2018一中自主招生试题附含答案解析
- 格式:doc
- 大小:421.57 KB
- 文档页数:11
2018-2019年最新乌鲁木齐市第一中学自主招生考试数学模拟精品试卷(第一套)考试时间:90分钟总分:150分一、选择题(本题有12小题,每小题3分,共36分)下面每小题给出的四个选项中,只有一个是正确的,请你把正确选项前的字母填涂在答题卷中相应的格子内.注意可以用多种不同的方法来选取正确答案.1.下列事件中,必然事件是( )A.掷一枚硬币,正面朝上B.a是实数,|a|≥0C.某运动员跳高的最好成绩是20.1米D.从车间刚生产的产品中任意抽取一个,是次品2、如图是奥迪汽车的标志,则标志图中所包含的图形变换没有的是()A.平移变换 B.轴对称变换 C.旋转变换 D.相似变换3.如果□×3ab=3a2b,则□内应填的代数式( )A.ab B.3ab C.a D.3a4.一元二次方程x(x-2)=0根的情况是( )A.有两个不相等的实数根B.有两个相等的实数根C.只有一个实数根D.没有实数根5、割圆术是我国古代数学家刘徽创造的一种求周长和面积的方法:随着圆内接正多边形边数的增加,它的周长和面积越来越接近圆周长和圆面积,“割之弥细,所失弥少,割之又割,以至于不可割,则与圆周合体而无所失矣”。
试用这个方法解决问题:如图,⊙的内接多边形周长为3 ,⊙的外切多边形O周长为3.4,则下列各数中与此圆的周长最接近的是()AB.10D6、今年5月,我校举行“庆五四”歌咏比赛,有17位同学参加选A拔赛,所得分数互不相同,按成绩取前8名进入决赛,若知道某同学分数,要判断他能否进入决赛,只需知道17位同学分数的()A.中位数 B.众数 C.平均数 D.方差7.如图,数轴上表示的是某不等式组的解集,则这个不等式组可能是( )A.Error!B. Error!C.Error!D.Error!8.已知二次函数的图象(0≤x≤3)如图所示,关于该函数在所给自变量取值范围内,下列说法正确的是( )A.有最小值0,有最大值3B.有最小值-1,有最大值0C.有最小值-1,有最大值3D.有最小值-1,无最大值9.如图,矩形OABC的边OA长为2 ,边AB长为1,OA在数轴上,以原点O为圆心,对角线OB的长为半径画弧,交正半轴于一点,则这个点表示的实数是( )A.2.5 B.2 C. D.23510.广场有一喷水池,水从地面喷出,如图,以水平地面为x轴,出水点为原点,建立平面直角坐标系,水在空中划出的曲线是抛物线y =-x2+4x(单位:米)的一部分,则水喷出的最大高度是( )水平面主视方向A .4米B .3米C .2米D .1米11、两个大小不同的球在水平面上靠在一起,组成如图所示的几何体,则该几何体的左视图是( )(A )两个外离的圆 (B )两个外切的圆(C )两个相交的圆 (D )两个内切的圆12.已知二次函数y =ax 2+bx +c (a ≠0)的图象如图所示,有下列结论:①b 2-4ac >0;②abc >0;③8a +c >0;④9a +3b +c <0.其中,正确结论的个数是( )A .1B .2C .3D .4二、填空题(本小题有6小题,每小题4分,共24分)要注意认真看清题目的条件和要填写的内容,尽量完整地填写答案13.当x ______时,分式有意义. 13-x14.在实数范围内分解因式:2a 3-16a =________.15.在日本核电站事故期间,我国某监测点监测到极微量的人工放射性核素碘-131,其浓度为0.0000963贝克/立方米.数据“0.0000963”用科学记数法可表示为________.16.如图,C 岛在A 岛的北偏东60°方向,在B 岛的北偏西45°方向,则从C 岛看A 、B 两岛的视角∠ACB =________.17.若一次函数y =(2m -1)x +3-2m 的图象经过 一、二、四象限,则m 的取值范围是________.18.将一些半径相同的小圆按如图所示的规律摆放,请仔细观察,第 n 个图形有________个小圆. (用含 n 的代数式表示)三、解答题(本大题7个小题,共90分)19.(本题共2个小题,每题8分,共16分)(1).计算:(-1)0+sin45°-2-1 201118。
2018年山东省枣庄实验高中自主招生数学试卷一、选择题(本大题共10小题,每小题3分,共30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.请将正确答案的选项填到二卷答题纸的指定位置处)1.如图,数轴上点A表示数a,则|a﹣1|是()A.1B.2C.3D.﹣22.若关于x的一元二次方程kx2﹣2x﹣1=0有两个不相等的实数根,则实数k的取值范围是()A.k>﹣1B.k>﹣1且k≠0C.k<﹣1D.k<﹣1或k=03.在公园内,牡丹按正方形种植,在它的周围种植芍药,如图反映了牡丹的列数(n)和芍药的数量规律,那么当n=11时,芍药的数量为()A.84株B.88株C.92株D.121株4.某校美术社团为练习素描,他们第一次用120元买了若干本资料,第二次用240元在同一商家买同样的资料,这次商家每本优惠4元,结果比上次多买了20本.求第一次买了多少本资料?若设第一次买了x本资料,列方程正确的是()A.﹣=4B.﹣=4C.﹣=4D.﹣=45.如图,某工厂有甲、乙两个大小相同的蓄水池,且中间有管道连通,现要向甲池中注水,若单位时间内的注水量不变,那么从注水开始,乙水池水面上升的高度h与注水时间t之间的函数关系图象可能是()A.B.C.D.6.如图在水平地面上有一幢房屋BC与一棵树DE,在地面观测点A处测得屋顶C与树稍的仰角分别是45°与60°,∠DCA=90°,在屋顶C处测得∠DCA=90°,若房屋的高BC=5米,则高DE的长度是()A.6米B.6米C.5米D.12米7.某单位组织职工开展植树活动,植树量与人数之间关系如图,下列说法不正确的是()A.参加本次植树活动共有30人B.每人植树量的众数是4棵C.每人植树量的中位数是5棵D.每人植树量的平均数是5棵8.如图,在矩形ABCD中,AB=4,AD=2,分别以点A、C为圆心,AD、CB为半径画弧,交AB 于点E,交CD于点F,则图中阴影部分的面积是()A.4﹣2πB.8﹣C.8﹣2πD.8﹣4π9.如图,是由若干个相同的小立方体搭成的几何体的俯视图和左视图.则小立方体的个数可能是()A.5或6B.5或7C.4或5或6D.5或6或710.如图,在平面直角坐标系中,△ABC的顶点坐标为A(﹣1,1)、B(0,﹣2)、C(1.0),点P(0,2)绕点A旋转180得到点P1,点P1绕点B旋转180°得到点P2,点P2绕点C旋转180°得到点P3,点P3绕点A旋转180°得到点P4,…,按此作法进行下去,则点P2018的坐标为()A.(2,﹣4)B.(0,4)C.(﹣2,﹣2)D.(2,﹣2)二、填空题(本大题共5小题,每小题5分,共25分,把答案填到二卷答题纸的指定位置处)11.若实数a满足a2﹣2a﹣1=0,则2a3﹣7a2+4a﹣2018=12.学校“百变魔方”社团准备购买A、B两种魔方.已知购买2个A种魔方和6个B种魔方共需130元,购买3个A种魔方和4个B种魔方所需款数相同,则购买一套魔方(A、B两种魔方各1个)需元.13.如图,在平面直角坐标系中,正方形OABC的顶点O与坐标原点重合,其边长为2,点A、点C 分别在x 轴、y 轴的正半轴上,函数y =2x 的图象与CB 交于点D ,函数y =(k 为常数,k ≠0)的图象经过点D ,与AB 交于点E ,与函数y =2x 的图象在第三象限内交于点F ,连接AF 、EF ,则△AEF 的面积为 .14.如图,已平行四边形OABC 的三个顶点A 、B 、C 在以O 为圆心的半圆上,过点C 作CD ⊥AB ,分别交AB 、AO 的延长线于点D 、E ,AE 交半圆于点F ,连接CF ,若半圆O 的半径为12,则阴影部分的周长为 .15.庄子说:“一尺之椎,日取其半,万世不竭”.这句话(文字语言)表达了古人将事物无限分割的思想,用图形语言表示为图1,按此图分割的方法,可得到一个等式(符号语言):1=+++…++….图2也是一种无限分割:在△ABC 中,∠C =90°,∠B =30°,过点C 作CC 1⊥AB 于点C 1,再过点C 1作C 1C 2⊥BC 于点C 2,又过点C 2作C 2C 3⊥AB 于点C 3,如此无限继续下去,则可将利△ABC 分割成△ACC 1、△CC 1C 2、△C 1C 2C 3、△C 2C 3C 4、…、△C n ﹣2C n ﹣1∁n 、….假设AC =2,这些三角形的面积和可以得到一个等式是 .三、解答题(共7道题,合计65分,解答应写出文字说明、证明过程或推演步骤,并把答案写在二卷答题纸的指定位置处)16.(7分)先简化,再求值:(),其中x=2,y=.17.(8分)从共享单车,共享汽车等共享出行到共享充电宝,共享雨伞等共享物品,各式各样的共享经济模式在各个领域迅速普及应用,越来越多的企业与个人成为参与者与受益者.根据国家信息中心发布的《中国分享经济发展报告2017》显示,2016年我国共享经济市场交易额约为34520亿元,比上年增长103%;超6亿人参与共享经济活动,比上年增加约1亿人.如图是源于该报告中的中国共享经济重点领域市场规模统计图:(1)请根据统计图解答下列问题:①图中涉及的七个重点领域中,2016年交易额的中位数是亿元.②请分别计算图中的“知识技能”和“资金”两个重点领域从2015年到2016年交易额的增长率(精确到1%),并就这两个重点领域中的一个分别从交易额和增长率两个方面,谈谈你的认识.(2)小宇和小强分别对共享经济中的“共享出行”和“共享知识”最感兴趣,他们上网查阅了相关资料,顺便收集到四个共享经济领域的图标,并将其制成编号为A,B,C,D的四张卡片(除编号和内容外,其余完全相同)他们将这四张卡片背面朝上,洗匀放好,从中随机抽取一张(不放回),再从中随机抽取一张,请用列表或画树状图的方法求抽到的两张卡片恰好是“共享出行”和“共享知识”的概率(这四张卡片分别用它们的编号A,B,C,D表示)18.(9分)鄂州某个体商户购进某种电子产品的进价是50元/个,根据市场调研发现售价是80元/个时,每周可卖出160个,若销售单价每个降低2元,则每周可多卖出20个.设销售价格每个降低x元(x为偶数),每周销售量为y个.(1)直接写出销售量y个与降价x元之间的函数关系式;(2)设商户每周获得的利润为W元,当销售单价定为多少元时,每周销售利润最大,最大利润是多少元?(3)若商户计划下周利润不低于5200元的情况下,他至少要准备多少元进货成本?19.(9分)在四边形ABCD中,∠B+∠D=180°,对角线AC平分∠BAD.(1)如图1,若∠DAB=120°,且∠B=90°,试探究边AD、AB与对角线AC的数量关系并说明理由.(2)如图2,若将(1)中的条件“∠B=90°”去掉,(1)中的结论是否成立?请说明理由.(3)如图3,若∠DAB=90°,探究边AD、AB与对角线AC的数量关系并说明理由.20.(10分)服装店准备购进甲乙两种服装,甲种每件进价80元,售价120元;乙种每件进价60元,售价90元,计划购进两种服装共100件,其中甲种服装不少于65件.(1)若购进这100件服装的费用不得超过7500,则甲种服装最多购进多少件?(2)在(1)条件下,该服装店在5月1日当天对甲种服装以每件优惠a(0<a<20)元的价格进行优惠促销活动,乙种服装价格不变,那么该服装店应如何调整进货方案才能获得最大利润?21.(10分)(1)阅读理解:如图①,在△ABC中,若AB=10,AC=6,求BC边上的中线AD的取值范围.解决此问题可以用如下方法:延长AD到点E使DE=AD,再连接BE(或将△ACD绕着点D逆时针旋转180°得到△EBD),把AB、AC,2AD集中在△ABE中,利用三角形三边的关系即可判断.中线AD的取值范围是;(2)问题解决:如图②,在△ABC中,D是BC边上的中点,DE⊥DF于点D,DE交AB于点E,DF交AC于点F,连接EF,求证:BE+CF>EF;(3)问题拓展:如图③,在四边形ABCD中,∠B+∠D=180°,CB=CD,∠BCD=140°,以C为顶点作一个70°角,角的两边分别交AB,AD于E、F两点,连接EF,探索线段BE,DF,EF之间的数量关系,并加以证明.22.(12分)如图,抛物线y=(x﹣3)2﹣1与x轴交于A,B两点(点A在点B的左侧),与y 轴交于点C,顶点为D.(1)求点A,B,D的坐标;(2)连接CD,过原点O作OE⊥CD,垂足为H,OE与抛物线的对称轴交于点E,连接AE,AD,求证:∠AEO=∠ADC;(3)以(2)中的点E为圆心,1为半径画圆,在对称轴右侧的抛物线上有一动点P,过点P作⊙E的切线,切点为Q,当PQ的长最小时,求点P的坐标,并直接写出点Q的坐标.2018年山东省枣庄实验高中自主招生数学试卷参考答案与试题解析一、选择题(本大题共10小题,每小题3分,共30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.请将正确答案的选项填到二卷答题纸的指定位置处)1.【分析】根据数轴上A点的位置得出a表示的数,利用绝对值的意义计算.【解答】解:根据数轴得:a=﹣2,∴|a﹣1|=|﹣2﹣1|=|﹣3|=3,故选:C.【点评】此题考查了数轴,以及绝对值,熟练掌握绝对值的意义是解本题的关键.2.【分析】利用一元二次方程的定义和判别式的意义得到k≠0且△=(﹣2)2﹣4k•(﹣1)>0,然后其出两个不等式的公共部分即可.【解答】解:根据题意得k≠0且△=(﹣2)2﹣4k•(﹣1)>0,解得k>﹣1且k≠0.故选:B.【点评】本题考查了根的判别式:一元二次方程ax2+bx+c=0(a≠0)的根与△=b2﹣4ac有如下关系:当△>0时,方程有两个不相等的实数根;当△=0时,方程有两个相等的实数根;当△<0时,方程无实数根.3.【分析】根据题目中的图形,可以发现其中的规律,从而可以求得当n=11时的芍药的数量.【解答】解:由图可得,芍药的数量为:4+(2n﹣1)×4,∴当n=11时,芍药的数量为:4+(2×11﹣1)×4=4+(22﹣1)×4=4+21×4=4+84=88,故选:B.【点评】本题考查规律型:图形的变化类,解答本题的关键是明确题意,发现题目中图形的变化规律.4.【分析】由设第一次买了x本资料,则设第二次买了(x+20)本资料,由等量关系:第二次比第一次每本优惠4元,即可得到方程.【解答】解:设他上月买了x本笔记本,则这次买了(x+20)本,根据题意得:﹣=4.故选:D.【点评】此题考查了由实际问题抽象出分式方程.找到关键描述语,找到合适的等量关系是解决问题的关键.5.【分析】根据特殊点的实际意义即可求出答案.【解答】解:因为该做水池就是一个连通器.开始时注入甲池,乙池无水,当甲池中水位到达与乙池的连接处时,乙池才开始注水,所以A、B不正确,此时甲池水位不变,所有水注入乙池,所以水位上升快.当乙池水位到达连接处时,所注入的水使甲乙两个水池同时升高,所以升高速度变慢.在乙池水位超过连通部分,甲和乙部分同时升高,但蓄水池底变小,此时比连通部分快.故选:D.【点评】主要考查了函数图象的读图能力.要能根据函数图象的性质和图象上的数据分析得出函数的类型和所需要的条件,结合实际意义得到正确的结论.6.【分析】首先解直角三角形求得表示出AC,AD的长,进而利用直角三角函数,求出答案.【解答】解:如图,在Rt△ABC中,∠CAB=45°,BC=6m,∴AC==5(m);在Rt△ACD中,∠CAD=60°,∴AD==10(m);在Rt△DEA中,∠EAD=60°,DE=AD•sin60°=5,答:树DE的高为5米.故选:C.【点评】此题主要考查了解直角三角形的应用,熟练应用锐角三角函数关系是解题关键.7.【分析】A、将人数进行相加,即可得出结论A正确;B、由种植4棵的人数最多,可得出结论B 正确;C、由4+10=14,可得出每人植树量数列中第15、16个数为5,即结论C正确;D、利用加权平均数的计算公式,即可求出每人植树量的平均数约是4.73棵,结论D错误.此题得解.【解答】解:A、∵4+10+8+6+2=30(人),∴参加本次植树活动共有30人,结论A正确;B、∵10>8>6>4>2,∴每人植树量的众数是4棵,结论B正确;C、∵共有30个数,第15、16个数为5,∴每人植树量的中位数是5棵,结论C正确;D、∵(3×4+4×10+5×8+6×6+7×2)÷30≈4.73(棵),∴每人植树量的平均数约是4.73棵,结论D不正确.故选:D.【点评】本题考查了条形统计图、中位数、众数以及加权平均数,逐一分析四个选项的正误是解题的关键.8.【分析】用矩形的面积减去半圆的面积即可求得阴影部分的面积.【解答】解:∵矩形ABCD,∴AD=CB=2,∴S阴影=S矩形﹣S半圆=2×4﹣π×22=8﹣2π,故选:C.【点评】本题考查了扇形的面积的计算及矩形的性质,能够了解两个扇形构成半圆是解答本题的关键,难度不大.9.【分析】易得这个几何体共有2层,由俯视图可得第一层立方体的个数,由左视图可得第二层最多和最少小立方体的个数,相加即可.【解答】解:由俯视图易得最底层有4个小立方体,由左视图易得第二层最多有3个小立方体和最少有1个小立方体,那么小立方体的个数可能是5个或6个或7个.故选:D.【点评】本题考查了由三视图判断几何体,也体现了对空间想象能力方面的考查.如果掌握口诀“俯视图打地基,主视图疯狂盖,左视图拆违章”就更容易得到答案.注意俯视图中有几个正方形,底层就有几个小立方体.10.【分析】画出P1~P6,寻找规律后即可解决问题.【解答】解:如图所示,P1(﹣2,0),P2(2,﹣4),P3(0,4),P4(﹣2,﹣2),P5(2,﹣2),P6(0,2),发现6次一个循环,∵2018÷6=336…2,∴点P2018的坐标与P2的坐标相同,即P2018(2,﹣4),故选:A.【点评】本题考查坐标与图形的性质、点的坐标等知识,解题的关键是循环探究问题的方法,属于中考常考题型.二、填空题(本大题共5小题,每小题5分,共25分,把答案填到二卷答题纸的指定位置处)11.【分析】由题意可得a2=2a+1,代入代数式可求值.【解答】解:∵a2﹣2a﹣1=0∴a2=2a+1∴2a3﹣7a2+4a﹣2018=2a(2a+1)﹣7(2a+1)+4a﹣2018=4a2+2a﹣14a﹣7+4a﹣2018=4(2a+1)﹣8a﹣2025=﹣2021故答案为:﹣2021【点评】本题考查了代数式求值,个体代入是本题的关键.12.【分析】设A种魔方的单价为x元/个,B种魔方的单价为y元/个,根据“购买2个A种魔方和6个B种魔方共需130元,购买3个A种魔方和4个B种魔方所需款数相同”,即可得出关于x、y的二元一次方程组,解之即可得出结论.【解答】解:设A种魔方的单价为x元/个,B种魔方的单价为y元/个,根据题意得:,解得:.答:购买一套魔方(A、B两种魔方各1个)需35元.故答案为:35.【点评】本题考查了二元一次方程组的应用,解题的关键是找准等量关系,列出关于x、y的二元一次方程组.13.【分析】根据正方形的性质,以及函数上点的坐标特征可求点D的坐标为(1,2),根据待定系数法可求反比例函数表达式,进一步得到E、F两点的坐标,过点F作FG⊥AB,与AB的延长线交于点G,根据两点间的距离公式可求AE=1,FG=3,再根据三角形面积公式可求△AEF的面积.【解答】解:∵正方形OABC的边长为2,∴点D的纵坐标为2,即y=2,将y=2代入y=2x,得x=1,∴点D的坐标为(1,2),∵函数y=的图象经过点D,∴2=,解得k=2,∴反比例函数的表达式为y=,∴E(2,1),F(﹣1,﹣2);过点F作FG⊥AB,与BA的延长线交于点G,∵E(2,1),F(﹣1,﹣2),∴AE=1,FG=2﹣(﹣1)=3,∴△AEF的面积为:AE•FG=×1×3=,故答案为.【点评】本题主要考查了待定系数法求函数解析式,以及正方形的性质,解题的关键是求得D、E、F点的坐标.14.【分析】根据菱形的判定定理得到四边形OABC为菱形,得到∴△COF为等边三角形,求出∠OCF=60°,根据弧长公式求出的长,根据直角三角形的性质求出EF、CE,得到答案.【解答】解:∵四边形OABC为平行四边形,OA=OC,∴四边形OABC为菱形,∴BA=BC,∴∠CFA=∠COA,∵BC∥AF,∴∠A=∠CFA,∴∠A=∠COA,又∠A+∠COA=180°,∴∠A=60°,∴∠COF=60°,∴△COF为等边三角形,∴∠OCF=60°,∴的长==4π,∵CD⊥AB,∠BDC=60°,∴∠BCD=30°,∴∠ECO=90°,又∠COE=60°,∴∠E=30°,∴OE=2OC=24,∴EF=12,EC==12,∴阴影部分的周长=12+12+4π,故答案为:12+12+4π.【点评】本题考查的是弧长的计算,掌握弧长公式:l=是解题的关键.15.【分析】先根据AC=2,∠B=30°,CC1⊥AB,求得S=;进而得到=△ACC1×,=×()2,=×()3,根据规律可知=×()n﹣1,再根据S=AC×BC=×2×2=2,即可得到等式.△ABC【解答】解:如图2,∵AC=2,∠B=30°,CC1⊥AB,∴Rt△ACC1中,∠ACC1=30°,且BC=2,∴AC1=AC=1,CC1=AC1=,=•AC1•CC1=×1×=;∴S△ACC1∵C1C2⊥BC,∴∠CC1C2=∠ACC1=30°,∴CC2=CC1=,C1C2=CC2=,∴=•CC2•C1C2=××=×,同理可得,=×()2,=×()3,…∴=×()n﹣1,=AC×BC=×2×2=2,又∵S△ABC∴2=+×+×()2+×()3+…+×()n﹣1+…∴2=.故答案为:2=.【点评】本题主要考查了图形的变化类问题,解决问题的关键是找出图形哪些部分发生了变化,是按照什么规律变化的,通过分析找到各部分的变化规律后直接利用规律求解.探寻规律要认真观察、仔细思考,善用联想来解决这类问题.三、解答题(共7道题,合计65分,解答应写出文字说明、证明过程或推演步骤,并把答案写在二卷答题纸的指定位置处)16.【分析】先根据分式的混合运算顺序和运算法则化简原式,再将x、y的值代入计算可得.【解答】解:原式=[﹣]÷=(﹣)•=[﹣]•=•=﹣,当x=2,y=时,原式=﹣=﹣=﹣.【点评】本题主要考查分式的混合运算﹣化简求值,解题的关键是掌握分式的混合运算顺序和运算法则.17.【分析】(1)根据图表将2016年七个重点领域的交易额从小到大罗列出来,根据中位数的定义即可得;(2)将(2016年的资金﹣2015年的资金)÷2015年的资金可分别求得两领域的增长率,结合增长率提出合理的认识即可;(3)画树状图列出所有等可能结果,根据概率公式求解可得.【解答】解:(1)由图可知,2016年七个重点领域的交易额分别为70、245、610、2038、3300、7233、20863,2016年交易额的中位数是2038亿元,故答案为:2038;(2)“知识技能”的增长率为:×100%=205%,“资金”的增长率为:≈109%,由此可知,“知识技能”领域交易额较小,其增长率最高,达到200%以上,其发展速度惊人.(3)画树状图为:共有12种等可能的结果数,其中抽到“共享出行”和“共享知识”的结果数为2,所以抽到“共享出行”和“共享知识”的概率==.【点评】本题主要考查条形统计图、折线统计图和列表法与树状图法求概率,根据条形图得出解题所需数据及画树状图列出所有等可能结果是解题的关键.18.【分析】(1)根据题意,由售价是80元/个时,每周可卖出160个,若销售单价每个降低2元,则每周可多卖出20个,可得销售量y个与降价x元之间的函数关系式;(2)根据题意结合每周获得的利润W=销量×每个的利润,进而利用二次函数增减性求出答案;(3)根据题意,由利润不低于5200元列出不等式,进一步得到销售量的取值范围,从而求出答案.【解答】解:(1)依题意有:y=10x+160;(2)依题意有:W=(80﹣50﹣x)(10x+160)=﹣10(x﹣7)2+5290,因为x为偶数,所以当销售单价定为80﹣6=74元或80﹣8=72时,每周销售利润最大,最大利润是5280元;(3)依题意有:﹣10(x﹣7)2+5290≥5200,解得4≤x≤10,则200≤y≤260,200×50=10000(元).答:他至少要准备10000元进货成本.【点评】此题主要考查了二次函数的应用以及一元二次方程的应用等知识,正确利用销量×每个的利润=W得出函数关系式是解题关键.19.【分析】(1)结论:AC=AD+AB,只要证明AD=AC,AB=AC即可解决问题;(2)(1)中的结论成立.以C为顶点,AC为一边作∠ACE=60°,∠ACE的另一边交AB延长线于点E,只要证明△DAC≌△BEC即可解决问题;(3)结论:.过点C作CE⊥AC交AB的延长线于点E,只要证明△ACE是等腰直角三角形,△DAC≌△BEC即可解决问题;【解答】解:(1)AC=AD+AB.理由如下:如图1中,在四边形ABCD中,∠D+∠B=180°,∠B=90°,∴∠D=90°,∵∠DAB=120°,AC平分∠DAB,∴∠DAC=∠BAC=60°,∵∠B=90°,∴,同理.∴AC=AD+AB.(2)(1)中的结论成立,理由如下:以C为顶点,AC为一边作∠ACE=60°,∠ACE的另一边交AB延长线于点E,∵∠BAC=60°,∴△AEC为等边三角形,∴AC=AE=CE,∵∠D+∠ABC=180°,∠DAB=120°,∴∠DCB=60°,∴∠DCA=∠BCE,∵∠D+∠ABC=180°,∠ABC+∠EBC=180°,∴∠D=∠CBE,∵CA=CE,∴△DAC≌△BEC,∴AD=BE,∴AC=AD+AB.(3)结论:.理由如下:过点C作CE⊥AC交AB的延长线于点E,∵∠D+∠B=180°,∠DAB=90°,∴DCB=90°,∵∠ACE=90°,∴∠DCA=∠BCE,又∵AC平分∠DAB,∴∠CAB=45°,∴∠E=45°.∴AC=CE.又∵∠D+∠ABC=180°,∠D=∠CBE,∴△CDA≌△CBE,∴AD=BE,∴AD+AB=AE.在Rt△ACE中,∠CAB=45°,∴,∴.【点评】本题考查四边形综合题、等边三角形的性质、等腰直角三角形的判定和性质、全等三角形的判定和性质等知识,解题的关键是学会添加常用辅助线,构造全等三角形解决问题,属于中考常考题型.20.【分析】(1)设甲种服装购进x件,则乙种服装购进(100﹣x)件,然后根据购进这100件服装的费用不得超过7500元,列出不等式解答即可;(2)首先求出总利润W的表达式,然后针对a的不同取值范围进行讨论,分别确定其进货方案.【解答】解:(1)设购进甲种服装x件,由题意可知:80x+60(100﹣x)≤7500 解得:x≤75答:甲种服装最多购进75件.(2)设总利润为w元,因为甲种服装不少于65件,所以65≤x≤75,W=(40﹣a)x+30(100﹣x)=(10﹣a)x+3000方案1:当0<a<10时,10﹣a>0,w随x的增大而增大,所以当x=75时,w有最大值,则购进甲种服装75件,乙种服装25件;方案2:当a=10时,所有方案获利相同,所以按哪种方案进货都可以;方案3:10<a<20时,10﹣a<0,w随x的增大而减小,所以当x=65时,w有最大值,则购进甲种服装65件,乙种服装35件.【点评】本题考查了一元一次方程的应用,不等式组的应用,以及一次函数的性质,正确利用x 表示出利润是关键.21.【分析】(1)延长AD至E,使DE=AD,由SAS证明△ACD≌△EBD,得出BE=AC=6,在△ABE中,由三角形的三边关系求出AE的取值范围,即可得出AD的取值范围;(2)延长FD至点M,使DM=DF,连接BM、EM,同(1)得△BMD≌△CFD,得出BM=CF,由线段垂直平分线的性质得出EM=EF,在△BME中,由三角形的三边关系得出BE+BM>EM即可得出结论;(3)延长AB至点N,使BN=DF,连接CN,证出∠NBC=∠D,由SAS证明△NBC≌△FDC,得出CN=CF,∠NCB=∠FCD,证出∠ECN=70°=∠ECF,再由SAS证明△NCE≌△FCE,得出EN=EF,即可得出结论.【解答】(1)解:延长AD至E,使DE=AD,连接BE,如图①所示:∵AD是BC边上的中线,∴BD=CD,在△BDE和△CDA中,,∴△BDE≌△CDA(SAS),∴BE=AC=6,在△ABE中,由三角形的三边关系得:AB﹣BE<AE<AB+BE,∴10﹣6<AE<10+6,即4<AE<16,∴2<AD<8;故答案为:2<AD<8;(2)证明:延长FD至点M,使DM=DF,连接BM、EM,如图②所示:同(1)得:△BMD≌△CFD(SAS),∴BM=CF,∵DE⊥DF,DM=DF,∴EM=EF,在△BME中,由三角形的三边关系得:BE+BM>EM,∴BE+CF>EF;(3)解:BE+DF=EF;理由如下:延长AB至点N,使BN=DF,连接CN,如图3所示:∵∠ABC+∠D=180°,∠NBC+∠ABC=180°,∴∠NBC=∠D,在△NBC和△FDC中,,∴△NBC≌△FDC(SAS),∴CN=CF,∠NCB=∠FCD,∵∠BCD=140°,∠ECF=70°,∴∠BCE+∠FCD=70°,∴∠ECN=70°=∠ECF,在△NCE和△FCE中,,∴△NCE≌△FCE(SAS),∴EN=EF,∵BE+BN=EN,∴BE+DF=EF.【点评】本题考查了三角形的三边关系、全等三角形的判定与性质、角的关系等知识;本题综合性强,有一定难度,通过作辅助线证明三角形全等是解决问题的关键.22.【分析】(1)根据二次函数性质,求出点A、B、D的坐标;(2)如何证明∠AEO=∠ADC?如答图1所示,我们观察到在△EFH与△ADF中:∠EHF=90°,有一对对顶角相等;因此只需证明∠EAD=90°即可,即△ADE为直角三角形,由此我们联想到勾股定理的逆定理.分别求出△ADE三边的长度,再利用勾股定理的逆定理证明它是直角三角形,由此问题解决;(3)依题意画出图形,如答图2所示.由⊙E的半径为1,根据切线性质及勾股定理,得PQ2=EP2﹣1,要使切线长PQ最小,只需EP长最小,即EP2最小.利用二次函数性质求出EP2最小时点P的坐标,并进而求出点Q的坐标.【解答】方法一:(1)解:顶点D的坐标为(3,﹣1).令y=0,得(x﹣3)2﹣1=0,解得:x1=3+,x2=3﹣,∵点A在点B的左侧,∴A(3﹣,0),B(3+,0).(2)证明:如答图1,过顶点D作DG⊥y轴于点G,则G(0,﹣1),GD=3.令x=0,得y=,∴C(0,).∴CG=OC+OG=+1=,∴tan∠DCG=.设对称轴交x轴于点M,则OM=3,DM=1,AM=3﹣(3﹣)=.由OE⊥CD,易知∠EOM=∠DCG.∴tan∠EOM=tan∠DCG==,解得EM=2,∴DE=EM+DM=3.在Rt△AEM中,AM=,EM=2,由勾股定理得:AE=;在Rt△ADM中,AM=,DM=1,由勾股定理得:AD=.∵AE2+AD2=6+3=9=DE2,∴△ADE为直角三角形,∠EAD=90°.设AE交CD于点F,∵∠AEO+∠EFH=90°,∠ADC+∠AFD=90°,∠EFH=∠AFD(对顶角相等),∴∠AEO=∠ADC.(3)解:依题意画出图形,如答图2所示:由⊙E的半径为1,根据切线性质及勾股定理,得PQ2=EP2﹣1,要使切线长PQ最小,只需EP长最小,即EP2最小.设点P坐标为(x,y),由勾股定理得:EP2=(x﹣3)2+(y﹣2)2.∵y=(x﹣3)2﹣1,∴(x﹣3)2=2y+2.∴EP2=2y+2+(y﹣2)2=(y﹣1)2+5当y=1时,EP2有最小值,最小值为5.将y=1代入y=(x﹣3)2﹣1,得(x﹣3)2﹣1=1,解得:x1=1,x2=5.又∵点P在对称轴右侧的抛物线上,∴x1=1舍去.∴P(5,1).∵△EQ2P为直角三角形,∴过点Q2作x轴的平行线,再分别过点E,P向其作垂线,垂足分别为M点和N点.由切割线定理得到Q2P=Q1P=2,EQ2=1设点Q2的坐标为(m,n)则在Rt△MQ2E和Rt△Q2NP中建立勾股方程,即(m﹣3)2+(n﹣2)2=1①,(5﹣m)2+(n ﹣1)2=4②①﹣②得n=2m﹣5③将③代入到①得到m1=3(舍,为Q1)m2=再将m=代入③得n=,∴Q2(,)此时点Q坐标为(3,1)或(,).方法二:(1)略.(2)∵C(0,),D(3,﹣1),∴KCD=,∵OE⊥CD,∴K CD×K OE=﹣1,∴K OE=,∴l OE:y=x,把x=3代入,得y=2,∴E(3,2),∵A(3﹣,0),D(3,﹣1),∴K EA==,∵K AD=,∴K EA×K AD=﹣1,∴EA⊥AD,∠EHD=∠EAD,∵∠EFH=∠AFD,∴∠AEO=∠ADC.(3)由⊙E的半径为1,得PQ2=EP2﹣1,要使切线长PQ最小,只需EP长最小,即EP2最小,设点P坐标为(x,y),EP2=(x﹣3)2+(y﹣2)2,∵y=(x﹣3)2﹣1,∴(x﹣3)2=2y+2,∴EP2=2y+2+(y﹣2)2=(y﹣1)2+5,∴当y=1时,EP2有最小值,将y=1代入y=(x﹣3)2﹣1得:x1=1,x2=5,又∵点P在对称轴右侧的抛物线上,∴x1=1舍去,∴P(5,1),显然Q1(3,1),∵Q1Q2被EP垂直平分,垂足为H,∴K Q1Q2×K EP=﹣1,∴K EP==﹣,K Q1Q2=2,∵Q1(3,1),∴l Q1Q2:y=2x﹣5,∵l EP:y=﹣x+,∴x=,y=,∴H(,),∵H为Q1Q2的中点,∴H x=,H Y=,∴Q2(x)=2×﹣3=,Q2(Y)=2×﹣1=,∴Q2(,).【点评】本题是二次函数压轴题,涉及考点众多,难度较大.第(2)问中,注意观察图形,将问题转化为证明△ADE为直角三角形的问题,综合运用勾股定理及其逆定理、三角函数(或相似形)求解;第(3)问中,解题关键是将最值问题转化为求EP2最小值的问题,注意解答中求EP2最小值的具体方法.。
2018-2019年最新南海区石门中学自主招生语文模拟精品试卷(第一套)(满分:100分考试时间:90分钟)③小屋在山的怀抱中,犹如在花蕊中一般,慢慢地花蕊绽开了一些,好像山后退了一些。
④当花瓣微微收拢,那就是夜晚来临了。
⑤小屋的光线既富于科学的时间性,也富于浪漫的文学性。
A.①③②④⑤ B.①④③②⑤ C.⑤③②①④ D.⑤③②④①二、阅读下面古诗文,完成7—14题。
(24分,7—12每题2分)勾践自会稽归七年,拊循其士民,欲用以报吴。
大夫逄同谏曰:“今夫吴兵加齐、晋,怨深于楚﹑越,名高天下,实害周室,德少而功多,必淫自矜。
为越计,莫若结齐,亲楚,附晋,以厚吴。
吴之志广,必轻战。
是我连其权,三国伐之,越承其弊,可克也。
”勾践曰:“善。
”其后四年。
吴士民罢弊,轻锐尽死于齐﹑晋。
而越大破吴,因而留围之三年,吴师败,越遂复栖吴王于姑苏之山。
吴王使公孙雄肉袒膝行而前,请成越王曰:“孤臣夫差敢布腹心,异日尝得罪于会稽,夫差不敢逆命,得与君王成以归。
今君王举玉趾而诛孤臣,孤臣惟命是听,意者亦欲如会稽之赦孤臣之罪乎?”勾践不忍,欲许之。
范蠡曰:“会稽之事,天以越赐吴,吴不取。
今天以吴赐越,越其可逆天乎?且夫君王蚤朝晏罢,非为吴邪?谋之二十二年,一旦而弃之,可乎?且夫天与弗取,反受其咎。
君忘会稽之厄乎?”勾践曰:“吾欲听子言,吾不忍其使者。
”范蠡乃鼓进兵,曰:“王已属政于执事,使者去,不者且得罪。
”吴使者泣而去。
勾践怜之,乃使人谓吴王曰:“吾置王甬东,君百家。
”吴王谢曰:“吾老矣,不能事君王!”遂自杀。
选自《史记·越王勾践世家》7.下列加点词语解释不正确的一项是( )A.越承其弊,可克也。
克:战胜 B.越遂复栖吴王于姑苏之山 栖:占领C.越其可逆天乎 逆:违背 D.吾老矣,不能事君王 事:侍奉8.下列加点词语古今意义相同的是( )A.今天以吴赐越 B.使者去,不者且得罪 C.谋臣与爪牙之士,不可不养而择也 D.微夫人之力不及此9.下列加点词语的用法和意义相同的一组是( )A.①德少而功多,必淫自矜 ②鼓瑟希,铿尔,舍瑟而作B.①得与君王成以归 ②王好战,请以战喻C.①亦欲如会稽之赦孤臣之罪 ②邻国之民不加少D.①异日尝得罪于会稽 ②吾长见笑于大方之家10.下列加点词语属于谦称的是( )A.吾欲听子言 B.君忘会稽之厄乎? C.君王举玉趾而诛孤臣 D.孤臣夫差敢布腹心11.下列句子,全都表现勾践具有仁慈之心的一项是( )①孤臣惟命是听②勾践不忍,欲许之。
2018年浙江省温州市苍南中学自主招生数学试卷一、选择题(本大题共8小题,每小题5分,共40分.在每小题给出的四个选项中,只有一项是符合题目要求的)1.(5分)有一正方体,六个面上分别写有数字1,2,3,4,5,6,有三个人从不同的角度观察的结果如图.如果记6的对面的数字为a,2的对面的数字为b,那么a+b的值为()A.3B.7C.8D.112.(5分)已知函数y=2018﹣(x﹣m)(x﹣n),并且a,b是方程2018﹣(x﹣m)(x﹣n)=0的两个根,则实数m,n,a,b的大小关系可能是()A.m<a<b<n B.m<a<n<b C.a<m<b<n D.a<m<n<b 3.(5分)如图,矩形ABCD中,AB=CD=x,AD=BC=y,把它折叠起来,使顶点A与C 重合,则折痕PQ的长度为()A.B.C.D.4.(5分)已知关于x的不等式组恰有三个整数解,则实数a的取值范围是()A.﹣3<a<﹣2B.﹣3≤a<﹣2C.﹣3<a≤﹣2D.﹣3≤a≤﹣2 5.(5分)设x、y、z是两两不等的实数,且满足下列等式:,则x3+y3+z3﹣3xyz的值是()A.0B.1C.3D.条件不足,无法计算6.(5分)某粮店用一架不准确的天平(两臂长不相等)称大米.某顾客要购买10kg大米,售货员先将5kg砝码放入天平左盘,置大米于右盘,平衡后将大米给顾客;然后又将5kg砝码放入天平右盘,置大米于左盘,平衡后再将大米给顾客.售货员的这种操作方式结果使()A.粮店吃亏B.顾客吃亏C.粮店和顾客都不吃亏D.不能确定7.(5分)设P是高为h的正三角形内的一点,P到三边的距离分别为x,y,z(x≤y≤z).若以x,y,z为边可以组成三角形,则z应满足的条件为()A.h≤z h B.h≤z h C.h≤z h D.8.(5分)已知y=x3+ax2+bx+c,当x=5时,y=50;x=6时,y=60;x=7时,y=70.则当x=4时,y的值为()A.30B.34C.40D.44二、填空题(本题有10个小题,每小题6分,共60分)9.(6分)方程x2﹣3|x﹣1|﹣1=0所有解的和为.10.(6分)设a、b、c、d、e的值均为0、1、2中之一,且a+b+c+d+e=6,a2+b2+c2+d2+e2=10,则a3+b3+c3+d3+e3的值为.11.(6分)设a,b为两个不相等的实数,且满足2a2﹣5a=2b2﹣5b=1,则ab3+a3b的值是12.(6分)已知(2x﹣1)9=a0+a1x+a2x2+……+a9x9,则a1+a2+……+a8+a9的值为.13.(6分)已知四边形ABCD是正方形,且边长为2,延长BC到E,使CE=﹣,并作正方形CEFG,(如图),则△BDF的面积等于.14.(6分)向一个三角形内加入2015个点,加上原三角形的三个点共计2018个点.用剪刀最多可以剪出个以这2018个点为顶点的三角形.15.(6分)如图,直角△ABC中,∠ABC=90°,∠A=20°,△ABC绕点B旋转至△A'BC'的位置,此时C点恰落在A'C'上,且A'B与AC交于D点,那么∠BDC=度.16.(6分)已知:对于正整数n,有,若某个正整数k满足,则k=.17.(6分)用f(n)表示组成n的数字中不是零的所有数字乘积,例如:f(5)=5,f(29)=18,f(207)=14.则f(1)+f(2)+……+f(200)=.18.(6分)设x1、x2是方程x2﹣6x+a=0的两个根,以x1、x2为两边长的等腰三角形只可以画出一个,则实数a的取值范围是.三、解答题(本大题共3题,共50分.解答应写出文字说明、证明过程或演算步骤)19.(16分)甲、乙两个粮库原来各存有整袋的粮食,如果从甲库调90袋到乙库,则乙库存粮是甲库的2倍;如果从乙库调若干袋到甲库,则甲库存粮是乙库的6倍.问甲库原来最少存粮多少袋?20.(16分)如图,函数的图象交y轴于M,交x轴于N,点P是直线MN上任意一点,PQ⊥x轴,Q是垂足,设点Q的坐标为(t,0),△POQ的面积为S(当点P与M、N重合时,其面积记为0).(1)试求S与t之间的函数关系式;(2)在如图所示的直角坐标系内画出这个函数的图象,并利用图象求使得S=a(a>0)的点P的个数.21.(18分)已知二次函数y=ax2+bx+c的图象经过点(﹣2,0),且对一切实数x,都有2x ≤ax2+bx+c≤x2+2成立.(1)当x=2时,求y的值;(2)求此二次函数的表达式;(3)当x=t+m时,二次函数y=ax2+bx+c的值为y1,当x=时,二次函数y=ax2+bx+c 的值为y2,若对一切﹣1≤t≤1,都有y1<y2,求实数m的取值范围.2018年浙江省温州市苍南中学自主招生数学试卷参考答案与试题解析一、选择题(本大题共8小题,每小题5分,共40分.在每小题给出的四个选项中,只有一项是符合题目要求的)1.(5分)有一正方体,六个面上分别写有数字1,2,3,4,5,6,有三个人从不同的角度观察的结果如图.如果记6的对面的数字为a,2的对面的数字为b,那么a+b的值为()A.3B.7C.8D.11【分析】由图一和图二可看出1的对面的数字是5;再由图二和图三可看出3的对面的数字是6,从而2的对面的数字是4.【解答】解:从3个小立方体上的数可知,与写有数字1的面相邻的面上数字是2,3,4,6,所以数字1面对数字5,同理,立方体面上数字3对6.故立方体面上数字2对4.则a=3,b=4,那么a+b=3+4=7.故选:B.【点评】本题考查灵活运用正方体的相对面解答问题,立意新颖,是一道不错的题.解题的关键是按照相邻和所给图形得到相对面的数字.2.(5分)已知函数y=2018﹣(x﹣m)(x﹣n),并且a,b是方程2018﹣(x﹣m)(x﹣n)=0的两个根,则实数m,n,a,b的大小关系可能是()A.m<a<b<n B.m<a<n<b C.a<m<b<n D.a<m<n<b 【分析】首先把方程化为一般形式,由于a,b是方程的解,根据根与系数的关系即可得到m,n,a,b之间的关系,然后对四者之间的大小关系进行讨论即可判断.【解答】解:由2018﹣(x﹣m)(x﹣n)=0变形得(x﹣m)(x﹣n)=2018,∴x﹣m>0,x﹣n>0或x﹣m<0,x﹣n<0,∴x>m,x>n或x<m,x<n,∵a,b是方程的两个根,将a,b代入,得:a>m,a>n,b<m,b<n或a<m,a<n,b>m,b>n,观察选项可知:a<b,m<n,只有D可能成立.故选:D.【点评】本题考查了抛物线与x轴的交点,根与系数的关系,难度较大,关键是对m,n,a,b大小关系的讨论是此题的难点.3.(5分)如图,矩形ABCD中,AB=CD=x,AD=BC=y,把它折叠起来,使顶点A与C 重合,则折痕PQ的长度为()A.B.C.D.【分析】由翻折可得到QP垂直平分AC,那么AQ=QC,易证△APO≌△CQO,再利用勾股定理求出AP的长,进而利用菱形的面积等于对角线乘积的一半,求出PQ的长即可.【解答】解:∵A,C两点关于PQ对称,所以AO=CO,∵AC⊥QP,从而∠AOP=∠QOC=90°,∵四边形ABCD是矩形,∴AB∥DC,∴∠APQ=∠PQC.∴△APO≌△CQO,∴CQ=AP,由PQ⊥AC且平分AC,可知AQ=CQ.∴四边形AQCP是菱形,设AP=a,则AQ=a,DQ=x﹣a,在Rt△ADQ中,利用勾股定理可知:a2=y2+(x﹣a)2,∴整理得:2ax=x2+y2,解得a=,菱形AQCP的面积为:PQ•AC=CQ•AD,∴PQ×=×y,整理得:PQ×=×y,解得:PQ=.故选:A.【点评】此题主要考查了翻折变换的性质和矩形的性质以及菱形的判定与性质等知识,遇到折叠变换问题注意找出翻折的边得出对应相等,再利用勾股定理求出,这是此类问题常用解题思路.4.(5分)已知关于x的不等式组恰有三个整数解,则实数a的取值范围是()A.﹣3<a<﹣2B.﹣3≤a<﹣2C.﹣3<a≤﹣2D.﹣3≤a≤﹣2【分析】首先熟练解得每个不等式,再根据它恰有三个整数解,分析出它的整数解,进而求得实数a的取值范围.【解答】解:由①,得x≥a+1;由②,得x<2.根据题意,得它的三个整数解只能是﹣1,0,1,所以﹣2<a+1≤﹣1,解得﹣3<a≤﹣2.故选:C.【点评】此题考查了不等式组的解法,同时能够根据它的整数解正确分析其字母的取值范围.5.(5分)设x、y、z是两两不等的实数,且满足下列等式:,则x3+y3+z3﹣3xyz的值是()A.0B.1C.3D.条件不足,无法计算【分析】由二次根式有意义可知x﹣z≥0,x3(y﹣x)3≥0,x3(z﹣x)3≥0,可得x=0,y=﹣z.代入代数式即可求解.【解答】解:依题意得:,解得x=0,∵,∴,∴y=﹣z∴把x=0,y=﹣z代入x3+y3+z3﹣3xyz得:原式=(﹣z)3+z3=0故选:A.【点评】此题考查了二次根式的有意义时被开方数是非负数的性质与不等式组解集的求解方法.此题比较难,注意仔细分析.6.(5分)某粮店用一架不准确的天平(两臂长不相等)称大米.某顾客要购买10kg大米,售货员先将5kg砝码放入天平左盘,置大米于右盘,平衡后将大米给顾客;然后又将5kg 砝码放入天平右盘,置大米于左盘,平衡后再将大米给顾客.售货员的这种操作方式结果使()A.粮店吃亏B.顾客吃亏C.粮店和顾客都不吃亏D.不能确定【分析】此题要根据天平的有关知识来解答,即在此题中天平的臂长不等,这是此题的关键.【解答】解:由于天平的两臂不相等,故可设天平左臂长为a,右臂长为b(不妨设a>b),先称得的大米的实际质量为m1,后称得的大米的实际质量为m2由杠杆的平衡原理:bm1=a×5,am2=b×5,解得m1=,m2=则m1+m2=+下面比较m1+m2与10的大小:(求差比较法)因为(m1+m2)﹣10=+﹣10=>0又因为a≠b,所以(m1+m2)﹣10>0,即m1+m2>10这样可知称出的大米质量大于10kg,商店吃亏.故选:A.【点评】此题学生要利用物理知识来求解,所以学生平时在学习时要各科融汇贯通.7.(5分)设P是高为h的正三角形内的一点,P到三边的距离分别为x,y,z(x≤y≤z).若以x,y,z为边可以组成三角形,则z应满足的条件为()A.h≤z h B.h≤z h C.h≤z h D.【分析】如图,连接AP,BP,CP,先利用S△ABC=S△APC+S△BPC+S△APB,找出x,y,z 与h的关系,再运用三角形三边关系可得z<h,由x≤y≤z可得z≥h,即可求出z 应满足的条件.【解答】解:如图,PE=x,PF=y,Pq=Q=z,连接AP,BP,CP,∵S△ABC=S△APC+S△BPC+S△APB,∴BC•h=AC•x+BC•y+AB•z,∵△ABC为等边三角形,∴AB=BC=AC,∴BC•h=BC(x+y+z),即x+y+z=h,∵以x,y,z为边可以组成三角形,∴x+y>z,∴2z<h,即z<h,又∵x≤y≤z,∴z≥(x+y+z),即z≥h,∴h≤z h.故选:B.【点评】本题主要考查了三角形边角关系,解题的关键是利用S△ABC=S△APC+S△BPC+S△APB,找出x,y,z与h的关系.8.(5分)已知y=x3+ax2+bx+c,当x=5时,y=50;x=6时,y=60;x=7时,y=70.则当x=4时,y的值为()A.30B.34C.40D.44【分析】将x、y的值分别代入y=x3+ax2+bx+c,转化为关于a、b、c的方程,求出a、b、c的值,再把x=4代入,求出y的值.【解答】解:把x=5,y=50;x=6,y=60;x=7,y=70代入y=x3+ax2+bx+c,得,解得;代入y=x3+ax2+bx+c得:y=x3﹣18x2+117x﹣210,把x=4代入y=x3﹣18x2+117x﹣210得:y=43﹣18×42+117×4﹣210=64﹣288+468﹣210=34,故选:B.【点评】本题通过建立关于a,b,c的三元一次方程组,求得a、b、c的值后而求解.二、填空题(本题有10个小题,每小题6分,共60分)9.(6分)方程x2﹣3|x﹣1|﹣1=0所有解的和为﹣1.【分析】含有绝对值的方程,一般要分两种情况进行解答,即当x﹣1≥0和x﹣1≤0两种情况分别求出方程的解,再求出所有解得和即可.【解答】解:若x≥1,则x﹣1≥0,原方程可变为:x2﹣3(x﹣1)﹣1=0,即:x2﹣3x+2=0,解得:x1=1,x2=2,若x≤1,则x﹣1≤0,原方程可变为:x2+3(x﹣1)﹣1=0,即:x2+3x﹣4=0,解得:x1=1,x2=﹣4,所有解得和为:1+2﹣4=﹣1故答案为:﹣1【点评】考查一元二次方程的解法、绝对值的意义、因式分解等知识,掌握绝对值方程分情况讨论是正确解答的关键.10.(6分)设a、b、c、d、e的值均为0、1、2中之一,且a+b+c+d+e=6,a2+b2+c2+d2+e2=10,则a3+b3+c3+d3+e3的值为18.【分析】根据条件可判断a、b、c、d、e中得0、1、2分别有几个,即可解答本题.【解答】解:由题可得,a+b+c+d+e=6,a2+b2+c2+d2+e2=10有且只有一种情况可满足上述两式,即a、b、c、d、e中有2个2,2个1,1个0∴a3+b3+c3+d3+e3=23+23+1+1+0=18故答案为18.【点评】本题主要考查整式的性质,了解整式的性质是解答本题的关键.11.(6分)设a,b为两个不相等的实数,且满足2a2﹣5a=2b2﹣5b=1,则ab3+a3b的值是【分析】ab3+a3b=ab(a2+b2),由题可得a2+b2的值,再根据(a+b)2与a2+b2的差得到ab的值,从而解得此题.【解答】解:2a2﹣5a=1 2b2﹣5b=1 两式相减得:2(a2﹣b2)﹣5(a﹣b)=0 且a≠b,∴a+b=两式相加得:2(a2+b2)﹣5(a+b)=2 得a2+b2=∵(a+b)2﹣(a2+b2)=2ab=∴ab=∴ab3+a3b=ab(a2+b2)==故答案为﹣【点评】本题主要考查一元二次方程,掌握一元二次方程得求解方法是本题得关键12.(6分)已知(2x﹣1)9=a0+a1x+a2x2+……+a9x9,则a1+a2+……+a8+a9的值为2.【分析】令x=0与x=1,分别求出相应的值,代入计算即可求出所求.【解答】解:当x=0时,a0=﹣1,当x=1时,a0+a1+a2+……+a8+a9=1,则a1+a2+……+a8+a9=2,故答案为:2【点评】此题考查了代数式求值,熟练掌握运算法则是解本题的关键.13.(6分)已知四边形ABCD是正方形,且边长为2,延长BC到E,使CE=﹣,并作正方形CEFG,(如图),则△BDF的面积等于2.【分析】根据正方形的性质可知三角形BDC为等腰直角三角形,由正方形的边长为2,表示出三角形BDC的面积,四边形CDFE为直角梯形,上底下底分别为小大正方形的边长,高为小正方形的边长,利用梯形的面积公式表示出梯形CDFE的面积,而三角形BEF 为直角三角形,直角边为小正方形的边长及大小边长之和,利用三角形的面积公式表示出三角形BEF的面积,发现四边形CDEF的面积与三角形EFB的面积相等,所求△BDF 的面积等于三角形BDC的面积加上四边形CDFE的面积减去△EFB的面积即为三角形BDC的面积,进而得到所求的面积.【解答】解:∵四边形ABCD是正方形,边长为2,∴BC=DC=2,且△BCD为等腰直角三角形,∴△BDC的面积=BC•CD=×2×2=2,又∵正方形CEFG,及正方形ABCD,∴EF=CE,BC=CD,由四边形CDFE的面积是(EF+CD)•EC,△EFB的面积是(BC+CE)•EF,∴四边形CDFE的面积=△EFB的面积,∴△BDF的面积=△BDC的面积+四边形CDFE的面积﹣△EFB的面积=△BDC的面积=2.故答案为:2.【点评】此题考查了正方形的性质,以及三角形的面积求法,解答此类题时注意不规则图形的面积可以转化为一些规则图形,或已知面积的图形的面积的和或差来计算.根据题意得到四边形CDFE的面积=△EFB的面积是解本题的关键.14.(6分)向一个三角形内加入2015个点,加上原三角形的三个点共计2018个点.用剪刀最多可以剪出4031个以这2018个点为顶点的三角形.【分析】当一个点的时候是3个三角形,2个点的时候是5个三角形,3个点的时候是7个三角形,依次算下去,就有公式2n+1;故2018个点时,有2×2015+1个三角形.【解答】解:加入1个点时有3个以这4个点为顶点的三角形;加入2个点时有5个以这5个点为顶点的三角形;加入3个点时有7个以这6个点为顶点的三角形;则加入n个点时有2n+1个以这(n+3)个点为顶点的三角形;故2018个点时,有2×2015+1=4031个.故答案为:4031.【点评】本题考查了规律探索,首先应找出图形哪些部分发生了变化,是按照什么规律变化的,通过分析找到各部分的变化规律后直接利用规律求解.探寻规律要认真观察、仔细思考,善用联想来解决这类问题.15.(6分)如图,直角△ABC中,∠ABC=90°,∠A=20°,△ABC绕点B旋转至△A'BC'的位置,此时C点恰落在A'C'上,且A'B与AC交于D点,那么∠BDC=60度.【分析】想办法求出∠DBC,∠DCB即可解决问题.【解答】解:∵∠A=∠A′=20°,∠ABC=∠A′BC′=90°,∴∠C′=∠ACB=70°,∵BC=BC′,∴∠BCC′=∠C′=70°,∴∠CBC′=40°,∴∠DBC=50°,∴∠BDC=180°﹣50°﹣70°=60°,故答案为60.【点评】本题考查旋转变换,三角形内角和定理等知识,解题的关键是熟练掌握基本知识,属于中考常考题型.16.(6分)已知:对于正整数n,有,若某个正整数k满足,则k=8.【分析】读懂规律,按所得规律把左边所有的加数写成的形式,把互为相反数的项结合,可使运算简便.【解答】解:∵,∴+,即1﹣,∴,解得k=8.故答案为:8.【点评】解答此题的关键是读懂题意,总结规律答题.17.(6分)用f(n)表示组成n的数字中不是零的所有数字乘积,例如:f(5)=5,f(29)=18,f(207)=14.则f(1)+f(2)+……+f(200)=4233.【分析】根据题意可以得到规律:一位数结果为个位数,两位数结果为十位数×个位数,三位数为百位数×个位数.据此规律解决此题即可.【解答】解:f(1)+f(2)+f(3)+…+f(200)=(1+2+3…+9)+1×(1+2+3…+9)+2×(1+2+3…+9)+3×(1+2+3…+9)+…+9×(1+2+3…+9)+(1+2+3…+9+1)=(1+2+3…+9)×(1+1+2+3…+9)+46=2(45×46+46)+1=4233故答案为:4233【点评】本题考查了数字变化类问题,解题的关键是仔细地观察题目并从中总结规律,利用总结的规律进行计算即可.18.(6分)设x1、x2是方程x2﹣6x+a=0的两个根,以x1、x2为两边长的等腰三角形只可以画出一个,则实数a的取值范围是0<a≤8.【分析】方程有两个根,设x1≤x2,则可以根据求根公式用含a的式子表示x,由x1>0,x2>0,则0<a≤9,再分两种情况讨论,即根据x1=x2,和x1≠x2,等腰三角形只能作一个,只要较小的根作底,较大的根为要,再根据三边关系得出不等式求解,再综合得出答案,确定a的取职范围.【解答】解:设x1,x2为方程两根,且x1≤x2,则x1=3﹣,x2=3+,∵x1>0,x2>0,∴0<a≤9,(1)当x1=x2时,即△=9﹣a=0,a=9时,可以画无数个等腰三角形,(2)当x1≠x2时,∵x1≤x2∴以x2为腰为等腰三角形必有一个,而等腰三角形只有一个,故不存在以x2为底,x1为腰的三角形,∴2x1≤x2,∴6﹣2≤3+,≥1,∴0<a≤8,综上所述:当0<a≤8时只有一个等腰三角形.故答案为:0<a≤8.【点评】考查一元二次方程的解法、根的判别式、一元一次不等式的解集以及等腰三角形的性质等知识,准确的理解题意是解决问题的关键.三、解答题(本大题共3题,共50分.解答应写出文字说明、证明过程或演算步骤)19.(16分)甲、乙两个粮库原来各存有整袋的粮食,如果从甲库调90袋到乙库,则乙库存粮是甲库的2倍;如果从乙库调若干袋到甲库,则甲库存粮是乙库的6倍.问甲库原来最少存粮多少袋?【分析】两个关系式为:(甲库存粮﹣90)×2=乙库存粮+90;甲库存粮+若干袋粮=(乙库存粮﹣若干袋粮)×6,进而得到相应的最小整数解即可.【解答】解:设甲库原来存粮a袋,乙库原来存粮b袋,依题意可得2(a﹣90)=b+90(1);再设乙库调c袋到甲库,则甲库存粮是乙库的6倍,即a+c=6(b﹣c)(2);由(1)式得b=2a﹣270 (3),将(3)代入(2),并整理得11a﹣7c=1620,由于又a、c是正整数,从而有≥1,即a≥148;并且7整除4(a+1),又∵4与7互质,∴7整除a+1.∴a+1最小为154,∴a最小是153.答:甲库原来最少存粮153袋.【点评】解决问题的关键是读懂题意,找到关键描述语,进而找到所求的量的等量关系.注意本题需求得最小的整数解.20.(16分)如图,函数的图象交y轴于M,交x轴于N,点P是直线MN上任意一点,PQ⊥x轴,Q是垂足,设点Q的坐标为(t,0),△POQ的面积为S(当点P与M、N重合时,其面积记为0).(1)试求S与t之间的函数关系式;(2)在如图所示的直角坐标系内画出这个函数的图象,并利用图象求使得S=a(a>0)的点P的个数.【分析】本题要根据题意把各种情况都讨论出来,同时把△POQ的面积表示出来.(2)要根据题意列式整理分析,在根据解析式画出图象.【解答】解:解法1:(1)①当t<0时,OQ=﹣t,PQ=,∴S=;②当0<t<4时,OQ=t,PQ=,∴S=;③当t>4时,OQ=t,PQ=,∴S=;④当t=0或4时,S=0;于是,S=(6分);(2)S=下图中的实线部分就是所画的函数图象.(12分)观察图象可知:当0<a<1时,符合条件的点P有四个;当a=1时,符合条件的点P有三个;当a>1时,符合条件的点P只有两个.(15分)解法2:(1)∵OQ=|t|,PQ=,∴S=(4分)(2)=(6分)以下同解法1.【点评】本题考查一次函数有关分情况讨论的问题,解题中要注意对各种情况做出准确分析,尤其是t值做好取值范围的分段,21.(18分)已知二次函数y=ax2+bx+c的图象经过点(﹣2,0),且对一切实数x,都有2x ≤ax2+bx+c≤x2+2成立.(1)当x=2时,求y的值;(2)求此二次函数的表达式;(3)当x=t+m时,二次函数y=ax2+bx+c的值为y1,当x=时,二次函数y=ax2+bx+c 的值为y2,若对一切﹣1≤t≤1,都有y1<y2,求实数m的取值范围.【分析】(1)可令x=2,可得4≤4a+2b+c≤4,即有4a+2b+c=4;(2)通过图象过一点点(﹣2,0)得到4a﹣2b+c=0,由x=2得4a+2b+c=4,再将b、c都有a表示.不等式2x≤ax2+bx+c≤x2+2对一切实数x都成立可转化成两个一元二次不等式即恒成立,即可解得;(3)当﹣1≤t≤1时,y1﹣y2<0,可得3t2+(8+8m)t+4m2+16m<0 恒成立.设W=3t2+(8+8m)t+4m2+16m,则,由此求得t的范围.【解答】解:(1)解:∵不等式2x≤ax2+bx+c≤x2+2对一切实数x都成立,∴当x=2时也成立,即4≤4a+2b+c≤4,即有y=4;(2)根据二次函数y=ax2+bx+c的图象经过点(﹣2,0),可得4a﹣2b+c=0 ①,又f(2)=4,即4a+2b+c=4 ②.由①②求得b=1,4a+c=2,∴y=ax2+x+2﹣4a,∴2x≤ax2+x+2﹣4a≤x2+2,即恒成立,∴,解得:,∴c=2﹣4a=1,二次函数的表达式为.(3)∵当﹣1≤t≤1时,y1<y2,即:y1﹣y2<0,即﹣<0.整理得:3t2+(8+8m)t+4m2+16m<0,∵当t=1或﹣1时均成立,∴,整理得:解得:,∴【点评】本题考查了二次函数与不等式恒成立问题,以及二次函数的性质,赋值法(特殊值法)可以使问题变得比较明朗,它是解决这类问题比较常用的方法.。
2018年温州瓯海中学提前招生模拟考试数学试题(满分120分,考试时间:100分钟)第Ⅰ卷(选择题)评卷人得分一、选择题(共10小题,满分40分,每小题4分)1.对于两个数,M=2018×20 192 019,N=2019×20182 018.则()A.M=N B.M>N C.M<N D.无法确定2.(2017•芜湖一中自主招生)已知,,则的值为()A.3 B.4 C.5 D.63.(2015•黄冈中学自主招生)已知实数x、y、z满足x2+y2+z2=4,则(2x﹣y)2+(2y﹣z)2+(2z﹣x)2的最大值是()A.12 B.20 C.28 D.364.(2017•延平区校级自主招生)设方程(k+1)x2+2x+1=0的两根为x1、x2,若+2,则满足条件的整数k的值有()A.无数个B.﹣2,﹣1,0 C.﹣1,0 D.﹣2,05.(2017•余姚中学自主招生)如图,在Rt△ABC中,∠C=90°,AC=3,以AB 为一边向三角形外作正方形ABEF,正方形的中心为O,且OC=4,那么BC 的长等于()A.3B.5 C.2D.第5题第7题第9题6.(2017•江阴中学自主招生)对于方程x2﹣2|x|+2=m,如果方程实根的个数为3个,则m的值等于()A.1 B.C.2 D.2.57.如图,已知直线y=﹣x+3分别与x轴,y轴交于A,B两点,与双曲线y=交于E,F两点.若AB=3EF,则k的值是()A. B.2 C.D.8.(2017•奉化中学自主招生)在△ABC中,AB=AC=a,BC=b,∠A=36°,记m=,则m、n、p的大小关系为()A.m>n>p B.p>m>n C.n>p>m D.m=n=p9.(2014•成都七中自主招生)如图,在边长为1的正方形ABCD中,E、F分别为线段AB、AD、上的动点,若以EF为折线翻折,A点落在正方形ABCD所在的A′点的位置,那么A'所有可能位置形成的区域面积为()A.B.C.﹣1 D.﹣110.(2015•慈溪中学自主招生)如图,已知长方体ABCD﹣A1B1C1D1,AB=2,AD=1,AA1=2,P是棱A1B1上任意一点,Q是侧面对角线AB1上一点,则PD1+PQ 的最小值是()A.3 B.C.D.1+第10题第Ⅱ卷(非选择题)请点击修改第Ⅱ卷的文字说明评卷人得分二、填空题(共5小题,满分25分,每小题5分)11.多项式6x3﹣11x2+x+4可分解为.12.设的整数部分为x,小数部分为y,则的值为.13.(2018•枣庄八中自主招生)已知有理数x满足:,若|3﹣x|﹣|x+2|的最小值为a,最大值为b,则ab=.14.正方形ABCD的中心为O,面积为1989cm2.P为正方形内一点,且∠OPB=45°,PA:PB=5:14.则PB=.第14题第15题15.(2017•奉化中学自主招生)如图,AB为半圆的直径,C是半圆弧上任一点,正方形DEFG的一边DG在直线AB上,另一边DE过△ABC的内切圆圆心I,且点E在半圆弧上,已知DE=9,则△ABC的面积为.评卷人得分三、解答题(共5小题,满分55分)16.(8分)(2016•杭州中国美院附中自主招生)如图,点E是菱形ABCD对角线CA的延长线上任意一点,以线段AE为边作一个菱形AEFG,且∠EAG=∠BAD,连接EC,GD.(1)求证:EB=GD;(2)若∠DAB=60°,AB=2,AG=,求GD的长.第16题17.(10分)(2017•芜湖一中自主招生)方程x2﹣kx+k﹣2=0有两个实数根x1,x2,且0<x1<1,2<x2<3,求k的取值范围.18.(10分)(2016•黄冈中学自主招生)如图,四边形ABCD为正方形,⊙O过正方形的顶点A和对角线的交点P,分别交AB、AD于点F、E.(1)求证:DE=AF;(2)若⊙O的半径为,AB=,求的值.第18题19.(12分)(2016•邯郸一中自主招生)如图,在平面直角坐标系xoy中,直线y=x+3交x轴于A点,交y轴于B点,过A、B两点的抛物线y=﹣x2+bx+c交x 轴于另一点C,点D是抛物线的顶点.(1)求此抛物线的解析式;(2)点P是直线AB上方的抛物线上一点,(不与点A、B重合),过点P作x 轴的垂线交x轴于点H,交直线AB于点F,作PG⊥AB于点G.求出△PFG的周长最大值;(3)在抛物线y=﹣x2+bx+c上是否存在除点D以外的点M,使得△ABM与△ABD的面积相等?若存在,请求出此时点M的坐标;若不存在,请说明理由.第19题20.(15分)(2017•奉化中学自主招生)如图,在直角坐标系中,⊙M外接于矩形OABC,AB=12,BC=16,点A在x轴上,点C在y轴上.(1)写出点A、B、C及M的坐标;(2)过点C作⊙M的切线交x轴于点P,求直线PC的解析式;(3)如果E为线段PC上一动点(运动时不与P、C重合),过点E作直线EF 交PA于点F.①直线EF将四边形PABC的周长平分,设E点的纵坐标为t,△PEF的面积为S,求S关于t的函数关系式,并求自变量t的取值范围;②是否存在直线EF将四边形PABC的周长和面积同时平分?若能,请求出直线EF的解析式;若不能,请说明理由.第20题2018年温州瓯海中学提前招生模拟考试数学试题参考答案与试题解析一、选择题(共10小题,满分40分,每小题4分)1.对于两个数,M=2018×20 192 019,N=2019×20 182 018.则()A.M=N B.M>N C.M<N D.无法确定【解析】M=2018×(20 190 000+2019)=2018×20 190 000+2018×2019=2018×2019×10000+2018×2019=2019×20180 000+2018×2019,N=2019×(20 180 000+2018)=2019×20180 000+2019×2018,所以M=N.故选:A.2.已知,,则的值为()A.3 B.4 C.5 D.6【解析】∵a===+2,b==﹣2,∴a2+b2=(a﹣b)2+2ab=42+2×(5﹣4)=18,∴==5,故选:C.3.已知实数x、y、z满足x2+y2+z2=4,则(2x﹣y)2+(2y﹣z)2+(2z﹣x)2的最大值是()A.12 B.20 C.28 D.36【解析】∵实数x、y、z满足x2+y2+z2=4,∴(2x﹣y)2+(2y﹣z)2+(2z﹣x)2=5(x2+y2+z2)﹣4(xy+yz+xz)=20﹣2[(x+y+z)2﹣(x2+y2+z2)]=28﹣2(x+y+z)2≤28∴当x+y+z=0时(2x﹣y)2+(2y﹣z)2+(2z﹣x)2的最大值是28.故选:C.4.设方程(k+1)x2+2x+1=0的两根为x1、x2,若+2,则满足条件的整数k的值有()A.无数个B.﹣2,﹣1,0 C.﹣1,0 D.﹣2,0【解析】∵方程(k+1)x2+2x+1=0有实数根,∴,解得:k≤0且k≠﹣1.∵方程(k+1)x2+2x+1=0的两根为x1、x2,∴x1+x2=﹣,x1x2=.∵+2,即k+1+2≥﹣k﹣1,解得:k≥﹣2,∴﹣2≤k≤0且k≠﹣1,∴满足条件的整数k为﹣2或0.故选:D.5.如图,在Rt△ABC中,∠C=90°,AC=3,以AB为一边向三角形外作正方形ABEF,正方形的中心为O,且OC=4,那么BC的长等于()A.3B.5 C.2D.【解析】如图,作EQ⊥x轴,以C为坐标原点建立直角坐标系,CB为x轴,CA为y轴,则A(0,3).设B(x,0),由于O点为以AB一边向三角形外作正方形ABEF的中心,∴AB=BE,∠ABE=90°,∵∠ACB=90°,∴∠BAC+∠ABC=90°,∠ABC+∠EBQ=90°,∴∠BAC=∠EBQ,在△ABC和△BEQ中,∴△ACB≌△BQE(AAS),∴AC=BQ=3,BC=EQ,设BC=EQ=x,∴O为AE中点,∴OM为梯形ACQE的中位线,∴OM=,又∵CM=CQ=,∴O点坐标为(,),根据题意得:OC=4=,解得x=4,则BC=5.故选:B.6.对于方程x2﹣2|x|+2=m,如果方程实根的个数为3个,则m的值等于()A.1 B.C.2 D.2.5【解析】原方程可化为x2﹣2|x|+2﹣m=0,解得|x|=1±,∵若1﹣>0,则方程有四个实数根,∴方程必有一个根等于0,∵1+>0,∴1﹣=0,解得m=2.故选:C.7.如图,已知直线y=﹣x+3分别与x轴,y轴交于A,B两点,与双曲线y=交于E,F两点.若AB=3EF,则k的值是()A. B.2 C.D.【解析】作FH⊥x轴,EC⊥y轴,FH与EC交于D,如图,∵直线y=﹣x+3分别与x轴,y轴交于A,B两点,∴A点坐标为(3,0),B点坐标为(0,3),OA=OB,∴△AOB为等腰直角三角形,∴AB=OA=3,∴EF=AB=,∴△DEF为等腰直角三角形,∴FD=DE=EF=1,设F点横坐标为t,代入y=﹣x+3,则纵坐标是﹣t+3,则F的坐标是:(t,﹣t+3),E点坐标为(t+1,﹣t+2),∴t(﹣t+3)=(t+1)•(﹣t+2),解得t=1,∴E点坐标为(2,1),∴k=2×1=2.8.在△ABC中,AB=AC=a,BC=b,∠A=36°,记m=,则m、n、p的大小关系为()A.m>n>p B.p>m>n C.n>p>m D.m=n=p【解析】作底角B的角平分线交AC于D,易推得△BCD∽△ABC,所以=,即CD=,AD=a﹣=b(△ABD是等腰三角形)因此得a2﹣b2=ab,∴n====m,p====m,∴m=n=p.故选:D.9.如图,在边长为1的正方形ABCD中,E、F分别为线段AB、AD、上的动点,若以EF为折线翻折,A点落在正方形ABCD所在的A′点的位置,那么A'所有可能位置形成的区域面积为()A.B.C.﹣1 D.﹣1【解析】如图,以EF为折线翻折,A点落在正方形ABCD所在的A′点的位置,那么A′所有可能位置形成的图形是图中阴影部分.∴S阴=2•S扇形BAC﹣S正方形ABCD=﹣1,故选:D.10.如图,已知长方体ABCD﹣A1B1C1D1,AB=2,AD=1,AA1=2,P是棱A1B1上任意一点,Q是侧面对角线AB1上一点,则PD1+PQ的最小值是()A.3 B.C.D.1+【解析】将正方形展开,取A1B1C1D1及ABB1A1两个面,过点D1作D1Q⊥AB1于点Q,D1Q交A1B1于点P,此时PD1+PQ取最小值D1Q.∵ABB1A1为正方形,∴∠D1AQ=45°.在Rt△D1QA中,AD1=AA1+A1D1=3,∠D1QA=90°,∠D1AQ=45°,∴D1Q=sin∠D1AQ•AD1=.故选:B.二、填空题(共5小题,满分25分,每小题5分)11.多项式6x3﹣11x2+x+4可分解为(x﹣1)(3x﹣4)(2x+1).【解析】6x3﹣11x2+x+4,=6x3﹣6x2﹣5x2+x+4,=6x2(x﹣1)﹣(5x2﹣x﹣4),=6x2(x﹣1)﹣(x﹣1)(5x+4),=(x﹣1)(6x2﹣5x﹣4),=(x﹣1)(3x﹣4)(2x+1).12.设的整数部分为x,小数部分为y,则的值为5.【解析】∵==,而0<<1,∴x=2,y=,∴=4+×2×+()2=4++=5.故答案为5.13.已知有理数x满足:,若|3﹣x|﹣|x+2|的最小值为a,最大值为b,则ab=5.【解析】解不等式:不等式两边同时乘以6得:3(3x﹣1)﹣14≥6x﹣2(5+2x)去括号得:9x﹣3﹣14≥6x﹣10﹣4x移项得:9x﹣14﹣6x+4x≥3﹣10即7x≥7∴x≥1∴x+2>0,当1≤x≤3时,x+2>0,则|3﹣x|﹣|x+2|=3﹣x﹣(x+2)=﹣2x+1则最大值是﹣1,最小值是﹣5;当x>3时,x+2>0,则|3﹣x|﹣|x+2|=x﹣3﹣(x+2)=x﹣3﹣x﹣2=﹣5,是一定值.总之,a=﹣5,b=﹣1,∴ab=5故答案是:5.14.正方形ABCD的中心为O,面积为1989cm2.P为正方形内一点,且∠OPB=45°,PA:PB=5:14.则PB=42cm.【解析】连接OA,OB,∵正方形ABCD的中心为O,∠OPB=45°,∴∠OAB=∠OPB=45°,∠OBA=45°,∴O,P,A,B四点共圆,∴∠APB=∠AOB=180°﹣45°﹣45°=90°,在△PAB中由勾股定理得:PA2+PB2=AB2=1989,由于PA:PB=5:14,设PA=5x,PB=14x,(5x)2+(14x)2=1989,解得:x=3,∴PB=14x=42.故答案为:42cm.15.如图,AB为半圆的直径,C是半圆弧上任一点,正方形DEFG的一边DG 在直线AB上,另一边DE过△ABC的内切圆圆心I,且点E在半圆弧上,已知DE=9,则△ABC的面积为81.【解析】设⊙I切AC与M,切BC于N,半径为r,则AD=AM,CM=CN=r,BD=BN,r=(AC+BC﹣AB),∵AB为半圆的直径,∴∠ACB=90°,∴AB2=AC2+BC2,∴AD•DB=AM•BN=(AC﹣r)(BC﹣r)=[AC﹣(AC+BC﹣AB)][BC﹣(AC+BC ﹣AB)]=(AC﹣BC+AB)(AB+BC﹣AC)=(AB2﹣AC2﹣BC2+2AC•BC)=AC•BC,由射影定理得AD•DB=DE2=81,∴S△ABC=AC•BC=81,故答案为:81.三.解答题(共5小题,满分55分)16.如图,点E是菱形ABCD对角线CA的延长线上任意一点,以线段AE为边作一个菱形AEFG,且∠EAG=∠BAD,连接EC,GD.(1)求证:EB=GD;(2)若∠DAB=60°,AB=2,AG=,求GD的长.【解析】(1)证明:∵∠EAG=∠BAD,∴∠EAG+∠GAB=∠BAD+∠GAB,∴∠EAB=∠GAD,在△AEB和△AGD中,∵∴△AEB≌△AGD,∴EB=GD;(2)连接BD交AC于点P,则BP⊥AC,∵∠DAB=60°,∴∠PAB=30°,∴BP=AB=1,AP==,AE=AG=,∴EP=2,∴EB==,∴GD=.17.方程x2﹣kx+k﹣2=0有两个实数根x1,x2,且0<x1<1,2<x2<3,求k的取值范围.【解析】∵方程x2﹣kx+k﹣2=0有两个实数根x1,x2,且0<x1<1,2<x2<3,∴二次函数y=x2﹣kx+k﹣2如图所示,∴x=0,y=k﹣2>0;x=1,y=1﹣k+k﹣2<0;x=2,y=4﹣2k+k﹣2<0;x=3,y=9﹣3k+k﹣2>0,而△=k2﹣4(k﹣2)=(k﹣2)2+4>0,∴2<k<3.5,即k的取值范围为2<k<3.5.18.如图,四边形ABCD为正方形,⊙O过正方形的顶点A和对角线的交点P,分别交AB、AD于点F、E.(1)求证:DE=AF;(2)若⊙O的半径为,AB=,求的值.【解析】(1)证明:连接EP、FP,如图,∵四边形ABCD为正方形,∴∠BAD=90°,∠BPA=90°∴∠FPE=90°,∴∠BPF=∠APE,又∵∠FBP=∠PAE=45°,∴△BPF≌△APE,∴BF=AE,而AB=AD,∴DE=AF;(2)连EF,∵∠BAD=90°,∴EF为⊙O的直径,而⊙O的半径为,∴EF=,∴AF2+AE2=EF2=()2=3①,而DE=AF,DE2+AE2=3;又∵AD=AE+ED=AB,∴AE+ED=②,由①②联立起来组成方程组,解之得:AE=1,ED=或AE=,ED=1,所以:或.提示:(1)连接EF、EP、FP,可证明△AEP≌△BFP(2)设:AE=x,ED=AF=y可得:和x2+y2=3,解得x=,y=1或x=1,y=,所以:或.19.如图,在平面直角坐标系xoy中,直线y=x+3交x轴于A点,交y轴于B 点,过A、B两点的抛物线y=﹣x2+bx+c交x轴于另一点C,点D是抛物线的顶点.(1)求此抛物线的解析式;(2)点P是直线AB上方的抛物线上一点,(不与点A、B重合),过点P作x 轴的垂线交x轴于点H,交直线AB于点F,作PG⊥AB于点G.求出△PFG的周长最大值;(3)在抛物线y=﹣x2+bx+c上是否存在除点D以外的点M,使得△ABM与△ABD的面积相等?若存在,请求出此时点M的坐标;若不存在,请说明理由.【解析】(1)∵直线AB:y=x+3与坐标轴交于A(﹣3,0)、B(0,3),代入抛物线解析式y=﹣x2+bx+c中,∴∴抛物线解析式为:y=﹣x2﹣2x+3;(2)∵由题意可知△PFG是等腰直角三角形,设P(m,﹣m2﹣2m+3),∴F(m,m+3),∴PF=﹣m2﹣2m+3﹣m﹣3=﹣m2﹣3m,△PFG周长为:﹣m2﹣3m+(﹣m2﹣3m),=﹣(+1)(m+)2+,∴△PFG周长的最大值为:.(3)点M有三个位置,如图所示的M1、M2、M3,都能使△ABM的面积等于△ABD的面积.此时DM1∥AB,M3M2∥AB,且与AB距离相等,∵D(﹣1,4),∴E(﹣1,2)、则N(﹣1,0)∵y=x+3中,k=1,∴直线DM1解析式为:y=x+5,直线M3M2解析式为:y=x+1,∴x+5=﹣x2﹣2x+3或x+1=﹣x2﹣2x+3,∴x1=﹣1,x2=﹣2,x3=,x4=,∴M1(﹣2,3),M2(,),M3(,).20.如图,在直角坐标系中,⊙M外接于矩形OABC,AB=12,BC=16,点A 在x轴上,点C在y轴上.(1)写出点A、B、C及M的坐标;(2)过点C作⊙M的切线交x轴于点P,求直线PC的解析式;(3)如果E为线段PC上一动点(运动时不与P、C重合),过点E作直线EF 交PA于点F.①直线EF将四边形PABC的周长平分,设E点的纵坐标为t,△PEF的面积为S,求S关于t的函数关系式,并求自变量t的取值范围;②是否存在直线EF将四边形PABC的周长和面积同时平分?若能,请求出直线EF的解析式;若不能,请说明理由.【解析】(1)A(16,0),B(16,12),C(0,12),M(8,6).(2)连接CM.∵CM是圆半径,PC是切线,∴PC⊥CM,K PC×K CM=﹣1,解得K PC=,由点斜式写出解析式为y=x+12.(3)①作EN⊥x轴于N.根据(2)中的直线解析式求得P(﹣9,0).则PC=15.则四边形ABCP的周长是15+9+16+16+12=68.又点E的纵坐标是t,则PE=t,∵直线EF将四边形PABC的周长平分,则PF=34﹣t,则S=×t(34﹣t)=﹣+17t∵点E为PC上一动点(运动时不与P、C重合),∴0<t<12,∵点F在PA上,∴0<PF≤AP,∵OP=9,OA=16,∴AP=25,∴0<PF≤25,∵PF=34﹣t,∴0<34﹣t≤25,∴7.2≤t<27.2∵0<t<12∴7.2≤t<12即:S=×t(34﹣t)=﹣+17t(7.2≤t<12);②因为四边形ABCP的面积=×(16+16+9)×12=246.若把四边形的面积等分,则S=123.有﹣+17t=123,此方程无实数根,故不存在直线EF将四边形PABC的周长和面积同时平分.。
第 1 页 共 4 页2018 年自主招生考试数学模拟试题(满分:120 分时间:120 分钟)一、选择题。
(每小题 4 分,共 24 分)1. 如图是以△ABC 的边AB 为直径的半圆O ,点C 恰好在半圆上,过C 作CD ⊥AB 交AB 于D.已知cos ∠ACD=,BC=4,则AC 的长为()A.1B. C.3 D.第 1 题图第 3 题图第 5 题图第 6 题图2. 满足(x 2-x -1)3-x =1 的所有实数 x 的个数为( )A.3B.4C.5D.63. 如图,在方格纸中,随机选择标有序号①②③④⑤中的一个小正方形涂黑,与图中阴影部分 构成轴对称图形的概率是( )A. B. C. D.20 - 14 = 14. 已知正整数 x , y ,则 x 2 y 3 的解(x , y )共有()组.A.1B.2C.3D.45. 如图,已知正方形 ABCD ,顶点 A (1,3)、B (1,1)、C (3,1)规定“把正方形 ABCD 先沿 x 轴翻折,再向左平移 1 个单位”为一次变换,如此这样,连续经过 2018 次变换后,正方形ABCD 的对角线交点 M 的坐标变为( )A.(-2017,2)B.(-2017,-2)C.(-2016,-2)D.(-2016,2)6.抛物线 y =ax 2+bx +c 交 x 轴于 A (-1,0),B (3,0),交 y 轴的负半轴于 C ,顶点为 D.下列 结论:①2a +b =0;②2c <3b ;③当 m ≠1 时,a +b <am 2+bm ;④当△ABD 是等腰直角三角形时,则a=;⑤当△ABC 是等腰三角形时,a 的值有3 个.其中正确的有()A.①③④B.①②④C.①③⑤D.③④⑤第 2 页共 4 页二、填空题。
(每小题4 分,共24 分)7.若a 是一元二次方程x 2 -x-1=0的一个根,则代数式a4 - 2a +1a5的值是.8.我国古代有这样一道数学问题:“枯木一根直立地上,高二丈,周三尺,有葛藤自根缠绕而上,五周而达其顶,问葛藤之长几何?”题意是:如图所示,把枯木看作一个圆柱体,因一丈是十尺,则该圆柱的高为20 尺,底面周长为3 尺,有葛藤自点A 处缠绕而上,绕五周后其末端恰好到达点B 处,则问题中葛藤的最短长度是尺.第8 题图第10 题图第12 题图9.已知实数a,b 满足a+ | a - 2 |=(1-a)(b - 2) 2 +b 2 + 2 ,则a+b 的值为.10.如图,A、B 两点在反比例函数y =k1 的图像上,C、D 两点在反比例函数y =k2 的图像x x上,AC、BD 均与y 轴平行AC 交x 轴于点E,BD 交x 轴于点F,AC=2,BD=3,EF=5,则k 2 -k1= .11.已知a,b,c,d,e为互不相等的有理数,且| a -b |=| b -c |=| c -d |=| d -e |= 3 ,则| a -e |= .12.如图,AB 是半圆的直径,点O 为圆心,OA=5,弦AC=8,OD⊥AC,垂足为E,交⊙O 于D,连接BE.设∠BEC=α,则sinα的值为.三、解答题。
⼀中⾃主招⽣试卷带答案银川⼀中2017年中考⾃主招⽣考试数学试卷(时间:90分钟卷分:120分)⼀、选择题(每题5分,共50分)1、银川⼀中为有效开展“阳光体育”活动,计划购买篮球和⾜球共50个,购买资⾦不超过3000元。
若每个篮球80元,每个⾜球50元,则篮球最多可购买()A.16个B.17个C.33个D.34个答案详解A正确率: 64%, 易错项: B解析:本题主要考查⼀元⼀次不等式的应⽤。
设篮球可以购买个,那么⾜球则购买了个。
买篮球总共花元,买⾜球共花元,由资⾦不超过列不等式,解得,由于必须取整数则篮球最多买个。
故本题正确答案为A。
2、若关于的⽅程有实数根,则实数的取值范围是()。
A: B: 且C: D:正确率: 43%, 易错项: B解析:本题主要考查⼆次函数与⼀元⼆次⽅程的联系。
记,那么原⽅程有实数根就转化为该函数与轴图象有交点即可。
当时,函数为⼀条直线必与轴有交点,即原⽅程有实数根,可取排除B选项。
当时,函数为⼆次抛物线,根据判别式定理时,函数必与轴相交,即时原⽅程有实数根。
综上为所求。
故本题正确答案为C。
3、已知等腰三⾓形的周长是10,底边长y是腰长x的函数,则下列图象中,能正确反映y 与x之间函数关系的图象是( )A B C D答案详解解:由题意得,,所以,,由三⾓形的三边关系得,,解不等式①得,,解不等式②的,,所以,不等式组的解集是,正确反映y与x之间函数关系的图象是D选项图象.所以D选项是正确的.4、⼀个⼏何体的主视图和俯视图如图所⽰,若这个⼏何体最多有个⼩正⽅体组成,最少有个⼩正⽅体组成,则等于()。
A: B: C: D:答案详解C正确率: 68%, 易错项: B解析:本题主要考查三视图。
包含最多⼩正⽅形和最少⼩正⽅形的⽴体⼏何图如下:由下图可知,,故。
5、⼀个圆锥的侧⾯积是底⾯积的倍,则圆锥侧⾯展开图的扇形的圆⼼⾓是()。
A: B: C: D:答案详解A正确率: 49%, 易错项: B解析:本题主要考查圆锥和⼏何体的平⾯展开图。
2018-2019年最新玉山一中自主招生语文模拟精品试卷(第一套)(满分:100分考试时间:90分钟)③小屋在山的怀抱中,犹如在花蕊中一般,慢慢地花蕊绽开了一些,好像山后退了一些。
④当花瓣微微收拢,那就是夜晚来临了。
⑤小屋的光线既富于科学的时间性,也富于浪漫的文学性。
A.①③②④⑤ B.①④③②⑤ C.⑤③②①④ D.⑤③②④①二、阅读下面古诗文,完成7—14题。
(24分,7—12每题2分)勾践自会稽归七年,拊循其士民,欲用以报吴。
大夫逄同谏曰:“今夫吴兵加齐、晋,怨深于楚﹑越,名高天下,实害周室,德少而功多,必淫自矜。
为越计,莫若结齐,亲楚,附晋,以厚吴。
吴之志广,必轻战。
是我连其权,三国伐之,越承其弊,可克也。
”勾践曰:“善。
”其后四年。
吴士民罢弊,轻锐尽死于齐﹑晋。
而越大破吴,因而留围之三年,吴师败,越遂复栖吴王于姑苏之山。
吴王使公孙雄肉袒膝行而前,请成越王曰:“孤臣夫差敢布腹心,异日尝得罪于会稽,夫差不敢逆命,得与君王成以归。
今君王举玉趾而诛孤臣,孤臣惟命是听,意者亦欲如会稽之赦孤臣之罪乎?”勾践不忍,欲许之。
范蠡曰:“会稽之事,天以越赐吴,吴不取。
今天以吴赐越,越其可逆天乎?且夫君王蚤朝晏罢,非为吴邪?谋之二十二年,一旦而弃之,可乎?且夫天与弗取,反受其咎。
君忘会稽之厄乎?”勾践曰:“吾欲听子言,吾不忍其使者。
”范蠡乃鼓进兵,曰:“王已属政于执事,使者去,不者且得罪。
”吴使者泣而去。
勾践怜之,乃使人谓吴王曰:“吾置王甬东,君百家。
”吴王谢曰:“吾老矣,不能事君王!”遂自杀。
选自《史记·越王勾践世家》7.下列加点词语解释不正确的一项是( )A.越承其弊,可克也。
克:战胜 B.越遂复栖吴王于姑苏之山 栖:占领C.越其可逆天乎 逆:违背 D.吾老矣,不能事君王 事:侍奉8.下列加点词语古今意义相同的是( )A.今天以吴赐越 B.使者去,不者且得罪 C.谋臣与爪牙之士,不可不养而择也 D.微夫人之力不及此9.下列加点词语的用法和意义相同的一组是( )A.①德少而功多,必淫自矜 ②鼓瑟希,铿尔,舍瑟而作B.①得与君王成以归 ②王好战,请以战喻C.①亦欲如会稽之赦孤臣之罪 ②邻国之民不加少D.①异日尝得罪于会稽 ②吾长见笑于大方之家10.下列加点词语属于谦称的是( )A.吾欲听子言 B.君忘会稽之厄乎? C.君王举玉趾而诛孤臣 D.孤臣夫差敢布腹心11.下列句子,全都表现勾践具有仁慈之心的一项是( )①孤臣惟命是听②勾践不忍,欲许之。
2018-2019年最新安徽肥东县第一中学自主招生考试英语模拟精品试卷(第一套)考试时间:120分钟总分:150分第I卷(选择题,共100分)第一节:单项填空(共25小题,每小题1分,满分25分)1. —When did the terrible earthquake in YaNan happen?—It happened ________ the morning of April 20, 2013.A. onB. atC. inD. /2. Our teacher told us ________ too much noise in class.A. to makeB. makeC. not to makeD. not make3. Here is your hat. Don’t forget______ when you __________.A. to put it on, leaveB. to wear it, leaveC. to wear it, will leaveD. putting it on, will leave4. The baby is sleeping. You _____ make so much noise.A. won’tB. mustn’tC. may notD. needn’t5. Since you are _____ trouble, why not ask _________ help?A. in, forB. in, toC. with, forD. with, to6. It’s about___________kilometers from Nanchong to Chengdu.A. two hundredsB. two hundreds ofC. two hundredD. two hundred of7. It is six years since my dear uncle ________China.A. leftB. has leftC. is leftD. had left8. —How long _______ you _______ the bicycle?—About two weeks.A. have, hadB. have, boughtC. did, buyD. have, have9. The Yellow River is not so ________ as the Yangtze River.A. longerB. longC. longestD. a long10. Mrs.Green usually goes shopping with ________ umbrella in ________ summer.A. a;theB. an; /C. the; aD. /;/11. At first, I was not too sure if he could answer the question. However, ____,he worked it out at last with the help of his friend.A. to my angerB. to my surpriseC. in other wordsD. ina word12. —Must I stay here with you?—No, you ______.You may go home, but you _____ go to the net bar (网吧).A. mustn't; needn'tB. needn't; mustn'tC. must; needD. need; must13. I ______ the newspaper while my mother _____TV plays yesterday evening.A. was reading; was watchingB. was reading; watchedC. read; was watchingD. read; would watch14. It's a rule in my class that our classroom ________ before 6:00 p. m.every day.A. be able to cleanB. should be cleaningC. must cleanD. must be cleaned15. —Tom wants to know if you ________ a picnic next Sunday.—Yes. But if it ________, we'll visit the museum instead.A. will have; will rainB. have; rainsC. have; will rainD. will have; rains16.—Would you mind looking after my dog while I'm on holiday?—________.A. Of course notB. Yes. I'd be happy toC. Not at all. I've no timeD. Yes, please17. Many students didn’t realize the importance of study _______they left school.A. whenB. untilC. afterD. unless18. My father _______ to Shanghai. He _______ for over 2 months.A. has been, has leftB. has gone, has goneC. has gone, has been awayD. has been, has gone19. They are your skirts. Please __________.A. put it awayB. put out itC. put them awayD. put them out20. —Please read every sentence carefully. you are, mistakes you’ll make.—Thank you for your advice.A. The more carefully; the fewerB. The more careful; the lessC. The more carefully; the lessD. The more careful; the fewer21. My friend is coming today but he didn’t tell me _______.A. when did the train arriveB. how did the train arriveC. when the train arrivedD. how the train arrived22. I felt it hard to keep up with my classmate s. But whenever I wantto _______, my teacher always encourages me to work harder.A. go onB. give upC. run awayD. give back23. —________ fine weather it is today!—Let's go for a picnic.A. WhatB. HowC. What aD. How a24. — Mary, you’re going to buy an apartment here, aren’t you?—Yes, but I can’t_______an expensive one.A. spendB. costC. payD. afford25. —Would you like to drink coffee or milk?—_________. Please give me some tea.A. NeitherB. BothC. EitherD. None第二节:完形填空(共20小题,每小题1分,满分20分)(A)Big schoolbags have been a serious problem for students for a long time.Maybe your schoolbag is too __26__ to carry, and it troubles you a lot __27__ you want to find a book out to read. Now an etextbook will __28__ you.It is said that etextbooks are going to be __29__ in Chinese middle schools.An etextbook, in fact, is a small __30__ for students.It is much __31__ than a usual schoolbag and easy to carry. Though it is as small as a book, it can __32__ all the materials (材料) for study.The students can read the text page by page on the __33__, take notes with the pointer (屏写笔). Or even “__34__” their homework to their teachers by sending emails. All they have to do is to press a button.Some people say etextbooks are good, but some say they may be __35__ for the students' eyes. What do you think of it?26.A.light B. heavy C. useful D. comfortable27.A.till B. after C. before D. when28.A.trouble B. prevent C. help D. understanded B. kept C. invented D. lent B. radio C. pen D. computer31.A.heavier B. lighter C. cheaper D. brighter32.A.hold B. build C. discover D. practice33.A.blackboard B. desk C. screen D. card34.A.find out B. hand in C. get back D. give back35.A.helpful B. famous C. good D. bad(B)Food is very important. Everyone needs to _36_ _well if he/she wants to have a strong body. Our minds also need a kind of food. This kind of food is__ 37 __.We begin to get a knowledge even when we are very young. Small children are __38__ in everything around them. They learn __39 __while they are watching and listening. When they are getting older, they begin to ___ 40__ story books, science books…anything they like. When they find something new, they have to ask questions and__41___ to find out the answers.What is the best ___42___to get knowledge? If we learn___43___ourselves, we will get the most knowledge, If we are__44___getting answers from others and don’t ask why, we will never learn more and understand___45_.36. A. sleep B. read C. drink D. eat37. A. sport B. exercise C. knowledge D. meat38. A. interested B. interesting C. weak D. meat39. A. everybody B. something C. nothing D. anything40. A. lend B. write C. think D. read41. A. try B. wait C. think D. need42. A. place B. school C. way D. road43. A. in B. always C. to D. by44. A. seldom B. always C. certainly D. sometimes45.A.harder B. much C. well D. better第三节:阅读理解(共25小题,每小题2分,满分50分)AFamous Museums_______ .A. BeijingB. LondonC. New YorkD. The USA47. New York Museum is America’s largest museum on American__________.A. areaB. historyC. collectionsD. buildings48. The Palace Museum. Which is in the center of Beijing, is also called“Forbidden City(紫禁城)” in China. It lies in __________.A.Chang’an StreetB. New Oxford StreetC. BerlingD. Chestnut Street49. According to the form, if you want to see ancient Chinese collections,you can visit ____ at most.A. one museumB. two museumsC. three museumsD. four museums50. Which of the following is TRUE according to the information above?A. Each ticket for the Palace Museum costs the same in the whole year.B. You don’t have to pay for tickets if you visit New York Museum on Monday.C. British Museum lies in Chestnut street, London.D. New York Museum is the largest in the world.BIn recent years, more and more people like to keep pets such as a dog, a cat, a monkey and other animals. But usually people would accept tame(温顺的) and loyal(忠诚的) animals as pets rather than dangerous ones such as a lion,a tiger or a snake.People love pets and take good care of them. The owners usually regard pets as good friends and some even consider them as members of the family. Although they are not human beings(人类), their behavior sometimes is better than human beings, for they are always loyal to their owners. There are always many stories about brave and smart pets. We often hear that a pet dog saved the owner's life or traveled thousands of miles to return home. Such stories often make pets more lovely.Some pets can also be trained to help people with some special work. For example, trained dogs can help the blind to walk and trained dogs and pigs can even help police to find where drugs are easily.But pets are sometimes trouble-makers. Some pets like dogs or snakes may hurt people without any warning. Some people may become ill after being hurt because of the virus carried by the pets. If they are not taken good care of, they will become very dirty and easily get ill. So pets are helpful to us but keeping pets is not an easy job.51. What animals are thought to be dangerous as pets?A. Cats.B. Dogs.C. Snakes.D. Monkeys52. Which of the following statements is TRUE about pets?A. All the pets are considered as family members.B. Pets always behave better than human beings.C. Sometimes some pets can protect their owners.D. Pets like traveling far away from home.53. Why do people train pets according to the passage?A. To make them more clever.B. To make them more lovely.C. To find drugs for the blind.D. To do some special work.54. What can we learn from the last paragraph?A. Pets often hurt strange people.B. Pets can live well with the virus.C. Pets are dirty and dangerous.D. Pets should be looked after well.55. What is the best title for the passage?A. Training Pets.B. Keeping Pets.C. Cleaning Pets.D.Loving Pets.CFrom Feb. 8 to Mar. 1 is our winter holiday. I think everybody did a lot in the holiday. But it seems that I did nothing and it was my most unlucky holiday.I spent a lot of time on my homework. Every morning my mother woke me up early and I had breakfast in a hurry. Then I had to do my homework almost the whole day! I’m not a very slow person but the homework was too heavy!I was also unlucky when playing. During the Spring Festival, I played fireworks but my finger was hurt because I was careless to light the fireworks. I began to fear playing with fireworks from then.I was still unlucky on my friend’s party. On my friend’s birthday, unusually I woke up at 10:50 because my parents went to visit my grandmother early in the morning. The party would start in 10 minutes! So I hurried to my friend’s home without breakfast. I returned very late that day and when I got home, my parents were very angry with me.Another worrying thing was my weight. Last term, I was 46 kg but nowI am 51 kg! I have to consider losing weight!56. How long did the winter holiday last?A. two monthsB. one monthC. 4 weeksD. 22 days57. The writer got up early every day during the holiday because ______.A. he had to finish homeworkB. he had to have breakfastC. he was a very slow personD. his mother was in a hurry58. He hurt his finger because of ________.A. the Spring FestivalB. his carelessnessC. the light of fireworksD. his fear of playing59. Why were the writer’s parents angry with him?A. Because he got up too late.B. Because he missed breakfast.C. Because he was late for the party.D. Because he came back home too late.60. What did the writer want to tell us in the passage?A. He had an unlucky holiday.B. He had too much homework.C. His parents were very strict.D. He planned to lose weight.DSteven Jobs, the designer of Apple Computer, was not clever when he was in school.At that time, he was not a good student and he always made troubles with his schoolmates.When he went into college, he didn't change a lot.Then he dropped out.But he was full of new ideas.After he left college, Steven Jobs worked as a video game designer.He worked there for only several months and then he went to India.He hoped that the trip would give him some new ideas and give him a change in life.Steven Jobs lived on a farm in California for a year after he returned from India.In 1975, he began to make a new type of computer.He designed the Apple Computer with his friend in his garage.He chose the name “Apple” just because it could help him to remember a happy summer he once spent in an apple tree garden.His Apple Computer was such a great success that Steven Jobs soon became famous all over the world.61.Steven Jobs was not a good student in school because he ________.A. never did his lessonsB. was full of new ideasC. always made troubles with his schoolmatesD. dropped out62.Did Steven Jobs finish college?A. Yes, he did.B. No, he didn't.C. No, he didn't go into college.D. We don't know.63.Steven Jobs designed his new computer ______.A. in IndiaB. with his friendC. in a pear tree gardenD. by himself64.Steven Jobs is famous for his ________ all over the world.A. new ideasB. appleC. Apple ComputerD. video games65.From this passage we know ________.A. Steven Jobs didn't finish his studies in the college because he hatedhis schoolmatesB. Steven Jobs liked traveling in India and CaliforniaC. Steven Jobs liked trying new things and making new ideas become trueD. Steven Jobs could only design video gamesEIf you go into the forest with friends, stay with them. If you don't, you may get lost. If you get lost, this is what you should do. Sit down and stay where you are. Don't try to find your friends. Let them find you. You can help them find you by staying in one place. There is another way to help your friends or other people to find you. You can shout or whistle (吹口哨) three times. Stop. Then shout or whistle three times again. Any signal given three times is a call for help.Keep up shouting or whistling. Always three times together. When people hear you, they will know that you are not just making a noise for fun. They will let you know that they have heard your signal. They will give you two shouts or two whistles. When a signal is given twice, it is an answer to a call for help.If you don't think that you will get help before night comes, try to make a little house with branches .Make yourself a bed with leaves and grass.When you need some water, you have to leave your little branch house to look for it. Don't just walk away .Pick off small branches and drop them as you walk in order to go back again easily.66.If you get lost in the forest, you should ________.A. walk around the forest to find your friendsB. stay in one place and give signalsC. climb up a tree and wait for your friends quietlyD. shout as loudly as possible67.Which signal is a call for help?A. Shouting one time as loudly as you can.B. Crying twice.C. Shouting or whistling three times together.D. Whistling everywhere in the forest.68.When you hear two shouts or two whistles, you know that ________.A. someone finds something interestingB. people will come and help youC. someone needs helpD. something terrible will happen69.Before night comes, you should try to make a little house with ________.A. stoneB. earthC. leaves and grassD. branches70.Which of the following is the best title?A. Getting Water in the ForestB. Spending the Night in the ForestC. Surviving (生存) in the ForestD. Calling for Help in the Forest 第四节:补全对话,从方框内7个选项中选择恰当的5个句子完成此对话(共5分)John: Hi, Karl. You were not here, in your class yesterday afternoon. What was wrong?Karl: 71________John: Sorry to hear that.72Karl: Much better. The fever is gone. But I still cough and I feel weak. John: 73Karl: Yes, I have. I went to the doctor’s yesterday afternoon. The doctor gave me some medicine and asked me to stay in bed for a few days. John: 74Karl: Because I’m afraid I’ll miss more lessons and I’ll be left behind. John: Don’t worry. Take care of yourself. 75第Ⅱ卷(非选择题,共50分)一、根据句意及所给提示,补全单词或用单词、固定短语、固定搭配的正确形式填空(10分)76. Many athletes won gold medals in the Olympics, they are our national h_____.77. Tom didn’t finish _____________( write) his test because he ran out of the time.78. The girl is making a model doll ___________ (care).79. The boy felt __________(困倦的) in class because he stayed up late last night.80. So Terrible! The airplane ______________(起飞) five minutes ago.81.I don't think students should be (允许)to bring mobile phones to school.82.I find it useless to spend much time (解释)it to him.83. She prefers keeping silent to (争吵)with others.84. It is important for us to be (有信心的)of doing everything.85. The doctor operated on the patient (成功)yesterday.二、汉译英, 一空一词(共5小题,每小题2分,计10分)86. 他默默地在雨中行走,浑身上下都被淋湿。
2018-2019年最新自贡市第一中学自主招生语文模拟精品试卷(第一套)(满分:100分考试时间:90分钟)③小屋在山的怀抱中,犹如在花蕊中一般,慢慢地花蕊绽开了一些,好像山后退了一些。
④当花瓣微微收拢,那就是夜晚来临了。
⑤小屋的光线既富于科学的时间性,也富于浪漫的文学性。
A.①③②④⑤ B.①④③②⑤ C.⑤③②①④ D.⑤③②④①二、阅读下面古诗文,完成7—14题。
(24分,7—12每题2分)勾践自会稽归七年,拊循其士民,欲用以报吴。
大夫逄同谏曰:“今夫吴兵加齐、晋,怨深于楚﹑越,名高天下,实害周室,德少而功多,必淫自矜。
为越计,莫若结齐,亲楚,附晋,以厚吴。
吴之志广,必轻战。
是我连其权,三国伐之,越承其弊,可克也。
”勾践曰:“善。
”其后四年。
吴士民罢弊,轻锐尽死于齐﹑晋。
而越大破吴,因而留围之三年,吴师败,越遂复栖吴王于姑苏之山。
吴王使公孙雄肉袒膝行而前,请成越王曰:“孤臣夫差敢布腹心,异日尝得罪于会稽,夫差不敢逆命,得与君王成以归。
今君王举玉趾而诛孤臣,孤臣惟命是听,意者亦欲如会稽之赦孤臣之罪乎?”勾践不忍,欲许之。
范蠡曰:“会稽之事,天以越赐吴,吴不取。
今天以吴赐越,越其可逆天乎?且夫君王蚤朝晏罢,非为吴邪?谋之二十二年,一旦而弃之,可乎?且夫天与弗取,反受其咎。
君忘会稽之厄乎?”勾践曰:“吾欲听子言,吾不忍其使者。
”范蠡乃鼓进兵,曰:“王已属政于执事,使者去,不者且得罪。
”吴使者泣而去。
勾践怜之,乃使人谓吴王曰:“吾置王甬东,君百家。
”吴王谢曰:“吾老矣,不能事君王!”遂自杀。
选自《史记·越王勾践世家》7.下列加点词语解释不正确的一项是( )A.越承其弊,可克也。
克:战胜 B.越遂复栖吴王于姑苏之山 栖:占领C.越其可逆天乎 逆:违背 D.吾老矣,不能事君王 事:侍奉8.下列加点词语古今意义相同的是( )A.今天以吴赐越 B.使者去,不者且得罪 C.谋臣与爪牙之士,不可不养而择也 D.微夫人之力不及此9.下列加点词语的用法和意义相同的一组是( )A.①德少而功多,必淫自矜 ②鼓瑟希,铿尔,舍瑟而作B.①得与君王成以归 ②王好战,请以战喻C.①亦欲如会稽之赦孤臣之罪 ②邻国之民不加少D.①异日尝得罪于会稽 ②吾长见笑于大方之家10.下列加点词语属于谦称的是( )A.吾欲听子言 B.君忘会稽之厄乎? C.君王举玉趾而诛孤臣 D.孤臣夫差敢布腹心11.下列句子,全都表现勾践具有仁慈之心的一项是( )①孤臣惟命是听②勾践不忍,欲许之。
2018-2019年最新山东省菏泽第一中学初升高自主招生物理模拟精品试卷(第一套)一.单项选择题(共15小题,每题3分,共45分)1.“大黄鸭”来到中国,下列能正确表示“大黄鸭”在水中所成倒影的是( )A B C D2.关于四季常见的自然现象,下面说法正确的是( )A.春雨是汽化现象 B.夏露是液化现象 C.秋霜是凝固现象 D.冬雪是升华现象3.自行车是非常方便的交通工具,它运用了许多科学知识.下列说法中错误的是( ) A.车轮上刻有凹凸不平的花纹是为了增大摩擦力 B.用力蹬脚踏板,自行车前进是因为受到地面的摩擦力 C.在水平地面上运动的自行车不用踩也会前进是因为自行车的惯性 D.上坡前用力猛踩几下是为了增大自行车的惯性4.关于电磁波的以下说法正确的是( ) A.在电磁波普中频率最高的电磁波是γ射线 B.手机信号所使用的电磁波频率为900Hz,其波长大约为3.3m C.电视信号利用了电磁波的能量特征D. X射线断层扫描照像(CT)是利用了电磁波的能量特征5.关于光现象,下面说法正确的是( ) A.开凿大山隧道时,用激光引导掘进方向是运用光的直线传播道理 B.当物体表面发生漫反射时,光线射向四面八方,不遵守光的反射定律 C.渔民叉鱼时,将鱼叉对准看到的“鱼”叉去可以叉到鱼 D.太阳光是由红、黄、蓝三种色光组成的6.下列有关热现象的解释正确的是( )A.炒菜时满屋子的香味说明分子间存在斥力B.给自行车打气时,打气筒内活塞向下压缩气体,气体分子间引力做正功C.物体从外界吸收热量,温度一定升高D.内燃机压缩冲程,汽缸内燃气温度上升,将内能转化为机械能7.丹麦物理学家奥斯特首先通过实验发现电流周围存在磁场.如图所示,实验时要在通电直导线下方放一个小磁针,通过小磁针的偏转来判断电流是否在其周围空间激发磁场及激发的磁场的方向.为使实验效果尽量明显,下列有关直导线AB放置方向的有关叙述正确的是( )A.直导线AB应该东西方向水平放置B.直导线AB应该南北方向水平放置C.直导线AB应该东南方向水平放置D.直导线AB应该东北方向水平放置8.某学习小组对一辆在平直公路上做直线运动的小车进行观测研究.他们记录了小车在某段时间内通过的路程和所用的时间,并根据记录的数据绘制了路程与时间关系图象,如图所示,根据图象可以判断( ) A.2s~5s内,小车的平均速度是0.4 m/s B.0~7s内,小车的平均速度是1.5m/s C.2s~5s内,小车受到的合力不为零 D.5s~7s内,小车受到的合力为零9.某测量仪器及其控制电路如图所示.仪器的等效内阻为90 Ω,正常工作电流范围为100~300 mA之间,控制电路电源电压恒为30V.控制电路由两个滑动变阻器R1和R2串联组成,为了在不同环境下快速而准确地调节仪器以达到正常工作电流,应选用哪一组变阻器?( )A.R1=100 Ω,R2=10 Ω B.R1=200 Ω,R2=20 ΩC.R1=100 Ω,R2=100 Ω D.R1=200 Ω,R2=200 Ω10.以下关于飞机正常飞行及失速下降的科学分析中正确的是()A.飞机正常飞行时机翼上方的空气流速大于下方空气流速导致飞机上下表面受压力差向上克服飞机重力B. 飞机正常飞行时机翼下方的空气流速大于下方空气流速导致飞机上下表面受压力差向上克服飞机重力C.飞机失速状态下机翼上、下方的空气流速不同导致的压力差大于飞机重力D. 飞机失速状态下机翼上、下方的空气流速不同导致的压力差等于飞机重力11.将一m = 50 kg的长方体木箱放置于电梯水平地面上的可以显示压力大小的压力传感器上随电梯一起由1楼道32楼,下面关于电梯上升过程中的相关分析正确的是()A.电梯匀速上升阶段木箱受到的重力和木箱对传感器的压力是一对平衡力,传感器示数为500 NB. 电梯匀速上升阶段木箱对传感器的压力和传感器对木箱的支持力是一对平衡力,传感器示数为500 NC..电梯加速上升阶段木箱对传感器的压力和传感器对木箱的支持力是一对相互作用力,传感器示数大于500 ND. 电梯减速上升阶段木箱受到的重力和传感器对木箱的支持力是一对平衡力,传感器示数为500 N12.凸透镜是一种基本光学元件,在生活中有广泛的应用.下列说法正确的是( ) A.用放大镜观察报纸上的小字时,应将报纸放在凸透镜的一倍焦距与二倍焦距之间,这时报纸上的字成正立、放大的虚像 B.某人去医院检查发现是远视眼,医生建议他佩戴装有凸透镜的眼镜加以矫正 C.照相机照全身照时,应该让照相机与人的距离大于照相机镜头的二倍焦距,人在相机中所成像是正立、缩小的实像 D.小孔成像、海市蜃楼、彩虹的形成以及凸透镜成像都是由光的折射形成的13.如图所示,三个相同的容器内水面高度相同,甲容器内只有水,乙容器内有木块漂浮在水面上,丙容器中悬浮着一个小球,则下列四种说法正确的是( ) A.三个容器对水平桌面的压力相等 B.三个容器中,丙容器对水平桌面的压力最大 C.如果向乙容器中加入盐水,木块将下沉 D.如果向丙容器中加入酒精,小球受到的浮力不变14.芷晴走到电动扶梯(电梯)前,发现电梯上没有站人时运行较慢,当她站到电梯上时又快了很多.她了解到电梯是由电动机带动运转的,电梯的控制电路中安装了力敏电阻(力敏电阻受到压力时,阻值会发生变化),控制电梯运动快慢的模拟电路如图所示.以下分析合理的是( ) A.电梯没有站人时,电磁铁的衔铁与触点1接触,电阻R连入电路中,电路消耗的总功率更大 B.电梯上站人后,压敏电阻的阻值减小,电磁铁的磁性变强,使衔铁接触触点2,电动机消耗的低昂率增大 C.调整弹簧的长度、软硬等不能有效防止因儿童单独上扶梯导致扶梯突然加速引发的潜在危险D.调整R的大小不能有效改变人站上扶梯后电梯速度的该变量15.如图所示的电路,电源电压恒为U,闭合开关S,将滑动变阻器的滑片P 由a端逐渐向右移动至b端,测得当滑片位于某两个不同位置时电流表A示数分别为I1、I2,电压表V1示数分别为U1、U3,电压表V2示数分别为U2、U4.则下列说法正确的是( )A.U1-U3=U4-U2B.U1-U3<U4-U2C.=R 2-R 3 C.=R 2+R 3 U 4-U 2I 2-I 1 U 1-U 3I 2-I 1 二、作图题(共4小题,每小题3分,共12分)16.如图所示,轻杆AC 一端用光滑铰链固定于墙壁上的A 点,另一端用细线系一电灯,从杆上点B 拉一细绳系于墙壁上的D 点使轻杆处于水平,请做出此杆模型的支点O 、作用于B 、C 两点的细绳和电线的作用力的力臂、.l B l C17.如图所示,空间中有一通电螺线管,在其过轴线的竖直面上放置4个可以360°旋转的小磁针a 、b 、c 、d ,其中小磁针c 位于螺线管内部,其余小磁针在螺线管外部。
2018-2019年最新常德市桃源一中自主招生考试数学模拟精品试卷(第一套)考试时间:90分钟总分:150分一、选择题(本题有12小题,每小题3分,共36分)下面每小题给出的四个选项中,只有一个是正确的,请你把正确选项前的字母填涂在答题卷中相应的格子内.注意可以用多种不同的方法来选取正确答案.1.下列事件中,必然事件是( )A.掷一枚硬币,正面朝上B.a是实数,|a|≥0C.某运动员跳高的最好成绩是20.1米D.从车间刚生产的产品中任意抽取一个,是次品2、如图是奥迪汽车的标志,则标志图中所包含的图形变换没有的是()A.平移变换 B.轴对称变换 C.旋转变换 D.相似变换3.如果□×3ab=3a2b,则□内应填的代数式( )A.ab B.3ab C.a D.3a4.一元二次方程x(x-2)=0根的情况是( )A.有两个不相等的实数根B.有两个相等的实数根C.只有一个实数根D.没有实数根5、割圆术是我国古代数学家刘徽创造的一种求周长和面积的方法:随着圆内接正多边形边数的增加,它的周长和面积越来越接近圆周长和圆面积,“割之弥细,所失弥少,割之又割,以至于不可割,则与圆周合体而无所失矣”。
试用这个方法解决问题:如图,⊙的内接多边形周长为3 ,⊙的外切多边形O周长为3.4,则下列各数中与此圆的周长最接近的是()AB.10D6、今年5月,我校举行“庆五四”歌咏比赛,有17位同学参加选A拔赛,所得分数互不相同,按成绩取前8名进入决赛,若知道某同学分数,要判断他能否进入决赛,只需知道17位同学分数的()A.中位数 B.众数 C.平均数 D.方差7.如图,数轴上表示的是某不等式组的解集,则这个不等式组可能是( )A.Error!B. Error!C.Error!D.Error!8.已知二次函数的图象(0≤x≤3)如图所示,关于该函数在所给自变量取值范围内,下列说法正确的是( )A.有最小值0,有最大值3B.有最小值-1,有最大值0C.有最小值-1,有最大值3D.有最小值-1,无最大值9.如图,矩形OABC的边OA长为2 ,边AB长为1,OA在数轴上,以原点O为圆心,对角线OB的长为半径画弧,交正半轴于一点,则这个点表示的实数是( )A.2.5 B.2 C. D.23510.常德市桃源一中广场有一喷水池,水从地面喷出,如图,以水平地面为x轴,出水点为原点,建立平面直角坐标系,水在空中划出的曲线是抛物线y=-x2+4x(单位:米)的一部分,则水喷出的最大高度是( )水平面主视方向A .4米B .3米C .2米D .1米11、两个大小不同的球在水平面上靠在一起,组成如图所示的几何体,则该几何体的左视图是( )(A )两个外离的圆 (B )两个外切的圆(C )两个相交的圆 (D )两个内切的圆12.已知二次函数y =ax 2+bx +c (a ≠0)的图象如图所示,有下列结论:①b 2-4ac >0;②abc >0;③8a +c >0;④9a +3b +c <0.其中,正确结论的个数是( )A .1B .2C .3D .4二、填空题(本小题有6小题,每小题4分,共24分)要注意认真看清题目的条件和要填写的内容,尽量完整地填写答案13.当x ______时,分式有意义. 13-x14.在实数范围内分解因式:2a 3-16a =________.15.在日本核电站事故期间,我国某监测点监测到极微量的人工放射性核素碘-131,其浓度为0.0000963贝克/立方米.数据“0.0000963”用科学记数法可表示为________.16.如图,C 岛在A 岛的北偏东60°方向,在B 岛的北偏西45°方向,则从C 岛看A 、B 两岛的视角∠ACB =________.17.若一次函数y =(2m -1)x +3-2m 的图象经过 一、二、四象限,则m 的取值范围是________.18.将一些半径相同的小圆按如图所示的规律摆放,请仔细观察,第 n 个图形有________个小圆. (用含 n 的代数式表示)三、解答题(本大题7个小题,共90分)19.(本题共2个小题,每题8分,共16分)(1).计算:(-1)0+sin45°-2-1 201118。
2018-2019年最新湖北荆门市龙泉中学自主招生考试英语模拟精品试卷(第一套)考试时间:120分钟总分:150分第I卷(选择题,共100分)第一节:单项填空(共25小题,每小题1分,满分25分)1. —When did the terrible earthquake in YaNan happen?—It happened ________ the morning of April 20, 2013.A. onB. atC. inD. /2. Our teacher told us ________ too much noise in class.A. to makeB. makeC. not to makeD. not make3. Here is your hat. Don’t forget______ when you __________.A. to put it on, leaveB. to wear it, leaveC. to wear it, will leaveD. putting it on, will leave4. The baby is sleeping. You _____ make so much noise.A. won’tB. mustn’tC. may notD. needn’t5. Since you are _____ trouble, why not ask _________ help?A. in, forB. in, toC. with, forD. with, to6. It’s about___________kilometers from Nanchong to Chengdu.A. two hundredsB. two hundreds ofC. two hundredD. two hundred of7. It is six years since my dear uncle ________China.A. leftB. has leftC. is leftD. had left8. —How long _______ you _______ the bicycle?—About two weeks.A. have, hadB. have, boughtC. did, buyD. have, have9. The Yellow River is not so ________ as the Yangtze River.A. longerB. longC. longestD. a long10. Mrs.Green usually goes shopping with ________ umbrella in ________ summer.A. a;theB. an; /C. the; aD. /;/11. At first, I was not too sure if he could answer the question. However, ____,he worked it out at last with the help of his friend.A. to my angerB. to my surpriseC. in other wordsD. ina word12. —Must I stay here with you?—No, you ______.You may go home, but you _____ go to the net bar (网吧).A. mustn't; needn'tB. needn't; mustn'tC. must; needD. need; must13. I ______ the newspaper while my mother _____TV plays yesterday evening.A. was reading; was watchingB. was reading; watchedC. read; was watchingD. read; would watch14. It's a rule in my class that our classroom ________ before 6:00 p. m.every day.A. be able to cleanB. should be cleaningC. must cleanD. must be cleaned15. —Tom wants to know if you ________ a picnic next Sunday.—Yes. But if it ________, we'll visit the museum instead.A. will have; will rainB. have; rainsC. have; will rainD. will have; rains16.—Would you mind looking after my dog while I'm on holiday?—________.A. Of course notB. Yes. I'd be happy toC. Not at all. I've no timeD. Yes, please17. Many students didn’t realize the importance of study _______they left school.A. whenB. untilC. afterD. unless18. My father _______ to Shanghai. He _______ for over 2 months.A. has been, has leftB. has gone, has goneC. has gone, has been awayD. has been, has gone19. They are your skirts. Please __________.A. put it awayB. put out itC. put them awayD. put them out20. —Please read every sentence carefully. you are, mistakes you’ll make.—Thank you for your advice.A. The more carefully; the fewerB. The more careful; the lessC. The more carefully; the lessD. The more careful; the fewer21. My friend is coming today but he didn’t tell me _______.A. when did the train arriveB. how did the train arriveC. when the train arrivedD. how the train arrived22. I felt it hard to keep up with my classmate s. But whenever I wantto _______, my teacher always encourages me to work harder.A. go onB. give upC. run awayD. give back23. —________ fine weather it is today!—Let's go for a picnic.A. WhatB. HowC. What aD. How a24. — Mary, you’re going to buy an apartment here, aren’t you?—Yes, but I can’t_______an expensive one.A. spendB. costC. payD. afford25. —Would you like to drink coffee or milk?—_________. Please give me some tea.A. NeitherB. BothC. EitherD. None第二节:完形填空(共20小题,每小题1分,满分20分)(A)Big schoolbags have been a serious problem for students for a long time.Maybe your schoolbag is too __26__ to carry, and it troubles you a lot __27__ you want to find a book out to read. Now an etextbook will __28__ you.It is said that etextbooks are going to be __29__ in Chinese middle schools.An etextbook, in fact, is a small __30__ for students.It is much __31__ than a usual schoolbag and easy to carry. Though it is as small as a book, it can __32__ all the materials (材料) for study.The students can read the text page by page on the __33__, take notes with the pointer (屏写笔). Or even “__34__” their homework to their teachers by sending emails. All they have to do is to press a button.Some people say etextbooks are good, but some say they may be __35__ for the students' eyes. What do you think of it?26.A.light B. heavy C. useful D. comfortable27.A.till B. after C. before D. when28.A.trouble B. prevent C. help D. understanded B. kept C. invented D. lent B. radio C. pen D. computer31.A.heavier B. lighter C. cheaper D. brighter32.A.hold B. build C. discover D. practice33.A.blackboard B. desk C. screen D. card34.A.find out B. hand in C. get back D. give back35.A.helpful B. famous C. good D. bad(B)Food is very important. Everyone needs to _36_ _well if he/she wants to have a strong body. Our minds also need a kind of food. This kind of food is__ 37 __.We begin to get a knowledge even when we are very young. Small children are __38__ in everything around them. They learn __39 __while they are watching and listening. When they are getting older, they begin to ___ 40__ story books, science books…anything they like. When they find something new, they have to ask questions and__41___ to find out the answers.What is the best ___42___to get knowledge? If we learn___43___ourselves, we will get the most knowledge, If we are__44___getting answers from others and don’t ask why, we will never learn more and understand___45_.36. A. sleep B. read C. drink D. eat37. A. sport B. exercise C. knowledge D. meat38. A. interested B. interesting C. weak D. meat39. A. everybody B. something C. nothing D. anything40. A. lend B. write C. think D. read41. A. try B. wait C. think D. need42. A. place B. school C. way D. road43. A. in B. always C. to D. by44. A. seldom B. always C. certainly D. sometimes45.A.harder B. much C. well D. better第三节:阅读理解(共25小题,每小题2分,满分50分)AFamous Museums_______ .A. BeijingB. LondonC. New YorkD. The USA47. New York Museum is America’s largest museum on American__________.A. areaB. historyC. collectionsD. buildings48. The Palace Museum. Which is in the center of Beijing, is also called“Forbidden City(紫禁城)” in China. It lies in __________.A.Chang’an StreetB. New Oxford StreetC. BerlingD. Chestnut Street49. According to the form, if you want to see ancient Chinese collections,you can visit ____ at most.A. one museumB. two museumsC. three museumsD. four museums50. Which of the following is TRUE according to the information above?A. Each ticket for the Palace Museum costs the same in the whole year.B. You don’t have to pay for tickets if you visit New York Museum on Monday.C. British Museum lies in Chestnut street, London.D. New York Museum is the largest in the world.BIn recent years, more and more people like to keep pets such as a dog, a cat, a monkey and other animals. But usually people would accept tame(温顺的) and loyal(忠诚的) animals as pets rather than dangerous ones such as a lion,a tiger or a snake.People love pets and take good care of them. The owners usually regard pets as good friends and some even consider them as members of the family. Although they are not human beings(人类), their behavior sometimes is better than human beings, for they are always loyal to their owners. There are always many stories about brave and smart pets. We often hear that a pet dog saved the owner's life or traveled thousands of miles to return home. Such stories often make pets more lovely.Some pets can also be trained to help people with some special work. For example, trained dogs can help the blind to walk and trained dogs and pigs can even help police to find where drugs are easily.But pets are sometimes trouble-makers. Some pets like dogs or snakes may hurt people without any warning. Some people may become ill after being hurt because of the virus carried by the pets. If they are not taken good care of, they will become very dirty and easily get ill. So pets are helpful to us but keeping pets is not an easy job.51. What animals are thought to be dangerous as pets?A. Cats.B. Dogs.C. Snakes.D. Monkeys52. Which of the following statements is TRUE about pets?A. All the pets are considered as family members.B. Pets always behave better than human beings.C. Sometimes some pets can protect their owners.D. Pets like traveling far away from home.53. Why do people train pets according to the passage?A. To make them more clever.B. To make them more lovely.C. To find drugs for the blind.D. To do some special work.54. What can we learn from the last paragraph?A. Pets often hurt strange people.B. Pets can live well with the virus.C. Pets are dirty and dangerous.D. Pets should be looked after well.55. What is the best title for the passage?A. Training Pets.B. Keeping Pets.C. Cleaning Pets.D.Loving Pets.CFrom Feb. 8 to Mar. 1 is our winter holiday. I think everybody did a lot in the holiday. But it seems that I did nothing and it was my most unlucky holiday.I spent a lot of time on my homework. Every morning my mother woke me up early and I had breakfast in a hurry. Then I had to do my homework almost the whole day! I’m not a very slow person but the homework was too heavy!I was also unlucky when playing. During the Spring Festival, I played fireworks but my finger was hurt because I was careless to light the fireworks. I began to fear playing with fireworks from then.I was still unlucky on my friend’s party. On my friend’s birthday, unusually I woke up at 10:50 because my parents went to visit my grandmother early in the morning. The party would start in 10 minutes! So I hurried to my friend’s home without breakfast. I returned very late that day and when I got home, my parents were very angry with me.Another worrying thing was my weight. Last term, I was 46 kg but nowI am 51 kg! I have to consider losing weight!56. How long did the winter holiday last?A. two monthsB. one monthC. 4 weeksD. 22 days57. The writer got up early every day during the holiday because ______.A. he had to finish homeworkB. he had to have breakfastC. he was a very slow personD. his mother was in a hurry58. He hurt his finger because of ________.A. the Spring FestivalB. his carelessnessC. the light of fireworksD. his fear of playing59. Why were the writer’s parents angry with him?A. Because he got up too late.B. Because he missed breakfast.C. Because he was late for the party.D. Because he came back home too late.60. What did the writer want to tell us in the passage?A. He had an unlucky holiday.B. He had too much homework.C. His parents were very strict.D. He planned to lose weight.DSteven Jobs, the designer of Apple Computer, was not clever when he was in school.At that time, he was not a good student and he always made troubles with his schoolmates.When he went into college, he didn't change a lot.Then he dropped out.But he was full of new ideas.After he left college, Steven Jobs worked as a video game designer.He worked there for only several months and then he went to India.He hoped that the trip would give him some new ideas and give him a change in life.Steven Jobs lived on a farm in California for a year after he returned from India.In 1975, he began to make a new type of computer.He designed the Apple Computer with his friend in his garage.He chose the name “Apple” just because it could help him to remember a happy summer he once spent in an apple tree garden.His Apple Computer was such a great success that Steven Jobs soon became famous all over the world.61.Steven Jobs was not a good student in school because he ________.A. never did his lessonsB. was full of new ideasC. always made troubles with his schoolmatesD. dropped out62.Did Steven Jobs finish college?A. Yes, he did.B. No, he didn't.C. No, he didn't go into college.D. We don't know.63.Steven Jobs designed his new computer ______.A. in IndiaB. with his friendC. in a pear tree gardenD. by himself64.Steven Jobs is famous for his ________ all over the world.A. new ideasB. appleC. Apple ComputerD. video games65.From this passage we know ________.A. Steven Jobs didn't finish his studies in the college because he hatedhis schoolmatesB. Steven Jobs liked traveling in India and CaliforniaC. Steven Jobs liked trying new things and making new ideas become trueD. Steven Jobs could only design video gamesEIf you go into the forest with friends, stay with them. If you don't, you may get lost. If you get lost, this is what you should do. Sit down and stay where you are. Don't try to find your friends. Let them find you. You can help them find you by staying in one place. There is another way to help your friends or other people to find you. You can shout or whistle (吹口哨) three times. Stop. Then shout or whistle three times again. Any signal given three times is a call for help.Keep up shouting or whistling. Always three times together. When people hear you, they will know that you are not just making a noise for fun. They will let you know that they have heard your signal. They will give you two shouts or two whistles. When a signal is given twice, it is an answer to a call for help.If you don't think that you will get help before night comes, try to make a little house with branches .Make yourself a bed with leaves and grass.When you need some water, you have to leave your little branch house to look for it. Don't just walk away .Pick off small branches and drop them as you walk in order to go back again easily.66.If you get lost in the forest, you should ________.A. walk around the forest to find your friendsB. stay in one place and give signalsC. climb up a tree and wait for your friends quietlyD. shout as loudly as possible67.Which signal is a call for help?A. Shouting one time as loudly as you can.B. Crying twice.C. Shouting or whistling three times together.D. Whistling everywhere in the forest.68.When you hear two shouts or two whistles, you know that ________.A. someone finds something interestingB. people will come and help youC. someone needs helpD. something terrible will happen69.Before night comes, you should try to make a little house with ________.A. stoneB. earthC. leaves and grassD. branches70.Which of the following is the best title?A. Getting Water in the ForestB. Spending the Night in the ForestC. Surviving (生存) in the ForestD. Calling for Help in the Forest 第四节:补全对话,从方框内7个选项中选择恰当的5个句子完成此对话(共5分)John: Hi, Karl. You were not here, in your class yesterday afternoon. What was wrong?Karl: 71________John: Sorry to hear that.72Karl: Much better. The fever is gone. But I still cough and I feel weak. John: 73Karl: Yes, I have. I went to the doctor’s yesterday afternoon. The doctor gave me some medicine and asked me to stay in bed for a few days. John: 74Karl: Because I’m afraid I’ll miss more lessons and I’ll be left behind. John: Don’t worry. Take care of yourself. 75第Ⅱ卷(非选择题,共50分)一、根据句意及所给提示,补全单词或用单词、固定短语、固定搭配的正确形式填空(10分)76. Many athletes won gold medals in the Olympics, they are our national h_____.77. Tom didn’t finish _____________( write) his test because he ran out of the time.78. The girl is making a model doll ___________ (care).79. The boy felt __________(困倦的) in class because he stayed up late last night.80. So Terrible! The airplane ______________(起飞) five minutes ago.81.I don't think students should be (允许)to bring mobile phones to school.82.I find it useless to spend much time (解释)it to him.83. She prefers keeping silent to (争吵)with others.84. It is important for us to be (有信心的)of doing everything.85. The doctor operated on the patient (成功)yesterday.二、汉译英, 一空一词(共5小题,每小题2分,计10分)86. 他默默地在雨中行走,浑身上下都被淋湿。
2018-2019年最新珠海一中自主招生考试英语模拟精品试卷(第一套)考试时间:120分钟总分:150分第I卷(选择题,共100分)第一节:单项填空(共25小题,每小题1分,满分25分)1. —When did the terrible earthquake in YaNan happen?—It happened ________ the morning of April 20, 2013.A. onB. atC. inD. /2. Our teacher told us ________ too much noise in class.A. to makeB. makeC. not to makeD. not make3. Here is your hat. Don’t forget______ when you __________.A. to put it on, leaveB. to wear it, leaveC. to wear it, will leaveD. putting it on, will leave4. The baby is sleeping. You _____ make so much noise.A. won’tB. mustn’tC. may notD. needn’t5. Since you are _____ trouble, why not ask _________ help?A. in, forB. in, toC. with, forD. with, to6. It’s about___________kilometers from Nanchong to Chengdu.A. two hundredsB. two hundreds ofC. two hundredD. two hundred of7. It is six years since my dear uncle ________China.A. leftB. has leftC. is leftD. had left8. —How long _______ you _______ the bicycle?—About two weeks.A. have, hadB. have, boughtC. did, buyD. have, have9. The Yellow River is not so ________ as the Yangtze River.A. longerB. longC. longestD. a long10. Mrs.Green usually goes shopping with ________ umbrella in ________ summer. A. a;theB. an; /C. the; aD. /;/11. At first, I was not too sure if he could answer the question. However, ____,he worked it out at last with the help of his friend.A. to my angerB. to my surpriseC. in other wordsD. in a word12. —Must I stay here with you?—No, you ______.You may go home, but you _____ go to the net bar (网吧).A. mustn't; needn'tB. needn't; mustn'tC. must; needD. need; must13. I ______ the newspaper while my mother _____TV plays yesterday evening.A. was reading; was watchingB. was reading; watchedC. read; was watchingD. read; would watch14. It's a rule in my class that our classroom ________ before 6:00 p. m.every day.A. be able to cleanB. should be cleaningC. must cleanD. must be cleaned15. —Tom wants to know if you ________ a picnic next Sunday.—Yes. But if it ________, we'll visit the museum instead.A. will have; will rainB. have; rainsC. have; will rainD. will have; rains16.—Would you mind looking after my dog while I'm on holiday?—________.A. Of course notB. Yes. I'd be happy toC. Not at all. I've no timeD. Yes, please17. Many students didn’t realize the importance of study _______they left school.A. whenB. untilC. afterD. unless18. My father _______ to Shanghai. He _______ for over 2 months.A. has been, has leftB. has gone, has goneC. has gone, has been awayD. has been, has gone19. They are your skirts. Please __________.A. put it awayB. put out itC. put them awayD. put them out20. —Please read every sentence carefully. you are, mistakes you’ll make.—Thank you for your advice.A. The more carefully; the fewerB. The more careful; the lessC. The more carefully; the lessD. The more careful; the fewer21. My friend is coming today but he didn’t tell me _______.A. when did the train arriveB. how did the train arriveC. when the train arrivedD. how the train arrived22. I felt it hard to keep up with my classmate s. But whenever Iwant to _______, my teacher always encourages me to work harder.A. go onB. give upC. run awayD. give back23. —________ fine weather it is today!—Let's go for a picnic.A. WhatB. HowC. What aD. How a24. — Mary, you’re going to buy an apartment here, aren’t you?—Yes, but I can’t_______an expensive one.A. spendB. costC. payD. afford25. —Would you like to drink coffee or milk?—_________. Please give me some tea.A. NeitherB. BothC. EitherD. None第二节:完形填空(共20小题,每小题1分,满分20分)(A)Big schoolbags have been a serious problem for students for a long time.Maybe your schoolbag is too __26__ to carry, and it troubles you a lot __27__ you want to find a book out to read. Now an etextbook will __28__ you.It is said that etextbooks are going to be __29__ in Chinese middle schools.An etextbook, in fact, is a small __30__ for students.It is much __31__ than a usual schoolbag and easy to carry. Though it is as small as a book, it can __32__ all the materials (材料) for study.The students can read the text page by page on the __33__, take notes with the pointer (屏写笔). Or even “__34__” their homework to their teachers by sending emails. All they have to do is to press a button.Some people say etextbooks are good, but some say they may be __35__ for the students' eyes. What do you think of it?26.A.light B. heavy C. useful D. comfortable27.A.till B. after C. before D. when28.A.trouble B. prevent C. help D. understanded B. kept C. invented D. lent B. radio C. pen D. computer31.A.heavier B. lighter C. cheaper D. brighter32.A.hold B. build C. discover D. practice33.A.blackboard B. desk C. screen D. card34.A.find out B. hand in C. get back D. give back35.A.helpful B. famous C. good D. bad(B)。
银川一中2017年中考自主招生考试数学试卷(时间:90分钟卷分:120分)一、选择题(每题5分,共50分)1、银川一中为有效开展“阳光体育”活动,计划购买篮球和足球共50个,购买资金不超过3000元。
若每个篮球80元,每个足球50元,则篮球最多可购买()A.16个B.17个C.33个D.34个答案详解A正确率: 64%, 易错项: B解析:本题主要考查一元一次不等式的应用。
设篮球可以购买个,那么足球则购买了个。
买篮球总共花元,买足球共花元,由资金不超过列不等式,解得,由于必须取整数则篮球最多买个。
故本题正确答案为A。
2、若关于的方程有实数根,则实数的取值范围是()。
A: B: 且 C: D:答案详解C正确率: 43%, 易错项: B解析:本题主要考查二次函数与一元二次方程的联系。
记,那么原方程有实数根就转化为该函数与轴图象有交点即可。
当时,函数为一条直线必与轴有交点,即原方程有实数根,可取排除B选项。
当时,函数为二次抛物线,根据判别式定理时,函数必与轴相交,即时原方程有实数根。
综上为所求。
故本题正确答案为C。
3、已知等腰三角形的周长是10,底边长y是腰长x的函数,则下列图象中,能正确反映y 与x之间函数关系的图象是( )A B C D答案详解解:由题意得,,所以,,由三角形的三边关系得,,解不等式①得,,解不等式②的,,所以,不等式组的解集是,正确反映y与x之间函数关系的图象是D选项图象.所以D选项是正确的.4、一个几何体的主视图和俯视图如图所示,若这个几何体最多有个小正方体组成,最少有个小正方体组成,则等于()。
A: B: C: D:答案详解C正确率: 68%, 易错项: B解析:本题主要考查三视图。
包含最多小正方形和最少小正方形的立体几何图如下:由下图可知,,故。
5、一个圆锥的侧面积是底面积的倍,则圆锥侧面展开图的扇形的圆心角是()。
A: B: C: D:答案详解A正确率: 49%, 易错项: B解析:本题主要考查圆锥和几何体的平面展开图。
设圆锥底面半径为,则,又设圆锥母线长为,则,因为,即,所以,所以,即扇形弧长为整个圆周长的,所以。
故本题正确答案为A。
6、某共享单车前公里元,超过公里的,每公里元,若要使使用该共享单车的人只花元钱,应该要取什么数()。
A: 平均数 B: 中位数 C: 众数 D: 方差答案详解B正确率: 42%, 易错项: C解析:本题主要考查数据的分析。
根据中位数的定义,其值总是将所有数据按从小到大依次排列后,处于最中间的那个数(或中间两个数的平均数)。
因而当取中位数时,必有一半的数据在该值以下,即共享单车的人只花元钱。
故B项符合题意。
故本题正确答案为B。
7、我们知道:四边形具有不稳定性.如图,在平面直角坐标系中,边长为2的正方形ABCD 的边AB在x轴上,AB的中点是坐标原点O,固定点A,B,把正方形沿箭头方向推,使点D 落在y轴正半轴上点D'处,则点C的对应点C'的坐标为( )A. B. C. D.答案详解解:, , ,,,,所以D选项是正确的.8、如图,在中,,,以BC的中点O为圆心作圆,分别与AB,AC相切于D,E两点,则的长为( )A. B. C. D.答案详解解:连接OE、OD,设半径为r,∵分别与AB,AC相切于D,E两点,,,是BC的中点,是中位线,, ,同理可知:, ,,由勾股定理可知,,故选(B)9、如图,学校环保社成员想测量斜坡旁一棵树的高度,他们先在点处测得树顶的仰角为,然后在坡顶测得树顶的仰角为,已知斜坡的长度为,的长为,则树的高度是()。
A: B: C: D:答案详解B正确率: 49%, 易错项: A解析:本题主要考查角的概念及其计算。
因为,则,所以,又在中,,所以,则,所以,则,所以。
故本题正确答案为B 。
10、如图,已知的顶点坐标分别为、、。
若二次函数的图象与阴影部分(含边界)一定有公共点,则实数的取值范围是( )。
A:B:C:D:答案详解C正确率: 49%, 易错项: A解析:本题主要考查二次函数的图象与性质。
二次函数与轴的交点为,因为点的坐标是,所以对称轴时,二次函数的图象与阴影部分(含边界)一定有公共点,所以。
故本题正确答案为C 。
二、填空题(每题5分,共20分) 11、己知,,求(2x+y )2+(x+y)(x-y)-5x(x-y)=12、如图,用同样大小的黑色棋子按如图所示的规律摆放:则第⑦个图案有 个黑色棋子.答案详解解:第一个图需棋子1,第二个图需棋子,第三个图需棋子,第四个图需棋子,…第n个图需棋子枚.所以第⑦个图形有19颗黑色棋子.故答案为:19;13、如图,在矩形中,,将沿折叠,使点恰好落在对角线上处,则的长是。
答案详解C正确率: 55%, 易错项: B解析:本题主要考查矩形。
因为四边形是矩形,所以,由折叠可得,所以,,,在中,,,根据勾股定理得:,即,设,则有,根据勾股定理得:,解得:(负值舍去),则。
故本题正确答案为C。
14、如图,抛物线的对称轴为直线,与轴的一个交点在和之间,其部分图象如图所示,则下列结论:;;;(为实数);点,,是该抛物线上的点,则,正确的个数有。
答案详解B正确率: 0%解析:本题主要考查二次函数的图象与性质。
①项,由对称轴公式知,所以结论①正确;②项,由对称轴的性质知,而,故结论②正确;③项,由图象知,又,即,则,故结论正确;④项,要使对于任意恒成立,只需要证明对于任意恒成立,又,即证恒成立,尽管,但时该式可以为0,不总是对于任意永远小于0,故结论④错误;⑤项,由抛物线性质知,离对称轴最近的点值越大或越小,显然,离对称轴最近,其次是,然后是,所以,故结论⑤错误;综上所述,故正确个数有个。
故本题正确答案为B。
三、解答题(共50分)15(10分)为了了解同学们每月零花钱的数额,校园小记者随机调查了本校部分同学,根据调查结果,绘制出了如下两个尚不完整的统计图表。
请根据以上图表,解答下列问题:(1)填空:这次被调查的同学共有_____人,_____,_____ 。
(2)求扇形统计图中扇形的圆心角度数。
(3)该校共有学生人,请估计每月零花钱的数额在范围的人数。
答案详解(1),,。
(2),即扇形统计图中扇形的圆心角为。
(3),即每月零花钱的数额在范围的人数为。
16、(12分)收发微信红包已成为各类人群进行交流联系,增强感情的一部分,下面是甜甜和她的双胞胎妹妹在六一儿童节期间的对话。
请问:(1)年到年甜甜和她妹妹在六一收到红包的年增长率是多少?(2)年六一甜甜和她妹妹各收到了多少钱的微信红包?答案详解(1)设年到年甜甜和她妹妹在六一收到红包的年增长率是,依题意得:,解得,(舍去)。
答:年到年甜甜和她妹妹在六一收到红包的年增长率是。
(2)设甜甜在年六一收到微信红包为元,依题意得:,解得,所以(元)。
答:甜甜在年六一收到微信红包为元,则她妹妹收到微信红包为元。
17、如图,是的直径,点在的延长线上,平分交于点,且,垂足为点。
(1)求证:直线是的切线。
(2)若,,求弦的长。
答案详解(1)证明:连结,如图所示,因为平分,所以,因为,所以,所以,所以,因为,所以,所以直线是的切线。
(2)因为,所以,因为,所以,所以,所以,所以,所以,所以,,设,则,在中,根据勾股定理可知,所以,所以。
18、如图,抛物线经过点,交轴于点:(1)求抛物线的解析式(用一般式表示)。
word 格式整理版范文范例 学习指导(2)点 为轴右侧抛物线上一点,是否存在点使,若存在请直接给出点 坐标;若不存在请说明理由。
(3)将直线绕点顺时针旋转,与抛物线交于另一点,求的长。
答案详解(1)因为抛物线交轴于点,,那么,横坐标可以看做是抛物线方程等于时的两个根。
即有根,。
则,解得。
所以抛物线解析式为。
(2)由知,。
在抛物线解析式中令,得,即,代入抛物线中,且在轴右侧,得或或。
(3)根据两直线夹角公式知,,而,且,解得或。
因为,所以直线的解析式,与抛物线联立得方程,即,。
根据直线距离坐标公式知,。