模拟卷2参考答案

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认证通用基础模拟试卷带参考答案-2

认证通用基础模拟试卷带参考答案-2

【单选题】1. 合格评定功能法中进行确定活动之一,其核心技术是()。

A. 评价技术B. 查阅方法C.面谈技巧D.观察能力参考答案:A题目解析:《审核概论》P143评价活动是合格评定功能法中的关键活动之一,而评价技术是审核的核心技术。

【单选题】2.WTO/TBT协议规定了各成员国合格评定的原则的目的是为了()。

A.增加各成员国的贸易壁垒B.提升合格评定活动对贸易的影响C.减少合格评定活动对贸易的负面影响D.提高合格评定活动的约束性参考答案:C题目解析:《合格评定基础》P9三、合格评定的原则为减少合格评定活动对贸易的负面影响,WTO/TBT协议规定了各成员国合格评定的原则。

【单选题】3.审核通常可包括:审核活动启动、审核活动的准备、审核活动的实施、审核报告的编制、审核的完成等活动,()不属于审核阶段活动内容。

A.审核发现的汇总分析B.审核的后续活动C.审核过程和结果的复核D.审核方案的监视和评审参考答案:B题目解析:《审核概论》P68审核通常可划分为审核的启动、审核活动的准备、审核活动的实施、审核报告的编制与分发、审核的完成、审核后续活动的实施等六个阶段。

审核后续活动是在审核完成之后进行的活动,不属于审核阶段的内容。

【单选题】4.关于检验机构的内部管理要求描述错误的是()。

A.管理方式B是按照GB/T 19001要求建立并保持管理体系B.检验机构管理方式A至少包括文件控制、记录控制、管理评审、内部审核、纠正措施、预防措施C.检验机构必须按照GB/T 19000要求建立并保持管理体系D.检验机构应建立并保持能持续满足管理方式A或者管理方式B要求之一的管理体系参考答案:C题目解析:《合格评定基础》P182:(二)检验机构管理方式B:一个检验机构已经按照GB/T 19001要求建立并保持管理体系,且有能力支持并证实其满足本准则的要求,则符合管理体系条款的要求。

P180:检验机构管理体系方式A应包括:管理体系文件、文件控制、记录控制、管理评审、内部审核、纠正措施、预防措施。

2023年陕西中考语文全真模拟卷2

2023年陕西中考语文全真模拟卷2

2023年陕西中考语文全真模拟卷2(本卷满分120分,考试时间为150分钟。

)一、积累与运用。

(17分)1.下列各组词语中,加点字的读音全都正确的一组是()(2分)A.侮.辱(wǔ)田垄.(lǒng)木讷.(nà)混.为一谈(hùn)B.着.落(zhuó)模.样(mú)门框.(kuāng)潜.移默化(qián)C.憎.恶(zēng)镌.刻(juān)炽.热(chì)不可捉摸.(mō)D.殷.红(yīn)黝.黑(yǒu)镂.空(lòu)妇孺.皆知(rǔ)【答案】C【详解】本题考查字音辨析。

A.木讷.(nà)——nè;B.门框.(kuāng)——kuàng;D.殷.红(yīn)——yān,妇孺.皆知(rǔ)——rú;故选C。

2.下列各组词语中,汉字书写全都正确的一组是()(2分)A.流盼琢磨任劳任怨怪诞不经B.贮立温驯精溢求精力不暇供C.田畴报歉巧妙绝纶大庭广众D.署光譬如悲天悯人美不盛收【答案】A【详解】本题考查字形。

B. 贮立——伫立,精溢求精——精益求精;C. 报歉——抱歉,巧妙绝纶——巧妙绝伦;D.署光——曙光,美不盛收——美不胜收;故选A。

3.默写。

(6分)在(1)—(7)题中任选5题,在(8)—(10)中任选1题.(1)此中有真意,______________。

(陶渊明《饮酒》(其五))(2)潮平两岸阔,______________。

(王湾《次北固山下》)(3)______________。

鬓微霜,又何妨!(苏轼《江城子·密州出猎》)(4)伤心秦汉经行处,______________。

(张养浩《山坡羊·潼关怀古》)(5)几处早莺争暖树,______________。

(白居易《钱塘湖春行》)(6)______________,以光先帝遗德。

(诸葛亮《出师表》)(7)______________,不知口体之奉不若人也。

2022~2023学年高一年级数学上册期末备考模拟试卷(2)【含答案】

2022~2023学年高一年级数学上册期末备考模拟试卷(2)【含答案】

期末模拟试卷(2)一、单选题: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.已知全集{}4U x x =∈≤N ,集合{1,},{1,2,4}A m B ==.若(){0,2,3}U A B = ð,则m =().A .4B .3C .2D .02.已知命题“R x ∀∈,214(2)04x a x +-+>”是假命题,则实数a 的取值范围为().A .(][),04,-∞+∞U B .[]0,4C .[)4,+∞D .()0,43.函数()log 14a y x =-+的图像恒过定点P ,点P 在幂函数()y f x =的图像上,则(4)f =().A .16B .8C .4D .24.函数()2log 21f x x x =+-的零点所在区间为().A .10,2⎛⎫ ⎪⎝⎭B .1,12⎛⎫ ⎪⎝⎭C .31,2⎛⎫⎪⎝⎭D .3,22⎛⎫ ⎪⎝⎭5.函数e 1()cos e 1x x f x x -=⋅+的图像大致为().A .B .C .D .6.牛顿冷却定律描述物体在常温环境下的温度变化:如果物体的初始温度为0T ,则经过一定时间t 分钟后的温度T 满足()012tha a T T T T ⎛⎫-=- ⎪⎝⎭,h 称为半衰期,其中a T 是环境温度.若25a T =℃,现有一杯80℃的热水降至75℃大约用时1分钟,那么水温从75℃降至45℃,大约还需要().(参考数据:lg 20.30≈,lg11 1.04≈)A .9分钟B .10分钟C .11分钟D .12分钟7.函数()()214tan πcos f x x x =--的最大值为().A .2B .3C .4D .58.定义在R 上的函数()f x 满足()()()()0,2x f x f x f x f -+==-,且当[]0,1x ∈时,()2f x x =.则函数()72y f x x =-+的所有零点之和为().A .7B .14C .21D .28二、多选题:本题共4小题,每小题5分,共20分.在每小题给出的选项中,有多项符合题目要求,全部选对的得5分,部分选对的得2分,有选错的得0分.9.下列函数中,最小正周期为π,且在0,2π⎛⎫⎪⎝⎭上单调递增的是().A .sin 2y x =B .tan y x =C .sin y x =D .tan y x =10.设正实数m ,n 满足2m n +=,则下列说法正确的是().A .11m n+的最小值为2B .mn 的最大值为1C 的最大值为4D .22m n +的最小值为5411.已知函数()2sin 213f x x π⎛⎫=-+ ⎪⎝⎭,则下列说法正确的是().A .()()f x f x π+=B .6f x π⎛⎫+ ⎪⎝⎭的图象关于原点对称C .若125012x x π<<<,则()()12f x f x <D .对1x ∀,2x ,3,32x ππ⎡⎤∈⎢⎣⎦,有()()()132f x f x f x +>成立12.已知()y f x =奇函数,()(2)f x f x =-恒成立,且当01x 时,()f x x =,设()()(1)g x f x f x =++,则().A .(2022)1g =B .函数()y g x =为周期函数C .函数()y g x =在区间(2021,2022)上单调递减D .函数()y g x =的图像既有对称轴又有对称中心三、填空题:本大题共4小题,每小题5分,共20分.把答案填写在答题卡相应位置上.13.已知正实数a ,b 满足2a b +=,则24a ab+的最小值是______.14.已知函数()223,02ln ,0x x x f x x x ⎧+-≤=⎨-+>⎩,方程()f x k =有两个实数解,则k 的范围是____.15.已知函数()sin ,06f x x πωω⎛⎫=+> ⎪⎝⎭,若5412f f ππ⎛⎫⎛⎫= ⎪ ⎪⎝⎭⎝⎭且()f x 在区间5,412ππ⎛⎫ ⎪⎝⎭上有最小值无最大值,则ω=_______.16.若函数22sin 2,0()2,()()2,0x a x x f x g x a R x a x -+≥⎧==∈⎨+<⎩,对任意1[1,)x ∈+∞,总存在2x R ∈,使12()()f x g x =,则实数a 的取值范围___________四、解答题:本大题共6小题,共70分.第17题10分,第18至22题均12分.解答应写出文字说明、证明过程或演算步骤.17.在①22{|1}1x A x x -=<+,②{||1|2}A x x =-<,③23{|log }1xA x y x -==+这三个条件中任选一个,补充在横线上,并回答下列问题.设全集U =R ,_____,22{|0}.B x x x a a =++-<(1).若2a =,求()()U UC A C B ;(2).若“x A ∈”是“x B ∈”的充分不必要条件,求实数a 的取值范围.18.已知关于x 的不等式2tan 0x θ-+≥对x ∈R 恒成立.(1).求tan θ的取值范围;(2).当tan θ取得最小值时,求22sin 3sin cos 1θθθ++的值.19.已知函数()π2sin 226f x x ⎛⎫=++ ⎪⎝⎭.(1).若()3f α=,且()0,πα∈,求α的值;(2).若对任意的ππ,42x ⎡⎤∈⎢⎥⎣⎦,不等式()3f x m >-恒成立,求实数m 的取值范围.20.某地区的一种特色水果上市时间11个月中,预测上市初期和后期会因供不应求使价格呈连续上涨态势,而中期又将出现供大于求使价格连续下跌,现有三种价格模拟函数:①()x f x p q =⋅;②2()1f x px qx =++;③()sin(44f x A x B ππ=-+(以上三式中,,,p q A B 均为非零常数,且1q >)(1).为准确研究其价格走势,应选哪种价格模拟函数,为什么?(2).若(3)8,(7)4,f f ==求出所选函数()f x 的解析式,为保证果农的收益,打算在价格在5元以下期间积极拓宽外销渠道,请你预测该水果在哪几个月份要采用外销策略?(注:函数的定义域是[]0,10,其中0x =表示1月份,1x =表示2月份, ,以此类推)21.已知函数41()log 2x a x f x +=(01)且a a >≠.(1).试判断函数()f x 的奇偶性;(2).当2a =时,求函数()f x 的值域;(3).已知()g x x =-[][]124,4,0,4x x ∀∈-∃∈,使得12()()2f x g x ->,求实数a的取值范围.22.已知函数2()1(0).f x ax x a =++>(1).若关于x 的不等式()0f x <的解集为(3,)b -,求a ,b 的值;(2).已知1()422x xg x +=-+,当[]1,1x ∈-时,(2)()x f g x ≤恒成立,求实数a 的取值范围;(3).定义:闭区间1212[,]()x x x x <的长度为21x x -,若对于任意长度为1的闭区间D ,存在,,|()()|1m n D f m f n ∈-≥,求正数a 的最小值.期末模拟试卷02参考答案一、单选题:本题共8小题,每小题5分,共40分.1.A 【详解】因为{}{}40,1,2,3,4U x x =∈≤=N ,又(){0,2,3}U A B = ð,所以{}1,4A B = ,即1A ∈且4A ∈,又{1,}A m =,所以4m =;故选A2.A 【详解】若“R x ∀∈,214(2)04x a x +-+>”是真命题,即()21Δ24404a =--⨯⨯<,解得04a <<,所以若该命题是假命题,则实数a 的取值范围为(][),04,-∞+∞U .故选A.3.A 【详解】当2x =时,log 144a y =+=,所以函数()log 14a y x =-+恒过定点(2,4)记()m f x x =,则有24m =,解得2m =,所以2(4)416f ==.故选A4.B【详解】函数()2log 21f x x x =+-在()0+∞,上单调递增,1102f ⎛⎫=- ⎪⎝⎭<,()110f =>,由零点存在性定理可得,函数()2log 21f x x x =+-零点所在区间为1,12⎛⎫⎪⎝⎭.故选B.5.A 【详解】函数定义域是R ,e 1e e 1()cos()c )11e os (x x xxf x x x f x -----=⋅-==-++,函数为奇函数,排除BD ,当02x π<<时,()0f x >,排除C .故选A .6.B【详解】由题意,25a T =℃,由一杯80℃的热水降至75℃大约用时1分钟,可得()11752580252h ⎛⎫-=- ⎪⎝⎭,所以11501025511h ⎛⎫== ⎪⎝⎭,又水温从75℃降至45℃,所以()1452575252th⎛⎫-=- ⎪⎝⎭,即12022505th⎛⎫== ⎪⎝⎭,所以11110222115tt thh ⎡⎤⎛⎫⎛⎫⎛⎫⎢⎥=== ⎪ ⎪ ⎪⎢⎥⎝⎭⎝⎭⎝⎭⎣⎦,所以10112lg 22lg 2120.315log 101051lg111 1.04lg 11t -⨯-===≈=--,所以水温从75℃降至45℃,大约还需要10分钟.故选B.7.B 【详解】()()22222sin cos 4tan tan 4tan 1tan 23cos x x f x x x x x x+=--=---=-++,当tan 2x =-时,()f x 取得最大值,且最大值为3,故选B8.B【详解】()f x 是奇函数.又由()()2f x f x =-知,()f x 的图像关于1x =对称.()()()()()()()4131322f x f x f x f x f x +=++=-+=--=-+()()()()2f x f x f x =---=--=,所以()f x 是周期为4的周期函数.()()()()()()()()211112f x f x f x f x f x f x +=++=-+=-=-=--,所以()f x 关于点()2,0对称.由于()()27207x y f x x f x -=-+=⇔=,从而求函数()f x 与()27x g x -=的图像的交点的横坐标之和.而函数()27x g x -=的图像也关于点()2,0对称.画出()y f x =,()27x g x -=的图象如图所示.由图可知,共有7个交点,所以函数()72y f x x =-+所有零点和为7214⨯=.故选B9.BCD【详解】A ,sin 2y x =,2T ππω==,由0,2x π⎛⎫∈ ⎪⎝⎭,得()20,x π∈,函数在区间0,2π⎛⎫ ⎪⎝⎭上不单调,故A 错误;B ,tan y x =最小正周期为π且在0,2π⎛⎫ ⎪⎝⎭上单增,故B 正确;C ,sin y x =最小正周期为π且在0,2π⎛⎫⎪⎝⎭上单增,故C 正确;D ,tan y x =,最小正周期为π,且在0,2π⎛⎫⎪⎝⎭上单调递增,故D 正确;故选BCD.10.AB 【详解】∵0,0,2m n m n >>+=,∴()1111111222222n m m n m n m n m n ⎛⎛⎫⎛⎫+=++=++≥+= ⎪ ⎪ ⎝⎭⎝⎭⎝当且仅当n m m n =,即1m n ==时等号成立,故A 正确;2m n +=≥ 1mn ≤,当且仅当1m n ==时,等号成立,故B正确;22224⎡⎤≤+=⎢⎥⎣⎦ ,2,当且仅当1m n ==时等号成立,最大值为2,故C 错误;()22222m n m n ++≥=,当且仅当1m n ==时等号成立,故D 错误.故选AB 11.ACD【详解】∵函数()2sin 213f x x π⎛⎫=-+ ⎪⎝⎭的周期22T ππ==,所以()()f x f x π+=恒成立,故A 正确;又2sin 216f x x π⎛⎫+=+ ⎪⎝⎭,所以2sin 11663f πππ⎛⎫+=+ ⎪⎝⎭,2sin 11663f πππ⎛⎫⎛⎫-+=-+= ⎪ ⎪⎝⎭⎝⎭,所以6666f f ππππ⎛⎫⎛⎫+≠--+ ⎪ ⎪⎝⎭⎝⎭,所以6f x π⎛⎫+ ⎪⎝⎭的图象不关于原点对称,故B 错误;当50,12x π⎛⎫∈ ⎪⎝⎭时,2,332x πππ⎛⎫-∈- ⎪⎝⎭,所以函数()2sin 213f x x π⎛⎫=-+ ⎝⎭在50,12π⎛⎫ ⎪⎝⎭上单调递增,故C 正确;因为,32x ππ⎡⎤∈⎢⎣⎦,所以22,333x πππ⎡⎤-∈⎢⎥⎣⎦,sin 213x π⎛⎫≤-≤ ⎪⎝⎭,()1,3f x ⎤∴∈⎦,又)213+>,即min max 2()()f x f x >,所以对123,,[,],32x x x ππ∀∈有132()()()f x f x f x +>成立,故D 正确.故选ACD.12.BCD【详解】因为()(2)f x f x =-,所以()(2)f x f x -=+,又()f x 为奇函数,故()()(2)(2)(2)f x f x f x f x f x -=-=--=-=+,利用(2)(2)f x f x -=+,可得()(4)f x f x =+,故()f x 的周期为4;因为()f x 周期为4,则()g x 的周期为4,又()f x 是奇函数,所以(2022)(50542)(2)(2)(3)(2)(1)(1)1g g g f f f f f =⨯+==+=+-=-=-,A 错误,B 正确;当01x 时,()f x x =,因为()f x 为奇函数,故10x -≤<时,()f x x =,因为()(2)f x f x =-恒成立,令021x ≤-≤,此时,(2)2f x x -=-,则21x ≥≥,()(2)2f x f x x =-=-,故02x ≤≤时,,01()2,12x x f x x x ≤≤⎧=⎨-<≤⎩,令21x -≤<-,即12x <-≤,则()2()f x x f x -=+=-,即()2f x x =--;令10x -≤<,即01x <-≤,则()()f x x f x -=-=-,即()f x x =;令23x <<,即32x -<-<-,120x -<-<,(2)2()f x x f x -=-=所以(),112,13f x x xx x⎪=-≤≤⎨⎪-<≤⎩,根据周期性()y g x=在(2021,2022)x∈上的图像与在(1,2)x∈相同,所以,当12x≤<,即213x≤+<时,()()(1)22(1)32g x f x f x x x x=++=-+-+=-,故()g x在(1,2)x∈上单调递减,C正确;由()f x是周期为4的奇函数,则(2)()(2)f x f x f x+=-=-且(1)(1)f x f x-=-+,所以(1)(1)(2)(1)(2)()(1)()g x f x f x f x f x f x f x g x-=-+-=----=++=,故()g x关于12x=对称,()(3)()(1)(3)(4)()(1)(1)()0g x g x f x f x f x f x f x f x f x f x+-=+++-+-=++-+-=,所以()g x关于3,02⎛⎫⎪⎝⎭对称,D正确.故选BCD三、填空题:本大题共4小题,每小题5分,共20分.13.3+【详解】242422222133a b a b a b b aa ab a ab a b a b a b++++=+=+=+=+++≥++(当且仅当2b aa b=,即42a b=-=时等号成立).所以24a ab+的最小值为3+ 14.{}()43,--+∞【详解】由题意可知,直线y k=与函数()f x的图象有两个交点,作出直线y k=与函数()f x的图象如图所示:由图象可知,当4k=-或3k>-时,直线y k=与函数()f x的图象有两个交点.因此,实数k的取值范围是{}()43,--+∞.15.4或10【详解】∵f(x)满足5412f fππ⎛⎫⎛⎫=⎪ ⎪⎝⎭⎝⎭,∴541223xπππ+==是f(x)的一条对称轴,∴362kπππωπ⋅+=+,∴13kω=+,k∈Z,∵ω>0,∴1,4,7,10,13,ω=⋯.当5,412xππ⎛⎫∈ ⎪⎝⎭时,5,646126xπππππωωω⎛⎫+∈++⎪⎝⎭,要使()f x在区间5,412ππ⎛⎫⎪⎝⎭上有最小值无最大值,则:31624624355321262ππππωωππππω⎧≤+<⎪⎪⇒≤<⎨⎪<+⎪⎩或57285224627593521262ππππωωππππω⎧≤+<⎪⎪⇒≤<⎨⎪<+⎪⎩,此时ω=4或10满足条件;区间5,412ππ⎛⎫⎪⎝⎭的长度为55312412126πππππ-=-=,当13ω 时,f(x)最小正周期22136Tπππω=<,则f(x)在5,412ππ⎛⎫⎪⎝⎭既有最大值也有最小值,故13ω 不满足条件.综上,ω=4或10.16.14a<或322a≤≤【详解】因2()2xf x-=在[1,)+∞上单调递增,则有min1()(1)2f x f==,于是得()f x在[1,)+∞上的值域是1[,)2+∞,设()g x的值域为A,1212在上的值域包含于()g x 的值域”,从而得1[,)2A +∞⊆,0x <时,2()2g x x a =+为减函数,此时()2g x a >,0x ≥时,()sin 2g x a x =+,此时2||()2||a g x a -≤≤+,当122a <,即14a <时,1[,)2A +∞⊆成立,于是可得14a <,当122a ≥,即14a ≥时,要1[,)2A +∞⊆成立,必有0x ≥,()[2,2]g x a a ∈-+满足22122a aa ≤+⎧⎪⎨-≤⎪⎩,即232a a ≤⎧⎪⎨≥⎪⎩,从而可得322a ≤≤,综上得14a <或322a ≤≤,所以实数a 的取值范围是14a <或322a ≤≤.四、解答题:本大题共6小题,共70分.第17题10分,第18至22题均12分.17.【详解】(1).若选①:222213{|1}{|0}{|0}{|13}1111x x x x A x x x x x x x x x --+-=<=-<=<=-<<++++,若选②:{|12}{|212}{|13}A x x x x x x =-<=-<-<=-<<若选③:()(){}233{|log }0|31011xxA x y x x x x x x ⎧⎫--===>=-+>=⎨⎬++⎩⎭{|13}x x -<<,()22{|0}{|()10}{|(2)(1)0}B x x x a a x x a x a x x x ⎡⎤=++-<=++-<=+-<⎣⎦,所以{|2<1}B x x =-<,{|13}U C A x x x =≤-≥或,{|21}U C B x x x =≤-≥或,故()()U U C A C B ⋃=1{}1|x x x ≤-≥或.(2).由(1)知{|13}A x x =-<<,()22{|0}{|()10}B x x x a a x x a x a ⎡⎤=++-<=++-<⎣⎦,因为“x A ∈”是“x B ∈”的充分不必要条件,①若(1)a a -<--,即12a >,此时{|(1)}B x a x a =-<<--,所以1,3(1)a a -≥-⎧⎨≤--⎩等号不同时取得,解得4a ≥.②若(1)a a -=--,则B =∅,不合题意舍去;③若(1)a a ->--,即12a <,此时{|(1)}B x a x a =--<<-,1(1),3a a-≥--⎧⎨≤-⎩解得3a ≤-.综上所述,a 的取值范围是(][),34,-∞-⋃+∞.18.【详解】(1).不等式2tan 0x θ-+≥对x ∈R 恒成立,则0∆≤,即24tan 0θ-≤,tan 2θ≥,则tan θ的取值范围为[2,)+∞(2).由(1)知tan θ的最小值为2,则22sin 3sin cos 1θθθ++22223sin 3sin cos cos sin cos θθθθθθ++=+223tan 3tan 1126119tan 1415θθθ++++===++.19.【详解】(1).因为()3f α=,所以π2sin 2236α⎛⎫++= ⎪⎝⎭,即1sin 262απ⎛⎫+= ⎪⎝⎭,又由()0,πα∈,得132666απππ<+<,所以π5π266α+=,解得π3α=.(2).对ππ,42x ⎡⎤∈⎢⎥⎣⎦,有2ππ7π2366x ≤+≤,所以1sin 226απ⎛⎫-≤+ ⎪⎝⎭()12f x ≤≤所以要使()3f x m >-对任意的ππ,42x ⎡⎤∈⎢⎣⎦恒成立,只需()min 3f x m >-,所以31m -<,解得4m <.故所求实数m 的取值范围为(),4-∞.的图象不具备先上升,后下降,再上升的特点,不符合题意,对于③,当0A >时,函数()sin()44f x A x B ππ=-+在[0,3]上的图象是上升的,在[3,7]上的图象是下降的,在[7,11]上的图象是上升的,满足题设条件,应选③.(2).依题意,84A B A B +=⎧⎨-+=⎩,解得2,6A B ==,则[]()2sin()6,0,10,N 44f x x x x ππ=-+∈∈,由2sin()6544x ππ-+<,即1sin()442x ππ-<-,而[]0,10,N x x ∈∈,解得{0,6,7,8}x ∈,所以该水果在第1,7,8,9月份应该采取外销策略.21.【详解】(1).()f x 的定义域为R ,4114()log log ()22x xa a x x f x f x --++-===,故()f x 是偶函数.(2).当2a =时,22411()log log (2)22x x x x f x +==+,因为20x >,所以1222x x +≥,所以()1f x ≥,即()f x 的值域是[1,)+∞.(3).“[][]124,4,0,4x x ∀∈-∃∈,使得12()()2f x g x ->”等价于min min ()()2g x f x <-.22()111)1g x x =-=--=--,所以min ()(1)1g x g ==-.令函数12[),0,)(2x x x h x +∈=+∞,对12,[0,)x x ∀∈+∞,当12x x >时,有211212121212*********()()2222(22)(10222222x x x x x x x x x x x x x x h x h x --=+--=-+=-->⋅⋅,所以()h x 在[0,)+∞上单调递增.于是,当1a >时,()f x 在[0,4]单调递增,故min ()(0)log 2a f x f ==,所以log 221a ->-,解得2a <,即a 的范围为12a <<;当01a <<时,()f x 在[0,4]单调递减,故min 257()(4)log 16a f x f ==,所以257log 2116a->-,无解.综上:a 的取值范围为(1,2).22.【详解】(1).∵不等式()0f x <解集为(3,)b -,则2()10f x ax x =++=的根为3,b -,且3b -<,∴11033a b b a a>-=-+=-,,,解得2392a b ==-,.(2).令1,22112x t =⎡⎤∈⎢⎥⎣⎦,若(2)()x f g x ≤,即2214112a t t t t++≤-+,则242a t t -≤-,∵22y t t =-的开口向上,对称轴为1t =,则22y t t =-在1,12⎡⎤⎢⎥⎣⎦单调递减,在(]1,2单调递增,且1|1t y ==-,∴41a -≤-,即03a <≤,故实数a 的取值范围为(]0,3.(3).2()1(0)f x ax x a =++>的开口向上,对称轴为12x a =-,∵211x x -=,根据二次函数的对称性不妨设121x x a+≥-,则有:当112x a≥-时,()f x 在12[,]x x 上单调递增,则可得()()()2222212221111()()1111211f x f x ax x ax x a x x ax a ⎡⎤-=++-++=+-+=++≥⎣⎦,即12112a a a ⎛⎫⨯-++≥ ⎪⎝⎭,解得1a ≥;当12x a <-,即22x a >-时,()f x 在1,2x a -⎪⎢⎣⎭上单调递减,在2,2x a -⎢⎥⎣⎦上单调递增,则可得()222222111()()111242f x f ax x a x a a a ⎛⎫⎛⎫--=++--=+≥ ⎪ ⎪⎝⎭⎝⎭,∵211211x x x x a -=⎧⎪⎨+≥-⎪⎩,则21122x a +≥,∴114a ≥,即4a ≥;综上所述:4a ≥,故正数a 的最小值为4.。

2024年全国硕士研究生招生考试《英语二》模拟测试卷(2)

2024年全国硕士研究生招生考试《英语二》模拟测试卷(2)

2024年全国硕士研究生招生考试《英语二》模拟测试卷(2)下列每小题的四个选项中,只有一项是最符合题意的正确答案,多选、错选或不选均不得分。

材料题根据以下材料,回答1-20题Holiday eating gets a bad rap. Around New Years, we’re hit with calls to clean up our wayward diets by eliminating 1and counting calories. But psychologists and epidemiologists alike 2 that all this fuss could do more harm than good.The social aspect of the holidays can make it difficult to 3 your usual diet—it’s hard to 4 another cookie or glass of wine when everyone around you is partaking—so it’s understandable if you feel like you must have caused your body harm. But research suggests that festive meals come with their own set of health 5. A 2017 study published in Adaptive Human Behavior and Physiology found that people who ate more meals in social 6 were more likely to feel better about themselves and have a wider social network—characteristics that, as the study authors point out, are 7with happiness, wellbeing, and lower risk of illness. Evening meals involving alcohol were the most likely to 8 feelings of warmth and bonding.Sometimes, participating in these social situations involves 9caloric, fatty, or sugary foods and drinks. Bioethicists at Johns Hopkins argue that those foods, too, have health value. “Sharing food is a way to express love, forge relationships, and 10 bonds,”they wrote in an article published in the Kennedy Institute of Ethics Journal. “What we eat expresses our personal and 11 identities.”Our cultural obsession over whether foods are healthy or not is 12universal. Psychologists at the University of Pennsylvania recently asked 947 Indian, French, and American participants to sort a list of foods in whatever 13they deemed mostappropriate. Their results showed that 14French and Indian respondents generally chose to sort foods into 15 groupings like “food vs. drink,” most Americans chose to 16 foods by whether they deemed them healthy or unhealthy.Researchers found that the French were the most likely to associate food with pleasure, and the least likely to associate food with health. Americans were on the 17 end of the spectrum. That’s significant because French people, on average, have lower rates of heart disease and live around four years longer. That’s not to say that French diets are the key to health and 18, but a(n) 19 on the pleasure of food rather than its health-value certainly doesn’t seem to 20.1. 【完形填空】第1题答案是A. weedsB. treatsC. drugsD. staples正确答案:B2. 【完形填空】第2题的答案是A. denyB. demandC. cautionD. promise正确答案:C3. 【完形填空】第3题的答案是A. diversifyB. improveC. balanceD. maintain正确答案:D4. 【完形填空】第4题的答案是A. turn down第 2 页共 21 页B. eat upC. indulge inD. settle for正确答案:A5. 【完形填空】第5题的答案是A. restrictionsB. benefitsC. regulationsD. risks正确答案:B6. 【完形填空】第6题的答案是A. classesB. termsC. settingsD. relations正确答案:C7. 【完形填空】第7题的答案是A. comparedB. presentedC. associatedD. equipped正确答案:C8. 【完形填空】第8题的答案是A. suppressB. exhaustC. exploitD. trigger正确答案:D9. 【完形填空】第9题的答案是A. preparingB. consumingC. limitingD. avoiding正确答案:B10. 【完形填空】第10题的答案是A. reinforceB. diminishC. aggravateD. disconnect正确答案:A11. 【完形填空】第11题的答案是A. nationalB. culturalC. privateD. group正确答案:D12. 【完形填空】第12题的答案是A. in partB. far fromC. in essenceD. as usual正确答案:B13. 【完形填空】第13题的答案是A. combinationB. situationC. directionD. manner正确答案:D第 4 页共 21 页14. 【完形填空】第14题的答案是A. sinceB. untilC. whileD. unless正确答案:C15. 【完形填空】第15题的答案是A. neutralB. binaryC. openD. sequential正确答案:A16. 【完形填空】第16题的答案是A. purchaseB. processC. categorizeD. evaluate正确答案:C17. 【完形填空】第17题的答案是A. oppositeB. identicalC. severeD. moderate正确答案:A18. 【完形填空】第18题的答案是A. prosperityB. freedomC. wealthD. longevity正确答案:D19. 【完形填空】第19题的答案是A. banB. focusC. attemptD. attack正确答案:B20. 【完形填空】第20题的答案是A. hurtB. careC. matterD. hold正确答案:A下列每小题的四个选项中,只有一项是最符合题意的正确答案,多选、错选或不选均不得分。

河南省九师联盟2024届高考英语试题模拟卷(2)含解析

河南省九师联盟2024届高考英语试题模拟卷(2)含解析

河南省九师联盟2024届高考英语试题模拟卷(2)注意事项1.考生要认真填写考场号和座位序号。

2.试题所有答案必须填涂或书写在答题卡上,在试卷上作答无效。

第一部分必须用2B 铅笔作答;第二部分必须用黑色字迹的签字笔作答。

3.考试结束后,考生须将试卷和答题卡放在桌面上,待监考员收回。

第一部分(共20小题,每小题1.5分,满分30分)1.It is obvious to the students _____________they should get well prepared for their future.A.as B.thatC.which D.whether2.public bicycles with a mobile app is more convenient for users.A.To unlock B.Unlock C.Unlocked D.Unlocking3.We packed all the hooks in wooden boxes so that they damaged.A.don’t get B.won’t getC.didn’t get D.wouldn’t get4.This car is important to our family. We would repair it at our expense _______ it break down within the first year. A.could B.wouldC.might D.should5.After the fire,________________ would otherwise be a cultural center is now reduced to a pile of ashes.A.that B.itC.what D.which6.It is difficult for any of us to eat better, exercise more, and sleep enough,______ we know we should.A.because B.even ifC.unless D.before7.— Thank you very much for giving me a hand when I was in trouble.—Don’t mention it. I only did what anyone else _______ in my place.A.must do B.could have doneC.would do D.can have done8.They are smiling. There ______ much trouble solving the problem.A.couldn’t be B.mustn’t beC.can’t have been D.mustn’t have been9.The Japanese people keep up cheerful spirits ________ the world that they can get over the crisis caused by the terrible tsunami(海啸).A.being convinced B.convinced C.to convince D.having convinced10.The expert points out the phenomenon that cream goes bad faster than butter______ its structure rather than itschemical composition.A.lives up to B.gets down toC.comes down to D.stands up to11.These new books are a very welcome _________ to the school library.A.addition B.arrival C.attitude D.audience12.— Y ou could always put the decision off a little bit longer.— __________ If I leave it much longer I might miss my chance.A.That’s reasonable.B.Isn’t it a good idea?C.Do you think so? D.I can’t agree more.13.—-What difference will it make _____we shall go to the concert on Tuesday or Saturday?—They offer a discount on weekdays.A.that B.whenC.if D.why14.Instead of making choices for their children, liberal parents usually say, “Go where you ________ .”A.will B.shouldC.can D.must15.Under good treatment, many patients are beginning to ________ and will soon recover.A.turn up B.catch upC.pick up D.show up16.Anyone with an eye on the employment situation knew the assessment about economic recovery _______ just around the corner was correct.A.being B.to be C.was D.having been17.The same boiling water softens the potato and hardens the egg. It’s about ________you’re made of, not the circumstances.A.that B.whatC.how D.who18.Y ou have a big mouth, Tom. You have told everybody the secret.A.can’t B.mustn’tC.shouldn’t D.mightn’t19.By the time I saw the angry expression on his face, I ________ exactly what I was having to face.A.knew B.had known C.would know D.have known20.________ has greater potential than flammable ice being mined from underneath the South China Sea when it comes to a global energy revolution.A.nothing B.neitherC.no one D.none第二部分阅读理解(满分40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。

2023届湖南省重点大学附属中学高三下学期模拟试卷(二)物理试题及答案

2023届湖南省重点大学附属中学高三下学期模拟试卷(二)物理试题及答案

湖南重点大学附中2023届模拟试卷(二)物理注意事项:1.答卷前,考生务必将自己的姓名、准考证号填写在答题卡上。

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,将本试题卷和答题卡一并交回。

第Ⅰ卷一、单项选择题(本题共7小题,每小题4分,共28分。

每小题给出的四个选项中,只有一个选项是符合题目要求的)1.下列说法正确的是A.图(a)中,观看3D电影时佩戴特殊的眼镜是运用了双缝干涉的原理B.图(b)叫牛顿环,是光的衍射现象C.图(c)水中的气泡看上去特别明亮,是由于光发生了折射引起的D.图(d)所示光导纤维的内芯折射率比外套大2,如图甲所示是用光照射某种金属时逸出的光电子的最大初动能随入射光频率的变化图像(直线与横轴的交点的横坐标为4.29,与纵轴的交点的纵坐标为0.5),如图乙所示是氢原子的能级图,下列说法正确的是A.根据该图像能求出普朗克常量B.该金属的逸出功为0.5eVC.该金属的极限频率为D.用能级的氢原子跃迁到能级时所辐射的光照射该金属能使该金属发生光电效应3.很多智能手机都有加速度传感器,能通过图像显示加速度情况。

用手掌托着手机,打开加速度传感器,手掌从静止开始迅速上下运动,得到如图所示的竖直方向上加速度随时间变化的图像,该图像以竖直向上为正方向,重力加速度的大小,由此可判断出A.手机可能离开过手掌B. 手机在时刻运动到最高点C.手机在时刻改变运动方向D.手机在~时间内,受到的支持力先减小再增大4.如图,光滑绝缘水平面上,由1、2、3三个带电量均为+q、质量均为m的相同金属小球,均用长为L的三根绝缘细绳连接,A、B、C分别为三根绝缘细绳的中点,O为三角形中心,现选取无穷远处为电势零点(已知单个点电荷q周围空间的电势,r为到点电荷的距离),则下列说法正确的是A.O点的电场强度不为零,且方向向上B.若长度L可调节,则A、O两点的电势可能相等C.系统的总电势能为D.系统的总电势能为5.如图所示,一理想自耦变压器线圈AB绕在一个圆环形闭合铁芯上,左端输入正弦交流电压,L1为相同的灯泡,其电阻均为且恒定不变,定值电阻的阻值为灯泡阻值的。

《保教知识与能力》模拟卷(二)

《保教知识与能力》模拟卷(二)

《保教知识与能力》模拟卷(二)一、单项选择题(本大题共10小题,每小题3分,共30分)1. 操作工人从机器的运转声音中能够辨别出是否有故障,而非此行业的人则是无法做到的。

这表明知觉具有()。

A.活动性B. 选择性C. 恒常性D. 理解性2. 蒙眼游戏结束后,孩子们摘掉了眼罩,感觉自己眼前一亮,白茫茫一片,慢慢才能看清。

这体现了感觉规律中的()。

A.适应B. 对比C. 明适应D. 暗适应3. 明明和闹闹在一起玩的时候发生了争执,他们来找老师评理。

原因是两个人面对面站着,明明说小汽车在自己的左边,球在右边;而闹闹说球在左边,小汽车在右边。

两人谁也不让谁。

明明和闹闹更符合哪一年龄阶段?()A.小班B. 中班C. 大班D. 学前班4. 2 岁的小宝不会自己穿鞋,可偏要自己穿;不会自己喝水,可偏要自己喝。

这反映了()。

A.动作的发展B.自我意识的发展C. 情感的发展D. 认知的发展5.“虽然人们的发展遵循着特定的规律,但在发展过程中,有的人颜色感知能力强,有的人音色辨别能力强。

”这反映的是发展的()。

A.连续性与阶段性B. 方向性和顺序性C. 普遍性和差异性D. 不平衡性6.看到明明推倒了亮亮,并获得了亮亮的小手枪,丽丽可能在以后尝试使用这个方法,这属于()。

A.直接强化B.替代性强化C.自我强化D.负强化7. 陶行知先生说:“要把教育和知识变成空气一样,弥漫于宇宙,洗荡于乾坤,普及众生,人人有得呼吸。

”下列选项中哪个理论更好地体现了这一教育观点?()A.小先生制B. 整个教学法C. 教学做合一D. 艺友制师范教育8.小二班的明明小朋友,经常把鸡蛋、苹果、梨、南瓜等归为一类,问她为什么把它们分到一起,她回答说:“因为都是圆形的”。

明明小朋友是依据()进行的分类。

A.感知特征B.概念C.生活情境D.功能和用途9. 学会评价自己和同伴的作品,培养儿童的审美能力。

这是()美术欣赏的教育内容。

A.托班B. 小班C. 中班D. 大班10. 神经系统是学前期发展最早且发育最快的系统。

大学英语四级模拟试卷二及参考答案

大学英语四级模拟试卷二及参考答案

⼤学英语四级考试全真预测试卷 Model Test TwoPart I Writing(30 minutes)Directions: For this part, you are allowed 30 minutes to write a composition on the topic Should Smoking Be Completely Banned. You should write at least 120 words following the outline given below in Chinese:1. 有⼈赞同完全禁⽌吸烟,理由是……2. 有⼈不赞同完全禁⽌吸烟,理由是……3. 我的看法。

Should Smoking Be Completely BannedPart II Reading Comprehension (Skimming and Scanning)(15 minutes)Directions: In this part, you will have 15 minutes to go over the passage quicklyand answer the questions on Answer Sheet 1.For questions 1-7,choose the best answer from the four choices marked [A],[B],[C]and [D]. For questions 8-10,completethe sentences with the information given in the passage.Space Our Future in Space: It Has Already Begun!We are all space travelers. But we’ve stayed close to home until now. One day,we may leave our “mother ship”Earth to make our home among the stars.A giant, spherical “spaceship”, about 8,000 miles in diameter, is speedingthrough the solar system right now. It is cruising at an incredible 66,600 milesper hour.It’s not a giant, Star Wars mother ship. It’s spaceship Earth, the home ofover four billion people. This water coated spaceship has been traveling throughthe universe for about five billion years. Only within the past 25 years, however,have some of its passengers broken free of Earth’s gravityBut 25 years from now, many people, including you, might live in an orbitingspace station 200 miles above the Earth.Space CitiesScientists have already designed special space factories. These factories will take advantage of the absence of gravity (zero gravity) to produce everything from life saving drugs to perfect ball bearings.Other scientists have designed space colonies, complete with farms, schools,and artificial day and night. Hundreds, or even thousands, of people will live, work,play—even go toschool, far above the Earth.Our conquest of space, of course, has already begun. We have explored part of the Moon, sent robot spaceships onto the surface of Venus and Mars, and aimed space probes past the planets of Jupiter and Saturn.Last June, one robot ship, Pioneer 10, left our solar system forever. Andastronauts from both the Soviet Union and the United States have lived in spacestations.The conquest of space, without question, is one of the greatest adventures human beings have ever set out on. But it may be more than a great adventure. Some scientiststhink the conquest of space may be a necessity for survival of the human species.We are tearing up more and more of the Earth to get raw materials for industry.And we are polluting the air and water as we manufacture products that we need or want. Almost everything that seems to make our lives more comfortable, and fromelectricity to pesticides, uses up or alters a piece of our planet’s natural environment.Why Go into Space?Yet our solar system is full of resources. The moon is chockfull of valuable metals. So are the asteroids, the small, rocky, planet like bodies orbiting the sun most of them between Mars and Jupiter. These metals, if we can get them, could be used to build factories and space stations.Also, in space, there is no atmosphere to filter out the sun’s energy. There is plenty of solar energy to be turned into electricity for manufacturing, for creating comfortable living conditions.Getting away from Earth has other advantages, too. Modern industry uses manykinds of metal alloys (mixtures of metal thatare better for certain purposes thanpure metals). Yet some metal alloys either can’t be made or are very expensive to make on Earth because of gravity. For instance, certain metals don’t mix well onEarth. But in zero gravity, molten (hot, liquid) metals mix more evenly. This is because there is no gravity to pull the heavier metals down, while the lighter ones float on top.From space, too, we can look down on the Earth and study the atmosphere, its weather, and the effects of air pollution.And because there is no strong gravity to break free from, our future homes away from Earth will be convenient starting points for travel to distant planets.But, while going into space might solve some problems, outer space can also be a dangerous place. For example, in outer space, we have to protect ourselves from the dangers of ultraviolet light and cosmic rays. Ultraviolet light from the sun can give us bad sunburns right here on Earth. Yet, Earth’s atmosphere screens out most of that harmful radiation. Cosmic rays are tiny high energy particles from outer space. Again, the Earth shields us from most of them.At Home in Space?But in space, without special protection, we would be exposed to much stronger radiation from ultraviolet light and cosmic rays. Also, in the zero gravity of outer space, our bones will lose calcium and become weaker. This will be more of a problem the longer people stay out in space. Doctors are looking for a way to keep our bones from losing calcium in outer space. And a small spaceship just might “drive you batty” after a while. But even on a short trip in outer space, you might not feelas well as you’d like to. Space travel could make you seasick!Yet, these risks won’t keep people from going into space. Eventually, an Earth like environment will be built in space. And they will be populated by people with many different interests: medicine, construction, farming, teaching, mining, and so on.The next hundred years will be filled with other worldly adventures, exciting scientific discoveries, and danger, as humans leave Earth—perhaps forever.Aging in SpaceSuppose a space traveler is moving at a velocity of 186,200 miles per second.For every hour that passes for him, 30 hours pass on Earth. If he travels for a year in this fashion (having accelerated instantaneously) and then turns around and comes back at this speed (having turned around instantaneously), he will find that while he has seemed to himself to have traveled two years, the men on Earth would claim he had been absent for 30 years.Suppose the space traveler had left at the age of 30, leaving behind a twin brother also aged 30. When he returned he would be 32, but his stay at home twinbrother would be 60. That is why the “clock paradox”, is sometimes called the “twin paradox”.Of course it takes quite a long while to accelerate to a high speed, and a long while to make a turn and head back again, so conditions aren’t quite as clear cut as just described.1.The giant, spherical spaceship mentioned in the passage is.[A]the outer space[B]a man made spaceship[C]the planet Earth[D]the Star Wars mothe ship2.Some persons have traveled into outer space after conquering within the past 25years.[A]the universe[B]Earth’s gravity[C]the earth[D]outer space3.We have explored or sent robot spaceships to the following space except.[A]the moon[B]Venus[C]Jupiter[D]Mars4.Why is the conquest of space more than a great adventure?[A]Because it is full of challenges for human beings.[B]Because it may be necessary for human beings to survive.[C]Because it is the greatest adventure in human history.[D]Because it is more exciting than any other adventures.5.The moon and the asteroids are alike with respect to their .[A]size and moving ways[B]comfortable living conditions[C]rich and valuable metals[D]solar energy6.Why can’t ultraviolet light scorch our skin on Earth as seriously as it does in space places?[A]Because the Earth’s atmosphere can make ultraviolet light less harmful.[B]Because ultraviolet can’t reach the Earth at all.[C]Because the Earth is far away from those planets radiating ultraviolet light.[D]Because other space places is near from those planets radiating ultravioletlight.7.In spite of many risks, scientists will finally build in space suitable for humans to live.[A]an environment without ultraviolet light[B]a lot of homes[C]an Earth like environment[D]an environment with atmosphere8.The reason some metal alloys can’t be made on Earth is that the heavier metals together with the lighter ones.9.In space, there is no atmosphere to filter out the sun’s energy. There is plentyof solar energy to be turned into, for creating comfortable living conditions.10.According to the author, will be caused to a man in gravity free space.Part III Listening Comprehension(35 minutes)Section ADirections: In this section, you will hear 8 short conversations and 2 long conversations. At the end of each conversation, one or more questions will be askedabout what was said. Both the conversation and the questions will be spoken only once. After each question there will be a pause. During the pause, you must read the four choices marked [A], [B], [C]and [D], and decide which is the bestanswer. Then mark the corresponding letter on Answer Sheet 2 with a single line through the centre.11.[A]Tennis equipment.[B]Volleyball equipment.[C]Football equipment.[D]Basketball equipment.12.[A]He must meet his teacher.[B]He must attend a class.[C]He must go out with his girlfriend.[D]He must stay at school to finish his homework.13.[A]It’s not as good as it was.[B]It’s better than it used to be.[C]It’s better than people say.[D]It’s even worse than people say.14.[A]Because he doesn’t like football.[B]Because Maria fell ill.[C]Because he didn’t have the time.[D]Because Maria can’t stand football.15.[A]A temporary job.[B]A permanent job.[C]Some money for the vacation.[D]Some money for the university fees.16.[A]The woman did most of the talking.[B]The man did most of the talking.[C]The woman was wearing a black sweater.[D]The man and the woman had dark hair.17.[A]A sunny day. [B]A raincoat.[C]An attractive hut. [D]A lovely hat.18.[A]Librarian and student. [B]Operator and caller.[C]Boss and secretary.[D]Customer and repairman. Questions 19 to 22 are based on the conversation you have just heard.19.[A]The benefits of strong business competition.[B]A proposal to lower the cost of production.[C]Complaints about the expense of modernization.[D]Suggestions concerning new business strategies.20.[A]It costs much more than its worth.[B]It should be brought up to date.[C]It calls for immediate repairs.[D]It can still be used for a long time.21.[A]The personnel manager should be fired for inefficiency.[B]A few engineers should be employed to modernize the factory.[C]The entire staff should be retrained.[D]Better educated employees should be promoted.22.[A]Their competitors have long been advertising on TV.[B]TV commercials are less expensive.[C]Advertising in newspapers alone is not sufficient.[D]TV commercials attract more investments.Questions 23 to 25 are based on the conversation you have just heard.23.[A]Searching for reference material.[B]Watching a film of the 1930s’.[C]Writing a course book.[D]Looking for a job in a movie studio.24.[A]It’s too broad to cope with. [B]It’s a bit outdated.[C]It’s controversial.[D]It’s of little practical value.25.[A]At the end of the online catalogue.[B]At the Reference Desk.[C]In the New York Times.[D]In the Reader’s Guide to Periodical Literature.Section BDirections: In this section, you will hear 3 short passages. At the end of each passage, you will hear some questions. Both the passage and the questions will be spoken only once. After you hear a question, you must choose the best answer from the four choices marked [A], [B], [C]and [D].Then mark the corresponding letter on Answer Sheet 2 with a single line through the centre.Passage OneQuestions 26 to 28 are based on the passage you have just heard.26.[A]The liberation movement of British women.[B]Rapid economic development in Britain.[C]Changing attitudes to family life.[D]Reasons for changes in family life in Britain.27.[A]Because millions of men died in the war.[B]Because women had proved their worth.[C]Because women were more skillful than men.[D]Because factories preferred to employ women.28.[A]The concept of “the family”as a social unit.[B]The attitudes to birth control.[C]The attitudes to religion.[D]The ideas of authority and tradition.Passage TwoQuestions 29 to 31 are based on the passage you have just heard.29.[A]Synthetic fuel. [B]Solar energy.[C]Alcohol.[D]Electricity.30.[A]Air traffic conditions. [B]Traffic jams on highways.[C]Road conditions.[D]New traffic rules.31.[A]Go through a health check. [B]Take little luggage with them.[C]Arrive early for boarding. [D]Undergo security checks.Passage ThreeQuestions 32 to 35 are based on the passage you have just heard.32.[A]Washing plates. [B]Clearing tables.。

2023届上海市上海中学高三模拟卷练习二英语试卷(含答案)

2023届上海市上海中学高三模拟卷练习二英语试卷(含答案)

高考英语上海卷模拟试卷(二)考生注意:1.考试时间120分钟,试卷满分140分。

2.本考试设试卷和答题纸两部分。

第I卷(共90分)I. Listening Comprehension (25 分)Section A (每题1分,共10分)Directions: In Section A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversations and the questions will be spoken only once. After you hear a conversation and the question about it, read the four possible answers on your paper, and decide which one is the best answer to the question you have heard.II. Grammar and Vocabulary (每题1 分,共20 分)Section ADirections: After reading the passage below, fill in the blanks to make the passage coherent and grammatically correct. For the blanks with a given word, fill in each blank with the proper form of the given word; for the other blanks, used one word that best fits each blank.An implant that can cool nerves to block pain signals has been unveiled by researchers who say the device could offer an alternative to drugssuch as opioids (类鸦片活性肽).The team behind the device say it could bring benefits for management of acute pain such as (21)_______ experienced after amputations (截肢)or nerve grafts."We are optimistic that this represents a very promising starting point for an engineering approach (22)_______ treating pain,” said Prof John Rogers of Northwestern University in the US, a co-author of the research. But he cautioned that it might be some time (23) _______ they were available to patients. "As with any implantable device, the regulatory process can be slow, typically (24) _______ (involve) much more extensive animal model studies over a period of years," he said.Writing in the journal Science, the team report that the device to block pain signals, which (25) _______(test) only on rats so far, involves a pump, external control system and an implant made from a soft, rubbery substance. The latter forms a sealed collection of tiny channels which form a twisting path in the part of the implant that sits around the target nerve like a cuff.When liquid coolant and dry nitrogen flow through the implant, the liquid causes a drop in temperature. An electronic sensor in the device allows the temperature at the nerve (26) _______(keep) constant."All body processes are based on metabolic chemical reactions, motions of ions and flows of fluids--all (27) _______slow down as a result ofcooling,” said Rogers. “ The net effect when cooling is applied to a nerve is in blocking of electrical signals."Among their experiments, the team tracked two rats with an injury, recording over a three- week period the minimum force that (28)_______be applied to the hind paw to cause the animal to retract (缩回)it. This data was then compared against three rats who were similarly injured but also had the implant. The results suggest bouts of cooling of the injured nerve from 37℃ to 10℃led to (29) _______(severe) pain, with a sevenfold increase in the force that could be applied to the paw. The team say the implant’s benefits include (30)_______, in contrast to opioids, it is not addictive. As the implant is made with water-soluble and biocompatible materials, it can break down in the body after use. The implant could be inserted as an extension of the patient’s initial surgery.Section BDirections: Complete the following passage by using the -words in the box. Each word can only be used once. Note that there is one more word than you need.A. attachedB. bottle-fedC. confusionD. invisibleE. originallyF. orphanedG. partneringH. procedureI. reproducedJ. subjectsK. unintentionallySaving Baby BearsReacting to the auditory assault of barking dogs, shouts and rifle blasts, a 168-pound American black bear shot out and hightailed (迅速逃走)it into the woods off a logging road.His sister, weighing in at 135 pounds, took a little more time to overcome her fear and (31) _______ before she, too, ran for the trees and away from the humans who had driven more than 100 miles to witness the bears’ return to the wild.The cubs were the 106th and 107th (32) _______ or injured bears to be raised or treated at the Progressive Animal Welfare Society (PAWS) Wildlife Center in Washington state, then released months later in the same general area where they were (33) _______found. Fitted with GPS collars and tattooed with identification numbers on their gums, the bears are also among the latest (34) _______of a long-term research study being conducted by Rich Beausoleil and Lindsay Welfelt, both biologists and bear and cougar specialists.The siblings were only two weeks old when a forest worker and his dog (35) _______ disturbed their den in February 2020, scaring away their mother. She never returned.The worker contacted the WDFW, an agency (36) _______with PAWS to rehabilitate sick, injured wild animals before releasing them back into their natural habitats."Their eyes and ears weren't open, and their teeth hadn't evenerupted,Jennifer Convy, PAWS senior director of wildlife, said of the cubs, which each weighed less than 2 pounds when they arrived and were the youngest ever to be raised at the center.Though (37) _______at first, the cubs weren't cuddled. Instead, their caregivers wore bear- scented bear suits during feedings once the cubs' eyes had opened. At PAWS, the staff and volunteers take pains to be (38)_______to all the animals in their care."They don't see us, ever. We don't talk to them. We don't name the animals because we don't want our staff or volunteers to get (39)_______,“ Convy said. After more than a year, the bear siblings were released.The wildlife biologists are studying how the bears raised at PAWS fare after their release, compared to their wild-reared brethren. One cub released in 2017 had her first litter of cubs in January."We've been to her den several times." Beausoleil said. "She (40)_______ and had cubs of her own ... This was kind of a turning point for us."After all, the whole point of PAWS' rehabilitation "is to protect and perpetuate (使持续) the species," he said.III. Reading Comprehension (45 分)Section A (每题1分,共15分)Directions: For each blank in the following passage there are four wordsor phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.The Other da Vinci CodeFor centuries, two of the most intriguing questions about Leonardo da Vinci's "Mon a Lisa" were "Who " and "When ”A(n) (41) _______made at Heidelberg University in 2005 pretty much answered both. A note written in a manuscript in the library (42) _______the account of da Vinci's first biographer, Giorgio Vasari: that the sitter was a merchanfs wife, Lisa Gherardini. The note also helped date the masterpiece to between 1503 and 1506.A(n) (43)_______ mystery-"Where "- is still in dispute. But on June 3rd a French engineer, Pascal Cotte, declared that he and a collaborator had (44) _______the landscape in the background of the painting. Arguments had (45)_______ been made for stretches of countryside in the Marche region and between Milan and Genoa. During a presentation in Vinci, near Florence, Mr Cotte maintained that the artist was more plausibly depicting a part of his native Tuscany 一one that keenly interested him at the time. According to this theory, da Vinci represented the area not as it was, but as, in an unrealised scheme, he (46) _______ it to be.Mr Cotte, who was asked by the Louvre (where the "Mona Lisa" hangs) to create a digital image of the painting, is the inventor of themultispectral camera: a device that can detect not only the drawing below the (47) _______of an oil painting, but also, where they exist, intermediate layers of work. It was among these, under what appears to be a pointed rock, that he found a(n) (48) _______sketch showing that da Vinci meant it to represent a castellated tower.The landscape of the "Mona Lisa" also includes a huge overhanging cliff. That is (49)_______ to one that da Vinci included in a sketch of a fortress (堡垒)contested by Pisa and Florence in the war that flared between them in 1503 (around the time he was painting Gherardini). The (50)_______ with the nearby cliff ——and a tower, known as the Caprona tower 一all overlook the river Amo as it snakes from Florence to Pisa. All three also feature in drawings made by da Vinci to illustrate a plan about which, says Mr Cotte, he became “(51) _______”.This involved diverting the Amo to (52) _______Pisa's water supply and give Florence an outlet to the Mediterranean. In the early 1500s, with the two citystates at war, the idea was under active consideration. Mr Cotte argues that a(n) (53) _______ winding through desolate countryside at the right of the "Mona Lisa" is too wide to be a road, as some have speculated, and is(54) _______the driedup bed of the Amo as da Vinci envisaged (设想)it once his plan had been adopted.It never was. But if Mr Cotte's theory is right, it might just explain why Gherardini, a Florentine, exhibits such a contented, if mysterious, (55)_______。

2024-2025学年湖北省八年级(上)期中语文模拟试卷(2)

2024-2025学年湖北省八年级(上)期中语文模拟试卷(2)

2024-2025学年湖北省八年级(上)期中语文模拟试卷(2)试卷满分:100分考试时间:120分钟日期:2024.11 姓名:班级:得分:一、积累运用(30分)1.(2分)下列加点字注音完全正确的一项是()A.溃.退(kuì)浩瀚.(hàn)黝.黑(yǒu)炽.热(chì)B.要塞.(sài)镌.刻(juān)滞.留(zhì)教诲.(huǐ)C.颁.发(bān)畸.形(qí)骤.雨(zhòu)胆怯.(qiè)D.翘.首(qiào)不辍.(chuò)匿.名(nì)绯.红(fēi)2.(2分)下列书写全部正确的一项是()A.锐不可挡抑扬顿挫待人接物匿名 B.张惶失措荡然无存永垂不朽懊丧C.眼花缭乱名副其实粗制滥造凛冽 D.锲而不舍振耳欲聋丰功伟绩管辖3.(2分)下列句子中加下划线成语使用不当的一项是()A.中国白手起家,一切从零开始,终于圆了航母舰载机着舰这一强军梦。

B.航母舰载战斗机着舰的一幕真是惊心动魄。

C.科研人员殚精竭虑,使我国的无人战机在当代天空叱咤风云。

D.中国军人展示出震耳欲聋、蓬勃向上的“中国力量”。

4.(2分)下列说法正确的一项是()A.《藤野先生》选自鲁迅的《呐喊》集,追忆了作者的恩师藤野严九郎的事迹,以及自己弃医从文的原因。

B.消息是迅速、简要地报道新近发生的重大事件的一种新闻体裁。

它的正文结构通常是按重要性递减的原则安排的,即所谓的“倒金字塔结构”。

C.郦道元是我国战国时魏国地理学家,所撰《水经注》既是古代地理名著,又具有较高的文学价值。

李白的《早发白帝城》、吴均的《与朱元思书》等后代山水作品都深受其影响。

D.律诗是近体诗的一种,分五律、七律等。

律诗的四联分别是首联、颈联、颔联、尾联,每首律诗的二三两联必须是对偶的。

5.(2分)给下面句子排序,衔接最恰当的一项是()春节,是全球华人寄予乡愁的节日,是中华文化的传播媒介,更是古老又现代、海纳百川又活力充沛的中国最鲜活的一张“名片”。

2024年广东省普通高中学业水平合格性考试数学模拟卷(二)(含答案)

2024年广东省普通高中学业水平合格性考试数学模拟卷(二)(含答案)

数 2024年第一次广东省普通高中学业水平合格性考试模拟卷(二)学位号填写在答题卡上。

用2B 铅笔将试卷类型(A )填涂在答题卡相应位置上。

将条形码横贴在答题卡右上角“条形码粘贴处”本试卷共22小题,满分150分。

考试用时90分钟。

注意事项:1. 答卷前,考生务必用黑色字迹的钢笔或签字笔将自己的姓名、考生号、考场号和座。

2. 选择题每小题选出答案后,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,然后再写上新的答案;不准使用铅笔和涂改液。

不按以上要求作答的答案无效。

4. 考生必须保持答题卡的整洁。

考试结束后,将试卷和答题卡一并交回。

─、选择题:本大题共12小题,每小题6分,共72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.已知全集U =R ,集合{}|13Ax x =<<,则CC UU AA =( )A .{|1x x <或3}x >B .{}|3x x ≥C .{|1x x ≤或3}x ≥D .{}|1x x ≤2.下列函数中,在区间(0,+∞)上是减函数的是( )A .y =x 2B .y =1x C .y =2x D .y =lg x 3. 已知角α的终边过点()1,2P −,则tan α等于( )A. 2B. 12−C. 2−D.124.函数lg y x =+的定义域是( )A .{1x x >或}0x <B .{}01x x <<C .{1x x ≥或}0x ≤D .{}01x x <≤5.已知R a ∈,则“1a >”是“11a<”的( ) A .充分不必要条件B .必要不充分条件6.不等式(2x −1)(x +2)>0的解集是(A .){2x x <−∣,或12x>B .12∣ >xx C .122xx−<<∣ D .{2}xx <−∣ 7.已知平面向量a =(-2,4),b =(n ,6),且a ∥b ,则n =( )A. 3 B .2C .1D .-18.已知,0x y >且xy =36,则x y +的最小值为( )A. B .4C .6D .129. 要得到函数4y sin x =−(3π)的图象,只需要将函数4y sin x =的图象( )A. 向左平移12π个单位 B. 向右平移12π个单位 C. 向左平移3π个单位 D. 向右平移3π个单位10. 已知函数()122,0,log ,0,x x f x x x ≤= > 则()()2f f −=( )A. -2B. -1C. 1D. 211.如图1,在正方体1111ABCD A B C D −中,E ,F 分别是AB ,AD 的中点,则异面直线1B C 与EF 所成的角的大小为( ) A .90° B .60°C .45°D .30°12. 某同学计划2023年高考结束后,在A ,B ,C ,D ,E 五所大学中随机选两所去参观,则A 大学恰好被选中的概率为( ) A.45B.35C.25 D. 15二、填空题:本大题共6小题,每小题6分,共36分。

2023年普通高等学校招生全国统一考试新高考仿真模拟卷数学(二)答案

2023年普通高等学校招生全国统一考试新高考仿真模拟卷数学(二)答案

2023年普通高等学校招生全国统一考试·仿真模拟卷数学(二)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,只有一个选项是符合题目要求的.1.已知集合{}2A x x x=≤,(){}2log1B x y x ==-,则A B ⋃=()A.[)1,+∞B.[)0,∞+C.(0,1)D.[]0,1【答案】B 【解析】【分析】分别化简集合,A B ,根据并集的定义求解.【详解】{}2A x x x=≤ ∴不等式2x x ≤的解集是集合A又因为(){}21001,01x x x x x A x x ≤⇒-≤⇒≤≤∴=≤≤又(){}2log 1x y x =- ,所以满足函数()2log 1y x =-中x 的范围就是集合B所以{}1011x x B x x ->⇒>∴=>所以{}{}{}[)01100,A B x x x x x x ∞⋃=≤≤⋃>=≥=+故选:B2.已知复数()()2i 1i z a =+-为纯虚数,则实数=a ()A.12-B.23-C.2D.2-【答案】D 【解析】【分析】根据复数乘法计算方法化简复数,结合纯虚数的概念求值即可.【详解】()()()2i 22i 1i i 2i 2i 2a a a a z a ==-++++---=,因为复数z 为纯虚数,所以2020a a -≠⎧⎨+=⎩,即2a =-.故选:D3.在正方形ABCD 中,M 是BC 的中点.若AC m = ,AM n = ,则BD =()A.43m n -B.43m n+ C.34m n -D.34m n+【答案】C 【解析】【分析】作图,根据图像和向量的关系,得到2()22BC AC AM m n =-=-和AB AC BC =- 222m m n n m =-+=-,进而利用BD BC CD BC AB =+=- ,可得答案.【详解】如图,AC m =,AM n =,且在正方形ABCD 中,AB DC=12AC AM MC BC -==,2()22BC AC AM m n ∴=-=- , AC AB BC =+,AB AC BC ∴=- 222m m n n m =-+=- ,∴BD BC CD BC AB =+=-= 22234m n n m m n--+=- 故选:C4.已知40.5=a ,5log 0.4b =,0.5log 0.4c =,则a ,b ,c 的大小关系是()A.b a c >>B.a c b >>C.c a b >>D.a b c>>【答案】C 【解析】【分析】利用指数函数,对数函数单调性,找出中间值0,1,使其和,,a b c 比较即可.【详解】根据指数函数单调性和值域,0.5x y =在R 上递减,结合指数函数的值域可知,()()400,0.50,10.5a ∈==;根据对数函数的单调性,5log y x =在(0,)+∞上递增,则55log 0.4log 10b =<=,0.5log y x =在(0,)+∞上递减,故0.50.5log 0.4log 0.51c =>=,即10c a b >>>>,C 选项正确.故选:C5.端午佳节,人们有包粽子和吃粽子的习俗.四川流行四角状的粽子,其形状可以看成一个正四面体.广东流行粽子里放蛋黄,现需要在四角状粽子内部放入一个蛋黄,蛋黄的形状近似地看成球,当这个蛋黄的表面积是9π时,则该正四面体的高的最小值为()A.4 B.6C.8D.10【答案】B 【解析】【分析】根据题意分析可知,当该正四面体的内切球的半径为32时,该正四面体的高最小,再根据该正四面体积列式可求出结果.【详解】由球的表面积为9π,可知球的半径为32,依题意可知,当该正四面体的内切球的半径为32时,该正四面体的高最小,设该正四面体的棱长为a 3a =,根据该正四面体积的可得2163334a a ⨯⨯=21334324a ⨯⨯⨯,解得a =.所以该正四面体的高的最小值为66633a =⨯=.故选:B6.现有一组数据0,l ,2,3,4,5,6,7,若将这组数据随机删去两个数,则剩下数据的平均数大于4的概率为()A.514 B.314C.27D.17【答案】D 【解析】【分析】先得到删去的两个数之和为4时,此时剩下的数据的平均数为4,从而得到要想这组数据随机删去两个数,剩下数据的平均数大于4,则删去的两个数之和要小于4,利用列举法得到其情况,结合组合知识求出这组数据随机删去两个数总共的情况,求出概率.【详解】0,l ,2,3,4,5,6,7删去的两个数之和为4时,此时剩下的数据的平均数为284482-=-,所以要想这组数据随机删去两个数,剩下数据的平均数大于4,则删去的两个数之和要小于4,有()()()()0,1,0,2,0,3,1,2四种情况符合要求,将这组数据随机删去两个数,共有28C 28=种情况所以将这组数据随机删去两个数,剩下数据的平均数大于4的概率为41287=.故选:D7.在棱长为3的正方体1111ABCD A B C D -中,O 为AC 与BD 的交点,P 为11AD 上一点,且112A P PD =,则过A ,P ,O 三点的平面截正方体所得截面的周长为()A. B.C.+D.+【答案】D 【解析】【分析】根据正方体的性质结合条件作出过A ,P ,O 三点的平面截正方体所得截面,再求周长即得.【详解】因为112A P PD =,即11113D P A D = ,取11113D H D C =uuuu r uuuu r,连接11,,PH HC A C ,则11//HP AC ,又11//AC AC ,所以//HP AC ,所以,,,,A O C H P 共面,即过A ,P ,O 三点的正方体的截面为ACHP ,由题可知APCH ===,PH =,11A C =,所以过A ,P ,O 三点的平面截正方体所得截面的周长为+.故选:D.8.不等式15e ln 1-≥+x a xx x对任意(1,)x ∈+∞恒成立,则实数a 的取值范围是()A.(,1e]-∞- B.(2,2e⎤-∞-⎦C.(,4]-∞- D.(,3]-∞-【答案】C 【解析】【分析】分离参数,将15e ln 1-≥+x a x x x 变为41e ,1ln x x xa x x---≤>,然后构造函数,即将不等式恒成立问题转化为求函数的最值问题,利用导数判断函数的单调性,求最值即可.【详解】由不等式15e ln 1-≥+x a xx x 对任意(1,)x ∈+∞恒成立,此时ln 0x >,可得41e ,1ln x x xa x x---≤>恒成立,令41e ,1ln x x x y x x ---=>,从而问题变为求函数41e ,1ln x x x y x x---=>的最小值或范围问题;令1()e x g x x -=-,则1()e 1x g x -'=-,当1x <时,1()e 10x g x -'=-<,当1x >时,1()e 10x g x -'=->,故1()e (1)0x g x x g -=-≥=,即1e x x -≥,所以4411ln 4ln 1e e e e 4ln x x x x x x x x ------=⋅=≥-,()*,当且仅当4ln 1x x -=时取等号,令()4ln 1h x x x =--,则44()1x h x x x-'=-=,当4x <时,()0h x '<,当>4x 时,()0h x '>,故min ()(4)34ln 40h x h ==-<,且当x →+∞时,()4ln 1h x x x =--也会取到正值,即4ln 1x x -=在1x >时有根,即()*等号成立,所以41e 4ln 4ln x x x x x x x---≥--=-,则41e 4ln x x xx---≥-,故4a ≤-,故选:C【点睛】本题考查了不等式的恒成立问题,解法一般是分离参数,构造函数,将恒成立问题转化为求函数最值或范围问题,解答的关键是在于将不等式或函数式进行合理的变式,这里需要根据式子的具体特点进行有针对性的变形,需要一定的技巧.二、选择题:本题共4小题,每小题5分,共20分.在每小题给出的选项中,有多项符合题目要求.全部选对的得5分,有选错的得0分,部分选对的得2分.9.在平面直角坐标系中,圆C 的方程为22210x y y +--=,若直线1y x =-上存在一点M ,使过点M 所作的圆的两条切线相互垂直,则点M 的纵坐标为()A.1B.C.1- D.【答案】AC 【解析】【分析】首先可根据圆的方程得出圆心与半径,然后根据题意得出点M 、圆心以及两个切点构成正方形,最后根据2MC =以及两点间距离公式即可得出结果.【详解】22210x y y +--=化为标准方程为:()2212x y +-=,圆心()0,1C ,,因为过点M 所作的圆的两条切线相互垂直,所以点M 、圆心以及两个切点构成正方形,2MC =,因为M 在直线1y x =-上,所以可设(),1M a a -,则()22224MCa a =+-=,解得:2a =或0a =,所以()2,1M 或()0,1M -,故点M 的纵坐标为1或1-.故选:AC.10.已知函数()()πsin 0,0,2f x A x A ωϕωϕ⎛⎫=+>><⎪⎝⎭的部分图象如图所示,若将()f x 的图象向右平移()0m m >个单位长度后得到函数()()sin 2g x A x ωϕ=-的图象,则m 的值可以是()A.π4B.π3C.4π3D.9π4【答案】AD 【解析】【分析】根据函数图象可确定A 和最小正周期T ,由此可得ω,结合π26f ⎛⎫= ⎪⎝⎭可求得ϕ,从而得到()(),f x g x 的解析式,根据()()f x m g x -=可构造方程求得()ππ4m k k =-∈Z ,由此可得m 可能的取值.【详解】由图象可知:2A =,最小正周期5ππ4π126T ⎛⎫=⨯-=⎪⎝⎭,2π2T ω∴==,ππ2sin 263f ϕ⎛⎫⎛⎫∴=+= ⎪ ⎪⎝⎭⎝⎭,()ππ2π32k k ϕ∴+=+∈Z ,解得:()π2π6k k ϕ=+∈Z ,又π2ϕ<,π6ϕ∴=,()π2sin 26f x x ⎛⎫∴=+ ⎪⎝⎭,()π2sin 23g x x ⎛⎫=- ⎪⎝⎭,()()π2sin 226f x m x m g x ⎛⎫-=-+= ⎪⎝⎭ ,()ππ22π63m k k ∴-+=-+∈Z ,解得:()ππ4m k k =-∈Z ,当0k =时,π4m =;当2k =-时,9π4m =.故选:AD.11.大衍数列来源于《乾坤谱》中对易传“大衍之数五十”的推论,主要用于解释中国传统文化中的太极衍生原理,数列中的每一项都代表太极衍生过程.已知大衍数列{}n a 满足10a =,11,,,n n na n n a a n n +++⎧=⎨+⎩为奇数为偶数,则()A.34a =B.221n n a a n +=++C.221,,2,2n n n a n n ⎧-⎪⎪=⎨⎪⎪⎩为奇数为偶数D.数列(){}1nn a -的前2n 项和的最小值为2【答案】ACD 【解析】【分析】当2n k =时,2122k k a a k +=+,当21n k =-时,2212k k a a k -=+,联立可得21214k k a a k +--=,利用累加法可得22122k a k k +=+,从而可求得221,2,2n n n a n n ⎧-⎪⎪=⎨⎪⎪⎩为奇数为偶数,在逐项判断即可.【详解】令k *∈N 且1k ≥,当2n k =时,2122k k a a k +=+①;当21n k =-时,221212112k k k a a k a k --=+-+=+②,由①②联立得21214k k a a k +--=.所以315321214,8,,4k k a a a a a a k +--=-=-= ,累加可得()22112114844222k k k k a a a k k k+++-==+++=⨯=+ .令21k n +=(3n ≥且为奇数),得212n n a -=.当1n =时10a =满足上式,所以当n 为奇数时,212n n a -=.当n 为奇数时,()21112n nn aa n ++=++=,所以22n n a =,其中n 为偶数.所以221,2,2n n n a n n ⎧-⎪⎪=⎨⎪⎪⎩为奇数为偶数,故C 正确.所以233142a -==,故A 正确.当n 为偶数时,()22222222n nn n aa n ++-=-=+,故B 错误.因为()()222212211222n n n n a a n ----=-=,所以(){}1nna -的前2n 项和21234212nn nSa a a a a a -=-+-++-+()()121222212n n n nn +=⨯+⨯++⨯=⨯=+ ,令()1n c n n =+,因为数列{}n c 是递增数列,所以{}n c 的最小项为1122c =⨯=,故数列(){}1nna -的前2n 项和的最小值为2,故D 正确.故选:ACD.【点睛】数列求和的方法技巧(1)倒序相加:用于等差数列、与二项式系数、对称性相关联的数列的求和.(2)错位相减:用于等差数列与等比数列的积数列的求和.(3)分组求和:用于若干个等差或等比数列的和或差数列的求和.12.已知抛物线()220y px p =>的准线为:2l x =-,焦点为F ,点(),P P P x y 是抛物线上的动点,直线1l 的方程为220x y -+=,过点P 分别作PA l ⊥,垂足为A ,1PB l ⊥,垂足为B ,则()A.点F 到直线1l 的距离为655B.2p x +=C.221p px y ++的最小值为1 D.PA PB +的最小值为655【答案】ABD 【解析】【分析】对于A ,用点到直线的距离公式即可判断;对于B ,利用抛物线的定义即可判断;对于C ,利用基本不等式即可判断;对于D ,利用抛物线的定义可得到PA PB PF PB BF +=+≥,接着求出BF 的最小值即可【详解】由抛物线()220y px p =>的准线为:2l x =-可得抛物线方程为28y x =,焦点为()2,0F ,对于A ,点F 到直线1l的距离为655d ==,故A 正确;对于B ,因为(),P P P x y 在抛物线上,所以利用抛物线的定义可得2P PF x =+,即2p x +=,故B 正确;对于C ,因为(),P P P x y 在抛物线上,所以28,0p p p y x x =≥,所以211221144111818888p p p pp p p p x x x x y x x x +=+=+=+++++1788≥=,当且仅当38p x =时,取等号,故C 错误;对于D ,由抛物线的定义可得PA PF =,故PA PB PF PB BF +=+≥,当且仅当,,P B F 三点共线时,取等号,此时1BF l ⊥,由选项A 可得点F 到直线1l的距离为5d =,故PA PB +的最小值为655,故D正确,故选:ABD三、填空题:本题共4小题,每小题5分,共20分.13.已知sin 3cos 0αα+=,则tan 2α=______.【答案】34##0.75【解析】【分析】利用已知等式可求得tan α,由二倍角正切公式可求得结果.【详解】由sin 3cos 0αα+=得:sin 3cos αα=-,sin tan 3cos ααα∴==-,22tan 63tan 21tan 194ααα-∴===--.故答案为:34.14.函数()()ln 211f x x x =++-的图象在点()()0,0f 处的切线方程是______.【答案】310x y --=【解析】【分析】求导函数,可得切线斜率,求出切点坐标,运用点斜式方程,即可求出函数()f x 的图象在点()()0,0f 处的切线方程.【详解】()()ln 211f x x x =++-,∴2()121f x x '=++,则(0)213f '=+=,又()ln 201(0)011f =⨯++-=-Q ,∴切点为()0,1-,∴函数()()ln 211f x x x =++-的图象在点()0,1-处的切线方程是()130,y x +=-即310x y --=.故答案为:310x y --=.15.2名老师带着8名学生去参加数学建模比赛,先要选4人站成一排拍照,且2名老师同时参加拍照时两人不能相邻.则2名老师至少有1人参加拍照的排列方法有______种.(用数字作答)【答案】3024【解析】【分析】分两种情况讨论:①若只有1名老师参与拍照;②若2名老师都拍照.利用计数原理、插空法结合分类加法计数原理可求得结果.【详解】分以下两种情况讨论:①若只有1名老师参与拍照,则只选3名学生拍照,此时共有134284C C A 2688=种排列方法;②若2名老师都拍照,则只选2名学生拍照,先将学生排序,然后将2名老师插入2名学生所形成的空位中,此时,共有222823C A A 336=种排列方法.综上所述,共有26883363024+=种排列方法.故答案为:3024.16.已知A ,B 是双曲线22:124x y C -=上的两个动点,动点P 满足0AP AB += ,O 为坐标原点,直线OA 与直线OB 斜率之积为2,若平面内存在两定点1F 、2F ,使得12PF PF -为定值,则该定值为______.【答案】【解析】【分析】设()()1122(,),,,,P x y A x y B x y ,根据0AP AB += 得到122x x x =-,122y y y =-,根据点A ,B 在双曲线22124x y -=上则22212212416,248y x y x -=-=,代入计算得22220x y -=,根据双曲线定义即可得到12PF PF -为定值.【详解】设()()1122(,),,,,P x y A x y B x y ,则由0AP AB += ,得()()()112121,,0,0x x y y x x y y --+--=,则122x x x =-,122y y y =-,点A ,B 在双曲线22124x y -=上,222211221,12424x y x y ∴-=-=,则22212212416,248y x y x -=-=()()222212122222x y x x y y ∴-=---()()()2222121212121212828442042x x x x y y y y x x y y =+--+-=--,设,OA OB k k 分别为直线OA ,OB 的斜率,根据题意,可知2OA OBk k ⋅=,即12122y y x x ⋅=,121220y y x x ∴-=22220x y ∴-=,即2211020x y -=P ∴在双曲线2211020x y -=上,设该双曲线的左、右焦点分别为12,F F ,由双曲线定义可知||12||||PF PF -为定值,该定值为.故答案为:.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.在ABC 中,角,,A B C 的对边分别是,,a b c ,()()()0a c a c b b a -++-=.(1)求C ;(2)若c =ABC 的面积是2,求ABC 的周长.【答案】(1)π3.(2).【解析】【分析】(1)将()()()0a c a c b b a -++-=化为222a b c ab +-=,由余弦定理即可求得角C .(2)根据三角形面积求得2ab =,再利用余弦定理求得3a b +=,即可求得答案.【小问1详解】由题意在ABC 中,()()()0a c a c b b a -++-=,即222a b c ab +-=,故2221cos 22a b c C ab +-==,由于(0,π)C ∈,所以π3C =.【小问2详解】由题意ABC 的面积是32,π3C =,即133sin ,2242ABC S ab C ab ab ===∴= ,由c =2222cos c a b ab C =+-得2223()6,3a b ab a b a b =+-=+-∴+=,故ABC 的周长为a b c ++=.18.已知数列{}n a 满足,()*1232311112222n n a a a a n n +++⋅⋅⋅+=∈N .(1)求数列{}n a 的通项公式;(2)若()21n n b n a =-,记n S 为数列{}n b 的前n 项和,求n S ,并证明:当2n ≥时,6n S >.【答案】(1)2nn a =(2)()12326n n S n +=-+【解析】【分析】(1)利用递推式相减得出2n n a =,并验证首项符合通项,最后得出答案;(2)错位相减法求前n 项和【小问1详解】1232311112222n n a a a a n ++++= ,①则()12312311111122222n n a a a a n n --++++=-≥ ,②①-②得11(2)2n n a n =≥,则2(2)n n a n =≥,当n =1时,由①得1112a =,∴1122a ==,∴2n n a =.【小问2详解】易得()212nn b n =-,()123123512222n n S n =⋅+⋅+∴+-⋅+ ,①()21341232522212n n S n +=⋅+⋅+⋅+∴+- ,②②-①得()()34112122222n n n S n ++=--++++- ()()21228212n n n +++=----()12326n n +=-+,故()12326n n S n +=-+,当2n ≥时,()12320n n +->6n S ∴>19.如图,四棱锥P ABCD -中,平面APD ⊥平面ABCD ,APD △为正三角形,底面ABCD 为等腰梯形,AB //CD ,224AB CD BC ===.(1)求证:BD ⊥平面APD ;(2)若点F 为线段PB 上靠近点P 的三等分点,求二面角F AD P --的大小.【答案】(1)证明见解析;(2)π4【解析】【分析】(1)先用几何关系证明π3A ∠=,然后根据余弦定理求出BD ,结合勾股定理可得BD AD ⊥,最后利用面面垂直的性质定理证明;(2)过P 作PG AD ⊥,垂足为G ,结合面面垂直的性质先说明可以在G 处为原点建系,然后利用空间向量求二面角的大小.【小问1详解】取AB 中点E ,连接CE ,根据梯形性质和2AB CD =可知,CD //AE ,且CD AE =,于是四边形ADCE 为平行四边形,故2CE AD BE CB ====,则CEB 为等边三角形,故π3A CEB ∠=∠=,在ABD △中,由余弦定理,222π2cos 1648123BD AB AD AB AD =+-⨯⨯=+-=,故BD =,注意到22212416BD AD AB +=+==,由勾股定理,π2ADB ∠=,即BD AD ⊥,由平面APD ⊥平面ABCD ,平面APD 平面ABCD AD =,BD ⊂平面ABCD ,根据面面垂直的性质定理可得,BD ⊥平面APD .【小问2详解】过P 作PG AD ⊥,垂足为G ,连接EG ,由平面APD ⊥平面ABCD ,平面APD 平面ABCD AD =,PG ⊂平面PAD ,根据面面垂直的性质定理,PG ⊥平面ABCD ,APD △为正三角形,PG AD ⊥,故AG GD =(三线合一),由AE EB =和中位线性质,GE //BD ,由(1)知,BD ⊥平面APD ,故GE ⊥平面APD ,于是,,GA GE GP 两两垂直,故以G 为原点,,,GA GE GP 所在直线分别为,,x y z 轴,建立如图所示的空间直角坐标系.由(1)知,BD ⊥平面APD ,又BD //y 轴,故可取(0,1,0)m =为平面APD的法向量,又P,(B -,根据题意,2BF FP = ,设(,,)F x y z,则()()1,2,,x y z x y z +-=--,解得12323,,333F ⎛- ⎝⎭,又(1,0,0)A ,(1,0,0)D -,(2,0,0)DA = ,42323,,333FA ⎛=-- ⎝⎭ ,设平面FAD 的法向量(,,)n a b c = ,由00n DA n FA ⎧⋅=⎪⎨⋅=⎪⎩ ,即0423230333a a =⎧⎪⎨--=⎪⎩,于是(0,1,1)n =- 为平面FAD 的法向量,故2cos ,2m n m n m n⋅=== ,二面角大小的范围是[]0,π,结合图形可知是锐二面角,故二面角F AD P --的大小为π420.为落实体育总局和教育部发布的《关于深化体教融合,促进青少年健康发展的意见》,某校组织学生参加100米短跑训练.在某次短跑测试中,抽取100名女生作为样本,统计她们的成绩(单位:秒),整理得到如图所示的频率分布直方图(每组区间包含左端点,不包含右端点).(1)估计样本中女生短跑成绩的平均数;(同一组的数据用该组区间的中点值为代表)(2)由频率分布直方图,可以认为该校女生的短跑成绩X 服从正态分布()2,N μσ,其中μ近似为女生短跑平均成绩x ,2σ近似为样本方差2s ,经计算得,2 6.92s =,若从该校女生中随机抽取10人,记其中短跑成绩在[]12.14,22.66以外的人数为Y ,求()1P Y ≥.2.63≈,随机变量X 服从正态分布()2,N μσ,则()0.6827P X μσμσ-<≤+=,()220.9545P X μσμσ-<<+=,()330.9974P X μσμσ-<<+=,100.68270.0220≈,100.95450.6277≈,100.99740.9743≈.【答案】(1)17.4(2)0.3723【解析】【分析】(1)结合频率分布直方图中求平均数公式,即可求解.(2)根据已知条件,可知,217.4, 6.92μσ==,即可求出212.14,222.66μσμσ-=+=,结合正态分布的对称性以及二项分布的概率公式,即可求解.【小问1详解】估计样本中女生短跑成绩的平均数为:()120.02140.06160.14180.18200.05220.03240.02217.4⨯+⨯+⨯+⨯+⨯+⨯+⨯⨯=;【小问2详解】该校女生短跑成绩X 服从正态分布()17.4,6.92N ,由题可知217.4, 6.92μσ==, 2.63σ=≈,则212.14,222.66μσμσ-=+=,故该校女生短跑成绩在[]12.14,22.66以外的概率为:1(12.1422.66)10.95450.0455P X -≤≤=-=,由题意可得,~(10,0.0455)Y B ,10(1)1(0)10.954510.62770.3723P Y P Y ≥=-==-≈-=.21.已知椭圆()2222:10x y C a b a b +=>>的左焦点为F ,右顶点为A ,离心率为22,B 为椭圆C 上一动点,FAB 面积的最大值为212+.(1)求椭圆C 的方程;(2)经过F 且不垂直于坐标轴的直线l 与C 交于M ,N 两点,x 轴上点P 满足PM PN =,若MN FP λ=,求λ的值.【答案】(1)2212x y +=;(2)λ=.【解析】【分析】(1)由题意可得22c e a ==,121()22a c b ++=,再结合222a b c =+可求出,a b ,从而可求出椭圆的方程;(2)由题意设直线MN 为1x ty =-(0t ≠),1122(,),(,)M x y N x y ,设0(,0)P x ,将直线方程代入椭圆方程中化简利用根与系数的关系,然后由PM PN =可得0212x t =-+,再根据MN FP λ=可求得结果.【小问1详解】因为椭圆的离心率为2,所以2c e a ==,因为FAB面积的最大值为12+,所以121()22a cb ++=,因为222a bc =+,所以解得1a b c ===,所以椭圆C 的方程为2212x y +=;【小问2详解】(1,0)F -,设直线MN 为1x ty =-(0t ≠),1122(,),(,)M x y N x y ,不妨设12y y >,设0(,0)P x ,由22112x ty x y =-⎧⎪⎨+=⎪⎩,得22(2)210t y ty +--=,则12122221,22t y y y y t t -+==++,所以12y y -==,因为PM PN =,所以2222101202()()x x y x x y -+=-+,所以222212102012220x x x x x x y y --++-=,所以12120121212()()2()()()0x x x x x x x y y y y +---+-+=,所以12120121212(11)()2()()()0ty ty ty ty x ty ty y y y y -+----+-+=,因为120y y -≠,所以12012(2)2()0t ty ty x t y y +--++=,所以20222222022t t t x t t t ⎛⎫--+= ⎪++⎝⎭,所以20222222022t x t t --+=++,解得0212x t =-+,因为MN FP λ=,所以222MN FP λ=,0λ>,所以222212120()()(1)x x y y x λ-+-=+,222212120()()(1)ty ty y y x λ-+-=+2222120(1)()(1)t y y x λ+-=+,所以22222222288(1)(1)(2)(2)t t t t t λ+++=++,化简得28λ=,解得λ=±,因为0λ>,所以λ=22.已知函数()()1ln R 1x f x x m m x -=-⋅∈+.(1)当1m =时,判断函数()f x 的单调性;(2)当1x >时,()0f x >恒成立,求实数m 的取值范围.【答案】(1)()f x 在()0,∞+上是单调递增的(2)2m ≤【解析】【分析】(1)对()f x 求导,从而确实()f x '为正及()f x 的单调性;(2)令()()()1(m )ln 1R x x x m x g =+--∈,然后分2m ≤和m>2两种情况讨论()g x 的单调性及最值,即可得答案.【小问1详解】当1m =时,()1ln 1x f x x x -=-+,定义域为()0,∞+()()()()()2222212111121x x x f x x x x x x x +-+'=-==+++,所以()0f x ¢>,所以()f x 在()0,∞+上是单调递增的.【小问2详解】当1x >时,()()1ln R 1x f x x m m x -=-⋅∈+,()0f x >等价于()()()()1ln 1g m x x x m x R =+--∈,则()0g x >,1g ()ln 1x x m x '=++-,令()1ln 1m h x x x =++-,则22111()x h x x x x-'=-=,当1x >时,()0h x '>,则()g x '在()1,+∞上是单调递增的,则()(1)2g x g m ''>=-①当2m ≤时,()0g x '>,()g x 在()1,+∞上是单调递增的,所以()(1)0g x g >=,满足题意.②当m>2时,(1)20g m '=-<,(e )e 1e 10m m m g m m --'=++-=+>,所以0(1,e )mx ∃∈,使00()g x '=,因为()g x '在()1,+∞上是单调递增的所以当0(1,)x x ∈时,()0g x '<,所以()g x 在0(1,)x 上是单调递减的,又(1)0g =,即得当0(1,)x x ∈时,()(1)0g x g <=,不满足题意.综上①②可知:实数m 的取值范围2m ≤.。

小升初数学模拟卷(2)及答案

小升初数学模拟卷(2)及答案

小升初数学模拟卷(二)一、“相信你的能力!"请你耐心填一填。

(本题共26分,每小题2分)1、在○里填上“<”、“>”、或“=”。

999○100141○616.53○6.5302米○18分米2、2.125精确到百分位约是(),把0.59万改写成以“一”为单位的数,写作()。

3、85=()÷8=10:()=()%=()小数4、把下面的各数按要求填在适当的圈里。

52201300723516886947324335能被2整除的数奇数5、2.4元=()元()角5千克230克=()千克6、73的分数单位是(),它有()个这样的单位。

7、()吨的92是12吨,50米的20%是()米。

8、一个平行四边形的高是15分米,底比高少31,这个平行四边形的面积是()平方分米。

9、前进小学六年级有200个学生,其中有120个女生,男生与女生的人数的最简整数比是(),比值是()。

10、上海到北京的距离大约是900千米。

在一幅中国地图上,量得上海到北京的图上距离是15厘米,那么这幅地图的比例尺是()。

11、自2006年1月1日起个人所得税标准由800元改为1600元,即工资超过1600元的那部分按20%缴纳税金。

李老师每月工资是1800元,那么李老师每月应缴纳税金()元。

12、如右图所示,把底面直径8厘米的圆柱切成若干等分,拼成一个近似的长方体。

这个长方体的表面积比原来增加80平方厘米,那么长方体的体积是()立方厘米。

13、甲用1000元人民币购买了一手股票,随即他将这手股票转卖给了乙,获利10%,而后来乙又将这手股票转给了甲,但乙损失了10%,最后甲按乙卖给甲的价格的90%将这手股票卖给了乙。

甲在上述股票交易中()[选填“盈利”或“亏本”]()元。

二、“惊慕你的判断"请你判一判。

你认为对的,请在每小题的后面括号里打上“√”,错的打上“×”。

(本题共5分,每小题1分)14、自然数都有它的倒数。

2023年北京第二次普通高中学业水平合格性考试语文仿真模拟试卷02(解析版)

2023年北京第二次普通高中学业水平合格性考试语文仿真模拟试卷02(解析版)

机密★本科目考试启用前2023年北京市第二次普通高中学业水平合格性考试语文仿真模拟试卷02(答案在最后)一、本大题共 3 小题,共 14分。

1.在横线上默写原句。

(6 分)(1)《登泰山记》中“____________,____________”两句描写了太阳将出时的景色,在天地相接的地方,有一线云层显现出奇异的颜色,很快又变为五彩缤纷。

(2)《子路、曾皙、冉有、公西华侍坐》中,孔子没有直接让弟子言志,而是先用温和自谦的话打消学生的顾虑,为他们创造一个轻松、亲切、活跃的环境。

他说:“_______________,__________________。

”(3)《桂枝香·金陵怀古》一词中,表现词人登高望远,所看到的深秋时节金陵最具典型特征的景物的句子:________,___________。

【答案】(1)极天云一线异色须臾成五采(2)以吾一日长乎尔毋吾以也(3)千里澄江似练翠峰如簇2.下列有关文学常识的表述,不正确...的一项是(4分)A.李白诗歌想象雄奇、情思奔涌,诗风豪放飘逸;杜甫诗艺精湛、众体皆工,诗风沉郁顿挫。

严羽《沧浪诗话》高度评价道:“论诗以李杜为准。

”B.《聊斋志异》是清代文学家蒲松龄创作的文言短篇小说集,以奇异的故事批判当时社会的腐败黑暗,被誉为“写鬼写妖高人一等,刺贪刺虐入木三分”。

C.韩愈是“唐宋八大家”之首,其他七位分别为柳宗元、范仲淹、欧阳修、苏洵、苏轼、苏辙、王安石。

他们反对骈文,提倡古文,主张“文以明道”。

D.新乐府运动是中唐诗人白居易、元稹倡导的一场诗歌革新运动,强调自创新题、咏写时事,主张“文章合为时而著,歌诗合为事而作”。

【答案】C【详解】本题考查学生了解并掌握常见的文学常识的能力。

C.“‘唐宋八大家’……其他七位分别为柳宗元、范仲淹、欧阳修、苏洵、苏轼、苏辙、王安石”错。

“范仲淹”应为“曾巩”。

故选C。

3.《红楼梦》第十七、十八回《大观园试才题对额荣国府归省庆元宵》中,宝玉、元春等人曾为大观园中的主要景观命名。

马克思模拟试卷二及参考答案

马克思模拟试卷二及参考答案

模拟试卷二及参考答案一、单项选择题(每题1 分,共20 分)1 .作为中国共产党和社会主义事业指导思想的马克思主义是指()A .不仅指马克思恩格斯创立的基本理论、基本观点和学说的体系,也包括继承者对它的发展B .无产阶级争取自身解放和整个人类解放的学说体系C .关于无产阶级斗争的性质、目的和解放条件的学说D .列宁创立的基本理论、基本观点和基本方法构成的科学体系2 .爱因斯坦指出:“哲学可以被认为是全部科学之母”,这说明()A .哲学是一切科学之科学B .哲学是各门学科的知识基础C .哲学对各门具体科学的研究具有指导作用D .哲学与具体科学是整体与局部的关系3 .哲学发展的历史表明,各种唯心主义派别之间的差异和矛盾,常常有利于唯物主义的发展,这一事实说明()A .矛盾一方克服另一方促使事物发展B .矛盾一方的发展可以为另一方的发展提供条件C .矛盾双方中每一方的自身矛盾,可以为另一方的发展所利用D .矛盾可以向自己的对立面转化5 .把科学的实践观第一次引人认识论是()A .费尔巴哈哲学的功绩B .黑格尔哲学的功绩C .马克思主义哲学的功绩D .康德哲学的功绩6 .列宁说:“没有革命的理论,就不会有革命的运动”,这句话应理解为()A .革命运动是由革命理论派生的B .革命理论是革命运动的基础C .革命理论对革命实践具有最终决定作用D .革命理论对革命实践具有重要指导作用7 .列宁对辩证唯物主义物质范畴的定义是通过()A .物质和意识的关系界定的B .哲学与具体科学的关系界定的C .主体和客体的关系界定的D .一般和个别的关系界定的8 . “人的思维是否具有真理性,这并不是一个理论的问题,而是一个实践的问题。

人应该在实践中证明自己思维的真理性,即自己思维的现实性和力量,亦即自己思维的此岸性。

”这一论断说明了()A .实践是认识的来源和动力B .实践是检验认识是否具有真理性的唯一标准C .实践检验真理不需要理论指导D .认识活动与实践活动具有同样的作用和力量9 . 1633 年,伽利略因宣传‘日心说”被教廷判处终身监禁。

模拟卷02-2023高三生物对接新高考全真模拟试卷2(解析版)

模拟卷02-2023高三生物对接新高考全真模拟试卷2(解析版)

2023高三生物对接新高考全真模拟试卷2一、选择题:本题共6小题,每小题6分,共36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.新冠病毒(SARS-CoV-2)和肺炎双球菌均可引发肺炎,两者结构不同。

新冠病毒是具外套膜的正链单股RNA病毒,其遗传物质是目前所有RNA病毒中最大的,该病毒在宿主细胞内的增殖过程如图所示,a~e表示相应的生理过程。

下列有关叙述错误的是( )A.新冠病毒必须寄生在活细胞中才能繁殖B.e过程需要成熟的mRNA、tRNA、氨基酸、ATP、核糖体等C.新冠病毒在宿主细胞内形成子代的过程可以体现中心法则的全过程D.a~e 过程均存在A-U的形成和解开【答案】C【解析】病毒都必须寄生在活细胞中才能繁殖,繁殖过程中蛋白质合成所需要的物质、能量和工具均由宿主细胞提供;新冠病毒为RNA病毒,其繁殖过程只涉及到RNA的复制和翻译,并未涉及中心法则全过程;a~e 过程均存在A-U的形成和解开。

故C错误。

解析:病毒都必须寄生在活细胞中才能繁殖,繁殖过程中蛋白质合成所需要的物质、能量和工具均由宿主细胞提供;新冠病毒为RNA病毒,其繁殖过程只涉及到RNA的复制和翻译,并未涉及中心法则全过程;a~e 过程均存在A-U的形成和解开。

故C错误。

2.A和a,B和b为一对同源染色体上的两对等位基因。

有关有丝分裂和减数分裂叙述正确的是()A. 多细胞生物体内都同时进行这两种形式的细胞分裂B. 减数分裂的两次细胞分裂前都要进行染色质DNA的复制C. 有丝分裂的2个子细胞中都含有Aa,减数分裂Ⅰ的2个子细胞中也可能都含有AaD. 有丝分裂都形成AaBb型2个子细胞,减数分裂都形成AB、Ab、aB、ab型4个子细胞【答案】C【解析】A、生物体内有些特殊的细胞如生殖器官中的细胞既能进行有丝分裂又能进行减数分裂,但体细胞只进行有丝分裂,A错误;B、减数分裂只在第一次分裂前进行染色质DNA的复制,B错误;C、有丝分裂得到的子细胞染色体组成与亲代相同,得到的2个子细胞中都含有Aa;减数分裂Ⅰ若发生交叉互换,则得到的2个子细胞中也可能都含有Aa,C 正确;D、有丝分裂得到的子细胞染色体组成与亲代相同,都形成AaBb型2个子细胞;因A和a,B和b为一对同源染色体上的两对等位基因,若不考虑交叉互换,则减数分裂能得到两种类型(AB、ab或Ab、aB)的子细胞,D错误。

全国自考(语言学概论)模拟试卷2(题后含答案及解析)

全国自考(语言学概论)模拟试卷2(题后含答案及解析)

全国自考(语言学概论)模拟试卷2(题后含答案及解析)题型有:1. 单项选择题 2. 多项选择题 3. 名词解释 4. 简答题 5. 分析题6. 论述题单项选择题1.下列关于语言起源的学说中,注意到语言符号任意性的是( )A.神授说B.社会契约说C.摹声说D.感叹说正确答案:B解析:“社会契约说”认为原始人类起初没有语言,后来为了相互交际.就通过彼此约定,规定了各种事物的名称,就产生了语言。

该观点注意到了语言的社会属性和语言符号的任意性。

2.语言谱系分类的层级体系中,最小的类别是( )A.语系B.语支C.语群D.语族正确答案:C解析:语系、语族、语支、语群这种层级分类体系反映了原始基础语(母语)随着社会的分化而不断分化的历史过程和结果。

其中语系是谱系分类中最大的类,语群是最小的类。

3.下列语言中不属于汉藏语系的有( )A.壮语B.苗语C.维吾尔语D.傣语正确答案:C解析:维吾尔语属阿尔泰语系突厥语族西匈语支,不属于汉藏语系。

4.下列关于亲属语言的表述,不正确的一项是( )A.亲属语言是社会完全分化的产物B.亲属语言是一种语言的地域变体C.亲属语言具有历史同源关系D.语音对应关系是亲属语言的重要标志正确答案:B解析:社会的地域分化会导致语言的地域分化,在一种语言的内部形成不同的地域方言,如果社会由不完全分化走向完全分化,那么,方言也会由语言的地域变体转变为独立的语言。

这些语言之间具有历史同源关系,叫做亲属语言,它们之间存在着语音对应关系。

5.下列汉语中的词属于半音译半意译的是( )A.啤酒B.剑桥C.卡通D.酒吧正确答案:B6.用本族语言的语素逐个对译外语原词的语素而产生的词是( )A.音译词B.仿译词C.意译词D.借词正确答案:B解析:仿译词是用本族语言的语素逐个时译外语原词的语素造成的词,这种词不仅把原词的词义翻译过来,而且保持了原词的内部构成方式。

7.下列关于“语言转用”的表述,不正确的一项是( )A.语言转用是语言统一的重要形式B.双语现象并非一定导致语言转用C.语言转用以民族融合为必要条件D.语言间的密切接触是语言转用的重要条件正确答案:C解析:语言转用虽然同民族的融合关系密切,但由于语言并不是民族最根本的特征,因而语言转用了,并不等于不同民族也融合了。

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《旅游文化》模拟卷二参考答案
一、填空题:(每空格1 分,共20分)
1、《茶经》首次把饮茶当做一种艺术过程来看,是中国乃至世界上的第一步茶学专著,它的作者是陆羽。

2、蒙古族人称奶食为白食,肉食为红食。

3、莫干山与江西庐山、河南鸡公山、河北北戴河并称中国四大避暑胜地。

4、浙江四大藏书楼是杭州文澜阁、宁波天一阁、温州瑞安玉海阁、湖州南浔嘉业堂。

5、“金华双绝”为金华火腿、金华斗牛。

6.陶器的发明标志着烹饪技术的第一次飞跃,人类真正进入了烹饪的时代。

7. 我国历史上最早的一篇烹饪理论文章是:《吕氏春秋·本味篇》。

8、浙江境内有8大水系,分别是钱塘江、瓯江、曹娥江、灵江、甬江、飞云江、鳌江、东西苕溪。

9、戏曲包括戏剧和曲艺两个部分。

10、三月三是畲族的传统节日。

11、浙江三雕黄杨木雕、东阳木雕、青田石雕。

12、道家把正月十五、七月十五和十月十五分别成为上元、中元和下元,和称“三元”。

13、轩辕黄帝的三大行宫是仙都山、黄山、庐山。

14、中国古氏绘画可分为五类:宫廷绘画、文人绘画、宗教绘画、市民绘画和民间绘画。

15、一般把山水诗词分成山水诗,近体诗山水诗,新诗体山水诗,山水词等。

16、四大佛教名山是五台山、峨眉山、普陀山、九华山。

二、名词解释:(每小题2分,共10分)
17、楹联:楹联又叫楹贴,对联,对子,是悬挂或粘贴在门框,墙上的联语。

18、左祖右社:左祖右社是封建王朝传统的祭祀格局,左面为太庙,右面为社稷坛。

19、漏景:透过漏窗的窗隙,可见园外或院内的美景,这叫做漏景。

20、绍兴三乌:乌干菜、乌蓬船、乌毡帽。

21、前朝后寝:所谓“前朝”,即为帝王上朝治政、举行大典之处。

所谓“后寝”,即帝王
与后妃们生活居住的地方。

三、选择题: (每小题2分,共30分)
22、什么酒是中国最具代表性的蒸馏酒-------------------------------------------------(B )
A、黄酒
B、白酒
C、果酒
D、高粱酒
23、以糖提鲜,以盐提香是哪类官府菜的特色----------------------------------------(B )
A、孔府菜
B、谭家菜
C、随园菜
D、红楼菜
24、龙虎斗是哪个菜系的名菜-------------------------------------------------------------(C )
A、川菜
B、鲁菜
C、粤菜D淮扬菜
25、云南西双版纳地区的少数民族习惯喝哪类茶-------------------------------------(B )
A、擂茶
B、烤茶
C、咸茶
D、嚼茶
26、“自言长官如灵运,能使江山似永嘉”是描写----------------------- (C )
A、富春江
B、曹娥江
C、楠溪江
D、新安江
27、被华东56位地质学家堪称为“中国一线天之最”的是---------------(D)
A、天目山
B、仙都山
C、方岩
D、江郎山
28、有“大树华盖闻九州”之称的景观位于-----------------------------(D )
A、大奇山
B、莫干山
C、雁荡山
D、天目山
29、六朝吴均称此地为“奇山异水、天下独绝”的是---------------------(A )
A、富春江
B、瓯江
C、曹娥江
D、新安江
30、张择端的《清明上河图》反应了代都市饮食市场的形成与发展空前繁荣的景象。

--------------------------------------------------------( C )
A、明
B、清
C、宋
D、元
31、“驴打滚”是著名的风味小吃---------------------------------------------(B )
A天津、B北京、C山东、D陕西
32、与历史上著名的1983年“芝加哥”博览会,1900巴黎博览会和1927年费城博览会并称为国际庆典的是------------------------------------------------ ( C )
A、杭州西湖国际烟花大会
B、中国梁祝婚俗节
C、中国西湖博览会
D、中国农民旅游节
33、下列哪部作品是唐代山水游记的代表作-------------------------------------- ( A )
A《永州八记》B《醉翁亭记》C《岳阳楼记》D《石钟山记》
34、《随园食单》的作者是清代文人------------------------------------------------- (C )
A.李渔
B.李调元C袁枚D徐珂
35、茶文化盛于哪个朝代----------------------------------------------------------------(B)
A、唐朝
B、宋朝
C、元朝D明朝
36、中国最大的淡水湖---------------------------------------------------------- (C )
A.千岛湖
B.太湖
C.鄱阳湖
D.青海湖
四、多项选择题(每小题2分,共30分)
37、自然景观中属丹霞地貌的景点是。

(B C )
A、武夷山、千山
B、武夷山、穿岩
C、方岩、江郎山
D、庐山、武当山
38、关于雪窦山的描述正确的是。

(B C D )
A、雪窦寺始建于南宋
B、有“四明第一山”之称
C、雪窦山为全国重点风景名胜区
D、它由江镇、雪窦山、亭下湖组成
39、西湖群山外围以石英砂岩为主,海拔较高,如。

(A D )
A、天竺山
B、玉泉山
C、南高峰
D、北高峰
40、以下属于浙江风味小吃的是( A D )
A、五芳斋粽子
B、生煎馒头
C、南翔小笼包
D、丁莲芳千张包子
41、下列属于畲族婚嫁中最具有特色的仪式(ABCD )
A、迎亲路上拦赤郎
B、长夜对歌闹吉祥
C、溜筷和衔十斤饭
D、拜堂少两拜
42、畲族主要分布在(ABD )
A、福建
B、浙江
C、江西
D、广东
43、下列作者与作品都对应正确的是(ABCD )
A、柳宗元——《石涧记》
B、曾巩——《黑池记》C苏轼——《石钟山记》
D、王安石——《游褒禅山记》
44、下列属于再加工茶的是--------------------------------------------(ABCD)
A、花茶
B、果味茶
C、含茶饮料
D、药用保健茶
45、下面哪几种是浙江风味小吃-----------------------------------(BC )
A、郭汤圆
B、嘉兴鲜肉粽
C、猫耳朵
D、叉烧包
46、下列几类名菜中那几个是浙江的名菜-------------------------( AC )
A、东坡肉
B、佛跳墙 C 、龙井虾仁D、天下第一菜
47、下列哪几种属于十大名茶之列---------------------------------(ABCD)
A、祁红
B、云南普洱茶
C、西湖龙井
D、君山银针
48、下列属于四大菜系的有哪几项?(ABC )
A、川菜
B、鲁菜
C、粤菜
D、淮扬菜
49、京杭大运河流经的省市是。

(AD )
A、河北
B、河南
C、山西
D、山东
E、江西
50、我国著名的奇特泉有。

(ACDE )
A、云南大理蝴蝶泉
B、山东青岛崂山泉
C、四川广元含羞泉
D、湖北当阳珍珠泉
E、湖南石门鱼泉
51、中国茶道精神包括(ABE )
A、儒
B、道 C 、墨 D 、荀 E 、释
五、综合题:(10分)
52、中国古典园林的游览方法与游览建筑的方法有所不同,游览古典园林讲究“游”“停”的结合。

a.从“游”的角度来说是什么? b.从“停”的角度来说是什么?
c.从心理学的角度来说,游园有哪几条路线?
答:
a、“游”:可顺着路、径、廊的走向漫步游览和观赏(3分)
b、“停”:遇到园中亭、榭、厅、堂等重要建筑物,最好驻步停留,以便细加观赏。

(3分)
c、心理学的角度看,园林中的游览路线有两条:登山越水的路或径;进廊游览。

(4分)
六、论述题:(15分)
53、你认为丽水应该如何利用“文化节庆”这个平台发展旅游?谈谈你的想法,要求观点鲜明(3分),论据充分(3分),结合实际(3分),论述条理清晰(3分)具有说服力(3分)。

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