2019-2020学年重庆市新高考模拟实战卷(三)
重庆市綦江县2019-2020学年高考第三次模拟数学试题含解析
重庆市綦江县2019-2020学年高考第三次模拟数学试题一、选择题：本题共12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．空间点到平面的距离定义如下：过空间一点作平面的垂线，这个点和垂足之间的距离叫做这个点到这个平面的距离．已知平面α，β，λ两两互相垂直，点A α∈，点A 到β，γ的距离都是3，点P 是α上的动点，满足P 到β的距离与P 到点A 的距离相等，则点P 的轨迹上的点到β的距离的最小值是（ ） A ．33- B ．3C ．33- D ．32【答案】D 【解析】 【分析】建立平面直角坐标系，将问题转化为点P 的轨迹上的点到x 轴的距离的最小值，利用P 到x 轴的距离等于P 到点A 的距离得到P 点轨迹方程，得到()26399y x =-+≥，进而得到所求最小值.【详解】如图，原题等价于在直角坐标系xOy 中，点()3,3A ，P 是第一象限内的动点，满足P 到x 轴的距离等于点P 到点A 的距离，求点P 的轨迹上的点到x 轴的距离的最小值． 设(),P x y ，则()()2233y x y =-+-，化简得：()23690x y --+=，则()26399y x =-+≥，解得：32y ≥， 即点P 的轨迹上的点到β的距离的最小值是32. 故选：D . 【点睛】本题考查立体几何中点面距离最值的求解，关键是能够准确求得动点轨迹方程，进而根据轨迹方程构造不等关系求得最值.2．在直角ABC ∆中，2C π∠=，4AB =，2AC =，若32AD AB =，则CD CB ⋅=u u u r u u u r （ ）A ．18-B ．63-C ．18D ．3【解析】 【分析】在直角三角形ABC 中，求得12AC cos CAB AB ∠== ，再由向量的加减运算，运用平面向量基本定理，结合向量数量积的定义和性质：向量的平方即为模的平方，化简计算即可得到所求值． 【详解】在直角ABC ∆中，2C π∠=，4AB =，2AC =，，12AC cos CAB AB ∠==， 若32AD AB =u u u v u u u v ，则2CD CB AD AC AB AC AD AB AD AC AC AB AC ⋅=-⋅-=⋅-⋅-⋅+u u u v u u u v u u u v u u u v u u u v u u u v u u u v u u u v u u u v u u u v u u u v u u u v u u u v （）（）223322AB AB AC AC AB AC =-⋅-⋅+u u u v u u u v u u u v u u u v u u u v u u u v 3511642418222=⨯-⨯⨯⨯+=． 故选C. 【点睛】本题考查向量的加减运算和数量积的定义和性质，主要是向量的平方即为模的平方，考查运算能力，属于中档题．3．我国古代有着辉煌的数学研究成果，其中的《周髀算经》、《九章算术》、《海岛算经》、《孙子算经》、《缉古算经》，有丰富多彩的内容，是了解我国古代数学的重要文献．这5部专著中有3部产生于汉、魏、晋、南北朝时期．某中学拟从这5部专著中选择2部作为“数学文化”校本课程学习内容，则所选2部专著中至少有一部是汉、魏、晋、南北朝时期专著的概率为（ ） A ．35B ．710C ．45D ．910【答案】D 【解析】 【分析】利用列举法，从这5部专著中选择2部作为“数学文化”校本课程学习内容，基本事件有10种情况，所选2部专著中至少有一部是汉、魏、晋、南北朝时期专著的基本事件有9种情况，由古典概型概率公式可得结果. 【详解】《周髀算经》、《九章算术》、《海岛算经》、《孙子算经》、《缉古算经》，这5部专著中有3部产生于汉、魏、晋、南北朝时期．记这5部专著分别为,,,,a b c d e ，其中,,a b c 产生于汉、魏、晋、南北朝时期．从这5部专著中选择2部作为“数学文化”校本课程学习内容，基本事件有,,,,,,,,,,ab ac ad ae bc bd be cd ce de 共10种情况，所选2部专著中至少有一部是汉、魏、晋、南北朝时期专著的基本事件有,,,,,,,,,ab ac ad ae bc bd be cd ce ，共9种情况，所以所选2部专著中至少有一部是汉、魏、晋、南北朝时期专著的概率为910m P n ==．故选D ． 【点睛】本题主要考查古典概型概率公式的应用，属于基础题，利用古典概型概率公式求概率时，找准基本事件个数是解题的关键，基本亊件的探求方法有 (1)枚举法：适合给定的基本事件个数较少且易一一列举出的；(2)树状图法：适合于较为复杂的问题中的基本亊件的探求.在找基本事件个数时，一定要按顺序逐个写出：先11(,)A B ，12(,)A B …. 1(,)n A B ，再21(,)A B ，22(,)A B …..2(,)n A B 依次31(,)A B 32(,)A B ….3(,)n A B … 这样才能避免多写、漏写现象的发生.4．512a x x x x ⎛⎫⎛⎫+- ⎪⎪⎝⎭⎝⎭的展开式中各项系数的和为2，则该展开式中常数项为 A ．-40 B ．-20C ．20D ．40【答案】D 【解析】令x=1得a=1.故原式=511()(2)x x x x +-．511()(2)x x x x+-的通项521552155(2)()(1)2r r r r r r r r T C x x C x ----+=-=-，由5-2r=1得r=2,对应的常数项=80，由5-2r=-1得r=3,对应的常数项=-40，故所求的常数项为40 ，选D解析2.用组合提取法,把原式看做6个因式相乘，若第1个括号提出x,从余下的5个括号中选2个提出x ，选3个提出1x ；若第1个括号提出1x ，从余下的括号中选2个提出1x，选3个提出x. 故常数项=223322335353111(2)()()(2)X C X C C C X X X X⋅⋅-+⋅-⋅=-40+80=405．已知P 与Q 分别为函数260x y --=与函数21y x =+的图象上一点，则线段||PQ 的最小值为( )A ．65B ．CD ．6【答案】C 【解析】 【分析】利用导数法和两直线平行性质，将线段||PQ 的最小值转化成切点到直线距离. 【详解】可知抛物线21y x =+存在某条切线与直线260x y --=平行，则2k =，设抛物线21y x =+的切点为()200,1x x +，则由2y x '=可得022x =，01x ∴=，所以切点为(1,2)，则切点(1,2)到直线260x y --=的距离为线段||PQ 的最小值，则min ||PQ ==. 故选：C. 【点睛】本题考查导数的几何意义的应用，以及点到直线的距离公式的应用，考查转化思想和计算能力. 6．已知α、,22ππβ⎛⎫∈- ⎪⎝⎭，αβ≠，则下列是等式sin sin 2αβαβ-=-成立的必要不充分条件的是（ ） A ．sin sin αβ> B ．sin sin αβ< C ．cos cos αβ> D ．cos cos αβ<【答案】D 【解析】 【分析】构造函数()sin h x x x =-，()sin 2f x x x =-，利用导数分析出这两个函数在区间,22ππ⎛⎫- ⎪⎝⎭上均为减函数，由sin sin 2αβαβ-=-得出sin sin 2ααββ-=-，分0α=、02πα-<<、02πα<<三种情况讨论，利用放缩法结合函数()y h x =的单调性推导出02παβ-<<<或02πβα<<<，再利用余弦函数的单调性可得出结论. 【详解】构造函数()sin h x x x =-，()sin 2f x x x =-， 则()cos 10h x x '=-<，()cos 20f x x '=-<，所以，函数()y f x =、()y h x =在区间,22ππ⎛⎫- ⎪⎝⎭上均为减函数，当02x π-<<时，则()()00h x h >=，()()00f x f >=；当02x π<<时，()0h x <，()0f x <.由sin sin 2αβαβ-=-得sin sin 2ααββ-=-. ①若0α=，则sin 20ββ-=，即()00fββ=⇒=，不合乎题意；②若02πα-<<，则02πβ-<<，则()()sin sin 2sin h h αααβββββ=-=->-=，此时，02παβ-<<<，由于函数cos y x =在区间,02π⎛⎫- ⎪⎝⎭上单调递增，函数sin y x =在区间,02π⎛⎫- ⎪⎝⎭上单调递增，则sin sin αβ<，cos cos αβ<；③若02πα<<，则02πβ<<，则()()sin sin 2sin h h αααβββββ=-=-<-=，此时02πβα<<<，由于函数cos y x =在区间0,2π⎛⎫ ⎪⎝⎭上单调递减，函数sin y x =在区间0,2π⎛⎫ ⎪⎝⎭上单调递增，则sin sin αβ>，cos cos αβ<.综上所述，cos cos αβ<. 故选：D. 【点睛】本题考查函数单调性的应用，构造新函数是解本题的关键，解题时要注意对α的取值范围进行分类讨论，考查推理能力，属于中等题.7．如图在直角坐标系xOy 中，过原点O 作曲线()210y x x =+≥的切线，切点为P ，过点P 分别作x 、y 轴的垂线，垂足分别为A 、B ，在矩形OAPB 中随机选取一点，则它在阴影部分的概率为（ ）A ．16B ．15C ．14D ．12【答案】A 【解析】 【分析】设所求切线的方程为y kx =，联立()201y kx k y x ⎧=>⎨=+⎩，消去y 得出关于x 的方程，可得出0∆=，求出k 的值，进而求得切点P 的坐标，利用定积分求出阴影部分区域的面积，然后利用几何概型概率公式可求得所求事件的概率.设所求切线的方程为y kx =，则0k >， 联立()201y kx k y x ⎧=>⎨=+⎩，消去y 得210x kx -+=①，由240k ∆=-=，解得2k =，方程①为2210x x -+=，解得1x =，则点()1,2P ， 所以，阴影部分区域的面积为()1232100111233S x x dx x x x ⎛⎫=+-=-+= ⎪⎝⎭⎰， 矩形OAPB 的面积为122S '=⨯=，因此，所求概率为16S P S =='. 故选：A. 【点睛】本题考查定积分的计算以及几何概型，同时也涉及了二次函数的切线方程的求解，考查计算能力，属于中等题.8．已知集合{}2|2150A x x x =-->，{}|07B x x =<<，则()R A B ðU 等于( )A ．[)5,7-B ．[)3,7-C ．()3,7-D ．()5,7-【答案】B 【解析】 【分析】解不等式确定集合A ，然后由补集、并集定义求解． 【详解】由题意{}2|2150A x x x =-->{|3x x =<-或5}x >，∴{|35}R A x x =-≤≤ð，(){|37}R A B x x =-≤<U ð．故选：B. 【点睛】本题考查集合的综合运算，以及一元二次不等式的解法，属于基础题型． 9．某四棱锥的三视图如图所示，记S 为此棱锥所有棱的长度的集合，则（ ）.A ．，且SB ．，且SC ．22S ∈，且23S ∉ D ．22S ∈，且23S ∈【答案】D 【解析】 【分析】首先把三视图转换为几何体，根据三视图的长度，进一步求出个各棱长. 【详解】根据几何体的三视图转换为几何体为：该几何体为四棱锥体， 如图所示：所以：2AB BC CD AD DE =====，22AE CE ==，22(22)223BE =+=.故选：D.. 【点睛】本题考查三视图和几何体之间的转换，主要考查运算能力和转换能力及思维能力，属于基础题. 10．已知数列{}n a 为等比数列，若a a a 76826++=，且a a 5936⋅=，则a a a 768111++=（ ） A ．1318B ．1318或1936C ．139D ．136【答案】A 【解析】 【分析】根据等比数列的性质可得25968736a a a a a ⋅=⋅==，通分化简即可.【详解】由题意，数列{}n a 为等比数列，则25968736a a a a a ⋅=⋅==，又a a a 76826++=，即68726a a a +=-，所以，()()76877786867678777683636261113636a a a a a a a a a a a a a a a a a a a +⋅++⋅-⋅+⋅+⋅++===⋅⋅⋅⋅，()277777777773626362636263626133636363618a a a a a a a a a a +⋅-+⋅-+⋅-⋅=====⋅⋅⋅⋅.故选：A. 【点睛】本题考查了等比数列的性质，考查了推理能力与运算能力，属于基础题.11．如图，这是某校高三年级甲、乙两班在上学期的5次数学测试的班级平均分的茎叶图，则下列说法不正确的是（ ）A ．甲班的数学成绩平均分的平均水平高于乙班B ．甲班的数学成绩的平均分比乙班稳定C ．甲班的数学成绩平均分的中位数高于乙班D ．甲、乙两班这5次数学测试的总平均分是103 【答案】D 【解析】 【分析】计算两班的平均值，中位数，方差得到ABC 正确，两班人数不知道，所以两班的总平均分无法计算，D 错误，得到答案. 【详解】由题意可得甲班的平均分是104，中位数是103，方差是26.4； 乙班的平均分是102，中位数是101，方差是37.6，则A ，B ，C 正确. 因为甲、乙两班的人数不知道，所以两班的总平均分无法计算，故D 错误. 故选：D . 【点睛】本题考查了茎叶图，平均值，中位数，方差，意在考查学生的计算能力和应用能力.12．已知双曲线()222210,0x y a b a b-=>>的左、右焦点分别为12F F 、，圆222x y b +=与双曲线在第一象限内的交点为M ，若123MF MF =．则该双曲线的离心率为 A ．2 B ．3C 2D 3【答案】D本题首先可以通过题意画出图像并过M 点作12F F 垂线交12F F 于点H ，然后通过圆与双曲线的相关性质判断出三角形2OMF 的形状并求出高MH 的长度，MH 的长度即M 点纵坐标，然后将M 点纵坐标带入圆的方程即可得出M 点坐标，最后将M 点坐标带入双曲线方程即可得出结果。
2019-2020学年重庆市第九十五中学高三语文三模试卷及答案
2019-2020学年重庆市第九十五中学高三语文三模试卷及答案一、现代文阅读（36分）(一)现代文阅读I(9分)阅读下面的作品，完成各题。
生沈从文北京城什刹海南头，煤灰土新垫就一片场坪，白日照着。
一个年过六十的老人扛了一对大傀儡走来，到了场坪，四下望人，似乎很明白这不是玩傀儡的地方，但无可奈何地停了下来。
这老头子把傀儡坐在场中烈日下，轻轻咳着，调理着嗓子。
他除了那对脸儿一黑一白简陋呆板的傀儡以外，什么都没有！看的人也没有。
他把那双发红小眼睛四方瞟着，场坪位置既不适宜，天又那么热，若无什么花样做出来，绝不能把闲人引过来。
老头子便望着坐在坪里傀儡中白脸的一个，亲昵地低声打着招呼，也似乎正用这种话安慰他自己。
“王九，不要着急，慢慢的会有人来的，咱们呆一会儿，就玩个什么给爷们看看，玩得好，还愁爷们不赏三枚五枚？玩得好，爷们回去还会说：王九赵四摔跤多扎实，六月天大日头下扭着蹩着搂着，还不出汗！可不是，天那么热，你也不累，好汉子！”来了一个人，把花条子衬衣下角长长的拖着，作成北京城大学生特有的样子。
老头子瞥了这学生一眼，便微笑着，以为帮场的“福星”来了，全身作成年轻人灵便姿势，膀子向上向下摇着，一面自言自语地说话，亲昵得如同家人父子应对：“王九，你瞧，先生可来了。
好，咱们动手，先生不会走的。
你小心，别让赵四小子扔倒。
先生帮咱们绷个场面，看你摔赵四这小子，先生准不走。
”于是他把傀儡扶起，整理傀儡身上的破旧长衫，又从衣下取出两只假腿来，把它缚在自己裤带上，再把傀儡举起，弯着腰，钻进傀儡所穿衣服里面去，用衣服罩好了自己，且把两只手套进假腿里，改正了两只假腿的位置，开始在灰土坪里扮演两人殴打的样子。
他移动着傀儡的姿势，跳着，蹿着，有时又用真脚去捞那双用手套着的假脚，装作掼跤盘脚的动作。
他既不能看清楚头上的傀儡，又不能看清楚场面上的观众，表演得却极有生气。
大学生忧郁地笑了，而且远远的另一方，有人注意到了这边空地上的情形，就被这情形引起了好奇兴味，第二个人跑来了。
2019-2020学年重庆市南开中学校高三英语三模试题及参考答案
2019-2020学年重庆市南开中学校高三英语三模试题及参考答案第一部分阅读（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、D四个选项中选出最佳选项AFor some people, there’s no better companion than mans best friend-a dog. This four-legged pet can bring comfort and joy and provide much- needed exercise for you when it needs walkies! This probably explains why dog ownership increased last year because people spent more time at home during he CovID-I9 lockdown.However, as demand for a new dog increased, so did the price tag. Popular breeds, such as Cockapoos and Cocker Spaniels, saw even sharper price increases, and puppies have been selling for $3,000 or more.Animal welfare charities fearthat high prices could encourage puppy farming, smuggling (走私) or dog theft. An investigation found some breeders have been selling puppies and kittens on social media sites--something charities have called “extremely irresponsible”.But despite some new owners purchasing a dog legally, maybe from a rescue center or registered breeder, they’ve proved to be ill-prepared for life with a new pet, and the pet itself has found it hard tocome to terms withlife in a new home.Looking to the future, there are concerns about the welfare of these much-loved pets. Lan Alkin manager of the Oxfordshire Animal Sanct uary in the UK, notes: “At the moment, the dogs are having a great time, but separation anxiety could still surface when people go back to work.” And Cliare Calder from the UKs Dogs Trust rescue charity says, “The economic situation also means that some people may find they can’t afford to look aftera dog.” The message is not to buy a dog in haste and to pick one that fits into our lifestyle.1. The greater demand for dogs can cause the following problems except ________.A. illegal trade of dogsB. less dog farmingC. high prices of dogsD. online sale of dogs2. What does the underlined phrase"come to terms with"in paragraph 4 mean?A. Fit in withB. Go in forC. Make up for.D. End up with3. What can we learn from the last paragraph?A. Despite the problems, dogs are living happily.B. The writer has a positive attitude towards dogs future.C. Experts are worried that dogs will be unaffordable to people.D. The writer advises people to think twice before keeping dogs as pets.BAt Aizo Chuo Hospital in Japan, employees greet newcomers, guide patients to and from the surgery area, and print out maps of the hospital for confused visitors. They don’t take lunch breaks or even get paid. Why? They’re robots!Robots have long worked in factories, helping to build cars and electronic appliances. But today’s robots don’t just do the jobs of people－they actually look and act a lot like people.Kansei, arobot from Japan, has a plastic face covering 19 movable parts. The robot can make 36 facial expressions in response to different words. Kansei shakes in fear at the word “war” and smiles when it hears the word “dinner”.Researchers in Europe are going even further with iCub, a “baby” robot. They are teaching it to speak and hold conversations.The ability to interact is crucial for robots that will one day work closely with humans says robotics professor ChrisAtkeson. “ This will require robots to understand what you say and how you are feeling and respond with appropriate emotions, ” he told WR News.Japanese scientist Minoru Asada agrees. He is building a robot called CB2 that acts like a real baby. “ Right now, it only goes, ̒Ah, ah. ̓But as we develop its learning function, it will start saying more complex sentences and moving on its own, ” Asada says. “ Next-generation robots need to be able to learn and develop by themselves.”Intelligent robot will become more important in the future, as populations age and the number of human workers declines in many countries. “ We’re going to have many more old people and not enough young people to care for them,” says robot researcher Matthew Mason. “ Technology can help the old people live at home longer, instead of going to nursing homes.”4. According to the passage, what jobs have robots already performed?A. Giving advice, answering customer questions and planning events.B. Producing factory goods, building cars and greeting customers.C. Greeting customers, producing factory goods and performing surgery.D. Building cars, driving passengers and providing directions.5. The second paragraph in this passage is mainly about?A. To explain how a robot works.B. To define what a robot is.C. To describe the functions of modern robots.D. To predict the future uses of robots.6. How does the Kansei robot react on the word “fire”?A. Use languages to warn nearby humans.B Back up its memory files.C. Activate an automatic fire alarm.D. Produce a worried look on his face.7. In Asada’s opinion, the next step for robots will be to develop_______.A. the ability to learn independentlyB. the ability to understand human commandsC. the capacity to interact with humansD. the willingness to work togetherCHidden beneath the surface in the roots of Earths astonishing and diverse plant life, there exists a biological superhighway linking together the members of the plant kingdom in what researchers call the "wood wide web".The network is comprised of thin threads of fungus (真菌) that grow outwards underground up to a few meters from its partnering plant, meaning that all of the plant life within a region likely connected to one another. The partnership is beneficial for both parties involved, plants provide carbohydrates (碳水化合物) to the funguses and in exchange, the funguses aid in gathering water and providing nutrients to its partnering plant.A study conducted by Rensen Zeng of theSouthChinaAgriculturalUniversityfound that this also allowed for plants to warn one another of potential harm. The study showed Broad Beans used the fungal network to spy on one another for upcoming danger.Like our Internet, this fungal connectivity is also full of crime. Some plants, such as Golden Marigolds have been found to release poisons into the network to slow down the growth of surrounding plants in the fight for water and light. Other plants, such as the Phantom Orchid, do not have the chlorophyll (叶绿素) and must get the necessary nutrients from surrounding plants.Research suggests that animals such as insects and worms may be able to detect slight exchanges of nutrients through the network, allowing them to more easily find delicious roots to feed on; however, this has not been conclusively made clear in experimentation. The more we learn about this phenomenon, the more ourunderstanding of the plant life of our planet will continue to change. Perhaps one day, we may be able to map out these complex networks entirely.8. What is the function of the first paragraph?A.To explain the aim of the web.B. To introduce the main topic.C. To give definition of diverse life.D. To show the importance of plants.9. The criminal behavior of plants can be seen as a way to________.A. compete for survivalB. gather more waterC. take in sunlightD. break natural rules10. What does the last paragraph suggest?A. Animals can also feed on the fungus.B. Nutrient exchanges are too slight to detect.C. No experiment can prove the phenomenon.D. More needs to be done to work out the network.11. Which can be the best title for the text?A. The Partnership between PlantsB. The Unknown Roots of the EarthC. The Superhighway Linking the PlantsD. The Mysterious Map Changing the WorldDIn the old days, when you had to drive to a movie theater to get some entertainment, it was easy to see how your actions could have an impact（影响）on the environment. After all, you were jumping into your car, driving across town, coughing out emissions（产生排放）and using gas all the way. But now that we're used to staying at home and streaming movies, we might get a littleproud. After all, we're just picking up our phones and maybe turning on the TV. You're welcome. Mother Nature.Not so fast, says a recent report from the French-based Shift Project. According to "Climate Crisis: The Unsustainable Use of Online Video", digital technologies are responsible for 4% of greenhouse gas emissions, andthat energy use is increasing by 9% a year. Watching a half-hour show would cause 1. 6 kilograms of carbon dioxide emissions. That's like driving 6. 28 kilometers. And in the European Union, the Eureca project found that data centers（where videos are stored）there used 25% more energy in 2017 compared to just three years earlier, reports the BBC.Streaming is only expected to increase as webecome more enamored ofour digital devices（设备）and the possibility of enjoying entertainment where and when we want it increases. Online video use is expected to increase by four times from 2017 to 2022 and account for 80% of all Internet traffic by 2022. By then, about 60% of the world's population will be online.You're probably not going to give up your streaming services, but there're things you can do to help lessen the impact of your online use, experts say. For example, according to Lutz Stobbe, a researcher from the Fraunhofer Institute in Berlin, we have no need to upload 25 pictures of the same thing to the cloud because it consumes energy every time. If instead you delete a few things here and there, you can save energy. Moreover, it's also a good idea to stream over Wi-Fi, watch on the smallest screen you can, and turn off your Wi-Fi in your home if you're not using your devices.12. What topic is the first paragraph intended to lead in?A. The environmental effects of driving private cars.B. The improvements on environmental awareness.C. The change in the way people seek entertainment.D The environmental impacts of screaming services.13. What does the underlined phrase become more enamored of" in paragraph 3 probably mean?A. Get more skeptical of.B. Become more aware of.C. Feel much crazier about.D. Get more worried about.14. What can we infer about the use of streaming services?A. It is being reduced to protect the planet.B. Its environmental effects are worsening.C. It is easily available to almost everyone.D. Its side effects have drawn global attention.15. Which of the following is the most environmentally-friendly?A. Watching downloaded movies on a mobile phone.B. Downloading music on a personal computer.C. Uploading a lot of images of the same thing.D. Playing online games over mobile networks.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
重庆市万州区2019-2020学年高考第三次模拟数学试题含解析
重庆市万州区2019-2020学年高考第三次模拟数学试题一、选择题：本题共12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．已知函数()sinx12sinxf x =+的部分图象如图所示，将此图象分别作以下变换，那么变换后的图象可以与原图象重合的变换方式有（ ）①绕着x 轴上一点旋转180︒; ②沿x 轴正方向平移; ③以x 轴为轴作轴对称;④以x 轴的某一条垂线为轴作轴对称. A ．①③ B ．③④C ．②③D ．②④【答案】D 【解析】 【分析】计算得到()()2f x k f x π+=，22f x f x ππ⎛⎫⎛⎫-=+ ⎪ ⎪⎝⎭⎝⎭，故函数是周期函数，轴对称图形，故②④正确，根据图像知①③错误，得到答案. 【详解】()sin 12sin xf x x=+，()()()()sin 2sin 212sin 212sin x k x f x k f x x k x πππ++===+++，k Z ∈， 当沿x 轴正方向平移2,k k Z π∈个单位时，重合，故②正确；co sin 2212co s s s 12in2x f x x x x πππ⎛⎫- ⎪⎛⎫⎝⎭-== ⎪+⎛⎫⎝⎭+- ⎪⎝⎭，co sin 2212co s s s 12in2x f x x x x πππ⎛⎫+ ⎪⎛⎫⎝⎭+== ⎪+⎛⎫⎝⎭++ ⎪⎝⎭，故22f x f x ππ⎛⎫⎛⎫-=+⎪ ⎪⎝⎭⎝⎭，函数关于2x π=对称，故④正确；根据图像知：①③不正确； 故选：D .【点睛】本题考查了根据函数图像判断函数性质，意在考查学生对于三角函数知识和图像的综合应用.2．已知双曲线2222:1(0,0)x y C a b a b-=>>的右焦点为F ，过右顶点A 且与x 轴垂直的直线交双曲线的一条渐近线于M 点，MF 的中点恰好在双曲线C 上，则C 的离心率为（ ） A1 B．CD【答案】A 【解析】 【分析】设(,)M a b ，则MF 的中点坐标为(,)22a c b+，代入双曲线的方程可得,,a b c 的关系，再转化成关于,a c 的齐次方程，求出ca的值，即可得答案. 【详解】双曲线2222:1(0,0)x y C a b a b-=>>的右顶点为(,0)A a ，右焦点为(c,0)F ，M 所在直线为x a =，不妨设(,)M a b ，∴MF 的中点坐标为(,)22a cb +.代入方程可得2222221a c b a b +⎛⎫⎛⎫ ⎪ ⎪⎝⎭⎝⎭-=， ∴22()544a c a +=，∴2240e e +-=，∴1e =（负值舍去）. 故选：A. 【点睛】本题考查双曲线的离心率，考查函数与方程思想、转化与化归思想，考查逻辑推理能力和运算求解能力，求解时注意构造,a c 的齐次方程.3．执行下面的程序框图，若输出的S 的值为63，则判断框中可以填入的关于i 的判断条件是（ ）A ．5i ≤B ．6i ≤C ．7i ≤D ．8i ≤【答案】B 【解析】 【分析】根据程序框图，逐步执行，直到S 的值为63，结束循环，即可得出判断条件. 【详解】 执行框图如下： 初始值：0,1S i ==，第一步：011,112S i =+==+=，此时不能输出，继续循环； 第二步：123,213S i =+==+=，此时不能输出，继续循环； 第三步：347,314S i =+==+=，此时不能输出，继续循环； 第四步：7815,415S i =+==+=，此时不能输出，继续循环； 第五步：151631,516S i =+==+=，此时不能输出，继续循环； 第六步：313263,617S i =+==+=，此时要输出，结束循环； 故，判断条件为6i ≤. 故选B 【点睛】本题主要考查完善程序框图，只需逐步执行框图，结合输出结果，即可确定判断条件，属于常考题型. 4．某几何体的三视图如图所示，则该几何体的体积是（ ）A ．53π B ．43π C ．223π+D ．243π+【答案】A 【解析】 【分析】观察可知，这个几何体由两部分构成,：一个半圆柱体，底面圆的半径为1，高为2；一个半球体，半径为1，按公式计算可得体积。
重庆市万州区2019-2020学年第三次高考模拟考试数学试卷含解析
重庆市万州区2019-2020学年第三次高考模拟考试数学试卷一、选择题：本题共12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．如图，在平面四边形ABCD 中，满足,AB BC CD AD ==，且10,8AB AD BD +==，沿着BD 把ABD 折起，使点A 到达点P 的位置，且使2PC =，则三棱锥P BCD -体积的最大值为（ ）A ．12B ．2C ．23D ．163【答案】C 【解析】 【分析】过P 作PE BD ⊥于E ，连接CE ，易知CE BD ⊥，PE CE =，从而可证BD ⊥平面PCE ，进而可知1833P BCD B PCE D PCE PCE PCE V V V S BD S ---=+=⋅=V V ，当PCE S V 最大时，P BCD V -取得最大值，取PC 的中点F ，可得EF PC ⊥，再由2112PCE S PC EF PE =⋅=-V PE 的最大值即可.【详解】在BPD △和BCD V 中，PB BC PD CD BD BD =⎧⎪=⎨⎪=⎩，所以BPD BCD V V ≌，则PBD CBD ∠=∠，过P 作PE BD ⊥于E ，连接CE ，显然BPE BCE V V ≌，则CE BD ⊥，且PE CE =， 又因为PE CE E =I ，所以BD ⊥平面PCE ， 所以1833P BCD B PCE D PCE PCE PCE V V V S BD S ---=+=⋅=V V ， 当PCE S V 最大时，P BCD V -取得最大值，取PC 的中点F ，则EF PC ⊥， 所以2112PCE S PC EF PE =⋅=-V 因为10,8PB PD BD +==，所以点P 在以,B D 为焦点的椭圆上（不在左右顶点），其中长轴长为10，焦距长为8，所以PE 的最大值为椭圆的短轴长的一半，故PE 22543-=， 所以PCE S ∆最大值为2，故P BCD V -的最大值为8223⨯1623=.故选：C.【点睛】本题考查三棱锥体积的最大值，考查学生的空间想象能力与计算求解能力，属于中档题. 2．在ABC V 中，内角A ，B ，C 所对的边分别为a ，b ，c ，D 是AB 的中点，若1CD =，且1sin 2a b A ⎛⎫-⎪⎝⎭()()sin sin c b C B =+-，则ABC V 面积的最大值是( ) A ．155B ．15C 15D ．155【答案】A 【解析】 【分析】根据正弦定理可得()()12a b a c b c b ⎛⎫-=+- ⎪⎝⎭，求出cos C ，根据平方关系求出sin C .由2CD CA CB =+u u u r u u u r u u u r 两端平方，求ab 的最大值，根据三角形面积公式in 12s S ab C =，求出ABC V 面积的最大值. 【详解】ABC V 中，()()1sin sin sin 2a b A c b C B ⎛⎫-=+- ⎪⎝⎭，由正弦定理可得()()12a b a c b c b ⎛⎫-=+-⎪⎝⎭，整理得22212c a b ab =+-， 由余弦定理2222cos c a b ab C =+-，得()115cos ,0,,sin 44C C C π=∈=Q . Q D 是AB 的中点，且1CD =，()()222,2CD CA CB CDCA CB ∴=+∴=+u u u r u u u r u u u r u u u ru u u r u u u r ，即22242CD CA CB CA CB =++u u u r u u u r u u u r u u u r u u u r g ，即222211542cos 2222b a ba C a b ab ab ab ab =++=++≥+=， 85ab ∴≤，当且仅当a b =时，等号成立. ABC ∴V 的面积1181515sin 225S ab C =≤⨯所以ABC V面积的最大值为5. 故选：A . 【点睛】本题考查正、余弦定理、不等式、三角形面积公式和向量的数量积运算，属于中档题. 3．已知复数21iz i=+，则z =（ ） A ．1i + B ．1i -CD ．2【答案】C 【解析】 【分析】根据复数模的性质即可求解. 【详解】21i z i=+Q, |2||1|i z i ∴===+, 故选：C 【点睛】本题主要考查了复数模的性质，属于容易题.4．已知向量(,1)a m =r ，(1,2)b =-r ，若(2)a b b -⊥r r r ，则a r 与b r夹角的余弦值为（ ）A．13-B．13C．65-D．65【答案】B 【解析】 【分析】直接利用向量的坐标运算得到向量2a b -r r 的坐标，利用(2)=0a b b -⋅r r r 求得参数m ，再用cos ,||||a ba b a b ⋅〈〉=r rr r r r计算即可. 【详解】依题意，2(2,3)a b m -=+-r r ， 而(2)=0a b b -⋅r r r， 即260m ---=， 解得8m =-，则cos ,||||a b a b a b ⋅〈〉===r rr r r r 故选：B.本题考查向量的坐标运算、向量数量积的应用，考查运算求解能力以及化归与转化思想.5．已知双曲线2222:1(0,0)x y C a b a b-=>>的左、右焦点分别为1F ，2F ，点P 是C 的右支上一点，连接1PF 与y 轴交于点M ，若12||FO OM =（O 为坐标原点），12PF PF ⊥，则双曲线C 的渐近线方程为（ ）A ．3y x =±B ．y =C ．2y x =±D ．y =【答案】C 【解析】 【分析】利用三角形1OMF ∆与2PF F ∆相似得122PF PF =，结合双曲线的定义求得,,a b c 的关系，从而求得双曲线的渐近线方程。
2019-2020学年重庆市田家炳中学高三英语三模试卷及答案解析
2019-2020学年重庆市田家炳中学高三英语三模试卷及答案解析第一部分阅读（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、D四个选项中选出最佳选项AEver wonder why there are so many people polluting the earth? Ever say to yourself：Hey, I wish that I could do more to help the environment? Have you ever thought about trying to help the earth but never really did it? Well, here are some pretty easyand skillful ideas for that green - earth desire inside you.●Turn off your computer. By leaving it on all day you are creating more CO2than a regular passenger would, driving to and from work in one day.● Ride your bike or carpool (合用汽车).Obviously, youare creating less CO2which will help the ozone(臭氧).● Make a garden. Even simply grow some plants in your kitchen, which will help produce more oxygen while eating up some of that evil CO2.● Buy local groceries. It creates less impact on the environment. Besides, you're supporting your local farmers.● Recycle. You had to see this coming. But you have no idea how much you are helping the environment by simply reusing a water bottle instead of buying a huge pack at the store.● Don't run the water while brushing. It saves you money and helps the water resources.● Open the curtains. Natural light is much prettier and it will keep the energy usage down.● Rechargeable batteries. You have no idea how much it takes to get rid of batteries. Do yourself a favor. Save some money and some energy.1. If you don't want to create more CO2, you may_______.A. turn off your computer or open the curtainsB. turn off your computer or ride your bike or carpoolC. make a garden or open the curtainsD. use rechargeable batteries or make a garden2. Which of the following can best describe the function of the first paragraph?A. Main body.B. Argument.C. Lead - in.D. Conclusion.3. The main idea of the passage is about________.A. the importance of environmental protectionB. some ways about how to prevent pollutionC. some suggestions about how to save energyD. some suggestions about environmental protectionBVietnammade preparations for theLunar New Year with a fish release on Thursday. The tradition involves releasing fish called carps into rivers and lakes in the country a week before the holiday known as Tet officially begins.The tradition comes from an ancient story of three “Kitchen Gods”. The three, two males and a female, take a ride on the brightly color1 ed carp at the end of the year. They go to Heaven to meet with the Jade Emperor, the God who rules there. The “Kitchen Gods” report news about the families they represent to seek the emperor's care and protection. Their efforts help to keep the kitchen fires burning, and families happy and healthy.As the Gods go to Heaven, families clean their houses in preparation to celebrate Tet. A clean house, Vietnamese believe, will bring luck in the new year. On Lunar New Year's Eve, the Gods will return to Earth and their duties in the kitchen of the house.Thursday's event inVietnamthis year was more controlled than ever before because of COVID-19. It is spreading in several northern areas of the country. However,Vietnamhas increased contact examination, mass testing, and quarantine measures to slow the spread. The aggressive action has limited infections and deaths in the country.“Vietnamese will still follow the tradition of releasing the fish, but COVID-19 has made people keep a safe distance,” said Tran Van Toi as he released a carp from a plastic bag atHanoi'sWestLake.This year, due to COVID-19, there were fewer people releasing fish there, but there was a major change in attitude towards plastic bags. After years of persuasion, now they don't throw the plastic bags into the water anymore but collect them to be recycled.4. What does paragraph 2 mainly talk about?A. Kitchen fires inVietnam.B. Traditional customs ofVietnam.C. A report on family activities inVietnam.D. A tale about Vietnamese Lunar New Year.5. What's the author's attitude towards theVietnam's actions to control COVID-19?A. Critical.B. Positive.C. Doubtful.D. Unconcerned.6. What was the change about fish release inVietnamthis year?A. It required few lake fish.B. It was more interesting and diverse.C. It was more environmentally friendly.D. It required more people and plastic bags.7. What can be the best title for the text?A. Vietnamese Respect Gods During the Festive TimeB. Vietnamese Mark Lunar New Year with Fish ReleaseC. Vietnamese Lunar New Year Celebrations Are FunnyD. Traditional Lunar New Year Promotes Animal ProtectionCMany cars in advertisements and on exhibition in the United States are red, blue or green, but almost 75 percent of new cars sold in the United States are black, white, silver orgray.Les Jackson is a reporter who writes about cars. He says the color1 s of cars Americans choose do not show dirt. He says that means the owners wash their cars less in order to save money. And he notes some areas that are suffering from water shortages do not permit people to wash their cars often.Dan Benton works for a company called Axalta, which makes supplies for international car makers. He says white cars are often sold more expensive than cars of other color1 s. And he notes that white cars “absorb(吸收)less energy” than cars of other color1 s. This means temperatures inside them are lower in warmer areas. Benton also says research at Monash University in Australia suggests that there is a lower risk of crashes during the day for white cars compared with darker ones.Car buyers in other countries also like white. Jane Harrington works for PPG Industries, a company that makes paint for cars. She said in China, buyers say white makes a small car look bigger.About 11 percent of cars sold in North America are red and 8 percent are blue. Green has become less popular. Benton notes that in the mid-1990s green was the most popular color1 in North America. Today, green is hard to find.Sometime in the future, people may not have to choose the color1 of their cars —— technology may let owners change their cars’ paint color1 anytime.8. What can we learn from Paragraph 2?A. Most Americans don’t like red cars.B. People in America are not allowed to wash their cars.C. Many people prefer to choose white cars in America.D. Americans may consider the cost of cleaning when choosing cars.9. Why do many people choose white cars?A. They are much cheaper than cars of other color1 s..B. They are much safer while crashing.C. They are bigger than cars of other color1 s.D. They are more comfortable inside in warmer areas.10. What do we know from the text?A. Les Jackson is a member of Axalta.B. Most Americans rarely wash their cars.C. PPG Industries mainly produces cars in China.D. Green cars were once popular in North America.11. What does the text mainly tell us?A. Choices of car color1 sB. How to buy a good car.C. Differences of car color1 s.D. Popular car color1 s in history.DThis year researchers expect the world to snap 1.35 trillion photographs, or about 3.7 billion per day. All those pixels (像素) take up a lot of room if they are stored on personal computers or s phones, which is one reason why many people store their images in the cloud. But unlike a hard on drive which can be encrypted to protect its data, cloud storage users have to trust that a tech platform will keep their private pictures safe. Now a team of Columbia University computer scientists has developed a tool to encrypt (加密) images stored on many popular cloud services while allowing authorized users to browse and display their photographs as usual.Malicious (恶意的) attempts to access or leak cloud-based photographs can expose personal information. In November 2019, for example, a bug in the popular photograph storage app Google Photos mistakenly shared some users' private videos with strangers. Security experts also worry about employees at cloud storage companies on purpose accessing users' images.So the Columbia researchers came up with a system called Easy Secure Photos (ESP), which they presented at a recent conference. “We wanted to see if we could make it possible to encrypt data while using existing services,” says computer scientist Jason Nieh, one of the developers of ESP. “Everyone wants to stay with Google Photos and not have to register on a new encrypted-image cloud storage service.”To overcome this problem, they created a tool that preserves blocks of pixels but moves them around to effectively hide the photograph. First, ESP's algorithm (算法) divides a photograph into three separate files, eachone containing the image's red, green or blue color1 data. Then the system hides the pixel blocks around among these three files (allowing a block from the red file, for instance, to hide out in the green or blue ones). But the program does nothing within the pixel blocks, where all the image processing happens. As a result, the files remain unchanged images but end up looking like grainy black-and-white ones to anyone who accesses them without the decryption (解密) key.12. What's probably the main purpose for people to store images in the cloud?A. To save storage room.B. To make photos beautiful.C. To try a new storage way.D. To keep their privacy safe.13. Why might employees in cloud storage companies be distrusted by experts?A. They sell users' passwords.B. They have invented new tools.C. They often let out personal information.D. They may steal a glance at users' images.14. What's the advantage of ESP?A. It can provide clear images.B. It can decrease the upload time.C. It can classify images automatically.D. It can encrypt data on the original platform.15. What does paragraph 4 mainly talk about?A. Method of decryption.B. Image-processing technique.C. Separate files of images.D. Data analysisof color1 s.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
2019-2020学年重庆市合川中学高三语文三模试卷及答案解析
2019-2020学年重庆市合川中学高三语文三模试卷及答案解析一、现代文阅读（36分）(一)现代文阅读I(9分)阅读下面的文章，完成下面小题。
我远来是为的这一湖水①刚下过一场雨，蒙自的老街湿漉漉的。
一线阳光就在这时划过来，南湖上又是一片光鲜。
如果没有穿梭往来的汽车，会让人想起多少年前的5月，同现在差不多的天气，南湖边霎时出现了一群来自全国各地的学生，男的衣衫整洁，女的裙裾飘摇。
一下子涌来这么多文化人，让人有一种震惊和欣喜，来的是西南联大的学生啊。
一个享誉世界的文化讲堂也在南湖边开启，主讲者有陈寅恪、冯友兰、钱穆、闻一多、朱自清、沈从文……②1938年，在北中国已经放不下一张书桌的情况下，北大、清华、南开三校南迁昆明，组建了西南联大。
由于昆明校舍不敷，边城蒙自便暂时接纳了联大的文学院、法学院。
政府尽心，绅士尽力，把联大师生安顿在风景秀丽的南湖边，让他们住进最好的房子。
③这样，西南联大师生的生活就与美丽的南湖融在了一起。
每天，师生上下课经过南湖东堤，课余在湖边读书、唱歌、诵诗，在湖里畅游，在亭上探讨，青春的气息弥漫水中。
鱼翔浅底，鸟儿扑飞，田田莲叶拨弄着微风。
南湖，一时成了联大师生感情的依托、诗情的沃土。
想起诗人周定一的《南湖短歌》，那首诗感情淋漓，淋漓得让人涌泪：我远来是为的这一园花。
你问我的家吗？我的家在辽远的蓝天下。
我远来是为的这一湖水。
我走得有点累，让我枕着湖水睡一睡。
让湖风吹散我的梦，让落花堆满我的胸，让梦里听一声故国的钟。
……我在这小城里学着异乡话，你问我的家吗？我的家在辽远的蓝天下。
④武庙街的颐楼，是蒙自十分有特色的民居。
楼高势险，古榕成荫，湖光山色，尽收眼底，十分幽雅闲静，被作为了联大女生的宿舍。
入夜，山风刮来，呜呜噓噓，如怨如诉，女生们总是长久不能成眠。
家乡、亲人、故都，无不随风而来，于是，她们将颐楼叫成了“听风楼”。
听风楼，听的是“风在吼，马在叫，黄河在咆哮”吗？听的是“大风起兮云飞扬，威加海内兮归故乡”吗？⑤虽然美丽的南湖给了师生们暂时的宁静，但是在那个烽火连三月的年代，宁静中又奔涌着激情。
重庆市綦江县2019-2020学年第三次高考模拟考试数学试卷含解析
重庆市綦江县2019-2020学年第三次高考模拟考试数学试卷一、选择题：本题共12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()(1)2z i i =++（i 是虚数单位），则复数z 在复平面内对应的点位于（ ）A ．第一象限B ．第二象限C ．第三象限D ．第四象限【答案】A【解析】【分析】将z 整理成a bi +的形式，得到复数所对应的的点，从而可选出所在象限.【详解】解：221()()2313z i i i i i =++=++=+，所以z 所对应的点为()1,3在第一象限. 故选:A.【点睛】本题考查了复数的乘法运算，考查了复数对应的坐标.易错点是误把2i 当成1进行计算.2．已知集合2{|1}A x x =<,2{|log 1}B x x =<，则A ．{|02}AB x x ⋂=<<B ．{|2}A B x x ⋂=<C ．{|2}A B x x ⋃=<D ．{|12}A B x x =-<<U【答案】D【解析】【分析】【详解】因为2{|1}{|11}A x x x x =<=-<<,2{|log 1}{|02}B x x x x =<=<<,所以{|01}A B x x =<<I ,{|12}A B x x =-<<U ,故选D ．3．若函数f(x)＝a |2x －4|(a>0，a≠1)满足f(1)＝19，则f(x)的单调递减区间是( ) A ．(－∞，2]B ．[2，＋∞)C ．[－2，＋∞)D ．(－∞，－2]【答案】B【解析】由f(1)=得a 2=,∴a=或a=-(舍),即f(x)=(.由于y=|2x-4|在(-∞,2]上单调递减,在[2,+∞)上单调递增,所以f(x)在(-∞,2]上单调递增,在[2,+∞)上单调递减,故选B.4．已知函数()2x f x x a =+⋅，()ln 42xg x x a -=-⋅，若存在实数0x ，使()()005f x g x -=成立，则正数a 的取值范围为（ ）A ．(]01，B ．(]04，C ．[)1+∞，D ．(]0,ln2 【答案】A【解析】【分析】 根据实数0x 满足的等量关系，代入后将方程变形0000242ln 5x x a a x x -⋅+⋅=+-，构造函数()ln 5h x x x =+-，并由导函数求得()h x 的最大值；由基本不等式可求得00242x x a a -⋅+⋅的最小值，结合存在性问题的求法，即可求得正数a 的取值范围.【详解】函数()2x f x x a =+⋅，()ln 42x g x x a -=-⋅，由题意得()()0000002ln 425x x f x g x x a x a --=+⋅-+⋅=， 即0000242ln 5x x a a x x -⋅+⋅=+-， 令()ln 5h x x x =+-，∴()111x h x x x -'=-=， ∴()h x 在()01，上单调递增，在()1+∞，上单调递减， ∴()()14max h x h ==，而000024222424x x x x a a a --⋅+⋅≥⋅⋅=，当且仅当00242x x -=⋅，即当01x =时，等号成立，∴44a ≤，∴01a <≤.故选：A.【点睛】本题考查了导数在求函数最值中的应用，由基本不等式求函数的最值，存在性成立问题的解法，属于中档题.5．已知正四面体A BCD -外接球的体积为86π，则这个四面体的表面积为（ ）A ．183B ．163C ．143D ．123【答案】B【解析】【分析】设正四面体ABCD 的外接球的半径R ，将该正四面体放入一个正方体内，使得每条棱恰好为正方体的面对角线，根据正方体和正四面体的外接球为同一个球计算出正方体的棱长，从而得出正四面体的棱长，最后可求出正四面体的表面积．【详解】将正四面体ABCD 放在一个正方体内，设正方体的棱长为a ，如图所示，设正四面体ABCD 的外接球的半径为R ，则34863R ππ=，得6R =．因为正四面体ABCD 的外接球和正方体的外接球是同一个球，则有3a=226R =，∴a=22 ．而正四面体ABCD 的每条棱长均为正方体的面对角线长，所以，正四面体ABCD 的棱长为2a=2224⨯=，因此，这个正四面体的表面积为2341634a ⨯=． 故选：B ．【点睛】本题考查球的内接多面体，解决这类问题就是找出合适的模型将球体的半径与几何体的一些几何量联系起来，考查计算能力，属于中档题．6．已知三棱柱111ABC A B C -的所有棱长均相等，侧棱1AA ⊥平面ABC ，过1AB 作平面α与1BC 平行，设平面α与平面11ACC A 的交线为l ，记直线l 与直线,,AB BC CA 所成锐角分别为αβγ，，，则这三个角的大小关系为( )A ．αγβ>>B ．αβγ=>C ．γβα>>D ．αβγ>=【答案】B【解析】【分析】 利用图形作出空间中两直线所成的角，然后利用余弦定理求解即可.【详解】如图，1111111，D C CC C E AC ==，设O 为11A C 的中点，1O 为11C E 的中点，由图可知过1AB 且与1BC 平行的平面α为平面11AB D ，所以直线l 即为直线1AD ，由题易知，11，D AB O CB ∠∠的补角，1D AC ∠分别为αβγ，，，设三棱柱的棱长为2，在1D AB ∆中，1125225，，D B AB AD === 2212542555cos cos 10102225D AB α+-∠==∴=⨯⨯； 在1O BC ∆中，111125，，O B BC OC = (221541155cos cos 1010225O CB β+-∠==-∴=⨯⨯； 在1D AC ∆中，114225，，CD AC AD ===，155cos cos 25D AC α∠==∴= cos cos cos ，αβγαβγ=<∴=>Q .故选：B【点睛】本题主要考查了空间中两直线所成角的计算，考查了学生的作图，用图能力，体现了学生直观想象的核心素养.7．过椭圆()2222:10x y C a b a b+=>>的左焦点F 的直线过C 的上顶点B ，且与椭圆C 相交于另一点A ，点A 在y 轴上的射影为A '，若34FO AA ='，O 是坐标原点，则椭圆C 的离心率为（ ） A．2 B．3 C ．12 D．2【答案】D【解析】【分析】求得点B 的坐标，由34FO AA ='，得出3BF FA =u u u r u u u r ，利用向量的坐标运算得出点A 的坐标，代入椭圆C 的方程，可得出关于a 、b 、c 的齐次等式，进而可求得椭圆C 的离心率.【详解】由题意可得()0,B b 、(),0F c -. 由34FO AA ='，得34BF BA =，则31BF FA =，即3BF FA =u u u r u u u r . 而(),BF c b =--u u u r ，所以,33c b FA ⎛⎫=-- ⎪⎝⎭u u u r ，所以点4,33b A c ⎛⎫-- ⎪⎝⎭. 因为点4,33b A c ⎛⎫-- ⎪⎝⎭在椭圆2222:1x y C a b +=上，则22224331b c a b ⎛⎫⎛⎫-- ⎪ ⎪⎝⎭⎝⎭+=， 整理可得2216899c a ⋅=，所以22212c e a ==，所以2e =. 即椭圆C故选：D.【点睛】本题考查椭圆离心率的求解，解答的关键就是要得出a 、b 、c 的齐次等式，充分利用点A 在椭圆上这一条件，围绕求点A 的坐标来求解，考查计算能力，属于中等题.8．以下两个图表是2019年初的4个月我国四大城市的居民消费价格指数（上一年同月100=）变化图表，则以下说法错误的是（）（注：图表一每个城市的条形图从左到右依次是1、2、3、4月份；图表二每个月份的条形图从左到右四个城市依次是北京、天津、上海、重庆）A．3月份四个城市之间的居民消费价格指数与其它月份相比增长幅度较为平均B．4月份仅有三个城市居民消费价格指数超过102C．四个月的数据显示北京市的居民消费价格指数增长幅度波动较小D．仅有天津市从年初开始居民消费价格指数的增长呈上升趋势【答案】D【解析】【分析】采用逐一验证法，根据图表，可得结果.【详解】A正确，从图表二可知，3月份四个城市的居民消费价格指数相差不大B正确，从图表二可知，4月份只有北京市居民消费价格指数低于102C正确，从图表一中可知，只有北京市4个月的居民消费价格指数相差不大D错误，从图表一可知上海市也是从年初开始居民消费价格指数的增长呈上升趋势故选：D【点睛】本题考查图表的认识，审清题意，细心观察，属基础题.9．已知a>0，b>0，a+b =1，若α=11a ba bβ+=+，，则αβ+的最小值是（）A．3 B．4 C．5 D．6 【答案】C根据题意，将a 、b 代入αβ+，利用基本不等式求出最小值即可.【详解】∵a>0，b>0，a+b=1， ∴211111152a b a b ab a b αβ+=+++=+≥+=+⎛⎫ ⎪⎝⎭， 当且仅当12a b ==时取“＝”号． 答案：C【点睛】 本题考查基本不等式的应用，“1”的应用，利用基本不等式求最值时，一定要正确理解和掌握“一正，二定，三相等”的内涵：一正是首先要判断参数是否为正；二定是其次要看和或积是否为定值（和定积最大，积定和最小）；三相等是最后一定要验证等号能否成立，属于基础题.10．已知函数()f x 的定义域为[]0,2，则函数()()282x g x f x =+-的定义域为（ ）A ．[]0,1B ．[]0,2 C ．[]1,2D ．[]1,3 【答案】A【解析】 试题分析：由题意，得022{820x x ≤≤-≥，解得01x ≤≤，故选A ． 考点：函数的定义域．11．如图，平面四边形ACBD 中，AB BC ⊥，AB DA ⊥，1AB AD ==，2BC =，现将ABD △沿AB 翻折，使点D 移动至点P ，且PA AC ⊥，则三棱锥P ABC -的外接球的表面积为（ ）A ．8πB ．6πC ．4πD ．823【答案】C由题意可得PA ⊥面ABC ，可知PA BC ⊥，因为AB BC ⊥，则BC ⊥面PAB ，于是BC PB ⊥.由此推出三棱锥P ABC -外接球球心是PC 的中点，进而算出2CP =，外接球半径为1，得出结果.【详解】解：由DA AB ⊥，翻折后得到PA AB ⊥，又PA AC ⊥，则PA ⊥面ABC ，可知PA BC ⊥．又因为AB BC ⊥，则BC ⊥面PAB ，于是BC PB ⊥，因此三棱锥P ABC -外接球球心是PC 的中点．计算可知2CP =，则外接球半径为1，从而外接球表面积为4π．故选：C.【点睛】本题主要考查简单的几何体、球的表面积等基础知识；考查空间想象能力、推理论证能力、运算求解能力及创新意识，属于中档题．12．一小商贩准备用50元钱在一批发市场购买甲、乙两种小商品，甲每件进价4元，乙每件进价7元，甲商品每卖出去1件可赚1元，乙商品每卖出去1件可赚1.8元.该商贩若想获取最大收益，则购买甲、乙两种商品的件数应分别为（ ）A ．甲7件，乙3件B ．甲9件，乙2件C ．甲4件，乙5件D ．甲2件，乙6件 【答案】D【解析】【分析】由题意列出约束条件和目标函数，数形结合即可解决.【详解】设购买甲、乙两种商品的件数应分别x ，y 利润为z 元，由题意*4750,,,x y x y N +≤⎧⎨∈⎩ 1.8z x y =+， 画出可行域如图所示，显然当5599y x z =-+经过(2,6)A 时，z 最大. 故选：D.【点睛】 本题考查线性目标函数的线性规划问题，解决此类问题要注意判断x ，y 是否是整数，是否是非负数，并准确的画出可行域，本题是一道基础题.二、填空题：本题共4小题，每小题5分，共20分。
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2019-2020学年重庆市新高考模拟实战卷（三）
数学
考试时间：120分钟
学校:___________班级:___________姓名:___________学号:___________
一、单项选择题: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.
1.设z的共轭复数为，若，，则在复平面内，复数z对应的点位于
A. 第一象限
B. 第二象限
C. 第三象限
D. 第四象限
z ⎯⎯⎯z +=−1z ⎯⎯⎯z ==i z ⎯⎯⎯2.已知全集，集合，，则图中阴影部分所表示的集合为
A.［-2,1)
B.［-2,1］
C.［-2,0)∪(1,2］
D.［-2,0］∪[1,2]
U =R M ={x |y =}x −1
‾‾‾‾‾√N ={x |−4≤0}x 23.设，，，则实数a，b，c的大小关系正确的是A.a>b>c
B.c>a>b
C.b>c>a
D.c>b>a
a =log 213
b =3－0.5
c =2sin 14.锐角中，角A，B，C所对的边长分别为a，b，c，若bcosC=bsinC=2ccosB=2，则的面积为
A. 3
B.
△ABC △ABC 23‾√
C. D. 6
33‾
√5.设函数与对应的零点分别是，，下面结论正确的是A.B.C.D.y =ln |x |−12e −x y =2x +ln |x |−32
x 1x 2<<0
x 1x 20<<x 1x 2
0<<x 2x 1
<<0
x 2x 16.在对任意的实数x,若［x］表示不超过x的最大整数,例如［0.5］=0，［-1.45］=-2，则“［x］+［y］≤1”是“”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 ［x +［y ≤1］2］2
7.2 000多年前，古希腊数学家最先开始研究圆锥曲线，并获得了大量的成果.对圆锥曲线进行系统研究成就最突出的是古希腊数学家阿波罗尼奥斯.他的著作《圆锥曲线论》与欧几里得的《几何原本》同被誉为古代希腊几何的登峰造极之作.在《圆锥曲线论》中指出：用一个与圆锥的母线平行且不过圆锥顶点的平面截圆锥，截得的交线为抛物线．如图是一个母线长和底面圆直径均为2的木质圆锥模型，点A，B是底面圆周上的两个点，且，过线段AB作与圆锥的母线平行的截面(截面与圆锥轴线所成的角与母线和轴线的夹角相等)，截得的抛物线C的焦准距（焦点到准线的距离）为
A. B. C. D. AB =3‾√14341
2328.定义在(0,1)上的函数恒成立，则a的取值范围是
A. (0,2］
B. ［2,+∞)
C. (－∞,2］
D. (－∞,2) f (x )=lnx −
+1<0ax x +1二、多项选择题:本大题共4小题,每小题5分,共20分.在每小题给出的四个选项中,有多项符合题目要求.全部选对的得5分,选对但不全的得3分,有选错的得0分.
9.为迎接2020年冬奥会，北京市对本市所有小区的交通、安全、环保、卫生、管理五项指标进行检查评分.每项评分最低分0分，最高分100分，每个小区的综合分为这五项得分之和,并规定综合分大于等于300分为合格.为了解各小区迎检的效果，在迎检的小区中随机抽取26个小区进行考核评分.根据考核评分结果，绘制交通得分与安全得分散点图、交通得分与综合分散点图如下，下列各选项中说法正确的是
A.交通得分与小区综合分成正相关
B.交通得分与安全得分的相关系数小于交通得分与小区综合分的相关系数
C.该市26个小区的交通平均得分比安全平均得分低
D.该市26个小区安全得分的中位数估计为80
10.设函数，下列说法正确的是
A.函数f(x)的图象是中心对称图形
B.函数f(x)在(1,+∞)上有最小值1
C.函数f(x+1)－1是一个奇函数
D.对任意的x≠1，都有f (x )=x x −1f (x )+f ()=1
1x 11.如图所示，三棱柱ABC-A'B'C'的底面是边长为2的等边，侧棱AA'⊥底面ABC，点P是棱AC上的动点，直线B'P与平面A'C'B交于点Q，设三棱柱ABC-A'B'C'的高为h，下列说法正确
△ABC
的是
A.点Q是B'P的中点
B.∠A'BC'的范围是(0°,90°］
C.存在点P满足B'P⊥A'C'
D.三棱锥Q-A'B'C'的体积恒为
ℎ3√612.已知椭圆E的方程为，左、右焦点分别为，，点是椭圆E上的一个动
点，下列说法正确的是
A.B.C.D.+=1x 24y 2F 1F 2P (,)x 0y 0|P |≤2+F 13‾
√∠P ≤90°
F 1F 20≤<2x 0y 0→=2−|→||
PF 1−→−PF 2−→−PF 1−→−PF 2−→−三、填空题:本大题共4小题,每小题5分,共20分.把答案填在题中的横线上.13.若展开式中第二项的二项式系数为8，则展开式中所有偶数项的系数的和为____.
(1－x )n 14.已知数列是项数为2 019的一个等差数列，若数列的所有项的
和为673，则数列所有项的和是____.
{}a n {}(1≤k ≤673,k ∈)a 3k －1N ∗{}a n 15.用一个里面填充透明材料的半球形容器和一个正方体大理石做的装饰品如图所示，大理石在
半球内的四个顶点均在容器表面上，且大理石的上表面与容器口所在平面平行.经测量在透明材料外的长方体高为5 cm，上表面的正方形边长为 12 cm，则半球形容器的体积为____．cm 316.如图所示，是边长为2的等边三角形，以顶点C为圆心作圆C与边AB相切，MN是圆C 的一条动直径，点M在点N的左侧.当AM与圆C相切时，的值为____；的最
大值为____．
△ABC ·AM −→−BN −→−·AM −→−BN −→−
四、解答题:本大题共6小题,共70分.解答应写出必要的文字说明、证明过程或演算步骤.
17.在①，②，③这三个条件中任选一个，补充在横线中，并解答下面的问题．
若数列是有限数列，且满足，则称数列为“对称数列”，如项数为 5的一个“对称数列”：-3，1，7，1，-3．已知是一个项数为51的“对称数列”，，等差数列满足____．
Ⅰ.写出数列中满足的项数n的取值范围和数列的通项公式（无需写演算步骤）；Ⅱ.求数列中所有满足的项的和．
=n b n =2n −21b n 4n −163b ={}a n =（i ∈，1≤i ≤n ）a i a n +1－i N +{}a n a n =4n −43（1≤n ≤26）a n b n （n ∈）N +{}a n ≥0a n {}a n a n ≥a n b n 18.设，，函数f（x）是偶函数．
Ⅰ.求φ的值；
Ⅱ.求函数的值域．
f (x )=2sin φcos )x +φ)+sinx ，x ∈R φ∈[0,π)−[f (x +)]π122[f (x −)]π4219.如图1，点E，F分别是边长为4的正方形ABCD边AB，BC的中点，沿EF减去得到一个五边形AEFCD，然后沿AC将该五边形折起，得到如图2所示的四棱锥D-AEFC，且.Ⅰ.在图2中，过点F作平面α，满足平面α⊥CF，画出平面α与平面AEFC和平面ACD的交线，证平面α⊥CF；
Ⅱ.求平面CFD和平面AED所成的锐二面角的余弦值．
△EFB DF =23‾√20.某市教育局为了解直属学校党员同志学习“学习强国”的效果，举行“挑战答题”竞赛活动．为了解本次竞赛成绩的情况，从中随机抽取了50名党员同志的成绩（单位：分，满分为100分）作为样本进行统计，得到如图1所示的频率分布直方图，并发现这50人中，其成绩低于 70分的有24人，这 24人成绩数据茎叶图如图2所示，以样本的频率作为总体的概率．
Ⅰ.求出频率分布直方图中的a，b，c的值；
Ⅱ.在抽取的样本中，从竞赛成绩在80分以上（含 80分）的党员同志中随机抽取2名党员，求所抽取的 2名党员来自同一分数段的概率；
Ⅲ.在(Ⅱ)的条件下，用X表示所抽取的2名党员竞赛成绩在90分以上（含90分），求X的分布列及数学期望．
21.已知圆C上的一段弧长为，这段弧所对的圆心角为60°，圆心C的坐标是（3,1）.Ⅰ.求圆C的标准方程；
Ⅱ.设点A（1,7），B（5,9），点P是圆C上的一个动点，求使最小的点P的坐标．π3
|PA +|PB |2|222.已知函数，．Ⅰ.设h（x）=f（x）-g（x），求函数h（x）的单调区间；
Ⅱ.点，是函数f（x）图象上的两个动点，，O为直角坐标系的原点，设直线OP，OQ和PQ的斜率分别为，和k，证明：.
f (x )=xlnx
g (x )=12
x 2P （,）x 1y 1Q （,）x 2y 2>x 2x 1k 1k 2+1<k <+1k 1k 2。