【原创】2021届高三上学期期中试卷 试卷 教师版
数学丨山东省菏泽市2021届高三上学期期中考试数学试卷及答案
保密★启用前2020－2021学年度第一学期期中考试高三数学试题(B)本试卷共4页，共150分，考试时间120分钟。
一、单项选择题：本题共8小题，每小题5分，共40分。
在每小题给出的四个选项中，只有一项是符合题目要求的。
1.全集U ＝{x|－1≤x<3}，集合A ＝{x|－1≤x ≤2}，则U A ＝A.{x|－1≤x<2}B.{x|2<x<3}C.{x|2≤x<3}D.{x|x<－1或x>2}2.己知复数z ＝1＋i ，z 为z 的共轭复数，则1z z + A.32i + B.12i + C.132i - D.132i + 3.下列函数中，既是偶函数又在(0，＋∞)上单调递减的是 A.y ＝x －2B.y ＝2－x C.y ＝|lnx| D.y ＝xsinx4.已知tan α＝2，则sin(α－4π)sin(α＋4π)＝ A.－310 B.－35 C.310 D.35 5.《九章算术》中《方田》章有弧田面积计算问题，术曰：以弦乘矢，矢又自乘，并之，二而一。
其大意是弧田面积计算公式为：弧田面积＝12(弦×矢＋矢×矢)。
弧田是由圆弧(弧田弧)和以圆弧的端点为端点的线段(弧田弦)围成的平面图形，公式中的“弦”指的是弧田弦的长，“矢”指的是弧田所在圆的半径与圆心到弧田弦的距离之差，现有一弧田，其弧田弦AB 等于6米，其弧田弧所在圆为圆O ，若用上述弧田面积计算公式算得该弧田的面积为72平方米，则sin ∠AOB ＝A.34B.725C.1225D.24256.在△ABC 中，AB AC 2AD +=，AE 2DE 0+=，若EB xAB yAC =+，则A.x ＋2y ＝0B.2x ＋y ＝0C.x －2y ＝0D.2x －y ＝07.函数f(x)＝Asin(ωx ＋φ)(其中A>0，ω>0，|φ|<2π)的图象如图所示，为了得到f(x)的图象，只需将g(x)＝Asin ωx 图象A.向左平移4π个单位长度 B.向右平移4π个单位长度 C.向左平移12π个单位长度 D.向右平移12π个单位长度 8.定义域为(－2π，2π)的函数f(x)满足f(x)＋f(－x)＝0，其导函数为f'(x)，当0<x<2π时，有f'(x)cosx ＋f(x)sinx<0成立，则关于x 的不等式2f(4π)·cosx 的解集为 A.(－2π，－4π)∪(4π，2π)B.(4π，2π) C.(－4π，0)∪(0，4π) D.(－4π，0)∪(4π，2π) 二、多项选择题：本题共4小题，每小题5分，共20分。
2021年高三上学期期中检测语文试卷 含解析
2021年高三上学期期中检测语文试卷含解析一、基础知识题（共1小题）1.阅读下面一段文字，完成1—5题。
时间愈久，愈爱这一室虚白，像画面上的大片留白，情味隽永。
世界至繁，天地至简，这小小的一室，容得下一个人的万千思绪。
坐在这简单的四壁之间，心无旁骛，独品一刻之(闲/静)。
我们需要的生活，其实比想象的更加简单，所谓“良田万顷，日食一升；广厦千间，夜眠七尺”。
身无长物，是一种让人（羡慕/艳羡）的状态。
也许我们本来就无须为太多的念头埋单，美丽的风景，看过就好。
庞杂的愿望中往往夹杂着太多的不切实际，付账时常常随着别人的风向，有时忘了自己的。
加法生活里充满了太多多余的对比和（奢想/向往），不如试试减法生活，如这四壁白雪，保留自然给予的一点天真和质朴。
放轻松，抛开重负，世界还是一样美好。
1．文中加横线字的注音或字形有错误的一项是（）A．隽（juàn）永万千思绪B．给（jǐ）予心无旁骛C．身无长（zhǎng）物付账D．广厦(shà)千间埋单2．依次选用文中括号里的词语，最恰当的一项是（）A．闲羡慕向往B．静羡慕奢想C．静艳羡向往D．闲艳羡奢想3．下列各句中，标点符号使用正确的一句是（）A．要解决长假出行拥堵问题：一是修路架桥并开发更多的旅游资源；二是落实好带薪休假制度，让错峰出游成为现实，事实上，后者远比前者效果更快更好。
B．1962年拍摄的电影《甲午风云》中，管带邓世昌高高屹立在舰桥上，双手握拳高举、头戴“暖式”顶戴花翎——民族英雄的壮烈形象打动了几代中国观众。
C．节日“堵”在路上、“人在囧途”的尴尬、诸多不文明的旅游行为、以“千元大虾”为典型的消费猫腻……黄金周虽已结束，但一个个“假日印记”却格外鲜明。
D．书展7天的热闹过后，剩下的358天怎么办？对于上海书展来说，如何“热热闹闹一周，长长久久一年”？成为书展组委会着力思考的问题。
4．下列各句中，加横线的成语使用正确的一项是（）A．中国药学家屠呦呦获“诺奖”实至名归，这无疑是一件值得庆祝的好事，不只是因为得奖，更因为这是一个造福众生的伟大贡献。
【数学】北京市海淀区2021届高三上学期期中考试考试题(解析版)
北京市海淀区2021届高三上学期期中考试考数学试题第一部分（选择题 共40分）一、选择题共10小题，每小题4分，共40分.在每小题列出的四个选项中，选出符合题目要求的一项.1. 已知集合{|30}A x x =-≤，{0,2,4}B =，则A B =（ ）A. {0，2}B. {0，2，4}C. {}3x x ≤D. {}03x x ≤≤【答案】A【解析】集合{|30}{|3}A x x x x =-≤=≤，{0,2,4}B =，则A B ={}0,2故选：A.2. 已知向量(,2)a m =，(2,1)b =-. 若//a b ，则m 的值为（ ） A. 4 B. 1C. -4D. -1【答案】C【解析】因为//a b ，所以40m --=，解得4m =- 故选：C.3. 命题“0x ∃>，使得21x ≥”的否定为（ ） A. 0x ∃>，使得21x < B. 0x ∃≤，使得21x ≥ C. 0x ∀>，都有21x < D. 0x ∀≤，都有21x <【答案】C【解析】命题“0x ∃>，使得21x ≥”的否定为“0x ∀>，都有21x <” 故选：C4. 设a ，b R ∈，且0a b <<，则（ ）A.11a b< B.b a a b> C.2a b+> D.2b a a b+> 【答案】D 【解析】0a b <<，11a b∴>，故A 错；0a b <<，22a b∴>，即220,0b a ab -<>，可得220b a b a a b ab --=<，b a a b ∴<，故B 错；0a b <<，02a b +∴<0>，则2a b+<，故C 错；0a b <<，0,0b a a b ∴>>，2b a a b +>=，等号取不到，故D 正确；故选：D.5. 下列函数中，是偶函数且在区间(0,)+∞上为增函数的是（ ） A. 2ln y x = B. 3||y x =C. 1y x x=-D. cos y x =【答案】B 【解析】对于A ，2ln y x =的定义域为(0,)+∞，故不是偶函数，故A 错误；对于B ，()3f x x =的定义域为R ，关于原点对称，且()()33f x x x f x -=-==，∴3y x =是偶函数，且根据幂函数的性质可得在(0,)+∞上为增函数，故B 正确；对于C ，()1f x x x=-的定义域为{}0x x ≠，关于原点对称，且()()11f x x x f x x x ⎛⎫-=--=--=- ⎪-⎝⎭，故1y x x =-是奇函数，故C 错误； 对于D ，cos y x =在(0,)+∞有增有减，故D 错误. 故选：B.6. 已知函数()ln 4f x x x =+-，在下列区间中，包含()f x 零点的区间是（ ） A. （0，1） B. （1，2）C. （2，3）D. （3，4） 【答案】C【解析】函数()ln 4f x x x =+-，是增函数且为连续函数， 又f （2）ln2240=+-<，f （3）ln3340=+->，可得()()230f f <所以函数()ln 4f x x x =+-包含零点的区间是(2,3)． 故选：C ．7. 已知数列{}n a 的前n 项和为n S ，且1(),2,3,n n S a n ==，则2020a =（ ）A. 0B. 1C. 2020D. 2021【答案】A【解析】当1n =时，11a S =，当2n ≥时，11n n n n n a S S a a --=-=-， 所以10n a -=，即1220200a a a ==⋅⋅⋅==， 故选：A.8. 已知函数sin()y A x ωϕ=+的部分图象如图所示，将该函数的图象向左平移()0t t >个单位长度，得到函数()y f x =的图象若函数()y f x =为奇函数，则t 的最小值是（ ）A.12πB.6π C.4π D.3π 【答案】B【解析】由图象可得6x π=时，函数sin()y A x ωϕ=+的函数值为0，即()6k k Z ωπϕπ+=∈，()6k k Z ωπϕπ∴=-+∈，sin()6y A x k ωπωπ∴=-+，将此函数向左平移()0t t >个单位得，()sin ()6f x A x t k ωπωπ⎡⎤=+-+⎢⎥⎣⎦，又因为()f x 为奇函数，11()6t k k k Z ωπωππ∴-+=∈，11(,)6k kt k Z k Z ππω-∴=+∈∈，因为0t >min 6t π∴=．故选：B ．9. 设x ，y 是实数，则“01x <<，且01y <<”是“22log log 0x y +<”的（ ） A. 充分而不必要条件 B. 必要而不充分条件 C. 充分必要条件 D. 既不充分也不必要条件【答案】A【解析】】若“01x <<，且01y <<”,则01xy <<，2222log log log log 10x y xy +=<=， 所以“01x <<，且01y <<”是“22log log 0x y +<充分条件；若22log log 0x y +<，则2222log log log log 10x y xy +=<=，可得01xy <<，但得不出“01x <<，且01y <<”，如116x =，2y =可得22log log 0x y +<，所以 22log log 0x y +<得不出“01x <<，且01y <<”，所以“01x <<，且01y <<”是“22log log 0x y +<充分不必要条件； 故选：A.10. 对于函数()f x ﹐若集合()(){}0,x x f x f x >=-中恰有k 个元素，则称函数()f x 是“k 阶准偶函数”.若函数21,()2,xx a f x x x a ⎧⎛⎫≤⎪ ⎪=⎨⎝⎭⎪>⎩是“2阶准偶函数”，则a 的取值范围是（ ） A. (),0-∞ B. [)0,2C. [)0,4D. [)2,4【答案】B【解析】根据题意，函数21,()2,xx af x x x a ⎧⎛⎫≤⎪ ⎪=⎨⎝⎭⎪>⎩是“2阶准偶函数”，则集合()(){}0,x x f x f x >=-中恰有2个元素.当0a <时，函数21,()2,xx a f x x x a ⎧⎛⎫≤⎪ ⎪=⎨⎝⎭⎪>⎩有一段部分为2,y x x a =>，注意的函数2y x 本身具有偶函数性质，故集合()(){}0,x x f x f x >=-中不止有两个元素，矛盾，当0a >时，根据“2阶准偶函数”的定义得()f x 的可能取值为2x 或12x⎛⎫ ⎪⎝⎭，()f x -为122-⎛⎫= ⎪⎝⎭xx ，故当122xx ⎛⎫= ⎪⎝⎭，该方程无解，当22x x =，解得2x =或4x =，故要使得集合()(){}0,x x f x f x >=-中恰有2个元素，则需要满足2a <，即02a <<；当0a =时，函数21,0()2,0xx f x x x ⎧⎛⎫≤⎪ ⎪=⎨⎝⎭⎪>⎩，()f x 的取值为2x ，()f x -为122-⎛⎫= ⎪⎝⎭xx ，根据题意得22x x =满足恰有两个元素，故0a =满足条件. 综上，实数a 的取值范围是[)0,2. 故选：B.第二部分（非选择题 共110分）二、填空题共5小题，每小题5分，共25分. 11. 若复数(1)z i i =+，则||z = _______.【解析】由题意得：2(1)1z i i i i i =+=+=-+，所以z ==12. 已知tan 24πα⎛⎫-= ⎪⎝⎭，则tan α=________. 【答案】-3.【解析】因为tan 24πα⎛⎫-= ⎪⎝⎭，所以tan 12tan 31tan ααα-=⇒=-+．13. 已知等差数列{}n a 的前n 项和为n S ．若19a =，公差2d =-，则n S 的最大值为_______. 【答案】25 【解析】19a =，2d =-，912112na n n令0n a ≥，解得112n ≤，又*n N ∈，则15n ≤≤ n S 的最大值为554592252S故答案为：25.14. 在边长为2的正三角形ABC 中，M 是BC 的中点，D 是线段AM 的中点. ①若BD xBA yBC =+，则x y +=_______； ②BD BM ⋅= _______.【答案】 (1). 34(2). 1 【解析】①M 是BC 的中点，∴12BMBC ， D 是AM 的中点，∴11112224BD BA BM BA BC =+=+， 12x ∴=，14y =，故34x y +=． ②ABC ∆是边长为2的正三角形，M 是BC 的中点，AM BC ∴⊥，且1BM =，∴2cos 1BD BM BD BM DBM BM ⋅=⋅⋅∠==．故答案：34，1．15. 唐代李皋发明了“桨轮船”，这种船是原始形态的轮船，是近代明轮航行模式之先导.如图，某桨轮船的子的半径为3m ，它以1rad/s 的角速度逆时针旋转.轮子外边沿有一点P ，点P 到船底的距离是H （单位：m ），轮子旋转时间为t （单位：s ）. 当0t =时，点P 在轮子的最高点处.①当点P 第一次入水时，t =__________；②当t t =0时，函数()H t 的瞬时变化率取得最大值，则0t 的最小值是________. 【答案】 (1).23π (2). 32π【解析】（1）当0t =时，点P 在轮子最高点处，由图可知，轮船距离船底1m ，半径3m ，设为r ，则cos 13cos 4,0H r t r t t =++=+≥，当点P 第一次入水时，水面高 2.5m ，即2.5H =，代入3cos 4H t =+得，1cos 2t =-，第一次入水即在满足1cos 2t =-的情况下满足现实条件0t ≥后可取的最小值，23t π=（2）瞬时变化率取得最大值，即'()H t 最大，'()3sin H t t =-，当3sin 3t -=时，瞬时变化率取得最大值，此时，0t 的最小值为32π 故答案为：①23π；②32π三、解答题共6小题，共85分.解答应写出文字说明、演算步骤或证明过程. 16. 在△ABC 中，sin 2sin B C =，3cos 4A =.（1）若△ABC ，求c 的值； （2）求ac的值. 解：（1）由正弦定理得：sin sin b cB C=. 因为sin 2sin B C =，所以2b c =.因为3cos4A=，0Aπ<<，所以27sin1cosA A=-=，因为S=211sin2sin22S bc A c A==⨯⨯=，所以24c=，所以2c=；（2）由（1）知2b c=，因为3cos4A=，所以222222232cos4424a b c bc A c c c c=+-=+-⨯=，所以a=，所以ac=17. 已知等差数列{}n a满足59a=，3922a a+=.（1）求{}n a的通项公式；（2）等比数列{}n b的前n项和为n S，且11b a=，再从条件①、条件②、条件③这三个条件中任选择两个作为已知条件，求满足2020nS<的n的最大值.条件①：312b a a=+；条件②：37S=；条件③：1n nb b+>.解：（1）设等差数列{}n a的公差为d，则()11na a n d+-=，因为59a=，3922a a+=，所以1492102ta da d+=⎧⎨+=⎩，解得：112ad=⎧⎨=⎩所以21na n=-；（2）（I）选择①②设等比数列{}n b的公比为q，因为11b a=，312b a a=+，所以11b=，34b=，因为37S=，所以23132b S b b=--=，所以212b q b ==，所以1(1)211n n n b q S q-==--， 因为2020n S <，所以212020n -≤， 所以10n ≤，即n 的最大值为10. （II ）选择①③设等比数列{}n b 的公比为q ， 因为11b a =，312b a a =+， 所以11b =，34b =，所以2314b q b ==，2q =±， 因为1n n b b +>，所以2q，所以1(1)211n n n b q S q-==--， 因为2020n S <，所以212020n -<， 所以10n ≤.即n 的最大值为10. 选择②③设等比数列{}n b 的公比为q 因为37S =，11b =， 所以217q q ++=. 所以2q，或3q =-.因为1n n b b +>，所以2q.所以1(1)211n n n b q S q-==-- 因为2020n S <，所以212020n -< 所以10n ≤.即n 的最大值为10.18. 已知函数2()(23)x f x e x x =-. （1）求不等式()0f x >的解集；（2）求函数()f x 在区间[0,2]上的最大值和最小值. 解：（1）因为0x e >，由()2(0)23xf x e x x =->，得2230x x ->.所以0x <或32x >. 所以不等式()0f x >的解集为{|x 0x <或32x ⎫>⎬⎭； （2）由()223()xf x e x x =-得：2()(23)x f x e x x '=+-()()231xex x =+-.令()0f x '=，得1x =，或32x =-（舍）. ()f x 与()f x '在区间[0，2]上的情况如下：所以当1x =时，()f x 取得最小值()1f e =-； 当2x =时，()f x 取得最大值()222f e =.19. 已知函数π()2sin 6f x x ⎛⎫=+⎪⎝⎭. （1）求()f x 的单调递减区间；（2）设π()()6g x f x f x ⎛⎫=- ⎪⎝⎭. 当[0,]x m ∈时，()g x 的取值范围为0,2⎡⎣，求m 的最大值.解：（1）令322262πππk πx k π+≤+≤+，k Z ∈. 所以42233ππk πx k π+≤≤+，()k Z ∈.所以函数()f x 的单调递减区间42,2()33k k k Z ππππ⎡⎤++∈⎢⎥⎣⎦. （2）()()4sin sin 66g x f x f x x x ππ⎛⎫⎛⎫=-=+ ⎪ ⎪⎝⎭⎝⎭14cos sin 2x x x ⎫=+⎪⎝⎭22cos sin x x x =+cos2)sin 2x x =-+2sin 23x π⎛⎫=-+ ⎪⎝⎭因为0x m ≤≤， 所以22333x m πππ-≤-≤-.因为()g x 的取值范围为0,2⎡+⎣，所以sin 23x π⎛⎫- ⎪⎝⎭的取值范围为2⎡⎤-⎢⎥⎣⎦所以42233m πππ≤-≤. 解得：55126m ππ≤≤. 所以m 的最大值为56π.20. 已知三次函数32()324f x ax ax a =-++.（1）当1a =-时，求曲线()y f x =在点(3,(3))f 处的切线方程； （2）若函数()f x 在区间(,3)a a +上具有单调性，求a 的取值范围； （3）当0a >时，若122x x +>，求12()()f x f x +的取值范围.解：由()32324f x ax ax a =-++可得：2()363(2)f x ax ax ax x '=-=-（1）当1a =-时，(3)2f =-，(3)9f '=-.所以曲线( )y f x =在点()()3,3f 处的切线方程为925y x =-+.（2）由已知可得0a ≠①当0a >时，令()0f x '=得0x =，22x =.()f x 与()f x '在区间(),-∞+∞_上的情况如下：因为()f x 在(),3a a +上具有单调性，所以2a ≥.②当0a <时，()f x 与()'f x 在区间(),-∞+∞上的情况如下：因为()f x 在(),3a a +上具有单调性， 所以30a +≤，即3a ≤-. 综上所述，a 的取值范围是(][),32,-∞-+∞.（3）先证明：()()12 4f x f x +≥.由（2）知，当0a >时，()f x 的递增区间是(),0-∞，()2,+∞，递减区间是（0，2）. 因为122x x +>，不妨设12x x ≤，则21>x . ①若10x ≤，则2122x x >-≥.所以()()()()12112444f x f x f x f x a +>+-=+>. ②若1>0x ，因为21>x ，所以()()12()()224f x f x f f +≥+=，当且仅当122x x ==时取等号.综上所述，12())4(f x f x +≥.再证明：12()()f x f x +的取值范围是[4,)+∞.假设存在常数()4m m ≥，使得对任意122x x +>，()()12f x f x m +≤.取12x =，且22x >+则 ()()3222222324f f x ax ax a+=+-++2222222()()222()224ax x a x a x m =+-+-+>-+>，与()()12f x f x m +≤矛盾.所以12()()f x f x +的取值范围是[4,)+∞.21. 已知{}n a 是无穷数列，1a a =，2a b =且对于{}n a 中任意两项i a ，()j a i j <在{}n a 中都存在一项(2)k a j k j <<，使得2k j i a a a =-. （1）若3a =，5b =求3a ； （2）若0a b ，求证：数列{}n a 中有无穷多项0；（3）若ab ，求数列{}n a 的通项公式.解：（1）取1i =，2j =，则存在24)k a k <<(，使得3212a a a =-，即3212a a a =-. 因为13a a ==，25a b ==，所以32127a a a =-=.（2）假设{}n a 中仅有有限项为0，不妨设0m a =，且当n m >时，n a 均不为0，则2m ≥.取1i =，j m =，则存在2)k a m k m <<(，使得120k m a a a =-=，与0k a ≠矛盾．（3）①当a b <时，首先证明数列{}n a 是递增数列，即证*n N ∀∈，1n n a a +<恒成立． 若不然，则存在最小的正整数0n ，使得001n n a a +≥，且012 n a a a <<<.显然02n ≥.取0j n =，1i =，2，…，01n -，则存在00(2k a n k n <<），使得02k n i a a a =-．因为00000121222n n n n n a a a a a a a -->->>->，所以012n a a -，022n a a -，…，0012n n a a --这01n -个不同数恰为01n a +，02n a +，…，021n a -这01n -项.所以001n n a a +>与001n n a a +≤矛盾. 所以数列{}n a 是递增数列.再证明： (1)()n a a n b a =+--，1,2,3,n= 记,d b a =- 即证(1)n a a n d =+-，1,2,3,n=当1,2n =时，结论成立.假设存在最小的正整数0,m 使得 (1)n a a n d =+-对任意01n m ≤≤恒成立， 但010,m a a m d +≠+则02m ≥. 取0j m =，1,2,i =，01m -，则存在()002k a m k m <<，使得02k m i a a a =-因为数列{}n a 是递增数列， 所以00012121m m m a a a a a +-<<<<<<.所以0600121222m m m m a a a a a a --<<-<-.因为0012m m a a --，…022m a a -，012m a a -这01m -个数恰为01m a +，02m a +，…021m a -这01m -项.所以()()004110002212m m m a a a a m d a m d a m d +-=-=+--+-=+⎡⎤⎡⎤⎣⎦⎣⎦， 与10n m a a m d +≠+矛盾.所以 (1)()n a a n b a =+--，1,2,3,n=②当a b >时，令n n b a =-，1,2,3,n =，则1b a =-，2b b =-，且12<b b .对于{}n b 中任意两项i b ，()j b i j <，的因为对任意i a ，()j a i j <，存在(2),k a j k j <<使得2k j i a a a =-， 所以()2k j i a a a -=---，即存在(2),k b j k j <<使得2k j i b b b =-. 因此数列{}n b 满足题设条件.由① 可知(1)()n b a n a b =-+--，1,2,3,,n =所以(1)()n a a n b a =+--，1,2,3,n =综上所述，(1)()n a a n b a =+--，1,2,3,n =经检验，数列{}n a 满足题设条件.。
2021届华中师范大学第一附属中学高三英语上学期期中试卷及答案
2021届华中师范大学第一附属中学高三英语上学期期中试卷及答案第一部分阅读（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A、B、C、D四个选项中选出最佳选项AArtificial intelligence (Al) is practically everywhere today. There are so many products out there which use Al. Some are being developed, some are already in use, and some failed and are being improved, so it’s very difficult to name a few of them and regard them as the best.ViIt is an Al personal trainer which is mainly concerned with fitness and coaching. It, however, requires the use of bio-sensing earphones and other fitness tracking equipment! It can play your favourite music while you work out and all you have to worry about is the exercise you're doing.Deep TextDo you ever wonder how an ad appears suddenly just when you are looking for something similar? This is because of Deep Text. It uses real-time consumer information to produce data which in turn is used to target consumers. Thus, if you search online for flight tickets fromBangaloretoDelhi, it is very likely that an ad relating to hotels inDelhiwill soon follow.Hello EggIf you live alone and miss your mother because you always miss your breakfast or don’t know what to eat for dinner, then Hello Egg is exactly what you are looking for. A very healthy choice of the 2-minute noodles and oats, Hello Egg provides you with a detailed weekly meal plan about the needs of your body. It is truly a modern AI-powered home cooking tool for the young.WordsmithYou can put Mr. Smith into your Microsoft Excel using their free API, and let it write up detailed analysis of the stories behind your numbers. It can produce detailed reports on thousands of pages of spreadsheets in seconds.1. What can we learn about Vi from the text?A. It is an AI music player.B. It is a bio-sensing earphone.C. It doesn't work without bio-sensing earphones.D. It can make you more energetic while you work out.2. Which can help you improve cooking skill?A. Deep Text.B. Vi.C. Wordsmith.D. Hello Egg.3. What can Wordsmith do for us?A. Produce a detailed report.B. Provide us with a detailed meal plan.C. Book a ticket ahead of time.D. Offer us information on hotels for traveling.BOwning a dog is associated with a significantly lower risk of heart disease and death, according to a comprehensive new study published by a team of Swedish researchers on Friday in the journal Scientific Reports.The scientists followed 3.4 million people over the course of 12 years and found that adults who lived alone and owned a dog were 33 percent less likely to die during the study than adults who lived alone without dogs. In addition, the single adults with dogs were 36 percent less likely to die from heart disease.“Dog ownership was especiallyprominentas a protective factor in persons living alone, which is a group reported previously to be at higher risk of heart disease and death than those living in a multi-person household,” Mwenya Mubanga, a Ph.D. student at Uppsala University in Uppsala, Sweden, and the lead junior author of the study, said in a statement announcing its findings. The link between dog ownership and lower mortality(死亡率)was less pronounced in adults who lived either with family members or partners, but still present, according to the study. “Perhaps a dog may stand in as an important family member in the single households,” Mubanga added. “Another interesting findingwas that owners of dogs which were intended originally for hunting were most protected.”The study, which is the largest to date on the health relations of owning a dog, suggested that some of the reasons dog owners may have a lower risk of mortality and heart disease were because dog owners walk more. “These kind of epidemiological (流行病学的)studies look for associations in large populations but do not provide answers on whether and how dogs could protect their owners from heart disease,” Tove Fall, a senior author of the study and a professor at Uppsala University, said in a statement“We know that dog owners in general have a higher level of physical activity, which could be one explanation to the observed results,” Fall added. “Other explanations include an increased well-being and social contacts or effects of the dog on the bacterial microbiome(微生物菌群) in the owner.” Fall added that because all participantsof dog owners in Sweden or other “European populations with similar culture regarding dog ownership.”4. Why did the researchers do the study related to 3.4 million people’s health and the dogs?A. To help Europeans,B. To find their association.C. To protect unhealthy adults.D. To reduce risk of heart disease.5. What does the underlined word “prominent” probably mean in Para.3?A. Universal.B. Confusing.C. Appealing.D. Important6. What’s the main idea of the text?A. Adults living with dogs are less likely to die.B. Swedish people are very fond of animal pets.C. Keeping a dog is a popular and healthy hobby.D. Owning dogs reduces the risk of heart disease.7. What’s the writer’s attitude towards owning a dog?A. Positive.B. Negative.C. Objective.D. Contradictory.CFor fishermen and sailors in the seaside town of Shangpan in Linhai, Zhejiang province, where recently 12 stranded（搁浅的）whales were found, it is not unusual to spot whales or dolphins in the sea — they would always call them haizi, or “son of the ocean”. And if the local fishermen spot such intelligent sea creatures in need, they will save them from danger without hesitation.In the recent rescue of melon-headed whales, they tried their best to help as a 37-second video of a young man sleeping in the water holding a whale soon went viral online, with netizens applauding his constant efforts and caring heart in trying to keep the whale alive. “The melon-headed whale knew I was trying to save it and would be more cooperative and wouldn’t move.” said Lu Wenhui, a 21-year-old diver from Hangzhou Changqiao Polar Ocean Park, who held up the whale’s head so it could breathe easily. Lu had been staying in the water for 10 hours by 4 a.m. Wednesday after coming for the rescue work on Tuesday, when the stranded whales were first spotted.“Whales are mammals and need to breathe in air — that’s why I had to hold its head, to ensure it couldbreathe smoothly the whole time,” Lu said. The stranded whales, after the struggling of being transported, were worn out when they arrived at the fish farm and might have drowned if they were left unattended.The reason why the whales were stranded is still not clear. And these stranded whales were returned to the ocean because they are not adapted to artificial breeding environments on land.8. What do local fishermen usually do after finding haizi in danger?A. Ignore them.B. Feed them.C. Transport them.D. Protect them.9. Why does the author describe Lu’s rescue work?A. To introduce the whales in detail.B. To voice his views on the diver.C. To praise the efforts made by rescuers.D. To explain the value of the whales.10. What can we say about the melon-headed whales according to the text?A. They are rarely seen by locals.B. They are used to breathing in the sea.C. They aren’t suited to artificial surroundings.D. They are fond of swimming near the seaside.11. Which of the following can be the best title for the text?A. Melon-headed Whales Were StrandedB. Rescuers Went All out to Help WhalesC. Fishermen Transported Stranded WhalesD. Witnesses Voiced Opinions on Rescue WorkDThey are smart. They know how to steal. They know how to find food. They know how to intimidate(恐吓) . Who are they? They are macaque monkeys. They have taken over the old city ofLopburiinThailand. About 8,400 of them are in the center of the city. They roam(漫游) neighborhoods in groups. Dozens of businesses in Lopburi are closing. They include a music school, a gold shop, a barber, a cellphone store and a movie theatre. The Buddhist culture believes reducing the number of monkeys would disturb spiritual well-being.The monkeys were not always such a hazard. They attracted tourists. Buddhists thought feeding them was a good deed. Now times have changed. Recently, the coronavirus made things worse. There are fewer tourists, which means that travelers give less food to the monkeys. Over the years, the monkeys moved into empty buildings. They trashed whatever they came across. They ripped(扯掉) antennas and windshield wipers off parkedcars. What happens when monkeys come into contact with humans? An observer said that years ago the monkeys were fewer, biggerand healthier. Their fur was shiny and thick. They kept to the temples，as well as the ruins of the ancient Khmer civilization.Then tourists came with easy and unhealthy food. Along with bananas and citrus(柑橘), the macaques feasted on junk food. An observer said, “The monkeys are never hungry. They are just like children who eat too much KFC. ” Compared with the monkeys of the forest, their urban counterparts have less muscle. They have more hypertension and blood disease. Their fur has thinned. Some have gone bald. With so much food available，they have more time to breed and to give birth. Their population has exploded. “These monkeys were here before us,” a man said. A juvenile macaque tugged(拽) his trousers demanding a treat. “We have to adapt to them，not the other way around. ”12. Why are macaque monkeys so popular in Lopburi inThailand?A. Because they are smart and know how to steal.B. Because they have taken over the old city ofLopburiinThailand.C. Because they are symbols of spiritual well-being in the Buddhist culture.D. Because 8,400 of them are in the centre of the city and roam everywhere.13. In what way has tourism influenced the macaque monkeys?A. The monkeys are bigger and healthier.B. The monkeys like eating KFC food.C. The monkeys have suffered from more diseases.D. The monkeys are subject to birth control.14. What does the underlined word “counterparts” in the last paragraph refer to?A. Monkeys.B. Tourists.C. Buddhists.D. Children.15. It can be inferred from what the man said that ________.A. monkeys are ancestors, so humans should adapt to monkeysB. humans should regard monkeys with awe and respectC. monkeys should be forced to follow rulesD. humans should give monkeys whatever they want第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
2021届哈尔滨市第一二二中学高三语文上学期期中考试试卷及参考答案
2021届哈尔滨市第一二二中学高三语文上学期期中考试试卷及参考答案一、现代文阅读（36分）(一)现代文阅读I(9分)阅读下面的文字，完成下面小题。
智县委贾大山他姓智，那时人们不叫他智书记，而是叫他智县委。
夜晚，我和我的伙伴们打戏院门口的电灯泡玩儿，看谁能打中。
我正瞄准，忽然有人揪住我脑后的小辫儿，伙伴们立即就跑散了。
扭头一看，揪我小辫儿的是个生人，中等个儿，白净脸儿，穿一身灰军装，戴一副眼镜。
我挣脱他的手，撒腿就跑，他一把又揪住我的小辫儿，揪得好疼。
我骂他的娘，他也不理睬，终于把我揪到父亲面前去了。
“这是你的小孩？”生硬的外地口音。
“噢，是我的孩子。
”父亲是个买卖人，开着一个杂货铺，一向胆小怕事。
一见那人，赶忙捻亮罩子灯，显得很惊慌：“智县委，请坐……”我也一惊，他就是智县委！智县委没有坐，眼睛忽然盯住桌上的一片字纸。
那是我写的一篇大楷：一去二三里烟村四五家亭台六七座八九十枝花“哪个写的？”他问。
“他写的。
”父亲指指我说，“瞎画。
”他立刻瞅定我，脸上竟然有了喜色，眼镜也显得明亮了。
“几岁了？”“十岁了。
”“十岁了还留小辫儿？”父亲赶忙解释：当地的风俗孩子们留小辫儿，要留到十二岁，成人。
智县委听了哈哈大笑，说：“好哇，那就留着吧！”智县委夸了一番我的毛笔字，就和父亲说起话来。
他问父亲的年龄、籍贯，又问这个小铺多大资本，生意如何，拿多少税。
我站在他的背后，并不注意他们的谈话，眼睛一直注视着他的衣襟下面露出的那块红布——那是一把盒子，真家伙！他和父亲谈着话，忽然仰着头，望着货架子说：“怎么，连个字号也没有？”“没有。
”父亲笑着说，“小本买卖，还值得立字号？”“怎么不值得？”智县委好像生气了，脸色红红的，说，“城里买卖家，哪个没字号？‘亨茂号’‘文兴成’‘荣泰昌’‘广顺正’，都有字号嘛！你也赶快立个字号！”父亲想了一下，说：“叫‘贾家小铺’？”“不好，小气！”“叫‘万宝店’？”“也不好，俗气！”父亲就笑了：“智县委赏个名儿吧！”“‘复兴成’，怎么样？共产党保护民族工商业，一切都要复兴的，你也要复兴嘛！”父亲说好，行，不错。
2021届江苏省南京市三校联盟高三年级上学期期中联考数学试题(教师版含答案)
绝密★启用前江苏省南京市三校（溧水二高、秦淮中学、天印中学）2021届高三年级上学期期中质量联考检测数学试题(教师版含答案)2020年11月注意事项：1．答卷前，考生务必将自己的姓名､考生号等填写在答题卡指定的位置上．2．回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑．如需改动，用橡皮擦干净后，再选涂其他答案标号．回答非选择题时，将答案写在答题卡上．写在本试卷上无效．3．考试结束后，将本试卷和答题卡一并交回．一、选择题：本题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的．1．设集合M＝[1，3)，N＝(2，5]，则M∩N＝(B) A．[1，5] B．(2，3) C．[1，2) D．(3，5]2．已知i是虚数单位,设复数a＋b i＝2－i2＋i,其中a,b∈R,则a＋b的值为(D)A．75B．－75C．15D．－153．从5名同学中选若干名分别到图书馆、食堂做志愿者,若每个地方至少去2名,则不同的安排方法共有(B)A ．20种B ．50种C ．80种D ．100种4．中国古代数学著作《算法统宗》中有这样一个问题：“三百七十八里关,初行健步不为难,次日脚痛减一半,六朝才得到其关,要见次日行里数,请公仔细算相还．”你的计算结果是 (D)A ．80里B ．86里C ．90里D ．96里5．若正数a 是一个不等于1的常数,则函数y ＝log a x 与函数y ＝x a(x ＞0)在同一个坐标系中的图象可能是(C)6．设a ＝0.32.1,b ＝2.10.3,c ＝log 0.32.1,d ＝log 2.10.3,则a ,b ,c ,d 的大小关系为(C) A ．a ＞b ＞c ＞d B ．d ＞c ＞b ＞a C ．b ＞a ＞c ＞d D ．b ＞a ＞d ＞c7．在平面直角坐标系xOy 中,已知圆C ：x 2＋y 2＝9及圆C 内的一点P (1,2),圆C 的过点P 的直径为MN ,若线段AB 是圆C 的所有过点P 的弦中最短的弦,则(→AM －→BN )·→AB 的值为(B) A ．8 B ．16 C ．4 D ．4 38．设f (x )是定义在R 上的函数,g (x )＝f (x ＋1)．若函数g (x )满足下列条件：①g (x )是偶函数；②g (x )在区间[0,＋∞)上是增函数；③g (x )有一个零点为2,B DC。
2021届福建省福州市八县(市)一中高三上学期期中联考数学试题(教师版含解析)
2020-2021学年度第一学期八县(市)一中期中试卷高中三年数学科试卷一､选择题(每小题5分，共40分，在每小题给出的四个选项中，只有一项是符合题目要求的)1. 已知集合{}{}2|560|22128xA x Z x xB x =∈--≤=<<，，则A B =( )A. {}|16x x <≤B. {}23456，，，， C. {}|16x x ≤≤ D. {}10123456-，，，，，，， 【答案】B 【解析】 【分析】先求出集合A ，B ，再求两集合的交集即可【详解】解：由2560x x --≤得16x -≤≤，由于x ∈Z ， 所以{}{}2|5601,0,1,2,3,4,5,6A x Z x x =∈--≤-=，由22128x <<，得17x <<，所以{}{}|2212817xB x x x =<<=<< 所以A B ={}23456，，，，， 故选：B2. 已知p ：“函数221y x ax =++在(1,)+∞上是增函数”，q ：“2a >-”，则p 是q 的( ) A. 充分不必要条件 B. 必要不充分条件 C. 充分必要条件 D. 既不充分也不必要条件【答案】A 【解析】 【分析】先求出命题p 对应的a 的取值范围，利用集合的包含关系即可判断. 【详解】由函数221y x ax =++在(1,)+∞上是增函数，因为221y x ax =++的对称轴为x a =-，开口向上，所有1a -≤，即1a ≥-，{}1a a ≥- {}2x a >-，∴p 是q 的充分不必要条件.故选：A.【点睛】结论点睛：本题考查充分不必要条件的判断，一般可根据如下规则判断： (1)若p 是q 的必要不充分条件，则q 对应集合是p 对应集合的真子集； (2)若p 是q 的充分不必要条件，则p 对应集合是q 对应集合的真子集； (3)若p 是q 的充分必要条件，则p 对应集合与q 对应集合相等；(4)若p 是q 的既不充分又不必要条件，则q 对应的集合与p 对应集合互不包含．3. 已知函数()f x 是定义在R 上的偶函数，且函数()f x 在[0,)+∞上是减函数，如果()31f =-，则不等式()110f x -+≥的解集为( )A. ](2-∞，B. [)2，+∞C. []24-，D. []14， 【答案】C 【解析】 【分析】根据题意可得()f x 在[0，)+∞上为减函数，结合奇偶性以及()31f =-可得(|1|)f x f ⇒-|1|3x -，解出x 的取值范围，即可得答案．【详解】函数()f x 是定义在R 上的偶函数，且函数()f x 在[0，)+∞上是减函数， 所以()f x 在(,0)-∞上是增函数，由f (3)1=-，则不等式(1)10(1)1(1)f x f x f x f -+⇒--⇒-(3)(|1|)f x f ⇒-(3)|1|3x ⇒-， 解之可得24x -，故不等式的解集为[2-，4]． 故选：C ．【点睛】将奇偶性与单调性综合考查一直是命题热点，解这种题型往往是根据函数在所给区间上的单调性，根据奇偶性判断出函数在对称区间上的单调性(偶函数在对称区间上单调性相反，奇函数在对称区间单调性相同)，然后再根据单调性列不等式求解.4. 下图是一个正方体的展开图，则在该正方体中( )A. 直线AB 与直线CD 平行B. 直线AB 与直线CD 相交C. 直线AB 与直线CD 异面垂直D. 直线AB 与直线CD 异面且所成的角为60°【答案】D 【解析】 【分析】首先画出正方体的展开图的立体图，从而得到直线AB 与直线CD 为异面直线，再求异面直线所成角即可得到答案.【详解】正方体的展开图的立体图形如图所示：由图知：直线AB 与直线CD 为异面直线，故A ，B 错误；连接CE ，DE ，因为//AB CE ，所以DCE ∠或其补角为异面直线AB 与CD 所成角. 又因为DCE 为等边三角形，所以60DCE ∠=.所以直线AB 与直线CD 异面且所成的角为60°，故C 错误，D 正确. 故选：D【点睛】本题主要考查异面直线成角问题，属于简单题.5. 记n S 为正项等比数列{}n a 的前n 项和，若2415S S ==，，则7S =( ). A. 710S = B. 723S =C. 7623S =D. 71273S =【答案】D 【解析】 【分析】利用等比数列前n 项和公式列出方程组，求出首项和公比，由此能求出这个数列的前7项和． 【详解】n S 为正项等比数列{}n a 的前n 项和，21S =，45S =，∴21410(1)11(1)51q a q qa q q ⎧⎪>⎪⎪-⎪=⎨-⎪⎪-⎪=-⎪⎩，解得113a =，2q ，771(12)1273123S -∴==-．故选：D ．6. 已知0042m n m n >>+=，，，则41m n+的最小值为( ) A. 36 B. 16C. 8D. 4【答案】C 【解析】 【分析】 巧用“1”拼凑()41141=42m n m n m n ⎛⎫+++ ⎪⎝⎭，应用基本不等式即得结果. 【详解】0042m n m n >>+=，，，()411411=4=82126m n m n m n m m n n ⎛⎫⎛⎫∴+++++ ⎪ ⎪⎝⎭⎝⎭18=82⎛≥+ ⎝，当且仅当16=n m m n 时即11,4m n ==时等号成立，故41m n+的最小值为8. 故选：C.7. 已知函数()sin()f x x ωϕ=+(0>ω，||2ϕπ<)，其图像相邻两条对称轴之间的距离为4π，将函数()y f x =的图像向左平移316π个单位后，得到的图像关于原点对称，那么函数()y f x =的图像( ) A. 关于点,016π⎛⎫-⎪⎝⎭对称 B. 关于点,016π⎛⎫⎪⎝⎭对称 C. 关于直线4x π=对称D. 关于直线4πx =-对称 【答案】A 【解析】根据函数()f x 图像的相邻两条对称轴之间的距离为4π，可求得()f x 的周期T ，进而可求得ω的值，根据平移后图像关于原点对称，利用正弦函数图像与性质，即可求得ϕ的值，分别求得()f x 的对称中心、对称轴的表达式，逐一分析选项，即可得答案.【详解】因为函数()f x 图像相邻两条对称轴之间的距离为4π， 所以24T π=，即2T π=，所以24Tπω==，即()sin(4)f x x ϕ=+， 将函数()y f x =的图像向左平移316π个单位后，得到函数3sin[4()]16y x πϕ=++的图像，且其关于原点对称， 所以3416k πϕπ⨯+=()k ∈Z ，又||2ϕπ<，令k =1， 解得4πϕ=，即()sin(4)4f x x π=+，令4,()4x k k Z ππ+=∈，解得,()416k x k Z ππ=-∈，即对称中心为(,0)416k ππ- 令k =0，则一个对称中心为,016π⎛⎫- ⎪⎝⎭，故A 正确，B 错误； 令4,()42x k k Z πππ+=+∈，解得,()416k x k Z ππ=+∈，即对称轴为,()416k x k Z ππ=+∈，故C 、D 错误， 故选：A【点睛】本题考查正弦型函数的图像与性质，解题的关键在于，根据两对称轴间距离，分析图像，可求得ω的值，再根据平移后图像，求得ϕ的值，再求解即可；易错点为平移后解析式为3sin[4()]16y x πϕ=++，注意平移要对x 进行加减，考查分析理解，计算求值的能力，属中档题.8. 已知可导函数()f x 的定义域为(,0)-∞，其导函数()'f x 满足()2()0xf x f x '->，则不等式2(2020)(2020)(1)0f x x f +-+-<的解集为( )A. (,2021)-∞-B. (2021,2020)--C. (2021,0)-D. (2020,0)-【答案】B【分析】由题可得当(,0)x ∈-∞时，()2()0xf x f x '->，进而构造函数2()()f x g x x=，可判断()g x 在(,0)-∞上的单调性，进而可将不等式转化为(2020)(1)g x g +<-，利用()g x 的单调性，可求出不等式的解集．【详解】解：构造2()()(0)f x g x x x =<，则243()2()()2()()x f x x f x xf x f x g x x x ''⋅-⋅-'==，因为()2()0xf x f x '->，则()0g x '<∴函数()g x 在(,0)-∞上是减函数，∵不等式2(2020)(2020)(1)0f x x f +-+-<，且()2(1)(1)(1)1f g f --==--，等价于()()()()()2220201120201f x f g x +-<=-+-，即为(2020)(1)g x g +<-，所以2020120200x x +>-⎧⎨+<⎩，解得20212020x -<<-．故选：B【点睛】本题考查函数单调性的应用，构造函数2()()f x g x x=是解决本题的关键，属于中档题． 二､多选题(每小题5分，共20分，在每小题给出的选项中，有多项符合题目要求，全部选对的5分，有选错的得0分，部分选对得3分)9. 已知复数z 满足(2i)i z -=(i 为虚数单位)，复数z 的共轭复数为z ，则( ) A. 3||5z =B. 12i5z +=-C. 复数z 的实部为1-D. 复数z 对应复平面上的点在第二象限【答案】BD 【解析】 【分析】因为复数z 满足(2i)i z -=，利用复数的除法运算化简为1255z i =-+，再逐项验证判断. 【详解】因为复数z 满足(2i)i z -=，所以()(2)1222(2)55i i i z i i i i +===-+--+所以5z ==，故A 错误；1255z i =--，故B 正确； 复数z 的实部为15- ，故C 错误； 复数z 对应复平面上的点12,55⎛⎫- ⎪⎝⎭在第二象限，故D 正确. 故选：BD【点睛】本题主要考查复数的概念，代数运算以及几何意义，还考查分析运算求解的能力，属于基础题. 10. 已知(2,4),(4,1),(9,5),(7,8)A B C D ，如下四个结论正确的是( ) A. AB AC ⊥；B. 四边形ABCD 为平行四边形；C. AC 与BD 夹角的余弦值为145； D. 85AB AC +=【答案】BD 【解析】 【分析】求出向量,,,AB AC DC BD 坐标，再利用向量的数量积、向量共线以及向量模的坐标表示即可一一判断. 【详解】由(2,4),(4,1),(9,5),(7,8)A B C D ，所以()2,3AB =-，()7,1AC =，()2,3DC =-， ()3,7BD =, 对于A ，143110AB AC ⋅=-=≠，故A 错误；对于B ，由()2,3AB =-，()2,3DC =-，则AB DC =， 即AB 与DC 平行且相等，故B 正确；对于C ，cos ,14550AC BD AC BD AC BD⋅===，故C 错误；对于D ，()||9,2AB AC +=-=D 正确； 故选：BD【点睛】本题考查了向量的坐标运算、向量的数量积、向量模的坐标表示，属于基础题. 11. 在ABC 中，角A ，B ，C 的对边分别是a ，b ，c，若222sin a a b c ab C =+-=，cos sin a B b A c +=，则下列结论正确的是( )A. tan 2C =B. 4A π=C. b =D.ABC 的面积为6【答案】ABD 【解析】 【分析】利用余弦定理，结合题意，可求得tan C的值，根据cos sin a B b A c +=，利用正弦定理边化角，可求得A∠的值，利用正弦定理及面积公式，可求得b 的值及ABC 的面积，即可得答案. 【详解】因为222sin a b c ab C +-=，所以222sin sin cos 222a b c ab C C C ab ab +-===， 所以sin tan 2cos CC C==，故A 正确； 因为cos sin a B b A c +=，利用正弦定理可得sin cos sin sin sin A B B A C +=， 因为()C A B π=-+，所以sin sin[()]sin()C A B A B π=-+=+，所以sin cos sin si sin()sin cos cos sin n A A B B A B A B A B ++==+， 即sin sin cos sin B A A B = 因为(0,)B π∈，所以sin 0B ≠， 所以tan 1A =，又(0,)A π∈， 所以4A π=，故B 正确；因为tan 2C =，(0,)C π∈所以sin 55C C ==所以sin sin()sin cos cos sin 252510B AC A C A C =+=+=+=， 因为sin sin a b A B=，所以31010sin1032 sin2a BbA⨯===，故C错误；1125sin10326225△==⨯⨯⨯=ABCS ab C，故D正确；故选：ABD【点睛】本题考查正弦定理、余弦定理解三角形，三角形面积的求法，解题的关键在于灵活应用正余弦定理及面积公式，考查计算化简的能力，属中档题.12. 已知直三棱柱111ABC A B C-中，AB BC⊥，1AB BC BB==，D是AC的中点，O为1A C的中点.点P是1BC上的动点，则下列说法正确的是()A. 当点P运动到1BC中点时，直线1A P与平面111A B C5B. 无论点P在1BC上怎么运动，都有11A P OB⊥C. 当点P运动到1BC中点时，才有1A P与1OB相交于一点，记为Q，且113PQQA=D. 无论点P在1BC上怎么运动，直线1A P与AB所成角都不可能是30°【答案】ABD【解析】【分析】构造线面角1PA E∠，由已知线段的等量关系求1tanEPPA EAE∠=的值即可判断A的正误；利用线面垂直的性质，可证明11A P OB⊥即可知B的正误；由中位线的性质有112PQQA=可知C的正误；由直线的平行关系构造线线角为11B A P ∠，结合动点P 分析角度范围即可知D 的正误 【详解】直三棱柱111ABC A B C -中，AB BC ⊥，1AB BC BB ==选项A 中，当点P 运动到1BC 中点时，有E 为11B C 的中点，连接1A E 、EP ，如下图示即有EP ⊥面111A B C∴直线1A P 与平面111A B C 所成的角的正切值：1tan EPPA E AE∠= ∵112EP BB =，22111152AE A B B E BB =+= ∴15tan 5PA E ∠=，故A 正确选项B 中，连接1B C ，与1BC 交于E ，并连接1A B ，如下图示由题意知，11B BCC 为正方形，即有11B C BC ⊥而AB BC ⊥且111ABC A B C -为直三棱柱，有11A B ⊥面11B BCC ，1BC ⊂面11B BCC ∴111A B BC ⊥，又1111A B B C B =∴1BC ⊥面11A B C ，1OB ⊂面11A B C ，故11BC OB ⊥同理可证：11A B OB ⊥，又11A B BC B ⋂=∴1OB ⊥面11A BC ，又1A P ⊂面11A BC ，即有11A P OB ⊥，故B 正确选项C 中，点P 运动到1BC 中点时，即在△11A B C 中1A P 、1OB 均为中位线∴Q 为中位线的交点 ∴根据中位线的性质有：112PQ QA =，故C 错误选项D 中，由于11//A B AB ，直线1A P 与AB 所成角即为11A B 与1A P 所成角：11B A P ∠结合下图分析知：点P 在1BC 上运动时当P 在B 或1C 上时，11B A P ∠最大为45°当P 在1BC 中点上时，11B A P ∠最小为23arctan302>=︒ ∴11B A P ∠不可能是30°，故D 正确 故选：ABD【点睛】本题考查了利用射影定理构造线面角，并计算其正弦值；利用线面垂直证明线线垂直；中位线的性质：中位线交点分中位线为1：2的数量关系；由动点分析线线角的大小三､填空题(本题共4小题，每小题5分，共20分)13.若cos 410πθ⎛⎫-=⎪⎝⎭，则sin 2θ=________． 【答案】45-【解析】【分析】由题意利用诱导公式、二倍角的余弦公式，求得结果．【详解】解：若cos()4πθ-= 则214sin 2cos(2)2cos ()12124105ππθθθ=-=--=⨯-=-， 故答案为：45-． 【点睛】本题主要考查诱导公式、二倍角的余弦公式的应用，属于基础题．14. 已知数列{}n a 的前n 项和231n S n n =--，则n a =__________.【答案】31242n n a n n -=⎧=⎨-≥⎩，， 【解析】【分析】由11,1,2n n n S n a S S n -=⎧=⎨-≥⎩进行求解即可 【详解】解：当1n =时，111313a S ==--=-，当2n ≥时，22131[(1)3(1)1]24n n n S n n n n a n S --=-------==-，当 1n =时，1242a -=-≠，所以31242n n a n n -=⎧=⎨-≥⎩，，， 故答案为：31242n n a n n -=⎧=⎨-≥⎩，， 15. 在三棱锥P ABC -中，平面PAB 垂直平面ABC，PA PB AB AC ====120BAC ∠=︒，则三棱锥P ABC -外接球的表面积为_________.【答案】52π【解析】【分析】如图，过点A 在面PAB 内作AQ AB ⊥交PAB △的外接圆于点Q ，平面PAB 垂直平面ABC ，两平面的交线为AB ，利用正弦定理和勾股定理，构造出Rt QHA ，然后，利用勾股定理得出2222(2)(2)448524R h r R =+=+==，进而求解可得 【详解】如图，过点A 在面PAB 内作AQ AB ⊥交PAB △的外接圆于点Q ，平面PAB 垂直平面ABC ，两平面的交线为AB ，AQ AB ⊥，AQ ⊂面PAB ，AQ ∴⊥面ABC ，PAB △的外接圆直径为234QB ==，222QA QB AB ∴=-=，而2h QA ==， ABC 中，23AB AC ==120BAC ∠=︒，30ACB ∴∠=︒，设底面ABC 的外接圆半径为r ，则243sin AB r BCA==∠R ，则有2222(2)(2)448524R h r R =+=+==，球的表面积为2452S R ππ==故答案为：52π【点睛】关键点睛：解题关键在于，构造直角三角形Rt QHA ，利用勾股定理得出2222(2)(2)448524R h r R =+=+==，进而求解16. 函数()f x 满足()()11f x f x +=-，当1x >时，()ln x f x x=，若()()2240f x mf x m -+=有8个不同的实数解，则实数m 的取值范围是______. 【答案】()24,22e e ⎛⎫ ⎪ ⎪-⎝⎭【解析】【分析】利用导数分析函数()y f x =在区间()1,+∞上的单调性与极值，由题意可知，函数()y f x =的图象关于直线1x =对称，数形结合可知关于t 的二次方程2240t mt m -+=有两个大于e 的实根，利用二次方程根的分布可得出关于m 的不等式组，由此可解得实数m 的取值范围.【详解】当1x >时，()ln x f x x=，()2ln 1ln x f x x -'=. 当1x e <<时，()0f x '<，此时函数()y f x =单调递减；当x e >时，()0f x '>，此时函数()y f x =单调递增.所以，函数()y f x =在x e =处取得极小值()f e e =，又()()11f x f x +=-，则函数()y f x =的图象关于直线1x =对称，令()t f x =，作出函数()t f x =的图象如下图所示：由于关于x 的方程()()2240f x mf x m -+=有8个不同的实数解，则关于t 的二次方程2240t mt m -+=有两个大于e 的实数根，由二次方程根的分布可得224160240m m m e e me m ⎧∆=->⎪>⎨⎪-+>⎩，解得()2422e m e <<-. 综上所述，实数m 的取值范围是()24,22e e ⎛⎫ ⎪ ⎪-⎝⎭.故答案为：()24,22e e ⎛⎫ ⎪ ⎪-⎝⎭.【点睛】本题考查利用方程根的个数求参数，考查了导数的应用以及一元二次方程根的分布，考查数形结合思想的应用，属于较难题.四､解答题(本题共6小题，共70分，解答应写出文字说明､证明过程或演算步骤.)17. 已知n S 是数列{}n a 的前n 项和，且233n n S a +=(1)求{}n a 的通项公式；(2)设3311log log n n n b a a ，求数列{}n b 的前n 项和n T . 【答案】(1)3n n a =；(2)1n n T n =+. 【解析】【分析】(1)利用1n n n a S S -=-可得{}n a 是以3为首项，3为公比的等比数列，即可求出通项公式；(2)利用裂项相消法可求.【详解】解：(1)1n =时，11233S a +=，11a S =，13a ∴=，2n ≥时，因为()312n n S a =-，所以()11312n n S a --=-. 相减得()132n n n a a a -=-，所以13n n a a -=. 所以{}n a 是以3为首项，3为公比的等比数列，所以1133n n n a a -=⋅=，即{}n a 通项公式为3n n a =.(2)由(1)可得()33111log log 1n n n b a a n n +==+111n n =-+. 所以12111111......12231n n T b b b n n ⎛⎫⎛⎫⎛⎫=+++=-+-++- ⎪ ⎪ ⎪+⎝⎭⎝⎭⎝⎭1111n n n =-=++. 【点睛】方法点睛：数列求和的常用方法：(1)对于等差等比数列，利用公式法可直接求解；(2)对于{}n n a b 结构，其中{}n a 是等差数列，{}n b 是等比数列，用错位相减法求和；(3)对于{}+n n a b 结构，利用分组求和法；(4)对于11n n a a +⎧⎫⎨⎬⎩⎭结构，其中{}n a 是等差数列，公差为d ，则111111n n n n a a d a a ++⎛⎫=- ⎪⎝⎭，利用裂项相消法求和.18. 在①()3cos cos cos sin C a B b A c C +=，②sin sin 2A B a c A +=，③()()2sin 2sin 2sin a b A b a B c C -+-=这三个条件中任选一个，补充在下列问题中，并解答. 已知ABC 的角A ，B ，C 对边分别为,,a b c ，3c =，而且___________.(1)求C ∠；(2)求ABC 周长的范围.【答案】条件选择见解析；(1)3π；(2)(2333⎤⎦，. 【解析】【分析】 (1)选①：由条件结合正弦定理可得()3cosCsin A B sinCsinC +=，即3tanC =，得出答案.选②：由条件结合诱导公式、正弦定理和二倍角公式可得122C sin =，从而得出答案. 选③：由条件结合正弦定理可得222a b c ab +-=，再根据余弦定理可得答案.(2)由(1)结合余弦定理可得223a b ab +-=,利用均值不等式可得周长的最大值，再利用三角形中两边之和大于第三边可得出答案.【详解】解：(1)选①：由正弦定理得()3cosC sinAcosB sinBcosA sinCsinC +=()3cosCsin A B sinCsinC +=因为0sinC tanC ≠∴=， 因为()03C C ππ∈∴=，，选②： 由正弦定理得2CsinAsin sinCsinA π-=， 因为02222c C C sinA cos sinC sin cos ≠∴==，因为02C cos ≠，所以122C sin =， 因为()03C C ππ∈∴=，，选③：因为()()2sin 2sin 2sin a b A b a B c C -+-=，所以()()2222a b a b a b c -+-=，即222a b c ab +-=， 所以2221cos 22a b c C ab +-==， 因为0C π<<，所以3C π=； (2)由(1)可知：3C π=，在ABC 中，由余弦定理得222cos 3a b ab C +-=，即223a b ab +-=，所以()()223334a b a b ab ++-=≤，所以a b +≤a b =时等号成立，所以a b c ++≤ABC 周长的最大值为又因为a b c +>=ABC 周长的取值范围为(【点睛】关键点睛：本题考查利用正弦定理和余弦定理解三角形，解题的关键是利用正弦定理进行边化角，第(2)问中结合(1)的结果，利用余弦定理得到223a b ab +-=，先配方再利用均值不等式()()223334a b a b ab ++-=≤求出+a b 的范围，最后三角形中两边之和大于第三边得到三角形周长的范围，属于中档题.19. 已知如图①，在菱形ABCD 中，60A ∠=︒且2AB =，E 为AD 的中点，将ABE △沿BE 折起使2AD =，得到如图②所示的四棱锥A BCDE -.(1)求证：平面ABE ⊥平面ABC ；(2)若P 为AC 的中点，求二面角P BD A --的余弦值.【答案】(1)证明见解析；(2)17. 【解析】【分析】(1)利用题中所给的条件证明AE ED ⊥，BE DE ⊥，因为//BC DE ，所以BC BE ⊥，BC AE ⊥，即可证明BC ⊥平面ABE ，进一步可得面面垂直；(2)先证明AE ⊥平面BCDE ，以E 为坐标原点，EB ，ED ，EA 的方向分别为x 轴，y 轴，z 轴，建立如图所示的空间直角坐标系，求出平面PBD 的一个法向量m ，平面BDA 的一个法向量n ，利用向量的夹角公式即可求解【详解】解：(1)在图①中，连接BD ，如图所示：因为四边形ABCD 为菱形，60A ∠=︒，所以ABD △是等边三角形.因为E 为AD 的中点，所以BE AE ⊥，BE DE ⊥.又2AD AB ==，所以1AE DE ==.在图②中，2AD =222AE ED AD +=，即AE ED ⊥.因为//BC DE ，所以BC BE ⊥，BC AE ⊥.又BE AE E =，AE ，BE ⊂平面ABE .所以BC ⊥平面ABE .又BC ⊂平面ABC ，所以平面ABE ⊥平面ABC .(2)由(1)知，AE DE ⊥，AE BE ⊥.因为BE DE E ⋂=，BE ，DE ⊂平面BCDE .所以AE ⊥平面BCDE .以E 为坐标原点，EB ，ED ，EA 的方向分别为x 轴，y 轴，z 轴，建立如图所示的空间直角坐标系：则()0,0,0E ，()0,0,1A ，)3,0,0B ，()3,2,0C ，()0,1,0D . 因为P 为AC 的中点，所以3122P ⎛⎫ ⎪ ⎪⎝⎭. 所以31,1,22PB ⎛⎫=-- ⎪ ⎪⎝⎭，31,0,22PD ⎛⎫=-- ⎪ ⎪⎝⎭. 设平面PBD 的一个法向量为(),,m x y z =，由00PB m PD m ⎧⋅=⎨⋅=⎩得31023102x y z x z ⎧--=⎪⎨⎪-=⎪⎩. 令3z =，得1x =-，3y =-(133m =-，. 设平面BDA 的一个法向量为()111n x y z =，，.因为()31BA =-，，，()011AD =-，， 由00BA n AD n ⎧⋅=⎨⋅=⎩得1111300x z y z ⎧-+=⎪⎨-=⎪⎩ 令11x =，3z =3y =(133n =，，则1cos ,777m nm n m n ⋅===-⨯⨯， 所以二面角P BD A --的余弦值为17. 【点睛】思路点睛：证明面面垂直的思路(1)利用面面垂直的定义，(不常用)(2)利用面面垂直的判定定理；(3)利用性质：//αβ，βγαγ⊥⇒⊥.20. 如图，有一生态农庄的平面图是一个半圆形，其中直径长为2km ，C ､D 两点在半圆弧上满足AD BC =，设COB θ∠=，现要在此农庄铺设一条观光通道，观光通道由,,AB BC CD 和DA 组成.(1)若6πθ=，求观光通道l 的长度；(2)用θ表示观光通道的长l ，并求观光通道l 的最大值； 【答案】(1)观光通道长(2362km +；(2)当3πθ=时，观光通道长l 的最大值为5km . 【解析】【分析】(1)由6πθ=，得6OCD ODC π∠=∠=，然后在OCD ，OCB ，OAD △利用余弦定理求出,,CD BC AD的长，从而可得结果；(2)作OE BC ⊥，垂足为E ，在直角三角形OBE 中，sin sin 22BE OB θθ==，则有2sin 2BC AD θ==，同理作OF CD ⊥，垂足为F ，cos cos CF OC θθ==，即：2cos CD θ=，从而24sin 2cos 2l θθ=++，然后利用三角函数的性质可得结果【详解】(1)因为6πθ=，所以6OCD ODC π∠=∠=在OCD 中，利用余弦定理可得，2211211cos 33CD π=+-⨯⨯⨯=，所以3CD =同理62232BC AD -==-= 所以观光通道长2362l km =++-(2)作OE BC ⊥，垂足为E ，在直角三角形OBE 中，sinsin 22BE OB θθ==， 则有2sin 2BC AD θ==，同理作OF CD ⊥，垂足为F ，cos cos CF OC θθ==，即：2cos CD θ=，从而有：22124sin 2cos 4sin 4sin 44sin 522222l θθθθθ⎛⎫=++=-++=--+ ⎪⎝⎭ 因为02πθ⎛⎫∈ ⎪⎝⎭，，所以当3πθ=时，l 取最大值5， 即观光通道长l 的最大值为5km .【点睛】关键点点睛：此题考查余弦定理的应用，解题的关键是把,,CD BC AD 用含θ的式子表示，然后利用三角恒等变换公式转化为同角的三角函数求解，解题时要注意θ的取值范围21. 已知函数()ax f x xe =的极值为1e-. (1)求a 的值并求函数()f x 在1x =处的切线方程；(2)已知函数()()0mx lnx g x e m m=->，存在()0x ∈+∞，，使得()0g x ≤成立，求m 得最大值. 【答案】(1)1a =，切线方程为：2y ex e =-；(2)最大值为1e . 【解析】【分析】(1)利用切线方程的公式求解即可(2)将问题转化为mx me lnx ≤，经过放缩得mx lnx mxe xlnx lnxe ≤=，转化成()()f mx f lnx ≤，再利用导数判断()f x 的最值情况，进而可求得最终答案【详解】解：(1)()f x 定义域为R因为()()1ax f x e ax ='+若0a =则()f x 在R 上单调递增，无极值，不合题意，舍去若0a ≠则令()0f x '=得1x a =-所以11f a e ⎛⎫-=- ⎪⎝⎭解得1a = 经检验，1a =符合题意.因为切线斜率()()11112f e e =+='又因为()1f e =所以切点为()1e ， 所以切线方程为：()21y e x e =-+即切线方程：2y ex e =-(2)因为存在()0x ∈+∞，，使得()0g x ≤成立 则mx lnx e m≤ 即mx me lnx ≤即mx lnx mxe xlnx lnxe ≤=即mx lnx mxe lnxe ≤即()()f mx f lnx ≤(*)由(1)得()()1x f x e x '=+所以()f x 在区间()1-∞-，上单调递减，在区间()1，-+∞上单调递增 因为00mx m x me lnx >>≤，，所以0lnx >，所以1x >即0mx >且0lnx >所以存在()1x ∈+∞，使得()()f mx f lnx ≤ 所以存在()1x ∈+∞，使得mx lnx ≤即()1lnx m x x≤∈+∞， 令()lnx s x x=所以()max m s x ⎡⎤≤⎣⎦ 因为()210lnx s x x '-==得x e = 所以()s x 在区间()1e ，上单调递增，在区间()e +∞，单调递减 所以()s x 的最大值为()1s e e =所以1m e≤又因为0m >，所以10m e <≤ 所以m 的最大值为1e 【点睛】关键点睛：解题的关键在于放缩得mx lnx mxe xlnx lnxe ≤=，把问题转化为()()f mx f lnx ≤，考查学生的转化化归和放缩的运用，属于难题22. 已知函数()()()2ln 1002x f x ax a x x -=+->≥+，. (1)当12a =时，讨论函数()y f x =的单调性； (2)若不等式()1f x ≥在[0,)x ∈+∞时恒成立，求实数a 的取值范围；(3)证明：()()*11111ln 1357212n n N n ++++<+∈+. 【答案】(1)在区间()02，上单调递减；在区间()2+∞，上单调递增；(2)[1，)+∞；(3)证明见解析. 【解析】【分析】(1)求出()f x 的导数，根据导数的正负即可判断单调性；(2)求出()f x 的导数，根据a 的范围讨论单调性，求出()f x 的最小值，满足()min 1f x ≥即可求出a 的取值范围；(3)由(2)可知当1a =时，不等式()1f x >在(0,)x ∈+∞时恒成立，可得11[ln(1)ln ]122k k k <+-+，即可得证. 【详解】解：(1)当12a =时，()()()221142122212x f x x x x -'=⋅-=+++， 当()0,2x ∈时，()0f x '<，()f x 单调递减；当()2,x ∈+∞时，()0f x '>，()f x 单调递增；所以()y f x =在区间()0,2上单调递减；在区间()2,+∞上单调递增；(2)()2224441(2)(1)(2)a ax a f x ax x ax x +-'=-=++++， 当1a ≥时，()0f x '≥，∴函数()y f x =在[)0+∞，上单调递增；当01a <<时，由()0f x '>可得x >∴函数在⎡⎫∞⎪⎢⎪⎣⎭上单调递增，在⎡⎢⎣上单调递减； ①当1a ≥时，函数()y f x =在[)0+∞，上单调递增， ()()01f x f ∴≥=，即不等式()1f x ≥，在[)0x ∈+∞，时恒成立，②当01a <<时，函数在0⎡⎢⎣上单调递减，存在00x ⎡∈⎢⎣使得()()001f x f <=，所以不合题意，舍去. 综上可知实数a 的取值范围为[)1,+∞；(3)由(2)得当1a =时，不等式()1f x >在(0,)x ∈+∞时恒成立， 即2ln(1)2x x x +>+，12ln(1)12k k ∴+>+，*()k N ∈. 即11[ln(1)ln ]122k k k <+-+， ∴11(ln 2ln1)32<-，11(ln3ln 2)52<-，11(ln 4ln3)72<-，11[ln(1)ln ]212n n n ⋯<+-+， 将上述式子相加可得()()()111111ln 1ln1ln 13572122n n n ++++<+-=++ 原不等式得证. 【点睛】结论点睛：本题考查不等式的恒成立问题，一般按如下规则转化，(1)对于[],x a b ∀∈，都有()f x m ≥恒成立，则()min f x m ≥；(2)对于[],x a b ∀∈，都有()f x m ≤恒成立，则()max f x m ≤.。
2021届太原市进山中学高三语文上学期期中考试试卷及答案
2021届太原市进山中学高三语文上学期期中考试试卷及答案一、现代文阅读（36分）(一)现代文阅读I(9分)阅读下面的文字，完成下面小题。
材料一作为“新动能”之一，共享经济有光明的前景。
共享单车、共享停车、共享租屋等就是已走近普通大众的共享产品。
“共享+”百花齐放，这是好事，然而，也有一些所谓的“共享”，因为缺少有效的管理手段和盈利模式，不仅浪费社会资源，也透支了“共享”的公信力，成了“伪共享”。
比如，有人把“共享雨伞”投放到地铁口、公交站台等交通节点，结果半个月后悉数消失；有人在城市街头投放“共享马扎”，但管理跟不上，最终被市政部门清理搬走；有人推出“共享睡眠舱”，但安全、消防、卫生都不过关，最终被管理部门叫停……共享经济是指在所有权不变的情况下，权利人将使用权进行临时性转移，以提高资源利用率，权利人也能从中获益。
其本质是整合闲散资源，盘活存量经济，减少浪费，避免新的资源开掘。
“伪共享”之所以没市场，就在于它们并未真正从消费者的需求出发来解决生产与生活的痛点，却因制造“虚假需求”，造成了不必要的资源浪费，成了“非共享、不经济”。
“共享项目”只有摆脱资本热捧与营销炒作，走向理性、合规、审慎，在分享平台上做深、做透，才能找到真正清晰的发展焦点和盈利路径。
有人预测，未来几年中国共享经济还将保持40%的高速增长。
（摘编自《“共享经济”越是被看好，越不能滥用》，2017.10.27《人民日报》）材料二中国的共享服务正在向汽车领城扩大。
预计到2020年，中国共享汽车的市场规模将从2016年的4亿元扩大至93亿元，在中国，共享单车等服务迅速普及，但共享汽车面临诸多课题。
大型汽车厂商由于担心汽车销量下滑，也对共享汽车持谨慎姿态。
但中国本土汽车厂商可能在共享汽车领域加强攻势。
中国约3亿人有驾照，而私家车保有量约1.5亿辆，单纯计算有1.5亿人有驾照但没车。
有问卷调查显示，77%的受访者“愿意尝试共享汽车服务”。
（摘编自《中国共享汽车服务业快马加鞭》2017.12.20“日本经济新闻网”）材料三五种出行方式在五个纬度上的示意图材料四近日，摩拜单车宣布与日本最大移动社交网络公司LINE达成战略协作。
2021年高三上学期期中试卷(语文)
2021年高三上学期期中试卷（语文）A．辟．邪／开辟．新路解．数／押解．症．结／症．状B．作．坊／作．客他乡囤．积／粮囤．粘．性／粘．贴C．着．凉／十分着．急字帖．／碑帖．角．逐／角．色D．载．体／载．歌载舞中．肯／中．伤狼藉．／慰藉．2．下列各句中，加点成语熟语使用正确的是 ( )A．《士兵突击》成功播出后，王宝强因为饰演许三多而声名鹊起，成为了一位炙手可热．．．．的新星。
B．最近几年，我国内地学生对香港地区的入学趋之若鹜．．．．。
这说明国内考生的传统观念正在逐渐发生变化。
C．为了加快城市现代化建设的步伐，我市“城市总体规划修编计划”启动，新一轮城市总体规划呼之欲出．．．．。
D．随着公众对政府行为和社会事务的关注度与参与度的不断提高，一些地方己开始征收卡拉OK版权费，这对著作权人来说是大快人心．．．．的好事。
3．下列各句中，，沒有语病的一句是( )A．加强对青少年的教育，增强青少年识别利抵制各种错误倾向的能力，是目前我国中小学思想品德教育工作的当务之急。
B．“十一”黄金周期间，奥林匹克中心区将推出奥运图片展览、奥运规划建设成果展、闭幕式部分道具展、奥运精品图片展、中国故事小屋、“福娃流动秀”和“北京礼物-xx 北京旅游商品展卖周”等系列活动。
C．今年的沙尘暴提高了干部群众对植树造林重要性的认识，仅一个春季全县就植树造林3700多万棵，其中防风林和果树10500棵。
D．国家质检总局近日紧急在全国开展了婴幼儿奶粉三聚氰胺含量专项检查。
为保证乳制品安全，有关部门采取了进一步加强监管。
4．依次填入下面横线上的句子，与上文衔接最恰当的一组是( )那个春天的黄昏，当满树繁花无意间闯入我的眼帘，我的心不禁为之震颤，我惊诧三年的时间，树天天在，花年年开，。
一连好几天，我带着一种愧疚的心情站在阳台上望它们，望它们于无声处悄然散发着生命的芬芳。
美丽的花树，寂寞的花树，使我领悟了一句话：。
①可我竟然从未留意过这近在咫尺的美②可对于这近在咫尺的美，我竟然从未留意过③美丽如河流，越深越无声④美丽如醇酒，越久越香浓A．①③B．②④C．①④D．②③二、阅读下面的文字，完成5-7题。
湖北省武汉市华师一附中2021届高三上学期期中考试化学试题 Word版含答案
华中师大一附中 2020—2021 学年度上学期高三期中检测化学试题考试时间：90 分钟 试卷满分：100 分可能用到的相对原子质量：H 1 B11 C 12 N 14 O 16一、选择题：本题共 10 小题，每小题 2 分，每小题只有一个选项符合题意。
1．关于“新型冠状病毒肺炎”的认识，下列说法正确的是A ．《新型冠状病毒肺炎诊疗方案》第六版增加“新型冠状病毒在相对封闭的环境中长时间暴露于高浓度气溶胶情况下存在经气溶胶传播的可能。
”此处气溶胶中的分散质是空气B ．乙醇、过氧化氢、过氧乙酸、84 消毒液、二氧化氯泡腾片消毒原理相同C ．医用口罩的原料聚丙烯属于合成纤维D ．酒精是良好的消毒杀菌剂，大量饮用高度白酒可以预防“新型冠状病毒”2．共建“一带一路”符合国际社会的根本利益，彰显人类社会的共同理想和美好追求。
下列贸易商品中，主要成分不属于高分子的是 中国丝绸乌克兰葵花籽油埃及长绒棉泰国乳胶枕A B C D2验证实验(夹持装置略去)。
下列说法正确的是A ．用 KMnO 4 和浓盐酸制备 Cl 2 时，可选①和②B ．③用于 Cl 2 收集时，a 为进气口C ．④中加入 NaBr 溶液和 CCl 4，通入 Cl 2 反应后振荡、静置，下层溶液呈橙红色D ．⑤中通入 Cl 2 使品红溶液褪色后，取出溶液加热可恢复红色4．捕获二氧化碳生成甲酸过程如图所示。
N A 为阿伏加德罗常数的值，下列说法正确的是A ．在捕获过程中，若有 22.4 L CO 2 参与反应，则在反应过程中 CO 2 分子只拆开了 N A 对电子对B ．已知N(C 2H 5)3 性质与NH 3 类似，0.1mol 该物质与盐酸恰好完全反应生成的盐溶液中，[N(C 2H 5)3H]+的数目为 0.1N AC ．100 g 46％的甲酸水溶液中所含的氧原子数目为 5N AD ．该反应是氧化还原反应，每生成 0.5 mol 甲酸，转移的电子数为 2N A 5．银－Ferrozine 法检测甲醛(HCHO)的原理如下(在原电池中完成氧化银与甲醛的反应)。
2021年高三语文上学期期中教学质量检测试题
2021年高三语文上学期期中教学质量检测试题本试卷分第I卷和第II卷两部分，共8页。
满分150分。
考试用时150分钟。
注意事项:1.答题前，考生务必用0.5毫米黑色签字笔将自己的姓名、座号、准考证号、学校和科类填写在答题卡和试卷规定的位置上。
2.第I卷每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦千净后，在选涂其他答案标号。
答案不能答在试卷上。
3.第II卷必须用0.5毫米黑色签字笔作答，答案必须写在答题卡各题目指定区域内相应的位置.不能写在试卷上;如需改动，先划掉原来的答案.然后再写上新的答案;不能使用第I卷(共36分)一、(每小题3分.共15分)1、下列词语加点的字，每对读音都不相同的一项是()A.牵强．/勉强．着．迷/软着．路氯．气/渌水荡漾B.请帖．/字帖．薄．饼/户口簿．冠．军/夺得桂冠．C.璎路．/洛．阳警．戒/擎．天柱对峙．/．恃．才傲物D.烘焙．/培．养咀嚼．/咬文嚼．字肖．像/惟妙惟肖．2、下列词语中.没有错别字的一项是()A.欠账风采亲和力和言悦色B.蜚声抱负出洋像委曲求全C.安详陨落汀书机兼容并蓄D.婆姿谜语白内瘴寥若晨星3、下列句子中标点符号使用没有惜误的一项是()A.去都江堰，一进入成灌高速公路，“上善若水”的巨型横幅扑面而来，这是指水吗，是褒扬都江堰吗，还是借水喻人，弘扬一种文化精神?B.社会主义现代化建设，离不开法治的引领和规范，中华民族的伟大复兴，离不开法治的保障和支撑。
C.这一带在古代就是一个“少草木，多风沙”的地方。
(《汉书·匈奴传>)D.据说明代士大夫案上总摆两部书:一部《庄子》，一部《水浒传》。
士大夫们认为《水浒传》笔法好，写人“明镜照物，妍媸毕露”。
4、下列各句中，加点的词语使用正确的一项是()A.望着那悠远．．的夜空中闪烁的群星，他又回想起那有趣的童年，回想起小时候妈妈讲给他的许多动人的故事。
B、类似的事情在我们身边也不止．．一次地发生，为什么至今没有引起我们的重视呢?C、香港“占中”非法集会已经持续超过了四个星期，他们绑架民意，一意孤行，在普选问题上危言危行．．．．，破坏了政府运作、法治根基，会造成短期甚至长期的政治不稳定。
2021届华南师范大学附属中学高三英语上学期期中考试试卷及参考答案
2021届华南师范大学附属中学高三英语上学期期中考试试卷及参考答案第一部分阅读（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A、B、C、D四个选项中选出最佳选项AOver the years, I’ve been guilty of hastily shutting the front door to many strangers when they came knocking with the intention of selling things. But earlier this year, around Easter time, a dear friend of mine had an experience that changed my mind and perception towards these “intruders”.Linda, who recently moved to a new neighbour hood, had been housebound all week suffering from a severe case of flu when, early one morning, there was the terrible knock on her front door. Peering out the window, she saw two young ladies demurely (端庄地) holding leaflets and a TV set model. She knew of a large TV set establishment nearby and assumed a message was about to be gently delivered. “This is the last thing I need today,” she muttered to herself and hesitantly opened the door.Coughing, she poked her head out and impatiently informed them she wasn’t the slightest bit interested in any TV whatsoever. Furthermore, she added, she was feeling quite unwell and abruptly shut the door. The ladies politely turned and left in silence.A few hours later, another knock. Linda glanced out the window and to her surprise, there were the same two ladies, back again. Really annoyed this time, she opened the door ready togive them a piece of her mind. Before she could speak, with concerned looks on their faces, these women handed Linda what looked to be a dish, saying, “We’re so sorry we disturbed you earlier. We thought you may like this homemade chicken soup. Hopefully it might make you feel a little better.”Linda was taken aback by this lovely gesture. Being new to the area, she didn’t know many neighbours or people who could assist her with shopping or errands (跑腿) so the gesture was all the more appreciated. She could only smile sheepishly and sincerely thanked them. With that, the ladies left.After she related this story to me, I thought about how touching this deed really was. Especially in today’s world where sadly, kindness and thoughtfulness seem so rare. I also realized that thinking about others and showing kindness is the real essence of love.1. What does the expression “give them a piece of her mind”in Paragraph 4 mean?A. Scold them severely.B. Drive them away.C. Refuse them directly.D. Speak out her idea.2. What can be learned about Linda from the passage?A. Linda always hastily shut the front door to many people.B. Linda was not a little interested in any TV whatsoever.C. Linda was very grateful to the ladies for their chicken soup.D. Linda would buy their TV set in reward for the ladies’ chicken soup.3. What’s the best title for the passage?A. A Chicken SoupB. Soup for the soulC. The Real Essence of LoveD. A Friend’s experienceBA new study finds almost one third of the world’s population is overweight. Since 1980, obesity rates(肥胖率)in children and adults have doubled in 73 countries, making people more worried. And rates are increasing in many other countries. Obesity is increasing faster in children than adults in many nations, including Algeria, Turkey and Jordan, the report said. But the world's weight problem is growing in both rich and poor countries alike. Researchers say an increasing number of people are dying of related health problems in what they called a "disturbing global public health problem."Researchers studied health information from 1980 through 2015. They examined obesity rates, average weight gain in 195 countries. They found that obesity rates are three times greater among youth and young adults in countries like China, Brazil and India. Almost 108 million children and more than 600 million adults were found to be overweight. Egypt had the highest number of overweight adults in 2015. Vietnam had the least. In the same year, the United States had the highest number of overweight children, and Bangladesh had the least.Yet hunger remains a problem in many areas. The United Nations said that almost 800 million people, including 300 million children, go to bed hungry each night. Experts said poor diets and lack of physical activity are mainly to blame for the rising numbers of overweight people.Growing populations have led to rising obesity rates in poor countries. Often, poor people will eat processed(加工的) foods instead of choosing a diet rich in vegetables.The London-based Overseas Development Institute studied the price of food in five countries: Britain, Brazil, China, Mexico and South Korea. It found that the cost of processed foods like ice creamand hamburgers has fallen since 1990. But the cost of fresh fruits and vegetables has gone up.4. What do we know about obesity in the world?A. It is worryingB. It is given no attentionC. It has been successfully solvedD. It is more serious among old people5. Which country has the most overweight children in 2015?A. AmericaB. VietnamC. EgyptD. Bangladesh6. What is one of the reasons why people get overweight?A. HungerB. Poor dietC. Diets with vegetablesD. Eating too much fruit7. Which ofthe following can be the best title for the text?A. It’s Important to Have a Balanced DietB. How Can We Lose Weight and Keep Fit?C. Is It Necessary for Us to Eat Fresh Vegetables?D. Nearly One Third of People in the World Are Overweight.CThe Native American of northern California were highly skilled at basketry, using the reeds, graeses, barks, and roots they found around them to fashion articles of all sorts and sizes-not only trays, containers, and cooking pots, but hats, boats, fish traps, baby carriers, and ceremonial objects.Of all these experts, none excelled the Pomo-a group who lived on or near the coast during the 1800's, and whose descendants continue to live in parts of the same region to this day. They made baskets three feet in diameter and othersno bigger than a thimble (顶针). The Pomo people were masters of decoration. Some of their baskets were completely covered with shell pendants；others with feathers that made the baskets’ surfaces as soft as the breasts of birds. Moreover, the Pomo people made use of more weaving techniques than did their neighbors. Most groups made al their basketwork by twining--the twisting of a flexible horizontal material, called a weft, around stiffer vertical strands of material, the warp. Others depended primarily on coiling-a process in which a continuous coil of stiff material is held in the desired shape with tight wrapping of flexible strands. Only the Pomo people used both processes with equal ease and frequency. In addition, they made use of four distinct variations on the basic twining process, often employing more than one of them in a single article.Although a wide variety of materials was available, the Pomo people used only a few. The warp was alwaysmade of willow, and the most commonly used weft was sedge root, a woody fiber that could easily be separated into strands no thicker than a thread. For color1 , the Pomo people used the bark of red-bud for their twined work and dyed bullrush root for black in coiled work. Though other materials were sometimes used, these four were the staples in their finest basketry.If the basketry materials used by the Pomo people were limited, the designs were amazingly varied. Every Pomo basket maker knew how to produce from fifteen to twenty distinct patterns that could be combined in a number of different.8. The word “fashion” in paragraph 1 is closest in meaning to ______.A. maintainB. organizeC. tradeD. create9. What is the author's main point in paragraph 2?A. The neighbors of the Pomo people tried to improve on the Pomo basket weaving techniques.B. The Pomo people were the most skilled basket weavers in their region.C. The Pomo people learned their basket weaving techniques from other Native Americans.D. The Pomo baskets have been handed down for generations.10. According to the passage, the relationship between red-bud and twining is most similar to the relationship between ______.A. bullrush and coilingB. weft and warpC. willow and feathersD. sedge and weaving11. Which of the following statements about Pomo baskets can be best inferred from the passage?A. Baskets produced by other Native Americans were less varied in design than those of the Pomo.B. Baskets produced by Pomo weaves were primarily for ceremonial and religious purposes.C. There were a very limited number of basket-making materials available to the Pomo people.D. The basket-making production of the Pomo people has been increasing over the years.DAs we all know, there are plenty of different parks to visit in theUK. All theme parks inBritainhave cafes, restaurants, picnic areas and gift shops, so you'll still have plenty to see and do when you and the kids have been on enough rides. There are usually smaller “funfair“ rides and games as well, so younger children won't get bored. Several theme parks also have other attractions next to them, e. g. water parks often open all year round, unlike the theme parks.Whenever you are inBritain, there's likely to be a theme park within one or two hours,drive, bus ride ortrain journey. Several theme parks even have accommodation(膳宿)so you can stay for a day or two if you want to make a trip into a short holiday.Prices forUKtheme parks vary considerably; some have an entrance price which allows you to go on all the rides, while in others you have to pay for every ride individually. It can also make a difference whether you go during peak time or not. For example, tickets always cost more during school holidays and weekends than they do during the weekdays.Theme parks always get very busy during the summer months, so if you don't like crowds ifs usually a good idea to go earlier or later in the year!If you're thinking of visiting aUKtheme park, it's worth having a look for special offers on tickets. Products such as chocolate bars and cereals sometimes have " buy one get one free" offers on theme park tickets, so keep a look out in shops and supermarkets.12. This passage mainly talks about all the following EXCEPT.______.A. things to doB. prices for theme parksC. rules to obeyD. special offers13. If you go to the theme park during the weekdays, you'll probably.______.A. have to spend moreB. save some moneyC. win a big prizeD. get something free14. According to the passage, what should you do if you are tired of crowds in the theme park?A. Avoid the busiest months.B. Go earlier or later in the daytime.C. Choose one with few visitors.D. Go there when no one is in it.15. The best title for the text would be ______.A. What to Do in the Theme ParkB. Theme Parks in theUKC. Visiting the Theme ParkD. Introduction to Famous Theme Parks第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
2021届宜昌市夷陵中学高三语文上学期期中试卷及参考答案
2021届宜昌市夷陵中学高三语文上学期期中试卷及参考答案一、现代文阅读（36分）(一)现代文阅读I(9分)阅读下面的文字，完成下面小题。
娱乐价值与社会价值是词之为体的双重价值“诗言志”是我国诗歌发展史中的一个传统的观念。
就《诗经》而言，它在古代诸侯外交中是被用来“言志”的，这就赋予了《诗经》以神圣的光环。
“言志”的功能与古代诗人自觉的社会责任感、道德感共同铸就了诗歌的严肃面目。
综观历代诗歌，伟大的诗人无不以胸襟和抱负而彪炳千秋，如屈原用诗歌歌唱对国家的热爱和对君主的忠贞，杜甫用诗歌抒发兼济天下的博爱精神，李白用诗歌表达对仕途的渴望与无奈。
与诗相比，词就轻松得多。
它是用来助兴的，这就决定了它必须是轻松、愉快、悦人的。
再加上歌妓的女性身份，更使词有一种软媚的情态。
如身为一朝重臣的欧阳修，在作诗与作词时的两种截然不同的面目，不单单是其人格分裂使然，也是诗与词两诗体的不同体性造成的。
词的这种娱乐性的天职，使得欧阳修等词人在作词时可以抛却思想内容方面的顾忌，而专注于艺术技巧的锤炼。
从史的角度来看，词的思想性的发展是比不上其艺术性的发展之快、之深、之精的。
宋代的词选大多数反映了这种倾向，这是由其娱乐价值所决定的。
然而，任何文学样式都是对社会现实的一种反映形式，词亦然。
南宋时期，民族矛盾渐趋激化，一些有识之士开始用词来抒写这种民族激情，爱国词人大量涌现。
爱国词篇放射出耀眼光芒，使得选词者不得不睁眼面对现实，于是《花庵词选》以比较新的面目在词坛上出现了。
在《花庵词选》中，辛弃疾一人就入选四十三首。
他那些被张炎等认为“非雅词也，于文章余暇，戏弄笔墨，为长短句之诗耳”的慷慨激昂的爱国词作，纷纷入选。
这表明了爱国词的地位得到了词坛的承认，一部分词论家已经开始重视词的思想性了。
黄昇介绍张元幹时，特别指出：“绍兴戊午之和，胡澹庵上书乞斩时相，坐谪新州。
仲宗以词送行，后并得罪。
”并全录了他送胡铨、寄李纲的两首《贺新郎》。
黄昇评论吴激在金国所作的两首感怀故国的词《春从天上来》《青衫湿》时说：“右二曲皆精妙凄惋，惜无人拈出，今录入选，必有能知其味者。
2021届上海宜川中学高三英语上学期期中试卷及答案解析
2021届上海宜川中学高三英语上学期期中试卷及答案解析第一部分阅读（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A、B、C、D四个选项中选出最佳选项ACharlie Thorne and the Last Equationby Stuart GibbsThe CIA is on a task to find an equation (方程式) called Pandora, which could destroy the world if the wrong people get it. For help, they turn to Charlie, a 12-year-old girl who's as smart as Albert Einstein. People who like action-packed mysteries will enjoy reading this exciting book.AstroNutsby Jon Scieszka and Steven WeinbergIn AstroNuts, the Earth has been destroyed by humans for thousands of years. Four animals set out from Mount Rushmore, the headquarters (总部) of NNASA. Their task is to find a new planet fit for human life. Finally, they discover one: Plant Planet. The story's theme (主题) is simple: Don't harm the planet. Readers who love fantasy will enjoy AstroNuts.Stargazingby Jen WangChristine hears that Moon, who's new in town, is the kind of kid who beats people up for fun. But Moon and her mum come to live with Christine's family, and the two kids become best friends. Moon even shares a big secret with Christine. Stargazing is based on author Jen Wang's experiences as a child. The story is about the power of friendship and how people are able to change.Roll with Itby Jamie SumnerRoll with It is a story about a 12-year-old girl named Ellie. She has difficulty walking on her own and uses a wheelchair. When Ellie and her mum move to another state to take care of Ellie's grandpa, she must learn to navigate (处理) a new school and new friendships. This page-turner is a must-read for everyone. It's a heartwarming story that really shows the value of familyand how being different is special.1. Which book tells readers to protect the place we live in?A.AstroNutsB.Stargazing.C.Roll with It.D.Charlie Thorne and the Last Equation.2. What makes Stargazing different from the other three books?A. It talks about friendship.B. It tells stories about animals.C. It contains lots of scientific knowledge.D. It was written according to the author's experiences.3. What happened to Ellie?A. She had difficulty in making friends.B. She had an accident which left her in a wheelchair.C. She went to a new school and had to start all over again.D. She lost her mum and was taken care of by her grandpa.BNASA has flown an aircraft on another planet for the first time. On Monday, April 19, Ingenuity, a 1. 8-kilogram drone (无人机) helicopter, took off from the surface of Mars, flew up about threemeters, then hovered for 40 seconds and took photos above with one of its two cameras. The historic moment was livestreamed on YouTube. “We can now say that human beings have flown a drone helicopter on another planet,” said MiMi Aung, the Ingenuity Mars Helicopter project manager at NASA's Jet Propulsion Laboratory, at a press conference. “We, together, flew at Mars, and we, together, now have our Wright brothers moment,” she added, referring to the first powered airplane flight on Earth in 1903.The details: The flight was a significant technical challenge, thanks to Mars' s bone-chilling temperatures (nights can drop down to -130 °F / -90 °C) and its incredibly thin atmosphere — just 1% the density of Earth' s. That meant Ingenuity had to be light, with rotor blades (螺旋桨) that were bigger and faster than would be needed to achieve liftoff on Earth (although the gravity on Mars, which is only about one-third of Earth's, worked in its favor). The flight had originally been scheduled to take place on April 11 but was delayed by software issues.Why it's significant: Beyond being a significantmilestonefor Mars exploration, the flight will also pave the way for engineers to think about new ways to explore other planets. Future drone helicopters could help astronauts by scoping out locations, exploring inaccessible areas, and capturing images. Ingenuity will also help inform the design of Dragonfly, a car-size drone that NASA is planning to send to Saturn's moon Titan in 2027.What's next: In the next few weeks, Ingenuity will conduct four more flights, each lasting up to 90 seconds. Each one is designed to further push the limits of Ingenuity's capabilities. Ingenuity is only designed to last for 30 Martian days, and is expected to stop functioning aroundMay 4. Its final resting place will be in the Jezero Crater as NASA moves on to the main focus of its mission: Getting the Perseverance rover (毅力号火星探测器) to study Mars for evidence of life.4. Why does MiMi Aung mention Wright brothers?A. To honor the historic contribution of Wright brothers.B. To make a comparison between Wright brothers and herself.C. To highlight the importance of Ingenuity's Martian exploration.D. To remind the public of the first powered airplane flight on Earth.5. What caused challenges for Ingenuity designers?A. The delay of software tests.B. The force of gravity on Mars.C. Mars's temperatures and atmosphere.D. Technical difficulty in liftoff on Earth.6. What does the underlined word “milestone” in Paragraph 3 probably mean?A. A great victory.B. A difficult task.C. A roadside stone.D. A historic event.7. Which of the following can be a suitable title for the text?A. Mars Drone Helicopter Makes Historic First FlightB. Modern Wright Brothers Moment: Mars ExplorationC. First Drone Helicopter Spots Evidence of Life on MarsD. Ingenuity: Mars Drone Flies to Take Incredible PicturesCYou run into the grocery store to quickly pick up your item. You grab what you need and head to the front of the store. After quickly sizing up the check-out lines, you choose the one that looks fastest. You chose wrong. People getting in other lines long after you have already checked out and headed to the parking lot. Why does this seem to always happen to you?Well, as it turns out, it's just math that is working against you. A grocery store tries to have enough employees at the checkout lines to get all their customers through with minimum delay. But sometimes, like on aSunday afternoon, they get super busy. Because most grocery stores don't have the physical space to add more checkout lines, their system becomes overburdened. Some small interruption — a price check, a particularly talkative customer — will have downstream effects, holding up the entire line behind them.If there are three lines at the store, these delays will happen randomly at different registers (收银台). Think about the probability. The chances of your line being that fastest one are only one in three, which means you have a two-thirds chance of not being in the fastest line. So it's not just in your mind: Another line is probably moving faster thanyours.Now, mathematicians have come up with a good solution, which they call queuing theory, to this problem: Just make all customers stand in one long snaking line, called a serpentine line, and serve each person at the front with the next available register. With three registers, this method is about three times faster on average than the more traditional approach. This is what they do at most banks, Trader Joe's, and some fast-food places. With a serpentine line, a long delay at one register won't unfairly punish the people who lined up behind it. Instead, it will slow everyone down a little bit.8. What phenomenon is described in the first paragraph?A. Queuing in a line.B. A shopping experience.C. A rush in the morning.D. Cutting in a line.9. According to the article, what may cause delays in checking out?A. The lack of employees in the grocery store.B. Some unexpected delays of certain customers.C. The increasing items bought by customers.D. A worsening shopping system of the store.10. What is the solution given by mathematicians?A. Employing more workers for checking out.B. Limiting the number of queuing people.C. Making only one line available.D. Always standing in the same line.11. What's the principle behind the queuing theory?A. To pursue the maximum benefit.B. To leave success or failure to luck.C. To avoid the minimum loss.D. To spread the risk equally among everyone.DA maverick describes a person who thinks independently. A maverick refuses to follow the customs or rules of a group to which he or she belongs. In the US, a maverick is often admired for his or her free spirit, although others who belong to the maverick’s group may not like the maverick’s independent ways.But where did the word “maverick” come from?Early in the 1800s, a man named Samuel Augustus Maverick settled down in Texas, which was a place of wide-open land, rich soil, cattle ranches(牛场) and cowboys. As the years passed, Mr. Maverick increased his property(财产) in Texas. Before long, he owned huge pieces of land that were good for raising cattle. But he had no cattle. He wasn’t a rancher.One day, a man came to Samuel Maverick to pay him an old debt. But the man didn’t have enough money. So he offered Mr. Maverick 400 head of cattle. Mr. Maverick accepted them, but he didn’t really want them. He simply put the cattle on his land to eat and care for themselves.It was not long before the cows reproduced(繁殖). The calves grew and had more calves. Soon, hundreds of cows and calves moved freely across Samuel Maverick’s land. They also moved across the land of nearby ranch owners.It was a tradition among ranchers in the West to put a mark of ownership on newborn calves. They burned the name of their ranch into the animal’s skin with a hot iron. The iron made a clear mark called a “brand”. Brands allowed ranchers to easily see who owned which cattle.Samuel Maverick refused to brand his calves. “Why should I?” he asked. If all the other cattle owners branded theirs, then those without a brand belonged to him.And this is how the word “maverick” entered the American language. It meant a calf without a brand. As time passed, the word “maverick” took on a wider meaning. It came to mean a person who was too independent to follow even his or her own group.12. Why did the man give Samuel Maverick 400 head of cattle?A. To get some money.B. To return what he owed him.C. To buy some of his land.D. To ask him to raise them.13. How could the ranchers easily know who the cattle belonged to?A. Through the brand on the cattle.B. Through the name of the cattle.C. Through the appearance of the cattle.D. Through the land on which the cattle stayed.14. What can we learn about Samuel Augustus Maverick from the text?A. He was born in Texas.B. He took good care of all his cattle.C. He didn’t really want to accept the cattle.D. He followed the tradition of ranchers in the West.15. What is the text mainly about?A. How to become an independent thinker.B. “Maverick” means a calf without a brand.C. The life story of Samuel Augustus Maverick.D. How the word “maverick” got into American English.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
2021届上海华东师大二附中高三英语上学期期中试卷及答案
2021届上海华东师大二附中高三英语上学期期中试卷及答案第一部分阅读（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A、B、C、D四个选项中选出最佳选项ASan Francisco Bay Area is a great place if you're a sports fan as you'll find several events all year round and plenty of team pride. If you are anywhere close to the area during a game,these fantastic sports events are here for you.San Francisco Giants BaseballThe San Francisco Giants baseball team plays in SF at Oracle Park. This is a fun ballpark because it's always packed with great energy and offers views of the bay. It's one of the most popular San Francisco sports events. The Giants are part of the National League West Division. Since their arrival here in 1958,they have been World Series Champions three times.Golden State Warriors BasketballThe fan base of the Golden State Warriors distributes the whole San Francisco Bay Area as this region's only NBA team.Their regular season runs from late October through mid-April, and all home games are played at the Chase Center in San Francisco.In total, the Warriors has won six NBA championships.San Francisco 49ers FootballThe 49ers are San Francisco's NFL team, though they have recently moved to Levi's Stadium in Santa Clara, about an hour south of SF. The football team was named for the prospectors (探矿者) who arrived in the area in 1849 for the Gold Rush. They’ve won 5 Super Bowl championships, all between 1981 and 1994.San Jose Sharks HockeyThe San Jose Sharks represent the Bay Area in hockey (冰球).They were founded in 1991 as the only Bay Area team to compete in the NHL. Sharks fans love going to these San Francisco sports events at the SAP Center,which they call the Shark Tank,located about an hour southeast of SF.1.Where can a sports fan have a good view of the area?A.The Oracle Park.B.The Chase Center.C.Levi's Stadium.D.The SAP Center2.Which team has claimed the most titles according to the text?A.The Giants.B.The Golden State Warriors.C.The 49ers.D.The San Jose Sharks.3.Where is the passage probably taken from?A.A book review.B.A news report.C.A science fiction.D.A tourist magazine.BFew people are aware that Waterloo Bridge, crossed by thousands of daily commuters (每天长途上下班的人) and celebrated as a London landmark, was mainly built by a female workforce.There are no written or photographic records ofthe women who built the bridge since the construction company that built it closed in the 1980s and with it went all the records. What’s left is anecdotal evidence, kept alive by the tourist boat skippers who have called it theLadiesBridge.During the Second World War, with the majority of the active male labor-force away at the front, women increasingly took on traditionally male roles. By 1944, 25,000 women were working in the construction industry, carrying out back-breaking tasks. These women were permitted to carry out this exhausting and dangerous work on the basis that it would only be temporary and that their pay would be lower than that of men. So the surge (激增) in women working in construction and engineering did not continue after the war.September sees a series of events celebrating the unknown work of the large female workforce. Musician Claudia Molitor has created a 45-minute musical entitled “theSingingBridge”, which runs at Somerset House from Sept9th. to Sept25th. In late September, the “Light up the Ladies Bridge” event saw the National Theatre’s fly tower lit up by large scale photographic projections (投影片) of female construction workers working in construction during World War II.Let’s all celebrate the women who have helped to build the cities in which we live.4. Why do few people know about the builders ofWaterlooBridge?A. The records of their work have been lost.B. Female workers received little attention.C. The construction took place long ago.D. Few people know about the bridge.5. What can we learn from the fourth paragraph?A. There are less female workers inLondon.B.WaterlooBridgeis popular among artists.C. The story ofWaterlooBridgeis better known.D. Many works of art were created during World War II.6. Why didLondonwomen do the construction work?A. It was a tradition inLondon.B. They wanted to get a steady job.C. Mostmen had gone to the battlefield.D. The company offered them better pay.7. Which can be the best title for the text?A. Female Workers on the RiseB. A Hidden Treasure inLondonC. ANew LondonLandmarkD. The Story behind theLadiesBridgeCDid you know that the average child has heard the word "no" over 20,000 times before they turn the age of three? Ironically, it is also around this time that children begin to develop enough personal character to refuse to obey. The "terrible twos" are categorized by a lack of understanding. Somewhere between three and four, children begin to acquire the skills to reason. It is during this time they watch how other children and adults reason. If we're not careful, the children will watch us model a world of "NOs".By the time a person turns eighteen, how many times have they been told no? I haven't found any studies that even attempted to track this statistic, but I'm sure if the number is 20,000 by three, then at eighteen that number has multiplied. You can do the math.Anyway, I think I know why we say no. We say no to protect. We say no to direct. We say no to stop potential confusion. However, do we sometimes say no just for the sake of saying no? Do we say no because we have internalized(内在化)all of the "NOs" we've heard over the years and we feel it is finally our time to say no to someone else?The internalized no can damage the growth process of dreams in infancy as quickly as it can weaken a three-year-old. And we wonder why we run intopeople with big, un-accomplished dreams who have a bit of a chip on their shoulder. They have to take on the 20,000 NOs. However, the thing that keeps them going is the possibility of the power of ONE YES! Just as it only takes one book to make a writer a Pulitzer Prize Winner, it only takes one word to change the course of your day. That word is YES!8. What do we know about two- year- olds?A. They understand well.B. They often say no to others.C. They think logically.D. They don't do all they're told.9. What effect does saying no have on children?A. They lose all their dreams.B. They aren't easy to succeed.C. They don't make mistakes.D. They never say yes to others.10. Which word best describes the author's attitude to a world of "NOs"?A. Tolerant.B. Disapproving.C. Favorable.D. Carefree.11. What is the text?A. A how-to guide.B. A survey report.C. An opinion essay.D. A short story.DAt the World Economic Forum last month, President Trump drew claps when he announced the United States would respond to the forum's proposal to plant one trillion(万亿) trees to fight climate change. The trillion-tree idea won wide attention last summer after a study published in the journal Science concluded thatplanting so many trees was “the most effective climate change solution to date”.If only it were true. But it isn't. Planting trees would slow down the planet's warming, but the only thing that will save us and future generations from paying a huge price in dollars, lives and damage to nature is rapid and considerable reductions in carbon release from fossil fuels, to net zero by 2050.Focusing on trees as the big solution to climate change is a dangerous diversion(偏离). Worse still, it takes attention away from those responsible for the carbon release that are pushing us toward disaster. For example, in the Netherlands, you can pay Shell an additional 1 euro cent for each liter of regular gasoline you put in your tank, to plant trees to balance the carbon release from your driving. That's clearly no more than disaster slightly delayed. The only way to stop this planet from overheating is through political, economic, technological and social solutions that end the use of fossil fuels.There is no way that planting trees, even across a global area the size of theUnited States, can absorb the huge amounts of fossil carbon released from industrial societies. Trees do take up carbon from the atmosphere as they grow. But this uptake merely replaces carbon lost when forests were cleared in the first place, usually long ago. Regrowing forests where they once grew can undo some damage done in the past, but even a trillion trees can't store enough carbon to head off dramatic climate changes this century.In a sharp counter argument to last summer's Paper in Science, Justin Gillis wrote in the same journal inOctober that the study's findings were inconsistent with the dynamics of the global carbon cycle. He warned that “the claimthat global tree restoration(复原) is our most effective climate solution is simply scientifically incorrect and dangerously misleading”.12. What do we know about the trillion-tree idea?A. It was published in a journal.B. It was proposed last summer.C. It was put forward by Trump.D. It drew lots of public attention.13. What is paragraph 3 mainly about?A. A drawback of the tree planting strategy.B. An example of balancing carbon release.C. An anecdote of making a purchase at Shell.D. A responsibility for politicians and economists.14. What was Justin Gillis's attitude towards global tree restoration?A. Indifferent.B. Opposed.C. Hesitant.D. Supportive.15. What is the best title for the text?A. Contradictory Ideas on Tree Planting.B. A Trillion Trees Come to the Rescue.C. Planting Trees Won't Save the World.D. The Best Solution to Climate Change.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。
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考场号
【原创】2021 届高三上学期期中试卷 数学试卷
注意事项： 1． 答 题 前 ， 先 将 自 己 的 姓 名 、 准 考 证 号 填 写 在 试 题 卷 和 答 题 卡 上 ， 并 将 准 考 证 号 条 形
码粘贴在答题卡上的指定位置。 2． 选 择 题 的 作 答 ： 每 小 题 选 出 答 案 后 ， 用 2B 铅 笔 把 答 题 卡 上 对 应 题 目 的 答 案 标 号 涂
9．已知
的展开式中各项系数之和为 ，第二项的二项式系数为 ，则（ ）
A．
B．
C．展开式中存在常数项 D．展开式中含 项的系数为 【答案】ABD
【解析】令 ，得
的展开式中各项系数之和为
，所以
，
选项 A 正确；
的展开式中第二项的二项式系数为
，所以
，
的展开式的通项公式为
，
，选项 B 正确；
令
，则
，所以展开式中不存在常数项，选项 C 错误；
的外接球表面积为 （即 ）与 是相交直线，故该说法错误；
B 项，由已知可得
，
又平面
平面
，所以
平面
，
在矩形
中，
的面积
，
又
，所以三棱锥
所以该说法正确；
的体积
C 项，由
平面
，得
，
又
，所以
平面
，所以
D 项，由题意可得四边形
为矩形，连接 ，
， ，所以该说法正确；
则矩形
外接圆的圆心为 的中点 ，且
，
过作
与点 ，连接 ， ，
A． B． C．
D．
【答案】A
【解析】
，
，
因为
，所以
．
3．为了评估某家快递公司的服务质量，某评估小组进行了客户满意度调查，从该公司参与调查的客
户中随机抽取 名客户的评分，评分均在区间
上，分组为
，
，
，
，
，其频率分布直方图如图所示．规定评分在 分以下表示对该公司的服务质量
不满意，则这 名客户中对该公司的服务质量不满意的客户的人数为（ ）
令
，则
，所以展开式中含 项的系数为
，选项 D 正确．
10．已知函数 导函数，函数
的图象的一条对称轴为直线 ，则下列说法正确的是（ ）
， 为函数
的
A．直线
是
图象的一条对称轴 B．
的最小正周期为
C．
是
【答案】BD
图象的一个对称中心 D．
的最大值为
【解析】因为
的图象的一条对称轴为直线
，
所以
，
，所以
，
，
又
，所以
，
由于
，所以
恒成立，故 A 不具有性质 ；
对于 B，函数
的定义域为 ，取
，
，则
，
所以
，所以
成立，故 B 具有性质 ；
对于 C，函数
的定义域为
，当
，
时，
，
由于
，所以
，易知
在
上单调递增，
所以 对于 D，函数
恒成立，故 C 不具有性质 ；
的定义域为 ，易知
为奇函数，
取
，则
，所以
，
，
所以
成立，故 D 具有性质 ．
，所以
，所以
，
所以 ，
，且 B、D 正确．
，所以
的最大值为 ，最小正周期为 ，故 A、C 错误，
11．如图，直接三棱柱
，
为等腰直角三角形，
，且
，
， 分别是 ， 的中点， ， 分别是 ， 上的两个动点，则（ ）
A． 与 一定是异面直线
B．三棱锥
的体积为定值
C．直线 与 所成角为 D．若 为 的中点，则四棱锥 【答案】BCD 【解析】A 项，当 ， 重合时，
第Ⅰ卷
一 、单项 选择 题：本题 共 8 小题 ，每小 题 5 分 ，共 40 分．在每 小题 给出 的四 个选 项中 ，只 有一项是符合题目要求的．
1．若复数
为纯虚数，则实数 的值为（ ）
A． B． C． D． 【答案】D
【解析】由
为纯虚数，
可得
，解得
．
2．已知集合
，
取值组成的集合为（ ）
，若
，则 的可能
A． B．
C．
D．
【答案】C
【解析】由题意知，
，
当
时，
，函数
在 上单调递增，没有两个不同的零点；
当
时，
，得
，
，
，函数
在
上单调递增；
，
，函数
在
上单调递减，
故
在
处取得最小值，
所以
，得
，
所以 的取值范围为
．
二 、多项 选择 题：本题 共 4 小题 ，每小 题 5 分 ，共 20 分．在每 小题 给出 的选 项中 ，有多 项 符合题目要求．全部选对的得 5 分，部分选对的得 3 分，有选错的得 0 分．
则
，
，
，故
，
所以 就是四棱锥 故外接球的表面积
的外接球的球心，所以外接球半径
，
，故该说法正确．
12．若 存在两 个不相 等的实 数 ， ，使 ， ，
均在函数
的定义域内，且满足
，则称函数 具有性质 ，下列函数具有性质 的是（ ）
A．
B．
C．
D．
【答案】BD
【解析】对于 A，因为函数
的定义域为 ，
，
所以
第Ⅱ卷
三、填空题：本大题共 4 小题，每小题 5 分．
A． B． C． D． 【答案】A
【解析】由频率分布直方图可知，评分在区间
上的频率为
，
所以评分在区间
上的客户有
（人），
即对该公司的服务质量不满意的客户有 人．
4．已知定义在 上的奇函数
在
上单调递减，且
，
，则 ， ， 的大小关系是（ ）
A．
B．
C．
【答案】A
【解析】因为定义在单调递减且
所以
，
又
，所以
，
而 5．已知四边形
，所以
，所以
．
中， ， 分别为 ， 的中点，
，则
（）
，若 ，
，
， ，若
A． B． C． D． 【答案】A
【解析】依题意，可知四边形
为直角梯形，
，
，
且
，
，
所以
．
6．已知在正方体
中， ， 分别为 ， 上的点，且满足
，
，则异面直线 与 所成角的余弦值为（ ）
A．
B． C． D．
黑，写在试题卷、草稿纸和答题卡上的非答题区域均无效。 3． 非 选 择 题 的 作 答 ： 用 签 字 笔 直 接 答 在 答 题 卡 上 对 应 的 答 题 区 域 内 。 写 在 试 题 卷 、 草
稿纸和答题卡上的非答题区域均无效。 4． 考 试 结 束 后 ， 请 将 本 试 题 卷 和 答 题 卡 一 并 上 交 。
【答案】A
【解析】取线段 上一点 ，使
，连接 ， ，如图所示，
因为 所以 又
，
，所以
，
，
，
，所以易知
为异面直线 与 所成的角．
设该正方体的棱长为 ，则
所以在
中，
，
，
，
所以
．
7．已知双曲线
的渐近线分别为 ， ，点 是 轴上与坐标原点 不重合
的一点，以 为直径的圆交直线 于点 ， ，交直线 于点 ， ，若 该双曲线的离心率是（ ）
，则
A． 或 【答案】C
B． C．
或
D．
【解析】由题意，不妨设
，
，
设
，则
，
设
，由
，得
，
由对称性知，
，且线段 被 平分．
如图，设 与 交于点 ，则
由于 为直径，所以
，
则
，
，连接 ， ，
由 因为
，得
，
，
，所以
或
，即
或
．
又
，所以
或
．
当
时，
，则
当
时，
，则
，离心率
；
，离心率
．
8．若函数
恰有两个不同的零点，则实数 的取值范围为（ ）