高三考前限时训练

高三英语限时训练一、阅读理解ARainforests are disappearing at a surprisingly fast speed, largely due to human development over the past few centuries. Once covering 14% of land on Earth, rainforests now make up only 6%. Since 1947, the total area of rainforests has probably been reduced by more than half, to about 6.2 to 7.8 million square kilometers.Many biologists expect rainforests will lose 5-10% of their species every ten years. Rapid deforestation could cause many important rainforest habitats to disappear completely in the next hundred years.Such rapid habitat loss is due to the fact that 100 acres of rainforest are cleared every minute for agricultural and industrial development. In the Pacific Northwest’s rainforests, logging companies cut down trees for wood used in building while paper industries use the wood for pulp. In the Amazon rainforest, large agricultural industries clear large parts of forests for growing crops. In the Congo rainforest, roads and other infrastructure (基础建设) development have reduced habitat and cut off living areas for many rainforest species. Throughout both the Amazon and Congo, logging operations clear-cut to build roads or for other purposes. Some rainforests are threatened by large hydroelectric power projects. Development is encroaching on rainforest habitats from all sides.Economic inequalities increase this rapid deforestation. Many rainforests are located in developing countries with economies based on natural resources. Wealthy nations drive need for products, and economic development increases energy use. These needs encourage local governments to develop rainforest at a slow speed. Poor people who live on or near these lands are also motivated to improve their lives by changing forests into farmland.1. What feeling does the author want to convey by listing figures?A．Frustrated. B．Disgusted. C．Inspired. D．Concerned.2. What caused the reduction of habitat for many rainforest species in Congo? A．The demand for numerous products.B．The motivation from local governments.C．The construction of fundamental facilities.D．The cooperation with developing countries.3. What does the underlined part “encroaching on” in paragraph 3 mean?A．Tearing up. B．Relying on. C．Holding back. D．Benefiting from.4. Which of the following can be the best title for the text?A．Changes of Rainforests B．Threats to Rainforests C．Comments on Rainforests D．Projects in RainforestsBAlexis, 17, sat quietly in the passenger seat of her dad’s car, looking out of the window. Then a deer came into view about 200 yards in front of them. “Dad, there’s a deer there!” Alexis said.As the car moved closer, Alexis heard a scream and saw an arm fly up near the deer’s head. Alexis reali zed the deer was attacking a woman. Sue, a 44-year-old mother, had been out for her morning run. The deer followed her and edged closer. “I knew I was in trouble,” Sue says. She went to pick up a stick for self-defense, and the deer charged. It lifted her with its antlers (鹿角) and threw her into the air. Sue could feel blood flowing down her leg.When Alexis and her father pulled up, the deer was throwing Sue like a doll. Alexis looked into the woman’s terrified eyes, and before her father had even stopped the car, the teenager jumped quickly out of the car and ran toward the deer. “I was kicking it to get its attention,” she says. Then her father, who had followed his daughter, tried to push the deer away from the women.Alexis helped Sue into the car, and then applied a piece of cloth to Sue’s injured leg. “We’re going to get you to a hospital,” Alexis said. Then she heard her father shout loudly. He had been knocked to the ground. Alexis took hold of a hammer from the car and ran to where her father lay on his back. She beat the deer’s head and neck, but the blows didn’t scare it away. “I was losing faith,” she says. “A couple more strikes, Alexis,” said her father. “You can do it.” Turning the hammer around, Alexis closed her eyes and beat the deer’s neck with all her strength. When she opened her eyes, the deer was running away. Alexis got in the driver’s seat and sped toward the nearest hospital.5. What was Sue doing when she was attacked by the deer?A．She was going to school. B．She was taking exercise.C．She was resting by the road. D．She was watering plants.6. What does the underlined word “edged” in paragraph 2 mean?A．Changed. B．Cut. C．Moved. D．Attacked.7. What did Alexis do to save her father and Sue?A．She pushed the deer away. B．She hit the deer with her feet. C．She drove the car to hit the deer. D．She beat the deer with a hammer.8. Which of the following words can best describe Alexis?A．Brave. B．Cruel. C．Energetic. D．Strong.CThis new packaging material looks like conventional plastic, but it’s actually made from fish waste and algae(海藻)—unlike plastic, if it escapes into the ocean, a fish could eat it safely. The material, called MarinaTex, is the winner of this year’s James Dyson Award.“It began with my desire to work with waste,” says Lu cy Hughes, a recent graduate from University of Sussex, who once visited a fish processing factory to see the massive quantity of waste generated by the industry and found new ways to use it. She focused on fish skins and scales(鳞片). “When I had it in my h ands, I realizedthis has got potential,” she says. “It s super strong and flexible.” Hughes spent months experimenting with fish waste in her kitchen, running more than 100 experiments to find a binder(粘合剂) and a process that could hold together theprote ins in the fish skins and scales.“I had a lot of failed attempts. A lot of things either were too easily broken or too soft or easily went bad,” she says. She finally found a type of algae that can be locally got.The material can be used to replace oil-based plastic in packaging such as bags or the half-transparent windows used to show products in boxes. As her research continues, Hughes says that she’ll be studying how MarinaTex works and how long it could be used to store food.Unlike some “composable” p lastics, which need to be processed in an industrial composting(堆肥) facility to break down, the new material can break down in four to six weeks in a backyard compost bin. “All the ingredients are food-grade safe,” Hughes says. “So, this breaks down much l ike a piece of food would break down.；9. Why did MarinaTex win this year s James Dyson Award?A．It doesn’t go into the ocean.B．It is environmentally-friendly.C．It doesn’t produce any waste.D．It doesn’t look like conventional plastic.；10. What is the main idea conveyed in paragraph 2?A．Where Hughes got inspiration for the binder.B．How Hughes developed MarinaTex.C．What advantages MarinaTex has.D．Where the binder can be used.；11. What is yet to be found out?A．How long it takes for MarinaTex to break down.B．How long MarinaTex can be used to store food.C．Whether MarinaTex can replace oil-based plastic.D．Whether MarinaTex is better than half-transparent windows.；12. Where is the text most likely to appear?A．In a science fiction. B．In a government report.C．In a science magazine. D．In a government announcement.二、完形填空As a community worker, I came into serious conflict with my community last week, and things completely fell apart between me and my community. As a result, I 13 to step away from them, blocked them on social media and did my best to 14 about them. However, the issues continued to 15 me, and even had a bad 16 on other areas of my life.Yesterday, I was 17 to see this community and held a 18 because of a major work event. I was still in fear of the issues and felt 19 when I first saw the people I 20 with. Despite all of this, the strangest but most 21 thinghappened to me. I felt something that I hadn’t felt in a long time, peace and forgiveness.At the meeting, there were no 22 or conflicts with anyone. It was 23 that a lot of people had 24 from the issues. I realized that there wasn’t anythingto worry about anymore. No one 25 bad will on me, and no one was holding anger against me anymore. There was peace, and room for people to forgive and move on.When you 26 others, you not only free them and find peace with them, but yourself as well. Now, I only wish them the best life that they can have, and the most 27 they could ever have. And then, I move on.13. A．pretended B．afforded C．decided D．refused14. A．forget B．argue C．inquire D．worry15. A．cheat B．bother C．alarm D．attack16. A．habit B．plan C．influence D．relationship17. A．designed B．supposed C．allowed D．forced18. A．record B．position C．meeting D．party19. A．ashamed B．disappointed C．excited D．embarrassed20. A．competed B．conflicted C．connected D．communicated21. A．wonderful B．sad C．awful D．careful22. A．movements B．achievements C．advertisements D．arguments23. A．impossible B．difficult C．obvious D．strange24. A．based on B．called on C．carried on D．moved on25. A．registered B．wished C．revised D．decorated26. A．represent B．forgive C．respect D．monitor27. A．happiness B．career C．resource D．pattern三、七选五阅读Camping is a fun and exciting experience, but to stay organized and safe, youmust carefully prepare for the trip.28 . If it is just you or your family, things will become easier. However, if you are taking a group of friends, you’d better make sure you get insurance information and health information before you start. If someone is injured on the trip, the insurance information will make a huge difference in the care they receive.29 . For example, for those who are suffering basic diseases, make sure they have enough medicines available.Then decide how long you will stay and figure out where you will be campin g. If you do not know this, you will go through the trouble of purchasing and packing atent when you were supposed to sleep in a cabin.30 . The food should be enough for three meals and an optional snack per day, as well as lots of water. Basically, ensure food safety. Foods like dairy products and meat go bad easily. 31 .At last, gather all of the other things on the “Things You’ll Need" list and try to pack them in a small, lightweight bag. 32 . Materials to deal with burns, cuts,and bumps should be prepared in it.A．Health information is also importantB．Stay calm when an accident happensC．Make sure they are fresh and safe to eatD．And never forget to pack up a first aid kitE．Firstly, decide with whom you are campingF．Taking exercise every day is a good way to keep healthyG．Of course, you’d better pack an appropriate amount of food四、语法填空阅读下面短文，在空白处填入1个适当的单词或括号内单词的正确形式。

高考二轮复习限时训练（一）（时间：60分钟）班级 姓名 得分一、填空题：本大题共12小题，每小题5分，共60分。

1、复数ii4321+-在复平面上对应的点位于第__ 象限. 2、命题“2,220x R x x ∃∈++≤”的否定是3、设{}{}=⋂+==∈==B A x y y x B R x x y y A 则,2|),(,,|2 4、已知x 、y 的取值如下表：从散点图分析，y 与x 线性相关，且回归方程为 0.95y x a =+，则a = 5、若椭圆的焦距长等于它的短轴长，则椭圆的离心率等于____ ____。

6、如果执行右面的程序框图，那么输出的S = 7、把函数4cos()3y x π=++1的图象向左平移ϕ个单位，所得的图象对应的函数为偶函数，则ϕ的最小正值为8、如果实数x ,y 满足x 2+y 2=1,则（1+xy ）(1－xy )的最小值为9、已知实数x ，y 满足条件5003x y x y x -+⎧⎪+⎨⎪⎩≥≥≤，i z x y =+（i 为虚数单位），则|12i |z -+的最小值是 ．10、一枚半径为1的硬币随机落在边长为3的正方形所在平面内，且硬币一定落在正方形内部或与正方形有公共点，则硬币与正方形没有公共点的概率是11、若函数f (x )=e x -2x-a 在R 上有两个零点，则实数a 的取值范围是12、设函数)0](,[,321)1ln()(2>-∈+-+=t t t x x e x x f x ，若函数的最大值是M ，最小值是m ，则M+m=二、解答题：本大题共2小题，共30分。

13、（本小题满分15分）在ABC ∆中，角A 、B 、C 的对边分别为,,a b c ，已知向量33(cos ,sin ),22A A m = (cos ,sin ),22A An =且满足m n +=（Ⅰ）求角A 的大小；（Ⅱ）若,b c +=试判断ABC ∆的形状。

14、（本小题满分15分）已知圆C ：224x y +=.（1）直线l 过点()1,2P ，且与圆C 交于A 、B 两点，若||AB =l 的方程;（2）过圆C 上一动点M 作平行于x 轴的直线m ，设m 与y 轴的交点为N ，若向量OQ OM ON =+，求动点Q 的轨迹方程，并说明此轨迹是什么曲线.高考二轮复习限时训练（一）1、三2、2,220.x R x x ∀∈++>3、∅4、2.6 5 6、25507、32π 8、43 9 10、 π+21111、()+∞-,2ln 2212、6 二、解答题： 13、（1）3A π=———————————————————7分（2）ABC ∆为直角三角形。

高三英语限时训练一总分65分第一部分：阅读理解（共两节，满分30分）第一节（共7个小题：每小题2.5分，满分17.5分）AWhat do the random, scribbled(潦草的)drawings crowding the margins(页边空白)of most high school students’ papers mean? When a student is caught doodling(乱画)in class, he will probably be criticized for daydreaming. But doodling while listening can help with remembering details, rather than implying that the mind is wandering, according to a study published in the scientific journal Applied Cognitive Psychology.In an experiment conducted b y the Medical Research Council’s Cognition and Brain Sciences Unit in Cambridge,40 subjects were asked to listen to a two-minute tape giving several names of people and places. Half of the participants were asked to shade in shapes on a piece of paper at the same time, without paying attention to neatness, while the rest were given no such instructions. After the tape had finished, all participants in the study were asked to recall the names of people and places. The doodlers recalled on average 7.5 names of people and places, compared to only 5.8 by the non-doodlers.“If someone is doing a boring task, like listening to a dull telephone conversation, they may start to daydream.” said study researcher, Professor Jackie Andrade, of the School of Psychology, U niversity of Plymouth. “Daydreaming distracts them from the task, resulting in poorer performance. A simple task, like doodling, may be enough tostop daydreaming without affecting performance on the main task.”“In psychology, tests of memory or attention will often use a second task to selectively block a particular mental process. If that process is important for the main task, then performance will be weakened. But my research suggests that in everyday life doodling may be something we do because it helps to keep us on track with a boring task, rather than being an unnecessary distraction(分心)that we should try to resist doing.” said Andrade.Dan Ware, a social study teacher, used to consider doodling a distraction from learning, but after teaching kids with all personality types he learned scribbling away during lectures helps certain students remember more information. “In my first few years of teaching, I thought, ‘Well, this kid isn’t paying attention. He’s daydreaming.’ But I had some really powerful experiences with students and came to understand in many cases that was their way of focusing, and those students were probably paying more attention than other students.” Ware said.1. What do we know about the participants involved in the experiment?A. Some were asked to note down the information neatly.B. Some were asked to memorize the names they would hear.C. Some were instructed to listen to the tape with full attention.D. Some were instructed to make random drawings on paper.2. Which of the following will both Jackie Andrade and Dan Ware agree with?A. Doodling helps some people focus.B. Doodling makes a dull task interesting.C. Students who doodle perform poorly.D. Students who doodle lack concentration.3. What is the best title of the text?A. Daydreaming Can Sharpen Study SkillsB. Doodling Can Help Memory RecallC. A Wandering Mind Improves ProductivityD. Distractions Harm Academic PerformanceBShyness is the cause of much unhappiness for a great many people. Shy people are anxious and self-conscious; that is, they are concerned about their own appearance and actions too much. Negative thoughts are constantly occurring in their minds: What kind of impression am I making? Do they like me? Do I sound stupid? Am I wearing unattractive clothes?It is obvious that such uncomfortable feelings must affect people unfavorably. A person’s self-concept is reflected in the way he or she behaves and the way a person behaves affects other people’s reactions. In general, the way people think about themselves has a deep effect on all areas of their lives.Shy people, who have low respect, are likely to be passive and easily influenced by others. They need faith that they are doing ＂the right thing＂. Shy people are very sensitive to criticism. It makes them feel inferior(自卑). They also find it difficult to be pleased by praises because they believe they are unworthy of praise. A shy person may respond to a praise with a statement like this one: ＂You’re just saying that to make me feel good. I kn ow it’s not true.＂It is clear that, while self-awareness is a healthy quality, overdoing it is harmful.Can shyness be completely got rid of, or at least reduced? Fortunately, people can overcome shyness with determination since shyness goes hand in hand with lack ofself-respect. It is important for people to accept their weaknesses as well as their strengths. Each one of us has his or her own characteristics. We are interested in our own personal ways. The better we understand ourselves, the easier it becomes to live up to our chances for a rich and successful life.4. The first paragraph is mainly about ____________. .A. the cause of shynessB. the effect of shyness on peopleC. the feelings of shy peopleD. the questions in the minds of shy people5. According to the writer, self-awareness is ____________.A. harmful to peopleB. a weak point of peopleC. the cause of unhappinessD. a good characteristic6. What is the shy people’s reaction to praise?A. They are pleased by it. B They feel it is not true.C. They are very sensitive to it.D. They feel they are worthy of it.7. We can learn from the passage that shyness ____________.A. blocks our chances for a successful lifeB. helps us to live up to our full developmentC. enables us to understand ourselves betterD. has nothing to do with lack of self-respect第二节七选五（共5小题，每小题2.5分，满分12.5分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项，选项中有两项为多余选项。

2024届高三二轮复习“8+3+3”小题强化训练(1)一、选择题：本题共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．1对两个具有线性相关关系的变量x 和y 进行统计时，得到一组数据1,0.3 ,2,4.7 ,3,m ,4,8 ，通过这组数据求得回归直线方程为y=2.4x -2，则m 的值为()A.3B.5C.5.2D.6【答案】A【解析】易知x =1+2+3+44=52,y =13+m4，代入y =2.4x -2得13+m 4=2.4×52-2⇒m =3.故选：A2已知m ，n 表示两条不同直线，α表示平面，下列说法正确的是()A.若m ⎳α,n ⎳α,则m ⎳nB.若m ⊥α，n ⊂α，则m ⊥nC.若m ⊥α，m ⊥n ，则n ⎳αD.若m ⎳α，m ⊥n ，则n ⊥α【答案】B【解析】线面垂直，则有该直线和平面内所有的直线都垂直，故B 正确.故选：B3已知向量a ，b 满足a =3，b =23，且a ⊥a +b，则b 在a 方向上的投影向量为()A.3B.-3C.-3aD.-a【答案】D【解析】a ⊥a +b ，则a ⋅a +b =a 2+a ⋅b =9+a ⋅b =0，故a ⋅b=-9，b 在a 方向上的投影向量a ⋅b a 2⋅a =-99⋅a =-a.故选：D .4若n 为一组从小到大排列的数1，2，4，8，9，10的第六十百分位数，则二项式3x +12xn的展开式的常数项是()A.7B.8C.9D.10【答案】A【解析】因为n 为一组从小到大排列的数1，2，4，8，9，10的第六十百分位数，6×60%=3.6，所以n =8，二项式3x +12x8的通项公式为T r +1=C r 8⋅3x 8-r ⋅12x r =C r 8⋅12 r⋅x8-r 3-r，令8-r 3-r =0⇒r =2，所以常数项为C 28×12 2=8×72×14=7，故选：A5折扇是我国古老文化的延续，在我国已有四千年左右的历史，“扇”与“善”谐音，折扇也寓意“善良”“善行”.它常以字画的形式体现我国的传统文化，也是运筹帷幄、决胜千里、大智大勇的象征(如图1).图2是一个圆台的侧面展开图(扇形的一部分)，若两个圆弧DE ，AC 所在圆的半径分别是3和6，且∠ABC =120°，则该圆台的体积为()A.5023π B.9π C.7π D.1423π【答案】D【解析】设圆台上下底面的半径分别为r 1,r 2，由题意可知13×2π×3=2πr 1，解得r 1=1，13×2π×6=2πr 2，解得：r 2=2，作出圆台的轴截面，如图所示：图中OD =r 1=1,O A =r 2=2，AD =6-3=3，过点D 向AP 作垂线，垂足为T ，则AT =r 2-r 1=1，所以圆台的高h =AD 2-AT 2=32-1=22，则上底面面积S 1=π×12=π，S 2=π×22=4π，由圆台的体积计算公式可得：V =13×(S 1+S 2+S 1⋅S 2)×h =13×7π×22=142π3，故选：D .6已知函数f x =x 2-bx +c (b >0,c >0)的两个零点分别为x 1,x 2，若x 1,x 2,-1三个数适当调整顺序后可为等差数列，也可为等比数列，则不等式x -bx -c≤0的解集为()A.1,52B.1,52C.-∞,1 ∪52,+∞D.-∞,1 ∪52,+∞ 【答案】A【解析】由函数f x =x 2-bx +c (b >0,c >0)的两个零点分别为x 1,x 2，即x 1,x 2是x 2-bx +c =0的两个实数根据，则x 1+x 2=b ,x 1x 2=c 因为b >0,c >0，可得x 1>0,x 2>0，又因为x 1,x 2,-1适当调整可以是等差数列和等比数列，不妨设x 1<x 2，可得x 1x 2=-1 2=1-1+x 2=2x 1 ，解得x 1=12,x 2=2，所以x 1+x 2=52,x 1x 2=1，所以b =52,c =1，则不等式x -b x -c ≤0，即为x -52x -1≤0，解得1<x ≤52，所以不等式的解集为1,52.故选：A .7已知双曲线C ：x 2a 2-y 2b2=1a >0,b >0 的左、右焦点分别为F 1，F 2，M ，N 为双曲线一条渐近线上的两点，A 为双曲线的右顶点，若四边形MF 1NF 2为矩形，且∠MAN =2π3，则双曲线C 的离心率为()A.3B.7C.213D.13【答案】C【解析】如图，因为四边形MF 1NF 2为矩形，所以MN =F 1F 2 =2c (矩形的对角线相等)，所以以MN 为直径的圆的方程为x 2+y 2=c 2.直线MN 为双曲线的一条渐近线，不妨设其方程为y =bax ，由y =b a x ,x 2+y 2=c 2,解得x =a y =b ，或x =-a ,y =-b , 所以N a ,b ，M -a ,-b 或N -a ,-b ，M a ,b .不妨设N a ,b ，M -a , -b ，又A a ,0 ，所以AM =a +a 2+b 2=4a 2+b 2，AN =a -a 2+b 2=b .在△AMN 中，∠MAN =2π3，由余弦定理得MN 2=AM 2+AN 2-2AM AN ⋅cos 2π3，即4c 2=4a 2+b 2+b 2+4a 2+b 2×b ，则2b =4a 2+b 2，所以4b 2=4a 2+b 2，则b 2=43a 2，所以e =1+b 2a2=213.故选:C .8已知a =ln 1.2e ,b =e 0.2,c =1.2e 0.2，则有()A.a <b <cB.a <c <bC.c <a <bD.c <b <a【答案】C【解析】令f x =e x -ln x +1 -1,x >0，则f x =e x -1x +1.当x >0时，有e x >1,1x +1<1，所以1x +1<1，所以，f (x )>0在0,+∞ 上恒成立，所以，f (x )在0,+∞ 上单调递增，所以，f (x )>f (0)=1-1=0，所以，f (0.2)>0，即e 0.2-ln1.2-1>0，所以a <b令g x =e x -x +1 ,x >0，则g x =e x -1在x >0时恒大于零，故g x 为增函数，所以x +1ex <1,x >0，而a =ln 1.2e =1+ln1.2>1，所以c <a ，所以c <a <b ，故选：C二、选择题：本题共3小题，每小题6分，共18分.在每小题给出的选项中，有多项符合题目要求.全部选对的得6分，部分选对的得部分分，有选错的得0分.9已知函数f x =sin 2x +3π4 +cos 2x +3π4，则()A.函数f x -π4 为偶函数 B.曲线y =f x 对称轴为x =k π,k ∈ZC.f x 在区间π3,π2单调递增D.f x 的最小值为-2【答案】AC【解析】f x =sin 2x +3π4 +cos 2x +3π4=sin2x cos 3π4+sin 3π4cos2x +cos2x cos 3π4-sin2x sin3π4=-22sin2x +22cos2x -22cos2x -22sin2x =-2sin2x ，即f x =-2sin2x ，对于A ，f x -π4 =-2sin 2x -π2=2cos2x ，易知为偶函数，所以A 正确；对于B ，f x =-2sin2x 对称轴为2x =π2+k π,k ∈Z ⇒x =π4+k π2,k ∈Z ，故B 错误；对于C ，x ∈π3,π2 ,2x ∈2π3,π ，y =sin2x 单调递减，则f x =-2sin2x 单调递增，故C 正确；对于D ，f x =-2sin2x ，则sin2x ∈-1,1 ，所以f x ∈-2,2 ，故D 错误；故选：AC10设z 为复数，则下列命题中正确的是()A.z 2=zz B.若z =(1-2i )2，则复平面内z对应的点位于第二象限C.z 2=z 2D.若z =1，则z +i 的最大值为2【答案】ABD【解析】对于A ，设z =a +bi ，故z =a -bi ，则z 2=a 2+b 2，zz =(a +bi )(a -bi )=a 2+b 2，故z 2=zz成立，故A 正确，对于B ，z =(1-2i )2=-4i -3，z =4i -3，显然复平面内z对应的点位于第二象限，故B 正确，对于C ，易知z 2=a 2+b 2，z 2=a 2+b 2+2abi ，当ab ≠0时，z 2≠z 2，故C 错误，对于D ，若z =1，则a 2+b 2=1，而z +i =a 2+(b +1)2=2b +2，易得当b =1时，z +i 最大，此时z +i =2，故D 正确.故选：ABD11已知菱形ABCD 的边长为2，∠ABC =π3．将△DAC 沿着对角线AC 折起至△D AC ，连结BD ．设二面角D -AC -B 的大小为θ，则下列说法正确的是()A.若四面体D ABC 为正四面体，则θ=π3B.四面体D ABC 的体积最大值为1C.四面体D ABC 的表面积最大值为23+2D.当θ=2π3时，四面体D ABC 的外接球的半径为213【答案】BCD【解析】如图，取AC 中点O ，连接OB ,OD ，则OB =OD ，OB ⊥AC ,OD ⊥AC ，∠BOC 为二面角D AC -B 的平面角，即∠BOC =θ．若D ABC 是正四面体，则BD =BC ≠BO ，△OBD 不是正三角形，θ≠π3，A 错；四面体D ABC 的体积最大时，BO ⊥平面ACD ，此时B 到平面ACD 的距离最大为BO =3，而S △ACD=34×22=3，所以V =13×3×3=1，B 正确；S △ABC =S △DAC =3，易得△BAD ≅△BCD ，S △BAD=S △BCD=12×22sin ∠BCD =2sin ∠BCD ，未折叠时BD =BD =23，折叠到B ,D 重合时，BD =0，中间存在一个位置，使得BD =22，则BC 2+D C 2=BD 2，∠BCD =π2，此时S △BAD=S △BCD=2sin ∠BCD 取得最大值2，所以四面体D ABC 的表面积最大值为23+2 ，C 正确；当θ=2π3时，如图，设M ,N 分别是△ACD 和△BAC 的外心，在平面AOD 内作PM ⊥OD ，作PN ⊥OB ，PM ∩PN =P ，则P 是三棱锥外接球的球心，由上面证明过程知平面OBD 与平面ABC 、平面D AC 垂直，即P ,N ,O ,M 四点共面，θ=2π3，则∠PON =π3，ON =13×32×2=33，PN =ON tan π3=33×3=1，PB =PN 2+BN 2=12+233 2=213为球半径，D 正确．故选：BCD ．三、填空题：本题共3小题，每小题5分，共15分.12设集合M =x log 2x <1 ，N =x 2x -1<0 ，则M ∩N =．【答案】x 0<x <12【解析】因为log 2x <1=log 22，所以0<x <2，即M =x log 2x <1 =x 0<x <2 ，因为2x -1<0，解得x <12，所以N =x 2x -1<0 =x x <12，所以，M ∩N =x 0<x <12 .故答案为：x 0<x <12 13已知正项等比数列a n 的前n 项和为S n ，且S 8-2S 4=6，则a 9+a 10+a 11+a 12的最小值为.【答案】24【解析】设正项等比数列a n 的公比为q ，则q >0，所以，S 8=a 1+a 2+a 3+a 4+a 5+a 6+a 7+a 8=a 1+a 2+a 3+a 4+q 4a 1+a 2+a 3+a 4 =S 41+q 4 ，则S 8-2S 4=S 4q 4-1 =6，则q 4>1，可得q >1，则S 4=6q 4-1，所以，a 9+a 10+a 11+a 12=q 8a 1+a 2+a 3+a 4 =S 4q 8=6q 8q 4-1=6q 4-1+1 2q 4-1=6q 4-1 2+1+2q 4-1 q 4+1=6q 4-1 +1q 4-1+2 ≥62q 4-1 ⋅1q 4-1+2 =24，当且仅当q 4-1=1q 4-1q >1 时，即当q =42时，等号成立，故a 9+a 10+a 11+a 12的最小值为24.故答案为：2414已知F 为拋物线C :y =14x 2的焦点，过点F 的直线l 与拋物线C 交于不同的两点A ，B ，拋物线在点A ,B 处的切线分别为l 1和l 2，若l 1和l 2交于点P ，则|PF |2+25AB的最小值为.【答案】10【解析】C :x 2=4y 的焦点为0,1 ，设直线AB 方程为y =kx +1，A x 1,y 1 ,B x 2,y 2 .联立直线与抛物线方程有x 2-4kx -4=0，则AB =y 1+y 2+2=k x 1+x 2 +4=4k 2+4.又y =14x 2求导可得y =12x ，故直线AP 方程为y -y 1=12x 1x -x 1 .又y 1=14x 21，故AP :y =12x 1x -14x 21，同理BP :y =12x 2x -14x 22.联立y =12x 1x -14x 21y =12x 2x -14x 22可得12x 1-x 2 x =14x 21-x 22 ，解得x =x 1+x 22，代入可得P x 1+x 22,x 1x 24 ，代入韦达定理可得P 2k ,-1 ，故PF =4k 2+4.故|PF |2+25AB=4k 2+4+254k 2+4≥24k 2+4 ×254k 2+4=10，当且仅当4k 2+4=254k 2+4，即k =±12时取等号.故答案为：102024届高三二轮复习“8+3+3”小题强化训练(2)一、选择题：本题共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．1抛物线y =12x 2的焦点坐标为()A.18,0B.12,0 C.0,18D.0,12【答案】D 【解析】由y =12x 2可得抛物线标准方程为：x 2=2y ，∴其焦点坐标为0,12 .故选：D .2二项式3x 2-1x 47的展开式中常数项为()A.-7B.-21C.7D.21【答案】A 【解析】二项式3x 2-1x47的通项公式为Tr +1=C r 7⋅3x 27-r⋅-1x4r=Cr 7⋅-1 r⋅x14-14r 3，令14-14r 3=0⇒r =1，所以常数项为C 17⋅-1 =-7，故选：A3已知集合A =x log 2x ≤1 ，B =y y =2x ,x ≤2 ，则()A.A ∪B =BB.A ∪B =AC.A ∩B =BD.A ∪(C R B )=R【答案】A【解析】由log 2x ≤1，则log 2x ≤log 22，所以0<x ≤2，所以A =x log 2x ≤1 =x 0<x ≤2 ，又B =y y =2x ,x ≤2 =y 0<y ≤4 ，所以A ⊆B ，则A ∪B =B ，A ∩B =A .故选：A ．4若古典概型的样本空间Ω=1,2,3,4 ，事件A =1,2 ，甲：事件B =Ω，乙：事件A ,B 相互独立，则甲是乙的()A.充分不必要条件B.必要不充分条件C.充分必要条件D.既不充分也不必要条件【答案】A【解析】若B =Ω，A ∩B =1,2 ，则P A ∩B =24=12，而P A =24=12，P B =1，所以P A P B =P A ∩B ，所以事件A ,B 相互独立，反过来，当B =1,3 ，A ∩B =1 ，此时P A ∩B =14，P A =P B =12，满足P A P B =P A ∩B ，事件A ,B 相互独立，所以不一定B =Ω，所以甲是乙的充分不必要条件.故选：A5若函数f x =ln e x -1 -mx 为偶函数，则实数m =()A.1B.-1C.12D.-12【答案】C【解析】由函数f x =ln e x -1 -mx 为偶函数，可得f -1 =f 1 ，即ln e -1-1 +m =ln e -1 -m ，解之得m =12，则f x =ln e x -1 -12x (x ≠0)，f -x =ln e -x -1 +12x =ln e x -1 -x +12x =ln e x -1 -12x =f x故f x =ln e x -1 -12x 为偶函数，符合题意.故选：C6已知函数y =f (x )的图象恰为椭圆C :x 2a 2+y 2b2=1(a >b >0)x 轴上方的部分，若f (s -t )，f (s )，f (s +t )成等比数列，则平面上点(s ，t )的轨迹是()A.线段(不包含端点) B.椭圆一部分C.双曲线一部分D.线段(不包含端点)和双曲线一部分【答案】A【解析】因为函数y =f (x )的图象恰为椭圆C :x 2a 2+y 2b2=1(a >b >0)x 轴上方的部分，所以y =f (x )=b ⋅1-x 2a2(-a <x <a )，因为f (s -t )，f (s )，f (s +t )成等比数列，所以有f 2(s )=f (s -t )⋅f (s +t )，且有-a <s <a ,-a <s -t <a ,-a <s +t <a 成立，即-a <s <a ,-a <t <a 成立，由f 2(s )=f (s -t )⋅f (s +t )⇒b ⋅1-s 2a 22=b ⋅1-(s -t )2a 2⋅b ⋅1-(s +t )2a 2，化简得：t 4=2a 2t 2+2s 2t 2⇒t 2(t 2-2a 2-2s 2)=0⇒t 2=0，或t 2-2a 2-2s 2=0，当t 2=0时，即t =0，因为-a <s <a ，所以平面上点(s ，t )的轨迹是线段(不包含端点)；当t 2-2a 2-2s 2=0时，即t 2=2a 2+2s 2，因为-a <t <a ，所以t 2<a 2，而2a 2+2s 2>a 2，所以t 2=2a 2+2s 2不成立，故选：A7若tan α+π4=-2，则sin α1-sin2α cos α-sin α=()A.65B.35C.-35D.-65【答案】C【解析】因为tan α+π4 =tan α+tan π41-tan αtan π4=tan α+11-tan α=-2，解得tan α=3，所以，sin α1-sin2αcos α-sin α=sin αsin 2α+cos 2α-2sin αcos α cos α-sin α=sin αcos α-sin α 2cos α-sin α=sin αcos α-sin 2α=sin αcos α-sin 2αcos 2α+sin 2α=tan α-tan 2α1+tan 2α=3-91+9=-35.故选：C .8函数f x =2ln xx,x >0sin ωx +π6,-π≤x ≤0，若2f 2(x )-3f (x )+1=0恰有6个不同实数解，正实数ω的范围为()A.103,4B.103,4 C.2,103D.2,103【答案】D【解析】由题知，2f 2x -3f x +1=0的实数解可转化为f (x )=12或f (x )=1的实数解，即y =f (x )与y =1或y =12的交点，当x >0时，f x =2ln xx ⇒f (x )=21-ln x x 2所以x ∈0,e 时，f (x )>0，f x 单调递增，x ∈e ,+∞ 时，f (x )<0，f x 单调递减，如图所示：所以x =e 时f x 有最大值：12<f (x )max =2e<1所以x >0时，由图可知y =f (x )与y =1无交点，即方程f (x )=1无解，y =f (x )与y =12有两个不同交点，即方程f (x )=12有2解当x <0时，因为ω>0，-π≤x ≤0，所以-ωπ+π6≤ωx +π6≤π6，令t =ωx +π6，则t ∈-ωπ+π6,π6则有y =sin t 且t ∈-ωπ+π6,π6，如图所示：因为x >0时，已有两个交点，所以只需保证y =sin t 与y =12及与y =1有四个交点即可，所以只需-19π6<-ωπ+π6≤-11π6，解得2≤ω<103．故选：D二、选择题：本题共3小题，每小题6分，共18分.在每小题给出的选项中，有多项符合题目要求.全部选对的得6分，部分选对的得部分分，有选错的得0分.9已知复数z 1,z 2是关于x 的方程x 2+bx +1=0(-2<b <2,b ∈R )的两根，则下列说法中正确的是()A.z 1=z 2B.z 1z 2∈R C.z 1 =z 2 =1D.若b =1，则z 31=z 32=1【答案】ACD【解析】Δ=b 2-4<0，∴x =-b ±4-b 2i 2，不妨设z 1=-b 2+4-b 22i ，z 2=-b2-4-b 22i ，z 1=z 2，A 正确；z 1 =z 2 =-b 22+4-b 222=1，C 正确；z 1z 2=1，∴z 1z 2=z 21z 1z 2=z 21=b 2-22-b 4-b 22i ，b ≠0时，z 1z 2∉R ，B 错；b =1时，z 1=-12+32i ，z 2=-12-32i ，计算得z 21=-12-32i =z 2=z 1 ，z 22=z 1=z 2 ，z 31=z 1z 2=1，同理z 32=1，D 正确．故选：ACD ．10四棱锥P -ABCD 的底面为正方形，P A 与底面垂直，P A =2，AB =1，动点M 在线段PC 上，则()A.不存在点M ，使得AC ⊥BMB.MB +MD 的最小值为303C.四棱锥P -ABCD 的外接球表面积为5πD.点M 到直线AB 的距离的最小值为255【答案】BD【解析】对于A ：连接BD ，且AC ∩BD =O ，如图所示，当M 在PC 中点时，因为点O 为AC 的中点，所以OM ⎳P A ，因为P A ⊥平面ABCD ，所以OM ⊥平面ABCD ，又因为AC ⊂平面ABCD ，所以OM ⊥AC ，因为ABCD 为正方形，所以AC ⊥BD .又因为BD ∩OM =O ，且BD ，OM ⊂平面BDM ，所以AC ⊥平面BDM ，因为BM ⊂平面BDM ，所以AC ⊥BM ，所以A 错误；对于B ：将△PBC 和△PCD 所在的平面沿着PC 展开在一个平面上，如图所示，则MB +MD 的最小值为BD ，直角△PBC 斜边PC 上高为1×56，即306，直角△PCD 斜边PC 上高也为1×56，所以MB +MD 的最小值为303，所以B 正确；对于C ：易知四棱锥P -ABCD 的外接球直径为PC ，半径R =12PC =1222+12+12=62，表面积S =4πR 2=6π，所以C 错误；对于D ：点M 到直线AB 距离的最小值即为异面直线PC 与AB 的距离，因为AB ⎳CD ，且AB ⊄平面PCD ，CD ⊂平面PCD ，所以AB ⎳平面PCD ，所以直线AB 到平面PCD 的距离等于点A 到平面PCD 的距离，过点A 作AF ⊥PD ，因为P A ⊥平面ABCD ，所以P A ⊥CD ,又AD ⊥CD ,且P A ∩AD =A ,故CD ⊥平面P AD ，AF ⊂平面P AD ，所以AF ⊥CD ，因为PD ∩CD =D ，且PD ，CD ⊂平面PCD ，所以AF ⊥平面PCD ，所以点A 到平面PCD 的距离，即为AF 的长，如图所示，在Rt △P AD 中，P A =2，AD =1，可得PD =5，所以由等面积得AF =255，即直线AB 到平面PCD 的距离等于255，所以D 正确，故选：BCD .11今年是共建“一带一路”倡议提出十周年.某校进行“一带一路”知识了解情况的问卷调查，为调动学生参与的积极性，凡参与者均有机会获得奖品.设置3个不同颜色的抽奖箱，每个箱子中的小球大小相同质地均匀，其中红色箱子中放有红球3个，黄球2个，绿球2个；黄色箱子中放有红球4个，绿球2个；绿色箱子中放有红球3个，黄球2个，要求参与者先从红色箱子中随机抽取一个小球，将其放入与小球颜色相同的箱子中，再从放入小球的箱子中随机抽取一个小球，抽奖结束.若第二次抽取的是红色小球，则获得奖品，否则不能获得奖品，已知甲同学参与了问卷调查，则()A.在甲先抽取的是黄球的条件下，甲获得奖品的概率为47B.在甲先抽取的不是红球的条件下，甲没有获得奖品的概率为1314C.甲获得奖品的概率为2449D.若甲获得奖品，则甲先抽取绿球的机会最小【答案】ACD【解析】设A 红，A 黄，A 绿，分别表示先抽到的小球的颜色分别是红、黄、绿的事件，设B 红表示再抽到的小球的颜色是红的事件，在甲先抽取的是黄球的条件下，甲获得奖品的概率为：P B 红∣A 黄 =P B 红A 黄 P A 黄=27×4727=47，故A 正确；在甲先抽取的不是红球的条件下，甲没有获得奖品的概率为：P B 红 ∣A 红 =P A 红 B 红 P A 红 =P A 黄B 红 +P A 绿B 红 P A 红 =27×37+27×1247=1328，故B 错误；由题意可知，P A 红 =37,P A 黄 =27,P A 绿 =27,P B 红∣A 红 =37,P B 红∣A 黄 =47，P B 红∣A 绿 =12，由全概率公式可知，甲获得奖品的概率为：P =P A 红 P B 红∣A 红 +P A 黄 ⋅P B 红∣A 黄 +P A 绿 ⋅P B 红∣A 绿 =37×37+27×47+27×12=2449，故C 正确；因为甲获奖时红球取自哪个箱子的颜色与先抽取小球的颜色相同，则P A 红∣B 红 =P A 红 ⋅P B 红∣A 红 P B 红=37×37×4924=38，P A 黄∣B 红 =P A 黄 ⋅P B 红∣A 黄P B 红=27×47×4924=13，P A 绿∣B 红 =P A 绿 ⋅P B 红∣A 绿 P B 红 =27×12×4924=724，所以甲获得奖品时，甲先抽取绿球机会最小，故D 正确.故选：ACD .三、填空题：本题共3小题，每小题5分，共15分.12已知△ABC 的边BC 的中点为D ，点E 在△ABC 所在平面内，且CD =3CE -2CA ，若AC =xAB +yBE，则x +y =．【答案】11【解析】因为CD =3CE -2CA ，边BC 的中点为D ，所以12CB=3BE -BC +2AC ，因为12CB =3BE -3BC +2AC ，所以52BC =3BE +2AC ，所以52BC =52AC -AB =3BE +2AC ，所以5AC -5AB =6BE +4AC ，即5AB +6BE =AC ，因为AC =xAB +yBE ，所以x =5，y =6，故x +y =11.故答案为：1113已知圆锥母线长为2，则当圆锥的母线与底面所成的角的余弦值为时，圆锥的体积最大，最大值为．【答案】①.63②.16327π【解析】设圆锥的底面半径为r ，圆锥的母线与底面所成的角为θ,θ∈0,π2 ，易知cos θ=r 2.圆锥的体积为V =13πr 2⋅4-r 2=43πcos 2θ⋅2sin θ=8π3cos 2θ⋅sin θ=8π31-sin 2θ sin θ令x =sin θ,x ∈0,1 ，则y =1-sin 2θ sin θ=-x 3+x ，y =-3x 2+1当y >0时，x ∈0,33，当y<0时，x ∈33,1 ，即函数y =-x 3+x 在0,33 上单调递增，在33,1上单调递减，即V max =8π333-33 3 =163π27，此时cos θ=1-323 =62.故答案为：62；163π2714已知双曲线C ：x 2-y 23=1的左、右焦点分别为F 1，F 2，右顶点为E ，过F 2的直线交双曲线C 的右支于A ，B 两点(其中点A 在第一象限内)，设M ，N 分别为△AF 1F 2，△BF 1F 2的内心，则当F 1A ⊥AB 时，AF 1=；△ABF 1内切圆的半径为．【答案】①.7+1##1+7②.7-1##-1+7【解析】由双曲线方程知a =1,b =3,c =2，如下图所示：由F 1A ⊥AB ，则AF 1 2+AF 2 2=F 1F 2 2=16，故AF 1 -AF 2 2+2AF 1 AF 2 =16，而AF 1 -AF 2 =2a =2，所以AF 1 AF 2 =6，故AF 2 2+2AF 2 -6=0，解得AF 2 =7-1，所以AF 1 =7+1，若G 为△ABF 1内切圆圆心且F 1A ⊥AB 可知，以直角边切点和G ,A 为顶点的四边形为正方形，结合双曲线定义内切圆半径r =12AF 1 +AB -BF 1 =12AF 1 +AF 2 +BF 2 -BF 1所以r =1227+BF 2 -BF 1 =1227-2 =7-1；故答案为：7+1，7-1；2024届高三二轮复习“8+3+3”小题强化训练(3)一、选择题：本题共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．1有一组按从小到大顺序排列数据：3，5，x ，8，9，10，若其极差与平均数相等，则这组数据的中位数为()A.7B.7.5C.8D.6.5【答案】B【解析】依题意可得极差为10-3=7，平均数为163+5+x +8+9+10 =1635+x ，所以1635+x =7，解得x =7，所以中位线为7+82=7.5.故选：B .2已知集合A =x x -1 >2 ，B =x log 4x <1 ，则A ∩B =()A.3,4B.-∞,-1 ∪3,4C.1,4D.-∞,4【答案】A【解析】由x -1 >2，得x <-1或x >3，所以A =x x <-1或x >3 ，由log 4x <1，得0<x <4，所以B =x 0<x <4 ，所以A ∩B =x 3<x <4 .故选：A .3已知向量a =(2,0)，b =sin α,32，若向量b 在向量a 上的投影向量c =12,0 ，则|a +b |=()A.3B.7C.3D.7【答案】B【解析】由已知可得，b 在a 上的投影向量为a ⋅b |a |⋅a |a |=2sin α2×2(2,0)=(sin α,0)，又b 在a 上的投影向量c =12,0 ，所以sin α=12，所以b =12,32，所以a +b =52,32 ，所以|a +b |=52 2+322=7.故选：B .4如图是两个底面半径都为1的圆锥底面重合在一起构成的几何体，上面圆锥的侧面积是下面圆锥侧面积的2倍，AP ⊥AQ ，则PQ =()A.74B.262C.52D.3【答案】C【解析】设两圆锥的高OP =x ，OQ =y ，则AP =x 2+1，AQ =y 2+1，由AP ⊥AQ ，有AP 2+AQ 2=PQ 2，可得x 2+1+y 2+1=x +y 2，可得xy =1，又由上下圆锥侧面积之比为2:1，即π×1×P A =2×π×1×QA ，可得P A =2QA ，则有x 2+1=2y 2+1，即x 2=4y 2+3，代入y =1x整理为x 4-3x 2-4=0，解得x =2(负值舍)，可得y =12，OP =x +y =2+12=52．故选：C ．5已知Q 为直线l :x +2y +1=0上的动点，点P 满足QP=1,-3 ，记P 的轨迹为E ，则()A.E 是一个半径为5的圆B.E 是一条与l 相交的直线C.E 上的点到l 的距离均为5D.E 是两条平行直线【答案】C【解析】设P x ,y ，由QP=1,-3 ，则Q x -1,y +3 ，由Q 在直线l :x +2y +1=0上，故x -1+2y +3 +1=0，化简得x +2y +6=0，即P 轨迹为E 为直线且与直线l 平行，E 上的点到l 的距离d =6-112+22=5，故A 、B 、D 错误，C 正确.故选：C .6已知x +1 x -1 5=a 0+a 1x +a 2x 2+a 3x 3+a 4x 4+a 5x 5+a 6x 6，则a 1+a 3的值为()A.-1B.1C.4D.-2【答案】C【解析】在x +1 x -1 5=a 0+a 1x +a 2x 2+a 3x 3+a 4x 4+a 5x 5+a 6x 6中，而x +1 x -1 5=x x -1 5+x -1 5，由二项式定理知x -1 5展开式的通项为T r +1=C r 5x 5-r (-1)r ，令5-r =2，解得r =3，令5-r =3，r =2，故a 3=C 35(-1)3+C 25(-1)2=0，同理令5-r =1，解得r =4，令5-r =0，解得r =5，故a 1=C 45(-1)4+C 55(-1)5=4，故a 1+a 3=4.故选：C7已知P 为抛物线x 2=4y 上一点，过P 作圆x 2+(y -3)2=1的两条切线，切点分别为A ，B ，则cos ∠APB 的最小值为()A.12B.23C.34D.78【答案】C【解析】如图所示：因为∠APB =2∠APC ，sin ∠APC =AC PC=1PC，设P t ,t 24，则PC 2=t 2+t 24-3 2=t 416-t 22+9=116t 2-4 2+8，当t 2=4时，PC 取得最小值22，此时∠APB 最大，cos ∠APB 最小，且cos ∠APB min =1-2sin 2∠APC =1-21222=34，故C 正确.故选：C8已知函数f x ,g x 的定义域为R ,g x 为g x 的导函数且f x +g x =3,f x -g 4-x =3，若g x 为偶函数，则下列结论一定成立的是()A.f -1 =f -3B.f 1 +f 3 =65C.g 2 =3D.f 4 =3【答案】D【解析】对于D ，∵g x 为偶函数，则g x =g -x ，两边求导可得g x =-g -x ，则g x 为奇函数，则g 0 =0，令x =4，则f 4 -g 0 =3，f 4 =3，D 对；对于C ，令x =2，可得f 2 +g 2 =3f 2 -g 2 =3 ，则f 2 =3g 2 =0 ，C 错；对于B ，∵f x +g x =3，可得f 2+x +g 2+x =3，f x -g 4-x =3可得f 2-x -g 2+x =3，两式相加可得f 2+x +f 2-x =6，令x =1，即可得f 1 +f 3 =6，B 错；又∵f x +g x =3，则f x -4 +g x -4 =f x -4 -g 4-x =3，f x -g 4-x =3，可得f x =f x -4 ，所以f x 是以4为周期的函数，所以根据以上性质不能推出f -1 =f -3 ，A 不一定成立.故选：D二、选择题：本题共3小题，每小题6分，共18分.在每小题给出的选项中，有多项符合题目要求.全部选对的得6分，部分选对的得部分分，有选错的得0分.9下列结论正确的是()A.若a <b <0，则a 2>ab >b 2B.若x ∈R ，则x 2+2+1x 2+2的最小值为2C.若a +b =2，则a 2+b 2的最大值为2D.若x ∈(0,2)，则1x +12-x ≥2【答案】AD【解析】因为a 2-ab =a (a -b )>0，所以a 2>ab ，因为ab -b 2=b (a -b )>0，所以ab >b 2，所以a 2>ab >b 2，故A 正确；因为x 2+2+1x 2+2≥2的等号成立条件x 2+2=1x 2+2不成立，所以B 错误；因为a 2+b 22≥a +b 2 2=1，所以a 2+b 2≥2，故C 错误；因为1x +12-x =12(x +2-x )1x +12-x =122+2-x x +x 2-x ≥12(2+2)=2，当且仅当1x =12-x，即x =1时，等号成立，所以D 正确.故选：AD10若函数f x =2sin 2x ⋅log 2sin x +2cos 2x ⋅log 2cos x ，则()A.f x 的最小正周期为πB.f x 的图像关于直线x =π4对称C.f x 的最小值为-1D.f x 的单调递减区间为2k π,π4+2k π ,k ∈Z【答案】BCD【解析】由sin x >0,cos x >0得f x 的定义域为2k π,π2+2k π ,k ∈Z .对于A ：当x ∈0,π2时，x +π∈π,32π 不在定义域内，故f x +π =f x 不成立，易知f x 的最小正周期为2π，故选项A 错误；对于B ：又f π2-x =2cos 2x ⋅log 2cos x +2sin 2x ⋅log 2sin x =f x ，所以f x 的图像关于直线x =π4对称，所以选项B 正确；对于C ：因为f x =sin 2x ⋅log 2sin 2x +cos 2x ⋅log 2cos 2x ，设t =sin 2x ，所以函数转化为g t =t ⋅log 2t +1-t ⋅log 21-t ,t ∈0,1 ,g t =log 2t -log 21-t ，由g t >0得，12<t <1.g t <0得0<t <12.所以g t 在0,12 上单调递减，在12,1 上单调递增，故g (t )min =g 12=-1，即f (x )min =-1，故选项C 正确；对于D ：因为g t 在0,12 上单调递减，在12,1 上单调递增，由t =sin 2x ，令0<sin 2x <12得0<sin x <22，又f x 的定义域为2k π,π2+2k π ,k ∈Z ，解得2k π<x <π4+2k π,k ∈Z ，因为t =sin 2x 在2k π,π4+2k π 上单调递增，所以f x 的单调递减区间为2k π,π4+2k π ,k ∈Z ，同理函数的递增区间为π4+2k π,π2+2k π ,k ∈Z ，所以选项D 正确.故选：BCD .11已知数列a n 的前n 项和为S n ，且2S n S n +1+S n +1=3，a 1=α0<α<1 ，则()A.当0<α<13-14时，a 2>a 1B.a 3>a 2C.数列S 2n -1 单调递增，S 2n 单调递减D.当α=34时，恒有nk =1S k -1 <54【答案】ACD【解析】由题意可得：S n +1=32S n +1，a 1=α，由S n +1=32S n +1可知：S n +1=1⇔S n =1，但S 1=α∈0,1 ，可知对任意的n ∈N *，都有S n ≠1，对于选项A ：若0<α<13-14，则a 2-a 1=S 2-2a 1=32α+1-2α=3-2α-4α22α+1=4α+1+13 13-14-α2α+1>0，即a 2>a 1，故A 正确；对于选项B ：a 3-a 2=S 3-2S 2+S 1=6α+32α+7-62α+1+α=α-1 4α2+32α+39 2α+1 2α+7<0，即a 3<a 2，故B 错误.对于选项C ：因为S n +1-1=-2S n -1 2S n +1，S n +1+32=3S n +32 2S n +1，则S n +1-1S n +1+32=-23⋅S n -1S n +32，且S 1-1S 1+32=α-1α+32<0，可知S n -1S n+32是等比数列，则S n -1S n +32=α-1α+32⋅-23n -1，设A =α-1α+32<0，t =232n -2，可得S 2n =3-3At 3+2At =3253+2At -1 ，S 2n -1=1+32At 1-At =521-At-32，因为At =A 232n -2，可知A 23 2n -2 为递增数列，所以数列S 2n -1 单调递增，S 2n 单调递减，故C 正确；对于选项D ：因为S n +1=32S n +1，S n +1-34=32S n +1-34=33-2S n 42S n +1，由S 1=α=34，可得S 2-34>0，即S 2>34，则S 2≤65，即34<S 2≤65；由34<S 2≤65，可得S 3-34>0，即S 3>34，则S 3<65，即34<S 3<65；以此类推，可得对任意的n ∈N *，都有S n ≥S 1=α=34，又因为S n +1-1S n -1=22S n +1，则S n +1-1 ≤22α+1S n -1 =45S n -1 ，所以∑nk =1S k -1 ≤541-45 n <54，故D 正确.故选：ACD .三、填空题：本题共3小题，每小题5分，共15分.12在(1+ax )n (其中n ∈N *,a ≠0)的展开式中，x 的系数为-10，各项系数之和为-1，则n =.【答案】5【解析】由题意得(1+ax )n 的展开式中x 的系数为aC 1n =-10，即an =-10，令x =1，得各项系数之和为(1+a )n =-1，则n 为奇数，且1+a =-1，即得a =-2,n =5，故答案为：513已知椭圆C :x 2a 2+y 2b2=1a >b >0 的左、右焦点分别F 1，F 2，椭圆的长轴长为22，短轴长为2，P 为直线x =2b 上的任意一点，则∠F 1PF 2的最大值为.【答案】π6【解析】由题意有a =2，b =1，c =1，设直线x =2与x 轴的交点为Q ，设PQ =t ，有tan ∠PF 1Q =PQ F 1Q=t3，tan ∠PF 2Q =PQ F 2Q=t ，可得tan ∠F 1PF 2=tan ∠PF 2Q -∠PF 1Q =t -t31+t23=2t t 2+3=2t +3t ≤2t 23t =33，当且仅当t =3时取等号，可得∠F 1PF 2的最大值为π6.故答案为：π614已知四棱锥P -ABCD 的底面为矩形，AB =23，BC =4，侧面P AB 为正三角形且垂直于底面ABCD ，M 为四棱锥P -ABCD 内切球表面上一点，则点M 到直线CD 距离的最小值为.【答案】10-1【解析】如图，设四棱锥的内切球的半径为r ，取AB 的中点为H ，CD 的中点为N ，连接PH ，PN ，HN ，球O为四棱锥P-ABCD的内切球，底面ABCD为矩形，侧面P AB为正三角形且垂直于底面ABCD，则平面PHN截四棱锥P-ABCD的内切球O所得的截面为大圆，此圆为△PHN的内切圆，半径为r，与HN，PH分别相切于点E，F，平面P AB⊥平面ABCD，交线为AB，PH⊂平面P AB，△P AB为正三角形，有PH⊥AB，∴PH⊥平面ABCD，HN⊂平面ABCD，∴PH⊥HN，AB=23，BC=4，则有PH=3，HN=4，PN=5，则△PHN中，S△PHN=12×3×4=12r3+4+5，解得r=1.所以，四棱锥P-ABCD内切球半径为1，连接ON.∵PH⊥平面ABCD，CD⊂平面ABCD，∴CD⊥PH，又CD⊥HN，PH,HN⊂平面PHN，PH∩HN=H，∴CD⊥平面PHN，∵ON⊂平面PHN，可得ON⊥CD，所以内切球表面上一点M到直线CD的距离的最小值即为线段ON的长减去球的半径，又ON=OE2+EN2=10.所以四棱锥P-ABCD内切球表面上的一点M到直线CD的距离的最小值为10-1.故答案为：10-12024届高三二轮复习“8+3+3”小题强化训练(4)一、选择题：本题共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．1已知双曲线的标准方程为x 2k -4+y 2k -5=1，则该双曲线的焦距是()A.1B.3C.2D.4【答案】C【解析】由双曲线方程可知a 2=k -4,b 2=5-k ，所以c 2=k -4+5-k =1，c =1，2c =2．故选：C2在等比数列a n 中，a 1+a x =82，a 3a x -2=81，前x 项和S x =121，则此数列的项数x 等于()A.4B.5C.6D.7【答案】B【解析】由已知条件可得a 1+a x =82a 3a x -2=a 1a x =81，解得a 1=1a x =81 或a 1=81a x =1 .设等比数列a n 的公比为q .①当a 1=1，a x =81时，由S x =a 1-a x q 1-q =1-81q1-q=121，解得q =3，∵a x =a 1q x -1=3x -1=81，解得x =5；②当a 1=81，a x =1时，由S x =a 1-a x q 1-q =81-q 1-q =121，解得q =13，∵a x =a 1q x -1=81×13x -1=35-x =1，解得x =5.综上所述，x =5.故选：B .3对任意实数a ，b ，c ，在下列命题中，真命题是()A.“ac 2>bc 2”是“a >b ”的必要条件B.“ac 2=bc 2”是“a =b ”的必要条件C.“ac 2=bc 2”是“a =b ”的充分条件D.“ac 2≥bc 2”是“a ≥b ”的充分条件【答案】B【解析】对于A ，若c =0，则由a >b ⇏ac 2>bc 2，∴“ac 2>bc 2”不是“a >b ”的必要条件，A 错．对于B ，a =b ⇒ac 2=bc 2，∴“ac 2=bc 2”是“a =b ”的必要条件，B 对，对于C ，若c =0，则由ac 2=bc 2，推不出a =b ，“ac 2=bc 2”不是“a =b ”的充分条件对于D ，当c =0时，ac 2=bc 2，即ac 2≥bc 2成立，此时不一定有a ≥b 成立，故“ac 2≥bc 2”不是“a ≥b ”的充分条件，D 错误，故选：B ．4已知m 、n 是两条不同直线，α、β、γ是三个不同平面，则下列命题中正确的是()A.若m ∥α，n ∥α，则m ∥nB.若α⊥β，β⊥γ，则α∥βC.若m ∥α，m ∥β，则α∥βD.若m ⊥α，n ⊥α，则m ∥n【答案】D【解析】A选项：令平面ABCD为平面α，A1B1为直线m，B1C1为直线n，有：m∥α，n∥α，但m∩n=B1，A错误；B选项：令平面ABCD为平面β，令平面B1BCC1为平面α，令平面A1ABB1为平面γ，有：α⊥β，β⊥γ，而α⊥β，B错误；C选项：令平面ABCD为平面α，令平面A1ABB1为平面β，C1D1为直线m，有：m∥α，m∥β，则α∥β，而α⊥β，C错误；D选项：垂直与同一平面的两直线一定平行，D正确.故选：D5将甲、乙等8名同学分配到3个体育场馆进行冬奥会志愿服务，每个场馆不能少于2人，则不同的安排方法有()A.2720B.3160C.3000D.2940【答案】D【解析】共有两种分配方式，一种是4:2:2，一种是3:3:2，故不同的安排方法有C48C24C222!+C38C35C222!A33=2940.故选：D6若抛物线y2=4x与椭圆E:x2a2+y2a2-1=1的交点在x轴上的射影恰好是E的焦点，则E的离心率为()A.2-12 B.3-12 C.2-1 D.3-1【答案】C【解析】不妨设椭圆与抛物线在第一象限的交点为A，椭圆E右焦点为F，则根据题意得AF⊥x轴，c2=a2-a2-1=1，则c=1，则F1,0，当x=1时，y2=4×1，则y A=2，则A1,2，代入椭圆方程得12a2+22a2-1=1，结合a2-1>0，不妨令a>0；解得a=2+1，则其离心率e=ca=12+1=2-1，故选：C.7已知等边△ABC 的边长为3，P 为△ABC 所在平面内的动点，且|P A |=1，则PB ⋅PC 的取值范围是()A.-32,92B.-12,112C.[1,4]D.[1,7]【答案】B【解析】如下图构建平面直角坐标系，且A -32,0 ，B 32,0 ，C 0,32，所以P (x ,y )在以A 为圆心，1为半径的圆上，即轨迹方程为x +322+y 2=1，而PB =32-x ,-y ,PC =-x ,32-y ，故PB ⋅PC =x 2-32x +y 2-32y =x -34 2+y -34 2-34，综上，只需求出定点34,34 与圆x +322+y 2=1上点距离平方范围即可，而圆心A 与34,34 的距离d =34+32 2+34 2=32，故定点34,34与圆上点的距离范围为12,52，所以PB ⋅PC ∈-12,112.故选：B 8设a 、b 、c ∈0,1 满足a =sin b ，b =cos c ，c =tan a ，则()A.a +c <2b ，ac <b 2B.a +c <2b ，ac >b 2C.a +c >2b ，ac <b 2D.a +c >2b ，ac >b 2【答案】A【解析】∵a 、b 、c ∈0,1 且a =sin b ，b =cos c ，c =tan a ，则c =tan a =tan sin b ，先比较a +c =sin b +tan sin b 与2b 的大小关系，构造函数f x =sin x +tan sin x -2x ，其中0<x <1，则0<sin x <1，所以，cos1<cos sin x <1，则f x =cos x +cos xcos 2sin x -2=cos x -2 cos 2sin x +cos x cos 2sin x，令g x =cos x -1-12x 2 ，其中x ∈0,1 ，则g x =x -sin x ，令p x =x -sin x ，其中0<x <1，所以，p x =1-cos x >0，所以，函数g x 在0,1 上单调递增，故g x >g 0 =0，所以，函数g x 在0,1 上单调递增，则g x =cos x -1-12x 2 >0，即cos x >1-12x 2，因为x ∈0,1 ，则0<sin x <sin1，所以，cos sin x >1-12sin 2x =1-121-cos 2x =121+cos 2x ，所以，cos 2sin x >141+cos 2x 2，因为cos x -2<0，所以，cos x -2 cos 2sin x +cos x <14cos x -2 1+cos 2x 2+cos x=14cos 5x -2cos 4x +2cos 3x -4cos 2x +5cos x -2 =14cos x -1 3cos 2x +cos x +2 <0，所以，对任意的x ∈0,1 ，f x =cos x -2 cos 2sin x +cos xcos 2sin x <0，故函数f x 在0,1 上单调递减，因为b ∈0,1 ，则f b =sin b +tan sin b -2b <f 0 =0，故a +c <2b ，由基本不等式可得0<2ac ≤a +c <2b (a ≠c ，故取不了等号)，所以，ac <b 2，故选：A .二、选择题：本题共3小题，每小题6分，共18分.在每小题给出的选项中，有多项符合题目要求.全部选对的得6分，部分选对的得部分分，有选错的得0分.9某大学生做社会实践调查，随机抽取6名市民对生活满意度进行评分，得到一组样本数据如下：88、89、90、90、91、92，则下列关于该样本数据的说法中正确的是()A.均值为90B.中位数为90C.方差为2D.第80百分位数为91【答案】ABD【解析】由题意可知，该组数据的均值为x =88+89+90+90+91+926=90，故A 正确；中位数为90+902=90，故B 正确；方差为s 2=1688-90 2+89-90 2+90-90 2×2+91-90 2+92-90 2 =53，故C 错误；因为6×80%=4.8，第80百分位数为91，故D 正确.故选：ABD .10设M ，N ，P 为函数f x =A sin ωx +φ 图象上三点，其中A >0，ω>0，ϕ <π2，已知M ，N 是函数f x 的图象与x 轴相邻的两个交点，P 是图象在M ，N 之间的最高点，若MP 2+2MN ⋅NP=0，△MNP 的面积是3，M 点的坐标是-12,0 ，则()A.A =2B.ω=π2C.φ=π4D.函数f x 在M ，N 间的图象上存在点Q ，使得QM ⋅QN <0【答案】BCD【解析】MP 2+2MN ⋅NP =MP 2-2NM ⋅NP =MP 2-2NM ⋅12NM =T 4 2+A 2 -T 22=A 2-3T 216=0，而S △MNP =AT 4=3，故A =3，T =4=2πω，ω=π2，A 错误、B 正确；-12⋅π2+φ=k π，φ=k π+π4(k ∈Z )，而ϕ <π2，故φ=π4，C 正确；显然，函数f x 的图象有一部分位于以MN 为直径的圆内，当Q 位于以MN 为直径的圆内时，QM⋅QN<0，D 正确，故选:BCD .11设a 为常数，f (0)=12,f (x +y )=f (x )f (a -y )+f (y )f (a -x )，则()．A .f (a )=12B .f (x )=12成立C f (x +y )=2f (x )f (y )D .满足条件的f (x )不止一个【答案】ABC 【解析】f (0)=12,f (x +y )=f (x )f (a -y )+f (y )f (a -x )对A ：对原式令x =y =0，则12=12f a +12f a =f a ，即f a =12，故A 正确；对B ：对原式令y =0，则f x =f x f a +f 0 f a -x =12f x +12f a -x ，故f x =f a -x ，对原式令x =y ，则f 2x =f x f y +f y f x =2f x f y =2f 2x ≥0，故f x 非负；对原式令y =a -x ，则f a =f 2x +f 2a -x =2f 2x =12，解得f x =±12，又f x 非负，故可得f x =12，故B 正确；对C ：由B 分析可得：f x +y =2f x f y ，故C 正确；对D ：由B 分析可得：满足条件的f x 只有一个，故D 错误.故选：ABC .三、填空题：本题共3小题，每小题5分，共15分.12在复平面内，复数z =-12+32i 对应的向量为OA ，复数z +1对应的向量为OB ，那么向量AB 对应的复数是．。

高三英语45分钟限时训练三班级_____________ 姓名_____________ 得分_____________第一节单项填空（共l5小题；每小题l分，满分l5分）1．The World Expo is like _________great stage for cultural exchange, bringing people into closer contact with _________ rest of the world．A．a; 不填B．the; 不填 C．a; the D．the; the2．It’s a common scene in the film Tangshan Earthquake that families cry helplessly over _________ is left of their homes.A. whatB. whichC. thereD. it3．When _________a job, you should highlight your experience and skills．A．applying to B．applied to C．applying for D．applied for4．There_________a trend that 3D films will become more popular ever since the film Avatar hit the market．A．has arisen B．arose C．is arising D．will arise5．It is the discovery of Cao Cao's tomb, _________ was made in Anyang, Henan Province,________ has received widespread attention.A．where; which B．where; that C．that; which D．which; that 6．The accident that happened at an oilfield_________ the southeast coast of the USA has caused great damage to the environment.A．away B．from C．off D．beyond7．If you take sleeping pills for a long time, you will_________them.A．be addicted to B．be attracted to C．be devoted to D．be exposed to8．—Did Linda see the traffic accident.—No, no sooner _________ than it happened.A．had she gone B．she had gone C．has she gone D．she has gone9．After the Yushu earthquake, to make sure that rescue work could be carried out as smoothly as possible, Premier Wen Jiabao his visit to Brunei.A．put away B．put off C．put up D．put down10．—You didn’t take the high-speed train with the leading letter "G" to Shanghai?—. I still took the "D" train, which saved me more than 50 yuan．A．Yes, I didn't B．No, I did C．Yes, I did D．No, I didn't 11．Fireworks, ________ of gun powder or flash powder _______ in a tight paper tube, make the big flash or boom effect.A. consisting…wrappingB. consisting…wrappedC. consisted…wrappedD. being consisted….wrapped12．You should always remember that, in the years to come, you have some people _______ company, the parents who have sacrificed to raise you and provide you with the best education possible.A. keepB. keepingC. keptD. to keep13．Many young people in the West are expected to leave ______ could be life’s most important decision --- marriage---almost entirely up to luck.A .as B. that C. which D. what14．In a typical business situation, women, _______, find themselves bumping their heads on “glass ceiling”, a barrier to their career which is not obvious to the observer, impossible to break through.A. as they might tryB. try as they mightC. hardly as they tryD. as they try very hard 15．Playing QQ’s “Happy Farm” by planting, watering, fertilizing, s praying, harvesting and selling of virtual vegetables, fruits and flowers, we can find out ________ all the excitement is about.A. thatB. ifC. whatD. when第二节完形填空（共20小题，每小题1分，满分20分）Children model themselves largely on their parents. They do so mainly through identification. Children identify 16 a parent when they believe they have the qualities and feelings that are 17 of that parent. The things parents do and say---and the 18 they do and say to them--therefore strongly influence a child's 19 . However, parents must consistently behave like the type of 20 they want their child to become.A parent's actions 21 affect the self-image that a child forms 22 identification. Children who see mainly positive qualities in their 23 will likely learn to see themselves in a positive way. Children who observe chiefly 24 qualities in their parents will have difficulty 25 positive qualities in themselves. Children may 26 their self-image, however, as they become increasingly 27 by peers groups standards before they reach 13.Isolated(孤立的) events, 28 dramatic(突然的) ones, do not necessarily have a permanent 29 on a child's behavior. Children interact such events according to their established attitudes and previous training. Children who know they are loved can, 30 , accept the divorce of their parents' or a parent's early 31 . But if children feel unloved, they may interpret such events 32 a sign of rejection or punishment.In the same way, all children are not influenced 33 by toys and games, reading matter, and television programs. 34 in the case of a dramatic change in family relations, the 35 of an activity or experience depends on how the child interprets it.16. A. to B. with C. around D. for17. A. informed B. characteristic C. conceived D. indicative18. A. gesture B. expression C. way D. extent19. A. behavior B. words C. mood D. reactions20. A. person B. humans C. creatures D. adult21. A. in turn B. nevertheless C. also D. as a result22. A. before B. besides C. with D. through23. A. eyes B. parents C. peers D. behaviors24. A. negative B. cheerful C. various D. complex25. A. see B. seeing C. to see D. to seeing26. A. modify B. copy C. give up D. continue27. A. mature B. influenced C. unique D. independent28. A. not B. besides C. even D. finally29. A. idea B. wonder C. stamp D. effect30. A. luckily B. for example C. at most D. theoretically31. A. death B. rewards C. advice D. teaching32. A. as B. being C. of D. for33. A. even B. at all C. alike D. as a whole34. A. Even B. Since C. Right D. As35. A. result B. effect C. scale D. cause第三节阅读理解（共８小题，每小题2分，满分１６分）ＡCities need to put efforts to deal with pollution and clear the air.Chinese cities will need to put efforts to clear up the sky when a new department to improve regional air quality is set up by 2015, according to the latest plan released by the State Council．Besides the existing pollution control program for S02, regional emission caps（区域排放上限）for other certain chemicals will be established in the three key air polluting areas— rhe Yangtze River Delta, Pearl River Delta and the Beijing-Tianjin-Hebei region．Coal-consumption caps will also be tried out in some areas, according to the plan．The plan is aimed at dealing with regional air pollution—such as acid rain and smog—which have become increasingly obvious in China in recent years and caused a severe threat to people's health, Zhang Lijun, vice-minister of environmental protection, said in an interview on Monday.The air quality in a city affects the regions nearby because pollutants（污染物）can travel through the atmosphere, said Chai Fahe, vice-director of Chinese Research Academy of Environmental Sciences．“So efforts to reduce air pollution in a single city, targeting a certain pollutant will not oe enough," Chai said．Zhang said the country's major industrial districts—the Yangtze River Delta, Pearl River Delta and the Beijing-Tianjin-Hebei region—have recorded more than 100 misty days annually in recent years．These three regions, home to at least 200 million people, occupy only 6．3 percent of the country's area but consume 40 percent of the country's coal and produce half of its steel, according to official figures．Studies also show that the visibility（能见度）in eastern regions of China has dropped by 7 to 15 km compared to that in the early 1960s, as a result of air pollution．Liaoning province, the Shandong Peninsula, Wuhan in Hubei province and its surrounding area, the Changsha-Zhuzhou-Xiangtan region in Hunan province, the Chengdu-Chongqing region, and the western coast of the Taiwan Straits are also listed as areas to carry out such regional air pollution control programs, according to the plan．The new plan also requires an improved air quality system, which will measure the pollution levels．"The current system, which only measures some major pollutants, cannot reflect the true picture．" said Chai．36．It will not be enough to reduce air pollution in a single city because ______.A．air pollution has been becoming more and more serious in recent yearsB．air pollution like acid rain and smog is threatening people's healthC．air pollutants in a city can travel in the sky and affect nearby areasD．air pollutants stay in fixed places over the cities37．Suppose the visibility in eastern regions of China was 130 km in the early 1960s, the visibility there nowadays is .A．123 to 115 km B．137 to 145 km C．123 to 145 km D．115 to 137 km 38．The main purpose of the passage is .A．to inform readers of the damage caused by air pollutionB．to introduce the new plan to control air pollutionC．to provide official figures of air pollution levelsD．to point out the regions affected by air pollution39．What does the underlined phrase "the true picture" in the last paragraph mean?A．The major pollutants．B．The key polluted areas．C．The major polluted cities．D．The pollution levels．ＢPete took his seat this day on his usual bench in Union Square．Every Thanksgiving Day for nine years Pete had taken this same seat exactly at one o'clock, and each time the same pleasant thing had happened．But this time Pete had come here more from habit than from hunger．Certainly today Pete was not hungry．Two rich old ladies gave him a Thanksgiving dinner of everything he could eat．Pete sat on the bench now, hardly able to move．The buttons on his ragged shirt and coat were about to burst．The November breeze and the first light fall of snow felt cool and pleasant to his face．He happened to look to the left and there in the distance he saw the old Gentleman coming toward him．He wanted to get up and run, but he was so full of food he stayed right there．Every Thanksgiving Day for nine years, the Old Gentleman had come there, found Pete on this same bench, and then taken him to a restaurant and bought him a Thanksgiving dinner．It was a kind of tradition which the Old Gentleman, who had no family and lived alone, had tried to continue．"How do you do' "said the Old Gentleman．" Glad to see that the changes of another year have permitted you to move in health through this beautiful world．"Each time the Old Gentleman had said exactly this same thin? It was part of the tradition．Pete, too, began to feel as though he himself was now a part of the tradition, and he therefore did not have the courage to tell the old man that he had already eaten．This dinner seemed to mean so much to the Old Gentleman."T hank you, sir, " said Pete at last. “I’ll go with you gladly. I’m very hungry, sir.” Together the Old Gentleman and Pete walked south to the same restaurant where each year Pete had his Thanksgiving dinner. The Old Gentleman seemed pleased and happy. When the waiter brought dish after dish of food to Pete, the Old Gentlenman sat quietly and smiled．Under the circumstances, Pete had to eat．It was part of the tradition, and so he ate like a hero, although when he entered the restaurant even the smell of food almost made him sick．At last Pete leaned back with the battle won．" Thank you, sir, " he said, with some effort, " for a fine dinner．"They parted as they did each year at the door, the Old Gentlemen going south, Pete north．Around the corner, Pete stopped for a moment, felt a terrible pain in his stomach, and then fell to the sidewalk unconscious．A little time later an ambulance came．In the hospital they discovered that he had had an attack of indigestion （消化不良）.An hour later, another ambulance brought the Old Gentleman to the same hospital．At first they thought it was also indigestion but later one of the nurses said, "That nice old gentleman over there—you wouldn't think that it was a case of hunger．"40．Where did Pete and the Old Gentleman meet every Thanksgiving Day?A．In the restaurant．B．In Union Square．C．At the Old Gentleman's house．D．At the hospital．41．Which of the following sentences in the passage doesn't show Pete was full?A．Pete sat on the bench now, hardly able to move．B．The buttons on his ragged shirt and coat were about to burst．C．Glad to see that the changes of another year have permitted you to move in health through this beautiful world．D．When he entered the restaurant even the smell of food almost made him sick．42．The underlined sentence "I'm very hungry, sir．" in Paragraph 5 implies that .A．Pete didn't have the heart to refuse the offerB．Pete was so greedy that he wanted to have another dinnerC．Pete was suffering from hungerD．Pete enjoyed having the Thanksgiving dinner for free43．Which of the following words can best describe the Old Gentleman?A．Self-centred. B．Modest. C．Stubborn．D．Generous CACAD CAABD BDDBCBBCAA CDBAB ABCDB AACDBCABD BCAD。

2023届高三英语晚自习限时训练一、阅读理解AParis is the most wooded city in Europe, and around one quarter of its area is taken up by parks, gardens and other green spaces.Jardin des Plantes: over 28 hectaresThis impressive botanical garden is home to four museums, including the Natural History Museum, which houses a variety of African animals. The gardens themselves boast thousands of species of plants. This is a lovely place to relax and explore the beautiful surroundings.Parc des Buttes-Chaumont: over 24 hectaresIf you want to get off the tourist trail and hang out with true Parisians, this park is the perfect spot for a relaxing picnic or a leisurely wander. It is in the 19th district and is one of the largest parks in Paris. Cross its lake on the suspension bridge, or admire the waterfalls and attractive temples.Jardin du Luxembourg: over 25 hectaresThese beautiful gardens were first laid out in 1612 as the gardens of the Luxembourg Palace. They include a geometric forest, French and English style gardens, as well as a large pond. There is so much to see and do with activities especially for children, such as slides, rides and puppet shows, as well as leisure and sporting activities for adults. The gardens are home to more than one hundred sculptures.Le Parc de la Villette: 55 hectaresThis urban cultural park is so much more than just a park as it is home to numerous music venues, exhibition spaces and cafes. It is also home to no less than 12 gardens, which all follow different themes. The Passenger Gardens, for example, which are built on what was once the sheep market, support a variety of ecosystems, demonstrating the importance of biodiversity.1. Which green space probably suits students who are interested in botany?A．Jardin des Plantes. B．Parc des Buttes-Chaumont.C．Jardin du Luxembourg. D．Le Parc de la Villette.2. Which is special about Jardin du Luxembourg?A．They used to be royal gardens. B．They are Parisian favourite gardens. C．They feature American style gardens. D．They house a variely of African animals.3. What can visitors do in Le Parc de la Villette?A．Visiting museums. B．Attending concerts.C．Admiring waterfalls. D．Watching puppet shows.BA team of researchers at ETH Zurich has the perfect long-lasting solution to our fog problem. They have developed a very thin and gold-based transparent coating (涂层) that can convert sunlight into heat, which can be applied to glass and other surfaces to prevent them from fogging.The special coating is developed using titanium oxide and gold particles. It selectively absorbs infrared radiations (红外线照射) from the sunlight and creates a heating effect that is powerful enough to keep fog away from the surface of an object.The coating basically employs heat to stop fogging. The coating absorbs a large part of the infrared radiation, which causes it to heat up-by up to 8℃.Previously, products like anti-fogging sprays create a very thin film of water on the surface to remove fog. The problem with such surfaces is pollution. Along with water, the sprays also attract dirt, dust, oil, and various other dirty substances, and just a little dirt on the surface makes it useless. However, the new coating repels (排斥) water. Furthermore, the coating is significantly thinner, which makes it more transparent as well as flexible. It is heated passively and requires, during daytime, no additional energy source.Gold might be expensive, but the researchers stress that their coating requires solittle that the material costs remain low. Their coating is produced with standardized and readily scalable methods, all cost-effectively. However, although this product can work at very low levels of solar irradiation, it does rely on a certain amount of light.The researchers will develop the coating further for other applications. In the process, they will examine whether other metals work just as well as gold. There is no need to fear. However, this would cause a car or a building to heat up more in the summer. The researchers have already filed a patent. Hopefully, this groundbreaking product will soon be available on the market.；4. How does the new coating work to remove fog?A．By reflecting part of sunlight selectively.B．By keeping cool air away from the object.C．By creating a thin film of water on the surface.D．By absorbing infrared radiations to heat the coating.；5. Why does the author mention previous anti-fogging products?A．To stress the advantages of the new coating.B．To introduce the inspiration behind the new coating.C．To prove the difficulty of creating anti-fogging products.D．To explain the reason for the anti-fogging coating being popular .；6. What is the limitation of the new coating?A．It is considerably expensive.B．It depends on sunlight to work.C．It takes too much time to produce.D．It is difficult to clean and maintain.；7. What do the researchers plan to do next?A．Apply for a patent for the invention.B．Develop the product for more practical uses.C．Explore the environmental impact of the coating.D．Seek a cheaper material as an alternative to gold.CAn exhibition of vivid photographs and a restored documentary give fresh insight into the Antarctic explorer, who died a century ago.One hundred years ago, the leader of the last great expeditio n of the heroic age of polar exploration died from a heart attack as his ship, Quest, headed for Antarctica. The announcement of the death of Ernest Shackleton on 30 January 1922 was greeted with an outpouring of national grief.This was the man, after all, who had saved the entire crew of his ship Endurance —which had been crushed and sunk by ice in 1915 — by making a daring trip in a tiny open boat over 750 miles of polar sea to raise the alarm at a whaling station in South Georgia.It remains one of the greatest rescue stories of modern history and led to the idolising of Shackleton in the United Kingdom, a reputation that survived undamaged for the rest of the century. As his contemporary Raymond Priestley, the geologist and Antarctic explorer, later put it: “When disaster strikes and all hope is gone, get down on your knees and pray for Shackleton.”And here and now in 2022, his death is being marked with an elaborately illustrated exhibition —Shackleton’s legacy and the power of early Antarctic pho tography — which opens at the Royal Geographical Society (RGS), and which includes a range of images and artefacts from his expeditions. Additionally, a digitally remade version of South, a documentary film of Shackleton’s 1914-16 Endurance expedition, is being screened at the British Film Institute.The film and most of the exhibition’s finest images are the handiwork of Frank Hurley, who sailed with Shackleton and who was one of the 20th century’s greatest photographers and film-makers. Both film and exhibition feature striking camera work and provide vivid accounts of the hardships that Shackleton and his men endured as they headed off to explore Antarctica.Even after he survived the great expedition, he still longed for another trip to Antarctica, and after long negotiations set sail in Quest, from England, with the aim of circumnavigating (环航) Antarctica, Shackleton was by now very ill and had suffered at least one heart attack. On 2 January 1922, he wrote in his diary: “I grow old and tired but must a lways lead on.” Three days later he had a major heart attack and died a few hours later. He is buried on South Georgia, scene of his greatest triumph.“Shackleton was an inspirational leader. He had an innate sense of what was possible and achievable. He also had a huge personality but led by example. At the same time, he was sensitive to the needs of the individuals he was leading. For example, after Endurance broke up, his men had lost their protection and shelter. Their social fabric had been destroyed. There would have been disagreement. Yet Shackleton succeeded in keeping them together and made sure they survived.”8. People were overcome with grief when Ernest Shackleton died because .A．it was a huge pity that such a brave explorer should ha ve died from a heart attack B．he was the man that wrote about one of the greatest rescue stories of modern history C．he came to his entire crew’s rescue and symbolised hope in extreme circumstances D．there was no one to pray to anymore when disaster came and there was no hope9. What can we learn about the exhibition?A．It presents Shackleton’s 1914-16 Endurance expedition with powerful Antarctic photos.B．It celebrates the 100th anniversary of the great explorer Ernest Shackleton’s birth. C．It consists of vivid photographs, artefacts, and documentaries of Ernest Shackleton.D．It is created by Frank Hurley, who witnessed Shackleton’s heroic acts with his own eyes.10. Which of the following is NOT true about Ernest Shackleton according to the passage? A．He was the leader of a heroic exploration to the South pole, who died from a heart attack off shore.B．He saved the crew members of the sunken Endurance by travelling to raise the alarmin a tiny boat.C．He is universally recognised as the greatest Antarctic explorer who has enjoyed enduring fame.D．He was inspirational, practical, responsible, sensitive towards his men, but had a strong character.11. What does “Their social fabric had been destroyed. ” in the last paragraph most probably mean?A．What they wore would not be accepted by others upon returning.B．They could no longer socialise with others even if they went back.C．The ship could not keep them together even if they survived.D．They could not function socially as they had when there was shelter.二、七选五阅读Do you want lots of success in school and in your friendship? If you want to learn how to balance your friendship and study, read on.• Talk about your goals for school.Share what academic success looks like for both of you. 12 Before the semester starts, sit down and discuss what you’d both like to achieve.• Share tips and tricks with each other.Exchange study techniques to learn how to be more successful. Maybe you know some learning methods like making a list of key ideas and reviewing them one hour a day. Chat about how you’ll both try to master your subjects. 13• Remind each other to take your health seriously.Put your wellness above everything else to succeed in school. 14 Get at least 8 hours of rest. Make sure to eat nutritious meals that will keep you energized. When you take care of yourselves, you’ll both reduce stress and improve your academic performances.• 15Remember that you both can overcome any challenge s with enough effort. In orderto get the most out of your studies, keep a “growth mindset”, an optimistic outlook that lets you see any difficult topic or task as an opportunity to learn.• Prioritize (优先安排) exams and finals.16 When you and your friend are in the middle of a finals week or have a huge assignment to prepare for, that should come first. Block out times on your schedules that need all your focus.A．Keep positive attitudes about school.B．Fight for any challenges enthusiastically.C．Your methods will be always encouraging.D．Give each other space during more intense periods.E．It’s important to be clear about what your goals are.F．You’ll lift each other up and get an extra edge in class.G．So you can both be at your best physically and mentally.三、完型填空When I was 13, I climbed my first mountain near my home. I was overweight thenand 17 when I reached the top. But I loved the 18 .Two years ago, my friend Mel Olsen and I drove to Oregon to climb 11,240-foot Mount Hood.As we went higher, the trail (山路) grew 19 and steeper. At around 9 a.m., we reached an ice step. Confident I was safe, I put my full 20 on it. Suddenly, I heard a 21 , and the ice broke off the step, right under my foot.In a second, I fell backward. Soon, I came to a stop on a flat slope. I 22 my whole body, feeling 23 that I was fine, except that I was suffering from a sharp painin my left 24 and couldn’t move.I 25 for help, and immediately Mel and other climbers came to assist me and called Mountain Rescue. Then I was loaded into a(n) 26 and taken to a hospital,27 I stayed for four days. The doctors told me it would be a year before I could 28 again, but I was back on the trails within six months.The fall has made me more 29 . One slip (滑倒) on a mountain can change everything. But the experience also made me 30 as a person. Since the accident, I’ve climbed another 60 mountains. I’m not going to let one fall 31 me from doing my favorite thing.17. A．out of danger B．out of trouble C．out of balance D．out of breath18. A．opportunity B．memory C．challenge D．mountain19. A．wider B．flatter C．narrower D．nearer20. A．body B．strength C．foot D．weight21. A．cry B．crack C．crash D．call22. A．calmed B．looked C．tested D．checked23. A．fortunate B．satisfied C．grateful D．confident24. A．arm B．shoulder C．leg D．eye25. A．yelled B．asked C．begged D．arranged26. A．car B．ambulance C．truck D．shelter27. A．which B．that C．where D．when28. A．jump B．move C．drive D．climb29. A．positive B．cautious C．frightened D．anxious30. A．grow B．develop C．feel D．act31. A．protect B．ban C．addict D．keep四、语法填空阅读下面短文，在空白处填入1个适当的单词或括号内单词的正确形式。

一. 词汇语法巩固练习1.Indeed, as time speeds by, I've had some of my most ____________ (memory)experiences working for CT, and made lifelong friends along the way.2.He often complains that the results of machine translation are usually in an____________ (nature) form of language and sometimes just plain weird.3.It is often believed however, that competition and cooperation are in____________ (oppose) to each other.4.He is well qualified for the job, but he does lack ____________ (person) (= he isa boring person).5.Being a great team player depends on your comfortable ____________(participate) in every process of such a team, including knowledge exchange.6.One of my business partners has many interesting new ideas, but he's a little____________ (practice), and often they don't hold water.7.This was a ____________ (predict) reaction, given the bitter hostility between thetwo countries.8.Then, my friends, I recommend we rest and get ourselves ready to move quickly.When war comes, we cannot be caught ____________ (prepare).9.Giving a man a single fish can only cope with his meal. But if you taught him theskill of fishing, it would be ____________ (profit) to him for lifetime.10.The best remedy is either to limit the use of the offending substance or to takeextra precautions, like wearing ____________ (protect) clothing.11.You could ____________ (reason) argue that the store should replace thesegoods.ck of exercise is also a risk factor for heart disease but it's ____________(relate) small when compared with the others.13.He received the award in recognition ____________ his success over the pastyear.14.This is ____________ my father has taught me—to always face difficulties andhope for the best.15.While they are rare north of 88°, there is evidence ____________ they range allthe way across the Arctic, and as far south as James Bay in Canada.16.My uncle is the owner of a restaurant close to ____________ I live.17.I soon realized ____________ my mom's greatest strength was taking care ofthose in need. That's ____________ I wanted to become a doctor.18.I don't remember ____________ it first started annoying me—her hands pushingmy hair that way. Finally, one night, I shouted at her，“Don't do that anymore—your hands are too rough！”19.When we got a call ____________ (say) she was short­listed, we thought it was ajoke.20.Modern methods of tracking polar bear populations have been employed onlysince the mid­1980s, and are expensive ____________ (perform) consistently over a large area.二. 根据句意或提示完成句子1. 我写这封信是想分享我对这个问题的看法，我认为可能会对你有帮助。

选择题限时训练（1）1、如图，墙上有两个钉子a 和b ，它们的连线与水平方向的夹角为45°，两者的高度差为l 。

一条不可伸长的轻质细绳一端固定于a 点，另一端跨过光滑钉子b 悬挂一质量为m 1的重物。

在绳子距a 端2l 得c 点有一固定绳圈。

若绳圈上悬挂质量为m 2的钩码，平衡后绳的ac 段正好水平（如图虚线所示），则重物和钩码的质量比12m m 为（ ） A.5 B. 52C. 2D.2 2、如图所示，绷紧的水平传送带始终以恒定速率1v 运行。

初速度大小为2v 的小物块从与传送带等高的光滑水平地面上的A 处滑上传送带。

若从小物块滑上传送带开始计时，小物块在传送带上运动的t v -图像（以地面为参考系）如图乙所示。

已知12v v >，则（ ）A. t 2时刻，小物块离A 处的距离达到最大B. t 2时刻，小物块相对传送带滑动的距离达到最大C. 0~ t 2时间内，小物块受到的摩擦力方向先向右后向左D. 0~ t 3时间内，小物块始终受到大小不变的摩擦力作用3、我国和欧盟合作正式启动伽利略卫星导航定位系统计划，这将结束美国全球卫星定位系统（GPS ）一统天下的局面。

据悉，“伽利略”卫星定位系统将由30颗轨道卫星组成，卫星的轨道高度为2.4×104km ，倾角为560，分布在3个轨道面上，每个轨道面部署9颗工作卫星和1颗在轨备份卫星，当某颗工作卫星出现故障时可及时顶替工作。

若某颗替补卫星处在略低于工作卫星的轨道上，则这颗卫星的周期、速度、加速度和角速度与工作卫星相比较，以下说法中正确的是（ ）A ．替补卫星的周期大于工作卫星的周期，速度大于工作卫星的速度B ．替补卫星的周期大于工作卫星的周期，速度小于工作卫星的速度C ．替补卫星的加速度大于工作卫星的加速度，角速度大于工作卫星的角速度D ．替补卫星的加速度小于工作卫星的加速度，角速度小于工作卫星的角速度4．如图示，空间存在着一簇关于y 轴对称电场线，O 是坐标原点，M 、N 、P 、Q 是以O为圆心的一个圆周上的四个点，其中M 、N 在y 轴上，Q 点在x 轴上，则（ ）A ．OM 间的电势差大于NO 间的电势差B ．M 点的电势比P 点的电势高C ．一正电荷在O 点时的电势能等于在Q 点时的电势能D ．将一负电荷由M 点移到P 点，电场力做正功x y M NO P Q v v 2 v 1 -v 1 v 2 t 1 t 3 t 2 t 甲乙 A5、在匀强磁场中，一矩形金属线框绕与磁感线垂直的转轴匀速转动，如图1所示。

高三限时训练四答案12.C; 解析：认为母亲的心情“痛苦与无奈”分析不恰当。

应该是对儿子在极为紧张的情况下,还用这种方法表达孝心感到十分欣慰。

13.(1)在情节结构上:设置悬念,引出下文,为下文情节做铺垫;(2)在人物塑造上:表达母亲对儿子的深情,侧面表现了军人在家也是一个好儿子;(3)从主题表达上:通过母亲在风雪中的焦灼与失望,表现军人付出的巨大牺牲,赞美了军人的无私奉献。

14.①从内容上讲是合理的:“回家过年”的故事的主体,主体部分写母亲盼儿回家过年,结尾部分写儿子的照片“回家过年”。

②从结构上讲是合理的:“回家过年”的情节贯穿全文,使叙事结构严谨。

③从人物形象的塑造上讲是合理的:母亲盼儿回家过年,表现的是浓浓的母爱;儿子只有照片回家过年,表现的是军人为国奉献的精神。

④从主题上讲是合理的:“回家过年”的愿望与最终未能回家的结局,形成矛盾,突出了军人为了国家舍小家的精神之伟大。

15.首先由闲书引出“中国艺术史是一部悲催的血泪史和心灵的放逐史”的观点;接着,例证不幸与放逐成全了一个个艺术大家;最后,得出艺术史是高尚者的墓志铭的结论。

16.①理论的包装无法提升艺术自身的价值;②应努力提升艺术本身的品质;③表达出对不甘寂寞追求浮华影响的艺术家们的讽刺与批判。

17.①不甘寂寞的折腾往往掩盖了艺术本身的品质;②不甘寂寞的折腾意味着背离艺术本身;③不甘寂寞的折腾会助长作假蒙骗的坏风气。

18.B; B项以偏概全,原文第3段中强调成功的原因是“很大程度上在于”。

19.AE; B项要“让历史节目获得观众的认可,要从孩子做起,照顾到孩子的兴趣和接受能力”无从说起,错,让历史节目获得观众的认可,真实又轻松好看是主要因素;C项“时效性”错,多角度引用体现了新闻的真实性客观性;D项《国家宝藏》主要是带动博物馆旅游热和文博、文旅产业发展,在“各领域、各个方面”错。

20.①深度挖掘优秀传统文化资源,在文化的内核和创新的表达中找到结合点,让诗词、文物、书法等传统文化“活起来”,变得更有温度,更易于接受、感知、亲近。