2017年军考数学真题《历年军考真题系列》
2017年解放军军考数学真题及参考答案
2017年士兵高中军考数学真题解放军军考数学真题，解放军士兵考军校资料，解放军2017数学，德方军考，解放军军考真题，解放军军考资料德方军考寄语 首先预祝你2018年军考取得好成绩！军考真题的参考意义巨大，希望你好好利用这份军考真题。
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一、单项选择（每小题4分，共36分）．1. 设集合A={y|y=2x ，x ∈R}，B={x|x 2﹣1＜0}，则A ∪B=（ ） A ．（﹣1，1） B ．（0，1） C ．（﹣1，+∞） D ．（0，+∞）2. 已知函数f （x ）=a x +log a x （a ＞0且a≠1）在[1，2]上的最大值与最小值之和为（log a 2）+6，则a 的值为（ ） A ． B ．C ． 2D ．43. 设a b 、是向量，则||=||a b 是|+|=|-|a b a 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4．已知421353=2,4,25a b c ==，则（ ）A ．b<a<cB ．a<b<cC ．b<c<aD ． c<a<b5. 设F 为抛物线C ：y 2=3x 的焦点，过F 且倾斜角为30°的直线交C 于A ，B 两点，O 为坐标原点，则△OAB 的面积为（ ） A ．B ．C ．D ．6. 设数列{a n }是首项为a 1、公差为-1的等差数列，S n 为其前n 项和，若S 1，S 2，S 4成等比数列，则a 1=（ ） A ．2 B ． C ．﹣2D ．﹣7. 袋中共有15个除了颜色外完全相同的球，其中有10个白球，5个红球．从袋中任取2个球，所取的2个球中恰有1个白球，1个红球的概率为（ ） A ．B ．C ．D ．18. 已知A ，B ，C 点在球O 的球面上，∠BAC=90°，AB=AC=2．球心O 到平面ABC 的距离为1，则球O 的表面积为（ ） A ．12π B ．16π C ．36π D ．20π9. 已知2017ln f x x x =+（）（），0'2018f x =（），则0x =（ ） A. 2eB.1C. ln 2D. e二、填空题（每小题4分，共32分）10. 设向量，，且，则m=．11.设tanα，tanβ是方程x2﹣3x+2=0的两个根，则tan（α+β）的值为．12. 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．13. 已知函数f（x）=，则f（f（））= ．14. 在的展开式中x7的项的系数是．15. 我国第一艘航母“辽宁舰”在某次舰载机起降飞行训练中，有5架“歼﹣15”飞机准备着舰，如果甲、乙两机必须相邻着舰，而丙、丁两机不能相邻着舰，那么不同的着舰方法数是_______。
2017年优秀士兵提干考试之数学运算备考专练3
【答案与解析】：D。请注意，本题问的是亮的灯。4÷6=66.7%。
6、一瓶挥发性药物，每天挥发 5 毫升，15 天后挥发了全部的 75%，假如每天挥发的速 度不变，余下的几天能挥发完?( ) A. 4 B.5 C. 6 D. 7
【答案与解析】 ： B。 5×15÷75%=100ml 这瓶药物共 100ml， 100-5×15=25ml， 剩下 25ml， 25÷5=5 天。 7、有两个相同的正方体，每个正方体的六个面上分别标有数字 1、2、3、4、5、6。将 两个正方体放到桌面上，向上的一面数字之和为偶数的有多少种情形?( )。 A.9 B.12 C.18 D.24
1、甲、乙两人在银行都有存款，已知甲有存款 160 元，若甲取出存款的 75%，乙取出 存款的 1/3，则甲的余款是乙的一半，那么乙原来在银行存款( )元。 A.120 B.100 C.150 D.200
【答案与解析】：A。本题的计算式为：160×
1 2 =40；40×2=80；80÷ =120(元)。 4 3
【答案与解析】：C。依题意可知：2a=1,b+5=0,c-3=0，则 a=0，b=-5，c=3。则 a+b+c=0+3-5=-2，故正确答案为 C。 4、正方形边长扩大四倍，那么面积扩大( )。 A.4 倍 B.8 倍 C.16 倍 D.64 倍
2017年士兵提干考试分析推理：数量关系练习题13
2017年士兵提干考试分析推理：数量关系练习题13关键词：士兵提干考试大学生士兵张为臻分析推理提干考试练习数量关系1、1，2，6，15，31，()A.46B.61C.66D.56答案：D【解析】2-1=12，6-2=22，15-6=33，31-15=42，56-31=52。
2、165，140，124，()，111A.135B.150C.115D.200答案：C【解析】前一项减后一项得到25，16，9，4(平方数列)，括号内应填115。
3、1，9，36，100，()A.81B.27C.125D.225答案：D【解析】后一项减前一项得到8，27，64，125(立方数列)，括号内应填225。
4、1，3，4，8，16，()A.26B.24C.32D.16答案：C【解析】从第三项开始，后一项都是前面所有项之和，即1+3=4，1+3+4=8，1+3+4+8=16，故第六项为1+3+4+8+16=32。
5、6，24，60，120，()A.186B.200C.210D.220答案：C【解析】6=23-2，24=33-3，60=43-4，120=53-5，210=63-6。
6、2，5，11，23，47，()A.71B.67C.97D.95答案：D【解析】此题中前五项的差分别为3，6，12，24，构成首项为3，公比为2的等比数列，因此第六项与第五项的差应为48，故第六项为48+47=95。
张为臻博客7、80，62，45，28，()A.20B.18C.12D.9答案：D【解析】80=92-1；62=82-2；45=72-4；28=62-8；9=52-16。
8、1，3，11，123，()A.15131B.146C.16768D.96543答案：A【解析】此数列的规律为：前一项的平方加2得到下一项，依此规律，答案为A。
9、能被15和12整除的最小正整数是()。
A.60B.120C.180D.30答案：A【解析】从答案选项入手，显然A能被15和12整除，然后查看比A选项小的数，D选项虽然比A选项小，但30不能被12整除，故答案为A。
军考数学真题《历年军考真题系列》
2013年军考数学真题《历年军考真题系列》历年军考真题系列之2013年军队院校招生士兵高中数学真题关键词：军考真题，德方军考，军考试题，军考资料，士兵高中，军考数学考 生 须 知 1.本试题共八大题，考试时间150分钟，满分150分。
2.将单位、姓名、准考证号分别填写在试卷及答题纸上。
3.所有答案均写在答题纸上，写在试卷上的答案一律无效。
4.考试结束后，试卷及答题纸全部上交并分别封存。
一、（36分）选择题，本题共有9个小题，每个小题都给出代号为A 、B 、C 、D 的四个结论，其中只有一个结论是正确的，将正确的结论代号写在答题纸指定位置上，选对得4分，选错、不选或多选一律得0分．1．已知集合P= {x │x (x -l)≥0，x ∈R ），Q={x │101x >-，x ∈R ），则P∩Q 等于( )．A ．∅B ．{x│x≥1，x ∈R ）C ．{x│x>1,x ∈R)D ．{x│x≥1或x<0,x ∈R)2．已知A·B·C≠0，则“A 、B 、C 三者符号相同”是“方程A x 2 +By 2 =C 表示椭圆”的( )． A ．充要条件 B ．充分不必要条件C ．必要不充分条件D ．既不充分也不必要条件 3．设11221logtan 70,log sin 25,()cos 252a b c =︒=︒=︒，则有( )．A ．a <b <cB ．b <c <aC ．a <c <bD ．c <b <a4．设S n 是等差数列{a n }的前n 项和，S 5 =3(a 2 +a 8)，则53a a的值为( )． A ．16 B ．13 C ．35 D ．565．函数3sin cos y x x=+的一个单调增区间是( )．A ．7,66ππ⎡⎤⎢⎥⎣⎦B ．4,33ππ⎡⎤⎢⎥⎣⎦C ．5,66ππ⎡⎤-⎢⎥⎣⎦D ．233ππ⎡⎤-⎢⎥⎣⎦6．已知向量a =（2，-1，3），b =(-4，2，x )，c =（1，-x ，2），若(a + b ) ⊥c ，则x =( )． A ．4 B ．-4 C ．2 D ．-2 7．设双曲线22221(0,0)x y a b a b-=>>的渐近线与抛物线y=x 2+1相切，则双曲线的离心率等于( )． A 3 B 5 C ．2 D 68．在正三棱柱ABC –A 1B 1C 1中，若12AB BB =，则AB 1与C 1B 所成角的大小为( )． A. 90° B.60° C.75° D. 105° 9．曲线y=x 4上的点到直线x -2y-l=0的距离的最小值是( )‘ A ．12B 5C ．58D 5二、（32分）填空题，本题共有8个小题，每个小题4分，只要求给出结果，并将结果写在答题纸指定位置上． 1·设函数22,1(),1x x f x x x -⎧<=⎨≥⎩，若f (x )>4，则x 的取值范围为 ．2．2cos10sin 20cos 20︒-︒=︒． 3．直线l 将圆x 2+ y 2-2x -4y=0平分，且l 不通过第四象限，则直线l 的斜率的取值范围是____． 4．甲、乙两人从4门课程中各选修2门，则甲、乙两人所选修的课程中恰有1门相同的选法有 种（用数字作答）． 5．若321()nxx+的展开式中只有第6项的二项系数最大，则常数项的值为 ．6．抛物线的顶点在原点，对称轴为坐标轴，且焦点在直线3x -4y-12=0上，则抛物线的方程 为 ．7．21n n =+ ．8．正四棱锥P -ABCD 的五个顶点在同一球面上，若正四棱锥的底面边长为4，侧棱长为26则此球的体积为 ．三、（16分）计算题，本题共有2个小题． 1．（6分）解不等式：213x x +-<2．（10分）在△ABC 中，BC=a ，AC=b ，且a 、b 是方程22320xx -+=的两根，2cos （A+B ）=1(1)求角C 的度数； (2)求AB 的长； (3)求△ABC 的面积．四．（12分）成等差数列的三个正数的和等于15，并且这三个数分别加上2、5、13后成为等比数列{b n }中的b 3，b 4，b 5（1）求数列{b n }的通项公式；（2）设数列{b n }的前n 项和为S n ，求证：数列54nS ⎧⎫+⎨⎬⎩⎭是等比数列．五、（12分）有一种舞台灯，外形是正六棱柱ABCDEF – A 1B 1C 1D 1E 1F 1，在每一个侧面上（不在棱上）安装5只颜色各异的彩灯，假如每只灯正常发光的概率是0.5，若一个面上至少有3只灯发光，则不需要维修，否则需要更换这个面．假定更换一个面需要100元，用ξ表示维修时所更换的面数。
消防士兵考军校真题试卷：数学部分(二)
消防士兵考军校真题试卷：数学部分（二）关键词：消防考军校 真题试卷 京忠教育 军考数学 消防考试资料参考公式（三角函数的积化和差公式）()()1sin cos sin sin 2αβαβαβ=⎡++-⎤⎣⎦()()1cos sin sin sin 2αβαβαβ=⎡+--⎤⎣⎦ ()()1cos cos cos cos 2αβαβαβ=⎡++-⎤⎣⎦()()1sin sin cos cos 2αβαβαβ=-⎡+--⎤⎣⎦ 一、单项选择题（共60分，每小题5分）1．设{(,)|4}P x y x y =+=，{(,)|2}Q x y x y =-=，则P Q = （ ）. A ．{3,1} B ．(3,1) C ．{(3,1)}D ．{3,1}x y ==2．函数242y x x =-+-在区间[3,4]上的最大值是（ ）. A ．2 B ．2- C ．1-D ．13．在等比数列{}n a 中，12100a a +=，3420a a +=，那么56a a +=（ ）. A ．2 B ．4 C ．10D ．54．如果关于x 的不等式250x a -…的正整数解是1，2，3，4，5，那么实数a 的取值范围是（ ）. A ．125180a <… B ．125a … C ．125a >D ．180a <5．已知两点(4,1)A ，(7,3)B -，则与向量AB反方向的单位向量是（ ）.A ．34(,)55-B ．34(,)55-C ．43(,)55-D ．43(,)55-6．五人站成一排，其中甲，乙，丙必须相邻，且甲必须站在乙、丙的中间，则不同的排法有（ ）种. A ．6 B ．12 C ．18D ．247．若直线340ax y +-=与圆22410x y x ++-=相切，则a 的值为（ ）.A ．6±B ．2±C ．8±D ．1±8．若角α，β满足αβ-π<<<π，则αβ-的取值范围是（ ）. A ．(2,0)-π B ．(2,2)-ππ C ．(0,)πD ．3(,)22ππ-- 9．下列命题中的真命题是（ ）. A ．垂直于同一条直线的两条直线平行 B ．平行于同一条直线的两个平面平行 C ．垂直于同一条直线的两个平面平行 D ．垂直于同一平面的两个平面平行10．若函数122log (2log )y x =-的值域是(0,)+∞，那么它的定义域是（ ）.A ．(0,2)B ．(2,4)C ．(0,4)D ．(0,1)11．函数2sin()34y x π=+，x R ∈的单调递增区间是（ ）.A ．3[2,2],44k k k πππ+π+∈ZB ．[(21),2],k k k -ππ∈ZC ．[2,2],2k k k ππ+π+π∈ZD ．3[2,2],44k k k πππ-π+∈Z 12．双曲线与椭圆221259x y +=有公共的焦点，若它们的离心率的和为145，则双曲线的方程为（ ）.A ．221124x y -=B ．221412y x -=C ．221412x y -=D ．221124y x -=二、填空题（本大题共6个小题，每小题5分，共30分）13．若集合2{|300}P x x x =+-=，集合{|30}T x mx =+=，且T P ⊆，则由实数m 的可取值组成的集合为14．2835()3x x-展开式中，整式的项是前项.15．在等差数列{}n a 中，若123989910050a a a a a a ++++++= ，则299a a +=.16．求值：1sin10= .17．若奇函数()y f x =在R 上单调递减，且2()()f m f m >-，则实数m 的取值范围是. 18．如图，在正三棱柱111ABC A B C -中，底面边长为2，侧棱长为3，则1BB 与平面11AB C 所成的角是.三、解答题（本大题共5小题，满分60分. 其中19小题10分，20～22小题每小题12分，23小题14分. 解答应写出文字说明、证明过程或演算步骤） 19．（10分）已知3tan 4α=，1tan()3αβ-=-，求tan()αβ+的值.20．（12分）已知函数3()log (01,0)3ax bf x a a b x b+=>≠>-且. （1）求()f x 的定义域；（7分）（2）讨论()f x 在(,)3b+∞上的单调性.（5分）21．（12分）设二次方程2*110()n n a x a x n N +-+=∈有两个实根αβ和，且满足43ααββ-+=，17a =. （1）试用n a 表示1n a +；（6分）（2）求证：{2}n a +是等比数列；（3分） （3）求数列{}n a 的通项公式.（3分）22．（12分）已知双曲线2212y x -=与点(2,1)P ，过P 作直线l 与双曲线交于A 、B 两点，若点P 为AB 的中点，求直线AB 的方程.23．（14分）如图所示，已知四棱锥P ABCD -的底面是边长为a 的菱形. 120ABC ∠= ，PC ABCD ⊥平面，PC a =，E 为PA 的中点.（1）求证：平面EBD ABCD ⊥平面；（8分）（2）求二面角A BE D --的大小.（6分）。
2017年军考英语真题《历年军考真题系列》
历年军考真题系列之2017年军队院校招生士兵高中军考英语真题关键词：军考真题，德方军考，军考试题，军考资料，士兵高中，军考英语一、选择填空（共15分，每小题1分）1. —We could invite John and Barbara to the Friday night party.—Yes, ____ ? I ’ll give them a call right now.A.why notB. what forC. whyD. what2. Try ____ she might, Sue couldn ’t get the door open.A. ifB. whenC. sinceD. as3. Planning so far ahead ____ no sense-so many things will have changed by next year.A. madeB. is makingC. makesD. has made4. I wasn ’t sure if he was really interested or if he ____ polite.A. was just beingB. will just beC. had just beenD. would just be5. The lawyer rarely wears anything other than jeans and a T-shirt ____ the season.A. whateverB. whereverC. wheneverD. however6. I like getting up very early in summer. The morning air is so good ____.A. to be breathedB. to breatheC. breathingD. being breathed7. —Have you known Dr. Jackson for a long time?—Yes, since she ____ the Chinese Society.A.has joinedB. joinsC. had joinedD. joined8. We all know that, ____, the situation will get worse.A. not if dealt carefully withB. if not carefully dealt withC. if dealt not carefully withD. not if carefully dealt with9. I smell something ____ in the kitchen. Can I call you back in a minute?A. burningB. burntC. being burntD. to be burnt10. Does this meal cost $50? I ____ something far better than this!A. preferB. expectC. suggestD. suppose11. The prize will go to the writer ____ story shows the most imagination.A. thatB. whichC. whoseD. what12. They ____ have arrived at lunchtime but their flight was delayed.A. willB. canC. mustD. should13. It is generally accepted that ____ boy must learn to stand up and fight like ____man.A. a; aB. a; theC. the; theD. a; /14. It is important to pay your electricity bill on time, as late payments may affect your ____.A. conditionB. incomeC. creditD. status15. —It was a wonderful trip. So, which city did you like better, Paris or Rome?—____. There were good things and bad things about them.A. It’s hard to sayB. I didn’t get itC. You must be kiddingD. Couldn’t be better二、阅读理解（共40分，每小题2分）Passage 1How can you remember a song from your childhood to this day? Why do your teachers use songs to teach you English? It seems there is a scientific reason for this.Researchers are now studying the relationship between music and remembering a foreign language. They find that remembering words in a song is the best way to remember even the most difficult language.“Singing could be a new way of learning a foreign language. The brain likes to remember things when they are used in a catchy and meaningful way”, said Dr, Karen Ludke. The findings may help those who have difficulties learning foreign languages. On his blog, Dr. Ludke writes, “A listen-and-repeat singing method can support foreign language learning, and opens the door for future research in this area.”Many language teachers know the value of using music and singing. Hua Zhuying, a teacher at a Chinese language school in Washington, D.C. depends heavily on songs in teaching Chinese. She says, “I use music all the time to t each children Chinese. For little kids usually we use the music. Not only does it work, but it is fun for kids.”“Sometimes, I think if I were taught English that way, maybe I could speak much better English than now,” Hua Zhuying adds.Our brain likes music, especially for remembering. So, if you’re still struggling in learning a language, why not try singing it out ?16. According to the passage, the best way to remember a foreign language is to .A. listen to the teacher carefullyB. copy the words many times.C. remember words in a songD. read and write more17.The underlined word “catchy”probably means “____”A.infamousB. boringC. impressiveD.emotional18.Dr. Ludke believes that foreign language learning can be supported by .A. writing songs with the languageB. listening to all kinds of famous musicC. using the listen-and-repeat singing methodD. reading the lyrics again and again19.From the passage, we know that Hua Zhuying .A. is interested in writing English songsB. teaches Chinese in an American schoolC. teaches children English by using musicD. is a researcher in a language school20.The passage mainly tells us that .A. many researchers realize the importance of language studyingB. many language teachers know the value of famous musicC. your brain remembers a language better if you sing it.D. you will never learn a language well unless you can singPassage 2Mass media are tools of communication. Mass media can be divided into two groups: print media and electronic media. By print media, we mean books, newspapers and magazines. Electronic media include television, computer, radio and movie. Mass media allow us to record and pass information rapidly to a large, scattered audience. They extend our ability to talk to each other by helping us overcome barriers caused by time and space.There are various ways in which mass media make daily life easier for us. Firstly, they inform and help us keep a watch on our world. They gather and pass on information we would be unlikely or unable to obtain on their own.Secondly, mass media help us to arrange our time and life. What we talk about and what we think about are greatly influenced by the media. When people get together, they tend to talk about certain happenings in newspapers or on TV. Because we are exposed to different points of view through different kinds of media every day, we are able to evaluate all sides of a certain issue.Thirdly, the media are used to persuade people. A good example is advertisements through the media. Newspapers, magazines and TV are filled with all kinds of colorful, persuasive advertisements. Though many advertisements may not say openly that they want you to buy a certain product, they describe their products in such a way that you may want to buy them.Fourthly, the media also entertain. All of the media make efforts to try to entertain their audience. For instance, even though the newspaper is a prime medium of information, it also contains entertainment features. Television, motion pictures, fiction books and some radio stations and magazines are devoted mainly to entertainment. It is estimated that in the future, the entertainment function of mass media will become even more important than it is now.21. What makes it possible for people living in different places to communicate with each other?A. Printed mediaB. Mass mediaC. Electronic mediaD. Computers22. Which of the following functions of mass media is Not mentioned?A. To make people well informed about the worldB. To amuse and entertain peopleC. To help people arrange their time and lifeD. To give people a sense of honor23. Certain matters in newspapers or on TV tend to be talked about when people get together because ____.A. people are curious about themB. people are influenced by those mass mediaC. it is fashionable for people to do soD. it is easy for people to communicate in this way24. How does advertisement make people purchase certain goods according to the passage?A. By giving an attractive account of the goods.B. By asking people to buy them.C. By forcing people to buy them.D. By giving people something extra.25. Which of the following media is mainly devoted to information according to the passage?A. TVB. MagazinesC. Motion PicturesD. NewspapersPassage 3When international aid is given, steps must be taken to ensure (确保)that the aid reaches the people for whom it is intended. The way to achieve this may not be simple. It is very difficult for a nation to give help directly to people in another nation. The United Nations Organization (UNO) could undertake to direct the distribution of aid. Here however rises the problem of costs. Also tied with this is time. Perhaps the UNO could set up a body of devoted men and women in every country who can speedily distribute aid to victims of floods and earthquakes.More than the help that one nation can give to another during a disaster; it would be more effective to give other forms of help during normal times. A common proverb says, “Give me a fish and I eat for day, teach me to fish and I eat for a lifetime.”If we follow this wise saying, it would be right to teach people from less developed nations to take care of themselves. For example, a country could share its technology with another. This could be in simple areas like agriculture or in more complex areas like medical and health care or even in building satellites. Even small country is able to help less developed nations. Sometimes what is take for granted, like the setting up of a water purification plant or the administration of a school, could be useful for countries which are looking about to solve common problems. It does not cost much to share such simple things. Exchange students could be attached for a number of months or years and learn the required craft whileon the site. They can then take their knowledge back to their homelands and if necessary come back from time to time to clear doubts or to update themselves. Such aid will be truly helpful and there is no chance of it being temporary or of it falling into the wrong hands.Many countries run extensive courses in all sorts of skills. It will not cost much to include deserving foreigners in these courses. Besides giving effective help to the countries concerned, there is also the build-up of friendships to consider. Giving direct help by giving materials may be effective in the short run and must continue to be given in the event of emergencies. However, in the long run what is really effective would be the sharing of knowledge.26. According to the author, how could international aid reach the victims in time?A. By solving the cost problemsB. By solving the transportation problemsC. By relying on the direct distribution of the UNO.D. By setting up a body of devoted people in every country.27. What does the author try to express in the underlined sentence?A. Providing food is vital . Learning to fish is helpfulC. Teaching skills is essentialD. Looking after others is important.28. The second paragraph is developed mainly ____.A.by contrastB.by processC.by comparisonD.by example29. Which aid is likely to fall into the wrong hands?A. A medical team.B. An exchange program.C. Financial support.D. A water plant.30. What can we infer about international aid from the passage?A. It is facing difficultiesB. It is unnecessary during normal timesC. It should be given in the form of materialsD. It has gained support from developed countriesPassage 4Human needs seem endless. When a hungry man gets a meal, he begins to think about an overcoat; when a manager gets a new sports car, a big house and pleasure boats dance into view.The many needs of mankind might be made up of several levels. When there is money enough to satisfy one level of needs, another level appears.The first and most basic level of needs involves food. Once this level is satisfied, the second level of needs, clothing and some sort of shelter, appears. By the end of World WarⅡ, these needs were satisfied for a great majority of Americans. Then a third level appeared. It included such items as automobiles and new houses.By 1957 or 1958 this third level of needs was fairly well satisfied. Then, in the late 1950s, a fourth level of needs appeared: the “life-enriching” l evel. While the other levels involve physical satisfaction, that is, the feed in comfort, safety, and transportation, this level stresses mental needs for recognition, achievement, and happiness. It includes a variety of goods and services, many of which c ould be called “luxury” items. Among them are vacation trips, the best medical and dental care, and recreation. Also included here are fancy goods and the latest fashions.On the fourth level, a lot of money is spent on services, while on the first three levels more is spent on goods. Will consumers raise their sights to a fifth level of needs as their income increases, or will they continue to demand luxuries and personal services on the fourth level?A fifth level would probably involve needs that can be achieved best by community action. Consumers may be spending more on taxes to pay for government action against disease, ignorance, crime, and prejudice. After filling our stomachs, our clothes closets, our garages, our teeth and our minds, we now may seek to ensure the health, safety, and35. The author tends to think that the fifth level .A．would be little better than the fourth levelB．may be a lot more desirable than the first fourC．may ask consumers to spend less on taxesD．would be attainable before the government takes actions三、完形填空（共15分，每小题1分）new travel experiences.Eco-tourism has many benefits. First, all the money spent by the tourists is used to protect the important environmentalbecause of a focus on more sustainable(不破坏生态环境的，可持续的)travel. 50 this, local people's living-standards have improved. Many others are now following Peru's example and using eco-tourism to preserve their environment for the future generations.36. A. finally B. usually C. suddenly D. roughly37. A. pleasure B. satisfaction C. popularity D. freedom38. A. approving B. confirming C. enriching D. supporting39. A. respectful B. aware C. uncertain D. independent40. A. rich B. curious C. lazy D. normal41. A. use B. recycle C. copy D. restrict42. A. challenging B. disappointing C. positive D. risky43. A. success B. adventure C. tool D. symbol44. A. polite B. sincere C. doubtful D. grateful45. A. mines B. factories C. schools D. hotels46. A. ban B. result C. decline D. appearance47. A. close B. safe C. friendly D. active48. A. influences B. examines C. balances D. improves49. A. environment B. law C. economy D. hosts50. A. But for B. According to C. In spite of D. In addition to四、翻译（共20分，汉译英每小题3分，英译汉每小题2分）51.每个人都能做与他体力和脑力相适应的工作。
2017年士兵考军校复习资料综合试卷.doc
1.下列函数既是奇函数，又是增函数的是( )A.y=log2 x| B・ y二x3+x C. y=3x D. y=x 3【分析】A： y=log2|x|是偶函数B： y=x+x3既是奇函数又是增函数.C：非奇非偶函数D： y=x 3是奇函数，但是在(0, +°°), ( - 0)递减函数，从而可判断【解答】解:A： y=log2|x|是偶函数B： y=x+x3既是奇函数又是增函数.C： y二丁非奇非偶函数D： y=x 3是奇函数，但是在(0, +°°), ( - 0)递减函数故选B.【点评】本题主要考察了函数的奇偶性及函数的单调性的判断，属于基础试题2.奇函数f (x)在区间［1, 4］上为减函数，则它在区间［-4, -1］上( )A.是减函数B.是增函数C.无法确定D.不具备单调性【分析】先根据函数单调性的定义，在区间［・4, -1］±任取X1，X2,且设岀大小关系，则・X1、-x2e ［l, 4］,根据奇函数f (x)在区间［1, 4］上为减函数，达到比较f (X1)与f (x2)的大小，从而判断函数在区间［-4, -1］±的单调性.【解答】解:Vf (x)为奇函数/.f ( - X)= - f(X),Vxi， X2^ ［ ~ 4, - 1］,且X1<X2Vf (x)区间［1, 4］上单调递减，/• 4^ - X］> - X2MI,/.f ( - Xi)<f ( - X2),/.f(Xi)>f (x2)・・・f (x)在区间［・4,上单调减.故选A.【点评】本题主要考查函数奇偶性和单调性的应用以及应用单调性的定义判断单调性的方法，体现了转化的思想.属基础题.3.下列函数是奇函数的是( )A.y二x - 1 B・ y二2x? 一3 C・ y=x3 D. y=2x【分析】根据函数的图象判断.【解答】解：A、D两项图象既不关于y轴对称，也不关于原点对称，所以它们不是奇函数.B项图彖关于y轴对称，所以它是偶函数.故选C.【点评】掌握基木初等函数的图象，解题时方便快捷！4.中共中央办公厅、国务院办公厅下发的《关于健全和完善村务公开和民主管理制度的意见》，对落实农民群众的知情权、决策权、参与权、监督权（〃四权〃）提出了许多新的明确的政策措施，为农民群众依法维护自己的民主权利提供了依据和保障。
2017士兵考军校复习资料
1．给出以下四个命题：①若ab≤0，则a≤0或b≤0；②若a＞b则am2＞bm2；③在△ABC中，若sinA=sinB，则A=B；④在一元二次方程ax2+bx+c=0中，若b2﹣4ac＜0，则方程有实数根．其中原命题、逆命题、否命题、逆否命题全都是真命题的是（）A．① B．② C．③ D．④【分析】根据题意，分别写出每个命题的逆命题、否命题和逆否命题，再判断它们的真假．【解答】解：对于①，原命题是：若ab≤0，则a≤0或b≤0，是真命题，逆命题是：若a≤0或b≤0，则ab≤0，是假命题，否命题是：若ab＞0，则a＞0或b＞0，是假命题，逆否命题是：若a＞0且b＞0，则ab＞0，是真命题；对于②，原命题是：若a＞b，则am2＞bm2，是假命题，逆命题是：若am2＞bm2，则a＞b，是真命题，否命题是：若a≤b，则am2≤bm2，是真命题，逆否命题是：若am2≤bm2，则a≤b，是假命题，对于③，原命题是：在△ABC中，若sinA=sinB，则A=B，是真命题，逆命题是：在△ABC中，若A=B，则sinA=sinB，是真命题，否命题是：在△ABC中，若sinA≠sinB，则A≠B，是真命题，逆否命题是：在△ABC中，若A≠B，则sinA≠sinB，是真命题；对于④，原命题是：在一元二次方程ax2+bx+c=0中，若b2﹣4ac＜0，则方程有实数根，是假命题，逆命题是：在一元二次方程ax2+bx+c=0中，若方程有实数根，则b2﹣4ac＜0，是假命题，否命题是：在一元二次方程ax2+bx+c=0中，若b2﹣4ac≥0，则方程无实数根，是假命题，逆否命题是：在一元二次方程ax2+bx+c=0中，若方程无实数根，则b2﹣4ac≥0，是假命题；综上，以上命题中，原命题、逆命题、否命题、逆否命题全都是真命题的是③．故选：C．【点评】本题考查了四种命题之间的关系，解题时应明确四种命题的语言叙述是什么，它们之间的真假关系是什么，是综合题．2．命题“若a＞﹣3，则a＞﹣6”以及它的逆命题、否命题、逆否命题中，假命题的个数为（）A．1 B．2 C．3 D．4【分析】根据四种命题的关系写出答案即可．【解答】解：在命题的四种形式中原命题和逆否命题互为逆否命题，同真同假，否命题和逆命题互为逆否命题同真同假．∵命题“若a＞﹣3，则a＞﹣6”为真命题；逆命题是假命题，∴命题的逆否命题为真命题，故选B．【点评】此题考查了四种命题的关系，熟练掌握他们之间的关系是解本题的关键．3．（2016春•晋中校级期中）考察以下列命题：①命题“lgx=0，则x=1”的否命题为“若lgx≠0，则x≠1”②若“p∧q”为假命题，则p、q均为假命题③命题p：∃x∈R，使得sinx＞1；则￢p：∀x∈R，均有sinx≤1④“x＞2”是“＜”的充分不必要条件则真命题的个数为（）A．1 B．2 C．3 D．4【分析】①根据否命题的定义进行判断，②根据复合命题真假关系进行判断，③根据特称命题的否定是全称命题进行判断，④根据充分条件和必要条件的定义进行判断．【解答】解：①命题“lgx=0，则x=1”的否命题为“若lgx≠0，则x≠1”，正确，故①正确，②若“p∧q”为假命题，则p、q至少有一个为假命题，故②错误③命题p：∃x∈R，使得sinx＞1；则￢p：∀x∈R，均有sinx≤1，正确，故③正确，④当x＞2时，＜成立，当x＜0时，＜成立，但x＞2不成立，故“x＞2”是“＜”的充分不必要条件，故④正确，故正确是①③④，故选：C【点评】本题主要考查命题的真假判断，涉及的知识点较多，综合性较强，但难度不大．。
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历年军考真题系列之
2017年军队院校招生士兵高中军考数学真题
关键词：军考真题，德方军考，军考试题，军考资料，士兵高中，军考数学
一、单项选择（每小题4分，共36分）．
1. 设集合A={y|y=2x ，x ∈R}，B={x|x 2﹣1＜0}，则A ∪B=（ ）
A ．（﹣1，1）
B ．（0，1）
C ．（﹣1，+∞）
D ．（0，+∞）
2. 已知函数f （x ）=a x +log a x （a ＞0且a≠1）在[1，2]上的最大值与最小值之和为（log a 2）+6，则a 的值为（ ）
A ．
B ．
C ．2
D ．4
3. 设a b 、是向量，则||=||a b 是|+|=|-|a b a 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 4．已知4
21353=2,4,25a b c ==，则（ ）
A ．b<a<c
B ．a<b<c
C ．b<c<a
D ． c<a<b
5. 设F 为抛物线C ：y 2=3x 的焦点，过F 且倾斜角为30°的直线交C 于A ，B 两点，O 为坐标原点，则△OAB 的面积为（ ）
A ．
B ．
C ．
D ．
6. 设数列{a n }是首项为a 1、公差为-1的等差数列，S n 为其前n 项和，若S 1，S 2，S 4成等比数列，则a 1=（ ）
A ．2
B ．
C ．﹣2
D ．﹣
7. 袋中共有15个除了颜色外完全相同的球，其中有10个白球，5个红球．从袋中任取2个球，所取的2个球中恰有1个白球，1个红球的概率为（ ）
A ．
B ．
C ．
D ．1
8. 已知A ，B ，C 点在球O 的球面上，∠BAC=90°，AB=AC=2．球心O 到平面ABC 的距
离为1，则球O 的表面积为（ ）
A ．12π
B ．16π
C ．36π
D ．20π
9. 已知2017ln f x x x =+（）（）
，0'2018f x =（），则0x =（ ） A. 2e B.1
C. ln 2
D. e 二、填空题（每小题4分，共32分）
10. 设向量，，且，则m= ． 11. 设tanα，tanβ是方程x 2﹣3x+2=0的两个根，则tan （α+β）的值为 ．
12. 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 ．
13. 已知函数f （x ）=，则f （f （））= ．
14. 在的展开式中x 7的项的系数是 ．
15. 我国第一艘航母“辽宁舰”在某次舰载机起降飞行训练中，有5架“歼﹣15”飞机准备着舰，
如果甲、乙两机必须相邻着舰，而丙、丁两机不能相邻着舰，那么不同的着舰方法数是_______。
16. 在极坐标系中，直线ρcosθ﹣ρsinθ﹣1=0与圆ρ=2cosθ交于A ，B 两点，则|AB|=_______．
17. 已知n 为正偶数，用数学归纳法证明
时，若已假设n=k （k≥2，k 为偶数）
时命题为真，则还需要用归纳假设再证n= 时等式成立．
三、解答题（共7小题，共82分，解答题应写出文字说明、演算步骤或证明过程）
18.（本小题8分）对任意实数x ，不等式﹣9＜22361
x px x x +--+＜6恒成立，求实数p 的取值范围。
19.（本小题12分）
20、（12分）已知数列{a n}中，a1=1，二次函数f（x）=a n•x2+（2﹣n﹣a n+1）•x的对称轴为x=．
（1）试证明{2n a n}是等差数列，并求{a n}通项公式；
（2）设{a n}的前n项和为S n，试求使得S n＜3成立的n值，并说明理由．
21、（10分）已知5只动物中有1只患有某种疾病，需要通过化验血液来确定患病的动物．血液化验结果呈阳性的即为患病动物，呈阴性即没患病．下面是两种化验方法：
方案甲：逐个化验，直到能确定患病动物为止．
方案乙：先任取3只，将它们的血液混在一起化验．若结果呈阳性则表明患病动物为这3只中的1只，然后再逐个化验，直到能确定患病动物为止；若结果呈阴性则在另外2只中任取1只化验．
（Ⅰ）求依方案甲所需化验次数不少于依方案乙所需化验次数的概率；
（Ⅱ）ξ表示依方案乙所需化验次数，求ξ的期望．
22、（12分）已知函数f（x）=ax+bsinx，当时，f（x）取得极小值．
（1）求a，b的值；
（2）设直线l：y=g（x），曲线S：y=f（x）．若直线l与曲线S同时满足下列两个条件：
①直线l与曲线S相切且至少有两个切点；
②对任意x∈R都有g（x）≥f（x）．则称直线l为曲线S的“上夹线”．试证明：直线l：y=x+2为曲线S：y=ax+bsinx“上夹线”．
23、（14分）已知圆M：x2+（y﹣4）2=4，点P是直线l：x﹣2y=0上的一动点，过点P作圆M的切线PA，PB，切点为A，B．
（1）当切线PA的长度为时，求点P的坐标；
（2）若△PAM的外接圆为圆N，试问：当P在直线l上运动时，圆N是否过定点？若存在，求出所有的定点的坐标；若不存在，说明理由．
（3）求线段AB长度的最小值．
24、（14分）如图，在四棱柱ABCD﹣A1B1C1D1中，侧棱AA1⊥底面ABCD，AB⊥AC，AB=1，AC=AA1=2，AD=CD=，且点M和N分别为B1C和D1D的中点．
（Ⅰ）求证：MN∥平面ABCD
（Ⅱ）求二面角D1﹣AC﹣B1的正弦值；
（Ⅲ）设E为棱A1B1上的点，若直线NE和平面ABCD所成角的正弦值为，求线段A1E 的长．。