全国100所名校最新高考模拟示范卷卷(四)
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全国100所名校2024届高三模拟示范卷(四)理科综合全真演练物理试题(基础必刷)学校:_______ 班级:__________姓名:_______ 考号:__________(满分:100分时间:75分钟)总分栏题号一二三四五六七总分得分评卷人得分一、单项选择题(本题包含8小题,每小题4分,共32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)(共8题)第(1)题亚里士多德在其著作《物理学》中说:一切物体都具有某种“自然本性”,物体由其“自然本性”决定的运动称之为“自然运动”,而物体受到推、拉、提、举等作用后的非“自然运动”称之为“受迫运动”.伽利略、笛卡尔、牛顿等人批判的继承了亚里士多德的这些说法,建立了新物理学;新物理学认为一切物体都具有的“自然本性”是“惯性”.下列关于“惯性”和“运动”的说法中不符合新物理学的是()A.一切物体的“自然运动”都是速度不变的运动——静止或者匀速直线运动B.作用在物体上的力,是使物体做“受迫运动”即变速运动的原因C.可绕竖直轴转动的水平圆桌转得太快时,放在桌面上的盘子会向桌子边缘滑去,这是由于“盘子受到的向外的力”超过了“桌面给盘子的摩擦力”导致的D.竖直向上抛出的物体,受到了重力,却没有立即反向运动,而是继续向上运动一段距离后才反向运动,是由于物体具有惯性第(2)题如图烟雾自动报警器的探测器中装有放射性元素镅241,其衰变方程为。
下列说法正确的是( )A.是光子,不具有能量B.是粒子,有很强的电离本领C.冬天气温较低,镅241的衰变速度会变慢D.镅241衰变过程要吸收能量,故比的原子核更稳定第(3)题铅球掷出后,在空中运动的轨迹如图所示。
a、b、c、d、e为轨迹上5个点,c为轨迹的最高点,不计空气阻力,下列说法正确的是()A.球运动到c点时重力的功率最小,但不为零B.铅球从点运动到点和从点运动到点速度变化方向相同C.铅球从点到点的过程比从点到点的过程中速度变化快D.铅球从点运动到点,合外力做的功大于重力势能的减少量第(4)题一定质量的理想气体的压强、内能的变化与气体体积的温度的关系是()A.如果保持其体积不变,温度升高,则气体的压强增大,内能增大B.如果保持其体积不变,温度升高,则气体的压强增大,内能减少C.如果保持其温度不变,体积增大,则气体的压强减小,内能增大D.如果保持其温度不变,体积增大,则气体的压强减小,内能减少第(5)题如图所示,一束复色光经半圆形玻璃砖分成a、b两束。
全国100所名校最新高考模拟示范卷(四)--语文(150分钟150分)第Ⅰ卷(选择题共36分)一、(18分,每小题3分)1.下列词语中,加点的字读音全都正确的一组是A.症结(zhēng) 折本(shé) 粘液(nián) 生拉硬拽(zhuài)B.曾祖(zēng) 包扎(zhā) 佣金(yòng) 长吁短叹(xū)C.肖像(xiāo) 炫目(xuàn) 反省(xǐng) 悄然无声(qiǎo)D.卡片(kǎ) 铺张(pū) 理发(fà) 街头巷尾(xiàng)2.下列词语中,没有错别字的一组是A.船舷赌博粗犷绿草如茵B.馋言辨别班师书写潦草C.抉择斗殴部署振惊中外D.札记荧屏恬静英雄倍出3.依次填入下列各句横线处的词语,最恰当的一组是①这位82岁的老人虽已是之年.但他精神矍铄,思路清晰。
②人类探测火星十分艰难,是成功入轨后也并不能高枕无忧。
③演员王月在沉寂两年后,引起演艺圈的轰动。
A.古稀虽然东山再起B.耄耋纵然东山再起C.古稀纵然死灰复燃D.耄雀虽然死灰复燃4.下列各句中,标点符号使用正确的一项是A.当卖油翁对他的箭术“睨之”时,陈尧咨(后世称之为“陈康肃公”)愤然曰:“尔安敢轻吾射!”——傲慢且缺乏自制力。
B.洛夫、谢冕两位老先生就当下诗歌边缘化、庸俗化等问题展开了深入探讨,笔者以为,两位先生关于诗歌价值的“对话”令人深思。
C.英国著名历史学家帕金森通过长期调查研究,写了一本“帕金森定律”,这本书中详尽阐述了机构人员膨胀的原因及后果。
D.为什么这些豪华的建筑物,都成了这些组织的“陵墓”呢?以中国传统文化来解释,有这两种可能,一是滥用民力,二是风水不好。
5.下列各旬中,没有语病的一项是A.读《论语》之前,不妨我们找中国近代史有关《论语》的章节来读读,可帮助我们了解《论语》的基本内容及其影响。
B.水资源短缺严重制约着我国西部地区经济的发展和繁荣,能否做好水的开源节流工作是西部大开发成功的关键。
全国100所名校2024届高三模拟示范卷(四)理科综合物理试题学校:_______ 班级:__________姓名:_______ 考号:__________(满分:100分时间:75分钟)总分栏题号一二三四五六七总分得分评卷人得分一、单项选择题(本题包含8小题,每小题4分,共32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)(共8题)第(1)题下列物理量中属于矢量的是A.温度B.重力势能C.电流强度D.电场强度第(2)题如图是一长软细绳,用手握住其一端上下抖动形成的某一时刻的部分波形,绳上a、b两点位置如图所示,相距l。
假设抖动频率不变,不考虑能量损失,根据上述信息能确定以下物理量值的是( )A.波长B.波速C.周期D.振幅第(3)题如图所示,不可伸长的轻绳一端固定在悬点A处,另一端与一小球相连。
现使小球在水平面内绕O点做匀速圆周运动。
测得A至小球球心的距离为L,A、O之间的距离为H。
已知当地重力加速度为g,则小球的运动周期T为( )A.B.C.D.第(4)题双缝干涉实验时,用黄光作为入射光照射双缝,在光屏上出现干涉图样。
改用蓝光再次进行实验,看到的图样( )A.条纹变宽,间距均匀B.条纹变宽,间距不均匀C.条纹变窄,间距均匀D.条纹变窄,间距不均匀第(5)题如图所示,有3根光滑杆AC、BC和BD,其端点正好在同一个竖直的圆周上,A为最高点,D为最低点。
现有一穿孔的小球,分别穿过3根杆从杆的顶端静止滑下,从A到C、B到C、B到D的时间分别计为、、。
下列判断正确的是( )A.B.C.D.第(6)题如图所示,三根长度均为L的轻细绳α、b、c组合系住一质量分布均匀且带正电的小球m,球的直径为,绳b、c与天花板的夹角,空间中存在平行于纸面竖直向下的匀强电场,电场强度,重力加速度为g,现将小球拉开小角度后由静止释放,则( )A.若小球在纸面内做小角度的左右摆动,则周期为B.若小球做垂直于纸面的小角度摆动,则周期为C.摆球经过平衡位置时合力为零D.无论小球如何摆动,电场力都不做功第(7)题如图甲所示,将一块平板玻璃放置在另一块平板玻璃上,在一端加入四张纸片,形成一个空气薄膜。
全国100所名校最新高考模拟示范卷文基/理基试题(理科)(四)单项选择题(75题,共150分。
在每小题给出的四个选项中,只有一项最符合题意。
)1.在玉米种子萌发初期,胚细胞内不可能...存在的代谢途径是A.葡萄糖一二氧化碳+水B.葡萄糖一淀粉C.氨基酸一蛋白质D.葡萄糖一纤维素2.用花药离体培养出马铃薯单倍体植株,当它进行减数分裂时,观察到染色体两两配对形成12对,据此现象可推知产生花药的马铃薯是A.三倍体B.二倍体C.四倍体D.六倍体3.菜田里需经常施加氮肥,但原野的草地却不需要,其主要原因是A.蔬菜会被收割B.草能利用空气中的氮C.草在含氮量较低的泥土中生长最好D.原野的草具有防风固沙的作用4.下列关于细胞中蛋白质的叙述中正确的是①构成染色体②构成“膜结构”③主要的能源物质④多数酶是蛋白质⑤各种激素是蛋白质⑥调节细胞代谢⑦组成维生素A.①②④⑦B.①②④⑤C.②③④⑥D.①②④⑥5.下列各项中,除哪一项外,均包含遗传信息A.染色体B.线粒体C.质粒D.病毒的衣壳6.高原地区生活垃圾和废物难以通过环境的自净作用解决,因为温度过低导致生态系统中某种成分数量很少。
该成分是A.生产者B.消费者C.分解者D.无机环境7.下图表示某池塘中几类生物在一昼夜内释放或吸收CO2量的变化,下列有关叙述中不正..确.的是A.曲线A可以代表池塘中腐生生物呼出CO2量的变化B.曲线B可以代表池塘中藻类吸收或放出CO2量的变化C.2 h左右A类生物CO2释放量下降与气温低有关D.18 h左右B类生物不进行呼吸作用8.2006年度诺贝尔生理学或医学奖授予了两名美国科学家安德鲁·费里和克拉格·米洛,以表彰他们发现了RNA干扰现象。
对这一新发现的理解。
错误的是A.RNA能抑制基因的表达B.RNA会干扰生物体本身的RNA“信使”功能C.RNA干扰现象可用于治疗某些疾病D.这些RNA可不以DNA为模板转录而来9.甲品系的小鼠和乙品系的小鼠的皮肤同时移植于A小鼠身上。
全国100所名校最新高考模拟示范卷卷(四)数学(理科)数 学(理科)本试卷共4页,21小题,满分150分.考试用时120分钟.注意事项:1.答卷前,考生务必用黑色字迹的钢笔或签字笔将自己的姓名和考生号、试室号、座位号填写在答题卡上.用2B 铅笔将试卷类型(A )填涂在答题卡相应位置上.2.选择题每小题选出答案后,用2B 铅笔把答题卡上对应题目选项的答案信息点涂黑,如需改动,用橡皮擦干净后,再选涂其它答案,答案不能答在试卷上.3.非选择题必须用黑色字迹钢笔或签字笔作答,答案必须写在答题卡各题目指定区域内相应位置上;如需改动,先划掉原来的答案,然后再写上新的答案;不准使用铅笔和涂改液.不按以上要求作答的答案无效.4.作答选做题时,请先用2B 铅笔填涂选做题的题号对应的信息点,再将答案填写在对应题号的横线上。
漏涂、错涂、多涂的,答案无效.5.考生必须保持答题卡的整洁.考试结束后,将试卷和答题卡一并交回. 参考公式:锥体的体积公式13V Sh =,其中S 是锥体的底面积,h 是锥体的高. 如果事件A 、B 互斥,那么()()()P A B P A P B +=+. 如果事件A 、B 相互独立,那么()()()P A B P A P B ⋅=⋅.如果事件A 在一次试验中发生的概率是p ,那么在n 次独立重复试验中恰好发生k 次的概率()()C 1n kk kn n P k p p -=-.一、选择题:本大题共8小题,每小题5分,满分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.已知全集U =R ,集合{}22A x x =-<<,{}220B x x x =-≤,则AB =A .()0,2B .(]0,2C .[)0,2D .[]0,22.某赛季,甲、乙两名篮球运动员都参加了11赛得分的情况用如图1所示的茎叶图表示,则甲、乙两名运动员的中位数分别为A .19、13B .13、19C .20、18D .18、203.已知函数2log ,0,()2,0.x x x f x x >⎧=⎨≤⎩若1()2f a =,则a =A .1- BC .1-D .1或4.直线20ax y a -+=与圆229x y +=的位置关系是A .相离B .相交C .相切D .不确定 5.在区间[]0,1上任取两个数,a b ,方程220x ax b ++=的两根均为实数的概率为 A .18 B .14 C .12 D .346.已知a ∈R ,则“2a >”是“22a a >”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7.抽气机每次抽出容器内空气的60%,要使容器内剩下的空气少于原来的0.1%,则至少要抽(参考数据:lg 20.3010=,lg30.4771=)A .15次B .14次C .9次D .8次8.在ABC ∆所在的平面上有一点P ,满足PA PB PC AB ++=,则PBC ∆与ABC ∆的面积之比是 A .13 B .12 C .23 D .34二、填空题:本大题共7小题,每小题5分,满分30分.本大题分为必做题和选做题两部分.(一)必做题:第9、10、11、12题是必做题,每道试题考生都必须做答.9.若复数()()2563i z m m m =-++-是实数,则实数m = . 10.已知3cos 5α=,则cos 2α= . 11.根据定积分的几何意义,计算x =⎰.12.按如图2所示的程序框图运算. 若输入8x =,则输出k = ;若输出2k =,则输入x 的取值范围是 . (注:“1=A ”也可写成“1:=A ”或“1←A ”,均表示赋值语句)(二)选做题:第13、14、15题是选做题,考生只能选做二题,三题全答的,只计算前两题的得分.13.(坐标系与参数方程选做题)在极坐标系中,过点4π⎛⎫⎪⎝⎭作圆4sinρθ=的切线,则切线的极坐标方程是.14.(不等式选讲选做题)若a、b、c∈R,且222236a b c++=,则a b c++的最小值是.15.(几何证明选讲选做题)在平行四边形ABCD中,点E在边AB上,且:1:2AE EB=,DE与AC交于点F,若AEF∆的面积为62cm,则ABC∆的面积为2cm.三、解答题:本大题共6小题,满分80分.解答须写出文字说明、证明过程和演算步骤.16.(本小题满分12分)已知函数()sin cosf x a x b x=+的图象经过点,03π⎛⎫⎪⎝⎭和,12π⎛⎫⎪⎝⎭.(1)求实数a和b的值;(2)当x为何值时,()f x取得最大值.17.(本小题满分12分)某计算机程序每运行一次都随机出现一个二进制的六位数123456N n n n n n n=,其中N的各位数中,161n n==,kn(k=2,3,4,5)出现0的概率为23,出现1的概率为13,记123456n n n n n nξ=+++++,当该计算机程序运行一次时,求随机变量ξ的分布列和数学期望(即均值).18.(本小题满分14分)如图3所示,在边长为12的正方形11AA A A''中,点,B C在线段AA'上,且3AB=,4BC=,作1BB1AA分别交11A A'、1AA'于点1B、P,作1CC 1AA ,分别交11A A '、1AA '于点1C 、Q ,将该正方形沿1BB 、1CC 折叠,使得1A A ''与1AA 重合,构成如图4所示的三棱柱111ABC A B C -.(1)在三棱柱111ABC A B C -中,求证:AB ⊥平面11BCC B ;(2)求平面APQ 将三棱柱111ABC A B C -分成上、下两部分几何体的体积之比; (3)在三棱柱111ABC A B C -中,求直线AP 与直线1AQ 所成角的余弦值.19.(本小题满分14分)已知数列}{n a 中,51=a 且1221n n n a a -=+-(2n ≥且*n ∈N ).(1)若数列2n na λ+⎧⎫⎨⎬⎩⎭为等差数列,求实数λ的值; (2)求数列}{n a 的前n 项和n S .20.(本小题满分14分)已知函数()xf x e x =-(e 为自然对数的底数). (1)求函数()f x 的最小值;(2)若*n ∈N ,证明:1211n nn nn n e n n n n e -⎛⎫⎛⎫⎛⎫⎛⎫++++< ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭⎝⎭⎝⎭. 21.(本小题满分14分)已知抛物线L :22x py =和点()2,2M ,若抛物线L 上存在不同两点A 、B 满足AM BM +=0.(1)求实数p 的取值范围;(2)当2p =时,抛物线L 上是否存在异于A 、B 的点C ,使得经过A 、B 、C 三点的圆和抛物线L 在点C 处有相同的切线,若存在,求出点C 的坐标,若不存在,请说明理由.全国100所名校最新高考模拟示范卷卷(四)数学(理科)数学(理科)试题参考答案及评分标准说明:1.参考答案与评分标准指出了每道题要考查的主要知识和能力,并给出了一种或几种解法供参考,如果考生的解法与参考答案不同,可根据试题主要考查的知识点和能力比照评分标准给以相应的分数.2.对解答题中的计算题,当考生的解答在某一步出现错误时,如果后继部分的解答未改变该题的内容和难度,可视影响的程度决定后继部分的得分,但所给分数不得超过该部分正确解答应得分数的一半;如果后继部分的解答有较严重的错误,就不再给分.3.解答右端所注分数,表示考生正确做到这一步应得的累加分数.4.只给整数分数,选择题和填空题不给中间分.一、选择题:本大题主要考查基本知识和基本运算.共8小题,每小题5分,满分40分.8.由PA PB PC AB ++=,得PA PB BA PC+++=0,即2PC AP =,所以点P 是CA 边上的第二个三等分 点,如图所示.故23PBC ABC S BC PC S BC AC ∆∆⋅==⋅.二、填空题:本大题主要考查基本知识和基本运算.本大题共7小题,每小题5分,满分30分.其中第12题第一个空2分,第二个空3分.9.3 10.725-11.3π 12.4;(]28,57 13.cos 2ρθ= 14. 15.72三、解答题:本大题共6小题,满分80分.解答须写出文字说明、证明过程和演算步骤. 16.(本小题满分12分)(本小题主要考查特殊角的三角函数、三角函数的性质等基础知识,考查运算求解能力) 解:(1)∵函数()sin cos f x a x b x =+的图象经过点,03π⎛⎫⎪⎝⎭和,12π⎛⎫⎪⎝⎭,∴sin cos 0,33sin cos 1.22a b a b ππππ⎧+=⎪⎪⎨⎪+=⎪⎩即10,221.b a +=⎪⎨⎪=⎩解得1,a b =⎧⎪⎨=⎪⎩.(2)由(1)得()sin f x x x =12sin 2x x ⎛⎫= ⎪ ⎪⎝⎭2sin 3x π⎛⎫=- ⎪⎝⎭.∴当sin 13x π⎛⎫-= ⎪⎝⎭,即232x k πππ-=+, 即526x k ππ=+()k ∈Z 时,()f x 取得最大值2. 17.(本小题满分12分)(本小题主要考查随机变量的分布列及其数学期望等基础知识,考查运算求解能力等) 解:ξ的可能取值是2,3,4,5,6.∵161n n ==,∴()4042162C 381P ξ⎛⎫=== ⎪⎝⎭, ()31412323C 3381P ξ⎛⎫==⋅= ⎪⎝⎭,()22241284C 3327P ξ⎛⎫⎛⎫==⋅= ⎪⎪⎝⎭⎝⎭, ()3341285C 3381P ξ⎛⎫==⋅= ⎪⎝⎭, ()444116C 381P ξ⎛⎫=== ⎪⎝⎭. ∴ξ的分布列为∴ξ的数学期望为16322481102345681818181813E ξ=⨯+⨯+⨯+⨯+⨯=.18.(本小题满分14分)(本小题主要考查空间几何体中线面的位置关系,面积与体积,空间向量等基础知识,考查空间想象能力和运算求解能力)(1)证明:在正方形11AA A A''中,∵5A C AA AB BC ''=--=, ∴三棱柱111ABC A B C -的底面三角形ABC 的边5AC =. ∵3AB =,4BC =,∴222AB BC AC +=,则AB BC ⊥.∵四边形11AA A A''为正方形,11AA BB ,∴1AB BB ⊥,而1BCBB B =,∴AB ⊥平面11BCC B . (2)解:∵AB ⊥平面11BCC B ,∴AB 为四棱锥A BCQP -的高.∵四边形BCQP 为直角梯形,且3BP AB ==,7CQ AB BC =+=,∴梯形BCQP 的面积为()1202BCQP S BP CQ BC =+⨯=, ∴四棱锥A BCQP -的体积1203A BCQP BCPQ V S AB -=⨯=,由(1)知1B B AB ⊥,1B B BC ⊥,且AB BC B =,∴1B B ⊥平面ABC .∴三棱柱111ABC A B C -为直棱柱, ∴三棱柱111ABC A B C -的体积为111172ABC A B C ABC V S BB -∆=⋅=.故平面APQ 将三棱柱111ABC A B C -分成上、下两部分的体积之比为722013205-=.(3)解:由(1)、(2)可知,AB ,BC ,1BB 两两互相垂直.以B 为原点,建立如图所示的空间直角坐标系B xyz -, 则()3,0,0A ,()13,0,12A ,()0,0,3P ,()0,4,7Q , ∴(3,0,3)AP =-,1(3,4,5)AQ =--, ∴1111cos ,5AP AQ AP AQ AP AQ ⋅<>==-,∵异面直线所成角的范围为0,2π⎛⎤⎥⎝⎦,∴直线AP 与1AQ 所成角的余弦值为15. 19.(本小题满分14分) (本小题主要考查等比数列、递推数列等基础知识,考查综合运用知识分析问题和解决问题的能力) 解:(1)方法1:∵51=a ,∴22122113a a =+-=,33222133a a =+-=. 设2n n na b λ+=,由}{n b 为等差数列,则有3122b b b +=. ∴321232222a a a λλλ+++⨯=+.∴13533228λλλ+++=+. 解得 1λ=-.事实上,1111122n n n n n n a a b b +++---=-()111212n n n a a ++=-+⎡⎤⎣⎦()1112112n n ++⎡⎤=-+⎣⎦1=,综上可知,当1λ=-时,数列2n na λ+⎧⎫⎨⎬⎩⎭为首项是2、公差是1的等差数列. 方法2:∵数列2n na λ+⎧⎫⎨⎬⎩⎭为等差数列, 设2n n na b λ+=,由}{n b 为等差数列,则有122n n n b b b ++=+(*n ∈N ). ∴12122222n n n n n n a a a λλλ+++++++⨯=+.∴1244n n n a a a λ++=--()()121222n n n n a a a a +++=---()()12221211n n ++=---=-.综上可知,当1λ=-时,数列2n na λ+⎧⎫⎨⎬⎩⎭为首项是2、公差是1的等差数列. (2)由(1)知,()1111122n n a a n --=+-⨯, ∴()121nn a n =+⋅+.∴()()()()12122132121121n nn S n n -⎡⎤=⋅++⋅+++⋅+++⋅+⎣⎦.即()1212232212n n n S n n n -=⋅+⋅++⋅++⋅+.令()1212232212n n n T n n -=⋅+⋅++⋅++⋅, ① 则()23122232212n n n T n n +=⋅+⋅++⋅++⋅. ②②-①,得()()12312222212n n n T n +=-⋅-+++++⋅12n n +=⋅.∴()11221n n n S n n n ++=⋅+=⋅+.20.(本小题满分14分)(本小题主要考查函数的导数、最值、等比数列等基础知识,考查分析问题和解决问题的能力、以及创新意识)(1)解:∵()x f x e x =-,∴()1xf x e '=-.令()0f x '=,得0x =.∴当0x >时,()0f x '>,当0x <时,()0f x '<.∴函数()xf x e x =-在区间(),0-∞上单调递减,在区间()0,+∞上单调递增.∴当0x =时,()f x 有最小值1.(2)证明:由(1)知,对任意实数x 均有1x e x -≥,即1xx e +≤.令k x n=-(*,1,2,,1n k n ∈=-N ),则01k n ke n-<-≤,∴1(1,2,,1)nnkkn k e e k n n --⎛⎫⎛⎫-≤==- ⎪ ⎪⎝⎭⎝⎭.即(1,2,,1)nk n k e k n n --⎛⎫≤=- ⎪⎝⎭.∵1,nn n ⎛⎫= ⎪⎝⎭∴(1)(2)211211n nn nn n n n e e e e n n n n -------⎛⎫⎛⎫⎛⎫⎛⎫++++≤+++++ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭.∵(1)(2)2111111111n n n e eeee e e e e ----------+++++=<=---, ∴ 1211n nn nn n e n n n n e -⎛⎫⎛⎫⎛⎫⎛⎫++++< ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭⎝⎭⎝⎭.21.(本小题满分14分)(本小题主要考查直线与圆锥曲线等基础知识,考查数形结合的数学思想方法,以及推理论证能力、运算求解能力)解法1:(1)不妨设A 211,2x x p ⎛⎫ ⎪⎝⎭,B 222,2x x p ⎛⎫ ⎪⎝⎭,且12x x <, ∵AM BM +=0,∴2212122,22,222x x x x p p ⎛⎫⎛⎫--+--= ⎪ ⎪⎝⎭⎝⎭0.∴124x x +=,22128x x p +=.∵()21222122x x x x ++>(12x x ≠),即88p >,∴1p >,即p 的取值范围为()1,+∞.(2)当2p =时,由(1)求得A 、B 的坐标分别为()0,0、()4,4.假设抛物线L 上存在点2,4t C t ⎛⎫⎪⎝⎭(0t ≠且4t ≠),使得经过A 、B 、C 三点的圆和抛物线L 在点C 处有相同的切线.设经过A 、B 、C 三点的圆的方程为220x y Dx Ey F ++++=,则2420,4432,1641616.F D E F tD t E F t t ⎧=⎪++=-⎨⎪++=--⎩整理得 ()()3441680t E t E ++-+=. ① ∵函数24x y =的导数为2x y '=, ∴抛物线L 在点2,4t C t ⎛⎫ ⎪⎝⎭处的切线的斜率为2t , ∴经过A 、B 、C 三点的圆N 在点2,4t C t ⎛⎫ ⎪⎝⎭处的切线斜率为2t . ∵0t ≠,∴直线NC 的斜率存在.∵圆心N 的坐标为,22D E ⎛⎫-- ⎪⎝⎭, ∴242122t E t D t +⨯=-+,即()()324480t E t E ++-+=. ② ∵0t ≠,由①、②消去E ,得326320t t -+=.即()()2420t t -+=.∵4t ≠,∴2t =-.故满足题设的点C 存在,其坐标为()2,1-.解法2:(1)设A ,B 两点的坐标为1122()()A x y B x y ,,,,且12x x <。
2023届江西省100所名校最新模拟示范卷高三高考全国统一考试英语卷(四)学校:___________姓名:___________班级:___________考号:___________一、阅读理解Support a kindergarten in Bali and teach children important life skills to give them a head-start for school.Typical dayThe program usually runs for around 90 minutes in the morning with the first half of the project centred around play activities with the children and the second half based in the kinder garten classroom supporting the local teacher with activities to keep the children engaged.The most important thing is that you engage the children, get them excited about coming to kindergarten and learning new things. Your role is not limited to just teaching and you are actively encouraged to get involved in other areas such as arts and crafts, physical education and helping local staff in their day-to-day role.In the afternoon, volunteers will have the chance to work at our after-school government approved community program me with the younger children of the Tabanan community. Volunteers will be expected to plan and prepare activities to engage in with the children.What’s included●Accommodation: volunteer house●Meals: breakfast & dinner●Airport pickup●In-country support●V olunteer handbook●Regular program inspectionHighlights●Travel to North Bali in your free time●Motivate children to actively learn important life lessons in a charming way●Free-time activities: water sports, swimming, concert/music, diving / snorkling, climbing, hiking, museum/ opera, yoga / meditation●Discover traditional Balinese markets to learn about the local cuisine, traditionaljewelry and clothingRequirementsMinimum age: 18Criminal background check: requiredEducation requirements: English at high school level1.What is the main task as a member of the program?A.Keeping the children safe on campus.B.Getting the children interested in schooling.C.Helping improve the school environment in Bali.D.Designing various activities for the local community.2.What is provided for the volunteers?A.Air tickets.B.Three meals.C.International support.D.A place to live.3.Which can be one of the exciting points of the project?A.Diving with the children.B.Getting close to rare animals. C.Exploring the local market.D.Experiencing special festivals.A father from Alabama feels favored after his twin sons’ quick thinking helped save his life last month. Brad Hassig was doing underwater exercise at his home swimming pool, something he said he’d done numerous times before. “We were just swimming. The boys were having fun. I like to do just some calming, breathing exercises in the waters, which involves just sitting underwater,” Hassig said. “I don’t ever remember finishing it.”Hassig’s 10-year old twin sons Bridon and Christian, as well as an 11 year old neighbor named Sam, were in the pool with him, enjoying the water, when they noticed something was wrong with their dad, who had turned a blue color. They quickly jumped into action, dragging their 185-pound father above water and toward the side of the pool.“They weigh 80 pounds around, and I weigh 185 pounds. So they should’ t have been able to physically do what they did,” Hassig said. “I mean everything just went as perfectly as it probably had to have.”Although the boys had no formal CPR (心肺复苏术) training, they remembered what they had seen in the movies. They started chest compressions and mouth-to-mouth and Christian and Sam also ran for help after the boys couldn’t unlock their dad’s phone.When Hassig came to life, he said he heard his sons calling out to him. “I hear the boys saying, you know, ‘Daddy, come back!’ ‘Daddy, you have to be OK!’ ” he recalled.Hassig and his sons’ experience calls to mind another recent close call when synchronized swimmer Anita Alvarez lost consciousness during competition and had to be pulled to the surface by her coach, Andrea Fuentes.Since the incident, Hassig has told others to never swim alone and the family urges people to learn how to do CPR. They’re planning a local, community-wide CPR event for kids and adults to get proper training.4.What were the boys doing when their dad got into trouble?A.They were swimming for fun.B.They were calling their neighbour.C.They were making preparations for swimming.D.They were cleaning the home swimming pool.5.What difficulty did the boys have when saving their father?A.They couldn’t pull their father.B.They couldn’t use their father’s phone.C.They didn’t know how to perform CPR.D.They were too frightened to run for help.6.What does Hassig advise people to do after the incident?A.Do underwater exercise before swimming.B.Make sure to swim with companions for safety.C.Never swim alone before getting the formal training.D.Turn to a swimming coach when meeting a similar situation.7.Which of the following can best describe the boys?A.Professional and kind.B.Cooperative and generous. C.Enthusiastic and smart.D.Courageous and calm.Thousands of emperor penguins pack together on the ice of Atka Bay in Antarctica, mostly unaware that among them lives a 3-foot-tall autonomous robot called ECHO. The birds occasionally notice the unmanned and remote controlled ground vehicle out of curiosity but quickly move on from the object, which acts like a mobile antenna(天线)for an observatory monitoring about 300 of them each year.Penguins dominate the South Pole, but the climate crisis could threaten their very existence. A study published last year reported 98% of the emperor penguin population could all but disappear by 2100 due to the impact of climate crisis in Antarctica. “As top predators, emperor penguins serve as ideal species to study in an unsteady ecosystem,” said Zitterbart, associate scientist at the Woods Hole Oceanographic Institution.Surprisingly little is known about these penguins because Antarctica isn’t the easiest place for scientists to access. Although it’s crucial to learn more about the penguins and their ecosystem, Zitterbart and his team didn’t want to introduce a harmful human footprint in an already vulnerable environment or negatively affect the colony.A successful trial run of ECHO this year is already showing how that may be possible.Since 2017, Zitterbart and other researchers have been tagging 300 penguin chicks with a system similar to how dogs and cats are microchipped. But the small sensors worn by the penguins don’t have their own power supply, so they can only be read from about a meter or two away.That’s where ECHO comes in. The robot acts like a receiving station with wireless receivers, automatically collecting data from the penguins’ sensors. With ECHO, the researchers don’t miss out on a chance to collect data when the birds return to the colony to feed their chicks and no longer have to search through a crowd of 20,000 birds to find the tagged ones because ECHO picks up on them automatically.Tracking the penguins allows the team to determine where the penguins go when they dive off the sea ice into the ocean and understand their food hunting strategies. “In the next stage, we will extend ECHO’s data collection to include penguins’ reproductive behaviors that scientists haven’t been able to collect before,” said Zitterbart.8.What’s emperor penguins’ reaction to ECHO?A.Defensive B.Frightened.C.Undisturbed.D.Unfriendly. 9.What does “that” in paragraph 4 refer to?A.Cutting carbon dioxide emission in Antarctica.B.Studying penguins without polluting the land.C.Having access to more knowledge about Antarctica.D.Involving more scientists in studying emperor penguins.10.What is the team likely to study about emperor penguins in the future?A.How they produce young.B.Where they search for food.C.When they dive into the deep sea.D.Why they can survive the extreme cold.11.What can be the best title for the text?A.Emperor penguins are dying out.B.Uncover more mystery of penguins.C.Technology brings life back to Antarctica.D.Meet the robot in the Antarctic penguin colony.Interest in sleep tourism is increasing, with a number of establishments focusing their attention on those suffering from sleep deprivation.Over the past 12 months, Park Hyatt New York has opened the Bryte Restorative Sleep Suite, a 900-square-foot suite filled with sleep-enhancing amenities(设施),while Rosewood Hotels &. Resorts recently launched a collection of retreats called the Alchemy of Sleep, which are designed to “promote rest”, and Swedish manufacturer Hastens established the world’s first Hastens Sleep Spa Hotel, a 15-room boutique hotel a year later.Dr Rebecca Robbins, a sleep researcher and co-author of the book Sleep for Success! believes this shift has been a long time coming, particularly with regards to hotels. “When it comes down to it, travelers book hotels for a place to sleep,” she says, before pointing out that the hotel industry has primarily been focused on things that actually detract from sleep in the past. “People often associate travel with luxury meals, postponing their bed times, the attractions and the things you do while you’re traveling, really almost at the cost of sleep,” she adds.According to Dr Robbins, travel experiences centered around “healthy sleep strategies” that aim to supply guests with the tools they need to improve their sleep can be hugely beneficial, provided a reputable medical or scientific expert is involved in some way to help to deter- mine whether there may be something else at play.Mandarin Oriental in Geneva has taken things a step further by teaming up with CENAS, a private medical sleeping clinic in Switzerland, to curate a three-day program that studies guests’ sleeping patterns in order to identify potential sleeping disorders.As sleep tourism continues to grow, Dr Robbins says she’s looking forward to seeing “who really continues to pioneer and think creatively about this space”, stressing that thereare countless means that haven’t been fully explored yet when it comes to travel and the science of sleep.12.What is the main purpose of paragraph 2?A.To list the consequences of sleep deprivation.B.To prove the popularity of some hotels.C.To highlight the importance of quality sleep.D.To provide evidence for the rise of sleep tourism.13.What does the underlined phrase “detract from” in paragraph 3 probably mean? A.Benefit.B.Influence.C.Measure.D.Analyze. 14.What is special about Mandarin Oriental in Geneva?A.It sets up a private medical sleeping clinic in its hotel.B.It suggests some good and scientific sleeping bedding.C.It provides special tour route for people with sleeping disorders.D.It offers professionally medical help in sleeping problems.15.What does Dr Robbins probably think of the sleep tourism?A.It’s promising.B.It meets some doubts.C.It is facing bottleneck.D.It needs to be systematized.二、七选五Becoming more optimistic can help you see people, situations and tasks with a more positive outlook. ___16___However, optimism is actually an ability that individuals can choose to develop. Here are some steps to become more optimistic.Keep a gratitude journal. Write every day in a gratitude journal. A gratitude journal is a place where you regularly write things you feel grateful for. The things you’re grateful for can be small, such as a good meal or a sunny day.___17___Consistently recognizing the things you’re grateful for can help you refocus on the positives in your life and foster a more optimistic perspective.Do activities you enjoy, Spend time on hobbies or interests that naturally boost your mood or make you laugh. Pursuing activities you enjoy can help you reduce stress levels.___18___Limit your consumption of the news. ___19___ However, most news mediums discussnegative topics or subjects from a negative angle. So consuming too much news can dampen your outlook on life. So consider setting a time limit for how much news you consume regularly. Choosing to only receive your news from a few reputable sources can also help you reduce the amount of news you read or watch daily.___20___ Make actionable plans when facing challenging situations. Some people might think that optimists only perceive the good in all situations. However, optimistic people do recognize difficulties. Instead of dwelling on the negative aspects of a situation, optimists make plans to alter, improve or overcome those challenges.A.Take action against the negative.B.Spend time with optimistic people.C.Optimists are often interested in trying new things.D.It’s important to stay informed about the world around you.E.Some people might believe that people are born with optimism.F.They can also be larger or more complex, such as a loved one or a good job.G.When you feel more joyful, it can become easier to look at situations positively.三、完形填空From the time I went to flea markets with Daddy, I have been ___21___ about antiques. Yesterday’s treasures ground us in what’s lasting and true.Then I heard about a ___22___ antiques venue; the tiny Texas town Round Top. Antiques dealer and show promoter Emma Lee Turney invited the best dealers to ___23___ their antiques for one week. Some 6,000 vintage devotees(古董爱好者) ___24___ there.So in 1995, I went ___25___ to see what all the hit was about. Highway 237, a two lane country road, was ___26___ with vendor(摊贩)tents. I hadn’t gone four yards when I___27___ a small brown-and yellow teapot. Majolica! I’d only seen the European china in magazines. Two tents down, a seen-better-days farm table ___28___ me. I felt a kinship(亲切感)with its ___29___ top and carved initials. “Chips and dents(凹痕)are where the____30____ is,” the vendor said. I rubbed shoulders with shoppers and left with great satisfaction.I loved it all so much, and I ____31____ five years ago.Emma Lee Turney ____32____ this year, but she lived to see her idea become a global____33____ ,with thousands of antiques vendors. And to be ____34____ the mayor of Round Top, one candidate, Mark Massey, raised a campaign ____35____ —Keep Round Top, Round Top. The sign welcoming visitors says Round Top’s population is 90, but the ____36____ during Antiques Week in April and October ____37____ to 90,000.I’d come for vintage ____38____. But there’s something that couldn’t be put in the____39____: the assurance that there’s nothing that can’t be repurposed for greater glory. The magic of Round Top ____40____ me. Ninety-one people can’t be wrong! 21.A.cautious B.crazy C.optimistic D.particular 22.A.normal B.super C.cultural D.potential 23.A.show off B.hand over C.bring back D.pick out 24.A.arose B.lived C.assisted D.gathered 25.A.privately B.consequently C.occasionally D.eagerly 26.A.connected B.tied C.packed D.challenged 27.A.spotted B.lost C.polished D.sold 28.A.called to B.went after C.focused on D.responded to 29.A.regular B.worn C.special D.smooth 30.A.expert B.evidence C.designer D.story 31.A.recovered B.celebrated C.returned D.suffered 32.A.arrived B.died C.sponsored D.settled 33.A.attraction B.goal C.faith D.conflict 34.A.recognized B.honored C.elected D.awarded 35.A.partner B.site C.slogan D.reminder 36.A.coverage B.cost C.position D.attendance 37.A.swells B.refers C.sticks D.relates 38.A.turns B.finds C.trips D.plays 39.A.mind B.show C.town D.suitcase 40.A.includes B.awakens C.refreshes D.fulfills四、用单词的适当形式完成短文阅读下面短文,在空白处填入1个适当的单词或括号内单词的正确形式。
全国100所名校2024届高三模拟示范卷(四)理科综合高效提分物理试题(基础必刷)学校:_______ 班级:__________姓名:_______ 考号:__________(满分:100分时间:75分钟)总分栏题号一二三四五六七总分得分评卷人得分一、单项选择题(本题包含8小题,每小题4分,共32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)(共8题)第(1)题均匀带电圆环电量为、半径为R,位于坐标原点O处,中轴线位于x轴上。
已知若规定无穷远处电势为零,真空中点电荷周围某点的电势可表示为,其中k为静电力常量,q为点电荷的电荷量,r为该点到点电荷的距离;若场源是多个点电荷,电场中某点的场强(电势)为各个点电荷单独在该点产生的场强(电势)的叠加。
取无穷远处电势为零,轴线上一点P的坐标为,下列说法正确的是()A.当时,P点的场强大小为零、电势为零B.当时,P点的场强大小约为,电势约为C.将一带负电的试探电荷从O点沿x轴正方向移动过程中,该试探电荷所受的电场力一直减小D.将一带正电的试探电荷从O点沿x轴正方向移动过程中,该试探电荷的电势能一直增大第(2)题从塔顶同一位置由静止先后释放两个小球A、B,不计空气阻力。
从释放B球开始计时,在A球落地前,两球之间的距离随时间t变化的图像为( )A.B.C.D.第(3)题如图所示,均匀分布有负电荷的橡胶圆环A和金属圆环B为同心圆,保持金属圆环位置固定,让橡胶圆环绕圆心O在金属圆环的平面内沿顺时针方向从静止开始加速转动,下列判断正确的是( )A.金属圆环B中的感应电流沿顺时针方向B.金属圆环B中的感应电流越来越大C.金属圆环B有收缩趋势D.金属圆环B有沿顺时针转动的趋势第(4)题如图所示,一根跨过光滑轻滑轮的轻质细绳两端各系一个小球a、b,b球的质量是a球的2倍,用手托住b球,a球静止于地面,细绳刚好被拉紧,突然松手,b球落地后立刻静止,a球上升的最大高度为h,则松手前b球距地面的高度为( )A.B.C.D.第(5)题如图所示,可视为质点的小球A、B分别同时从倾角为37°的光滑斜面顶端水平抛出和沿斜面下滑,平抛初速度大小为,下滑初速度未知,两小球恰好在斜面底端相遇,重力加速度,,,则( )A.斜面长B.B球初速度C.相遇前,A、B两球始终在同一高度D.相遇前两小球最远相距第(6)题如图所示,一个竖直圆盘转动时,固定在圆盘上的小圆柱带动一个T型支架在竖直方向振动,T型支架下面连接一个弹簧和小球组成的振动系统,小球浸没在水中,当小球振动稳定时( )A.小球振动的频率与圆盘转速无关B.小球振动的振幅与圆盘转速无关C.圆盘的转速越大,小球振动的频率越大D.圆盘的转速越大,小球振动的振幅越大第(7)题在xOy竖直平面内存在沿y轴正方向的匀强电场和垂直于平面向外的匀强磁场,现让一个质量为m,电荷量为q的带正电小球从O点沿y轴正方向射入,已知电场强度大小为,磁感应强度大小为B,小球从O点射入的速度大小为,重力加速度为g,则小球的运动轨迹可能是( )A.B.C.D.第(8)题如图所示,餐厅服务员托举菜盘给顾客上菜.若菜盘沿水平向左加速运动,则( )A.手对菜盘的摩擦力方向向右B.手对菜盘的作用力等于菜盘的重力C.菜盘对手的作用力方向斜向右下D.菜盘对手的作用力方向斜向左下评卷人得分二、多项选择题(本题包含4小题,每小题4分,共16分。
全国 100 所名校最新高考模拟示范卷·理科综合卷(四)化学试卷可能用到的相对原子质量:H-1 C-12 N-14 O-16 Na-23 Al-27 S-32Cl-35.5C. X 与 Y 形成的化合物 X2Y 溶于水所得溶液在常温下 pH< 7 D.Y 与 Z 形成的化合物 Y2Z2中存在极性键和非极性键11. NaOH标准溶液的配制和标定,需经过NaOH溶液配制、基准物质邻苯二甲酸氢钾()的称量以及用NaOH溶液滴定等操作。
下列有关说法正确的是()K-39 Mn-55 Ga-70 As-75Ag-108一.选择题(本题共13 小题,每小题 6 分,在每小题给出的四个选项中,只有一项是符合题目要求的。
)7.明代医学家李时珍所著《本草纲目》中记载了氯化亚汞的制法:“用水银一两,白矾二两,食盐一两,同研不见星,铺于铁器内,以小乌盆覆之⋯⋯封固盆口。
以炭打二炷香取开,则粉升于盆上矣。
”该过程涉及的反应为12Hg + 4KAl(SO4) 2 + 12NaCl + 3O2 = 2K2SO4 + 6Na2SO4 + 2Al 2O3+ 6Hg 2Cl 2。
该过程中不涉及的操作是()A. 研磨B.蒸馏C.加热D.升华 A. 用图甲所示操作转移 NaOH溶液到容量瓶中8. “分子机器”设计和合成有着巨大的研究潜力,蒽醌分子光控反应示意图如图所示。
下列有 B.用图乙所示装置准确称得 2.14g 邻苯二甲酸氢钾固体关说法正确的是() C.用图丙所示操作排去碱式滴定管尖嘴处的气泡D.用图丁所示装置以 NaOH待测溶液滴定邻苯二甲酸氢钾溶液12.钠离子电池具有成本低、能量转换效率高、寿命长等优点。
某种钠离子电池如下图所示,利用钠离子在正负极之间的嵌脱过程实现充放电,该钠离子电池的工作原理为。
下列说法不正确的是()A. X 分子结构中含有 3 个苯环B. Y 分子中所有原子可能共平面C. Y 不能与 H2发生加成反应D. X 与 Y 互为同分异构体9. 设 N A为阿伏加德罗常数的值。
全国100所名校2024届高三模拟示范卷(四)理科综合全真演练物理试题学校:_______ 班级:__________姓名:_______ 考号:__________(满分:100分时间:75分钟)总分栏题号一二三四五六七总分得分评卷人得分一、单项选择题(本题包含8小题,每小题4分,共32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)(共8题)第(1)题下列物理量及对应的国际单位符号,正确的是( )A.速度,B.加速度,m/s C.动能,W D.力,N第(2)题下列说法中正确的是( )A.单晶体有固定的熔点,多晶体没有固定的熔点B.1kg氧气在0℃时分子热运动的平均动能比0.1kg氧气在27℃时分子热运动的平均动能小C.在阳光照射下的教室里,眼睛直接看到的空气中尘粒的运动属于布朗运动D.地面上自由滚动的足球逐渐停下来,违反了能量守恒定律第(3)题原子处于磁场中,某些能级会发生劈裂。
某种原子能级劈裂前后的部分能级图如图所示,相应能级跃迁放出的光子分别设为①②③④。
若用光①照射某金属表面时能发生光电效应,且逸出光电子的最大初动能为E k,则( )A.②光子的频率大于④光子的频率B.①光子的动量与③光子的动量大小相等C.用②照射该金属一定能发生光电效应D.用④照射该金属逸出光电子的最大初动能小于E k第(4)题2022年3月23日,“天宫课堂”第二课开讲,“太空教师”翟志刚、王亚平、叶光富在中国空间站再次为广大青少年带来一堂精彩的太空科普课,其中有一个实验是王亚平在太空拧毛巾,拧出的水形成一层水膜,附着在手上,像手套一样,晃动也不会掉。
形成这种现象的原因,下列说法正确的是( )A.在空间站水滴不受重力B.水和手发生浸润现象,且重力影响很小C.水和手发生不浸润现象,且重力影响很小D.在空间站中水的表面张力变大,使得水“粘”在手上第(5)题2023年5月,中国科学技术大学团队利用纳米纤维与合成云母纳米片研制出一种能适应极端环境的纤维素基纳米纸材料。
全国 100 所名校最新高考模拟示范卷·理科综合卷(四)化学试卷可能用到的相对原子质量:H-1 C-12 N-14 O-16 Na-23 Al-27 S-32Cl-35.5C. X 与 Y 形成的化合物 X2Y 溶于水所得溶液在常温下 pH< 7 D.Y 与 Z 形成的化合物 Y2Z2中存在极性键和非极性键11. NaOH标准溶液的配制和标定,需经过NaOH溶液配制、基准物质邻苯二甲酸氢钾()的称量以及用NaOH溶液滴定等操作。
下列有关说法正确的是()K-39 Mn-55 Ga-70 As-75Ag-108一.选择题(本题共13 小题,每小题 6 分,在每小题给出的四个选项中,只有一项是符合题目要求的。
)7.明代医学家李时珍所著《本草纲目》中记载了氯化亚汞的制法:“用水银一两,白矾二两,食盐一两,同研不见星,铺于铁器内,以小乌盆覆之⋯⋯封固盆口。
以炭打二炷香取开,则粉升于盆上矣。
”该过程涉及的反应为12Hg + 4KAl(SO4) 2 + 12NaCl + 3O2 = 2K2SO4 + 6Na2SO4 + 2Al 2O3+ 6Hg 2Cl 2。
该过程中不涉及的操作是()A. 研磨B.蒸馏C.加热D.升华 A. 用图甲所示操作转移 NaOH溶液到容量瓶中8. “分子机器”设计和合成有着巨大的研究潜力,蒽醌分子光控反应示意图如图所示。
下列有 B.用图乙所示装置准确称得 2.14g 邻苯二甲酸氢钾固体关说法正确的是() C.用图丙所示操作排去碱式滴定管尖嘴处的气泡D.用图丁所示装置以 NaOH待测溶液滴定邻苯二甲酸氢钾溶液12.钠离子电池具有成本低、能量转换效率高、寿命长等优点。
某种钠离子电池如下图所示,利用钠离子在正负极之间的嵌脱过程实现充放电,该钠离子电池的工作原理为。
下列说法不正确的是()A. X 分子结构中含有 3 个苯环B. Y 分子中所有原子可能共平面C. Y 不能与 H2发生加成反应D. X 与 Y 互为同分异构体9. 设 N A为阿伏加德罗常数的值。
全国100所名校最新高考模拟示范卷卷(四)数学(理科)数学(理科)本试卷共4页,21小题,满分150分.考试用时120分钟.注意事项:1.答卷前,考生务必用黑色字迹的钢笔或签字笔将自己的姓名和考生号、试室号、座位号填写在答题卡上.用2B 铅笔将试卷类型(A )填涂在答题卡相应位置上.2.选择题每小题选出答案后,用2B 铅笔把答题卡上对应题目选项的答案信息点涂黑,如需改动,用橡皮擦干净后,再选涂其它答案,答案不能答在试卷上.3.非选择题必须用黑色字迹钢笔或签字笔作答,答案必须写在答题卡各题目指定区域内相应位置上;如需改动,先划掉原来的答案,然后再写上新的答案;不准使用铅笔和涂改液.不按以上要求作答的答案无效.4.作答选做题时,请先用2B 铅笔填涂选做题的题号对应的信息点,再将答案填写在对应题号的横线上。
漏涂、错涂、多涂的,答案无效.5.考生必须保持答题卡的整洁.考试结束后,将试卷和答题卡一并交回. 参考公式:锥体的体积公式13V Sh =,其中S 是锥体的底面积,h 是锥体的高. 如果事件A 、B 互斥,那么()()()P A B P A P B +=+. 如果事件A 、B 相互独立,那么()()()P A B P A P B ⋅=⋅.如果事件A 在一次试验中发生的概率是p ,那么在n 次独立重复试验中恰好发生k 次的概率()()C 1n kk kn n P k p p -=-.一、选择题:本大题共8小题,每小题5分,满分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.已知全集U =R ,集合{}22A x x =-<<,{}220B x x x =-≤,则A B =A .()0,2B .(]0,2C .[)0,2D .[]0,22.某赛季,甲、乙两名篮球运动员都参加了11赛得分的情况用如图1所示的茎叶图表示,则甲、乙两名运动员 图1的中位数分别为A .19、13B .13、19C .20、18D .18、203.已知函数2log ,0,()2,0.x x x f x x >⎧=⎨≤⎩若1()2f a =,则a =A .1-B.1-.1或4.直线20ax y a -+=与圆229x y +=的位置关系是 A .相离B .相交C .相切D .不确定5.在区间[]0,1上任取两个数,a b ,方程220x ax b ++=的两根均为实数的概率为 A .18 B .14 C .12 D .346.已知a ∈R ,则“2a >”是“22a a >”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7.抽气机每次抽出容器内空气的60%,要使容器内剩下的空气少于原来的0.1%,则至少要抽(参考数据:lg 20.3010=,lg30.4771=) A .15次 B .14次C .9次D .8次8.在ABC ∆所在的平面上有一点P ,满足PA PB PC AB ++=,则PBC ∆与ABC ∆的面积之比是 A .13B .12C .23D .34二、填空题:本大题共7小题,每小题5分,满分30分.本大题分为必做题和选做题两部分. (一)必做题:第9、10、11、12题是必做题,每道试题考生都必须做答.9.若复数()()2563i z m m m =-++-是实数,则实数m =. 10.已知3cos 5α=,则cos 2α=. 11.根据定积分的几何意义,计算x =⎰.12.按如图2所示的程序框图运算. 若输入8x =,则输出k =;若输出2k =,则输入x 的取值范围是.(注:“1=A ”也可写成“1:=A ”或“1←A ”,均表示赋值 语句)(二)选做题:第13、14、15题是选做题,考生只能选做二题,三题全答的,只计算前两题的得分. 13.(坐标系与参数方程选做题)在极坐标系中,过点4π⎛⎫⎪⎝⎭作圆4sin ρθ=的切线,则切线的极坐标方程是.14.(不等式选讲选做题)若a 、b 、c ∈R ,且222236a b c ++=,则a b c ++的最小值是. 15.(几何证明选讲选做题)在平行四边形ABCD 中,点E 在边AB 上,且:1:2AE EB =,DE 与AC 交于点F ,若AEF ∆的面积为62cm ,则ABC ∆的面积为2cm .三、解答题:本大题共6小题,满分80分.解答须写出文字说明、证明过程和演算步骤. 16.(本小题满分12分) 已知函数()sin cos f x a x b x =+的图象经过点,03π⎛⎫⎪⎝⎭和,12π⎛⎫⎪⎝⎭. (1)求实数a 和b 的值;(2)当x 为何值时,()f x 取得最大值. 17.(本小题满分12分)某计算机程序每运行一次都随机出现一个二进制的六位数123456N n n n n n n =,其中N 的各位数中,161n n ==,k n (k =2,3,4,5)出现0的概率为23,出现1的概率为13,记123456n n n n n n ξ=+++++,当该计算机程序运行一次时,求随机变量ξ的分布列和数学期望(即均值).18.(本小题满分14分)如图3所示,在边长为12的正方形11AA A A ''中,点,B C 在线段AA '上,且3AB =,4BC =,作1BB1AA分别交11A A '、1AA '于点1B 、P ,作1CC 1AA ,分别交11A A '、1AA '于点1C 、Q ,将该正方形沿1BB 、1CC 折叠,使得1A A ''与1AA 重合,构成如图4所示的三棱柱111ABC A B C -.(1)在三棱柱111ABC A B C -中,求证:AB ⊥平面11BCC B ;(2)求平面APQ 将三棱柱111ABC A B C -分成上、下两部分几何体的体积之比; (3)在三棱柱111ABC A B C -中,求直线AP 与直线1AQ 所成角的余弦值.19.(本小题满分14分)已知数列}{n a 中,51=a 且1221n n n a a -=+-(2n ≥且*n ∈N ).(1)若数列2n na λ+⎧⎫⎨⎬⎩⎭为等差数列,求实数λ的值; (2)求数列}{n a 的前n 项和n S .20.(本小题满分14分)已知函数()xf x e x =-(e 为自然对数的底数). (1)求函数()f x 的最小值;(2)若*n ∈N ,证明:1211n n n nn n e n n n n e -⎛⎫⎛⎫⎛⎫⎛⎫++++< ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭⎝⎭⎝⎭.21.(本小题满分14分)已知抛物线L :22x py =和点()2,2M ,若抛物线L 上存在不同两点A 、B 满足AM BM +=0.(1)求实数p 的取值范围;(2)当2p =时,抛物线L 上是否存在异于A 、B 的点C ,使得经过A 、B 、C 三点的圆和抛物线L 在点C 处有相同的切线,若存在,求出点C 的坐标,若不存在,请说明理由.全国100所名校最新高考模拟示范卷卷(四)数学(理科)数学(理科)试题参考答案及评分标准说明:1.参考答案与评分标准指出了每道题要考查的主要知识和能力,并给出了一种或几种解法供参考,如果考生的解法与参考答案不同,可根据试题主要考查的知识点和能力比照评分标准给以相应的分数.2.对解答题中的计算题,当考生的解答在某一步出现错误时,如果后继部分的解答未改变该题的内容和难度,可视影响的程度决定后继部分的得分,但所给分数不得超过该部分正确解答应得分数的一半;如果后继部分的解答有较严重的错误,就不再给分.3.解答右端所注分数,表示考生正确做到这一步应得的累加分数.4.只给整数分数,选择题和填空题不给中间分.一、选择题:本大题主要考查基本知识和基本运算.共8小题,每小题5分,满分40分.8.由PA PB PC AB ++= ,得PA PB BA PC +++=0,即2PC AP =,所以点P 是CA 边上的第二个三等分点,如图所示.故23PBC ABC S BC PC S BC AC ∆∆⋅==⋅.二、填空题:本大题主要考查基本知识和基本运算.本大题共7小题,每小题5分,满分30分.其中第12题第一个空2分,第二个空3分.9.3 10.725-11.3π12.4;(]28,57 13.cos 2ρθ=14..72三、解答题:本大题共6小题,满分80分.解答须写出文字说明、证明过程和演算步骤. 16.(本小题满分12分)(本小题主要考查特殊角的三角函数、三角函数的性质等基础知识,考查运算求解能力) 解:(1)∵函数()sin cos f x a x b x =+的图象经过点,03π⎛⎫⎪⎝⎭和,12π⎛⎫⎪⎝⎭,∴sin cos 0,33sin cos 1.22a b a b ππππ⎧+=⎪⎪⎨⎪+=⎪⎩即10,221.b a +=⎪⎨⎪=⎩解得1,a b =⎧⎪⎨=⎪⎩.(2)由(1)得()sin f x x x =12sin 2x x ⎛⎫= ⎪ ⎪⎝⎭2sin 3x π⎛⎫=- ⎪⎝⎭.∴当sin 13x π⎛⎫-= ⎪⎝⎭,即232x k πππ-=+, 即526x k ππ=+()k ∈Z 时,()f x 取得最大值2. 17.(本小题满分12分)(本小题主要考查随机变量的分布列及其数学期望等基础知识,考查运算求解能力等) 解:ξ的可能取值是2,3,4,5,6.∵161n n ==,∴()4042162C 381P ξ⎛⎫=== ⎪⎝⎭,()31412323C 3381P ξ⎛⎫==⋅= ⎪⎝⎭,()22241284C 3327P ξ⎛⎫⎛⎫==⋅=⎪ ⎪⎝⎭⎝⎭,()3341285C 3381P ξ⎛⎫==⋅= ⎪⎝⎭,()444116C 381P ξ⎛⎫=== ⎪⎝⎭. ∴ξ的分布列为∴ξ的数学期望为16322481102345681818181813E ξ=⨯+⨯+⨯+⨯+⨯=.18.(本小题满分14分)(本小题主要考查空间几何体中线面的位置关系,面积与体积,空间向量等基础知识,考查空间想象能力和运算求解能力)(1)证明:在正方形11AA A A''中,∵5A C AA AB BC ''=--=, ∴三棱柱111ABC A B C -的底面三角形ABC 的边5AC =. ∵3AB =,4BC =,∴222AB BC AC +=,则AB BC ⊥.∵四边形11AA A A ''为正方形,11AA BB ,∴1AB BB ⊥,而1BC BB B = , ∴AB ⊥平面11BCC B . (2)解:∵AB ⊥平面11BCC B ,∴AB 为四棱锥A BCQP -的高.∵四边形BCQP 为直角梯形,且3BP AB ==,7CQ AB BC =+=,∴梯形BCQP 的面积为()1202BCQP S BP CQ BC =+⨯=, ∴四棱锥A BCQP -的体积1203A BCQP BCPQ V S AB -=⨯=,由(1)知1B B AB ⊥,1B B BC ⊥,且AB BC B = , ∴1B B ⊥平面ABC .∴三棱柱111ABC A B C -为直棱柱, ∴三棱柱111ABC A B C -的体积为111172ABC A B C ABC V S BB -∆=⋅=.故平面APQ 将三棱柱111ABC A B C -分成上、下两部分的体积之比为722013205-=.(3)解:由(1)、(2)可知,AB ,BC ,1BB 两两互相垂直.以B 为原点,建立如图所示的空间直角坐标系B xyz -, 则()3,0,0A ,()13,0,12A ,()0,0,3P ,()0,4,7Q ,∴(3,0,3)AP =-,1(3,4,5)AQ =-- , ∴1111cos ,5AP AQ AP AQ AP AQ ⋅<>==-, ∵异面直线所成角的范围为0,2π⎛⎤⎥⎝⎦,∴直线AP 与1AQ 所成角的余弦值为15. 19.(本小题满分14分) (本小题主要考查等比数列、递推数列等基础知识,考查综合运用知识分析问题和解决问题的能力) 解:(1)方法1:∵51=a ,∴22122113a a =+-=,33222133a a =+-=. 设2n n na b λ+=,由}{n b 为等差数列,则有3122b b b +=. ∴321232222a a a λλλ+++⨯=+.∴13533228λλλ+++=+. 解得1λ=-.事实上,1111122n n n n n n a a b b +++---=-()111212n n n a a ++=-+⎡⎤⎣⎦()1112112n n ++⎡⎤=-+⎣⎦1=,综上可知,当1λ=-时,数列2n na λ+⎧⎫⎨⎬⎩⎭为首项是2、公差是1的等差数列. 方法2:∵数列2n na λ+⎧⎫⎨⎬⎩⎭为等差数列, 设2n n na b λ+=,由}{n b 为等差数列,则有122n n n b b b ++=+(*n ∈N ). ∴12122222n n n n n n a a a λλλ+++++++⨯=+.∴1244n n n a a a λ++=--()()121222n n n n a a a a +++=---()()12221211n n ++=---=-.综上可知,当1λ=-时,数列2n na λ+⎧⎫⎨⎬⎩⎭为首项是2、公差是1的等差数列. (2)由(1)知,()1111122n n a a n --=+-⨯, ∴()121nn a n =+⋅+.∴()()()()12122132121121n nn S n n -⎡⎤=⋅++⋅+++⋅+++⋅+⎣⎦ .即()1212232212n n n S n n n -=⋅+⋅++⋅++⋅+ . 令()1212232212n n n T n n -=⋅+⋅++⋅++⋅ ,①则()23122232212nn n T n n +=⋅+⋅++⋅++⋅ .②②-①,得()()12312222212n n n T n +=-⋅-+++++⋅12n n +=⋅.∴()11221n n n S n n n ++=⋅+=⋅+.20.(本小题满分14分)(本小题主要考查函数的导数、最值、等比数列等基础知识,考查分析问题和解决问题的能力、以及创新意识)(1)解:∵()x f x e x =-,∴()1xf x e '=-.令()0f x '=,得0x =.∴当0x >时,()0f x '>,当0x <时,()0f x '<.∴函数()xf x e x =-在区间(),0-∞上单调递减,在区间()0,+∞上单调递增.∴当0x =时,()f x 有最小值1.(2)证明:由(1)知,对任意实数x 均有1x e x -≥,即1xx e +≤.令k x n =-(*,1,2,,1n k n ∈=-N ),则01k n k e n-<-≤,∴1(1,2,,1)nnkkn k e e k n n --⎛⎫⎛⎫-≤==- ⎪ ⎪⎝⎭⎝⎭. 即(1,2,,1)nk n k e k n n --⎛⎫≤=- ⎪⎝⎭ . ∵1,nn n ⎛⎫= ⎪⎝⎭∴(1)(2)211211n n n nn n n n e e e e n n n n -------⎛⎫⎛⎫⎛⎫⎛⎫++++≤+++++ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭. ∵(1)(2)2111111111n n n e eeee e e e e ----------+++++=<=--- , ∴ 1211n n n nn n e n n n n e -⎛⎫⎛⎫⎛⎫⎛⎫++++< ⎪ ⎪ ⎪ ⎪-⎝⎭⎝⎭⎝⎭⎝⎭.21.(本小题满分14分)(本小题主要考查直线与圆锥曲线等基础知识,考查数形结合的数学思想方法,以及推理论证能力、运算求解能力)解法1:(1)不妨设A 211,2x x p ⎛⎫ ⎪⎝⎭,B 222,2x x p ⎛⎫ ⎪⎝⎭,且12x x <, ∵AM BM +=0 ,∴2212122,22,222x x x x p p ⎛⎫⎛⎫--+--= ⎪ ⎪⎝⎭⎝⎭0.∴124x x +=,22128x x p +=.∵()21222122x x x x ++>(12x x ≠),即88p >,∴1p >,即p 的取值范围为()1,+∞.(2)当2p =时,由(1)求得A 、B 的坐标分别为()0,0、()4,4.假设抛物线L 上存在点2,4t C t ⎛⎫⎪⎝⎭(0t ≠且4t ≠),使得经过A 、B 、C 三点的圆和抛物线L 在点C 处有相同的切线.设经过A 、B 、C 三点的圆的方程为220x y Dx Ey F ++++=,则2420,4432,1641616.F D E F tD t E F t t ⎧=⎪++=-⎨⎪++=--⎩整理得()()3441680t E t E ++-+=.① ∵函数24x y =的导数为2x y '=, ∴抛物线L 在点2,4t C t ⎛⎫ ⎪⎝⎭处的切线的斜率为2t , ∴经过A 、B 、C 三点的圆N 在点2,4t C t ⎛⎫ ⎪⎝⎭处的切线斜率为2t . ∵0t ≠,∴直线NC 的斜率存在.∵圆心N 的坐标为,22D E ⎛⎫-- ⎪⎝⎭, ∴242122t E t D t +⨯=-+,即()()324480t E t E ++-+=.② ∵0t ≠,由①、②消去E ,得326320t t -+=.即()()2420t t -+=.∵4t ≠,∴2t =-.故满足题设的点C 存在,其坐标为()2,1-.解法2:(1)设A ,B 两点的坐标为1122()()A x y B x y ,,,,且12x x <。