浙江省绍兴市第一中学高二英语上学期期末考试试题
浙江省绍兴市柯桥区2023-2024学年五年级第一学期期末学业评价测试语文试题(含答案)

绍兴市柯桥区2023年学业评价测试五年级语文第一学期时间:100分钟满分:100分一、书写展示(3分)全卷用黑色钢笔或水笔书写,字迹端正,如需更正,请用修改符号。
请依照示例,用楷书书写下面的唐诗,注意诗句要居中,上下、左右文字要对齐。
二、积累与运用(29分)(一)根据语境写词语。
(8分)秋日huánɡhūn(),门前的小溪轻唱着,流向miǎn qiǎnɡ()可以看见的淡淡远山。
柿树上的果子bú jì qí shù(),正偷吃的鸟儿被我一惊,dǎn qiè()地飞走了。
这一番shī qínɡ huà yì()的景象是城里人所 xī hɑn()的吧。
树下的窗上透出一圈hónɡyùn(),母亲正lù xù()把饭菜端上桌。
我轻轻推开了门……(二)选择题。
(14分)1.下列加点字读音完全正确的一组是()A.削弱(xuē)直奔(bēn)嗜好(shì)待一会儿(dāi)B.强逼(qiánɡ)冠军(ɡuàn)床铺(pù)水浒传(zhuàn)C.间隔(jiàn)湖畔(pàn)美差(chāi)默而识之(zhì)D.教诲(huǐ)脊背(jǐ)纸屑(xiè)落在后边(là)2.下列词语书写没有错误的一组是()A.隐敝冀中夜幕降临兵荒马乱B.崩塌妨碍诡计多端难以置信C.牛郎追逐金壁辉煌直言不讳D.煤碳誊写不能自己应接不暇3.下列加点字词意思相同的一组是()A.敏而好学敏以求之者也B.漫浪诵读浪花淘尽英雄C.学而不厌喜新厌旧D.生气勃勃九州生气恃风雷4.下列加点词语或俗语使用最恰当的一项是()A.林升看到南宋权贵们不顾内忧外患,只图享乐,愤而写下《题临安邸》。
B.王老师吩咐我们每天打开天窗说亮话,多通风,免得细菌感染。
2025届浙江省绍兴市上虞实验中学数学八年级第一学期期末质量检测试题含解析

2025届浙江省绍兴市上虞实验中学数学八年级第一学期期末质量检测试题量检测试题注意事项:1.答卷前,考生务必将自己的姓名、准考证号、考场号和座位号填写在试题卷和答题卡上。
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考试结束后,请将本试卷和答题卡一并交回。
一、选择题(每题4分,共48分)1.25的平方根是( ) A .±5 B .5 C .±5 D .52.长度分别为a ,2,4的三条线段能组成一个三角形,则a 的值可能是( ) A .1 B .2 C .3 D .63.下面是“北”“比”“鼎”“射”四个字的甲骨文,其中不是轴对称图形的是( ) A . B . C . D .4.在同一平面直角坐标系中,直线()2y k x k =-+和直线y kx =的位置可能是( ) A . B .C .D .5.若长度分别为,3,5a 的三条线段能组成一个三角形,则a 的值可以是( )A .1B .2C .3D .86.已知直线MN EF ∥,一个含30角的直角三角尺()ABC AB BC >如图叠放在直线MN 上,斜边AC 交EF 于点D ,则1∠的度数为( )A .30B .45︒C .50︒D .60︒7.如图,△ABC 中,AB =AC ,AB 的垂直平分线交AC 于P 点,若AB =6,BC =4,△PBC的周长等于( )A .10B .12C .14D .168.如图,在△ABC 和△ADE 中,∠BAC =∠DAE =90°,AB =AC ,AD =AE ,C ,D ,E 三点在同一条直线上,连接BD ,则下列结论错误的是( )A .△ABD ≌△ACEB .∠ACE+∠DBC =45° C .BD ⊥CE D .∠BAE+∠CAD =200°9.在平面直角坐标系中,点(4,2)-关于y 轴对称的点的坐标是( )A .(4,2)B .(4,2)-C .(4,2)--D .(4,2)-10.下列长度的三条线段中,能组成三角形的是( )A .1, 2, 3cm cm cmB .2,3,5cm cm cmC .3, 4, 7cm cm cmD .4,5,8cm cm cm11.点P(2,-3)所在的象限是( )A .第一象限B .第二象限C .第三象限D .第四象限 12.立方根等于本身的数是( )A .-1B .0C .±1D .±1或0二、填空题(每题4分,共24分)13.若()1,1A m n +-与点()-3,2B 关于y 轴对称,则()2019m n +的值是___________;14.在实数范围内,使得3x +有意义的x 的取值范围为______.15.计算:(a-b )(a 2+ab+b 2)=_______.16.比较大小:3_____8.(填“>”、“<“、“=“)17.化简:129=_______________.18.如图,将ABC ∆沿着过AB 中点D 的直线折叠,使点A 落在BC 边上的1A 处,称为第1次操作,折痕DE 到BC 的距离记为1h ,还原纸片后,再将ADE ∆沿着过AD 中点1D 的直线折叠,使点A 落在DE 边上的2A 处,称为第2次操作,折痕11D E 到BC 的距离记为2h ,按上述方法不断操作下去…经过第2020次操作后得到的折痕20192019D E 到BC 的距离记为2020h ,若11h =,则2020h 的值为______.三、解答题(共78分)19.(8分)如图,直线l 与m 分别是ABC ∆边AC 和BC 的垂直平分线,它们分别交边AB 于点D 和点E .(1)若10AB =,则CDE ∆的周长是多少?为什么?(2)若125ACB ︒∠=,求DCE ∠的度数.20.(8分)已知O 为原点,点(8,0)A 及在第一象限的动点(),P x y ,且12x y +=,设OPA ∆的面积为S .(1)求S 关于x 的函数解析式;(2)求x 的取值范围;(3)当12S =时,求P 点坐标;(4)画出函数S 的图象.21.(8分)如图:AE=DE ,BE=CE ,AC 和BD 相交于点E ,求证:AB=DC22.(10分)如图所示,△ABD 和△BCD 都是等边三角形,E 、F 分别是边AD 、CD 上的点,且DE =CF ,连接BE 、EF 、FB .求证:(1)△ABE ≌△DBF ;(2)△BEF 是等边三角形.23.(10分)如图(1),一架云梯AB 斜靠在一竖直的墙上,云梯的顶端A 距地面15米,梯子的长度比梯子底端B 离墙的距离大5米.(1)这个云梯的底端B 离墙多远?(2)如图(2),如果梯子的顶端下滑了8m (AC 的长),那么梯子的底部在水平方向右滑动了多少米?24.(10分)如图1,AC BC =,CD CE =,ACB DCE α∠=∠=,AD 、BE 相交于点M ,连接CM .()1求证:BE AD =;()2求AMB ∠的度数(用含α的式子表示);()3如图2,当90α=时,点P 、Q 分别为AD 、BE 的中点,分别连接CP 、CQ 、PQ ,判断CPQ 的形状,并加以证明.25.(12分)我国边防局接到情报,近海处有一可疑船只A 正向公海方向行驶,边防部迅速派出快艇B 追赶(如图1) .图2中12,l l 分别表示两船相对于海岸的距离S (海里)与追赶时间t (分)之间的关系.根据图象问答问题:(1)①直线1l 与直线2l 中 表示B 到海岸的距离与追赶时间之间的关系; ②A 与B 比较 速度快;③如果一直追下去,那么B ________ (填 “能”或“不能")追上A ;④可疑船只A 速度是 海里/分,快艇B 的速度是 海里/分;(2)1l 与2l 对应的两个一次函数表达式111s k t b =+与222s k t b =+中12,k k 的实际意义各是什么?并直接写出两个具体表达式.(3)15分钟内B 能否追上A ?为什么?(4)当A 逃离海岸12海里的公海时,B 将无法对其进行检查,照此速度,B 能否在A 逃入公海前将其拦截?为什么?26.如图,已知在△ABC 中,AB =AC ,D 是BC 边上任意一点,E 在AC 边上,且AD =AE .(1)若∠BAD =40°,求∠EDC 的度数;(2)若∠EDC =15°,求∠BAD 的度数;(3)根据上述两小题的答案,试探索∠EDC 与∠BAD 的关系.参考答案一、选择题(每题4分,共48分)1、C25,再根据平方根定义求出即可.255,5的平方根是±525±5故选C.【点睛】本题考查了对平方根和算术平方根的应用,主要考查学生对平方根和算术平方根的定义的理解能力和计算能力,难度不大.2、C【分析】根据三角形三边关系定理得出4-2<a<4+2,求出即可.【详解】由三角形三边关系定理得:4﹣2<a<4+2,即2<a<6,即符合的只有1.故选:C.【点睛】此题考查三角形三边关系定理,能根据定理得出5-1<a<5+1是解题的关键,注意:三角形的两边之和大于第三边,三角形的两边之差小于第三边.3、B【解析】根据轴对称的定义,逐一判断选项,即可得到答案.【详解】A是轴对称图形,不符合题意,B不是轴对称图形,符合题意,C 是轴对称图形,不符合题意,D 是轴对称图形,不符合题意,故选B .【点睛】本题主要考查轴对称图形的定义,掌握轴对称图形的定义,是解题的关键. 4、C【分析】根据一次函数的性质,对k 的取值分三种情况进行讨论,排除错误选项,即可得到结果.【详解】解:由题意知,分三种情况:当k >2时,y=(k-2)x+k 的图象经过第一、二、三象限;y=kx 的图象y 随x 的增大而增大,并且l 2比l 1倾斜程度大,故B 选项错误,C 选项正确;当0<k <2时,y=(k-2)x+k 的图象经过第一、二、四象限;y=kx 的图象y 随x 的增大而增大,A 、D 选项错误;当k <0时,y=(k-2)x+k 的图象经过第二、三、四象限,y=kx 的图象y 随x 的增大而减小,但l 1比l 2倾斜程度大.∴直线()2y k x k =-+和直线y kx =的位置可能是C.故选:C .【点睛】本题考查了一次函数图象与系数的关系:对于y =kx +b (k 为常数,k ≠0),当k >0,b >0,y =kx +b 的图象在一、二、三象限;当k >0,b <0,y =kx +b 的图象在一、三、四象限;当k <0,b >0,y =kx +b 的图象在一、二、四象限;当k <0,b <0,y =kx +b 的图象在二、三、四象限.5、C【分析】根据三角形三边关系可得5﹣3<a <5+3,解不等式即可求解.【详解】由三角形三边关系定理得:5﹣3<a <5+3,即2<a <8,由此可得,符合条件的只有选项C ,故选C .【点睛】本题考查了三角形三边关系,能根据三角形的三边关系定理得出5﹣3<a <5+3是解此题的关键,注意:三角形的两边之和大于第三边,三角形的两边之差小于第三边. 6、D【分析】首先根据直角三角形的性质判定∠A=30°,∠ACB=60°,然后根据平行的性质得出∠1=∠ACB.【详解】∵含30角的直角三角尺()ABC AB BC >∴∠A=30°,∠ACB=60°∵MN EF ∥∴∠1=∠ACB=60°故选:D.【点睛】此题主要考查直角三角形以及平行的性质,熟练掌握,即可解题.7、A【分析】先根据等腰三角形的性质得出AC =AB =6,再根据线段垂直平分线的性质得出AP =BP ,故AP +PC =AC ,由此即可得出结论.【详解】解:∵△ABC 中,AB=AC ,AB=6,∴AC=6,∵AB 的垂直平分线交AC 于P 点,∴BP+PC=AC ,∴△PBC 的周长=(BP+PC )+BC=AC+BC=6+4=1.故选:A .【点睛】本题考查的是线段垂直平分线的性质,三角形的周长计算方法,熟知线段垂直平分线上任意一点,到线段两端点的距离相等是解答此题的关键.8、D【分析】根据SAS 即可证明△ABD ≌△ACE ,再利用全等三角形的性质以及等腰直角三角形的性质即可一一判断.【详解】∵∠BAC =∠DAE =90°,∴∠BAC +∠CAD =∠DAE +∠CAD ,即∠BAD =∠CAE .在△BAD 和△CAE 中,∵AB AC BAD CAE AD AE ∠∠=⎧⎪=⎨⎪=⎩,∴△BAD ≌△CAE (SAS ),∴BD =CE ,故A 正确; ∵△ABC 为等腰直角三角形,∴∠ABC =∠ACB =45°,∴∠ABD +∠DBC =45°. ∵△BAD ≌△CAE ,∴∠ABD =∠ACE ,∴∠ACE +∠DBC =45°,故B 正确.∵∠ABD+∠DBC=45°,∴∠ACE+∠DBC=45°,∴∠DBC+∠DCB=∠DBC+∠ACE +∠ACB=90°,则BD⊥CE,故C正确.∵∠BAC=∠DAE=90°,∴∠BAE+∠DAC=360°﹣90°﹣90°=180°,故D错误.故选D.【点睛】本题考查了全等三角形的判定和性质,等腰直角三角形的性质等知识,解题的关键是正确寻找全等三角形解决问题,属于中考常考题型.9、C【分析】直接利用关于y轴对称点的性质得出答案.【详解】解:点(4,-2)关于y轴对称的点的坐标是:(-4,-2).故选:C.【点睛】此题主要考查了关于y轴对称点的性质,正确记忆横纵坐标的符号是解题关键.10、D【分析】根据三角形三边关系定理:①三角形两边之和大于第三边,②三角形的两边之差小于第三边,逐个判断即可.【详解】A、1+2=3,不符合三角形三边关系定理,故本选项错误;B、2+3=5,不符合三角形三边关系定理,故本选项错误;C、3+4=7,不符合三角形三边关系定理,故本选项错误;D、4+5>8,符合三角形三边关系定理,故本选项正确;故选:D.【点睛】本题主要考查了三角形的三边关系:用两条较短的线段相加,如果大于最长的那条线段就能够组成三角形.11、D【解析】析:应先判断出所求的点的横纵坐标的符号,进而判断点P所在的象限.解答:解:∵点P的横坐标为正,纵坐标为负,∴点P(2,-3)所在象限为第四象限.故选D.12、D【分析】根据立方根的定义得到立方根等于本身的数.【详解】解:∵立方根是它本身有3个,分别是±1,1.故选:D .【点睛】本题主要考查了立方根的性质.对于特殊的数字要记住,立方根是它本身有3个,分别是±1,1.立方根的性质:(1)正数的立方根是正数. (2)负数的立方根是负数.(3)1的立方根是1.二、填空题(每题4分,共24分)13、1【分析】根据关于y 轴对称的点,纵坐标相同,横坐标互为相反数,可得m 、n 的值,代入计算可得答案.【详解】由点()11A m n +-,与点()32B -,的坐标关于y 轴对称,得: 13m +=,12n -=,解得:2m =,1n =-,∴20192019()(21)1m n +=-=.故答案为:1.【点睛】本题考查了关于y 轴对称的点的坐标,解决本题的关键是掌握好对称点的坐标规律:关于x 轴对称的点,横坐标相同,纵坐标互为相反数;关于y 轴对称的点,纵坐标相同,横坐标互为相反数;关于原点对称的点,横坐标与纵坐标都互为相反数.14、3x ≥-【分析】直接利用二次根式有意义的条件分析得出答案.有意义,则1+x≥0,解得:x≥-1.故答案为:x≥-1.【点睛】本题考查二次根式有意义的条件,正确把握二次根式的定义是解题关键.15、a 3-b 3【分析】根据多项式乘以多项式法则进行计算即可求解.【详解】3222322233()()=a a b ab a b ab b a b a b a ab b ++---=--++故答案为:33a b -【点睛】本题考查了多项式乘以多项式法则,先用一个多项式的每一项分别乘另一个多项式的每一项,再把所得的积相加.16、>【分析】利用估算法比较两实数的大小.<<,∴2<3,∴3.故答案是:>.【点睛】本题考查实数的大小比较,正确对无理数进行估算是解题关键.17、3【分析】根据分数指数幂的定义化简即可. 【详解】解:1293== 故答案为:3【点睛】本题主要考查了分数指数幂的意义,熟知分数指数幂意义是解题关键.18、2019122-【分析】根据中点的性质及折叠的性质可得DA=DA ₁=DB,从而可得∠ADA ₁=2∠B,结合折叠的性质可得.,∠ADA ₁=2∠ADE,可得∠ADE=∠B,继而判断DE// BC,得出DE 是△ABC 的中位线,证得AA ₁⊥BC,AA ₁=2,由此发现规律:01 2122h =-=-₁同理21122h =-3211122222h =-⨯=-…于是经过第n 次操作后得到的折痕Dn-1 En-1到BC 的距离1122n n h -=-,据此求得2020h 的值. 【详解】解:如图连接AA ₁,由折叠的性质可得:AA ₁⊥DE, DA= DA ₁ ,A ₂、A ₃…均在AA ₁上又∵ D 是AB 中点,∴DA= DB ,∵DB= DA ₁ ,∴∠BA ₁D=∠B ,∴∠ADA ₁=∠B +∠BA ₁D=2∠B,又∵∠ADA ₁ =2∠ADE ,∴∠ADE=∠B∵DE//BC,∴AA ₁⊥BC ,∵h ₁=1∴AA ₁ =2, ∴01 2122h =-=-₁ 同理:21122h =-; 3211122222h =-⨯=-; …∴经过n 次操作后得到的折痕D n-1E n-1到BC 的距离1122n n h -=- ∴20202019122h =-【点睛】 本题考查了中点性质和折叠的性质,本题难度较大,要从每次折叠发现规律,求得规律的过程是难点.三、解答题(共78分)19、(1)10;(2)70DCE ︒∠=【分析】根据垂直平分线定理即可推出CD AD =,同理CE BE =,即CDE ∆的周长为10由垂直平分线定理可得ACD A ∠=∠,BCE B ∠=∠,再根据三角形内角和定理2CDE A ∠=∠,2CDE A ∠=∠即22180DCE A B ︒∠+∠+∠=,再由三角形外角和定理得125DCE A B ︒∠+∠+∠= ,即可计算出70DCE ︒∠=.【详解】解:(1)CDE ∆的周长为10∵l 是AC 的垂直平分线∴CD AD =同理CE BE =∴CDE ∆的周长10CD DE CB AD DE BE AB =++=++==(2)∵l 是AC 的垂直平分线∴ACD A ∠=∠同理BCE B ∠=∠∴2CDE A ∠=∠,2CDE A ∠=∠∵180DCE CDE CED ︒∠+∠+∠=①∴22180DCE A B ︒∠+∠+∠=∵125DCE ACD BCE ACB ︒∠+∠+∠=∠=∴125DCE A B ︒∠+∠+∠=②联立①②,解得:70DCE ︒∠=【点睛】本题考查垂直平分线和三角形的内角和定理,熟练掌握垂直平分线定理推出CDE ∆=AB 是解题关键.20、(1)S =−4x +48;(2)0<x <12;(3)P (1,3);(4)见解析.【分析】(1)根据三角形的面积公式即可得出结论;(2)根据(1)中函数关系式及点P 在第一象限即可得出结论;(3)把S =12代入(1)中函数关系即可得出x 的值,进而得出y 的值;(4)利用描点法画出函数图象即可.【详解】解:(1)∵A 点和P 点的坐标分别是(8,0)、(x ,y ),∴S =12×8×y =4y . ∵x +y =12,∴y =12−x .∴S =4(12−x )=48−4x ,∴所求的函数关系式为:S=−4x+48;(2)由(1)得S=−4x+48>0,解得:x<12;又∵点P在第一象限,∴x>0,综上可得x的取值范围为:0<x<12;(3)∵S=12,∴−4x+48=12,解得x=1.∵x+y=12,∴y=12−1=3,即P(1,3);(4)∵函数解析式为S=−4x+48,∴函数图象是经过点(12,0)(0,48)但不包括这两点的线段.所画图象如图:【点睛】本题考查的是一次函数的应用,根据题意得到函数关系式,并熟知一次函数的图象和性质是解答此题的关键.21、见详解.【详解】由SAS可得△ABE≌△DCE,即可得出AB=CD.∵AE=DE,BE=CE,∠AEB=∠CED(对顶角相等),∴△ABE≌△DCE(SAS),∴AB=CD.22、(1)详见解析;(2)详见解析.【分析】(1)根据等边三角形的性质及SAS推出△ABE≌△DBF即可;(2)根据全等三角形的性质得出BE=BF,∠ABE=∠DBF,求出∠EBF=60°,根据等边三角形的判定推出即可.【详解】证明:(1)∵△ABD和△BCD都是等边三角形,∴∠ABD=∠A=∠BDF=60°,AB=AD=DB=CD,∵DE=CF,∴AE=DF,在△ABE和△DBF中,AB DBA BDF AE DF=⎧⎪∠=∠⎨⎪=⎩∴△ABE≌△DBF(SAS);(2)∵△ABE≌△DBF,∴BE=BF,∠ABE=∠DBF,∴∠EBF=∠EBD+∠DBF=∠EBD+∠ABE=∠ABD=60°,∴△BEF是等边三角形.【点睛】本题主要考查全等三角形的判定及性质,等边三角形的判定及性质,掌握全等三角形和等边三角形的判定方法和性质是解题的关键.23、(1)这个云梯的底端B离墙20米;(2)梯子的底部在水平方向右滑动了4米. 【解析】(1)由题意得OA=15米,AB-OB=5米,根据勾股定理OA2+OB2=AB2,可求出梯子底端离墙有多远;(2)由题意得此时CO=7米,CD=AB=25米,由勾股定理可得出此时的OD,继而能和(1)的OB进行比较.【详解】解:(1)设梯子的长度为米,则云梯底端B 离墙为米。
浙江省绍兴市诸暨中学 2020年高二英语联考试题含解析

浙江省绍兴市诸暨中学 2020年高二英语联考试题含解析一、选择题1. _____ development of nuclear weapon is ____ threat to ___ mankind.A. The; / ;theB. The; a; /C. A; /; theD. A; a; / 参考答案:b略2. The total income of the family, _______, 500 yuan a month, can never cover the livingcost.A. for exampleB. that isC. which isD. however参考答案:B3. E-mail, as well as telephones, _______an important part in our daily communication.A. play B.have played C.are playing D. is playing参考答案:D4. Wind power is an ancient source of energy ____we may return in the near future.A. on whichB. by whichC. to whichD. from which参考答案:C知识点:介词与关系代词连用解析:选C,根据定语从句中的谓语动词return后应跟to sth。
先行词为energy,故选C。
技巧点拨:观动词,判介词。
5. --- Where was it _____ you interviewed the famous actor yesterday?--- It was in the hotel _____ he stayed.A. that; whereB. which; thatC. that; thatD. which; where参考答案:A6. -- How do you like living here ?-- Fine , thank you . I just feel ______A. freeB. at homeC. numbD. at risk参考答案:B7. After a whole afternoon’s heated discussion, they finally ________ to us what had been decided.A. toldB. declaredC. statedD. made参考答案:B略8. --- Could you do me a favor and take these books to my office?--- Yes, ______.A. for pleasureB. I couldC. my pleasureD. with pleasure 参考答案:D9. While hosting the Academy Awards, Billy Crystal always keeps a toothbrush with him__________ he started practicing stand-up as a child, he told himself jokes in front of the mirror, brushing his teeth.A. because whenB. because thatC. whenD. as参考答案:A【详解】考查原因状语从句和时间状语从句。
2020年浙江省绍兴市柯桥中学高二英语月考试题含解析

2020年浙江省绍兴市柯桥中学高二英语月考试题含解析 一、 选择题 1. Mike seldom asks __________ questions in class,but today he raised __________ fairly good one.
A./; a B.the; a C./; the D./; / 参考答案: A 略 2. As we all know no countries would ________ being controlled by another country. A. submit B. submitting to C. submitted to D. submit to
参考答案:
D 3. —What do you usually do in the afternoon, Mrs. Smith?
— My afternoon ______ includes doing some shopping and walking my dogs. A. appointment B. action C. routine D. reservation 参考答案: C 4. Her eyes __________ on her plate and in a short time, she was busy ________. A. were fixed, eating B. fixed, to eat C. were fixed, to eat D. fixed, eating 参考答案:
A 5. -What do you think of Mrs Smith ?
- She is ______ as you could ever meet. A. an as kind lady B. as a kind lady C. as kind a lady D. kind as a lady 参考答案: C
浙江省绍兴市第一中学2023-2024学年高一下学期创新班期中考试数学试卷

浙江省绍兴市第一中学2023-2024学年高一下学期创新班期中考试数学试卷学校:___________姓名:___________班级:___________考号:___________一、单选题1.若复数z 满足z=i (2+z )(i 为虚数单位),则z=( )A .1+iB .1-iC .-1+iD .-1-i2.如图所示,梯形A B C D ''''是平面图形ABCD 用斜二测画法得到的直观图,2A D ''=,1A B B C ''''==,则平面图形ABCD 中对角线AC 的长度为( )AB C D .53.已知样本数据12100,,,x x x L 的平均数和标准差均为4,则数据121001,1,,1x x x ------L 的平均数与标准差分别为( )A .54-,B .516-,C .416,D .44,4.一个圆锥的侧面展开图是半径为1的半圆,则此圆锥的内切球的表面积为( ) A .π B .π2 C .π3 D .π45.光源(3,2,1)P 经过平面Oyz 反射后经过(1,6,5)Q ,则反射点R 的坐标为( )A .75(0,,)22B .(0,4,3)C .97(0,,)22D .(0,5,4)6.若4,2145,,,的第 p 百分位数是4,则 p 的取值范围是( )A .(]4080,B .[)4080,C .[]40,80D .()40,807.如图是棱长均相等的多面体EABCDF ,其中四边形ABCD 是正方形,点P Q M N ,,,分别为DE ,AB ,AD ,BF 的中点,则异面直线PQ 与MN 所成角的余弦值为( )A .13B .12 C .23 D .348.在正方体1111ABCD A B C D -中,点M N ,分别是直线CD AB ,上的动点,点 P 是△11AC D 内的动点(不包括边界),记直线1D P 与MN 所成角为θ,若θ的最小值为π3,则1D P 与平面11AC D 所成角的正弦的最大值为( )A B C D二、多选题9.在12件同类产品中,有9件正品和3件次品,从中任意抽出3件产品,设事件A “3件产品都是次品”,事件B “至少有1件是次品”,事件C “至少有1件是正品”,则下列结论正确的是( )A .A 与C 为对立事件B .B 与C 不是互斥事件 C .A B A =ID .()()1P B P C +=10.在某市高三年级举行的一次模拟考试中,某学科共有20000人参加考试.为了了解本次考试学生成绩情况,从中抽取了部分学生的成绩(成绩均为正整数,满分为100分)作为样本进行统计,样本容量为n ,按照[)[)[)[)[]506060707080809090100,,,,,,,,,的分组作出频率分布直方图如图所示,其中,成绩落在区间[)5060,内的人数为16.则( )A .图中0.016x =B .样本容量1000n =C .估计该市全体学生成绩的平均分为71.6分D .该市要对成绩前25%的学生授予“优秀学生”称号,则授予“优秀学生”称号的学生考试成绩大约至少为77.25分11.勒洛四面体是一个非常神奇的“四面体”,它能在两个平行平面间自由转动,并且始终保持与两平面都接触,因此它能像球一样来回滚动.勒洛四面体是以正四面体的四个顶点为球心,以正四面体的棱长为半径的四个球的公共部分,如图所示,若正四面体ABCD 的棱长为a .则( )A .能够容纳勒洛四面体的正方体的棱长的最小值为aB .勒洛四面体能够容纳的最大球的半径为1a ⎛ ⎝⎭C .勒洛四面体中过A B C ,,三点的截面面积为(212π4aD .勒洛四面体的体积3312V a ⎛⎫∈ ⎪ ⎪⎝⎭三、填空题12.从含有6件次品的50件产品中任取4件,观察其中次品数,这个试验的样本空间Ω=. 13.如图,甲乙做游戏,两人通过划拳(剪刀、石头、布)比赛决胜谁首先登上第3个台阶,并规定从平地开始,每次划拳赢的一方登上一级台阶,输的一方原地不动,平局时两人都上一个台阶.如果一方连续赢两次,那么他将额外获得上一级台阶的奖励,除非已经登上第3个台阶,当有任何一方登上第3个台阶时游戏结束,则游戏结束时恰好划拳3次的概率为.14.在三棱锥 A BCD -中,二面角 A BD C --的大小为π 3, BAD CBD ∠∠=,2BD BC ==,则三棱锥外接球表面积的最小值为.四、解答题15.已知复数22(232)(32)z m m m m i =--+-+,(其中i 为虚数单位)(1)当复数z 是纯虚数时,求实数m 的值;(2)若复数z 对应的点在第三象限,求实数m 的取值范围.16.如图,在四棱锥P ABCD -中,底面ABCD 是以2为边长的菱形,且120BAD ∠=︒,PB PD =,M 为PC 的中点.(1)求证:平面PBD ⊥平面PAC ;(2)若PC =PD 与平面AMD 所成角的正弦值.17.为了了解学生躯干、腰、髋等部位关节韧带和肌肉的伸展性、弹性等,某学校对在校1500名学生进行了一次坐位体前屈测试,采用按学生性别比例分配的分层随机抽样抽取75人,已知这1500名学生中男生有900人,且抽取的样本中男生的平均数和方差分别为13.2和13.36,女生的平均数和方差分别为15.2和17.56.(1)求样本中男生和女生应分别抽取多少人;(2)求抽取的总样本的平均数,并估计全体学生的坐位体前屈成绩的方差.18.如图,已知直角三角形ABC 的斜边//BC 平面α,A 在平面α上,AB ,AC 分别与平面α成30o 和45o 的角,6BC =.(1)求BC 到平面α的距离;(2)求平面ABC 与平面α的夹角.19.如图,四棱锥S ABCD -的底面是平行四边形,平面α与直线AD SA SC ,,分别交于点,,P Q R ,且AP SQ CR AD SA CS==,点M 在直线SB 上运动,在线段CD 上是否存在一定点N ,使得其满足:(i )直线//MN α;(ii )对所有满足条件(i )的平面α,点M 都落在某一条长为m 的线段上,且m SB =若存在,求出点N 的位置;若不存在,说明理由.。
浙江省绍兴市高级中学2019-2020学年高二下学期质量检测英语试卷(含解析)

高二英语第一部分阅读理解(共两节,满分30分)第一节(共10小题;每小题2分,满分20分)阅读下列短文,从每题所给的A、B、C和D四个选项中,选出最佳选项。
AHumans and elephants don't always get along. In Africa and Asia, elephants damage public facilities, farmers' crops, and natural habitat critical for other species. If an elephant becomes a problem, humans may kill it. So local people are trying to find better ways to keep elephants away from sensitive places. Now, researchers think they have a good tool to discourage elephants: honeybee."Elephants strongly dislike being bitten in the trunk, so it's an extremely sensitive organ. Imagine being bitten in your nose by a bee, and multiply that a few thousand times." Mark Wright,a professor of entomology at the University of Hawaii at Manoa.Kenyan farmers have long known that if they hang live beehives around their farms, they will drive elephants away. "The problem is how do you manage hundreds of thousands of hives, if you have a huge area to protect? And then the idea came up to look at the alarm, which are chemicals that the bees release to launch an attack on something that's robbing the hive." Because elephants have a highly developed sense of smell, Wright and his team thought they might be able to use the bees' chemical alarm signal to affect elephant behavior.So the researchers tested out chemicals that imitate honeybees' alarm. In Greater Kruger National Park in South Africa, the team hung white socks treated with the mixture as well as untreated control socks around an elephant watering hole, and then observed the animals' behavior. "Elephants are really curious organisms, they'll walk along and see something like a sock hangingon a branch in the environment, and they'll inspect it, and if it's not treated with the alarm, they'llpick it up, throw it around, try to taste it, things like that. And in the case of the treated ones, we found that they would show immediate signs of unease or uncertainty, and they'd eventually backoff from them. Very seldom would they go pick them up and play with them."The study is in the journal Current Biology.Next up, the researchers plan to refine the mixture, in order to get an even stronger avoidance response from elephants. That way, these remarkable animals will stay far away from whereverthey should not be going—and both elephants and humans can coexist a bit more peacefully.1. According to the author, what would people do if an elephant continually destroys the farm?A. They would drive it away.B. They would beat it hard.C. They may force it to live in a poor area.D. They may kill it without mercy.2.The chemicals can successfully drive elephants away because of _________.A. elephants’ sensitive trunkB. elephants’ well developed sense of smellC. elephants’ powerful hearingD. elephants’ huge curiosity3. According to paragraph 5, when an elephant find a sock with chemicals, it feels________.A. pleasantB. curiousC. nervousD. thrilled4. Where might you read the text?A. In a scientific magazine.B. On the school notice board.C. At an exhibition.D. In a newspaper.5. What does the underlined word beehives mean?A. 蜂王B. 蜂窝C.蜂鸣器D. 黄蜂BSelf-driving vehicles(车辆) will rely on cameras, sensors and artificial intelligence (人工智能) to recognize and respond to road and traffic conditions, but sensing is the most effective for objects and movement in the neighborhood of the vehicle. Not everything important in a car’s environment will be caught by the vehicl e’s camera. Another vehicle approaching at high speed ontoo late. This is why vehicle-to-vehiclea collision (碰撞) track might not be visible until it’scommunication is undergoing rapid development. Our research shows that cars will need to be able to chat and cooperate on the road, although the technical challenges are still great.The usages of vehicle-to-vehicle communication contain vehicles driving together in a row,as well as safety messages about nearby emergency vehicles. Vehicles could alert each other to avoid collisions or share notices about passers-by and bicycles. From as far as several hundred metres away, vehicles could exchange messages with one another or receive information from roadside units(RSUs) about nearby incidents or dangerous road conditions through 4G network. Ahigh level of AI seems required for such vehicles, not only to self-drive from A to B, but also toreact intelligently to messages received. Vehicles will need to plan, reason, strategize and adapt according to information received in real time and to carry out cooperative behaviors. For example,a group of autonomous vehicles might avoid a route together because of possible risks, or a vehicle could decide to drop someone off earlier due to messages received, a foreseen crowding ahead.Further applications(应用) of vehicle-to-vehicle communication are still being researched, including how to perform cooperative behavior.6.What is the first paragraph mainly about?A.The reasons for the accidents by self-driving vehicles.B.The research about how to use self-driving vehicles.C.The reasons for developing communication between self-driving vehicles.D.The importance of artificial intelligence of self-driving vehicles.?7.What does the underlined word “alert” mean in Paragraph 2A.change. B.tell. C.ignore. D.warn.8.What can we learn about roadside units (RSUs)?A.They classify the vehicles on the road. B.They serve as efficient information stations. C.They take over the passing vehicles. D.They can change bad road conditions. 9.What is the best title for the text?A.When do vehicles communicate?B.Vehicle-to-vehicle communication is comingC.The reasons why a high level of AI is importantD.What do applications for vehicle-to-vehicle communication need?10. What is the writing style of the passage?A.argumentation议论文B.narration记叙文C.exposition说明文D.practical writing应用文第二节(共5小题;每小题2分,满分10分)根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。
浙江省绍兴市诸暨市诸暨中学暨阳分校2023-2024学年高二下学期期中考试语文试卷(含答案)
高二语文试题考生注意:1.本试题卷分四部分,全卷共8页,满分150分,考试时间150分钟。
2.答题前,在答题卷指定区域填写班级、姓名、考场、座位号及准考证号(填涂)。
3.请考生按规定用笔将所有试题的答案填写在答题卷上。
2024.4一、现代文阅读(35分)(一)现代文阅读Ⅰ(本题共5小题,19分)阅读下面的文字,完成下面小题。
材料一:生态文学以生态整体主义或生态整体观作为指导考察自然与人的关系,它对人类所有与自然有关的思想、态度和行为的判断标准是:是否有利于生态系统的整体利益,即生态系统和谐、稳定和可持续地自然存在。
不把人类作为自然界的中心、不把人类的利益作为价值判断的终极度,并不意味着生态文学蔑视人类或者反人类;恰恰相反,生态灾难的恶果和生态危机的现实使生态文学家认识到,只有把生态系统的整体利益作为根本前提和最高价值,人类才有可能真正有效地消除生态危机,而凡是有利于生态系统整体利益的,最终也一定有利于人类的长远利益或根本利益。
生态文学对自然与人的关系的考察和表现主要包括:自然对人的影响(物质的和精神的两个方面),人类在自然界的地位,自然整体以及自然万物与人类的关系,人对自然的征服、控制改造、掠夺和摧残,人对自然的保护和对生态平衡的恢复与重建,人对自然的赞美和审美,人重返和重建与自然的和谐等。
在表现自然与人的关系时,生态文学特别重视人对自然的责任义务,急切地呼吁保护自然万物和维护生态平衡,热情地赞美为生态整体利益而做出的自我牺牲,生态文学不能忽视生态责任,它把人类对自然的责任作为文本的主要伦理取向。
生态文学表现的是自然与人的关系,而落点却在人类的思想、文化、经济、科技、生活方式、社会发展模式上。
探寻和揭示造成生态灾难的社会根源,使得生态文学具有了显著的文明批判的特点。
许多作家对人类中心主义、征服和统治自然观、发展至上论等思想观念,对破坏生态平衡的自然改造、竭泽而渔地榨取自然资源的经济发展、违反自然规律和干扰自然进程的科技创造、严重污染自然的工业化和农业现代化等许许多多的思想、文化、社会现象提出了严厉的批判。
浙江省绍兴市诸暨市诸暨中学2023-2024学年化学高二第一学期期末达标测试试题含解析
浙江省绍兴市诸暨市诸暨中学2023-2024学年化学高二第一学期期末达标测试试题考生须知:1.全卷分选择题和非选择题两部分,全部在答题纸上作答。
选择题必须用2B铅笔填涂;非选择题的答案必须用黑色字迹的钢笔或答字笔写在“答题纸”相应位置上。
2.请用黑色字迹的钢笔或答字笔在“答题纸”上先填写姓名和准考证号。
3.保持卡面清洁,不要折叠,不要弄破、弄皱,在草稿纸、试题卷上答题无效。
一、选择题(每题只有一个选项符合题意)1、分别取50mL 0.5mol/L的盐酸溶液与50mL 0.55mol/L的氢氧化钠溶液进行中和热测定实验。
下列操作和说法不正确的是A.稍过量的氢氧化钠是确保盐酸完全反应B.向盛装盐酸的烧杯中小心缓慢地加入氢氧化钠溶液C.仪器A的名称是环形玻璃搅拌棒D.在实验过程中,把温度计上的酸用水冲洗干净后再测量氢氧化钠溶液的温度2、电解NO制备NH4NO3的工作原理如图所示,为使电解产物全部转化为NH4NO3,需要补充物质M。
下列有关分析或说法中正确的是A.a 应接直流电源的正极B.补充的物质M是NH3C.阳极的电极反应式:NO+5e-+6H+=NH4++H2OD.电路中通过5mol 电子时,理论上最多可处理22.4L NO气体(标准状况)3、构成金属晶体的基本微粒A.分子B.阴离子和自由电子C.阳离子和阴离子D.阳离子和自由电子4、由水电离产生的c(H+)为1×10-12mol/L的溶液中,能大量共存的离子组是A .Na +、Cl -、NO 3-、SO 42-B .K +、Fe 3+、I -、SO 42-C .K +、Cl -、NH 4+、S 2-D .Na +、Ca 2+、Cl -、HCO 3-5、某烃的相对分子质量为86,如果分子中含有3个3CH -、2个2CH --和1个CH --,则该结构的烃的一氯取代物最多可能有(不考虑立体异构())A .9种B .6种C .5种D .4种6、氮氧化物具有不同程度的毒性,利用构成电池方法既能实现有效消除氮氧化物的排放,减轻环境污染,又能充分利用化学能,发生反应 6NO 2+ 8NH 3= 7N 2+12H 2O ,装置如图所示。
绍兴市第一中学初中英语七年级下册期末基础卷(含答案解析)
一、选择题1.---_________do you usually go to school, Simon?---By bike.A.What B.When C.How C解析:C【详解】句意:——西蒙,你通常怎样去上学?——骑自行车。
考查特殊疑问词辨析。
A. What什么,对于事物的提问;B. When什么时候,对于时间的提问;C. How如何、怎样,对于方式的提问。
根据答语By bike.可知上文是问“西蒙,你通常怎样去上学?”所以应用how提问,故答案选C。
2.Most of the students are having a great time in the Art Festival.A.singing B.sang C.sing D.to sing A解析:A【解析】句意:大多数学生们在艺术节上唱歌玩得非常开心。
singing唱歌,动名词形式;sang是sing的过去式;sing唱歌,动词原形;to sing动词不定式。
句中使用了句型have a great time doing sth.做某事很开心,由这个句型可知应选A。
3.Tom is good at the violin and he is good old people.A.play; for B.to play; at C.playing; with C解析:C【解析】句意:汤姆很擅长拉小提琴,而且他与老人相处的很好。
考查固定搭配。
be good at doing sth.擅长做某事;be good with与…相处的好;故答案选C。
4.There are no_________ now, so we can’t ___________ the river.A.boats; cross B.boats; across C.train; across A解析:A【解析】句意:现在没有船了,所以我们不能过河。
考查名词和动词。
根据so we can’t ____ the river.“我们不能过河”可知是没有船了,所以第一个空应填boats,排除C选项;第二个空由题干可知句中缺少谓语动词,cross是动词,“穿过、越过”,相当于go across;across是介词,意思是“穿过、越过”,所以应填动词cross,前面有情态动词can’t,所以这里用原形形式;故答案选A。
浙江省绍兴市柯桥中学高二数学上学期第一次段考试卷(含解析)
2015-2016学年浙江省绍兴市柯桥中学高二(上)第一次段考数学试卷一、选择题1.(3分)(2014秋•台州期末)直线x+y+3=0的倾斜角是()A.πB.πC. D.2.(3分)(2014秋•台州期末)两平行直线3x﹣4y﹣3=0和6x﹣8y+5=0之间的距离是()A. B. C. D.3.(3分)(2015•青羊区校级模拟)如果实数x,y满足等式(x﹣2)2+y2=3,那么的最大值是()A. B. C. D.4.(3分)(2014春•淮南期末)若直线经过点P(1,1)和点Q(2,t+),其中t>0,则该直线的倾斜角的取值范围是()A.(0,]B.[,)C.(,]D.[,π)5.(3分)(2015秋•绍兴校级月考)圆x2+y2﹣(4m+2)x﹣2my+4m2+4m+1=0的圆心在直线x+y ﹣4=0上,那么圆的面积为()A.9πB.πC.2πD.由m的值而定6.(3分)(2011•湖北校级模拟)已知点A(﹣3,﹣4),B(6,3)到直线l:ax+y+1=0的距离相等,则实数a的值等于()A. B.﹣ C.﹣或﹣D.或7.(3分)(2012•安徽)若直线x﹣y+1=0与圆(x﹣a)2+y2=2有公共点,则实数a取值范围是()A.[﹣3,﹣1]B.[﹣1,3]C.[﹣3,1]D.(﹣∞,﹣3]∪[1,+∞)8.(3分)(2014•城厢区校级一模)已知圆C:x2+y2+mx﹣4=0上存在两点关于直线x﹣y+3=0对称,则实数m的值()A.8B.﹣4C.6D.无法确定9.(3分)(2015•南昌模拟)设两条直线的方程分别为x+y+a=0,x+y+b=0,已知a,b是方程x2+x+c=0的两个实根,且0≤c≤,则这两条直线之间的距离的最大值和最小值分别是()A., B., C., D.,10.(3分)(2009•上海)过圆C:(x﹣1)2+(y﹣1)2=1的圆心,作直线分别交x、y正半轴于点A、B,△AOB被圆分成四部分(如图),若这四部分图形面积满足S|+S IV=S||+S|||则直线AB有()A.0条B.1条C.2条D.3条二、填空题11.(3分)(2015秋•上海校级期中)a、b、c是两两不等的实数,则经过P(b,b+c)、C (a,c+a)两点的直线的倾斜角为.12.(3分)(2016春•苏州期末)求过两点A(0,4),B(4,6)且圆心在直线x﹣2y﹣2=0上的圆的标准方程.13.(3分)(2015秋•绍兴校级月考)已知圆C:x2﹣(1+a)x+y2﹣ay+a=0,若圆C与x轴相切,则圆C的方程为.14.(3分)(2015秋•绍兴校级月考)过点P(1,2)的直线l与圆C:(x+3)2+(y﹣4)2=36交于A,B两点,C为圆心,当∠ACB最小时,直线l的方程是.15.(3分)(2015秋•绍兴校级月考)已知直线: x+y=1(a,b为给定的正常数,θ为参数,θ∈[0,2π))构成的集合为S,给出下列命题:①当θ=时,S中直线的斜率为;②S中的所有直线可覆盖整个坐标平面.③当a=b时,存在某个定点,该定点到S中的所有直线的距离均相等;其中正确的是(写出所有正确命题的编号).三、解答题16.(14分)(2014秋•台州期末)已知两直线x﹣2y+4=0和x+y﹣2=0的交点为P,直线l 过点P且与直线5x+3y﹣6=0垂直.(Ⅰ)求直线l的方程;(Ⅱ)求直线l关于原点对称的直线方程.17.(10分)(2014秋•烟台期末)设直线l的方程为(a+1)x+y+2﹣a=0(a∈R).(1)若l在两坐标轴上的截距相等,求l的方程;(2)若l不经过第二象限,求实数a的取值范围.18.(12分)(2015秋•江西校级期中)已知圆C经过点A(2,﹣1),和直线x+y=1相切,且圆心在直线y=﹣2x上.(Ⅰ)求圆C的方程;(Ⅱ)已知直线l经过原点,并且被圆C截得的弦长为2,求直线l的方程.19.(14分)(2016春•安徽校级月考)如图,函数f(x)=x+的定义域为(0,+∞).设点P是函数图象上任一点,过点P分别作直线y=x和y轴的垂线,垂足分别为M,N.(1)证明:|PM|•|PN|为定值;(2)O为坐标原点,求四边形OMPN面积的最小值.20.(15分)(2015秋•成都期中)已知以点C(t,)(t∈R,t≠0)为圆心的圆与x轴交于点O、A,与y轴交于点O、B,其中O为原点.(Ⅰ)求证:△AOB的面积为定值;(Ⅱ)设直线2x+y﹣4=0与圆C交于点M、N,若丨OM丨=丨ON丨,求圆C的方程;(Ⅲ)在(Ⅱ)的条件下,设P、Q分别是直线l:x+y+2=0和圆C的动点,求丨PB丨+丨PQ丨的最小值及此时点P的坐标.2015-2016学年浙江省绍兴市柯桥中学高二(上)第一次段考数学试卷参考答案与试题解析一、选择题1.(3分)(2014秋•台州期末)直线x+y+3=0的倾斜角是()A.πB.πC. D.【分析】先求出直线的斜率,再求直线的倾斜角.【解答】解:∵直线x+y+3=0的斜率k=﹣,∴直线x+y+3=0的倾斜角α=.故选:A.【点评】本题考查直线的倾斜角的求法,是基础题,解题时要注意直线的性质的合理运用.2.(3分)(2014秋•台州期末)两平行直线3x﹣4y﹣3=0和6x﹣8y+5=0之间的距离是()A. B. C. D.【分析】利用平行线间的距离公式求解.【解答】解:两平行直线3x﹣4y﹣3=0和6x﹣8y+5=0之间的距离:d==.故选:A.【点评】本题考查两平行线间的距离的求法,是基础题,解题时要认真审题,注意距离公式的合理运用.3.(3分)(2015•青羊区校级模拟)如果实数x,y满足等式(x﹣2)2+y2=3,那么的最大值是()A. B. C. D.【分析】表示圆上动点与原点O连线的斜率,画出满足等式(x﹣2)2+y2=3的图形,由数形结合,我们易求出的最大值.【解答】解:满足等式(x﹣2)2+y2=3的图形如图所示:表示圆上动点与原点O连线的斜率,由图可得动点与B重合时,此时OB与圆相切,取最大值,连接BC,在Rt△OBC中,BC=,OC=2易得∠BOC=60°此时=故选D【点评】本题考查的知识点是简单线性规划,分析出表示圆上动点与原点O连线的斜率,是解答本题的关键.4.(3分)(2014春•淮南期末)若直线经过点P(1,1)和点Q(2,t+),其中t>0,则该直线的倾斜角的取值范围是()A.(0,]B.[,)C.(,]D.[,π)【分析】利用直线的斜率公式和均值定理求解.【解答】解:∵直线经过点P(1,1)和点Q(2,t+),其中t>0,∴直线的斜率k==t+﹣1≥2﹣1=1.∴该直线的倾斜角的取值范围是[,).故选:B.【点评】本题考查直线的倾斜角的取值范围的求法,是基础题,解题时要认真审题,注意均值定理的合理运用.5.(3分)(2015秋•绍兴校级月考)圆x2+y2﹣(4m+2)x﹣2my+4m2+4m+1=0的圆心在直线x+y ﹣4=0上,那么圆的面积为()A.9πB.πC.2πD.由m的值而定【分析】由圆的方程求出圆心坐标,代入直线方程求出m的值,求出圆的方程后并配方求圆的半径,代入圆的面积求解即可.【解答】解:∵圆的方程是:x2+y2﹣(4m+2)x﹣2my+4m2+4m+1=0,∴圆心坐标是(2m+1,m),∵圆心在直线x+y﹣4=0上,∴2m+1+m﹣4=0,解得m=1,则圆的方程是:x2+y2﹣6x﹣2y+9=0,即(x﹣3)2+(y﹣1)2=1,∴半径r=1,圆的面积S=πr2=π,故选:B.【点评】本题考查由圆的一般式方程求圆心和半径的方法:公式法和配方法,属于基础题.6.(3分)(2011•湖北校级模拟)已知点A(﹣3,﹣4),B(6,3)到直线l:ax+y+1=0的距离相等,则实数a的值等于()A. B.﹣ C.﹣或﹣D.或【分析】因为A和B到直线l的距离相等,根据点到直线的距离公式列出关于a的方程,求出方程的解即得到a的值.【解答】解:由题意知点A和点B到直线l的距离相等得到=,化简得6a+4=﹣3a﹣3或6a+4=3a+3解得a=﹣或a=﹣.故选C【点评】此题考查学生灵活运用点到直线的距离公式化简求值,做题时注意两种情况.7.(3分)(2012•安徽)若直线x﹣y+1=0与圆(x﹣a)2+y2=2有公共点,则实数a取值范围是()A.[﹣3,﹣1]B.[﹣1,3]C.[﹣3,1]D.(﹣∞,﹣3]∪[1,+∞)【分析】根据直线x﹣y+1=0与圆(x﹣a)2+y2=2有公共点,可得圆心到直线x﹣y+1=0的距离不大于半径,从而可得不等式,即可求得实数a取值范围.【解答】解:∵直线x﹣y+1=0与圆(x﹣a)2+y2=2有公共点∴圆心到直线x﹣y+1=0的距离为∴|a+1|≤2∴﹣3≤a≤1故选C.【点评】本题考查直线与圆的位置关系,解题的关键是利用圆心到直线的距离不大于半径,建立不等式.8.(3分)(2014•城厢区校级一模)已知圆C:x2+y2+mx﹣4=0上存在两点关于直线x﹣y+3=0对称,则实数m的值()A.8B.﹣4C.6D.无法确定【分析】因为圆上两点A、B关于直线x﹣y+3=0对称,所以直线x﹣y+3=0过圆心(﹣,0),由此可求出m的值.【解答】解:因为圆上两点A、B关于直线x﹣y+3=0对称,所以直线x﹣y+3=0过圆心(﹣,0),从而﹣+3=0,即m=6.故选C.【点评】本题考查圆的性质和应用,解题时要认真审题,仔细解答.9.(3分)(2015•南昌模拟)设两条直线的方程分别为x+y+a=0,x+y+b=0,已知a,b是方程x2+x+c=0的两个实根,且0≤c≤,则这两条直线之间的距离的最大值和最小值分别是()A., B., C., D.,【分析】利用方程的根,求出a,b,c的关系,求出平行线之间的距离表达式,然后求解距离的最值.【解答】解:因为a,b是方程x2+x+c=0的两个实根,所以a+b=﹣1,ab=c,两条直线之间的距离d=,d2==,因为0≤c≤,所以≤1﹣4c≤1,即d2∈[,],所以两条直线之间的距离的最大值和最小值分别是,.故选:A.【点评】本题考查平行线之间的距离的求法,函数的最值的求法,考查计算能力.10.(3分)(2009•上海)过圆C:(x﹣1)2+(y﹣1)2=1的圆心,作直线分别交x、y正半轴于点A、B,△AOB被圆分成四部分(如图),若这四部分图形面积满足S|+S IV=S||+S|||则直线AB有()A.0条B.1条C.2条D.3条【分析】由圆的方程得到圆心坐标和半径,根据四部分图形面积满足S|+S IV=S||+S|||,得到S IV﹣S II=SⅢ﹣S I,第II,IV部分的面积是定值,所以三角形FCB减去三角形ACE的面积为定值即SⅢ﹣S I为定值,所以得到满足此条件的直线有且仅有一条,得到正确答案.【解答】解:由已知,得:S IV﹣S II=SⅢ﹣S I,由图形可知第II,IV部分的面积分别为S正方形OECF﹣S扇形ECF=1﹣和S扇形ECF=,所以,S IV﹣S II为定值,即SⅢ﹣S I为定值,当直线AB绕着圆心C移动时,只可能有一个位置符合题意,即直线AB只有一条.故选B.【点评】此题考查学生掌握直线与圆的位置关系,会求三角形、正方形及扇形的面积,是一道综合题.二、填空题11.(3分)(2015秋•上海校级期中)a、b、c是两两不等的实数,则经过P(b,b+c)、C(a,c+a)两点的直线的倾斜角为.【分析】由直线经过P(b,b+c)、C(a,c+a)两点,能求出直线AB的斜率,从而能求出直线AB的倾斜角.【解答】解:∵直线经过P(b,b+c)、C(a,c+a)两点,∴直线AB的斜率k==1,∴直线AB的倾斜角α=;故答案为:.【点评】本题考查直线的倾斜角的求法,是基础题.解题时要认真审题,仔细解答,注意合理地进行等价转化.12.(3分)(2016春•苏州期末)求过两点A(0,4),B(4,6)且圆心在直线x﹣2y﹣2=0上的圆的标准方程(x﹣4)2+(y﹣1)2=25 .【分析】由圆心在直线x﹣2y﹣2=0上,可设圆心坐标为(2b+2,b),再根据圆心到两点A (0,4)、B(4,6)的距离相等,求出b的值,可得圆心坐标和半径,从而求得圆的标准方程.【解答】解:由于圆心在直线x﹣2y﹣2=0上,可设圆心坐标为(2b+2,b),再根据圆过两点A(0,4),B(4,6),可得[(2b+2)﹣0]2+(b﹣4)2=[(2b+2)﹣4]2+(b ﹣6)2,解得b=1,可得圆心为(4,1),半径为=5,故所求的圆的方程为(x﹣4)2+(y﹣1)2=25,故答案为:(x﹣4)2+(y﹣1)2=25.【点评】本题主要考查圆的标准方程的求法,求出圆心的坐标,是解题的关键,属于中档题.13.(3分)(2015秋•绍兴校级月考)已知圆C:x2﹣(1+a)x+y2﹣ay+a=0,若圆C与x轴相切,则圆C的方程为.【分析】把圆的一般方程化为标准形式,根据圆心到x轴的距离等于半径,求得a的值,可得圆C的方程.【解答】解:圆C:x2﹣(1+a)x+y2﹣ay+a=0,即+=,由圆C与x轴相切,可得||=,求得a=1,故圆C的方程为,故答案为:.【点评】本题主要考查圆的一般方程的特征,直线和圆的位置关系,属于基础题.14.(3分)(2015秋•绍兴校级月考)过点P(1,2)的直线l与圆C:(x+3)2+(y﹣4)2=36交于A,B两点,C为圆心,当∠ACB最小时,直线l的方程是2x﹣y=0 .【分析】当直线AB与直线CP垂直时,∠ACB最小,由M与C的坐标求出直线CP的斜率,利用两直线垂直时斜率的乘积为﹣1求出直线AB的斜率,由P坐标与求出的斜率即可得出此时直线l的方程.【解答】解:圆C:(x+3)2+(y﹣4)2=36的圆心坐标C为(﹣3,4),∵P(1,2),∴k CP==﹣,∴k AB=2,则此时直线l的方程为y﹣2=2(x﹣1),即2x﹣y=0.故答案为:2x﹣y=0【点评】此题考查了直线与圆的位置关系,涉及的知识有:圆的标准方程,点到直线的距离公式,难度不大,属于基础题.15.(3分)(2015秋•绍兴校级月考)已知直线: x+y=1(a,b为给定的正常数,θ为参数,θ∈[0,2π))构成的集合为S,给出下列命题:①当θ=时,S中直线的斜率为;②S中的所有直线可覆盖整个坐标平面.③当a=b时,存在某个定点,该定点到S中的所有直线的距离均相等;其中正确的是③(写出所有正确命题的编号).【分析】①当θ=时,S中直线的斜率为k=﹣;②(0,0)不满足方程x+y=1,S中的所有直线不可能覆盖整个平面;③当a=b时,方程变为xsinθ+ycosθ=a,存在定点(0,0),该定点到S中的所有直线的距离均相等.【解答】解:①当θ=时,直线方程变为S中直线的斜率为k=﹣,故①错误;②∵(0,0)不满足方程x+y=1,∴S中的所有直线不可覆盖整个平面,故②错误;③当a=b时,方程为xsinθ+ycosθ=a,存在定点(0,0),该定点到S中的所有直线的距离均相等,故③正确.故答案为:③.【点评】本题考查命题的真假判断与应用,考查直线系方程的应用,要明确直线系中直线的性质,结合三角函数的性质,判断各个命题的正确性,是中档题.三、解答题16.(14分)(2014秋•台州期末)已知两直线x﹣2y+4=0和x+y﹣2=0的交点为P,直线l 过点P且与直线5x+3y﹣6=0垂直.(Ⅰ)求直线l的方程;(Ⅱ)求直线l关于原点对称的直线方程.【分析】(Ⅰ)联立方程组可得交点P的坐标,由垂直关系可得直线的斜率,可得点斜式方程,化为一般式即可;(Ⅱ)由题意和对称性可得(0,﹣2)在要求的直线上,斜率为,同(Ⅰ)可得.【解答】解:(Ⅰ)联立方程组,解得,∴直线x﹣2y+4=0和x+y﹣2=0的交点P(0,2),又∵直线5x+3y﹣6=0的斜率为,∴直线l的斜率为,∴直线l的方程为y﹣2=(x﹣0),化为一般式可得3x﹣5y+10=0;(Ⅱ)由题意和对称性可得直线l上的点P(0,2)关于原点的对称点(0,﹣2)在要求的直线上,由对称可得要求的直线与l平行,故斜率也为,∴直线l关于原点对称的直线方程为y+2=x,化为一般式可得3x﹣5y﹣10=0【点评】本题考查直线的一般式方程和垂直关系,涉及直线的对称性,属中档题.17.(10分)(2014秋•烟台期末)设直线l的方程为(a+1)x+y+2﹣a=0(a∈R).(1)若l在两坐标轴上的截距相等,求l的方程;(2)若l不经过第二象限,求实数a的取值范围.【分析】(1)先求出直线l在两坐标轴上的截距,再利用 l在两坐标轴上的截距相等建立方程,解方程求出a的值,从而得到所求的直线l方程.(2)把直线l的方程可化为 y=﹣(a+1)x+a﹣2,由题意得,解不等式组求得a的范围.【解答】解:(1)令x=0,得y=a﹣2.令y=0,得(a≠﹣1).∵l在两坐标轴上的截距相等,∴,解之,得a=2或a=0.∴所求的直线l方程为3x+y=0或x+y+2=0.(2)直线l的方程可化为 y=﹣(a+1)x+a﹣2.∵l不过第二象限,∴,∴a≤﹣1.∴a的取值范围为(﹣∞,﹣1].【点评】本题考查直线在坐标轴上的截距的定义,用待定系数法求直线的方程,以及确定直线位置的几何要素.18.(12分)(2015秋•江西校级期中)已知圆C经过点A(2,﹣1),和直线x+y=1相切,且圆心在直线y=﹣2x上.(Ⅰ)求圆C的方程;(Ⅱ)已知直线l经过原点,并且被圆C截得的弦长为2,求直线l的方程.【分析】(Ⅰ)设出圆心的坐标为(a,﹣2a),利用两点间的距离公式表示出圆心到A的距离即为圆的半径,且根据圆与直线x+y=1相切,根据圆心到直线的距离等于圆的半径列出关于a的方程,求出方程的解得到a的值,确定出圆心坐标,进而求出圆的半径,根据圆心和半径写出圆的标准方程即可.(Ⅱ)分类讨论,利用被圆C截得的弦长为2,求出直线的斜率,即可求直线l的方程.【解答】解:(Ⅰ)设所求圆心坐标为(a,﹣2a)由条件得=,化简得a2﹣2a+1=0,∴a=1,∴圆心为(1,﹣2),半径r=∴所求圆方程为(x﹣1)2+(y+2)2=2(Ⅱ)①当直线l的斜率不存在时,直线l的方程为x=0,此时直线l被圆C截得的弦长为2,满足条件.②当直线l的斜率存在时,设直线l的方程为y=kx,由题得=1,解得k=﹣,∴直线l的方程为y=﹣x.综上所述:直线l的方程为x=0或y=﹣x.【点评】本题考查了直线与圆的位置关系,涉及的知识有两点间的距离公式,点到直线的距离公式,圆的标准方程,当直线与圆相切时,圆心到直线的距离等于圆的半径,常常利用此性质列出方程来解决问题.19.(14分)(2016春•安徽校级月考)如图,函数f(x)=x+的定义域为(0,+∞).设点P是函数图象上任一点,过点P分别作直线y=x和y轴的垂线,垂足分别为M,N.(1)证明:|PM|•|PN|为定值;(2)O为坐标原点,求四边形OMPN面积的最小值.【分析】(1)根据条件,设出P的坐标,求出|PM|•|PN|,判断是否为定值即可.(2)根据条件将四边形OMPN分解为两个三角形OPM和OPN,分别表示出两个三角形的面积,利用基本不等式的性质进行求最值.【解答】解:(1)设P的坐标为(m,n)(m>0),则有n=m+,即有n﹣m=,由点到直线的距离公式得|PM|==,|PN|=m,即|PM|•|PN|=1,即|PM|•|PN|为定值1;(2)由题意可设M(t,t),知N(0,n),由PM与直线y=x垂直,知k PM=﹣1,即=﹣1,又n=m+解得t=m+,故|OM|=t=m+,∴S△OPM=•|OM|•|PM|=(m+)•=(),S△OPN=•|ON|•|PN|==(m2+),∴S OMPN=S△OPM+S△OPN=(2+m2+)≥(2+2)=,当且仅当m2=,即m=1时等号成立,故四边形面积有最小值+1.【点评】本题主要考查曲线和方程,以及点到直线的距离公式的应用,利用基本不等式是解决本题的关键,涉及的知识点较多,综合性较强,运算量较大.20.(15分)(2015秋•成都期中)已知以点C(t,)(t∈R,t≠0)为圆心的圆与x轴交于点O、A,与y轴交于点O、B,其中O为原点.(Ⅰ)求证:△AOB的面积为定值;(Ⅱ)设直线2x+y﹣4=0与圆C交于点M、N,若丨OM丨=丨ON丨,求圆C的方程;(Ⅲ)在(Ⅱ)的条件下,设P、Q分别是直线l:x+y+2=0和圆C的动点,求丨PB丨+丨PQ丨的最小值及此时点P的坐标.【分析】(Ⅰ)根据题意写出圆C的方程,整理后分别令y=0与x=0求出对应的x与y的值,确定出A与B坐标,求出三角形AOB面积,即可得证;(Ⅱ)根据|OM|=|ON|,得到O在MN的中垂线上,设MN中点为H,得到CH与MN垂直,进而确定出C,H,O共线,求出直线OC斜率,得到t的值确定出圆心C坐标,即可得到圆C 的方程;(Ⅲ)找出B关于x+y+2=0的对称点B′坐标,利用三角形两边之和大于第三边求出|PB|+|PQ|的最小值,以及此时直线B′C的方程,即可求出交点P坐标.【解答】解:(Ⅰ)由题设知,圆C的方程为(x﹣t)2+(y﹣)2=t2+,化简得x2﹣2tx+y2﹣y=0,当y=0时,x=0或2t,则A(2t,0);当x=0时,y=0或,则B(0,),∴S△AOB=|OA|•|OB|=×|2t|×||=4为定值;(II)∵|OM|=|ON|,∴原点O在MN的中垂线上,设MN的中点为H,则CH⊥MN,∴C、H、O三点共线,则直线OC的斜率k===,∴t=2或t=﹣2,∴圆心C(2,1)或C(﹣2,﹣1),∵当圆方程为(x+2)2+(y+1)2=5时,直线2x+y﹣4=0到圆心的距离d>r,此时不满足直线与圆相交,故舍去;∴圆C的方程为(x﹣2)2+(y﹣1)2=5;(Ⅲ)点B(0,2)关于直线x+y+2=0的对称点为B′(﹣4,﹣2),则|PB|+|PQ|=|PB′|+|PQ|≥|B′Q|,又B′到圆上点Q的最短距离为|B′C|﹣r=﹣=3﹣=2,∴|PB|+|PQ|的最小值为2,直线B′C的方程为y=x,则直线B′C与直线x+y+2=0的交点P的坐标为(﹣,﹣).【点评】此题考查了圆的标准方程,两点间的距离公式,对称的性质,三角形的三边关系,以及两直线的交点坐标,熟练掌握公式及性质是解本题的关键.。
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浙江省绍兴市第一中学2014-2015学年高二英语上学期期末考试试题第一部分:听力(共10小题;每小题1分,满分10分)第一节听下面5段对话。
每段对话后有一道小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1.What do we know about the man?A.He doesn’t like chocolate.B.He had a cold last week.C.He doesn’t want to eat ice cream.2.Who is the woman?A. A journalist.B. A teacher.C. A writer.3.What does the woman mean?A.She doesn’t want to go to the game.B.She will only go if Tammy can come.C.The man should go with Tammy instead.4.How many people are in the group?A. Four.B. Six.C. Eight.5.What time is it now?A. 8:45a.m.B. 8:50a.m.C. 10:00a.m.第二节听下面2段对话。
每段对话后有2至4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。
听每段对话前,你将有5秒钟的时间阅读各个小题;听完后,各小题将给出5秒钟的作答时间。
每段对话读两遍。
听第6段材料,回答第6、7题。
6.What does the woman ask the man about first?A. Some text messages.B. Some voicemails.C. Some phone calls.7.Why didn’t the man text the woman?A. He didn’t take his phone with him.B. He got distracted(分心) by something else.C. He was phoning his friends.听第7段材料,回答第8至10题。
8.What does the woman say about Coke and beer?A. They’re not healthy.B. They don’t taste good.C. She’ll buy some next time.9.What do we know about Almond Breeze?A. It’s from China.B. It looks more like juice.C. It’s good for human body.10.What did the man drink this morning?A. Coffee.B. Water.C. Juice.11.---Well, guess what! I came across my primary school teacher when I was visiting the Frick Collection in Manhattan.--- Really? ________.A. It’s really a piece of cake.B. It’s a small world.C. It’s a long time.D. It’s a once-in-a-lifetime chance.12. Some experts say ____ birthrate of China, after the one-child policy is relaxed,can see ___ modest increase in future.A. the, aB. a, aC. a, /D. the, /13. To keep healthy, Professor Johnson________ cycling as a regular form of exercise after he retired.A. took upB. caught onC. carried outD. made for14. _____ to sunlight for too much time will do harm to one’s skin.A. Being exposedB. Having exposedC. ExposedD. After being exposed15. Find information in today’s world is not difficult. The _____ is how you can tell if the information is reliable or not.A. abilityB. competitionC. challengeD. knowledge16. If a ball is placed near a fire, the ball will _______.A. exploreB. extendC. exchangeD. expand17. Lin Tao, who is said to have settled in the UAS recently, ______ in my neighborhood for four years.A. livedB. has livedC. had livedD. was living18. The impatient driver received a ticket for speeding. He _______ so fast.A. mustn’t have drivenB. wouldn’t have drivenC. couldn’t have driveD. shouldn’t have driven20. To make your argument more ________, you should put forward more supporting information.A. convincedB. convincingC. to be convincedD. to be convincing21. Only when the war was over in 1945 ______ able to get back to work.A. was heB. did heC. could heD. could he be22. He has worked out exactly what his income will have to be during the first six months to _____ his costs.A. spareB. coverC. saveD. spend23. It’s our presence rather than our presents ____ really counts.A. whichB. whatC. whyD. that24. It’s greatly ______ that Arthur gave back the money he found.A. in his honorB. to his creditC. for his referenceD. in his power25. Sometimes to get out of difficulty, one must change his way of thinking, or simply change his ____ towards a problem.A. emotionB. approachC. purposeD. sense26. Miss Green won’t be available at 9 tomorrow morning, for she ____ a speech at that time.A. will be deliveringB. deliveredC. has deliveredD. was delivering27. He _____ to participate in the school debate, but on second thoughts, he gave up the idea.A. intendedB. has intendedC. had intendedD. would intend28. ---I’m afraid this question is beyond me.---______. Tell me what you think. There’s no wrong answer.A. You are jokingB. Come onC. By no meansD. All right29. Maintaining(维持) peace and stability of the South China Sea and the East China Sea will certainly ____ a more promising Asian-Pacific area.A. contribute toB. belong toC. compare toD. devote to30. The printer is of good quality. _____ it break down within the first year, we would repair it at our expense.A. WouldB. CouldC. MightD. Should第二节完形填空(共20小题;每小题1.5分,满分30分)In 1972, I returned to Miami Beach High School to speak to the drama class. Afterward I asked the drama teacher 31 any of my English teachers are still there. Irene Roberts, he tells me, is in the class 32 down the hall.I was no one special in Miss Roberts’ class — just another student who did okay work. I don’t recall any one special bit of wisdom she passed on. Yet I cannot forget her 33 for language, for ideas and for her students. I 34 now, many years later, that she is the perfect example of a 35 teacher. I’d like to say something to her, I say, but I don’t want to 36 her from a class. Nonsense, he says, she’ll be 37 to see you.And there in the hallway, this lovely woman, now nearing 43 age, this teacher who doesn’t remember me, begins to weep; and she encircles me in her arms.44 this moment, I begin to sense that everything I will ever know, everythingI will ever pass to my students, is an inseparable part of a legacy(遗产) of our ancestors.Irene Roberts holds me 45 in her arms and through her tears whispers 46 my cheek, “Thank you.” And then, with the briefest of looks into my forgotten face, she 47 back into her classroom, 48 to what she has done thousands of days through all the years of my 49 .Looking back, maybe those were, 50 , just the right words to say to Irene Roberts. Maybe they are the very words I would like to speak to all those teachers through my life, the very words I would like spoken to me one day by some returning student: “I want you to know you were important to me.”31. A. that B. if C. as D. when32. A. just B. almost C. nearly D. about33. A. kindness B. respect C. friendship D. devotion34. A. know B. learn C. realize D. believe35. A. selfish B. self-confident C. self-conscious D. selfless36. A. call B. force C. push D. pull37. A. upset B. delighted C. interested D. bored38. A. brings B. takes C. fetches D. introduces39. A. bows B. raises C. rises D. puts40. A. thought B. brain C. attention D. memory41. A. announce B. speak C. deliver D. tell42. A. useful B. important C. hopeful D. astonishing43. A. retirement B. enjoyment C. employment D. amusement44. A. Reminding B. Explaining C. Remembering D. ForgettingB. fortunatelyC. calmlyD. briefly45. A.particularly46. A. against B. with C. off D. beyond47. A. escapes B. gathers C. disappears D. fails48. A. longs B. continues C. goes D. returns49. A. absence B. function C. explanation D. focus50. A. or rather B. in addition C. as usual D. after all第三部分阅读理解(共15小题;每小题2分,满分30分)AThere it is - your selfie.Over the past year, "selfie" has become a well-known term across the globe. This August the Oxford dictionary added the word to their online dictionary and define it as: "A photograph that one has taken of oneself, typically with a smartphone or webcam and uploaded to a social media website."Today it's not difficult to find social networking pages full of photos people have taken of themselves and their friends. And selfie culture has become especially relevant for young people. As many as 91 percent of teenagers have posted photos of themselves online, according to a recent survey by the US Pew Research Center.So what are the reasons for the rise of selfie culture?"The craziness about the selfie celebrates regular people," Pamela Rutledge, a professor at the Massachusetts School of Professional Psychology, told Vogue magazine. "There are many more photographs available now of real people than models."Posting selfies also allows you to control your image online. "I like having the power to choose how I look, even if I'm making a funny face:' Samantha Barks, 19, a high school student in the US, told Vogue.In addition to self-expression and documentation (记录), selfies "allow for a close friendship for long-distance friends, because you can see each other's faces every day", wrote Casey Miller at The Huffington Post.But US psychologist Jill Weber is concerned that selfies might lead to socialproblems. "There's a danger that your self-esteem (自尊心) may start to be tied to the comments and '1ikes' you get when you post a selfie, and they aren't based on who you are - they're based on what you look like," Weber told Vogue. "When you get nothing or a negative response, your confidence can decrease."51. Why does the writer mention the fact that "selfie" was added to an online dictionary?A. To inform readers where to find the meaning of the word.B. To recommend the dictionary to readers.C. To show that the online dictionary is widely used.D. To indicate that the word has become very popular.52. According to the passage, "selfie" mainly involves _______.A. young peopleB. modelsC. studentsD. celebrities53. How many reasons are mentioned for the rise of selfie?A.2.B.3.C.4.D.5.54, Many people post photos of themselves online to _________.A. make fun of themselvesB. become powerfulC. beautify themselvesD. choose how they look55. In Jill Weber's opinion, selfies may cause one to _________.A. be cheatedB. feel discouragedC. lose moneyD. succeed more easilyBClose your eyes for a minute and imagine what life would be like if you had a hundred dollars less. Also imagine what it would be like spending the rest of your life with you eyes closed. Imagine having to read this page, not with your eyes but with your fingertips.With existing medical knowledge and skills, two-thirds of the world’s 42 million blind should not have to suffer. Unfortunately, rich countries posses most of this knowledge, while developing countries do not.ORBIS is an international non-profit organization which operates the world’s only flying teaching eye hospital. ORBIS intends to help fight blindness worldwide. Inside a DC-8 aircraft, there is a fully-equipped teaching hospital with television studio and classroom. Doctors are taught the latest techniques of bringing sight back to people there. Project ORBIS also aims at promoting peaceful cooperation among countries.For just US$38,you can help one person see; for $380 you can bring sight to 10 people; $1,300 helps teach a doctor new skills; and for $13,000 you can provide a training program for a group of doctors who can make thousands of blind people seeagain. Your money can open their eyes to the world. Please help ORBIS improve the quality of life for so many people less fortunate than ourselves.56.The first paragraph is intended to ______.A. introduce a new way of readingB. advise the public to lead a simple lifeC. direct the public’s attention to the blindD. encourage the public to use imagination57.What do we learn about existing medical knowledge and skills in the world?A.They are enough.B. They are not equally distributed(分布,分配).C. They have not been updated.D. They have benefited most of the blind.58.ORRIS aims to help the blind by ______.A.teaching medical studentsB. training doctors and nursesC. running flying hospitals globallyD. setting up non-profit organization59.What can be the best title for the passage?A. ORRIS in ChinaB. Fighting BlindnessC. Sight-seeing TechniquesD. ORRIS Flying HospitalCYou may have heard of the man who decided to repair the roof of his house. To be safe, he tied a rope around his waist and threw the other end of it over the top of the house. He asked his son to tie it to something secure. The boy fastened the safety rope to the bumper(保险杠) of their car parked in the driveway. It seemed like a good idea at the time. But a little while later, his wife, unaware of the rope securing her husband, she started the car and drove away. Imagine what the result would be.This story, factual or not, points to a great truth. It is a truth about where we place our security; about those things to which we’ve tied our safety lines. What is your rope tied to? Think about it. What do you depend on to keep you from disaster? Is your rope tied to a good job? Is it tied to a relationship with somebody you rely on? Is it tied to a company or an organization?Thoroughly exhausted, she thought that maybe she should be listening for rescuers instead of crying. So she grew still and listened. In the silence around her, the only sound she heard was the beating of her own heart. It occurred to her then that at least she was still alive and, amazingly enough, unhurt.She thought about her situation. In the stillness, fear abandoned her and afeeling of indescribable peace and happiness flooded in, the likes of which she had never before known. It was an experience that was to change her life forever.In the deepest part of her being, Susan realized a remarkable truth. She realized she had nothing to fear. Amazingly, whether or not she was ever rescued, whether she ever made it out alive, she sensed she had nothing to fear.For the first time in her life she understood that her true security did not depend on those things in which she had placed her trust. It lay deep within. And also for the first time, she knew what it was to be content in all circumstances. She realized that whether she had plenty or hardly enough, somehow she would be all right. She just knew it. She later wrote, “Before the quake I appeared very successful, but my life was out of balance. I wasn’t happy because I had been making money and always wanting more. My home, my job, my clothes, a relationship — I thought they were my security. It took an earthquake and losing everything I owned for me to discover that my security had been with me all along…There’s a power within us that we can depend upon no matter what is happening around us.”She had tied her rope to the wrong things. It took a disaster for her to understand that those things are not trustworthy. So she let go of the rope and discovered peace. She found that her true security was a power within- dependable and sure.What is your rope tied to? And what would happen if you found the courage to let go of it?60. In Paragraph 1, the writer tells the story of the man to___________.A. describe a sceneB. introduce a topicC. of fer an argumentD. satisfy the readers’ curiosity61. If the story of the man was true, the most probable result would be that___________.A. the bumper of the car broke upB. the house was destroyed by the carC. the man was badly injured or even killedD. the wife was scolded by her husband62. Why was Susan crying in the darkness?A. Because she was afraid of staying in the dark.B. Because she wanted to be heard by the rescuers.C. Because she was frightened and felt helpless.D. Because she had tied her rope to the wrong thing.63. What did Susan feel after she thought about her situation?A. She felt she was still alive.B. She felt very peaceful.C. She felt completely hopeless.D. She wanted to change her life forever.64. How did Susan like interpersonal relationship and income before the earthquake?A. They were meaningless.B. They were very important.C. They were unnecessary.D. They were among some choices.65. Which of the following should be the best title?A. An Experience from an Earthquake.B. A Rope Tied to a Car.C. Please Let Go of Your Trust.D. What to Depend on.第四部分写作(共两节,满分20分)第一节:短文改错(共10小题;每小题1分,满分10分)文中共有10处错误,每句中最多有两处。