-=-++-+222211*********()()2()102(1)124x y x y x x y y x x x =+++-+=-+-=- 又由12MN AB =，则224AB MN =，可得220001244[(1)]x x y -=-+，化简得220019()24x y -+=， ∴点00(,)N x y 的轨迹是以1(,0)2为圆心、半径等于32的圆C 3， ∵M 在圆C 3内，∴ MN 的最小值即是半径减去M 到圆心1(,0)2的距离， 即min 31122MN r d =-=-=，故选A ． 【点睛】本题主要考查了圆的方程及性质的应用，以及点圆的最值问题，其中解答中根据圆的性质，求得N 点的轨迹方程，再利用点与圆的位置关系求解是解答的关键，着重考查了推理与运算能力，属于中档试题．二、多项选择题（本题共2小题，每小题5分，共10分.全部选对得5分，部分选对得3分，有选错的得0分）11.已知α、β是两个不同的平面，m 、n 是两条不同的直线，下列说法中正确的是（ ）A. 若m α⊥，//m n ，n β⊂，则αβ⊥B. 若//αβ，m α⊥，n β⊥，则//m nC. 若//αβ，m α⊂，n β⊂，则//m nD. 若αβ⊥，m α⊂，n αβ=I ，m n ⊥，则m β⊥【答案】ABD【解析】【分析】根据线面的位置关系对每个选项进行判断．【详解】由m α⊥，//m n ，得n α⊥，又由n β⊂，得αβ⊥，A 正确；由//αβ，m α⊥，得m β⊥，又由n β⊥，得//m n ，B 正确；若//αβ，m α⊂，n β⊂，,m n 可能平行也可能是异面直线，C 错误；由面面垂直的性质定理知D 正确．故选：ABD ．【点睛】本题考查空间线面间的平行与垂直关系，掌握直线、平面间平行垂直的判定定理的性质定理是解题关键．12.设有一组圆k C ：()()224132x k y k k -++-=（*k N ∈）.下列四个命题中真命题的是（ ） A. 存在一条定直线与所有的圆均相切B. 存在一条定直线与所有的圆均相交C. 存在一条定直线与所有的圆均不相交D. 所有圆均不经过原点【答案】BD【解析】【分析】由圆与圆的位置关系判断A ．由圆心所在直线判断B ，由圆半径可能无穷大，判断C ，代入原点坐标确定方程是否有整数解判断D ．【详解】圆心为(1,3)k C k k -，半径为2k r =，1(0,3)C ，1r =2(1,6)C ，2r =12C C ==<=1C 与圆2C 是内含关系，因此不可能有直线与这两个圆都相切，从而A 错误；易知圆心在直线3(1)y x =+上，此直线与所有圆都相交，B 正确；若k 取无穷大，则所有直线都与圆相交，C 错；将(0,0)代入圆方程得224(1)92k k k -+=，即2410212k k k -+=，等式左边是奇数，右边是偶数，因此方程无整数解，即原点不在任一圆上，D 正确．故选：BD ．【点睛】本题考查直线与圆的位置关系，考查圆与圆的位置关系，掌握反证法，特殊值法，综合性较高．三、填空题（本题共4小题，每小题5分，其中第15题第一空2分，第二空3分；共20分）13.若直线()2540a x y +-+=与()2210x a y +--=互相垂直，则a 的值是__________.【答案】4-.【解析】【分析】由垂直的条件求解．【详解】∵已知两直线垂直，∴2(25)(2)0a a +--=，解得4a =-．故答案为：－4．【点睛】本题考查两直线垂直的条件，属于基础题．14.在四面体ABCD 中，E ，F 分别是AB ，CD 的中点.若BD ，AC 所成的角为60°，且1BD AC ==，则EF 的长为__________.【答案】12【解析】【分析】 取BC 中点G ，可证EGF ∠（或其补角）是BD ，AC 所成的角，分类计算．【详解】取BC 中点G ．连接,GE GF ，∵E ，F 分别是AB ，CD 的中点，∴//,//EG AC GF BD ，1122GE BD ==，1122GF BD ==， ∴BD ，AC 所成的角是EGF ∠（或其补角）， 若60EGF ∠=︒，则12EF GE ==，若120EGF ∠=︒，则12sin 6022EF GF =︒=⨯=，故答案为：12或2．【点睛】本题考查异面直线所成的角，解题时要注意通过平行线作出异面直线所成角时，对应的角或其补角是异面直线所成的角，因此可分类讨论．15.2020年是中国传统的农历“鼠年”，有人用3个圆构成“卡通鼠”的形象，如图：()0,3Q -是圆Q 的圆心，圆Q 过坐标原点O ；点L 、S 均在x 轴上，圆L 与圆S 的半径都等于2，圆S 、圆L 均与圆Q 外切.已知直线l 过点O .（1）若直线l 与圆L 、圆S 均相切，则l 截圆Q 所得弦长为__________；（2）若直线l 截圆L 、圆S 、圆Q 所得弦长均等于d ，则d =__________.【答案】 (1). 3 (2).125【解析】【分析】圆L 与圆S 关于原点对称，直线l 过原点，只要与一个圆相切，必与另一圆相切．求出圆L 与圆S 的圆心坐标，（1）求出切线方程后，求出Q 到切线l 的距离后由勾股定理得弦长．（2）设出直线l 方程，由三个弦长相等得直线方程，从而可得弦长d ．【详解】由题意圆L 与圆S 关于原点对称，设(,0)(0)S a a >23=+，4a =，即(4,0)S ，∴(4,0)L -．（1）设l 方程为y kx =，即0kx y -=2=得k =，由对称性不妨取k =l方程为y x =，0x -=，圆心Q 到l2=，∴弦长为3=； （2）同（1）设直线l 方程为0kx y -=，点Q 到直线l，直线截圆Q得弦长为d ==S 到直线l，直线截圆S得弦长为d ===，解得2421k =，∴125d ==． 故答案为：3；125． 【点睛】本题考查直线与圆的位置关系，考查直线与圆相交弦长问题．求出圆心到直线的距离，用勾股定理求得弦长是求圆弦长的常用方法．16.在锐角ABC V 中，2BC =，sin sin 2sin B C A +=，则BC 边上的中线AD 的长的取值范围是__________.【答案】2⎭【解析】【分析】由正弦定理化角为边，由余弦定理求出中线长（用三边表示），然后根据已知条件求出b 的范围，结合二次函数性质得bc 的范围，从而得中线取值范围．【详解】因为sin sin 2sin B C A +=，由正弦定理得2b c a +=，又2a =，所以4b c +=，由余弦定理得2222cos b AD CD AD CD ADC =+-⋅∠，2222cos c AD BD AD BD ADB =+-⋅∠，又cos cos ADB ADC ∠=-∠，12BD CD a ==， 所以2222122b c AD a +=+，所以AD === 又4b c +=，即4c b =-，因为ABC V 是锐角三角形，∴222222222b c a b a c a c b ⎧+>⎪+>⎨⎪+>⎩，所以222222(4)44(4)(4)4b b b b b b ⎧+->⎪+>-⎨⎪-+>⎩，解得3522b <<，∴2215(4)4(2)4(,4]4bc b b b b b =-=-=--+∈，AD ≤<故答案为：． 【点睛】本题考查了正弦定理，余弦定理，二次函数的性质的综合应用，解题时利用余弦定理建立中线与三角形边长之间的关系是基础，利用锐角三角形求出b 的取值范围是解题关键．四、解答题（本题共6小题，其中第17题10分，其他每题12分，共70分；解答应写出文字说明、证明过程或演算步骤）17.在△ABC 中，a =7，b =8，cos B = –17． ∴Ⅰ）求∠A ∴∴Ⅱ）求AC 边上的高． 【答案】(1) ∠A =π3 (2) AC【解析】分析：（1）先根据平方关系求sin B ,再根据正弦定理求sin A ，即得A ∠∴∴2）根据三角形面积公式两种表示形式列方程11sin 22ab C hb =，再利用诱导公式以及两角和正弦公式求sin C ，解得AC 边上的高∴ 详解：解∴∴1）在△ABC 中，∵cos B =–17∴∴B ∴∴π2∴π∴∴∴sin B=由正弦定理得sin sin a b A B =⇒ 7sin A=AB ∴∴π2∴π∴∴∴A ∴∴0∴π2∴∴∴∴A =π3∴ ∴2）在∴ABC 中∴∴sin C =sin∴A +B ∴=sin A cos B +sin B cos A1172⎛⎫-+ ⎪⎝⎭∴ 如图所示，在△ABC 中，∵sin C =h BC ∴∴h =sin BC C ⋅=7=∴∴AC∴点睛：解三角形问题，多为边和角的求值问题，这就需要根据正、余弦定理结合已知条件灵活转化边和角之间的关系，从而达到解决问题的目的. 18.在如图所示五面体ABCDEF 中,四边形ABCD 为菱形,且60,22,//,DAB EA ED AB EF EF AB M ∠=︒====为BC 中点.(1)求证:FM ∕∕平面BDE ;(2)若平面ADE ⊥平面ABCD ,求F 到平面BDE 的距离. 【答案】(1)见解析(2) 【解析】【详解】(1)取BD 中点O ,连接,OM OE ,因为,O M 分别为,BD BC 的中点,所以//OM CD ,且12OM CD =,因为四边形ABCD 为菱形,所以//,CD AB CD ⊄又平面,ABFE AB ⊂平面ABFE ,所以//CD 平面ABFE .因为平面ABFE I 平面,CDEF EF CD =⊂平面CDEF , 所以CD EF ∕∕.又2AB CD ==,所以12EF CD =.所以四边形OMFE 为平行四边形∴所以//MF OE .又OE ⊂平面BDE ∴且MF ⊄平面BDE ,所以//MF 平面BDE .(2)由(1)得//FM 平面BDE ,所以F 到平面BDE 的距离等于M 到平面BDE 的距离. 取AD 的中点H ,连接,EH BH ,因为四边形ABCD 为菱形,且60,2DAB EA ED AB EF ∠====o,所以,EH AD BH AD ⊥⊥因为平面ADE ⊥平面ABCD ,平面ADE I 平面ABCD AD =,所以EH ⊥平面,ABCD EH BH ⊥,因为EH BH ==,所以BE =所以122BDES ==V , 设F 到平面BDE 的距离为h ,又因为11422BDM BCD S S ===V V , 所以由E BDM M BDE V V --=,得113232h =⨯⨯,解得5h = 即F 到平面BDE的距离为5∴ 19.已知以点P 为圆心的圆经过点A （－1，0）和B （3，4），线段AB 的垂直平分线交圆P 于点C 和D ，且|CD |＝,（1）求直线CD 的方程； （2）求圆P 的方程.【答案】（1）x ＋y －3＝0（2）圆P 的方程为（x ＋3）2＋（y －6）2＝40或（x －5）2＋（y ＋2）2＝40 【解析】 【分析】（1）求出AB 中点坐标和直线CD 的斜率，即得直线CD 的方程；（2）设圆心P （a ，b ），求出,a b 的值，即得圆P 的方程.【详解】（1）由题意知，直线AB 的斜率k ＝1，中点坐标为（1，2）. 所以1CD k =-.则直线CD 的方程为y －2＝－（x －1）， 所以直线CD 的方程为x ＋y －3＝0.（2）设圆心P （a ，b ），则由点P 在CD 上得a ＋b －3＝0.①又因为直径|CD |＝，所以|P A |＝ 所以（a ＋1）2＋b 2＝40.② 由①②解得36a b =-⎧⎨=⎩或52a b =⎧⎨=-⎩所以圆心P （－3，6）或P （5，－2）.所以圆P 的方程为（x ＋3）2＋（y －6）2＝40或（x －5）2＋（y ＋2）2＝40.【点睛】本题主要考查直线和圆的方程的求法，考查直线和圆的位置关系的求法，意在考查学生对这些知识的理解掌握水平.20.如图，AB 是O e 的直径，PA 垂直于O e 所在的平面，C 是圆周上不同于A ，B 的一动点.（1）证明：PBC V 是直角三角形；（2）若2PA AB ==，且当直线PC 与平面ABC 时，求直线AB 与平面PBC 所成角的正弦值.【答案】（1）证明见解析；（2）3【解析】 【分析】（1）由PA ABC ⊥平面，得BC PA ⊥，再有BC AC ⊥，这样可由线面垂直的判定定理得线面垂直，从而得证线线垂直，即得证结论；（2）过A 作AH PC ⊥于H ，由（1）可证AH PBC ⊥平面，从而有ABH ∠是直线AB 与平面PBC 所成的角，求出此角正弦值即可．【详解】（1）证明∴AB 是O e 的直径，C 是圆周上不同于A ，B 的一动点.∴BC AC ⊥， ∴PA ABC ⊥平面，∴BC PA ⊥，又PA AC A =I ，PA ，AC PAC ⊂平面， ∴BC PAC ⊥平面，∴BC PC ⊥， ∴BPC △是直角三角形.（2）如图，过A 作AH PC ⊥于H ，∴BC PAC ⊥平面， ∴BC AH ⊥，又PC BC C ⋂=，PC ，BC PBC ⊂平面， ∴AH PBC ⊥平面，∴ABH ∠是直线AB 与平面PBC 所成的角， ∴PA ABC ⊥平面，∴PCA ∠即是PC 与平面ABC 所成的角，∴tan PAPCA AC∠==又2PA =，∴AC =∴在Rt PAC △中，3AH ==，∴在Rt ABH △中，3sin 2AH ABH AB ∠===，即直线AB 与平面PBC 【点睛】本题考查证明线线垂直，考查直线与平面所成的角，求线面角时一般可作出平面的垂直，得出直线与平面所成的角，在三角形中计算即可，即通常所说的作证算三步． 21.已知方程（2＋λ）x －（1＋λ）y －2（3＋2λ）＝0与点P （－2，2）.（1）证明：对任意的实数λ，该方程都表示直线，且这些直线都经过同一定点，并求出这一定点的坐标；（2）证明：该方程表示的直线与点P 的距离d 小于【答案】（1）证明见解析；直线经过的定点为M （2，－2）（2）证明见解析 【解析】 【分析】（1）变形得到2x －y －6＋λ（x －y －4）＝0，得到方程26040x y x y --=⎧⎨--=⎩计算得到答案.（2）易知d ≤|PM |=PM 与直线垂直时，直线方程为x －y －4＝0.，而直线系不能表示此直线，故得证.【详解】（1）解显然2＋λ与－（1＋λ）不可能同时为零，故对任意的实数λ，该方程都表示直线. ∵方程可变形为2x －y －6＋λ（x －y －4）＝0，∴26040x y x y --=⎧⎨--=⎩ 解得22x y =⎧⎨=-⎩故直线经过的定点为M （2，－2）.（2）证明：易知d ≤|PM |=PM 与直线垂直时，等号成立 此时对应的直线方程是y ＋2＝x －2，即x －y －4＝0.但直线系方程唯独不能表示直线x －y －4＝0，∴d <.【点睛】本题考查了直线过定点，点到直线的距离范围，确定直线系不能表示x －y －4＝0是解题的关键.22.已知直线220x y -+=与圆C ：2240x y y m +-+=. （1）求圆C 的方程；（2）过原点O 作圆C 的两条切线，与函数2y x =的图象相交于M 、N 两点（异于原点），证明：直线MN 与圆C 相切；（3）若函数2y x =图象上任意三个不同的点P 、Q 、R ，且满足直线PQ 和PR 都与圆C 相切，判断线QR 与圆C 的位置关系，并加以证明.【答案】（1）()2221x y +-=（2）证明见解析；（3）直线QR 与圆C 相切；证明见解析； 【解析】 【分析】（1）化圆方程为标准方程，得圆心坐标和半径，求出圆心到直线的距离，用表示出弦长，从而求得m ，得圆方程；（2）求出过原点的圆C 的两条切线方程，然后求得两条切线与抛物线的交点坐标后可得证； （3）设()2,P a a，()2,Q b b ，()2,R c c ，由此写出直线,,PQ PR QR 的方程，由直线,PQ PR 与圆相切得出,,a b c 的关系，可得221a b c a +=-；2231a bc a-=-，然后可证直线QR 也与圆相切． 【详解】（1）解：圆C ：2240x y y m +-+=，可化为圆()2224x y m +-=-+，圆心到直线的距离d =，∴，∴224m +=-+⎝⎭， ∴3m =，∴圆C 的方程为()2221x y +-=；（2）证明：设过原点O 的切线方程为y kx =，即0kx y -=，1=，∴k =∴设过原点O 的切线方程为y =，与函数2y x =，联立可得3x y ==，∴3y =与圆C 相切；（3）解：设()2,P a a，()2,Q b b ，()2,R c c ，可得22PQb a ka b b a-==+-， 直线PQ 的方程为()()2y a a b x a -=+-，即为()y a b x ab =+-，同理可得，直线PR 的方程为()y a c x ac =+-， 直线QR 的方程为()y b c x bc =+-， ∴直线PQ 和PR 都与圆C 相切，1=1=，即为()2221230b a ab a --+-=，()2221230c a ac a --+-=，即有b ，c 为方程()2221230x a ax a --+-=的两根， 可得221a b c a +=-；2231a bc a-=-， 由圆心到直线QR222211111a a a a ---===+-，则直线QR 与圆C 相切.【点睛】本题考查直线与圆相交弦长问题，考查直线与圆的位置关系，掌握用几何方法求弦长和判断直线与圆的位置关系是解题基础．。
江苏省泰州中学2021-2022学度高一上学期期中考试英语试卷
江苏省泰州中学2021-2022学度高一上学期期中考试英语试卷高一英语期中考试试题注意事项：1、本试卷共分两部分，第Ⅰ卷为选择题，第Ⅱ卷为非选择题。
2、所有试题的答案均填写在答题纸上（选择题部分使用答题卡的学校请将选择题的答案直截了当填涂到答题卡上），答案写在试卷上的无效。
第I卷（选择题共85分）第一部分：听力（共两节，满分20分）第一节（共5小题，每小题1分，满分5分）听下面5段对话。
每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。
听完每段对话后，你都有10秒钟的时刻来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1．Who is probably the woman?A. Steve’s mother.B. A teacher.C. A student.2. What is David going to do?A. Catch a train home.B. Do his homework.C. Go to a park.3. When does the film end?A. At 7:14 p.m.B. At 8:14 p.m.C. At 8:40 p.m.4. What color is the woman’s dress?A. Blue.B. White.C. Purple.5. What is probably the local（当地的）weather?A. Sunny and hot.B. Cold and cloudy.C. Sunny and windy.第二节（共15小题：每小题1分，满分15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，并标在试卷的相应位置。
听每段对话或独白前，你将有时刻阅读各个小题，每小题5秒钟；听完后，各小题将给出5秒钟的作答时刻。
每段对话或独白读两遍。
听第6段材料，回答第6至7题。
数学卷·2019届江苏省泰州中学高一下学期期中考试(2017.04)
江苏省泰州中学2016—2017学年度第二学期高一数学质量检测2017.3一、填空题：（本大题共14小题，每小题5分，共70分）1. 已知直线1:10l ax y --=，若直线1l 的倾斜角为3π，则a = . 2. 两条平行线1:342l x y +=与1:47l ax y +=的距离为 .3.已知数列{}n a 的前n 项和为21n S n =+，则15a a += .4.在ABC ∆中，角,,A B C 的对边分别为,,a b c ，若120,2,A a b ===，则B = .5．数列{}n a 的通项公式()()1121n n a n +=--，则它的前100项之和为 .6.设n S 是等差数列{}n a 的前n 项和，()7193S a a =+则的54a a 值为 . 7.ABC ∆的内角A,B,C 的对边分别为,,a b c，面积ABC S ∆=ABC ∆外接圆的半径为c = .8.已知等比数列n 的前项和为n S ，若32:3:2S S =，则公比q = .9.在ABC ∆中，角,,A B C 的对边分别为,,a b c，已知2,sin ,a b B C == sin 2C = . 10.直线1:2l y x =与直线()2:00l ax by c abc ++=≠相互垂直，当,,a b c 成等差数列时，直线12,l l 与y 轴围成的三角形的面积S = .11.各项均为正数的等比数列{}n a 中，211a a -=，当取5a 最小值时，数列{}n a 的通项公式n a = .12.在ABC ∆中，已知1,2,b c AD ==是A ∠的平分线，3AD =则C ∠= . 13．n S 是等差数列{}n a 的前n 项和，若2142n n S n S n +=+，则35a a = .14.在锐角三角形ABC 中，角A,B,C 的对边分别为,,a b c ，且满足22b a ac -=，则11tan tan A B+的取值范围为 .二、解答题：本大题共6小题，共90分.解答应写出必要的文字说明或推理、验算过程.15.（本题满分14分）（1）已知直线l 经过点()4,1P ，且在两坐标轴上的截距相等，求直线l 的方程；（2）已知直线l 经过点()3,4P ，且直线l 的倾斜角为()90θθ≠，若直线l 经过另外一点()cos ,sin θθ，求此时直线l 的方程.16.（本题满分14分）在ABC ∆中，,,a b c 分别为角A,B,C 的对边，若cos 3,cos 1,a B b A ==且6A B π-=. （1）求边c 的长；（2）求角B 的大小.17.（本题满分14分）已知ABC ∆的顶点()5,1A ，AB 边上的中线CM 所在的直线方程为250x y --=，AC 边上的高BH 所在直线的方程为250x y --=.（1）求直线BC 的方程；（2）求直线BC 关于CM 的对称直线方程.18.（本题满分16分）如图，在半径为2，圆心角为2π的扇形金属材料中剪出一个四边形MNPQ ，其中M,N 两点分别在半径OA,OB 上，P,Q 两点在弧AB 上，且OM=ON,MN//PQ.（1）若M,N 分别是OA,OB 的中点，求四边形MNPQ 的面积的最大值；（2）若PQ=2，求四边形MNPQ 的面积的最大值.19.（本题满分16分）已知数列{}n a 的首项为2，前n 项和为n S ，且()1112.41n n n n N a a S *+-=∈-. （1）求2a 的值；（2）设1n n n na b a a +=-，求数列{}n b 的通项公式； （3）若(),,,,,m p r a a a m p r N m p r *∈<<成等比数列，试比较2,p mr 的大小.20.（本题满分16分）已知n 为正整数，数列{}n a 满足0n a >，()221410n n n a na ++-=，设数列{}n b 满足2n n n a b t= （1）求证：数列为等比数列； （2）若数列{}n b 是等差数列，求实数t 的值；（3）若数列{}n b 是等差数列，前n 项和为n S ，对任意的n N *∈，均存在m N *∈，使得242211816n n a S a n b -=成立，求满足条件的所有整数1a 的值.。
江苏省泰州中学2019-2020学年高一上学期期中考试英语试题
江苏省泰州中学2019—2020学年度第一学期中考试高一英语命题人：乔霖琳审核人：刘慧敏第一部分听力（共两节，满分30分）做题时，先将答案标在试卷上，录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。
第一节（共5小题，每小题1.5分，满分7.5分）听下面5段对话。
每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。
听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题，每段对话仅读一遍。
1.When will the speakers probably study together?A.At four o’clock.B.At six o’clock.C.At eight o’clock.2.Which restaurant will the speakers probably go to?A.Mario’s.B.Luigi’s.C.Gino’s.3.What does the man want to do tonight?A.Watch football on TV.B.Buy some books.C.Go to a basketball game.4.Who might Mary be?A.The woman’s dog.B.The man’s daughter.C.The man’s neighbor.5.What is the woman doing?A.Working.B.Apologizing.C.Expressing her thanks.第二节（共15小题，每小题15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，并标在试卷的相应位置。
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。
2020-2021学年江苏省泰州市姜堰区姜堰中学高一下学期开学考试英语试题 Word版含答案
2020-2021学年江苏省泰州市姜堰区姜堰中学高一下学期开学考试英语试题(考试时间：120分钟总分150分）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. How does the woman check the weather?A. She uses her phone.B. She listens to the radio.C. She watches television.2. What will the man do tonight?A. Practice debating.B. Study for an exam.C. Watch a basketball game.3. Why doesn’t the man use the method the woman gave?A. He doesn’t like it.B. He doesn’t have it yet.C. He doesn’t understand it.4. What is more important in the man’s opinion?A. Teamwork.B. Star players.C. Game time.5. Where are the speakers?A. At a store.B. At a bank.C. At a park.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，并标在试卷的相应位置。
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。
江苏省泰州中学2019-2020学年高一上学期期中考试英语试题(解析版)
● Simple and user-friendly design.
● Take a picture of your food to easily track meals.
● The app uses artificial intelligence to recognize what’s in a photo. Easily add the size of your food and extras like barbecue sauce to keep an exact record.
● Track weight regularly, as well as track total calories eaten, burned and left each day..
Disadvantages:
● The app is based on the principles of calories and doesn’t consider food quality.
A. The woman’s dog.B. The man’s daughter.C. The man’s neighbor.
5. What is the woman doing?
A. Working.B. Apologizing.C. Expressing her thanks.
第二节（共15小题，每小题15分，满分22.5分）
A. He will be sent an email.
B He will receive some money.
C. He will be given more information.
13. What will the man do next?
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第 1 页 共 18 页 江苏省泰州中学2016-2017学年高一下学期期中考试 英语试题 试卷分第I卷（选择题）和第Ⅱ卷（非选择题）两卷，满分120分，考试时间120分钟。 第I卷将正确的选项填涂在答题卡上的相应位置上，第Ⅱ卷做在答题卡上。 第I 卷（共75分） 第一部分：听力（共两节，满分20分） 第一节 （共5小题；每小题1分，满分5分） 听下面5段对话。每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。每段对话仅读一遍。 1. What color key does the man need to get into the house? A. Blue. B. Red. C. White. 2. Where are the speakers talking? A. At the man’s house. B. In the woman’s garage. C. In an office building. 3. Who is the woman talking with? A. Her student. B. A postman. C. Her boss. 4. What will the speakers do next? A. Wait for their parents. B. Go swimming. C. Go home. 5. What are the speakers mainly talking about? A. Cooking skills. B. Their colleagues. 第 2 页 共 18 页
C. Weekend plans. 第二节 （共15小题；每小题1分，满分15分） 听下面5段对话或独白。每段对话或独白后有几个小题，从题中做给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。听每段对话或独白前，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 does the woman ask the man to do? A. Look for some toys. B. Take care of a baby. C. Take her to the office. 7. What’s the probable relationship between the speakers? A. A couple. B. Neighbors. C. Co-workers. 听第7段材料，回答第8、9题。 8. What has the man been doing lately? A. Designing a website. B. Looking for a job at Twitter. C. Chatting with friends on Face book. 9. How does the man sound? A. Worried. B. Surprised. C. Confident. 听第8段材料，回答第10至12题。 10. Which apartment does the woman live in? A. 15B. B. 16B. C. 17B. 11. Why is the woman having trouble opening her apartment door? A. They key won’t work. 第 3 页 共 18 页
B. The hallway is dark. C. She is at the wrong door. 12. How did the man help the woman? A. He told her where to get a new key. B. He opened the elevator for her. C. He explained the situation to her. 听第9段材料，回答第13至16题。 13. Who might Richie be? A. The woman’s boss. B. The man’s team leader. C. The woman’s teammate. 14. Why does the man want to join the woman’s team? A. He likes field work. B. He thinks he could help a lot. C. He doesn’t get along with Paul. 15. When did the woman have a meeting with Richie? A. Last month. B. Last week. C. Yesterday. 16. What will the woman tomorrow? A. The man will put a new team together. B. The speakers will go to Edmonson. C. The woman will talk with Paul. 听第10段材料，回答第17至20题。 17. Where are escape rooms already very popular? A. In the big cities of China. B. In rural areas of China. C. All over China. 18. What does the speaker say about the escape rooms? A. They are easier than they seem. 第 4 页 共 18 页
B. Each one has a similar subject. C. They’re quite difficult. 19. What must plays do before they can open the door? A. Figure out a series of puzzles. B. Find a key before time is up. C. Answer some questions about math or history. 20. What theme is mentioned by the speaker? A. Communicating with animals. B. Solving crimes. C. Looking for treasure. 第二部分：英语知识运用（共两节，满分55分） 第一节 单项填空（共15小题，每小题1分，满分15分） 从A、B、C、D四个选项中，选出可以填入空白处的最佳选项，并在答题卡上将该项涂黑。 21. He just stood at distance, wondering if it was real goodbye. A. a, a B. a, the C. the, the D. the, a 22. In Britain, packets of cigarettes come with a government health warning them. A. attach to B. attaching to C. attached to D. to attach to 23. With time going by, we have come to realize that creativity is it takes to keep a nation highly competitive. A. what B. how C. why D. that 24. , please start the work tomorrow. A. If you are convenient B. If it is convenient for you C. If it is convenient of you D. If you have convenience 25. You her more help, even though you were really very busy. A. ought to have given B. must have given C. might give D. ought to give 26. Icy roads are dangerous for drivers it’s hard to know what is going happen 第 5 页 共 18 页
and accidents take place easily. A. even if B. as if C. in that D. so that 27. I have sent Miss Green an invitation to our party, but I don’t have the slightest idea she will accept it. A. whether B. that C. how D. why 28. Brian has been studying really hard all the time; he has got an A in each of his subjects. A. instead B. anyhow C. thus D. rather 29. ---Are you satisfied with the 20 students’ performance? ---Yes, though haven’t passed the road test. A. all B. none C. those D. ones 30. After the earthquake, a team which fifteen volunteers and ten doctors was sent to search for and save survivors. A. made up of B. consisted in C. made of D. consisted of 31. Mr. Black, company I have been working donates a lot of money to Project Hope every year. A. which B. in which C. whose D. for whose 32. himself in doing research all day, Jerry didn’t even have time to time to have a meal with his family. A. Occupied B. Occupying C. Being occupied D. To occupy 33. Tom’s mother kept telling him she should work harder, but didn’t help. A. he B. she C. it D. they 34. Our school was moved to Xiangjiang New District in 2014, and the new campus is the old one. A. three times as the size of B. more larger three times than C. three times larger than that of D. three times as large as 35. ---Oh, my God! This week I got fired and got my wallet stolen. ---Poor guy! . But maybe your good luck is on the way. A. Failure is the mother of success B. It never rains but it pours