2020年济宁市高三第一次模拟考试题及答案

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2020届济宁市第一中学高三英语一模试卷及参考答案

2020届济宁市第一中学高三英语一模试卷及参考答案

2020届济宁市第一中学高三英语一模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AStudents, teachers, and local community members are strongly encouraged to register online to receive real-time information of emergency events fromColumbiaUniversity.Text message warnings will only be used in rare cases where ongoing events causean immediate threat or have a significant influence. Possible situations include severe weather conditions, emergency campus closures, crimes in progress that may endanger the community, and major transportation interruptions.Read instructions on how to sign up for emergency notifications(通知).ColumbiastudentsColumbiastudents can easily register for text message notifications. Simply enter Student Services Online, click on "Text Message Enrollment" and add your cellphone number. Students can register parents or family members by following the instructions for the public below.ColumbiateachersColumbiateachers can register for text message notifications by following the step by step instructions below:● Register with your UNI and password .● Select "Faculty & Staff"at the top of the page.● Select "Contact Details".● Click the "+" under "Phone".● Select "Campus Alert 1".● Enter your mobile number.To receive emergency information on additional mobile phones, you may follow the above instructions for "Campus Alert2" and "Campus Alert 3" for a total of three.Community/General PublicCommunity/General Public interested in receiving emergency information fromColumbiacan sign up by clicking on the "Register" button on the registration page and entering their email and mobile number. Users will receive confirmation code(密码)on their phone and will need to confirm their account via email.Users can choose not to use the Emergency Text Alert System at any time by texting STOP to 226787, calling226789 or sending an email tolr27682@.1.Which is a possiblesituation where a text message will be sent?A.The temperature will drop slightly tomorrow.B.The campus will be closed during Christmas.C.The main railway system of the city is interrupted.D.A bank robber is being sentenced in the court of the city.2.Which step comes before the others when teachers register?A.Selecting Campus Alert.B.Entering mobile numbers.C.Clicking "+" under "Phone".D.Selecting "Faculty& Staff".3.How can users stop receiving text messages?A.By sending an email.B.By texting STOP to 226789.C.By managing information online.D.By making a phone call to 226787.BDengue is a very painful illness spread by mosquitoes. In severe cases, dengue can even be deadly. Dengue is a serious disease affecting people in around 120 countries. It can cause high fevers, headaches, and severe pain. It’s caused by a virus spread by bites from mosquitoes. Therefore, dengue is more common in warm areas. Every year, roughly 390 million people get dengue, and as many as 25,000 die from it.Now scientists seem to have found a way to protect humans from dengue by first protecting mosquitoes. Dengue fever is caused by a virus. Though it may seem strange to think of it this way, the mosquitoes that spread the dengue virus are also infected with it. But the virus doesn’t seem to hurt the mosquitoes.Wolbachia is a kind of bacteria commonly found in many insects. In some insects, Wolbachia can keep some viruses fromduplicatingthemselves, which is how viruses grow inside a body. Wolbachia isn’t naturally found in mosquitoes. But by infecting these mosquitoes with Wolbachia, scientists can keep the mosquitoes from catching the dengue virus. Even better, the young mosquitoes coming from the eggs of the infected mosquitoes also carry Wolbachia.Researchers working with the World Mosquito Program (WMP) ran a 27-month study in Yogyakarta, Indonesia. They split a 10-square-mile area up into 24 smaller areas. In half of the areas, the scientists did nothing. In the other half, they set out containers of eggs from mosquitoes that had Wolbachia. They did this every twoweeks for just 4 to 6 months.Ten months later, 80% of the mosquitoes in the treated areas carried Wolbachia. The researchers report the number of dengue cases in the treated areas was reduced by 77% and that the number of people needing hospital care for dengue dropped by 86%.Because the results of the experiment were so good, the WHO has placed Wolbachia-infected mosquito eggs in all parts of Yogyakarta and surrounding areas. The WHO says that within a year, their efforts will protect 2.5 million people against dengue and that their efforts will be turned into a program that can be repeated worldwide.4. What kind of disease is dengue?A. It is likely to cause death.B. It causes no pain but fevers.C. It happens less often in hot areas.D. It hurts both people and mosquitoes.5. The underlined word “duplicating” in paragraph 3 most probably means “________”.A. worsening the harm ofB. expanding the size ofC. increasing forces ofD. making copies of6. What can be inferred about the method from the figures listed in paragraph 5?A. Its wide use.B. Its effectiveness.C. Its complexity.D.Its easy operation.7. What’s the WHO’s attitude towards the method?A. Ambiguous.B. Positive.C. Tolerant.D. Skeptical.CWhere doyou usually put your toothbrush?Do you keep it in the bathroom? How’s your toothbrush looking these days? Even if you can’t see it with a naked eye, experts say it may be saturated(使饱和)with millions of toilet germs!Dr. Charles Oerba, a germ expert, is amicrobiology professor at the University of Arizona. He says there are approximately 3 million bacteria per square inch in most toilet bowls, and every time you flush it without closing the lid, those millions of bacteria droplets spray into the air as far as twenty feet away and dirty everything in their path. And a common victim is your poor toothbrush, usually, left out on the bathroom sink, right?So, what do we do? Dr. Gerba says it’s easy. Close the toilet lid before you flush—that’ll greatly cut downthe germs, which will otherwise float in the air. And wash your toothbrush every few days in mouthwash or peroxide to get rid of any germs hiding in it. You can even put it through the dishwasher to sanitize(消毒)it. And always store your toothbrush in a closed cabinet.Here’s one more tip from Dr. Gerba, who says our kitchen sink is probably dirtier than our toilet. “If an alien came from space and studied the bacterial counts, he probably would conclude he should wash his hands in your toilet and go to the bathroom in your sink.” He says that’s because the kitchen sink is a great place where E. coli(大肠杆菌)to live and grow since it’s wet and damp. Bacteria feed on the food that people put down the drain or—that’s left on dishes in the sink. To reset your sink’s bacteria count back to zero, you’d better regularly wash it with hot water and sanitize yoursink with special chemicals. In fact, you may want to do it every day or before preparing dinner.8. What is the purpose of the text?A. To show how to brush your teeth.B. To tell people the importance of health.C. To warn people of the invisible germs.D. To introduce a microbiology professor.9. What can we learn from Paragraph 2?A. Bathroom sinks are the dirtiest places.B. Bacteria are bad for people’s health.C. Why bacteria spread through the air.D. How bacteria spread in the bathroom.10. What does the underlined word“that”in Paragraph 4 refer to?A. The food.B. The toothbrush.C. The sink.D. The chemical.11. Why does Dr. Gerba mention the example of an alien?A. To tell us a fiction story of an alien studying bacteria.B. To show our kitchen sink may be dirtier than our toilet.C. To teach us how to reset sink’s bacteria count back to zero.D. To prove coli prefers to live in the kitchen and the drain.DAfter a year at sea, 16-year-old Laura Dekker can finally say, “Missionaccomplished!” Last month, she finished a daring trip around the world aboard her 38-foot boat, Guppy. Dekker, who is from theNetherlands, traveled more than 30,000 miles all by herself. She is the youngest person ever to sail around the globe alone.Dekker had wanted to lake on this challenge when she was even younger. She first tried to set sail at the age of 13,but a court in theNetherlandsstopped her. They said that she was too young to make such a risky trip by herself. But Dekker insisted she had the navigation skills and patience of an adult sailor.She finally took off on January 20, 2011. During her trip, Dekker battled loneliness, storms, and worries about pirates. But she also got to surf, scuba dive, and started a new hobby: playing the flute. Although Dekker didn’t spend all of her time at sea—she stopped at ports along the way—she did spend her 16lh birthday on the open ocean. To celebrate, she ate doughnuts for breakfast.But Dekker didn’t sail into the record books. Guinness World Records and the World Sailing Speed Record Council no longer recognize records for “youngest” sailors. They dropped the category in 2009 to discourage children fromattempting such dangerous feats (壮举). But that didn’t stop Dekker,who was born on a yacht during a seven-year world voyage undertaken by her parents.Dekker doesn’t mind that she won’t hold an official record. She says it was a personal goal, and she is happy she achieved it.“I am not disappointed at all that Guinness World Record won’t recognize my attempt.” Dekker wrote on her website. “I did not start on my trip to achieve any record…I did it just for myself.”12. Dekker wasn’t allowed to sail at the age of 13 because .A. people were concerned about her safetyB.she didn’t learn any sailing skills wellC. she had to continue her study at schoolD. she didn’t have enough patience for long trips13. What does the underlined word “accomplished” in Paragraph 1 mean?A. Failed.B. Completed.C. Continued.D. Started.14. In 2009 records for youngest sailors were canceled in order to .A. set a higher sailing standard for teen sailorsB. stop children entering Guinness World RecordsC. encourage parents to sail with their childrenD. prevent children making dangerous attempts15. What could be the best title for this passage?A. A New Guinness World Record in SailingB. A New Sailing Standard for Teen SailorsC. A Teen Girl Sailing Alone Aroundthe WorldD. The Youngest Sailor in Guinness World Records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。

2020年高考模拟山东省济宁市嘉祥一中高考数学第一次模拟试卷 含解析

2020年高考模拟山东省济宁市嘉祥一中高考数学第一次模拟试卷 含解析

2020年高考数学第一次模拟测试试卷一、选择题1.若全集U=R,集合A={x∈Z|x2<16},B={x|x﹣1≤0},则A∩(∁U B)=()A.{x|1≤x<4}B.{x|1<x<4}C.{1,2,3}D.{2,3}2.复数z满足,则|z|=()A.2i B.2C.i D.13.已知向量=(3,﹣4),=(6,﹣3),=(2m,m+1).若,则实数m的值为()A.B.C.﹣3D.﹣4.函数f(x)=的部分图象是()A.B.C.D.5.“a<﹣1”是“∃x0∈R,a sin x0+1<0”的()A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分也不必要条件6.若,则a+2b的最小值为()A.6B.C.3D.7.已知圆C:x2+y2﹣10y+21=0与双曲线的渐近线相切,则该双曲线的离心率是()A.B.C.D.8.已知正三棱锥S﹣ABC的侧棱长为4,底面边长为6,则该正三棱锥外接球的表面积是()A.16πB.20πC.32πD.64π二、多项选择题9.已知a,b,c,d均为实数,则下列命题正确的是()A.若a>b,c>d,则ac>bdB.若ab>0,bc﹣ad>0,则C.若a>b,c>d,则a﹣d>b﹣cD.若a>b,c>d>0,则10.已知α,β是两个不重合的平面,m,n是两条不重合的直线,则下列命题正确的是()A.若m∥n,m⊥α,则n⊥αB.若m∥α,α∩β=n,则m∥nC.若m⊥α,m⊥β,则α∥βD.若m⊥α,m∥n,n∥β,则α∥β11.如图,在四边形ABCD中,AB∥CD,AB⊥AD,AB=2AD=2DC,E为BC边上一点,且,F为AE的中点,则()A.B.C.D.12.已知函数f(x)是定义在R上的奇函数,当x<0时,f(x)=e x(x+1),则下列命题正确的是()A.当x>0时,f(x)=﹣e﹣x(x﹣1)B.函数f(x)有3个零点C.f(x)<0的解集为(﹣∞,﹣1)∪(0,1)D.∀x1,x2∈R,都有|f(x1)﹣f(x2)|<2三、填空题13.在△ABC中,内角A,B,C的对边分别为a,b,c,若,b2+c2﹣a2=bc,则tan B=.14.我国古代的天文学和数学著作《周髀算经》中记载:一年有二十四个节气,每个节气晷(guǐ)长损益相同(晷是按照日影测定时刻的仪器,晷长即为所测量影子的长度),夏至、小署、大暑、立秋、处暑、白露、秋分、寒露、霜降、立冬、小雪、大雪是连续十二个节气,其日影子长依次成等差数列,经记录测算,夏至、处暑、霜降三个节气日影子长之和为16.5尺,这十二节气的所有日影子长之和为84尺,则夏至的日影子长为尺.15.已知抛物线y2=2px(p>0)的焦点为F(4,0),过F作直线l交抛物线于M,N两点,则p=,的最小值为.16.设函数f(x)在定义域(0,+∞)上是单调函数,∀x∈(0,+∞),f[f(x)﹣e x+x]=e,若不等式f(x)+f'(x)≥ax对x∈(0,+∞)恒成立,则实数a的取值范围是.四、解答题:本题共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,内角A,B,C所对的边分别为a,b,c.已知a sin A=4b sin B,ac=(a2﹣b2﹣c2)(Ⅰ)求cos A的值;(Ⅱ)求sin(2B﹣A)的值18.已知数列{a n}的前n项和S n=3n2+8n,{b n}是等差数列,且a n=b n+b n+1.(Ⅰ)求数列{b n}的通项公式;(Ⅱ)令c n=,求数列{c n}的前n项和T n.19.如图,三棱柱ABC﹣A1B1C1中,AB⊥侧面BB1C1C,已知∠BCC1=,BC=1,AB =C1C=2,点E是棱C1C的中点.(1)求证:C1B⊥平面ABC;(2)在棱CA上是否存在一点M,使得EM与平面A1B1E所成角的正弦值为,若存在,求出的值;若不存在,请说明理由.20.为了解某地区初中学生的体质健康情况,统计了该地区8所学校学生的体质健康数据,按总分评定等级为优秀,良好,及格,不及格.良好及其以上的比例之和超过40%的学校为先进校.各等级学生人数占该校学生总人数的比例如表:学校比例等级学校A学校B学校C学校D学校E学校F学校G学校H优秀8%3%2%9%1%22%2%3%良好37%50%23%30%45%46%37%35%及格22%30%33%26%22%17%23%38%不及格33%17%42%35%32%15%38%24%(Ⅰ)从8所学校中随机选出一所学校,求该校为先进校的概率;(Ⅱ)从8所学校中随机选出两所学校,记这两所学校中不及格比例低于30%的学校个数为X,求X的分布列;(Ⅲ)设8所学校优秀比例的方差为S12,良好及其以下比例之和的方差为S22,比较S12与S22的大小.(只写出结果)21.已知椭圆C:3x2+4y2=12.(Ⅰ)求椭圆C的离心率;(Ⅱ)设A,B分别为椭圆C的左右顶点,点P在椭圆C上,直线AP,BP分别与直线x=4相交于点M,N.当点P运动时,以M,N为直径的圆是否经过x轴上的定点?试证明你的结论.22.已知函数f(x)=m sin(1﹣x)+lnx.(1)当m=1时,求函数f(x)在(0,1)的单调性;(2)当m=0且时,,求函数g(x)在(0,e]上的最小值;(3)当m=0时,有两个零点x1,x2,且x1<x2,求证:x1+x2>1.参考答案一、单项选择题:本题共8小题,每小题5分,共40分.在每小题给出的四个选项中,只有一项是符合题目要求的.1.若全集U=R,集合A={x∈Z|x2<16},B={x|x﹣1≤0},则A∩(∁U B)=()A.{x|1≤x<4}B.{x|1<x<4}C.{1,2,3}D.{2,3}【分析】可以求出集合A,B,然后进行补集和交集的运算即可.解:A={x∈Z|﹣4<x<4}={﹣3,﹣2,﹣1,0,1,2,3},B={x|x≤1},∴∁U B={x|x>1},A∩(∁U B)={2,3}.故选:D.2.复数z满足,则|z|=()A.2i B.2C.i D.1【分析】根据已知条件,先求出复数z的代数形式,代入模长公式即可.解:依题意,因为复数z满足,所以z===i,所以|z|=1,故选:D.3.已知向量=(3,﹣4),=(6,﹣3),=(2m,m+1).若,则实数m的值为()A.B.C.﹣3D.﹣【分析】先求得得==(3,1),再由,则这两个向量的坐标对应成比例,解方程求得实数m的值,可得结论.解:由题意可得==(3,1),若,则这两个向量的坐标对应成比例,即,解得m=﹣3,故选:C.4.函数f(x)=的部分图象是()A.B.C.D.【分析】根据题意,由排除法分析:分析可得f(x)为奇函数,排除B,结合函数的解析式可得当0<x<1时,f(x)<0,排除C,当x>1时,f(x)>0,排除D;据此即可得答案.解:根据题意,f(x)=,其定义域为{x|x≠0},又由f(﹣x)==﹣f(x),即函数f(x)为奇函数,排除B,当0<x<1时,ln|x|=lnx<0,x3>0,则有f(x)<0,排除C,当x>1时,ln|x|=lnx>0,x3>0,则有f(x)>0,排除D,故选:A.5.“a<﹣1”是“∃x0∈R,a sin x0+1<0”的()A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分也不必要条件【分析】设f(x)=a sin x+1,分类求得函数的值域,由∃x0∈R,a sin x0+1<0求得a的范围,可知“a<﹣1”是“∃x0∈R,a sin x0+1<0”的不必要条件;取,当a<﹣1时,a sin x0+1<0成立,说明“a<﹣1”是“∃x0∈R,a sin x0+1<0”的充分条件.解:必要性:设f(x)=a sin x+1,当a>0时,f(x)∈[1﹣a,1+a],∴1﹣a<0,即a >1;当a<0时,f(x)∈[1+a,1﹣a],∴1+a<0,即a<﹣1.故a>1或a<﹣1;充分性:取,当a<﹣1时,a sin x0+1<0成立.∴“a<﹣1”是“∃x0∈R,a sin x0+1<0”的充分不必要条件.故选:A.6.若,则a+2b的最小值为()A.6B.C.3D.【分析】,变形log3(2a+b)=1+log3ab,可得a,b>0,+=3,可得a+2b=(a+2b)(+)=(5++),利用基本不等式的性质即可得出.解:,∴log3(2a+b)=1+log3ab,∴2a+b=3ab,a,b>0.化为:+=3.则a+2b=(a+2b)(+)=(5++)≥(5+2×2)=3,当且仅当a=b=1时取等号.故选:C.7.已知圆C:x2+y2﹣10y+21=0与双曲线的渐近线相切,则该双曲线的离心率是()A.B.C.D.【分析】由双曲线的标准方程写出渐近线方程,利用圆心到切线的距离d=r,列方程求出离心率e=的值.解:双曲线﹣=1的渐近线方程为bx±ay=0,圆C:x2+y2﹣10y+21=0化为标准方程是:x2+(y﹣5)2=4,则圆心C(0,5)到直线bx﹣ay=0的距离为d=r;即==2,解得=,即双曲线的离心率是e=.故选:C.8.已知正三棱锥S﹣ABC的侧棱长为4,底面边长为6,则该正三棱锥外接球的表面积是()A.16πB.20πC.32πD.64π【分析】正棱锥的外接球的球心在顶点向底面做投影所在的直线上,先求底面外接圆的半径,再由勾股定理求锥的高,由勾股定理求出外接球的半径,由球的表面积公式求出表面积.解:如图所示:由正棱锥得,顶点在底面的投影是三角形ABC的外接圆的圆心O',接圆的半径r,正三棱锥的外接球的球心在高SO'所在的直线上,设为O,连接OA得,:r=,∴r=2,即O'A=2,所以三棱锥的高h===6,由勾股定理得,R2=r2+(R﹣h)2,解得:R=4,所以外接球的表面积S=4πR2=64π.故选:D.二、多项选择题:本题共4小题,每小题5分,共20分.在每小题给出的选项中,有多项符合题目要求,全部选对的得5分,部分选对的得3分,有选错的得0分.9.已知a,b,c,d均为实数,则下列命题正确的是()A.若a>b,c>d,则ac>bdB.若ab>0,bc﹣ad>0,则C.若a>b,c>d,则a﹣d>b﹣cD.若a>b,c>d>0,则【分析】利用不等式的基本性质,或者反例判断选项的正误即可.解:若a>b>0,c>d>0,则ac>bd,所以A不正确;若ab>0,bc﹣ad>0,可得,即﹣>0,所以B正确;若a>b,c>d,则a+c>b+d,即a﹣d>b﹣c,所以C正确;若a>b,c>d>0,则.不正确,反例a=1,b=﹣1,c=﹣2,d=﹣3,显然,,所以D不正确.故选:BC.10.已知α,β是两个不重合的平面,m,n是两条不重合的直线,则下列命题正确的是()A.若m∥n,m⊥α,则n⊥αB.若m∥α,α∩β=n,则m∥nC.若m⊥α,m⊥β,则α∥βD.若m⊥α,m∥n,n∥β,则α∥β【分析】利用空间线面、面面位置关系的判定即可得出结论.解:A.由m∥n,m⊥α,则n⊥α,正确;B.由m∥α,α∩β=n,则m与n的位置关系不确定;C.由m⊥α,m⊥β,则α∥β正确D.由m⊥α,m∥n,n∥β,则α⊥β,因此不正确.故选:AC.11.如图,在四边形ABCD中,AB∥CD,AB⊥AD,AB=2AD=2DC,E为BC边上一点,且,F为AE的中点,则()A.B.C.D.【分析】利用向量的加法法则,先用,进而表示出.解:由AB=2AD=2DC知:∵,∴==,故A选项正确.又∵,∴===,故B选项正确.∵,∴=,故C正确.∵==,D不正确.故选:ABC.12.已知函数f(x)是定义在R上的奇函数,当x<0时,f(x)=e x(x+1),则下列命题正确的是()A.当x>0时,f(x)=﹣e﹣x(x﹣1)B.函数f(x)有3个零点C.f(x)<0的解集为(﹣∞,﹣1)∪(0,1)D.∀x1,x2∈R,都有|f(x1)﹣f(x2)|<2【分析】函数f(x)是定义在R上的奇函数,当x<0时,f(x)=e x(x+1),设x>0时,﹣x<0,可得f(x)=﹣f(﹣x)=e﹣x(x﹣1),x=0时,f(0)=0.当x<0时,f(x)=e x(x+1),f′(x)=)=e x(x+2),可得x=﹣2时,函数f(x)取得极小值,进而判断出结论.解:函数f(x)是定义在R上的奇函数,当x<0时,f(x)=e x(x+1),设x>0时,﹣x<0,f(﹣x)=e﹣x(﹣x+1),∴f(x)=﹣f(﹣x)=e﹣x(x﹣1),x=0时,f(0)=0.因此函数f(x)有三个零点:0,±1.当x<0时,f(x)=e x(x+1),f′(x)=)=e x(x+2),可得x=﹣2时,函数f(x)取得极小值,f(﹣2)=.可得其图象:f(x)<0时的解集为:(﹣∞,﹣1)∪(0,1).∀x1,x2∈R,都有|f(x1)﹣f(x2)|≤|f(0+)﹣f(0﹣)|<2.因此BCD都正确.故选:BCD.三、填空题:本题共4小题,每小题5分,共20分.13.在△ABC中,内角A,B,C的对边分别为a,b,c,若,b2+c2﹣a2=bc,则tan B=4.【分析】先由余弦定理求出cos A的值,结合正弦定理进行化简即可.解:由b2+c2﹣a2=bc得cos A===,则sin A=,若,则+==1,即+=1,得=,得tan B=4,故答案为:4.14.我国古代的天文学和数学著作《周髀算经》中记载:一年有二十四个节气,每个节气晷(guǐ)长损益相同(晷是按照日影测定时刻的仪器,晷长即为所测量影子的长度),夏至、小署、大暑、立秋、处暑、白露、秋分、寒露、霜降、立冬、小雪、大雪是连续十二个节气,其日影子长依次成等差数列,经记录测算,夏至、处暑、霜降三个节气日影子长之和为16.5尺,这十二节气的所有日影子长之和为84尺,则夏至的日影子长为1.5尺.【分析】根据题意列等式,再用等差数列的通项公式和求和公式去求解,即得.解:由题意知为单调递增的等差数列,设为a1,a2,…,a12,公差为d,,代入得,联立方程解得a1=1.5,故答案为:1.5.15.已知抛物线y2=2px(p>0)的焦点为F(4,0),过F作直线l交抛物线于M,N两点,则p=8,的最小值为.【分析】先有焦点坐标求出p,再讨论当直线l的斜率不存在时,求出答案,当直线l的斜率存在时,根据韦达定理和抛物线的定义即可求出+=,代入,根据基本不等式即可求最小值解:抛物线y2=2px的焦点F,因为F(4,0),∴=4⇒p=8⇒y2=16x;当直线l的斜率不存在时,直线l为x=4,由,可得M(4,8),N(4,﹣8),∴|MF|=|NF|=8,∴=﹣=;当直线l的斜率存在时,设过点F作直线l的方程为y=k(x﹣4),不妨设M(x1,y1),N(x2,y2),由,消y可得k2x﹣(16+8k2)x+16k2=0,∴x1+x2=8+,x1x2=16,∴|MF|=x1+=x1+4,|NF|=x2+=x2+4,∴+=+===.∴=﹣4(﹣)=+﹣1≥2﹣1=.(当且仅当|NF|=6时等号成立).故答案为:8,.16.设函数f(x)在定义域(0,+∞)上是单调函数,∀x∈(0,+∞),f[f(x)﹣e x+x]=e,若不等式f(x)+f'(x)≥ax对x∈(0,+∞)恒成立,则实数a的取值范围是{a|a ≤2e﹣1}.【分析】由已知可得f(x)=e x﹣x+t,且f(t)=e t,进而可求t及f(x),然后代入已知不等式,结合恒成立与最值求解的相互转化可求.解:令t=f(x)﹣e x+x,所以f(x)=e x﹣x+t,因为f(x)在定义域(0,+∞)上是单调函数,∀x∈(0,+∞),f[f(x)﹣e x+x]=e,故t为常数且f(t)=e t=e,所以,t=1,f(x)=e x﹣x+1,f′(x)=e x﹣1因为f(x)+f'(x)≥ax对x∈(0,+∞)恒成立,所以2e x≥(a+1)x对x∈(0,+∞)恒成立,即a+1对x∈(0,+∞)恒成立,令g(x)=,x>0,则g′(x)=,当x>1时,g′(x)>0,g(x)单调递增,当0<x<1时,g′(x)<0,g(x)单调递减,故当x=1时,函数取得最小值g(1)=2e,故a+1≤2e即a≤2e﹣1.故答案为:{a|a≤2e﹣1}.四、解答题:本题共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,内角A,B,C所对的边分别为a,b,c.已知a sin A=4b sin B,ac=(a2﹣b2﹣c2)(Ⅰ)求cos A的值;(Ⅱ)求sin(2B﹣A)的值【分析】(Ⅰ)由正弦定理得a sin B=b sin A,结合a sin A=4b sin B,得a=2b.再由,得,代入余弦定理的推论可求cos A的值;(Ⅱ)由(Ⅰ)可得,代入a sin A=4b sin B,得sin B,进一步求得cos B.利用倍角公式求sin2B,cos2B,展开两角差的正弦可得sin(2B﹣A)的值.【解答】(Ⅰ)解:由,得a sin B=b sin A,又a sin A=4b sin B,得4b sin B=a sin A,两式作比得:,∴a=2b.由,得,由余弦定理,得;(Ⅱ)解:由(Ⅰ),可得,代入a sin A=4b sin B,得.由(Ⅰ)知,A为钝角,则B为锐角,∴.于是,,故.18.已知数列{a n}的前n项和S n=3n2+8n,{b n}是等差数列,且a n=b n+b n+1.(Ⅰ)求数列{b n}的通项公式;(Ⅱ)令c n=,求数列{c n}的前n项和T n.【分析】(Ⅰ)求出数列{a n}的通项公式,再求数列{b n}的通项公式;(Ⅱ)求出数列{c n}的通项,利用错位相减法求数列{c n}的前n项和T n.解:(Ⅰ)S n=3n2+8n,∴n≥2时,a n=S n﹣S n﹣1=6n+5,n=1时,a1=S1=11,∴a n=6n+5;∵a n=b n+b n+1,∴a n﹣1=b n﹣1+b n,∴a n﹣a n﹣1=b n+1﹣b n﹣1.∴2d=6,∴d=3,∵a1=b1+b2,∴11=2b1+3,∴b1=4,∴b n=4+3(n﹣1)=3n+1;(Ⅱ)c n========6(n+1)•2n,∴T n=6[2•2+3•22+…+(n+1)•2n]①,∴2T n=6[2•22+3•23+…+n•2n+(n+1)•2n+1]②,①﹣②可得﹣T n=6[2•2+22+23+…+2n﹣(n+1)•2n+1]=12+6×﹣6(n+1)•2n+1=(﹣6n)•2n+1=﹣3n•2n+2,∴T n=3n•2n+2.19.如图,三棱柱ABC﹣A1B1C1中,AB⊥侧面BB1C1C,已知∠BCC1=,BC=1,AB =C1C=2,点E是棱C1C的中点.(1)求证:C1B⊥平面ABC;(2)在棱CA上是否存在一点M,使得EM与平面A1B1E所成角的正弦值为,若存在,求出的值;若不存在,请说明理由.【分析】(1)推导出BC1⊥BC,AB⊥BC1,由此证明C1B⊥平面ABC;(2)以B为原点,BC,BC1,BA分别为x,y,z轴,建立空间直角坐标系,设在棱CA 上存在点M,使得EM与平面A1B1E所成角的正弦值为,由=λ,λ∈[0,1],求出的坐标及平面A1B1E的法向量,利用法向量求出EM与平面A1B1E所成角的正弦值,列方程求出λ的值即可.【解答】(1)证明:∵BC=1,CC1=2,∠BCC1=,∴BC1=,∴BC2+BC12=CC12,得BC1⊥BC,又AB⊥侧面BB1C1C,∴AB⊥BC1,又AB∩BC=B,∴C1B⊥平面ABC;(2)以B为原点,BC,BC1,BA分别为x,y,z轴,建立空间直角坐标系,则B(0,0,0),A(0,0,2),B1(﹣1,,0),A1(﹣1,,2),E(,,0),C(1,0,0).则=(﹣,,0),=(0,0,2).设平面A1EB1的法向量为=(x,y,z),则,令x=1,求得=(1,,0).假设在棱CA上存在一点M(a,b,c),使得EM与平面A1B1E所成角的正弦值为,不妨设=λ,λ∈[0,1].又=(a﹣1,b,c),=(﹣1,0,2),∴,∴M(1﹣λ,0,2λ),∴=(﹣λ,﹣,2λ),又平面A1B1E的法向量为=(1,,0),则EM与平面A1B1E所成角的正弦值为:|cos<,>|===,化简得69λ2﹣38λ+5=0,解得λ=或λ=.∴在棱CA上存在点M,使得EM与平面A1B1E所成角的正弦值为.此时=或.20.为了解某地区初中学生的体质健康情况,统计了该地区8所学校学生的体质健康数据,按总分评定等级为优秀,良好,及格,不及格.良好及其以上的比例之和超过40%的学校为先进校.各等级学生人数占该校学生总人数的比例如表:学校学校学校B学校C学校D学校E学校F学校G学校H比例A等级优秀8%3%2%9%1%22%2%3%良好37%50%23%30%45%46%37%35%及格22%30%33%26%22%17%23%38%不及格33%17%42%35%32%15%38%24%(Ⅰ)从8所学校中随机选出一所学校,求该校为先进校的概率;(Ⅱ)从8所学校中随机选出两所学校,记这两所学校中不及格比例低于30%的学校个数为X,求X的分布列;(Ⅲ)设8所学校优秀比例的方差为S12,良好及其以下比例之和的方差为S22,比较S12与S22的大小.(只写出结果)【分析】(Ⅰ)8所学校中有四所学校学生的体质健康测试成绩达到良好及其以上的比例超过40%,即可得出从8所学校中随机取出一所学校,该校为先进校的概率.(Ⅱ)8所学校中,学生不及格率低于30%的学校有学校B、F、H三所,所以X的取值为0,1,2.利用超几何分布列即可得出随机变量X的分布列.(Ⅲ)经过计算即可得出S12与S22的关系.解:(Ⅰ)8所学校中有四所学校学生的体质健康测试成绩达到良好及其以上的比例超过40%,所以从8所学校中随机取出一所学校,该校为先进校的概率为.(Ⅱ)8所学校中,学生不及格率低于30%的学校有学校B、F、H三所,所以X的取值为0,1,2.,所以随机变量X的分布列为:X012P(Ⅲ)S12=S22.21.已知椭圆C:3x2+4y2=12.(Ⅰ)求椭圆C的离心率;(Ⅱ)设A,B分别为椭圆C的左右顶点,点P在椭圆C上,直线AP,BP分别与直线x=4相交于点M,N.当点P运动时,以M,N为直径的圆是否经过x轴上的定点?试证明你的结论.【分析】(Ⅰ)将方程化成标准方程,可得a,b,c进而求出离心率;(Ⅱ)分两种方法解题,由题意求出A,B的坐标,设直线AP,BP与x=4联立求出M,N的坐标,设x轴一点Q,使得=0,求出Q的坐标,即为定点.解:(Ⅰ)由得,那么a2=4,b2=3,所以c2=a2﹣b2=1,解得a=2,c=1所以离心率;(Ⅱ)解法一:A(﹣2,0),B(2,0),设P(x0,y0),则,直线AP的方程:,令x=4,得,从而M点坐标为,直线BP的方程:,令x=4,得,从而N点坐标为,设以MN为直径的圆经过x轴上的定点Q(x1,0),则MQ⊥NQ,由得,由①式得,代入②得,解得x1=1或x1=7,所以以MN为直径的圆是否经过x轴上的定点(1,0)和(7,0),解法二:A(﹣2,0),B(2,0)设P(x0,y0),则,,设直线AP的方程:y=k(x+2),令x=4,得y M=6k,从而M点坐标为(4,6k),则直线BP的方程:,令x=4,得,从而N点坐标为,设以MN为直径的圆经过x轴上的定点Q(x1,0),则MQ⊥NQ,由得,可得,解得x1=1或x1=7,所以以MN为直径的圆经过x轴上的定点(1,0)和(7,0).22.已知函数f(x)=m sin(1﹣x)+lnx.(1)当m=1时,求函数f(x)在(0,1)的单调性;(2)当m=0且时,,求函数g(x)在(0,e]上的最小值;(3)当m=0时,有两个零点x1,x2,且x1<x2,求证:x1+x2>1.【分析】(1)将m=1代入f(x)中,然后求导判断f(x)在(0,1)上的单调性;(2)由条件求出g(x)的解析式,然后求导判断g(x)在(0,e]上的单调性,再求出其最小值;(3)求出个零点x1,x2,得到,构造函数,根据函数的单调性证明即可.解:(1)当m=1时,f(x)=sin(1﹣x)+lnx,则f'(x)=﹣cos(1﹣x)+,当x∈(0,1),f'(x)在(0,1)上单调递减,∴f'(x)>f(1)=0,∴当x∈(0,1)时,f(x)在(0,1)上单调递增.(2)当m=0时,(,0<x≤e),则=,∵,∴g'(x)<0,∴g(x)在(0,e]上单调递减,∴.(3)当m=0时,,∵x1,x2是函数的两个零点,∴,,.两式相减,可得,即,∴,∴,.令(0<t<1),则.记,则.∵0<t<1,∴F'(t)>0恒成立,∴F(t)<F(1),即.∴,故x1+x2>1.。

2020届济宁市第一中学高三英语模拟试题及答案

2020届济宁市第一中学高三英语模拟试题及答案

2020届济宁市第一中学高三英语模拟试题及答案第一部分阅读(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AWelcome to Oxford University MuseumsAshmolean MuseumEstablished in 1683, the Ashmolean Museum is the oldest museum in the UK and one of the oldest in the world. It houses the University’s extensive collections of art and antiquities, ranging back over four millennia.Location: Beaumont Street Tel: 01865278000Open: Tue. Sun. 10: 00-17: 00.Charge: Admission is free; special exhibitions are ticketed and a charge may applyNote: For group bookings Tel:01865278015Oxford University Museum of Natural HistoryThe University Museum of Natural History houses the University’s collections of zoological, entomological, paleontological and mineral specimens. With 4. 5 million specimens it is the largest collection of its type outside of the national collections.Location: Parks Road Tel: 01865 272950Open: 10: 00-17: 00 dailyCharge: Admission is freeNote: Groups must book in advanceMuseum of the History of ScienceThe Museum of the History of Science is housed in the world’s oldest surviving purpose-built museum building. It contains the world’s finest collection of historic scientific instruments.Location: Broad Street Tel: 01865277280Open: Tue.Sun.12: 00-17:00Charge: Admission is freeNote: Booking required for groups of 15 or morePitt Rivers MuseumThe Pitt Rivers Museum holds one of the world’s finest collections of anthropology and archaeology, withobjects from every continent and from throughout human history.Location: Parks Road enter via the Oxford University Museum of Natural HistoryTel:01865270927Open: Tue Sun. and Bank Holiday Mondays: 10: 00-16: 30Charge: Admission is freeNote: Groups must book in advance1.If a group of 20 students want to visit the oldest museum in the UK, they should call_______A.01865277280B.01865278015C.01865270927D.018652729502.Which of the museums can visitors go to any day of the week?A.Ashmolean Museum.B.Museum of the History of Science.C.Oxford University Museum of Natural History.D.Pitt Rivers Museum3.Which of the following is true according to the passage?A.Ashmolean Museum is on the Broad Street.B.Pitt Rivers Museum contains the world’s finest collection of historic scientific instruments.C.We don’t have to book in advance if our group want to visit the University Museum of Natural History.D.We can enter the Pitt Rivers Museum through the Oxford University Museum of Natural History.BI dropped out of college after my first year. Three years later, I returned to college after having been stuck in a dead-end job, working at a department store. I saw school as my way out. But I quickly found myself up against the same problems that had caused me to give up before. I was in over my head with college-level algebra (代数) and a heavy workload of reading and writing homework. In addition, I was still unsure of my career (职业) direction。

2020年山东省济宁市高考物理一模试卷 (有详解)

2020年山东省济宁市高考物理一模试卷 (有详解)

2020年山东省济宁市高考物理一模试卷一、单选题(本大题共8小题,共24.0分)1.关于固体和液体,下列说法正确的是A. 晶体都有确定的几何形状B. 玻璃、蜂蜡、蔗糖、橡胶都是非晶体C. 液体的表面张力是分子力作用的表现D. 对于一定的液体和一定材质的管壁,管的内径越粗,液体能达到的高度越高2.下列说法正确的是()A. 汤姆孙成功解释了氢原子光谱B. 卢瑟福发现了电子C. 法拉第发现了电磁感应定律D. 牛顿成功解释了光电效应3.如图所示,虚线AB和CD分别为椭圆的长轴和短轴,相交于O点,两个等量异种点电荷分别处于椭圆的两个焦点M、N上,下列说法中正确的是()A. A、B两处电势、场强均相同B. C、D两处电势、场强均相同C. 在虚线AB上O点的场强最大D. 带正电的试探电荷在O处的电势能小于在B处的电势能4.一木块在水平桌面上保持静止,下列说法中正确的是()A. 木块所受的重力就是地球对物体产生的吸引力B. 木块对桌面的压力就是木块受到的重力,施力物体是地球C. 木块对桌面的压力是弹力,是由于木块发生形变而产生的D. 木块保持静止是由于木块对桌面的压力与桌面对木块的支持力保持平衡5.如图所示,四个粗细均匀的线框中通以大小相等的恒定电流I,且固定于纸面内,线框区域存在垂直纸面向里的匀强磁场,磁感应强度均为B.甲线框三条边长度均为a,倾斜的两条边与水平边夹角均为75°;乙线框两条竖直边与水平边长度均为a;丙线框是边长为a的正三角形线框;丁线框是半径为a、圆心角为90°的圆弧.关于四个线框受到的安培力大小,下列说法正确的是()A. 甲线框受到的安培力大小为BIaB. 丁线框受到的安培力大小为√2BIaC. 丙线框受到的安培力大小为3BIaD. 乙线框受到的安培力大小为3BIa6.一理想变压器原、副线圈的匝数比为5:1,副线圈电路中仅接入一个11Ω的电阻;原线圈电路中接有理想电流表。

2020年济宁市高三第一次模拟考试题及答案

2020年济宁市高三第一次模拟考试题及答案

发展旅游观光农业,农业附加值提高。
(答出2点即可4分)
17.(13分)阅读图文资料,完成下列要求。 民勤县位于甘肃省武威市北部,石羊河下游,
沙漠、戈壁、剥蚀山地和盐碱滩地占全县面积 91%,人口集中分布在占全县面积9%的绿洲上 (如图)。脆弱的生态环境以及水、土资源的 匮乏是制约民勤县发展的重要因素。 为了摆 脱水、土资源的制约,缓解日益恶化的生态环 境,该县实行了“关井压田”、水资源“总量控制, 定额管理”等政策,加强对水、土资源的管理。
16.(14分)阅读图文资料,完成下列要求。 意大利以山地丘陵为主(图 1),却是世界上的传统农业大国和农业强国。意大利重视农业的多种经
营,形成了众多特色鲜明的农业地域(图2)和多元化的产业结构(谷物、蔬果、乳畜等多元发展)。
农业区位: 意大利农业生产历来崇尚对自然的尊重,通过施用有机肥、生物农药,规范种养标准等措施发展有 机农业,自现已然成条为欧件洲头:号气有机候食、品生地产国形。、近年地来壤,意、大利水利文用乡村丰富的自然资源,将乡村变成具
3.图示研究区内,行政村地名占比最大的是 A.以地貌命名(如塬、梁、峁)
√B.以水文命名(如沟、湾、畔) C.以植被命名(如柳、芦、麻) D.以颜色命名(如红、黑、青)
4.定边县北部分布有大量沙地,却是以水文 命名的行政村密集区,原因可能是
A.该地地下水丰富,多以泉、井等命名
√B.历史上气候变化,使地表水环境变化
低高 热冷
13.图示时刻最可能是 A.4月16日2:00 B. 6月16日2:00
√C. 8月16日15:00
D. 10月16日15:00
湖陆风是指在较大湖面与陆地之间形成的以24小时为周期的局地性环流现象。密云水 库位于北京市密云县城北的燕山丘陵之中,面积约为 180 平方千米。下图示意密云水库 研究区某日某时刻(背景天气较稳定)近地面气温与风场分布。据此完成13~15题。

济宁市2020年高考模拟考试语文参考答案及评分标准

济宁市2020年高考模拟考试语文参考答案及评分标准

济宁市2020年高考模拟考试语文参考答案及评分标准一、现代文阅读(35分)1.D(A根据语言景观的定义,语言景观指的应是各类标牌之上的语言;B因果关系表述不当;C信息功能是显性的,象征功能是隐性的)2.C(“关键因素”表述错误)3.B(不属于语言景观)4.①应该如何界定语言景观这一概念?(1分)语言景观研究的价值和意义表现在哪些方面?(1分)②如何评价近年来的语言景观研究?(或“请谈谈语言景观研究的成绩和不足。

”)(1分)③未来如何推进语言景观研究?(1分)5. ①选取具有代表性的研究地点。

②广泛搜集资料,针对不同类别做好信息标注。

③拓宽研究视角,挖掘象征意义,不能仅局限在语言使用规范等方面。

(6分。

每点2分)6.B(“冷漠麻木、自私自利”不准确)7.C(“意在消除人物身上的个性色彩”不当)8.①一方面,士兵身上有着一般的人性,他爱自己的孩子和兄弟,渴望战争结束,回到家乡。

②另一方面,他深受法西斯思想毒害,认为杀人是结束战争的正确方式,以致冷酷残暴,泯灭了人性和良知。

(4分,每点2分。

意思答对即可。

如有其他答案,只要言之成理,可酌情给分)9.①婴儿是小说的中心故事展开的关键,推动着情节的发展。

地下室内外两条线索,都是围绕婴儿的命运变化逐步展开并推进的。

②婴儿是小说刻画人物形象的凭借。

女人伟大的母爱以及她的痛苦、挣扎和死亡,士兵的残暴冷酷和人性上的矛盾,都是通过他们对待婴儿的态度来体现的。

③婴儿是小说主旨的集中体现。

小说揭露战争、反省人性的主旨,主要是通过婴儿的悲惨命运具体形象地揭示出来的。

(6分,每点2分。

意思答对即可。

如有其他答案,只要言之成理,可酌情给分)二、古代诗文阅读(35分)10. C(京兆控制陇蜀,诸王贵籓分布左右,民杂羌戎,尤号难治。

希宪讲求民病,抑强扶弱)11. B(庐指古人在父母或老师死后,为守丧而在墓旁构筑居住的屋舍)12. C(由原文“忠臣良臣,何代无之,顾人主用不用尔”可见,廉希宪是说任何一个朝代都会有忠臣良臣,只是君主用或不用罢了,并未举荐任何人)13.(1)廉希宪彻底查办阿合马的事情,把他的情况上报,杖责阿合马,革去了他所统领的职权,并归属主管该事务的官员。

【附加15套高考模拟试卷】山东省济宁市2020届高三第一次模拟考试数学(理)试题含答案

【附加15套高考模拟试卷】山东省济宁市2020届高三第一次模拟考试数学(理)试题含答案

山东省济宁市2020届高三第一次模拟考试数学(理)试题一、选择题:本题共12小题,每小题5分,共60分。

在每小题给出的四个选项中,只有一项是符合题目要求的。

1.已知函数()ln 2f x x x x a =-+,若函数()y f x =与()()y ff x =有相同的值域,则a 的取值范围是( ) A .1,12⎛⎤ ⎥⎝⎦ B .(],1-∞ C .31,2⎡⎫⎪⎢⎣⎭ D .[)1,+∞ 2.已知幂函数()a f x x =的图象过点13,3⎛⎫ ⎪⎝⎭,则函数()()()21g x x f x =-在区间1,22⎡⎤⎢⎥⎣⎦上的最小值是( )A .1-B .0C .2-D .323.已知点()mod N n m ≡表示N 除以m 余n ,例如()71mod6≡,()133mod5≡,则如图所示的程序框图的功能是( )A .求被5除余1且被7除余3的最小正整数B .求被7除余1且被5除余3的最小正整数C .求被5除余1且被7除余3的最小正奇数D .求被7除余1且被5除余3的最小正奇数4.已知各项均为正数的等差数列{}n a 的公差为2,等比数列{}n b 的公比为-2,则( )A .14n n a a b b --=B .14n n a a b b -=C .14n n a a b b --=-D .14nn a a b b -=- 5.设函数()322ln f x x ex mx x =-+-,记()()f xg x x =,若函数()g x 至少存在一个零点,则实数m 的取值范围是( )A .21,e e ⎛⎤-∞+ ⎥⎝⎦ B .210,e e ⎛⎤+ ⎥⎝⎦C .21e ,e ⎛⎫++∞ ⎪⎝⎭ D .2211e ,e e e ⎛⎤--+ ⎥⎝⎦ 6.已知x ∈R,sin 3cos x x -=tan2x =( )A .43B .34C .34-D .43-7.设[]x 表示不超过x 的最大整数(如5[2]2,[]14==),对于给定的*n N ∈,定义(1)([]1)(1)([]1)x n n n n x C x x x x -⋅⋅⋅-+=-⋅⋅⋅-+,[1,)x ∈+∞,则当3[,3)2x ∈时,函数8x C 的值域是( ) A .16[,28]3 B .16[,56)3 C .28(4,)[28,56)3⋃ D .1628(4,](,28]33⋃8.若直线y=2x 上存在点(x ,y )满足约束条件30230x y x y x m +-≤⎧⎪--≤⎨⎪≥⎩,则实数m 的最大值为A .-1B .1C .32 D .29.将函数()2sin 6f x x π⎛⎫=- ⎪⎝⎭的图象上各点的纵坐标保持不变,横坐标扩大到原来的2倍,再把所得函数图象向右平移4π个单位,得到函数()g x 的图象,则函数()g x 图象的一条对称轴的方程为( ) A .4x π= B .1912x π= C .1312x π= D .6x π= 10.若正数,m n 满足21m n +=,则11m n +的最小值为 A.3+B.3C.2+ D .311.等比数列{a n }中,11,28aq ==,则4a 与8a 的等比中项是( ) A .±4 B .4 C .14±D .14 12.如图,四棱锥P ABCD -的底面为矩形,矩形的四个顶点A ,B ,C ,D 在球O 的同一个大圆上,且球的表面积为16π,点P 在球面上,则四棱锥P ABCD -体积的最大值为( )A .8B .83C .16D .163二、填空题:本题共4小题,每小题5分,共20分。

2020年山东省济宁市高考化学一模试卷 (含答案解析)

2020年山东省济宁市高考化学一模试卷 (含答案解析)

2020年山东省济宁市高考化学一模试卷一、单选题(本大题共12小题,共28.0分)1.化学与社会、生产、生活紧切相关.下列说法正确的是()A. 棉花和木材的主要成分都是纤维素,蚕丝和人造丝的主要成分都是蛋白质B. 石油干馏可得到石油气、汽油、煤油、柴油等C. 变质的油脂有特殊难闻气味,是因为油脂发生了氧化反应D. 制作快餐盒的聚苯乙烯塑料是易降解塑料2.下列实验装置正确且能达到实验目的的是()A. 用装置甲制取少量NH3B. 用装置乙验证SO2的还原性C. 用装置丙配制100mL0.10mol·L−1的硫酸溶液D. 用装置丁分离水和乙醇3.已知某有机物的结构简式如图所示,下列说法不正确的是()A. 分子式为C15H18B. 能使酸性KMnO4溶液褪色,且是氧化反应C. 1mol该物质最多和2molH2加成D. 苯环上的一氯化物有4种4.设N A为阿伏加德罗常数的值,下列说法正确的是()A. 某密闭容器中盛有0.1molN2和0.3molH2,在一定条件下充分反应,转移电子的数目为0.6N AB. 常温下,1L pH=9的CH3COONa溶液中,发生电离的水分子数为1×10−9N AC. 14.0gFe发生吸氧腐蚀生成Fe2O3⋅xH2O,电极反应转移的电子数为0.5N AD. 标准状况下,2.24L丙烷含有的共价键数目为1.1N A5.最新报道:科学家首次用X射线激光技术观察到CO与O在催化剂表面形成化学键的过程。

反应过程的示意图如下:下列说法正确的是()A. CO和O生成CO2的反应是吸热反应B. 在该过程中,CO断键形成C和OC. CO和O生成了具有极性共价键的CO2D. 状态I→状态Ⅲ表示CO与O2反应的过程6.如图是利用一种微生物将废水中的有机物(如淀粉)和废气NO的化学能直接转化为电能,下列说法中一定正确的是()A. 质子透过阳离子交换膜由右向左移动B. 电子流动方向为N→Y→X→MC. M电极反应式:(C6H10O5)n+7nH2O−24ne−=6nCO2↑+24nH+D. 当M电极微生物将废水中16.2g淀粉转化掉时,N电极产生134.4LN2(标况下)7.金刚烷是具有类似樟脑气味的无色晶体,其衍生物在医药方面有着重要的用途。

山东省济宁市2020届高三试题第一次模拟考英语试题word版含答案

山东省济宁市2020届高三试题第一次模拟考英语试题word版含答案

试卷类型:A山东省济宁市2020届高三试题第一次模拟考英语试题第一部分阅读(共两节,满分50分)第一节(共15小题;每小题2.5分,满分37.5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。

AWhistler, B.C.Beloved by tourists and Canadians alike, Whistler increased its international profile after the success of the 2010 Winter Olympics. With its close proximity (临近) to Vancouver—it’s only two hours north of the city—Whistler offers the ideal winter playground for all ages. Challenge yourself on its snow-covered mountains, enjoy some retail therapy in its 200 shops, or sample the finest cuisine in 90+ restaurants and bars.QuebecFor more than 60 years, Que bec’s world famous Winter Carnival has been one of Canada’s must-visit winter destinations. With evening parades, snow tubing, skating, snow rafting, outdoor BBQs, ice sliding, s now sculpture competitions and dance parties, there’s something for every taste and age. And be sure to keep your eyes peeled for the fete’s main man—Bonhomme Carnaval—Que bec’s beloved snowman and Carnival ambassador.Nova ScotiaYou can’t get more Canadia n than maple syrup, and Sugar Moon Farm in Nova Scotia has tasty fun in store for you! Learn all about the art of maple syrup—and sample the yummy results. Sugar Moon Farm offers warm hospitality along with their sugar camp tours, all-day brunch, maple-tasting experiences and hiking trails. Foodies will be thrilled to take part in Sugar Moon’s Chef Nights when the finest chefs create delicate dinners for a hungry andappreciative audience.SaskatchewanFor a true frontier experience, enjoy a breathtaking journey through the Canadian wild by dog sled. Saskatchewan’s Sundogs Sled Excursions lets you ride along on a trip of a lifetime. Each winter from late November to mid-March, Sundogs takes visitors deep into the beautiful and unspoiled environs of Anglin Lake by dog sled. Sundogs offers a host of amazing activities including puppy camps, half or full day trips, and overnight camping.1.What is Whistler, B.C. known for worldwide?A. The success of the 2010 Winter Olympics.B. Its close proximity to Vancouver.C. Its snow-covered mountains.D. The finest cuisine in restaurants and bars.2.Which is suggested if one has a sweet tooth?A.Whistler,B.C. B. Saskatchewan. C. Nova Scotia.D. Quebec.3. What can be expected in Saskatchewan?A. A snow sculpture competition.B. A maple-tasting experience.C. A cross-country hiking trip.D. A thrilling snow sled ride.BThe expression, “everybody’s doing it,” is very much at the center of the concept of peer pressure. It is a social influence applied on an individual in order to get that person to act or behave in a similar way as a larger group.People are social creatures by nature, and so it is hardly surprising that some part of their self-respect comes from the approval of others. This instinct explains why the approval of peers, or the fear of disapproval, is such a powerful force in many people’s lives. This instinct drives people to dress one way at home and another way at work, or to answer a simple “fine” when a stranger asks “how are you?” even if it is not necessarily true.For certain individuals, seeking social acceptance is so important that it becomes an addiction. Teens and young adults may feel forced to smoke, or drink alcohol, which might encourage criminal behavior. Mature adults may sometimes feel pressured to cover up illegal activity at the company where they work, or end up in debt because they are unable to hold back the desire to buy a house or car that they can’t afford in an effort to keep up with the peers.However, peer pressure is not always negative. A student whose friends are good at academics may be urged to work harder and get good grades. Players on a sports team may feel driven to play harder in order to help the team win. This type of influence can also get a friend off smoking, or to help an adult take up a good habit or drop a bad one.Although peer pressure is sometimes quite obvious, it can also be so subtle that a person may not even notice that it is affecting his or her behavior. For this reason, when making important decisions, simply going with an instinct is risky. Instead, people should seriously consider why they feel drawn to taking a particular action, or it is simply because everyone else is doing the same thing.4. What does the underlined word “instinct” in paragraph 2 probably mean?A. A natural tendency.B. An acquired ability.C. A popular idea.D. An obvious mistake.5. What is paragraph 4 mainly about?A. Potential causes of peer pressure.B. Possible positive effects of peer pressure.C. Negative consequences of peer pressure.D. Tips for coping with peer pressure.6. What is the author’s suggestion when making a decision?A. Simply go with an instinct.B. Ask for advice from adults.C. Listen to the inner motivation.D. Follow in the footsteps of others.7. What is the author’s attitude to peer pressure?A. Objective.B. Ambiguous.C. Critical.D. Indifferent.CThe worst outbreak of desert locusts (蝗虫) in decades is presently underway in the Horn of Africa. It is the biggest of its kind in 25 years for Ethiopia and Somalia –and the worst Kenya has seen for 70 years.What we are seeing in East Africa today is unlike anyt hing we’ve seen in a very long time. Its destructive potential is enormous, and it’s taking place in a region where farmers need every gram of food to feed themselves and their families. Most of the countries hardest hit are those where millions of people are already vulnerable (脆弱的) or in serious humanitarian need, as they endure the impact of violence, drought, and floods.We have acted quickly to respond to this outbreak. The primary method of battling locusts is the aerial spraying of pesticides (杀虫剂). FAO’s “Locust Watch” service explains that “although giant nets, flamethrowers, lasers, and huge vacuums have been proposed in the past, these are not in use for locust control. People and birds often eat locusts but usually not enough to significantly reduce population levels over large areas.”The UN’s Office for the Coordination of Humanitarian Affairs has released $10 million from its Central Emergency Relief Fund to fund a huge scale-up in aerial operations to manage the outbreak.But the window to contain this crisis is closing fast. We only have until the beginning of March to bring this infestation under control as that is when the rain and planting season begins. If left unchecked – and with expected additional rains – locust numbers in East Africa could increase 500 times by June.We must act now to avoid a full-blown catastrophe. And we will. At the same time, weneed to pay attention to a bigger picture. This is not the first time the Greater Horn of Africa has seen locust outbreak approach this scale, but the current situation is the worst in decades. This is linked to climate change. Warmer seas mean more tropical storms, generating the perfect breeding conditions for locusts.8.What is implied in paragraph 2?A. People in East Africa are suffering drought.B. People in East Africa are going through floods.C. The locust outbreak will cause crop failure.D. The locust outbreak is worsening locals’ life.9. What can we learn from the last paragraph?A. We need a bigger picture to study the disaster.B. It is the second outbreak of locusts in East Africa.C. It is the largest outbreak of locusts ever in history.D. The outbreak of locusts is fueled by global warming.10. What is the purpose of the text?A. To analyse and compare.B. To inform and call for.C. To argue and discuss.D. To introduce and assess.11. Where does the text probably come from?A. A guidebook.B. A health magazine.C. A news report.D. A chemistry paper.DScientists say they have developed a system that uses machine learning to predict when and where lightning will strike. Researchers report the system is able to predict lightning strikes up to 30 minutes before they happen within a 30-kilometer area.Lightning is a strong burst of electricity in the atmosphere. Since it carries an extremely powerful electrical charge, it can be destructive and deadly. European researchers haveestimated that between 6,000 and 24,000 people are killed by lightning worldwide each year. For this reason, climate scientists have long sought to develop methods to predict lightning.The system tested in the experiments uses a combination of data from weather stations and machine learning methods. The researchers developed a prediction model that was trained to recognize weather conditions that were likely to cause lightning.The model was created with data collected over a 12-year period from 12 Swiss weather stations in cities and mountain areas. The data, related to four main surface conditions: air pressure, air temperature, relative humidity and wind speed, was placed into a unique machine learning algorithm (算法), which compared it to records of lightning strikes. Researchers say the algorithm was then able to learn the conditions under which lightning happens.The researchers test-ran the system several times. They found that the system made predictions that proved correct almost 80 percent of the time. “It can now be used anywhere,” the Swiss Federal Institute of Technology said in a statement.The researchers plan to keep developing the technology in partnership with a European effort that aims to create a lightning protection program. The effort is called the European Laser Lightning Rod project. Scientists working on the project are experimenting with a laser technology that could someday control lightning activity, transferring lightning charges from clouds to the ground. They hope that such technology can one day be used as protection against lightning strikes. Possible uses could be at stations, airports or places where large crowds gather.12.Why was the system developed?A. To meet kids’ curiosity about lightning.B. To show the power of lightning.C. To keep track of lightning deaths.D. To protect people from lightning.13. What is special about the system?A. It was based on a number of samples worldwide.B. It adopted a different machine learning algorithm.C. It used an effective method of collecting data.D. It required low cost of predicting lightning.14. What will the researchers do next with the laser technology?A. Monitor lightning activity.B. Prevent the occurrence of lightning.C. Direct energy from lightning.D. Generate electricity with lightning.15. What can be a suitable title for the text?A. A System of Controlling LightningB. A Method of Forecasting LightningC. A Theory of Employing LightningD. A Model of Creating Lightning第二节(共5小题;每小题2.5分,满分12.5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。

山东省济宁市2020届高三高考模拟考试数学试题 含答案

山东省济宁市2020届高三高考模拟考试数学试题 含答案

·. �. i' :· (川_::l ,济宁市12020年高考模拟考试2020.05注意事项:1.答题前考生务必将自己的姓名、准考证号在答题卡上涂写清楚:2:每小题选出答案后i 用2B 铅笔把答蝠卡上对应届臼的答案标号涂黑,如需改动,角橡皮擦干净后,再选涂其他答案标号,在试题卷上作答无效.一、单项选择题:本题共8小题,每小题5分,共40分在每小题给出.的四个选项中,只有一项符l合题目要求J I.已知集合A=lx1χ2 -,2x-3 <01,B = !xl 2.r �_!_I ,贝�"xeB ”是“xeA ”的2 A.充分不必要条件l B.必要不充分条件c .充耍条件D .既不充分也不必要条件2. i 是虚数单位,复数z =旦土i.cα>0),若lz:l= 1,则α=1 -2i8..• A ÷,,} .,.川B .1 · ' ·’’C.2 D.33.双曲线Z-·-乙=λ(λ>0)的渐近线方程为.4 2 .忌’A.y =土扫马 B.r = ±-fl-x 1 , : c . r 可2x D.y =个I .L 4已知α=ln 言,b =肘,c =logλ则α,b ,c 的大小顺序为A.α>b >cB. b >α>cC. c >α>b D .b >c >a5.已知(x -2)(x +m)5 =α6元6+α.5X .5+…+α,x+町,m 为常数,若α。

=2,则αs =A.-7B.-2C.3D.76.《丸章算术》是我国古代内容极为丰富的数学名著,书中有如下问题:“今有仓,广三丈,袤四丈五尺,容粟一万剧,问高几何?”其意思为:“今有一个长方体的粮仓,宽3丈,长4丈5尺,可装粟一万剧,问该粮仓的高是多少?”已知I斟粟的体积为2.7立方尺,一丈为10尺,该粮仓的外接球的体积是()立方丈•. I A. 133.一-,,,.4 B .旦2臂48” .n ·c .i33、/i33何4 .. D. 133打育48 ,,,.7如图,在MBC中,LBAC=f ,AD =2DB,P 为CD 上一点,且满足AP=m.AC +担,若AC=--i ’-→ 3,AB =4,贝UAP ·en 的值为’ • ' .cA .-3 B.13--12D.上12'Bc .1312 A 高三数学试题第1页(共4·页)数学试题8.已知π是一个三位正整数,若n 的十位数字大于个位数字〉百位数字大于十位数字}则称n为三位递增数.已知α,b,cel0,1,2,3,例,设事件A为“由α,b,c 组成三位正整数”.事件B为“由α,b,c 组成三位正整数为递增数’·.则P(BIA)= 2…A .+B .上10 c 2 ·252-5’且-呵’』D 二、多项选择题:本题共4小题,每小题5分,共20分.在每小题给出的选项中,有多项符合题目要求.全部选对的得5分,部分选对的得3分,有选错的得0分.9.下列说法中正确的是A.对具有线性相关关系的变量x,.有一组观测数据(x i ,Y i )(i =l,2,…,8),其线性回归方程是归卡+ι且引+乌+句+. +x g =2(y. +r 2�川B.正态分布N(1,9)在区间(-1,0)和(2,3)上取值的概率相等c.若两个随机变盘的线性相关性越强,则相关系数r的值越接近于1D .若一组数据1,α,2,3的平均数是2,则这组数据的众数和中位数都是210.已知α,β是两个不同平面,m ,n 是两条不同直线i则下列命题中正确的是A.如果mJ.n,m J.α,nJ.β,那么α土β B.如果me α,α//卢,那么ml/{3c.如果αnβ=l,ml /α,ml 徊,那么m//l D.如果mJ.n,m J.α,nll/3,那么α4β11.已知函数f (x)=cos(2x _:!!..) -2sin(.x +立)cos ( x + :!!.. ) ( x E R ),现给出下列四个命题,其34 4 中正确的是A.函数J (功的最小正周期为2作B .函数J(付的最大值为1C函数f(x )在[-f.f ]上单调递增D .将函数J(功的图象向左平移立个单位长度,得到的函数解析式为g (x ),= s in 2x 12 12.已知抛物线E::i:2=付的焦点为F,圆C:泸+(y -1 )2 =16与抛物线E交于A,B 两点,点P 为劣弧AB 上不同于A,B 的一个动点,过点P作平行于y轴的直线l交抛物线E于点N,则下列四个命题中正确的是A.点P 的纵坐标的取值范围是(2./3,5)B .IPNI + INFI 等于点P 到抛物线准线的距离c.困C的圆心到抛物线准线的距离为2D. b.PF N 周长的取值范围是(8,10)高三数学试题第2页(共4页)王4填空题本题共4b题,每小题分,今io分;...川-·, i l:"、、.、·;I 'l ' i , ,, . 3.}已知向盘a=:C 卡ψ,6}.:b '=2\x,)满足a/lb 典中元eR ;"那么lbl =,。

【附28套精选模拟试卷】山东省济宁市2020届高三第一次模拟考试数学(理)试题及答案

【附28套精选模拟试卷】山东省济宁市2020届高三第一次模拟考试数学(理)试题及答案

山东省济宁市2020届高三第一次模拟考试数学(理)试题及答案 本试卷分第I 卷和第Ⅱ卷两部分,共5页.满分150分.考试用时120分钟,考试结束 后,将试卷和答题卡一并交回.注意事项:1.答卷前,考生务必用0.5毫米黑色签字笔将自己的姓名、座号、准考证号填写在答题纸上.2.第I 卷每小题选出答案后,用2B 铅笔把答题纸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.答案不能答在试题卷上.3.第Ⅱ卷必须用0.5毫米黑色签字笔作答,答案必须写在答题纸各题指定区域内相应的位置,不能写在试题卷上;如需改动,先划掉原的答案,然后再写上新的答案;不准使用涂改液、胶带纸、修正带.不按以上要求作答的答案无效.参考公式:如果事件A 、B 互斥,那么P(A+B)=P(A)+P(B)如果事件A 、B 独立,那么P(AB)=P(A)·P(B)第I 卷(选择题 共50分)一、选择题:本大题共10小题,每小题5分,共50分.在每小题给出的四个选项中,只有一项是符合题目要求的.1.已知复数211i z i +=+(i 为虚数单位),则复数z 在复平面内对应的点位于 A.第一象限 B.第二象限 C.第三象限 D.第四象限2.已知集合{}211,3402x A x B x x x A B ⎧⎫⎪⎪⎛⎫=<=-->⋂⎨⎬ ⎪⎝⎭⎪⎪⎩⎭,则等于A.{}0x x >B. {}0x x x <-1>或C.{}4x x >D. {}4x x -1≤≤ A.88 88B.90 89C.89 88D.89 90 4.若点(),P x y 满足线性约束条件20220,40x y x y z x y y -≤⎧⎪-+≥=+⎨⎪≥⎩则的最大值为A.1B.2C.3D.45.给出命题p :直线()3102110ax y x a y ++=+++=与直线互相平行的充要条件是3a =-;命题q :若210mx mx --<恒成立,则40m -<<.关于以上两个命题,下列结论正确的是A.命题“p q ∧”为真B. 命题“p q ∨”为假C.命题“p q ∧⌝”为真D. 命题“p q ∨⌝”为真 6.在△ABC 中,角A 、B 、C 的对边分别是a 、b 、c.若sin sin 3sin sin .aA c C a C bB +-=则角B 等于A.56πB.23π C.3π D.6π 7.函数()1ln f x x x ⎛⎫=- ⎪⎝⎭的图象大致是8.已知向量()()11,1,1,2,0,0,//a m n b m n a b m n=-=>>+其中若,则的最小值是 A.22 B.322+ C.42 D.32 9.设()ln f x x =,若函数()()g x f x ax =-在区间(]0,3上有三个零点,则实数a 的取值范围是A.10,e ⎛⎫ ⎪⎝⎭B.ln 3,3e ⎛⎫ ⎪⎝⎭C.ln 30,3⎛⎤ ⎥⎝⎦ D.ln 31,3e ⎡⎫⎪⎢⎣⎭ 10.已知12,F F 是双曲线()222210,0x y a b a b-=>>的两个焦点,点P 是该双曲线和圆2222x y a b +=+的一个交点,若1221sin 2sin PF F PF F ∠=∠,则该双曲线的离心率是A.10B.5C.10D.10 第II 卷(非选择题 共100分)二、填空题:本大题共5小题,每小题5分,共25分.11.函数1lg 123x y x ⎛⎫=-+- ⎪⎝⎭的定义域是 ▲ . 12.阅读如图所示的程序框图,若输出()f x 的范围是2,2⎡⎤⎣⎦,则输入实数x 的范围应是 ▲ .13.已知在正方体1111ABCD A B C D -中,点E 是棱11A B 的中点,则直线AE 与平面11BDD B 所成角的正弦值是 ▲ .14.若()()()()()234525012345411111x x a a x a x a x a x a x a +=+-+-+-+-+-,则 = ▲ .15.设区域Ω是由直线0,=1x x y π==±和所围成的平面图形,区域D 是由余弦曲线y=cosx 和直线x=0,x=π和y=1±所围成的平面图形,在区域Ω内随机抛掷一粒豆子,则该豆子落在区域D 的概率是 ▲ .三、解答题:本大题共6小题,共75分.解答应写出文字说明,证明过程或演算步骤.16.(本小题满分12分)已知函数()3sin cos .34f x x x π⎛⎫=++ ⎪⎝⎭ (I )当,36x ππ⎡⎤∈-⎢⎥⎣⎦时,求函数()f x 的值域; (II )将函数()y f x =的图象向右平移3π个单位后,再将得到的图象上各点的横坐标变为原的12倍,纵坐标保持不变,得到函数()y g x =的图象,求函数()g x 的表达式及对称轴方程. 17.(本小题满分12分)如图,已知斜三棱柱ABC 111A B C -的底面是正三角形,点M 、N 分别是1111B C A B 和的中点,112,60AA AB BM A AB ===∠=o .(I )求证:BN ⊥平面111A B C ;(II )求二面角1A AB M --的余弦值.18.(本小题满分12分)甲、乙、丙三位同学彼此独立地从A 、B 、C 、D 、E 五所高校中,任选2所高校参加自主招生考试(并且只能选2所高校),但同学甲特别喜欢A 高校,他除选A 校外,在B 、C 、D 、E 中再随机选1所;同学乙和丙对5所高校没有偏爱,都在5所高校中随机选2所即可.(I )求甲同学未选中E 高校且乙、丙都选中E 高校的概率;(II )记为甲、乙、丙三名同学中未参加E 校自主招生考试的人数,求的分布列及数学期望.19.(本小题满分12分)在等比数列{}121342,,n a a a a a a =+中,已知,且成等差数列.(I )求数列{}n a 的通项公式n a ;(II )设数列{}2n n a a -的前n 项和为2,nn n n S b S =记,求数列{}n b 的前n 项和n T . 20.(本小题满分13分) 已知抛物线214x y =的焦点与椭圆()2222:10x y C a b a b +=>>的一个焦点重合,12F F 、是椭圆C 的左、右焦点,Q 是椭圆C 上任意一点,且12QF QF ⋅u u u r u u u u r 的最大值是3.(I )求椭圆C 的标准方程;(II )过右焦点2F 作斜率为k 的直线l 与椭圆C 交于M 、N 两点,在x 轴上是否存在点(),0P m ,使得PM 、PN 为邻边的平行四边形是菱形?如果存在,求出m 的取值范围;如果不存在,请说明理由.21.(本小题14分)设函数()()2ln f x ax x a R =--∈.(I )若()()(),f x e f e 在点处的切线为20,x ey e a --=求的值;(II )求()f x 的单调区间;(III )当()0.x x f x ax e >0-+>时,求证:高考模拟数学试卷第Ⅰ卷注意事项:第Ⅰ卷共12小题,每小题5分,共60分。

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济宁市高三第一次模拟考试 试题及答案
2020.3.20
广西荔浦县生产的砂糖桔,是国家地理标志农产品。 近年来随着树冠覆膜技术(树冠上方覆盖塑料薄膜,如图 所示)的推广使用,砂糖桔留树时间延长,实现了错峰上 市,经济效益显著提高。据此完成1~2题。
1.荔浦砂糖桔覆膜的最
佳时间是
√A.8月 C.12月
意大利农业生产历来崇尚对自然的尊重,通过施用有机肥、生物农药,规范 种养标准等措施发展有机农业,现已成为欧洲头号有机食品生产国。近年来, 意大利利用乡村丰富的自然资源,将乡村变成具有教育、游憩、文化等多种 功能的生活空间,发展现代化的旅游观光农业,有机农业和旅游观光农业已 逐渐成为其农业发展的新动向。 近年来,虽然面临着从业人口老龄化、自然 灾害多发等问题,意大利农业产值仍在增加,2016年其农业产值高达529.02 亿欧元,位居欧盟第二(图3)。
√ A.洋底泥沙上涌固结
C.草原腐殖质的堆积
B.当地岩石的风化物 D.风力的搬运和沉积
√ 9.北极洲草原所需的水分主要来自 A.暖季雨水 B.积雪融水 C.冰层融水 水

10.猛犸象灭绝的原因最可能是
D.上涌的海
囊谦县位于青海省最南端,
该县已发现的29处盐泉,是世 界上已发现矿化度最高的盐泉 群。下图为囊谦地区盐泉形成 过程示意图。据此完成11~12 题。
B. 9月 D.1月
2.砂糖桔实现了错峰上市,主 要得益于树冠覆膜技术可以
①增加树冠内温度,防止果 品受冻害 ②改善果园小气 候,提高果品品质 ③增加树冠内湿度,提高果
√ 树抗病能力 ④减弱雨水的影
响,保证果品质量
某科研小组,对下图研究区域内行政村的地名进行了调查。 调查表明,该区域内行政村名与自然环境关系密切。据此完成3~ 4题。
C.地下水溶滤的岩盐中的
盐分
D.高山沉积物的盐分积累
湖陆风是指在较大湖面与陆地之间形成的以24小时为周期的局 地性环流现象。密云水库位于北京市密云县城北的燕山丘陵之中, 面积约为 180 平方千米。下图示意密云水库研究区某日某时刻 (背景天气较稳定)近地面气温与风场分布。据此完成13~15题。
高低
15. 6月10日,若天
气晴朗,下列四地点中
最先看到日出的是
√ A.①
B.②
C.③
D.④
16.(14分)阅读图文资料,完成下列要求。 意大利以山地丘陵为主(图 1),却是世界上的传统农业大国和农业强国。
意大利重视农业的多种经营,形成了众多特色鲜明的农业地域(图2)和多元 化的产业结构(谷物、蔬果、乳畜等多元发展)。
11.推测囊谦地区地下岩盐层的形成
√过程是
A.地壳隆起—海退成湖—湖水蒸发— 盐分沉淀 B.地壳下沉—河水成湖—湖水蒸发— 盐分沉淀 C.地壳隆1起2—.图淡中水盐流泉出中—盐盐类度物变质大的— 盐分析出来源是
√ D.地壳下A沉.泉—水形蒸成发海后洋盐—类盐物分质沉积淀—
隆起抬升累 B.封存在地层中的海水
功能的生活空间,发展现代化的旅游观光农业,有机农业和旅游观光农业已
逐渐成为其农业发展的新动向。 近年来,虽然面(临1着)从分业析人意口老大龄利化农、业自多然 灾害多发等问题,意大利农业产值仍在增加,2种01经6年营其的农有业利产条值高件达。52(9.602分)
3.图示研究区内,行政村地名占 比最大的是
√A.以地貌命名(如塬、梁、峁)
B.以水文命名(如沟、湾、畔) C.以植被命名(如柳、芦、麻) D.以颜色命名(如红、黑、青)
4.定边县北部分布有大量沙地, 却是以水文命名的行政村密集区,
√原因可能是
A.该地地下水丰富,多以泉、井 等命名
市辖区,是我国的行政 区划之一,属县级行政区, 受地级市、直辖市管辖,是 城市市区的组成部分。随着 城市的发展,政府通过行政 区划调整来增设市辖区。有 学者将我国城市市辖区的空 间结构划分为圈层式、组合 式、并排式和独立式等几种 基本模式(如下图)。据此 完成5~7题。
14.观测发现图示时刻 水库北侧与南侧的风速 存在较明显的差异,其 中风速较大的一侧及其 成因是
√A.北侧 湖风与山风相
互叠加 B. 北侧 湖风 与谷风相互叠加 C.南侧 湖风与山风相
湖陆风是指在较大湖面与陆地之间形成的以24小时为周期的局 地性环流现象。密云水库位于北京市密云县城北的燕山丘陵之中, 面积约为 180 平方千米。下图示意密云水库研究区某日某时刻 (背景天气较稳定)近地面气温与风场分布。据此完成13~15题。
低高 热冷
13.图示时刻最 可能是 A.4月16日2:00
√B. 6月16日2:00
C. 8月16日1成的以24小时为周期的局 地性环流现象。密云水库位于北京市密云县城北的燕山丘陵之中, 面积约为 180 平方千米。下图示意密云水库研究区某日某时刻 (背景天气较稳定)近地面气温与风场分布。据此完成13~15题。
16.(14分)阅读图文资料,完成下列要求。
意大利以山地丘陵为主(图 1),却是世界上的传统农业大国和农业强国。
农业区位: 意大利重视农业的多种经营,形成了众多特色鲜明的农业地域(图2)和多元
化的产业结构(谷物、蔬果、乳畜等多元发展)。
意大利自农然业生条产历件来:崇尚气对候自然、的尊地重形,通、过地施用壤有机、肥水、生文物农药,规范 种养标社准等会措施因发素展有:机市农业场,现、已技成为术欧、洲头交号有通机、食品政生策产国、。近年来, 劳动力 意大利利用乡村丰富的自然资源,将乡村变成具有教育、游憩、文化等多种
√度
C.地形起伏状况
D.经济发展水
约一万年前,北冰洋被完整的冰原覆盖,形成“气候性陆 地”,有的学者称之为北极洲。北极洲的冰层上面覆盖着黄土层, 进而演化成为一片辽阔的草原,生活着身披厚密长毛的猛犸象 等大型食草动物。随着环境巨变,这些史前动物逐渐灭绝。据 此完成8~10题。
8.北极洲的黄土层形成于
5.研究表明,城市市辖区空间结构
的演化具有一定规律性,其演化路
径可能为
√A.并排式——组合式——圈层式——
独立式
B.独立式——并排式——组合式——
圈层式
C.独立式——并排式——圈层式——
组合式
√ D.并排式——组合式——独立式——
圈层式
6.我国东部地区城市较西部更易增
设市辖区,其主要影响因素是
A.政策支持力度 B.交通便利程
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