江门市新会第一中学2012届高二级研究性学习档案袋

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江门市2012年普通高中毕业班先进备课组推荐表

江门市2012年普通高中毕业班先进备课组推荐表

江门市2012年普通高中毕业班先进备课组推荐表
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江门市2012届普通高中高三调研测试(理综)

江门市2012届普通高中高三调研测试(理综)

江门市2012届普通高中高三调研测试理科综合本试题卷共10页,满分300分,考试时间150分钟。

注意事项:1.答题前,务必将自己的姓名、准考证号填写在答题卡规定的位置上。

2.做选择题时,必须用2B铅笔将答题卷上对应题目的答案标号涂黑,如需改动,用橡皮擦干净后,再选涂其它答案标号。

3.非选择题必须使用黑色字迹钢笔或签字笔,将答案写在答题卡规定的位置上。

4.所有题目必须在答题卡上作答,在试题卷上作答无效。

5.考试结束后,将答题卡交回。

可能用到的相对原子质量:Hl C12 N14 016 C135.5 Mn55 Fe56一、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,选错或不选的得O分。

1.近期,一种可耐受大多数抗生素的超级细菌“NDM-1”在许多国家爆发。

下列有关超级细菌的叙述正确的是A. 在细胞分裂前先进行DNA复制B.遗传物质主要分布在染色体上C.核糖体分布在细胞质基质中和内质网上D. 细胞壁可用纤维素酶、果胶酶除去,且不影响细胞活性2.下列关于线粒体、叶绿体和细胞核的叙述,不正确的是A.都具有DNA、RNA和双层膜结构B.细胞质基质中的[H]可以进入线粒体C.细胞核控制合成的蛋白质可以进入线粒体和叶绿体D.叶绿体合成的葡萄糖可以进入线粒体中进行氧化分解3.祖母通过儿子传给孙子的染色体条数为A.0条B. 23条 C.0-22条 D. 0-23条4.提出生物膜流动镶嵌模型的科学家是A. 欧文顿 B.罗伯特森C.桑格和尼克森 D. 施旺和施莱登5.下列选项中,前者随后者变化的情况与右图不相符的是A. 抗利尿激素——饮水量B.胰岛素浓度——血糖浓度C.促甲状腺激素浓度——甲状腺激素浓度6.HIV能通过细胞表面的CD4(一种受体蛋白)识别T细胞(如图甲),如果给AIDS患者大量注射用CD4修饰过的红细胞,红细胞也会被HIV识别、入侵(如图乙)。

江门一中2012年报刊征订通知

江门一中2012年报刊征订通知

江门一中2012年报刊征订通知全体教职工:2012年报刊杂志征订工作开始了,为了做好这次征订工作,现将有关事项通知如下:1.请各位教职工在限额范围内填写好报刊杂志征订单。

2.限额如下:(均以在编教职工计算)高级教师300元;一级教师200元;二级教师150元;其他教师及教职工100元(其他职称按相应的教师职称计算)。

3.如需自费订阅其他报刊,请在自费订阅刊物一栏中填写,所需金额待办理订阅手续完毕后由财务室在订阅人的帐户中扣除。

4.个人公费订阅报刊的总金额须在第二条标明的限额以内,剩余部分不作为个人自费订阅之用。

如果订阅总金额超出订阅人的限额,学校将视情况将订阅人的刊物少订一个季度。

5.杂志订期可选择全年、半年或某季度;报纸订期可选全年、半年、季度或某一个月。

6.学校规定公费订阅的刊物必须与教职工所教的科目或从事的工作内容相关,不能用公费订阅婴儿、幼儿、小学、初中类刊物。

例如:《婴儿画报》、《发现号》、《启蒙》、《少年科学画报》、《智力》、《大灰狼画报》等。

7.想将报刊投回家的, 请注明家庭地址。

(投回学校,寒暑假学校会封存杂志,开学后才发放)8.在书本目录未发放之前,各位教职工可上中国邮政报刊订阅网查阅有关订价。

/index.do或者 9.请教职工们于10月30日前填写好以下表格,以工会小组为单位收集好交图书馆。

(自己交到图书馆二楼也可)。

特此通知。

江门一中图书馆2011年10月14日江门一中报刊征订单报刊订阅人: 职称:公费订阅的刊物自费订阅的刊物邮发代号报刊名称订期订价邮发代号报刊名称订期订价投递地址:投递地址:公费报刊订价合计:自费报刊订价合计:。

表1 2012年高一调研测试成绩单 - 副本

表1 2012年高一调研测试成绩单 - 副本

2012年高一调研测试市、区学校姓名学号数学英语总分市直江门市第一中学陈金科1701010101120.00114.50234.5市直江门市第一中学崔励彬1701010102109.0092.50201.5市直江门市第一中学邓森耀1701010103136.00111.50247.5市直江门市第一中学方子丹1701010104100.00108.50208.5市直江门市第一中学冯超铭170101010598.0095.50193.5市直江门市第一中学冯耀文1701010106106.00116.50222.5市直江门市第一中学冯梓凡1701010107116.00107.00223市直江门市第一中学何浩儒1701010108116.0068.50184.5市直江门市第一中学何源俊170101010992.00113.50205.5市直江门市第一中学胡璐瑶170101011093.00120.00213市直江门市第一中学黄冠烽1701010111101.00101.50202.5市直江门市第一中学黄锦钊1701010112111.00115.50226.5市直江门市第一中学黎天鹏1701010114104.00123.00227市直江门市第一中学黎子杰170101011582.0084.50166.5市直江门市第一中学李宝珊170101011677.00121.25198.25市直江门市第一中学李方舆1701010117125.00113.00238市直江门市第一中学李嘉雯1701010118133.00134.00267市直江门市第一中学李淑怡1701010119122.00122.50244.5市直江门市第一中学李晓明1701010120115.00115.00230市直江门市第一中学李晓璇1701010121119.00107.00226市直江门市第一中学梁嘉敏170101012299.00125.00224市直江门市第一中学梁嘉雯170101012390.00124.75214.75市直江门市第一中学梁巧彤1701010124107.00117.50224.5市直江门市第一中学林翠萍1701010125104.00112.00216市直江门市第一中学林杰辉170101012675.00105.00180市直江门市第一中学刘心宇1701010128124.00125.50249.5市直江门市第一中学刘悦170101012992.00123.25215.25市直江门市第一中学卢伟杰170101013076.00123.00199市直江门市第一中学吕梦星1701010131101.00114.00215市直江门市第一中学满印怡1701010132110.0085.50195.5市直江门市第一中学欧阳慧婷170101013382.00120.00202市直江门市第一中学欧阳瑶珠1701010134108.00126.50234.5市直江门市第一中学区钊雄1701010136102.0099.00201市直江门市第一中学容嘉乾1701010137115.0097.50212.5市直江门市第一中学容浚铭1701010138114.00110.00224市直江门市第一中学阮嘉茵1701010139112.00121.50233.5市直江门市第一中学韦梅1701010140103.00122.50225.5市直江门市第一中学吴阜珊1701010141113.00129.50242.5市直江门市第一中学吴健豪170101014299.00103.50202.5市直江门市第一中学吴若莹170101014380.0087.50167.5市直江门市第一中学吴志平170101014485.00124.50209.5市直江门市第一中学伍兆富170101014588.00118.00206市直江门市第一中学肖俊东1701010146109.0097.50206.5市直江门市第一中学谢耀威1701010147101.00110.00211市直江门市第一中学许一颖1701010148101.00127.50228.5市直江门市第一中学许紫荆1701010149111.00130.50241.5市直江门市第一中学杨盛霖1701010150120.00121.00241市直江门市第一中学叶铭熙1701010151110.00120.50230.5市直江门市第一中学尹瑞霞1701010152102.00139.50241.5市直江门市第一中学曾孟儒1701010153127.0091.50218.5市直江门市第一中学张妙莹170101015494.00105.00199市直江门市第一中学赵静诗170101015574.00131.75205.75市直江门市第一中学赵镇1701010156101.0093.50194.5市直江门市第一中学钟韵炫1701010157113.00120.00233市直江门市第一中学周汉宁170101015890.00118.00208市直江门市第一中学周昕宇1701010159100.0094.25194.25市直江门市第一中学庄培鑫1701010160113.00133.00246市直江门市第一中学庄思吕1701010161103.00108.00211市直江门市第一中学岑瑞仪1701010201101.00121.25222.25市直江门市第一中学陈宏浩1701010202119.00122.50241.5市直江门市第一中学陈焕铭1701010203121.00114.50235.5市直江门市第一中学陈家立1701010204108.00113.50221.5市直江门市第一中学陈佩莹1701010205122.00120.50242.5市直江门市第一中学陈韵荷1701010206117.00100.25217.25市直江门市第一中学冯梓豪170101020774.00120.50194.5市直江门市第一中学甘玉婷170101020899.00127.00226市直江门市第一中学古名辉1701010209113.0096.00209市直江门市第一中学郭蕊1701010210114.0095.50209.5市直江门市第一中学何洁仪1701010211131.00112.50243.5市直江门市第一中学胡楚夷170101021296.00120.00216市直江门市第一中学黄斐然1701010213112.0097.00209市直江门市第一中学黄俊铭170101021494.0087.50181.5市直江门市第一中学黄雄韬170101021567.0099.00166市直江门市第一中学黄秀云170101021678.00109.50187.5市直江门市第一中学黄仲玲170101021797.00105.50202.5市直江门市第一中学黄卓超1701010218109.0092.50201.5市直江门市第一中学孔繁鑫170101021979.00108.00187市直江门市第一中学李劼聪170101022098.00113.00211市直江门市第一中学李桥铭170101022181.0069.50150.5市直江门市第一中学李歆柱1701010222109.00121.50230.5市直江门市第一中学李英华1701010223102.00118.00220市直江门市第一中学李振君170101022592.0098.00190市直江门市第一中学梁健聪1701010226104.00108.00212市直江门市第一中学梁乐赋1701010227100.0095.50195.5市直江门市第一中学梁伟中170101022897.00121.00218市直江门市第一中学梁曦敏170101022997.0098.50195.5市直江门市第一中学梁咏芝170101023086.00107.50193.5市直江门市第一中学廖声威1701010231121.00121.00242市直江门市第一中学林靖怡170101023277.0096.00173市直江门市第一中学林汝奂1701010233109.00112.50221.5市直江门市第一中学刘建明1701010235107.00102.50209.5市直江门市第一中学刘婉玲1701010236100.00113.50213.5市直江门市第一中学刘勇1701010237125.00124.50249.5市直江门市第一中学龙瑞芳1701010238106.00105.00211市直江门市第一中学罗浚乘1701010239108.00109.00217市直江门市第一中学马胡喜子1701010240113.00112.00225市直江门市第一中学聂年德1701010241117.00122.00239市直江门市第一中学区焕明1701010242122.00100.50222.5市直江门市第一中学全晓慧170101024394.00126.50220.5市直江门市第一中学宋庆莹1701010245125.00109.50234.5市直江门市第一中学谭乐文170101024695.00126.00221市直江门市第一中学谭昭宏1701010247101.00105.00206市直江门市第一中学王锐文1701010248113.00118.50231.5市直江门市第一中学温嘉明170101024951.0090.50141.5市直江门市第一中学伍锦玲1701010251115.00118.00233市直江门市第一中学夏明德1701010252119.00106.00225市直江门市第一中学谢秀贞1701010253106.00118.50224.5市直江门市第一中学叶乘风1701010254111.00103.50214.5市直江门市第一中学叶琦欣170101025584.00110.50194.5市直江门市第一中学余文杰1701010256136.0067.00203市直江门市第一中学张德均1701010257102.00101.00203市直江门市第一中学赵嘉琪1701010258101.00124.00225市直江门市第一中学钟伟燊1701010259111.00110.50221.5市直江门市第一中学周淑妍170101026078.00105.00183市直江门市第一中学周志豪1701010261129.0091.00220市直江门市第一中学陈嘉忻1701010262113.00134.50247.5市直江门市第一中学陈健宁1701010301136.00138.50274.5市直江门市第一中学陈伟聪1701010302128.00139.50267.5市直江门市第一中学陈熙临1701010303135.00133.50268.5市直江门市第一中学关慧哲1701010304142.00146.50288.5市直江门市第一中学何家盛1701010305136.00140.00276市直江门市第一中学胡耀森1701010306139.00130.50269.5市直江门市第一中学黄晓明1701010307125.00140.00265市直江门市第一中学黄学知1701010308126.00134.00260市直江门市第一中学黄羽翔1701010309137.00133.00270市直江门市第一中学李德庸1701010310124.00139.00263市直江门市第一中学梁瑞峰1701010311124.00118.00242市直江门市第一中学林文辉1701010312129.00133.50262.5市直江门市第一中学刘乃新1701010313129.00141.50270.5市直江门市第一中学卢奥威1701010314132.00129.50261.5市直江门市第一中学罗观璋1701010315139.00137.00276市直江门市第一中学罗梓浩1701010316137.00137.5027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广东省江门市新会第一中学2023-2024学年高二下学期期末考试数学试题

广东省江门市新会第一中学2023-2024学年高二下学期期末考试数学试题

广东省江门市新会第一中学2023-2024学年高二下学期期末考试数学试题一、单选题1.已知ξ的分布列为设25ηξ=-,则()η=E ( ) A .23B .56C .196 D .322.已知直线y kx m =+(m 为常数)与圆224x y +=交于点M N ,,当k 变化时,若||MN 的最小值为2,则m =A .1±B .C .D .2±3.已知等差数列{}n a ,等比数列{}n b ,满足794a a +=,261027b b b =,则3813483a a ab b ++=+( ).A .14B .12C .2D .44.若曲线ln x ay x+=在点(1,)a 处的切线与直线:50l x y ++=平行,则实数=a ( ) A .12B .1C .32D .25.今天是星期天,则713天后是( ) A .星期五B .星期六C .星期天D .星期一6.某单位五一放假,安排甲、乙等五人值班五天,每人值班一天.若甲、乙都至少需要三天的连休假期,则不同的值班安排共有( ) A .60种B .66种C .72种D .78种7.袋中装有10个球,其中3个黑球、7个白球,从中依次取两球(不放回),则第二次取到的是黑球的概率为( ) A .29B .310 C .13D .7108.已知函数2()ex x f x =,下列关于()f x 的四个命题,其中是假命题是( )A .函数()f x 在[]0,1上是增函数B .函数()f x 的最小值为0C .如果[]0,x t ∈时,max 24()ef x =,则t 的最小值为2 D .函数()f x 有2个零点二、多选题9.给出下列命题,其中正确的命题有( )A .两个变量的线性相关性越强,则相关系数r 越大B .在103x⎛⎝的展开式中,各项系数和与所有项二项式系数和相等C .将4名老师分派到两个学校支教,每个学校至少派1人,则共有14种不同的分派方法D .公共汽车上有10位乘客,沿途5个车站,乘客下车的可能方式有510种10.已知双曲线()222:102x y E a a -=>的左、右焦点别为1F ,2F ,过点2F 的直线l 与双曲线E的右支相交于,P Q 两点,则( )A.若E 的两条渐近线相互垂直,则a =B .若EE 的实轴长为1 C .若1290F PF ∠=︒,则124PF PF ⋅=D .当a 变化时,1F PQ V 周长的最小值为11.如图,在棱长为1的正方体1111ABCD A B C D -中,E 、F 分别为棱11A D 、1AA 的中点,G 为线段1B C 上一个动点,则( )A .三棱锥1A EFG -的体积为定值B .存在点G ,使平面//EFG 平面1BDCC .当点G 与1B 重合时,二面角1G EF A --的正切值为D .当点G 为1B C 中点时,平面EFG三、填空题12.已知12,F F 分别为椭圆2211810x y +=的左、右焦点,P 为椭圆上一点且122PF PF =,则12PF F △的面积为.13.甲、乙两人争夺一场羽毛球比赛的冠军,比赛为“三局两胜”制.如果每局比赛中甲获胜的概率为23,乙获胜的概率为13,则在甲获得冠军的情况下,比赛进行了三局的概率为.14.已知*a ∈N ,函数()e 0x af x x =->恒成立,则a 的最大值为.四、解答题15.某班级有60名同学参加了某次考试,从中随机抽选出5名同学,他们的数学成绩x 与物理成绩y 如下表:数据表明y 与x 之间有较强的线性相关性.(1)利用表中数据,求y 关于x 的经验回归方程,并预测该班某同学的数学成绩为90分时的物理成绩;(2)在本次考试中,规定数学成绩达到125分为数学优秀,物理成绩达到100分为物理优秀. 若该班的数学优秀率与物理优秀率分别为50%和60%,且所有同学中数学优秀但物理不优秀的同学共有6人,请你完成下面的22⨯列联表,依据小概率值0.005α=的独立性检验,能否认为数学成绩与物理成绩有关联?参考公式及数据:()()()1122211ˆnniii ii i nniii i x x y y x y nxybx x xnx ====---==--∑∑∑∑,ˆˆay bx =-,5154900i i i x y ==∑,()5211000ii x x =-=∑,()()()()()22n ad bc ab c d a c b d χ-=++++,其中n a b c d =+++.下表是2χ独立性检验中几个常用的小概率值和相应的临界值.16.已知{}n a 数列的前n 项和为()*,23n n n S S a n n =-∈N .(1)证明:数列{}3n a +是等比数列,并求出数列{}n a 的通项公式; (2)设()33n n nb a =+,求数列{}n b 的前n 项和n T ; (3)数列{}n a 中是否存在三项,它们可以构成等差数列?(直接写出结论,不要求证明). 17.如图所示,在三棱锥-P ABC 中,已知PA ⊥平面ABC ,平面PAB ⊥平面PBC .(1)证明:BC ⊥平面PAB ;(2)若6PA AB ==,3BC =,在线段PC 上(不含端点),是否存在点D ,使得二面角B AD C --D 的位置;若不存在,说明理由. 18.面试是求职者进入职场的一个重要关口,也是机构招聘员工的重要环节.某科技企业招聘员工,首先要进行笔试,笔试达标者进入面试,面试环节要求应聘者回答3个问题,第一题考查对公司的了解,答对得2分,答错不得分,第二题和第三题均考查专业知识,每道题答对得4分,答错不得分.(1)若一共有100人应聘,他们的笔试得分X 服从正态分布()60,144N ,规定72X ≥为达标,求进入面试环节的人数大约为多少(结果四舍五入保留整数);(2)某进入面试的应聘者第一题答对的概率为23,后两题答对的概率均为45,每道题是否答对互不影响,求该应聘者的面试成绩Y 的数学期望.附:若()2~,X N μσ(0σ>),则()0.683P X μσμσ-<<+≈,()220.954P X μσμσ-<<+≈,()330.997P X μσμσ-<<+≈. 19.已知函数()()()ln 101axf x x a x =+->+. (1)若1x =是函数()f x 的一个极值点,求a 的值; (2)若()0f x ≥在[)0,∞+上恒成立,求a 的取值范围; (3)证明:20242024e 2023⎛⎫> ⎪⎝⎭(e 为自然对数的底数).。

广东省江门市新会第一中学2023-2024学年高二上学期第二次测验英语试题

广东省江门市新会第一中学2023-2024学年高二上学期第二次测验英语试题

广东省江门市新会第一中学2023-2024学年高二上学期第二次测验英语试题学校:___________姓名:___________班级:___________考号:___________一、阅读选择Experiencing cultureCHI’s Academic Year Program (AYP) is your chance to study in America and spend either an academic year or a semester living with a volunteer host family.The requirements for studying in America*High school students who are aged 15 to 18*A score of at least 212 on the ELTiS (English Language Test for International Students) examApplicationContact a partner agency in your country or a nearby one. We can also send you a list of partners near you. They will set you up with an application and let you know what you need to submit in order to be considered.Admission and visaOur Admission Department reviews and accepts your application. We issue and send your agency the Form DS-2019, which is the official US Department of State document that allows you to apply for and receive your J-1 visa. Once you receive the DS-2019, you can make your visa appointment, When the visa is granted(授予), you are set to travel to America.HomestayMeanwhile, our Field and Operations staff work hard to find you the perfect family and school as soon as possible! Once they find the right host family for you, you can start to get to know your host family by email.Depart to the US and meet your family and friendsYour agency will book your flight according to your school start date. When you arrive, we will give you a student orientation(迎新会)and introduce you to other exchange students and the surrounding community. When school starts, you will begin the process of cultural exchange.Contact us: 1-800-432-4643; 1-415-459-5397Location: 255 W End Avenue San Rafael, CA 949011.Who can take part in this homestay project?A.15-year-old high school students with 210 on the ELTiS exam.B.17-year-old high school students with 213 on the ELTiS exam.C.20-year-old college students with 220 on the ELTiS exam.D.18-year-old college student with 218 on the ELTiS exam.2.What should applicants do before making a visa appointment?A.Gain approval from their school’s Admission Department.B.Get a letter of introduction from their teacher.C.Contact the host family via email.D.Fill and send the Form DS-2019.3.What awaits the students when they arrive in the US?A.A student orientation.B.A party with their host family.C.A volunteer activity in the community.D.A meeting with the Field and Operations staff.“It’s incredible. I never thought my dream would come true so soon,” Chinese ballet dancer Chun Wai Chan recalled the scene when he got the news in May that he was promoted to the principal dancer with New York City Ballet (NYCB), the company’s first Chinese and fourth Asian principal dancer in 74 years.Born in Huizhou City, South China’s Guangdong Province in 1992, Chan has become attached to dance since childhood. However, his parents preferred him to be a doctor or a lawyer in the future. The uncompromising boy then wrote a seven-page letter to his parents describing his determination to study dance.Thanks to the sincere letter, Chan finally gained the support of his family, and at the age of 12, he was admitted to Guangzhou Arts School, marking the start of his dream-seeking journey. Chan’s first turning point in life came at the age of 18, when he was a finalist in the 2010 Prix de Lausanne, which earned him a full scholarship to study with Houston Ballet’s second company, Houston Ballet II. Two years later, Chan joined Houston Ballet and became a principal in 2017. There, he gained a reputation as a confident and sensitive performer. In 2020, he appeared as a contestant on the Chinese television show Dance Smash on HunanSatellite TV, which gained him a large following.Chan returned to New York last year, and was finally promoted to the principal dancer with the NYCB in May this year. He attributes his success to his passion, hard work, concentration, perseverance and the pursuit of excellence.In China, Chan’s success has become a source of pride. News of his promotion to principal dancer was widely circulated, and he has been featured repeatedly in the Chinese media, under headlines like “The Ballet Knight” and “After Dance Smash, he conquered New York.”After performances, audience members sometimes tell Chan that they have never seen Asian dancers in leading roles. He has been moved to hear young dancers of color say his example has given them hope for their own careers. “I used to think I danced just for myself,” he said. “Now I’m dancing for my family, for the audience, for the whole dance community.”4.What can we learn about Chan?A.He is the first Asian principal dancer.B.He is fond of dancing when he was young.C.His parents have supported him from the beginning.D.He gained the popularity in the 2010 Prix de Lausanne.5.What does the underlined word “uncompromising” in paragraph 2 mean?A.Fearless.B.Cautious.C.Thoughtful.D.Determined. 6.Which of the following can best describe Chan?A.Generous and considerate.B.Passionate and devoted.C.Warm-hearted and ambitious.D.Perseverant and talented7.Chan changed his opinion about the significance of dancing because ________.A.he got a lot of praises from the audience B.he had been promoted to principal dancer C.he realized his dance inspired other dancers D.he received persistent support from his familyHorror movies are designed to cause seemingly bad emotions — shock, fear and disgust (厌恶). Yet many people want to sit through those films. Why? Clasen, who directs the Recreational Fear Lab at Aarhus University in Denmark, made a survey of more than 250 American horror fans with his team, and the responses of those participants revealed three types of fans.The first type felt great and were in high spirits. The second type reported some negative reactions to scary movies, and some even had nightmares at first, but they explained that they were feeling like they learned something about themselves at last. The third type of fans seemed to use horror to deal with bad feelings and events in real life. They supported statements such as “Watching horror movies makes me realize that everything in my own life is OK.”According to previous studies, thrill seekers, who often seek out new and intense experiences, tend to be horror fans. Men also seemed to be slightly bigger horror fans than women. Clasen’s teammates didn’t think so. “They are stereotypes (老套),” they said. Men are less likely to admit in surveys that they are afraid of horror. Likewise, researchers think it’s unfair to paint all horror fans as thrill seekers. What may influence how much people like horror is something called the need for affections (情感). “People differ on that — how much emotion they want in their lives,” Clasen explained.After watching a horror movie, you could be left unsettled and anxious, or you have nightmares, but it may actually help you become more prepared in real life. Horror fans reported lower levels of distress (忧虑) during the pandemic, for instance. “That’s because horror movies let people practise negative emotions in a safe setting,” Clasen said. “That way, when it comes to scary situations in real life, you know how to react quickly and correctly.”Good or not, scary movies aren’t for everyone. And that’s OK. “Monster movies are good,” sociologist Margee Ker r said, “because you know that they’re not real.” “If you are not ready to watch a horror movie, there are other ways to test yourself. One is writing your own horror-movie story. If you are writing a story about something that you are afraid of, you are taking control of it,” she explained. “That can be really helpful.”8.How did the second type of horror fans respond after watching horror movies?A.The movies made them feel worried about their real life for some time.B.The movies stopped the fans from having unpleasant dreams.C.The fans could have a better understanding of themselves.D.The fans felt fantastic and were in high spirits.9.What do the researchers mainly want to express in paragraph 3?A.The reason why people need affections.B.The reason why people love horror.C.The concept of thrill seekers.D.The features of horror fans.10.Why is the pandemic mentioned in paragraph 4?A.Horror movies can help people deal with scary situations quickly and correctly.B.Horror movies can make people have a false judgement about scary situations.C.Horror movies can help people get rid of negative emotions in real life.D.Horror movies can make people unsettled and anxious in real life.11.How can creating a horror story affect people?A.They can bring their sense of excitement under control.B.They can bring their sense of fear under control.C.They can take control of their sense of sadness.D.They can take control of their sense of pride.A lot of businesses are started by people who have hobbies or special talents and want to turn these interests into a business. But switching from a hobby to a real business requires business know-how and investments, which stops many people. However, an increasing number of ambitious business owners have found a way to take a first step. They are taking their dreams and talents on the road—in vans and lorries, and even horse trucks.Beginning on a small scale has its advantages, the most important of which is the relatively modest size of start-up costs. These costs, which consist primarily of, the vehicle and any required equipment, usually come to about only £5,000-£10,000. With such tightly controlled costs, mobile businesses often break even in a year or two; in contrast, success comes to brick-and-mortar businesses much more slowly, and they often fail within the first two years. In short, mobile businesses are relatively low-risk.The first entrepreneurs (企业家) to do this have been in the food business. The mobile food vendor trend in the US and Europe has spread to other parts of the globe. The success of these street food vans has inspired other entrepreneurs to consider starting out on wheels.UK entrepreneurs could take inspiration from the US where, today, there are vehicles that sell flowers, shoes, clothes and all kinds of specialty food items. There are also mobile businesses that provide services, such as hair styling, pet grooming and repair of high-tech devices. Mobile retail is not without problems, however. Weather, the increasing price of fuel, and just finding a place to set up are all challenges that mobile entrepreneurs have to deal with. These business owners, however, feel the advantages outweigh the disadvantages. In the US, the mobile food business alone generates an average annual revenue of $857 million.Once convinced that their business has achieved enough success, some successfulmobile entrepreneurs move on to a brick-and-mortar business. Others are satisfied to stay mobile. As stand-alone businesses or as extensions of shops, mobile retail has huge future potential in these economic times.12.What’s the major cause for the rise of mobile business?A.High profit.B.Strong interest.C.Low investment.D.Convenient transport.13.How does the author prove the advantages of the mobile business in paragraph 2?A.By quoting a famous saying.B.By giving an exampleC.By providing an explanation.D.By making a comparison.14.Which of the following will make it harder for the owners to run the business?A.The vehicle care.B.The business permit.C.The terrible storm.D.The insufficient supply.15.What’s the author’s attitude towards the future of mobile business?A.Negative.B.Positive.C.Unconcerned.D.Neutral.There were times when I thought success was about creating great work; times when I believed success was about making more money; and times when I thought success was aboutusually describe success as combination of achievements and happiness. For me, success is about having fun and joy along the journey of creating the results you desire. Many habits help one achieve success. 17 .Practicing self-awarenessPracticing self-awareness uses your ability to make the right choices for yourself. The best way to learn about yourself is to keep asking yourself difficult questions and answering them. 18 . Do this every day, write them down, and think carefully about them regularly.Choosing the child-like attitude19 , and that's very true. First, kids don't over-think and always see the world as it is. We can act faster and live happier if we stop over-thinking. Second, kids don't stop playing. What do kids do when they fall down? They get back up quickly and start playing again. And that's exactly what many of us need to achieve in order to get what we want in life.Finding joy in hard work, challenges and setbacksLife never goes as smoothly as we want it to. 20 .If you're a businessman, don't wait for customers to come knocking at your door. Go out, share your product, and get customers for your business.Accept that success is not easy. Then, be the other group -- the group who finds joy in challenges and setbacks(挫折),the group who loves the things they do,and the group who puts in the hard work and gets devoted to the process.A.The questions require you to think deeplyB.None of the above sayings of success are wrongC.Don't wish for more challenges, wish for more joyD.These are what help us hold on and bring us pure joyE.Hoping life gets easier, you will find it easy to be successfulF.Instead of hoping that life gets easier, hope that you will get betterG.There is a saying that kids are here to remind adults about how to live二、完形填空Simone is sitting in front of me and crying. She looks away, embarrassed. “Sorry, doctor,”she is overwhelmed by a stranger's 34 .One of the administration staff in the outpatient clinic has just won a $20 Marks and Spencer voucher and given it to Simone on a whim so she could 35 her daughter, who often plays in the department while her mother sees me.That sudden,impulsive act of kindness and generosity from a stranger has made all the difference to Simone.21.A.mounting up B.taking up C.summing up D.brightening up 22.A.only B.instead C.rather D.else 23.A.trend B.complaint C.need D.anxiety 24.A.domestic B.financial C.severe D.physical 25.A.examined B.joined C.linked D.restricted 26.A.disturbing B.warning C.fueling D.abusing 27.A.until B.since C.before D.after 28.A.bills B.taxes C.discounts D.sponsors 29.A.opposed B.determined C.committed D.recognized 30.A.afterwards B.therefore C.however D.moreover 31.A.denied B.withdrew C.continued D.stopped 32.A.mentioned B.referred C.distributed D.adjusted 33.A.limited B.sad C.ripe D.global 34.A.disability B.poverty C.kindness D.suggestion 35.A.amuse B.handle C.affect D.treat三、翻译词义匹配We’re told that writing is dying. Typing on keyboards and screens curbs written communication today. Learning cursive (草书), joined-up handwriting was once obligatory in schools. But now, not so much. Countries such as Finland have dropped joined-up handwriting lessons in schools to stick up for typing courses. And in the U.S., the requirement to learm cursive has been left out of core standards since 2013. A few U.S. states still place value on formative cursive education, such as Arizona, but they’re not the mayority.Some experts point out that writing lessons can have indirect benefts. Anne Trubek, author of The History and Uncertain Future of Handwriting, argues that such lessons canreinforce a skill called automaticity. Thats when you’ve perfected a task, and can do it almost without thinking, granting you extra mental bandwidth to think about or do other things while you’re doing the task. In this sense, Trubek likens handwriting to driving. “Once you have driven for a while, you don’t consciously think‘Step on gas now’ or ‘Turn the steering wheel a bit’,” she explains. 36.A.abandons B.dominates C.enters D.absorbs 37.A.compulsory B.opposite C.crucial D.relevant 38.A.get through B.make up for C.side with D.put up with 39.A.allowing B.getting C.bringing D.coming‘40.A.eventually B.constantly C.dramatically D.intentionally四、短文填空五、语法填空阅读下面短文,在空白处填如1个适当的单词或括号内单词的正确形式。

3新会一中2016届“研究性学习”结题报告评价表

3新会一中2016届“研究性学习”结题报告评价表
15
12
9
6
3
15
专家签名:
指导教师签名:
评价结果
等级说明:A等90分以上;B等80分以上;C等70分以上;D等60分以上;E等60分一下(为不合格)
总分
等次
A
94
新会一中2016届“研究性学习”结题报告评价表
课题:新会圭峰山的地理环境及旅游资源2014年9月
评价内容
评价指标
等级
得分
A
B
C
D
E
结题报告的内容
课题选择具有现实性和可行性
10
8
6
4
2
9
课题研究的思路清楚,观点的阐述能与具体材料相结合
10
8
6
4
2
9
结题报告格式规范,文字表述清晰、流畅,中心突出。
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8
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4
2
8Hale Waihona Puke 课题研究的成果研究目的明确,方法恰当,实现了预期目标
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12
9
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3
14
研究成果观点明确,论据充分、结论正确。
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6
3
13
研究有自己的见解,有个人的切身体验
10
8
6
4
2
10
报告的陈述与答辩
报告陈述人仪态大方,语句清楚、流畅,报告陈述条理清晰,图文并茂。
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9
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15
课题组成员能正确回答有关专家及指导教师的提问,且回答得有理有据。

广东省江门市高二数学上学期期末试题新人教A版

广东省江门市高二数学上学期期末试题新人教A版

新会一中2012-2013学年度第一学期期末考试 高二级数学(理科)试卷(选修2-1)本试卷共4页,共21题,本卷必做题满分100分,附加题10分,考试时间为120分钟. 注意事项:1.答卷前,考生务必用黑色字迹的钢笔或签字笔将自己的姓名和学号填写在答题卡上,并用2B 铅笔填涂答题卡上相应的信息点.2.选择题每小题选出答案后,用2B 铅笔把答题卡上对应题目选项的答案信息点涂黑,如需改动,用橡皮擦干净后,再选涂其他答案,答案不能答在试卷上.3.非选择题必须用黑色字迹钢笔或签字笔作答,答案必须写在答题卡各题目指定区域内相应位置上;如需改动,先划掉原来的答案,然后再写上新的答案;不准使用铅笔和涂改液。

不按以上要求作答的答案无效.第一卷: 选择题一、选择题(本大题10小题,每小题3分,共30分)1.命题:“若12<x ,则11<<-x ”的逆否命题是( )A.若12≥x ,则11-≤≥x x ,或 B.若11<<-x ,则12<xC. 若1,1-≤≥x x 且,则12≥x D.若11-≤≥x x ,或,则12≥x2.条件p :1>x ,1>y ,条件q :2>+y x ,1>xy ,则条件p 是条件q 的 ( ) A .充分而不必要条件 B .必要而不充分条件 C .充要条件 D .既不充分也不必要条件3.有金盒、银盒、铅盒各一个,只有一个盒子里有肖像.金盒上写有命题p :肖像在这个盒子里;银盒上写有命题q :肖像不在这个盒子里;铅盒上写有命题r :肖像不在金盒里.p 、q 、r 中有且只有一个是真命题,则肖像在( ) A .金盒里 B .银盒里 C .铅盒里 D .在哪个盒子里不能确定4.等轴双曲线22122x y -=的离心率为( ) A 2 B 2 C 22 D 225.抛物线y x 82=的准线方程是 ( )A . 2-=yB .2-=xC . 2=yD . 2=x6.在同一坐标系中,方程12222=+y b x a 与02=+by ax )0(>>b a 的图象大致是( )7.已知P N M ,,为不共线的三点,对空间中任意一点O ,若4113412OQ OM ON OP =--u u u r u u u u r u u u r u u u r,则Q P N M ,,, 四点( )A .不一定共面B .一定不共面C .一定共面D .无法判断8.已知向量)0,1,1(=a ,)2,0,1(-=b ,且b a k +与b a -2互相垂直,则k 等于( ) A.1 B.51 C. 53D. 579.在棱长为1的正方体ABCD —A 1B 1C 1D 1中,M 和N 分别为A 1B 1和BB 1的中点,那么异面直线AM 与CN 所成角的余弦值是( )A .52-B .52C .1010-D .101010.已知(1,2,3)OA =u u u v ,(2,1,2)OB =u u u v ,(1,1,2)OP =u u u v,点Q 在直线OP 上运动,则当•取得最小值时,点Q 的坐标为( )A.)314321(,, B. )323221(,, C. )383434(,, D. )373434(,,第二卷: 非选择题二、填空题(本大题共4小题,每小题4分,共16分)11.已知空间向量=(2,-3,t ),=(-3,1,-4),若·=1-,则实数t =________. 12.命题p :“任意素数都是奇数”,则p 的否定为:__________________________.13.已知椭圆192522=+y x 的两焦点分别为21,F F ,若椭圆上一点P 到1F 的距离为6,则点P 到2F 的距离为______________.14.当用反证法证明来命题:“若0||||=+b a ,则0==b a ”时,应首先假设“______________”成立.三、解答题(共7道题,前6题共54分,全体考生必做题。

新会一中2009——2010学年第一学期

新会一中2009——2010学年第一学期

新会一中、葵城中学2013—2014学年度暑假教师读书情况汇报表姓名黄运华科组历史级组高一书(刊)名陶行知著作作(编)者陶行知出版社出版日期读书心得(800字以上,不另页,可双面打印)陶行知,先行后知,知行合一,他是一名留洋的教育家,在深刻了解了中国的文化和社会现实的基础上,所提出的教育学说,既强调了教育的显示功能,又关住了教育的终极目的,是适合中国国情的。

他的理论和实践,应该成为中国教育血液的重要成分。

他的“捧着一颗心来,不带半根草去”的敬业精神让人感动,我也为之折服。

在读到《教学合一》这节时,陶行知先生提出了教学要合一的观点,有三个理由。

第一,先生的责任不在教,而在教学生学;第二,新的法子必须根据学的法子;第三,先生不但要拿他教的法子和学生学的法子联络,并须和他自己的学问联系起来。

简而言之,一,先生的责任在教学生学。

二,先生教的法子必须根据学的法子。

三,先生须一面教一面学。

我仔细阅读,细细品味,联系我的教学,我感觉陶先生所提出的“教学合一”的观点很有道理。

《学生的精神》中提到三点;(一)学生求学必须具有科学的精神;(二)要改造社会必具有委婉的精神。

(三)应付环境必具有坚强人格和百折不回的精神。

我想说说我自己的感受。

现在的学生正如陶公所说容易“自满”,自己刚刚对这个知识点有一定的领悟,就沾沾自喜,但真正实践做题时,却无从下手。

学生对学习缺少一定的目标,很多学生不知道自己到学校来接受教育是为了什么。

每当找学生谈话时,我便会问“你准备读完干什么?”学生的回答:“不知道。

”他们从来很少会想过我以后的路该怎么走?会是什么样?这就导致学生在学习上不想下功夫,更不愿意吃苦,对于周围的环境学生很少从自身找原因,而是把更多的原因归纳在外界的环境上。

例如,学生作业未完成,当问其原因时,学生回答无非这几种“我不知道什么时候交作业?”“我不会做。

”“我不知道做哪道题?”“我忘记交作业啦!”“课代表交作业时没告诉我!”诸如此类的冠冕堂皇的理由让人无可奈何,更多的时候我在问自己:“我们的学生到底是怎么了?是什么原因让他们变成这样?”回首这些年的教学,我在不断的告诉自己:你一定要学会去转变角色了,你现在已经是一名教师了,陶行知先生的《教育名著》中师范生的第一变——孙悟空,也是这样告诉我的,只有先知道怎样做一个好学生,才能培养出来许多好学生。

江门市2012年普通高中高二调研测试理科数学试题及答案

江门市2012年普通高中高二调研测试理科数学试题及答案

江门市2012年普通高中高二调研测试数 学(理科)本试卷共21题,满分150分,测试用时120分钟.不能使用计算器. 参考公式:锥体的体积公式h S V 31=,其中S 是锥体的底面积,h 是锥体的高. 方差公式])()()[(1222212x x x x x x ns n -++-+-=.一、选择题:本大题共8小题,每小题5分,满分40分。

在每小题给出的四个选项中,只有一项是符合题目要求的.⒈复数i i Z ) 43(+=(其中i 为虚数单位)在复平面上对应的点位于A .第一象限B .第二象限C .第三象限D .第四象限 ⒉等差数列{}n a 的前n 项和记为n S ,若n a n 29-=,则使得n S 最大的序号=nA .2B .3C .4D .5⒊如图1,空间四边形OABC 中,点M 在OB 上,且MB OM 2=,点N 为AC 的中点。

若a OA =,b OB =,c OC =,则= MNA . 21 21 32c b a ++-B . 21 21 21c b a -+C . 21 32 32c b a -+D . 21 32 21c b a +-⒋若542sin -=θ,532cos =θ,则角θ的终边所在的直线是A .0247=+y xB .0247=-y xC .0724=+y xD . 0724=-y x ⒌设α、β是两个不同的平面, l 是一条直线,以下命题中,正确的是A .若α⊥l ,βα⊥,则β⊂lB .若α//l ,βα//,则β⊂lC .若α⊥l ,βα//,则β⊥lD .若α//l ,βα⊥,则β⊥l ⒍已知离散型随机变量X 的 的分布列如右表,则=EXA .9.0B .0.1C ⒎阅读图2的程序框图,若输入m A .2 B .3⒏将正偶数按下表排列则2012所在的位置是A .第252行第3列B .第252行第4列C .第251行第3列D .第251行第4列二、填空题:本大题共7小题,考生作答6小题,每小题5分,满分30分.(一)必做题(9~13题)⒐在等比数列{}n a 中,45=a ,67=a ,则_______9=a .⒑对于曲线C :11422=-+-k y k x ,给出下面四个命题: ①曲线C 不可能表示椭圆; ②当41<<k 时,曲线C 表示椭圆; ③若曲线C 表示双曲线,则1<k 或4>k ; ④若曲线C 表示焦点在x 轴上的椭圆,则251<<k . 其中所有正确命题的序号为__ _ __ .⒒若)4 , 1(~N X ,6826.0)31(=≤<-X P ,________)3(=>X P . ⒓已知命题p :R x ∈∀,0222>++x x .它的否定p ⌝:______________________. ⒔已知x 、y 满足⎪⎩⎪⎨⎧≥++≤≥+-0305k y x x y x ,且y x z 2+=的最小值为3-,则常数____=k .(二)选做题(14~15题,考生只能从中选做一题)⒕82)2(xx -的展开式中,4x 的系数是_ _ __ (用数字表示).⒖曲线x e y =,e y =,0=x 围成的图形的面积_______=S (用数字表示).三、解答题:本大题共6小题,满分80分.解答须写出文字说明、证明过程和演算步骤.B 1已知函数x xx f sin 32sin 2)(2+=,R x ∈. ⑴求)3(π-f 的值;⑵求函数)(x f 的单调递增区间.⒘(本小题满分13分)如图3,正方体1111D C B A ABCD -的棱长为2,E 、F 分别是AB 、BC 的中点. ⑴求多面体1111D C B A AEFCD -的体积; ⑵求F D 1与平面EF B 1所成角的余弦值.⒙(本小题满分13分)已知某同学上学途中必须经过三个交通岗,且在每一个交通岗遇到红灯的概率均为31,假设他在3个交通岗是否遇到红灯是相互独立的,用随机变量ξ表示该同学上学途中遇到红灯的次数.⑴求该同学在第一个交通岗遇到红灯,其它交通岗未遇到红灯的概率; ⑵若2≥ξ,则该同学就迟到,求该同学不迟到...的概率; ⑶随机变量ξ的数学期望和方差. ⒚(本小题满分14分)设) , (1+n n n a a P (+∈N n )是直线l :12+=x y 上的点列,其中1P 是 l 与y 轴的交点. ⑴求1a ,2a⑵用数学归纳法证明:对于一切正整数n ,121-=-n n a ;⑶记直线 l 与直线n a x =、1+=n a x 、x 轴围成的梯形的面积为n b (+∈N n ),试求数列{}n b 的前n 项和n S .已知圆C 的方程为422=+y x ,过点)4 , 2(M 作圆C 的两条切线,切点分别为A 、B ,直线AB 恰好经过椭圆T :12222=+by a x (0>>b a )的右顶点和上顶点.⑴求椭圆T 的方程;⑵记过点M 的两条切线中经过第二象限的为l ,P 是椭圆T 上任意一点,P 到直线为 l 的距离为d ,求d 的取值范围.21.(本小题满分14分)已知函数x e x f =)(. ⑴证明:当10<≤x 时,xe x -≤11; ⑵若函数3)(|)(1|)(-+--=x af x f x h (0>a 是常数)在区间]3ln , 3ln [-上有零点,求a 的取值范围.评分参考一、选择题 BCDD CABA 二、填空题⒐9 ⒑③④(只填1个正确的给3分;若填有错误的,则为0分) ⒒ ⒓R x ∈∃0,022020≤++x x (0x 全写成x 不扣分;≤写成<给3分) ⒔0 ⒕1120 ⒖1 三、解答题⒗⑴)3sin(3)6(sin 2)3(2πππ-+-=-f ……1分,)23(3)21(22-⨯+-⨯=……3分,1-=……5分⑵x x x f sin 3cos 1)(+-=……7分,1)6sin(2+-=πx ……9分,解不等式22622πππππ+≤-≤-k x k ……10分,得32232ππππ+≤≤-k x k (Z k ∈)……11分,所以)(x f 的单调递增区间为]322 , 32[ππππ+-k k (Z k ∈)……12分(是否包含区间端点不扣分;Z k ∈写一次即可)⒘⑴82221111=⨯⨯=-D C B A ABCD V ……1分,B B S Sh V BEF BEF B 131311⨯⨯==∆-……2分,312112131=⨯⨯⨯⨯=……3分,所以,多面体1111D C B A AEFCD -的体积32311111=-=--BEF B D C B A ABCD V V V ……4分 ⑵以1B 为原点,11C B 、11A B 、B B 1分别为x 轴、y 轴、z 轴建立空间直角坐标系……5分,则)0 , 0 , 0(1B ,)0 , 2 , 2(1D ,)2 , 1 , 0(E ,)2 , 0 , 1(F ……6分,设平面EF B 1的一个法向量为) , , ( c b a n =,则⎪⎩⎪⎨⎧=⋅=⋅0 01F B n EF n ……8分,即⎩⎨⎧=+=-020c a b a……9分,取1=c ,则)1 , 2 , 2( --=n ……10分,||||, cos 111F D n F D n F D n ⋅>=<……11分,98=……12分,F D 1与平面EF B 1所成角的余弦值917, sin cos 1>=<=F D n θ……13分。

广东省江门市新会第一中学2024-2025学年高二上学期11月期中考试数学试题

广东省江门市新会第一中学2024-2025学年高二上学期11月期中考试数学试题

广东省江门市新会第一中学2024-2025学年高二上学期11月期中考试数学试题一、单选题1.在简单随机抽样中,某一个个体被抽到的可能性()A .与第几次抽样有关,第一次抽到的可能性大一些B .与第几次抽样无关,每次抽到的可能性都相等C .与第几次抽样有关,最后一次抽到的可能性要大些D .与第几次抽样无关,每次都是等可能抽取,但各次抽取的可能性不一定2.某烟花爆竹厂从20万件同类产品中随机抽取了100件进行质检,发现其中有5件不合格,那么请你估计该厂这20万件产品中合格产品约有()A .1万件B .18万件C .19万件D .20万件310y ++=的倾斜角为α,在y 轴上的截距为b ,则()A .5π6α=,1b =B .2π3α=,1b =-C .2π3α=,1b =D .5π6α=,1b =-4.如图,在平行六面体1111ABCD A B C D -中,M 为11A C 与11B D 的交点,若AB a=,AD b = ,1AA c = ,则下列向量中与BM相等的向量是()A .1122-++ a b cB .1122a b c++C .1122a b c--+ D .1122a b c-+ 5.给定数组5,4,3,5,3,2,2,3,1,2,则错误的是()A .中位数为3BC .众数为2和3D .第85百分位数为46.已知向量(),a x y =,()1,2b =- ,从6张大小相同分别标有号码1,2,3,4,5,6的卡片中,有放回地抽取两张,x 、y 分别表示第一次、第二次抽取的卡片上的号码,则满足0a b ⋅>的概率是()A .112B .34C .15D .167.设()()2,3,1,2A B -,若点(),P x y 在线段AB 上,则1y x+的取值范围是()A .[]2,3-B .()2,3-C .][(),23,∞∞--⋃+D .()(),23,-∞-⋃+∞8.在空间直角坐标系Oxyz 中,(1,1,1)A ,(0,1,0)B ,(0,0,1)C ,点H 在平面ABC 内,则当点O 与H 间的距离取最小值时,点H 的坐标是()A .211,,333⎛⎫ ⎪⎝⎭B .211,,333⎛⎫- ⎝⎭C .111,,333⎛⎫- ⎪⎝⎭D .111,,333⎛⎫-- ⎪⎝⎭二、多选题9.如图所示,下列频率分布直方图显示了三种不同的分布形态.图(1)形成对称形态,图(2)形成“右拖尾”形态,图(3)形成“左拖尾”形态,根据所给图作出以下判断,正确的是()A .图(1)的平均数=中位数=众数B .图(2)的众数<中位数<平均数C .图(2)的众数<平均数<中位数D .图(3)的平均数<中位数<众数10.下列事件中,,A B 是相互独立事件的是()A .一枚硬币掷两次,A =“第一次为正面”,B =“第二次为反面”B .袋中有2个白球,2个黑球,不放回地摸两球,A =“第一次摸到白球”,B =“第二次摸到白球”C .掷一枚骰子,A =“出现点数为奇数”,B =“出现点数为3或4”D .掷一枚骰子,A =“出现点数为奇数”,B =“出现点数为偶数”11.如图,在棱长为2的正方体1111ABCD A B C D -中,P 为线段1B C 上的动点,则()A .当12B P PC =时,3AP =B .直线1A P 与BD 所成的角不可能是π6C .若1113B P BC = ,则二面角11B A P B --D .当12B P PC =时,点1D 到平面1A BP 的距离为23三、填空题12.经过()()0,2,1,0A B -两点的直线的方向向量为()1,k ,求k 的值为.13.如图,在平行六面体1111ABCD A B C D -中12,3,4AD AB AA ===,90BAD ∠=︒,1160A AB A AD ∠=∠=,则1AC =.14.由1,2,3,…,1000这1000个正整数构成集合A ,先从集合A 中随机取一个数a ,取出后把a 放回集合A ,然后再从集合A 中随机取出一个数b ,则13a b >的概率为.四、解答题15.已知ABC V 的顶点分别为(2,4)A ,(7,1)B -,(6,1)C -.(1)求BC 边的中线AD 所在直线的方程;(2)求BC 边的垂直平分线DE 的方程.16.为鼓励青年大学生积极参与暑期社会实践,某高校今年暑假组织返乡大学生积极参与了当地的暑假社区儿童托管服务.现抽样调查了其中100名大学生,统计他们参加社区托管活动的时间(单位:小时),并将统计数据制成如图所示的频率分布直方图.另外,根据参加社区托管活动的时间从长到短按3:4:3的比例分别被评为优秀、良好、合格.(1)求m 的值,并估计该校学生在暑假中参加社区托管活动的时间的平均数(同一组中的数据用该组区间的中点值作代表);(2)试估计至少参加多少小时的社区托管活动,方可以被评为优秀.17.如图,在四棱锥P ABCD -中,底面ABCD 是正方形,侧棱PD ⊥底面,,ABCD PD DC E =是PC 的中点,作EF PB ⊥交PB 于点F .(1)求证://PA 平面EDB ;(2)求证:PB ⊥平面EFD ;(3)求平面CPB 与平面PBD 的夹角的大小.18.甲、乙、丙三人结伴去游乐园玩射击游戏,其中甲射击一次击中目标的概率为12,甲、乙两人各射击一次且都击中目标的概率为16,乙、丙两人各射击一次且都击中目标的概率为19,且任意两次射击互不影响.(1)分别计算乙,丙两人各射击一次击中目标的概率;(2)求甲、乙、丙各射击一次恰有一人击中目标的概率;(3)若乙想击中目标的概率不低于99100,乙至少需要射击多少次?(参考数据:lg 20.3010≈,lg30.4771≈)19.在空间直角坐标系O xyz -中,已知向量(),,u a b c = ,点()0000,,P x y z .若直线l 以u为方向向量且经过点0P ,则直线l 的标准式方程可表示为()0000x x y y z z abc a b c---==≠;若平面α以u为法向量且经过点0P ,则平面α的点法式方程表示为()()()0000a x x b y y c z z -+-+-=.(1)已知直线l 的标准式方程为1111x z -+=,平面1α的点法式方程可表示为50y z +-+=,求直线l 与平面1α所成角的余弦值;(2)已知平面2α的点法式方程可表示为2310x y z ++-=,平面外一点()1,2,1P ,求点P 到平面2α的距离;(3)(i )若集合{(,,)|||||2,||1}M x y z x y z =+≤≤,记集合M 中所有点构成的几何体为S ,求几何体S 的体积:(ii )若集合{(,,)|||||2,||||2,||||2}N x y z x y y z z x =+≤+≤+≤.记集合N 中所有点构成的几何体为T ,求几何体T 相邻两个面(有公共棱)所成二面角的大小。

广东省新会一中2012-2013学年高一上学期第二次测验数学试题

广东省新会一中2012-2013学年高一上学期第二次测验数学试题

新会一中2012-2013学年第一学期高一 数学科第二次数学测验本试卷共4页,共23题,满分100分,附加题10,考试用时80分钟. 注意事项:1.答卷前,考生务必用黑色字迹的钢笔或签字笔将自己的姓名和学号填写在答题卡上. 2.选择题每小题选出答案后,用2B 铅笔把答题卡上对应题目选项的答案信息点涂黑,如需改动,用橡皮擦干净后,再选涂其他答案,答案不能答在试卷上.3.非选择题必须用黑色字迹钢笔或签字笔作答,答案必须写在答题卡各题目指定区域内相应位置上;如需改动,先划掉原来的答案,然后再写上新的答案;不准使用铅笔和涂改液.不按以上要求作答的答案无效.第一部分 选择题(共30分)一、选择题:本大题共10小题,每小题3分,满分30分.在每小题给出的四个选项中,只有一项是符合题目要求的.1.已知全集{}1,2,3,4,5,6,7U =,{}2,4,5A =,则U A =ð( ). A.∅ B.{}2,4,6 C.{}1,3,6,7 D.{}1,3,5,72.已知函数f(x)= 2(1)x x x ⎧⎨+⎩,0,0x x ≥< ,则(2)f -=( ).A.1B.2C.3D.4 3.下列四个图象中,不是y 关于x 的函数的图象是( ).4.如果指数函数y=(2)x a -在x ∈R 上是减函数,则a 的取值范围是( ). A.a >2 B.a <3 C.2<a <3D.a >35.下列幂函数中过点(0,0),(1,1)的偶函数是( ).A.B.C.D.A .12y x = B .4y x = C .2y x -= D .13y x = 6.函数()23x f x =-的零点所在区间为( )A .(-1,0)B .(0,1)C .(1,2)D .(2,3)7.函数2()2f x x ax a =-+在区间(,1)-∞上有最小值,则a 的取值范围是( ). A.1a < B.1a ≤ C.1a > D.1a ≥8.某人2003年1月1日到银行存入一年期存款a 元,若按年利率为x ,并按复利计算,到2008年1月1日可取回的款共( ).A.a(1+x)5元 B.a(1+x)6元 C.a(1+x 5)元 D.a(1+x 6)元 9. 已知函数)(x f 是定义在(,0)(0,)-∞+∞上的偶函数,当0x >时,函数()f x 与函数2,x y x R =∈互为反函数...,则当0<x 时,=)(x f ( ) A.2log ()x - B.2log x - C.2x - D.2x-10.已知(3),1()log ,1a a x a x f x x x --<⎧=⎨≥⎩是(-∞,+∞)上的增函数,那么a 的取值范围是( ).A.(1,+∞)B.(-∞,3)C.[32,3) D.(1,3)第二部分 非选择题(共80分)二、填空题:本大题共6小题,每小题3分,满分18分. 11. 函数ln(1)y x =-的定义域为____________________. 12.已知函数(),mf x x x=+ 且此函数图象过点(1,5),实数m 的值为 . 13.若2510a b ==,则11a b+= . 14.若函数1y ax =+在区间(0,1)内恰有一个零点,则实数a 的取值范围是 . 15.已知集合{(,)|2},{(,)|4}M x y x y N x y x y =+==-=,那么集合MN = .16.已知函数()f x 满足:当x ≥4时,()f x =1()2x;当x <4时()f x =(1)f x +,则2(2log 3)f += .三、解答题:本大题共7小题,满分62分.解答须写出文字说明,证明过程或演算步骤. 17.(本题8分)设U R =,{|24}A x x =-≤<,{|8237}B x x x =-≥-,求A B ,()U A B ð,U A ð,()()U UA B 痧.18.(本题8分)化简:212lg2lg52⨯+⨯+19.(本题9分)已知21()21x x f x -=+,用函数单调性的定义证明()f x 在R 上是增函数.20.(本题9分)某租赁公司拥有汽车100辆. 当每辆车的月租金为3000元时,可全部租出. 当每辆车的月租金每增加50元时,未租出的车将会增加一辆.租出的车每辆每月需要维护费150元,未租出的车每辆每月需要维护费50元.(1)当每辆车的月租金定为3600元时,能租出多少辆车?(2)当每辆车的月租金定为多少元时,租赁公司的月收益最大?最大月收益是多少?21.(本题9分) 试推导出换底公式:log log log c a c bb a=(0a >,且1a ≠;0c >,且1c ≠;0b >)..22.(本题9分)已知函数()log (1),()log (1)a a f x x g x x =+=-(01)a a >≠且. (1)求函数()()f x g x +的定义域;(2)判断函数()()f x g x +的奇偶性,并说明理由; (3)求使()(2)0f x g x ->成立的x 的集合.23.(附加题,本题10分)已知a 是实数,函数()2222f x ax x a =+--,如果函数()x f y =在区间[]1,1-上有零点,求a 的取值范围.2012-2013学年第一学期高一级二次数学测验答案一、选择题:本大题共10小题,每小题3分,满分30分. CBBCB CAAAC二、填空题:本大题共6小题,每小题3分,满分18分.11.(1,)+∞; 12.4; 13.1; 14.(,1)-∞-; 15.{}(3,1)-; 16.124三、解答题:本大题共7小题,满分62分.解答须写出文字说明,证明过程或演算步骤. 17. 解:{|24}A x x =-≤<,{|8237}{|3}B x x x x x =-≥-=≤ 2分∴{}4A B x x =< 3分U R =∴{}()4U A B x x =≥ð 5分{}{}2,4,3U UA x x xB x x =<-≥=>或痧 7分∴()()U UA B 痧{}4x x =≥ 8分18. 解:原式=2112(lg2)lg2lg5(lg 22⨯++分=211lg 2lg2lg5(lg 1)22+- 4分 =2111lg 2lg2lg5lg21222+-+ 5分 =1lg 2(lg 2lg51)12+-+ 7分 =1lg 2(lg101)10112-+=+=. 8分 19. 证明:设任意12,x x R ∈,且12x x <,则 1分 121212*********(22)()()2121(21)(21)x x x x x x x x f x f x ----=-=++++. 5分由于12x x <,从而1222x x <,即12220x x -<. 7分 ∴ 12()()0f x f x -<,即12()()f x f x <. 8分∴ ()f x 为R 上的增函数. 9分 20. 解:(1)月租金定为3600元时,未租出的车辆数为:3600300050- =12, 2分所以这时租出了88辆车. 3分 (2)设每辆车的月租金定为x 元, 4分 则月收益为30003000()(100)(150)505050x x f x x --=---⨯, 6分 整理得:221()16221000(4050)3070505050x f x x x =-+-=--+(30008000)x ≤≤. 7分所以,当x=4050时,()f x 最大,其最大值为f (4050)=307050. 8分即当每辆车的月租金定为4050元时,租赁公司的月收益最大,最大收益为307050元. 9分21. 证明:设log c b m =,log c a n =,log a b p =, 2分 则m c b =,n c a =,p a b =. 4分 从而()n p m c b c ==,即np m =. 6分 由于log log 10c c n a =≠=,则mp n=. 8分 所以,log log log c a c bb a=. 9分 22. 解:(1)()()log (1)log (1)a a f x g x x x +=++- , 若要式子有意义,则{1010x x +>-> , 1分即1x >.所以所求定义域为{}1x x >. 3分 (2)设()()()F x f x g x =+,因为()()()F x f x g x =+的定义域为{}1x x >,所以()()()F x f x g x =+是非奇非偶函数。

广东省江门市新会第一中学2023-2024学年高二下学期期中考试数学答案

广东省江门市新会第一中学2023-2024学年高二下学期期中考试数学答案

新会第一中学2023-2024学年高二下学期期中考试数学答案单项选择:DDBA BBBC多项选择:ABD ABC ABD 填空题:12.28 13.14. 390 1.D 【详解】通项公式为,当时,,所以项的系数为80.故选:D 2.D 【详解】,由得,又函数定义域为,当时,,递减,当时,,递增,因此是函数的极小值点.故D .3、【答案】B 【解析】分两类:一是取出1本画册,3本集邮册,此时赠送方法有种;二是取出2本画册,2本集邮册,此时赠送方法有种.故赠送方法共有10种.4. 【答案】A 【详解】试题分析:因为函数y=f (x )的导函数在区间[a ,b]上是增函数,所以对任意的a <x 1<x 2<b ,有也即在a,x 1,x 2,b 处它们的斜率是依次增大的.所以A 满足上述条件,对于B 存在使,对于C 对任意的a <x 1<x 2<b ,都有,对于D 对任意的x [a ,b],不满足逐渐递增的条件,故选A .5.【答案】B 【详解】记事件该家族某位成员出现性状,事件该家族某位成员出现性状,则,,,则,又因为,则,故所求概率为.6、【答案】B 【解析】因为每人至少一张,且分给同一人的多张票必须连号,又分给甲、乙、丙、丁四个人,则在座位号、、、、、的五个空位插3个板子,有种,然后再分给甲、乙、丙、丁四个人,有种,所以不同的分法种数为,7、【答案】B 解:抛掷一枚质地均匀的硬币100次,设硬币正面向上次数为,则,所以,,,且,,因为,516515C 2rrr r T x -+=2r =232235C 280T x x ==2x 144C =246C =∈:E A :F B ()415P E =()215P F =()710P E F = ()()3110P E F P E F =-= ()()()()P E F P E P F P EF =+- ()()()()110P EF P E P F P E F =+-= ()()()11531048P EF P F E P E ==⨯=1234563510C=4424A =1024240⨯=X 1~100,2X B ⎛⎫ ⎪⎝⎭()1100502E X np ==⨯=()()11110012522D X np p ⎛⎫=-=⨯⨯-= ⎪⎝⎭()2,X N μσ:()50E X μ==()225D X σ==(22)0.9545P X μσμσ-≤≤+≈所以利用正态分布近似估算硬币正面向上次数超过60次的概率为, 8.【答案】C 参变分离得,,设,得,,设,,求导讨论单调性,可得9.【答案】ABD 【详解】对于A ,若,则,故A 正确;对于B ,若,则,故B 正确;对于C ,若,则,故C 错误;对于D ,若,则,故D 正确. 故选:ABD 10.【答案】ABC 【解析】∵已知的展开式中第5项的二项式系数最大,当时,第5项的二项式系数为,此时和第4项二项式系数同为最大,符合题意。

2023-2024学年广东省江门市新会一中高二(下)期末数学试卷(含答案)

2023-2024学年广东省江门市新会一中高二(下)期末数学试卷(含答案)

2023-2024学年广东省江门市新会一中高二(下)期末数学试卷一、单选题:本题共8小题,每小题5分,共40分。

在每小题给出的选项中,只有一项是符合题目要求的。

1.已知ξ的分布列为ξ1234P 161613m设η=2ξ−5,则E(η)=( )A. 12B. 13C. 23D. 322.已知圆C:x2+y2=4,直线l:y=kx+m,若当k的值发生变化时,直线被圆C所截的弦长的最小值为2,则m的取值为( )A. ±2B. ±2C. ±3D. ±33.已知等差数列{a n},等比数列{b n},满足a7+a9=4,b2b6b10=27,则a3+a8+a13b4b8+3=( )A. 14B. 12C. 2D. 44.若曲线y=lnx+ax在点(1,a)处的切线与直线l:x+y+5=0平行,则实数a=( )A. 12B. 1 C. 32D. 25.今天是星期天,则137天后是( )A. 星期五B. 星期六C. 星期天D. 星期一6.某单位五一放假,安排甲、乙等五人值班五天,每人值班一天.若甲、乙都至少需要三天的连休假期,则不同的值班安排共有( )A. 60种B. 66种C. 72种D. 78种7.袋中装有10个球,其中3个黑球、7个白球,从中依次取两球(不放回),则第二次取到的是黑球的概率为( )A. 29B. 310C. 13D. 7108.已知函数f(x)=x2e x,下列关于f(x)的四个命题,其中是假命题是( )A. 函数f(x)在[0,1]上是增函数B. 函数f(x)的最小值为0C. 如果x∈[0,t]时,f(x)max=4e2,则t的最小值为2D. 函数f(x)有2个零点二、多选题:本题共3小题,共18分。

在每小题给出的选项中,有多项符合题目要求。

9.给出下列命题,其中正确的命题有( )A. 两个变量的线性相关性越强,则相关系数r越大B. 在(3x−1x)10的展开式中,各项系数和与所有项二项式系数和相等C. 将4名老师分派到两个学校支教,每个学校至少派1人,则共有14种不同的分派方法D. 公共汽车上有10位乘客,沿途5个车站,乘客下车的可能方式有105种10.已知双曲线E:x2a2−y22=1(a>0)的左、右焦点分别为F1,F2,过点F2的直线l与双曲线E的右支相交于P,Q两点,则( )A. 若E的两条渐近线相互垂直,则a=2B. 若E的离心率为3,则E的实轴长为1C. 若∠F1PF2=90°,则|PF1|⋅|PF2|=4D. 当a变化时,△F1PQ周长的最小值为8211.如图,在棱长为1的正方体ABCD−A1B1C1D1中,E、F分别为棱A1D1、AA1的中点,G为线段B1C上一个动点,则( )A. 三棱锥A1−EFG的体积为定值B. 存在点G,使平面EFG//平面BDC1C. 当点G与B1重合时,二面角G−EF−A1的正切值为22D. 当点G为B1C中点时,平面EFG截正方体所得截面的面积为334三、填空题:本题共3小题,每小题5分,共15分。

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