2014徐汇高三二模

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2014年徐汇高三二模试卷分析

2014年徐汇高三二模试卷分析

起承上启下的作用,承接上文论证了充分的物质能带来 幸福感是错误的,引出下文对能否找到幸福的思考。
承上启下,承接上文只有充分获得物质满足才能体会幸 福感的误区,引出下文我们是否还能找到幸福。内容上 指出了物质条件与财富与个体幸福感并不构成正相关, 解释了中:分析段落在文中的作用 参考答案: 承接上文对幸福感认识的误区,从主观感受与客观福 祉未必对等的角度,解释了中国人幸福感迷失的原因; 又进一步引出下文“如今还能不能找到幸福”的问题。 主要问题: 只从概括本段的主要内容角度答题,未关注其与上下 文之间的关系。 只出现“承上启下”,并未结合具体内容明确承接上 文什么内容,引出下文什么内容,或对承上启下的内 容理解有偏差。 只关注本段与上文和下文之间的关系,忽略了其本身 的内容。
答案示例:
作者用大量的描写,描绘出一幅四月天内生机 勃勃景象。又运用大量修辞,夸张写出林子内 的生机勃勃。这些手法使文章更加生动,有感 染力。 作者从视觉角度进行描写,体现青杠林的环境 优美;“青杠树把它们的绿荫投在我身上”运 用拟人的修辞,写出树长的茂盛。 运用夸张,表现林子在清明后的快速生长;将 对青杠林的感觉比喻为鱼儿在溪水中,表现作 者的舒畅,对四月天的热爱;
第16题考点:理解诗歌的内容。 主要问题在于: 1、审题不清,这个比较突出。忽略题干中“变”一词, 简单分析词中主人公的情绪以及作者的情感,不能够展 现出主人公情绪变化的几个层次。故而失分比较严重, 虽然分析了很多,但是停留于“愉快”这个层面。另一 处审题不清,就是题干中的“下片”被忽视,回答时结 合的是上片内容。 2、不能够理解“不道吴……几行斑”这句词句的含义, 从而影响答题。 3、分析比较笼统,不能结合词中的词语或句子加以具 体分析,用许多套话、术语。 4、回答过简或过繁。有的十几个字,一笔带过,不加 分析;有的缺乏概括力,洋洋洒洒。

2014届 徐汇区 高三化学二模试卷及答案-推荐下载

2014届 徐汇区 高三化学二模试卷及答案-推荐下载

C.放入装有铁粉的透气的小纸袋
二、选择题(本题共 36 分,每小题 3 分,只有一个正确选项。)
6.右图是一种有机物的比例模型,该模型代表的有机物可能是
A.饱和一元醇
C.羧酸
7.下列微粒的存在最能说明碘可能呈现金属性的是
A.IBr
B.I2O5
B.羟基酸
D.饱和一元醛
8.在试管中注入某红色溶液,给试管加热,溶液颜色逐渐变浅,则原溶液可能是
A.钠
4.下列有机物系统命名正确的是
A.2-甲基-氯丙烷
C.2,4,4-三甲基戊烷
B.铝
B.氧化剂
D.既不是氧化剂又不是还原剂
C.溴
B.2-甲基-3-丁烯
D.2,3-二甲基-4-乙基己烷
5.除去密封食品包装盒内的氧气可延长食品的保质期,下列措施不能延长食品保质期的是
A.适当添加含酚类结构单元的天然化合物 B.对食物充氮包装
搅拌,再加入过量的 NaOH 溶液,则溶液中物质的量基本不变的阴离子是
A.NO3-
B.S2-
11.已知粗碘中含有的 IBr 和 ICl 受热都会升华,若先在粗碘中加入下列物质中的一种再加
热,就可制得纯净的碘。该物质是
A. KI
B.H2O
12.实验室中某些气体的制取、收集及尾气处理装置如图所示(省略夹持和净化装置)。仅
三、选择题(本题共 20 分,每小题 4 分,每小题有一个或两个正确选项。只有一个正确选项 的,多选不给分;有两个正确选项的,选对一个给 2 分,选错一个,该小题不给分,答案 涂写在答题卡上。)
18.某浓 H2SO4 的标签上写明含 H2SO4 96~98.3%,欲得其准确的值,有效的方法是
A.测 pH 值

2014年上海市徐汇区二模卷试题

2014年上海市徐汇区二模卷试题

2014年上海市徐汇区、松江区、金山区高考化学二模试卷一、选择题(本题共10分,每小题2分,只有一个正确选项).1.(2分)(2014•松江区二模)将CO2转化成有机物可有效实现碳循环.下列反应中,最节能的是()A.CO2+3H2CH3OH+H2OB.6CO2+6H2O C6H12O6+6O2C.CO2+CH4CH3COOHD.2CO2+6H2CH2=CH2+4H2O2.(2分)(2014•松江区二模)吸进人体内的氧有2%转化为加速人体衰老的氧化性极强的活性氧,若Na2SeO3能清除人体内活性氧,则Na2SeO3的作用是()A.还原剂B.氧化剂C.既是氧化剂又是还原剂 D.既不是氧化剂又不是还原剂3.(2分)(2014•松江区二模)下列元素的单质,工业上不需要用电解法制取的是()A.钠B.铝C.溴D.氯4.(2分)(2014•松江区二模)下列有机物系统命名正确的是()A.2﹣甲基﹣氯丙烷B.2﹣甲基﹣3﹣丁烯C.2,4,4﹣三甲基戊烷D.2,3﹣二甲基﹣4﹣乙基己烷5.(2分)(2014•松江区二模)除去密封食品包装盒内的氧气可延长食品的保质期,下列措施不能延长食品保质期的是()A.适当添加含酚类结构单元的天然化合物B.对食物充氮包装C.放入装有铁粉的透气的小纸袋D.放入装有生石灰透气的小纸袋二、选择题(本题共36分,每小题3分,只有一个正确选项.)6.(3分)(2014•松江区二模)如图是一种有机物的比例模型,该模型代表的有机物可能是()A.饱和一元醇B.羟基酸C.羧酸酯D.饱和一元醛7.(3分)(2014•松江区二模)下列微粒的存在最能说明碘可能呈现金属性的是()A.IBr B.I2O2C.I3+D.I3﹣8.(3分)(2014•松江区二模)在试管中注入某红色溶液,给试管加热,溶液颜色逐渐变浅,则原溶液可能是()A.滴有酚酞的Na2CO3溶液B.溶有SO2的品红溶液C.滴有石蕊的AlCl3溶液D.滴有酚酞的饱和Ca(OH)2溶液9.(3分)(2014•松江区二模)下列物质都具有较强的吸水性.欲使稀硝酸成为浓硝酸,不能使用的试剂是()A.浓硫酸B.五氧化二磷C.无水硫酸镁D.甘油10.(3分)(2014•松江区二模)某溶液中含有NO3﹣、S2﹣、AlO2﹣、SO32﹣四种阴离子,若向其中加入过量的盐酸,微热并搅拌,再加入过量的NaOH溶液,则溶液中物质的量基本不变的阴离子是()A.NO3﹣B.S2﹣C.AlO2﹣D.SO32﹣11.(3分)(2014•松江区二模)已知粗碘中含有的IBr和ICl受热都会升华,若先在粗碘中加入下列物质中的一种再加热,就可制得纯净的碘.该物质是()A.KI B.H2O C.Zn D.NaCl12.(3分)(2013•安徽)实验室中某些气体的制取、收集及尾气处理装置如图所示(省略)D 浓盐酸MnO2Cl2NaOH溶液A.A B.B C.C D.D13.(3分)(2015•潍坊模拟)将5.4g Al投入200.0mL 2.0mol•L﹣1的某溶液中有氢气产生,充分反应后有金属剩余.该溶液可能为()A.HNO3溶液B.Ba(OH)2溶液C.H2SO4溶液D.HCl溶液14.(3分)(2014•松江区二模)下列事实中不能说明氯的单质分子是双原子分子的是()A.在标准状况下,氯单质为气态B.在标准状况下,71g氯气所占的体积约为22.4LC.电解饱和食盐水,阴极和阳极分别逸出气体的体积大致相等D.含0.2mol NaOH的溶液,恰好与2.24L(标准状况下)氯气反应15.(3分)(2014•松江区二模)1mol过氧化钠与2mol碳酸氢钠固体混合后,在密闭的容器中加热充分反应,排出气体物质后冷却,残留的固体物质是()A.Na2CO3B.Na2O2 Na2CO3C.NaOH Na2CO3 D.Na2O2 NaOH Na2CO316.(3分)(2014•松江区二模)将通有H2S的导气管伸入充满氧气的集气瓶内点燃,描述反应过程的曲线错误的是()A.B.C.D.17.(3分)(2014•松江区二模)烷烃分子中去掉2个氢原子形成一个双键是吸热反应,1,3﹣环己二烯失去2个氢原子生成苯是放热反应.下列说法正确的是()A.1,3﹣环己二烯加氢是吸热反应B.1,3﹣环己二烯比苯稳定C.苯比1,3﹣环己二烯稳定D.苯加氢生成环己烷是吸热反应三、选择题(本题共20分,每小题4分,每小题有一个或两个正确选项.只有一个正确选项的,多选不给分;有两个正确选项的,选对一个给2分,选错一个,该小题不给分.)18.(4分)(2014•松江区二模)实验室进行下列实验,在选择实验装置时应当考虑冷凝回流装置的是()A.葡萄糖与银氨溶液共热制银镜B.用蒸馏烧瓶从石油中提取汽油C.苯与浓硝酸、浓硫酸共热制硝基苯D.乙醇、丁酸与浓硫酸共热制丁酸乙酯19.(4分)(2014•松江区二模)如图是工业制纯碱的部分物质转化示意图,下列推测错误的是()A.若是氨碱法,溶液c可在转化流程中循环利用B.若是联碱法,则L的主要成分是NH3C.M可在转化流程中循环利用D.X是NH3,Y是CO220.(4分)(2014•松江区二模)在O.1mol/L的NaHS03溶液中粒子关系为:c(Na+)>c (HSO3﹣)>c(SO32﹣)>c(H2SO3),向NaHS03溶液中慢慢地加入少量的NaOH溶液,发生的主要反应的离子方程式为()A.SO32﹣+H2O⇌HSO3﹣+OH﹣B.HSO3﹣+H2O⇋H2SO3+OH﹣C.HSO3﹣+OH﹣═SO32﹣+H20 D.H2SO3+OH﹣═HSO3﹣+H2021.(4分)(2014•松江区二模)在含有n mol FeI2的溶液中通入Cl2,有x mol Cl2发生反应.下列说法正确的是()A.当x≤时,反应的离子方程式为:2Fe2++Cl2→2Fe3++2Cl﹣B.当x≥n时,反应的离子方程式为:2Fe2++2I﹣+2Cl2→2Fe3++I2+4Cl﹣C.当Fe2+和I﹣都被氧化时,x与n的关系为x>nD.当x=n时,反应后氧化产物的物质的量为n mol22.(4分)(2014•松江区二模)将气体A2和B2以等物质的量充入密闭容器,一定条件下发生反应生成C气体.平衡时测得c(A2)=0.58mol/L,c(B2)=0.16mol/L,c(C)=0.84mol/L,则C的分子式为()A.AB B.AB2C.A2B D.A2B3四、(本题12分)23.(6分)(2014•松江区二模)液化石油气中常存在少量有毒气体羰基硫(COS),必须将其脱除以减少环境污染和设备腐蚀.完成下列填空.(1)写出羰基硫的电子式,羰基硫分子属于(选填“极性”、“非极性”)分子.(2)下列能说明碳与硫两元素非金属性相对强弱的是.a.相同条件下水溶液的pH:Na2CO3>Na2SO4b.酸性:H2SO3>H2CO3c.CS2中碳元素为+4价,硫元素为﹣2价(3)羰基硫在水存在时会缓慢水解生成H2S,对钢铁设备产生电化学腐蚀.写出正极的电极反应式,负极的反应产物为(填化学式).24.(6分)(2014•松江区二模)为除去羰基硫,工业上常采用催化加氢转化法,把羰基硫转化为H2S再除去:COS+H2═CO+H2S(1)已知升高温度,会降低羰基硫的转化率.则升高温度,平衡常数K,反应速率(均选填“增大”、“减小”、“不变”).(2)若反应在恒容绝热密闭容器中进行,能说明该反应已达到平衡状态的是.a.容器内气体密度保持不变b.容器内温度保持不变c.c(H2)=c(H2S)d.υ(H2)正=υ(H2S)正(3)已知该反应的平衡常数很大,说明.五、(本题共12分)25.(12分)(2014•松江区二模)铁及其化合物在生产、生活中存在广泛用途,完成下列填空.(1)如图1所示为铁元素在周期表中的信息,方格中“55.85”的意义为:.铁原子核外有种运动状态不同的电子.(2)二茂铁[Fe(C5H5)2]是一种有机金属化合物,熔点172℃,沸点249℃,易升华,难溶于水易溶于有机溶剂.二茂铁属于晶体;测定表明二茂铁中所有氢原子的化学环境都相同,则二茂铁的结构应为图2中的(选填“a”或“b”).(3)绿矾(FeSO4•7H2O)可用于治疗缺铁性贫血,其水溶液露置于空气中会变质,写出发生变质反应的离子方程式.(4)无水FeCl3是水处理剂,遇潮湿空气即产生白雾,易吸收空气中的水分成为结晶氯化铁(FeCl3•6H2O).制备无水FeCl3的试剂是.(5)高铁酸盐也是常用的水处理剂.高铁酸钠(Na2FeO4)可用如下反应制备:2FeSO4+6Na2O2→2Na2FeO4+2Na2O+2Na2SO4+O2↑若生成2mol Na2FeO4,则反应中电子转移的物质的量为mol.(6)高铁酸盐可将水体中的Mn2+氧化为MnO2进行除去,若氧化含Mn2+1mg的水体样本,需要1.2mg/L高铁酸钾L.六、(本题共12分)26.(12分)(2014•松江区二模)某研究性学习小组通过下列实验探究SO2能否与BaCl2溶液反应生成BaSO3沉淀,并制备硫酸铜晶体.完成下列填空.甲同学用装置I进行实验,加热反应物至沸腾,发现BaCl2溶液中出现白色沉淀,且白色沉淀不溶于盐酸.(1)白色沉淀是.(2)甲同学就白色沉淀的生成原因提出了两种假设,这两种假设可能是:①;②;乙同学设计了改进装置Ⅱ进行实验,检验甲同学提出的假设(夹持装置和A中加热装置已略,气密性已检验).①打开弹簧夹,通入N2一段时间后关闭弹簧夹;②滴加一定量浓硫酸,加热A,一段时间后C中未见沉淀生成.(3)操作①的目的是,洗气瓶B中的试剂是.(4)该实验能否检验甲同学上述两种假设中的任意一种,理由是.丙同学向反应后的蓝色溶液中加入足量的CuO,过滤后将滤液制成硫酸铜晶体(5)若无需再进行第三次称量,则a的数值范围应为.(6)若加热后坩埚与固体总质量为18.620g,计算x的实测值(保留二位小数),相对误差为.七、(本题共12分)27.(12分)(2014•松江区二模)为了回收利用钢铁生产过程中产生的SO2,工业上采用软锰矿(主要成分MnO2)脱硫同时制取硫酸锰技术,其流程示意图如图1:已知:浸出液的pH接近4,其中的金属离子主要是Mn2+,还含有少量的Fe2+.完成下列填空.(1)写出浸出过程中主要反应的化学方程式:.(2)浸出过程的副反应之一是部分SO2被氧化为硫酸,致使浸出液的pH下降,这将(“有利”、“不利”)于软锰矿浆继续吸收SO2.欲消除生成的硫酸,试剂A最好是.a.MnCO3b.MnO2c.CaO d.CaCO3(3)操作I的目的是除去浸出液中的Fe2+,MnO2在氧化Fe2+的同时还,使Fe3+沉淀.检验过滤后溶液中是否含有Fe3+的操作是.(4)已知大于27℃时,MnSO4•H2O溶解度随温度上升而明显下降,则操作II的过程为:、、洗涤、干燥.工业上为了充分利用锰元素,在流程中可循环使用.(5)通过煅烧MnSO4•H2O 可制得生产软磁铁氧体材料的Mn x O4,图2是煅烧MnSO4•H2O 时温度与剩余固体质量变化曲线.该曲线中A段所表示物质的化学式为;Mn x O4中x=.八、(本题共10分)28.(10分)(2014•松江区二模)碳酸二甲酯(DMC)是一种低毒性的绿色化学品,可用于代替高毒性的光气(COCl2)作羰基化试剂.DMC的合成流程如下.完成下列填空.已知:RCOOR′+R″OH→RCOOR″+R′OH(1)写出反应类型:反应①反应③(2)石油工业上获得C2H4的操作名称为.(3)写出结构简式:DMC;X(4)已知物质Y与DMC互为同分异构体,Y的水溶液呈酸性,在一定条件下2mol Y能生成1mol分子中含六元环结构的有机物,则Y的结构简式为:(5)DMC与双酚()在一定条件下可生成芳香族聚碳酸酯,写出反应的化学方程式:.九、(本题共12分)29.(12分)(2014•松江区二模)苯巴比妥是安眠药的成分,化学式为C12H12N2O3,分子结构中有两个六元环:如下图是以A为原料合成苯巴比妥的流程示意图.完成下列填空.已知:①有机物D、E中亚甲基(﹣CH2﹣)的氢原子受羰基影响活性较高,容易发生如下反应:②(1)芳香烃A与HCl反应后,制取B还需进行反应的类型依次为.(2)一种酯与B互为同分异构体,且苯环上只有一个取代基,该酯同分异构体有种,写出其中一种结构的系统命名.(3)写出D转化为E的化学方程式:(4)苯巴比妥G的结构简式:(5)E与CO(NH2)2在一定条件下合成的高分子结构简式:(6)已知:,请设计合理方案以B的同系物为原料合成(用合成路线流程图表示为:A B…目标产物).十.(本题共14分)30.(6分)(2014•松江区二模)将CO2通入NaOH溶液中,所得产物随通入的CO2的物质的量的不同而不同.完成下列填空.(1)250mL 1mol/L的NaOH溶液,最多可吸收CO2L(标准状况下).(2)向250mL 2mol/L的NaOH溶液中通入一定量CO2,溶液增重4.4g,将所得溶液蒸干,计算所得固体中各成分的物质的量.31.(8分)(2014•松江区二模)向未知浓度、体积为V L的NaOH溶液中缓缓通入一定体积(标准状况下)的CO2,充分反应后,在减压低温的条件下蒸发溶液,得到白色固体.(1)若反应中CO2和NaOH均无剩余,反应后向溶液中加入过量的澄清石灰水生成m1 g 白色沉淀.①根据以上数据,用代数式表示CO2的体积V(CO2)=.②根据以上数据,推理计算出NaOH溶液的浓度范围.(2)写出确定NaOH浓度的实验步骤,并用代数式表示NaOH溶液的浓度.(设计实验方案时,只能用题中提供的CO2和NaOH溶液,不得使用其它化学试剂.)实验步骤:.c(NaOH)=.。

上海市徐汇、松江、金山三区2014届高三二模化学试卷(带解析)

上海市徐汇、松江、金山三区2014届高三二模化学试卷(带解析)

上海市徐汇、松江、金山三区2014届高三二模化学试卷(带解析)1.将CO2转化成有机物可有效实现碳循环。

下列反应中,最节能的是A.CO223OH +H2OB.6CO2 + 6H26H12O6 + 6O2C.CO243COOHD.2CO2 + 6H2=CH2 + 4H2O【答案】B【解析】试题分析:光合作用是自然界中碳循环的反应,不需要再人为提供能源,所以是最好、最节能的,因此答案选B。

考点:考查自然界中碳循环2.吸进人体内的氧有2%转化为加速人体衰老的氧化性极强的活性氧,若Na2SeO3能清除人体内活性氧,则Na2SeO3的作用是A.还原剂B.氧化剂C.既是氧化剂又是还原剂 D.既不是氧化剂又不是还原剂【答案】A【解析】试题分析:“活性氧”氧化性极强,服用含硒元素(Se)的化合物亚硒酸钠(Na2SeO3),能消除人体內的活性氧,表明Na2SeO3能将“活性氧”还原,“活性氧”为氧化剂,Na2SeO3为还原剂,故选A。

考点:考查氧化还原反应的判断3.下列元素的单质,工业上不需要用电解法制取的是A.钠 B.铝 C.溴 D.氯【答案】C【解析】试题分析:A、金属钠在工业上采用电解氯化钠的方法来获得,故A错误;B、金属铝在工业上采用电解氧化铝的方法来获得,故B错误;C、溴的提取是利用氧化还原法从海水中提取,不属于电解法,故C正确;D、氯气在工业上用电解饱和食盐水来获得,故D错误,答案选C。

考点:考查金属冶炼、电解原理4.下列有机物系统命名正确的是A.2-甲基-氯丙烷 B.2-甲基-3-丁烯C.2,4,4-三甲基戊烷 D.2,3-二甲基-4-乙基己烷【答案】D【解析】试题分析:A、卤代烃的命名,必须指出卤素原子的位置,故A错误;B、双键编号不是最小,正确命名是:3-甲基-1-丁烯,故B错误;C、支链编号之和必须最小,正确命名是:2,2,4-三甲基戊烷,故C错误;D、符合烷烃命名原则,主链最长,主链编号之和最小等,故D 正确,答案选D。

2014年徐汇区语文高三二模卷

2014年徐汇区语文高三二模卷

2014年徐汇区语文二模考试卷一、阅读80分(一)阅读下文,完成第1—6题。

(18分)①幸福感是一种主观体验,是个体依据自己没定的标准对其生活质量所作的整体评价与感受。

尽管人的幸福感各有差异,但幸福感的存在是有科学依据的。

②对进化论的解读使我们发现,人类的形态在进化过程中发生了剧烈的变化。

由于直立行走使骨盆变得狭短,人类的婴儿比其他动物的婴儿史加脆弱,更需要保护和关怀,这决定了人类必须关照自己的后代,否则很难生存。

人类的毛发少而裸露的皮肤多,拥有发达的触觉,需要接触,需要抚摸。

人类还有巨大的脑,人脑皮质相对于身体特质的比例居动物之首,因而人类是唯一具有灵性的动物,能进行心灵的沟通和对话。

可见,进化决定了人类的本性是善良、友爱、相互支持的,具有获取幸福感的基础。

③一些新的研究进一步提示了幸福感产生的生机机制。

④通过对大脑进行核磁共振的扫描,科学家发现积极的情感对应的是大脑前额叶左边的皮层活动,消极情感则对应同一个区域右边的皮层。

扫描还显示,脑部所分泌的0xytocin能抑制杏仁核区域(脑部处理恐惧和危险的部位)和纹状体区域(脑部凭借过去的回馈来指引未来行为的地带)的活动,从而降低人的防卫心和恐惧感,让人忘掉过去的目痛苦,更相信他人。

0xytocin在中国被翻译成“催产素”,但它实则是一种男女皆有的神经激索,不仅可在女性生育时促进产道的收缩,还有更多其他的心理影响,尤其对于抑郁症和孤独症患者会有很大帮助。

⑤科学家们继而发现迷走神经与人类积极的体验和亲社会行为有关。

迷走神经支配呼吸、消化系统的绝大部分器官,以及心脏的感觉、运动和腺体的分泌,所以人一旦遇到激动、兴奋的事件,心胸便比较开阔。

当人在进行利他服务的时候,迷走神经是张开的,会感到特别舒畅、痛快。

⑥此外,大脑伏隔核中的多巴胺和阿片受体两种神经递质也对幸福感起到重要的作用:多巴胺系统帮助人去“寻求”,阿片系统则负责“咂摸..”。

这也意味着幸福知觉包含了不同的阶段。

【VIP专享】2014届上海市徐汇区、金山区、松江区高三二模化学试卷(官方版)

【VIP专享】2014届上海市徐汇区、金山区、松江区高三二模化学试卷(官方版)

C.放入装有铁粉的透气的小纸袋
二、选择题(本题共 36 分,每小题 3 分,只有一个正确选项。)
6.右图是一种有机物的分子模型,图中的“棍”代表单键或双键,该模型代表的有机物可能是
A.饱和一元醇
C.羧酸
7.下列微粒的存在最能说明碘可能呈现金属性的是
A.IBr
B.I2O5
B.羟基酸
Dห้องสมุดไป่ตู้饱和一元醛
C.I3+
8.在试管中注入某红色溶液,给试管加热,溶液颜色逐渐变浅,则原溶液可能是
A.滴有酚酞的 Na2CO3 溶液 C.滴有石蕊的 AlCl3 溶液
D.放入装有生石灰透气的小纸袋
D.I-
B.溶有 SO2 的品红溶液 D.滴有酚酞的饱和 Ca(OH)2 溶液
1
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徐汇区2014年高三语文二模试卷综述

徐汇区2014年高三语文二模试卷综述

2013学年第二学期徐汇区学习能力诊断卷高三年级语文学科2014.4考生注意:1.本考试设试卷和答题卷两部分,试卷包括试题与答题要求,所有答题必须涂(选择题)或写(非选择题)在答题卷上,做在试卷上一律不得分。

2.答题卷与试卷在试题编号上是一一对应的,答题时应特别注意,不能错位。

3.考试时间150分钟。

试卷满分150分。

一阅读80分(一)阅读下文,完成第1——6题。

(18分)①幸福感是一种主观体验,是个体依据自己设定的标准对其生活质量所作的整体评价与感受。

尽管人的幸福感各有差异,但幸福感的存在是有科学依据的。

②对进化论的解读使我们发现,人类的形态在进化过程中发生了剧烈的变化。

由于直立行走使骨盆变得狭短,人类的婴儿比其他动物的婴儿更加脆弱,更需要保护和关怀,这决定了人类必须关照自己的后代,否则很难生存。

人类的毛发少而裸露的皮肤多,拥有发达的触觉,需要接触,需要抚摸。

人类还有巨大的脑,大脑皮质相对于身体特质的比例居动物之首,因而人类是唯一具有灵性的动物,能进行心灵的沟通和对话。

可见,进化决定了人类的本性是善良、友爱、相互支持的,具有获取幸福感的基础。

③一些新的研究进一步揭示了幸福感产生的生理机制。

④通过对大脑进行核磁共振的扫描,科学家发现积极的情感对应的是大脑前额叶左边的皮层活动,消极情感则对应同一个区域右边的皮层。

扫描还显示,脑部所分泌的Oxytocin能抑制杏仁核区域(脑部处理恐惧和危险的部位)和纹状体区域(脑部凭借过去的回馈来指引未来行为的地带)的活动,从而降低人的防卫心和恐惧感,让人忘掉过去的痛苦,更相信他人。

Oxytocin在中国被翻译成“催产素”,但它实则是一种男女皆有的神经激素,不仅可在女性生育时促进产道的收缩,还有更多其他的心理影响,尤其对于抑郁症和孤独症患者会有很大帮助。

⑤科学家们继而发现迷走神经与人类积极的体验和亲社会行为有关。

迷走神经支配呼吸、消化系统的绝大部分器官,以及心脏的感觉、运动和腺体的分泌,所以人一旦遇到激动、兴奋的事件,心胸便比较开阔。

2014上海徐汇高三语文二模参考答案

2014上海徐汇高三语文二模参考答案

2013学年第二学期徐汇区高三年级语文学科学习能力诊断卷参考答案及评分标准一阅读 80分(一)18分1. 仔细辨别、品味,享受幸福(2分)2. 大脑前额叶皮层活动和催产素分别反应或左右着人的幸福感。

(2分)3. 幸福感与人的健康关系密切,而且幸福并不虚幻。

(大意对即可,2分)4. 承接上文对幸福感认识的误区,从主观感受与客观福祉未必对等的角度,解释了中国人幸福感迷失的原因;又进一步引出下文“如今还能不能找到幸福”的问题。

(3分)5. D(3分)6. 文章先后从生理学和心理学两个角度对“幸福感”加以说明。

生理学角度,首先以进化论的观点,解释人体形态上的进化使人类具有了获取幸福感的基础;其次结合脑科学的发现,进一步说明了幸福感产生的生理机制。

在此之后,文章再从心理学角度,说明幸福感同人的生理及心理健康密切相关。

(6分)(二)20分7. 生动形象地表现了四月岷山的春色在不知不觉中逐渐变浓的过程。

(2分)8. 四月天里,生命已经萌发,处处生机勃勃,充满着变化的可能;此时勃发的生命也是新鲜稚嫩的,略带青涩,尚不成熟。

(2分)9. 转折并推进文意,强调四月没有永恒的美,写出了四月的春意在带来生机的同时却又变化万端、转瞬即逝,为结尾抒发感慨作了铺垫。

(3分)10.C(3分)11.随着时代的发展,人类身上某些最本真、最新鲜、最充满活力的东西在不经意之间被成熟、理性、世故甚至颓废所取代,作者对此流露出淡淡的伤感情绪。

(4分)12.可以从修辞(反复、比喻、拟人)、白描、描写抒情结合、多角度描写、动静相衬、虚实相生等方面加以赏析。

(6分)(三)6分(超过6空,按前6空评分)13.⑴出则无敌国外患者⑵屋舍俨然⑶澄江一道月分明⑷又恐琼楼玉宇⑸宫阙万间都做了土⑹岩上无心云相逐⑺予尝求古仁人之心⑻恐美人之迟暮(四)8分14.柳永(1分)15.B(3分)16.阴晴不定的天气喻示着姑娘们情绪的阴晴不定。

“撩”花“拨”柳的动作,写出了她们一览春色的急切心情。

2014上海徐汇、松江、金山区高考地理二模试题(详细解析版).

2014上海徐汇、松江、金山区高考地理二模试题(详细解析版).

2014届上海徐汇区、松江区、金山区高三地理二模试卷(120分钟完卷 满分为150分)全卷包括两大题,第一大题为选择题,第二大题为综合分析题。

一、选择题(共60分,每小题2分,每小题只有一个正确答案)(一)由我国科学家主持的南海第二次大洋钻探,首次钻取了南海形成时期的岩石样本,为精确确定南海深海海盆扩张始末的年代提供了可靠证据。

1.研究南海海盆扩张始末所需的岩石类型是 A .玄武岩B .花岗岩C .沉积岩D .变质岩 答案:A解析:海盆扩张地区通常是板块生长区,地壳厚度薄弱, 地球内部岩浆在强大的压力下在此处喷涌出地表,形成 新的地壳。

在喷出岩中最多的为玄武岩。

因此,要研究 海盆扩张,只需要研究此处玄武岩的性态。

故选A 。

2.南海所属的板块和其附近的板块边界类型是 A .南极洲板块;生长边界 B .太平洋板块;消亡边界 C .印度洋板块;生长边界D .亚欧板块;消亡边界 答案:D解析:从板块分布图上可以看到,南海所属板块为亚欧板块,此处为消亡边界。

故选D 。

3.此次南海大洋钻探的应用价值包括①了解洋流和潮汐等海水运动为南海海运安全保驾护航 ②有利于勘探开发该区域丰富的油气和深海金属矿资源 ③对海底沉积物中微生物群进行研究以准确了解海底环境 ④为地震海啸等灾害寻找原因以提高灾害预警的准确性 A .①②③ B .②③④ C .①③④ D .①②④答案:B解析:海底勘探对洋流及潮汐的研究的没有作用。

②③④对,故选B 。

(二)人口金字塔的类型不仅能显示一个国家的人口结构与人口发展阶段,同时也能反映该国社会经济发展的程度。

读人口金字塔图,回答问题:全球局部地区板块分布示意图南海第二次大洋钻探预设井位分布甲 乙 丙 丁4.上述四种类型的人口金字塔图,若按人口发展阶段的前后排列,其正确的顺序是A.甲乙丙丁B.甲丙丁乙C.甲乙丁丙D.甲丁丙乙答案:D解析:按照人口增长理论,人口的发展一般要经历增长阶段,稳定阶段和缩减阶段,其中增长阶段又可分为快速增长阶段和减速增长阶段,在图中对应的人口金字塔分别为甲丁丙乙。

2014届上海徐汇二模作文(7篇)(选择了不同层次的典型范文)

2014届上海徐汇二模作文(7篇)(选择了不同层次的典型范文)

徐汇区作文《汉书·艺文志》中有“安其所习,毁所不见”一语,大意是指:人往往安于自己所习惯、所熟悉的事物,对从没见过,或未能直接、间接经验过的事物,则常常予以否定。

对这种生活中普遍存在的现象,你有怎样的联想和思考?请选取一个角度写一篇文章,谈谈你的想法。

要求:⑴题目自拟;⑵全文不少于800字;⑶不要写成诗歌。

有容乃大(一类上:67)犹记得房龙《宽容序言》中的无知山谷,人们安于山谷中平静却落后的生活,对于开拓者所言的那个从未目见知晓的谷外世界,予以了强烈的抗拒与否定。

可见,无论是于《汉书》“安其所习,毁所不见”还是于《宽容序言》都向我们揭示了一个现象:人往往安于自己所习惯,所熟悉的事物,对未曾目见和经历的则常常予以否定。

然,世界发展的根源在于旧事物的灭亡,新事物的产生,若想寻求世界的进步与发展,我们必须以更积极开放的姿态接纳新生事物,有容才能成其大。

“安其所习,毁所不见”是生活是普遍存在的现象。

究其原因有三。

其一便是对结果好坏的未知。

历来,社会的变革都将会遭到保守派的层层阻隔,原因很简单,人们无法预见变革的后果,当现时的安稳与革新的未知安放于天秤两端,人们手中的砝码宁愿偏向虽有弊病的当下,也不愿投向未知的变革。

但我们需要看到先秦时代,彻底的商鞅变法使曾经积贫积弱边陲小国蜕变为强秦;反观六国之变法,多在变法者死后就又回复了过去的轨道,不思新变,终为秦灭。

其二是对权威和现时利益者的维护。

爱迪生的直流发电曾为其带来巨大的财富与利益,因此在明知特斯拉交流电更为优异的情况下,仍对其进行强烈的打压与否定。

而当如今交流电的普遍使用,爱迪生的自私与浅陋显露无遗。

最后是人们惯性思维的(殆)怠惰,这份惰性让我们停止思考,将创新的思考转为对旧有的维护,如此对创新进步的伤害之大让人唏嘘不已。

不管是出于对未知的求称心理和对权威利益的奉行,还是人们所先具有的怠惰都将毫无疑问的阻碍社会的发展和进步,因此对于未知的新事物,宽容接纳是上策,切莫用否定将真理的萌芽抑掐在初始阶段;然后要打破权威,不能因为旧有保守者们的维护和否定,我们自身就加以不经思考的否定;要勇于实践,实践是检验真理的唯一标准,对于未知的新生事物通过实践才能确实判断优劣。

上海市徐汇、松江、金山三区2014届高三学习能力诊断(二模)物理试题.pdf

上海市徐汇、松江、金山三区2014届高三学习能力诊断(二模)物理试题.pdf

须将原选项用橡皮擦去,重新选择。第四、第五和第六大题的作答必须用蓝色或黑色的钢笔或圆珠笔写在答题纸上与试
卷题号对应的位置(作图可用铅笔)。
3、第30、3l、32、33题要求写出必要的文字说明、方程式和重要的演算步骤。只写出最后答案,而未写出主要演
算过程的,不能得分。有关物理量的数值计算问题,答案中必须明确写出数值和单位。
起以大小为gtanθ的加速度向左做匀加速直线运动,两者保持相对静止,则运动过程中,小物块受力的示意图为
二.单项选择题(共24分,每小题3分。每小题只有一个正确选项)
9.
10.A)2(B)4(C)6(D)8
11.A)h=H(B)h<(C)h=(D)<h<H
12.S1闭合;当红灯亮起时,S2闭合。若要在汽车违章时进行拍摄,且摄像仪需要一定的工作电压,能实现此功能
一直增大一直减小(C)先减小后增大(D)先增大后减小
15.x轴上的电势分布如图所示,由图可以判断
(A)x=2m处电场强度可能为零
(B)x=2m处电场方向一定沿x轴正方向
(C)沿x轴正方向,电场强度先增大后减小
(D)某负电荷沿x轴正方向移动,电势能始终增大
16.COD处于匀强磁场中,磁场方向垂直导轨平面,电阻不计。一均匀金属棒M在外力F作用下,以恒定速率v由O点
轨间存在一匀强磁场,磁场方向垂直于导轨向上,磁感应强度B2=1T。两根与倾斜导轨垂直的金属杆M、N被固定在导轨
上,M、N的质量均为m=1kg,电阻均为R=0.5Ω,杆与水平导轨间的动摩擦因数为(=0.4。现将M杆从距OO′边界
x=10m处静止释放,已知M杆到达OO′边界前已开始做匀速运动。当M杆一到达OO′边界时,使N杆在一平行导轨向下的
时间为Δt。

2014年上海市松江区、徐汇区、金山区高考数学二模试卷(文科)含详解

2014年上海市松江区、徐汇区、金山区高考数学二模试卷(文科)含详解

2014年上海市松江区、徐汇区、金山区高考数学二模试卷(文科)一.填空题:(本题满分56分,每小题4分)1.(4分)已知集合A={x|<0},B={x|x2﹣2x﹣3≥0,x∈R},则A∩B=.2.(4分)直线x+y+1=0的倾斜角的大小为.3.(4分)函数y=cos(2x+)的单调递减区间是.4.(4分)函数的值域.5.(4分)设复数z满足i(z+1)=﹣3+2i,则=.6.(4分)某学校高一、高二、高三共有2400名学生,为了调查学生的课余学习情况,拟采用分层抽样的方法抽取一个容量为120的样本.已知高一有760名学生,高二有840名学生,则在该学校的高三应抽取名学生.7.(4分)函数的最小正周期T=.8.(4分)已知函数f(x)=arcsin(2x+1),则f﹣1()=.9.(4分)如图在直三棱柱ABC﹣A1B1C1中∠ACB=90°,AA1=2,AC=BC=1,则异面直线A1B与AC所成角的余弦值是.10.(4分)已知实数x、y满足不等式组,则z=3x+4y的最大值是.11.(4分)若(1﹣)n(n∈N*,n>1)的展开式中x﹣4的系数为a n,则(++…+)=.12.(4分)如图,三行三列的方阵中有9个数a ij(i=1,2,3;j=1,2,3),从中任取三个数,则这三个数位于不同行不同列的概率是.(结果用分数表示)13.(4分)对于集合A={a1,a2,…,a10},定义集合S={x|x=a i+a j,1≤i<j≤10},记集合S中的元素个数为S(A).若a1,a2,…,a10是公差大于零的等差数列,则S(A)=.14.(4分)如图所示,在边长为2的正六边形ABCDEF中,动圆Q的半径为1,圆心在线段CD(含端点)上运动,P是圆Q上及内部的动点,设向量=m+n(m、n为实数),则m+n的最大值为.二.选择题:(本题满分20分,每小题5分)15.(5分)命题p:a≥1;命题q:关于x的实系数方程x2﹣2x+a=0有虚数解,则p是q的()A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分也不必要条件16.(5分)已知直线l⊥平面α,直线m⊂平面β,有下面四个命题,其中正确命题是①α∥β⇒l⊥m②α⊥β⇒l∥m③l∥m⇒α⊥β④l⊥m⇒α∥βA.①与②B.①与③C.②与④D.③与④17.(5分)在△ABC中,角A、B、C的对边分别是a、b、c,且∠A=2∠B,则等于()A.B.C.D.18.(5分)函数图象上存在不同的三点到原点的距离构成等比数列,则以下不可能成为公比的数是()A.B.C.D.三.解答题:(本大题共5题,满分74分)19.(12分)如图示,给出的是某几何体的三视图,其中正视图与侧视图都是边长为2的正三角形,俯视图为半径等于1的圆.试求这个几何体的侧面积与体积.20.(14分)如图所示,某旅游景点有一座风景秀丽的山峰,山上有一条笔直的山路BC和一条索道AC,小王和小李打算不坐索道,而是花2个小时的时间进行徒步攀登.已知∠ABC=120°,∠ADC=150°,BD=1(千米),AC=3(千米).假设小王和小李徒步攀登的速度为每小时1200米,请问:两位登山爱好者能否在2个小时内徒步登上山峰.(即从B点出发到达C点)21.(14分)已知椭圆x2+2y2=a2(a>0)的一个顶点和两个焦点构成的三角形的面积为4.(1)求椭圆C的方程;(2)已知直线y=k(x﹣1)与椭圆C交于A、B两点,若点M(,0),求证•为定值.22.(16分)定义:对于函数f(x),若存在非零常数M,T,使函数f(x)对于定义域内的任意实数x,都有f(x+T)﹣f(x)=M,则称函数f(x)是广义周期函数,其中称T为函数f(x)的广义周期,M称为周距.(1)证明函数f(x)=x+(﹣1)x(x∈Z)是以2为广义周期的广义周期函数,并求出它的相应周距M的值;(2)试求一个函数y=g(x),使f(x)=g(x)+Asin(ωx+φ)(x∈R)(A、ω、φ为常数,A>0,ω>0)为广义周期函数,并求出它的一个广义周期T 和周距M;(3)设函数y=g(x)是周期T=2的周期函数,当函数f(x)=﹣2x+g(x)在[1,3]上的值域为[﹣3,3]时,求f(x)在[﹣9,9]上的最大值和最小值.23.(18分)一个三角形数表按如下方式构成(如图:其中项数n≥5):第一行是以4为首项,4为公差的等差数列,从第二行起,每一个数是其肩上两个数的和,例如:f(2,1)=f(1,1)+f(1,2);f(i,j)为数表中第i行的第j个数.(1)求第2行和第3行的通项公式f(2,j)和f(3,j);(2)证明:数表中除最后2行以外每一行的数都依次成等差数列;(3)求f(i,1)关于i(i=1,2,…,n)的表达式.2014年上海市松江区、徐汇区、金山区高考数学二模试卷(文科)参考答案与试题解析一.填空题:(本题满分56分,每小题4分)1.(4分)已知集合A={x|<0},B={x|x2﹣2x﹣3≥0,x∈R},则A∩B={x|﹣5<x≤﹣1}.【考点】1E:交集及其运算.【专题】51:函数的性质及应用;5J:集合.【分析】利用分式不等式和一元二次不等式分别求出集合A和B,由此能求出A ∩B.【解答】解:∵集合A={x|<0}={x|﹣5<x<2},B={x|x2﹣2x﹣3≥0,x∈R}={x|x≤﹣1或x≥3},∴A∩B={x|﹣5<x≤﹣1}.故答案为:{x|﹣5<x≤﹣1}.【点评】本题考查集合的交集的运算,是基础题,解题时要认真审题,注意分式不等式和一元二次不等式的合理运用.2.(4分)直线x+y+1=0的倾斜角的大小为.【考点】I2:直线的倾斜角.【专题】5B:直线与圆.【分析】化直线的一般式方程为斜截式,求出直线的斜率,由倾斜角的正切值等于斜率求倾斜角.【解答】解:由x+y+1=0,得,∴直线x+y+1=0的斜率为,设其倾斜角为θ(0≤θ<π),则,∴θ=.故答案为:.【点评】本题考查直线的倾斜角,考查直线倾斜角与斜率的关系,是基础题.3.(4分)函数y=cos(2x+)的单调递减区间是.【考点】HA:余弦函数的单调性.【专题】57:三角函数的图像与性质.【分析】根据余弦函数的单调性的性质即可得到结论.【解答】解:由2kπ≤2x+≤2kπ+π,即kπ﹣≤x≤kπ,k∈Z故函数的单调减区间为,故答案为:.【点评】本题主要考查余弦函数的单调性的求法,要求熟练掌握三角函数的图象和性质.4.(4分)函数的值域.【考点】34:函数的值域.【专题】11:计算题.【分析】因为自变量x是正数,所以可以用基本不等式进行求解:,所以函数的最小值为,当且仅当x=,即x=时取到等号,由此可得函数的值域.【解答】解:当x>0时,f(x)=x+,当且仅当x=,即x=时取到等号,因此该函数的值域为[2,+∞).故答案为:[2,+∞).【点评】本题考查了函数值域的求法,利用函数解析式的特点选择合适的方法求解函数的值域.利用基本不等式求值域是解决函数值域问题的一种方法,关键要用到基本不等式的放缩办法,要注明等号成立的条件.5.(4分)设复数z满足i(z+1)=﹣3+2i,则=1﹣3i.【考点】A5:复数的运算.【专题】5N:数系的扩充和复数.【分析】利用复数的运算法则即可得出.【解答】解:∵复数z满足i(z+1)=﹣3+2i,∴﹣i•i•(z+1)=﹣i(﹣3+2i),化为z+1=2+3i,化为z=1+3i,∴=1﹣3i.故答案为:1﹣3i.【点评】本题考查了复数的运算法则,属于基础题.6.(4分)某学校高一、高二、高三共有2400名学生,为了调查学生的课余学习情况,拟采用分层抽样的方法抽取一个容量为120的样本.已知高一有760名学生,高二有840名学生,则在该学校的高三应抽取40名学生.【考点】B3:分层抽样方法.【专题】5I:概率与统计.【分析】由所给的学校的总人数和要抽取的样本容量,得到每个个体被抽到的概率,做出高三年级的人数,乘以概率得到结果.【解答】解:∵某高中共有学生2400人,采用分层抽样法抽取容量为120的样本,∴每个个体被抽到的概率是=,高三年级有2400﹣760﹣840=800人∴要在高三抽取800×=40人,故答案为:40.【点评】本题考查分层抽样方法,本题解题的关键是做出每个个体被抽到的概率,用这个概率乘以指定年级的人数,就可以得到这个年级要抽取的样本数.7.(4分)函数的最小正周期T=π.【考点】GL:三角函数中的恒等变换应用;H1:三角函数的周期性;O1:二阶矩阵.【专题】57:三角函数的图像与性质.【分析】先利用二阶矩阵化简函数式f(x),再把函数y=f(x)化为一个角的一个三角函数的形式,然后求出它的最小正周期.【解答】解:函数=(sinx+cosx)(﹣sinx+cosx)﹣2sinxcos(π﹣x)=cos2x+sin2x=sin(2x+),它的最小正周期是:T==π.故答案为:π【点评】本题考查三角函数的周期性及其求法,二倍角的正弦,考查计算能力,是基础题.8.(4分)已知函数f(x)=arcsin(2x+1),则f﹣1()=.【考点】4R:反函数.【专题】11:计算题.【分析】欲求,只需令arcsin(2x+1)=求出x的值,根据原函数与反函数之间的关系可得结论.【解答】解:令arcsin(2x+1)=即sin=2x+1=解得x=故答案为:【点评】本题主要考查了反函数,以及反函数求值和三角形函数的运算,属于基础题.9.(4分)如图在直三棱柱ABC﹣A1B1C1中∠ACB=90°,AA1=2,AC=BC=1,则异面直线A1B与AC所成角的余弦值是.【考点】LM:异面直线及其所成的角.【专题】11:计算题.【分析】先通过平移将两条异面直线平移到同一个起点A1,得到的锐角或直角就是异面直线所成的角,在三角形中再利用余弦定理求出此角的余弦值.【解答】解:∵A1C1∥AC,∴异面直线A1B与AC所成角为∠BA1C1,易求,∴.故答案为:【点评】本小题主要考查异面直线所成的角,考查空间想象能力、运算能力和推理论证能力,属于基础题.10.(4分)已知实数x、y满足不等式组,则z=3x+4y的最大值是20.【考点】7C:简单线性规划.【专题】11:计算题.【分析】先根据约束条件画出可行域,再利用几何意义求最值,z=3x+4y中z 表示在y轴上的截距,要求z得最大值,只需求出可行域直线在y轴上的截距最大值即可.【解答】解:作出不等式组表示的平面区域如图所示,由z=3x+4y得y=x+,则z表示直线在y轴上的截距,截距越大,z越大,要求z的最大值,则只要求解直线y=x+的截距的最大值结合图形可知,当直线z=3x+4y过点C(0,5)时,在y轴上截距最大,z最大由此时z取得最大值20故答案为:20【点评】本题主要考查了简单的线性规划,以及利用几何意义求最值,属于基础题.11.(4分)若(1﹣)n(n∈N*,n>1)的展开式中x﹣4的系数为a n,则(++…+)=2.【考点】DA:二项式定理.【专题】5P:二项式定理.【分析】先求出二项式展开式的通项公式,再令x的幂指数等于﹣4,求得r的值,即可求得展开式中的x﹣4的系数a n,再用裂项法求得++…+的值,从而求得所给式子的值.【解答】解:(1﹣)n(n∈N*,n>1)的展开式的通项公式为T r+1=•(﹣1)r•x﹣2r,令﹣2r=﹣4,r=2,故展开式中x﹣4的系数为a n==,∴==2(﹣).则(++…+)=2(+++…+)=2(1﹣)=2,故答案为:2.【点评】本题主要考查二项式定理的应用,二项式系数的性质,二项式展开式的通项公式,求展开式中某项的系数,用裂项法进行数列求和,属于中档题.12.(4分)如图,三行三列的方阵中有9个数a ij(i=1,2,3;j=1,2,3),从中任取三个数,则这三个数位于不同行不同列的概率是.(结果用分数表示)【考点】C6:等可能事件和等可能事件的概率.【专题】11:计算题;5I:概率与统计.【分析】从9个数中任取3个数共有C93=84种取法,若三个数分别位于不同的三行,有三种方法;若三个数分别位于不同的三列,满足要求的选法共有6种,综合可得答案.【解答】解:从9个数中任取3个数共有C93=84种取法,三个数分别位于三行或三列的情况有6种;∴所求的概率为=.故答案为:.【点评】本题主要考查计数原理和组合数公式的应用、概率的计算公式,属于基础题.13.(4分)对于集合A={a1,a2,…,a10},定义集合S={x|x=a i+a j,1≤i<j≤10},记集合S中的元素个数为S(A).若a1,a2,…,a10是公差大于零的等差数列,则S(A)=17.【考点】12:元素与集合关系的判断;83:等差数列的性质.【专题】54:等差数列与等比数列.【分析】由已知条件,利用等差数列的性质,用列举法能求出S(A).【解答】解:∵集合A={a1,a2,…,a10},定义集合S={x|x=a i+a j,1≤i<j≤10},a1,a2,…,a10是公差大于零的等差数列,∴集合S中的元素有:a1+a2,a1+a3,a1+a4,a1+a5,,a1+a7,a1+a8,a1+a9,a1+a10,a2+a10,a3+a10,a4+a10,,a6+a10,a7+a10,a8+a10,a9+a10,共17个,∴S(A)=17.故答案为:17.【点评】本题考查集合中元素个数的求法,是基础题,解题时要认真审题,注意等差数列的性质的灵活运用.14.(4分)如图所示,在边长为2的正六边形ABCDEF中,动圆Q的半径为1,圆心在线段CD(含端点)上运动,P是圆Q上及内部的动点,设向量=m+n(m、n为实数),则m+n的最大值为5.【考点】9H:平面向量的基本定理.【专题】5A:平面向量及应用.【分析】连接AE会发现它与AB垂直,所以构造,将条件中的代入,便会得到,而,所以经过化简就可得到.同样的办法你会得到,显然得到的这两式需相加便经过化简得到m+n=,而这正好是在方向上的投影,所以求这个投影的最大值即可,而投影的最大值,通过图形就能得到.【解答】解:如图所示,.∴==6n ①同理,②①+②得:;∵,∴.∵=.∴,其几何意义就是在上的投影.∴求m+n的最大值就转化为求在上投影最大值.从图形上可以看出:当点Q和D点重合时,在上的投影取到最大值5.【点评】本题需注意的是构造两组数量级,将求m+n的最大值转化为求在方向上投影的最大值.二.选择题:(本题满分20分,每小题5分)15.(5分)命题p:a≥1;命题q:关于x的实系数方程x2﹣2x+a=0有虚数解,则p是q的()A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分也不必要条件【考点】29:充分条件、必要条件、充要条件.【专题】5L:简易逻辑.【分析】根据复数的有关性质,利用充分条件和必要条件的定义进行判断.【解答】解:若关于x的实系数方程x2﹣2x+a=0有虚数解,则判别式△<0,即8﹣4a<0,解得a>2,∴p是q的必要不充分条件,故选:B.【点评】本题主要考查充分条件和必要条件的判断,利用复数的性质是解决本题的关键.16.(5分)已知直线l⊥平面α,直线m⊂平面β,有下面四个命题,其中正确命题是①α∥β⇒l⊥m②α⊥β⇒l∥m③l∥m⇒α⊥β④l⊥m⇒α∥βA.①与②B.①与③C.②与④D.③与④【考点】LP:空间中直线与平面之间的位置关系;LQ:平面与平面之间的位置关系.【专题】14:证明题.【分析】①α∥β⇒l⊥m,可由线面垂直的性质进行判断;②α⊥β⇒l∥m,可以由面面垂直的性质进行判断;③l∥m⇒α⊥β面面垂直的判定定理进行判断;④l⊥m⇒α∥β,可由面面平行的判定定理进行判断.【解答】解:对于①l⊥α,α∥β,m⊂β⇒l⊥m正确;对于②l⊥α,m⊂β,α⊥β⇒l∥m;l与m也可能相交或者异面;对于③l∥m,l⊥α⇒m⊥α,又因为m⊂β则α⊥β正确;对于④l⊥m,l⊥α则m可能在平面α内,也可能不在平面α内,所以不能得出α∥β;综上所述①③正确,故选:B.【点评】本题考查平面与平面之间的位置关系,考查空间想像能力及组织材料判断面面间位置关系的能力,属于基本题型.17.(5分)在△ABC中,角A、B、C的对边分别是a、b、c,且∠A=2∠B,则等于()A.B.C.D.【考点】HP:正弦定理.【专题】11:计算题;35:转化思想.【分析】先根据三角形的内角和以及∠A=2∠B把所求问题转化,再结合正弦定理即可得到答案.【解答】解:∵A+B+C=π,A=2B,∴===.再结合正弦定理得:=.故选:A.【点评】本题主要考查正弦定理的应用.解决本题的关键在于根据三角形的内角和以及∠A=2∠B把所求问题转化.18.(5分)函数图象上存在不同的三点到原点的距离构成等比数列,则以下不可能成为公比的数是()A.B.C.D.【考点】87:等比数列的性质.【专题】11:计算题.【分析】根据平面几何切割线定理:从圆外一点做圆的切线和割线,则切线长是割线与它的圆外部分的比例中项.假设存在,则可计算出公比的范围,从而可下结论.【解答】解:根据平面几何切割线定理:从圆外一点做圆的切线和割线,则切线长是割线与它的圆外部分的比例中项.鉴于此,从原点作该半圆的切线,切线长为:,设割线与半圆的另外两个交点到原点的距离分别是a和b,则b=aq2,且ab=(aq)2=3,所以aq=;所以q=,当,则;当时,考查四个选项,只有B选项不符合上述范围故选:B.【点评】本题的考点是等比关系的确定,主要课程等比数列的定义,等比中项及切割线定理,属于基础题.三.解答题:(本大题共5题,满分74分)19.(12分)如图示,给出的是某几何体的三视图,其中正视图与侧视图都是边长为2的正三角形,俯视图为半径等于1的圆.试求这个几何体的侧面积与体积.【考点】L!:由三视图求面积、体积.【专题】5F:空间位置关系与距离.【分析】由已知中的三视图,可以分析出该几何体是一个底面半径为1,母线长为2的圆锥,进而可得到圆锥的高为,代入圆锥的体积公式和表面积公式,可得答案.【解答】解:根据几何体的三视图知,原几何体是以半径为1的圆为底面,母线长为2的圆锥则圆锥的高为的圆锥.…3分=πrl=2π,…7分则它的侧面积S侧体积.…11分【点评】本题考查的知识点是由三视图求体积和表面积,其中根据已知的三视图判断出几何体的形状及底面半径,母线长等几何量是解答的关键.20.(14分)如图所示,某旅游景点有一座风景秀丽的山峰,山上有一条笔直的山路BC和一条索道AC,小王和小李打算不坐索道,而是花2个小时的时间进行徒步攀登.已知∠ABC=120°,∠ADC=150°,BD=1(千米),AC=3(千米).假设小王和小李徒步攀登的速度为每小时1200米,请问:两位登山爱好者能否在2个小时内徒步登上山峰.(即从B点出发到达C点)【考点】HU:解三角形.【专题】12:应用题;58:解三角形.【分析】先利用正弦定理,求出AD,再在△ADC中,由余弦定理,求出DC,即可得出结论.【解答】解:由∠ADC=150°知∠ADB=30°,由正弦定理得,所以,.﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣(4分)在△ADC中,由余弦定理得:|AC|2=|AD|2+|DC|2﹣2|AD|•|DC|cos150°,即,即DC2+3•DC﹣6=0,解得(千米),﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣(10分)所以|BC|≈2.372(千米),﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣(12分)由于2.372<2.4,所以两位登山爱好者能够在2个小时内徒步登上山峰.﹣﹣﹣(14分)【点评】本题考查解三角形的实际应用,考查正弦定理、余弦定理,考查学生的计算能力,正确运用正弦定理、余弦定理是关键.21.(14分)已知椭圆x2+2y2=a2(a>0)的一个顶点和两个焦点构成的三角形的面积为4.(1)求椭圆C的方程;(2)已知直线y=k(x﹣1)与椭圆C交于A、B两点,若点M(,0),求证•为定值.【考点】K4:椭圆的性质.【专题】5D:圆锥曲线的定义、性质与方程.【分析】(1)由已知条件推导出,,由此能求出椭圆方程.(2)由,得(2k2+1)x2﹣4k2x+2k2﹣8=0,设A(x1,y1),B(x2,y2),利用韦达定理结合已知条件能证明为定值.【解答】(本题满分(14分);第(1)小题(6分),第(2)小题8分)解:(1)设椭圆的短半轴为b,半焦距为c,则,由c2=a2﹣b2,得,由,解得a2=8,b2=4,∴椭圆方程为.(6分)(2)由,得(2k2+1)x2﹣4k2x+2k2﹣8=0,(8分)设A(x1,y1),B(x2,y2),由韦达定理得:,∴=,====,∴为定值.(14分)【点评】本题考查椭圆方程的求法,考查向量的数量积为定值的证明,解题时要认真审题,注意椭圆的简单性质的合理运用.22.(16分)定义:对于函数f(x),若存在非零常数M,T,使函数f(x)对于定义域内的任意实数x,都有f(x+T)﹣f(x)=M,则称函数f(x)是广义周期函数,其中称T为函数f(x)的广义周期,M称为周距.(1)证明函数f(x)=x+(﹣1)x(x∈Z)是以2为广义周期的广义周期函数,并求出它的相应周距M的值;(2)试求一个函数y=g(x),使f(x)=g(x)+Asin(ωx+φ)(x∈R)(A、ω、φ为常数,A>0,ω>0)为广义周期函数,并求出它的一个广义周期T 和周距M;(3)设函数y=g(x)是周期T=2的周期函数,当函数f(x)=﹣2x+g(x)在[1,3]上的值域为[﹣3,3]时,求f(x)在[﹣9,9]上的最大值和最小值.【考点】3Q:函数的周期性.【专题】51:函数的性质及应用.【分析】(1)由已知条件推导出f(x+2)﹣f(x)═2,由此证明函数f(x)=x+(﹣1)x(x∈Z)是广义周期函数,它的周距为2.(2)设g(x)=kx+b(k≠0),由=,推导出f(x)是广义周期函数,并能求出并求出它的一个广义周期T和周距M.(3)由f(x+2)﹣f(x)=﹣4,知f(x)是广义周期函数,且T=2,M=﹣4,由此能求出f(x)在[﹣9,9]上的最大值和最小值.【解答】(本题满分(16分);第(1)小题(4分),第(2)小题(5分),第(3)小题7分)(1)证明:∵f(x)=x+(﹣1)x(x∈Z),∴f(x+2)﹣f(x)=[(x+2)+(﹣1)x+2]﹣[x+(﹣1)x]=2,(非零常数)∴函数f(x)=x+(﹣1)x(x∈Z)是广义周期函数,它的周距为2.(4分)(2)解:设g(x)=kx+b(k≠0),则f(x)=kx+b+Asin(ωx+φ)∵=(非零常数)∴f(x)是广义周期函数,且.(9分)(3)解:∵f(x+2)﹣f(x)=﹣2(x+2)+g(x+2)+2x﹣g(x)=﹣4,∴f(x)是广义周期函数,且T=2,M=﹣4.(10分)设x1,x2∈[1,3]满足f(x1)=﹣3,f(x2)=3,由f(x+2)=f(x)﹣4得:f(x1+6)=f(x1+4)﹣4=f(x1+2)﹣4﹣4=f(x1)﹣4﹣4﹣4=﹣3﹣12=﹣15,又∵f(x+2)=f(x)﹣4<f(x),∴f(x)在区间[﹣9,9]上的最小值是x在[7,9]上获得的,而x1+6∈[7,9],∴f(x)在[﹣9,9]上的最小值为﹣15.(13分)由f(x+2)=f(x)﹣4,得f(x﹣2)=f(x)+4,∴f(x2﹣10)=f(x2﹣8)+4=f(x2﹣6)+4+4=…=f(x2)+20=23,又∵f(x﹣2)=f(x)+4>f(x),∴f(x)在区间[﹣9,9]上的最大值是x在[﹣9,﹣7]上获得的,而x2﹣10∈[﹣9,﹣7],f(x)在[﹣9,9]上的最大值为23.(16分)【点评】本题考查广义周期函数的证明,考查广义周期函数的求法,考查函数的最大值的求法,解题时要认真审题,注意等价转化思想的合理运用.23.(18分)一个三角形数表按如下方式构成(如图:其中项数n≥5):第一行是以4为首项,4为公差的等差数列,从第二行起,每一个数是其肩上两个数的和,例如:f(2,1)=f(1,1)+f(1,2);f(i,j)为数表中第i行的第j个数.(1)求第2行和第3行的通项公式f(2,j)和f(3,j);(2)证明:数表中除最后2行以外每一行的数都依次成等差数列;(3)求f(i,1)关于i(i=1,2,…,n)的表达式.【考点】F1:归纳推理.【专题】5M:推理和证明.【分析】1)根据等差数列和等比数列的定义即可求出相应的通项公式,(2)由已知,第一行是等差数列,若成等差数列求出公差d即可.(3)根据条件建立方程关系即可求出f(i,1)的表达式.【解答】解:(1)f(2,j)=f(1,j)+f(1,j+1)=2f(1,j)+4=8j+4(j=1,2,…,n﹣1),f(3,j)=f(2,j)+f(2,j+1)=2f(2,j)+8=2(8j+4)+8=16j+16(j=1,2,…,n﹣2)(2)由已知,第一行是等差数列,假设第i(1≤i≤n﹣3)行是以d i为公差的等差数列,则由f(i+1,j+1)﹣f(i+1,j)=[f(i,j+1)+f(i,j+2)]﹣[f(i,j)+f(i,j+1)]=f(i,j+2)﹣f(i,j)=2d i(常数)知第i+1(1≤i≤n﹣3)行的数也依次成等差数列,且其公差为2d i.综上可得,数表中除最后2行以外每一行都成等差数列(3)由于d1=4,d i=2d i﹣1(i≥2),所以,,所以f(i,1)=f(i﹣1,1)+f(i﹣1,2)=2f(i﹣1,1)+d i﹣1由得f(i,1)=2f(i﹣1,1)+2i,于是,即,又因为,所以,数列是以2为首项,1为公差的等差数列,所以,,所以f(i,1)=(i+1)•2i(i=1,2,…n)【点评】本题主要考查等差数列和等比数列的综合应用,考查学生的运算能力,综合性较强,运算量较大.。

上海市徐汇区2014届高三第二次测试物理试卷全国通用-一轮复习

上海市徐汇区2014届高三第二次测试物理试卷全国通用-一轮复习

ABv 0徐汇区2013学年高三物理第二次测试试卷(考试时间120分钟,满分150分)第I 卷(共64分)本卷分单项选择题和多项选择题,共20小题,单项选择题每小题给出的四个答案中只有一个是正确的,选对得3分;多项选择题每小题给出的四个答案中,有二个或三个以上是正确的.选对得4分;选对但不全,得2分;有选错或不答的,得0分. (一)单项选择题 ( 每小题3分,共24分,每小题只有一个正确选项。

) 1.物体做匀速圆周运动过程中,始终在变化的物理量是( ) (A )转速 (B )角速度 (C )周期(D )线速度 2.下列物理量的单位属于国际单位制基本单位的是( )(A )速度(B )力(C )摄氏温度(D )热力学温度3.某物质的密度为ρ,摩尔质量为μ,阿伏伽德罗常数为N A ,则单位体积该物质中所含的分子个数为( )(A )N A ρ (B )N Aμ(C )ρN Aμ(D )μN Aρ4.物体做竖直上抛运动时,在任意相同时间间隔内,速度的变化量( )(A )大小相同、方向相同 (B )大小相同、方向不同 (C )大小不同、方向不同(D )大小不同、方向相同5.秋千的吊绳有些磨损,在摆动过程中,吊绳最容易断裂的时候是秋千()(A )在下摆过程中(B )在上摆过程中 (C )摆到最高点时 (D )摆到最低点时6.消防车停于水平地面上,在缓缓放水的过程中,若车胎不漏气,胎内气体温度不变,不计气体分子势能,则胎内气体( ) (A )分子平均动能减少(B )对外界做正功 (C )分子对胎内壁单位面积上的撞击力增大(D )内能增加7.如图所示,某人向对面的山坡上水平抛出两个质量相等的石块,分别落到A 、B 两处。

不计空气阻力,则落到B 处的石块( )(A )初速度大,运动时间长 (B )初速度大,运动时间短 (C )初速度小,运动时间长 (D )初速度小,运动时间短 8.如图,甲、乙两颗卫星以相同的轨道半径分别绕质量为M 和2M的行星做匀速圆周运动,则() (A )甲的线速度比乙的大 (B )甲的角速度比乙的大 (C )甲的运行周期比乙的小 (D )甲的向心加速度比乙的小(二)单项选择题 ( 每小题3分,共24分,每小题只有一个正确选项。

2014年徐汇区高三数学二模(理)

2014年徐汇区高三数学二模(理)

2013学年第二学期徐汇区学习能力诊断卷高三年级数学学科(理科) 2014.4一.填空题:(本题满分56分,每小题4分) 1.已知集合2|05x A x x -⎧⎫=<⎨⎬+⎩⎭,{}2|230,B x x x x R =--≥∈,则=B A ____________. 2.直线10x +=的倾斜角的大小是____________. 3.函数cos 24y x π⎛⎫=+ ⎪⎝⎭的单调递减区间是____________. 4.函数()22y x x x=+≥的值域是____________. 5.设复数z 满足()132i z i +=-+,则z =____________.6.某学校高一、高二、高三共有2400名学生,为了调查学生的课余学习情况,拟采用分层抽样的方法抽取一个容量为120的样本.已知高一有820名学生,高二有780名学生,则在该学校的高三应抽取____________名学生. 7.函数()()sin cos cos 2sin cos sin x x x f x xx xπ+-=-的最小正周期T =____________.8.已知函数)12(arcsin )(+=x x f ,则=-)6(1πf____________. 9.如图,在直三棱柱111ABC A B C -中,0190,2,1ACB AA AC BC ∠====,则异面直线1A B 与AC 所成角的余弦值是____________.10.若()211,1nn N n x *⎛⎫-∈> ⎪⎝⎭的展开式中4-x 的系数为n a ,则23111lim n n a a a →∞⎛⎫+++⎪⎝⎭=____________.11.在极坐标系中,定点A (2,),2π点B 在直线0sin cos =+θρθρ上运动,则点A 和点B 间的最短距离为____________.12.如图,三行三列的方阵中有9个数(123123)ij a i j ==,,;,,,从中任取三个数,111213212223313233a a a a a a a a a ⎛⎫ ⎪ ⎪ ⎪⎝⎭ 则至少有两个数位于同行或同列的概率是____________. (结果用分数表示) 13.如图所示,在边长为2的正六边形ABCDEF 中,动圆Q 的半径为1,圆心在线段CD (含端点)上运动,P 是圆Q 上及内部的动点,设向量(,AP mAB nAF m n =+为实数),则m n +的最大值为____________.14.对于集合12{,,,}n A a a a =⋅⋅⋅(*,3)n N n ∈≥,定义集合,1}{i j x a a i j n S x =+≤<≤=,记集合S 中的元素个数为()S A .若12,,,n a a a ⋅⋅⋅是公差大于零的等差数列,则()S A =____________.二.选择题:(本题满分20分,每小题5分)15.已知直线⊥l 平面α,直线m ⊆平面β,给出下列命题,其中正确的是-------------( ) ①m l ⊥⇒βα// ②m l //⇒⊥βα ③βα⊥⇒m l // ④βα//⇒⊥m l A .②④ B. ②③④ C. ①③ D. ①②③16.在ABC ∆中,角C B A 、、的对边分别是c b a 、、,且B A ∠=∠2,则BB3sin sin 等于-------( ) A .c a B .b c C .abD .c b17.函数y =...成为公比的数是---------------------------------------------------------------------------------- ( )A .23 B .21 C .33D .3 18.设圆O 1和圆O 2是两个相离的定圆,动圆P 与这两个定圆都相切,则圆P 的圆心轨迹可能是 ①两条双曲线;②一条双曲线和一条直线;③一条双曲线和一个椭圆.以上命题正确的是--( )A .① ③B .② ③C .① ②D .① ② ③三.解答题:(本大题共5题,满分74分)19.(本题满分12分,第(1)小题6分,第(2)小题6分)如图,△ABC 中,090=∠ACB ,030=∠ABC ,3=BC ,在三角形内挖去一个半圆(圆心O 在边BC 上,半圆与AC 、AB 分别相切于点C 、M ,与BC 交于点N ),将△ABC 绕直线BC 旋转一周得到一个旋转体.(1)求该几何体中间一个空心球的表面积的大小;(2)求图中阴影部分绕直线BC 旋转一周所得旋转体的体积.20.(本题满分14分)如图所示,某旅游景点有一座风景秀丽的山峰,山上有一条笔直的山路BC 和一条索道AC ,小王和小李打算不坐索道,而是花2个小时的时间进行徒步攀登.已知0120ABC ∠=,0150ADC ∠=,1BD =(千米),3AC =(千米).假设小王和小李徒步攀登的速度为每小时1200米,请问:两位登山爱好者能否在2个小时内徒步登上山峰. (即从B 点出发到达C 点)21.(本题满分14分;第(1)小题6分,第(2)小题8分)已知椭圆2222(0)x y a a +=>的一个顶点和两个焦点构成的三角形的面积为4. (1)求椭圆C 的方程;(2)已知直线)1(-=x k y 与椭圆C 交于A 、B 两点,试问,是否存在x 轴上的点(),0M m ,使得对任意的k R ∈,MA MB ⋅为定值,若存在,求出M 点的坐标,若不存在,说明理由. 22.(本题满分16分;第(1)小题4分,第(2)小题5分,第(3)小题7分)定义:对于函数()f x ,若存在非零常数,M T ,使函数()f x 对于定义域内的任意实数x ,都有()()f x T f x M +-=,则称函数()f x 是广义周期函数,其中称T 为函数()f x 的广义周期,M 称为周距.(1)证明函数()()()1xf x x x Z =+-∈是以2为广义周期的广义周期函数,并求出它的相应周距M 的值;(2)试求一个函数()y g x =,使()()()()s i n fx g x A x x R ωϕ=++∈(A ωϕ、、为常数,0,0A ω>>)为广义周期函数,并求出它的一个广义周期T 和周距M ;(3)设函数()y g x =是周期2T =的周期函数,当函数()()2f x x g x =-+在[]1,3上的值域为[]3,3-时,求()f x 在[]9,9-上的最大值和最小值.ACBD23.(本题满分18分,第(1)小题3分,第(2)小题9分,第(3)小题6分) 一个三角形数表按如下方式构成(如图:其中项数5n ≥):第一行是以4为首项,4为公差的等差数列,从第二行起,每一个数是其肩上两个数的和,例如:()()()2,11,11,2f f f =+;(),f i j 为数表中第i 行的第j 个数.(1) 求第2行和第3行的通项公式()2,f j 和()3,f j ;(2) 证明:数表中除最后2行外每一行的数都依次成等差数列,并求(),1f i 关于i (1,2,,i n =)的表达式;(3)若()()(),111i f i i a =+-,11i i i b a a +=,试求一个等比数列()()1,2,,g i i n =,使得()()()121123n n S b g b g b g n =+++<,且对于任意的11,43m ⎛⎫∈ ⎪⎝⎭,均存在实数λ ,当n λ>时,都有n S m >.()()()()()()()()()()1,11,21,11,2,12,22,13,13,2,1f f f n f n f f f n f f n f n ---2013学年第二学期徐汇区学习能力诊断卷 高三年级数学学科(理科)参考答案及评分标准 2014.4一.填空题:(本题满分56分,每小题4分) 1.(]5,1-- 2.56π 3.()3,88k k k Z ππππ⎡⎤-+∈⎢⎥⎣⎦ 4.[)3,+∞5.13i - 6.40 7.π 8.14- 9 10.211 12.141313.5 14.23n - 二.选择题:(本题满分20分,每小题5分)15.C 16.D 17.B 18.C三.解答题:(本大题共5题,满分74分) 19.(本题满分12分,第(1)小题6分,第(2)小题6分)解:(1)连接OM ,则OM AB ⊥,设OM r =,则OB r =, 在BMO ∆中,1sin2OM ABC OB ∠===,所以3r =--------------------------(4分) 所以2443S r ππ==.-----------------(6分)(2)ABC ∆中,90ACB ∠=,30ABC ∠=,BC =,1AC ∴=,-------------------------------(8分)232314141(3333327V V V AC BC r ππππ∴=-=⨯⨯-=⨯=圆锥球.(12分) 20.(本题满分14分)解:由0150ADC ∠=知030ADB ∠=,由正弦定理得001sin 30sin120AD =,所以,AD =---------------------------------------(4分) 在ADC ∆中,由余弦定理得:2222cos150AC AD DC AD DC =+-⋅,即222032cos150DC DC =+-,即2360DC DC +⋅-=,解得 1.372DC =≈(千米), -----------------------------------------------(10分)2.372BC ∴≈(千米),--------------------------------------------------------------------(12分) 由于2.372 2.4<,所以两位登山爱好者能够在2个小时内徒步登上山峰.---(14分) 21.(本题满分14分;第(1)小题6分,第(2)小题8分)解:(1)设椭圆的短半轴为b ,半焦距为c ,则222a b =,由222c a b =-得222222a a c a =-=, 由4221=⨯⨯c b 解得4,822==b a ,则椭圆方程为14822=+y x . ----------(6分) (2)由22(1)28y k x x y =-⎧⎨+=⎩得2222(21)4280,k x k x k +-+-= 设1122(,),(,),A x y B x y 由韦达定理得:,1282,12422212221+-=+=+k k x x k k x x MA MB ∴⋅=221122121212(,)(,)()(1)(1)x m y x m y x x m x x m k x x -⋅-=-+++--=22221212(1)()()k x x m k x x k m +-++++=22222222284(1)()2121k k k m k k m k k -+-+++++=()22254821m k m k ++-++,----------------(10分) 当5416m +=,即114m =时,MA MB ⋅=167-为定值,所以,存在点11(,0)4M使得MA MB ⋅为定值(14分).22.(本题满分16分;第(1)小题4分,第(2)小题5分,第(3)小题7分) 解:(1)()()()1xf x x x Z =+-∈,∴()()()()()222112x x f x f x x x +⎡⎤⎡⎤+-=++--+-=⎣⎦⎣⎦,(非零常数) 所以函数()()()1xf x x x Z =+-∈是广义周期函数,它的周距为2.-----(4分)(2)设()()0g x kx b k =+≠,则()()sin f x kx b A x ωϕ=+++()2f x f x πω⎛⎫+- ⎪⎝⎭()222sin sin k k x b A x kx b A x πππωϕωϕωωω⎡⎤⎛⎫⎛⎫=+++++-+++=⎡⎤⎪ ⎪⎢⎥⎣⎦⎝⎭⎝⎭⎣⎦(非零常数) 所以()f x 是广义周期函数,且22,k T M ππωω==.-----------------( 9分)(3)()()()()()222224f x f x x g x x g x +-=-++++-=-,所以()f x 是广义周期函数,且2,4T M ==- .------------------------------------------(10分)设[]12,1,3x x ∈满足()()123,3f x f x =-=, 由()()24f x f x +=-得:()()()()111164424444431215f x f x f x f x +=+-=+--=---=--=-,又()()()24f x f x f x +=-<知道()f x 在区间[]9,9-上的最小值是x 在[]7,9上获得的,而[]167,9x +∈,所以()f x 在[]9,9-上的最小值为15-.--------------------( 13分)由()()24f x f x +=-得()()24f x f x -=+得:()()()()222210846442023f x f x f x f x -=-+=-++==+=,又()()()24f x f x f x -=+>知道()f x 在区间[]9,9-上的最大值是x 在[]9,7--上获得的,而[]2109,7x -∈--,所以()f x 在[]9,9-上的最大值为23.-----------------------(16分) 23.(本题满分18分;第(1)小题3分,第(2)小题9分,第(3)小题6分.) 解:(1)()()()()()2,1,1,121,4841,2,,1f j f j f j f j j j n =++=+=+=-()()()()()()3,2,2,122,8284816161,2,,2f j f j f j f j j j j n =++=+=++=+=-.----(3分) (2)由已知,第一行是等差数列,假设第()13i i n ≤≤-行是以i d 为公差的等差数列, 则由()()()()()()1,11,,1,2,,1f i j f i j f i j f i j f i j f i j ++-+=+++-++⎡⎤⎡⎤⎣⎦⎣⎦()(),2,2i f i j f i j d =+-=(常数)知第()113i i n +≤≤-行的数也依次成等差数列,且其公差为2i d .综上可得,数表中除最后2行以外每一行都成等差数列;------------(7分) 由于()114,22i i d d d i -==≥,所以11422i i i d -+=⋅=,所以1(,1)(1,1)(1,2)2(1,1)i f i f i f i f i d -=-+-=-+,由12i i d -=,得(),1f i 2(1,1)2i f i =-+,------------------------------------------------------------------------------------ (9分)于是()()1,11,1122i i f i f i --=+ , 即()()1,11,1122i i f i f i ---=, 又因为()11,14222f ==,所以,数列(),12i f i ⎧⎫⎨⎬⎩⎭是以2为首项,1为公差的等差数列, 所以,()(),12112if i i i =+-=+,所以()(),112i f i i =+⋅(1,2,,i n =). -------------------(12分)(3)()()(),111i f i i a =+-(),11211i i f i a i ⇒=+=++ , ()()11111111221212121i i i i i ii i b a a +++⎛⎫⇒===- ⎪++++⎝⎭, 令()2i g i =1111111()2221212121i i i i i i i b g i ++⎛⎫⇒=-⨯=- ⎪++++⎝⎭,-----------------(14分) 2231111111212121212121n n n S +⎛⎫⎛⎫⎛⎫⇒=-+-++- ⎪ ⎪ ⎪++++++⎝⎭⎝⎭⎝⎭11113213n +=-<+. -------(15分) n S m >111321n m +⇔->+111132133n m m +-⇔<-=+, 11,43m ⎛⎫∈ ⎪⎝⎭10134m ⇒<-<,132113n m +⇒+>-23log 1113n m ⎛⎫⇒>-- ⎪-⎝⎭,令λ=23log 113m ⎛⎫-⎪-⎝⎭,则当n λ>时,都有n S m >,∴适合题设的一个等比数列为()2i g i =.-------------------------------------------------------(18分)。

【2014徐汇区等三区二模】上海市徐汇、松江、金山三区2014届高三学习能力诊断历史试题Word版含答案

【2014徐汇区等三区二模】上海市徐汇、松江、金山三区2014届高三学习能力诊断历史试题Word版含答案

2013学年第二学期徐汇区学习能力诊断卷高三年级历史学科2014.4 (考试时间120分钟,试卷总分150分)考生注意:本试卷分为选择题与非选择题,所有试题答案按照试题对应编号写在答题纸上,本试题卷不作为阅卷和评分依据。

第I卷选择题(共75分)以下每小题2分,共60分,每题只有一个正确的选项。

1 .史家描述第二次世界大战中各国领袖的性格:丘吉尔“有想象力但不稳定”,罗斯福“天真而喜欢喊口号”,他们并不比“往往情绪失常而暴怒”的希特勒更明智。

对于这种说法,以下理解何者较为适当?()A.这位史家的说法是对历史事实的客观陈述,应该可以采信B.这种说法只是这位史家个人的片面的主观意见,不足采信C.这是史家从特定角度评论历史人物的结果,可以作为参考D.可见当时各国领导人都不明智,因而掀起第二次世界大战2.工具的进步是生产力发展的重要标志之一。

下列图示中的工具按材质出现先后顺序排列应该是()①带木柄的石斧②铁锄③青铜斧A.①②③B.②③①C.①③②D.③①②3 .文字的出现是人类文明的标志之一,下列哪项不属于亚非文明古国为人类文明发展所作出的重大贡献()A.甲骨文B.象形文字C.楔形文字D.希腊字母4.历史学家钱穆说:“严格说来,要到秦汉才是中国历史上正式有统一政府。

秦汉以前的中国,只可说是一种封建的统一。

”他做出这一判断的主要依据是秦汉时期实行了()A.郡国并行制B.三公九卿制C.郡县制D.皇帝制5.“感身世之戮辱,传畸人于春秋,虽背《春秋》之义,固不失为史家之绝唱,无韵之《离骚》矣”。

这是鲁迅先生对我国古代一部重要史学著作的赞叹。

这部史书在叙述历史时采用()A.以人物为中心展开B.以事件之始末记述C.以时序为主线写作D.以地域之联系综述6.仔细观察右面图示。

图示中所反映的这种内外形势与格局得以最终解决是由于()A.设置河西四郡、郡国并行制B.三次北击匈奴、内外朝体制C.设置河西四郡、颁布推恩令D.三次北击匈奴、颁布推恩令7.秦隋都是二世而亡的短命王朝,但却是一个承上启下、继往开来的重要时期,其最重要的共同举措是()A.完成了国家统一B.创建了中央官制C.修建了巨大工程D.完善了法律制度8.国学大师陈寅恪先生认为:“华夏民族之文化,历数千年之演进,造极于赵宋之世。

上海市徐汇、松江、金山三区2014届高三学习能力诊断(二模)生物试题 Word版含答案

上海市徐汇、松江、金山三区2014届高三学习能力诊断(二模)生物试题 Word版含答案

上海市徐汇松江金山三区2014届高三学习能力诊断(二模)生命科学学科2014.4考生注意:试卷满分150分,考试时间为120分钟。

本考试分设为试卷和答题纸。

试卷包括两部分,第一部分全部为选择题,考生应将代表正确答案的小方格用铅笔涂黑。

第二部分为综合题,答案必须全部写在答题纸上。

一、选择题(共60分。

每小题只有一个正确选项)1.人体内的下列平衡不受激素调控的是A.水的平衡B.盐的平衡C.pH的平衡D.血糖的平衡2. 吞噬细胞不参与A.非特异性免疫B.体液免疫C.细胞免疫D.合成并分泌抗体3. 炭疽杆菌能产生两种成分的蛋白质内毒素。

科学家将该菌的大型DNA分子破坏,结果该菌仍能产生内毒素。

请分析推断,应该破坏炭疽杆菌的哪个结构,才能使它从根本上不再产生内毒素?A.核糖体B.质粒C.内质网D.拟核4. 正常情况下,下列细胞中最可能不含等位基因的是5.下列关于艾滋病毒、大肠杆菌、胚胎干细胞的比较,正确的是A.都有细胞结构B.都具有增殖能力C.都属于原核生物D.都含有核糖体6. 下列四个图中,能正确反映生长素浓度与芽位置关系的是(侧芽1、侧芽2、侧芽3分别表示距顶芽的距离由近到远)7.下表是某中年男子血液化验单中的部分数据:据所学知识判断下列叙述正确的是AB.该男子可能患有高血脂,应不吃脂肪,多吃糖类食物C.该男子可能患有地方性甲状腺肿,细胞代谢速率偏低D.血浆的生化指标应保持稳定,否则将引起代谢紊乱8.右图所示为真核细胞蛋白合成过程中必需的两种物质(甲、乙),下列有关叙述中正确的是A.遗传信息只位于甲上B.乙由三个碱基组成C.甲、乙均具有特异性D.乙中不存在碱基互补配对9.下图表示细胞膜部分功能结构模式图。

据图分析,下列说法不正确...的是A.功能①在生命起源过程中具有重要作用B.功能②表示的运输方式均消耗能量C.激素调控生命活动与图中功能③有一定关系D.相邻的植物细胞可通过功能④进行通讯10. 细胞自噬是指细胞内受损、变性、衰老的蛋白质或细胞器运输到溶酶体内并进行降解的过程。

2014上海徐汇高三语文二模参考答案

2014上海徐汇高三语文二模参考答案

2013学年第二学期徐汇区高三年级语文学科学习能力诊断卷参考答案及评分标准一阅读 80分(一)18分1. 仔细辨别、品味,享受幸福(2分)2. 大脑前额叶皮层活动和催产素分别反应或左右着人的幸福感。

(2分)3. 幸福感与人的健康关系密切,而且幸福并不虚幻。

(大意对即可,2分)4. 承接上文对幸福感认识的误区,从主观感受与客观福祉未必对等的角度,解释了中国人幸福感迷失的原因;又进一步引出下文“如今还能不能找到幸福”的问题。

(3分)5. D(3分)6. 文章先后从生理学和心理学两个角度对“幸福感”加以说明。

生理学角度,首先以进化论的观点,解释人体形态上的进化使人类具有了获取幸福感的基础;其次结合脑科学的发现,进一步说明了幸福感产生的生理机制。

在此之后,文章再从心理学角度,说明幸福感同人的生理及心理健康密切相关。

(6分)(二)20分7. 生动形象地表现了四月岷山的春色在不知不觉中逐渐变浓的过程。

(2分)8. 四月天里,生命已经萌发,处处生机勃勃,充满着变化的可能;此时勃发的生命也是新鲜稚嫩的,略带青涩,尚不成熟。

(2分)9. 转折并推进文意,强调四月没有永恒的美,写出了四月的春意在带来生机的同时却又变化万端、转瞬即逝,为结尾抒发感慨作了铺垫。

(3分)10.C(3分)11.随着时代的发展,人类身上某些最本真、最新鲜、最充满活力的东西在不经意之间被成熟、理性、世故甚至颓废所取代,作者对此流露出淡淡的伤感情绪。

(4分)12.可以从修辞(反复、比喻、拟人)、白描、描写抒情结合、多角度描写、动静相衬、虚实相生等方面加以赏析。

(6分)(三)6分(超过6空,按前6空评分)13.⑴出则无敌国外患者⑵屋舍俨然⑶澄江一道月分明⑷又恐琼楼玉宇⑸宫阙万间都做了土⑹岩上无心云相逐⑺予尝求古仁人之心⑻恐美人之迟暮(四)8分14.柳永(1分)15.B(3分)16.阴晴不定的天气喻示着姑娘们情绪的阴晴不定。

“撩”花“拨”柳的动作,写出了她们一览春色的急切心情。

2014年 高三物理二模徐汇卷答案

2014年 高三物理二模徐汇卷答案

徐汇区2013学年第二学期高三年级物理学科学习能力诊断题参考答案和评分标准(一)单项选择题 每小题2分,共16分.(二)单项选择题. 每小题3分,共24分.(三)多项选择题. 每小题4分,共16分.(四)填空题. 每小题4分,共20分.21.14,H +O →He +N 111784214722A .10,3 22B .7∶1,7/4923.2,2⨯10-2 24.mg ,0.25πmgl 25.6,4(五)实验题.共24分. 26.(4分)2NxaAV ⋅ 27.(6分)(1)封闭气柱的的长度l 0(2分),)(12.00x l x H ∆+∆=(2分) (2)>(2分) 28.(7分)(1)22)Δ(2=t L d a (2分),(2)BCD (4分,部分答对得2分) (3)偏大 (1分)29.(7分)(1)2V (1分),0.5Ω(2分) (2)小于(1分) 大于(1分) (3)电动势相同(1分),内阻偏大(1分) (六)计算题. 共50分.30.(10分)(1)U 形管内两边水银面的高度差为Δh =25cm A 中气体p A1=p 0+p Δh =75+25=100cmHgB 中为大气,所以活塞产生的压强p 塞=25cmHg (1分)闭合阀门后,两部分气体温度降低,压强均减小且A 处降低较多,活塞会下移(1分) 设此时U 形管表示的是A 中压强,p A2=p 0-p Δh =75-25=50cmHg (1分) 对A 中气体222111T SL p T S L p A A A A =,解得L A2=72cm >49cm假设不成立。

说明此时U 形管表示的应该是B 中压强 p B2=50cmHg (1分)则A 中气体压强 p A2=p B2+p 塞=50+25=75cmHg (1分) 对A 中气体222111T SL p T S L p A A A A =(1分) 解得L A2=48cm (1分) 活塞离容器底部高度L ′=L A2=48cm(判断、讨论共2分,得出48cm 的解题过程共3分) (2)对B 中气体2B221B11T L p T L p B B =(1分) 2A221A10))T L H p T L H p B -=-(((1分),解得H =75cm (2分)31.(13分)(1)小球自由下落进入小孔用时s 5.01025.1221=⨯==g h t (1分) 这段时间内装置在做匀减速运动,其加速度41==g μa m/s 2(1分)小球与小孔的水平距离为m 35.04215.052122111=⨯⨯+⨯=+=t a t v x t (1分)(2)小球下落到Q 板时速度为零从最高点到最低点:电W W E K +=∆G (1分) mg (h +d )-Eqd =0(1分)3109)(⨯=+=qdd h mg E N/C (1分)(3)总势能变化量:J 50.=+=Δ+Δ=ΔEqd mgd E E E P -电势(公式2分,结论1分) (4)小球在电场中的加速度5.1204.05.0==-=m mg Eq a m/s 2(1分)小球在电场中运动的时间ada s t 22=22=2=0.8s (1分) 小球进入电场后,装置的加速度57.4=)+=2MEq Mg μa (m/s 2(1分)当小球返回到P 板时,装置的速度2.1==221t a v v t m/s (1分)32.(12分)(1)mθmg θF a sin cos 1-=(2分)m θmg θF s s a v B )sin cos (221-==(2分)(2)继续上滑的加速度a 2=g sin θ+μg cos θ(1分)若mg sin θ<μmg cos θ,即μ>tan θ,物体停在BC 段,(1分)s θμθmg θmg μF s θμθmg θmg θF s a v s s s s )cos (sin cos )()cos (sin sin cos 222B 2++=+-+=+=+=总(2分)若μ<tan θ,物体停在水平段(1分)从A 点至停下,21f f F K W W W E ++=∆(1分)3222cos 2cos 0mgs μa v θmg μθFs B--=(1分)223cos cos a θv mg μθFs s B-= )cos (sin )sin cos (cos 2cos 3θμθmg θmg θF θs mg μθFs s +--=(1分)33.(15分)(1)M 杆在斜轨道滑行过程中开始做匀速运动R v d B mg M230sin 222=︒(1分),解得m /s 5=M v (1分) 22130sin Mmv Q mgx =-︒(1分) ,解得Q =37.5J M 杆产生的焦耳热Q M =Q 21=18. 75J (1分)(2)N 杆下滑时做匀加速运动ma R at d B mg F =++230sin 2220(2分),解得16+=t F (1分) (3)N 杆在斜轨道滑行中,M 杆在水平面上作减速运动M Nma Rv L B B mg μN μf =+==)2(221(1分)代入得t a M 24+=(1分) M 杆作加速度逐渐增大的减速运动 作出M 杆加速度与时间的图像,如图,可知图中围成的面积是M 杆速度的变化量,当M 杆停止时,速度的改变量大小为5m/s 。

2014届上海徐汇二模作文(7篇)(选择了不同层次的典型范文)

2014届上海徐汇二模作文(7篇)(选择了不同层次的典型范文)

徐汇区作文《汉书·艺文志》中有“安其所习,毁所不见”一语,大意是指:人往往安于自己所习惯、所熟悉的事物,对从没见过,或未能直接、间接经验过的事物,则常常予以否定。

对这种生活中普遍存在的现象,你有怎样的联想和思考?请选取一个角度写一篇文章,谈谈你的想法。

要求:⑴题目自拟;⑵全文不少于800字;⑶不要写成诗歌。

有容乃大(一类上:67)犹记得房龙《宽容序言》中的无知山谷,人们安于山谷中平静却落后的生活,对于开拓者所言的那个从未目见知晓的谷外世界,予以了强烈的抗拒与否定。

可见,无论是于《汉书》“安其所习,毁所不见”还是于《宽容序言》都向我们揭示了一个现象:人往往安于自己所习惯,所熟悉的事物,对未曾目见和经历的则常常予以否定。

然,世界发展的根源在于旧事物的灭亡,新事物的产生,若想寻求世界的进步与发展,我们必须以更积极开放的姿态接纳新生事物,有容才能成其大。

“安其所习,毁所不见”是生活是普遍存在的现象。

究其原因有三。

其一便是对结果好坏的未知。

历来,社会的变革都将会遭到保守派的层层阻隔,原因很简单,人们无法预见变革的后果,当现时的安稳与革新的未知安放于天秤两端,人们手中的砝码宁愿偏向虽有弊病的当下,也不愿投向未知的变革。

但我们需要看到先秦时代,彻底的商鞅变法使曾经积贫积弱边陲小国蜕变为强秦;反观六国之变法,多在变法者死后就又回复了过去的轨道,不思新变,终为秦灭。

其二是对权威和现时利益者的维护。

爱迪生的直流发电曾为其带来巨大的财富与利益,因此在明知特斯拉交流电更为优异的情况下,仍对其进行强烈的打压与否定。

而当如今交流电的普遍使用,爱迪生的自私与浅陋显露无遗。

最后是人们惯性思维的(殆)怠惰,这份惰性让我们停止思考,将创新的思考转为对旧有的维护,如此对创新进步的伤害之大让人唏嘘不已。

不管是出于对未知的求称心理和对权威利益的奉行,还是人们所先具有的怠惰都将毫无疑问的阻碍社会的发展和进步,因此对于未知的新事物,宽容接纳是上策,切莫用否定将真理的萌芽抑掐在初始阶段;然后要打破权威,不能因为旧有保守者们的维护和否定,我们自身就加以不经思考的否定;要勇于实践,实践是检验真理的唯一标准,对于未知的新生事物通过实践才能确实判断优劣。

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2013学年第二学期徐汇区学习能力诊断卷高三年级英语学科2014.4第I 卷I. Listening ComprehensionSection ADirections: In Section A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversations and the questions will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard.1. A. Visit the woman. B. Go to an interview.C. Attend a lecture.D. Work in the office.2. A. At a post office. B. At a fast-food restaurant.C. At a booking office.D. At a check-in desk.3. A. One hour later. B. Thirty minutes later.C. Twenty minutes later.D. Around ten minutes.4. A. $8. B. $12. C. $16. D. $20.5. A. There is something wrong with the yoga class.B. John and Tom are good friends.C. Tom has attended a yoga class.D. The woman may have dialed a wrong number.6. A. What to take up as a hobby. B. How to keep fit.C. How to handle pressure.D. What to play with.7. A. Classmates. B. Lecturers. C. Strangers. D. Relatives.8. A. He has made a careful plan of writing it. B. He hasn’t got a whole pictur e of it.C. He has quit writing it.D. He is seeking for a publisher now.9. A. A head waiter. B. A shop assistant. C. A customer. D. A receptionist.10. A. He’s not surprised at the woman’s playing at a concert tomorrow.B. He won’t give the wo man a surprise at the concert tomorrow.C. He will also play at the concert tomorrow with the woman.D. He understands the woman’s nervous feeling and thinks it normal.Section BDirections: In Section B, you will hear two short passages, and you will be asked three questions on each of the passages. The passages will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one would be the best answer to the question you have heard.Questions 11 through 13 are based on the following passage.11. A. In small cups. B. In a boat on a river.C. At the bottom of a river.D. On food stored in the ground.12. A. It came from a river. B. It was extremely pure.C. It was found in large quantities.D. It was dug up from a deep mine.13. A. An early means of travel on rivers. B. An ancient method of fishing in rivers.C. An important discovery at a village.D. The way of trading foods in old times.Questions 14 through 16 are based on the following passage.14. A. She enjoyed removing others’ drinks. B. She became more and more forgetful.C. She preferred to do everything by herself.D. She wanted to keep her house in good order.15. A. She is happy to clean windows. B. She loves to clean used windows.C. She is fond of clean used windows.D. She likes clean windows as my mother did.16. A. My mother often made us confused.B. My family members had a poor memory.C. My mother helped us to form a good habitD. My wife was surprised when she visited my mother.II. Grammar and VocabularySection ADirections: After reading the passages below, fill in the blanks to make the passages coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, use one word that best fits each blank.(A)The California state assembly recently approved the so-called Paris Hilton bill, which prevents dogs 25______ occupying th e driver’s seat in a moving vehicle. The bill passed 26______ Democrats wanted to make an example of Miss Hilton. They think she is a “little rich girl” who always gets her way; plus, her daddy is a big contributor to the Republican Party. “The judge 27______ have put her in jail for four months,” said one Democrat when Paris got only a four-day jail sentence for driving while drunk.28______ (drive) around town with her little dog Lovey hanging out the driver's window, Paris is a familiar sight throughout Los Angeles. Pictures of her and Lovey are common in newspapers and magazines.“That dog is a deadly threat to everyone on the streets and the sidewalks,” said assembly leader Fabian Nunez. “We wrote this bill to protect the public. There’s no telling 29______ the dog might cause her to drive into a crowd of pedestrians. 30______ Paris wants to be with her dog, let her chauffeur do the driving. The state assembly is responsible for protecting people, and we take that responsibility seriously.”The assembly bill passed by a vote of 44 to 11. The state senate(参议院) 31______(expect) to approve the bill, and Governor Schwarzenegger has promised to sign it. “I love dogs,” he said, “but when it comes to 32______ (protect) the people of California, dogs will have to take a back seat.”The eleven assembly members who objected to the bill were all Republicans. “While Californians,” said Republican Tom Ridge, “are being attacked daily by murderers, rapists, and muggers, who do the Democrats protect citizens from—a 33______ (spoil) little girl and her dog!”(B)He almost killed somebody, but one minute changed his life. This beautiful story comes from Sherman Rogers’ book, Foremen: Leaders or Drivers? .In his true-life story, Rogers illustrates the importance of effective relationships. During his college years, Rogers spent 34______ summer in an Idaho(爱达荷州) logging camp. When the superintendent had to leave for a few days, he put Rogers in charge.“35______ if the men refuse to follow my orders?” Rogers asked. He though t of Tony, an immigrant worker who roared all day, giving the other men a hard time.“Fire them,” the superintendent said. Then, as if 36______ (read) Roger’s mind, he added, “I suppose you think you are going to fire Tony if you get the chance. I’d feel badly about that. I have been logging for 40 years. Tony is the most reliable worker I’ve ever had. I know he hates everybody and everything. But he comes in first and leaves last. There has not been an accident for eight years on the hill 37______ he work s.”Rogers took over the next day. He went to Tony and spoke to him.“Tony, do you know I’m in charge here today?” Tony grunted(发哼声). “I was going to fire you the first time we fought, but I want you to know I’m not,” he told Tony, adding what the superintendent 38______ (say).When he finished, Tony dropped the shovelful of sand he had held and tears streamed down his face.“Why he no tell me dat eight years ago?”That day Tony worked harder than ever before --- and he smiled! He later said to Rogers, “My first foreman who ever say, ‘Good work, Tony’ and it made me feel like Christmas.”Rogers went back to school after that summer. Twelve years later he met Tony again who was now superintendent for railroad construction for one of 39______ (large) logging companies in the West. Rogers asked him how he came to California and happened to have such success. Tony replied, “If it not be for the one minute you talk to me back in Idaho, I keel somebody someday. One minute change my whole life.”Effective managers know the importance of taking a moment to point out 40______ a worker is doing well. But what a difference a minute of affirmation can make in any relationship!Section BDirections: Complete the following passage by using the words in the box. Each word can only be“Here’s a nice bit of gossip!” Do I have your attention? Probably. We listen, but 41______ we often feel terrible with ourselves. That’s the problem with gossip: it’s something that as a social species we are primed to enjoy, but it can also be 42______ and harmful.Not all gossip is bad. Small talk establishes relationships and 43______ the other person that our intentions are friendly. So gossip, in the sense of exchanging bits and pieces of news about ourselves and others, can be perfectly 44______. If I say to y ou, “Let’s meet for coffee and have a bit of a gossip,” I’m inviting you to a social 45______ in which two people chew the fat. There is nothing wrong with that: life would be very dull if we were unable to talk about what goes on around us.But it’s not that simple. If we say that somebody is a gossip, we do not mean that he or she enjoys gentle social chat: it carries a crueler 46______. A true gossip enjoys spreading stories about other people --- stories in which others do not usually come out 47______. The gossip is one who spreads bad gossip; good gossip is still fine, but it’s not what gossips spread. Thedistinction between good and bad gossip is not always clear. It would be easy if we could 48______ the two by saying that bad gossip is just about people; but innocent gossip may be about people too. The best way to tell the difference is to look at the intention behind the remarks. Bad gossip 49______ itself in its desire to make the 50______ of the story look foolish. It also intrudes on their privacy. So we all know the difference.Now then, did you hear about ...III. Reading ComprehensionSection ADirections: For each blank in the following passages there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.“What are you going to be when you grow up?” is a question that you may have been asked. You may not even know there are a variety of geography-related jobs.The Association of American Geographers lists nearly 150 different geography jobs. So, if you are interested in people, places, and environments, consider a job in geography. Your work will not be limited to maps—it might range from 51 data to planning projects, or making decisions about the environment.Processing Geographic DataA geographer’s main activity is analyzing geographic information to answer geographic questions. Jobs processing geographic data begin, of course, with collecting the information. One on-the-ground job in data collection is that of a surveyor. Surveyors 52 and measure the land directly. They may mark boundaries, study the shape of the land, or even help find sewer(下水道) and water systems beneath the Earth. High-tech information-gathering jobs include working with Geographic Information Systems (GIS) data. Some examples of these jobs include remote sensing specialists and GIS analysts. Data analysis jobs require the ability to think 53 , high-level computer skills, and a college education.Once data have been processed, a geographer may study the information to use in planning projects such as a new urban area, a(n) 54 evacuation(撤退) plan, or the placement of a new highway. Planners can also help determine how to make a neighborhood a better place to live. These jobs, too, require good critical thinking, writing, and computer skills, as well as a college education. Planners are 55 to the success of a community.Advising Businesses and GovernmentAbout half of jobs using geography are in business and government. All kinds of businesses use geographic information to help build and 56 their operations. A location analyst studies an area to find the best location for a client. The client might be a large retail store chain that wants to know which location would be best for opening a new store. The location analyst can study GIS reports on such elements as transportation networks or population in an area and give the business owners the positive and 57 points about a location being considered.In 1967, the Mexican government was looking for a location to create a new international tourist resort. They used location analysts to find an area that had good beaches and was easy to reach from the United States. The 58 was Cancún, today one of the world’s most desirab le vacation sites.Businesses connected with natural 59 such as forests also rely on geographers. Geographers help them understand the relationship between their business and the environment where their business is located.In 1967, Cancún was a sm all island on Mexico’s Caribbean coast. It had white sand beaches, many birds and mangrove(红树) trees, but few people. After it was selected as a resort site, it was quickly 60 . Today, Cancún has more than 100 hotels and 500,000 permanent residents. Many work in the tourist industry that serves the millions of visitors who come each year from all over the world.Physical and Human GeographyPhysical geographers are sometimes called earth scientists. Some study such topics as geomorphology (地形学), that is, the study of how the shape of the Earth 61 . Others study weather and climate. Still others study water, the oceans, soils, or ecology. Jobs in these fields require 62 scientific training.Some geographers study economic, political, and 63 issues as they relate to place or region. Human geographers are usually hired by government agencies to analyze a specific problem. These geographers work 64 with political scientists, economists, and sociologists. Together, they provide possible solutions to problems from many different aspects of life in an area. And, of course, geographers teach the subject at all levels of 65 , from elementary schools to universities. But no matter what geography jobs people might hold, they are always trying to answer the basic geographic questions: “Where are things located?” and “Why are they there?”51. A. performing B. analyzing C. appreciating D. downloading52. A. map B. mine C. shape D. honour53. A. creatively B. critically C. wildly D. moderately54. A. disaster B. radioactivity C. excursion D. vacation55. A. qualified B. determined C. valuable D. feasible56. A. survive B. expand C. manage D. boom57. A. negative B. effective C. depressive D. profitable58. A. scenery B. destination C. result D. foundation59. A. potentials B. histories C. resources D. sciences60. A. specialized B. polluted C. governed D. transformed61. A. proceeds B. stretches C. extends D. changes62. A. peculiar B. special C. reasonable D. enthusiastic63. A. psychological B. contemporary C. religious D. social64. A. closely B. peacefully C. loyally D. sensitively65. A. demonstration B. revolution C. examination D. educationSection BDirections: Read the following passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.(A)The Dangers of DietingThanks to our modern lifestyle, with more and more time spent sitting down in front of computers than ever before, the number of overweight people is at a new high. As people crazily search for a solution to this problem, they often try some of the popular fad(时尚) diets being offered. Many people see fad diets as harmless ways of losing weight, and they are grateful to have them. Unfortunately, not only don’t fad diets usua lly do the trick, but they can actually bedangerous for your health.Although permanent weight loss is the goal, few are able to achieve it. Experts estimate that 95 percent of dieters return to their starting weight, or even add weight. While the irresponsible or unwise use of fad diets can bring some initial results, long-term results are very rare. Nonetheless, people who are bored with the difficulties of changing their eating habits often turn to fad diets. Rather than being moderate, fad diets involve extreme dietary changes. They advise eating only one type of food, or they prohibit other types of foods entirely. This results in a situation where a person’s body doesn’t get all the vitamins and other things that it needs to stay healthy.One popular fad diet recommends eating lots of meat and animal products, while nearly eliminating carbohydrates(碳水化合物). A scientific study from Britain found that this diet is very high in fat. According to the study, the increase of damaging fats in the blood can lead to heart disease and, in extreme cases, kidney failure. Furthermore, diets that are too low in carbohydrates can cause the body to use its own muscle for energy. The less muscle you have, the less food you use up, and the result is slower weight loss.Veteran(老兵) dieters may well ask at this point, “What is the ideal diet?” Well, to some extent, it depends on the individual. A United States government agency has determined that to change your eating habits requires changing your psychology of eating, and everyone has a different psychology. That being said, the British study quoted above recommends a diet that is high in carbohydrates and high in fiber, with portions of fatty foods kept low. According to the study, such a diet is the best for people who want to stay healthy, lose weight, and keep that weight off. And, any dieting program is best undertaken with a doctor’s supervision.66. After losing weight by dieting, what usually happens to people?A. They have kidney failure.B. They gain the weight back again.C. They keep the weight off.D. They have less muscle.67. Which of the following best expresses the essential information in paragraph 3?A. Bored people turn to fad diets, which, being too extreme, don’t give the body everything it needs.B. People are bored with fad diets and turn to diets which provide what the body needs.C. People prefer fad dieting to moderate dieting because it requires fewer foods to give the body what it needs.D. Fad diets give boring people the moderate dietary changes they need to get all the required vitamins.68. Which is not mentioned as an effect of the meat and animal product diet?A. Heart disease.B. Slower weight loss.C. Psychological changes.D. Kidney failure.69. According to the passage, why does the ideal diet depend on the individual?A. The less muscle you have, the less food you use up.B. Everyone can gain the weight back.C. Everyone has a different psychology.D. Everyone likes different foods.(B)Charlie Bell became chief execut ive of McDonald’s in April. Within a month doctors told him that he had colorectal cancer. After stock market hours on November 22nd, the fast-food firm said he had resigned; it would need a third boss in under a year. Yet when the market opened, its share price barely dipped then edged higher. After all, McDonald’s had, again, shown how to act swiftly and decisively in appointing a new boss.Mr. Bell himself got the top job when Jim Cantalupo died of a heart attack hours before he was due to address a con vention of McDonald’s franchisees(获特许经营联营店者). Mr. Cantalupo was a McDonald’s veteran brought out of retirement in January 2003 to help remodel the firm after sales began falling because of dirty restaurants, indifferent service and growing concern about junk food. He devised a recovery plan, backed by massive marketing, and promoted Mr. Bell to chief operating officer. When Mr. Cantalupo died, a rapidly convened(召集) board confirmed Mr. Bell, a 44-year-old Australian already widely seen as his heir apparent, in the top job. The convention got its promised chief executive’s address, from the firm’s first non-American leader.Yet within weeks executives had to think about what to do if Mr. Bell became too ill to continue. Perhaps Mr. Bell had the same thing on his mind: he usually introduced Jim Skinner, the 60-year-old vice-chairman, to visitors as the “steady hand at the wheel”. Now Mr. Skinner, an expert on the firm’s overseas operations, becomes chief executive, and Mike Roberts, head of its American operations, joins the board as chief operating officer.Is Mr. Roberts now the new heir apparent? Maybe. McDonald’s has brought in supposedly healthier choices such as salads and toasted sandwiches worldwide and, instead of relying for most of its growth on opening new restaurants, has turned to upgrading its 31,000 existing ones. America has done best at this; under Mr. Roberts, like-for-like sales there were up by 7. 5% in October on a year earlier.The new team’s task is to keep the revitalization plan on cours e, especially overseas, where some American brands are said to face political hostility from consumers. This is a big challenge. Is an in-house succession(交替、继承) the best way to tackle it? Mr. Skinner and Mr. Roberts are both company veterans, having joined in the 1970s. Some recent academic studies find that the planned succession of a new boss from within, such as Mr. Bell and now (arguably) Mr. Roberts, produces better results than looking hastily, or outside, for one. McDonald’s smooth handling of its serial misfortunes at the top certainly seems to prove the point. Even so, everyone at McDonald’s must be hoping that it will be a long time before the firm faces yet another such emergency.70. The main reason for the constant change at the top of McDonald is _______.A. the constant change of its share priceB. the board’s failure to reach an agreementC. the falling salesD. the physical problems of the chief executives71. The underlined phrase “heir apparent” (in Paragraph 2) in the article most probabl y means someone who _______.A. has the same ideas, aims and style with a personB. has the same right to receive the family titleC. is appointed as an executive of a companyD. is likely to take over a person’s position when that person leaves72. Which of the following was NOT a cause of the falling sales of McDonald?A. The change of the chief executive.B. People’s concern about junk food.C. Dirty restaurant.D. Indifferent service.73. In terms of succession at the top, McDonald_______.A. has had to made rather hasty decisionsB. prefers to appoint a new boss from withinC. acts in a quick and unreasonable wayD. surprises all the people with its decisions(C)It is evident that there is a close connection between the capacity to use language and the capacities covered by the verb “to think”. Indeed, some writers have identified thinking with using words: Plato coined the saying, “In think¬ing the soul is talking to itself”; J. B. Watson reduced thinking to inhibited(拘谨的) speech located in the minute(微小的) movements or tensions of the physiological mechanisms involved in speaking; and although Ryle is careful to point out that there are many senses in which a person is said to think and in which words are not in evidence, he has also said that say¬ing something in a specific frame of mind is thinking a thought.Is thinking reducible to, or dependent upon, language habits? It would seem that many thinking situations are hardly distinguishable from the skillful use of language, although there are some others in which language is not involved. Thought cannot be simply identified with using language. It may be the case, of course, that the non-linguistic skills involved in thought can only be acquired and developed if the learner is able to use and understand language. However, this question is one which we cannot hope to answer in this book. Obviously being able to use language makes for a considerable development in all one’s capacities but how precisely this comes about we cannot say.At the common-sense level it appears that there is often a distinction between thought and the words we employ to communicate with other people. We often have to struggle hard to find words to capture what our thinking has already grasped, and when we do find words we sometimes feel that they fail to do their job properly. Again when we report or describe our thinking to other people we do not merely report unspoken words and sentences. Such sentences do not always occur in thinking, and when they do they are merged with vague imagery and the hint of unconscious or subliminal(潜意识)activities going on just out of rage. Thinking, as it happens, is more like struggling, striving, or searching for something than it is like talking or reading.Again the study of speech disorders due to brain injury disease suggests that patients can think without having adequate control over their language. Some patients, for example, fail to find the names of objects presented to them and are unable to describe simple events which they witness; they even find it difficult to interpret long written notices. But they succeed in playing games of chess or draughts. They can use the concepts needed for chess playing or draught playing but are unable to use many of the concepts in ordinary language. How they manage to do this we do not know. Presumably human beings have various capacities for thinking situations which are likewise independent of language.74. According to the theory of "thought" devised by J. B. Watson, thinking is_______.A. talking to the soulB. concealed speechC. speaking nonverballyD. a non-linguistic behavior75. What does the author think about the relationship between language and thinking?A. The ability to use language enhances one’s capacity of thinking.B. Words and thinking match more often than not.C. Thinking never goes without language.D. Language and thinking are generally distinguishable.76. According to the author, when we intend to describe our thoughts, we______.A. merely report internal speechB. have to search for proper words in the way we readC. are overwhelmed with vague imageryD. sometimes are not able to find appropriate words77. Why are patients with speech disorders able to think without having adequate control of language?A. They use different concepts.B. They do not think linguistically.C. It still remains an unsolved mystery.D. Thinking is independent of language.Section CDirections: Read the passage carefully. Then answer the questions or complete the statements in the fewest possible words.Space Travel and Science FictionSpace travel and science fiction have long been connected. Early science fiction writers such as Jules Verne inspired scientists and engineers to develop new space technologies. Writers of science fiction, as well as creators of science fiction TV shows and movies, often study the latest scientific concepts and use or adapt them to help expose what future space travel, space ships, and space stations might look like. And while many of their predictions have come true, many others have not.Jules Verne (1828-1905) was a French author. He was a pioneer of science fiction. In his novels From the Earth to the Moon (1865) and Around the Moon (1870), a kind of space ship is fired from a 900-foot-long cannon(大炮) at the moon. On their journey, the three travelers are deprived of gravity at one point and float around their small ship. When landing on the moon, rockets are used to slow the ship down.Given the year in which he was writing, Verne’s predictions were very good. The size of his space ship is about the size of the first one to go to the moon, the Apollo, minus its large rockets. Both Verne’s ship and the Apollo carried three people into space. Furthermore, rockets were indeed used by the Apollo to slow its descent. However, V erne’s ship, by analogy(相似) with a gun, shot his travelers into space, which never could have worked. The intense pressure of such an event would cause great physiological damage to the crew.During the first half of the 20th century, science fiction novels and comic books were widely distributed in the United States. Their portrayal(描写) of space travel was less far-fetched thanVerne’s. Pictures began showing astronauts in space suits, as writers realized that exposure of human beings to space was deadly. Ideas of other planets were still often wrong, though. A 1928 drawing of the surface of a moon of the planet Jupiter shows it covered in plant life. Only later was it discovered that other than Earth, the planets and moons around our sun are without life as we know it. Drawings in the early 20th century showed very large space ships and stations. They were like floating cities. Writers at the time knew that trips to other stars would take hundreds of years. Those who left Earth would die on the journey; their descendants would arrive. Some writers avoided this problem by using the concept of suspended animation—a deep sleep in which a person doesn't grow older. Such travelers would awaken at the end of their journey, hundreds of years in the future. It is not impossible that these ideas could become reality one day. After the American space program had begun, the television show Star Trek became very popular. It follows the adventures of a large space ship with over 400 crew members that flies around the universe at speeds faster than light. This is a wonderful dream, but it will probably never come to fulfillment. The laws of physics tell us that it is quite impossible for any object to travel faster than the speed of light.It’s clear to see that science fic tion has progressed along with science and technology. Science fiction writers continue to study scientific concepts and to use them to portray the future. Looking back at their ideas, some were correct and cannot be distinguished from today’s reality; many others were nothing but fiction. But, they have always inspired new generations of humans to dream of someday going into space.78. According to the passage, what did early science fiction writers do to help reveal what future space travel might look like?79. If the modern spaceship were shot into space by analogy with a gun, the astronauts in it would ___________________________.80. The space travelers in early science fictions didn’t wear space suits, because ___________________________________________________.81. According to the passage, suspended animation was a concept used in science fiction to _______________________________(Note: Answer the questions or complete the statements in No More Than Twelve Words.)第II 卷I. TranslationDirections: Translate the following sentences into English, using the words given in the brackets.1. 20世纪末电子产品发生了巨大的变化。

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