2019-2020高一上学期期中考试试卷及答案
北京八中2019-2020高一上期中测试含答案

北京八中2019-2020学年度第一学期期中练习题年级:高一科目:生物考试时间90分钟,满分100分一、单选题(1-30每小题1分,31-40每小题2分,共50分)1.细胞学说揭示了A.植物细胞与动物细胞的区别B.生物体结构的统一性C.细胞为什么能产生新细胞D.认识细胞经历了曲折过程2.从生命系统的结构层次看,人体所具有的最低层次和最高层次分别是A.生物大分子和细胞B.细胞和生物圏C.细胞和个体D.个体和生态系统3.美国细胞生物学家成尔逊曾经说过:“每一个生物科学问题的答案都必须在细胞中寻找”。
他作出这一结论的理由最可能是A.新细胞可以从老细胞中产生B.细胞是一切生物体结构和功能的基本单位C.细胞是一个完全独立的单位,独立完成生命活动D.生物的生命活动是在细胞内或细胞参与下完成的4.生态系统存在着从细胞到生物圈各个不同的结构层次。
下列相关叙述错误的是A.细胞是基本的生命系统B.草履虫可以看作是基本的生命系统C.植物体和动物体共有的生命系统层次有细胞、组织、器官、个体D.生态系统中存在非生命的物质和成分,不属于生命系统5.蓝细菌与酵母菌在结构上的最明显区别是A.前者无细胞结构B.前者无成形的细胞核C.后者有核糖体D.后者有细胞壁6.2017年中秋节,一位忠糖尿病的韩奶奶因食用“无糖月併”而被“甜晕”,还好抢救及时脱离危险。
部分商家广告语存在科学性错误,容易误导消费者。
你认为下列描述正确的是A.某品牌无糖饼干没有甜味,属于无糖食品B.某品牌口服液含有丰富的N、P、Zn等微量元素C.某地大棚菜,天然种植,不含任何化学元素,是真正的绿色食品D.某品牌鱼肝油,含有丰富的维生素有助于宝宝骨骼健康7.植物细胞中重要的多糖是A.葡萄糖B蔗糖和麦芽糖 C.纤维素和淀粉 D.糖元和淀粉8.给低血糖休克病人静脉注射50%的葡萄糖溶液,其主要目的是A.供给全面营养B.供给能源C.维持细胞内液浓度D供给水分9.碳元素是构成细胞的最基本元素,对此最有说服力的解释是A.碳在细胞的各种化合物中含量最多B.碳在自然界中含量最为丰富C.细胞的各种化合物中都含有碳D.碳链构成了有机物的基本骨架10.下列选项中,属于动植物细胞共有的糖类是A葡萄糖、核糖、脱核糖 B.葡萄糖、淀粉和果糖C.粉、脱氧核糖、乳糖D.麦芽糖、果糖、乳糖11.下列关于细胞中化合物的叙述中,不正确的是A.细胞中无机盐含量很少且大多数是以离子的形式存在B.脂肪、磷脂、胆固醇、性激素、维生素D都属于脂质C.动物乳汁中的乳糖和植物细胞中的纤维素都属于多糖D.新陈代谢越旺盛,细胞中的自由水与结合水的比值越高12.已知Mn2+是许多酶的活化剂,如它能激活硝酸还原酶,使缺Mn2+的植物无法利用硝酸盐。
精品解析:上海市金山中学2019-2020学年高一上学期期中考试英语试题(解析版)

2. A. $6B. $ 9C. $ 15D. $18
3. A. A salesmanB.A lawyerC. A journalistD. An employer
4. A. In the hospital.B. At the airport.C. In the school.D. In the prison.
Questions 11 through 13 are based on the following passage.
11. A. Solar system.B. English people.C.A leaf.D. The sky
12. A. Fox Talbot.B. Larry Schaaf.C. Thomas Wedgwood, D. New York.
The British often disagrees about the best way___6___(educate)their children. Many people say that comprehensive schools help more children to succeed because they provide everybody___7___similar opportunities in a fairer way. Another view, though, is that more intelligent children, especially___8___from poor homes are better supported at grammar schools. Now, the government plans to open new grammar schools___9___almost two million children will go to the same type of school that Paul attended.
黑龙江省哈师大附中2019-2020学年高一上学期期中物理试卷 (含答案解析)

黑龙江省哈师大附中2019-2020学年高一上学期期中物理试卷一、单选题(本大题共9小题,共36.0分)1.关于时间和时刻,下列说法正确的是()A. 物体在5s时就是指物体在5s末时,指的是时刻B. 物体在5s时就是指物体在5s初时,指的是时刻C. 物体在5s内就是指物体在4s末到5s末的这1s时间D. 物体在第5s内就是指物体在4s末到5s初的这1s的时间2.下列说法正确的是()A. 木块放在桌面上受到一个向上的弹力,这是由于木块发生微小形变而产生的B. 质量均匀分布、形状规则的物体的重心可能在物体上,也可能在物体外C. 重力的方向总是垂直接触面向下的D. 由磁铁间存在相互作用可知:力可以离开物体而单独存在3.关于摩擦力,下列说法正确的是()A. 人走路前进时,地面给人的摩擦力阻碍人前进B. 擦黑板时,静止的黑板受到的摩擦为滑动摩擦力C. 人握竖直杆向上爬,杆给人的摩擦力向下D. 摩擦力的大小一定与接触面处的压力成正比4.在变速运动中对瞬时速度大小的理解,正确的是()A. 表示物体在某一时刻运动的快慢程度B. 表示物体在某段时间内运动的快慢程度C. 表示物体经过某一路程运动的快慢程度D. 表示物体经过某段位移的运动的快慢程度5.甲、乙两物体从同一位置开始做直线运动的图象如图所示,下列说法正确的是()A. 若y表示位移,则0~t1时间内甲的位移小于乙的位移B. 若y表示位移,则t=t1时甲的速度大于乙的速度C. 若y表示速度,则t=t1时甲的加速度等于乙的加速度D. 若y表示速度,则t=t1时甲、乙两物体相遇6.如图所示,用细线将一小球悬挂在光滑墙壁上,小球的质量为m,若增加细线的长度,以下说法中正确的是()A. 细线的拉力变大B. 墙壁对球的支持力变大C. 墙壁对球的支持力变小D. 细线和墙壁对球的作用力的合力变小7.取一根长2m左右的细线、5个铁垫圈和一个金属盘.在线端系上第一个铁垫圈,隔12cm再系一个,以后铁垫圈之间的距离分别为36cm、60cm、84cm,如图所示.站在椅子上,向上提起线的上端,让线自由垂下,且第一个铁垫圈紧靠放在地面上的金属盘内.松手后开始计时,若不计空气阻力,则第2、3、4、5各铁垫圈()A. 落到盘上的声音时间间隔越来越长B. 落到盘上的声音时间间隔相等C. 依次落到盘上的速率之比为1:√2:√3:2D. 依次落到盘上所用的时间之比为1:(√2−1):(√3−√2):(2−√3)8.有一种“傻瓜”相机的曝光时间(快门从打开到关闭的时间)是固定不变的.为了估测相机的曝光时间,有位同学提出了下述实验方案:他从墙面上A点的正上方与A相距H=1.5m处,使一个小石子自由落下,在小石子下落通过A点后按动快门对小石子照相得到如图所示的照片,由于小石子的运动,它在照片上留下一条模糊的径迹CD.已知每块砖的平均厚度约为6cm,从这些信息估算该相机的曝光时间最近于(取g=10m/s2)()A. 0.5sB. 0.06sC. 0.02sD. 0.008s9.甲、乙两物体在同一条直线上,甲以v=6m/s的速度作匀速直线运动,在某时刻乙以a=3m/s2的恒定加速度从静止开始运动,则()A. 在2s内甲、乙位移一定相等B. 在2s时甲、乙速率一定相等C. 在2s时甲、乙速度一定相等D. 在2s内甲、乙位移大小一定相等二、多选题(本大题共5小题,共20.0分)10.一物体由静止开始以恒定加速度下落,经过时间1s落至地面,落地时速度是9m/s.下列说法中正确的是()A. 物体下落高度为4.5mB. 物体下落高度为4.9mC. 物体下落的加速度为9 m/s2D. 物体下落的加速度为9.8m/s211.如图所示为某物体做直线运动的v−t图象,关于物体在前4s的运动情况,下列说法中正确的是()A. 物体始终向同一方向运动B. 物体的加速度大小不变,方向与初速度方向相同C. 物体在前2s内做减速运动,物体在后2s内做加速运动D. 物体4s末回到了原出发点12.一个小球从某高度作自由落体运动,掉进下方一个深水池中.已知小球受水的阻力与速度大小成正比,不计小球受水的浮力.当小球进入水中后运动的v−t图象,可能正确的有()A. B.C. D.13.有一质点从t=0开始由原点出发,其运动的速度−时间图象如图所示,则()A. t=1s时,质点离原点的距离最大B. t=2s时,质点离原点的距离最大C. t=2s时,质点回到原点D. t=4s时,质点回到原点14.对于匀变速直线运动的速度与时间关系式v=v0+at可以作以下的理解()A. v0是时间间隔t开始的速度,v是时间间隔t结束时的速度,它们均是瞬时速度B. v一定大于v0C. at可以是在时间间隔t内速度的增加量,也可以是时间间隔t内速度的减少量,在匀加速直线运动中at为正值,在匀减速直线运动中at为负值D. a与匀变速直线运动的v−t图象的倾斜程度无关三、实验题(本大题共2小题,共18.0分)15.在如图所示“探究弹力和弹簧伸长关系”的实验中.让刻度尺(分度值是1mm)的零刻度与弹簧上端相平,在弹簧下端挂1个钩码,静止时弹簧长度为______ cm.在弹簧下端不断增加钩码个数,记下每增加一个钩码后对应的钩码总重和弹簧的总长度,已知每个钩码质量为50g,挂4个钩码,静止时,弹簧弹力为______ N(g=9.8m/s2).要得到挂不同钩码时弹簧的伸长量,还需测量的是______ .在坐标纸上作出弹力与弹簧伸长量的图象,可分析出它们的关系.16.某同学利用打点计时器测量福州的重力加速度,某次实验得到的一段纸带如图所示,O、A、B、C、D为相邻的五个点,测得OA=5.5mm,OB=14.9mm,OC=28.3mm,OD=45.2mm,打下相邻两个点间的时间间隔为0.02s.①用逐差法算出福州的重力加速度g=______ m/s2(结果保留三位有效数字)②通过查阅资料发现福州的重力加速度标准值为9.79m/s2,比较①的结果发现两者并不相等,除了读数误差外,你认为产生误差的其它主要原因可能是______ .(只要求写出一种原因)四、计算题(本大题共3小题,共30.0分)17.如图所示,木杆长5m,上端固定在某一点,由静止放开后让它自由落下(不计空气阻力),木杆通过悬点正下方20m处圆筒AB,圆筒AB长为5m,g取10m/s2,则:(1)木杆通过圆筒的上端A所用的时间t1是多少⋅(2)木杆通过圆筒AB所用的时间t2是多少⋅18.如图14所示,质量M=2kg的木块套在水平固定杆上,并用轻绳与质量m=1kg的小球相连,今用跟水平方向成60°角的力F=10√3N拉着小球并带动木块一起向右匀速运动,运动中M、m 的相对位置保持不变,g=10m/s2。
山东省济南2019-2020学年高一上学期期中考试语文试卷含答案

济南2019—2020学年高一上学期期中考试语文试题(2019.11)本试卷共10页,22小题,满分150分,考试用时150分钟。
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。
2.作答选择题时,选出每小题答案后,用2B铅笔把答题卡上对应题目选项的答案信息点涂黑;如需改动,用橡皮擦干净后,再选涂其他答案。
答案不能答在试题卷上。
3.非选择题必须用黑色签字笔作答,答案必须写在答题卡各题目指定区域内相应位置上;如需改动,先划掉原来的答案,然后再写上新答案;不准使用铅笔和涂改液。
不按以上要求作答无效。
一、现代文阅读(36分)(一)论述类文本阅读(本题共3小题,9分)阅读下面的文字,完成1~3题。
孔子认为,修身是从政的前提,欲正人先正己,人君只有加强自身道德修养,以身作则,百姓才会效法学习,社会风气也才会随之改善,社会治理才会出现事半功倍的效果。
所以孔子说:“七教者,治民之本也,教定是正矣。
上者,民之表也,表正则何物不正。
”孔子认为,“七教”是人君治国理民的根本,只有人君坚持德教,整个社会才能和谐有序。
“三至”则体现了孔子礼乐治国的为政理念。
孔子认为,明王只有达到“至礼不让”“至赏不费”“至乐无声”这三种境界,才可以对外征伐。
明王所征伐的对象必定是无道之君,是为了吊民罚罪,必然会得到所征伐之地百姓的拥护和支持。
孔子认为,明王所进行的征伐是战无不胜的,在他看来,征伐有时是必要的,而且是必须的。
孔子还在《大戴礼记·用兵》中把战争分为两种,即“圣人之用兵”和“后世贪者之用兵”,这两种战争的动机和后果是截然不同的。
孔子还认为,明王要达到天下大治,并不意味着一定要劳苦和耗费。
所以《大戴礼记·主言》载:“孔子曰:‘内修七教而上不劳,外行三至而财不费,此之谓明主之道也。
’”这里所提到的“不劳不费”是孔子“无为而治”政治理想的具体化。
文中还提到“昔者舜左禹而右皋陶,不下席而天下治”,这也体现了早期儒家“无为而治”的施政理念。
浙江省温州市2019-2020学年高一上学期期中化学试卷 (含答案解析)

浙江省温州市2019-2020学年高一上学期期中化学试卷一、单选题(本大题共25小题,共50.0分)1.下列物质中,按照只有氧化性、只有还原性、既有氧化性又有还原性的顺序排列的一组是()A. F2、K、HClB. S、Al、H2C. NO2、Na、Br2 D. O2、SO2、H2O2.下列物质中,属于非电解质的是()A. 铜丝B. 氯化钠C. 浓硫酸D. 乙醇3.某城市以“液氯泄漏”作应急预警演习,演习时,下列防护措施合理的是()①及时向上风口和地势较高处转移疏散人群;②及时戴上用浓烧碱溶液湿润过的口罩;③用高压水枪向泄露口附近喷洒稀碱液并设法控制氯气泄漏;④及时戴上用稀纯碱溶液湿润过的口罩;⑤及时用毛巾包裹活性炭捂住口鼻;⑥就近向地下商场转移.A. ②③④⑤B. ①②④⑤C. ①③④⑤D. ①③④⑥4.下列关于钠的说法不正确的是()A. 金属钠和氧气反应,条件不同,产物不同B. 钠钾合金通常状况下呈液态,可作原子反应堆的导热剂C. 钠的化学活泼性很强,少量的钠可保存在煤油中D. 由于钠比较活泼,所以它能从溶液中置换出金属活动顺序表中钠后面的金属5.完全沉淀等物质的量浓度的NaCl、MgCl2、AlCl3溶液中的Cl−,消耗等物质的量浓度的AgNO3溶液的体积比为1∶2∶3,则上述溶液的体积比为()A. 1∶1∶1B. 6∶3∶2C. 3∶2∶1D. 9∶3∶16.在配制一定物质的量浓度的硫酸溶液时,下列错误操作可使所配制溶液浓度偏高的是()A. 浓硫酸稀释后马上转移至容量瓶B. 溶解搅拌时有液体飞溅C. 定容时仰视容量瓶瓶颈刻度线D. 摇匀后见液面下降,再加水至刻度线7.有关Na2CO3和NaHCO3的叙述中正确的是()A. Na2CO3溶液和NaHCO3溶液可用澄清石灰水鉴别B. NaHCO3比Na2CO3热稳定性强C. 相同质量的Na2CO3和NaHCO3与足量盐酸作用时,产生的气体质量相同D. 在饱和的碳酸钠溶液中通入二氧化碳溶液变浑浊8.下列各组离子一定能大量共存的是()A. 在无色溶液中:NH4+、Fe2+、SO42−、CO32−B. 在含大量Ba2+的溶液中:NH4+、Na+、Cl−、OH−C. 在强碱性溶液中:Na+、K+、SO42−、HCO3−D. 在强酸性溶液中:K+、Fe2+、Cl−、SO42−9.亚硝酸盐可将正常的血红蛋白氧化成高铁血红蛋白,失去携氧能力,是人体组织出现缺氧现象.美蓝是亚硝酸盐中毒的有效解毒剂.下列说法不正确的是()A. 在中毒过程中血红蛋白被氧化B. 中毒时亚硝酸盐发生氧化反应C. 药品美蓝应具有还原性D. 解毒时高铁血红蛋白被还原10.下列说法正确的是()A. 酸、碱、盐都是电解质B. 电解质都是易溶于水的物质C. 氧化物都是电解质D. 氢化物都不是电解质11.实验室欲用Na2CO3·10H2O晶体配制1mol/L的Na2CO3溶液100mL,下列说法正确的是()A. 要完成实验需称取10.6gNa2CO3·10H2O晶体B. 本实验需用到的仪器是天平、药匙、玻璃棒、烧杯、100mL容量瓶C. 配制时若容量瓶不干燥,含有少量蒸馏水会导致浓度偏低D. 定容时俯视刻度会导致浓度偏高12.ClO2是一种杀菌消毒效率高、二次污染小的水处理剂。
江苏省徐州市2019~2020学年度高一第1学期期中考试数学试题及参考答案解析

2019~2020学年度江苏省徐州市高一第一学期期中数学试卷一、选择题(本大题共12小题)1.已知集合A={1,3,5},B={3,5,7},则A∩B=( )A.3,5,B.C.D.2.函数f(x)=+ln(1-x)的定义域为( )A. B. C. D.3.已知幂函数f(x)的图象过点(2,16),则f(3)=( )A.27B.81C.12D.44.函数f(x)=a x+1+2(a>0且a≠1)的图象恒过定点( )A. B., C. D.5.设a=logπ3,b=π0.3,c=log0.3π,则( )A. B. C. D.6.已知函数,则的值是( )A.27B.C.D.7.已知函数f(x)=ax5-bx3+cx-3,f(-3)=7,则f(3)的值为( )A.13B.C.7D.8.函数y=(a>1)的图象的大致形状是( )A. B. C. D.9.已知y=f(x)是定义在R上的奇函数,当x<0时,f(x)=x+2,那么不等式2f(x)-1<0的解集是( )A. B.或C. D.或10.已知函数f(x)=x2•(a+)是R上的奇函数,则实数a=( )A. B. C. D.111.若函数f(x)=a x-a-x(a>0且a≠1)在R上为减函数,则函数的单调递增区间( )A. B. C. D.12.若函数f(x)=|lg x|-()x+a有2个零点,则实数a的取值范围是( )A. B. C. D.二、填空题(本大题共4小题)13.已知集合A={-2,0,1,3},B={x|-<x<},则A∩B的子集个数为______.14.若函数f(x)=lg x+x-3的零点在区间(k,k+1),k∈Z,则k=______.15.若函数f(x)=的值域为R,则实数a的范围是______.16.已知函数y=x+有如下性质:常数a>0,那么函数在(0,]上是单调减函数,在[,+∞)上是单调增函数.如果函数f(x)=|x+-m|+m在区间[1,4]上的最小值为7,则实数m的值是______.三、解答题(本大题共6小题)17.计算:(1);(2)2lg5+lg8+lg5•lg20+(lg2)2.18.已知集合A={x|3≤3x≤27},B={x|1<log2x<2}.(1)分别求A∩B,(∁R B)∪A;(2)已知集合C={x|2a<x<a+2},若C⊆A,求实数a的取值范围.19.已知函数f(x)是定义在(-4,4)上的奇函数,满足f(2)=1,当-4<x≤0时,有f(x)=.(1)求实数a,b的值;(2)求函数f(x)在区间(0,4)上的解析式,并利用定义证明函数f(x)在(0,4)上的单调性.20.某公司生产一种化工产品,该产品若以每吨10万元的价格销售,每年可售出1000吨,若将该产品每吨分价格上涨x%,则每年的销售数量将减少mx%,其中m为正常数,销售的总金额为y万元.(1)当m=时,该产品每吨的价格上涨百分之几,可使销售总金额最大?(2)当x=10时,若能使销售总金额比涨价前增加,试设定m的取值范围.21.已知函数f(x)=x|x-a|+x(a∈R)(1)若函数f(x)是R上的奇函数,求实数a的值;(2)若对于任意x∈[1,2],恒有f(x)≥2x2,求实数a的取值范围;(3)若a≥2,函数f(x)在区间[0,2]上的最大值为4,求实数a的值.22.已知函数f(x)=lg(m+),m∈R.(1)当m=-1时,求函数f(x)的定义域;(2)若函数g(x)=f(x)+2x lg2有且仅有一个零点,求实数m的取值范围;(3)任取x1,x2∈[t,t+2],若不等式|f(x1)-f(x2)|≤1对任意t∈[1,2]恒成立,求实数m的取值范围.答案和解析1.【参考答案】C【试题分析】解:∵集合A={1,3,5},B={3,5,7},∴A∩B={3,5}.故选:C.利用交集定义直接求解.本题考查交集的求法,考查交集定义等基础知识,考查运算求解能力,是基础题.2.【参考答案】B【试题分析】解:要使f(x)有意义,则,解得,∴f(x)的定义域为.故选:B.可看出,要使得f(x)有意义,则需满足,解出x的范围即可.本题考查了函数定义域的定义及求法,对数函数的定义域,考查了计算能力,属于基础题.3.【参考答案】B【试题分析】解:设幂函数f(x)=xα,又f(x)过点(2,16),∴2α=16,解得α=4,∴f(x)=x4,∴f(3)=34=81.故选:B.用待定系数法求出f(x)的解析式,再计算f(3)的值.本题考查了幂函数的定义与应用问题,是基础题.4.【参考答案】D【试题分析】解:由x+1=0,解得x=-1,此时y=1+2=3,即函数的图象过定点(-1,3),故选:D.根据指数函数过定点的性质,直接领x+1=0即可得到结论本题主要考查指数函数过定点问题,利用指数幂等于0是解决本题的关键.5.【参考答案】D【试题分析】解:0=logπ1<logπ3<logππ=1,π0.3>π0=1,log0.3π<log0.31=0,∴b>a>c.故选:D.容易得出,从而得出a,b,c的大小关系.考查对数函数、指数函数的单调性,以及增函数和减函数的定义.6.【参考答案】B【试题分析】解:∵∴=f(-3)=故选B.由已知中的函数的解析式,我们将代入,即可求出f()的值,再代入即可得到的值.本题考查的知识点是分段函数的函数值,根据分析函数的解析式,由内到外,依次代入求解,即可得到答案.7.【参考答案】B【试题分析】解:∵函数f(x)=ax5-bx3+cx-3,f(-3)=7,令g(x)=ax5-bx3+cx,则g(-3)=10,又g(x)为奇函数,∴g(3)=-10,故f(3)=g(3)-3=-13,故选:B.令g(x)=ax5-bx3+cx,则g(-3)=10,又g(x)为奇函数,故有g(3)=-10,故f(3)=g(3)-3.本题考查函数的奇偶性的应用,求函数值,令g(x)=ax5-bx3+cx,求出g(3)=-10,是解题的关键.8.【参考答案】C【试题分析】解:当x>0时,y=a x,因为a>1,所以函数y=a x单调递增,当x<0时,y=-a x,因为a>1,所以函数y=-a x单调递减,故选:C.根据函数的单调性即可判断.本题考查了函数图象和识别,关键掌握函数的单调性,属于基础题9.【参考答案】B【试题分析】解:因为y=f(x)为奇函数,所以当x>0时,-x<0,根据题意得:f(-x)=-f(x)=-x+2,即f(x)=x-2,当x<0时,f(x)=x+2,代入所求不等式得:2(x+2)-1<0,即2x<-3,解得x<-,则原不等式的解集为x<-;当x≥0时,f(x)=x-2,代入所求的不等式得:2(x-2)-1<0,即2x<5,解得x<,则原不等式的解集为0≤x<,综上,所求不等式的解集为{x|x<-或0≤x<}.故选:B.根据f(x)为奇函数,得到f(-x)=-f(x),设x大于0,得到-x小于0,代入已知的解析式中化简即可求出x 大于0时的解析式,然后分两种情况考虑,当x小于0时和x大于0时,分别把所对应的解析式代入所求的不等式中,得到关于x的两个一元一次不等式,求出不等式的解集的并集即为原不等式的解集.此题考查了其他不等式的解法,考查了函数奇偶性的应用,是一道基础题.10.【参考答案】A【试题分析】解:根据题意,函数f(x)=x2•(a+)是R上的奇函数,则有f(-x)=-f(x),即(-x)2(a+)=-(x2•(a+),变形可得:a+=-(a+),则有2a=-1,即a=-;故选:A.根据题意,由函数奇偶性的定义可得f(-x)=-f(x),即(-x)2(a+)=-(x2•(a+),变形分析可得a的值,即可得答案.本题考查函数的奇偶性的性质以及应用,关键是掌握函数奇偶性的定义,属于基础题.11.【参考答案】C【试题分析】解:∵函数f(x)=a x-a-x(a>0且a≠1)在R上为减函数,则0<a<1.则函数的单调递增区间,即y=x2+2x-3在y>0时的减区间.由y=x2+2x-3>0,求得x<-3,或x>1.再利用二次函数的性质可得,y=x2+2x-3在y>0时的减区间为(-∞,-3),故选:C.复合函数的单调性,指数函数、二次函数的性质,先判断0<a<1,本题即求y=x2+2x-3在y>0时的增区间,再利用二次函数的性质得出结论.本题主要考查复合函数的单调性,指数函数、二次函数的性质,属于中档题.12.【参考答案】B【试题分析】解:原函数转化为f(x)=|lg x|-()x+a,|lg x|=()x-a,函数有2个零点,相当于y=|lg x|与y=()x-a有两个交点,根据图象:当x=1时,y=()x-a的值-a>0即可所以a∈(-∞,).故选:B.原函数转化为f(x)=|lg x|-()x+a,|lg x|=()x-a,根据图象:当x=1时,y=()x-a的值-a>0即可.把零点问题转换为两个函数的交点问题,考察图象法的应用,中档题.13.【参考答案】8【试题分析】解:∵A={-2,0,1,3},B={x|-<x<},∴A∩B={-2,0,1},∴A∩B的子集个数为:23=8个.故答案为:8.进行交集的运算求出A∩B,从而得出A∩B的元素个数,进而可得出A∩B的子集个数.本题考查了描述法、列举法的定义,交集的运算,集合子集个数的计算公式,考查了计算能力,属于基础题.14.【参考答案】2【试题分析】解:因为函数y=lg x与y=x-3都是定义域上的增函数,所以函数f(x)=lg x+x-3也为定义域上的增函数.因为f(2)=lg2+2-3<lg10+2-3=0,f(3)=lg3+3-3>0,所以由零点存在性定理可得函数f(x)=lg x+x-3的近似解在区间(2,3)上,所以k=2.故答案为:2.确定函数f(x)=lg x+x-3也为定义域上的增函数.计算f(2)=lg2+2-3<lg10+2-3=0,f(3)=lg3+3-3>0,由零点存在性定理可得函数f(x)=lg x+x-3的近似解在区间(2,3)上,即可得出结论.本题考查零点存在性定理,考查学生的计算能力,比较基础.15.【参考答案】[0,+∞)【试题分析】解:x≤1时,f(x)≤2+a;x>1时,f(x)=(x-a)2+1-a2,∴①a>1时,f(x)≥1-a2,且f(x)的值域为R,∴2+a≥1-a2,解得a∈R,∴a>1;②a≤1时,f(x)>(1-a)2+1-a2=2-2a,且f(x)的值域为R,∴2+a≥2-2a,解得a≥0,∴0≤a≤1,∴综上得,实数a的范围是[0,+∞).故答案为:[0,+∞).根据f(x)的解析式得出,x≤1时,f(x)≤2+a;x>1时,f(x)=(x-a)2+1-a2,从而得出:a>1时,f(x)≥1-a2,进而得出2+a≥1-a2;a≤1时,f(x)>2-2a,进而得出2+a≥2-2a,从而解出a的范围即可.本题考查分段函数值域的求法,配方求二次函数值域的方法,考查计算能力,属于中档题.16.【参考答案】6【试题分析】解:设t=在[1,2]上单调递减,在[2,4]上单调递增,所以t∈[4,5],问题化为y=|t-m|+m在区间[4,5]上的最小值为7,当m>5时,y min=y(5)=m-5+m=7,m=6;当m∈[4,5]时,y min=y(m)=m=7(舍去);当m<4时,y min=y(4)=4-m+m=7,不成立.故答案为:6.换元将问题化为绝对值函数在闭区间上的最小值问题,根据对称轴在闭区间的右侧、中间、左侧分三类讨论即可.本题是一个经典题目,通过换元将问题化为绝对值函数在闭区间上的最小值问题,接下来根据对称轴在闭区间的右侧、中间、左侧分三类讨论即可.17.【参考答案】解:(1)原式==4-4+3-π-1+π=2.(2)原式=2lg5+2lg2+lg5•(lg2+1)+(lg2)2=2+lg2(lg5+lg2)+lg5=2+lg2+lg5=3.【试题分析】(1)利用指数幂的运算性质即可得出.(2)利用对数的运算性质及其lg2+lg5=1即可得出.本题考查了指数幂与对数的运算性质,考查了推理能力与计算能力,属于基础题.18.【参考答案】解:(1)因为A={x|3≤3x≤27}={x|1≤x≤3},B={x|1<log2x<2}={x|2<x<4},所以A∩B={x|2<x≤3},从而(C R B)∪A={x|x≤3或x≥4}.(2)当2a≥a+2,即a≥2时C=∅,此时C⊆A,符合条件;当2a<a+2,即a<2时,C≠∅,要使C⊆A,只需即.故要使C⊆A,实数a的取值范围是{a|a≥2或}.【试题分析】(1)求出集合A,B,由此能求出A∩B和(C R B)∪A.(2)当2a≥a+2,即a≥2时C=∅,符合条件;当2a<a+2,即a<2时,C≠∅,要使C⊆A,只需由此能求出实数a的取值范围是.本题考查交集、补集、并集的求法,考查交集、补集、并集定义等基础知识,考查运算求解能力,是基础题.19.【参考答案】解:(1)∵函数f(x)是定义在(-4,4)上的奇函数,∴f(0)=0,即,∴b=0,又因为f(2)=1,所以f(-2)=-f(2)=-1,即,所以a=1,综上可知a=1,b=0,(2)由(1)可知当x∈(-4,0)时,,当x∈(0,4)时,-x∈(-4,0),且函数f(x)是奇函数,∴,∴当x∈(0,4)时,函数f(x)的解析式为,任取x1,x2∈(0,4),且x1<x2,则=,∵x1,x2∈(0,4),且x1<x2,∴4-x1>0,4-x2>0,x1-x2<0,于是f(x1)-f(x2)<0,即f(x1)<f(x2),故在区间(0,4)上是单调增函数.【试题分析】(1)根据f(x)是定义在(-4,4)上的奇函数及-4<x≤0时的f(x)解析式即可得出b=0,并可求出f(-2)=-1,从而可得出,求出a=1;(2)根据上面知,x∈(-4,0)时,,从而可设x∈(0,4),从而得出,从而得出x∈(0,4)时,,然后根据函数单调性的定义即可判断f(x)在(0,4)上的单调性:设任意的x1,x2∈(0,4),且x1<x2,然后作差,通分,提取公因式,然后判断f(x1)与f(x2)的大小关系即可得出f(x)在(0,4)上的单调性.本题考查了奇函数的定义,奇函数在原点有定义时,原点处的函数值为0,求奇函数在对称区间上的解析式的方法,以及函数的单调性,考查了推理能力和计算能力,属于基础题.20.【参考答案】解:(1)由题设,当价格上涨x%时,每年的销售数量将减少mx%,销售总金额y=10(1+x%)•1000(1-mx%)=-mx2+100(1-m)x+10000().当时,y=[-(x-50)2+22500],当x=50时,y max=11250.即该产品每吨的价格上涨50%时,销售总金额最大.(2)当x=10时,若能使销售总金额比涨价前增加,能使销售总金额增加,则存在使y>10×10000,由得,所以m<10.由y>10×10000,即-100m+1000(1-m)+10000>10000亦即,所以.故若能使销售总金额比涨价前增加,m的取值范围设定为.【试题分析】(1)得出y关于x的函数,根据二次函数的性质求出结论;(2)根据题意列不等式得出m的范围.本题考查了函数解析式,函数最值的计算,考查不等式的解法,属于中档题.21.【参考答案】解:(1)∵f(x)是奇函数,∴f(-1)=-f(1),∴-|-1-a|-1=-(1•|1-a|+1)∴-|1+a|-1=-|1-a|-1,∴|1+a|=|1-a|,∴a=0,当a=0时,f(x)=x•|x|+x是奇函数,∴a=0;(2)任意的x∈[1,2],f(x)≥2x2恒成立,∴x|x-a|+x≥2x2恒成立,∴|x-a|+1≥2x恒成立,∴|x-a|≥2x-1恒成立, ∵x∈[1,2],∴2x-1∈[1,3],2x-1>0,∴x-a≥2x-1恒成立或x-a≤-2x+1恒成立,∴a≤-x+1恒成立或a≥3x-1恒成立,而-x+1∈[-1,0],3x-1∈[2,5],∴a≤-1或a≥5;(3)∵a≥2,x∈[0,2],∴x-a≤0,∴|x-a|=-(x-a),∴f(x)=x[-(x-a)]+x=-x2+(a+1)x,开口向下,对称轴为x=≥,①当,即2≤a≤3时,f(x)max=f()==4,∴a=3或a=-5(舍),②当>2,即a>3时,f(x)max=f(2)=-4+2a+2=2a-2=4,∴a=3,又a>3,矛盾,综上a=3.【试题分析】(1)由奇函数的性质f(-x)=-f(x),进而求解;(2)x∈[1,2],2x-1∈[1,3],2x-1>0,f(x)≥2x2等价于x-a≥2x-1恒成立或x-a≤-2x+1恒成立,进而求解;(3))∵a≥2,x∈[0,2],∴x-a≤0,∴f(x)=x[-(x-a)]+x=-x2+(a+1)x,进而比较对称轴与区间端点的关系求解;(1)考查奇函数的性质,去绝对值号;(2)考查不等式恒成立的转化,得出x-a≥2x-1恒成立或x-a≤-2x+1恒成立,是突破本题的关键点;(3)考查不等式在特定区间上的最值问题,将不等式恒成立转化为二次函数在特定区间上的最值.22.【参考答案】解:(1)当m=-1时,,要使函数f(x)有意义,则需,即2x<2,从而x<1.故函数f(x)的定义域为{x|x<1};(2)若函数g(x)=f(x)+2x lg2有且仅有一个零点,即有且仅有一个根,亦即,即,即m(2x)2+2•2x-1=0有且仅有一个根.令2x=t>0,则mt2+2•t-1=0有且仅有一个正根,当m=0时,2•t-1=0,,即x=-1,成立;当m≠0时,若△=4+4m=0即m=-1时,t=1,此时x=0成立;若△=4+4m>0,需,即m>0,综上,m的取值范围为[0,+∞)∪{-1};(3)若任取x1,x2∈[t,t+2],不等式|f(x1)-f(x2)|≤1对任意t∈[1,2]恒成立,即f(x)max-f(x)min≤1对任意t∈[1,2]恒成立,因为在定义域上是单调减函数,所以,,即,即,,所以,即,又有意义,需,即,所以,t∈[1,2],.所以m的取值范围为.【试题分析】(1)将m=-1代入f(x)中,根据,解不等式可得f(x)的定义域;(2)函数g(x)=f(x)+2x lg2有且仅有一个零点,则可得方程m(2x)2+2•2x-1=0有且仅有一个根,然后求出m的范围;(3)由条件可得f(x)max-f(x)min≤1对任意t∈[1,2]恒成立,求出f(x)的最大值和最小值代入该式即可得到m 的范围.本题考查了函数定义域的求法,函数的零点判定定理和不等式恒成立问题,考查了分类讨论思想和转化思想,属难题.。
福建省泉州市2019-2020学年高一上学期期中化学试卷1(含答案解析)

福建省泉州市2019-2020学年高一上学期期中化学试卷1一、单选题(本大题共19小题,共47.5分)1.将4gNaOH溶解在10ml水中,再稀释成1L,稀释后溶液的物质的量浓度是()A. 1mol⋅L−1B. 0.1mol⋅L−1C. 0.001mol⋅L−1D. 10mol⋅L−12.下列物质中,属于电解质的是()A. 铜丝B. MgCl2C. 稀盐酸D. 蔗糖3.摩尔是()A. 国际单位制中的一个基本物理量B. 表示物质质量的单位C. 计量微观粒子的物质的量的单位D. 描述宏观世界的4.下列化学方程式中,不能用离子方程式Ba2++SO42−=BaSO4表示的是()A. Ba(NO3)2+H2SO4=BaSO4↓+2HNO3B. BaCl2+Na2SO4=BaSO4↓+2NaClC. BaCO3+H2SO4=BaSO4↓+H2O+CO2↑D. BaCl2+H2SO4=BaSO4↓+2HCl5.下列说法正确的是()A. SO2属于酸性氧化物,CaO属于碱性氧化物B. 盐酸,硝酸都属于无氧酸C. 氢氧化钠,氢氧化铁都是可溶性碱D. CO,CO2都是酸性氧化物6.同温同压下,xg的甲气体和yg的乙气体占有相同的体积,则x:y的值不能表示()A. 同温同体积下,等质量的甲与乙的压强之比B. 等质量的乙与甲的分子个数之比C. 甲与乙的相对分子质量之比D. 同温同压下,甲与乙的密度之比7.下列每组内的物质,按单质、化合物、混合物顺序排列的是()A. 冰、生石灰、天然气B. 碘酒、纯碱、胆矾C. 金刚石、烧碱、纯盐酸D. 液态氧、食醋、沼气8.下列离子反应不能发生的是()A. NaOH+KCl=NaCl+KOHB. AgNO3+NaCl=AgCl↓+NaNO3C. Na2CO3+Ca(OH)2=CaCO3↓+2NaOHD. NaOH+NH4NO3=NaNO3+NH3·H2O9.与100mL 0.4mol⋅L−1NH4NO3溶液中的NO3−浓度相同的是()A. 400 mL 0.1mol⋅L−1的KNO3溶液B. 100 mL 0.4mol⋅L−1的Ba(NO3)2溶液C. 100 mL 0.2mol⋅L−1的Mg(NO3)2溶液D. 200 mL 0.1mol⋅L−1的Fe(NO3)3溶液10.如果1g水中含有n个氢原子,则阿伏加德罗常数是()A. n mol−1B. 9n mol−1C. 2n mol−1D. n mol−111.在实验室用自来水制取蒸馏水的实验中,下列说法错误的是()A. 烧瓶中要放入碎瓷片以防止暴沸B. 温度计的水银球应放在支管口处,不能插入液面C. 冷凝水应该是下进上出,与蒸汽的流向相反D. 开始蒸馏时,应该先加热,再开冷凝水;蒸馏完毕,应该先关冷凝水再撤酒精灯12.下列各组中的离子,能在溶液中大量共存的是A. Na+、NH4+、CO32−、OH−B. Mg2+、H+、SO42−、Cl−C. K+、H+、NO3−、ClO−D. Ba2+、Na+、HCO3−、OH−13.下列关于0.2mol·L−1K2SO4溶液的正确说法是()A. 该溶液中所含K+、SO42−总数为0.6N A B. 500mL该溶液中含有0.1N A个K+C. 任意体积的该溶液中,K+浓度均是0.4mol·L−1 D. 2L该溶液中SO42−浓度是0.4mol·L−114.有关化学实验的下列操作中,一般情况下不能相互接触的是()A. 过滤操作中,玻璃棒与三层滤纸B. 过滤操作中,漏斗颈与烧杯内壁C. 过滤操作中,玻璃棒与烧杯口D. 胶头滴管向试管中滴加液体时,滴管尖端与试管内壁15.下列离子方程式正确的是()A. 大理石与醋酸反应:CO32−+2CH3COOH=2CH3COO−+H2O+CO2↑B. 高锰酸钾与浓盐酸制氯气的反应:MnO4−+4Cl−+8H+=Mn2++2Cl2↑+4H2OC. 漂白液吸收少量二氧化硫气体:SO2+H2O+ClO−=SO42−+Cl−+2H+D. 氢氧化钠溶液与过量的碳酸氢钙溶液反应:OH−+Ca2++HCO3−=CaCO3↓+H2O16.300mLAl2(SO4)3溶液中,含Al3+为1.62g,在该溶液中加入0.1mol/L Ba(OH)2溶液300mL,反应后溶液中SO42−的物质的量浓度为()A. 0.4mol/LB. 0.3mol/LC. 0.2mol/LD. 0.1mol/L17.2molNaOH配成200mL溶液,其物质的量浓度为()A. 2mol/LB. 10 mol/LC. 0.2mol/LD. 0.1mol/L18.下列说法正确的是()A. 摩尔是一种国际基本物理量B. 在同温同压下,相同体积的任何气体所含分子数都相同C. 1mol氧气的质量为32g.mol−1D. 标准状况下气体的摩尔体积为22.4L19.下列关于胶体的叙述不正确的是()A. 胶体区别于其他分散系的本质特征是光线透过胶体时,胶体中可发生丁达尔效应B. 在Fe(OH)3胶体中逐滴滴入稀硫酸时出现的现象是先沉淀,后沉淀溶解C. 渗析就是用半透膜把分子或离子从胶体溶液分离出来达到净化胶体的目的D. 将饱和FeCl3溶液滴入沸水中,并继续加热煮沸得到红褐色液体即可制备Fe(OH)3胶体二、双选题(本大题共1小题,共2.5分)20.VmL密度为ρg⋅mL−1的某溶液,含有相对分子质量为 M 的溶质m g,该溶液中溶质的质量分数为ω,物质的量浓度为c mol⋅L−1,则下列关系式正确的是()A. m=Vρω1000B. c=1000ρωMC. ω=cM1000ρ% D. c=1000mVM三、填空题(本大题共2小题,共20.0分)21.可用于分离或提纯物质的方法有:过滤、蒸发结晶、蒸馏、萃取、分液、洗气。
辽宁省大连市2019-2020学年高一上学期期中考试英语试卷(有答案)

辽宁省大连市2019-2020学年高一上学期期中考试英语试卷(时间:100分钟总分:150分)注意事项:1、答卷前,考生务必将自己的姓名、准考证号填写在答题卡上;并将条形码粘贴在指定区域。
2、第Ⅰ卷每小题选出答案后,用2B铅笔把答题卡上对应题目的答案号涂黑。
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第Ⅰ卷第一部分阅读理解第一节(共10小题;每小题2.5分,满分25分)阅读下列短文,从每题所给的四个选项(A、B、C和D)中,选出最佳选项。
AIn America, when the eighth graders leave the middle school, they are often worried about moving to the high school. It is a hard time for them, but it can be an exciting one as well. To make it easier, students need to get familiar with their school.Even though they’ve done this already in the middle school, it’s still important to find where their classrooms are. Most schools take students to the high school for a visit at the end of their eighth grade year. And, most schools also have a freshman(新生) speech for parents and students before the first year begins. Sometimes the main office will give away a map of the school. This can help students and parents to find different places in the school.One way to know your high school more and make new friends is to join a club or play a sport. Schools often offer chances to join different clubs and sports. Fall activities begin in August before the first day of school. If you are not interested in sports, there are clubs for any interest, such as drama, dance, chess, photography, community service, etc. These clubs allow students from different grades. By joining an activity, students can find new friendships, not to mention improving their chances in future college applications. The school office will have a list of activities offered at the school and information on how to join them.What’s more, to have a great start to a high school year, students can write down some of their worries, and ask for help from teachers and school workers.1. According to the text, most students feel ________ when entering the high school.A. excitedB. nervousC. lonelyD. disappointed2. Which of the following is TRUE about the school clubs?A. There are fewer students from higher grades in the school clubs.B. The school clubs in high school only welcome sports fans.C. Summer activities begin before school starts.D. The school office can help new students join school clubs.3. What is the purpose of this text?A. To tell the new students how to learn in high schoolB. To introduce new schoolsC. To give the teachers some adviceD. To give advice to new high school studentsBIf you're reading this, then you're probably studying English. Maybe you speak a couple of other languages too. But what are the British like when it comes to learning languages?According to a survey published by the European Commission, the British are officially the worst language learners in Europe-62 percent of them can't speak any other language apart from their own! While 38 percent of Britons speak at least one foreign language, only 18 percent speak two.However, the European Union data show that 56 percent speak at least one foreign language and 28 percent speak at least two. Moreover, 51 percent of EU citizens can have a conversation in English.Learning a foreign language is not a popular choice at schools in Britain; children start studying a foreign language at the age of 11 and many give up completely at 14. So why don't young people continue with languages at school? Research suggests that students think that it is more difficult to get good grades in languages than in other subjects such as science or history.The government is now looking for different ways to improve language learning. One idea is to start much earlier and introduce foreign languages from the age of five. Another plan is to give schoolchildren more choices and expand(增加) the range of languages taught to include Arabic, Mandarin and Urdu.Mandarin is predicted to become the second most popular foreign language learned in UK schools. Gareth from Wales says, "I am learning Chinese and find it fun." Another student, Thomas from London, says, "Just saying that I learn Mandarin impresses people. Even having a very basic level gives you an advantage." Brighton College headmaster Richard Cairns says, "One of my key tasks is to ensure that pupils are equipped for the realities of the 21st century. One of those realities is that China has the fastest-growing economy in the world."It may be a challenging task to change the Brits' attitude to learning languages but the government is determined to try.4. What does the survey published by the European Commission indicate?A. Britons aren't fond of speaking other languages.B. Britons are the people who are excellent in languages.C. More than half of EU citizens can only speak one language.D. English is a second language to half of EU citizens.5. Why don't British students pick up foreign languages?A. Because they don't want to make an effort.B. Because they have too many subjects to learn.C. Because English is the most widely spoken language.D. Because they feel other languages boring and useless.6. What does the passage mainly tell?A. British students are poor in language learning.B. British students are losing their advantages of learning languages.C. There is no need for the British to learn other languages.D. The British government is encouraging language learning.CFamily fights are never pleasant. Whether you are dealing with a loud sister or a stubborn mom or dad, the situation can be bad. At least that’s what we have learned to expect.What if it could be different? What if family fights and conflicts could be turned into opportunities to become closer? What if problems could be solved with everyone walking away feeling more deeply cared for and loved?Family fighting is, on one level, about power. Someone is telling someone else what to do. In a short moment, emotions(情绪) can be brought out and the fight is on. You can go down that old road or you can try something different.In any fight, you have a need to be listened to and understood. People raise their voices and shout at each other because they’re not being listened to. It’s totally unconscious. You want your mom or dad to listen and understand. But you always seem to forget the person you’re fighting with has the same need to be listened to. So if you can put aside your need for a few minutes, the situation will change quickly. Give the other person some time to express hisor her feelings, and you will earn your turn to be listened to.And while you are listening, you have to listen for the right information. The best way to do this in emotional situations is to forget most of the words because many people can’t express what they are feeling with words alone, especially when they are upset.But if the situation starts to get out of hand, it would be wise to just calm down and leave the problem until later because many fights happen at the wrong time. The situation will die down quickly, usually in less than 30 seconds. You can either move on to problem solving or reach an agreement to work on it at a better time.This way, everyone will feel respected and listened to, even when there is conflict. Emotions will be understood and respected, and it will bring your family closer together.7. From the first 3 paragraphs, we can conclude that ______.A. people had better avoid family fights.B. family members fight to control each otherC. family fights can be changed into something positiveD. when family members fight, it’s worse as they know each other’s weakness8. In the author’s opinion, the key to dealing with family fights is to ______.A. satisfy the need to be listened to firstB. make sure that you’re the first to be listen ed toC. pay no attention to what others say in emotional situationsD. stop quarrelling when you feel you are giving orders9. When the situation gets out of control, it’s best to ______.A. wait for a suitable moment to speakB. let the other person express their feelingsC. work on a written agreement firstD. move on to problem solving10. The main purpose of this article is to ______.A. tell readers that family fights are not that badB. teach reader how to listen in family fightsC. remind readers to show respect in family fightsD. give advice on dealing with family fights第二节(共5小题;每小题2分,满分10分)根据短文内容,从短文后的选项中选出能填入空白处的最佳选项,选项中有两项为多余选项。
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第一卷(选择题)第一部分英语知识运用(满分50分)第一节语音知识(每小题1分)。
从A、B、C、D四个选项中,找出其划线部分与所给单词的划线部分读音相同的选项。
⒈learn A. pear B. dear C. wearD. early⒉though A. southern B. think C. bathroomD. south⒊wanted A. pleased B. worked C. stayedD. deserted⒋exchange A. get B. peg C. vegetable D.glass⒌dead A. stream B. death C. teachD. great第二节语法和词汇知识(每小题1分)。
从A、B、C、D四个选项中,选出可以入空白处的最佳选项。
⒈“What is your good friend like?” “He is ____ .”A. richB. poorC. honestD. ill⒉Your composition is well-written ____ a few spellingmistakes.A. exceptB. except forC. except thatD. besides⒊____ catch the first bus, she got up earlier that morning.A. In order toB. So as toC. In order thatD. So that⒋ He likes Chinese, ____ his brother likes English.A. whichB. whatC. whileD. whether⒌ I was about to go to bed ____ the telephone rang.A. whileB. asC. at the timeD. when⒍ What ____ it is to have a cold drink in hot weather!A. funnyB. funC. a funD. funs⒎ The teacher asked the students ____ any noise in class.A. don’t makeB. to not makeC. not to makeD. notmake⒏ With so many people ____ in English every day, it willbecome more and more important to have ____ English.A. communicated, good knowledge ofB.communicating, a good knowledge ofC. communicating good knowledgeD. tocommunicate, a good knowledge at⒐ Some Chine se students have ____ English well.A. difficulty in learningB. difficulty tolearnC. difficulties learnedD. difficulties tolearn⒑ The boy wrote a ____ article last week.A. two –thousand-wordsB.two-thousands-wordC. two- thousands-wordsD.two-thousand-word⒒ Many people watched the ____ TV broadcast of the football match.A. liveB. aliveC. livingD. lively⒓ I’m going to Europe tomorrow and all my friends will ____ at the airport.A. see off meB. see me offC. see of meD.see me of⒔ It was in a poor family ____ the great scientist was born.A. whenB. thatC. whichD. how⒕ I ____ my success to my parents and my teachers.A. oweB. ownC. shareD. pay⒖ The man ____ leg broke in a match used to be a football player.A. whichB. whoC. whoseD. whom第三节完形填空(每小题1. 5分)。
阅读下面短文,从短文后各题所给的四个选项中,选出可以填入空白处的最佳选项。
When one takes shoes to a shop to be mended, he is given a ticket with a number on it. Then, _1___ his shoes are ready, he goes back to the shop, gives the ticket to the shoemaker, gets his shoes and ____2 __ for them.One day Mrs Smith gave her husband a pair of her shoes which needed ____3 ___ and asked him to ____4 ___ them to the shop. Mr Smith did so, and put the ___5 ___ for them in his pocket.He went back four days ___6 ___ to get the shoes, ___7___ when he went into the shop, he was not able to find his ticket ____ 8____ , and the shoemaker did not want to give him the ___9 ___ until he got the ticket. “How do I know that the shoes are ___ 10___ unless you give me the ticket?” he said. “If I give them to you now, somebody ____11___ may come into my shop with the ticket tomorrow, and then I shan’t be a ble to give him the shoes.” Mrs Smith needed the shoes very much, so her husband ___12____ for a moment and then went out to his ____13____ , which was at the side of the road ____14___ the shop. He ____15___ the door, and whistled(耳语)to his wife’s small ___16 ___, which was sitting on the back seat. Then he went back into the shop with the dog and ____17 ___ it, “Get the shoes!” the dog began to ___18 ____ around the shop, and soon it found Mrs Smith’s sho es andbrought one of them to Mr Smith, then the ___ 19 ___. “That should prove(证明)that they are my wife’s,.” said Mr Smith. The shoemaker ___ 20___ . Then he gave the shoes to Mr Smith.⒈ A. before B. when C. since D. unless⒉ A. cares B. looks C. pays D. waits⒊ A. making B. mending C. buying D. selling⒋ A. show B. change C. find D. take⒌ A. ticket B. paper C. bill D. list⒍ A. late B. later C. ago D. long⒎ A. or B. because C. and D. but⒏ A. with his son B. again C. in his office D. slowly⒐ A. shoes B. idea C. dog D. answer⒑ A. black B. beautiful C. yours D. whose⒒ A. old B. young C. other D. else ⒓ A. worried B. stood C. thought D. hoped ⒔ A. car B. bus C. boat D. train⒕ A. inside B. round C. up D. outside⒖ A. opened B. closed C. broke D. built ⒗ A. clock B. dog C. hen D. cat⒘ A. talked B. replied C. answered D. said to ⒙ A. hear B. taste C. smell D. see ⒚ A. other B. others C. ticket D. pairs⒛ A. surprised B. cried out C. laughed D.frightened第二部分阅读理解(满分45分)第一节阅读下列短文,从每题所给的四个选项(A、B、C和D) 中,选出最佳选项。
(每小题2分)(A)When we think of Hollywood, we think of films and famous film stars. They are part of Hollywood’s history. Today people make films in other places, too. Not all famous film stars live in Hollywood. But Hollywood is still a very special part in Los Angeles, California.You can easily see where Hollywood is in Los Angeles. There is a big sign on the hills. It says “HOLLYWOOD”. The white letters are fifty feet tall. You can see the sign from far away. The Hollywood sign is a famous Hollywood landmark (标记) in Los Angeles. Many postcards show this famous Hollywood landmark.In the hills of Hollywood, there is also the Hollywood Bowl. This is an open-air theatre. It is one of the biggest open-air theatres in the world. It has seventeen thousand seats and a very different stage (舞台). The designer (设计师) of the stage was a great American called Frank Lloyd Wright. You can listen to different kinds of concerts at the Hollywood Bowl.⒈ The best title for the article should be “____”.A. The History of HollywoodB. HollywoodC. Famous Films and Film StarsD. FamousHollywood Sign⒉ We learn from the reading that ____ .A. Hollywood is one part of Los AngelesB. Hollywood is now a place only for travelersC. all the films are made In HollywoodD. “HOLLYWOOD” can just be seen from far away⒊ The word “special” in the last sen tence of the first paragraphmeans ____ .A. unusualB. famousC. oldD. new⒋ At the Hollywood Bowl ____ .A. all postcards are madeB. only 7000 peoplecan get togetherC. people can enjoy musicD. you can hardlylisten to all kinds of concerts(B)In China, more and more middle school students are getting shorter sleeping time than before. More students sleep less than nine hours every night, because they have much homework to do. Some homework is given by their teachers, and some by their parents. Also, some students don’t knowhow to save time. They are not careful enough while they do their homework, so it takes them a lot of time. Some students spend too much time watching TV or playing computer games. They stay up very late. Some students have to get up early every morning on weekdays to get to school on time by bus of by bike . It can be a long way from home to school. It will probably make them sleepy during school hours.Schools and parents should cut down some of the homework so that our children can enjoy more than nine hours of sleep every night for their health. For children, they should make the most best use of their time. When they have enough time for sleeping .They will find it much better for both their study and health.⒌ What make most students sleep less than nine hours everynight?A. Doing housework.B. Playing computergames.C. Doing homework.D. Watching TV.⒍ Who else gives children some homework to do except theirteachers according to the article?A. Their parents.B. Other students.C. Nobody else.D. They themselves.⒎ How many hours of sleep is good for students’ heal th?A. Only 9 hours.B. Over 9 hours.C. 8 hours.D. Less than 9 hours.⒏ Which is the following is NOT true?A. Some students are careless that they spent much timeon the homework.B. All the students stay up late for their homework.C. Sch ools and parents should take care of children’shealth.D. Children should be given less homework.(C)Dear Mom and Dad,I’m afraid I have some very bad news for you. I have been very naughty and the school master is very angry with me. She is going to write to you. You must come and take me away from here. She does not want me in the school any longer.The trouble started last night when I was smoking a cigarette in bed. As I was smoking, I heard footsteps coming towards the room. I did not want a teacher to catch me smoking, so I threw the cigarette away. Unfortunately, the cigarette fell into the waste--paper basket. It caught fire. There was a curtain near the waste --paper basket. It caught fire too. Soon the whole room was burning. The master phoned the fire brigade (消防队). The school is a long way from the town and beforethe fire brigade arrived, the whole school was on fire. The master said that the fire was all my fault and I must pay for the damage. She will send you a bill for about a million dollars.I’m very so rry for this.uch love,arah⒐ What did Sarah write to his parents for?A. He was tired of studying, and wanted to leave school.B. He told his parents what he had done in school.C. He thought he was old enough to help them dosomething.D. He missed them a lot for they hadn’t seen each other fora long time.⒑Where did Sarah smoke?A. In the classroom.B. In the street.C. In the dormitory.D. In the Dinning hall. ⒒Why was the school master angry with him?A. Because he was too naughty.B. Because he wastoo lazy.C. Because he caused a big fire.D. Because hestole something at school.⒓According to the passage, the students should not ____ .A. smoke because smoking is dangerous to themB. dance because dancing would waste a lot of moneyC. make friends because making friends would take themabsentD. wear expensive clothes because they would waste a lotof money(D)Now I’d like to talk to you a bout your final exam. The exam will be held next Thursday, the last day of the exam week. Remember to bring two or three pens in case you run out of ink. And unlike the midterm exam, this test will not include multiple-choice questions; it will consist (由…组成) entirely of essays (文章). You’ll have to answer three of the five essay questions. The exam will be comprehensive (理解性的), which means you’ll be responsible (负责) for all of the subject matters we covered in class this term. I would suggest you review your midterm exam as well as textbooks and your class notes. The research project will count as 20 percent and the midterm exam 30 percent. I’ll be in my office almost all day next Tuesday. If you run into any problems, please drop in. Gook luck to you and I’ll see you on Tuesday.⒔The final exam take place on a ____ .A. TuesdayB. WednesdayC. ThursdayD. Friday⒕What will be included in the exam?A. There will be only multiple-choice questions.B. The exam will contain both multiple-choice and essayquestions.C. The exam will have an oral and written section.D. There will be only essay questions.⒖Why does the teacher call the exam comprehensive?A. Because it will be easy to understand.B. Because it will cover topics from a variety (多种多样) ofsubjects.C. Because students will be tested on all the materialdiscussed in class.D. Because students must complete all parts of it.⒗The underlined phrase run into probably means ____ .A. meet somebody unexpectedlyB. go intoC. come across (偶然遇见)D. comeup against something with force(E)In every language, there are words or phrases that have some interesting stories behind them. The word “sandwich”, for example, is very common in English. If we want to know thestory behind it, we must know something about an English nobleman (贵族) called Sandwich.Sandwich lived in the 18th century. He liked to play cards for money. He often played for twenty four hours, and not even stop to have his meal. He ordered his servants to bring some meat and bread, and he played while eating. He put the meat between two pieces of bread, and he held the food in his left hand while he played with his right hand. People liked Sandwich’s idea and began to eat bread and meat in this way. From the name of the man Sandwich, we have the word “sandwich” today.⒘Sandwich is very popular food today, which comes from ____ .A. a man’s nameB. people’s ideaC. a servant’s nameD. a nobleman’s play⒙What did Sandwich play cards for?A. Pleasure.B. Living.C. Food.D. Pounds.⒚Which of the following can best explain the reason why Sandwich did not stop to have his meals?A. He played cards with his right hand.B. He put the meat between two pieces of bread.C. He played cards for twenty four hours.D. He played cards while eating.⒛The best title for the passage should be “____”.A. How Sandwich Played CardsB. How the Word“Sandwich” Came into UseC. A Bright IdeaD. An Interesting Story 第二节根据对话内容,从对话后的选项中选出能填入空白处的最佳选项。