2012年佛山市普通高中高三教学质量检测
广东省佛山市普通高中2012届高三历史上学期教学质量检测(1)试题新人教版

2012年佛山市普通高中高三教学质量检测(一)文科综合历史部分试题12.钱穆认为,中国古代史“前一段落为秦以前的封建政治,后一段落为秦以后之郡县政治”。
其中“封建政治”是A.封邦建国 B.宗法世袭制C.礼乐制度 D.封建社会13.合理的历史联系是历史学习的常用方法之一,右图空白a、d相对应处,应补填上A.开埠通商、入超B.开埠通商、商品市场与原料产地C.鸦片走私、自然经济解体D.鸦片走私、出超14.竹枝词是风土诗的一种,保存了大量有价值的第一手资料。
如要研究近代上海开埠后的商业活动,可选用的竹枝词是A.“邑志详陈旧土风,镇升为县百年中。
田家妇女帮农作,镇市夫男晓女工。
蓬首晨兴遥入市,归家手挈米和盐。
”B.“南北分开两市忙,南为华界北洋场。
有城不若无城富,第一繁华让此方。
”C.“菜圃瓜畦拓百弓,杭家村北夕阳红。
楝花倒覆三间屋,酒幔低垂十里风。
”D.“北口山头民力忙,耕牛无恙乐平康。
一鞭残照吹横笛,蚕豆青青麦半黄。
”15.某图书目录有如下内容:上古茫昧无稽考……周末诸子并起创教考……诸子创教改制考……诸子改制托古考……汉武帝后儒教一统考。
据此判断该图书是A.《海国图志》 B.《新学伪经考》 C.《孔子改制考》 D.《变法通议》16.20世纪以来,中国历代政府都重视中学历史教育。
下列历史教育目标,分别颁布于1904年、1912年、1936年和1956年。
其中1912年的是A.“多讲本朝仁政,俾知历圣德泽之深厚,以养成国民自强之志气,忠爱之性情”B.“了解劳动人民是历史的创造者,是历史的主人”C.“明于民族之进化,社会变迁,邦国之盛衰,尤宜注意于政体之沿革,与民国建立之本”D.“特别说明其(指中国)历史上之光荣,及近代所受列强侵略之经过,与其原因,以激发学生民族复兴之思想”17.1958年5月27日,村民阿牛在日记中写到“广播里面说全村的人都要集中起来,学习中央的最新理论指示。
”是因为当时在A.开展“三大改造”运动 B.开展“大跃进”与人民公社化运动C.庆祝“一五”计划超额完成 D.庆祝“八大”胜利召开18.下图是美国军费开支曲线图,根据该图获取的历史信息不正确的是A.导致阶段A军费上升的主要原因是朝鲜战争爆发B.阶段B时期的中美处于敌视和对抗状态C.阶段C军费下降的主要原因是美国实行全球战略收缩D.阶段D时期的中美结束对抗,两国关系开始实现正常化19.西方学者麦马虹在总结西方从古希腊至今两千多年来的幸福观时指出:荷马时期,幸福就是幸运;公元前五世纪的古希腊,幸福等同于;启蒙时期,幸福就是及时行乐。
2012~2013学年佛山市普通高中教学质量检测高三物理答案

2013年佛山市普通高中高三教学质量检测(一)理科综合参考答案选择题答案一、本题共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,错选、不选得0 分.二、本题共9小题,每小题6分,共54分.在每小题给出的四个选项中,有两个选项符合题目要求,全选对的得6分,只选一个且正确的得3分,错选、不选得0 分.物理非选择题答案34.(1)(共8分) ①125(2分) ;②B.方向(1分),11.38~11.42(2分); C.如左下图所示(3分)。
(2)(共10分)①C ,B ;②完整电路如右上图所示;③A 端;④8.19×10-2W(每空2分,连图正确2分) 35.(18分)解:(1)(共6分)小球由静止沿电场方向加速 从M 到N :212qE l m v =(2分)进入磁场时的速度 v =(2分)速度方向与电场方向一致,与x 轴垂直 (2分) (2)(共6分)带电小球进入磁场做匀速圆周运动,轨迹如图所示。
由几何关系可得cos 60R R l +︒=即圆周运动的半径23R l =(2分)由2vqvB m R= (2分)解得 m v B q R== (2分)(3)(共6分)由平抛运动规律:212h gt = (2分)飞离桌面飞行时间t =(2分)水平飞行距离x vt === (2分)x36.(18分)解:(1)(共7分)a 球从A 滑到C :21(5sin 37 1.8)cos 37502a m g R R m g R m v μ⨯︒+-︒⨯=- (2分)解得,a v =(1分)a 、b 球在C 发生弹性碰撞动量守恒:a a b m v m v m v '=+ (1分) 222111222a ab m v m v m v '=+ (1分)(缺上两式,无论结论是否正确,不给相应的得分) a 、b球碰后交换速度,0,ab v v '== (2分)(2) (共11分)①要使小球不离轨道,并从水平轨道EF 滑出,在竖直圆轨道最高点P :2P v mg mR≤;P v ≥(1分)因ab 两球碰撞无能量损失,故由A 到P :21(5sin 37 1.8)cos 375202P m g R R m g R m g R m g R m v μμ'⨯︒+-︒⨯-⨯-=- (1分)得23'0.9225R R R ≤= (1分)②让b 球恰好不脱离轨道且能够滑回D ,则b 球上滑到四分之一圆轨道的Q 处速度为零(1分),因ab 两球碰撞无能量损失,故由A 到Q :21(5sin 37 1.8)cos 37502Q m g R R m g R m g R m gR m v μμ'⨯︒+-︒⨯-⨯-=- (1分)得' 2.3R R > (1分) ③若 2.5R R '=,则b 球必定将滑回D ,设其能滑过CD 轨道与a 球碰撞,且a 球到达B 点,在B 点的速度为B v ,则由A 经CD 再返回B :21(5sin 37 1.8)cos 3752 1.802B m g R R m g R m g R m g R m v μμ⨯︒+-︒⨯-⨯-⨯=-(1分)解得:0B v = (1分)故ab 两球将在A 、Q 之间做往返运动,最终a 球停在C 处。
佛山市普通高中2012届高三教学质量检测(一)(理)

4.“关于x 的不等式220x ax a -+>的解集为R ”是“01a ≤≤”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6.已知点P 是抛物线24x y =上的一个动点,则点P 到点(2,0)M 的距离与点P 到该抛物线准线的距离之和的最小值为A .172B .5C .22D .92 8.对于非空集合,A B ,定义运算:{|,}A B x x A B x A B ⊕=∈∉且,已知}|{},|{d x c x N b x a x M <<=<<=,其中d c b a 、、、满足a b c d +=+,0ab cd <<,则=⊕N MA. (,)(,)a d b cB.(,][,)c a b dC. (,][,)a c d bD.(,)(,)c a d b9.某学校三个社团的人员分布如下表(每名同学只参加一个社团)学校要对这三个社团的活动效果进行抽样调查,按分层抽样的方法从社团成员中抽取30人,结果合唱社被抽出12人,则a =_______________.11.已知不等式组02,20,20x x y kx y ≤≤⎧⎪+-≥⎨⎪-+≥⎩所表示的平面区域的面积为4,则k 的值为__________. 13.对任意实数b a ,,函数|)|(21),(b a b a b a F --+=,如果函数2()23,f x x x =-++ ()1g x x =+,那么函数()()(),()G x F f x g x =的最大值等于 .18.(本题满分13分) 佛山某学校的场室统一使用“佛山照明”的一种灯管,已知这种灯管使用寿命ξ(单位:月)服从正态分布2(,)N μσ,且使用寿命不少于12个月的概率为0.8,使用寿命不少于24个月的概率为0.2.(1)求这种灯管的平均使用寿命μ;(2)假设一间功能室一次性换上4支这种新灯管,使用12个月时进行一次检查,将已经损坏的灯管换下(中途不更换),求至少两支灯管需要更换的概率.19.(本题满分12分)已知圆221:(4)1C x y -+=,圆222:(2)1C x y +-=,动点P 到圆1C ,2C 上点的距离的最小值相等.(1)求点P 的轨迹方程;(2)点P 的轨迹上是否存在点Q ,使得点Q 到点(22,0)A -的距离减去点Q 到点(22,0)B 的距离的差为4,如果存在求出Q 点坐标,如果不存在说明理由.20.(本题满分14分)设a R ∈,函数()ln f x x ax =-. 合唱社 粤曲社 书法社 高一45 30 a 高二 15 10 20(1) 若2a =,求曲线()y f x =在()1,2P -处的切线方程;(2) 若()f x 无零点,求实数a 的取值范围;(3) 若()f x 有两个相异零点12,x x ,求证: 212x x e ⋅>.21.(本题满分14分)设*N n ∈,圆n C :222(0)n n x y R R +=>与y 轴正半轴的交点为M ,与曲线y x =的交点为1(,)n N y n,直线MN 与x 轴的交点为(,0)n A a .(1)用n 表示n R 和n a ;(2)求证:12n n a a +>>;(3)设123n n S a a a a =++++,111123n T n =++++,求证:27352n n S n T -<<.答案 4、A ;6、B ;8、C9.30 11.1 13. 318.(本题满分13分)解:(1)∵2(,)N ξμσ,(12)0.8P ξ≥=,(24)0.2P ξ≥=,∴(12)0.2P ξ<=,显然(12)(24)P P ξξ<=> ……………3分 由正态分布密度函数的对称性可知,1224182μ+==, 即每支这种灯管的平均使用寿命是18个月; …………………5分(2)每支灯管使用12个月时已经损坏的概率为10.80.2-=, ……………6分假设使用12个月时该功能室需要更换的灯管数量为η支,则(4,0.2)B η, ………10分 故至少两支灯管需要更换的概率1(0)(1)P P P ηη=-=-=0413********.80.80.2625C C =--⨯=(写成≈0.18也可以). …………………13分19.(本题满分13分)解:(1)设动点P 的坐标为(,)x y ,圆1C 的圆心1C 坐标为(4,0),圆2C 的圆心2C 坐标为(0,2), …………………2分因为动点P 到圆1C ,2C 上的点距离最小值相等,所以12||||PC PC =, …………3分 即2222(4)(2)x y x y -+=+-,化简得23y x =-,…………………4分因此点P 的轨迹方程是23y x =-; ……………………5分(2)假设这样的Q 点存在,因为Q 点到(22,0)A -点的距离减去Q 点到(22,0)B 点的距离的差为4,所以Q 点在以(22,0)A -和(22,0)B 为焦点,实轴长为4的双曲线的右支上,即Q 点在曲线221(2)44x y x -=≥上, ……………………9分 又Q 点在直线:23l y x =-上, Q 点的坐标是方程组2223144y x x y =-⎧⎪⎨-=⎪⎩的解,………11分 消元得2312130x x -+=,21243130∆=-⨯⨯<,方程组无解,所以点P 的轨迹上不存在满足条件的点Q . …………………13分20.(本题满分14分)解:方法一在区间()0,+∞上,11()ax f x a x x-'=-=. ………1分(1)当2a =时,(1)121f '=-=-,则切线方程为(2)(1)y x --=--,即10x y ++= …………3分(2)①若0a <,则()0f x '>,()f x 是区间()0,+∞上的增函数,(1)0f a =->Q ,()(1)0a a a f e a ae a e =-=-<,(1)()0a f f e ∴⋅<,函数()f x 在区间()0,+∞有唯一零点. …………6分②若0a =,()ln f x x =有唯一零点1x =. …………7分③若0a >,令()0f x '=得: 1x a=. 在区间1(0,)a 上, ()0f x '>,函数()f x 是增函数; 在区间1(,)a +∞上, ()0f x '<,函数()f x 是减函数;故在区间()0,+∞上, ()f x 的极大值为11()ln1ln 1f a a a =-=--. 由1()0,f a <即ln 10a --<,解得:1a e>. 故所求实数a 的取值范围是1(,)e +∞. …………9分方法二、函数()f x 无零点⇔方程ln x ax =即ln x a x =在()0,+∞上无实数解 ……4分 令ln ()x g x x =,则21ln ()x g x x -'= 由()0g x '=即21ln 0x x-=得:x e = …………6分 在区间(0,)e 上, ()0g x '>,函数()g x 是增函数;在区间(,)e +∞上, ()0g x '<,函数()g x 是减函数;故在区间()0,+∞上, ()g x 的极大值为1()g e e=. …………7分 注意到(0,1)x ∈时,()(),0g x ∈-∞;1x =时(1)0g =;()1,x ∈+∞时,1()0,g x e ⎛⎤∈ ⎥⎝⎦故方程ln x a x=在()0,+∞上无实数解⇔1a e >. 即所求实数a 的取值范围是1(,)e +∞. …………9分 [注:解法二只说明了()g x 的值域是1,e ⎛⎤-∞ ⎥⎝⎦,但并没有证明.] (3) 设120,x x >>12()0,()0,f x f x ==Q 1122ln 0,ln 0x ax x ax ∴-=-=1212ln ln ()x x a x x ∴+=+,1212ln ln ()x x a x x -=-原不等式21212ln ln 2x x e x x ⋅>⇔+>12()2a x x ⇔+>121212ln ln 2x x x x x x -⇔>-+1122122()ln x x x x x x -⇔>+ 令12x t x =,则1t >,于是1122122()2(1)ln ln 1x x x t t x x x t -->⇔>++. ……12分 设函数2(1)()ln 1t g t t t -=-+(1)t >, 求导得: 22214(1)()0(1)(1)t g t t t t t -'=-=>++ 故函数()g t 是()1,+∞上的增函数, ()(1)0g t g ∴>= 即不等式2(1)ln 1t t t ->+成立,故所证不等式212x x e ⋅>成立. ………………14分 21.(本题满分14分)解: (1)由点N 在曲线y x =上可得11(,)N n n, ……………………1分 又点在圆n C 上,则2221111(),n n n n R R n n n n ++=+==, ……………………2分 从而直线MN 的方程为1n nx y a R +=, ……………………4分 由点11(,)N n n 在直线MN 上得: 111n nna n R +=⋅,将1n n R n +=代入 化简得: 1111n a n n =+++. ……………………6分 (2) 1111,11n n +>+>,*11,112n n N a n n∴∀∈=+++> ……………7分 又111111,1111n n n n +>++>+++, 11111111111n n a a n n n n +∴=+++>+++=++ ……………………9分 (3)先证:当01x ≤≤时,1(21)112x x x +-≤+≤+. 事实上, 不等式1(21)112x x x +-≤+≤+22[1(21)]1(1)2x x x ⇔+-≤+≤+ 22212(21)(21)114x x x x x ⇔+-+-≤+≤++ 222(223)(21)04x x x ⇔-+-≤≤ 后一个不等式显然成立,而前一个不等式2001x x x ⇔-≤⇔≤≤. 故当01x ≤≤时, 不等式1(21)112x x x +-≤+≤+成立. 1111(21)112n n n∴+-≤+<+, ……………………11分 1113221122n a n n n n∴+⋅≤=+++<+(等号仅在n =1时成立) 求和得: 32222n n n n T S n T +⋅≤<+⋅ 273252n n S n T -∴<≤< ……………………14分。
2012佛山二模文科综合

3.非选择题必须用黑色字迹的钢笔或签字笔作答,答案必须写在答卷上各题目指定 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.
2012 年佛山市普通高中高三教学质量检测(二)
理科综合能力测试
本试卷共 12 页,满分 300 分.考试时间 150 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考号填写在答
题卡上.用 2B 铅笔将答题卡试卷类型(A)填涂在答题卡上,并在答题卡右上角的“试室 号”和“座位号”栏填写试室号、座位号,将相应的试室号、座位号信息点涂黑.
A.与 DNA 复制相关的酶基因 B.与转录相关的酶基因 C.与逆转录相关的酶基因 D.与翻译相关的酶基因 3.如图所示为人工培养的肝细胞中 DNA 含量随时间的变化曲线,据图判断正确的是
A.细胞周期时长为 14 小时 B.染色体数量倍增发生在 I 段 C.II 段可以看到圆柱状或杆状的染色体 D.基因突变最可能发生于 III 段 4.尿崩症患者因某种激素分泌不足,每天尿量可多达十几升,且尿液的渗透压低于正常人, 这最可能是下列哪种器官或组织病变所致? A.下丘脑一垂体 B.肾上腺 C.肾脏 D.胰岛 5.用矮壮素处理可以抑制植株增高,培育出矮化的水仙,提高观赏价值。与矮壮素的作用 效果正好相反的植物激素是 A.乙烯利 B.赤霉素 C.2,4 一 D D.脱落酸
2012.4
对全部高中资料试卷电气设备,在安装过程中以及安装结束后进行高中资料试卷调整试验;通电检查所有设备高中资料电试力卷保相护互装作置用调与试相技互术关,通系电1,力过根保管据护线0生高不产中仅工资22艺料22高试可中卷以资配解料置决试技吊卷术顶要是层求指配,机置对组不电在规气进范设行高备继中进电资行保料空护试载高卷与中问带资题负料22荷试,下卷而高总且中体可资配保料置障试时23卷,23调需各控要类试在管验最路;大习对限题设度到备内位进来。行确在调保管整机路使组敷其高设在中过正资程常料1工试中况卷,下安要与全加过,强度并看2工且55作尽22下可2都能护1可地关以缩于正小管常故路工障高作高中;中资对资料于料试继试卷电卷连保破接护坏管进范口行围处整,理核或高对者中定对资值某料,些试审异卷核常弯与高扁校中度对资固图料定纸试盒,卷位编工置写况.复进保杂行护设自层备动防与处腐装理跨置,接高尤地中其线资要弯料避曲试免半卷错径调误标试高方中等案资,,料要编5试求写、卷技重电保术要气护交设设装底备备4置。高调、动管中试电作线资高气,敷料中课并3设试资件且、技卷料中拒管术试试调绝路中验卷试动敷包方技作设含案术,技线以来术槽及避、系免管统不架启必等动要多方高项案中方;资式对料,整试为套卷解启突决动然高过停中程机语中。文高因电中此气资,课料电件试力中卷高管电中壁气资薄设料、备试接进卷口行保不调护严试装等工置问作调题并试,且技合进术理行,利过要用关求管运电线行力敷高保设中护技资装术料置。试做线卷到缆技准敷术确设指灵原导活则。。:对对在于于分调差线试动盒过保处程护,中装当高置不中高同资中电料资压试料回卷试路技卷交术调叉问试时题技,,术应作是采为指用调发金试电属人机隔员一板,变进需压行要器隔在组开事在处前发理掌生;握内同图部一纸故线资障槽料时内、,设需强备要电制进回造行路厂外须家部同出电时具源切高高断中中习资资题料料电试试源卷卷,试切线验除缆报从敷告而设与采完相用毕关高,技中要术资进资料行料试检,卷查并主和且要检了保测解护处现装理场置。设。备高中资料试卷布置情况与有关高中资料试卷电气系统接线等情况,然后根据规范与规程规定,制定设备调试高中资料试卷方案。
佛山市普通高中2012届高三教学质量检测(一)(理)

4.“关于x 的不等式220x ax a -+>的解集为R ”是“01a ≤≤”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6.已知点P 是抛物线24x y =上的一个动点,则点P 到点(2,0)M 的距离与点P 到该抛物线准线的距离之和的最小值为A .172B .5C .22D .92 8.对于非空集合,A B ,定义运算:{|,}A B x x A B x A B ⊕=∈∉ 且,已知}|{},|{d x c x N b x a x M <<=<<=,其中d c b a 、、、满足a b c d +=+,0ab cd <<,则=⊕N MA. (,)(,)a d b cB.(,][,)c a b dC. (,][,)a c d bD.(,)(,)c a d b9.某学校三个社团的人员分布如下表(每名同学只参加一个社团)学校要对这三个社团的活动效果进行抽样调查,按分层抽样的方法从社团成员中抽取30人,结果合唱社被抽出12人,则a =_______________.11.已知不等式组02,20,20x x y kx y ≤≤⎧⎪+-≥⎨⎪-+≥⎩所表示的平面区域的面积为4,则k 的值为__________. 13.对任意实数b a ,,函数|)|(21),(b a b a b a F --+=,如果函数2()23,f x x x =-++ ()1g x x =+,那么函数()()(),()G x F f x g x =的最大值等于 .18.(本题满分13分) 佛山某学校的场室统一使用“佛山照明”的一种灯管,已知这种灯管使用寿命ξ(单位:月)服从正态分布2(,)N μσ,且使用寿命不少于12个月的概率为0.8,使用寿命不少于24个月的概率为0.2.(1)求这种灯管的平均使用寿命μ;(2)假设一间功能室一次性换上4支这种新灯管,使用12个月时进行一次检查,将已经损坏的灯管换下(中途不更换),求至少两支灯管需要更换的概率.19.(本题满分12分)已知圆221:(4)1C x y -+=,圆222:(2)1C x y +-=,动点P 到圆1C ,2C 上点的距离的最小值相等.(1)求点P 的轨迹方程;(2)点P 的轨迹上是否存在点Q ,使得点Q 到点(22,0)A -的距离减去点Q 到点(22,0)B 的距离的差为4,如果存在求出Q 点坐标,如果不存在说明理由.20.(本题满分14分) 合唱社 粤曲社 书法社 高一45 30 a 高二 15 10 20设a R ∈,函数()ln f x x ax =-.(1) 若2a =,求曲线()y f x =在()1,2P -处的切线方程;(2) 若()f x 无零点,求实数a 的取值范围;(3) 若()f x 有两个相异零点12,x x ,求证: 212x x e ⋅>.21.(本题满分14分)设*N n ∈,圆n C :222(0)n n x y R R +=>与y 轴正半轴的交点为M ,与曲线y x =的交点为1(,)n N y n,直线MN 与x 轴的交点为(,0)n A a .(1)用n 表示n R 和n a ;(2)求证:12n n a a +>>;(3)设123n n S a a a a =++++ ,111123n T n =++++ ,求证:27352n n S n T -<<.答案 4、A ;6、B ;8、C9.30 11.1 13. 318.(本题满分13分)解:(1)∵2(,)N ξμσ ,(12)0.8P ξ≥=,(24)0.2P ξ≥=,∴(12)0.2P ξ<=,显然(12)(24)P P ξξ<=> ……………3分 由正态分布密度函数的对称性可知,1224182μ+==, 即每支这种灯管的平均使用寿命是18个月; …………………5分(2)每支灯管使用12个月时已经损坏的概率为10.80.2-=, ……………6分假设使用12个月时该功能室需要更换的灯管数量为η支,则(4,0.2)B η , ………10分故至少两支灯管需要更换的概率1(0)(1)P P P ηη=-=-=0413********.80.80.2625C C =--⨯=(写成≈0.18也可以). …………………13分19.(本题满分13分)解:(1)设动点P 的坐标为(,)x y ,圆1C 的圆心1C 坐标为(4,0),圆2C 的圆心2C 坐标为(0,2), …………………2分因为动点P 到圆1C ,2C 上的点距离最小值相等,所以12||||PC PC =, …………3分 即2222(4)(2)x y x y -+=+-,化简得23y x =-,…………………4分因此点P 的轨迹方程是23y x =-; ……………………5分(2)假设这样的Q 点存在,因为Q 点到(22,0)A -点的距离减去Q 点到(22,0)B 点的距离的差为4,所以Q 点在以(22,0)A -和(22,0)B 为焦点,实轴长为4的双曲线的右支上,即Q 点在曲线221(2)44x y x -=≥上, ……………………9分 又Q 点在直线:23l y x =-上, Q 点的坐标是方程组2223144y x x y =-⎧⎪⎨-=⎪⎩的解,………11分 消元得2312130x x -+=,21243130∆=-⨯⨯<,方程组无解,所以点P 的轨迹上不存在满足条件的点Q . …………………13分20.(本题满分14分)解:方法一在区间()0,+∞上,11()ax f x a x x-'=-=. ………1分 (1)当2a =时,(1)121f '=-=-,则切线方程为(2)(1)y x --=--,即10x y ++= …………3分(2)①若0a <,则()0f x '>,()f x 是区间()0,+∞上的增函数,(1)0f a =->Q ,()(1)0a a a f e a ae a e =-=-<,(1)()0a f f e ∴⋅<,函数()f x 在区间()0,+∞有唯一零点. …………6分②若0a =,()ln f x x =有唯一零点1x =. …………7分③若0a >,令()0f x '=得: 1x a=. 在区间1(0,)a 上, ()0f x '>,函数()f x 是增函数; 在区间1(,)a +∞上, ()0f x '<,函数()f x 是减函数;故在区间()0,+∞上, ()f x 的极大值为11()ln1ln 1f a a a =-=--. 由1()0,f a <即ln 10a --<,解得:1a e>. 故所求实数a 的取值范围是1(,)e +∞. …………9分方法二、函数()f x 无零点⇔方程ln x ax =即ln x a x =在()0,+∞上无实数解 ……4分 令ln ()x g x x =,则21ln ()x g x x -'= 由()0g x '=即21ln 0x x-=得:x e = …………6分 在区间(0,)e 上, ()0g x '>,函数()g x 是增函数;在区间(,)e +∞上, ()0g x '<,函数()g x 是减函数;故在区间()0,+∞上, ()g x 的极大值为1()g e e=. …………7分 注意到(0,1)x ∈时,()(),0g x ∈-∞;1x =时(1)0g =;()1,x ∈+∞时,1()0,g x e⎛⎤∈ ⎥⎝⎦ 故方程ln x a x=在()0,+∞上无实数解⇔1a e >.即所求实数a 的取值范围是1(,)e +∞. …………9分[注:解法二只说明了()g x 的值域是1,e ⎛⎤-∞ ⎥⎝⎦,但并没有证明.] (3) 设120,x x >>12()0,()0,f x f x ==Q 1122ln 0,ln 0x ax x ax ∴-=-= 1212ln ln ()x x a x x ∴+=+,1212ln ln ()x x a x x -=-原不等式21212ln ln 2x x e x x ⋅>⇔+>12()2a x x ⇔+>121212ln ln 2x x x x x x -⇔>-+1122122()ln x x x x x x -⇔>+ 令12x t x =,则1t >,于是1122122()2(1)ln ln 1x x x t t x x x t -->⇔>++. ……12分 设函数2(1)()ln 1t g t t t -=-+(1)t >, 求导得: 22214(1)()0(1)(1)t g t t t t t -'=-=>++ 故函数()g t 是()1,+∞上的增函数, ()(1)0g t g ∴>= 即不等式2(1)ln 1t t t ->+成立,故所证不等式212x x e ⋅>成立. ………………14分 21.(本题满分14分)解: (1)由点N 在曲线y x =上可得11(,)N n n, ……………………1分 又点在圆n C 上,则2221111(),n n n n R R n n n n ++=+==, ……………………2分 从而直线MN 的方程为1n nx y a R +=, ……………………4分 由点11(,)N n n 在直线MN 上得: 111n nna n R +=⋅,将1n n R n +=代入 化简得: 1111n a n n=+++. ……………………6分 (2) 1111,11n n +>+> ,*11,112n n N a n n∴∀∈=+++> ……………7分又111111,1111n n n n +>++>+++ , 11111111111n n a a n n n n +∴=+++>+++=++ ……………………9分 (3)先证:当01x ≤≤时,1(21)112x x x +-≤+≤+. 事实上, 不等式1(21)112x x x +-≤+≤+ 22[1(21)]1(1)2x x x ⇔+-≤+≤+ 22212(21)(21)114x x x x x ⇔+-+-≤+≤++ 222(223)(21)04x x x ⇔-+-≤≤ 后一个不等式显然成立,而前一个不等式2001x x x ⇔-≤⇔≤≤. 故当01x ≤≤时, 不等式1(21)112x x x +-≤+≤+成立. 1111(21)112n n n∴+-≤+<+, ……………………11分 1113221122n a n n n n∴+⋅≤=+++<+(等号仅在n =1时成立) 求和得: 32222n n n n T S n T +⋅≤<+⋅ 273252n n S n T -∴<≤< ……………………14分。
2012年佛山市普通高中高三教学质量检测(

2012年佛山市普通高中高三教学质量检测(二)4月18日数学(理科)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1.设全集}5,4,3,2,1{=U ,集合}2,1{=A ,}3,2{=B ,则=⋂)(B C A U ( ) A .}5,4{ B .}3,2{ C . }1{ D .}2{2.设向量a 、b 满足1||=a ,2||=b ,0)(=-⋅b a a ,则a 与b 的夹角为( ) A .︒30 B .︒60 C .︒90 D .︒1203.若0≥x ,0≥y ,且12=+y x ,则232y x +的最小值是( ) A .2 B .43 C .32D .04.已知a 、b 是实数,则“1||||<+b a ”是“21||<a 且21||<b ”的( ) A .充分不必要条件 B .必要不充分条件C .充要条件D .既不充分又不必要条件5.函数xxy sin =(),0()0,(ππ⋃-∈x )的图象是( ).6.已知直线m 、l 与平面α、β、γ满足l =⋂γβ,α//l ,α⊂m ,γ⊥m ,下A .B . D .C .列命题一定正确的是( )A .γα⊥且m l ⊥B .γα⊥且β//mC .β//m 且m l ⊥D . βα//且γα⊥7. 如图所示为函数)sin(2)(ϕω+=x x f(0>ω,πϕ≤≤0A 、B 之间的距离为5,则=)1(f ( )A .2 B .3C .3- D . 2-8.已知函数)(x f M 的定义域为实数集R ,满足⎩⎨⎧∉∈=M x Mx x f M 01)(,(M 是R 的非空真子集),在R 有两个非空真子集A 、B ,且φ=⋂B A ,则1)()(1)()(+++=⋃x f x f x f x F B A B A 的值域为( )A .]32,0( B .}1{ C .}1,32,21{ D .]1,31[二.填空题:本大题7小题,每小题5分,满分30分 (一)必做题(9——13题)9.设i 为虚数单位,则5)1(i +的虚部为 .10.设x 、y 满足约束条件⎪⎩⎪⎨⎧≤-≥-≥1020y x y x x ,则y x z 2+=的最大值是 .11.抛掷一枚质地均匀的骰子,所得点数的样本空间为}6,5,4,3,2,1{=S ,令事件}5,3,2{=A ,事件}6,5,4,2,1{=B ,则)|(B A P 的值为 . 12.直线x y 2=和圆122=+y x 交于A 、B 两点,以Ox 为始边,OA 、OB 为终边的角分别为α、β,则)sin(βα+的值为 .13.已知等比数列}{n a 的首项为2,公比为2,则=⋅⋅⋅+nn a a a a a a a a 211.(二)选做题(14——15题,考生只能从中选做一题) 14.(坐标系与参数方程)在极坐标系中,已知射线3πθ=(0≥ρ)与曲线1C :θρsin 4=的异于极点的交点为A ,与曲线2C :θρsin 8=的异于极点的交点为B ,则=||AB .15.(几何证明选讲)如图,圆中两条弦AB 与CD 相交 于点F ,E 是AB 延长线上一点,且2==CF DF , 1:2:4::=BE FB AF ,若CE 与圆相切,则线段CE 的长为 .三.解答题:本大题共6小题,满分80分,解答须写出文字说明、证明过程或演算步骤 16.(本题满分12分)在四边形ABCD 中,2=AB ,4==CD BC ,6=AD ,︒=∠+∠180C A . (1)求AC 的长;(2)求四边形ABCD 的面积.17.(本题满分12分)空气质量指数5.2PM (单位:3/m g μ)表示每立方米空气中可入肺颗粒物某市2012年3月8日—4月7日(30天)对空气质量指数5.2PM 进行监测,获得数据得到如下条形图:(1)估计该城市一个月内空气质量类别为良的概率;(2)在上述30个监测数据中任取2个,设X 为空气质量类别为优的天数,求X 的分布列.A18.(本题满分14分)如图所示的四棱锥ABCD P -中,⊥PA 底面ABCD ,四边形ABCD 中,AD AB ⊥,AD BC //,2===BC AB PA ,4=AD ,E 为PD 的中点,F 为PC 的中点. (1)求证:⊥CD 平面PAC ; (2)证明://BF 平面ACE ;(3)求直线PD 与平面PAC 所成的角的正弦值. 19.(本题满分14分)已知椭圆E :12222=+by a x (0>>ba )的一个焦点为)0,3(1-F ,而且过点)21,3(H . (1)求椭圆E 的方程; (2)设椭圆E 的上下顶点分别为1A 、2A ,P 是椭圆上异于1A 、2A 的任意一点,直线1PA 、2PA 分别交x 轴于点N 、M ,若直线OT 与过点M 、N 的圆相P A F EB C D 1510一级二级三级四级级别切,切点为T ,证明:线段OT 的长为定值,并求出该定值.20.(本题满分14分)记函数1)1()(-+=n n x x f (2≥n ,*N n ∈)的导函数为)(x f n ',函数nx x f x g n -=)()(.(1)讨论)(x g 的单调区间和极值;(2)若实数0x 和正数k 满足:)()()()(1010k f k f x f x f n n n n ++='' ,求证:k x <<00.21.(本题满分14分)设曲线C :122=-y x 上的点P 到点),0(n n a A 的距离的最小值为n d ,若00=a ,12-=n n d a (*N n ∈).(1)求数列}{n a 的通项公式; (2)求证:222644212125331++-+++<+++n n n n a a a aa a a a a a a a ; (3)是否存在常数M ,对*N x ∈∀,都有不等式M a a a n<+++33231111 成立?说明理由.2012年佛山市普通高中高三教学质量检测(一)答案一.选择题:ABCC DBBA二.填空题:9.30 10.π2 11.1 12.49 13.3 14.213- 15.49π三.解答题: 16.(1)3π=B ,71])(cos[cos =-+=B C B C ;----------6分 (2)734sin =C ,利用正弦定理得8=c ,5=b ,310sin 21==∆B ac S ABC17.(2)方法一:利用向量;方法二:取FA 的中点G ,AB 的中点M ,连CM ,GM ,CG ,则面//CGM 面BEF ,AMG ∠是A CM G --的平面角,332cos 222=⋅-+=∠GM AM AG GM AM AMG ---------7分18.(1)1822412=+=μ;---------5分 (2)每支灯管使用12个月时已经损坏的概率为2.08.01=-,假设使用12个月时该功能室需要更换的灯管数量为η,则η服从)2.0,4(B ,至少两支灯管需要更换的概率6251132.08.08.01)1()0(11314404=⨯--==-=-=C C P P P ηη.19.(1)点P 的轨迹方程是32-=x y ;---------5分(2)点Q 在以)0,22(-A 、)0,22(B 为焦点,实轴长为4的双曲线的右支上,即点Q 的轨迹方程为422=-y x (2≥x ),⎩⎨⎧=--=43222y x x y ,得0131232=+-x x ,0<∆,方程组无解,所以不存在满足条件的点Q .-----------13分20.(1)01=++y x ;--------3分(2)函数)(x f 无零点⇔方程ax x =ln ,即xxa ln =在),0(∞+上无解,研究x x x g ln )(=,实数a 的取值范围是),1(∞+e;---------9分 (3)0ln 11=-ax x ,0ln 22=-ax x ,)(ln ln 2121x x a x x +=+,)(ln ln 2121x x a x x -=-,⇔>+⇔>+⇔>⋅2)(2ln ln 2121221x x a x x e x x 112ln 2ln 2ln ln 212121212121+->⇔+->⇔+>--t t t x x x x x x x x x x x x (21x x t =,1>t ),研究函数112ln )(+--=t t t t g ,易知)(t g 在),1(∞+上是增函数,0)1()(=>g t g ,所以 112ln +->t t t 成立,所以221e x x >⋅.-------14分21.(1)易得nn R n 1+=,n n a n 1111+++=;---------6分(2)2111111=+>+++=nn a n , 11111111111+=+++++>+++=n n a n n n n a ,21>>+n n a a ;---------9分 (3)证明不等式nn n 211111)12(1+<+≤-+,从而得 n n n a n n 2321111122+<+++=≤⋅+,求和得n n n T n S T n 23222+<≤+,所以 23257<-<n n T n S .----------14分。
2012年佛山市普通高中高三教学质量检测(二)理综及答案
2012年佛山市普通高中高三教学质量检测(二)理科综合化学试题一、单项选择题:本大题共6小题,每小题4分。
共24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. 下列哪种物质是用右图所示的运输方式进入细胞的?()A.CO2B.甘油C.氨基酸D.胰岛素2. 关于组成细胞的有机物,下列叙述正确的是()A.盘曲和折叠的肽链被解开时,蛋白质的特定功能可能会发生改变B.RNA与DNA的分子结构很相似,由四种核苷酸组成,但不能储存遗传信息C.葡萄糖和蔗糖都可以被细胞直接用于呼吸作用D.组成细胞膜的脂质是磷脂,不含胆固醇3. 甲至丁为二倍体生物卵巢中的一些细胞分裂图,有关判断正确的是()A.若图中所示细胞分裂具有连续性,则顺序依次为乙→丙→甲→丁B.甲、乙、丙细胞中含有的染色体组数目依次为4、2、1C.若乙的基因组成为AAaaBBbb,则丁的基因组成为AaBbD.乙是次级卵母细胞,丁可能为卵细胞4. 下列有关种群和群落的叙述不.正确的是()A.种间关系属于群落水平的研究内容B.随着时间的推移,弃耕的农田可能演替成森林C.群落中动物的垂直分层现象与植物有关D.出生率和死亡率是种群最基本的数量特征5. 某同学绘制的生态系统概念图如下,下列叙述不.正确的是()A.①表示生态系统的组成成分B.③越复杂,生态系统的抵抗力稳定性越强C.④中可能有微生物D.该图漏写了生态系统的某项功能6. 下列有关实验的叙述正确的是()A.经甲基绿染色的口腔上皮细胞,可在高倍镜下观察到蓝绿色的线粒体B.在噬菌体侵染细菌的实验中,用35S 标记噬菌体的蛋白质C.用于观察质壁分离与复原的洋葱表皮细胞也可以用来观察有丝分裂D.可用标志重捕法精确地得到某地野驴的种群密度7.下列说法不正确...的是A.天然气的主要成分是甲烷B.蛋白质、糖类物质都能发生水解反应C.煤的干馏是化学变化,石油的分馏是物理变化D.乙醇、乙烯、乙醛都可被酸性高锰酸钾溶液氧化8.下列离子反应方程式正确的是A.氨水吸收过量的SO2:OH-+SO2=HSO3-B.FeSO4溶液被氧气氧化:4Fe2++O2+2H2O=4Fe3++4OH-C.NaAlO2溶液中加入过量稀硫酸:AlO2-+H++H2O=Al(OH)3↓D.Cl2与稀NaOH溶液反应:Cl2+2OH-=Cl-+ClO-+ H2O9.N A为阿伏加德罗常数,下列叙述正确的是A.22.4L NH3中含氮原子数为N AB.1 mol Na2O2与水完全反应时转移的电子数为N AC.1 L 0.1mol·L-1碳酸钠溶液的阴离子总数小于0.1 N AD.1 mol O2和2 mol SO2在密闭容器中充分反应后的分子数等于2N A10.下列说法正确的是A.用盐析法分离NaCl溶液和淀粉胶体B.工业制硫酸的吸收塔中用水吸收SO3C.加足量的稀盐酸可除去BaCO3固体中少量的BaSO4D.向硝酸银稀溶液中逐滴加入稀氨水至白色沉淀恰好溶解,即得银氨溶液11.A、B、C为短周期元素,A的最外层电子数是次外层的3倍,B是最活泼的非金属元素,C的氯化物是氯碱工业的原料,下列叙述正确的是A.A是O,B是ClB.A、B、C的原子半径大小关系是:A>C>BC.B的气态氢化物比A的稳定D.向AlCl3溶液中加过量C的最高价氧化物对应水化物可得白色沉淀12.有关右图的说法正确的是移向Cu极A.构成原电池时溶液中SO 24B.构成原电池时Cu极反应为:Cu﹣2e-=Cu2+C.构成电解池时Fe极质量既可增也可减D.a和b分别接直流电源正、负极,Fe极会产生气体二. 双项选择题:本大题共2小题,每小题6分,共12分。
佛山市2012年普通高中高三教学质量检测一
16.请联系全文,简要概述百合花的特点。(4分) ①美丽、芬芳、妩媚;②清纯高雅(超凡脱俗、不附庸风 雅、天然淡泞)、矜持含蓄;③羸弱轻柔却具有顽强的生 命力和激进的灵魂。 17.⑤段划线句子使用的是什么修辞手法?请分析其表 达效果。(5分) 比喻(博喻)。(1分)作者连用四个喻体形象写出百合
花的美丽,“月下的霜雪”从整体上写出花朵颜色的乳白、
动向泰国警方投案自首,至此,震惊东南亚的“10.5湄公河惨案”
(A、变幻莫测:变化很多,不能预料。B、形容事情错综复杂, 不易识别;C、莫衷一是:不能决定哪个是对的。形容意见分歧, 没有一致的看法。D、呼之欲出:形容人像画得逼真,似乎叫一声 就会从画中走出来。泛指文学作品中人物的描写十分生动。这里属 于望文生义,可改为“水落石出”)
句式杂糅,“原因是……”与“由……所导致的” 杂糅)
4.把下列句子组成语意连贯的语段,排序最恰当的一项 是( C ) ①正如人是由动物进化而来的一样,动物和人都是 有情绪与情感的 ②在论述人类的艺术时,人们常说艺术是发展、继 承和创新的 ③而人类的艺术就起源于人类对情绪与情感作用的 认识与需要 ④艺术是用来表现人的情绪与对某种事物的情感 ⑤然而,如果我们把人类的艺术看作一个整体,人 类的艺术不是上帝的作品,而是从动物的“艺术”进化而 来 ⑥这是艺术的一般功能,同时这也是艺术具有其他 各种功能的基础 A.④①⑤③⑥② B.①⑤④⑥②③
动物的鸣叫声与下列因素有关:①求偶;②群体 的大小;③对象的不同;④对陌生世界的好奇; ⑤争取食物; ⑥环境噪声。
21.作者以“你能听到我吗”为题有何作用?请 结合全文分析。(6分)
①用询问的语气写出了动物想要在周围的噪声中 突出自己,从而使读者产生阅读的兴趣。 ②用第二人称对话的方式,拉近了与读者的距离, 富有亲切感; ③含蓄地表达希望人类减少噪声污染,给动物提 供一个安静的环境的主旨。
2012年佛山市普通高中高三教学质量检测(二)文科数学试题答案4.19
2012年佛山市普通高中高三教学质量检测(二)数学试题(文科)参考答案和评分标准一、选择题:本大题共10小题,每小题5分,满分50分.11.e12.(1,)-+∞13.114.15三、解答题:本大题共6小题,满分80分,解答须写出文字说明、证明过程或演算步骤.16.(本题满分12分)解析:(1)由条形统计图可知,空气质量类别为良的天数为16天,-------------------------1分所以此次监测结果中空气质量类别为良的概率为1683015=.-------------------------4分(2)样本中空气质量级别为三级的有4天,设其编号为1234,,,a a a a;样本中空气质量级别为四级的有2天,设其编号为12,b b.-------------------------6分则基本事件有:12(,)a a,13(,)a a,14(,)a a,11(,)a b,12(,)a b;23(,)a a,24(,)a a,21(,)a b,22(,)a b;34(,)a a,31(,)a b,32(,)a b;41(,)a b,42(,)a b;12(,)b b.共15个.(注:基本事件可以用树状图或列表等均正确)-------------------------9分其中至少有1天空气质量类别为中度污染的情况为:11(,)a b,12(,)a b,21(,)a b,22(,)a b,31(,)a b,32(,)a b,41(,)a b,42(,)a b,12(,)b b.共9个.所以至少有1天空气质量类别为中度污染的概率为93155=.-------------------------12分17.(本题满分12分)解析:(1)在Rt ABD∆中,DB,6ABDπ∠=,∴1AD=,3DABπ∠=,-------------------------3分又∵1DF=,∴ADF∆为等边三角形,故3ADFπ∠=,-------------------------5分2∴6πα=. -------------------------6分解法二、在BDF ∆中,sinsin 66DF DB ππα=⎛⎫+ ⎪⎝⎭,-------------------------2分即sin 62πα⎛⎫+=⎪⎝⎭, -------------------------3分 注意到03πα<<,所以63ππα+=,即6πα=. -------------------------6分坐标法可相应给分.(2)∵//AD BC ,AD DB ⊥,∴DB BC ⊥,则0AD DB ⋅=,0DB BE ⋅=, -------------------------8分∵AE AD DB BE =++, -------------------------10分 ∴22()3AE DB AD DB BE DB AD DB DB BE DB DB ⋅=++⋅=⋅++⋅==. -------------------12分 18.(本题满分14分)解析:(1)∵四边形ABCD 中,AB AD ⊥,//BC AD , ∴四边形ABCD 为直角梯形, ∴11()(24)2622ABCD S BC AD AB =+⨯=⨯+⨯=,-------------------2分 又PA ⊥底面ABCD ,∴四棱锥P ABCD -的高为PA , ∴1162433P ABCD ABCD V S PA -=⋅=⨯⨯=.-------------------4分 (注:只要结果对,就给满分)(2)∵PA ⊥底面ABCD ,CD ⊂平面ABCD ,∴PA CD ⊥, -------------------------6分 又∵直角梯形ABCD 中,AC ==CD =∴222AC CD AD +=,即AC CD ⊥, -------------------------8分又PA AC A =,∴CD ⊥平面PAC ; -------------------------9分文科试题参考答案 第 3 页 共 7 页(3)不存在.-------------------------10分 下面用反证法说明:假设存在点M (异于点C )使得//BM 平面PAD . 在四边形ABCD 中,//BC AD , ∵AD Ì平面PAD ,BC ⊄平面PAD , ∴//BC 平面PAD ,∵BM ⊂平面PBC ,BC ⊂平面PBC ,BC BM B =,∴平面//PBC 平面PAD而平面PBC 与平面PAD 相交,矛盾. -------------------------14分 (注1:有反证法思想的给1分; 注2:证明面面平行不严密扣1分)19.(本题满分14分) 解析:(1)由题意知,当10x =时, 28u =, ∴22158528(10)48k =--+,解得2k =,-------------------------3分 ∴年销售利润221585(6)(6)[2()]48y x u x x =-⋅=-⋅--+,-------------------------5分 (6,)x ∈+∞.-------------------------6分(2)由(1)知,221585(6)[2()]48y x x =-⋅--+ 2(6)(22118)x x x =-⋅-++,∴222118(6)(421)y x x x x '=-+++--+ -------------------------8分26(1118)6(2)(9)x x x x =--+=---,(6,)x ∈+∞ 令0y '=得,12x =(舍去)或29x =, -------------------------10分 当(6,9)x ∈时,0y '<;当(9,)x ∈+∞时,0y '>; -------------------------12分 因此,9x =是利润函数的极大值点,也是最大值点.故max 135y =万元.-------------------------13分 所以当售价为9元时,最大年利润为135万元. -------------------------14分20.(本题满分14分)解析:(1)解法1:由题意得222231341a b a b ⎧-=⎪⎪⎨⎪+=⎪⎩,-------------------------1分4解得24a =,-------------------------2分; 21b =.-------------------------3分所以椭圆的方程为2214x y +=. -------------------------4分 解法2:椭圆的两个焦点为1(F,2F ,由椭圆的定义可得12712||||422a HF HF =+==+=,(1分) ∴2a =,22221b =-=,(正确解出一个给1分)所以椭圆的方程为2214x y +=. -------------------------4分(2)解法1:由(1)知1(0,1)A ,2(0,1)A -,设00(,)P x y ,-------------------------5分直线1PA 的方程为0011y y x x --=,令0y =,得001N x x y -=-;-------------------------7分 直线2PA 的方程为0011y y x x ++=,令0y =,得001N xx y =+;-------------------------9分 设圆G 的圆心为00001((),)211x x h y y -+-, -------------------------10分则22222000000000011[()]()2111411x x x xx r h h y y y y y =--+=+++-++-,-------------------------11分22200001()411xx OG h y y =-++-,22222220000000011()()411411x x x x OT OG r h h y y y y =-=-+-+-+-+-2021x y =-, -----12分 而220014x y +=,即22004(1)x y =-,-----13分 所以220204(1)4(1)y OT y -==-,所以||2OT =.即线段OT 的长为定值2. -------------------------14分 解法2:由(1)知1(0,1)A ,2(0,1)A -,设00(,)P x y ,------------------------5分文科试题参考答案 第 5 页 共 7 页直线1PA 的方程为0011y y x x --=,令0y =,得001N x x y -=-;-------------------------7分直线2PA 的方程为0011y y x x ++=,令0y =,得001N xx y =+;-------------------------9分则20002000||||||||111x x x OM ON y y y -⋅=⋅=-+-,而220014x y +=,即22004(1)x y =-,∴||||4OM ON ⋅=;取线段MN 的中点Q ,连接,,GQ GM GO .||r GM =,2222222()()OT OG GM OQ QG MQ QG =-=+-+22(||)(||||)OQ MQ OQ MQ OQ MQ =-=+-||||4OM ON =⋅=∴||2OT =.即线段OT 的长为定值2. -------------------------14分 (注:可不用证明切割线定理) 解法3:过点P 作PR x ⊥轴于点R .设00(,)P x y ,则12||||1OA OA ==,0||||OR x =,0||||RP y =,------------------------5分 由2PRM A OM ∆∆得2||||||||PR RM A O OM =,即00||||||1||y x OM OM -=,则00||||||1x OM y =+; -------8分 由1PRNAON ∆∆得1||||||||PR RN AO On =,即00||||||1||y ON x ON -=,则00||||1||x ON y =-; ---------11分而220014x y +=,即22004(1)x y =-,-------------------------12分 ∴20002000||||||||4||11||1x x x OM ON y y y ⋅=⋅==+--.-----------13因直线OT 与圆G 相切,且OMN 是圆G 的一条割线,所以2||||4OT OM ON =⋅=,所以||2OT =.即线段OT 的长为定值2. -------------------------14分621.(本题满分14分)解析:(1)设点(,)P x y ,则221x y -=,∴n PA =分==-------------------------2分 ∵y ∈R ,∴当2n a y =时,n PA 取最小值n d,且n d =-------① ------------------3分又1n n a -,∴1n n a +=,1n n d +=,将1n n d +=1n +=-------------------------4分 两边平方得:2212n n a a +-=,又00a =,212a =,-------------------------5分 故数列{}2n a 是首项212a =公差为2的等差数列,∴22n a n =.∵10n n a ->,∴n a = -------------------------6分(2)由点到直线距离公式得n t =,-------------------------7分∴n t ==,=<0+>,∴n t =-------------------------9分 ∴1211()112123n t tt n n+++=+++++-+,-------------------------10分<=-------------------------12分∴112+<L ,故111(212+<L,-------------------------13分∴121122nt t t+++=+<L.-------------------------14分文科试题参考答案第7 页共7 页。
广东省佛山市2012届高三教学质量检测一文综试题版
广东省佛山市2012年普通高中高三教学质量检测(一)文科综合试题本试卷共10页,41小题,满分300分。
考试时间150分钟注意事项:1.答卷前,考生务必填涂好答题卷和选择题答题卡的相关信息。
2.选择题每小题选出答案后,用2B铅笔把选择题答题卡上对应题目选项的答案信息点涂黑。
3.非选择题必须用黑色字迹钢笔或签字笔作答,答案必须写在答题卷各题目指定内相应位置上;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。
一、选择题:本题共35小题,每小题4分,共140分。
每小题给出的四个选项中,只有一个选项符合题目要求。
1.下列四图(图1)中能够正确表示地理事物和现象之间关系的是()图1A.①B.②C.③D.④2.我国现行流通人民币20元的背面图案(图2)取景于广西桂林,该景观()A.天然植被为落叶阔叶林B.山体的基岩是岩浆岩C.山体连绵,是褶皱山D.受流水化学和物理作用图2 图33.读某区域流域图(图3),能导致该区域海岸线变迁的人类活动是()A.发展海水养殖业B.营造沿海防护林C.大量开采地下水D.河流中上游修建水库读“我国某两省份第六次人口普查数据统计表”,答4—5题。
4.由表中信息直接可以反映出甲省比乙省()A.城市化水平更高B.人口素质更高C.老龄化问题更突出D.人口合理容量更大5.2011年夏季,乙省某城市多次出现严重内涝现象。
它的形成与下列人类活动无关的是()A.城市空气中尘埃多,增加暴雨形成机率B.城市人工绿地面积增多,使地下水位上升C.城市“热岛效应”,使大气对流运动增强D.城市建设使地面硬化,地表水下渗能力弱6.我国房价变动引起社会关注,图4中影响保障性住房布局的主导因素是()A.土地价格和交通条件B.环境因素和土地价格C.交通条件和环境因素D.土地利用效益和地租支付能力图4 图5泰国兰花生产条件优越。
2011年1月,泰国农业部推出2011―2016泰国兰花发展战略计划。
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2012年佛山市普通高中高三教学质量检测(二)英语2012.4本试卷共11页,满分135分,考试时间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考号填写在答题卡上.用2B铅笔将答题卡试卷类型(A)填涂在答题卡上,并在答题卡右上角的“试室号”和“座位号”栏填写试室号、座位号,将相应的试室号、座位号信息点涂黑。
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,答案答在试题卷上无效。
3、非选择题必须用黑色字迹的钢笔或签字笔作答,答案必须写在答卷上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。
4、考试结束后,将答卷和答题卡一并交回。
I 语言知识及应用(共两节,满分45分)第一节完形填空(共15小题;每小题2分,满分30分)阅读下面短文,掌握其大意,然后从1~15各题所给的A、B、C和D项中,选出最佳选项,并在答题卡上将该项涂黑。
Every summer, hundreds of thousands of students travel to other countries looking for work and adventure. Most of the opportunities are in 1 work. The pay is usually poor, but most people work 2 for the thrill of travel. You can pick grapes in France, entertain kids on American summer camps, and, of course, there are always 3 in hotels and restaurants.But it is not as easy as it used to be to find work. Unless you speak the language of the country well, there will be very 4 openings. For example, when you arrive to wash dishes in a restaurant in Paris, the owner will 5 you to speak French. British students only have a language 6 for jobs in the USA and Australia.Not every one 7 the experience. Sarah James was once responsible for forty American children in Europe. During the 8 , one child lost his passport; four children were lost in Madrid for a whole day; the whole group was thrown out of one hotel because of the 9 they made. Sarah says, “It really was a 24-hour-a-day job since the kids never 10 ! And the pay was awful. It wasn’t worth it.”The trouble is that 11 expect to have an easy time of it. After all, they see it as a 12 . In practice, though, they have to work hard. At the same time, all vacation work is casual work, and jobs are 13 only when the hotel, the restaurant, or the campsite is busy. But students have few employment 14 . As soon as the holiday season finishes, companies will get rid of them. And if their employer doesn’t like them, they’ll be 15 , too.1. A. seasonal B. mental C. professional D. formal2. A. hard B. voluntarily C. abroad D. continuously3. A. customers B. visitors C. jobs D. parties4. A. good B. new C. attractive D. few5. A. teach B. expect C. allow D. forbid6. A. program B. lesson C. advantage D. exam7. A. has B. enjoys C. forgets D. remembers8. A. trip B. flight C. discussion D. ceremony9. A. promise B. progress C. complaint D. noise10. A. cried B. studied C. slept D. helped11. A. children B. students C. employers D. parents12. A. job B. lesson C. holiday D. shame13. A. countless B. available C. interesting D. boring14. A. experiences B. rules C. plans D. rights15. A. dismissed B. charged C. fined D. punished第二节语法填空(共10小题; 每小题1.5分,满分15分)阅读下面短文,按照句子结构的语法性和上下文连贯的要求,在空格处填入一个适当的词或使用括号中词语的正确形式填空,并将答案填写在答题卷标号为16~25的相应位置上。
The International Red Cross and Red Crescent(弯月) Museum was opened in Geneva in 1988. 16 tells the story of men and women who, in the course of the major events of the last 150 years, have given assistance 17 victims of wars and natural disasters.The organization 18 (found) in 1863, and was based on an idea by a Swiss businessman called Henry Dunant. He had witnessed too many 19 (die) and wounds at the Battle of Solferino in Italy four years 20 (early), in which 40,000 people were killed, wounded or missing. He had seen the lack of medical services and the great suffering of many of the wounded, 21 simply died from lack of care. The International Red Cross/Red Crescent exists 22 (help) the victims of conflicts and disasters regardless of their nationality.23 symbol of the organization was originally just the red cross. It has no religious significance; the founders 24 the movement adopted it in honor of Switzerland. However, the original symbol, the red cross, could hurt Muslim soldiers’feelings, 25 a second symbol, the red crescent, was used. Both are now official symbols.II 阅读(共两节,满分50分)第一节阅读理解(共20小题;每小题2分,满分40分)阅读下列短文,从每题所给的A、B、C和D项中,选出最佳选项,并在答题卡上将该项涂黑。
AKelly Reeves was getting ready for a trip when her phone slipped into a sink full of water. Panic moment! She quickly picked up the wet phone and tried to turn it on, but nothing worked. Her first reaction? She got dressed, drove to the nearest store, and bought a new model at full price.A new study finds that fear of losing your phone is a common illness. About 66 percent of those surveyed suffer from nomophobia or “no mobile phone phobia”. Interestingly, more women worry about losing their phone than men.Fortunately, there’s a solution.The first step is to figure out if you have nomophobia. Checking your phone too often is one thing, but the true sign of a problem is that you can’t conduct business or go about your routine when the fear becomes so severe.Do you go to unusual lengths to make sure you have your phone? That’s another sign of a problem. If you find you check your phone plenty of times per hour, or a total of an hour per day, there may be a problem.Some of the treatments are similar to those for treating anxiety attacks: Leaving the phone behind and not checking e-mail or text messages, and then learning to tolerate the after anxiety. Even if this leads to a high level of worry and stress, the solution is to push through the fear and learn to deal with not having your phone.Of course, there are also technological alternatives. Luis Levy, a co-founder at Novy PR, says he uses an application called Cerberus that can automatically track the location of his phone. To find it, he can just go to a Web site and see the phone’s location.He also insures his phone through a service called Asurion. The company’s description of its product reads like a prescription for anxiety: “60 million phones are lost, stolen or damaged each year. You’ll have complete peace of mind knowing that your phone is protected and you can quickly reconnect with family, friends and work, as soon as the very next day!”26. Why does the author mention Kelly’s experience in the first paragraph?A. To introduce the topic for discussion.B. To inform us that mobile phones are useful.C. To warn us that we should be careful.D. To tell us we should get phones ready for a trip.27. The underlined word “nomophobia” in Paragraph 2 means ________.A. Habits of using mobile phones.B. Fear of losing mobile phones.C. Eagerness for new mobile phones.D. Independence of mobile phones.28. Which of the following is a way to treat nomophobia?A. Avoiding using phone for some timeB. Learning more about modern technology.C. Protecting one’s phone against any damage.D. Not using a mobile phone in one’s daily work.29. Why can the service called Asurion help to treat nomophobia?A. It lets you know other people also lose their phones.B. It will give you a new phone through insurance.C. It enables you to reconnect with your acquaintance.D. It gives you a prescription to treat nomophobia.30. What is the passage mainly about?A. Attitude toward mobile phone.B. New mobile phone technology.C. Disadvantages of mobile phone.D. Solutions to nomophobiaBCarbon monoxide(一氧化碳) poisoning kills and injures many people and animals around the world. The gas has been a problem since people first began burning fuels to cook food or to create heat. It is a problem in all parts of the world that experience cold weather.Carbon monoxide is called the silent killer because people do not know it is in the air. The gas has no color. It has no taste. It has no smell. It does not cause burning eyes. And it does not cause people to cough. But it is very deadly. It robs the body of its ability to use oxygen.Carbon monoxide decreases the ability of the blood to carry oxygen to body tissues. It does this by linking with the blood. When the gas links with the blood, the blood is no longer able to carry oxygen to the tissues that need it.Damage to the body can begin very quickly from large amounts of carbon monoxide. How quickly this happens depends on the length of time a person is breathing the gas and the amount of the gas he or she breathes in.Carbon monoxide poisoning has warning signs. But people have to be awake to recognize them. Small amounts of the gas will cause a person’s head to hurt. He or she may begin to feel tired. The person may feel sick. The room may appear to be turning around. The person may have trouble thinking clearly. People develop severe head pain as the amount of gas continues to enter their blood. They will begin to feel very tired and sleepy. They may have terrible stomach pains.Medical experts say carbon monoxide affects people differently. For example, a small child will experience health problems or die much quicker than an adult will. The general health of the person or his or her age can also be important. An older adult with health problems may suffer the effects of carbon monoxide more quickly than a younger person with no health problems. People with heart disease may suffer chest pains. They may begin to have trouble breathing.31. Why is carbon monoxide called the silent killer?A. Because it tastes and smells good.B. Because it is not easily noticed.C. Because it kills and injures people.D. Because it always harm people.32. How does carbon monoxide harm people?A. It makes people’s blood unable to move.B. It decreases the amount of blood in the body.C. It makes body tissues full of blood.D. It makes the blood less able to carry oxygen.33. When people breathe in small amounts of the gas, they may ________.A. feel a little dizzyB. suffer a severe headacheC. go around in the roomD. have a terrible stomachache34. Which of the following about carbon monoxide poisoning is true?A. Adults are affected more seriously than children.B. Young people are more severely affected than old people.C. People in poor health may have more severe consequences.D. People with heart problem only suffer from chest pains.35. The purpose of the passage is to ________.A. warn people not to burn fuels to keep warm in the winterB. list the damages that carbon monoxide brings to peopleC. give advice on how to avoid carbon monoxide poisoningD. introduce some knowledge about carbon monoxide poisoningCMy wife Julie and I were out on the road that runs around where we live. Two dog walkers passed by and we heard one of them say, “Never seen that dog around here before.”We looked down the hill and saw an old black dog stumbling(蹒跚) painfully up the hill. We bent down, talked gently to the dog and patted it. I checked and there was a collar. There was a phone number on the collar, but no one answered.The dog was painfully thin. It didn’t have many teeth left and, well, it just seemed done.Julie ran home to get some of our dog’s food while I tried to encourage the dog to come along with me. Julie got home and back again when I was only half way there. So, we sat down on the sidewalk while our new friend made short work of the food.Eventually we got her home. We gave her a blanket, more food, kept her warm and wondered what to do next.After trying for many times we got a response from the number. A lady came around with a bunch of flowers for us. She explained that Tara had been her father’s dog. She was very old and should probably be put to sleep, but the lady just couldn’t bring herself to do it.So, Tara was safely returned home.Here is the truth of the story:Actually Julie and I were out that morning because I was leaving. She was trying to persuade me to come back, but I wasn’t hearing anything that made that sound likely.I was about to turn and go when an old, worn out dog walked between us and collapsed. Suddenly we had something more important than our problem to worry about. There was a creature in need right before us and we had to work together to help it.We did help it. And here am I writing the story in my own home, in my own family.In the song “Love Is Not a Fight” Warren Barfield talks about marriage. At one point he sings, “And if we try to leave, may God send angels to guard the door.”Sometimes angels come disguised (伪装) as dogs.36. Which of the following statements is true?A. The two dog walkers didn’t like the dog.B. The old black dog was sick and weak and couldn’t walk.C. After finding the dog, the writer phoned its owners.D. Most of the dog’s teeth had been pulled out by someone.37. We can infer from Paragraph 4 that ________.A. Julie bought some food for the dogB. the writer had his own pet dogC. the dog didn’t go with the writerD. the dog didn’t eat any of the food38. After they got the dog home, the couple ________.A. decided to raise the dogB. took good care of the dogC. found out that its owner was a ladyD. found it hard to get to sleep39. The underlined part “our problem” in the passage refers to the fact that ________.A. the writer didn’t want to take Julie’s adviceB. Julie disagreed with the writer’s travel on businessC. an old and worn out dog appeared in front of themD. the couple had some trouble with their marriage40. Which of the following is the best title for the passage?A. A Helpful Couple .B. An Angel Dog.C. Saving the Dog.D. A Famous Song.DDo you know what Fear and Faith have in common? Fear believes in a negative future. Faith believes in a positive future. Both believe in something that has not yet happened. So I ask you, if neither the positive nor negative future has happened yet then why not choose to believe in the positive future?I believe during these challenging times we have a choice between two roads: the positive road and the negative road. And our bus can’t b e on two roads at the same time. So we have to make a choice and this choice determines our belief about the future and the attitude and actions we bring to the present.I’m not saying we shouldn’t have any fear. There are times when fear is a gift. A heal thy dose(剂量) of fear causes us to examine our situation and plan for the future. It moves us to smell the cheese and expect change. When used wisely it allows us to manage risk and make better decisions. Some fear is good.However, what I have observed lately is a supersized, huge dose of fear that is spreading the hearts and minds of far too many people. This oversized fear is causing leaders and their organizations to either act unreasonably or to not act at all. They are either hurrying in a million different directions because of fear. In both cases, fear is leading them to take the negative road to failure.The answer is the positive road paved (铺满) with faith and a belief that your best days are not behind you but ahead of you. With this belief you make the right choices and actions today that will create your positive future tomorrow. You stay calm, focused and committed to your purpose. You look for ways to save money and cut costs without making unreasonable fear-based decisions that sacrifice your future success. You identify opportunities in the midst of the challenges and you focus on solutions instead of problems.Your faith and belief in a positive future leads to powerful actions today. The future has not happened yet and you have a say in what it looks like by the way you think and act. Fear or Faith.The choice is yours.41. According to the passage, we can know that ________.A. Fear and Faith have nothing in commonB. both Fear and Faith trust something in the futureC. neither negative nor positive future will happenD. people usually choose to believe in the positive future42. What does the writer mean by saying “our bus can’t be on two roads”?A. We usually choose the negative roadB. We should choose the positive roadC. We must choose either a positive or negative roadD. Our belief about the future depends on our choice43. Which of the following about “Fear” is true?A. We should have fear because it is a gift.B. Reasonable fear helps to plan for future.C. Fear can bring cheese and risk to people.D. Leaders usually have oversized fear.44. If we have faith in future, we will ________.A. have the best days ahead of usB. be committed to saving moneyC. stay away from any challengesD. take powerful actions today45. What does the writer intend to tell us in the passage?A. Fear is closely related to faith.B. Life is full of fear and faith.C. We should have faith in a positive future.D. Wrong decisions sacrifice our future success.第二节信息匹配(共5小题;每小题2分,满分10分)阅读下列应用文及相关信息,并按照要求匹配信息。