浙江省名校新高考研究联盟(Z20联盟)2020届高三上学期第一次联考 英语 含答案
浙江省名校新高考研究联盟（Z20联盟）2020届第一次联考 英语试题卷 考生须知： 1.本试题卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题），满分为150分，考试时间为120分钟。 2.用黑色签字笔将学校、班级、姓名、考号分别填写在答题卷和机读卡的相应位置上。 第Ⅰ卷（选择题部分） 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。 例：Hour much is the shirt? A.￡19.15. B. ￡9.15 C.￡9.18 答案是B。 1.What is the woman planning to do? A. Search for the new tie. B. Paint the shelf. C.Fix the shelf 2.What can we learn from the conversation? A. The man can't drive well. B. The car has broken down. C.They are on the wrong way 3. When does the conversation take place? A. On Friday. B. On Saturday. C. On Sunday. 4. Who is the man? A. A teacher. B.A doctor. C. A patient 5. What does the man think about the price of the car? A. Acceptable B.Too high C.Unbelievable 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。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。 听第6段材料，回答第6～7题。 6. What does the man think of his job? A. Tiring. B. Boring. C.He doesn’t like it. 7. What is his job? A. A secretary. B. A clerk C.A manager. 听第7段材料，回答第8～10题。 8.Why was the woman on the underground train? A.She was going to work. B.She was coming home from work. C.She was traveling to another city. 9. What did the woman do when the robber pointed a knife at her? A. She held her handbag tightly. B. She caught hold of his knife. C. She asked the passengers for help. 10. Who caught the robber when he was running away? A. Two policemen. B. The woman herself. C. Two other passengers. 听第8段材料，回答第11～14题。 11. Why didn't the woman buy the book? A. Because it's too expensive. B. Because she can't buy it anywhere. C. Because she has already got one. 12. Why did the man suggest that the woman read the book? A. The professor has written it. B. The professor uses it in his course. C. It is about sociology. 13. Why does the woman have problems getting the book from the library? A. It is in great demand. B. It was sold out already. C. It isn't owned by the library. 14. How does the woman react to Tom's idea? A. She thinks it ridiculous. B. She wonders if she can afford it. C. She thinks it a good idea. 听第9段材料，回答第15～17题。 15. Where will the man have Thanksgiving? A. At his own home. B. At the woman's home. C. At his parents' home. 16. What are the traditional dishes at the woman's home on Thanksgiving? A. Turkey, sweet potatoes and apple pie. B. Turkey, salty potatoes and pumpkin pie. C. Turkey, sweet potatoes and pumpkin pie. 17. What are the men in the woman's family doing while the women arc cooking? A. They are chatting in the room. B. They are watching football games. C. They are helping the women cook. 听第10段材料，回答第18～20题。 18. What caused the traffic to stop? A.The storm. B. The snow. C. The wind. 19. What will the weather be like on Saturday? A.It will probably be rainy in the evening. B. It will be fine all day, C. It will be windy in the afternoon. 20. What is the season now? A. Summer. B. Spring. C. Winter. 第二部分：阅读理解(共两节，满分35分) 第一节：(共10小题；每小题2.5分，满分25分) 阅读下列短文，从每题所给的A、B、C和D四个选项中，选出最佳选项，并在答题纸上将该项涂黑。 A I used to be crazy about the hunting season. The excitement of waiting for a prey(猎物)and the pride of showing off the kill fascinated me. However, everything changed after that cold morning. Early on that day of the late fall, I set off alone for the woods, packing a gun, a bottle of hot coffee and three thick sandwiches. After finding the fresh deer's tracks in the snow, I settled down behind a little bush. I sat there for about an hour. It was then that I saw him. A deer, a big beautiful deer! There was no cover nearer to him than 30 yards. Surely I couldn't miss! I waited for him to realize I was there. I waited for him to be shocked and run away. But he fooled me completely. He came towards me! He was curious, I suppose, or maybe lie was stupid---how else can you explain it? Well, that deer walked right up to where I was sitting. Then he stopped and looked at me! What happened next is hard to believe, but it's true. And it all seemed quite natural. Just as when a friendly young deer comes near you, I reached up and scratched his head. And he liked to be scratched. In fact, he practically asked for more. Then, I fed him my sandwich! Yes, I know what a deer eats, but that deer ate my sandwich. Well, he finally went his way, down the hill and up the deer trail. Shoot him? Not me. You wouldn't have either, not after that. I just watched him go. When I was about half way back, I heard two shots, followed by a dull slam(撞击)a few seconds later. Those two shots usually mean a kill. I had forgotten there were other hunters that day. Those hunters would never know they could have scratched his head. 21. Why didn't the author kill the deer? A. He preferred to shoot a shy deer. B. He was fooled by the tricky deer. C. He was sympathetic for the deer. D. He was too shocked to shoot the deer. 22. What most probably happened to the deer in the end? A. Other hunters shot the deer to death. B. Other hunters scratched the deer's head too. C. The deer managed to escape from being shot. D. The deer would become friends with the author. 23. What's the best title of this text? A. A Hunting in Late Fall. B. A Lovely Deer. C. The Cruel Killing. D. The Last Hunting. B People have grown taller over the last century, with South Korean women shooting up by more than 20cm on average, and Iranian men gaining 16.5cm. A global study looked at the average height of 18-year-olds in 200 countries 1914 and 2014. The results show that while Swedes were the tallest people in the world in 1914, Dutch men have risen from l2th place to claim top spot with an average height of 182.5cm. Larvian women, meanwhile, rose from 28th place in 1914 to become the tallest in the world a century later, with an average height of 169.8cm. James Bentham, a co-author of the research says the global trend is likely but once you average over whole populations, genetics plays a less key role," he added. But while height has increased around the world, the trend in many countries of north and sub-Saharan Africa causes concern, says Elio Riboli of Imperial College. While height increased in Uganda and Niger during the early 20th century, the trend has reversed in recent years, with height decreasing among 18-year-olds. "One reason for these decreases in height is the economic situation in the 1980s," said Professor Alexander. The nutritional and health problems that followed the policy of structural adjustment, he says, led to many children and teenagers failing to reach their full potential in terms of height. Bentham believes the global trend of increasing height has important implications. "How tall we are now is strongly influenced by the environment we grew up in," he said. "If we give children the best possible start in life now, they will be healthier and more productive for decades to come." 24. What can be learned from Paragraph I? A. The increase in women's height is much bigger than men's in the last century.
浙江省名校新高考研究联盟(Z20联盟)2020届高三上学期第一次联考+化学+Word版含答案
绝密★考试结束前（高三署假返校联考）浙江省名校新高考研究联盟（Z20联盟）2020届第一次联考化学试题卷本试题卷分选择题和非选择题两部分，共8页，满分100分，考试时间90分钟。
可能用到的相对原子质量：H1 C12 N14 O16 Na23 Mg24 S32 Cl35.5 K39 Fe56 Zn65选择题部分一、选择题（本大题共16小题，每小题3分，共48分。
每小题列出的四个备选项中只有一个是符合题目要求的，不选、多选、错选均不得分）1.化学与社会、环境密切相关，下列说法正确的是A.光催化还原水制氢比电解水制氢更节能环保、更经济B.CO2、NO2或SO2都会导致酸雨的形成C.人造棉、蚕丝、棉花、涤纶的主要成分都是纤维素D.向汽油中添加乙醇后，该混合燃料的热值不变2.下列表示不正确．．．的是O B.乙烷的球棍模型：A.中子数为10的氧原子：188C.Mg2＋的结构示意图：D.次氯酸的电子式：3.下列有关物质性质与用途具有对应关系的是A.AgBr淡黄色难溶于水，可用于制作感光材科B.SiO2熔点高硬度大，可用于制光导纤维C.Al2O3是两性氧化物，可用作耐高温材料D.NH3沸点高易液化，可用作致冷剂4.下列说法正确的是A.准确量取25.00mL的液体可选用移液管、量筒或滴定管等量具B.实验过程中若皮肤不慎沾上少量酸液，应先用大量水冲洗，再用饱和碳酸氢钠溶液洗，最后再用水冲洗C.用分液漏斗分液时要经过振荡、放气、静置后，从上口倒出上层液体，在打开旋塞，将下层液体从下口放出D.在中和热测定实验中，盐酸和NaOH溶液的总质量m g，反应前后体系温度变化为t，反应液的比热容为cJ•g－1•°C－1，则生产1mol水放出的热量为5.在给定条件下，下列选项所示的物质间转化均能实现的是A.2323()()()CO NaCl aq NaHCO s Na CO s ∆−−−→−−→ B.()22()()()()NaOH aq CuCl aq Cu OH s Cu s ∆−−−−→−−−→葡萄糖 C.2226()()()MgCl H O s MgCl s Mg s ∆⋅−−→−−−−→HCl 熔融电解 D.233()()()O NH g NO g HNO g ∆−−−−→−−→2H O催化剂， 6.右表为元素周期表的一部分。
2020届浙江省名校新高考研究联盟(Z20联盟)高三12月第二次联考英语试卷(有答案)
浙江省名校新高考研究联盟(Z20联盟)2020届第二次联考英语试卷考生须知：1.本试卷分第I卷(选择题)和第II卷(非选择题)。
满分为150分，考试时间为120分钟。
2.请用黑色签字笔将学校、班级、姓名、考号分别填写在答题卷和机读卡的相应位置上。
第I卷(选择题部分)第一部分：听力(共两节，20小题，每小题1.5分，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纸上。
第一节：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's the man going to do first?A. Wash his hands.B. Do some typing.C. Move the sofa.2. What color does the man like best?A. Light green.B. Dark blue.C. Brown.3. How much does one ticket cost?A. $20.B. $40.C. $80.4. What is the conversation about?A. A robbery.B. A video phone.C. A phone problem.5. How does the man find his living place?A. Too crowded.B. Quite noisy.C. All right.第二节：听下面5段对话或独白。
每段对话或独白后有2至4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5秒钟的时间阅读各个小题；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。
高三试卷英语-浙江省名校新高考研究联盟(Z20联盟)2023届高三第二次联考英语试卷及参考答案
Z20名校联盟（浙江省名校新高考研究联盟）2023届高三第二次联考英语试题卷第I卷第一部分：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 girl like about her parents?A.They are kind.B.They are rich.C.They have influence.2.How much does the keyboard cost now?A.$50.B.$40.C.$20.3.Why did the boy refuse the girl's offer?A.He has a doctor's appointment.B.He has gone to the beach earlier.C.He needs to go to see his family.4.What are the speakers doing?A.Making dinner.B.Shopping for a party.C.Ordering some food.5.What does the man want to know?A.How to understand expressions.B.How to make desserts.C.How to solve problems.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。
浙江省Z20名校联盟(浙江省名校新高考研究联盟)2025届高三第一次联考+数学答案
Z20名校联盟（浙江省名校新高考研究联盟）2025届高三第一次联考数学参考答案一、单选题：本题共8小题，每小题5分，共40分.在每小题给出的选项中，只有一项是符合题目要求的. 1．【答案】B【解析】{|12}A x x =-≤≤，3{|}2B x x =<，所以A B =3{|1}2x x -≤<，故选B.2．【答案】D【解析】347232(2)(150)6C x x x=--，故选D.3．【答案】D【解析】113137951913()13131242999133()2a a S a S a a a +===⨯=+，故选D. 4．【答案】C【解析】由分布列的11136a ++=，得12a =，所以11111()1233266E X =⨯+⨯+⨯=,所以14(21)2()13E X E X +=+=，故选C.5．【答案】B若函数()f x 在(1,)+∞上单调递增，则1210a a ⎧≤⎪⎨⎪-≥⎩，得1a ≤，所以“2a ≤”是“函数()f x 在(1,)+∞上单调递增”的必要不充分条件. 6．【答案】C【解析】因为(0,1)x ∈，则，得666x πππωω<+<+，由题意得3262πππω<+≤，得433ππω<≤，故选C. 7．【答案】A【解析】将圆台母线延长交于点S ，得圆锥1SO ，作圆锥1SO 的轴截面如右图，设底面直径2AB R =，由条件知11cos 3SAO ∠=，得3SA R =，1SO =，设内切球半径为r ，则12OT OO OO r ===，所以3SO r =，那么14SO r ==，则R =，2O 为1SO 的中点，CD 为SAB ∆的中位线，1A于是内切球的体积3143V r π=，圆台的体积2232171772483243V R SO r r r πππ=⋅⋅=⋅⋅=,所以圆台与其内切球的体积比为2174V V =，故选A.8．【答案】C【解析】令()()()0h x f x g x =-=，即2(1)1=cos 22xa x ax π---，整理得方程21cos 2xax a π+-=在(1,1)-上有解，记2()1,()cos 2xF x ax aG x π=+-=，即函数(),()F x G x 图像有公共点，如图，0()1G x <≤，当0a ≤时，2()11F x ax a =+-≤-， 显然函数(),()F x G x 图像无公共点，当0a >时，由(),()F x G x 图像的对称性，得(0)(0)(1)(1)F G F G ≤⎧⎨>⎩，即11210a a -≤⎧⎨->⎩，解得122a <≤，故选C.二、多选题：本题共3小题，共18分.在每小题给出的选项中，有多项符合题目要求. 9．【答案】BCD【解析】若1c =，2510a b ==，则25log 10,log 10a b ==，显然25log 10log 101a b c =≠+=+，故A 不正确.因为,,0a b c >，显然a b c >>，故B 正确.设2510a b c t ===，得2510log ,log ,log a t b t c t ===，则111log 2,log 5,log 10t t t a b c ===，所以111log 2log 5log 10t t t a b c+=+==，故C 正确. 对于D ，1144(4)()(14)9b aa b c a b c c a b a b+=++=+++≥，故D 正确. 10．【答案】AC【解析】如图，选项A ，直线2y x =，d CM ==，||2||AB AM ===A 正确.对于选项B ，CA CB ⋅u u r u u r||||cos cos CA CB ACB ACB =⋅∠=∠因为点A, B 不重合，所以cos 1ACB ∠<，故B 不正确. 对于选项C ，||||(||||)(||||)OA OB OM MA OM MA ⋅=+- 222222||||||()OM MA OC d r d =-=---22||1OC r =-=，故C 正确.对于选项D ，如图线段AB 中点M 满足OM CM ⊥，M 的轨迹是以OC 为直径的圆（圆C 内部部分），所以轨迹长为12222π⋅=，故D 不正确. 11．【答案】ACD【解析】当2n =时，22(cos )1cos22sin 22cos f x x x x =-==-，那么2()222f x x =-≤，故A 正确.对于B ，当3n =时，(cos )[cos()]1cos3()f x f x x ππ-=-=-- 1cos(33)x π=--1cos3x =+,三、填空题：本题共3小题，共15分.12．【答案】11314【解析】设双曲线得右焦点为F '，BF m =，则4AF m =，连结'AF ，'BF ，则'2BF m a =+，42AF m a '=+, 在BFF '∆中，BFF '∠=60︒，由余弦定理得222(2)42m a m c mc +=+-，整理得2222(2)c a m a c -=+ ①在AFF '∆中，120AFF '∠=︒，由余弦定理得222(42)1648m a m c mc +=++，整理得22(42)c a m a c -=- ②① ②两式相除得，222a c c+=，得65a c =，所以渐近线方程为5y x =±.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．（本小题满分13分） 【解析】 （1）连结AQ ，因为PM //平面ABC ，PM ⊂平面ADQ又平面ADQ 与平面ABC 相交于AQ ，所以//PM AQ ， 因为P 是AD 的中点，所以M 是DQ 中点. ┄┄6 （2）方法一，因为AD ⊥底面BCD ，BC CD ⊥，如图建立坐标系， ┄┄7(2,0,0)D ，(0,2,0)B ，(2,0,2)A ，(0,1,0)Q ， (2,1,0)DQ =-uuu r ，(2,0,2)CA =uu r ，(0,2,0)CB =uu r, 则平面ABC 的法向量为(1,0,1)=-n ，┄┄9 所以cos ,DQ DQ >=DQ ⋅<=⋅u u u ru u u r u u u rn n n，┄┄12 因此直线DQ 与平面ABC . ┄┄13 方法二，取AC 中点N ，因为DA DC =，所以DN AC ⊥, 因为AD ⊥底面BCD ，所以AD BC ⊥，又BC CD ⊥，则BC ⊥平面ACD ， ┄┄7 所以BC DN ⊥ 所以DN ⊥平面ABC ，于是DQN ∠即为所求，┄┄9DN DQ┄┄11因此sin DN DQN DQ ∠=. ┄┄13 方法三，设D 到平面ABC 的距离为d ，1242333A BCDBCD V AD S -∆=⋅=⨯=， ┄┄8易知12ABC S BC AC ∆=⋅= ┄┄9所以1433A BCD D ABC ABC V V d S --∆==⋅==，得d11因此直线DQ 与平面ABC 所成角的正弦值d DQ ┄┄13P B16．（本小题满分15分） 【解析】（1）由sin sin cos =22sin a c A CB c C--=, ┄┄2 则2sin cos =sin sin sin()sin sin cos sin cos sin C B A C B C C B C C B C -=+-=+- 整理得sin sin cos sin cos sin()C B C C B B C =-=-,则C B C =-，即2B C =， ┄┄5由3A π=，得233B C C π+==，则24,99C B ππ==. ┄┄7（2）由ABC ∆是锐角三角形知2232B C B C C ππ⎧=<⎪⎪⎨⎪+=>⎪⎩，得64C ππ<<， ┄┄9则cos C << ┄┄11 由正弦定理得sin c b B =，得sin 4sin 28cos sin sin c B Cb C C C===, ┄┄13因此b <<┄┄1517．（本小题满分15分） 【解析】（1）由条件得12c e a ==，即2a c =，则b，┄┄2所以12OM a c ==，2max 1()()2BMP S b a c ∆=+1c =，┄┄4因此椭圆E 的方程为22143x y +=. ┄┄6 （2）设直线PQ ：(1)y k x =+，1122(,),(,)P x y Q x y ，BP uu r11(2,)x y =-，BQ uu u r 22(2,)x y =-,与椭圆联列方程得22(1)3412y k x x y =+⎧⎨+=⎩，得2222(34)84120k x k x k +++-=， 则221212228412,3434k k x x x x k k -+=-=++, ┄┄8所以BP BQ ⋅uu r uu u r212121212(2(2(2(2(1(1x x y y x x k x x =--+=--+++）））））） ┄┄1022222222121222(1)(412)8(2)(1)(2)()443434k k k k k x x k x x k k k k+--=++-+++=-++++ 2227634k k ==+，┄┄13 得26k =，k =PQ 的方程为1)y x =+. ┄┄1518．（本小题满分17分） 【解析】（1）()()(1)ln (1)ln(1)g x f x f x x x x x =+-=+--，01x <<,令'()1ln ln(1)1ln ln(1)g x x x x x =+---=--， ┄┄1令'()0g x =，得12x =， ┄┄2 当1(0,)2x ∈时，'()0g x <，当1(,1)2x ∈时，'()0g x >，所以()g x 有极小值1()ln 22g =-，无极大值. ┄┄5（2）()1ln 0f x x '=+=，得1x e=，易知()f x 在1(0,)e 上递减，在1(,)e +∞上递增，结合()f x 的图象，由题意得()0f e e ae bb ==+⎧⎨≥⎩，得0b e ea =-≥，1a ≤. ┄┄9于是21(1)()24e ab ea a e a =-=--+，故max ()4eab =. ┄┄11（3）先证明左边：作差()()ln ln ln ln ln f n f m n n m m n m m mm n m n m---+-=-- (ln ln )ln 1nn n m n m n n m m m-==-- ┄┄12令1n t m=>，(ln ln )ln ln 111n n m t tt n m t t-==---， 令()ln 1h t t t t =-+，'()1ln 1ln h t t t =+-=，当1t >时，'()0h t >，函数()h t 在(1,)+∞上是增函数，所以()ln 1(1)0h t t t t h =-+>=，因此ln 1t t t >-，所以ln 11t t t >-，即()()ln 1f n f m m n m -->-，故()()ln 1f n f m m n m->+-. （或者利用1ln 1t t >-，得ln 111t t>-） ┄┄15 对于右边()()ln ln ln ln ln f n f m n n m m n n m nn n m n m---+-=-- (ln ln )1ln1m n m n n n m m m-==--. 令1n t m =>，(ln ln )m n m n m --ln 11tt =<-， （利用ln 1t t <-，得ln 11tt <-）即()()ln 1f n f m n n m--<-，故()()ln 1f n f m n n m -<+-.综上得()()ln 1ln 1f n f m m n n m-+<<+-. ┄┄17（证出任何一边得4分）19．（本小题满分17分） 【解析】（1）由于{}n x 是等比数列，则212n n n x x x ++=，且12,,0nn n x x x ++≠，0a ≠, ┄┄2 由条件得21n n n x ax ax +=-，所以22111()()n n n n n n x ax ax x ax ax +++-=-，则1n n ax a ax a +-=-，即n x =21n nn x ax ax +=-，得01n a x x a+== ┄┄4 所以101a a +<<，即110a-<<，得1a <-. ┄┄6 （2）①由1a =-知21n n n x x x +=-+，11111=11n n n nn x x x x x +=+--（）， 则11111n n nx x x +-=-， ┄┄8 因为210n n n x x x +-=-<，所以数列{}n x 是递减数列， 于是012n x x ≤=，111121n n nx x x +-=≤-； ┄┄10 又110n n nx x x +=->，所以1,n n x x +同号，那么n x 与0x 同号，即0n x >， 于是111111n n n x x x +-=>-，因此11112n nx x +<-≤. ┄┄12 ②由21nn n x x x +=-，得2011012n i n n i x x x x ++==-=-∑, ┄┄14 因为1112n n x x +-≤，所以10112(1)24n n n x x +≤++=+，则1124n x n +≥+，┄┄16 所以21111122242(2)ni n i n x x n n +=+=-≤-=++∑. ┄┄17。
2020年6月浙江省Z20联盟-英语参考答案
浙江省名校新高考研究联盟（Z20联盟）2020届第三次联考英语参考答案第一部分：听力部分（30分）1-5 BBAAC 6-10 CACAB 11-15 CCBCA 16-20 BCAAB第二部分：阅读理解（35分）21-23 CCB 24-26 DCA 27-30 CDBA 31-35 EAGFB第三部分：语言运用（45分）完形填空36-40 ACBAD 41-45 BCADB 46-50 DCABB 51-55 CDDAC语法填空56. ending 57. successful 58. be spent 59. growth 60. an61. on 62. effectively 63. will provide 64. to see 65. Whether第四部分：写作部分（40分）应用文写作Dear George,I’m terribly sorry that I’m not able to be your guide for the one-day tour this weekend, for I’ve signe d up for the voluntary work at a local hospital.I highly recommend the West Lake in Hangzhou as your destination. The picturesque scenery there in this season must be a feast for your eyes. You’ll also be attracted by the historical and cultural sites an d the legends behind. Considering there is a long way to go, please wear comfortable sneakers. Don’t forget to apply some sunscreen.Wish you an enjoyable weekend.Yours,Li Hua读后续写Paragraph 1Unfortunately, she soon came again, with her shoes in hands and preparing for an outing. “Don’t even ask, Meghan. Not today.” I knew her plan too well: First th e shoes. Then the stroller. And pretty soon we were in the park. The way she always did. Having sensed my impatience with her, Meghan put down the shoes, tugged at my sleeve with her free hands and pleaded using her hard-to-understand baby talk. “No, Meghan,” I said again. “Not now. Go away and leave me alone.” Finally, she left. She made no further attempt to bother me. Then I tried to concentrate on the test paper without interference.Paragraph 2Out of the corner of my eye I could see the little girl sobbing because I didn’t have time for her.wondering if you would care to join me.” Hearing this, she laughed excitedly and began the frantic search for her shoes. I knew I should put aside my work and go right at the moment—while bib-bibs still sparked wonder and while my girl thought that a walk with her father was a gift beyond measure.听力原文：Text 1M: Well, I think maybe you could watch me play basketball tomorrow afternoon.W: I’d love to, but I have to go to the denti st.Text 2W: Which platform does the 4:30 train leave from?M: Platform No. 6. Wait. It’s Platform No. 5. The 5:30 train leaves from Platform No. 6.Text 3M: I bought some snacks for us to have on the trip.W: OK, great. But I’m allergic to chocolate and nuts.M: Oh, never mind. I have brought some fruit.Text 4W: I have this huge essay that’s due tomorrow, and I haven’t even started it.M: You’re in luck. There’s supposed to be a big snowstorm tonight. I bet school will get canceled and you’ll get an e xtra day.W: Really? That would be fantastic.Text 5M: It’s already 6:30 pm now. Let’s call it a day.W: But I haven’t finished my work yet. I’m afraid I cannot leave now.M: Then I’ll go without you. See you tomorrow.Text 6W: Nancy said she’s moving to Russia.M: Isn’t she in South Africa?W: No. She is in Kenya, and now she is going to Russia.M: She doesn’t speak Russian, does she?W: A little. She’s been studying it for about three months. I think she’s going to enroll in a language class at the university when she gets there.M: Where’s she going to live, Moscow?W: I’m not sure. She’s talked a lot about St. Petersburg.M: Well, I wish I could go to some other country, too, just have a visit.W: Then you’d better start saving your pennies. I’ve been doing this for my traveling plan for a long time. Text 7W: Ooops, I’m glad that the lecture’s over. Alan, can I borrow your notes? I missed many points.M: Sorry, but I didn’t take notes.W: Are you kidding? This lecture is important.M: Actually, I don’t know how. It’s already hard for me to follow the instructor’s fast speed, let alone takenotes.W: Did you read the materials that he handed out last time?M: No.W: You’d better read it or spend 5 to 10 minutes skimming it before the lecture. It will hel p you get the main idea.M: I see. But there is still so much information in a lecture.W: Of course you have to select the most important points. Don’t write down every word the instructor says.Pay attention to how he organizes the materials, and you may want to add your own headings.M: Well, where is my pen? Ah, here it is. And? Anything else?W: Oh, yes. Sometimes there will be gaps in your notes, you can exchange notes with a classmate to fill them in right after the class while your memory is fresh. I was trying to do that just now.M: Eh.W: Hey! What are you doing? You are not looking at me!M: Me? Ha! I’m taking notes, of course.Text 8W: I’d like to reserve two tickets on Saturday, the 10th.M: For which movie, ma’am?W: Um.…for Dark and Stormy Night, please.M: I’m sorry, that show is sold out.W: Well, how about on the 11th?M: Yes, ma’am, we do have tickets for that show.W: Are there any seats left in the middle?M: Yes, but they’re not next to each other. If you want them together, we still have seats in the front and at the back.W: The front row sounds good. How much are the tickets?M: ＄13 in total.W: Okay. Do I have to pay in cash?M: Well, credit card is also acceptable.W: Uh, just a second. Here is my card.Text 9W: Good morning, Mr. Bourne. I assume you have a nice weekend.M: Not really. My wife and my daughter were arguing the whole weekend about God knows what, and my son came down with the flu, so we had to miss a really important football game.W: Oh, is your son feeling any better now?M: He’ll be fine in a few days. So, what’s on the schedule for today?W: You’ve got a 10: 00 a m. meeting with Mr. Peterson. And the car will pick you up downstairs at noon to take you to Charmaines. You’ll have lunch there with Ms. Smith. She wan ts to discuss the details about your business cooperation before she leaves for Paris this evening.M: All right. What about the afternoon?W: So far, it looks like you just have the regular 3: 00 p. m. managers’ meeting and some e-mails to respond to.M: That’s relief. Maybe I’ll even make it out of there on time today.W: It’s not very likely considering your daily routine, but I’ll keep my fingers crossed.Text 10M: In the age of mobile payments, what we really need to survive is a smartphone. But one company is going to make your traditional wallets cool again, by bringing them into the digital age. Armenian firm V olterman’s Smart wallet takes security to a new level. It’s fitted with different technology. In a smart wallet, there is an alarm, a GPS tracker, and even a camera. If your phone and wallet become separated, the alarm will warn you so that you don’t leave it behind. And if your wallet is lost or stolen, the GPS tracker can help you reach it. But perhaps the most unusual feature of the V olterman smart wallet is the built-in front-facing camera. If someone dares to open your wallet without your permission, the little camera will take a picture and send it to your mobile phone. Once your wallet is stolen, you will find it easily by receiving the p icture of the thief. The V olterman is really the world’s most powerful smart wallet. It’s also lightweight and thin, making it perfect for travel. The high-tech wallet comes in three models, with the cheapest cost expected to be $169.。
Z20 名校联盟(浙江省名校新高考研究联盟)2023 届高三第一次联考参考答案
Z20名校联盟（浙江省名校新高考研究联盟）2023届高三第一次联考数学参考答案（后附评分细则）一、单选题（共8小题，每小题5分，共40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）8．解法一：不妨设()()1,0,2,0a b =−=，(),c x y =，因为12c a c b −=−，=即2240x x y ++=，由图可知，向量c b −与a 夹角的最大值是6π. 解法二：∵2c a c b −=−，∴2c b b a c b −+−=−，又∵2b a =−，∴()23c b a c b −−=−， 则()()()222469c b a c b a c b ⎡⎤−−⋅−+=−⎢⎥⎣⎦， 即()()28120c b a c b −−⋅−+=，即()()2128c b a c b−+⋅−=，所以()()()()22212123cos ,288c b a c b c b a c b a c bc bc b−⋅−−+<−>==≥=−−−， 向量c b −与a 夹角的最大值是6π.二、多选题（本大题共4小题，每小题5分共20分.每小题列出的四个选项中有多个是符合题目要求的，全部选对得5分，部分选对得2分，有选错的得0分）11．解析：如图，过A 、B 作准线1y =−的垂线，垂足分别为H 、G ，设线段AB 的中点为C ，C 在准线上的射影为D .当线段AB 为通径时长度最小为24p =，故A 正确；y xOac bc b −因为1212AB x x k p+==，故B 正确； 因为直线1y =−为抛物线准线，由抛物线定义可知弦AB 的中点到准线的距离CD 等于()11||||||22BG AH AB +=， 故圆与直线1y =−相切，所以点M 在该圆的圆上或者圆外，故C 错误；由题意(0,1)M −，设211(,)4x A x ，222(,)4x B x ，直线AB 方程为1y mx =+， 则214y mx x y=+⎧⎪⎨=⎪⎩可得2440x mx −−=，所以12124,4x x m x x +==−， 2212121122111144,44MA MB x x x x k k x x x x ++==+==+,1212121212121211044444MA MB x x x xx x x x x x k k x x x x ++++∴+=+++=+=−=，所以直线MA 与直线MB 的斜率互为相反数，直线倾斜角互补，所以∠AMO =∠BMO ， 故D 正确（D 选项也可用平面几何三角形相似得到）， 故选：ABD.12．解析：∵ln ()x f x x =，∴21ln ()x f x x −'=，()f x ∴在(0,e)上单调递增，在(e,)+∞上单调递减， 又∵2211ln ln e ex x x k x ==， ∴当0k >时，要使12x x +越小，则取21e 1x x =→，故有121x x +>，故A 正确； 又21e x x 与均可趋向于+∞，故B 错误；当0k <，21e x x =，且1(0,1)x ∈，1211ln 1x x x x ∴+=+<，故C 正确； 21e e kk x k x ⋅=，令()e ,0k g k k k =<，'()(1)e k g k k =+， ()g k ∴在(,1)−∞−单调递减，在(1,0)−单调递增，1()(1)eg k g ∴≥−=−，故D 正确，故选：ACD.三、填空题（本大题有4小题，单空每空4分，多空每空3分，共20分） 13．π；14．122n +−；15．63；16．132a −±=．16．解析：直线l 的方程可化为()3230a x y x y −−++−=，由23030x y x y +−=⎧⎨−−=⎩，解得直线l 的恒过定点()2,1−，又点C 到直线l 的距离为d ==，因为2211sin 2=222ABC S r BCA r r ∆=∠≤=⇒， 则当ABC ∆的面积最大为2时，ABC ∆为等腰直角三角形， 圆心C到直线l的距离为d =解得 a =四、解答题（本大题有6小题，共70分．解答应写出必要的文字说明，证明过程或演算步骤） 17．解： （1）()()3sin cos cosbA C c aB −=−，)()sin cos sin sincos BA C C AB ∴−=−，()sinsin sin cos A B B C A B =+−sin sin sin cos A B A A B =−， sin 0,cos 1,A B B ≠+=即有1sin(),62B π+=7(,),666B πππ+∈23B π∴=； 5分（2）若选①O 为ABC ∆的重心，111sin 3324OAC BAC S S ac B ∆∆===； 10分若选②O 为ABC ∆的内心，∵2222cos 49b ac ac B =+−=，∴7b =， 设内切圆半径为r ，则有1()24ABC a b c r S ∆++==， 则有2r =，此时124OAC S br ∆==； 10分若选③O 为ABC ∆的外心，∵2222cos 49b a c ac B =+−=，∴7b =，设外接圆半径为R ，则2R sin b B =，解得 R 3=，如图，23AOC π∠=, AB CE FD O ABCO此时,21R sin 2OACSAOC =∠=. 10分18．解： （I=N n *∈且2n ≥），∴n a =∴当2n ≥时，1n n S S −−∴=+，又∵0n a >0，1(2)n =≥，∴数列1==为首项，公差为1的等差数列，1(1)1n n =+−⨯=，所以2n S n =. 4分 ∴当2n ≥时，121n a n n n =+−=−，又∵11a =满足上式，∴数列{}n a 的通项公式为21n a n =−. 6分 另解：当2n ≥时，221(1)21n n n a S S n n n −=−=−−=−， 当1n =时，11a =，满足上式，所以{}n a 的通项公式为21n a n =−. 6分 （II ）当2n ≥时，221111114441n a n n n n ⎛⎫==− ⎪−−−⎝⎭， 故22211111111111111141223144n a a n n n ⎛⎫⎛⎫++=⨯−+−++−=⨯−< ⎪ ⎪−−−⎝⎭⎝⎭， 所以对,2n N n *∈≥，都有222111114n a a ++<−−. 12分 19．解：（I ）方法一：延长,CB DA 交于点F ，连接PF ，在CDF ∆中， ∵BD 是ADC ∠的平分线，且BD BC ⊥， ∴点B 是CF 的中点，又∵E 是PC 的中点，∴BE ∥PF ，又PF ⊂平面PAD ，BE ⊄平面PAD ，∴直线BE ∥平面PAD . 6分方法二：取CD 的中点为G ，连接GE ， ∵E 为PC 的中点，∴GE ∥PD ， 又PD ⊂平面PAD ，GE ⊄平面PAD ，F P AB CD E∴GE ∥平面PAD ，① 又在四边形ABCD 中，2AD =，4BD =，AB =则90,60BAD BDA BDC ∠=∠=∠=,又因为BD BC ⊥，G 为CD 的中点，所以60DBG BDA ∠=∠=,所以AD ∥BG ，可得BG ∥平面PAD ，②由①②得平面BEG ∥平面PAD ，又BE ⊂平面BEG ，BE ⊄平面PAD ，∴直线BE ∥平面PAD .（II ）在ABD ∆中，2AD =，4BD =，AB =则90BAD ∠=，即BA AD ⊥，由已知得60BDC BDA ∠=∠=，8CD =，又平面PAD ⊥平面ABCD ，BA ⊂平面ABCD ，所以BA ⊥平面PAD ，即BA PA ⊥， 所以PAD ∠为二面角P AB D −−的的平面角，所以60PAD ∠=， 又2PA AD ==，所以PAD ∆为正三角形，取AD 的中点为O ，连OP ，则OP AD ⊥，OP 如图建立空间直角坐标系，则()(()1,0,0,1,23,0,,1,0,0,A B C D P −−, 所以()()()1,0,3,2,23,0,DP BD DC ==−−=−,设()()111222,,,,,m x y z n x y z ==分别为平面PBD 和平面PCD 的法向量，则 0m DP m BD ⎧⋅=⎪⎨⋅=⎪⎩，即1111020x x ⎧+=⎪⎨−−=⎪⎩，取11y =−，则()3,1,1m =−−，n DP n DC ⎧⋅=⎪⎨⋅=⎪⎩，即222204430x x ⎧+=⎪⎨−+=⎪⎩，取21y =，则()3,1,1n =−，所以3cos ,5m n m n m n⋅==⋅， 则平面PBD 和平面PCD 所成夹角的余弦值为35. 12分20．解： （I ）由题意得45670.20.30.40.55.5,0.3544x y ++++++====，又4170.560.450.340.8.22i ii x y==⨯+⨯+⨯+⨯=∑，PA B CDE G∴4148.24 5.50.350.5i ii x y x y =−⋅=−⨯⨯=∑∵42222217654126,ii x==+++=∑ ∴4222141264 5.55ii xx ==−−⨯=∑∴41422140.5ˆ0.154i ii ii x y xybxx ==−===−∑∑， 所以0.35ˆˆ0.1 5.50.2a y bx=−=−⨯=−， 故得y 关于x 的线性回归方程为0.10.2y x =−. 5分 （II ）（ⅰ）将8x =代入0.10.20.180.20.6y x =−=⨯−=，估计该省要发放补贴的总金额为0.610000.5300⨯⨯=（万元） 7分（ⅱ）设小浙、小江两人中选择考研的的人数为X ，则X 的所有可能值为0，1，2；2(0)(1)(23)352P X p p p p ==−−=−+，2(1)(1)(31)(23)661P X p p p p p p ==−−+−=−+−， 2(2)(31)3P X p p p p ==−=−，∴()()()222()0352********E X p p p p p p p =⨯−++−+−⨯+−⨯=−，5(0.5)0.5(41)0.758E X p p =⨯−≤⇒≤， 1031113p p ∴≤−≤∴≤≤，，1385p ∴≤≤,故p 的取值范围为15,38⎡⎤⎢⎥⎣⎦. 12分注：p 的取值范围未取等不符不扣分 21．解： （I）因为c e a ==222243c a a b ==+，即223a b =，又点(在双曲线()2222:10,0x y C a b a b−=>>图象上，所以22921a b −=，即229213b b−=，解得221,3b a ==，所以双曲线22:13x C y −=. 4分（II ）由已知点,A B 在以OP 为直径的圆22220000224x y x y x y +⎛⎫⎛⎫−+−= ⎪ ⎪⎝⎭⎝⎭上，又点,A B 在221x y +=上，则有方程组2222000022,2241,x y x y x y x y ⎧+⎛⎫⎛⎫−+−=⎪ ⎪ ⎪⎨⎝⎭⎝⎭⎪+=⎩ 解得直线AB 的方程为001x x y y +=， 设直线AB与渐近线,y y x ==的交点分别为,M N ，由001,,x x y y y +=⎧⎪⎨⎪⎩解得M ，由001,,x x y y y x +=⎧⎪⎨=⎪⎩解得N ，所以2200313MN x y ==−， 又点O 到直线AB的距离为d =，则三角形MON的面积222200001113112233S MN d x y x y =⋅=⨯=−−， 又因为220013x y −=，所以201833S y =+0=，由已知S =，解得203y =，即0y =，因为点P在双曲线右支上，解得0x =，即点(P或(P . 12分22．解： （I ）当22e a =时，()22211ln ln 1e e f x x x x x x x x ⎛⎫=−−=−− ⎪⎝⎭， 要证()0f x ≤，即证21ln 10ex x −−≤，设()21ln 1,0eg x x x x =−−>，令()2110eg x x '=−=，解得2e x =，所以()g x 在()20,e 上递增，在()2e ,+∞上递减， 则()()2222max1e ln e1e 0eg x g ==−−⨯=， 所以()0g x ≤，即21ln 10ex x −−≤成立， 所以()0f x ≤成立. 5分（II ） 因为对任意的0,()x H x >在(0,)+∞上单调递减，所以()0H x '≤恒成立，即e ln 1x x x a x−−≤在(0,)+∞上恒成立，解法一：令e ln 1()(0)x x x F x x x −−=>，则22e ln ()x x xF x x +'=， 令2()e ln x h x x x =+，则()21()2e 0xh x x x x'=++>， 所以()h x 在(0,)+∞上为增函数，又因为11e2e 21e (1)e 0,1e 10e eh h −⎛⎫=>=−=−< ⎪⎝⎭， 所以01,1e x ⎛⎫∃∈ ⎪⎝⎭，使得()00h x =，即0200e ln 0x x x +=， 当00x x <<时，()0h x <，可得()0F x '<，所以()F x 在()00,x 上单调递减； 当0x x >时，()0h x >，可得()0F x '>，所以()F x 在()0,x +∞上单调递增， 所以()000min00e ln 1()x x x F x F x x −−==，由0200e ln 0x x x +=，可得01ln 000000ln 111e ln ln e x x x x x x x x ⎛⎫=−== ⎪⎝⎭，令()e x t x x =，则()001ln t x t x ⎛⎫= ⎪⎝⎭，又由()(1)e 0x t x x '=+>，所以()t x 在(0,)+∞上单调递增， 所以001lnx x =，可得00ln x x =−，所以001e x x =，即00e 1x x =， 所以()0000min000e ln 111()1x x x x F x F x x x −−+−====，即得1a ≤. 12分解法二： 先证e 1x x ≥+（0x ≥），设函数()e 1x h x x =−−，令()e 10xh x '=−=，解得0x =， ∴()h x 在[)0,+∞上单调递增，∴()()00h x h ≥=，即e 1x x ≥+成立. 设()ln k x x x =+（0x >）， ∵()110k x x'=+>，∴()k x 在()0,+∞上单调递增， ∵()1110,110e e k k ⎛⎫=−+<=> ⎪⎝⎭，∴存在()00,x ∈+∞，使得00ln 0x x +=.令e ln 1()(0)x x x F x x x−−=>， 则()ln ln e e ln 1e ln 1ln 1ln 11x x x x x x x x x F x x x x+−−−−++−−==≥=， 当ln 0x x +=时，即0x x =时，取等号. ∴()min 1F x =，即得1a ≤. 12分Z20名校联盟（浙江省名校新高考研究联盟）2023届高三第一次联考数学试卷阅卷细则13-16．（每题5分，共20分）以数值正确为准， 注：第16题给出一个正确数值得3分. 17．（本题满分10分） （Ⅰ）5分1、有正确结论，23B π=，有过程，5分（无过程，3分） 2、无正确结论，找得分点：○1 ()sin sin sin cos A B B C A B =+− ， 2分○21sin()62B π+=，2分 ○323B π=，1分 （Ⅱ）5分1、有正确结论，有过程，5分（无过程，3分）2、无正确结论，找得分点：①1sin 2ABCS ac B ==2分 15334OACABCSS ==，3分 ② ∵2222cos 49b a c ac B =+−=，∴7b =，2分解得内切圆半径2r =，2分124OAC S br ∆==，1分③∵2222cos 49b a c ac B =+−=，∴7b =，2分解得R 3=，，2分解得21sin 2OAC S R AOC ∆=∠，1分 18．（本题满分12分）（Ⅰ）6分1、有正确结论，得21n a n =−，有过程，6分（无过程，2分）2、无正确结论，找得分点：○1n a =2分○22nS n =，2分 ○321na n =−，2分 （Ⅱ）6分1、有正确证明过程，6分（无过程，不得分）2、证明有误，找得分点： ①221111114441n a n n n n ⎛⎫==− ⎪−−−⎝⎭，3分 ②22211111111111111141223144n a a n n n ⎛⎫⎛⎫++=⨯−+−++−=⨯−< ⎪ ⎪−−−⎝⎭⎝⎭，3分 19．（本题满分12分）（Ⅰ）6分1、有证明过程，6分（无过程，不得分）2、证明有误，找得分点：方法一：○1BD BC ⊥，2分 ○2BE ∥PF ，2分 ○3直线BE ∥平面PAD ，2分 方法二：○1取CD 的中点为G ，GE ∥PD ，2分 ○2AD ∥BG ，2分 ○3由平面BEG ∥平面PAD 得直线BE ∥平面PAD ，2分（Ⅱ）6分1、有正确结论35，有过程，6分（无过程，3分） 2、无正确结论，找得分点：①60PAD ∠=，1分②有建系思想，1分○3 求出法向量()3,1,1m =−−，()3,1,1n =−，2分 （法向量计算错误但有法向量计算公式的给1分）④解得余弦值为35，2分（结论错误但有法向量夹角计算公式的给1分） 其他证法酌情给分20．（本题满分12分）（Ⅰ）5分1、有正确结论：0.10.2y x =−，有过程，5分（无过程，2分）2、无正确结论，找得分点：①∵4148.24 5.50.350.5i i i x y x y =−⋅=−⨯⨯=∑， 4222141264 5.55i i xx ==−−⨯=∑，∴41422140.5ˆ0.154i ii i i x y xy b xx ==−===−∑∑，3分 ②0.35ˆˆ0.1 5.50.2a y bx=−=−⨯=−，1分 ③得0.10.2y x =−，1分（Ⅱ）7分（ⅰ）1、有正确结论：300万元，有过程，2分（无过程，1分）2、无正确结论，找得分点：将8x =代入0.10.20.180.20.6y x =−=⨯−=，1分（ⅱ）1、有正确结论：300万元，有过程，5分（无过程，2分）2、无正确结论，找得分点：①2(0)(1)(23)352P X p p p p ==−−=−+，2(1)(1)(31)(23)661P X p p p p p p ==−−+−=−+−，2(2)(31)3P X p p p p ==−=−，()()()222()0352********E X p p p p p p p =⨯−++−+−⨯+−⨯=−，3分 ②解1358p ≤≤，2分（1358p <≤或1358p ≤<或1358p <<均得2分） 21．（本题满分12分）（Ⅰ）4分1、有正确结论：双曲线22:13x C y −=，有过程，4分（无过程，2分） 2、无正确结论，找得分点：①得223a b =， 1分②点(代入()2222:10,0x y C a b a b −=>>，得22921a b−=， 1分 ③解得221,3b a ==，双曲线22:13x C y −=， 2分 （Ⅱ）8分1、有正确结论：点(P或(P ，有过程，8分（无过程，3分，只写出一个坐标的扣1分）2、无正确结论，找得分点：①解得直线AB 的方程为001x x y y +=， 1分②由001,,x x y y y +=⎧⎪⎨=⎪⎩解得M ，1分由001,,3x x y y y x +=⎧⎪⎨=⎪⎩解得N ，1分③2200313MN x y ==−， 点O 到直线AB的距离为d =，三角形MON 的面积222200001113112233S MN d x y x y =⋅=⨯=−−0，3分 ○4点(P或(P ，2分 本小题其他解法酌情给分22．（本题满分12分）（Ⅰ）5分找得分点累加：①要证()0f x ≤，即证21ln 10e x x −−≤，1分 ②设()21ln 1,0e g x x x x =−−>，得()g x 在()20,e 上递增，在()2e ,+∞上递减，2分 ③()()2222max 1e ln e 1e 0e g x g ==−−⨯=，即21ln 10e x x −−≤成立，2分（Ⅱ）7分1、有正确结论：1a ≤，有过程，7分（无过程，2分）2、无正确结论，找得分点：①由()0H x '≤恒成立，得e ln 1x x x a x −−≤，2分 ②令e ln 1()(0)x x x F x x x−−=>，得()F x 在()00,x 上单调递减；在()0,x +∞上单调递增，2分③01,1e x ⎛⎫∃∈ ⎪⎝⎭，使得0200e ln 0x x x +=，1分 ④求得()0000min 000e ln 111()1x x x x F x F x x x −−+−====，即1a ≤，2分 本小题其他解法酌情给分。
2020届浙江省浙南名校联盟高三上学期第一次联考试题英语
绝密★考试结束前2019学年浙南名校联盟高三年级第一学期第一次联考英语学科试题考生须知：1.本卷满分150分，考试时间120分钟；2.答题前，在答题卷指定区域填写班级、姓名、试场号、座位号及准考证号并填涂相应数字；3.所有答案必须写在答题卷上，写在试卷上无效；4.考试结束后，只需上交答题卷。
第I卷(选择题部分)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，并标在试卷的相应位置，听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. What does the man offer to do?A. Go to 7th street.B. Show another shirt.C. Call another branch.2. What is the cause of the woman's quietness?A. The violent film.B. Her tiredness.C. The crowded theater.3. How does the man know about animals?A. From books.B. On TV.C. Through the Internet.4. Where are the speakers?A. At a shop.B. In a restaurant.C. At home.5. What does the woman do?A. A nurse.B. A waitress.C. A saleswoman.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，各小题给出5秒钟的作答时间。
2020届浙江省名校新高考研究联盟(Z20联盟)高三12月第二次联考英语试题答案
•明显遗漏主要内容，写了一些无关内容，原因可能是未理解试题要求； •语法结构单调、词汇项目有限； •较多语法结构或词汇方面的错误，影响对写作内容的理解； •缺乏语句间的连接成分，内容不连贯； 信息未能传达给读者。
0分 未能传达给读者任何信息：内容太少，无法评判；写的内容均与所要求内容无关或所写内容无法看
some relevant information. My primary concern is the detailed schedule. Could you inform me of the arrangement of activities in
advance, so that I can make some preparations. Besides, will we be put up in school dorms or host families? Personally, I prefer the latter, a good opportunity to expose myself to local culture.
浙江省名校新高考研究联盟（Z20 联盟）2020 届第二次联考 英语参考答案 第 3 页 共 6 页
•语法结构单调，词汇项目很有限，有较多语法结构和词汇方面的错误，严重影 响意思了的表达。 •缺乏句间的连接成分，全文内容不连贯。 0 分 白卷，内容太少无法评判或所写内容与所提供内容无关。 三、One possible version
I highly expect the experience to be a perfect mix of learning and fun, which will definitely turn out
rewarding. Looking forward to your reply.
Z20英语答案
Z20名校联盟（浙江省名校新高考研究联盟）2024届高三第二次联考英语参考答案第一部分：听力（每小题1.5分，满分30分）15 BCABA 610 CABBC 1115 AACBA 1620 CCCBB第二部分：阅读理解（每小题2.5分，满分50分）2123 BAA 2427 BCDD 2831 DCBA 3235 BCDC3640 GCEAF第三部分：语言运用（共两节，满分30分）第一节：完形填空（每小题1分，满分15分）4145 DBACA 4650 CDBCA 5155 CDBDA第二节：语法填空（每小题1分，满分15分）56.what 57. traditionally 58. passing 59. selected 60. and 61.is designed 62. circulation 63. have piloted 64. to / among 65. the 第四部分：写作（共两节，满分40分）第一节：应用文写作应用文评分标准一、各档次的给分范围和要求第五档（13~15分）一覆盖了所有内容要点,表述清楚、合理。
一使用了多样并且恰当的词汇和语法结构,可能有个别小错,但完全不影响理解。
一有效地使用了语句间衔接手段,全文结构清晰,意义连贯。
完全达到了预期的写作目的。
第四档（10~12分）一覆盖了所有内容要点,表述比较清楚、合理。
一使用了比较多样并且恰当的词汇和语法结构,可能有些许错误,但不影响理解。
一比较有效地使用了语句间衔接手段,全文结构比较清晰,意义比较连贯。
达到了预期的写作目的。
第三档（7~9分）一覆盖了大部分内容要点,有个别地方表述不够清楚、合理。
一使用了简单的词汇和语法结构,有一些错误或不恰当之处,但基本不影响理解。
一基本有效地使用了语句间衔接手段,全文结构基本清晰,意义基本连贯。
基本达到了预期的写作目的。
第二档（4~6分）一遗漏或未清楚表述一些内容要点,或一些内容与写作目的不相关。