华南师大附中、广东实验中学、广雅中学三校2017届高三广州一模后联合适应性考试(数学理) 推荐
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高考模拟检测12本试卷分第Ⅰ卷(阅读题)和第Ⅱ卷(表达题)两部分。
共150分,考试时间150分钟。
第Ⅰ卷(阅读题,共70分)甲必考题一、现代文阅读(9分,每小题3分)阅读下面的文字,完成1~3题。
孔子所处的春秋时代,正值“富者田连阡陌,贫者亡立锥之地”。
基于对这一现象的担忧,为了维护社会安定,巩固封建统治,孔子提出了如下思想:“丘也闻有国有家者,不患寡而患不均,不患贫而患不安。
盖均无贫,和无寡,安无倾。
”(《论语·季氏》)一些人把“寡”解为土地财产少当是误读,所谓“寡”类似于老子“小国寡民”中的“寡民”。
孔子的“贫”是指土地和财产少。
孔子的“均”应有两解:一是“平均”,如《荀子·君道》“以礼分施,均遍而不偏”;二是“公平”,相当于《管子·形势》所言“天公平而无私,故美恶莫不覆;地公平而无私,故小大莫不载”。
这段话经常被当作儒家宣传平均主义的经典表达加以批判。
有的学者指出,社会财富分配的公正合理,确实是社会稳定的前提之一,但孔子“患不均”之正确,并不能证明“不患贫”之合理;“均无贫”不合事实,“不均才贫”更不合逻辑;“均”是分配形式,“贫”是财富多少,财富分配之均与不均,与财富之多少没有关系。
“不患贫”是危险的,因为“贫”正是“不安”的首要原因;物贫则人争,争则不安;人所争者,正是这不足之物。
“均无贫”彻底的没逻辑,其极端形式是人人都是零,也就不是贫;而人人都是零,均是绝对均了,但贫还是贫。
这番言论虽不乏深刻,但存在诸多误解。
不能脱离特定的语境加以解释。
要知道,孔子这段话是针对季康子掌权者之流通过对外战争攻伐等不义手段掠夺土地和人口而讲的,不是泛泛而论。
他分别把“均”“和”“安”和“贫”“寡”“倾”当作两种价值,“不患寡而患不均,不患贫而患不安”表面上看是只追求平均、安定而排斥土地和人口的增加,其实不然,他不过是强调通过采用“均”“和”“安”的手段和途径达到解决“贫”“寡”“倾”的问题。
2017-2018学年广东省华南师大附中、实验中学、广雅中学、深圳高中等四校联考高三(上)期末化学试卷一、选择题(共7小题,每小题6分,满分42分)1.(6分)关于自然资源的利用,下列说法不正确的是( )A.海水资源丰富,可从海水中提取溴、镁、铀、重水等物质B.工业上利用还原剂高温下将铝土矿冶炼为金属铝C.石油通过化学变化可得到合成聚乙烯的原料乙烯D.煤的气化可得到CO 和H2,再利用CO 和H2 可合成甲醇2.(6分)4─苯基─1─丁烯()是合成降压药福辛普利钠的中间体,关于它的说法下列正确的是( )A.和萘()互为同分异构体B.所有的碳原子不可能都处于同一平面C.沸点比苯高,易溶于水D.可发生取代、氧化、加聚反应3.(6分)用图所示装置(熔融CaF2﹣CaO 作电解质)获得金属钙,并用钙还原TiO2 制备金属钛。
下列说法正确的是( )A.电解过程中,Ca2+向阳极移动B.若用铅蓄电池作该装置的供电电源,石墨电极连接的是铅蓄电池的Pb 电极C.阴极的电极反应式为:TiO2+4e=Ti+2O2D.理论上,若生成1mol Ti,电路中通过了4mol 电子的电量4.(6分)某无色溶液中只大量存在Na+、Mg2+、Fe3+、CO3、SO4、Cl﹣中的几种,某学生进行了下列实验:①取少量原溶液,向其中滴加足量氯水,有无色无味气体产生,再将溶液分为2 份;②向其中一份溶液中加入AgNO3溶液,有白色沉淀产生;③向另一份溶液中加入酸化BaCl2溶液,有白色沉淀产生。
原溶液中一定大量存在的离子组是( )A.Na+、Fe3+、SO42﹣B.Na+、SO42﹣、CO32﹣C.Mg2+、Cl﹣、SO 42﹣D.CO 32﹣、Cl﹣、SO 42﹣5.(6分)有关下列装置说法正确的是( )A.用装置①制备SO2B.用装置②检验和收集SO2C.用装置③分离NaCl 和I2固体D.用装置④测定溶液中c(H+)6.(6分)W、X、Y、Z、Q 为原子序数依次增大的短周期主族元素,且原子最外层电子数之和为21,W 的内层电子数是最外层电子数的 2 倍,X 与Y 同族,Y、Z 同周期且相邻,X 的一种核素在考古时常用来鉴定文物的年代,下列叙述不正确的是( )A.最高价氧化物水化物酸性:X>Z>YB.W 的单质在空气中受热生成氧化物C.简单氢化物的热稳定性:Q>ZD.Y 的最高价氧化物晶体中只含有共价键7.(6分)室温时,下列叙述中正确的是( )A.pH=3 的醋酸、盐酸溶液中:c(CH3COO﹣)>c(Cl ﹣)B.已知电离常数:K(HF)>K(CH3COOH),pH=11 的NaF 与CH3COONa 溶液中:<C.0.1mol/L 氨水的pH=a,0.01mol/L 氨水的pH=b,则a﹣b>1D.已知某温度下,K sp(AgCl)=1.56×10﹣10,K sp(Ag2CrO4)=1×10﹣12,AgCl、Ag2CrO4两者的饱和溶液相比,后者的c(Ag+)大二、解答题(共3小题,满分43分)8.(14分)过氧化钙(CaO2)是一种白色晶体,不溶于醇类、乙醚等;能潮解,难溶于水,可与水缓慢反应;275℃时开始分解;易与酸反应。
2017届高三综合测试理科综合试题 (2017-5-23) 注意事项:1. 答卷前,请务必用黑色字迹的钢笔或签字笔将自己的班级、姓名和考号填写在答题卡和答卷上。
2. 选择题在选出答案后,用2B 铅笔把答题卡上对应题目选项的答案涂黑。
如需改动,用橡皮擦干净后,再选涂其他答案,答案不能答在试卷上。
3. 非选择题必须用黑色字迹钢笔或签字答作答,答案必须写在答卷各题目指定区域内相应位置上;如需改动,先划掉原来的答案,再写上新答案;不准使用铅笔和涂改液,不按要求作答的答案无效。
4. 考生必须保持答题卡的整洁,考试结束后,将答卷和答题卡一并交回。
[可能用到的相对原子质量:H —1 C —12 O —18 S —32 Fe —56 Cu —64]第Ⅰ卷二、选择题(本大题共8小题,每小题6分。
在每小题给出的四个选项中,14~17题只有一项符合题目要求。
18~21题有多项符合题目要求。
全部选对的得6分,选对但不全的得3分。
有选错的得0分)14. 在光滑水平面上,a 、b 两球沿水平面相向运动。
Iv 两球间距小于或等于L 时,则相互作用力为零,两球沿水平平面相向运动。
当两球间距小于或等于L 时,则相互作用力为零,两球在相互作用区间运动时始终未接触。
两球运动的t v -图像如图所示,则A .a 球质量小于b 球质量B .1t 时刻两小球间距最小C .20t -时间内,两小球间距逐渐减小D .30t -时间内,b 球所受排斥力方向始终与运动方向相反15. 如图所示,图形区域内有垂直于纸面向里的匀强磁场,一个带电粒子以速度v 从A 点沿直径AOB 方向射入磁场,经过△t 时间从C 点射出磁场,OC 与OB 成︒60角。
现将带点粒子的速度变为3v ,仍从A 点沿原方向射入磁场,不计重力,则粒子在磁场中的运动时间变为A.2△tB.21△tC.3△tD.31△t 16.如图所示,一个小型水电站,其交流发电机的输出电压1U 一定,通过理想升压变压器1T 和理想降压变压器2T 向远处用户供电,输电线的总电阻为R 。
2017届高三上学期期末华附、省实、广雅、深中四校联考理科数学考试参考答案一、选择题:每小题5分,满分60分.1.C 2.B 3.A 4.C 5.D 6.C7.B 8.A 9.B 10.A 11.A 12.D二.填空题:本大题共4小题,每小题5分。
13.20-; 14.8; 15.]63,0[π+; 16. 93.三.解答题:解答应写出文字说明,证明过程或演算步骤。
17.(本小题满分12分) 法一解:(1)在等比数列}{n a 中,因为21748≠=S S ,所以1≠q ,所以q q a S n n --=1)1(1…………………1分由⎪⎪⎩⎪⎪⎨⎧=--==--=3401)1(201)1(818414q q a S q q a S ,两式相除得:1714=+q ,且0>q ,所以34,21==a q ……………4分所以3223411+-==n n n a 。
…………………5分(2)当1=n 时,222112===b T b当2≥n 时,由n n T b 21=+,得n n n n n b T T b b 2)(211=-=--+,即)2(31≥=+n b b n n ,…………6分 所以⎩⎨⎧≥⋅==-2,321,12n n b n n ,⎩⎨⎧≥⋅==-2,321,12n n n nb n n …………………8分 2103236341-⋅++⋅+⋅+=n n n H12132363433-⋅++⋅+⋅+=n n n H两式相减得:12232)333(2422--⋅-+++++-=-n n n n H …………………9分1)21(3323131232)3331(2111122--=⋅---=⋅-++++=-----n n n n n n n n …………………11分所以2132121+⋅-=-n n n H …………………12分 法二当2≥n ,232-⋅=n n n nb =213]21)1[(3)21(------n n n n ,所以:++-+-+= )323325()321323(11201n H 213]21)1[(3)21(------n n n n13)21(211--+-=n n 2132121+⋅-=-n n对应步骤相应给分。
【关键字】试卷华南师大附中2017届高三综合测试(一)数学(理科)第Ⅰ卷一、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.设全集,集合,则()A.B.C.D.2.命题“”的否定是()A.B.C.D.3.设,则下列关系中正确的是()A.B.C.D.4.设,则“”是“”的()A.充分不必要条件B.必要不充分条件C.充分必要条件D.既不充分也不必要条件5.已知,则()A.B.C.D.6.由曲线,直线及轴所围成图形的面积是()A.B.4 C.D.67.已知函数在单调递减,则的取值范围是()A.B.C.D.8.函数与在同一直角坐标系下的图象大致是()A.B.C.D.9.已知在偶函数,且在单调递减,若,则的解集为()A.B.C.D.10.已知函数,则的大小关系为()A.B.C.D.11.下列命题中是假命题的是()A.,使是幂函数,且在上递减B.函数的值域为,则C.关于的方程至少有一个负根的弃要条件是D.函数与函数的图像关于直线对称12.已知函数是定义在上的以4为周期的函数,当时,,其中.若函数的零点个数是5,则的取值范围为()A.B.C.D.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.函数的定义域为____________.14.已知集合,若,则实数的所有可能取值的集合为____________.15.若,且,则__________.16.过函数图像上一个动点作函数的切线,则切线倾斜角的范围是__________.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.(本小题满分10分)已知集合.(1)分别求;(2)已知集合,若,求实数的取值集合.18.(本小题满分12分)已知实数满足,其中实数满足.(1)若,且为真,求实数的取值范围;(2)若是的必要不充分条件,求实数的取值范围.19.(本小题满分12分)函数是定义在实数集上的奇函数.(1)若,试求不等式的解集;(2)若且在上的最小值为-2,求的值.20.(本小题满分12分)某山区外围有两条相互垂直的直线型公路,为进一步改善山区的交通现状,计划修建一条连接两条公路和山区边界的直线型公路.记两条相互垂直的公路为12l l 、,山区边界曲线为C .计划修建的公路为l ,如图所示,,M N 为C 的两个端点,测得点M 到12l l 、的距离分别为5千米和40千米,点N 到12l l 、的距离分别为20千米和2.5千米,以12l l 、所在直线分别为,x y 轴,建立平面直角坐标系xOy .假设曲线C 符合函数2ay x b=+(其中,a b 为常数)模型. (1)求,a b 的值;(2)设公路l 与曲线C 相切于P 点,P 的横坐标为t . ①请写出公路l 长度的函数解析式()f t ,并写出其定义域; ②当t 为何值时,公路l 的长度最短?求出最短长度. 21.(本小题满分12分)已知定义为R 的函数()f x 满足下列条件:①对任意的实数,x y 都有:()()()1f x y f x f y +=+-;②当0x >时,()1f x >.(1)求()0f ;(2)求证:()f x 在R 上为增函数;(3)若()67,3f a =≤-,关于x 的不等式()()223f ax f x x -+-<对任意[)1,x ∈-+∞恒成立,求实数a 的取值范围. 22.(本小题满分12分) 已知函数()ln x mf x ex +=-.(1)设1x =是函数()f x 的极值点,求m 并讨论()f x 的单调性;(2)设0x x =是函数()f x 的极值点,且()0f x ≥恒成立,求m 的取值范围(其中常数a 满足ln 1a a =).参考答案一、选择题二、填空题13. ()2,+∞ 14. {}1,0,1-30,,24πππ⎡⎫⎡⎫⎪⎪⎢⎢⎣⎭⎣⎭三、解答题:17.解:(1)∵3327x ≤≤,即13333x ≤≤,∴13x ≤≤,∴{}|13A x x =≤≤,...........2分 ∵2log 1x >,即22log log 2x >,∴2x >,∴{}|2B x x =>,..................3分 当C 为空集时,1a ≤,当C 为非集合时,可得13a <≤,综上所述3a ≤........................................10分 18.解:(1)对:p 由22430x ax a -+<得()()30x a x a --<, 因为0a >,所以3a x a <<............................... 2分 当1a =时,解得13x <<,即p 为真时,实数x 的取值范围是13x <<. 又q 为真时实数x 的取值范围是23x <≤...................4分 若p q ∧为真,则p 真且q 零点,所以实数x 的取值范围是()2,3................................. 6分 (2)p 是q 的必要不充分条件 ,即q p ⇒,且p q ≠,设(){}(){}|,|A x p x B x q x ==,则B A ≠...................8分 又(]()2,3,,3B A a a ==;所以有233a a ≤⎧⎨≤⎩解得12a ≤≤,所以实数a 的取值范围是(]1,2................12分 19.解:(1)∵()f x 是定义在R 上的奇函数,∴()00f =,∴10k -=,∴1k =....................2分 ∵()10f >,∴10a a->,又0a >且1a ≠,∴1a >....................4分易知()f x 在R 上单调递增,原不等式化为:()()224f x x f x +>-, ∴1x >或4x <-,∴不等式的解集为{}|14x x x ><-或...................6分 (2)∵()312f =,∴132a a -=,……22320a a --=,∴122a a ==-或(舍去) ∴()()()()22222222222222xx x x x x x x g x m m ----=+--=---+....................9分 令()22xxt f x -==-,∵1x ≥,∴()312t f ≥=,∴()()222222g t t mt t m m =-+=-+-, 当32m ≥时,当t m =时,()min 17324g t m =-=-,∴2m =,当32m <时,当32t =时,()min 17324g t m =-=-,解得253122m =>,舍去,综上可知2m =............12分20.解:(1)由题意知,点,M N 的坐标分别为()()5,40,20,2.5.将其分别代入2a y x b =+,得4025 2.5400aba b⎧=⎪⎪+⎨⎪=⎪+⎩,解得10000a b =⎧⎨=⎩.................4分(2)①由(1)知,()21000520y x x =≤≤,则点P 的坐标为21000,t t ⎛⎫⎪⎝⎭, 设在点P 处的切线l 交,x y 轴分别交于,A B 点,32000y x'=-, 则l 的方程为()2310002000y x t t t -=--,由此得233000,0,0,2t A B t ⎛⎫⎛⎫⎪ ⎪⎝⎭⎝⎭.故()[]5,20f t t ==∈...............8分 ②设()624410g t t t ⨯=+,则()6516102g t t t⨯'=-,令()0g t '=,解得t =当(t ∈时,()0g t '<,()g t 是减函数;当()20t ∈时,()()0,g t g t '>是增函数.从而,当t =()g t 有极小值,也是最小值,所以()min 300g t =,此时()min f t =答:当t =l 的长度最短,最短长度为千米................12分21.解:(1)令0x y ==,恒等式可变为()()()00001f f f +=+-,解得()01f =............1分(2)任取12x x <,则210x x ->,由题设0x >时,()1f x >,可得()211f x x ->, ∵()()()1f x y f x f y +=+-,∴()()()()()212112111f x f x x x f x f x x f x =+-=+-->⎡⎤⎣⎦, 所以()f x 是R 上增函数................................4分 (3)由已知条件有:()()()22221f ax f x x f ax x x -+-=-+-+,故原不等式可化为:()2213f ax x x -+-+<,即()2122f x a x ⎡⎤-++-<⎣⎦,而当*n N ∈时,()()()()()()()()()1112212331311f n f n f f n f f n f nf n =-+-=-+-=-+-==--,所以()()6615f f =-,所以()12f =,故不等式可化为()()2121f x a x f ⎡⎤-++-<⎣⎦,由(2)可知()f x 在R 上为增函数,所以()2121x a x -++-<,即()2130x a x -++>在[)1,x ∈-+∞上恒成立,令()()213g x x a x =-++,即()min 0g x >成立即可.①当112a +<-,即3a <-时,()g x 在[)1,x ∈-+∞上单调递增,则()()()min 11130g x g a =-=+++>解得5a >-,所以53a -<<-,②当112a +≥-即3a ≥-时,有()()2min 111130222a a a g x g a +++⎛⎫⎛⎫==-++> ⎪ ⎪⎝⎭⎝⎭解得11a --<<-,而13-<-,所以31a -≤<-, 综上,实数a 的取值范围是(]5,3--......................12分22.解:(1)()()1,0x m f x e x x+'=->,因为1x =是函数()f x 的极值点, 所以()1110mf e +'=-=,所以1m =-,所以()11x f x e x-'=-.................2分当01x <<时,1101,1x e x -<<-<-,所以()0f x '<,当1x >时,111,10x e x->-<-<,所以()0f x '>,所以()f x 在()0,1单调递减,在()1,+∞单调递增............................5分 (2)()()1,0x m f x e x x +'=->,设()1x m g x e x +=-,则()210x m g x e x+'=+>, 所以()g x 在()0,+∞单调递增,即()f x '在()0,+∞单调递增.由于0x x =是函数()f x 的极值点,所以0x x =是()0f x '=在()0,+∞的唯一零点, 所以00001,ln x mex m x x +=+=-.............................6分 由于00x x <<时,()()00f x f x ''<=;当0x x >时,()()00f x f x ''>=, 所以函数()f x 在()00,x 单调递减,在()0,x +∞单调递增.....................8分且函数()f x 在0x x =处取得最小值,所以()()000001ln x mf x f x e x x m x +≥=-=++, 因为()0f x ≥恒成立,所以0010x m x ++≥.....................9分 ∴00001ln x m x x x +≥-=+,即001ln x x ≥.又因为ln 1a a =,故可解得0x a ≤...............................11分 所以00,ln ln x a x a -≥--≥-,所以00ln ln m x x a a =--≥--,即m 的取值范围是[)ln ,a a --+∞..................................12分此文档是由网络收集并进行重新排版整理.word 可编辑版本!。
2017届高三上学期期末华附、省实、广雅、深中四校联考英语本试卷分选择题和非选择题两部分,共8页,满分120分,考试用时120分钟。
注意事项:1.答卷前,考生务必用2B铅笔在“准考证号”处填涂准考证号.用黑色字迹的钢笔或签字笔将自己姓名、班级、考场号、座位号、准考证号填写在答题卷指定区域内.2.选择题每小题选出答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑;如需改动,用橡皮擦干净后,再选涂其它答案;不能答在试卷上.3.非选择题必须用黑色字迹的钢笔或签字笔作答,答案必须写在答题卡各题目指定区域内的相应位置上;如需改动,先划掉原来的答案,然后再写上新的答案;不准使用铅笔和涂改液.不按以上要求作答的答案无效.4.考生必须保持答题卡的整洁.第I卷选择题(共70分)第一部分阅读理解(共两节,满分40分)第一节阅读下列短文,从每题所给的四个选项(A、B、C和D)中,选出最佳选项,并在答题卡上将该项涂黑.(共15小题;每小题2分,满分30分)ATop Five Events for Kids This AutumnAutumn in London means crisp leaves, deep puddles, pumpkins and a whole new variety of fun for kids. The leaves, the conkers and the rain are falling; our picnic days are over. Never fear, there's still plenty of family fun to be had in London's top five museums for kids.London's Night Owls SleepoverExplore the Museum of London by torchlight at these special after-hours events for ages seven to eleven.Pillows and PaintbrushesKids aged seven to thirteen can embark on a very arty overnight at the Dulwich Picture Gallery on its first ever sleepover.Michael Rosen's Bear Hunt, Chocolate Cake and Bad ThingsMuch-loved children's author Michael Rosen is behind this interactive family exhibition which reveals the inspiration behind his writing. Younger children will explore a Bear Hunt trail, while older kids can go on a mission to find facts which they can use for inspiration to write their very own poem to take home.Free Range Urban KidsChildren aged three to five can be dropped off for a forest adventure that might include den building, mini-beast-hunting, foraging, wildlife spotting, singing and free play (tots aged two to three can attend with a parent) .The Comedy of Errors"The Comedy of Errors" returns to the National Theatre, this time in a new production forages eight to twelve by Ben Power. It should be a short, sweet and fun version of the tale of two sets of identical twins that get separated and then reunited.1.Kim has two children, aged 4 and 11 respectively, so he will probably chooseA. Free Range Urban KidsB. Pillows and PaintbrushesC. London's Night Owls SleepoverD. Michael Rosen's Bear Hunt, Chocolate Cake and Bad Things2.In Free Range Urban Kids, children can enjoy the following EXCEPT .A.wildlife-hunting B.foragingC.den building D.singing3.Which of the following statements is true?A.Ben Power is a children's writer.B.Michael Rosen is an inspiring painter.C.Kids aged 8-12 can join in Pillows and Paintbrushes.D.Kids who explore the Museum of London should carry a torchlight.BIt was my first day at school in London and I was half-excited and half-frightened. On my way to school I wondered what sort of questions the other boys would ask me and rehearsed all the answers: "I am nine years old. I was born here but I haven't lived here since I was two. I was living in Farley. It's about thirty miles away. I came back to London two months ago." I also wondered if it was the custom for boys to fight strangers like me, but I was tall for my age. I hoped they would decide not to risk it.No one took any notice of me before school. I stood in the center of the playground, expecting someone to say "hello", but no one spoke to me. When a teacher called my name and told me where my classroom was, one or two boys looked at me but that was all the curiosity my arrival aroused.My teacher was called Mr. Jones. There were 42 boys in the class, so I didn't stand out there, either, until the first lesson of the afternoon. Mr. Jones was very fond of Charles Dicker and he had decided to read aloud to us from David Copperfield, but first he asked several boys . they knew Dickens' birthplace, but no one guessed right. A boy called Brian, the biggest in the class, said: "Timbuktu", and Mr. Jones went red in the face. Then he asked me. I said "Portsmouth", and everyone stared at me because Mr. Jones said I was right. This didn't make me very popular, of course. "He thinks he's clever," I heard Brian sayAfter that, we went out to the playground to play football. I was in Brian's team, and he obviously bad Dickens in mind because he told me to go in goal. No one ever wanted to be the goalkeeper. "He's big enough and useless enough," Brian said when someone asked him why he had chosen me.I suppose Mr. Jones remembered Dickens, too, because when the game was nearly over, Brian pushed one of the players on the other team, and he gave them a penalty. As the boy kicked the ball hard along the ground to my night, I threw my self down instinctively and saved itAll my team crowded round me. My bare knees were grazed and bleeding. Brian took out a handkerchief and offered it to me."Do you want to join my gang (team) ?" he said.At the end of the day, I was no longer a stranger.4. Which of the following statements is true about the writer?A.He was born in Farley and moved to London two months ago.B.He was taller than most of his classmates because he was older.C.He knew what sorts of questions his classmates would ask and he had prepared answers.D.He looked forward to his new school life but were afraid that the boys might not be friendly 5.We can learn from the passage that .A.boys were usually unfriendly to new studentsB.the writer was not greeted as he had expectedC.Brian envied the writer because the teacher thought he was cleverD.the writer was chosen as the goalkeeper because he was useless and big6. The underlined part "I didn't stand out" in paragraph 3 means that the writer was notA.noticeable B.welcomeC.important D.outstanding7.The writer was offered a handkerchief becauseA.he threw him self down and saved the goalB.his bare knees were bleedingC.he was beginning to be acceptedD.he joined Brian's gangCPublicity offers several benefits. There are not costs for message time or space. An ad in prime-time television may cost $250,000 to $5,000,000 or more per minute, whereas a five-minute report on a network newscast would not cost anything. Publicity reaches a mass audience within a short time and new products or company policies are widely known Credibility about messages is high, because they are reported in independent media. A newspaper review of a movie has more credibility than an ad in the same paper, because the reader associates independence with objectivity (客观现实). Similarly, people are more likely to pay attention to news reports than to ads. For example, Women's Wear Daily has both fashion reports and advertisements. Readers spend time reading the stories, but they skim through the ads. Furthermore, there may be 10 commercials during a half-hour television program or hundreds of ads in a magazine. Feature stories are much fewer in number and stand out clearly.Publicity has some significant limitations. A firm has little control over messages, their timing, their placement, or their coverage by a given medium. It may issue detailed news releases but find only portions mentioned by the media, and media have the ability to be much more critical than a firm would like.For example, in 1982, Procter & Gamble faced a massive publicity problem over the meaning of its 123-year-old company logo. To fight this negative publicity, the firm had a spokespersonappear on Good Morning America to disprove the rumor (谣言). The false rumors were temporarily put to rest. However, in 1985, publicity became so trouble-making that Procter & Gamble decided to remove the logo from its products.A firm may want publicity during certain periods, such as when a new product is introduced or new store opened, but the media may not cover the introduction or opening until after the time it would aid the firm. Similarly, media determine the placement of a story; it may follow a report on crime or political scandal. Finally, the media decide whether to cover a story at all and the amount of coverage to be devoted to it.8.All of the following advantages of publicity are mentioned EXCEPTA. time savingB. money savingC.profitability D.credibility9.What does "it 'in paragraph 3 refer to?A.A giver medium B.A firm C.Publicity D.Coverage 10.Which of the following statements is true?A.A five-minute report via the media would not cost people any moneyB.Procter & Gamble fighted against another company on company logo.C.Ads appear less believable merely because there are too many of them.D.Timing of publicity can influence the efficiency of promoting products.11.What's the author's attitude towards publicity?A.Objective. B.Doubtful. C.Passive. D.Supportive.DSunny countries are often poor. A shame, then, that solar power is still quite expensive. Eight19, a British company by Cambridge University, has, however, invented a novel way to get round this. In return for a deposit of around $10 it is supplying poor Kenyan families with a solar cell able to generate 2.5 watts of electricity, a battery that can deliver a three amp (安培) current to store this electricity, and a lamp whose bulb is a light-emitting diode(二极管). The firm thinks that this system, once the battery is fully charged, is enough to light two small rooms and to power a mobile-phone charger for seven hours. Then, next day, it can be put outside and charged back up again.The trick is that, to be able to use the electricity, the system's keeper must buy a scratch car d—for as little as a dollar-on which is printed a reference number. The keeper sends this reference, plus the serial number of the household solar unit, by SMS to Eight19. The company's server will respond automatically with an access code to the unit.Users may consider that they are paying an hourly rate for their electricity. In fact, they are paying off the cost of the unit. After buying around $80 worth of scratch cards - which Eight19 expects would take the average family around 18 months - the user will own it. He will then have the option of continuing to use it for nothing, or of trading it in for a bigger one, perhaps driven by a 10-watt solar cell.In that case, be would go then through the same process again, paying off the additional cost of the upgraded kit at a slightly higher rate. Users would therefore increase their electricity supplysteadily and affordably.According to Eight19's figures, this looks like a good deal for customers. The firm believes the average energy-starved Kenyan spends around $10 a month on oil - enough to fuel a couple of smoky lamps - plus $2 on charging his mobile phone in the market-place. Regular users of one of Eight19's basic solar units will spend around half that, before owning it completely. Meanwhile, as the cost of solar technology falls, it should get even cheaper.12.Where can the users find the reference number needed if they want to use electricity?A. On the battery.B. On the scratch card.C. On the solar unit.D. In the automatic message.13.How much would users pay for the cell and scratch cards before they own a 2.5-watt solar cell?A. Around $10.B. Around $80.C. Around $90.D. Around $180.14.Which of the following statements is true according to the passage?A. Users pay for the eletricity by the hour.B. Users can get a bigger unit free of charge.C. The reference number on the scratch card is the access code to the solar unit.D. The price of the solar units may decrease in the future according to Eight19.15. What might be the most suitable title for the passage?A. Solar Energy: Starting from Scratch.B. Eight19: a creative British Company.C. Kenyan Families: Using Solar Energy for Free.D. Poor Countries: Beginning to Use Solar Energy.第二节根据短文内容,从短文后的选项中选出能填入空白处的最佳选项,并在答题卡上将该项涂黑。
2019届华附、省实、广雅三校 广州一模后联合适应性考试理科数学 2019.3.21一、选择题:本大题共8小题,每小题5分,满分40分。
在每小题给出的四个选项中,只有一项是符合题目要求的。
1.设集合{1,2,3,4},{1,2},{2,4},()U U A B C A B ===⋃=则 ( )A .}2{B .}3{DC .}4,2,1{D.}4,1{2.已知函数()12f x x =-,若3(log 0.8)a f =,131[()]2b f =,12(2)c f -=,则( )A .a b c <<B .b c a <<C .c a b <<D .a c b << 3.下列命题不正确...的是 A .如果一个平面内的一条直线垂直于另一个平面内的任意直线,则两平面垂直; B .如果一个平面内的任一条直线都平行于另一个平面,则两平面平行; C .如果两条不同的直线在一平面内的射影互相垂直,则这两条直线垂直;D .如果一条直线和一个平面平行,经过这条直线的平面和这个平面相交,那么这条直线和交线平行x( ))2222(解:(,)P m n 则222211()10am a b b m -+-=而这方程有一个根为x a =则另一个概为 从另一个角度椭圆越扁则离心率越大才有可能 典型的坐标系好题6在直三棱柱111A B C ABC -中,2BAC π∠=,11AB AC AA ===. 已知G与E分别为11A B 和1CC 的中点,D与F分别为线段AC 和AB 上的动点(不包括端点). 若GD EF ⊥,则线段DF 的长度的取值范围为A . 1⎫⎪⎭ B.1, 25⎡⎫⎪⎢⎣⎭ C. 1,⎡⎣ D. 解:建立空间直角坐标系。
1122(,0,0),(0,,0),(0,1,),(0,1,)F f D d E GB. A.由于12021,01)DG EF f d f DF d ⋅=→+=<<→=<< 7. 袋内有8个白球和2个红球,每次从中随机取出一个球,然后放回1个白球,则第4次恰好取完所有红球的概率为A. 0.0324 B .0.0434 C.0.0528 D.0.0562解:9989882121211010101010101010101010100.0434P =⨯⨯⨯+⨯⨯⨯+⨯⨯⨯=要分步。
高考模拟检测2本试卷分第Ⅰ卷(阅读题)和第Ⅱ卷(表达题)两部分。
共150分,考试时间150分钟。
第Ⅰ卷(阅读题,共70分)甲必考题一、现代文阅读(9分,每小题3分)阅读下面的文字,完成1~3题。
礼仪文化与社会主义核心价值观蒋璟萍中国礼仪文化博大精深,蕴含着丰富的思想内涵,代表着社会的价值观念。
从基本框架看,礼仪文化蕴含着善良、和谐、秩序的价值观念。
第一,追求善良的价值观念。
从个人的视角看,礼仪是一种“向善”的价值追求。
孟子认为,“辞让之心,礼之端也”,它同“恻隐之心”“羞恶之心”“是非之心”皆为“善端”,由此形成仁、义、礼、智“四德”,进而达到至善。
礼仪表现出善良的人性,雍容大度的仪态,彬彬有礼的行为,庄重诚敬的仪式,表明与人为善的态度,善良宽容的心灵。
第二,追求和谐的价值观念。
从社会的视角看,礼仪是一种“为和”的价值追求。
儒家强调,“礼之用,和为贵”。
礼仪是以建立和谐关系为目标的行为规范,礼让包含对自我的克制,对他人的理解,体现以礼待人的尊重,乐群贵和的美德。
第三,追求秩序的价值观念。
从国家的视角看,礼仪是一种“有序”的价值追求。
“礼,经国家,定社稷,序民人,利后嗣者也”,“国无礼则不宁”。
孔子以礼为基础,提出一套完整的规范体系,进而建立有条不紊的社会秩序。
礼仪是经世治国的道德秩序,人们遵循礼仪规范,各就其位,各司其职,国家就会秩序井然。
中国礼仪文化根植于传统社会,不可避免地带有封建思想的印迹。
然而,遵循“取其精华,去其糟粕”的原则,挖掘其中的合理内核,就可以提炼一些超越时空的价值观念。
将礼仪文化中蕴含的向善、为和、有序的价值观念,放进现代文化的框架下进行科学的阐释,就可以融入社会主义核心价值观,使之与友善、和谐、文明、法制等现代价值观念相联系,成为涵养核心价值观的重要源泉。
中国礼仪文化有着深刻的价值观内涵,传统礼仪文化教育也有着其独特的优势。
为此,我们应该充分发挥礼仪文化教育的优势,将核心价值观教育和礼仪文化教育有机地结合起来,以礼仪文化教育促进社会主义核心价值观教育的落实。
高考模拟检测7本试卷分第Ⅰ卷(阅读题)和第Ⅱ卷(表达题)两部分。
共150分,考试时间150分钟。
第Ⅰ卷(阅读题,共70分)甲必考题一、现代文阅读(9分,每小题3分)阅读下面的文字,完成1~3题。
民谣谚语,是人们世世代代的生活经验的高度总结与卓越智慧的长期积淀,有着极为丰富、深厚的内涵,是极富特色的传统民间文化的形式之一。
两宋时期的民间谣谚中,有大量的倡导廉洁、颂扬廉政的内容。
宋人认为:为官之要,曰廉曰勤。
岳飞更是直率地指出:文臣不爱钱,武臣不惜死,天下太平矣。
至于名相范仲淹,则其境界更高。
他提出:先天下之忧而忧,后天下之乐而乐;清心做官,莫营私利。
这些至理名言,后多成为人们的口头禅,转化为歌谣民谚。
另外,当时还有劝廉、守廉的民谣。
例如,世常羞贫贱,贫贱非我羞;人人为官莫敛财;住世一日,则做一日好人,居官一日,则做一日好事;否则尔俸尔禄,民膏民脂,下民易虐,上天难欺。
这些或诗或歌的民谣,代表了广大民众希望官场廉洁、官员廉正、不奢华、莫敛财的良好愿望。
据传,当时民歌谣谚中还广为流传着宋真宗写的《劝学诗》:“富家不用买良田,书中自有千钟粟。
安房不用架高梁,书中自有黄金屋。
娶妻莫恨无良媒,书中自有颜如玉。
出门莫恨无随人,书中车马多如簇。
男儿欲遂平生志,六经勤向窗前读。
”诗歌既有劝人勤学苦读的用意,也将读书、做官、发财之间的关系写得清楚明白。
但不论怎么说,学而优则仕,总比不学无术、只顾投机钻营和贪赃枉法要好。
因此,时人又唱道:谁不爱,黄金屋;谁不羡,千钟粟。
但他们也深知“儿孙自有儿孙福,何必为官去贪污”;且“富贵有余乐,贫贱不堪忧”。
两宋时期,大量的民谣谚语表达了人们对贪官污吏的愤恨,尤其是对北宋末年童贯、蔡京、朱勔等人的切齿之恨。
如贪官朱勔受宠,仅家奴即有百余人封官赐爵。
其中受金带者至有数十人。
当时民谣说:金腰带,银腰带,赵家世界,朱家坏!具有讽刺意味的是,这些以贪污勒索的钱财而换取的腰带到手后,又被高价转卖,从而大获其利。
2017届华附、省实、广雅三校理科数学一、选择题:本大题共8小题,每小题5分,满分40分。
在每小题给出的四个选项中,只有一项是符合题目要求的。
1.设集合{1,2,3,4},{1,2},{2,4},()U U A B C A B ===⋃=则 ( )A .}2{B .}3{DC .}4,2,1{D.}4,1{2.已知函数()12f x x =-,若3(log 0.8)a f =,131[()]2b f =,12(2)c f -=,则( )A .a b c <<B .b c a <<C .c a b <<D .a c b <<3.下列命题不正确...的是 A .如果一个平面内的一条直线垂直于另一个平面内的任意直线,则两平面垂直; B .如果一个平面内的任一条直线都平行于另一个平面,则两平面平行; C .如果两条不同的直线在一平面内的射影互相垂直,则这两条直线垂直;D .如果一条直线和一个平面平行,经过这条直线的平面和这个平面相交,那么这条直线和交线平行x1)<的图象的大致形状是 ( )5. 设A 1、A 2为椭圆)0(12222>>=+b a by a x 的左右顶点,若在椭圆上存在异于A 1、A 2的点P ,使得02=⋅PA ,其中O 为坐标原点,则椭圆的离心率e 的取值范围是( ) A 、)21,0( B 、 )22,0( C 、)1,21( D 、)1,22( 解:(,)P m n 则222211()10ama b b m -+-=而这方程有一个根为x a =则另一个概为2220ab a b a e -<<→>从另一个角度椭圆越扁则离心率越大才有可能典型的坐标系好题6在直三棱柱111A B C ABC -中,2BAC π∠=,11AB ACAA ===. 已知G与E分别为11A B 和1CC 的中点,D与F分别为线段AC和AB 上的动点(不包括端点). 若GD EF ⊥,则线段DF 的长度的取值范围为A . 1⎫⎪⎭ B.1, 25⎡⎫⎪⎢⎣⎭C. 1,⎡⎣D. 解:建立空间直角坐标系。
1122(,0,0),(0,,0),(0,1,(0,1,)F f D d E G 由于12021,01)DG EF f d f DF d ⋅=→+=<<→=<<7. 袋内有8个白球和2个红球,每次从中随机取出一个球,然后放回1个白球,则第4次恰好取完所有红球的概率为A. 0.0324 B .0.0434 C.0.0528 D.0.0562解:9989882121210.0434P =⨯⨯⨯+⨯⨯⨯+⨯⨯⨯=要分步。
8.任意a 、R b ∈,定义运算⎪⎩⎪⎨⎧>-≤⋅=*.0 , ,0, ab b a ab b a b a ,则xe x xf *=)(的A.最小值为e - B .最小值为e 1- C.最大值为e1- D.最大值为e解:xe x xf *=)(()()01x f x xe x ''==→=-此时为最小值二、填空题:本大题共7小题,每小题5分,满分30分。
本大题分为必做题和选做题两部分.(一)必做题:第9、10、11、12、13题为必做题,每道试题考生都必须作答。
9. 若框图(图1)所给程序运行的结果20102009>s ,那么 判断框中可以填入的关于k 的判断条件是_ ____. 10.已知定义域为R 的函数()f x 满足2()(2)242f x f x x x++=-+,②(1)(1)f x f x +-- 4(2)x =-,若1(1),,()2f t f t --成等差数列,则t 的值为 .解:2()(2)242f x f x x x ++=-+则1x x →-则得2(1)2f x x x +=-得2,3x =11.若对一切θ∈R ,复数(cos )(2sin )i z a a θθ=++-的模不超过2,则实数a 的取值范围为 . 解:22cos ()(2)12sin x a x a y a y a θθ=-⎧→-+-=⎨=-⎩所以得到原心到原点的距离1≤即可112.设O 点在ABC ∆内部,且有230OA OB OC ++=,则ABC ∆的面积与AOC ∆的面积的比为 .13.记集合}6,5,4,3,2,1,0{=T ,⎭⎬⎫⎩⎨⎧=∈+++=4,3,2,1,77774433221i T a a a a a M i ,将M 中的元素按从大到小顺序排列,则第2005个数是 .(二)选做题:第14、15题为选做题,考生只能选做一题,两题全答的,只计前一题的得分 14.(几何证明选讲选做题)如图,半径为2的⊙O 中,90AOB ∠=︒,D 为OB 的中点,AD 的延长线交⊙O 于点E ,则线段DE 的长为_______.15.(坐标系与参数方程选做题)曲线C 的极坐标方程2cos ρθ=,直角坐标系中的点M 的坐标为(0,2),P 为曲线C 上任意一点,则MP 的最小值是 .三、解答题:本大题共6小题,满分80分.解答须写出文字说明、证明过程和演算步骤.16.(本小题12分)已知()2312sin 2cos 3cos sin 322+⎪⎭⎫⎝⎛-+-=πωωωωx x x x x f (其中0>ω)的最小正周期为π.(1) 求()x f 的单调递增区间;(2) 在ABC ∆中,c b a ,,分别是角C B A ,,的对边,已知(),1,2,1===A f b a 求角C .17.(本小题满分12分) 在甲、乙等7个选手参加的一次演讲比赛中,采用抽签的方式随机确定每个选手的演出顺序(序号为1,2,……7),求:(1)甲、乙两个选手的演出序号至少有一个为奇数的概率; (2)甲、乙两选手之间的演讲选手个数ξ的分布列与期望.解:(2)甲、乙两选手不一定在1与2号的可以其它号只要捆扎一起26!7!(0)P ⨯=1525!57!21(1)C P ⨯⨯==即中间插入一个后把三个捆扎一起全排列即18.(本小题14分)如图2,在四面体ABOC 中,,120,,0=∠⊥⊥AOB OB OC OA OC 且.1===OC OB OA (1)设P 为AC 的中点,证明:在AB 上存在一点Q ,使OA PQ ⊥,并计算AQAB的值; (2)求二面角B AC O --的平面角的余弦值.26!7!(0)P ⨯=19.(本小题14分)在平面直角坐标系xoy 中,给定三点4(0,),(1,0),(1,0)3A B C -, 点P 到直线BC 的距离是该点到直线AB ,AC 距离的等比中项。
(Ⅰ)求点P 的轨迹方程;(Ⅱ)若直线L 经过ABC ∆的内心(设为D ),且与P 点的轨迹恰好有3个公共点,求L 的斜率k 的取值范围。
20.(本小题14分)已知,αβ是方程24410()x tx t R --=∈的两个不等实根,函数22()1x tf x x -=+的定义域为[],αβ。
(Ⅰ)求()max ()min ()g t f x f x =-; (Ⅱ)证明:对于(0,)(1,2,3)2i u i π∈=,若123sin sin sin 1,u u u ++=123111(tan )(tan )(tan )g u g u g u ++<则。
APBCO图221.(本小题14分)(I )已知数列{}n a 满足111,2n n a a a n +==+()1,2,3n =L ,{}n b 满足11b =,21n n n b b b n +=+ ()1,2,3n =L,求证:1112n k =≤<。
. (II) 已知数列{}n a 满足:a 1=1且)2(213221≥=---n a a n n n 。
设m ∈N +,m ≥n ≥2,证明(a n +n 21)1(m-n+1)≤mm 12-2017届华附、省实、广雅三校 广州一模后联合适应性考试数学试题(理科)参考答案和评分标准一、选择题:(每题5分,共40分)9.2010<k 10.2或3 11.⎡⎢⎣⎦ 12.3 13.396240114 15. 1三、解答题:本大题共6小题,满分80分,解答须写出文字说明、证明过程或演算步骤.16.解:(1)())3sin 1cos 21cos 222122f x x x x πωωω⎛⎫=-++--+ ⎪⎝⎭162cos 2cos 232sin 23+⎪⎭⎫ ⎝⎛---=πωωωx x x 132sin 2+⎪⎭⎫ ⎝⎛-=πωx …………2分1,22,0,===∴>=ωπωπωπT T()132sin 2+⎪⎭⎫ ⎝⎛-=∴πx x f …………4分故递增区间为Z k k k ∈⎥⎦⎤⎢⎣⎡+-125,12ππππ …………6分 (2)()1132sin 2=+⎪⎭⎫⎝⎛-=πA A f 032sin =⎪⎭⎫ ⎝⎛-∴πA52333A πππ-<-<Q πππ=-=-∴32A 或032A即6A π=或32A π=又,,B A b a <∴<故32A π=舍去,∴6A π=. …………9分由B b A a sin sin =得,22sin =B 4π=∴B 或43π=B , 若4π=B ,则127π=C . 若43π=B ,则12π=C . …………12分注意:没有说明 "52333A πππ-<-<Q "扣两分17.解:(1)设A 表示“甲、乙的演出序号至少有一个为奇数”,则A 表示 “甲、乙的演出序号均为偶数”.由等可能性事件的概率计算公式得()()76112723=-=-=C C A P A P .…………4分(2)ξ的可能取值为5,4,3,2,1,0,…………5分(),726027===C P ξ (),2155127===C P ξ (),2144227===C P ξ (),2133327===C P ξ (),2122427===C P ξ ().2111527===C P ξ…………8分从而ξ的分布列为…………10分所以,3521152124213321422151720=⨯+⨯+⨯+⨯+⨯+⨯=ξE . …………12分18.解法一:(1)在平面OAB 内作ON OA ⊥交AB 于N ,连接NC .…………1分 又OA OC ⊥, OA ONC ∴⊥平面N C O N C ⊂ 平面, O AN C ∴⊥。
取Q 为AN 的中点,则NC PQ // P Q O A ∴⊥ …………4分 在等腰AOB ∆中,120AOB ∠=, 30OAB OBA ∴∠=∠=在AON Rt ∆中, 30OAN ∠=, 12ON AN AQ ∴==…………4分 在ONB ∆中, 1209030NOB NBO ∠=-==∠,.N B O N A Q ∴== …………5分3ABAQ ∴=…………8分 (2)连接 PNPO ,由OC OA ⊥,OC OB ⊥知:OC OAB ⊥平面. 又OAB ON 平面⊂, OC ON ∴⊥又由ON OA ⊥,ON AOC ⊥平面. 又 AO C AC 平面⊂,AC ON ⊥∴ 又 P 是AC 的中点,OC OA =∴O O N O P O P AC =⋂⊥,,PO N AC 平面⊥∴,PO N PN 平面⊂,PN AC ⊥∴O P N ∴∠为二面角O AC B --的平面角 …………10分在等腰COA Rt ∆中,1OC OA ==,2OP ∴=在AON Rt ∆中,tan 30ON OA ==∴在PON Rt ∆中, PN ==. …………12分cos PO OPN PN ∴∠===…………14分 解法二:在平面AOB 中,过点O ,作OA ON ⊥交AB 于N ,取O 为坐标原点,分别以OA ,ON ,OC 所在的直线为x 轴,y 轴,z 轴,建立空间直角坐标系O xyz - (如图所示) …………1分则1(1,0,0),(0,0,1),(2A C B - P 为AC 中点,11(,0,)22P ∴ …………2分设 ((0,1)),AQ AB λλ=∈3(2AB =- .131(,).222PQ OQ OP λ∴=-=--33(1,0,0)((1,0),22OQ OA AQ λλ∴=+=+-=- N,PQ OA ⊥ 0=⋅∴ 即13022λ-=,13λ=. …………6分所以存在点1(2Q 使得 PQ OA ⊥ 且3AB AQ =. …………8分 (2)记平面ABC 的法向量为123(,,)n n n n =,则由n CA ⊥ ,n AB ⊥ ,且(1,0,1)CA =-,得13230302n n n n -=⎧⎪⎨-=⎪⎩, 故可取n =() …………10分又平面OAC 的法向量为 (0,1,0)e =. …………11分cos ,n e ∴≥=…………13分 二面角O AC B --的平面角是锐角,记为θ,则cos θ=…………14分19.解:(Ⅰ)直线AB 、AC 、BC 的方程依次为44(1),(1),033y x y x y =+=--=。