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15年15 第一篇_15年高考真题——理科数学(新课标II卷)

2015年普通高等学校招生全国统一考试新课标II卷

数学（理科）

一．选择题：共12小题，每小题5分，共60分。在每个小题给出的四个选项中，只有一项是符合题目

要求的一项。

1．已知集合A2,1,0,1,2，Bx|x1x20，则AB( ) （A）1,0 （B）0,1 （C）1,0,1 （D）0,1,2

2．若a为实数且2aia2i4i，则a( ) （A）1 （B）0 （C）1 （D）2

3．根据下面给出的2004年至2013

年我国二氧化硫排放量（单位：万吨）柱形图。以下



结论不正确的是（ ） （A）逐年比较，2008年减少二氧化硫排放量的效果最显著 （B）2007年我国治理二氧化硫排放显现 （C）2006年以来我国二氧化硫年排放量呈减少趋势 （D）2006年以来我国二氧化硫年排放量与年份正相关

4．等比数列an满足a13，a1a3a521，则a3a5a7( ) （A）21 （B）42 （C）63 （D）84

1log22xx15．设函数fx，则x1

x12

f2flog212 ( )

（A）3 （B）6 （C）9 （D）12

6．一个正方体被一个平面截去一部分后，剩余部分的三视图如右图，则截去部分体积与剩余部分体积的比值为( ) （A） （B）7 （C）6 （D）5

7．过三A1,3，B4,2，C1,7的圆交于y轴于M,N两点，则|MN|( ) （A）26 （B）8 （C）46 （D）10

8．右边程序抗土的算法思路源于我国古代数 学名著《九章算术》中的“更相减损术”。 执行该程序框图，若输入a,b分别为14,18，则输出的

输入

a,b

a( ) （A）0 （B）2 （C）4 （D）14

AOB900，9．设A,B是球O的球面上两点，C为该球面上的动点，若三棱锥OABC体积的最

大值为36，则球O的表面积为( )

36 （B）64 （C）144 （D）256 （A）

10．如图，长方形ABCD的边AB2，BC1，O是

a = ab

a > b

a ≠ b

输出a

b = ba

AB的中点，点P沿着边BC，CD与DA运动，记

BOPx，将动点P到A,B两点距离之和表示为x的函

数fx，则fx的图像大致为（ ）

11．已知A,B为双曲线E的左，右顶点，点M在E上，ABM为等腰三角形，且顶角为120，则E的离心率为( ) （A

（B）2 （C

（D

12．设函数fx是奇函数fxxR的导函数，f10，当x0时，

xfxfx0，则使得fx0成立的x的取值范围是( ) （A）,10,1

（B）1,01, （C）,11,0 （D）0,11,

二．填空题：本大题共四小题，每小题5分。



13．设向量a,b不平行，向量ab与a2b平行，则实数________。

xy10

14．若x,y满足约束条件x2y0，则zxy的最大值为__________。

x2y20

15．ax1x的展开式中x的奇数次幂项的系数之和为32，则a__________。

4

16．设Sn是数列an的前n项和，且a11，an1SnSn1，则Sn __________。

三.解答题：解答应写出文字说明，证明过程或演算步骤。

17．（本小题满分12分）ABC中，D是BC上的点，AD平分BAC，ABD是ADC

面积的2倍。⑴求

sinB；⑵若AD

1，DCBD和AC的长。

sinCA地区

4

56789

B地区

18．（本小题满分12分）某公司为了解用户对其产品的满意度，从

A,B两地区分别随机调查了20个用

户，得到用户对产品的满意度评分如右表所示。⑴根据两组数据完成两地区用户满意度评分的茎叶图，并通过茎叶图比较两地区满意度评分的平均值及分散程度（不要求计算出具体值，得出结论即可）；⑵根据用户满意度评分，将用户的满意度从低到高分为三个不等级（右表所示）。记事件C：“A地区用户的满意度等级高于B地区用户的满意度等级”。假设两地区用户的评价结果相互独立。根据所给数据，以事件发生的频率作为相应事件发生的概率，求C的概率。

19．（本小题满分12分）长方体ABCDA，AB16，BC10，1BC11D1中（如图）点E,F分别在AA1ED1F4，AA18，1B1,DC11上，过点E,F的平面与此长方体的面相交，交线围成一个正方形。⑴在图中画出这个正方形（不必说明画法和理由）；⑵求直线AF与平面所成的角的正弦值。

20．（本小题满分12分）已知椭圆C：直线l不过原点O且不平行于坐9xymm0，【15年15】

2

2

2

D1

A1

ED

A

F

B1

C1

CB

标轴，l与C有两个交点A,B，线段AB的中点为M。⑴证明：直线OM的斜率与l的斜率的乘积为定值；⑵若l过点3,m，延长线段OM与C交于点P，四边形OAPB能否为平行四边形？若能，求此时l的斜率；若不能，说明理由。

21．（本小题满分共12分）设函数fxemxx2mx。⑴证明：fx在,0单调递减，在0,单调递增；⑵若对于任意x1,x21,1，都有|fx1fx2|e1，求m的取值范围。

请考生在第（22）、（23）、（24）三题中任选一题作答。注意：只能做所选定的题目。如果多做，则按所做的第一个题目计分，作答时请用2B铅笔在答题卡上将所选题号后的 方框涂黑。

22．（本小题满分10分）如图，O为等腰三角形ABC内一点，⊙O与ABC的底边BC交于M,N两点，与底边上的高

AD交于点G，且与AB,AC分别相切于E,F两点。⑴证明：

EF//BC；⑵若AG等于⊙O的半径，

且AEMN求

四边形EBCF的面积。

BC

xtcos23．（本小题10分）在直角坐标系xOy中，曲线C1：（t为参数，t0），

ytsin

其中0，在以O为极点，x轴正半轴为极轴的极坐标系中，曲线C2：2sin，C

3：

。⑴求C2与C3交点的直角坐标；⑵若C1与C2相交于点A，C1与C3相交于

点B，求|AB|的最大值。

24．（本小题满分10分）设a,b,c,d均为正数，且abcd，证明：⑴若abcd，



是|ab||cd|的充要条件。

2015年普通高校招生全国统考数学试卷新课标II卷解答

一．ABDBC DCBCB DA

二．13．2；14．2；15．3；16．n 17．解：⑴由题SABD

SABD

11

ABADsinBAD，SADCACADsinCAD。因为22

sinBAC1

； 2SADC，BADCAD，所以AB2AC。由正弦定理可得

sinCAB2

⑵因SABD:SADC【15年15】

BD:DC，故BDABD和ADC中，由余弦定理可知

AB2AD2BD22ADBDcosADB，AC2AD2DC22ADDCcosADC，故AB22AC23AD2BD22DC26。由⑴知AB2AC，所以AC1。

18．解：⑴两地区用户满意度评分的茎叶图如右下表所示。通过茎叶图可以看出：A地区用户满意度评分的平均值高于B地区用户满意度评分的平均值，A地区用户满意度评分比较集中，B地区用户满意度评分比较分散；

⑵记CA1为事件“A地区用户满意度等级为满意或非常满意”，CA2为事件“A地区用户满意度等级为非常满意”，CB1为事件“B地区用户满意度等级为不满意”，

CB2为事件“B地区用户满意度等级为满

意”，则CA1与CB1独立，CA2与CB2独立，CB1与CB2互斥，CCB1CA1CB2CA2，故

PCPCB1CA1CB2CA2PCB1PCA1PCB2PCA2。由题CA1,CA2,CB1,CB2

16410816410

,,,，故PCA1，PCA2，PCB1，202020202020208101684

PCB20.48。 ，故PCz2020202020FC1D1

19．解：⑴交线围成的正方形EHGF如图所示；

A1⑵作EMAB于M，则AMAB11E4，

yG

EMAA18。因EHGF为正方形，故EH10，

C

A

MHB

x于是MH6，所以AH10。以D

为坐标原点，DA,DC,DD1的方向为x,y,z轴的正方向，建立空间直角坐标系Dxyz，则

发生的频率分别为



A10,0,0，H10,10,0，E10,4,8，F0,4,8，FE10,0,0，HE0,6,8。



10x0nFE0设nx,y,z为平面EHGF的法向量，则，即。可取

6y8z0nHE0



|AFn|，所以AF与平面n0,4,3。又AF10,4,8，故|cosAF,n||AF||n|EHGF

所成角的正弦值为。

20．解：⑴设直线l：ykxbk0,b0，Ax1,y1,Bx2,y2，MxM,yM，

ykxbx1x2kb2222

k9x2kbxbm0x由2得，故，M222

2k99xym

yMkxMb

9byM9

k，因此，即kOMk9。所以直线OM的斜率与l的

OM

k29xMk

15年15 第二篇_15年高考真题——理科数学(浙江卷)

2015年普通高等学校招生全国统一考试数学试卷（浙江卷）

一．选择题：本大题共8小题，每小题5分，共40分，在每小题给出的四个选项中 只有一项是符合题目要求的。

2 1．已知集合Px|x2x0，Qx|1x2，则ðRPQ（ ） 

（A）0,1 （B）0,2 （C）1,2 （D）1,2

2．某几何体的三视图如图所示（单位：cm），则该几何体的体

积是（ ）

（A）8cm （B）12cm （C）3332340cm （D）cm3 33

3．已知an是等差数列，公差d不为零，前n项和是Sn，若

a3,a4,a8成等比数列，则（ ） （A）a1d0，dSn0

（B）a1d0，dSn0 （C）a1d0，dSn0 （D）a1d0，dSn0

4．命题“nN，fnN且fnn”的否定形式是（ ） 

（A）nN，fnN且fnn （B）nN，fnN或fnn 

（C）n0N，fn0N且fn0n0 （D）n0N，fn0N或fn0n0 

5．如图，设抛物线y4x的焦点为F，不经过焦点的直线

上有三个不同的点A,B,C，其中点A,B在抛物线上，点C在y轴2

上，则BCF与ACF的面积之比是（ ） （A）|BF|1 |AF|1

|BF|1|BF|21|BF|21（B） （C） （D） |AF|1|AF|21|AF|21

6．设A,B是有限集，定义dA,BcardABcardAB，其中cardA表示有限集A中的元素个数，命题①：对任意有限集A,B，“AB”是“dA,B0”的充分必要条件；命题②：对任意有限集A,B,C，dA,CdA,BdB,C。则（ ）

（A）命题①和命题②都成立 （B）命题①和命题②都不成立

（C）命题①成立，命题②不成立 （D）命题①不成立，命题②成立

7．存在函数fx满足，对任意xR都有（ ）

（A）fsin2xsinx （B）fsin2xx2x

22（C）fx1|x1| （D）fx2x|x1| 

D是AB的中点，8．如图，已知ABC，沿直线CD将ACD

折成ACD，所成二面角ACDB的平面角为，则（ ）

（A）ADB （B）ADB

（C）ACB （D）ACB

二．填空题：本大题共7小题，多空题每题6分，单空题每题4分，共36分。

x2

y21的焦距是_____________。 9．双曲线2

2x3x1x10．已知函数fx，则ff3fx的最小

2lgx1x1

值是________。

11．函数fxsinxsinxcosx1的最小正周期是2

是_________________。

12．若alog23，则22aa________。

13．如图，三棱锥ABCD中，ABACBDCD3，

ADBC2，点M,N分别是AD,BC的中点，则异面直线

AN,CM所成的角的余弦值是________。

14．若实数x,y满足xy1，则|2xy2||6x3y|的最小值是________。 22

51 15．已知e1,e2是空间单位向量，e1e2，若空间向量b满足be12，be2，22|bxe1ye2||bx0e1y0e2|1x0,y0R，且对于任意x,yR，则x0，

y0，|b|________。

三．解答题：本大题共5小题，共74分．解答应写出文字说明、证明过程或演算步骤。

16．（本题满分14分）在ABC中，内角A,B,C所对边分C1

D

A1B112别为a,b,c。已知A，bac。⑴求tanC的值；42

⑵若ABC的面积为7，求b的值。 22

C

17．（本题满分15分）如图，在三棱柱ABC

A1B1C1中，

AB

BAC900，ABAC2，A1A4，A1在底面ABC的射影为BC的中点，D为B1C1的中点。⑴证明：A1D平面A1BC；⑵求二面角A1BDB1的平面角的余弦值。

18．（本题满分15分）已知函数fxx2axba,bR，记Ma,b是|fx|在区间1,1上的最大值。⑴证明：当|a|2时，⑵当a,b满足Ma,b2，Ma,b2；

求|a||b|的最大值。 x2y21上两个不同 19．（本题满分15分）已知椭圆21对称。⑴求实数m的取值范2

围；⑵求AOB面积的最大值（O为坐标原点）。 的点A,B关于直线ymx

2220．（本题满分15分）已知数列an满足a12且an1anannN，数列an

的前n项和为Sn，证明：⑴1San11 nnN。2nN；⑵an12n2n2n1

2015年普通高校招生全国统考数学试卷浙江卷解答

一．CCBDA ADB

二．9．2，y237x；10．0

，3；11．，k,kkZ；288

12

．；13．7；14．3；15．1，2

，16．解：⑴由ba又由A2212112c及正弦定理得sin2Bsin2C，故cos2BsinC。222

4，即BC3，得cos2Bsin2C2sinCcosC，解得tanC2； 4

⑵由tanC

2得sinC

cosC，又sinBsinACsinC，

4故sinB

1，又A，bcsinA

3，故bc，，由正弦定理得c42103

故b3。

17．⑴设E为BC的中点，连A1E,AE。由题A1E平面ABC，故A1EAE。因ABAC，故AEBC，从而AE平面A1BC。由D,E

分别B1C1,BC的中点，得DE//B1B且DEB1B，从而A1C1DB1

F

DE//A1A且DEA1A，所以A1AED为平行四边形，故

A1D//AE。又AE平面A1BC，故A1D平面A1BC；

⑵作A1FBD于F，连B1F，

由题AEEBACEB

A1EAA1EB900，得A1BA1A4。由A1DB1D，A1BB1B，得

A1DBB1DB。由A1FBD，得B1FBD，因此A1FB1为二面角A1BDB1的

0平面角。由A，，得BDAAB4DAB

90D11FB1F3，由11

余弦定理得cosA1FB1即为所求。

aaa218．解：⑴由fxxb，得对称轴为直线x，由|a|2，得2242

a||1，故fx在1,1上单调，因此Ma,bmax|f1|,|f1|。当a2时，2

由f1f12a4，故4|f1||f1|，因此max|f1|,|f1|2，即Ma,b2；当a2时，由f1f12a4，故4|f1||f1|，因此max|f1|,|f1|2，即Ma,b2。综上，当|a|2时，Ma,b2；

⑵由Ma,b2得|1ab||f1|2，|1ab||f1|2，故|ab|3，

|ab|ab0|ab|3，由|a||b|，得|a||b|3。当a2，b1时，

|ab|ab0

|a||b|3，且|x22x1|在1,1的最大值为2，即M2,12，故|a||b|的最大值为3。

19．解：⑴由题知m0，可设直线AB：y

22221xb，代入椭圆方程并整理得mx2m2x4mbx2mb10。因直线AB与椭圆2y21有两个不同的交点，

2mbm2b,2故8mm2mb0 ①。将AB中点M2代入直线方程

m2m22222

1m22ymx得b

②。由①②得或； mm

22m233

⑵令t13|AB|2t1，则

0,，且O到AB的距离

2m2

1，AOB的面积S

t|AB|d221为dt2当且仅当t2时，等号成立，故

AOB。

20．解：⑴由题an1anan20，即an1an，故an1。由an1an1an1得2

an1an11an21a1a10，故0anan2； an1

⑵由题an2anan1，故Sna1an1 ①。由1a1，从而n1,2，即2an11an1a11a=n和1n2得，an1anan1an1

11111112，故n2n，因此an1nN ②， an1anan1a12n1n2

S11nnN。 2n2n2n1由①②得

15年15 第三篇_15年考研英语试卷

15年考研英语一及答案

Section 1 Use of English

Directions:

Read the following text. Choose the best word(s) for each numbered blank and mark [A], [B],

[C] or [D] on ANSWER SHEET 1. (10 points)

Though not biologically related, friends are as "related" as fourth cousins, sharing about 1% of genes. That is 1 a study published from the University of California and Yale University in the Proceedings of the National Academy of Sciences, has 2 .

The study is a genome-wide analysis conducted 3 1932 unique subjects which 4 pairs of unrelated friends and unrelated strangers. The same people were used in both 5 .While 1% may seem 6 , it is not so to a geneticist. As James Fowler, professor of medical genetics at UC San Diego, says, "Most people do not even 7 their fourth cousins but somehow manage to select as friends the people who 8 our kin."

The study 9 found that the genes for smell were something shared in friends but not genes for immunity. Why this similarity in olfactory genes is difficult to explain, for now. 10 Perhaps, as the team suggests, it draws us to similar environments but there is more 11 it. There could be many mechanisms working in tandem that 12 us in choosing genetically similar friends 13 than "functional kinship" of being friends with 14 !One of the remarkable findings of the study was that the similar genes seem to be evolving 15 than other genes. Studying this could help 16 why human evolution picked pace in the last 30,000 years, with social environment being a major 17 factor.

The findings do not simply corroborate people's 18 to befriend those of similar 19 backgrounds, say the researchers. Though all the subjects were drawn from a population of European extraction, care was taken to 20 that all subjects, friends and strangers were taken from the same population. The team also controlled the data to check ancestry of subjects. Section II Reading Comprehension

1、What

2、Concluded

3、On

4、Compared

5、Samples

6、Insignificant

7、Know

8、Resemble

9、Also

10、Perhaps

11、To

12、Drive

13、Ratherthan

14、Benefits

15、Faster

16、understand

17、Contributory

18、Tendency

19、Ethnic

20、see

Part A

Directions:

Read the following four texts. Answer the questions below each text by choosing [A], [B], [C] or [D]. Mark your answers on ANSWER SHEET 1. (40 points)

TEXT 1

King Juan Carlos of Spain once insisted"kings don't abdicate, they die in their sleep." But embarrassing scandals and the popularity of the republicans left in the recent Euro-elections have forced him to eat his words and stand down. So, does the Spanish crisis suggest that monarchy is seeing its last days? Does that mean the writing is on the wall for all European royals, with their magnificent uniforms and majestic lifestyles?

The Spanish case provides arguments both for and against monarchy. When public opinion is particularly polarized, as it was following the end of the France regime, monarchs can rise above "mere" polities and "embody" a spirit of national unity.

It is this apparent transcendence of polities that explains monarchy's continuing popularity as heads of state. And so, the Middle East expected, Europe is the most monarch-infested region in the world, with 10 kingdoms (not counting Vatican City and Andorra). But unlike their absolutist counterparts in the Gulf and Asia, most royal families have survived because they allow voters to avoid the difficult search for a non-controversial but respected public figure.

Even so, kings and queens undoubtedly have a downside. Symbolic of national unity as they claim to be, their very history-and sometimes the way they behave today-embodies outdated and indefensible privileges and inequalities. At a time when Thomas Piketty and other economists are warming of rising inequality and the increasing power of inherited wealth, it is bizarre that wealthy aristocratic families should still be the symbolic heart of modern democratic states.

The most successful monarchies strive to abandon or hide their old aristocratic ways. Princes and princesses have day-jobs and ride bicycles, not horses (or helicopters). Even so, these are wealthy families who party with the international 1%, and media intrusiveness makes it increasingly difficult to maintain the right image.

While Europe's monarchies will no doubt be smart enough to survive for some time to come, it is the British royals who have most to fear from the Spanish example.

It is only the Queen who has preserved the monarchy's reputation with her rather ordinary (if well-heeled) granny style. The danger will come with Charles, who has both an expensive taste of lifestyle and a pretty hierarchical view of the world. He has failed to understand that monarchies have largely survived because they provide a service-as non-controversial and non-political heads of state. Charles ought to know that as English history shows, it is kings, not republicans, who are the monarchy's worst enemies.

21. According to the first two paragraphs, King Juan Carlos of Spain

[A]eased his relationship with his rivals.

[B]used to enjoy high public support.

[C]was unpopular among European royals.

[D]ended his reign in embarrassment.

22. Monarchs are kept as head of state in Europe mostly

[A]to give voters more public figures to look up to.

[B]to achieve a balance between tradition and reality.

[C]owing to their undoubted and respectable status.

[D]due to their everlasting political embodiment.

23. Which of the following is shown to be odd, according to Paragraph 4?

[A] The role of the nobility in modern democracies.

[B] Aristocrats' excessive reliance on inherited wealth.

[C] The simple lifestyle of the aristocratic families.

[D] The nobility's adherence to their privileges.

24. The British royals "have most to fear" because Charles

[A]takes a tough line on political issues.

[B]fails to change his lifestyle as advised.

[C]takes republicans as his potential allies.

[D]fails to adapt himself to his future role.

25. Which of the following is the best title of the text?

[A]Carlos, Glory and Disgrace Combined

[B]Charles, Anxious to Succeed to the Throne

[C]Charles, Slow to React to the Coming Threats

[D]Carlos, a Lesson for All European Monarchs

21.Dended his reign in embarrassment.

22. C owing to the undoubted and respectable status

23. A the role of the nobility in modern democracy

24. B fails to change his lifestyle as advised.

25. D Carlos, a lesson for all Monarchies

TEXT 2

Just how much does the Constitution protect your digital data? The Supreme Court will now consider whether police can search the contents of a mobile phone without a warrant if the phone is on or around a person during an arrest.

California has asked the justices to refrain from a sweeping ruling, particularly one that upsets the old assumptions that authorities may search through the possessions of suspects at the time of their arrest. It is hard, the state argues, for judges to assess the implications of new and rapidly changing technologies.

The court would be recklessly modest if it followed California's advice. Enough of the implications are discernable, even obvious, so that the justice can and should provide updated guidelines to police, lawyers and defendants.

They should start by discarding California's lame argument that exploring the contents of a smartphone- a vast storehouse of digital information is similar to say, going through a suspect's purse .The court has ruled that police don't violate the Fourth Amendment when they go through the wallet or pocketbook, of an arrestee without a warrant. But exploring one's smartphone is more like entering his or her home. A smartphone may contain an arrestee's reading history, financial history, medical history and comprehensive records of recent correspondence. The development of "cloud computing." meanwhile, has made that exploration so much the easier.

But the justices should not swallow California's argument whole. New, disruptive technology sometimes demands novel applications of the Constitution's protections. Orin Kerr, a law professor, compares the explosion and accessibility of digital information in the 21st century with

the establishment of automobile use as a digital necessity of life in the 20th: The justices had to specify novel rules for the new personal domain of the passenger car then; they must sort out how the Fourth Amendment applies to digital information now.

26. The Supreme court, will work out whether, during an arrest, it is legitimate to

[A] search for suspects' mobile phones without a warrant.

[B] check suspects' phone contents without being authorized.

[C] prevent suspects from deleting their phone contents.

[D] prohibit suspects from using their mobile phones.

27. The author's attitude toward California's argument is one of

[A] tolerance.

[B] indifference.

[C] disapproval.

[D] cautiousness.

28. The author believes that exploring one's phone content is comparable to

[A] getting into one's residence.

[B] handing one's historical records.

[C] scanning one's correspondences.

[D] going through one's wallet.

29. In Paragraph 5 and 6, the author shows his concern that

[A] principles are hard to be clearly expressed.

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