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2018届山东省枣庄八中第一学期高三年级期中考试试卷与答案

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{正文}2018届山东省枣庄八中第一学期高三年级期中考试英语试题第 I 卷（选择题，共100分）第一部分听力（共两节，满分30分）做题时，先将答案标在试卷上。

录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节（共5小题;每小题1．5分，满分7．5分）听下面5段对话。

每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项。

听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

例：How much is the shirt?A．£ 19．15．B．£ 9．18．C．£ 9．15．答案是C。

1．How much is a ticket to New Jersey?A．$ 21．B．$ 42．C．$ 50．2．How does the woman feel?A．Annoyed．B．Confused．C．Excited．3．What is the conversation mainly about?A．Bad neighborhoods．B．A loud noise．C．Chinese New Year 4．What does the woman mean?A．They should get more butter．B．She doesn’t want to ask Andy again．C．The butter is not in its usual place．5．What place are the speakers looking for?A．A cafe．B．A bookstore．C．A bank．第二节（共15小题;每小题1．5分，满分22．5分）听下面5段对话或独白。

每段对话或独白后有几个小题，从题中所给的A、B、C三个选项中选出最佳选项。

听每段对话或独白前，你将有时间阅读各个小题，每小题5秒钟;听完后，各小题将给出5秒钟的作答时间。

每段对话或独白读两遍。

听第6段材料，回答第6、7题。

6．What is the woman looking for?A．An information office．B．A shoe repair shop．C．A police station．7．What is the Town Guide probably according to the man?A．A map．B．A newspaper．C．A brochure．听第7段材料，回答第8、9题。

8．What is the woman’s opinion about the art field trip?A．It is very worthwhile．B．It is a waste of money．C．It is a waste of time．9．How will the woman get to Hong Kong?A．By train．B．By air．C．By bus．听第8段材料，回答第10至12题。

10．What does the man want to do?A．Buy office supplies．B．Buy his father a gift．C．Get a birthday gift．11．What’s the relationship between the speakers?A．Co-workers．B．Boss and employee．C．Shop owner and customer．12．What does the woman suggest?A．A trip to the office supplies store．B．A couple of shopping trips．C．A shopping trip downstairs．听第9段材料，回答第13至16题。

13．Why does the woman need the job?A．To learn to start a business．B．To support herself through college．C．To gain some work experience．14．Where exactly will the woman work if she gets the job?A．At the cashier’s desk．B．Between the shop shelves．C．In the manager s office．15．How is the woman s performance at school?A．Poor．B．Excellent．C．Average．16．What will be the woman^ working hours if she gets the job?A．8:30 a．m．— 6:10 p．m．B．9:00 a．m．—6:00 p．m．C．8:30 a．m．—10:00 p．m．听第10段材料，回答第17至20题。

17．Where is the speaker making the announcement?A．In an outdoor market．B．In a grocery store．C．In a restaurant．18．Who is Donna Jones?A．A customer．B．A weatherman．C．A manager．19．Where should one go if he is afraid of small spaces or crowds?A．To the restrooms．B．To the customer service area．C．To the parking lot．20．What should an employee of the store do?A．Help customers with their things．B．Keep an eye on the weather．C．Lock up the money．第二部分阅读理解（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A、B、C和D四个选项中，选出最佳选项。

AWhere to see amazing colours around the worldLake DistrictAs you wander along Windermere to the centre of Ambleside，autumn colours are abundant．Looking across the water，it seems that the treeline is alight with red and orange leaves．As you round the bend on to the road into town，you are faced with a hillside that celebrates autumn at its best．CANADA and USA Dorset Lookout Tower，near TorontoBefore you reach Dwight，climb the Dorset Lookout Tower，142 metres above the Lake of Bays．From up here，the canopy（树冠）is an autumnal quilt reaching to the horizon，refreshing the eyes and invigorating the spirit．Great Smoky Mountains，Tennessee／North CarolinaFall in this national park is amazing．Just outside quaint Townsend，Tennessee，it is a bikeable and driveable 11-mile loop（环形车道），but be sure to explore the cabins and fields on foot．Late October and early November is my favoufite time to visit；the colours are great and the weather is cool enough to allow nice walks and hikes．Kamikochi national park，HonshuIn central Honshu，three hours’drive north—east of Nagoya，is one of the most charming places in the Japanese Alps．The mountain slopes are covered with forests of larch and beech that blaze scarlet and orange in the autumn．Wisps of smoke hang in the still air above Yakedake volcano，and the glass surface of Taisho Pond perfectly reflects the trees and snow-dusted mountains．21．What season scenery does the text describe?A．Spring．B．Summer．C．Autumn．D．Winter．22．Where would you go if you like to ride a bike?A．Lake District．B．Great Smoky Mountains，Tennessee／North Carolina．C．CANADA and USA Dorset Lookout Tower．D．Kamikochi national park，Honshu．23．What is different from others about Kamikochi national park?A．People can enjoy colorful leaves from up．B．The scenery here is the most charming．C．You can appreciate the snow scene．D．No smoke can be seen above the volcano．BYou’re swallowing a fruit salad and everything’s fine．Then suddenly your tongue feels itchy．The corners of your mouth start to sting．Before you know it，your entire mouth feels burnt and irritated．What’s going on?Is there pineapple or kiwi fruit in that salad? They both contain specific enzymes（生化酶）called bromelain（sourced from pineapple）and actinidain（found in kiwis，as well as mangos, papayas and bananas in lower amounts）that break down proteins，including the ones in your mouth．Literally，if you eat pineapple or kiwi fruit，the enzymes will eat the surface of your tongue in return．Wow，isn’t that dangerous for my tongue?!Bromelain and actinidain are so potent that they are actually used commercially to tenderize meat．But，usually，the enzymes are not harmful to humans．It’s important， however，to be aware of the difference between a normal reaction and an allergic one．The actinidain in kiwis，in particular，is a common allergen，so if you experience swelling，rashes，vomiting or unusual pain，see a doctor!But it is strange that canned pineapple doesn’t hurt my mouth at all．Well，that’s because in the canning process，the pineapple is heated up to kill any bacteria （and enzymes）．But the canned stuff is often full of preservatives and added sugar．We definitely recommend flesh pineapple，in small amounts，paired with nutritious delicious dishes like a beef burger．It’ll minimize the itch !24．Actinidain can be found in various fruits EXCEPT_________．A．banana B．kiwi C．mango D．pineapple 25．Enzymes can___________________．A．make some people vomitB．be totally harmlessC．make our tongues tenderD．break down our tongues26．Why did the author prefer fresh pineapple to canned pineapple?A．Because flesh pineapple can minimize the itch．B．Because enzymes in fresh pineapple can break down protein．C．Because canned pineapple has lost enzymes during the canning process．D．Because fresh pineapple is free of preservatives and added sugar．27．We might read this passage from a____________．A．science fiction B．cook brochureC．botany journal D．biology textbookCAs part of the “Facebook generation”, we have grown up in a world where access to the Internet is at our fingertips and we’re able to connect with people at the touch of a button Nowadays, it’s become increasingly common to offer guests the wifi password along with a cup of tea when they come round to visit．We’ve been becoming addicted to our smartphones．Of course, smartphones are undeniably useful and have revolutionized our lives in so many ways．However, it seems a shame that these addictive pieces of technology are slowly beginning to take over our lives．It is so easy to become dependent on our phones, whether it’s to find the way around with the map application or even just check the time．As a result, we’re becoming more anti-social-choosing to consult the Internet rather than talk with others face-to-face．Even when we’re socializing, it seems that most of us struggle without our phones at arm’s reach．Phones were invented to keep us more connected, and indeed they have done just that．We can now get in touch by an endless list of social media outlets, yet this means we’re starting to lack genuine connect with each other．People are connecting more and more across screens rather than in person, even when it comes to relationship—dating websites are more popular than ever before．Recently, an artist released a series of images depicting people in social situations and edited out their smartphones and tables．The result was a collection of powerful images that have, perhaps ironically, gone viral．They force people to wake up to the fact that we are no longer interacting properly with each other．28．What do people nowadays provide their guests with according to the passage?A．A smartphone．B．The wifi password．C．A tablet．D．A tea party．29．What’s the main idea of the second paragraph?A．Smartphones are beginning to take over our lives．B．We should consult the Internet for knowledge．C．We haven’t been addicted to our smartphones．D．Smartphones have revolutionized our lives in many ways．30．Which of the followings is NOT the consequence of smartphone addiction?A．Giving away our wifi password．B．Connecting more across screens．C．Becoming more and more anti-social．D．Depending on phones to know anything．31．The artist released the images so as to show that ．A．the images are as powerful as the virusB．the contents of the images are improperC．dating websites are more and more popular than beforeD．We rely on human-machine interaction too muchDVirtual reality（虚拟现实）（VR），a kind of fantastic technology，is finally here．It is the wonderful world created by computers．If you want to experience it，you will have to depend on a headset．But before you rush out to buy a headset，you’d better take notice of experts’ warnings．“These devices are designed mostly for adult use，”said Dr．Justin Asgarpour，Vancouver optometrist（whose job is to examine people’s eyes and to sell glasses）．Little research has been done about the long-term effects of VR on children，but makers have not designed the VR headsets for young eyes．VR headset maker Oculus Rift states that younger users may suffer from discomfort or ill health effects，and they are in a very important period of visual development．“What a study has suggested is that a VR device can actually cause what we call amblyopia，potential lazy eyes，in these children with pre-existing Visual conditions，”Dr．Asgarpour said．Even children without underlying eye conditions could experience problems，like temporary double vision．Adults should be careful too，according to the doctor，who says longer virtual reality experiences can lead to balance problems．You should be careful of your surroundings and be careful not to trip over wires．Those with heart conditions are advised not to play．But every coin has two sides．There are some positive health uses of virtual reality．The technology has been used to treat conditions like fear of heights，and it’s also being used to help treat addictions．“What we’re trying to do is take people into a dru g-related setting and let them experience cravings（extremely strong wishes for something），”Dr．Zach Rosenthal with Duke University said．He and his team take addicts inside a virtual house and use a series of tones to condition them to reduce their cravings．Scientists say that a lot more research needs to be done about the long-term health effects of VR，but in the meantime you had better spend only a limited time on VR and make sure your children use it as little as possible．32．What’s one of the purposes of this passage?A．To introduce the uses of VR headset．B．To warn children not to use VR．C．To inform people of the advantages of VR．D．To advertise the VR headset．33．What problems can the use of the VR headset bring to children?A．Hearing problems．B．Balance problems．C．Visual problems．D．Addiction problems．34．Why is Dr．Zach Rosenthal’s research mentioned?A．To prove VR has beneficial effects．B．To explain how VR works．C．To reduce people’s cravings．D．To prove the bad effects of using the headset．35．What does the author think of VR?A．VR will not be popular any longer．B．VR does no harm to adults．C．VR has no benefits to people．D．VR’s health effects need more researches．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。

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配套K12山东省枣庄市第八中学东校区2018-2019学年高一数学上学期期中试题(扫描版)

配套K12山东省枣庄市第八中学东校区2018-2019学年高一数学上学期期中试题(扫描版)

山东省枣庄市第八中学东校区2018-2019学年高一数学上学期期中试题（扫描版）高一数学试题参考答案一、选择题DADCC ABDBA CB二、填空题13. 3 ； 14.(1)x x +； 15.413a <<； 16. ①④⑤ 三、解答题17 解：（I ）原式=220.533492()()(0.008)28125---+⨯ 472259925=-+⨯， -------3分 15233=-+=， -------5分 （II ）原式=2(lg 2)lg 20lg 5+⨯ 2(lg 2)(1lg 2)(1lg 2)=++- --------8分 22(lg 2)1(lg 2)=+- 1= --------10分18. 解：（1） 若0a =，集合{}{}|11|11A x a x a x x =-<<+=-<<， ……………2分 {}|03B x x =<<．则=I A B {}|11x x -<<{}|03<<I x x {}|01x x =<<； ……6分（2） 若A B ⊆，则10,13,a a -≥⎧⎨+≤⎩ 即12a ≤≤，所以实数a 的取值范围是12a ≤≤． ……………………………12 19（本小题满分12分）解：（Ⅰ）()f x ∵是奇函数，()()f x f x -=-∴． 即2211ax b ax b x x -++=-++，ax b ax b -+=--，0b =∴． ……………………3分 2()1ax f x x =+∴，又12()25f =，1221514a =+∴，1a =，2()1x f x x =+∴． ……………………6分 （Ⅱ）任取12,(1,1)x x ∈-，且12x x <， 1212121222221212()(1)()()11(1)(1)x x x x x x f x f x x x x x ---=-=++++， ………8分 ∵1211x x -<<<，∴1211x x -<<，∴1210x x ->，又120x x -<，2110x +>，2210x +>，∴12()()0f x f x -<，12()()f x f x <，∴()f x 在区间(1,1)-上是增函数． ……………………12分20. 解： (1)由题意，得y ＝⎩⎪⎨⎪⎧ 0.1x ，0<x ≤15，1.5＋2log 5x －，x >15. …………6分(2)∵x ∈(0,15]时，0.1x ≤1.5，又y ＝5.5>1.5，∴x >15， ------------------9分 所以1.5＋2log 5(x －14)＝5.5，x ＝39. ---------------11分 老张的销售利润是39万元． …………12分21.解：（1）由题设知101_0x x +>⎧⎨>⎩，解得11x -<<，所以函数()h x 的定义域为(-1,1)[]()log (1)log 1()log (1)log(1)()a a a h x x x x x h x -=-++--=-++=Q ∴函数()h x 是偶函数. ………… …6分（1） 由()()f x g x >，即log (1)log (1)a a x x +>-①当10,01,10,1011,x a x x x x +>⎧⎪<<->-<<⎨⎪+<-⎩有解得 ②当10,1,10,111,x a x x x x +>⎧⎪>-><<⎨⎪+>-⎩有解得0 ………… …10分 所以，当01a <<时，不等式的解集是{}|10x x -<<；当1a >时，不等式的解集是{}|01x x << ………… …12分22.解：（1）函数1122-=-+=+x b x f y )(过定点（-2,0），2p q +=- …2分（2）],[,1131-∈=x b 时，()⎥⎦⎤⎢⎣⎡∈⎪⎪⎭⎫ ⎝⎛=33131,xx f 222233133123132a a a x af x f y x x x -+⎥⎥⎦⎤⎢⎢⎣⎡-⎪⎪⎭⎫ ⎝⎛=+⎪⎪⎭⎫ ⎝⎛-⎪⎪⎭⎫ ⎝⎛=+-=)()]([ 当31<a 时，32928a a g -=)(； 当331≤≤a 时，23a a g -=)(； 当3>a 时，a a g 612-=)(.故()⎪⎪⎪⎩⎪⎪⎪⎨⎧>-≤≤-<-=3612331331329282a a a a a a a g ,,, ………… …8分 （2）假设存在满足题意的n m ,由3>>n m 且()a a g 612-=在），（∞+3上是减函数 又)(a g 的定义域为],[m n ，值域为],[22m n 所以⎩⎨⎧=-=-22612612m n n m ………… …9分 两式相减得))(()(n m n m n m -+=-6由3>>n m 知6=+n m ， ………… …11分 这与3>>n m 矛盾所有满足题意的n m ,不存在 ………… …12分。

2018届山东省枣庄市枣庄一中高三月考理科数学试题及答案

2018届山东省枣庄市枣庄一中高三月考理科数学试题及答案

2018届山东省枣庄市枣庄一中高三月考数学试题（理）试题分第Ｉ卷和第Ⅱ卷两部分。

满分150 分，考试时间120分钟。

第Ⅰ卷 （选择题 共50分）一、选择题：本大题共l0个小题，每小题5分，共50分。

每小题给出的四个选项中只有一项是符合题目要求的。

1．设集合3.022},032|{=≤-=m x x x P ，则下列关系中正确的是A ．P m ⊆B ．P m ∉C ．P m ∈}{D ．}{m ≠⊂P2．已知复数2z i =-，则z z ⋅的值为（ ） A ．5 B C ．3D3．已知数列{}n a 是等差数列，若91130a a +<，10110a a ⋅<，且数列{}n a 的前n 项和n S 有最大值，那么n S 取得最小正值时n 等于 A ．20 B ．17 C ．19D ．214．若随机变量ξ的分布列如下：那么()54E ξ+等于ξ1 2 4 P0.40.30.3A ．15B ．11C ．2.2D ．2.35．设n m ,是两条不同的直线，βα,是两个不同的平面，下列命题中正确的是A ．若βα⊥, ,α⊂m ,β⊂n 则n m ⊥B ．若α⊥m ，m ∥n ，n ∥β ，则βα⊥C ．若n m ⊥，,α⊂m ,β⊂n 则βα⊥D ．若,//ββα⊂m,αm ,β⊂n 则n m // 6．将5名学生分配到甲、乙两个宿舍，每个宿舍至少安排2名学生，那么互不相同的安排方法的种数为 A ．10 B ．20 C ．30D ．407．为了得到函数()⎪⎭⎫⎝⎛-=32sin 2πx x f 的图象，只要将2sin y x =的图象上所有的点A ．向右平移3π个单位长度，再把所得各点的横坐标缩短到原来的12倍，纵坐标不变B ．向右平移3π个单位长度，再把所得各点的横坐标伸长到原来的2倍，纵坐标不变C ．向右平移6π个单位长度，再把所得各点的横坐标缩短到原来的12倍，纵坐标不变D ．向右平移6π个单位长度，再把所得各点的横坐标伸长到原来的2倍，纵坐标不变8．已知四棱锥ABCD P -的三视图如图，则四棱锥ABCD P -的全面积为A ．53+B ．52+C ．5D ．49．阅读如图所示的程序框图，运行相应的程序，则程序运行后输出的结果是A ．7B ．9C ．10D ．1110．定义域为R 的函数()f x 满足(2)2()f x f x +=，当[0,2)x ∈时，23||2,[0,1),()1(),[1,2),2x x x x f x x -⎧-∈⎪=⎨-∈⎪⎩若当[4,2)x ∈--时，函数21()42t f x t ≥-+恒成立，则实数t 的取值范围为A ．23t ≤≤B ．13t ≤≤C ．14t ≤≤D ．24t ≤≤第II 卷（非选择题，共100分）二、填空题（本大题共5小题，每小题5分，共25分）11．已知函数,0,)21(0,)(21⎪⎩⎪⎨⎧≤>=x x x x f x则=-)]4([f f ________． 12．在区间[]1,2-上随机取一个实数x ，则事件“122x ≤≤”发生的概率为________．13．若某一离散型随机变量ξ的概率分布如下表，且() 1.5E ξ=，则a b -的值为________．14．已知n⎛ ⎝展开式的第4项为常数项，则=n ________．15．()f x 是定义在D 上的函数，若存在区间[]m n D ⊆，，使函数()f x 在[]mn ，上的值域恰为[]km kn ，，则称函数()f x 是k 型函数。

山东省枣庄市第八中学东校区2018届高三10月月考化学试题(重点班)缺答案

山东省枣庄市第八中学东校区2018届高三10月月考化学试题(重点班)缺答案

秘密★启用前试卷类型：A枣庄八中东校2018届高三1，2班月考化学测试题（测试时间：90分钟，满分100分)可能用到的相对原子质量:H-1 C—12 O—16 Cu-64第Ⅰ卷（选择题,共46分）一、选择题（本题共17小题,1---5每小题2分，6--—-17每小题3分，共46分；每小题只有一个符合题目要求）1．实验发现，298K时，在FeCl3酸性溶液中加入少量锌粒后，Fe3+立即被还原成Fe2+.某化学兴趣小组根据该实验事实设计了如下图所示的原电池装置.下列有关说法中正确的是( )2．A,B，C，D,E是同一周期的五种主族元素,A和B的最高价氧化物对应的水化物显碱性，且碱性B＞A;C,D两种元素对应的气态氢化物的稳定性C＞D，E是这五种元素中原子半径最小的，则它们的原子序数由小到大的顺序为（)。

A．B，A，C,D，E B．A,E，B，C，D C．E，B,A，C，D D．B，A，D,C,E3．现将足量的AgCl分别放入下列物质中，AgCl的溶解度由大到小的顺序是（）①20mL 0．01mol/L KCl溶液②30mL 0．02mol/L CaCl2溶液③40mL 0．03mol/L HCl溶液④10mL 蒸馏水⑤50mL 0．05mol/L AgNO3溶液A．④＞①＞③＞②＞⑤ B．①＞②＞③＞④＞⑤C．⑤＞④＞②＞①＞③ D．④＞③＞⑤＞②＞①4．物质的量浓度相同时，下列既能与NaOH溶液反应、又能跟盐酸溶液反应的溶液中pH最大的是（)A．Na2CO3溶液B．NH4HCO3溶液C．NaHCO3溶液D．NaHSO4溶液5．700℃时，向容积为2 L的密闭容器中充入一定量的CO和H2O，发生反应:CO(g）+H2O（g)CO2（g)+H2(g)，反应过程中测定的部分数据见下表（表中t1＞t2)。

反应时间/min n（CO)/molH2O/mol0 1.200。

60t10.80t20。

20下列说法正确的是（）A.反应在t1 min内的平均速率为v(H2)=0.40/t1 mol·L—1·min-1B.保持其他条件不变，起始时向容器中充入0。

山东省枣庄市第八中学东校区2018届高三1月月考数学(理)试题

山东省枣庄市第八中学东校区2018届高三1月月考数学(理)试题

山东省枣庄市第八中学东校区2018届高三1月月考数学（理科）第Ⅰ卷（共60分）一、选择题（共大题共12个小题,每小题5分,共60分.在每小题给出的四个选项中，只有一项是符合题目要求的）1.若复数11iz i+=-，z 为z 的共轭复数，则2017()z =（ ） A.i B.i -C.20172i - D 20172i2.已知全集U R =，集合{}260A x x x =--≤，401x B xx ⎧-⎫=≤⎨⎬+⎩⎭，那么集合()A C B ⋃⋂=（ ）A.[2,4)-B.(1,3]-C.[2,1]--D.[1,3]-3.ln 2a =，125b -=，201cos 2c xdx π=⎰的大小关系为（ ） A.b c a << B.b a c << C.a b c <<D.c b a <<4.已知双曲线222:1(0)4x y C a a -=>的一条渐近线方程为230x y +=，1F ，2F 分别是双曲线C 的左、右焦点，点P 在双曲线C 上，且17PF =，则2PF =（ ） A.1B.13C.4或10D.1或135.将函数21()cos cos 2f x x x x =+-的图象向左平移312π个单位长度后得到()y g x =的图象，则()g x 在[,]123ππ-上的值域为（ ） A.1[,1]2-B.1[1,]2-C.1[]22-D.1[22-6.已知()f x 为奇函数，函数()f x 与()g x 的图象关于直线1y x =+对称，若(1)4g =，则(3)f -=(A .2-B.2C.1-D.47.某空间几何体的三视图如图所示，则该几何体的体积为（ ）A.73 B .83π- C.83D.73π- 8.已知函数1()ln 1f x x x =--，则()y f x =的图象大致为（ ）9.已知数列{}n a ，{}n b 满足*2log ,n n b a n N =∈，其中{}n b 是等差数列，且920094a a =，则1232017...b b b b ++++=（ ）10.在Rt △ABC 中，90BCA ∠=︒，1CA CB ==，P 为AB 边上的点，且AP AB λ=.若CP AB PA PB ≥，则λ的最大值是（ ）C.111.已知点,M N 是抛物线24y x =上不同的两点，F 为抛物线的焦点，且满足23MFN π∠=，弦MN 的中点P 到直线1:16l y =-的距离记为d ，若22MN d λ= ，则λ的最小值为（ ）A.3C.1D.412.已知3()32(0)f x x x m m =-++>，在区间[0,2]上存在三个不同的实数,,a b c ，使得以(),(),()f a f b f c 为边长的三角形是直角三角形，则m 的取值范围是（ ）A.4m >+B.02m <<+C.44m -<<+.04m <<+第Ⅱ卷（非选择题，共90分）二、填空题（本题共4小题，每题5分，共20分）13.已知数列{}n a 为等比数列，n S 是它的前n 项和.若2312a a a = ，且4a 与72a 的等差中项为54，则5S = . 14.若,,,A B C D 四个人站成一排照相，,A B 相邻的排法总数为k ，则二项式(1)kx k+的展开式中含2x 的项的系数为 .15.已知变量,x y 满足约束条件,2,6,x y y x x y ≤⎧⎪≤⎨⎪≤+⎩则2z x y =-的取值范围是 .16.下列说法中错误的是 .（填序号）①命题“1212,,x x M x x ∃∈≠，有1221[()()]()0f x f x x x -->”的否定是“1212,,x x M x x ∀∉≠”，有1221[()()]()0f x f x x x --≤”； ②已知0a >，0b >，1a b +=，则23a b+的最小值为5+； ③设,x y R ∈，命题“若0xy =，则220x y +=”的否命题是真命题； ④已知2:230p x x +->，1:13q x>-，若命题()q p ⌝∧为真命题，则x 的取值范围是(,3)(1,2)[3,)-∞-⋃⋃+∞.三、解答题 （本大题共70分.解答应写出文字说明、证明过程或演算步骤.）17.已知向量(cos ,1)a x =-，1,)2b x =-，函数()()2f x a b a =+-（1）求函数()f x 的最小正周期及单调递增区间；（2）在ABC △中，内角,,A B C 的对应边分别为,,a b c ，已知函数()f x 的图象经过点1(,)2A ，,,b a c 成等差数列，且9AB AC = ，求a 的值.18.某厂生产甲、乙两种产品每吨所需的煤、电和产值如下表所示：但国家每天分配给该厂的煤、电有限，每天供煤至多56吨，供电至多450千瓦，问该厂如何安排生产，使得该厂日产值最大？最大日产值为多少？ 19.已知数列{}n a 与{}n b 满足*112()()n n n n a a b b n N ++-=-∈. （1）若11a =，35n b n =+，求数列{}n a 的通项公式；（2）若16a =，*2()n n b n N =∈且22n n a n λλ>++对于一切*n N ∈恒成立，求实数λ的取值范围.20.如图1，在ABC △中，2AC =，90ACB ∠=︒，30ABC ∠=︒，P 为AB 边的中点，现把△ACP 沿CP 折叠，使其与BCP △构成如图2所示的三棱锥A BCP -，且AB =（1）求证：平面ACP ⊥平面BCP ； （2）求平面ABC 与平面ABP 夹角的余弦值.21.已知右焦点为F 的椭圆222:1(3x y M a a +=>与直线y =,P Q 两点，且PF QF ⊥.（1）求椭圆M 的方程；（2）O 为坐标原点，A ，,B C 是椭圆M 上不同的三点，并且O 为ABC △的重心，试探究ABC △的面积是否为定值.若是，求出这个定值；若不是，说明理由.22.已知函数21()2f x x =，()ln g x a x =. （1）若曲线()()y f x g x =-在1x =处的切线方程为6250x y --=，求实数a 的值；（2）设()()()h x f x g x =+，若对任意两个不等的正数12,x x ，都有()()12122h x h x x x +>-恒成立，求实数a 的取值范围；（3）若在[1,]e 上存在一点0x ，使得()()()00001()f x g x g x f x ''+<+'成立，求实数a 的取值范围.福建省莆田第九中学2017-2018学年高二上学期第二次月考(12月)试题数学(文)参考答案一、选择题15-：BDACB610-：ABDBC 1112-：AD二、填空题13.3114.112415.(,3]-∞-16.①④三、解答题17.解（1）∵21()()22cos 2sin 226f x a b a a a b x x x π⎛⎫=+-=+-==+ ⎪⎝⎭ ∴()f x 的最小正周期22T ππ==. 由222()262k x k k Z πππππ-≤+≤+∈得()36k x k k Z ππππ-≤≤+∈.∴()f x 的单调递增区间为,()26k k k Z ππππ⎡⎤++∈⎢⎥⎣⎦. （2）由1()sin 262f A A π⎛⎫=+= ⎪⎝⎭，得 2266A k πππ+=+或522()66A k k Z πππ+=+∈. 又0A π<<，∴3A π=.∵,,b a c 成等差数列，∴2a b c =+.∵1cos 92AB AC b A bc === ，∴18bc =，由余弦定理，得222221()4cos 111223612b c a a a a A bc +--==-=-=-，∴a =.18.解：设该厂每天安排生产甲产品x 吨，乙产品y 吨，日产值为z ，可得812z x y =+，其中,x y 满足约束条件735620504500,0x y x y x y +≤⎧⎪+≤⎨⎪≥≥⎩，作出可行域如图所示将直线:812l z x y =+进行平移，由图可知当直线l 经过可行域上的点M 时，直线在y 轴上的截距最大，目标函数z 同时达到最大值 解方程组73562050450x y x y +=⎧⎨+=⎩∴z 的最大值为min 85127124z =⨯+⨯=答：该厂每天安排生产甲产品5吨，乙产品7吨，可得日产值为z 的最大值为124万元. 19.解：（1）因为()112n n n n a a b b ++-=-，35n b n =+所以()1122(3835)6n n n n a a b b n n ++-=-=+--=，所以{}n a 是等差数列，首项为11a =，公差为6，即65n a n =-.（2）因为2n n b =，所以()1112222n n n n n a a +++-=-=当2n ≥时，()()()121112211...22...2622n n n n n n n n a a a a a a a a -+---=-+-++-+=++++=+；当1n =时，16a =，符合上式，所以122n n a +=+.由22nn a n λλ>++，得1121222n n n n nλ+++>=+.又11110222n n n n n n +++-+=≤，所以当1,2n =时，122n n n++取得最大值34， 故实数λ的取值范围为3,4⎛⎫+∞⎪⎝⎭. 20.证明：（1）如图1，取CP 得中点O ，连接AO 并延长交BC 于点E ，在ABC △中，因为2AC =，90ACB ∠=︒，30ABC ∠=︒，P 为AB 边的中点，所以ACP 是正三角形，所以AE CP ⊥，且AO =1CE =,CB =由折叠过程可知，在图2中，AO CP ⊥，OE CP ⊥，如图2，连接OB ，在OCB △中，由余弦定理得(2221217OB =+-⨯⨯︒=，所以22210AO OB AB +==，所以AO OB ⊥.又因为AO CP ⊥，CP OB O ⋂=，所以AO PCB ⊥平面，又因为AO ACP ⊂平面，所以平面ACP ⊥平面CPB.（2）因为AO ⊥平面CPB ，且OC OE ⊥，所以可建立如图二所示的空间直角坐标系.则(0,0,0)O ，()100C ，，，(A ，()1,0,0P -，()B -，(=AB -，(1,0,AC =.设平面ABC 的一个法向量为(,,)m x y z =，则由0,0,m AB m AC ⎧=⎪⎨=⎪⎩得m . 同理可求得平面ABP的一个法向量为()11n =-，. 设所求角为θ，则所求角的余弦值()cos cos ,13m n θ===. 20.解：（1）设(,0)F c，(P t，则,7Q t ⎛- ⎝⎭. ∵点P 在椭圆M 上，∴22317t a +=，∴2247t a =.①∵PF QF ⊥，∴771t c t c =---- ，即2297c t -=-.②由①②得224977c a -=-. 又∵223a c -=，∴24a =，∴椭圆M 的方程为22143x y +=.（2）当直线AB 的斜率存在时，设()11,A x y ，()22,B x y ，直线AB 的方程为(0)y kx m m =+≠.由221,43,x y y kx m ⎧+=⎪⎨⎪=+⎩得()2223484120k x kmx m +++-= ∴122834km x x k -+=+，2122412=34m x x k -+，122634m y y k +=+. 由0> ，得2234k m +>.∵O 为ABC △的重心，∴()2286,3434kmm OC OA OB k k -⎛⎫=-+=⎪++⎝⎭， ∴2286,3434kmm C k k -⎛⎫⎪++⎝⎭. ∵点C 在椭圆M 上，∴2222863434143km m k k -⎛⎫⎛⎫ ⎪ ⎪++⎝⎭⎝⎭+=，化简得22443m k =+（经验证，满足0> ）.∵AB =点C 到直线AB的距离d =（d 是原点O 到直线AB 的距离的3倍），∴26192342ABCm S AB d k ====+△. 当直线AB 的斜率不存在时，3AB =，3d =，92ABC S =△. ∴ABC △的面积为定值92. 21.解：（1）由()21()ln 2y f x g x x a x =-=-，得()a y x x x'=-. 由题意，13a -=，所以2a =-. （2）()()()21ln 2h x f x g x x a x =+=+. 因为对任意两个不等的正数12,x x ，都有()()12122h x h x x x ->-恒成立，设12x x >，则()()()12122h x h x x x ->-即()()112222h x x h x x ->-恒成立.问题等价于函数()()2F x h x x =-，即()21ln 22F x x a x x =+-在()0,+∞上为增函数，所以()20aF x x x'=+-≥在()0,+∞上恒成立.即22a x x ≥-在()0,+∞上恒成立.所以()2max21a x x ≥-=，即实数a 的取值范围是[1,)+∞.（3）不等式()()()()00001f x g x g x f x ''+<-'等价于00001ln ax a x x x +<-，整理得0001ln 0ax a x x +-+<. 设()1ln am x x a x x+=-+， 由题意知，在[1,]e 上存在一点0x ，使得()00m x <.()()()2222111(1)1x a x a a x ax a m x x x x x--++--+'=--==. 因为0x >，所以10x +>，令()0m x '=，得1x a =+. ①当11a +≤，即0a ≤时，()m x 在[1,]e 上单调递增. 只需()120m a =+<，解得2a <-.②当11a e <+≤即01a e <≤-时，()m x 在1x a =+处取最小值. 令()()11ln 110m a a a a +=+-++<即()11ln 1a a a ++<+，可得()11ln 1(*)a a a++<+. 令1t a =+，即1t e <≤，不等式()*可化为1ln 1t t t +<-. 因为1t e <≤，所以不等式左端大于1，右端小于等于1，所以不等式不能成立. ③当1a e +>，即1a e >-时，()m x 在[1,]e 上单调递减，只需()10a m e e a e +=-+<，解得211e a e +>-.综上所述，实数a 的取值范围是()21,2,1e e ⎛⎫+-∞-⋃+∞ ⎪-⎝⎭.。

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