2021年陜西專升本院校

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1、2021年陜西專升本院校 篇一：2021年陜西統(tǒng)招專升本英語真題 2021年陜西省普通高等教育專升本招生考試試題 大學英語 Ⅰ.Vocabulary and Structure(40 points) Directions:In this part,there are 40 incomplete sentences.For each sentence there are 4 choice marked A，B，C，D.You should decide on the best choice and mark the corresponding let

2、ter on the Answer Sheet. A.bare B.empty C.blank D.vacant A.source B.resource C.birth D.origin context. A.approachB.solution C.manner D.road 4.A.That B.WithC.It D.What books. A.helpB.helpingC.helpsD.to help large increase in food price. A.strengthB.sup

3、portC.agreementD.vote A.reminds me ofB.reminds me toC.remembers me of D.remember me to A.twice as many asB.as twice many as C.as twice much asD.twice as much as language. A.accumulate B.collect C.assembleD.gather of life:driving everywhere. A.enviousB.hopefulC.pl

4、easedD.happy 11.One of the requirements for a fire is that the material A.is heated B.will be heated C.be heated D.would be heated A.transfer B.adjust C.direct D.add A.can write B.could have written C.could write D.have written 14.With the development of industry,this region wi

5、ll surely . A.develop B.profitC.succeedD.thrive I have relatives. A.whichB.neverthelessC.whereD.when 16.-Could I borrow your dictionary? -I’d get it for you I could remember who last borrowed it. A.except thatB.if onlyC.only ifD.unless 17.The terrorists to blow up the plane if thei

6、r demands were not met. A.pretended B.determined C.threatenedD.proceeded 18.It is generally thought to be of importance to a man that he himself. A.knewB.knowC.knowsD.must know 19.My friends us into going swimming. A.persuadedB.told C.invited D.suggested 20.The total cultivat

7、ed area is 13,000 acres, 10,000 acres are irrigated fields. A.whichB.of which C.in thatD.of that 21.There was a large crowd in the square against the war. A.protectingB.protestingC.preventing D.promoting 22.I could not persuade him to accept it,make him see the importance of it.

8、 A.if only I could notB.no more than I could C.or I could notC.nor could I 23.Mr.Green said his clients our samples by the end of last month. A.didn’t receive B.hadn’t received C.haven’t received D.don’t receive 24.Language belongs to each if us,to the flower-seller to the profe

9、ssor. A.as far as B.as much as C.as many as D.as long as 25.Because of the reduction of air pollution,this city now is a good place. A.where to live B.which to live C.to live D.to be live 26.Only when it is correct in every detail. A.his model can really runB.can his model re

10、ally run C.his model really can runD.can really his model run 27.He is to win.No one else in the race stands a chance. A.boundB.liableC.probableD.apt 28.I continued to study the discouragement I hand received. A.despite of B.despite C.in spite D.in spite that 29.It’s the

11、in this country to go out and pick flowers on the first day of spring. A.normalB.habitC.custom D.use 30.Without the friction between their feet and the ground,people wouldbe able to walk. A.in no timeB.on any account C.by all means D.in no way 31.He didn’t allowin his room;actua

12、lly he did not allow his family at all. A.to smoke;to smoke B.smoking;to smoke C.to smoke;smoking D.smoking;smoking 32.With such poor he really needs glasses. A.visionB.viewC.sense D.scene 33.the plan carefully,he rejected it. A.To have considered B.To consider C.Ha

13、ving considered D.Considering 34.Finding it difficult toto the climate in the city,he decided to move to the north. A. fit B.adopt C.suit D.adapt 35.Our public transportation system is notfor the needs of the people.We need more buses and subways. A.completeB.adequateC.normalD.good 36.He

14、 apologized having to leave so early. 篇二：2021陜西專升本考點 -連續(xù)知識擴展 第三節(jié) 連續(xù) 題型一：討論分段函數(shù)在分段點的連續(xù)性 解題提示 （一）兩種情形： 1.已知分段函數(shù)，研究其在分段點的連續(xù)性； 2.已知分段函數(shù)在分段點連續(xù)，反求函數(shù)關系式中的參數(shù)． （二）方法 在分段點x0處連續(xù)? f(x0?0)?f(x0?0)?f(x0)． ?cosx,x?0?例1設f(x)??sinx，試問f(x)在x?0處是否連續(xù)？ ,x?0??x f(x)?li

15、m?cosx?1 解 由于f(0?0)?lim?x?0x?0 f(x)?lim? f(0?0)?lim?x?0x?0sinx?1 x f(0)?1 即有f(0?0)?f(0?0)?f(0),f(x)在點x?0連續(xù)． ?1?xsinx?a,x?0 ?例2 已知f(x)??b,x?0(a,b為常數(shù))，問a,b為何值時，f(x)在x?0處連續(xù)． ?1?xsin,x?0x? 1 x?0x?0x 1f(x)?limx?sin?0 f(0?0)?limx?0?x?0?xf(x)?lim?(sinx?a)?1?

16、a 解由于f(0?0)?lim? 若要 f(x)在x?0處連續(xù)，必有f(0?0)?f(0?0)?f(0) 即1?a?0?b 解得 a??1,b?0． 題型二：確定函數(shù)的間斷點并進行分類 解題提示 函數(shù)y?f(x)的間斷點，是指滿足下列三個條件之一的點： 1f(x)在點x0處無定義； 2在x0點limf(x)不存在； x?x0 3在x0點limf(x)存在，但limf(x)?f(x0)． x?x0x?x0 求函數(shù)y?f(x)的間斷點也就是尋找滿足三個條件之一的點． x2?1例 求函數(shù)f(

17、x)?2的間斷點，并確定其類型． x?x?2 解 所給函數(shù)在點x??1,2沒有意義，因此x??1,2是所給函數(shù)的間斷點． 由于在x??1點有 f(?1?0)?lim?f(x)?lim?x??1x??1(x?1)(x?1)x?12?lim?? x??1(x?1)(x?2)x?23 f(?1?0)?lim?f(x)?lim?x??1x?1x?12? x?23 f(?1?0)?f(?1?0)， 故x??1是第一類間斷點且為可去間斷點． 在x?2點有 x2?1 f(2?0)?limf(x)?lim?? x?2?x?2?(x?1)(x?2)

18、 x2?1 f(2?0)?limf(x)?lim?? x?2?x?2?(x?1)(x?2) 故x?2是第二類間斷點且為無窮間斷點． 題型三：有關閉區(qū)間上連續(xù)函數(shù)的命題的證明 解題提示 其基本步驟是：把已知方程右端項全部移到左端，令其為f(x)，再由題設確定區(qū)間?a,b?，然后驗證f(a)?f(b)?0即可推證在a與b之間至少存在一個根． 例 設f(x)在?a,b?上連續(xù)，且f(a)?a，f(b)?b，證明f(x)?x在?a,b?內至少有一實根 證令F(x)?f(x)?x，則F(x)在?a,b?上連續(xù)， F(a)?f(

19、a)?a?0， F(b)?f(b)?b?0，則F(a)?F(b)?0 由零點定理知，在(a,b)內至少存在一點?，使F(?)?0，即f(?)?? 也就是說方程f(x)?x在區(qū)間(a,b)內至少有一個根． 篇三：2021陜西統(tǒng)招專升本考點-函數(shù)的概念 第一節(jié) 函數(shù) 一、概念 （一）定義 （二）常見形式 1.由一個解析式表示 2.分段函數(shù) 在定義域內的不同點集內由不同（段）的數(shù)學表達式表示的函數(shù)稱為分段函數(shù)． 3.隱函數(shù) 如果函數(shù)的對應法則是由方程F(x,y)?0給出，則稱y為x的隱函數(shù)．

20、4.參數(shù)方程表示的函數(shù) 如果x與y的關系通過第三個變量聯(lián)系起來，如 ??x??(t) ?y??(t) 5.復合函數(shù) y是u的函數(shù)：y?f(u)，而u又是x的函數(shù)：u??(x) 6.反函數(shù) 對數(shù)函數(shù)y?logax與指數(shù)函數(shù)y?ax互為反函數(shù)，其定義域和值域互相對應，一個函數(shù)的定義域恰好是另一個函數(shù)的值域 （三）初等函數(shù)（圖像，定義域，值域） （四）性質 （單調性、奇偶性、周期性、有界性） 二、考試題型 題型一：求函數(shù)定義域 （1）y?x?2?12x?12?ln(5?x)；（2）y??x?arcsin． x?

21、37 解（1）要使y有意義，x應滿足x?2?0,x?3?0,5?x?0，即x?2,x?3且x?5，其公共部分為?2,3??(3,5)，故所求定義域為D??2,3??(3,5)． ?16?x2?0(1)?（2）要使y有意義，必須?2x?1 ?7?1(2)? 由（1）式解得?4?x?4；解（2）式得?7?2x?1?7，即?3?x?4． 于是，所給函數(shù)的定義域為D???3,4????4,4????3,4?． 例2 已知函數(shù)f(x)的定義域為?0,1?，求f(2x?3)的定義域． 解 f(x)的定義域為?0,1?，所以f(2x?3)的定

22、義域為0?2x?3?1，即3?x?2． 2 題型二：復合函數(shù)求法 類型一：已知f(x)和g(x)的表達式，求函數(shù)f?g(x)?的表達式 方法：只需用g(x)替換f(x)中的x即可． 類型二：已知f?g(x)?的表達式，反求f(x)的表達式 方法一：令g(x)?u，解出x??(u)，求出f(u)的表達式，再將u換成x即得f(x)方法二：將f?g(x)?的表達式湊成g(x)的表達式，再將g(x)換成x即得f(x)的表達式 例1設f(x)?x ?x2，求f(x?1),f?f(x)?． 解 把g(x)?x?1代替f(x)中的x

23、，得 f(x?1)?x?1 ?(x?1)2?x?1x?2x?22 再把g(x)?f(x)代替f(x)中的x可得 x f?f(x)??f(x) ?f(x)22x?x?? 22x?2x1?1?x2 例2 已知f(1?xx?1)?2，求f(x) xx1?11?x1?u，得x?解令，于是f(u)??u2?u 12xu?1()u?1 再將u換成x，得f(x)?x?x． 例3. f(sinx)?cosx，求f(x) 222222解f(sinx)?1?sinx，另sinx?t，得f(t)?1?t，于是f( x)?1? x 《2021年陜西專升本院?！?