最新卓越内测试卷

七年级下册英语卓越配套试卷As a student in Grade 7, the English Excellence Supplementary Workbook has been a valuable resource for meto enhance my language skills. The workbook, which covers a range of topics from basic grammar to reading comprehension, has been instrumental in helping me consolidate my knowledge and improve my performance in school.The workbook is divided into several sections, each focusing on a different aspect of English language learning. The first section, Grammar Focus, provides clear explanations and examples of key grammar points, followedby practice exercises to help me apply what I have learned. The Vocabulary section introduces new words and phrases, along with example sentences and exercises to help me memorize and use them correctly.The Reading Comprehension section is particularly helpful, as it exposes me to different types of texts and challenges me to understand and analyze them. The exercises in this section range from simple comprehension questionsto more complex tasks like summarizing or inferring information. These exercises have not only improved myreading skills but have also enhanced my critical thinking abilities.The workbook also includes a Writing section, which guides me through the process of writing different types of essays. From writing simple narratives to persuasive essays, this section has helped me develop my writing skills and confidence. The exercises provide valuable feedback and suggestions for improvement, which I have found very helpful.In addition to the workbook's structured approach to learning, I have also found it to be a highly engaging resource. The texts are interesting and relevant, and the exercises are challenging but not overwhelming. The workbook's user-friendly layout and clear instructions make it easy to follow, and the included answer key provides convenient reference for self-checking and revision.Studying with the English Excellence Supplementary Workbook has been a rewarding experience for me. Not only has it helped me improve my English skills, but it has also fostered a love for learning and a sense of accomplishment as I complete each section. The workbook's comprehensivecoverage of English language topics and its focus on practical application have made it an invaluable tool for my academic success.**七年级下册英语卓越配套试卷分析与学习心得** 作为七年级的学生，英语卓越配套试卷对于提升我的语言能力起到了重要的作用。

第1页共11页交大附中2022学年第二学期高一年级数学卓越考2023.3一、选择题1．已知,αβ∈R ，则“()sin sin 2αβα+=”是“()2πk k βα=+∈Z ”的（）．A．充分而不必要条件B．必要而不充分条件C．充分必要条件D．既不充分也不必要条件2．设α是第三象限角，则下列函数值一定为负数的是（）．A．cos 2αB．tan2αC．sin2αD．cos2α3．对于给定的实数a ，不等式()2110ax a x +--<的解集可能是（）．A．11x x a ⎧⎫<<⎨⎬⎩⎭B．{}1x x ≠-C．{}1x x <-D．R4．若1tan 3α=-，则222ππcos sin 332sin cos cos ααααα⎛⎫⎛⎫-+- ⎪ ⎪⎝⎭⎝⎭+的值为（）．A．103B．53C．23D．103-5．在ABC △中，内角A ，B ，C 的对边分别为a ，b ，c22cos32BB +=，cos cos sin sin 6sin BC A Bb c C+=，则ABC △的外接圆的面积为（）．A．12πB．16πC．24πD．64π6．阻尼器是一种以提供阻力达到减震效果的专业工程装置．我国第一高楼上海中心大厦的阻尼器减震装置，被称为“定楼神器”，如图1．由物理学知识可知，某阻尼器的运动过程可近似为单摆运动，其离开平衡位置的位移()y m 和时间()t s 的函数关系为()()sin 0,πy t ωϕωϕ=+><，如图2，若该阻尼器在摆动过程中连续三次到达同一位置的时间分别为1t ，2t ，()31230t t t t <<<，且122t t +=，235t t +=，则在一个周期内阻尼第2页共11页器离开平衡位置的位移大于0.5m 的总时间为（）．A．13s B．23s C．1sD．43s 7．函数()()ππsin 0,,22f x A x A ωϕωϕ⎛⎫=+>>-<< ⎪⎝⎭的部分图象如图所示，则ω，ϕ的值分别是（）．A．2，π3-B．2，π6-C．4，π6-D．4，π38．己知函数()πsin 3f x x ⎛⎫=+⎪⎝⎭，给出下列结论：①π3f x ⎛⎫-⎪⎝⎭为奇函数；②π2f ⎛⎫⎪⎝⎭是()f x 的最大值；③把函数sin y x =的图象上所有点向左平移π3个单位长度，可得到函数()y f x =的图象．其中所有正确结论的序号是（）．A．①B．①③C．②③D．①②③9．已知函数()sin 22sin 1f x x x =+-，则()f x 在[]0,2023πx ∈上的零点个数是（）．A．2023B．2024C．2025D．202610．设函数()ln 21ln 21f x x x =++-，则()f x （）．第3页共11页A．是偶函数,且在1,2⎛⎫+∞⎪⎝⎭单调递增B．是奇函数，且在11,22⎛⎫-⎪⎝⎭单调递减C．是偶函数，且在1,2⎛⎫-∞- ⎪⎝⎭单调递增D．是奇函数，且在1,2⎛⎫-∞-⎪⎝⎭单调递增11．已知函数()()π2sin 04f x x ωω⎛⎫=-> ⎪⎝⎭在区间[]0,2π上存在零点，且函数()f x 在区间[]0,2π上的值域为2M ⎡⎤⊆⎣⎦，则ω的取值范围是（）．A．13,42⎡⎤⎢⎥⎣⎦B．13,84⎡⎤⎢⎥⎣⎦C．14,83⎡⎤⎢⎥⎣⎦D．1,18⎡⎤⎢⎥⎣⎦12．已知函数()πsin 3f x x ω⎛⎫=+⎪⎝⎭，0ω>对任意3π0,8x ⎛⎫∈ ⎪⎝⎭都有()12f x >，则当ω取到最大值时，()f x 的一个对称中心为（）．A．π,08⎛⎫⎪⎝⎭B．3π,016⎛⎫⎪⎝⎭C．π,02⎛⎫⎪⎝⎭D．3π,04⎛⎫⎪⎝⎭13．定义在R 上的奇函数()f x ，满足102f ⎛⎫=⎪⎝⎭，且在()0,+∞上单调递减，则不等式()()()0f x f x x x --<--的解集为（）．A．102102x x x ⎧⎫<<<⎭<⎨⎩-⎬或B．1122x x x ⎧⎫<->⎨⎬⎩⎭或C．10221x x x <⎧⎫<<⎨⎩⎭-⎬或D．12102x x x -<⎧⎫>⎨⎩⎭<⎬或14．“数缺形时少直观，形少数时难入微”，数和形是函数的神和形两方面、在数学的学习研究中，常用函数的图象来研究函数的性质，也常用函数的解析式来琢磨函数的图象的特征．若下图为()y f x =的大致图第4页共11页象，则函数()y f x =的解析式最可能为（）．A．()ln x f x e x =⋅B．()ln x f x e x =⋅C．()ln xf x e x=⋅D．()ln x f x e x-=⋅15．在ABC △中，“ABC △是锐角三角形”是“tan tan 1A B >”的（）．A．充分不必要条件B．必要不充分条件C．充要条件D．既不充分也不必要条件16．已知定义在R 上的函数()f x 满足()()12f x f x +=，当(]0,1x ∈时，()1sin π4f x x =-．若对任意(],x m ∈-∞，都有()2f x ≥-，则m 的取值范围是（）．A．9,4⎛⎤-∞ ⎥⎝⎦B．7,3⎛⎤-∞ ⎥⎝⎦C．5,2⎛⎤-∞ ⎥⎝⎦D．8,3⎛⎤-∞ ⎥⎝⎦二、填空题17．己知函数()ln 1f x x x =+-，则不等式()0f x <的解集是______．18．已知1sin cos 5αα+=，()0,πα∈，则()()sin 1cos 1αα-+=______．19．函数()()πsin 03f x x ωω⎛⎫=+> ⎪⎝⎭的部分图象如图所示，则ω=______．20．已知实数x ，y 满足221x y xy ++=，则222x y +的最大值为______．21．已知ABC △的内角A ，B ，C 所对的边分别为a ，b ，c ，若222222sin sin sin A C b c a B b c a --=+-，则tan C 的取值范围为______．22．己知π,0,2x y ⎛⎫∈ ⎪⎝⎭，且tan tan tan sin sin 1x y x y x +-≤，则()2221x y --的最大值为______．第5页共11页三、解答题23．在ABC △中，内角A ，B ，C ，所对的边分别为a ，b ，c ，且cos cos c a b c B b Cb c a+++=-．（1）求C ；（2若角C 的内角平分线与AB 边交于点D ，且2CD =，求4b a +的最小值．24．若函数()y f x =满足()3π2f x f x ⎛⎫=+ ⎪⎝⎭且()ππ44f x f x x ⎛⎫⎛⎫+=-∈ ⎪ ⎪⎝⎭⎝⎭R ，则称函数()y f x =为“M 函数”．（1）试判断4sin3y x =是否为“M 函数”，并说明理由；（2）函数()f x 为“M 函数”，且当π,π4x ⎡⎤∈⎢⎥⎣⎦时，sin y x =，求()y f x =的解析式，并写出在3π0,2⎡⎤⎢⎥⎣⎦上的单调增区间；（3）在（2）条件下，当π5π,22x ⎡⎤∈-⎢⎥⎣⎦，关于x 的方程()f x a =（a 为常数）有解，记该方程所有解的和为S ，求S ．第6页共11页参考答案一、选择题1.B; 2.B; 3.B; 4.A;5.B; 6.D;7.A;8.B;9.B;10.A;11.B;12.C 13.D;14.B;15.C;16.B;15.在ABC △中，“ABC △是锐角三角形”是“tan tan 1A B >”的（）．A．充分不必要条件B．必要不充分条件C．充要条件D．既不充分也不必要条件由tan tan A B >1得tan 0,tan 0A B >>,即,A B 为锐角,()()tan tan tan tan tan 01tan tan A B C A B A B A Bπ+=--=-+=->-又所以C 为锐角,由此可知ABC ∆是锐角三角形,则必要性成立;由ABC ∆是锐角三角形,则C 为锐角,从而tan 0C >,即()tan 0A B -+>,则tan tan 01tan tan A BA B+<-,又,A B 也是锐角,故有1tan tan 0A B -<,即tan tan 1A B >,所以充分性成立，16．已知定义在R 上的函数()f x 满足()()12f x f x +=，当(]0,1x ∈时，()1sin π4f x x =-．若对任意(],x m ∈-∞，都有()32f x ≥-，则m 的取值范围是（）．A．9,4⎛⎤-∞ ⎥⎝⎦B．7,3⎛⎤-∞ ⎥⎝⎦C．5,2⎛⎤-∞ ⎥⎝⎦D．8,3⎛⎤-∞ ⎥⎝⎦因为当(]0,1x ∈时,()1sin 4f x x π=-,所以()1,04f x ⎡⎤∈-⎢⎥⎣⎦;第7页共11页当(]1,2x ∈时,(]()()10,1,21x f x f x -∈=-=()11sin 1,022x π⎡⎤⎡⎤--∈-⎣⎦⎢⎥⎣⎦(](]()()()[]2,3,20,1,42sin 21,0;x x f x f x x π⎡⎤∈-∈=-=--∈-⎣⎦当时()()78sin 2;233x x x π⎡⎤--=-==⎣⎦令得或舍若对任意(],x m ∈-∞,意有()2f x ≥-,则m 的取值范围是7,3⎛⎤-∞ ⎥⎝⎦.二、填空题17.()10，；18.252-；19.23;20.3322+;21.()10，22.2222+-ππ22.己知π,0,2x y ⎛⎫∈ ⎪⎝⎭，且tan tan tan sin sin 1x y x y x +-≤，则()2221x y --的最大值为______．2222+-ππ因tan tan tan sin sin 1x y x y x +- ,则1sin 1tan sin cos tan sin tan tan 22x y y x x x x x ππ+⎛⎫⎛⎫+=+=-+- ⎪ ⎪⎝⎭⎝⎭因函数sin ,tan y x y x ==均在0,2π⎛⎫⎪⎝⎭上单调递增,则函数tan sin y x x =+在0,2π⎛⎫⎪⎝⎭上单调递增,故有:02x y π<+<,第8页共11页设x y m +=,其中02m π<<,则()()()()()()()222222222212142222121x y m y y y m y m y m m m --=---=-+-+-⎡⎤=---+--⎣⎦ 当且仅当2y m =-时取等号,则此时022m π<-<,得222m ππ-< ,又函数()()221f m m =-在2,12π⎛⎫-⎪⎝⎭时单调递减,在1,2π⎛⎤⎥⎝⎦时单调递增,222f f ππ⎛⎫⎛⎫-= ⎪ ⎪⎝⎭⎝⎭,则()()22212222f m m fπππ⎛⎫=-=-+ ⎪⎝⎭ 此时2,22y x ππ=-=-三、解答题23．在ABC △中，内角A ，B ，C ，所对的边分别为a ，b ，c ，且cos cos c a b c B b Cb c a+++=-．（1）求C ；（2若角C 的内角平分线与AB 边交于点D ，且2CD =，求4b a +的最小值．(1)设ABC ∆外接圆的半径为R ,由正弦定理得:()cos cos 2sin cos 2sin cos 2sin 2sin c B b C R C B R B C R B C R A a +=+=+==则cos cos c a b c B b C b c a +++=-可化为c a b ab c a++=-整理得222a b c ab +-=-,由余弦定理得22212cos ,0,.2223a b c ab C C C ab ab ππ+--===-<<=又所以(2)由BCD ∆和ACD ∆的面积之和等于ABC ∆的面积,得1112sin sin 232323CD a CD b absin πππ⋅+⋅=可得22ab b a =+,即1112a b +=.第9页共11页则()(1144242525218a b b a b a a b b a ⎛⎫⎛⎫+=++=⨯++≥⨯+=⎪ ⎪⎝⎭⎝⎭当且仅当41112a b b aa b ⎧=⎪⎪⎨⎪+=⎪⎩,即6,3b a ==时,等号成立.故4b a +的最小值为18.24．若函数()y f x =满足()3π2f x f x ⎛⎫=+ ⎪⎝⎭且()ππ44f x f x x ⎛⎫⎛⎫+=-∈ ⎪ ⎪⎝⎭⎝⎭R ，则称函数()y f x =为“M 函数”．（1）试判断4sin3y x =是否为“M 函数”，并说明理由；（2）函数()f x 为“M 函数”，且当π,π4x ⎡⎤∈⎢⎥⎣⎦时，求()y f x =的解析式，并写出在3π0,2⎡⎤⎢⎥⎣⎦上的单调增区间；（3）在（2）条件下，当π5π,22x ⎡⎤∈-⎢⎥⎣⎦，关于x 的方程()f x a =（a 为常数）有解，记该方程所有解的和为S ，求S．(1)()4sin 3f x x ⎛⎫=⎪⎝⎭不为M 函数,理由如下：()()()424sin sin ,2323344sin ,,sin 323f x x x f x x f x f x f x ππππ⎛⎫⎛⎫⎛⎫-=-=- ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎛⎫⎛⎫=∴-≠∴= ⎪ ⎪⎝⎭⎝⎭不是M的函数(2) 函数()f x 对任意的实数x 满足()32f x f x π⎛⎫=+⎪⎝⎭,∴函数()f x 的周期为32T π=,第10页共11页753,4242x x πππππ≤≤∴≤-≤,()32f x f x π⎛⎫∴=+ ⎪⎝⎭,()()()()33,,sin ,sin cos ,42275,cos 42x f x x f x f x x x f x f x xππππππ⎡⎤⎛⎫⎛⎫∈=∴=-=-= ⎪ ⎪⎢⎥⎣⎦⎝⎭⎝⎭⎡⎤∴=⎢⎣⎦时在时的解析式为(3)由(2)知()f x 在75,42ππ⎡⎤⎢⎣⎦时的解析式为()()()cos ,,2442sin cos ,2273,422433cos sin 22f x x x x f x f x x x x x f x f x x x ππππππππππππππ=-≤≤∴≤-≤⎛⎫⎛⎫∴=-=-= ⎪ ⎪⎝⎭⎝⎭≤≤∴-≤-≤⎛⎫⎛⎫∴=-=-=- ⎪ ⎪⎝⎭⎝⎭ (),24,47,475,42cosx x sinx x f x sinx x cosx x ππππππππ⎧-≤≤⎪⎪⎪≤≤⎪∴=⎨⎪-≤≤⎪⎪⎪≤≤⎩作出函数()f x 的图象,如图,2O π-∣关于x 的方程()(f x a a =为常数)有解等价于()y f x =与y a =的图象有交点,由图可知当0a =时,方程()(f x a a =为常数)有3个解,其方程所有解的和为5322S ππππ=-++=当22a<<或1a=时,方程()(f x a a=为常数）有4个解,其方程有解的和为:2144,44Sπππ=+=当22a=时,方程()(f x a a=为常数)有6个解，其方程有解的和为:27146, 4444Sπππππ=+++=当212a<<时,方程()(f x a a=为常数)有8个解，其方程有解的和为：2214148,,3,06,.44442 S S a a πππππππ=+++====综上第11页共11页。

卓越联盟2021届新高考省份高三9月份检测英语试题及答案卓越联盟2021届新高考省份高三年级9月份检测试题英语试卷（本试卷满分150分，考试时间：120分钟）注意事项：1.答卷前：先将自己的姓名、准考证号填写在试卷和答题卡上，并将准考证条码粘贴在答题卡上指定位置。

2.选择题，每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑。

3.非选择题，用0.5mm黑色签字笔写在答题卡上对应的答题区域，写在非答题区域无效。

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 will the woman do this afternoon?A. Go shopping.B. Visit friends.C. Attend classes.2. What is the weather like now?A. Sunny.B. Cloudy.C. Rainy.3. Where does the conversation probably take place?A. In a library.B. In a bookstore.C. In a classroom.4. How does the man probably feel when hearing the news?A. Pleased.B. Nervous.C. Annoyed.5. What does the man advise the woman to do?A. Put her article in his office.B. Have her article improved.C. Have her article published.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。

2017届卓越学校一模测试英语（问卷）本试卷共四大题，8页，满分110分。

考试时间120分钟。

注意事项：1、答卷前，考生务必在答题卡上用黑色字迹的钢笔或签字笔填写自己的考生号、姓名、试室号、座位号，再用2B 铅笔把对应这两个号码的标号涂黑。

2、选择题每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑。

如需要改动，用橡皮擦干净后，再选涂其他答案。

3、非选择题必须用黑色字迹的钢笔或签字笔作答，答案必须写在答题卡各题目指定区域的相应位置上；如需要改动，先划掉原来的答案，然后再写上新的答案，改动的答案也不能超出指定的区域；不准使用铅笔、圆珠笔和涂改液。

不按以上要求作答的答案无效。

4、考生必须保持答题卡的整洁，考试结束，将本试卷和答题卡一并交回。

一、语法选择（共15小题；每小题1分，满分15分）阅读下面短文，按照句子结构的语法性和上下文连贯的要求，从1～15各题所给的A、B、C和D项中选出最佳选项，并在答题卡上将该项涂黑。

Is watching TV a waste of time? Here is a story happened to teenager Jake Deham.Jake Deham ___ 1 ___ with his family in the US when he fell over and lost one of his skis. His family didn’t know that he had a problem. They kept on skiing. When they got to the foot of the mountain, there was no sign of Jake.Jake couldn’t find his ski anywhere. In the end, he decided to take off ___ 2 ___ ski and walk down the mountain. But he couldn’t work out the right way to go.It was now getting dark and he was a long way from any place of ___ 3 ___. He knew ___ 4 ___ that he might die that night in the cold temperatures. But Jake kept calm. At home, Jake watched a lot of programmes about ___ 5 ___ in difficult situations. He remembered ___ 6 ___ advice from these programmes and knew he should build a hole in the snow. He made a hole ___ 7 ___ the wind couldn’t blow into it. Outside his hole, the temperature fell to a dangerous-15℃ that night, but inside it Jake was safe from the cold.After the long evening passed, Jake began ___ 8 ___ his way out. He had to get down the mountain. He thought of the words ___ 9 ___ the TV programmes always said,“ If you are lost, you should find someone else’ s tracks(足迹) through the snow and follow ___ 10 ___.” “I wanted to live my life.” Remembered Jake. “So I got up and I found some ski tracks and I followed those.” He walked and walked and finally saw lights.…His mum was very happy when she heard his son ___ 11 ___. To everyone’s___ 12 ___, Jake didn’t even have to go to hospital. He got through the terrible experience ___ 13 ___ any injuries.So, the next time someone says ___ 14 ___ watching TV is a waste of time, think of Jake. Sometimes TV ___ 15 ___ save your life!1. A. is skiing B. skied C. was skiing D. were skiing2. A. others B. another C. other D. the other3. A. safe B. safely C. safer D. safety4. A. clear B. clearly C. clearness D. clearer5. A. lived B. lives C. living D. live6. A. a B. an C. the D. /7. A. and B. but C. or D. so8. A. to think B. thinks C. thought D. to thinking9. A. what B. which C. when D. who10. A. they B. them C. their D. themselves11. A. has saved B. was saved C. is saved D. was saving12. A. surprised B. surprising C. surprise D. surprisedly13. A. with B. of C. about D. without14. A. that B. whether C. what D. when15. A. can B. should C. must D. need二、完形填空（共10小题；每小题1.5分，满分15分）阅读下面短文，掌握其大意，然后从16～25各题所给的A、B、C和D项中，选出最佳选项，并在答题卡上将该项涂黑。