最新北师大版必修一Unit1单元测试题(含答案解析)

单元综合检测(一)Ⅰ.阅读理解(共15小题；每小题2分，满分30分)AVolunteering abroad is a great way to help others，improve your skills and become a more confident，independent person.If you are interested in volunteering abroad，here are some places to go.AustraliaIf you seek a life on the beach，in a great climate with friendly people，Australia might be your best bet.With thousands of volunteers already signing up to leave for Australia during the holidays，it will be a very popular option.BrazilThis vast country has some great opportunities on offer and is especially popular this year.A lot of volunteer work is available in Brazil centers on conservation and ecological projects.With the world's largest rainforest in Brazil，of which huge part is in danger，there are lots of different jobs you can do.KenyaKenya is the destination for those seeking an African experience.Generally volunteers in Kenya work in orphanages (孤儿院) or with local children living in slums (贫民窟)．There is a great demand for volunteers in Africa because many people live in poverty，so if you truly feel like making a difference to a community，Kenya should be at the top of your list.Costa RicaGolden sandy beaches，clear coastlines and beautiful cities are what Costa Rica is all about.Because of its small size，you can see a lot of the country in just a short time.You can really make the most of your time as a volunteer.A lot of work is needed with wildlife and ecological work，so if you want to look after turtles on the beach，Costa Rica could be just the place for you.【语篇解读】本文介绍的是到海外去做志愿者的四个不同的国家的人所从事的不同事情的详细信息。

北师大版高中英语必修一Unit 1 Life styles单元测试卷（附解析）单元达标检测(一) Lifestyles授提示单独成册对应学生用书第71页【说明】本试卷满分150分。

考试时间120分钟。

第一部分听力(共两节，满分30分)第一节(共5小题；每小题15分，满分75分)听下面5段对话。

每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。

听完每段对话后，你都有10秒钟的时间回答有关小题和阅读下一小题。

每段对话仅读一遍。

1．What does the man like best?A．Apples B．Bananas C．Pears2．What does the woman mean?A．Lucy is not going to Jenny’s birthday partyB．Lucy likes surprising othersC．Lucy is often late3．Where do the speakers want to go?A．The supermarketB．The post officeC．The cinema4．Where is Mr Lee?A．In his officeB．In the meeting roomC．On the playground5．What was the woman’s plan for tonight?A．To see a movie with the manB．To visit the man’s grandparentsC．To have classes with the man。

2023届高考北师大版英语必修一单元练习一和答案2023届高考北师大版英语必修一单元练习一Ⅰ.阅读理解(建议用时15′)A[2023·昆明调研]Last Thursday, Michael and Linda stood behind large food trucks distributing meals to 4,000 homeless people for their wedding reception on the border town of Kilis. The couple had decided that instead of hosting their friends and family for a traditional feast reception, they would feed the victims from an earthquake-stricken area.The idea came from the bridegroom's father, Ted, who volunteers for a Turkish relief organization. For the past few years, the organization has distributed daily meals to thousands of peoplewho've suffered from natural disasters. He approached a representative of the organization and suggested that the family cover part of the costs of feeding them for the day.Then he told his son, who was surprised by the suggestion, but soon won over. When he told that to the bride, she was really shocked but finally accepted because in southeastern Turkey there is a real culture of sharing with people in need. They love to share their food, their table and everything they have. And afterwards she was quite amazed about it. So, they arrived at the distribution center on Thursday to spend the day serving food and taking photographs with their grateful recipients (接受者).On Tuesday evening, the newly married couple were still pleased with their decision to quit a personal celebration for one with a greater good. “It's like sharing a dinner with your friends and family who have this kind of thing on a daily basis-or sharingsomething with people who don't even have the most basic things，” Michael said. “Hopefully， this will also give the start for other wedding dinners to be held here with our brothers and sisters in need.”本文讲述的是一对新婚夫妇，在结婚日，放弃庆祝自己的婚礼，而去受灾的地区分发食物给受害者。

最新北师大版高中英语必修一单元测试题及答案全套A第一单元检测(A)(时间:120分钟满分:150分)第一部分听力(共两节,满分30分)第二部分阅读理解(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下面短文,从每题所给的A、B、C和D四个选项中,选出最佳选项。

ADuring my first day at senior high school,I met my new teacher.She looked at me and said,“I am happy to have an international student in my class.” I was very excited.One day,I came to the class about three minutes late,and she told me to write my name on a piece of paper and,beside it,the word “late”.In that way she could know I was late for the first time.At first,I was embarrassed and angry.Then I did not complain because she taught me an important lesson:time is very valuable(珍贵的).I learned another important lesson during one of my classes—the day of the first exam.After I finished my exam,I did not review my answers because it was the end of the exam time.One week later,she returned my exam paper.I found that one of my answers was wrong.Even though the only choices were A,B and C,I chose D instead of C.I tried to explain my mistake to her,“I marked D though I knew C was the right answer.” However,she refused my request and told me,“I do believe you,but I am not going to forgive your mistake.This will encourage you to be more careful before you turn your papers in.” I was not expecting that from her,but she was completely right.21The teacher told the writer to write his name to .A.force him to work hardB.test his writing styleC.make him angry“Then I did not complain because she taught me an important lesson:time is very 可知,老师这样做的目的是告诉他时间的重要性。

一、选择题1．下列表示正确的个数是（ ） （１）{}{}2100;(2)1,2;(3){(,)}3,435x y x y x y +=⎧∉∅∅⊆=⎨-=⎩；（４）若A B ⊆则A B A =A ．０B ．１C ．２D ．３2．设集合2{|}A x x x =<，2}6{|0B x x x =+-<，则A B =（ ）A ．（0，1）B ．()()3,01,2-⋃C ．（－3，1）D ．()()2,01,3-⋃3．由实数x ，﹣x ，|x | ） A ．2个B ．3个C ．4个D ．5个4．已知集合A 、B 均为非空集合，定义{*|()A B x x A B =∈⋃且}()x A B ∉⋂，若{}1,0,1,2,3A =-，{}2|1,B x x t t A ==+∈，则集合*A B 的子集共（ ）A ．64个B ．63个C ．32个D ．31个5．已知集合{}1,2,3,4,5,6U =，集合A 、B 是U 的子集，且A B U ⋃=，A B ⋂≠∅．若{}3,4=UAB ，则满足条件的集合A 的个数为（ ）A ．7个B ．8个C ．15个D ．16个6．已知集合{}2|230A x x x =--≤，集合{}||1|3B x x =-≤，集合4|05x C x x -⎧⎫=≤⎨⎬+⎩⎭，则集合A ，B ，C 的关系为（ ）A ．B A ⊆B ．A B =C ．C B ⊆D ．A C ⊆7．已知集合2{|120}A x x x =--≤, {|211}B x m x m =-<<+.且A B B =，则实数m 的取值范围为 ( ) A ．[-1，2)B ．[-1，3]C ．[-2，+∞)D ．[-1，+∞)8．已知集合123,,A A A 满足: {}*123|19A A A x N x =∈≤≤,且每个集合恰有3个元素,记()1,2,3i A i =中元素的最大值与最小值之和为()1,2,3i M i =,则123M M M ++的最小值为（ ） A ．21B ．24C ．27D ．309．集合{}*|421A x x N =--∈，则A 的真子集个数是（ ） A ．63B ．127C ．255D ．51110．设集合1{|0}x A x x a-=≥-，集合{}21B x x =->，且B A ⊆，则实数a 的取值范围是 （） A ．1a ≤B ．3a ≤C ．13a ≤≤D ．3a ≥11．已知全集U =R ，集合(){}{}20,1A x x x B x x =+<=≤，则图中阴影部分表示的集合是（ ）A ．()2,1-B ．[][)1,01,2-C ．()[]2,10,1--D ．0,112．已知集合{0,1,2,3,4},{|21,}A B x x n n A ===+∈，则A B 等于（ ）A ．{}1,3,5B ．{}3C ．{}5,7,9D ．{}1,3二、填空题13．对非空有限数集12{,,,}n A a a a =定义运算“min”：min A 表示集合A 中的最小元素．现给定两个非空有限数集A ，B ，定义集合{|,,}M x x a b a A b B ==-∈∈，我们称min M 为集合A ，B 之间的“距离”，记为AB d ．现有如下四个命题：①若min min A B =，则0AB d =；②若min min A B >，则0AB d >；③若0AB d =，则A B ⋂≠∅；④对任意有限集合A ，B ，C ，均有AB BC AC d d d +． 其中所有真命题的序号为__________．14．已知{2}A x x x =->，{|(3)(3)0}B x x x x =-+>，则AB =________15．定义有限数集A 中的最大元素与最小元素之差为A 的“长度”，如：集合1{1,2,4}A =的“长度”为3，集合{}23A =的“长度”为0.已知集合{1,2,3,4,5,6}U =，则U 的所有非空子集的“长度”之和为_________.16．若使集合{}2()(6)(4)0,A k x kx k x x Z =---≥∈中元素个数最少，则实数k 的取值范围是 ________.17．已知集合{}10,A x ax x R =+=∈，集合{}2280B x x x =--=，若A B ⊆,则a 所有可能取值构成的集合为______________18．已知集合{|11},{|01}A x a x a B x x =-<<+=<<若A B φ⋂=，实数a 的取值范围是______． 19．设全集{|35}Ux x =-≤≤，集合1{|||1},{|0}2A x xB x x =≤=>+，则()UC A B ⋂=_____________．20．若集合{}|121A x m x m =+<≤-，{}|25B x x =-≤<,若()()R R C A C B ⊇，则m 的取值范围是_____________．三、解答题21．已知集合A ={x |12x -≤≤}，B ={x |123m x m +≤≤+} （1）当m =1时，求AB ；（2）若B A ⊆，求实数m 的取值范围 22．已知全集为R ，集合{}503x A x R x -=∈>+，()2{|21050}B x R x a x a =∈-++≤． （1）若RB A ⊆，求实数a 的取值范围；（2）从下面所给的三个条件中选择一个，说明它是RB A ⊆的什么条件(充分必要性)．①[)7,10a ∈-；②(]7,10a ∈-；③(]6,10a ∈．23．已知集合A ＝{x |2≤x <7}，B ＝{x |3<x <10}，C ＝{x |x a ≤}． (1)求A ∪B ，(∁R A )∩B ；(2)若A ∩C ≠∅，求a 的取值范围．24．已知0a ≠，集合{}2|60A x x x =--<，{}2|280B x x x =+-≥，{}22|430C x x ax a =-+<，且()RC A B ⊆.求实数a 的取值范围.25．已知集合{|123}A x a x a =+≤≤+，{}2|7100B x x x =-+-≥. （1）已知3a =，求集合()R A B ；（2）若B A ⊆，求实数a 的范围．26．关于x 的不等式22(21)(2)0x a x a a -+++->，223()0x a a x a -++<的解集分别为M 和N（1）试求M 和N ；（2）若M N ⋂=∅，求实数a 的取值范围．【参考答案】***试卷处理标记，请不要删除一、选择题 1．D 解析：D 【解析】选项（1）中元素与空集的关系是不属于，正确；（2）空集是非空集的子集正确；（3）集合前后不相等，一个是方程的根构成的集合，有一个元素，一个是两个实数构成的集合，故不正确；（4）根据集合子集的意义知若A B ⊆则AB A =正确.2．B解析：B 【分析】化简集合A ，B ，根据交集运算即可求值. 【详解】因为2{|}A x x x =<(,0)(1,)=-∞⋃+∞，26{|}(32)0,B x x x =+-<=-所以()()3,01,2A B ⋂=-⋃. 故选：B 【点睛】本题主要考查了一元二次不等式的解法，集合的运算，属于中档题.3．A解析：A 【分析】根据绝对值的定义和开平方、立方的方法,应对x 分0,0,0x x x >=<三种情况分类讨论,根据讨论结果可得答案. 【详解】当0x >时，0x x x ===-<，此时集合共有2个元素，当0x =时，0x x x ====-=，此时集合共有1个元素，当0x <时，0x x -===>，此时集合共有2个元素，综上所述，此集合最多有2个元素. 故选：A . 【点睛】本题考查了元素与集合关系的判断及根式的化简求值,其中解答本题的关键是利用分类讨论思想,对x 分三种情况进行讨论,是基础题.4．C解析：C 【分析】先求集合B ，再求并集、交集、补集，最后根据元素确定子集个数. 【详解】因为{}2|1,{1,2,5,10}B x x t t A ==+∈=， 所以{}{}1,0,1,2,3510,1,2,AB A B =-=，，*{1,0,3,5,10}A B ∴=-因此集合*A B 的子集有5232=个， 故选：C 【点睛】本题考查并集、交集、补集定义以及子集个数，考查综合本分析求解能力，属基础题.5．C解析：C 【分析】由题意知3、4B ∉，则集合A 的个数等于{}1,2,5,6非空子集的个数，然后利用公式计算出集合{}1,2,5,6非空子集的个数，即可得出结果. 【详解】由题意知3、4B ∉，且集合A 、B 是U 的子集，且A B U ⋃=，A B ⋂≠∅， 则AB 为集合{}1,2,5,6的非空子集，因此，满足条件的集合A 的个数为42115-=.故选C. 【点睛】本题考查集合个数的计算，一般利用列举法将符合条件的集合列举出来，也可以转化为集合子集个数来进行计算，考查化归与转化思想的应用，属于中等题.6．D解析：D 【分析】根据一元二次不等式的解法可求出集合A ，根据绝对值不等式的解法可求出集合B ，根据分式不等式的解法可求出集合C ，从而可得出集合A ，B ，C 间的关系. 【详解】解：由于{}{{}2|23013A x x x x x =--≤=-≤≤，{}{}|1324B x x x x =-≤=-≤≤， {}4|0545x C x x x x -⎧⎫=≤=-<≤⎨⎬+⎩⎭，可知，A C ⊆. 故选：D. 【点睛】本题考查一元二次不等式、绝对值不等式和分式不等式的解法，以及集合间的关系，考查计算能力.7．D解析：D 【分析】 先求出集合A ,由A B B =,即B A ⊆,再分B φ=和B φ≠两种情况进行求解.【详解】由2120x x --≤,得34x -≤≤. 即[3,4]A =-. 由AB B =,即B A ⊆.当B φ=时，满足条件,则211m m -≥+解得2m ≥.当B φ≠时，要使得B A ⊆，则12121314m m m m +>-⎧⎪-≥-⎨⎪+≤⎩.解得：12m -≤<.综上满足条件的m 的范围是：1m ≥-. 故选：D. 【点睛】本题主要考查集合的包含关系的判断及应用，以及集合关系中的参数范围问题，考查分类讨论思想，属于中档题.8．C解析：C 【分析】 求出{}{}*123|191,2,3,4,5,6,7,8,9A A A x N x =∈≤≤=，由题意列举出集合123,,A A A ，由此能求出123M M M ++的最小值. 【详解】 由题意可知，{}{}*123|191,2,3,4,5,6,7,8,9A A A x N x =∈≤≤=123,,A A A 各有3个元素且不重复，当{}13,4,5A =，{}22,6,7A =，{}31,8,9A =时，123M M M ++取得最小值，此时最小值为12357927+++++=，故选C 【点睛】本题主要考查集合中的元素运算，解题的关键是理解题中满足的条件，属于中档题.9．B解析：B 【分析】先求得{}*|421A x x N =--∈的元素个数,再求真子集个数即可.【详解】由{}*|421A x x N=--∈,则421x --为正整数.则21x -可能的取值为0,1,2,3,故210,1,2,3x -=±±±,故x 共7个解.即{}*|421A x x N =--∈的元素个数为7故A 的真子集个数为721127-= 故选：B 【点睛】本题主要考查集合中元素个数的求解与知识点：元素个数为n 的集合的真子集有21n -个. 属于基础题型.10．C解析：C 【解析】 【分析】先求出集合B ，比较a 与1的大小关系，结合B A ⊆，可求出实数a 的取值范围. 【详解】解不等式21x ->，即21x -<-或21x ->，解得1x <或3x >，{1B x x ∴=<或}3x >.①当1a =时，{}1A x x =≠，则B A ⊆成立，符合题意； ②当1a <时，{A x x a =<或}1x ≥，B A ⊄，不符合题意；③当1a >时，{1A x x =≤或}x a >，由B A ⊆，可得出3a ≤，此时13a .综上所述，实数a 的取值范围是13a ≤≤. 故选：C. 【点睛】本题考查集合之间关系的判断，涉及分式、绝对值不等式的解法，解分式不等式一般要转化为整式不等式，有参数时，一般要分类讨论．11．C解析：C 【分析】由集合描述求集合,A B ，结合韦恩图知阴影部分为()()U C A B A B ⋂⋂⋃，分别求出()U C A B 、()A B ⋃，然后求交集即可.【详解】(){}20{|20}A x x x x x =+<=-<<，{}1{|11}B x x x x =≤=-≤≤，由图知：阴影部分为()()U C A B A B ⋂⋂⋃，而{|10}A B x x ⋂=-≤<，{|21}A B x x ⋃=-<≤，∴(){|1U C A B x x ⋂=<-或0}x ≥，即()(){|21U C A B A B x x ⋂⋂⋃=-<<-或01}x ≤≤，故选：C 【点睛】本题考查了集合的基本运算，结合韦恩图得到阴影部分的表达式，应用集合的交并补混合运算求集合.12．D解析：D 【分析】首先求得集合B ，然后进行交集运算即可.【详解】由题意可得：{}1,3,5,7,9B =，则{}1,3A B =.故选D . 【点睛】本题主要考查集合的表示方法，交集的定义与运算等知识，意在考查学生的转化能力和计算求解能力.二、填空题13．①③【分析】根据题意可得①③正确通过举反例可得②④错误【详解】对于结论①若则中最小的元素相同故①正确；对于结论②取集合满足但故②错误；对于结论③若则中存在相同的元素则交集非空故③正确；对于结论④取集解析：①③ 【分析】根据题意可得①③正确，通过举反例可得②④错误. 【详解】对于结论①，若min min A B =，则A ，B 中最小的元素相同，故①正确；对于结论②，取集合{}1,2A =，{}0,2B =，满足min min A B >，但0AB d =，故②错误；对于结论③，若0AB d =，则,A B 中存在相同的元素，则交集非空，故③正确； 对于结论④，取集合{}1,2A =，{}2,3B =，{}3,4C =，可知0AB d =，0BC d =，1AC d =，则AB BC AC d d d +≥不成立，故④错误. 故答案为：①③．14．【分析】先分别求解集合中元素的所满足的不等式再由交集的定义求解即可【详解】由题因为解得则因为解得或则或所以故答案为:【点睛】本题考查集合的交集运算考查含根式的不等式的运算考查解高次不等式 解析：{|30}-<<x x【分析】先分别求解集合中元素的所满足的不等式,再由交集的定义求解即可 【详解】由题,因为20xx >-≥⎪⎩,解得1x <,则{}|1A x x =<, 因为()()330x x x -+>,解得30x -<<或3x >,则{|30B x x =-<<或}3x >, 所以{}|30A B x x ⋂=-<<,故答案为:{|30}-<<x x 【点睛】本题考查集合的交集运算,考查含根式的不等式的运算,考查解高次不等式15．201【分析】根据集合长度的定义可将集合的非空子集分六类分别计算可求出答案【详解】集合有6个元素非空子集有个①集合长度为0的子集有：；②集合长度为1的子集有：；③集合长度为2的子集有：；④集合长度为解析：201 【分析】根据集合“长度”的定义，可将集合U 的非空子集分六类，分别计算可求出答案. 【详解】集合U 有6个元素,非空子集有62163-=个，①集合“长度”为0的子集有：{}{}{}{}{}{}1,2,3,4,5,6； ②集合“长度”为1的子集有：{}{}{}{}{}1,2,2,3,3,4,4,5,5,6； ③集合“长度”为2的子集有：{}{}{}{}1,3,2,4,3,5,4,6,{}{}{}{}1,2,3,2,3,4,3,4,5,4,5,6；④集合“长度”为3的子集有：{}{}{}1,4,2,5,3,6,{}{}{}1,2,4,1,3,4,2,3,5，{}{}{}2,4,5,3,4,6,3,5,6,{}{}1,2,3,4,2,3,4,5,{}3,4,5,6；⑤集合“长度”为4的子集有：{}{}1,5,2,6,{}{}{}1,2,5,1,3,5,1,4,5,{}{}{}2,3,6,2,4,6,2,5,6,{}{}{}1,2,3,5,1,2,4,5,1,3,4,5,{}{}{}2,3,4,6,2,3,5,6,2,4,5,6,{}2,3,4,5,6,{}1,2,3,4,5；⑥集合“长度”为5的子集有：{}1,6,{}1,2,6,{}1,3,6,{}1,4,6,{}1,5,6,{}1,2,3,6,{}1,2,4,6,{}1,2,5,6,{}1,3,4,6,{}1,3,5,6,{}1,4,5,6{1,3,4,5,6},{1,2,4,5,6},{1,2,3,5,6},{1,2,3,4,6},{1,2,3,4,56},.U 的所有非空子集的“长度”之和为061528312416516201⨯+⨯+⨯+⨯+⨯+⨯=. 故答案为：201. 【点睛】本题考查新定义，要求读懂题意并结合已有知识、能力进行理解，根据新定义进行计算、推理、迁移，新定义问题要把握两点：一是掌握问题原型的特点及其问题解决的思想方法；二是根据问题情境的变化，通过思考，合理进行思想方法的迁移.16．【分析】首先讨论的取值解不等式；再由集合的元素个数最少推出只有满足若集合的元素个数最少由集合只需求的最大值即可再由集合中只需即可求解【详解】由题知集合内的不等式为故当时可得；当时可转化为或因为所以不 解析：[]3,2--【分析】首先讨论k 的取值，解不等式；再由集合A 的元素个数最少，推出只有k 0<满足，若集合A 的元素个数最少，由k 0<，集合A =64x Z k x k ⎧⎫∈+≤≤⎨⎬⎩⎭，只需求6k k +的最大值即可，再由集合A 中x ∈Z ，只需654k k-<+<-即可求解. 【详解】由题知集合A 内的不等式为2(6)(4)0,kx k x x Z ---≥∈，故 当0k =时，可得{}4A x Z x =∈<； 当0k >时， 2(6)(4)0kx k x ---≥可转化为24060x kx k -≥⎧⎨--≥⎩ 或24060x kx k -≤⎧⎨--≤⎩，因为64k k <+， 所以不等式的解集为{4x x ≤或6x k k ⎫≥+⎬⎭，所以A ={4x Z x ∈≤或6x k k ⎫≥+⎬⎭当k 0<时，由64k k +<，所以不等式的解集为64x k x k ⎧⎫+≤≤⎨⎬⎩⎭，所以A =64x Z k x k ⎧⎫∈+≤≤⎨⎬⎩⎭，此时集合A 的元素个数为有限个. 综上所述，当0k ≥时，集合A 的元素个数为无限个，当k 0<时，集合A 的元素个数为有限个，故当k 0<时，集合A 的元素个数最少，且当6k k+的值越大，集合A 的元素个数越少，令6()f k k k =+（k 0<），则26()1f k k'=-，令()0f k '= 解得k =()f k在(,-∞内单调递增，在()内单调递减，所以max ()(f k f ==-又因为x ∈Z ，54-<-<-，所以当654k k-≤+<-，即32k -≤≤-时， 集合A =64x Z k x k ⎧⎫∈+≤≤⎨⎬⎩⎭中元素的个数最少，故32k -≤≤- 故答案为：[]3,2-- 【点睛】本题主要考查集合的运算和解不等式，综合性比较强.17．【分析】先化简集合利用分类讨论和即可求出构成的集合【详解】由可得:即:解得或故:由可得:当时方程无实数解此时满足当时方程的实数解为故:由可得:或解得或的所有取值构成的集合为:故答案为:【点睛】本题主 解析：11{0,,}24- 【分析】先化简集合B ,利用A B ⊆,分类讨论=0a 和0a ≠,即可求出构成a 的集合.【详解】 由{}2280B x x x =--=可得:2280x x --= 即:()()240x x +-=解得2x =-或4x = 故:{}2,4B =- {}10,A x ax x R =+=∈由10ax += 可得:1ax =-当0a =时,方程1ax =-无实数解,此时A =∅,满足A B ⊆当0a ≠时,方程1ax =-的实数解为1x a =-,故:1{}A a=- 由A B ⊆可得:12a -=-或14a-= 解得12a =或14a =- a 的所有取值构成的集合为:11{0,,}24-.故答案为:11{0,,}24-.【点睛】本题主要考查了集合间的基本关系以及一元二次方程的解法,要注意集合A 是集合B 的子集时,集合A 有可能是空集. 18．【分析】由根据集合的交集的运算得到或即可求解【详解】由题意集合因为则满足或解得或即实数的取值范围是故答案为：【点睛】本题主要考查了集合的运算以及利用集合的交集求参数其中解答中熟记集合交集运算列出相应 解析：(][),12,-∞-⋃+∞【分析】由A B φ⋂=，根据集合的交集的运算，得到11a -≥或10a +≤，即可求解.【详解】由题意，集合{|11},{|01}A x a x a B x x =-<<+=<<，因为A B φ⋂=，则满足11a -≥或10a +≤，解得2a ≥或1a ≤-，即实数a 的取值范围是(][),12,-∞-⋃+∞.故答案为：(][),12,-∞-⋃+∞.【点睛】本题主要考查了集合的运算，以及利用集合的交集求参数，其中解答中熟记集合交集运算，列出相应的不等式是解答的关键，着重考查了推理与运算能力，属于基础题. 19．【分析】解绝对值不等式求得集合然后求得其补集解分式不等式求得集合由此求得【详解】由解得所以由解得所以故填：【点睛】本小题主要考查集合交集和补集的概念和运算考查绝对值不等式和分式不等式的解法属于基础题 解析：(2,1)(1,5]--【分析】解绝对值不等式求得集合A ，然后求得其补集.解分式不等式求得集合B ，由此求得()U C A B ⋂.【详解】 由1x ≤解得11x -≤≤，所以[)(]3,11,5U C A =--⋃.由102x >+解得2x >-，所以()U C A B ⋂(2,1)(1,5]=--.故填：(2,1)(1,5]--.【点睛】本小题主要考查集合交集和补集的概念和运算，考查绝对值不等式和分式不等式的解法，属于基础题. 20．【分析】由进行反推可分为集合和集合两种情况进行分类讨论【详解】由进行反推若则解得成立由可知集合因应满足解得综上所述故答案为：【点睛】本题考查根据集合的补集与包含关系求解参数问题是中档题型在处理此类题 解析：(),3-∞【分析】由()()R R C A C B ⊇进行反推，可分为集合A =∅，和集合A ≠∅两种情况进行分类讨论【详解】由()()R R C A C B ⊇进行反推，若A =∅，则121m m +≥-，解得2m ≤，成立 由A ≠∅可知，集合{}|121U A x x m x m =≤+>-或，{}|25U B x x x =<-≥或因()()R R C A C B ⊇，应满足12215211m m m m +≥-⎧⎪-<⎨⎪->+⎩，解得()2,3m ∈综上所述，(),3m ∈-∞故答案为：(),3-∞【点睛】本题考查根据集合的补集与包含关系求解参数问题，是中档题型，在处理此类题型中，易错点为忽略端点处等号取不取得到的问题，解题时要特别仔细三、解答题21．（1）{}2；（2）1,2⎛⎤-∞ ⎥⎝⎦. 【分析】（1）根据集合的交集运算求解即可；（2）讨论集合B 是否为空集，根据包含关系列出不等式，即可得出实数m 的取值范围.【详解】（1）当m =1时，B ={x |2≤x ≤5}，因此A B ={2} （2）A B ⇔B A ⊆，则①当B =∅时，即123m m +>+，即2m <-，符合题意②当B ≠∅时，要满足B A ⊆，则12311232m m m m +≤+⎧⎪+≥-⎨⎪+≤⎩2212m m m ⎧⎪≥-⎪⇒≥-⎨⎪⎪≤-⎩122m ⇒-≤≤- 综上所述，当B A ⊆时，实数m 的取值范围时1(,2)2,2⎡⎤-∞-⋃-⎢⎥⎣⎦=1,2⎛⎤-∞ ⎥⎝⎦ 【点睛】本题考查交集的运算，同时也考查了利用集合的包含关系求参数，解题的关键就是对含参集合分空集和非空集合两种情况讨论，考查分类讨论思想的应用，属于中档题. 22．（1）610a -≤≤；（2）答案见解析.【分析】()1先求集合A ，B ，A R ，再由R B A ⊆得到a 的不等式，解得即可；()2结合()1利用充分必要条件的定义逐一判定．【详解】解：()1集合5|0(3)(5,)3x A x R x -⎧⎫=∈>=-∞-⋃+∞⎨⎬+⎩⎭， 所以[]35R A =-，，集合()()()2{|21050}{|250}B x R x a x a x R x a x =∈-++≤=∈--≤，若R B A ⊆， 只需352a -≤≤，所以610a -≤≤．()2由()1可知的充要条件是[]610a ∈-，， 选择①，则结论是既不充分也不必要条件；选择②，则结论是必要不充分条件；选择③，则结论是充分不必要条件．【点睛】关键点睛，利用集合关系求参数范围，求集合A ，B ，A R ，再由RB A ⊆得到a 的不等式，进而利用a 的范围，判定充分必要条件，属于中档题． 23．(1) {x |2≤x <10}, {x |7≤x <10};(2) 2a ≥【分析】（1）根据交、并、补集的运算分别求出A ∪B ，（∁R A ）∩B ；（2）根据题意和A∩C≠∅，即可得到a 的取值范围．【详解】解：(1)因为A ＝{x |2≤x <7}，B ＝{x |3<x <10}，所以A ∪B ＝{x |2≤x <10}．因为A ＝{x |2≤x <7}，所以∁R A ＝{x |x <2，或x ≥7}，则(∁R A )∩B ＝{x |7≤x <10}．(2)因为A ＝{x |2≤x <7}，C ＝{x |x a ≤}，且A ∩C ≠∅，所以2a ≥所以a 的取值范围为2a ≥．【点睛】高考对集合知识的考查要求较低，均是以小题的形式进行考查，一般难度不大，要求考生熟练掌握与集合有关的基础知识．纵观近几年的高考试题，主要考查以下两个方面：一是考查具体集合的关系判断和集合的运算．解决这类问题的关键在于正确理解集合中元素所具有属性的含义，弄清集合中元素所具有的形式以及集合中含有哪些元素．二是考查抽象集合的关系判断以及运算．24．22,00,33a ⎡⎫⎛⎤∈-⎪ ⎢⎥⎣⎭⎝⎦. 【分析】先化简集合,A B ,求出R AB ,再对a 分类讨论，根据()RC A B ⊆得解.【详解】 {}{}2|60|23A x x x x x =--<=-<<，{}{2|2804B x x x x =+-≥=≤-或}2x ≥，∴{}|42R B x x =-<<，则(){}|22R A B x x =-<<，又∵{}()(){}22|430|30C x x ax a x x a x a =-+<=--<，∵0a ≠，∴当0a >时，{}|3C x a x a =<<，当0a <时，{}|3C x a x a =<<.∵()R C A B ⊆，∴0232a a a >⎧⎪≥-⎨⎪≤⎩或0322a a a <⎧⎪≥-⎨⎪≤⎩， 解得203a <≤或203a -≤<. 所以实数a 的取值范围是22,00,33a ⎡⎫⎛⎤∈-⎪ ⎢⎥⎣⎭⎝⎦. 【点睛】本题主要考查一元二次不等式的解法，考查集合的关系和运算，意在考查学生对这些知识的理解掌握水平. 25．（1）(){|24}R A B x x ⋂=≤<（2）1a =【分析】化简集合B ，（1）计算3a =时集合A ，根据补集与交集的定义；（2）由题意得出A ≠∅，根据包含关系，列出关于a 的不等式，求出实数a 的取值范围．【详解】集合{|123}A x a x a =+≤≤+ {}{}22|7100|7100{|25}B x x x x x x x x =-+-≥=-+≤=≤≤；（1）当3a =时，{|49}A x x =≤≤{| 4 R A x x ∴=<或9}x >则(){|24}R A B x x ⋂=≤<（2）因为B A ⊆，{|25}B x x =≤≤，所以A ≠∅，则1232a a a +≤+⇒≥-并且由B A ⊆，得12235a a +≤⎧⎨+≥⎩，解得1a = 综上，实数a 的取值范围是1a =．【点睛】本题主要考查了交集，并集的运算以及根据包含关系求参数范围，属于中档题. 26．（1）(,1)(2,)M a a =-∞-⋃++∞，集合N 见解析；（2）[1,2]-.【分析】（1）对两个不等式进行因式分解，分类讨论即可得解；（2）结合（1）的结论进行分类讨论求解.【详解】（1）22(21)(2)0x a x a a -+++->即()()()120x a x a ---+>所以(,1)(2,)M a a =-∞-⋃++∞；223()0x a a x a -++<即()()20x a x a --<当1a >或0a <时，2(,)N a a =；当01a <<时，2(,)N a a =；当1a =或0a =时，N =∅；（2）分类讨论：当1a =或0a =时，N =∅，符合题意；当01a <<时，2(,)N a a =，M N ⋂=∅， 即212a a a a ≥-≤+⎧⎨⎩，2102a a a a -+≥≤+⎧⎨⎩恒成立，所以01a <<符合题意； 当1a >或0a <时，212a a a a ≥-≤+⎧⎨⎩解得：12a -≤≤，所以[)(]1,01,2a ∈-， 综上所述：[1,2]a ∈-【点睛】此题考查求二次不等式的解集，关键在于准确进行因式分解并分类讨论，根据两个集合的交集为空集求参数的取值范围，考查分类讨论思想.。

第一单元测评(时间:120分钟满分:120分)第二部分阅读(共两节,满分50分)第一节(共15小题;每小题2.5分,满分37.5分)阅读下列短文,从每题所给的A、B、C、D四个选项中选出最佳选项。

AAre you a high school student aged 15 to 18?Then consider spending your summer at Harvard,where you can explore a variety of college-level courses,and live and learn with students from different countries.TWO CHOICES TO CHOOSE FROM:●The choice to live on campus,commute (走读),or study online.●The freedom to schedule your days.续表For students living on campus,the average cost of attending the program is$11,900,which includes fees for two 4-credit courses ($3,050 for each course) or one 8-credit course ($6,100),room and meal fees,and a health insurance fee.For those attending as a commuting or online student,the cost ranges from $3,050 to $6,100 depending on the number of courses taken.【语篇解读】本文是一篇说明文。

文章主要介绍了哈佛大学为高中生提供的两种夏季学习项目。

最新北师大版高中英语必修一单元测试题及答案全套A第一单元检测(A)(时间:120分钟满分:150分)第一部分听力(共两节,满分30分)第二部分阅读理解(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下面短文,从每题所给的A、B、C和D四个选项中,选出最佳选项。

ADuring my first day at senior high school,I met my new teacher.She looked at me and said,“I am happy to have an international student in my class.” I was very excited.One day,I came to the class about three minutes late,and she told me to write my name on a piece of paper and,beside it,the word “late”.In that way she could know I was late for the first time.At first,I was embarrassed and angry.Then I did not complain because she taught me an important lesson:time is very valuable(珍贵的).I learned another important lesson during one of my classes—the day of the first exam.After I finished my exam,I did not review my answers because it was the end of the exam time.One week later,she returned my exam paper.I found that one of my answers was wrong.Even though the only choices were A,B and C,I chose D instead of C.I tried to explain my mistake to her,“I marked D though I knew C was the right answer.” However,she refused my request and told me,“I do believe you,but I am not going to forgive your mistake.This will encourage you to be more careful before you turn your papers in.” I was not expecting that from her,but she was completely right.21The teacher told the writer to write his name to .A.force him to work hardB.test his writing styleC.make him angry“Then I did not complain because she taught me an important lesson:time is very 可知,老师这样做的目的是告诉他时间的重要性。

Unit 1-writing&viewing&reading基础能力检测【匹配单词词性及汉意】()1.junior A．n.幻灯片()2.formal B．adv.向前；进展()3.adapt C．n.某一时刻()4.moment D．n.报告；陈述，说明()5.presentation E．adj.正式的，官方的()6.slide F．adj.低年级的()7.forward G．vi.(使)适应【选择句中短语汉意】()2.They were almost completely cut off from the outside world.()3.How can a leader adapt to a changing environment?()4.As a result，costs will be reduced by as much as 90%.()5.He could not come up with a proper answer.()6.Thanks to his effort，it is more successful than we have expected. 【衍生词汇】1．adj.正式的，官方的→informal adj.非正式的2．vi.(使)适应；改编→adaptation n．适应；改编3．present vt.提出；出现；介绍；赠送→n．报告；陈述，说明4．adj.低年级的→senior adj.(反义词)高年级的；年长的【补全短语及应用】**补全1．be different 不同于2．adapt sth. 适应某事3．the moment 此刻，目前，眼下4．look forward 盼望5． a result 结果6．come up 提出7．play an important part ... 在……起重要作用8．fact 事实上9．cut 切断；使隔绝10．thanks 幸亏；由于**应用1．the bad weather，the match had been canceled.2．He good ideas for the product promotion.3．He quickly the climate in the high mountains.4．he doesn't have a penny to his name.5．We welcoming you and your party.6．The floods have our supplies.【补全句子及应用】**补全【单句语法填空】1．Today they had sent back all the pictures and (幻灯片)，she thought.2．I always ask how much time I have to make my (报告)．3．Her daughter studies in the third grade in a high school.4．The boss had an discussion with me this morning，I feel really down.5．They have had to themselves to a war economy.6．I suggested she should learn to adapt (her)to her new campus life.7．It is said the boy grew up to be a noted scientist.8．I didn't wake up I heard the alarm clock.9．They gave the first-class (entertain)to us.10．Where were you when the power was cut ?11．She is senior everyone else in the company.12．I look forward to (come)to work every day.13．The sales manager will give a (present) on the new products.14．He won't go there with us unless (invite)．15．Let's leave the matter aside the moment.【补全句子】1．直到太阳从东方升起，我们才出发。

Unit 1 Life Choices 单元测试第一部分听力（共两节，满分30分）第一节（共5小题；每小题1.5分，满分7.5分）听下面5段对话。

每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1. What did the man do yesterday?A. He went to school.B. He went to the doctor’s.C. He went to a drug store.2. Why doesn’t the man take the sweater?A. It’s too costly.B. It doesn’t suit him.C. It doesn’t look great.3. How will the speakers get there?A. In Tom’s car.B. By bus.C. By train.4. What are the speakers talking about?A. What the woman did yesterday.B. A movie the woman watched.C. Yesterday’s weather.5. What time is it now?A. 9: 50.B. 10: 00.C. 10: 10.第二节（共15小题；每小题1.5分，满分22.5分）听下面5段对话或独白。

每段对话或独白后有几个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。

听每段对话或独白前，你将有时间阅读各个小题，每小题5秒钟；听完后，各小题将给出5秒钟的作答时间。

每段对话或独白读两遍。

听第6段材料，回答第6、7题。

6. What did the man think of the supper?A. Great.B. Just so so.C. Terrible.7. Where did the speakers have supper tonight?A. At the man’s.B. At the woman’s.C. At a restaurant.听第7段材料，回答第8、9题。

最新北师大版高中英语必修一单元检测试卷（共3套试卷附答案）单元质量检测(一)(时间：100分钟满分：120分本卷共4页)Ⅰ.阅读理解(共15小题；每小题2分，满分30分)阅读下列短文，从每题所给的四个选项(A、B、C和D)中，选出最佳选项。

AA long time ago, I was returning from my friend's house. My father parked the car in front of our house. We were opening the gate when suddenly we noticed two puppies run onto our lawn.They were differently colored — one was brown as a cup of coffee with its tail and ear hair dipped in black while the other puppy was a biscuit color with the tip of its tail dipped in white and its forehead had a white spot. And that's why we named them Coffee and Biscuit. Coffee is female and Biscuit is male. In other words they both are twins by a minute. Well, since then they have been with us.Now here comes the shocking part. Coffee has been with us only for one year and 11 months. Guess why? Coffee will soon be littering. At this time it is known that dogs have a sense of temper and start biting people that come near them. And so with a lot of arguing I let my parents take Coffee to a farmer's far away. They tried to assure me that Coffee would have a great time. But Biscuit and I couldn't stand the pain of losing a friend and a sister.After a month, the night before Christmas I made my Christmas wish: I wished that Coffee would come back to me on Santa's sleigh (雪橇). The next morning, I woke up to hear the singing of birds. I jumped out of bed and ran towards the Christmas tree. There were almost 100 presents. We started toopen our gifts. In the end there was just one remaining. It was wrapped in a green gift paper and tied with a red ribbon. I opened it up and guessed who jumped out. It was my lost Coffee. She looked absolutely beautiful. My parents explained that she had been scratching the front door last night!1．The story mainly tells ________．A．a lovely and little pet was recoveredB．the author gave up his petC．the pet can bring a whole world of happinessD．people should protect dogs2．The features of the dogs are that ________．A．Coffee's ear is brownB．Biscuit likes to eat biscuits terriblyC．Coffee has a black tailD．Biscuit's forehead has a white spot3．The author had got Coffee lost ________.A．because it often fought with BiscuitB．because his parents disliked a dog at homeC．because it began attacking peopleD．because it get lost when going out4．What can we infer from the passage?A．Coffee will be thrown away after Christmas.B．His parents will never give Coffee away again.C．The author is always worried about Coffee.D．His parents hardly bought presents for the author.B“Children who eat three or more hamburgers a week are more likely to suffer from asthma (哮喘)，” researchers say. However, they say that the risk may not be caused by the meat­rich diet itself, but by a lack of fresh fruit and vegetables.The study looked into 50，000 children from 20 countries. Parents were asked about their children's eating habits and whether they had ever suffered from asthma. The effects of their diet seemed to be different depending on where they lived. Fruit and vegetables are more helpful in less affluent parts of the world, while eating lots of fish was more helpful in richer countries. Eating at least three hamburgers a week was linked to a greater risk of asthma, but only in wealthier countries.Dr Gabriele Nagel said different things may cause asthma in different parts of the world. “This gives us more understanding of how asthma affects different people, and its effects in developing as well as developed countries，” she said.High levels of unhealthy fat in hamburgers could increase the risk of asthma. However, the researchers said children who ate several hamburgers a week were likely to have other unhealthy lifestyle habits as well.Researchers said the study helped us understand how asthma and diet are connected.“Previous studies have shown that a diet­rich in fruit and vegetables can help to reduce a child's risk of developing asthma symptoms，” Dr Elaine Vickers said.“Our advice to parents is to make sure that children have a healthy diet and also get plenty of exercise.”5．What can we learn about the study from the passage?A．Fat in hamburgers doesn't increase the risk of asthma.B．Children who like hamburgers don't like exercise.C．Eating three hamburgers a month could cause asthma.D．The parents of fifty thousand children were interviewed.6．What did Dr Elaine Vickers advise parents to do?A．To help children have healthy lifestyle.B．To keep children away from hamburgers.C．To have dinner with children at home.D．To make sure children take exercise as much as possible.7．The passage is mainly about ________．A．asthma affects people in wealthy placesB．hamburgers raise a child's risk of asthmaC．parents can influence children's healthD．more and more children become fatterCThe Key to Your HealthA car needs gas to run and your body also needs food to work for you. Eating the right kind of foodis very important. It can help your body grow strong to take care of what you eat.There are four main food groups altogether. The dairy group has food like milk, cheese and sour milk. The other three groups are the meat and fish group, the fruit and vegetable group, and the bread and rice group. Each meal should have at least one food from all four main groups. With all these food together, you will be given enough energy during the day.It is easy to get into bad eating habits. You may eat your breakfast in a hurry to get to school on time. Or you may not have time for a good lunch. It may seem easy to finish your supper with fish and chips all the time. But you will find yourself tired in these days and you can not think quickly.Watching what you eat will help your body remain healthy and strong. It is also good to take some exercise. It will help you eat more if you take a walk or play games in the open air. Having a good eating habit with some exercise is the key to your health.8．Which of the following diets do you think is the best one?A．Milk, bread, cabbages and beef.B．Eggs, tomatoes and chicken.C．Corn, fish, cream and pork.D．Rice, beancurd, apples, fish and chicken.9．Which of the following is a good eating habit?A．Eating fish and chips for supper all the time.B．Finishing your lunch in a very short time.C．Going to school without any breakfast.D．Having at least one food from all four groups in each meal.10．In this passage the writer mainly tells us that ________．A．every person needs food to grow wellB．taking exercise can keep your body strongC．the right kind of food with exercise will keep you healthyD．enough energy helps people think more quickly11．The underlined word “dairy” in the second paragraph means “________”．A．a farm where cows are keptB．the food made out of cows such as milk and butterC．the shop that sells milk and butterD．a place where milk products are madeDDo you know what Street Sense is? It's a newspaper in Washington D．C. about homeless people and problems that affect them. Homeless, or formerly homeless, Washingtonians write many of the articles. The newspaper's business model is based on homeless sellers who sell the newspaper. You can hear them call out “Street Sense for sale！” near subway entrances, lunch places and other areas aroundthe city.The Street Sense newspaper is housed in an office in a Christian church in Washington. Every other Wednesday about fourteen thousand copies are printed. The newspaper expresses the thoughts and experiences of people who call the streets home.Four staff members work at Street Sense, and two of them are paid. The staff members write the first two pages of the paper. Interns — students working as part of their studies — and volunteers help. Homeless writers provide the rest of the material. This includes poems, stories and essays.Street Sense provides training for the homeless people who want to become part of the sales team. After the training, each student is given ten free copies of Street Sense. Once those are sold, trainees become real salespeople. They buy papers for thirty­five cents each and sell them for a dollar.Lisa Gillespie is the managing editor of Street Sense. She says the newspaper plays a part in homeless people's lives that other media cannot. With the help of the newspaper, a lot of homeless people have become confident again, and their lives have also been improved a lot.12．What does the writer mainly tell us in Paragraph 1?A．Something about the Street Sense.B．Homeless people in Washington D．C.C．How Street Sense solves homeless people's problems.D．Street Sense is very interesting.13．How often is Street Sense printed?A．Once a week.B．Twice a week.C．Every two weeks. D．Once a month.14．What do the staff members mainly do at Street Sense?A．They sell newspapers along the street.B．They write the first two pages of the material.C．They edit the newspaper after they receive articles.D．They provide the last two pages of the material.15．What can we infer from this passage?A．If you are one of the trainees of Street Sens e, you can get ten dollars from selling all of your newspapers.B．There are too many homeless people in Washington D．C. for the government to help.C．Most articles of Street Sense are about Washingtonians' lives.D．If you become real salespeople of Street Sense, you can get thirty­five cents from a copy of the newspaper.Ⅰ.任务型阅读(共5小题；每小题2分，满分10分)根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。

一、选择题1．已知集合{}11M x Z x =∈-≤≤，{}Z (2)0N x x x =∈-≤，则如图所示的韦恩图中的阴影部分所表示的集合为（ ）A ．{}0,1B ．{}1,2-C ．{}1,0,1-D ．1,0,1,22．由实数x ，﹣x ，|x |2x 33x - ） A ．2个B ．3个C ．4个D ．5个3．设有限集合A =123{,,,}n a a a a ，则称123A n S a a a a =++++为集合A 的和.若集合M ={x ︳2,N ,6x t t t *=∈<},集合M 的所有非空子集分别记为123,,,k P P P P ，则123k P P P P S S S S ++++=（ ）A ．540B ．480C ．320D ．2804．已知}{|21M x x =-<<，3|0x N x x ⎧-⎫=≤⎨⎬⎭⎩，则M N ⋂=（ ） A ．()0,1 B ．[)0,1C ．(]1,3D ．[]0,35．已知集合302x A xx ⎧⎫+⎪⎪=⎨⎬-⎪⎪⎩⎭，{}B y y m =<，若A B ⊆，则实数m 的取值范围为( ) A ．()2∞+， B ．[)2∞+，C ．()3∞-+，D ．[)3∞-+，6．已知集合2{|120}A x x x =--≤, {|211}B x m x m =-<<+.且AB B =，则实数m 的取值范围为 ( ) A ．[-1，2)B ．[-1，3]C ．[-2，+∞)D ．[-1，+∞)7．集合{}*|421A x x N =--∈，则A 的真子集个数是（ ） A ．63B ．127C ．255D ．5118．若集合{}2|560A x x x =-->，{}|21xB x =>，则()R C A B =（ ）A ．{}|10x x -≤<B ．{}|06x x <≤C ．{}|20x x -≤<D ．{}|03x x <≤9．设集合{}2110P x x ax =++>，{}2220P x x ax =++>，{}210Q x x x b =++>，{}2220Q x x x b =++>，其中a ，b ∈R 下列说法正确的是（ ） A ．对任意a ，1P 是2P 的子集；对任意的b ，1Q 不是2Q 的子集 B ．对任意a ，1P 是2P 的子集；存在b ，使得1Q 是2Q 的子集 C ．存在a ，使得1P 不是2P 的子集；对任意的b ，1Q 不是2Q 的子集 D ．存在a ，使得1P 不是2P 的子集；存在b ，使得1Q 是2Q 的子集10．已知集合A ，B 是实数集R 的子集，定义{},A B x x A x B -=∈∉，若集合1113A y y x x ⎧⎫==≤≤⎨⎬⎩⎭，，{}21,12B y y x x ==--≤≤，则B A -=（ ）A ．[]1,1-B ．[)1,1-C ．[]0,1D ．[)0,111．设{}|22A x x =-≥，{}|1B x x a =-<，若A B =∅，则a 的取值范围为（ ） A ．1a <B ．01a <≤C ．1a ≤D ．03a <≤12．已知函数()f x =M ，()ln(1)g x x =+的定义域为N ，则()R MC N =（ ）A ．{|1}<x xB ．{|1}x x ≥C ．φD ．{|11}x x -≤<二、填空题13．已知{}A x x =>，{|(3)(3)0}B x x x x =-+>，则AB =________14．已知集合(){}22112|2103x P x Q x x x m ⎧-⎫=-=-+-⎨⎬⎩⎭≤，≤,其中m >0,全集U =R .若“Ux P ∈”是“∈Ux Q ”的必要不充分条件,则实数m 的取值范围为__________.15．已知集合A ={x |x ≥2}，B ={x ||x ﹣m |≤1}，若A ∩B =B ，则实数m 的取值范围是______． 16．已知集合{}10,A x ax x R =+=∈，集合{}2280B x x x =--=，若A B ⊆,则a 所有可能取值构成的集合为______________17．设集合{}[1,2),0M N x x k =-=-≤,若M N ⋂=∅，则实数k 的取值范围为_______.18．已知集合{|11},{|01}A x a x a B x x =-<<+=<<若A B φ⋂=，实数a 的取值范围是______．19．对于集合M ，定义函数1()1M x M f x x M ∈⎧=⎨-∉⎩，对于两个集合M 、N ，定义集合{|()()1}M N M N x f x f x *=⋅=-,用()Card M 表示有限集合M 所含元素的个数,若{1,2,4,8}A =,{2,4,6,8,10}B =,则能使()()Card X A Card X B *+*取最小值的集合X 的个数为________.20．任意两个正整数x 、y ，定义某种运算⊗：()()x y x y x y x y x y +⎧⊗=⎨⨯⎩与奇偶相同与奇偶不同，则集合{(,)|6,,}M x y x y x y =⊗=∈*N 中元素的个数是________三、解答题21．设集合{}{}222280,430A x x x B x x ax a =+-<=-+= （1）若x A ∈是x B ∈的必要条件，求实数a 的取值范围；（2）是否存在实数a ，使A B ϕ⋂≠成立？若存在，求出实数a 的取值范围；若不存在，请说明理由.22．设集合{}227150A x x x =+-≤，{}122B x a x a =-<<． （Ⅰ）若B =∅，求实数a 的取值集合； （Ⅱ）若A B ⊆，求实数a 的取值集合．23．已知集合A ={x |12x -≤≤}，B ={x |123m x m +≤≤+} （1）当m =1时，求AB ；（2）若B A ⊆，求实数m 的取值范围24．已知集合A ＝{x |2≤x <7}，B ＝{x |3<x <10}，C ＝{x |x a ≤}． (1)求A ∪B ，(∁R A )∩B ；(2)若A ∩C ≠∅，求a 的取值范围．25．已知集合2{|320}A x ax x =-+=，其中a 为常数，且a R ∈． （1）若A 中至少有一个元素，求a 的取值范围； （2）若A 中至多有一个元素，求a 的取值范围．26．已知不等式()210x a x a -++≤的解集为A ，不等式2103x x +≤-的解集为B . （1） 当3a =时，求A B ；（2）若不等式的解集A B ⊆，求实数a 的取值范围.【参考答案】***试卷处理标记，请不要删除一、选择题 1．B 解析：B 【分析】阴影部分可以用集合M N 、表示为()()M N C M N ⋃⋂，故求出M N 、、M N ⋃，M N ⋂即可解决问题． 【详解】解：由题意得，{}1,0,1M =-，{}0,1,2N ={}1,0,1,2M N ⋃=-，{}0,1M N ⋂=阴影部分为()(){}1,2M N C M N ⋃⋂=-故选B 【点睛】本题考查用韦恩图表示的集合的运算，解题时要能用集合的运算表示出阴影部分．2．A解析：A 【分析】根据绝对值的定义和开平方、立方的方法,应对x 分0,0,0x x x >=<三种情况分类讨论,根据讨论结果可得答案. 【详解】当0x >时，0x x x ===-<，此时集合共有2个元素，当0x =时，0x x x ====-=，此时集合共有1个元素，当0x <时，0x x -===>，此时集合共有2个元素，综上所述，此集合最多有2个元素. 故选：A . 【点睛】本题考查了元素与集合关系的判断及根式的化简求值,其中解答本题的关键是利用分类讨论思想,对x 分三种情况进行讨论,是基础题.3．B解析：B 【分析】求出{2,4.6.8.10}M =后，分别求出含有2,4,6,8,10的子集个数，然后可求得结果. 【详解】{2,4.6.8.10}M =，其中含有元素2的子集共有4216=个，含有元素4的子集共有4216=个，含有元素6的子集共有4216=个，含有元素8的子集共有4216=个，含有元素10的子集共有4216=个， 所以123k P P P P S S S S ++++(246810)16480=++++⨯=.故选：B 【点睛】本题考查了对新定义的理解能力，考查了集合的子集个数的计算公式，属于基础题.4．A【分析】根据分式不等式的解法，求得{}03N x x =<≤，再结合集合的交集的运算，即可求解. 【详解】由题意，集合{}3|003x N x x x x ⎧-⎫=≤=<≤⎨⎬⎭⎩， 又由}{|21M x x =-<<，所以{}()010,1M N x x ⋂=<<=. 故选：A. 【点睛】本题主要考查了集合交集的概念及运算，以及分式不等式的求解，其中解答中正确求解集合N 是解答的关键，着重考查运算与求解能力.5．B解析：B 【分析】求出集合A ，由A B ⊆，结合数轴，可得实数m 的取值范围. 【详解】 解不等式302x x +≤-，得32x -≤<，[)3,2A ∴=-. A B ⊆，可得2m ≥.故选：B . 【点睛】本题考查集合间的关系，属于基础题.6．D解析：D 【分析】 先求出集合A ,由A B B =,即B A ⊆,再分B φ=和B φ≠两种情况进行求解.【详解】由2120x x --≤,得34x -≤≤. 即[3,4]A =-. 由AB B =,即B A ⊆.当B φ=时，满足条件,则211m m -≥+解得2m ≥.当B φ≠时，要使得B A ⊆，则12121314m m m m +>-⎧⎪-≥-⎨⎪+≤⎩.解得：12m -≤<.综上满足条件的m 的范围是：1m ≥-.【点睛】本题主要考查集合的包含关系的判断及应用，以及集合关系中的参数范围问题，考查分类讨论思想，属于中档题.7．B解析：B 【分析】先求得{}*|421A x x N =--∈的元素个数,再求真子集个数即可.【详解】由{}*|421A x x N=--∈,则421x --为正整数.则21x -可能的取值为0,1,2,3,故210,1,2,3x -=±±±,故x 共7个解.即{}*|421A x x N =--∈的元素个数为7故A 的真子集个数为721127-= 故选：B 【点睛】本题主要考查集合中元素个数的求解与知识点：元素个数为n 的集合的真子集有21n -个. 属于基础题型.8．B解析：B 【解析】 【分析】求得集合{|1A x x =<-或6}x >，{}|0B x x =>，根据集合运算，即可求解，得到答案． 【详解】由题意，集合{}2|560{|1A x x x x x =-->=<-或6}x >，{}{}|21|0x B x x x =>=>，则{}|16R C A x x =-≤≤，所以(){}|06R C A B x x =<≤.故选B ． 【点睛】本题主要考查了集合的混合运算，其中解答中正确求解集合,A B ，结合集合的运算求解是解答的关键，着重考查了推理与运算能力，属于基础题．9．B解析：B 【分析】先证得1P 是2P 的子集，然后求得b 使1Q 是2Q 的子集，由此确定正确选项．【详解】对于1P 和2P ，由于210x ax ++>时222110x ax x ax ++=+++>，所以1P 的元素，一定是2P 的元素，故对任意a ，1P 是2P 的子集；对于1Q 和2Q ，根据判别式有140440b b -<⎧⎨-<⎩，即1b >时，12Q Q R ==，满足1Q 是2Q 的子集，也即存在b ，使得1Q 是2Q 的子集. 故选： B. 【点睛】方法点睛：该题主要考查子集的判断，解题方法如下：（1）利用子集的概念，可以判断出1P 的元素，一定是2P 的元素，得到对任意a ，1P 是2P 的子集；（2）利用R 是R 的子集，结合判别式的符号，存在实数1b >时，有12Q Q R ==，得到结果.10．B解析：B 【分析】先根据题意得{}13A y y =≤≤，{}13B y y =-≤≤，再根据集合运算即可得答案. 【详解】解：根据题意得{}111133A y y x y y x ⎧⎫==≤≤=≤≤⎨⎬⎩⎭，， {}{}21,1213B y y x x y y ==--≤≤=-≤≤，再根据集合的运算得}{11B A y y -=-≤<. 故选：B. 【点睛】本题考查集合的运算，函数值域的求解，考查运算能力，是中档题.11．C解析：C 【分析】解集绝对值不等式求得,A B ，结合A B =∅求得a 的取值范围.【详解】由22x -≥得22x -≤-或22x -≥，解得0x ≤或4x ≥，所以(][),04,A =-∞⋃+∞， 由1x a -<得1a x a -<-<，解得11a x a -<<+，所以()1,1B a a =-+. 当0a ≤时，B =∅，AB =∅，符合题意.当0a >时，由于A B =∅，所以1014a a -≥⎧⎨+≤⎩，解得01a <≤.综上所述，a 的取值范围是1a ≤. 故选：C 【点睛】本小题主要考查绝对值不等式的解法，考查根据交集的结果求参数的取值范围.12．A解析：A 【解析】 【分析】根据函数定义域的求法求得,M N ，再求得()R M C N .【详解】由210x ->解得11x -<<，由10x +>解得1x >-.所以{}|1R C N x x =≤-，故()R MC N ={|1}<x x ，故选A.【点睛】本小题主要考查函数定义域的求法，考查集合补集和并集的运算，属于基础题.二、填空题13．【分析】先分别求解集合中元素的所满足的不等式再由交集的定义求解即可【详解】由题因为解得则因为解得或则或所以故答案为:【点睛】本题考查集合的交集运算考查含根式的不等式的运算考查解高次不等式 解析：{|30}-<<x x【分析】先分别求解集合中元素的所满足的不等式,再由交集的定义求解即可 【详解】 由题,因为220x x x ⎧->⎪⎨-≥⎪⎩,解得1x <,则{}|1A x x =<,因为()()330x x x -+>,解得30x -<<或3x >,则{|30B x x =-<<或}3x >, 所以{}|30A B x x ⋂=-<<, 故答案为:{|30}-<<x x 【点睛】本题考查集合的交集运算,考查含根式的不等式的运算,考查解高次不等式14．【分析】解出集合PQ 根据充分条件和必要条件关系得出两个集合的包含关系即可求出范围【详解】由题：是的必要不充分条件即PQ 解不等式所以0PQ 所以解得：故答案为：【点睛】此题考查根据充分条件和必要条解析：9m ≥【分析】解出集合P ，Q ，根据充分条件和必要条件关系得出两个集合的包含关系即可求出范围. 【详解】 由题：“Ux P ∈”是“∈Ux Q ”的必要不充分条件,UQUP ，即P Q ，解不等式1123x --≤，12123x --≤-≤， 646x -≤-≤，210x -≤≤所以[]1122,103x P x ⎧-⎫=-=-⎨⎬⎩⎭≤， (){}()()()(){}22|210|110Q x x x m x x m x m =-+-=-+--≤≤，m >0,P Q ， 所以11012m m +≥⎧⎨-≤-⎩，解得：9m ≥.故答案为：9m ≥ 【点睛】此题考查根据充分条件和必要条件判断集合的包含关系求解参数范围，关键在于准确判断两个集合的包含关系，列出不等式组求解.15．3+∞）【分析】先求出集合再利用交集定义和不等式性质求解【详解】∵集合解得∴实数m 的取值范围是故答案为：【点睛】本题考查实数的取值范围的求法解题时要认真审题注意不等式性质的合理运用是基础题解析：[3，+∞） 【分析】先求出集合B ，再利用交集定义和不等式性质求解． 【详解】∵集合{|2}A x x =≥，{|||1}{|11}B x x m x m x m =-≤=-≤≤+，A B B =，12m ∴-≥，解得3m ≥，∴实数m 的取值范围是[)3,+∞． 故答案为：[)3,+∞． 【点睛】本题考查实数的取值范围的求法，解题时要认真审题，注意不等式性质的合理运用，是基础题.16．【分析】先化简集合利用分类讨论和即可求出构成的集合【详解】由可得:即:解得或故:由可得:当时方程无实数解此时满足当时方程的实数解为故:由可得:或解得或的所有取值构成的集合为:故答案为:【点睛】本题主解析：11{0,,}24-【分析】先化简集合B ,利用A B ⊆,分类讨论=0a 和0a ≠,即可求出构成a 的集合. 【详解】由{}2280B x x x =--=可得:2280x x --= 即:()()240x x +-= 解得2x =-或4x = 故:{}2,4B =- {}10,A x ax x R =+=∈ 由10ax += 可得:1ax =-当0a =时,方程1ax =-无实数解,此时A =∅,满足A B ⊆ 当0a ≠时,方程1ax =-的实数解为1x a =-,故:1{}A a=- 由A B ⊆可得:12a -=-或14a -= 解得12a =或14a =-a 的所有取值构成的集合为:11{0,,}24-.故答案为:11{0,,}24-. 【点睛】本题主要考查了集合间的基本关系以及一元二次方程的解法,要注意集合A 是集合B 的子集时,集合A 有可能是空集.17．【分析】首先求得集合N 然后确定实数k 的取值范围即可【详解】由题意可得：结合可知实数k 的取值范围是：故答案为：【点睛】本题主要考查交集的运算由集合的运算结果求参数取值范围的方法等知识意在考查学生的转化 解析：{}|1k k <-【分析】首先求得集合N ，然后确定实数k 的取值范围即可. 【详解】由题意可得：{}|N x x k =≤，结合M N ⋂=∅可知实数k 的取值范围是：1k <-. 故答案为：{}|1k k <-． 【点睛】本题主要考查交集的运算，由集合的运算结果求参数取值范围的方法等知识，意在考查学生的转化能力和计算求解能力.18．【分析】由根据集合的交集的运算得到或即可求解【详解】由题意集合因为则满足或解得或即实数的取值范围是故答案为：【点睛】本题主要考查了集合的运算以及利用集合的交集求参数其中解答中熟记集合交集运算列出相应 解析：(][),12,-∞-⋃+∞【分析】由A B φ⋂=，根据集合的交集的运算，得到11a -≥或10a +≤，即可求解.【详解】由题意，集合{|11},{|01}A x a x a B x x =-<<+=<<，因为A B φ⋂=，则满足11a -≥或10a +≤，解得2a ≥或1a ≤-，即实数a 的取值范围是(][),12,-∞-⋃+∞.故答案为：(][),12,-∞-⋃+∞.【点睛】本题主要考查了集合的运算，以及利用集合的交集求参数，其中解答中熟记集合交集运算，列出相应的不等式是解答的关键，着重考查了推理与运算能力，属于基础题. 19．【分析】通过定义可以用集合中的补集来解释再根据取最小值时所满足的条件最后可以求出集合的个数【详解】因为所以有要想最小只需最大且最小要使最小则有所以集合是集合和集合子集的并集因此集合的个数为个故答案为 解析：8【分析】通过定义可以用集合中的补集来解释，再根据()()Card X A Card X B *+*取最小值时所满足的条件,最后可以求出集合X 的个数.【详解】因为{|()()1}M N M N x f x f x *=⋅=-,所以有()M N M N C M N *=⋂,要想()Card X A *最小,只需()Card X A ⋂最大,且()Card X A ⋃最小,要使()()Card X A Card X B *+*最小, 则有A B X A B ⋂⊆⊆⋃,{}{}1,2,4,6,8,10,2,4,8A B A B ⋃=⋂=,所以集合X 是集合{}2,4,8和集合{}1,6,10子集的并集,因此集合X 的个数为328=个.故答案为：8【点睛】本题考查了新定义题,考查了集合与集合之间的关系,考查了数学阅读能力.20．【分析】根据正整数的奇偶讨论的不同取值情况:若一奇一偶则取;若都是奇数或都是偶数则取列举出所有可能即可【详解】集合若一奇一偶则取此时所有个数为此时共有4个;若都是偶数则取此时所有个数为此时共有2个;解析：9【分析】根据正整数的奇偶,讨论x y 、的不同取值情况:若一奇一偶,则取6xy =;若都是奇数或都是偶数,则取6x y +=,列举出所有可能即可.【详解】集合{(,)|6,,}M x y x y x y =⊗=∈*N若x y 、一奇一偶,则取6xy =,此时所有个数为16x y =⎧⎨=⎩,23x y =⎧⎨=⎩,32x y =⎧⎨=⎩,61x y =⎧⎨=⎩,此时(),x y 共有4个;若x y 、都是偶数,则取6x y +=,此时所有个数为24x y =⎧⎨=⎩,42x y =⎧⎨=⎩,此时共(),x y 有2个; 若x y 、都是奇数,则取6x y +=,此时所有个数为15x y =⎧⎨=⎩,33x y =⎧⎨=⎩, 51x y =⎧⎨=⎩此时(),x y 共有3个;综上可知,满足条件的元素共有9个.故答案为:9【点睛】本题考查了新定义运算与集合的综合应用,注意分析题意并正确理解新定义是解决此类问题的关键,属于中档题.三、解答题21．（1）4233a -<<；（2）存在，42a -<<. 【分析】（1）x A ∈是x B ∈的必要条件可转化为B A ⊆，建立不等式求解即可；（2）假设A B ⋂≠∅，建立不等关系，有解则存在，无解则不存在.【详解】 {}42A x x =-<<，()(){}30B x x a x a =--=（1）由已知得：B A ⊆ 42432a a -<<⎧∴⎨-<<⎩ 4233a ⇒-<<， 即实数a 的取值范围4233a -<<， （2）假设存在a 满足条件， 则42a -<<或432a -<<，42a ∴-<<即存在42a -<<使A B ⋂≠∅.【点睛】本题主要考查了根据集合的包含关系求参数的取值范围，考查了必要条件，属于中档题. 22．（Ⅰ）14a ≤；（Ⅱ）{}3a a >. 【分析】（Ⅰ）由空集的意义知，当且仅当212a a ≤-时，集合B 中无任何元素，解不等式即可得实数a 的取值范围；（Ⅱ）根据A B ⊆，得到a 的取值范围，即可得到结论．【详解】解：∵集合{}()(){}2327150235052A x x x x x x x x ⎧⎫=+-≤=-+≤=-≤≤⎨⎬⎩⎭， （Ⅰ）∵B =∅，∴{}122x a x a -<<=∅，∴212a a ≤-，解得14a ≤， （Ⅱ）∵A B ⊆，则集合B ≠∅,所以212a a >-，则14a >∴1253322a a a -<-⎧⎪⇒>⎨>⎪⎩∴实数a 的取值集合为{}3a a >．【点睛】本题考查解二次不等式，根据集合的包含关系求参数的范围，属于中档题. 23．（1）{}2；（2）1,2⎛⎤-∞ ⎥⎝⎦. 【分析】（1）根据集合的交集运算求解即可；（2）讨论集合B 是否为空集，根据包含关系列出不等式，即可得出实数m 的取值范围.【详解】（1）当m =1时，B ={x |2≤x ≤5}，因此A B ={2} （2）A B ⇔B A ⊆，则①当B =∅时，即123m m +>+，即2m <-，符合题意②当B ≠∅时，要满足B A ⊆，则12311232m m m m +≤+⎧⎪+≥-⎨⎪+≤⎩2212m m m ⎧⎪≥-⎪⇒≥-⎨⎪⎪≤-⎩122m ⇒-≤≤-综上所述，当B A ⊆时，实数m 的取值范围时1(,2)2,2⎡⎤-∞-⋃-⎢⎥⎣⎦=1,2⎛⎤-∞ ⎥⎝⎦ 【点睛】本题考查交集的运算，同时也考查了利用集合的包含关系求参数，解题的关键就是对含参集合分空集和非空集合两种情况讨论，考查分类讨论思想的应用，属于中档题. 24．(1) {x |2≤x <10}, {x |7≤x <10};(2) 2a ≥【分析】（1）根据交、并、补集的运算分别求出A ∪B ，（∁R A ）∩B ；（2）根据题意和A∩C≠∅，即可得到a 的取值范围．【详解】解：(1)因为A ＝{x |2≤x <7}，B ＝{x |3<x <10}，所以A ∪B ＝{x |2≤x <10}．因为A ＝{x |2≤x <7}，所以∁R A ＝{x |x <2，或x ≥7}，则(∁R A )∩B ＝{x |7≤x <10}．(2)因为A ＝{x |2≤x <7}，C ＝{x |x a ≤}，且A ∩C ≠∅，所以2a ≥所以a 的取值范围为2a ≥．【点睛】高考对集合知识的考查要求较低，均是以小题的形式进行考查，一般难度不大，要求考生熟练掌握与集合有关的基础知识．纵观近几年的高考试题，主要考查以下两个方面：一是考查具体集合的关系判断和集合的运算．解决这类问题的关键在于正确理解集合中元素所具有属性的含义，弄清集合中元素所具有的形式以及集合中含有哪些元素．二是考查抽象集合的关系判断以及运算．25．（1）9,8⎛⎤-∞ ⎥⎝⎦；（2）{}90,8⎡⎫⋃+∞⎪⎢⎣⎭【分析】（1）对a 分类讨论：0a =，解出即可判断出是否满足题意．0a ≠时，A 中至少有一个元素，满足0∆，解得a 范围即可得出．（2）对a 分类讨论：0a =，直接验证是否满足题意．0a ≠时，由A 中至多有一个元素，可得0∆≤，解得a 范围即可得出．【详解】解：（1）0a =，由320x -+=，解得23x =，满足题意，因此0a =． 0a ≠时，A 中至少有一个元素，∴980a ∆=-，解得98a ，0a ≠． 综上可得：a 的取值范围是9,8⎛⎤-∞ ⎥⎝⎦．（2）0a =，由320x -+=，解得23x =，满足题意，因此0a =． 0a ≠时，A 中至多有一个元素，∴980a ∆=-，解得98a． 综上可得：a 的取值范围是{}90,8⎡⎫⋃+∞⎪⎢⎣⎭．【点睛】 本题考查了集合的性质、一元二次方程的实数根与判别式的关系，考查了分类讨论方法、推理能力与计算能力，属于中档题．26．（1）{}|13A B x x ⋂=≤<（2）132a -≤< 【分析】 先求解不等式,可得1|32B x x ⎧⎫=-≤<⎨⎬⎩⎭, （1）当3a =时,{}|13A x x =≤≤,再由交集的定义求解即可；（2）由A B ⊆,判断a 与集合B 的端点的位置即可.【详解】由题,因为()210x a x a -++≤,则()()10x a x --≤, 因为2103x x +≤-,即()()213030x x x ⎧+-≤⎨-≠⎩,所以132x -≤<,即集合1|32B x x ⎧⎫=-≤<⎨⎬⎩⎭, （1）当3a =时,()()310x x --≤,解得13x ≤≤,即{}|13A x x =≤≤,所以{}|13A B x x ⋂=≤<（2）由题,当1a <时,{}|1A x a x =≤≤；当1a ≥时,{}|1A x x a =≤≤,因为A B ⊆,所以132a -≤< 【点睛】本题考查集合的交集运算,考查已知集合的包含关系求参数问题,考查解一元二次不等式和分式不等式.。

最新北师大版高中英语必修一各单元测试题（全册共3个单元附答案）目录单元达标检测(一)Lifestyles单元达标检测(二)Heroes单元达标检测(三)Celebration单元达标检测(一)Lifestyles【说明】本试卷满分150分。

考试时间120分钟。

第一部分听力(共两节，满分30分)第一节(共5小题；每小题1.5分，满分7.5分)听下面5段对话。

每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1．What does the man like best?A．Apples.B．Bananas.C．Pears.2．What does the woman mean?A．Lucy is not going to Jenny's birthday party.B．Lucy likes surprising others.C．Lucy is often late.3．Where do the speakers want to go?A．The supermarket.B．The post office.C．The cinema.4．Where is Mr. Lee?A．In his office.B．In the meeting room.C．On the playground.5．What was the woman's plan for tonight?A．To see a movie with the man.B．To visit the man's grandparents.C．To have classes with the man.第二节(共15小题；每小题1.5分，满分22.5分)听下面5段对话或独白。

每段对话或独白后有几个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。

北师大版必修一Unit1单元测试题(含答案解析)北师大版必修一Unit1单元测试题(含答案解析)第一部分听力（共三节，满分30分）做题时，先将答案标在试卷上。

录音内容结束后，你将有两分钟的时间将试卷上答案转涂到答题卡上。

第一节（共5 小题；每小题 1.5 分，满分7.5 分）听下面5段对话。

每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

例：How much is the shirt?A. ￡19.15.B. ￡9.15.C. ￡9.18.答案是B。

1.What does the man mean?A.Of course, you canB.No, please don’tC.Never2.What does the husband tell his wife to do?A.To buy a new dress.B.To decide by herself quickly.C.To wear an evening dress.3.What are they doing?A.Making an appointment(约会).B.Deciding what to doC.Checking the time.4.How many pills should the man take in24 hours?A.4 pillsB.24 pillsC.16 pills5.Why isn’t Jack present?A.Because he wasn’t invited.B.Because he changed hi mind.C.Because he forgot the invitation.第二节（共15小题；每小题 1.5 分，满分22.5 分）听下面 5 段对话或独白。

每段对话或独白后有几个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。

第一单元测评(时间:120分钟满分:150分)第一部分听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。

每段对话后有一个小题,从题中所给的 A、B、C 三个选项中选出最佳选项。

听完每段对话后,你都有 10 秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

1.What is the woman?A.A student.B.A teacher.C.A saleswoman.2.What will the man do on Friday?A.See a film.B.Attend a lecture.C.Go to the theater.3.Which character is the woman’s favourite in Friends?A.Monica.B.Phoebe.C.Joey.4.What is the man complaining about?A.His career.B.His health.C.His kid.5.Who will the man celebrate the festival with?A.His brother.B.His parents.C.His grandparents.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。

每段对话或独白后有几个小题,从题中所给的 A、B、C 三个选项中选出最佳选项。

听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。

每段对话或独白读两遍。

听第6段材料,回答第6、7题。

6.What’s the woman doing?A.Making an invitation.B.Asking for help.C.Giving advice.7.What does the man plan to do today?A.Play tennis.B.Go on a trip.C.Help Emily study.听第7段材料,回答第8、9题。

【6份】高一英语北师大版必修 1单元练习含答案Unit1 Lifestyles单元练习1（Lesson1&2）一、单词拼写(单句首字母填空)（共15小题；共15分）1. Chinese students always c about the stress of schoolwork.2. At that moment, he p not to think about the future.3. My car is in u need of repair.4. He doesn't like his job because it is very s .5. People who love peace want to live a p life.6. She is s from loss of memory.7. The missile is guided by r control.8. I'd like to see all the legal d concerning the sale of this land before we sign this paper.9. He studied chemistry and then s to biology.10. For example, you might o a group of animal-lovers among your friends or classmates.11. The p of the water caused the wall of the dam to crack.12. Eating just one carrot a day can r the possibility of getting the serious eye disease by 40%.13. If you want to keep fit, you'd better go to Dr. Smith, who is an e on losing weight.14. The doctor suggests that people who have a high blood pressure have a d low in salt.15. He has a r lifestyle now. All he does is fishing.二、单词拼写(根据中文提示拼写单词)（共15小题；共15分）16. Still a lot of people in Africa (遭受) from hunger and poverty.17. I (更喜欢) classical music to pop music.18. He slowed the bleeding by applying (压力) to the wounds until the police and ambulance arrived.19. After a career as a journalist, she was looking forward to a happy and (宁静的) retirement.20. He (抱怨) to the police of the boys stealing his apples.21. The report called for (紧急的) action to reduce lead in petrol.22. It's (轻松的) to listen to music after a day's work.23. The sky stretched out clear and bright above our heads as we approached the (遥远的) village.24. I can't (忍受) being laughed at.25. According to the (文件), all the people left the city 200 years ago.26. The government is helping to solve the (社会) problems.27. He became an (知识丰富的) language teacher.28. Wang Ping flashed a (开关) on a computer screen, and a table and some chairs rose from under the floor as if by magic.29. What is a good teacher? First of all, he or she should (组织) his/her lessons well.30. The government will be looking to (减少) the unemployment rate.三、翻译(根据中文提示完成句子)（共10小题；共20分）31. 你应该接受我的建议，开始存钱。

北师大版英语必修一第一单元测试题(含答案解析)第二部分英语知识运用第一节单项填空(共15小题, 满分15分)从A、B、C、D四个选项中，选出可以填入空白处的最佳选项，并在答题卡上将该项涂黑。

21.I can’t stand_____ with Jane in the same company.A.works B worked C.working D.to work22.Rather than_____on a crowed bus, he always prefers____a bicycles.A ride, rideB riding, rideC ride, to rideD to ride, riding23.After he retired from office, Rogers ___ painting for a while, but soon lost interest.A took upB save upC kept upD drew up24.I’d like to study law at university____ my cousin prefers geography.A thoughB asC whileD for25.—The last one___ pays the meal--- Agreed.A arrivedB arrivesC to arriveD arriving26.The farmers___ a lot ___ the flood.A sufferd….fromB Sufferd…..ofC will suffer….forD Suffer…..to27.The hurricane（飓风）damaged many houses and business buildings, ___, it caused 20 deaths.A or elseB thereforeC after allD besides28.--- Be careful when you cross this very busy street, if not, you may___ hurt by a car.A haveB becomeC getD run29.He often complains the weather.A.aboutB.forC.withD.to30.John shut everybody out of the kitchen___ he could prepare his grand surprise for the evening party.A whichB whenC so thatD as if31.It took us much time to get there.It was ____trip.A.a two hoursB.a two-hourC.two-hourD.two hours32.The man seldom has time ____.A.for a funB.for funC.for funsD.for a funs33. The boy was caught ___________in the shop.A. stealingB.stealC. was stealingD. stole34.Tom often____himself as a gifted singer.A.thinksB.thinks ofC.thinks aboutD.thinks over35. --- Are you still busy?----Yes, I___ my work, and it won’t take long.A just finishB am just finishingC have just finishD just going to finish第二节完形填空(共20小题, 满分30分)阅读下面短文，掌握其大意，然后从第36至第55小题所给的A、B、C、D四个选项中，选出最佳选项。

Unit 1 LifestylesLesson 1 A Perfect Day？I.单词拼写（共12小题；每小题1分，满分12分）1. The novel is written from p experience.2. It is most u that the patient should be sentto hospital.3. He has a r lifestyle now. All he does isfishing.4. Chinese students always c about thestress of schoolwork.5. The businessman leads a s life: too manyplaces to go, too many people to see and toomany bills to pay, etc.6. Nowadays, many people have a (便携式的) computer to do their work.7. The (文件) he passed me just now isvery important.8. All the people ran out of the building when thefire （警报）went off.9. There is something wrong with the(遥控) control, so you can't watch TV tonight.10. He went out in order to r his mind.11. I bought a Walt Disney c for my son tocelebrate his birthday.12. I’m b .Let’s go to the cinema.II.短语翻译（共10小题；每小题1分，满分10分）1．打开_______________2．下楼梯______________3．转变，转换频道______________4．换好衣服______________5．占据______________6．be filled with______________7．be ready for…______________8．switch off______________9．complain about sth．______________ 10．get bored____________III.用所给短语的适当形式填空（共10小题；每小题1分，满分10分）1.Don't you think these boxes of yours aretoo much space？2.I didn't until someone knocked at thedoor.3.It's time for you to ,Tom. The sun hasrisen high in the sky.4. The alarm as soon as the thief got intothe front garden.5. His speech was long and boring, and almost all ofus it.6. He all the lights as he left the room to saveelectricity.7. This play is too boring. Will you pleaseto CCTV 5 to watch the live football match?8. The warm-hearted girl led us into a large hallwhich passengers.9. All the friends are waiting downstairs, so don’ tspend too much time .10. Many students always _______ too muchhomework.IV.单项填空（共18小题；每小题1分，满分18分）1.---There is still a copy of the book in the library.Will you go and borrow ?---No, I'd rather buy in the bookstore.A. it; oneB. one; oneC. one; itD. it; it2. I really don't like art, I find his workimpressive.A.AsB. Since C．If D．While3. Without your help，I the exam last term．A．would have failed in B．wouldn’t failC．will have failed in D．failed in4. I have to go to work by taxi because my carat the garage．A．will be repaired B．is repairedC．is being repaired D．has been repaired5.—The water to waste．—Oh，it’s my fault．I forgot to turn it off．A．is running B．has runC．will have run D．will run6. I want to know when he for New York．A．has left B．is leavingC．had left D．would leave7.She was the first in our class the mold：instead of getting a job in education，she decided to become an accountant.A．to break B．breakingC．broke D．broken8. —It shouldn’t long to clear up after theparty if we all volunteer to help．—That’s right．Many hands make light work．A．take B．spendC．cost D．pay9.The young woman got to the cinema two hours ahead of time she could get a good position．A．so that B．in caseC．in order D．now that10. I’ll take the story where I finishedyesterday．A．up B．overC．out D．in11.The number of deaths from heart disease will bereduced greatly if people to eat morefruit and vegetables．A．persuade B．will be persuadedC．be persuaded D．are persuaded12.I don’t like to watch this．There’s a betterprogramme than this on CCTV 6.Please .A．turn back B．turn onC．switch off D．switch over13.Be careful when you cross this very busystreet．If not，you may run over by a car．A．have B．becomeC．get D．turn14.---Have you got any job offers?---No. I .A. waitedB. had been waitingC. have waitedD. am waiting15. ---- What's that noise?---Oh, I forgot to tell you. The newmachine .A. was testedB. will be testedC. is being testedD. has been tested16. We tried to find a table for seven, but they wereall .A. given awayB. kept awayC. taken upD. used up17. The message is very important, so it is ______tobe sent as soon as possible.A. supposeB. supposingC. supposedD. to suppose18. Please fill_____ the blanks _____proper words.A. up; withB. in; withC. up; ofD. in; ofV.短文填词(共10小题；每小题1分，满分10分)阅读P8和P9课文内容，根据所给提示，在每个空格内填入一个适当的英语单词。