9. 吉林省长春市普通高中2017年高三年级质量检测(二)试题及答案

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吉林省长春市普通高中2017届高三第二次模拟考试英语试题(详解答案]

吉林省长春市普通高中2017届高三第二次模拟考试英语试题(详解答案]

第Ⅰ卷第Ⅰ卷第一部分第一部分 听力(1-20小题)在笔试结束后进行。

小题)在笔试结束后进行。

注意事项:英语听力共两节,共20小题;每小题1.5分,满分30分。

分。

第一节(共5小题;每小题1.5分,满分7.5分)分)听下面5段对话。

每段对话后有一个小题,从题中所给的A 、B 、C 三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

和阅读下一小题。

每段对话仅读一遍。

例:How much is the shirt? A.£19.15. B.£9.15. C.£9.18. 答案是B 。

1.Where does the conversation probably take place? A.In a library. B.In a laboratory. C.In a restaurant. 2.What does the man mean ? A.He knows Thomas’ birth date.B.The woman is good with dates. C.He has trouble remembering dates. 3.What is the weather like now? A.Fine B.Hot C.Rainy. 4.What will the speakers do tomorrow afternoon? A.Go to a park. B.Go shopping. C.Eat out with Joe. 5.What is the man complaining about? A.His work. B.The weather. C.The noise from neighbors. 第二节(共15小题;每小题1.5分,满分22.5分)分)听下面5段对话或独白。

每段对话或独白后有几个小题,从题中多给的A 、B 、C 三个选项中选出最佳选项,选项中选出最佳选项,并标在试卷的相应位置。

吉林省长春市普通高中2017届高三下学期第二次模拟考试英语试题(图片版)

吉林省长春市普通高中2017届高三下学期第二次模拟考试英语试题(图片版)

长春市普通高中2017届高三质量监测(二)英语参考答案及评分参考说明:本试题满分150分。

其中听力30分,笔试120分。

第I卷第一部分听力(共两节,共20小题;每小题1.5分,满分30分)听力原文:Text 1W: May I take your order now?M: Thank you, but I'm waiting for someone to join me.Text 2W: Do you know when Thomas was born?M: Don’t ask me. I’m not good with dates.Text 3W: What bad weather!M: Don’t worry. The rain won’t last long. The radio says it’ll be fine later on.Text 4W: This is a beautiful park. Shall we come again tomorrow?M: Don’t we have to go shopping?W: We can do that in the afternoon, and come here in the morning.M: OK. And let’s bring Joe with us.Text 5M: Excuse me, madam. I am trying to do some work now. I’m afraid your children are making too much noise.W: I’m sorry. But you know it's diffi cult to keep boys quiet.Text 6W: Hello, John!M: Hello, Betty! Long time no see! Where have you been all this time?W: I was in France for a business study. I stayed there for about three months and came back only yesterday.M: Oh, I see. Did you visit any interesting places there?W: Not many. Every day I had to stay inside all day long. I didn’t have much free time to do any sightseeing, so I was only shown around the city of Paris.M: That’s a pity.W: I’ll have many chances to go to France in the futu re, so I can travel around the country some other time.M: I hope I’ll get the chance to go too.Text 7M: Did you watch the basketball game between the US and Spain?W: No, I missed it. Was it exciting?M: Yes, it was a close game. Finally, the American team beat the Spanish.W: They must have had a hard time. By the way, which is your favorite basketball team?M: The Chinese team, of course. What about you?W: In fact, I m not so interested in basketball. My favorite sport is ping-pong.M: Really? I like ping pong too. There is going to be a ping pong match in our class this weekend.Why not come with us then? We can watch it together.W: OK. Many of my classmates like ping pong. Maybe, we can have a game between classes. M: Good idea! Let's talk about it on Thursday. We will need a whole day to prepare.Text 8M: Hello, Cathy. How are you today? I heard you were ill last week.W: I'm fine now. Thank you, Peter.M: What was the matter? Nothing serious I hope.W: Oh, no. I had a bad cold and had to stay in bed for two days.M: I’m glad you’re better. Anyway, what about your friend, Ann? I heard she was ill, too.W: She was ill, but she’s all right now.M: Everybody seems to have a cold now. One day is hot and the next day is cold.W: And very windy, too. Tha t’s why I'm wearing a sweater today. What do you think of it? Pretty, isn’t it?M: It certainly is. It must have cost a lot. Where did you buy it?W: Oh, my mother bought it for me. It was on sale. It was very cheap.M: Well, Cathy. I must say, it suits you very well.W: Really? Thank you.Text 9W: I thought you went to work. Why are you still home, Daniel?M: I didn’t want to tell you, but I lost my job last week.W: Uh-oh! What about the rent? How are you going to pay it?M: Um... I was hoping you’d help me out a little, Emily. Isn’t that what older sisters are for?W: Yeah, I guess they are…but I can’t pay my little brother’s rent forever. We’ll have to move to a cheaper place soon if you don’t get another job. What are you planning to do?M: I don’t kno w! I thought my job at the theater was perfect.W: You only sold tickets to people and helped them find their seats.M: Right, but I got to watch great performances for free!W: Come on! You could be doing something better than that! I know you’re a great cook and you’ve always wanted to open your own sandwich shop. Why don’t you try to do that?M: Where would I begin? I don't know anything about running a business.W: Well, I do. I’ve been running my online store for five years. I know all the legal stuff you have to do to get started. And my friend Mike just closed his coffee house, and it’s the perfect space for your sandwich shop. It has a nice little kitchen in it already.M: That sounds perfect! You're a great sister!Text 10I used to dislike television. I thought that people spent too much time watching it. A lot of myfriends, however, always talked about the sports programs and films on it. They never read any books or went out in the evenings. So I refused to buy a TV set.Last year I turned 60 a nd I retired from my job. My son bought me a TV set. “It will make you learn some of the latest news,” he said. It’s quite true. I’ve watched all the news programs. I know far more about the world now. And I read more books, too. In fact, I think I may follow one of the Open University TV courses next year. Perhaps I’ll get a degree when I'm 65.There’s only one problem. I get quite angry when people interrupt my favorite programs. My friends don’t understand that I can change at 60.【参考答案】1.C 【命题立意】考查学生对所听内容简单推断的能力。

2017年吉林省长春高考数学二模试卷(文科) 含解析

2017年吉林省长春高考数学二模试卷(文科) 含解析

2017年吉林省长春市高考数学二模试卷(文科)一、选择题(本大题包括12小题,每小题5分,共60分,每小题给出的四个选项中,只有一项是符合题目要求的,请将正确选项涂在答题卡上)1.已知集合A={0,1,2},B={y|y=2x,x∈A}则A∩B=() A.{0,1,2} B.{1,2} C.{1,2,4} D.{1,4}2.已知复数z=1+i,则下列命题中正确的个数为()①;②;③z的虚部为i;④z在复平面上对应点在第一象限.A.1 B.2 C.3 D.43.下列函数中,既是奇函数又在(0,+∞)单调递增的函数是()A.y=e x+e﹣x B.y=ln(|x|+1)C.D.4.圆(x﹣2)2+y2=4关于直线对称的圆的方程是( )A.B.C.x2+(y﹣2)2=4 D.5.堑堵,我国古代数学名词,其三视图如图所示.《九章算术》中有如下问题:“今有堑堵,下广二丈,袤一十八丈六尺,高二丈五尺,问积几何?”意思是说:“今有堑堵,底面宽为2丈,长为18丈6尺,高为2丈5尺,问它的体积是多少?”(注:一丈=十尺).答案是()A.25500立方尺B.34300立方尺C.46500立方尺D.48100立方尺6.某游戏设计了如图所示的空心圆环形标靶,图中所标注的一、二、三区域所对的圆心角依次为,,,则向该标靶内投点,该点落在区域二内的概率为()A.B. C. D.7.在△ABC中,D为三角形所在平面内一点,且,则=( )A.B. C. D.8.运行如图所示的程序框图,则输出结果为( )A.1008 B.1009 C.2016 D.20179.关于函数,下列叙述有误的是()A.其图象关于直线对称B.其图象可由图象上所有点的横坐标变为原来的倍得到C.其图象关于点对称D.其值域是[﹣1,3]10.如图是民航部门统计的2017年春运期间十二个城市售出的往返机票的平均价格以及相比去年同期变化幅度的数据统计图表,根据图表,下面叙述不正确的是()。

吉林省长春市普通高中2017届高三下学期第二次模拟考试语文试题

吉林省长春市普通高中2017届高三下学期第二次模拟考试语文试题

长春市普通高中2017届高三质量监测(二)语文本试题卷共10页,22题。

全卷满分150分。

考试用时150分钟。

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

孔府档案是围绕孔子直系后裔历代衍圣公的活动所形成的文书档案,也是我国现存数量最多、收藏最完整、内容最丰富、涵盖时间最长的私家档案文献。

因档案中保存了衍圣公与明清以来中央和地方机构之间事务往来的大量文书资料,使其又兼具官方档案的性质。

孔府档案表明,居住在孔府的衍圣公凭借大宗主的地位,在家族中建立了严密的宗族组织和管理机构,并通过修宗谱、订族规等方式统管全国各地的孔氏族人,孔氏家族宗族体系之完整、宗法制度之完善、祖训族规之完备,是其他宗族很难比拟的。

孔子世家谱汇集了分散在全国80余处支派的谱系衍变信息和流寓朝鲜半岛的孔氏族人的世系信息,其对于考察孔氏宗族繁衍,迁移、发展和影响等,具有重大参考价值。

崇儒尊孔是历代统治者巩固和强化统治秩序的手段,孔子直系后裔也因之被扶植成为拥有部分政治和经济特权的世袭贵族。

朝廷与与孔氏贵族之间有着相互依存的共同利益,这在孔府档案中都有较深刻的反映。

明清帝王或亲赴辟雍诣学观礼,或临幸阙里释奠孔子,或遣子派官致祭庙林;对孔子后裔或优免差徭,或置官封爵,或赐土赐民。

这固然表明国家对孔子学说的尊崇和对孔子后裔的优待,但也是出于强化国家思想的需要。

孔府是中国历史上持续时间最长的贵族地主庄园,保存了成序列的土地文书,包括不下10万件的各种土地执照、纳税和过割凭证等。

这些文献信息,为研究明清以来的地权分配和转移、土地买卖和经营、租佃制度及其变迁,以及农业耕作制度等经济史问题提供了翔实而可靠的材料,对探索中国古代基层社会实态和演变轨迹具有重重要价值。

孔府司房日用账簿、日收支款项账簿等,也为探究明清及民国时期基层社会的商业贸易网络、物价和生活水平及其变迁等,提供了全面而原始的记录。

在中华优秀传统文化传承和弘扬越来越受到重视的今天,孔府档案的价值也日渐凸显,它不仅保留了作为道德规范存在的族规家训,还记载了族人弘扬孝道、敦睦宗族、彰显忠义的言行事迹,其中可资弘扬家风、传承家训、承继家史的资料,无疑是文化传承的重要思想宝库。

吉林省吉林市2017届高三下学期教学质量检测(二)英语试题解析(原卷版)

吉林省吉林市2017届高三下学期教学质量检测(二)英语试题解析(原卷版)

吉林省吉林市2017届高三第二次模拟考试英语试题第一部分听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

例:How much is the shirt?A. £19.15.B. £9.18. C £9.15.答案是C。

1. Where are the speakers?A. On a plane.B. On a bus.C. On a ship2. What time is it now?A. 7:00.B. 7:25.C. 7:30.3. What does the man mean?A. He is too busy to help her.B. His hands are holding something.C. He wants to move the sofa all by himself.4. Who is the woman?A. Mr. Johnson‟s secretary.B. Mr. Johnson‟s wife.C. Mr. Johnson‟s mother.5. How does the man feel?A. Worried.B. Excited.C. Unconcerned.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。

每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。

每段对话或独白读两遍。

听第6段材料,回答第6至8题。

6. Why did Mary‟s parents make her stay at home yesterday evening?A. To let her do her homework.B. To let her take care of her baby sister.C. To let her watch TV.7. What did Mary do yesterday evening?A. She watched boxing on TV.B. She watched a movie about boxing.C. She went to a concert.8. What did John do last night?A. He watched boxing on TV.B. He went to the cinema.C. He went to a concert.听第7段材料,回答第9至11题。

吉林省吉林市普通中学2017届高三数学毕业班第二次调研测试试题理(扫描版)

吉林省吉林市普通中学2017届高三数学毕业班第二次调研测试试题理(扫描版)

吉林省吉林市普通中学2017届高三数学毕业班第二次调研测试试题理(扫描版)吉林市普通中学2016—2017学年度高中毕业班第二次调研测试数 学(理科)参考答案与评分标准一、选择题:本大题共12题,每小题5分,共60分,在每小题给出的四个选项中,只有一个是符合题目二、填空题:本大题共4小题,每小题5分,共20分.把答案填在答题卡的相应位置. 13. [0,2];2n 三、解答题17解:(1)由图象知A=1, 54(),2126T πππω=-== ----------------------------------------------------3分将点(,1)6π代入解析式得sin()1,3πϕ+=因为||2πϕ<,所以6πϕ=所以()sin(2)6f x x π=+ --------------------------------------------------------------------------5分(2)由(2)cos cos a c B b C -=得: (2sin sin )cos sin cos A C B B C -= 所以2sin cos sin(),2sin cos sin A B B C A B A =+=因为(0,)A π∈,所以sin 0A ≠,所以12cos ,,233B B A C ππ==+= -------------------------------8分25()sin(),0,263666A f A A A πππππ=+<<<+<,所以1sin()(,1]62A π+∈所以1()(,1]22A f ∈ ------------------------------------------------------------------------10分18.(本小题满分12分)解:(Ⅰ)设数列{a n }的公比为q ,当1q =时,符合条件,133a a ==,a n =3 -----------------------------------2分当1q ≠时,21313(1)91a q a q q ⎧=⎪⎨-=⎪-⎩所以21213(1)9a q a q q ⎧=⎪⎨++=⎪⎩,解得1112,2a q ==- ----5分 1112()2n n a -=⨯-综上:a n =3或1112()2n n a -=⨯- ---------------------------------------------------6分注:列方程组21211139a q a a q a q ⎧=⎪⎨++=⎪⎩求解可不用讨论 (Ⅱ)证明:若a n =3,则b n =0,与题意不符;222231112()3()22n n n a ++=⨯-=⨯,222233log log 22n n n b n a +=== -----------------8分 14111(1)1n n n c b b n n n n +===-++ ----------------------------------------------------10分123111111(1)()()1122311n c c c c n n n ++++=-+-++-=-<++ ---------12分19.(本小题满分12分)解 (Ⅰ) 由题意可知,这20名工人年龄的众数是30, --------------------------------2分这20名工人年龄的平均数为x =120(19+3×28+3×29+5×30+4×31+3×32+40)=30,------------------------------4分(Ⅱ) 这20名工人年龄的茎叶图如图所示:------------------------------------------7分(Ⅲ) 记年龄为24岁的三个人为A 1,A 2,A 3;年龄为26岁的三个人为B 1,B 2,B 3则从这6人中随机抽取2人的所有可能为{A 1,A 2},{A 1,A 3},{A 2,A 3},{A 1,B 1},{A 1,B 2}, {A 1,B 3},{A 2,B 1},{A 2,B 2},{A 2,B ,3},{A 3,B 1},{A 3,B 2},{A ,3,B 3},{B 1,B 2},{B 1,B 3},{B 2,B 3}共15种。

2017届吉林省长春市普通高中高三质量监测(二)理科数学试题及答案

2017届吉林省长春市普通高中高三质量监测(二)理科数学试题及答案

长春市普通高中2017届高三质量监测(二)数 学(理 科)一、选择题(本大题共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、已知集合{}0x x P =≥,1Q 02x xx ⎧+⎫=≥⎨⎬-⎩⎭,则()R Q P = ð( ) A .(),2-∞ B .(],1-∞- C .()1,0- D .[]0,22、复数12i i--的共轭复数对应的点位于( )A .第一象限B .第二象限C .第三象限D .第四象限3、已知随机变量ξ服从正态分布()21,σN ,若()20.15ξP >=,则()01ξP ≤≤=( )A .0.85B .0.70C .0.35D .0.15 4、已知:p 函数()f x x a =+在(),1-∞-上是单调函数,:q 函数()()log 1a g x x =+(0a >且1a ≠)在()1,-+∞上是增函数,则p ⌝成立是q 成立的( )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5、若x ,y 满足约束条件5315153x y y x x y +≤⎧⎪≤+⎨⎪-≤⎩,则35x y +的取值范围是( )A .[]13,15-B .[]13,17-C .[]11,15-D .[]11,17-6、一个几何体的三视图如图所示,则该几何体的体积为( )A .163B .203C .152D .1327、已知平面向量a ,b 满足a = ,2b = ,3a b ⋅=-,则2a b += ( )A .1 B . C .4D .8、下面左图是某学习小组学生数学考试成绩的茎叶图,1号到16号同学的成绩依次为1A 、2A 、⋅⋅⋅⋅⋅⋅、16A ,右图是统计茎叶图中成绩在一定范围内的学生人数的算法流程图,那么该算法流程图输出的结果是( )A .6B .10C .91D .929、已知函数()1cos cos 22f x x x x =+,若将其图象向右平移ϕ(0ϕ>)个单位后所得的图象关于原点对称,则ϕ的最小值为( ) A .6π B .56π C .12πD .512π10、设m ,R n ∈,若直线()()1120m x n y +++-=与圆()()22111x y -+-=相切,则m n +的取值范围是( ) A .(),22⎡-∞-++∞⎣ B .(),⎡-∞-+∞⎣C .22⎡-+⎣ D .(][),22,-∞-+∞11、若()F ,0c 是双曲线22221x y a b-=(0a b >>)的右焦点,过F 作该双曲线一条渐近线的垂线与两条渐近线交于A ,B 两点,O 为坐标原点,∆OAB 的面积为2127a ,则该双曲线的离心率e =( )A .53B .43C .54D .8512、设数列{}n a 的前n 项和为n S ,且121a a ==,(){}2n n nS n a ++为等差数列,则n a =( ) A .12n n- B .1121n n -++ C .2121n n -- D .112n n ++二、填空题(本大题共4小题,每小题5分,共20分.)13、62x ⎛ ⎝的展开式中常数项为 .14、已知0a >且曲线y x a =与0y =所围成的封闭区域的面积为2a ,则a = .15、正四面体CD AB 的外接球半径为2,过棱AB 作该球的截面,则截面面积的最小值为 . 16、已知函数()f x 为偶函数且()()4f x f x =-,又()235,01222,12x x x x x f x x -⎧--+≤≤⎪=⎨⎪+<≤⎩,函数()12xg x a ⎛⎫=+ ⎪⎝⎭,若()()()F x f x g x =-恰好有4个零点,则a 的取值范围是 .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、(本小题满分12分)在C ∆AB 中,tan 2A =,tan 3B =. ()1求角C 的值;()2设AB =C A . 18、(本小题满分12分)根据某电子商务平台的调查统计显示,参与调查的1000位上网购物者的年龄情况如下图显示.()1已知[)30,40、[)40,50、[)50,60三个年龄段的上网购物者人数成等差数列,求a,b的值;()2该电子商务平台将年龄在[)30,50之间的人群定义为高消费人群,其他的年龄段定义为潜在消费人群,为了鼓励潜在消费人群的消费,该平台决定发放代金券,高消费人群每人发放50元的代金券,潜在消费人群每人发放100元的代金券,现采用分层抽样的方式从参与调查的1000位上网购物者中抽取10人,并在这10人中随机抽取3人进行回访,求此三人获得代金券总和X的分布列与数学期望.19、(本小题满分12分)如图,在四棱锥CDP-AB中,PA⊥平面CDAB,D2PA=AB=A=,四边形CDAB满足DAB⊥A,C//DB A且C4B=,点M为CP中点,点E为C B边上的动点,且C λBE=E.()1求证:平面D A M⊥平面CPB;()2是否存在实数λ,使得二面角DP-E-B的余弦值为23?若存在,试求出实数λ的值;若不存在,说明理由.20、(本小题满分12分)在C ∆AB 中,顶点()1,0B -,()C 1,0,G 、I 分别是C ∆AB 的重心和内心,且G//C I B. ()1求顶点A 的轨迹M 的方程;()2过点C 的直线交曲线M 于P 、Q 两点,H 是直线4x =上一点,设直线C H 、PH 、Q H 的斜率分别为1k ,2k ,3k ,试比较12k 与23k k +的大小,并加以证明. 21、(本小题满分12分)设函数()()()1ln 1f x ax x bx =-+-,其中a 和b 是实数,曲线()y f x =恒与x 轴相切于坐标原点. ()1求常数b 的值;()2当01x ≤≤时,关于x 的不等式()0f x ≥恒成立,求实数a 的取值范围;()3求证:10000.41000.5100011001100001000e ⎛⎫⎛⎫<< ⎪⎪⎝⎭⎝⎭.请考生在第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. 22、(本小题满分10分)选修4-1:几何证明选讲 如图,过点P 作圆O 的割线PBA 与切线PE ,E 为切点,连接AE ,BE ,∠APE 的平分线与AE ,BE 分别交于点C ,D ,其中30∠AEB = .()1求证:D DD CE PB P ⋅=B PAP ;()2求C ∠P E 的大小. 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x y O 中,曲线1C的参数方程为21x y ⎧=⎪⎨=-+⎪⎩(t 为参数),以原点O 为极点,以x 轴正半轴为极轴,建立极坐标系,曲线2C 的极坐标方程为ρ=.()1求曲线1C 的普通方程与曲线2C 的直角坐标方程;()2试判断曲线1C 与2C 是否存在两个交点,若存在,求出两交点间的距离;若不存在,说明理由.24、(本小题满分10分)选修4-5:不等式选讲 设函数()212f x x x a a =++-+,R x ∈. ()1当3a =时,求不等式()7f x >的解集;()2对任意R x ∈恒有()3f x ≥,求实数a 的取值范围.长春市普通高中2017届高三质量监测(二)数学(理科)参考答案及评分标准一、选择题(本大题包括12小题,每小题5分,共60分)1.D2.A3.C4.C5.D6.D7.B8.B9.C 10.A 11.C 12.A简答与提示:1. 【命题意图】本题主要考查集合交集与补集的运算,属于基础题.【试题解析】D 由题意可知{|1Q x x =-≤或2}x >,则{|12}Q x x =-<≤R ð,所以{|02}P Q x x =≤≤R ð. 故选D.2. 【命题意图】本题考查复数的除法运算,以及复平面上的点与复数的关系,属于基础题.【试题解析】A131255ii i-=--,所以其共轭复数为3155i +. 故选A.3. 【命题意图】本题考查正态分布的概念,属于基础题,要求学生对统计学原理有全面的认识.【试题解析】C (01)(12)0.5(2)0.35P P P ξξξ==->=≤≤≤≤. 故选C. 4. 【命题意图】本题借助不等式来考查命题逻辑,属于基础题. 【试题解析】C 由p 成立,则1a ≤,由q 成立,则1a >,所以p ⌝成立时1a >是q 的充要条件.故选C.5. 【命题意图】本题主要考查线性规划,是书中的原题改编,要求学生有一定的运算能力. 【试题解析】D 由题意可知,35x y +在(2,1)--处取得最小值,在35(,)22处取得最大值,即35[11,17]x y +∈-.故选D.6. 【命题意图】本题通过正方体的三视图来考查组合体体积的求法,对学生运算求解能力有一定要求.【试题解析】D 该几何体可视为正方体截去两个三棱锥,所以其体积为41138362--=. 故选D.7. 【命题意图】本题考查向量模的运算.【试题解析】B|2|+==a b . 故选B.8. 【命题意图】本题考查学生对茎叶图的认识,通过统计学知识考查程序流程图的认识,是一道综合题. 【试题解析】B 由算法流程图可知,其统计的是数学成绩大于等于90的人数,所以由茎叶图知:数学成绩大于等于90的人数为10,因此输出结果为10. 故选B.9. 【命题意图】本题主要考查三角函数的图像和性质,属于基础题.【试题解析】C 由题意()sin(2)6f x x π=+,将其图像向右平移ϕ(0)ϕ>个单位后解析式为()sin[2()]6f x x πϕ=-+,则26k πϕπ-=,即212k ππϕ=+()k ∈N ,所以ϕ的最小值为12π. 故选C.10. 【命题意图】本题借助基本不等式考查点到直线的距离,属于中档题.【试题解析】A由直线与圆相切可知||m n +=理得1mn m n =++,由2()2m n mn +≤可知211()4m n m n ++≤+,解得(,2[2)m n +∈-∞-++∞ . 故选A.11. 【命题意图】本题主要考查双曲线的几何性质,结合着较大的运算量,属于难题.【试题解析】C 由题可知,过I 、III 象限的渐近线的倾斜角为θ,则tan b aθ=,222tan 2ab a bθ=-,因此△OAB 的面积可以表示为3222112tan 227a b a a a a b θ⋅⋅==-,解得34b a=,则54e =. 故选C.12. 【命题意图】本题是最近热点的复杂数列问题,属于难题. 【试题解析】A 设(2)n n n b nS n a =++,有14b =,28b =,则4n b n =, 即(2)4n n n b nS n a n =++= 当2n ≥时,1122(1)(1)01n n n n S S a a n n ---++-+=-所以12(1)11n n n n a a n n -++=-,即121n n a a n n -⋅=-,所以{}n a n 是以12为公比,1为首项的等比数列,所以11()2n n a n -=,12n n n a -=. 故选A.二、填空题(本大题包括4小题,每小题5分,共20分)13.60 14.4915.83π 16.192,8⎛⎫⎪⎝⎭简答与提示: 13. 【命题意图】本题主要考查二项式定理的有关知识,属于基础题.【试题解析】由题意可知常数项为2246(2)(60C x =. 14. 【命题意图】本题考查定积分的几何意义及微积分基本定理,属于基础题.【试题解析】由题意32223aa x ==⎰,所以49a =.15. 【命题意图】球的内接几何体问题是高考热点问题,本题通过求球的截面面积,对考生的空间想象能力及运算求解能力进行考查,具有一定难度.【试题解析】由题意,面积最小的截面是以AB 为直径,可求得AB =,进而截面面积的最小值为283ππ=.16. 【命题意图】本题主要考查数形结合以及函数的零点与交点的相关问题,需要学生对图像进行理解,对学生的能力提出很高要求,属于难题.【试题解析】由题意可知()f x 是周期为4的偶函数,对称轴为直线2x =. 若()F x 恰有4个零点,有(1)(1)(3)(3)g f g f >⎧⎨<⎩,解得19(2,)8a ∈.17. (本小题满分12分)【命题意图】本小题主要考查两角和的正切公式,以及同角三角函数的应用,并借助正弦定理考查边角关系的运算,对考生的化归与转化能力有较高要求. 【试题解析】解:(1) +,tan tan()A B C C A B π+=∴=-+(3分)tan 2,tan 3,tan 1,4A B C C π==∴=∴=(6分)(2)因为tan 3B =sin 3sin 3cos cos B B B B⇒=⇒=,而22sincos 1B B +=,且B 为锐角,可求得sin B =.(9分)所以在△ABC 中,由正弦定理得,sin sin AB AC B C =⨯=.(12分)18. (本小题满分12分)【命题意图】本小题主要考查统计与概率的相关知识、离散型随机变量的分布列以及数学期望的求法. 本题主要考查数据处理能力.【试题解析】(1)由图可知0.035a =,0.025b =. (4分)(2) 利用分层抽样从样本中抽取10人,其中属于高消费人群的为6人,属于潜在消费人群的为4人. (6分)从中取出三人,并计算三人所获得代金券的总和X , 则X 的所有可能取值为:150,200,250,300.363101(150)6C P X C ===,21643101(200)2C C P X C ===, 12643103(250)10C C P X C ===, 343101(300)30C P X C ===,(10分) 且1131150200250300210621030EX =⨯+⨯+⨯+⨯=. (12分)19. (本小题满分12分)【命题意图】本小题主要考查立体几何的相关知识,具体涉及到线面以及面面的垂直关系、二面角的求法及空间向量在立体几何中的应用. 本小题对考生的空间想象能力与运算求解能力有较高要求.【试题解析】解:(1) 取PB 中点N ,连结MN 、AN ,M 是PC 中点,1//,22MN BC MN BC ∴==,又//BC AD ,//,MN AD MN AD ∴=,∴四边形ADMN 为平行四边形,AP AD AB AD ⊥⊥ ,AD ∴⊥平面PAB ,AD AN ∴⊥,AN MN ∴⊥AP AB = ,AN PB ∴⊥,AN ∴⊥平面PBC ,AN ⊂ 平面ADM ,∴平面ADM ⊥平面PBC . (6分)(2) 存在符合条件的λ.以A 为原点,AB 方向为x 轴,AD 方向为y 轴,AP 方向为z 轴,建立空间直角坐标系A xyz -,设(2,,0)E t ,(0,0,2)P ,(0,2,0)D ,(2,0,0)B从而(0,2,2)PD =- ,(2,2,0)DE t =-,则平面PDE 的法向量为1(2,2,2)n t =-,又平面DEB 即为xAy 平面,其法向量2(0,0,1)n =,则1212122cos ,3||||n n n n n n ⋅<>===⋅, 解得3t =或1t =,进而3λ=或13λ=.(12分) 20. (本小题满分12分) 【命题意图】本小题主要考查直线与圆锥曲线的综合应用能力,具体涉及到轨迹方程的求法,椭圆方程的求法、直线与圆锥曲线的相关知识. 本小题对考生的化归与转化思想、运算求解能力都有很高要求. 【试题解析】解:(1) 已知11(||||||)||||22ABC A S AB AC BC r BC y ∆=++⋅=⋅,且||2BC =,||3A y r =,其中r 为内切圆半径,化简得:||||4AB AC +=,顶点A 的轨迹是以B C 、为焦点,长轴长为4的椭圆(去掉长轴端点),其中2,1,a c b ===进而其方程为22143x y +=(0)y ≠.(5分)(2) 1232k k k =+,以下进行证明:当直线PQ 斜率存在时,设直线:(1)PQ y k x =-且11(,)P x y ,22(,)Q x y ,(4,)H m联立22143(1)x y y k x ⎧+=⎪⎨⎪=-⎩可得2122834k x x k +=+,212241234k x x k -=+. (8分)由题意:13m k =,1214y m k x -=-,2324y m k x -=-.11212312()(4)()(4)(4)(4)y m x y m x k k x x --+--+=--21212121212882(5)()2424224()1636363m k kx x m k x x mk m mk x x x x k ++-+++====-+++当直线PQ 斜率不存在时,33(1,),(1,)22P Q -,231332222333m m m k k k -++=+== 综上可得1232k k k =+. (12分) 21. (本小题满分12分)【命题意图】本小题主要考查函数与导数的综合应用能力,具体涉及到用导数来描述原函数的单调性、极值以及函数零点的情况. 本小题对考生的逻辑推理能力与运算求解有较高要求. 【试题解析】解:(1) 对()f x 求导得:1()ln(1)1ax f x a x b x-'=-++-+,根据条件知(0)0f '=,所以101b b -=⇒=. (3分)(2) 由(1)得()(1)ln(1)f x ax x x =-+-,01x ≤≤1()ln(1)11axf x a x x-'=-++-+22(1)(1)21()1(1)(1)a a x ax ax a f x x x x -+--++''=-+=-+++. ① 当12a ≤-时,由于01x ≤≤,有221()()0(1)a a x a f x x ++''=-≥+,于是()f x '在[0,1]上单调递增,从而()(0)0f x f ''≥=,因此()f x 在[0,1]上单调递增,即()(0)0f x f ≥=而且仅有(0)0f =;②当0a ≥时,由于01x ≤≤,有221()0(1)ax a f x x ++''=-<+,于是()f x '在[0,1]上单调递减,从而()(0)0f x f ''≤=,因此()f x 在[0,1]上单调递减,即()(0)0f x f ≤=而且仅有(0)0f =;③当102a -<<时,令21min{1,}a m a+=-,当0x m ≤≤时,221()()0(1)a a x a f x x ++''=-≤+,于是()f x '在[0,]m 上单调递减,从而()(0)0f x f ''≤=,因此()f x 在[0,]m 上单调递减,即()(0)0f x f ≤=而且仅有(0)0f =.综上可知,所求实数a的取值范围是1(,]2-∞-.(8分)(3) 对要证明的不等式等价变形如下:2110000100010000.41000.55210001100111()()(1)(1)100001000100001000e e ++<<⇔+<<+ 所以可以考虑证明:对于任意的正整数n,不等式215211(1)(1)n n e n n+++<<+恒成立. 并且继续作如下等价变形 2152112111(1)(1)()ln(1)1()ln(1)52n n e n n n n n n +++<<+⇔++<<++211(1)ln(1)0()5111(1)ln(1)0()2p n n nq n n n ⎧++-<⎪⎪⇔⎨⎪++->⎪⎩对于()p 相当于(2)中21(,0)52a =-∈-,12m =情形,有()f x 在1[0,]2上单调递减,即()(0)0f x f ≤=而且仅有(0)0f =.取1x n=,当2n ≥时,211(1)ln(1)05nn n++-<成立;当1n =时,277(1)ln 21ln 210.710555+-=-<⨯-<.从而对于任意正整数n 都有211(1)ln(1)05n n n++-<成立.对于()q 相当于(2)中12a =-情形,对于任意x ∈[0,1],恒有()0f x ≥而且仅有(0)0f =. 取1x n=,得:对于任意正整数n 都有111(1)ln(1)02n n n++->成立. 因此对于任意正整数n ,不等式215211(1)(1)n n e n n+++<<+恒成立.这样依据不等式215211(1)(1)n n e n n+++<<+,再令10000n =利用左边,令1000n = 利用右边,即可得到10000.41000.5100011001()()100001000e <<成立.(12分) 22. (本小题满分10分)【命题意图】本小题主要考查平面几何的证明,具体涉及到弦切角定理以及三角形 相似等内容. 本小题重点考查考生对平面几何推理能力.【试题解析】解:(1) 由题意可知,EPC APC ∠=∠,PEB PAC ∠=∠, 则△PED ∽△PAC ,则PE PD PAPC=,又PE ED PBBD=,则ED PB PD BD PAPC⋅=. (5分)(2) 由EPC APC ∠=∠,PEB PAC ∠=∠,可得CDE ECD ∠=∠,在△ECD 中,30CED ∠= ,可知75PCE ∠= . (10分) 23. (本小题满分10分) 【命题意图】本小题主要考查极坐标系与参数方程的相关知识,具体涉及到极坐标方程与平面直角坐标方程的互化、利用直线的参数方程的几何意义求解直线与曲线交点的距离等内容. 本小题考查考生的方程思想与数形结合思想,对运算求解能力有一定要求.【试题解析】解:(1) 对于曲线1C 有1x y +=,对于曲线2C 有2214x y +=.(5分)(2) 显然曲线1C :1x y +=为直线,则其参数方程可写为21x y ⎧=⎪⎪⎨⎪=-⎪⎩(为参数)与曲线2C :2214x y +=联立,可知0∆>,所以1C 与2C 存在两个交点,由12t t +=,1285t t =,得21||d t t =-==. (10分)24. (本小题满分10分)【命题意图】本小题主要考查不等式的相关知识,具体涉及绝对值不等式及不等式证明等内容. 本小题重点考查考生的化归与转化思想.【试题解析】解:(1)当3a =时,()174,2135,22341,2x x f x x x x ⎧-≤⎪⎪⎪=<<⎨⎪⎪-≥⎪⎩所以()7f x >的解集为{}02x x x <>或 (5分) (2)()2122121f x x a x a x a x a a a =-+-+≥-+-+=-+由()3f x ≥恒成立,有13a a -+≥,解得2a ≥所以a 的取值范围是[)2,+∞ (10分)。

吉林省长春市2017届高三下学期第二次模拟考试英语试题 Word版含答案

吉林省长春市2017届高三下学期第二次模拟考试英语试题 Word版含答案

第Ⅰ卷第一部分听力(1-20小题)在笔试结束后进行。

注意事项:英语听力共两节,共20小题;每小题1.5分,满分30分。

第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

例:How much is the shirt?A.£19.15.B.£9.15.C.£9.18.答案是B。

1.Where does the conversation probably take place?A.In a library.B.In a laboratory.C.In a restaurant.2.What does the man mean ?A.He knows Thomas’ birth date.B.The woman is good with dates.C.He has trouble remembering dates.3.What is the weather like now?A.FineB.HotC.Rainy.4.What will the speakers do tomorrow afternoon?A.Go to a park.B.Go shopping.C.Eat out with Joe.5.What is the man complaining about?A.His work.B.The weather.C.The noise from neighbors.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。

每段对话或独白后有几个小题,从题中多给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。

吉林省普通中学2017届高三毕业班第二次调研测试-英语-Word版含答案

吉林省普通中学2017届高三毕业班第二次调研测试-英语-Word版含答案

吉林市普通中学2016—2017学年度高中毕业班第二次调研测试英语本试卷分第I卷(选择题)和第II卷(非选择题)。

第I卷1至10页,第II卷10至12页。

共150分.考试时间120分钟。

注意事项:请按照题号顺序在答题纸上各题目的答题区域内作答,超出答题区域书写的答案无效;在草稿纸、试题卷上答题无效.第I卷第一部分:听力(共两节, 满分30分)第一节(共5小题;每小题1。

5分, 满分7.5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置.听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题.每段对话仅读一遍。

1. When will the man meet with Nick?A。

On May 12th。

B. On May 14th。

C. On May 15th。

2. What does the woman want the man to do?A。

Get her a disk.B. Repair a computer。

C. Help to deal with a document.3. How long has the man been working?A。

For seven hours。

B。

For half an hour. C. For seven and a half hours。

4. What are the speakers mainly talking about?A。

A supermarket。

B。

A new store. C。

A piece of furniture.5. What weather does the woman dislike?A. Foggy。

B。

Cloudy。

C。

Sunny.第二节(共15小题;每小题1。

5分,满分22。

5分)听下面5段对话或独白。

每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置.听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题给出5秒钟的作答时间。

【吉林省长春】2017学年高考二模数学年(文科)试题

【吉林省长春】2017学年高考二模数学年(文科)试题

吉林省长春市2017年高考二模理科数学试卷答 案一、选择题1~5.BCDDC 6~10.BACDA 11~12.BA 二、填空题13.211e 22+14.91 15.1 080 16.2 三、解答题17.解:(1)由题可知1112(23())n n a n a *+=--∈N , 从而有13n n b b +=,11112a b =-=, 所以{}n b 是以1为首项,3为公比的等比数列. (2)由(1)知113n b -=,从而1132n n a -=+,13131(3)312log log n n n c n --+==>-,有12(1)01212n n T c c c n n n ->+++=++⋯+⋅⋅⋅-=,所以(1)2n n n T ->.18.解:(1)根据统计数据做出列联表如下:经计算7.287 6.635k ≈>,因此可以在犯错误概率不超过1%的前提下,认为抗倒伏与玉米矮茎有关. (2)(ⅰ)按照分层抽样的方式抽到的易倒伏玉米共4株,则X 的可能取值为0,1,2,3,4.416420(0)P X ==,13416420(1)P X ==,22416420(2)P X ==,31416420(3)P X ==,44420(4)P X ==即X 的分布列为:416416416(ⅱ)在抗倒伏的玉米样本中,高茎玉米有10株,占5,即每次取出高茎玉米的概率均为25,设取出高茎玉米的株数为ξ,则~(505,)2B ξ,即2E 50205np ξ==⨯=,23D (1)501255np p ξ=-=⨯⨯=.19.解:(1)证明:因为AD BCD BC BCD AD BC ⊥⊂⊥平面,平面,所以,又AC BC ACAD A BC ACD BC ABC ABC ACD ⊥=⊥⊂⊥因为,,所以平面,平面,所以平面平面.(2)由已知可得CD =如图所示建立空间直角坐标系,由已知(0,0,0)(0,2,0)C B A ,,,D ,1)2E .有31()2CE =,(3,0,1)CA = (3,0,0)CD =,设平面ACE 的法向量(,,z)n x y =,有001002z n CA n CE y z +=⎧=⎪⎨=++=⎪⎩,令1x =,得(1,0,n =-, 设平面CED 的法向量(,,z)m x y =,有01002m CD m CE y z ⎧=⎧=⎪⎪⎨⎨=++=⎪⎪⎩, 令y=1,得(0,1,2)m =-,二面角A CE D --的余弦值||23cos ||||25n m n m θ===.20.解:(1)联立方程有,2402x y py⎧+=⎪⎨=⎪⎩,有280y p -+=,由于直线与抛物线相切,得22832048p p p y x ====-,,所以.(2)假设存在满足条件的点(,0)(0)M m m >,直线l :x ty m =+,有28x ty m y x=+⎧⎨=⎩,2880y ty m -=-,设1221(,)(,)A x B y y x ,,有121288y yt y y m +==-,,22222111||()(1)AM x m y t y =-+=+,22222222||()(1)BM x m y t y =-+=+,22222222212111114()||||(1)(1)(1)4t mAM BM t y t y t m ++=+=+++,当4m =时,2211||||AM BM +为定值,所以(4,0)M . 21.解:(1)()1af x x a x'=+--,因为()f x 存在极值点为1,所以(1)02201f a a '=-==,即,,经检验符合题意,所以1a =. (2)()1(1)(1)(0)a af x x a x x x x'=+--=+-> ①0()()(0,0)a f x f x '≤+∞>当时,恒成立,所以在上为增函数,不符合题意; ②0()0a f x x a '>==当时,由得,()00()0()x a f x x a f x f x ''>><<<当时,,所以为增函数,当时,,所为减函数,所以当x a =时,()f x 取得极小值()f a又因为()f x 存在两个不同零点12x x ,,所以()0f a <,即()211ln 02a a a a a +--<整理得1ln 12a a >-,作()y f x =关于直线x a =的对称曲线()(2)g x f a x =-,令2()()()(2)()22lna x h x g x f x f a x f x a x a x-=-=--=--222222()220(2)()a a h x a x x x a a '=-+=-+≥---+所以()(0,2)h x a 在上单调递增,不妨设1222221()()0()(2)()()x a x h x h a g x f a x f x f x <<>==->=,则,即, 又因为212(0,)(0,)a x a x a -∈∈,,且()f x 在(0,)a 上为减函数,故211222a x x x x a +<->,即,又1ln 12a a >-,易知1212a x x >>+成立,故.22.解:(1)由22(3sin )12ρθ+=得22143x y +=,该曲线为椭圆.(2)将1tcos tsin x y αα=+⎧⎨=⎩代入22143x y +=得224cos 6cos 9()0t t αα-+-=,由直线参数方程的几何意义,设12||||||||t PB PA t ==,,1226cos t t 4cos αα-+=-,1229t t 4cos α-=-,所以122127|||||t t |4cos 2PA PB α+=-==-,从而24cos 7α=,由于(0)2πα∈,,所以cos α.23.解:(1)令24x ||1||6245x x x y x -+≤⎧⎪==<<⎨-⎩+⎪≥+-,-1,-1x 5,x 5,可知|||65|1x x ++-≥,故要使不等式|||m 5|1x x ++-≤的解集不是空集,有6m ≥.(2)证明:由a b ,均为正数,则要证a b b a a b a b ≥,只需证1a b b a a b --≥,整理得()1a b ab-≥,由于当a b ≥时,0a b -≥,可得1a ba b -⎛⎫≥ ⎪⎝⎭,当0a b a b <-<时,,可得1a ba b -⎛⎫> ⎪⎝⎭,可知a b ,均为正数时1a ba b -⎛⎫≥ ⎪⎝⎭,当且仅当a b =时等号成立, 从而a b b a a b a b ≥成立.吉林省长春市2017年高考二模理科数学试卷解 析一、选择题1.【考点】交集及其运算.【分析】由题意求出集合B ,由交集的运算求出A∩B . 【解答】解:由题意可知,集合A={0,1,2},则B={}2A xy y x =∈,={1,2,4},所以A∩B={1,2}, 故选:B .2.【考点】复数求模.【分析】利用复数的模、共轭复数、虚部与复数与平面内点的对应关系即可判断出正误. 【解答】解:∵复数z=1+i , ②z = ②1z i =-,正确; ③z 的虚部为1;④z 在复平面上对应点(1,1)在第一象限. 可得:①②④正确,③错误. 故选:C .3.【考点】奇偶性与单调性的综合.【分析】根据函数的单调性和奇偶性判断即可. 【解答】解:对于A .B 选项为偶函数,排除, C 选项是奇函数,但在(0,+∞)上不是单调递增函数. 故选:D .4.【考点】关于点、直线对称的圆的方程. 【分析】求出圆(x ﹣2)2+y 2=4的圆心关于直线y =对称的坐标,即可得出结论. 【解答】解:设圆(x ﹣2)2+y 2=4的圆心关于直线y =对称的坐标为(a ,b ),则312222b a b a ⎧=-⎪⎪-⎨+⎪=⎪⎩, ∴a=1,∴圆(x ﹣2)2+y 2=4的圆心关于直线y x =对称的坐标为( ,从而所求圆的方程为()(2214x y -+=.故选D .5.【考点】由三视图求面积、体积.【分析】由三视图得到几何体为横放的三棱柱,底面为直角三角形,利用棱柱的体积公式可求.【解答】解:由已知,堑堵形状为棱柱,底面是直角三角形,其体积为12018625=46502⨯⨯⨯立方尺.故选C .6.【考点】向量在几何中的应用.【分析】利用三角形以及向量关系,求解三角形的面积即可. 【解答】解:由已知,在△ABC 中,D 为三角形所在平面内一点,且11AD AB AC 32=+,点D 在AB 边的中位线上,且为靠近BC 边的三等分点处,从而有ABD ABC 1S S 2=△△,ABC 1S S 3=△ACD △,ABC ABC 111S 1S S 236⎛⎫=--= ⎪⎝⎭△BCD △△,有S 1S 3=△BCD △ABD .故选:B .7.【考点】程序框图.【分析】由已知,S=0﹣1+2﹣3+4+…﹣2015+2016=1008,即可得出结论 【解答】解:由已知,S=0﹣1+2﹣3+4+…﹣2015+2016=1008. 故选A .8.【考点】正弦函数的对称性.【分析】利用正弦函数的图像和性质,判断各个选项是否正确,从而得出结论.【解答】解:关于函数2sin 314y x π⎛⎫=++ ⎪⎝⎭,令4x π=-,求得y=﹣1,为函数的最小值,故A 正确;由2sin 14y x π⎛⎫=++ ⎪⎝⎭图像上所有点的横坐标变为原来的13倍,可得2sin 34y x π⎛⎫=+ ⎪⎝⎭的图像,故B 正确;令11x 12π=,求得y=1,可得函数的图像关于点1112π⎛⎫⎪⎝⎭,1对称,故C 错误;函数的值域为[﹣1,3],故D 正确, 故选:C .9.【考点】进行简单的合情推理.【分析】根据折线的变化率,得到相比去年同期变化幅度、升降趋势,逐一验证即可. 【解答】解:由图可知D 错误.故选D . 10.【考点】几何概型. 【分析】求出扇形AOC 的面积为34π,扇形AOB 的面积为3π,从而得到所求概率. 【解答】解:设OA=3,则,由余弦定理可求得AOP=30°,所以扇形AOC 的面积为34π,扇形AOB 的面积为3π,从而所求概率为31434ππ=. 故选A .11.【考点】双曲线的简单性质.【分析】利用已知条件求出a ,b 求出双曲线方程,利用双曲线的定义转化求解三角形的最小值即可.【解答】解:双曲线C 的渐近线方程为,一个焦点为0F (,,可得22a 43b =,a 2cb ==,双曲线方程为22143y x -=,设双曲线的上焦点为F',则|PF|=|PF'|+4,△PAF 的周长为|PF|+|PA|+|AF|=|PF'|+4+|PA|+3,当P 点在第一象限时,|PF'|+|PA|的最小值为|AF'|=3, 故△PAF 的周长的最小值为10. 故选:B .12.【考点】利用导数研究函数的单调性.【分析】令F (x )=f (x )+2x ,求出导函数F'(x )=f'(x )+2>0,判断F (x )在定义域内单调递增,由f(1)=1,转化()2log 313f 1xx -<--为()22log 31233f log 1x x -+-<,然后求解不等式即可.【解答】解:令F (x )=f (x )+2x ,有F'(x )=f'(x )+2>0,所以F (x )在定义域内单调递增,由f (1)=1,得F (1)=f (1)+2=3,因为()2log 313f 1xx -<--等价于()22log 31233f log 1x x-+-<,令2t log 31x=-,有f (t )+2t <3,则有t <1,即2log 311x -<,从而312x-<,解得x <1,且x≠0.故选:A . 二、填空题13.【考点】定积分.【分析】根据定积分的计算法则计算即可.【解答】解:11ex dx x ⎛⎫+ ⎪⎝⎭⎰=222111111ln ln ln122222ex x e e e ⎛⎫⎛⎫+=+-+=+ ⎪ ⎪⎝⎭⎝⎭故答案为:21122e +.14.【考点】归纳推理.【分析】由三角形数组可推断出,第n 行共有2n ﹣1项,且最后一项为n2,所以第10行共19项,最后一项为100,即可得出结论.【解答】解:由三角形数组可推断出,第n 行共有2n ﹣1项,且最后一项为n2, 所以第10行共19项,最后一项为100,左数第10个数是91. 故答案为91.15.【考点】排列、组合的实际应用.【分析】根据题意,求甲、乙两人至少一人参加,则分2种情况讨论:①、若甲乙同时参加,②、若甲乙有一人参与,分别求出每种情况下的情况数目,由分类计数原理计算可得答案, 【解答】解:根据题意,分2种情况讨论: ①若甲乙同时参加,先在其他6人中选出2人,有C 62种选法, 选出2人进行全排列,有A 22种不同顺序, 甲乙2人进行全排列,有A 22种不同顺序,甲乙与选出的2人发言,甲乙发言中间需恰隔一人,有2种情况,此时共有2226222C A A 120=种不同顺序,②若甲乙有一人参与,在甲乙中选1人,有C 21种选法,在其他6人中选出3人,有C 63种选法, 选出4人进行全排列,有A 44种不同情况,则此时共有134264C C A 960=种,从而总共的发言顺序有1080种不同顺序. 故答案为:1080.16.【考点】球内接多面体.【分析】由正弦定理可求出三角形PBC 外接圆半径为,F 为BC 边中点,求出,利用勾股定理结论方程,求出四棱锥P ﹣ABCD 外接球半径.【解答】解:由已知,设三角形PBC 外接圆圆心为O1,由正弦定理可求出三角形PBC,F 为BC 边中点,求出11O F=2, 设四棱锥的外接球球心为O ,外接球半径的平方为221BD O F 42⎛⎫+= ⎪⎝⎭,所以四棱锥外接球半径为2.故答案为2. 三、解答题17.【考点】数列与不等式的综合;数列的求和. 【分析】(1)利用数列的递推关系式推出()111223N n n a a n *+⎛⎫=∈ ⎪-⎝⎭-,然后证明{n b }是以1为首项,3为公比的等比数列.(2)求出13n n b -=,化简1132n n a -=+,推出131313log 312n n n o c l g n --⎛⎫=+>=- ⎪⎝⎭,然后通过数列求和,证明结果.18.【考点】离散型随机变量的期望与方差;独立性检验;离散型随机变量及其分布列. 【分析】(1)利用已知条件写出2×2列联表即可. (2)(i )按照分层抽样的方式抽到的易倒伏玉米共4株,则X 的可能取值为0,1,2,3,4;求出概率即可得到即X 的分布列.(ii )设取出高茎玉米的株数为ξ,判断概率满足ξ~B (50,),然后求解期望与方差. 19.【考点】二面角的平面角及求法;平面与平面垂直的判定.【分析】(1)证明AD ⊥BC ,AC ⊥BC ,推出BC ⊥平面ACD ,然后证明平面ABC ⊥平面ACD . (2)建立空间直角坐标系,求出相关点的坐标,求出平面ACE 的法向量,平面CED 的法向量,利用空间向量的数量积求解二面角A ﹣CE ﹣D 的余弦值. 20.【考点】直线与抛物线的位置关系. 【分析】(1)联立方程有,,通过△=0,求出p=4,即可求解抛物线方程.(2)假设存在满足条件的点M (m ,0)(m >0),直线l :x=ty+m ,有,y 2﹣8ty ﹣8m=0,设A (x 1,y 1),B (x 2,y 2),利用韦达定理弦长公式,化简求解即可.21.【考点】利用导数研究函数的单调性;利用导数研究函数的极值. 【分析】(1)求出,利用f (x )存在极值点为1,结合f'(1)=0,求出a .(2)求出,通过①当a≤0时,②当a >0时,判断函数的单调性求出函数的极值,所以当x=a 时,f (x )取得极小值f (a ),利用f (x )存在两个不同零点x 1,x 2,f (a )<0,作y=f (x )关于直线x=a 的对称曲线g (x )=f (2a ﹣x ),令h (x )=g (x )﹣f (x )=f (2a ﹣x )﹣f (x ),求出导数,利用函数的单调性,最值推出结果.22.【考点】参数方程化成普通方程;简单曲线的极坐标方程.【分析】(1)利用极坐标与直角坐标的关系化简曲线C 1的极坐标方程为普通方程;(2)对参数方程x ,y 代入椭圆方程,然后根据直线参数方程的几何意义,设|PA|=|t 1|,|PB|=|t 2|,结合韦达定理得到所求.23.【考点】不等式的证明;绝对值不等式的解法.【分析】(1)利用绝对值不等式推出|x+1|+|x﹣5|≥6,转化不等式|x+1|+|x﹣5|≤m的解集不是空集,推出m即可;(2)利用分析法,集合指数函数的性质,推出结果即可.。

2017年吉林省长春市普通高中高考英语二模试卷(附答案详解)

2017年吉林省长春市普通高中高考英语二模试卷(附答案详解)

2017年吉林省长春市普通高中高考英语二模试卷一、阅读理解(本大题共15小题,共30.0分)AGrand City Tour with Griffith Park Observatory &Hollywood SignLos Angeles,Hollywood,homes of the movie stars-it's all here on this Grand Tour!Your tour guide takes great pride in offering amazing customer service,too.Hotel pickups are available at almost any hotel in Hollywood,Downtown LA,Mid-Wilshire,West Hollywood and North Hollywood.From:69 US DollarsMovie Star Homes GPS Self-Guided Bike TourWant the flexibility of taking a"whenever you feel like it"Hollywood tour?Here's the perfect solution.Explore and experience the best of West Hollywood and Beverly Hills on your own time,at your own pace!Cycle past legendary homes and all the places where the rich and famous live,shop,eat and play.And your tour is guided by GPS with an iPad fixed to the bicycle! 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Even last year,I ended up hiking up Mt.Kilimanjaro.I would never have been that adventurous before I started my 30-day challenges.I also figured out that if you really want something badly enough,you can do anything for 30 days.Have you ever wanted to write a novel?Tens of thousands of people try to write their won 50,000-word novel in 30 days.It turns out that all you have to do is write 1,667 words a day for a month.So I did.So here's my question to your:What are you waiting for?The next 30days are going to pass whether you like it or not,so why not think about something you have always wanted to try and give it a shot!4.The underlined sentence in Para.1 means ______ .A. I felt my life was unchanged and boringB. I didn't like following others'footstepsC. I thought that I was in troubleD. It seemed that my car broke down5.The author did all the following things every day to change EXCEPT ______ .A. take a photoB. cycle to workC. write over 1000 wordsD. stick to computer games6.The author of the passage aims to ______ .A. talk about his own experiencesB. encourage people to try something newC. introduce a new philosophyD. show people the challenges in life7.Which of the following can be the best title for the passage?______A. Would you like to change yourself?B. Fed up with your life?C. Ways to challenge yourself.D. Try something new for 30 days.CLife in the Internet age is lonely-or is it?That's what experts in human interaction are debating after a new Stanford University survey has been published.According to the study,the more time people spend online,the less they can spare for real-life relationships with family and friends.The researchers asked 113 people about the Web's influence on daily activities.36% of those people are online for more than five hours a week."As Internet use becomes more widespread,it will have an increasingly isolating (孤立的)effect on society,"says Robert Kraut,one of the researchers.Scholars and Web lovers criticized the study for stretching its data to make the"isolating"point.While 13% of regular Web users admitted the loss of time with loved ones,60% reported watching less TV.The survey also shows that E-mail is the most popular online activity,If some of webheads (网虫)spend what was once passive TV time keeping company with friends via E-mails,"that's a move toward greater connectedness,"says Paul Resnick,a professor at the University of Michigan.This isn't the first claim that the Web should be criticized,A 1998 report monitored73Pittsburgh-area families'Net use for a year.People who used the Internet more"talked less to family members and reported being lonelier and more depressed."says Robert Kraut."It's true that there have been big declines in social connectedness over the past decades,but those declines began before the Internet was invented,"says Thomas Putnam.As Amitay Etzioni says,the Internet gives us a different kind of social life-not better or worse than before,but just different.8.Who claimed that the Web had negative influence?______A. Robert Kraut.B. Paul Resnick.C. Thomas Putnam.D. Amitay Etzioni.9.The underlined word"This"in Para.4 refers to ______ .A. the study conducted by Stanford UniversityB. the survey made by the University of MichiganC. the conclusion in a report written in 1998D. the opinion expressed in Bowling Alone10.From the passage we learn that ______ .A. watching TV used to take time away from staying onlineB. 36% of web users spend more than five hours a week onlineC. the Web was blamed more than once for causing an isolating effectD. the Web has the same influence as telephones and televisions11.The passage mainly discusses ______ .A. how we can make a better use of the InternetB. whether the Internet causes an isolating effectC. how declines in social connectedness appearD. what a different life the Internet brings to us.DFood production does great harm to our environment.There are many procedures involved in the manufacture of food that result in greenhouse gases and other pollutants.Some procedures require the consumption of large amounts of fossil fuels,such as the transportation and storage of food products.Other factors that cause great damage to our environment include the overuse of fresh water.The production of beef is more damaging to the environment than that of any other food we consume,Raising large numbers of cattle requires the production of large amounts of food forthe animals.It's estimated that producing one pound of beef requires seven pounds of feed. Land use is also a problem.If the cattle are free-range cattle,large areas of land are required for them to live on.This has led to disastrous forest cutting and the loss of rare plants and animal species,particularly in tropical rain forests in Central and South America.Another problem specific to beef production is methane emissions(甲烷排放).Although many people are aware of the damaging effects of carbon dioxide,they don't realize methane's global warming potential is 25times worse,making it a more dire problem. Unfortunately,beef consumption is growing rapidly.This is the result of simple supply and demand factors.Specifically,there are two main causes of demand that are encouraging the production of more supply.First,the increase in the world population means there are more people to consume meat.The second factor is socioeconomic advancement.As citizens in developing nations become financially stable,they can afford to buy more meat. Therefore,one way to reduce the greenhouse gas emissions is for people around the world to significantly cut down on the amount of beef they eat.12.Which of the following is TRUE?______A. Raising free-range cattle is eco-friendly.B. People cut down trees for animal habitats.C. Producing beef can damage the environment.D. Carbon dioxide causes far more harm than methane.13.The underlined word"dire"in Para.4 means ______ .A. urgentB. commonC. typicalD. avoidable14.Beef production is growing rapidly because ______ .A. more people are in demand of beefB. developing countries raise more cattleC. more land is available to raise cattleD. the cost of raising cattle is relatively low15.The author writes this article to ______ .A. describe the booming of the beef productionB. emphasize the advantages of beef productionC. condemn the deforestation of the rain forestD. argue for a decrease in beef consumption.二、阅读七选五(本大题共5小题,共10.0分)The Case For and Against Homework Home work is typically defined as any tasks"assigned to students by school teachers that are meant to be carried out during nonschool hours".Homework has been a constant topic of believed that homework helped create disciplined minds.(1) Since then,arguments for and against homework have continued to multiply.A number of studies have been conducted on homework.Two analyses by Cooper and colleagues are the most comprehensive.They concluded that the relationship between homework and student achievement was found to be positive.(2)(3) The authors of How Homework Is Hurting Our Children and What We Can Do About It criticized both the quantity and quality of homework.They provided evidence that too much homework harms student's health and family time.They insisted that teachers reduce the amount of homework and avoid homework over breaks and holidays.One of the more controversial issues in the homework debate is the amount of time students should spend on homework.Researchers have offered various recommendations.(4) Another question regarding homework is the extent to which schools should involve parents.Some studies have reported minimal positive effects or even negative effects for parental involvement.(5)Finally,assignments should cause students and their parents or other family members to become engaged in conversations and thus extend the students' learning.A.Inappropriate homework may produce little or no benefit.B.Teachers are to blame for providing too much homework.C.Doing homework causes improved academic achievement.D.Although the research support for homework is powerful,the case against homework is popular.E.By 1940,growing concern that homework affected other home activities caused a reaction against it.F.Experts recommend parents receive clear rules and teachers do not expect parents to act as experts.G.For example,5 to 10 minutes per subject might be appropriate for 4th graders,while 30 to 60 minutes for high school students.16. A. A B. B C. C D. D E.E F.F G. G17. A. A B. B C. C D. D E.E F.F G. G18. A. A B. B C. C D. D E.E F.F G. G19. A. A B. B C. C D. D E.E F.F G. G20. A. A B. B C. C D. D E.E F.F G. G三、完形填空(本大题共20小题,共30.0分)When my husband Ian died in 2014,ten days after being diagnosed with cancer,I was completely broken up.He was just 54.We'd planned to(21)together and live the years to the fullest.At first,the(22)of spending the following years of my life without him left me feeling(23).I was in deep(24)having lost him.But as time passed,I began to(25)that life is so precious,and none of us can(26)what's around the comer.I realized I could sit at home and(27)Ian,(28)I could make the most of every moment of my life.Three months(29)he died,a friend mentioned a choir run by a(30)Tenovus Cancer Care,which was for anyone(31)by cancer.I've(32)loved to sing,so I agreed,That decision really has changed my(33).We rehearse (排练)once a week and perform(34)to raise money.We sing inspiring songs like"You've Got a Friend"which really(35)all of us.Spending time with the choir,I have not only made a whole new friendship circle,but I feel so(36),which has been fantastic for my(37)wellbeing.I(38)money for the charity too,including travelling a long way to Machu Picchu,where I scattered (播撒)some of Ian's ashes,I know Ian would(39)my decision to look after myself.Living a good life,and finding ways to enjoy my retirement,is the best(40)I can offer him.21. A. die B. retire C. live D. sing22. A. feeling B. thought C. sense D. hope23. A. lost B. tired C. excited D. relaxed24. A. trouble B. breath C. shock D. regret25. A. wonder B. understand C. forget D. remember26. A. change B. doubt C. predict D. leave27. A. mourn for B. wait for C. look after D. talk with28. A. and B. so C. for D. or29. A. before B. after C. since D. as30. A. charity B. company C. club D. hospital31. A. killed B. defeated C. scared D. affected32. A. hardly B. seldom C. always D. often33. A. mind B. life C. dream D. goal34. A. dances B. plays C. services D. concerts35. A. amuse B. interest C. push D. impress36. A. skillful B. cautious C. secure D. positive37. A. mental B. physical C. social D. economical38. A. spend B. save C. raise D. earn39. A. approve of B. care aboutC. be opposed toD. be sorry about40. A. award B. experience C. comfort D. help四、语法填空(本大题共1小题,共10.0分)41.阅读下面材料,在空白处填入适当的内容(一个单词)或括号内单词的正确形式.On the first day of her work,Sally found that a class full of problems was waiting forher.She was told six teachers (1) (quit)before her.When she walked into the classroom,it was chaos:two boys were fighting in a corner,yet (2) rest of the class seemed not to notice them;some girls were chatting and some were running about with paper,food packages and other garbage (3) (leave)everywhere.Sally walked onto the platform, (4) (pick)up a piece of chalk and wrote on theblackboard:"Rule 1:We are family!"All students stopped (5) (look)at her.And she continued with Rule 2,Rule 3…In the following weeks,Sally worked out 10class rules and posted them (6) the walls of the classroom.She patiently explained all the rules to the students and required everyone to follow them. (7) (surprise),Sally was not driven out like the former teachers;(8) ,she won respect from the students.Over the years,she witnessed gradual changes in the class.At the graduation ceremony,just (9) sheexpected,she was very proud to stand with a class of care,manners and (10) (confident).五、短文改错(本大题共1小题,共15.0分)42.第一节短文改错假设英语课上老师要求同桌之间交换修改作文,请你修改你同桌写的以下作文.文中共有10处错误,每句中最多有两处错误.错误涉及一个单词的增加、删除或修改.增加:在缺词处加一个漏字符号(∧),并在其下面写出该加的词.修改:在错的词下划一横线,并在该词下面写出修改后的词.注意:1.每处错误及其修改均仅限一词;2.只允许修改10处,多者(从第11处起)不计分.Born in America,Thomas Edison was a great scientist and an inventor.He was oncethought to be a boy whom was not worth educating.In fact,he was a man full ofimaginations.I admire Edison a lot because his great contributions to the world.He is said to havingmore than 1,000inventions.In his lifetime,he were always eager to know how things worked.He was also very diligent and work day and night.And this explained why he had so much great inventions.Success can never be easy achieved.Probably I can't be another Edison himself,but I will work hard to be a worthy person.六、书面表达(本大题共1小题,共25.0分)43.假设你是李华,你班同学在今天的英语课上就"高中生是否应该在学校使用智能手机"进行了课堂讨论,同学们的观点不一.请你根据下表内容,为本次课堂讨论结果撰写一篇英文报道稿,内容应该包括:1.活动简介;2.支持理由;3.支持者理由;4.你的观点.注意:1.词数100左右;2.可适当加入细节,以使行文连贯;3.文章开头已经给出,不计入总词数.Smart phones are becoming more and more of a necessity for high school students.___ _______________.答案和解析1.【答案】【解析】21.A,细节理解题,根据第二部分"Movie Star Homes GPS Self-Guided Bike Tour"中"Explore and experience the best of West Hollywood and Beverly Hills on your own time,at your own pace!",可知答案为A.22.C,细节理解题,根据第三部分"Amazing Scavenger Quest in Los Angeles"中"ONLY $44 FOR UP TO 5 PEOPLE!"和"Offered at a fixed price for a group as opposed to per person pricing",可知答案为C.23.D,细节理解题,根据第四部分"Disneyland Excursion from Los Angeles"内容,可知答案为D.本文是一篇广告类应用文,介绍了四个在洛杉矶市可以参加的一日游项目,内容包括活动特色的介绍,行程安排及费用等信息.解答任务型阅读理解题,首先对原文材料迅速浏览,掌握全文的主旨大意.因为阅读理解题一般没有标题,所以,速读全文,抓住中心主旨很有必要,在速读的过程中,应尽可能多地捕获信息材料.其次,细读题材,各个击破.掌握全文的大意之后,细细阅读每篇材料后的问题,弄清每题要求后,带着问题,再回到原文中去寻找、捕获有关信息.最后,要善于抓住每段的主题句,阅读时,要有较强的针对性.对于捕获到的信息,要做认真分析,仔细推敲,理解透彻,只有这样,针对题目要求,才能做到稳、准.4.【答案】【解析】24-27.ADBD24:A 词义猜测题,根据后文"so I decided to follow in the footsteps of the great American philosopher,Morgan Spurlock,and try something new for 30 days."(因此我决定跟随美国伟大的哲学家Morgan Spurlock的脚步,并且尝试某个新的事物30天)可知作者认为他的生活毫无改变,并且无聊,才不满足于现状,想尝试一些新事物,故答案为A.25:D,细节理解题,根据第二段中"take a picture"和"bikes to work",第四段中"all you have to do is write 1,667words a day for a month.So I did.",可知作者做了拍照、骑车去上班和写东西,没有提及D选项,故答案为D.26:B,推理判断题,作者整篇文章的目的就是鼓励读者去尝试一下自己想做的事情,尤其是最后一段"What are you waiting for?",可知答案为B.27:D,主旨大意题,整篇文章作者强调的就是用30天尝试一些新的事情,可知答案为D.本文是一篇记叙文,讲述通过自己的亲身尝试,告诉读者30天刚好是这么一段合适的时间去养成一个新的习惯或者改掉一个习惯.本文是一篇人生感悟类阅读,题目涉及多道细节理解题,做题时结合原文和题目有针对性找出相关语句进行仔细分析,结合选项选出正确答案.推理判断也是要在抓住关键句子的基础上合理的分析才能得出正确的答案.8.【答案】【解析】28-31.AACB28题答案:A 细节理解题,根据文章第二段的."As Internet use becomes more widespread,it will have an increasingly isolating (孤立的) effect on society,"says Robert Kraut,one of the researchers.(一位名叫Robert Kraut的研究者说道"随着网络使用变得越来越普及,它将会对社会拥有越来越大的孤立作用")可知本题答案为A选项.29题答案:A 词义猜测题,本句话意为"这并不是批评网络(的消极影响)的第一个论断."This指代的是Stanford University开展的调查研究.可知答案为A.30题答案:C 推理判断题,由文章第四段第一句话可知,因为网络使人们之间的联系减少,不止一次有人就此现象进行调查研究,可知答案为C.31题答案:B 主旨大意题,文章以一项调查研究入题,展开关于网络是否导致"社会孤立效应"的辩论,可知答案为B.本文是一篇议论文.讲述的是对于斯坦福大学进行的一项关于网络是否导致"社会孤立效应"的研究,社会各界展开的讨论.本文是一篇科教类阅读,题目涉及多道细节理解题,做题时结合原文和题目有针对性找出相关语句进行仔细分析,结合选项选出正确答案.推理判断也是要在抓住关键句子的基础上合理的分析才能得出正确的答案.12.【答案】【解析】32.C.细节理解题.由第二段The production of beef is more damaging to the environment than that of any other food we c onsume可知,生产牛肉特别损害环境,故选C.33.A.词义猜测题.由前一句Although many people are aware of the damaging effects of carbon dioxide,they don't realize methane's global warming potential is 25times worse,虽然很多人都知道二氧化碳的破坏性影响,但是他们没有意识到甲烷的全球变暖潜能是二氧化碳的25倍,这是一个更可怕的事情,故选A.34.A.细节理解题.由倒数第二段the increase in the world population means there are more people to consume meat,可知人口的增长导致需要更多的牛肉,故选A.35.D.主旨归纳题.由最后一段one way to reduce the greenhouse gas emissions is for people around the world to significantl y cut down on the amount of beef they eat.可知作者希望可以使人们降低牛肉的消费来减少温室气体排放,故选D.本文讲述了生产牛肉给环境带来的危害,希望可以使人们降低牛肉的消费.本篇文章主要是细节理解方面的题型,需要考生通过跳读迅速找到对应的句子进行详细理解,句子的理解可以结合上下文意思,通过逻辑推理得到.16.【答案】【解析】36-40.ECDGF36.E 考查上下文联系.根据"Since then,arguments for and against homework have continued to multiply"可知,后来越来越多的人担心作业影响家庭活动,因此引起了争论.故选E.37.C 考查细节理解.根据"They concluded that the relationship between homework and student achievement was found to be positive"可知,做家庭作业会提高学业成绩.故选C.38.D 考查上下文联系.根据"They provided evidence that too much homework harms student's health and family time"可知,虽然种种原因都是支持布置家庭作业的,但是还是有人反对写家庭作业.故选D.39.G 考查推理判断.根据"One of the more controversial issues in the homework debate is the amount of time students should spend on homework"可推断出,低级的学生适合布置花费较少时间的作业,而高中学生需要布置花费较多时间的作业.故选G.40.F 考查上下文联系.根据"Some studies have reported minimal positive effects or even negative effects for parental inv olvement"可知,父母也产生了一些消极的影响,所以父母也要遵守规则.故选F.本文是一篇议论文,主要阐述了学生对家庭作业的不同看法.七选五"这样的题型,主要目的在于"考查考生对文章的整体内容和结构以及上下文逻辑意义的理解和掌握.解题时最主要的两个步骤就是1.理清文章的逻辑和结构,2.在所给的原文中找出关键词或者说是线索词.文章的整体思路能帮助你在答案中筛选出符合逻辑的选项,关键词则能帮助你确定更多的细节,排除相近的选项.这些技巧多练练就能孰能生巧.21.【答案】【解析】41--45:BBACB 46--50:CADBA 51--55:DCBDD 56--60:DACAC 41.B 考查动词.根据上文提到了作者丈夫去世时的年龄,可以推断出之前二人曾计划过一起退休后的生活,可知答案为B.42.B 考查名词.此处句意是"一想到未来二十年没有丈夫的老年生活,就觉得很迷茫.",可知答案为B.43.A 考查形容词.此处句意是"一想到未来二十年没有丈夫的老年生活,就觉得很迷茫.",可知答案为A.44.C 考查名词.此处句意为"丈夫的突然去世让我很受打击",可知答案为C.45.B 考查动词.此处句意为"随着时间的推移,我开始意识到了生命的珍贵",可知答案为B.46.C 考查动词.此处句意为"我们当中没有人能预测即将发生什么",可知答案为C 47.A 考查动词短语.此处句意为"待在家里哀悼他",可知答案为A.48.D 考查连词.此处表示作者的不同选择:是呆在家中悲伤,还是走出去积极面对生活,可知答案为D.49.B 考查连词.此处表达在作者丈夫去世之后,可知答案为B.50.A 考查名词.根据54空后"toraise money"及58空后"for the charity",可知答案为A.51.D 考查非谓语动词.这个机构的目的在于帮助像作者这样有亲人因癌症去世而生活受到影响的人,可知答案为D.52.C 考查副词.此处表示作者喜欢唱歌,所以参加慈善排练活动,可知答案为C.53.B 考查名词.参加慈善歌唱的决定改变了作者的生活,即把作者从悲伤之中拉了出来,可知答案为B.54.D 考查名词.此处应与歌唱和排练相呼应,可知答案为D.55.D 考查动词.从"inspiringsongs"可以知道这些歌曲都是打动人的,可知答案为D.56.D 考查形容词.此处意为"排练不仅使我有了新的朋友圈,而且我感觉情绪积极.",可知答案为D.57.A 考查形容词.根据上文可知此处应表示心理健康,可知答案为A.58.C 考查动词.作者除了跟合唱团一起为慈善尽力,自己也通过个人活动来筹集善款(raisemoney),可知答案为C.59.A 考查动词短语.作者相信丈夫会支持她好好生活,可知答案为A.60.C 考查名词.作者认为,她好好享受生活,就是对丈夫最好的慰藉,可知答案为C.本文主要讲述了主人公在失去丈夫后的迷茫以及如何走出阴霾积极面对生活的心路历程.本文是一个人物故事类完形填空,题目涉及多道细节理解题,做题时结合原文和题目有针对性的找出相关语句进行仔细分析,结合选项选出正确答案.推理判断题也是要在抓住关键句子的基础上合理的分析才能得出正确答案,切忌胡乱猜测,一定要做到有理有据.41.【答案】【小题1】had quit(ted)【小题2】The【小题3】left【小题4】picked【小题5】to look【小题6】on/onto【小题7】Surprisingly【小题8】instead【小题9】as【小题10】confidence【解析】61.had quit(ted)考查时态.由句子Sally found that a class full of problems was waiting for her.可见是过去时,她被告知之前已经有六个老师离开这个班级,这应该发生在她接手这个班级之前,所以应该用过去完成时,故答案had quit/quited.62.The 考查固定短语.根据文章内容两个男孩在打架,剩下的学生好像没有注意到他们.the rest of是英语的固定搭配"剩下的,剩余的".63.left 考查with的复合结构. with paper,food packages and other garbage____(leave) everywhere.此结构中paper,food packages and other garbage 与leave之间是被动关系,所以用过去分词left做宾语补足语.64.picked 考查时态.Sally walked onto 走上讲台,(pick) up a piece of chalk 拿起一支粉笔and wrote on the blackboard在黑板上写字,这三个动作是连续发生的,都应该用过去时,所以答案填picked.65.to look 考查固定短语.根据句意:同学们停下来看,stop to do 是英语的固定短语"停下来做某事".66.on/onto 考查介词.根据句意:把它们贴在教室的墙上.on/onto是介词"在…的上面".67.Surprisingly 考查副词.根据下文:莎丽没有像以前的老师那样被赶出去,对于赶走了六名老师的这样一个班级,应该是令人惊讶的事,所以填副词surprisingly. 68.instead考查副词.根据句意:相反她获得了学生们的尊敬,instead是副词"相反,反之".69.as 考查定语从句.根据句意:正如她所期望的,她很骄傲地站在这群学生中,as 关系代词引导定语从句,并在定语从句中做宾语,指代后文整个句子的意思.70.confidence考查名词.关心、礼貌和自信. care,manners and confidence这是三个并列名词,所以用名词confidence.本文写了一名老师和一群很难管教的学生的故事.老师用十条班规改变了整个班级和学生的精神面貌.语法填空是通过语篇在语境中考查语法知识的运用能力,在解题前应快速浏览短文掌握大意,在读懂短文的基础上,结合短文提供的特定的语言环境去逐句分析.要解决好语法填空,离不开坚实的语法知识,有了坚实的语法知识才能对语言进行正确的分析和判断,从而答对题目.42.【答案】Born in America,Thomas Edison was a great scientist and an inventor.He was once thought to be a boy whom was not worth educating.In fact,he was a man full of imaginations.I admire Edison a lot because∧his great contributions to the world.He is said to having more than 1,000inventions.In his lifetime,he were always eager to know how things worked.He was also very diligent and work day and night.And this explained why he had so much great inventio ns.Success can never be easy achieved.Probably I can't be another Edison himself,but I will work hard to be a worthy person.详解:1.去掉an 考查冠词.根据句意:Thomas Edison是一位伟大的科学家和发明家.表示一个人身兼两种身份,只用一个不定冠词.2.whom改为who 考查定语从句.分析句子结构可知这是一个定语从句,先行词boy 指人,在从句中作主语,用关系代词who.3.imaginations改为imagination 考查名词.imagination是不可数名词,没有复数形式.4.加of 考查词语用法.because of意为"因为",是介词短语,后接名词或代词作宾语;because意为"因为",是连词,后接句子.此处用介词,因为后面接的是名词短语.5.having改为have 考查动词不定式.sb./sth is/was said to do.是一个句型,意为"据说某人/某物…",to是不定式符号.6.were改为was 考查主谓一致.主语he是单数,谓语动词也用单数was.7.work改为worked 考查动词时态.全文讲述的是已发生的事情,用一般过去时态.8.加many 考查形容词.inventions是复数,用many修饰,much修饰不可数名词.9.easy改为asily 考查副词.achieve是动词,要用副词修饰.10.himself改为myself 考查代词.根据句意:可能我自己不可能成为另一个爱迪生.用反身代词myself.【解析】本文主要介绍爱迪生是一位伟大的科学家和发明家.他对世界作出了伟大的贡献,作者也希望像他一样,努力工作成为一个有价值的人.高考短文改错题的形式有说明文,短文故事,书信等,具有很强的实用性.短文的内容和语言都符合高中学生的实际,从表面上看类似一篇学生习作.首先,通读全文,了解短文大意,把握全篇的时态、人称及行文逻辑,在通读全文时把一些容易的错误先改好,再进行逐句改错.其次,要进行逐个句子的改错.这是要对文中的词法、句法和语篇着重分析和特别注意.最后把改好的短文再阅读一遍,检查答案是否正确,感觉是否还有不妥之处,最终形成定稿.43.【答案】Smart phones are becoming more and more of a necessity for high school students.Therefore,today in our English class a discussion was held about whether we senior school students should use our smart phones at school.Opinions are different.【解析】Smart phones arebecoming more and more of a necessity for high school students.Therefore,today in our English class a discussion was held about whether we senior schoolstudents should use our smart phones at school.(高分句型一)Opinions are different.(引出话题)Students who arein favor of it consider it convenient to contact their parents when necessary.(高分句型二)Besides,they point out it is smart phones that give them access to theInternet and thus benefit their study.(高分句型三)While others hold the opposite opinion.They find it easy for some students to be addicted to playing the smart phones,which has a negative effect on their studies.Worse still it can alsodiscourage students'will.(阐述各自不同看法)。

吉林省长春市普通高中2017届高三质量监测(二)历史试题及参考答案【人教版】〖必修三册、选修一二三四〗

吉林省长春市普通高中2017届高三质量监测(二)历史试题及参考答案【人教版】〖必修三册、选修一二三四〗

吉林省长春市普通高中2017届高三质量检测(二)历史试题24.中国古代官职包括职官的名称、职权范围和品级地位等内容。

秦汉以后,官职复杂多变,每一细微变化都折射出相关政治变迁。

据此判断,下列选项正确的是A.秦汉“以吏治天下”,品位等级从属于职位的色彩浓厚B.魏晋官僚“世族化”,品位等级因素与职位的关系不大C.唐宋“入仕皆授官阶”,说明已基本摆脱品位的影响D.明清以职能官职为主,禹城市经济功能的增强紧密相关25.明朝规定,佃农见田主“不论齿序,并如少事长之礼”,仅亲戚间例外;但在一些土地荒芜,招募劳动力较难的地方,还常有佃户“刁悍成风”、地主“吞声茹苦”一类记载。

这一现象说明A.佃农逐步获得独立地位B.人地矛盾影响租佃关系C.政府决策脱离地方实际D.租佃关系改变社会结构26.春秋时期思想家老子主张“为无为,则无不治”,“勿伐”、“不为”、“不言”、“无欲”、“无兵”、“无味”、“我有三宝,持有保之。

一曰慈,二曰俭,三曰不敢为天下先。

”、“圣人处无为之事,行不言之教。

”下列选项对材料中信息理解最准确的是A.无为是指时有时无的一般的人类活动B.老子的主张主要是对普通人的道德要求C.体现想社会管理方式的设想D.是老子对现实的不满及消极避世的心态27.一位驻中国的外交官在1872年写道:中国正在迅速地成为一个令人生畏的对手,整个官僚阶级都决心恢复中国的国际地位。

兵工厂和造船厂的产量给人以深刻的印象,中国建造的军舰不久就将达到欧洲的最高水平。

这位外交官的陈述A.代表了西方人对中国的发展态度B.可佐证中国军事近代化的起步C.肯定了他对中国逐渐崛起的担忧D.肯定了民族资本主义快速发展28.中国大规模海外移民出现于1840年之后,晚于国际人口迁移(1500年前后),原因有①中国卷入资本主义世界市场的时间较晚②“普天之下,莫非王土”的天下观促成自大的心理③《中英天津条约》增开十处通商口岸④建立在农业社会基础上的伦理观的影响A.①②③B.②③④C.①②④D.①③④29.1935年瓦窑堡会议通过的决议中,决定将“工农共和国”改为“人民共和国”,提出红军行动的战略方针是“把国内战争同民族战争结合起来”。

长春市2017届高三质量监测(二)文科数学

长春市2017届高三质量监测(二)文科数学

数学试题卷(文科) 第3页(共 4 页)
19. (本小题满分 12 分)
A
已知三棱锥 A− BCD 中,△ ABC 是等腰直角三角形,
且 AC ⊥ BC, BC = 2, AD ⊥ 平面 BCD, AD =1.
E
(1) 求证:平面 ABC ⊥ 平面 ACD ;
(2) 若 E 为 AB 中点,求点 A 到平面 CED 的距离.
B
D
20. (本小题满分 12 分)
C
已知抛物线 C : y2 = 2 px( p 0) 与直线 x − 2 y + 4 = 0 相切.
(1) 求该抛物线的方程;
(2) 在 x 轴正半轴上,是否存在某个确定的点 M ,过该点的动直线 l 与抛物线 C 交

A, B 两点,使得 |
1 AM
|2
+
|
1 BM
已知数列 {an} 满足
a1
=
3 2
,
an+1
=
3an
−1(n
N+
)
.
(1)
若数列{bn}满足 bn
= an
−1 2
,求证:{bn}是等比数列;
(2) 求数列{an}的前 n 项和 Sn .
18. (本小题满分 12 分) 为了打好脱贫攻坚战,某贫困县农科院针对玉 米种植情况进行调研,力争有效地改良玉米品
P(K2≥k) 0.15 0.10 0.05 0.025 0.010 0.005
k
2.072 2.706 3.841 5.024 6.635 7.879
(K2 =
n(ad − bc)2
,其中 n = a + b + c + d )

吉林省长春市高三第二次质量监测理综物理试题Word版含解析

吉林省长春市高三第二次质量监测理综物理试题Word版含解析

长春市普通高中2017届高三质量监测(二)理科综合能力测试(物理部分)二、选择题:本题共8小题,每小题6分。

在每小题给出的四个选项中,第14~18题只有一项符合题目要求,第19~21题有多项符合题目要求。

全部选对的得6分,选对但不全的得3分,有选错的得0分。

14.下列说法正确的是A .物体抵抗运动状态变化的性质是惯性B .物体如果受到力的作用,运动状态一定发生变化C .带电的导体球总可以看作是一个位于其球心的点电荷D .力的单位“牛顿”是国际单位制中的基本单位15.一个质点做匀加速直线运动,在时间间隔t 内发生的位移为x ,动能变为原来的4倍,该质点运动的加速度大小为 A. 232t x B .252t x C .22t x D .23t x 16.如图所示,水平面上的小车内固定一个倾角为30°的光滑斜面,平行于斜面的细绳一端固定在车上,另一端系着一个质量为m 的小球,小球和小车均处于静止状态。

如果小车在水平面上向左加速且加速度大小不超过a 1时,小球仍能够和小车保持相对静止;如果小车在水平面上向右加速且加速度大小不超过a 2时,小球仍能够和小车保持相对静止。

则a 1和a 2 的大小之比为A .1:3B .3:3C .3:1D .1:317.如图所示,水平放置的平行板电容器间存在竖直向下的匀强电场,现有比荷相同的甲、乙两种带电粒子从电容器中央O 点分别沿OP 、OQ 以相同的速率水平射出,不计粒子重力及粒子间的相互作用,粒子在运动过程中均未与电容器两极板接触。

则以下对粒子运动的描述正确的是A .若甲、乙电性相同,则粒子离开电容器时速度一定相同B .若甲、乙电性相反,则粒子离开电容器时侧向位移一定相同C .若甲、乙电性相反,则粒子在电容器运动过程中电场力做功一定相等D .若甲、乙电性相反,则粒子离开电容器时速度偏转角大小一定相等18.1990年5月18日,经国际小行星中心批准,中科院紫金山天文台将国际编号为,密度与地球近似相等,则该小行星与地球的第一宇宙速度之比约为19.如图所示,边长为L 、匝数为N 、电阻不计的正方形线圈abcd 在磁感应强度为B 的匀强磁场中绕垂直于磁感线的转轴OO ′以恒定角速度ω转动。

吉林省吉林市普通中学2017届高三毕业班第二次调研测试数学(文)试题 Word版含答案

吉林省吉林市普通中学2017届高三毕业班第二次调研测试数学(文)试题 Word版含答案

吉林省普通中学2016-2017学年度高中毕业班第二次调研测试 数学(文科)第Ⅰ卷(选择是 共60分)一、选择题:本大题共12个小题,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.已知集合{}1 2 3 4A =,,,,集合{}3 4 5 6B =,,,,则集合A B 真子集的个数为( ) A.1B.2C.3D.42.已知复数21iz i+=+,则复数z 在复平面内对应的点在( ) A.第一象限B.第二象限C.第三象限D.第四象限3.命题“[]0 1m ∀∈,,12m x x+≥”的否定形式是( ) A.[]10 1 2m m x x∀∈+<,,B.[]10 1 2m m x x ∃∈+≥,, C.()()100 2m m x x∃∈-∞+∞+≥ ,,, D.[]10 1 2m m x x∃∈+<,, 4.阅读如图所示的程序框图,运行相应的程序,则输出S 的值为( )A.10-B.6C.14D.185.抛物线24x y =上一点A 的纵坐标为4,则点A 与抛物线焦点的距离为( )A.5B.46.若 x y ,满足约束条件02323x x y x y ≥⎧⎪+≥⎨⎪+≤⎩,则z x y =-的最小值是( )A.0B.3-C.32D.37.{}n a 是公差不为0的等差数列,满足22224567a a a a +=+,则该数列的前10项和10S =( ) A.10- B.5- C.0D.58.双曲线()222210 0x y a b a b-=>>,的一条渐近线与圆(()2211x y -+-=相切,则此双曲线的离心率为( ) A.29.若将函数()sin 2cos 2f x x x =+的图象向右平移ϕ个单位,所得图象关于y 轴对称,则ϕ的最小正值是( ) A.8πB.4πC.38πD.34π 10.某几何体的三视图如下图,若该几何体的所有顶点都在一个球面上,则该球面的表面积为( )A.4πB.283πC.443πD.20π11.在等腰直角ABC △中,AC BC =,D 在AB 边上且满足:()1CD tCA t CB =+-,若60ACD ∠=︒,则t 的值为( )1-12.设函数()'f x 是奇函数()()f x x R ∈的导函数,()20f -=,当0x >时,()()'03xf x f x +>,则使得()0f x >成立的x 的取值范围是( )A.()() 20 2-∞- ,,B.()()2 0 2 -+∞ ,,C.()() 2 2 2-∞-- ,,D.()()0 2 2 +∞ ,, 第Ⅱ卷(非选择题 共90分)二、填空题(每题5分,满分20分,将答案填在答题纸上)13.设函数()122 11log 1x x f x x x -⎧≤⎪=⎨->⎪⎩,,,则()1f f -=⎡⎤⎣⎦ . 14.已知2a = ,2b = ,a 与b 的夹角为45︒,且b a λ- 与a垂直,则实数λ= .15.给出下列命题:①若函数()y f x =满足()()11f x f x -=+,则函数()f x 的图象关于直线1x =对称;②点()2 1,关于直线10x y -+=的对称点为()0 3,; ③通过回归方程 y bxa =+ 可以估计和观测变量的取值和变化趋势; ④正弦函数是奇函数,()()2sin 1f x x =+是正弦函数,所以()()2sin 1f x x =+是奇函数,上述推理错误的原因是大前提不正确. 其中真命题的序号是 .16.设n S 为数列{}n a 的前n 项和,若()()()*21212nnn n n n a a n N +-⋅=+-⋅∈,则10S = .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.已知函数()()sin 0 2f x M x M πωϕϕ⎛⎫=+>< ⎪⎝⎭,的部分图象如图所示.(1)求函数()f x 的解+析式;(2)在ABC △中,角 A B C ,,的对边分别是 a b c ,,,若()2cos cos a c B b C -=,求2A f ⎛⎫⎪⎝⎭的取值范围.18.已知{}n a 是公比不等于1的等比数列,n S 为数列{}n a 的前n 项和,且33a =,39S =.(1)求数列{}n a 的通项公式; (2)设2233log n n b a +=,若14n n n c b b +=,求数列{}n c 的前n 项和n T . 19.某车间20名工人年龄数据如下表:(1)求这20名工人年龄的众数与平均数;(2)以十位数为茎,个位数为叶,作出这20名工人年龄的茎叶图; (3)从年龄在24和26的工人中随机抽取2人,求这2人均是24岁的概率.20.如图,在四棱锥PABCD -中,底面ABCD 是边长为2的正方形,E ,F 分别为PC ,BD 的中点,平面PAD ⊥平面ABCD ,且PA PD AD ==.(1)求证:EF ∥平面PAD ; (2)求三棱锥C PBD -的体积.21.已知椭圆()222210x y a b a b +=>>,左、右焦点分别为12 F F ,,左顶点为A ,11AF =.(1)求椭圆的方程;(2)若直线l 经过2F 与椭圆交于 M N ,两点,求11F M F N ⋅的取值范围.22.设函数()()ln f x x b x =+,已知曲线()y f x =在点()()1 1f ,处的切线与直线20x y +=垂直.(1)求b 的值;(2)若函数()()()01x f x g x e a a x ⎛⎫=-≠ ⎪+⎝⎭,且()g x 在区间()0 +∞,上是单调函数,求实数a 的取值范围.吉林省普通中学2016-2017学年度高中毕业班第二次调研测试数学(文科)参考答案与评分标准一、选择题1-5:CDDBA 6-10:BCACB 11、12:AB二、填空题13.1-15.②③ 16.27283三、解答题17.解:(1)由图象知1A =,54(),2126T πππω=-==, 将点(,1)6π代入解+析式得sin 13πϕ⎛⎫+= ⎪⎝⎭,因为2πϕ<,所以6πϕ=,所以()sin 26f x x π⎛⎫=+ ⎪⎝⎭.(2)由()2cos cos a c B b C -=得:()2sin sin cos sin cos A C B B C -=, 所以()2sin cos sin A B B C =+,2sin cos sin A B A =,因为()0 A π∈,,所以sin 0A ≠,所以1cos 2B =,3B π=,23A C π+=, sin 26A f A π⎛⎫⎛⎫=+ ⎪ ⎪⎝⎭⎝⎭,203A π<<,5666A πππ<+<,所以1sin 162A π⎛⎫⎛⎤+∈ ⎪ ⎥⎝⎭⎝⎦,,所以1 122A f ⎛⎫⎛⎤∈ ⎪ ⎥⎝⎭⎝⎦,.18.解:(1)设数列{}n a 的公比为q ,1q ≠,()21313191a q a q q ⎧=⎪-⎨=⎪-⎩,所以()2121319a q a q q ⎧=⎪⎨++=⎪⎩,解得112a =,12q =-. 11122n n a -⎛⎫=⨯- ⎪⎝⎭(2)222231112322n nn a ++⎛⎫⎛⎫=⨯-=⨯ ⎪⎪⎝⎭⎝⎭,222233log log 22n n n b n a +===,()1411111n n n c b b n n n n +===-++, 12311111111223111n n c c c c n n n n ⎛⎫⎛⎫⎛⎫++++=-+-++-=-= ⎪ ⎪ ⎪+++⎝⎭⎝⎭⎝⎭……. 19.解:(1)由题意可知,这20名工人年龄的众数是30,这20名工人年龄的平均数为: ()119328329530431332403020x =+⨯+⨯+⨯+⨯+⨯+=. (2)这20名工人年龄的茎叶图如图所示:(3)记年龄为24岁的三个人为123 A A A ,,;年龄为26岁的三个人为123 B B B ,,,则从这6人中随机抽取2人的所有可能为:{}{}{}{}{}1213231112 A A A A A A A B A B ,,,,,,,,,, {}{}{}{}{}1321222331 A B A B A B A B A B ,,,,,,,,,, {}{}{}{}{}3233121323 A B A B B B B B B B ,,,,,,,,,共15种. 满足题意的有{}{}{}121323 A A A A A A ,,,,,3种, 故所求的概率为31155P ==. 20.(1)证明:连接AC ,则F 是AC 的中点,E 为PC 的中点, 故在CPA △中,EF PA ∥,且PA ⊂平面PAD ,EF ⊄平面PAD , ∴EF ∥平面PAD .(2)取AD 的中点M ,连接PM , ∵PA PD =, ∴PM AD ⊥,又平面PAD ⊥平面ABCD ,平面PAD 平面ABCD AD =, ∴PM ⊥平面ABCD ,∴1111222233223C PBD P BCD BCD V V S PM --==⋅=⨯⨯⨯⨯⨯=△.21.解:(1)设()1 0F c -,,()2 0F c ,,∴1c a a c ⎧=⎪⎨⎪-=⎩,1a c ⎧=⎪⎨=⎪⎩. ∴2221b a c =-=,∴2212x y +=.(2)当直线l 斜率存在时,设()11 M x y ,,()22 N x y ,,直线l 为:()1y k x =-,代入2212x y +=, 得:()222112x k x +-=,整理得:()2222124220k x k x k +-+-=,由题意0∆>.所以2122421k x x k +=+,21222221k x x k -=+,所以()()()()21211221212121 1 111F M F N x y x y x x x x k x x ⋅=+⋅+=++++--,, ()()()()()222222221212222241111112121k k k x x k x x k k k k k k -=++-+++=++-++++()22222799217172222121221k k k k k +--===-+++, 因为2121k +≥,所以1271 2F M F N ⎡⎫⋅∈-⎪⎢⎣⎭ ,.当直线l 斜率不存在时:22112x x y =⎧⎪⎨+=⎪⎩,y =,∴ 1 M ⎛ ⎝, 1 N ⎛- ⎝,,所以1272 2 2F M F N ⎛⎛⋅=⋅= ⎝⎝ ,, 综上:1271 2F M F N ⎡⎤⋅∈-⎢⎥⎣⎦ ,.22.解:(1)曲线()y f x =在点()()1 1f ,处的切线斜率为2,所以()'12f =, 又()'ln 1bf x x x=++,即ln112b ++=,所以1b =. (2)由(1)知,()()ln 1x x x f x g x e a e x ae x ⎛⎫=-=-⎪+⎝⎭, 所以()()1'ln 0x g x a x e x x ⎛⎫=-+> ⎪⎝⎭,若()g x 在()0 +∞,上为单调递减函数,则()'0g x ≤在()0 +∞,上恒成立,即1ln 0a x x -+≤,所以1ln a x x≥+, 令()()1ln 0h x x x x =+>,则()22111'x h x x x x-=-+=, 由()'0h x >,得1x >,()'0h x <,得01x <<,故函数()h x 在(]0 1,上是减函数,在[)1 +∞,上是增函数, 则1ln x x+∞→,()h x 无最大值,()'0g x ≤在()0 +∞,上不恒成立, 故()g x 在()0 +∞,不可能是单调减函数, 若()g x 在()0 +∞,上为单调递增函数,则()'0g x ≥在()0 +∞,上恒成立, 即1ln 0a x x -+≥,所以1ln a x x ≤+,由前面推理知,()1ln h x x x=+的最小值为1, ∴1a ≤,故a 的取值范围是(] 1-∞,.吉林市普通中学2016—2017学年度高中毕业班第二次调研测试数 学(文科)参考答案与评分标准一、选择题:本大题共12题,每小题5分,共60分,在每小题给出的四个选项中,只有一个是符合题目要求。

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