2018届高三高考模拟考试试卷2

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2018届东北三省四市高三高考第二次模拟考试理科综合物理试卷

2018届东北三省四市高三高考第二次模拟考试理科综合物理试卷SYS201806270701一、选择题详细信息1. 难度：简单下列说法正确的是A. 太阳辐射的能量主要来自太阳内部的核裂变反应B. 黑体辐射的实验规律可用光的波动性解释C. 一束光照到某金属上，不能发生光电效应，是因为该束光的频率低于极限频率D. 氢原子从较低能级跃迁到较高能级时，棱外电子的动能增大，势能减小SYS20180627070详细信息2. 难度：中等比邻星是离太阳系最近（距离太阳4.2光年）的一颗恒星。

据报道：2016年天文学家在比邻星的宜居带发现了一颗岩石行星——比邻星b，理论上在它的表面可以维持水的存在，甚至有可能拥有大气层。

若比邻星b绕比邻星的公转半径是地球绕太阳的公转半径的p倍，比邻星b绕比邻星的公转周期是地球绕太阳的公转周期的a倍，则比邻星与太阳的质量的比值为A. B. C. D.SYS20180627070详细信息3. 难度：简单一物体沿直线运动，用x表示运动位移，用t表示运动时间。

从t=0时刻开始计时，物体与r的图象如图所示，图线斜率的绝对值为k，则以下说法正确的是A. 物体做匀速直线运动，速度大小等于kB. 物体做变减速直线运动，加速度均匀减小C. 物体做匀减速直线运动，加速度大小等于kD. 物体做匀减速直线运动，加速度大小等于2kSYS20180627070详细信息4. 难度：中等如图所示，一个理想变压器原线圈的匝数为50匝，副线圈的匝数为100匝，原线圈两端接在光滑的水平平行导轨上，导轨间距为0.4m。

导轨上垂直于导轨有一长度略大于导轨间距的导体棒，导轨与导体棒的电阻忽略不计，副线圈回路中电阻R1=5Ω，R2=15Ω．图中交流电压表为理想电压表。

导轨所在空间有垂直于导轨平面、磁感应强度大小为1T的匀强磁场。

导体棒在水平外力的作用下运动，其速度随时间变化的关系式为：，则下列说法中正确的是A. R1的功率为0.2 WB. 电压表的示数为C. 水平外力为恒力D. 变压器铁芯中磁通量变化率的最大值为0.04 Wb/sSYS201806270702二、不定项选择题详细信息5. 难度：中等如图所示，真空中有一边长为a的等边三角形ABC，P点是三角形的中心。

2018届黄浦区高考数学二模和参考答案

黄浦区2018年高考模拟考数学试卷(完卷时间：120分钟满分：150分) 2018.4考生注意：1．每位考生应同时收到试卷和答题卷两份材料，解答必须在答题卷上进行，写在试卷上的解答一律无效； 2．答卷前，考生务必将姓名等相关信息在答题卷上填写清楚，并在规定的区域贴上条形码； 3．本试卷共21道试题，满分150分；考试时间120分钟．一、填空题（本大题共有12题，满分54分）考生应在答题纸相应编号的空格内直接填写结果，每个空格填对前6题得4分、后6题得5分，否则一律得零分.1．已知集合{}{}1,2,31,A B m ==，，若3m A -∈，则非零实数m 的数值是． 2．不等式|1|1x ->的解集是．3．若函数()f x 是偶函数，则该函数的定义域是．4．已知ABC ∆的三内角A B C 、、所对的边长分别为a b c 、、，若2222sin a b c bc A =+-，则内角A 的大小是．5．已知向量a 在向量b 方向上的投影为2-，且3b =，则a b ⋅= ．(结果用数值表示)6．方程33log (325)log (41)0x x⋅+-+=的解x = ．7．已知函数2sin cos 2()1cos x x f x x-=，则函数()f x 的单调递增区间是．8．已知α是实系数一元二次方程22(21)10x m x m --++=的一个虚数根，且||2α≤，则实数m 的取值范围是．9．已知某市A 社区35岁至45岁的居民有450人，46岁至55岁的居民有750人，56岁至65岁的居民有900人．为了解该社区35岁至65岁居民的身体健康状况，社区负责人采用分层抽样技术抽取若干人进行体检调查，若从46岁至55岁的居民中随机抽取了50人，试问这次抽样调查抽取的人数是人． 10．将一枚质地均匀的硬币连续抛掷5次，则恰好有3次出现正面向上的概率是．(结果用数值表示) 11．已知数列{}n a 是共有k 个项的有限数列，且满足11(2,,1)n n nna a n k a +-=-=-，若1224,51,0k a a a ===，则k = ．12．已知函数2()(02)f x ax bx c a b =++<<对任意R x ∈恒有()0f x ≥成立，则代数式(1)(0)(1)f f f --的最小值是．二、选择题（本大题满分20分）本大题共有4题，每题有且只有一个正确答案，考生应在答题卷的相应编号上，将代表答案的小方格涂黑，选对得5分，否则一律得零分．13．在空间中，“直线m ⊥平面α”是“直线m 与平面α内无穷多条直线都垂直 ”的答( )．(A )充分非必要条件 (B )必要非充分条件 (C )充要条件 (D )非充分非必要条件14．二项式40的展开式中，其中是有理项的项数共有答( ). (A ) 4项 (B ) 7项 (C ) 5项 (D ) 6项15．实数x y 、满足线性约束条件3,0,0,10,x y x y x y +≤⎧⎪≥≥⎨⎪-+≥⎩则目标函数23w x y =+-的最大值是答( )．(A ) 0 (B ) 1 (C ) 2- (D ) 316．在给出的下列命题中，是假命题的是答( )． (A )设O A B C 、、、是同一平面上的四个不同的点，若(1)(R)OA m OB m OC m =⋅+-⋅∈，则点A B C 、、必共线(B )若向量a b 和是平面α上的两个不平行的向量，则平面α上的任一向量c 都可以表示为(R)c a b λμμλ=+∈、，且表示方法是唯一的(C )已知平面向量OA OB OC 、、满足||||(0)OA OB OC r r ==>|=|，且0OA OB OC ++=，则ABC ∆是等边三角形(D )在平面α上的所有向量中，不存在这样的四个互不相等的非零向量a b c d 、、、，使得其中任意两个向量的和向量与余下两个向量的和向量相互垂直三、解答题（本大题满分76分）本大题共有5题，解答下列各题必须在答题卷的相应编号规定区域内写出必要的步骤．17.（本题满分14分）本题共有2个小题，第1小题满分4分，第2小题满分10分．在四棱锥P A B -中，P A A B ⊥平面，,,1,AB AD BC AD BC ⊥=045CD CDA =∠=．(1)画出四棱锥P ABCD -的主视图；(2)若PA BC =，求直线PB 与平面PCD 所成角的大小．(结果用反三角函数值表示)18.（本题满分14分）本题共有2个小题，第1小题满分6分，第2小题满分8分．某企业欲做一个介绍企业发展史的铭牌，铭牌的截面形状是如图所示的扇形环面(由扇形OAD 挖去扇形OBC 后构成的)．已知10,(010)OA OB x x ==<<米米，线段BA CD 、线段与弧BC 、弧AD 的长度之和为30米，圆心角为θ弧度． (1)求θ关于x 的函数解析式；(2)记铭牌的截面面积为y ，试问x 取何值时，y 的值最大？并求出最大值．19.（本题满分14分）本题共有2个小题，第1小题满分6分，第2小题满分8分．已知动点(,)M x y 到点(2,0)F 的距离为1d ，动点(,)M x y 到直线3x =的距离为2d，且12d d =. (1)求动点(,)M x y 的轨迹C 的方程； (2)过点F 作直线:(2)(0)l y k x k =-≠交曲线C 于P Q 、两点，若OPQ ∆的面积OPQ S ∆(O 是坐标系原点)，求直线l 的方程.20.（本题满分16分）本题共有2个小题，第1小题满分4分，第2小题满分6分，第3小题满分6分．已知函数22, 10,()=1, 0 1.x x f x x x --≤<⎧⎨-≤≤⎩(1) 求函数()f x 的反函数1()fx -；(2)试问:函数()f x 的图像上是否存在关于坐标原点对称的点，若存在，求出这些点的坐标；若不存在，说明理由； (3)若方程()|()240f x f x ax +---=的三个实数根123x x x 、、满足:123x x x <<，且32212()x x x x -=-，求实数a 的值．21.（本题满分18分）本题共有3个小题，第1小题满分3分，第2小题满分6分，第3小题满分9分．定义：若数列{}n c 和{}n d满足*10,0,N nn n c d n +>>=∈且c ，则称数列{}n d 是数列{}n c 的“伴随数列”.已知数列{}n b 是数列{}n a 的伴随数列，试解答下列问题： (1)若*(N )nn b a n =∈，1b {}n a 的通项公式n a ；(2)若*11(N )n n n b b n a +=+∈，11b a 为常数，求证：数列2n n b a ⎧⎫⎛⎫⎪⎪⎨⎬ ⎪⎝⎭⎪⎪⎩⎭是等差数列； (3)若*1N )n nb n +=∈，数列{}n a 是等比数列，求11a b 、的数值．黄浦区2018年高考模拟考数学试卷参考答案和评分标准2018.4说明：1．本解答仅列出试题的一种解法，如果考生的解法与所列解答不同，可参考解答中的评分精神进行评分．2．评阅试卷，应坚持每题评阅到底，不要因为考生的解答中出现错误而中断对该题的评阅，当考生的解答在某一步出现错误，影响了后继部分，但该步以后的解答未改变这一题的内容和难度时，可视影响程度决定后面部分的给分，这时原则上不应超过后面部分应给分数之半，如果有较严重的概念性错误，就不给分．一、填空题. 1．2 2．(,0)(2,)-∞+∞ 3．[2,2]- 4．4π5．6- 6．27．3[,],Z 88k k k ππππ-+∈ 8．3(4- 9．140 10．51611．50 12．3.二、选择题．13．()A 14．()B 15．()D 16．()D三、解答题． 17．（本题满分14分）本题共有2个小题，第1小题满分4分，第2小题满分10分．解 (1)主视图如下：(2) 根据题意，可算得1,2AB AD ==. 又1PA BC ==，按如图所示建立空间直角坐标系，可得，(0,0,0),(1,0,0),(1,1,0),(0,2,0),(0,0,1)A B C D P . 于是，有(1,0,1),(1,1,0),(0,2,1)PB CD PD =-=-=- . 设平面PCD 的法向量为(,,)n x y z =，则0,0,n CD n PD ⎧⋅=⎪⎨⋅=⎪⎩即0,20.x y y z -+=⎧⎨-=⎩令2z =，可得1,1y x ==，故平面PCD 的一个法向量为(1,1,2)n =.设直线PB 与平面PCD 所成角的大小为θ，则||3sin 6||||n PB n PB θ⋅==. 所以直线PB 与平面PCD 所成角的大小为arcsin 6.18．（本题满分14分）本题共有2个小题，第1小题满分6分，第2小题满分8分．解 (1)根据题意，可算得弧BC x θ=⋅(m )，弧10AD θ=(m ). 又30BA CD BC CD +++=弧弧，于是，10101030x x x θθ-+-+⋅+=，所以，210(010)10x x x θ+=<<+.(2) 依据题意，可知22111022OAD OBC y S S x θθ=-=⨯-扇扇化简，得2550yx x =-++25225()24x =--+. 于是，当52x =(满足条件010x <<)时，max 2254y =(2m ).答所以当52x =米时铭牌的面积最大，且最大面积为2254平方米.19．（本题满分14分）本题共有2个小题，第1小题满分6分，第2小题满分8分．解 (1)结合题意，可得12|3|d d x ==-.又12d d =3=，化简得 22162x y +=. 因此，所求动点(,)M x y 的轨迹C 的方程是22162x y +=. (2) 联立方程组221,62(2),x y y k x ⎧+=⎪⎨⎪=-⎩得2222(13)121260k x k x k +-+-=.设点1122(,)(,)P x y Q x y 、，则2122212212,13126,130.k x x k k x x k ⎧+=⎪+⎪-⎪=⎨+⎪∆>⎪⎪⎩于是，弦||PQ == 点O 到直线l的距离d =.由OPQS ∆== 42210k k -+=，解得1k =±，且满足0∆>，即1k =±都符合题意. 因此，所求直线的方程为2020x y x y --=+-=或.20．（本题满分16分）本题共有3个小题，第1小题满分4分，第2小题满分6分，第3小题满分6分．解 (1)22, 10,()=1, 0 1.x x f x x x --≤<⎧⎨-≤≤⎩∴当10x -≤<时，()2,0()2f x x f x =-<≤且.由2y x =-，得12x y =-，互换x y 与，可得11()(02)2f x x x -=-<≤. 当01x ≤≤时，2()1,()0f x x f x =-≤≤且-1.由21y x =-，得x =x y 与，可得1()10)f x x -=-≤≤.11, 0<2,2() 10.x x f x x -⎧-≤⎪∴=-≤≤(2) 答函数图像上存在两点关于原点对称.设点00000(,)(01)(,)A x y x B x y <≤--、是函数图像上关于原点对称的点，则00()()0f x f x +-=，即200120x x -+=，解得001(1,)x x ==舍去，且满足01x <≤ .因此，函数图像上存在点1,2(12)A B -和关于原点对称.(3) 考察函数()y f x =与函数y =当12x -≤≤-时，有()f x ≥4240x ax ---=，解得 2+2x a =-，且由21+22a -≤-≤-，得02a ≤≤.当1x <≤时，有()f x <240ax -=，化简得 22(4)40a x ax ++=，解得24=0+4a x x a =-，或(当02a ≤≤时，24024aa -<-<+). 于是，123224,,024ax x x a a =-=-=++. 由32212()x x x x -=-，得22442=2(+)+442a a a a a -++，解得32a -±=.因为312a -=<-，故32a --=不符合题意，舍去；02a <=<，满足条件.因此，所求实数a =21．（本题满分18分）本题共有3个小题，第1小题满分3分，第2小题满分6分，第3小题满分9分．解 (1)根据题意，有*10,0,N n n n a b a n +>>=∈且.由*(N )nn b a n =∈，1b =111n a a b +====*N n ∈.所以n a =，*N n ∈．证明 (2)*11(N )n n n b b n a +=+∈，*10,0,N n n n a b a n +>>=∈且，∴11nn b a ++==11n n b a ++=*N n ∈．∴22111n n n n b b a a ++⎛⎫⎛⎫-= ⎪ ⎪⎝⎭⎝⎭，*N n ∈．∴数列2n n b a ⎧⎫⎛⎫⎪⎪⎨⎬ ⎪⎝⎭⎪⎪⎩⎭是首项为211b a ⎛⎫ ⎪⎝⎭、公差为1的等差数列．解(3)*1N )n n b n +=∈，*10,0,N n n n a b a n +>>=∈且，*N n n a b n <+≤∈，得11n a +<.{}n a 是等比数列，且0n a >，设公比为(0)r r >，则1*1(N )n n a a r n -=∈.∴当1r >，即lim n n a →∞→+∞，与11n a +<≤矛盾．因此，1r >不成立. 当01r <<，即lim 0n n a →∞→，与11n a +<01r <<不成立.∴1r =，即数列{}n a 是常数列，于是，1n a a =(11a <≤).*11(N )n n b n +∴=∈. 100n b b >∴>，，数列{}n b 也是等比数列，设公比为(0)q q >，有11n n b b q +=.2n a +∴=可化为222221111111(1)2(1)0(1n n b a q a b q a a a --+-=<≤，*N n ∈.2222422111111111(1)0,20,(1)0,4(2)0b a a b a a a b a ->≠->∆=-≥，∴关于x 的一元二次方程22222111111(1)2(1)0b a x a b x a a --+-=有且仅有两个非负实数根.一方面，n q (*N n ∈)是方程22222111111(1)2(1)0b a x a b x a a --+-=的根；另一方面，若1(0)q q ≠>，则无穷多个互不相等的234,,,,,,n q q q q q 都是该二次方程的根.这与该二次方程有且仅有两个非负实数根矛盾！1q ∴=，即数列{}n b 也是常数列，于是，1n b b =，*N n ∈.∴由*1N )n nb n +=∈，得1a =把1a =1n a +=解得1b11a b ⎧=⎪∴⎨=⎪⎩ ．。

广东省化州市2018届高三语文上学期第二次模拟考试试题(扫描版)

广东省化州市2018年高考第二次模拟考试语文试题化州市2018年高考第二次模拟语文试题参考答案一、现代文阅读（35 分）（一）论述类文本阅读（本题共3小题，9 分）1. C（C 选项中的“人们平常表现出的彬彬有礼其实是端架子、装腔作势的假象”和原文中“在一个公开的场合”这一信息是存在出入的。

选项有意模糊了其范围和程度，从而存在错误。

）2. A（A 选项中的因果关系不当。

王国维的评价中并没有指出，词“要眇宜修”的特点对其境界的限制，只是客观地表述诗和词的不同特点。

3. A（A 选项中“诗人文士主观上并不想要把思想志意完全容纳进去”，意思是诗人文士仍然是希望把思想志意部分地融入，这一判断显然存在问题。

）（二）文学类文本阅读（本题共3小题，14 分）4．C（A 项“自豪”是“我”当时的真实心理感受，并非反语。

B 项儿媳的看法不完全是偏见。

D 项“情节跌宕起伏”“始终在轻快而诙谐的笔调中发展”与原文不符，“如一只优美婉转的小夜曲”也评价不当，“随想曲”与“小夜曲”是两种不同曲风，本文叙述平静、自然，更近）似“随想曲”。

“静谧、清新”1 分，结（2 分，5．（1）特点：新月迷蒙，柔柳低拂，营造出静谧、清新的氛围。

合原文分析1分。

意思相近即可）（2）作用：扣题，交代故事发生的背景——乡下月夜（环境方面）；引出下文，为组长赞美乡下空气作铺垫（情节上）；景中寓情，烘托出“我”有幸陪组长坐在小河边时的喜悦之情（2 分，每答出1点给1分）。

（人物）（1）第一次听组长赞美乡下的自然条件时，充满自豪，满怀邀请组长孙女下乡玩的热情。

6．“我”的热情（2）听组长担心乡下卫生条件却又忧虑孙女不能像乡下人那样吃上新鲜东西时，冷却。

“我”进而想到自己和组长的巨大差距，感到失落，心生怨气。

（3）“我”感到悲凉。

（答出1 点（4）组长对“我”进行“传统教育”并对“我”的埋怨毫不理解时，给1分，答出2点得3分，答出3点给5分。

如有其他答案，只要言之成理，也可给分）（三）实用类文本阅读（本题共3小题，12 分）“首次对变味的‘收视率’开战”错误，原文是“首次明确将《美人私房菜》收视不佳7.A（归结为‘制作方没有购买收视数据……’”。

广东省化州市2018年第二次高考模拟考试语文试卷

广东省化州市2018年高考第二次模拟考试语文试题一、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。

词之所以微妙是什么缘故呢？因为词在初起的时候，本来就是那些个诗人文士写给美丽的歌女去唱歌的歌词，没有想把我的思想怀抱理想志意都写到词里边去。

他最初本来没有这种用心，没有这种想法。

写美丽女子的爱情，就是写美丽女子的爱情。

可是，奇妙的事情就是在这里发生的。

每一个人都是带着自己的思想文化教养性格的种种不同的背景的，不管你是欣赏诗歌，还是创作诗歌，都无法避免，你就是你，你无论是解说，无论是创作，都带着自己的背景在其中了。

微妙的事情发生在哪里呢？就是在这些个诗人文士，当他用游戏笔墨为了娱宾遣兴给歌女写歌词的时候，无法避免地把自己的性格思想，在不知不觉之中，隐意识的，自己完全都不知道的，流露表现在爱情的歌词中去了。

这是很微妙的一件事情。

而且，我们有一句俗话说，观人于揖让，不若观人于游戏。

我们要观察一个人，看他揖让进退，他在一个公开的场合，表现得彬彬有礼——当然，因为那是要给大家看的。

可是，当他游戏的时候，因为不用端着架子，不用装腔作态了，所以，不知不觉反而把他更真的自我表现出来了。

词就是有这样一种微妙的作用。

就是说他本来没有要写自己理想志意的用心，只是给美丽的歌女，写一些漂亮的爱情的歌词。

可是他不知不觉地就把他最深隐的本质，这不是拿腔作调说出来的什么伦理道德，而是他自己真正的感情人格的最基本的本质，无意之中，不注意之间就流露表现出来了。

词既有这样一种微妙的作用，因此，王国维在他的《人间词话》里，就为词的性质下了这样几句定义，他说：“词之为体，要眇宜修，能言诗之所不能言，而不能尽言诗之所能言。

诗之境阔，词之言长。

”他说词这种文学体式是“要眇宜修”。

“要眇宜修”出于《楚辞·九歌·湘君》“美要眇兮宜修”，是写湘水上的一个神灵具有一种要眇宜修的美。

这是一种什么样的美呢？《楚辞》有王逸的注解《楚辞章句》，说“要眇”是“好貌”，是一种美好的样子。

2018-2019年广西南宁市二模：南宁市2018届高三第二次模拟考试数学(文)试题-附答案精品

青霄有路终须到，金榜无名誓不还！
2018-2019年高考备考
广西南宁市2018届第二次模拟考试
文科数学
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{}|8U x x =≤，集合{}
2|80A x x x =-≤，则U C A =（） A ．(),8-∞ B ．(],0-∞ C ．(),0-∞ D ．∅
2.下列命题正确的是（）
A ．命题“若αβ=，则sin sin αβ=”的逆否命题为真命题
B ．命题“若a b <，则22
ac bc ≤”的逆命题为真命题
C ．命题“0,50x x ∀>>”的否定是“000,50x x ∃≤≤”
D ．“1x <-”是“()ln 20x +<”的充分不必要条件 3.已知tan 3α=，则
sin 21cos 2αα
=+（） A ．-3 B ．13- C ．13 D ．3 4.已知向量b 在向量a 方向上的投影为2，且1a =，则a b = （）
A ．-2
B ．-1 C. 1 D ．2
5.若点P 为圆22
1x y +=上的一个动点，点()()1,0,1,0A B -为两个定点，则PA PB +的最大值是（）
A ．2
B ．22 C. 4 D ．42
6.《九章算术》中对一些特殊的几何体有特定的称谓，例如：将底面为直角三角形的直三棱柱称为堑堵.将一堑堵沿其一顶点与相对的棱刨开，得到一个阳马（底面是长方形，且有一条侧棱与底面垂直的四棱锥）和一个鳖臑（四个面均匀直角三角形的四面体）.在如图所示的堑堵111ABC A B C -中，。

2018年广东省高考英语听说模拟考试题2

【做答案】2018年广东省高考英语听说模拟考试题2PART A Reading AloudFish are masters of the waters. And sailfish are the fastest of them all. Their speed makes them one of the ocean’s most fearsome predators. Off the coast of Mexico, 30 sailfish have surrounded a ball of sardines. To catch their prey requires more than speed alone. In the tightest turns, fins maintain stability. Their sickle-shaped tail powers them forwards, and that extraordinary dorsal fin helps intimidate their prey. To the naked eye, the action is too fast, slowed, their challenge becomes clear. Just picking a target is hard enough.PART B Role Play角色：你是辅导员。

任务：（1）和你所辅导的学生讨论有关考试的话题；（2）根据谈话内容回答其他同学的提问。

生词：counselor 顾问cram 死记硬背hastily 仓促地录像原文：M: You look better today, Mandy. How did your test go?W: Much better than the did yesterday.M: Did you pass?W: I not only passed my test, but I did quite well. I’m so happy.M: You should be. You worked really hard last night preparing for it.W: Thanks for helping me with it, Vincent. If you hadn’t pushed me to do my best, I wouldn’t able to pass.M: You don’t have to thank me. It’s just a part of my job, I’m your counselor. Well, when are your final exams?W:I find my finals in two months.下面请你用英语提出三个问题。

房山区2018届高三二模数学试题及答案(官方版)

房山区2018年高考第二次模拟测试试卷数学（文）本试卷共5页，150分。

考试时长120分钟。

考生务必将答案答在答题纸上，在试卷上作答无效。

考试结束后，将本试卷和答题纸一并交回。

第一部分（选择题共40分）一、选择题共8小题，每小题5分，共40分。

在每小题列出的四个选项中，选出符合题目要求的一项。

（1）设集合{|2},{|03}A x x B x x =≤=<<，则AB =（A ）{}2≤x x （B ） {|3}x x < （C ）{|23}x x << （D ）{|23}x x ≤< （2）设复数 iz 1i =-+，则复数z 在复平面内对应的点位于（A ）第一象限（B ）第二象限（C ）第三象限（D ）第四象限（3）下列函数中，在区间(2,)+∞上为增函数的是（A ）3xy =- （B ）12y x =- （C ） 2(2)y x =-- （D ）12log y x = （4）已知实数,x y 满足10,0,0,+-≥⎧⎪≥⎨⎪≥⎩x y x y的取值范围是（A ）()01，（B ）(]01，（C ）[)1+∞，（D）+⎫∞⎪⎪⎭（5）将函数sin y x =的图象上所有点的横坐标扩大到原来的2倍，纵坐标不变，再把所得的图象上的所有点向右平移2个单位长度，得到的图象所对应的函数解析式为(A)sin(22)=-y x (B) sin(22)=+y x (C) 1sin(1)2=+y x (D) 1sin(1)2=-y x （6）已知某几何体的三视图如图所示，则该几何体的最长棱为俯视图左视图（A ）4 （B ）22 （C ）7 （D ）2（7）12+>“”x x是1>“”x 的（A ）充要条件（B ）充分而不必要条件（C ）必要而不充分条件（D ）既不充分也不必要条件（8）正方形ABCD 的边长为1，点E 在边AB 上，点F 在边BC 上，11,24AE BF ==．动点P 从E 出发沿直线向F 运动，每当碰到正方形的边时反弹，反弹时反射角等于入射角，当点P 第一次碰到E 时，P与正方形的边碰撞的次数为（A ）3 （B ）4 （C ）6 （D ）8第一部分（非选择题共110分）二、填空题共6小题，每小题5分，共30分。

2018届黄浦区高考英语二模试卷和参考答案

黄浦区2017 学年度第二学期期终教学质量监控测试高三英语试卷2018.04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。

所有答題必须涂（选择题）或写（非选择题）在答题纸上，做在试卷上一律不得分。

3. 答題前，务必在答題纸上填写准考证号和姓名，并将核对后的条形码貼在指定位置上，在答題纸反面清楚地填写姓名。

第I 卷( 共100 分)I. Listening Comprehension Section ADirections: In Section A, you will hear ten short conversations between two speakers. At the endof 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.1. A. In a fruit store B. In a gym C. At a restaurant D. At a watch shop2. A. 4:30 B. 5:00 C. 5:10 D. 5:153. A. Boss and secretary B. Nurse and patientC. Salesman and customerD. Teacher and student4. A. The tickets are more expensive B. The tickets told online are cheaperC. It is difficult to get tickets on the spotD. It ‘ s better to buy tickoefftlsi ne5. A. He wants to be a musician in the futureB. He shows more interest in English learningC. He displays great music talent in the exhibitionD. He doesn ‘ t make enough efforts in English learning6. A. He wants to get some sleep B. He needs time to write a paperC. He has a literature class to attendD. He is troubled by his sleep problem7. A. It looks old B. It looks newC. It doesn ‘ t need paintingD. It doesn ‘ t run well8. A. Extremely dull B. Hard to understandC. Lacking a good storyD. Not worth seeing twice9. A. Plan his budget carefully B. Buy a gift for his motherC. Ask someone else for adviceD. Give her more information10. A. She didn ‘ t like telling jokes B. She went to school after 9 a.m.C. She may not have gone to school todayD. She may have been late for schoolSection BDirections:In Section B, you will hear two short passages and one longer conversation, and you will be asked several questions on each of the passages and the conversation. The passages and theconversation will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one would be the best answer to the question you have heard.Questions 11 through 13 are based on the following dialogue11. A. He found his TV was broken B. He missed a good TVC. He felt bored with the TV programD. He lost his meal tickets in the cafeteria12. A. He lost fifty dollars B. His time was wastedD. His brain wasn ve‘rytactive D. He watched just one program13. A. Part of the brain is not in useB. Part of the brain becomes more passiveC. It takes longer to process visual informationD. It processes complex information less activelyQuestions 14 through 17 are based on the following dialogue14. A. From the newspaper B. From her classmateC. From her friendsD. From the man15. A. Plant more trees in the school yard B. Organize a picnic on ThursdayD. Build a parking lot for students D. Protect the natural beauty on campus16. A. Attend a meeting B. Attend a classC. Visit her friendsD. Go to the parking lot17. A. Lend her pen to the man B. Go to the administrationC. Support the students action unionD. Give out the handoutsQuestions 18 through 20 are based on the following dialogue18. A. There are numerous languages in the existenceB. Most public languages are essentially vagueC. People differ greatly in their ability to communicateD. Big gaps exist between private and public languages19. A. It is a sign of human intelligence B. It improves with constant practiceC. It is something we are born withD. It varies from person to person20. A. How various languages are related to each otherB. How children learn to use language in particular waysC. How private languages are developed from public onesD. How people of different ages create their own languagesII. Grammar and Vocabulary 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, use one word that best fits each blank.Wildlife secrets of Nigeria's last wildernessResearchers from Chester Zoo, working with the Nigeria National Park Service, surveyed over1,000 square kilometres of the national park. Known （21）its mountain rainforests, savannah woodlands and rolling grasslands, it is home to some of West Africa's most endangered animals.The cameras （22）(spot) some animals that have never been recorded before in the area and others, like chimps, （23）are rarely seen. Stuart Nixon, the Africa Field Programme Co-ordinator at Chester Zoo, said confirmation of the locations of chimps was an important discovery." Gashaka's been regarded for many years as （24）(have)the biggest population of this Nigeria-Cameroon chimp, which is the rarest chimp subspecies,"he said." We consider it the most important population - that's really （25）we need to count it and see what the status of the chimp is right now - that will ultimately affect what we know about this subspecies elsewhere."The chimp （26） (endanger) across its range in Cameroon and Nigeria. Its total population is down to fewer than 9,000 individuals, of which about 1,000 are thought （27） (live)to live within the borders of the national park. "It's an incredible tool to use these camera traps and to reveal that this park - which is a （28）(forget)forgotten wilderness, really, for Nigeria- still has a really important reservoir of important species for Nigeria and Africa in general," said Stuart Nixon.Chester Zoo is funding guards for the rangers and providing training in wildlife onitoring and protection. "This work is helping us learn more about the secrets of one of our last wilderness areas and we must continue to work together to ensure （29）survival for future generations. "（30）all this beauty were lost it would be a terrible tragedy for all."Section BDirections: Complete the following passage by using the words in the box. Each word can only beWhy India's Pink City' is a Photographer's HeavenThe city of Jaipur is one of India's wonders. It 31 some of the country's most decorative royal palaces-elegant structures designed hundreds of years ago that still attract visitors today. Largely built in the 1700s, Jaipur is surrounded by a city wall and several 32 castles. Considered as a commercial center, it was ahead of its time due to the use of grid iron (网格状) city planning.A romantic dusty pink type-which has 33 the city since 1876, after it was painted pink towel come Queen Victoria's husband, Prince Albert-gives Jaipur its 34 as the "Pink City". This 35 beauty is what first brought HongKong-based photographers Victor Cheng and Samantha Wong to Jaipur.Walking in glass skyscrapers for century-old royal palacesand historic castles, the pair-who have 130,000 Instagram followers between them-said that the images they 36 in Jaipur received a lot of response online. "A lot of our followers hadn't seen this side of India, so we're happy we were able to show this side of the country." Cheng said.For the photographers, one of the city's most fascinating features is the light pink coloring of itsbuildings. “The first gate y ou see when you enter are pink, “ s aid Wang “Once you ‘ re through, everything around you varies in different 37 of the color--from bright pinks to reddish browns."The building is a(n) 38 of the City Palace, and its windows allowed royal women to observe street life without appearing in public. One of Cheng's most striking photos shows a straight front of the building and its hundreds of windows. The building's lively coloring also pushed Cheng to take a different 39 to editing than with images of other cities. "I toned down my usual editing process because the pink was so bright in reality," he said, "I wanted the photos to40 the actual color I was seeing myself and to maintain its tone."III. Reading Comprehension Section ADirections：For each blank in the following passages there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.Concerns about the harm caused by “too much” screen time – particularly when it is spent on social media — are widespread. But working out what a “healthy”41 might be is far from easy.Some negative experiences on social media— like 42 how your appearance compares to others— do affect some children. However, this does not mean that technology use in 43 is harmful and it is difficult to make claims about how it will affect different people.Consider the picture painted by a UNICEF review of existing research into the effects of digital technology on children‘s 44 comforts, including happiness, mental health and social life. Rather than stating that social media was harmful, it suggested a more 45 effect.The UNICEF report highlighted a 2017 study that examined 120,000 UK 15-year-olds. Among those teenagers who were the lightest users, it was found that increasing the time spent using technology was linked to 46 comfort — possiblybecause it was important for keeping up friendships. 47 , among the heaviest users of technology, any increase in time was linked to lower levels of comfort. Overall, the UNICEF study suggested that some screen time could be good for children‘s mental health.A broader look at evidence provided by some other high quality studies again suggests the story is not 48 . An early study in 2013 looked at how the television and video game habits of 11,000 UK five-year-olds affected them two years later. It is one of few studies actually 49 the effects of technology over time. It suggested that, compared with children who watched one hour of television or less on a weekday, a small increase in conduct problems was seen among those who watched more than three hours each day. Playing e lectronic g ames, however, was not seen as leading to a greater 50 of friendship or emotional problems.So how much time should our children spend looking at screens? It i s difficult to be 51 as different people spend time online in such different ways. A useful comparison might be with sugar. Broadly speaking, people 52 that too much sugar can be bad for your health. But the effect it might have can depend on many factors, from the type of sugar to the person and the amount. We would not 53 trust anyone who claims to predict how someone is affected by consuming one gram of sugar. The same could be said for 54 usage: the outcomes depend on so many factors that only very 55 predictions are possible.41. A. amount B. comparison C. experience D. medium42. A. accounting for B. boasting of C. commenting on D. worrying about43. A. general B. particular C. private D. public44. A. domestic B. material C. physical D. psychological45. A. complex B. dramatic C. harmless D. predictable46. A. improved B. maximum C. relative D. small47. A. As a rule B. In contrast C. On the whole D. Worse still48. A. convincing B. definite C. probable D. true49. A. estimating B. experiencing C. reducing D. tracing50. A. connection B. power C. promotion D. risk51. A. balanced B. independent C. precise D. subjective52. A. agree B. forget C. object D. remember53. A. equally B. readily C. reluctantly D. weakly54. A. emotion therapy B. social media C. TV broadcasting D. video game55. A. confident B. optimistic C. rough D. wildSection BDirections: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.(A)The cold northern wind here in the streets of Petersburg strengthens my nerves and fills me with delight. I cannot think of the Pole as cold and empty; in my imagination it is a region of beauty and delight. Who knows what strange landscapes and creatures we may find there. I shall satisfy my curiosity with the sight of an unknown part of the world-and walk where no man has before. Thinking of it, I feel the same joy a child feels when he sails his little boat on a voyage of discovery up his native river.This voyage was the favorite dream of my early years. My education was neglected, yet I was passionately fond of reading. Uncle Thomas's library contained only books about exploration, which I read day and night. Finally my thoughts comes to the idea of making a voyage of discovery.Six years have passed sinceI decided on the present voyage. I can, even now, remember the hour when I committed myself to this great enterprise. I began by making my body used to hardship. I went on whale hunting voyages to the North Sea; I voluntarily endured cold, hunger, thirst, and lack of sleep. I often worked harder than the common sailors during the day. Then, at nights, I studied mathematics, the theory of medicine, and sciences of practical importance for a seagoing adventurer. Twice I took jobs as an officer on a Greenland whaling ship. I felt a little proud when my captain asked me to remain with the ship, so valuable did he consider my services. And now, do I not deserve to achieve some great task? My life might have been passed inease and comfort, but I preferred glory to every pleasure that wealth placed in my path.56. What does the author think of the Pole? A.It reminds him of his childhood.B. It must be a region full of surprises.C. It would fulfil his dream to be an adventurer.D. It's too cold a destination with almost nothing.57. To realize his childhood dream, the author got .A. physically prepared by experiencing great sufferingB. spiritually prepared by gaining captain's recognitionC. academically prepared by reading books on explorationD. financially prepared by serving on a whale hunting ship.58. According to the passage, the author is definitely a person full of .A. curiosityB. fancyC. perseveranceD.pride(B)Subject Art & Design, Craft & Creative, BeautyDelivery method OnlineStudy level Professional development, Short. AccreditedRef FACE-GUARDPrice￡30, was ￡299 . use code: GUARD90Face Painting Academy DiplomaStart a career in Face Painting or simply learn for fun. Do you have a love for entertaining people? Are you artistic and want to impress people with a new skill?Have you ever thought about doing a course in face painting so you can earn fantastic money?If so then with this course you could become a qualified face painter just like hundreds of other people who have taken our courses. For a one-off fee一(次性付款), you can study online and complete the diploma in about 28 hours.The comprehensive syllabus (教学大纲) is supported by 16 instructional videos so you can learn all the designs with ease, andyou will learn a wide range of designs including dog, rabbit and spider man. With 14 modules to cover, you can become an accomplished face painter.Your qualification will be recognized and can be checked for validity by all of your future clients too !Take a step in the right direction and get your Face Painting Academy Diploma today.￡30, was ￡299, use code: GUARD90Module 1 Your Introduction to Becoming a Face PainterModule 2 The Equipment and Materials You Will Need for Face Painting Module 3 Health & Safety and Risk AssessmentsModule 4 Starting / Running Your Own BusinessModule 5 Pricing and CostsModule 6 Marketing Your Business& Social MediaModule 7 The Do's and Don'ts and What to Do If Your Business Doesn't Go Well Module 8 How to do a Dog /Cat Face Paint DesignModule 9 How to do a Butterfly / Dolphin Face Paint DesignModule 10 How to do a Monkey / Frog Face Paint Design Module 11 How to do a Rabbit / Swan Face Paint Design Module 12 How to do a Tiger / Dinosaur Face Paint Design Module 13 How to do a Spiderman / Batman Face Paint DesignModule 14 How to do a Minnie Mouse /Princess Face Paint Design59. The course is intended mainly for those .A. keen on showing off new skillsB. eager to get an academy diplomaC. interested in learning face paintingD. equipped with a unique taste for art60. You can save￡if signing up for the course now.A. 30B.269C. 299D. 32961. Which of the following statements is TRUE about the course?A. It is presented both online and offline.B. It provides not only lessons on business.C. The diploma can be obtained in one day.D. Some clients will be invited to examine your qualification.62. In which module are you likely to learn how to advertise your business?A. Module 4.B. Module 6.C. Module 7.D. Module 10.(C)All across America, students are anxiously finishing their "What I Want To Be .." college application essays, advised to focus on STEM (Science, Technology, Engineering, and Mathematics) by experts and parents who insist that's the only way to become workforce ready. But two recent studies of workplace success contradict the traditional wisdom about "hard skills". Google originally set its hiring systems to sort for computer science students with top grades from top science universities. In 2013, Google decided to test its hiring theory by quickly dealing with large amounts hiring, firing, and promotion data collected since the company's establishment. Project Oxygen shocked everyone by concluding that, among the eight most important qualities of Google's top employees, STEM capability comes in dead last. The seven top characteristics of success at Google are all soft skills: being a good coach; communicating and listening well; possessing comprehension into others, being supportive of one's colleagues; being a good critical thinker and problem solver, and being able to make connections across complex ideas.Those characteristics sound more like what one gains as an English or theater major than as a programmer. Could it be that top Google employees were succeeding despite their technical training, not because of it? After bringing in more experts to dive even deeper into the data, the company enlarged its previous hiring practices to include humanities majors, artists, and even the MBAs (Master of Business Administration).Project Aristotle, a study released by Google this past spring, further supports the importance of soft skill seven in high-tech environments. Project Aristotle analyzes data on inventive andproductive teams. Google takes pride in its A-teams, assembled with top scientists, each with the most specialized knowledge and able to throw down one creative idea after another. Its data analysis revealed, however, that the company's most important and productive new ideas come from B-teams comprised of employees who don't always have to be the smartest people in the room.Project Aristotle shows that the best teams at Google exhibit a range of soft skills: equality, generosity, curiosity toward the ideas of your teammates, understanding, and emotional intelligence. And topping the list: emotional safety. To succeed, each and every team member must feel confident speaking up and making mistakes. They must know they are being heard.STEM skills are vital to the world we live in today, but technology alone, as Steve Jobs famously insisted, is not enough. We desperately need those who are educated to the human, cultural, and social as well as the computational.63. The underlined word: “contradict” most probably means ““.A. add toB. back upC. bring aboutD. conflict with64. Google conducted the studies of workplace success in order to .A. determine what makes a workplace-ready studentB. check whether its hiring system serves the purposeC. prove soft skills are more important than hard onesD. impress its competitors with the employees ‘ excellence65. What can be inferred from Project Aristotle?A. Emotional safety enables people to express themselves freely.B. Listening and hearing helps develop problem-solving abilities.C. Learning from mistakes doesn‘t necessarily m ean improvement.D. Those without specialized knowledge can also make inventions.66. Which of the following is the best title for the passage?A. STEM skills our society needs for better educationB. The principal focus students have on application essaysC. The surprising thing Google learned about its employeesD. The soft skills Google programmers lack for career growthSection CDirections: Read the following passage.Fill in each blank with a proper sentence given in the box. Each sentence can be used only once. Note that there are two more sentences than you need.Car washes have been automated for decades, but companies developing fully autonomous vehicles must rely on a human touch to keep their cars and trucks in working condition.67 For example, soap residue or water spots could effectively "blind" an autonomous car. A traditional car wash's heavy brushes could jar the vehicle's sensors, disrupting their calibration and accuracy. Even worse, sensors, which can cost over $100,000, could be broken.68 Dirt, dead bugs, bird droppings or water spots can impact the vehicle's ability to drive safely.Avis, which has years of experience managing large fleets of rental cars, has been tasked with cleaning and refueling the self-driving van fleet of Waymo, the self-driving arm of Google's parent company. Avis modified three of its branches in the Phoenix area to tend to the Chrysler Pacifica vans." There are special processes that definitely require a lot more care and focus, and you have to clean [the vans] quite often,"69 But other self-driving car companies such as Toyota, Aptiv, Drive.AI and Uber described to CNN that they use microfiber cloths along with rubbing alcohol, water or glass cleaner for manual cleanings.70 This should alleviate some need for manual cleaning. But because autonomous vehicles can have dozens of sensors, Seeva CEO Diane Lansinger doesn't imagine products like this will be able to clean every camera, radar or LIDAR, a laser sensor that most experts see as essential for self-driving vehicles.IV ．Summary WritingDirections: Read the following passage.Summarize the main idea and the main point(s) of the passagein no more than 60 words. Use your own words as far as possible.In one of my earliest memories, I‘m drawing. I don’t remember what the picture is supposed to be, but I remember the mistake. My marker slips, an unintentional line appears and my lip trembles. The picture has long since disappeared. But that feeling of deep frustration, even shame, stays with me.That‘s the thing about perfectionism, a crucially self-defeating way to move through the world. It makes you better at your career and relationships and life in general. Culturally, we often see perfectionism as a positive.But the disadvantage of perfectionism isn’t just that it holds you back from being your most successful, productive-self. Perfectionistic tendencies have been linked to a laundry list of clinical issues: depression and anxiety (even in children), self-harm, social anxiety disorder, eating disorders, and most damaging of all, suicide.“Based upon the 60-odd studies that we‘ve done,the higher the perfectionism is, the more psychological disorders you’re going to suffer,”says York St John University‘s Hill.“Factors often labeled ‘healthy’ perfectionism, like striving for excellence, aren‘t actually perfectionism at all. They’re just conscientiousness（尽职尽责）–which explains why people with those tendencies often have different outcomes in studies. Perfectionism isn ‘t defined by working hard or setting high goals. It‘s that critical inner voice.”Take the student who works hard and gets a poor mark. If she tells herself: “I’m disappointed,but it‘s o kay; I‘m still a good person overall,” th at‘s healthy. If the message is: “I‘m a failure. I‘m not good enough, “that‘s perfectionism.Perfectionists can make smooth sailing into a storm, a brief ill wind into a category-five hurricane. Eventually, the behaviours perfectionists adapt, actually, do make them more likely to fail.V. TranslationDirections: Translate the following sentences into English, using the words given in the brackets.72.他一直提醒自己不要对他人太苛刻。

【解析】重庆市巴蜀中学2018届高三高考适应性月考(二)理综生物试题含解析

巴蜀中学2018 届高考适应性月考卷(二)理科综合生物—、选择题1. 组成生物体的细胞既具有多样性，又具有统一性。

下列叙述正确的是A. 不同物种的细胞所含有机分子的结构都不同B. 同一个体的不同细胞，其元素含量都相同C. 原核细胞与真核细胞的区别之一在于DNA分子在细胞内的主要存在形式D. 细胞学说揭示了细胞的统一性和多样性，以及生物体结构的统一性【答案】C【解析】不同物种的细胞所含有机分子有糖类、脂质、蛋白质和核酸等，糖类和脂质在不同物种的细胞中结构没有特异性，各细胞都相同，A错误；同一个体的不同细胞，其元素种类大致相同，而含量相差较大，B错误；DNA在原核细胞中存在于拟核和质粒中，而DNA在真核细胞中主要存在于细胞核中的染色体上，故C正确；细胞学说并未揭示细胞的多样性，D错误。

考点：细胞中的元素和化合物、细胞学说的内容点睛：本题要求掌握有关细胞结构的相关基础知识，考查考生把握所学知识的要点和知识之间内在联系的能力。

2. 下列选项中，一般而言H/0比值前者高于后者的是A. 葡萄糖麦芽糖B. DNA RNAC. 自由水结合水D. 糖类脂质【答案】B【解析】葡萄糖的分子式为C6H12O6，麦芽糖的分子式为C12H23O11，由此可知葡萄糖的H/O比值低于麦芽糖，A错误；DNA比RNA的氧原子少，故DNA中H/O比值比RNA高，B正确；自由水和结合水都是H2O，自由水的H/O比值等于结合水的，C错误；一般而言，糖类的氧含量高于脂质，故糖类的H/O比值比脂质的低，D错误。

考点：细胞中化合物的元素组成点睛：解决该题的关键在于熟记各种化合物的元素组成及特点3. 下列有关生物体内化合物或结构的叙述，正确的是A. 盐析后的蛋白质分子，其空间结构变得伸展、松散B. 核酸的唯一功能是携带和传递遗传信息C. 多糖具有多样性的原因就在于葡萄糖数目不同D. 念珠藻细胞中含有藻蓝素和叶绿素，是自养生物【答案】D【解析】盐析后的蛋白质分子，其结构并没有变化，A错误；核酸包括DNA和RNA，如rRNA 是构成核糖体的成分，B错误；多糖有纤维素、淀粉和糖原，其结构不同的原因在于葡萄糖数目不同及其空间结构不同，C错误；念珠藻属于蓝藻，其细胞中含有藻蓝素和叶绿素，能进行光合作用，是自养生物，D正确。

2018学年度高三二模理综试卷及答案

2018学年度高三二模理综试卷及答案可能用到的相对原子质量H～1 O～16 C～12 N～14 S～32 F～19 Cl～35.5 Br～80 I～127 Si～28 Na～23 K～39 Ca～40 Mg～24 Al～27 Fe～56 Cu～64 Ag～108 Zn～65 Ba～137 Mn～55 Pb～207第Ⅰ卷(选择题共126分)一、选择题：本大题共13小题，每小题6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．用15N标记含有100个碱基对的DNA分子（其中有腺嘌呤60个），该DNA分子在含14N 的培养基中连续复制4次。

下列有关判断错误的是A．含有15N的DNA分子有两个B．含有14N的DNA分子占总数的7／8 C．第四次复制消耗胞嘧啶脱氧核苷酸320个D．复制结果共产生16个DNA分子2．现用洋葱根尖进行如下处理并完成实验：①用蒸馏水培养根尖，观察有丝分裂；②将正常培养的装置移到冰箱中再培养一段时间，观察并统计有丝分裂中期的染色体数；③在蒸馏水中加入一定量的甲苯培养根尖，观察有丝分裂中期染色体的形态和数目。

下列相关叙述合理的是A．实验①应先在低倍镜下找到长方形细胞，再换用高倍镜观察染色体变化特点B．实验②是为了验证“低温可能是形成多倍体的诱因”，其原理是低温抑制着丝点分裂C．实验③的目的可能是研究甲苯对染色体结构和数目的影响，但还需设计对照组D．实验③依据的原理之一是“致癌物质会使细胞发生基因突变，原癌基因被激活”3．在一个新鲜萝卜中挖一凹槽，在凹槽中放入浓盐水。

一段时间后，萝卜变软，凹槽中水分增多。

下列图示的实验与此实验原理差异最大的是（）4. 溶菌酶是存在于眼泪和白细胞中的酶，有杀菌功能，整个分子大致呈球形，故称为球蛋白(如图)。

下列关于溶菌酶的说法，不正确的有( )①溶菌酶从细胞进入泪液不穿过生物膜②溶菌酶是由两条多肽链共同组成的③双缩脲试剂5%NaOH溶液与1%CuSO4溶液等体积混合后与溶菌酶反应呈紫色④溶菌酶的空间结构与其功能密切关系⑤溶菌酶的杀菌作用属于特异性免疫A.一项B.两项C.三项D.四项5.如图表示处于平衡状态的某生物种群因某些外界环境变化导致种群中生物个体数量改变时的四种情形，下列有关叙述中，不正确．．．的是A．若图①所示为草原生态系统中某人工种群，则a点后的变化可能原因是过度放牧B．若图②所示为培养液中的酵母菌种群，则b点后的变化可能原因是天敌的减少C．图③中c点后发生的变化表明生态系统的自我调节能力有一定限度D．图④曲线可用于指导海洋捕捞，维持资源种群数量在K/2左右6．下图为人体内某些生理过程示意图解。

2018届青浦区高考英语二模试卷和参考答案及听力原文.doc

青浦区2017学年高三年级第二次学业质量调研测试英语学科试卷2018.04（时间120分钟，满分140分）考生注意：1．本试卷共13页。

满分140分。

考试时间120分钟。

2．答题前，考生务必在答题卡（纸）上用钢笔或水笔清楚填写姓名、准考证号，并用铅笔正确涂写准考证号。

3．答案必须全部涂写在答题卡（纸）上。

如用铅笔答题，或写在试卷上也一律不给分。

I. Listening ComprehensionSection ADirections: 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.1. A. To the bank. B. To the museum. C. To the clinic. D. To the restaurant.2. A. 10:07. B. 10:30. C. 11:00. D. 11:07.3. A. U se the woman’s phone. B. Pay for the phone call.C. Get some change from Jane.D. Go and look for a pay phone.4. A. He likes to wear clean clothes.B. He changes his job frequently.C. He is careless about his appearance.D. He is ashamed of his present condition.5. A. Whether he has time on Saturday.B. Whether he can get access to the concert.C. Whether the tickets will be too expensive.D. Whether the woman is available on Saturday.6. A. They are complaining. B. They are bargaining.C. They are negotiating.D. They are arguing.7. A. The window smells of fresh paint. B. The man will clean the air-conditioner.C. She’d like to have the window open.D. She prefers keeping the air-conditioner on.8. A. They are entertaining. B. They are time-consuming.C. They are boring.D. They are rewarding.9. A. She is fully engaged. B. She doesn’t like get-together.C. She has some paper to bury.D. She questions the man’s purpose.10. A. His notebook is missing.B. His handwriting is difficult to read.C. He wasn’t in class this morning either.D. He’s already lent his notes to someone else.Section BDirections: In Section B, you will hear two short passages and one longer conversation, and you will be asked several questions on each of the short passages and the longer conversation. The short passages and the longer conversation will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one is the best answer to the question you have heard.Questions 11 through 13 are based on the following passage.11. A. Photos of polar bears.B. Photos heavily edited.C. Photos of fine quality to appear in print.D. Photos reflecting sc ientists’ working life.12. A. A year’s personal subscription to Nature is another award to winners.B. Photos can be uploaded to Twitter with the tag ScientistAtWork.C. Five winning photos will be published in one issue of Nature.D. All entries have to be sent to photocompetition@.13. A. To encourage people to work with scientists.B. To attract people to participate in a contest.C. To increase the sales of the magazine.D. To teach people how to take pictures.Questions 14 through 16 are based on the following passage.14. A. Because no team invited him to play. B. Because he stopped training for a while.C. Because he appeared to be too tired.D. Because he suffered from a disease.15. A. 8.5. B. 19.2. C. 44. D. 53.16. A. Bosh’s great basketball career in NBA.B. The regular training Bosh took in NBA.C. Bosh’s repeated desire to return to NBA.D. The reason why Bosh couldn’t play at NBA.Questions 17 through 20 are based on the following conversation.17. A. He is training to fly a single engine plane. B. He hopes to be an international pilot.C. He is not sure about his future plan.D. He works for Air Canada.18. A. From his friends. B. From his father.C. From his training.D. From his university.19. A. By attending special schools.B. By getting a private license first.C. By getting into an airline company first.D. By passing a test for commercial license.20. A. His flying hours in total. B. His interview performance.C. His university degrees.D. His private pilot license.II. Grammar and VocabularySection 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, use one word that best fits each blank.The kindness of Kiwi Lotto (乐透彩票) winnersWe’ve all dreamed of winning the Lotto but what actually happens when our numbers come up? The answer is rather heart-warming. New research out today from Lotto New Zealand reveals three-quarters of Powerball winners think of sharing (21) ______ spending.One lucky Powerball winner from Tauranga was even thinking of others before his numbers came up. “A week before I won Lotto, I saw an ambulance (22) ______ side bore the name of the donor, and I thought, ‘if I ever win Lotto, that’s what I will do’ and then 10 days’ later I won,” he said. Having won $5.5 mill ion dollars, he’s now spending a part of the money on two ambulances for his local hospital. “They’re going to say ‘(23) ______ (donate) by a Lotto winner’ on the side.I hope that it will inspire others to pay it forward if they ever find (24) ______ in a fortunate position like I have.”“My life (25) ______ (save) many years ago by a St John ambulance and it’s a marvellous feeling repaying back that kindness.” It was this kind of behaviour (26) ______ led to Lotto NZ’s inspiring true story of a Kiwi man who won $15 million and, honouring a promise made long ago, went halves with his mate.2017 was the (27) ______ (lucky) year on record for Powerball winners. “Last year, there were 19 different Powerball winners — (28) ______ most of them had in common was the desire to share their good fortune.” said Emilia Mazur, General Manager Corporate Communications.“Another Tauranga man won $10 million with Powerball in July and once he got over theshock of winning, his first thought was his community and he has since shared some of his winnings to upgrade its facilities.”Group players are natural sharers —not only (29) ______ they share the winnings among themselves but they also then want to help out others.“Everyone is just so happy, it’s created an amazing sense of freedom.” said one of the group leaders Tina. “For me personally, (30) ______ (know) how much of a significant difference you have made to your family and your circle of friends, I have a feeling that I have never experienced. It’s an unquantifiable feeling —it’s magic.”Section BDirections: Fill in each blank with a proper word chosen from the box. Each word can be used only once. Note that there is one word more than you need.Swiss village bans tourists from taking photos because it’s too beautifulLocated in the Swiss Alps, near the famous resort of St. Moritz, the commune of Bergün is one of the most beautiful mountain villages in Europe. So beautiful, in fact, that photos of it shared on social media may make people feel depressed that they can’t visit, so local authorities banned tourists from taking photos.It may sound like a joke, but it’s actually a new law adopted by the Bergün village council and __31__ by its mayor. To discourage visitors from taking photos in Bergün, they plan to __32__ a symbolic €5 fine for those caught breaking the new rules.According to a statement by the Bergün tourism authority, “It is scientifically proven that beautiful holiday photos on social media make the viewers unhappy because they cannot be there themselves.”However, it seems unlikely that Bergün’s new law was really thought of as a way to __33__ Facebook or Instagram users of the depressing experience of seeing the beauty of the village. As the news went __34__ online, many assumed that it was actually a clever marketing __35__, a theory that was at least partially confirmed by the village’s director of tourism, Marc-Andrea Barandun.“In the background of course the idea is that everyone is talking about Bergün,” he told The Local. “So it’s a combination of both —we made the law and also there’s some marketing aim behind it.”To show that they were serious about the law, Bergün authorities __36__ photos of the village from its Facebook and Twitter accounts, and declared their intention to delete them from the Bergün website too.If the new law was just a __37__ strategy, it actually worked wonders, as people started __38__ more photos of the beautiful mountain village soon after news of the photo ban started making news headlines.A few days ago, Mayor Nicolay __39__ the debatable photo ban in a video, where he offeredpeople visiting Bergün a special __40__ to take photos. He reminded them to think twice before sharing the pictures online, though, as they could be making their friends depressed.III. Reading ComprehensionSection ADirections: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.A cliché is a phrase that has been used so many times that it comes out of the mouth or the computer without stirring up a wave in the mind of the speaker, the typist, the listener or the reader. The word was part of the technical term of the French printing trade in the 19th century, the name for a plate used in the printing process, and it is still used with that meaning in English and other languages. By the middle of the same century, the word was being used in French, shortly followed by English, as a simile (比喻) for __41__ used expressions.Clichés can be __42__ according to whether they were originally idioms, similes and proverbs, expressions from trades or __43__ phrases.Many idioms have been so universally overused that they have been __44__ — phrases like far and wide, by leaps and bounds or safe and sound. Our second category could be similes and proverbs that now fall off the __45__ with little meaning, similes like as cool as a cucumber, which __46__ around 400 years.A large category is __47__ from the terms of trades and professions, sports and games, and other national concerns. Many are __48__ clichés, as is fitting for the British, as an island nation, with examples like to leave a sinking ship, to know the ropes, to stick to one’s g uns.Our last broad category of cliché might be phrases which were __49__ when they were first coined, but have become ineffective through constant use. When a football manager, asked how he felt about the __50__ of his team, said that he was as sick as a parrot. Since then, it has been so overused that it has lost its __51__. To explore every avenue and to leave no stone unturned are two political clichés of this class. No politician with any sensitivity for language could use either of those phrases __52__, yet you hear them still, all the time.No doubt we could specify the classes of clichés into further subdivisions until the cows come home. But there is no need to. We all agree that clichés are to be __53__ by careful writers and speakers at all times, don’t we? Well, actually, no, not I. Life, and language, are so full of clichés that silence will hold the position if you __54__ us the use of cliché. So many millions of people have spoken and written clichés so __55__ that it is almost impossible to find ideas and phrases that have not been used many times before.41. A. occasionally B. frequently C. technically D. grammatically42. A. confirmed B. quoted C. inferred D. classified43. A. invented B. customized C. recognized D. underlined44. A. highlighted B. tailored C. weakened D. enriched45. A. nose B. eyes C. lips D. forehead46. A. dates back B. catches on C. takes shape D. gives out47. A. detected B. drawn C. excluded D. initiated48. A. remote B. temperate C. urban D. oceanic49. A. boring B. striking C. entertaining D. annoying50. A. expense B. punishment C. defeat D. age51. A. origin B. shine C. statue D. humour52. A. seriously B. fluently C. flexibly D. properly53. A. adjusted B. adapted C. adopted D. avoided54. A. deny B. allow C. forbid D. promise55. A. casually B. decently C. reluctantly D. ceaselesslySection BDirections: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.(A)Free to SoarOne windy spring day, I observed young people having fun using the wind to fly their kites. Multicolored creations of varying shapes and sizes filled the skies like beautiful birds racing and dancing. As the strong winds blew against the kites, a string kept them in check.Instead of blowing away with the wind, they arose against it to achieve great heights. They shook and pulled, but the string and the tail kept them attached, facing upward and against the wind. The kites struggled and kept being dragged behind, facing upward and against the wind. As the kites struggled and trembled against the string, they seemed to say, “Let me go! Let me go! I want to be free!” They soared beautifully even as they fought the restriction of the string. Finally, one of the kites succeeded in breaking loose. “Free at last,” it seemed to say. “Free to fly with the wind.”Yet freedom from restriction simply put it at the mercy of a cruel breeze. It flew ungracefully to the ground and landed in a messed mass of weeds and string against a dead bush. “Free at last”, free to lie powerless in the dirt, to be blown helplessly along the ground, and to stop lifeless against the first obstruction.How much like kites we sometimes are. The heaven gives us misfortune and limitations, rules to follow from which we can grow and gain strength. Restriction is a necessary companion to the winds of opposition. Some of us resist the rules so hard that we never soar to reach the heights we might have obtained. We keep part of the order and never rise high enough to get our tails off the ground.Let us each rise to the great heights, recognizing that some of the restrictions that we may be annoyed at are actually the steadying force that helps us improve and achieve.56. According to paragraph 2, “Let me go!” is said by _______.A. the kiteB. the windC. the birdD. the flyer57. Which of the following words has the meaning closest to the underlined word “obstruction” in paragraph 3?A. destructionB. miracleC. observationD. obstacle58. According to the writer, in what way is man similar to kites?A. Man will never reach the desirable height unless he breaks some rules.B. Man can be empowered by difficulties to go further and higher.C. Man desires freedom but only a few will succeed.D. Man is limited by his surroundings.59. By telling the story of kite flying, the writer wants to share the lesson that _______.A. rules are made to be brokenB. flying a kite involves skills and patienceC. sometimes difficulties can be potential blessingsD. it’s no use complaining about the difficulties we encounter(B)Products▼Tel:(855)776-7763 Get a Demo Login Sign Up Free ProProfsKnowledgebase Tour Pricing Solutions Integrations Blog Clients Examples Help Knowledge Base Software That Answers Questions Instantly Create help sites, knowledge bases, user guides, manuals, wikis & moreGet, share & apply knowledgeOffer awesome support 24/7With a self-service online knowledge base, customers can find instant answers to their questions. This means fewer support tickets and less workload for your agents. Likewise, support agents and employees can also find instant answers by searching the internal knowledge base for any questions about company best practices, policies and more.Access anytime, anywhereProProfs ensures your documentation fits all screen sizes and is accessible on multiples devices including iPhone, iPad, Android, iOs etc. There’s also no extra effort or coding at your end. All your documentation including tables, fonts (字体) sizes, large images and screenshots are automatically resized to fit smaller screens.Collaborate to create & share knowledgeBring different teams together to create and share knowledge on a central platform. Define roles and permissions for your team members to control who can access what in your knowledge base. Build private knowledge bases, accessible only to authorized users, with advanced authoring60. By searching the internal knowledge base, you can _______.A. access the Help CenterB. enjoy user-friendly manualsC. create a secure resource centerD. learn the company’s practices and policies61. The Knowledge Base Software can offer the following functions EXCEPT that _______.A. users of ProProfs can read the manuals on their mobile phonesB. the service that ProProfs offers to its users is accessible around the clockC. with ProProfs, users can readjust the size of their documentation to their screensD. different users can be grouped together on a central platform to share knowledge62. How does the software ProProfs tighten its security?A. By authorizing different users.B. By creating a central platform.C. By offering single sign-on settings.D. By building private knowledge bases.(C)The largest-ever study of the link between city walkability and high blood pressure has been held up as evidence of the “invisible value of urban design” in improving long-term health outcomes, say researchers.The study of around 430,000 people aged between 38 and 73 and living in 22 UK cities found significant associations between the increased walkability of a neighborhood, lower blood pressure and reduced risk of high blood pressure among its residents.The outcomes remained consistent even after adjustments for socio-demographic (社会人口统计学), lifestyle and changing physical environment factors, though the protective effects were particularly pronounced among participants aged between 50 and 60, women, and those residing in higher density and poor neighborhoods.The paper was published in the International Journal of Hygiene and Environmental Health this week. With high blood pressure a major risk factor for chronic (慢性的) and particularly heart diseases, researchers at the University of Hong Kong and Oxford University said the findings demonstrated the need to take notice of the health-influencing factor in urban design.“With the increasing pace of urbanization an d demographic shifts towards an ageing population, we become more likely to suffer from chronic diseases,” said Dr. Chinmoy Sarkar, an assistant professor at the Healthy High Density Cities Lab of the University of Hong Kong and lead author of the study. “The action taken to improve public health must consider the invisible value of urban planning and design.“We are spending billions of pounds in preventing and curing heart diseases — if we are ableto invest in creating healthy cities through small changes in the design of our neighborhoods to make them more activity-friendly and walkable, then probably, we will have significant savings in future healthcare expenses.”To measure a neighborhood’s activity-promoting potential, researchers developed a set of index of walkability containing relevant urban elements, including residential and retail (零售) density, public transport, street-level movement, and distance to attractive destinations.Poorly designed spaces generally reduced walking and physical activity, promoting the lifestyles of long time sitting down and not moving; and were harmful to social interactions, and as such associated with poorer mental and physical health.Because walkability was “based on the underlying design of the city”, said Sarkar, cities could be modified or designed to encourage it. “Such investments in healthy d esign are likely to bring in long-term gains as they are enduring and common.”63. By considering “invisible value of urban design”, people can _______.A. reduce the ageing populationB. slow down the pace of urbanizationC. promote activity-friendly and walkable citiesD. invest in preventing and curing heart diseases64. What can be inferred from the passage?A. A set of index is essential to ensure that urban design promotes walkability.B. Walkable cities can lower blood pressure and the risk of high blood pressure.C. Chronic diseases are becoming common due to people’s neglect of their health.D. Middle-aged women living in poor areas are less likely to benefit from increased walkability.65. All of the following are the undesirable consequences of poorly-designed neighborhoods EXCEPT _______.A. failing healthB. unhealthy lifestyleC. fewer social interactionsD. fewer neighborhoods66. According to Dr. Chinmoy Sarkar, _______.A. cities should encourage the residents to engage in social interactionsB. the design of our neighborhoods should meet people’s needs for retail densityC. money invested in creating healthy cities is money saved in future healthcare expensesD. chronic diseases will be common because of our lifestyle and the physical environmentSection CDirections: Read the following passage. Fill in each blank with a proper sentence given in the box. Each sentence can be used only once. Note that there are two more sentences than you need.Make traditional treasures come aliveThe Palace Museum Director Shan Jixiang delivered a cultural heritage speech on Feb 27 in Beijing, which was co-organized by the Beijing Diplomatic Service Bureau and Beijing Housing Service Corporation for Diplomatic Missions. 67On the theme The World of the Palace Museum and the Palace Museum of the World, the 64-year-old director shared his ideas about how to make traditional treasures come alive again. During the speech, which lasted two and a half hours, Shan touched on topics including upgrading museum infrastructure（基础设施）, restoring cultural sites, digitalizing online museums, setting up restoration hospitals, providing better visitor experiences and promoting the Palace Museum’s cultural items.“The abundant collection of cultural objects at the Palace Museum is the inspiration for the creative souvenirs and cultural items available,” Shan said. “68 ” Throughout 2017, the total sales of Palace Museum’s cultural items have been more than 1 billion yuan ($158million). Explaining the huge success of Palace Museum’s cultural souvenirs, Shan said: “The museum opened a shop on the e-commerce website Taobao in 2008, but sales remained neither high nor low for years, as more than 80 percent of the souvenirs sold in stores in the past were not related to our museum.” “T herefore, I wanted to change the situation. Now, souvenirs from the Palace Museum cover almost every aspect of life. After all, what matters to a museum is not how many visitors they have, but how close they are to people’s daily lives.”69 Around 200 “doctors” are employed to analyze, examine, detect flaws or damage in ancient objects and restore them using more than 100 pieces of specialized equipment, including 3-D printers and scanners. The restoration hospital covers 13,000 square meters and boasts the nation’s most advanced restoration workshops.John Aquilina, Malta’s ambassador to China said that Shan’s speech showed a totally different Palace Museum to foreign people. “China enjoys a lon g and profound culture and many of the national treasures have been preserved at the Palace Museum. It is no easy task to preserve them well. 70 ”IV. Summary WritingDirections: Read the following passage. Summarize the main idea and the main point(s) of the passage in no more than 60 words. Use your own words as far as possible.Ban the Bag!Standing in line at the grocery store last week, I watched the woman in front of me buy a tube of toothpaste. As the clerk placed her purchase in a plastic bag, I couldn’t help wondering howlong it would take for that bag to end up in the trash. Then I noticed the big purse the woman was carrying and wondered why she had needed a plastic bag at all.People have come to rely on plastic bags as everything from shopping bags for groceries to trash-can bags. Although plastic bags can be recycled, only about one percent of those used in the United States are. Instead, after helping people transport items from one place to another, most are thrown away. They end up in landfills, where it can take a plastic bag up to a thousand years to decay. Some bags end up elsewhere in the environment, sticking to trees and fences, blocking rivers and oceans, or floating along city sidewalks.Plastic bags harm the environment in several ways. First, they break down into particles that pollute our soil and water. Because most plastic bags are made of polyethylene, a product derived from crude oil (原油) or natural gas, they waste nonrenewable resources. Plastic bags can also harm animals. Scientists estimate that more than one million sea animals, including whales, seabirds, and turtles, die each year from intaking or becoming stranded in plastic.People all over the world are starting to recognize the problems associated with plastic bags. Countries such as China, South Africa, Switzerland, and Uganda are taking action and banning the bags. Other nations, including Italy and Ireland, have been trying to restrict the use of plastic bags by taxing them. In the United States more and more communities are ridding themselves of plastic bags. Now more and more people are also purchasing inexpensive, reusable bags and using them when they shop. If we all take this simple step, we can be a part of a “green” revolution.V. TranslationDirections: Translate the following sentences into English, using the words given in the brackets. 72．全世界掀起了一股学中文的潮流。

2018届高考模拟试卷二参考答案 .doc

2018届高考模拟试卷二参考答案一、填空题：本大题共14小题，每小题5分，共70分． 1. 0 . 由{}0,1A B ⋂=,可得21x =，所以，0x =2. 1. 法一：由()(1i)1i (1)(1)i z a a a =+-=++-，所以z =222(1)(1)2a a ++-=，所以21a =，即1a =±，所以20162016()()1ai i ==法二：由(1i)1i 2z a =+-=，所以212a +=，所以21a =，即1a =±，所以20162016()()1ai i ==.3. 45-. 因为tan 2=α，所以，22220162sin cos 2tan 4sin(2)sin 23sin cos 1tan 5παααααααα-=-=-=-=-++. 4. 600. 设高二女生人数为x 人，所以，0.192000x=，即380x =，所以，高三人数为 2000-650-370-380=600人。

5.()1,3-. 根据偶函数的性质，可得2323x x -<-<，从而可得13x -<<，从而不等式的解集为()1,3-.6. 6. 根据算法流程图， 2112(13)12(1333)6(31)201713k k k s --=++++==-≥-，所以6k =故输出结果为6. 7.34. 所有基本事件共12个：(2,1)--，(2,0)-，(2,1)-，(2,2)-，(1,1)--，(1,0)-，(1,1)-，(1,2)-， (0,1)-，(0,0)，(0,1)，(0,2). 其中，b a A B -∈的事件共有9个，分别为(2,1)--，(2,0)-，(1,1)--，(1,0)-，(1,1)-，(0,1)-，(0,0)，(0,1)，(0,2).所以，概率93()124P E ==. 8.1008. 显然数列{}n a 中通项0n a ≠，由1111n n n n n n a a a a a a --++-=-可得，1111n n n n n n n n a a a aa a a a -+-+⋅⋅=-- 两边取倒数可得：111111n n n n a a a a -+-=-,所以1n a ⎧⎫⎨⎬⎩⎭是等差数列,首项1112a =,公差d =11122-=，所以()1111222n nn a =+-=,即2n a n =，所以，由20172n a a =可得2222016n =⨯，所以1008n =. 9. 73π.()sin 2sin()3f x x x a x a π=-=+-，函数在区间[]0,2π上恰有三个零点x 1，x 2，x 3，则a =令sin()3x π+=，所以233x k πππ+=+或者233x k ππππ+=+-，所以2x k π=或者23x k ππ=+，所以10x =，23x π=，32x π=，即12373x x x π++=.10.22143x y +=.依题意知()21,0F ，设()11,M x y ，由椭圆的定义可得253MF =，由抛物线定义得21513MF x =+=，即123x =，将123x =代入抛物线方程得1y =，进而由2222231a b ⎛⎫ ⎪⎝⎭⎝⎭+=及221a b -=，解得224,3a b ==，故椭圆1C 的方程为22143x y +=.11.102m -≤<.法一：由题意得：当0m ≥时，函数2()222f x x mx =+-的对称轴02m -≤，且(0)1f =-，所以，此时()f x 在[]0,1上至多有一个零点，而()2f x mx =+在()1,+∞没有零点.所以，0m ≥不符合题意．当0m <时，函数2()221f x x mx =+-的对称轴02m->，且(0)1f =-，所以，此时()f x 在[]0,1 上至多有一个零点，而()2f x mx =+在()1,+∞至多有一个零点，若()f x 在[)0,+∞有且只有2个零点，则要求012221020m m m ⎧<-≤⎪⎪+-≥⎨⎪+>⎪⎩，解之可得102m -≤<.综上：102m -≤<法二：由题意得：x ＝0不是函数f (x )的零点.当0＜x ≤1时，由f (x )＝0，得12m x x=-，此时函数12y x x =-在(]0,1上单调递减，从而1122y x x =-≥-，所以，当m ≥－12时，f (x )在(]0,1上有且只有一个零点，当x ＞1时，由f (x )＝0，得2m x =-，此时函数2y x=-在()1,+∞上单调递增，从而()22,0y x=-∈-，所以，当－2＜m ＜0时，f (x )在()1,+∞上有且只有一个零点，若()f x 在[)0,+∞有且只有2个零点，则要求1220m m ⎧≥-⎪⎨⎪-<<⎩，解之可得102m -≤<.综上，102m -≤<.12.32.令2,2(0,0)x y m x y n m n +=+=>>，则问题转化为6,m n +≤求41m n+的最小值，而41()()9m n m n ++≥，即41932m n m n +≥≥+故知最小值为32.13.5.以AB 所在直线为x 轴，过点A 作垂直于直线AB 所在的直线为y 轴，建立如图所示的直角坐标系.设BM CN BCCD==λ（0≤λ≤1），所以，BM λ=，2CN λ=，所以，(2)2M λ+，)23,225(λ-N ，所以，2535444AM AN λλλλ⋅=-+-+2225(1)6λλλ=--+=-++，因为[01]λ∈，，所以，[25]AM AN ⋅∈，，所以AM AN ⋅的取值范围是]52[，，即最大值为5.14.1a ≥.仅考虑函数()f x在0x >时的情况，可知3312()12x x x f x x x x ⎧-<⎪=⎨-⎪⎩，，≥函数()f x 在2x =时，取得极大值16．令31216x x -=，解得，4x =．作出函数的图象（如右图所示）．函数()f x 的定义域为[0,]m ，值域为2[0]am ，，分为以下情况考虑：（1）当02m <<时，函数的值域为2[0(12)]m m -，，有22(12)m m am -=，所以12a m m =-，因为02m <<，所以4a >；（2）当24m ≤≤时，函数的值域为[016]，，有216am =，所以216a m=，因为24m ≤≤，所以14a ≤≤；（3）当4m >时，函数的值域为2[0(12)]m m -，，有22(12)m m am -=，所以12a m m =-，因为4m >，所以1a >；综上所述，实数a 的取值范围是1a ≥．二、解答题15．（11sin()62C π-=，因为()0,C π∠∈，所以5,666C πππ⎛⎫-∈- ⎪⎝⎭，所以66C ππ-=或56π,即3C π=或π（舍去）.（2）因为2sin cR C=，所以24R =, 要使三角形周长最大，即要求a b +最大.所以，2(sin sin )4(sin sin())3a b R A B A A π+=+=++14(sin sin ))26A A A A π=+=+因为20,3A π⎛⎫∈ ⎪⎝⎭，所以，当3A π=时，a b +有最大值.此时，ABC∆为等边三角形，c =所以12ABCS=⨯=.16．（1）连AC交BD于O，连CO；因为AB∥CD，2AB DC=，所以2AO CO=，又因为2EM CM=，所以，AE∥MO，又因为AE⊄面BDM，MO⊂面BDM，所以AE∥面BDM.（2）设1DC=，因为DC⊥BC，1BC=，所以BD，在梯形ABCD中，//AB CD，所以45ABD BDC︒∠=∠=，因为2AB DC=，所以在ABD∆中，由余弦定理知AD因为AB=2，所以AD2+BD2=AB2，所以∠ADB=90°，所以，AD⊥BD，因为平面ADEF⊥平面ABCD，BD⊥AD，平面ADEF∩平面ABCD=AD，BD⊂面ABCD 所以BD⊥平面ADEF，因为BD⊂平面BDM，所以平面BDM⊥平面ADEF.17.（1）过O作直线OE AB⊥于E，则10,OE=设,EOAα∠=则3,(),442EOBπππαα∠=-<<故310tan,10tan(),4AE BEπαα==-3sin()3sin410tan tan()10()34cos cos()4ABπαπαααπαα-=+-=+-310sin4,3cos cos()4ππαα=⋅-又31cos cos()cos()sin(2)424ππαααααα⋅-=⋅+=-，由42ππα<<，得32(,),444πππα-∈故max32cos cos()44παα⋅-=，当且仅当32,428πππαα-==时取等号．此时，AB有最小值为1)．即两出入口之间距离的最小值为1) .（2）由题意可知直线AB是以O为圆心，10为半径的圆O的切线，根据题意，直线AB与圆C要相离，其临界位置为直线AB与圆C相切，设切点为F此时直线AB 为圆C 与圆O 的公切线．因为，出入口A 在古建筑群和市中心O 之间，如图，以O 为坐标原点，以CO 所在的直线为x 轴，建立平面直角坐标系xoy 由CF=5，OE=10，因为圆O 的方程为22100x y +=，圆C 的方程为22(30)25x y ++=，设直线AB 的方程为(0)y kx t k =+>，则10,(1)5,(2)==，所以，（1）/（2）得230t k t =-+，所以20t k =或60t k =，所以此时(20,0)A -或(60,0)A -（舍去），此时20OA =，又由（1）知当//AB ON时，OA =综上，(60,).OA ∈+∞即设计出入口A 离市中心O的距离在到20km 之间时，才能使高架道路及其延伸段不经过保护区. 18．（1）设点P (x ，y )，x 2 + y 2 = 4，P A = (x - a)2 + (y - 2)2，PB = (x - m)2 + (y - 1)2，因为PAPB= k ，所以(x –a )2 + (y –2)2 = k 2[(x –m )2 + (y –1)2]，又x 2 + y 2 = 4，化简得2ax + 4y – a 2 – 8 = k 2(2mx + 2y – m 2 – 5)，因为P 为圆O 上任意一点，所以⎩⎨⎧2a = 2mk24 = 2k2a2 + 8 = k2(m2 + 5)，又m > 0，k > 0，解得⎩⎨⎧k = 2a = 2m = 1，所以常数k = 2．（2）法一：设M (x 0，y 0)，M 是线段NE 的中点，N (2x 0 – 2，2y 0 – t )，又MN 在圆C 上，即关于x ，y 的方程组⎩⎨⎧x02 + y02 = 1(2x0 -2)2 + (2y0 - t)2 = 1有解，化简得⎩⎨⎧x02 + y02 = 18x0 + 4t y0 - t2 - 7 = 0有解，即直线n ：8x + 4t y –t 2– 7 = 0与圆C ：x 2 + y 2 = 1有交点，则d o -n =|t2 + 7|64 + 16t2≤1，化简得：t 4 – 2t 2 – 15 ≤0，解得t ∈[5，5]．法二：设过E 的切线与圆C 交于切点F ，EF 2 = EM ·EN ，又M 是线段NE 的中点，所以EN = 2MN ，EM = MN ，所以EF 2 = 2MN 2，又EF 2 = EO 2 – OF 2 = 22 + t 2 – 1 = t 2 + 3，所以MN ≤ 2，t 2 + 3 ≤ 8，所以t ∈[-5，5]．19．(1)由已知，得f '(x )1221x a x=---+，据题意，f '(1) = 0，得到1a =-.所以2()ln f x x x x =-++， f '(x )(21)(1)121x x x xx+-+=-++=.由0x >，令f '(x )0>，得01x <<，令f '(x )0<，得1x >，所以函数()f x 在1x =处取得极值，所以1a =-， ()f x 的单调增区间为(0)，1，()f x 的单调减区间为(1+)，∞.(2)257()()ln 22x x g x f x b x x b =-+=-++-，(0,2016)x ∈.则g '(x ) 7122x x =-++，令g '(x )0=，得2x =，负舍.当02x <<时，g '(x )0>，g (x )在(02)，上递增，当22016x <<时，g '(x )0<，g (x )在(22016)，上递减，所以函数5()()2g x f x b x =-+在区间(0,2016)上只有一个零点，等价于(2)0g =，解得ln23b =+. (3) 由条件可得2ln ()x kh x x x x=-- 因为12()()0h x h x ==，所以2211222ln 2ln x x x x -=-令2()2ln x x x ϕ=-，所以222(1)()2x x x x x-'ϕ=-=当01x <<时，()0x 'ϕ>，当1x >时，()0x 'ϕ<，所以()x ϕ在()0,1上递增，在()1,+∞上递减，所以()x ϕ在1x =处有极大值，所以1201x x <<< 令()()()2s x x x =--ϕϕ，()0,1x ∈， ()()242440222s x x x x x '=->-=-+-⎛⎫⎪⎝⎭()s x 在()0,1上单调递增，()()10s x s <=有()()21x x =ϕϕ()12x <-ϕ，因为，()x ϕ在()1,+∞上递减，且211,21x x >->所以211222x x x x >-⇒+>． 20．（1）①因为211112a a a a =+∆=-，322114a a a a =+∆=-，且{}n a 为等比数列. 所以2213a a a =⋅，即211111()()24a a a -=-，解得113a =.当113a =时，当2n ≥时，1n n a a -=∆+……111111()1()11122()13321()2n n a a --⎡⎤---⎢⎥⎣⎦+∆+=+=⋅---. 1n =适合上式，所以{}n a 为等比数列，即113a =.②因为n m a a -=1n a -∆+……m a +∆11()1()21122[()()]13221()2m n m n m -⎡⎤---⎢⎥⎣⎦==⋅-----所以||n m a a -=211|()()|322n m ⋅---211[()()]322n m ≤⋅+41()32m ≤⋅，令41()32m t ⋅≤，则24log 3m t ≥，故可取k 不小于24log 3t的正整数，则对任意,,n m k n N m N **>≥∈∈，||n m a a -41()32m t ≤⋅≤.（2）因为n a ∆=21n a -∆+ (12)1113(13)2(1)13n a a n a --+∆+∆=--+∆-131222n n a =-++∆231222n n a =-+-. 由23-20n n a ∆=>知 {}n a ∆递增，所以4n a a ≥对n N *∈恒成立当且仅当满足23234300a a a a a a ∆=-≤⎧⎨∆=-≥⎩，即22070a a ≤⎧⎨+≥⎩，解得2-70a ≤≤. 所以2a 的取值范围是[7,0].-2018届高考模拟试卷一参考答案一、填空题（本大题共14小题，每小题5分，共70分．请把答案填写在答卷规定的横线上）1.22.四3.284.35.8π 6.a ＞2 7.6π 8.54 9.6π10.3π11.448 12.2 13.24 14.()5333，二、解答题（本大题共6小题，共90分．解答时应写出文字说明、证明过程或演算步骤） 15．（本小题满分14分）如图，在几何体中，四边形ABCD 为菱形，对角线AC 与BD 的交点为O ，四边形DCEF为梯形，EF ∥CD ，FB FD =．（1）若2CD EF =，求证：OE ∥平面ADF ；（2）求证：平面ACF ⊥平面ABCD ．【解析】（Ⅰ）证明：取AD 的中点G ，连接OG 、FG ，因为O 为对角线AC 与BD 的交点，则O 为AC 中点，所以OG ∥CD ，且12OGCD =．又因为EF ∥CD ，且2CD EF =，所以OG ∥EF ，OG EF =，则四边形OGFE 为平行四边形，----------3分所以OE ∥FG ．又因为FG ⊂平面ADF ，OE ⊄平面ADF ，OE ∥FG ，所以OE ∥平面ADF ；-------------------------------------------------------------------6分（Ⅱ）证明：因为四边形ABCD 为菱形，所以OC BD ⊥，--------------------------7分又因为FB FD =，O 是BD 的中点，所以OF BD ⊥，------------------8分又有OFOC O OF =⊂，平面ACF ，OC ⊂平面ACF ,所以BD ⊥平面ACF ，----------------------------------------------12分又因为BD ⊂平面ABCD ，所以平面ACF⊥平面ABCD ．----------------------------------------14分16．（本小题满分14分）已知函数()2sin()cos 6f x x x π=-.（1）求函数()f x 的最大值和最小正周期；（2）设ABC ∆的角A B C ，，的对边分别为a b c ，，，且c =，1()2f C =，若sin 2sin B A =，求边a ，b 的值.【解析】（Ⅰ）因为)2()2sin()cos 612cos cos 2cos cos 1cos 2221sin(2)62f x x xx x x x x x x x x ππ=-=-=-+=-=---------------------------------------------------------------------4分当且仅当,3x k k Z ππ=+∈时，max 1()2f x =--------------------------------------6分最小正周期分别为和22T ππ==.------------------------------------------------7分（Ⅱ）因为11()sin(2)622f C C π=--=，即sin(2)16C π-=，因为0C π<<，所以 112666C πππ-<-<，于是262C ππ-=，即3C π=.------------------------------10分因为sin 2sin B A =，由正弦定理得2b a =，-------------------------------------12分由余弦定理得2222cos3c a b ab π=+-，即2212a b ab +-=，联立22212b aa b ab =⎧⎨+-=⎩，解得24a b =⎧⎨=⎩.-------------------------------------------14分17．（本小题满分14分）在平面直角坐标系xOy 中，椭圆C ：22221(0)x y a b a b+=>>，且点在椭圆C 上.（1）求椭圆C 的方程；-（2）设P 为椭圆上第一象限内的点，点P 关于原点O 的对称点为A ，点P 关于x 轴的对称点为Q ，设PD PQ λ=，直线AD 与椭圆C 的另一个交点为B ，若PA ⊥PB ，求实数λ的值.【解析】17.解:(1)因为点222，在椭圆C 上，则222112a b+=，------------------------------1分又椭圆C 的离心率为32，可得32ca，即32ca ，所以2222223124bacaa a ，代入上式，可得22221a a +=，解得24a ，故22114ba .所以椭圆C 的方程为2214x y += ...............................................................................................5分(2)设P (x 0，y 0)，则A (-x 0，-y 0)，Q (x 0，-y 0). 因为=λ,则(0，y D -y 0)=λ(0，-2y 0)，故y D =(1-2λ)y 0.所以点D 的坐标为(x 0，(1-2λ)y 0). ..................................................................................................7分设B (x 1，y 1)，221222*********210101010114414PB BAx x y y y y y y k k x x x x x x x x...............................9分又0000121BA ADy y y k k x x x故001441PBBAx k k y .----------------------------------------------------------------------11分又PA ⊥PB ，且0PAx k y ， D QBPxAOy第17题所以1PB PA k k ，即0000141x y x y ，解得34. 所以34....................................................................................................................................14分 18．（本小题满分16分）一块圆柱形木料的底面半径为12cm ，高为32cm ，要将这块木料加工成一只毛笔筒，在木料一端正中间掏去一个小圆柱，使小圆柱与原木料同轴，并且掏取的圆柱体积是原木料体积的三分之一，设小圆柱底面半径为r cm ，高为h cm ，要求笔筒底面的厚度超过2cm ．（1）求r 与h 的关系，并指出r 的取值范围；（2）笔筒成形后进行后续加工，要求笔筒上底圆环面、桶内侧面、外表侧面都喷上油漆，其中上底圆环面、外表侧面喷漆费用均为a （元/ cm 2），桶内侧面喷漆费用为2a （元/cm 2），而桶内底面铺贴金属薄片，其费用是7a （元/ cm 2）（其中a 为正常数）． ①将笔筒的后续加工费用y （元）表示为r 的函数；②求出当r 取何值时，能使笔筒的后续加工费用y 最小，并求出y 的最小值．【解析】（Ⅰ）据题意，221(1232)3r h ππ=⋅⋅，所以23248h r ⨯=，----------------------3分因为322h ->，所以30h <即2324830r ⨯<，解得r >----------------------------------------------------------5分又012r <<，所以125r <<；----------------------------------------------------------6分（Ⅱ）①据题意，笔筒的后续加工费用22272(2)(1221232)y a r a rh a r πππππ=++⋅-⋅+⋅⋅，整理得2226412763248641276y a r a rh a a r a r a rππππππ=++⨯⨯=+⋅+⨯ 232326(152)a r rπ⨯=++，定义域为；----------------------11分 ②由①知，33/22323286(2)12r y a r a r rππ⨯-=-=⋅，令/0y =得8(,12)5r =∈，由表知，当8r =时，y 取极小值即最小值2064a π.------------------------15分答：当8r cm =时，能使笔筒的后续加工费用y 最小，最小值为2064a π元．----16分19．（本小题满分16分）已知数列{}n a 中，首项11a =，2a a =，12()n n n a k a a ++=+对任意正整数n 都成立，数列{}n a 的前n 项和为n S ．（1）若12k =，且18171S =，求实数a 的值；（2）是否存在实数k ，使数列{}n a 是公比不为1的等比数列，且任意相邻三项n a ，1n a +，2n a +按某顺序排列后成等差数列．若存在，求出所有的k 的值；若不存在，请说明理由；（3）若12k =-，求n S （用a ，n 表示）．【解析】（Ⅰ）当12k =时，由12()n n n a k a a ++=+得121()2n n n a a a ++=+，即211n n n n a a a a +++-=-，所以数列{}n a 为等差数列，--------------------1分公差为211d a a a =-=-，数列{}n a 的前n 项和为(1)(1)2n n n S n a -=+⋅-，由18171S =得18(181)17118(1)2a -=+⋅-，解得2a =；---------------------------------------------------------3分（Ⅱ）设数列{}n a 为等比数列，则其公比为21a q a a ==，1n n a a -=，1n n a a +=，12n n a a ++=． 1︒若1n a +为等差中项，则122n n n a a a ++=+即112n n n a a a -+=+，解得1a =，与已知不符，舍去； 2︒若n a 为等差中项，则122n n n a a a ++=+即112n n n a a a -+=+，即220a a +-=，解得2a =-或1a =（舍），此时由12()n n n a k a a ++=+得11()n n n a k aa -+=+即2(1)a k a =+，故2215a k a ==-+；3︒ 若2n a +为等差中项，则212n n n a a a ++=+即112n n n a a a +-=+，即2210a a --=，解得12a =-或1a =（舍），仿2︒得2215a k a ==-+．---------------------------------------------------8分综上，满足要求的实数k 有且仅有一个，25k =-；---------------------------------9分（Ⅲ）当12k =-时，121()2n n n a a a ++=-+，所以211()n n n n a a a a ++++=-+，于是32n n a a +++=211()n n n n a a a a +++-+=+．----------------------------------------11分1︒ 当n 为偶数时，123456112(1)()()()()()22n n n n n a S a a a a a a a a a a -+=++++++++=+=； ---------------------------------------------------------------------------------13分2︒ 当n 为奇数时，1234511231()()()()2n n n n S a a a a a a a a a a --=+++++++=++ 11211[()]1(1)22n n a a a a --=+⋅-+=-+（2n ≥），当1n =时，也适合该式，所以11(1),2(1),2n n a n S n a n -⎧-+⎪⎪=⎨+⎪⎪⎩为奇数为偶数．-----------------------------------------------16分20．（本小题满分16分）已知函数1()ln f x a x x=+（0a ≠）. （1）求函数()f x 的单调区间；（2）若存在两条直线1y ax b =+，2y ax b =+（12b b ≠）都是曲线()y f x =的切线，求实数a 的取值范围；（3）若{}|()0(0,1)x f x ⊆≤，求实数a 的取值范围.【解析】（Ⅰ）/2211()a ax f x x x x-=-=（0x >）. 当0a <时，/()0f x <，()f x 的递减区间为(0,)+∞；----------------------------1分当0a >时，由/()0f x =得1x a=，列表得：所以，函数()f x 的递减区间为1(0,)a ，递增区间为1(,)a+∞；-----------------------4分（Ⅱ）因为存在两条直线1y ax b =+、2y ax b =+（12b b ≠）都是曲线()y f x =的切线，所以/()f x a =至少有两个不等的正根，-----------------------------------------------5分令/21()ax f x a x-==，得210ax ax -+=，记其两个根为1x 、2x （12x x <），则2124010a a x x a ⎧∆=->⎪⎨=>⎪⎩，解得4a >，------------------------------------------------------------------------------------7分而当4a >时，曲线()y f x =在点11(,())x f x 、22(,())x f x 处的切线分别为11()y ax f x ax =+-、22()y ax f x ax =+-，设()()F x f x ax =-（0x >），由2//1222()()1()()a x x x x ax ax F x f x a x x----+-=-==知，当12x x x <<时，/()0F x >即()F x 在区间12[,]x x 上是单调函数，因此12()()F x F x ≠，所以11()y ax f x ax =+-、22()y ax f x ax =+-不重合，即1y ax b =+、2y ax b =+（12b b ≠）是曲线()y f x =的两条不同的切线，故4a >；----------------10分（Ⅲ）当0a <时，函数()f x 是(0,)+∞内的减函数，因为11111()ln()10aaaaf ea e e e---=+=-<，而1(0,1)ae-∉，不符合题意；----------------------------------------------------------12分当0a >时，由（Ⅰ）知()f x 的最小值为1()ln (1ln )f a a a a a a=-+=-.1︒若1()0f a>即0a e <<时，{}|()0(0,1)x f x φ≤=⊆，所以0a e <<符合题意；2︒若1()0f a =即a e =时，{}1|()0(0,1)x f x e ⎧⎫≤=⊆⎨⎬⎩⎭，所以a e =符合题意；3︒若1()0f a <即a e >时，101a <<，而(1)10f =>，函数()f x 在1(,)a+∞内递增，所以当1x ≥时，()0f x >，又因为()f x 的定义域为(0,)+∞，所以{}|()0(0,1)x f x ≤⊆，符合题意.综上，实数a 的取值范围为(0,)+∞.----------------------------------------------16分。

XXX2018届高三第二次模拟考试理科综合试题Word版

XXX2018届高三第二次模拟考试理科综合试题Word版2018年XXX第二次高考模拟考试理科综合能力测试，考试时间为150分钟。

本试卷分为第I卷（选择题）和第Ⅱ卷（非选择题）两部分，其中第Ⅱ卷第33～38题为选考题，其它题为必考题。

考生作答时，应将答案填写在答题卡上，本试卷上的答题无效。

考试结束后，应将本试卷和答题卡一并交回。

注意事项：1.答题前，考生应在答题卡上填写自己的姓名和准考证号，并认真核对条形码上的姓名和准考证号，将条形码粘贴在答题卡的指定位置上。

2.选择题应使用2B铅笔填涂，如需改动，应用橡皮擦干净后再重新填涂；非选择题应使用0.5毫米的黑色中性（签字）笔或碳素笔书写，字体工整、笔迹清楚。

3.请按照题号在各题的答题区域（黑色线框）内作答，超出答题区域书写的答案无效。

4.保持卡面清洁，不折叠，不破损。

5.做选考题时，考生应按照题目要求作答，并用2B铅笔在答题卡上把所选题目对应的题号涂黑。

可能用到的原子量：H-1、C-12、N-14、O-16、Na-23、Al-27、Si-28、P-31、S-32、Cl-35.5、K-39、Ca-40、Fe-56、Cu-64一、选择题（共13小题，每小题6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.细胞相关的叙述，正确的是A.蓝藻细胞的能量来源于其线粒体有氧呼吸过程B.细胞膜两侧的离子浓度差是通过协助扩散实现的C.分泌蛋白先经过高尔基体再经过内质网分泌到细胞外D.细胞凋亡是由基因决定的细胞自动结束生命的过程2.生物实验的叙述，正确的是A.用洋葱鳞片叶内表皮为材料也能观察到质壁分离B.检测氨基酸的含量可用双缩脲试剂进行显色C.观察细胞有丝分裂要用8%的盐酸处理实验材料D.苏丹Ⅲ染色后的花生子叶细胞中可观察到红色颗粒3.某生物体（2n=4）正常的细胞分裂示意图，有关叙述正确的是A.该细胞处于减数第二次分裂后期B.若染色体①有基因A，则④一定有基因aC.若图中的②表示X染色体，则③表示Y染色体D.该细胞产生的子细胞中无同源染色体4.人体内环境与稳态的叙述，错误的是A.发生局部炎症反应时的肿胀，是由于组织中的Na浓度增加所致B.肾小管细胞能够选择性表达抗利尿激素受体基因5.葡萄糖转运载体（GLUT）有多个成员，其中对胰岛素敏感的是GLUT4.GLUT1～3分布于全身所有组织细胞，它们的生理功能不受胰岛素的影响，其生理意义在于维持细胞对葡萄糖的基础转运量。

2018年上海市虹口区高考英语二模试卷

2018年上海市虹口区高考英语二模试卷2018年上海市虹口区高考英语二模试卷II. Grammar and VocabularySection 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， use one word that best fits each blank.1.阅读下面短文，在空白处填入1个适当的单词或括号内单词的正确形式。

Nook’s arrival, Good or Bad?Booklovers, most of them, will tell you（1）______ a pleasure it is to lend a favorite read toa friend—the novel you stayed up all night to get to the end of; the travel book that madeyou feel（2）______ you yourself were on a train ride through India. For a while it seemed that e-book users were to be denied this pleasure of lending to friends. You could buy a book or magazine for your r eading device, but you couldn’t lend it out.But now, with the Nook, the US book chain Barnes and Noble’s response to Amazon’s Kindle, electronic read ers will be able to have their latest literary enthusiasm（3）______（press）on their friends, just like readers of physical books can. You simply email the book from your Nook and your friend can read it for two weeks,（4）______（use）any device with the Barnes & amp; Noble e-book reader software. It’s a big improvement from previous e-book readers.The Nook offers other features too. You read in black andwhite on the main screen, just like with Kindle. The difference is （5）______ on the lower part of the device there’s a color touch screen,（6）______ allows you to browse through a book or magazine, but goes black whe n you’re not using it so that you save power.（7）______ exciting thing about the Nook is that it offers Wi-Fi, arguably a big advance on previous e-book readers. Customers in the United States can use the Internet connection （8）______（read）whole e-books at Barnes and Noble’s hundreds of bookstores for free. None of Barnes and Noble’s competitors can come close to this.But the Nook, ironically,（9）______（turn out）to be a money-loser for Barnes and Noble, or at least a job-loser for Barnes and Noble’s employee s. According to Marian Maneker atThe Big Money Website,（10）______ the Nook is successful it might take sales from the company’s bookstores, eventua lly forcing their closure and the loss of thousands of jobs.【答案】（1）what,（2）as if/ though,（3）pressed,（4）using,（5）that,（6）which,（7）Another,（8）to read,（9）has turned out/turns out,（10）if【考点】固定用法高三说明文语法填空句子基础知识动词时态代词表语从句非限制性定语从句状语从句过去分词动名词从属连词【解析】本文属于说明文阅读，主要介绍了一种新的电子书阅读器NooK。

2018届江苏高考数学模拟试题(2)数学之友

2018届江苏高考数学模拟试题（2）数学I 注意事项考生在答题前请认真阅读本注意事项及答题要求1．本试卷共4页，包含填空题（第1题～第14题）、解答题（第15题～第20题）．本卷满分为160分，考试时间为120分钟．考试结束后，请将答题卡交回．2．答题前，请您务必将自己的姓名、准考证号用0.5毫米黑色墨水的签字笔填写在试卷及答题卡的规定位置．3．作答试题，必须用0.5毫米黑色墨水的签字笔在答题卡上指定位置作答，在其它位置作答一律无效．4．如需作图，须用2B 铅笔绘、写清楚，线条、符号等须加黑、加粗．参考公式：球体的体积公式：V ＝334R π，其中为球体的半径．一、填空题（本大题共14小题，每小题5分，计70分.不需写出解答过程，请把答案写在答题纸的指定位置上）1．已知集合},02/{2R x x x x M ∈=+=，},02/{2R x x x x N ∈≤-=，则=N M ▲．2．已知复数z 满足＝i ，其中i 为虚数单位，则复数z 的虚部为▲．3．某校共有400名学生参加了一次数学竞赛，竞赛成绩的频率分布直方图如图所示．成绩分组为[50，60)，[60，70)，…，[90，100]，则在本次竞赛中，得分不低于80分的人数为▲．4．在标号为0，1，2，4的四张卡片中随机抽取两张卡片，则这两张卡片上的标号之和为奇数的概率是▲．5．运行如图所示的流程图，则输出的结果S 是▲．6．已知等差数列{a n }的前n 项和为S n ．若S 15＝30，a 7＝1，则S 10的值为．7．已知()y f x =是R 上的奇函数，且0x >时，()1f x =，则不等式2()(0)f x x f -<的解集为▲．8．在直角坐标系xOy 中，双曲线x 2－＝1的左准线为l ，则以l 为准线的抛物线的标准方程是▲．9．四面体ABCD 中，AB ⊥平面BCD ，CD ⊥平面ABC ，且1c mA BB C C D ===，则四面体ABCD 的外接球的表面积为▲2cm .10.已知0πy x <<<，且tan tan 2x y =，1sin sin 3x y =，则x y -=▲. 11．在平面直角坐标系xOy 中，若直线l ：20x y +=与圆C ：22()()5x a y b -+-=相切，且圆心C 在直线l 的上方，则ab 的最大值为▲．(第3题)12．正五边形ABCDE的边长为AE AC ⋅的值为▲．13．设0a ≠，e 是自然对数的底数，函数2,0,(),0x ae x x f x x ax a x ⎧-≤⎪=⎨-+>⎪⎩有零点，且所有零点的和不大于6，则a 的取值范围为▲． 14．若对任意实数x 和任意θ∈[0，]，恒有(x +2sin θcos θ)2+(x +a sin θ+a cos θ)2≥，则实数a 的取值范围是▲．二、解答题（本大题共6小题，计90分.解答应写出必要的文字说明，证明过程或演算步骤，请把答案写在答题卡的指定区域内） 15．(本小题满分14分)如图，在直角坐标系xOy 中，角α的顶点是原点，始边与x 轴正半轴重合，终边交单位圆于点A ，且(,)62ππα∈.将角α的终边按逆时针方向旋转3π，交单位圆于点B ，记A (x 1，y 1)，B (x 2，y 2). （1）若113x =，求2x ；（2）分别过A ，B 作x 轴的垂线，垂足依次为C ，D ，记△AOC 的面积为S 1，△BOD 的面积为S 2，若122S S =，求角α的值. .16.（本小题满分14分）如图，在直三棱柱ABC -A 1B 1C 1中，AC ⊥BC ，BC =BB 1，D 为AB 的中点.（1）求证：BC 1∥平面A 1CD ；（2）求证：BC 1⊥平面AB 1C . 17．(本小题满分14分)某生物探测器在水中逆流行进时，所消耗的能量为n E cv T =，其中v 为探测器在静水中行进时的速度，T 为行进时的时间（单位：小时），c 为常数，n 为能量次级数．如果水的速度为4km/h ，该生物探测器在水中逆流行进200km ．（1）求T 关于v 的函数关系式；（2）(i)当能量次级数为2时，求该探测器消耗的最少能量； (ii)当能量次级数为3时，试确定v 的大小，使该探测器消耗的能量最少．18．(本小题满分16分)如图，椭圆22:143x y C +=的右焦点为F ，右准线为l ，过点F 且与x 轴不重合的直线交椭圆于A ，B 两点，P 是AB 的中点，过点B 作BM ⊥l 于M ，连AM 交x 轴于点N ，连PN . （1）若165AB =，求直线AB 的倾斜角；（2）当直线AB 变化时，求PN 长的最小值. 19．(本小题满分16分)设函数()e ()x f x ax a a =-+∈R ，其图象与x 轴交于1(0)A x ，，2(0)B x ，两点，且x 1＜x 2．（1）求a 的取值范围；（2）证明：0f '<（()f x '为函数()f x 的导函数）；（3）设点C 在函数()y f x =的图象上，且△ABC 为等腰直角三角形，t =，求(1)(1)a t --的值．20．(本小题满分16分)已知数列{n a }满足*111,||,.n n n a a a p n N +=-=∈（1）若{n a }是递增数列，且12,3,23a a a 成等差数列，求p 的值；（2）若12p =，且{21n a -}是递增数列，{2n a }是递减数列，求数列{n a }的通项公式．数学Ⅱ（附加题）一个特征向量．C ．选修4—4：坐标系与参数方程已知点P 是曲线C ：⎩⎨⎧==θθsin 3cos 2y x （θ为参数，πθπ2≤≤）上一点，O 为原点．若直线OP 的倾斜角为3π，求点P 的直角坐标． D ．选修4—5：不等式选讲已知实数x ，y ，z 满足x +y+z =2，求22232z y x ++的最小值．(第21题A)【必做题】第22题、第23题，每题10分，共计20分．请在答卷卡．．．指定区域内．．．．．作答．解答应写出文字说明、证明过程或演算步骤． 22．（本小题满分10分）某小组共10人，利用暑期参加义工活动，已知参加义工活动次数为1，2，3的人数分别为3，3，4，现从这10人中选出2人作为该组代表参加座谈会．（1）记“选出2人参加义工活动的次数之和为4”为事件A ，求事件A 的发生的概率；（2）设X 为选出2人参加义工活动次数之差的绝对值，求随机变量X 的分布列和数学期望．23．（本小题满分10分）在集合{A =1，2，3，4，…，2n }中，任取m （m n ≤，m ，n ∈N *）元素构成集合m A ．若m A 的所有元素之和为偶数，则称m A 为A 的偶子集，其个数记为()f m ；若m A 的所有元素之和为奇数，则称m A 为A 的奇子集，其个数记为()g m ．令()()()F m f m g m =-．（1）当2n =时，求(1)F ，(2)F ，的值；（2）求()F m ．2018高考数学模拟试题（2）数学I 答案一、填空题答案 1.{0}2.33.1204.215.216.－57.（0，1）8.y 2＝2x9.3π10.3π11.258解：因为直线l ：20x y +=与圆C ：22()()5x a y b -+-=相切，又因为圆心C 在直线l 的上方，所以20a b +>，所以25a b +=，52a b =+≥所以ab 的最大值为258． 12.6解：利用在上的投影得，221=⋅=6． 13．()[]6,40, ∞-解：①0<a0≤x 时，01e )(<-=x a x 'f ，所以)(x f 在)0(,-∞单调递减，且0)0(<=a f ，所以)(x f 在)0(,-∞有一个小于0的零点．0>x 时，)(x f 在)0(+∞,单调递增，因为1)1(=f ，所以)(x f 在)0(+∞,有一个小于1的零点．因此满足条件． ②0>a（1）1≤0a <时，)(x f 在)0(,-∞单调递减，0)0(>=a f ，所以)(x f 在(]0,∞-上没有零点．又因为042<-=∆a a ，故)(x f 在)0(+∞,上也没有零点.因此不满足题意．（2）41<<a 时，)(x f 在⎪⎭⎫ ⎝⎛∞-a 1ln ,上单调递减，在⎪⎭⎫⎝⎛01ln ,a上单调递增， 0ln 11ln >+=⎪⎭⎫⎝⎛a a f ，所以)(x f 在(]0,∞-上没有零点．又因为042<-=∆a a ，故)(x f 在)0(+∞,上也没有零点.因此不满足题意．（3）4=a 时，⎩⎨⎧>+--=04404)(2x x x x x e x f x ,≤ ,，)(x f 在(]0,∞-上没有零点，零点只有2，满足条件．（4）4>a 时，)(x f 在(]0,∞-上没有零点，在)0(+∞,上有两个不相等的零点，且和为a ，故满足题意的范围是64≤a <．综上所述，a 的取值范围为()[]6,40, ∞-． 14.a ≤或a ≥解：因为222()2a b a b -+≥对任意a 、b 都成立，所以，(x +2sin θcos θ)2+(x +a sin θ+a cos θ)2≥(2sin θcos θ-a sin θ-a cos θ)2， (2sin θcos θ-a sin θ-a cos θ)2≥，即对任意θ∈[0，]，都有132sin cos 2sin cos a θθθθ++≥+或132sin cos 2sin cos a θθθθ+-≤+，因为132sin cos 512sin cos sin cos 2sin cos θθθθθθθθ++=++⋅++,当θ∈[0，]时，1sin cos θθ≤+≤所以72a ≥，同理a ≤.因此，实数a 的取值范围是a ≤或a ≥．二、解答题答案15.解：（1）由三角函数定义，1cos x α=，2cos()3x πα=+，因为(,)62ππα∈，1cos 3α=，所以sin 3α==. 211cos()cos 3226x πααα-=+=-=.（2）依题意，1sin y α=，2sin()3y πα=+，所以111111cos sin sin 2224S x y ααα==⋅=，)322sin(41-)3sin()3cos(2121222παπαπα+=++-==y x S ,依题意，2sin 22sin(2)3παα=-+，化简得cos20α=，因为62ππα<<，则23παπ<<，所以22πα=，即4πα=.16.证明：（1）在直三棱柱ABC -A 1B 1C 1中，CC 1⊥平面A 1B 1C 1，四边形ACC 1A 1为矩形，设AC 1∩A 1C =G ，则G 为AC 1中点，D 为AB 中点，连DG ，则DG ∥BC 1. 因为DG ⊂平面A 1CD ，BC 1⊄平面A 1CD ，所以BC 1∥平面A 1CD.（2）由（1）四边形BCC 1B 1为矩形，又BC =BB 1，则四边形BCC 1B 1为正方形，所以BC 1⊥B 1C ，由（1）CC 1⊥平面ABC ，所以CC 1⊥AC ，又AC ⊥BC ，则AC ⊥平面BCC 1B 1，AC ⊥BC 1，因此，BC 1⊥平面AB 1C .17.解：（1）由题意得，该探测器相对于河岸的速度为200T，又该探测器相对于河岸的速度比相对于水的速度小4km/h ，即4v -，所以200T=4v -，即2004T v =-，4v >；（2）(ⅰ)当能量次级数为2时，由（1）知22004v E c v =⋅-，4v >， 3200c =（当且仅当1644v v -=-即8v =km/h 时，取等号）（9分）(ⅱ)当能量次级数为3时，由（1）知32004v E c v =⋅-，4v >，所以222(6)2000(4)v v E c v -'=⋅=-得6v =，当6v <时，0E '<；当6v >时，0E '>，所以当6v =时，min E 21600c =．答：(ⅰ)该探测器消耗的最少能量为3200c ； (ⅱ)6v =km/h 时，该探测器消耗的能量最少． 18.解（1）显然)0,1(,21,3,2F e b a ===，当AB ⊥x 轴时，易得221635b AB a ==≠，不合题意.所以可设AB 的方程为(1)(0)y k x k =-≠，与椭圆方程联立得2222(43)84120k x k x k +-+-=，设A （x 1，y 1）,B （x 2，y 2）,则212221228,4341243k x x k k x x k ⎧+=⎪⎪+⎨-⎪=⎪+⎩，因此2212(1)16435k k +=+，解得k =，所以直线AB 的倾斜角等于60o 或120o . （2）因为椭圆的右准线的方程为4x =，由（1），当AB 不垂直于x 轴时，点211(4,(1)),(,(1))M k x A x k x --，所以直线AM 的方程为12111()(1)()4k x x y k x x x x ---=--，令y =0，得1121254N x x x x x x --=- 2211221212412205454343k k x x k k x x x x ----++==--=1121255()522x x x x x -+=-. 当AB ⊥x 轴时，易得52N x =，所以无论AB 如何变化，点N 的坐标均为5(,0)2.因此，当AB ⊥x 轴时，PN 取最小值，PN min =53122-=.19.解（1）()e x f x a '=-．若0a ≤，则()0f x '>，则函数()f x 是单调增函数，这与题设矛盾．所以0a >，令()0f x '=，则ln x a =．当ln x a <时，()0f x '<，()f x 是单调减函数；当ln x a >时，()0f x '>，()f x 是单调增函数. 于是当ln x a =时，()f x 取得极小值．因为函数()e ()x f x ax a a =-+∈R 的图象与x 轴交于两点1(0)A x ，，2(0)B x ，(x 1＜x 2)，所以(ln )(2ln )0f a a a =-<，即2e a >．. 此时，存在1ln (1)e 0a f <=>，；存在33ln ln (3ln )3ln a a f a a a a a >=-+，3230a a a >-+>，又由()f x 在(ln )a -∞，及(ln )a +∞，上的单调性及曲线在R 上不间断，可知2e a >为所求取值范围.（2）因为1212e 0e 0xx ax a ax a ⎧-+=⎪⎨-+=⎪⎩，，两式相减得2121e e x x a x x -=-．记21(0)2x x s s -=>，则()121221212221e e e e 2(e e )22x x x x x x s sx x f s x x s ++-+-'⎡⎤=-=--⎣⎦-，设()2(e e )s s g s s -=--，则()2(ee )0ssg s -'=-+<，所以()g s 是单调减函数，则有()(0)0g s g <=，而122e 02x x s+>，所以()1202x x f +'<．又()e x f x a '=-是单调增函数，且122x x +>所以0f '<．（3）依题意有e 0ix i ax a -+=，则(1)e 0i x i a x -=>⇒112i x i >=（，）．于是122ex x +=，在等腰三角形ABC 中，显然C =90°，所以12012()2x x x x x +=∈，，即00()0y f x =<，由直角三角形斜边的中线性质，可知2102x x y -=-，所以21002x x y -+=，即1221212e ()022x x x x a x x a +--+++=，所以2112()022x x a x x a -+++=，即2112(1)(1)[(1)(1)]022x x a x x ----+-+=．因为110x -≠，则()2211111110212x x x a x ----++=-，t =，所以221(1)(1)022a at t t -++-=，即211a t =+-，所以(1)(1) 2.a t --= 20.解：（1）因为{n a }是递增数列，所以n n n p a a =-+1，又11=a ，1,1232++=+=p p a p a ，因为12,3,23a a a 成等差数列，所以p p p p p a a a =+++=++=223123,333144,34，解得0,31==p p ，当0=p ，01=-+n n a a ，与{n a }是递增数列矛盾，所以31=p . （2）因为{21n a -}是递增数列，所以01212>--+n n a a ，于是()+-+n n a a 212()0122>--n n a a ① 由于1222121-<n n ，所以122212-+-<-n n n n a a a a ② 由①②得()0122>--n n a a ，所以()122121222121----=⎪⎭⎫ ⎝⎛=-n n n n n a a ③ 因为{2n a }是递减数列，所以同理可得0212<-+n n a a ，()nn nnn a a 21222122121++-=⎪⎭⎫⎝⎛-=-.④ 由③④得()nn nn a a 2111++-=-，所以()()()123121--++-+-+=n n n a a a a a a a a()()()123122121211--++-+-+=n n()11213134211211211---+=+⎪⎭⎫⎝⎛--⋅+=n nn ，所以数列{n a }的通项公式为()1213134--+=n nn a ．数学Ⅱ答案21．【选做题】答案 A ．选修4—1：几何证明选讲解：连结OC ，BE ．因为AB 是圆O 的直径，所以BE ⊥AE ．因为AB ＝8，BC ＝4，所以OB ＝OC ＝BC ＝4，即△OBC 为正三角形．所以∠BOC ＝60?．又直线l 切⊙O 与于点C ，所以OC ⊥l ．因为AD ⊥l ，所以AD ∥l ．所以∠BAD ＝∠BOC ＝60?．在Rt △BAE 中，因为∠EBA ＝90?－∠BAD ＝30°，所以AE ＝AB ＝4． B ．选修4—2：矩阵与变换解：矩阵M 的特征多项式为f (λ)＝＝(λ－1)(λ－x )－4．因为λ1＝3是方程f (λ)＝0的一个根，所以(3－1)(3－x )－4＝0，解得x ＝1．由(λ－1)(λ－1)－4＝0，得λ＝－1或3，所以λ2＝－1．设λ2＝－1对应的一个特征向量为α＝，则从而y ＝－x ．取x ＝1，得y ＝－1，所以矩阵M 的另一个特征值为－1，对应的一个特征向量为α＝． C ．选修4—4：坐标系与参数方程解：由题意得,曲线C 的普通方程为22143x y +=(1)00sin 2≤⇒≤⇒≤≤y θπθπ 直线OP的方程为y =(2)A D(第21题A)联立(1)(2)得55xy⎧=⎪⎪⎨⎪=⎪⎩（舍）或55xy⎧=-⎪⎪⎨⎪=-⎪⎩所以点P的坐标为(D．选修4—5：不等式选讲解：由柯西不等式可知22222221)1](23)z x y z ++⋅≤++++,所以2222()24231111123x y zx y z++++≥=++,当且仅当1112,114,116===zyx时取等号.【必做题】答案22.解：（1）由已知有P(A)=C31C41+C32C102=13,所以事件A发生的概率为13.（2）随机变量X的所有可能的取值为0，1，2P(X=0)=C32+C32+C42C102=415；P(X=1)=C31C31+C31C41C102=715；P(X=2)=C31C41C102=415.所以随机变量X的分布列为23.解：（1）当2n =时，集合为{1,2,3,4}．当1m =时，偶子集有{2}，{4}，奇子集有{1}，{3}，(1)2f =，(1)2g =，(1)0F =；当2m =时，偶子集有{2,4}，{1,3}，奇子集有{1,2}，{1,4}，{2,3}，{3,4}，(2)2f =，(2)4g =，(2)2F =-；（2）当m 为奇数时，偶子集的个数0224411()C C C C C C C C m m m m n n n nn n n n f m ---=++++，奇子集的个数11330()C C C C C C m m m n n n nn n g m --=+++，所以()()f m g m =，()()()0F m f m g m =-=．当m 为偶数时，偶子集的个数022440()C C C C C C C C m m m m n n n nn n n n f m --=++++，奇子集的个数113311()C C C C C C m m m n n n nn n g m ---=+++，所以()()()F m f m g m =-0112233110C C C C C C C C C C C C m m m m m m n n n n n n n nn n n n ----=-+-+-+．一方面，01220122(1)(1)(C C C C )[C C C (1)C ]n n n n n n nn n n n n n n n x x x x x x x x +-=++++-+-+-，所以(1)(1)n n x x +-中m x 的系数为0112233110C C C C C C C C C C C C m m m m m m n n n n n n n nn n n n -----+-+-+；另一方面，2(1)(1)(1)nnnx x x +-=-，2(1)nx -中mx 的系数为22(1)C mm n-，故()F m =22(1)C m m n-．综上，22(1)C , ()0,m mn m F m m ⎧⎪-=⎨⎪⎩为偶数，为奇数．。

四川省成都市2018届高三语文二模试卷 Word版含解析

2018年四川省成都市高考语文二模试卷一、阅读下面的文字，完成小题．1．（9分）阅读下面的文字，完成小题。

①比较文明是个新课题。

之所以新，是因为我们刚开始习惯以世界为关心对象。

②比较文明的目的是理解文明。

个体研究固然有其合理性，然而无视整体的结果一定是盲人摸象。

比较研究是整体下的个体与个体的比较，或个体与整体关系的钩沉。

从中国的视角来看世界以及从世界的维度理解中国，是出和入、间离和沉浸的关系，是辩证发展的关系，是螺旋上升的关系。

有了“从周边看中国”，对中国的认识才会更完整，更接近真实。

③比较文明研究是有伦理的。

有些比较无益甚或有害，伦理就是约束。

比如“汤显祖”和“莎士比亚”，比人生轨迹，没什么好比，但在创造文学形象的维度比较，不但有益而且必需。

有些比较无意义，并非学术前提出了问题，而是回答不了“然后呢”这个问题。

④比较文明是有方法的。

文明离不开人，物作为文明的见证，也很重要。

物是实实在在的，罐就是罐，盆就是盆。

当年瑞典人安特生在西北见到彩陶就一眼看出它和世界各地彩陶文化的联系，后来我们批判“中国文明西来说”，把世界维度下的中国彩陶变成了中国自己的彩陶。

彩陶西来与否和中国的正当性其实毫无关系。

⑤文明比较有宏观和微观两个层面。

制度比较属于宏观的文明比较。

因为有文献，制度很容易比较，但也有缺陷，一是文献资料有太多不可信的成分。

二是制度大概未必属于文明的范畴，它更像是文明的副产品，其本质是制度的制定者对受制度约束的人实践优势的工具。

如今的民主制度在柏拉图那里未必多么理想，同理，当年的“君臣父子”也未必全然“反动”。

⑥微观的文明比较主要在这两个范畴：从具体的物到生活方式、思维方式；从核心价值观念到具体的物。

前者受时空约束，它是已然的，找出异同作出解释，能改变的很有限。

后者则有超越时空的启发和引领意义：罗马人的沃伦杯与两千年后的中国电影《霸王别姬》同样在叙述欲望与爱的故事；静穆而伟大的希腊石头与北齐佛造像都是人类最伟大的气质与精神的外化，存在于所有不朽的艺术品之中；敦煌的鹿王本生与圣艾格尼斯杯主题都是牺牲……⑦文明的比较，核心在于解释文明的演进，以达成文化认同与自信。

2018年安徽省合肥市高考数学二模试卷(理科)(解析版)

当 x≤2 时，由 y＝﹣2x﹣3＝1，解得：x＝﹣2，
第 8 页（共 23 页）
综上可得若输出的结果为 1，则输入 x 的值为 3 或﹣2，故选：A． 7．（5 分）小李从网上购买了一件商品，快递员计划在下午 5：00﹣6：00 之间送货上门，
已知小李下班到家的时间为下午 5：30﹣6：00．快递员到小李家时，如果小李未到家，则快递员会电话联系小李．若小李能在 10 分钟之内到家，则快递员等小李回来；否则，就将商品存放在快递柜中．则小李需要去快递柜收取商品的概率为（）
求 m 的取值范围．
第 6 页（共 23 页）
2018 年安徽省合肥市高考数学二模试卷（理科）
参考答案与试题解析
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．（5 分）已知复数 z 满足 z（• 1﹣2i）＝（i i 是虚数），则复数 z 在复平面内对应的点在（）
到的绵是（）
A．174 斤
B．184 斤
C．191 斤
D．201 斤
6．（5 分）执行如图所示的程序框图，若输出的结果为 1，则输入 x 的值为（）
第 1 页（共 23 页）
A．3 或﹣2
B．2 或﹣2
C．3 或﹣1
D．﹣2 或﹣1 或 3
7．（5 分）小李从网上购买了一件商品，快递员计划在下午 5：00﹣6：00 之间送货上门，
对称
12．（5 分）已知点 I 在△ABC 内部，AI 平分∠BAC，述条件的所有△ABC，下列说法正确的是（）
，对满足上
A．△ABC 的三边长一定成等差数列 B．△ABC 的三边长一定成等比数列 C．△ABI，△ACI，△CBI 的面积一定成等差数列 D．△ABI，△ACI，△CBI 的面积一定成等比数列二、填空题（每题 5 分，满分 20 分，将答案填在答题纸上）