高二上学期月考试卷月考卷
湖北省十堰市六校教学体2024-2025学年高二上学期9月月考物理试卷

湖北省十堰市六校教学体2024-2025学年高二上学期9月月考物理试卷一、单选题1.两个质量不同的物体,如果它们的()A.动能相等,则质量大的动量小B.动能相等,则动量大小也相等C.动量大小相等,则动能也相等D.动量大小相等,则质量大的动能小2.某电路如图所示,已知电池组的总内阻r=1 Ω,外电路电阻R=5 Ω,理想电压表的示数U=3.0 V.则电池组的电动势E等于()A.3.0 V B.3.6 V C.4.0 V D.4.2 V3.将一电源与一电阻箱连接成闭合回路,测得电阻箱所消耗功率P与电阻箱读数R变化的曲线如图所示,由此可知()A.电源最大输出功率可能大于45WB.电源内阻一定等于5ΩC.电源电动势为45VD.电阻箱所消耗功率P最大时,电源效率大于50%4.一质量为2kg的物块在合外力F的作用下从静止开始沿直线运动。
F随时间t变化的图线如图所示,则()A .t =1s 时物块的速率为1m/sB .t =2s 时物块的动量大小为2kg·m/sC .t =3s 时物块的动量大小为5kg·m/sD .t =4s 时物块的速度为零5.如图所示电路中,电源电动势为E ,内阻为r ,电流表A 、电压表1V 、2V 、3V 均为理想电表,1R 为定值电阻,2R 为滑动变阻器。
闭合开关S ,当2R 的滑动触头P 由上端向下滑动的过程中( )A .电压表1V 、2V 的示数增大,电压表3V 的示数不变;B .电流表A 示数变大,电压表3V 的示数变小;C .电压表2V 示数与电流表A 示数的比值不变;D .电压表3V 示数变化量与电流表A 示数变化量比值的绝对值不变。
6.某同学用一个微安表(量程1 mA ,内阻900 Ω)、电阻箱R 1和电阻箱R 2组装一个多用电表,有电流10 mA 和电压3 V 两挡,改装电路如图所示,则R 1、R 2应调到多大阻值( )A .R 1=100 Ω,R 2=3000 ΩB .R 1=10 Ω,R 2=210 ΩC .R 1=100 Ω,R 2=210 ΩD .R 1=10 Ω,R 2=3000 Ω7.关于电源的电动势,下面说法正确的是( )A .电动势是表征电源把其它形式的能转化为电能本领的物理量B .电动势在数值上等于电路中通过2C 电量时电源提供的能量C .电源的电动势跟电源的体积有关,跟外电路有关D .电动势有方向,因此电动势是矢量二、多选题8.电饭锅工作时有两种状态:一种是锅内水烧干前的加热状态,另一种是锅内水烧干后保温状态,如图所示是电饭锅电路原理示意图,S 是用感温材料制造的开关。
山西省部分学校2024-2025学年高二上学期10月月考数学试题(含答案)

2024~2025学年高二10月质量检测卷数学(A 卷)考生注意:1.本试卷分选择题和非选择题两部分。
满分150分,考试时间120分钟。
2.答题前,考生务必用直径0.5毫米黑色墨水签字笔将密封线内项目填写清楚。
3.考生作答时,请将答案答在答题卡上。
选择题每小题选出答案后,用2B 铅笔把答题卡上对应题目的答案标号涂黑;非选择题请用直径0.5毫米黑色墨水签字笔在答题卡上各题的答题区域内作答,超出答题区域书写的答案无效,在试题卷、草稿纸上作答无效。
4.本卷命题范围:人教A 版选择性必修第一册第一章~第二章。
一、选择题:本题共8小题,每小题5分,共40分。
在每小题给出的四个选项中,只有一项是符合题目要求的。
1.已知直线经过,两点,则的倾斜角为()A.B.C.D.2.已知圆的方程是,则圆心的坐标是( )A. B. C. D.3.在长方体中,为棱的中点.若,,,则()A. B. C. D.4.两平行直线,之间的距离为( )B.3D.5.曲线轴围成区域的面积为( )l (A (B l 6π3π23π56πC 2242110x y x y ++--=C ()2,1-()2,1-()4,2-()4,2-1111ABCD A B C D -M 1CC AB a = AD b =1AA c = AM =111222a b c -+ 111222a b c ++12a b c-+12a b c++ 1:20l x y --=2:240l x y -+=y =xA. B. C. D.6.已知平面的一个法向量,是平面内一点,是平面外一点,则点到平面的距离是( )A. B.D.37.在平面直角坐标系中,圆的方程为,若直线上存在点,使以点为圆心,1为半径的圆与圆有公共点,则实数的取值范围是( )A. B.C. D.8.在正三棱柱中,,,为棱上的动点,为线段上的动点,且,则线段长度的最小值为( )A.2二、选择题:本题共3小题,每小题6分,共18分。
河南省平顶山市叶县高级中学2024-2025学年高二上学期9月月考英语试卷

河南省平顶山市叶县高级中学2024-2025学年高二上学期9月月考英语试卷一、听力选择题1.What did the woman buy for her mum?A.A hat.B.A coat.C.A T- shirt.2.What does the man like doing?A.Travelling alone.B.Joining a guided tour.C.Backpacking with friends. 3.Why is the woman broke at the end of the month?A.She likes shopping.B.She doesn't work hard.C.She earns little money. 4.What time will the man’s party probably start?A.At 7: 30 p.m.B.At 8: 00 p.m.C.At 11: 00 p.m.5.Where are the speakers probably?A.In a hospital.B.In the police office.C.On the street.听下面一段较长对话,回答以下小题。
6.What should the woman do to order checks?A.Wait in a line.B.Fill in a form.C.Check the mail.7.When will the woman probably get the check?A.In two days.B.In four days.C.In a week.听下面一段较长对话,回答以下小题。
8.What is the man’s attitude towards art class?A.Favourable.B.Unconcerned.C.Worried.9.What does the woman mean about talent?A.She wants to be a painter too.B.She knows how to draw and paint.C.She hopes she could have some kind of talent.10.What are the speakers mainly talking about?A.The man’s hobby.B.The talent of the woman.C.The woman’s favourite class.听下面一段较长对话,回答以下小题。
2023-2024学年高二上学期10月月考语文试卷(含解析)

2023-2024学年高二上学期10月月考语文试卷(含解析)一、现代文阅读(一)现代文阅读Ⅰ(14分)1.(14分)阅读下面的文字,完成下列各题。
材料一:大家都知中国人喜讲“中庸之道”,一般人以为中庸之道是指平易近人,不标新立异,不惊世骇俗,调和折衷,不走极端而言。
然此等乃通俗义,非正确义,《中庸》上说:“执其两端,用其中于民。
”无论何事都有两端,此两端,可以推到极处各成为一极端。
在此两极端间之中间都叫做中,此一“中”可以有甚长之距离。
所谓“中”,非折中之谓,乃指此两极端之全过程。
如言真善美,是此一极端,不真不善不美是那一极端。
但此真、善、美三分,只是西方人说法。
照中国人讲,此世界便是一真,不是伪,真伪不能对立。
若论美丑,此世界是浑沌中立,既非极美,也非极丑。
中国文化是人本位的,以人文主义为中心,看重了人的一面,则善、恶对立不能不辨。
但纵是一大圣人,亦不能说他已达到了百分之百的善。
纵是一大恶人,亦不能说他是百分之百的恶。
人只在善、恶两极端之“中道”上,既不在此极端,亦不在彼极端。
但必指出此两极端,始能显出此中道,始能在此中道上理论有根据、行为有目标,故说“执其两端,用其中于民”。
若非执其两端,则中道无可见。
真实可用者乃此中道,非其两端。
此乃中国人所讲“中庸之道”之正确意义。
在此等观念与意识之下,我认为中国文化尽可以借鉴西方文化,使中国文化更充实更光辉。
并不如一般人想法,保守了中国固有之旧,即不能吸收西方现代之新。
似乎大家总爱把一切事物作相反对立看,不肯把此等相反对立来作互通合一看。
所以我们中国所讲“执两用中的中庸之道”,此刻实该大大地再阐明。
这和我们当前所该采用的一切想法和做法实有很大的关系。
中国人惟其有此中庸之道,亦可使各不同的宗教信仰也一样调和融通起来。
佛教、回教、耶稣教来到中国,不仅和中国传统文化无冲突,在此诸大宗教之相互间都能和平共存,没有大冲突。
你看中国社会上,这里一个天主堂,那里一个和尚庙:母亲信佛教,儿子信耶稣。
湖南省长沙市雅礼中学2024-2025学年高二上学期10月月考语文试题(含答案)

雅礼集团2024下学期第一次月考试卷高二语文时量:150分钟分值:150分一、现代文阅读(35分)(一)现代文阅读1(本题共5小题,19分)阅读下列文字,完成下面小题。
我们不可像霍布斯那样,因为人没有任何善的观念,便认为人天生是恶人;因为人不知道什么是美德,便认为人是邪恶的;人从来不对他的同类效劳,因为他认为他对他们没有任何义务;人自认为他有取得自己所需之物的权利,因此便以为他自己是整个宇宙的唯一的主人。
诚然,霍布斯看出了现今的人们对自然的权利所作的种种解释的缺点,然而从他自己所作的解释中得出的结论就可看出,他的解释的着眼点也是错误的。
既然这位作者是根据他自己提出的原则进行推理的,他的论点就应该这样来表述:我们在自然状态中对保护我们自己的生存的关心,是丝毫不妨碍他人对保护他自己的生存的关心的,因此这个状态是有利于和平的,是适合于人类的。
然而他在书中所说的话却恰恰相反,因为他把为了满足许许多多欲望而产生的需要,与野蛮人为了保护自己的生存而产生的需要混为一谈了;其实,这些欲望乃是社会造成的,而且,正因为人的欲望丛生,才使法律成为必需的东西。
既然霍布斯认为恶人是一个强壮的孩子,那我们就要问:野蛮人是否也是一个强壮的孩子?如果我们承认他是一个强壮的孩子,那该得出什么样的结论呢?如果这个人强壮的时候也像他柔弱的时候那样依赖他人,那么,什么过分的事他干不出来呢?他的母亲如果不及时喂他奶,他就会打她;如果他觉得他的弟弟招他讨厌,他就会掐死他:如果别人碰撞了他或打扰了他,他就会咬别人的腿。
说自然状态中的人是强壮的,与说自然状态中的人需要依赖于人,这两种说法是矛盾的。
人只有在处于依赖状态的时候才是柔弱的:如果他无拘无束,不依赖他人的话,他早就是很强壮的了。
霍布斯没有看出:我们的法学家所说的阻碍野蛮人使用理智的原因,正好就是霍布斯本人所说的阻碍野蛮人滥用他们的官能的原因。
因此,我们认为野蛮人之所以不是恶人,其原因恰恰在于他不知道什么是善,因为防止他们作恶的,既不是智慧的发达、也不是法律的约束,而是欲念的平静和对恶事的无知:他们从对恶事的无知中得到的益处,比别人从对美德的认识中得到的益处多得多。
高二上上语文月考试卷

考试时间:120分钟满分:100分一、选择题(每小题2分,共20分)1. 下列词语中,字形、字音、词义完全正确的一项是()A. 沉鱼落雁振聋发聩轻歌曼舞B. 靡靡之音声东击西水落石出C. 惊弓之鸟破釜沉舟胸有成竹D. 气壮山河雕梁画栋美轮美奂2. 下列句子中,没有语病的一项是()A. 为了提高学生的综合素质,学校决定增设多个选修课程。
B. 这本书的内容丰富,文字优美,深受广大读者的喜爱。
C. 在这次比赛中,他的表现让人刮目相看,赢得了评委的一致好评。
D. 通过这次实践活动,我对环保有了更深刻的认识,也明白了保护环境的重要性。
3. 下列各句中,句式变换最恰当的一项是()原句:我国经济发展迅速,人民生活水平不断提高。
A. 我国经济发展迅速,人民生活水平不断提高,这使得我国在国际上的地位日益提高。
B. 由于我国经济发展迅速,所以人民生活水平不断提高,从而使得我国在国际上的地位日益提高。
C. 我国经济发展迅速,人民生活水平不断提高,使得我国在国际上的地位日益提高。
D. 我国经济发展迅速,人民生活水平不断提高,因此我国在国际上的地位日益提高。
4. 下列各句中,标点符号使用正确的一项是()A. “这个句子中的逗号使用得非常恰当。
”他说。
B. “我明白了,你的意思是……”他点了点头。
C. “你看看这个句子,这里的冒号用得对吗?”老师问。
D. “你的看法很有道理,我赞同你的观点。
”他微笑着说。
5. 下列各句中,修辞手法使用正确的一项是()A. 月亮悄悄地爬上了树梢,像一块银盘挂在天空中。
B. 那朵花在阳光下绽放,仿佛一位美丽的少女在翩翩起舞。
C. 这本书的内容非常丰富,像一座知识的宝库。
D. 夜晚的星空,像一张五彩斑斓的画卷。
6. 下列各句中,下列词语使用正确的一项是()A. 他的成绩一直名列前茅,是班级的佼佼者。
B. 这位老师教学方法独特,深受学生喜爱。
C. 那个城市的风景非常美丽,令人流连忘返。
D. 这本书的内容非常有趣,让人百读不厌。
湖北云学名校联盟2024-2025学年高二上学期10月月考数学试题(解析版)

2024年湖北云学名校联盟高二年级10月联考数学试卷一、选择题:本题共8小题,每小题5分,共40分.在每小题给出的四个选项中,只有一项考试时间:2024年10月15日15:00-17:00 时长:120分钟满分:150分是符合题目要求的.1. 已知i 为虚数单位,20253i 1i ++的虚部为( )A. i −B. iC. 1−D. 1【答案】C 【解析】【分析】根据复数乘方、乘法、除法运算法则结合复数的概念运算即可得出结果.【详解】根据复数的乘方可知()50620254i i i i =⋅=,则()()()()20253i 1i 3i 3i32i 12i 1i 1i1i 1i 2+−++−+====−+++−,其虚部为1−. 故选:C2. 已知一组数据:2,5,7,x ,10的平均数为6,则该组数据的第60百分位数为( ) A. 7 B. 6.5C. 6D. 5.5【答案】B 【解析】【分析】先根据平均数求x 的值,然后将数据从小到大排列,根据百分位数的概念求值. 【详解】因为2571065x ++++=⇒6x =.所以数据为:2,5,6,7,10.又因为560%3×=,所以这组数据的第60百分位数为:676.52+=. 故选:B3. 直线1l :20250ax y −+=,2l :()3220a x ay a −+−=,若12l l ⊥,则实数a 的值为( ) A 0 B. 1C. 0或1D.13或1 【答案】C.【分析】根据两直线垂直的公式12120A A B B +=求解即可. 【详解】因为1l :20250ax y −+=,2l :()3220a x ay a −+−=垂直, 所以()()3210a a a −+−=, 解得0a =或1a =,将0a =,1a =代入方程,均满足题意, 所以当0a =或1a =时,12l l ⊥. 故选:C .4. 为了测量河对岸一古树高度AB 的问题(如图),某同学选取与树底B 在同一水平面内的两个观测点C 与D ,测得15BCD ∠=°,30BDC ∠=°,48m CD =,并在点C 处测得树顶A 的仰角为60°,则树高AB 约为( )1.4≈1.7≈)A. 100.8mB. 33.6mC. 81.6mD. 57.12m【答案】D 【解析】【分析】先在BCD △中,利用正弦定理求出BC ,再在Rt ABC △中求AB 即可.【详解】在BCD △中,15BCD ∠=°,30BDC ∠=°,所以135CBD ∠=°,又48CD =,由正弦定理得:sin sin CD CBCBD CDB=∠∠⇒12CB=⇒CB =在Rt ABC △中,tan 60AB BC =°=24 1.4 1.7≈××57.12=. 故选:D5. 如果直线ax +by =4与圆x 2+y 2=4有两个不同的交点,那么点P (a ,b )与圆的位置关系是( ) A. P 在圆外 B. P 在圆上D. P 与圆的位置关系不确定 【答案】A 【解析】224a b ∴+,所以点(),a b 在圆外考点:1.直线与圆的位置关系;2.点与圆的位置关系6. 在棱长为6的正四面体ABCD 中,点P 与Q 满足23AP AB = ,且2CD CQ =,则PQ 的值为( )A.B.C.D.【答案】D 【解析】【分析】以{},,AB AC AD 为基底,表示出PQ,利用空间向量的数量积求模.【详解】如图:以{},,AB AC AD 为基底,则6AB AC AD ===,60BAC BAD CAD ∠=∠=∠=°,所以66cos 6018AB AC AB AD AC AD ⋅=⋅=⋅=××°=.因为()1223PQ AQ AP AC AD AB =−=+− 211322AB AC AD =−++. 所以22211322PQ AB AC AD =−++222411221944332AB AC AD AB AC AB AD AC AD =++−⋅−⋅+⋅ 169912129=++−−+19=.所以PQ =.故选:D7. 下列命题中正确的是( )A. 221240z z +=,则120z z ==; B. 若点P 、Q 、R 、S 共面,点P 、Q 、R 、T 共面,则点P 、Q 、R 、S 、T 共面;C. 若()()1P A P B +=,则事件A 与事件B 是对立事件; D. 从长度为1,3,5,7,9的5条线段中任取3条,则这三条线段能构成一个三角形的概率为310; 【答案】D 【解析】【分析】举反例说明ABC 不成立,根据古典概型的算法判断D 是正确的.【详解】对A :若1i z =,22z =,则221240z z +=,但120z z ==不成立,故A 错误; 对B :如图:四面体S PRT −中,Q 是棱PR 上一点,则点P 、Q 、R 、S 共面,点P 、Q 、R 、T 共面,但点P 、Q 、R 、S 、T 不共面,故B 错误; 对C :掷1枚骰子,即事件A :点数为奇数,事件B :点数不大于3, 则()12P A =,()12P B =,()()1P A P B +=,但事件A 、B 不互斥,也不对立,故C 错误; 对D :从长度为1,3,5,7,9的5条线段中任取3条,有35C 10=种选法, 这三条线段能构成一个三角形的的选法有:{}3,5,7,{}3,7,9,{}5,7,9共3种, 所以条线段能构成一个三角形的的概率为:310P =,故D 正确. 故选:D8. 动点Q 在棱长为3的正方体1111ABCD A B C D −侧面11BCC B 上,满足2QA QB =,则点Q 的轨迹长度为( )A. 2πB.4π3C.D.【解析】【分析】结合图形,计算出||BQ =,由点Q ∈平面11BCC B ,得出点Q 的轨迹为圆弧 EQF,利用弧长公式计算即得.【详解】如图,易得AB ⊥平面11BCC B ,因BQ ⊂平面11BCC B ,则AB BQ ⊥,不妨设||BQ r =,则||2AQ r =, ||3AB ==,解得r =又点Q ∈平面11BCC B ,故点Q 的轨迹为以点B EQF,故其长度为π2. 故选:D.二、选择题:本题共36分,共18分.在每小题给出的选项中,有多项符合题目要求.全部选对的得6分,部分选对的得部分分,有选错的得0分.9. 在平面直角坐标系中,下列说法正确的是( ) A. 若两条直线垂直,则这两条直线的斜率的乘积为1−;B. 已知()2,4A ,()1,1B ,若直线l :20kx y k ++−=与线段AB 有公共点,则21,32k∈−; C. 过点()1,2,且在两坐标轴上截距互为相反数的直线l 的方程为10x y −+=;D. 若圆()2214x y −+=上恰有3个点到直线y x b =+的距离等于1,则1b =−±. 【答案】BD 【解析】【分析】根据直线是否存在斜率判断A 的真假;数形结合求k 的取值范围判断B 的真假;根据截距的概念判断真假;转化为点(圆心)到直线的距离求b 判断D 的真假.【详解】对A :“若两条直线垂直,则这两条直线的斜率的乘积为1−”成立的前提是两条直线的斜率都存若两条直线1条不存在斜率,另一条斜率为0,它们也垂直.故A 是错误的. 对B :如图:对直线l :20kx y k ++−=⇒()21y k x −=−+,表示过点()1,2P −,且斜率为k −的直线, 且()422213APk −==−−,()121112BP k −==−−−, 由直线l 与线段AB 有公共点,所以:203k ≤−≤或102k −≤−<,即203k −≤≤或102k <≤,进而得:2132k −≤≤.故B 正确; 对C :过点()1,2,且在两坐标轴上截距互为相反数的直线l 的方程为10x y −+=或2y x =,故C 错误; 对D :“圆()2214x y −+=上恰有3个点到直线y x b =+的距离等于1”可转化为“圆心(1,0)到直线y x b =+的距离等于1”.1⇒1b =−±.故D 正确.故选:BD10. 如图所示四面体OABC 中,4OB OC ==,3OA =,OB OC ⊥,且60AOB AOC ∠=∠=°,23CD CB =,G 为AD 的中点,点H 是线段OA 上动点,则下列说法正确的是( )A. ()13OG OA OB OC =++ ;B. 当H 是靠近A 的三等分点时,DH ,OC ,AB共面;C. 当56OH OA = 时,GH OA ⊥ ;D. DH OH ⋅的最小值为1−.【答案】BCD 【解析】【分析】以{},,OA OB OC为基底,表示出相关向量,可直接判断A 的真假,借助空间向量共面的判定方法可判断B 的真假,利用空间向量数量积的有关运算可判断CD 的真假.【详解】以{},,OA OB OC 为基底,则3OA = ,4OB OC == ,6OA OB OA OC ⋅=⋅= ,0OB OC ⋅=.对A :因为23AD AC CD AC CB =+=+ ()23AC AB AC =+−2133AB AC +()()2133OB OA OC OA =−+−2133OA OB OC =−++ . 所以12OG OA AG OA AD =+=+ 121233OA OA OB OC =+−++111236OA OB OC =++ ,故A 错误;对B :当H 是靠近A 的三等分点,即23OH OA =时,DH AH AD =− 121333OA OA OB OC =−−−++221333OA OB OC =−− ,又AB OB OA =−,所以13DH AB OC − .故DH ,AB ,OC 共面.故B 正确;对C :因为HG OG OH OA AG OH =−=+− 1526OA AD OA =+−12152336OA OA OB OC OA =+−++− 111336OA OB OC =−++,所以:HG OA ⋅= 111336OA OB OC OA −++⋅ 2111336OA OB OA OC OA =−+⋅+⋅1119660336=−×+×+×=,所以HG OA ⊥ ,故GH OA ⊥,故C 正确;对D :设OH OA λ=,()01λ≤≤.因为:DH OH OD =−()OA OA AD λ=−+ 2133OA OA OA OB OC λ =−−++2133OA OB OC λ=−− .所以DH OH ⋅ 2133OA OB OC OAλλ =−−⋅()2233OA OA OB OA OCλλλ−⋅−⋅296λλ−,()01λ≤≤.当13λ=时,DH OH ⋅ 有最小值,为:1196193×−×=−,故D 正确. 故选:BCD11. 已知()2,3P 是圆C :22810410x y x y a +−−−+=内一点,其中0a >,经过点P 的动直线l 与C 交于A ,B 两点,若|AAAA |的最小值为4,则( ) A. 12a =;B. 若|AAAA |=4,则直线l 的倾斜角为120°;C. 存在直线l 使得CA CB ⊥;D. 记PAC 与PBC △的面积分别为PAC S ,PBC S ,则PAC PBC S S ⋅△△的最大值为8. 【答案】ACD 【解析】【分析】根据点()2,3P 在圆内,列不等式,可求a 的取值范围,在根据弦|AAAA |的最小值为4求a 的值,判断A 的真假;明确圆的圆心和半径,根据1l CP k k ⋅=−,可求直线AB 的斜率,进而求直线AB 的倾斜角,判断B 的真假;利用圆心到直线的距离,确定弦长的取值范围,可判断C 的真假;由三角形面积公式和相交弦定理,可求PAC PBC S S ⋅△△的最大值,判断D 的真假. 【详解】对A :由222382103410a +−×−×−+<⇒8a >. 此时圆C :()()2245x y a −+−=.因为过P 点的弦|AAAA |的最小值为4,所以CP=又CP =⇒12a =.故A 正确;对B :因为53142CP k −==−,1l CP k k ⋅=−,所以直线l 的斜率为1−,其倾斜角为135°,故B 错误; 对C :当|AAAA |=4时,如图:sin ACP ∠==,cos ACP ∠==41cos 1033ACB ∠=−=>, 所以ACB ∠为锐角,又随着直线AB 斜率的变化,ACB ∠最大可以为平角, 所以存在直线l 使得CA CB ⊥.故C 正确; 对D :如图:直线CP 与圆C 交于M 、N 两点,链接AM ,BN ,因为MAP BNP ∠=∠,APM NPB ∠=∠,所以APM NPB .所以AP MP NPBP=⇒(4AP BP MP NP ⋅=⋅=−+=.又1sin 2PACS PA PC APC APC =⋅⋅∠=∠ ,PBCS BPC =∠ ,且sin sin APC BPC ∠=∠.所以22sin PAC PBC S S PA PB APC⋅=⋅⋅∠ 28sin APC ∠8≤,当且仅当sin 1APC ∠=,即AB CP ⊥时取“=”.故D 正确. 故选:ACD【点睛】方法点睛:在求PAC PBC S S ⋅△△的最大值时,应该先结合三角形相似(或者蝴蝶定理)求出AP BP ⋅为定值,再结合三角形的面积公式求PAC PBC S S ⋅△△的最大值. 三、填空题:本题共3小题,每小题5分,共15分.12. 实数x 、y 满足224x y +=,则()()2243x y −++的最大值是______. 【答案】49 【解析】【分析】根据()()2243x y −++几何意义为圆上的点(),x y 与()4,3−距离的平方,找出圆上的与()4,3−的最大值,再平方即可求解.【详解】解:由题意知:设(),p x y ,()4,3A −,则(),p x y 为圆224x y +=上的点, 圆224x y +=的圆心OO (0,0),半径2r =, 则()()2243x y −++表示圆上的点(),p x y 与()4,3A −距离的平方,又因为max 27PA AO r=+=+=, 所以22max749PA==; 故()()2243x y −++的最大值是49. 故答案为:49.13. 记ABC 的三个内角A ,B ,C 的对边分别为a ,b ,c ,已知()cos2cos a B c b A =−,其中π2B ≠,若ABC 的面积S =,2BE EC = ,且AE = ,则BC 的长为______.【解析】【分析】利用正弦定理对()cos 2cos a B c b A =−化简,可得π3A =,再由三角形面积公式求出8bc =,根据题意写出1233AE AB AC =+,等式两边平方后,可求出,b c 的值,由余弦定理2222cos a b c bc A =+−,求出BC 的长.【详解】()cos 2cos a B c b A =−,由正弦定理可得:sin cos 2sin cos sin cos A B C A B A =−,sin cos cos sin 2sin cos A B A B C A +=, ()sin 2sin cos A B C A +=,()sin πC 2sin cos C A −=,sin 2sin cos (sin 0)C C A C >,即1cos 2A =,π3A =,1sin 2ABC S bc A == ,得8bc =, ∵2BE EC = ,∴1233AE AB AC =+ ,221233AE AB AC =+, 即2228144cos 3999c b bc A =++,由8bc =,解得42b c = = 或18b c = = , 根据余弦定理2222cos a b c bc A =+−,当42b c = =时,a =,此时π2B =,不满足题意, 当18b c = =时,a =..14. 如图,已知四面体ABCD 的体积为9,E ,F 分别为AB ,BC 的中点,G 、H 分别在CD 、AD 上,且G 、H 是靠近D 的三等分点,则多面体EFGHBD 的体积为______.【答案】72##3.5 【解析】 【分析】多面体EFGHBD 的体积为三棱锥G DEH −与四棱锥E BFGD −的体积之和,根据体积之比与底面积之比高之比的关系求解即可.【详解】连接ED ,EG ,因为H 为AAAA 上的靠近D 的三分点,所以13DH AD =, 因为E 为AAAA 的中点,所以点E 到AAAA 的距离为点B 到AAAA 的距离的一半, 所以16DEH BAD S S = , 又G 为CCAA 上靠近D 的三分点,所以点G 到平面ABD 的距离为点C 到平面ABD 的距离的13, 所以111119663182G DEH G BAD C BAD V V V −−−==×=×=, 1233BCD FCG BCD BCD BCD BFGD S S S S S S =−=−= 四边形, 所以2211933323E BFGD E BCD A BCD V V V −−−==×=×=, 所以多面体EFGHBD 的体积为17322G DEH E BFGD V V −−+=+=. 故答案为:72. 【点睛】关键点点睛:将多面体转化为两个锥体的体积之和,通过体积之比与底面积之比高之比的关系求解.四、解答题:本题共5小题,共77分.解答应写出文字说明、证明过程或演算步骤. 15. 在对某高中1500名高二年级学生的百米成绩的调查中,采用按学生性别比例分配的分层随机抽样抽取100人,已知这1500名高二年级学生中男生有900人,且抽取的样本中男生成绩的平均数和方差分别为13.2秒和13.36,女生成绩的平均数和方差分别为15.2秒和17.56.(1)求抽取的总样本的平均数;(2)试估计高二年级全体学生的百米成绩的方差.【答案】(1)14 (2)16【解析】【分析】(1)先确定样本中男生、女生的人数,再求总样本的平均数.(2)根据方差的概念,计算总样本的方差.【小问1详解】 样本中男生的人数为:100900601500×=;女生的人数为:1006040−=. 所以总样本的平均数为:6013.24015.214100x ×+×=. 【小问2详解】记总样本的方差为2s , 则()(){}22216013.3613.2144017.5615.214100s =×+−+×+− 16=. 所以,估计高二年级全体学生的百米成绩的方差为16.16. 在平面直角坐标系xOy 中,ABC 的顶点A 的坐标为()4,2−,ACB ∠的角平分线所在的直线方程为10x y −+=,AC 边上中线BM 所在的直线方程为220x y +−=. (1)求点C 的坐标;(2)求直线BC 的方程.【答案】(1)(3,4)C ;(2)72130x y −−=【解析】【分析】(1)设(,1)C m m +,则43(,)22m m M −+,代入220x y +−=,求解即可; (2)设直线BC 的方程为:340x ny n +−−=,在直线10x y −+=取点(0,1)P ,利用点P 到直线AC 的距离等于点P 到直线BC 的距离,求解即可.【小问1详解】解:由题意可知点C 在直线0x y −+=上, 所以设(,1)C m m +,所以AC 中点43(,)22m m M −+, 又因为点43(,)22m m M −+在直线220x y +−=上, 所以34202m m +−+−=,解得3m =, 所以(3,4)C ;【小问2详解】解:因为(3,4)C ,设直线BC 的方程为:340x ny n +−−=, 又因为(4,2)A −,所以直线AC 的方程为:27220x y −+=, .又因为ACB ∠的角平分线所在的直线方程为10x y −+=, 在直线10x y −+=取点(0,1)P ,则点P 到直线AC 的距离等于点P 到直线BC 的距离,=,整理得21453140n n ++=, 解得:72n =−或27n =−, 当72n =−时,所求方程即为直线AC 的方程, 所以27n =−, 所以直线BC 的方程为: 72130x y −−=. 17. 直三棱柱111ABC A B C −中,12AB AC AA ===,其中,,E F D 分别为棱111,,BC B A B C 的中点,已知11AF A C ⊥,(1)求证:AF DE ⊥;(2)设平面EFD 与平面ABC 的交线为直线m ,求直线AC 与直线m 所成角的余弦值.【答案】(1)证明见解析(2【解析】【分析】(1)取AB 的中点G ,连接1,EG A G 证得四边形ADEG 为平行四边形,得到1//DE A G ,利用1A AG ABF ≌,证得90AHG ∠= ,得到1AF A G ⊥,即可证得AF DE ⊥;(2)根据题意,证得11A C ⊥平面11ABB A ,得到1111A C A B ⊥,以A 为原点,建立空间直角坐标系,求得(0,2,0)AC = ,再取AC 的中点M ,延长,MB DF 交于点N ,得到直线AC 与直线m 所成角,即为直线AC 与直线EN 所成角,求得(4,1,0)N −,得到(3,2,0)EN =− ,结合向量的夹角公式,即可求解.【小问1详解】证明:取AB 的中点G ,连接1,EG A G ,因为E 的中点,可得//EG AC ,且12EG AC =, 又因为1//A D AC ,且112A D AC =,所以1//EG A D ,且1EG A D =, 所以四边形ADEG 平行四边形,所以1//DE A G ,在正方形11ABB A 中,可得1A AG ABF ≌,所以1A GA AFB ∠=∠, 因为90AFB AFB ∠+∠= ,所以190AFB A GA ∠+∠= ,AGH 中,可得90AHG ∠= ,所以1AF A G ⊥,又因为1//DE A G ,所以AF DE ⊥.【小问2详解】解:在直三棱柱111ABC A B C −中,可得1AA ⊥平面111A B C ,因为11AC ⊂平面111AB C ,所以111AA A C ⊥, 又因为11AF A C ⊥,且1AA AF A ∩=,1,AA AF ⊂平面11ABB A ,所以11A C ⊥平面11ABB A , 因为11A B ⊂平面11ABB A ,所以1111A C A B ⊥,即直三棱柱111ABC A B C −的底面为等腰直角三角形,以A 为原点,以1,,AB AC AA 所在的直线分别为,,x y z 轴,建立空间直角坐标系,如图所示,因为12AB AC AA ===,可得(0,0,0),(0,2,0)A C ,则(0,2,0)AC =, 为在取AC 的中点M ,连接,MB DM ,可得1//DM CC 且1DM CC =,因为11//BB DD 且11BB DD =,所以//BF DM ,且12BF DM =, 延长,MB DF 交于点N ,可得B 为MN 的中点,连接EN ,可得EN 即为平面DEF 与平面ABC 的交线,所以直线AC 与直线m 所成角,即为直线AC 与直线EN 所成角,又由(0,1,0),(2,0,0),(1,1,0)M B E , 设(,,)N x y z ,可得MB BN =,即(2,1,0)(2,,)x y z −=−, 可得4,1,0x y z ==−=,所以(4,1,0)N −,可得(3,2,0)EN =− ,设直线EN 与直线AC 所成角为θ,可得cos cos ,AC EN AC EN AC EN θ⋅=== 即直线AC 与直线m18. 已知圆C :22430x y y +−+=,过直线l :12y x =上的动点M 作圆C 的切线,切点分别为P ,Q .(1)当π3PMQ ∠=时,求出点M 的坐标; (2)经过M ,P ,C 三点的圆是否过定点?若是,求出所有定点的坐标;(3)求线段PQ 的中点N 的轨迹方程.【答案】(1)(0,0)或84(,)55(2)过定点(0,2)或42(,)55(3)22173042x y x y +−−+= 【解析】【分析】(1)点M 在直线l 上,设(2,)M m m ,由对称性可知30CMP ∠= ,可得2MC =,从而可得点M 坐标.(2)MC 的中点,12m Q m+,因为MP 是圆P 的切线,进而可知经过C ,P ,M 三点的圆是以Q 为圆心,以MC 为半径的圆,进而得到该圆的方程,根据其方程是关于m 的恒等式,进而可求得x 和y ,得到结果;(3)结合(2)将两圆方程相减可得直线PQ 的方程,且得直线PQ 过定点13,42R,由几何性质得MN RN ⊥,即点N 在以MR 为直径的圆上,进而可得结果.【小问1详解】(1)直线l 的方程为20x y −=,点M 在直线l 上,设(2,)M m m , 因为π3PMQ ∠=,由对称性可得:由对称性可知30CMP ∠= ,由题1CP =所以2MC =,所以22(2)(2)4+−=m m , 解之得:40,5==m m 故所求点M 的坐标为(0,0)或84(,)55. 【小问2详解】 设(2,)M m m ,则MC 的中点(,1)2m E m +,因为MP 是圆C 的切线, 所以经过,,C P M 三点的圆是以Q 为圆心,以ME 为半径的圆,故圆E 方程为:2222()(1)(1)22m m x m y m −+−−=+−化简得:222(22)0x y y m x y +−−+−=,此式是关于m 的恒等式,故2220,{220,x y y x y +−=+−=解得02x y = = 或4525x y = = , 所以经过,,C P M 三点的圆必过定点(0,2)或42(,)55.【小问3详解】 由()22222220,430x y mx m y m x y y +−−++= +−+=可得PQ :()22320mx m y m +−+−=,即()22230m x y y +−−+=, 由220,230x y y +−= −=可得PQ 过定点13,42R . 因为N 为圆E 的弦PQ 的中点,所以MN PQ ⊥,即MN RN ⊥,故点N 在以MR 为直径的圆上,点N 的轨迹方程为22173042x y x y +−−+=. 19. 四棱锥P ABCD −中,底面ABCD 为等腰梯形,224AB BC CD ===,侧面PAD 为正三角形;(1)当BD PD ⊥时,线段PB 上是否存在一点Q ,使得直线AQ 与平面ABCD所成角的正弦值为若存在,求出PQ QB 的值;若不存在,请说明理由. (2)当PD 与平面BCD 所成角最大时,求三棱锥P BCD −的外接球的体积.【答案】(1)存在;1.(2【解析】【分析】(1)先证平面PAD ⊥平面ABCD ,可得线面垂直,根据垂直,可建立空间直角坐标系,用空间向量,结合线面角的求法确定点Q 的位置.(2)根据PD 与平面BCD 所成角最大,确定平面PAD ⊥平面ABCD ,利用(1)中的图形,设三棱锥P BCD −的外接球的球心,利用空间两点的距离公式求球心和半径即可.【小问1详解】因为底面ABCD 为等腰梯形,224AB BC CD ===,所以60BAD ∠=°,120BCD ∠=°,30CBD ABD ∠=∠=°,所以90ADB ∠=°. 所以BD AD ⊥,又BD PD ⊥,,AD PD ⊂平面PAD ,且AD PD D = ,所以BD ⊥平面PAD .又BD ⊂平面ABCD ,所以平面PAD ⊥平面ABCD .取AD 中点O ,因为PAD △是等边三角形,所以PO AD ⊥,平面PAD ∩平面ABCD AD =,所以⊥PO 平面ABCD .再取AB 中点E ,连接OE ,则//OE BD ,所以OE AD ⊥.所以可以O 为原点,建立如图空间直角坐标系.则()0,0,0O ,()1,0,0A ,()1,0,0D −,()E ,()1,B −,(P ,()C −.(1,PB =−− .设PQ PB λ= ,可得)()1Q λλ−−所以)()1,1AQ λλ=−−− ,取平面ABCD 的法向量()0,0,1n = .因为AQ 与平面ABCD ,所以AQ nAQ n ⋅⋅ ,解得12λ=或5λ=(舍去). 所以:线段PB 上存在一点Q ,使得直线AQ 与平面ABCD ,此时1PQ QB =. 【小问2详解】当平面PAD ⊥平面ABCD 时, PD 与平面BCD 所成角为PDA ∠.当平面PAD 与平面ABCD 不垂直时,过P 做PH ⊥平面ABCD ,连接HD ,则PDH ∠为PD 与平面BCD 所成角,因为PH PO <,sin PH PDH PD ∠=,sin PO PDA PD∠=,s s n i i n PDA PDH ∠∠<,所以A PDH PD ∠∠<. 故当平面PAD ⊥平面ABCD 时,PD 与平面BCD 所成角最大.此时,设棱锥P BCD −的外接球球心为(),,G x y z ,GP GB GC GD R====,所以(()(()(()2222222222222222121x y z R x y z R x y z R x y z R ++= ++−+= ++−+=+++=,解得20133x y z R = = = = 所以三棱锥P BCD −的外接球的体积为:34π3V R ==. 【点睛】方法点睛:在空间直角坐标系中,求一个几何体的外接球球心,可以利用空间两点的距离公式,根据球心到各顶点的距离相等列方程求解..。
高二上期月考试题

高二语文月考试卷(共150分,把答案写在答题卡上)第Ⅰ卷(选择题,共30分)一、选择题(共12分,每小题3分)1.下列词语中加点的字,注音全都正确的一组是()A.敕.令(chì)两靥.(yè) 丝绦.(tiāo) 监.生(jiān)B.讪讪.. (shàn) 榫.头(sǔn)扪参.(shēn)湍.流(tuān)C.猿猱.(náo)贾.人(jiǎ)宵柝.(chè)洿.池(wū)D.孝悌.(dì)饿莩.(piǎo)跬.步(kuǐ)庠.序(xiáng)2.下列各组词语中,没有错别字的一组是()A.寒喧雕梁画栋膏粱出奇制胜B.缪种敛声屏气瞋视扪心自问C.崔嵬义愤填膺鱼凫春寒料峭D.编纂天崖海角边陲遍体磷伤3、下列各句中加点的成语使用正确的一项是()A.这些大学的一些学生语文水平实在低劣,往往让人贻笑大方....,影响学校的声誉。
B.这种药能治疗心脏病,又没有副作用,我们屡试不爽....,你还有什么可以怀疑的呢?C.刘白羽的《长江三峡》描绘了一幅巧夺天工....的自然图景,瞿塘峡的雄,巫峡的秀,西陵峡的险,充分展现了造化的神奇。
D.超级女声周笔畅唱功很好,专家评价说,她在歌唱事业上会很有前途,一定会成为明日黄花....。
4、下列各句中,没有..语病的一句是( )A.中午还是阳光灿烂,但到下午5时左右,老天突然变脸,市区狂风大作,天昏地暗。
据气象部门监测,这次特强沙尘暴瞬间风力达11级,地面能见度0米。
B.记者来到卧龙镇人民政府南侧的中国卧龙大熊猫博物馆前,只见这座被称为“中国唯一大熊猫博物馆”坐落在风景秀丽的山下,周围流水淙淙,绿树成荫。
C.世博园开园以来,无论是风和日丽还是刮风下雨,参观的人流一直络绎不绝,截至5月9日17时30分,累计检票人员入园已达19.59万人次。
D.昨天上午,一位老人突然晕倒在购物中心,后经迅速赶到的120急救中心医护人员以及商场保安,在场群众的救护下,老人得到及时抢救,最终脱离了危险。
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高二上第一次月考第一部分:英语知识运用(共两节,满分45分)第一节:单项填空(共15小题;每小题1分,满分15分)21._______ at the theater , he found the ticket a friend gave him was left at home.A. To arriveB. On arrivingC. While arrivedD. On arrived22. After studying in a medical college for five years, Jane ____her job as a doctor in the countryside .A .set out B. took over C. took up D. set up23. Follow the man and keep him in_ ___all the time.A. sightB. viewC. opinionD. purpose24. _____for the breakdown of the school computer network, Alice was in low spirits.A. BlamingB. BlamedC. To blameD. To be blamedhurried to the station only ____that the train had left.A. to findB. findingC. foundD. to be found26. While watching the TV, _________A. the doorbell rangB. the doorbell ringsC. we heard the doorbell ringD. we heard the doorbell rings27. ____straight on and you will see a churchA. GoB. GoingC. If you goD. When going28 .The teacher _____ his students______ five groups..A divid ed…into B. separated…from .C. separated…into… D. divided…fromUnited Kingdom of Great Britain _______ three main parts, England, Scotland and WalesA. is made upB. makes upC. consists ofD. is consisted of30. A quarrel _________, which made him ______ his family.A. was broken out; break awayB. broke out; break away fromC. was broken away; break downD. broke down; break out31. He _______ the enemy and was raised to the rank of general as a reward.A. wonB. hitC. stuckD. defeated32. -----Would you like to ______ us in celebrating John’s return from Africa tonight-----I’d like to, but I have to _______ a meeting.A. join; attendB. attend; joinC. take part in; attendD. join; join in33. Our English teacher is very strict _____ us and ______ his teaching.A. with; onB. with; inC. to; onD. to; in34. ________ that 10 dollars, I have another ten given by my mother.A. ExceptB. Apart fromC. Except for D . Except that35. -----_____ to the sun, the flower will become dry soon.-----What you said does make _____.A. Exposed; senseB. Exposing; senseC. Exposed; sensesD. Exposing; a sense第二节:完形填空一预招班做(共20小题;每小题分,满分30分)“Where would we be right now if I did not have the restaurant How else would I be able to provide for our family Who give me, a man without 36,a job” My father 37 say these things on our car rides to the stores that supply our 38.He would tell me stories of long, hard hours spent 39 in the hot kitchen and helping all the 40. Now I 41 late on weekends to welcome him when he comes home, 42 that he will tell me something43 .He always knows how to inspire me. I could never understand 44 he grew to be so wise without even having the opportunity to completehigh school.“Make sure you get a job 45 you have to wear a tie!” he’d always say with a smile. They were like a paintbrush(画笔) that began 46 this beautiful picture in my head that I have entitled(有资格的)my 47 in life.My father has worked day in and day out,with few days off, for 20 years, never 48 .He would explain how this is 49 he had to do to make my 50 better. My father worked so I 51 go to the best schools. He worked to supply me with 52 that would provide a job where I had to wear a tie.My father is a 53 in which I can only hope to catch a reflection(反省)of myself. He is the one I strive to model myself after. 54 , when I make a great name for myself as a doctor, someone will ask, “Where did you get such a gre at work ethic (道德) ”“My father!” I’ ll say, adjusting(调整) my 55 and I’ ll smile.36. A work B family C money D an education37. A used to B would C always D might38. A family B kitchen C restaurant D hotel39. A to work B work C works D working40. A friends B cooks C customers D waiters41. A hold up B keep up C stand up D stay up42. A wished B wishing C hoped D hoping43. A wonderful B interesting C big D wise44. A why B how C where D when45. A when B where C which D that46. A drawing B describing C imagining D painting47. A job B success C goal D future48. A complained B complains C complaining D complaint49. A what B that C which D why50. A job B life C days D future51. A would B could C should D should52. A opportunities B hopes C occasions D possibilities53. A model B mirror C hero D example54. A The other day B Another day C Some days D One day55. A tie B suit C dress D shirt完形填空二普通班做Michel is a young girl who works for the police 36 a handwriting expert (专家). She has helped 37 many criminals (罪犯) by using her special talents (天才).When she was fourteen, Michel was already 38 interested in the differences in her friends' 39 that she would spend hours 40 them. After 41 college she went to France for a 42 two-year class in handwriting at the School of Police Science.Michel says that it is 43 for people to hide their handwriting. She can discover 44 of what she needs to know simply 45 looking at the writing with her own eyes, 46 she also has machines 47 help her make 48 different kinds of paper and ink. This knowledge is often 49 great help to the police.Michel believes that handwriting is a good 50 of what kind of person the 51 is. "I wouldn't go out with a fellow 52 I didn't like his handwriting. " She says. But she 53 she fell in love with her future husband, a young policeman 54 she studied his handwriting. It is later proved to be 55 , however.36 A. with B. by C. like D. as37 A. search B. follow C. catch D. judge38 A. so B. too C. quite D. extra39 A. books B. letter C. tongues D. handwriting40 B. studying C. settling D. uncovering41 B. finishing C. starting D. stepping into42 A. powerful B. natural C. special D. common43 B. safe C. easy44 A. most B. nothing C. little D. sight45 A. with B. by C. of D. about46 A. so B. for C. thus D. but47 A. they B. in which C. that D. those48 A. up B. out C. for D. into49 A. of B. to C. with D. for50 A. test B. sign(标记) C. means D. habit51 A. thief B. criminal C. writer D. policeman52 A. whether B. unless C. if D. after53 A. adds B. tells C. repeats D. cries54 A. before B. after C. so D. and55 B. all right D. quite easy第三部分:阅读理解普通班做A、B、E、F篇;预招班A、B、C、D篇(共20小题;每小题2分,满分40分)A(预、普)Nearly everyone is shy in some ways. If shyness is making you uncomfortable, it may be time for a few lessons in self-confidence. You can build your confidence by following some suggestions from doctors and psychologists.Make a decision not to hold back in conversations. What you have to say is just as important as what other people say. And don’t turn down party invitations just because of your shyness. Prepare for yourself for being with others in groups. Make a list of the good qualities you have.Then make a list of ideas, experiences, and skills you would like to share with other people . I think about what you would like to say in advance. Then say it.If you start feeling self-conscious in a group, take a deep breath and focus your attention on other people, Remember, you are not alone. Other people are concerned about the impression they are making, too.No one ever gets over being shy completely, but most people do learn to live with their shyness. Even entertainers admit that they often feel shy. They work at fighting their shy feelings so that they can face the cameras and the public. Just making the effort to control shyness can have many rewards. But perhaps the best reason to fight shyness is to give other people a chance to know about you.would this article probably appearA.In a popular magazineB.On the top-line position of a newspaperC.In a science textbookD.In an encyclopedia(百科全书)57. The main purpose of the article is to _____.A. explain how shyness developedB .recommend ways of dealing with shynessC. persuade readers that shyness is naturalD. prove that shyness can be overcome58. Which of these can you conclude from reading the articleA.Shy people never have any funB.Entertainers choose their work to fight shynessC.The attempt to overcome shyness is always successfulD.The attempt to overcome shyness is always rewarding59. Who probably gives the suggestion for fighting shynessA. The author of the articleB. Shy men and womenC. Doctors and psychologistsD. Popular entertainersB(预、普)A new plan for getting children to and from school is being started by a local government in Eastern England. This could end the worries of many parents fearful for their children’s safety on the roads.Until now the local government have only been prepared to provide bus services for children living more than three miles from their school, or sometimes less if special reasons existed. Now it has been decided that if a group of parents ask for help in organizing transport they will be prepared to go ahead, as long as the arrangement will not lose money and children taking part will be attending their nearest school.The new plan is to be fired out this term for children living at Milton who attend Impington School. The children live just within the three-mile limit and the local government said in the past that they would not undertake to provide free transport to the school . But now they have agreed to offer a sum of money for a bus service from Milton to Impington and back, a plan which has the support of the school’s headmaster. Between 50 and 60 parents have said they would like their children to take part . Final calculations have still to be carried out, but a government official has said the cost to parents should be less than £20 a term. They have been able to arrange the service at a low cost because there is already an agreement with the bus company for a bus to take children who live further away to Impington . The same bus would now just make one more journey to pick up the Milton children. The official said they would get in touch with other groups of parents who in the past had asked if transport could be provided for their children , to see if they would like to take part in the new plan. is the aim of the planA.To prevent the students’ road accidentsB.To relieve the traffic pressureC.To save time for the parents and studentsD.To help the parents save money61. How can the local government arrange the new bus service at a low costA.By letting the bus run in the morning onlyB.By limiting the number of the studentsC.By getting the support from the headmasterD.By linking the new bus service with the existing one62. Which of the following is possible if the plan is carried outA.The bus company will make much more moneyB.The children can choose whatever school they likeC.The parents can get rid of their worriesD.The students in Impington School can have free bus rides63. This passage is most probably _____A. a personal letterB.an advertisementC. a headmaster’s reportD. a newspaper articleC(预)Every artist knows in his heart that he is saying something to the public. Not only does he want to say it well, but he wants it to be something that has not been said before. He hopes that the public will listen and understand ----he wants to teach them, and he wants them to learn fromhim.What visional artists like painters want to teach is easy to make out but difficult to explain, because painters translate their experience into shapes and colors, not words. They seem to feel that a certain choice of shapes and colors, out of the countless billions possible, is very interesting for them and worth showing to us. Without their work we should never have notice these particular shapes and colors ,or have felt the delight which they brought to the artists.Most artists take their shapes and colors from the world of nature and from human bodies in movement and at rest; their choices show that these aspects(方面)of the world are worth looking at, that they contain beautiful sights. Modern artists might say that they only choose subjects that provide an interesting pattern, that there is nothing more in it. Yet even they do not choose totally without thinking about the character of their subjects.If one painter chooses to paint a decaying(溃烂的) leg and another a lake in moonlight, each of them is directing our attention to a certain aspect of the world . Each painter is telling us something , showing us something , emphasizing something----- all of which means that, consciously or unconsciously, he is trying to teach us.is hard to explain what a painter is saying because _______.A. most painters do not express themselves wellpainter uses unusual words and phrasespainter use shapes and colors instead of wordspainters do not say anything65. Modern artists might say their choice of subjects ______.A.carries a message to the public provides interesting patternsno pattern or form D. teaches the public important truths66. The writer says that modern art contains______A. nothing but meaningless patterns aspects of the worldC. subjects chosen partly for their meaningD. completely meaningless subjectsD(预)Cole Bettles had been rejected by a number of universities when he received an e-mail from the University of California, San Diego, last month, congratulating him on his admission and inviting him to tour the campus. His mother booked a hotel in San Diego, and the 18-year-old Ojai high school senior arranged for his grandfather, uncle and other family members to meet them at the campus for lunch during the Saturday tour.“They were like ‘Oh my God, that’s so awesome (棒的)’,” Bettles said. Right before he got in bed, he checked his e-mail one last time and found another message saying the school had made a mistake and his application had been denied.In fact, all 28, 000 students turned away from UC San Diego, in one of the toughest college entrance seasons on record, had received the same incorrect message. The students’ hopes had been raised and then dashed (破灭) in a cruel twist that shows the danger of instant communications in the Internet age.UCSD admissions director Mae Brown called it an “administrative error” but refused to say who had made the mistake, or if those responsible would be disciplined (受训)。