江西省新余一中、樟树中学等六校2019-2020学年高一(创新班)下学期第二次联考英语试题含答案
2019-2020学年江西省新余一中、樟树中学等六校高一下学期第二次联考数学(理)试题(解析版)
2019-2020学年江西省新余一中、樟树中学等六校高一（创新班）下学期第二次联考数学（理）试题一、单选题1．已知集合{0A =，1，2，3}，2{|1B y y x ==+，}x R ∈，P A B =⋂，则P 的子集个数( ) A ．4 B ．6 C ．8 D ．16【答案】C【解析】求出集合B ，然后计算出集合P ，得出元素个数即可求出子集个数 【详解】{|1}B y y =，{0A =，1，2，3}；{1P AB ∴==，2，3}；P ∴的子集个数为：328=． 故选C ．【点睛】本题考查了求子集个数问题，较为基础2．已知1m ，12log a m =，12mb ⎛⎫= ⎪⎝⎭，12c m =，则（ ）A ．a b c <<B ．a c b <<C ．b a c <<D ．b c a <<【答案】A【解析】分别求出1m 时a ，b ，c 的范围，再根据a ，b ，c 的大小得到选项. 【详解】当1m 时，由对应函数的单调性可知，1122log log 10a m =<=，1112212m b ⎛⎫⎛⎫=< ⎪ ⎪⎝⎭⎝⎭=且201mb ⎛⎫= ⎪⎭>⎝，121c m =>，排序得a b c <<. 故选：A . 【点睛】本题考查了利用函数单调性比较函数值的大小，属于基础题.3．下列四个函数中，与函数()tan f x x =完全相同的是（ ）A ．22tan21tan 2x y x=- B ．1cot y x=C ．sin 21cos 2xy x=+D ．1cos 2sin 2xy x-=【答案】C【解析】先判断函数的定义域是否相同，再通过化简判断对应关系是否相同，从而判断出与()f x 相同的函数. 【详解】()f x 的定义域为|,2x x k k Z ππ⎧⎫≠+∈⎨⎬⎩⎭，A. 22tan 21tan2x y x =-，因为tan 12,22x x k k Z ππ⎧≠±⎪⎪⎨⎪≠+∈⎪⎩，所以,24,22x k k Z x k k Z ππππ⎧≠±+∈⎪⎪⎨⎪≠+∈⎪⎩， 定义域为{|22x x k ππ≠±或2,}x k k Z ππ≠+∈，与()tan f x x =定义域不相同；B. 1cot y x =，因为cos 0sin 0x x ≠⎧⎨≠⎩，所以,2,x k k Zx k k Zπππ⎧≠+∈⎪⎨⎪≠∈⎩， 所以定义域为,2k x x k Z π⎧⎫≠∈⎨⎬⎩⎭，与()tan f x x =定义域不相同； C. sin 21cos 2x y x =+，因为1cos20x +≠，所以定义域为|,2x x k k Z ππ⎧⎫≠+∈⎨⎬⎩⎭，又因为2sin 22sin cos tan 1cos 22cos x x xy x x x===+，所以与()tan f x x =相同； D. 1cos 2sin 2xy x-=，因为sin 20x ≠，所以2,x k k Z π≠∈，定义域为|,2k x x k Z π⎧⎫≠∈⎨⎬⎩⎭， 与()tan f x x =定义域不相同. 故选：C. 【点睛】本题考查与三角函数有关的相同函数的判断，难度一般.判断相同函数时，首先判断定义域是否相同，定义域相同时再去判断对应关系是否相同（函数化简），结合定义域与对应关系即可判断出是否是相同函数.4．已知函数()()log 130,1a y x a a =-+>≠且的图象恒过点P ，若角α的终边经过点P ，则2sin sin 2αα-的值等于（ ）A ．313 B ．513 C ．313-D ．513-【答案】C【解析】【详解】试题分析：由题意可得定点坐标为(2,3)P , 由三角函数的定义可得3tan 2α=, 所以222tan 2tan 3sin sin 21tan 13ααααα--==-+. 故选：C.【考点】对数函数的图象和性质及三角函数的定义与同角三角函数的关系. 5．已知数列{}n a 的前n 项和为n S ，若()2*12n n S S n n ++=∈N ，且1028a=，则2a =（ ） A ．－5 B ．－10 C ．12 D ．16【答案】C【解析】由题意利用递推关系式确定数列为隔项等差数列，然后结合10a 的值可得2a 的值. 【详解】由题意可得：212n n S S n ++=，()2121n n S S n -+=-，两式作差可得：()122142n n a a n n ++=-=-， ① 进一步有：()141246n n a a n n -+=--=-， ② ①-②可得：114n n a a +--=，故数列的偶数项为等差数列，且公差为4，据此可得：1024a a d =+，即：22844a =+⨯，解得：212a =. 故选C. 【点睛】给出n S 与n a 的递推关系，求a n ，常用思路是：一是利用1n n n S S a +-=转化为a n 的递推关系，再求其通项公式；二是转化为S n 的递推关系，先求出S n 与n 之间的关系，再求a n .6．已知等差数列{a n }首项为a ，公差为1，1n n na b a +=，若对任意的正整数n 都有b n ≥b 5，则实数a 的取值范围是（ ） A ．()(),43,-∞-⋃-+∞ B ．()4,3-- C ．()(),54,-∞-⋃-+∞ D ．()5,4--【答案】D【解析】直接利用数列的递推关系式求出函数的关系式，进一步利用函数的最小值和不等式的解法求出结果． 【详解】解：等差数列{a n }首项为a ，公差为1， 所以a n =a +n -1， 所以11n n n a a nb a a n ++==+-， 则111111n n n n a b a a a n +==+=++-， 若对任意的正整数n 都有b n ≥b 5， 所以（b n ）min =b 5=114a ++， 所以5456b b b b ⎧⎨⎩＜＜，解得-5＜a ＜-4． 故选D ． 【点睛】本题考查的知识要点：数列的通项公式的求法及应用，函数的关系式的应用，恒成立问题的应用，主要考查学生的运算能力和转换能力及思维能力，属于中档题．7．黄金三角形有两种，一种是顶角为36°的等腰三角形，另一种是顶角为108°的等腰三角形，例如，正五角星可以看成是由一个正五边形剪去五个顶角为108°的黄金三角形，如图所示，在黄金三角形ABC中，AB AC =根据这些信息，可得cos144︒=（ ）A ．154- B ．358+-C ．15+ D ．458+-【答案】C 【解析】先根据51AB AC -=求得cos36︒，再根据诱导公式得结果. 【详解】因为1108(180108)362ABC BAC ∠=∴∠=-= 因此11512cos362451ACAB ===- 从而51cos144cos364︒=-=- 故选：C 【点睛】本题考查诱导公式、等腰三角形性质，考查基本分析求解能力，属基础题. 8．在ABC 中，角A ，B ，C 所对的边分别为a ，b ，c ，若直线cos cos 0bx y A B ++=,cos cos 0ax y B A ++=平行，则ABC 一定是（ ）A ．锐角三角形B ．等腰三角形C ．直角三角形D ．等腰或者直角三角形【答案】C【解析】解法一根据直线的平行关系，结合正弦定理即可求得A 与B 的关系，根据直线平行又不重合的条件即可判断三角形形状；解法二根据直线平行关系得到cos cos 0b B a A -=，由余弦定理转化为边的表达式，进而利用因式分解可得a b 、的关系，根据平行又不重合的条件即可得三角形形状． 【详解】解法一：由两直线平行可得cos cos 0b B a A -=由正弦定理可知sin cos sin cos 0B B A A -=，即11sin 2sin 222A B = 又,(0,)A B π∈，且(0,)A B π+∈所以22A B =或22A B π=＋，即A B =或2A B π+=.若A B =，则a b =，cos cos A B =，此时两直线重合，不符合题意，舍去 故2A B π+=，则ABC 是直角三角形故选C.解法二：由两直线平行可得cos cos 0b B a A -=，由余弦定理得22222222b c a a c b a b bc ac+-+-⋅=⋅所以()()22222222a b c a b a c b +-=+- 所以()()()2222222cab a b a b -=+-所以()()222220a bab c -+-=所以a b =或222+=a b c若a b =，则两直线重合，不符合题意，故222+=a b c 则ABC 是直角三角形 故选C. 【点睛】本题考查了正弦定理与余弦定理在判断三角形形状中的应用，注意边角转化的应用，直线平行时不重合的条件限制，属于中档题．9．已知ABC 三内角,,A B C 的对边分别为,,a b c cos sin 0A a C +=，若角A 的平分线交BC 于D 点，且1AD =，则b c +的最小值为( )A ．2B ．C ．4D ．【答案】C【解析】由已知，易得120A =︒，再利用ABCABDACDSSS=+得到bc b c =+，即111b c+=，然后利用“1”的代换并结合基本不等式可得到答案. 【详解】cos sin 0A a C +=cos sin sin 0C A A C +=，因为(0,180)C ∈︒，sin 0C ≠，所以3cos sin 0A A +=，即tan 3A =-， 因为(0,180)A ∈︒，所以120A =︒. 如图，ABCABDACDSSS=+，所以111sin1201sin 601sin 60222bc c b ⋅︒=⋅⋅︒+⋅⋅︒， 所以bc b c =+，即111b c+=， ∴()112224b c b c b c b c c b c b ⎛⎫+⋅+=++≥+⨯=⎪⎝⎭，当且仅当c b =，bc b c =+，即2c b ==时，等号成立， 所以b c +的最小值为4. 故选：C.【点睛】本题考查正弦定理在解三角形中的应用，涉及到基本不等式求最值，考查学生的数学运算求解能力，是中档题.10．已知BC 是圆O 的直径，H 是圆O 的弦AB 上的一动点，10BC =，8AB =，则HB HC ⋅的最小值为( )A ．4-B ．25-C ．9-D ．16-【答案】D【解析】 以BC 所在的直线为x 轴，线段BC 的垂直平分线为y 轴，建立平面直角坐标系，设点,()H x y ，则(5,0),(5,0)B C -， 所以(5,),(5,)HB x y HC x y =---=--则22(5,)(5,)25HB HC x y x y x y ⋅=---⋅--=+-，又因为8AB =，且H 在弦AB 上一动点，所以22925x y ≤+≤，其中当取AB 的中点时取得最小值，所以92516HB HC ⋅=-=-，故选D ． 点睛：本题考查了平面向量的数量积与应用问题，解答的关键是建立适当的直角坐标系，表示出向量,HB HC 的坐标，再利用圆的性质求解，着重考查了学生分析问题和解答问题的能力，对于平面向量的运算问题，通常有两种方法：一是建立平面的基底，利用基底运算；二是建立适当的平面直角坐标系，转化为坐标运算即可．11．已知函数()14sin cos f x x x =-，下列结论错误的是( ) A ．()f x 的最小正周期为πB ．曲线()y f x =关于直线π4x =-对称C ．()f x 在π5π,412⎛⎫⎪⎝⎭上单调递增 D ．方程()2f x []π,π-上有4个不同的实根【答案】C【解析】化简得()2sin 21f x x =-，进而利用三角函数图象的平移变换及翻折变换，可作出()f x 的图象，结合四个选项可选出答案. 【详解】由题意，()14sin cos 2sin 21f x x x x =-=-，2sin 2y x =的图象向下平移1个单位可得到2sin 21y x =-的图象，将所得图象在x 轴下方的部分沿x 轴翻折，可得到()f x 的图象，如下图所示，由图可知()f x 的最小正周期为π，即A 正确；曲线()y f x =关于直线π4x =-对称，即B 正确；()f x 在π5π,412⎛⎫⎪⎝⎭上单调递减，即C 错误；方程()2f x =在[]π,π-上有4个不同的实根，即D 正确.故选：C.【点睛】本题考查三角函数的恒等变换，考查三角函数图象的平移变换与翻折变换，考查三角函数的性质，考查学生的推理能力，属于中档题.12．已知正方体1111ABCD A B C D -的棱长为1，P 是空间中任意一点，下列说法错误的个数是（ ）①若P 为棱1CC 中点，则异面直线AP 与CD 所成角的正切值为5；②若P 在线段1A B 上运动，则1AP PD +的最小值为622+；③若P 在半圆弧CD 上运动，当三棱锥P ABC -的体积最大时，三棱柱P ABC -外接球的表面积为2π；④若过点P 的平面α与正方形每条棱所成角相等，则α截此正方体所得截面面积的最大值为33A ．1个B ．2个C ．3个D ．4个【答案】A【解析】根据异面直线的夹角求解，棱锥外接球的求解，以及正方体截面的性质，对选项进行逐一分析即可. 【详解】对于①，如图所示，由//AB CD ，可知BAP ∠即为异面直线AP 与CD 所成的角． 设正方体的棱长为2，连接BP ，则在RT BAP 中，2AB =，2222215BP BC CP =+=+=5tan 2BP BAP AB ∠==，故①正确 对于②，将三角形1AA B 与四边形11A BCD 沿1A B 展开到同一个平面上，如图所示．由图可知，线段1AD 的长度即为1AP PD +的最小值． 在11AA D 中，利用余弦定理可得122AD =+，故②错误.对于③，如下图所示：当P 为CD 中点时，三棱锥P ABC -体积最大， 此时，三棱锥P ABC -的外接球球心是AC 中点，半径为22﹐其表面积为2π．故③正确.对于④﹐平面α与正方体的每条棱所在直线所成的角都相等，只需与过同一顶点的三条棱所成的角相等即可，如图所示：AP AR AQ==．则平面PQR与正方体过点A的三条棱所成的角相等．若点E，F，G，H，M，N分别为相应棱的中点，可得平面EFGHMN平行于平面PQR，且六边形EFGHMN为正六边形．正方体棱长为1，所以正六边形EFGHMN的边长为22，33故④正确．故选：A【点睛】本题考查异面直线夹角的求解，棱锥外接球表面积，正方体截面问题，属较难题.二、填空题13．已知,x y满足约束条件3442x yyx y+≥⎧⎪≤⎨⎪-≤⎩，若(0)Z ax y a=+>的最大值是16，则a 的值为_________.【答案】2【解析】画出满足约束条件可行域，求出A，B的坐标，由Z ax y=+得：y ax Z=-+，结合函数的图象确定直线y ax Z=-+过A时，Z最大，求出a的值即可．【详解】解：画出满足约束条件3442x yyx y+≥⎧⎪≤⎨⎪-≤⎩的平面区域，如图示：由2040x yy--=⎧⎨-=⎩，解得：(6,4)A，由34040x yy+-=⎧⎨-=⎩，解得：(0,4)B，当直线y ax Z=-+过(0,4)B时，416Z=≠由Z ax y=+得：y ax Z=-+，当直线y ax Z=-+过(6,4)A 时，Z最大，此时，6416a+=解得：2a=故答案为：2【点睛】本题考查了简单的线性规划问题，考查数形结合思想，属于中档题．14．已知向量()2,1a=-，()6,b x=，且//a b，则2a b-=_________.5【解析】先根据向量平行求出x，再根据向量模的定义求结果.【详解】//263a b x x∴=-∴=-2(2,1)5a b∴-=-=5【点睛】本题考查向量平行坐标表示、向量模的定义，考查基本分析求解能力，属基础题. 15．若偶函数()f x 的图像关于32x =对称，当30,2x ⎡⎤∈⎢⎥⎣⎦时，()f x x =，则函数20()()log ||g x f x x =-在[20,20]-上的零点个数是__________.【答案】26【解析】先确定函数()f x 周期，再根据周期作()y f x =与20log ||y x =图象，最后根据交点个数确定结果. 【详解】因为偶函数()f x 的图像关于32x =对称， 所以()(3),()()(3)()f x f x f x f x f x f x =-=-∴-=-∴()f x 周期为3 因为当30,2x ⎡⎤∈⎢⎥⎣⎦时，()f x x =，所以当3,02x ⎡⎤∈-⎢⎥⎣⎦时，()()f x f x x =-=-， 因此当3,32x ⎡⎤∈⎢⎥⎣⎦时，()(3)(3)f x f x x =-=--，因为20log 201,(20)(2)1f f ===所以当[]0,3x ∈时，有一个交点；当(3,6]x ∈时，有两个交点；当(6,9]x ∈时，有两个交点；当(9,12]x ∈时，有两个交点；当(12,15]x ∈时，有两个交点；当(15,18]x ∈时，有两个交点；当(18,20]x ∈时，有两个交点；从而当[0,20]x ∈时，共有16213+⨯=个交点；（原点不是交点） 根据偶函数对称性，当[20,20]x ∈-时，共有21326⨯=个交点；故答案为：26 【点睛】本题考查函数周期、偶函数性质、函数零点个数，考查数形结合思想方法，属基础题. 16．在ABC 中，()()3cos ,cos ,cos ,sin AB x x AC x x ==，则ABC 面积的最大值是____________ 【答案】34【解析】计算113sin 22624ABC S x π⎛⎫=--≤ ⎪⎝⎭△，得到答案. 【详解】()22211sin ,1cos ,22ABC S AB AC AB AC AB ACAB AC=⋅=⋅-△()()222222114cos 3cos sin cos 22AB AC AB ACx x x x=⋅-⋅=-+211133sin cos sin 222624x x x x π⎛⎫=-=--≤ ⎪⎝⎭， 当sin 216x π⎛⎫-=- ⎪⎝⎭时等号成立.此时262x ππ-=-，即6x π=-时，满足题意. 故答案为：34. 【点睛】本题考查了三角形面积的最值，向量运算，意在考查学生的计算能力和综合应用能力.三、解答题17．如图，在四棱锥P ABCD -中，底面ABCD 为菱形，2AB =,060BAD ∠=，面PAD ⊥面ABCD ,PAD ∆为等边三角形，O 为AD 的中点．(1)求证：AD ⊥平面POB ；(2)若E 是PC 的中点，求三棱锥P EDB -的体积． 【答案】（1）详见解析（2）12【解析】（1）由AD PO ⊥，AD BO ⊥结合线面垂直的判定即可得证； （2）由E 是PC 的中点，所以12P EDB P BCD V V --=，则将求三棱锥P EDB -的体积转化为求三棱锥P CDB -的体积，再由条件即可得解. 【详解】(1)证：因为O 为等边PAD ∆中边AD 的中点， 所以AD PO ⊥,又因为在菱形ABCD 中，060BAD ∠=， 所以ABD ∆为等边三角形，O 为AD 的中点， 所以AD BO ⊥,而PO BO O =,所以AD ⊥平面POB .(2)解：由(1)知AD PO ⊥，面PAD ⊥面ABCD ,所以PO ⊥底面ABCD ， 因为等边PAD ∆的边长为2，所以3PO =,易知BCD ∆为边长为2的等边三角形，所以三棱锥P BCD -的体积为：212134P BCD V -=⨯=, 因为E 是PC 的中点，所以1122P EDB P BCD V V --==， 所以三棱锥P EDB -的体积为12． 【点睛】本题考查了线面垂直的判定及三棱锥体积的求法，重点考查了空间想象能力及运算能力，属中档题.18．已知函数21()cos sin cos ,64f x x x x x R π⎛⎫=⋅+-+∈ ⎪⎝⎭. （1）求()f x 的最小正周期； （2）判断函数()f x 在,44ππ⎡⎤-⎢⎥⎣⎦上的单调性. 【答案】（1）π；（2）减区间为,46ππ⎡⎤--⎢⎥⎣⎦，增区间为,64ππ⎡⎤-⎢⎥⎣⎦.【解析】（1）先根据两角和正弦公式展开，再根据二倍角公式、辅助角公式化简，最后根据正弦函数性质求周期；（2）根据正弦函数性质求单调区间，再根据,44ππ⎡⎤-⎢⎥⎣⎦与单调区间关系讨论单调性. 【详解】（1）由题意，函数211()cos cos cos 224f x x x x x ⎛⎫=⋅+-+ ⎪ ⎪⎝⎭21111cos cos 2(1cos 2)224444x x x x x =⋅-+=-++()112cos 2sin 2,426x x x f x π⎛⎫=-=-∴ ⎪⎝⎭的最小正周期22T ππ==. （2）由（1）得1()sin 226f x x π⎛⎫=- ⎪⎝⎭， 因为,44x ππ⎡⎤∈-⎢⎥⎣⎦时，则2,22x ππ⎡⎤∈-⎢⎥⎣⎦，所以22,633x πππ⎡⎤-∈-⎢⎥⎣⎦，当22,632x πππ⎡⎤-∈--⎢⎥⎣⎦时，即,46x ππ⎡⎤∈--⎢⎥⎣⎦时，()f x 单调递减， 当2,623x πππ⎡⎤-∈-⎢⎥⎣⎦时，即,64x ππ⎡⎤∈-⎢⎥⎣⎦时，()f x 单调递增. 【点睛】本题考查两角和正弦公式、二倍角公式、辅助角公式、正弦函数性质，考查基本分析求解能力，属基础题.19．已知等比数列{a n }的公比q >1，且a 3+a 4+a 5=28，a 4+2是a 3，a 5的等差中项 （1）求数列{a n }通项公式；（2）求数列{()()1111n n n a a a ++++}的前n 项和T n .【答案】（1）12n na ；（2）n T 2121n n -=+.【解析】（1）由已知条件直接求得48a =，再利用58a q =，38a q=，列方程求q 和通项公式；（2）由（1）可知，()()()()1112112121nn n n n n n a b a a +-+==++++，利用裂项相消法求和.【详解】（1）由42a +是35,a a 的等差中项得35424a a a +=+， 所以34543428a a a a ++=+=， 解得48a =.由3520a a +=得18()20q q+=， 因为1q >，所以2q .所以12n na（2）记()()()()1112112121nn n n n n n a b a a +-+==++++则()()1112211221212121n n n n n nb ---⋅==-++++（）所以01122311111111122121212121212121n n n T -⎛⎫=-+-+-++-⎪++++++++⎝⎭1121222121n n n-⎛⎫=-= ⎪++⎝⎭。
2022年数学选择性必修1寒假必刷题专题04 直线和圆的方程(解答题)(解析版)
（1）若 l1 ⊥ l2 ，求实数 a 的值； （2）若 l1//l2 ，求实数 a 的值． 10．（四川省宜宾市第四中学 2020-2021 学年高二上学期开学考试数学（文））已知直线 l1 ：x + 3y − 5 =0 ，
直线 l2 ： ax − y + 4= 0(a ∈ R) ．
（1）求 BC 边上的中线所在直线的方程； （2）求 AB 边上的高线所在直线的方程．
6．（福建省普通高中 2019-2020 学年高二 1 月学业水平合格性考试）已知圆 O ： x2 + y2 = 8 ，点 P0 (−1,2) ，
直线 l 过点 P0 且倾斜角为α ．
1
（1）判断点 P0 与圆 O 的位置关系，并说明理由； （2）若α = 3π ，求直线 l 被圆 O 所戴得的弦 AB 的长．
24．（江西省南昌市第二中学 2020-2021 学年高二上学期第一次月考数学（文））已知以点 A(−1, 2) 为圆心
的圆与直线 m : x + 2 y + 7 =0 相切，过点 B (−2, 0) 的动直线 l 与圆 A 相交于 M，N 两点．
（1）求圆 A 的方程．
（2）当 MN = 2 19 时，求直线 l 方程．
18．（四川省资阳市 2019-2020 学年高一下学期期末）已知直线 l1 ： 2x − y +1 =0 和 l2 ： x − y − 2 =0 的交
点为 P ．
（1）若直线 l 经过点 P 且与直线 l3 ： 4x − 3y − 5 =0 平行，求直线 l 的方程； （2）若直线 m 经过点 P 且与 x 轴， y 轴分别交于 A ，B 两点，P 为线段 AB 的中点，求 OAB 的面积（其 中 O 为坐标原点）．
江西省新余一中2019-2020学年高一下学期第二次月考物理答案
新余市一中2019-2020学年高一年级第二次段考物理试题参考答案一、本题共10小题：每小题4分，共计40分。
在每小题给出的四个选项中，有一个或多个选项正确，二、填空题：本题共2小题，每空2分，共18分。
把答案填写在题后括号内或横线上。
11、 AD BC 0.6512、 3.67 6.87 6.73 > 阻力做功 机械能守恒三、计算题：本大题共4小题，共42分。
解答应写出必要的文字说明、方程式和重要演算步骤。
只写出最后答案的不能得分，有数值计算的题，答案中必须明确写出数值和单位。
13、【答案】（1（2）6mg 方向竖直向下【解析】（1）物块从A 点运动到B 点的过程中，由机械能守恒得221B mv mgh =，2分 解得gR v B 5=．………………………………………………2分（2）物块从B 至C 做匀速直线运动，所以gR v v B C 5==．…………1分 物块通过圆形轨道最低点C 时做圆周运动，由牛顿第二定律有：R vm mg F CN 2=-……1分所以mgF N 6=………………1分由牛顿第三定律可知，小物块对轨道的压力mg F F N N6=='方向竖直向下…………1分 14、【答案】（1）小球从抛出点O 到P 点的竖直高度h 为15m ；（2）抛出点O 到P 点的水平距离x 为； 【解析】(1)小球到达P 点时的竖直分速度为：0tan /y v v s θ== 由速度位移关系公式得：22y v gh = 解得：15h m = （2）飞行时间：y v t g==从O 到P 点的水平距离:0x v t ==.15、【答案】(1)加速 (2)212)(h R gR + (3)R n t gR -322224π 【解析】(1)飞船在B 点经椭圆轨道进入预定圆轨道时需要加速的 …………2分；(2)在地球表面重力有200R Mm Gg m = ① …………2分 根据牛顿第二定律有：A ma h R MmG =+21)(① …………2分 由①①式联立解得，飞船经过椭圆轨道近地点A 时的加速度大小为212)(h R gR a A +=……1分；(3)飞船在预定圆轨道上飞行时由万有引力提供向心力，有)(4)(22222h R Tm h R Mm G +=+π①………………2分 由题意可知，飞船在预定圆轨道上运行的周期为ntT =① ………… 2分 由①①①式联立解得：R n tgR h -=3222224π…………1分 16、【答案】①1m/s ①40N ①0.45m≤h≤0.8m 或h≥1.25m【解析】⑴小球恰能通过第二个圆形轨道最高点，有：Rv m mg 22= …………1分求得：gR v =2=1m/s ① …………1分⑵在小球从第一轨道最高点运动到第二圆轨道最高点过程中，应用动能定理有：212212121mv mv mgL -=-μ ① …………1分 求得：12212gL v v μ+==m/s …………1分在最高点时，合力提供向心力，即Rv m mg F N 21=+ ① …………1分求得：F N = 40N …………1分根据牛顿第三定律知，小球对轨道的压力为：F N ′=F N =40N ① ………… 1分(3)若小球恰好通过第二轨道最高点，小球从斜面上释放的高度为h 1，在这一过程中应用动能定理有：2211212mv mgR mgL mgh =--μ ① 求得：gv L R h 222211++=μ=0.45m …………1分若小球恰好能运动到E 点，小球从斜面上释放的高度为h 2，在这一过程中应用动能定理有： 00)(212-=+-L L mg mgh μ ①求得： h 2=0.8m …………1分 使小球停在DE 段，应有h 1≤h ≤h 2，即：0.45m≤h ≤0.8m 若小球能通过E 点，并恰好越过壕沟时，则有221gt d =得s gd t 4.02==① …………1分 t v x E =得s m v E /3= ① …………1分设小球释放高度为h 3，从释放到运动E 点过程中应用动能定理有：021)(2213-=+-E mv L L mg mgh μ ① 求得：g v L L h E2)(2211++=μ=1.25m …………1分即小球要越过壕沟释放的高度应满足：h ≥1.25m ………… 1分综上可知，释放小球的高度应满足：0.45m≤h ≤0.8m 或 h ≥1.25m ① ………… 1分。
江西省2019-2020学年下学期高一英语期末试卷精选汇编：阅读理解专题部分详解
阅读理解专题江西省新余一中、樟树中学等六校2019-2020学年高一（常规班）下学期期末联考英语试题第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的四个选项（A、B、和D）中，选出最佳选项。
ABefore you know it, the sound of sleigh(雪橇)bells will be heard on your rooftop. If you love the holidays but hate the cold, why not wear your coat and wool gloves this year and celebrate a warm Christmas?Beaufort, South CarolinaBeaufort is a small town with a huge heart and is located in the heart of the Lowcountry. It is close to Savannah and Charleston, and a short drive to the beaches of Hilton Head Island. Its location is great for visiting the larger cities and enjoying the sights and sounds of a small town. December temperatures average in the low 60s.Orlando, FloridaEnjoy everything in Orlando without the huge crowds of summer. Theme parks have special attractions this time of year, such as “Mickey's Very Merry Christmas”, but be aware that some of the special attractions require a separate ticket. Orlando temperatures average around 72 degrees in December.San Francisco, CaliforniaThe City by the Bay comes alive in December as it transforms into a wonderland of lights. Stay near Union Square so you are close to all the activities and can easily get around via the historic cable cars. San Francisco's December temperatures average in the high 50s.New Orleans, LouisianaCelebrate Christmas season in the Big Easy. The Oaks, visited by over 122,000 people last year, has one of the most beautiful Christmas exhibitions in the country. Many hotels in the area offer Papa Noel specials, making this a very affordable holiday destination. December temperatures average in the 60s.1. You can easily travel through San Francisco by_________ .A. bikeB. cable carC. taxiD. bus2. Spending the Christmas in New Orleans, you can .A. enjoy a wonderland of lightsB. avoid the huge crowdsC. enjoy Papa Noel specialsD. visit various theme parks3. What's the purpose of writing this passage?A. To inform readers of some popular sports events.B. To describe different celebrations during the Christmas.C. To introduce four places' special Christmas traditions.D. To introduce four places to spend a warm Christmas.BI’ve loved my mother’s desk since I was just tall enough to see above the top of it as mother sat doingletters. Standing by her chair, looking at the ink bottle, pens, and white paper, I decided that the act of writing must be the more wonderful thing in the world.Years later, during her final illness, mother kept different things for my sister and brother. “But the desk,” she’d said again, “it’s for Elizabeth.”I never saw her angry, never saw her cry. I knew she loved me; she showed it in action. But as a young girl, I wanted heart-to-heart talks between mother and daughter.They never happened. And a gulf opened between us. I was “too emotional(易动感情的)”. But she lived “on the surface”.As years passed I had my own family. I loved my mother and thanked her for our happy family. I wrote to her in careful words and asked her to let me know in any way she chose that she did forgive me.I posted the letter and waited for her answer. None came.My hope turned to disappointment, then little interest and, finally, peace— it seemed that nothing happened. I couldn’t be sure that the letter had even got to mother. I only knew that I had written it, and I could stop trying to make her into someone she was not.Now the present of her desk told, as she’d never been able to, that she was pleased that writing was my chosen work. I cleaned the desk carefully and found some papers inside —a photo of my father and a one-page letter, folded and refolded many times.Give me an answer, my letter asks, in any way you choose. Mother, you always chose the act that speaks louder than words.4. The passage shows that ______.A. mother was cold on the surface but kind in her heart to her daughterB. mother was too serious about everything her daughter had doneC. mother cared much about her daughter in wordsD. mother wrote to her daughter in careful words5. The word “gulf” in the passage means ______.A. deep understanding between the old and the youngB. different ideas between the mother and the daughterC. free talks between mother and daughterD. part of the sea going far in land6. What did mother do with her daughter’s letter asking forgiveness?A. She had never received the letter.B. For years, she often talked about the letter.C. She didn’t forgive her daughter at all in all her life.D. She read the letter again and again till she died.7. What’s the best title of the passage?A. My letter to MotherB. Mother and ChildrenC. My mother’s DeskD. Talks between Mother and MeCClowns (丑角) have been around for a long time. They not only make us laugh at the circus (马戏团) but they used to make the king and his followers laugh in court. The earliest of these fun-makers are ca lled “fools”. In Greek and Roman times these fools dressed up like clowns do today with painted faces and funny costumes (戏服). They danced around the stage and did things to make people laugh as clowns do in circuses today. By medieval times (中世纪) every court had its fools. Fools wore bright colours and had bells (大肚子) on their shoes and their hats.They were not really meant to be foolish and often they would say wise and sensible things to the king which nobody else was brave enough to say The king never got cross with the fool as this was his job. He was meant to be both wise and foolish and try to make the situation in court less serious.Of course people wanted to have their own kind of fool and shows developed in the 1500s in which fools (or Zanies) performed to make the crowd laugh. This kind of entertainment became so popular that it started a special kind of drama in Italy called Commedia dell´arte, which gradually turned into our comedy programmes today. When you watch the antics (滑稽剧) of Charlie Chaplin or Rowan Atkinson, you are watching a kind of comedy that is directly originated from the fools of ancient Rome and Greece.8. How do clowns entertain people?A. By their amusing talk and body language.B. By their amusing voices and body langauge.C. By their clothes, songs and the way they talk.D. By their clothes, make-up and the way they act.9. Where did fools first give performances?A. In different towns in Europe.B. In shows in ancient Greece and Rome.C. At fairs or market where there were a lot of people.D. in the building where law cases could be heard and judged.10. When did the Commedia dell´arte begin to develop?A. Before the fifth century.B. Around the tenth century.C. During the Middle Ages.D. In the sixteenth century.11. What does the underlined “got cross” in the second paragraph mean?A. got satisfiedB. got throughC. got angryD. got delightedDDo you sometimes find yourself drawing random patterns(图案)during meetings and lectures? Some people believe it shows you aren’t li stening or paying attention. However, there are studies whichclaim the opposite, and that doodling might actually be beneficial.Doodling seems to be popular. In the past, it was seen as a mindless activity and a product of the absentminded. However, in 2016, an article on the Harvard University website discussed the idea that random drawings may assist memory retention(保持)and concentration.The article stresses that 26 of 44 American presidents were known to doodle, with Ronald Reagan famous for drawing cowboys. A 2009 study conducted by psychologist Jackie Andrade asked 40 people to listen to a “dull and boring” voice message. Half of the group were asked to doodle and shade in a picture. Those who did were able to recall 29% more of the information contained in the message.Some experts believe that willingly accepting your creative and artistic sides during meetings may engage your mind. Jesse Prinz, a famous professor of philosophy, actively encourages his students to doodle. He believes that doodling i s the “attentional sweet spot”.And it might not just be good for your concentration. An artist known as Mr Doodle has made a career out of his scribbles(乱涂乱画).He sells his art online and also goes to people’s homes to decorate their walls.It seems that random drawing to aid concentration might be becoming more of a thing. People like Sunni Brown, an advocate for doodling, teach adults how to sketch and scribble in the workplace. So, the next time you see someone randomly drawing pictures during a meeting, it might be a sign that they are listening more than you think.12. Which phrase has the same meaning as doodling?A. A product of the absentminded.B. Random drawing.C. Scribbles in the workplace.D. Drawing on the walls.13. The author refers to Ronald Reagan in order to ________.A. inform readers of Reagan’s abilityB. provide some background informationC. introduce a new topic for further discussionD. give an example of the advantages of doodling14. Why does Jesse Prinz think that doodling is the “attentional sweet spot”？A. He thinks that it draws more attention.B. He thinks that it promises a good career.C. He believes that it helps to engage the mind.D. He believes that it can bring economic benefits.15. What is the author’s purpose in writi ng the passage?A. To state the benefits that you can get from random drawing.B. To provide guidance on leading a random drawing life.C To stress the need of distraction at work and in our life.D. To suggest a way of pursuing concentration in our lifelong career.江西省宜春市上高二中2019-2020学年高一英语下学期期末考试试题第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的四个选项（A、B、C和D）中，选出最佳选项，并在题卡上将该项涂黑。
【精准解析】江西省新余市第一中学2019-2020学年高一下学期第二次月考数学试题
新余一中2019-2020高一年级第二次段考数学答案一、单选题1.若点P (m ，n )(m ≠0)为角600︒终边上一点，则nm等于（ ） 333 D.12【答案】B 【解析】 【分析】利用任意角的三角函数的定义及诱导公式可求得tan 600n m︒=. 【详解】∵tan 600n m︒=， 又()tan 600tan 180360tan 603︒︒︒=︒⨯+==∴nm=3故选：B ．【点睛】本题考查任意角的三角函数的定义及诱导公式，属于容易题． 2.已知向量()cos ,sin a θθ=，()0,1b =-，π0,2θ⎛⎫∈ ⎪⎝⎭，则向量a 与向量b 的夹角为（ ）A. πθ-B. π2θ- C.π2θ+ D. θ【答案】C 【解析】 【分析】由题意，根据向量夹角公式的坐标表示，得到cos ,sin θ<>=-a b ，即可得出结果. 【详解】因为()cos ,sin a θθ=，()0,1b =-，所以()222cos ,sin cos sin 01θθθ<>==-++⋅-a b ，又π0,2θ⎛⎫∈ ⎪⎝⎭，所以,2a b πθ<>=+. 故选：C【点睛】本题主要考查求向量的夹角，熟记向量夹角公式即可，属于常考题型.3.《周脾算经》有记载：一年有二十四个节气，每个节气晷（gui ）长损益相同，晷是按照日影测定时刻的仪器，晷长即所测定的影子的长度，二十四节气及晷长变化如图所示，相邻两个节气晷长变化量相同，周而复始，若冬至晷长最长是一丈三尺五寸，夏至晷长最短是一尺五寸，（一丈等于10尺，一尺等于10寸），则秋分节气的晷长是（ ）A. 七尺五寸B. 二尺五寸C. 五尺五寸D. 四尺五寸【答案】A 【解析】 【分析】由题意从夏至到秋分到冬至的过程中晷长为等差数列，设为{}n a ，则夏至晷长为首项，冬至晷长为第13项，利用等差数列的通项公式即可得出．【详解】由题意从夏至到秋分到冬至的过程中晷长为等差数列，设为{}n a . 则115a =，13135a =,则公差131135151013112a a d --===-.秋分晷长为716156075a a d =+=+=. 所以秋分节气的晷长是七尺五寸 故选：A ．【点睛】本题考查了等差数列的通项公式，考查了推理能力与计算能力，属于中档题． 4.为了得到函数2sin 36x y π⎛⎫=+ ⎪⎝⎭，x ∈R 的图像，只需把函数2sin y x =，x ∈R 的图像上所有的点（ ） A. 向右平移6π个单位长度，再把所得各点的横坐标伸长到原来的3倍（纵坐标不变） B. 向左平移6π个单位长度，再把所得各点的横坐标伸长到原来的3倍（纵坐标不变） C. 向右平移6π个单位长度，再把所得各点的横坐标伸长到原来的13倍（纵坐标不变） D. 向左平移6π个单位长度，再把所得各点的横坐标伸长到原来的13倍（纵坐标不变） 【答案】B 【解析】 【分析】根据三角函数图象的平移变换和伸缩变换求解. 【详解】把函数2sin y x =向左平移6π个单位长度，得到2sin 6y x π⎛⎫=+ ⎪⎝⎭， 再把所得各点的横坐标伸长到原来的3倍（纵坐标不变）得到2sin 36x y π⎛⎫=+ ⎪⎝⎭. 故选：B【点睛】本题主要考查三角函数的图象变换，还考查了数形结合的思想方法，属于中档题.5.已知点()4,1A 和坐标原点O ，若点(),B x y 满足1133x y x y x y -≥-⎧⎪+≥⎨⎪-≤⎩，则OA OB ⋅的最大值是（ ） A. 11 B. 4 C. 1 D. 1-【答案】A 【解析】 【分析】设4z OA OB x y =⋅=+，作出不等式组所表示的可行域，平移直线4z x y =+，找出使得直线4z x y =+在y 轴上截距最大时对应的最优解，代入目标函数计算即可.【详解】设4z OA OB x y=⋅=+，作出不等式组1133x yx yx y-≥-⎧⎪+≥⎨⎪-≤⎩所表示的可行域如下图所示：联立10330x yx y-+=⎧⎨--=⎩，解得23xy=⎧⎨=⎩，即点()2,3A，平移直线4z x y=+，当该直线经过可行域的顶点A时，直线4z x y=+在y轴上的截距最大，此时z取最大值，即max42311z=⨯+=.故选：A.【点睛】本题考查线性目标函数的最值，一般通过平移直线找出最优解，考查数形结合思想的应用，属于基础题.6.若(sin)4cos2f x x=+，则(cos)f x=（）A. 4cos2x+ B. 4cos2x- C. 4sin2x- D.4sin2x+【答案】B【解析】【分析】用诱导公式转化．【详解】(cos )[sin()]4cos 2()42cos(2)42cos 222f x f x x x x πππ=-=+-=+-=-． 【点睛】本题考查求函数解析式，掌握诱导公式是解题关键．本题也可以先求出()f x ，再求解．7.设等比数列{}n a 的前n 项和为n S ，若39S =，636S =，则789(a a a ++= ) A. 144 B. 81C. 45D. 63【答案】B 【解析】 【分析】根据等比数列性质，得到关于3S ，63S S -，96S S -的新等比数列，求解出公比后，求出96S S -的值即可．【详解】由等比数列性质可知：3S ，63S S -，96S S -，……成等比数列，设公比为q 由题意得：6336927S S -=-= 2739q ⇒== 7899627381a a a S S ∴++=-=⨯=本题正确选项：B【点睛】解决本题的关键在于根据等比数列的性质得到：232,,,k kk k k S S S S S 依然成等比数列，从而快速求解此题．本题也可以利用等比数列的基本项1a 和q 来进行求解，但计算量较大．8.设点M 是线段BC 的中点，点A 在直线BC 外，若2BC =，AB AC AB AC +=-，则AM =（ ）A.12B. 1C. 2D. 4【答案】B 【解析】 【分析】||||AB AC AB AC +=-两边平方，可得0AB AC ⋅=，即AB AC ⊥，利用直角三角形斜边中线与斜边长度的关系，即可求出||AM .【详解】||||AB AC AB AC +=-，两边平方得，222222AB AB AC AC AB AB AC AC +⋅+=-⋅+，0,AB AC AB AC ∴⋅=∴⊥，M 是线段BC 的中点，1||||12AM BC ∴==. 故选:B【点睛】本题考查向量的模长以及向量的数量积运算，属于基础题. 9.已知函数()cos 26π⎛⎫=-⎪⎝⎭f x x ，把()y f x =的图象向左平移6π个单位得到函数()g x 的图象，则下列说法正确的是（ ）A. 3π⎛⎫=⎪⎝⎭g B. ()g x 的图像关于直线2x π=对称C. ()g x 的一个零点为,03π⎛⎫⎪⎝⎭D. ()g x 的一个单调减区间为5,1212ππ⎡⎤-⎢⎥⎣⎦ 【答案】D 【解析】 【分析】先把()f x 变为()cos 26f x x π⎛⎫=- ⎪⎝⎭，根据平移得到()g x 的解析式，从而可以讨论()g x 的相关性质.【详解】()cos 2cos 2666g x x x πππ⎡⎤⎛⎫⎛⎫=+-=+ ⎪ ⎪⎢⎥⎝⎭⎝⎭⎣⎦，所以5cos 36g ππ⎛⎫== ⎪⎝⎭，故A 不正确， 令2,6x k k z ππ+=∈，故对称轴方程为,212k x k z ππ=-∈，故B 错，令2,62x k k z πππ+=+∈，故对称中心的横坐标为,26k x k z ππ=+∈，故C 错， 因5,1212x ππ⎡⎤∈-⎢⎥⎣⎦，故 []20,6u x ππ=+∈，因cos y u =在[]0,π上是减函数，故()cos 26f x x π⎛⎫=- ⎪⎝⎭在5,1212ππ⎡⎤-⎢⎥⎣⎦上是减函数，故D 正确.综上，选D.【点睛】（1）平移变换有“左加右减”（水平方向的平移），注意是对自变量x 做加减，比如把()y f x =-的图像向右平移1个单位后，得到的图像对应的解析式为()()11y f x f x ⎡⎤=--=-+⎣⎦．（2）形如()()sin 2f x A x B ωϕ=++的正弦型函数，可根据复合函数的讨论方法求该函数的单调区间、对称轴方程和对称中心等．10.如图，在三角形OPQ 中，M 、N 分别是边OP 、OQ 的中点，点R 在直线MN 上，且(,)OR xOP yOQ x y R =+∈，则代数式2212x y x y +--+的最小值为（ ）A. 24-B. 3C.243【答案】C 【解析】 【分析】 本题首先可设λμOROM ON 并得出1λμ+=，然后根据M 、N 分别是边OP 、OQ 的中点得出12x y +=，最后将12yx 2212x y x y +--+. 【详解】因为点R 、M 、N 共线， 所以设λμOROM ON ，其中1λμ+=，因为M 、N 分别是边OP 、OQ 的中点， 所以11λμλμ22OROM ONOP OQ ，111λμ222x y ，12y x ， 2221112222y x yx x xx ， 21112224484xx x，故当14x =4， 故选：C.【点睛】本题考查向量的共线定理，主要考查平面向量的三点共线定理，考查通过配方法求最值，考查计算能力，考查化归与转化思想，是中档题.11.在平面直角坐标系xOy 中，已知向量10a b a b a b ==⋅=、，，，点Q 满足()2OQ a b =+，曲线{}1C P OP ==，区域{}|0，，Ω=<≤≤<P r PQ R r R 若C Ω为两段分离的曲线，则（ ） A. 13r R <<< B. 13r R <<≤ C. 13r R ≤<< D. 13r R <<<【答案】A 【解析】 【分析】不妨设(1,0),(0,1)a b ==，由{}1C P OP ==，所以点P 的轨迹表示一个单位圆， 又由{}|0P r PQ R r R Ω=<≤≤<，表示的平面区域为：以Q 为圆心，内径为r 外径为R 的圆环，根据CΩ为两端分离的曲线，则单位圆与圆环的内外均相交，利用圆与圆的位置关系，即可求解．【详解】由题意，平面直角坐标系xOy 中，已知向量10a b a b a b ==⋅=、，，，不妨设(1,0),(0,1)a b ==，则()2(2,OQ a b =+=，由{}1C P OP ==，所以点P 的轨迹表示一个单位圆，又由{}|0P r PQ R r R Ω=<≤≤<，表示的平面区域为：以Q 为圆心，内径为r 外径为R 的圆环， 若CΩ为两端分离的曲线，则单位圆与圆环的内外均相交，所以11OQ r R OQ -<<<+，因为2OQ =，所以13r R <<<． 故选A ．【点睛】本题主要考查了平面向量在几何问题中的应用，其中根据已知条件得到点P 的轨迹，以及{}|0P r PQ R r R Ω=<≤≤<，所表示的平面区域，结合圆与圆的位置关系求解是解答的关键，着重考查了转化思想，以及推理与运算能力，属于中档试题． 12.将函数()3cos 3f x x π⎛⎫=- ⎪⎝⎭的图象上的所有点的横坐标缩短为原来的12，纵坐标不变，再把所得的图象向左平移3π个单位长度，然后再把所得的图象向下平移1个单位长度，得到函数()g x 的图象，若()()1216g x g x =，且[]12,2,2x x ππ∈-，则122x x -的最大值为（ ） A.133π B.103π C. 52πD.256π 【答案】A 【解析】 【分析】根据三角函数平移变换,先求得()g x 的解析式.根据()()1216g x g x =,可知()()124g x g x ==-,即12cos 21,cos 2133x x ππ⎛⎫⎛⎫+=-+=- ⎪ ⎪⎝⎭⎝⎭.根据[]12,2,2x x ππ∈-可分别求得12x 的最大值和2x 的最小值,即可求得122x x -的最大值.【详解】根据平移变换将函数()3cos 3f x x π⎛⎫=- ⎪⎝⎭的图象上的所有点的横坐标缩短为原来的12,纵坐标不变,再把所得的图象向左平移3π个单位长度,然后再把所得的图象向下平移1个单位长度,可得()3cos 213g x x π⎛⎫=+- ⎪⎝⎭由()()1216g x g x =, 可知()()124g x g x ==- 即12cos 21,cos 2133x x ππ⎛⎫⎛⎫+=-+=- ⎪ ⎪⎝⎭⎝⎭[]12,2,2x x ππ∈-所以12111311132,,2,333333x x ππππππ⎡⎤⎡⎤+∈-+∈-⎢⎥⎢⎥⎣⎦⎣⎦123x π+的最大值为3π,223x π+的最小值为3π-则12x 的最大值为83π,2x 的最小值为53π- 所以122x x -的最大值为8513333πππ⎛⎫--= ⎪⎝⎭故选:A【点睛】本题考查了三角函数图象的平移变换,三角函数性质的综合应用,利用函数的最值求参数的取值情况,属于难题. 二、填空题13.已知数列{}n a 为等差数列，若159a a a π++=，则()28cos +a a 的值为_______． 【答案】12- 【解析】 【分析】先利用等差数列的性质求出53a π=，进而得2823a a π+=，再代入所求即可． 【详解】因为{}n a 为等差数列，且159a a a π++=， 由等差数列的性质得53a π=，所以2823a a π+=， 故()2821cos cos 32a a π⎛⎫+==- ⎪⎝⎭. 故答案为：12-. 【点睛】本题主要考查等差数列性质的应用．属于较易题.14.已知正项等比数列{}n a 满足28516a a a ，3520a a +=，若存在两项m a ，n a，使得32=，则14m n+的最小值为________． 【答案】34【解析】【分析】 先计算12n na 32=得到12m n +=，再利用均值不等式得到1434m n +≥. 【详解】正项等比数列{}n a2285551616a a a a a353204a a a +=⇒=12n n a -⇒=210322212m n m n +-=⇒=⇒+=144()()1414531212124n m m n m n m n m n ++++++==≥= 当8,4n m ==时等号成立.故答案为34【点睛】本题考查了数列的通项公式，均值不等式，综合性强，意在考查学生对于数列方法和均值不等式的综合应用.15.已知平面向量a 与b 的夹角为锐角，4a =，2b =，且b ta+的最小值为c 满足()()•0c a c b --=，则c 的取值范围为__________．【答案】3,73⎤+【解析】 分析】根据b ta +的最小值为可知,a b 的夹角为π3，画出向量对应的平面图形，建立平面直角坐标系，求得,a b 两点的坐标，设出c 的坐标，代入()()•0c a c b --=，求得c 坐标满足的方程，根据这个方程对应的曲线是圆，由圆上的点和原点的距离的最大值和最小值，求得c 的取值范围.【详解】画出图像如下图所示，其中OD BD ⊥，设,a OA b OB ==.由于b ta +的最小值为3，根据向量加法的几何意义可知3OD =,而2OB =，故π3∠=OBD ，3OD =.以O 为坐标原点，,OA OD 分别为,x y 轴建立如图所示的平面直角坐标系，()()4,0,1,3a b ==，设(),c x y =.由于()()•0c a c b --=，即()()4,1,30---=x y x y ，化简得2253322⎛⎫⎛⎫-+-= ⎪ ⎪ ⎪⎝⎭⎝⎭x y ，即c 对应的点在以53,22⎛⎫⎪ ⎪⎝⎭为圆心，半径为3的圆上，而c 表示圆上的点到原点的距离.圆心到原点的距离为2253722⎛⎫⎛⎫+= ⎪ ⎪ ⎪⎝⎭⎝⎭，故c 的取值范围是73,73⎡⎤-+⎣⎦.【点睛】本小题主要考查平面向量的坐标运算，考查平面向量加法的几何意义，考查建立平面直角坐标系的方法研究向量模的取值范围，考查化归与转化的数学思想方法、考查数形结合的数学思想方法，属于中档题.解题的关键点在于将c 的坐标满足的方程转化为圆的方程，将模的为题转化为圆上的点到原点距离来求解.16.给出以下五个结论： ①函数sin 3y x π⎛⎫=+⎪⎝⎭是偶函数； ②当0,2x π⎡⎤∈⎢⎥⎣⎦时，函数()2cos 26f x x π⎛⎫=+ ⎪⎝⎭的值域是⎡-⎣； ③等差数列{}n a 的前n 项和为n S ，若633S S =，则12953S S =； ④已知定义域为R 的函数()sin cos sin cos 22x xx x f x -+=-，当且仅当()222k x k k Z πππ<<+∈时，()0f x >成立.⑤函数224()sin (,)sin f x x x k k Z xπ=+≠∈的最小值4； 则上述结论中正确的是______（写出所有正确结论的序号）． 【答案】②③④ 【解析】 【分析】利用特殊值代入①中的解析式即可判断①；根据函数单调性及自变量取值范围，可判断②；讨论sin cos x x -的符号去绝对值，即可判断④；换元得()g t ，利用函数单调性即可判断⑤. 【详解】当3x π=与3x π=-时，代入①中的解析式所得函数值不相等，所以①错误；当0,2x π⎡⎤∈⎢⎥⎣⎦时，72,666x πππ⎡⎤+∈⎢⎥⎣⎦，由余弦函数图象可知()2cos 26f x x π⎛⎫=+ ⎪⎝⎭的值域是⎡-⎣，所以②正确；设12366391295,3,2,6,S 10,3S S a S a S S a S a a S =∴=∴-===∴=，故③正确； 当sin cos 0x x -≥时，()sin cos sin cos cos 22x x x xf x x +-=-=，当()2242k x k k Z ππππ+<<+∈时，()0f x >；当sin cos 0x x -<时，()sin cos cos sin sin 22x x x x f x x +-=-=，当()224k x k k Z πππ<<+∈时，()0f x >，综上，()222k x k k Z πππ<<+∈时，()0f x >，所以④正确. ⑤设2244sin ,(01),()()10t x t g t t g t t t'=<≤∴=+∴=-<，，所以函数g (t )在(]0,1上单调递减，所以函数的最小值为g (1)=5，所以该命题是假命题. 故答案为：②③④.【点睛】本题主要考查了三角函数图象与性质的综合应用，三角函数定义域与值域的求法，数列和函数最值问题.属于较难题. 三、解答题17.已知向量(),3a λ=，()2,4b =-. （1）若()2a b b +⊥，求λ；（2）若4λ=，求向量a 在b 方向上的投影cos a θ（其中θ是a 与b 的夹角）【答案】（1）11λ=；（2. 【解析】试题分析：（1）利于垂直数量积0求解即可； （2）利用向量数量积的几何一意义求解即可. 试题解析：（1）∵(),3a λ=，()2,4b =-，∴()222,10a b λ+=-， 又()2a b b +⊥，∴()20a b b +⋅=， ∴()()2224100λ-⨯-+⨯=，∴11λ=. （2）由4λ=，可知()4,3a =，()2,4b =-，∴4a b ⋅=，25b =，∴cos 25a b a bθ⋅===. 18.建设生态文明，是关系人民福祉，关乎民族未来的长远大计.某市通宵营业的大型商场，为响应节能减排的号召，在气温超过28℃时，才开放中央空调降温，否则关闭中央空调.如图是该市夏季一天的气温（单位：℃）随时间（t ≤≤024，单位：小时）的大致变化曲线，若该曲线近似的满足函数()()sin 0,0,y A t b A ωϕωϕπ=++>><关系.（1）求函数()y f x =的表达式；（2）请根据（1）的结论，判断该商场的中央空调应在本天内何时开启？何时关闭？ 【答案】（1）()()2248sin 024123f t t t ππ⎛⎫=+-≤≤ ⎪⎝⎭（2）上午10时开启，下午18时关闭. 【解析】 【分析】（1）根据函数图象可知周期T ，进而根据2T πω=求得ω的值；结合函数的最大值和最小值，可求得A ，代入最低点坐标()216,，即可求得ϕ，进而得函数()f t 的解析式． （2）根据题意，令2248sin 28123t ππ⎛⎫+->⎪⎝⎭，解不等式，结合t 的取值范围即可求得开启和关闭中央空调时间．【详解】（1）由图知，()214224T =-=， 所以224πω=，得12πω=.由图知，1632242b +==，321682A -==， 所以()8sin 2412f t t πϕ⎛⎫=++⎪⎝⎭. 将点()216,代入函数解析式得248sin 21612πϕ⎛⎫+⨯+= ⎪⎝⎭，得262k ππϕπ+=-，()k Z ∈即()223k k Z ϕππ=-∈ 又因为ϕπ<，得23ϕπ=-. 所以()()2248sin 024123f t t t ππ⎛⎫=+-≤≤⎪⎝⎭. （2）依题意，令2248sin 28123t ππ⎛⎫+->⎪⎝⎭， 可得21sin 1232t ππ⎛⎫->⎪⎝⎭， 所以()252261236k t k k Z ππππππ+<-<+∈ 解得：()24102418k t k k Z +<<+∈， 令0k =得，1018t <<，故中央空调应在上午10时开启，下午18时关闭.【点睛】本题考查了利用部分函数图象求三角函数解析式，三角函数在实际问题中的应用，属于基础题．19.已知数列{}n a 是等差数列，n S 是其前n 项和，若510S =，且13a +，2a ，1-成等比数列．（1）求数列{}n a 的通项公式； （2）设7=+n n b a ，若11n n n c b b +=，求数列{}n c 的前n 项和n T ． 【答案】（1）37n a n =-（2）9(1)+nn【解析】（1）设等差数列{}n a 的公差为d ，因为510S =，所以()15355102a a a +==，所以32a =，因为13a +，2a ，1-成等比数列，所以()2213a a =-+， 又232a a d d =-=-，13222a a d d =-=-， 所以()()22223d d -=--+，解得3d =，所以()()3323337n a a n d n n =+-=+-=-． （2）由（1）可得73773n n b a n n =+=-+=， 故()11111133191n n n c b b n n n n +⎛⎫===- ⎪⋅++⎝⎭， 所以()1111111111922319191n n T n n n n ⎡⎤⎛⎫⎛⎫⎛⎫⎛⎫=-+-++-=-= ⎪ ⎪ ⎪ ⎪⎢⎥+++⎝⎭⎝⎭⎝⎭⎝⎭⎣⎦． 20.已知a ，b ，1a b ==，且3a kb a kb +=-，其中0k ＞. (1)若a 与b 的夹角为60°，求k 的值；(2)记()f k a b =⋅，是否存在实数k ，使得()1f k tk ≥-对任意的[]1,1t ∈-恒成立?若存在，求出实数k 的取值范围；若不存在，请说明理由. 【答案】(1) 1k =；(2) 203k <≤． 【解析】 【分析】(1)由3a kb a kb +=-两边平方得，223a kb a kb +=-，展开即可求出k 的值； (2)根据223a kb a kb +=-，可求出()f k a b =⋅，再将()1f k tk ≥-变形为21104k kt k ++-≥，设()2114k t kt k ϕ+=+-，然后解不等式组()()1010ϕϕ⎧≥⎪⎨-≥⎪⎩，即可求出实数k 的取值范围．【详解】(1) 由3a kb a kb +=-得，223a kb a kb +=-，因为12a b ⋅=， 所以()2212312ka b k ka b k+⋅+=-⋅+，即()22131k kk k ++=-+，解得1k =．(2)由(1)可知，()2212312ka b k ka b k+⋅+=-⋅+，所以21()4k f k a b k+=⋅=，()1f k tk ≥-变形为21104k kt k ++-≥，设()2114k t kt kϕ+=+-，所以()0t ϕ≥对任意的[]1,1t ∈-恒成立，即有()()1010ϕϕ⎧≥⎪⎨-≥⎪⎩，2211041104k k k k k k ⎧++-≥⎪⎪⎨+⎪-+-≥⎪⎩，解得203k <≤ ． 【点睛】本题主要考查数量积的运算以及不等式恒成立问题的解法，意在考查学生的转化能力和数学运算能力，属于中档题．21.已知数列{}n a 为等差数列，35a =，713a =，数列{}n b 的前n 项和为n S ，且有21n n S b =-.（1）求{}n a 、{}n b 的通项公式；（2）若()1n n n c a b =--⋅，123n n T c c c c =++++，求使1230++⋅>n n T n 成立的n 的最小值.【答案】（1）()*21n a n n N =-∈，()1*2n nbn N -=∈；（2）5.【解析】 【分析】（1）设等差数列{}n a 的公差为d ，根据题意列方程组解出1a 和d 的值，利用等差数列的通项公式可求得{}n a 的通项公式，令1n =可求得1b 的值，令2n ≥，由21n n S b =-得出1121n n S b --=-，两式作差可推导出数列{}n b 为等比数列，确定该数列的首项和公比，可求得数列{}n b 的通项公式；（2）求得2nn c n =-⋅，利用错位相减法求得n T ，由不等式1230++⋅>n n T n 得出1232n +>，解此不等式即可得出正整数n 的最小值.【详解】（1）设等差数列{}n a 的公差为d ，由题意可得317125613a a d a a d =+=⎧⎨=+=⎩，解得112a d =⎧⎨=⎩，()()1112121n a a n d n n ∴=+-=+-=-.由于数列{}n b 的前n 项和为n S ，且有21n n S b =-. 当1n =时，11121b S b ==-，解得11b =；当2n ≥时，由21n n S b =-可得1121n n S b --=-，上述两式相减得122n n n b b b -=-，12n n b b -∴=，可得12nn b b -=， 所以，数列{}n b 是以1为首项，以2为公比的等比数列，11122n n n b --∴=⨯=；（2）()11222n n n n n c a b n n -=--⋅=-⋅=-⋅，1231222322n n T n ∴=-⨯-⨯-⨯--⨯，()23121222122n n n T n n +=-⨯-⨯---⨯-⨯，上式-下式得()()12311121222222212212n n n n n n T n n n +++--=-----+⨯=⨯-=-⨯+-，()1122n n T n +∴=-⋅-，()1111212222230n n n n n T n n n +++++⋅=-⋅-+⋅=->，即1232n +>，15n ∴+>，解得4n >. 因此，满足不等式1230++⋅>n n T n 成立的n 的最小值为5.【点睛】本题考查等差数列通项公式的求解，利用n S 求通项，同时也考查了错位相减法与数列不等式的求解，考查计算能力，属于中等题. 22.已知函数()sin()0,22f x x b ππωϕωϕ⎛⎫=++>-<<⎪⎝⎭的相邻两对称轴间的距离为2π，若将()f x 的图像先向左平移12π个单位，再向下平移1个单位，所得的函数()g x 为奇函数．（1）求()f x 的解析式；（2）若关于x 的方程23(())()20g x m g x +⋅+=在区间0,2π⎡⎤⎢⎥⎣⎦上有两个不等实根，求实数m 的取值范围．【答案】（1）()sin 216f x x π⎛⎫=-+ ⎪⎝⎭（2）{}(,5)26m ∈-∞--【解析】 【分析】（1）根据相邻两对称轴间的距离求出ω值，由函数图像的变换关系，求出函数()g x ，再结合()g x 是奇函数，即可求出参数； （2）设sin 2t x =，0,2x π⎡⎤∈⎢⎥⎣⎦，原方程在区间0,2π⎡⎤⎢⎥⎣⎦上有两个不等实根，转化为方程2320t mt ++=在[0,1)t ∈内仅有一个根，且另一个根1≠，转化一元二次方程根的分布求参数，或分离参数转化为对勾函数与直线交点横坐标范围，即可求解. 【详解】解：（1）由题意知()f x 的周期22T ππωω==⇒=，故()sin(2)f x x b ϕ=++，而()sin 21sin 21126g x x b x b ππϕϕ⎛⎫⎛⎫⎛⎫=+++-=+++- ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭为奇函数，则101b b -=⇒=，且20()66k k k Z ππϕπϕπ⨯++=⇒=-∈，而,22ππϕ⎛⎫∈-⎪⎝⎭，故6πϕ=-，因此()sin 216f x x π⎛⎫=-+ ⎪⎝⎭；（2）由（1）知()sin 2g x x =，题意等价于[]23sin 2sin 220x m x +⋅+=在区间0,2π⎡⎤⎢⎥⎣⎦上有两个不等实根，令sin 2t x =，0,2x π⎡⎤∈⎢⎥⎣⎦，则题意 ⇔方程2320t mt ++=在[0,1)t ∈内仅有一个根，且另一个根1≠．法一：令2()32h t t mt =++，则题意2240016m m⎧∆=-=⎪⇔⎨<-<⎪⎩或{(0)0(,5)(1)0h m h ≥⎧⇒∈-∞-⋃-⎨<⎩； 法二：显然0不是该方程的根，题意22323mt t m t y m t⇔-=+⇔-=+⇔=- 与23y t t=+的图像在(0,1)t ∈内仅有一个交点且另一个交点不为()1,5，由于对勾函数23y t t =+在0,3⎛ ⎝⎦上单减，在3⎫⎪⎪⎣⎭上单增，故有5m ->或m -=，因此{}(,5)26m ∈-∞--．【点睛】本题考查三角函数图像的变换关系，考查函数的零点分布常用的方法，一是直接研究函数与x 轴交点范围，结合零点存在性定理求出参数范围；二是转化为两个函数交点横坐标的范围.。
江西新余一中2019-2020学年高一下学期第二次段考物理试题 Word版含答案
新余一中2019--2020年度高一年级下学期物理测试卷本卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分。
满分100分，时间90分钟。
第Ⅰ卷(选择题共40分)一、选择题(共10小题，每小题4分，共40分，在每小题给出的四个选项中，第1～6小题只有一个选项符合题目要求，第7～10小题有多个选项符合题目要求，全部选对的得4分，选不全的得2分，有选错或不答的得0分)1．(2019·山东省济南一中高一下学期合格考)列车在通过桥梁、隧道的时候，要提前减速。
假设列车的减速过程可看作匀减速直线运动，下列与其运动相关的物理量(位移x、加速度a、速度v、动能E k)随时间t变化的图像，能正确反映其规律的是()2．2010年诺贝尔物理学奖授予英国曼彻斯特大学科学家安德烈·海姆和康斯坦丁·诺沃肖洛夫，以表彰他们在石墨烯材料方面的卓越研究。
石墨烯是目前世界上已知的强度最高的材料，它的发现使“太空电梯”缆线的制造成为可能，人类将有望通过“太空电梯”进入太空。
现假设有一“太空电梯”悬在赤道上空某处，相对地球静止，如图所示，那么关于“太空电梯”，下列说法正确的是()A．“太空电梯”各点均处于完全失重状态B．“太空电梯”各点运行周期随高度增大而增大C．“太空电梯”上各点线速度与该点离地球球心距离的开方成反比D．“太空电梯”上各点线速度与该点离地球球心距离成正比3．(2019·辽宁省沈阳二十中高一下学期期中)如图所示，一铁球用细线悬挂于天花板上，静止垂在桌子的边缘，细线穿过一光盘的中间孔，手推光盘在桌面上平移，光盘带动细线紧贴着桌子的边缘以水平速度v匀速运动，当光盘由A位置运动到图中虚线所示的B位置时，细线与竖直方向的夹角为θ，此时铁球()A．竖直方向速度大小为v cosθB．竖直方向速度大小为v sinθC．竖直方向速度大小为v tanθD．相对于地面速度大小为v4．(2018·吉林省实验中学高一下学期期末)如图所示，小球从静止开始沿光滑曲面轨道AB滑下，从B 端水平飞出，撞击到一个与地面呈θ＝37°的斜面上，撞击点为C 。
江西省新余一中、樟树中学等六校2019-2020学年高一下学期第二次联考语文试题
江西省六校2019-2020学年高一年级下学期第二次联考语文试卷樟树中学宜春中学高安二中丰城九中宜春九中新余一中命题一、现代文阅读(36分)（一）论述类文本闵读（本题共3小题，9分）阅读下面的文字，完成1~3题。
综观杜甫的全部七绝，可以发现他创作七绝的情绪状态与其他诗体的明显差别，在于大多数作于兴致较高、心情轻松甚至是欢愉的状态中。
这一特点目前尚未见研究者论及，却是考察杜甫七绝“别趣”的重要出发点。
与其情绪状态相应，杜甫七绝的抒情基调也多数是轻松诙谐、幽默风趣的。
由以上两点可以看出，杜甫对于七绝的表现功能有其独到的认识。
盛唐七绝在传统题材里充分展现了以浅语倾诉深情的特长，使七绝突破南朝初唐七绝含蕴浅狭的藩篱，固然达到了艺术的巅峰。
但七绝这种体式的表现潜能尚未充分得到开掘，杜甫发现了这种诗体还有适宜于表现多种生活情趣的潜力。
所以他很少用这种体式来抒发沉重悲抑的情绪，而是在七绝中呈现了沉郁顿挫的基本风格之外的另一面，让人更多地从中看到他性情中的放达、出默和风趣。
这种不同于盛唐的趣味追求，应当就是他七绝中的“别趣”所在。
而“别趣”的内涵可以从他对外物的体察和对内心的发掘两方面来看，二者交融在一起，不能截然区分。
杜甫在体察外物中发现的“别趣”大多是他在成都和夔州时期对日常生活中多种诗趣的敏锐体悟。
大致有三个方面。
其一是诗人善于捕捉自然景物和人居环境中的生机和处处可见的趣味。
如《绝句四首（其一）》：“堂西长笋别开门，堑北行椒却背村。
梅熟许同朱老吃，松高拟对阮生论。
”不但写出了诗人与朱阮二人的特殊交情，更藉梅、松与竹、椒合围，形成了一个封闭的小天地，突出了草堂与世隔绝的清幽之趣。
其二是在人际交往和应酬中的雅兴和逸趣。
如《从韦二明府续处觅绵竹》：“华轩蔼蔼他年到，绵竹亭亭出县高。
江上舍前无此物，幸分苍翠拂波涛。
”既夸赞韦明府县斋绵竹的茂盛，又预想将来自己舍前苍翠竹影在江中倒映的美景。
将希望赠竹说成幸“分”苍翠之色，已十分新颖，“拂”字更写出竹影在波涛中摇漾的动态，这就使讨要竹子一事显得优雅别致。
江西省新余市第一中学2019_2020学年高一英语下学期第二次月考试题
江西省新余市第一中学2019-2020学年高一英语下学期第二次月考试题第一部分：听力（共两节，满分30分）第一节 (共 5 小题；每小题 1.5 分，满分 7.5 分)听下面 5 段对话。
每段对话后有一个小题，从题中所给的 A、B、C 三个选项中选出最佳选项。
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。
例：How much is the shirt?A. ￡19.15.B. ￡9.15.C. ￡9.18.答案是B.1. What does the man ask the woman to do?A. Give her ID card to him.B. Move a table.C. Sign for a parcel.2. What does the woman think of cleaning the shirt?A. Easy.B. Time-wasting.C. Impossible.3. How does the woman most probably go to work?A. By car.B. By bike.C. By bus.4. What relation is Tom to the woman?A. Her teacher.B. Her agent.C. Her husband.5. What are the speakers mainly talking about?A. A picnic.B. The weather.C. A forecast.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。
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。
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。
【精准解析】江西省新余市第一中学2019-2020学年高一下学期第二次月考英语试题
新余市高一下学期第二次月考英语试题第Ⅰ卷命题人：高一英语组第一部分：听力（共两节，满分30分）第一节 (共 5 小题；每小题 1.5 分，满分 7.5 分)听下面 5 段对话。
每段对话后有一个小题，从题中所给的 A、B、C 三个选项中选出最佳选项。
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。
例：How much is the shirt? A.￡19.15. B. ￡9.15. C. ￡9.18. 答案是B. 1. What does the man ask the woman to do? A. Give her ID card to him. B. Move a table. C. Sign for a parcel. 2. What does the woman think of cleaning the shirt? A. Easy. B. Time-wasting. C. Impossible. 3. How does the woman most probably go to work? A. By car. B. By bike. C. By bus. 4. What relation is Tom to the woman? A. Her teacher. B. Her agent. C. Her husband. 5. What are the speakers mainly talking about? A. A picnic. B. The weather. C. A forecast. 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。
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。
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。
高一(创新班)下学期第二次联考英语试题
江西省新余一中、樟树中学等六校2020-2021学年高一（创新班）下学期第二次联考英语试题学校:___________姓名：___________班级：___________考号：___________一、阅读选择New Zealand Car RentalCustomers who demand superior service and value hire Apex rental cars. With a wide section of over 15 different types of vehicles including small economical cars, medium-sized and full-size sedans (轿车), station wagons (旅行车), 4WDs and people movers as well as 14 conveniently located branches throughout New Zealand, you can be sure to find a vehicle that is perfect for your travel needs.When you hire a car from Apex CarRentals, there are no hidden surprises. All rates include GST (消费税), comprehensive insurance and unlimited kilometers. Apex also offers a very affordable Collision Damage Waiver upgrade option which allows customers to reduce their legal responsibility for paying money for damage in the event of an accident to $0.With Apex there are no hidden costs, guaranteed:*No extra charges for one-way hires and free airport/hotel transfers*No booking or administration fees*No credit card processing fees*No extra driver fees*No surcharges for seniors and divers under 25*No hidden fees; all rates include GST, insurance and unlimited kilometersFree Ferry TicketsCrossing Cook Strait? Your vehicle ferry ticket could be free, and Apex can help you book passenger tickets. Check vehicle availability for more information.Apex CarRental Optional ExtrasWinter Extras1．Which of the following is an advantage of Apex CarRentals?A．It has big economical cars. B．It has 15 branches in New Zealand. C．It has more than 15 types of vehicles. D．It has a wide section of payment methods. 2．On what condition will you be charged extra fees?A．Hiring a driver. B．Purchasing insurance.C．Paying by credit card. D．Renting a child seat.3．What can we learn from the text?A．Apex can’t help you book passenger tickets.B．You need not pay to rent ski and snowboard racks or snow chains.C．There are extra charges for one-way hires and airport hotel transfers.D．Customers can pay no money for damage when hiring a car from Apex CarRentals.My best friend travelled to stay with my family last weekend. when she arrived, she went straight to the kitchen and, without asking, ate some dried fruit. She wasn’t being rude. I knew she would do this. We’ve known each other for almost 20 years. She can eat anything she wants from my kitchen. Indeed, I bought her favorite fruits and snacks at the shop that morning.Our long weekend together was simple. I was recovering from surgery and couldn't go to shopping malls, We passed the time running errands（差事）, but there's never been a quiet moment. We’ve lived in different cities for almost a decade, Reunions demand conversation.Our personalities are matched, to be sure，and a shared history is indescribably（难以形容地）valuable. We were competitors at high school before bonding. Then we discovered the many interests that we had in common. Our friendship developed itself quickly. We stayed companions and supported each other through law school and through our first jobs. Ours is afriendship for the ages.There is something special about friends who know everything about you. They are rare. They have seen your bright lights of achievements, the depths of desperation（绝望的境地）and the boring routine of the in-between. It's special to unpack feelings without wasting time filling in the blanks. As my long weekend shows, with such friends we don't have to “do”, but we simply have to “be”. We drop the act, the performance, the public version of ourselves, The special friendships are those which never fail to delight, whose continuation is worth the extra effort, despite distance and difference in our separate lives. I had the very great joy of this reminder last weekend. I'm lucky to have found this friend, and to see a future where her companionship remains. Being together is perfection.4．What can we infer from Para. 2?A．They talked constantly about their lives. B．They enjoyed running errands. C．They had a noisy weekend D．They quarreled at times.5．What does the author appreciate about their friendship?A．They help each other achieve in their fieldsB．They’re best friends despite different hobbiesC．They inspire each other to be their best selves.D．They' re comfortable just being themselves6．What message is conveyed in the last paragraph?A．Real friends are easy to makeB．Good friendships need devotionC．Special friendships may fail to delight.D．Distance and difference bring friendship7．Why does the author write this text?A．To express thanks to her best friendB．To share the friendship she treasures.C．To describe what her best friend is likeD．To explain how she offers help to her friendDrivers who drive a little too close to cyclists on the road could soon be caught on the spot. A new technology adopted by legal departments in Ottawa could help carry out legal distance between bikers and cars on the road.The device, which is fixed on a bicycle’s handlebars like a bike bell, uses sonar(声呐) technology to measure the distance between the bike and passing cars. The device will make a loud noise if the car is within one meter of the bike, the legal limit in the city of Ottawa, allowing the police rider to radio ahead to his colleagues so that the driver can be pulled over. “The safety of all road users is extremely vital, in cluding cyclists. These cycling changes are directed at encouraging cycling, promoting road safety, and sharing the road,” said Rob Wilkinson, coordinator of the Safer Roads Ottawa Program.The authorities started the program last week with a single sonar device. One police officer rode the bike bearing the device around the city on Tuesday to prove the effectiveness(有效性) of the technology. Within a few minutes of riding, the device was beeping, registering that two drivers had violated the one-meter distance requirement. The drivers were pulled over and given brochures informing them that they had broken the safe distance law.Wilkinson noted that the device is not currently being used to issue fines, which can go up to $110, and that there are no plans to use it for enforcement(执法)in the future. At this point, its main use is to spread awareness about the safe distance law, which was passed last September in an effort to encourage rider safety and reduce deadly crashes.8．What will happen if the safe distance is beyond the legal limit?A．The cyclist will soon be caught on the spot.B．The police will make the driver stop by the road.C．The driver will be arrested for driving too fast.D．The device will at once call the police of itself.9．What’s the main purp ose of using the device?A．To make the bicycle attractive. B．To encourage people to walk.C．To guarantee road safety. D．To warn drivers of danger. 10．Which of the following can replace the underlined word “beeping” in paragraph 3? A．Making a loud noise. B．Receiving an urgent message. C．Sending a stop signal. D．Radioing the police rider.11．What does Wilkinson say about the device?A．It is being developed at present. B．It still has room for improvement. C．It may be used to fine drivers later. D．It helps reduce traffic accidents.Japanese scientists and technology companies are coming up with new ways to deal with employee shortages in delivery service. How exactly? By introducing a robot that can deliver food to your home. A Japanese company ZMP has launched robot tests recently.The ZMP’s delivery robot in the process of testing is a red box that measurs109 cm and 133 cm in heights and lengths respectively. It is designed to carry up to 100 kilograms of anything with an approximate speed of 4 miles an hour. It has its own navigating system and a map, sensors (传感器) and cameras that allow it to self-dive. These, for now, are its main technical abilities. The next developments will be control of food temperature, and perhaps voice control and speaking abilities.The robot will bested together with a local sushi (寿司) delivery company Ride on Express Co. Upon making a sushi order, customers will receive codes on their smart phones that will allow them to unlock the robot and get their orders out. Similar testing attempts have been undertaken by Domino’s Pizza in Australia, where they tested a delivery robot a year ago.In any case, before robots are able to deliver food or any other goods to real customers, massive testing on public roads or in the public airspace will have to take place. While it is in the government’s best interest to make up for the shortage of delivery in the Japanese labor market, it is clear that self-driving machines of any kind will not be permitted on the roads until they are fully tested and proven safe enough for public roads. According to the experts, this could take another 3 to 5 years.In any case, robot delivery is not such a distant future after all, and recent developments show that there is significant market demand for such type of delivery.12．How do Japanese handle the employee shortages in delivery service?A．By introducing a robot that can cook food. B．By employing science and technology. C．By increasing the salary. D．By delivering more at a time. 13．What is the second paragraph mainly about?A．The basic data of delivery robots. B．The working principle of delivery robot. C．The introduction of delivery robots. D．The appearance of delivery robots. 14．What’s the attitude of Japanese government to delivery robots?A．Critical. B．In different. C．Worried. D．Interested. 15．What can be the best title for the text?A．Robot Delivery: New Tests in Japan B．Robot Delivery Coming to Use in Japan C．Employee Shortage in Delivery in Japan D．Latest Automatic Service: Robots二、七选五Find Your Self-worthDo you measure your personal self-worth based on what others think of you? Do you work really hard to make others like or support you? Do you lack confidence? 16．Increasing self-image will help build your confidence and your personal self-worth. So how?Pay attention to your body language. How you hold your physical body has an effect on how others see you and on how you personally feel. Similarly, how you feel often determines your body language and your outward behavior. 17．A simple way to make you feel good is to check your body language. Stand tall with your head held high, shoulders squared, back straight and legs solid; you will feel more in control.Develop your own measuring stick. 18．Don’t let others influence your view of success or happiness. Take time to recognize your achievements no matter how big or small they are. Keep track of your achievements and proud moments in a journal. If you keep a personal journal or even a work journal, start writing in the journal from back to front keeping track of all the little and big things that happen every day.19．Here’s a simple exercise you can do to gain a richer way of thinking on how you are seen by others. Take a close look at yourself and create a list of your unique strengths. Who are you? 20．Ask friends, families to share with you what they see as your key characteristics.Now you know that you’re more than you think you are. So, create a positive picture of your future and then confidently move forward.A．Know your worth.B．The two go hand in hand.C．What sets you apart from others?D．Do you beat yourself up with your own language?E.If your answer is yes, perhaps you just don’t consider yourself worthy.F.Know what’s important to yo u and take time to do what you value most.G.Besides, take a step towards something you really want by taking action.三、完形填空It was an early blow to my self-confidence. I was attending my first group meeting in the lab 21 a postdoc (博士后), and I was 22 that I could follow most of thediscussion. Then, in front of everyone, the professor turned to me and asked about my previous accomplishments. I 23 . As a PhD student. I had achieved lots to be 24 of. But 25 of those were in my home country of Brazil. Now I was in Washington D．C．and I didn’t know what the word “accomplishment” meant. All I could say was “I don’t know”.I had decided to do a postdoc abroad because I thought the training would 26 me a teaching position. I applied for it and received a Brazil in government fellowship to spend 18 months working abroad and got a position in a lab. It all seemed so easy—until I 27 started.In a new country, I 28 . After the 29 lab meeting incident, my confidence took another 30 . The paper I had finished painstakingly (费力地) did not 31 . I used to be invited to give talks. Now, I was asked to speak more 32 in case that others couldn’t follow me because of my accent. My confidence was 33 .After months of 34 , I reminded myself that I had made the 35 to succeed.I needed to do something to 36 my confidence. I thought a change of environment might be what I needed. So I made the 37 .In another new lab, I still felt not confident. 38 , I was determined to make the most of the time I had. There were more opportunities to communicate with other, which forced me to 39 more. It was awful at first, but with practice I began to feel more confident in my English. I proposed new projects. I began to receive 40 feedback (反馈) on my presentations. My confidence is back.21．A．for B．as C．like D．with 22．A．aware B．afraid C．glad D．amazed 23．A．suffered B．fell C．froze D．refused 24．A．thankful B．protective C．ashamed D．proud 25．A．any B．all C．none D．little 26．A．secure B．lead C．land D．supply 27．A．clearly B．frequently C．hardly D．actually 28．A．failed B．struggled C．crashed D．cried 29．A．exciting B．interesting C．entertaining D．embarrassing 30．A．hit B．loss C．failure D．pain 31．A．come about B．come on C．come out D．come in 32．A．naturally B．slowly C．quickly D．accurately33．A．drowned B．changed C．destroyed D．delayed 34．A．self-harm B．self-help C．self-rule D．self-doubt 35．A．promise B．mark C．permission D．comment 36．A．regain B．return C．review D．replace 37．A．reaction B．balance C．move D．plan 38．A．Otherwise B．Besides C．Then D．However 39．A．write B．read C．talk D．listen 40．A．positive B．disturbing C．subjective D．terrible四、用单词的适当形式完成短文阅读下面短文，在空白处填入1个适当的单词或括号内单词的正确形式。
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A. Hiring a driver.B. Purchasing insurance.
C. Paying by credit card.D. Renting a child seat.
23.What can we learn from the text?
When you hire a car from Apex CarRentals, there are no hidden surprises. All rates include GST (消费税), comprehensive insurance and unlimited kilometers. Apex also offers a very affordable Collision Damage Waiver upgrade option which allows customers to reduce their legal responsibility for paying money for damage in the event of an accident to $0.
A. In the station.B. At a park.C. In a restaurant.
听第8段材料，回答第10至12题。
10.How many coats are mentioned by the man?
A. Two.B. Three.C. Four.
11.What is the woman’s attitude to the man at last?
16.How many Pulitzer Prizes did Carl win for his poetry?
A. Three.B. Two.C. One.
听第10段材料，回答第17至20题。
17.What is the probable purpose of the passage?
A. To deal with concerns.B. To report a piece of news.C. To introduceBentleyCollege.
A. Apex can’t help you book passenger tickets.
B. You need not pay to rent ski and snowboard racks or snow chains.
C. There are extra charges for one-way hires and airport hotel transfers.
*No credit card processing fees
*No extra driver fees
*No surcharges for seniors and divers under 25
*No hidden fees; all rates include GST, insurance and unlimited kilometers
5.What are the speakers talking about?
A. Where to hold the party.B. Whether to go to the party.C. How to finish the homework.
第二节（共15小题；每小题1.5分，满分22.5分）
18.What will the students be educated well in atBentleyCollege?
A. Art.B. Business.C. Computer.
19.Which skill is not what Bentley students try to develop?
14.Which place did Carl fist travel to?
A. Iowa.B. Wisconsin.C. Chicago.
15.What are Carl’s stories mainly about?
A. Kings and queens.B. Talking animals.C. Modern buildings or vehicles.
Free Ferry Tickets
CrossingCook Strait? Your vehicle ferry ticket could be free, and Apex can help you book passenger tickets. Check vehicle availability for more information.
Apex CarRental Optional Extras
GPS navigation hire
$7.60/day
Affordable Roadside Assistance Cover
$4.00/day
4G portable wi-fi units hire
$12.00/day
Child and booster seats (车载的儿童座椅)
1.What does the woman want to do?
A. Take the ordered cake.B. Buy a birthday cake.C. Change the cake.
2.What could the man’s job be?
A. A cleaner.B. A hotel manager.C. A security guard.
听第7段材料，回答第8至9题。
8.What’s the matter with the woman?
A. Her car doesn’t work.B. There is no taxi there.C. She doesn’t catch the bus.
9.Where does the conversation most probably take place?
A. Glad.B. Patient.C. Annoyed.
12.Whats the probable relationship between the speakers?
A. Policewoman and citizen.B. Driver and passenger.C. Shop assistant and customer.
听第9段材料，回答第13至16题。
13.What did Carl do when he was in the eighth grade?
A. He moved to a new school.
B. He left school and started to work.
C. He had a more American-like name.
D. Customers can pay no money for damage when hiring a car from Apex CarRentals.
B
My best friend travelled to stay with my family last weekend. When she arrived, she went straight to the kitchen and, without asking, ate some dried fruit. She wasn’t being rude. I knew she would do this. We’ve known each other for almost 20 years. She can eat anything she wants from my kitchen. Indeed, I bought her favorite fruits and snacks at the shop that morning.
A. It has big economical cars.B. It has 15 branches inNew Zealand.
C. It has more than 15 types of vehicles.D. It has a wide section of payment methods.
With Apex there are nபைடு நூலகம் hidden costs, guaranteed:
*No extra charges for one-way hires and free airport/hotel transfers
*No booking or administration fees
A. Class Two.B. Class Three.C. Class Four.
7.What will the girl do next Saturday after o on?
A. Watch a game.B. Study in the classroom.C. Play basketball.
江西省六校2019-2020学年高一年级下学期第二次联考英语试卷
（时间：120分钟总分：150分）
第一部分听力（共两节，满分30分）
第一节短对话（共5小题；每小题1.5分，满分7.5分）
听下面5段对话。每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题，每段对话仅读一遍。
$12.00 each
Winter Extras
Ski and snowboard racks (搁物架) hire
$25.00 each
Snow chains hire
$25.00 each
21.Which of the following is an advantage of Apex CarRentals?