云南师大附中2021届月考(二)作文题目及讲评例文

2021届云南师范大学附属中学高三英语月考试卷及参考答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AShopping centers，stadiums and universities may soon have a new tool to help fight crime.ACaliforniacompany called Knightscope says its robots can predict and prevent crime. Knightscope says the goal is to reduce crime by half in areas the robots guard.William Santana Li is the chief executive officer of Knightscope. He says，"These robot security guards will change the world. Our planet has more than seven billion people on it. It's going to quickly get to nine billion people. The security equipmentthat we have globally is just not going to develop that fast. The company's Autonomous Data Machines can become the eyes and ears of law enforcement（执法）.""You want them to be machines plus humans. Let. the machines do the heavy and sometimes dangerous work and let the humans do the strategic decision-making work，so it's always working all together."The machines do not carry weapons but they have day and night video cameras which are able to turn 360 degrees and can also sense chemical and biological weapons.Some people may become concerned about their privacy, especially in connection with the video recordings. Some people may worry that such recordings will appear on the Internet. Eugene Volokh, a law professor at the UCLA School of Law, says the machines have to be used in the right way and it will be interesting to see how state laws deal with this kind of video.William Santana Li says there is a long waiting list for the robots in theUS. Workers in the company are working overtime to meet the demands of the market. At least 25 other countries are also interested in these robot security guards.1. What can this new tool do for humans?A. Make strategic decisions.B. Keep watching day and night.C. Carry heavy weapons.D. Stop crime autonomously.2. Why are some people worried about the new robots?A. Their privacy may be let out.B. The robots are very expensive.C. Robots will replace humans.D. They will be out of work soon.3. Which of the following can be the best title of the text?A. Robots Are Becoming More PopularB. Robots Contribute aLotto the WorldC. Robots Are in Great Demand NowD. Security Robots Could Help Cut CrimeBCigarettes aren’t just harmful when they’re being smoked. Even when cigarette ends go out and are cold, new research has found they continue to give off harmful chemicals in the air. In the first 24 hours alone, scientists say a used cigarette end will produce 14 percent of the nicotine (尼古丁) that an actively burning cigarette would produce.While most of these chemicals are released within a day of being put out,an analysis for the United States Food and Drug Administration (FDA) found the level of nicotine fell by just 50% five days later.“I was ly surprised,” since environmental engineer Dustin Poppendieck from the United States National Institute of Standards and Technology (NIST). “The numbers are significant and could have important impacts when cigarette ends are dealt with indoors or in cars. While much attention has been paid to the health influence of first-hand, second-hand and now third-hand smoking, it is not the case when it comes to the actual cigarette end of the matter.”To measure emissions (排放) from this forgotten thing, Poppendieck and his team placed 2,100 cigarettes that were recently put out inside a special room. Once the ends weresealed away, the team measured eight chemicals commonly produced by cigarettes, four of which the FDA have their eye on for being harmful or potentially so.After setting the room’s temperature, the researchers tested how emissions changed under certain conditions. When the air temperature of the room was higher, for instance, they noticed the ends produced these chemicals at higher rates. This finding might discourage those who want to leave ashtrays (烟灰缸) out for days ata time, especially in the heat.4. What do the researchers say about cigarette ends?A They contain little nicotine.B. They produce no nicotine five days later.C. They give off nicotine for days.D. They create as much nicotine as burning cigarettes.5. What do Poppendieck’s words suggest?A. First-hand smoking does most harm.B. The findings are within his expectation.C. Cigarettes should be dealt with indoors.D. Health influence of cigarette ends is ignored.6. Which word best describes the author’s attitude to not cleaning ashtrays for days?A. Unclear.B. Disapproving.C. Unconcerned.D. Puzzled.7. What can be the best title for the text?A. Used Cigarette Ends Release Harmful ChemicalsB. Cigarettes Are More Harmful While Being SmokedC. Research Found Reasons For Cigarette Ends’ HarmD. Cigarette Ends Produce More Chemicals in the HeatCAs a 51-year-old first-aid responder since 1984, Jeffrey never knows what type of situation he might walk into, or who he'll meet along the wayTen years into the job, Jeffrey received a call that reported that a man in his early 30s had fallen down in the Mall of America. When Jeffrey and his partner arrived at the scene, they found the young male face down on the ground. He had gone unconscious, making weak attempts to breathe. His wife stood beside him holding their small son in horror. They quickly rushed to calm the man to keep him under control and offer necessary first aid. After Jeffrey dropped the patient off at the neighboring hospital, he thought about the man and his family for a long time.Jeffrey thought he had experienced everything under the sun until one random visit to Office Max three years ago, where he met a man repeatedly walking back and forth while staring at him. As it turned out, the man was the patient he had saved 20 years earlier."You gave me 20 years more than I ever thought I'd have," the man said. He thanked Jeffrey repeatedly and told him he had someone he wanted him to meet. He stepped around the corner and reappeared with a 20-something-year-old man. Jeffrey instantly knew that it was the son he had seen standing by his mother allthose years ago"That day changed my life," Jeffrey said. "Before that, everything was about work…When I talk to my beginner-training class, I tell them you never know the effect you can have on someone's life."8. What did Jeffrey do with the young man?A. He cured the man at the scene.B. He took care of the man's wife and son.C. He only sent the man to hospital.D. He did what was needed9. What did Jeffrey think of the encounter with the man at Office Max?A. It was a common routine.B. It was troublesomeC. It was unbelievableD. It was a dangerous situation.10. Why was the man thankful to Jeffrey?A. Jeffrey helped bring up his little sonB. Jeffrey donated to support his family.C. Jeffrey's help gave him the present happy life.D. Jeffrey's kindness taught his son to be a new doctor.11. How did the meeting change Jeffrey's life?A. He was rewarded with much moneyB. He changed his attitude to his job.C. He got a promotion to be a team leader.D. He took up teaching work to train newcomers.DWho is a genius? This question has greatly interested humankind for centuries.Let's state clearly: Einstein was a genius. His face is almost the international symbol for genius. But we want to go beyond one man and explore the nature of genius itself. Why is it that some people are so much more intelligent or creative than the rest of us? And who are they?In the sciences and arts, those praised as geniuses were most often white men, of European origin. Perhaps this is not a surprise. It's said that history is written by the victors, and those victors set the standards for admission to the genius club. When contributions were made by geniuses outside the club—women, or people of a different color1 or belief—they were unacknowledged and rejected by others.A study recently published bySciencefound that as young as age six, girls are less likely than boys to say that members of their gender(性别)are “really, really smart.” Even worse, the study found that girls act on that belief:Around age six they start to avoid activities said to be for children who are “really, really smart.” Can our planet afford to have any great thinkers become discouraged and give up? It doesn't take a genius to know the answer: ly not.Here's the good news. In a wired world with constant global communication, we're all positioned to see flashes of genius wherever they appear. And the more we look, the more we will see that social factors(因素)like gender, race, and class do not determine the appearance of genius. As a writer says, future geniuses come from those with “intelligence, creativity, perseverance(毅力), and simple good fortune, who are able to change the world.”12. Whatdoes the author think of victors' standards for joining the genius club?A. They're unfair.B. They're conservative.C. They're objective.D. They're strict.13. What can we infer about girls from the study inScience?A. They think themselves smart.B. They look up to great thinkers.C. They see gender differences earlier than boys.D. They are likely to be influenced by social beliefs14. Why are more geniuses known to the public?A. Improved global communication.B. Less discrimination against women.C.Acceptance of victors' concepts.D. Changes in people's social positions.15. What is the best title for the text?A. Geniuses Think AlikeB. Genius Takes Many FormsC. Genius and IntelligenceD. Genius and Luck第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

云师大附中202届高考适应性月考（二）英语试题及答案秘密★启用前云南师大附中2021届高考适应性月考卷（二）英语试卷注意事项：1. 答题前，考生务必用黑色碳素笔将自己的姓名、准考证号、考场号、座位号在答题卡上填写清楚。

2. 每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑。

如需改动，用橡皮擦干净后，再选涂其他答案标号。

在试题卷上作答无效。

3. 考试结束后，请将本试卷和答题卡一并交回。

满分150分，考试用时120分钟。

第一部分听力（共两节，满分30分）注意，听力部分答题时请先将答案标在试卷上，听力部分结束前你将有两分钟的时间将答案转涂到答题卡上。

第一节（共5小题；每题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 is the man's wife?A. A writer.B. A housewife.C. A babysitter.2. What are the speakers talking about?A. The final exam.B. The holiday plan.C. Sights in Europe.3. What is the woman asking about?A. Moving tips.B. Good movers.C. Moving plans.4. How can we describe the lecturer?A. He is hard-working.B. He is well-off.C. He is honest.5. How did the woman feel about her son's return?A. Excited.B. Regretted.C. Shocked.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。

语文试卷注意事项：1.答题前，考生务必用黑色碳素笔将自己的姓名、准考证号、考场号、座位号在答题卡上填写清楚。

2.每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑。

如需改动，用橡皮擦干净后，弄选涂其他答案标号。

在试题卷上作答无效。

3.考试结束后，请将本试卷和答题卡一并交回。

满分150分，考试用时150分钟。

一、现代文阅读（35分）（一）论述类文本阅读（9分，每小题3分）阅读下面的文字，完成1-3题。

人工智能，新起点上再发力去年出尽风头让人惊叹的谷歌围棋人工智能“阿尔法狗（AlphaGo）”，5月将来到中国，在浙江乌镇与世界排名第一的中国棋手柯洁上演人机大战。

黑白子此起彼落之间，柯洁探寻的是已有几千年历史的围棋“真理”，而对“阿尔法狗”和它的发明者来说，比赢得比赛更重要的，是寻找人工智能的科学真理。

人工智能称得上是当前科技界和互联网行业最为热门的话题。

无论将其称作“下一个风口”“最强有力的创新加速器”“驱动未来的动力”，还是关于它会不会比人更聪明甚至取代人的各种争论，都在说明，人工智能又一次迎来了黄金发展期。

与以往几十年不同的是，这次人工智能的高潮，是伴随着生活和工作的应用而来，它是科技进步的水到渠成，也嵌入了十分广泛的生活场景。

因此也有科学家认为，“我们或许是和人工智能真正共同生活的第一代人”。

对大众来说，人工智能充满着科幻色彩；对科学家来说，人工智能可能是最受内心驱动、最具理想色彩的一门科学。

从1956年的美国达特茅斯会议算起，明确提出人工智能的概念并开始科学上的研究，到现在已有61年的历史，并经历过至少两个“冬天”。

一直到上世纪90年代，人工智能仍然走不出实验室。

人工智能遭遇的技术瓶颈，一方面有着时代的限制，另一方面也是由于人们对它的期待太高，一直梦想着的是创造出类似科幻电影《人工智能》中那个小机器人的形象——会找寻自我、探索人性，想成为一个真正意义上的人。

这也是一些人对人工智能既向往又恐惧的原因之一。

3 - 05 3 2 36 2 1⎛ 1⎫ 云南师大附中 2021 届高考适应性月考卷（二） 理科数学参考答案一、选择题（本大题共 12 小题，每小题 5 分，共 60 分）题号 1 2 3 4 5 6 7 8 9 10 11 12 答案 CBBADDBAADCC【解析】1． A = {x | x ≥ 0}，B = {x | x ≥ 3或x ≤ - 1} ，∴ A B = {x | x ≥ 3}，故选 C ． 2． z = 1 - 2i = - 1 - 3 i ， z =- 1 + 3i ，故选 B ．1 + i2 2 2 23．对于∀x ∈ R ，x 2 + ax + a ≥ 0 成立是真命题，∴ ∆ = a 2 - 4a ≤0 ，即0≤a ≤4 ，故选 B ． 4．由题意可知输出结果为 S = -1 + 2 - 3 + 4 - ⋅ ⋅ ⋅ + 8 = 4 ，故选 A ．5．∵ k l k l = -2 - m= -1，∴ m = -5 ，故选 D ．6． ⎛ ax + 1 ⎫6的展开式通项为Tr = C r (ax )6-r = C r a 6-r x 6-3r，令6 - 3r = 0 ，则有 r = 2 ， x 2 ⎪ r +1 6x 2 ⎪6 ⎝ ⎭ ⎝ ⎭∴ C 2a 4 = 15 ，即 a 4 = 1 ，解得 a = ± 1，故选 D ．616 16 27．由题意可得函数 g (x ) 的解析式为 g (x ) = 2 cos ⎡ω ⎛ x - π ⎫ + π ⎤= 2 cos ω x ，函数 g (x ) 的一个⎢ 4ω ⎪ 4 ⎥⎣ ⎝⎭ ⎦ 单调递减区间是 ⎡0 π ⎤⎡ π ⎤ ⎡ π ⎤ ⎡ π ⎤⎢ ， ⎥ ，若函数 y = g (x ) 在区间 ⎢0， ⎥ 上为减函数，则 ⎢0， ⎥ ⊆ ⎢0， ⎥ ，⎣ ω ⎦ ⎣ 3 ⎦ ⎣ 3 ⎦ ⎣ ω ⎦只要 π ≥ π，∴ ω≤3 ，则ω 的最大值为3 ，故选 B ．ω 38．由三视图知：几何体为四棱锥，且四棱锥的一条侧棱与底面垂直，如图 1，PA ⊥ 平面 ABCD ，PA = 2 ，AB = 2 ，AD = 4 ，BC = 2 ，经计算，PD = 2 ，PC = 2 ， DC = 2 ， ∴ PC ⊥ CD ， ∴ S = 1⨯ 2 ⨯ 2 = 2 ，S = 1 ⨯ 2 ⨯ 4 = 4 ，S = 1⨯ 2 ⨯ 2 = 2 △PAB ， 2 图 1 S = 1 ⨯ 2 2 ⨯ 2 = 2 ， △PAD 2 △PBC 2 △PCD2S = 1⨯ (2 + 4) ⨯ 2 = 6 ，∴ S = 12 + 2 + 2 ，故选 A ．ABCD 2表- 3 2 2 67 CA CB CO9．设△ABC 外接圆半径为 r ，三棱锥外接球半径为 R ，∵ AB = 2，AC = 3，∠BAC = 60︒ ，∴ BC 2 = AB 2 + AC 2 - 2AB AC cos 60︒ = 22 + 32 - 2 ⨯ 2 ⨯ 3⨯ 1= 7 ，∴ BC =，∴2r = 2BC =sin60︒7 = 2 21 ，∴ r = 21 ，由题意知， PA ⊥ 平面 ABC ，则将三棱锥补成三棱柱可得，3 3 2⎛ PA ⎫2 3 21 1010 40 R 2 = ⎪+ r 2 = 1 + = ，∴ S = 4πR 2= 4π ⨯ = π ，故选 A ． ⎝ 2 ⎭9 3 3 310 ．设 | PF 1 |= r 1，| PF 2 |= r 2 ，由椭圆的定义得： r 1 + r 2 = 2a ， ∵ △F 1PF 2 的三条边|PF 2|，| PF 1 |，| F 1F 2 | 成等差数列， ∴ 2r 1 = 2c + r 2 ，联立 r 1 + r 2 = 2a ， 2r 1 = 2c + r 2 ，解得 r = 2a + 2c ，r = 4a - 2c，由余弦定理得 ：(2c )2 = r 2 + r 2 - 2r rcos 60︒ ，将1 32 32a + 2c 4a - 2c 1 2 1 2⎛ 2a + 2c ⎫2r = ，r = 代入 (2c )2 = r 2 + r 2 - 2r r cos 60︒ 可得， 4c 2 = +1 3 231 2 1 2 3 ⎪ ⎝ ⎭⎛ 4a - 2c ⎫22a + 2c 4a - 2c 1 c ⎪- 2 ，整理得 ： 2c 2 + ac - a 2= 0 ，由 e = ，得 ⎝ 3 ⎭3 3 2 a2e 2 + e - 1 = 0 ，解得： e = 1或e = -1 （舍去），故选 D ．2 11 ．对于任意的 m ，n ∈ ⎡ 1 ，2⎤ ，都有 f (m ) ≥ g (n ) 成立， 等价于在 ⎡ 1 ，2⎤，函 数 ⎢⎣ 2 ⎥⎦⎢⎣ 2 ⎥⎦f (x ) ≥g (x ) ，g '(x ) = 3x 2- 4x =⎛ - 4 ⎫ ，g (x ) 在⎡ 1 4 ⎤ 上单调递减，在⎛ 4，2⎤ 上min max3x x 3 ⎪ ⎢ ， ⎥ ⎥ ⎝ ⎭ 单调递增，且 - 11 = g ⎛ 1 ⎫ < g (2) = -1 ， ∴ g (x ) ⎣ 2 3 ⎦= g (2) = -1 ⎝ 3 ⎦ ．在 ⎡ 1 ，2⎤上，82 ⎪ max⎢ 2 ⎥f (x ) = 2ax + x ln x ≥ -1 ⎝ ⎭ 恒成立， 等价于 2a ≥-x ln x - 1 = -ln x - 1x x ⎣ ⎦恒成立．设 h (x ) = -ln x - 1 ， h '(x ) =- 1 + 1 = 1 - x ， h (x ) 在⎡ 1 ， ⎤上单调递增，在 (1，2] 上单调递x x x 2 x 2 ⎢⎣ 21 减，所以 h (x )max = h (1) = -1，所以 a ≥ - 1 ，故选 C ．22 12．因为CA CB = (CO + OA ) (CO + OB ) = CO+ CO (OA + OB ) + OA OB ，由于圆O 的半径为 2 ， AB 是圆 O 的一条直径，所以 OA + OB = 0 ， OA OB = 2 ⨯ 2 ⨯ (-1) = -4 ，又∠POQ = 60︒ ，⎥ ⎦=2 2-4 =[(λ-1)OP +λOQ]2 -4 =(λ-1)2 OP+2(λ-1)λ OP OQ所以2⎡ ⎛ 1 ⎫2 3 ⎤ 1 +λ 2 OQ - 4 = 4(3λ 2 - 3λ + 1) - 4 = 4(3λ 2- 3λ) = 4 ⎢3 λ - ⎪ - ⎥ ，所以，当λ = 时，⎡ ⎛ 1 ⎫2 3 ⎤3⎣⎢ ⎝ 2 ⎭ ⎛ 3 ⎫ 4 ⎥⎦ 2 ⎢3 λ - 2 ⎪ - 4 ⎥ = - 4 ，故CA CB 的最小值为4 ⨯ - 4 ⎪ = -3 ，故选 C ．⎣⎢ ⎝ ⎭ ⎥⎦min⎝ ⎭ 二、填空题（本大题共 4 小题，每小题 5 分，共 20 分）题号13 14 15 16答案-148513⎡⎢ ln 2， 1 ⎫⎣6 3e ⎪ ⎭【解析】13．画出不等式组表示的可行域知， z = 2x + 3y 的最小值为 -14 ．14． a n +1 =3S n +1①， a n = 3S n -1 +1(n ≥2) ②，① - ②得： a n +1 = 4a n (n ≥2)，又a 1 =1，a 2 = 3a 1 +1 = 4，∴数列{a n } 是首项为 1，公比为4 的等比数列，∴ S 4 = 1 + 4 + 16 + 64 = 85 ．15．依题意知，平面区域 D 1 是一个边长为2 的正方形区域（包括边12 ⎛ 13 ⎫ 界），其面积为4 ，D = (1 - x )d x = x - x = 4，如图 2，2 ⎰-13 ⎪ 3⎝ ⎭ -14 点 M 恰好取自区域 D 的概率 P = 3 = 1 ．24 3图 216．由 g (x ) =| f (x ) | -3ax - 3a = 0 ，得| f (x ) |= 3ax + 3a = 3a (x + 1) ，设 y = 3a (x + 1) ，则直线过定点(-1，0) ，作出函数| f (x ) | 的图象（图象省略）．两函数图象有三个交点. 当3a ≤0 时，不满足条件；当3a > 0 时，当直线 y = 3a (x + 1) 经过点 (3，ln 4) 时，此时两函数图象有3 个交点，此时3a = ln 4 ，a = ln 2 ；当直线 y = 3a (x + 1) 与 y = ln(x + 1) 相切时，有两个交点，此时函数的4 6导数 f '(x ) = 1 ，设切点坐标为(m ，n ) ，则 n = ln(m + 1) ，切线的斜率为 f '(m ) = 1，x + 1 m + 1则切线方程为 y - ln(m + 1) =1 m + 1 (x - m ) ，即 y = 1 m + 1 x - mm + 1 + ln(m + 1) ，∵ 3a =1 m + 1且3a =- m + ln(m + 1) ，∴ 1 = - m + ln(m + 1) ，即 1 + m= ln(m + 1) = 1 ， 则m + 1 m + 1 m + 1 m + 1 m + 1m + 1 = e ，即 m = e - 1 ，则3a = 1 = 1 ，∴ a = 1，∴要使两个函数图象有3 个交点，则 ln 2 ≤a < 1 ．m + 1 e 3e6 3e 13 3 C 2 66 6 三、解答题（共 70 分．解答应写出文字说明，证明过程或演算步骤） 17．（本小题满分 12 分）解：（Ⅰ）因为(2b - c ) cos A - a cos C = 0 ，所以 2b cos A - c cos A - a cos C = 0 ，由正弦定理得 2sin B cos A - sin C cos A - sin A cos C = 0 ， 即2sin B cos A - sin( A + C ) = 0 ，又 A + C = π - B ，所以sin( A + C ) = sin B ， 所以sin B (2 cos A -1) = 0 ，在△ABC 中，sin B ≠ 0 ，所以2 cos A -1 = 0 ，所以 A = π．…………………………（6 分）3 （Ⅱ）由余弦定理得： a 2 = b 2 + c 2 - 2bc cos A = b 2 + c 2 - bc ， ∴4 ≥ 2bc - bc = bc ，∴ S = 1 bc sin A = 3 bc ≤ 3 ⨯ 4 = ，2 4 4当且仅当b = c 时“ = ”成立，此时△ABC 为等边三角形，∴△ABC 的面积 S 的最大值为 ．…………………………………………………（12 分）18．（本小题满分 12 分）解：（Ⅰ） 2 ⨯ 2 列联表补充如下：喜欢数学课程不喜欢数学课程合计 男生40 30 70 女生 35 15 50 合计75 45 120……………………………………………………………………………………………（3 分）2 120 ⨯ (40 ⨯15 - 30 ⨯ 35)2由题意得 K =≈ 2.057 ，………………………………………（5 分） 70 ⨯ 50 ⨯ 75 ⨯ 45 ∵ 2.057 < 2.706 ，∴没有90% 的把握认为喜欢数学课程与否与性别有关．…………（6 分）（Ⅱ）用分层抽样的方法抽取时，抽取比例是 6 = 2，则抽取男生30 ⨯ 2 15 = 4 人，抽取女生15 ⨯ 2 15 45 15= 2 人，所以 X 的分布列服从参数 N = 6，M = 2，n = 2 的超几何分布，……………………（8 分）C i C 2-iX 的所有可能取值为0，1，2 ，其中 P ( X = i ) = 2 4(i = 0，1，2) ．6C 0C 2 6 C 1 C 1 8 C 2C 0 1 由公式可得 P ( X = 0) = 2 4 = ， P ( X = 1) = 2 4 = ， P ( X = 2) = 2 4= ，2 15 2 15 2 15…………………………………………………………………………………………（10 分）C C C3 所以 X 的分布列为：X 0 1 2 P6 158 151 15所以 X 的数学期望为 E ( X ) = 0 ⨯ 6 + 1⨯ 8 + 2 ⨯ 1 = 2．……………………………（12 分）15 15 15 3 19．（本小题满分 12 分）（Ⅰ）证明：由已知，得 AC = ∵ BC = AD = 2 ， AB = 4 ，又 BC 2 + AC 2 = AB 2 ，∴ BC ⊥ AC ． 又 PA ⊥ 底面 ABCD ， BC ⊂ 平面 ABCD ， 则 PA ⊥ BC ，∵ PA ⊂ 平面 PAC ， AC ⊂ 平面 PAC ，且 PA AC = A ， ∴ BC ⊥ 平面 PAC ．= 2 ， ∵ BC ⊂ 平面 PBC ，∴平面 PBC ⊥ 平面 PAC ．………………………………………（6 分） （Ⅱ）解：以 A 为坐标原点，过点 A 作垂直于 AB 的直线为 x 轴， AB ，AP 所在直线分别为 y 轴， z 轴建立空间直角坐标系 A - xyz ，如图 3 所示． 则 A (0，0，0)，B (0，4，0)，P (0，0，3) ，因为在平行四边形 ABCD 中， AD = 2，AB = 4，∠ABC = 60︒ ， 则∠DAx = 30︒ ，∴ D ( 3，- 1，0) ．又 PE= λ(0 < λ < 1) ，知 E (0，4λ，3(1 - λ)) ． PB设平面 ADE 的法向量为 m = (x 1，y 1，z 1) ，⎧ ⎧ 图 3则⎪m AD = 0，即⎪ 3x 1 - y 1 = 0， ⎨ ⎪⎩m AE = 0， ⎨⎪⎩4λ y 1 + 3(1 - λ)z 1 = 0，⎛ 4 3λ ⎫取 x 1 = 1 ，则 m = 1， 3， ⎪ ．……………………………………………………（8 分）⎝ 3(λ - 1) ⎭设平面 PAD 的法向量为 n = (x 2，y 2，z 2 ) ，⎧⎪n AP = 0， ⎨ ⎧⎪3z 2 = 0， 即⎨ ⎪⎩n A D = 0， ⎩⎪ 3x 2 - y 2 = 0， AB 2 + BC 2 - 2 A B ⨯ BC ⨯ cos ∠ABC 则3⎝ ⎭ + = 2 ⎛ ⎫取 y 2 = 1 ，则 n = 3 ，1，0 ⎪ ．…………………………………………………………（10 分）⎝ ⎭ 若平面 ADE 与平面 PAD 所成的二面角为60︒ ，1 则cos 〈m ，n 〉 = cos 60︒ =1⨯ 3+ 3 3 ⨯1 + 0 = 1 ， 2 16λ 2 1 + 3 + 3(λ - 1)21 + 1 23⎛ λ ⎫29 = 2 ，即 λ - 1 ⎪ = 4，解得λ = 3 （舍去）或λ = 3．5于是，存在λ = 3，使平面 ADE 与平面 PAD 所成的二面角为60︒ ．………………（12 分）5 20．（本小题满分 12 分）解：由题意知函数的定义域为{x | x > 0} ， f '(x ) =- a + 1 = x - a．x x（Ⅰ）①当 a ≤0 时， f '(x ) > 0 ，所以函数 f (x ) 的单调递增区间是(0，+ ∞) ，无极值； ②当 a > 0 时，由 f '(x ) > 0 ，解得 x > a ，所以函数 f (x ) 的单调递增区间是(a ，+ ∞) ， 由 f '(x ) < 0 ，解得 x < a ，所以函数 f (x ) 的单调递减区间是(0，a ) ．所以当 x = a 时，函数 f (x ) 有极小值 f (a ) = -a ln a + a + 1 ．…………………………（6 分） （Ⅱ）由（Ⅰ）可知，①当 a ≤1 时，函数 f (x ) 在[1，e] 为增函数，∴函数 f (x ) 在[1，e] 上的最小值为 f (1) = a ln1 + 1 + 1 = 2 ，显然 2 ≠ 1 ，故不满足条件； ②当1 < a ≤e 时，函数 f (x ) 在[1，a ) 上为减函数，在[a ，e] 上为增函数，故函数 f (x ) 在[1，e] 上的最小值为 f (x ) 的极小值 f (a ) = -a ln a + a + 1=1，即a = e ，满足条件； ③当 a > e 时，函数 f (x ) 在[1，e] 为减函数，故函数 f (x ) 在[1，e] 上的最小值为 f (e) = a ln 1+ e + 1 = 1 ，即 a = e ，不满足条件．e 综上所述，存在实数 a = e ，使得函数f (x ) 在[1，e] 上的最小值为1 ．……………（12 分） 21．（本小题满分 12 分）解：（Ⅰ）设动点Q (x ，y )，A (x ，y ) ，则 N (x ，0) ，且 x 2 + y 2 = 8 ，①1又OQ = mOA + (1 - m )ON ，得 x 0 = x ，y 0 = my ，2代入①得动点Q 的轨迹方程为 x y 1 ．…………………………………………（4 分） 8 8m 24λ21 + 3(λ - 1)22 3 - 13 + = ⎨ 2 2y （Ⅱ）当 m = 时，动点Q 的轨迹曲线C 为 x y 1 ．2 8 4x 2 y 2直线l 的斜率存在，设为 k ，则直线l 的方程为 y = k (x + 4) ，代入得(1 + 2k 2 )x 2 + 16k 2 x + 32k 2 - 8 = 0 ， 由∆ = (16k 2 )2 - 4(1 + 2k 2 )(32k 2 - 8) > 0 ，+ = 1 ， 8 4解得- 2 < k < 2，②…………………………………………………………………（7 分）2 2设 E (x 1，y 1)，F (x 2，y 2 ) ，线段 EF 的中点G (x '，y ') ，' = x + x = - 8k 2 ' = ' + = 4k则 x 1 2，y k (x2 1 + 2k 24) 1 + 2k 2 ．由题设知，正方形Γ 在 y 轴左边的两边所在的直线方程分别为 y = x + 2，y = -x - 2 ，注意到点G 不可能在 y 轴右侧，则点G 在正方形Γ 内（包括边界）的条件是⎧ 4k 8k 2⎧ y '≤x ' + 2， ≤ - + 2， ⎪1 + 2k 2 1 + 2k 2 ⎨ y '≥ -x ' - 2，即⎨ 4k 8k 2⎩ ⎪ ≥ ⎪⎩1 + 2k 2 1 + 2k 2- 2， 解得 - 3 - 1≤k ≤3 - 1，此时②也成立． 2 2⎡ 3 - 1⎤于是直线l 的斜率的取值范围为 ⎢- 2 ， 2 ⎥ ．………………………………（12 分）⎣ ⎦ 22．（本小题满分 10 分）【选修 4−4：坐标系与参数方程】⎧x = 1 + 1 t ，解：（Ⅰ）直线l 的参数方程为： ⎪2 (t 为参数) ， ⎪ y = 1 +3 t ，曲线C 的直角坐标方程为： ⎪⎩x2+ 232 2= 1 ．………………………………………………（5 分） ⎧x = 1 + 1 t ，⎪ 2 x 2 2（Ⅱ）把直线l 的参数方程⎨1 3 代入曲线C 的方程 + y 3 = 1 中， ⎪ y = + t ， ⎛ 1 ⎫2⎛ 1 ⎩⎪ 2 2⎫2得 1 + t ⎪ + 3 + t ⎪ = 3 ，即10t 2 + (6 + 4)t - 5 = 0 ， ⎝ 2 ⎭ ⎝ 2 2 ⎭3⎝ ⎦ ⎣ ⎭ 设点 A ，B 所对应的参数分别为t ，t ，则t t = - 1，1 2 1 22∴ | PA | | PB |=| t | | t |=| t t |= - 1 = 1．……………………………………………（10 分）1 2 1 22 223．（本小题满分 10 分）【选修 4−5：不等式选讲】解：（Ⅰ）不等式 f (x )≤0 ，即| x - 2 | ≤| 2x + 1| ，即 x 2 - 4x + 4≤4x 2 + 4x + 1，3x 2 + 8x - 3≥ 0 ，解得 x 1x ≤ - 3 ， ≥ 或3所以不等式 f (x )≤0 的解集为⎧x x1x ≤ -⎫ ……………………………………（5 分）⎨ ≥ 或3 3⎬ .⎩⎭⎧x + 3，x < - 1 ，⎪ ⎪ （Ⅱ） f (x ) =| x - 2 | - | 2x + 1|= ⎪-3x + 1，- 1 ≤x ≤2， ⎨ ⎪⎪-x - 3，x > 2， ⎪⎩ 故 f (x ) 的最大值为 f ⎛ - 1 ⎫ = 5，2 ⎪ 2 ⎝ ⎭ 因为对于∀x ∈ R ，使 f (x ) - 2m 2≤4m 恒成立，所以 2m 2 + 4m ≥ 5，即4m 2 + 8m - 5≥ 0 ，2m 1 5⎛ 5 ⎤ ⎡ 1 ⎫解得 ≥ 2 或m ≤ - 2 ，∴ m ∈ -∞，- 2 ⎥ ⎢ 2，+ ∞⎪ ．……………………………（10 分） 2 2。

2021届云南师范大学附属中学高三英语月考试题及答案解析第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AConservation Volunteering in New ZealandWhether you are a student, professional or a retiree (退休者), anyone is welcome to make a difference and contribute to protecting some of the most beautiful islands in the world. Choose a suitable city and travelout to your conservation (保护) site to work with local people!Duration: 1-12 weeks Dates: Throughout the yearArrival day: Friday Return day: FridayRequirement: General level of fitness Age: 18+What will I be doing?Volunteer in New Zealand and enjoy conserving the environment through activities such as:·Tree planting·Walking trail construction·Protect native birds, insects, fish and penguins·Seed collection·Weed controlYou, and a group of up to 10 volunteers, will work under the guidance of a conservation team leader. Your team leader will give you regular safety instructions, inform you of the project aims and assist you with working effectively.No previous experience is necessary to join the project. All you need is a love of the environment and a fairly good level of fitness to help out!1.Who can sign up for this conservation volunteering project?A.A retired maths teacher.B.A primary school student.C.A scientist with heart disease.D.A businessman in a wheelchair.2.What can you do on the volunteer trip?A.Protect cultural sites and go shopping.B.Enjoy local sightseeing and go fishing.C.Protect weeds and build roads.D.Collect seeds and plant trees.3.From which is the text probably taken?A.A history book.B.A travel magazine.C.A research paper.D.A novel.BTo hear people talk about Internet friendships, you would think it was one giant web of cat-fishing and e-crime. While we all undoubtedly have to take measures to remain safe online, assuming every friendship or connection made on Instagram, Twitter or Facebook is cheating or insincere would be a mistake.As a woman who works in the creative industry, I have found real joy in seeking out a community I couldn’t find elsewhere, and making some great friends along the way. My first online friendship was on Twitter with my(now) best friend, during the university exam period. We exchanged study notes in dozens of direct messages, set a study date, and haven’t looked back since.Drawn to each other by similar circumstances, friendships online are similar to offline in that they tend to begin because of shared interest or common ground-maybe they’ve read the post on Instagram. Maybe they have the same taste in food or politics. Or maybe they just love memes too. If online friendships start similar to friendships offline, they grow in the same way, too. Often through mutual support: apart from calling a friend to congratulate him on that new job, you also re-tweet his jokes and praise his Instagram story.Despite my positive experiences when I tell people, most are still suspicious. Eyebrows are raised higher when I explain not only have I found a community online but have made friendships with people I meet face-to-face too. Actually, these are just as valid as other friendships, according to behavioural psychologist Jo Hemmings, who says online friendships can be real.So how do you know if people are there for the real you or just because you’re popular on Instagram? Hemmings has simple rules. She tells me “You have to equally feel comfortable that you’re getting something of each other instead of being used to enable something that isn’t friendship.”Therefore, if all a “friend” online is asking you to do is to promote their work or personal brand and rarely takes an interest in you, then there may be room to question the basis of the friendship. On that note it is worth remembering that just because someone has a lot of followers, it doesn’t necessarily mean they have lots of friends.4. What is most people’s attitude towards online friendship?A. Negative.B. Positive.C. Objective.D. Neutral.5. Why does the writer share her own experience in paragraph 2?A. To introduce the background information of the text.B. To convey the writer’s attitude and give the related example.C. To prove the likely risk for people to develop friendship online.D. To remind people of the various benefits of making friends online.6. How can online and offline friendships be deepened?A. They should be based on shared interest.B. They need to have common ground.C. They require support from each other.D. They can’t live without social media.7. According to the author, what’s the golden rule to make friends online?A. A friend to all is a friend to none.B. Without confidence there is no friendship.C. A friend without faults will never be found.D. Friendship cannot stand always on one side.COne of the most popular street food found inChinais no doubt the barbecue. A new program, called Chinese Barbecue, tells the story of this popular food cooked over hot coals on just about every street corner in cities and towns across the country. Barbecued meat is an important part of people’s nightlife.Shown on June 20, the program has had more than 25 million clicks on the video site . To find the most popular barbecue stalls (摊位), the production team travelled to more than 500 locations in 30 cities across the country. Some viewers compare Chinese Barbecue to Midnight Diner, a Japanese TV program telling stories from late night informal Japanese bars.“I’m happy to hear this comparison because Midnight Diner is a good program, and we share the same topic― night food,” Chinese Barbecue’s director Chen Yingjie said. “However, they are quite different.” He said that Midnight Diner focused more on food itself, though there was someconversation while people were eating.However, the night food scene of Chinese people means joy and a more lively atmosphere. People eating these barbecue snacks develop a feeling of connection, which can be a cure for loneliness.The barbecue, regarded as the most ordinary and common night street snack, is different from home-made food by mothers as that is a symbol of family and kinship. The barbecue is where you go to become connected to people in society. And unlike official business lunches, during which people are rather polite, the barbecue lets people relax with old friends and new friends, leaving a lasting impression of friendship.The world, as a whole, holds deep-rooted good feelings toward the barbecue, either for the taste or the warmth produced by fire. “What we should do is to present the Chinese barbecue just the way it is because with its special ingredients(食材)，ways of cooking and more importantly, the special environment and people, the world will recognize it and might fall in love with it just as we do.” said Chen.8. What do we know about Chinese Barbecue?A. It has been becoming very popular on the Internet.B. It is thought highly of by most of the foreign tourists.C. It mainly talks about the most famous Chinese food.D. It shows the color1 ful nightlife in large cities ofChina.9. In which way was Chinese Barbecue different from Midnight Diner?A. It showed more kinds of food.B. It focused more on the diners.C. It showed the eating habits.D. It was less popular than Midnight Diner.10. What is Paragraph 4 mainly about?A. The importance of the barbecue to family.B. The influence of the barbecue on people’s manners.C. The influence of the barbecue on people’s lifestyle.D. The role of the barbecue in people’s relationship.11. Why did Chen Yingjie choose the barbecue as the topic of the series?A. To show the feature of Chinese food.B. To research a special way of cooking.C. To help the world understandChina.D. To introduce the history of the barbecue.DIn a recent survey of 2000 Americans, housecleaning was shown to have some mood-boosting effects — but that doesn't mean everybody is willing to do it.The majority of respondents (受访者) said cleaning gave them a sense of accomplishment (65%) and helpedthem clear their mind (63%). Half of these adults said they are most often motivated to clean when they're happy. In fact, 63% of those surveyed find the experience of cleaning to be relaxing - even more so than getting fresh air (61%).But that's not the only reason people clean. A big 70% admitted that tidying their home was a way of putting off having to do other things, with the average procrastinator (拖延者) using that trick four times a week. The survey showed that 86% of respondents do feel on top of their housework, but the last deep clean of their kitchen happened over a week and a half ago. That's no surprise because the kitchen is most terrible of all.Conducted by OnePoll on behalf of DishFish, the survey investigated people's attitudes toward dirty dishes and how they get through tricky task. More than two-thirds of people (69%) let their dishes pile up between washings with 20% saying “always” letting them be placed in the sink, which left them feeling stressed. More than any other room, the kitchen was rated as “very difficult” to cope with. And most people enjoy cleaning their toilet or taking out the garbage more than washing dishes by hand.How do they get through it? 66% listen to music while they clean. 72% have a best-loved song that they sounded while tidying up their home, with “Uptown Funk,”“Read All About It” and “Work” being the three favorite tunes on America's cleaning playlist.12. What is the result of the survey?A. Housecleaning may contribute to a good mood to some extent.B. Housecleaning may strengthen people's willingness to volunteer.C. Housecleaning may cause anxiety and concern for some people.D. Housecleaning may improve people's motivation to other housework.13. What is the top reason why people undertake housecleaning?A. They can entertain themselves.B. They can take in fresh air.C. They get a sense of achievement.D. They can delay other things.14. What are respondents' attitudes to dirty dishes?A. Many would rather wash dishes than throw out the rubbish.B. Half are under pressure with dirty dishes lying in the sink.C. A quarter will let dirty dishes pile up after their meals.D. Most prefer cleaning their toiletto washing dishes by hand.15. What column does the text belong to?A. Feature Story.B. Family Life.C. Scientific Hotspot.D. Finance Focus.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

秘密★启用前2020届云南师大附中高考适应性月考语文试卷（二）注意事项：1.答题前,考生务必用黑色碳素笔将自己的姓名、准考证号、考场号、座位号在答题卡上填写清楚。

2.每小题选出答案后,用2B铅笔把答题卡上对应題目的答案标号涂黑。

如需改动,用橡皮擦干净后,再选涂其他答案标号。

在试题卷上作答无效。

3.考试结束后,请将本试卷和答题卡一并交回。

满分150分,考试用时150分钟。

一、现代文阅读(36分)(一)论述类文本阅读(本题共3小题,9分)阅读下面的文字,完成1～3题。

尽管不偏不倚的中和美是中国传统艺术追求的理想境界,平庸和单调却是艺术的敌人。

“中和”并不是简单的缺少变化的“同”,而是各种丰富的情感糅杂形成的冲淡含蓄的审美境界。

在整体的和谐中蕴藏着变化,变化中又体现着统一。

钱钟书充分意识到这一点,谈论文学艺术的时候始终贯穿着“和而不同”、一与多彼此统一的思想。

如果把事物比做一,那么,这个一中包含着无限的多。

古今中外的人们都曾参透过这个上帝创造世界的秘密。

为了能让我们有更清晣的认识,钱先生在《管锥编》里把古希謄和古中国放在一起,显示了他广泛搜罗中外例证的眼界和中西贯通的学术视野。

赫拉克利特(古希腊著名哲学家)说:“自然是由联合对立物造成最初和谐的……绘画在画面上混合着白色和黑色、黄色和红色的部分,从而造成与原物相似的形象。

音乐混合不同音调的高低、长音和短音,从而造成一个和谐的曲调。

”“结合物既是整个的,又不是整个的,既是协调的,又不是协调的,既是和谐的,又不是和谐的。

”无独有偶,中国古代也有过类似的论述,《左传》里说,声音妤比味道,有“一气、二体、三类、四物、五声、六律、七音、八风、九歌”彼此配合而成,“清浊、大小、长短、疾徐、哀乐、刚柔、迟速、高下”融合在一起,如果只有琴或瑟的一种声音,谁还愿意去听呢?古希腊的诗人还概括过一个有趣的现象,“争斗”有两种情况,一种是两善相争,一种是两恶相争,前者可以互补互利,后者则是互相残杀。

云南师大附中2021届高考适应性月考卷（二）英语参考答案第一部分听力（共两节，满分30分）1~5 ABBAA 6~10 CACBB 11~15 CCBBC 16~20 CAACB第二部分阅读理解（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）21~25 BABAD 26~30 DBBAC 31~35 CADCC第二节（共5小题；每小题2分，满分10分）36~40 EAFBG第三部分语言知识运用（共两节，满分45分）第一节（共20小题；每小题1.5分，满分30分）41~45 ACBDB 46~50 ABDCC 51~55 DABCA 56~60 DCABD第二节（共10小题；每小题1.5分，满分15分）61．However/Nevertheless 62．habits 63．Unlike 64．to breathe 65．a 66．swimming 67．which 68．partially 69．are equipped 70．larger 第四部分写作（共两节，满分35分）第一节短文改错（共10小题；每小题1分，满分10分）Tens of million of people carry their smartphones with them everywhere they go，included the①millions ②including bathroom．Our phones are always within their grasp，which means they are also constantly being③ourexposed to however bacteria we may have accumulated on our hands．Unfortunately，a new survey④whateverof 1,200 Americans and their cleaning habits find that 25 percent of them have never cleaned their⑤findsphones．Researchers estimate that a regularly use smartphone that has never been cleaned may hold英语参考答案·第1页（共11页）⑥usedup more bacteria than an entire household bathroom．According to∧survey conducted by the company ⑦⑧the/aVital Vio，that is common for smartphone users to come into contact with various forms of bacteria.⑨itSuddenly，eating a sandwich in your bathroom doesn’t sound so badly.⑩bad第二节书面表达（满分25分）【参考范文】NoticeAn exhibition of Han clothes will be held from 2：30 to 5：30 this Friday afternoon in the school hall．The purpose of this activity is to enrich students’knowledge of Han clothes and promote Chinese traditional culture．The exhibition will present Han clothes with different designs and decorations by means of some pictures and videos．With the help of craftsmen，visitors can also gain some hands-on experience of making a Han dress．What’s more，a fashion show on the stage can be a visual feast for everyone to appreciate the special charm of Han clothes．There will be a lot of fun．Everyone is welcome．Hope you will enjoy it!The Students’ Union【解析】第二部分阅读理解A【语篇导读】本文是应用文。

理科数学参考答案·第1页（共10页）云南师大附中2021届高考适应性月考卷（二）理科数学参考答案一、选择题（本大题共12小题，每小题5分，共60分）题号 1 2 3 4 5 6 7 8 9 10 11 12 答案D C A C B A A B C B D B 【解析】1．由题意知，[35)A =，，(46)B =，，所以(45)A B = ，，故选D ． 2．由题意知，iπe 1cos πisin π10+=++=，故选C ．3．原式cos 45cos15sin 45sin15cos(4515)cos302=︒︒+︒︒=︒-︒=︒=，故选A ． 4．由题意知，双曲线的右焦点为0)F ，双曲线的渐近线方程为2y x =±，即2y - 0=，所以点0)F 到渐近线的距离d ==，故选C ．5．由题意可知，九个儿子的年龄可以看成以老大的年龄1a 为首项，公差为3-的等差数列，所以1989(3)2072a ⨯+⨯-=，解得135a =，故选B ． 6．由茎叶图可知，甲年级的平均分主要集中在70多分，而且比较集中，而乙主要集中在80分以上，但是比较分散，故选A ．7．如图1，()()()CA CB CO OA CO OB CO OA =++=+2222()()()||||3CO OA CO OA CO OA -=-=-=，故选A ．8．圆的标准方程为22(3)(4)25x y -+-=，即圆是以(34)M ，为圆心，5为半径的圆，且由22(03)(44)925-+-=<，即点(04)P ，在圆内，则最短的弦是以(04)P ，为中点的弦，所以22592AC ⎛⎫=+ ⎪⎝⎭，所以8AC =，过(04)P ，最长的弦BD 为直径，所以10BD =，且AC BD ⊥，故而1402ABCD S AC BD == ，故选B ．图1理科数学参考答案·第2页（共10页）9．如图2，该正四面体可以看成边长为1的正方体六个面对角线组成的正四面体ABCD ，所以正四面体ABCD 的外接球，即为边长为1的，则24π3πS ==⎝⎭，故选C ．10．由2()sin cos f x x x =，所以22()sin()cos ()sin cos ()f x x x x x f x -=--=-=-，所以()f x 是奇函数；22(2π)sin(2π)cos (2π)sin cos ()f x x x x x f x +=++==，所以()f x 又是周期函数；22(π)sin(π)cos (π)sin cos ()f x x x x x f x -=--==，所以()f x 关于直线π2x =对称；22(2π)sin(2π)cos (2π)sin cos ()f x x x x x f x -=--=-=-，所以()f x 关于点(π0)，对称，即选项A ，C ，D 正确；又222222()(sin cos )sin (1sin )(1sin )f x x x x x x ==--32222sin (1sin )(1sin )12422327x x x --⎛⎫== ⎪⎝⎭ ≤，当且仅当sin x =，max ()f x =，故B 选项错误，故选B ．11．由题意知，令直线2px my =+，11()A x y ，，22()B x y ，，与抛物线C ：22y px =联立方程，消去x 得2220y pmy p --=，由韦达定理知：122y y pm +=，212y y p =-，如图3所示，过A ，B 分别作抛物线准线的垂线，垂足分别为A '，B '，记AB 的中点为I ，过I作抛物线准线的垂线，垂足为I '，由||||AB AA '=||2||BB II ''+=，所以以AB 为直径的圆与抛物线C 的准线相切，故A 正确；由12x x =212224p p p my my ⎛⎫⎛⎫++=⎪⎪⎝⎭⎝⎭，所以12122121212121111||||()222224x x p x x p p p p p p p AF BF x x x x x x x x +++++=+===⎛⎫⎛⎫+++++++ ⎪⎪⎝⎭⎝⎭12122212122()()2424x x p x x p p p p p p x x p x x ++++==+++++，故B 正确；由图，抛物线在第一象限的解析图2图3理科数学参考答案·第3页（共10页）式为y =，所以y '=，所以过点B抛物线的切线的斜率为1k =同理过点A抛物线的切线的斜率为2k =1212p k k =-=- ，所以两切线垂直，故C 正确；由1πtan 2m αα⎛⎫=≠ ⎪⎝⎭，所以12||||||AB AF BF x x p =+=++= 22122212()2222(1)21tan sin p m y y p pm p p m p αα⎛⎫++=+=+=+= ⎪⎝⎭； 如图，作OE 垂直AB 于E ，则22112sin 22sin 22sin AOBp p p S AB OE ααα=== △ ，当π2α=时，经检验AOB S =△22sin p α亦成立，故D 错误，故选D ． 12．由2ln 2ln 4ln e 1=>=，故①正确；由2ln2ln e ln 2e 2e>⇔>，考察函数ln x y x =，21ln xy x -'=，所以当(0e)x ∈，时，0y '>，即y 在(0e)，上单调递增，当(e )x ∈+∞，时，0y '<，即y在(e )+∞，上单调递减，所以e x =时，y 取到最大值1e ，所以ln2ln e 2e <，故②错误；令0.2log 0.4a =，2log 0.4b =，所以0.40.40.411log 0.2log 2log 0.41a b+=+==，所以a b ab +=，即0.220.22log 0.4log 0.4log 0.4log 0.4+= ，故③正确；由4372401219713=>=，所以133log 74>，由41328561=<32979131=，所以313log 134<，故④错误，故选B ． 二、填空题（本大题共4小题，每小题5分，共20分）【解析】13．约束条件所表示的线性区域，如图4所示，由00y y x x -=-，即区域中的点与原点O 的斜率，所以OA 的斜率即为yx的最大值，又有点A 的坐标为(32)，，则y x的最大值为23．图4理科数学参考答案·第4页（共10页）14．由3n n x x ⎛⎫- ⎪⎝⎭展开式的二项式系数为2n ，即264n =，所以6n =，则二项式为62x x ⎛⎫- ⎪⎝⎭，故展开式中的常数项为33362C 160x x ⎛⎫-=- ⎪⎝⎭．15．如图5甲，将等边ACD '△沿CD '向后旋转到与面A BCD ''共面，得到等边1A CD '△，则AP BP +的最小值即为图乙中线段1A B 的长，取A B '的中点I ，由题意知：等边ACD '△的边长为，A BCD ''是以1BC =，A B '=的矩形，所以1A B == 16．由题意知，2221cos ()2AB AC bc A b c a ==+- ，同理，2221()2BA BC a c b =+- ，2221()2CA CB a b c =+- ，故由已知，2222222222()3()b c a a c b a b c +-++-=+-，即22223a b c +=，由22222221(2)3cos 22a b a b a b c C ab ab+-++-==363a b b a =+=≥，所以sin 3C =，当且仅当::a b c =时取等号，所以sin C 的最大值是3． 三、解答题（共70分．解答应写出文字说明，证明过程或演算步骤）17．（本小题满分12分）解：（1）补充22⨯的列联表如下：更擅长理科 其他 合计男生 22 33 55 女生 9 36 45 合计3169100图5理科数学参考答案·第5页（共10页）所以22100(2236933)100334.628 3.841554531693123K ⨯⨯-⨯⨯==≈>⨯⨯⨯⨯，所以有95%的把握认为文理科偏向与性别有关．…………………………………………………（6分）（2）由题意可知，选取的5人中，有2人更擅长理科，3人不更擅长理科， 所以X 的可能取值为0，1，2，故022325C C 3(0)C 10P X ===，112325C C 3(1)C 5P X ===，202325C C 1(0)C 10P X ===， 所以X 的分布列为所以3314()012105105E X =⨯+⨯+⨯=． ……………………………（12分）18．（本小题满分12分）（1）证明：由等腰梯形22AB CD AD===，则60ABC ∠=︒， 又2AB BC =，所以AC BC ⊥①， 又PC BC ==，PB = 则222CB CP PB +=， 所以BC CP ⊥②， 又AC CP C = ③，由①②③知，BC ⊥平面APC ，所以平面APC ⊥平面ABC ．…………………………………………………………（6分）（2）解：如图6，取AB 的中点E ，连接DE ，CE ，AC ， 则AECD 为菱形，且60DAE ∠=︒， 则AC DE ⊥，记垂足为O ， 由（1）知，平面APC ⊥平面ABC ， 又PO AC ⊥，所以PO ⊥平面ABC ，同理，EO ⊥平面APC ，所以OA ，OE ，OP 两两垂直，图6理科数学参考答案·第6页（共10页）如图7，建立分别以OA ，OE ，OP 为x ，y ，z 轴的空间直角坐标系， 则6AC =，DO =，所以(300)A ，，，(30)B -，，(300)C -，，，(00P ，，所以(3BP =- ，，(60)BA =-，，(00)BC =-，，设平面ABP 的法向量为1111()n x y z =，，， 所以1100BA n BP n ⎧=⎪⎨=⎪⎩，，即111116030x x ⎧-=⎪⎨-+=⎪⎩，，令1y =，得111x z =⎧⎪⎨=⎪⎩，所以平面ABP的一个法向量为1(1n =； 设平面CBP 的法向量为2222()n x y z =，，，所以2200BC n BP n ⎧=⎪⎨=⎪⎩，，即2222030x ⎧-=⎪⎨-+=⎪⎩，，令2z =2210x y =-⎧⎨=⎩，， 所以平面CBP的一个法向量为2(10n =-，； 令二面角A PB C --为θ，有题意知θ为钝角，所以1212||cos 7||||n n n n θ=-==-， 所以二面角A PB C --的余弦值为7-． ………………………………（12分）19．（本小题满分12分） （1）解：由11a =，23a =， 所以123121(22)5a a a a a +=++=+，234231(32)7a a a a a +=++=+．图7理科数学参考答案·第7页（共10页）猜想：21n a n =-，证明：当2n =时，由11a =，23a =，故成立； 假设(2)n k k =≥时成立，即21k a k =-， 所以1111(2)212(1)1k k k k k a a a k k k a a -+-+=++=+=+-+，即当1n k =+时成立， 综上所述，21n a n =-． …………………………………………………（6分）（2）证明：由（1）知，2241(1)n a n =+，所以22222212444111(1)(1)(1)12n a a a n +++=++++++ (222111)121311n <++++---…11111324(1)(1)n n =++++⨯⨯-+… 111111111111232435211n n n n ⎛⎫=+-+-+-++-+- ⎪--+⎝⎭…11117112214n n ⎛⎫=++--< ⎪+⎝⎭，证毕．…………………………………（12分）20．（本小题满分12分） 解：（1）由题意知：1(2)2y k x x =≠-+，2(2)2yk x x =≠-， 由1234k k =- ，即3(2)224y y x x x =-≠±+- ， 整理得点()P x y ，的轨迹C 的方程为221(2)43x y x +=≠±．…………………………………………………………（4分）（2）假设在x 轴上存在点0(0)Q x ，，使得QA QB为定值．当直线l 的斜率存在时，设直线l 的方程为(1)(0)y k x k =-≠， 联立方程22143(1)x y y k x ⎧+=⎪⎨⎪=-⎩，，消去y 得2222(34)84120k x k x k +-+-=，理科数学参考答案·第8页（共10页）令11()A x y ，，22()B x y ，，则2122834k x x k +=+，212241234k x x k -=+ ，由101()QA x x y =- ，，202()QB x x y =-，，所以2102012102012()()()()(1)(1)QA QB x x x x y y x x x x k x x =--+=--+--2222120120(1)()()k x x x k x x k x =+-++++22002(58)1234x k x k-+-=++， 将0x 看成常数，要使得上式为定值，需满足05816x +=，即0118x =， 此时13564QA QB =- ；当直线l 的斜率不存在时，可得312A ⎛⎫ ⎪⎝⎭，，312B ⎛⎫- ⎪⎝⎭，，1108Q ⎛⎫⎪⎝⎭，所以3382QA ⎛⎫=- ⎪⎝⎭ ，，3382QB ⎛⎫=-- ⎪⎝⎭ ，，13564QA QB =- ，综上所述，存在1108Q ⎛⎫⎪⎝⎭，，使得QA QB 为定值．…………………………………………………………（12分）21．（本小题满分12分）解：（1）有题意知，()e e(ln )x h x x x x =-+，(0)x ∈+∞，， 所以，1e ()(1)e e 1(1)e x x h x x x x x ⎛⎫⎛⎫'=+-+=+- ⎪ ⎪⎝⎭⎝⎭，所以，当(01)x ∈，，()0h x '<，即()h x 在(01)，上单调递减， 当(1)x ∈+∞，，()0h x '>，即()h x 在(1)+∞，上单调递增， 故()(1)0h x h =≥，所以()h x 的最小值为0．…………………………………………………………（4分）（2）原不等式等价于e (ln )(2)1x x x x b x -+-+≥， 即e ln 1x x x x bx +--≥，在(0)x ∈+∞，上恒成立理科数学参考答案·第9页（共10页）等价于e ln 1x x x x b x +--≥，在(0)x ∈+∞，上恒成立．令e ln 1()x x x x t x x +--=，(0)x ∈+∞，，所以22e ln ()x x xt x x +'=，令2()e ln x x x x ϕ=+，则()x ϕ为(0)+∞，上的增函数， 又当0x →时，()x ϕ→-∞，(1)e 0ϕ=>，所以()x ϕ在(01)，存在唯一的零点0x ，即0200e ln 0x x x +=，由0200e ln 0x x x +=⇔001ln 0000ln 1e ln e x x x x x x ⎛⎫=-= ⎪⎝⎭，又有e x y x =在(0)+∞，上单调递增， 所以0001lnln x x x ==-，001e x x =， 所以0000min00e ln 1[()]()2x x x x t x t x x +--===，所以2b ≤． …………………………………………………………（12分）22．（本小题满分10分）【选修4−4：坐标系与参数方程】 解：（1）由222x y ρ=+，所以曲线C 的直角坐标方程为224x y +=，由2x t y =--⎧⎪⎨=⎪⎩，(t 为参数)，消去t 得直线l0y +-=．…………………………………………………………（5分）（2）由题意知，关于点(2P -，的直线l的参数方程为22t x y ⎧=--⎪⎪⎨⎪=+⎪⎩，，(t 为参数)，代入曲线C 的直角坐标方程得211270t t ++=，理科数学参考答案·第10页（共10页）又121108130∆=-=>，所以方程有两个不同的解1t ，2t ， 又12110t t +=-<，12270t t => ， 所以1200t t <<，， 有1t ，2t 的几何意义可知，1212121211111111||||||||27t t PA PB t t t t t t ⎛⎫++=+=-+=-= ⎪⎝⎭．…………………………………………………………（10分）23．（本小题满分10分）【选修4−5：不等式选讲】 （1）解：由绝对值三角不等式可知：()|1|2|3|1||3||13|2f x x x x x x x =-+--+--+-=≥|≥，当且仅当3x =时，两个不等式同时取等号， 所以()f x 的最小值2M =．……………………………………………………………（5分）（2）证明：由（1）知，2a b +=，则(1)(1)4a b +++=， 所以22(11)(11)11(1)2(1)21111a b a b a b a b +-+-+=+-+++-+++++211(11)11144a b a b ⎛⎫+++++ ⎪++⎝⎭⎝⎭==≥， 当且仅当1a b ==，不等式取等号，所以22111a b a b +++≥．…………………………………………………………（10分）。

3 - 02 3 6 2 7 1云南师大附中 2021 届高考适应性月考卷（二） 文科数学参考答案一、选择题（本大题共 12 小题，每小题 5 分，共 60 分）题号 1 2 3 4 5 6 7 8 9 10 11 12 答案 CBBCADDDAADC【解析】1． A = {x | x ≥ 0}，B = {x | x ≥ 3或x ≤ - 1} ，∴ A B = {x | x ≥ 3}，故选 C ． 2． z = 1 - 2i = - 1 - 3 i ， z =- 1 + 3i ，故选 B ．1 + i2 2 2 23．对于∀x ∈ R ，x 2 + ax + a ≥ 0 成立是真命题，∴ ∆= a 2 - 4a ≤0 ，即0≤a ≤4 ，故选 B ． 4．∵ tan α = 2 ，∴ cos α ≠ 0 ，∴ 2sin α - cos α = 2 tan α - 1 = 3，故选 C ．sin α + 2 c os α tan α + 2 45．由题意可知输出结果为 S = -1 + 2 - 3 + 4 - ⋅ ⋅ ⋅ + 8 = 4 ，故选 A ．6．∵ k l k l = -2 - m= -1，∴ m = -5 ，故选 D ．7．∵ a 1 + a 3 = 2a 2，a 4 + a 8 = 2a 6 ，又 a 2 + a 6 = 2a 4 ，∴ a 1 + a 3 + a 4 + a 8 = 2(a 2 + a 6 ) = 4a 4 = 32 ，故选 D ．8．画出不等式组表示的可行域知， z = 2x + 3y 的最小值为-14 ，故选 D ．9．由三视图知：几何体为四棱锥，且四棱锥的一条侧棱与底面垂直，如图，PA ⊥ 平面 ABCD ， PA = 2 ， AB = 2 ， AD = 4 ， BC = 2 ，经计算，PD = 2 ，PC = 2 ，DC = 2 ，∴ PC ⊥ CD ，∴ S = 1⨯2⨯2 = 2，S = 1 ⨯ 2 ⨯ 4 = 4 ，S = 1⨯ 2 ⨯ 2 = 2 △PAB， 2 S = 1 ⨯ 2 2 ⨯ 2 = 2 ， △PAD 2 △PBC 2 △PCD2S= 1⨯ (2 + 4) ⨯ 2 = 6 ，∴ S = 12 + 2 + 2 ，故选 A ．ABCD 2表10．设△ABC 外接圆半径为 r ，三棱锥外接球半径为 R ，∵ AB = 2，AC = 3，∠BAC = 60︒ ，∴ BC 2 = AB 2 + AC 2 - 2AB AC cos 60︒ = 22 + 32 - 2 ⨯ 2⨯ 3⨯ 1= 7 ，∴ BC =，∴ 2r =2BC =sin60︒- 3 5 3 2 2 67 = 2 21 ，∴ r = 21 ，由题意知， PA ⊥ 平面 ABC ，则将三棱锥补成三棱柱可得，3 3 2⎛ PA ⎫2 3 21 1010 40 R 2 = ⎪+ r 2 = 1 + = ，∴ S = 4πR 2= 4π ⨯ = π ，故选 A ． ⎝ 2 ⎭9 3 3 311 ．设 | PF 1 |= r 1，| PF 2 |= r 2 ， 由椭圆的定义得： r 1 + r 2 = 2a ， ∵ △F 1PF 2 的三条边|PF 2|，| PF 1 |，| F 1F 2 | 成等差数列， ∴ 2r 1 = 2c + r 2 ，联立 r 1 + r 2 = 2a ， 2r 1 = 2c + r 2 ，解得r = 2a + 2c ，r = 4a - 2c，由余弦定理得 ：(2c )2 = r 2 + r 2 - 2r r cos 60︒ ，将1 32 32a + 2c 4a - 2c 1 2 1 2⎛ 2a + 2c ⎫2r = ，r = 代入 (2c )2 = r 2 + r 2 - 2r r cos 60︒ 可得， 4c 2 = +1 3 231 2 1 2 3 ⎪ ⎝ ⎭⎛ 4a - 2c ⎫22a + 2c 4a - 2c 1 c ⎪ - 2，整理得 ： 2c 2 + ac - a 2= 0 ，由 e = ，得 ⎝ 3 ⎭3 3 2 a2e 2 + e -1 = 0 ，解得： e = 1或e = -1 （舍去），故选 D ．212．若至少存在一个 x ∈ ⎡1，1⎤ ，使得 f (x ) > g (x ) 成立，则 f (x ) - g (x ) > 0 在 x ∈ ⎡1，1⎤有解，⎢⎣ e ⎥⎦0 0 ⎢⎣ e ⎥⎦ 即 a ⎛ 1 - x ⎫ - 2 ln 1 + ax = a + 2 ln x > 0 在 x ∈ ⎡1，1⎤ 上有解，即 a > -2x ln x 在 x ∈ ⎡1，1⎤上x ⎪ x x ⎢⎣ e ⎥⎦ ⎢⎣ e ⎥⎦ ⎝ ⎭至少有一个 x 成立，令 h (x ) = -2x ln x ， h '(x ) = -2(ln x + 1) ，所以 h (x ) 在⎡1，1⎤上单调递减，则 h (x )min = h (1) = 0 ，因此 a > 0 ，故选 C ． 二、填空题（本大题共 4 小题，每小题 5 分，共 20 分）⎢⎣ e ⎥⎦题号13 1415 16答案7242 7 7⎡⎢ ln 2， 1 ⎫⎣6 3e ⎪ ⎭【解析】713． a + b = (m - 2，3)，2b = (-4，2) ，∵ (a + b ) ⊥ 2b ，∴ (m - 2) ⨯ (-4) + 3 ⨯ 2 = 0 ，∴ m = ．214 = 4a ，∴ b a 2= 15 ，∴双曲线的离心率e = = 4 ．a 2+ b 2b2 1 + a 2 22 7715．在△ABC 中，由余弦定理得 AC 2 = AB 2 + BC 2 - 2 AB BC cos B = 32 + 22 - 2 ⨯ 3 ⨯ 2 ⨯ 1= 7 ，2∴ AC = 2 ⨯ 3 7 ，由正弦定理得sin A = BC sin B = 2 = 21 ，∵ BC < AC ，∴ A < B = π ， AC 77 3 ∴ cos A =．16．由 g (x ) =| f (x ) | -3ax - 3a = 0 ，得| f (x ) |= 3ax + 3a = 3a (x + 1) ，设 y = 3a (x + 1) ，则直线过定点(-1，0) ，作出函数| f (x ) | 的图象（图象省略）．两函数图象有三个交点. 当3a ≤0 时，不满足条件；当3a > 0 时，当直线 y = 3a (x + 1) 经过点 (3，ln 4) 时，此时两函数图象有3 个交点，此时 3a =ln 4 ，a = ln 2；当直线 y = 3a (x + 1) 与 y = ln(x + 1) 相切时，有两个交点，此时函数的 4 6导数 f '(x ) = 1 ，设切点坐标为(m ，n ) ，则 n = ln(m + 1) ，切线的斜率为 f '(m ) = 1，x + 1 m + 1 则切线方程为 y - ln(m + 1) =1 m + 1 (x - m ) ，即 y = 1 m + 1 x - m m + 1 + ln(m + 1) ，∵ 3a =1m + 1且3a =- m + ln(m + 1) ，∴ 1 = - m + ln(m + 1) ，即 1 + m= ln(m + 1) = 1 ，则m + 1 m + 1 m + 1 m + 1 m + 1m + 1 = e ，即 m = e -1 ，则3a = 1 = 1 ，∴ a = 1，∴要使两个函数图象有3 个交点，则 ln 2 ≤a < 1 ．m + 1 e 3e6 3e 三、解答题（共 70 分．解答应写出文字说明，证明过程或演算步骤）17．（本小题满分 12 分）解：（Ⅰ）因为(2b - c ) cos A - a cos C = 0 ，所以 2b cos A - c cos A - a cos C = 0 ，由正弦定理得 2sin B cos A - sin C cos A - sin A cos C = 0 ， 即2sin B cos A - sin( A + C ) = 0 ，又 A + C = π - B ，所以sin( A + C ) = sin B ， 所以sin B (2 c os A -1) = 0 ，在△ABC 中，sin B ≠ 0 ，所以2 cos A - 1 = 0 ，所以 A = π．…………………………（6 分）33 3 3 （Ⅱ）由余弦定理得： a 2 = b 2 + c 2 - 2bc cos A = b 2 + c 2 - bc ， ∴4 ≥ 2bc - bc = bc ，∴ S = 1 bc sin A = 3 bc ≤ 3 ⨯ 4 = ，2 4 4当且仅当b = c 时“ = ”成立，此时△ABC 为等边三角形，∴△ABC 的面积 S 的最大值为 ．…………………………………………………（12 分）18．（本小题满分 12 分）解：（Ⅰ） 2 ⨯ 2 列联表补充如下：喜欢数学课程不喜欢数学课程合计 男生 40 30 70 女生 35 15 50 合计7545120……………………………………………………………………………………………（3 分）2 120 ⨯ (40 ⨯15 - 30 ⨯ 35)2由题意得 K =≈ 2.057 ，………………………………………（5 分） 70 ⨯ 50 ⨯ 75 ⨯ 45∵ 2.057 < 2.706 ，∴没有90% 的把握认为喜欢数学课程与否与性别有关．…………（6 分）（Ⅱ）用分层抽样的方法抽取时，抽取比例是 6 = 2，则抽取男生30 ⨯ 2 15 = 4 人，抽取女生15 ⨯ 2 15 45 15= 2 人．…………………………………（8 分）记抽取的女生为 A ，B ，抽取的男生为 a ，b ，c ，d ， 从中随机抽取2 名学生共有15 种情况：( A ，B )，( A ，a )，( A ，b )，( A ，c )，( A ，d )，(B ，a )，(B ，b )，(B ，c )，(B ，d )，(a ，b )，(a ，c )，(a ，d )，(b ，c )，(b ，d )，(c ，d ) ．其中至少有1 名是女生的事件为：( A ，B )，( A ，a )，( A ，b )，( A ，c )，( A ，d )，(B ，a )，(B ，b )，(B ，c )，(B ，d )， 有9 种情况． 记“抽取的学生中至少有1 名是女生”为事件 M ，则 P (M ) = 9 = 3．……………（12 分）19．（本小题满分 12 分）（Ⅰ）证明：由已知，得 AC =∵ BC = AD = 2 ， AB = 4 ，15 5= 2 ，AB 2 + BC 2- 2 A B ⨯ BC ⨯ cos ∠ABC又BC 2 +AC 2 =AB2 ，∴BC ⊥AC ．又PA ⊥底面ABCD ，BC ⊂平面ABCD ，则PA ⊥BC ，∵PA ⊂平面PAC ，AC ⊂平面PAC ，且PA AC =A ，∴BC ⊥平面PAC ．∵BC ⊂平面PBC ，∴平面PBC ⊥平面PAC ．………………………………………（6 分）（Ⅱ）线段PB 上存在一点E ，使得MN∥平面ACE ．证明：在线段PB 上取一点E ，使PE=3，连接ME，AE，EC，MN，PB 5∵PM=PE=3，∴ME∥AB ，且ME =3AB ，PA PB 5 5又∵CN∥AB ，且CN =3AB ，5∴CN∥ME ，且CN =ME ，∴四边形CEMN 是平行四边形，∴CE∥MN ，又CE ⊂平面ACE ，MN ⊄平面ACE ，∴MN∥平面ACE ．∴V = V =3V =1S BC =1⨯1⨯ 3 ⨯ 2 3 ⨯ 2 =6 3．……………（12 分）P -ACE E -PAC 5 B -PAC 5 △PAC 5 2 5 20．（本小题满分12 分）解：由题意知函数的定义域为{x | x > 0} ， f '(x) =-a+ 1 =x -a．x x（Ⅰ）当a = 1时， f '(x) =-1+ 1 =x - 1，x x当0 <x < 1时， f '(x) < 0 ，当x > 1 时， f '(x) > 0 ，所以函数f (x) 的单调递减区间是(0，1) ，单调递增区间是(1，+∞) ．所以当x =1 时，函数f (x) 有极小值f (1) = ln1 +1 +1 = 2 ，无极大值．………………（6 分）（Ⅱ）①当a≤1 时，函数f (x) 在[1，e] 为增函数，∴函数 f (x) 在[1，e] 上的最小值为f (1) =a ln1 + 1 + 1 = 2 ，显然2 ≠ 1 ，故不满足条件；②当1 <a≤e 时，函数f (x) 在[1，a) 上为减函数，在[a，e] 上为增函数，故函数f (x) 在[1，e] 上的最小值为f (x) 的极小值f (a) =-a ln a +a +1=1，即a = e ，满足条件；2 2 2 2 + =+ = ⎨ y 2 2 2③当a > e 时，函数 f (x ) 在[1，e] 为减函数，故函数 f (x ) 在[1，e] 上的最小值为 f (e) = a ln 1+ e + 1 = 1 ，即 a = e ，不满足条件．e 综上所述，存在实数 a = e ，使得函数f (x ) 在[1，e] 上的最小值为1 ．……………（12 分）21．（本小题满分 12 分）解：（Ⅰ）设动点Q (x ，y )，A (x ，y ) ，则 N (x ，0) ，且 x 2 + y 2 = 8 ，①1又OQ = mOA + (1 - m )ON ，得 x 0 = x ，y 0 = my ，2代入①得动点Q 的轨迹方程为 x y 1 ．…………………………………………（4 分） 8 8m 2（Ⅱ）当 m = 时，动点Q 的轨迹曲线C 为 x y 1 ．2 8 4x 2 y 2设直线l 的方程为 y = -x + b ，代入得3x 2 - 4bx + 2b 2 - 8 = 0 ，+ = 1 中， 8 4由∆ = (-4b )2 - 4 ⨯ 3 ⨯ (2b 2 - 8) > 0 ，∴ b 2 < 12 ，设 B (x 1，y 1)，D (x 2，y 2 ) ， x 1 + x 2 =4b，x x = 3 1 2 2b 2 - 8 3，……………………………（7 分） ∵点O 到直线l 的距离 d = | b |，| BD |= 2 (x + x )2 - 4x x = 4 12 - b 2 ，1 2 2 21 2 1 2 3 b 2 + 12 - b 2S △OBD = 2 d BD = 3b (12 - b )≤ 3= 2 ，2 当且仅当b 2 = 12 - b 2 ，即b 2 = 6 < 12 时取到最大值．∴ △OBD 面积的最大值为2 ．………………………………………………………（12 分）22．（本小题满分 10 分）【选修 4−4：坐标系与参数方程】⎧x = 1 + 1 t ，解：（Ⅰ）直线l 的参数方程为： ⎪2 (t 为参数) ，⎪ y = 1 + 3 t ，曲线C 的直角坐标方程为： ⎪⎩x2+ 232 2= 1 ．………………………………………………（5 分） ⎧x = 1 + 1 t ，⎪ 2 x 2 2（Ⅱ）把直线l 的参数方程⎨1 3 代入曲线C 的方程 + y 3 = 1 中， ⎪ y = + t ， ⎩⎪2 223 ⎝ ⎦ ⎣ ⎭ ⎪得⎛1 + 1 t ⎫ + 3⎛ 1⎫ 2 ⎪ + t = 3 ，即10t 2 + (6 + 4)t - 5 = 0 ， ⎝ 2 ⎭ ⎝ 2 2 ⎭设点 A ，B 所对应的参数分别为t ，t ，则t t = - 1，1 2 1 22∴ | PA | | PB |=| t | | t |=| t t |= - 1 = 1．……………………………………………（10 分）1 2 1 22 223．（本小题满分 10 分）【选修 4−5：不等式选讲】解：（Ⅰ）不等式 f (x )≤0 ，即| x - 2 | ≤| 2x + 1| ，即 x 2 - 4x + 4≤4x 2 + 4x + 1，3x 2 + 8x - 3≥ 0 ，解得 x 1x ≤- 3 ， ≥ 或3所以不等式 f (x )≤0 的解集为⎧x x1x ≤-⎫ ……………………………………（5 分）⎨ ≥ 或3 3⎬ .⎩⎭⎧x + 3，x < - 1 ，⎪ ⎪ （Ⅱ） f (x ) =| x - 2 | - | 2x + 1|= ⎪-3x + 1，- 1 ≤x ≤2， ⎨ ⎪⎪-x - 3，x > 2， ⎪⎩故 f (x ) 的最大值为 f ⎛ - 1 ⎫ = 5，2 ⎪ 2 ⎝ ⎭ 因为对于∀x ∈ R ，使 f (x ) - 2m 2≤4m 恒成立，所以 2m 2 + 4m ≥ 5，即4m 2 + 8m - 5≥ 0 ，2解得 m1 5⎛ 5 ⎤ ⎡ 1 ⎫≥ 2 或m ≤- 2 ，∴ m ∈ -∞，- 2 ⎥ ⎢ 2，+ ∞⎪ ．……………………………（10 分） 3 2 2 2。

2021年云南师大实验中学高三语文月考试卷及参考答案一、现代文阅读（36分）(一)现代文阅读I(9分)阅读下面的文字，完成下面小题。

朦胧的敬慕——悼念鲁迅先生萧乾①也许有人比我更怕死，我却不相信有比我更怕看死人的。

走在街上，我从没有胆子向寿衣铺里望望。

夜半，即便是从很远很远的地方飘来的僧器或诵经声，也必害得我用棉被厚厚包起头来，直像那是什么符咒一样。

②我曾见过三位死人，在我的记忆中，他们都将是我永不会忘记的。

而且，我还该陈说我都例外地不曾害怕过：一个黄昏，我的母亲死在我的怀抱里；小学时代，曾排着队去中央公园社稷堂瞻仰过孙中山先生的遗体；最近，在鲁迅先生灵前，我守了两天灵。

③扶着那绛色帏幔，职务使我看见了数千张陌生的但却诚笃的脸，一个个都像脚跟坠了铅球，那么轻又那么沉重地向灵堂踱着。

低垂的头，低垂的手，低垂的眉眼和心。

待踱到灵堂中央，冥冥中似有什么使他们肃然驻足了。

敬慕和哀悼如一双按住身体的手，他们的身子皆极自然地屈下了，然后噙了一汪眼泪，用手巾堵着嘴，仓皇地奔了出来。

④最感人的莫如一群小学生的吊唁。

在那近三十位小吊客中间，我特别留意一个衣服褴褛、腿下微跛的孩子，他肋下夹着的画册和石板说明了他是刚刚放学，如今正是回家或在街头玩耍的时候，然而他却结伴迢迢跑到了这里。

那个微跛的孩子，一拐一拐地，一直来到灵前，两只颇清秀的眼睛直直地凝视着鲁迅先生的遗体，然后，又放下肋下的画册，深深地鞠躬。

我不信作了那么些纪念周，他还不知道“三鞠躬”的礼数，然而，当我数到第三次以后，他仍向下屈着小小的腰身，他一连鞠了七个躬才红涨着脸，也红涨着眼睛，走出灵堂。

⑤如果稍换一个情况，我将忍不住笑出来，然而，我那时是用极大的崇敬的心情替他掀开帏幔，一直目送他走下殡仪馆的台阶。

⑥那个背影唤起我一点回忆。

十多年前的一个傍晚，如一切贪爱窗外景色的孩子一样，四点钟以后的时间对我来说变了滋味，变成鲜艳颜色。

然而我放下了玩具，和同伴沿着朱色皇城走了好长好长一段路去瞻仰一位“民国缔造者”的遗体。

2021届云南师大附中高三高考适应性月考数学(文)试题Word版含解析2021届云南师大附中高三高考适应性月考数学（文学）试题第ⅰ卷（共60分）一、多项选择题：这道主题有12道小题，每道小题5分，四道题各60分，只给出一个选项要求的.1.设a={0,1,2,4}，B=x？r | 1？十、4.然后是a？b＝（）a.｛1，2，3，4｝b.｛2，3，4｝c.｛2，4｝d.｛x|1?x?4｝【答案】c【解析】问题分析：ab？{0,12,4}{x1？x≤4}? {2,4}，所以选择 C.测试点：集合的交集运算。

2如果复数为Z？1.2I的共轭络合物是Z？A.Bi（a，B？R），其中I是虚单位，那么点（a，B）是（）ia（1.2）B.（-2，1）C.（1，-2）d.（2，1）[回答]B[分析]试题分析：≓ Z1.2i？？2.我∴Z2.i、所以选择B.i测试点：复数的计算12，且x为第四象限的角，则tanx的值等于（）13121255a、b、－c、d、－5512123.如果cosx？[答：]d[分析]555试题分析：∵x为第四象限的角，∴sinx??1?cos2x??，于是tanx?13??，故选d．12121313? 测试地点：商关系4.有3个不同的社团，甲、乙两名同学各自参加其中1个社团，每位同学参加各个社团的可能性相同，则这两位同学参加同一个社团的概率为（）A.1123b、c、d、3234【答案】a【解析】试题分析：记住这三个关联分别是a、B和C。

根据问题的意思，学生a和B有9种可能的情况参与社团，分别是（a，a），（a，B），（a，c），（B，a），（B，B），（B，c），（c，B），（c，c），而两个学生参与同一社团的人数是3，所以概率是测试点：概率31?，故选a．93??ex?1,x?05.已知函数f(x)??，若f(a)＝－1，则实数a的值为（）十、2，x？0a、2b、±1c。

1D。

1[答]C[分析]?a≤0，?a≤0，?a?0，?a?0，试题分析：∵?a?1a??，a?1，故选c．A.1a？2.1a？1.E1.测试点：函数值6.“0≤m≤l”是“函数f(x)?cosx?m?1有零点”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】试题分析：∵ f（x）？0 cosx？1.m、从0开始≤ M≤ 1, 0 ≤ 1.M≤ 1，然后呢？1.≤ 科斯≤ 1.所以函数f(x)?cosx?m?1有零点．反之，函数f(x)?cosx?m?1有零点，只需|m?1|≤1?0≤m≤2，故选a．试验场地：充分必要条件7.将某正方体工件进行切削，把它加工成一个体积尽可能大的新工件，新工件的三视图如图1所示，则原工件材料的利用率为〔材料的利用率新工件的体积（）原工件的体积a、7654b、c、d、8765【答案】c【解析】试题分析：如图1所示，建议将立方体的边长设置为1，切割部分为三角棱锥a？A1b1d1，其体积为1，则剩余零件（新工件）的体积为1，又正方体的体积65，故选c．6测试场地：三个视图8.在△abc中，|ab?ac|?|ab?ac|，ab=2,ac＝1，e,f为bc的三等分点，则aeaf＝a、8102526b，C，D，9999[答]B考点：向量的活动9.等比数列an?中，a1?2,a8?4，函数f(x)?x(x?a1)(x?a2)?a、2b、2c、2d、2【答案】c【解析】六9十二15（x？A8），然后f'（0）=（）试题分析：依题意，记g(x)?(x?a1)(x?a2)f?(0)?g(0)?a1a2(x?a8)，则f(x)?xg(x)，f?(x)?g(x)?xg?(x)，a8？（a1a8）4？212，所以C考点：等比数列的性质.10.在第九章算术中，四个面为直角三角形的四面体被称为乌龟，如图2所示。

2021年云南大学附属中学高三语文二模试题及答案解析一、现代文阅读（36分）(一)现代文阅读I(9分)阅读下面的文字，完成下面小题。

大师刘兆亮老傅拍过一组照片叫西湖边的一棵树：西湖边一棵桃树的春夏秋冬，人去湖空，人来湖拥，都安静得能听见风去风来，你说不出来个所以然的好，却让人体会到渗落到心底的妙。

老傅喜欢定点在一个位置拍植物，四季轮回，雷打不动，像是一把锁总喜欢待在门扣地方，而这里，恰恰是一扇门或一个城市的某种重要地带。

终于有一年，他拿出潜心之作——西湖边的一棵树，一举获得全球摄影最高奖的荷赛奖，领奖地点是在自行车的天堂城市——阿姆斯特丹。

于是，大家都开始管老傅叫大师，且把大师常挂在嘴上，见到本人叫，见到照片也叫。

老傅成为大师后，便在城市北郊布置了一个工作室，是旧民房改的，最大一间是冲洗照片的暗房。

边上原先有座大钢厂，刚迁走，那里原本是轰轰烈烈的大炼钢铁场景，如今复归自然。

颓败，荒芜，秋天的狗尾草，总能在风中把阳光撩拨得魅力缤纷。

老傅就是想借比桃树更小的植物拍出大意境。

他这次是站在路边的一片草地前，开发之前的荒芜，像烟雾一般缭绕。

老傅想，内心再静一些，再静一些，等这个地方再次喧闹起来，超越西湖边的一棵树的片子可能就出来了。

某天夜里的十点钟，他在暗房冲洗照片的手抖了一下，造成冲出的照片失去光影水准。

他托照片的食指是被暗房上顶一个沉重而笨拙的钢琴键给弄的。

楼上那户人家，恰好在暗房头顶，有架钢琴，每晚都在十点准时响起，老傅的安静被沉闷的琴声侵占了。

笨笨的钢琴声过后，便是“吱吱呀呀”的纤细二胡声，两种乐器交替，要到夜里12点才停歇。

而乐器声落，准有“嘣嘣嘣”的拖鞋声，老傅停下手里的活儿听楼上琴声之后的一来一回，两趟脚步声，一趟是碎步，心情有点儿小兴奋走出那种声音，一趟则是缓慢，充满思考与疲惫，两种脚步声都像是贴着老傅的头皮在走。

这种情况持续了大半年，老傅常待在暗房里，再望望窗外，感觉天上的星星，都是那些杂乱的音符飘上天空的。