【含高考模拟卷31套】江苏省扬州中学2020-2021学年高三下学期开学考试(2月)英语试题含解析
江苏省扬州中学2020届高三英语下学期2月开学考试试题
江苏省扬州中学2020学年高三年级开学考试英语试卷 2020.02(本卷分共五部分，满分120分。
考试时间120分钟。
)第一部分听力（共两节，每题1分，满分20分）第一节（共5小题；每小题1分，满分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. What is the man waiting to do?A. Enjoy some noodles.B. Eat some eggs.C. Drink some hot water.2. What does the man advise the woman to do?A. Give running another try.B. Try some of the other events.C. Talk with the P.E teacher.3. For what purpose did the woman choose to take Spanish?A. She wanted to be classmates with the man.B. She studied it when she was a little girl.C. She had studied a similar language before.4. What is Steve worried about?A. His football.B. His lamp.C. His desk.5. Where does the conversation probably take place?A. In a bookstore.B. In a game center.C. In a library.第二节（共15小题；每小题1分，满分15分）听下面5段对话或独白。
江苏省扬州中学2020┄2021届高三下学期开学考试2月 英语
第一卷（选择题，共85分）第一部分：听力（共两节，满分20分）第一节（共5小题；每小题1分，满分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. Where is the man’s brother？A. At home.B. In the hospital.C. In the office.2. What is the probable relationship between the two speakers？A. Husband and wife.B. Waiter and customer.C. Teacher and student.3. Who is the man？A. The woman’s husband.B. The woman’s boss.C. The woman’s teacher.4. What is the girl going to do next？A. Go to school.B. Have breakfast.C. Pack her school bag.5. What will the woman do at about ten o’clock？A. Have a meeting.B. Call Mr. Johnson again.C. Go to Siemens.第二节（共15小题；每小题1分，满分15分）听下面5段对话或独白。
每段对话或独白后有几个小题，从题中所给的A、B、C三个项中选出最佳选项，并标在试卷的相应位置。
听每段对话或独白前，你将有时间阅读各个小题，每小题5秒钟；听完后，各小题给出5秒钟的作答时间。
每段对话或独白读两遍。
听下面一段对话，回答第6和第7两个小题。
6. What does the woman want to find at first？A. A fruit market.B. A supermarket.C. A parking area.7. Where is the parking area？A. Behind the fruit market.B. In front of the supermarket.C. Between the fruit market and the supermarket.听下面一段对话，回答第8和第9两个小题。
江苏省扬州市扬州中学高三数学下学期开学考试试题苏教版
江苏省扬州中学2014～2015学年第二学期开学检测高三数学卷注意事项：所有试题的答案均填写在答题纸上，答案写在试卷上的无效 一、填空题：（本大题共14小题，每小题5分，共70分请将答案填入答题纸填空题的相应答题线上）1．已知集合{113}A =-，，，}5,3,1{=B ，则=B A ▲ ． 2．复数212a ii-+（i 是虚数单位）是纯虚数，则实数a 的值为 ▲ ． 3．右图是一个算法的流程图，则最后输出的S = ▲ ． 4．从1,3,5,7这4个数中一次随机地取2个数,则所取2个数的和小于9的概率是 ▲ ． 5．已知样本7,5,,3,4x 的平均数是5,则此样本的方差为 ▲ ． 6．已知函数()2sin()(0)6f x x πωω=->的最小正周期为π，则f (x )在[0,]2π上的单调递增区间为[a ，]b ，则实数a b += ▲ ．7．已知体积相等的正方体和球的表面积分别为1S ，2S ，则321)(S S 的值是 ▲ ．8. 抛物线212y x =-的准线与双曲线22162x y -=的两条渐近线所围成的三角形的面积等 于 ▲ ．9．已知32x ≥，则22211x x x -+-的最小值为 ▲ ．10．在平面直角坐标系xOy中，若曲线()sin cos f x x x =(a 为常数)在点(,())33f ππ处的切线与直线0132=++y x 垂直，则a 的值为 ▲ ．11．设等差数列{}n a 的前n 项和为n S ，且满足21n n a S An Bn +=++（0A ≠）则1B A-=___▲___．12．已知函数()f x 是定义在R 上的偶函数，且当0x ≥时，()|2|f x x x =-．若关于x 的方程2()()0(,)f x af x b a b R ++=∈恰有10个不同实数解，则a 的取值范围为 ___▲ ．13．在直角ABC ∆中，2,AB AC ==，斜边BC 上有异于端点两点B C 、的两点E F 、，且=1EF ，则AE AF ⋅的取值范围是 ▲ ． 14．已知三个正数,,a b c 满足3a b c a ≤+≤，223()5b a a c b ≤+≤，则2b ca-的最小值 是 ▲ ．二、解答题（本大题共6小题，共90分解答应写出文字说明、证明过程或演算步骤） 15（本小题满分14分）设平面向量a ＝(cos ,sin )x x，(cos )b x x =+，(sin ,cos )c αα=，x R ∈． （1）若a c ⊥，求cos(22)x α+的值；（2）若0α=，求函数()(2)f x a b c =⋅-的最大值，并求出相应的x 值．16（本小题满分14分）如图，在三棱柱111ABC A B C -中，D 为棱BC 的中点，1,AB BC BC BB ⊥⊥，111,AB AB BB ===求证：(1) 1A B⊥平面ABC ； (2)1A B ∥平面1AC D .17（本小题满分14分）如图，椭圆22122:1(0)x y C a b a b +=>>和圆2222:C x y b +=，已知椭圆1C过点，焦距为2.C(1) 求椭圆1C 的方程；(2) 椭圆1C 的下顶点为E ，过坐标原点O 且与坐标轴不重合的任意直线l 与圆2C 相交于点A B 、，直线EA EB 、与椭圆1C 的另一个交点分别是点P M 、.设PM 的斜率为1k ，直线l 斜率为2k ，求21k k 的值.18（本小题满分16分）在距A 城市45千米的B 地发现金属矿，过A 有一直线铁路AD ．欲运物资于A ，B 之间，拟在铁路线AD 间的某一点C 处筑一公路到B ．现测得BD =45BDA ∠=（如图）．已知公路运费是铁路运费的2倍，设铁路运费为每千米1个单位，总运费为y ．为了求总运费y 的最小值，现提供两种方案：方案一：设AC x =千米；方案二设BCD θ∠=．（1）试将y 分别表示为x 、θ的函数关系式()y f x =、()y g θ=；（2）请选择一种方案，求出总运费y 的最小值，并指出C 点的位置．19（本小题满分16分）已知数列{}n a 、{}n b 满足1=n b a ，110k k k k b b a a --=≠，其中2,3,,k n =，则称{}n b 为{}n a 的“生成数列”．（1）若数列12345a a a a a ，，，，的“生成数列”是1,2,3,4,5，求1a ；（2）若n 为偶数，且{}n a 的“生成数列”是{}n b ，证明：{}n b 的“生成数列”是{}n a ； （3）若n 为奇数，且{}n a 的“生成数列”是{}n b ，{}n b 的“生成数列”是{}n c ，…，依次将数列{}n a ，{}n b ，{}n c ，…的第(1,2,,)i i n =项取出，构成数列:,,,i i i i a b c Ω．探究：数列i Ω是否为等比数列，并说明理由．20（本小题满分16分）已知函数2()f x x ax b =++，()ln g x x =．（1）记()()()F x f x g x =-,求()F x 在[1,2]的最大值；（2）记()()()f x G xg x =，令4a m =-，24()b m m R =∈，当210<<m 时，若函数()G x 的3个极值点为123123,,()x x x x x x <<，（ⅰ）求证：321120x x x <<<<；（ⅱ）讨论函数()G x 的单调区间（用123,,x x x 表示单调区间）．高三第二学期期初联考数学附加题 (考试时间：30分钟 总分：40分)21.（［选做题］请考生在A 、B 、C 、D 四小题中任选两题作答，如果多做，则按所做的前两题记分．A ．（本小题满分10分，几何证明选讲）如图，P 是⊙O 外一点，PA 是切线，A 为切点，割线PBC 与⊙O 相交于点B ，C ，PC ＝2PA ，D 为PC 的中点，AD 的延长线交⊙O 于点E . 证明： AD ·DE ＝2PB 2.B ．（本小题满分10分，矩阵与变换）设矩阵12M x y ⎡⎤=⎢⎥⎣⎦，2411N ⎡⎤=⎢⎥--⎣⎦，若02513MN ⎡⎤=⎢⎥⎣⎦，求矩阵M 的特征值．C ．（本小题满分10分，坐标系与参数方程选讲）在平面直角坐标系xOy 中，已知直线l 的参数方程为：122x ty t =+⎧⎨=-⎩(t 为参数)．以坐标原点为极点，x 轴正半轴为极轴建立极坐标系，圆C 的极坐标方程为ρ＝2cos θ．直线l 与圆相交于A ，B 两点，求线段AB 的长．D ．（本小题满分10分，不等式选讲）已知实数z y x ,,满足123=++z y x ，求22232z y x ++的最小值.［必做题］第22题，第23题，每题10分，共计20分．解答时应写出文字说明、证明过程或演算步骤．22.（本小题满分10分)如图，四棱锥P －ABCD 中，底面ABCD 为矩形，PA ⊥平面ABCD ，AP ＝1，ADE 为线段PD 上一点，记PE PD λ=． 当12λ=时，二面角D AE C --的平面角的余弦值为23． （1）求AB 的长； （2）当13λ=时，求直线BP 与直线CE 所成角的余弦值．23.（本小题满分10分)已知数列{}n a 通项公式为11n n a AtBn -=++，其中,,A B t 为常数，且1t >，n N *∈．等D式()()()()1022020122022111x x b b x b x b x ++=+++++⋅⋅⋅++，其中()0,1,2,,20i b i =⋅⋅⋅为实常数．（1）若0,1A B ==，求1021n nn a b=∑的值；（2）若1,0A B ==，且()1011212222n n nn ab =-=-∑，求实数t 的值．高三第二学期期初联考数学参考答案 一、填空题1．{1,1,3,5}-； 2．4； 3．9； 4．23； 5．2； 6．3π； 7．6π； 8． 9．2； 10．23-；11．3； 12．(2,1)--； 13．11[,9)4； 14．185-．二、解答题15．解：（1）若a c ⊥，则0a c ⋅=， ………2分 即()cos sin sin cos 0,sin 0x x x ααα+=+= ………4分 所以()()2cos 2212sin 1x x αα+=-+=. ………6分（2）若()0,0,1c α==则………10分………12分所以max ()5,2()6f x x k k Z ππ==-∈. ………14分()()()()(()2cos ,sin cos 2cos cos sin sin 212sin 214sin 3f x a b c x x x x x x x x x x x π=⋅-=⋅+-=++-=-+⎛⎫=++ ⎪⎝⎭16．证明：（1）因为1111,,,AB BC BC BB AB BB B AB BB ABB ⊥⊥=⊂、平面，所以111BC ABB AB ABB ⊥⊂平面,又平面，所以1AB BC ⊥； ………3分又因为1111,AB A B BB AA ====，得22211AA AB A B =+，所以1A B AB ⊥. ………6分 又AB BC ABC ABBC B ⊂=、平面，，所以1A B ⊥平面ABC ； ………8分（2）连接1AC 交1AC 与点E ，连接DE ，在1A BC ∆中，D E 、分别为1BC AC 、的中点，所以1//DE A B ，又111,A B AC D DE AC D ⊄⊂平面平面，所以1A B ∥平面1AC D ．………14分17．解：（1）解法一：将点代入椭圆方程，解方程组，求得222,1a b ==，所以椭圆1C 的方程为2212x y +=． ………4分解法二：由椭圆的定义求得2a =，所以椭圆1C 的方程为2212x y +=． ………4分说明：计算错全错.（2）由题意知直线,PE ME 的斜率存在且不为0，PE EM ⊥， 不妨设直线PE 的斜率为(0)k k >，则:1PE y kx =-，由221,1,2y kx x y =-⎧⎪⎨+=⎪⎩得2224,2121,21k x k k y k ⎧=⎪⎪+⎨-⎪=⎪+⎩或0,1,x y =⎧⎨=-⎩BAC222421(,)2121k k P k k -∴++. ………6分用1k -去代k ，得22242(,)22k k M k k--++， ………8分 则2113PMk k k k-== ………10分由221,1,y kx x y =-⎧⎨+=⎩得2222,11,1k x k k y k ⎧=⎪⎪+⎨-⎪=⎪+⎩或0,1,x y =⎧⎨=-⎩22221(,)11k k A k k -∴++. ………12分则2212OAk k k k -==，所以2132k k =． ………14分评讲建议：此题还可以求证直线PM 恒过定点，求PME ∆面积的最大值.18.解：(1)在ABD ∆中，由余弦定理解得AD=63 ………2分 方案一：在ABC ∆中，222222227)36(7245cos 45245+-=-+=⋅-+=x x x A x x BC 2227)36(22)(+-+=+=∴x x BC AC x f ………5分方案二：在BCD ∆中，θθsin 2745sin sin 227==BC ，θθθθθsin )cos (sin 27)45sin(sin 227+=+= CD ， θθθθθθθsin cos 22736)sin cos sin sin 2(2763221)(-+=+-+=+-=⋅+⋅=BC CD AD BC AC g ………9分 (2)若用方案一，则8100)144(23)4572(4)(457222222222=+--+⇒+-=-⇒+-+=y x y x x x x y x x x y………11分 由0≥∆得327360891720)8100(3)144(222+≥⇒≥--⇒≥-+-y y y y y ………14分32736min +=∴y ，这时39363144-=-=yx ，C 距A 地)3936(-千米 ………16分若用方案二，则θθθθθ222sin cos 2127sin cos )cos 2(sin 27-=--='y ………11分)(θg 在↓)3,0(π，在↑),3(ππ32736232122736min +=-+=∴y ………14分 这时3πθ=，C 距A 地)3936(-千米 ………16分19．（1）解：151b a ==，4544520a a a =⨯⇒=同理，32131,10,55a a a ===. ………4分 （写对一个i a 得1分，总分4分） （2）证明：1n b b = 1212232311n n n nb b a a b b a a b b a a --=== ………7分∵n 为偶数，将上述n 个等式中第2，4，6，…，n 这2n个式子两边取倒数，再将这n 个式子相乘得：1234523451234112341111111n n n n nb b b b b a a a a a b b b b b b a a a a a a --⋅⋅⋅⋅=⋅⋅⋅⋅ ∴1n b a = ………9分因为1n a b =，11(2,3,,)k k k k a a b b k n --==所以根据“生成数列”的定义，数列{}n a 是数列{}n b 的“生成数列”. ………10分（3）证明：因为11(2,3,,)i ii i a a b i n b --==，所以111(2,3,,)i i i i b i n a a b --==.所以欲证i Ω成等差数列，只需证明1Ω成等差数列即可. ………12分对于数列{}n a 及其“生成数列”{}n b1n b b = 1212232311n n n nb b a a b b a a b b a a --===∵n 为奇数，将上述n 个等式中第2，4，6，…，1n -这12n -个式子两边取倒数，再将这n 个式子相乘得：12345123451123421123421111111n n n n nn n n n b b b b b b b a a a a a a a b b b b b b a a a a a a ------⋅⋅⋅⋅⋅=⋅⋅⋅⋅⋅ ∴21111n n n n n a b a bb a a a a =⇒==因为1n a b =，11(2,3,,)k k k k a a b b k n --==数列{}n b 的“生成数列”为{}n c ，因为22111111,nn n a b c c b b a c a ===⇒= 所以111,,a b c 成对比数列. 同理可证，111111,,;,,,b c d c d e 也成等比数列. 即 1Ω是等比数列.所以 i Ω成等差数列. ………16分20.解：（1）x b ax x X F ln )(2-++=（0>x ）x ax x x a x X F 1212)('2-+=-+= ………2分令0)('=x F ，得04821<+--=a a x ，04822>++-=a a x()()xx x x x X F 212)('--=………3分易知()()(){}2,1max max F F x F =而()()()()32ln 2ln 42121-+-=-++-++=-a b a b a F F 所以当32ln -≤a 时， ()()11max ++==b a F x F当32ln ->a 时，()()2ln 422max -++==b a F x F ………5分（2）（ⅰ）()xm mx x x G ln 4422+-=，()()xxm x m x x G 2ln 12ln 22'⎪⎭⎫⎝⎛-+-=令()12ln 2-+=x m x x h ，()222'xmx x h -= 又()x h 在()m ,0上单调减，在()+∞,m 上单调增，所以()()1ln 2min +==m m h x h 因为函数()x G 有3个极值点，所以01ln 2<+m 所以em 10<< ………7分所以当210<<m 时，()04ln 121ln 211ln 2<-=+<+=m m h ，()0121<-=m h 从而函数()x G 的3个极值点中，有一个为m 2，有一个小于m ，有一个大于1………9分 又321x x x <<，所以m x <<10，m x 22=，13>x 即2021x x <<，3212x m x <<=，故321120x x x <<<< ………11分 （ⅱ）当()1,0x x ∈时，()012ln 2>-+=xmx x h ，02<-m x ，则()0'<x G ，故函数()x G 单调减；当()21,x x x ∈时，()012ln 2<-+=xmx x h ，02<-m x ，则()0'>x G ，故函数()x G 单调增；当()1,2x x ∈时，()012ln 2<-+=xmx x h ，02>-m x ，则()0'<x G ，故函数()x G 单调减；当()3,1x x ∈时，()012ln 2<-+=xmx x h ，02>-m x ，则()0'<x G ，故函数()x G 单调减；当()+∞∈,3x x 时，()012ln 2>-+=xmx x h ，02>-m x ，则()0'>x G ，故函数()x G单调增；综上，函数()x G 的单调递增区间是()21,x x ()+∞,3x ，单调递减区间是()1,0x ()1,2x ()3,1x 。
2024江苏省扬州中学高三下学期开学考数学试题及答案
2023~2024学年度第二学期开学检测高三数学一、选择题: 本题共8小题, 每小题5分, 共40分．在每小题给出的四个选项中,只有一项是符合题目要求的。
题目要求．全部选对的得6分，部分选对的得部分分，有选错的得0分．四、解答题：本题共5小题，共77分．解答应写出文字说明、证明过程或演算步骤。
22:4C y x =在第一象限与．分别作直线():l x t t a =>的18．（本题满分17分）某城市的青少年网络协会为了调查该城市中学生的手机成瘾情况，对该城市中学生中随机抽出的200名学生进行调查，调查中使用了两个问题．问题1：你的学号是不是奇数？问题2：你是否沉迷手机？调查者设计了一个随机化装置，这是一个装有大小、形状和质量完全一样的50个白球和50个红球的袋子，每个被调查者随机从袋中摸取一个球（摸出的球再放回袋中），摸到白球的学生如实回答第一个问题，摸到红球的学生如实回答第二个问题，回答“是”的人往一个盒子中放一个小石子，回答“否”的人什么都不要做．由于问题的答案只有“是”和“否”，而且回答的是哪个问题也是别人不知道的，因此被调查者可以毫无顾虑地给出符合实际情况的答案．(1) 如果在200名学生中，共有80名回答了“是”，请你估计该城市沉迷手机的中学生所占的百分比．(2) 某学生进入高中后沉迷手机，学习成绩一落千丈，经过班主任老师和家长的劝说后，该学生开始不玩手机．已知该学生第一天没有玩手机，若该学生前一天没有玩手机，后面一天继续不玩手机的概率是0.8；若该学生前一天玩手机，后面一天继续玩手机的概率是0.5．①求该学生第三天不玩手机的概率P ；②设该学生第n 天不玩手机的概率为n P ，求n P ．19．（本题满分17分）已知函数()22(ln )(1),f x x a x a =--∈R ．(1) 当1a =时，求()f x 的单调区间；(2) 若1x =是()f x 的极小值点，求a 的取值范围．2023~2024学年度第二学期开学检测高三数学参考答案12t=，e eFC，10，，回答“是”记为事件B，则。
江苏省扬州中学2020届高三下学期开学考试(2月)语文.docx
江苏省扬州中学高三年级第二学期开学检测语文试卷一、语言文字运用（15分）1.依次填入下列横线处的词语，最恰当的一项是( )曹雪芹创造性地吸收和运用了中国古代诗歌、绘画等艺术手法，使小说充满了诗情画意。
这既表现在宝黛共读《西厢》、黛玉葬花、宝钗扑蝶等众多优美场景的构思中，也表现在人物形象的塑造上。
例如林黛玉________的倩影、________的眉眼、________的低泣，以及她所住的那个________的潇湘馆，使她在群芳云集的大观园中，独具一种“风流态度”。
A．静谧高雅幽怨含情哀婉缠绵纤细清丽B．静谧高雅哀婉缠绵幽怨含情纤细清丽C．纤弱清丽幽怨含情哀婉缠绵静谧高雅D．纤细清丽哀婉缠绵幽怨含情静谧高雅2. 下列各句中，没有语病的一项是( )A.如果发生疑似心脏病的胸痛，切忌不要盲目走动，以防止病情进一步发展甚至猝死，最有效的办法是立即静卧。
B.总结历史教训，需要深入的民族自省。
甲午战争失败的内因，正是清末腐朽至极的观念、制度、官吏所导致的。
腐至而殇，腐盛而败，朽极而亡。
C. 新兴国家群体崛起，世界格局发生变化，美国如果不能改变高高在上的霸主心态，不能转变传统对抗的零和思维，发展空间就会受限。
D.法国和平艺术节组委会是一个在全球颇有影响的民间艺术，每年夏季，都要组织全球上百个艺术团体到法国各地巡回演出。
3. 某服装店开业，老板的朋友送来四副贺联，其中最得体的一副是( )A．锦绣乾坤真事业经纶山海大文章B．春满柜台宾客至货盈橱架利源开C．丹青夺造化之工粉黛染山川之色D．愿将天上云霞色化作人间锦绣裳4.依次填入下面一段文字横线处的语句，衔接最恰当的一组是( )中国书法艺术是中华大地上土生土长、地地道道的民族传统艺术。
__________。
__________，__________。
__________，__________，__________，是中国传统文化的精粹体现和辉煌标本。
①它至今仍是从头至尾、从里到外②始终保持着地道的中国作风与中国气派③在生成和发展的过程中，它与中国传统文化始终难解难分④唯独书法艺术的情况不一样⑤当然，中国传统文化对古往今来中国的人文、历史乃至一切事物都有深刻的渗透与影响⑥但那影响毕竟在逐渐淡化A．⑤⑥③④①②B．⑤⑥③④②①C．③⑤⑥④①② D．③⑤⑥②①④5. 阅读下面这幅漫画，对它的寓意理解不贴切的一项是( )A．多行不义必自毙，人类最终将自食因无知造成的恶果。
2020-2021学年江苏扬州高三下数学高考模拟有答案
2020-2021学年江苏扬州高三下数学高考模拟一、选择题1. 已知集合A={(x,y)|x2+y2=1}，B={(x,y)|y=2x+1}，则集合A∩B中元素的个数为( )A.3B.2C.1D.02. 若复数z1=1+2i，复数z2=1−i，则|z1z2|=()A.6B.√10C.√6D.√23. 已知函数f(x)=x2−2x+m．若p：f(x)有零点；q:0<m≤1，则p是q的（）A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分也不必要条件4. 已知角α是第三象限角，则α2终边落在( )A.第一象限或第二象限B.第二象限或第三象限C.第二象限或第四象限D.第一象限或第三象限5. 已知曲线y=ae x+x ln x在点(1,ae)处的切线方程为y=2x+b，则( )A.a=e, b=−1B.a=e, b=1C.a=e−1, b=1D.a=e−1，b=−16. 在△ABC中，角A，B，C的对边分别是a，b，c，若a：b：c=4：3：2，则2sin A−sin Bsin2C=( )A.3 7B.57C.97D.1077. 函数y=ln|x|−x2的图象大致为( )A. B.C. D.8. 已知菱形ABCD 的边长为4, ∠ABC =60∘，E 是BC 的中点， DF →=−2AF →，则AE →⋅BF →=( )A.24B.−7C.−10D.−12二、多选题下列说法中，正确的命题是( )A.已知随机变量X 服从正态分布N (2,σ2)， P (X <4)=0.8，则P (2<X <4)=0.2B.线性相关系数r 越大，两个变量的线性相关性越强；反之，线性相关性越弱C.已知两个变量具有线性相关关系，其回归直线方程为y =a ̂+b ̂x ，若b ̂=2，x ¯=1，y ¯=3，则a ̂=1D.若样本数据2x 1+1 ，2x 2+1，…，2x 16+1的方差为8，则数据x 1，x 2，…，x 16的方差为2下列不等式不一定成立的是( ) A.若a >b ，则a 2>b 2 B.若a >b >0，则b a <b+ma+m C.若ab =4，则a +b ≥4 D.若ac 2>bc 2，则a >b函数f (x )=A sin (ωx +φ)(A >0,ω>0,|φ|<π2)的部分图像如图所示，将函数f (x )的图像向左平移π3个单位长度后得到y =g (x )的图像，则下列说法正确的是( )A.函数g (x )为奇函数B.函数g (x )的最小正周期为πC.函数g (x )的图像的对称轴为直线x =kπ+π6(k ∈Z ) D.函数g (x )的单调递增区间为[−5π12+kπ,π12+kπ](k ∈Z )如图，在四棱锥P −ABCD 中，底面ABCD 为菱形，∠DAB =60∘．侧面PAD 为正三角形，且平面PAD ⊥平面ABCD ，则下列说法正确的是（ ）A.在棱AD 上存在点M ，使AD ⊥平面PMBB.异面直线AD 与PB 所成的角为90∘C.二面角P −BC −A 的大小为45∘D.BD ⊥平面PAC 三、填空题若sin x =−23，则cos 2x =________.已知向量OA →=(3，−4)，OB →=(6，−3)，OC →=(2m ，m +1)，若AB →//OC →，则实数m 的值为________.函数f(x)=ax 2+(b −2a)x −2b 为偶函数，且在(0,+∞)单调递增，则f(x)>0的解集为________．函数f(x)的导函数为f′(x)，对任意x∈R，都有f′(x)>f(x)成立，若f(ln2)=2，则满足不等式f(x)>e x的x的范围是________.四、解答题已知命题p：“∀−1≤x≤1，不等式x2−x−m<0成立”是真命题．(1)求实数m的取值范围；(2)若q:−4<m−a<4是p的充分不必要条件，求实数a的取值范围．已知向量m→=(−1,cosωx+√3sinωx)，n→=(f(x),cosωx)，其中ω>0，m→⊥n→，又函数f(x)的图象任意两相邻对称轴间距为32π．(1)求ω的值；(2)设α是第一象限角，且f(32α+π2)=2326，求sin(α+π4)cos(π+2α)的值．已知函数f(x)=log a x(a>0，且a≠1），且f(3)=1.(1)求a的值，并写出函数f(x)的定义域；(2)设函数g(x)=f(1+x)−f(1−x)，试判断g(x)的奇偶性，并说明理由；(3)若不等式f(t⋅4x)≥f(2x−t)对任意x∈[1,2]恒成立，求实数t的取值范围．如图，四棱锥P−ABCD的底面ABCD为直角梯形，其中BA⊥AD，CD⊥AD，CD= AD=2AB，PA⊥底面ABCD，E是PC的中点．(1)求证：BE // 平面PAD；(2)若BE⊥平面PCD，求平面EBD与平面BCD夹角的余弦值．中学为研究学生的身体素质与体育锻炼时间的关系，对该校200名高三学生平均每天体育锻炼时间进行调查，如表：（平均每天锻炼的时间单位：分钟）将学生日均体育锻炼时间在[40，60）的学生评价为“锻炼达标”.(1)请根据上述表格中的统计数据填写下面2×2列联表；并通过计算判断，是否能在犯错误的概率不超过0.025的前提下认为"锻炼达标"与性别有关？(2)在”锻炼达标“的学生中，按男女用分层抽样方法抽出10人，进行体育锻炼体会交流，①求这10人中，男生、女生各有多少人？②从参加体会交流的10人中，随机选出2人作重点发言，记这2人中女生的人数为X，求X的分布列和数学期望.参考公式：K2=n(ad−bc)2，其中n=a+b+c+d.(a+b)(c+d)(a+c)(b+d)临界值表已知函数f(x)=ln x+ax2−3x(a∈R)．(1)若函数f(x)在点(1, f(1))处的切线方程为y=−2，求函数f(x)的极值；(2)若a=1时，对于任意x1，x2∈[1, 10]，当x1<x2时，不等式f(x1)−f(x2)>m(x2−x1)恒成立，求实数m的取值范围．x1x2参考答案与试题解析2020-2021学年江苏扬州高三下数学高考模拟一、选择题1.【答案】B【解析】利用点到直线的距离公式得到d<r，则圆x2+y2=1与直线y=2x+1相交，有两个公共点，，集合A∩B中元素的个数为2个.【解答】解：∵圆x2+y2=1的圆心坐标为O(0,0)，半径为r=1，O(0,0)到直线y=2x+1的距离为d=√22+12=√55<r=1，∴圆x2+y2=1与直线y=2x+1相交，有两个公共点，∴集合A∩B中元素的个数为2个.故选B.2.【答案】B【解析】直接利用复数的模等于模的乘积求解．【解答】解：∵z1=1+2i，z2=1−i，∴|z1z2|=|1+2i|⋅|1−i|=√5×√2=√10．故选B.3.【答案】B【解析】利用判别式大于等于0求得m的范围，然后结合充分必要条件的判定方法得答案．【解答】解：函数f(x)=x2−2x+m有零点，则Δ=4−4m≥0，即m≤1．∴p不能推出q，但q能够推出p，∴p是q的必要不充分条件．故选B.4.【答案】C【解析】先根据α所在的象限确定α的范围，进而确定α2的范围，进而看当k为偶数和为奇数时所在的象限．【解答】解：∵解：∵α是第三象限角，即2kπ+π<α<2kπ+32π，k∈Z．当k为偶数时，α2为第二象限角；当k为奇数时，α2为第四象限角．故选C．5.【答案】D【解析】此题暂无解析【解答】解：y′=ae x+ln x+1,∵曲线y=ae x+x ln x在点(1,ae)处的切线方程为y=2x+b，∴ae+ln1+1=2,解得a=e−1.∴切线方程为y=2x−1，解得b=−1.故选D.6.【答案】D【解析】此题暂无解析【解答】解：设a=4k，b=3k，c=2k，根据余弦定理可知：cos C=a2+b2−c22ab =21k224k2=78，根据正弦定理可知2sin A−sin Bsin2C =2a−b2c×cos C=2×4k−3k2×2k×78=107.故选D.7.【答案】A【解析】【解答】解：令f(x)=y=ln|x|−x2，定义域为(−∞, 0)∪(0, +∞)，且f(−x)=ln|x|−x2=f(x)，所以函数y=ln|x|−x2为偶函数，因此图象关于y轴对称，故排除B，D；当x >0时，设g(x)=ln x −x 2， g ′(x)=1x −2x ，当x ∈(0,√22)时，g ′(x)=1x −2x >0，所以g(x)=ln x −x 2在(0,√22)上单调递增，故排除C . 故选A .8.【答案】 D【解析】 此题暂无解析 【解答】解：由已知得AF →=13AD →，BE →=12BC →，AD →=BC →，所以AE →=AB →+12AD →，BF →=AF →−AB →=13AD →−AB →. 因为在菱形ABCD 中，∠ABC =60∘， 所以 ∠BAD =120∘.又因为菱形ABCD 的边长为4， 所以AB →⋅AD →=|AB →|⋅|AD →|cos 120∘ =4×4×(−12)=−8，所以AE →⋅BF →=(AB →+12AD →)⋅(13AD →−AB →) =−|AB →|2−16AB →⋅AD →+16|AD →|2=−16−16×(−8)+16×16=−12.故选D . 二、多选题 【答案】C,D【解析】由正态分布的性质可判断A ，由相关系数的概念可判断B ，由回归方程过样本中心(x ¯，y ¯)可判断C ，由方差的性质可判断D . 【解答】解：对于A 选项，随机变量X 服从正态分布N (2,σ2)， P (X <4)=0.8， 则P (2<X <4)=P (X <4)−P (X <2)=0.8−0.5=0.3，故A 错误；对于B 选项，因为线性相关系数绝对值越大，两个变量的线性相关性越强，故B 错误； 对于C 选项，因为回归方程过样本中D (x ¯,y ¯)，所以有3=a +2×1，解得a ̂=1，故C 正确；对于D 选项，由方差的性质D (aX +b )=a 2D (X )，可得，若样本数据2x 1+1 ，2x 2+1，…，2x 16+1的方差为8，则数据x 1，x 2，…，x 16的方差为822=2，故D 正确． 故选CD . 【答案】 A,B,C【解析】本题考查不等式，考查推理论证能力． 【解答】解：对于A ，当a =−1，b =−2时，a 2<b 2，故选项A 不一定成立； 对于B ，ba −b+ma+m =b(a+m)−a(b+m)a(a+m)=(b−a)ma(a+m)，因为a >b >0，所以b −a <0， 当a +m >0，m <0时，(b−a)ma(a+m)>0，即ba >b+ma+m ，故选项B 不一定成立； 对于C ，当a =−1，b =−4时，a +b =−5，故选项C 不一定成立； 对于D ，因为ac 2>bc 2，所以c 2>0，所以a >b ，故选项D 一定成立． 故选ABC ． 【答案】 B,D【解析】根据函数f (x )的部分函数图像得到f (x )=3sin (2x −π3)，即可得到将函数g (x )=3sin (2x +π3)，再结合选项逐一判定即可得解. 【解答】解：依题意，A =3，3T 4=5π12+π3=3π4，∴ T =π， ∴ ω=2，∴ f (x )=3sin (2x +φ). 又∵ 函数图像过点(5π12,3), ∴ 3=3sin (2×5π12+φ)， ∴ 5π6+φ=2kπ+π2(k ∈Z )， ∴ φ=2kπ−π3.又∵ |φ|<π2， ∴ φ=−π3，∴ f (x )=3sin (2x −π3).将函数f (x )的图象向左平移π3个单位长度，得g (x )=3sin (2x +π3)，显然g (x )不是奇函数，故A 错误； 函数g (x )=3sin (2x +π3)的最小正周期T =2π2=π，故B 正确；由2kπ−π2≤2π+π3≤2kπ+π2(k ∈Z )，可得−5π12+kπ≤x ≤π12+kπ(k ∈Z )， ∴ g (x )的单调递增区间为[−5π12+kπ,π12+kπ](k ∈Z )，故D 正确．故选BD .【答案】 A,B,C 【解析】根据线面垂直，异面直线所成角的大小以及二面角的求解方法分别进行判断即可． 【解答】解：对于A ，如图取AD 的中点M ，连结PM ，BM ，∵ 侧面PAD 为正三角形， ∴ PM ⊥AD ，又底面ABCD 是菱形，且∠DAB =60∘， ∴ 三角形ABD 是等边三角形， ∴ AD ⊥BM ，∴ AD ⊥平面PBM ，故A 正确， 对于B ，∵ AD ⊥平面PBM ，∴ AD ⊥PB ，即异面直线AD 与PB 所成的角为90∘，故B 正确， 对于C ，∵ 底面ABCD 为菱形，∠DAB =60∘,平面PAD ⊥平面ABCD ， ∴ BM ⊥BC ，则∠PBM 是二面角P −BC −A 的平面角， 设AB =1，则BM =√32，PM =√32， 在直角三角形PBM 中，tan ∠PBM =PMBM =1，即∠PBM =45∘，故二面角P −BC −A 的大小为45∘，故C 正确； 对于D ，∵ BD 与PA 不垂直，∴ BD 与平面PAC 不垂直，故D 错误. 故选ABC ． 三、填空题 【答案】19【解析】由已知条件利用二倍角的余弦公式计算即可得到结果. 【解答】解：由二倍角的余弦公式可得：cos 2x =1−2sin 2x =1−2(−23)2=1−89=19.故答案为：19. 【答案】−3【解析】 此题暂无解析 【解答】解：∵ 向量OA →=(3，−4)，OB →=(6，−3)， ∴ AB →=(3，1)，OC →=(2m ，m +1)， 由AB →//OC →可得：3m +3=2m ， 解得m =−3． 故答案为：−3. 【答案】{x|x <−2或x >2} 【解析】 此题暂无解析 【解答】解：由已知得f(x)为二次函数且对称轴为y 轴， ∴ a ≠0，−b−2a 2a=0，即b =2a ，∴ f(x)=ax 2−4a ．再根据函数在(0,+∞)上单调递增， 可得a >0．令f(x)=0，求得x =2或x =−2， 故由f(x)>0，可得x <−2或x >2， 故解集为{x|x <−2或x >2}． 故答案为：{x|x <−2或x >2}． 【答案】 (ln 2,+∞) 【解析】造函数g(x)=f(x)e x，利用导数可判断g(x)的单调性，再根据f(ln2)＝2，求得g(ln2)＝1，继而求出答案．【解答】解：∵∀x∈R，都有f′(x)>f(x)成立，∴f′(x)−f(x)>0，于是有(f(x)e x)′>0，令g(x)=f(x)e x，则有g(x)在R上单调递增，∵不等式f(x)>e x，∴g(x)>1，∵f(ln2)=2，∴g(ln2)=1，∴x>ln2.故答案为：(ln2,+∞).四、解答题【答案】解：(1)由题意命题p：“∀−1≤x≤1，不等式x2−x−m<0成立”是真命题，∴m>x2−x在−1≤x≤1恒成立，即m>(x2−x)max，x∈(−1, 1)，因为x2−x=(x−12)2−14，所以−14≤x2−x≤2，即m>2，所以实数m的取值范围是(2, +∞).(2)由p得，设A={m|m>2}，由q得，设B={m|a−4<m<a+4}，因为q:−4<m−a<4是p的充分不必要条件，所以q⇒p，但p推不出q，所以B⫋A，所以a−4≥2，即a≥6，所以实数a的取值范围是[6, +∞)．【解析】(Ⅰ)分离出m，将不等式恒成立转化为函数的最值，求出(x2−x)max，求出m的范围．(Ⅱ)设p对应集合A，q对应集合B，“q是p的充分不必要条件”即B⫋A，求出a的范围【解答】解：(1)由题意命题p：“∀−1≤x≤1，不等式x2−x−m<0成立”是真命题，∴m>x2−x在−1≤x≤1恒成立，即m>(x2−x)max，x∈(−1, 1)，因为x2−x=(x−12)2−14，所以−14≤x2−x≤2，即m>2，所以实数m的取值范围是(2, +∞).(2)由p得，设A={m|m>2}，由q得，设B={m|a−4<m<a+4}，因为q:−4<m−a<4是p的充分不必要条件，所以q⇒p，但p推不出q，所以B⫋A，所以a−4≥2，即a≥6，所以实数a的取值范围是[6, +∞)．【答案】解：(1)由题意得m →⋅n →=0，所以，f(x)=cos ωx ⋅(cos ωx +√3sin ωx) =1+cos 2ωx 2+√3sin 2ωx2=sin (2ωx +π6)+12.根据题意知，函数f(x)的最小正周期为3π.又ω>0， 所以ω=13.(2)由(1)知f(x)=sin (23x +π6)+12， 所以f(32α+π2)=sin (α+π2)+12=cos α+12=2326， 解得cos α=513.因为α是第一象限角，故sin α=1213， 所以sin (α+π4)cos (π+2α)=sin (α+π4)−cos 2α=√2−2(cos α−sin α)=1314√2.【解析】（1）利用向量的数量积，而二倍角公式以及两角和的正弦函数，化简数量积为sin (2ωx +π6)+12，利用周期求出ω的值．（2）设α是第一象限角，且f(32α+π2)=2326，化简方程为cos α=513，求出sin α=1213，利用两角和的正弦函数，诱导公式化简sin (α+π4)cos (π+2α)并求出它的值． 【解答】解：(1)由题意得m →⋅n →=0，所以，f(x)=cos ωx ⋅(cos ωx +√3sin ωx) =1+cos 2ωx +√3sin 2ωx=sin (2ωx +π6)+12.根据题意知，函数f(x)的最小正周期为3π. 又ω>0， 所以ω=13.(2)由(1)知f(x)=sin (23x +π6)+12，所以f(32α+π2)=sin(α+π2)+12=cosα+12=2326，解得cosα=513.因为α是第一象限角，故sinα=1213，所以sin(α+π4 )cos(π+2α)=sin(α+π4)−cos2α=√2−2(cosα−sinα)=1314√2.【答案】解：(1)f(3)=loga3=1，故a=3.f(x)=log3x定义域为(0,+∞).(2)g(x)=f(1+x)−f(1−x)，∴{1+x>0，1−x>0，∴−1<x<1，g(−x)=f(1−x)−f(1+x)=−g(x)，∴g(x)为奇函数.(3)f(x)=log3x，∴f(x)是单调递增函数，f(t⋅4x)≥f(2x−t)，∴(t⋅4x)≥(2x−t)>0，∴t(4x+1)≥2x，∴t≥2x4x+1=12x+12x.令y=2x+12,x∈[1,2]时该函数为增函数，∴y min=2+12=52，∴ t≥152=25.又∵2x−t>0，∴t<(2x)min=2.综上t∈[25,2).【解析】答案未提供解析。
江苏省扬州中学2020┄2021届高三8月开学考试 英语
江苏省扬州中学2021届高三8月开学考试英语8一.听力（共两节，满分20分）第一节听下面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. What is the man going to buy？A. $5.5 for a red.B. $13.6 for two green ones.C. $11 for two red ones.2. What is the feeling of the man？A. He felt sleepy.B. He is tired of listening.C. The work is important.3. What is the man going to do for his holiday？A. Stay at home.B. Collect stamps.C. Volunteer in the west.4. Where does the conversation probably take place？A. In a plane.B. In a train.C. In a restaurant.5. Why didn’t Mary sleep well？A. She had a headache.B. She had a stomachache.C. She was troubled by noise.第二节听下面5段对话或独白。
每段对话或独白后有几个小题，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。
听每段对话或独白前，你将有时间阅读各个小题，每小题5秒钟；听完后，各小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
听第6段材料，回答第6至8题。
6. When will the man go to see the doctor？A. ON Tuesday.B. On Wednesday.C. On Thursday.7. What’s wrong with the man？A. He was hit by a ladder.B. He broke his leg.C. He hurt his foot.8. Which statement of the following is TRUE？A. The appointment time is quite fit to the man.B. The appointment time isn’t quite fit to the man.C. The man can’t go to the hospital.听第7段材料，回答第9至11题。
2021-2022学年江苏省扬州市扬州中学高三下学期开学摸底考检测语文试卷带讲解
根据“就其方式而言，美育更加强调具象性与实践性。美育不同于德育、智育、体育、劳育，就其本质而言，美育更加强调感通性与情感性。”可知美育的方式。
根据“就其目标而言，美育更加强调和谐性与整体性。”可知美育的本质。
5.①材料一分别从美育的核心、本质、方式、目标等方面阐述了“美育何以育人”，
②在此基础上，材料二结合互联网时代的美育特点以及我国目前审美存在的问题，强调要建立对“美的标准”的重构探索。
③两则材料先后从是什么、为什么源自怎么办三个角度，围绕“美育”这一核心话题进行了论述。
【1题详解】
本题考查对文本相关内容的理解和分析的能力。
江苏省扬州中学2022年开学考试
高三语文试卷
2022.02
一、现代文阅读（35分）
（一）现代文阅读I（本题共5小题，19分）
阅读下面的文字，完成各题。
材料一：
美育，即审美教育，又被称为美感教育，也因为美本身所具有的特殊性，而成为区别于德育、智育、体育、劳育的一种独特的教育形态。美育范畴认识的科学性和定位的准确性，不仅是美育研究和美育工作开展的逻辑起点，更关涉到人才培养、课程设置、美育目标、体系建设等美育实施环节的依据和宗旨。
故选C。
【3题详解】
本题考查把握文章内容、分析作者观点态度的能力。
B项，“明代文学家杨慎认为《江南春》‘千里莺啼绿映红’原本应是‘十里莺啼绿映红’，理由是：千里莺啼谁能听见？千里红绿谁能看见呢？”错误。杨慎对“千里莺啼绿映红”中的理解不符合题干所说美育的通感性与情感性，杜牧写“千里”并非一定就是眼前之景，它是有想象的成分在里面的，“千里”虽然看不见，但在诗人的审美世界里，眼前的江南春色可通过想象得到合理的延伸，“有一些东西并不是眼可见、耳可听，而是需要彼此的心意相通”，而杨慎是用实证的方法理解，即杨慎是“理性把握对象的属性与特征”。
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江苏省扬州中学2020-2021学年高三下学期开学考试（2月）英语试题注意事项1．考生要认真填写考场号和座位序号。
2．试题所有答案必须填涂或书写在答题卡上，在试卷上作答无效。
第一部分必须用2B 铅笔作答；第二部分必须用黑色字迹的签字笔作答。
3．考试结束后，考生须将试卷和答题卡放在桌面上，待监考员收回。
第一部分（共20小题，每小题1.5分，满分30分）1．The matches of the FIFA Women’s World Cup will be played in 2019 all around France, whose men’s team _____ the 2018 World Cup.A．wins B．wonC．has won D．had won2．Sympathy for the rebels, the government claimed, is beginning to ______because of more and more harm they did to common people.A．fade B．decline C．fail D．collapse3．He insisted what he did _______ right and that anyone who broke laws ________．A．was ; be punished B．be ; was punishedC．was; was punished D．be ; be punished4．—Jack, my plane arrives at 8:30 pm when, I suppose, you ______ dinner.—But I can wait.A．will have B．have hadC．will have had D．are having5．—I will be a vice president in a year or two.—You can’t be serious!_______.A．I can’t make it B．I can’t help it C．I won’t tell a soul D．I wouldn’t bet on it 6．—What a mess! You are always so lazy!—I’m not to blame, mum. I am ________ you have made me.A．how B．what C．that D．who7．-- Did Jim come?-- I don’t know. He _______ while I was out.A．might have come B．might comeC．must have come D．should have come8．Nature is understandable in the sense ______ she will answer truly and reward with discoveries when we ask her questions via observation.A．that B．whereC．how D．what9．I like such houses with beautiful gardens in front, but I don’t have enough money to buy.A．it B．one C．that D．this10．Not until they left school________how much their teachers loved them and helped them.A．they realized B．did they realizedC．the would realized D．had they realized11．Professor Li ________ for his informative lecture, was warmly received by the students.A．known B．knowingC．having known D．to be known12．Join us and you will discover an environment ______ you can make the most of your skills and talents. A．that B．whereC．how D．what13．—What’s that noise?—Oh，I forget to tell you.The new machine________.A．is testing B．was being testedC．is being tested D．has been tested14．——What was wrong? Why didn’t you go to the picnic as scheduled?——I’m sorry. I _________ a seriously-injured old man to the hospital.A．would deliver B．deliveredC．had delivered D．was delivering15．Obama didn’t explain ______ any larger principles have guided him through the historic convulsions of the 2011 Arab Spring.A．what B．that C．where D．whether16．Egyptian President decided to ______on Friday afternoon after an 18-day campaign against him, ending his thirty-year rule.A．step down B．break in C．break down D．step in17．The book ______ through the air to Aunt Dede and she began to read it aloud.A．got B．pushed C．sailed18．In 2012, Sun Yang became the first Chinese man _____ an Olympic gold medal in swimming. A．winning B．to winC．having won D．being won19．---Where is the plane？I can't see it.---It went off its ________________ to keep away from the sudden storm.A．course B．roadC．flight D．direction20．An old lady came to the bus stop only the bus had gone.A．to run ; to find B．running;to find C．and ran ; finding D．running; finding第二部分阅读理解（满分40分）阅读下列短文，从每题所给的A、B、C、D四个选项中，选出最佳选项。
21．（6分）Conformity involves changing your behaviour to “fit in” or “go along” with the peoplearound you. In some cases, this social influence might involve agreeing with or acting like the majority of people in a specific group, or involve behaving in a particular way to be viewed as “normal” by the group. Actually, conformity is someth ing that happens regularly in our social world. Sometimes we’re aware of our behaviour, but in many cases it happens without much thought or awareness on our parts. And even in some cases we go along with things we disagree with or behave in ways we know w e shouldn’t.There are countless examples of conformity in life. For instance, a teenager dresses in a certain style because he wants to fit in with the rest of the teenagers in his social group. A woman reads a book for her book club and really enjoys it. When she attends her book club meeting, the other members all dislike the book. Rather than go against the group opinion, she simply agrees with the others that the book is terrible.Why do we conform? In many cases, looking to the rest of the group for clues for how we should behave can actually be helpful. Others might have greater knowledge or experience than we do, so following their lead can actually be instructive. And in some cases, we conform to the expectations of the group to avoid looking foolis h. This tendency can become especially strong in situations where we aren’t quite sure how to act. Additionally, there are some situations where we conform just in order to avoid punishments or gain rewards.There are many factors that influence conformity. For example, not knowing how to perform a difficult task makes people more likely to conform. Personal characteristics such as motivation to achieve and strong leadership abilities are linked with a decreased tendency to conform. And people are more likely to conform in situations that involve between three and five other people.1、What can we learn about conformity from the first paragraph?A．Actually it is rare for us to conform.B．Blind conformity should be criticized.C．Sometimes people tend to conform unconsciously.D．Conformity means changing your behaviour reluctantly.2、Which of the following is a case of conformity?A．Joining a certain organization.B．Following the ongoing fashion.C．Sharing something personal with others.D．Giving up your creative ideas cautiously.3、When are we most likely to conform according to the text?A．When we are at a loss.B．When we are rejected.C．When we feel threatened.D．When we feel uncomfortable.4、What can be inferred from the last paragraph?A．Conformity cannot be avoided by anyone.B．Conformity is linked to the size of the group.C．Conformity is a natural response to challenges.D．Conformity isn’t related to individual differences.22．（8分）While staring out of the window during a flight, not everyone will think carefully about the question why airplanes have rounded windows rather than square ones.Over the years, aerospace engineering has made huge steps in airplane technology, meaning planes can carry more passengers and go faster. The planes have also changed shape to increase safety—including the windows. As commercial air travel took off in the mid－20th century, airline companies began to fly at higher altitudes to lower their cost—the air density(密度) is lower up there, creating less drag(阻力)for airplanes. However, higher altitudes came with problems, like the fact human beings can’t r eally survive at 30，000 feet. To make that possible, the cabin was changed to a cylindrical(圆柱体) shape to support the pressure inside. But at first, plane builders left in the standard square windows and this expansion meant disaster. The de__Havilland__Comet came into fashion in the 1950s. With a closed cabin, it was able to go higher and faster than other aircraft.However, where there’s a corner, there’s a weak spot. Windows, having four corners, have four potential weak spots, making them likely to crash under stress—such as air pressure. By curving the window, the stress that would eventually break the window corner is distributed and the chance of it breaking is reduced. Rounded shapes are also stronger and resist deformation(变形), and can thus survive the extreme differences in pressure between the inside and outside of the aircraft.Fortunately, designers figured out the lack of design pretty quick. Now we have nice, rounded airplane windows that can resist the pressure of traveling altitude. It gives being able to gaze out of your window to the world from 35，000 feet a whole new outlook, doesn’t it?1、Why did airlines aim to fly at higher altitudes?A．To increase safety of the plane.B．To help the plane to take off.C．To save money for less drag.D．To carry more passengers and go faster.2、What does the underlined words “de Havilland Comet” in Paragra ph 2 refer to?A．A planet. B．A band. C．An aircraft. D．A design.3、Which is the advantage of the rounded window?A．It reduces the possibility of breaking up.B．It weakens the strength of air pressure.C．It increases the air pressure.D．It helps to survive the extreme weather.4、Where does this text come from?A．A newspaper on safe driving. B．A magazine on fashion design.C．A website on survival skills. D．A science book on flying.23．（8分）Surfing: Famous Beaches, Famous WavesNorth Shore, Oahu, Hawaii: Famous for being the birthplace of surfingWide, sandy beaches stretch nearly 20 miles along the Pacific Ocean. Between December and February, this surfing destination is suitable only for experienced surfers, as its big waves can reach 30 feet. In summer, the ocean can be almost completely flat, making it perfect for swimming.Huntington Beach, California: Famous for the US Open of Surfing competitionThis busy 8.5-mile-long beach attracts 8 million visitors a year for bodysurfing, boogie boarding （趴板冲浪）, and board surfing at every level, beginner to expert. At night, the beach’s fire pits （深坑）draw families as much as the waves do during the day. The best time for surfers is winter, when the swells can hit 15 feet.Jeffreys Bay, South Africa: Famous for being the setting of the classic movie The Endless Summer This area of the ocean may contain the most consistent waves on the planet, with some up to 10 feet. The best waves are between late May and late August. The beach sometimes closes because of sharks, but at other times, surfers are lucky enough to surf alongside dolphins.Tamarindo, Costa Rica: Famous for being featured in the movie The Endless Summer ⅡThis beach has waves up to 12 feet high, which are good for long-boarders or short-boarders, beginners or experts, with the best waves from April to July. Bodysurfing is not recommended because of offshore rocks. The laid-back atmosphere and nearly perfect year-round weather make it feel like the California beaches of the 1950s.1、The beach with the biggest waves is ______.A．Oahu’s North Shore B．Huntington BeachC．Jeffreys Bay D．Tamarindo2、Jeffreys Bay and Tamarindo are both known for ______.A．offshore rocks B．competitionsC．movies D．dolphins3、How are North Shore and Huntington Beach similar?A．Both beaches are the same length.B．Both are suitable for all surfing levels.C．The waves are both perfect for bodysurfing.D．Winter is the best time to surf at both beaches.4、What does the underlined word “laid-back” mean in English?A．Worried. B．Relaxed.C．Excited. D．Depressed.5、The Huntington Beach information differs from the others in that it includes ______.A．the height of the waveB．the weather conditionsC．the best time to visitD．the number of visitors each year24．（8分）For the business traveler who’s all about efficiency: check out these hotels that will get you in and out with a minimum trouble.When you’re pressed for time on a business trip, nothing can annoy you more than a slow hotelcheck-in process. On your next trip, try these hotels that offer a speedier check-in process.◆ Marriott Detroit AirportAnother option for business travelers in a hurry: Marriott is rolling out its mobile check-in app to 325 hotels this year, including the Marriott Detroit Airport hotel. (I’ve tested the app itself but not for a real visit quite yet.) here is the basic idea: you download the iPhone or Android app. The night before, you can “check-in” virtually. When you arrive, you get an alert that the room is ready and your key, which is already tied to your reservation, is waiting for you at the desk.◆Hyatt Regency MinneapolisI happened to stay at this hotel recently and liked how fast the kiosk check-in works. The kiosk asks you to insert your credit card, similar to an airport terminal. The whole process took about 3 minutes. When I left, I was equally impressed with the fast check-out: An agent meets you in the lobby with, an iPad and asks for an email to use for a receipt. The big advantage: you never have to wait in line.◆Radisson LaCrosseThe Radisson is trying to make the kiosk process even faster. At a few select hotels like the Radisson Lacrosse in Wisconsin，you use a mobile app to register and then receive a barcode by email or text. When you get to the kiosk, you can scan the barcode to get your key without any other steps required. It’s sup er fast. You can find this new check-in system at the Radisson hotels in Salt Lake City, Seattle, and Phoenix as well.1、Which two hotels offer a mobile app for customers to check in?A．Marriott Detroit Airport and Yotel New York .B．Marriott Detroit Airport and Radisson LaCrosse.C．Marriott Detroit Airport and Hyatt Regency Minneapolis.D．Hyatt Regency Minneapolis and Radisson LaCrosse.2、Which hotel will send you a receipt by email?A．Y otel New York. B．Marriott Detroit Airport.C．Radisson LaCrosse. D．Hyatt Regency Minneapolis.3、What is the best title for the passage?A．Checking out the hotels will make you in troubleB．Three hotels that will make your life easierC．Try these hotels that offer you comfortD．Hotels for the travelers25．（10分）Are you a true “cheapskate” (吝啬鬼)? You are if you do this when traveling: You always use a carry-on to avoid the checked-bag fee. You pack a lunch from home so you don’t pay for airline meals. True “cheapskates” also fly on the three cheapest days of the week (Tuesdays, W ednesdays and Saturdays).◇ For U. S. domestic travel, everyone wants to fly Fridays and Sundays and those are usually the most expensive days to fly.﹡The fares for Los Angeles-to-New York flights in September:• Friday & Sunday: $ 365• Tuesday & Wednes day: $ 288﹡The fares for Houston-to-New York flights in September:• Friday & Sunday: $ 302• Tuesday & Wednesday: $ 226◇ For transatlantic flights, the difference can be more startling and the rules a little looser.﹡Boston-to-Dublin fares from United Airlines:• Friday & Sunday: $ 581• Tuesday & Wednesday: $ 457﹡The fares for Washington-to-London flights in a week:• Friday & Sunday: $ 896• Tuesday & Wednesday: $ 732﹡It’s time for some Saturday fares now. This example features Chicago-to-Atlanta routes.• Friday & Sunday: $ 207• Saturday: $ 89﹡There were cheaper fares for this trip,but only if the Sunday flight departed at 5 am or earlier. Is the rule on the “cheapest three days to fly” etched in stone? No. It is usually true but not always. S o you should be flexible. After all it can save much money. And isn’t that what being a “cheapskate” is all about?1、According to Para. 1, what does a true “cheapskate” really do?A．Choose a luxury hotel to live in.B．Save money in every way during the trip.C．Fly on Fridays when traveling.D．Have their lunch on the plane.2、How much should you pay for a flight to New York from Houston on Tuesday?A．$ 226. B．$ 288.C．$ 302. D．$ 365.3、What is the purpose of the last paragraph?A．To recommend some cheap trips. B．To call on us to fly on Sundays.C．To introduce another topic. D．To explain a special case.第三部分语言知识运用（共两节）第一节（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A、B、C和D四个选项中，选出可以填入空白处的最佳选项．26．（30分）Adventure is in my blood. And I had been considering how I was going to 1 my high school graduation. I didn’t just want a small party in the backyard. I started thinking about doing a solo2 somewhere out of the ordinary. I took out maps and drew the 1,500-mile route along which I would be3 from the northernmost point in Norway to the southernmost section of Sweden. When I4 my plans with my dad, he5 as I thought he would. Because I get my6 spirits from him, he was7 it.I had only been away from my home three days now, but there was an inner 8 going on inside of me. Part of me was 9 and doubting whether I 10 could make it. The other part of me was ready to 11 to myself and my family that I could do it by myself.On the road, I met another bicyclist who was quite a bit older than I was. He had started his journey12 by bike at the southern part of Norway and had just finished. I could tell he had a great sense of13 . It encouraged me not to 14 .As I listened to my 15 artists on my MP4 player, I pedaled (踩踏板) with my feet. There was 16 around me for miles. 17 , that wasn’t entirely true. There were mosquitoes—millions of them. My arms were so dotted with 18 that they looked like a topographical map (地形图). But, however 19 it would be, nothing could stop my advance 20 the destination. As you know, adventure is in my blood.1、A．celebrate B．finish C．spend D．organize2、A．flight B．activity C．performance D．trip3、A．walking B．flying C．biking D．jogging4、A．provided B．shared C．exchanged D．compared5、A．agreed B．sighed C．teased D．obtained6、A．aggressive B．adventurous C．optimistic D．athletic7、A．in fear of B．in charge of C．in favor of D．in need of8、A．battle B．dilemma C．request D．discussion9、A．stubborn B．ambitious C．homesick D．astonished10、A．naturally B．really C．extremely D．obviously11、A．submit B．turn C．prove D．adapt12、A．alone B．practically C．patiently D．sincerely13、A．humor B．direction C．balance D．satisfaction14、A．calm down B．break down C．keep on D．give up15、A．personal B．favourite C．professional D．grateful16、A．nobody B．everybody C．anything D．everything17、A．Simply B．Actually C．Eventually D．Fortunately18、A．wounds B．cuts C．bites D．burns19、A．boring B．confusing C．complex D．tough20、A．from B．with C．in D．towards第二节（每小题1.5分，满分15分）阅读下面材料，在空白处填入1个适当的单词或括号内单词的正确形式。