黑龙江省实验中学2021届高三上学期10月月考英语试卷及答案
数学丨黑龙江省龙东联盟2025届高三10月月考数学试卷及答案
试题参考答案一．单选题1.【解析】选B.{|2}{|12}U A x x A B x x ==< ≤，≤ð，故选B.2.【解析】选C.0a <且0b <⇒0a b +<且0ab >，反之也成立，故选C.3.【解析】选C.12(43i)(i)=(4-3)+(4+3)i z z a a a ⋅=++为实数，所以430a +=所以43a =-，故选C.4.【解析】选D.因为|||2|-=+ab a b 平方得，21||2⋅=-a b b ，a 在b 方向上的投影向量为1||||2⋅⋅=-a b b b b b ，故选D.5.【解析】选A.53357S a a =⇒=，453623a a a a +=+=，所以616a =，所以63363a a d -==-，故选A. 6.【解析】选A.由102sin cos 2αα+=两边平方得2254sin 4sin cos cos 2αααα++=，所以4sin cos αα233cos 2α-=-所以2332sin 2(2cos 1)222ααα=-=所以3tan 24α=.故选A.7.【解析】选D.因为ln()ln ln ln ln 3333xy x y x y +==⋅故选D.8.【解析】选A.设零点为(01]t ∈，，则ln 0at b t ++=，()a b ，在直线ln 0xt y t ++=上，22a b +的几何意义为点()a b ，到原点距离的平方，其最小值为原点到直线ln 0xt y t ++=的距离d 的平方，222ln 1t d t =+，设22ln ()1t g t t =+，22222ln (12ln )()0(1)t t t t g t t t +-'=<+所以()g t 在(01]，单调递减，所以min ()(1)0g t g ==.故选A.二．多选题9.【解析】选AD.|||2i ||2|z z y y -==知A 对C 错，222222i z x xy y x y =+-≠+，故B 错，||||||z x y =+成立，故选AD.10.【解析】选ABD.由21((0)22n d d S n a n d =+-≠及二次函数的性质知A B ，为真，对D 知100a d <<，从而{}n S 是递减数列，对C ：1258--- ，，，，满足{}n S 是递减数列，但0n S <不恒成立，故选ABD .11.【解析】选BC.对A ：(0)1()1(0)2f f f π===，A 错，对B ，令sin x t =，21()sin sin 1f x x x =-++，210t t -++=则1sin [02]2t x x π==∈，，，有两个实根.B 对.对C ：232()sin cos f x x x =+，22()2sin cos 3cos sin f x x x x x '=-，令2()0f x '=即2cos sin 203x x ==，，2cos 3x =的两个根为123(0(2)22x x πππ∈∈，，，，sin 20x =的根为30222ππππ，，，，，所以2()f x 的极小值点为12x x π，，，C 对.对D ：22(2)()f x f x π+=，所以2()f x 为周期函数，但232()sin cos f x x x =+，232()sin cos f x x x π+=-，22()()f x f x π≠+，D 错.三.填空题12.【解析】0.()()f x f x -=特值()()f a f a -=即cos cos |2|a a a =-所以0a =.13.【解析】π.21cos 2cos 2x x +=与cos(2)4x π-的最小正周期相同，14.【解析】4+解1：设|+a b |x =，||-a b y θ=<，，a b >=，254cos [13]x x θ=+∈，，，254cos [13]y y θ=-∈，，且2210x y +=，设x y ϕϕ==，，其中sinϕ，则)4x y πϕ+=+，当4πϕ=，x y ==x y +取得最大值cos sinϕϕ==即3x =，1y =时x y +取得最小值4,所以最大值与最小值之和为4+解2：换元后，利用平行直线系和圆弧的位置关系四.解答题15.解：（1）由223n S n n =+得当1n =时，115a S ==，………………………………1分当2n ≥时，22123[2(1)3(1)]41n n n a S S n n n n n -=-=+--+-=+……3分所以41n a n =+…………………………………………………………………4分由34log 141n n a b n =+=+，所以3n n b =………………………………6分（2）由（1）知(41)3n n n a b n =+…………………………………………………7分125393(41)3n n T n =⨯+⨯+++ ①23135393(43)3(41)3n n n T n n +=⨯+⨯++-++ ②……………9分①-②得212154343(41)3n n n T n +-=+⨯++⨯-+⨯ ……………………10分119(132154(41)313n n n T n -+--=+-+⨯-），所以131(2322n n T n +=--⨯.…………………………………………13分16.解：（1）因为222sin C sin sin sin A B A B =+-222a b c ⇒+-=，…2分由余弦定理得2222cos 22a b c C ab +-==，(0)C π∈，，所以4C π=，…4分因为6sin 2B C =所以3sin 2B =，………………………………………6分因为(0)2B π∈，，所以3B π=…………………………………………………7分（2）512A B C ππ=--=……………………………………………………………8分sin sin()A B C =+624+=…………………………………………………10分sin sin sin a b c A B C ==得312a ==，62b c =………12分由213(31)sin 128ABC S ab C +===△，…………………………14分得2833c =.……………………………………………………………………15分（17）解：（1）因为()ln f x x x =-，所以()()ln a a g x f x x x x x =-=--，0x >，2221()1a x x a g x x x x-++'=-+=，………………………………………………………2分令2211()()24m x x x a x a =-++=--++①当14a -≤时，()0g x '≤恒成立，此时()g x 在(0)+∞，上单调递减；②当104a -<<时，()0m x >可得11411422x -+<<所以()g x 在1(0)2-，上单调递减，在11()22+，上单调递增，在114()2++∞，上单调递减；③当0a >时，()0m x >，可得114114022x -+<<<所以()g x 在114(0)2，上单调递增，在114()2+∞，上单调递减；……5分综上所述：当14a -≤时，()g x 的单调递减区间为(0)+∞，，无单调递增区间；当104a -<<时，()g x 的单调递减区间为114(0)2，和114()2+∞，单调递增区间为11()22+，；当0a >时，()g x 的单调递增区间为1(0)2，，单调递减区间为114()2++∞，；……………………………………………………………………7分（2）由()ln f x x x =-，1()x f x x-'=，由()0f x '>得01x <<，()0f x '<得1x >所以()f x 在(01)，上单调递增，在(1)+∞，上单调递减，所以max ()(1)1f x f ==-，所以min |()|1f x =，………………………………………10分设ln 1()2x g x x =+，则21ln ()x g x x -'=由()0g x '>得0e x <<，由()0g x '<得e x >，所以()g x 在(0e)，上单调递增，在(e )+∞，上单调递减，所以max ()g x =(e)g 111e 2=+<所以max min ()|()|g x f x <，…………………………………………………………………14分所以ln 1|()|2x f x x >+对任意的(0)+∞，恒成立.……………………………………15分18.解：（1）(0)1()e (0)1x g g x g ''==-=-，所以()g x 在(0(0))g ，处的切线方程为：(11y x =-+………………………………………………………………2分(1)1h b c =+-，2()1(1)1b h x h b x''=-=-，，所以()h x 在(1(1))h ，处切线方程为：(1)2y b x b c =-+-所以21b c -=，11b -=-6分即1(1)c a =≥所以c 的最小值为1.…………………………………………7分（2）()e x g x =-，则()e x g x '=-当ln (0)2a x ∈，时ln ()0()2a g x x '<∈+∞，，时()0g x '>所以()g x 在ln (0)2a ，上单调递减，在ln ()2a +∞，上单调递增，故min ln ln ()())22a a g x g ==-………………………………………………………9分()b h x x cx =+-，则()h x 在(0上单调递减，在)+∞上单调递增令()0h x =，即20x cx b -+=，24c b∆=-1.0∆>即c >(0+∞，）上()h x 的两个零点为12x x ，，同时它们恰好为()g x 的零点.12()0()0ln 102g x g x a ⎧⎪=⎪∴=⎨⎪⎪-<⎩即12122e e e x x a ⎧=⎪⎪=⎨⎪>⎪⎩又1212x x c x x b +==，，则2e 1e c ab a ⎧=>⎪⎨>⎪⎩，此时…11分1ln ln e e e a a a b a a a b a-++--=>，令1ln y a a a =-+，则21110y a a'=--<，y ∴递减且a →+∞时y →-∞，则2212e e e (0e )y -+∈，，故2212e e e e a b a -+->.…………………………………14分2.0∆≤即0c <≤时，在(0)+∞，上()0h x ≥，此时只需min ()0g x ≥即21e a ≤≤即可.此时，e e eb a b a aa -⋅=，令()e a a k a =，则10e a a k -'=≤，即k 在2[1e ]，递减，22e 1[e ]e k -∴∈，而e 1b >，故22e e e a b a -->.……………………………………………………………………16分综上所述，e a b a -的取值范围为22e (e )-+∞，………………………………………………17分（19）解：（1）设{}n a 的公差为d ，32318S a ==所以26a =，323a a d -==，3n a n =；……………………………2分由214b b q ==，313(1)141b q T q-==-，所以22520q q -+=，2q =或12q =（舍）所以2n n b =.……………………………………………………………………4分1132a b ==，所以1223c c ==，；2264a b ==，所以3446c c ==，3398a b ==，所以5689c c ==，；441216a b ==，所以7812c c =，16=.3574812c c c +=+==，所以1k =.………………………………………5分（2）221233(363)(222)222n n nn n n n M S T n ++=+=+++++++=+- …7分7231n n M b =-，即2133223212n n n n +++-=⋅-所以233222n n n +=⋅+，当1n =时符合，…………………………………………………8分令233222n n r n n =+-⋅-1234081826r r r r ====，，，，524r =，64r =-16622nn n r r n +-=+-⋅当4n ≥，10n n r r +-<所以123456r r r r r r <<<>>>所以有且只有1n =符合.…………………………………………………………11分（3）由2122122(36)(1)n n n n n n n n a b d c c c c -+++=-得1(96)2(1)(3)2(33)2n nn n n n d n n ++=-+111(1)()32(33)2n n n n n +=-++………………13分22231111()(32(313)2(313)2(323)2n E =-+++⨯⨯+⨯+⨯+）22111()3(2)23(21)2n n n n +-+++ ……………………………………15分21116(63)2n n +=-++16>-.………………………………………………17分。
山东新高考联合质量测评2024届高三10月联考英语试题含答案
山东省滨州市普通高中新高考大联考2023-2024学年高三上学期10月月考英语试题本卷满分150分，考试时间120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、座号和考生号填写在答题卡和试卷规定的位置上。
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第一部分听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B:C三个选项中选出最佳选项。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1.What does the man want to do?A.Dry his hair.B.Wash his hair.C.Have his hair cut.2.When is the documentary on?A.At5:00.B.At8:00.C.At9:00.3.What does the man tell the woman?A.The exams have been canceled.B.The exams will be held at another place.C.The building work will stop due to the exams.4.Where does the conversation most probably take place?A.In the woman's house.B.In an art gallery.C.In a store.5.What are the speakers talking about in general?A.Video games.B.Sales performance.C.Places of entertainment.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。
湖北省沙市中学2024届高三上学期10月月考英语
2023—2024学年度上学期2021级10月月考英语试卷命题人：朱茂林审题人：Nancy第一部分听力(共两节，满分30分)该部分分为第一、第二两节，注意：回答听力部分时，请先将答案标在试卷上。
听力部分结束前，你将有两分钟的时间将你的答案转涂到答题卡上。
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 does the man plan to do for the summer vacation?A．Stay in London. B．Go sightseeing. C．See his parents.2．Why did the woman change the reservation?A．She canceled the party. B．She got the date wrong. C．She put off her plan. 3．What happened to the man?A．He found a car key. B．He lost his car key. C．He got his key ring. 4．What are the speakers talking about?A．The coming tests. B．The stress in life. C．The learning methods. 5．What is the weather like now?A．Rainy. B．Stormy. C．Fine.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；听后，各小题将给出5秒钟的作答时间。
生物丨黑龙江省哈尔滨市第三中学2025届高三10月月考生物试卷及答案
哈三中2024-2025学年度上学期高三学年十月月考生物试题一、单选题（每题只有一个选项符合题意，1-10题每题2分，11-20题每题1分，共30分）1.婴儿的肠道上皮细胞可以吸收母乳中的免疫球蛋白，此过程不涉及（）A.消耗ATPB.载体蛋白协助C.受体蛋白识别D.细胞膜流动性2.关于大肠杆菌和豌豆的共同点，表述正确的是（）A.都是真核生物B.能量代谢都发生在线粒体中C.都能进行光合作用D.都具有核糖体3.秋冬季是呼吸道疾病高发的季节，溶菌酶含片可用于急慢性咽喉炎、口腔黏膜溃疡的治疗。
下列有关叙述，正确的是（）A.溶菌酶是在人体细胞的溶酶体中合成的B.溶菌酶可以对支原体起到杀伤作用C.溶菌酶对有细胞壁的全部微生物起作用D.临床上溶菌酶可与抗生素混合使用，以增强疗效4.下列不属于水在植物生命活动中的作用的是（）A.物质运输的良好介质B.保持植物枝叶挺立C.降低酶促反应活化能D.参与光合作用5.组成细胞的化学元素，常见的有20多种。
下列关于元素与化合物的叙述错误的是（）A.镁是叶绿体中参与光合作用的叶绿素的组成元素B.有氧呼吸时，NADH来自于葡萄糖和水的分解C.植物从土壤溶液中吸收的氮可以用于合成蛋白质、磷脂D.组成细胞的各种元素大多以离子形式存在6.下列有关细胞结构与功能的叙述错误的是（）A.卵细胞体积较大，与周围环境进行物质交换效率低B.染色质处于细丝状，有利于DNA完成复制、转录等生命活动C.在抗体的合成分泌过程中内质网会接收、释放囊泡D.白细胞的主要功能是吞噬病菌等，其寿命短于红细胞7.下列有关人体中化合物及生化反应的描述错误的是（）A.腺苷是人体重要的化合物，组成元素为C、H、O、NB.细胞中由氨基酸合成新的肽链过程伴随着ATP的水解C.固醇的合成与核糖体、内质网和高尔基体有关D.人体中的酶合成过程都需要RNA聚合酶的参与8.下列有关植物细胞能量代谢的描述正确的是（）A.含有两个特殊化学键的ATP是RNA的基本组成单位之一B.加入呼吸抑制剂可使细胞中ADP生成减少，ATP生成增加C.无氧条件下，丙酮酸转变为乳酸的过程中伴随有ATP的合成D.光下植物叶肉细胞中会发生ATP的合成和水解9.下列关于生物学实验的叙述，正确的是（）实验名称实验结论或原理A伞藻的嫁接实验证明了生物体形态结构的建成由细胞核决定B绿叶中色素的提取实验光合色素易溶于层析液C洋葱鳞片叶表皮细胞质壁分离实验原生质层两侧溶液存在浓度差及原生质层比细胞壁的伸缩性大D人鼠细胞融合实验细胞膜上的分子都可以运动10.为探究温度对绿藻光合作用和呼吸作用的影响，将绿藻培养液均分成4份，装入密闭培养瓶中，置于4种不同温度（已知t1＜t2＜t3＜t4)下培养，分别在光照和黑暗条件下测定密闭培养瓶中氧气的含量变化，得到如图所示数据。
江苏省淮阴中学2020┄2021届高三10月月考英语试题 Word版含答案
第一部分听力（共两节，满分20分）做题时，先将答案标在试卷上。
录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到客观题答题卡上。
第一节（共5小题；每小题1分，满分5分）听下面5段对话。
每段对话后有一个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。
听完每段对话后，你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. Where will the woman go first？A. To the beach.B. To the bank.C. To the bathroom.2. What does the woman mean？A. The man forgot to do his hair.B. The man forgot to put on a tie.C. The man didn’t dress properly.3. How does the woman probably feel？A. Annoyed.B. Hungry.C. Excited.4. Why didn’t the man answer the phone？A. He lost it.B. He didn’t hear it.C. His phone ran out of power.5. Who did the woman want to call？A. James.B. Drake.C. Daniel.第二节（共15小题；每小题1分，满分15分）听下面5段对话或独白，每段对话或独白后有几个小题，从题中所给的A、B、C三个选项中选出最佳选项，并标在试卷的相应位置。
听每段对话或独白前，你将有时间阅读各个小题，每小题5秒钟; 听完后，各小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
听第6段材料，回答第6、7题。
6. What does the man order？A. Hot dogs and fries.B. Burgers and fries.C. Sandwiches and sodas.7. How much does the man give the woman as a tip？A. Three dollars.B. Two dollars.C. One dollar.听第7段材料，回答第8、9题。
数学丨河南省2025届高三上学期10月联考(二)数学试卷及答案
2024-2025年度河南省高三年级联考（二）数学注意事项：1.答题前，考生务必将自己的姓名、考生号、考场号、座位号填写在答题卡上.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.4.本试卷主要考试内容：集合与常用逻辑用语，函数与导数，三角函数，平面向量，数列，不等式.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.已知集合{}21A x x =-<，{}3B x a x a =<<+.，若{}15A B x x =<< ，则a =（）A.0B.1C.2D.32.已知符号）（表示不平行，向量(1,2)a =--，(,7)b m m =+ .设命题:(0,)p m ∀∈+∞，a ）（b ，则（）A.:(0,)p m ⌝∃∈+∞，//a b，且p ⌝为真命题B.:(0,)p m ⌝∀∈+∞，//a b，且p ⌝为真命题C.:(0,)p m ⌝∃∈+∞，//a b，且p ⌝为假命题D.:(0,)p m ⌝∀∈+∞，//a b，且p ⌝为假命题3.若||0a b >>，则下列结论一定成立的是（）A.22a b ab> B.2211ab a b> C.33a b< D.a c c b->-4.已知等比数列{}n a 的前n 项和为n S ，且31S ma =，则“7m =”是“{}n a 的公比为2”的（）A.必要不充分条件B.充分不必要条件C.充要条件D.既不充分也不必要条件5.已知函数3()log f x x =，若0b a >>，且a ，b 是()f x 的图像与直线(0)y m m =>的两个交点对应的横坐标，则4a b +的最小值为（）A.2B.4C.6D.86.三角板主要用于几何图形的绘制和角度的测量，在数学、工程制图等领域被广泛应用.如图，这是由两块直角三角板拼出的一个几何图形，其中||||AB AC = ，||||BD BC =，0BD BC ⋅= .连接AD ，若AD x AB y AC =+，则x y -=（）A.1B.2D.327.若0a ≠，()2ππsin 066x ax bx c ⎛⎫-++≥ ⎪⎝⎭对[0,8]x ∈恒成立，则（）A.0a > B.0bc +> C.0c > D.16b c a-=-8.已知A 是函数()e 3xf x x =+图象上的一点，点B 在直线:30l x y --=上，则||AB 的最小值是（）A.72e 22e- B.3C. D.二、选择题：本题共3小题，每小题6分，共18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6分，部分选对的得部分分，有选错的得0分.9.设数列{}n a ，{}n b 的前n 项和分别为n S ，n T ，且3n an b =，则下列结论不正确的是（）A.若{}n a 是递增数列，则{}n S 是递增数列B.若{}n a 是递减数列，则{}n S 是递减数列C.若{}n a 是递增数列，则{}n T 是递增数列D.若{}n a 是递减数列，则{}n T 是递减数列10.已知(31)f x +为奇函数，(3)1f =，且对任意x ∈R ，都有(2)(4)f x f x +=-，则必有（）A.(11)1f =-B.(23)0f =C.(7)1f =- D.(5)0f =11.已知函数()sin sin 3f x x x =+，则（）A.()f x 的图象关于点(π,0)中心对称B.()f x 的图象关于直线π4x =对称C.()f x 的值域为⎡⎢⎣⎦D.()f x 在π3π,24⎡⎤⎢⎥⎣⎦上单调递增三、填空题：本题共3小题，每小题5分，共15分.12.在ABC △中，角A ，B ，C 的对边分别是a ，b ，c ，且1a =，3b =，1cos 3C =，则ABC △外接圆的面积是__________.13.已知某种污染物的浓度C （单位：摩尔/升）与时间t （单位：天）的关系满足指数模型(1)0ek t C C -=，其中0C 是初始浓度（即1t =时该污染物的浓度），k 是常数.第2天（即2t =）测得该污染物的浓度为5摩尔/升，第4天测得该污染物的浓度为15摩尔/升，若第n 天测得该污染物的浓度变为027C ，则n =__________.14.1796年，年仅19岁的高斯发现了正十七边形的尺规作图法.要用尺规作出正十七边形，就要将圆十七等分.高斯墓碑上刻着如图所示的图案.设将圆十七等分后每等份圆弧所对的圆心角为α，则162121tan2k k α==+∑__________.四、解答题：本题共5小题，共77分.解答应写出文字说明、证明过程或演算步骤.15.（13分）在ABC △中，角A ，B ，C 的对边分别是a ，b ，c ，4cos 5A =，2cos 3cos a C c A =.（1）求sin C 的值；（2）若3a =，求ABC △的周长.16.（15分）已知函数()sin()(0,0,0π)f x A x b A ωϕωϕ=++>><<的部分图象如图所示.（1）求()f x 的解析式；（2）求()f x 的零点；（3）将()f x 图象上的所有点向右平移π12个单位长度，得到函数()g x 的图象，求()g x 在7π0,12⎡⎤⎢⎥⎣⎦上的值域.17.（15分）已知函数3()33xx a f x ⋅=+，且()()66log 3log 122f f +=.（1）求a 的值；（2）求不等式()22310f x x +->的解集.18.（17分）已知函数2()(2)ln(1)2f x ax x x x =++--.（1）当0a =时，求()f x 的单调区间与极值；（2）当0x ≥时，()0f x ≤恒成立，求a 的取值范围.19.（17分）设数列{}n a 的前n 项和为n S ，若对任意的n +∈N ，都有2n n S kS =（k 为非零常数），则称数列{}n a 为“和等比数列”，其中k 为和公比.（1）若23n a n =-，判断{}n a 是否为“和等比数列”.（2）已知{}n b 是首项为1，公差不为0的等差数列，且{}n b 是“和等比数列”，2n b nc =，数列{}n c 的前n 项和为n T .①求{}n b 的和公比；②求n T ；③若不等式2134(1)22nn n n T m -+->--对任意的n +∈N 恒成立，求m 的取值范围.2024-2025年度河南省高三年级联考（二）数学参考答案1.C 由题意可得{}13A x x =<<.因为{}15A B x x =<< ，所以1,35a a ≥⎧⎨+=⎩，解得2a =.2.A :(0,)p m ⌝∃∈+∞，//a b ，当(7)2m m -+=-，即7m =时，//a b，所以p ⌝为真命题.3.B 当3a =，2b =-时，2218,12a b ab =-=，此时22a b ab <，则A 错误.因为||0a b >>，所以a b >，且0ab ≠，所以2210a b >，所以2211ab a b>，则B 正确.当2a =，1b =-时，338,1a b ==-，此时33a b >，则C 错误.当2a =，1b =，3c =时，1a c -=-，2c b -=，此时a c c b -<-，则D 错误.4.A 设{}n a 的公比为q ，则()23123111S a a a q q a ma =++=++=.因为10a ≠，所以21q q m ++=.由7m =，得217q q ++=，即260q q +-=，解得2q =或3q =-.由2q =，得7m =，则“7m =”是“{}n a 的公比为2”的必要不充分条件.5.B 由题意可得01a b <<<，1b a=，则44a b +≥，当且仅当42a b ==时，等号成立.故4a b +的最小值为4.6.A 如图，以A 为原点，AB ，AC的方向分别为x ，y 轴的正方向，建立直角坐标系，设1AB =，则(0,0)A ，(1,0)B ，(0,1)C ，故(1,0)AB = ，(0,1)AC =.作DF AB ⊥，交AB 的延长线于点F .设||1AB = ，则||||1BF DF ==，所以(2,1)D ，所以(2,1)AD = .因为AD x AB y AC =+，所以2,1x y ==，则1x y -=.7.B 因为[0,8]x ∈，所以πππ7π,6666x ⎡⎤-∈-⎢⎥⎣⎦.当[0,1)x ∈时，ππsin 066x ⎛⎫-< ⎪⎝⎭；当()1,7x ∈时，ππsin 066x ⎛⎫-> ⎪⎝⎭；当(7,8]x ∈时，ππsin 066x ⎛⎫-< ⎪⎝⎭.因为()2ππsin 066x ax bx c ⎛⎫-++≥ ⎪⎝⎭对[0,8]x ∈恒成立，所以1，7是20ax bx c ++=的两根，且0a <，则17,17,b ac a ⎧+=-⎪⎪⎨⎪⨯=⎪⎩故80b a =->，70c a =<，15b c a -=-，0b c a +=->.8.D由题意可得()(1)e x x f x +'=.设()()g x f x '=，则()(2)e xg x x '=+，当1x <-时，()0f x '<，当1x >-时，()0g x '>，()f x '单调递增.因为(0)1f '=，所以()(1)e 1xf x x '=+=，得0x =，此时(0,3)A ，故min ||AB ==.9.ABD当7n a n =-时，{}n a 是递增数列，此时{}n S 不是递增数列，则A 错误.当12n a n =-+时，{}n a 是递减数列，此时{}n S 不是递减数列，则B 错误.由{}n a 是递增数列，得{}n b 是递增数列，且0n b >，则{}n T 是递增数列，故C 正确.由{}n a 是递减数列，得{}n b 是递减数列，且0n b >，则{}n T 是递增数列，故D 错误.10.CD由(31)f x +为奇函数，可得(31)(31)f x f x -+=-+，则()f x 的图象关于点(1,0)对称.又(2)(4)f x f x +=-，所以()f x 的图象关于直线3x =对称，则()f x 是以8为周期的周期函数，所以(7)(3)1f f =-=-，(5)(1)0f f ==，(11)(3)1f f ==，(23)(7)1f f ==-，故选CD.11.ACD因为(π)(π)sin(π)sin 3(π)sin(π)sin 3(π)0f x f x x x x x ++-=++++-+-=，所以()f x 的图象关于点(π,0)中心对称，则A 正确.由题意可得()sin sin 32sin 2cos f x x x x x =+=，则ππππ2sin 2cos 2cos 2cos 4244f x x x x x ⎛⎫⎛⎫⎛⎫⎛⎫+=++=+ ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭，ππππ2sin 2cos 2cos 2cos 4244f x x x x x ⎛⎫⎛⎫⎛⎫⎛⎫-=--=- ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭，所以ππ44f x f x ⎛⎫⎛⎫+≠- ⎪ ⎪⎝⎭⎝⎭，所以()f x 的图象不关于直线π4x =对称，则B 错误.由题意可得3()2sin 2cos 4sin 4sin f x x x x x ==-.设sin [1,1]t x =∈-，则3()44y g t t t ==-+，故()22()124431g t t t '=-+=--.由()0g t '>，得3333t -<<；由()0g t '<，得313t -≤<-或313t <≤，则()g t 在31,3⎡⎫--⎪⎢⎣⎭和3,13⎛⎤⎥⎝⎦上单调递减，在33,33⎛⎫- ⎪⎝⎭上单调递增.因为(1)(1)0g g -==，38339g ⎛⎫-=- ⎪⎝⎭，38339g ⎛⎫=⎪⎝⎭，所以8383()99g t ⎡∈-⎢⎣⎦，即()f x 的值域是838399⎡-⎢⎣⎦，则C 正确.当π3π,24x ⎡⎤∈⎢⎥⎣⎦时，2sin 2t x ⎤=∈⎥⎣⎦.因为sin t x =在π3π,24⎡⎤⎢⎥⎣⎦上单调递减，且()g t在,13⎤⎥⎣⎦上单调递减，所以()f x 在π3π,24⎡⎤⎢⎥⎣⎦上单调递增，则D 正确.12.9π4由余弦定理可得22212cos 1921383c a b ab C =+-=+-⨯⨯⨯=，则c =因为1cos 3C =，所以22sin 3C =，则ABC △外接圆的半径32sin 2c R C ==，故ABC 外接圆的面积为29ππ4R =.13.7由题意可得030e 5,e 15,k kC C ⎧=⎨=⎩则2e 3k =，解得ln32k =.因为(1)00e 27k n C C -=，即3ln(1)200e 27n C C -=，所以ln 3(1)2e 27n -=，所以ln 3(1)ln 273ln 32n -==，解得7n =.14.15由题可知2π17α=，则222π11tan 1tan π217cos 17k k k α+=+=，则161616162211112π2π2π2cos 1cos 16cos 1717171tan 2k k k k k k k k α====⎛⎫==+=+ ⎪⎝⎭+∑∑∑∑.由161611π2π(21)π(21)π33πππ2sin cos sin sin sin sin 2sin 17171717171717k k k k k ==+-⎡⎤⋅=-=-=-⎢⎥⎣⎦∑∑，得1612πcos117k k ==-∑，故原式16115=-=.15.解：（1）因为4cos 5A =，且0πA <<，所以3sin 5A ==.因为2cos 3cos a C c A =，所以2sin cos 3sin cos A C C A =，所以342cos 3sin 55C C ⨯=⨯，即cos 2sin C C =.因为22sin cos 1C C +=，所以21sin 5C =.因为0πC <<，所以5sin 5C =.（2）由（1）可知3sin 5A =，4cos 5A =，5sin 5C =，25cos 5C =，则3254525sin sin()sin cos cos sin 55555B AC A C A C =+=+=⨯+⨯=.由正弦定理可得sin sin sin a b cA B C==，则sin sin a B b A ==sin sin a Cc A==，故ABC △的周长为3a b c ++=+.16.解：（1）由图可知3(1)22A --==，3(1)12b +-==，()f x 的最小正周期7ππ2π1212T ⎛⎫=-= ⎪⎝⎭.因为2π||T ω=，且0ω>，所以2ω=.因为()f x 的图象经过点π,312⎛⎫⎪⎝⎭，所以ππ2sin 2131212f ϕ⎛⎫⎛⎫=⨯++= ⎪ ⎪⎝⎭⎝⎭，即πsin 16ϕ⎛⎫+=⎪⎝⎭，所以ππ2π()62k k ϕ+=+∈Z ，即π2π()3k k ϕ=+∈Z .因为0πϕ<<，所以π3ϕ=.故π()2sin 213f x x ⎛⎫=++ ⎪⎝⎭.（2）令()0f x =，得π1sin 232x ⎛⎫+=- ⎪⎝⎭，则ππ22π()36x k k +=-∈Z 或π5π22π()36x k k +=-∈Z ，解得ππ4x k =-或7ππ()12k k -∈Z ，故()f x 的零点为ππ4k -或7ππ()12k k -∈Z .（3）由题意可得πππ()2sin 212sin 211236g x x x ⎡⎤⎛⎫⎛⎫=-++=++ ⎪ ⎪⎢⎝⎭⎝⎭⎣⎦.因为7π0,12x ⎡⎤∈⎢⎥⎣⎦，所以ππ4π2,663x ⎡⎤+∈⎢⎥⎣⎦.当ππ262x +=，即π6x =时，()g x 取得最大值π36g ⎛⎫= ⎪⎝⎭；当π4π263x +=，即7π12x =时，()g x 取得最小值7π112g ⎛⎫=- ⎪⎝⎭.故()g x 在7π0,12⎡⎤⎢⎥⎣⎦上的值域为1⎡⎤⎣⎦.17.解：（1）因为3()33x x a f x ⨯=+，所以221393(2)333933x x x x a a af x --+⨯-===+++，则33()(2)3333x xx a a f x f x a ⨯+-=+=++.又666log 3log 12log 362+==，所以()()66log 3log 12f f a +=，从而2a =.（2）由（1）可知236()23333x x x f x ⨯==-++，显然()f x 在R 上单调递增.因为1(0)2f =，所以由()22310f x x +->，可得()23(0)f x x f +>，则230x x +>，解得3x <-或0x >，故不等式()22310f x x +->的解集为(,3)(0,)-∞-+∞ .18.解：（1）当0a =时，2()2ln(1)2f x x x x =+--，其定义域为(1,)-+∞，则()222(2)22111x x x x f x x x x x ---+'=--==+++.当(1,0)x ∈-时，()0f x '>，()f x 的单调递增区间为(1,0)-，当(0,)x ∈+∞时，()0f x '<，()f x 的单调递减区间为(0,)+∞，故()f x 的极大值为(0)0f =，无极小值.（2）设1t x =+，[1,)t ∈+∞，2()(2)ln 1g t at a t t =+--+，[1,)t ∈+∞，则2()ln 2at a t t a tg -=+-+'.设()()h t g t '=，则222222()2a a t at a h t t t t --++-'=--=.设2()22m t t at a =-++-，则函数()m t 的图象关于直线4at =对称.①当2a ≤时，()m t 在[1,)+∞上单调递减.因为(1)240m a =-≤，所以2()220m t t at a =-++-≤在[1,)+∞上恒成立，即()0h t '≤在[1,)+∞上恒成立，则()h t 在[1,)+∞上单调递减，即()g t '在[1,)+∞上单调递减，所以()(1)0g t g ''≤=，所以()g t 在[1,)+∞上单调递减，则()(1)0g t g ≤=，即()0f x ≤在[0,)+∞上恒成立，故2a ≤符合题意.②当2a >时，()m t 在[1,)+∞上单调递减或在[1,)+∞上先增后减，因为(1)240m a =->，所以存在01t >，使得()00m t =.当()01,t t ∈时，()0m t >，即()0h t '>，所以()g t '在()01,t 上单调递增.因为(1)0g '=，所以()0g t '>在()01,t 上恒成立，所以()g t 在()01,t 上单调递增，则()0(1)0g t g >=，故2a >不符合题意.综上，a 的取值范围为(,2]-∞.19.解：（1）因为23n a n =-，所以121n a n +=-，所以12n n a a +-=.因为11a =-，所以{}n a 是首项为-1，公差为2的等差数列，则22n S n n =-，所以2244n S n n =-，所以222444422n n S n n n S n n n --==--.因为442n n --不是常数，所以{}n a 不是“和等比数列”.（2）①设等差数列{}n b 的公差为d ，前n 项和为n S ，则21(1)1222n n n d d S nb d n n -⎛⎫=+=+- ⎪⎝⎭，所以222(2)n S dn d n =+-.因为{}n b 是“和等比数列”，所以2n n S kS =，即222(2)22kd kd dn d n n k n ⎛⎫+-=+- ⎪⎝⎭，所以2,22,2kd d kd d k ⎧=⎪⎪⎨⎪-=-⎪⎩解得4,2,k d =⎧⎨=⎩即{}n b 的和公比为4.②由①可知12(1)21n b n n =+-=-，则212n n n c -=，所以35211232222n n n T -=++++ ，所以2352121112122222n n n n nT -+-=++++ ，所以235212121211122311111422222212nn n n n n n T -++⎡⎤⎛⎫⨯-⎢⎥ ⎪⎝⎭⎢⎥⎣⎦=++++-=- ，即2132344332n n n T ++=-⨯，所以21834992nn n T -+=-⨯.③设2121212134834348103429922992n n n n n n n n n n P T ----++++=-=--=-⨯⨯，12121103710345(1)092924n n n n nn n n P P ++-+++-=-⨯+⨯=>.不等式2134(1)22n n n n T m -+->--对任意的n +∈N 恒成立，即不等式(1)2n n P m >--对任意的n +∈N 恒成立.当n 为奇数时，()1min 23n m P P --<==-，则1m >；当n 为偶数时，()2min 122n m P P -<==-，则32m <.综上，m 的取值范围是31,2⎛⎫⎪⎝⎭.。
重庆市南开中学2021届高三英语10月月考试题（含解析）
重庆市南开中学2021届高三英语10月月考试题（含解析）新人教版第一节单项填空(共15小题；每题1分，总分值15分)21．——Which type of IPhone shall I buy?——，whatever they have in the shop．A．Everything B．Something C．Some one D．Any one【答案】【知识点】A2代词【答案解析】D。
解析：那个地址指代“任何一个苹果电话”，用any one，相当于any one of the Iphones in the shop. anything是任何东西，没有范围，something某个东西，some one某一个,应选D。
句意：--我买哪一种苹果电话？--任何一个，凡是他们店里有的。
【题文】22．——May I look at the menu for a little while?——Of course，，sir．A．don’t worry B．it doesn’t matterC．enjoy yourself D．take your time【答案】【知识点】A17 情景交际【答案解析】D。
解析：A．don’t worry 不用担忧B．it doesn’t matter. 没关系。
C．enjoy yourself 玩得高兴D．take your time. 不着急句意：——我能够再看一会菜单吗？——固然，不着急，先生。
【题文】23．At this time tomorrow over the Atlantic．A．we’re going to fly B．we’ll be flying ．C．we’ll fly D．we’re to fly【答案】【知识点】A11动词的时态与语态【答案解析】B。
解析：At this time tomorrow 明天的那个时候，要用以后进行时，will be doing。
2021届哈尔滨市第四十九中学校高三英语月考试卷及答案
2021届哈尔滨市第四十九中学校高三英语月考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ABored with your life? Dreaming of something different? I always wonder what life would be like400 kmabove my head. That's where the International Space Station orbits the earth, with six astronauts living and working on board, for months at a time.How do they sleep? They spend the night floating in a sleeping bag inside a small cubicle (小隔间) on the ceiling. American astronaut Sunita Williams explains, “It's like a little phone booth, but it's pretty comfortable and it doesn't matter if I turn overand sleep upside down. I don't have any sensation (感觉) in my head that tells me I'm upside down.”Brushing your teeth in a place where you can't have a tap or a sink can be a challenge. Can you imagine the mess that running water would make in zero gravity? Canadian astronaut Chris Hadfield uses a straw to put a big blob of water from a sealed bag onto an ordinary toothbrush and adds a little toothpaste which he has to swallow when he's done.Daily exercise is essential. The lack of gravity makes bones more fragile and muscles lose strength — so astronauts are encouraged to work out for at least two hours a day.The role of astronauts in the International Space Station is to act as lab technicians for scientists back on earth. So they spend their time maintaining their environment and performing and monitoring experiments in a confined space about the size of a Boeing 747. Almost every task is carefully planned by mission control — although most astronauts spend their first days losing things until they get used to sticking everything they use to the walls with Velcro, duct tape (强力胶带) or clips (夹子)．One of their most valued perks (额外待遇) is the view from “the office”, dominated by that gigantic blue ball down there, sitting in the darkness of space. Wow! Absolutely breathtaking!1. When they sleep upside down, the astronauts willnot get dizzy because ________.A. they don't feel itB. they sleep in the daytimeC. they sleep in special sleeping bagsD. they are trained to adapt to the conditions of weightlessness2. What parts become weak if astronauts don't exercise?A. Their teeth and bones.B. Their brains and bones.C. Their bones and muscles.D. Their teeth and muscles.3. What is the passage mainly about?A. The Problems We Met in Space.B. Living and Working in Space.C. How to Become an Astronaut.D. The International Space Station.BClara Daly was seated on an Alaska Airlines flight from Boston to Los Angeles when a flight attendant asked an urgent(紧急的) question over the loudspeaker: “Does anyone on board know American Body Language?” She knew she needed to help.Clara, 15 at the time, pressed the call button. The flight attendant came by and explained the situation. “We have a passenger on the plane who’s blind and deaf,” she said. The passenger seemed to want something, but he was traveling alone and the flight attendants couldn’t understand what he needed, according to PEOPLE magazine.Clara had been studying ASL for the past year to help with her dyslexia (阅读障碍) and knew she’d be able to spell on the man’s palm(手掌) by finger. So she unbuckled her seat belt, walked toward the front of the plane, and knelt by the aisle seat of Tim Cook, then 64. Gently taking his hand, she wrote, “How are you? Are you OK?” Cook asked for some water. When it arrived, Clara returned to her seat. She came by again a bit later because he wanted to know the time. On her third visit, she stopped and stayed for a while.“He didn’t need anything. He was lonely and wanted to talk,” Clara said. So for the next hour, that was what they did. She talked about her family and her plans for the future (she wants to be a politician). Cook told Clara how he had gradually become blind over time and shared stories of his days as a traveling salesman. Even though he couldn’t see her, she “looked attentively at his face with such kindness”, a passenger reported.“Clara was amazing,” a flight attendant told Alaska Airlines in a blog interview. “You could tell Cook was very excited to have someone he could speak to, and she was such a warm-hearted girl.” Cook’s reaction: “Best trip I’ve ever had.”Looking for ways to offer help? Start with this random(随时的) act of kindness that can change someone’s liferight now.4. The flight attendant asked an urgent question because ________.A. the passenger was traveling aloneB. the plane was in a dangerous situationC. the passenger asked for something suddenlyD. none of the flight attendants could communicate with the passenger5. Why did Clara talk about her plans for the future?A. Because the flight attendant asked her to do so.B. Because she needed topics to go on talking with Cook.C. Because Cook hoped to understand teenagers better.D. Because she wanted to show her dream for the future.6. Which of the following words can best describe Clara?A. Kind and caring.B. Warm-hearted and brave.C. careful and calm.D. opened-minded and confident.7. The passage is mainly written to ________.A. tell a touching story of an amazing girlB. show the great importance of American Body LanguageC. encourage readers to give a hand kindly and randomlyD. show how kind the flight attendant was to help CookCThe founder of Earth Day was Gaylord Nelson, a U.S. Senator fromWisconsin. During the late 1960s, Americans witnessed the uninvited side effects of high productivity. Factories and power plants were sending out smoke and industrial waste while Americans were using petrol for their massive(大量的) cars, making air pollution almostsynonymous withthe nation’s development.What moved Senator Nelson to action was the 1969 massive oil spill inCalifornia, the largest in theUnited Statesat that time. The spill proved to be an environmental nightmare as it had a significant effect on marine life, killing about 3,500 sea birds, as well as marine animals such as dolphins, elephant seals and sea lions, fueling public anger. Inspired by the student antiwar movement at that period of time, Nelson found it an appropriatetime to direct the energy of the students towards a fight for environmental protection. He decided that it was time to educate the Americans on the need to protect the environment. Thus Earth Day was born in 1970, and public environmentalawareness took centre stage.On 22ndApril 1970, millions of Americans took to the street and thousands of students marched to appeal for a healthy, sustainable environment. There was now a new synergy(协同作用) among different groups which had previously been fighting their causes related to the environment. Their fight for environmental conservation became so overwhelming that affected businesses were forced to follow environmental standards if they wanted to continue their operations.As it became more apparent that environmental issues were not just localized ones but a global concern, the year 1990 saw Earth Day reach out to many more around the world. Earth Day 1990 helped pave the way for the 1992 United Nations Earth Summit inRio de Janeiro, bringing together many nations for a united effort towards protecting the environment.8. Which of the following can replace the underlined phrase “synonymous with” in paragraph 1?A. familiar withB. opposite toC. different fromD. equal to9. Why did Nelson found Earth Day?A. To support students’ antiwar movement.B. To draw people’s attention to the seriousness of the oil spill.C. To arouse American’s awareness of environmental conservation.D. To educate Americans to protect marine life threatened by oil spill events.10. What can be inferred from the passage?A. Businesses would like to follow environmental standards.B. Earth Day united people to fight for environmental protection.C. It was the side effects of high productivity that led Nelson to take action.D. The 1992 United Nations Earth Summit made Earth Day known to more countries.11. The passage mainly talks about_______.A. how Earth Day came into beingB. why Earth Day was so significantC. who the founder of Earth Day wasD. what Earth Day meantto the worldDA 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。