2016年陕西专升本院校

篇一：2016年陕西统招专升本英语真题

2015年陕西省普通高等教育专升本招生考试试题

大学英语

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 letter 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.supportC.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.pleasedD.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 will 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 their 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 cultivated 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.

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 professor.

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 really 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 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;actually 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.Having 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 apologized having to leave so early.

篇二：2016陕西专升本考点 -连续知识扩展

第三节 连续

题型一：讨论分段函数在分段点的连续性

解题提示

（一）两种情形：

1.已知分段函数，研究其在分段点的连续性；

2.已知分段函数在分段点连续，反求函数关系式中的参数．

（二）方法

在分段点x0处连续? f(x0?0)?f(x0?0)?f(x0)．

?cosx,x?0?例1设f(x)??sinx，试问f(x)在x?0处是否连续？ ,x?0??x

f(x)?lim?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)在点x?0连续．

?1?xsinx?a,x?0

?例2 已知f(x)??b,x?0(a,b为常数)，问a,b为何值时，f(x)在x?0处连续．

?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?a 解由于f(0?0)?lim?

若要 f(x)在x?0处连续，必有f(0?0)?f(0?0)?f(0)

即1?a?0?b

解得 a??1,b?0．

题型二：确定函数的间断点并进行分类

解题提示

函数y?f(x)的间断点，是指满足下列三个条件之一的点：

1°f(x)在点x0处无定义；

2°在x0点limf(x)不存在； x?x0

3°在x0点limf(x)存在，但limf(x)?f(x0)． x?x0x?x0

求函数y?f(x)的间断点也就是寻找满足三个条件之一的点．

x2?1例 求函数f(x)?2的间断点，并确定其类型． x?x?2

解 所给函数在点x??1,2没有意义，因此x??1,2是所给函数的间断点．

由于在x??1点有 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是第一类间断点且为可去间断点．

在x?2点有 x2?1 f(2?0)?limf(x)?lim?? x?2?x?2?(x?1)(x?2)

x2?1 f(2?0)?limf(x)?lim??

x?2?x?2?(x?1)(x?2)

故x?2是第二类间断点且为无穷间断点．

题型三：有关闭区间上连续函数的命题的证明

解题提示 其基本步骤是：把已知方程右端项全部移到左端，令其为f(x)，再由题设确定区间?a,b?，然后验证f(a)?f(b)?0即可推证在a与b之间至少存在一个根．

例 设f(x)在?a,b?上连续，且f(a)?a，f(b)?b，证明f(x)?x在?a,b?内至少有一实根

证令F(x)?f(x)?x，则F(x)在?a,b?上连续，

F(a)?f(a)?a?0， F(b)?f(b)?b?0，则F(a)?F(b)?0

由零点定理知，在(a,b)内至少存在一点?，使F(?)?0，即f(?)??

也就是说方程f(x)?x在区间(a,b)内至少有一个根．

篇三：2016陕西统招专升本考点-函数的概念

第一节 函数

一、概念

（一）定义

（二）常见形式

1.由一个解析式表示

2.分段函数 在定义域内的不同点集内由不同（段）的数学表达式表示的函数称为分段函数．

3.隐函数 如果函数的对应法则是由方程F(x,y)?0给出，则称y为x的隐函数．

4.参数方程表示的函数 如果x与y的关系通过第三个变量联系起来，如

??x??(t)

?y??(t)

5.复合函数 y是u的函数：y?f(u)，而u又是x的函数：u??(x)

6.反函数 对数函数y?logax与指数函数y?ax互为反函数，其定义域和值域互相对应，一个函数的定义域恰好是另一个函数的值域

（三）初等函数（图像，定义域，值域）

（四）性质 （单调性、奇偶性、周期性、有界性）

二、考试题型

题型一：求函数定义域

（1）y?x?2?12x?12?ln(5?x)；（2）y??x?arcsin． x?37

解（1）要使y有意义，x应满足x?2?0,x?3?0,5?x?0，即x?2,x?3且x?5，其公共部分为?2,3??(3,5)，故所求定义域为D??2,3??(3,5)．

?16?x2?0(1)?（2）要使y有意义，必须?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，所给函数的定义域为D???3,4????4,4????3,4?．

例2 已知函数f(x)的定义域为?0,1?，求f(2x?3)的定义域．

解 f(x)的定义域为?0,1?，所以f(2x?3)的定义域为0?2x?3?1，即3?x?2． 2

题型二：复合函数求法

类型一：已知f(x)和g(x)的表达式，求函数f?g(x)?的表达式

方法：只需用g(x)替换f(x)中的x即可．

类型二：已知f?g(x)?的表达式，反求f(x)的表达式

方法一：令g(x)?u，解出x??(u)，求出f(u)的表达式，再将u换成x即得f(x)方法二：将f?g(x)?的表达式凑成g(x)的表达式，再将g(x)换成x即得f(x)的表达式

例1设f(x)?x

?x2，求f(x?1),f?f(x)?．

解 把g(x)?x?1代替f(x)中的x，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

再将u换成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

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