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2024届广东省惠州市惠阳高级中学全国高三模拟考试(六)数学试题

2024届广东省惠州市惠阳高级中学全国高三模拟考试(六)数学试题

2024届广东省惠州市惠阳高级中学全国高三模拟考试(六)数学试题请考生注意:1.请用2B 铅笔将选择题答案涂填在答题纸相应位置上,请用0.5毫米及以上黑色字迹的钢笔或签字笔将主观题的答案写在答题纸相应的答题区内。

写在试题卷、草稿纸上均无效。

2.答题前,认真阅读答题纸上的《注意事项》,按规定答题。

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. “角谷猜想”的内容是:对于任意一个大于1的整数n ,如果n 为偶数就除以2,如果n 是奇数,就将其乘3再加1,执行如图所示的程序框图,若输入10n =,则输出i 的( )A .6B .7C .8D .92.函数()1ln1xf x x-=+的大致图像为( ) A . B .C .D .3.已知双曲线2222:10,0()x y C a b a b-=>>的左、右顶点分别为12A A 、,点P 是双曲线C 上与12A A 、不重合的动点,若123PA PA k k =, 则双曲线的离心率为( ) A .2B .3C .4D .24.已知随机变量X 的分布列是X12 3P1213a则()2E X a +=( ) A .53B .73C .72D .2365.对于函数()f x ,若12,x x 满足()()()1212f x f x f x x +=+,则称12,x x 为函数()f x 的一对“线性对称点”.若实数a 与b 和+a b 与c 为函数()3xf x =的两对“线性对称点”,则c 的最大值为( )A .3log 4B .3log 41+C .43D .3log 41-6.如下的程序框图的算法思路源于我国古代数学名著《九章算术》中的“更相减损术”.执行该程序框图,若输入的a ,b 分别为176,320,则输出的a 为( )A .16B .18C .20D .157.若x ,y 满足约束条件40,20,20,x y x x y -+≥⎧⎪-≤⎨⎪+-≥⎩且z ax y =+的最大值为26a +,则a 的取值范围是( )A .[1,)-+∞B .(,1]-∞-C .(1,)-+∞D .(,1)-∞-8.已知12,F F 是椭圆和双曲线的公共焦点,P 是它们的-一个公共点,且1223F PF π∠=,设椭圆和双曲线的离心率分别为12,e e ,则12,e e 的关系为( ) A .2212314e e += B .221241433e e += C .2212134e e += D .221234e e +=9.《九章算术》是我国古代内容极为丰富的数学名著,书中有如下问题:“今有刍甍,下广三丈,袤四丈,上袤二丈,无广,高二丈,问:积几何?”其意思为:“今有底面为矩形的屋脊状的楔体,下底面宽3丈,长4丈,上棱长2丈,高2丈,问:它的体积是多少?”已知l 丈为10尺,该楔体的三视图如图所示,其中网格纸上小正方形边长为1,则该楔体的体积为( )A .10000立方尺B .11000立方尺C .12000立方尺D .13000立方尺10.若复数z 满足(1)12i z i +=+,则||z =( )A .22B .32C .102D .1211.一个正三角形的三个顶点都在双曲线221x ay +=的右支上,且其中一个顶点在双曲线的右顶点,则实数a 的取值范围是( ) A .()3,+∞B .)3,+∞C .(,3-∞-D .(),3-∞-12.已知向量(2,4)a =-,(,3)b k =,且a 与b 的夹角为135︒,则k =( ) A .9-B .1C .9-或1D .1-或9二、填空题:本题共4小题,每小题5分,共20分。

2023届辽宁省锦州市渤海大学附属高级中学高三第六次模拟考试英语试题(含解析)

2023届辽宁省锦州市渤海大学附属高级中学高三第六次模拟考试英语试题(含解析)

2023届辽宁省锦州市渤海大学附属高级中学高三第六次模拟考试英语试题学校:___________姓名:___________班级:___________考号:___________一、阅读理解Stuck at home with nothing to watch? Curious about China, but don’t know where to begin? Well, we’ve got you covered with this brand-new video series exploring Chinese culture. All you have to do is press play.● Dazu Rock CarvingsOne of China’s UNESCO World Heritage sites is hidden among the mountains on the outskirts of the southwestern city of Chongqing. Here, tens of thousands of sculptures collectively make up the Dazu Rock Carvings — considered one of the finest examples of China’s cave art, exhibiting the most sophisticated craftsmanship of Dazu Rock Carvings. The delicate Dazu Rock Carvings tell ancient, mystical stories.● Bamboo CarvingThe art of bamboo carving originated at the Qing imperial court. Today, the art lives on, appreciated for its historical value and elegance.Just 100 kilometers southwest of Beijing, in the Xiong’an New Area, a group of skilled artists are keeping this art alive. Carving bamboo requires focus, precision and elbow grease (重活). Watch the video and feast your eyes on the bamboo masterpieces.● Peach-stone CarvingIn the small town of Siyang in Jiangsu Province in eastern China, craftsmen create art on the tiny cores of the peaches.While most people think nothing of peach stones and just throw them away, they are a source of inspiration for these artists. The artist goes to great length to find the perfect core to fit his concept. Using tiny tools, they chip away at the cores to create intricate designs,Chinese people in the near future.4.Why does the author quote Leung Long-kong in Paragraph 2?A.To call on Chinese women to wear qipao in everyday life.B.To introduce the development of qipao in China.C.To emphasize the importance of qipao in China nowadays.D.To show that qipao is no longer as popular as it was.5.Which of the following best describes Mary Yu?A.Creative.B.Conservative.C.Cautious.D.Considerate 6.What does the underlined word “evolving” in Paragraph 7 refer to?A.passing B.withdrawing C.developing D.following 7.What can we infer from the passage about qipao?A.Qipao is an iconic sign in the fashion industry.B.Qipao is on its way back to the daily life of Chinese.C.Qipao enjoys a good reputation in the world.D.Qipao is seen as a symbol of wealth in modern China.What’s more important in determining life success-book smarts or street smarts? This question gets at the heart of an important debate contrasting the relative importance of cognitive(认知的)intelligence (CI) and emotional intelligence (EI).Cognitive intelligence is still recognized as an important element of success, particularly when it comes to academic achievements. People with high cognitive intelligence typically do well in school, often earn more money, and tend to be healthier in general.But today experts recognize that cognitive intelligence is not the only determining factor of life success. Instead, it is part of a complex range of influences-one that includes emotional intelligence. Many companies now provide emotional intelligence training and use emotional intelligence tests as part of the hiring process. Research has found that individuals with strong leadership potential also tend to be more emotionally intelligent, suggesting that high emotional intelligence is an important equality for business leaders and managers. According to a survey of hiring managers, almost 75% of the responders suggested that they valued an employee’s emotional intelligence more than his cognitive intelligence.Now that emotional intelligence is so important, can it be taught or strengthened? According to one meta-analysis that looked at the results of social and emotional learningprogrammes, the answer to that question is definitely yes. Strategies for teaching emotional intelligence include character education, modeling positive behaviours, encouraging people to think about how others are feeling, and finding ways to be more empathetic(感同身受的)towards others.All in all, life success is a result of many factors. Both cognitive intelligence and emotional intelligence play roles in overall success, as well as health, wellness, and happiness. Rather than focusing on which factors have a prior influence, the greatest benefit may lie in learning to improve skills in multiple areas. In addition to strengthening cognitive abilities, such as memory and mental focus, you can also acquire and improve social and emotional skills.8.What can we know about people with book smarts?A.They can debate with other people.B.They can deal with various situations.C.They can be outstanding in academic research.D.They can be good at gaining real life experience.9.Why does the author mention the data in Paragraph 3?A.To indicate the strictness of the hiring process.B.To prove the importance of emotional intelligence.C.To explain the result of emotional intelligence tests.D.To show the influence of cognitive intelligence on success.10.What can be learned concerning emotional intelligence?A.Evaluating how others feel.B.One’s extreme behavioursC.One’s academic performance.D.Controlling others’ emotions. 11.Which of the following is a suitable title for the text?A.Does book smarts matter?B.Is CI or El more important?C.What counts most in life?D.Mental health or physical health?New research suggests that a gene that governs the body’s biological (circadian) clock acts differently in males versus females and may protect females from heart disease. The study is the first to analyse circadian blood pressure rhythms(节奏)in female mice.The body’s circadian clock-the biological clock that organizes bodily activities over a 24-hour period—contributes to normal variations in blood pressure and heart function overthe course of the day. In most healthy humans, blood pressure dips(下降)at night. People who do not experience this temporary drop, called “non-dippers”, are more likely to develop heart disease. The circadian clock is made up of four main proteins (encoded by “clock genes”) that regulate close to half of all genes in the body, including those important for blood pressure regulation.Previous research has shown that male mice that are missing one of the four clock genes (PER1) become non-dippers and have a higher risk for heart and kidney disease. A research team studied the circadian response and blood pressure of female mice that lack PERI and compared them with a healthy female control group. On both low-and high-salt diets, both groups “kept an apparent circadian rhythm” of blood pressure, the researchers explained. Unlike the male mice in previous research, the females without PERI showed normal dips in blood pressure overnight.These results suggest that the lack of PER1 acts differently in males and females. The findings are consistent with research showing that women are less likely to be non-dippers than men of the same age. “This study represents an important step in understanding sex differences in the regulation of cardiovascular(心血管的)function by the circadian clock,” the researchers wrote.12.What does the new research find?A.Biological clock may protect males from heart disease.B.Biological blood pressure rhythms in female mice act normally.C.Biological clock organizes bodily activities over a 24-hour period.D.A gene controlling biological clock works differently between sexes.13.What role can circadian clock play according to the text?A.Helping males cure heart disease.B.Helping blood pressure vary normally.C.Contributing to abnormal variations in blood pressure.D.Making up four main proteins regulating almost half of all genes.14.The lack of PRRI ______A.has the same impact on males and femalesB.makes no difference to malesC.does harm to male’s healthD.is more likely to affect female’s health15.What would be a suitable title for the text? A.One clock gene is importantB.Women may benefit from body clock C.New study analyses blood pressure rhythms D.Blood pressure of healthy humans dips at night二、七选五E.Most universities are considered to perform better in at least one academic field.F.Therefore, research each university and its course guide in as much detail as possible.G.One of the biggest deciding factors in choosing a university is where you want to study.三、完形填空27.A.target B.turn C.block D.appointment 28.A.happens B.remains C.offers D.works 29.A.celebrating B.communicating C.arguing D.competing 30.A.encourage B.help C.defeat D.please 31.A.college B.destination C.business D.competition 32.A.next to B.behind C.around D.ahead of 33.A.beautiful B.grateful C.careful D.successful 34.A.advantage B.competitor C.position D.challenge 35.A.Live B.Drive C.Run D.Ride五、其他应用文46.你校将举办英语演讲比赛。

湖北省荆州市名校2024学年高三冲刺模拟(6)数学试题

湖北省荆州市名校2024学年高三冲刺模拟(6)数学试题

湖北省荆州市名校2024学年高三冲刺模拟(6)数学试题注意事项1.考试结束后,请将本试卷和答题卡一并交回.2.答题前,请务必将自己的姓名、准考证号用0.5毫米黑色墨水的签字笔填写在试卷及答题卡的规定位置. 3.请认真核对监考员在答题卡上所粘贴的条形码上的姓名、准考证号与本人是否相符.4.作答选择题,必须用2B 铅笔将答题卡上对应选项的方框涂满、涂黑;如需改动,请用橡皮擦干净后,再选涂其他答案.作答非选择题,必须用05毫米黑色墨水的签字笔在答题卡上的指定位置作答,在其他位置作答一律无效. 5.如需作图,须用2B 铅笔绘、写清楚,线条、符号等须加黑、加粗.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.已知实数0a b <<,则下列说法正确的是( ) A .c c a b> B .22ac bc < C .lna lnb <D .11()()22ab<2.i 为虚数单位,则32i 1i-的虚部为( )A .i -B .iC .1-D .13.复数12iz i=+的共轭复数在复平面内所对应的点位于( ) A .第一象限B .第二象限C .第三象限D .第四象限4.如图,在ABC 中,,(,),2AD AB BD xAB yAC x y R AD ⊥=+∈=,且12AC AD ⋅=,则2x y +=( )A .1B .23-C .13-D .34-5.在311(21)x x ⎛⎫++ ⎪⎝⎭展开式中的常数项为( ) A .1B .2C .3D .76.在满足04i i x y <<≤,i i y xi i x y =的实数对(),i i x y (1,2,,,)i n =⋅⋅⋅⋅⋅⋅中,使得1213n n x x x x -++⋅⋅⋅+<成立的正整数n 的最大值为( ) A .5 B .6C .7D .97.已知复数为纯虚数(为虚数单位),则实数( )A .-1B .1C .0D .28.中国古典乐器一般按“八音”分类.这是我国最早按乐器的制造材料来对乐器进行分类的方法,最先见于《周礼·春官·大师》,分为“金、石、土、革、丝、木、匏(páo )、竹”八音,其中“金、石、木、革”为打击乐器,“土、匏、竹”为吹奏乐器,“丝”为弹拨乐器.现从“八音”中任取不同的“两音”,则含有打击乐器的概率为( ) A .314B .1114C .114D .279.已知0a b >>,则下列不等式正确的是( ) Ab a <Bb a >C .abe b e a -<- D .abe b e a ->-10.已知定义在[)0,+∞上的函数()f x 满足1()(2)2f x f x =+,且当[)0,2x ∈时,2()2f x x x =-+.设()f x 在[)22,2n n -上的最大值为n a (*n N ∈),且数列{}n a 的前n 项的和为n S .若对于任意正整数n 不等式()129n k S n +≥-恒成立,则实数k 的取值范围为( )A .[)0,+∞B .1,32⎡⎫+∞⎪⎢⎣⎭C .3,64⎡⎫+∞⎪⎢⎣⎭D .7,64⎡⎫+∞⎪⎢⎣⎭11.已知1F 、2F 是双曲线22221(0,0)x y a b a b-=>>的左右焦点,过点2F 与双曲线的一条渐近线平行的直线交双曲线另一条渐近线于点M ,若点M 在以线段12F F 为直径的圆外,则双曲线离心率的取值范围是( ) A .(2,)+∞B.2)C.D.12.已知集合{|24}A x x =-<<,集合2560{|}B x x x =-->,则A B =A .{|34}x x <<B .{|4x x <或6}x >C .{|21}x x -<<-D .{|14}x x -<<二、填空题:本题共4小题,每小题5分,共20分。

2021-2022年高三下学期第六次模拟考试数学(理)试题含答案

2021-2022年高三下学期第六次模拟考试数学(理)试题含答案

2021年高三下学期第六次模拟考试数学(理)试题含答案一、选择题(本大题包括12小题,每小题5分,共60分,) 1.集合,,则( )A 、B 、C 、D 、 2.若复数,其中是虚数单位,则复数的模为 A . B .C .D .23.某学生在一门功课的22次考试中,所得分数如下茎叶图所示,则此学生该门功课考试分数的极差与中位数之和 为A .117B .118C .118.5D .119.5 4.已知,函数在上单调递减.则的取值范围是() A. B. C. D. 5.数列的前n 项和为,若,则( ) A. B. C.D.6.若程序框图如图所示,则该程序运行后输出的值是 A .B .C .D .7.设函数()log (01)a f x x a =<<的定义域为,值域为,若的最小值为,则实数a 的值为 A .B .或C .D .或8.设x ∈R ,向量a =(2,x ),b =(3,-2),且a ⊥b ,则|a -b |=A .5B .C .2D .6 9.二项式展开式中的系数是( )A .-14B .14C .-28D .28 10.在△ABC 中,若,,则b=( ) A .3 B .4 C.5 D .611.设函数11,(,2)()1(2),[2,)2x x f x f x x ⎧--∈-∞⎪=⎨-∈+∞⎪⎩,则函数的零点的个数为开始否 n =3n +1n 为偶数k =k +1 结束n =5,k =0 是 输出k n 否是A .4B .5C .6D .712.已知双曲线上一点,过双曲线中心的直线交双曲线于两点,记直线的斜率分别为,当最小时,双曲线离心率为( ) A . B . C D二、填空题(本大题包括4小题,每小题5分,共20分). 13.—个几何体的三视图如图所示(单位:m )则该几何体的体积为___.14.若整数..满足0700y x x y x -≥⎧⎪+-≤⎨⎪≥⎩,则的最大值为 . 15.向平面区域}10,20|),{(≤≤≤≤y x y x .内随机投入一点,则该点落在曲线⎪⎩⎪⎨⎧≤<-≤≤=)21(2)10(23x x x x y 下方的概率等于_______.16.若一个棱锥的底面是正多边形,并且顶点在底面的射影是底面的中心,这样的棱锥叫做正棱锥.已知一个正六棱锥的各个顶点都在半径为3的球面上,则该正六棱锥的体积的最大值为_____.三、解答题(本大题包括6小题,共70分,解答应写出文字说明,证明过程或演算步骤). 17.(本小题满分12分)已知二次函数的图像经过坐标原点,其导函数为,数列的前项和为,点均在函数的图像上. (Ⅰ)求数列的通项公式;(Ⅱ)设是数列的前项和, 求使得对所有都成立的最小正整数18.(本小题满分12分) A 、B 两个投资项目的利润率分别为随机变量X 1和X 2.根据市场分析,X 1和X 2的分布列分别为X 1 5% 10% P0.80.2X 2 2% 8% 12% P0.20.50.3(Ⅰ)在两个项目上各投资100万元,Y 1和Y 2分别表示投资项目A 和B 所获得的利润,求方差DY 1,DY 2;(Ⅱ)将万元投资A 项目,万元投资B 项目,表示投资A 项目所得利润的方差与投资B 项目所得利润的方差的和.求的最小值,并指C 1B 1A 1出x 为何值时,取到最小值.(注:)19.(本小题满分12分) 如图,在三棱柱中,侧面底面,, ,,为中点. (Ⅰ)证明:平面;(Ⅱ)求直线与平面所成角的正弦值;(Ⅲ)在上是否存在一点,使得平面?若存在,确定点的位置;若不存在,说明理由 20.(本小题满分12分)已知两定点,和定直线l :,动点在直线上的射影为,且. (Ⅰ)求动点的轨迹的方程并画草图;(Ⅱ)是否存在过点的直线,使得直线与曲线相交于, 两点,且△的面积等于?如果存在,请求出直线的方程;如果不存在,请说明理由 21.(本小题满分12分)已知函数,且.(Ⅰ)若曲线在点处的切线垂直于轴,求实数的值;(Ⅱ)当时,求函数的最小值;(Ⅲ)在(Ⅰ)的条件下,若与的图像存在三个交点,求的取值范围请考生在第22、23、24题中任选一...题.作答,如果多做,按所做第1题计分。

海南省2022-2023学年高三下学期高考全真模拟卷(六)语文试题(含答案)

海南省2022-2023学年高三下学期高考全真模拟卷(六)语文试题(含答案)

2022—2023学年海南省高考全真模拟卷(六)语文1.本试卷满分150分,测试时间150分钟,共8页。

2.考查范围:高考全部内容。

一、现代文阅读(35分)(一)现代文阅读Ⅰ(本题共5小题,19分)阅读下面的文字,完成1~5题。

材料一:作家梁晓声说,我想把从前的故事讲给年轻人听,让他们知道从前的中国是什么样子。

原著者以文学语言叙说了中国父辈的故事。

然后,我们在李路导演的电视剧《人世间》中,看到中国从苦难走向变革的半个世纪,看到颠簸于历史浪涛的百姓生活,历经悲欢离合。

小说和电视剧《人世间》的成功都源于尊重生活。

《人世间》叙说的是双重生活:一是中国的社会生活。

中国当代社会生活波澜壮阔,少有小桥流水与田园平静;它总是深刻地影响着所有人的命运,少有人能超然度外、徜徉桃花源。

《人世间》的故事从1969年展开,许多青年人的命运悄然扭转。

接着,它依次表现了恢复高考、知青返城、国企改革、经商热潮、棚区改建等重要社会事件。

这些事件是中国社会走过的路标,构成当代社会史,也构成中国百姓命运史。

中国社会是波涛汹涌的大海,每个人是颠簸于其间的小船,或者是一个不带救生圈的泳者,同呼吸共命运。

二是个人的日常生活。

社会生活背景是辽阔的,但编导紧紧聚焦于普通百姓的生活命运,没有戏剧化的强情节推动,更多的是生活琐碎的细节。

一个东北的工人家庭,一群“光字片”的棚区人,一出烟火气扑鼻而来的百姓生活之剧,就此徐徐展开。

比如开篇的第一集,周家面临着上山下乡的抉择。

周秉义下乡离家,周蓉不辞而别,五口之家就此分处三省四地。

父亲拿着洗印好的全家福照片说:“这或许是我们全家最后一张全家福了,难了!”社会生活是风,个人生活是草。

风吹草动,即通过细致描写个人命运的“草动”,来折射时代生活的“风吹”,这是编导对中国百姓生活的理解,也是整部《人世间》的艺术逻辑。

电视剧《人世间》采用家庭叙事结构。

家庭是中国社会生活的微型标本,蕴藏着中国人的生活哲理和情感。

2018-2019年高中化学重庆高三高考模拟测试试题【6】含答案考点及解析

2018-2019年高中化学重庆高三高考模拟测试试题【6】含答案考点及解析

2018-2019年高中化学重庆高三高考模拟测试试题【6】含答案考点及解析班级:___________ 姓名:___________ 分数:___________1.答题前填写好自己的姓名、班级、考号等信息2.请将答案正确填写在答题卡上一、选择题1.下列实验误差分析错误的是A.用润湿的pH试纸测稀碱溶液的pH,测定值偏小B.用容量瓶配制溶液,定容时俯视刻度线,所配溶液浓度偏小C.滴定前滴定管内无气泡,终点读数时有气泡,所测体积偏小D.测定中和反应的反应热时,将碱缓慢倒入酸中,所测温度值偏小【答案】B【解析】试题分析:A项润湿pH试纸,等于稀释溶液,碱溶液稀释后碱性减小,测定值偏小,正确;B项用容量瓶配制溶液,定容时俯视刻度线,溶液的液面低于刻度线,则所配溶液浓度偏大,错误;C项滴定前滴定管内无气泡,终点读数时有气泡,所测体积偏小,正确;D项测定中和反应的反应热时,操作要快,若将碱缓慢倒入酸中,热量损失多,则所测温度值偏小,正确。

考点:考查化学实验误差分析。

2.我国科学家在世界上第一次为一种名为“钴酞菁”的分子(直径为1.3纳米)恢复了磁性,“钴酞菁”分子结构和性质与人体的血红素及植物体内的叶绿素非常相似。

下列关于“钴酞菁”分子的说法中正确的是( )A.在水中所形成的分散系属悬浊液B.分子直径比Na+小C.在水中形成的分散系能产生丁达尔效应D.“钴酞菁”分子不能透过滤纸【答案】C【解析】试题分析:根据该种物质的分子直径判断,该物质在水中形成的分散系属于胶体,能产生丁达尔效应,能透过滤纸,分子直径大于钠离子直径,所以答案选C。

考点:考查对分散系的判断及胶体的性质3.为证明稀硝酸与铜反应产物中气体为NO ,设计如图实验(实验过程中活塞2为打开状态),下列说法中不正确的是A .关闭活塞1,加入稀硝酸至液面a 处B .在装置左侧稍加热可以加快稀硝酸与铜的反应速率C .通过关闭或开启活塞1可以控制反应的进行D .反应开始后,胶塞下方有无色气体生成,还不能证明该气体为NO 【答案】A【解析】如果关闭活塞1,稀硝酸加到一定程度后,左侧液面将不再升高,即不可能加到液面a 处,A 错误。

理科综合-【名校面对面】河南省三甲名校2024届高三校内模拟试题(六)-高中物理

理科综合-【名校面对面】河南省三甲名校2024届高三校内模拟试题(六)-高中物理

理科综合-【名校面对面】河南省三甲名校2024届高三校内模拟试题(六)-高中物理一、单选题 (共6题)第(1)题假设摩托艇受到的阻力的大小正比于它的速率。

如果摩托艇发动机的输出功率变为原来的2倍,则摩托艇的最大速率变为原来的( )A.4倍B.2倍C.倍D.倍第(2)题在如图所示的图像中,直线为某电源的路端电压与电流的关系图线,曲线为某一小灯泡的伏安特性曲线,曲线与直线的交点坐标为,该点的切线与横轴的交点坐标为,用该电源直接与小灯泡连接成闭合电路,由图像可知( )A.电源电动势为B.电源内阻为C.小灯泡接入电源时的电阻为D.小灯泡实际消耗的电功率为第(3)题如图所示,水平地面上放一质量为M的落地电风扇,一质量为m的小球固定在叶片的边缘,启动电风扇小球随叶片在竖直平面内做半径为r的圆周运动。

已知小球运动到最高点时速度大小为v,重力加速度大小为g,则小球在最高点时地面受到的压力大小为( )A.B.C.D.第(4)题如图所示为演示光电效应的实验装置,用光强相同、频率差距较大的单色光a、b分别照射光电管的K极,得到a的反向遏止电压比b的大,则两种单色光分别实验时,得到的电流的示数I和对应电压表的示数U关系图像,正确的是( )A.B.C.D.第(5)题一列简谐波在两时刻的波形如图中实践和虚线所示,由图可确定这列波的A.周期B.波速C.波长D.频率第(6)题如图所示,理想变压器输入端a、b间接入电压有效值恒定的交变电流,、为定值电阻,灯泡、的阻值恒定。

在滑动变阻器R的滑片从上端滑到下端的过程中,两个灯泡始终发光且工作电压在额定电压以内,则下列说法正确的是( )A.变亮,变亮B.变亮,变暗C.变暗,变暗D.变暗,变亮二、多选题 (共4题)第(1)题我国第三艘航母福建舰已正式下水,如图甲所示,福建舰配备了目前世界上最先进的电磁弹射系统。

图乙是一种简化的电磁弹射模型,直流电源的电动势为E,电容器的电容为C,两条相距L的固定光滑导轨,水平放置处于磁感应强度B的匀强磁场中。

2024届海南省高三高考全真模拟卷(六)地理试题及答案

2024届海南省高三高考全真模拟卷(六)地理试题及答案

2023-2024学年海南省高考全真模拟卷(六)地理1.本试卷满分100分,测试时间90分钟,共8页。

2.考查范围:高考全部内容。

一、选择题:本题共15小题,每小题3分,共45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。

自然地理环境作为人类活动生存的场所,对服饰产生重要影响。

我国安徽某地以山地丘陵地形为主,服饰种类繁多,人们上山时常穿用多层粗布密针纳成的长及膝盖的袜子。

据此完成1~2题。

1.为适应当地环境,该地裤装应()A.上下均紧B.上下均宽C.下宽上紧D.上宽下紧2.厚实的袜子长及膝盖主要是为了()①防止蛇虫叮咬②防止荆棘刮伤③保持腿部干燥④保持腿部温度A.①②B.②③C.①③D.③④区域经济发展在随时间演化过程中,根据区域经济发展的集聚程度和平衡程度可分为四种模式。

图1为区域经济协调发展状况的四种模式图。

据此完成3~4题。

图13.区域经济协调发展状况的四种模式一般的演变顺序是()A.①②③④B.②③①④C.③④②①D.④②③①4.与③模式相比,②模式()A.区域内部收入差异较大B.企业类型和数量比较多C.不同企业规模相差不大D.区域内部发展效率较低石林彝族自治县位于云南省昆明市东南远郊,喀斯特地貌发育典型。

近年来其生产空间、生活空间和生态空间“三生”空间用地转移频繁,对当地社会经济和环境都产生了一定的影响。

表1为该地2015-2020年“三生”空间用地转移面积表。

据此完成5~6题。

表1转移面积/km22015-2020年生产空间生活空间生态空间转出合计生产空间—12.4630.7343.19生活空间 2.76— 1.01 3.77生态空间45.81 1.35—47.16转入合计48.5713.8131.74—5.石林彝族自治县“三生”空间用地转换特征为()A.生活空间变化最大B.生态空间转入转出平衡C.生产空间增长最多D.“三生”空间动态转移多6.石林彝族自治县“三生”空间用地转移会使()A.地表景观破碎化减轻B.生态脆弱性增强C.地表景观多样化下降D.生物多样性增加锋的移动会引起锋前、锋后三小时内有明显变压(前后两特定时间的气压变化,如后一时刻的气压数值减前一时刻的气压数值的差值,大于零时叫正变压,小于零时叫负变压)。

2024年“圆梦杯”高三统一模拟考试(六)数学试卷

2024年“圆梦杯”高三统一模拟考试(六)数学试卷

数学试题 第 1 页(共 4 页)2024年普通高等学校招生“圆梦杯”统一模拟考试(六)数 学注意事项:1.答卷前,考生务必将自己的姓名、准考证号填写在答题卡上。

2.回答选择题时,选出每小题答案后,用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,再选涂其他答案标号。

回答非选择题时,将答案写在答题卡上,写在本试卷上无效。

3.考试结束后,将本试卷和答题卡一并交回。

一、选择题: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.已知3i z =−,则||z = A .0BCD2.若2()(1)2f x x ax =−+为偶函数,则a = A .2−B .1−C .0D .13.已知某商场在上半年的六个月中,每个月的销售额y (万元)与月份x (126,,,x = )满足线性回归方程ˆˆ12278..yx =+,则该商场上半年的总销售额为 A .160万元 B .176万元 C .180万元 D .192万元4.命题p :0xy >;命题q :||||||x y x y +=+,则命题p 是命题q 的 A .充要条件B .充分不必要条件C .必要不充分条件D .既不充分也不必要条件5.已知函数()ln f x x ax =+在区间(1,2)单调递增,则a 的最小值为 A .1−B .12−C .12D .16.在ABC △中,角A ,B ,C 所对的边分别为a ,b ,c ,ABC △的面积为S .若4a =,且2tan S A =,则22b c += A .24B .27C .32D .36数学试题 第 2 页(共 4 页)7.设椭圆2221(0)9:x y C b b +=>的离心率为23,左、右焦点分别为1F ,2F ,点A 在C上.若211cos 4AF F ∠=,则12AF F △的面积为 ABCD.8.已知集合{|010π}A θθ=<<,7π{|sin 3cos()}6B θθθ==+,则集合A B 中的元素个数为 A .15B .20C .30D .40二、选择题:本题共4小题,每小题5分,共20分。

2023年湖南省株洲市攸县一中高三化学模拟试卷(六)+答案解析(附后)

2023年湖南省株洲市攸县一中高三化学模拟试卷(六)+答案解析(附后)

2023年湖南省株洲市攸县一中高三化学模拟试卷(六)1. 2021年末,詹姆斯.韦伯红外线太空望远镜搭载火箭发射升空,将奔赴远离地球150万公里的第二拉格朗日点,肩负起观测宇宙形成后最初出现的星系、搜寻地外生命迹象等重任。

下列有关叙述错误的是( )A. 主镜材料为密度小,性能好的金属铍,表面的金涂层可提高红外线反射率B. 望远镜需要避免太阳辐射,制作遮阳帆的聚酰亚胺薄膜属于有机高分子材料C. 望远镜需要超低温环境,主动冷却器用氦制冷,氦的分子间作用力很弱,沸点极低D.望远镜工作的推进剂为和,本身的颜色会随温度升高变为红棕色2. 设为阿伏加德罗常数的值。

下列说法正确的是( )A.中含有的质子数目为B. 铅蓄电池中,当正极理论上增加时,电路中通过的电子数目为C.的溶液中,阴离子总数小于D.常温常压下,中所含的分子数目为3. 下列有关海水及海产品综合利用的离子方程式不正确的是( )A. 海水提镁中用石灰乳沉镁:B. 海水提溴中用水溶液富集溴:C. 氯碱工业中电解饱和食盐水:D. 海产品中提取碘单质:4. 炼铁工业中高炉煤气的一种新的处理过程如图所示,有关该过程的叙述错误的是( )A.可减少的排放 B. 和CaO均可循环利用C. 过程①和④的转化均为氧化还原反应D. 过程①的反应中有非极性键的断裂和生成5. 化学是实验的科学,下列有关实验设计能达到目的的是部分夹持装置已略去( )A B C D进行喷泉实验可吸收多余的HCl气体测定化学反应速率配制稀硫酸溶液A. AB. BC. CD. D6. 有机化合物X 与Y在一定条件下可反应生成Z,反应方程式如图:下列有关说法正确的是( )A. 有机物X 与Y生成Z的反应属于取代反应B. 有机物Y 分子中最多有9个原子在同一平面上C. 1molZ的芳香族同分异构体化合物可能与2molNa反应D. Z在酸性条件下水解生成和7.《Science》杂志报道了王浩天教授团队发明的制取的绿色方法,原理如图所示。

辽宁省沈阳市东北育才学校科学高中部2023-2024学年高三下学期第六次模拟考试数学含答案

辽宁省沈阳市东北育才学校科学高中部2023-2024学年高三下学期第六次模拟考试数学含答案

2023-2024学年度高三年级下学期第六次模拟考试数学学科试卷命题人:佟国荣刘娜王苗苗校对人:佟国荣刘娜王苗苗一.选择题: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.集合8|1,3A x x x ⎧⎫=-≤≤∈⎨⎬⎩⎭N ,集合{|03}B x x =<<,则A B = ()A.{|02}x x <≤B.{|12}x x ≤≤C.{1,2}D.{0,1,2}2.在ABC 中,点D 是BC 的中点,点M 是AC 的中点,点N 在线段AD 上并且12DN AN =,则MN = ()A.2136AB BC -B.2132AB BC -C.1163AB BC -D.1166AB BC -3.已知数列{}n a 的前n 项和为n S ,11a =,12n n S a +=,则4a =()A.274B.94C.278D.984.6sin 1x x θ⎛⎫-+ ⎪⎝⎭的展开式中4x 的系数为12,则cos 2θ=()A.14B.12-C.12D.345.洛书,古称龟书,是阴阳五行术数之源,在古代传说中有神龟出于洛水,其甲壳上心有此图象,结构是戴九履一,左三右七,二四为肩,六八为足,以五居中,五方白圈皆阳数,四黑点为阴数.如图,若从四个阴数和五个阳数中随机选取3个不同的数,其和等于15的概率是()A.221B.114C.328D.176.某班学生每天完成数学作业所需的时间的频率分布直方图如右图,为响应国家减负政策,若每天作业布置量在此基础上减少5分钟,则减负后完成作业的时间的说法中正确的是()A.减负后完成作业的时间的标准差减少25B.减负后完成作业的时间的方差减少25C.减负后完成作业的时间在60分钟以上的概率为12%D.减负后完成作业的时间的中位数为257.函数()y xf x =是定义在R 上的奇函数,且()f x 在区间[0,)+∞上单调递增,若关于实数t 的不等式()313log log 2(2)f t f t f ⎛⎫+> ⎪⎝⎭恒成立,则t 的取值范围是()A.10,(3,)3⎛⎫+∞ ⎪⎝⎭B.10,3⎛⎫ ⎪⎝⎭C.(9,)+∞ D.10,(9,)9⎛⎫+∞ ⎪⎝⎭8.已知抛物线()2:20C y px p =>的焦点为F ,准线为l ,A ,B 为C 上两点,且均在第一象限,过A ,B 作l 的垂线,垂足分别为D ,E .若1AB =,1sin 4DFE ∠=,则AFB △的外接圆面积为().A.16π15B.15π16C.14π15D.15π14二.选择题: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m ,n 为正整数且1n m >>,则()A.12C C C 2nnn n n +++= B.111C C C mm m n n n +++=-C.11C (1)C m m n n m n --=- D.11A A A mm m n nn m -++=10.已知抛物线C :2x my =的焦点为()0,1F ,点A ,B 为C 上两个相异的动点,则()A.抛物线C 的准线方程为1y =-B.设点()2,3P ,则AP AF +的最小值为4C.若A ,B ,F 三点共线,则AB 的最小值为2D.若60AFB ∠=︒,AB 的中点M 在C 的准线上的投影为N ,则MN AB ≤11.如图,在ABC 中,2B π∠=,AB =,1BC =,过AC 中点M 的直线l 与线段AB 交于点N .将AMN 沿直线l 翻折至A MN '△,且点A '在平面BCMN 内的射影H 在线段BC 上,连接AH 交l 于点O ,D 是直线l 上异于O 的任意一点,则()A.A DH A DC ''∠≥∠B .A DH A OH''∠≤∠C.点O 的轨迹的长度为6πD.直线A O '与平面BCMN 所成角的余弦值的最小值为13三.填空题:本题共3小题,每小题5分,共15分.12.已知复数z 满足22z z -==,则3z =__________.13.若ππ665πeecos 3t ααα+--⎛⎫-++= ⎪⎝⎭,ππ445πe e cos 4t βββ--⎛⎫-++=- ⎪⎝⎭,则()sin αβ+=______.14.若实数x ,y 满足(4x y ++=,则22x y +的最小值为________.四.解答题:本题共5小题,共77分.解答应写出文字说明、证明过程或演算步骤.15.如图,在数轴上,一个质点在外力的作用下,从原点O 出发,每次等可能地向左或向右移动一个单位,质点到达位置的数字记为X.(1)若该质点共移动2次,位于原点O 的概率;(2)若该质点共移动6次,求该质点到达数字X 的分布列和数学期望.16.如图,四棱柱1111ABCD A B C D -的底面ABCD是正方形,11AB AA A C ==⊥平面11BB D D.(1)求点1A 到平面ABCD 的距离;(2)若M 是线段1BB 上一点,平面MAC 与平面11BB D D 夹角的余弦值为105时,求1BM BB 的值.17.如图,动.双曲线的一个焦点为(0,F ,另一个焦点为P ,若该动双曲线的两支分别经过点(1,0),(1,0)M N -.(1)求动点P 的轨迹方程;(2)斜率存在且不为零的直线l 过点(1,0)M ,交(1)中P 点的轨迹于,A B 两点,直线(2)x t t =>与x 轴交于点D ,Q 是直线x t =上异于D 的一点,且满足AQ DQ ⊥.试探究是否存在确定的t 值,使得直线BQ 恒过线段DM 的中点,若存在,求出t 值,若不存在,请说明理由.18.已知数列{}n a 是正项等比数列,{}n b 是等差数列,且1122a b ==,24a b =,534a a =,(1)求数列{}n a 和{}n b 的通项公式;(2)[]x 表示不超过x 的最大整数,4n T 表示数列()221n n b ⎡⎤⎢⎥⎣⎦⎧⎫-⋅⎨⎬⎩⎭的前4n 项和,集合*422,N n n n T b A n n a λ++⎧⎫⋅⎪⎪=≤∈⎨⎬⎪⎪⎩⎭共有4个元素,求λ范围;(3)**21,N ,2,Nn n n n k k c a b n k k =-∈=⋅=∈⎩,求数列{}n c 的前2n 项和.19.给出下列两个定义:I.对于函数()y f x =,定义域为D ,且其在D 上是可导的,若其导函数定义域也为D ,则称该函数是“同定义函数”.II.对于一个“同定义函数”()y f x =,若有以下性质:①()()()f xg f x '=;②()()()f x h f x =',其中()(),yg x yh x ==为两个新的函数,()y f x '=是()y f x =的导函数.我们将具有其中一个性质的函数()y f x =称之为“单向导函数”,将两个性质都具有的函数()y f x =称之为“双向导函数”,将()y g x =称之为“自导函数”.(1)判断函数tan y x =和ln y x =是“单向导函数”,或者“双向导函数”,说明理由.如果具有性质①,则写出其对应的“自导函数”;(2)已知命题():p y f x =是“双向导函数”且其“自导函数”为常值函数,命题():(,0,1)x q f x k a k a a =⋅∈>≠R .判断命题p 是q 的什么条件,证明你的结论;(3)已知函数()()e axf x x b =-.①若()f x 的“自导函数”是y x =,试求a 的取值范围;②若1a b ==,且定义()()34e 3xI x f x kx kx =-+,若对任意][1,2,0,k x k ⎡⎤∈∈⎣⎦,不等式()I x c ≤恒成立,求c 的取值范围.2023-2024学年度高三年级下学期第六次模拟考试数学学科试卷命题人:佟国荣刘娜王苗苗校对人:佟国荣刘娜王苗苗一.选择题: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.集合8|1,3A x x x ⎧⎫=-≤≤∈⎨⎬⎩⎭N ,集合{|03}B x x =<<,则A B = ()A.{|02}x x <≤B.{|12}x x ≤≤C.{1,2}D.{0,1,2}【答案】C 【解析】【分析】列举法表示出集合A ,进而根据交集的概念即可求出结果.【详解】因为{}8|1,0,1,23A x x x ⎧⎫=-≤≤∈=⎨⎬⎩⎭N ,所以{}1,2A B = ,故选:C2.在ABC 中,点D 是BC 的中点,点M 是AC 的中点,点N 在线段AD 上并且12DN AN =,则MN = ()A.2136AB BC -B.2132AB BC -C.1163AB BC -D.1166AB BC -【答案】D 【解析】【分析】根据平面向量的线性运算计算即可.【详解】因为2AN ND =,所以23AN AD = ,又点D 是BC 的中点,点M 是AC 的中点,所以11,22AM AC BD BC ==,故()21211113232266MN AN AM AD AC AB BC AB BC AB BC ⎛⎫=-=-=+-+=- ⎪⎝⎭.故选:D.3.已知数列{}n a 的前n 项和为n S ,11a =,12n n S a +=,则4a =()A.274B.94C.278 D.98【答案】D 【解析】【分析】根据给定递推公式求出23,a a 即可计算作答.【详解】因数列{}n a 的前n 项和为n S ,11a =,12n n S a +=,则211111222a S a ===,321211113()(1)22224a S a a ==+=+=,43123111139()(1)222248a S a a a ==++=++=,所以498a =.故选:D4.6sin 1x x θ⎛⎫-+ ⎪⎝⎭的展开式中4x 的系数为12,则cos 2θ=()A.14B.12-C.12D.34【答案】C 【解析】【分析】根据乘法的运算法则,结合组合数的性质、二倍角的余弦公式进行求解即可.【详解】6sin 1x x θ⎛⎫-+ ⎪⎝⎭的展开式中4x 的系数可以看成:6个因式sin 1x x θ⎛⎫-+ ⎪⎝⎭中选取5个因式提供x ,余下一个因式中提供sin x x 或者6个因式sin 1x x θ⎛⎫-+ ⎪⎝⎭中选取4个因式提供x ,余下两个因式中均提供1,故4x 的系数为4566C C sin 12θ-⋅=,∴1sin 2θ=,∴211cos 212sin 1242θθ=-=-⨯=,故选:C5.洛书,古称龟书,是阴阳五行术数之源,在古代传说中有神龟出于洛水,其甲壳上心有此图象,结构是戴九履一,左三右七,二四为肩,六八为足,以五居中,五方白圈皆阳数,四黑点为阴数.如图,若从四个阴数和五个阳数中随机选取3个不同的数,其和等于15的概率是()A.221B.114C.328D.17【答案】A 【解析】【分析】先计算从四个阴数和五个阳数共9个数字中随机选取3个不同的数,总共有39C 种选法,再计算符合条件和等于15的三个数的种类,即可算出概率.【详解】从四个阴数和五个阳数共9个数字中随机选取3个不同的数,总共有3984C =种选法,其和等于15的三个数的种类共有8种,即:(图形中各横,各列,对角线所在的三个数字之和均为15).故其和等于15的概率是:828421=,故选:A .【点睛】本题考查古典概型的概率计算,运用分类和分步分别选出符合条件的种类,找出古典概型的分子和分母是关键,属于中等题.6.某班学生每天完成数学作业所需的时间的频率分布直方图如右图,为响应国家减负政策,若每天作业布置量在此基础上减少5分钟,则减负后完成作业的时间的说法中正确的是()A.减负后完成作业的时间的标准差减少25B.减负后完成作业的时间的方差减少25C.减负后完成作业的时间在60分钟以上的概率为12%D.减负后完成作业的时间的中位数为25【答案】D 【解析】【分析】根据给定的频率分布直方图求出x ,利用方差、标准差的意义判断AB ;求出减负前完成作业的时间在60分钟以上的概率及中位数判断CD.【详解】由频率分布直方图可得:200.025200.0065200.0032201x +⨯+⨯+⨯⨯=,解得0.0125x =,减负后每天作业布置量减少5分钟,则减负后完成作业的时间的平均数减少5分钟,而完成作业的时间波动大小不变,因此减负后完成作业的时间的标准差、方差不变,AB 错误;减负前完成作业时间在60分钟以上的频率为0.003400.12⨯=,减负后完成作业时间在60分钟以上的频率小于0.12,由此估计减负后完成作业的时间在60分钟以上的概率小于12%,C 错误;减负前,第一组的频率为0.0125200.25⨯=,第二组的频率为0.025200.5⨯=,则完成作业的时间的中位数在第二组的中间,即中位数为2040302+=(分钟),所以减负后完成作业时间的中位数为30525-=(分钟),D 正确.故选:D7.函数()y xf x =是定义在R 上的奇函数,且()f x 在区间[0,)+∞上单调递增,若关于实数t 的不等式()313log log 2(2)f t f t f ⎛⎫+> ⎪⎝⎭恒成立,则t 的取值范围是()A.10,(3,)3⎛⎫+∞ ⎪⎝⎭B.10,3⎛⎫ ⎪⎝⎭C.(9,)+∞ D.10,(9,)9⎛⎫+∞ ⎪⎝⎭【答案】D【解析】【分析】首先得出()f x 是偶函数,把不等式化为()3(log )2f t f >,结合函数的单调性与奇偶性,得到3log 2t >,求解不等式即可.【详解】因为函数()y xf x =是定义在R 上的奇函数,即()()xf x xf x --=-,当0x ≠时()()f x f x -=,又()0f 有意义,所以()f x 是定义域R 上的偶函数,又因为()f x 在区间[0,)+∞上单调递增,所以()313333(log )(log )(log )(log )2(log )22f t f t f t f t f t f +=+-=>,所以()3(log )2f t f >,即()()3log 2ft f >,所以3log2t >,则3log 2t >或3log 2t <-,解得9t >或109t <<,所以t 的取值范围是10,(9,)9⎛⎫+∞ ⎪⎝⎭.故选:D .8.已知抛物线()2:20C y px p =>的焦点为F ,准线为l ,A ,B 为C 上两点,且均在第一象限,过A ,B作l 的垂线,垂足分别为D ,E .若1AB =,1sin 4DFE ∠=,则AFB △的外接圆面积为().A.16π15B.15π16C.14π15D.15π14【答案】A 【解析】【分析】由抛物线的定义及平行线的性质可得2AFB DFE ∠=∠,结合同角三角函数的平方关系及二倍角公式可得15sin 8AFB ∠=,进而由正弦定理可求得结果.【详解】如图所示,由抛物线的定义可知AF AD =,BF BE =,所以BFE BEF EFO ∠=∠=∠,AFD ADF DFO ∠=∠=∠,所以DFE EFO DFO BFE AFD BAF DFE ∠=∠-∠=∠-∠=∠-∠,故2AFB DFE ∠=∠,易知DFE ∠为锐角,且由1sin 4DFE ∠=可知cos 4DFE ∠=,所以15sin 2sin cos 8AFB DFE DFE ∠=∠∠=.设AFB △的外接圆半径为R ,由正弦定理可知2sin AB R AFB=∠,又||1AB =,所以R =所以AFB △的外接圆面积为216ππ15R =.故选:A.二.选择题: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m ,n 为正整数且1n m >>,则()A.12C C C 2nnn n n +++= B.111C C C mm m n n n +++=-C.11C (1)C m m n n m n --=- D.11A A A mm m n nn m -++=【答案】BD 【解析】【分析】对A :借助二项式的展开式计算即可得;对B 、C 、D :结合排列数与组合数的计算公式计算即可得.【详解】对A :()01211C C C nn nn n n =+=+++ ,又0C 1n =,故A 错误;对B :()()()()()()1111!!CC1!!1!1!m m n nn n m n m m n m ++++-=-+-+--()()()()()()()()()()()()()1!!1!!C 1!!1!!1!!!!m nn n m n n n m n n m m n m m m n m m m n m m n m +-+-+=-===+-+-+--,故B 正确;对C :()()()()()()()()111!1!1!C 1!!1!!!!m n n m n m n m n m m m n m m n m -----===-----,()()!C !!m n m n m m n m =-,即11C (1)C mm n n m n --≠-,故C 错误;对D :()()()()()()11!1!!!A A !1!1!1!m m n n n m m n n n m n m n m n m n m n m --++⋅+⋅+=+==--+-+-+,()()11!A 1!m n n n m ++=-+,即11A A A mm m n n n m -++=,故D 正确.故选:BD.10.已知抛物线C :2x my =的焦点为()0,1F ,点A ,B 为C 上两个相异的动点,则()A.抛物线C 的准线方程为1y =-B.设点()2,3P ,则AP AF +的最小值为4C.若A ,B ,F 三点共线,则AB 的最小值为2D.若60AFB ∠=︒,AB 的中点M 在C 的准线上的投影为N ,则MN AB ≤【答案】ABD 【解析】【分析】对于A ,由抛物线的焦点可求出抛物线的准线方程,对于B ,过点A 作AK 垂直准线于K ,则AP AF AP AK +=+,从而可求出其最小值,对于C ,由抛物线的性质可判断,对于D ,过,A B 分别作11,AA BB 垂直准线,垂足分别为11,A B ,则由梯形中位线定理可得()()111122MN AA BB AF BF =+=+,然后在ABF △利用余弦定理结合基本不等式可判断【详解】对于A ,因为抛物线C :2x my =的焦点为()0,1F ,所以抛物线C 的准线方程为1y =-,所以A 正确,对于B ,由题意可得抛物线的方程为24x y =,则点()2,3P 在抛物线外,如图,过点A 作AK 垂直准线于K ,则AP AF AP AK +=+,当,,A P K 三点共线时,AP AK +取得最小值,最小值为4,所以B 正确,对于C ,由抛物线的性质可得当A ,B ,F 三点共线,且AB ⊥y 轴时,弦AB 最短为抛物线的通径24p =,所以C 错误,对于D ,过,A B 分别作11,AA BB 垂直准线,垂足分别为11,A B ,则由梯形中位线定理可得()()111122MN AA BB AF BF =+=+,设,AF a BF b ==,则()12MN a b =+,在ABF △中由余弦定理得22222cos 60()3AB a b ab a b ab =+-︒=+-,因为22a b ab +⎛⎫≤ ⎪⎝⎭,所以222231()3()()()44a b ab a b a b a b +-≥+-+=+,所以221()4AB a b ≥+,所以1()2AB a b MN ≥+=,当且仅当a b =时取等号,所以D 正确,故选:ABD11.如图,在ABC 中,2B π∠=,3AB =,1BC =,过AC 中点M 的直线l 与线段AB 交于点N .将AMN 沿直线l 翻折至A MN '△,且点A '在平面BCMN 内的射影H 在线段BC 上,连接AH 交l 于点O ,D 是直线l 上异于O 的任意一点,则()A.A DH A DC ''∠≥∠B.A DH A OH''∠≤∠C.点O 的轨迹的长度为6πD.直线A O '与平面BCMN 所成角的余弦值的最小值为13【答案】BCD 【解析】【分析】A 、B 选项结合线面角最小,二面角最大可判断;对于C ,先由旋转,易判断出MN AO ⊥,故其轨迹为圆弧,即可求解.对于D 求直线与平面所成角的余弦值,即求OH OH A O AO =',,32AMN ππθ⎛⎫∠=∈ ⎪⎝⎭,用θ表示,AO OH ,再结合三角恒等变换求出函数的最值即可【详解】依题意,将AMN 沿直线l 翻折至A MN '△,连接AA ',由翻折的性质可知,关于所沿轴对称的两点连线被该轴垂直平分,故AA MN '⊥,又A '在平面BCMN 内的射影H 在线段BC 上,所以A H '⊥平面BCMN ,MN ⊂平面BCMN ,所以A H MN '⊥,AA A H A '''⋂=,AA '⊂平面A AH ',A H '平面A AH'所以MN ⊥平面A AH '.AO ⊂平面A AH ',A O '⊂平面A AH ',A H '⊂平面A AH ',,,AO MN A O MN A H MN ''⊥⊥⊥,AOM ∴∠=90 ,且A OH '∠即为二面角A MN B '--的平面角对于A 选项,由题意可知,A DH '∠为A D '与平面BCMN 所成的线面角,故由线面角最小可知A DH A DC ''∠≤∠,故A 错误;对于B 选项,A OH '∠ 即为二面角A MN B '--的平面角,故由二面角最大可知A DH A OH ''∠≤∠,故B 正确;对于C 选项,MN AO ⊥ 恒成立,故O 的轨迹为以AM 为直径的圆弧夹在ABC 内的部分,易知其长度为1236ππ⨯=,故C 正确;对于D 选项,如下图所示设,32AMN ππθ⎛⎫∠=∈ ⎪⎝⎭,在AOM 中,AOM ∠=90 ,sin sin AO AM θθ∴==,在ABH 中,2B π∠=,3cos cos 3AB AH BAH πθ==∠⎛⎫- ⎪⎝⎭,所以3sin cos 3OH AH AO θπθ=-=-⎛⎫- ⎪⎝⎭,设直线A O '与平面BCMN 所成角为α,则3sin cos 333cos 11sin 3sin cos sin 2332OH AO θπθαπθπθθθ-⎛⎫- ⎪⎝⎭===-=-⎛⎫⎛⎫--+ ⎪ ⎪⎝⎭⎝⎭31313312≥=+,当且仅当523212πππθθ-=⇒=时取等号,故D 正确.故选:BCD .三.填空题:本题共3小题,每小题5分,共15分.12.已知复数z 满足22z z -==,则3z =__________.【答案】8-【解析】【分析】设i z a b =+,根据22z z -==得到方程组,求出1,3a b ==而求出3z .【详解】设i z a b =+,则22i z a b -=-+,所以()2222424a b a b ⎧+=⎪⎨-+=⎪⎩,解得1,a b ==当1,a b ==时,1=+z,故()222113i 2z =+=++=-+,()()322126i 8z =-++=-+=-;当1,a b ==1z =-,故()222113i 2z =-=-+=--,()()322126i 8z =---=-+=-故答案为:-813.若ππ665πeecos 3t ααα+--⎛⎫-++= ⎪⎝⎭,ππ445πe e cos 4t βββ--⎛⎫-++=- ⎪⎝⎭,则()sin αβ+=______.【答案】4【解析】【分析】运用诱导公式化简后构造函数()1e sin e =-+xxf x x ,研究其奇偶性,运用导数研究其单调性,依据奇偶性及单调性解方程即可.【详解】由ππ665πeecos 3t ααα+--⎛⎫-++= ⎪⎝⎭,ππ445πe e cos 4tβββ--⎛⎫-++=- ⎪⎝⎭得ππ663ππeecos 26t ααα+--⎛⎫-+++= ⎪⎝⎭,ππ443ππe e cos 24t βββ-⎛⎫-++-=- ⎪⎝⎭,即ππ66πeesin 6t ααα+--⎛⎫-++= ⎪⎝⎭,ππ44πe e sin 4t βββ-⎛⎫-+-=- ⎪⎝⎭.设()1e sin e =-+xx f x x ,定义域为R ,则()11esin())e e esin (xxx x x f x x f x ---=-+-=--=-所以()f x 是R 上的奇函数,又因为()1e cos cos 2cos 0e x x f x x x x =++≥+='+>,所以()f x 是R 上的单调增函数.又因为π6f t α⎛⎫+= ⎪⎝⎭,π4f t β⎛⎫-=- ⎪⎝⎭,所以πππ644f f f αββ⎛⎫⎛⎫⎛⎫+=--=- ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭,所以ππ64αβ+=-,即ππ46αβ+=-,所以()ππππ62sin sincos cos 46464αβ+=-=.故答案为:4-.14.若实数x ,y满足(4x y ++=,则22x y +的最小值为________.【答案】1-.【解析】【分析】由(4x y ++=,得出(2,(x y ++构成成等比数列,求得y x =-,进而结合二次函数的性质,即可求解.【详解】由题意,实数x ,y满足(4x y ++=,可得(2,(x y ++构成成等比数列,设公比为q,则22x q y q ⎧+=⎪⎨⎪=⎩,整理得2222244(2)x x q y q y ⎧⎛⎫+=-⎪ ⎪⎨⎝⎭⎪+=-⎩,解得11x q q y q q ⎧=-⎪⎪⎨⎪=-⎪⎩,可得y x =-,所以22x y +=222(1)11x x x -=--≥-,故22x y +的最小值为1-.故答案为:1-.【点睛】本题主要考查了最值问题的求解,其中解答中根据题设条件,合理转化为等比数列,利用等比数列的性质求解是解答的关键,着重考查了转化与化归思想,以及分析问题和解决问题的能力.四.解答题:本题共5小题,共77分.解答应写出文字说明、证明过程或演算步骤.15.如图,在数轴上,一个质点在外力的作用下,从原点O 出发,每次等可能地向左或向右移动一个单位,质点到达位置的数字记为X .(1)若该质点共移动2次,位于原点O 的概率;(2)若该质点共移动6次,求该质点到达数字X 的分布列和数学期望.【答案】(1)12;(2)分布列见解析,0.【解析】【分析】(1)由题意知质点移动2次的所有可能种数,再求出移动2次后在原点的所有可能种数,根据古典概型求解即可;(2)设向左移动的次数为随机变量Y ,易知1(6,)2Y B ,得出随机变量62X Y =-,由二项分布求出对应的概率,即可求出分布列,再由期望的性质求解X 的期望.【小问1详解】质点移动2次,可能结果共有224⨯=种,若质点位于原点O ,则质点需要向左、右各移动一次,共有12C 2=种,故质点位于原点O 的概率2142P ==.【小问2详解】质点每次移动向左或向右,设事件A 为“向右”,则A 为“向左”,故1()(2P A P A ==,设Y 表示6次移动中向左移动的次数,则1(6,)2Y B ,质点到达的数字62X Y =-,所以06611(6)(0)C ()264P X P Y =====,16613(4)(1)C ()232P X P Y =====,266115(2)(2)C ()264P X P Y =====,36615(0)(3)C ()216P X P Y =====,466115(2)(4)C ()264P X P Y =-====,56613(4)(5)C ()232P X P Y =-====,66611(6)(6)C ()264P X P Y =-====,所以X 的分布列为:X6-4-2-0246P164332156451615643321641()(62)2()626602E X E Y E Y =-=-+=-⨯⨯+=.16.如图,四棱柱1111ABCD A B C D -的底面ABCD是正方形,11AB AA A C ==⊥平面11BB D D.(1)求点1A 到平面ABCD 的距离;(2)若M 是线段1BB 上一点,平面MAC 与平面11BB D D夹角的余弦值为5时,求1BM BB 的值.【答案】(1)1(2)112BM BB =【解析】【分析】(1)连接AC 交BD 于点O ,先证平面11AA C C ⊥平面ABCD ,然后由面面垂直性质定理知1AO 即为所求,然后计算可得;(2)以O 为原点,分别以1,,OB OC OA分别为,,x y z 轴的正方向建立如图所示空间直角坐标系,利用平面夹角的向量公式列方程可解.【小问1详解】连接AC 交BD 于点O ,连接1AO .因为1A C ⊥平面11BB D D ,BD ⊂平面11BB D D ,所以1A C BD ⊥,因为底面ABCD 是正方形,所以BD AC ⊥,又1AC AC C ⋂=,1,A C AC ⊂平面11AA C C ,所以BD ⊥平面11AA C C .又BD ⊂平面ABCD ,所以平面11AA C C ⊥平面ABCD .因为1A C ⊥平面11BB D D ,1BB ⊂平面11BB D D ,所以11A C BB ⊥,又11//BB AA ,所以11A C AA ⊥.在1Rt AA C中,12AA AC ==,所以1AC =又O 为AC 的中点,所以1A O AC ⊥且11A O =,又平面11AA C C ⊥平面ABCD ,平面11AA C C 平面ABCD AC =,1A O ⊂平面11AA C C ,所以1A O ⊥平面ABCD .故点1A 到平面ABCD 的距离为11A O =.【小问2详解】以O 为原点,分别以1,,OB OC OA分别为,,x y z轴的正方向建立如图所示空间直角坐标系,则()()()()11,0,0,0,1,0,0,0,1,0,1,0B C A A -,()()()1110,1,1,1,1,0,0,1,1A C AB BB AA =-===,由(1)知,平面11BB D D 的一个法向量()110,1,1n A C ==-,设1(01)BMBB λλ=<<,则()()()10,,,1,1,,0,2,0BM BB AM AB BM AC λλλλλ===+=+=设()2,,n x y z =为平面MAC 的一个法向量,由()22100n AM x y z n AC y λλ⎧⋅=+++=⎪⎨⋅==⎪⎩ ,取x λ=,得()2,0,1n λ=- ,设平面MAC 与平面11BB D D 的夹角为θ,则有121210cos 5n n n n θ===,解得12λ=,即112BM BB =.17.如图,动.双曲线的一个焦点为(0,F ,另一个焦点为P ,若该动双曲线的两支分别经过点(1,0),(1,0)M N -.(1)求动点P 的轨迹方程;(2)斜率存在且不为零的直线l 过点(1,0)M ,交(1)中P 点的轨迹于,A B 两点,直线(2)x t t =>与x 轴交于点D ,Q 是直线x t =上异于D 的一点,且满足AQ DQ ⊥.试探究是否存在确定的t 值,使得直线BQ 恒过线段DM 的中点,若存在,求出t 值,若不存在,请说明理由.【答案】(1)22143x y +=(0)x ≠(2)存在,4t =【解析】【分析】(1)根据题意,结合双曲线的定义和椭圆的定义即可求解;(2)设直线l 的方程为:(1)y k x =-(0)k ≠,1122(,),(,)A x y B x y 12()x x ≠,根据题意得出()()121220y y k tk y y +-+=,然后将直线方程与圆锥曲线方程联立,利用韦达定理即可求解.【小问1详解】由题意以及双曲线定义可得:MP MF NF NP -=-,42MP NP NF MF MN ∴+=+=>=由椭圆的定义可知,点P 的轨迹是以,M N 为焦点,2,1a c ==的椭圆(不含短轴端点),其方程为22143x y +=(0)x ≠.【小问2详解】设直线l 的方程为:(1)y k x =-(0)k ≠,1122(,),(,)A x y B x y 12()x x ≠,则由0AQ DQ ⋅= ,知()1,Q t y ,所以()2112:y y BQ y y x t x t--=--,令0y =,得()1221y t x x t y y -=+-()*因点22(,)B x y 在直线l 上,所以22(1)y k x =-,变形得221y x k =+,代入()*式化简得()121212y y ky kty x k y y +-=-,若直线BQ 恒过线段DM 的中点1,02t +⎛⎫ ⎪⎝⎭,则有()12121212y y ky kty t k y y +-+=-,整理得()()121220y y k tk y y +-+=()**由22(1)143y k x x y =-⎧⎪⎨+=⎪⎩,得222(34)690k y ky k ++-=,所以212122269,3434k k y y y y k k +=-=-++代入()**整理得,()21860k k k tk ---=,解得4t =,所以存在4t =,即直线4x =,使得直线BQ 恒过线段DM 的中点.18.已知数列{}n a 是正项等比数列,{}n b 是等差数列,且1122a b ==,24a b =,534a a =,(1)求数列{}n a 和{}n b 的通项公式;(2)[]x 表示不超过x 的最大整数,4n T 表示数列()221n n b ⎡⎤⎢⎥⎣⎦⎧⎫-⋅⎨⎬⎩⎭的前4n 项和,集合*422,N n n n T b A n n a λ++⎧⎫⋅⎪⎪=≤∈⎨⎬⎪⎪⎩⎭共有4个元素,求λ范围;(3)**21,N ,2,N n n nn k k c a b n k k =-∈=⋅=∈⎩,求数列{}n c 的前2n 项和.【答案】(1)2n n a =,n b n=(2)353322λ<≤(3)12252241839n n S n +⎛⎫=+-⎝- ⎪⎭⋅【解析】【分析】(1)设出数列{}n a 公比,数列{}n b 公差,结合题意计算即可得;(2)由()()()()22222222434241443424144n n n n b b b b n n n n -----+=-----+=,即可得()424224,2n n n n n n n T b T n a +++⋅==,令()22n n n n D +=,由1n n D D +-的值,可得数列n D 的单调性,计算出前五项,即可得λ的取值范围;(3)分奇偶讨论后,分别借助错位相减法与裂项相消法求和计算即可得.【小问1详解】设数列{}n a 首项12a =,设公比()0q q >,设数列{}n b 首项11b =,设公差d ,∵53244a a a b =⎧⎨=⎩,即424213q q q d ⎧=⎨=+⎩,∴2q =,2q =-(舍去),1d =,∴2n n a =.n b n =;【小问2详解】()()()2222222222224123456784342414n n n n n T b b b b b b b b b b b b ---=--++--++⋅⋅⋅+--+,其中()()()()22222222434241443424144n n n n b b b b n n n n -----+=-----+=,∴()424224,2n n n nn n n T b T n a +++⋅==,集合()*2,N 2n n n A n n λ⎧⎫+⎪⎪=≤∈⎨⎬⎪⎪⎩⎭,设()22n n n n D +=,()()()21111323222n n n n n n n n n n D D ++++++-+-=-=,所以当1n =时,21D D >,当2n ≥时,234D D D >>>⋅⋅⋅.计算可得132D =,22D =,3158D =,432D =,53532D =,因为集合有4个元素,353322λ<≤;【小问3详解】**21,N ,2,Nn n n n k k c a b n k k =-∈=⋅=∈⎩,21232n n S C C C C =+++⋅⋅⋅+,设2462246222426222n n n A C C C C n =+++⋅⋅⋅+=⋅+⋅+⋅+⋅⋅⋅+⋅,()462224224222222n n n A n n +=⋅+⋅+⋅⋅⋅+-⋅+⋅,则()4246822222224382222222822214n n n n n A n n ++-⋅-=++++⋅⋅⋅+-⋅=+-⋅-222222322282822223333n n n n n +++⋅⎛⎫=-+-⋅=-+-⋅ ⎪⎝⎭,所以228222939n n A n +⎛⎫=+-⋅ ⎪⎝⎭,当n为奇数时,n C ==,设13521n n B C C C C -=+++⋅⋅⋅+12⎛⎛⎛=-++⋅⋅⋅+ ⎝⎝⎝12=所以12252241839n n n n S A B n +⎛⎫=+=+-⋅ ⎪⎝⎭.19.给出下列两个定义:I.对于函数()y f x =,定义域为D ,且其在D 上是可导的,若其导函数定义域也为D ,则称该函数是“同定义函数”.II.对于一个“同定义函数”()y f x =,若有以下性质:①()()()f x g f x '=;②()()()f x h f x =',其中()(),yg x yh x ==为两个新的函数,()y f x '=是()y f x =的导函数.我们将具有其中一个性质的函数()y f x =称之为“单向导函数”,将两个性质都具有的函数()y f x =称之为“双向导函数”,将()y g x =称之为“自导函数”.(1)判断函数tan y x =和ln y x =是“单向导函数”,或者“双向导函数”,说明理由.如果具有性质①,则写出其对应的“自导函数”;(2)已知命题():p y f x =是“双向导函数”且其“自导函数”为常值函数,命题():(,0,1)x q f x k a k a a =⋅∈>≠R .判断命题p 是q 的什么条件,证明你的结论;(3)已知函数()()e a xf x x b =-.①若()f x 的“自导函数”是y x =,试求a 的取值范围;②若1a b ==,且定义()()34e 3x I x f x kx kx =-+,若对任意][1,2,0,k x k ⎡⎤∈∈⎣⎦,不等式()I x c ≤恒成立,求c 的取值范围.【答案】(1)答案见解析(2)既不充分也不必要条件;证明见解析(3)452[e ,)3-+∞【解析】【分析】(1)由()tan f x x =和()ln f x x =,结合题设中函数的定义,即可得到答案;(2)由q 成立,得到()ln x f x ka a '=,设()ln g x x a =,得出()f x 为“单向导函数”,再设()ln x h x a =,得到()f x 为“双向导函数”,结合()g x 不是常值函数,求得p 不是q 的必要条件;再由p 成立,得到()(())f x g f x m '==,进而得出结论;(3)①由题意得到10a ax -=,求得0a =;②由题意求得()22(21)e 4x I x x kx k '=--+且1()02I '=,令()22(21)e 4x p x x kx k =--+,求得()24(e 2)x p x x k '=-,得到存在0x 使得02e 20x k -=,进而得到()p x 单调性,分类讨论,即可求解.【小问1详解】解:对于函数()tan f x x =,则()21tan '=+f x x ,这两个函数的定义域都是π{|π,Z}2x x k k ≠+∈,所以函数()f x 为“同定义域函数”,此时,()21g x x =+,由函数的定义,对于4πx =±,()(())f x h f x '=无法同时成立,所以()f x 为“单向导函数”,其“自导函数”为()21g x x =+,对于函数()ln f x x =,则()1f x x'=,因为这两个函数的定义域不同,所以不是“同定义函数”.【小问2详解】解:若q 成立,()x f x ka =,则()ln x f x ka a '=,设()ln g x x a =,则()(())f x g f x '=,所以()f x 为“单向导函数”,又设()ln x h x a=,则()(())f x h f x '=,所以()f x 为“双向导函数”,但()g x 不是常值函数,所以p 不是q 的必要条件;若p 成立,则()g x m =,所以()(())f x g f x m '==,所以()f x mx n =+,所以q 不成立,所以p 是q 的既不充分也不必要条件.【小问3详解】解:①由题意,()1()e a a x f x ax x b -'=+-,且1()e ()e a a x a x ax x b x b -+-=-,所以10a ax -=,所以0a =;②由题意()234(1)e 3x I x x kx kx =--+,所以()22(21)e 4x I x x kx k '=--+且1()02I '=,令()[][]22(21)e 4,0,,1,2x p x x kx k x k k =--+∈∈,可得()21e 30p k =->,且()224e 84(e 2)x xp x x kx x k '=-=-,因为2e 2x y k =-为单调递增函数,且20|120,|e 20kx x k y k y k ===-<=->,所以存在01ln 2(0,)2x k k =∈使得02e 20x k -=,且当0[0,]x x ∈时,()0p x '≤,()p x 单调递减;当0[,]x x k ∈时,()0p x '≥,()p x 单调递增,(i )当011ln 222x k ==时,即e 2k =,所以2min 0000()()(21)24(21)0p x p x x k kx k k x ==-⋅-+=--=,此时()0I x '≥,()I x 在[0,]x k ∈上单调递增,可得()()max I x I k =;(ii )当1k =时,(0)110p =-+=,此时()200min 1ln 2,(21)02x p x k x ==--<,所以当1[0,]2x ∈时,()0I x '≤,()I x 单调递减;当1[,1]2x ∈时,()0I x '≥,()I x 单调递增,又由()()()10I k I I =>,所以()()max I x I k =;(iii )当(1,2]k ∈且e 2k ≠时,()20min (21)0,(0)0p x k x p =--<>,所以函数()I x 在(0,1)上存在两个极值点,若011ln 222x k =>,即e 22k <≤时,极大值点为12;若011ln 222x k =<,即e 12k <<时,极大值点为11x 2<,则()max I x 为函数的极大值或()I k ,由当102x ≤≤时,()()23242414(1)e 10,(1)e 323x k I x x kx kx k I k k k k =--+≤-+≤=--+,令()[]2424(1)e ,1,23k t k k k k k =--+∈,则()[]2316(21)e 2,1,23k t k k k k k '=--+∈,设()[]2316(21)e 2,1,23k s k k k k k =--+∈,则()2224e 1624(e 4)20k ks k k k k k '=-+=-+>,所以()s k ,即()t k '单调递增,所以()()2161e 203t k t ''≥=-+>,所以()t k 单调递增,所以()()4522e 03t k t ≤=->,综上可得,()4max 52e 3I x c =-≤,所以实数c 的取值范围为452[e ,)3-+∞.【点睛】方法技巧:对于利用导数研究不等式的恒成立与有解问题的求解策略:1、通常要构造新函数,利用导数研究函数的单调性,求出最值,从而求出参数的取值范围;2、利用可分离变量,构造新函数,直接把问题转化为函数的最值问题.3、根据恒成立或有解求解参数的取值时,一般涉及分离参数法,但压轴试题中很少碰到分离参数后构造的新函数能直接求出最值点的情况,进行求解,若参变分离不易求解问题,就要考虑利用分类讨论法和放缩法,注意恒成立与存在性问题的区别.。

22023年高三数学对接新高考全真模拟试卷6(云南,安徽,黑龙江,山西,吉林五省通用)(解析版)

22023年高三数学对接新高考全真模拟试卷6(云南,安徽,黑龙江,山西,吉林五省通用)(解析版)

2023年高三数学对接新高考全真模拟试卷(06)(云南,安徽,黑龙江,山西,吉林五省通用)数学(新高考卷)(考试时间:120分钟 试卷满分:150分)一、 单项选择题: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.1.已知集合{}2230A x x x =+-<,{}310B x x =+>,则A B =( )A .133x x ⎧⎫-<<-⎨⎬⎩⎭B .113x x ⎧⎫-<<⎨⎬⎩⎭C .{}31x x -<<D .133x x ⎧⎫-<<⎨⎬⎩⎭【详解】A ,则为()A .20,10x x x ∃<-+≤B .20,10x x x ∃<-+>C .20,10x x x ∃>-+≤D .20,10x x x ∃>-+>【答案】C【分析】由全称命题的否定判定.【详解】由题意得p ⌝为20,10x x x ∃>-+≤. 故选:C3.意大利数学家斐波那契()17701250,以兔子繁殖为例,引入“兔子数列”:即1、1、2、3、5、8、13、21、34、55、89、144、233、,在实际生活中,很多花朵(如梅花,飞燕草,万寿简等)的瓣数恰是斐波那契数列中的数,斐波那契数列在物理及化学等领域也有着广泛得应用.已知斐波那契数列{}n a 满足:11a =,21a =,21n n n a a a ++=+,若2357959k a a a a a a a ++++++=,则k =( )A .2020B .2021C .59D .60【答案】D【解析】利用21n n n a a a ++=+化简得出235795960a a a a a a a ++++++=,即可得出结果.【详解】由于21n n n a a a ++=+,则2357959795945a a a a a a a a a a a +++++=++++++67959585960a a a a a a a ++++==+==,因此,60k =. 故选:D.4.下列四个函数中,以π为最小正周期且在区间,2ππ⎛⎫⎪⎝⎭上单调递增的函数是( )A .sin 2y x =B .cos y x =C .cos 2xy =D .tan y x =ππ⎛⎫A .8B .C .9D .6.定义在上的函数满足:对12,0,x x ∀∈+∞,且12x x ≠,都有()()2112120x f x x f x x x ->-成立,且()24f =,则不等式()2f x x>的解集为( )A .()4,+∞B .()0,4C .()0,2D .()2,+∞7.已知1F 、2F 分别为双曲线()2210,0x y a b a b -=>>的左、右焦点,且122b F F a=,点P为双曲线右支一点,I 为12PF F △的内心,若1212IPF IPF IF F S S S △△△成立,给出下列结论:∴点I 的横坐标为定值a ; ∴离心率e = ∴λ=; ∴当2PF x ⊥轴时,1230PF F ∠=︒.上述结论正确的是( ) A .∴∴ B .∴∴C .∴∴∴D .∴∴∴121212111,,2222IPF IPF IF F S PF r S PF r S c r =⋅=⋅=⋅⋅,1212IPF IF F S S △△,121112222PF r PF r c r λ⋅=⋅+⋅⋅⋅, 12151PF PF a --====,所以∴正确,8.《周髀算经》中给出了弦图,所谓弦图是由四个全等的直角三角形和中间一个小正方形拼成一个大的正方形,若图中直角三角形两锐角分别为α、β,且小正方形与大正方形面积之比为4:9,则()cos αβ-的值为( )A .59B .49C .23D .0符合题目要求.全部选对的得5分,部分选对的得3分,有选错的得0分.9.已知向量()1,2a =-,()1,b m =,则( ) A .若a 与b 垂直,则12m =B .若//a b ,则m 的值为2-C .若a b =,则2m =D .若3m =,则a 与b 的夹角为45°夹角的坐标表示计算判断C 、D ;【详解】解:因为()1,2a =-,()1,b m =,对于:若a 与b 垂直,则12a b ⋅=-+正确;:若//a b ,则12m -⨯=⨯,解得m B 正确; a b =,则()22121-+=+2m =±,故C 错误;:若3m =,则()1,3b =,设a 与b 的夹角为()222112322121a b a b⋅-⨯+⨯==-+⨯+,因为θABD10.如图,在正方体1111ABCD A B D -中,M ,N 分别是11的中点,则()A .四点A ,M ,N ,C 共面B .MN ∥CDC .1AD ∥平面1BCDD .若1MN =,则正方体1111ABCD A B C D -外接球的表面积为12π11.函数()3sin(2)f x x ϕ=+的部分图象如图所示,则下列选项中正确的有( )A .()f x 的最小正周期为πB .2π3⎛⎫⎪⎝⎭f 是()f x 的最小值C .()f x 在区间π0,2⎡⎤⎢⎥⎣⎦上的值域为33,22⎡⎤-⎢⎥⎣⎦D .把函数()y f x =的图象上所有点向右平移π12个单位长度,可得到函数3sin 2y x =的图象,π0,2x ⎡∈⎢⎣3sin(2=,函数y =A .不等式121x x >-的解集为1,12⎛⎫ ⎪⎝⎭B .已知x y z >>,且0x y z ++=,则xy xz >C .正数a ,b 满足191a b+=,若不等式2418a b x x m +≥-++-对任意实数x 恒成立,则实数m 的取值范围是(],6-∞D .若不等式23208kx kx +-<对一切实数x 都成立,则k 的取值范围为(]3,0-()2min 418a b x x m +≥-++-,因为()1999+=++=++102+10=16a ba ba b a b a b bab a≥⎛⎫⨯ ⎪⎝⎭当且仅当=4a ,12b =时取等号,所以241186x x m ≥-++-,()242x x m --≥-()2二、 填空题:本题共4小题,每小题5分,共20分 13.若1tan 3α=-,则3sin 2cos 2sin cos αααα+=-_______. 【答案】35【分析】利用同角三角函数的基本关系,分子、分母同除以【详解】将原式分子、分母同除以2cos 3tan cos 2αα=故答案为:35【点睛】本题考查了同角三角函数的基本关系、齐次式,属于基础题14.如图,四边形ABCD 为平行四边形,11,22AE AB DF FC ==,若AF AC DE λμ=+,则λμ-的值为_________.【分析】选取,AB AD 为基底将向量AF 进行分解,然后与条件对照后得到【详解】选取,AB AD 为基底,则13AF AD DF AB AD=+=+, 又()()122AF AC DE AB AD AB AD AB AD μλμλμλλμ⎛⎫⎛⎫=+=++-=++- ⎪ ⎪⎝⎭⎝⎭,将以上两式比较系数可得1λμ-=. 故答案为:1..如图,在棱长为4的正方体1111ABCD A B C D -中,E 为BC 的中点,点P 在线段1D 上,点Р到直线1CC 的距离的最小值为_______.则(0,4,0),(0,0,4),(2,4,0),(0,4,4)C D E C ,11(2,0,0),(0,0,4),(2,4,4)CE CC ED ===--, ,1(2,4,4)EP ED λλλλ==--, (2CP CE EP =+=-,向量CP 在向量1CC 上投影长为11||4||CP CC d CC λ⋅==而2||(22)CP λ=-,则点Р到直线1CC 的距离2221445||25()555h CP d λ=-=-+≥,当且仅当15=时取“=”,所以点Р到直线CC 45. 故答案为:45.新能源汽车是战略性新兴行业之一,发展新能源汽车是中国从汽车大国迈向汽车强国的必由之路,某汽车企业为了适应市场需求引进了新能源汽车生产设备,2019年该企业新能源汽车的销售量逐月平稳增长,1,2,3月份的销售量分别为1.2千台,1.4千台,1.8千台,为估计以后每个月的销售量,以这三个月的销售量为依据,用一个函数模拟汽车的月销售量y (单位:千台)和月份x 之间的函数关系,有以下两个函数模型可供选择: ∴2()(0)f x ax bx c a =++≠;∴()(0,1)x g x pq r q q ≠,如果4月份的销售量为2.3千台,选择一个效果较好的函数进行模拟,则估计5月份的销售量为________千台.17.在等差数列{}n a 中,已知 12318a a a ++=且45654a a a =++. (1)求{}n a 的通项公式;(2)设14n n n b a a +=⋅,求数列{}n b 的前n 项和n S .解:4n n b a =⋅11122n ⎡⎛⎛=-++ ⎢-⎝⎝⎣18.如图,在四边形ABCD 中,112CA CD AB ===,sin 5BCD ∠=.且______;在∴、∴、∴中选一个作为条件,解答下列问题;∴222AB AC BC AB AC +-=⋅;∴2sin ACB=;∴1AB AC ⋅=.(1)求四边形ABCD 的面积; (2)求sin D 的值.ACDSABCS,相加后求出四边形面积;∴:求出得到ACB ∠ACDS与ABCS,相加后求出四边形面积;SACDS,相加后求出四边形面积;ABC)先求出【详解】(S=ACDS=ABC故四边形AC在ABC中,B=︒所以30由余弦定理得:BC>结合0因为2+BC ACACDS =ABCS=故四边形∴:1AB AC ⋅=,cos AB AC BAC ⋅∠因为()0,πBAC ∠∈,所以因为12CA CD AB ==由余弦定理得:ACDS =ABCS=故四边形由图可知ACD 为锐角三角形,由余弦定理得:cos 0>,解得:学年推行“52+”课后服务.为缓解教师压力,在2021年9月10日教师节大会上该校就是否实行“弹性上下班”进行了调查.另外,为鼓舞广大教职工的工作热情,该校评出了十位先进教师进行表彰﹑并从他们中间选出三名教师作为教师代表在教师节大会上发言.(1)调查结果显示:有23的男教师和35的女教师支持实行“弹性上下班”制,请完成下列22⨯列联表﹒并判断是否有90%的把握认为支持实行“弹性上下班”制与教师的性别相关?(2)已知十位先进教师足按“分层抽样”的模式评选的,用X 表示三位发言教师的女教师人数,求随机变量X 的分布列和数学期望.参考公式:()()()()22()n ad bc K a b c d a c b d -=++++,其中n a b c d =+++.参考数据:将数据代入公式()()()()2n ad bc K a b c d a c b d -=++++,计算得2125 2.315 2.70654K =≈<, 据此可知没有90%的把握认为支持实行“弹性上下班”制与教师的性别相关. (2)依题意,在此十名优秀教师中男教师6人、女教师4人.若用X 表示三位发言教师的女教师人数,则X 的可能取值为:0,1,2,3,其概率分别为:()034631010;6C C P X C ⋅=== ()124631011;2C C P X C ⋅=== ()214631032;10C C P X C ⋅=== ()304631013;30C C P X C ⋅=== 随机变量X 的分布列如下: 随机变量X 的数学期望为:1316123210305EX =⨯+⨯+⨯=20.如图,在四棱锥P ABCD -中,底面ABCD 是矩形,侧棱P A ∴底面ABCD ,点E 为棱PD 的中点,1AB =,2AD AP ==.(1)求证:PB ∴平面ACE ;(2)求平面ACE 与平面P AB 夹角的余弦值;(3)若F 为棱PC 的中点,则棱P A 上是否存在一点G ,使得PC ∴平面EFG .若存在,求线段AG 的长;若不存在,请说明理由. ,,AB AD AP 所在直线分别为标系,求出平面ACE 的一个法向量为n ,利用向量法证明即可;)易得(0,2,0AD =的一个法向量,利用向量求出求解即可;与PC 不垂直,则不可能垂直平面1)因为底面是矩形, AD , 平面ABCD 平面ABCD1⎛⎫所以()()(1,2,0,0,1,1,1,0,AC AE PB ===-设平面ACE 的一个法向量为(),,n x y z =,200n AC x y n AE y z ⎧⋅=+=⎪⎨⋅=+=⎪⎩,即2x y y z =-⎧⎨=-⎩,令1y =,则()2,1,1n =--,又2020n PB ⋅=-++=PB ⊄平面ACE 所以//PB 平面ACE ;2)由(1)可知AB PA AB A =,所以AD ⊥平面PAB ,所以()0,2,0AD =是平面的一个法向量, 设平面PAB 与平面2cos ,6AD n AD n AD n⋅===⋅, 所以平面PAB 与平面ACE 的夹角的余弦值为)因为1,0,02EF ⎛⎫= ⎪⎝⎭,(1,2,PC =所以11002EF PC ⋅=+≠, EF 与PC 不垂直,EF ⊂平面EFG ,PC 不可能垂直平面EFG ,所以棱PA 上不存在点21.已知椭圆:)0(1:2222>>=+b a by a x E 的一个顶点为)1,0(A ,焦距为32.∴1∴求椭圆E 的方程;∴2∴过点(2,1)P -作斜率为k 的直线与椭圆E 交于不同的两点B ,C ,直线AB ,AC 分别与x 轴交于点M ,N ,当||2MN =时,求k 的值.【答案】(1)2214x y +=(2)4k =- 【解析】【分析】(1)依题意可得22212b c c a b =⎧⎪=⎨⎪=-⎩,即可求出a ,从而求出椭圆方程;(2)首先表示出直线方程,设()11,B x y 、()22,C x y ,联立直线与椭圆方程,消元列出韦达定理,由直线AB 、AC 的方程,表示出M x 、N x ,根据N M MN x x =-得到方程,解得即可; 【小问1详解】解:依题意可得1b =,2c =222c a b =-,所以2a =,所以椭圆方程为2214x y +=;【小问2详解】解:依题意过点()2,1P -的直线为()12y k x -=+,设()11,B x y 、()22,C x y ,不妨令1222x x -≤<≤,由()221214y k x x y ⎧-=+⎪⎨+=⎪⎩,消去y 整理得()()22221416816160k x k k x k k +++++=, 所以()()()222216841416160k kk k k ∆=+-++>,解得0k <,所以212216814k k x x k ++=-+,2122161614k kx x k+⋅=+, 直线AB 的方程为1111y y x x --=,令0y =,解得111M xx y =-,直线AC 的方程为2211y y x x --=,令0y =,解得221N xx y =-,所以212111N M x x MN x x y y =-=--- ()()2121121121x x k x k x =--++-++⎡⎤⎡⎤⎣⎦⎣⎦()()212122x x k x k x =+-++ ()()()()2121212222x x x x k x x +-+=++()()12212222x x k x x -==++, 所以()()122122x x k x x -=++, ()212124k x x x x ⎡⎤=+++⎣⎦ 22221616168241414k k k k k kk ⎡⎤⎛⎫++=+-+⎢⎥ ⎪++⎝⎭⎣⎦即()()22221616216841414k k k k k k k ⎡⎤=+-+++⎣⎦+ 整理得4k =,解得4k =-22.已知函数2()e .=--x f x ax b(1)记()()g x f x '=,讨论()g x 的单调性;(2)若对R x ∀∈,都有(1)()0x f x -≥,求实数a 的取值范围.。

2023届辽宁省沈阳市东北育才学校高中部高三第六次模拟英语考题

2023届辽宁省沈阳市东北育才学校高中部高三第六次模拟英语考题

阅读理解Each applicant to Harvard College is considered with great care. We consider each applicant to Harvard College as a whole person, and put enormous care into evaluating every application. We hope you will explore the information in this section to understand what we look for in our admissions process.How to ApplySubmit your application through the Common Application, the Coalition Application, or the Universal College Application. Each is treated equally by the Admissions Committee. Complete and submit your materials as soon as possible to ensure full and timely consideration of your application. View our Application Tips for steppromise【11】A.manage B.operate C.purchase D.pray【12】A.window B.door C.fence D.box【13】A.explored B.fastened C.ignored D.investigated【14】A.groceries B.vegetables C.snacks D.fruits【15】A.held up B.looked into C.checked out D.took over【16】A.thrilled B.touched C.frustrated D.ashamed【17】A.insisted B.determined C.expected rmed 【18】A.money B.check C.ice D.gift【19】A.personally B.secretly C.sacrificially D.excitedly 【20】A.appreciate B.deserve prehend D.accept 【答案】【1】C【2】A【3】B【4】D【5】A【6】D【7】B【8】C【9】A【10】C【11】D【12】B【13】C【14】A【15】B【16】D【17】A【18】D【19】C【20】B【解析】这是一篇记叙文。

海南省2023届高三高考全真模拟(六)历史试题(无答案)

海南省2023届高三高考全真模拟(六)历史试题(无答案)

海南省2023届高三高考全真模拟(六)历史试题学校:___________姓名:___________班级:___________考号:___________一、单选题1.战国时期,各诸侯国实施的编户齐民包括“分家"和“立户”两个过程。

所谓“分家”,商鞅在秦国变法时就明确要求“民有二男以上不分异者,倍其赋”;所谓“立户”,是指以家庭为单位将所有民众编制为户籍人口,使之负责纳税、服役、完成官府任务。

编户齐民的实施()A.减轻了农民的经济负担B.导致了家庭人口的急剧增加C.降低了苛政带来的影响D.有助于构建大一统国家形态二、未知2.孝文帝时期,韩显宗上书道:“使寺署有别,士庶异居”,“令伎作家习士人风礼,则百年难成;令士人儿童效伎作容态,则一朝可得。

是以士人同处,则礼教易兴;伎作杂居,则风俗难改”。

这一言论()A.强调独尊儒家文化B.说明了门阀政治的腐朽性C.意在加强礼乐教化D.强调了移风易俗的重要性三、单选题3.据统计,唐太宗贞观十三年(639年)至唐玄宗天宝元年(742年),全国户数年均增长率为10.59%,而长江中游的荆楚地区鄂岳等道则为13.8%,明显高于全国平均增长率。

这一现象()A.说明荆楚地区开始得到开发B.为经济重心的南移奠定基础C.反映南方社会环境优于北方D.得益于赋税制度的重大变革四、未知五、单选题5.明朝政府建立了以明朝为中心的朝贡贸易圈,羁縻海外诸国。

后来,占据了原来明朝朝贡国领土的西欧殖民者为追逐利益,除了在最初接触时伪装成朝贡国而试图进人朝贡贸易网络外,更多的仗恃武力走私贸易。

据此可知,当时()A.东西方国家实力发生逆转B.传统朝贡贸易体系受到冲击C.明朝主权遭到殖民者侵犯D.海禁政策导致东南倭患严重6.1846年,英使戴维斯请求在西藏定界通商,钦差大臣耆英即以此要求“殊与成约不符”,予以驳斥。

戴维斯又以前往天津为要挟,耆英认为“惟有坚守条约”,“即使驶往天津,所请亦不能允准”。

2021届天津市宁河区芦台第一中学高三下学期第六次模拟英语试卷(word版含答案)

2021届天津市宁河区芦台第一中学高三下学期第六次模拟英语试卷(word版含答案)

芦台一中2020~2021学年度高三第六次模拟英语试卷第一部分:英语知识运用(共两节;满分45分)第一节:单项选择(共15小题;每小题1分,满分15分)从A、B、C、D四个选项中,选出可以填入空白处的最佳选项。

1. —David, I lost your book at school. I suppose I pay for it.—Oh. ________. It was just an old book anyway.A. Take it easy.B. Forget it.C. By no means.D. You got it2. I have heard that it is a common _______ in that country to hold a big party to celebrate when a child turns 15.A. practiceB. methodC. evidenceD. effects3. As the rule says, every student _______ remain seated until their papers are collected.A. canB. mayC. shallD. will4. Hearing the telephone ring, she rushed out of the kitchen and left the water _______.A. boilB. boilingC. boiledD. to boil5. With the word PM2.5 _______ appearing in media reports, people pay greater attention to it and seek health tips for smoggy days.A. immediatelyB. consequentlyC. permanentlyD. constantly6. Hopefully in 2025 we will no longer be e-mailing each other, for we _______ more convenient electronic communication tools by then.A. have developedB. had developedC. will have developedD. developed7. —Can we finish packing these orders some other time?—__________? They are not urgent.A. What forB. Why notC. So whatD. Why bother8. ______ makes our school famous is ______ more than 90% of the students have been admitted to universities.A. What; thatB. Which; becauseC. That; whatD. What; because9. Lisa wouldn’t ______ the job any more. She had a big argument with her boss and resigned.A. come up withB. keep up withC. make up withD. put up with10. I’m sorry you have been waiting so long, but it will still be some time ______ you can get your passport.A. sinceB. tillC. afterD. before11. ______ their final medical check, the astronauts boarded their spacecraft.A. ReceivedB. Being receivedC. Having receivedD. To receive12. There are a great number of attractions in Tianjin, ______ I like the Ancient Cultural Street best.A. whichB. whereC. among themD. among which13. Generally, students’ inner motivation with high expectations from others ______ essential to their development.A. isB. areC. wasD. were14. Some experts think, _____ genes, intelligence also depend on an adequate diet, a good education and a nice home environment.A. instead ofB. apart fromC. except forD. far from15. He doesn’t think that the plan is practical, _____?A. does heB. doesn’t heC. is itD. isn’t it第二节:完形填空(共20小题,每小题1.5分,满分30分)阅读下面短文,掌握其大意,然后从16-35各题所给的A、B、C、D四个选项中,选出最佳选项。

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高考英语模拟题第I卷第一部分听力(共两节,满分30分)(略)第二部分阅读理解(共两节,满分60分)第一节(共15小题;每小题3分,满分45分)ATo avoid becoming the lead story on the evening news, be prepared. Before you head out on a hike, check the weather, take plenty of water, and make sure someone knows where you’ll be and when you’ll be back. Bring clothes to keep you warm when wet. Avoid cotton, which is easy to take in moisture(潮气).“To avoid getting lost, check often to make sure you’re still on the trail(路线),” says John Dill, a search-and-rescue worker at Yosemite National Park in California. “The minute you think you might not be on the trail, stop.” If you’re not alone, focusing on the needs of others can help you get rid of your own fears. Other keys to survival: staying observant and remembering to rest.The surest way to get out alive is to take basic precautions(预防措施), such as taking a set of survival equipment. It includes waterproof matches for starting a fire, a folding knife for cutting branches, and a plastic tarp(防水布) for making shelter.Generally, people who try to find their own way out get along worse than those who stay, says Richard N. Bradley, MD, of the American Red Cross. Find shelter before dark, and try to keep dry. Stay visible so that anyone searching can see you. In a wide-open area, make a signal with colorful tools, make a big X out of rocks, or dig a long shallow hole.You can go several days without eating, so in most cases, you’d better not search for food, since there are lots of poisonous plants in the wild, says Dr Bradley. Y ou cannot survive without water, so if you run out of water, it’s usually better to drink from a stream with suspect water than to go without.21. By saying “becoming the lead story on the evening news” in Para. 1, the author really means ______.A. you will be interviewed by the journalistB. you will become famous for your survival in a forestC. the news that you are lost will spread around quicklyD. your experience in the wild will be reported in the evening news22. According to the writer, how can you avoid getting lost when you are alone?A. Make sure you don’t leave the trail.B. Remember not to stop on the trail.C. Try to look for others’ help.D. Overcome the fear of getting lost.23. According to the passage, you may get into trouble if you ______.A. try to find food to fill your stomach in the wildB. stay in the shelter before darkC. remain where you get lost and wait for helpD. drink water from the stream24. Why does the author suggest digging a long shallow hole in a wide-open area when you get lost?A. To find some water to drink.B. To store some food you’ve brought along.C. To attract the rescuer’s attention.D. To use it as a shelter.BMany New Zealand children are being studied for sleep disorders(紊乱) after overloading on technology before bedtime. Specialists say computers and televisions disturb sleep patterns and cause learning, concentration and growth problems.Dr Alex Bartle, director of the Sleep Well Clinic, said that different sorts of technology were keeping children awake. Computer games and social networking sites such as Bebo and Facebook were worse than TV for exciting young minds because they were more interactive(互动的). They should be switched off at least an hour before bedtime. “To fall asleep, you need to have a certain amount of calmness in the brain,” said Bartle. “being exciteddoesn’t help those brainwaves settle.”Professor Philippe Gander, director of Massey University’s Sleep / Wake Research Centre, said poor sleep had been linked to learning, behavioral and growth problems.Bartle said parents should make sure that children were well-rested and not kept awake by the technology surrounding them. “They need to encourage kids to get out during the day. Watching TV or playing on computers as soon as they get home isn’t good from a sleep point of view. Parents need to be strong enough to remove TVs or computers.”In Britain, National Health Service data show almost 3,000 under-11s were referred to specialists for problems such as sleeplessness and sleep-walking(梦游) in 2007, up 26 percent from 2002.While there are no figures in New Zealand, experts have no doubt that the number of children in need of help is on the increase. “We have the same sorts of technology, the same lifestyles,” said Bartle.The Child Health Research Foundation in New Zealand has funded(资助) studies on sleep problems and is waiting for the results.25. What does “technology”mainly refer to according to the passage?A. pianos and guitarsB. televisions and computersC. learning machinesD.MP3s and MP4s26. According to Dr Alex Bartle’s opinion, we can learn that ______.A. Children should not watch TV or play online games in the eveningB. Watching TV does more harm to children than playing computer gamesC. Some parents fail to set a limit on the time their children spend on technologyD. About 3,000 New Zealand under-11s have sleep problems in 200727. Experts are sure that the number of New Zealand children with sleep problems is increasing because _____.A. New Zealand children play more online games than British childrenB. They have exact data from National Health ServiceC. New Zealand children are in the same situation as the British childrenD. They have exact data provided by hospitals28. What is the main idea of the passage?A. Children like computer games.B. Technology disturbs children’s sleep.C. The ways of dealing with sleep disorders.D. New Zealand parents should prevent children from watching TV.CTo “sacrifice” means to give up something in e xchange for something better. My husband and I decided to make a sacrifice and exchange city life and move to a seaside village with our children aged 9 and 3 years. We had a beautiful home in the city and plenty of money but little real safety because crime was on the increase every day. We never knew if we would all make it safely home at night. I hoped my children would be brought up in a peaceful and friendly environment.We were very happy in our new town, but life was very difficult economically. Income was based on the tourist seasons and jobs were few. Although many of the long- standing residents(居民) were well off in the town, it was a very competitive environment for newcomers. We had to adapt to the difficult new life over the past four years.Reading the newspapers from the big cities made us better believe that we had made the right choice. But it was not always easy to explain to the children why they could no longer have what they used to have.However, our little daughter proved to me she had the ability to think about problems in a reasonable way. The son of a visiting friend preferred riding in the back of our old little truck to his father’s big car.Our little daughter called me and said, “Mom, please help me to explain to James that he is wrong. He says that ‘Today, ifyou don’t have money you are nothing.’ I know that is not true. If you do have money you share it with your friends and if you don’t have money and they do, they share it with you. That is what makes us all rich.” It made me realize that the sacrifice was well worth it.29. Why did the family move away from the city?A. They were tired of the city life.B. The city was not safe enough.C. Pollution in the city was serious.D. Life in the city was expensive.30. What wa s the worst to the author’s family in their new town?A. They couldn’t get used to the weather.B. It was hard for them to get a steady and satisfying income.C. Their neighbours were richer than them.D. They had few friends in the new environment.31. The author realized that their sacrifice was well worth it because _______.A. they had a beautiful home in the seaside villageB. they made a lot of money in the tourist seasonsC. their children could receive good educationD. their children could form a good value concept in the new environmentD.2014 Good Neighbor Dream Vacation PackageStay close to the magic and make your dreams come true. Good Neighbor Hotels offer quality accommodations (住宿) and are located near the Disneyland Resort.Disney’s G ood Neighbor Dream Vacation Package includes:●Accommodations at a conveniently located Disneyland Resort Good Neighbor Hotel.●Disneyland Resort Park Hopper Souvenir ticket, valid (有效的) for same day entry into both Disneyland Park and Disney’s Californi a Adventure Park.●One Magic Morning admission (进入权) into Fantasyland in Disneyland Park, valid with Disneyland Resort Park Hopper souvenir tickets of 3 days or longer.●Get two extra days of Theme Park admission FREE when you buy a 3-day or longer Disneyland Resort Park. Hopper Bonus Ticket from August 7 to October 24, 2010.●Disney Character Calls — Receive a phone call from Mickey, Minnie or Goofy!●Mickey’s Toontown (卡通城) Morning Madness — an interactive Guest and Character experience.● Dream Coin — small collectible objects as souvenirs (one per package).All features of tickets, offers, events, age ranges, services, attractions and entertainment may be seasonal and changeable without notice.*Tickets may not be sold or transferred for commercial use. Offer may not be combined with other discounts or promotions.**Magic Morning allows admission into selected attractions at Disneyland Park before the park opens to the public. Based on availability, does not operate daily, subject to change without notice.***Operation of attractions, entertainment, stores and restaurants, and appearance of Characters may vary and change without notice. Each guest must have a valid multi-day (多天的) admission ticket from Walt Disney Travel Company in order to enter Disn eyland Park and Mickey’s Toontown.32. The biggest attraction of Good Neighbor Hotels may lie in its ____.A. locationB. serviceC. priceD. scenery33. Disney’s Good Neighbor Dream Vacation Package ____.A. offers free meals to touristsB. provides admission into Fantasyland unconditionallyC. includes a phone call from a Disney characterD. gives away several souvenirs to each guest34. If you want to enjoy Mickey’s Toontown Morning Madness, you should ____.A. have a more-than-one-day ticketB. arrive before the park opensC. know the plan is never changeableD. remember your discounts35.The purpose of this passage is to ________.A. introduce the favorable location of Good Neighbor Hotels.B. inform visitors of a free tour to Disney’s Californ ia Adventure ParkC. introduce Disneyland Park and Mickey’s Toontown.D. advertise the 2010 Good Neighbor Dream Vacation Package第二节(共5小题;每小题3分,满分15 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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