典型错误题例
3、数一数（共5分，每空1分）。
.
（1）一共有（ ）只小动物， （2）从左数 排第4 ， 排第（ ），
（3） 前面有（ ）只小动物， 后面有（ ）只小
动物。
（4）从右边起圈出3只小动物。
五．数一数，分一分
=
高中英语典型陷阱单项选择100题附答案
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导致无法扫描。 7、考生书写不规范。考生答题时，避免书写过意潦草或字迹太小，否则影响监考教师评阅。 8、答题卡损坏、撕破。考生答题时尽量使用垫板，如果答题卡出现损坏和撕破，考生应报告监考，更换答题卡。 9、作图题使用铅笔画图。作图题考生应使用0.5毫米黑色签字笔画图，如果没有把握，可先用铅笔画图，再用0.5毫米黑色签字笔描一遍。只用铅笔画图，扫描后图像不清楚影响评分。 10、使用其它纸张作答。根据《国家教育考试违规处理办法》（教育部第18号令）第五条第（八）款，使用规定以外的纸张答题属于考试违纪行为。
高中数学经典例题错题详解
高中数学经典例题、错 题详解
【例1】设M=｛1、2、3｝，N=｛e、g、h｝，从M至N的四种对应方式，其中是从M到N的映射是（） M N A M N B M N C M N D 映射的概念:设A、B是两个集合，如果按照某一个确定的对应关系f，是对于集合A中的每一个元素x，在集合B中都有一个确定的元素y与之对应，那么就称对应f：A→B为从集合A到集合B的一个映射。 函数的概念：一般的设A、B是两个非空数集，如果按照某种对应法则f，对于集合A中的每一个元素x，在集合B中都有唯一的元素y和它对应，这样的对应叫集合A到集合B的一个函数。（函数的本质是建立在两个非空数集上的特殊对应） 映射与函数的区别与联系： 函数是建立在两个非空数集上的特殊对应；而映射是建立在两个任意集合上的特殊对应；函数是特殊的映射，是数集到数集的映射，映射是函数概念的扩展，映射不一定是函数，映射与函数都是特殊的对应。 映射与函数（特殊对应）的共同特点：○1可以是“一对一”；○2可以是“多对一”；○3不能“一对多”；○4A中不能有剩余元素；○5B中可以有剩余元素。 映射的特点：（1）多元性：映射中的两个非空集合A、B，可以是点集、数集或由图形组成的集合等；（2）方向性：映射是有方向的，A到B的映射与B到A的映射往往不是同一个映射；（3）映射中集合A的每一个元素在集合B中都有它的象，不要求B中的每一个元素都有原象；（4）唯一性：映射中集合A中的任一元素在集合B中的象都是唯一的；（5）一一映射是一种特殊的映射方向性 上题答案应选 C 【分析】根据映射的特点○3不能“一对多”，所以A、B、D都错误；只有C完全满足映射与函数（特殊对应）的全部5个特点。 本题是考查映射的概念和特点，应在完全掌握概念的基础上，灵活掌握变型题。 【例2】已知集合A=R，B={(x、y)︱x、y∈R},f是从A到B的映射fx：→（x+1、x2）,（1）求2在B 中的对应元素；（2）（2、1）在A中的对应元素 【分析】（1）将x=2代入对应关系，可得其在B中的对应元素为（2+1、1）；（2）由题意得：x+1=2,x2=1 得出x=1，即（2、1）在A中的对应元素为1 【例3】设集合A={a、b}，B={c、d、e}，求：（1）可建立从A到B的映射个数（）；（2）可建立从B到A的映射个数（） 【分析】如果集合A中有m个元素，集合B中有n个元素，则集合A到集合B的映射共有n m 个；集合B到集合A的映射共有m n个，所以答案为23=9；32=8 【例4】若函数f(x)为奇函数，且当x﹥0时，f(x)=x-1,则当x﹤0时，有（） A、f(x) ﹥0 B、f(x) ﹤0 C、f(x)·f(-x)≤0 D、f(x)-f(-x) ﹥0 奇函数性质： 1、图象关于原点对称；? 2、满足f(-x) = - f(x)?； 3、关于原点对称的区间上单调性一致；? 4、如果奇函数在x=0上有定义,那么有f(0)=0；? 5、定义域关于原点对称（奇偶函数共有的）
高考英语典型陷阱题详解(时态)
高考英语典型陷阱题详解（时态） 1.“I _________ his telephone number.” “I have his number, but I _________ to brin g my phone book.” A.forget, forget B.forgot, forgot C.forget, forgot D.forgot, forget 【陷阱】容易误选A，认为“忘记”是现在的事。 【分析】仔细体会一下对话的语境：第一个人说“我忘记他的电话号码了”，这个“忘记”应该是现在的情况，即现在忘记了，要不然就没有必要同对方说此话了，故第一空应填for get；第二个人说“我有他的号码，但我忘记带电话本了”，这个“忘记”应该是过去的情况，即过去忘记带电话本，所以现在电话本不在身上(注意句中的转折连词but)，故第二空应填f orgot，即答案选应C。请再看一例： —Oh, I _________ where he lives. —Don’t you carry your address book? No, I _________ to bring it. A.forget, forget B.forgot, forgot C.forget, forgot D.forgot, forget 答案选C，理由同上。 2.Dear me! Just _________ at the time! I _________ no idea it was so late. A.look, have B.looking, had C.look, had D.looking, have 【陷阱】此题容易误选D，认为第一空用现在分词表伴随，第二空填have 的一般现在时，以保持与前面时态的一致性。 【分析】其实，此题应选C，第一空应填look, 因为这是祈使句的谓语；第二空应填had，因为前一句说“看看时间吧”，这一看当然知道了现在很迟的情形，“不知道这么迟了”显然应是“过去”的事，故应用一般过去时态，许多同学由于忽略这一隐含的语境而误选。 3.Mr Smith _________ a book about China last year but I don’t know whether he has finished it. A.has written B.wrote C.had written D.was writing 【陷阱】容易误选B或C。 【分析】此题应选D，这是由but I don’t know whether he has finished it 这一句话的语境决定的，全句意为“史密斯先生去年在写一本书，但我不知道他现在是否写完了”。有的同学可能由于受last year的影响而误选B。但若选B，则句子前半部分的意思则变为“史密斯先生去年写了一本书”，既然是“写了”，那么这与下文的“但我不知道他现在是否写完了”相矛盾。 4.He has changed a lot. He _________ not what he _________. A.is, is B.was, was C.is, was D.was, is 【陷阱】几个干扰项均有可能误选。 【分析】最佳答案选C，上文说“他”变化很大，即“他”现在不是过去的那个样子了，故第一空填is，第二空填was（其实第二空也可用used to be）。请看类例：
八年级数学经典错题分析
八年级错题集 1、如图11-1，,12,,ABE ACD B C ???∠=∠∠=∠指出对应边和另外一组对应角。 错解：对应边是AB 与AD ，AC 与AE ，BD 与CE ，另一组对应角是∠BAD 与∠CAE 。 错误原因分析：对全等三角形的表示理解不清，在全等三角形的表示中对应顶点的位置需 要对齐，不能根据对应顶点来确定对应角和对应边。同时对全等三角形中对应角与对应边之间的对应关系也没有理解，对应角所对的边应该是对应边，如∠2所对的边是AB ，∠1所对的边是AC ，因为∠1=∠2，即∠1与∠2是对应角，所以AB 与AC 是对应边。 正解：对应边是AB 与AC ，AE 与AD ，BE 与CD ，另一组对应角是∠BAD 与∠CAE 。 2、如图11-2，在ABD ACE ??和中，AB=AC ，AD=AE ，欲证ABD ACE ???，须补充的条 件是（ ）。 A 、∠B=∠C ； B 、∠D=∠E ； C 、∠BAC=∠DAE ； D 、∠CAD=∠DA E 。 错解：选A 或B 或D 。 错误原因分析：对全等三角形的判定定理（SAS ）理解不清，运用SAS 判定定理来证明两 三角形全等时，一定要看清角必须是两条对应边的夹角，边必须是夹相等角的两对应边。上题中AB 与AC ，AD 与AE 是对应边，并且AB 与AD 的夹角是∠BAD ，AC 与AE 的夹角是∠CAE,而∠B 与∠C ，∠D 与∠E 不是AB 与AC ，AD 与AE 的夹角，故不能选择A 或B 。∠CAD 与∠DAE 不是ABD ?和ACE ?中的内角，故不能选择D 。所以只有选择C ，因为∠BAC+∠CAD=∠DAE+∠CAD ，即：∠BAD=∠CAE 。 正解：选C 。 3、如图11-3所示，点0为码头，A ，B 两个灯塔与码头的距离相等，0A 、OB 为海岸线，一轮船离开码头，计划沿∠AOB 的平分线航行，在航行途中，测得轮船与灯塔A 和灯塔B 的距离相等，试问轮船航行是否偏离指定航线？ 错解：不能判断，因为应该是到角两边距离相等（即垂线段相等）的点才在角平分线上。 错误原因分析：生搬硬套“角的内部到角的两边的距离相等的点在角的平分线上”，而忽 略了角平分线的实质是所分得的两个角相等，本题由OA=OB ，轮船到两灯塔的距离相等，再加上已行的航线，可构造出一对全等三角形，从而可得到已行航线把∠AOB 分成相等的两个角，即没有偏离指定航线。 正解：没有偏离指定航线，如图11-4，依题意可得：OA=OB ，AC=BC ，OC=OC ，AOC BOC ???， ∴∠AOC=∠BOC ，即OC 平分∠AOB ，∴没有偏离指定航线。 4、如图11-5，,CAB DBA C D ∠=∠∠=∠，E 为AC 和BD 的交点，ADB ?与BCA ?全等吗？说明理由。 错解：ADB BCA ???。理由如下： ,, , () CAB DBA C D CBA DBA ADB BCA AAA ∠=∠∠=∠∴∠=∠∴???Q
高中英语典型陷阱题对比练习100例
1. (1) Mr. Smith did all he could ______ his family’s living standard. (2) Mr. Wang made up his mind to devote all he could ___ his oral English before going abroad. (3) He couldn’t believe that such a little boy could _____ his English with so little time. A. improve B. to improve C. improving D. to improving 2. (1) He was very careless, so I am afraid he _____ your umbrella by mistake yesterday. (2) Everything he ______ away from him before he returned to his hometown. (3) Everything ______ away from him before he returned to his hometown. A. took B. had been taken C. had had been taken D. had taken 3. (1) Before he went abroad, he looked forward ______ English as much as possible. (2) Before he went abroad, he spent as much time as he ______ English. (3) As their English teacher had gone back America, so as a result, they stayed only _____ French. A. could learning B. to learn C. to learning D. could learn 4. (1) You can not imagine what a silly mistake I ______yesterday. A. make B. made C. have made D. hade made (2) You can never imagine what great difficulty I have ______ your house. A. found B. finding C. to find D. for finding 5. (1) The person we spoke to _______ no answer at first. (2) The person was made _____ a speech at the meeting. A. making B. makes C. to make D. made 6. (1) He seemed _____ everything, but in fact he knew nothing. A. to know B. to have known C. to knowing D. knowing (2) The person we referred to ______ us a report yesterday. A. giving B. have given C. gave D. give 7. (1) The days we have been looking forward to ______ soon.
高中政治难点突破典型例题错误分析专题22对立统一新人教版
【挑战高考】2013高中政治难点突破+典型例题错误分析专题22 对 立统一新人教版 难点一矛盾就是对立统一 1．矛盾的同一性和斗争性的关系 同一性斗争性 区别含义 矛盾双方相互吸引、相互联结的属 性、趋势 矛盾双方相互排斥、相互分离的属性、 趋势 特点 相对的 有条件的 绝对的 无条件的 联系①二者相互联系、不可分离：a.同一性离不开斗争性，同一性以差别和对立为前提； b.斗争性离不开同一性，斗争性寓于同一性之中，并为同一性所制约；②矛 盾双方既对立又统一，由此推动事物的运动、变化和发展 【特别提醒】认为事物自身不包含对立面，事物没有矛盾的观点是形而上学的观点。相反，脱离统一的对立，是绝对的对立，也是形而上学的观点，都是错误的。要在统一中把握对立，在对立中看到统一。 2．注意区分日常生活中的矛盾和哲学上的矛盾 (1)日常生活用语中的“矛盾”与哲学意义上的“矛盾”既相互区别又相互联系。 (2)日常用语中的“矛盾”是指人与人之间的一些冲突和分歧，如邻里之间的争吵、同学之间的口角等。哲学意义上的“矛盾”则是指事物自身包含的既对立又统一的关系。 (3)日常生活中“矛盾”只是哲学意义上讲的矛盾的“对立”的一种表现形式而已。哲学
意义上讲的矛盾的“对立”是指矛盾双方相互排斥、相互分离的属性、趋势，是对一切具体矛盾斗争的共性的概括与反映。 可见，哲学意义上讲的“矛盾”包含了日常生活用语中的“矛盾”，但决不能把二者等同起来。 3．注意区分逻辑矛盾和辩证矛盾 (1)哲学上讲的矛盾是辩证矛盾。辩证矛盾是指客观事物本身所包含的既对立又统一的关系。辩证矛盾是事物本身所固有的，是客观存在的，需要人们正确认识。 (2)逻辑矛盾是指人们在叙述问题时违背同一律而出现的首尾不一、互相“打架”的现象。逻辑矛盾是思维混乱的表现，应该尽量避免。 例1、(2011·高考天津卷)“太极图”是中华文化的瑰宝，它是由黑白两个鱼形纹组成的圆形图案，俗称“阴阳鱼”(见下图)。该图看似简单，却包含着丰富的哲学意蕴。其内涵主要有( ) 难点二矛盾的普遍性与客观性 理解矛盾普遍性应注意的问题 (1)“事事有矛盾”是说每一事物内部包含着矛盾，而不是说每一事物同所有事物之间存在着矛盾。两个毫不相干的事物未处于统一体中，就不能构成现实的矛盾。只有在一定条件下，它们共处于统一体中时，才能构成矛盾关系。