2013年新课标1卷
2013年高考试题及解析:文科数学(新课标Ⅰ卷)

绝密★启封并使用完毕前2013年普通高等学校招生全国统一考试文科数学本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分。
第Ⅰ卷1至2页,第Ⅱ卷3至4页。
全卷满分150分。
考试时间120分钟。
注意事项:1. 本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分。
第Ⅰ卷1至3页,第Ⅱ卷3至5页。
2. 答题前,考生务必将自己的姓名、准考证号填写在本试题相应的位置。
3. 全部答案在答题卡上完成,答在本试题上无效。
4. 考试结束,将本试题和答题卡一并交回。
第Ⅰ卷一、 选择题共12小题。
每小题5分,共60分。
在每个小题给出的四个选项中,只有一项是符合题目要求的一项。
(1)已知集合A={1,2,3,4},B={x |x =n 2,n ∈A },则A ∩B= ( ) (A ){0} (B ){-1,,0} (C ){0,1} (D ){-1,,0,1} 【答案】A 【解析】【难度】容易【点评】本题考查集合之间的运算关系,即包含关系.在高一数学强化提高班上学期课程讲座1,第一章《集合》中有详细讲解,其中第02节中有完全相同类型题目的计算.在高考精品班数学(文)强化提高班中有对集合相关知识的总结讲解. (2)1+2i(1-i)2= ( ) (A )-1-12i(B )-1+12i(C )1+12i(D )1-12i【答案】B 【解析】【难度】容易【点评】本题考查复数的计算。
在高二数学(文)强化提高班下学期,第四章《复数》中有详细讲解,其中第02节中有完全相同类型题目的计算。
在高考精品班数学(文)强化提高班中有对复数相关知识的总结讲解。
(3)从1,2,3,4中任取2个不同的数,则取出的2个数之差的绝对值为2的概率是( )(A )12(B )13(C )14(D )16【答案】B【难度】容易【点评】本题考查几何概率的计算方法。
在高二数学(文)强化提高班,第三章《概率》有详细讲解,在高考精品班数学(文)强化提高班中有对概率相关知识的总结讲解。
2013年高考理科数学新课标1卷解析版

2013 年高考理科数学新课标1 卷解析版一、选择题(题型注释)1.已知集合 A={x|x2-2x >0},B={x| - 5 <x < 5},则 () A 、A ∩B= B 、 A B=R C、B AD 、A B【答案】 B ; 【解析】依题意Ax x 0或x 2 ,由数轴可知,选 B.【考点定位】 本题考查集合的基本运算,考查学生数形结合的能力 . 2.若复数 z 满足(3 -4i)z =|4 + 3i | ,则 z 的虚部为 ( )A 、- 4 (B )-【答案】 D ;45( C )4 (D )45【 解 析 】设z a bi , 故 ( 3 i 4 )a( b i ) 3a 3b i 4a i 4b 4, 所3i 以3b 4a 0 3a 4b 5,解得4 b.5【考点定位】 本题考查复数的基本运算,考查学生的基本运算能力.3.为了解某地区的中小学生视力情况,拟从该地区的中小学生中抽取部分学生进行调 查,事先已了解到该地区小学、初中、高中三个学段学生的视力情况有较大差异,而男 女生视力情况差异不大,在下面的抽样方法中,最合理的抽样方法是( ) A 、简单随机抽样 B 、按性别分层抽样C 、按学段分层抽样D 、系统抽样【答案】 C ;【解析】不同的学段在视力状况上有所差异,所以应该按照年段分层抽样 .【考点定位】 本题考查随机抽样,考查学生对概念的理解 .4.已知双曲线C:2 x 2a- 2 y2b=1(a >0, b >0) 的离心率为5 2,则 C 的渐近线方程为 ()A 、y=± 1 4x(B )y=± 1 3x(C )y=± 1 2x( D )y=±x【答案】 C ; 【 解 析 】e2 2c b 1aa5 2, 故2b2a1 4, 即b a1 2, 故 渐 近 线 方 程 为b 1 yxx.a 2【考点定位】 本题考查双曲线的基本性质,考查学生的化归与转化能力 .5.执行右面的程序框图,如果输入的t ∈[ -1,3] ,则输出的 s 属于 ()12 页1页,总试卷第A、[ -3,4] B 、[ -5,2] C、[ -4,3] D 、[ -2,5] 【答案】A;【解析】若t 1,1 ,则S 3t 3,3 ;若t1,3 ,2S 4t t 3,4 ;综上所述S3,4 .【考点定位】本题考查算法框图,考查学生的逻辑推理能力.6.如图,有一个水平放置的透明无盖的正方体容器,容器高8cm,将一个球放在容器口,再向容器内注水,当球面恰好接触水面时测得水深为6cm,如果不计容器的厚度,则球的体积为( )A、50033 B、866cm33 C、1372cm33 D、2048cm33cm【答案】A;【解析】作出该球轴截面的图像如下图所示,依题意BE 2,AE CE 4 ,设D E x ,故AD 2 x ,因为AD2 AE2 DE 2 ,解得x 3,故该球的半径AD 5 ,所以4 5003V R .3 3试卷第 2 页,总12 页【考点定位】本题考查球体的体积公式,考查学生的空间想象能力.7.设等差数列{a n} 的前n 项和为S n,S m-1=-2,S m=0,S m+1=3,则m=( ) A、3 B 、4 C 、5 D 、6【答案】C;【解析】m(m 1)a 2,a 3 ,故d 1;因为S m 0 ,故ma1 d 0 ,故m m 12m 1a ,因为12 a a 1 5 ,故m ma a 1 2a1 (2m 1)d (m 1) 2m 1 5,即m 5 .m m【考点定位】本题考查等差数列的基本公式,考查学生的化归与转化能力. 8.某几何函数的三视图如图所示,则该几何的体积为( )A、16+8 B 、8+8C、16+16 D 、8+16【答案】A;【解析】上半部分体积为V1 2 2 4 16 ,下半部分体积12V 2 4 8 ,22故总体积V2 16 8 .【考点定位】本题考查三视图以及简单组合体的体积计算,考查学生的空间想象能力. 9.设m 为正整数,(x +y) 2m展开式的二项式系数的最大值为a,(x +y) 2m+1 展开式的二项式系数的最大值为b,若13a=7b,则m=( )A、5 B 、6 C 、7 D、8【答案】B;【解析】ma C ,2mmb C ,因为2m 1m m13C 7C ,解得m=6.2m 2m 1【考点定位】本题考查二项式定理的应用以及组合数的计算,考查学生的基本运算能力.试卷第 3 页,总12 页10.已知椭圆2x 2a +2y 2b=1(a>b>0) 的右焦点为 F(3,0) ,过点 F 的直线交椭圆于A 、B两点。
2013年高考数学新课标全国卷Ⅰ试题及答案

绝密★启封2013年普通高等学校招生全国统一考试文科数学本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分。
第Ⅰ卷1至2页,第Ⅱ卷3至4页。
全卷满分150分。
考试时间120分钟。
注意事项:1. 本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分。
第Ⅰ卷1至3页,第Ⅱ卷3至5页。
2. 答题前,考生务必将自己的姓名、准考证号填写在本试题相应的位置。
3. 全部答案在答题卡上完成,答在本试题上无效。
4. 考试结束,将本试题和答题卡一并交回。
第Ⅰ卷一、选择题共8小题。
每小题5分,共40分。
在每个小题给出的四个选项中,只有一项是符合题目要求的一项。
(1)已知集合A={1,2,3,4},B={x|x=n2,n∈A},则A∩B= ( ) (A){0}(B){-1,,0}(C){0,1} (D){-1,,0,1}(2)错误!未找到引用源。
= ( )(A)-1 - 错误!未找到引用源。
i (B)-1 + 错误!未找到引用源。
i (C)1 + 错误!未找到引用源。
i (D)1 - 错误!未找到引用源。
i(3)从1,2,3,4中任取2个不同的数,则取出的2个数之差的绝对值为2的概率是()(A)错误!未找到引用源。
(B)错误!未找到引用源。
(C)错误!未找到引用源。
(D)错误!未找到引用源。
(4)已知双曲线C:错误!未找到引用源。
= 1(a>0,b>0)的离心率为错误!未找到引用源。
,则C的渐近线方程为()(A)y=±错误!未找到引用源。
x (B)y=±错误!未找到引用源。
x (C)y=±错误!未找到引用源。
x (D)y=±x(5)已知命题p:,则下列命题中为真命题的是:()(A) p∧q (B)¬p∧q (C)p∧¬q (D)¬p∧¬q(6)设首项为1,公比为错误!未找到引用源。
的等比数列{an }的前n项和为Sn,则()(A)Sn =2an-1 (B)Sn=3an-2 (C)Sn=4-3an(D)Sn=3-2an(7)执行右面的程序框图,如果输入的t∈[-1,3],则输出的s属于(A)[-3,4](B)[-5,2](C)[-4,3](D)[-2,5](8)O为坐标原点,F为抛物线C:y²=4x的焦点,P为C上一点,若丨PF丨=4,则△POF的面积为(A)2 (B)2(C)2(D)4(9)函数f(x)=(1-cosx)sinx在[-π,π]的图像大致为(10)已知锐角△ABC的内角A,B,C的对边分别为a,b,c,23cos²A+cos2A=0,a=7,c=6,则b= (A)10 (B)9 (C)8 (D)5(11)某几何函数的三视图如图所示,则该几何的体积为(A)18+8π(B)8+8π(C)16+16π(D)8+16π(12)已知函数f(x)= 若|f(x)|≥ax,则a的取值范围是(A)(-∞] (B)(-∞] (C)[-2,1] (D)[-2,0]第Ⅱ卷本卷包括必考题和选考题两个部分。
2013年高考真题——理综化学(新课标I卷)pdf解析版

【知识点】有机物的组成、结构以及性质。 【答案】 A 【解析】 A 项,根据香叶醇的结构简式可知其分子式为 C10H18O; B 项,香叶醇结构中存在 碳碳双键,故能和溴的四氯化碳溶液发生加成反应而使其褪色;C 项,香叶醇结构中存在碳 碳双键能和酸性高锰酸钾溶液发生氧化反应而使其褪色; D 项,香叶醇结构中存在羟基 ( OH) ,故能和活泼金属以及羧酸发生取代反应。
【解析】 ( 1) LiCoO2 中的氧元素是 2 价,Li 元素是 +1 价,所以 Co 元素是+3 价; ( 2)在 “正极碱浸”过程中,正极发生的反应应为正极材料中的铝箔与 NaOH 溶液之间的反应, (3)根据流程图可知, “酸 所以离子方程式为 2Al 2OH 6H2O 2Al OH 4 3H 2 ;
-3
所以成品的纯度 w=
0.236 g 100% 94.5% 0.25 g
【知识点】基态原点的电子排布式的书写、晶体结构、杂化轨道、化学键等知识。 【答案】
【解析】 (1 ) 硅原子核外有 14 个电子, 其基态原子的核外电子排布式为 1S 2S 2P 3S 3P , 对应能层分别别为 K、L、M,其中能量最高的是最外层 M 层。该能层有 s、p、d 三个能级, s 能级有 1 个轨道, p 能级有 3 个轨道, d 能级有 5 个轨道,所以共有 9 个原子轨道。硅原 子的 M 能层有 4 个电子( 3s23p2) ; ( 3)硅元素在自然界中主要以化合态(二氧化硅和硅酸 盐)形式存在; (3)硅晶体和金刚石晶体类似都属于原子晶体,硅原子之间以共价键结合。 在金刚石晶体的晶胞中,每个面心有一个碳原子(晶体硅类似结构) ,则面心位置贡献的原 子为 6
【知识点】物质的分类、化学方程式和离子方程式的数学、化学实验基本操作、化学计算等 知识。 【答案】
2013年全国新课标1卷

高考训练题三可能用到的相对原子质量:H 1 C 12 N14 O 16 Mg 24 S 32 K 39 Mn 557 •化学无处不在,与化学有关的说法不正确..的是A .侯氏制碱法的工艺过程中应用了物质溶解度的差异B .可用蘸浓盐酸的棉棒检验输送氨气的管道是否漏气C.碘是人体必需微量元素,所以要多吃富含高碘酸的食物D .黑火药由硫磺、硝石、木炭三种物质按一定比例混合制成&香叶醇是合成玫瑰香油的主要原料,其结构简式如下:下列有关香叶醇的叙述正确的是A .香叶醇的分子式为C IO H IB O B .不能使溴的四氯化碳溶液褪色C.不能使酸性高锰酸钾溶液褪色D.能发生加成反应不能发生取代反应9.短周期元素 W、X、Y、Z的原子序数依次增大,其简单离子都能破坏水的电离平衡的是2- x Z + + 、z3+ —3+ 2- + 2-A . W、X B. X、Y C. Y、Z D. X、Z10.银制器皿日久表面会逐渐变黑,这是生成了Ag2S 的缘故。
根据电化学原理可进行如下处理:在铝质容器中加入食盐溶液,再将变黑的的银器浸入该溶液中,一段时间后发现黑色会褪去。
下列说法正确的是A .处理过程中银器一直保持恒重B .银器为正极,Ag 2S被还原生成单质银C.该过程中总反应为 2AI + 3Ag 2S = 6Ag + Al 2S3D .黑色褪去的原因是黑色Ag2S转化为白色AgCl11.已知 K sp(AgCI)=1.56 1为-1°, K sp(AgBr)=7.7 10-13, K sp(Ag 2CrO4)=9.0 W-12。
某溶液中含有Cl-、Br-和 CrO:,浓度均为0.010mol ?L-1,向该溶液中逐滴加入0.010 mol ?L-1 的AgNO3溶液时,三种阴离子产生沉淀的先后顺序为A . Cl-、Br-、CrO:- B. CrO:、Br-、Cl-C. Br-、Cl-、CrO:D. Br-、CrO:、Cl-12.分子式为C5H10O2的有机物在酸性条件下可水解为酸和醇,若不考虑立体异构,这些酸和醇重新组合可形成的酯共有A. 15种 B . 28 种C. 32 种D. 40 种13•下列实验中,所采取的分离方法与对应原理都正确的是第□卷(58分)(一)必考题(共 43分)26.(13 分)醇脱水是合成烯烃的常用方法,实验室合成环己烯的反应和实验装置如下:可能用到的有关数据如下:合成反应:在a中加入20g环己醇和2小片碎瓷片,冷却搅动下慢慢加入 1 mL浓硫酸。
2013年高考理科数学全国新课标卷1(附答案)

2013年普通高等学校夏季招生全国统一考试数学理工农医类(全国卷I新课标)注意事项:1.本试题分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分,第Ⅰ卷1至3页,第Ⅱ卷3至5页.2.答题前,考生务必将自己的姓名、准考证号填写在本试题相应的位置.3.全部答案在答题卡上完成,答在本试题上无效.4.考试结束后,将本试题和答题卡一并交回.第Ⅰ卷一、选择题:本大题共12小题,每小题5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.(2013课标全国Ⅰ,理1)已知集合A={x|x2-2x>0},B={x|-5<x<5},则().A.A∩B=B.A∪B=RC.B⊆A D.A⊆B答案:B解析:∵x(x-2)>0,∴x<0或x>2.∴集合A与B可用图象表示为:由图象可以看出A∪B=R,故选B.2.(2013课标全国Ⅰ,理2)若复数z满足(3-4i)z=|4+3i|,则z的虚部为().A.-4 B.45-C.4 D.45答案:D解析:∵(3-4i)z=|4+3i|,∴55(34i)34i 34i(34i)(34i)55z+===+--+.故z的虚部为45,选D.3.(2013课标全国Ⅰ,理3)为了解某地区的中小学生的视力情况,拟从该地区的中小学生中抽取部分学生进行调查,事先已了解到该地区小学、初中、高中三个学段学生的视力情况有较大差异,而男女生视力情况差异不大.在下面的抽样方法中,最合理的抽样方法是().A.简单随机抽样B.按性别分层抽样C.按学段分层抽样D.系统抽样答案:C解析:因为学段层次差异较大,所以在不同学段中抽取宜用分层抽样.4.(2013课标全国Ⅰ,理4)已知双曲线C:2222=1x ya b-(a>0,b>0)的离心率为52,则C的渐近线方程为().A.y=14x±B.y=13x±C.y=12x±D.y=±x答案:C解析:∵52cea==,∴22222254c a bea a+===.∴a2=4b2,1 =2 ba±.∴渐近线方程为12b y x x a =±±.5.(2013课标全国Ⅰ,理5)执行下面的程序框图,如果输入的t ∈[-1,3],则输出的s 属于( ).A .[-3,4]B .[-5,2]C .[-4,3]D .[-2,5] 答案:A解析:若t ∈[-1,1),则执行s =3t ,故s ∈[-3,3). 若t ∈[1,3],则执行s =4t -t 2,其对称轴为t =2.故当t =2时,s 取得最大值4.当t =1或3时,s 取得最小值3,则s ∈[3,4]. 综上可知,输出的s ∈[-3,4].故选A.6.(2013课标全国Ⅰ,理6)如图,有一个水平放置的透明无盖的正方体容器,容器高8 cm ,将一个球放在容器口,再向容器内注水,当球面恰好接触水面时测得水深为6 cm ,如果不计容器的厚度,则球的体积为( ).A .500π3cm 3 B .866π3cm 3C .1372π3cm 3D .2048π3cm 3答案:A解析:设球半径为R ,由题可知R ,R -2,正方体棱长一半可构成直角三角形,即△OBA 为直角三角形,如图.BC =2,BA =4,OB =R -2,OA =R , 由R 2=(R -2)2+42,得R =5, 所以球的体积为34500π5π33=(cm 3),故选A. 7.(2013课标全国Ⅰ,理7)设等差数列{a n }的前n 项和为S n ,若S m -1=-2,S m =0,S m +1=3,则m =( ).A .3B .4C .5D .6 答案:C解析:∵S m -1=-2,S m =0,S m +1=3,∴a m =S m -S m -1=0-(-2)=2,a m +1=S m +1-S m =3-0=3. ∴d =a m +1-a m =3-2=1.∵S m =ma 1+12m m (-)×1=0,∴112m a -=-. 又∵a m +1=a 1+m ×1=3,∴132m m --+=. ∴m =5.故选C.8.(2013课标全国Ⅰ,理8)某几何体的三视图如图所示,则该几何体的体积为( ).A .16+8πB .8+8πC .16+16πD .8+16π 答案:A解析:由三视图可知该几何体为半圆柱上放一个长方体,由图中数据可知圆柱底面半径r =2,长为4,在长方体中,长为4,宽为2,高为2,所以几何体的体积为πr 2×4×12+4×2×2=8π+16.故选A. 9.(2013课标全国Ⅰ,理9)设m 为正整数,(x +y )2m 展开式的二项式系数的最大值为a ,(x +y )2m +1展开式的二项式系数的最大值为b .若13a =7b ,则m =( ).A .5B .6C .7D .8 答案:B解析:由题意可知,a =2C mm ,b =21C mm +, 又∵13a =7b ,∴2!21!13=7!!!1!m m m m m m ()(+)⋅⋅(+),即132171m m +=+.解得m =6.故选B. 10.(2013课标全国Ⅰ,理10)已知椭圆E :2222=1x y a b+(a >b >0)的右焦点为F (3,0),过点F 的直线交E于A ,B 两点.若AB 的中点坐标为(1,-1),则E 的方程为( ).A .22=14536x y + B .22=13627x y + C .22=12718x y + D .22=1189x y + 答案:D解析:设A (x 1,y 1),B (x 2,y 2),∵A ,B 在椭圆上,∴2211222222221,1,x y a b x y a b ⎧+=⎪⎪⎨⎪+=⎪⎩①②①-②,得1212121222=0x x x x y y y y a b (+)(-)(+)(-)+, 即2121221212=y y y y b a x x x x (+)(-)-(+)(-), ∵AB 的中点为(1,-1),∴y 1+y 2=-2,x 1+x 2=2,而1212y y x x --=k AB =011=312-(-)-,∴221=2b a . 又∵a 2-b 2=9,∴a 2=18,b 2=9.∴椭圆E 的方程为22=1189x y +.故选D. 11.(2013课标全国Ⅰ,理11)已知函数f (x )=220ln(1)0.x x x x x ⎧-+≤⎨+>⎩,,,若|f (x )|≥ax ,则a 的取值范围是( ).A .(-∞,0]B .(-∞,1]C .[-2,1]D .[-2,0] 答案:D解析:由y =|f (x )|的图象知:①当x >0时,y =ax 只有a ≤0时,才能满足|f (x )|≥ax ,可排除B ,C. ②当x ≤0时,y =|f (x )|=|-x 2+2x |=x 2-2x . 故由|f (x )|≥ax 得x 2-2x ≥ax .当x =0时,不等式为0≥0成立. 当x <0时,不等式等价于x -2≤a . ∵x -2<-2,∴a ≥-2. 综上可知:a ∈[-2,0].12.(2013课标全国Ⅰ,理12)设△A n B n C n 的三边长分别为a n ,b n ,c n ,△A n B n C n 的面积为S n ,n =1,2,3,….若b 1>c 1,b 1+c 1=2a 1,a n +1=a n ,b n +1=2n n c a +,c n +1=2n nb a +,则( ). A .{S n }为递减数列B .{S n }为递增数列C .{S 2n -1}为递增数列,{S 2n }为递减数列D .{S 2n -1}为递减数列,{S 2n }为递增数列 答案:B第Ⅱ卷本卷包括必考题和选考题两部分.第(13)题~第(21)题为必考题,每个试题考生都必须做答.第(22)题~第(24)题为选考题,考生根据要求做答.二、填空题:本大题共4小题,每小题5分.13.(2013课标全国Ⅰ,理13)已知两个单位向量a ,b 的夹角为60°,c =t a +(1-t )b .若b ·c =0,则t =__________.答案:2解析:∵c =t a +(1-t )b , ∴b ·c =t a ·b +(1-t )|b |2.又∵|a |=|b |=1,且a 与b 夹角为60°,b ⊥c , ∴0=t |a ||b |cos 60°+(1-t ),0=12t +1-t . ∴t =2.14.(2013课标全国Ⅰ,理14)若数列{a n }的前n 项和2133n n S a =+,则{a n }的通项公式是a n =__________. 答案:(-2)n -1解析:∵2133n n S a =+,① ∴当n ≥2时,112133n n S a --=+.②①-②,得12233n n n a a a -=-,即1n n aa -=-2. ∵a 1=S 1=12133a +,∴a 1=1.∴{a n }是以1为首项,-2为公比的等比数列,a n =(-2)n -1.15.(2013课标全国Ⅰ,理15)设当x =θ时,函数f (x )=sin x -2cos x 取得最大值,则cos θ=__________.答案: 解析:f (x )=sin x -2cos xx x ⎫⎪⎭,令cos αsin α=则f (x )α+x ),当x =2k π+π2-α(k ∈Z )时,sin(α+x )有最大值1,f (x )即θ=2k π+π2-α(k ∈Z ),所以cos θ=πcos 2π+2k α⎛⎫- ⎪⎝⎭=πcos 2α⎛⎫- ⎪⎝⎭=sin α=5=-16.(2013课标全国Ⅰ,理16)若函数f (x )=(1-x 2)(x 2+ax +b )的图像关于直线x =-2对称,则f (x )的最大值为__________.答案:16解析:∵函数f (x )的图像关于直线x =-2对称, ∴f (x )满足f (0)=f (-4),f (-1)=f (-3),即15164,0893,b a b a b =-(-+)⎧⎨=-(-+)⎩解得8,15.a b =⎧⎨=⎩∴f (x )=-x 4-8x 3-14x 2+8x +15. 由f ′(x )=-4x 3-24x 2-28x +8=0,得x 1=-2x 2=-2,x 3=-2易知,f (x )在(-∞,-2上为增函数,在(-22)上为减函数,在(-2,-2上为增函数,在(-2∞)上为减函数.∴f (-2=[1-(-22][(-22+8(-2+15]=(-8--=80-64=16.f (-2)=[1-(-2)2][(-2)2+8×(-2)+15] =-3(4-16+15) =-9.f (-2=[1-(-22][(-22+8(-2+15]=(-8++=80-64=16.故f (x )的最大值为16.三、解答题:解答应写出文字说明,证明过程或演算步骤.17.(2013课标全国Ⅰ,理17)(本小题满分12分)如图,在△ABC 中,∠ABC =90°,AB BC =1,P 为△ABC 内一点,∠BPC =90°.(1)若PB =12,求P A ; (2)若∠APB =150°,求tan ∠PBA . 解:(1)由已知得∠PBC =60°,所以∠PBA =30°.在△PBA 中,由余弦定理得P A 2=11732cos 30424+-︒=.故P A (2)设∠PBA =α,由已知得PB =sin α.在△PBAsin sin(30)αα=︒-,α=4sin α. 所以tan α,即tan ∠PBA18.(2013课标全国Ⅰ,理18)(本小题满分12分)如图,三棱柱ABC -A 1B 1C 1中,CA =CB ,AB =AA 1,∠BAA 1=60°.(1)证明:AB ⊥A 1C ;(2)若平面ABC ⊥平面AA 1B 1B ,AB =CB ,求直线A 1C 与平面BB 1C 1C 所成角的正弦值. (1)证明:取AB 的中点O ,连结OC ,OA 1,A 1B . 因为CA =CB ,所以OC ⊥AB . 由于AB =AA 1,∠BAA 1=60°, 故△AA 1B 为等边三角形, 所以OA 1⊥AB .因为OC ∩OA 1=O ,所以AB ⊥平面OA 1C . 又A 1C ⊂平面OA 1C ,故AB ⊥A 1C . (2)解:由(1)知OC ⊥AB ,OA 1⊥AB .又平面ABC ⊥平面AA 1B 1B ,交线为AB , 所以OC ⊥平面AA 1B 1B ,故OA ,OA 1,OC 两两相互垂直.以O 为坐标原点,OA 的方向为x 轴的正方向,|OA |为单位长,建立如图所示的空间直角坐标系O -xyz.由题设知A (1,0,0),A 1(0,3,0),C (0,0,B (-1,0,0).则BC =(1,0,1BB =1AA =(-10),1AC =(0,. 设n =(x ,y ,z )是平面BB 1C 1C 的法向量,则10,0,BC BB ⎧⋅=⎪⎨⋅=⎪⎩nn 即0,30.x x y ⎧+=⎪⎨-+=⎪⎩可取n =1,-1).故cos 〈n ,1AC 〉=11AC AC ⋅n n =. 所以A 1C 与平面BB 1C 1C 19.(2013课标全国Ⅰ,理19)(本小题满分12分)一批产品需要进行质量检验,检验方案是:先从这批产品中任取4件作检验,这4件产品中优质品的件数记为n .如果n =3,再从这批产品中任取4件作检验,若都为优质品,则这批产品通过检验;如果n =4,再从这批产品中任取1件作检验,若为优质品,则这批产品通过检验;其他情况下,这批产品都不能通过检验.假设这批产品的优质品率为50%,即取出的每件产品是优质品的概率都为12,且各件产品是否为优质品相互独立.(1)求这批产品通过检验的概率;(2)已知每件产品的检验费用为100元,且抽取的每件产品都需要检验,对这批产品作质量检验所需的费用记为X (单位:元),求X 的分布列及数学期望.解:(1)设第一次取出的4件产品中恰有3件优质品为事件A 1,第一次取出的4件产品全是优质品为事件A 2,第二次取出的4件产品都是优质品为事件B 1,第二次取出的1件产品是优质品为事件B 2,这批产品通过检验为事件A ,依题意有A =(A 1B 1)∪(A 2B 2),且A 1B 1与A 2B 2互斥,所以P (A )=P (A 1B 1)+P (A 2B 2)=P (A 1)P (B 1|A 1)+P (A 2)P (B 2|A 2)=41113161616264⨯+⨯=. (2)X 可能的取值为400,500,800,并且 P (X =400)=41111161616--=,P (X =500)=116,P (X =800)=14.所以X 的分布列为EX =111400+500+80016164⨯⨯⨯=506.25. 20.(2013课标全国Ⅰ,理20)(本小题满分12分)已知圆M :(x +1)2+y 2=1,圆N :(x -1)2+y 2=9,动圆P 与圆M 外切并且与圆N 内切,圆心P 的轨迹为曲线C .(1)求C 的方程;(2)l 是与圆P ,圆M 都相切的一条直线,l 与曲线C 交于A ,B 两点,当圆P 的半径最长时,求|AB |. 解:由已知得圆M 的圆心为M (-1,0),半径r 1=1;圆N 的圆心为N (1,0),半径r 2=3. 设圆P 的圆心为P(x ,y ),半径为R .(1)因为圆P 与圆M 外切并且与圆N 内切, 所以|PM |+|PN |=(R +r 1)+(r 2-R )=r 1+r 2=4.由椭圆的定义可知,曲线C 是以M ,N 为左、右焦点,长半轴长为2(左顶点除外),其方程为22=143x y +(x ≠-2). (2)对于曲线C 上任意一点P (x ,y),由于|PM |-|PN |=2R -2≤2,所以R ≤2,当且仅当圆P 的圆心为(2,0)时,R =2. 所以当圆P 的半径最长时,其方程为(x -2)2+y 2=4. 若l 的倾斜角为90°,则l 与y 轴重合,可得|AB |=若l 的倾斜角不为90°,由r 1≠R 知l 不平行于x 轴,设l 与x 轴的交点为Q ,则1||||QP RQM r=,可求得Q (-4,0),所以可设l :y =k (x +4).由l 与圆M =1,解得k =4±. 当k =4时,将4y x =代入22=143x y +, 并整理得7x 2+8x -8=0,解得x 1,2=47-±.所以|AB |2118|7x x -=.当4k =-时,由图形的对称性可知|AB |=187.综上,|AB |=|AB |=187.21.(2013课标全国Ⅰ,理21)(本小题满分12分)设函数f (x )=x 2+ax +b ,g (x )=e x (cx +d ).若曲线y =f (x )和曲线y =g (x )都过点P (0,2),且在点P 处有相同的切线y =4x +2.(1)求a ,b ,c ,d 的值;(2)若x ≥-2时,f (x )≤kg (x ),求k 的取值范围.解:(1)由已知得f (0)=2,g (0)=2,f ′(0)=4,g ′(0)=4. 而f ′(x )=2x +a ,g ′(x )=e x (cx +d +c ), 故b =2,d =2,a =4,d +c =4. 从而a =4,b =2,c =2,d =2.(2)由(1)知,f (x )=x 2+4x +2,g (x )=2e x (x +1). 设函数F (x )=kg (x )-f (x )=2k e x (x +1)-x 2-4x -2, 则F ′(x )=2k e x (x +2)-2x -4=2(x +2)(k e x -1). 由题设可得F (0)≥0,即k ≥1. 令F ′(x )=0得x 1=-ln k ,x 2=-2.①若1≤k <e 2,则-2<x 1≤0.从而当x ∈(-2,x 1)时,F ′(x )<0;当x ∈(x 1,+∞)时,F ′(x )>0.即F (x )在(-2,x 1)单调递减,在(x 1,+∞)单调递增.故F (x )在[-2,+∞)的最小值为F (x 1).而F (x 1)=2x 1+2-21x -4x 1-2=-x 1(x 1+2)≥0.故当x ≥-2时,F (x )≥0,即f (x )≤kg (x )恒成立.②若k =e 2,则F ′(x )=2e 2(x +2)(e x -e -2).从而当x >-2时,F ′(x )>0,即F (x )在(-2,+∞)单调递增. 而F (-2)=0,故当x ≥-2时,F (x )≥0,即f (x )≤kg (x )恒成立.③若k >e 2,则F (-2)=-2k e -2+2=-2e -2(k -e 2)<0. 从而当x ≥-2时,f (x )≤kg (x )不可能恒成立. 综上,k 的取值范围是[1,e 2].请考生在第(22)、(23)、(24)三题中任选一题做答.注意:只能做所选定的题目.如果多做,则按所做的第一个题目计分,做答时请用2B 铅笔在答题卡上将所选题号后的方框涂黑. 22.(2013课标全国Ⅰ,理22)(本小题满分10分)选修4—1:几何证明选讲如图,直线AB 为圆的切线,切点为B ,点C 在圆上,∠ABC 的角平分线BE 交圆于点E ,DB 垂直BE 交圆于点D .(1)证明:DB =DC ;(2)设圆的半径为1,BC CE 交AB 于点F ,求△BCF 外接圆的半径. (1)证明:连结DE ,交BC 于点G . 由弦切角定理得,∠ABE =∠BCE .而∠ABE =∠CBE ,故∠CBE =∠BCE ,BE =CE .又因为DB⊥BE,所以DE为直径,∠DCE=90°,由勾股定理可得DB=DC.(2)解:由(1)知,∠CDE=∠BDE,DB=DC,故DG是BC的中垂线,所以BG=2.设DE的中点为O,连结BO,则∠BOG=60°.从而∠ABE=∠BCE=∠CBE=30°,所以CF⊥BF,故Rt△BCF23.(2013课标全国Ⅰ,理23)(本小题满分10分)选修4—4:坐标系与参数方程已知曲线C1的参数方程为45cos,55sinx ty t=+⎧⎨=+⎩(t为参数),以坐标原点为极点,x轴的正半轴为极轴建立极坐标系,曲线C2的极坐标方程为ρ=2sin θ.(1)把C1的参数方程化为极坐标方程;(2)求C1与C2交点的极坐标(ρ≥0,0≤θ<2π).解:(1)将45cos,55sinx ty t=+⎧⎨=+⎩消去参数t,化为普通方程(x-4)2+(y-5)2=25,即C1:x2+y2-8x-10y+16=0.将cos,sinxyρθρθ=⎧⎨=⎩代入x2+y2-8x-10y+16=0得ρ2-8ρcos θ-10ρsin θ+16=0.所以C1的极坐标方程为ρ2-8ρcos θ-10ρsin θ+16=0. (2)C2的普通方程为x2+y2-2y=0.由2222810160,20x y x yx y y⎧+--+=⎨+-=⎩解得1,1xy=⎧⎨=⎩或0,2.xy=⎧⎨=⎩所以C1与C2交点的极坐标分别为π4⎫⎪⎭,π2,2⎛⎫⎪⎝⎭.24.(2013课标全国Ⅰ,理24)(本小题满分10分)选修4—5:不等式选讲已知函数f(x)=|2x-1|+|2x+a|,g(x)=x+3.(1)当a=-2时,求不等式f(x)<g(x)的解集;(2)设a>-1,且当x∈1,22a⎡⎫-⎪⎢⎣⎭时,f(x)≤g(x),求a的取值范围.解:(1)当a=-2时,不等式f(x)<g(x)化为|2x-1|+|2x-2|-x-3<0. 设函数y=|2x-1|+|2x-2|-x-3,则y =15,,212,1,236, 1.x x x x x x ⎧-<⎪⎪⎪--≤≤⎨⎪->⎪⎪⎩其图像如图所示.从图像可知,当且仅当x ∈(0,2)时,y <0. 所以原不等式的解集是{x |0<x <2}.(2)当x ∈1,22a ⎡⎫-⎪⎢⎣⎭时,f (x )=1+a . 不等式f (x )≤g (x )化为1+a ≤x +3.所以x ≥a -2对x ∈1,22a ⎡⎫-⎪⎢⎣⎭都成立. 故2a -≥a -2,即43a ≤. 从而a 的取值范围是41,3⎛⎤- ⎥⎝⎦.。
2013年高考真题——英语(新课标I卷)Word版含答案_1

2013年普通高等学校招生全国统一考试英语本试题分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)考试结束后,将本试卷和答题卡一并交回。
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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。
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1.What does the man want to do?A. Take photos.B. Buy a camera.C. Help the woman.2. What are the speakers talking about>A. A noisy nightB. Their life in townC. A place of living.3.Where is the man now?A. On his way.B. In a restaurant.C. At home4. What will Celia do?A.Find a player.B.Watch a game.C.Play basketball.5.What day is it when the conversation takes place?A.Saturday.B.Sunday.C.Monday.第二节(共15小题:每小题1.5分,满分22.5分)听下面5段对话或独白。
2013年高考真题——英语(新课标I卷)Word版含答案

第一节单项天空(共15小题;每小题1分,满分15分)21. -Wly, this is nothing but common vegetable soup!A.Let me see Bso it is CDon …t mention it D Neither do I22. .They might just have a place ______on the writing course一why don't you give it a try?A.LeaveB.LeftC. leavingD. to leave23. Try not to sough more than you can since it may cause problems to your lungs.A. checkB. allowC.stopD.help24. If we now to protect the environment, we‟ll live to regret it.A. hadn't actedB. haven't actedC. don't actD. won't act25. Tony can hardly boil an egg,still cook dinner.A.lessB.littleC.muchD.more26. Police have found appears to be the lost ancient statue.A. whichB. whereC. howD.what27.When I first met Bryan I didn‟t like him, but I my mind.A. have changedB. changeC. had changedD.would change28. The driver wanted to park his car near the roadside but was asked by}the police .A .not to doB. not to C. not do D. do not29.The door open , no matter how hard she pushed.A. shouldn‟tB. couldn'tC. wouldn'tD. mightn't30. At the last moment Tom-decided to a new character to make the story seem more likely.A. put upB. put inC. put onD. put off31. India attained independence in 1947, after long struggle.A.不填;aB. the; aC. an;不填D. an; the32.There‟s no way of knowing why one man makes an important discovery another man, also intelligent, fails.A. since B: if C. as D. while33.”You can‟t judge a book by its cover,”.A. as the saying goes oldB. goes as the old sayingC. as the old saying goesD. goes as old the saying34.It was a real race time to get the project done. Luckily, we made it.A. overB. byC. forD. against35. The sunlight is white and blinding, hard-edged shadows on the ground.A. throwingB.being thrownC.to throwD.to be thrown 第二节完形填空(共20小题:每小题1.5分,满分30分)I went to a group activity,“Sensitivity Sunday”which was to make us more 36the problem faced by disabled people,We where asked asked to “ 37 a disability”for sever hours one Sunday,Some member 38 chose the wheel chair,Other wore sound-blocking carplug(耳塞)or bilndfold(眼罩).Just sitting in the wheelchair was a 39 experience,I had never considered before how 40 it would be to use one ,As soon as I sat down my 41 made the chair begin to roll ,Its wheel were not 42 Then I wondered where to put my 43 ,It took me quite a while to get the metal footrest into 44 ,I took my first uneasy look at what was to be my only means of 45 for serveral hours,For disabled people,“adoping a wheelchair”is not a tempoarty(临时的) 46I tried to find a 47 position and thought it might be restful, 48 kind of nice to be 49 around for a while, Looking around,I 50 would have to handle the thing myself!My hands statred to ache as I 51 the heavy wheels,I came to know that controlling the 52 of the wheelchair as not going to be 53 task,My wheelchair experment was soon 54 It made a deep impression on me ,A few hours of “disability” gave me only a taste of the 55 ,both physical and mental,that disbled people must overcome36.A.curious about B. interested in C. aware on D. careful with37.A.cure B.prevent C.adopt D.analyze38.A.inserted B.strangely C.as usual D.like me39.A.learning B.working C.satrstying D.relaxing40.A.convening B.awkward C.boring D.exciting41.A.height B.force C.skill D.weight42.A.locked B.repaired C.powered D.grasped43.A.hands B.feet C.keys D.handles44.A.place B.action C.play D.effect45.A.operationB.ecommunication C.transportation D.production46.A.exploration B.edcation C.experiment D.entertainment47.A.flexble B. safe C. starting D. comfortable48.A. yet B. just C. still D. even49.A.shown B. pushed C. driven D. guided50.A.realized B. suggested C. agreed D. admitted51.A. lifted B. turned C. pressed D. seized52.A. path B. position C. direction D. way53.A. easy B. heavy C. major D. extra54.A. forgotten B. repeated C. conducted D. finished55.A. weaknesses B. challenges C. anxieties D. illnesses第三部分阅读理解(共两节,满分40分)ASome people will do just about anything to save money. And I am one of them .Take my family‟s last vacation .It was my six-year-old son‟s winter break form school ,and we were heading home form Fort Lauderdale after a weeklong trip. The flight wasoverbooked ,and Delta , the airline ,offered us $400 per person in credits to give up our seats and leave the next day .I had meeting in New York,So I had to get back . But that didn't mean my husband and my son couldn't stay. I took my nine-month-old and took off for home.The next day my husband and son were offered more credits to take an even later flight.Yes, I encouraged一okay, ordered-them to wait it out at the airport, to "earn" more Delta Dollars. Our total take: $1,600. Not bad, huh? Now some people may think I'm a bad mother and not such a great wife either. But as a big-time bargain hunter, I know the value of a dollar. And these days, a good deal is some-timething few of us can afford to pass up.I've made living looking for the best deals and exposing (揭露) the worst tricks .I have been the consumer reporter of NBC's Today show for over a decade. I have written a coupleof books including one titled Tricks of the Trade: A Consumer Survival Guide. And I really do what I believe in.I tell you this because there is no shame in getting your money‟s worth. I‟m also tightfisted when it comes to shoes, clothes for my children,and expensive restaurants.But I wouldn't hesitate to spend on a good haircut. It keeps its longer, and it's the first thing people notice. And I will also spend on a classic piece of furniture. Quality lasts.56. Why did Delta give the author's family credits?A. They took a later flight.B. They had early bookings.C. Their flight had been delayed.D. Their flight had been cancelled.57. What can we learn about the author?A. She rarely misses a good deal.B. She seldom makes a compromise.C. She is very strict with her childrenD. She is interested in cheap products.58. What does the author do?A. She's a teacher.B. She's a housewife.C. She's a media person.D. She's a businesswoman.59. What does the author want to tell us?A.How to expose bad tricks.B. How to reserve airline seats.C. How to spend money wisely,D. How to make a business deal.BThey baby is just one day old and has not yet left hospital. She is quiet but alert (警觉)。
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高三周考英语试卷(七)阅读理解(共两节)第一节(每小题3分)ASome people will do just about anything to save money. And I am one of them .Take my family’s last vacation .It was my six-year-old son’s winter break form school ,and we were heading home form Fort Lauderdale after a week-long trip. The flight was overbooked ,and Delta , the airline ,offered us $400 per person in credits to give up our seats and leave the next day .I had meeting in New York,So I had to get back . But that didn't mean my husband and my son couldn't stay. I took my nine-month-old and took off for home.The next day my husband and son were offered more credits to take an even later flight.Yes, I encouraged一okay, ordered-them to wait it out at the airport, to "earn" more Delta Dollars. Our total take: $1,600. Not bad, huh?Now some people may think I'm a bad mother and not such a great wife either. But as a big-time bargain hunter, I know the value of a dollar. And these days, a good deal is something few of us can afford to pass up.I've made a living looking for the best deals and exposing (揭露) the worst tricks .I have been the consumer reporter of NBC's Today show for over a decade. I have written a couple of books including one titled Tricks of the Trade: A Consumer Survival Guide. And I really do what I believe in.I tell you this because there is no shame in getting your money’s worth. I’m also tightfisted when it comes to shoes, clothes for my children,and expensive restaurants.But I wouldn't hesitate to spend on a good haircut. It keeps its longer, and it's the first thing people notice. And I will also spend on a classic piece of furniture. Quality lasts. 1. Why did Delta give the author's family credits?A. They took a later flight.B. They had early bookings.C. Their flight had been delayed.D. Their flight had been cancelled.2. What can we learn about the author?A. She rarely misses a good deal.B. She seldom makes a compromise.C. She is very strict with her childrenD. She is interested in cheap products.3. What does the author do?A. She's a teacher.B. She's a housewife.C. She's a media person.D. She's a businesswoman.4. What does the author want to tell us?A.How to expose bad tricks.B. How to reserve airline seats.C. How to spend money wisely,D. How to make a business deal.DThe National GalleryDescription:The National Gallery is the British national art museum built on the north side of Trafalgar Squaer in London It houses a diverse collection of more than 1300 examples.European art ranging from 13th-century religious paintings to more modern ones by Renoir and Van Gogh.The older collections of the gallery are reached through the main entrance while the more modern works in the East Wing are most easily reached from Trafalgar Square by a ground floor entranceLayout:The modern Sainsbury Wing on the western side of the building houses 13th-to15th-century paintings,and artists include Duccio,Uccello,Van Eyck,Lippi,Mantegna,Botticelli and Memling.The main West Wing houses 16th-century paintings ,and artists include Leonardo da Vinci,Cranach,Michelangelo,Raphael,Bruegel,Bronzino,Titan and V eronest.The North Wing houses 17th-century paintings,and artists include Caravaggio,Rubens,Poussin,Van Dyck,Velazquez,Claude and Vermeer.The East Wing houses 18th-to early 20th-century paintings,and artists include Canaletto,Goya,Turner,Constable,Renoir and Van GoghOpening Hours:The Gallery is open every day from 10am to 6pm(Fridays 10am to 9pm)and is free, but charges apply to some special exhibitions.Getting There:Nearest underground stations: Charing Cross(2-minute walk). Leicester Square(3-minute walk),Embankment(7-minute walk),and Piccadilly Circus(8-minute walk).5.In which century’s collection can you see religious paintings?A.The 13th B.The 17thC.The 18th D.The 20th6.Where are Leonardo da Vinci’s works shown?A.In the East Wing .B.In the main West Wing.C.In the Sainsbury Wing. D.In the North Wing.7.Which underground station is closest to the National Gallery?A.Piccadilly Circus .B.Leicester Square.C.Embankment. D.Charing Cross.第二节(共5小题:每小题3分,满分15分)根据短文内容,从短文后的选项中选出能填入空白处的最佳选项·选项中有两项为多余选项。
Business is the organized approach to providing customers with the goods and services they want.The word business also refers to an organization that provides these goods and services.Most business seek to make a profit(利润)-that is,they aim to achieve income that is more than the costs of operating the business. 8 Commonly called nonprofits,these organizations are primarily nongovernmental service providers.9Business management is a term used to describe the techniques of planning, direction,and control of the operations of a business. 10 One is the establishment (制定)of broad basic policies with respect to production; sales; the purchase of equipment, materials and and supplies; and accounting. 11 The third relates to the establishment of standards of work in all departments. Direction is concerned primarily with supervision(监管)and guidance by the management in authority. 12A. Control includes the use of records and reports to compare actual work with the set standards for work.B. In this connection there is the difference between top management and operative management.C.Examples of nonprofit business include such organizations as social service agencies and many hospitals.D. However, some businesses only seek to earn enough to cover their operating costsE. The second aspect relates to the application of these policies by departments.F. In the theory of business management, organization has two main aspects.G Planning in business management has three main aspects.第二节完形填空(共20小题:每小题1.5分,满分30分)阅读下面短文,从短文后各题所给的四个选项(A、B、C、D)中,选出可以填入空白处的最佳选项,并在答题卡上将该项涂黑I went to a group activity,“Sensitivity Sunday”which was to make us more 36the problem faced by disabled people,We were asked to “37 a disability”for sever hours one Sunday,Some member, 38 chose to use the wheel chair,Other wore sound-blocking earplugs(耳塞)or blindfolds(眼罩).Just sitting in the wheelchair was a 39 experience,I had never considered before how 40 it would be to use one ,As soon as I sat down my 41 made the chair begin to roll ,Its wheel were not 42 Then I wondered where to put my 43 ,It took me quite a while to get the metal footrest into 44 ,I took my first uneasy look at what was to be my only means of 45 for several hours,For disabled people,“adopting a wheelchair”is not a temporary(临时的) 46 .I tried to find a 47 position and thought it might be restful, 48 kind of nice to be 49 around for a while, Looking around,I 50 would have to handle the thing myself!My hands started to ache as I 51 the heavy wheels,I came to know that controlling the 52 of the wheelchair as not going to be 53 task,My wheelchair experiment was soon 54 It made a deep impression on me ,A few hours of “disability”gave me only a taste of the 55 ,both physical and mental,that disabled people must overcome.36.A.curious about B. interested in C. aware on D. careful with37.A.cure B.prevent C.adopt D.analyze38.A.inserted B.strangely C.as usual D.like me39.A.learning B.working C.satisfying D.relaxing40.A.convening B.awkward C.boring D.exciting41.A.height B.force C.skill D.weight42.A.locked B.repaired C.powered D.grasped43.A.hands B.feet C.keys D.handles44.A.place B.action C.play D.effect45.A.operation munication C.transportation D.production46.A.exploration cation C.experiment D.entertainment47.A.flexible B. safe C. starting D. comfortable48.A. yet B. just C. still D. even49.A.shown B. pushed C. driven D. guided50.A.realized B. suggested C. agreed D. admitted51.A. lifted B. turned C. pressed D. seized52.A. path B. position C. direction D. way53.A. easy B. heavy C. major D. extra 54.A. forgotten B. repeated C. conducted D. finished55.A. weaknesses B. challenges C. anxieties D. illnesses短文改错(共10小题;每小题1分,满分10分)I hardly remember my grandmother.She used to holding me on her knees and sing old songs.I was only four when she passes away.She is just a distant memory for me now.I remember my grandfather very much.He was tall,with broad shoulder and a beard that turned from black toward gray over the years.He had a deep voice,which set himself apart from others in our small town,he was strong and powerful.In a fact,he even scared my classmates away during they came over to play or do homework with me.However, he was the gentlest man I have never known.答题卡阅读(每题3分)完形(每题2分)改错(答在原题,共10分)。