西北师大附中高三第五次诊断考试

西北师大附中2017-2018学年高三第五次诊断考试英语考生注意：1.本试卷分第I卷（选择题）和第II卷（非选择题）两部分。

满分120分，考试时间100分钟。

2.答题前，考生务必将自己的姓名、考号写在答题卡和答卷密封线内相应的位置上。

考生要认真核对答题卡上粘贴的条形码的“准考证号、姓名、考试科目”与考生本人准考证号、姓名是否一致。

3.第I卷每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其它答案标号。

第II卷用黑色签字笔在答题卡上书写作答，在试卷卷上作答，答案无效。

第I卷第一部分: 阅读理解（共两节，满分40 分）第一节：(共15小题, 每小题2分, 满分30分)阅读下列短文，从每题所给的四个选项（A、B、C和D）中，选出最佳选项。

AIt was a hot, humid day. My brother Walt and I had decided that the only way to survive it would be to go swimming in a deep swimming hole across Mr. Blickez’s pasture(牧场) and through some woods.The only problem with our plan was that this pasture was guarded by a huge, mean Hereford bull. Mr. Blickez had told us that Elsie was the meanest bull in the township, maybe even the county, and we believed him. But the hotter it got, the more we thought there was something fishy about his claim. For one thing, we remembered Mr. Blickez liked telling tall tales; for another, Elsie seemed like an odd name for a bull.Finally, I talked Mom into asking permission for us to walk through the pasture, but then another problem surfaced. Mom said she would talk to Mr. Blickez if we would take our cousin Joanie along with us. Joanie was almost two years older than me and a head taller. In fact, I still had a headache from a quarrel with her that morning. “I’m not going swimming with that dumb girl cousin.” I told my mom.“Either Joanie goes with, or you stay home alone,” Mom said in her serious tone. I gave in and we set out. On our way across the pasture, Walt yelled suddenly. Elsie had approached him quietly and was licking(舔) his back. Joanie and I dove under the wire fence, but while I was on the ground I looked up and saw that Elsie wasn’t a big mean bull after all. She was go ing to keep licking my brother’s back as long as he stood still.We had many good days growing up and visiting our secret swimming hole guarded by the so-called “big mean bull”. And as it turned out, for a girl cousin, Joanie hasn’t been too bad. She’s been one of my best friends over the years.1. What’s the meaning of the underline word “fishy” in Paragraph 2?A. Funny.B. Interesting.C. Doubtful.D. Believable.2. What’s the second problem the author has to face?A. His cousin made jokes on him in his grade school.B. His mother insisted on his cousin going with him.C. He quarreled with his cousin and had a headache.D. His mother failed to ask permission for him.3. What does the author think of Elsie in the end?A. Aggressive.B. Unkind.C. Friendly.D. Bad-tempered.4. What’s the passage mainly about?A. The story of visiting the swimming hole.B. The bull guarding Mr. Blickez’s farm.C. How friendly the so-called mean bull was.D. How the author changed his attitude to Joanie.BImagine a mass of floating waste is two times the size of the state of Texas. Texas has a land area of more than 678 000 square kilometers. So it might be difficult to imagine anything twice as big. All together, this mass of waste flowing in the North Pacific Ocean is known as the Great Pacific Ocean Garbage Patch. It weighs about 3,500,000 tons. The waste includes bags, bottles and containers — plastic products of all kinds.The eastern part of the Great Pacific Ocean Garbage Patch is about l,600 kilometers west of California. The western part is west of the Hawaiian Islands and east of Japan. The area has been described as a kind of oceanic desert，with light winds and slow moving water currents. The water moves so slowly that garbage from all over the world collects there.In recent years, there have been growing concerns about the floating garbage and its effect on sea creatures and human health. Scientists say thousands of animals get trapped in the floating waste, resulting in death or injury. Even more die from a lack of food or water after swallowing pieces of plastic. The trash can also make animals feel full, lessening their desire to eat or drink.The floating garbage can also have harmful effects on people. There is an increased threat of infection of disease from polluted waste, and from eating fish that swallowed waste. Divers can also get trapped in the plastic.Its existence first gained public attention in l997. That was when racing boat captain and oceanographer Charles Moore and his crew sailed into the garbage while returning from a racing event. Five years earlier, another oceanographer learned of the trash after a shipment of rubber duck got lost at sea. Many of those toys are now part of the Great Pacific Ocean Garbage Patch.5. How did the writer introduce the topic of the passage?A. By giving an example.B. By listing the facts.C. By telling a story.D. By giving a comparison.6. What do we know about the Great Pacific Ocean Garbage Patch?A. It is made up of various kinds of plastic products.B. It is a solid mass of floating waste materials.C. It lies l60 000 kilometers east of California.D. It is described as a kind of oceanic desert.7. Which column can you find the passage on a newspaper?A. Sports and entertainment.B. Media and culture.C. Environment and society.D. Science and technology.8. The purpose of writing this passage is to____________.A. warn people of the danger to travel in the pacificB. analyze what caused the waste patch in the pacificC. give advice on how to recycle waste in the oceanD. introduce the Great Pacific Ocean Garbage PatchCAs the 2016 Presidential election is approaching, US high school students have become more interested in the election. They hope to have a voice in the process and, finally, have a say in deciding the next US president.“Even though I will not be old enough to vote in the next presidential election, my voice matters, as do the voices of every other individual in this country,” Wilentz (will only be 17 this November) said. “Discussions I may have with my peers(同龄人), debates with teachers and communication with my parents allow me to be heard and give my opinions and insight (洞察力).”Alain Jean, the 2008 African American vote director for Barack Obama in the state of Florida, said high school students should be paying attention to the presidential election because many of them will be of voting age by election time. She said candidates have views and positions on certain topics that may affect teens’ lives.Cypress Bay High j unior Olivia Ohayon said, “I believe it is my responsibility to discus s, and try to persuade my parents and other adults to concern themselves with the issues important to the younger generation when casting their vote for president, such as the cost of college education.”Not only do teenagers think their voices are important, but some also think parents have to consider the views of their children when casting their vote. According to Lucy Rimalower, a licensed marriage and family therapist (治疗专家), parents must consider whether their vote represents their household and the teens living in it.Jean said teenagers can also influence the election in other ways by getting involved in campaigns. She suggests that teenagers who actually believe in one of the candidates find the local headquarters (总部) for a campaign and volunteer. Campaign volunteers stuff(装满) envelopes, wave signs and go into neighborhoods to drop off literature (宣传材料) and try to get the word out.9. Which of the following statements is TRUE?A. Teenagers can influence the election by expressing their opinions.B. Most US high school students have the right to vote in the presidential election.C. Wilentz is interested in the election because he enjoys debates and discussions.D. Students below voting age are not allowed to participate in election campaigns.10. According to Alain Jean, ______________.A. teenagers are too young to take part in the electionB. the candidates’ views could influence students’ livesC. candidates should pay more attention to teenagers’ opinionsD. students should be prepared for voting from an early age11. According to Rimalower, parents should consider the views of their children during the election because _____________.A. they should respect their children’s right to voteB. the election process could help their children develop debating and thinking skillsC. their votes should represent not only themselves but also other family membersD. they could reduce the cost of education for their children12. What can we infer from the article?A. Parents do not like their children to participate in elections.B. Many people encourage teenagers’ involvement in the election process.C. Candidates who care more about the younger generation’s issues are more likely to win.D. Students who are interested in elections only care about themselves.DDate: Monday, 4 April 2016Time: 17:00 to 20:00We are a local business that has been operating for 25 years and we would love to introduce you to the wonders of the Yorkshire Dales underground.We will meet you at the parking place nearest the cave and directions will be given on booking. Everyone will change into the caving kit(全套衣服) at the road head and walk a short distance to the cave before beginning your adventure underground. The trip involves walking in a stream way, some stooping (俯身) and some crawling. There are some simple climbs, a few deep pools and a waterfall to deal with too. You just need a reasonable amount of fitness and flexibility to move through the cave as there is bending and some short crawling sections. Minimum Age is 8 years old. We are licensed to take children without a parent with them if necessary.The trips start at 5.00 pm and finish by 8.00 pm.What we supply: Caving suits, helmets, lights, belts, wellies (雨靴), and all safety equipment. What to Wear: Each person should wear warm clothing i.e. trousers (NOT jeans), two pairs of socks, underwear, two layers of upper-body wear (e.g. sweatshirt /pullover (套衫)/fleece (羊毛衣) etc), at least one of which should have full length arms. At certain times of the year hat, gloves and a scarf are also recommended.What to bring: A complete change of clothes and a towel as you will get WET, a bag for your wet clothes.What we will need from you before the event: A contact number for when you are in the area; the name(s) of participants and their feet size, approximate height and build (S, M, L, XL, XXL etc.).The age of any under 18s: We will provide you with a medical form or parental agreement form as appropriate on booking.Charges: Adult £25.00, Child £22.00.Other charges: Booking is essentialFor further information or to book, contact:Tel: 01729 824455Email: info@Web: 13. For the trip, you are advised to ______________.A. bring your own safety equipmentB. bring extra clothes to change intoC. exercise for a certain period of timeD. bring at least two T-shirts14. Teenagers under 18 years old ________________.A. should be accompanied by their parentsB. will be charged less than adultsC. will go on a different route specially prepared for themD. must have a signed medical form or parental agreement form15.Sophia, a 10-year-old girl, and her parents have booked the trip, how much should they pay?A. £72.00B. £75.00C. £47.00D. £69.00第二节（共5小题, 每小题2分，满分10分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。

陕西师范大学附中2025届高三第五次模拟考试语文试卷注意事项：1．答卷前，考生务必将自己的姓名、准考证号、考场号和座位号填写在试题卷和答题卡上。

用2B铅笔将试卷类型（B）填涂在答题卡相应位置上。

将条形码粘贴在答题卡右上角"条形码粘贴处"。

2．作答选择题时，选出每小题答案后，用2B铅笔把答题卡上对应题目选项的答案信息点涂黑；如需改动，用橡皮擦干净后，再选涂其他答案。

答案不能答在试题卷上。

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内相应位置上；如需改动，先划掉原来的答案，然后再写上新答案；不准使用铅笔和涂改液。

不按以上要求作答无效。

4．考生必须保证答题卡的整洁。

考试结束后，请将本试卷和答题卡一并交回。

1、阅读下面的文字，完成下面小题。

呼兰河传萧红冯歪嘴子把小孩搬到磨房南头那草棚子里去了。

那小孩哭的声音很大，好像他并不是刚刚出生，好像他已经长大了的样子。

那草房里吵得不得了，我想去看看。

这回那女人坐起来了，身上披着被子，很长的大辫子垂在背后，面朝里，坐在一堆草上不知在干什么，她一听门响，一回头。

我看出来了，她就是我们同院住着的老王家的大姑娘，我们都叫她王大姐。

这可奇怪，怎么就是她呢？她一回头几乎是把我吓了一跳。

我转身就想往家里跑。

跑到家里好赶快地告诉祖父，这到底是怎么回事？王大姐看是我，就先向我一笑，她有很大的脸孔，很尖的鼻子，每笑时，她的鼻梁上就皱了一堆的褶。

今天她的笑法还是和从前的一样，鼻梁处堆满了皱褶。

她是很能说能笑的人，她是很响亮的人，她和别人相见之下，她问别人：“你吃饭了吗？”那声音才大呢，好像房顶上落了喜鹊似的。

她的父亲是赶车的，她牵着马到井上去饮水，她打起水来，比她父亲打得更快，三绕两绕就是一桶。

别人看了都说：“这姑娘将来是个兴家立业好手！”她在我家后园里摘菜，摘完临走的时候，常常就折一朵马蛇菜花戴在头上。

她那辫子梳得才光呢，红辫根，绿辫梢，干干净净，又加上一朵马蛇菜花戴在鬓角上，非常好看。

西北师大附中2015届高三第五次诊断考试理科综合能力测试化学试题（A卷）本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分，其中第Ⅱ卷第33~40题为选考题，其他题为必考题。

考生作答时，将答案答在答题卡上，在本试卷上作答无效。

考试结束后，将答题卡交回。

可能用到的元素相对原子质量：相对原子质量： Li—7 Mg—24 Cu—64 Ca—40Pb—207 O—16 Cl—35.5第Ⅰ卷（选择题共126分）一、选择题：本题共13小题，每小题6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。

7.下列关于化学用语的表示正确的是（）A.CO2的电子式：O::C::Ol4的比例模型：C.含有18个中子的氯原子的核素符号：错误！未找到引用源。

D.硫原子的结构示意图：8. 下列说法错误的是（）A.明矾可用于海水的净化，但不能用于海水的淡化B.使用清洁能源是防止酸雨的重要的措施C.胃酸过多的病人均可服用含Al(OH)3或NaHCO3的药物进行治疗D.84消毒液可用来漂白某些衣物，为加快漂白速度，使用时可加入少量稀盐酸9. 甲苯苯环上的一个H原子被—C3H6Cl取代，形成的同分异构体有（不考虑立体异构）（）A．9种 B．12种 C．15种 D．18种10. 热激活电池可用作火箭、导弹的工作电源。

一种热激活电池的基本结构如图所示，其中作为电解质的无水LiCl-KCl混合物受热熔融后，电池即可瞬间输出电能。

该电池的总反应为PbSO4+2LiCl+Ca=CaCl2+Li2SO4+Pb。

下列有关说法正确的是（）A.正极反应式：Ca+2Cl－-2e－==CaCl2B.常温时，在正负极间接上电流表或检流计，指针不偏转C.放电过程中，Li+向负极移动D.每转移0.1mol电子，理论上生成20.7g Pb11. W、X、Y、Z是原子序数依次增大的四种短周期元素，甲、乙、丙、丁、戊是由其中的两种或三种元素组成的化合物，己是由Y元素形成的单质，常温下丙和己均为气体。

西北师大附中2015届高三第五次诊断考试数学 （理科） A 卷一、选择题（每小题5分，共60分，在每小题给出的四个选项中，只有一项符合要求.）1. 已知i 是虚数单位，则311i i -⎛⎫⎪+⎝⎭=（ ） A. 1 B. i C. i - D. 1-2. sin 3的取值所在的范围是（ ）A ．⎛ ⎝⎭B ．⎫⎪⎪⎝⎭C ．⎛⎫ ⎪ ⎪⎝⎭D ．1,2⎛⎫-- ⎪ ⎪⎝⎭3.在某次诊断考试中，某班学生数学成绩ξ服从正态分布()()2100,,0σσ>，若ξ在()80,120内的概率为0.8，则ξ落在()0,80内的概率为（ ） A. 0.05B. 0.1C. 0.15D.0.24.数列{n a }的前n 项和)(322+∈-=N n n n S n ，若p-q=5,则q p a a -= （ ） A. 10 B. 15 C. -5 D.205. 在△ABC 中, AB AC BA BC ⋅=⋅“” 是 AC BC =“”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6. 根据如下样本数据：得到了回归方程为ˆybx a =+，则 A. 0,0a b >> B. 0,0a b <>C. 0,0a b ><D. 0,0a b <<7.如图是一个四棱锥在空间直角坐标系xoz 、xoy 、yoz 三个平面上的正投影，则此四棱锥的体积为（ ）A ．94B ．32C ．64D ．168. 函数()1ln ||f x x x=+的图象大致为（ ）9．设函数y f x =在区间,a b 上的导函数为f x '，f x '在区间,a b 上的导函数为()f x ''，若在区间(),a b 上0)(<''x f 恒成立，则称函数()f x 在区间(),a b 上为“凸函数”；已知234236121)(x x m x x f --=在()1,3上为“凸函数”，则实数m 的取值范围是（ ） A ．31(,)9-∞ B ．31[,5]9 C ．)2,(--∞ D ．),2[+∞10. 已知函数()cos sin f x x x x =-，当][3,3x ππ∈-时，函数f （x ）的零点个数是（ ）A. 7B. 5C. 3D. 111. 已知双曲线22221(0,0)x y a b a b-=>>与函数0)y x =≥的图象交于点P . 若函数y =点P 处的切线过双曲线左焦点(1,0)F -，则双曲线的离心率是（ ）D.3212. 对于函数()f x ，若,,a b c R ∀∈，(),(),()f a f b f c 为某一三角形的边长，则称()f x 为“可构造三角形函数”。

西北师大附中2016届高三第五次诊断考试数学（文科）数学（文科）一、选择题（本大题共12小题，每小题5分，共60分）1．已知,m n R ∈，集合{}72,log A m =，集合{},B m n =，若{}0A B =I ， 则m n +=( )A ．1B ．2C ．4D ．8 2．已知复数z 满足25)43(=+z i ，则z 的共轭复数为（ ）A.34i -+B.34i --C.34i +D.34i -3.双曲线C :22221y x a b-=（0a >，0b >）的一条渐近线方程为2y x =，则C 的离心率是（ ）A B ．2 D 4.我国明朝程大位《算法统宗》中用一首诗歌形式提出了的一个数列问题：远望巍巍塔七层，红灯向下成倍增，共灯三百八十一，请问塔顶几盏灯？你算出塔顶有灯的盏数为 （ ）A. 5B. 4C. 3D. 25. 若变量y x ,满足约束条件⎪⎩⎪⎨⎧≤+-≤-+≥043041y x y x x ，则目标函数y x z -=3的最小值为( )A.4-B.34C.0D.4 6. 下列说法中正确的是（ ） A. “5x >”是“3x >”的必要条件B.命题“对2,10x R x ∀∈+>，”的否定是“2,10x R x ∃∈+≤”C.R m ∈∃，使函数)()(2R x mx x x f ∈+=是奇函数D.设,p q 是简单命题，若p q ∨是真命题，则p q ∧也是真命题. 7. 执行如图程序框图，如果输入4a =，那么输出的n 的值为（ ） A.2 B.3 C.4 D.5 8.某空间几何体的三视图如图所示， 则该几何体的体积等于（ ） A ．10 B ．15 C ．20 D ．309．将函数sin(2)y x ϕ=+的图象向左平移4π个单位后得到的函数图象关于点4(,0)3π成中心对称，那么||ϕ的最小值为（ ） A ．6πB ．4π C ．3π D ．2π 10.若直线20ax by -+=)0,0(>>b a 被圆222410x y x y ++-+=截得的弦长为4，则11a b+的最小值为（ ） A ．322+B ．322+C ．14D ．211.在ABC ∆中，角,,A B C 的对边分别为,,a b c ，且2c cos 2B a b ⋅=+，若ABC ∆的面积为2S =，则ab 的最小值为( ) A. 4 B. 12 C. 16 D. 2412. 已知函数()11,1,4ln ,1,x x f x x x ⎧+⎪=⎨⎪>⎩… 则方程()f x ax =恰有两个不同的实根时，实数a 的取值范围是A.10,4⎛⎫⎪⎝⎭ B.1,e 4⎡⎫⎪⎢⎣⎭ C.10,e ⎛⎫ ⎪⎝⎭ D.11,4e ⎡⎫⎪⎢⎣⎭二、填空题（本大题共4小题，每小题5分，共20分）13.已知函数,0,)21(0,)(21⎪⎩⎪⎨⎧≤>=x x x x f x则=-)]4([f f . 14. 若向量b a ,的夹角为3π，且1,2==b a ，则a 与b a 2+ 的夹角为.15. 已知正三棱柱111C B A ABC -的所有棱长都等于6，且各顶点都在同一球面上，则此球的表面积等于____.16. 已知函数),(3)(23R b a bx ax x x f ∈++-=，若函数)(x f 在]1,0[上单调递减，则22b a +的最小值为____.三、解答题（每小题12分，共60分.解答应写出必要的文字说明、证明过程或演算步骤） 17．（本小题满分12分）在等差数列{}n a 中，已知35a =，12749a a a +++=L . （1）求通项n a ； （2）若*11()n n n b n a a +=∈N ，设数列{}n b 的前n 项和为n S ，求n S18.（本小题满分12分）如图，四棱锥ABCD P -中，底面ABCD 是边长为3的菱形，ο60=∠ABC .⊥PA 面ABCD ，且3=PA .E 为PD 中点，F 在EF ADP棱PA 上，且1=AF .（Ⅰ）求证：//CE 面BDF ； （Ⅱ）求三棱锥BDF P -的体积.19. 近年来，我国许多省市雾霾天气频发，为增强市民的环境保护意识， 某市面向全市征召n 名义务宣传志愿者，成立环境保护宣传组织.现把该组织的成员按年龄分成5组：第1组[20,25)，第2组[25,30)，第3组[30,35)，第4组[35,40)， 第5组[40,45)，得到的频率分布直方图如图所示，已知第2组有35人. （1）求该组织的人数.（2）若从第3，4，5组中用分层抽样的方法抽取6名志愿者参加某社区的宣传活动，应从第3，4，5组各抽取多少名志愿者？（3）在（2）的条件下，该组织决定在这6名志愿者中随机抽取2名志愿者介绍宣传经验，求第3组至少有一名志愿者被抽中的概率.20．（本小题满分12分）已知椭圆C ：22221(0)x y a b a b+=>>3，右顶点(2,0)A 。

西北师大附中2015届高三第五次诊断考试 理科综合能力测试物理试题（A 卷）本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分，其中第Ⅱ卷第33~40题为选考题，其他题为必考题。

考生作答时，将答案答在答题卡上，在本试卷上作答无效。

考试结束后，将答题卡交回。

可能用到的元素相对原子质量：相对原子质量： Li —7 Mg —24 Cu —64 Ca —40 Pb —207 O —16 Cl —35.5第Ⅰ卷（选择题 共126分）二、选择题：本大题共8小题，每小题6分。

在每小题给出的四个选项中，14—18题只有一项符合题目要求。

19—21题有多项符合题目要求。

全选对的得6分，选对但不全的得3分，有错选的得0分。

14．下列说法中正确的是：A ．在探究求合力的方法的实验中利用了理想模型的方法B ．牛顿首次提出“提出假说，数学推理，实验验证，合理外推科学推理方法”C ．用点电荷来代替实际带电体是采用了等效替代的方法D ．奥斯特通过实验观察到电流的磁效应，揭示了电和磁之间的关系15.将一只皮球竖直向上抛出，皮球运动时受到空气阻力的大小与速度的大小成正比．下列描绘皮球在上升过程中加速度大小a 与时间t 关系的图象，可能正确的是()16．我国“玉兔号”月球车被顺利送抵月球表面，并发回大量图片和信息。

若该月球车在地球表面的重力为1G ，在月球表面的重力为2G 。

已知地球半径为1R ，月球半径为2R ，地球表面处的重力加速度为g ，则（ ）A ．“玉兔号”月球车在地球表面与月球表面质量之比为21G G B ．地球的质量与月球的质量之比为212221R G R GC ．地球表面处的重力加速度与月球表面处的重力加速度之比为12G G D ．地球的第一宇宙速度与月球的第一宇宙速度之比为2211R G R G17．如图所示，在水平地面上M 点的正上方某一高度处，将S 1球以初速度v 1水平向右抛出，同时在M 点右方地面上N 点处，将S 2球以初速度v 2斜向左上方抛出，两球恰在M 、N 连线的中点正上方相遇，不计空气阻力，则两球从抛出到相遇过程中( )A.初速度大小关系为v 1＝v 2 B ．速度变化量相等 C ．都是变加速运动D ．都不是匀变速运动18．如图甲所示，以速度v 逆时针匀速转动的足够长的传送带与水平面的夹角为θ。

甘肃省西北师大附中2023-2024学年高三第五次模拟考试生物试卷考生须知：1．全卷分选择题和非选择题两部分，全部在答题纸上作答。

选择题必须用2B铅笔填涂；非选择题的答案必须用黑色字迹的钢笔或答字笔写在“答题纸”相应位置上。

2．请用黑色字迹的钢笔或答字笔在“答题纸”上先填写姓名和准考证号。

3．保持卡面清洁，不要折叠，不要弄破、弄皱，在草稿纸、试题卷上答题无效。

一、选择题：（共6小题，每小题6分，共36分。

每小题只有一个选项符合题目要求）1．某研究小组为了研究一定浓度的IAA和GA1对黄瓜幼苗生长的影响，进行了相关实验。

实验设置（其中CK处理的为对照组）和结果如下表所示。

下列叙述错误的是（）不同处理对黄瓜幼苗生长的影响A．实验处理的对象是种子，处理前可将其浸泡一段时间B．根据一般实验设计的要求，表中对照组应该加入等量的配制其他三组溶液所用的溶剂C．从根体积、根干重和鲜重这些指标可以看出，GA1比IAA更能促进根的生长D．GA1和IAA在促进黄瓜幼苗根系生长方面表现为协同作用2．下列关于“观察洋葱表皮细胞的质壁分离及质壁分离复原”活动的叙述，正确的是（）A．本活动应选择洋葱内表皮作为实验材料B．第一次观察时中央大液泡把细胞质和细胞核都被挤到四周，紧贴着细胞壁C．吸水纸的作用是吸除滴管滴出的多余液体D．细胞处于质壁分离状态时，细胞液浓度可以大于外界溶液浓度3．红细胞中的血红蛋白可以与O2结合，随血液循环将O2运输至人体各处的细胞，供细胞生命活动利用。

下图为喜马拉雅登山队的队员们在为期110天的训练过程中随运动轨迹改变（虚线），红细胞数量变化过程。

以下相关叙述错误的是（）A．随海拔高度增加，人体细胞主要进行无氧呼吸B．血液中的O2以自由扩散方式进入组织细胞C．红细胞数量增加，利于增强机体携带氧的能力D．返回低海拔时，人体红细胞对高海拔的适应性变化会逐渐消失4．取未转基因的水稻（W）和转Z基因的水稻（T）各数株，分组后分别喷施蒸馏水、寡霉素和NaHSO3，24h后进行干旱胁迫处理（胁迫指对植物生长和发育不利的环境因素），下图为测得未胁迫组和胁迫组植株8h时的光合速率柱形图。

奋斗没有终点任何时候都是一个起点第5题图西北师大附中2016届高三第五次诊断考试数学（理科）5分，共60分.下列每小题所给选项只有一项符合题意，请将正确答案的序号填涂在3.在等差数列｛a n ｝中，已知S n 是其前n 项和，且a 1 a 《答题卡上） 1.已知R 是实数集, xl- 1 x,N y y，则 N C R MA. 1,2 B.0,2C.D. 1,22.m mi1为纯虚数”的A.充分但不必要条件B.必要但不充分条件C.充要条件D.既不充分也不必要条件A 30B. 30C. 15D.154.给出下列四个命题: 1P I ： x (0,),-P 2 : x (0,1),1og 』x 2log 1x ；3P 3: x (0,),1 1x(0,3), 2 3 210g l x .3俯视图.选择题（每小题 a 8 a 12 a 15其中真命题是A. P l , P3B.P1, P4C. P2, P3D. P2 , P45.某几何体的三视图如图所示，则其侧面的直角三角形的个数为A.1B.2C.3D.46.已知图象不间断函数f(x)是区间a,b上的单调函数，且在区间(a,b)上存在零点.下图是用二分法求方程f(x) 0近似解的程序框图，判断框内可以填写的内容有如下四个选择：① f(a)f(m) 0;② f(a)f(m) 0;③ f(b)f(m) 0;④ f(b)f(m) 0;其中能够正确求出近似解的是第6题图A.②④B.②③C.①③D.①④7.已知过定点2,0的直线与抛物线x2 y相交于Ax1,y1 ,B x2, y2两点，若、？2是方程A.1B.^C.^D.1概率为A. 1B. 1C. 1D.6 3 2uuur uuu 10.已知在△ ABC中，AB=1, BC=/6, AC=2 点。

为△ ABC勺外心，若AO sAB4 3 3 4A.(-,-)B.(-,-)C.2x xsin cos 0的两个不相等实数根，则tan的值是八1 - 1A. B. --C.2D.-28.若函数f (x) sin(2x )(1 I 2) 的图像关于直线x 一对称，且当12x i x2时，f (X I) f (x2), 则f (X I X2)2 29.已知圆C:x y 2x 10,直线l:3x 4y 12 0,圆C上任意一点P到直线l的距离小于2的uurtAC，则有序实数对(s,t )5-)D.(? 4)和区间［2b)/5 5 5 52 211.如图，5、F2是双曲线。

西北师大附中考生注意：1.本试卷分阅读理解、英语知识运用和写作三部分，满分150分，考试时间120分钟。

2.答题前，考生务必将自己的姓名、考号写在答题卡和答卷密封线内相应的位置上。

考高三英语生要认真核对答题卡上粘贴的条形码的“准考证号、姓名、考试科目”与考生本人准考证号、姓名是否一致。

3.回答选择题时，选出每小题答案后，用2B 铅笔把答题卡上对应题目的答案标号涂黑。

如需改动，用橡皮擦干净后，再选涂其它答案标号。

回答非选择题时，用黑色签字笔在答题卡上书写作答，在试卷卷上作答，答案无效。

第一部分:阅读理解（共两节，满分40分）第一节：(共15小题,每小题2分,满分30分)阅读下列短文，从每题所给的四个选项（A 、B 、C 和D ）中，选出最佳选项。

AA.Because of bad weather.B.Because of maintenance.C.Because of management.D.Because of funds.2.Which platform will you get off at when reach Hornsby?A.Platform4.B.Platform1.C.Platform6.D.Platform2.3.Which place will you most probably go through,if you leave Hornsby?A.Berowra.B.Asquith.C.Epping.D.Chatswood.BDr.Ofri’s new book,“Singular Intimacies:Becoming a Doctor at Bellevue,”recounts her experiences as a doctor at New York’s Bellevue Hospital.NPR’s Melissa Block,host of All Things Considered,recently spent a day at the hospital to get a sense of her world,through her relationships with her patients.Dr.Danielle Ofri is an attending physician in internal medicine at Bellevue.For her,poetry and literature are as much a part of the job as X-rays and pills.She’s written about her experiences there in the book,“Singular Intimacies:Becoming a Doctor at Bellewe”.It’s a collection of essays about learning to listen to the narrative of her patients.Dr.Ofri tries to keep an ear to the stories behind her patients’medical complaints.Answers to questions about family or jobs may not help with medical diagnosis,but conversations like these can help gain a patient’s trust,and they help the doctor,too.“At night,I recall our conversations, and wonder what else I could do for them.It makes me curious about them,”Dr.Ofri says,“so when I go back the next day,I’m more connected with them.And I think a connection has healing powers.Most of the patients brighten,when they talk about themselves and I think they actually feel better.”A good part of Dr.Ofri’s day is also spent overseeing the work of new doctors.The days are filled with jargon(行业术语)and medical shorthand.But Dr.Ofri also tries to inject another kind of language into the training poetry.She carves out five minutes or so each day to gather with her interns and read a poem.She calls it her“literary rounds”.Through these brief pauses in the day, she says she’s giving her students“a chance to let the other part of their brain flower a little bit”. And above all,“I’m just hoping the experience of doing that is helpful to train my students to listen more carefully to patients and be more empathetic towards them.”she said.4．Why did Melissa Block recently go to Bellevue Hospital?A．To make friends with Dr.Ofri.B．To receive medical treatment.C．To know about Dr.Ofri’s experiences.D．To collect essays on treatment.5．What does the third paragraph mainly focus on?A．The stories behind illnesses.B．The effect of family and jobs.C．The benefits of listening to patients.D．The healing powers of complaints.6．What does the underlined word“inject”in paragraph4mean?A．cut.B．change.C．accept.D．add.7．Why does Dr.Ofri hold“literary rounds”for new doctors?A．To encourage them to write books.B．To improve their humanistic quality.C．To prepare an entire career for them.D．To make their brain grow.CMany people think work meetings are a waste of time,and that might be because most meetings keep employees from working well.One survey of76companies found that productivity was71percent higher when meetings were reduced by40percent.Unnecessary meetings waste $37billion in salary hours a year in the U.S.alone.Many meetings occur without a specific reason.Another motive for meetings is what some scholars call the Mere Urgency effect,in which we engage in tasks—such as a meeting where each person recites what they’re working on,whether others need that information or not—to help us feel like we are accomplishing something actual.But the real problem with meetings is not lack of productivity—it’s unhappiness.When meetings are a waste of time,job satisfaction declines.And when job satisfaction declines, happiness in general falls.Thus,for a large population,eliminating meetings—or at least minimizing them—is one of the most straightforward ways to increase well-being.Nobody likes excessive and unproductive meetings.First,they generally increase fatigue. You have probably experienced a day of meetings,after which you are exhausted and haven’t accomplished much.Second,people tend to engage in“surface acting”(faking emotions that are deemed appropriate)during work meetings.Finally,researchers have found that the strongest predictor of meeting effectiveness is active involvement by the participants.If you are asking yourself,“Why am I here?”you are not likely to think that the meeting is a good use of your time —which is obviously bad for your work satisfaction.Taken together,the research on meetings shows that if you want to be happier at work(or want your employees to be happier),you should fight against the time-consuming,unproductive meetings at every opportunity.If there is one rule to remember about work meetings,it might be that they are a necessary evil.They are necessary as organizations need them for proper communication,but they are evil in that they are not irreplaceable,and should thus be used as little as possible for the sake of productivity and happiness.8．Why does the author mention the survey in Para1?A．To explain the survey.B．To introduce the topic.C．To stress the importance of meetings.D．To state the advantage of meetings. 9．Which of the following best explains“eliminating”underlined in Para3? A．reducing.B．hosting.C．increasing.D．avoiding. 10．What is mainly talked about in Para4?A．How to increase people’s job satisfaction.B．The importance and necessity of minimizing meetings.C．Why excessive and unproductive meetings lower job satisfaction.D．Active involvement by the participants indicates meeting effectiveness.11．Which is the most suitable title for the text?A．Work meetings—Evil but necessaryB．Work meetings—The less,the betterC．Work meetings—The more,the merrierD．Work meetings—Excessive and productiveDA routine dentist’s tool may help with more than just oral health.A new study has found that dental imaging scanners can serve as portable devices to track the growth of baby corals(珊瑚虫)—a crucial predictor of how reefs will fare(生长)during climate change.Heat stress can kill mature coral reefs and curtail their regrowth.“Growth,reproduction and survival are the main things that we’re always looking at in terms of how healthy reefs are,”says marine biologist Kate Quigley of the Australian Institute of Marine Science.By modeling baby corals in3-D,researchers can track how well they branch,develop complex shapes and reach reproducing size.If harsh water conditions make corals grow too slowly,a reef won’t recover.Corals this small are difficult to model in3-D;researchers can CT scan them,dip them in wax or laboriously put the measurements together from a commercial3-D scanner-but these methods can be slow and provide lower-resolution views.So one day,as Quigley’s dentist used a scanning device that focused light to create a detailed3-D tooth model,Quigley got an idea.If this device could examine the teeth in detail,she thought,it should be able to scan tiny living corals, too—both teeth and coral are calcium-based and wet.For a study published in Methods in Ecology and Evolution,Quigley tested the scanning device and found that it offered a cheap,easy and portable way to model baby corals significantly faster and at higher resolution than currently available techniques.This tool could let scientists more easily examine how coral species endure stress.“Baby corals are critical for reef restoration and recovery from disturbances like hurricanes and heat waves.Unfortunately,they are the stages we know least about because they are so difficult to measure accurately without great expense,”says Hawaii Institute of Marine Biology researcher Joshua Madin,who was not involved in the study.“This paper is a great example of taking a mature technology developed in another field and applying it to coral reef science.”12．Why is it important to study the baby corals?A．It helps to reduce heat stress.B．It helps to protect the coral reefs.C．It helps to prevent climate change.D．It helps to improve the marine environment. 13．What does the underlined word“curtail”in Paragraph2probably mean?A．guide.B．aid.C．limit.D．follow. 14．What makes it possible to use dental scanning device to scan the corals?A．The low price of the scanning devices.B．The similarity between corals and teeth.C．The excellent skills of the dentists and scientists.D．The assistance of other technologies such as CT scanning.15．Which can be a suitable title for the text?A．Improving Scanning TechnologyB．Fighting Against Climate ChangeC．Saving Coral Reefs with Dental TechD．Protecting Marine Life with Advanced Scanners第二节（共5小题；每小题2分，满分10分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。

西北师大附中2024届高三第五次诊断考试试题高三数学注意事项：1．答题前考生需将姓名、班级填写在答题卡指定位置上，并粘贴好条形码.2．回答选择题时，选出每小题答案后，用2B 铅笔把答题卡上对应题目的答案标号涂黑.如需改动，用橡皮擦干净后，再选涂其它答案标号.3．回答非选择题时，请使用0.5毫米黑色字迹签字笔将答案写在答题卡各题目的答题区域内，超出答题区域或在草稿纸、本试卷上书写的答案无效.4．保持卡面清洁，不要折叠、不要弄皱、弄破，不准使用涂改液、修正带、刮纸刀.一、单选题：本题共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．1.在复平面内，2i 33i z z +=+，其中i 是虚数单位，z 是z 的共轭复数，则复数z 的对应点位于（）A.第一象限B.第二象限C.第三象限D.第四象限【答案】D 【解析】【分析】利用复数的四则运算化简以及共轭复数的定义，结合复数的几何意义可得出结论.【详解】设i(,R)z a b a b =+∈，则共轭复数为i(,R)z a b a b =-∈，所以()()2i i i 3+3i a b a b -++=，所以()()22i 3+3i a b a b -+-=，所以2323a b a b -=⎧⎨-=⎩，解得11a b =⎧⎨=-⎩，所以1i z =-，故复数z 对应的点位于第四象限.故选：D.2.已知直线m 平面α，直线n ⊥平面β，则“m n ”是“αβ⊥”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【答案】A 【解析】【分析】根据题意，由空间中的线面关系，分别验证命题的充分性与必要性即可得到结果.【详解】因为直线m 平面α，直线n ⊥平面β，当m n 时，可得αβ⊥，即充分性满足；当αβ⊥时，,m n 不一定平行，有可能相交还有可能异面，故必要性不满足；所以“m n ”是“αβ⊥”的充分不必要条件.故选：A3.已知两个向量,a b满足1a b b ⋅== ，a b -= ，则a =r （）A.1B.C.D.2【答案】D 【解析】【分析】将a b -=.【详解】因为1a b b ⋅==，a b -= ，所以2232a a b b ⋅=-+，即222113a -⨯+= ，解得2=a 或2a =-r（舍去）.故选：D4.ABC 的内角A B C ､､所对的边分别为,1,2a b c a b A B ===､､，则c =（）A.2B.C.D.1【答案】A 【解析】【分析】由已知可得sin sin 2A B =，结合三角恒等变换，正弦定理可得2cos a b B =，由此可求A B C ､､，再结合勾股定理求c 即可.【详解】因为2A B =，所以sin sin 2A B =，故sin 2sin cos A B B =，由正弦定理可得sin sin a bA B=，所以2cos a b B =，又1a b ==，所以3cos 2B =，又()0,πB ∈，所以π6B =，π3A =，故π2πC A B =--=由勾股定理可得2224c a b =+=，所以2c =，故选：A.5.已知函数()()sin f x x ωϕ=+，如图,A B 是直线12y =与曲线()y f x =的两个交点，π13π,1624AB f ⎛⎫==- ⎪⎝⎭，则5π6f ⎛⎫= ⎪⎝⎭（）A.0B.12C.32D.32【答案】C 【解析】【分析】设1211,,,22A x B x ⎛⎫⎛⎫ ⎪ ⎪⎝⎭⎝⎭，依题可得，21π6x x -=，结合1sin 2x =的解可得()212π3x x ω-=，从而得到ω的值，再根据13π124f ⎛⎫=-⎪⎝⎭即可得2()sin 4π3f x x ⎛⎫=- ⎪⎝⎭，进而求得5π6f ⎛⎫⎪⎝⎭．【详解】设1211,,,22A x B x ⎛⎫⎛⎫ ⎪ ⎪⎝⎭⎝⎭，由π6AB =可得21π6x x -=，由1sin 2x =可知，π2π6x k =+或5π2π6x k =+，Z k ∈，由图可知，当0ω>时，()215π2ππ663x x ωϕωϕ+-+=-=，即()212π3x x ω-=，4ω∴=；当0ω<时，()125π2ππ663x x ωϕωϕ+-+=-=，即()122π3x x ω-=，4ω∴=-；综上：4ω=±；因为同一图象对应的解析式是一样的，所以此时不妨设4ω=，则()()sin 4f x x ϕ=+，因为13π13πsin 1246f ϕ⎛⎫⎛⎫=+=- ⎪ ⎪⎝⎭⎝⎭，则13π3π2π,62k k ϕ+=+∈Z ，解得2π2π,Z 3k k ϕ=-+∈，所以2π2()sin 42πsin 4π33f x x k x ⎛⎫⎛⎫=-+=- ⎪ ⎪⎝⎭⎝⎭，5π10π22π2πsin πsin 2πsin 633332f ⎛⎫⎛⎫⎛⎫∴=-=+== ⎪ ⎪ ⎝⎭⎝⎭⎝⎭．故选：C.6.过抛物线()220y px p =>焦点的直线l 交抛物线于,A B 两点，已知18AB =，线段AB 的垂直平分线交x轴于点()11,0M ，则p =（）A.2B.4C.6D.8【答案】B 【解析】【分析】设直线l 的方程为2px my =+，利用设而不求法求弦长AB 的表达式，再求线段AB 的垂直平分线，由条件列方程求,m p 可得结论.【详解】抛物线22y px =的焦点F 的坐标为,02p ⎛⎫⎪⎝⎭，由题意可知：直线l 的斜率不为0，但可以不存在，且直线l 与抛物线必相交，可设直线l 的方程为2px my =+，()()1122,,,A x y B x y，联立方程222y px p x my ⎧=⎪⎨=+⎪⎩，消去x 可得2220y pmy p --=，则21212,2y y pm y y p +==-，可得()()2121222118AB x x p m y y p p m =++=++=+=，即29pm p +=，设AB 的中点为()00,P x y ，则0y pm =，202p x pm =+，可知线段AB 的垂直平分线方程为22p y pm m x pm ⎛⎫-=--- ⎪⎝⎭，因为()11,0M 在线段AB 的垂直平分线上，则2112p pm m pm ⎛⎫-=---⎪⎝⎭，可得23112p pm +=，联立方程2293112pm p p pm ⎧+=⎪⎨+=⎪⎩，解得2454p m =⎧⎪⎨=⎪⎩，故选：B.7.如图，为球形物品设计制作正四面体、正六面体、正八面体形状的包装盒，最少用料分别记为123S S S ､､，则它们的大小关系为（）A.123S S S <<B.321S S S <<C.312S S S <<D.231S S S <<【答案】B 【解析】【分析】由题意包装盒的最少用料为球形物品的外切多面体，根据多面体的结构特征求出正四面体、正六面体、正八面体形状的包装盒的内切球半径与其表面积的关系，再进行比较.【详解】由题意包装盒的最少用料为球形物品的外切多面体，下面求正四面体、正六面体、正八面体形状的包装盒的内切球的半径与其表面积的关系.设球形物品的半径为R ，则正方体的棱长为2R ，表面积()2226224S R R ==；设正四面体的棱长为a，则正四面体的表面积为221344S a =⨯=，如图正四面体A BCD-，由正四面体的对称性与球的对称性可知内切球的球心在正四面体的高上，如图OG R=，底面等边三角形BCD的高32CE=，外接圆半径233323CG a a=⨯=，正四面体的高63AG a===，体积21131343V Saa R=⨯⨯=，所以21131343V Saa R=⨯⨯=，又21S=，所以a=，所以正四面体的表面积221S==；设正八面体的棱长为b，如图，在正八面体中连接AF，DB，CE，可得AF，DB，CE互相垂直平分，四边形BCDE为正方形，1222OD BD b==，在Rt AOD中，22AO b===，则该正八面体的体积231223232V b b'=⨯⨯⨯=，该八面体的表面积232384b S =⨯=，因为313S R V '=，即2313R ⨯⋅=，解得b =，所以)2223S ===，所以321S S S <<.故选：B.8.已知0.12e 1,,ln1.121a b c =-==，则（）A.b a c <<B.<<c a bC.c b a <<D.<<b c a【答案】D 【解析】【分析】构造函数，判断函数单调性，代入数值可比较大小.【详解】设()e 1x f x x =--，()e 1x f x '=-，(),0x ∈-∞时，()0f x '<，()f x 为减函数，()0,x ∈+∞时，()0f x '>，()f x 为增函数，所以()(0)0f x f ≥=，(0.1)0f >，即0.1e 10.1->.设()ln 1g x x x =-+，11()1x g x x x-'=-=，()0,1x ∈时，()0g x '>，()g x 为增函数，()1,x ∈+∞时，()0g x '<，()g x 为减函数，所以()(1)0g x g ≤=，(1.1)0g <，即ln1.10.1<，所以a c >.设()2()ln 12x h x x x =+-+，()()()22214()01212h x x x x x x '=-=>++++，()h x 为增函数，所以(0.1)(0)0h h >=，所以2ln1.121>，即c b >.故选：D二、多选题：本题共3小题，每小题6分，共18分．在每小题给出的选项中，有多项符合题目要求，全部选对的得6分，部分选对的得部分分，有选错的得0分．9.若集合A B B C = ，则一定有（）A .C B⊆ B.B C ⊆C.B A ⊆ D.A B⊆【答案】AC 【解析】【分析】根据A B A ⊆ 以及A B B ⊆ ，可得B C A ⋃⊆、B C B ⋃⊆、可得C B A ⊆⊆，结合选项即可求解.【详解】因为A B A ⊆ ，A B B C = ，所以B C A ⋃⊆，所以B A ⊆，C A ⊆，因为A B B ⊆ ，A B B C = ，所以B C B ⋃⊆，所以C B ⊆，所以C B A ⊆⊆，故选项A 、C 正确，B 、D 错误.故选：AC.10.已知函数()1221xx f x -=+，则下列说法正确的是（）A.函数()f x 单调递增B.函数()f x 值域为()0,2C.函数()f x 的图象关于()0,1对称D.函数()f x 的图象关于()1,1对称【答案】ABD 【解析】【分析】根据复合函数单调性的判断方法，即可判断A ，根据函数形式的变形，根据指数函数的值域，求解函数的值域，即可判断B ，根据对称性的定义，()2f x -与()f x 的关系，即可判断CD.【详解】()111222222212121x x x x x f x ---+-===-+++，函数22y t=-，121x t -=+，则1t >，又内层函数121x t -=+在R 上单调递增，外层函数22y t=-在()1,∞+上单调递增，所以根据复合函数单调性的法则可知，函数()f x 单调递增，故A 正确；因为1211x -+>，所以120221x -<<+，则1202221x -<-<+，所以函数()f x 的值域为()0,2，故B 正确；()2112422212221x x x x f x ----===+++，()()22f x f x -+=，所以函数()f x 关于点()1,1对称，故C 错误，D 正确.故选：ABD11.已知12,F F 分别为双曲线的左､右焦点，过1F 的直线交双曲线左､右两支于,A B 两点，若2ABF △为等腰直角三角形，则双曲线的离心率可以为（）A.1+ B.C.D.【答案】BC 【解析】【分析】利用等边三角形的性质，结合双曲线的定义，建立,a c 的等量关系式求解.【详解】如果2BAF ∠为直角，设2AF AB m ==，则2BF =，又122BF BF a -=，212AF AF a -=，所以122AF m =，由212AF AF a -=，则222m m a -=，得(4m a =+，在12AF F △中，2221212AF AF F F +=，即222242m m c ⎛⎫+= ⎪ ⎪⎝⎭，即((22222224442a ac ⎛⎫+++= ⎪ ⎪⎝⎭，化简得229c a=+e =如果2AF B ∠为直角，设2BF m =，则2AF m =，AB =，12AF m a =-，12BF m a =-+，因为122BF BF a -=，所以22a a -+=，故m =，在12AF F △中，由余弦定理可知()()2222428222c a a a ⎛⎫=-+--⋅⋅- ⎪ ⎪⎝⎭，整理得22412c a =，即23e =，所以e =B 正确；如果2ABF ∠为直角，则2AB BF =，122BF BF a -=，则12AF a =，又212AF AF a -=，所以24AF a =，22BF ==，()122BF a a =+=，在等腰直角12BF F △中，222212124BF BF F F c +==，即()()222224a c ++=，化简得225c a=+e =C 正确.故选：BC.【点睛】关键点睛：求解离心率的关键是结合题中的已知关系，找出,,a b c 之间的数量关系.三、填空题：本题共3小题，每小题5分，共15分．12.已知直线:21l y kx k =--与圆22:5C x y +=相切，则k =__________.【答案】2【解析】【分析】利用圆心到直线的距离等于半径列方程，解方程求得k 的值.【详解】直线l 的一般方程为210kx y k ---=，圆225x y +=的圆心C 的坐标为()0,0，半径r =，由于直线l 和圆C 相切，所以圆心C 到直线l 的距离等于半径，=解得2k =.故答案为：2.13.春暖花开季节，小王､小李､小张､小刘四人计划“五･一”去踏青，现有三个出游的景点：南湖､净月､莲花山，假设每人随机选择一处景点，在至少有两人去南湖的条件下有人去净月的概率为__________.【答案】23【解析】【分析】由古典概率结合条件概率的形式计算即可.【详解】至少有两人去南湖的情况有三种：两人去，三人去，四人去，其概率为21134422444C C C C 2C 33381+⨯+=，至少有两人去南湖且有人去净月的概率为23444C 3C 22381⨯+=，所以在至少有两人去南湖的条件下有人去净月的概率为222333=，故答案为：23.14.记表[](){},max x a b f x ∈示()f x 在区间[],a b 上的最大值，则[]{}20,1max x x x c ∈-+取得最小值时，c =__________.【答案】18##0.125【解析】【分析】根据题意，[]{}20,1max x x x c ∈-+取得最小值，即为()2f x x x c =-+在区间[]0,1上的最大值取得最小值，先用分段函数表示()f x 在区间[]0,1上的最大值，再根据图象求分段函数的最小值即可.【详解】[]{}20,1max x x x c ∈-+取得最小值，即为()2f x x x c =-+在区间[]0,1上的最大值取得最小值，因为()f x 的对称轴12x =，且()()01f f c ==，所以()f x 的最大值为1124f c ⎛⎫=-⎪⎝⎭或()()01f f c ==，当14c c -=时，即18c =，所以()max 1,81148c c f x c c ⎧⎛⎫ ⎪⎪⎪⎝⎭=⎨⎛⎫⎪-≤ ⎪⎪⎝⎭⎩＞，，当18c =时，()max f x 取最小值，最小值为18.故答案为：18.【点睛】关键点点睛：本题主要考查函数的最值，关键在于理解题意，[]{}20,1max x x x c ∈-+取得最小值，即为()2f x x x c =-+在[]0,1的最大值取得最小值，所以先要将()f x 的最大值表示出来，再用分段函数的性质即可.四、解答题：本题共5小题，共77分．解答应写出文字说明、解答过程或演算步骤．15.如图，在正三棱柱111ABC A B C -中，12,AA AB M ==为1BB 中点，点N 在棱11A B 上，112A N NB =.（1）证明：MC 平面1NAC ；（2）求锐二面角1M AC N --的余弦值.【答案】（1）证明见解析；（2）155.【解析】【分析】（1）解法1：作出辅助线，得到线线平行，进而得到线面平行；解法2：建立空间直角坐标系，写出点的坐标，求出平面的法向量，由0n CM ⋅= 证明出结论；（2）解法1：作出辅助线，得到MDE ∠即为二面角1M AC N --的平面角，求出各边长，求出锐二面角的余弦值；解法2：求出平面的法向量，得到平面的法向量，求出答案.【小问1详解】解法1：设11AC AC D ⋂=，则D 为1AC中点，1A M AN E ⋂=，连接DE ，延长AN 交1BB 延长线于F ，由112A N NB =得112AA B F =，11,,AA MF A E EM E ==为1A M 中点，MC DE ，DE ⊂平面1,NAC MC ⊄平面1NAC ，MC 平面1NAC ，解法2：取AC 中点O ，取11A C 中点1O ，连接1,OB OO ，因为111ABC A B C -为正三棱柱，所以1,,AC OB OO 两两垂直，以O 为坐标原点，1,,OB OC OO 所在直线分别为,,x y z 轴，建系如图，则()()())10,1,0,0,1,2,0,1,0,3,0,1A C C M -，())1231,,2,0,2,2,3,1,133N AC CM ⎛⎫-==- ⎪ ⎪⎝⎭，)134,,0,3,1,133C N AM ⎛⎫=-=⎪ ⎪⎝⎭ ，设平面1NAC 的一个法向量为(),,n x y z = ，则11220234033n AC y z n C N x y ⎧⋅=+=⎪⎨⋅=-=⎪⎩，令3y =，则(2,3,3,3x z n === ，0n CM MC ⋅=⊄ ，平面1NAC ，故MC 平面1NAC .【小问2详解】解法1：因为12AA AB ==，所以1AA AC =，故四边形11ACC A 为正方形，故1AC ⊥1AC ，且D 为1AC 中点，又22415AM AB BM =+=+=，2211115C M B C B M =+=，故1AM C M =，故DM ⊥1AC ，因为1A C DM D ⋂=，1,A C DM ⊂平面1MA C ，所以1AC ⊥平面1MA C ，因为DE ⊂平面1MA C ，所以1AC DE ⊥，所以MDE ∠即为二面角1M AC N --的平面角，又MC ===11122AD AC ===且11515,2222DE MC EM A M ====，DM ==2225531544cos 25DE DM EM MDE DE DM ∠+-+-==⋅，故锐二面角1M AC N --的余弦值为155.解法2：设平面1MAC 的一个法向量为(),,m a b c =，则12200m AC b c m AM b c ⎧⋅=+=⎪⎨⋅=++=⎪⎩ ，令1b =，则()1,0,0,1,1c a m =-==-，15cos ,5m n m n m n ⋅=== ，所以锐二面角1M AC N --的余弦值为5.16.某校研究性学习小组研究的课题是数学成绩与物理成绩的关系，随机抽取了20名同学期末考试中的数学成绩和物理成绩，如表1：表1：序号数学物理114495213090312479412085511069610782710380810262910067109875119868129577139459149265159057168858178570188555198052207554（1）数学120分及以上记为优秀，物理80分及以上记为优秀.（i）完成如下列联表；数学成绩物理成绩合计优秀不优秀优秀不优秀合计（ii ）依据0.01α=的独立性检验，能否认为数学成绩与物理成绩有关联？（2）从这20名同学中抽取5名同学的成绩作为样本，如表2：表2：数学成绩1301101008575物理成绩9069677054如图所示：以横轴表示数学成绩､纵轴表示物理成绩建立直角坐标系，将表2中的成对样本数据表示为散点图，观察散点图，可以看出样本点集中在一条直线附近，由此推断数学成绩与物理成绩线性相关.（i ）求样本相关系数r ；（ii ）建立物理成绩y 关于数学成绩x 的一元线性回归模型，求经验回归方程，并预测数学成绩120的同学物理成绩大约为多少？（四舍五入取整数）参考公式：（1）样本相关系数()()ni i x x y y r --=∑.（2）经验回归方程ˆˆˆy a bx =+；.()()()121ˆˆˆ,.ni i i n i i x x y y b ay bx x x ==--==--∑∑（3）()()()()22()n ad bc a b c d a c b d χ-=++++，其中n a b c d =+++.临界值表：α0.10.050.010.0050.001x α 2.706 3.841 6.6357.87910.828【答案】（1）（i ）答案见解析；（ii ）认为数学成绩与物理成绩有关联.（2）（i ）3337；（ii ）9961018537y x =+，81分【解析】【分析】（1）（i ）由表1可直接填写列联表；（ii ）根据列联表，计算2χ的值，结合临界值表可得出结论；（2）（i ）根据参考公式计算样本相关系数；（ii ）根据参考公式计算经验回归方程，并将120x =代入，预测该同学的物理成绩.【小问1详解】（i ）数学成绩物理成绩合计优秀不优秀优秀314不优秀21416合计51520（ii ）零假设0H ：数学成绩与物理成绩相互独立，即数学成绩与物理成绩无关联.()()()()222()20(31412)416515n ad bc a b c d a c b d χ-⨯⨯-⨯==++++⨯⨯⨯0.0120 6.667 6.6353α=≈>=依据0.01α=的独立性检验，推断0H 不成立，即认为数学成绩与物理成绩有关联.【小问2详解】（i ）由题意100,70x y ==，所以r ⨯+⨯-+⨯-+-⨯+-⨯-=33.37==（ii ）由题意()()()()()2222230201010315025ˆ1630100(15)(25)b ⨯+⨯-+⨯-+-⨯+-⨯-=+++-+-990991850185==，所以99610ˆ7010018537a y bx =-=-⨯=，所以经验回归方程为9961018537y x =+，当120x =时，996102986ˆ12080.7811853737y =⨯+=≈≈，所以物理成绩约为81分.17.已知1a ，函数()ln 1a f x ax x x =-+.（1）当1a =时，求()f x 的最小值；（2）若1x >时，()0f x <恒成立，求a 的取值范围.【答案】（1）0；（2）2a .【解析】【分析】（1）由已知可得()1ln 1ln f x x x =+-='，进而可求()f x 的单调区间；（2）求导得()()11ln a f x a x x-'=+-，令()11ln ,a g x x x -=+-进而求导()()211a g x a x x-'=--，分类讨论可求a 的取值范围.【小问1详解】当1a =时，()()ln 1,1ln 1ln f x x x x f x x x =-+=='+-，()()()0,1,0,x f x f x '∈<单调递减；()()()1,,0,x f x f x '∈+∞>单调递增；()min ()10f x f ==【小问2详解】()()()111ln 1ln a a f x a x ax a x x --=+-=+-'，设()()()1211ln ,1a a g x x x g x a x x--=+-=--'，①若1a =，由（1）知()()10f x f >=，不合题意；②若()()()211112,111a a a g x a x a x x x--⎡⎤<<=--='--⎣⎦，设()()()()12211,(1)0,a a h x a x h x a x h x --=--=--'<单调递减，()()11120h a a =--=->，令()()111000110,(1)a a h x a x x a ---=--==-，()()()()01,,0,0,x x h x g x g x ∈>'>单调递增，()()10g x g >=，()()0,f x f x '>单调递增，()()10f x f >=，不合题意；③()()()212,1,,10a a x g x a x x∞-≥∈+-'=-<，()g x 单调递减，()()()()10,0,g x g f x f x <=<'单调递减，()()10f x f <=；综上，2a ≥.18.已知椭圆2222:1(0)x y C a b a b +=>>过点()2,1M ，离心率为2.不过原点的直线:l y kx m =+交椭圆C 于,A B 两点，记直线MA 的斜率为1k ，直线MB 的斜率为2k ，且1214k k =.（1）求椭圆C 的方程；（2）证明：直线l 的斜率k 为定值；（3）求MAB △面积的最大值.【答案】（1）22182x y +=（2）证明见解析（3）max S =【解析】【分析】（1）根据离心率和过点M ，用待定系数法可求出椭圆C 的方程；（2）设出直线并与椭圆进行联立，用韦达定理表示出1214k k =，并进行化简，即可求出斜率定值；（3）根据弦长公式和点到直线的距离公式表示出三角形面积，将其转化为函数，再利用导数求出最大值.【小问1详解】依题意222222411c aa b b a c ⎧=⎪⎪⎪+=⎨⎪=-⎪⎪⎩，解得228,2a b ==，所以椭圆的标准方程为22182x y +=.【小问2详解】设直线l 方程为()()1122,0,,,,y kx m m A x y B x y =+≠，由22182y kx m x y =+⎧⎪⎨+=⎪⎩得()222418480k x kmx m +++-=，()222121222848Δ16820,4141km m k m x x x x k k --=+->+==++，()()()()121212121211112222kx m kx m y y k k x x x x +-+---=⋅=----()()()()2222222121221212224881(1)1(1)414148162444141m km k k m m k x x k m x x m k k m km x x x x k k --⋅+-⋅+-+-++-++==--++++++()()22224(1)12141244144k m m k m k m mk k -+---===++-++，解得12k =-.【小问3详解】由（2）得221,0,22402y x m m x mx m =-+≠-+-=，22Δ1640,4,22,0m m m m =-><-<<≠，()12252552AB x h m =-==-MAB △的面积(122S AB h m ==-=，()()3(2)2f m m m =-+，()()()2323(2)2(2)(2)44f m m m m m m =--++-=---'，令()0f m '＞，解得21m --＜＜，即()f m 在()2,1--上单调递增，令()0f m '＜，解得10m -＜＜或02m ＜＜，即()f m 在()10-，和()02，上单调递减，所以当1m =-时，取到最大值()127f -=，MAB △的面积max S =【点睛】关键点点睛：本题主要考查直线与椭圆的位置关系，解决直线与椭圆的综合问题，关键在于（1）注意题设中每一个条件，明确确定直线和椭圆的条件；（2）直线和椭圆联立得韦达定理，与弦长公式和点到直线距离公式的结合运用；（3）求最值时，要善于转化为函数关系，利用导数来求解.19.对于数列{}n a ，称{}Δn a 为数列{}n a 的一阶差分数列，其中()*1Δn n n a a a n +=-∈N .对正整数()2k k ≥，称{}Δk n a 为数列{}n a 的k 阶差分数列，其中()1111ΔΔΔΔΔk k k k n n n n a a a a ---+==-已知数列{}n a 的首项11a =，且{}1Δ2n n n a a +--为{}n a 的二阶差分数列.（1）求数列{}n a 的通项公式；（2）设(){}212,2n n b n n x =-+为数列{}n b 的一阶差分数列，对*n ∀∈N ，是否都有1C n i i n n i x a ==∑成立？并说明理由；（其中C in 为组合数）（3）对于（2）中的数列{}n x ，令2n n x x n t t y -+=，其中122t <<.证明：2122n n n i i y -=<-∑.【答案】（1）12n n a n -=⋅；（2）成立，理由见解析；（3）证明见解析.【解析】【分析】（1）由二阶差分数列的定义可得21Δ2Δn n n n a a a +--=，将21ΔΔΔn n n a a a +=-，可得122n n n a a +-=，构造等差数列即可求解；（2）由一阶差分数列的定义可得1n n n x b b n +=-=，要证1Cn i i n n i x a ==∑成立，即证121C 2C C 2n n n n n n n -+++=⋅ ，根据二项式定理即可证明；（3）作差可得22n nn n t t --<++，故()()111112222n n n i i i i i i i i y t t --====+<+∑∑∑，根据等比数列的求和公式即可证明.【小问1详解】因为{}1Δ2n n n a a +--为{}n a 的二阶差分数列，所以21Δ2Δn n n n a a a +--=，将21ΔΔΔn n n a a a +=-，代入得11Δ2ΔΔn n n n n a a a a ++--=-，整理得Δ2n n n a a -=，即122n n n a a +-=，所以111222n n n n a a ++-=.故数列2n n a ⎧⎫⎨⎬⎩⎭是首项为12，公差为12的等差数列，因此，()111222n n a n =+-⋅，即12n n a n -=⋅.【小问2详解】因为{}n x 为数列{}n b 的一阶差分数列，所以1nn n x b b n +=-=，故1Cn i i n n i x a ==∑成立，即为121C 2C C 2n n n n n n n -+++=⋅ .①当1n =时，①式成立；当2n ≥时，因为()110111112(11)C C C n n n n n n n n n ------⋅=⋅+=⋅+++ ，且11C C k k n n n k --=，所以①成立，故对*n ∀∈N 都有1C n i i n n i x a ==∑成立.【小问3详解】2n nn t t y -+=，因为122t <<，所以(2)1,2n n n t t ><，故()()()1222(2)10(2)n n n n n n n n t t t t t --⎡⎤+-+=-->⎣⎦，即22n n n n t t --<++，所以()()()111111221111222212222112n n n n n i i i i i i i i y t t --===⎡⎤⎛⎫- ⎪⎢⎥-⎝⎭⎢⎥=+<+=+-⎢⎥-⎢⎥⎣⎦∑∑∑()()2111121121222222222n n n n n n n --⎛⎫=-+-=-+<-⋅=- ⎪⎝⎭.【点睛】关键点点睛：涉及数列新定义问题，关键是正确理解给出的定义，由给定的数列结合新定义探求数列的相关性质，并进行合理的计算、分析、推理等方法综合解决.。