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篇1:人教版数学必修2第三章高考复习试题附有答案详解
1.双曲线的方程为=1(a>0,b>0),焦距为4,一个顶点是抛物线y2=4x的焦点,则双曲线的离心率e=( )
A.2 B. C. D.
2.已知F1,F2是椭圆的两个焦点,满足=0的点M总在椭圆内部,则椭圆离心率的取值范围是( )
A. (0,1) B. C. D.
3.设F为抛物线y2=4x的焦点,A,B,C为该抛物线上三点.若=0,则||+||+||=( )
A.9 B.6 C.4 D.3
4.已知抛物线y2=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=-2
5.已知A,B,P是双曲线=1上不同的三点,且A,B连线经过坐标原点,若直线PA,PB的斜率乘积kPA·kPB=,则该双曲线的离心率为( )
A.1 B.2 C. -1 D.-2
6.已知抛物线y2=4x的焦点为F,准线为l,经过F且斜率为的直线与抛物线在x轴上方的部分相交于点A,AKl,垂足为K,则AKF的面积是( )
A.4 B.3 C.4 D.8
7.过抛物线y2=2px(p>0)的焦点F作倾斜角为45°的直线交抛物线于A,B两点,若线段AB的长为8,则p= .
8.(湖南,文14)平面上一机器人在行进中始终保持与点F(1,0)的距离和到直线x=-1的距离相等.若机器人接触不到过点P(-1,0)且斜率为k的直线,则k的取值范围是 .
9.已知双曲线的中心在原点,且一个焦点为F(,0),直线y=x-1与其相交于M, N两点,线段MN中点的横坐标为-,求此双曲线的方程.
10.(2014安徽,文21)设F1,F2分别是椭圆E:=1(a>b>0)的左、右焦点,过点F1的直线交椭圆E于A,B两点,|AF1|=3|F1B|.
(1)若|AB|=4,ABF2的周长为16,求|AF2|;
(2)若cosAF2B=,求椭圆E的离心率.
11.已知点F是双曲线=1(a>0,b>0)的左焦点,点E是该双曲线的右顶点,过点F且垂直于x轴的直线与双曲线交于A,B两点,若ABE是直角三角形,则该双曲线的离心率是( )
A. B.2 C.1+ D.2+
12.(2014湖北,文8)设a,b是关于t的方程t2cosθ+tsinθ=0的两个不等实根,则过A(a,a2),B(b,b2)两点的直线与双曲线=1的公共点的个数为( )
A.0 B.1 C.2 D.3
13.已知椭圆C:=1(a>b>0)的离心率为,双曲线x2-y2=1的渐近线与椭圆C有四个交点,以这四个交点为顶点的四边形的面积为16,则椭圆C的方程为( )
A.=3 B.=1C.=-1D=-2
C.=1 D.=1
14.(2014江西,文20)如图,已知抛物线C:x2=4y,过点M(0,2)任作一直线与C相交于A,B两点,过点B作y轴的平行线与直线AO相交于点D(O为坐标原点).
(1)证明:动点D在定直线上;
(2)作C的任意一条切线l(不含x轴),与直线y=2相交于点N1,与(1)中的定直线相交于点N2,证明:|MN2|2-|MN1|2为定值,并求此定值.
15.已知点A(0,-2),椭圆E:=1(a>b>0)的离心率为,F是椭圆E的右焦点,直线AF的斜率为,O为坐标原点.
(1)求E的方程;
(2)设过点A的动直线l与E相交于P,Q两点,当OPQ的面积最大时,求l的方程.
参考答案
1.A 解析:抛物线y2=4x的焦点为(1,0),则在双曲线中a=1.又2c=4,c=2,e==2.
2.C 解析:设F1,F2为焦点,由题意知,点M的轨迹是以F1F2为直径的圆,
则c1或k<-1.
9.解:设双曲线的方程为=1(a>0,b>0),
则a2+b2=2=7.
由
消去y,得=1.
整理,得(b2-a2)x2+2a2x-a2-a2b2=0.(*)
由直线y=x-1与双曲线有两个交点知a≠b,
设M(x1,y1),N(x2,y2),
则x1和x2为方程(*)的根,
于是x1+x2=.
由已知得=-,
则=-,即5a2=2b2.
由得
故所求双曲线方程为=1.
10.解:(1)由|AF1|=3|F1B|,|AB|=4,
得|AF1|=3,|F1B|=1.
因为ABF2的周长为16,
所以由椭圆定义可得4a=16,
|AF1|+|AF2|=2a=8.
故|AF2|=2a-|AF1|=8-3=5.
(2)设|F1B|=k,则k>0,
且|AF1|=3k,|AB|=4k.
由椭圆定义可得|AF2|=2a-3k,|BF2|=2a-k.
在ABF2中,由余弦定理可得|AB|2=|AF2|2+|BF2|2-2|AF2|·|BF2|cosAF2B,
即(4k)2=(2a-3k)2+(2a-k)2-(2a-3k)·(2a-k),
化简可得(a+k)(a-3k)=0,
而a+k>0,故a=3k.
于是有|AF2|=3k=|AF1|,|BF2|=5k.
因此|BF2|2=|F2A|2+|AB|2,可得F1AF2A,
故AF1F2为等腰直角三角形.
从而c=a,所以椭圆E的离心率e=.
11.B 解析:将x=-c代入双曲线方程得A.
由ABE是直角三角形,得=a+c,
即a2+ac=b2=c2-a2,
整理得c2-ac-2a2=0.
∴e2-e-2=0,
解得e=2(e=-1舍去).
12.A 解析:可解方程t2cosθ+tsinθ=0,
得两根0,-.
不妨设a=0,b=-,
则A(0,0),B,
可求得直线方程y=-x,
因为双曲线渐近线方程为y=±x,
故过A,B的直线即为双曲线的一条渐近线,直线与双曲线无交点,故选A.
13.D 解析:因为椭圆的离心率为,
所以e=,c2=a2,a2=a2-b2.
所以b2=a2,即a2=4b2.
因为双曲线的渐近线为y=±x,代入椭圆得=1
即=1,
所以x2=b2,x=±b,y2=b2,y=±b.
则在第一象限的交点坐标为.
所以四边形的面积为4×b×b=b2=16.解得b2=5,
故椭圆方程为=1.
14.(1)证明:依题意可设AB方程为y=kx+2,代入x2=4y,得x2=4(kx+2),即x2-4kx-8=0.
设A(x1,y1),B(x2,y2),
则有x1x2=-8,
直线AO的方程为y=x;BD的方程为x=x2.
解得交点D的坐标为
注意到x1x2=-8及=4y1,
则有y==-2.
因此D点在定直线y=-2上(x≠0).
(2)解:依题设,切线l的斜率存在且不等于0,设切线l的方程为y=ax+b(a≠0),
代入x2=4y得x2=4(ax+b),
即x2-4ax-4b=0,
由Δ=0得(4a)2+16b=0,化简整理得b=-a2.
故切线l的方程可写为y=ax-a2.
分别令y=2,y=-2得N1,N2的坐标为N1,N2.
则|MN2|2-|MN1|2=+42-=8,
即|MN2|2-|MN1|2为定值8.
15.解:(1)设F(c,0),由条件知,,得c=.
又,所以a=2,b2=a2-c2=1.
故E的方程为+y2=1
(2)当lx轴时不合题意,故设l:y=kx-2,P(x1,y1),Q(x2,y2).
将y=kx-2代入+y2=1,
得(1+4k2)x2-16kx+12=0.
当Δ=16(4k2-3)>0,即k2>时,x1,2=.
从而|PQ|=|x1-x2|
=.
又点O到直线PQ的距离d=,
所以OPQ的面积SOPQ=d·|PQ|=.
设=t,则t>0,
SOPQ=.
因为t+≥4,当且仅当t=2,即k=±时等号成立,且满足Δ>0.
所以,当OPQ的面积最大时,l的方程为y=x-2或y=-x-2.
篇2:人教版数学必修2第三章高考复习试题附有答案详解
一、选择题
.下列函数中,与函数y=定义域相同的函数为( ).
A.y= B.y=
C.y=xex D.y=
解析 函数y=的定义域为{x|x≠0,xR}与函数y=的定义域相同,故选D.
答案 D
.若一系列函数的解析式相同,值域相同,但定义域不同,则称这些函数为“同族函数”,则函数解析式为y=x2+1,值域为{1,3}的同族函数有( ).
A.1个 B.2个 C.3个 D.4个
解析 由x2+1=1,得x=0.由x2+1=3,得x=±,所以函数的定义域可以是{0,},{0,-},{0,,-},故值域为{1,3}的同族函数共有3个.
答案 C
.若函数y=f(x)的定义域为M={x|-2≤x≤2},值域为N={y|0≤y≤2},则函数y=f(x)的图象可能是( ).解析 根据函数的定义,观察得出选项B.
答案 B
.已知函数f(x)=若a,b,c互不相等,且f(a)=f(b)=f(c),则abc的取值范围是( ).
A.(1,10) B.(5,6)
C.(10,12) D.(20,24)
解析 a,b,c互不相等,不妨设ag[f(x)]的x的值是________.
解析 g(1)=3,f[g(1)]=f(3)=1,由表格可以发现g(2)=2,f(2)=3,f(g(2))=3,g(f(2))=1.
答案 1 2.已知函数f(x)=则满足不等式f(1-x2)>f(2x)的x的取值范围是________.
解析 由题意有或解得-11时,函数g(x)是[1,3]上的减函数,此时g(x)min=g(3)=2-3a,g(x)max=g(1)=1-a,所以h(a)=2a-1;
当0≤a≤1时,若x[1,2],则g(x)=1-ax,有g(2)≤g(x)≤g(1);
若x(2,3],则g(x)=(1-a)x-1,有g(2)2x+m,即x2-3x+1>m,对x[-1,1]恒成立.令g(x)=x2-3x+1,则问题可转化为g(x)min>m,又因为g(x)在[-1,1]上递减, 所以g(x)min=g(1)=-1,故m<-1.
篇3:人教版数学必修2第三章高考复习试题附有答案详解
一、选择题
11.(文)(·重庆理,3)已知变量x与y正相关,且由观测数据算得样本平均数=3,=3.5,则由该观测数据算得线性回归方程可能为( )
A.=0.4x+2.3 B.=2x-2.4
C.=-2x+9.5 D.=-0.3x+4.4
[答案] A
[解析] 因为变量x和y正相关,所以回归直线的斜率为正,排除C、D;又将点(3,3.5)代入选项A和B的方程中检验排除B,所以选A.
(理)一个车间为了规定工时定额,需要确定加工零件所花费的时间,为此进行了8次试验,收集数据如下:
零件数x(个) 10 20 30 40 50 60 70 80 加工时间y(min) 62 68 75 81 89 95 102 108 设回归方程为y=bx+a,则点(a,b)在直线x+45y-10=0的( )
A.左上方 B.左下方
C.右上方 D.右下方
[答案] C
[解析] =45,=85,a+45b=85,
a+45b-10>0,故点(a,b)在直线x+45y-10=0的右上方,故选C.
12.(2014·沈阳市质检)某高校进行自主招生,先从报名者中筛选出400人参加笔试,再按笔试成绩择优选出100人参加面试.现随机调查了24名笔试者的成绩,如下表所示:
分数段 [60,65) [65,70) [70,75) [75,80) [80,85) [85,90) 人数 2 3 4 9 5 1 据此估计允许参加面试的分数线大约是( )
A.75 B.80 C.85 D.90
[答案] B
[解析] 由题可知,在24名笔试者中应选出6人参加面试.由表可得面试分数线大约为80.故选B.
13.(·陕西文,5)对一批产品的长度(单位:毫米)进行抽样检测,下图为检测结果的频率分布直方图.根据标准,产品长度在区间[20,25)上为一等品,在区间[15,20)和[25,30)上为二等品,在区间[10,15)和[30,35]上为三等品.用频率估计概率,现从该批产品中随机抽取1件,则其为二等品的概率是( )
A.0.09 B.0.20 C.0.25 D.0.45
[答案] D
[解析] 解法1:用样本估计总体.在区间[15,20)和[25,30)上的概率为0.04×5+[1-(0.02+0.04+0.06+0.03)×5=0.45.
解法2:由图可知,抽得一等品的概率P1=0.06×5=0.3;抽得三等品的概率为P3=(0.02+0.03)×5=0.25.故抽得二等品的概率为1-(0.3+0.25)=0.45.
14.(2014·江西理,6)某人研究中学生的性别与成绩、视力、智商、阅读量这4个变量之间的关系,随机抽查52名中学生,得到统计数据如表1至表4,则与性别有关联的可能性最大的变量是( )
A.成绩 B.视力 C.智商 D.阅读量
[答案] D
[解析] A中,K2==;
B中,K2==;
C中,K2==;
D中,K2==.
因此阅读量与性别相关的可能性最大,所以选D.
15.(文)某养兔场引进了一批新品种,严格按照科学配方进行喂养,四个月后管理员称其体重(单位:kg),将有关数据进行整理后分为五组,并绘制频率分布直方图(如图所示).根据标准,体重超过6kg属于超重,低于5kg的不够分量.已知图中从左到右第一、第三、第四、第五小组的频率分别为0.25、0.20、0.10、0.05,第二小组的频数为400,则该批兔子的总数和体重正常的频率分别为( )
A.1000,0.50 B.800,0.50
C.800,0.60 D.1000,0.60
[答案] D
[解析] 第二组的频率为1-0.25-0.20-0.10-0.05=0.40,所以兔子总数为=1000只,体重正常的频率为0.40+0.20=0.60.故选D.
(理)(2014·山东理,7)为了研究某药品的疗效,选取若干名志愿者进行临床试验.所有志愿者的舒张压数据(单位:kPa)的分组区间为[12,13),[13,14),[14,15),[15,16),[16,17],将其按从左到右的顺序分别编号为第一组,第二组,……,第五组.下图是根据试验数据制成的频率分布直方图.已知第一组与第二组共有20人,第三组中没有疗效的有6人,则第三组中有疗效的人数为( )
A.6 B.8 C.12 D.18
[答案] C
[解析] 第一、二两组的频率为0.24+0.16=0.4
志愿者的总人数为=50(人).
第三组的人数为:50×0.36=18(人)
有疗效的人数为18-6=12(人)
二、填空题
16.(2013·辽宁文,16)为了考察某校各班参加课外书法小组的人数,从全校随机抽取5个班级,把每个班级参加该小组的人数作为样本数据,已知样本平均数为7,样本方差为4,且样本数据互不相同,则样本数据中的最大值为________.
[答案] 10
[解析] 设5个班级中参加的人数分别为x1,x2,x3,x4,x5,则=7,
=4,即5个整数平方和为20,x1,x2,x3,x4,x5这5个数中最大数比7大,但不能超过10,因此最大为10,平方和
20=0+1+1+9+9=(7-7)2+(8-7)2+(6-7)2+(10-7)2+(4-7)2.
因此参加的人数为4,6,7,8,10,故最大值为10,最小值为4.
三、解答题
17.(文)(2014·重庆文,17)20名学生某次数学考试成绩(单位:分)的频率分布直方图如下:
(1)求频率分布直方图中a的值;
(2)分别求出成绩落在[50,60)与[60,70)中的学生人数;
(3)从成绩在[50,70)的学生中任选2人,求此2人的成绩都在[60,70)中的概率.
[分析] 由频率之和为1,求a,然后求出落在[50,60)和[60,70)中的人数,最后用列举法求古典概型的概率.
[解析] (1)组距为10,(2a+3a+6a+7a+2a)×10=200a=1,
a==0.005.
(2)落在[50,60)中的频率为2a×10=20a=0.1,
落在[50,60)中的人数为2.
落在[60,70)中的学生人数为3a×10×20=3×0.005×10×20=3.
(3)设落在[50,60)中的2人成绩为A1,A2,落在[60,70)中的3人为B1,B2,B3.
则从[50,70)中选2人共有10种选法,Ω={(A1,A2),(A1,B1),(A1,B2),(A1,B3),(A2,B1),(A2,B2),(A2,B3),(B1,B2),(B1,B3),(B2,B3)}
其中2人都在[60,70)中的基本事件有3个:(B1,B2),(B1,B3),(B2,B3),故所求概率p=.
(理)(2014·辽宁理,18)一家面包房根据以往某种面包的销售记录,绘制了日销售量的频率分布直方图,如图所示.
将日销售量落入各组的频率视为概率,并假设每天的销售量相互独立.
(1)求在未来连续3天里,有连续2天的日销售量都不低于100个且另1天的日销售量低于50个的概率;
(2)用X表示在未来3天里日销售量不低于100个的天数,求随机变量X的分布列,期望E(X)及方差D(X).
[解析] (1)设A1表示事件“日销售量不低于100个”,A2表示事件“日销售量低于50个”,B表示事件“在未来连续3天是有连续2天日销售量不低于100个且另一天销售量低于50个”,因此
P(A1)=(0.006+0.004+0.002)×50=0.6
P(A2)=0.003×50=0.15,
P(B)=0.6×0.6×0.15×2=0.108.
(2)X可能取的值为0,1,2,3,相应的概率为
P(X=0)=C·(1-0.6)3=0.064,
P(X=1)=C·0.6(1-0.6)2=0.288.
P(X=2)=C·0.62(1-0.6)=0.432.
P(X=3)=C·0.63=0.216.
分布列为
X 0 1 2 3 P 0.064 0.288 0.432 0.216 因为X~B(3,0.6)
所以期望E(X)=3×0.6=1.8,
方差D(X)=3×0.6×(1-0.6)=0.72.
18.(文)为加强中学生实践、创新能力和团队精神的培养,促进教育教学改革,郑州市教育局举办了全市中学生创新知识竞赛.某校举行选拔赛,共有200名学生参加,为了解成绩情况,从中选取50名学生的成绩(得分均为整数,满分为100分)进行统计.请你根据尚未完成的频率分布表,解答下列问题:
分组 频数 频率 一 60.5~70.5 a 0.26 二 70.5~80.5 15 c 三 80.5~90.5 18 0.36 四 90.5~100.5 b d 合计 50 e (1)若用系统抽样的方法抽取50个样本,现将所有学生随机地编号为000,001,002,…,199,试写出第二组第一位学生的编号;
(2)求出a、b、c、d、e的值(直接写出结果),并作出频率分布直方图;
(3)若成绩在85.5~95.5分的学生为二等奖,问参赛学生中获得二等奖的学生约为多少人.
[解析] (1)004
(2)a,b,c,d,e的值分别为13,4,0.30,0.08,1.
频率分布直方图如下:
(3)由样本中成绩在80.5~90.5的频数为18,成绩在90.5~100.5的频数为4,可估计成绩在85.5~95.5的人数为11人,故获得二等奖的学生约为×11=44人.
(理)(·山西省高考联合模拟)为了了解某年级1000名学生的百米成绩情况,随机抽取了若干学生的百米成绩,成绩全部介于13s与18s之间,将成绩按如下方式分成五组:第一组[13,14);第二组[14,15);……;第五组[17,18].按上述分组方法得到的频率分布直方图如图所示,已知图中从左到右的前3个组的频率之比为3?8?19,且第二组的频数为8.
(1)将频率当作概率,求调查中随机抽取了多少个学生的百米成绩;
(2)若从第一、五组中随机取出两个成绩,求这两个成绩的差的绝对值大于1秒的概率.
[解析] (1)设图中从左到右前3个组的频率分别为3x,8x,19x依题意,得3x+8x+19x+0.32×1+0.08×1=1,x=0.02,设调查中随机抽取了n个学生的百米成绩,则8×0.02=,n=50,调查中随机抽取了50个学生的百米成绩.
(2)百米成绩在第一组的学生数为3×0.02×1×50=3,记他们的成绩为a、b、c百米成绩在第五组的学生数有0.08×1×50=4,记他们的成绩为m、n、p、q,则从第一、五组中随机取出两个成绩,基本事件有{a,b}、{a,c}、{a,m}、{a,n}、{a,p}、{a,q}、{b,c}、{b,m}、{b,n}、{b,p}、{b,q}、{c,m}、{c,n}、{c,p}、{c,q}、{m,n}、{m,p}、{m,q}、{n,p}、{n,q}、{p,q},共21个
其中满足“成绩的差的绝对值大于1s”所包含的基本事件有{a,m}、{a,n}、{a,p}、{a,q}、{b,m}、{b,n}、{b,p}、{b,q}、{c,m}、{c,n}、{c,p}、{c,q},共12个,所以P==.
篇4:人教版数学必修2第三章高考复习试题
1.双曲线的方程为=1(a>0,b>0),焦距为4,一个顶点是抛物线y2=4x的焦点,则双曲线的离心率e=( )
A.2 B. C. D.
2.已知F1,F2是椭圆的两个焦点,满足=0的点M总在椭圆内部,则椭圆离心率的取值范围是( )
A. (0,1) B. C. D.
3.设F为抛物线y2=4x的焦点,A,B,C为该抛物线上三点.若=0,则||+||+||=( )
A.9 B.6 C.4 D.3
4.已知抛物线y2=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=-2
5.已知A,B,P是双曲线=1上不同的三点,且A,B连线经过坐标原点,若直线PA,PB的斜率乘积kPA·kPB=,则该双曲线的离心率为( )
A.1 B.2 C. -1 D.-2
6.已知抛物线y2=4x的焦点为F,准线为l,经过F且斜率为的直线与抛物线在x轴上方的部分相交于点A,AKl,垂足为K,则AKF的面积是( )
A.4 B.3 C.4 D.8
7.过抛物线y2=2px(p>0)的焦点F作倾斜角为45°的直线交抛物线于A,B两点,若线段AB的长为8,则p= .
8.(2014湖南,文14)平面上一机器人在行进中始终保持与点F(1,0)的距离和到直线x=-1的距离相等.若机器人接触不到过点P(-1,0)且斜率为k的直线,则k的取值范围是 .
9.已知双曲线的中心在原点,且一个焦点为F(,0),直线y=x-1与其相交于M, N两点,线段MN中点的横坐标为-,求此双曲线的方程.
10.(2014安徽,文21)设F1,F2分别是椭圆E:=1(a>b>0)的左、右焦点,过点F1的直线交椭圆E于A,B两点,|AF1|=3|F1B|.
(1)若|AB|=4,ABF2的周长为16,求|AF2|;
(2)若cosAF2B=,求椭圆E的离心率.
11.已知点F是双曲线=1(a>0,b>0)的左焦点,点E是该双曲线的右顶点,过点F且垂直于x轴的直线与双曲线交于A,B两点,若ABE是直角三角形,则该双曲线的离心率是( )
A. B.2 C.1+ D.2+
12.(2014湖北,文8)设a,b是关于t的方程t2cosθ+tsinθ=0的两个不等实根,则过A(a,a2),B(b,b2)两点的直线与双曲线=1的公共点的个数为( )
A.0 B.1 C.2 D.3
13.已知椭圆C:=1(a>b>0)的离心率为,双曲线x2-y2=1的渐近线与椭圆C有四个交点,以这四个交点为顶点的四边形的面积为16,则椭圆C的方程为( )
A.=3 B.=1C.=-1D=-2
C.=1 D.=1
14.(2014江西,文20)如图,已知抛物线C:x2=4y,过点M(0,2)任作一直线与C相交于A,B两点,过点B作y轴的平行线与直线AO相交于点D(O为坐标原点).
(1)证明:动点D在定直线上;
(2)作C的任意一条切线l(不含x轴),与直线y=2相交于点N1,与(1)中的定直线相交于点N2,证明:|MN2|2-|MN1|2为定值,并求此定值.
15.已知点A(0,-2),椭圆E:=1(a>b>0)的离心率为,F是椭圆E的右焦点,直线AF的斜率为,O为坐标原点.
(1)求E的方程;
(2)设过点A的动直线l与E相交于P,Q两点,当OPQ的面积最大时,求l的方程.
篇5:人教版数学必修2第三章高考复习试题
1.已知抛物线x2=ay的焦点恰好为双曲线y2-x2=2的上焦点,则a=( )
A.1 B.4 C.8 D.16
2.(辽宁,文8)已知点A(-2,3)在抛物线C:y2=2px的准线上,记C的焦点为F,则直线AF的斜率为( )
A.- B.-1 C.- D.-
3.抛物线y=-4x2上的一点M到焦点的距离为1,则点M的纵坐标是( )
A.- B.- C. D.
4.抛物线C的顶点为原点,焦点在x轴上,直线x-y=0与抛物线C交于A,B两点,若P(1,1)为线段AB的中点,则抛物线C的方程为( )
A.y=2x2 B.y2=2x C.x2=2y D.y2=-2x
5.已知抛物线C:y2=8x的焦点为F,准线与x轴的交点为K,点A在C上,且|AK|=|AF|,则AFK的面积为( )
A.4 B.8 C.16 D.32
6.以抛物线x2=16y的焦点为圆心,且与抛物线的准线相切的圆的方程为 .
7.已知抛物线x2=2py(p为常数,p≠0)上不同两点A,B的横坐标恰好是关于x的方程x2+6x+4q=0(q为常数)的两个根,则直线AB的方程为 .
8.已知F是抛物线C:y2=4x的焦点,A,B是C上的两个点,线段AB的中点为M(2,2),求ABF的面积.
9.已知一条曲线C在y轴右边,C上每一点到点F(1,0)的距离减去它到y轴距离的差都是1.
(1)求曲线C的方程;
(2)是否存在正数m,对于过点M(m,0),且与曲线C有两个交点A,B的任一直线,都有<0?若存在,求出m的取值范围;若不存在,请说明理由.
10.已知抛物线y2=2px,以过焦点的弦为直径的圆与抛物线准线的位置关系是( )
A.相离 B.相交 C.相切 D.不确定
11.设x1,x2R,常数a>0,定义运算“*”,x1*x2=(x1+x2)2-(x1-x2)2,若x≥0,则动点P(x,)的轨迹是( )
A.圆 B.椭圆的一部分
C.双曲线的一部分 D.抛物线的一部分
12.已知抛物线C:y2=8x的焦点为F,准线为l,P是l上一点,Q是直线PF与C的一个交点.若=4,则|QF|=( )
A. B.3 C. D.2
13.过抛物线x2=2py(p>0)的焦点作斜率为1的直线与该抛物线交于A,B两点,A,B在x轴上的正射影分别为D,C.若梯形ABCD的面积为12,则p= .
14.(2014大纲全国,文22)已知抛物线C:y2=2px(p>0)的焦点为F,直线y=4与y轴的交点为P,与C的交点为Q,且|QF|=|PQ|.
(1)求C的方程;
(2)过F的直线l与C相交于A,B两点,若AB的垂直平分线l'与C相交于M,N两点,且A,M,B,N四点在同一圆上,求l的方程.
15.已知抛物线C:y2=2px(p>0)的焦点为F,A为C上异于原点的任意一点,过点A的直线l交C于另一点B,交x轴的正半轴于点D,且有|FA|=|FD|.当点A的横坐标为3时,ADF为正三角形.
(1)求C的方程;
(2)若直线l1l,且l1和C有且只有一个公共点E,
证明直线AE过定点,并求出定点坐标;
ABE的面积是否存在最小值?若存在,请求出最小值;若不存在,请说明理由.参考答案及解析:1.C 解析:根据抛物线方程可得其焦点坐标为,双曲线的上焦点为(0,2),依题意则有=2,解得a=8.
2.C 解析:由已知,得准线方程为x=-2,
F的坐标为(2,0).
又A(-2,3),直线AF的斜率为k==-.故选C.
3.B 解析:抛物线方程可化为x2=-,其准线方程为y=.
设M(x0,y0),则由抛物线的定义,可知-y0=1y0=-.
4.B 解析:设A(x1,y1),B(x2,y2),抛物线方程为y2=2px,
则两式相减可得2p=×(y1+y2)=kAB×2=2,
即可得p=1,故抛物线C的方程为y2=2x.
5.B 解析:抛物线C:y2=8x的焦点为F(2,0),准线为x=-2,K(-2,0).
设A(x0,y0),过点A向准线作垂线AB垂足为B,则B(-2,y0).
|AK|=|AF|,
又|AF|=|AB|=x0-(-2)=x0+2,
由|BK|2=|AK|2-|AB|2,
得=(x0+2)2,即8x0=(x0+2)2,
解得A(2,±4).
故AFK的面积为|KF|·|y0|
=×4×4=8.
6.x2+(y-4)2=64 解析:抛物线的焦点为F(0,4),准线为y=-4,
则圆心为(0,4),半径r=8.
故圆的方程为x2+(y-4)2=64.
7.3x+py+2q=0 解析:由题意知,直线AB与x轴不垂直.
设直线AB的方程为y=kx+m,与抛物线方程联立,得x2-2pkx-2pm=0,
此方程与x2+6x+4q=0同解,
则解得
故直线AB的方程为y=-x-,
即3x+py+2q=0.
8.解:由M(2,2)知,线段AB所在的直线的斜率存在,
设过点M的直线方程为y-2=k(x-2)(k≠0).
由消去y,
得k2x2+(-4k2+4k-4)x+4(k-1)2=0.
设A(x1,y1),B(x2,y2),
则x1+x2=,
x1x2=.
由题意知=2,
则=4,解得k=1,
于是直线方程为y=x,x1x2=0.
因为|AB|=|x1-x2|=4,
又焦点F(1,0)到直线y=x的距离d=,所以ABF的面积是×4=2.
9.解:(1)设P(x,y)是曲线C上任意一点,
则点P(x,y)满足-x=1(x>0),
化简得y2=4x(x>0).
(2)设过点M(m,0)(m>0)的直线l与曲线C的交点为A(x1,y1),B(x2,y2).
设l的方程为x=ty+m.
由得y2-4ty-4m=0,
Δ=16(t2+m)>0,
于是
因为=(x1-1,y1),
=(x2-1,y2),
所以=(x1-1)(x2-1)+y1y2=x1x2-(x1+x2)+y1y2+1.
又<0,
所以x1x2-(x1+x2)+y1y2+1<0,③
因为x=,所以不等式可变形为
+y1y2-+1<0,
即+y1y2-[(y1+y2)2-2y1y2]+1<0.
将代入整理得m2-6m+1<4t2.
因为对任意实数t,4t2的最小值为0
所以不等式对于一切t成立等价于m2-6m+1<0,
即3-20),则FD的中点为.
因为|FA|=|FD|,
由抛物线的定义知3+,
解得t=3+p或t=-3(舍去).
由=3,解得p=2.
所以抛物线C的方程为y2=4x.
(2)由(1)知F(1,0).
设A(x0,y0)(x0y0≠0),D(xD,0)(xD>0),
因为|FA|=|FD|,
则|xD-1|=x0+1.
由xD>0得xD=x0+2,
故D(x0+2,0).
故直线AB的斜率kAB=-.
因为直线l1和直线AB平行,设直线l1的方程为y=-x+b,
代入抛物线方程得y2+y-=0,
由题意Δ==0,
得b=-.
设E(xE,yE),
则yE=-,xE=.
当≠4时,kAE==-,
可得直线AE的方程为y-y0=(x-x0),
由=4x0,整理可得y=(x-1),
直线AE恒过点F(1,0).
当=4时,直线AE的方程为x=1,过点F(1,0).
所以直线AE过定点F(1,0).
由知直线AE过焦点F(1,0),
所以|AE|=|AF|+|FE|=(x0+1)+=x0++2.
设直线AE的方程为x=my+1,
因为点A(x0,y0)在直线AE上,
故m=.
设B(x1,y1),
直线AB的方程为y-y0=-(x-x0),由于y0≠0,
可得x=-y+2+x0,
代入抛物线方程得y2+y-8-4x0=0.
所以y0+y1=-,
可求得y1=-y0-,
x1=+x0+4.
所以点B到直线AE的距离为
d=
==4.
则ABE的面积S=×4≥16,
当且仅当=x0,即x0=1时等号成立.
所以ABE的面积的最小值为16.
篇6:人教版数学必修2第三章高考复习试题
1.甲、乙两名篮球运动员每场比赛的得分情况用茎叶图表示如右:
则下列说法中正确的个数为( )
甲得分的中位数为26,乙得分的中位数为36;
甲、乙比较,甲的稳定性更好;
乙有的叶集中在茎3上;
甲有的叶集中在茎1,2,3上.
A.1 B.2 C.3 D.4
2.一组数据的平均数是4.8,方差是3.6,若将这组数据中的每一个数据都加上60,得到一组新数据,则所得新数据的平均数和方差分别是( )
A.55.2,3.6 B.55.2,56.4 C.64.8,63.6 D.64.8,3.6
3.某中学高三(2)班甲、乙两名学生自高中以来每次考试成绩的茎叶图如图,下列说法正确的是( )
A.乙学生比甲学生发挥稳定,且平均成绩也比甲学生高
B.乙学生比甲学生发挥稳定,但平均成绩不如甲学生高
C.甲学生比乙学生发挥稳定,且平均成绩比乙学生高
D.甲学生比乙学生发挥稳定,但平均成绩不如乙学生高
4.为了研究某药品的疗效,选取若干名志愿者进行临床试验.所有志愿者的舒张压数据(单位:kPa)的分组区间为[12,13),[13,14),[14,15),[15,16),[16,17],将其按从左到右的顺序分别编号为第一组,第二组,…,第五组.下图是根据试验数据制成的频率分布直方图.已知第一组与第二组共有20人,第三组中没有疗效的有6人,则第三组中有疗效的人数为( )
A.6 B.8 C.12 D.18
5.若某校高一年级8个班参加合唱比赛的得分如茎叶图所示,则这组数据的中位数和平均数分别是( )
A.91.5和91.5 B.91.5和92
C.91和91.5 D.92和92
6.某工厂对一批产品进行了抽样检测.下图是根据抽样检测后的产品净重(单位:克)数据绘制的频率分布直方图,其中产品净重的范围是[96,106],样本数据分组为[96,98),[98,100),[100,102),[102,104),[104,106],已知样本中产品净重小于100克的个数是36,则样本中净重大于或等于98克并且小于104克的产品的个数是( )
A.90 B.75 C.60 D.45
7.某赛季,甲、乙两名篮球运动员都参加了11场比赛,他们每场比赛得分的情况用右图所示的茎叶图表示,若甲运动员的中位数为a,乙运动员的众数为b,则a-b= .
8.为了调查某厂工人生产某种产品的能力,随机抽查了20位工人某天生产该产品的数量,产品数量的分组区间为[45,55),[55,65),[65,75),[75,85),[85,95],由此得到频率分布直方图如图,则由此估计该厂工人一天生产该产品数量在[55,70)的人数约占该厂工人总数的百分率是 .
9.(2014广东,文17)某车间20名工人年龄数据如下表:
年龄(岁) 工人数(人) 19 1 28 3 29 3 30 5 31 4 32 3 40 1 合计 20
(1)求这20名工人年龄的众数与极差;
(2)以十位数为茎,个位数为叶,作出这20名工人年龄的茎叶图;
(3)求这20名工人年龄的方差.
10.在发生某公共卫生事件期间,有专业机构认为该事件在一段时间没有发生大规模群体感染的标志为“连续10天,每天新增疑似病例不超过7人”.根据过去10天甲、乙、丙、丁四地新增疑似病例数据,一定符合该标志的是( )
A.甲地:总体均值为3,中位数为4
B.乙地:总体均值为1,总体方差大于0
C.丙地:中位数为2,众数为3
D.丁地:总体均值为2,总体方差为3
11.样本(x1,x2,…,xn)的平均数为,样本(y1,y2,…,ym)的平均数为),若样本(x1,x2,…, xn,y1,y2,…,ym)的平均数=α+(1-α),其中0<α<,则n,m的大小关系为( )
A.nm C.n=m D.不能确定
12.(2014课标全国,文18)从某企业生产的某种产品中抽取100件,测量这些产品的一项质量指标值,由测量结果得如下频数分布表:
质量指标
值分组 [75,85) [85,95) [95,105) [105,115) [115,125) 频数 6 26 38 22 8
(1)在答题卡上作出这些数据的频率分布直方图;
(2)估计这种产品质量指标值的平均数及方差(同一组中的数据用该组区间的中点值作代表);
(3)根据以上抽样调查数据,能否认为该企业生产的这种产品符合“质量指标值不低于95的产品至少要占全部产品80%”的规定?
参考答案
1.C 解析:由茎叶图可知乙的集中趋势更好,故错误,正确.
2. D 解析:每一个数据都加上60时,平均数也应加上60,而方差不变.
3.A 解析:从茎叶图可知乙同学的成绩在80~100分分数段的有9次,而甲同学的成绩在80~100分分数段的只有7次;再从题图上还可以看出,乙同学的成绩集中在90~100分分数段的最多,而甲同学的成绩集中在80~90分分数段的最多.故乙同学比甲同学发挥较稳定且平均成绩也比甲同学高.
4.C 解析:设样本容量为n,
由题意,得(0.24+0.16)×1×n=20,解得n=50.
所以第三组频数为0.36×1×50=18.
因为第三组中没有疗效的有6人,
所以第三组中有疗效的人数为18-6=12.
5.A 解析:按照从小到大的顺序排列为87,89,90,91,92,93,94,96.
有8个数据,中位数是中间两个数的平均数:=91.5,
平均数:
=91.5.
6.A 解析:样本中产品净重小于100克的频率为(0.050+0.100)×2=0.3,
又频数为36,样本容量为=120.
样本中净重大于或等于98克并且小于104克的产品的频率为(0.100+0.150+0.125)×2=0.75,
样本中净重大于或等于98克并且小于104克的产品的个数为120×0.75=90.
7.8 解析:由茎叶图可知,a=19,b=11,
a-b=8.
8.52.5% 解析:结合直方图可以看出:生产数量在[55,65)的人数频率为0.04×10=0.4,生产数量在[65,75)的人数频率为0.025×10=0.25,而生产数量在[65,70)的人数频率约为0.25×=0.125,所以生产数量在[55,70)的人数频率约为0.4+0.125=0.525,即52.5%.
9.解:(1)由图可知,众数为30.极差为:40-19=21.
(2)
1 9 2 888999 3 000001111222 4 0
(3)根据表格可得:
∴s2=[(19-30)2+3(28-30)2+3(29-30)2+5(30-30)2+4(31-30)2+3(32-30)2+(40-30)2]
=12.6.
10.D 解析:根据信息可知,连续10天内,每天的新增疑似病例不能有超过7的数,选项A中,中位数为4,可能存在大于7的数;同理,在选项C中也有可能;选项B中的总体方差大于0,叙述不明确,如果数目太大,也有可能存在大于7的数;选项D中,根据方差公式,如果有大于7的数存在,那么方差不会为3,故答案选D.
11.A 解析:由题意知样本(x1,…,xn,y1,…,ym)的平均数为,
又=α+(1-α),即α=,1-α=.
因为0<α<,所以0<,
即2n
篇7:英语必修一UNIT1-2知识点详解及练习教学总结(人教版英语高考复习)
Unit 1 知识点
一、知识点
1. be good to 对……友好
be good for 对……有益;be bad to…/be bad for…
I will be good to other people.我会善良的对待其他人.
It would be good for you to spend a holiday in the sun. 在有阳光的地方度假会给你带来很多好处。
The Olympics will be good for business. 奥运会的召开将有利于商业的发展。
be good at 擅长make good 有成就;成功as good as 实际上;几乎等于
a good deal 许多,大量 彻底的;完全的;痛快的to have a good drink 喝个痛快
2. add up 加起来
add up to 合计,总计
add… to 把……加到…… add to 增加
Add up your score and see how many points you get? 把你的分数加起来,看看得多少?
Some people can add up quite easily in their heads, but not all.
Good friends do not add up what they do for each other; instead they offer help when it is needed.
The figures add up to 270. 这些数字加起来是270。
You shouldn’t add fuel to the flame 你不应该火上加油
Fireworks added to the attraction of the festival night. 焰火使节日的夜晚更加生色。
The bad weather added to the shipwrecked sailors’ difficulties.恶劣的天气增加了失事船只的船员们的困难。
Your friend can not go until he finishes cleaning his bike.
not…until/till 意思是“直到…才”,表示主句谓语所表示的动作直到until状语所表示的时间才发生,主句的谓语动词表示的是动作的开始,动词既可以是延续性的,也可以是非延续性的。
They did not come back until eleven. 他们会在十一点后回来。
I did not notice it until yesterday.我一直到昨天才注重到它。
4. You had to pay to get it repaired
get sth done 使……完成/让某人做某事
5. You will ignore the bell and go somewhere quiet to calm your friend down.
I said hello to her, but she ignored me completely!
calm …down使平息, 使平静
calm down平息/平静下来
The crying child soon calmed down.哭闹的小孩不多一会就安静下来。
It was a long time before he managed to calm himself down. 过了很久他才努力使自己冷静下来。
We tried to calm him down, but he kept crying.
我们试图让他平静下来,但他仍不停地哭着。
6. Tell your friend that you are concerned about him. be concerned about关心,挂念
He was very concerned about his children's education. 他很关心他儿子的教育。
Please don’t be concerned about me.请别为我操心。
Why is she so concerned about his attitude to her work? 她为什么那么关注他对她的工作的看法?
7. Your friend has gone on holiday and asked you to take care of his dog.
go on holiday 度假
be on holiday 正在休假
What fun it will be when we all go on holiday together.我们大家一起去度假那可太有意思了.
take care of 爱护,照料
take care 注意,当心
You are not (physically) strong, so you may as well take care of your health. 你的体格不壮,因此最好注意健康。
8. While walking the dog, you were careless and it got loose. 在遛狗的时候,你一粗心松开了手中的狗链。
当while, when, before, after 等引导的时间状语从句中的主语与主句的主语一致时,可将从句中的主语和be动词省去。
walk sb home/ to a place: 为保证安全而陪某人去某地 It’s late ---- let me walk you home.
9. take one’s end-of-term exam 参加期末考试
10. 3) Your friend, who doesn’t work hard, asks you to help him cheat in the end-of-term exam.(非限制性定语从句)
cheat in the exam 考试作弊
11. look at someone else’s paper 看别人的试卷
12. make a list of reasons 列举一些原因
13. Do you want a friend whom you could tell everything to, like your deepest feelings and thoughts? 你想有一位无话不谈、能推心置腹的朋友吗?
14. go through遭遇;经历;熬过;用光(钱);获准,通过
It can go through the test of the time. 它能经受时间的考验.
She knew that she had got to go through all the difficulties with her family.
He would go through fire and water for his country. 他愿为国家赴汤蹈火。
15. hide away 躲藏;隐藏
16. I don’t want to set down a series of facts in a diary as most people do,…我不愿像大多数人一样在日记中记流水账,……
Why don't you set your ideas down on paper?
We have had a series of stormy days when we were on the island.
The police asked him to set down what he had seen in a report. 警察让他在报告中写下他所看见的事情。
16. I wonder if it’s because I haven’t been able to be outdoors for so long that I’ve grown so crazy about everything to do with nature. 我不知道这是不是因为我长久无法出门的缘故,我变得对一切与大自然有关的事物都无比狂热。
17. I can well remember that there was a time when a deep blue sky, the song of birds, moonlight and flowers could never have kept me spellbound. 我记得非常清楚,曾有一段时间,湛蓝的天空、鸟儿的歌唱、月光和鲜花,从未使我心醉神迷过。
18. I stayed awake on purpose until half past eleven one evening in order to have a good look at the moon for once by myself.
有一天晚上,我熬到11点半故意不睡觉,为的是独自好好看看月亮一次。
19. But as the moon gave far too much light, I didn't dare open a window.但是因为月光太亮了,我不敢打开窗户。
She speaks French far better than I, so I don't dare talk with her in French.
20. I happened to be upstairs at dusk when the window was open. 黄昏时我碰巧在楼上,那时窗户是开着的。
sth happen to sb 某人发生某事
What happened to him?
sb happen to do sth 某人碰巧做某事 正巧 it so happened that 。。。
It happened that he was seen by his father. = He happened to be seen by his father.
他碰巧被他父亲看见了。
As I was about to go out and search for him, he happened to come in. 正当我打算出去找他时,他恰巧进来。
The street lights go on at dusk. 街上的路段在傍晚时分亮起来。
21. It was the first time in a year and a half that I had seen the night face to face. 这是我一年半以来第一次目睹夜晚。
It is the first (second…etc) that… (从句谓语动词用现在完成时)
It was the first (second…etc) that… (从句谓语动词用过去完成时)
the first time 可作从属连词用,引导时间状语从句。The first time I saw her, my heart stopped.
It was the first time that I talked with a foreigner face to face.
I think we need a face-to-face talk so as to clear the misunderstanding.
I have often heard of her. Actually, I've never met her face to face.
22. in one’s power 处于……的控制之中
I have got him in my power. I can ask him to do anything I want. 我控制了他,我可以让他为我做任何事。
23. It’s no pleasure looking through these any longer because nature is one thing that really must be experienced.观看这些已不再是乐趣,因为大自然是你必须亲身体验的。
It’s no good/ use doing sth. 做某事时没用的。
24. She found it difficult to settle and calm down in the hiding place.
25. suffer from 患…病; 受…苦痛;遭受
Most of the important cities of the world suffer from traffic jam. 世界上大多数大城市都交通堵塞为患。
26. It was such fun to watch it run loose in the park.
27. I’ve got tired of looking nature through dirty curtains and dusty windows.
28. I need to pack up my things in the suitcase very quickly.
29. Mum asked her if she was very hot with so many clothes on.
with+名词/代词(宾格)+分词/形容词/介词短语/不定式/副词在句中常作伴随状语。动词形式的选择取决于宾语同动词之间的逻辑关系。
The murderer was brought in, with his hands tied behind his back.
30. have some trouble with sb or sth. 在……上遇到了麻烦
I have some trouble with my studies.
31. get along … with sb/sth. 与某人相处怎样/某事进展如何?
If you have some trouble (in) getting along with your friends, you can write to the editor and ask for advice. 如果你在和朋友的相处上有问题,你可以写信给编辑向他征求建议。
32. This has made me angry.
…he made her diary her best friend…
make 后接复合宾语,宾语补足语须用不带to 的不定式、形容词、过去分词、名词等。常见的有以下几种形式:
make sb. do sth.让(使)某人做某事。He was made to repeat it.(注意在被动句中,不定式前要加to)
make sb. /sth. +adj.使某人/物…We should do our best to make our country stronger and more beautiful.
make sb./ oneself +v-ed 让某人/自己被…When you speak, you should make yourself understood.
(4) make sb.+n. 使某人成为…
make it n. /adj.+(for sb.) to do sth. We made him leader of our team. (注意表示职位的名词前不加冠词)
He made it easy for us to understand the text.
33. I’m not good at communicating with people.
34. Although I tried to talk to my classmates, I still found it hard to make friends with them.
35. I do want to change this situation, but I don’t know how.
36. Mr. Jones lives alone and often feels lonely. 琼斯先生单独一人生活,常常感到孤独。
37. I would be grateful if you could give me some advice. 如果您给我提些建议,我会非常感谢的。 (I would be grateful if… 委婉客气提出请求)
38. join in discussions and show interest in other people’s ideas
39. It’s a good habit for you to keep a diary. 记日记对你来说是个好习惯。
40. Why not have a try?
41.True friends are like wine; the older, the better.
42. People are told that their actions should be as gentle as the wind that blows from the sea.
43. A friend in need is a friend indeed.
新课标英语必修 一---Unit 2 知识点
一、知识点
1. go to the pictures去看电影(美);go to the movies 去看电影(英)
2. …list the countries that use English as an official language 列举把英语用作官方语言的国家
3. the road to …通向……之路
4. at the end of在……末端,在……尽头,by the end最后(=finally)
5. because of 因为…… (注意和because 的区别)
Many beautiful fish are fast disappearing because of the severe pollution.因为污染严重,许多美丽的鱼类正在面临绝种。
An argument was inevitable because they disliked each other so much.
争论是不可避免的,因为他们彼此非常厌恶。
6. native English speakers 以英语作为母语的人
7. even if (= even thoug)即使,用来引导一个让步状语从句,后面既可用陈述语气,也可用虚拟语气,但是even if/even though,引导的从句中不用将来时。如:Even though/if it rains tomorrow, we will leave for Beijing.
8. come up 走上前来,走近,发生,出现 come up with 追上,赶上,提出
9. Actually all languages change and develop when cultures meet and communicate with each other.事实上,当不同文化相互交流渗透时,所有的语言都会有所发展、有所变化。
10. be different from… 与……不同
be different in … 在……不同
Most of my projects will be wildly different in performance from one night to the next.
我多数作品每天晚上的演奏风格都各不相同。
As we know, Britain English is a little different from American English.中所周知,英国英语和美国英语有点不同。
11. be based on 以……为基础The relationship between our two countries is based upon mutual respect. 两个国家的关系以相互尊重为基础。This book is based on a true story that happened in the 1930s. 这本书以发生在20世纪三十年代的真实故事为基础。 The reporter asked the writer who he based his character on. 记者问作家他作品的人物是以谁为原型的。
12. at present 目前,眼下be present at 在席;出席present sth to sb / present sb with sth把……推荐,呈现……for the present眼前;暂时present oneself 出席;到场
13. make (great/ good/better/full)use of
We have a lot of work to do, so we have to make good use of time.我们有很多工作要做,所以要好好利用时间。
14. The latter gave a separate identity to Amerian English speaking. 后者体现了美国英语的不同特色。
15. For example, India has a very large number of fluent English speakers because Britain ruled India from 1765 to 1947. 比如说, 印度拥有众多讲英语流利的人,这是应为英国于1765到1947年统治过印度。(A small number of friends came to help him when he was in trouble)
16. such as 例如
for example In this paragraph there are many nouns, such as boy, girl, and book. 这一段里面有很多名词,例如男孩、女孩和书本。Many great men have risen from poverty---Lincoln, for example. 许多伟人从贫困中崛起,例如林肯。You can take your research work for example.
你可以拿你的研究工作做个例子。
17. Today, the number of people learning English in China is increasing rapidly. 目前在中学习英语的人数正在迅速增长。
18. the largest number of 大多数的
China has the largest number of people.中国有着世界上最多的人。
19. It is not easy for a Chinese person to speak English as fluently as a native speaker. 中国人说英语很难像以英语为母语的人说英语那么流利。
20. One reason is that English has a large vocabulary. 一个原因是英语有很大的词汇量。
21. different English speaking countries 不同的说英语的国家
22. sing sb a song = sing a song for sb
23. turn off
turn on
turn up
turn down
24. hold on 坚持住,握住不放;(打电话时)不挂断,等-会hold on to vt. 拉住(抓牢)
25. believe it or not 信不信由你
26. those who reported the news were expected to speak excellent English 人们期望新闻播音员所说的英语是最好的英语
27. … you will hear the difference in the way(that/ in which) people speak. 你会听出人们在说话时的差异。
28. play a role/ part (in) 在…中担任角色;在…中起作用;扮演一个角色;参与
play an important role/ part 在…中起重要作用
Deng Xiaoping played an important part in developing the economy in China.邓小平在中国经济的发展过程中起着重要作用。
29. from one place to another 从一个地方到另一个地方
30. the same …as… 与……一样
31. … they still recognize and understand each other’s dialects. ……他们仍然能够辨别、理解彼此的方言。
32. No problem.没问题
33. a nice fall day = a lovely autumn day
34. at the top of…在…顶上,在最高位,
at the bottom of 在……底部
35. keep fit
保持健康
You need exercise and keep fit.你需要运动和保持体形。
36. build up逐渐积聚,集结;逐步建立;增进,增强
bring up 教养,养育;提出
37. When you learn English, try to have fun with the language. 当学英语的时候,努力找出语言的乐趣。
38. Visitors are requested not to take photos in the museum. 博物馆要求参观的旅客不得在馆内拍照。
39. by candle light 借助于烛光
40. be satisfied with…对……感到满意,满足于
Never be satisfied with just a little success. 不要有一点成绩就满足。
41. She suggested using CDs to listen to English songs and learn English expressions, watching the news and interviews on CCTV 9, and trying to listen to native speakers.她建议用CD来听英语歌曲和学习英语短语,看新闻和中央电视台9套访谈,努力听以英语为母语的人说话。
It is suggested that ...有人提议... I suggest that ...我觉得[认为]
I suggested you do what he says. 我建议你按照他说的去做。
I suggest you not go tomorrow. 我想你明天还是不要去了。
His pale face suggested that he was in bad health. 他苍白的脸色暗示了他身体不好。
42. at sea在海上 当海员 迷惑, 茫然by sea乘船,经海路
by the sea
在海边, 在海岸边 in the sea在海里
on the sea 在海上
beyond/over the sea在海外
She tried to understand the instructions, but she was completely at sea.
她费尽力气想看懂那些说明文字,却全然不知所云。
43. according to … 按照…… He lives according to her means他按他的方式生活
二、练习
一)单项选择
1 Mr. Huang will ________ in the movement.
A. play a leading part B. take parts C. play leading partD. take a part
2. We discussed where to go for a whole morning, but we decided to stay at home_____.
A. at the end B. by the end C. in the end D. on end
3. _____ of the students who took part in the military training is 450.
A. A number B. A lot C. Lots D. The number
4. Sometimes ________ English is quite different from _______ English in many ways.
A. speaking, writing B. spoken, written
C. speaking, written D. spoken, writing
5. Can you tell me if you have found the key ________ your car.
A. for B. to C. about D. by
6. When we visited Zhou Zhuang again ten years later, we found it changed so much that we could hardly ________ it.
A. rememberB. think about C. believe D. recognize
7. It is so nice to hear from her, _______, we last met more than 30 years ago.
A. what’s more B. that’s to say C. in other words D. believe it or not
8. They lived a hard life and were often made _______ for over ten hours a day.
A. work B. to work C. to working D. worked
9. Do you have any difficulty ________?
A. on listening B. to listening C. for listening D. with listening
10. Please tell me the way you thought of _______ the garden.
A. take care of B. to take care of C. taking care of D. to take care
11. Can you explain how it _______ that you missed the morning classes?
A. came across B. came toC. came up D. came about
12. China Daily is ____ a newspaper, for it can also serve as a useful textbook for English study.
A. more than B. more or less C. less than D. more and more
13. The leader of the factory told us that very little _______ was made of the waste material in the past.
A. costB. value C. use D. matter
14. The reason _____ being late for the meeting was ____ his little son fell ill this morning.
A. for, that B. why, thatC. for, because D. why, because
15. You made the same mistake for _____ second time, dropping ____ “n” in the word “government”.
A. a, the B.a , a C. the, anD. a, an
16. The president, together with his bodyguards, _____ to the nuclear station _____ there was an accident 20 minutes ago.
A. have come, which B. came, in which
C. has come, where D. came, in where
17.“Not all of the dinosaurs were dangerous”. This sentence means ____ .
A. none of the dinosaurs were dangerous.
B. all of the dinosaurs were not dangerous.
C. few of the dinosaurs were dangerous.
D. no dinosaurs were dangerous
18. -He asked Tom, “Have you finished your homework?”
-He asked Tom ______ .
A. if had he finished his homework.
B. whether he had finished his homework.
C. if he had finished homework.
D. if you had finished your homework.
19. He realized she was crying ________ what he had sad.
A. because B. because of C. as D. since
20. ------You haven’t done it well.------ But I tried my best and did it _______ the way _______ I think is the best.
二)根据首字母提示完成句子。
1.Mr. Wu doesn't like the W ______-way of life, so he came back to our hometown last year.
2.She has a very large v____________ and she can read English novels now.
3.His n___________ language is not Chinese but he can speak it fluently.
4.Everyone went to see the film, me i_______________.
5.What the British call “petrol” the Americans call “g____________”.
6.The UN is an i__________organization that tries to solve problems between countries
7.In this part of the city, you can see ancient and m_______ buildings next to each other.
8.It is the duty of a g_______________ to provide education for the children of its countries.
9.I love working abroad and meeting people from different c__________.
10.We are concerned about the p__________ situation in the Middle East.
11.A_________, I’ve known Barbara for years since we were babies.
12.Jet Li has played lots of leading r_______ in Kong Fu films.
13.Though I hadn’t seen Lily for ten years, I r_______ her voice immediately I picked up the phone.
14. Judging from his a_______, he must be from North East of China next to each other.
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