单元质量评估(三)
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单元质量评估(三)第5、6章(90分钟100分)一、选择题(本题共25小题,每小题2分,共50分)1.基因突变是生物变异的根本来源。
下列关于基因突变特点的说法正确的是()A.无论是低等还是高等生物都可能发生突变B.突变对生物的生存往往是有利的C.突变只能定向形成新的等位基因D.生物在个体发育的特定时期才可发生突变2.(2011·邵阳高一检测)下列有关基因重组的说法不正确的是()A.基因重组发生在有性生殖形成配子时B.基因工程的原理是基因重组C.基因重组可以产生新的基因型D.基因重组可以产生很多新基因,是变异的重要来源3.(2011·邯郸高一检测)下列属于染色体变异的是()①花药离体培养后长成的植株②染色体上DNA碱基对的增添、缺失③非同源染色体的自由组合④四分体中非姐妹染色单体之间相应部位的交叉互换⑤21三体综合征患者细胞中的21号染色体有3条A.①④⑤B.②④C.②③④D.①⑤4.多基因遗传病是一类在人群中发病率较高的遗传病。
以下疾病属于多基因遗传病的是()①哮喘病②猫叫综合征③抗维生素D佝偻病④青少年型糖尿病A.①②B.①④C.②③D.③④5.如图是某植物正常的体细胞(图示细胞中所有的染色体)。
试判断该植物可能的基因型及细胞中所含的染色体组数是()A.ABCd,4B.AaaaBBBb,4C.AaBbCcDd,8D.BBBbDddd,86.下列说法错误的是()A.人类基因组计划测定的染色体有24条B.遗传咨询包括推算后代患病风险率C.家系分析法是人类遗传病研究最基本的方法D.21三体综合征发生的根本原因是减数第一次分裂同源染色体未分开7.(2011·大同高一检测)下列关于育种优点的叙述,不正确的是()A.多倍体较二倍体茎秆粗大,果实种子大B.杂交育种能产生新基因C.人工诱变育种能提高变异频率D.利用单倍体育种能明显缩短育种年限8.基因工程中常作为基因的运载体的一组是()A.质粒、线粒体、噬菌体B.染色体、叶绿体、线粒体C.质粒、噬菌体、动植物病毒D.细菌、噬菌体、动植物病毒9.在调查人群中的遗传病时,有许多注意事项,下列说法错误的是()A.调查的人群要随机抽样B.调查的人群基数要大,基数不能太小C.要调查群体发病率较低的单基因遗传病D.要调查患者至少三代之内的所有家庭成员10.(2011·泰安高一检测)已知某个核基因片段碱基排列如图所示:①—GGCCTGAAGAGAAGA—②—CCGGACTTCTCTTCT—若该基因由于一个碱基被置换而发生改变,氨基酸序列由“—脯氨酸—谷氨酸—谷氨酸—赖氨酸—”变成“—脯氨酸—谷氨酸—甘氨酸—赖氨酸—”。
单元质量评估(三)第4章(90分钟100分)一、选择题(共25小题,每小题2分,共50分)1.某同学在无菌培养基上接种酵母菌后,不久就出现了酵母菌菌落,该菌落属于( )A.种群B.群落C.细菌斑D.生态系统2.你认为下列调查动物种群密度的方法中,正确的是( )A.可采用样方法调查草地中跳蝻的种群密度B.可采用样方法调查草地中野兔的种群密度C.用标志重捕法调查土壤中小动物的种群密度D.用标志重捕法调查土壤中昆虫卵的种群密度3.(2011·临沂高二检测)下列有关种群和群落的研究和分析中,正确的是( )A.田鼠在被捉过一次后更难捕捉,统计得到的种群密度比实际数值要低B.对培养液中酵母菌进行计数采用的是抽样检测法C.对小动物的种群密度进行调查时,不能采用标志重捕法,一定要采用样方法D.某同学决定以校园内的池塘为对象进行物种丰富度调查,应在同一水层取样4.(2011·北京高考)在生态学研究中,下列方法与研究目的不相符的是( )A.给海龟安装示踪器调查其洄游路线B.给大雁佩戴标志环调查其迁徙路线C.用样方法研究固着在岩礁上的贝类的种群关系D.用标志重捕法调查达乌尔黄鼠的丰富度5.下列各项中,不属于对种群数量特征描述的是( )A.我国的人口将逐渐步入老龄化阶段B.第六次人口普查中,成都的人口出生率为6.77%C.橡树种子散布能力差,常在母株周围形成集群D.由于薇甘菊的入侵,松树种群的死亡率比较高6.据预测,我国人口数量在2050年将达到最高值,作出这一预测的依据是( )A.各年龄期人口的比例B.现有人口数量密度C.出生率、死亡率和迁入率、迁出率D.人口性别比例7.如图表示的是四个不同种群中不同年龄期的个体所占的比例,其中种群密度可能越来越小的是( )8.长期生活在某环境中的种群,其数量变化最可能的是( )A.呈“J”型或“S”型曲线增长B.增长、波动C.波动、稳定D.增长、下降9.种群是生态研究的一个重要单位,有关种群的叙述正确的是( )A.种群是指一个生态系统中同种生物所有成熟个体的总和B.一个呈“S”型增长的种群中,种群增长率在各个阶段是不同的C.种群中各年龄期的个体数目比例适中,这样的种群正处于发展阶段D.合理密植农作物时,其数量可以大于K值10.下列几组图中,能正确表示生存斗争、种间斗争、种内斗争和竞争关系的是( )11.如图表示处于平衡状态的某生物种群因某些外界环境变化导致种群中生物个体数量改变时的四种情形,下列对产生这些变化的原因分析,正确的是( )(1)若图①所示为草原生态系统中的某种群,则a点后变化的可能原因是过度放牧(2)若图②所示为某发酵罐中的菌种数量,则b点后变化的原因可能是增加营养供应(3)图③中c点后发生的变化表明生态系统遭到了破坏(4)图④曲线可用于指导海洋捕捞A.只有一种说法正确B.只有两种说法正确C.只有三种说法正确D.四种说法都正确12.下列对探究培养液中酵母菌种群数量动态变化实验的叙述,错误的是( )A.从试管中吸出培养液之前应振荡试管,有利于正确计数B.先将培养液滴在计数板上,轻盖盖玻片防止气泡产生C.当一个小方格中酵母菌数量过多时,应将培养液按比例稀释处理D.压在方格边上的酵母菌,只计数相邻两边及其顶角的个体数13.有一种紫腹巨蚊,当它处于幼虫状态时,专以毒蚊幼虫为食,当它发育成成虫后,又以吸食竹类植物的叶汁花浆为生。
单元质量评估(三)1.按劳分配是社会主义公有制经济中个人消费品分配的基本原则。
实行按劳分配的直接原因是() A.生产资料公有制B.生产力的发展水平C.多种分配方式并存D.社会主义条件下人们劳动的性质和特点2.居民购买国债所得属于按资本要素分配取得的收入,下列对这一分配方式认识正确的是() A.它是提高居民收入的主要途径B.它是个人消费品分配的主要方式C.它是与市场经济相适应的收入分配方式D.它是由生产资料公有制决定的分配方式3.刘月的爸妈承包集体土地种粮年收入1.5万元,收购、销售蔬菜年收入4万元,房屋出租年收入5千元。
其中属于按劳分配的收入为()A.6万元B.5.5万元C.4万元D.1.5万元4.广东某国家控股的药业有限公司与三位博士合办一家新的有限责任公司,原公司注册300万元拥有70%的股份,三位年轻博士以“无形资产”——知识拥有30%的股份,这种合办既实现了科技成果的产业化,又实现了个人技术与国有资产的结合。
这三位博士取得的收入属于()A.按劳分配所得B.按劳动要素分配所得C.按生产要素分配所得D.按劳分配与按生产要素分配所得5.从2012年1月1日起,南京市事业单位统一实施绩效工资,并同步推进和实施事业单位养老保险制度。
绩效工资是指根据工作成绩和劳动效率确定的工资。
事业单位实施绩效工资()①与社会主义公有制经济中个人消费品分配的基本原则相吻合②有利于调动事业单位劳动者的积极性,提高服务质量③有利于增加事业单位劳动者的工资收入④实现了按劳分配和按生产要素分配的统一A.①②B.②③C.②④D.③④6.下列对我国分配方式理解不正确的是()①劳动收入属于按劳分配,非劳动收入属于按生产要素分配②按劳分配体现效率优先,其他分配方式体现兼顾公平③按劳分配为主体、多种分配方式并存④按劳分配存在于公有制经济中,其他分配方式存在于非公有制经济中A.①②③④B.①②③C.②③④D.①②④7.“十二五”规划纲要提出,要“努力实现居民收入增长和经济发展同步、劳动报酬增长和劳动生产率提高同步”。
单元质量评估(三)第三章(90分钟 100分)一、选择题(本题包括16小题,每小题3分,共48分)1.(双选)硫酸锶(SrSO4)在水中的沉淀溶解平衡曲线如下。
下列说法正确的是( )A.温度一定时,K sp(SrSO4)随c(SO42-)的增大而减小B.三个不同温度中,313 K时K sp(SrSO4)最大C.283 K时,图中a点对应的溶液是不饱和溶液D.283 K下的SrSO4饱和溶液升温到363 K后变为不饱和溶液2.喷墨打印机墨汁的pH为7.5~9.0。
当墨汁喷在纸上时,生成不溶于水的固体。
由此可知( )①墨汁偏碱性②墨汁偏酸性③纸张偏酸性④纸张偏碱性A.①③B.②④C.①④D.②③3.在25 ℃时,用蒸馏水稀释1 mol·L-1氨水至0.01 mol·L-1,随溶液的稀释,下列各项中始终保持增大趋势的是( )A.32c(OH ) c(NH H O)- B.c(H )c(OH )+- C.324c(NH H O)c(NH )+D.c(OH -)4.有①、②、③三瓶体积相等,浓度都是1 mol ·L -1的HCl 溶液,将①加热蒸发至体积减小一半;在②中加入少量CH 3COONa 固体(加入后溶液仍呈强酸性);③不作改变。
然后以酚酞作指示剂,用NaOH 溶液滴定上述三种溶液,所消耗NaOH 溶液的体积是( )A.①=③>②B.③>②>①C.③=②>①D.①=②=③5.(2012·烟台高二检测)相同体积的pH=3的强酸溶液和弱酸溶液分别跟足量的镁完全反应,下列说法中正确的是( ) A.弱酸溶液产生较多的氢气 B.强酸溶液产生较多的氢气 C.两者产生等量的氢气 D.无法比较两者产生氢气的量6.已知次氯酸是比碳酸还弱的酸,氯水中存在平衡Cl 2+H 2OHCl+HClO ,HClOH ++ClO -。
达平衡后,要使HClO 浓度增大,可加入( )A.H 2SB.Na 2CO 3C.盐酸D.NaOH7.(双选)(2012·徐州高二检测)下列溶液中有关离子浓度的说法可能正确的是( )A.NH4Cl溶液:c(Cl-)>c(NH4+)>c(OH-)>c(H+)B.Na2CO3溶液中:c(Na+)+c(H+)=c(HCO3-)+c(OH-)+c(CO32-)C.显酸性的CH3COOH与CH3COONa的混合溶液:c(CH3COO-)>c(Na+)>c(H+)>c(OH-)D.在NaHCO3溶液中:c(Na+)=c(HCO3-)+c(CO32-)+c(H2CO3)8.(2012·苏州高二检测)二元酸H2A在水中发生电离:H2A====H++HA-和HA-H++A2-,则下列几种物质的水溶液中关系式不正确的是( )A.NaHA:c(Na+)>c(HA-)>c(OH-)>c(H+)B.NaHA:c(Na+)=c(A2-)+c(HA-)C.Na2A:c(Na+)>c(A2-)>c(OH-)>c(H+)D.H2A:c(H+)=c(HA-)+2c(A2-)+c(OH-)9.在蒸发皿中加热蒸干并灼烧(低于400 ℃)下列的溶液,可以得到该物质的固体是( )A.氯化镁溶液B.硫酸亚铁溶液C.碳酸氢钠溶液D.硫酸铝溶液10.常温下,一定浓度的某溶液,由水电离产生的c(OH-)=10-4mol·L-1,则该溶液中的溶质可能是( )A.Al2(SO4)3B.CH3COONaC.NaOHD.KHSO411.已知当NH4Cl溶液的浓度小于0.1 mol·L-1时,其pH>5.1。
单元质量评估(三)第三章 圆锥曲线与方程 (120分钟 150分)一、选择题(本大题共12小题,每小题5分,共60分,在每小题给出的四个选项中,只有一项是符合题目要求的)1.(2011·湖南高考)设双曲线222x y 1a 9-=(a>0)的渐近线方程为3x ±2y=0,则a 的值为( )(A)4 (B)3 (C)2 (D)12.若椭圆的对称轴为坐标轴,长轴长与短轴长的和为18,焦距为6,则椭圆的方程为( )(A)22x y 1916+=(B)22x y 12516+=(C)2222x y x y 1125161625+=+=或(D)以上都不对3.(2011·许昌高二检测)已知抛物线y 2=2px(p >0),过其焦点且斜率为1的直线交抛物线于A 、B 两点,若线段AB 的中点的纵坐标为2,则该抛物线的准线方程为( )(A)x=1 (B)x=-1 (C)x=2 (D)x=-24.中心在原点,焦点在x 轴上的双曲线的一条渐近线经过点(4,-2),则它的离心率为( )225.(2011·广东高考)设圆C 与圆x 2+(y-3)2=1外切,与直线y=0相切,则C 的圆心轨迹为( )(A)抛物线 (B)双曲线 (C)椭圆 (D)圆6.连接抛物线x 2=4y 的焦点F 与点M(1,0)所得的线段与抛物线交于点A ,设点O 为坐标原点,则三角形OAM 的面积为( )32327.椭圆2222x y 1a b+=(a >b >0)的左焦点为F ,A(-a ,0)和B(0,b)是两个顶点,若F到直线AB 的距离为7,则椭圆的离心率为( )12 (D)458.(2011·张家界高二检测)椭圆2222x y 1a b += (a >b >0)则双曲线2222x y 1a b -=的离心率为( )(A)54329.以正方形ABCD 的相对顶点A 、C 为焦点的椭圆,恰好过正方形四边的中点,则该椭圆的离心率为( )10.设抛物线y 2=8x 的焦点为F ,准线为l ,P 是抛物线上的一点,PA ⊥l ,A 为垂足,如果直线AF 的斜率为|PF|=( )(A)11.(2011·新课标全国高考)已知直线l 过抛物线C 的焦点,且与C 的对称轴垂直,l 与C 交于A ,B 两点,|AB|=12,P 为C 的准线上的一点,则△ABP 的面积为( )(A)18 (B)24 (C)36 (D)4812.(2011·兰州高二检测)连接双曲线2222x y 1a b -=(a >0,b >0)与2222y x 1b a-=(a >0,b >0)的四个顶点的四边形面积为S 1,连接其四个焦点的四边形面积为S 2,则12S S 的最大值是( )(A)2 (B)4 (C)14(D)12二、填空题(本大题共4小题,每小题4分,共16分,请把正确的答案填在题中的横线上)13.(2011·大庆高二检测)已知抛物线y 2=4x 焦点F 恰好是双曲线2222x y 1a b-=的右焦点,且双曲线一条渐近线过点(23a 2,b),则该双曲线的渐近线方程为 .14.以双曲线22y x 13-=的右焦点为圆心,离心率为半径的圆的方程是 .15.在一椭圆中以焦点F 1,F 2为直径的圆,恰好过短轴的两顶点,则此椭圆的离心率e = .16.(2011·新课标全国高考)在平面直角坐标系xOy 中,椭圆C 的中心为原点,焦点F 1,F 2在x 过F 1的直线l 交C 于A ,B 两点,且△ABF 2的周长为16,那么C 的方程为 .三、解答题(本大题共6小题,共74分,解答时应写出必要的文字说明、证明过程或演算步骤)17.(12分)双曲线的离心率等于2,且与椭圆22x y 1259+=有相同的焦点,求此双曲线的标准方程.18.(12分)某抛物线形拱桥跨度是20米,拱高4米,在建桥时每隔4米需用一支柱支撑,求其中最长的支柱的长.19.(12分)在直角坐标系中,以M(-1,0)为圆心的圆与直线相切. (1)求圆M 的方程;(2)已知A(-2,0)、B(2,0),圆内动点P 满足|PA|·|PB|=|PO|2,求PA PB的取值范围.20.(12分)过点(0,4),斜率为-1的直线与抛物线y 2=2px(p >0)交于两点A 、B ,如果弦|AB|的长度为. (1)求p 的值;(2)求证:OA ⊥OB(O 为原点).21.(12分)(2011·孝感高二检测)已知椭圆22x y 14+=的左、右顶点分别为A 、B ,曲线E 是以椭圆的中心为顶点,B 为焦点的抛物线. (1)求曲线E 的方程;(2)直线l 与曲线E 交于不同的两点M 、N.当AM AN≥17时,求直线l 的倾斜角θ的取值范围.22.(14分)(2011·北京高考)已知椭圆G :2222x y 1a b +=(a >b >0)右焦点为(0),斜率为1的直线l 与椭圆G 交于A ,B 两点,以AB 为底边作等腰三角形,顶点为P(-3,2). (1)求椭圆G 的方程; (2)求△PAB 的面积.答案解析1.【解析】选C.由222x y 1a 9-=可得到双曲线的渐近线方程为y=〒3a x,又已知双曲线的渐近线方程为3x 〒2y=0,根据直线重合的条件可得到a=2. 2. 【解析】选C.由题意2a 2b 182c 6+=⎧⎨=⎩,又a 2=b 2+c 2. 解得:a=5,b=4,∴椭圆方程为2222x y x y 11.25161625+=+=或3.【解析】选B.设A(x 1,y 1),B(x 2,y 2),则由211222y 2px y 2px ⎧=⎪⎨=⎪⎩,得:(y 1-y 2)(y 1+y 2)=2p(x 1-x 2), 即121212y y 2px x y y -=-+,∴1=2p4,∴p=2, ∴抛物线的准线为x=-1.4.【解析】选D.易知双曲线的渐近线方程为y=-12x , ∴=5.【解析】选A.由题意,C 的圆心到点(0,3)与直线y=-1的距离相等,由抛物线的定义知C 的圆心轨迹为抛物线,故选A.6.【解析】选B.由题意得F 的坐标为(0,1).又点M(1,0),故线段MF 所在直线的方程为x+y=1(0≤x ≤1). 解2x y 1(0x 1)x 4y+=≤≤⎧⎨=⎩, 得交点A 的坐标为,∴S △OAM =12〓1〓327.独具【解题提示】由点F 到直线AB出关系式,消去b 后构造关于c a的方程,解方程求c a,即离心率e. 【解析】选C.直线AB 的方程为bx-ay+ab=0, F(-c ,0),∴7, ∴5a 2-14ac+8c 2=0, ∴8e 2-14e+5=0,解得:e=12.8.【解析】选B.由题意椭圆中=∴22b 1a 4=, ∴双曲线中=9.【解析】选D.如图所示,令|AC|=2c , 则,|EC|=2c ,2a =, ∴e=c a ==10.独具【解题提示】解答本题可借助直线AF 的倾斜角为120°的几何性质将条件转化,在等边三角形PAF 中可得|PF|.【解析】选B.如图所示: ∵直线AF 的斜率为∴∠AFK=60°, ∴∠PAF=60°, 又|PA|=|PF|,∴△APF 为等边三角形, 在Rt △AKF 中,|FK|=4, ∴|AF|=8,∴|PF|=8.11.独具【解题提示】确定点P 到直线AB 的距离d ,利用S △ABP =12|AB|·d 求面积. 【解析】选C.设抛物线方程为y 2=2px ,则点C(p 2,0),在方程中,令x=p 2,则y=〒6,即36=p 2,得p=6,∴y 2=12x ,∴点P 到直线AB 的距离为p=6, ∴S △ABP =12|AB|·6=36.12.【解析】选D.易知S 1=2ab ,S 2=2(a 2+b 2), ∴()122222S 2ab ab ab 1S a b 2ab 22a b ==≤=++,当且仅当a=b 时取等号. 13.【解析】由222a b 1b b 3a a 2⎧+=⎪⎪⎨=⎪⎪⎩,得:2a 3b ⎧=⎪⎪⎨⎪=⎪⎩ ∴双曲线的渐近线方程为y=〒2x. 答案:y=14.【解析】,∴右焦点为(2,0),e=c a=2,∴圆的方程为(x-2)2+y 2=4. 答案:(x-2)2+y 2=415.【解析】由题意b=c ,∴a 2=b 2+c 2=2c 2,∴e=c a2=.答案:216.独具【解题提示】△ABF 2的周长为4a ,求得a 的值,再由离心率求得c 的值,可得椭圆的方程.【解析】由△ABF 2的周长=4a=16,得a=4,即c a 进而c=所以a 2=16,b 2=a 2-c 2=16-8=8,∴C 的方程为22x y 1168+=.答案:22x y 1168+=17.【解析】∵椭圆22x y 1259+=的焦点坐标为(-4,0)和(4,0),则可设双曲线方程为2222x y 1a b-=(a >0,b >0),∵c=4,又双曲线的离心率等于2,即c a=2,∴a=2.∴b 2=c 2-a 2=12.故所求双曲线方程为22x y 1.412-=18.【解析】以拱顶为原点,水平线为x 轴,建立坐标系,如图,由题意知, |AB|=20,|OM|=4,A 、B 坐标分别为(-10,-4)、(10,-4),设抛物线方程为x 2=-2py ,将A 点坐标代入, 得100=-2p 〓(-4),解得p=12.5,于是抛物线方程为x 2=-25y , 由题意知E 点坐标为(2,-4),E ′点横坐标也为2, 将2代入得y=-0.16,从而|EE ′|=(-0.16)-(-4)=3.84, 故最长支柱长应为3.84米.19.【解析】(1)依题意,圆M 的半径等于圆心M(-1,0)到直线的距离, 即,∴圆M 的方程为(x+1)2+y 2=4.(2)设P(x ,y),由|PA|·|PB|=|PO|2,得22x y =+,即x 2-y 2=2.PA PB =(-2-x ,-y)·(2-x ,-y)=y 2+x 2-4=2x 2-6,由()2222x y 2x 1y 4⎧-=⎪⎨++=⎪⎩,解得:11x 22--+=舍), ∵点P 在圆M内,∴x ≤, ∴2≤x2,∴-2≤2x2-6, ∴PA PB的取值范围是[-2).独具【误区警示】本题易出现的错误是利用点在圆内,求出-2<y <2或-3<x<1,从而求出PA PB 的范围为[-2,6)或 [-6,12),错误的原因是忽视了借助图像准确确定x 的范围,致使所求范围过大.20.【解析】(1)直线AB 的方程为y=-x+4,由2y x 4y 2px =-+⎧⎨=⎩,得:y 2+2py-8p=0,设A(x 1,y 1),B(x 2,y 2),∴y 1+y 2=-2p ,y 1y 2=-8p ,∴AB ==解得:p=2.(2)由(1)可知OA OB =x 1x 2+y 1y 2=2212y y 44 -8〓2=16-16=0, ∴OA ⊥OB.21.【解析】(1)依题意得:A(-2,0),B(2,0),∴曲线E 的方程为y 2=8x. (2)由)2y x 1y 8x⎧=-⎪⎨=⎪⎩得:kx 2-(2k+8)x+k=0. 由()222k 84k 0,k 0⎧∆=+-⎪⎨⎪⎩>>∴k >0, 设M(x 1,y 1),N(x 2,y 2), 则x 1+x 2=2k 8k+,x 1x 2=1. ∴AM AN =(x 1+2,y 1)·(x 2+2,y 2) =(x 1+2)(x 2+2)+y 1y 2=(k+1)x 1x 2+(2-k)(x 1+x 2)+4+k=161k+≥17,∴0<k ≤1,∴θ∈(0,4π].22.独具【解题提示】(1)利用a 、b 、c 的关系及离心率求出a ,b ,代入标准方程;(2)联立直线方程与椭圆方程,然后利用根与系数的关系,设而不求,整体代入.【解析】(1)由已知得c a = ,解得又b 2=a 2-c 2=4,所以椭圆G 的方程为22x y 1.124+=(2)设直线l 的方程为y=x+m , 由22y x mx y 1124=+⎧⎪⎨+=⎪⎩得,4x 2+6mx+3m 2-12=0① 设A ,B 的坐标分别为(x 1,y 1),(x 2,y 2)(x 1<x 2),AB 的中点为E(x 0,y 0),则x 0=12x x 2+=-3m4,y 0=x 0+m=m4.因为AB 是等腰△PAB 的底边,所以PE ⊥AB.所以PE 的斜率m24k 13m34-==--+,解得m=2.此时方程①为4x 2+12x=0,解得x 1=-3,x 2=0,所以y 1=-1,y 2=2.所以此时,点P(-3,2)到直线AB :x-y+2=0的距离2=所以△PAB 的面积S=19AB d 22= .。
单元评估检测(三) 牛顿运动定律(90分钟100分)一、选择题(此题共10小题,每一小题4分,共40分)1.关于牛顿第一定律的理解错误的答案是( )A.牛顿第一定律反映了物体不受外力的作用时的运动规律B.不受外力作用时,物体的运动状态保持不变C.在水平地面上滑动的木块最终停下来,是由于没有外力维持木块运动D.飞跑的运动员,由于遇到障碍而被绊倒,这是因为他受到外力作用迫使他改变原来的运动状态【解析】选C。
牛顿第一定律描述的是物体不受外力作用时的状态,即总保持匀速直线运动状态或静止状态,A、B正确;牛顿第一定律揭示了力和运动的关系,力是改变物体运动状态的原因,而不是维持物体运动状态的原因,在水平地面上滑动的木块最终停下来,是由于摩擦阻力的作用而改变了运动状态,飞跑的运动员,遇到障碍而被绊倒,是因为他受到外力作用而改变了运动状态,C错误,D正确。
2.(2019·杭州模拟)如下的运动图象中,物体所受合外力不为零的是 ( )【解析】选D。
选项A表示物体静止,其合外力为零;选项B、C表示物体做匀速直线运动,其合外力为零;选项D表示物体做匀变速直线运动,其合外力不为零,所以此题应选D。
3.如下列图,一同学在水平桌面上将三个形状不规如此的石块成功叠放在一起,保持平衡,如下说法正确的答案是( )A.石块b对a的支持力与a受到的重力是一对相互作用力B.石块c对b的支持力一定等于b受到的重力C.石块c受到水平桌面向左的摩擦力D.石块c对b的作用力一定竖直向上【解析】选D。
石块b对a的支持力与其对a的静摩擦力的合力,跟a受到的重力是平衡力,A 错误;石块c对b的支持力与石块c对b摩擦力的合力一定等于石块a和b受到的重力,B错误;以三块作为整体研究,如此石块c不会受到水平桌面的摩擦力,C错误;选取石块a、b作为整体研究,根据平衡条件,如此石块c对b的作用力与其重力平衡,如此石块c对b的作用力一定竖直向上,D正确。
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单元质量评估(三)Unit 3(120分钟150分)第I卷第一部分听力(共两节,满分30分)第一节(共5小题;每小题1. 5分,满分7. 5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1.Why is the man going to London?A.To live there・B・ To visit a friend.C. To work there.2.What is the woman looking for?A.Her camera.B. Her coat・C. Her plane ticket.3.When will the man go to the store?A.This Monday.B. Tomorrow.C. This weekend.4.Where does the conversation most probably take place?C. In the classroom.5.How much can one student save in the gallery?A・$2・ B. $8. C. $10.【听力材料】Text 1W: I heard you9re leaving.M: Yeah, I am moving to London. @Tve got a new job・W: Fll miss you.Text 2M: Oh, my God! Sophie, ifs already 9: 30. We have to get to the airport before 10 o,clock.W: But where is my camera? I can't find it anywhere. I need it to take pictures while we are in Shanghai.Text 3W: Do you have any special plans for 5 your weekend?M: Fm going to the store to buy a huge HDTV. I like watching sports programs. The screen of my present TV set is so small.Text 4W: Will you give me that book, please? ' The one in the second shelf! Here is my card. M: Here you are. You must be collecting information about the experiment, too.Text 5W: My dear, I hardly believe the admission price of the gallery is ~ ten dollars perperson. Ifs pretty expensive for a single exhibit.M: But if we have student cards, we can get in for two dollars・W: Really? Lefs have a try.答案:CACBB第二节(共15小题;每小题1. 5分,满分22. 5分)听下面5段对话。
Unit3单元质量评估(三)(新人教版选修7)Ⅱ. 单项填空(共15小题;每小题1分,满分15分)21. Teenagers should keep in mind that there’s something more important than ______ fashion for them to seek after in ______ life.A. /; theB. /; /C. the; /D. the; the22. Tom was so lucky tha t a fisherman who was fishing nearby ______ him ______ of the seawater after he fell off his boat.A. helped; outB. picked; outC. let; outD. blocked; out23. He urged ______ us the need for patience in doing this job.A. toB. onC. ofD. with24. Global food prices ______ very sharp increase during the year 2010 and they are continuing to rise further.A. observedB. witnessedC. provedD. supplied25. It is useless to attempt to______ from every danger: some risks must be taken.A. fleeB. hideC. hesitateD. run26. It’s not good manners for a lady to speak almost without ______ for breath; it’s very necessary to listen to others’ opinion.A. pausingB. standingC. predictingD. opposing27. It was several minutes before I was ______ of what happened.A. tiredB. accusedC. awareD. ashamed28. If I correct someone, I will do it with as much good humor and self-restraint as if I were the one ______.A. to be correctedB. correctingC. having being correctedD. being corrected29. The lost car of the Lees was found ______ in the woods off the highway.A. vanishedB. abandonedC. scatteredD. rejected30. The lack of eco-friendly habits among the public is thought to be a major ______ of global climate change.A. resultB. causeC. warningD. reflection31. The meeting is beginning at 8 o’clock. ______, let’s make some preparations for it.A. WhenB. MeanwhileC. WhileD. During32. As she was late again,Mary was scared ______ her boss because she was scared ______.A. to face; to be punishedB. to face; of being punishedC. of facing; of being punishedD. of facing; to be punished33. —I’m looking forward to your ______ here.—Sorry, I’m too busy and haven’t even a minute to ______.A. come; spareB. come; sparingC. coming; sparingD. coming; spare34. His laboratory has become the______, which was attacked by the animal rights groups for he had had drug trials on monkeys.A. exampleB. handleC. faceD. target35. All the leaders were ______ leave the hall when the explosion happened.A. able toB. about toC. sure toD. likely toⅢ. 完形填空(共20小题;每小题1. 5分,满分30分)A study, published in January in the US magazine, shows many people believe 36 are dangerous and will always try to hurt37 even kill humans. In fact, 94 percent of the world’s 400 species are harmless to humans. A shark 38 at the National Aquarium in Baltimore, US, proves this. Visitors can 39 young sharks, see their eggs deve lop and watch dozens of different species swim 40 around a huge tank. Most people41 to realize that shark attacks don’t happen very often. Humans are more42 to be killed by lightning than by a shark. And to make this point clear, the museum has43 a special touching pool for children. There, kids can learn, from an early age,44 sharks. They can watch them develop inside their eggs and feel the skin of the older swimmers. “People fear what they don’t know,” said Nancy Hotchkiss, an organizer of the exhibition, which runs45 December. “Sharks have been around for 400 million years and play an important role in the ocean’s food 46 . We want people to discover that sharks are47 animals that need our respect and 48 . ” A shocking 100 million sharks are killed every year around the world by humans. Scientists found that almost all recorded shark species have fallen 49 half in the past 8 to 15 years. Thousands of sharks are hunted in Asia for special foods,50 shark fin(鱼翅)soup. And many others get 51 in nets, while fishermen are hunting other fish.52 half of all sharks are smaller than one meter long. “Some fishing methods are actually 53 the ocean,” said Dave Schofield, the manager of the aquarium’s ocean health programme. “The fis hermen throw them away 54 rubbish. It is a worrying situation and some areas have put55 in place to protect these special fish. ”36. A. sharks B. fish C. animals D. beings37. A. and B. or C. but D. so38. A. talk B. filmC. TV playD. exhibition39. A. touch B. kill C. hurt D. buy40. A. hard B. frequentlyC. difficultyD. smoothly41. A. want B. begin C. fail D. attempt42. A. possible B. likelyC. probableD. lively43. A. stand up B. look upC. set upD. get up44. A. fear B. fearsC. not to fearD. to fear45. A. until B. when C. while D. as46. A. factory B. chain C. ship D. producer47. A. amazing B. happy C. shy D. cold48. A. punishment B. protectionC. preventionD. pollution49. A. by B. to C. in D. from50. A. for B. toC. for exampleD. such as51. A. to catch B. caught C. catching D. catch52. A. Rather than B. More thanC. LessD. Few53. A. cleaned B. cleaning outC. cleanD. cleaned54. A. as B. looking likeC. likeD. as if55. A. up B. rules C. value D. measuresⅣ. 阅读理解(共20小题;每小题2分,满分40分)(A)In October on the east coast of Australia migrating humpback whales (座头鲸) pass very close to the shore. With the help of Max Egan,we went to filmthem near Byron Bay. It was extremely exciting to spot the little puffs (水柱) of spray on the horizon. When migrating,the whales can be traveling at speeds of about 12 km/hour and can dive under anywhere between five and forty minutes, but with many years of experience Max usually guided our boat to exactly the right place to provide us with amazing sighting. As the whales break the surface like submarines (潜水艇),the water spills off their broad dark backs and huge tails. Then, with a powerful downstroke, they dive back into the depths of the sea, leaving just a smooth glassy patc h of water or “footprint” on the surface to indicate where they had been.On one day we had traveled up and down the shore and far out to sea in search of whales, but the horizon was deserted. There was no report of whales in the area so we decided to test out Max’s new, highly sensitive hydrophone (水下听音器) and the result was astonishing. As soon as the hydrophone was in the water, we found ourselves listening to the most beautiful voices. What may have seemed like a vast empty ocean was filled with the singing of whales.It is the males which sing, as they migrate between feeding and breeding grounds. No one knows for sure what the purpose of these songs is, whether they are love songs to the females or warnings to other males to stay away.It was thought that each population of humpbacks had its own song which remained constant but recent findings suggest that the whales appreciate a memorable tune and quickly adopt any new songs they hear.56. Which finding about whales is beyond the writer’s expectation?A. The little puffs.B. The songs of whales.C. The traveling speed of whales.D. The depth of the whale’s dive.57. What do we know about Max?A. He likes whales very much.B. He is a brave scientist on whales.C. He is an experienced man in finding whales.D. He is an expert in making hydrophones.58. The writer’s main purpose of looking for whales is ______.A. carrying out research on whalesB. studying the beautiful songs of whalesC. making films about whalesD. hunting whales for research59. What’s the WRONG opinion abou t whales before according to the passage?A. Each group of humpbacks had the song of their own.B. Whales can’t sing to each other.C. Whales aren’t easy to find.D. Whales can’t dive in the sea as long as forty minutes.(B)Dolphins(海豚)are not fish, but warm-blooded animals. Theylive in groups, and speak to each other in their own language. In thisway they are like other animals,such as bees and birds. Butdolphins are very different from almost all land animals. Their brainis nearly the same size as our own and they live a long time—at least twenty or thirty years.Like some animals, dolphins use sound to help them find their way around. They also make these sounds to talk to each other and to help them find food. We now know they do no t use their ears to receive these sounds,but the lower part of the mouth, called the jaw.Strangely, dolphins seem to like man, and for thousands of years there have been stories about the dolphin and its friendship with people.There is a story about sailors in the 19th century. In a dangerous part of the sea off the coast of New Zealand,they learnt to look for a dolphin called Jack. From 1871 to 1903,Jack met every boat in the area and showed it the way. Then in 1903 a passenger on a boat call ed The Penguin shot and wounded Jack. He recovered and for nine years more continued to guide all ships through the area—except for The Penguin.Today, some people continue to kill dolphins,but many countries of the world now protect them and in these places it is against the law to kill them.60. Dolphins are different from many other animals because they ______.A. live in groupsB. have large brainsC. are warm-bloodedD. have their own language61. Which of the following does the dolphin use to he lp it find its way around?A. Its nose.B. Its ears.C. Its mouth.D. Its eyes.62. Why did the sailors off the coast of New Zealand look for Jack?A. He was lonely and liked to be with people.B. They enjoyed playing with him.C. He was seriously wounded.D. They wanted his help.63. By telling the story of Jack the writer wanted to show that ______.A. dolphins are friendly and cleverB. people are cruel to animalsC. Jack is different from other dolphinsD. dolphins should be protected by law(C)Michael Fish may soon be replaced as a weather forecaster bysomething truly fishier—the shark (鲨鱼).Research by a British biology student suggests that sharkscould be used to predict storms.Lauren Smith,24,is close to completing her study on shark’sability to sense pressure.If her studies prove the theory,scientists may be able to monitor the behaviour of sharks to predict bad weather.Miss Smith had previously studied the behaviour of lemon sharks in theBahamas.She then used their close relatives,lesser spotted dogfish,for further research at Aberdeen University.Her work—thought to be the first of its kind to test the pressure theory—resulted from the observation that juvenile blacktip sharks off Flori da moved into deeper water ahead of a violent storm in 2001.Miss Smith said: “I’ve always been crazy about traveling and diving and this led me to an interest in sharks. ”“I was delighted to have been able to research in the area for my degree. I know there’s so much more we need to understand—but it certainly opens the way to more research. ”It has been discovered that a shark senses pressure using hair cells in its balance system.At the Bimini Shark Lab in the Bahamas, Miss Smith fixed hi-tech sensors to sharks to record pressure and temperature,while also tracking them using GPS (Global Positioning System) technology.In Aberdeen, she was able to study the effects of tidal(潮汐的) and temperature changes on dogfish—none of which were harmed. She also used a special lab which can mimic(模拟) oceanic pressure changes caused by weather fronts.She is due to complete her study and graduate later this year. She says she will be looking for a job which will give her the chance to enri ch her experience of shark research.64. The passage is most probably taken from ______.A. a short-story collectionB. a popular science magazineC. a research paperD. a personal diary65. What do we learn from the first four paragraphs of the passage?A. Sharks may be used to predict bad weather.B. Sharks’ behaviour can be controlled.C. Michael Fish is not qualified for his job.D. Lauren Smith will become a weather forecaster.66. Lauren Smith conducted her research by ______.A. removing hai r cells from a shark’s balance systemB. measuring the air pressure of weather frontsC. recording sharks’ body temperatureD. monitoring sharks’ reaction to weather changes67. What is the passage mainly about?A. A popular way of forecasting weather.B. A new research effort in predicting storms.C. Biologists’ interest in the secrets of sharks.D. Lauren Smith’s devotion to scientific research.(D)The Giant Squids(鱿鱼)in the Deep OceanJapanese scientists have taken the first photographs of oneof the most mysterious creatures in the deep ocean—the giantsquid.Until now the only information about the behavior of thecreatures which measure up to 18 meters(59 feet) in length hasbeen based on dead or dying squid washed up on shore or captured in commercial fishing nets.But Tsunemi Kubodera of the National Science Museum,and Kyoichi Mori of the Ogasawara Whale Watching Association, both in Tokyo have ca ptured the first images of Architeuthis attacking bait(鱼饵)900 meters below the surface in the cold, dark waters of the North Pacific. “We show the first wild images of a giant squid in its natural environment,”they said in a report in the journal Proceedings B of the Royal Society.Little is known about the creatures because it has been so difficult tolocate and study them alive. Large ships and specialist equipment,which is costly, are needed to study deep sea environments.The Japanese scientists found the squid by following sperm whales,the most effective hunters of giant squids,as they gathered to feed between September and December in the deep waters off the coast of the Ogasawara Islands in the North Pacific. They used a remote long-line camera and depth logging system to capture the giant squid in the ocean depths.The most dramatic character of giant squids is the pair of extremely long tentacles(触须),distinct from the eight shorter arms. The long tentacles make up to two-thirds of the length of the dead specimens(标本,样本)to date. The giant squid appears to be a much more active meat-eating animal than researchers thought before.68. The passage mainly tells us that ______.A. scientists captured a giant squid aliveB. scientists captured a giant squid on cameraC. giant squids are spe cial meat-eating animalsD. giant squids mainly live in the deep sea69. According to the passage,we can infer that the underlined word “Architeuthis” should refer to ______.A. a scientistB. the sperm whaleC. a big shipD. the giant squid70. According to the passage, giant squids ______.A. like living in the cold and deep watersB. mainly feed on the dead fishC. like playing, using their tentaclesD. have only eight arms around the mouth71. According to the passage, the scientists located the giant squid _____.A. through a remote long-line cameraB. by using depth logging systemC. by following sperm whalesD. by using the bait to attract it(E)A child who has once been pleased with a tale likes, as a rule, tohave it retold in almost the same words, but this should not leadparents to treat printed fairy stories as formal texts. It is always muchbetter to tell a story than read it out from a book, and, if a parent canproduce what, in the actual circumstances(情况) of the time and theindividual child, is an improvement on the printed text,that’s muchbetter.A charge made against fairy tales is that they harm the children by frightening them or making them sad thinking. To prove the latter,one would have to show in a controlled ex periment that children who had read fairy stories were more often guilty of cruelty than those who had not. As to fears, there are, I think, some cases of children being strongly terrified by some fairy stories. This also arises from the child who has hear d the story once. However,familiarity with the story by repetition turns the pain of fear into the pleasure of a fear faced and mastered.There are also people who are opposed to fairy stories in that they are not true, that giants, witches, two-headed dragons, magic carpets, etc. do not exist; and that, instead of being fond of the strange side in fairy tales, the child should be taught how to adapt to reality by studying history and mechanics. I find such people, I must say so strange that I do not k now how to argue with them. If their cases were sound, the world should be full of mad men attempting to fly from New York to Philadelphia on a stick or covering a telephone with kisses in the belief that it was their beloved girlfriend.No fairy story has ever claimed(声称) to be description of the real world and no clever child has ever believed that it is.72.The author considered that a fairy story is more effective ______.A. when a story is repeated without any changeB. when we treat a story as a jokeC. when the parent makes some changes about a storyD. when the story is set in the present time73.According to the passage, a child can feel scared when a story is ____.A. in a realistic settingB. heard for the first timeC. repeated too oftenD. told in a different way74.What is the advantage claimed for repeating fairy stories to young children?A. Making them less fearful.B. Developing their power of memory.C. Making them believe there is nothing to be afraid of.D. Encouraging them not to have stran ge beliefs.75.According to the last paragraph, we can know ______.A. fairy stories are still being made upB. there are some misunderstandings about fairy talesC. people try to update old fairy storiesD. there is more concern for children’s fears now adays第Ⅱ卷(非选择题)Ⅴ. 书面表达(共1题;满分35分)假设你是李华,在出国交流学习期间,到当地一家电影院看了一场精彩的电影,但在观看电影过程中发现了一些问题。
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单元质量评估(三)第三章概率(120分钟150分)一、选择题(本大题共12小题,每小题5分,共60分,在每小题给出的四个选项中,只有一项是符合题目要求的)1.下列事件中,随机事件的个数是( )①如果a,b是实数,那么b×a=a×b;②某地6月6日刮风下雨;③当x是实数时,x2≥0;④一个电影院某天的上座率超过50%.(A)1个 (B)2个 (C)3个 (D)4个2.(2011·江苏高考改编)从1,2,3,4这四个数中一次随机取两个数,则其中一个数是另一个的两倍的概率是( )(A)18(B)14(C)13(D)123.(2011·顺义高一检测)在一个袋子中装有分别标注数字1,2,3,4,5的五个小球,这些小球除标注的数字外完全相同.现从中随机取出2个小球,分两次抽取,每次取后放回,则取出的小球标注的数字之和为3的倍数的概率是( )(A)925 (B)825(C)725(D)6254.从装有两个红球和两个黑球的口袋内任取两个球,那么互斥而不对立的两个事件是( )(A)“至少有一个黑球”与“都是黑球”(B)“至少有一个黑球”与“至少有一个红球”(C)“恰有一个黑球”与“恰有两个黑球”(D)“至少有一个黑球”与“都是红球”5.某产品分一、二、三级,其中只有一级是正品,若生产中出现正品的概率是0.97,二级品的概率为0.02,那么出现二级品或三级品的概率是( )(A)0.01 (B)0.02 (C)0.03 (D)0.046.(2011·南京高一检测)从写上0,1,2,…,9的十张卡片中,有放回地每次抽一张,连抽两次,则两张卡片数字各不相同的概率是( )(A) 910 (B) 1100 (C) 190(D)1 7.如图,在一个边长为a 、b(a>b>0)的矩形内画一梯形,梯形上、下底分别为13a 与12a ,高为b.向该矩形内随机投一点,则所投的点落在梯形内部的概率为( )(A) 13 (B) 12 (C) 25 (D) 5128.如图所示,墙上挂有一边长为a 的正方形木板,它的四个角的空白部分都是以正方形的顶点为圆心,半径为a 2的圆弧,某人向此板投镖,假设每次都能击中木板,且击中木板上每个点的可能性都一样,则他击中阴影部分的概率是( )(A) 14π- (B) 4π (C) 18π- (D)与a 的取值有关 9.抛掷两枚骰子,则两枚骰子点数之和不大于4或不小于11的概率为( )(A) 14 (B) 13 (C)34 (D) 2510.下列说法不正确的是( )(A)在一场乒乓球比赛前,裁判员利用抽签器来决定由谁先发球,这是公平的(B)如果某种彩票中奖的概率为110,那么买10张彩票不一定能中奖 (C)不可能事件的概率不一定为0(D)必然事件的概率一定为111.甲、乙两人玩数字游戏,先由甲任想一个数字记为a ,再由乙猜甲刚才想的数字,把乙想的数字记为b ,且a,b ∈{1,2,3,4,5,6},若|a-b|≤1,则称“甲、乙心有灵犀”.现任意找两个人玩这个游戏,得出他们“心有灵犀”的概率为( )(A) 29 (B) 49 (C) 59 (D) 7912.从4名男生和2名女生中任选3人参加演讲比赛.则所选3人中至少有1名女生的概率为( )(A) 15 (B) 25 (C) 35 (D) 45二、填空题(本大题共4小题,每小题5分,共20分,请把正确答案填在题中横线上)13.在一个池塘里随机地捕捞到m 条鱼,作标记后放回,又随机地捕捞到n 条鱼,其中作过标记的有a 条,估计池塘中有______条鱼.14.(2011·秦皇岛高一检测)今有四张卡片上分别写有“好”、“好”、“学”、“习”四个字,现将其随机排成一行,则恰好排成“好好学习”的概率是_____.15.如图,四边形ABCD 为矩形,BC=1,以A 为圆心,1为半径作四分之一个圆弧DE ,在圆弧DE 上任取一点P ,则直线AP 与线段BC 有公共点的概率是 ______.16.有下列问题:(1)甲、乙两射手同时射击一目标,甲的命中率为0.65,乙的命中率为0.60,则目标被命中的概率等于0.65+0.60=1.25(2)一射手命中靶的内圈的概率是0.25,命中靶的其余部分的概率是0.50,则目标被命中的概率等于0.75(3)两人各掷一枚硬币,“同时出现正面”的概率可以算得为212.由于“不出现正面”是上述事件的对立事件,所以它的概率等于1-212=34 以上说法中正确的命题为_______.(请把正确的命题的序号都填上)三、解答题(本大题共6小题,共70分,解答时应写出必要的文字说明、证明过程或演算步骤)17.(10分)张华的朋友小明自远方来,小明乘火车、轮船、汽车、飞机来的概率分别为0.3、0.2、0.1、0.4.试求:(1)小明乘火车或乘飞机来的概率;(2)小明不乘轮船来的概率;(3)如果小明来的概率为0.5,请问小明有可能是乘何种交通工具来的.18.(12分)(2011·海淀高一检测)在甲、乙两个盒子中分别装有号为1、2、3、4的四个小球,现从甲、乙两个盒子中各取出1个小球,每个小球被取出的可能性相等.(1)求取出两个小球标号恰好相同的概率;(2)求取出的两个小球的标号至少有一个大于2的概率.19.(12分)某人欲从某车站乘车出差,已知该站发往各站的客车均每小时一班,求此人等车时间不多于10分钟的概率.20.(12分)(2011·天津高考)编号分别为A1,A2,…,A16的16名篮球运动员在某次训练比赛中的得分记录如下:(1)将得分在对应区间内的人数填入相应的空格:(2)从得分在区间[20,30)内的运动员中随机抽取2人,①用运动员编号列出所有可能的抽取结果;②求这2人得分之和大于50的概率.21.(12分)某高校在2010年的自主招生考试成绩中随机抽取了100名学生的笔试成绩,按成绩分组,得到的频率分布如下图所示.(1)请求出频率分布表中①、②处应填的数据;(2)为了能选拔最优秀的学生,高校决定在笔试成绩高的第3、4、5组中用分层抽样法抽取6名学生进入第二轮面试,问第3、4、5组每组各抽取多少名学生进入第二轮面试?(3)在(2)的前提下,学校决定在6名学生中随机抽取2名学生接受A考官进行的面试,求第4组有一名学生被考官A面试的概率.22.(12分)设有关于x的一元二次方程x2+2ax+b2=0.(1)若a是从0,1,2,3四个数中任取的一个数,b是从0,1,2三个数中任取的一个数,求上述方程有实根的概率.(2)若a是从区间[0,3]任取的一个数,b是从区间[0,2]任取的一个数,求上述方程有实根的概率.答案解析1.【解析】选B.根据在一定条件下有可能发生,也有可能不发生的事件为随机事件知②④为随机事件,故选B.2.【解析】选C.从1,2,3,4这四个数中一次随机取两个数,共有(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)6个基本事件,其中一个数是另一个的两倍的有(1,2),(2,4)2个基本事件,所以其中一个数是另一个的两倍的概率是2163=.3.【解析】选B.随机取出2个小球得到的结果数有5×5=25种.取出的小球标注的数字之和为3或6或9的结果为(1,2),(1,5),(2,1),(2,4),(3,3),(4,2),(5,1),(5,4)共8种,故所求答案为B.4.【解析】选C.由互斥、对立事件的定义知A、B不是互斥事件,C是互斥事件但不对立,D是对立事件.5.【解析】选C.由对立事件的概率公式可得1-0.97=0.03.6.【解析】选A.总的基本事件为{(x,y)|0≤x≤9,0≤y≤9,x∈N,y∈N},基本事件总数n=100,记A=“两张卡片数字各不相同”,A=“两张卡片数字相同”={(0,0),(1,1),…,(9,9)},事件A含基本事件数m=10,所以P(A)=1-P(A)=m1091n10010=-=.7.【解析】选D.记事件“所投的点落在梯形内部”为A,由几何概型得P(A)=1 2(13a+12a)×b/ab=512.8.【解析】选A.阴影部分的面积=边长为a的正方形的面积-半径为a2的圆的面积=a2-π(a2)2=44-πa2.所以击中阴影部分的概率为:P=224a44a4-π-π==阴影部分的面积正方形的面积.9.独具【解题提示】解答本题可先用图表列出所有可能的结果,然后根据图表统计两枚骰子点数之和不大于4或不小于11的基本事件个数,进而求出相应的概率.【解析】选A.抛掷两枚骰子,所有可能的结果有36种,记A=“点数之和不大于4”,则A包含的基本事件为(1,1)(1,2)(1,3)(2,1)(2,2)(3,1)共6个,B=“点数之和不小于11”,B包含的基本事件为(5,6)(6,5)(6,6)共3个,且A、B互斥,因此所求概率为P(A+B)=63136364+=.10.【解析】选C.对于A:因为抽签上抛后,红圈朝上与绿圈朝上的概率均是0.5,因此任何一名运动员猜中的概率都是0.5,也就是每个运动员取得先发球权的概率都是0.5.故A正确;买10张彩票相当于做10次试验,因为每次试验的结果都是随机的,即每张彩票可能中奖也可能不中奖,因此,10张彩票中可能没有一张中奖,也可能有一张、两张乃至多张中奖,故B正确,由不可能事件、必然事件的概念知C错误,D正确.11.独具【解题提示】可用列表法得出所有的基本事件.【解析】选B.∵a,b∈{1,2,3,4,5,6}∴a-b所有可能的结果如图所示,共有36种不同结果,而|a-b|≤1的结果共有16种.因此P(|a-b|≤1)=164=.36912.【解析】选A.从编号为男1、2、3、4号和女5、6号的6个人中选3人的方法数有(1,2,3)、(1,2,4)、(1,2,5)、(1,2,6)、(1,3,4)、(1,3,5)、(1,3,6)、(1,4,5)、(1,4,6)、(1,5,6)、(2,3,4)、(2,3,5)、(2,3,6)、(2,4,5)、(2,4,6)、(2,5,6)、(3,4,5)、(3,4,6)、(3,5,6)(4,5,6)、共20种方法.方法1 所选3人都是男生的情况有(1,2,3)、(1,2,4)、(2,3,4)、(1,3,4)共4种方法,故所选3人都是男生的概率为41=205“所选3人中至少有1名女生”的对立事件为“所选3人都是男生”,故所求事件的概率为14155-=.方法2 “所选3人中至少有1名女生”是两互斥事件“所选3人中恰有1名女生”、“所选3人中恰有2名女生”的和事件,所选3人中恰好有1名女生的情况有(1,2,5)、(1,2,6)、(2,3,5)、(2,3,6)、(3,4,5)、(3,4,6)、(1,3,5)、(1,3,6)、(1,4,5)、(1,4,6)、(2,4,5)、(2,4,6)共12种方法,所选3人中恰好有2名女生的情况有(1,5,6)、(2,5,6)、(3,5,6)、(4,5,6)4种情况,故所选3人中至少有1名女生的概率为124420205+=. 独具【方法技巧】解决概率综合题的方法:解答概率综合题时,一般“大化小”,即将问题划分为若干个彼此互斥事件,然后运用加法公式求解或采用其对立事件求解,运用“正难则反”的思想进行转化,以简化解答过程.13.【解析】设池塘中有x 条鱼,则应有n a x m =, 解得x=mn a. 答案: mn a 14.独具【解题提示】可用树图法不重不漏写出所有的基本事件.【解析】总的基本事件为:好好学习,好好习学,好学好习,好习好学,好学习好,好习学好,学好好习,学好习好,学习好好,习好好学,习好学好,习学好好,共12个,故恰好排成“好好学习”的概率为112. 答案: 112 15.【解析】如图,连结AC 交圆弧DE 为点M,当点P 在圆弧ME 上时,直线AP 与线段BC 有公共点.则直线AP 与线段BC 有公共点的概率为BAM 301BAD 903∠︒==∠︒的度数的度数.答案:1316.【解析】(1)不正确.因为甲命中目标与乙命中目标两事件不互斥.(2)正确.因为命中靶的内圈和命中靶的其余部分是互斥事件.(3)不正确.因为“不出现正面”与“同时出现正面”不是对立事件,故其概率和不为1.答案:(2)17.【解析】(1)记“小明乘火车来”为事件A1,“小明乘轮船来”为事件A2,“小明乘汽车来”为事件A3,“小明乘飞机来”为事件A4,这四个事件不可能同时发生,故它们彼此互斥.故P(A1∪A4)=P(A1)+P(A4)=0.3+0.4=0.7.即小明乘火车或乘飞机来的概率为0.7.(2)P(A)=1-P(A2)=1-0.2=0.8.2即小明不乘轮船来的概率为0.8.(3)由于0.3+0.2=0.5,0.1+0.4=0.5,1-(0.3+0.2)=0.5,1-(0.1+0.4)=0.5,故小明有可能是乘火车或轮船来;也有可能是乘汽车或飞机来.18.【解析】利用树状图可以列出从甲、乙两个盒子中各取出1个球的所有可能结果:可以看出,试验的所有可能结果有16种,且每种结果都是等可能的.(1)所取两个小球上的标号为相同整数的结果有1-1,2-2,3-3,4-4共4种.根据古典概型公式,所求概率P=41164=. 即取出的两个小球上的标号为相同整数的概率为14.(2)记事件“取出两个小球上的标号至少有一个大于2”为A ,所取出的两个小球上的标号都不大于2的结果有1-1,1-2,2-1,2-2,共4种. 故P (A )=41164=,P(A)=1-P(A )=34. 即取出的两个球上的标号至少有一个大于2的概率为34.19.独具【解题提示】到站等车的时刻X 是随机的,可以是0到60之间的任何一刻,并且是等可能的.【解析】设A={等待的时间不多于10分钟},我们所关心的事件A 恰好是到站等车的时刻位于[50,60]这一时间段内,因此由几何概型的概率公式,得P(A)=60501606-=,即此人等车时间不多于10分钟的概率为16. 独具【方法技巧】转化与化归思想的应用在许多实际问题中,其几何概型特征并不明显,要能将它们转化为几何概型,并正确应用几何概型的概率计算公式解决问题.如与时间有关的等候问题、约会问题,与数域有关的点集问题等等. 20.【解析】(1)4,6,6(2)①得分在区间[20,30)内的运动员编号为A 3,A 4,A 5,A 10,A 11,A 13.从中随机抽取2人,所有可能的抽取结果有:{A 3,A 4},{A 3,A 5},{A 3,A 10},{A 3,A 11},{A 3,A 13},{A 4,A 5}, {A 4,A 10},{A 4,A 11},{A 4,A 13},{A 5,A 10},{A 5,A 11},{A 5,A 13},{A 10,A 11},{A 10,A 13},{A 11,A 13},共15种.②“从得分在区间[20,30)内的运动员中随机抽取2人,这2人得分之和大于50”(记为事件B)的所有可能结果有:{A 4,A 5},{A 4,A 10},{A 4,A 11},{A 5,A 10},{A 10,A 11},共5种.所以P(B)=51.15321.【解析】(1)由题可知,第2组的频数为0.35×100=35(人),=0.300.第3组的频率为30100(2)因为第3、4、5组共有60名学生,所以利用分层抽样在60名学生中抽取6名学生,每组抽取的学生人数分别为:第3组:30×6=3(人);60×6=2(人);第4组:2060×6=1(人);第5组:1060所以第3、4、5组分别抽取3人、2人、1人.(3)设第3组的3位同学为A1,A2,A3,第4组的2位同学为B1,B2,第5组的1位同学为C1,则从6名同学中抽取2名同学有15种如下可能:(A1,A2),(A1,A3),(A1,B1),(A1,B2),(A1,C1),(A2,A3),(A2,B1),(A2,B2),(A2,C1),(A3,B1),(A3,B2),(A3,C1),(B1,B2),(B1,C1),(B2,C1).其中第4组的2位同学B1,B2有一位同学入选的有:(A1,B1),(A1,B2),(A2,B1),(A2,B2),(A3,B1),(A3,B2),(B1,C1),(B2,C1)8种可能,所以第4组.有一位同学入选的概率为81522.独具【解题提示】(1)为古典概型问题,(2)为几何概型问题.【解析】设事件A为“方程x2+2ax+b2=0有实根”.当a>0,b>0时,方程x2+2ax+b2=0有实根的充要条件为a≥b.(1)基本事件共12个:(0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,0),(2,1),(2,2),(3,0),(3,1),(3,2).其中第一个数表示a的取值,第二个数表示b的取值.事件A中包含9个基本事件,事件A发生的概率为P(A)=93124=.(2)试验的全部结果所构成的区域为{(a,b)|0≤a≤3,0≤b≤2}.构成事件A的区域为{(a,b)|0≤a≤3,0≤b≤2,a≥b}.所以所求的概率为2132222323⨯-⨯=⨯.独具【误区警示】本题中不少考生在区域约束条件不等式列式过程中出现错误.导致平面区域的面积求解出现偏差,在几何概型问题的分析中,试验的构成区域决定着概率运算的正确性,因而列式时要注意范围列式中边界值的确定依据.。
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单元质量评估(三)(90分钟100分)一、选择题(每小题3分,共24分)1.函数y=中,自变量x的取值范围是( )A.x≥-1B.x>2C.x>-1且x≠2D.x≥-1且x≠2【解析】选D.根据题意得:解得x≥-1且x≠2,故选D.2.均匀地向一个瓶子注水,最后把瓶子注满.在注水过程中,水面高度h随时间t 的变化规律如图所示,则这个瓶子的形状是下列的( )【解析】选B.∵水面高度开始增加得慢,后来增加得快,∴容器下面粗,上面细.故选B.3.如图,在平面直角坐标系中,点A的坐标为(0,3),△OAB沿x轴向右平移后得到△O′A′B′,点A的对应点A′在直线y=x上,则点B与其对应点B′间的距离为( )A. B.3 C.4 D.5【解析】选C.由平移可知,O′A′=OA=3,∴点A′的纵坐标为3,把y=3代入y=x,得x=4,∴OO′=4,即平移距离为4,∴点B与其对应点B′间的距离为4.4.(2014·乐山模拟)关于抛物线y=-(x+1)2-1,下列结论错误的是( )A.顶点坐标为(-1,-1)B.当x=-1时,函数值y的最大值为-1C.当x<-1时,函数值y随x值的增大而减小D.将抛物线向上平移1个单位,再向右平移1个单位,所得抛物线的解析式为y=-x2【解析】选C.A.抛物线y=-(x+1)2-1的顶点坐标为(-1,-1),所以A选项的结论正确;B.对于抛物线y=-(x+1)2-1,由于a=-1<0,所以x=-1时,函数值y的最大值为-1,所以B选项的结论正确;C.对于抛物线y=-(x+1)2-1,由于a=-1<0,当x<-1时,函数值y随x值的增大而增大,所以C选项的结论错误;D.将抛物线向上平移1个单位,再向右平移1个单位,所得抛物线的解析式为y=-(x+1-1)2-1+1=-x2,所以D选项的结论正确.故选C.5.某种子公司以一定价格销售“黄金1号”玉米种子,如果一次购买10kg以上(不含10kg)的种子,超过10kg的那部分种子的价格将打折,并依此得到付款金额y(单位:元)与一次购买种子数量x(单位:kg)之间的函数关系如图所示.下列说法:①一次购买种子数量不超过10kg时,销售价格为5元/kg;②一次购买30kg种子时,付款金额为100元;③一次购买10kg以上种子时,超过10kg的那部分种子的价格打五折;④一次购买40kg种子比分两次购买且每次购买20kg种子少花25元钱.其中正确的个数是( )A.1个B.2个C.3个D.4个【解析】选D.一次购买种子数量不超过10kg时,销售价格50÷10=5元/kg,①正确;一次购买10kg以上种子时,超过10kg的那部分种子的销售价格为(150-50)÷(50-10)=2.5元/kg,所以购买30kg的金额为5×10+2.5×20=100元,②正确;超过10kg的那部分种子的销售价格为2.5元/kg,所以超过10kg的那部分种子的价格打五折,③正确;一次购买40kg种子的金额为5×10+2.5×30=125元,分两次购买且每次购买20kg种子的金额为2×(5×10+2.5×10)=150元,150-125=25元,④正确.6.(2014·泸州模拟)如图,已知双曲线y=(k<0)经过直角三角形OAB斜边OA的中点D,且与直角边AB相交于点C.若点A的坐标为(-6,4),则△AOC的面积为( )A. 12B.9C.6D.4【解析】选B.∵OA的中点是D,点A的坐标为(-6,4),∴D(-3,2),∵双曲线y=经过点D,∴k=-3×2=-6,∴S△BOC=|k|=3.又∵S△AOB=×6×4=12,∴S△AOC=S△AOB-S△BOC=12-3=9.7.二次函数y=ax2+bx+c的图象如图所示,则下列关系式错误的是( )A.a>0B.c>0C.b2-4ac>0D.a+b+c>0【解析】选D.由抛物线开口向上,知a>0;当x=0时,y=c,抛物线与y轴的交点在正半轴上,故c>0;抛物线与x轴有两个交点,即ax2+bx+c=0有两个不相等的实数根,故b2-4ac>0;由图象知,当x=1时,y=a+b+c<0.8.某棵果树前x年的总产量y与x之间的关系如图所示,从目前记录的结果看,前x年的年平均产量最高,则x的值为( )A.3B.5C.7D.9【解析】选C.若果树前x年的总产量y与x在图中对应P(x,y)点,则前x年的年平均产量即为直线OP的斜率,由图易得当x=7时,直线OP的斜率最大,即前7年的年平均产量最高,x=7.故选C.二、填空题(每小题4分,共24分)9.如图,把“QQ”笑脸放在直角坐标系中,已知左侧眼A的坐标是(-2,3),嘴唇C 点的坐标为(-1,1),则将此“QQ”笑脸向右平移3个单位后,右侧眼B的坐标是.【解析】因为点A,B关于点C所在的纵向的网格线对称,根据对称性可知右眼的坐标为(0,3),再将右眼向右平移3个单位后的点B的坐标为(3,3).答案:(3,3)10.(2014·椒江区一模)若点P(m,n)在一次函数y=2x-1的图象上,则m-n= .【解析】∵点P(m,n)在一次函数y=2x-1的图象上,∴n=2m-1,∴2m-n=1,∴m-n=.答案:11.一次函数y=(m+2)x+1,若y随x的增大而增大,则m的取值范围是. 【解析】由题意得m+2>0,解这个不等式,得m>-2.答案:m>-212.二次函数y=-x2+bx+c的图象如图所示,则一次函数y=bx+c的图象不经过第象限.【解析】∵二次函数y=-x2+bx+c的对称轴在y轴的右侧,∴a,b异号,又∵a=-1<0,∴b>0;∵二次函数的图象与y轴的交点在y轴正半轴,∴c>0;∴一次函数y=bx+c过第一、二、三象限,不过第四象限.答案:四13.如果一个正比例函数的图象与反比例函数y=的图象交于A(x1,y1),B(x2,y2)两点,那么(x2-x1)(y2-y1)的值为.【解析】∵A(x1,y1),B(x2,y2)两点是正比例函数的图象与反比例函数y=的图象的交点,∴x1y1=x2y2=6,x1=-x2,y1=-y2,∴(x2-x1)(y2-y1)= x2y2-x2y1-x1y2+ x1y1=x2y2+x2y2+x2y2+x1y1=4×6=24.答案:2414.如图,正方形ABCD的边长为4,E,F分别是BC,CD上的两个动点,且AE⊥EF.则AF的最小值是.【解析】根据题意,若AF最小,则CF最大.设BE=x,CF=y,∵∠B=∠AEF=90°, 则∠BAE+∠AEB=∠FEC+∠AEB=90°,∴∠BAE=∠FEC,∴△ABE∽△ECF,得到=,有=,化简y==-(x-2)2+1,所以当x=2时,y有最大值为1,此时DF最小为3,由勾股定理得AF==5.答案:5三、解答题(共52分)15.(8分)已知二次函数图象的顶点坐标为(1,-1),且过原点(0,0),求该函数解析式.【解析】∵二次函数图象的顶点坐标为(1,-1),∴可设为y=a(x-1)2-1,当x=0时,y=0,∴0=a(0-1)2-1,a=1,所以函数解析式为y=(x-1)2-1.16.(8分)如图,等边三角形ABC放置在平面直角坐标系中,已知A(0,0),B(6,0),反比例函数的图象经过点C.(1)求点C的坐标及反比例函数的解析式.(2)将等边△ABC向上平移n个单位,使点B恰好落在双曲线上,求n的值.【解析】(1)过点C作CH⊥x轴,垂足为H.∵△ABC是等边三角形,∴AH=AB=3,∴CH==3,∴C(3,3).设反比例函数的解析式为y=,∴k=xy=9,即y=.(2)∵将等边△ABC向上平移n个单位,使点B恰好落在双曲线上,∴设此时的点B坐标为(6,n),∴6n=9,解得n=.17.(10分)周末,小明骑自行车从家里出发到野外郊游,从家出发1 h后到达南亚所(景点),游玩一段时间后按原速前往湖光岩.小明离家1 h50min后,妈妈驾车沿相同的路线前往湖光岩,如图是他们离家的路程y(km)与小明离家时间x(h)的函数图象.(1)求小明骑车的速度和在南亚所游玩的时间.(2)若妈妈在出发后25min时,刚好在湖光岩门口追上小明,求妈妈驾车的速度及CD所在直线的函数解析式.【解析】(1)小明骑车的速度为:20km/h,在南亚所游玩的时间为1h.(2)设妈妈驾车的速度为xkm/h,则×x=20+×20,解得x=60,点C的坐标为.设CD所在直线的解析式为:y=kx+b,所以解得k=60,b=-110,所以CD所在直线的解析式为:y=60x-110.18.(12分)如图,在平面直角坐标系中,直线y=x-2与y轴相交于点A,与反比例函数y=在第一象限内的图象相交于点B(m,2).(1)求反比例函数的解析式.(2)将直线y=x-2向上平移后与该反比例函数的图象在第一象限内交于点C,且△ABC的面积为18,求平移后的直线的函数解析式.【解析】(1)∵点B(m,2)在直线y=x-2的图象上,∴m-2=2,∴m=4,即B(4,2).∵点B在反比例函数y=的图象上,∴k=4×2=8,∴y=.(2)如图,过点C作CE⊥y轴,垂足为E,过点B作x轴、y轴的垂线,垂足分别为G,F.设C(x,y).∵直线y=x-2与y轴相交于点A,∴A(0,-2).S△ABC=S△AOD+S矩形OFBG+S梯形BCEF-S△ACE-S△BDG=OA〃OD+8+〃EF-CE〃AE-DG〃BG=×2×2+8+-x-×2×2=4-2x+2y,且y=,S△ABC=18, ∴4-2x+2×=18,即x2+7x-8=0.∴x1=1,x2=-8(不合题意,舍去),∴C(1,8),∵平移后的直线与直线y=x-2平行,∴设平移后直线的解析式为:y=x+b,把点C(1,8)代入得:b=7,∴y=x+7.19.(14分)某大学生利用暑假40天社会实践参与了一家网店的经营,了解到一种成本为20元/件的新型商品在第x天销售的相关信息如表所示.圆学子梦想铸金字品牌- 11 - ,q=30+x; ,q=20+(1)请计算第几天该商品的销售单价为35元/件.(2)求该网店第x天获得的利润y关于x的函数解析式.(3)这40天中该网店第几天获得的利润最大?最大利润是多少?【解析】(1)①对于q=30+x,当q=35时,30+x=35,解得x=10,在1≤x≤20范围内;②对于q=20+,当q=35时,20+=35,解得x=35,在21≤x≤40范围内.综上所述,当第10天或第35天时,该商品的销售单价为35元/件.(2)①当1≤x≤20时,y=(50-x)=-x2+15x+500;②当21≤x≤40时,y=(50-x)=-525.(3)①y=-x2+15x+500=-(x-15)2+612.5,由于-<0,抛物线开口向下,且1≤x≤20,所以当x=15时,y最大=612.5(元);②y=-525,越大(即x越小)y的值越大,由于21≤x≤40,所以当x=21天时,y最大=1250-525=725(元),综上所述,这40天中该网店第21天获得的利润最大,最大利润是725元.关闭Word文档返回原板块。