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高三上学期理科综合第11周周测试题

高三上学期理科综合第11周周测试题

湛江市第二十一中学高三上学期理科综合第11周周测试题相对原子质量:H 1 C 12 N 14 O 16 S 32 I 127 Cl 35.5 Al 27 Ca 40 Cu 64Fe 56 K 39 Mg 24 Na 23 Zn 65 Sn 119 Ag 108一、单项选择题:本大题共16小题,每小题4分,共64分。

在每小题给出的四个选项中,只有一个选项符合题目要求,选对的得4分,选错或不答的得0分。

1、有一位高中生,在实验中观察发现了一种新的单细胞生物,若想确定该生物的所属类型,则该同学进行分析判断要参考的依据应该有①有无细胞壁②有无膜结构③有无核糖体④有无核膜A.①③ B.②③ C.①④ D.②④2、以下有关基因重组的叙述,正确的是A.非同源染色体的自由组合能导致基因重组 B.姐妹染色单体间相同片段的交换导致基因重组C.基因重组导致纯合体自交后代出现性状分离 D.基因重组也就是基因的自由组合3、下列有关实验的描述不科学的是A.还原糖和蛋白质的检测实验中分别需水浴加热和不需加热B.提取绿叶中的色素时碳酸钙的作用是防止色素被破坏C.视野中观察到处于质壁分离状态的细胞,不能据此判断该细胞正在失水D.低温诱导多倍体形成的实验是先剪下根尖再放到低温环境中去诱导培养4、某农科所通过下图所示的育种过程培育成高品质的糯性水稻。

下列有关叙述正确的是A.该育种过程中运用的遗传学原理是基因突变B.图示的育种方式包括杂交育种和单倍体育种C.人工条件下,选用卵细胞培育yR效果更好D. a的操作比b简单,所以可以明显缩短了育种年限5、下列各项中与植物激素或植物生长调节剂的调节功能相符的一项是A.在形成无子番茄过程中生长素改变了细胞的染色体数目B.乙烯广泛存在于植物多种组织中,主要作用是促进果实的发育。

C.利用赤霉素处理大麦,可以使大麦种子无须发芽就可以产生α-淀粉酶D.赤霉素引起植株增高的原因主要是促进了细胞的分裂6、下列关于细胞器的描述正确的是①溶酶体内含有多种水解酶,能分解衰老、损伤的细胞②动植物细胞都有两个互相垂直排列的中心粒③用高倍镜观察叶绿体可选用黑藻幼叶④所有酶、抗体、激素都在核糖体上合成⑤衰老细胞中的线粒体功能增强⑥植物细胞有丝分裂末期细胞板周围分布较多的高尔基体⑦减数第一次分裂染色体联会不需要线粒体提供能量A.①③⑥ B.①③⑦ C.②④⑦ D.④⑤⑦7.下列说法正确的是A.235U原子中,核内中子数与核外电子数的差值为14392B .纯碱、42CuSO 5H O 和生石灰分别属于盐、混合物和氧化物C .凡是能电离出离子的化合物都是离子化合物D .3NH 、硫酸钡和水分别属于非电解质、强电解质和弱电解质 8.下列实验或操作正确的是9.下列叙述正确的是A .粗铜精炼时,把粗铜接在电源的负极B .充电时,把铅蓄电池的负极接在外电源的正极C .镀锡铁破损后铁更易被腐蚀D .碱性氢氧燃料电池工作时,负极反应为:22O 2H O 4e -++4OH -10.下列有关有机物的说法正确的是A .乙醇汽油是多种烃的混合物B .煤是由有机物和无机物组成的,干馏可制得甲苯C .油脂、蔗糖、蛋白质水解均可得到含羧基的化合物D .工业制乙烯的方法是用乙醇脱水11.下表为第2、3周期某些元素性质的数据。

高三上学期理科综合第14周周测试题

高三上学期理科综合第14周周测试题

湛江市第二十一中学高三上学期理科综合第14周周测试题一、单项选择题:本大题共16小题,每小题4分,共64分。

在每小题给出的四个选项中,只有一个选项符合题目要求,选对的得4分,选错或不答的得0分。

1.蛋白质与DNA是细胞内重要的化合物,以下描述正确的是A.合成的场所均含有磷脂分子B.均参与叶绿体、线粒体、核糖体等细胞器的活动C.DNA通过复制、转录、翻译来控制蛋白质的合成D.只有蛋白质和DNA也可构成生物2.下列有关细胞生命历程的叙述,正确的是A.细胞凋亡对于维持人体内部环境的稳定有重要作用B.胡萝卜叶肉细胞脱分化形成愈伤组织后不具全能性C.有丝分裂过程中,细胞不会发生染色体数目的改变D.减数分裂过程中,细胞中的遗传物质平均分配至子细胞3.PM2.5是指大气中直径小于或等于2.5微米的颗粒物,富含大量的有毒、有害物质,很容易入人肺进入血液。

PM2.5今年成为广东空气污染指数重要标准。

下列推测不.合理的是A.颗粒物如有硅尘入肺可能会破坏吞噬细胞的溶酶体膜,释放水解酶破坏细胞结构B.过敏病人在PM2.5超标的空气中会发病,是因为该颗粒中有相关的过敏原C.PM2.5含量升高主要是人类活动的影响D.PM2.5的颗粒中的一些酸性物质进入人体血液会导致其pH成酸性4.小宁和小静为同卵双胞胎,下列说法不.正确的是A.自然条件下同卵双胞胎和人工胚胎分割技术产生的结果类似B.小宁患某种常染色体显性遗传病,小静正常,则可能是小宁胚胎发生了基因突变C.小宁和小静在表现型上的差异主要是由基因重组导致的D.若二人母亲为伴X显性遗传病患者,在其怀孕期间最好对胎儿进行基因检测5.下列有关叙述正确的是:A、物质借助载体蛋白质进行的跨膜运输,叫做协助扩散B、酶是活细胞产生具有调节作用的有机物C、植物生长调节剂的使用效果与浓度、使用方法、使用时期等有密切关系。

D、细胞进行无氧呼吸不需要水的参与,也没有二氧化碳的释放。

[来6.下列关于生物技术叙述正确的是A.为快速繁殖良种牛,可对受精卵进行分割移植B.利用基因工程得到的转基因马铃薯与普通马铃薯属于同一物种C.PCR技术中使用的DNA聚合酶与人体细胞中的DNA聚合酶完全相同D.胰蛋白酶处理桑椹胚获得的细胞可直接移植到同期发情受体的子宫中继续发育7.下列说法正确的是A.硫酸、纯碱、醋酸钠和生石灰分别属于酸、碱、盐和氧化物B.蔗糖、硫酸钡、水分别属于非电解质、弱电解质和弱电解质C.Mg、Al、Cu可以分别用置换法、直接加热法和电解法冶炼得到D.由热化学方程式:C(s)+O2(g)==CO2 (g) △H1;2CO(g)+O2(g)==2CO2(g) △H2;C (s)+1/2O2(g)==CO(g) △H3;则△H3=△H1—1/2△H28.短周期金属元素甲~戊在元素周期表中的相对位置如右表所示。

2024届广东省佛山市禅城区高三上学期统一调研测试物理试卷(一)

2024届广东省佛山市禅城区高三上学期统一调研测试物理试卷(一)

2024届广东省佛山市禅城区高三上学期统一调研测试物理试卷(一)一、单项选择题(本题包含8小题,每小题4分,共32分。

在每小题给出的四个选项中,只有一项是符合题目要求的)(共8题)第(1)题如图为理想自耦变压器,在A、B间输入电压有效值恒定的交变电流,开始时滑片位于线圈的中点G,滑片位于滑动变阻器R的中点,电流表和为理想电表,下列说法正确的是( )A.若仅将滑片向上滑动,则、的示数均变小B.若仅将滑片向上滑动,则、的示数均变大C.若仅将滑片向上滑动,则的示数变大,的示数变小D.若仅将滑片向上滑动,则的示数变小,的示数变大第(2)题如图是甲、乙、丙、丁四个物体的物理量随时间变化关系的图像,下列说法正确的是( )A.甲物体的机械能一定守恒B.乙物体受到的合力越来越大C.丙物体相同时间内的速度变化量相同D.丁物体的速度越来越大第(3)题一列简谐横波沿x轴正方向传播,周期为T,时的波形如图所示。

时()A.质点a速度方向沿y轴负方向B.质点b沿x轴正方向迁移了1mC.质点c的加速度为零D.质点d的位移为-5cm第(4)题使用如图所示的卸货装置从高处卸下货物时,先将质量为m的货物放置在倾角为α、长为L的粗糙木板上端,货物开始加速下滑的同时,自动液压杆启动并逐渐缩短,液压杆装置最终完全缩回到地面以下,货物以较小的速度v水平向右滑出木板,完成卸货。

已知重力加速度大小为g。

下列说法正确的是( )A.木板对货物做功为B.木板对货物做功为C.摩擦力对货物做功为D.支持力对货物不做功第(5)题如图为某一电场的电场线,M、N、P为电场线上的三个点,M、N是同一电场线上两点,下列判断正确的是()A.M、N、P三点中M点的场强最大B.M、N、P三点中N点的电势最高C.负电荷在N点的电势能大于在M点的电势能D.正电荷从M点自由释放,电荷将沿电场线运动到N点第(6)题2023年11月27日20时02分,摄影爱好者成功拍摄到中国空间站“凌月”(空间站从图中a点沿虚线到b点)的绝美画面,整个“凌月”过程持续时间为t=0.5s。

高三上学期第九周周测历史试题 Word版含答案

高三上学期第九周周测历史试题 Word版含答案

高三第三次周测试题第九周2020.11.4 出题人:一.选择题,在每小题给出的四个选项中,只有一个选项是符合题目要求的。

每小题4分,共60分。

1.周天子分封诸侯时要举行授土授民仪式,既要给受封者颁赐礼器以作凭证,又规定受封者应“以供王职”“法则周公”,并将这些载入文告,公布于世。

这一做法旨在()A.明确周王与诸侯的隶属关系B.突出周天子权力的至高无上C.缓和周王与诸侯之间的矛盾D.推动诸侯为周王朝开疆拓土2.据史书记载,唐文宗太和二年(公元828年),政府征集江南造水车匠赴京,“所成水车,分发畿内诸县令,令依样制造,以广溉种”。

这反映出当时()A.江南地区农业得到一定发展B.灌溉农业已经普及C.北方生产力水平落后于南方D.经济重心加速南移3.周敦颐、邵雍、程颢、程颐、张载并称北宋五子,是理学的开创者。

下图是北宋五子学术生涯的主要情况示意图,所示情况反映出当时()A 文化教育的均衡发展B 文化重心的南移已经完成C 南北文化交流的加强D 经济重心发展程度决定文化水平4. 利玛窦在《中国札记》中说,虽然我们已经说过中国的政府形式是君主制……它在一定程度上是贵族政体……如果没有与内阁大臣磋商或考虑他们的意见,皇帝本人对国家大事就不能作出最后的决定……所有这类文件都必须先由大臣审阅然后呈交给皇帝。

利玛窦认为中国明朝时()A.君主制不是中国政府的主要形式B.内阁一定程度上制约着君权C.君主与内阁大臣有同等的决策权D.内阁掌握着明朝的行政大权5.光绪十四年(1888年),维新人士谭嗣同在著名学者刘人熙的指导下,开始认真研究黄宗羲、王夫之、顾炎武等人的著作。

谭嗣同这一研究的主要意图是()A.追溯政治变革的历史依据B.汲取朴素的民主思想精华C.糅合民权思想否定君权论D.形成独特的激进政治理念6.武昌起义后,各省代表联合会在汉口英租界举行,与会代表成分十分复杂,革命派与立先派占有几乎相等的席位,另有少数旧官僚和封建士绅。

河南省天一大联考2024-2025学年高三上学期阶段性检测(二)生物试题(含答案)

河南省天一大联考2024-2025学年高三上学期阶段性检测(二)生物试题(含答案)

大联考2024—2025学年高中毕业班阶段性测试(二)生物学考生注意:1.答题前,考生务必将自己的姓名、考生号填写在试卷和答题卡上,并将考生号条形码粘贴在答题卡上的指定位置。

2.回答选择题时,选出每小题答案后,用铅笔把答题卡对应题目的答案标号涂黑。

如需改动,用橡皮擦干净后,再选涂其他答案标号。

回答非选择题时,将答案写在答题卡上。

写在本试卷上无效。

3.考试结束后,将本试卷和答题卡一并交回。

一、选择题:本题共13小题,每小题2分,共26分。

在每小题给出的四个选项中,只有一项是符合题目要求的。

1.机体中的生物大分子不能及时被正常降解而贮积,会引起细胞、组织及器官功能的障碍,这主要与下列哪种细胞器的功能障碍有关()A.内质网B.高尔基体C.溶酶体D.核糖体2.奶茶作为年轻人喜爱的新式茶饮,由茶、奶和糖等原材料制成,含有多种多样的化学物质。

下列有关奶茶中的物质检测的说法,错误的是()A.若奶茶和斐林试剂水浴加热后不出现砖红色,则说明奶茶中不含糖类B.奶茶中的蛋白质加热变性后,仍可以与双缩脲试剂产生紫色反应C.向奶茶中加入适量的苏丹Ⅲ染液后,奶茶中的脂肪被染成橘黄色D.向奶茶中加入适量的碘液,通过显色反应检测奶茶中是否含有淀粉3.腺苷是一种神经递质,在海马结构中的作用最强。

腺苷激酶(ADK)通过使腺苷磷酸化转变为AMP来降低组织中腺苷的水平,控制中枢神经系统中腺苷的代谢,调节细胞外腺苷的水平。

目前已证实腺苷具有抗癫痫作用,下列有关说法错误的是()A.组成腺苷的化学元素有C、H、O、NB.腺苷分子彻底水解能获得两种不同的产物C.腺苷磷酸化的产物AMP可作为合成DNA的原料D.降低体内ADK的活性对癫痫的发作有明显的抑制作用4.近年来研究发现,原核细胞也存在细胞骨架,人们已经在细菌中发现了FtsZ、MreB和CreS这3种重要的细胞骨架蛋白。

下列有关说法错误的是()A.细菌合成FtsZ、MreB和CreS时直接在细胞质基质中对蛋白质进行加工B.FtsZ、MreB和CreS等蛋白锚定并支撑着线粒体、核糖体等多种细胞器C.高温破坏FtsZ、MreB和CreS的空间结构,从而使其功能不可逆地丧失D.细胞骨架与物质运输、能量转化、信息传递、细胞分裂等生命活动密切相关5.低温处理下,植物细胞能通过改变细胞液渗透压增强其抗寒性。

高三数学上学期周练试卷(十)文(含解析)-人教版高三全册数学试题

高三数学上学期周练试卷(十)文(含解析)-人教版高三全册数学试题

2014-2015学年某某省某某外国语学校高三(上)周练数学试卷(文科)(十)一.选择题1.在复平面内,复数对应的点位于()A.第一象限B.第二象限C.第三象限D.第四象限2.下列说法正确的是()A.若a∈R,则“<1”是“a>1”的必要不充分条件B.“p∧q为真命题”是“p∨q为真命题”的必要不充分条件C.若命题p:“∀x∈R,sinx+cosx≤”,则¬p是真命题D.命题“∃x0∈R,使得x02+2x0+3<0”的否定是“∀x∈R,x2+2x+3>0”3.设S n是等差数列a n的前n项和,若,则=()A.B.C.D.4.若△ABC为锐角三角形,则下列不等式中一定能成立的是()A.log cosC>0 B.log cosC>0C.log sinC>0 D.log sinC>05.把函数图象上各点的横坐标缩短到原来的倍(纵坐标不变),再将图象向右平移个单位,那么所得图象的一条对称轴方程为()A.B.C.D.6.某几何体的三视图如图所示,则其侧面积为()A.B.C.D.7.对任意非零实数a,b,若a⊗b的运算规则如图的程序框图所示,则(3⊗2)⊗4的值是()A.0 B.C.D.98.设实数x,y满足约束条件,则u=的取值X围是()A.[,] B.[,] C.[,] D.[,]9.若函数f(x)=ax3+bx2+cx+d(a,b,c>0)在R上是单调函数,则的取值X围为()A.(4,+∞)B.(2+2,+∞)C.[4,+∞)D.[2+2,+∞)10.(5分)在区间[1,5]和[2,4]分别取一个数,记为a,b,则方程表示焦点在y轴上且离心率小于的椭圆的概率为()A.B.C.D.11.已知函数f(x)=|x+a|(a∈R)在[﹣1,1]上的最大值为M(a),则函数g(x)=M(x)﹣|x2﹣1|的零点的个数为()A.1个B.2个C.3个D.4个12.过双曲线﹣=1(a>0,b>0)的一个焦点F引它到渐近线的垂线,垂足为M,延长FM交y轴于E,若=2,则该双曲线离心率为()A.B.C.D.313.已知P、M、N是单位圆上互不相同的三个点,且满足||=||,则的最小值是()A.﹣B.﹣C.﹣D.﹣114.设函数y=f(x)的定义域为D,若函数y=f(x)满足下列两个条件,则称y=f(x)在定义域D上是闭函数.①y=f(x)在D上是单调函数;②存在区间[a,b]⊆D,使f(x)在[a,b]上值域为[a,b].如果函数f(x)=为闭函数,则k的取值X围是()A.(﹣1,﹣] B.[,1﹚C.(﹣1,+∞)D.(﹣∞,1)二.填空题15.(5分)(2014某某二模)已知||=2,||=2,||=2,且++=,则++=.16.设,若当且仅当x=3,y=1时,z取得最大值,则k的取值X围为.17.(5分)(2014某某一模)已知点P是椭圆=1(x≠0,y≠0)上的动点,F1,F2为椭圆的两个焦点,O是坐标原点,若M是∠F1PF2的角平分线上一点,且=0,则|的取值X围是.18.对于定义在区间D上的函数f(X),若存在闭区间[a,b]⊊D和常数c,使得对任意x1∈[a,b],都有f(x1)=c,且对任意x2∈D,当x2∉[a,b]时,f(x2)<c恒成立,则称函数f(x)为区间D上的“平顶型”函数.给出下列说法:①“平顶型”函数在定义域内有最大值;②函数f(x)=x﹣|x﹣2|为R上的“平顶型”函数;③函数f(x)=sinx﹣|sinx|为R上的“平顶型”函数;④当t≤时,函数,是区间[0,+∞)上的“平顶型”函数.其中正确的是.(填上你认为正确结论的序号)三.解答题19.(12分)(2014正定县校级三模)已知△ABC是半径为R的圆内接三角形,且2R(sin2A ﹣sin2C)=(a﹣b)sinB.(1)求角C;(2)试求△ABC面积的最大值.20.(12分)(2014某某二模)某公司研制出一种新型药品,为测试该药品的有效性,公司选定2000个药品样本分成三组,测试结果如表:分组A组B组C组药品有效670 a b药品无效80 50 c已知在全体样本中随机抽取1个,抽到B组药品有效的概率是0.35.(1)现用分层抽样的方法在全体样本中抽取360个测试结果,问应在C组抽取样本多少个?(2)已知b≥425,c≥68,求该药品通过测试的概率(说明:若药品有效的概率不小于90%,则认为测试通过).21.(12分)(2015某某模拟)已知几何体A﹣BCED的三视图如图所示,其中俯视图和侧视图都是腰长为4的等腰直角三角形,正视图为直角梯形.(1)求此几何体的体积V的大小;(2)求异面直线DE与AB所成角的余弦值;(3)试探究在DE上是否存在点Q,使得AQ⊥BQ并说明理由.22.(12分)(2014春雁峰区校级月考)在平面直角坐标系xOy中,已知中心在坐标原点且关于坐标轴对称的椭圆C1的焦点在抛物线C2:y2=﹣4x的准线上,且椭圆C1的离心率为.(1)求椭圆C1的方程,(2)若直线l与椭圆C1相切于第一象限内,且直线l与两坐标轴分别相交与A,B两点,试探究当三角形AOB的面积最小值时,抛物线C2上是否存在点到直线l的距离为.23.(12分)(2014某某校级模拟)已知函数f(x)=lnx+x2﹣ax(a为常数).(1)若x=1是函数f(x)的一个极值点,求a的值;(2)当0<a≤2时,试判断f(x)的单调性;(3)若对任意的a∈(1,2),x0∈[1,2],使不等式f(x0)>mlna恒成立,某某数m的取值X围.2014-2015学年某某省某某外国语学校高三(上)周练数学试卷(文科)(十)参考答案与试题解析一.选择题1.在复平面内,复数对应的点位于()A.第一象限B.第二象限C.第三象限D.第四象限【分析】利用复数的运算法则、几何意义即可得出.【解答】解:复数==﹣i﹣1对应的点(﹣1,﹣1)位于第三象限,故选:C.【点评】本题考查了复数的运算法则、几何意义,考查了推理能力与计算能力,属于基础题.2.下列说法正确的是()A.若a∈R,则“<1”是“a>1”的必要不充分条件B.“p∧q为真命题”是“p∨q为真命题”的必要不充分条件C.若命题p:“∀x∈R,sinx+cosx≤”,则¬p是真命题D.命题“∃x0∈R,使得x02+2x0+3<0”的否定是“∀x∈R,x2+2x+3>0”【分析】利用充要条件的定义,可判断A,B,判断原命题的真假,进而根据命题的否定与原命题真假性相反,可判断C,根据存在性(特称)命题的否定方法,可判断D.【解答】解:若“<1”成立,则“a>1”或“a<0”,故“<1”是“a>1”的不充分条件,若“a>1”成立,则“<1”成立,故“<1”是“a>1”的必要条件,综上所述,“<1”是“a>1”的必要不充分条件,故A正确;若“p∧q为真命题”,则“p,q均为真命题”,则“p∨q为真命题”成立,若“p∨q为真命题”则“p,q存在至少一个真命题”,则“p∧q为真命题”不一定成立,综上所述,“p∧q为真命题”是“p∨q为真命题”的充分不必要条件,故B错误;命题p:“∀x∈R,sinx+cosx=sin(x+)≤”为真命题,则¬p是假命题,故C 错误;命题“∃x0∈R,使得x02+2x0+3<0”的否定是“∀x∈R,x2+2x+3≥0”,故D错误;故选:A.【点评】本题以命题的真假判断为载体,考查了充要条件,命题的否定等知识点,是简单逻辑的简单综合应用,难度中档.3.设S n是等差数列a n的前n项和,若,则=()A.B.C.D.【分析】由题意可得 S3、S6﹣S3、S9﹣S6、S12﹣S9也成等差数列,由此可得 S6=S9+S3①,S12=3S9﹣3S6+S3②,再由可得 S12=S6③,利用①、②、③化简可得的值.【解答】解:∵S n是等差数列a n的前n项和,∴S3、S6﹣S3、S9﹣S6、S12﹣S9也成等差数列,∴S6﹣2S3=S9﹣2S6+S3,∴S6=S9+S3①.同理可得,S12﹣2S9+S6=S9﹣2S6+S3,即 S12=3S9﹣3S6+S3②.而由可得 S12=S6③.由①、②、③化简可得S3=S9,∴=,故选:C.【点评】本题主要考查等差数列的性质的应用,属于中档题.4.若△ABC为锐角三角形,则下列不等式中一定能成立的是()A.log cosC>0 B.log cosC>0C.log sinC>0 D.log sinC>0【分析】由锐角三角形ABC,可得1>cosC>0,0<A<,0<B<,,利用正弦函数的单调性可得sinB>sin(﹣A)=cosA>0,再利用对数函数的单调性即可得出.【解答】解:由锐角三角形ABC,可得1>cosC>0,0<A<,0<B<,,∴0<<B<,∴sinB>sin(﹣A)=cosA>0,∴1>>0,∴>0.故选:B.【点评】本题考查了锐角三角形的性质、锐角三角函数函数的单调性、对数函数的单调性等基础知识与基本技能方法,属于中档题.5.把函数图象上各点的横坐标缩短到原来的倍(纵坐标不变),再将图象向右平移个单位,那么所得图象的一条对称轴方程为()A.B.C.D.【分析】先对函数进行图象变换,再根据正弦函数对称轴的求法,即令ωx+φ=即可得到答案.【解答】解:图象上各点的横坐标缩短到原来的倍(纵坐标不变),得到函数;再将图象向右平移个单位,得函数,根据对称轴处一定取得最大值或最小值可知是其图象的一条对称轴方程.故选A.【点评】本小题综合考查三角函数的图象变换和性质.图象变换是考生很容易搞错的问题,值得重视.一般地,y=Asin(ωx+φ)的图象有无数条对称轴,它在这些对称轴上一定取得最大值或最小值.6.某几何体的三视图如图所示,则其侧面积为()A.B.C.D.【分析】从三视图可以推知,几何体是四棱锥,底面是一个直角梯形,一条侧棱垂直底面,易求侧面积.【解答】解:几何体是四棱锥,底面是一个直角梯形,一条侧棱垂直底面.且底面直角梯形的上底为1,下底为2,高为1,四棱锥的高为1.四个侧面都是直角三角形,其中△PBC的高PB===故其侧面积是S=S△PAB+S△PBC+S△PCD+S△PAD==故选A【点评】本题考查三视图求面积、体积,考查空间想象能力,是中档题.7.对任意非零实数a,b,若a⊗b的运算规则如图的程序框图所示,则(3⊗2)⊗4的值是()A.0 B.C.D.9【分析】由框图知,a⊗b的运算规则是若a≤b成立,则输出,否则输出,由此运算规则即可求出(3⊗2)⊗4的值【解答】解:由图a⊗b的运算规则是若a≤b成立,则输出,否则输出,故3⊗2==2,(3⊗2)⊗4=2⊗4==故选C.【点评】本题考查选择结构,解题的关键是由框图得出运算规则,由此运算规则求值,此类题型是框图这一部分的主要题型,也是这几年对框图这一部分考查的主要方式.8.设实数x,y满足约束条件,则u=的取值X围是()A.[,] B.[,] C.[,] D.[,]【分析】作出不等式组对应的平面区域,利用数形结合将目标函数进行转化,利用直线的斜率结合分式函数的单调性即可得到结论.【解答】解:作出不等式组对应的平面区域如图:则对应的x>0,y>0,则u==,设k=,则u==,由图象可知当直线y=kx,经过点A(1,2)时,斜率k最大为k=2,经过点B(3,1)时,斜率k最小为k=,即.∴,,∴,即,即≤z≤,故选:C【点评】本题主要考查线性规划的应用,利用目标函数的几何意义,结合数形结合是解决本题的关键,综合性较强,运算量较大.9.若函数f(x)=ax3+bx2+cx+d(a,b,c>0)在R上是单调函数,则的取值X围为()A.(4,+∞)B.(2+2,+∞)C.[4,+∞)D.[2+2,+∞)【分析】利用导数求解,由函数f(x)=ax3+bx2+cx+d(a,b,c>0)在R上是单调函数,可得f′(x)>0恒成立,找出a,b,c的关系,再利用基本不等式求最值.【解答】解:∵函数f(x)=ax3+bx2+cx+d(a,b,c>0)在R上是单调函数,∴f′(x)≥0在R上恒成立,即3ax2+2bx+c≥0恒成立,即△=4b2﹣12ac≤0 即b2≤3ac,∴==++2≥2+2≥4.故选C.【点评】考查利用导数即基本不等式的解决问题的能力,把问题转化为恒成立问题解决是本题的关键,应好好体会这种问题的转化思路.10.(5分)在区间[1,5]和[2,4]分别取一个数,记为a,b,则方程表示焦点在y轴上且离心率小于的椭圆的概率为()A.B.C.D.【分析】根据椭圆的性质结合椭圆离心率,求出a,b满足的条件,求出对应的面积,结合几何概型的概率公式进行求解即可.【解答】解:∵在区间[1,5]和[2,4]分别取一个数,记为a,b,∴,若方程表示焦点在y轴上且离心率小于,则,由e=<得c<a,平方得c2<a2,即a2﹣b2<a2,即b2>a2,则b>a或b a(舍),即,作出不等式组对应的平面区域如图:则F(2,2),E(4,4),则梯形ADEF的面积S==4,矩形的面积S=4×2=8,则方程表示焦点在y轴上且离心率小于的椭圆的概率P=,故选:C.【点评】本题主要考查几何概型的概率的计算,根据椭圆的性质求出a,b的条件,求出对应的面积,利用数形结合是解决本题的关键.11.已知函数f(x)=|x+a|(a∈R)在[﹣1,1]上的最大值为M(a),则函数g(x)=M(x)﹣|x2﹣1|的零点的个数为()A.1个B.2个C.3个D.4个【分析】求出M(a)的解析式,根据函数g(x)=M(x)﹣|x2﹣1|的零点,即函数M(x)=与函数y=|x2﹣1|交点的横坐标,利用图象法解答.【解答】解:∵函数f(x)=|x+a|(a∈R)在[﹣1,1]上的最大值为M(a),∴M(a)=,函数g(x)=M(x)﹣|x2﹣1|的零点,即函数M(x)=与函数y=|x2﹣1|交点的横坐标,由图可得:函数M(x)=与函数y=|x2﹣1|有三个交点,故函数g(x)=M(x)﹣|x2﹣1|有3个零点,故选:C【点评】本题考查函数图象的作法,熟练作出函数的图象是解决问题的关键,属中档题.12.过双曲线﹣=1(a>0,b>0)的一个焦点F引它到渐近线的垂线,垂足为M,延长FM交y轴于E,若=2,则该双曲线离心率为()A.B.C.D.3【分析】先利用FM与渐近线垂直,写出直线FM的方程,从而求得点E的坐标,利用已知向量式,求得点M的坐标,最后由点M在渐近线上,代入得a、b、c间的等式,进而变换求出离心率【解答】解:设F(c,0),则c2=a2+b2∵双曲线﹣=1(a>0,b>0)的渐近线方程为y=±x∴垂线FM的斜率为﹣∴直线FM的方程为y=﹣(x﹣c)令x=0,得点E的坐标(0,)设M(x,y),∵=2,∴(x﹣c,y)=2(﹣x,﹣y)∴x﹣c=﹣2x且y=﹣2y即x=,y=代入y=x得=,即2a2=b2,∴2a2=c2﹣a2,∴=3,∴该双曲线离心率为故选C【点评】本题考查了双曲线的几何性质,求双曲线离心率的方法,向量在解析几何中的应用13.已知P、M、N是单位圆上互不相同的三个点,且满足||=||,则的最小值是()A.﹣B.﹣C.﹣D.﹣1【分析】由题意可得,点P在MN的垂直平分线上,不妨设单位圆的圆心为O(0,0),点P (0,1),点M(x1,y1),则点N(﹣x1,y1),由得=,求出最小值.【解答】解:由题意可得,点P在MN的垂直平分线上,不妨设单位圆的圆心为O(0,0),点P(0,1),点M(x1,y1),则点N(﹣x1,y1),﹣1≤y1<1∴=(x1,y1﹣1),=(﹣x1,y1﹣1),.∴===2﹣,∴当y1=时的最小值是故选:B.【点评】本题主要考查两个向量的数量积公式,二次函数的性质,属于中档题.14.设函数y=f(x)的定义域为D,若函数y=f(x)满足下列两个条件,则称y=f(x)在定义域D上是闭函数.①y=f(x)在D上是单调函数;②存在区间[a,b]⊆D,使f(x)在[a,b]上值域为[a,b].如果函数f(x)=为闭函数,则k的取值X围是()A.(﹣1,﹣] B.[,1﹚C.(﹣1,+∞)D.(﹣∞,1)【分析】若函数f(x)=为闭函数,则存在区间[a,b],在区间[a,b]上,函数f(x)的值域为[a,b],即,故a,b是方程x2﹣(2k+2)x+k2﹣1=0(x,x≥k)的两个不相等的实数根,由此能求出k的取值X围.【解答】解:若函数f(x)=为闭函数,则存在区间[a,b],在区间[a,b]上,函数f(x)的值域为[a,b],即,∴a,b是方程x=的两个实数根,即a,b是方程x2﹣(2k+2)x+k2﹣1=0(x,x≥k)的两个不相等的实数根,当k时,,解得﹣1<k≤﹣.当k>﹣时,,无解.故k的取值X围是(﹣1,﹣].故选A.【点评】本题考查函数的单调性及新定义型函数的理解,解题时要认真审题,仔细解答,注意挖掘题设中的隐含条件,合理地进行等价转化.二.填空题15.(5分)(2014某某二模)已知||=2,||=2,||=2,且++=,则++= ﹣12 .【分析】把++=两边平方,变形可得++=(),代入数据计算可得.【解答】解:∵++=,∴平方可得(++)2=2,∴+2(++)=0,∴++=()=(4+8+12)=﹣12故答案为:﹣12【点评】本题考查平面向量数量积的运算,由++=两边平方是解决问题的关键,属中档题.16.设,若当且仅当x=3,y=1时,z取得最大值,则k的取值X围为(﹣,1).【分析】作出不等式对应的平面区域,利用线性规划的知识,确定目标取最优解的条件,即可求出a的取值X围.【解答】解:作出不等式对应的平面区域如图:由z=kx﹣y得y=kx﹣z,要使目标函数z=kx﹣y仅在x=3,y=1时取得最大值,即此时直线y=kx﹣z的截距最小,则阴影部分区域在直线y=kx﹣z的上方,目标函数处在直线x+2y﹣5=0和x﹣y﹣2=0之间,而直线x+2y﹣5=0和x﹣y﹣2=0的斜率分别为﹣,和1,即目标函数的斜率k,满足﹣<k<1,故答案为:(﹣,1).【点评】本题主要考查线性规划的应用,利用数形结合是解决线性规划题目的常用方法.根据条件目标函数z=kx﹣y仅在点A(3,1)处取得最大值,确定直线的位置是解决本题的关键.17.(5分)(2014某某一模)已知点P是椭圆=1(x≠0,y≠0)上的动点,F1,F2为椭圆的两个焦点,O是坐标原点,若M是∠F1PF2的角平分线上一点,且=0,则|的取值X围是.【分析】延长PF2、F1M,交与N点,连接OM,利用等腰三角形的性质、三角形中位线定理和椭圆的定义,证出|OM|=||PF1|﹣|PF2||.再利用圆锥曲线的统一定义,化简得||PF1|﹣|PF2||=|x0|,利用椭圆上点横坐标的X围结合已知数据即可算出|的取值X围.【解答】解:如图,延长PF2、F1M,交与N点,连接OM,∵PM是∠F1PF2平分线,且=0可得F1M⊥MP,∴|PN|=|PF1|,M为F1F2中点,∵O为F1F2中点,M为F1N中点∴|OM|=|F2N|=||PN|﹣|PF2||=||PF1|﹣|PF2||设P点坐标为(x0,y0)∵在椭圆=1中,离心率e==由圆锥曲线的统一定义,得|PF1|=a+ex0,|PF2|=a﹣ex0,∴||PF1|﹣|PF2||=|a+ex0﹣a+ex0|=|2ex0|=|x0|∵P点在椭圆=1上,∴|x0|∈[0,4],又∵x≠0,y≠0,可得|x0|∈(0,4),∴|OM|∈故答案为:【点评】本题求两点间的距离的取值X围,着重考查了椭圆的定义、等腰三角形的性质、三角形中位线定理和椭圆的简单几何性质等知识,属于中档题.18.对于定义在区间D上的函数f(X),若存在闭区间[a,b]⊊D和常数c,使得对任意x1∈[a,b],都有f(x1)=c,且对任意x2∈D,当x2∉[a,b]时,f(x2)<c恒成立,则称函数f(x)为区间D上的“平顶型”函数.给出下列说法:①“平顶型”函数在定义域内有最大值;②函数f(x)=x﹣|x﹣2|为R上的“平顶型”函数;③函数f(x)=sinx﹣|sinx|为R上的“平顶型”函数;④当t≤时,函数,是区间[0,+∞)上的“平顶型”函数.其中正确的是①④.(填上你认为正确结论的序号)【分析】根据题意,“平顶型”函数在定义域内某个子集区间内函数值为常数c,且这个常数是函数的最大值,但是定义并没有指出函数最小值的情况.由此定义再结合绝对值的性质和正弦函数的图象与性质,对于四个选项逐个加以判断,即得正确答案.【解答】解:对于①,根据题意,“平顶型”函数在定义域内某个子集区间内函数值为常数c,且这个常数是函数的最大值,故①正确.对于②,函数f(x)=x﹣|x﹣2|=的最大值为2,但不存在闭区间[a,b]⊊D和常数c,使得对任意x1∈[a,b],都有f(x1)=2,且对任意x2∈D,当x2∉[a,b]时,f(x2)<2恒成立,故②不符合“平顶型”函数的定义.对于③,函数f(x)=sinx﹣|sinx|=,但是不存在区间[a,b],对任意x1∈[a,b],都有f(x1)=2,所以f(x)不是“平顶型”函数,故③不正确.对于④当t≤时,函数,,当且仅当x∈[0,1]时,函数取得最大值为2,当x∉[0,1]且x∈[0,+∞)时,f(x)=<2,符合“平顶型”函数的定义,故④正确.故答案为:①④.【点评】本题以命题真假的判断为载体,着重考查了函数的最值及其几何意义、带绝对值的函数和正弦函数的定义域值域等知识点,属于中档题.三.解答题19.(12分)(2014正定县校级三模)已知△ABC是半径为R的圆内接三角形,且2R(sin2A ﹣sin2C)=(a﹣b)sinB.(1)求角C;(2)试求△ABC面积的最大值.【分析】(1)根据正弦定理,已知等式中的角转换成边,可得a、b、c的平方关系,再利用余弦定理求得cosC的值,可得角C的大小;(2)根据正弦定理算出c=R,再由余弦定理c2=a2+b2﹣2abcosC的式子,结合基本不等式找到边ab的X围,利用正弦定理的面积公式加以计算,即可求出△ABC面积的最大值.【解答】解:(1)∵2R(sin2A﹣sin2C)=(a﹣b)sinB,∴根据正弦定理,得a2﹣c2=(a﹣b)b=ab﹣b2,可得a2+b2﹣c2=ab∴cosC===,∵角C为三角形的内角,∴角C的大小为(2)由(1)得c=2Rsin=R由余弦定理c2=a2+b2﹣2abcosC,可得2R2=a2+b2﹣ab≥2ab﹣ab=(2﹣)ab,当且仅当a=b时等号成立∴ab≤=()R2∴S△ABC=absinC≤()R2=R2即△ABC面积的最大值为R2【点评】本题给出三角形的外接圆半径为R,在已知角的关系式情况下,求三角形面积最大值.着重考查了三角形的外接圆、正余弦定理和基本不等式求最值等知识,属于中档题.20.(12分)(2014某某二模)某公司研制出一种新型药品,为测试该药品的有效性,公司选定2000个药品样本分成三组,测试结果如表:分组A组B组C组药品有效670 a b药品无效80 50 c已知在全体样本中随机抽取1个,抽到B组药品有效的概率是0.35.(1)现用分层抽样的方法在全体样本中抽取360个测试结果,问应在C组抽取样本多少个?(2)已知b≥425,c≥68,求该药品通过测试的概率(说明:若药品有效的概率不小于90%,则认为测试通过).【分析】(1)利用抽样的性质先求出a,再根据样本总个数得出b+c=500,从而根据分层抽样的特点确定应在C组抽取样本多少个;(2)列举(b,c)的所有可能性,找出满足b≥425,c≥68,情况,利用古典概型概率公式计算即可.【解答】解:(1)∵,∴a=700∵b+c=2000﹣670﹣80﹣700﹣50=500∴应在C组抽取样本个数是个.(2)∵b+c=500,b≥425,c≥68,∴(b,c)的可能性是(425,75),(426,74),(427,73),(428,72),(429,71),(430,70),(431,69),(432,68)若测试通过,则670+700+b≥2000×90%=1800∴b≥430∴(b,c)的可能有(430,70),(431,69),(432,68)∴通过测试的概率为.【点评】本题考查分层抽样的性质,古典概型概率公式的应用,属于中档题.21.(12分)(2015某某模拟)已知几何体A﹣BCED的三视图如图所示,其中俯视图和侧视图都是腰长为4的等腰直角三角形,正视图为直角梯形.(1)求此几何体的体积V的大小;(2)求异面直线DE与AB所成角的余弦值;(3)试探究在DE上是否存在点Q,使得AQ⊥BQ并说明理由.【分析】(1)由该几何体的三视图知AC⊥面BCED,且EC=BC=AC=4,BD=1,则体积可以求得.(2)求异面直线所成的角,一般有两种方法,一种是几何法,其基本解题思路是“异面化共面,认定再计算”,即利用平移法和补形法将两条异面直线转化到同一个三角形中,结合余弦定理来求.还有一种方法是向量法,即建立空间直角坐标系,利用向量的代数法和几何法求解.(3)假设存在这样的点Q,使得AQ⊥BQ.解法一:通过假设的推断、计算可知以O为圆心、以BC为直径的圆与DE相切.解法二:在含有直线与平面垂直垂直的条件的棱柱、棱锥、棱台中,也可以建立空间直角坐标系,设定参量求解.这种解法的好处就是:1、解题过程中较少用到空间几何中判定线线、面面、线面相对位置的有关定理,因为这些可以用向量方法来解决.2、即使立体感稍差一些的学生也可以顺利解出,因为只需画个草图以建立坐标系和观察有关点的位置即可.以C为原点,以CA,CB,CE所在直线为x,y,z轴建立空间直角坐标系.设满足题设的点Q存在,其坐标为(0,m,n),点Q在ED上,∴存在λ∈R(λ>0),使得=λ,解得λ=4,∴满足题设的点Q存在,其坐标为(0,,).【解答】解:(1)由该几何体的三视图知AC⊥面BCED,且EC=BC=AC=4,BD=1,∴S梯形BCED=×(4+1)×4=10∴V=S梯形BCED AC=×10×4=.即该几何体的体积V为.(3分)(2)解法1:过点B作BF∥ED交EC于F,连接AF,则∠FBA或其补角即为异面直线DE与AB所成的角.(5分)在△BAF中,∵AB=4,BF=AF==5.∴cos∠ABF==.即异面直线DE与AB所成的角的余弦值为.(7分)解法2:以C为原点,以CA,CB,CE所在直线为x,y,z轴建立空间直角坐标系.则A(4,0,0),B(0,4,0),D(0,4,1),E(0,0,4)∴=(0,﹣4,3),=(﹣4,4,0),∴cos<,>=﹣∴异面直线DE与AB所成的角的余弦值为.(3)解法1:在DE上存在点Q,使得AQ⊥BQ.(8分)取BC中点O,过点O作OQ⊥DE于点Q,则点Q满足题设.(10分)连接EO、OD,在Rt△ECO和Rt△OBD中∵∴Rt△ECO∽Rt△OBD∴∠EOC=∠OBD∵∠EOC+∠CEO=90°∴∠EOC+∠DOB=90°∴∠EOB=90°.(11分)∵OE==2,OD==∴OQ===2∴以O为圆心、以BC为直径的圆与DE相切.切点为Q∴BQ⊥CQ∵AC⊥面BCED,BQ⊂面CEDB∴BQ⊥AC∴BQ⊥面ACQ(13分)∵AQ⊂面ACQ∴BQ⊥AQ.(14分)解法2:以C为原点,以CA,CB,CE所在直线为x,y,z轴建立空间直角坐标系.设满足题设的点Q存在,其坐标为(0,m,n),则=(﹣4,m,n),=(0,m﹣4,n)=(0,m,n﹣4),=(0,4﹣m,1﹣n)∵AQ⊥BQ∴m(m﹣4)+n2=0①∵点Q在ED上,∴存在λ∈R(λ>0)使得=λ∴(0,m,n﹣4)=λ(0,4,m,1﹣n)⇒m=,n=②②代入①得(﹣4)()2=0⇒λ2﹣8λ+16=0,解得λ=4∴满足题设的点Q存在,其坐标为(0,,).【点评】本小题主要考查空间线面关系、面面关系、二面角的度量、几何体的体积等知识,考查数形结合、化归与转化的数学思想方法,以及空间想象能力、推理论证能力和运算求解能力.22.(12分)(2014春雁峰区校级月考)在平面直角坐标系xOy中,已知中心在坐标原点且关于坐标轴对称的椭圆C1的焦点在抛物线C2:y2=﹣4x的准线上,且椭圆C1的离心率为.(1)求椭圆C1的方程,(2)若直线l与椭圆C1相切于第一象限内,且直线l与两坐标轴分别相交与A,B两点,试探究当三角形AOB的面积最小值时,抛物线C2上是否存在点到直线l的距离为.【分析】(1)由题意设椭圆C1的方程,(a>b>0),且,由此能求出椭圆C1的方程.(2)设直线l的方程为y=kx+m(k<0,m>0)由,得(3+4k2)x2+8kmx+4m2﹣12=0,由此利用根的判别式、韦达定理、点到直线距离公式、弦长公式能推导出抛物线C2上不存在点到直线l的距离为.【解答】解:(1)∵椭圆C1的焦点在抛物线C2:y2=﹣4x的准线上,且椭圆C1的离心率为.∴椭圆焦点在x轴上,设椭圆C1的方程:,(a>b>0),且,解得a=2,b=,∴椭圆C1的方程为.(2)∵直线l与椭圆C1相切于第一象限内,∴直线l的斜率存在且小于零,设直线l的方程为y=kx+m(k<0,m>0)由,得(3+4k2)x2+8kmx+4m2﹣12=0,由题可知,△=0,∴m2=4k2+3,当即时上式等号成立,此时,直线l为设点D为抛物线C2上任意一点,则点D到直线l的距离为,利用二次函数的性质知,∴抛物线C2上不存在点到直线l的距离为.【点评】本题考查椭圆方程的求法,考查当三角形面积最小时满足条件的点是否存在的判断与求法,解题时要认真审题,注意根的判别式、韦达定理、点到直线距离公式、弦长公式的合理运用.23.(12分)(2014某某校级模拟)已知函数f(x)=lnx+x2﹣ax(a为常数).(1)若x=1是函数f(x)的一个极值点,求a的值;(2)当0<a≤2时,试判断f(x)的单调性;(3)若对任意的a∈(1,2),x0∈[1,2],使不等式f(x0)>mlna恒成立,某某数m的取值X围.【分析】(1)求导数,利用极值的定义,即可求a的值;(2)当0<a≤2时,判断导数的符号,即可判断f(x)的单调性;(3)问题等价于:对任意的a∈(1,2),不等式1﹣a>mlna恒成立.即恒成立.【解答】解:.(1)由已知得:f'(1)=0,∴1+2﹣a=0,∴a=3.…(3分)(2)当0<a≤2时,f′(x)=因为0<a≤2,所以,而x>0,即,故f(x)在(0,+∞)上是增函数.…(8分)(3)当a∈(1,2)时,由(2)知,f(x)在[1,2]上的最小值为f(1)=1﹣a,故问题等价于:对任意的a∈(1,2),不等式1﹣a>mlna恒成立.即恒成立记,(1<a<2),则,…(10分)令M(a)=﹣alna﹣1+a,则M'(a)=﹣lna<0所以M(a),所以M(a)<M(1)=0…(12分)故g'(a)<0,所以在a∈(1,2)上单调递减,所以即实数m的取值X围为(﹣∞,﹣log2e].…(14分)【点评】本题考查导数知识的综合运用,考查函数的极值,考查函数的单调性,考查恒成立问题,正确分离参数是关键.。

2021-2022年高三上学期英语周测卷(一) 含答案

2021-2022年高三上学期英语周测卷(一) 含答案

2021年高三上学期英语周测卷(一)含答案II. Grammar and Vocabulary(26%)SectionADirections: Read the following two passages. Fill in the blanks to make the passage coherent. For the blanks with a given word, fill in each blank with the proper form of the given word. For the other blanks, fill in each blank with one proper word. Make sure that your answers are grammatically correct.ANo trip to South Korea is plete without a visit to its fascinating theme parks. For those who assume that amusement parks are just playgrounds teeming with kids, South Korea’s theme parks are sure (25)______(change) their minds.With their charming garden plantations, hot springs and exciting rides, they offer travellers a romantic and exciting getaway. Within the appropriate driving distance from the Korean capital Seoul, there are three theme parks worth (26)______ (visit)—Lotte World, Everland and Seoul Land.Everland, the (27)______(large) theme park in South Korea, covering various areas like Festival World, Caribbean Bay and Speedway. It is such (28)______ huge park that you will have to plan in advance where you are investing your time there.Lotte World, (29)_____ designer was determined to create a wonderland for fun-seekers, consists of Lotte World Adventure, and movie theatres. You (30)______ also take part in activities such as ice-skating or bowling there.(31)______ travellers want, South Korea is there to provide. Hop on aheart-stopping ride, amuse yourself in one of the theme parks or simply enjoy a (32)_____(relax) hot spring bath. e and visit South Korea now!(B)Some of young soldiers who had recently joined the army were being trained in modern ways of fighting. One of the lessons they should take was (33) ______ an unarmed man could trick an armed enemy, take his weapon away and have him (34) ______ (arrest). First one of their two instructors took a knife away from the other, using only his bare hands, and then he took a gun away from him in the same way.After the lesson, and before they went on to train the young soldiers to do these things themselves, the two instructors asked them a number of questions to see how well they had understood what (35) ______ (show). One of the questions was this, “Well, you now know (36) ______ an unarmed man can do against a man with a gun. Imagine that you (37) ______ (guard) a bridge at eight one night, and that you have a gun. Suddenly you see an unarmed enemy soldier (38) ______ (e) towards you, and what will you do?”The young soldier (39) ______ had to answer this question thought carefully for a few seconds (40) ______ he answered, and then said, “Well after what I have just seen, I think that the first thing I would do would be to get rid of my gunas quickly as I could so that the unarmed enemy soldier couldn’t take it from me and kill me with it!”Section BDirections: plete the following passage by using the words in the box. Each word can be used only once. Note that there is one word more than you need.If this summer you pay a visit to Milan, the fashion center of Italy, make sure you’re not caught eating ice cream in the streets after midnight as doing so is now ___41___.A new law was passed by Milan’s city council banning the sale of take-away food and drinks after midnight in some districts which are famous for their nightlife ___42___. The purpose of this unusual move is, according to the city council, to discourage night gathering in downtown areas.The law inevitably has given rise to a number of protests, accusing that the government has ___43___ people’s normal lives. However, if you take into consideration the country’s ___44___ economy and its high unemployment rate, the local government’s fear of ‘night assembling’may be reasonable.In fact, Milan’s law is only the strangest of a host of restrictions on nightlife that have ___45___ up in European cities recently. Madrid’s citycenter was declared a ‘low-noise zone’ last September and the city council has been refusing to ___46___ bar and club licenses ever since.Why do European cities deal so strictly with nightlife? It may be because Europe’s population is getting older and can no longer ___47___ late night activities within the neighborhood.In the past, bars and clubs bloomed in European city centers, which were ___48___ to working class populations. But gradually, these people began to move out of the city centers and into the suburbs. Only the wealthy and the upper-class people can afford to live in ___49___ centers now. But these people don’t go to bars and clubs to socialize. Instead, they consider fun-seekers who wander in their neighborhoods annoying. They also worry that bars and clubs will make their neighborhoods less ___50___ and devalue their housing property.III. Reading prehension(47%)Section ADirections:For each blank in the following passage 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.Many of us have characteristics that are inherited from our families. Maybewe have our father’s eyes or our grandmother’s hair color – __51__ features that result from the genes we inherit. But there are other elements of our personalities such as behaviors and viewpoints that we pick up by __52__ our parents. Perhaps we have picked up our mother’s love of cleaning or our dad’s sense of humor. Now, researchers at the University of Michigan believe there may be a specific viewpoint we can thank our parents for: the way we __53__ work.The researchers referred to previous studies that identified three main ways of viewing work.First, there are job-oriented (以求职为目的的)people, those who __54__ to see their job as simply a way to make money. They __55__ to clocking out(打卡下班)every day and pursuing fun activities __56__ the office. If you have a job-oriented father, you may view work this way. __57__, if you grew up close to your mother you probably will not hold this viewpoint. One possible __58__ cited by the study: Youngsters that grow up close to their mothers are less likely to view work as just a job.Second, there are career-oriented(追求事业型的)people who see their job as a place to find __59__ and gain a sense of __60__. These are the people who don’t __61__ working overtime. In fact, some people love their jobs so much that they bee workaholics! They are more fortable in the office than at home. According tothe study, being close to a career- oriented father while growing up means there’s a good chance you’ll share his perspective. Oddly, having a mother with this viewpoint seems to have little __62__.Third, there are calling-oriented workers – folks who view their job as a way to make a (n) __63__ impact upon the world. They are more __64__ with improving the world around them than earning a large salary. In the study, those who expressed a strong calling (天职)orientation came from homes where both parents were calling-oriented. This suggests that adolescents need the support of both parents in order to have the confidence to put __65__ first and career success second.The good news is –we still have choice. Whether we share our parents’views of work or not, we can still find a career that suits us.51. A. medical B. physical C. biological D. mental52. A. stimulating B. neglecting C. implying D. modeling53. A. view B. think C. evaluate D. ment54. A. attempt B. manage C. tend D. offer55. A. pay attention B. are opposed C. look forward D. are devoted56. A. outside B. inside C. upside D.downside57. A. Otherwise B. Therefore C. However D. Moreover58. A. observation B. explanation C. negotiation D. instruction59. A. salary B. pany C. direction D. achievement60. A. trust B. identity C. despair D. urgency61. A. like B. resist C. mind D. consider62. A. influence B. evidence C. performance D. justice63. A. objective B. negative C. subjective D. positive64. A. familiar B. concerned C. anxious D. bined65. A. personal ideals B. economic profitsC. professional achievementsD. global developmentSection BDirections:Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices markedA, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.(A)On “Super Bowl Sunday”, millions of Americans are glued to their TVs. They are eating pizza, chicken wings and chips and cheering every move. They’re watching the Super Bowl.Why are Americans so crazy about American football? Well, it is more exciting than other sports. One team can lose possession of the ball in a minute, which may allow their opponents to make a touchdown(触地得分). Then that team may win the game unexpectedly.The Super Bowl also entertains its audience with a great halftime show. The football field is turned into a stage. Then an impressive performance of dancing and singing with special effects occurs.Since the Super Bowl is the most-watched TV program in America, mercial airtime is also very expensive. Big money is also spent on mercials. They draw the viewers’attention and advertise their image or products during the mercial break.After football season, the “March Madness”begins. The NCAA (National Collegiate Athletic Association) petitions begin with 68 men’s collegebasketball teams. They play until the field is reduced to the “Final Four”. The winning team bees the national champion. The NBA (National Basketball Association) All-Star game is also held in February. The best players from all the teams play in this game.Baseball is no doubt American’s national sport. From grandpas to young kids, whole families go to ball games together. They wear their favorite team’s caps or even carry their mascot(吉祥物). Our family went to watch Wang Chien-Ming play in D.C. once. We ate hot dogs, waved flags and sang during the 7th inning(棒球的一局) stretch.If you are not a sports fan yet, e and pick a sport or a team. Go to a ball game with your family, and cheer your team on. Sports are definitely a part of American culture one should not miss.66. According to the passage, the Super Bowl is _____.A. a well-received American football gameB. an expensive American sportC. an impressive TV performanceD. a most-watched TV series67. Which of the following may be one of the reasons for the popularity of the Super Bowl?A. Its results are too exciting to meet viewers’ expectations.B. People can eat pizza, chicken wings and chips when watching it.C. Its mercials are expensive enough to draw viewer s’ attention.D. The performances during its halftime show are appealing to viewers.68. Which of the following is TRUE according to the passage?A. Americans spend a large amount of money on ball games.B. American people of all ages like going to watch baseball games.C. All the American stars take part in the national basketball game.D. There are important national ball games in America almost every month.69. The passage is mainly written to _____.A. inform readers of the three popular sports in AmericaB. teach readers how to understand the sports culture in AmericaC. encourage readers to fit in with American culture through sportsD. show readers the importance of sports in young people’s life in America(B)Getting to work by bicycle has never been more ernments and nations are transforming their cities and highways to meet the needs of this new generation of cycling enthusiasts , and in a world where green alternatives (选择) are the new must-have, average citizens are eagerly seizing their chance to help make their streets and their bodies cleaner than ever before.Bee an EBTC Member-Click Here.The Edmonton Bicycle and Touring Club ( EBTC) is a recreationalnot-for-profit volunteer-run group devoted to promoting its members to cooperatively run bicycling trips during the spring, summer and fall, cross-country skiing trips in the winter, and social events all the year round. We wele both road bike and mountain bike riders !Benefits of Joining EBTC :·Opportunity to create the kinds of events you like: enjoy dozens of events all year round, organized by members like you !·Participation in the local cycling munity.·Use of club bicycle tools on tour and library materials.·Discounts at various bicycle shops.Why Cycle with a Group?·It's more FUN !·It encourages you to e out often and get healthy exercise without going to a gym.·Develop skills and gain through the experience of others.·Meet new and interesting people with a mon interest in cycling.·Meet a fun-loving, energetic and different group of individuals who enjoy a healthy lifestyle.·Enjoy the outdoors experience with the panionship and security of a group.·Group atmosphere provides challenge to strong cyclists and support and confidence to the novice .For more info on the EBTC:Phone the Club Hotline at 780-424-2453 (780-424-BIKE)E-mail: jprimeau@ edmontonbicycle.Mail us at : Edmonton Bicycle & Touring ClubP. O. Box 5xxGarneau Postal Stn.Edmonton, AB T6G 2T5 Canada70. The aim of EBTC is ______ .A. to call for governments to produce more bicyclesB. to make money by organizing bicycling tripsC. to popularize cycling by attracting more membersD. to look for young volunteers for sporting events71.What's the advantage of cycling with a group? ______A. Participating in whatever events.B. Meeting more new people.C. Having bicycles free of charge.D. Enjoying discounts in a gym.72.The underlined word "novice" in the passage probably means "________ .A. professional cyclistB. inexperienced riderC. event organizerD. healthy individual73.Apart from the website, how many other ways can people gain information from EBTC?A. Two.B. Three.C. Four.D. Five.(C)The Tourist Trade Contributes Absolutely Nothing to Increasing Understanding between Nations) The tourist trade is booming. With all this ing and going, you’d expect greater understanding to develop between the nations of the world. Not a bit of it! Superb systems of munication by air, sea and land make it possible for us to visit each other ’s countries at a moderate cost. What was once the ‘grand tour ’, reserved for only the very rich, is now within everybody ’s grasp? The package tour and chartered flights are not to be sneered at. Modern travelers enjoy a level of fort which the lords and ladies on grand tours in the old days couldn’t have dreamed of. But what ’s the sense of this mass exchange of populations if the nations of the world remain basically ignorant of each other? Many tourist organizations are directly responsible for this state of affairs. They deliberately set out to protect their clients from too much contact with the local population. The modern tourist leads a cosseted, sheltered life. He lives at international hotels, where he eats his international food and sips his international drink while he gazes at the natives from a distance. Conducted tours to places of interest are carefully censored. The tourist is allowed to see only what the organizers want him to see and no more. A strict schedule makes itimpossible for the tourist to wander off on his own; and anyway, language is always a barrier, so he is only too happy to be protected in this way. At its very worst, this leads to a new and hideous kind of colonization. The summer quarters of the inhabitants of the cite universitaire: are temporarily reestablished on the island of Corfu. Blackpool is recreated at Torremolinos where the traveler goes not to eat paella, but fish and chips.The sad thing about this situation is that it leads to the persistence of national stereotypes. We don’t see the people of other nations as they really are, but as we have been brought up to believe they are. You can test this for yourself. Take five nationalities, say, French, German, English, American and Italian. Now in your mind, match them with these five adjectives: musical, amorous, cold, pedantic, native. Far from providing us with any insight into the national characteristics of the peoples just mentioned, these adjectives actually act as barriers. So when you set out on your travels, the only characteristics you notice are those which confirm your preconceptions. You e away with the highly unoriginal and inaccurate impression that, say, ‘Anglo-Saxons are hypocrites’ of that ‘Latin peoples shout a lot ’. You only have to make a few foreign friends to understand how absurd and harmful national stereotypes are. But how can you make foreign friends when the tourist trade does its best to prevent you?Carried to an extreme, stereotypes can be positively dangerous. Wild generalizations stir up racial hatred and blind us to the basic fact—how triteit sounds! – That all people are human. We are all similar to each other and atthe same time all unique.74.The passage mainly wants to tell us________A.tourism contributes nothing to increasing understanding between nations.B.Tourism is tiresome.C.Conducted tour is dull.D. tourism really does something to one’s country.75.What is the author's attitude toward tourism?A.apprehensive.B.negative.C.critical.D. appreciative.76.The word”cosseted” (para 2 line 3) probably means_____A.over-protectedB.variedC.limitedD.luxurious77.What is ‘grand tour’ now?A. moderate cost.B. local sight-seeing is investigated by the tourist organization.C.people enjoy the first-rate forts.D.everybody can enjoy the ‘grand tour’.Section CDirections: Read the passage carefully. Then answer the questions or plete the statements in the fewest possible words.Yes, but what did we use to do before there was television? How often we hear statements like this! Television hasn’t been with us all that long, but we are already beginning to forget what the world was like without it. Before we admitted the one-eyed monster into our homes, we never fond it difficult to occupy our spare time. We used to enjoy civilized pleasures. For instance, we used to have hobbies, we used to entertain our friends and be entertained by them, we used to go outside for our amusements to theatres, cinemas, restaurants and sporting events. We even used to read books and listen to music and broadcast talks occasionally. All that belongs to the past. Now all our free time is regulated by the goggle box. We rush home or gulp down our meals to be in time for this or that programme. We have even given up sitting at table and having a leisurely evening meal, exchanging the news of the day. A sandwich and a glass of beer will do – anything, providing it doesn’t interfere with the programme. The monster demands and obtains absolute silence and attention. If any member of the family dares to open his mouth during aprogramme, he is quickly silenced.Whole generations are growing up addicted to the telly. Food is left uneaten, homework undone and sleep is lost. The telly is a universal pacifier. It is now standard practice for mother to keep the children quiet by putting them in the living-room and turning on the set. It doesn’t matter that the children will watch rubbishy mercials or spectacles of sadism and violence – so long as they are quiet.There is a limit to the amount of creative talent available in the world. Every day, television consumes vast quantities of creative work. That is why most of the programmes are so bad: it is impossible to keep pace with the demand and maintain high standards as well. When millions watch the same programmes, the whole world bees a village, and society is reduced to the conditions which obtain in preliterate munities. We bee utterly dependent on the two most primitive media of munication: pictures and the spoken word.Television encourages passive enjoyment. We bee content with second-hand experiences. It is so easy to sit in our armchairs watching others working. Little by little, television cuts us off from the real world. We get so lazy, we choose to spend a fine day in semi-darkness, glued to our sets, rather than go out into the world itself. Television may be s splendid medium of munication, but itprevents us from municating with each other. We only bee aware how totally irrelevant television is to real living when we spend a holiday by the sea or in the mountains, far away from civilization. In quiet, natural surroundings, we quickly discover how little we miss the hypnotic tyranny of King Telly.78.What does a mother usually do to keep her children quiet?_______________________________________________________________________ ___79.what’s the author’s suggestion to forget TV?_________________________________________80.&81Please smmerize at least two harms of TV80.__________________________________________________________________________ __81.__________________________________________________________________________ __第II卷I. Translation (22%)Directions:Translate the following sentences into English, using the words given in the brackets.1.这位明星因为吸毒被警方逮捕。

高三上学期第二次周考(地理)试题含答案

高三上学期第二次周考(地理)试题含答案

高三上学期第二次周考(地理)(考试总分:100 分)一、选择题组(本题共计12小题,总分60分)1.(4分)在每小题给出的四个选项中,只有一个是符合题目要求的)西班牙的耶罗岛地形崎岖,全年温和,多地形雨。

下图为耶罗岛地理位置示意图。

据此回答1~2题:1.耶罗岛地形雨主要分布在岛屿的()A.东南部B.东北部C.西北部D.西南部2.耶罗岛全年温和,气候并不炎热的重要原因是常年()A.受东北信风影响B.受副热带高气压带影响C.受洋流影响D.受人类影响2.(6分)读“大洋表层海水温度、盐度、密度随纬度变化示意图”,据此完成3—5题。

3.图中①②③三条曲线代表海水性质对应正确的是( )A.①海水盐度B.①海水温度C.②海水密度D.③海水温度4.与南纬60°附近相比,赤道附近的表层海水( )A.温度低、盐度低、密度小B.温度高、盐度低、密度大C.温度高、盐度高、密度小D.温度低、盐度高、密度大5.海洋蕴藏着丰富的资源。

北半球副热带近岸地区( )A.海水温度适宜的海域可发展海水养殖业B.海盐资源丰富,可直接加工C.海盐资源丰富,不能进行海水淡化D.随着海水深度增加,海洋生物数量增多3.(6分)每年秋季,美国加利福尼亚州(下称加州)易出现山林火灾。

如图为2018年11月加州山火分布示意图,此次山火造成了巨大的人员伤亡和财产损失。

“圣安娜风”特指秋冬季节从大盆地向西南扫过加州的一种风,对所经过的山林火灾影响极大。

据此回答6~8题:6.形成“圣安娜风”的气压或风带是()A.陆地冷高压B.副热带高压C.盛行西风D.极地东风7.到达加利福尼亚州着火点的“圣安娜风”的性质是()A.低温干燥B.高温干燥C.低温湿润D.高温湿润8.秋季,美国加利福尼亚州林区多火灾的主要原因是()①多雷电天气②多受焚风效应影响③多枯枝落叶④多家庭燃木取暖A.①③B.②③C.①④D.②④4.(4分)下图为我国某区域某时段霜冻线变化及该时段北京天气状况图。

江西省赣州市石城中学2020届高三物理上学期第五次周考试题[含答案]

江西省赣州市石城中学2020届高三物理上学期第五次周考试题[含答案]

江西省赣州市石城中学2020届高三物理上学期第五次周考试题考试时间:60分钟总分:100分本次考试范围:1.1-2.6 下次考试范围:1.1-3.2一、选择题(9×6=54分,第1~6小题只有一个选项符合题目要求,第7~9小题有多个选项符合题目要求,全部选对的得6分,选不全的得3分,有选错或不答的得0分)1.如图所示,小物块原来静止在固定的粗糙斜面上,现施加水平向右的推力F,F的大小由零逐渐增大,直到小物块刚要开始滑动为止。

则在此过程中,小物块所受的( )A.合外力可能增大 B.斜面的支持力可能不变C.斜面的摩擦力可能先增大后减小 D.斜面的摩擦力一定先减小后增大2.(错题再现)如图位于水平桌面上的物块P,由跨过定滑轮的轻绳与物块Q相连,从滑轮到P和到Q的两段绳都是水平的,已知Q与P之间以及P与桌面之间的动摩擦因数都是μ,两物体的质量都是m,滑轮轴上的摩擦都不计。

若用一水平向右的力F拉P使它做匀速运动,则F的大小为( )A.3μmg B.4μmg C.5μmg D.6μmg3.日常生活中,我们在门下缝隙处塞紧一个木楔(侧面如图所示),往往就可以把门卡住。

有关此现象的分析,下列说法正确的是()A.木楔对门的作用力大于门对木楔的作用力,因而能将门卡住B.门对木楔作用力的水平分量等于地面对木楔摩擦力的大小C.只要木楔的厚度合适都能将门卡住,与顶角θ的大小无关D.只要木楔对门的压力足够大就能将门卡住,与各接触面的粗糙程度无关4.如图所示,穿在一根光滑的固定杆上的小球A、B连接在一条跨过定滑轮的细线两端,杆与水平面成θ角,不计所有摩擦,当两球静止时,OA绳与杆的夹角为θ,OB绳沿竖直方向。

则正确的说法是( )A.A可能受到两个力作用 B.B可能受到三个力作用C.绳子对A的拉力大于对B的拉力 D.A、B 的质量之比为1:tanθ5.如图所示的风向测试仪上有五个浮球彼此用轻绳连接,悬挂于空中。

若假设风力水平稳定,测试仪发生倾斜,绳与竖直方向的夹角为30°,设每个浮球的质量为m 。

河北省沧州市任丘市第一中学南校区2022-2023学年高三上学期周测物理试题 (1)

河北省沧州市任丘市第一中学南校区2022-2023学年高三上学期周测物理试题 (1)

任丘一中南校区高三物理周测试题第Ⅰ卷(选择题,共48分)一、选择题(本题共10小题,共48分。

第1~6题每题4分,只有一项符合题目要求;第7~10题有多个选项符合题目要求,全部选对得6分,选对但不全得3分,错选得0分).1.一直导线平行于通电螺线管的轴线放置在螺线管的上方,如图所示,如果直导线可以自由地运动且通以由a 到b 的电流,则关于导线ab 受磁场力后的运动情况,下列说法正确的是().A .从上向下看顺时针转动并靠近螺线管B .从上向下看顺时针转动并远离螺线管C .从上向下看逆时针转动并远离螺线管D .从上向下看逆时针转动并靠近螺线管2.欧姆在探索导体的导电规律的时候,没有电流表,他利用小磁针的偏转检测电流,具体的做法是:在地磁场的作用下,处于水平静止的小磁针上方,平行于小磁针水平放置一直导线,当该导线中通有电流的时候,小磁针就会发生偏转;当通过该导线的电流为I 时,发现小磁针偏转了30°,由于直导线在某点产生的磁场与通过直导线的电流成正比,当他发现小磁针偏转了60°时,通过该导线的电流为()A .3IB .2I C.3I D .I 3.如图所示,圆形区域内有一垂直纸面的匀强磁场,P 为磁场边界上的一点.有无数带有同样电荷、具有同样质量的粒子在纸面内沿各个方向以相同的速率通过P 点进入磁场.这些粒子射出边界的位置均处于边界的某一段圆弧上,这段圆弧的弧长是圆周长的13.将磁感应强度的大小从原来的B 1变为B 2,结果相应的弧长变为原来的一半,则B 2/B 1等于()A.2B.3C .2D .34.如图所示,一段长方体形导电材料,左右两端面的边长都为a 和b ,内有带电量为q 的某种自由运动电荷.导电材料置于方向垂直于其前表面向里的匀强磁场中,内部磁感应强度大小为B.当通以从左到右的稳恒电流I时,测得导电材料上、下表面之间的电压为U,且上表面的电势比下表面的低、由此可得该导电材料单位体积内自由运动电荷数及自由运动电荷的正负分别为().A.IB|q|aU,负 B.IB|q|aU,正C.IB|q|bU,负 D.IB|q|bU,正5.如图所示,两个完全相同的木模质量均为m,通过三根轻质竖直细线对称连接,放在水平面上呈“互”字型静置,上方木模呈现悬浮效果,这是利用了建筑学中的“张拉整体”的结构原理。

2025届江苏无锡高三上学期期中考试物理试题+答案

2025届江苏无锡高三上学期期中考试物理试题+答案

无锡市2024年秋学期高三期中教学质量调研测试物理2024.11命题单位:江阴市教师发展中心 制卷单位:无锡市教育科学研究院注意事项考生在答题前请认真阅读本注意事项及各题答题要求1.本试卷共6页,满分为100分,考试时间为75分钟.考试结束后,请将本试卷和答题卡一并交回.2.答题前,请务必将自己的姓名、准考证号用0.5毫米黑色墨水的签字笔填写在试卷及答题卡的规定位置.3.请认真核对监考员在答题卡上所粘贴的条形码上的姓名、准考证号与本人是否相符. 4.作答选择题,必须用2B 铅笔将答题卡上对应选项的方框涂满、涂黑;如需改动,请用橡皮擦干净后,再选涂其他答案.作答非选择题,必须用0.5毫米黑色墨水的签字笔在答题卡上的指定位置作答,在其他位置作答一律无效.5.如需作图,必须用2B 铅笔绘、写清楚,线条、符号等须加黑、加粗.一、单项选择题(本题共11小题,每小题4分,共44分.每小题只有一个选项符合题意).1.某同学在列车车厢的顶部用细线悬挂一个小球,列车在平直轨道行驶过程中,细线偏过一定角度并相对车厢保持静止,如图所示.则列车( )A .匀速运动B .匀变速运动C .向左运动D .向右运动2.主动降躁耳机能主动收集周围环境中的躁声信号,并产生相应的抵消声波,取得最佳降躁效果的抵消声波与收集到的躁声具有不同的( )A .频率B .振幅C .相位D .波长3.如图是单摆做阻尼振动的位移—时间图像,摆球在P 与N 时刻具有相同的( )A .速度B .加速度C .重力势能D .机械能4.质量为1m 和2m 的两个物体在光滑的水平面上正碰,其位置—时间图像如图所示.则( )A .碰撞后,2m 的速率大于1m 的速率B .碰撞后,2m 的动量等于1m 的动量C .碰撞后,2m 的动量小于1m 的动量D .1m 小于2m5.把一个小球放在玻璃漏斗中,晃动漏斗,可使小球在短时间内沿光滑的漏斗壁在高度为h 的水平面内做匀速圆周运动,如图所示.若h 越大,则小球( )A .对侧壁的压力越大B .加速度越小C .角速度越小D .线速度越小6.伽利略创造地把实验、假设和逻辑推理相结合,此科学方法促进了人类科学认识的发展.利用如图所示的装置做如下实验:小球从左侧斜面上的O 点由静止释放后沿斜面向下运动,并沿右侧斜面上升.斜面上先后铺垫三种粗糙程度逐渐减低的材料时,小球沿右侧斜面上升到的最高位置依次为1、2、3.根据三次实验结果的对比,可以得到的最直接的结论是( )A .如果斜面光滑,小球将上升到与O 点等高的位置B .如果小球不受力,它将一直保持匀速运动或静止状态C .如果小球受到力的作用,它的运动状态将发生改变D .小球受到的力一定时,质量越大,它的加速度越小7.鹊桥二号中继星的成功发射,为嫦娥六号在月球背面的探月任务提供地月间中继通讯.如图所示,鹊桥二号采用周期为T 的环月椭圆冻结轨道,远月点B 距月心为近月点A 距月心距离的9倍,CD 为椭圆轨道的短轴,关于该卫星的说法正确的是( )A .从C 经B 到D 的运动时间为0.5T B .在A 、B 两点的速度大小之比为3:1C .在A 、B 两点的加速度大小之比为9:1D .在地球表面附近的发射速度大于7.9km/s 小于11.2km/s8.2024年第11号台风“摩羯”于9月6日在海南文昌登陆,登陆时中心附近最大风力有17级以上,造成具大破坏.已知11级台风的风速范围为28.5m/s~32.6m/s ,17级台风的风速范围为56.1m/s~61.2m/s .若台风迎面垂直吹向一固定的交通标志牌,则17级台风对该标志牌的作用力大小约为11级台风的( ) A .2倍 B .4倍 C .8倍 D .16倍9.2024年,中国选手陈艺文获得巴黎奥运会跳水项目女子3米板金牌.运动员(可视为质点)从跳板起跳后 运动速度—时间关系图像如图所示,0t =时刻跳板恢复水平,运动员向上跳离跳板,忽略空气阻力,重力加速度210m/s g =,关于图像中12t t 、和1v 不可求的是( )A .1tB .2tC .1vD .都不可求10.如图所示,水平圆盘上P 点有一滴粘稠液体,使圆盘绕竖直中心轴缓慢转动起来,则液体在圆盘上留下的痕迹示意图可能为( )A .B .C .D .11.如图所示,竖直轻弹簧固定在水平地面上,一铁球从弹簧的正上方h 高处由静止释放.以铁球释放点为原点,竖直向下为正方向,分别用y 、k E 和E 表示铁球的位移、动能和机械能.不计空气阻力,弹簧在弹性限度内.关于铁球从释放到最低点的过程中,其k E y −或E y −图像可能正确的是( )A .B .C .D .二、非选择题:本题共556分.其中第13题~第16题解答时请写出必要的文字说明、方程式和重要的演算步骤,只写出最后答案的不能得分;有数值计算时,答案中必须写出数值和单位.12.(15分)如图(1),用图示装置测量滑块沿斜面下滑的加速度.将长直木板B 支成斜面,挡光片P 固定在滑块C 上且两者前端平齐,光电门G 固定在木板上,导光孔与木板上的位置A 相齐.让滑块C 的前端与斜面上的位置O 相齐,滑块由静止释放,测出OA 的距离x 、挡光片的宽度x ∆及挡光时间t ,算出挡光片通过光电门的平均速度v ,即可测出滑块下滑的加速度.图(1) 图(2)(1)用游标卡尺测量挡光片的宽度,正确的操作应为图(2)中的____________图(选填“a ”或“b ”). (2)用v 作为滑块到达A 处时的瞬时速度,得到的加速度____________(填“偏大”或“偏小”);(3)换用不同宽度的挡光片,重复实验,测出挡光片的宽度x ∆及挡光时间t ,得出不同挡光片通过光电门的平均速度v ,图(3)中已描出几个点.根据图象,滑块前端过A 处的瞬时速度大小为____________m/s (结果保留三位有效数字),加速度大小为____________2m/s (结果保留一位有效数字). (4)有同学认为新方案每次都改变了研究对象的总质量,因此实验没有重复性,你认为这一观点正确吗?简要说明理由.图(3)13.(6分)为了清理堵塞河道的冰凌,实施投弹爆破,飞机在河道上空高180m H =处以水平速度0216km/h v =匀速飞行,投放炸弹并击中目标,不计空气阻力,g 取210m/s .求:(1)炸弹刚脱离飞机到击中目标所飞行的水平距离; (2)炸弹击中目标时的速度大小.14.(8分)如图所示,一质量为m 的小球,用长为L 的轻绳静止悬挂于O 点的正下方P 点.小球在水平恒力F 的作用下,从P 点运动到Q 点,立即撤去F .已知F mg =,30θ=°,重力加速度为g .求:(1)水平拉力F 做的功W ; (2)小球回到P 点时绳子的拉力T .15.(12分)某仓库运货的简化装置如图所示,配重和电动机连接小车的缆绳均平行于斜坡.装满货物的小车从静止开始加速运动距离L ,再以速度v 沿斜坡匀速上行,关闭电动机后,小车又沿斜坡上行一段距离速度为零.卸货后,给小车一个向下的初速度,小车沿斜坡刚好匀速下行.已知配重质量为m ,小车质量2m ,车上货物质量为30m ,重力加速度为g ,小车运动时受到的摩擦阻力与车及车上货物总重力成正比,比例系数0.25k =,配重始终未接触地面,不计电动机自身机械摩擦及缆绳质量.求:(1)斜坡倾角的正弦值sin θ; (2)关闭电动机后小车上行的距离d ; (3)加速上行阶段电动机对小车做的功W .16.(15分)如图甲所示,上表面光滑的固定平台上有A 、B 两物体,A 与一轻弹簧相连,以初速度0v 向B 运动.从弹簧接触B 到与B 分离过程A 、B 的v t −图像如图乙所示.已知从0t =到0t t =时间内,A 运动的距离为000.73v t .完全分离后B 滑上静止在光滑地面上与平台等高的木板C ,C 由水平粗糙轨道和1/4光滑圆弧轨道组成、两者相切,圆弧轨道半径232v R g=,水平轨道长度208v L g =.已知B 、C 质量均为m ,A 、B 可视为质点,不计空气阻力,重力加速度g .求:图甲图乙m;(1)A物体质量A(2)为使B物体能进入圆弧轨道,且在上升阶段不脱离C,则B与C的水平轨道间的动摩擦因数μ满足的条件;x∆.(3)A、B碰撞过程中,弹簧压缩量的最大值m无锡市2024年秋学期高三期中教学质量调研测试答案物理一、单项选择题:每小题只有一个选项符合题意(本大题11小题,每小题4分,共44分).1 2 3 4 5 6 7 8 9 10 11 BCCDCADBBCC二、非选择题:本题共5题,共56分.其中第13题~第16题解答时请写出必要的文字说明、方程式和重要的演算步骤,只写出最后答案的不能得分;有数值计算时,答案中必须明确写出数值和单位.12.(15分,每空3分)(1)a (2)偏大 (3)0.252、0.07 (4)不正确.(1分)由牛顿运动定律,sin cos mg mg ma θµθ−=, 得运动加速度sin cos ag g θµθ−,与质量无关.(2分) 13.(6分)解:(1)炸弹脱离飞机到击中目标在空中做平抛运动, 竖直方向212H gt =(1分) 水平方向0x v t = (1分)代入数据得:360m x = (1分) (2)22y v gH = (1分)v= (1分)代入数据得:/s v = (1分) 14.(8分)解:(1)Lsin W F θ= (2分) 代入已知得:0.5W mgL = (1分) (2)设小球回到P 点速度为v ,小球从P 点到Q 点再回到P 点,根据动能定理有:212W mv =(2分) 在P 点:2v T mg m L−= (2分)代入已知,得:2T mg = (1分)15.(12分)解:(1)小车沿斜坡匀速下行,对小车sin m g km g T θ=+车车 (1分)对配重有T m g =配 (1分)将2m m m m ==配车、和0.25k =代入,得:sin 0.75θ= (1分) (2)关闭电动机后,设小车上行的加速度大小为a ,对小车sin m g km g T m a θ′+−=总总总 (1分) 对配重T m g m a ′−=配配 (1分) 22v ad = (2分)将32m m m m ==配总、和0.25k =代入,得:23362v d g= (1分) (3)对小车、货物和配重系统, 由功能关系得:()21sin 2W km gL m m v m gL mgL θ−=++−总总总 (3分) 将320.25m m k ==总、代入,得:23116.5W mgL mv =+ (1分) 16.(15分)解:(1)A 、B 碰撞,系统动量守恒,由图乙有()04A A v m v m m =+ (2分) 解得:3A mm =(1分) (2)从弹簧接触B 到与B 分离,设分离时A 物体速度A ,v B 物体速度B v ,有A A AB m v m v mv =+ 2220111222A A A Bm v m v mv =+ 解得:00.5B v v = (2分)若B 物块恰好运动到圆弧轨道的最低点,此时两者共速,则对B 与C 整体由水平方向动量守恒12B mv mv = (1分) 由能量守恒定律221111222B mv mv mgL µ=×+(1分) 解得:10.5µ=若B 物块恰好运动到与圆弧圆心等高的位置,此时两者共速,则对B 与C 整体由水平方向动量守恒22B mv mv = (1分) 由能量守恒定律222211222B mv mv mgL mgR µ=×++ (1分) 解得:20.25µ=综上所述B 与C 的水平轨道间的动摩擦因数μ的取值范围为0.250.5µ≤< (1分) (3)弹簧接触B 后,弹簧压缩过程中,A 、B 动量守恒,有01133A B mv mv mv =+ (1分) 对方程两边同时乘以时间t ∆,有01133A B mv tmv t mv t ∆=∆+∆ (1分) 00~t 之间,根据位移等于速度在时间上的累积,可得001133A B mv t mx mx =+ (1分) 将A 000.73x v t =代入,得000.09B x v t = (1分) 则弹簧压缩量的最大值m A B 000.64x x x v t ∆=−=. (1分)。

2022-2023学年辽宁省沈阳市第二中学高三上学期周练习化学试卷

2022-2023学年辽宁省沈阳市第二中学高三上学期周练习化学试卷

沈阳二中2022-2023学年上学期周练(9-24)高三(23届)化学试题命题人:高三化学组 审校人:高三化学组说明:1.测试时间:60分钟 总分80分2.客观题涂在答题卡上,主观题写在答题纸相应位置上3.可能用到的相对原子量:H:1 C:12 0:16 Cu:64第Ⅰ卷(36分)一、选择题(本题包括12小题,每小题3分,共36分)1.中华诗词中蕴含着丰富的化学知识。

下列关于诗词的分析错误的是A.“章山之铜,所谓丹阳铜也、今有白铜,盖点化为之,非其本质”,纯铜比白铜硬度大熔点低B.“梨花院落溶溶月,柳絮池塘淡淡风”,句中柳絮的主要成分属于糖类C.“遍身罗绮者,不是养蚕人”,句中的罗绮不可用沸水浸泡D.“朝坛雾卷,曙岭烟沉”,雾有丁达尔现象是因为胶体粒子对光有散射作用2.我国自主研制的核电技术成果“华电一号”其主要核燃料铀238得到一个中子后经过2次β衰变成为钚239:U 23892+n 10→U 23992,U 23992+e 01-→p 23993N ,p 23993N →u 23994P +e 01-。

下列有关说法错误的是 A.U 23892和U 23992互为同位素 B.p 23993N 和u 23994P 化学性质不相同C.U 23992的中子数为145 D.U 23892经过三步化学反应得到u 23994PA为阿伏加德罗常数的值,下列说法正确的是A. 12g NaHSO 4固体中含有A 个阳离子 与2混合后的分子数目为N A412CH 含有中子数为3N A2和2于密闭容器中充分反应后,HI 分子总数为A4.化学与社会、生活、生产密切相关。

下列事实与解释一致的是A.能使KI淀粉试纸变蓝的溶液:Na+、NH4+、S2-、SO42-B. 能使甲基橙变红的溶液:K+、Na+、Cl-、S2O32-H=14的溶液:K+、Na+、Cl-、C1O-D.能与金属铝反应生成H2的溶液:K、Ca2+、NO3-、HCO3-6.下列由废铁屑制取无水Fe2(SO4)3的实验原理与装置不能达到实验目的的是A.用装置甲除去废铁屑表面的油污B. 用装置乙溶解废铁屑制Fe2(SO4)3C. 用装置丙过滤得到Fe2(SO4)3溶液D.用装置丁蒸干溶液获得Fe2(SO4)37.下列方程式不能准确解释相应实验现象的是A. MnO2和浓盐酸共热产生黄绿色气体:MnO2+4H+ +2Cl-∆=Mn2++Cl2↑+2H20B. Na加入滴有酚酞的水中,溶液变红:2Na+2H20=2Na++20H-+H2↑C.加热Fe和S的混合物生成黑色固体:2Fe+3S ∆=Fe2S3D.加热蓝色的CuCl2溶液,溶液变绿:[Cu(H2O)4]2+(aq,蓝色)+4Cl-(aq)[CuCl4]2-(aq,黄色)+4H2O(1) ΔH>08.从硫化物中提取单质梯(Sb)是先在高温下将硫化物转化为氧化物,再用碳还原:①2Sb2S3+3O2+6Fe→Sb4O6+6FeS ②Sb4O6+6C→4Sb+6CO↑关于反应①、②的说法正确的是A.反应①②中的氧化剂分别是Sb2S3、Sb4O6B.反应①中每生成了3mol FeS时,共转移6mol电子C.反应②说明高温下Sb的还原性比C强D. 每生成4mol Sb时,反应①与反应②中还原剂的物质的量之比为4:39. X、Y、Z、M、Q五种短周期元素,原子序数依次增大。

江西省赣州市石城中学2020届高三物理上学期第五次周考试题[带答案]

江西省赣州市石城中学2020届高三物理上学期第五次周考试题[带答案]

江西省赣州市石城中学2020届高三物理上学期第五次周考试题考试时间:60分钟总分:100分本次考试范围:1.1-2.6 下次考试范围:1.1-3.2一、选择题(9×6=54分,第1~6小题只有一个选项符合题目要求,第7~9小题有多个选项符合题目要求,全部选对的得6分,选不全的得3分,有选错或不答的得0分)1.如图所示,小物块原来静止在固定的粗糙斜面上,现施加水平向右的推力F,F的大小由零逐渐增大,直到小物块刚要开始滑动为止。

则在此过程中,小物块所受的()A.合外力可能增大 B.斜面的支持力可能不变C.斜面的摩擦力可能先增大后减小 D.斜面的摩擦力一定先减小后增大2.(错题再现)如图位于水平桌面上的物块P,由跨过定滑轮的轻绳与物块Q相连,从滑轮到P 和到Q的两段绳都是水平的,已知Q与P之间以及P与桌面之间的动摩擦因数都是μ,两物体的质量都是m,滑轮轴上的摩擦都不计。

若用一水平向右的力F拉P使它做匀速运动,则F的大小为( )A.3μmg B.4μmg C.5μmg D.6μmg3.日常生活中,我们在门下缝隙处塞紧一个木楔(侧面如图所示),往往就可以把门卡住。

有关此现象的分析,下列说法正确的是()A.木楔对门的作用力大于门对木楔的作用力,因而能将门卡住B.门对木楔作用力的水平分量等于地面对木楔摩擦力的大小C.只要木楔的厚度合适都能将门卡住,与顶角θ的大小无关D.只要木楔对门的压力足够大就能将门卡住,与各接触面的粗糙程度无关4.如图所示,穿在一根光滑的固定杆上的小球A、B连接在一条跨过定滑轮的细线两端,杆与水平面成θ角,不计所有摩擦,当两球静止时,OA绳与杆的夹角为θ,OB绳沿竖直方向。

则正确的说法是( )A.A可能受到两个力作用 B.B可能受到三个力作用C.绳子对A的拉力大于对B的拉力 D.A、B 的质量之比为1:tanθ5.如图所示的风向测试仪上有五个浮球彼此用轻绳连接,悬挂于空中。

若假设风力水平稳定,测试仪发生倾斜,绳与竖直方向的夹角为30°,设每个浮球的质量为m。

生物丨江苏省扬州中学2022-2023学年高三上学期10月双周练生物试卷及答案

生物丨江苏省扬州中学2022-2023学年高三上学期10月双周练生物试卷及答案

江苏省扬州中学2022-2023学年第一学期高三10月周测试题2022.10一、单项选择题:本大题共14小题,每小题2分,共28分。

在每题给出的四个选项中,只有一项是最符合题意的。

(请将所有选择题答案填到答题卡的指定位置中。

)1.下列关于元素和无机盐与人体健康的描述,正确的是()A.Mg是构成人体的微量元素,缺乏时会出现一定程度的贫血B.P是组成细胞膜、细胞核的重要成分,缺乏时影响细胞增殖C.细胞外液渗透压的90%以上来自Ca2+,血液中Ca2+太低会抽搐D.I是甲状腺激素的组成成分,缺乏时会引起神经系统兴奋性增强2.二硫键“—S—S—”是蛋白质中连接两条肽链之间的一种化学键。

如图是由280个氨基酸组成的某蛋白质的结构图,对其叙述正确的是()。

A.该蛋白质至少有2个游离的羧基B.形成该蛋白质的过程中脱去了277个水分子C.该蛋白质至少有280个氨基D.该蛋白质的功能由氨基酸的数量、种类、排列顺序三方面决定3.蛋白质分选是蛋白质依靠自身信号序列,从起始合成部位定向转运到功能发挥部位的过程。

下图示为人体某种线粒体蛋白的分选途径,下列说法错误的是()A.结构A是蛋白质通道,结构B是受体B.信号序列发挥定向功能与翻译过程几乎是同时进行的C.参与③过程的酶是肽酶D.信号序列和受体是蛋白质分选的物质基础4.下图是花生种子在发育、萌发过程中,糖类和脂肪干重的变化曲线。

下列分析正确的是()A.干重相等的可溶性糖和脂肪,可溶性糖所贮存的能量较大B.由图可见,种子发育中,可溶性糖可转变为脂肪,种子需要的N、P、K增加C.种子萌发时,可溶性糖是种子生命活动的直接能源物质D.种子萌发时,脂肪酶的活性很高,可降低反应的活化能5.下列关于不同生物细胞结构和功能的叙述正确的是()A.大肠杆菌和草履虫都有复杂的生物膜系统B.酵母菌和乳酸菌都可以通过有丝分裂增殖C.颤藻和伞藻合成蛋白质的场所都是核糖体D.根瘤菌和硝化细菌都能将无机物合成有机物6.如图所示,生物膜上的ATP可通过VNUT蛋白进人囊泡,再经囊泡运输分泌到细胞外,作为信号分子调节生命活动。

重庆市南开中学2024-2025学年高三上学期第一次开学质量检测物理(含解析)

重庆市南开中学2024-2025学年高三上学期第一次开学质量检测物理(含解析)

物理试题一、单项选择题:本题共7小题,每小题4分,共28分。

在每小题给出的四个选项中,只有一项是符合题目要求的。

1.关于功和功率,下列说法正确的是()A.功是矢量,功的正负表示方向B.恒力做功的大小一定与物体的运动路径有关C.某时刻的瞬时功率可能等于一段时间内的平均功率D.发动机的实际功率总等于额定功率2.关于物理学的发展史,下列说法正确的是()A.开普勒分析第谷的观测数据之后认为天体的运动是最完美、最和谐的匀速圆周运动B.第谷分析开普勒的观测数据之后发现了行星运动的规律,并总结概括为三大定律C.牛顿发现了万有引力定律,并测出了引力常量的大小D.人们利用万有引力定律不仅能够计算天体的质量,还能发现未知天体3.近年来,我国在机器人领域的研究取得重大进展,逐步实现从“中国制造”到“中国智造”。

一个机器人的速度-时间图像(v-t图)如图所示,在0~10s内,下列说法正确的是()A.机器人做匀减速运动B.机器人在0~10s内的位移大于55mC.机器人的加速度逐渐减小D.机器人可能做曲线运动4.2024年8月1日,我国在西昌卫星发射中心使用长征三号乙运载火箭成功将互联网高轨卫星02星发射升空。

这是长征系列运载火箭的第529次飞行。

如图所示,最初该卫星在轨道1上绕地球做匀速圆周运动,后来卫星通过变轨在轨道2上绕地球做匀速圆周运动。

下列说法正确的是()A .卫星在轨道2上运行的周期小于在轨道1上运行的周期B .卫星在轨道2上运行的速度大于在轨道1上运行的速度C .卫星在轨道2上运行的角速度小于在轨道1上运行的角速度D .卫星在轨道2上运行的线速度大于地球的第一宇宙速度5.如图所示,用一根轻绳系着一个可视为质点的小球,轻绳的长度为L 。

最初小球静止在圆轨迹的最低点A 点,现在A 点给小球一个初速度,使其在竖直平面内沿逆时针方向做圆周运动,已知B 点与圆心O 等高,C 点是圆轨迹的最高点,重力加速度为g 。

不计一切阻力,下列说法正确的是()A .小球做的是匀变速曲线运动B .若要使得小球做完整的圆周运动,小球运动到CC .若小球无法做完整的圆周运动,则小球可能在C 点脱离圆轨迹D .若小球无法做完整的圆周运动,则小球可能在B 点和C 点之间的某一点脱离圆轨迹6.如图所示,光滑的圆锥体顶部有一根杆,一根不可伸长的轻绳一端与杆的上端相连,另一端与一个质量为的小球(视为质点)相连,其中轻绳的长度为。

高三上学期理科综合第2周周测试题

高三上学期理科综合第2周周测试题

湛江市第二十一中学高三上学期理科综合第2周周测试题一、单项选择题:本大题共16小题,每小题4分,共64分。

在每小题给出的四个选项中,只有一个选项符合题目要求,选对的得4分,选错或不答的得0分。

1.下图为某二倍体生物体内的一组细胞分裂示意图,据图分析正确的是()A.图②产生的子细胞一定为精细胞B.图中属于体细胞有丝分裂这一过程的有①③⑤C.图示5个细胞均含有同源染色体D.该生物的体细胞中均含有2个染色体组2、下列有关光合作用和细胞呼吸的叙述中,正确的是()A.光合作用和细胞呼吸过程中产生的[H]都能与氧气结合生成水B.适宜的光照条件下叶绿体和线粒体合成ATP都需要氧气C.CO2的固定发生在叶绿体中,葡萄糖分解成CO2的过程只发生在线粒体中D.动物细胞的无氧呼吸产生的是乳酸,没有CO23.某小组进行观察洋葱根尖分生组织细胞有丝分裂的实验,下列关于该实验的叙述正确的是A.盐酸和酒精混合液主要起固定作用B.碱性染料吡罗红(派洛宁)可用于染色体染色C.观察到分裂末期细胞内细胞板向四周扩展形成新的细胞壁D.细胞内染色体的存在状态可作为判断有丝分裂各时期的依据4.用高倍显微镜观察洋葱根尖细胞的有丝分裂。

下列描述正确的是A.处于分裂间期和中期的细胞数目大致相等B.视野中不同细胞的染色体数目可能不相等C.观察处于分裂中期的细胞,可清晰看到赤道板和染色体D.细胞是独立分裂的,因此可选一个细胞持续观察它的整个分裂过程5.如图是表示某生物有丝分裂过程中核DNA含量变化的曲线,下列关于曲线各段的描述不正确的是A.若E 表示的是分裂末期,则e 段的核DNA 含量和c 、d 段的相同B.若B 到E 是分裂期,则b 段是DNA 复制时期C.若A 是分裂间期,B 到E 是分裂期,则ab 段比cf 段经历的时间短D.若A 到E 是一个完整的细胞周期,则c 段表示的是分裂前期6、细胞学家发现人类的一种神经细胞中DNA 含量是a ,在同一种器官中可以找到DNA 含量为0.5a 、a 、2a 的细胞,这种器官应该是A.骨髓B.肝脏C.卵巢D.皮肤7.下列说法正确的是 ( )A .水煤气是通过煤的液化得到的燃料B .乙酸和乙酸乙酯都能与氢氧化钠溶液反应C .淀粉、蛋白质、脂肪和葡萄糖都可发生水解反应D .葡萄糖、麦芽糖和蔗糖都能与新制氢氧化铜反应生成Cu 2O8.若N A 表示阿伏加德罗常数的值,下列叙述正确的是 ( )A .56 g Fe 2+被氧化时,需要得到N A 个电子B .2 L 0.3 mol ·L -1 Na 2S 溶液中,S 2-离子数是0.6 N A 个C .1 mol C 4H 10含有非极性键数为3N A 个D .N A 个CO 2分子和N A 个O 2分子的体积相同9.下列各组离子在指定的环境中能大量存在的是 ( )A .在pH=1的溶液中:SiO 32-、NO 3-、Na +、I -B .由水电离出的c(H +)=1×10-12mol/L 的溶液中:K +、Na +、Cl -、HCO 3-C .在能使甲基橙变红色的溶液中:NH 4+、Fe 2+、NO 3-、Cl -D .在能使石蕊试纸变蓝色的溶液中:Na +、K +、S 2-、CO 32-10.短周期元素X 、Y 、Z 、W 、Q 在元素周期表中的相对位置如图1所示。

江苏省南通市2024-2025学年高三上学期期初测试地理试题含答案

江苏省南通市2024-2025学年高三上学期期初测试地理试题含答案

江苏省如皋2024-2025学年度高三年级测试地理试题(答案在最后)一、单选题:共23题,每题2分,共46分。

每题只有一个选项最符合题意。

广东与江苏是我国的经济大省,两省在人口结构方面存在较大差异。

下表为“2020年第七次人口普查广东、江苏两省主要人口指标对比”。

据此完成下面小题。

省份常住人口/万人)0-14岁人口比重/%15-64岁人口比重/%65岁及以上人口比重/%出生率‰自然增长率‰流动人口/万人广东12624.0018.8572.578.5810.28 5.585206.62江苏8477.2615.2168.5916.20 6.660.172366.381.导致广东与江苏老龄化程度差异较大的主要因素是()A .自然条件B .人口基数C .人口流动D .地方政策2.广东省自然增长率与江苏差异明显,主要的原因是()A .医疗卫生条件好B .育龄人口比重大C .二孩政策发布早D .经济发展水平高3.相对于广东,江苏目前急需完善()A .社会养老保障体系B .省内交通网络建设C .环境污染治理法规D .城乡户籍管理制度农村人口老龄化程度高于城市的现象称作老龄化“城乡倒置”。

下图为2000~2020年我国四个省市区65岁以上的城镇和农村人口比例,其中柱状图为城镇,折线代表农村。

完成下面小题。

4.四省市中老龄化“城乡倒置”起步最早、目前最严重的是()A.北京市B.黑龙江省C.福建省D.新疆维吾尔自治区5.出现这种现象的主要诱因是()A.城市化占用土地B.农业实现机械化C.农村的耕地流转D.城乡收入差距大新型城镇化是实现共同富裕的关键。

自2014年新型城镇化成为国家战略以来,中国城镇化持续快速演进。

下图为2001—2018中国4个超大城市(现阶段常住人口大于1000万人)的常住与户籍人口变化,完成下面小题。

6.总体上四个超大城市常住人口数增加最快的时段是()A.2001-2005B.2005-2010C.2010-2014D.2014-20187.下列关于新型城镇化战略实施后的影响叙述正确的是()A.a城户籍人口变化趋势反转B.b城户籍人口数量止步不前C.c城常住人口增速突然放缓D.d城常住人口数量迎来突增8.未来我国超大城市的新型城镇化应致力于()①严控非户籍人口流入②推进流动人口市民化③扩大公共服务覆盖率④提高自然资源开发率A.①②B.①④C.②③D.②④伴随乡村人口迁出,部分地区出现“空心村”。

江苏省南通市如皋市2024-2025学年高三上学期期初调研测试英语试卷

江苏省南通市如皋市2024-2025学年高三上学期期初调研测试英语试卷

江苏省南通市如皋市2024-2025学年高三上学期期初调研测试英语试卷一、阅读理解The High School Creator Expo is a melting pot of hands-on activities, interactive demonstrations, and educational workshops that bring the wonders of science and technology to life.Highlights: * Engage with cutting-edge technologies such as virtual reality, augmented reality, and robotics. Explore how these technologies are shaping our world and inspiring new ways of learning, creating, and communicating.* Participate in workshops led by industry experts and educators, where you'll learn how to code, design apps, build circuits, and create your own tech projects. These sessions are designed to spark your creativity and foster a love for problem-solving.* Meet professionals from various fields of technology, including software development engineering, data science, and cybersecurity. They’ll share their experiences, offer advice on career paths, and answer your questions, helping you chart your tech-driven future.* Showcase your skills in coding competitions, robotics challenges, and other tech-themed events. Compete against fellow students, win prizes, and gain valuable experience that can help you stand out in college applications and future job interviews.* Connect with like-minded students, educators, and industry professionals. Make new friends, build your network, and discover potential mentors who can guide you on your tech journey.The High school Creator Expo is not just about learning about technology, its about empowering you to become the innovators, creators, and problem-solvers of tomorrow. Mark your calendars and get ready to be inspired, educated, and entertained at the High School Creator Expo! 1.Who is the Expo intended for?A.University staff.B.Tech enthusiasts.C.Business owners.D.Technology tutors2.How can one benefit from the Expo?A.Promoting their products B.Having job interviews.C.Getting career insights.D.Shaping the world.3.Where is the text probably taken from?A.A school newspaper.B.An academic articleC.A technology lecture.D.A project handbookAt just 5 years old, little Selah noticed that many of her fellow kindergarten classmates were struggling with their ABCs and didn’t know how to read, so she wanted to do something. The determined girl enlisted (争取) the help of her parents and The Empowered Readers Literacy Project was born.After listening to their daughter’s concerns, Khalil and Nicole Thompson researched the issue and found some shocking information about literacy here in America including the fact that 85% of children in the juvenile (少年的) prison system are functionally illiterate. “All these crazy statistics for literacy blew our minds,” says Khalil Thompson. “The statistics said some prisons forecast the number of prison cells they are going to build in the future based on third-grade reading test scores.”Despite the awesome nature of the problem they were facing, the couple was spurred on by Selah’s resolution, “We really, should just give away 20 hundred thousand books,” Nicole remarks, recalling her daughter’s words, The family got to work—starting initiatives such as Project 500 and their Holiday Book Drive.As they began, one of the main issues the family found, contributing to illiteracy, was representation—children are, not as excited about reading when they don’t find themselves reflected in the pages of the book. And after realizing that very few of the characters in the stories she’d read looked like her. Selah then decided to write a book of her own that features a heroine who looks like her and shares her same interests and sense of adventure. In November 2019, Selah’s first book Penelope the Private Princess, was published, and the sequel (续集) just released in December 2020.“There is so mush inspiration behind the fact that this is a child that wrote this story and came up with this” says Khalil. “You can reclaim your story, do your own story, and do all the things that adults at a at your young age and you should really know that you have power in your voice and ideas.”4.Why does Khalil mention the practice of prisons Para.2?A.To prove the accuracy of his statistics.B.To stress the importance of early reading.C.To indicate the consequences of illiteracy.D.To predict the trend of future juvenile crime.5.What can we learn from the third paragraph?A.The words of Selah forced the couple to work.B.The parents were encouraged by their daughter.C.The number of the book needed is too big to reach.D.The awesome nature of the problem surprised the couple.6.What’s the problem Nicole discovered in kid’s books?A.Their plots were uninteresting.B.The illustrations were not lifelike.C.Kids were represented insufficiently.D.The colors applied were unattractive. 7.What’s the main character like in Penelope the Pirate Princess?A.Wealthy and powerful.B.Naughty and troublesome.C.Talented and sensitive.D.Determined and adventurous.High employment, falling inflation (通货膨胀), and economic growth. On the surface, the US economy currently seems to be doing great. But as it turns out, things are not so rosy for the average working American.According to a March 6 survey conducted by the Savings website, 47 percent of parents provide some degree of financial support to their adult children to help deal with the cost of living struggles.Gen Zers (defined in the survey as aged 18 -27) were found to be the most likely generation to get financial help from their parents, receiving on average 1,515 dollars (about 10.950 an) per month. Still, many older parents were found to be supporting their Millennial (aged 28 -43) or Generation X (aged 44 -59) adult children, accounting for 21 percent of parents providing support. The most common expenses cited were groceries, tuition and health insurance, among others, reported The Guardian.But why are so many US adults, both young and old, struggling to “leave the nest”? Onemajor reason is soaring food prices. According to the US Department of Agriculture, food prices are expected to increase by nearly 2.5 percent across the board in 2024.Meanwhile, stable housing remains an elusive dream for many. The average rate for a 30-year, fixed rate mortgage (抵押贷款) was almost 8 percent in October 2023, said the Mortgage a Bankers Association. An extremely expensive mortgage means that many Americans are priced out of affording their own home.The final piece of the puzzle is falling salaries. Previous generations were able to count on graduating from college as the key to landing a job that paid well. But now, young US graduates have found that they earn less than their parents. According to a CNBC analysis, college graduate salaries have decreased more than 10 percent in four decades after adjusting for inflation. In other words, steadily declining salaries make it even harder for young Americans to cope with steadily rising costs.AS Gen Zers in the US struggle under financial pressures, so do the older generations supporting them, as their generosity puts their prospects for retirement at risk. The Savings website discovered that working parents were found to contribute “2.4 times more to support adult children than they contribute to their retirement accounts each month”.Essentially, while the US economy remains afloat, we might witness multiple generations of Americans sinking in their own lifeboats in the years ahead.8.What did the March 6 survey find out about adult children in the US?A.Nearly half of them are financially dependent.B.Over half of Gen Zers are financially struggling.C.Generation X often offer their parents financial aid.D.A majority of them rely on their parents for education.9.What does the underlined word “elusive” in paragraph 5 most probably mean?A.Easy to realize.B.Difficult to achieve.C.Often misunderstood.D.Commonly sought after.10.What can be inferred from the article about people now in the US?A.Government policies favor older generations.B.College education guarantees a well-paid job.C.Salaries for graduates remain stable in four decades.D.Financial pressures affect both the young and the old.11.What is the author’s attitude toward the future prospects of Americans?A.Reserved.B.Optimistic.C.Negative.D.Wait-and-see.The human brain is a powerful natural computer that has evolved over tens of thousands of years. But what would you get if you combined this biological computer with areal one?Researchers at Indiana University Bloomington in the US have linked human brain-like tissues to an electronic chip, which can perform simple computer tasks. Their creation, named Brainoware, is part of a growing field called biological computing that might one day an do better than current computers, according to Nature.The brain-like tissues the researchers used, also, known as brain organoids (类器官) weren’t part of a living person’s brain. They were “grown” from human stem cells, which are capable of producing different types of cells and forming body tissues. It took two months for the stem cells to mature into an organoid.The researchers then placed one organoid onto an electronic chip with thousands of electrodes (电极). Although organoids are much simpler and smaller than an actual brain, they act similarly to human brains, such as responding to electrical signals, which is What our brain does all the time. These responses lead to changes in the brain, which fuel our ability to learn. To test Brainoware, the researchers used it for voice recognition by training it with 240 Japanese recordings spoken by eight speakers. They found that the organoid reacted different toward each voice. With an accuracy of 78 percent, it successfully identified the speakers by showing different neural activities.What’s special about Brainoware is that researchers can take advantage of the organoid’s complexity without understanding its cell networks, according to Daily Mail. In other words, scientists don’t need to know exactly how the organoid works in order to use it.According to the researchers' work published in December, combining organoids and electronic chips could increase the speed and efficiency of AI in the future. Also, such models can be used to study human brains, according to Nature. Using Brainoware to model and study neurological disorders, such as Alzheimer s disease, is one example. It could also be used to test the effects of different treatments for such diseases. “That’s where the promise is: using the to oneday hopefully replace animal models of the brain,” Arti Ahluwalia, a researcher in Taly, told Nature.12.What do we know about the tissues used by the researchers?A.They were artificially constructed.B.They formed a complete human brain.C.They were harvested from dead human bodies.D.They were cultured from specific human cells.13.What success did Brinoware achieve in the voice recognition test?A.Distinguishing the voices between speakers.B.Understanding the content of the recordings.C.Showing preferences for some of the voices.D.Correctly matching all the speakers with their voices.14.What makes Brainoware special, according to Daily Mail?A.Its complexity is easy to learn and understand.B.It opens up new possibilities for studying human brains.C.Its cell networks are the same as those of human brains.D.No in-depth knowledge of organoids is needed to use it.15.What does the last paragraph focus on regarding Brainoware?A.Its underlying problems.B.Its potential applications.C.Its influence on AI technology.D.Its contribution to animal welfare.Does the sound of rain really help you fall asleep?You’ve likely heard that the pitter pattering of raindrops can help people fall asleep. There’s also research that supports the idea that rain sounds are a useful sleep aid. Here’s what to know.16Rain sounds can mask other sharper noises in your environment. Think of a dripping faucet (水龙头) in the other room in utter silence; it could be very annoying. With a masking background of wind in the leaves, a soft rain, or some gentle waves playing on a loop on a speaker, the drip would never be noticed. 17 Rain sounds might literally hit the right frequenciesBy masking background sounds, rain tracks might seem like they’re just ordinary white noise, butthey are not. 18 They are named after colors. Rain sounds are considered pink noise, which is characterized by louder sounds at a low frequency and softer sounds at high frequencies. Some research suggests that pink noise in particular can improve sleep in healthy individuals. But this may not be the case for everyone.Research suggests that nature has a calming effectIn addition to helping people fall asleep, rain sounds can be straight-up relaxing. 19 Evidence points to psychological benefits, including increased cognitive capacity and reduced anxiety and stress. People exposed to nature sounds in a 2017 study had an increased parasympathetic response, meaning that the calming part of their nervous system was activated by these tracks. Should you try rain sounds?Rain or might help some people but here nor for everyone. 20 One person’s gentle rain is another one’s flood worry or need to go to the bathroom. Similarly, gentle rain with distant thunder is calming for some and alarming for others. Will they work for you?There’s only one way to find out! There’s only one way to find out!A.Rain noise hides other sounds.B.Rain noises sometimes disrupt seep.C.So rain sounds can create a calming background noise.D.White noise is an umbrella term for a variety of sounds.E.This may boil down to the fact that they’re a natural sound.F.It’s worth experimenting with the frequency to see if it can work for you.G.Different sounds will bring out different responses depending on the individual.二、完形填空I was 68 when I first stepped onstage as a stand-up comedian in 2003. That first time was21 . I had no idea what the audience would make of me. In New York, comedians are 22 young men. It’s hard enough to 23 as a woman, but even harder as someone 24 enough to be their grandmother.As I stepped onstage that first time, people were clearly surprised to 25 me. I introduced myself, stated my age and tried my first joke. There was 26 . But by the time I got to the third line, people were finally laughing properly. I was hooked.Even today, I’m 27 before I step out. My first 28 is, “Hello, everybody. I’m 88 and three-quarters years old. I’m telling you that 29 I don’t make it all the way through the show.” Then people laugh and I relax.After a show, people young and old come up to me saying what a(n) 30 I am. They say they want to be as active as I am when they reach my 31 . I’m pleased that I can 32 others that, as an older person, you don’t need to take a back seat in life and you have got more to 33 .Last year, I was recognized as the oldest female stand-up comedian in the world. But I have no 34 of stopping my comedy work. I hope to still be performing when I’m 100. My goal is to still be up there making people 35 .21.A.scary B.precious C.familiar D.private 22.A.relatively B.absolutely C.eventually D.typically 23.A.taken in B.put up C.break through D.note down 24.A.lucky B.old C.wise D.careful 25.A.help B.see C.check D.offer 26.A.silence B.bonus C.blessing D.thrill 27.A.excited B.grateful C.nervous D.skillful 28.A.surprise B.strategy C.relief D.line29.A.as if B.in case C.even though D.in that 30.A.inspiration B.recreation C.crisis D.frustration 31.A.peak B.audience C.goal D.age 32.A.suggest B.challenge C.show D.relax 33.A.prove B.publish C.oppose D.please 34.A.mission B.principle C.intention D.burden 35.A.mature B.laugh C.learn D.win三、语法填空阅读下面短文,在空白处填入1个适当的单词或括号内单词的正确形式。

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数学理科一、选择题:本大题共12小题,每小题5分,共60分.在每小题给出的四个选项中,选择一个符合题目要求的选项.1.把复数z 的共轭复数记作z ,i 为虚数单位,若i z +=1,则(2)z z +⋅=A .42i -B .42i +C .24i +D .4 2.已知集合{}11M =-,,11242x N x x +⎧⎫=<<∈⎨⎬⎩⎭Z ,,则M N = ( ) A .{}11-,B .{}1-C .{}0D .{}10-,3.下列说法正确的是 A .函数xx f 1)(=在其定义域上是减函数 B .两个三角形全等是这两个三角形面积相等的必要条件C .命题“R x ∈∃,220130x x ++>”的否定是“R x ∈∀,220130x x ++<” D .给定命题q p 、,若q p ∧是真命题,则p ⌝是假命题 4.函数sin 2cos 263y x x ππ⎛⎫⎛⎫=+++ ⎪ ⎪⎝⎭⎝⎭的最小正周期和最大值分别为( )A .π,1B .πC .2π,1D .2π5.给出下列三个等式:()()()f xy f x f y =+,()()()f x y f x f y +=,()()()1()()f x f y f x y f x f y ++=-,下列函数中不满足其中任何一个等式的是( )A .()3xf x =B . ()tan f x x =C .2()log f x x =D .()sin f x x =6.一个几何体的三视图如图所示,其正视图和侧视图都是底边长分别为2和4,腰长为4的等腰梯形,则该几何体的侧面积是A .π6B .π12C .π18D .π247.我们定义若函数)(x f 为D 上的凹函数须满足以下两条规则:(1)函数在区间D 上的任何取值有意义;(2)对于区间D 上的任意n 个值n x x x ,,,21 ,总满足)()()()(2121n x x x nf x f x f x f n n +++≥+++ ,那么下列四个图象中在]2,0[π上满足凹函数定义的是(A)8.若2013(2)x -220130122013a a x a x a x =++++ ,则02420121352013a a a a a a a a ++++=++++A .201320133131+-B .201320133131+--C .201220123131+-D .201220123131+--9.若双曲线)0,0(12222>>=-b a b y a x 的一个焦点到一条渐近线的距离等于其焦距的41,则该双曲线的渐近线方程是A .02=±y xB .02=±y xC .03=±y xD .03=±y x 10.位于坐标原点的一个质点P 按下列规则移动:质点每次移动一个单位;移动的方向为向上或向右,并且向上、向右移动的概率都是12,质点P 移动五次后位`于点(23),的概率是( )A .212⎛⎫ ⎪⎝⎭B .2231C 2⎛⎫ ⎪⎝⎭C .3231C 2⎛⎫⎪⎝⎭D .312231C C 2⎛⎫⎪⎝⎭二、填空题:本大题共4小题,每小题4分,共16分.答案须填在题中横线上.11.设O 是坐标原点,F 是抛物线22(0)y px p =>的焦点,A 是抛物线上的一点,FA与x 轴正向的夹角为60,则OA为2p . 12.已知向量)3,2(=a ,)2,1(=b ,且b a ,满足)()(b a b a -⊥+λ,则实数=λ___35-____.13.设D 是不等式组21023041x y x y x y +⎧⎪+⎪⎨⎪⎪⎩≤,≥,≤≤,≥表示的平面区域,则D 中的点()P x y ,到直线10x y +=距离的最大值是14.与直线20x y +-=和曲线221212540x y x y +---=都相切的半径最小的圆的标准方程是 22(2)(2)2x y -+-= .15.函数log (3)1a y x =+-(01)a a >≠且,的图象恒过定点A ,若点A 在直线10mx ny ++=上,其中0mn >,则12m n+的最小值为 8 . 三、解答题:本大题共6小题,共74分.解答应写出文字说明、证明过程或演算步骤. 16.(本小题满分12分)在△ABC 中,三个内角分别为A ,B ,C ,已知4π=A ,54cos =B . (1)求cosC 的值;(2)若BC=10,D 为AB 的中点,求CD 的长.17.(本小题满分12分)如图,矩形ABCD 和梯形BEFC 所在平面互相垂直,BE//CF ,BC ⊥CF ,3=AD ,EF=2,BE=3,CF=4. (1)求证:EF ⊥平面DCE ;(2)当AB 的长为何值时,二面角C EF A --的平面角的大小为︒60.16.【解析】(1)因为54cos =B ,且),0(π∈B ,=-=B B 2cos 1sin 53,则)cos(cos B A C --=π+=-=B B cos 43cos )43cos(ππB sin 43sin π 10253225422-=⨯+⨯-=. (2)由(1)可得=∠-=∠ACB ACB 2cos 1sin 1027)102(12=--=. 由正弦定理得ACB ABA BC ∠=sin sin ,即10272210AB =,解得AB=14. 因为在△BCD 中,721==AB BD , ⋅⋅-+=BD BC BD BC CD 222237541072107cos 22=⨯⨯⨯-+=B , 所以37=CD . 17.【解析】(1)由题易知在△BCE 中,3==AD BC ,BE=3, 所以3222=+=BE BC EC ,又在△FCE 中,==162CF 22CE EF +,所以 EF ⊥CE , 因为平面ABCD ⊥平面EFCB ,DC ⊥BC ,所以DC ⊥平面EFCB , 又EF ⊂平面EFCB ,所以DC ⊥EE ,又DC EC=C ,所以EF ⊥平面DCE .(2) 法一过点B 作BH ⊥EF 交FE 的延长线于点H ,连接AH . 由平面ABCD ⊥平面BEFC ,又平面ABCD 平面BEFC=BC ,AB ⊥BC ,所以AB ⊥平面BEFC ,从而AB ⊥EF ,又因为BH ⊥EF ,BH AB=B ,所以EF ⊥平面ABH . 又AH ⊂平面ABH ,所以EF ⊥AH ,所以∠AHB 为二面角C EF A --的平面角. 在Rt △CEF 中,因为EF=2,CF=4,所以∠CFE=︒60,因为BE ∥CF ,所以∠BEH=∠CFE=︒60. 又在Rt △BHE 中,BE=3,所以233233sin =⨯=∠⋅=BEH BE BH , 由二面角C EF A --的平面角的大小为︒60,得∠AHB=︒60, 在Rt △ABH 中,解得293233tan =⨯=∠⋅=AHB BH AB . 所以当29=AB 时,二面角C EF A --的平面角的大小为︒60. (2)法二 由题知,平面ABCD ⊥平面BEFC ,又平面ABCD 平面BEFC=BC ,DC ⊥BC ,则DC ⊥平面BEFC .又CF ⊥BC ,则BC ,CD ,CF 两两垂直,以点C 为坐标原点,CB ,CF 和CD 所在直线分别作为x 轴,y 轴和z 轴,建立如图所示的空间直角坐标系xyz C -.设)0(>=a a AB ,则)0,0,0(C ,),0,3(a A ,)0,0,3(B ,)0,3,3(E ,)0,4,0(F , 从而)0,1,3(-=EF ,),3,0(a AE -=.设平面AEF 的法向量为),,(z y x n =, 由0=⋅n EF ,=⋅n AE 0得,⎩⎨⎧=-=+-0303az y y x ,取1=x ,则3=y ,az 33=, 即平面AEF 的二个法向量为)33,3,1(an =. 不妨设平面EFCB 的法向量为),0,0(a BA =,由条件,得||,cos |BA n =><21274332=+=a ,解得29=a .所以当29=AB 时,二面角C EF A --的平面角的大小为︒60. 18.(本题满分12分)某学生参加某高校的自主招生考试,须依次参加A 、B 、C 、D 、E 五项考试,如果前四项中有两项不合格或第五项不合格,则该考生就被淘汰,考试即结束;考生未被淘汰时,一定继续参加后面的考试。

已知每一项测试都是相互独立的,该生参加A 、B 、C 、D 四项考试不合格的概率均为12,参加第五项不合格的概率为23(1)求该生被录取的概率; (2)记该生参加考试的项数为X ,求X 的分布列和期望.18.解:(1)若该生被录取,则前四项最多有一项不合格,并且第五项必须合格记A={前四项均合格}B={前四项中仅有一项不合格}则P(A)=4121()(1)2324∙-=…………………………………………………………2分P(B)=311121(1)(1)422316C ⨯-⨯-=………………………………………………4分 又A 、B 互斥,故所求概率为 P=P(A)+P(B)=115241648+=…………………………………………………………………………………5分(2)该生参加考试的项数ξ可以是2,3,4,5.111(2)224P X ==⨯=,121111(3)(1)2224P X C ==-⨯⨯=1231113(4)(1)()22216P X C ==-⨯⨯=,1135(5)1441616P X ==---=…………………………………9分……………………10分113557()234544161616E X =⨯+⨯+⨯+⨯= …………………………………………12分19.(本小题满分12分) 已知各项均不相等的等差数列{}n a 的前5项和为513235,1,1,1S a a a =+++成等比数列。

(1)求数列{}n a 的通项公式;(2)设n T 为数列1n S ⎧⎫⎨⎬⎭⎩的前n 项和,问是否存在常数m ,使12(2)n n nT n n n ⎡⎤=+⎢⎥++⎣⎦,若存在,求m 的值;若不存在,说明理由。

20.(本题满分13分)已知椭圆)0(12222>>=+b a b y a x 的离心率为22,且过点2(. (1)求椭圆的标准方程;(2)四边形ABCD 的顶点在椭圆上,且对角线AC 、BD 过原点O ,若22ab k k BDAC -=⋅,(i) 求⋅的最值.(ii) 求证:四边形ABCD20解:(1)由题意22==a c e ,12422=+b a ,又222c b a +=,……………………………………………2分解得4,822==b a ,椭圆的标准方程为14822=+y x .……………………………………………………………4分 (2)设直线AB 的方程为m kx y +=,设),(),,(2211y x B y x A联立⎩⎨⎧=++=8222y x m kx y ,得0824)21(222=-+++m kmx x k ()2222244(12)(28)8840km k m k m ∆=-+-=-+>() ----------①⎪⎩⎪⎨⎧+-=+-=+22212212182214k m x x k km x x ……………………………………………………………6分2122-=-=⋅a b k k OBOA 212121-=∴x x y y2222212121421822121k m k m x x y y +--=+-⋅⋅-=-=∴ ……………………………………………………………7分2212122121)())((m x x km x x k m kx m kx y y +++=++==222222142182m kkm km k m k ++-++-222812m k k -=+ …………………………………………………8分22222218214kk m k m +-=+--∴2228)4(k m m -=--∴ 2242k m ∴+= …………………………………………………………9分(i)2121y y x x +=⋅2222222222844424421212121212m m m k k k k k k---+-=-===-+++++ 2242O A O B ∴-=-≤⋅<当k =0(此时22=m 满足①式),即直线AB 平行于x 轴时,⋅的最小值为-2.又直线AB 的斜率不存在时2OA OB ⋅=,所以⋅的最大值为2. …………………………………11分 (ii)设原点到直线AB 的距离为d ,则22442)4(16642||218242142||4)(2||1||||121||212222222222212212122=+-=--=+--⎪⎭⎫ ⎝⎛+-=-+=+⋅-⋅+=⋅=∆m k mm m k m k m k km m x x x x m k m x x k d AB S AOB 284==∴∆AO B ABCD S S 四边形.即,四边形ABCD 的面积为定值…………………………………………………21、(本小题满分14分)已知函数22af(x)aln x x(a)x=-++≠(I)若曲线y f(x)=在点(1,1f()))处的切线与直线20x y-=垂直,求实数a的值;(Ⅱ)讨论函数f(x)的单调性;(Ⅲ)当0a(,)∈-∞时,记函数f(x)的最小值为g(a),求证:g(a)≤-e-4.。

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