常德市一中2021届高三年级第三次月考答案
2021年湖南省常德市临第三中学高三数学理月考试题含解析
2021年湖南省常德市临第三中学高三数学理月考试题含解析一、选择题:本大题共10小题,每小题5分,共50分。
在每小题给出的四个选项中,只有是一个符合题目要求的1. 设直线与函数的图象分别交于点M、N,则当达到最小时的值为__________.A. B. C. 1 D. 2参考答案:C2. 双曲线(a>0,b>0)的渐近线为等边三角形OAB的边OA、OB所在直线,直线AB过焦点,且|AB|=2,则双曲线实轴长为()A.B.C.D.3参考答案:D【考点】KC:双曲线的简单性质.【分析】利用双曲线方程以及渐近线的性质求出a,b关系式,通过|AB|=2,求出c,然后求解a即可得到结果.【解答】解:双曲线(a>0,b>0)的渐近线为等边三角形OAB的边OA、OB所在直线,可得,直线AB过焦点,且|AB|=2,可得c=,则,解得a=.则双曲线实轴长为:3.故选:D.3. 已知x,y满足,则z=2x-y的最大值为lax-Y}-3,(A) 2 (B)1 (C) -1 (D) 3参考答案:A4. 已知集合,则等于A. B. C. D.参考答案:D试题分析:,故答案为D考点:集合的交集5. 已知F是双曲线E:﹣=1(a>0,b>0)的右焦点,过点F作E的一条渐近线的垂线,垂足为P,线段PF与E相交于点Q,记点Q到E的两条渐近线的距离之积为d2,若|FP|=2d,则该双曲线的离心率是()A.B.2 C.3 D.4参考答案:B【考点】双曲线的简单性质.【分析】E上任意一点Q(x,y)到两条渐近线的距离之积为d1d2===d2,F(c,0)到渐近线bx﹣ay=0的距离为=b=2d,求出可求双曲线的离心率.【解答】解:E上任意一点Q(x,y)到两条渐近线的距离之积为d1d2===d2,F(c,0)到渐近线bx﹣ay=0的距离为=b=2d,∴,∴e==2,故选B.【点评】本题考查双曲线的离心率,考查点到直线距离公式的运用,属于中档题.6. 设全集, ,,则A. B. C.D.参考答案:【知识点】集合及其运算A1【答案解析】B 由则,,所以故选B。
2020-2021学年常德市一中高三语文月考试卷及参考答案
2020-2021学年常德市一中高三语文月考试卷及参考答案一、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。
补锅人补的是什么干亚群每隔一段时间,母亲会把家里做饭的铁锅拎出去,拿一把铲子“咻咻”地刨。
锅灰清除后,饭煮得特别快。
可锅像人一样是有年纪的,上了年纪的锅就像老人一样会豁嘴。
锅是慢慢老的,从渗一滴水开始,一点点地漏。
这时候,锅与人一样等候补锅人的出现。
如果补锅人不出现,锅就一直耷拉着耳朵,补锅人不来,锅就没办法被端到灶上,就像犯了错误的小学生,不经老师允许不能坐到座位上去。
锅渗水了,不能煮饭。
母亲心急火燎赶到集市上买了一口新锅。
新锅烧开的第一锅水是不能喝的,带着一股生铁味。
锅冷却后,母亲拿着一片月牙形的砂轮,不停地锉,新锅大张着嘴巴,发出“啊嘎啊嘎”的声音。
新锅的饭菜里有一股生铁味,吃在嘴里涩涩的。
所以,母亲宁愿补一口旧锅。
补锅人是外乡人,他的脸灰扑扑的,似乎因为整天跟锅打交道,那些锅灰都腻着他。
补锅人年龄不好猜,看上去过于沧桑,脸上的皱纹像机耕路一样。
补锅虽不是力气活,但有时会看到他流汗,那些汗纵横流淌,流过沟沟壑壑后才滴下来,落到地上分成八辫。
那些汗是热出来的,他得把一些敲烂的锅铁放进炉中,待锅铁熔化了,用铁水补锅。
补锅人用一把小榔头,围着锅的破洞轻轻敲,一些经不起敲的铁锈碎片纷纷坠落。
这是补锅的第一道程序,像一位医生的清创术,得把周围坏死的组织清理掉。
与坏死的组织不同的是,那些碎片有再生的价值,它们还得回到原来的位置。
补锅人把铁锈片收集起来,放进炉中。
有人戏称补锅人是锅的再生父母。
补锅人有耐心,不急不躁拉着风箱。
只有我们按捺不住,个个伸长脖子去瞧。
补锅人看见了,腾出一只手冲我们摇摆,叼着烟的嘴巴含含糊糊发出一串声音。
我们的好奇心并没有因为他的阻止而消失,反而与他炉底的火一样旺起来。
这时有几只破旧的铁锅被人拎过来。
来人咨询锅还能不能补,补的话多少钱。
补锅人像一位老中医,一丝不苟地执行望、闻、问、切的程序,捏捏,敲敲,瞧瞧,对破锅一一做出诊断,有的可以小补,有的大补,但大补的工钱有些贵,所以他劝人别补了。
2021届湖南省常德市一中高三上学期第三次月考数学试题(解析版)
2021届湖南省常德市一中高三上学期第三次月考数学试题一、单选题1.设x ∈R ,则2x >的一个充分不必要条件是( ) A .1x < B .1x >C .1x >-D .3x >【答案】D【分析】根据充分不必要条件的定义判断. 【详解】1x <显然不能得出2x >,不充分;1x >时,若32x =,不能得出2x >,不充分; 1x >-时,若0x =,不能得出2x >,不充分,若3x >,则一定有2x >,充分性满足,且当2x >时,若52x =,则不能得出3x >,不必要也满足.只有D 是充分不必要条件. 故选:D .2.设直线l 不在平面α内,直线m 在平面α内,则下列说法正确的是( ) A .直线l 与直线m 没有公共点 B .直线l 与直线m 异面 C .直线l 与直线m 至多一个公共点 D .直线l 与直线m 不垂直【答案】C【分析】利用空间中直线与直线的性质以及直线与平面的性质进行判断选项即可 【详解】对于A ,直线l 不在平面α内,直线m 在平面α内,但是,直线l 与m 可以相交,所以,A 错;对于B ,直线l 不在平面α内,直线m 在平面α内,但是,直线l 与m 可以相交也可以平行,所以,B 错;对于C ,直线l 不在平面α内,直线m 在平面α内,则直线l 与直线m 只可以平行或者相交,不可能重合,所以,直线l 与直线m 至多一个公共点,C 正确;对于D ,直线l 不在平面α内,直线m 在平面α内,则当直线l 垂直于平面α时,直线l 与直线m 垂直,所以,D 错误;故选:C3.已知幂函数()()22322n nf x n n x-=+-()n Z ∈的图象关于y 轴对称,且在()0,∞+上是增函数,则n 的值为( ) A .3- B .1C .1-D .3-或1【答案】A【分析】根据函数是幂函数得2221+-=n n ,求得3n =-或1,再检验是否符合题意即可. 【详解】()f x 是幂函数,2221n n ∴+-=,解得3n =-或1,当3n =-时,()18=f x x 是偶函数,关于y 轴对称,在()0,∞+单调递增,符合题意, 当1n =时,()2f x x -=是偶函数,关于y 轴对称,在()0,∞+单调递减,不符合题意,3n ∴=-.故选:A.4.欧拉公式i e cos isin θθθ=+把自然对数的底数e ,虚数单位i ,三角函数cos θ和sin θ联系在一起,充分体现了数学的和谐美,被誉为“数学的天桥”,若复数z 满足()i i i ez π+⋅=,则z =( )A .1B .2C D【答案】B【分析】由新定义化为复数的代数形式,然后由复数的除法运算求出z 后再求模. 【详解】由题意(1)cos sin 1(1)(1)i i i i i i z e ii i i i i πππ--====+++-+-+--111222i i -+==-,∴2z ==. 故选:B .【点睛】本题考查复数的新定义,考查复数的除法运算和求复数的模,解题关键是由新定义化i e π为代数形式,然后求解.5.设b R ∈,数列{}n a 的前n 项和3nn S b =+,则( )A .{}n a 是等比数列B .{}n a 是等差数列C .当1b ≠-时,{}n a 是等比数列D .当1b =-时,{}n a 是等比数列【答案】D【分析】根据n S 与n a 的关系求出n a ,然后判断各选项.【详解】由题意2n ≥时,111(3)(3)23n n n n n n a S S b b ---=-=+-+=⨯,13n na a +=(2)n ≥, 113a Sb ==+,若212333a a b⨯==+,即1b =-,则{}n a 是等比数列,否则不是等比数列,也不是等差数列, 故选:D .【点睛】关键点点睛:本题考查等比数列的定义.在由1n n n a S S -=-求通项时,2n ≥必须牢记,11a S =它与(2)n a n ≥的求法不相同,因此会影响{}n a 的性质.对等比数列来讲,不仅要求3423a a a a ==,还必须满足3212a a a a =. 6.为得到函数6sin 23y x π⎛⎫=- ⎪⎝⎭的图象,只需要将函数6cos 2y x =的图象( ) A .向右平行移动6π个单位 B .向左平行移动6π个单位 C .向右平行移动512π个单位 D .向左平行移动512π个单位 【答案】C【分析】将目标函数的解析式变形为56sin 26cos 236y x x ππ⎛⎫⎛⎫=-=- ⎪ ⎪⎝⎭⎝⎭,利用三角函数图象的平移规律可得出结论. 【详解】将目标函数的解析式变形为56sin 26cos 26cos 233212y x x x ππππ⎡⎤⎛⎫⎛⎫⎛⎫=-=--=- ⎪ ⎪ ⎪⎢⎥⎝⎭⎝⎭⎝⎭⎣⎦,因此,为了得到函数6sin 23y x π⎛⎫=- ⎪⎝⎭的图象,只需要将函数6cos 2y x =的图象向右平行移动512π个单位. 故选:C.7.若平面区域30230230x y x y x y +-≥⎧⎪--≤⎨⎪-+≥⎩夹在两条斜率均为1的平行直线之间,则这两条平行直线间的距离的最小值为( )A .35B .2C .35D .5【答案】B【分析】利用线性规划的性质,作图求解即可【详解】画出不等式组的平面区域如题所示,由230{30x y x y -+=+-=得(1,2)A ,由230{30x y x y --=+-=得(2,1)B ,由题意可知,当斜率为1的两条直线分别过点A 和点B 时,两直线的距离最小,即22(12)(21)2AB =-+-=. 故选B .【点睛】关键点睛:解题的关键在于根据不等式组,作出相关图像,进而求解,属于基础题 8.设'()f x 是函数()f x 的导函数,且'()2()()f x f x x R >∈,12f e ⎛⎫= ⎪⎝⎭(e 为自然对数的底数),则不等式2(ln )f x x <的解集为( ) A .0,2e ⎛⎫ ⎪⎝⎭B .)eC .1,2e e ⎛⎫⎪⎝⎭D .2e e ⎛⎝ 【答案】B【分析】构造函数F (x )=()2xf x e,求出导数,判断F (x )在R 上递增.原不等式等价为F (lnx )<F (12),运用单调性,可得lnx <12,运用对数不等式的解法,即可得到所求解集.【详解】可构造函数F (x )=()2xf x e,F′(x )=()()22222()x xx f x e f x e e -=()()2'2xf x f x e-,由f′(x )>2f (x ),可得F′(x )>0,即有F (x )在R 上递增. 不等式f (lnx )<x 2即为()2f lnx x<1,(x >0),即()2lnxf lnx e<1,x >0.即有F (12)=12f e⎛⎫⎪⎝⎭=1,即为F (lnx )<F (12),由F (x )在R 上递增,可得lnx <12,解得0<x. 故不等式的解集为(0), 故选B .【点睛】利用导数解抽象函数不等式,实质是利用导数研究对应函数单调性,而对应函数需要构造. 构造辅助函数常根据导数法则进行:如()()f x f x '<构造()()xf xg x e=, ()()0f x f x '+<构造()()xg x e f x =, ()()xf x f x '<构造()()f xg x x=, ()()0xf x f x '+<构造()()g x xf x =等 9.已知函数()()tan 0,02f x x ωϕωϕπ⎛⎫=+><< ⎪⎝⎭的图象关于点,06π⎛⎫ ⎪⎝⎭成中心对称,且与直线y a =的两个相邻交点间的距离为2π,则下列叙述正确的是( ) A .函数()f x 的最小正周期为π B .函数()f x 图象的对称中心为(),06k k Z π⎛⎫π+∈ ⎪⎝⎭C .函数()f x 的图象可由tan 2y x =的图象向左平移6π得到 D .函数()f x 的递增区间为(),2326k k k Z ππππ⎛⎫-+∈ ⎪⎝⎭ 【答案】D【分析】利用正切函数的性质进行判断.【详解】由()f x 的图象与直线y a =的两个相邻交点间的距离为2π,知函数最小正周期是2π,A 错; 由此得22πωπ==,又由,06π⎛⎫⎪⎝⎭是其图象对称中心得,26k πϕπ⨯+=或262k ππϕπ⨯+=+,k Z ∈,又02πϕ<<,所以6π=ϕ. 所以()tan(2)6f x x π=+,由262k x ππ+=得412k x ππ=-,对称中心是(,0),412k k Z ππ-∈,B 错; tan 2y x =的图象向左平移6π得到tan 2()tan(2)()63y x x f x ππ=+=+≠,C 错;由2262k x k πππππ-<+<+,得2326k k x ππππ-<<+,即函数的增区间是(),2326k k k Z ππππ⎛⎫-+∈⎪⎝⎭,D 正确. 故选:D .【点睛】本题考查正切型函数的图象与性质,掌握正切函数的性质是解题关键.二、多选题10.在平面直角坐标系xOy 中,如图放置的边长为2的正方形ABCD 沿x 轴滚动(无滑动滚动),点D 恰好经过坐标原点,设顶点(),B x y 的轨迹方程是()y f x =,则对函数()y f x =的判断正确的是( )A .函数()y f x =是奇函数B .对任意的x ∈R ,都有()()44f x f x +=-C .函数()y f x =的值域为0,22⎡⎣D .函数()y f x =在区间[]6,8上单调递增【答案】BCD【分析】根据正方形的运动,得到点(),B x y 的轨迹,作出对应函数图像,根据图像,即可得出结果.【详解】由题意,当42x -≤<-时,顶点(),B x y 的轨迹是以点(2,0)A -为圆心,以2为半径的14圆; 当22x -≤<时,顶点(),B x y 的轨迹是以点(0,0)D 为圆心,以22为半径的14圆; 当24x ≤<时,顶点(),B x y 的轨迹是以点(2,0)C 为圆心,以2为半径的14圆; 当46x ≤<,顶点(),B x y 的轨迹是以点(4,0)A 为圆心,以2为半径的14圆,与42x -≤<-的形状相同,因此函数()y f x =在[]4,4-恰好为一个周期的图像;所以函数()y f x =的周期是8; 其图像如下:A 选项,由图像及题意可得,该函数为偶函数,故A 错;B 选项,因为函数的周期为8,所以(8)()f x f x +=,因此(4)(4)f x f x +=-;故B 正确;C 选项,由图像可得,该函数的值域为0,22⎡⎣;故C 正确;D 选项,因为该函数是以8为周期的函数,因此函数()y f x =在区间[]6,8的图像与在区间[]2,0-图像形状相同,因此,单调递增;故D 正确; 故选:BCD.【点睛】本题主要考查分段函数的应用,熟记函数的性质,灵活运用数形结合的思想求解即可,属于常考题型.11.《九章算术》中“勾股容方”问题:“今有勾五步,股十二步,问勾中容方几何?”魏晋时期数学家刘徽在其《九章算术注》中利用出入相补原理给出了这个问题的一般解法:如图1,用对角线将长和宽分别为b 和a 的矩形分成两个直角三角形,每个直角三角形再分成一个内接正方形(黄)和两个小直角三角形(朱、青).将三种颜色的图形进行重组,得到如图2所示的矩形,该矩形长为+a b ,宽为内接正方形的边长d .由刘徽构造的图形可以得到许多重要的结论,如图3.设D 为斜边BC 的中点,作直角三角形ABC 的内接正方形对角线AE ,过点A 作AF BC ⊥于点F ,则下列推理正确的是( )①由图1和图2面积相等得abd a b=+; ②由AE AF ≥2222a b a b++≥; ③由AD AE ≥222112a b a b+≥+; ④由AD AF ≥可得222a b ab +≥. A .①B .②C .③D .④【答案】ABCD【分析】根据图1,图2面积相等,可求得d 的表达式,可判断A 选项正误,由题意可求得图3中,,AD AE AF 的表达式,逐一分析B 、C 、D 选项,即可得答案. 【详解】对于①:由图1和图2面积相等得()S ab a b d ==+⨯,所以abd a b=+,故①正确;对于②:因为AF BC ⊥,所以221122a b a b AF ⨯⨯=+,所以22AF a b =+,设图3中内接正方形边长为t ,根据三角形相似可得a t t ab -=,解得abt a b=+, 所以22abAE t a b==+,因为AE AF ≥,所以a b ≥+2a b +≥,故②正确; 对于③:因为D 为斜边BC的中点,所以2AD =, 因为AD AE ≥,所以2a b≥+211a b≥+,故③正确; 对于④:因为AD AF ≥,所以2≥,整理得:222a b ab +≥,故④正确; 故选:ABCD【点睛】解题的关键是根据题意及三角形的性质,利用几何法证明基本不等式,求得,,AD AE AF 的表达式,根据图形及题意,得到,,AD AE AF 的大小关系,即可求得答案,考查分析理解,计算化简的能力.12.对于函数()f x 和()g x ,设(){}0x f x α∈=,(){}0x g x β∈=,若存在α,β使得1αβ-≤,则称()f x 与()g x 互为“零点相邻函数”.若函数()12x f x e x -=+-与()23g x x ax a =--+互为“零点相邻函数”,则实数a 的取值可以是( ) A .2 B .73C .3D .4【答案】ABC【分析】首先确定函数()f x 的零点,然后结合新定义的知识得到关于a 的等式,分离参数,结合函数的单调性确定实数a 的取值范围即可. 【详解】很明显函数()12x f x e x -=+-是R 上的单调递增函数,且()10f =,据此可知1α=,结合“零点相邻函数”的定义可得11β-≤,则02β≤≤, 据此可知函数()23g x x ax a =--+在区间[]0,2上存在零点,即方程230x ax a --+=在区间[]0,2上存在实数根,整理可得:()22134121121224x x x x a x x x x ++===++++-++--+,根据对勾函数的性质,很明显函数()()4121h x x x =++-+在区间[]0,1上单调递减,在[]1,2上单调递增,所以,()()0373,2h h ==,(1)2h =则函数()h x 的值域为[]2,3, 据此可知实数a 的取值范围是[]2,3. 故选:ABC【点睛】“新定义”主要是指即时定义新概念、新公式、新定理、新法则、新运算五种,然后根据此新定义去解决问题,有时还需要用类比的方法去理解新的定义,这样有助于对新定义的透彻理解.但是,透过现象看本质,它们考查的还是基础数学知识,所以说“新题”不一定是“难题”,掌握好三基,以不变应万变才是制胜法宝.三、填空题13.给出下列三个函数:①1y x =;②sin y x =;③e x y =,则直线12y x b =+(b R ∈)不能作为函数_______的图象的切线(填写所有符合条件的函数的序号). 【答案】①【分析】分别求得三个函数的导数,由导数的几何意义,解方程可得不满足题意的函数.【详解】直线12y x b =+的斜率为k =12,对于①1y x =,求导得:'21y x =-,对于任意x≠0,21x -=12无解,所以,直线12y x b =+不能作为切线;对于②sin y x =,求导得:'1cos 2y x ==有解,可得满足题意;对于③xy e =,求导得:'12x y e ==有解,可得满足题意;故答案为①【点睛】本题考查导数的运用:求切线的斜率,考查导数的运算,以及方程思想、运算能力,属于中档题.14.已知扇形的周长为8cm ,圆心角为2,则扇形的面积为___________ 【答案】4【解析】试题分析:由题意可得28211{{4242422r l r S lr l r l +==∴∴==⨯⨯===【解析】扇形面积15.已知向量a 和b 满足|||2|2,||1a a b a b =-=-=,则⋅=a b ________.【答案】1【分析】将已知|22a b -=∣,||1a b -=两边平方,并结合||2a =,利用平面向量的数量积的运算法则,消项求得a b ⋅的值. 【详解】∵向量a 和b 满足|||22a a b =-=∣,||1a b -=.∴22442a a b b -⋅+=;① 2221a a b b -⋅+=,②22a =.③联立①②③可得:1a b ⋅=. 故答案为:1.【点睛】本题考查平面已知平面向量的模求向量的数量积,属基础题.关键在于熟练掌握向量的数量积的运算法则.16.在平行四边形ABCD 中,22AB =3BC =,且2cos A =,以BD 为折痕,将BDC ∆折起,使点C 到达点E 处,且满足AE AD =,则三棱锥E ABD -的外接球的半径为_________. 【答案】132【分析】由题意可得BD BC AD ==,折起后可得三棱锥的对棱相等,放在长方体中,设长方体的长宽高,根据长方体的对角线长等于球的直径,即可求得球的半径. 【详解】在ABD △中,由22AB =3BC =,且2cos 3A =,平行四边形中,可得BC AD =,由余弦定理可得2222cos BD AB AD AB AD A =+-⋅, 即(22223BD =+-22239=,解得3BD =, 折起后,AE AD =,可得3AE BD ==,3AD BE ==,且22AB ED == 所以三棱锥的三组对棱长相等,可将四面体ABED 放在长方体中,如图所示, 设长方体的相邻三棱长分别为,,x y z ,外接球半径为R ,则222222998x yy zz x⎧+=⎪+=⎨⎪+=⎩,则22213x y z++=,即213R=,所以132R=,所以四面体E ABD-外接球的半径为132.故答案为:132.【点睛】与球有关的几何体的内切与外接问题的求解策略:1、由球的定义确定球心:如果是内切球,球心到切点的距离相等且为半径;如果是外接球,球心到接点的距离相等且为半径;2、构造长方体或正方体确定球心或半径:如:三条侧棱两两垂直的正三棱锥,四个面都是直角三角形的三棱锥,三组对棱相等的三棱锥,三个侧面两两垂直的三棱锥等,都可放置在一个长方体或正方体内,结合长方体或正方体的性质求解.四、解答题17.在①7c=,1cos7A=-,②1cos8A=,9cos16B=.这两个条件中任选一个,补充在下面问题中:在ABC∆中,它的内角A,B,C 的对边分别为a,b,c,已知11a b+=,.求a,b的值.【答案】答案见解析.【分析】若选①给出角A的余弦值,可用余弦定理求解;若选②,给出两个角的余弦,可求出对应角的正弦,用正弦定理可求解.【详解】选择条件①7c=,1cos7A=-,11a b+=,2222cosa b c bc A=+-∴()222117a a=-+-()121177a⎛⎫-⋅⋅-⎪⎝⎭,∴8a=,3b=选择条件②1cos8A=,9cos16B=,A,B()0,π∈,∴sin8A==,sin16B==由正弦定理得:sin sina bA B=,∴816=∴6a=,5b=.【点睛】本题考查正余弦定理解三角形,属于中档题方法点睛:(1)若给出一个角的余弦值和边长,可用余弦求解;(2)若给两个角的正弦或余弦,应转化为两角的正弦,应用正弦定理求解. 18.已知等差数列{}n a的前n项和为n S,且11a=,515S=,公比大于1的等比数列{}nb满足:2420b b+=,38b=.(1)求n a,n b;(2)令n n nc a b=⋅,求数列{}n c的前n项和n T.【答案】(1)n a n=,2nnb=;(2)()1122nnT n+=-⋅+.【分析】(1)利用等差数列和等比数列的性质,直接设出基本量,列方程求解即可;(2)利用错位相减求和法求解即可【详解】(1)等差数列{}n a的前n项和为n S,且11a=,515S=,设公差为d∴53515S a==,得33a=,∴()3112a ad-==,得n a n=;对于公比大于1的等比数列{}n b,由2420b b+=,38b=,设公比为q得32438820bb b b q qq q+=+=+=,化简得,282080q q-+=,整理得(2)(21)0q q--=,得2q或12q=,又因为公比大于1,所以2q,则312824bbq===,所以,2nnb=(2)由(1)得n a n=,2nnb=,2nn n nc a b n=⋅=⋅,23222322nnT n=+⋅+⋅++⋅,①23412222322nnT n+=+⋅+⋅++⋅,②-①②得,231112(222)2222n n n nnT n n+++-=++++-⋅=--⋅;则()1122n n T n +=-⋅+【点睛】关键点睛:通过等差等比性质列方程求出通项n a ,n b ;然后,利用错位相减求和法求解,主要考查学生的运算能力,属于中档题 19.函数()26cos3sin 3(0)2xf x x ωωω=+->在一个周期内的图象如图所示,A 为图象的最高点,B 、C 为图象与x 轴的交点,且ABC ∆为正三角形.(1)求ω的值及函数()f x 的值域; (2)若()0835f x =,且0102,33x ⎛⎫∈- ⎪⎝⎭,求()01f x +的值.【答案】(2)ω4π=,函数的值域为23,23⎡-⎣;(2)65.【分析】(1)将函数()f x 化简整理,根据正三角形ABC 的高为23ω,进而可得其值域;(2)由()083f x 0πx π4 sin 435⎛⎫ ⎪⎝⎭+=,再由010233x ⎛⎫∈ ⎪⎝⎭-,求出0cos 43xππ⎛⎫ ⎪⎝⎭+,进而可求出结果.【详解】(1)由已知可得()2633332xf x cos sin x cos x sin x ωωωω+-==233sin x πω⎛⎫=+ ⎪⎝⎭,又正三角形ABC 的高为23BC 4=, 所以函数()f x 的最小正周期428T ⨯==,即28πω=,得ω4π=,函数()f x 的值域为2323⎡⎣-,.(2)因为()083f x (1)得()00πxπ8323sin 435f x ⎛⎫ ⎪⎝⎭=+=, 即0πx π4sin 435⎛⎫⎪⎝⎭+=, 由010233x ⎛⎫∈ ⎪⎝⎭-,,得0,4322x ππππ⎛⎫+∈ ⎪⎝⎭-, 即20πx π4cos 1435⎛⎫⎛⎫- ⎪⎪⎝⎭⎝⎭+==35, 故()00πx ππ123sin 443f x ⎛⎫+⎪⎝⎭=++ 0πx ππ23sin 434⎡⎤⎛⎫ ⎪⎢⎥⎝⎭⎣⎦=++0023sin cos cos sin 434434x xππππππ⎡⎤⎛⎫⎛⎫= ⎪ ⎪⎢⎥⎝⎭⎝⎭⎦+++4232762355⎛⎫=⨯⨯⨯= ⎪ ⎪⎝⎭+. 【点睛】本题主要考查三角函数的图象和性质,熟记正弦函数的性质即可求解,属于基础题型.20.如图,已知四棱锥的侧棱PD ⊥底面ABCD ,且底面ABCD 是直角梯形,AD ⊥CD ,AB ∥CD ,AB=AD=12CD=2,点M 在侧棱上.(1)求证:BC ⊥平面BDP ;(2)若侧棱PC 与底面ABCD 所成角的正切值为12,点M 为侧棱PC 的中点,求异面直线BM 与PA 所成角的余弦值. 【答案】(1)证明见解析;(2310【解析】试题分析:(1)证明,BD BC PD BC ⊥⊥,即可证明BC ⊥平面BDP ;(2)取PD 中单为N ,并连结,AN MN ,则PAN ∠即异面直线BM 与PA 所成的角,在PAN ∆中,利用余弦定理,即求出一年直线BM 与PA 所成角的余弦值.试题解析:(1)证明:由已知可算得D C 22B =B =, ∴BD 2+BC 2=16=DC 2,故BD ⊥BC ,又PD ⊥平面ABCD ,BC ⊂平面ABCD ,故PD ⊥BC , 又BD∩PD=D ,所以BC ⊥平面BDP ;(2)解:如图,取PD 中点为N ,并连结AN ,MN ,BM ∥AN , 则∠PAN 即异面直线BM 与PA 所成角;又PA ⊥底面ABCD , ∴∠PCD 即为PC 与底面ABCD 所成角, 即1tan CD 2∠P =,∴1D CD 22P ==,即,又5AN =,22PA =,则在△PAN 中,222310cos 2AP +AN -PN ∠PAN ==AP ⋅AN , 即异面直线BM 与PA 所成角的余弦值为310.【解析】直线与平面垂直的判定与证明;异面直线所成的角.21.已知函数()()4log 41xf x mx =+-是偶函数,函数()42x xng x +=是奇函数. (1)求m n +的值; (2)设()()12h x f x x =+,若()()4log 21g h x h a >+⎡⎤⎡⎤⎣⎦⎣⎦对任意4log 3x ≥恒成立,求实数a 的取值范围. 【答案】(1)12m n +=-;(2)1,32⎛⎫- ⎪⎝⎭. 【分析】(1)根据偶函数的定义,求m 的值,根据奇函数若在原点有意义,则必满足()00g =,求n 的值,从而求得m n +;(2)求参数的恒成立问题转化为求最值问题,本题形如()f x a >恒成立,转化为()min f x a >恒成立,即转化为求()min f x ,从而求得a 的取值范围.【详解】(1)由()f x 是偶函数,得()()f x f x -= 即()()()()()444log 41log 411log 41xx x f x mx m x mx --=++=++-=+-化简得:()1mx m x -=-,故12m =由()g x 为奇函数,且定义域为R ,所以()00g =,即004012n n +=⇒=-,经检验,1n =-符合题意; 综上,可得12m n +=-(2)∵()()()41log 412x h x f x x =+=+,∴()()44log 21log 22h a a +=+⎡⎤⎣⎦ 又()()4log 21g h x h a >+⎡⎤⎡⎤⎣⎦⎣⎦对4log 3x ∀≥恒成立,即()()4min log 22g h x a >+⎡⎤⎣⎦对4log 3x ∀≥恒成立,下面求()min g h x ⎡⎤⎣⎦, 又()()4log 41xh x =+,在区间[)4log 3,+∞上是增函数()()4log 31h x h ∴≥=又()41222x x xxg x --==-在区间[)1,+∞上是增函数, ()()()4min 3g log 312g h x h g ∴===⎡⎤⎡⎤⎣⎦⎣⎦ 由题意,得32224122032210a a a a ⎧+<⎪⎪⎪+>⇒-<<⎨⎪+>⎪⎪⎩所以实数a 的取值范围是:1,32⎛⎫- ⎪⎝⎭.【点睛】本题考查利用函数奇偶性求参数,及函数恒成立求参数问题,在解函数恒成立问题时,往往转化为最值问题求解,考查学生的转化与化归思想与计算能力,属于中档题.22.已知函数()sin xf x e x =.(e 是自然对数的底数)(1)求()f x 的单调递减区间;(2)记()()g x f x ax =-,0<<3a ,试讨论()g x 在()0,π上的零点个数.(参考数据:2 4.8e π≈) 【答案】(1)372,244k k ππππ⎛⎫++⎪⎝⎭()k Z ∈;(2)答案见解析. 【分析】(1)先求出导数()f x ',再解()0f x '<,结合三角函数的性质可解得; (2)求出()()sin cos xg x ex x a '=+-,令()()h x g x '=,由导数的知识求得()h x 的单调性,然后通过讨论()()0,,2g g g ππ⎛⎫'''⎪⎝⎭的正负确定()g x 的单调性,确定其零点个数.【详解】解:(1)()sin xf x e x =,定义域为R .()()sin cos x f x e x x '=+sin 4x x π⎛⎫=+⎪⎝⎭. 由()0f x '<,解得sin 04x π⎛⎫+< ⎪⎝⎭,可得222()4k x k k Z πππππ+<+<+∈解得372244k x k ππππ+<<+()k Z ∈.∴()f x 的单调递减区间为372,244k k ππππ⎛⎫++⎪⎝⎭()k Z ∈. (2)由已知()sin xg x e x ax =-,∴()()sin cos xg x e x x a '=+-,令()()h x g x '=,则()2cos xh x e x '=.()0,x π∈,∴当0,2x π⎛⎫∈ ⎪⎝⎭时,()0h x '>;当,2x ππ⎛⎫∈ ⎪⎝⎭时,()0h x '<,∴()h x 在0,2π⎛⎫ ⎪⎝⎭上单调递增,在,2ππ⎛⎫⎪⎝⎭上单调递减,即()g x '在0,2π⎛⎫ ⎪⎝⎭上单调递增,在,2ππ⎛⎫⎪⎝⎭上单调递减.()01g a '=-,202g e a ππ⎛⎫'=-> ⎪⎝⎭,()0g e a ππ'=--<.①当10a -≥时,即01a <≤时,()00g '≥,∴0,2x ππ⎛⎫∃∈ ⎪⎝⎭,使得()00g x '=,∴当()00,x x ∈时,()0g x '>;当()0,x x π∈时,()0g x '<,∴()g x 在()00,x 上单调递增,()0,x π上单调递减.()00g =,∴()00g x >.又 ()0g a ππ=-<,∴由零点存在性定理可得,此时()g x 在()0,π上仅有一个零点.②若13a <<时,()010g a '=-<,又()g x '在0,2π⎛⎫ ⎪⎝⎭上单调递增,在,2ππ⎛⎫⎪⎝⎭上单调递减,而202g e a ππ⎛⎫'=-> ⎪⎝⎭, ∴10,2x π⎛⎫∃∈ ⎪⎝⎭,2,2x ππ⎛⎫∈ ⎪⎝⎭,使得()10g x '=,()20g x '=,且当()10,x x ∈、()2,x x π∈时,()0g x '<;当()12,x x x ∈时,()0g x '>.∴()g x 在()10,x 和()2,x π上单调递减,在()12,x x 上单调递增.()00g =,∴()10g x <.2230222g e a e πππππ⎛⎫=->-> ⎪⎝⎭,∴()20g x >.又()0g a ππ=-<,由零点存在性定理可得,()g x 在()12,x x 和()2,x π内各有一个零点,即此时()g x 在()0,π上有两个零点.综上所述,当01a <≤时,()g x 在()0,π上仅有一个零点;当13a <<时,()g x 在()0,π上有两个零点.【点睛】关键点睛:本题考查利用导数判断函数的单调性、求极值等问题,考查转化思想、分类讨论思想的综合应用,涉及构造函数、多次求导等方法,本题的关键是对()0g '的正负性的讨论,以及零点存在原理的应用.。
2020-2021学年湖南省常德市高三(上)月考化学试卷(Word+答案)
2020-2021学年湖南省常德市高三(上)月考化学试卷一、选择题1.在无色透明强酸性溶液中,能大量共存的离子组是()A.K+、MnO4﹣、SO42﹣B.Na+、Cl﹣、CO32﹣C.Zn2+、Al3+、Cl﹣D.Na+、Fe2+、SO42﹣2.在常温、常压下,取下列四种气态烃质量相等,分别在足量的氧气中燃烧,消耗氧气最多的是()A.C4H10B.C3H8C.C2H6D.CH43.设N A表示阿伏加德罗常数的值,下列叙述正确的是()A.100g 46%甲酸(HCOOH)水溶液中所含的氧原子为5N AB.标准状况下,11.2L Cl2溶于水,转移电子数为N AC.2g氢气所含原子数目为N AD.100mL 18.4mol•L﹣1浓硫酸与足量铜加热反应,生成SO2的分子数为0.92N A4.浓度为0.50mol/L的某金属阳离子M n+的溶液10.00mL,与0.80mol/L的NaOH溶液12.50mL完全反应,生成沉淀,则n等于()A.1B.2C.3D.45.某溶液中有S2﹣、SO32﹣、Br﹣、I﹣四种阴离子各0.1mol.现通入Cl2,则通入Cl2的体积(标准状况)和溶液中相关离子的物质的量的关系图正确的是()A.B.C.D.6.配制0.1mol•L﹣1NaCl溶液不需要用到的仪器是()A.B.C.D.7.如图表示一些物质或概念间的从属关系,下表中不能满足如图关系的是()X Y Z如金属单质单质纯净物A苯的同系物芳香烃芳香族化合物B胶体分散系混合物C电解质离子化合物化合物D碱性氧化物金属氧化物氧化物A.A B.B C.C D.D8.在给定条件下,下列选项所示的物质间转化均能实现的是()A.NaHCO3(s)Na2CO3(s)NaOH(aq)B.Al(s)NaAlO2(aq)Al(OH)3(s)C.AgNO3(aq)[Ag(NH3)2]+(aq)Ag(s)D.Fe2O3(s)Fe(s)FeCl3(aq)9.将5mol•L﹣1的盐酸10mL稀释到200mL,再取出5mL,这5mL溶液的物质的量浓度是()A.0.05 mol•L﹣1B.0.25 mol•L﹣1C.0.10 mol•L﹣1D.0.50 mol•L﹣110.在t℃时,将agNH3完全溶于水,得到V mL溶液,假设该溶液的密度为ρg•cm﹣1,质量分数为ω,其中含NH4+的物质的量为b moL.下列叙述中正确的是()A.溶质的质量分数为ω=×100%B.溶质的物质的量浓度c=mol•L﹣1C.溶液中c(OH﹣)=mol•L﹣1+c(H+)D.上述溶液中再加入VmL水后,所得溶液的质量分数大于0.5ω11.等质量的CO2和SO2相比较,下列结论正确的是()A.它们的分子数目之比是11:16B.它们的氧原子数目之比为2:3C.它们的分子数目之比为16:11D.它们所含原子数目之比为16:1112.下列说法正确的是()A.2﹣丁醇发生消去反应产物有2 种B.卤代烃都能发生水解反应C.醇在Cu/△条件下都能氧化成醛D.在酸性条件下水解产物是和C2H5OH13.下列关于指定粒子构成的几种描述中正确的是()A.614C与612C是不同的核素,所以由这两种原子构成的石墨化学性质不同B.H2O和D2O是同分异构体C.H3O+与OH﹣具有相同的电子数D.37Cl与39K具有相同的中子数14.下列解释事实的方程式正确的是()A.用FeCl3溶液制作铜质印刷线路板:2Fe3++Cu═Cu2++2Fe2+B.Al片溶于NaOH溶液中产生气体:2Al+2OH﹣═2AlO2﹣+H2↑C.用难溶的MnS除去MnCl2溶液中含有的Pb2+:MnS(s)+Pb2+(aq)═PbS(s)+Mn2+(aq)D.向银氨溶液中滴加乙醛后水浴加热,出现银镜:CH3CHO+Ag(NH3)2OH CH3COONH4+Ag↓+3NH3+H2O 15.水热法制备Fe3O4纳米颗粒的反应:3Fe2++2S2O32﹣+O2+xOH﹣═Fe3O4+S4O62﹣+2H2O,下列说法正确的是()A.反应方程式中x=4B.此反应中,氧化剂为O2,将Fe元素全部氧化C.每生成1 mol Fe3O4,该反应共转移电子4 molD.反应中S元素的价态未发生变化二、解答题16.已知A、B、C为中学化学中常见的单质。
精品解析:湖南省常德市第一中学2021届高三第三次月水平检测化学试题(解析版)
常德市一中2021届高三第三次月水平检测试卷化学时量:90分钟满分:100分可能用到的相对原子质量:H 1 C 12 N 14 O 16 S 32 Fe 56 Cl 35.5 Mn 55一、选择题:本题共15小题,每小题2分,共30分。
每小题只有一个选项符合题意。
1. 面对突如其来的新型冠状病毒,越来越多人意识到口罩和医用酒精的重要作用,医用口罩由三层无纺布制成,无纺布的主要原料是聚丙烯树脂。
下列说法错误..的是A. 医用口罩能有效预防新型冠状病毒传染B. 聚丙烯树脂属于合成有机高分子材料C. 医用酒精中乙醇的体积分数为95%D. 抗病毒疫苗冷藏存放的目的是避免蛋白质变性【答案】C【解析】【详解】A.医用口罩能够有效阻挡病毒的进入人体感染,故A不符合题意;B.聚丙烯树脂为合成树脂材料,属于合成有机高分子材料,故B不符合题意;C.医用酒精中乙醇的体积分数为75%,故C符合题意;D.疫苗中主要成分为蛋白质,若温度过高,蛋白质会发生变性,故D不符合题意;故答案为:C。
2. 下列有关化学用语的表示正确的是A. 医用“钡餐”的化学式:BaCO3B. C1-的结构示意图:C. NaHCO3在水中的电离方程式:NaHCO3=Na++H++CO32-D. N2的电子式:【答案】D【解析】【详解】A.医用“钡餐”为硫酸钡,化学式为BaSO4,故A错误;B.从Cl原子到C1-,质子数不变,所以C1-的结构示意图为,故B错误;C.NaHCO 3在水中的电离方程式为NaHCO3=Na++HCO3-,HCO3-H++CO32-,故C错误;D.N2的电子式为,故D正确;故选D。
3. 化学与生产生活密切相关,下列有关说法不正确的是A. 大量使用含磷洗衣粉会造成生活污水的富营养化B. 用未经处理的电镀厂废水灌溉农田,易造成土壤重金属污染C. 臭氧-生物活性炭用于自来水深度处理,利用了活性炭的还原性D. 纳米铁粉可将地下水中的NO3-转化为N2,是因为纳米铁具有还原性【答案】C【解析】【详解】A.含大量氮磷的废水能造成水体富营养化,故A正确;B.电镀厂废水中含有重金属离子,用未经处理的电镀厂废水灌溉农田,易造成土壤重金属污染,故B正确;C.臭氧-生物活性炭用于自来水深度处理,利用了臭氧的强氧化性和活性炭的吸附性,故C错误;D.NO3-转化为N2化合价降低,作氧化剂,则需要还原剂,所以纳米铁粉作还原剂,具有还原性,故D正确;答案选C。
湖南省常德市一中2021届高三上学期第三次月考水平检测考试数学试题参考答案
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2025届湖南省常德市第一中学高三第三次模拟考试语文试卷含解析
2025届湖南省常德市第一中学高三第三次模拟考试语文试卷注意事项:1.答卷前,考生务必将自己的姓名、准考证号、考场号和座位号填写在试题卷和答题卡上。
用2B铅笔将试卷类型(B)填涂在答题卡相应位置上。
将条形码粘贴在答题卡右上角"条形码粘贴处"。
2.作答选择题时,选出每小题答案后,用2B铅笔把答题卡上对应题目选项的答案信息点涂黑;如需改动,用橡皮擦干净后,再选涂其他答案。
答案不能答在试题卷上。
3.非选择题必须用黑色字迹的钢笔或签字笔作答,答案必须写在答题卡各题目指定区域内相应位置上;如需改动,先划掉原来的答案,然后再写上新答案;不准使用铅笔和涂改液。
不按以上要求作答无效。
4.考生必须保证答题卡的整洁。
考试结束后,请将本试卷和答题卡一并交回。
1、阅读下面的文字,完成下面小题。
兵爷的心愿柴双政嫩娃谈了三个女朋友兵爷都不满意。
嫩娃是兵爷的亲孙子。
兵爷是嫩娃对爷爷的敬称。
爷爷是抗战老兵,身高一米九零,即使活到九十五岁,拄着拐杖走路也是腰板挺直。
嫩娃把爷爷称兵爷,是调侃中满含敬佩。
嫩娃是兵爷对孙子的爱称。
嫩娃是名牌大学学生,学计算机,身高一米六五,走路拖泥带水。
兵爷把孙子叫嫩娃,是疼爱中略有不满。
大二开始,嫩娃就陆陆续续往家里领女朋友。
领回家,先直奔爷爷房间让兵爷过目。
兵爷总会把嫩蛙的女朋友上上下下左左右右打量三五个来回,然后并不说话,而是用见面礼的金额表达自己的判分。
嫩娃领回的第一个是四川姑娘,细胳膊细腿,小巧玲珑。
嫩娃用不断搓手的肢体语言表达自己的满意。
但是兵爷的见面礼是十元!嫩娃领回的第二个是江浙姑娘,吴侬软语,莺莺燕燕。
嫩娃用佝偻着腰,围着姑娘不停地转圈的夸张方式向兵爷宣示自己的欢心.可是爷爷的见面礼依然是十元!嫩娃领回的第三个是东北姑娘,一听到姑娘的家乡,兵爷一个鲤鱼打挺就从床上弹了起来,可一打量那姑娘,兵爷就不停地咳嗽。
见面礼就当然还是十元.这次,嫩娃不再像前两次那样调侃着问,兵爷,这个妹纸肿么了?大大的打赏有木有?而是奇怪了,结巴了,愤怒了:“兵!兵!兵爷啊,你还不满意啊,她可是我们的班花,我们全班都说她长得像章子怡!”张什么?兵爷没有听懂。
2021年常德市一中高三英语第三次联考试题及参考答案
2021年常德市一中高三英语第三次联考试题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ABook reading is certainly one of the most absorbing habits. For young adults who love to read, finding some good books to read is very essential. Writing a book review can help you to improve your language and writing skills.The Book ThiefListed onThe New York Times Children’s Best Seller List for over 100 weeks, The Book Thief by Markus Zusak is the story of a young girl in the Nazi camps set during World War II. So, if you love history and wish to learn how the life was during Adolf Hitler’s time, read this historic book.The Diary of Young GirlEven Anne Frank can not have imagined that her personal diary written during World War II would become such a popular book. It’s a must read that describes the situation of a family in the evils of wars through the eyes of a teenager.Animal FarmAnimal Farm is one of the most popular books by George Orwell. It is just a reflection of the Stalin and World War II period that has been so creatively presented in this book. It is an interesting example of how literature can be used to present conditions common in the society.Adventures of Huckleberry FinnMark Twain’s Adventures of Huckleberry Finn is one of the great American novels in history, and is certainly a great pick for young adults. Young Huck Finn and his mischief along with the color1 ful description of people around theMississippi Rivermake this novel a great book to read.1.Which book describes the author’s own experiences according to this passage?A.The Book ThiefB.The Diary of Young GirlC.Animal FarmD.Adventures of Huckleberry Finn2.What do the first three books have in common?A.All of them are about wars.B.All of them are about farms.C.All of them are intended for history lovers.D.All of them were written during World War II.3.The purpose of this passage is to _________.A.instruct youngsters how to improve skillsB.tell youngsters some wonderful reading habitsC.introduce several good books to youngstersD.give youngsters advice on writing a book reviewBTOKYO—Japanese Prime Minister Yoshihide Suga told the media on Monday if any places hosting events of the Tokyo Olympics and Paralympics declare a state of emergency due to the COVID-19 epidemic during the games, the events will continue to beheld but without spectators (观众). With one month to go before the games are due to begin on July 23, Suga is again showing his administration’s determination to hold the Olympic Games as planned, despite so much pressure from various parties urging it tocancel the event.Although the Japanese government regards the Tokyo Olympics as an important opportunity to improve its soft power, the Japanese people’s enthusiasm for the Games has been continuously dented (挫伤) since they were postponed last year. The resurgence (再猖獗) of the novel coronavirus in some places is Japan in recent months has cast a shadow over people’s confidence that the Olympics will not give rise to new clusters (群) of infections, and there are fears that the Games will provide new channels for the virus’ global transmission.Some torchbearers from Japan have withdrawn from the Olympic torch relay in the country. And the latest survey indicates only 34 percent of Japanese people support holding the games as scheduled. Predictably, the Suga administration will do all it can to try to ensure the games go ahead. But it remains to be seen whether it can stand the tests of the uncertainties related to epidemic prevention and control that might happen during the Games.Since it has not yet got the virus under control at home, the people have reasons to question is ability to deal with the prevention and control work when large numbers of participants will be flocking to Japan from around the world in a short time. It is to be hoped that Japan can draw lessons from the organization of epidemic prevention and control work during the ongoing UEFA European Championship, carry out strict epidemic prevention and control measures, and be prepared for emergencies to guarantee the safety and success of the Olympics at this special time.It should be a common wish of the whole world that the Tokyo Olympics can become a stage showing unityand resolve of human beings in their fight against the virus. That willendowthe games with special meaning beyond sports.4. What is the second paragraph mainly about?A. The virus’ global transmission.B. People’ worry about the infections.C. The resurgence of the novel coronavirus.D. The benefit of holding the Tokyo Olympics.5. How do about one third of Japanese people like holding the games as planned?A. Uncertain.B. Negative.C. Approving.D. Indifferent.6. Which of the following words can replace the underlined word “endow” in the last paragraph?A. Compare.B. Equip.C. Provide.D. Charge.7. What can be the best title for the news report?A. Japan can ensure Olympics go aheadB. Olympics big test for Japanese governmentC. Japanese people’s enthusiasm for the GamesD. Japan to carry out strict epidemic prevention during the GamesCSleep problems in early childhood may be linked to the development of certain mental health disorders in adolescence, according to a new research.A study of 7,155 children in theUnited Kingdomfound that waking up frequently during the night and irregular sleep routines as babies and toddlers was linked to psychotic experiences in children aged 12 and 13. Also, children who slept for shorter periods at night were more likely to be associated with borderline personality disorder at ages 11 and 12.The research, published in the journalJAMA Psychiatry, was the first time possible links between early childhood sleep problems and adolescent psychotic experiences and borderline personality disorder (BPD) symptoms have been examined.“We know from previous research that persistent nightmares in children have been associated with both psychosis and borderline personality disorder,” said lead author Isabel at theInstituteofMental Healthat theUniversityofBirmingham.“But nightmares don’t tell the whole story. We’ve found that, in fact, a number of behavioral sleep problemsin childhood can point towards these problems in adolescence,” she said.Adolescence, typically defined as the ages between 10 and 19, is a key period in human development because of brain and hormonal changes, anditis now thought to be when many mental health problems start.Prior research inAustraliafound that babies with persistent severe sleep problems in their first year were at greater risk for anxiety and emotional issues in later childhood. Sleep problems in children and adolescents have been shown to predict the development of various emotional and behavioral problems, including depression, anxiety, attention-deficit hyperactivity disorder, risk-taking and aggression. However, findings have been inconsistent, especially when based on objective measurements of sleep, rather than parental reports.Sleep and mental health are closely connected in adults, with sleep problems increasing the risk for developing particular mental illnesses as well as resulting in mental health issues.8. What can we know from paragraph 3?A. Childhood sleep problems probably cause youth mental health problems.B. Severe sleep problems in childhood lead to emotional issues.C. Persistent nightmares in children are closely related to psychosis.D. The research has not been known to the public.9. What does “it” refer to in paragraph 6?A. Adolescence.B. Key period.C. Human developmentD. Hormonal change.10. According to the Australian research, older children who had long-term serious sleep issues at one year old are more likely to ________.A. grow slowlyB. develop attention-deficit hyperactivity disorderC. do badly in lessonsD. suffer from anxiety11. Which of the following can be the best title for the text?A. A Study About Childhood Sleep Problems Is Under ProgressB. Pay Attention to Children Who Lack SleepC. Childhood Sleep Issues Linked to Adolescent Mental Health ProblemsD. How to Help Children Develop a Good Sleep HabitDIf you struggle to fall asleep quickly, you’re not alone! Fortunately, thereare plenty of solutions you can try. With a few changes, you can fall asleep fast every night!Keep your room dark. Turn off all the lights above your head when you go to bed. Any bright light can make you believe it is too early in the day for sleep. If you want to read or write before bed, try using a small book light. Now that blue lights can keep you awake, red ones are a great choice.If you can, keep noise in and around your room the lowest at night. If you have an old clock that ticks loudly and keeps you awake, replace it with a silent one. If you share your home with anyone else, request that they keep noises like talking, music, or TV shows at the lowest while you are trying to sleep. It is difficult to fall asleep if you live near a busy road or hear other boring sounds after bedtime. You could get a white noise machine or play recordings of nature sounds, like waves or whales’ singing. You could also listen to soft, relaxing music.Read a book in bed if you have difficulty in falling asleep. Staying in bed doing nothing when you’re having trouble falling asleep may keep you wide awake. While reading in bed may be slightly harmful to your eyes, it can distract (分散) you from your thoughts and help you feel sleepy. But remember to read from a print book rather than something with a screen. The light from electronic screens can keep you awake.Lowering your body temperature helps you sleep, so set the room temperature between 15.5℃-21℃could do the trick.12. What color1 book light should you choose toread before bed?A. Red.B. Blue.C. White.D. Orange.13. What is the author’s attitude towards reading in bed?A. Doubtful.B. Worried.C. Favorable.D. Uncaring.14. What can we inferred from the text?A. Reading on cellphones sometimes helps you fall asleep.B. The lower your temperature while sleeping is, the better.C. Bright lights are better for your reading before going to bed.D. Playing recordings like birds’ singing can improve your sleep.15. How does the author organize the text?A. By givingexamples.B. By asking questions.C. By offering suggestions.D. By listing research results.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年常德市一中高三英语第三次联考试卷及答案
2021年常德市一中高三英语第三次联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AThese wonderful NYC attractions offer pay — what — you — wish days, free entry hours/days and other great stuff for local families.Staten IslandZooThere are plenty of creatures who call NYC home—the Staten Island Zoo is one of them. Once you’ve finished learning about the wildlife in the animal nursery, reptile (爬行动物) side rooms, horse barn and other areas of the attraction, make sure to mark your schedule for fun seasonal happenings, such as the Easter Egg Games and the scary, crazy Halloween Shows.Entry on Wednesdays is by suggested donation; children aged two and under free.Children’s Museum of the ArtsThe Children’s Museum of the Arts welcomes 135,000 little visitors each year through its doors. Once inside, the whole family can enjoy interactive programs, exhibitions (展览) and events that celebrate the changed power of the arts on youngsters and grown-ups alike.Pay-as-you-wish Thursdays, 3-6 p. m.Wave HillEveryone needs a few hours of calm now and then-kids included-and you’d be hard- pressed to find a more peaceful spot within city limits than Wave Hill the broad grounds located above the river, covering 28 acres of public gardens, plus woodlands and grasses to wander. Jump in on nature walks, story times and family art projects often led by local artists and free with general admission.Pay — as — you — wish Tuesdays and Saturdays,9 a. m — noon.New York Hall of ScienceNaturally, kids love it when the New York Hall of Science pleases them with neat exhibits and fun hands-on activities. The museum’s playground is themost attractivetochildren A tube slide (管道滑梯) will give little ones the knowledge on science topics, while the climbing area mirrors a giant spider web. There are also wind pipes, metal drums, sand- boxes and much more. What better way to make the mostout of science?Free entry Sep-Jun on Fridays, 2 — 5 p. m,and Sundays, 10 —11 a. m.1. What can children do in Staten Island Zoo?A. Feed injured animals.B. Join in seasonal activities.C. Build a home for creatures.D. Deal with the donations to the zoo.2. What do Children’s Museum of the Arts and Wave Hill have in common?A. They both have peaceful spots.B. They both are located by a river.C. They both have public gardens.D. They both have activities about arts.3. Which place can be free of charge for all?A. Wave Hill.B.Staten IslandZoo.C. New York Hall of Science.D. Children’s Museum of the Arts.BIt all happened one afternoon in Carl's backyard a few years ago. We had just finished playing stickball, and I was about to go home.“Wait a minute,” Carl yelled. He ran into his house and cameback with a book for me to take home and read. All he said was, “See if you like it.”I said ly nothing. I kept the book for a couple of weeks and then returned it unread. Carl never asked me if I liked it or not. During the following two years Carl lent me three more books. Each time I returned them unread.A few days after graduating from high school, Carl asked, “Benny, which college are you going to?”“I'm not going to college.” I said.“Why not?” he asked.“Because my father can't afford the tuition (学费).” I answered.“Is that it?” Carl asked. “Yes,” I said.I lied. I had no intention of going back to school now that I was out. The following day, Carl knocked on my door and handed me a check for seventy-five dollars from his father along with the bookMartin Eden.“I think that should do it.” he said.Once again I was in shock. I was working full-time in my brother's bakery. I attended two classes atWayneUniversitypart-time at night. Halfway through the semester, after receiving failing grades on exams and essays, I decided there was no way I would ever become a good student and get satisfactory grades. I dropped out of college.One day, curious, I picked up the book, thinking Carl was trying to tell me something. Despite difficulty, I pressed on. By the time I finished the book, I understood why: the main character, Martin Eden, had my own poor educational background, but managed to educate himself and become a published author.4. What do we know about the author?A. He often told lies.B. He quit school unwillingly.C. He had thought little of education before.D. He became a published author.5. What kind of person is Carl?A. Caring.B. Emotional.C. Stubborn.D. Cautious.6. What message does Carl want to convey?A. Reading makes a rich man.B. Reading is the journey of the soul.C. Reading makes a person better known.D.Readingopens up new opportunities.7. What is the best title for the text?A. A strong desire for collegeB. A wish for better educationC. A wise friendD. A wise bookCWhat a day! I started at my new school this morning and had the best time. I made lots of new friends and really liked my teachers. I was nervous the night before, but I had no reason to be. Everyone was so friendly and polite. They made me feel at ease. It was like I'd been at the school for a hundred years!The day started very early at 7:00 am. I had my breakfast downstairs with my mom. She could tell that I was very nervous. Mom kept asking me what was wrong. She told me I had nothing to worry about and that everyone was going to love me. If they didn't love me, Mom said to send them her way for a good talking to. I couldn't stoplaughing.My mom dropped me off at the school gates about five minutes before the bell. A little blonde girl got dropped off at the same time and started waving at me. She ran over and told me her name was Abigail. She was very nice and we became close straight away. We spent all morning together and began to talk to another girl called Stacey. The three of us sat together in class all day and we even made our way home together! It went so quickly. Our teacher told us that tomorrow we would really start learning and developing new skills.I cannot wait until tomorrow and feel as though I am really going to enjoy my time at my new school. I only hope that my new friends feel the same way too.8. How did the author feel the night before her new school?A. Tired.B. ConfidentC. Worried.D. homesick9. What did the author think of her mother’s advice?A. Clear.B. Funny.C. OptionalD. Respectable10. What happened on the author's first day of school?A. She met many nice people.B. She had a hurried breakfast.C. She learned tome new skills.D. She arrived at school very early.11. What can we infer about Abigail?A. She disliked Stacey.B. She was shy and quiet.C. She got on well with the author.D. She was an old friend of the author.DA dog spentthe lastfour years of his life waitingat a crossroad in the Thai city ofKhon Kaenas if waiting for someone. People originally thought the dog had been abandoned, but then realized that he looked healthy, so people asked around about him. It turned out that the dog had indeed been spending most of his time around that crossroad, but a woman had been coming round regularly to bring him food and water.One day, while photographing the dog everyone called Leo, a reporter met the woman who had been taking care of him. She had come to drop off some food. After learning the story about the dog and the woman, the reporter decided to share the story on social media. The post soonwent viraland the photos of Leo got shared hundreds of times. And the photos reached the eyes of Leo’s former old owner.Nang Noi Sittisarn, a 64-year-old woman fromThailand’sRoiEtProvince, almost had a heart attack when her daughter showed her a photo of the beloved dog named BonBon she had lost during a car trip. When she learned that he had been waiting for her in the same spot for the last four years,her heart melted(融化).Auntie Noi told her daughter to drive her to where the dog was waiting. When she got there and called his name. BonBon,the poor dog started wiggling(扭动)his tailand came to her,but when she tried to take him home with her, he was unwilling to follow. She didn’ t want to force the dog to come with her so she agreed to leave him with his new master. However, she and her daughter will come to visit him regularly.12.Why did the dog look healthy after separation from his former owner?A. He walked around the crossroad constantly.B. He was kept at a woman’s home all the time.C. A local reporter brought him food and water.D. A woman looked after him on a regular basis.13. What does the underlined phrase “went viral” in paragraph 2 probably mean?A. Changed surprisingly.B. Spread quickly.C. Appeared gradually.D. Fell directly.14. How did Nang Noi Sittisam feel about the dog's waiting for her?A. Shocked.B. Regretful.C. Touched.D. Proud.15. What can we mainly learn from the story about the dog?A. Unbelievable success is worth waiting for.B. We should adjust ourselves to environments.C. We need to learn to be faithful and thankful.D. No one knows the result until the last minute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年常德市一中高三英语三模试卷及参考答案
2021年常德市一中高三英语三模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AConservation Volunteering in New ZealandWhether you are a student, professional or a retiree (退休者), anyone is welcome to make a difference and contribute to protecting some of the most beautiful islands in the world. Choose a suitable city and travelout to your conservation (保护) site to work with local people!Duration: 1-12 weeks Dates: Throughout the yearArrival day: Friday Return day: FridayRequirement: General level of fitness Age: 18+What will I be doing?Volunteer in New Zealand and enjoy conserving the environment through activities such as:·Tree planting·Walking trail construction·Protect native birds, insects, fish and penguins·Seed collection·Weed controlYou, and a group of up to 10 volunteers, will work under the guidance of a conservation team leader. Your team leader will give you regular safety instructions, inform you of the project aims and assist you with working effectively.No previous experience is necessary to join the project. All you need is a love of the environment and a fairly good level of fitness to help out!1.Who can sign up for this conservation volunteering project?A.A retired maths teacher.B.A primary school student.C.A scientist with heart disease.D.A businessman in a wheelchair.2.What can you do on the volunteer trip?A.Protect cultural sites and go shopping.B.Enjoy local sightseeing and go fishing.C.Protect weeds and build roads.D.Collect seeds and plant trees.3.From which is the text probably taken?A.A history book.B.A travel magazine.C.A research paper.D.A novel.BOne weekend I went toBuffaloto talk at a writers' conference organized by a group of women writers. The women were serious about their writing skills, and the articles they had written were solid and useful. They asked me to take part in a radio talk show earlier in the week to publicize the conference-they would be with the host in the studio and I would be on a telephone linking from my apartment inNew York.The appointed evening arrived, and my phone rang, and the host came on and greeted me. He said he had three lovely ladies in the studio with him and he was eager to find out what we all thought of the present state of literature and what advice we had for all his listeners who were members of the literati and had literary ambitions themselves.This hearty introduction dropped like a stone among us, and none of the three lovely ladies said anything, which I thought was the proper response.The silence lengthened, and finally I said, “I think we should stop mentioning the words literature and literary and literati. We're here to talk about the skills of writing.” Iknew that the host had been given information about what kind of writers we were and what we wanted to discuss. But he had no other preparation. "Tell me what insights do you have about the literary experience inAmericatoday?” Silence also greeted this question.He didn’t know what to do with that, and he began to mention the names of authors like Ernest Hemingway and Saul Bellow and William Styron, whom we surely regarded as literary giants. We said those writers didn't happen to be our models, and we mentioned people like Lewis Thomas and Joan Didion and Gary Wills, whom he hadn't heard of. We explained that these were writers we admired. “But don't you want to write anything literary?” our host said We were speechless.It was one of the all-time upset radio talk shows.4. What do we know about the talk show?A. It was organized by women writers.B. It was publicized at the conference.C. The author went toBuffaloto take part in it.D. The author participated in it inNew York.5. What does the underlined sentence in paragraph 2 mean?A. The introduction struck us heavily with a stone.B. The introduction received embarrassing response.C. The introduction increased the listeners' interest.D. The introduction carried the host's praise for us.6. What was the author's reaction when the host mentioned the three great literary- giants?A. Excited.B. Inspired.C. Uninterested.D. Satisfied.7. Who may be the author's model?A. Joan Didion.B. Ernest Hemingway.C. Saul Bellow.D. William Styron.CAt the foot of the Tianmu Mountain in Zhejiang, a homestay (民宿) is attracting travelers from far and wide, which has won architectural (建筑学的) medal at the 2021 German iF Design Awards.The owners of the homestay are a couple in their late 30s who decided to return to their hometown three years ago. Li Xiumei used to be in charge of a division at a company in Hangzhou, and her husband was a sales director. It was an ordinary situation where Li’s husband was on business trips a lot and Li worked overtime on weekends. City life sometimes is not easy.In 2018, they quit jobs and went back to Dongtianmu village, which lies in a forest of bamboo. The first time they drove into the village was one late afternoon. The cooking smoke was rising from the foot of the mountain, which gave them a very different feeling form thecity.The homestay was built beside her husband’s old countryside house. The old house is preserved (保留), while a brand-new building was built on its side and the whole site is made up of for courtyards. It has been updated to have a hall, a tea room, a kitchen, a dining room. Japanese cherry trees are planted in the east courtyard. A swimming pool is placed in the west courtyard, with a bar located on one side.Li and her husband love gardening and music, and their new home gives them enough space to continue their interests and relax in the heart of nature. Li wants to share the quiet country life, so she makes her new home a homestay. In 2019, the homestay became an online hit after guests shared their experiences on social media. “The longer I stay here, the more I feel it was the right choice to come back, and this is more meaningful than making money,” Li says.8. How did Li feel about city life?A. Satisfied.B. Tired.C. Attractive.D. Noisy.9. What impressed the couple when first driving to the village?A. The smoke of cooking.B. The forest of bamboo.C. The smell of the village.D. The feeling of loneliness.10. What can we infer about the homestay from paragraph 4?A. It is ancient and broken.B. It can hold many guests.C. It has been rebuilt bythe couple.D. It must have been carefully designed.11. What’s more meaningful than earning money according to Li?A. Continuing their music dream.B. Staying at the old house.C. Living in the countryside.D. Developing the economy of cities.DWho is a genius? This question has greatly interested humankind for centuries.Let's state clearly: Einstein was a genius. His face is almost the international symbol for genius. But we want to go beyond one man and explore the nature of genius itself. Why is it that some people are so much more intelligent or creative than the rest of us? And who are they?In the sciences and arts, those praised as geniuses were most often white men, of European origin. Perhaps this is not a surprise. It's said that history is written by the victors, and those victors set the standards for admission to the genius club. When contributions were made by geniuses outside the club—women, or people of a different color1 or belief—they were unacknowledged and rejected by others.A study recently published bySciencefound that as young as age six, girls are less likely than boys to say that members of their gender(性别)are “really, really smart.” Even worse, the study found that girls act on that belief: Around age six they start to avoid activities said to be for children who are “really, really smart.” Can our planet afford to have any great thinkers become discouraged and give up? It doesn't take a genius to know the answer: ly not.Here's the good news. In a wired world with constant global communication, we're all positioned to see flashes of genius wherever they appear. And the more we look, the more we will see that social factors(因素)like gender, race, and class do not determine the appearance of genius. As a writer says, future geniuses come from those with “intelligence, creativity, perseverance(毅力), and simple good fortune, who are able to change the world.”12. Whatdoes the author think of victors' standards for joining the genius club?A. They're unfair.B. They're conservative.C. They're objective.D. They're strict.13. What can we infer about girls from the study inScience?A. They think themselves smart.B. They look up to great thinkers.C. They see gender differences earlier than boys.D. They are likely to be influenced by social beliefs14. Why are more geniuses known to the public?A. Improved global communication.B. Less discrimination against women.C.Acceptance of victors' concepts.D. Changes in people's social positions.15. What is the best title for the text?A. Geniuses Think AlikeB. Genius Takes Many FormsC. Genius and IntelligenceD. Genius and Luck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年常德市一中高三生物三模试卷及答案
2021年常德市一中高三生物三模试卷及答案一、选择题:本题共15小题,每小题2分,共30分。
每小题只有一个选项符合题目要求。
1. 分析一个DNA分子时,发现30%的脱氧核苷酸含有腺嘌呤,由此可知该分子中一条链上鸟嘌呤含量的最大值可占此链碱基总数的( )A.20%B.30%C.40%D.70%2. 如图所示为人体某类免疫过程的模式图,下列相关叙述正确的是()A. 该图示体现了人体免疫系统的监控和清除功能B. 细胞A在特异性免疫过程中识别病菌具有特异性C. 物质I和II都是免疫活性物质,都能与病菌特异性结合D. 细胞A和B在体液免疫和细胞免疫中都发挥作用3. 将在暗处生长的燕麦胚芽鞘尖端与琼脂块一起放置,几小时后,再将琼脂块放在去除尖端的胚芽鞘一侧,一段时间后观察胚芽鞘生长情况(如下图)。
下列叙述错误的是A. 重力的作用导致琼脂块甲中的生长素浓度低于乙B. 胚芽鞘放置琼脂块乙的一侧细胞伸长生长比另一侧快C. 琼脂块甲中的生长素浓度高,对胚芽鞘的生长有抑制作用D. 将琼脂块放在去除尖端的胚芽鞘顶端后,有无光照对结果无影响4. 当体表痛和内脏痛共用一个中间神经元时(如下图),神经中枢无法判断刺激究竟来自内脏还是体表,但由于神经中枢更习惯于识别体表的信息,常将内脏痛误认为是体表痛,这种现象称为牵涉痛。
下列有关叙述正确的是()A.图中a、b、c组成一个完整的反射弧B.在牵涉痛形成过程中a没有发生兴奋C.牵涉痛产生过程中兴奋在a、b之间双向传递D.牵涉痛为条件反射,神经中枢位于大脑皮层5. 由许多氨基酸缩合而成的肽链,经过盘曲折叠才能形成具有一定空间结构的蛋白质。
下列有关蛋白质结构多样性原因的叙述,错误的是A.组成肽链的化学元素不同B.肽链盘曲折叠方式不同C.组成蛋白质的氨基酸排列顺序不同D.组成蛋白质的氨基酸种类和数量不同6. 下列属于相对性状的是()A.人的直发和黑发B.桃的无毛果皮和有毛果皮C.兔的白毛与狗的黑毛D.李子的红色果实和番茄的黄色果实7. 用显微镜的一个目镜分别与4个物镜组合来观察某一细胞装片。
2021年常德市一中高三生物三模试卷及参考答案
2021年常德市一中高三生物三模试卷及参考答案一、选择题:本题共15小题,每小题2分,共30分。
每小题只有一个选项符合题目要求。
1. 微生物与人类的关系极为密切。
目前,微生物已经在食品、医疗等许多领域得到了广泛的应用。
请回答下列有关微生物的问题:(1)制作葡萄酒的过程中,需将温度控制在__________℃。
家庭制作泡菜时无须刻意灭菌,原因是__________________。
变酸果酒表面的菌膜、泡菜坛内的白膜相关的微生物分别是______________________________。
(2)采用平板划线法分离菌种时,在培养基表面共划__________个区域时,需要6次灼烧接种环。
对菌种进行临时保藏时,试管中的固体培养基常呈斜面状,目的是___________。
此外,试管口通常塞有棉花,其作用有_______________________。
(3)土壤中的某些微生物可以利用空气中的二氧化碳作为碳源。
若要设计实验验证某细菌属于该类微生物,请简要写出实验思路、预期结果和结论:___________________________________。
2. 英国医生塞达尼・任格在对离体蛙心进行的实验中发现,用不含钙和钾的生理盐水灌注蛙心,其收缩不能维持;用含有少量钙和钾的生理盐水灌注蛙心时,蛙心可持续跳动数小时。
该实验说明钙盐和钾盐()A. 对维持细胞的形态有着重要作用B. 是细胞中某些复杂化合物的重要组成部分C. 为蛙心的持续跳动提供能量D. 对维持生物体的生命活动有重要作用3. 下列关于群落结构和群落演替的叙述,正确的是()A.荒漠环境恶劣,动植物稀少,生活在该环境内的生物不存在群落结构B.若两个群落的物种数量相同、物种种类相同,它们一定是同一种群落C.若森林生物群落中乔木全部被砍伐,则草本植物生长得不一定会更好D.群落演替的结果是物种丰富度逐渐增加,所以退耕后的农田总会形成森林4. 以下实例,能证明微量元素是生命活动必需的是( )A.人体缺硒——克山病B.植物缺镁——白化苗C.动物缺钙——抽搐D.植物缺钾一易倒伏5. 下列属于组成细胞的微量元素的是()A. Fe、Mn、Zn、MgB. Zn、Cu、Mn、KC. N、S、Cu、MoD. Zn、Cu、B、Mn6. 取一红色牡丹的2个大小相同、生理状态相似的花瓣细胞,将它们分别置于℃和℃两种溶液中,测得细胞中液泡直径的变化如图所示。
2021年常德市一中高三英语三模试题及答案解析
2021年常德市一中高三英语三模试题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ACanadais one of the most beautiful countries in the world. Here are 4 attractive places worth your visit.ChurchillChurchill is a town with the nickname "Polar Bear Capital of the World”, where tourists can safely view polar bears from special vehicles in the autumn and winter. Thousands of beluga whales, which move into the warmer waters of theChurchill Riverduring July and August, are a major summer attraction. Churchill is also a destination for bird watchers from late May until August.Niagara FallsNiagara Fallsis a group of three waterfalls, crossing the border betweenCanadaand theUnited States. The largest of the three is Horseshoe Falls, also known asCanadian Falls. Niagara Falls illumination(彩灯)is a must for any visitor! Every night of the year, the three waterfalls are illuminated in color1 s creating an attractive scene that can be viewed from near and far.VancouverVancouverisCanada's third-largest city, always named as one of the top five worldwide cities for its comfortable environment and quality of life.Vancouverhas an active nightlife scene, whether its food and dining, or bars and nightclubs. From mid-June to early July, the Vancouver International Jazz Festival features 300 concerts, including a free opening Downtown Jazz Weekend.OttawaThe capital ofCanadais situated on the banks of theOttawa Riverand has a lot ofEnglish buildings in it. It is a beautiful city which has the Parliament buildings on the banks and English influenced houses and parks around. There are museums and art galleries that will give you a complete knowledge of the English culture there. It is really the heart ofCanada. So if you are a history and art loverOttawais the best choice for your visit inCanada.1. If you want to watch birds, which place will you choose to visit?A. Churchill.B.Niagara Falls.C. Vancouver.D.Ottawa.2. What is the best season for visitingVancouver?A. Spring.B. Summer.C. Autumn.D. Winter.3. What doNiagara FallsandVancouverhave in common?A. They are both famous for natural scenery.B. The best visiting time are both at nights.C. They are both located inCanadaentirely.D. The tickets there are both free at weekends.BMost animals living in crowded conditions have particularly strong immune systems, so it long puzzled researchers that honeybees do not.Part of the answer, discovered in 2015, is that queen bees vaccinate their eggs by moving parts of proteins from disease-causing pathogens to them before they are laid. These act as antigens totriggerthe development of a protective immune response in the developing young. But that observation raises the question of how the queen receives her antigen supply in the first place? Dr. Harwood wondered if the nurse bees were taking in parts of pathogens and passing them to royal jelly they were producing while eating the food brought to the hive.To test this idea, he teamed up with a group at theUniversityofHelsinki, inFinland, led by Dr Heli Salmela. Together, they collected about 150 nurse bees and divided them among six queenless mini hives equipped with baby bees to look after. Instead of honey, they fed the nurses on sugar water, and for three of the hives they added P. larvae, a bacterium causing a hive-killing disease, to the sugar water.In this case, to stop such an infection happening, Dr Harwood and Dr Salmela heat-treated the pathogens and so killed them in advance. They also labelled the dead bacteria with a fluorescent dye, to track them easily. And, sure enough, it was confirmed that parts of P. larvae were getting into royal jelly released by those bees which had been fed with the sugar water containing that.All told, these findings suggest that nurse bees are indeed, through their royal jelly, passing antigens onto the queen for vaccinating her eggs. They also mean the nurses are vaccinating baby bees as well, because baby bees, too, receive royal jelly for the first few days after they come out.4. What does the underlined word “trigger" in Paragraph 2 probably mean?A. Cut out.B. Set off.C. Slow down.D. Put off.5. Which is the main experimental subject in Paragraph 3?A. Queen bees.B. Nurse bees.C. Bee eggs.D. Baby bees.6. Why was P. larvae added to the sugar water?A. To test if it would cause a hive-killing disease.B. To check how the bacterium would affect the hive.C. To see whether the target bees would favor the taste.D. To confirm the bees would pass pathogens to royal jelly.7. What is the text mainly about?A. How bees multiply.B. How antigens function.C. How bees get vaccinated.D. How immune system works.CNow most of the workers work from 9 am to 5 pm. However, according to the global Internet survey done by the UK Sleep Council, thesiesta(午睡)was the right idea all along. The UK Sleep Council called on the country'sbosses to end nine-to-five working in favor of more flexible hours. They believe what would really increase the workers' productivity is a nice afternoon nap, rather than those bonuses.Forty-one percent of the 12,000 people who responded to the council's survey said they were most productive in the morning, while 38 percent said theyhit their stridein the evening. "This means most of them cannot fully pay attention to what they do in the middle of the day," said sleep expert Dr. Chris Idzikowskii. "We must conclude from this survey that the traditional nine-to-five working day does not suit most workers." He suggested that allowing workers to follow their natural sleeping habits would actually benefit employers by allowing them to expand their working hours and be more productive.Fortunately, being a college lecturer, I don't have to go to work everyday. I only work three days a week, but during the three days I work really long hours and have no time for a little siesta. I'm usually so tired and sleepy in the afternoon, which really affects thevitality(活力)of my classes.I think Dr. Chris Idzikowskii's idea is worthwhile. When people have flexible working hours they could reach their highest productivity. On top of that, flexible working hours mean thatpeople don't have to work all at the same time. That way we could avoid traffic jams. Therefore, it's really killing two birds with one stone!8. What can improve the workers' productivity, according to the UK Sleep Council?A. More bonuses.B. The flexible working time.C. Working for long hours.D. Working in a relaxing way.9. What does the underlined part probably mean in Paragraph 2?A. Were most sleepy.B. Were most flexible.C. Worked at their own pace.D. Worked at their best.10. What did the UK Sleep Council's survey find?A. Few people are suitable to work at noon.B. People are more productive in the morning.C. Some people like to expand their working hours.D. More and more people prefer to work in the evening.11. Why does the author support Dr. Chris Idzikowskii's idea?A. It could solve most of the traffic problems.B. He finds Dr. Chris Idzikowskii respectable.C. He thinks the idea can benefit the society.D. It allows him to work for fewer hours.DA new look for technology, Solestrom’s new high-tech swimsuits promise to stand out all summer long. First in their new products is Solestrom’s new SmartSwimCMUV Smart Bikini featuring a smart UV meter.The bikini collects UV information through a smart fabric belt and reports the UV index to the wearer with 0.01 exactness. The electronic parts are neatly built into the removable belt, and can be worn even underwater. Next in the list is a lower cost cousin, the SmartSwimCMUV Index Detector Bikini, which has UV sensitive beads that change color1 with the level of UV intensity. The reading gives more of a range rather than an exact number, but for those who simply need to know if the UV is low, moderate or high, this bikini fits the bill.It is now available in Banana Split, more color1 s may become available later in the season. And finally, what could be better than a beach tote with built in energy source? No need to suffer dead batteries in your MP3, just plug them into your Solar Powered Beach Tote for on-the-go mobile charging. This beach tote combines fashion and functionality all in one, lightweight and roomy beach accessory. A built in solar panel charges fully most small electronics in only 2-3 hours of sunlight. Pauline Butler, Media Relations Manger at Solestrom states “the concept of blending fashion and technology is growing at amazing rate, and we are right on the leading edge. Our products are new, creative and meet the need of the young and environmentally conscious crowd.’’Solestrom’s SmartSwinirCMproducts retail from $98. 99- $189. 99,and can be found in their online store, Solestrom. com.12. What is the passage mainly about?A. Famous Solestrom and its Products.B. Fashionable and Functional Beach Tote.C. Smart swimsuits and Solar Powered Beach Tote.D. Where and How to Buy SmartSwirn and the Beach Tote.13. Why are UV sensitive beads used in the SmartSwirnCMUV Index Detector Bikini?A. To give an exact number.B. To know if the UV is low, moderate or high.C. To change color1 with the level of UV intensity.D. To improve the equality of the UV Index Detector Bikini.14. What can we know about Solar Powered Beach Tote?A. Its parts are fixed in the belt.B. It combines lightweight and functionality.C. It can tell the weaver the level of UV intensity.D. It saves people from worrying about dead batteries in their MP3.15. Who will prefer the new products mentioned in the text?A. The young people and the environmentalists.B. People who like to travel all over the world.C. People who love to buy goods online.D. People interested in the latest bag.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年常德市一中高三生物第三次联考试题及参考答案
2021年常德市一中高三生物第三次联考试题及参考答案一、选择题:本题共15小题,每小题2分,共30分。
每小题只有一个选项符合题目要求。
1. 蛋白质是生命活动的主要体现者,下列叙述错误的是()A.蛋白质的特定功能都与其特定的结构有关B.唾液淀粉酶进入胃液后将不再发挥催化作用C.细胞膜上的某些蛋白质起着物质运输的作用D.蛋白质的结构一旦改变将失去生物学活性2. 在我国西部大开发的战略中,“保护天然林”和“退耕还林(草)”是两项重要内容,采取这两项措施的首要目标是()A.开展生态旅游B.发展畜牧业C.增加木材产量D.改善生态环境3. 心脏的搏动受交感神经和副交感神经的控制,其中副交感神经释放乙酰胆碱,作用于心肌细胞膜上的M 型受体,使心肌细胞的收缩受到抑制,心率减慢;交感神经释放的去甲肾上腺素可以和心肌细胞膜上的β-肾上腺素受体结合,使心率加快。
但交感神经和副交感神经对心脏的作用强度不是等同的,利用心得安和阿托品进行如下实验(心得安是β-肾上腺素受体的阻断剂,阿托品是M型受体的阻断剂)。
对两组健康青年分别注射等量的阿托品和心得安各4次,给药次序和测得的平均心率如图所示。
有关叙述正确的是()A. 乙酰胆碱与M型受体结合,使得心肌细胞的静息电位绝对值减小B. 注射阿托品后交感神经的作用减弱,副交感神经作用加强C. 每一组的每位健康青年共进行了8次心率的测定D. 副交感神经对心跳的抑制作用远超过交感神经对心跳的促进作用4. 鳞翅目昆虫黄粉蝶的性别决定方式为ZW型,其幼体的黑斑和白斑受Bb基因的控制,皮肤不透明和透明受E、e基因的控制。
一对幼体皮肤不透明而有黑斑的雌雄个体交配,F1中黑斑个体与白斑个体的比例为3∶l,皮肤不透明与透明个体的比例为2∶1,且皮肤透明的幼体全为雌性。
下列有关叙述错误的是()A.B、b和E、e两对等位基因遵循自由组合定律B.E、e基因位于性染色体上,且基因型Z E Z E致死C.F1中皮肤不透明而有白斑的雄性幼体所占的比例为1/8D.F1中皮肤透明的幼体全为雌性,则雄性亲本基因型为BbZ E Z e5. 细胞核由核膜、染色质、核仁、核孔组成,下列有关叙述正确的是()A.核孔实现了细胞间的信息交流B.染色质主要由DNA和蛋白质组成C.核仁与DNA合成以及线粒体的形成有关D.核膜是单层膜,把核内物质与细胞质分开6. 下列四种化合物中,属于构成生物蛋白质的氨基酸是()A. B.C. D.7. 如图是人体细胞中三种重要有机物A、C、E的元素组成及相互关系图,下列叙述正确的是()A. 图中X、Y所指的元素都为N、PB. 有机物A彻底水解的产物是磷酸、核糖、含氮碱基C. 有机物E多样性的原因是b的种类、数量、排列顺序、空间结构不同D. 过程∶叫做脱水缩合,细胞结构∶的名称叫糖被8. 接种疫苗是预防传染病最有效、最经济的方式。
2021年湖南省常德市三岔河镇中学高三物理月考试题含解析
2021年湖南省常德市三岔河镇中学高三物理月考试题含解析一、选择题:本题共5小题,每小题3分,共计15分.每小题只有一个选项符合题意1. (单选)我国于2013年12月发射了“嫦娥三号”卫星,该卫星在距月球表面高度为h的轨道上做匀速圆周运动,其运行的周期为T;卫星还在月球上软着陆。
若以R表示月球的半径,忽略月球自转及地球对卫星的影响。
则()A.“嫦娥三号”绕月运行时的向心加速度为B.月球的第一宇宙速度为C.由题给条件不可求出物体在月球表面自由下落的加速度D.“嫦娥三号”用降落伞逐渐减小降落速度,最终实现软着陆参考答案:B2. (单选)对一些实际生活中的现象,某同学试图从惯性角度加以解释,其中正确的是A.采用大功率的发动机后,某些一级方程式赛车的速度甚至能超过某些老式螺旋桨飞机的速度.这表明通过科学手段能使小质量的物体获得大的惯性B.射出枪膛的子弹在运动相当长一段距离后连一件棉衣也穿不透,这表明它的惯性小了C.货运列车运行到不同的车站时,经常要摘下或加挂一些车厢,这会改变它的惯性D.摩托车转弯时,车手一方面要控制适当的速度,另一方面要将身体稍微向里倾斜,通过调控人和车的惯性达到安全行驶的目的参考答案:C 解析:A、采用了大功率的发动机后,一级方程式赛车的速度甚至能超过某些老式螺旋桨飞机的速度,不是由于使小质量的物体获得大惯性,是由于功率增大的缘故.故A错误.B、射出枪膛的子弹在运动一段距离后连一件棉衣也穿不透,是由于速度减小了,不是由于惯性减小,子弹的惯性没有变化.故B错误.C、货车运行到不同的车站时,经常要摘下或加挂一些车厢,质量改变了,会改变它的惯性.故C正确.D、摩托车转弯时,车手一方面要控制适当的速度,另一方面要将身体稍微向里倾斜,改变向心力,防止侧滑,而人和车的惯性并没有改变.故D错误.故选C3. (单选)如图所示,两根平行放置、长度均为L的直导线a和b,放置在与导线所在平面垂直的匀强磁场中,当a导线通有电流强度为I,b导线通有电流强度为2I,且电流方向相反时,a导线受到磁场力大小为F1,b导线受到的磁场力大小为F2,则a通电导线的电流在b导线处产生的磁感应强度大小为()参考答案:C4. 2009年3月1日16时13分,“嫦娥一号”完成了“受控撞月”行动,探月一期工程完美落幕。
2021年常德市一中高三生物三模试卷及答案
2021年常德市一中高三生物三模试卷及答案一、选择题:本题共15小题,每小题2分,共30分。
每小题只有一个选项符合题目要求。
1. 下图为生长素浓度对某种植物根部生长发育的影响(设图1中c点为10一10mol·L一1)曲线和将该植物水平放置一段时间后的生长状态示意图(图2)。
下列相关两图示分析错误的是()A.图1在a-d的浓度范围内生长素都均能促进根部细胞的生长B.图1在a-e的浓度范围内生长素都均能促进茎部细胞的生长C.若此时图2中的茎部A点的生长素浓度为e,则B点的生长素浓度可能为fD.若此时图2中的根部C点的生长素浓度为c,则D点的生长素浓度可能为d2. 某生物兴趣小组为研究渗透现象,将一个长颈漏斗的漏斗口外密封上一层玻璃纸(水分子可以自由透过,蔗糖分子不能通过),往漏斗内注入蔗糖溶液(S1),然后将漏斗浸人盛有蔗糖溶液(S2)的烧杯中,使漏斗管内外的液面高度相等。
过一段时间后,出现如图所示的现象。
下列相关叙述错误的是()A.开始注入的蔗糖溶液S1的浓度大于S2的浓度B.当漏斗内外的高度差达到平衡时,仍有水分子透过玻璃纸C.若向漏斗中加入蔗糖分子,则再平衡时m变小D.若吸出漏斗中高出烧杯液面的溶液,再次平衡时m将变小3. 下图是细胞膜的模式图,下列有关说法正确的是A. 细胞膜的基本骨架是②B. ②的种类和数量决定了细胞膜功能的复杂程度C. 具有②的一侧为细胞膜内侧D. 不同细胞膜中②的种类不同4. 下列有关细胞内物质和结构的比值关系,正确的是A. 细胞内结合水/自由水的比值,种子萌发时比休眠时高B. 人体细胞内O2/CO2的比值,线粒体内比细胞质基质低C. 相同形状的细胞,其细胞表面积/体积的比值与细胞大小呈正相关D. 适宜条件下光合作用过程中C5/C3的比值,停止供应CO2后比停止前的低5. 下列与细胞有关的叙述,正确的是()A.T2噬菌体不含有膜包被的细胞核,因此属于原核细胞B.细胞生物的遗传物质均为DNAC.细胞学说揭示了动物细胞和植物细胞的区别D.心肌细胞是高度分化的细胞,其细胞膜不具有流动性6. 下图是外源性过敏原引起哮喘的示意图。
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数学(参考答案)
一、选择题: 1. D 2. C 3. A 4. B 5. D 6.C 7. B
8. B
二、选择题:9. BCD 12. ABC
10. AD
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即异面直线 BM 与 PA 所成角的余弦值为 3 10 . 10
(建系法略)
21.(本题满分 12 分)解:(1)由 g0() 0 有 n 1,又由 f (x) f (x) 有 m 1 ,则 m n 1
g
(
a 2
)
0
2
a
3
三、填空题:本题共 4 小题,每小题 5 分,共 20 分。
13. ①
14. 4
15. 1
16. 13 2
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 . 【 解 析 】 选 择 条 件 ① c 7, cos A 1 , a b 11 , 7
11 a ,a 6,b 5 . 57
8 16
18. 解:(1) an n , bn 2n
(2)由题有 cn n 2n ,则
Tn 1 21 2 22 3 23 (n 1) 2n1 n 2n
①
2Tn 1 22 2 23 3 24 (n 1) 2n n 2n1 ②
① - ②得: Tn 21 22 23 2n n 2n1 2n1 2 n 2n1
则Tn (n 1) 2n1 2 . 19.解: f (x) 3(cos x 1) 3 sin x 3 2 3 sin( x )
3
(1)由 f (x) 的最大值为 2 3 有: ABC 的高为 2 3 ,故边长为 4,则 f (x) 的周期
4
x0
) 3
] 4
2
3[
2 2
sin( 4
x0
3
)
2 2
cos( 4
x0
3
)]
76 5
.
20.(1)证明:由已知可算得 BD BC 2 2 , BD2 BC2 16 DC2 ,
故 BD BC ,又 PD 平面ABCD , BC 平面 ABCD ,故 PD BC ,
又 BD PD D ,所以 BC 平面 BDP ;
法一: g(x) 在[0, 2] 上有零点 g(x) 0 在[0, 2] 上有解 a x2 3 在[0, 2] 上有解 x 1
y a 与 y x2 3 在[0, 2] 上的图象有交点 2 a 3 x 1
g(0) 0
法二:注意到
g
(x)
必过点(-1,4),则零点
b
[0,
2]
有一个零点,
即此时 g x 在 0, 上有两个零点.
综上所述,当 0 a 1 时, g x 在 0, 上仅有一个零点;当1 a 3 时, g x 在 0, 上有两
个零点.
2
②若1
a
3
时,
g0
1
a
0
,又∵
g
x
在
0 , 2
上单调递增,在
2
,
上单调递减,
而
g
2
e2
a
0
,从而
gx
在 0,
上图象大致如右图.∴ x1
0 , 2
,
x2
2
,
,使
得 g x1 0 ,g x2 0 ,且当 x 0,x1 、x x2, 时,g x 0 ;当 x x1,x2
1
(2)解:如图,取 PD 中点为 N,并连结 AN,MN,易证明 BM // AN ,
P
则 PAN 即异面直线 BM 与 PA 所成角; 又 PA 底面 ABCD ,PCD 即为 PC 与底面 ABCD 所成角,
N
M
即 tanPCD 1 , PD 1 CD 2 ,即 PN 1 PD 1,
2
2
22.(本题满分 12 分)解:(1) f x ex sin x ,定义域为 R .
f x ex sin x cos x
2ex
sin
x
4
.
由
f
x
0
解得
sin
x
4
0 ,解得 2k
3 4
x
7 4
2k
(kZ
).
∴
f
x
的单调递减区间为
3 4
2k,7 4
2k
(
k
Z
).
(2)由已知 g(x) ex sin x ax ,∴ g x ex sin x cos x a .令 h x g x ,则 h x 2ex cos x .
T 8 ,且 f (x) 2 3 sin( x ) 的值域为[2 3, 2 3] ;
4
43
⑵由
f
(x0 )
83 5
有
sin(
4
x0
3
)
4 5
,且
4
x0
3
(
2
,
2
)
则
cos(
4
x0
3
)
3 5
,
f (x0 1) 2
3 sin[ 4 (x0 1) 3 ] 2
3 sin[(
2
,
,使得
g
x0
0
,∴当
x
0,x0
时,
g x 0 ;当 x x0, 时,g x 0 ,∴ g x 在 0,x0 上单调递增,在 x0, 上单
调递减.∵ g 0 0 ,∴ g x0 0 .又∵ g a 0 ,∴由零点存在性定理可得,此
时 g x 在 0, 上仅有一个零点.