【全国百强校word】(衡水金卷调研卷)2018年普通高等学校招生全国统一考试模拟试题英语五(有答案)

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语文-衡水金卷2018年普通高等学校招生全国统一考试模拟(调研卷)(二)试题(解析版)

语文-衡水金卷2018年普通高等学校招生全国统一考试模拟(调研卷)(二)试题(解析版)

衡水金卷2018年普通高等学校招生全国统一考试模拟(调研卷)(二)语文试题一、现代文阅读(35分)(一)论述类文本阅读(本题共3小题,9分)阅读下面的文字,完成1~3题。

我国的民间刺绣历史非常悠久,可以追溯到战国时期,已经具有两千多年历史,它是经过人类文明进步,不断进化与发展而凝聚起来的艺术精髓。

在历史的长河中,民间刺绣的图案慢慢形成了独特的风格,展示出了深犀的文化底蕴、独特的民俗风情与民族特征。

我国的民间刺绣来自于社会群众之间,民间刺绣作品上的图案具有非常浓厚的象征意义,表达出不同民族祈福求祥的信息。

民间刺绣图案象征符号把表层结构与深层意义有机地结合为一体,体现出民间艺人在图案构思中自然与直接的纯朴观念与愿望。

比如最为常见的“鲤鱼跳龙门”,在作品上活灵活现地展示出鲤鱼在激流中跃起,跳向龙门,表达出前途无量、步步高升的祝福。

民间刺绣图案除了部分文字之外,大多数的图案内容都是我国民间传统中的吉祥物,吉祥物有着深厚的文化内涵,它比直接的话言表达具有更深层次的寓意。

动物作为民间刺绣图案中的象征符号之一,每种动物符号都能够体现出文化内涵。

比如蝙蝠,作为民间吉祥物的一种,它所表达出的是“蝠”与福、富的谐音,因此,蝙蝠通常象征着福气与富贵,使用蝙蝠这种象征符号通常所表达的是福到吉祥、富贵满堂的寓意。

还有很多民间传说与神话中的动物,比如龙。

龙在民间是种神圣的灵物,县有至高无上的权威与能力,被视为吉祥之物,龙的传人就是非常有特色的象征符号。

人们非常喜欢龙的象征符号,望子成龙寓意为希望子孙能够成为栋梁人才。

在民间刺绣图案中的植物象征符号主要包括花草类与树木类,一种或者多种植物相搭配组合,形成一种具有文化寓意的象征符号。

人们通过对大自然中各种植物的观察与了解,熟悉并总结出植物的特点与性情,在观赏的同时,还根据植物的属性或者同音、谐音来表达出吉祥祝福的象征符号。

比如梅、兰、竹、菊通常被寓为四君子,这四种植物传递着自强不息、淡泊名利、坚贞高雅的内涵。

【全国省级联考word】河北省衡水金卷2018年普通高等学校招生全国统一考试模拟(调研卷)试题(

【全国省级联考word】河北省衡水金卷2018年普通高等学校招生全国统一考试模拟(调研卷)试题(

7.化学与生产和生活密切相关,下列说法不正确的是()A.钢铁在潮湿的空气中更容易生锈,其主要原因是形成了原电池B.从海带、海藻中提取碘单质必须通过化学反应才能实现C.用纳米铁粉除去污水中的Cu2+、Hg2+等重金属离子是利用了铁的还原性D.一次性纸杯的内层材料主要成分是聚氯乙烯8.以Al2O3为原料(含杂质Fe2O3)制备药品级的氢氧化铝的一种工艺流程如下,下列说法不正确的是()A.操作Ⅰ中用到的玻璃仪器有漏斗、烧杯、玻璃棒B.NaHCO3溶液可用足量氨水代替C.碱溶过程中发生反应的离子方程式为Al2O3+2OH-=2AlO2-+H2OD.加入NaHCO3溶液发生反应的离子方程式为AlO2-+HCO3-+H2O=Al(OH)3↓+CO32-9.已知甲在一定条件下可转化为乙和丙,转化关系如图所示,下列说法正确的是()A.丙与是同分异构体B.丙与乙酸是同系物C.乙中所有碳原子一定位于同一平面上D.丙能发生加聚反应、取代反应和氧化反应10.25℃时,在一体积为10 L的绝热密闭容器中充入1molH2与1moI2(g),反应中能量变化曲线如图所示。

测得平衡时H2的物质的量为0.5mol,已知H-H键的键能为436kJ·mol-1,I-I键的键能为151 kJ·mol-1,H-I键的键能为b kJ·mol-1。

下列说法不正确的是()A.25℃时.反应H2(g)+I2(g)2HI(g)的平衡常数K=4B.当容器中压强保持不变时,H2(g)+I2(g)2HI(g)达到平衡状态C.b=299D.H2(g)+I2(g)2HI(l) △H=(a-11) kJ·mol-111.设N A为阿伏加德罗常数的值,下列叙述正确的是()A.将1molFe 与足量碘加热反应,转移电子数目为3N AB.84g溶有聚乙烯的环已烷中含有的C-H键数目小于12N AC.标准状况下,11.2 LCH3Cl所含的分子数目为0.5N AD.1L0.5mol·L-1的Na2S溶液中,阳离子与阴离子数目之和为1.5N A12.下表中,陈述Ⅰ、Ⅱ均正确,并且两者之间具有因果关系的是()13.常温下,向2 0mL0.1 mol·L-1的Na2CO3溶液中逐滴加入40mL0.1 mol·L-1的盐酸,溶液的pH逐渐降低,下列说法确的是()A.完全反应后.溶液呈中性B.当加入20 mL盐酸时,溶液中NaHCO3的物质的量为0.002 molC.20mL0.1 mol·L-1Na2CO3溶液中:c(Na+)+c(H+)=c(CO32-)+c(HCO3-)+c(OH-)D.20 mL0.1 mol·L-1Na2CO3溶液中:c(H+)+2c(H2CO3-)+c(HCO3-)=c(OH-)26.(14 分)某合作小组的同学为了测定某铜的硫化物(Cu x S y)的组成,在实验室中利用如图装置(夹持装置略去)进行实验探究。

【全国省级联考word】衡水金卷2018年普通高等学校招生全国统一考试模拟(调研卷)试题(二)理综生物试题

【全国省级联考word】衡水金卷2018年普通高等学校招生全国统一考试模拟(调研卷)试题(二)理综生物试题

衡水金卷2018年普通高等学校招生全国统一考试模拟(调研卷)试题(二)理综生物试题一、选择题:在下列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.蓝球藻和小球藻在细胞结构等方面存在着较大的差异。

下列对于这两种生物的叙述正确的是A.均含有叶绿素和类胡萝卜素B.均在叶绿体基质中完成二氧化碳的固定和还原C.细胞膜均有控制内外物质交换的作用D.有丝分裂过程中均会以一定方式形成细胞壁2.下列做法对实验结论的得出不会产生明显影响的是A.进行酵母菌计数时从静置的试管中吸取培养液滴在计数室内B.用洋葱鳞片叶内表皮作为材料观察植物细胞有丝分裂C.测定种子萌发初期呼吸速率变化的实验在光照条件下进行D.探索生长素类似物促进插条生根的最适浓度时不设蒸馏水处理的组别3.如图为人体免疫过程的示意图。

下列与此图相关的说法错误的是A.图中的同有免疫应答属于第二进防线B.图中淋巴细胞接受抗原刺激后大部分分化为记忆细胞C.图中接受抗原刺激的淋巴细胞不可能是浆细胞D.图中过程能说明免疫系统具有防卫功能4.神经系统、内分泌系统和免疫系统之间可以通过信息分子相互联系,如图显示的是三者之间的部分联系。

下列相关叙述正确的是A.神经末梢释放的信息分子进入免疫细胞内部发挥作用B.激素作为信息分子通过体液传送作用于免疫细胞C.图中信息分子的化学本质均为蛋白质D.免疫细胞表面的受体均属于信息分子5.艾滋病病毒(HIV)是一种逆转录病毒,HIV的宿主细胞主要是T钿胞,病毒进入人体后在宿主细胞内经一系列过程形成新的病毒。

下列与HIV增殖有关的叙述中正确的是A.HIV进入淋巴细胞内的只是病毒的RNAB.逆转录出的DNA分子中A+T与C+G的数目相等C.A与U的配对只发生在转录过程中D.HIV增殖时所用原料全部来自宿主细胞6.下列有关基因突变的叙述正确的是A.基因突变可能改变基因中密码子的种类或顺序B.密码子的简并性可以减少有害突变对机体造成的危害C.癌症的发生是几个正常基因突变成了原癌基因和抑癌基因D.一个基因中不同的碱基均可改变说明細突变具有随机性二、非选择题:29.(10分)生长素(吲哚乙酸)在植物体内的代谢过程如图1所示,图中的吲哚乙酸氧化酶是一种含铁蛋白。

【全国百强校Word】衡水金卷2018届全国高三大联考理数试题

【全国百强校Word】衡水金卷2018届全国高三大联考理数试题

衡水金卷2018届全国高三大联考理科第Ⅰ卷一、选择题:本大题共12个小题,每小题5分,在每小题给出的四个选项中,只有一项是符合题目要求的.1. 已知集合2{|540}M x x x =-+≤,{|24}xN x =>,则 ( ) A .{|24}M N x x =<<I B .M N R =U C .{|24}M N x x =<≤I D .{|2}M N x x =>U2. 记复数z 的虚部为Im()z ,已知复数5221iz i i =--(i 为虚数单位),则Im()z 为( ) A .2 B .-3 C .3i - D .33. 已知曲线32()3f x x =在点(1,(1))f 处的切线的倾斜角为α,则222sin cos 2sin cos cos ααααα-=+( ) A .12 B .2 C .35 D . 38- 4. 2017年8月1日是中国人民解放军建军90周年,中国人民银行为此发行了以此为主题的金银纪念币,如图所示是一枚8克圆形金质纪念币,直径22mm ,面额100元.为了测算图中军旗部分的面积,现用1粒芝麻向硬币内投掷100次,其中恰有30次落在军旗内,据此可估计军旗的面积大约是( ) A .27265mm π B .236310mm π C.23635mm π D .236320mm π5. 已知双曲线C :22221(0,0)x y a b a b-=>>的渐近线经过圆E :22240x y x y +-+=的圆心,则双曲线C 的离心率为( )A .5B .52C.2 D .2 6. 已知数列{}n a 为等比数列,且2234764a a a a =-=-,则46tan()3a a π⋅=( ) A .3- B .3 C.3± D .33- 7. 执行如图的程序框图,若输出的S 的值为-10,则①中应填( )A .19?n <B .18?n ≥ C. 19?n ≥ D .20?n ≥8.已知函数()f x 为R 内的奇函数,且当0x ≥时,2()1cos f x e m x =-++,记2(2)a f =--,(1)b f =--,3(3)c f =,则a ,b ,c 间的大小关系是( )A .b a c <<B .a c b << C.c b a << D .c a b <<9. 已知一几何体的三视图如图所示,俯视图是一个等腰直角三角形和半圆,则该几何体的体积为( )A .23π+ B .12π+ C.26π+ D .23π+ 10. 已知函数()2sin()(0,[,])2f x x πωϕωϕπ=+<∈的部分图象如图所示,其中5||2MN =.记命题p :5()2sin()36f x x ππ=+,命题q :将()f x 的图象向右平移6π个单位,得到函数22sin()33y x ππ=+的图象.则以下判断正确的是( )A.p q ∧为真B.p q ∨为假C.()p q ⌝∨为真D.()p q ∧⌝为真11.抛物线有如下光学性质:过焦点的光线经抛物线反射后得到的光线平行于抛物线的对称轴;反之,平行于抛物线对称轴的入射光线经抛物线反射后必过抛物线的焦点.已知抛物线24y x =的焦点为F ,一条平行于x 轴的光线从点(3,1)M 射出,经过抛物线上的点A 反射后,再经抛物线上的另一点B 射出,则ABM ∆的周长为 ( )A .712612+ B .926+ C. 910+ D .832612+ 12.已知数列{}n a 与{}n b 的前n 项和分别为n S ,n T ,且0n a >,2*63,n n S a a n N =+∈,12(21)(21)n n n a n a a b +=--,若*,n n N k T ∀∈>恒成立,则k 的最小值是( ) A .71 B .149 C. 49 D .8441第Ⅱ卷本卷包括必考题和选考题两部分.第13~21题为必考题,每个试题考生都必须作答.第22~23题为选考题,考生根据要求作答.二、填空题:本大题共4小题,每题5分.13.已知在ABC ∆中,||||BC AB CB =-u u u r u u u r u u u r ,(1,2)AB =u u u r,若边AB 的中点D 的坐标为(3,1),点C 的坐标为(,2)t ,则t = .14. 已知*1()()2nx n N x-∈的展开式中所有项的二项式系数之和、系数之和分别为p ,q ,则64p q +的最小值为 .15. 已知x ,y 满足3,,60,x y t x y π+≤⎧⎪⎪≥⎨⎪≥⎪⎩其中2t π>,若sin()x y +的最大值与最小值分别为1,12,则实数t 的取值范围为 .16.在《九章算术》中,将四个面都为直角三角形的三棱锥称之为鳖臑(bie nao ).已知在鳖臑M ABC -中,MA ⊥平面ABC ,2MA AB BC ===,则该鳖臑的外接球与内切球的表面积之和为 . 三、解答题 :解答应写出文字说明、证明过程或演算步骤.17. 已知函数21()cos 3sin()cos()2f x x x x ππ=+-+-,x R ∈. (Ⅰ)求函数()f x 的最小正周期及其图象的对称轴方程;(Ⅱ)在锐角ABC ∆中,内角A ,B ,C 的对边分别为a ,b ,c ,已知()1f A =-,3a =,sin sin b C a A =,求ABC ∆的面积.18. 如图,在四棱锥E ABCD -中,底面ABCD 为直角梯形,其中//,CD AB BC AB ⊥,侧面ABE ⊥平面ABCD ,且222AB AE BE BC CD =====,动点F 在棱AE 上,且EF FA λ=. (1)试探究λ的值,使//CE 平面BDF ,并给予证明; (2)当1λ=时,求直线CE 与平面BDF 所成的角的正弦值.19. 如今我们的互联网生活日益丰富,除了可以很方便地网购,网上叫外卖也开始成为不少人日常生活中不可或缺的一部分.为了解网络外卖在A 市的普及情况,A 市某调查机构借助网络进行了关于网络外卖的问卷调查,并从参与调查的网民中抽取了200人进行抽样分析,得到下表:(单位:人)(Ⅰ)根据以上数据,能否在犯错误的概率不超过0.15的前提下认为A 市使用网络外卖的情况与性别有关? (Ⅱ)①现从所抽取的女网民中利用分层抽样的方法再抽取5人,再从这5人中随机选出3人赠送外卖优惠卷,求选出的3人中至少有2人经常使用网络外卖的概率②将频率视为概率,从A 市所有参与调查的网民中随机抽取10人赠送礼品,记其中经常使用网络外卖的人数为X ,求X 的数学期望和方差.参考公式:22()()()()()n ad bc K a b c d a c b d -=++++,其中n a b c d =+++.参考数据:20()P K k ≥0.050 0.010 0.001 0k3.8416.63510.82820. 已知椭圆C :22221(0)x y a b a b +=>>的左、右焦点分别为点1F ,2F ,其离心率为12,短轴长为23.(Ⅰ)求椭圆C 的标准方程;(Ⅱ)过点1F 的直线1l 与椭圆C 交于M ,N 两点,过点2F 的直线2l 与椭圆C 交于P ,Q 两点,且12//l l ,证明:四边形MNPQ 不可能是菱形.21. 已知函数,()(1)(,)xf x e a x b a b R =-+-∈其中e 为自然对数的底数. (Ⅰ)讨论函数()f x 的单调性及极值;(Ⅱ)若不等式()0f x ≥在x R ∈内恒成立,求证:(1)324b a +<. 请考生在22、23两题中任选一题作答,如果多做,则按所做的第一题记分.22.选修4-4:坐标系与参数方程在平面直角坐标系xOy 中,已知曲线C 的参数方程为cos ,sin x t y αα=⎧⎨=⎩(0t >,α为参数).以坐标原点O 为极点,x 轴的正半轴为极轴,取相同的长度单位建立极坐标系,直线l 的极坐标方程为2sin()34πρθ+=.(Ⅰ)当1t =时,求曲线C 上的点到直线l 的距离的最大值; (Ⅱ)若曲线C 上的所有点都在直线l 的下方,求实数t 的取值范围. 23.选修4-5:不等式选讲 已知函数()21|1|f x x x =-++. (Ⅰ)解不等式()3f x ≤;(Ⅱ)记函数()()|1|g x f x x =++的值域为M ,若t M ∈,证明:2313t t t+≥+.衡水金卷2018届全国高三大联考理科参考答案及评分细则一、选择题1-5: CBCBA 6-10:ACDAD 11、12:BB二、填空题13. 1 14. 16 15. 57[,]66ππ16. 2482ππ- 三、解答题17. 解:(1)原式可化为,21()cos 3sin cos 2f x x x =--, 1cos 231sin 2222x x +=--, sin(2)sin(2)66x x ππ=-=--, 故其最小正周期22T ππ==,令2()62x k k Z πππ-=+∈,解得()23k x k Z ππ=+∈,即函数()f x 图象的对称轴方程为,()23k x k Z ππ=+∈. (2)由(1),知()sin(2)6f x x π=--, 因为02A π<<,所以52666A πππ-<-<. 又()sin(2)16f A A π=--=-,故得262A ππ-=,解得3A π=.由正弦定理及sin sin b C a A =,得29bc a ==. 故193sin 24ABC S bc A ∆==. 18.(1)当12λ=时,//CE 平面BDF . 证明如下:连接AC 交BD 于点G ,连接GF . ∵//,2CD AB AB CD =,∴12 CG CD GAAB==.∵12EF FA=,∴12EF CGFA GA==.∴//GF CE.又∵CE⊄平面BDF,GF⊂平面BDF,∴//CE平面BDF.(2)取AB的中点O,连接EO.则EO AB⊥.∵平面ABE⊥平面ABCD,平面ABE I平面ABCD AB=,且EO AB⊥,∴EO⊥平面ABCD.∵//BO CD,且1BO CD==,∴四边形BODC为平行四边形,∴//BC DO.又∵BC AB⊥,∴//AB DO.由,,OA OD OE两两垂直,建立如图所示的空间直角坐标系Oxyz.则(0,0,0)O,(0,1,0)A,(0,1,0)B-,(1,0,0)D,(1,1,0)C-,(0,0,3)E. 当1λ=时,有EF FA=u u u r u u u r,∴可得13(0,,)22F.∴(1,1,0)BD=u u u r,(1,1,3)CE=-u u u r,33(1,,)22BF=u u u r.设平面BDF的一个法向量为(,,)n x y z=r,则有0,0,n BDn BF⎧⋅=⎪⎨⋅=⎪⎩r u u u rr u u u r即0,330,22x yy z+=⎧⎪⎨+=⎪⎩令3z=,得1y=-,1x=.即(1,1,3)n =-r.设CE 与平面BDF 所成的角为θ,则sin |cos |CE n θ=<⋅>=u u u r r |113|1555--+=⨯. ∴当1λ=时,直线CE 与平面BDF 所成的角的正弦值为15. 19.解:(1)由列联表可知2K 的观测值,2()()()()()n ad bc k a b c d a c b d -=++++2200(50405060) 2.020 2.07211090100100⨯-⨯=≈<⨯⨯⨯.所以不能在犯错误的概率不超过0.15的前提下认为A 市使用网络外卖情况与性别有关. (2)①依题意,可知所抽取的5名女网民中,经常使用网络外卖的有6053100⨯=(人), 偶尔或不用网络外卖的有4052100⨯=(人). 则选出的3人中至少有2人经常使用网络外卖的概率为2133233355710C C C P C C =+=. ②由22⨯列联表,可知抽到经常使用网络外卖的网民的频率为1101120020=, 将频率视为概率,即从A 市市民中任意抽取1人, 恰好抽到经常使用网络外卖的市民的概率为1120. 由题意得11~(10,)20X B , 所以1111()10202E X =⨯=;11999()10202040D X =⨯⨯=. 20. 解:(1)由已知,得12c a =,3b =,又222c a b =-, 故解得224,3a b ==,所以椭圆C 的标准方程为22143x y +=. (2)由(1),知1(1,0)F -,如图,易知直线MN 不能平行于x 轴. 所以令直线MN 的方程为1x my =-,11(,)M x y ,22(,)N x y .联立方程2234120,1,x y x my ⎧+-=⎨=-⎩,得22(34)690m y my +--=, 所以122634m y y m +=+,122934y y m -=+. 此时221212(1)[()]MN m y y y y =++-, 同理,令直线PQ 的方程为1x my =+,33(,)P x y ,44(,)Q x y ,此时342634m y y m -+=+,342934y y m -=+, 此时223434(1)[()4]PQ m y y y y =++-. 故||||MN PQ =.所以四边形MNPQ 是平行四边形.若MNPQ Y 是菱形,则OM ON ⊥,即0OM ON ⋅=u u u u r u u u r,于是有12120x x y y +=. 又1212(1)(1)x x my my =--,21212()1m y y m y y =-++,所以有21212(1)()10m y y m y y +-++=,整理得到22125034m m --=+, 即21250m +=,上述关于m 的方程显然没有实数解, 故四边形MNPQ 不可能是菱形.21.解:(1)由题意得'()(1)xf x e a =-+.当10a +≤,即1a ≤-时,'()0f x >,()f x 在R 内单调递增,没有极值. 当10a +>,即1a >-, 令'()0f x =,得ln(1)x a =+,当ln(1)x a <+时,'()0f x <,()f x 单调递减; 当ln(1)x a >+时,'()0f x >,()f x 单调递增,故当ln(1)x a =+时,()f x 取得最小值(ln(1))1(1)ln(1)f a a b a a +=+--++,无极大值. 综上所述,当1a ≤-时,()f x 在R 内单调递增,没有极值;当1a >-时,()f x 在区间(,ln(1))a -∞+内单调递减,在区间(ln(1),)a ++∞内单调递增,()f x 的极小值为1(1)ln(1)a b a a +--++,无极大值.(2)由(1),知当1a ≤-时,()f x 在R 内单调递增,当1a =-时,(1)3024b a +=<成立. 当1a <-时,令c 为1-和11ba -+中较小的数,所以1c ≤-,且11bc a-≤+.则1x e e -≤,(1)(1)a c b -+≤--+.所以1()(1)(1)0xf c e a c b e b b -=-+-≤---<, 与()0f x ≥恒成立矛盾,应舍去.当1a >-时,min ()(ln(1))f x f a =+=1(1)ln(1)0a b a a +--++≥, 即1(1)ln(1)a a a b +-++≥,所以22(1)(1)(1)ln(1)a b a a a +≤+-++.令22()ln (0)g x x x x x =->,则'()(12ln )g x x x =-.令'()0g x >,得0x e <<,令'()0g x <,得x e >,故()g x 在区间(0,)e 内单调递增, 在区间(,)e +∞内单调递减. 故max ()()ln 2eg x g e e e e ==-=, 即当11a e a e +=⇒=-时,max ()2eg x =. 所以22(1)(1)(1)ln(1)2ea b a a a +≤+-++≤. 所以(1)24b a e+≤.而3e <, 所以(1)324b a +<.22.解:(1)直线l 的直角坐标方程为30x y +-=.曲线C 上的点到直线l 的距离,|cos sin 3|2d αα+-==|2sin()3|42πα+-, 当sin()14πα+=-时,max |23|23222d ++==,即曲线C 上的点到直线l 的距离的最大值为2322+.(2)∵曲线C 上的所有点均在直线l 的下方,∴对R α∀∈,有cos sin 30t αα+-<恒成立, 即21cos()3t αϕ+-<(其中1tan t ϕ=)恒成立,∴213t +<.又0t >,∴解得022t <<,∴实数t 的取值范围为(0,22).23.解:(1)依题意,得3,1,1()2,1,213,,2x x f x x x x x ⎧⎪-≤-⎪⎪=--<<⎨⎪⎪≥⎪⎩于是得1,()333,x f x x ≤-⎧≤⇔⎨-≤⎩或11,223,x x ⎧-<<⎪⎨⎪-≤⎩或1,233,x x ⎧≥⎪⎨⎪≤⎩解得11x -≤≤.即不等式()3f x ≤的解集为{|11}x x -≤≤.(2)()()|1|g x f x x =++=|21||22|x x -++≥|2122|3x x ---=,当且仅当(21)(22)0x x -+≤时,取等号,∴[3,)M =+∞. 原不等式等价于2331t t t -+-,22233(3)(1)t t t t t t t -+--+==.∵t M ∈,∴30t -≥,210t +>. ∴2(3)(1)0t t t -+≥. ∴2313t t t +≥+.。

衡水金卷2018年普通高等学校招生全国统一考试模拟(调研卷)试题(二)理综试题+版含答案

衡水金卷2018年普通高等学校招生全国统一考试模拟(调研卷)试题(二)理综试题+版含答案

2.下列对实验试剂及其实验效果的分析,错误的是
A. 细胞膜的通透性与所使用盐度的浓度及处理时间有关
B.茎段的生根数和生根长度与所使用的 NAA 浓度有关 C.染色体数目加倍的细胞所占的比例与固定液处理的时间有关
D.洋葱鳞片叶外表皮细胞质壁分离的程度与外界下列与此图相关的说法错误的是
A.C 膜可以为质子交换膜 B. 阴极室的电极反应式为 2H 2O-4e- =O2 ↑ +4H+ C.可用铁电极替换阴极的石墨电极 D. 每转移 2mole-,阳极室中 c(Ca2+)降低 1mol/L 12.下列实验的操作、现象、结论或解释均合理的是 A. 酸碱中和滴定时,未用标准液润洗滴定管可导致最终计算结果偏大
2018 年普通高等学校招生全国统一考试模拟试题理综(二)
理科综合试题
一、选择题
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下列对于这两种生物的叙述正确的
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C.细胞膜均有控制内外物质交换的作用 D.有丝分裂过程中均会以一定方式形成细胞壁
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0.01NA B. 一定条件下, 一定量的氧气通过 Na 单质后,Na 单质增重 3.2 g,转移电子数目为 0.4NA C.0.1mol/L 的 CH3COONa 溶液中所含碳原子总数为 0.2NA

【全国百强校】衡水金卷2018年普通高等学校招生全国统一考试 分科综合卷 理科数学(二)模拟试题(解析版)

【全国百强校】衡水金卷2018年普通高等学校招生全国统一考试 分科综合卷 理科数学(二)模拟试题(解析版)

2018年普通高等学校招生全国统一考试模拟试题理数(二)第Ⅰ卷(共60分)一、选择题:本大题共12个小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的.1. 已知集合)B. C. D.【答案】B故选:B2. )【答案】C由题意知:,解得:故选:C3. ( )D.【答案】D故选:D4. 的渐近线与抛物线)B. C. D.【答案】D,得到:故选:D点睛:解决椭圆和双曲线的离心率的求值及范围问题其关键就是确立一个关于a,b,c的方程或不等式,再根据a,b,c的关系消掉b得到a,c的关系式,建立关于a,b,c的方程或不等式,要充分利用椭圆和双曲线的几何性质、点的坐标的范围等.5. 袋中装有4个红球、3个白球,甲、乙按先后次序无放回地各摸取一球,在甲摸到了白球的条件下,乙摸到白球的概率是()【答案】B【解析】用A表示甲摸到白球,B故选:B6. 《算法统宗》是中国古代数学名著,由程大位所著,其中记载这样一首诗:九百九十九文钱,甜果苦果买一千,四文钱买苦果七,十一文钱九个甜,甜苦两果各几个?请君布算莫迟疑!其含义为:用九百九十九文钱共买了一千个甜果和苦果,其中四文钱可以买苦果七个,十一文钱可以买甜果九个,请问究竟甜、苦果各有几个?现有如图所示的程序框图,()343657【答案】B即若按全是甜果来算钱超出文,一个苦果和一个甜果差价位则p故选:B在区间)B. C. D.【答案】C【解析】内,得:,可知两个交点关于对称,故两个零点的和,.故选:C8. 为真命题,则实数)A. 2B. 3C. 4D. 5【答案】A2,故由存在性的意义知. 2.故选:A9. 已知某几何体的三视图如图所示,则该几何体的体积为()【答案】B【解析】由三视图,可知该几何体为一个半圆柱与一个三棱锥结合而成的(如图所示).半圆柱的底面半径为1,侧棱长为2,个侧面是全等的等腰三角形,腰长为22的等边三角形,因此故选:B点睛:三视图问题的常见类型及解题策略(1)由几何体的直观图求三视图.注意正视图、侧视图和俯视图的观察方向,注意看到的部分用实线表示,不能看到的部分用虚线表示.(2)由几何体的部分视图画出剩余的部分视图.先根据已知的一部分三视图,还原、推测直观图的可能形式,然后再找其剩下部分三视图的可能形式.当然作为选择题,也可将选项逐项代入,再看看给出的部分三视图是否符合.(3)由几何体的三视图还原几何体的形状.要熟悉柱、锥、台、球的三视图,明确三视图的形成原理,结合空间想象将三视图还原为实物图.10.,运动过程种,点与平面的距离保持不变,运动的路程关系,则此函数图象大致是()A. B. C. D.【答案】CN,计算得:同理,当N为线段AC或的中点时,计算得符合C项的图象特征.故选:C11. 两点,,则直线的斜率为()【答案】D.,,由韦达定理得,得,带入整理,得故选:D12. 已知函数的取值范围是()C.【答案】C时,a值:解得:②再求a值:,解得:一的解,此时,把的图象是由的图象向左平移1个单位,再向上平移a个单位(或向下平移-a个单位),由时,,可知的取值范围是故选:C点睛:已知函数有零点求参数取值范围常用的方法和思路(1)直接法:直接根据题设条件构建关于参数的不等式,再通过解不等式确定参数范围;(2)分离参数法:先将参数分离,转化成求函数值域问题加以解决;(3)数形结合法:先对解析式变形,在同一平面直角坐标系中,画出函数的图象,然后数形结合求解.第Ⅱ卷(共90分)二、填空题(每题5分,满分20分,将答案填在答题纸上)13. _________.【解析】设当.故答案为:14. __________.【解析】如图,阴影部分即为不等式表示的区域,在直线为1,最大值为过点.5的取值范围为点睛:线性规划问题,首先明确可行域对应的是封闭区域还是开放区域、分界线是实线还是虚线,其次确定目标函数的几何意义,是求直线的截距、两点间距离的平方、直线的斜率、还是点到直线的距离等等,最后结合图形确定目标函数最值取法、值域范围.最大时,.,取等号,∴∠C的最大值为75°,此时sinC=,故答案为:16. 3位逻辑学家分配10枚金币,因为都对自己的逻辑能力很自信,决定按以下方案分配:(1)抽签确定各人序号:1,2,3;(2)1号提出分配方案,然后其余各人进行表决,如果方案得到不少于半数的人同意(提出方案的人默认同意自己方案),就按照他的方案进行分配,否则1好只得到2枚金币,然后退出分配与表决;(3)再由2号提出方案,剩余各人进行表决,当且仅当不少于半数的人同意时(提出方案的人默认同意自己方案),才会按照他的提案进行分配,否则也将得到2枚金币,然后退出分配与表决;(4)最后剩的金币都给3号.每一位逻辑学家都能够进行严密的逻辑推理,并能很理智的判断自身的得失,1号为得到最多的金币,提出的分配方案中1号、2号、3号所得金币的数量分别为__________.【答案】9,0,1【解析】先看一下个人的利益最大化:①3号:如果1号的方案被否定,此时剩余金币有8枚,那么2号的方案必然是2号8枚,3号0枚,然后2号方案不低于半数通过,②由①的分析可知,只要1号的分配方案分配给3号的金币数量多于0,3号就会同意,方案就会通过,所以1号的利益最大化的分配方案是1号,2号,3号所得金币数量分别是9,0,1.故答案为:9,0,1三、解答题(本大题共6小题,共70分.解答应写出文字说明、证明过程或演算步骤.)17. 已知数列(1)的通项公式;(2).【答案】【解析】试题分析:(1,作差易得:(2的值.试题解析:(1)两式相减得2..(2)两式相减得,所以点睛:用错位相减法求和应注意的问题(1)要善于识别题目类型,特别是等比数列公比为负数的情形;(2)在写出“S n”与“qS n”的表达式时应特别注意将两式“错项对齐”以便下一步准确写出“S n-qS n”的表达式;(3)在应用错位相减法求和时,若等比数列的公比为参数,应分公比等于1和不等于1两种情况求解.18. 某校高三年级有1000且所有得分都是整数.(1)求全班平均成绩;(2)计算得分超过141的人数;(精确到整数)(3)甲同学每次考试进入年级前1004次考试,100名的次数,写.【答案】人;(3)见解析.【解析】试题分析:(1)(2),从而计算出得分超过141的人数;(3)0,1,2,3,4,计算出相应的概率值,利用公式即可算得期望与方差.试题解析:(1)故141分以上的人数为.0,1,2,3,4,,,故的分布列为19. 已知在直角梯形中,,折起至.(1)(2),当二面角.【答案】(1)见解析;(2【解析】试题分析:(1)要证平面平面,转证平面即可;(2)建立空间直角坐标系计算平面的法向量,利用二面角为45°建立等量关系求出的值............................试题解析:(1)(2)由(1)方向分别为立如图所示的空间直角坐标系.的一个法向量为,,则.,∴.20.(1)(2)若是,求出此定值,若不是,请说明理由.【答案】(1(2【解析】试题分析:(1)可得M(﹣2,2λ),N(﹣2+4λ,2)Q(x,y P的轨迹方程为;(2)设直线的斜率为,把代入椭圆方程,化简整理得利用韦达定理易得四边形GFHE,试题解析:(1)求得,,整理得的轨迹为第二象限的椭圆,由对称性可知曲线(2,当直线的斜率为,把程,化简整理得.∴∵,,∴.当直线斜率不存在或为零时,∴为定值点睛:求定值问题常见的方法①从特殊入手,求出定值,再证明这个值与变量无关.②直接推理、计算,并在计算推理的过程中消去变量,从而得到定值.21. 已知函数.(1)有极值点,求证:必有一个极值点在区间(2)【答案】(1)见解析;(2)见解析.【解析】试题分析:(1)易知,设,(2)对任意,即证.试题解析:(1),有极值点,或,有两个零点,且有一个在区间..(2),∴只需证,∴当时,为增函数,,∴当时,为增函数,∴原不等式成立.22. 在平面直角坐标系(1)(2)2,交曲线两点,若.【答案】(1(2【解析】试题分析:(1把极坐标方程化为直角坐标方程;(2(为参数,,利用韦达定理可得.的斜率.试题解析:(1)(2),的方程,整理得..同向共线.由,得.23.((2)证明:【答案】(1)最小值为9;(2)见解析.【解析】试题分析:(1(2累加即可得结果.试题解析:(1)由柯西不等式,得时,取等号.的最小值为9.(2)由同理得,.,.。

【完整版】衡水金卷2018年普通高等学校招生全国统一考试模拟试题(三)文综地理试题

【完整版】衡水金卷2018年普通高等学校招生全国统一考试模拟试题(三)文综地理试题

衡水金卷2018年普通高等学校招生全国统一考试模拟〔调研卷〕〔三〕文综地理试题考前须知:1.本试卷分第一卷〔选择题〕和第二卷〔非选择题〕两局部。

答题前,考生务必将自己的姓名、考生号填写在答题卡上。

2. 答复第一卷时,选出每题的答案后,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。

写在试卷上无效。

3. 答复第二卷时,将答案填写在答题卡上,写在试卷上无效。

4. 考试完毕,将本试卷和答题卡一并交回。

第一卷本卷共35小题。

每题4分,共140分。

在每个小题给出的四个选项中,只有一项为哪一项符合题目要求的。

净初级消费力(NPP)是生态系统在一段时间内所固定的碳总量,是由光合作用所产生的有机质总量扣除自养呼吸后的剩余局部。

农业植被净初级消费力代表了农田生态系统通过光合作用固定大气中CO2的才能,决定了农田土壤中可获得的有机质的含量。

根据每种作物的枯燥系数、收获指数和根冠比等指标,将我国划分为9个农业区。

下面两幅图分别表示我国20年农作物的NPP分布(单位:TgC)和农作物NPP密度分布(单位gC/m2)。

据此完成下面小题。

1. 据统计左图中1、3、5农业区产生的NPP相加占到全国NPP总量的一半以上,其主要原因是些区域A. 机械化程度高B. 农业消费条件好C. 单产量高D. 种植面积大2. 关于右图中不同区域农作物NPP密度说法正确的选项是A. 1区域作物NPP密度与积温呈负相关B. 5区域作物NPP密度与日照时数呈负相关C. 6区域作物NPP密度与降水呈正相关D. 8区域作物NPP密度与降水呈正相关3. 改革开放以来,5农业区产生的NPP总体呈现下降趋势,其原因和影响结果可能是A. 农田被占粮食产量增加B. 水田转为蔬菜地饮食构造发生改变C. 农田被占粮食产量下降D. 水田转为旱地农业构造发生改变【答案】1. D 2. D 3. C【解析】1. 读左图可知,我国地势的第三级阶梯分布着1、3、5农业区,第三级阶梯地形以平原为主,种植面积大,故1、3、5农业区产生的NPP相加能占到全国NPP总量的一半以上。

(衡水金卷调研卷)2018年普通高等学校招生全国统一考试模拟试题二(解析版)

(衡水金卷调研卷)2018年普通高等学校招生全国统一考试模拟试题二(解析版)

(衡水金卷调研卷)2018年普通高等学校招生全国统一考试模拟试题英语二本试题卷共8页。

全卷满分120分,考试用时100分钟。

第一部分阅读理解(百强校英语解析团队专供)(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下面短文,从每题所给的A、B、C和D四个选项中,选出最佳选项。

AMovie Nights at the Museum brings you classic movies in a classic location. Each film has been chosen for its connection to an area of our knowledge.We start this movie season with ocean-inspired cinema to celebrate the arrival of the noble blue whale in the Museum’s typical hall, the site for the series.So get your popcorn, take your seat, and settle in for Movie Nights at the Museum.JawsDid you know that great whites actually find the taste of human flesh not tasty? However, this legendary Steven Spielberg thriller sees a giant great white shark cause destruction on the shores of a New England beach town in the mid-1970s.Free WillyDid you know that the killer whale is not a whale? It’s actually a dolphin. In this heart-warming 90s classic, orphan Jesse makes friends with a trapped killer whale and does whatever it takes to return him to his family and ocean home.Finding NemoContrary to popular belief, fish are actually good at remembering things. Follow young clownfish Nemo, taken unexpectedly from his Great Barrier Reef home, and his father and forgetful partner who go on a brave journey to find him, in this charming Disney adventure.The Little MermaidIn this classic Disney tale of a mermaid princess who dreams of becoming human, Ariel falls in love with a handsome prince, much to the sadness of her father and long-suffering friends. But did you know that a fish is actually brown and flat, with both eyes on one side of its body?Ticket: £ 28, Member: £ 25.1. In which movie do fish memorize things well?A. Jaws.B. Free Willy.C. Finding Nemo.D. The Little Mermaid2. Who helps a whale return to his home?A. Ariel.B. Jesse.C. Nemo.D. Steven.3. Who will most likely be interested in the passage?A. A scientist on farming.B. A director of war movies.C. A teenager liking ocean life.D. A farmer hating cruel animals.【答案】1. C 2. B 3. C【解析】试题分析:本文是一篇说明文。

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本试题卷共8页。

全卷满分120分,考试用时100分钟。

第一部分阅读理解(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下面短文,从每题所给的A、B、C和D四个选项中,选出最佳选项。

AHave you ever been to France before? It is not only a country of great food, fashion and art. It‟s also home to the most influential painters in the world.Edouard ManetHe was one of the first artists to paint modern life. He began to paint in his own style, but still used some of Couture‟s techniques like thick lines and dark colors. He was greatly influenced by Claude Monet and Bert he Morisot, which can be seen in his use of light shades. Most of his paintings had scenes of daily life on the streets of Paris. His works include Olympia and The Absinthe Drinker.Camille PissarroIn his early years, Pissarro painted scenes of a river or a path from memory. After meeting Claude Monet and Paul Cezanne, who painted in a more realistic style, he changed his course to Impressionism. During his career, he experimented with various styles, and finally formed his own one. His works include Old Market at Rouen and Sunset at St. Charles.Vincent van GoghHe had a huge influence on art in the 20th century. His early works were most painted in somber tones. However, influenced by Monet, Pissarro, and Bernard, he adopted brighter colors in his works, and started creating his own techniques. Although he had produced more than 2,000 works of art, the artist sold only one painting during his lifetime —Red Vineyard at Arles. His works include The Potato Eaters, Starry Night and Bedroom in Arles.Claude MonetHe was the founder of the Impressionist movement and completely changed the French painting in the 19th century. Although he first started by selling charcoal caricatures(木炭讽刺画)in Paris, he soon started painting with oil after meeting Eugene Boudin, who taught him to use oil paints and also encouraged him to paint outdoors. And then he painted with his own style. His works include Impression, Sunrise and The Water Liles.1. What can we learn about Edouard Manet‟s paintings?A. They reflected the changes of life.B. They were mainly about daily life.C. They were all painted in bright colors.D. They were painted in Morisot‟s style.2. Which painting was sold by Vincent van Gogh in person?A. The Potato Eaters.B. Bedroom in Arles.C. Red Vineyard at Arles.D. Starry Night.3. What‟s the common point of the four painters from the text?A. All of them were given many awards in their life.B. All of them were taught by some famous painters.C. All of them had a good taste in delicious food.D. All of them had their unique styles in painting.BFinding true love can be pretty tough for a lot of people, but a lady from a fairly well-known San Francisco advertising agency seems to think money helps. She is offering $10,000 to any of her friends who can introduce her to her Mr. Right. She wants to find her future husband through this way.The unnamed husband seeker who sent out the email had just finished reading the best-selling book named Lean In. It was 11 p. m. on a Sunday night and she realized this was the second self-help book she had read in the month. She was still single. Things were not looking fine, but there was hope for her still. If the book had taught her anything, it was that she needed to take a more positive role in finding love. After all, if she wanted to get a better job, she wouldn‟t just sit outside an employer‟s building and wait for someone to offer it to her, so why should finding a husband be any different? But instead of going out and meeting new people she decided to write an email to all her friends, offering to give them $10,000 on her wedding day if any of them managed to introduce her to her future husband.“I am writing you today because I‟ve decided to make an aggressive action plan on finding the man that I get to hang out with fo rever,” the woman writes in her email. “Introducing me to my husband is just not high on your to-do list. But I think I have an idea that might change that…” You guessed it, and this is where she offers to reward her“closest friends” with cold hard cash.“I will personally give ten thousand dollars to the friend who introduces me to my husband.Here is how the program works:Step 1: You set me up on a date with a man.Step 2: I marry that man.Step 3: I give you $10,000 on my wedding day.I know you‟re thi nking that this is nuts. Just plain crazy. …You can find a husband without giving $10,000.‟ Well for starters, thank you! I‟m happy.”4. What does the lady offer $10,000 to any of her friends for?A. Celebrating the fact that she has made a decision to find a husband.B. Checking the power of money among her circle of friends.C. Encouraging her friends to help find her Mr. Right.D. Sharing her happiness of having found true love.5. What does the underlined word “nuts” mean in the last paragraph?A. deliciousB. sensibleC. angryD. foolish6. What‟s the purpose of the author‟s mentioning getting a better job in Paragraph 2?A. To stress the importance of finding a good job.B. To stress the importance of taking a positive attitude.C. To show that waiting patiently is necessary to get a job.D. To state that we need to be patient before a job is offered.7. What kind of person do you think the lady is?A. Adventurous.B. Imaginative.C. Considerate.D. Polite.CTaxi-booking app Uber agreed to sell its business in China to Didi Chuxing. The two firms had been fierce competitors, but Didi Chuxing had controlled the Chinese market with an 87% share.Uber China launched in 2014, but it had failed to make any profit for a long time. Cheng Wei, founder and chief executive of Didi Chuxing, said the two companies had learned a great deal from each other over the past two years in China. He added that the deal would set the mobile transportation industry on a healthier path of growth at ahigher level. As part of the deal, Mr. Cheng would join the board of Uber, while Uber chief executive Travis Kalanick would also join Didi‟s board.Uber‟s China business would own its separate branding while US-based Uber Technologies would hold about 17.5% in the combined company. Didi Chuxing is backed by Chinese Internet giants Tencent and Alibaba.Uber had been struggling to break into the Chinese market despite having Chinese search engine Baidu as an investor. Last February, the company admitted it was los ing more than $1 billion a year in China. “Funding their Chinese dreams was becoming too expensive for Uber,” Duncan Clark, chairman of Beijing-based consultancy BDA, told the BBC. Travis Kalanick said, “As a businessman, I‟ve learned that being successful is about listening to your head as well as following your heart.”The fierce competition had led both companies to spend much more on their journeys. The combination is likely to see fewer such subsidies(补贴). “One thing to watch carefully is how quickly c onsumers feel the impact as subsidies are withdrawn.” Mr. Clark added.The deal with Didi Chuxing came just days after China had agreed to provide a legal framework for taxi-ordering apps. Both Uber and Didi welcomed the decision. The new rules took effect last November and could, among other things, forbid such platforms to operate below cost.8. According the second paragraph, what can we know?A. Being successful is about listening to your head and following your heart.B. The deal would make the mobile transportation industry grow much faster.C. Didi Chuxing had learnt more in China than Uber over the past two years.D. Mr. Cheng would be working as a member of the board of Uber as planned.9. What is the best title of the passage?A. Uber sold Chinese business to Didi ChuxingB. Using Didi Chuxing brings more subsidiesC. Listen to your head and follow your heartD. The new rules look effect last November10. What is the impact of the fierce competition between Uber and Didi?A. Uber dominated the Chinese market with an 87% share.B. China provided a legal framework for taxi-ordering apps.C. Funding their Chinese dreams became expensive for Uber.D. Chinese search engine Baidu became an investor of Uber‟s.11. The passage is probably taken from a website about ________.A. appsB. politicsC. economyD. technologyDYou get anxious if there‟s no wi-fi in the hotel or mobile phone signal up the mountain. You feel upset if your phone is getting low on power, and you secretly worry things will go wro ng at work if you‟re not there. All these can be called “always on” stress caused by smart phone addiction.For some people, smart phones have liberated them from the nine-to-five work. Flexible working has given them more autonomy(自主权)in their working lives and enabled them to spend more time with their friends and families. For many others though, smart phones have become tyrants(暴君)in their pockets, never allowing them to turn them off, relax and recharge their batteries.Pittsburgh-based developer Kevin Holesh was worried about how much he was ignoring his family and friends in favour of his iPhone. So he developed an app — Moment — to monitor his usage. The app enables users to see how much time they‟re spending on the device and set up warnings if the usage limits are breached(突破). “Moment‟s goal is to promote balance in your life,” his website explains. “Some time on your phone, some time off it enjoying your loving family and friends around you.”Dr. Christine Grant, an occupational(职业的)psychologist a t Coventry University, said, “The effects of this …always on‟ culture are that your mind is never resting, and you‟re not giving your body time to recover, so you‟re always stressed. And the more tired and stressed we get, the more mistakes we make. Physical and mental health can suffer.”And as the number of connected smart phones is increasing, so is the amount of data. This is leading to a sort of decision paralysis(瘫痪)and is creating more stress in the workplace because people have to receive a broader range of data and communications which are often difficult to manage. “It actually makes it more difficult to make decisions and many do less because they‟re controlled by it all and fell they can never escape the office,” said Dr. Christine Grant.12. Wha t‟s the first paragraph mainly about?A. The popularity of smart phones.B. The progress of modern technology.C. The signs of “always on” stress.D. The cause of smart phone addiction.13. Kevin Holesh developed “Moment” to ________.A. research how people use their mobile phonesB. help people control their use of mobile phonesC. make people love parents and friends aroundD. increase the fun of using mobile phones14. What‟s Dr. Christine Grant‟s attitude towards “always on” culture?A. Confused.B. Positive.C. Doubtful.D. Critical.15. According to the last paragraph, a greater amount of data means ________.A. we will become less productiveB. we can make a decision more quicklyC. we will be equipped with more knowledgeD. we can work more effectively第二节(共5小题;每小题2分,满分10分)根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。

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