尖子生训练题四(必修三第一、二章)
八年级尖子生训练题(4)答案.docx
八年级尖子生训练题（4）答案见下页3、2«-180°4, 45°或135°5、心4注意:一次函数图象是直线，但直线不一定是一次函数。
如直线y + 2 = 0, x-3 = 06、60°或120°7、②8、20BADCB BDCDA CB1、解（1） Si=100t 分）（2）①VS2=kt+b,依题意得t=9时，S2=0,……（4分）②（解法一）由①得，S2=-80t+720令 &=S2,得 100t=-80t+720,解得 t=4 ..... （9 分）当 t<4 时，S2>Si , .,.S2-S I<288即（-80t+720） -100t<288 , -180tV-432180t>432,解得 t>2. 4........................ （12 分）在两车相遇之前，当2.4<t<4时，两车的距离小于288千米。
............. （13分）（解法二）由①得，S2=-80t+720,令 t=0, ...$2=720,即王红所乘汽车的平均速度为— =80 （千米/时） ........................... （8分）9设两辆汽车ti小时后相遇,.•.100ti+80ti=720,解得切=4 .................. （9分）又设两车在相遇之前行驶t2小时后，两车之距小于288千米，则有 720- （100t2+80t2） <288 ...... （11 分）解得：t2>2. 4 .... （12 分）...在两车相遇之前，当2.4<t<4时，两车的距离小于288千米。
.............. （13分）2、解（2）第二组由甲地出发首次到达乙地所用的时间为：（3Vt=2, S2=560 j9k + 0 = 0\2k+b = 56Qk = -800 =720（7分）（11 分）8 + [2X （8 + 2） + 2] = 8 + 10 = 0.8 （小时）第二组由乙地到达丙地所用的时间为：2 + [2x （8 + 2） + 2] = 2 + 10 = 0.2 （小时）（3）根据题意得A 、B 的坐标分别为（0.8, 0） 和（1, 2）,设线段AB 的函数关系式为：S 2=kt + b,根据题意得：解得：r=i°2 = k+b[b = -8 .••图中线段AB 所表示的S2与t 间的函数关系式为：&=皿一8,自变量t 的取值范围是：0.8<?<1. 3、 解：（1） 4 分钟，40 升（各一分） （2） y=40-19 （x-15） =-19x+325 , （3 分）2 升（1分） 4、(1) y = \-x (2) -：OP = t, :.Q 点的横坐标为①当 0 <Lf<l,即 0 <f< 2 时，QM②当-t>l,即 时，QM = l--t2 2 214分-r-l I ，诊 2.5> a 2a 6a 7 7 （7a ） X10 m 26注意：⑴书写数学符号语言一定要规范!⑵在不会引起误会情况下，角尽量用匕1、匕2、匕3、匕4、… 形式表达，或用表示角顶点的一个字母表示，如匕A 、ZB> ZC> ZD> -o3分答案见下页京=冲7*T久7 #》09D兰\-7曲-（嘴-从）*二（dj - "尸。
河南省三门峡市陕州中学高三下学期尖子生专题训练(三)英语试题 含答案
下期高三尖子生专题训练英语试卷第二部分: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)阅读下列短文，从每题所给的四个选项（A,B,C和D）中，选出最佳选项，并在答题卡上将该项涂黑。
AI clearly remember the day when my little brother was born: January 1, 1994. One of my favorite family photos shows me lying with my mother in the hospital bed, happy but un-aware of the small, sleeping baby in the background.I stayed with my grandparents for the weekend of my brother’s birth, excited about my new brother but not yet aware of what having a little brother would mean. I didn’t truly realize what was going on until we were in the hospital room at Duke University----coincidentally (碰巧地), the same room in which I had been born two years earlier.When I looked at my brother for the first time, I felt a mixture of fear and interest. Little did know that small, pink creature would grow up to be one of my favorite people in the world.In reality, though I am two years older than my brother, I am more often than not the real baby in the family. I am very lacking when it comes to common sense. Instructions constantly confuse me and I frequently find myself totally puzzled by things like knowing how to start the washing machine or manage the storage settings on my iPhone.That’s where Gibson comes in. The poor kid has had to guide me through more tasks than I would care to admit, but he never complains. Though I should probably be told to figure it out myself, he always comes through.I’m envious of his ability to readily answer the ever-present, “What do you want to do with your life?” question at family gatherings. “Be a doctor,” he says----a solid answer, completely opposite to my shaky one, “Well, I’m an English major, so...”My brother truly is my best friend. No one understands me better, and there isn’t anyone else I would want to be stuck with in our family. I may not have a clear idea of where I’m headed, but he is stuck with me.21. How did the author feel when he saw his brother for the first time?A. Angry and sad.B. Excited and moved.C. Curious and scared.D. Happy and interested.22. What does the underlined sentence in the fourth paragraph mean?A. My brother is the real baby in the family.B. In fact, I seem to be less mature than my little brother.C. My brother gets more love from the family than I do.D. I am growing more slowly than my brother.23. What does the brother often help the author do?A. Help him deal with many daily tasks.B. Help him with his studies.C. Give him advice on how to choose a major.D. Comfort him when he is in a bad mood.24. What is the best title for this text?A. My Strange Family GatheringsB. The Real Baby in the FamilyC. Stuck with Me----My Not-So-Little Little BrotherD. The Feeling of Having a BrotherBCan dogs and cats live in perfect harmony in the same home? People who are thinking about adopting a dog as a friend for their cats are worried that they will fight. A recent research has found a new recipe of success. According to the study, if the cat is adopted before the dog, and if they are introduced when still young (less than 6 months for cats, a year for dogs), it is highly probable that the two pets will get along swimmingly. Two-thirds of the homes interviewed reported a positive relationship between their cat and dog.However, it wasn't all sweetness and light. There was a reported coldness between the cat and dog in 25% of the homes, while aggression and fighting were observed in 10% of the homes. One reason for this is probably that some of their body signals were just opposite. For example, when a cat turns its head away it signals aggression, while a dog doing the same signals submission.In homes with cats and dogs living peacefully, researchers observed a surprising behavior. They are learning how to talk each other's language. It is a surprise that cats can learn how to talk ‘dog’ and dogs can learn how to talk ‘Cat’.What's interesting is that both cats and dogs have appeared to develop their intelligence. They can learn how to read each other's body signals, suggesting that the two may have more in common than we previously suspected. Once familiar with each other's presence and body language, cats and dogs can play together, greet each other nose to nose, and enjoy sleeping together on the sofa. They can easily share the same water bowl and in some cases groom (梳理) each other.The significance of the research on cats and dogs may go beyond pets ─ to people who don't get along, including neighbors, colleagues at work, and even world superpowers. If cats and dogs can learn to get along, surely people have a good chance.25. Some cats and dogs may fight when ________.A. they misunderstood each other's signalsB. they look away from each otherC. they are cold to each otherD. they are introduced at an early age26. What is found surprising about cats and dogs?A. They eat and sleep each other.B. They learn to speak each other's language.C. They observe each other's behaviors.D. They know something from each other's voices.27. It is suggested in Paragraph 4 that cats and dogs ________.A. have common interestsB. have a common body languageC. are less different than was thoughtD. are less intelligent than was expected28. What can we human beings learn from cats and dogs?A. We should live in peace with animals.B. We should know more about animals.C. We should learn to live in harmony.D. We should learn more body languages.CLost wallets which contain a picture of a baby are more likely to be returned to their owners, scientists have discovered.Researchers left 240 wallets on the streets of Edinburgh last year to see how many were returned to their owners. Some of the wallets contained one of four photographs—a baby, a cute little dog, a family and a portrait of an elderly couple. Other wallets contained a card suggesting the owner had recently made a charity donation, while a control sample(对照组) contained nothing at all.Professor Richard Wiseman, a psychologist who was in charge of the experiment, said 42percent of the wallets were posted back in total. Those containing the picture of the infant, or the baby were most likely to bring about an honest reaction from the finder, with 88 percent being returned, followed by those containing pictures of the little dog at 53 percent. Of those featuring the family snapshot, 48 percent were sent to the return address and only 28 percent of those with the picture of the elderly couple. Wallets containing the charity cards and the control sample were least likely to be returned, with rates of 20 and 15 percent.Prof Wiseman, of the University of Hertfordshire, said, “The baby kicked off a caring feeling in people, which is not surprising from an evolutionary perspective(角度). We were amazed by the high percentage of wallets that came back.”The wallets were planted at random about a quarter of a mile apart. Snapshots were inserted into a clear plastic window inside the wallets, none of which contained money.29. What’s the main idea of the passage?A. Researchers left 240 wallets on the streets of Edinburgh last year.B. Professor Richard Wiseman carried out an experiment with 240 wallets.C. Lost wallets containing a picture of a baby are more likely to be returned.D. Wallets with contain nothing are mostly returned to their owners.30. The passage is made believable by .A. telling storiesB. using figuresC. asking questionsD. giving examples31. The underlined phrase “kicked off” in Paragraph 4 most probably means “.”A. causedB. showedC. avoidedD. accepted32. It can be inferred that .A. people who returned the wallets are interested in psychologyB. dogs are more respected in western countries than the elderlyC. Professor Wiseman had expected the result before the experimentD. the result of the experiment can be different if wallets contain moneyDThe UK remains one of Chinese students’ top three destination for studying abroad. 58,810 Chinese undergraduates started studying in Britain in January 2015. Let’s take a look at the UK universities that attracted the most Chinese students last year, according to the Cultural and Education Section of the British Consulate-General(总领事馆)The University of LiverpoolIt is one of the great universities of research and innovation that attracts students and experts from around the world. The university has produced nine Nobel Prizewinners. It currently has 22,000 students, including 7,000 international students on campus. Around 3,200 Chinese students were enrolled last year.The University of BirminghamIt was established by Queen Victoria in 1900 and was the UK’s first “redbrick” university. It is one of the world’s top 100 universities and a member of the Russell Group. More than 2,500Chinese students were enrolled last year.The University of NottinghamIt is positioned as one of Britain’s leading research universities. The university reached a landmark in its long list of academic achievements in 2003 when Sir Peter Mansfield, a professor there, was awarded the Nobel Prize for Medicine. It is one of the world’s top 100 universities and a member of the Russell Group. Around 2, 800 Chinese students were enrolled last year.The University of SheffieldLocated in England’s fourth-largest city, it dates back to 1828. It is one of the world’s top 100 universities and a member of the Russell Group. The university has launched a project to help the city’s businesses break into the Chinese market by working with its Chinese-speaking graduates. Around 2,500 Chinese students were enrolled last year.33. How many Chinese students were enrolled in the “redbrick” university mentioned in the textlast year?A. Around 3,200B. More than 2,500.C. Around 2, 800D. Around 2, 500.34. What do we know about the universities mentioned in the passage?A. Queen Victoria set up the University of Nottingham in 1900.B. International students play a huge role at the University of Birmingham.C. The University of Liverpool has the most of Nobel Prizewinners in four universities.D. The University of Sheffield has set up a project to help Chinese enter the market in theUK.35. Which university does not belong to the Russell Group?A. The University of LiverpoolB. The University of BirminghamC. The University of Nottingham.D. The University of Sheffield.第二节（共5小题；每小题2分，满分10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。
广东省始兴县风度中学高三数学 尖子生培优训练资料4 文
广东省始兴县风度中学高三数学（文）尖子生培优训练资料2、已知两个等比数列{a n}，{b n}，满足a1＝a(a>0)，b1－a1＝1，b2－a2＝2，b3－a3＝3.(1)若a＝1，求数列{a n}的通项公式；(2)若数列{a n}唯一，求a的值．3、某企业在第1年初购买一台价值为120万元的设备M ，M 的价值在使用过程中逐年减少，从第2年到第6年，每年初M 的价值比上年初减少10万元；从第7年开始，每年初M 的价值为上年初的75%.(1)求第n 年初M 的价值a n 的表达式；(2)设A n ＝a 1＋a 2＋…＋a nn.若A n 大于80万元，则M 继续使用，否则须在第n 年初对M更新．证明：须在第9年初对M 更新．4、等比数列{a n}中，a1，a2，a3分别是下表第一、二、三行中的某一个数，且a1，a2，a3中的任何两个数不在下表的同一列．第一列第二列第三列第一行3210第二行6414第三行9818(1)求数列{a n}(2)若数列{b n}满足：b n＝a n＋(－1)n ln a n，求数列{b n}的前n项和S n.培优资料4答案2、解答 ：(1)设{a n }的公比为q ，则b 1＝1＋a ＝2，b 2＝2＋aq ＝2＋q ，b 3＝3＋aq 2＝3＋q 2，由b 1，b 2，b 3成等比数列得(2＋q )2＝2(3＋q 2)，即q 2－4q ＋2＝0，解得q 1＝2＋2，q 2＝2－2，所以{a n }的通项公式为a n ＝(2＋2)n －1或a n ＝(2－2)n －1.(2)设{a n }的公比为q ，则由(2＋aq )2＝(1＋a )(3＋aq 2)，得aq 2－4aq ＋3a －1＝0，(*)由a ＞0得Δ＝4a 2＋4a ＞0，故方程(*)有两个不同的实根，由{a n }唯一，知方程(*)必有一根为0，代入(*)得a ＝13.3、【解答】 (1)当n ≤6时，数列{a n }是首项为120，公差为－10的等差数列． a n ＝120－10(n －1)＝130－10n ；当n ≥6时，数列{a n }是以a 6为首项，公比为34的等比数列，又a 6＝70，所以a n ＝70×⎝ ⎛⎭⎪⎫34n －6.因此，第n 年初，M 的价值a n 的表达式为a n ＝⎩⎪⎨⎪⎧130－10n ，n ≤6，70×⎝ ⎛⎭⎪⎫34n －6，n ≥7.(2)设S n 表示数列{a n }的前n 项和，由等差及等比数列的求和公式得 当1≤n ≤6时，S n ＝120n －5n (n －1)， A n ＝120－5(n －1)＝125－5n ； 当n ≥7时，由于S 6＝570，故S n ＝S 6＋(a 7＋a 8＋…＋a n )＝570＋70×34×4×⎣⎢⎡⎦⎥⎤1－⎝ ⎛⎭⎪⎫34n －6＝780－210×⎝ ⎛⎭⎪⎫34n －6， A n ＝780－210×⎝ ⎛⎭⎪⎫34n －6n，因为{a n }是递减数列，所以{A n }是递减数列．又A 8＝780－210×⎝ ⎛⎭⎪⎫3428＝824764>80，A 9＝780－210×⎝ ⎛⎭⎪⎫3439＝767996<80，所以须在第9年初对M 更新．综上所述，S n＝⎩⎪⎨⎪⎧3n＋n2ln3－1，n 为偶数，3n－n －12ln3－ln2－1，n 为奇数.。
2021-2022年高三下学期尖子生专题训练生物试题
2021-2022年高三下学期尖子生专题训练生物试题一.选择题(每小题6分共36分)1. 有一位同学做根尖有丝分裂实验，在显微镜中观察到的图像如下图所示。
造成这种情况的原因可能是( )①取材位置不合适②取材时间不合适③制片时压片力量不合适④解离时间不合适⑤视野选择不合适A．②③ B．②⑤ C．①②⑤ D．①③④2．下列关于研究材料、方法及结论的叙述,错误的是（）A. 孟德尔以豌豆为研究材料,采用人工杂交的方法,发现了基因分离与自由组合定律B. 摩尔根等人以果蝇为研究材料,通过统计后代雌雄个体眼色性状分离比,认同了基因位于染色体上的理论C. 赫尔希与蔡斯以噬菌体和细菌为研究材料,通过同位素示踪技术区分蛋白质与 DNA,证明DNA是遗传物质D. 沃森和克里克以DNA大分子为研究材料,采用 X 射线衍射的方法,破译了全部密码子3．为研究高光强对移栽幼苗光合色素的影响,某同学用乙醇提取叶绿体色素,用石油醚进行纸层析,右图为滤纸层析的结果(Ⅰ、Ⅱ、Ⅲ、Ⅳ为色素条带)。
下列叙述不正确的是（）A. 强光照导致了该植物叶绿素含量降低B. 类胡萝卜素含量增加有利于该植物抵御强光照C. 色素Ⅲ、Ⅳ吸收光谱的吸收峰波长不同D. 画滤液线时,滤液在点样线上只能画一次4．为达到实验目的，必须在碱性条件下进行的实验是（）A、利用双缩脲试剂检测生物组织中的蛋白质B、测定胃蛋白酶分解蛋白质的最适温度C、利用重铬酸钾检测酵母菌培养液中的酒精D、观察植物细胞的质壁分离和复原5．下列是以酵母菌为材料进行的实验，有关叙述错误的是（）A．探究酵母菌的呼吸方式，可用溴麝香草酚蓝检测产生的CO2B．用酵母菌发酵酿制果酒，选择酸性重铬酸钾检测产生的酒精C．探究酵母菌种群数量变化，应设空白对照排除无关变量干扰D．用稀释涂布平板法培养计数，应选择有30～300菌落数的平板6．下列有关生物学实验的叙述，正确的是（）A.叶绿体色素滤液细线浸入层析液，可导致滤纸条上色素带重叠B.低温诱导大蒜根尖时间过短，可能导致难以观察到染色体加倍的细胞C.用显微镜观察洋葱根尖装片时，需保持细胞活性以便观察有丝分裂过程D.将洋葱表皮放入0.3g/mL蔗糖溶液中，水分交换平衡后制成装片观察质壁分离过程二.非选择题(共54分)7.（8分）为探究影响光合作用强度的因素，将同一品种玉米苗置于25℃条件下培养，实验结果如图所示。
河南省三门峡市陕州中学高三下学期尖子生专题训练数学
2015-2016学年河南省三门峡市陕州中学高三（下）尖子生专题训练数学试卷（文科）（四）一、选择题：本大题共18个小题，每小题5分，共90分.1．已知全集U=R ，A=，B={x|lnx ＜0}，则A∪B=（ ）A ．{x|﹣1≤x ≤2}B ．{x|﹣1≤x ＜2}C ．{x|x ＜﹣1或x ≥2}D ．{x|0＜x ＜2}2．已知双曲线的离心率为，则双曲线的渐近线方程为（ ）A ．y=±2xB ．C ．D ．3．已知直线l 过抛物线C 的焦点，且与C 的对称轴垂直．l 与C 交于A ，B 两点，|AB|=12，P 为C 的准线上一点，则△ABP 的面积为（ ） A ．18 B ．24 C ．36 D ．484．O 为坐标原点，F 为抛物线C ：y 2=4x 的焦点，P 为C 上一点，若|PF|=4，则△POF 的面积为（ ）A ．2B ．2C ．2D ．45．设F 1、F 2是椭圆的左、右焦点，P 为直线x=上一点，△F 2PF 1是底角为30°的等腰三角形，则E 的离心率为（ ）A ．B ．C ．D ．6．等轴双曲线C 的中心在原点，焦点在x 轴上，C 与抛物线y 2=16x 的准线交于点A 和点B ，|AB|=4，则C 的实轴长为（ ）A ．B ．C ．4D ．87．设椭圆C ： =1（a ＞b ＞0）的左、右焦点分别为F 1、F 2，P 是C 上的点，PF 2⊥F 1F 2，∠PF 1F 2=30°，则C 的离心率为（ ）A ．B ．C ．D ．8．设F 为抛物线C ：y 2=3x 的焦点，过F 且倾斜角为30°的直线交于C 于A ，B 两点，则|AB|=（ ）A ．B ．6C ．12D ．79．若点P在抛物线y=x2上，点Q（0，3），则|PQ|的最小值是（）A．B．C．3 D．10．曲线y=e x在点（2，e2）处的切线与坐标轴所围三角形的面积为（）A． e2B．2e2C．e2D． e211．用min{a，b，c}表示a，b，c三个数中的最小值，设f（x）=min{2x，x+2，10﹣x}（x≥0），则f（x）的最大值为（）A．4 B．5 C．6 D．712．已知函数y=f（x）的周期为2，当x∈时 f（x）=x2，那么函数y=f（x）的图象与函数y=|lgx|的图象的交点共有（）A．10个B．9个C．8个D．1个13．已知函数f（x）=|lgx|，若0＜a＜b，且f（a）=f（b），则a+2b的取值范围是（）A．B．C．（3，+∞）D． B．（﹣∞，1] C． D．16．当0＜x≤时，4x＜log a x，则a的取值范围是（）A．（0，）B．（，1）C．（1，）D．（，2）17．已知三棱锥S﹣ABC的各顶点都在一个半径为r的球面上，球心O在AB上，SO⊥底面ABC，，则球的体积与三棱锥体积之比是（）A．πB．2πC．3πD．4π18．如图正方体ABCD﹣A1B1C1D1的棱长为2，线段B1D1上有两个动点E、F，且EF=，则下列结论中错误的是（）A．AC⊥BEB．EF∥平面ABCDC．三棱锥A﹣BEF的体积为定值D．△AEF的面积与△BEF的面积相等二、填空题：本大题共6个小题，每小题5分，共30分.19．设抛物线C：y2=4x的焦点为F，直线l过F且与C交于A，B两点．若|AF|=3|BF|，则l 的斜率为．20．已知抛物线C的顶点坐标为原点，焦点在x轴上，直线y=x与抛物线C交于A，B两点，若P（2，2）为AB的中点，则抛物线C的方程为．21．过椭圆的右焦点作一条斜率为2的直线与椭圆交于A、B两点，O 为坐标原点，则△OAB的面积为．22．已知F是双曲线C：x2﹣=1的右焦点，P是C的左支上一点，A（0，6）．当△APF周长最小时，该三角形的面积为．23．已知H是球O的直径AB上一点，AH：HB=1：2，AB⊥平面α，H为垂足，α截球O所得截面的面积为π，则球O的表面积为．24．已知正四棱锥O﹣ABCD的体积为，底面边长为，则以O为球心，OA为半径的球的表面积为．三、解答题：本大题共3小题，共30分，解答应写出必要的文字说明、证明及演算步骤.25．在△ABC中，角A，B，C的对边分别为a，b，c，已知bcos2+acos2=c．（Ⅰ）求证：a，c，b成等差数列；（Ⅱ）若C=，△ABC的面积为2，求c．26．如图，已知四棱锥P﹣ABCD中，侧面PAD是边长为2的正三角形，底面ABCD为菱形，∠DAB=60°．（Ⅰ）证明：PB⊥AD；（Ⅱ）若PB=3，求四棱锥P﹣ABCD的体积．27．已知椭圆E的中心在坐标原点，且抛物线x2=﹣4y的焦点是椭圆E的一个焦点，以椭圆E的长轴的两个端点及短轴的一个端点为顶点的三角形的面积为6．（Ⅰ）求椭圆E的方程；（Ⅱ）若斜率为的直线l与椭圆E交于不同的两点A、B，又点C（，2），求△ABC面积最大时对应的直线l的方程．2015-2016学年河南省三门峡市陕州中学高三（下）尖子生专题训练数学试卷（文科）（四）参考答案与试题解析一、选择题：本大题共18个小题，每小题5分，共90分.1．已知全集U=R，A=，B={x|lnx＜0}，则A∪B=（）A．{x|﹣1≤x≤2} B．{x|﹣1≤x＜2} C．{x|x＜﹣1或x≥2} D．{x|0＜x＜2} 【考点】并集及其运算．【分析】求出A与B中不等式的解集，分别确定出A与B，找出两集合的并集即可．【解答】解：由A中不等式变形得：≤0，即（x+1）（x﹣2）＜0，且x﹣2≠0，解得：﹣1≤x＜2，即A={x|﹣1≤x＜2}，由B中不等式变形得：lnx＜0=ln1，得到0＜x＜1，即B={x|0＜x＜1}，则A∪B={x|﹣1≤x＜2}，故选：B．2．已知双曲线的离心率为，则双曲线的渐近线方程为（）A．y=±2x B．C．D．【考点】双曲线的简单性质．【分析】由离心率的值，可设，则得，可得的值，进而得到渐近线方程．【解答】解：∵，故可设，则得，∴渐近线方程为，故选C．3．已知直线l过抛物线C的焦点，且与C的对称轴垂直．l与C交于A，B两点，|AB|=12，P 为C的准线上一点，则△ABP的面积为（）A．18 B．24 C．36 D．48【考点】直线与圆锥曲线的关系．【分析】首先设抛物线的解析式y2=2px（p＞0），写出次抛物线的焦点、对称轴以及准线，然后根据通径|AB|=2p，求出p，△ABP的面积是|AB|与DP乘积一半．【解答】解：设抛物线的解析式为y2=2px（p＞0），则焦点为F（，0），对称轴为x轴，准线为x=﹣∵直线l经过抛物线的焦点，A、B是l与C的交点，又∵AB⊥x轴∴|AB|=2p=12∴p=6又∵点P在准线上∴DP=（+||）=p=6∴S△ABP=（DP•AB）=×6×12=36故选C．4．O为坐标原点，F为抛物线C：y2=4x的焦点，P为C上一点，若|PF|=4，则△POF的面积为（）A．2 B．2C．2D．4【考点】抛物线的简单性质．【分析】根据抛物线方程，算出焦点F坐标为（）．设P（m，n），由抛物线的定义结合|PF|=4，算出m=3，从而得到n=，得到△POF的边OF上的高等于2，最后根据三角形面积公式即可算出△POF的面积．【解答】解：∵抛物线C的方程为y2=4x∴2p=4，可得=，得焦点F（）设P（m，n）根据抛物线的定义，得|PF|=m+=4，即m+=4，解得m=3∵点P在抛物线C上，得n2=4×3=24∴n==∵|OF|=∴△POF的面积为S=|OF|×|n|==2故选：C5．设F1、F2是椭圆的左、右焦点，P为直线x=上一点，△F2PF1是底角为30°的等腰三角形，则E的离心率为（）A．B．C．D．【考点】椭圆的简单性质．【分析】利用△F2PF1是底角为30°的等腰三角形，可得|PF2|=|F2F1|，根据P为直线x=上一点，可建立方程，由此可求椭圆的离心率．【解答】解：∵△F2PF1是底角为30°的等腰三角形，∴|PF2|=|F2F1|∵P为直线x=上一点∴∴故选C．6．等轴双曲线C的中心在原点，焦点在x轴上，C与抛物线y2=16x的准线交于点A和点B，|AB|=4，则C的实轴长为（）A．B．C．4 D．8【考点】圆锥曲线的综合．【分析】设等轴双曲线C：x2﹣y2=a2（a＞0），y2=16x的准线l：x=﹣4，由C与抛物线y2=16x的准线交于A，B两点，，能求出C的实轴长．【解答】解：设等轴双曲线C：x2﹣y2=a2（a＞0），y2=16x的准线l：x=﹣4，∵C与抛物线y2=16x的准线l：x=﹣4交于A，B两点，∴A（﹣4，2），B（﹣4，﹣2），将A点坐标代入双曲线方程得=4，∴a=2，2a=4．故选C．7．设椭圆C： =1（a＞b＞0）的左、右焦点分别为F1、F2，P是C上的点，PF2⊥F1F2，∠PF1F2=30°，则C的离心率为（）A．B．C．D．【考点】椭圆的简单性质．【分析】设|PF2|=x，在直角三角形PF1F2中，依题意可求得|PF1|与|F1F2|，利用椭圆离心率的性质即可求得答案．【解答】解：设|PF2|=x，∵PF2⊥F1F2，∠PF1F2=30°，∴|PF1|=2x，|F1F2|=x，又|PF1|+|PF2|=2a，|F1F2|=2c∴2a=3x，2c=x，∴C的离心率为：e==．故选A．8．设F为抛物线C：y2=3x的焦点，过F且倾斜角为30°的直线交于C于A，B两点，则|AB|=（）A．B．6 C．12 D．7【考点】抛物线的简单性质．【分析】求出焦点坐标，利用点斜式求出直线的方程，代入抛物线的方程，利用根与系数的关系，由弦长公式求得|AB|．【解答】解：由y2=3x得其焦点F（，0），准线方程为x=﹣．则过抛物线y2=3x的焦点F且倾斜角为30°的直线方程为y=tan30°（x﹣）=（x﹣）．代入抛物线方程，消去y，得16x2﹣168x+9=0．设A（x1，y1），B（x2，y2）则x1+x2=，所以|AB|=x1++x2+=++=12故选：C9．若点P在抛物线y=x2上，点Q（0，3），则|PQ|的最小值是（）A．B．C．3 D．【考点】抛物线的简单性质．【分析】由已知条件，设P（x，y），利用两点间距离公式，求出|PQ|，由此利用配方法能求出|PQ|的最小值．【解答】解：设P（x，y），∵Q（0，3），∴|PQ|===≥，∴|PQ|的最小值是．故选：B．10．曲线y=e x在点（2，e2）处的切线与坐标轴所围三角形的面积为（）A． e2B．2e2C．e2D． e2【考点】利用导数研究曲线上某点切线方程．【分析】欲切线与坐标轴所围成的三角形的面积，只须求出切线在坐标轴上的截距即可，故先利用导数求出在x=2处的导函数值，再结合导数的几何意义即可求出切线的斜率．最后求出切线的方程，从而问题解决．【解答】解析：依题意得y′=e x，因此曲线y=e x在点A（2，e2）处的切线的斜率等于e2，相应的切线方程是y﹣e2=e2（x﹣2），当x=0时，y=﹣e2即y=0时，x=1，∴切线与坐标轴所围成的三角形的面积为：S=×e2×1=．故选D．11．用min{a，b，c}表示a，b，c三个数中的最小值，设f（x）=min{2x，x+2，10﹣x}（x ≥0），则f（x）的最大值为（）A．4 B．5 C．6 D．7【考点】函数的最值及其几何意义．【分析】在同一坐标系内画出三个函数y=10﹣x，y=x+2，y=2x的图象，以此作出函数f（x）图象，观察最大值的位置，通过求函数值，解出最大值．【解答】解：10﹣x是减函数，x+2是增函数，2x是增函数，令x+2=10﹣x，x=4，此时，x+2=10﹣x=6，如图：y=x+2 与y=2x交点是A、B，y=x+2与 y=10﹣x的交点为C（4，6），由上图可知f（x）的图象如下：C为最高点，而C（4，6），所以最大值为6．故选：C12．已知函数y=f（x）的周期为2，当x∈时 f（x）=x2，那么函数y=f（x）的图象与函数y=|lgx|的图象的交点共有（）A．10个B．9个C．8个D．1个【考点】对数函数的图象与性质；函数的周期性．【分析】根据对数函数的性质与绝对值的非负性质，作出两个函数图象，再通过计算函数值估算即可．【解答】解：作出两个函数的图象如上∵函数y=f（x）的周期为2，在上为减函数，在上为增函数∴函数y=f（x）在区间上有5次周期性变化，在、、、、上为增函数，在、、、、上为减函数，且函数在每个单调区间的取值都为，再看函数y=|lgx|，在区间（0，1]上为减函数，在区间B．（﹣∞，1] C． D．【考点】其他不等式的解法．【分析】由函数图象的变换，结合基本初等函数的图象可作出函数y=|f（x）|的图象，和函数y=ax的图象，由导数求切线斜率可得l的斜率，进而数形结合可得a的范围．【解答】解：由题意可作出函数y=|f（x）|的图象，和函数y=ax的图象，由图象可知：函数y=ax的图象为过原点的直线，当直线介于l和x轴之间符合题意，直线l 为曲线的切线，且此时函数y=|f（x）|在第二象限的部分解析式为y=x2﹣2x，求其导数可得y′=2x﹣2，因为x≤0，故y′≤﹣2，故直线l的斜率为﹣2，故只需直线y=ax的斜率a介于﹣2与0之间即可，即a∈故选：D16．当0＜x≤时，4x＜log a x，则a的取值范围是（）A．（0，）B．（，1）C．（1，）D．（，2）【考点】对数函数图象与性质的综合应用．【分析】由指数函数和对数函数的图象和性质，将已知不等式转化为不等式恒成立问题加以解决即可【解答】解：∵0＜x≤时，1＜4x≤2要使4x＜log a x，由对数函数的性质可得0＜a＜1，数形结合可知只需2＜log a x，∴即对0＜x≤时恒成立∴解得＜a＜1故选 B17．已知三棱锥S﹣ABC的各顶点都在一个半径为r的球面上，球心O在AB上，SO⊥底面ABC，，则球的体积与三棱锥体积之比是（）A．πB．2πC．3πD．4π【考点】球内接多面体．【分析】求出三棱锥的体积，再求出球的体积即可．【解答】解：如图，⇒AB=2r，∠ACB=90°，BC=，∴V三棱锥=，V球=，∴V球：V三棱锥=．18．如图正方体ABCD﹣A1B1C1D1的棱长为2，线段B1D1上有两个动点E、F，且EF=，则下列结论中错误的是（）A．AC⊥BEB．EF∥平面ABCDC．三棱锥A﹣BEF的体积为定值D．△AEF的面积与△BEF的面积相等【考点】空间中直线与直线之间的位置关系．【分析】连结BD，则AC⊥平面BB1D1D，BD∥B1D1，点A、B到直线B1D1的距离不相等，由此能求出结果．【解答】解：连结BD，则AC⊥平面BB1D1D，BD∥B1D1，∴AC⊥BE，EF∥平面ABCD，三棱锥A﹣BEF的体积为定值，从而A，B，C正确．∵点A、B到直线B1D1的距离不相等，∴△AEF的面积与△BEF的面积不相等，故D错误．故选：D．二、填空题：本大题共6个小题，每小题5分，共30分.19．设抛物线C：y2=4x的焦点为F，直线l过F且与C交于A，B两点．若|AF|=3|BF|，则l的斜率为±．【考点】抛物线的简单性质．【分析】由题意设出直线AB的方程，联立直线和抛物线方程，求出A，B的横坐标，由|AF|=3|BF|得到x1=3x2+2，代入A，B的坐标得答案．【解答】解：由y2=4x，得F（1，0），设AB所在直线方程为y=k（x﹣1），联立y2=4x，得k2x2﹣（2k2+4）x+k2=0．设A（x1，y1），B（x2，y2），结合|AF|=3|BF|，解方程得：x1=+，x2=﹣．再由|AF|=3|BF|，得x1+1=3（x2+1），即x1=3x2+2，∴+=3（﹣）+2，解得：k=±．故答案为：±．20．已知抛物线C的顶点坐标为原点，焦点在x轴上，直线y=x与抛物线C交于A，B两点，若P（2，2）为AB的中点，则抛物线C的方程为y2=4x ．【考点】抛物线的标准方程．【分析】先根据题意设出抛物线的标准方程，与直线方程联立消去y，利用韦达定理求得x A+x B 的表达式，根据AB中点的坐标可求得x A+x B的，继而p的值可得．【解答】解：设抛物线方程为y2=2px，直线与抛物线方程联立求得x2﹣2px=0∴x A+x B=2p∵x A+x B=2×2=4∴p=2∴抛物线C的方程为y2=4x故答案为：y2=4x21．过椭圆的右焦点作一条斜率为2的直线与椭圆交于A、B两点，O为坐标原点，则△OAB的面积为．【考点】直线与圆锥曲线的综合问题．【分析】将椭圆与直线方程联立：，得交点，进而结合三角形面积公式计算可得答案．【解答】解：由题意知，解方程组得交点，∴．答案：．22．已知F是双曲线C：x2﹣=1的右焦点，P是C的左支上一点，A（0，6）．当△APF周长最小时，该三角形的面积为12．【考点】双曲线的简单性质．【分析】利用双曲线的定义，确定△APF周长最小时，P的坐标，即可求出△APF周长最小时，该三角形的面积．【解答】解：由题意，设F′是左焦点，则△APF周长=|AF|+|AP|+|PF|=|AF|+|AP|+|PF′|+2 ≥|AF|+|AF′|+2（A，P，F′三点共线时，取等号），直线AF′的方程为与x2﹣=1联立可得y2+6y﹣96=0，∴P的纵坐标为2，∴△APF周长最小时，该三角形的面积为﹣=12．故答案为：12．23．已知H是球O的直径AB上一点，AH：HB=1：2，AB⊥平面α，H为垂足，α截球O所得截面的面积为π，则球O的表面积为．【考点】球的体积和表面积．【分析】本题考查的知识点是球的表面积公式，设球的半径为R，根据题意知由与球心距离为R的平面截球所得的截面圆的面积是π，我们易求出截面圆的半径为1，根据球心距、截面圆半径、球半径构成直角三角形，满足勾股定理，我们易求出该球的半径，进而求出球的表面积．【解答】解：设球的半径为R，∵AH：HB=1：2，∴平面α与球心的距离为R，∵α截球O所得截面的面积为π，∴d=R时，r=1，故由R2=r2+d2得R2=12+（R）2，∴R2=∴球的表面积S=4πR2=．故答案为：．24．已知正四棱锥O﹣ABCD的体积为，底面边长为，则以O为球心，OA为半径的球的表面积为24π．【考点】球的体积和表面积；棱锥的结构特征．【分析】先直接利用锥体的体积公式即可求得正四棱锥O﹣ABCD的高，再利用直角三角形求出正四棱锥O﹣ABCD的侧棱长OA，最后根据球的表面积公式计算即得．【解答】解：如图，正四棱锥O﹣ABCD的体积V=sh=（×）×OH=，∴OH=，在直角三角形OAH中，OA===所以表面积为4πr2=24π；故答案为：24π．三、解答题：本大题共3小题，共30分，解答应写出必要的文字说明、证明及演算步骤.25．在△ABC中，角A，B，C的对边分别为a，b，c，已知bcos2+acos2=c．（Ⅰ）求证：a，c，b成等差数列；（Ⅱ）若C=，△ABC的面积为2，求c．【考点】数列与三角函数的综合；正弦定理；余弦定理的应用．【分析】（Ⅰ）利用正弦定理以及两角和与差的三角函数，三角形的内角和，化简求解即可．（Ⅱ）利用三角形的面积以及余弦定理化简求解即可．【解答】解：（Ⅰ）证明：由正弦定理得：即，∴sinB+sinA+sinBcosA+cosBsinA=3sinC…∴sinB+sinA+sin（A+B）=3sinC∴sinB+sinA+sinC=3sinC…∴sinB+sinA=2sinC∴a+b=2c…∴a，c，b成等差数列．…（Ⅱ）∴ab=8…c2=a2+b2﹣2abcosC=a2+b2﹣ab=（a+b）2﹣3ab=4c2﹣24．…∴c2=8得…26．如图，已知四棱锥P﹣ABCD中，侧面PAD是边长为2的正三角形，底面ABCD为菱形，∠DAB=60°．（Ⅰ）证明：PB⊥AD；（Ⅱ）若PB=3，求四棱锥P﹣ABCD的体积．【考点】棱柱、棱锥、棱台的体积；空间中直线与直线之间的位置关系．【分析】（Ⅰ）取AD的中点E，连接PE，BD，BE，推导出BE⊥AD，PE⊥AD，从而AD⊥面PBE，由此能证明AD⊥PB．（Ⅱ）作PO⊥BE于E，PO⊥面ABCD，求出，由此能求出四棱锥P﹣ABCD的体积．【解答】证明：（Ⅰ）取AD的中点E，连接PE，BD，BE，∵底面ABCD为菱形，∠DAB=60°，∴△ABD为正三角形，又∵E为AD的中点，∴BE⊥AD，∵侧面PAD为正三角形，E为AD的中点，∴PE⊥AD，∴AD⊥面PBE，∴AD⊥PB．…解：（Ⅱ）由（Ⅰ）AD⊥面PBE，得面ABCD⊥面PBE，作PO⊥BE于O，PO⊥面ABCD，∵侧面PAD为边长等于2的正三角形、△ABD为正三角形，E为AD的中点，∴，又∵PB=3，设PB的中点为F，，…∴，∴∠EBP=30°，∴，…∴四棱锥P﹣ABCD的体积…27．已知椭圆E的中心在坐标原点，且抛物线x2=﹣4y的焦点是椭圆E的一个焦点，以椭圆E的长轴的两个端点及短轴的一个端点为顶点的三角形的面积为6．（Ⅰ）求椭圆E的方程；（Ⅱ）若斜率为的直线l与椭圆E交于不同的两点A、B，又点C（，2），求△ABC面积最大时对应的直线l的方程．【考点】椭圆的简单性质．【分析】（Ⅰ）由抛物线方程求得焦点坐标，求得c，由三角形的面积公式可知，根据椭圆的性质，a2=b2+c2，即可求得a和b的值，求得椭圆方程；（Ⅱ）求得直线方程，并将直线方程代入椭圆方程，由韦达定理求得求得x1+x2及x1•x2，由弦长公式求得丨AB丨，根据点到直线的距离公式，求得d，根据三角形的面公式及基本不等式的性质即可求得m的值，求得直线方程．【解答】解：（Ⅰ）设椭圆E（a＞b＞0），由抛物线的焦点是椭圆E的一个焦点得：，由椭圆的性质可知：a2=b2+c2，∴5=a2﹣b2，，即ab=6，∴a2b2=36，即（b2+5）b2=36，（b2+9）（b2﹣4）=0，b2=4a2=9，∴椭圆…（Ⅱ）设，A（x1，y1），B（x2，y2），与，联立得：9x2+6mx+2m2﹣18=0，△=36m2﹣36（2m2﹣18）＞0，可知：m2＜18，由韦达定理可知：，…，，到的距离，…当m2=9即m=±3时，S最大，对应的直线l的方程为…2016年10月28日。
河南省三门峡市陕州中学2016届高三下学期尖子生专题训练(四)英语试题 Word版含答案
2015-2016学年下期高三尖子生专题训练（四）英语试卷第二部分阅读理解（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的四个选项（A、B、C和D）中，选出最佳选项，并在答题卡上将该项涂黑。
AJust finishing loading his car, my husband Jerry went back to the kitchen. “I guess filling my thermos（保温杯）is all I have left to do, and then I’m off.” Down the drain went the hot water that had been warming his thermos.With a nostalgic（怀旧的）look on his face, he softly said: “Anytime I fill my thermos I think of your mom.” The tone in his voice was so tender as h e spoke of my mother, who is sadly no longer with us.“Oh honey, I know she’s watching and listening right now with a big smile on her face, ”I responded, tears in my eyes.One morning years ago when we were leaving, mother watched as Jerry was to fill his thermos. Mom then casually said, ‘Jerry, if you filled your thermos with hot water and let it sit a while, your coffee would stay hot for longer then.” With her simple suggestion, his face showed that he was shocked by his own lack of common sense, well, of course it would!He poured out the small amount of coffee already in the thermos and refilled it with boiling hot water. Then he poured the water and refilled it with hot coffee. Mom smiled as Jerry did so.Words of wisdom from a loving mother-in-law made an impact that will never be forgotten. She never dreamed her words would be a treasured and everlasting memory in the heart of her son-in-law.The lasting memories don’t necessarily come from major moments in life. Memories are often made during those small, ordinary times. Life is about small moments we share with friends and loved ones.21. Which word best describes the author’s feeling in Paragraph 3? ____A. TouchedB. ConcernedC. ConfusedD. Interested22. The advice the mother gave Jerry was to ___.A. fill the thermos with a little coffee firstB. fill thermos with coffee before with warm water.C. warm the thermos up before filling it with coffeeD. think of her whenever filling the thermos with coffee.23. The underlined part “don’t necessarily” is similar in meaning to ________.A. without any doubtB. possibly but uncertainlyC. certainly and absolutelyD. out of the question24. The message that the story intends to convey is ______.A. how valuable a mother’s advice is to her children.B. that lasting memories come from small thingsC. what a wise woman the mother is to her son-in-lawD. why we should respect our parents and listen to them.BMen are spending more and more time in the kitchen encouraged by celebrity (名人) chefs like Gordon Ramsay and Jamie Oliver，according to a report from Oxford University.The effect of the celebrity role models，who have given cooking a more manly picture，has combined with a more general drive towards sexual equality and men now spend more than twice the amount of time preparing meals than they did in 1961.According to the research by Prof. Jonatahn Gershuny，who runs the Centre for Time Research at Oxford，men now spend more than half an hour a day cooking，up from just 12 minutes a day in 1961.Prof.Gershuny said，“The man in the kitchen is part of a much wider social trend.There has been 40 years of sexual equality，but there is another 40 years probably to come.”Women，who a generation ago spent nearly two hours a day cooking，now spend just one hour and seven minutes—a great fall，but they still spend far more time in the kitchen than men.Some experts have named these men in aprons（围裙）as “Gastrosexuals (men using cooking skills to impress friends)”，who have been inspired to pick up a kitchen knife by the success of Ramsay，Oliver as well as other male celebrity chefs such as Hugh Fearnley－Whittingstall，Marco Pierre White and Keith Floyd.“I was married in 1974.When my father came to visit me a few weeks later，I was wearing an apron when I opened the door.He laughed，” said Prof.Gershuny.“That would never happen now.”Two－thirds of adults say that they come together to share at least three times a week，even if it is not necessarily around a kitchen or dining room table. Prof. Gershuny pointed out that the family meal was now rarely eaten by all of its members around a table—with many “family meals” in fact taken on the sofa in the sitting room，and shared by family members.“The family meal has changed a lot，and few of us eat—as I did when I was a child—at least two meals a day together as a family. But it has survived in a different format.”25.What is one reason behind the trend that men spend more time cooking than before?A. The improvement of cooks’ status.B. The influence of popular female chefs.C. The change of female's view on cooking.D. The development of sexual equality campaign.26.How did Prof. Gershuny see the family meal according to the passage?A. It has become a thing of the past.B. It is very different from what it used to be.C. It shouldn't be advocated in modern times.D. It is beneficial to the stability of the family.27.Which is the best title for the passage?A. The Changes of Family MealsB. Equality between Men and WomenC. Cooking into a New Trend for MenD. Cooking—a Thing of the Past for WomenCThanks to this new hands free suitcase, carrying around heavy luggage may soon become a thing of the past. Designed by Israeli company NUA Robotics, this “smart”suitcase can follow you everywhere you go.The carry-on suitcase, currently a prototype(雏形) , connects to a smartphone app via bluetooth. It has a built-in camera sensor that can “see” you and follow you around on flat surfaces like airport floors. It comes with an anti-theft alarm to prevent someone carrying it away when you’re not looking, and it has a backup battery that you can use to charge all your devices.“It can follow and carry things for people around while communicating with their smartphone, and avoiding obstacles, ”explained Alex Libman, founder of NUA Robotics. “We’re combining sensor network, computer vision, and robotics．So if you download our app, press the ‘follow me’ button, the luggage recognizes its user and knows to follow and communicate.”NUA is still testing the device and trying improve features like speed and customization, but they hope to make the suitcase available to custom ers in a year’s time. If it proves successful, they want to use the Bluetooth pairing technology to automate lots of other devices，like shopping carts at the supermarket. These devices are especially meant to be useful to the physically disabled and elderly. “Any object can be smart and robotic,” Libman told Mashable. “We want to bring robots into everyday life.”The Israeli tech company made it clear that the finished product will look just like a regular suitcase. They’re planning to partner with a carry-on luggage maker, since the device that makes it “smart” weighs just 2.5 pounds and can be fixed on old-fashion luggage.With NUA'S smart suitcase, carrying around heavy luggage isn’t a burden. But how their invention will handle tough obstacles like stairs? It will be interesting to wait.28. According to the passage the “smart” suitcase_______．A. it can go everywhere it would like to goB. is on sale now in the marketC. can make our hands free during travelingD. was designed by Alex Libman29. If you want to use this free hand suitcase，you must_____．A. download the appB. avoid obstaclesC. look at it all the timeD. carry a camera30. What can we learn about NUA?A. It was a company which was founded by Mashable．B. It has improved the features of the device up to now．C. It wants to bring more smart and robotic devices into daily life．D. It will make the suitcase available to customers in a month．31. Where is the article probably from?A. A textbook．B. A poster．C. A biography．D. A website．DWith autumn on the horizon, the season’s first Oscar competitors are here. Already, the Hollywood Reporter has predicted that the often-snubbed(冷落的) Leonardo DiCaprio could win. Although before we get swept up in predictions, it’s nice to remember that little gold men are not the reason we love the movies. They shouldn’t be the primary reason for acting in them either.Bill Murray knows this first hand. Murray was positioned to win his first-ever Oscar Academy Award for 2004’s “Lost in Translation”, but he lost it to Sean Penn. Years later, Murray explained how he took the defeat in stride(从容应对), and his words should act as a blueprint for anyone who has failed:“I went through it once before and it’s nice to get nominated and win s ome prizes. You get to go to dinners and tell little stories and so forth. Then you get to get dressed up. And then you get to be on TV, which is sweet. And you can either win or lose. Well you’re not supposed to say ‘lose’ when you talk about the Oscar. Y ou are supposed to say ‘not chosen’. But I later realized that I had gotten a little caught up in the possibility of winning. So, shame on myself for getting caught in it. But I won a lot of the prizes for Lost in Translation. So I thought it didn’t seem u nnatural to expect that I would be rewarded just one more time. So when it didn’t happen, I thought, ‘Well, that’s kind of funny.’ But to get prizes is not why you work.”Murray isn’t alone in those feelings. No mention of actors without Oscars would be c omplete without recognition of Leonardo DiCaprio, who said, “I don’t think I ever expected anything like an Oscar ever, to tell you the truth. That is not my motivation when I do these roles. I really am motivated by being able to work with great people an d create a body of work that I can look back and be proud of.”Perhaps neither Murray nor DiCaprio will ever be introduced as an Academy Award-winning actor, but winning an Oscar for acting shouldn’t be the point. Here are some other amazing actors who never won an Oscar, even if you think they did: Julianne Moore, Brad Pitt, Johnny Depp, Tom Cruise, Harrison Ford, Will Smith. And they have proved that countless times before.32.We can learn about Bill Murray from the passage that_________.A.he felt ashamed tha t he didn’t win the Oscar.B.he never expected himself to win the Oscar.C.he had already won many other awards before he failed in Oscar.D.he is the first actor who had been predicted to win the Oscar but failed.33. Leonard DiCaprio is motivated to act different roles because _________A.he wants to win the Oscar.B. he can cooperate with great people.B.he wants others to take pride in him. D. he is not the only actor who hasn’twon the Oscar.34. The underlined word “that” in the last paragraph refers to “_________”.A.Neither Murray nor DiCaprio will won an Oscar.B.Winning an Oscar for acting shouldn’t be the point.C.Some amazing actors have never won an Oscar.D.You think that some actors should have won an Oscar.35. What is the writer’s attitude to Leonardo’s idea on winning an Oscar?A. Approving.B. CriticalC. opposing.D. Neutral.第二节（共5小题；每小题2分，满分10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。
选择性必修三Unit1 高一英语尖子生选拔卷(人教版2019选择性必修三)(原卷版)
选择性必修三Unit 1提分小卷（考试时间：30分钟试卷满分：60分）一、词汇过关：本题共20个小题，每小题1分，共20分。
1.（北京高考完形填空改编）If they run out of certain daily necessities, Wilson will drive to the nearest store and p_________what's needed.2. Confucius is regarded as one of the most_______________（有影响的）figures in Chinese history.3.（全国Ⅲ卷语法填空改编）In ancient China lived an artist whose paintings were almost lifelike. The artist's_______________（名誉）had made him proud.4.Over the past 30 years evidence has_____________（出现）that the forests have been affected by climate change.5.—The hill village remains _______（原始的）and unspoiled（未遭破坏的）.—That explains why people desire to explore such a remote place.6.Relatives and friends gave their best wishes to the b______________and bridegroom.7.Tom was c____________by Mr Smith because he didn't arrive on time.8.Rodin is thought to have made a big contribution to modern_______________(雕塑）9.Calligraphy, which has a long history,is_____________(视觉的）art related to writing.10.The young could suffer __________（永久的）hearing loss if they listen to music at full volume for more than five hours a week.11.The actor many people were fond of gave a great performance,but he was very _______（谦逊的）.12.They suggested building a ____________（纪念碑）in memory of the heroes who died in the rescue activity.13.Many of us regard reading as an __________（投资）in ourselves, so it is natural that we want to learn something useful in return for our effort.14.China's e____________into the World Trade Organization makes a big difference to the world economy.15.It's also helpful to learn something about the museum in advance，such as its___________（展览会）and activities.16. He has gained global____________（赞誉）for his artistic talent.17.（河北衡水中学模考）The mountain's deep valleys and diverse wildlife will___________（确保）you wonderful scenery and a fantastic hiking experience.18.（浙江高考阅读理解改编）The number of trees larger than two feet across has___________（减少）by 50 percent on more than 46, 000 square miles of California forests. 19.I felt lucky to be chosen as a____________（代表）of our school and join in the cultural tour of Pompeii.20.Not all works of art are _____________(永久的）, though.二、课文改编语法填空：本题共10个小题，每小题2分，共20分。
2021-2022年高三化学下学期尖子生专题训练试题(四)
2021-2022年高三化学下学期尖子生专题训练试题（四）可能用到的相对原子质量：C 12 H 1 O 16 S 32 N 14 Fe 56 Mg 24 Mn 55一、选择题（单项选择，每题4分，共56分）1、xx年，我国的航天事业取得了巨大的成就，航天员翟志刚顺利实现了太空行走。
下列说法不正确的是（）A．在生活舱中，可用过氧化钠制备氧气B．金属镁可用作制造飞机、火箭的重要材料C．在航天试验用的仪器仪表中大量使用了硅等非金属材料D．返回舱外层的隔热瓦使用的是金属材料2、化学与生活、社会发展息息相关，下列说法不正确的是( )A．“霾尘积聚难见路人”，雾霾所形成的气溶胶有丁达尔效应B．“熬胆矾铁釜，久之亦化为铜”，该过程发生了置换反应C．“青蒿一握，以水二升渍，绞取汁”，屠呦呦对青蒿素的提取属于化学变化D．古剑“沈卢”“以剂钢为刃，柔铁为茎干，不尔则多断折”，剂钢指的是铁的合金3、下列说法不正确的是( )A．Na与H2O的反应是熵增的放热反应，该反应能自发进行B．饱和Na2SO4溶液或浓硝酸均可使蛋白质溶液产生沉淀，但原理不同C．FeCl3和MnO2均可加快H2O2分解，同等条件下二者对H2O2分解速率的改变相同D．Mg(OH)2固体在溶液中存在平衡：Mg(OH)2(s)Mg2＋(aq)＋2OH－(aq)，该固体可溶于NH4Cl溶液4、下列设计的实验方案能达到实验目的的是( )A. 制备Fe(OH)3胶体：0.1 mol·L－1 FeCl3溶液与0.3 mol·L－1 NaOH溶液等体积混合B. 验证溴原子：取C2H5Br少许与NaOH溶液共热，冷却后滴加AgNO3溶液C. 检验Fe2＋：取少量待测液，向其中加入适量新制氯水，再滴加KSCN溶液D. 比较AgCl和AgI的溶解度：向2 mL 0.1 mol·L－1 NaCl溶液中滴加数滴同浓度AgNO3溶液，然后再滴入KI溶液5、下列叙述正确的是( )A．将Na2SO3和KMnO4溶液加热蒸干后分别得到Na2SO4和KMnO4固体B．往稀氨水中加水，c（H＋）·c（NH3·H2O）c（NH＋4）的值变小C．c(Na＋)相同的三种溶液①CH3COONa、②NaClO、③Na2CO3中pH大小：①＞②＞③D．25 ℃时Cu(OH)2在水中的溶解度大于其在Cu(NO3)2溶液中的溶解度6、室温下，用0.100 mol·L－1 NaOH溶液分别滴定20.00 mL0.100 mol·L－1的盐酸和醋酸，滴定曲线如图所示。
高考数学-下期高三尖子生专题训练(四).docx
2015-2016学年下期高三尖子生专题训练（四）（理科）数学试题时间：120分钟 满分：150分一、选择题(本大题共12小题，每小题5分，共60分；) 1．复数212ii+-的共轭复数是 A.35i - B.35i C.i - D.i2．若f (x )＝x 2－2x －4ln x ，则f ′(x )>0的解集为( )A ．(0，＋∞)B ．(－1,0)∪(2，＋∞)C ．(2，＋∞)D ．(－1,0)3．在等差数列{}811162n a a a =+中，，则数列前9项之和9S 等于（ ）A ． 24B ．48C ．72D ．1084．已知a 与b 均为单位向量，其夹角为θ，有下列四个命题：p 1：|a ＋b |＞1⇔θ∈⎣⎢⎡⎭⎪⎫0，2π3；p 2：|a ＋b |＞1⇔θ∈⎝ ⎛⎦⎥⎤2π3，πp 3：|a －b |＞1⇔θ∈⎣⎢⎡⎭⎪⎫0，π3；p 4：|a －b |＞1⇔θ∈⎝ ⎛⎦⎥⎤π3，π.其中的真命题是( )A ．p 1，p 4B ．p 1，p 3C ．p 2，p 3D ．p 2，p 45．若函数22()(sin cos )2cos f x x x x m =++-在0,2π⎡⎤⎢⎥⎣⎦上有零点，则m 的取值范围为（ ）A. 1,22⎡+⎣B. []1,2-C. 1,22⎡-⎣D. []1,36．如图，四棱锥S －ABCD 的底面为正方形，SD ⊥底面ABCD ，则下列结论中不正确．．．的是( ) A ．AC ⊥SBB ．AB ∥平面SCDC ．SA 与平面SBD 所成的角等于SC 与平面SBD 所成的角 D ．AB 与SC 所成的角等于DC 与SA 所成的角7． 已知324log 0.3log 3.4log 3.615,5,,5a b c ⎛⎫=== ⎪⎝⎭则( )．A ．a b c >>B ．b a c >>C ．a c b >>D ．c a b >>8．设⎩⎨⎧<+≥-=)10()],6([)10(,2)(x x f f x x x f 则)5(f 的值为（ ）A ．10B ．11C ．12D ．139．一个几何体的三视图如图所示，其中正视图是一个正三角形，则这个几何体的外接球的表面积为 （ ）A ．23πB ．83πC ．43D ．163π10．已知函数()sin cos f x x a x =+的图象的一条对称轴是53x π=，则函数()sin cos g x a x x =+ 的最大值是（ ）A ．223B 23C ．43D 2611．设1e 、2e 分别为具有公共焦点1F 、2F 的椭圆和双曲线的离心率，P 是两曲线的一个公共点，且满足1212P F P F F F +=u u u r u u u u r u u u u r ,122212e e +2 B.2 2 D.112．设()f x 是定义在R 上的增函数，且对于任意的x 都有(1)(1)0f x f x -++=恒成立. 如果实数m n 、满足不等式组22(623)(8)03f m m f n n m ⎧-++-<⎨>⎩，那么22m n +的取值范围是A.(3, 7)B.(9, 25)C.(13, 49)D. (9, 49)二、填空题(本大题共4小题，每小题5分，共20分，将答案填写在题中横线上．) 13．由曲线y x =2y x =-及y 轴所围成的图形的面积为________;14．在ABC ∆中，60,3,C AB AB =︒=边上的高为4,3则AC+BC= .15．若点P 在曲线C 1：221169x y -=上，点Q 在曲线C 2：(x －5)2＋y 2＝1上，点R 在曲线C 3：(x ＋5)2＋y 2＝1上，则 | PQ |－| PR | 的最大值是 ．16．设f (x )＝a sin2x ＋b cos2x ，其中a ，b ∈R ，ab ≠0.若f (x )≤⎪⎪⎪⎪⎪⎪f ⎝ ⎛⎭⎪⎫π6对一切x ∈R 恒成立，则①f ⎝⎛⎭⎪⎫11π12＝0；②⎪⎪⎪⎪⎪⎪f ⎝ ⎛⎭⎪⎫7π10<⎪⎪⎪⎪⎪⎪f ⎝ ⎛⎭⎪⎫π5；③f (x )既不是奇函数也不是偶函数； ④f (x )的单调递增区间是⎣⎢⎡⎦⎥⎤k π＋π6，k π＋2π3(k ∈Z )． ⑤存在经过点(a ，b )的直线与函数f (x )的图像不相交． 以上结论正确的是________(写出所有正确结论的编号)．三、解答题(本大题共6小题，共70分，解答应写出文字说明、证明过程或演算步骤) 17．在△ABC 中，角A ，B ，C 的对边分别是a ，b ，c ，已知sin C ＋cos C ＝1－sin C2.(1)求sin C 的值；(2)若a 2＋b 2＝4(a ＋b )－8，求边c 的值．18．已知等差数列}{n a 的公差不为零，且53=a ，521,,a a a 成等比数列．（Ⅰ）求数列}{n a 的通项公式；（Ⅱ）若数列}{n b 满足21123222n n n b b b b a -++++=L ，求数列}{n b 的前n 项和n T ． 19．如图所示，在三棱柱ABC －A 1B 1C 1中，H 是正方形AA 1B 1B 的中心，AA 1＝22，C 1H ⊥平面AA 1B 1B ，且C 1H ＝ 5. (1)求异面直线AC 与A 1B 1所成角的余弦值； (2)求二面角A －A 1C 1－B 1的正弦值； (3)设N 为棱B 1C 1的中点，点M 在平面AA 1B 1B 内，且MN ⊥平面A 1B 1C 1，求线段BM 的长．20．已知平面内一动点P 到点F (1,0)的距离与点P 到y 轴的距离的差等于1.(1)求动点P 的轨迹C 的方程；(2)过点F 作两条斜率存在且互相垂直的直线l 1，l 2，设l 1与轨迹C 相交于点A ，B ，l 2与轨迹C 相交于点D ，E ，求AD →·EB →的最小值．21．已知函数()1(0,)x f x e a x a e =-->为自然对数的底数. ⑴求函数()f x 的最小值；⑵若()f x ≥0对任意的x ∈R 恒成立，求实数a 的值；⑶在⑵的条件下，证明：121()()()()(*)1n n n n n n e n n n n n e -++⋅⋅⋅++<∈-N 其中. 请考生在22、23、24三题中任选一题作答，如果多做，则按所做的第一题计分. 作答时请写清题号.（22）（本小题满分10分）选修4-1：几何证明选讲如图，点A 是以线段BC 为直径的圆O 上一点，AD ⊥BC 于点D,过点B 作圆O 的切线， 与CA 的延长线交于点E ，点G 是AD 的中点，连接CG 并延长与BE 相交于点F,延长AF 与CB 的延长线相交于点P. （Ⅰ）求证：BF=EF; （Ⅱ）求证：PA 是圆O 的切线.23）(本小题满分10分）选修4-4：坐标系与参数方程在直角坐标系xoy 中，l 是过定点P （4,2）且倾斜角为α的直线；在极坐标系（以坐标原点O为极点，以x 轴非负半轴为极轴，取相同长度单位）中，曲线C 的极坐标方程为4cos ρθ=.（Ⅰ）写出直线l的参数方程，并将曲线C的极坐标方程化为直角坐标方程；（Ⅱ）若曲线C与直线l相交于不同两点M、N,求PM PN+的取值范围. 24) (本小题满分10分）选修4-5：不等式选讲已知关于x的不等式2324-++≥+的解集为A.x a x x（Ⅰ）若a=1，求A;（Ⅱ）若A=R,求a的取值范围.参考答案一、选择题；（1）C 2，C 3.D 4. A 5. A. 6. D 7. C 8. B 9. B 10. A 11. A 12. C 二、填空题;13.16314.11①③三、解答题17．【解答】 (1)由已知得sin C ＋sin C2＝1－cos C ，即sin C2⎝ ⎛⎭⎪⎫2cos C2＋1＝2sin 2C2，由sin C 2≠0得2cos C 2＋1＝2sin C 2，即sin C 2－cos C 2＝12，两边平方得：sin C ＝34.(2)由sin C 2－cos C 2＝12＞0得π4＜C 2＜π2，即π2＜C ＜π，则由sin C ＝34得cos C ＝－74，由a 2＋b 2＝4(a ＋b )－8得：(a －2)2＋(b －2)2＝0，则a ＝2，b ＝2.由余弦定理得c 2＝a 2＋b 2－2ab cos C ＝8＋27，所以c ＝7＋1. 18. （1）解：在等差数列中，设公差为)0(≠d d ，2152a a a =Q ，∴2333)()2)(2(d a d a d a -=+-， ……2分化简得01052=-d d ，2=∴d ……4分 122)3(5)3(3-=-+=-+=∴n n d n a a n ……6分（2）解：1123242n n n b b b b a -++++=L ①1123112422n nn n n b b b b b a -+++++++=L ②②-①得： 221=⋅+n n b ，n n b -+=∴112 ……8分当1=n 时，111==a b ⎩⎨⎧=≥=∴-1,,12,22n n b n n ……10分2213--=∴n n T ……12分19.【解答】 方法一：如图所示，建立空间直角坐标系，点B 为坐标原点．依题意得A (22，0,0)，B (0,0,0)，C (2，－2，5)，A 1(22，22，0)，B 1(0,22，0)，C 1(2，2，5)． (1)易得＝(－2，－2，5)，＝(－22，0,0)，于是cos 〈，〉＝＝43×22＝23.所以异面直线AC 与A 1B 1所成角的余弦值为23. (2)易知＝(0,22，0)，＝(－2，－2，5)．设平面AA 1C 1的法向量m ＝(x ，y ，z )， 则即⎩⎨⎧－2x －2y ＋5z ＝0，22y ＝0.不妨令x ＝5，可得m ＝(5，0，2)．同样地，设平面A 1B 1C 1的法向量n ＝(x ，y ，z )，则即⎩⎨⎧－2x －2y ＋5z ＝0，－22x ＝0.不妨令y ＝5，可得n ＝(0，5，2)．于是cos 〈m ，n 〉＝m ·n |m|·|n|＝27·7＝27，从而sin 〈m ，n 〉＝357.所以二面角A －A 1C 1－B 1的正弦值为357.(3)由N 为棱B 1C 1的中点，得N ⎝ ⎛⎭⎪⎫22，322，52.设M (a ，b,0)，则＝⎝ ⎛⎭⎪⎫22－a ，322－b ，52.由MN ⊥平面A 1B 1C 1，得即⎩⎪⎨⎪⎧⎝ ⎛⎭⎪⎫22－a ·(－22)＝0，⎝ ⎛⎭⎪⎫22－a ·(－2)＋⎝ ⎛⎭⎪⎫322－b ·(－2)＋52·5＝0.解得⎩⎪⎨⎪⎧a ＝22，b ＝24，故M ⎝⎛⎭⎪⎫22，24，0. 因此＝⎝⎛⎭⎪⎫22，24，0， 所以线段BM 的长||＝104. 20.【解答】 设动点P 的坐标为(x ，y )，由题意有(x －1)2＋y 2－|x |＝1.化简得y 2＝2x ＋2|x |.当x ≥0时，y 2＝4x ；当x <0时，y ＝0.所以，动点P 的轨迹C 的方程为y 2＝4x (x ≥0)和y ＝0(x <0)． (2)由题意知，直线l 1的斜率存在且不为0，设为k ， 则l 1的方程为y ＝k (x －1)．由⎩⎪⎨⎪⎧y ＝k (x －1)，y 2＝4x 得k 2x 2－(2k 2＋4)x ＋k 2＝0.设A (x 1，y 1)，B (x 2，y 2)，则x 1，x 2是上述方程的两个实根，于是x 1＋x 2＝2＋4k2，x 1x 2＝1.因为l 1⊥l 2，所以l 2的斜率为－1k.设D (x 3，y 3)，E (x 4，y 4)，则同理可得 x 3＋x 4＝2＋4k 2，x 3x 4＝1. 故·＝(＋)·(＋) ＝·＋·＋·＋· ＝||·||＋||·||＝(x 1＋1)(x 2＋1)＋(x 3＋1)(x 4＋1)＝x 1x 2＋(x 1＋x 2)＋1＋x 3x 4＋(x 3＋x 4)＋1＝1＋⎝ ⎛⎭⎪⎫2＋4k 2＋1＋1＋(2＋4k 2)＋1＝8＋4⎝⎛⎭⎪⎫k 2＋1k2≥8＋4×2k 2·1k2＝16.当且仅当k 2＝1k2，即k ＝±1时，·取最小值16.21.解：（1）由题意0,()xa f x e a'>=-， 由()0xf x e a '=-=得l n x a =. 当(,l n)x a ∈-∞时, ()0f x '<；当(l n,)x a ∈+∞时，()0f x '>. ∴()f x 在(,l n )a -∞单调递减，在(l n ,)a +∞单调递增.即()f x 在l n x a =处取得极小值，且为最小值，其最小值为l n (l n )l n 1l n 1.af a e a a a a a =--=-- （4分）（2）()0f x ≥对任意的x ∈R 恒成立，即在x ∈R 上，m i n ()0f x ≥. 由（1），设()l n 1.g a a aa =--，所以()0g a ≥.由()1l n 1l n 0g a a a '=--=-=得1a =. ∴()g a 在区间(0,1)上单调递增，在区间(1,)+∞上单调递减，∴()g a 在1a =处取得极大值(1)0g =.因此()0g a ≥的解为1a =，∴1a =.（8分）（3）由（2）知，因为1a =，所以对任意实数x 均有1xe x --≥0，即1xx e +≤.令kx n=- (*,0,1,2,3,1)n k n ∈=-N …,，则01kn k e n - <-≤. ∴(1)()k n n kn k e e n - --=≤.∴(1)(2)21121()()()()1n n n n n n n n e e e e n n n n-------+++++++++≤…… 1111111ne e e e e ----=<=---. （12分） （22）证明：（Ⅰ） 因为BC 是圆O 的直径，BE 是圆O 的切线，所以EB BC ⊥.又因为AD BC ⊥，所以AD BE ∥，可知BFC DGC ∽△△， FEC GAC ∽△△，所以BF CF EF CFDG CG AG CG==，，所以BF EFDG AG=. 因为G 是AD 的中点，所以DG AG =，所以F 是BE 的中点，BF EF =. …………（5分） （Ⅱ）如图，连接AO AB ，，因为BC 是圆O 的直径，所以90BAC ∠=°．在Rt BAE △中，由（Ⅰ）知F 是斜边BE 的中点， 所以AF FB EF ==，所以FBA FAB ∠=∠. 又因为OA OB =，所以ABO BAO ∠=∠． 因为BE 是圆O 的切线，所以90EBO ∠=°.因为90EBO FBA ABO FAB BAO FAO ∠=∠+∠=∠+∠=∠=°，所以PA 是圆O 的切线.……………………………………………………………………（10分）（23）解：（Ⅰ）直线l 的参数方程为4cos ,(2sin x t t y t αα=+⎧⎨=+⎩为参数).………………………（2分）因为4cos ρθ=，所以24cos ρρθ=，所以曲线C 的直角坐标方程为224x y x +=.…………………………………………………………………………………………………（4分）（Ⅱ）将4cos ,2sin x t y t αα=+⎧⎨=+⎩代入22:4C x y x +=中，得24(sin cos )40t t αα+++=，则有2121216(sin cos )160,4(sin cos ),4,t t t t ∆αααα⎧=+->⎪+=-+⎨⎪=⎩………………………………………………………（6分） 所以sin cos 0αα>.又[0,π)α∈，所以π0,2α⎛⎫∈ ⎪⎝⎭， 1212||||||||()t t t PN t PM +=-++==π4(sin cos )4ααα⎛⎫+=+ ⎪⎝⎭，………（8分）由ππ3π,444α⎛⎫+∈ ⎪⎝⎭πsin 14α⎛⎫<+ ⎪⎝⎭„，所以||||(4,PM PN +∈.………（10分） （24）解：（Ⅰ）当3x -„时,原不等式化为3224x x --+…, 得3x -„； 当132x -<„时,原不等式化为424x x -+…,得30x -<„； 当12x >时，原不等式化为3224x x ++…,得2x …， 综上，{|0A x x =„或2}x ….………………………………………………………………（5分） （Ⅱ）当240,x +„即2x -„时,|2||3|024x a x x -+++厖成立，当240,x +>.即2x >-时, |2||3||2|324x a x x a x x -++=-+++…，得1x a +…或13a x -„, 所以12a +-„或113a a -+„,得2a -„. 综上，a 的取值范围为(],2-∞-.…………………………………………………………（10分）。
六年级尖子生综合训练4
六年级尖子生综合训练4work Information Technology Company.2020YEAR六年级尖子生综合训练（4）31、（1）n条直线，最多把平面分成几个部分？（2）n个平面，最多把空间分成几个部分？32、用n张2×1的纸片，去覆盖一张2×n的棋盘，有多少种不同的方法a n求a10的值。
33、从1、2、3、…、100这100个数中任意挑出51个数字，证明在这51个数中，一定：（1）有2个数互质；（2）有2个数的差为50；（3）有8个数，它们的最大公约数大于1。
34、求证：可以找到一个各位数字都是4的自然数，它是1996的倍数。
35、有17个科学家，其中每一个人与其他所有人通信，他们的通信仅讨论三个题目，且每两个科学家之间只讨论一个题目。
求证：至少有三个科学家相互之间讨论同一问题。
36、线段AB上有1998个点（包括A、B两点），将点A染成红色，点B 染成蓝色，其余各点染成红色或蓝色。
这时，图中共有1997条互不重叠的线段。
问：两个端点颜色相异的小线段的条数是奇数还是偶数为什么37、有n名（n≥3）选手参加的一次乒乓球循环赛中，没有一个全胜的。
问：是否能够找到三名选手A、B、C，使得A胜B，B胜C，C胜A？38、有三堆石子，每堆分别有1998、998、98粒。
现在对这三堆石子进行如下的“操作”：每次允许从每堆中拿掉一个或相同个数的石子，或从任一堆中取出一些石子放入另一堆中。
按上述方法进行“操作”，能否把这三堆石子都取光？如能，请设计一种取石子的方案；如不能，请说明理由。
39、在八边形的8个顶点上是否可以分别记上数1，2，…，8，使得任意三个相邻的顶点上的数的和大于13？40、如右图，正方体的8个顶点处标注的数字为a 、b 、c 、d 、e 、f 、g 、h ，其中每个数都等于相邻3个顶点处的数的和的31。
求（a ＋b ＋c ＋d ）－（e ＋f ＋g ＋h ）的值。
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a
．右图为“细胞直接与内环境进行物质交换”的图解，其中②④⑤为细胞外液。
下列相关叙述中，错误的是( )
．图中①和③细胞具体的内环境分别是血液和组织液
．某人皮肤烫伤后，出现了水泡，该水泡内的液体主要是指．右图为人体内环境三种主要组成成分之间的相互转化模式图，．人体中内环境稳态所包含的内容不包括( )
．细胞质基质渗透压的相对稳定
依次为组织液、淋巴、细胞内液
．右图表示人运动前后血液中乳酸浓度的变化曲线，那么BC
．下列有关内环境和稳态的说法，不正确的是( )
．血浆中的蛋白质含量明显高于淋巴和组织液
渗透压等保持相对稳定是细胞正常代谢
的方式是协助运输
．由于血液中缓冲物质的调节作用，使内环境维持在相对稳
B、C、D各代表体液中的哪一类？血浆
________、细胞内液________。
相对于内环境来说，相当于外界环境的是A、。
假如①为淀粉，则所形成的②从肠腔进入B
的运输，②可达到身体各部分的组织细胞，
有见到他。
次日在农田里发现他时，已经死亡多时。
据估计，下图中曲线能反映赵某(a)和高某(b)内环境渗透压变化趋势
______________。
高某的细胞内液的渗透压的变化趋势是_______________
高某血压下降的原因是_____________________________
赵某的死亡说明
p
量的Na2CO3溶液(或乳酸)
(3)试管1 pH变化明显，pH明显降低(或升高)，2、3试管pH 变化不明显试管1 pH变化明显，pH明显升高(或降低)，2、3试管pH变化不明显
(4)动物血浆与缓冲液一样具有缓冲作用，说明其成分中含有缓冲物质。