重庆市第一中学2020届高三英语3月月考试题(PDF)

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重庆一中2020-2021学年度初2023届第一次月考试题(图片版无答案)

重庆一中2020-2021学年度初2023届第一次月考试题(图片版无答案)

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重庆市巴蜀中学2020届高三高考适应性月考(七)英语试题 PDF版含答案

重庆市巴蜀中学2020届高三高考适应性月考(七)英语试题 PDF版含答案
Nakushi 没有告诉她父母是因为知道告诉父母只会受到责备而不会有任何帮助。故选 C。 25.B 事实细节题。根据第二段中信息,可知 Kapil 是因为有学习障碍而被学校拒绝入学。
故选 B。
英语参考答案·第 3 页(共 12 页)
26.C 推理判断题。根据第三段“Inside one of the schools,some of the classrooms have low benches and desks.In others,the little girls sit on the floor,books in their laps.”等 内容可知,在印度公立学校的教学设施很差,不令人满意。故选 C。
第二节 【语篇导读】本篇是说明文。对于读者如何做一个有责任感的人提出建议。
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【语篇导读】本篇是社科文。介绍了美国的浪漫主义思潮的开始及特点。 28.B 事实细节题。根据第二段第三句“This was a time of growth and expansion and their
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英语参考答案·第 2 页(共 12 页)
【解析】
Li Hua
第二部分 阅读理解
第一节 A

2021届重庆市第一中学校高三上学期第三次月考数学试题(解析版)

2021届重庆市第一中学校高三上学期第三次月考数学试题(解析版)

2021届重庆市第一中学校高三上学期第三次月考数学试题一、单选题1.复数z 满足21iz i=-,则复数z 的虚部为()A .﹣1B .1C .iD .﹣i【答案】B【分析】利用复数的除法运算化简211ii i=-+-,再利用复数的代数形式求出结果.【详解】解:∵()()()()2121211112i i i i i z i i i i ++====-+--+,则复数z 的虚部为1.故选:B .【点睛】本题考查复数的除法运算.复数的除法运算关键是分母“实数化”,其一般步骤如下:(1)分子、分母同时乘分母的共轭复数;(2)对分子、分母分别进行乘法运算;(3)整理、化简成实部、虚部分开的标准形式.2.已知集合{}22,A xx x Z =<∈∣,则A 的真子集共有()个A .3B .4C .6D .7【答案】D【分析】写出集合{1,0,1}A =-,即可确定真子集的个数.【详解】因为{}22,{1,0,1}A xx x Z =<∈=-∣,所以其真子集个数为3217-=.故选:D.【点睛】本题考查集合的真子集个数问题,属于简单题.3.已知某圆锥的母线长为4,底面圆的半径为2,则圆锥的全面积为()A .10πB .12πC .14πD .16π【答案】B【分析】首先求得底面周长,即侧面展开图的扇形弧长,然后根据扇形的面积公式即可求得侧面积,即圆锥的侧面积,再求得圆锥的底面积,侧面积与底面积的和就是全面积.【详解】底面周长是:2×2π=4π,则侧面积是:14π48π2⨯⨯=,底面积是:π×22=4π,则全面积是:8π+4π=12π.故选B .【点睛】本题考查了圆锥的全面积计算,正确理解圆锥的侧面展开图与原来的扇形之间的关系是解决本题的关键,理解圆锥的母线长是扇形的半径,圆锥的底面圆周长是扇形的弧长.4.为了衡量星星的明暗程度,古希腊天文学家喜帕恰斯在公元前二世纪首先提出了星等这个概念.星等的数值越小,星星就越亮;星等的数值越大它的光就越暗.到了1850年,由于光度计在天体光度测量的应用,英国天文学家普森又提出了亮度的概念,天体的明暗程度可以用星等或亮度来描述.两颗星的星等与亮度满足()12212.5lg lg m m E E -=-,其中星等为k m 的星的亮度为(1,2)k E k =.已知“心宿二”的星等是1.00,“天津四”的星等是1.25,则“心宿二”的亮度大约是“天津四”的()倍.(当||x 较小时,2101 2.3 2.7x x x ≈++)A .1.27B .1.26C .1.23D .1.22【答案】B【分析】把已知数据代入公式计算12E E .【详解】由题意211 1.25 2.5(lg lg )E E -=-,12lg0.1E E =,∴0.1212101 2.30.1 2.70.1 1.257 1.26E E =≈+⨯+⨯=≈.故选:B .【点睛】本题考查数学新文化,考查阅读理解能力.解题关键是在新环境中抽象出数学知识,用数学的思想解决问题.5.向量,a b 满足||1a = ,a 与b 的夹角为3π,则||a b - 的取值范围为()A .[1,)+∞B .[0,)+∞C .1,2⎡⎫+∞⎪⎢⎣⎭D .3,2⎫+∞⎪⎢⎪⎣⎭【答案】D【分析】把||a b -用数量积表示后结合函数的性质得出结论.【详解】22222||()2121cos 3a b a b a a b b b b π-=-=-⋅+=-⨯⨯+ 21b b -+= 2134423b ⎛⎫=+≥⎪⎝⎭- ,所以3||2a b -≥ .1||2b = 时取得最小值.故选:D .【点睛】本题考查平面向量的模,解题关键是把模用向量的数量积表示,然后结合二次函数性质得出结论.6.已知三棱锥P ABC -,过点P 作PO ⊥面,ABC O 为ABC ∆中的一点,,PA PB PB PC ⊥⊥,PC PA ⊥,则点O 为ABC ∆的()A .内心B .外心C .重心D .垂心【答案】D【分析】连接AO 并延长交BC 于一点E ,连接PO ,由于PA ,PB ,PC 两两垂直可以得到PA ⊥面PBC ,而BC ⊂面PBC ,可得BC ⊥PA ,由PO ⊥平面ABC 于O ,BC ⊂面ABC ,PO ⊥BC ,可得BC ⊥AE ,同理可以证明CO ⊥AB ,又BO ⊥AC .故O 是△ABC 的垂心.【详解】连接AO 并延长交BC 于一点E ,连接PO ,由于PA ,PB ,PC 两两垂直可以得到PA ⊥面PBC ,而BC ⊂面PBC ,∴BC ⊥PA ,∵PO ⊥平面ABC 于O ,BC ⊂面ABC ,∴PO ⊥BC ,∴BC ⊥平面APE ,∵AE ⊂面APE ,∴BC ⊥AE ;同理可以证明CO ⊥AB ,又BO ⊥AC .∴O 是△ABC 的垂心.故选D .【点睛】本题主要考查了直线与平面垂直的性质,解题时要注意数形结合,属于基本知识的考查.7.设sin5a π=,b =,2314c ⎛⎫= ⎪⎝⎭,则()A .a c b <<B .b a c <<C .c a b<<D .c b a<<【答案】C【分析】借助中间量1和12比较大小即可.【详解】解:由对数函数y x =在()0,∞+单调递增的性质得:1b =>=,由指数函数12xy ⎛⎫= ⎪⎝⎭在R 单调递减的性质得:2413311142212c ⎛⎫⎛⎫⎛⎫= ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭=<=,由三角函数sin y x =在0,2π⎛⎫ ⎪⎝⎭上单调递增的性质得1sin sin 562a ππ=>=.所以c ab <<.故选:C.【点睛】本题考查利用函数的单调性比较大小,考查运算能力,化归转化思想,是中档题.本题解题的关键在于借助中间量1和12,尤其在比较a 与c 的大小时,将c 变形得24331142c ⎛⎫⎛⎫= ⎪ =⎪⎝⎭⎝⎭,进而与12比较大小是重中之核心步骤.8.已知三棱锥P ABC -的四个顶点均在同一个确定的球面上,且BA BC ==,2ABC π∠=,若三棱锥P ABC -体积的最大值为3,则其外接球的半径为()A .2B .3C .4D .5【答案】A【分析】由题意分析知三棱锥P ABC -体积的最大时,P ,O ,O '共线且O P '⊥面ABC ,P 在大于半球的的球面上,根据棱锥体积公式求得||O P ',进而应用勾股定理求外接球的半径.【详解】由题意知:AC 中点O '为面ABC 外接圆圆心,若外接球球心为O ,半径为R ,三棱锥P ABC -体积的最大时,P ,O ,O '共线且O 在P ,O '之间,∴1||33P ABC ABC V S O P -'=⋅⋅= ,1||||32ABC S BA BC =⋅⋅= ,即||3O P '=,||||32AC O C '==,所以()22222'|||'|33O C OC OO R R =-=--=,解得2R =,故选:A【点睛】关键点点睛:理解三棱锥P ABC -体积的最大时P 的位置及与球心、底面外接圆圆心的关系,结合棱锥体积公式、勾股定理求球体半径.二、多选题9.设m 、n 是两条不同的直线,α、β是两个不同的平面,下列命题中错误..的是()A .若,,//m n m n αβ⊂⊂,则//αβB .若,m n m α⊂⊥,则n α⊥C .若,m n αα^Ì,则m n ⊥D .若//,,m n αβαβ⊂⊂,则//m n【答案】ABD【分析】根据空间线、面关系,结合空间关系相关图例以及线线、线面、面面间的平行、垂直判定与性质,即可知选项的正误.【详解】A :,,//m n m n αβ⊂⊂,α、β不一定平行,错误.B :,m n m α⊂⊥,n 不一定垂直于α,错误.C :由线面垂直的性质:,m n αα^Ì,则必有m n ⊥,正确.D ://,,m n αβαβ⊂⊂,m 、n 不一定平行,错误.故选:ABD10.下列函数中,在(0,1)内是减函数的是()A .||12x y ⎛⎫= ⎪⎝⎭B .212log y x =C .121=+y x D .2log sin y x=【答案】ABC【分析】根据复合函数的单调性判断确定选项中各函数是否为减函数即可.【详解】A :1(2t y =为减函数,||t x =在(0,1)上为增函数,所以||12x y ⎛⎫= ⎪⎝⎭为减函数;B :12log y t =为减函数,2t x =在(0,1)上为增函数,所以212log y x =为减函数;C :1y t =为减函数,21t x =+在(0,1)上为增函数,所以121=+y x 为减函数;D :2log y t =为增函数,sin t x =在(0,1)上为增函数,所以2log sin y x =为增函数;故选:ABC【点睛】结论点睛:对于复合函数的单调性有如下结论1、内外层函数同增或同减为增函数;2、内外层函数一增一减为减函数;11.下列关于函数1()2sin 26f x x π⎛⎫=+⎪⎝⎭的图像或性质的说法中,正确的为()A .函数()f x 的图像关于直线83x π=对称B .将函数()f x 的图像向右平移3π个单位所得图像的函数为12sin 23y x π⎛⎫=+ ⎪⎝⎭C .函数()f x 在区间5,33ππ⎛⎫-⎪⎝⎭上单调递增D .若()f x a =,则1cos 232a x π⎛⎫-=⎪⎝⎭【答案】AD 【分析】令1262x k πππ+=+得到对称轴,即可判断A ;根据平移变换知识可判断B ;求出其单调增区间即可判断C ;利用配角法即可判断D.【详解】对于A ,令1262x k πππ+=+()k ∈Z ,解得22()3x k k Z ππ=+∈,当1k =时,得83x π=,故A 正确;对于B ,将函数()f x 的图像向右平移3π个单位,得112sin[()]2sin 2362y x x ππ=-+=,故B 错误;对于C ,令122()2262k x k k Z πππππ-+<+<+∈4244()33k x k k Z ππππ⇒-+<<+∈,故C 错误;对于D ,若12sin()26x a π+=,则11cos()sin[()]23223x x πππ-=+-=1sin()262ax π+=,故D 正确.故选:AD【点睛】方法点睛:函数()sin (0,0)y A x B A ωϕω=++>>的性质:(1)max min =+y A B y A B =-,.(2)周期2π.T ω=(3)由()ππ2x k k +=+∈Z ωϕ求对称轴(4)由()ππ2π2π22k x k k -+≤+≤+∈Z ωϕ求增区间;由()π3π2π2π22k x k k +≤+≤+∈Z ωϕ求减区间.12.定义在(0,)+∞上的函数()f x 的导函数为()'f x ,且()()f x f x x'<,则对任意1x 、2(0,)x ∈+∞,其中12x x ≠,则下列不等式中一定成立的有()A .()()()1212f x x f x f x +<+B .()()()()21121212x xf x f x f x f x x x +<+C .()1122(1)x x f f <D .()()()1212f x x f x f x <【答案】ABC【分析】构造()()f x g x x=,由()()f x f x x '<有()0g x '<,即()g x 在(0,)+∞上单调递减,根据各选项的不等式,结合()g x 的单调性即可判断正误.【详解】由()()f x f x x '<知:()()0xf x f x x'-<,令()()f x g x x =,则()()()20xf x f x g x x '-='<,∴()g x 在(0,)+∞上单调递减,即122112121212()()()()()g x g x x f x x f x x x x x x x --=<--当120x x ->时,2112()()x f x x f x <;当120x x -<时,2112()()x f x x f x >;A :121()()g x x g x +<,122()()g x x g x +<有112112()()x f x x f x x x +<+,212212()()x f x x f x x x +<+,所以()()()1212f x x f x f x +<+;B:由上得21121212()()()()x f x x x x f x x x -<-成立,整理有()()()()21121212x xf x f x f x f x x x +<+;C :由121x >,所以111(2)(1)(2)(1)21x x x f f g g =<=,整理得()1122(1)x x f f <;D :令121=x x 且121x x >>时,211x x =,12111()()()()g x g x f x f x =,12()(1)(1)g x x g f ==,有121()()g x x g x >,122()()g x x g x <,所以无法确定1212(),()()g x x g x g x 的大小.故选:ABC【点睛】思路点睛:由()()f x f x x '<形式得到()()0xf x f x x'-<,1、构造函数:()()f x g x x =,即()()()xf x f x g x x'-'=.2、确定单调性:由已知()0g x '<,即可知()g x 在(0,)+∞上单调递减.3、结合()g x 单调性,转化变形选项中的函数不等式,证明是否成立.三、填空题13.若一个球的体积为323π,则该球的表面积为_________.【答案】16π【解析】由题意,根据球的体积公式343V R π=,则343233R ππ=,解得2R =,又根据球的表面积公式24S R π=,所以该球的表面积为24216S ππ=⋅=.14.设向量a ,b 不平行,向量a b λ+ 与2a b + 平行,则实数λ=_________.【答案】12【解析】因为向量a b λ+ 与2a b + 平行,所以2a b k a b λ+=+ (),则{12,k k λ==,所以12λ=.【解析】向量共线.15.一般把数字出现的规律满足如图的模型称为蛇形模型:数字1出现在第1行;数字2,3出现在第2行;数字6,5,4(从左至右)出现在第3行;数字7,8,9,10出现在第4行,依此类推,则第21行从左至右的第4个数字应是____________.【答案】228【分析】由题知,第n 行有n 个数字,奇数行从右至左由小变大,偶数行从左至右由小变大,则前20行共有20(120)123202102+++++==L 个数字,第21行最左端的数为21021231+=,从左到右第4个数字为228.【详解】观察数据可知,第n 行有n 个数字,奇数行从右至左由小变大,偶数行从左至右由小变大,则前20行共有20(120)123202102+++++==L 个数字,第21行最左端的数为21021231+=,所以第21行从左到右第4个数字为228.故答案为:228.【点睛】关键点睛:本题考查合情推理、数列的前n 项和,解题关键要善于观察发现数据特征,考查了学生的逻辑思维能力、数据处理能力、运算求解能力,综合性较强,属于较难题型.四、双空题16.已知等比数列{}n a 的公比为q ,且101a <<,20201a =,则q 的取值范围为______;能使不等式12121110m m a a a a a a ⎛⎫⎛⎫⎛⎫-+-++-≤ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭ 成立的最大正整数m =______.【答案】(1,)+∞4039【分析】根据已知求得1a 的表达式,由此求得q 的取值范围.根据12121110m m a a a a a a ⎛⎫⎛⎫⎛⎫-+-++-≤ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭ 成立列不等式,化简求得m 的取值范围,从而求得最大正整数m .【详解】由已知201911201911a qa q =⇒=,结合101a <<知2019101q <<,解得1q >,故q 的取值范围为(1,)+∞.由于{}n a 是等比数列,所以1n a ⎧⎫⎨⎬⎩⎭是首项为11a ,公比为1q 的等比数列.要使12121110m m a a a a a a ⎛⎫⎛⎫⎛⎫-+-++-≤ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭ 成立则1212111m ma a a a a a +++≤+++ 即()111111111m m a q a q q q⎛⎫-⎪-⎝⎭≤--,将120191a q=代入整理得:40394039m q q m ≤⇒≤故最大正整数4039m =.故答案为:(1,)+∞;4039【点睛】本小题主要考查等比数列的性质,考查等比数列前n 项和公式,属于中档题.五、解答题17.在四棱柱1111ABCD A B C D -中,底面ABCD 是等腰梯形,M 是线段AB 的中点,1160,22,2,DAB AB CD DD C M ∠=︒====(1)求证:1//C M 平面11A ADD ;(2)求异面直线 CM 与1DD 所成角的余弦值.【答案】(1)证明见解析;(2)14.【分析】(1)易得1111//,C D MA C D MA =,则四边形11AMC D 为平行四边形,得到11//C M D A ,再利用线面平行的判定定理证明.(2)由//CM DA ,将异面直线CM 与1DD 成的角,转化为 DA 与1DD 相交所成的角,然后在1ADD ,利用余弦定理求解.【详解】(1)因为四边形ABCD 是等腰梯形,且2AB CD =,所以//AB DC .又由M 是AB 的中点,因此//CD MA 且CD MA =.如图所示:连接1AD ,在四棱柱1111ABCD A B C D -中,因为1111//,CD C D CD C D =,可得1111//,C D MA C D MA =,所以四边形11AMC D 为平行四边形.因此11//C M D A ,又1C M ⊄平面11A ADD ,1D A ⊂平面11A ADD ,所以1//C M 平面11A ADD .(2)因为//CM DA ,所以异面直线CM 与1DD 成的角,即为 DA 与1DD 相交所成的直角或锐角,在1ADD中,1C M =,所以111,2AD AD DD ===,由余弦定理可得:22211111cos 24AD DD AD ADD AD DD +-∠==-⋅,所以异面直线CM 和1DD 余弦值为14.【点睛】方法点睛:判断或证明线面平行的常用方法:(1)利用线面平行的定义,一般用反证法;(2)利用线面平行的判定定理(a ⊄α,b ⊂α,a ∥b ⇒a ∥α),其关键是在平面内找(或作)一条直线与已知直线平行,证明时注意用符号语言的叙述;(3)利用面面平行的性质定理(α∥β,a ⊂α⇒a ∥β);(4)利用面面平行的性质(α∥β,a ⊄β,a ∥α⇒a ∥β).18.已知数列{}n a 满足:13a =,且对任意的n *∈N ,都有1,1,n n a a +成等差数列.(1)证明数列{}1n a -等比数列;(2)已知数列{}n b 前n 和为n S ,条件①:()1(21)n n b a n =-+,条件②:11n n n b a +=-,请在条件①②中仅选择一个条件作为已知条件.............来求数列{}n b 前n 和n S .【答案】(1)证明见解析;(2)答案不唯一,具体见解析.【分析】(1)由条件得121n n a a +=-,利用等比数列定义可得证.(2)选条件①得(21)2nn b n =+,选条件②得1(1)()2nn b n =+⋅利用错位相减法可得解.【详解】(1)由条件可知112n n a a ++=,即121n n a a +=-,∴()1121n n a a +-=-,且112a -=∴{}1n a -是以112a -=为首项,2q =为公比的等比数列,∴12nn a -=,∴()21nn a n N*=+∈(2)条件①:()1(21)(21)2nn n b a n n =-+=+,123325272(21)2nn S n =⋅+⋅+⋅+++⋅ 23412325272(21)2n n S n +=⋅+⋅+⋅+++⋅利用错位相减法:123413222222222(21)2nn n S n +-=⋅+⋅+⋅+⋅++⋅+⋅- 118(12)6(21)212n n n S n -+--=++⋅--化简得()12(21)2n n S n n N +*=-+∈条件②:11(1)()12nn n n b n a +==+⋅-231111234(1)2222n nS n =⋅+⋅+⋅+++⋅ 234111111234(1)22222n n S n +=⋅+⋅+⋅+++⋅ 利用错位相减法:23411111111(1)222222n n n S n +=++++-+⋅ 1111[1()]11421(1)12212n n n S n -+-=+-+⋅-化简得()13(3)(2n n s n n N *=-+∈【点睛】错位相减法求和的方法:如果数列{}n a 是等差数列,{}n b 是等比数列,求数列{}n n a b 的前n 项和时,可采用错位相减法,一般是和式两边同乘以等比数列{}n b 的公比,然后作差求解;在写“n S ”与“n qS ”的表达式时应特别注意将两式“错项对齐”以便下一步准确写出“n n S qS -”的表达式19.已知椭圆C 的两个焦点分别为12(1,0),(1,0)F F -,短轴的两个端点分别为12,B B .且122B B =.(1)求椭圆C 的标准方程;(2)过点2F 的直线l 与椭圆C 相交于P ,Q 两点,且11F P FQ ⊥ ,求直线l 的方程.【答案】(1)2212x y +=;(2)10x +-=,或10x -=.【分析】(1)由题干条件可得c 和b 的值,进而求出2a 的值,从而求出椭圆方程;(2)首先考虑斜率不存在的情况,不符合题意;当斜率存在时,联立方程,可得()22121222214,2121k k x x x x k k -+=⋅=++,又110F P FQ ⋅= ,向量坐标化可得()()()2221212111110k x x k x x k F P FQ ⋅--==++++uuu r uuu r ,代入1212,x x x x +⋅,化简,即可求出k 的取值,从而求出直线方程.【详解】解(1)由条件可知:1c =,又122B B =,所以1b =,则22a =,所以椭圆C 的方程为2212x y +=(2)当直线l 的斜率不存在时,其方程为1x =,不符合题意;当直线l 的斜率存在时,设直线l 的方程为(1)y k x =-,22(1)12y k x x y =-⎧⎪⎨+=⎪⎩得()()2222214210k x k x k +-+-=,()2810k ∆=+>,设()()1122,,,P x y Q x y ,则()22121222214,2121k k x x x x k k -+=⋅=++,()()1111221,,1,F P x y F Q x y =+=+ ,∵110F P FQ ⋅= ,即()()()()()22212121212111110x x y y k x x k x x k +++=+--+++=,即()()()222222221411()102121k k kk k k k -+--++=++化简得:2201172k k =+-解得217,77k k ==±.故直线l的方程为10x +-=,或10x --=.【点睛】方法点睛:(1)将向量转化为坐标的关系;(2)联立直线和椭圆,求出两根之和,两根之积;(3)将两根之和和两根之积代入坐标关系中,解出k .20.已知()cossin 222x x x f x ⎛⎫=+ ⎪⎝⎭,记ABC 的内角,,A B C 的对边分别为,,a b c .(1)求()f B 的取值范围;(2)当4a =,433b =,且()f B 取(1)中的最大值时,求ABC 的面积.【答案】(1)30,12⎛+ ⎝⎦;(2)833或433【分析】(1)利用公式对函数化简,根据B 角的范围,求函数值域.(2)由(1)求出B 的大小,利用正弦定理和三角形面积公式即可求出结果.【详解】(1)2()cossin sin cos 222222x x x x x x f x ⎛⎫=+=+ ⎪⎝⎭13(cos 1)3sin sin 2232x x x π+⎛⎫=+=++ ⎪⎝⎭因为B 为三角形的内角,所以(0,)B π∈所以4,333B πππ⎛⎫+∈ ⎪⎝⎭,所以3()0,12f B ⎛∈+ ⎝⎦(2)34()11,,23333f B B B ππππ⎛⎫⎛⎫=++=+∈ ⎪ ⎝⎭⎝⎭,,326B B πππ∴+==,由正弦定理得:4343sin 1sin sin sin 22a b A A B A =⇒=⇒=()0,,3A A ππ∈∴=,或23A π=,若3A π=,则2C π=,183sin 23ABC S ab C ==若23π=A ,则6π=C,1sin 23==ABC S ab C 【点睛】本题考查了三角恒等变换、正弦定理和三角形面积公式等基本数学知识,考查了数学运算能力和逻辑推理能力,属于中档题目.21.在直三棱柱111ABC A B C -中,112,120,,AB AC AA BAC D D ==∠=分别是线段11,BC B C 的中点,过线段AD 的中点P 作BC 的平行线,分别交,AB AC 于点,M N .(1)证明:平面1A MN ⊥平面11ADD A ;(2)求二面角1A A M N --的余弦值.【答案】(1)证明见解析;(2)155.【分析】(1)根据线面垂直的判定定理即可证明MN ⊥平面ADD 1A 1;又MN ⊂平面A 1MN ,所以平面A 1MN ⊥平面ADD 1A 1;(2)建立空间坐标系,利用向量法求出平面的法向量,利用向量法进行求解即可.【详解】(1)证明:∵AB=AC ,D 是BC 的中点,∴BC ⊥AD ,∵M ,N 分别为AB ,AC 的中点,∴MN ∥BC ,∴MN ⊥AD ,∵AA 1⊥平面ABC,MN ⊂平面ABC ,∴AA 1⊥MN ,∵AD,AA 1⊂平面ADD 1A 1,且AD∩AA 1=A ,∴MN ⊥平面ADD 1A 1∴,又MN ⊂平面A 1MN ,所以平面A 1MN ⊥平面ADD 1A 1;(2)设AA 1=1,如图:过A 1作A 1E ∥BC ,建立以A 1为坐标原点,A 1E ,A 1D 1,A 1A 分别为x ,y ,z 轴的空间直角坐标系如图:则A 1(0,0,0),A(0,0,1),∵P 是AD 的中点,∴M ,N 分别为AB ,AC 的中点.则31,,122M ⎛⎫ ⎪ ⎪⎝⎭,31,,122N ⎛⎫- ⎪ ⎪⎝⎭,则131,,122A M ⎛⎫= ⎪ ⎪⎝⎭,()10,0,1A A =,)NM = ,设平面AA 1M 的法向量为(),,m x y z=,则100m AM m A A ⎧⋅=⎪⎨⋅=⎪⎩,得10220x y z z ++=⎨⎪=⎩,令1x =,则y =,则()1,m =,同理设平面A 1MN 的法向量为(),,n x y z=,则100n A M n NM ⎧⋅=⎨⋅=⎩,得310220x y z ++=⎪⎨⎪=⎩,令2y =,则1z =-,则()0,2,1n =-,则()15cos ,5m n m n m n ⋅===-⋅,∵二面角A-A 1M-N 是锐二面角,∴二面角A-A 1M-N 的余弦值是155.【点睛】本题主要考查直线垂直的判定以及二面角的求解,建立空间直角坐标系,利用向量法进行求解,综合性较强,运算量较大.22.已知21()(1)2xf x e ax b x =---.其中常数 2.71828e ≈⋅⋅⋅⋅⋅⋅.(1)当2,4a b ==时,求()f x 在[1,2]上的最大值;(2)若对任意0,()a f x >均有两个极值点()1212,x x x x <,(ⅰ)求实数b 的取值范围;(ⅱ)当a e =时,证明:()()12f x f x e +>.【答案】(1)max ()1f x e =-;(2)(ⅰ)1b >;(ⅱ)证明见解析.【分析】(1)由题得2()4(1)x f x e x x =---,()24x f x e x '=--,()2x f x e ''=-,由[1,2]x ∈,可得()0f x ''>,即()'f x 在[1,2]上单增,且2(2)80f e -'=<,即()0f x '<,可知()f x 在[1,2]上单减,求得max ()(1)1f x f e ==-.(2)(ⅰ)利用两次求导可得(,ln )x a ∈-∞时,()'f x 单减;(ln ,)x a ∈+∞时,()'f x 单增,再由()f x 有两个极值点,知(ln )ln 0f a a a a b =--<',即ln b a a a >-恒成立,构造函数()ln g a a a a =-,利用导数求其最大值,可得实数b 的取值范围;(ⅱ)设()()(2),(1)h x f x f x x ''=--<,求导可得()h x 在(,1)-∞单增,得到()(2)f x f x ''<-,可得()()112f x f x ''<-,()()122f x f x ''->,结合()'f x 在(1,)+∞上单增,可得()()122f x f x >-,得到()()()()2222122222222x x f x f x f x f x e e ex ex e -+>-+=+-+-,构造22()22x x M x e e ex ex e -=+-+-,(1)x >,再利用导数证明()2(1)M x M e >=,即可得到()()12f x f x e+>【详解】(1)由2,4a b ==得,2()4(1)x f x e x x =---,求导()24x f x e x '=--,()2x f x e ''=-,[1,2]x ∈ ,2[,]x e e e ∴∈,20x e ∴->,即()0f x ''>()f x '∴在[1,2]上单增,且2(2)80f e -'=<,即[1,2]x ∀∈,()0f x '<,()f x ∴在[1,2]上单减,max ()(1)1f x f e ∴==-.(2)(ⅰ)求导()x f x e ax b '=--,因为对任意0,()a f x >均有两个极值点12,x x ,所以()0f x '=有两个根,求二阶导()x f x e a ''=-,令()0f x ''=,得ln x a=当(,ln )x a ∈-∞时,()0f x ''<,()'f x 单减;当(ln ,)x a ∈+∞时,()0f x ''>,()'f x 单增,由()0f x '=有两个根12,x x ,知(ln )ln 0f a a a a b =--<',即ln b a a a >-对任意0a >都成立,设()ln g a a a a =-,求导()ln g a a '=-,令()0g a '=,得1a =,当(0,1)x ∈时,()0g a '>,()g a 单增;当(1,)x ∈+∞时,()0g a '<,()g a 单减,max (()1)1g g a =∴=,1b ∴>又0,,()ba b f e x f x a -⎛⎫''-=>→+∞→+∞ ⎪⎝⎭Q ,所以实数b 的取值范围是:1b >.(ⅱ)当a e =时,()x f x e ex b '=--,()x f x e e ''=-,令()0f x ''=,得1x =当(,1)x ∈-∞时,()0f x ''<,()'f x 单减;当(1,)x ∈+∞时,()0f x ''>,()'f x 单增,又12,x x 是()0f x '=的两根,且12x x <,121,1x x <∴>,121x ∴->设()()(2),(1)h x f x f x x ''=--<,即22(2)2()2,(1)xxx xe ex b ee x b e e ex e x h x --⎡⎤=-=-------+<⎣⎦,则2()2220x x h x e e e e e -=+->-='()h x ∴在(,1)-∞单增,()(1)0h x h ∴<=,即()(2)f x f x ''<-又11,x <,()()112f x f x ''∴<-,()()122f x f x ''∴->又()f x ' 在(1,)+∞上单增,122x x ∴->,即1222x x x <-<,又()f x 在()12,x x 上单减,()()122f x f x ∴>-()()()()2222122222222x x f x f x f x f x e e ex ex e-∴+>-+=+-+-令22()22x x M x e e ex ex e -=+-+-,(1)x >则2()22x x M x e e ex e -'=--+,2()20x x M x e e e -''=+-≥()M x '∴在(1,)+∞单增,且(1)0M '=,()0M x '∴>,故()M x 在(1,)+∞单增又21x > ,()2(1)M x M e ∴>=,即()()12f x f x e+>【点睛】方法点睛:本题考查利用导数研究函数的单调性,求极值,最值,以及证明不等式,证明不等式的方法:若证明()()f x g x <,(,)x a b ∈,可以构造函数()()()F x f x g x =-,如果()0F x '<,则()F x 在(,)a b 上是减函数,同时若()0F a ≤,由减函数的定义可知(,)x a b ∈时,有()0F x <,即证明了()()f x g x <,考查学生的函数与方程思想,化归与转化思想,考查逻辑思维能力与推理论证能力,属于难题.。

2020届雅礼中学高三第3次月考试卷答案(英语)

2020届雅礼中学高三第3次月考试卷答案(英语)
英语参考பைடு நூலகம்案雅礼版!O
&$3%!上下文串连!作者在第一段提出了自己的建议!下面解释自己的理由!首先作者建议用认真)理性的 态度考虑这个建议被接受后的结果!根据/+-20=-F;+的含义也可判断出!
&"3$!名词辨析!从文章第一句作者就表明自己要提出建议#因此选28@@+2.>0=!如果这样的建议被接受! &%3$!短 语 辨 析!2>.-/08=9 固 定 短 语#此 句 的 主 语 还 是 6-1>;>+2#根 据 下 文 的 .0@+.*+/-6.+/9>==+/-=9
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也不行!+K+=表示强调! ''3$!名词辨析!针对上文的写不出句子#指出学习写作的途径是通过阅读! '(3%!名词辨析!指阅读的另一种形式...朗读! ')3"!上下文串连!阅读的那种安静时间可以变成讲故事的时间#这个时间一结束#电视网络系统就必须提供
好的节目来把人们再吸引回来!
'*3%!短语辨析!401+8:<>.*"提出)提供&#不能用被动语态,401+-4/022"偶遇&,401+-F08."发生#产生&, 401+08."出版#出现&!

重庆市巴蜀中学2020届高考英语适应性月考试题(七)答案

重庆市巴蜀中学2020届高考英语适应性月考试题(七)答案

巴蜀中学2020届高考适应性月考卷(七)英语参考答案第一部分听力(共两节,满分30分)1~5 ACBAB 6~10 BCACA 11~15 ABCBB 16~20 ABCBA第二部分阅读理解(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)21~25 BADCB 26~30 CBBCC 31~35 DBAAC第二节(共5小题;每小题2分,满分10分)36~40 CBDEA第三部分语言知识运用(共两节,满分45分)第一节(共20小题;每小题1.5分,满分30分)41~45 DABBD 46~50 ACDAC 51~55 ADBAB 56~60 CDACA第二节(共10小题;每小题1.5分,满分15分)61.lasting 62.is expected 63.another 64.truly 65.despite66.Elections 67.effective 68.who 69.involve 70.held 第四部分写作(共两节,满分35分)第一节短文改错(共10小题;每小题1分,满分10分)When I was twelve,I went to a soccer field for the first time.I found∧hard to drag myself①itaway from it.But all I could do were make a paper one instead,so I did n’t have enough money to buy1②was ③because/fora football.Every day on my way home,I couldn’t help kick whatever on the road could be kicked.④kickingSeveral weeks late,I felt I was on top of the world.While I got home,a real black and white ball was⑤later ⑥Whenwaiting for me on my desk,in which was my birthday present from my parents.I was very wild with⑦⑧sojoy that I didn’t know what to say.I simply rushed out with a ball and playing with other kids until I⑨the ⑩playedwas completely tired out.第二节书面表达(满分25分)【参考范文】Dear Emily,Having known that you take interest in the Spring Festival in China,I am overwhelmed with eagerness to write to brief you on it and enclosed in my letter are two pictures specially taken during this Spring Festival.The Spring Festival is the grandest festival in China,when China is dominated by iconic red lanterns,loud fireworks,and massive banquets.The first picture was about my family members setting off fireworks on New Year’s Eve.It is believed that2doing so helps keep away evil spirits in general.The other was taken of our family reunion dinner,presenting a table brimful of various dishes.On that particular occasion,fish will be the main dish,as it is commonly held that this will help realize your wishes for the New Year.I do hope what I mentioned above will give you a better understanding of the Spring Festival and that you will have the chance to experience the Spring Festival atmosphere here yourself.Looking forward to your reply.Yours sincerely,Li Hua【解析】第二部分阅读理解第一节A【语篇导读】本篇是应用文。

重庆市第一中学2020-2021学年高三上期第三次月考英语试题及答案

重庆市第一中学2020-2021学年高三上期第三次月考英语试题及答案

2020年重庆一中高2021届高三上期第三次月考英语测试试题卷第一部分听力(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三个选项中选岀最佳选项。

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。

1. Where does the conversation probably take place?A. In a bank.B. In a restaurant.C. In a hospital.2. What will the woman most probably do this Friday?A. Do some shopping.B. Watch a movie.C. Organize a sales promotion.3. What does the woman think of the trip to Indonesia?A. Safe.B. Dangerous.C. Exciting.4. How much does the man make in a year?A. £49,000.B. £50,000.C. £ 60,000.5. What are the speakers mainly talking about?A. The man's travel experience.B. A flight to Romania.C. Family members.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。

听下面一段对话,回答第6和第7两个小题。

6. Who bought the Christmas tree?A. The woman's grandpa.B. The woman's mother.C. The woman’s father.7. How old is the tree?A. 40 years old.B. 75 years old.C. 80 years old.听下面一段对话,回答第8和第9两个小题。

重庆市第一中学2020届高三3月模拟考试英语试题-含答案

重庆市第一中学2020届高三3月模拟考试英语试题-含答案

秘密★启用前【考试时间:3月29日15:00—17:00】2020年重庆一中高2020级高三下期3月月考英语试题卷2020.3.29英语试题卷共9页。

满分150分。

考试时间120分钟。

注意事项:1.答题前,务必将自己的姓名、准考证号填写在答题卡规定的位置上。

2.答选择题时,必须使用2B铅笔将答题卡上对应题目的答案标号涂黑,如需改动,用橡皮擦擦干净后,再选涂其他答案标号。

3.答非选择题时,必须使用0.5毫米黑色签字笔,将答案书写在答题卡规定的位置上。

4.所有题目必须在答题卡上作答,在试题卷上答题无效。

第Ⅰ卷Ⅰ.听力部分(共二节,每小题1分,满分20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。

1.What will the man do in the job interview?A.Keep smiling.B.Take notes.C.Speak loudly.2.What is the probable relationship between the speakers?A.Teacher and student.B.Doctor and patient.C.Husband and wife.3.What does the woman advise the man to do?A.Never complain about failure.B.Never care about exam results.C.Keep on trying hard.4.Why is the baby crying?A.He wishes to stay with his parents.B.He desires to drink milk.C.He wants to sleep.5.How much should a couple with two children pay for the tickets?A.15dollars.B.30dollars.C.60dollars.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。

2020届高三第二学期第一月考英语英语科试题

2020届高三第二学期第一月考英语英语科试题

2020届高三第二学期第一月考英语科试题第一部分听力(共两节,满分30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。

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 weather like today?A. Cool.B. Warm.C. Hot.2. What does the man want?A. A one-dollar bill.B. Some coins.C. A parking place.3. What are the speakers discussing?A. Whether to hire more workers.B. How to improve production.C. When to put the new orders.4. Where are the speakers?A. In a park.B. In a lift.C. On a bus.5. What has the woman done?A. She has changed her car for a new one.B. She has taken a new car for a test drive.C. She has finished recording a radio program.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。

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。

听下面一段对话,回答第 6 和第7 两个小题。

6. When will the man leave the hotel?A. On October 22nd.B. On October 25th.C. On October 26th.7. How much will the man pay for the room?A. $ 70.B. $ 280.C. $400.听下面一段对话,回答第8 和第9 两个小题8. Why didn’t the man choose the £12.50 tickets?A. The tickets weren’t enough.B. The time was not suitable.C. The seats were at the back.9. Which circus tickets did the man decide to buy?A. The £15.00 ones.B. The £15.50 ones.C. The £17.50 ones.听下面一段对话,回答第10 至第12 三个小题。

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秘密★启用前【考试时间:3月29日15:00—17:00】2020年重庆一中高2020级高三下期3月月考英语试题卷2020.3.29英语试题卷共9页。

满分150分。

考试时间120分钟。

注意事项:1.答题前,务必将自己的姓名、准考证号填写在答题卡规定的位置上。

2.答选择题时,必须使用2B铅笔将答题卡上对应题目的答案标号涂黑,如需改动,用橡皮擦擦干净后,再选涂其他答案标号。

3.答非选择题时,必须使用0.5毫米黑色签字笔,将答案书写在答题卡规定的位置上。

4.所有题目必须在答题卡上作答,在试题卷上答题无效。

第Ⅰ卷Ⅰ.听力部分(共二节,每小题1分,满分20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。

1.What will the man do in the job interview?A.Keep smiling.B.Take notes.C.Speak loudly.2.What is the probable relationship between the speakers?A.Teacher and student.B.Doctor and patient.C.Husband and wife.3.What does the woman advise the man to do?A.Never complain about failure.B.Never care about exam results.C.Keep on trying hard.4.Why is the baby crying?A.He wishes to stay with his parents.B.He desires to drink milk.C.He wants to sleep.5.How much should a couple with two children pay for the tickets?A.15dollars.B.30dollars.C.60dollars.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、7题。

6.When will the speakers meet?A.At7:00tonight.B.At7:00pm on Friday.C.At7:30pm tomorrow.7.What are the speakers talking about?A.Eating together.B.Attending a concert.C.Making a cake.听第7段材料,回答第8至10题。

8.What does the woman want to do?A.Live a slow life.B.Get herself occupied.C.Make herself promoted.9.How does the woman feel at the moment?A.Upset.B.Puzzled.C.Bored.10.What does the man advise the woman to do?A.Enjoy the free time.B.Give him a hand.C.Work hard.听第8段材料,回答第11至13题。

11.What was the man doing last night?A.Buying clothes.B.Watching a movie.C.Reading newspapers.12.What did the critic think of the film?A.It was worth seeing ten times.B.It deserved the award it had won.C.It was one of the best films in a decade.13.What can we learn from the conversation?A.The man will see the film tonight.B.The woman shares the critic’s view.C.The film will be the best of the year.听第9段材料,回答第14至17题。

14.What does the woman think of the jacket?A.It is cheap and fine.B.It is not expensive or cheap.C.It is cool and expensive.15.What is the game rule according to the woman?A.Shooting for an item.B.Shooting for more points.C.Shooting to choose your favourite.16.What made the woman disappointed?A.Gaining nothing.B.Wasting too much money.C.Failing to get what she desired.17.What did the woman think was good?A.The sales skills.B.Her playing skills.C.The shooting outcome.听第10段材料,回答第18至20题。

18.Which is NOT mentioned in the text?A.Racial diversity.B.Varied religions.C.Civil wars.19.What is typical of American life?plex.B.Mobile.fortable.20.What has influenced American society according to the text?A.Its laws.B.Its history.C.Its location.第二部分:阅读理解(共两节,满分40分)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,并在答题卡上将该项涂黑。

ADIFFERENT PACKS,SAME GREAT QUALITYTownsville,Australia’s largest farming families,enjoyed the most exceptional pasture in Australia.The people at Townsville blend dairy goodness and clever ideas to bring you great tasting innovative products,which have been shipped to America, Canada and exported to Austria.Like our lightest one—instant skim(脱脂的)milk powder that makes10litres of delicious skim milk.So you never run out of milk again.DIRECTIONS FOR USE:TO MIX1CUP(250ML)•Add1/4cup(25g)of instant Townsville Skim Milk Powder to1/2cup of hot or cold water.•Add more water to make up1cup.•For best result always add powder to water.•For richer taste and extra nutrition,use additional powder to suit your personal taste or preference.•To improve the natural taste of this product,we recommend storing the liquid milk in the refrigerator overnight before consumption.NOTICE:•Store powder in a cool,dry place.•Keep refrigerated at or below4degrees centigrade no more than3days.•Once opened,store in an airtight container.21.In which country is Townsville—a world-famous dairy product brand made?A.Australia.B.Austria.C.Canada.D.America.22.What should a user do if he/she wants the milk to taste a natural taste?A.Add some sugar to the milk.B.Remove the cream from the milk.C.Put it in the fridge before using it.D.Increase its amount in the liquid milk.23.Where are you suggested putting the powder once it is opened?A.In a plastic bowl.B.In a warm room.C.In a watertight jar.D.In a closed container.BOld Mrs.Lynn was working in the cottage,hanging the washed clothes on the line.What she wasn’t aware was that some children were hiding in a nearby tree watching her every move.They were sure that she was a witch and wanted to find the evidence.They watched nervously as she took a broomstick to sweep the dirt from her stone steps.But much to their disappointment,she didn’t get on the broomstick and fly off.The old lady only looked up when her hen began to make sounds loudly—signaling that she had laid an egg in the nest on the top of the haystack.(干草堆).The old lady put aside her broomstick and walked to the haystack,followed by Michael,a black cat she had rescued from a fox trap.With only three legs,it was hard for Michael to keep up with his mistress.The cat was proof for the children that only a witch would own a black cat with three legs!Accidentally,she tripped and crashed to the ground.The children were in horror.“Should we go and help her?”asked Mia.“What if it’s a trick?”replied Patrick.“She probably knows we’re here.Witches know things like that!”After thinking for a while,Julia said,“Anyway,we should go and check whether she is all right.”Approaching prudently,they could see a wound on the old lady’s forehead.She had knocked her head on a stone and was unconscious.“Go and get Dad,”Mia yelled to her brothers.“Tell him about the accident.”Later,in the hospital,the old lady smiled her thanks.“I was so lucky that you lovely children happened to be passing when I fell.I must have yelled quite loudly.”The children exchanged guilty glances,but were very pleased that she was not a witch after all!24.Mrs.Lynn stopped sweeping when_______.A.her doorstep became very cleanB.she heard the hen making sounds loudlyC.she noticed the children in the treeD.her cat Michael managed to get her attention25.Why was Patrick not willing to help Mrs.Lynn when she fell?A.He thought that she could be cheating them.B.He was afraid of the three-legged black cat.C.He did not think that she was hurt in the fall.D.He knew he and the others shouldn’t have been in her tree.26.Which of the definitions is closest in meaning to the underlined word“prudently”?A.Slowly.B.Hurriedly.C.Carefully.D.Quietly.27.What is the main idea of the story?A.Constant dropping wears away a stone.B.Never judge a book by its cover.C.A friend in need is a friend indeed.D.A good medicine tastes bitter.CUN Humanitarian Chief Mark Lowcock today released US$15million from the Central Emergency Response Fund(CERF)to help fund global efforts to contain the COVID-19virus.The announcement came as the World Health Organization(WHO)upgraded the global risk of the coronavirus outbreak to“very high”–its top level of risk assessment.The WHO has said there is still a chance of containing the virus if its chain of transmission is broken.The UN funding has been released to the WHO and the United Nations Children’s Fund(UNICEF).It will fund essential activities including being aware of the spread of the virus,investigating cases,and putting national laboratories into use.The WHO has called for US$675million to fund the fight against coronavirus. There is a window of opportunity to contain the spread of the virus if countries take strong measures to detect cases early,isolate and care for patients,and trace contacts.Emergency Relief Coordinator and Under-Secretary-General for Humanitarian Affairs,Mark Lowcock said,“We do not yet see evidence that the virus is spreading freely.As long as that’s the case,we still have a chance of containing it,by strengthening surveillance,conducting thorough outbreak investigations to identify contacts and applying appropriate measures to prevent further spread.”“This announcement from the UN’s Emergency Fund will help countries with fragile health systems improve their detection and response operations.It has the potential to save the lives of millions of vulnerable people.”Tedros Adhanom Ghebreyesus,WHO Director-General,said,“The potential spread of the virus to countries with weaker health systems is one of our biggest concerns.These funds will help support these countries get ready for detecting and isolating cases,protecting their health workers,and treating patients with dignity and appropriate care.This will help us save lives and push back the virus.”“At this critical moment,every effort must be made to push back against the outbreak,”said UNICEF Executive Director Henrietta Fore.“These funds will support our global efforts to promote weaker health systems and inform children, pregnant women and families about how to protect themselves.”28.We can learn from the passage that__________.A.WHO referred to the COVID-19as the medium level of risk assessmentB.The virus is still hard to contain even if timely measures are takenC.The fund will help people with weak healthD.The fund released by UN is much less than WHO originally demanded29.According to the passage,the fund can be used in the following aspectsexcept________.A.The examination of the people concernedB.The monitor of the the spread of the virusC.The operation of the national laboratoriesD.The establishment of the mobile hospital30.What is Tedros Adhanom Ghebreyesus’attitude toward the fight against the virus?A.Optimistic.B.Skeptical.C.Cautious.D.Ambiguous.31.In which part of newspaper can you most probably read the passage?A.Travel.B.Health.C.Business.D.Lifestyle.DE-cigarettes lead to as many lung diseases as tobacco products,a new study has found.The report from the University of North Carolina at Chapel Hill compared saliva(唾液)samples from tobacco smokers,e-cigarette smokers and nonsmokers. Researchers found that e-cigarette smokers were likely to develop dangerous proteins associated with lung diseases.The study adds to a growing body of evidence proving that e-cigarettes might not be the perfect alternative smokers addicted to tobacco are looking for.Last year a Surgeon General’s report claimed that the use of e-cigarettes among a certain group of people jumped900percent from2011to2015and more studies were carried out to research their side effects.That same year,the FDA put e-cigarettes in the tobacco products the administration monitors.Previous research from UCLA has proven that e-cigarettes can cause lifelong damage to one’s heart,and that one puff(吸一口烟)of an e-cigarette is all it takes to increase one’s risk of having a heart attack.For the new study UNC researchers observed15e-cigarette users,14cigarette smokers and15nonsmokers.The study revealed that e-cigarette smokers have raised levels of NET-related proteins in their airways,the increased levels of which can lead to lung illnesses,making it difficult for patients to breathe.Study author Dr Mehmet Keismer said,“There is confusion about whether e-cigarettes are‘safer’than cigarettes because the potential adverse effects of e-cigarettes are only beginning to be studied.Our results suggest that e-cigarettes might be just as bad as cigarettes.”Dr Keismer also stressed that e-cigarettes come with their own harmful risks along with those linked to tobacco,which challenges the concept that switching from cigarettes to e-cigarettes is a healthier alternative.A citizen named Dyclna said,“The e-cigarettes probably make you cough less, but nobody gives a warning about your lungs.For producers,it’s just a money thing —it’s just about getting profits from it.The problem is that our government just stands there with its hands behind the back.”32.Why did the FDA list e-cigarettes into monitored products?A.It might realize the harm of them.B.It wanted to improve their quality.C.It aimed to reduce their illegal sales.D.It might be warned by the government.33.What harm does the increased level of NET do to people?A.Leading to weight gain.B.Spoiling people’s appetite.C.Making breathing hard.D.Speeding people’s heartbeat.34.What does Dyclna mainly want to express?A.E-cigarettes are safer than regular ones.B.E-cigarettes can only benefit few people.C.Producers make a high profit from e-cigarettes.ernments are to blame for e-cigarettes’consumption.35.What is the best title for the text?A.E-cigarettes:A new way of smoking.B.E-cigarettes:A better way than tobacco?C.E-cigarettes:The harmful proteins it produces.D.E-cigarettes:The urgent need of government control.第二节(共5小题;每小题2分,满分10分)In our fast world of phones,fax machines and computers,the old-fashioned art of letter writing is at risk of disappearing altogether.36There is the excitement of its arrival,the pleasure of seeing who it is from and,finally,the enjoyment of the contents.Letter writing has been part of my life for as long as I can remember.It probably began with the little notes I would write to my mother.My mother,also,always insisted I write my own thank-you letters for Christmas and birthday presents.37.When I left home at18to train as a doctor in London,I would write once a week, and so would my mother.Occasionally my father would write and it was always a joy to receive his long,amusing letters.38Of course,we also made phone calls but it is the letters that I remember most.There were also letters from my boyfriend.In our youth he had to work or study away at some time and we were only able to stay in touch by letter.39I found that I could often express myself more easily in writing than by talking.I love the letters that come with birthday or Christmas cards.Notes are appreciated,but how much better to have a year’s supply of news!And it’s better still when it’s an airmail envelope with beautiful stamps.40Like my mother before me,I insist they write their own thank-you letters. My daughter writes to me little letters,just as I did to my mother.I strongly urge readers not to allow letter writing to become another“lost art”.A.Poor handwriting can spoil(破坏)your enjoyment of a letter.B.I am pleased that my children are carrying on the tradition.C.We had been writing to each other for a long time but never met.D.It didn’t matter how short or untidy they were as long as they were letters.E.But instead of harming the relationship,letter writing seemed to improve them.F.Yet,to me,receiving a letter cannot be matched by any form of communication.G.The letters from him contained just everyday events concerning my parents and their friends.第三部分英语知识运用(共三节,满分70分)第一节完形填空(共20小题;每小题1.5分,满分30分)阅读下面短文,从短文后各题所给四个选项(A、B、C和D)中,选出可以填入空白处的最佳选项,并在答题卡上将该项涂黑。

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