Removed_2014广州市小联盟试卷以及答案
广州小升初小联盟语文真题及答案.doc
2013 年广州小升初小联盟语文真题及答案．年广州市民办初中（五校联盟）新生入学检测（语文）2013注意事项： 60 全卷共两大大页，六大题，时间： 100 分。
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13 分）一、词语积累（ 1 、给加点的字注音或按拼音写汉字。
）（3）（2）瞻仰（）稠（1 密（）．．）（）脍炙人口（（4）出类拔萃．．）（（7）à（ú（5）f）晓6)荡yng旁征）引 bó（答案填 (2、下面加点字读音完全正确的一项是 )( )(2 在前面的括号里分）ch(di zh 立伫 A、（ù）腼腆视怒目嗔（ǎn) ．．． ně）2B、鞭笞（ t à）栖居（得益彰（ zh．．．āng）C、皎洁（ ji ǎo）蜿蜒(y qì）án)相三缄其口（ji ．．．ān）D、禀告（ bǐn）朔风 (s ù）绮丽风光（ y．．．ǐ））3、下列各组没有错别字的一项是（（答案填在前面的括号里）（2分）A、啜泣隐秘物竞天择因地治宜B 、狼籍阔绰捉襟见肘头晕目眩好高鹜远契而不舍枯洞、悲呛 C 义愤填膺魅力相形见绌 D、闪烁、下列的四字词语与描述的事物不对应的一项是 4 分）(2( )（答案填在前面的括号里）火伞高涨A、季节：莺歌燕舞——春天霜天红叶——秋天——夏天舂秋鼎盛 B、年龄：鹤发童颜——老年——中年两小无猜——童年不绝如缕、声音：响切云霄——响亮 C——雄浑万籁俱静——安静杏花烟雨、地理：赤县神州——中国D ——江南百草黄边——边塞 3二、按要求回答问题。
（17 分）画；1、“噬”字用部首查字法应先查部，再査用音序检字法应先査字母。
（3 分）、判断下列句子变换是否恰当，在括号内用“√”或2“×”标识。
（ 4 分））根据创作者霍夫曼的形容，“大黄鸭”无国界之（ 1 分，可以放轻松、治疗心灵。
2014广州小升初面试模拟题(5套)
第一套题题目一、自我介绍 Could you introduce yourself in English ?（10%）题目二、回答问题：（15%）1.你学习成绩怎样？2. 小升初想考哪所中学？3. 中国现任主席是谁？题目三、英语对话（10%）（问两遍）1. What time do you usually have lunch?2. How long does it take you to go to school from your home?3. What did you do for your mother on Mother’s Day last year?4. Which subject do you think is easier, English or maths?题目四、一分钟演讲题：评价一下爸爸和妈妈的一次冲突（10%）题目五、数学题（5%（提供稿纸和笔，稿纸没带走的扣两个桶里共盛水40斤，若把第一桶里的水倒7斤到第2个桶里，两个桶里的水就一样多，则第一桶有______斤水．题目一、自我介绍 Would you please tell us something about you in English? （10%）题目二、回答问题：（15%）1.你父母生日是哪一天？2. 小升初想考哪所中学？3. 美国现任总统是谁？题目三、英语对话（10%）（问两遍）1. What’s the date today?2. What did you have for breakfast today?3. How often do you go traveling with you family?4. Which do you prefer, tea or milk ?题目四、一分钟演讲题：我的业余生活（10%）题目五、数学题（5%）（两分钟内完成）（提供稿纸和笔，稿纸没带走的扣良好教养分）一个周长为20厘米的大圆内有许多小圆，这些小圆的圆心都在大圆的一个直径上．则小圆的周长之和为______厘米．题目一、自我介绍 Could you introduce yourself in English? （10%）题目二、回答问题：（15%）1.你父母是做什么工作的？2. 小升初想考哪所中学？3. 火警电话是什么？题目三、英语对话（10%）（问两遍）1. How do you go to school everyday?2. What did you have for breakfast this morning?3. How often do you help with the housework?4. Which sport do you like better, basketball or football?题目四、一分钟演讲：当朋友误会你了怎么办（10%）题目五、数学题（5%（提供稿纸和笔，稿纸没带走的扣某次数学竞赛，试题共有10道，每做对一题得8分，每做错一题倒扣5分．小宇最终得41分，他做对______题．题目一、自我介绍 Would you please tell us something about you in English? （10%）题目二、回答问题：（15%）1.你的特长是什么，能表演一下吗2. 小升初想考哪所中学？3. 110是什么电话号码？题目三、英语对话（10%）（问两遍）1. What day is today?2. How many students are there in your class?3. How did you go to school yesterday?4. Which sports do you like better, swimming or running?题目四、一分钟演讲题：谈谈你最尊敬的一个人（10%）题目五、数学题（5%（提供稿纸和笔，稿纸没带走的扣某工厂，三月比二月产量高20％，二月比一月产量高20％，则三月比一月高______％．题目一、自我介绍 Could you introduce yourself in English? （10%）题目二、回答问题：（15%）1.你的邻桌是个怎样的人，说说他（她）的优缺点2. 小升初想考哪所中学？3. 中国的全称是什么？题目三、英语对话（10%）（问两遍）1. What was the weather like yesterday?2. How often do you watch TV?3. How long did it take you to go back home from school yesterday?4. What sport do you think is more popular, football or table tennis?题目四、一分钟演讲：心目中的中学（10%）题目五、数学题（5%（提供稿纸和笔，稿纸没带走的扣20名乒乓球运动员参加单打比赛，两两配对进行淘汰赛，要决出冠军，一共要比赛______场．。
2014年广州市一模数学试题答案(文科发排稿2)
数学（文科）试题参考答案及评分标准 第 1 页 共 9 页2014年广州市普通高中毕业班综合测试（一）数学（文科）试题参考答案及评分标准说明：1．参考答案与评分标准给出了一种或几种解法供参考，如果考生的解法与参考答案不同，可根据试题主要考查的知识点和能力比照评分标准给以相应的分数．2．对解答题中的计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的得分，但所给分数不得超过该部分正确解答应得分数的一半；如果后继部分的解答有较严重的错误，就不再给分．3．解答右端所注分数，表示考生正确做到这一步应得的累加分数．4．只给整数分数，选择题和填空题不给中间分．一、选择题：本大题考查基本知识和基本运算．共10小题，每小题，满分50分．二、填空题：本大题考查基本知识和基本运算，体现选择性．共5小题，每小题，满分20分．其中14~15题是选做题，考生只能选做一题．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． 16．（本小题满分）（本小题主要考查古典概型等基础知识，考查化归与转化的数学思想方法，以及数据处理能力与应用意识）（1）解：记“从6瓶饮料中任意抽取1瓶，抽到没过保质期的饮料”为事件A ，从6瓶饮料中中任意抽取1瓶，共有6种不同的抽法．因为6瓶饮料中有2瓶已过保质期，所以事件A 包含4种情形． 则()4263P A ==． 所以从6瓶饮料中任意抽取1瓶，抽到没过保质期的饮料的概率为23． （2）解法1：记“从6瓶饮料中随机抽取2瓶，抽到已过保质期的饮料”为事件B ，随机抽取2瓶饮料，抽到的饮料分别记为x ，y ，则),(y x 表示第一瓶抽到的是x ，第二瓶抽到的是y ，则),(y x 是一个基本事件．由于是随机抽取，所以抽取到的任何基本事件的概率相等．不妨设没过保质期的饮料为1，2，3，4，数学（文科）试题参考答案及评分标准 第 2 页 共 9 页已过保质期的饮料为a ，b ，则从6瓶饮料中依次随机抽取2瓶的基本事件有：()1,2，()1,3，()1,4，()1,a ，()1,b ，()2,1，()2,3，()2,4，()2,a ，()2,b ， ()3,1，()3,2，()3,4，()3,a ，()3,b ，()4,1，()4,2，()4,3，()4,a ，()4,b ， (),1a ，(),2a ，(),3a ，(),4a ，(),a b ，(),1b ，(),2b ，(),3b ，(),4b ，(),b a ．共30种基本事件．由于2瓶饮料中有1瓶已过保质期就表示抽到已过保质期的饮料，所以事件B 包含的基本事件有：()1,a ，()1,b ，()2,a ，()2,b ，()3,a ，()3,b ，()4,a ，()4,b ，(),1a ，(),2a ， (),3a ，(),4a ，(),a b ，(),1b ，(),2b ，(),3b ，(),4b ，(),b a ．共18种基本事件． 则183()305P B ==． 所以从6瓶饮料中随机抽取2瓶，抽到已过保质期的饮料的概率为35． 解法2：记“从6瓶饮料中随机抽取2瓶，抽到已过保质期的饮料”为事件B ， 随机抽取2瓶饮料，抽到的饮料分别记为x ，y ，则),(y x 是一个基本事件．由于是随机抽取，所以抽取到的任何基本事件的概率相等．不妨设没过保质期的饮料为1，2，3，4， 已过保质期的饮料为a ，b ，则从6瓶饮料中随机抽取2瓶的基本事件有：()1,2，()1,3，()1,4，()1,a ，()1,b ，()2,3，()2,4，()2,a ，()2,b ，()3,4， ()3,a ，()3,b ，()4,a ，()4,b ，(),a b ．共15种基本事件．由于2瓶饮料中有1瓶已过保质期就表示抽到已过保质期的饮料，所以事件B 包含的基本事件有：()1,a ，()1,b ，()2,a ，()2,b ，()3,a ，()3,b ，()4,a ，()4,b ，(),a b ．共9种基本事件． 则93()155P B ==． 所以从6瓶饮料中随机抽取2瓶，抽到已过保质期的饮料的概率为35．数学（文科）试题参考答案及评分标准 第 3 页 共 9 页（本小题主要考查三角函数图象的周期性与单调性、同角三角函数的基本关系、三角函数的化简等知识，考查化归与转化的数学思想方法，以及运算求解能力）解：（1）因为函数()sin cos f x x a x =+的图象经过点π03⎛⎫- ⎪⎝⎭，，所以03f π⎛⎫-= ⎪⎝⎭． 即ππsin cos 033a ⎛⎫⎛⎫-+-= ⎪ ⎪⎝⎭⎝⎭．即02a+=．解得a =（2）由（1）得，()sin f x x x =+12sin 2x x ⎛⎫= ⎪ ⎪⎝⎭2sin cos cos sin 33x x ππ⎛⎫=+ ⎪⎝⎭π2sin 3x ⎛⎫=+ ⎪⎝⎭．所以函数()x f 的最小正周期为2π． 因为函数sin y x =的单调递增区间为2,222k k ππ⎡⎤π-π+⎢⎥⎣⎦()k ∈Z ， 所以当πππ2π2π232k x k -≤+≤+()k ∈Z 时，函数()x f 单调递增， 即5ππ2π2π66k x k -≤≤+()k ∈Z 时，函数()x f 单调递增．所以函数()x f 的单调递增区间为5ππ2π,2π66k k ⎡⎤-+⎢⎥⎣⎦()k ∈Z ．数学（文科）试题参考答案及评分标准 第 4 页 共 9 页（本小题主要考查空间线面关系、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） （1）证明：连结11B D ，BD ，因为四边形1111A B C D 是正方形，所以1111AC B D ⊥． 在正方体1111ABCD A B C D -中，1DD ⊥平面1111A B C D ，11AC ⊂平面1111A B C D ，所以111AC DD ⊥．因为1111B D DD D = ，11B D ，1DD ⊂平面11BB D D ， 所以11AC ⊥平面11BB D D ．因为EF ⊂平面11BB D D ，所以11EF AC ⊥． （2）解：取1C C 的中点H ，连结BH ，则BH AE ．在平面11BB C C 中，过点F 作FG BH ，则FG AE ． 连结EG ，则A ，E ，G ，F 四点共面．因为11122CH C C a ==，11133HG BF C C a ===， 所以1C G 116C C CH HG a =--=．故当1C G 16a =时，A ，E ，G ，F 四点共面．（3）解：因为四边形EFBD 是直角梯形，所以几何体ABFED 为四棱锥A EFBD -．因为()2113222EFBDa a BF DE BD S ⎛⎫+ ⎪+⎝⎭===，点A 到平面EFBD的距离为12h AC ==，所以231153312236A EFBD EFBD V S h a a a -==⨯⨯=． 故几何体ABFED 的体积为3536a ．1D ABCD EF 1A1B1C 1D ABCDEF 1A1B 1CG H数学（文科）试题参考答案及评分标准 第 5 页 共 9 页（本小题主要考查等差数列、分组求和等知识，考查化归与转化的数学思想方法，以及运算求解能力和创新意识）解：（1）因为等差数列{}n a 的首项为10，公差为2，所以()1012n a n =+-⨯， 即28n a n =+． 所以62n n nb a n =-22n n =-． （2）由（1）知()()2228n n b a n n n -=--+()(24822n n n n ⎡⎤⎡⎤=--=+-+⎣⎦⎣⎦，因为526<+，所以当5n ≤时，n n a b >，当5n >时，n n b a >． 所以{}max ,n n n c a b =228,5,2, 5.n n n n n +≤⎧=⎨->⎩当5n ≤时，123n n S c c c c =++++ 123n a a a a =++++ ()10121428n =+++++()10282n n ++=⨯29n n =+．当5n >时，123n n S c c c c =++++()()12567n a a a b b b =+++++++()()()()()222225956267278282n n ⎡⎤=+⨯+-⨯+-⨯+-⨯++-⨯⎣⎦ ()()2222706782678n n ⎡⎤=+++++-++++⎣⎦()()()()22222222265701231234522n n n+-⎡⎤=+++++-++++-⎢⎥⎣⎦()()()()1217055656n n n n n ++⎡⎤=+--+-⎢⎥⎣⎦数学（文科）试题参考答案及评分标准 第 6 页 共 9 页3211545326n n n =--+． 综上可知，n S 2329,5,11545,5.326n n n n n n n ⎧+≤⎪=⎨--+>⎪⎩20．（本小题满分）（本小题主要考查函数的极值、函数的导数、函数的零点与单调性等知识，考查数形结合、化归与转化、分类与讨论的数学思想方法，以及运算求解能力、抽象概括能力与创新意识） 解：（1）因为()32693f x x x x =-+-，所以()23129f x x x '=-+()()313x x =--．令'()0f x =，可得1x =或3x =． 则'(),()f x f x 在R 上的变化情况为：所以当1x =时，函数()f x 有极大值为1，当3x =时，函数()f x 有极小值为3-． （2）假设函数()f x 在()3,+∞上存在“域同区间”[],s t ()3s t <<，由（1）知函数()f x 在()3,+∞上单调递增．所以()(),.f s s f t t =⎧⎪⎨=⎪⎩即3232693,693.s s s s t t t t ⎧-+-=⎪⎨-+-=⎪⎩ 也就是方程32693x x x x -+-=有两个大于3的相异实根． 设32()683g x x x x =-+-()3x >，则2()3128g x x x '=-+． 令()g x '0=，解得123x =<，223x =+>． 当23x x <<时，()g x '0<，当2x x >时，()g x '0>，所以函数()g x 在区间()23,x 上单调递减，在区间()2,x +∞上单调递增．数学（文科）试题参考答案及评分标准 第 7 页 共 9 页因为()3 60g =-<，()()230g x g <<，()5120g =>， 所以函数()g x 在区间()3,+∞上只有一个零点．这与方程32693x x x x -+-=有两个大于3的相异实根相矛盾，所以假设不成立． 所以函数()f x 在()3,+∞上不存在“域同区间”．21．（本小题满分）（本小题主要考查直线的斜率、双曲线的方程、直线与圆锥曲线的位置关系等知识，考查数形结合、化归与转化、函数与方程的数学思想方法，以及推理论证能力和运算求解能力） （1）解：设双曲线E 的半焦距为c ，由题意可得2254.c a c a ⎧=⎪⎨⎪=+⎩解得a =．（2）证明：由（1）可知，直线2533a x ==，点()23,0F ．设点5,3P t ⎛⎫⎪⎝⎭,()00,Q x y ，因为220PF QF = ，所以()0053,3,03t x y ⎛⎫----= ⎪⎝⎭．所以()00433ty x =-． 因为点()00,Q x y 在双曲线E 上，所以2200154x y -=，即()2200455y x =-． 所以20000200005533PQ OQy t y y ty k k x x x x --⋅=⋅=-- ()()2002004453453553x x x x ---==-．所以直线PQ 与直线OQ 的斜率之积是定值45．数学（文科）试题参考答案及评分标准 第 8 页 共 9 页（3）证法1：设点(),H x y ，且过点5,13P ⎛⎫⎪⎝⎭的直线l 与双曲线E 的右支交于不同两点()11,M x y ，()22,N x y ，由（2）知()2211455y x =-，()2222455y x =-． 设PM MH PN HN λ==，则,.PM PN MH HN λλ⎧=⎪⎨=⎪⎩ ． 即()()1122112255,1,1,33,,.x y x y x x y y x x y y λλ⎧⎛⎫⎛⎫--=--⎪⎪ ⎪⎝⎭⎝⎭⎨⎪--=--⎩整理，得()()()1212121251,31,1,1.x x y y x x x y y y λλλλλλλλ⎧-=-⎪⎪⎪-=-⎨⎪+=+⎪+=+⎪⎩①②③④由①×③，②×④得()()22221222221251,31.x x x y y y λλλλ⎧-=-⎪⎨⎪-=-⎩⑤⑥将()2211455y x =-，()2222455y x =-代入⑥，得2221224451x x y λλ-=⨯--． ⑦ 将⑤代入⑦，得443y x =-． 所以点H 恒在定直线43120x y --=上．证法2：依题意，直线l 的斜率k 存在． 设直线l 的方程为513y k x ⎛⎫-=- ⎪⎝⎭，由2251,31.54y k x x y ⎧⎛⎫-=- ⎪⎪⎪⎝⎭⎨⎪-=⎪⎩消去y 得()()()22229453053255690kxk k x k k -+---+=．因为直线l 与双曲线E 的右支交于不同两点()11,M x y ，()22,N x y ，数学（文科）试题参考答案及评分标准 第 9 页 共 9 页则有()()()()()()()22222122212290053900455690,3053,95425569.954k k k k k k k x x k k k x x k ⎧⎪∆=-+--+>⎪⎪-⎪+=⎨-⎪⎪-+⎪=⎪-⎩设点(),H x y ，由PM MH PN HN=，得112125353x x x x x x --=--． 整理得()()1212635100x x x x x x -+++=．将②③代入上式得()()()()()2222150569303553100954954k k x k k x k k -++--+=--． 整理得()354150x k x --+=． ④ 因为点H 在直线l 上，所以513y k x ⎛⎫-=- ⎪⎝⎭． ⑤ 联立④⑤消去k 得43120x y --=． 所以点H 恒在定直线43120x y --=上．（本题（3）只要求证明点H 恒在定直线43120x y --=上，无需求出x 或y 的范围．）①② ③。
2014广州一模英语试题&答案(已校对)
广州市2014届普通高中毕业班综合测试（英语）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）In America, if you are invited to a wedding, baby shower, bar mitzvah(成年礼)or other celebrations, you're expected to bring a gift. Usually, it should be modest in 1 , about$25.For a wedding, the bride will often have "registered" a list of gifts at a local department store, indicating the items she 2 ．When you buy a registered item, tell the store that you're doing this, so the couple doesn't receive the 3 gift twice. For a baby shower, bring a gift 4 for a newborn baby. For a bar mitzvah, bring a gift appropriate for a 13-year-old boy. Because they are such important occasions, gifts for bar mitzvahs tend to be more 5 , for example, a gold-plated pen. 6 the pen by carving the boy's full name will be appreciated.If you wish to give a gift to American friends, choose something that is 7 to your country. It needn't be valuable or 8 , just typical of your homeland. 9 include a book about your country, an inexpensive souvenir, or something else that reflects your 10 .Young children who like collecting will probably be very 11 with a set of your country's coins or stamps. Items that are 12 in your country but difficult to find abroad are also good.If staying with an American family, a good way of expressing your thanks is to take them to a form of 13 , such as a basketball game or a concert.When giving gifts to a business acquaintance, don't give anything too personal, 14 to a woman. A scarf or a hat is ok, but other types of 15 are not. Something appropriate for the office is best.1. A. size B. value C. weight D. appearance2. A. prefers B. owns C. uses D. imagines3. A. first B. best C. same D. similar4. A. general B. suitable C. demanding D. expensive5. A. modest B. cheerful C. normal D. formal6. A. Personalizing B. Replacing C. Designing D. Changing7. A. convenient B. appropriate C. unique D. beneficial8. A. colorful B. rare C. heavy D. nice9. A. Opportunities B. Expectations C. Inventions D. Possibilities10. A. character B. interest C. culture D. progress11. A. annoyed B. impressed C. amused D. puzzled12. A. limited B. banned C. common D. priceless13. A. education B. discussion C. exercise D. entertainment14. A. directly B. especially C. merely D. deliberately15. A. clothing B. perfume C. jewelry D. equipment笫二节语法填空（共10小题；每小题1.5分，满分15分）While thousands of college students headed for warm climates to enjoy sun and fun during their week off from classes, seven local students had other plans.The Northern Essex Community College( NECC) students and one of their teachers spent part of their spring break in New York City, helping repair an area 16. (destroy) by the hurricane.“I wanted to see for myself what happened,” said Terry. “I couldn't imagine 17. it is like to lose your home and everything that you know and the 18. (power) effect the hurricane had on those people. I wanted to do something, to understand their feeling ofhelplessness.”The group headed into Brooklyn's Red Hook district, which was hit hard by the hurricane. There they met people from other parts of the country, 19. had also volunteered to help. Together, those volunteers and the NECC students 20. (work) to clear rubbish out of a three-story building. They put on protective suits and gloves 21. they entered the building.Inside the building, the students saw nothing but broken walls and doors and pieces of the building 22. (lie) all over the place.The students returned to school with 23. sense of achievement, a feeling that 24.______ helped people in need. It was remarkable how a community lost so much and was still able to recover, and this left the deepest impression 25. the students.Ⅱ阋读(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）AI once met a well-known botanist at a dinner party. I had never talked with a botanist before, and I found him very interesting. I sat there absorbed and listened while he spoke of unusual plants and his experiments (he even told me astonishing facts about the simple potato). I had a small indoor garden of my own -- and he was good enough to tell me how to solve some of my problems.As I said, we were at a dinner party. There must have been a dozen other guests, but I broke an important rule of politeness. I ignored everyone else and talked for hours to the botanist.Midnight came. I said good night to everyone and departed. The botanist then turned to our host and said many nice things about me, including that I was a “most interesting conversationalist ”.An interesting conversationalist? I had said hardly anything at all. I couldn't have said anything if I had wanted to without changing the subject, for I didn't know any more about plants than I knew about sharks. But I had done this one thing: I had listened carefully. I listened because I was really interested. And he felt it. Naturally that pleased him. That kind of listening is one of the best ways to show respect to others, and it makes them feel great too. “Few human beings," wrote Jack Woodford in Strangers in Love "can resist the sweet effect of rapt attention.” I went even further t han that. I was “sincere in my admiration and generous in my praise”.I told him that I had been hugely entertained and instructed. I told him I wished I had his knowledge. I told him that I should love to wander the fields with him. What's more, it was all true.And so I had him thinking of me as a good conversationalist when, in reality, I had only beena good listener and had encouraged him to talk.26. From Paragraph l, we can learn that the writer_________A. was deeply moved by the botanist's talkB. was amazed by what he was hearingC. was not in a comfortable situationD. behaved politely and properly27. Which of the following does the writer describe as a rule of politeness at dinner parties?A. Avoiding discussions about politics and religion.B. Listening carefully to what another guest says.C. Arriving and leaving at the appropriate time.D. Giving attention to all those in attendance.28. The underlined expression "rapt attention" in Paragraph 4 is closest in meaningto__________A. full understandingB. strong interestC. great uncertaintyD. little curiosity29. According to the writer, which of the following is an important characteristic of a good conversationalist?A. Listening attentively and encouraging the other side to continue.B. Encouraging the other side by sharing his/her own opinions.C. Promising a future meeting for more communication.D. Expressing respect by nodding his/her head.30. What is the purpose of the passage?A. To prove the writer is an interesting conversationalist.B. To share an interesting experience at a dinner party.C. To explain what makes a good conversationalist.D. To show that botanists can be really talkative.BA British dog-lover has invented a high-tech way of feeding his pet by Twitter(推特，流行社交网络)．Computer expert Nat Morris, 30, has designed a system to give his pet a "tweet treat" by sending him a Twitter message.His dog Toby: gets some delicious dog biscuits from a computer-controlled food machine whenever Nat sends a message to “@ feedtoby “.Nat often works away from home and isn't always able to feed Toby by hand. But his new invention allows Nat to feed his dog from anywhere in the world.Nat said, "Toby absolutely loves it. At first he didn't know what was going on. Now he sits underneath the machine, wagging his tail and waiting for the food to drop. "Nat fills the food machine with small pieces of dog biscuits, but not too many in case four-year-old Toby gets too many messages. And Nat has even equipped his house with an online camera so he can see Toby enjoying the food at his home.But one problem is that friends and family have been so amazed w ith the “tweet treat" machine that they have started sending tweets to Toby too. So Nat has had to restrict feeding time to make sure Toby doesn't turn into Tubby.“People have been sending him tweets at all hours of the day, so I had to limit it to betwe en 9 a. m. and 9 p. m. I'm thinking of doing an updated one which can measure his weight before he is fed, just to make sure he's not putting on too much puppy fat, “explained Nat.How Nat's Twitter Feeder works:When a message is sent to @ feedtoby, it is received by a mini-computer that is linked to the food machine.When the mini-computer receives the message, a bell rings and Toby comes running over and sits in front of the feeding machine. Next, the machine's motor pulls open a trap door which releases a serving of food.The doggy biscuits then drop into Toby's food bowl. Finally a digital camera takes a photo of him and sends it back to Nat on Twitter - so he knows Toby has been fed.31. Nat has invented a high-tech way to feed his dog because he________.A. wants his friends to feed TobyB. has very strong computing skillsC. is often too busy to feed his dogD. doesn't like to feed Toby by hand32. Why has Nat decided to limit the feeding machine's operating time?A. He doesn't want Toby to get too fat.B. He fears the machine will run out of food.C. He wants his friends to stop feeding Toby.D. He doesn't want Toby to be woken up at night.33. It can be learned from the passage that Toby_________A. sits beneath his feeder all day longB. is now used to being fed by machineC. doesn't know what happens to the feederD. no longer receives tweets from Nat's friends34. Which of the following shows the correct order of how the Twitter Feeder works?a. The bell goes off.b. A message is sent to @ feedtoby.c. The mini-computer gets the message.d. The digital camera takes a photo of Toby and sends it to Nat.e. The motor starts to work and opens the door to release dog food.A. a, b, c, e, d.B. b, c, e, a, d.C. b, c, a, e, d.D. c, b, a, d, e35. In which section of the newspaper would you most probably find this passage?A. Technology.B. Health.C. Environment.D. Style.CNo one knows why we dream, but some dreams might be connected to the mental processes that help us learn. In a recent study, scientists found a connection between dreams and better memory in people learning a new skill.So perhaps one way to learn something new is to practice, practice, practice -- and then sleep on it."I was very surprised by this finding,” said Robert Stickgold, a Harvard University scientist who led the study.In the study, 100 college students each spent an hour on a computer, trying to get through amaze (谜宫). The maze was difficult, and the study participants had to start from a different place each time they tried -- making it even more difficult.Then, for the first 90 minutes of a five-hour break, half of the participants were required to stay awake while half were asked to sleep. Participants who stayed awake were asked to describe their thoughts. Participants who slept were asked to describe any dream they had.Stickgold and his colleagues wanted to know about NREM, or non-REM sleep. REM stands for “rapid eye movement, “which is what happens during REM sleep. This period of sleep often brings strange dreams to a sleeper, although dreams can happen in both kinds of sleep. Stickgold wanted to know what people were dreaming about when their eyes weren't moving, during NREM sleep. Other studies have found a connection between NREM brain activity and learning ability.Four of the 50 people who slept said their dreams were about the maze. Later, when these four people tried the computer maze again, they were able to complete it faster.Stickgold believes the dream itself doesn't help a person learn -- it's the other way around. He suspects that such dreams are caused by the brain processes associated with learning.All the maze-dreamers had done the task poorly the first time, which makes Stickgold wonder if the NREM dreams show up when a person finds a new task particularly difficult. People who had other dreams, or people who didn't sleep, didn't show the same improvement.36. In the first stage of the study, the participants were asked to________A. design a maze on computerB. find their way out of a mazeC. decide where to begin a mazeD. remember a location in a maze37. What happened to the participants during the break?A. Half of them were woken up when they started to dream.B. Half of them were asked to dream about the maze.C. All of them were asked to describe their thoughts.D. Half of them were asked to sleep for 90 minutes.38. What can we learn from the passage?A. Everyone will dream about a new skill after learning it.B. Stickgold was the first to study dreams and learning.C. During NREM sleep, people usually don't dream.D. Unusual dreams often occur during REM sleep.39. According to the last paragraph, before sleeping the maze-dreamers________A. found it difficult to do the mazeB. were greatly interested in the mazeC. were mostly slow and poor thinkersD. completed the maze faster than others40. Which of the following statements best summarizes the study's conclusion?A. Dreams have a role in learning.B. Dreams have no basis in reality.C. Dreams are important for health.D. Dreams are the best way to study.DThe recent publication of autobiographies by two of Britain's greatest scientists, biologist Richard Dawkins and physicist Stephen Hawking is a wonderful opportunity to compare and contrast these two remarkable men. Surprisingly, they have rather more in common than we think.Most striking is the similarity in their backgrounds. They were born in the early 1940s to middle class families -- not wealthy but comfortably off, with a strong commitment to academic excellence and public service. Both families were keen to send their boys to Oxford University --and both succeeded, Dawkins studying zoology and Hawking physics.Neither man has a very positive view of his early university life. Hawking describes the attitude at Oxford in the 1950s and 1960s as very anti-work, "You were supposed to either be brilliant without effort or fail. Hard work was looked down upon by students and we all pretended that nothing was worth making an effort for. “He estimates that he studied for no more than an hour a day as an undergraduate student (本科生).Undergraduate life was somewhat more rewarding for Dawkins. Like Hawking, he wasn't particularly hard-working and never attended his lectures. But he found Oxford's system of weekly essay-based lessons with an academic tutor useful, "It was really only the tutorial system that educated me.”For both men, scientific life really got going as postgraduates after 1962. Dawkins, who remained at Oxford, describes brilliantly the academic competition among the postgraduate students, which he believed helped push him to develop the ideas that formed the basis of his most famous book, The Selfish Gene. This volume transformed scientific thinking aboutDarwinian evolution.Hawking, on the other hand, moved to Cambridge University after graduation, where his research into the universe would eventually make him the most famous physicist since Albert Einstein. He writes movingly about the disease which progressively crippled his entire body, leaving him unable to move and only able to communicate using a computer controlled by his eyes. Although communication is slow - he can write only 3 words a minute using the machine - his illness has not affected his mind or his research on space-time and the origins of the universe.Each book is recommended individually as a personal introduction to an important scientific thinker. Read together, they provide a superb background to the academic and social climate of postwar British research.41. Which of the following describes a similarity in Hawking's and Dawkins' backgrounds?A. They were both from wealthy families.B. They studied the same subject in university.C. They graduated from the same secondary school.D. They both came from families that valued good education.42. Why did Hawking study very little as an undergraduate student?A. He preferred doing his own research and experiments.B. Students considered it inappropriate to study too much.C. The materials discussed in lectures were very easy for him.D. He was more interested in making friends with his classmates.43. According to Dawkins, what helped him develop his most important ideas?A. His hard work as an undergraduate.B. The support he received from his family.C. The excellent tutors at Oxford University.D. The competition from other postgraduate students.44. What can we reasonably infer about the two scientists from the passage?A. Dawkins worked much harder than Hawking as an undergraduate.B. Hawking is more respected by the scientific community.C. They knew each other during their studies at Oxford.D. Hawking has experienced more physical difficulties.45. What is the function of the last paragraph?A. To state which book the writer prefers.B. To recommend the reviewed books to readers.C. To summarize the achievements of the two scientists.D. To suggest the order in which the books should be read.笫二节信息匹配（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。
广州市小升初小联盟广雅中学二中珠江六中考试英语卷及答案
⼴州市⼩升初⼩联盟⼴雅中学⼆中珠江六中考试英语卷及答案2011年⼴州市三所民办初中新⽣⼊学检测试题（⼴雅荔湾、⼴雅⽩云、⼆中应元、⼆中苏元、六中珠江）⼀、语⾳A)找出下列每组单词划线部分有⼏种读⾳，把答案的字母写在相应的括号中(A ⼀种读⾳、B两种读⾳、C三种读⾳、D四种读⾳)l. A. bird B. her C. purple D. worker2. A .music B. lung C. full D. put3. A. turns B. jumps C. draws D. throws4 A. bone B. cost C. money D. woman5.A. expensive B. enjoy C. wet D. tennisB)从以下四个选项中找出划线部分与所给单词划线部分读⾳相同的选项1．know A. down B. brown C. how D. window 2．name A. have B. table C. watch D. class3．clothes A. with B. think C. thank D. three4．here A. their B. chair C. there D. near5．yellow A. my B. boy C. your D. bye⼆、单项选择’l. -Are these storybooks ____?-No, they are ____.A. yours; her B your; hers C yours; his D your; his2. Stop talking! You ____ make noises in the library. We ____ keep quite.A can't; mustB mustn't; shouldC shouldn't; canD can’t; should3. The little baby was born ____ July 7h, 2009.A onB atC of D. in4 -What animals can you find on the farm?-I can find some ____ , many ____ and cows.A. gooses; sheepsB. geese; sheeps C goose; sheep D geese; sheep5.-Would you like to go swimming with me?- ____ .A Thank youB Sure, I'd love toC Fine. I like toD No. I don't want to6.-Do you often play___ piano?- No. I prefer playing ____ football.A. the; /B.／; theC. the, aD. a, the7. My sister doesn't like getting up early. I don't like, ____ .A. tooB. alsoC. eitherD. or8. He works ____ in his class.A .harder B. hardest C. the most hard D. the harder9.-Hello, may I speak to Mr. Wang?- ____ Mr. Wang. Who's ____？A. I 'm; thatB. I'm; thisC. This is; thatD. That's; you10. I think he had better ____ his homework on time.A. finishesB. finishedC. finishingD. finish11. – Tom, what's wrong with you? You don't look____. - I feel so ____.A. well; sadB. good; sadlyC. well; sadlyD. goodly; sad12.-Dad, can I ____ your car?- Of course. But you mustn't ____ it to your friends again.A borrow; lend B. lend; borrow C. borrowed; lent D. lent; borrowed13.- Mary ____ your jacket. It’s cold today.- Don't worry. I’m ____ a sweater. I feel warm.A put on; wearB wear; put onC wear; putting onD put on; wearing14.-Which floor do you____? -The ____ floor.A. live; ninethB. live on; ninthC. live on; ninethD. live; ninth15. I don't like these skirts. Can you show me____?A. other oneB. another onesC. another oneD. the other one16. Why don't you ____ to the cinema with us.A. go B to go C going D. goes17. - ____ have you been like this?- For two days.A. How longB. How oftenC. How soonD. How many18.When you're learning foreign language, use it, ____ you will lose it.A. butB. orC. thenD. and19. There ____ a hook and two pens on the desk just now.A. wasB. wereC. isD. are20. -Who did you stay with last night?- ——A. The Greens'B. The GreensC. GreenD. The Green's三、完形填空。
2014广东省广州市高考数学一模试卷(理科)(含解析)
2014年广东省广州市高考数学一模试卷（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．（5分）已知i 是虚数单位，若2(i)34i m +=-，则实数m 的值为（ ）．A ．2-B ．2±C ．D ．2【答案】A【解答】解：∵2(i)34i m +=-， ∴222i i 34i m m ++=-， 即22i 134i m m +-=-， ∴22413m m =-⎧⎨-=⎩，解得2m =-，故选A ．2．（5分）在ABC △中，角A ，B ，C 所对的边分别为a ，b ，c ，若2C B =，则cb为（ ）． A ．2sin CB ．2cos BC ．2sin BD ．2cos C【答案】B【解答】解：在ABC △中， ∵2C B =，∴sin sin 22sin cos C B B B ==， 即2cos c b B =，则2cos cB b=． 故选B ．3．（5分）圆22(1)(2)1x y -+-=关于直线y x =对称的圆的方程为（ ）．A ．22(2)(1)1x y -+-=B ．22(1)(2)1x y ++-=C ．22(2)(1)1x y ++-=D ．22(1)(2)1x y -++=【答案】A【解答】解：∵点(,)P x y 关于直线y x =对称的点为(,)P y x '， ∴(1,2)关于直线y x =对称的点为(2,1)，∴圆22(1)(2)1x y -+-=关于直线y x =对称的圆的方程为22(2)(1)1x y -+-=． 故选A ．4．（5分）若函数()f x R ，则实数a 的取值范围为（ ）．A ．(2,2)-B ．(,2)(2,)-∞-+∞UC ．]([,22,)-∞-+∞UD ．[]2,2-【答案】D【解答】解：函数()f x R ， 则210x ax ++≥恒成立，即240a ∆=-≤，解得22a -≤≤，即实数a 的取值范围是[]2,2-，故选D ． 5．（5分）某中学从某次考试成绩中抽取若干名学生的分数，并绘制成如图的频率分布直方图．样本数据分组为[5060)，，[60,70)，[70,80)，[80,90)，[90,100]．若用分层抽样的方法从样本中抽取分数在[80,100]范围内的数据16个，则其中分数在[90,100]范围内的样本数据有（ ）．A ．5个B ．6个C ．8个D ．10个【答案】B【解答】解：由频率分布直方图知：抽取分数在[80,100]范围内的频率为(0.0250.015)100.4+⨯=， 又在[80,100]范围内的数据有16个，∴样本容量16400.4==个， ∵分数在[90,100]范围内的频率为0.015100.15⨯=， ∴在[90,100]范围内的频数为0.15406⨯=个． 故选B ．6．（5分）已知集合{|A x x =∈Z 且3}2x∈-Z ，则集合A 中的元素个数为（ ）． A ．2B ．3C ．4D ．5【答案】C【解答】解：∵{|A x x =∈Z 且{}3}=1,1,3,52x∈--Z ， ∴集合A 中的元素有4个． 答案C ．7．（5分）设a r ，b r 是两个非零向量，则使||||a b a b ⋅=r r r r成立的一个必要非充分条件是（ ）．A ．a b =r rB ．a b r r ⊥C ．(0)a b λλ=>r rD ．a b r r ∥【答案】D【解答】解：∵a r ，b r 是两个非零向量，则||||a b a b ⋅=r r r r ， ∴||||cos ,||||a b a b a b a b ⋅==r r r r r r r r ，∴cos ,1a b =r r，∴,0a b =r r． ∴a b r r ∥． a r ，b r 是两个非零向量，则使||||a b a b ⋅=r r r r成立的一个必要非充分条件是a b r r ∥．故选D ． 8．（5分）设a ，b ，m 为整数(0)m >，若a 和b 被m 除得的余数相同，则称a 和b 对模m 同余，记为(mod )a b m ≡．若0122202020202020C C 2C 2C 2a =+⋅+⋅++⋅L ，(mod10)a b ≡，则b 的值可以是（ ）． A ．2011 B ．2012 C ．2013 D ．2014【答案】A【解答】解：∵0122202020202020C C 2C 2C 2a =+⋅+⋅++⋅L ，2020122202020202012312C 2C 2C +==++++L （），∴203a =．∵13个位是3，23个位是9，33个位是7，43个位是1，53个位是3，L ∴203个位是1，若(mod10)a b ≡，则b 的个位也是1．故选A ．二、填空题：本大题共5小题，考生作答6小题，每小题5分，满分25分．（一）必做题（9～13题） 9．（5分）若不等式||1x a -<的解集为{}3|1x x <<，则实数a 的值为__________． 【答案】2【解答】解：∵||1x a -<， ∴11x a -<-<， ∴11a x a -<<+，∴不等式||1x a -<的解集为{}1|1x a x a -<<+， ∵不等式||1x a -<的解集为{}3|1x x <<， ∴11a -=且13a +=， 解得：2a =． 故答案为：2．10．（5分）执行如图的程序框图，若输出7S =，则输入*()k k ∈N 的值为__________．【答案】3【解答】解：由程序框图知，程序第一次运行1n =，11021S -=+=； 第二次运行112n =+=，1123S =+=； 第三次运行3n =，121227S =++=． ∵输出7S =，∴程序运行终止时3n =， 又不满足条件n k <时输出S ， ∴3k =，故答案为：3． 11．（5分）一个四棱锥的底面为菱形，其三视图如图所示，则这个四棱锥的体积是__________．正主()视图侧左()视图俯视图【答案】4【解答】解：由三视图知几何体为四棱锥，且四棱锥的一条侧棱垂直于底面， 3， ∵底面为菱形，对角线互相垂直平分，∴底面面积124142S =⨯⨯⨯=，∴几何体的体积14343V =⨯⨯=．故答案为：4．12．（5分）设α为锐角，若π3cos 65α⎛⎫+= ⎪⎝⎭，则πsin 12α⎛⎫-= ⎪⎝⎭ __________．【解答】解：∵α为锐角，π3cos 65α⎛⎫+= ⎪⎝⎭为正数，∴π6α+是锐角，π4sin 65α⎛⎫+= ⎪⎝⎭，∴πππsin sin 1264αα⎡⎤⎛⎫⎛⎫-=+- ⎪ ⎪⎢⎥⎝⎭⎝⎭⎣⎦，ππππsin cos cos sin 6464αα⎛⎫⎛⎫=+-+ ⎪ ⎪⎝⎭⎝⎭4355=-=，．13．（5分）在数列{}n a 中，已知11a =，111n n a a ++=-，记n S 为数列{}n a 的前n 项和，则2014S =__________． 【答案】20112-【解答】解：∵11a =，111n n a a ++=-， ∴212a =-，312112a =-=-⎛⎫-+ ⎪⎝⎭，411(2)1a =-=-+，512a =-，L∴数列{}n a 是以3为周期的数列， 又3123131222S a a a =++=--=-，∴20142013201432013201167111222S S a ⎛⎫=+=⨯-+=-+=- ⎪⎝⎭．故答案为：20112-．三、选做题（14～15题，考生只能从中选做一题）（坐标系与参数方程选做题） 14．（5分）在极坐标系中，直线(sin cos )a ρθθ-=与曲线2cos 4sin ρθθ=-相交于A ，B 两点，若||AB =a 的值为__________． 【答案】1-或5-【解答】解：直线(sin cos )a ρθθ-=即0x y a -+=；曲线2cos 4sin ρθθ=-即22cos 4sin ρρθρθ=-，即22 240x y x y ++=-，即22(1)(2)5x y -++=，表示以(1,2)C -设圆心到直线的距离为d ，则d再根据点到直线的距离公式可得d解得1a =-，或5a =-， 故答案为：1-或5-．（几何证明选讲选做题）15．如图，PC 是圆O 的切线，切点为C ，直线PA 与圆O 交于A 、B 两点，APC ∠的平分线分别交弦CA ，CB 于D ，E 两点，已知3PC =，2PB =，则PEPD的值为__________．【答案】2 3【解答】解：作直线CF，连结BF，∴CF PC⊥，∴90PCB BCF∠+∠=︒，∵CF是直径，∴90BCF F∠+∠=︒，∴PCB F∠=∠，∵F A∠=∠，∴PCB A∠=∠，∴PCB PAC△∽△，∴23 PC PBPA PC==，∵PCE PCB A∠=∠=∠，CPE APD∠=∠，∴PCE PAD△∽△，∴23 PE PCPD PA==．故答案为：23．四、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．16．（12分）已知函数()sin cosf x x a x=+的图象经过点π,03⎛⎫-⎪⎝⎭．（1）求实数a的值．（2）设2()()2[]g x f x=-，求函数()g x的最小正周期与单调递增区间．【答案】见解析．【解答】解：（1）∵函数()sin cos f x x a x =+的图象经过点π,03⎛⎫- ⎪⎝⎭，∴π03f ⎛⎫-= ⎪⎝⎭，即ππsin cos 033a ⎛⎫⎛⎫-+-= ⎪ ⎪⎝⎭⎝⎭，即02a =，解得a（2）由（1）得()sin f x x x =． ∴2()()2[]g x f x =-2(sin )2x x =-22sin cos 3cos 2x x x x =++-2cos2x x +122cos22x x ⎫=+⎪⎪⎝⎭ ππ2sin 2cos cos2sin 66x x ⎛⎫=+ ⎪⎝⎭ π2sin 26x ⎛⎫=+ ⎪⎝⎭．∴函数的最小正周期为2ππ2=． ∵函数sin y x =的单调递增区间为ππ2π,2π()22k k k ⎡⎤-+∈⎢⎥⎣⎦Z ，令πππ2π22π262k x k -++≤≤，k ∈Z ，求得ππππ36k x k -+≤≤，∴函数的单调递增区间为πππ,π()36k k k ⎡⎤-+∈⎢⎥⎣⎦Z ．17．（12分）甲，乙，丙三人参加某次招聘会，假设甲能被聘用的概率是25，甲，丙两人同时不能被聘用的概率是625，乙，丙两人同时能被聘用的概率是310，且三人各自能否被聘用相互独立．（1）求乙，丙两人各自能被聘用的概率．（2）设ξ表示甲，乙，丙三人中能被聘用的人数与不能被聘用的人数之差的绝对值，求ξ的分布列与均值（数学期望）． 【答案】见解析．【解答】解：（1）记甲，乙，丙各自能被聘用的事件分别为1A ，2A ，3A ， 由已知1A ，2A ，3A 相互独立，且满足113232()56[1()][1()]253()()10P A P A P A P A P A ⎧=⎪⎪⎪--=⎨⎪⎪=⎪⎩解得21()2P A =，33()5P A =． ∴乙，丙各自能被聘用的概率分别为12，35． （2）ξ的可能取值为1，3． ∵123123(3)()()P P A A A P ξ==+，123123)))[)][)][)(((1(1(1(]P A P A P A P A P A P A =+---213312525525=⨯⨯+⨯⨯ 625=． ∴619(1)1(3)12525P P ξξ==-==-=． ∴ξ的分布列为∵1963713252525E ξ=⨯+⨯=．18．（14分）如图，在棱长为a 的正方体1111ABCD A B C D -中，点E 是棱1D D 的中点，点F 在棱1B B 上，且满足12B F FB =． （1）求证：11EF AC ⊥．（2）在棱1C C 上确定一点G ，使A ，E ，G ，F 四点共面，并求此时1C G 的长． （3）求平面AEF 与平面ABCD 所成二面角的余弦值．D ABCE F A 1B 1D 1C 1【答案】见解析．【解答】（1）证明：连结11B D ，BD ， ∵四边形1111A B C D 是正方形，∴1111B D AC ⊥．在正方体1111ABCD A B C D -中，∵1DD ⊥平面1111A B C D ，11AC ⊂平面1111A B C D ， ∴111AC DD ⊥．∵1111B D DD D =I ，11B D ，1DD ⊂平面11BB D D ， ∴11AC ⊥平面11BB D D ． ∵EF ⊂平面11BB D D ， ∴11EF AC ⊥．（2）解：以点D 为坐标原点，以DA ，DC ，1DD 所在的直线分别为x 轴，y 轴，z 轴， 建立如图的空间直角坐标系，则(,0,0)A a ，1,(0,)A a a ，10,(,)C a a ，10,0,2E a ⎛⎫ ⎪⎝⎭，1,,3F a a a ⎛⎫ ⎪⎝⎭，∴11(,,0)AC a a =-u u u u r ，1,,6EF a a a ⎛⎫=- ⎪⎝⎭u u u r ． 设(0,,)G a h ，∵平面11ADD A ∥平面11BCC B ，平面11ADD A I 平面AEGF AE =， 平面11BCC B I 平面AEGF FG =，∴存在实数λ，使得FG AE λ=u u u r u u u r． ∵1,0,2AE a a ⎛⎫=- ⎪⎝⎭u u u r ，1,0,3FG a h a ⎛⎫=-- ⎪⎝⎭u u u r ，∴11,0,,0,32a h a a a λ⎛⎫⎛⎫--=- ⎪ ⎪⎝⎭⎝⎭．∴1λ=，56h a =．∴115166C G CC CG a a a =-=-=．∴当116C G a =时，A ，E ，G ，F 四点共面．（3）解：由（1）知1,0,2AE a a ⎛⎫=- ⎪⎝⎭u u u r ，10,,3AF a a ⎛⎫= ⎪⎝⎭u u u r ．设(,,)n x y z =r是平面AEF 的法向量，则00n AE n AF ⎧⋅=⎪⎨⋅=⎪⎩r u u u r r u u u r，即102103ax az ay az ⎧-+=⎪⎪⎨⎪+=⎪⎩取6z =，则3x =，2y =-． 所以(3,2,6)n =-r是平面AEF 的一个法向量． 而1(0,0,)DD a =u u u u r是平面ABCD 的一个法向量，设平面AEF 与平面ABCD 所成的二面角为θ，则6cos 7θ==．故平面AEF 与平面PQ 所成二面角的余弦值为67．C 1D 1B 1A 1F EC B AD19．（14分）已知等差数列{}n a 的首项为10，公差为2，等比数列{}n b 的首项为1，公比为2，*n ∈N ． （1）求数列{}n a 与{}n b 的通项公式．（2）设第n 个正方形的边长为{}min ,n n n C a b =，求前n 个正方形的面积之和n S ．（注：{}min ,a b 表示a 与b 的最小值．） 【答案】见解析．【解答】解：（1）因为等差数列{}n a 的首项为10，公差为2，所以10(1)2n a n =+-⨯，即28n a n =+．因为等比数列{}n b 的首项为1，公比为2，所以112n n b -=⨯，即12n n b -=．（2）因为110a =，212a =，314a =，416a =，518a =，620a =，11b =，22b =，34b =，48b =，516b =，632b =．易知当5n ≤时，n n a b >．下面证明当6n ≥时，不等式n n b a >成立．方法1：①当6n =时，616623220268b a -==>=⨯+=，不等式显然成立．②假设当(6)n k k =≥时，不等式成立，即1228k k ->+．则有12222(28)2(1)8(26)2(1)8k k k k k k -=⨯+=++++>++>．这说明当1n k =+时，不等式也成立．综合①②可知，不等式对6n ≥的所有整数都成立．所以当6n ≥时，n n b a >．方法2：因为当6n ≥时112(28)(11)(28)n n n n b a n n ---=-+=+-+ 01211111(C C C C )(28)n n n n n n -----=++++-+L 012321111111(C C C C C C )(28)n n n n n n n n n n ---------+++++-+≥ 0121112(C C C )(28)n n n n ---=++-+ 236(4)(6)0n n n n n =-=-+->-， 所以当6n ≥时，n n b a >．所以5n ≤时，22222222123123n n nS c c c c b b b b =++++=++++L L 024222222n -=++++L1414n-=-1(41)3n =-． 当5n >时，2222123n nS c c c c =++++L ， 22222212567()()n b b b a a a =+++++++L L 52221(41)464)(74()3[(])4n =-+++++++L 2223414678(67)1[(6(5))]n n n =+++++++++-L L222222[()(34141212532(67)64(]5))n n n =++++++++++++--L L L(1)(21)(6)(5)3414553264(5)62n n n n n n +++-⎡⎤=+-+⨯+-⎢⎥⎣⎦3242421867933n n n =++-． 综上可知，321(41),53424218679,533n n n S n n n n ⎧-⎪⎪=⎨⎪++->⎪⎩≤．20．（14分）已知双曲线222:1(0)4x y E a a -=>的中心为原点O ，左，右焦点分别为1F ，2F ，，点P 是直线23a x =上任意一点，点Q 在双曲线E 上，且满足220PF QF ⋅=u u u u r u u u u r ． （1）求实数a 的值．（2）证明：直线PQ 与直线OQ 的斜率之积是定值．（3）若点P 的纵坐标为1，过点P 作动直线l 与双曲线右支交于不同两点M ，N ，在线段MN 上取异于点M ，N 的点H ，满足||||||||PM MH PN HN =，证明点H 恒在一条定直线上． 【答案】见解析．【解答】（1）解：设双曲线E 的半焦距为c ，由题意可得224c a c a ⎧=⎪⎨⎪=+⎩，解得a（2）证明：由（1）可知，直线2533a x ==，点2)(3,0F ． 设点5,3P t ⎛⎫ ⎪⎝⎭，00)(,Q x y ， 因为220PF QF ⋅=u u u u r u u u u r ，所以0053,(3,)03t x y ⎛⎫--⋅--= ⎪⎝⎭， 所以004(3)3ty x =-． 因为点00)(,Q x y 在双曲线E 上，所以2200154x y -=，即22004(5)5y x =-． 所以220000002200000044(5)(3)4535555333PQ OQ x x y t y y ty k k x x x x x x -----⋅=⋅===---． 所以直线PQ 与直线OQ 的斜率之积是定值45． （3）证明：设点(,)H x y ，且过点5,13P ⎛⎫ ⎪⎝⎭的直线l 与双曲线E 的右支交于不同两点11)(,M x y ，22)(,N x y ，则22114520x y -=，22224520x y -=，即22114(5)5y x =-，22224(5)5y x =-． 设||||||||PM MH PN HN λ==，则PM PN MH HNλλ⎧=⎪⎨=⎪⎩u u u u r u u u r u u u u r u u u u r ． 即1122112255,1,133(,)(,)x y x y x x y y x x y y λλ⎧⎛⎫⎛⎫--=--⎪ ⎪ ⎪⎝⎭⎝⎭⎨⎪--=--⎩， 整理，得121212125(1)31(1)(1)x x y y x x x y y y λλλλλλλλ⎧-=-⎪⎪⎪-=-⎨⎪+=+⎪+=+⎪⎩①②③④由①×③，②×④得2221222212x x y y λλ⎧-=⎪⎨⎪-⎩将22114(5)5y x =-，22224(5)5y x =-代入⑥， 得2221224451x x y λλ-=⨯--．⑦ 将⑤代入⑦，得443y x =-． 所以点H 恒在定直线43120x y --=上．21．（14分）已知函数2()(21)e x f x x x -=+（其中e 为自然对数的底数）． （1）求函数()f x 的单调区间．（2）定义：若函数()h x 在区间[](,)s t s t <上的取值范围为[],s t ，则称区间[],s t 为函数()h x 的“域同区间”．试问函数()f x 在(1,)+∞上是否存在“域同区间”？若存在，求出所有符合条件的“域同区间”；若不存在，请说明理由．【答案】见解析．【解答】解：（1）因为2()(21)e x f x x x -=+，所以22()(22)e (21)e (1)e (1)(1)e x x x x f x x x x x x x '=-++==+---． 当1x <-或1x >时，()0f x '>，即函数()f x 的单调递增区间为(,1)-∞-和(1,)+∞． 当11x -<<时，()0f x '<，即函数()f x 的单调递减区间为(1,1)-． 所以函数()f x 的单调递增区间为(,1)-∞-和(1,)+∞，单调递减区间为(1,1)-． （2）假设函数()f x 在(1,)+∞上存在“域同区间”,1)[](s t s t <<， 由（1）知函数()f x 在(1,)+∞上是增函数，所以()()f s s f t t =⎧⎨=⎩即22(1)e (1)e s t s s t t ⎧-⋅=⎪⎨-⋅=⎪⎩， 也就是方程2(1)e x x x -=有两个大于1的相异实根．设2()(1)e (1)x g x x x x --=>，则2()(1)e 1x g x x -'=-． 设2()()(1)e 1x h x g x x '==--，则2()(21)e x h x x x '=+-． 因为在(1,)+∞上有()0h x '>，所以()h x 在(1,)+∞上单调递增． 因为(1)10h =-<，2(2)3e 10h =->，即存在唯一的0(1,2)x ∈，使得0)(0h x =．当0)(1,x x ∈时，()()0h x g x '=<，即函数()g x 在0(1,)x 上是减函数； 当0(),x x ∈+∞时，()()0h x g x '=>，即函数()g x 在0(),x +∞上是增函数． 因为(1)10g =-<，0)((1)0g x g <<，2(2)e 20g =->， 所以函数()g x 在区间(1,)+∞上只有一个零点．这与方程2(1)e x x x -=有两个大于1的相异实根相矛盾，所以假设不成立． 所以函数()f x 在(1,)+∞上不存在“域同区间”．故答案为：（1）函数()f x 的单调递增区间为(,1)-∞-和(1,)+∞，单调递减区间为(1,1)-． （2）函数()f x 在(1,)+∞上不存在“域同区间”．。
广州市小联盟英语真题
广州市小联盟英语真题一、语音知识，找出与所给单词划线部分读音相同的词。
1、thank A these B brother C both D without2、where A whole B what C who D whose3、look A football B food C school D.tooth4、care A here B hear C fair D clear5、bread A please B sheep C tea D head6、cake A make B map C sad D cat7、music A cut B bus C use D put8、Christmas A match B chair C child D character9、now A grow B flower C follow D bowl10、helped A played B planned C killed D worked二、单项选择。
1.Reading aloud is____useful way for us to learn English well.A aB theC anD\2.My sister likes playing violin while my brother likes playing basketball.A the;theB the;\ C.\;the D a;a3.My mum always gets up___six_____Monday morning and cooks breakfast for me.A at;onB in;atC on;atD at;in4.Would you like to drink____,Joe?A AnythingB nothingC somethingD somethings5.—Mary,____book is on the desk,but I can't find______.A Your;yoursB yours;mineC you;meD Your;mine6.If it_____rain tomorrow,we_____to have a picnicA isn't,goB doesn't;goC isn't;will go D.doesn't will go7.----You look beautiful in white tonight.Sally!----__________.A My pleasureB You're welcome C.Thank you D That's OK.8.Though he is an American boy,he can speak Chinese _______.A wellB goodC betterD nice9._________the black board and listen to me carefully, boys and girls.A look upB Look atC Look downD Look after10.It rained all day,______I had to stay inside.A soB butC becauseD too11.I have two friends in the new class.One is Jack, ________is Liu Mei.A another oneB the otherC the othersD other one12.He often_______English after________his homework on weekends.A study;finishB studied;finishedC studies;finishingD studying;finishing13.----_______do you go shopping every month?----Sometimes.A How oftenB How manyC How muchD How long14.There are more than____________students in our school now.A two thousandsB two thousands ofC two thousandD thousand of15.The stadium________people who are watching a basketball game.A are full ofB fill withC is fill withD is full of16.He asked______________________.A when did the match beginB when does the match beginC when the match beganD when the match will begin.17.The little boy is_________in_________football.A interest;playB interests;playsC interesting;playingD interested;playing18.My mummy_________book at nine o'clock last night.A is seeingB was readingC is lookingD was watching19.---Is the man over there Mr Cool?---No,it______behim.He has gone to Shanghai.A Can'tB mustn'tC canD must三、完型填空。
2014广州一模(文数)【含答案--全WORD--精心排版】
2014年广州市普通高中毕业班综合测试（一）数学（文科）参考公式：()()22221211236n n n n ++++++=()*n ∈N ． 一、选择题：1．函数()()ln 1f x x =+的定义域为（ ）A ．(),1-∞-B ．(),1-∞C ．()1,-+∞D ．()1,+∞ 2．已知i 是虚数单位，若()2i 34i m +=-，则实数m 的值为（ ） A ．2- B ．2±C ．D ．23．在△ABC 中，角A ，B ，C 所对的边分别为a ，b ，c ，若2C B =，则cb为（ ） A ．2sin C B ．2cos B C ．2sin B D ．2cos C4．圆()()22121x y -+-=关于直线y x =对称的圆的方程为（ ）A ．()()22211x y -+-= B ．()()22121x y ++-= C ．()()22211x y ++-= D ．()()22121x y -++= 5．已知1x >-，则函数11y x x =++的最小值为（ ） A ．1- B ．0 C ．1 D ．2 6．函数()21xf x x =+的图象大致是（ ）7．已知非空集合M 和N ，规定{}M N x x M x N -=∈∉且，那么()M M N --等于（ ） A ．MN B ．M N C ．M D ．N8．任取实数a ，b ∈[]1,1-，则a ，b 满足22a b -≤的概率为（ ） A ．18 B ．14 C ．34 D ．789．设a ，b 是两个非零向量，则使a b =a b 成立的一个必要非充分条件是（ ） A ．=a b B ．ab C ．⊥a b D ．λ=a b ()0λ>10．在数列{}n a 中，已知11a =，()11sin2n n n a a ++π-=，记n S 为数列{}n a 的前n 项和，则2014S =（ ）（一）必做题（11～13题）11．执行如图1的程序框图，若输入=3k ，则输出S 的值为 ．12．一个四棱锥的底面为菱形，其三视图如图2所示，则这个四棱锥的体积是 ．13．由空间向量()1,2,3=a ，()1,1,1=-b 构成的向量集合{},A k k ==+∈Z x x a b ，则向量x 的模x 的最小 值为 ．（二）选做题（14～15题，考生只能从中选做一题）14．（坐标系与参数方程选做题）在极坐标系中，直线()sin cos a ρθθ-=与曲线2cos 4sin ρθθ=-相交于A ，B 两点，若AB =a 的值为 ．15．（几何证明选讲选做题）如图3，PC 是圆O 的切线，切点为C ，直线PA与圆O 交于A ，B 两点，APC ∠的平分线分别交弦CA ，CB 于D ，E两点，已知3PC =，2PB =，则PEPD的值为 ． 三、解答题：本大题共6小题，满分80分．解答须写出文字说明、证明过程和演算步骤． 16．（本小题满分12分）已知某种同型号的6瓶饮料中有2瓶已过保质期． （1）从6瓶饮料中任意抽取1瓶，求抽到没过保质期的饮料的概率； （2）从6瓶饮料中随机抽取2瓶，求抽到已过保质期的饮料的概率．侧（左）视图图2俯视图P图317．（本小题满分12分）已知函数()sin cos f x x a x =+的图象经过点π03⎛⎫- ⎪⎝⎭，． （1）求实数a 的值；（2）求函数()x f 的最小正周期与单调递增区间．18．（本小题满分14分）如图4，在棱长为a 的正方体1111ABCD A B C D -中，点E 是棱1D D 的中点，点F 在棱1B B 上，且满足12B F FB =．（1）求证：11EF AC ⊥；（2）在棱1C C 上确定一点G ，使A ，E ，G ，F 四点共面，并求此时1C G 的长；（3）求几何体ABFED 的体积． 1D ABCDEF1A1B1C 图419．（本小题满分14分） 已知等差数列{}n a 的首项为10，公差为2，数列{}n b 满足62n n nb a n =-，*n ∈N ． （1）求数列{}n a 与{}n b 的通项公式；（2）记{}max ,n n n c a b =，求数列{}n c 的前n 项和n S ．（注：{}max ,a b 表示a 与b 的最大值．）20．（本小题满分14分） 已知函数()32693f x x x x =-+-．（1）求函数()f x 的极值；（2）定义：若函数()h x 在区间[],s t ()s t <上的取值范围为[],s t ，则称区间[],s t 为函数()h x 的“域同区间”．试问函数()f x 在()3,+∞上是否存在“域同区间”？若存在，求出所有符合条件的“域同区间”；若不存在，请说明理由．21．（本小题满分14分）已知双曲线E ：()222104x y a a -=>的中心为原点O ，左，右焦点分别为1F ，2F ，离，点P 是直线23a x =上任意一点，点Q 在双曲线E 上，且满足220PF QF =．（1）求实数a 的值；（2）证明：直线PQ 与直线OQ 的斜率之积是定值；（3）若点P 的纵坐标为1，过点P 作动直线l 与双曲线右支交于不同两点M ，N ，在线段MN 上取异于点M ，N 的点H ，满足PM MHPN HN=，证明点H 恒在一条定直线上．2014年广州市普通高中毕业班综合测试（一）数学（文科）试题参考答案及评分标准一、选择题：CABAC ABDBC二、填空题：11. 7 12. 4 13.14. 1-或5- 15.2316．解：记“从6瓶饮料中任意抽取1瓶，抽到没过保质期的饮料”为事件A ，从6瓶饮料中中任意抽取1瓶，共有6种不同的抽法．因为6瓶饮料中有2瓶已过保质期，所以事件A 包含4种情形．则()4263P A ==． 所以从6瓶饮料中任意抽取1瓶，抽到没过保质期的饮料的概率为23． （2）解法1：记“从6瓶饮料中随机抽取2瓶，抽到已过保质期的饮料”为事件B ，随机抽取2瓶饮料，抽到的饮料分别记为x ，y ，则),(y x 表示第一瓶抽到的是x ，第二瓶抽到的是y ，则),(y x 是一个基本事件．由于是随机抽取，所以抽取到的任何基本事件的概率相等．不妨设没过保质期的饮料为1，2，3，4， 已过保质期的饮料为a ，b ，则从6瓶饮料中依次随机抽取2瓶的基本事件有：()1,2，()1,3，()1,4，()1,a ，()1,b ，()2,1，()2,3，()2,4，()2,a ，()2,b ，()3,1，()3,2，()3,4，()3,a ，()3,b ，()4,1，()4,2，()4,3，()4,a ，()4,b ，(),1a ，(),2a ，(),3a ，(),4a ，(),a b ，(),1b ，(),2b ，(),3b ，(),4b ，(),b a ．共30种基本事件．由于2瓶饮料中有1瓶已过保质期就表示抽到已过保质期的饮料，所以事件B 包含的基本事件有：()1,a ，()1,b ，()2,a ，()2,b ，()3,a ，()3,b ，()4,a ，()4,b ，(),1a ，(),2a ，(),3a ，(),4a ，(),a b ，(),1b ，(),2b ，(),3b ，(),4b ，(),b a ．共18种基本事件．则183()305P B ==．所以从6瓶饮料中随机抽取2瓶，抽到已过保质期的饮料的概率为35． 解法2：记“从6瓶饮料中随机抽取2瓶，抽到已过保质期的饮料”为事件B ，随机抽取2瓶饮料，抽到的饮料分别记为x ，y ，则),(y x 是一个基本事件．由于是随机抽取，所以抽取到的任何基本事件的概率相等．不妨设没过保质期的饮料为1，2，3，4， 已过保质期的饮料为a ，b ，则从6瓶饮料中随机抽取2瓶的基本事件有：()1,2，()1,3，()1,4，()1,a ，()1,b ，()2,3，()2,4，()2,a ，()2,b ，()3,4，()3,a ，()3,b ，()4,a ，()4,b ，(),a b ．共15种基本事件．由于2瓶饮料中有1瓶已过保质期就表示抽到已过保质期的饮料，所以事件B 包含的基本事件有：()1,a ，()1,b ，()2,a ，()2,b ，()3,a ，()3,b ，()4,a ，()4,b ，(),a b ．共9种基本事件．则93()155P B ==． 所以从6瓶饮料中随机抽取2瓶，抽到已过保质期的饮料的概率为35． 17．解：（1）因为函数()sin cos f x x a x =+的图象经过点π03⎛⎫- ⎪⎝⎭，，所以03f π⎛⎫-= ⎪⎝⎭．即ππsin cos 033a ⎛⎫⎛⎫-+-= ⎪ ⎪⎝⎭⎝⎭．即022a +=．解得a =（2）由（1）得，()sin f x x x =12sin 2x x ⎛⎫= ⎪⎪⎝⎭2sin cos cos sin 33x x ππ⎛⎫=+ ⎪⎝⎭π2sin 3x ⎛⎫=+ ⎪⎝⎭． 所以函数()x f 的最小正周期为2π．因为函数sin y x =的单调递增区间为2,222k k ππ⎡⎤π-π+⎢⎥⎣⎦()k ∈Z ， 所以当πππ2π2π232k x k -≤+≤+()k ∈Z 时，函数()x f 单调递增，即5ππ2π2π66k x k -≤≤+()k ∈Z 时，函数()x f 单调递增．所以函数()x f 的单调递增区间为5ππ2π,2π66k k ⎡⎤-+⎢⎥⎣⎦()k ∈Z ． 18．（1）证明：连结11B D ，BD ，因为四边形1111A B C D 是正方形，所以1111AC B D ⊥． 在正方体1111ABCD A B C D -中，1DD ⊥平面1111A B C D ，11AC ⊂平面1111A B C D ， 所以111AC DD ⊥．因为1111B D DD D =，11B D ，1DD ⊂平面11BB D D ，所以11AC ⊥平面11BB D D ．因为EF ⊂平面11BB D D ，所以11EF AC ⊥．（2）解：取1C C 的中点H ，连结BH ，则BH AE ．在平面11BB C C 中，过点F 作FG BH ，则FGAE ．连结EG ，则A ，E ，G ，F 四点共面． 因为11122CH C C a ==，11133HG BF C C a ===，所以 1C G 116C C CH HG a =--=．故当1C G 16a =时，A ，E ，G ，F 四点共面． （3）解：因为四边形EFBD 是直角梯形，所以几何体ABFED 为四棱锥A EFBD -．因为()211322212EFBDa a BF DE BD S a ⎛⎫+ ⎪+⎝⎭===，点A 到平面EFBD的距离为122h AC a ==，所以231153336A EFBD EFBD V S h a -===．故几何体ABFED 的体积为3536a ．19．解：（1）因为等差数列{}n a 的首项为10，公差为2，所以()1012n a n =+-⨯，即28n a n =+． 所以62n n nb a n =-22n n =-． （2）由（1）知()()2228n n b a n n n -=--+()(24822n n n n ⎡⎤⎡⎤=--=+-+⎣⎦⎣⎦，因为526<+<，所以当5n ≤时，n n a b >，当5n >时，n n b a >．所以{}max ,n n n c a b =228,5,2, 5.n n n n n +≤⎧=⎨->⎩当5n ≤时，123n n S c c c c =++++123n a a a a =++++()10121428n =+++++()10282n n ++=⨯29n n =+．1DABCDEF 1A1B1C 1DABCD EF 1A1B 1C G H当5n >时，123n n S c c c c =++++()()12567n a a a b b b =+++++++()()()()()222225956267278282n n ⎡⎤=+⨯+-⨯+-⨯+-⨯++-⨯⎣⎦()()2222706782678n n ⎡⎤=+++++-++++⎣⎦()()()()22222222265701231234522n n n +-⎡⎤=+++++-++++-⎢⎥⎣⎦()()()()1217055656n n n n n ++⎡⎤=+--+-⎢⎥⎣⎦3211545326n n n =--+．综上可知，n S 2329,5,11545,5.326n n n n n n n ⎧+≤⎪=⎨--+>⎪⎩20．解：（1）因为()32693f x x x x =-+-，所以()23129f x x x '=-+()()313x x =--． 令'()0f x =，可得1x =或3x =．则'(),()f x f x 在R 上的变化情况为：所以当1x =时，函数()f x 有极大值为1，当3x =时，函数()f x 有极小值为3-．（2）假设函数()f x 在()3,+∞上存在“域同区间”[],s t ()3s t <<，由（1）知函数()f x 在()3,+∞上单调递增．所以()(),.f s s f t t =⎧⎪⎨=⎪⎩即3232693,693.s s s s t t t t ⎧-+-=⎪⎨-+-=⎪⎩也就是方程32693x x x x -+-=有两个大于3的相异实根．设32()683g x x x x =-+-()3x >，则2()3128g x x x '=-+．令()g x '0=，解得123x =-，223x =>．当23x x <<时，()g x '0<，当2x x >时，()g x '0>，所以函数()g x 在区间()23,x 上单调递减，在区间()2,x+∞上单调递增．因为()3 60g =-<，()()230g x g <<，()5120g =>，所以函数()g x 在区间()3,+∞上只有一个零点．这与方程32693x x x x -+-=有两个大于3的相异实根相矛盾，所以假设不成立．所以函数()f x 在()3,+∞上不存在“域同区间”．21．（1）解：设双曲线E 的半焦距为c ，由题意可得22 4.c a c a ⎧=⎪⎨⎪=+⎩解得a =． （2）证明：由（1）可知，直线2533a x ==，点()23,0F ．设点5,3P t ⎛⎫⎪⎝⎭,()00,Q x y ，220PF QF =，5⎛⎫422x y 4∴2000020*******PQ OQy t y y ty k k x x x x --⋅=⋅=--()()2002004453453553x x x x ---==-．∴直线PQ 与直线OQ 斜率之积是定值45．（3）证法1：设点(),H x y ，且过点5,13P ⎛⎫⎪⎝⎭的直线l 与双曲线E 的右支交于不同两点()11,M x y ，()22,N x y ，由（2）知()2211455y x =-，()2222455y x =-．设PM MH PN HN λ==，则,.PM PN MH HN λλ⎧=⎪⎨=⎪⎩．即()()1122112255,1,1,33,,.x y x y x x y y x x y y λλ⎧⎛⎫⎛⎫--=--⎪⎪ ⎪⎝⎭⎝⎭⎨⎪--=--⎩整理，得()()()1212121251,31,1,1.x x y y x x x y y y λλλλλλλλ⎧-=-⎪⎪⎪-=-⎨⎪+=+⎪+=+⎪⎩①②③④由①×③，②×④得()()22221222221251,31.x x x y y y λλλλ⎧-=-⎪⎨⎪-=-⎩⑤⑥将()2211455y x =-，()2222455y x =-代入⑥，得2221224451x x y λλ-=⨯--． ⑦，将⑤代入⑦，得443y x =-．所以点H 恒在定直线43120x y --=上． 证法2：依题意，直线l 的斜率k 存在．设直线l 的方程为513y k x ⎛⎫-=- ⎪⎝⎭，由2251,31.54y k x x y ⎧⎛⎫-=- ⎪⎪⎪⎝⎭⎨⎪-=⎪⎩ 消去y 得()()()22229453053255690k x k k x k k -+---+=．因为直线l 与双曲线E 的右支交于不同两点()11,M x y ，()22,N x y ，则有()()()()()()()22222122212290053900455690,3053,95425569.954k k k k k k k x x k k k x x k ⎧⎪∆=-+--+>⎪⎪-⎪+=⎨-⎪⎪-+⎪=⎪-⎩ 设点(),H x y ，由PM MH PN HN =，得112125353x x x x x x --=--．整理得()()1212635100x x x x x x -+++=．将②③代入上式得()()()()()2222150569303553100954954k k x k k x k k -++--+=--．整理得()354150x k x --+=． ④，因为点H 在直线5⎛⎫ ①② ③。
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2014广州小升初小联盟语文试卷答案
一、字词积累
1、按拼音写汉字（6分）
（1）湖pàn（）（2）xùn( )情枉法（3）tǎng( )若 (4) è（）然（5）hú( )囵吞枣（6）瘟yì( )
2、下列词语中加点的字，每一对的读音都不相同的一组是（）(3分) A.友谊\毅力沾染\玷污渺茫\藐视媳妇\偷袭．．．．．．．．
B.憧憬\美景枢纽\殴打堤岸\提纲澄清\驰骋．．．．．．．．
C.醒目\省事谥号\隘口绊倒\湖畔琼浆\糨糊．．．．．．．．
D.险峻\竣工阻塞\边塞驱赶\屈服船舶\停泊．．．．．．．．
3. “赖"字用部首检字法，先查部首，再查画；用音序检字法应先查字母。
4.下列词语中共有4个错别字，请找出来并改正(4分)
（1）走投无路（2）得意忘型（3）买牍还珠 (4) 束手无策
（5）因才施教（6）责无旁代（7）寻物启事（8）众志成城
二、文学文化常识积累
1.1995年联合国教科文组织将每年的4月23日定为“世界读书日”，是为纪念《堂.吉诃德》的作者和《哈姆雷特》的作者。
2.古人四兄弟从大到小排列的次序：、、、。
3.我喜欢写的《水浒传》，对《水浒传》中武松的故事，鲁
智深的故事，吴用的故事倒背如流。
三、古诗文积累
1.古诗文默写
(1) 1911年，清华大学定《周易》中的名句为其校训,后来减缩成八字校
训，那就是“ ，。
”
(2) 古往今来，关于读书，我们的先辈感慨万千，孔子的“ ，思而不学则殆”折射出辩证的光芒；陈寿的“读书百遍， ”
强调要多读书才能有所得；西汉刘向的“书犹药也， ”都出了
读书之于健康的重要性；诗圣杜甫的“ ，下笔如有神”则认为阅读
与写作密不可分。
这些都值得我们借鉴。
2.古诗鉴赏
游山西村
莫笑农家腊酒浑，丰年留客足鸡豚。
萧鼓追随春社近，衣冠简朴古风存。
从今若许闲乘月，拄杖无时夜叩门。
诗歌颔联是千古流传的名句，请你默写在诗中的横线上，并说说我们在什
么情况下会使用这两句诗。
四、语言积累 1.
将下列句子改为“被”字句
热心网友将为了兑现承诺与小猪亲吻的一小学副校长评为“中国好校长”。
2.
用学过的修改符号，在原句上修改下面的片段。
林敏出席了市英语演讲比赛，获得了一等奖。
同学们向他高兴地祝贺，老师夸奖他说：“你真是太棒了，我为你感到骄傲！”
3.依次填入文中横线上的语句，与上下文衔接最恰当的一组是（）
一锤起来就发狠了，忘情了，没命了！百十个斜背响鼓的后生，如百十块被强震不断击起的石头，狂舞在你的面前。
骤雨一样，；旋风一样，；乱蛙一
样，；火花一样，；斗虎一样，。
黄土高原上，爆出一场多么壮阔、多么豪放、多么火烈的舞蹈哇！——安塞腰鼓
①强健的风姿②是飞扬的流苏③是闪射的瞳仁④是急促的鼓点⑤
是蹦跳的脚步
A.⑤②④①③
B.④②⑤③①
C.⑤③④②①
D.④①②③⑤
五、阅读下面一篇文章，按要求回答问题。
①我靠在昏暗、潮湿的小矮墙上，我正在思考我的人生——我是一只猪。
我们祖辈都一样，有着极其悲惨的宿命。
②宿命，宿命。
③我一生不愁吃，不愁穿，我只需要吃好睡好，每天重复着同样的事，日复一日??看着同伴们渐渐膘肥的身材，我却始终壮不起来，我有着自己的思想。
我不愿臣服于我的宿命，因为我深知，我的自由之日便是我受死之时。
④夜幕降临，我做了一个梦，梦见了我被屠夫送进了屠宰场。
那儿陈列着一行一行的猪头，一列一列的猪肉，我明白，我们身上的猪头、猪肉、猪肠通通不是我们的，除了我们自己的灵魂。
我一个激灵醒了过来。
⑤夜色朦胧，我站在猪圈门口，挥洒冷汗。
同伴一个一个都睡着了，我沐浴着从断崖处吹来的山风，我的灵魂在内心躁动不安。
我似乎看见了狼在断崖上嗥叫。
我也情不自禁，摆正了姿态，嗥叫了一声，同伴们瑟瑟发抖，痴迷的眼神望着我，显得很害怕。
他们一定以为我患上了猪瘟，或者以为我是一只披着猪皮的狼。
我没有解释什么，因为我只是顺从我内心的灵魂罢了。
⑥我决定要逃跑。
⑦“什么?”同伴们瞪着眼睛惊恐地看着我，此刻的我站在猪圈门口，继续沐浴我的清风。
我知道他们不会理解我，就像人们不理解韩寒一样，我和他一样，都是叛逆的少年。
就像汪国真说的：要输就输给追求，要嫁就嫁给幸福。
我的内心似乎更坚定了。
⑧明天，明天。
⑨红艳的晚霞如期而至，浓薄的雾气没忘记给大地披上一层灰色的荧幕。
我趁着夜色，拱开了猪圈门，玩命似的往断崖飞奔，迎面来的是凉爽的清风。
逃跑时我回头望了一眼，看到了我的母亲，从她的眼神里我看到的肯定的目光，似乎是我做了她想做却又不敢做的事情。
⑩断崖的清风不像猪圈，猪圈里的风夹杂着湿气和污臭。
我摆正了狼的英姿，不断嗥叫，因为我战胜了我的宿命，我顺从了我内心的灵魂。
?心里有个声音一直在嘀咕：“你战胜了你自己的宿命，多么值得啊!” 1.请给这篇短文拟个合适的标题： 2.请简要概括③④段的主要内容（30字以内）
3.文中两次提到“内心的灵魂”，它在文中主要是指什么？
4.在猪圈中的母亲肯定地看“我”离去，她肯定了“我”什么？她自己为什么不跟随我的脚步？
5.练习生活实际，想想有没有像此猪一样的人，请举例说明并谈谈你的感受。
六、作文
有人统计过，一个人的生命中起码有三分之一的时间花在了“等待”。
等待，有时漫长，譬如在某一领域有所建树，譬如真爱的到来；有时也短暂，譬如考试成绩的揭晓。
新的一天的开端。
你有没有等待过？对等待有什么特别的感受？请写一篇与“等待”有关的文章，题目自拟。
要求不少于400字，不能出现考生姓名和学校名称。