【2015红桥二模 全科】天津市红桥区2015届高三第二次模拟考试 扫描版试题Word版答案
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高三政治答案(2014、05)1.C2.D3.B4.C5.B6.A7.A8.B9.D 10.C 11.A12.(1)上表反映了最近10年来,我国农民收入实现了大幅度提高,收入结构明显改善,农民收入来源日益多元化,农民家庭经营性收入占比下降,工资性收入和转移性收入占比增加较多。
(6分)通过发展农业生产,调整农村经济结构,支持农村二、三产业,努力增加农民经营性收入;通过引导农民进城务工,保护农民工合法权益,努力增加农民工资性收入;通过完善市场体系,保障农民在土地流转、产权改革等过程中的财产权益,拓宽投资渠道,努力增加农民财产性收入;通过加大财政支持力度,提高转移支付比例,完善农村社保和农业生产支持体系,努力增加农民转移性收入。
(8分)(2)(10分)新自由主义的主张 对我国深化农村改革的借鉴之处反对政府过多干预,主张政府发挥市场决定作用,更好发挥政府作用作用最小化、市场自由化反对公有制,主张私有化 坚持和完善农村基本经营制度,坚持农村土地集体所有,鼓励多种所有制经济参与农村改革13.(1)这是由党和政府的性质、宗旨决定的;(4分)这是化解当前社会矛盾、解决形式主义、官僚主义、享乐主义和奢靡之风等作风建设中存在的突出问题的需;(4分)弘扬“马上就办”工作作风,有利于提高行政效能和服务水平,提高坚持党的群众观点、贯彻党的群众路线的自觉性。
(2分)(2)①改革是社会主义制度的自我完善和发展,当上层建筑适合经济基础状况时,它会促进经济基础的巩固和完善,当它不适合经济基础状况时,会阻碍经济基础的发展和变革。
(4分)②价值判断往往会因人而异,人们认识事物的角度不同,对其评价也会不同。
我们想事情、做工作,想的对不对,做得好不好,要以人民拥护不拥护、人民赞成不赞成为标准。
(4分)(从意识自觉选择性角度分析也可酌情给分)③实践是检验认识正确与否的唯一标准。
是否应该成立“马上就办办公室”及其效果要靠最终的实践来检验。
(4分) 14.(1)答案示例:认为将“继续恶化”并不可收拾:量变引起质变,中日关系持续恶化,民众相互缺乏好感,最终将引起两国关系的“不可收拾”。
高三数学(理)答案(2014、05)一、选择题:本大题共8小题,每小题5分,满分40分.二、填空题:本大题共6小题,每小题5分,满分30分. 9.45-10.4 11.(1,+∞) 12.9 13.4 14.2700 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. 15.(本小题满分13分) (Ⅰ)∵cos cos A b B a =,由正弦定理得cos sin cos sin A BB A=,即sin 2sin 2A B = ………2 ∴A B =或2A B π+=(舍去),23C π∠=,则6A B π== …………..4 (Ⅱ)x x A x x f 22cos sin )2sin()(+-+=)32sin(3π+=x (8)πωπ==2T (10)∵∈x ]6,12[ππ-,则32326πππ≤+≤x ………………………….11 而正弦函数sin y x =在[,]62ππ上单调递增,在2[,]23ππ上单调递减∴函数()f x ,即函数()f x 在[,]62ππ上的值域为. (13)16.(本小题满分13分) (Ⅰ)玩具A 为正品的概率约为4032841005++=. (1)玩具B 为正品的概率约为4029631004++=. ………………2 (Ⅱ)解:(ⅰ)随机变量X 的所有取值为90,45,30,15-. ………………3 433(90)545P X ==⨯=; 133(45)5420P X ==⨯=; 411(30)545P X ==⨯=; 111(15)5420P X =-=⨯=. (7)所以,随机变量X 的分布列为: (8)3311904530(15)66520520EX =⨯+⨯+⨯+-⨯=. (10)(ⅱ)设生产的5件玩具B 中正品有n 件,则次品有5n -件. 依题意,得 5010(5)140n n --≥, 解得 196n ≥. 所以 4n =,或5n =. ………………11 设“生产5件玩具B 所获得的利润不少于140元”为事件A , 则 445531381()C ()()444128P A =⨯+=.………………13 17.(本小题满分13分) 解法一:因为 90PAD ∠=︒, 所以PA AD ⊥.又因为侧面PAD ⊥底面ABCD , 且侧面PAD底面ABCD AD =,所以 PA ⊥底面ABCD . 又因为90BAD ∠=︒,所以AB ,AD ,AP 两两垂直. ……………………………………1 分别以AB ,AD ,AP 为x 轴,y 轴,z 轴建立空间直角坐标系,如图.设2AD =,则(0,0,0)A ,(1,0,0)B ,(1,1,0)C ,(0,2,0)D ,(0,0,1)P . (Ⅰ)(0,0,1)AP =,(1,1,0)AC =,(1,1,0)CD =-,所以 0AP CD ⋅=,0AC CD ⋅=,所以AP ⊥CD ,AC ⊥CD . 又因为APAC A =, 所以CD ⊥平面PAC . (4)(Ⅱ)在PA 上存在中点E ,使得//BE 平面PCD证明如下:侧棱PA 的中点是E ,则1(0, 0, )2E ,1(1, 0, )2BE =-.设平面PCD 的一个法向量是(,,)x y z =n ,则0,0.CD PD ⎧⋅=⎪⎨⋅=⎪⎩n n因为(1, 1, 0)CD =-,(0, 2,1)PD =-,所以0,20.x y y z -+=⎧⎨-=⎩ 取1x =,则(1, 1, 2)=n .所以1(1, 1, 2)(1, 0, )02BE ⋅=⋅-=n , 所以BE ⊥n .因为BE ⊄平面PCD ,所以//BE 平面PCD . ………………………………8 (Ⅲ)由已知,AB ⊥平面PAD ,所以(1, 0, 0)AB =为平面PAD 的一个法向量.由(Ⅱ)知,(1, 1, 2)=n 为平面PCD 的一个法向量. 设二面角A PD C --的大小为θ,即二面角A PD C -- ………………………………13 解法二:(Ⅰ)因为 90PAD ∠=︒,所以PA AD ⊥.又因为侧面PAD ⊥底面ABCD ,且侧面PAD底面ABCD AD =,所以PA ⊥底面ABCD .而CD ⊂底面ABCD ,所以PA ⊥CD . 在底面ABCD 中,因为90ABC BAD ∠=∠=︒,12AB BC AD ==, 所以 AC CD AD ==,∴222AD CD AC =+ ∴AC ⊥CD . 又因为PAAC A =, 所以CD ⊥平面PAC . (4)(Ⅱ)在PA 上存在中点E ,使得//BE 平面PCD ,证明如下:取PD 的中点F , 连结BE ,EF ,FC , 则//EF AD ,且12EF AD =. 由已知90ABC BAD ∠=∠=︒, 所以//BC AD . 又12BC AD =, 所以//BC EF ,且BC EF =,所以四边形BEFC 为平行四边形,所以//BE CF . 因为BE ⊄平面PCD ,CF ⊂平面PCD ,所以//BE 平面PCD . ……………8 (Ⅲ)取AD 中点G ,连结CG ,则 CG ⊥AD .又因为平面ABCD ⊥平面PAD , 所以 CG ⊥平面PAD . 过G 作GH PD ⊥于H , 连结CH ,∴CH PD ⊥.所以GHC ∠是二面角A PD C --的平面角.设2AD =,则1PA AB CG DG ====, DP =. 在PAD ∆中,GH DG PA DP =,所以GH =. 所以tan CG GHC GH ∠==,cos GHC ∠= 即二面角A PD C --………………………………13 18.(本小题满分13分)(Ⅰ)由已知得2b =2a =1d +, 3b =5a 14d =+, 2b =14a 113d =+, (1)由于{}n b 为等比数列,所以2324b b b =⋅.∴2(14)d +=(1)(113)d d ++, 0,2d d >∴=. (2)∴21n a n =- . (3)又2b =2a =3,3b = 5a =9 , (4)∴数列{n b }的公比为3, ..................5 ∴n b =3⋅23n -=13n -. (6)(Ⅱ)由11c b +22c b +…+n nc b =1n a + , (1) 当1n =时,11c b =2a =3, ∴1c =3. (7)当1n >时,11c b +22c b +…+11n n c b --= n a , (2) 由(1)-(2)得nnc b =1n a +-n a =2 , (9)∴n c =2n b =2⋅13n -,(2)n ≥ ……………10 ∴n c =13,123,2n n n -=⎧⎨⋅≥⎩ ..................11 ∴2014321......c c c c ++++=3+2⋅3+2⋅23+...+2⋅20133 (12)=1+2⋅03+2⋅3+2⋅23+…+2⋅20133=1+2⋅31312014--=20143 (13)19.(本小题满分14分)(Ⅰ)由题意可设椭圆C 的方程为22221(0)x y a b a b+=>>,(,0)F c .由题意知解得b =,1c =. (3)故椭圆C 的方程为22143x y +=,离心率为12. (5)(Ⅱ)以BD 为直径的圆与直线PF 相切.证明如下:由题意可设直线AP 的方程为(2)y k x =+(0)k ≠. (6)则点D 坐标为(2, 4)k ,BD 中点E 的坐标为(2, 2)k . (7)由22(2),143y k x x y =+⎧⎪⎨+=⎪⎩得2222(34)1616120k x k x k +++-=. (8)设点P 的坐标为00(,)x y ,则2021612234k x k--=+. 所以2026834k x k -=+,00212(2)34k y k x k =+=+. ……………………………10 因为点F 坐标为(1, 0), 当12k =±时,点P 的坐标为3(1, )2±,点D 的坐标为(2, 2)±. 直线PF x ⊥轴,此时以BD 为直径的圆22(2)(1)1x y -+=与直线PF 相切. (11)⎧⎪⎨⎪⎩2221222, .a b a a b c ⋅⋅===+当12k ≠±时,则直线PF 的斜率0204114PF y k k x k==--. 所以直线PF 的方程为24(1)14ky x k=--. 点E 到直线PF 的距离d 2||k . 又因为||4||BD k = ,所以1||2d BD =. 故以BD 为直径的圆与直线PF 相切.综上得,当直线AP 绕点A 转动时,以BD 为直径的圆与直线PF 相切.………14 20.(本小题满分14分)(Ⅰ)当5=a 时,由xe x x x g )35()(2-+-=得,e g =)1( (1)x e x x x g )23()('2++-=,故切线斜率为e g 4)1('= (2)所以切线方程为:e ex y 34-= .................................4 (Ⅱ)根据题意m 大于)(x f 在]2,[+t t 上的最小值即可. (5)1ln )('+=x x f (6) (7)①当et 1≥时,在区间]2,[+t t 上)(x f 为增函数, 所以t t t f x f ln )()(min == (8)②当e t 10<<时,在区间)1,(e t 上)(x f 为减函数,在区间),1(e e上)(x f 为增函数 所以e e f x f 1)1()(min -== (9)综上,当e t 1≥时,t t m ln >;e t 10<<时,em 1->(Ⅲ)由)(2)(x f e x g x =得,3ln 22-+-=ax x x x ,xx x a 3ln 2++= (10)第 11 页 共 11 页 令)(x h x x x 3ln 2++=,22)1)(3(321)('x x x x x x h -+=-+= (12)231)1(-+=e e e h ,4)1(=h ,23)(++=e ee h 0224)1()(<+-=-ee e h e h …………………………………………………………13 ee a 324++≤<∴ (14)。
1-11题,每题4分,共44分。
DCCDA DDBBB A12.(16分)(1)旅游、文化等(一点即可得2分)(2)传承民族文化和文明(2分)(3)丰富了人文旅游景点(提高了人文旅游资源的开发价值);突出了人文旅游景观特色(增强了人文旅游景点的吸引力);增加了游客(增加旅游业的收入)。
(6分)(4)(6分)13.(18分)(1)③(2分)(2)甲(2分)南北纬30°-40°的大陆西岸(2分)甲地气温年较差比乙地小(2分)甲地夏季降水量比乙地少(甲地降水季节变化比乙地大)(2分)(3)沟通大西洋东岸地区与太平洋西岸地区(亚欧大陆东西岸)之间的联系(2分)(4)①地理位置优越,交通便捷;②经济发达,商贸往来频繁;③历史悠久,旅游资源丰富;④人口众多,城市规模大。
(答出任三项即可得6分)14.(22分)(1)低山丘陵(2分)(2)在地转偏向力的作用下,海浪向北偏折,使北岸易于遭受侵蚀;夏秋季节,盛行东南风,北岸成为迎风岸,也易遭受海浪的冲刷。
(4分)(3)沿高速公路、铁路等交通线分布,沿城市分布。
(4分)(4)自西南向东北(2分)水量大,水位季节变化大,无结冰期,汛期长等。
(4分)(5)地处我国东部沿海;滨海平原面积较大;气候温暖湿润。
海岸线曲折,多港湾;大陆架宽阔,海洋资源(海洋生物资源、海底矿产资源、海洋旅游资源)丰富。
(三点即可得6分高三历史答案Ⅱ卷共3题,共56分。
12.(18分)(1)中国丝绸制作精良,质量上乘,中国丝绸深受古罗马人的喜爱追捧;运输的曲折艰难。
(3分)(2)状况:长期居于领先地位;近代以来虽有下降,但到1870年仍居世界第一位。
(2分)领先原因:中国古代农业、手工业发达,长期领先世界并远销海外;新航路的开辟和殖民扩张使世界市场出现并得到拓展。
(3分)下降原因:鸦片战争以后,西方列强的经济侵略,瓦解着中国的自然经济,客观促进商品经济的发展,手工业品和农产品在商品流通中的比重加大,国内外市场不断扩大,到1870年虽有下降仍居世界第一。
2015年天津市红桥区高考化学二模试卷一、选择题(共6小题,每小题6分,满分36分)1.(6分)化学与生产、生活密切相关,下列说法错误的是()A.铝需经过特别处理才具有抗腐蚀能力B.洁厕剂和漂白精不能混合使用,否则易产生有毒气体C.Li是最轻的金属也是活动性很强的金属,是制造电池的理想物质D.只要符合限量,“食用色素”、“亚硝酸盐”可以作为某些食品的添加剂2.(6分)解释下列操作或事实的离子方程式正确的是()A.把钠投入水中:2Na+2H2O═2Na++2OH﹣+O2↑B.把铜片放入氯化铁溶液中:Cu+Fe3+═Cu2++Fe2+C.铝制容器不能长期存放碱性食物:2Al+2OH﹣═2AlO2﹣+H2↑D.向海带灰浸出液中加入硫酸、双氧水:2I﹣+2H++H2O2═I2+2H2O3.(6分)在实验室进行物质制备,下列从原料及试剂分别制取相应的最终产物的设计中,理论上正确、操作上可行、经济上合理的是(少量杂质忽略不计)()A.C CO CO2Na2CO3B.CaO Ca(OH)2溶液NaOH溶液C.Fe Fe2(SO4)3溶液D.Cu Cu(NO3)2溶液Cu(OH)24.(6分)双酚A作为食品、饮料包装和奶瓶等塑料制品的添加剂,能导致人体内分泌失调,对儿童的健康危害更大.下列有关双酚A的叙述不正确的是()A.双酚A的分子式是C15H16O2B.双酚A的核磁共振氢谱显示氢原子数之比是1:2:2:3C.反应①中,1mol双酚A最多消耗2 mol Br2D.反应②的产物中只有一种官能团5.(6分)常温下,下列表述正确的是()A.pH=2的CH3COOH溶液的浓度小于pH=2的盐酸的浓度B.向NaHC03溶液中加入少量NaOH固体,可抑制HCO3﹣的水解,使c(HCO﹣)增大C.Na2C03溶液中:即c(Na+)>C(CO32﹣)>C(OH﹣)=C(HCO3﹣)>C(H+)D.将任意量的CH3COOH溶液与NaOH溶液混合,其混合溶液均存在:C(Na+)+c(H+)=C(CH3COO﹣)+C(OH﹣)6.(6分)常温下,将Cl2缓慢通入水中至饱和,然后向所得饱和氯水中滴加0.1mol•L﹣1的NaOH溶液.整个实验进程中溶液的pH变化曲线如图所示,下列叙述正确的是()A.实验进程中可用pH试纸测定溶液的pHB.c点所示溶液中:c(Na+)=2c(ClO﹣)+c(HClO)C.向a点所示溶液中通入SO2,溶液的酸性和漂白性均增强D.由a点到b点的过程中,溶液中减小二、非选择题(共64分)7.己知X、Y、Z、W为短周期四种常见元素,其中X是原子半径最小的元素,Y原子的最外层电子数是内层电子数二倍,Z能和X形成两种常见的液态化合物,W元素的单质既能与盐酸反应也能与NaOH溶液反应.元素Q是地壳中含量第二位的金属元素.试回答下列问题:(1)请写出Y元素在周期表中的位置:.(2)请写出X2Z2的电子式.(3)请写出Y元素的最高价氧化物与少量NaOH溶液反应的离子方程式.A、B、C、D、E五种物质(或离子)均含有同一种元索,它们之间有如图所示的转化关系:(4)若A为元素W的单质,B、C均属于盐类,D是一种白色沉淀.若B水溶液呈酸性,C水溶液呈碱性,则B溶液呈酸性的原因是(用离子方程式表示)(5)若A为元素Q的单质,B、C均属于盐中的阳离子,若B溶液呈黄色,与硫氰化钾溶液混合后显红色,D、E均为沉淀,且D在空气中能转化成E,B→C的离子方程式为.D→E的化学方程式为.选用氢碘酸与E反应实现E→C的转化,该反应的离子方程式为.8.A是一种重要的食用香料.存在于玉米、葡萄等多种食品中.主要用于配制奶油、乳品、酸奶和草莓等香精,也可用于有机合成.以A为原料制备阴离子树脂M和新型聚酯材料N 等的合成路线如图:已知:Ⅰ.RCHO+OHCR′Ⅱ.R﹣CH═CHOH不稳定,很快转化为R﹣CH2﹣CHO(1)反应Ⅰ的反应类型是.(2)B中官能团的名称是.(3)D与新制Cu(OH)2反应的化学方程式是.(4)下列关于E的说法正确的是.a.不溶于水b.能与H2发生还原反应c.能被重铬酸钾酸性溶液氧化d.与NaOH醇溶液共热发生消去反应(5)M的结构简式是.(6)反应Ⅱ的化学方程式是.(7)A的属于酯类的同分异构体有种,其中只含一个甲基的同分异构的结构简式是:.9.焦亚硫酸钠(Na2S2O5)是一种常用的抗氧化剂.某研究小组对焦亚硫酸钠进行如下研究:(1)采用如图装置(实验前已除尽装置内的空气)制取Na2S2O5.装置Ⅱ中有Na2S2O5析出,发生的化学反应方程式为:Na2SO3+SO2=Na2S2O5.①装置Ⅰ中产生气体的化学方程式为.②要从装置Ⅱ中获得已析出的晶体,可采取的分离方法是.③装置Ⅲ用于处理尾气,可选用的最合理装置(夹持仪器已略去)为(填序号).(2)【查阅资料】Na2S2O5溶于水即生成NaHSO3.①NaHSO3溶液显酸性.请用化学用语和必要的文字解释原因;证明该结论可采用的实验方法是(填序号).a.测定溶液的pH b.加入Ba(OH)2溶液c.加入盐酸d.加入品红溶液e.用蓝色石蕊试纸检测②检验Na2S2O5晶体在空气中已被氧化的实验方案是.(3)焦亚硫酸钠(Na2S2O5)在酸性条件下可将工业废水中的Cr2O72﹣还原为Cr3+.①写出该反应的离子方程式.②若处理Cr2O72﹣浓度为1×10﹣3mol/L的工业废水1L,需用Na2S2O5固体mg.10.氮的氢化物NH3、N2H4等在工农业生产、航空航天等领域有广泛应用.(1)液氨作为一种潜在的清洁汽车燃料已越来越被研究人员重视.它在安全性、价格等方面较化石燃料和氢燃料有着较大的优势.氨的燃烧实验涉及下列两个相关的反应:①4NH3(g)+5O2(g)=4NO(g)+6H2O(l)△H1②4NH3(g)+6NO(g)=5N2(g)+6H2O(l)△H2则反应4NH3(g)+3O2(g)=2N2(g)+6H2O(l)△H=.(请用含有△H1、△H2的式子表示)(2)合成氨实验中,在体积为3L的恒容密闭容器中,投入4mol N2和9mol H2在一定条件222mol NH3(g)中的化学键消耗的能量.①则T1T2(填“>”、“<”或“=”)②在T2K下,经过10min达到化学平衡状态,则0~10min内H2的平均速率v(H2)=,平衡时N2的转化率α(N2)=.③下列图象1分别代表焓变(△H)、混合气体平均相对分子质量()、N2体积分数φ(N2)和气体密度(ρ)与反应时间的关系,其中正确且能表明该可逆反应达到平衡状态的是.(3)某N2H4(肼或联氨)燃料电池(产生稳定、无污染的物质)原理如图2所示.①M区发生的电极反应式为.②用上述电池做电源,用图3装置电解饱和氯化钾溶液(电极均为惰性电极),设饱和氯化钾溶液体积为500mL,当溶液的pH值变为13时(在常温下测定),若该燃料电池的能量利用率为80%,则需消耗N2H4的质量为g(假设溶液电解前后体积不变).2015年天津市红桥区高考化学二模试卷参考答案与试题解析一、选择题(共6小题,每小题6分,满分36分)1.(6分)(2015•红桥区二模)化学与生产、生活密切相关,下列说法错误的是()A.铝需经过特别处理才具有抗腐蚀能力B.洁厕剂和漂白精不能混合使用,否则易产生有毒气体C.Li是最轻的金属也是活动性很强的金属,是制造电池的理想物质D.只要符合限量,“食用色素”、“亚硝酸盐”可以作为某些食品的添加剂【分析】A.常温下,铝在空气中与氧气反应,其表面生成的氧化铝薄膜具有抗腐蚀能力;B、次氯酸钠、浓盐酸相遇会产生有毒气体氯气;C、Li密度小,较活泼,单位质量提供的电能较大;D、正确使用食品添加剂对人体健康有益.【解答】解:A.常温下,铝在空气中与氧气反应,其表面生成一层致密的氧化铝薄膜,具有抗腐蚀能力,从而防止铝进一步氧化,故A错误;B、漂白精的主要成分为次氯酸钠,洁厕精的主要成分是浓盐酸,两者相遇会产生有毒气体氯气,故B正确;C、Li密度小,位于周期表第ⅠA族,较活泼,且单位质量提供的电能较大,是制造电池的理想物质,故C正确;D、正确使用食品添加剂对人体健康有益,只要符合限量,“食用色素”、“亚硝酸盐”可以作为某些食品的添加剂,故D正确.故选A.【点评】本题考查化学与科学、技术、社会、环境等问题,题目难度不大,侧重对基础知识的巩固,注意对基础知识的理解掌握.2.(6分)(2015•红桥区二模)解释下列操作或事实的离子方程式正确的是()A.把钠投入水中:2Na+2H2O═2Na++2OH﹣+O2↑B.把铜片放入氯化铁溶液中:Cu+Fe3+═Cu2++Fe2+C.铝制容器不能长期存放碱性食物:2Al+2OH﹣═2AlO2﹣+H2↑D.向海带灰浸出液中加入硫酸、双氧水:2I﹣+2H++H2O2═I2+2H2O【分析】A.不符合反应客观事实;B.电荷不守恒;C.原子个数不守恒;D.双氧水具有强的氧化性,能够氧化碘离子.【解答】解:A.钠与水反应生成氢氧化钠和氢气,离子方程式:2Na+2H2O═2Na++2OH﹣+H2↑,故A错误;B.把铜片放入氯化铁溶液中,离子方程式:Cu+2Fe3+═Cu2++2Fe2+,故B错误;C.铝能够与碱性物质发生反应,离子方程式:2Al+2OH﹣+2H2O=2AlO2﹣+3H2↑,故C错误;D.向海带灰浸出液中加入硫酸、双氧水,发生氧化还原反应,离子方程式:2I﹣+2H++H2O2═I2+2H2O,故D正确;故选:D.【点评】本题考查了离子方程式的书写,明确反应实质是解题关键,注意离子方程式书写应遵循客观事实、遵循原子个数、电荷数守恒规律,题目难度不大.3.(6分)(2015•红桥区二模)在实验室进行物质制备,下列从原料及试剂分别制取相应的最终产物的设计中,理论上正确、操作上可行、经济上合理的是(少量杂质忽略不计)()A.C CO CO2Na2CO3B.CaO Ca(OH)2溶液NaOH溶液C.Fe Fe2(SO4)3溶液D.Cu Cu(NO3)2溶液Cu(OH)2【分析】A.一氧化碳是有毒的气体,碳在氧气中燃烧生成的进行实验时要进行尾气处理,且一氧化碳是可燃性气体,不纯时加热或点燃引起爆炸;B.理论上正确,操作较为简便,利用碳酸钠获得氢氧化钠,经济上也合理;C.铁在氧气中燃烧生成四氧化三铁;D.银比铜要贵重,经济上不合理.【解答】解:A.碳不完全燃烧可生成一氧化碳,一氧化碳高温下与氧化铜反应生成铜和二氧化碳,二氧化碳与氢氧化钠反应生成碳酸钠;理论上正确,一氧化碳是有毒的气体,碳在氧气中燃烧生成的进行实验时要进行尾气处理,且一氧化碳是可燃性气体,不纯时加热或点燃引起爆炸,操作上较为复杂,故A错误;B.氧化钙与水反应可生成Ca(OH)2,氢氧化钙与碳酸钠反应可生成氢氧化钠;理论上正确,操作也只需要加入液体较为简便,利用碳酸钠获得氢氧化钠,经济上也合理,故B正确;C.铁在氧气中燃烧生成四氧化三铁而不能生成三氧化二铁,因此该方案在理论上就是错误的,故C错误;D.铜与硝酸银反应可生成硝酸铜,硝酸铜与氢氧化钠可生成氢氧化铜沉淀;理论上正确,操作上也较简便,银比铜要贵重,利用硝酸银、氢氧化钠制取氢氧化铜,经济上不合理,故D错误;故选B.【点评】本题考查实验制备方案设计与评价、绿色化学等知识,题目难度中等,是理论知识联系生活实际的体现、理论知识对实际的指导;注意实验方案设计必须满足理论正确、操作简单可行、经济节约、绿色环保.4.(6分)(2015•红桥区二模)双酚A作为食品、饮料包装和奶瓶等塑料制品的添加剂,能导致人体内分泌失调,对儿童的健康危害更大.下列有关双酚A的叙述不正确的是()A.双酚A的分子式是C15H16O2B.双酚A的核磁共振氢谱显示氢原子数之比是1:2:2:3C.反应①中,1mol双酚A最多消耗2 mol Br2D.反应②的产物中只有一种官能团【分析】双酚A中含有酚羟基,且﹣OH邻位有H原子,可与溴发生取代反应,含有苯环,可发生加成反应,根据有机物结构简式判断含有的元素种类和原子个数,可确定分子式,结合结构的对称性判断含有的H原子种类和个数.【解答】解:A.有结构简式可知分子中含有15个C原子、16个H原子、2个O原子,则分子式为C15H16O2,故A正确;B.两个甲基相同,两个苯酚基相同,根据苯环的对称可知有4种不同的H,H原子数之比是2:4:4:6=1:2:2:3,故B正确;C.双酚A中,2个酚羟基有4个邻位H可被取代,则1mol双酚A最多消耗4molBr2,故C错误;D.加成反应后,产物的官能团只有﹣OH,故D正确.故选C.【点评】本题考查有机物的结构和性质,题目难度不大,本题注意把握有机物官能团的性质,为解答该题的关键,易错点为B,注意H原子种类的判断.5.(6分)(2015•红桥区二模)常温下,下列表述正确的是()A.pH=2的CH3COOH溶液的浓度小于pH=2的盐酸的浓度B.向NaHC03溶液中加入少量NaOH固体,可抑制HCO3﹣的水解,使c(HCO﹣)增大C.Na2C03溶液中:即c(Na+)>C(CO32﹣)>C(OH﹣)=C(HCO3﹣)>C(H+)D.将任意量的CH3COOH溶液与NaOH溶液混合,其混合溶液均存在:C(Na+)+c(H+)=C(CH3COO﹣)+C(OH﹣)【分析】A、醋酸是弱电解质而盐酸是强电解质完全电离,所以氢离子浓度相同的两溶液醋酸的浓度大;B、向NaHC03溶液中加入少量NaOH固体,以促进碳酸氢根离子的电离;C、应考虑碳酸氢根离子继续水解生成氢氧根离子并且水电离产生氢氧根离子;D、电荷守恒分析解答.【解答】解:A、醋酸是弱电解质而盐酸是强电解质完全电离,所以氢离子浓度相同的两溶液醋酸的浓度大,故A错误;B、向NaHC03溶液中加入少量NaOH固体,以促进碳酸氢根离子的电离,所以c(HCO﹣)减小,故B错误;C、应考虑碳酸氢根离子继续水解生成氢氧根离子并且水电离产生氢氧根离子,所以离子浓度大小为:c(Na+)>C(CO32﹣)>C(OH﹣)>C(HCO3﹣)>C(H+),故C错误;D、电荷守恒可知:混合溶液均存在:C(Na+)+c(H+)=C(CH3COO﹣)+C(OH﹣),故D 正确;故选D.【点评】本题考查离子浓度的大小,学生要学会影响盐的水解平衡移动的因素以及离子浓度大小比较的方法等知识,难度不大.6.(6分)(2015•红桥区二模)常温下,将Cl2缓慢通入水中至饱和,然后向所得饱和氯水中滴加0.1mol•L﹣1的NaOH溶液.整个实验进程中溶液的pH变化曲线如图所示,下列叙述正确的是()A.实验进程中可用pH试纸测定溶液的pHB.c点所示溶液中:c(Na+)=2c(ClO﹣)+c(HClO)C.向a点所示溶液中通入SO2,溶液的酸性和漂白性均增强D.由a点到b点的过程中,溶液中减小【分析】整个过程发生的反应为Cl2+H2O═HCl+HClO、HCl+NaOH═NaCl+H2O、HClO+NaOH ═NaClO+H2O,根据溶液的酸碱性结合电荷守恒和物料守恒分析解答.【解答】解:A、溶液中含有次氯酸,具有漂白性,不能用pH试纸测量,故A错误;B、c点溶液呈中性,则c(H+)=c(OH﹣),根据电荷守恒得c(H+)+c(Na+)=c(Cl﹣)+c(Cl0﹣)+c(OH﹣),所以c(Na+)=c(Cl﹣)+c(ClO﹣),根据物料守恒得c(Cl﹣)=c (ClO﹣)+c(HClO),所以c(Na+)=c(HClO)+2c(ClO﹣),故B正确;C、氯气具有氧化性,能将二氧化硫氧化为硫酸,自身被还原为盐酸,所得到的溶液不再具有漂白性,故C错误;D、由a点到b点的过程是氯气的溶解平衡Cl2+H2O⇌H++Cl﹣+HClO向右进行的过程,酸性逐渐增强,氢离子浓度逐渐增大,溶液中增大,故D错误.故选B.【点评】本题综合考查氯气的性质,侧重于学生的分析能力的考查,注意理解溶液中的溶质及其性质是解本题关键,根据物料守恒得c(Cl﹣)=c(ClO﹣)+c(HClO),为易错点,难度中等.二、非选择题(共64分)7.(2015•红桥区二模)己知X、Y、Z、W为短周期四种常见元素,其中X是原子半径最小的元素,Y原子的最外层电子数是内层电子数二倍,Z能和X形成两种常见的液态化合物,W元素的单质既能与盐酸反应也能与NaOH溶液反应.元素Q是地壳中含量第二位的金属元素.试回答下列问题:(1)请写出Y元素在周期表中的位置:第二周期ⅣA族.(2)请写出X2Z2的电子式.(3)请写出Y元素的最高价氧化物与少量NaOH溶液反应的离子方程式CO2+OH﹣=HCO3﹣.A、B、C、D、E五种物质(或离子)均含有同一种元索,它们之间有如图所示的转化关系:(4)若A为元素W的单质,B、C均属于盐类,D是一种白色沉淀.若B水溶液呈酸性,C水溶液呈碱性,则B溶液呈酸性的原因是Al3++3H2O⇌Al(OH)3+3H+(用离子方程式表示)(5)若A为元素Q的单质,B、C均属于盐中的阳离子,若B溶液呈黄色,与硫氰化钾溶液混合后显红色,D、E均为沉淀,且D在空气中能转化成E,B→C的离子方程式为2Fe3++Fe ═3Fe2+.D→E的化学方程式为4Fe(OH)2+O2+2H2O═4Fe(OH)3.选用氢碘酸与E反应实现E→C的转化,该反应的离子方程式为2Fe(OH)3+6H+2I═2Fe2++I2+6H2O.【分析】已知X、Y、Z、W为短周期四种常见元素,其中X是原子半径最小的元素,则X 为H元素;Y原子的外层电子数是内层电子数二倍,原子只能有2个电子层,最外层电子数为4,故Y为C元素;Z能和X形成两种常见的液态化合物,则Z为O元素;W元素的单质既能与盐酸反应能与NaOH溶液反应,则W为Al,元素Q是地壳中含量第二位的金属元素,应为Fe,以此解答(1)(2)(3);(4)A为金属单质,B、C均属于盐类,D是一种白色沉淀.若B的水溶液呈酸性,C的水溶液呈碱性,则A为Al,B为AlCl3,C为NaAlO2,D为Al(OH)3;(5)若A为Fe,B溶液呈黄色,与硫氰化钾溶液混合后显红色,B含有Fe3+,C含有Fe2+,D、E均为沉淀,且D在空气中能转化成E,由转化关系可知D为Fe(OH)2,E为Fe(OH)3.【解答】解:已知X、Y、Z、W为短周期四种常见元素,其中X是原子半径最小的元素,则X为H元素;Y原子的外层电子数是内层电子数二倍,原子只能有2个电子层,最外层电子数为4,故Y为C元素;Z能和X形成两种常见的液态化合物,则Z为O元素;W元素的单质既能与盐酸反应能与NaOH溶液反应,则W为Al,元素Q是地壳中含量第二位的金属元素,应为Fe,(1)Y为C元素,位于周期表第二周期ⅣA族,故答案为:第二周期ⅣA族;(2)X2Z2为H2O2,H2O2的电子式为,故答案为:;(3)Y为C元素,对应的最高价氧化物为二氧化碳,与少量NaOH溶液反应生成碳酸氢钠,反应的离子方程式为CO2+OH﹣=HCO3﹣,故答案为:CO2+OH﹣=HCO3﹣;(4)A、B、C、D、E五种物质(或离子)均含有同一种元素,A为金属单质,B、C均属于盐类,D是一种白色沉淀.若B的水溶液呈酸性,C的水溶液呈碱性,则A为Al,B为AlCl3,C为NaAlO2,D为Al(OH)3,B为AlCl3,为强酸弱碱盐,水解呈酸性,离子方程式为Al3++3H2O⇌Al(OH)3+3H+,故答案为:Al3++3H2O⇌Al(OH)3+3H+;(5)A、B、C、D、E五种物质(或离子)均含有同一种元素,若A为Fe,B溶液呈黄色,与硫氰化钾溶液混合后显红色,B含有Fe3+,C含有Fe2+,D、E均为沉淀,且D在空气中能转化成E,由转化关系可知D为Fe(OH)2,E为Fe(OH)3,B→C的离子方程式为2Fe3++Fe ═3Fe2+,D→E的化学方程式为4Fe(OH)2+O2+2H2O═4Fe(OH)3,HI与Fe(OH)3反应,反应的离子方程式为2Fe(OH)3+6H++2I﹣═2Fe2++I2+6H2O,故答案为:2Fe3++Fe═3Fe2+;4Fe(OH)2+O2+2H2O═4Fe(OH)3;2Fe(OH)3+6H++2I﹣═2Fe2++I2+6H2O.【点评】本题考查无机物的推断,注意利用信息及转化关系图推断各物质是解答的关键,习题综合性较强,涉及热化学反应、离子反应及溶液中物料守恒、电荷守恒的考查,题目难度中等.8.(2015•红桥区二模)A是一种重要的食用香料.存在于玉米、葡萄等多种食品中.主要用于配制奶油、乳品、酸奶和草莓等香精,也可用于有机合成.以A为原料制备阴离子树脂M和新型聚酯材料N等的合成路线如图:已知:Ⅰ.RCHO+OHCR′Ⅱ.R﹣CH═CHOH不稳定,很快转化为R﹣CH2﹣CHO(1)反应Ⅰ的反应类型是加成反应.(2)B中官能团的名称是羟基.(3)D与新制Cu(OH)2反应的化学方程式是CH3CHO+2Cu(OH)2+NaOHCH3COONa+Cu2O↓+3H2O.(4)下列关于E的说法正确的是c.a.不溶于水b.能与H2发生还原反应c.能被重铬酸钾酸性溶液氧化d.与NaOH醇溶液共热发生消去反应(5)M的结构简式是.(6)反应Ⅱ的化学方程式是.(7)A的属于酯类的同分异构体有4种,其中只含一个甲基的同分异构的结构简式是:HCOOCH2CH2CH3.【分析】A中不饱和度==1,所以含有一个双键,A能和氢气发生加成反应生成B,B能和HIO4发生反应,结合题给信息I知,B中含有醇羟基,D和氢气发生加成反应生成E,D为醛,则E为乙醇结构简式为CH3CH2OH,D为CH3CHO,B为CH3CH(OH)CH(OH)CH3,A为CH3COCH(OH)CH3,B和F发生酯化反应生成N,F结构简式为;A发生消去反应生成H,H结构简式为CH3COCH=CH2,H发生加聚反应生成M,M结构简式为,据此分析解答.【解答】解:A中不饱和度==1,所以含有一个双键,A能和氢气发生加成反应生成B,B能和HIO4发生反应,结合题给信息I知,B中含有醇羟基,D和氢气发生加成反应生成E,D为醛,则E为乙醇结构简式为CH3CH2OH,D为CH3CHO,B为CH3CH (OH)CH(OH)CH3,A为CH3COCH(OH)CH3,B和F发生酯化反应生成N,F结构简式为;A发生消去反应生成H,H结构简式为CH3COCH=CH2,H发生加聚反应生成M,M结构简式为,(1)通过以上分析知,反应I为A发生加成反应生成B,所以其反应类型是加成反应,故答案为:加成反应;(2)B为CH3CH(OH)CH(OH)CH3,其官能团名称是羟基,故答案为:羟基;(3)D为CH3CHO,D与新制氢氧化铜悬浊液发生氧化反应,反应方程式为CH3CHO+2Cu (OH)2+NaOH CH3COONa+Cu2O↓+3H2O,故答案为:CH3CHO+2Cu(OH)2+NaOH CH3COONa+Cu2O↓+3H2O;(4)为乙醇结构简式为CH3CH2OH,a.乙醇易溶于水,故a错误;b.乙醇不含不饱和键,所以不能与H2发生还原反应,故b错误;c.乙醇含有醇羟基,能被重铬酸钾酸性溶液氧化而使重铬酸钾溶液变色,故c正确;d.乙醇不能和NaOH的醇溶液发生消去反应,但可以在浓硫酸作催化剂、170℃条件下发生消去反应,故d错误;故选c;(5)通过以上分析知,M结构简式为,故答案为:;(6)B和F在浓硫酸作催化剂、加热条件下发生酯化反应,反应方程式为,故答案为:;(7)A为CH3COCH(OH)CH3,A属于酯类的同分异构体中,有HCOOCH2CH2CH3、HCOOCH(CH3)2、CH3COOCH2CH3、CH3CH2COOCH3,所以有四种同分异构体;只含一个甲基的同分异构的结构简式为HCOOCH2CH2CH3,故答案为:4;HCOOCH2CH2CH3.【点评】本题考查有机物推断,为高考高频点,侧重考查学生分析、推断能力,根据反应条件、题给信息、流程图中部分物质结构式采用正逆结合的方法进行推断,难点是同分异构体种类的判断,要考虑碳链异构、官能团位置结构等.9.(2015•红桥区二模)焦亚硫酸钠(Na2S2O5)是一种常用的抗氧化剂.某研究小组对焦亚硫酸钠进行如下研究:(1)采用如图装置(实验前已除尽装置内的空气)制取Na2S2O5.装置Ⅱ中有Na2S2O5析出,发生的化学反应方程式为:Na2SO3+SO2=Na2S2O5.①装置Ⅰ中产生气体的化学方程式为Na2SO3+H2SO4=Na2SO4+SO2↑+H2O.②要从装置Ⅱ中获得已析出的晶体,可采取的分离方法是过滤.③装置Ⅲ用于处理尾气,可选用的最合理装置(夹持仪器已略去)为d(填序号).(2)【查阅资料】Na2S2O5溶于水即生成NaHSO3.①NaHSO3溶液显酸性.请用化学用语和必要的文字解释原因HSO3﹣⇌H++SO32﹣,HSO3﹣电离程度大于水解程度,所以溶液显酸性;﹣+H20⇌H2SO3+OH证明该结论可采用的实验方法是ae(填序号).a.测定溶液的pH b.加入Ba(OH)2溶液c.加入盐酸d.加入品红溶液e.用蓝色石蕊试纸检测②检验Na2S2O5晶体在空气中已被氧化的实验方案是取少量Na2S2O5晶体于试管中,加入适量水溶解,滴加盐酸,振荡,再滴加氯化钡溶液,有白色沉淀生成.(3)焦亚硫酸钠(Na2S2O5)在酸性条件下可将工业废水中的Cr2O72﹣还原为Cr3+.①写出该反应的离子方程式2Cr2O72﹣+3S2O52﹣+10H+=4Cr3++6SO42﹣+5H2O或2Cr2O72﹣+6HSO3﹣+10H+=4Cr3++6SO4+8H2O.②若处理Cr2O7浓度为1×10mol/L的工业废水1L,需用Na2S2O5固体285mg.【分析】(1)①由装置Ⅱ中发生的反应可知,装置Ⅰ中产生的气体为SO2,亚硫酸钠与硫酸反应生成硫酸钠、二氧化硫与水;②装置Ⅱ中获得已析出的晶体,分离固体与液态,应采取过滤操作;③装置Ⅲ用于处理尾气,吸收为反应的二氧化硫,应防止倒吸,且不能处于完全密闭环境中;(2)①NaHSO3溶液中HSO3﹣的电离程度大于水解程度,溶液呈酸性,故检验溶液呈酸性即可;②Na2S2O5晶体在空气中易被氧化为Na2SO4,用盐酸、氯化钡溶液检验样品中是否含有硫酸根即可;(3)①焦亚硫酸钠(Na2S2O5)在酸性条件下可将工业废水中的Cr2O72﹣还原为Cr3+.本身被氧化为硫酸根离子,依据原子守恒和电荷守恒配平书写离子方程式;②依据化学方程式中物质的定量关系,结合离子方程式2Cr2O72﹣+3S2O52﹣+10H+=4Cr3++6SO42﹣+5H2O列式计算Na2S2O5物质的量,n=得到质量.【解答】解:(1)①由装置Ⅱ中发生的反应可知,装置Ⅰ中产生的气体为SO2,亚硫酸钠与硫酸反应生成硫酸钠、二氧化硫与水,反应方程式为Na2SO3+H2SO4=Na2SO4+SO2↑+H2O,故答案为:Na2SO3+H2SO4=Na2SO4+SO2↑+H2O;②装置Ⅱ中获得已析出的晶体,分离固体与液态,应采取过滤进行分离,故答案为:过滤;③a.装置应将导管深入氨水中可以吸收二氧化硫,但为密闭环境,装置内压强增大易产生安全事故,故错误;b.该装置吸收二氧化硫能力较差,且为密闭环境,装置内压强增大易产生安全事故,故错误;c.该装置不能吸收二氧化硫,所以无法实现实验目的,故错误;d.该装置中氢氧化钠与二氧化硫反应,可以吸收,且防止倒吸,故正确;故答案为:d;(2)①NaHSO3溶液中HSO3﹣的电离程度大于水解程度,HSO3﹣⇌H++SO32﹣,HSO3﹣+H20⇌H2SO3+OH﹣,溶液呈酸性,测定溶液的pH,可以确定溶液酸碱性,酸性溶液可以使湿润蓝色石蕊试纸变红,所以能用测定溶液pH值、湿润的蓝色石蕊试液检验,而加入Ba(OH)2溶液、HCl溶液、品红溶液都不能说明溶液呈酸性;故答案为:HSO3﹣⇌H++SO32﹣,HSO3﹣+H20⇌H2SO3+OH﹣电离程度大于水解程度,所以溶液显酸性;ae;②Na2S2O5中S元素的化合价为+4价,因此会被氧化为为+6价,即晶体在空气中易被氧化为Na2SO4,用盐酸、氯化钡溶液检验样品中是否含有硫酸根即可,实验方案为:取少量Na2S2O5晶体于试管中,加入适量水溶解,滴加盐酸,振荡,再滴加氯化钡溶液,有白色沉淀生成,故答案为:取少量Na2S2O5晶体于试管中,加入适量水溶解,滴加盐酸,振荡,再滴加氯化钡溶液,有白色沉淀生成;(3)①焦亚硫酸钠(Na2S2O5)在酸性条件下可将工业废水中的Cr2O72﹣还原为Cr3+.本身被氧化为硫酸根离子,反应的离子方程式为:2Cr2O72﹣+3S2O52﹣+10H+=4Cr3++6SO42﹣+5H2O 或2Cr2O72﹣+6 HSO3﹣+10H+=4Cr3++6SO42﹣+8H2O;故答案为:2Cr2O72﹣+3S2O52﹣+10H+=4Cr3++6SO42﹣+5H2O或2Cr2O72﹣+6 HSO3﹣+10H+=4Cr3++6SO42﹣+8H2O;②依据反应定量关系列式计算,处理Cr2O72﹣浓度为1×10﹣3mol/L的工业废水1L中Cr2O72﹣物质的量=1×10﹣3mol/L×1L=10﹣3mol;2Cr2O72﹣+3S2O52﹣+10H+=4Cr3++6SO42﹣+5H2O2 310﹣3mol 1.5×10﹣3mol需用Na2S2O5固体质量=1.5×10﹣3mol×190g/mol=285×10﹣3g=285mg,故答案为:285.【点评】本题考查物质的制备实验、实验方案设计、物质含量的测定、氧化还原反应滴定等,难度中等,明确实验原理是解本题关键,根据物质的性质分析解答,注意元素化合物知识的积累和灵活运用.10.(2015•红桥区二模)氮的氢化物NH3、N2H4等在工农业生产、航空航天等领域有广泛应用.(1)液氨作为一种潜在的清洁汽车燃料已越来越被研究人员重视.它在安全性、价格等方面较化石燃料和氢燃料有着较大的优势.氨的燃烧实验涉及下列两个相关的反应:①4NH3(g)+5O2(g)=4NO(g)+6H2O(l)△H1②4NH3(g)+6NO(g)=5N2(g)+6H2O(l)△H2则反应4NH3(g)+3O2(g)=2N2(g)+6H2O(l)△H=.(请用含有△H1、△H2的式子表示)(2)合成氨实验中,在体积为3L的恒容密闭容器中,投入4mol N2和9mol H2在一定条件222mol NH3(g)中的化学键消耗的能量.①则T1<T2(填“>”、“<”或“=”)②在T2K下,经过10min达到化学平衡状态,则0~10min内H2的平均速率v(H2)=0.1mol•L﹣1•min﹣1,平衡时N2的转化率α(N2)=25%.③下列图象1分别代表焓变(△H)、混合气体平均相对分子质量()、N2体积分数φ(N2)和气体密度(ρ)与反应时间的关系,其中正确且能表明该可逆反应达到平衡状态的是BC.。
天津市红桥区2015届高三下学期3月月考高三地理答案1-11题,每题4分,共44分DACCA BCCCA A12.(14分)(1)特点:各月都有地质灾害发生(2分),(夏季多,冬季少)5~8月相对集中(2分),其中6月份发生次数最多(2分)。
(2)亚热带季风气候(2分)冬季少雨,夏季多雨(2分)降水多,且多暴雨(2分)副热带高气压(2分)13.(每空2分,共24分)(1)丘陵(2分)海拔较低,500米以下(2分);起伏较小,坡度和缓(2分)(2)约为1300米左右(2分)(3)28.5(2分)(4)冬季温和,夏季高温(4分)(5)流量丰富;流量季节变化大;水位季节变化大;流速较缓;无结冰期;汛期较长。
(答出其中三项即可)(6分)(6)冬春遇寒潮(低温)天气,农业生产易遭受冻害(2分)冬春降水少,春温回升快,农业生产易遭受旱灾(2分)14.(每空2分,共18分)天津市红桥区2015届高三下学期3月月考历史扫描版高三历史(2015、3)Ⅱ卷共3题,共56分。
12.(22分)(1)意大利最早出现资本主义萌芽,古典文化遗存丰厚;(2分)德意志处于四分五裂的状态,罗马教廷的精神控制与经济掠夺;(2分)法国资本主义经济发展程度较高,典型的君主专制激发了资产阶级的强烈反抗。
(2分)(2)鼓励追求财富;(2分)主张适应现世生活,提倡个人奋斗;(2分)反对政府干预经济。
(2分)(13.(20分)(1)“二程”把天理和伦理道德直接联系起来,朱熹认为三纲五常就是天理,强调“存天理,灭人欲”;(2分)程朱理学在宋以后成为居于统治地位的官方哲学,并成为地方教化的工具;(2分)女性贞节观是程朱理学的体现,受到地方推崇和不断强化,从而成为地方道德标准。
(2分)(2)20世纪初的中国社会正发生着深刻变化。
经济上,中国民族资本主义经济发展,为反封建斗争和社会进步奠定了物质基础,让更多妇女可以走出家庭并实现经济独立。
(3分)政治上,辛亥革命推翻了君主专制制度,民主共和观念深入人心。
高三数学(文史类)第Ⅰ卷一、选择题(在每小题给出的四个选项中,只有一项是符合题目要求的) 1、已知集合2{|1},{|0}A x x B x x x =<=-≤,则AB =A .{|11}x x -≤≤B .{|01}x x ≤≤C .{|01}x x <≤D .{|01}x x ≤<2、盒子装有形状、大小完全相同的5个球,其中红色球3个,黄色球2个,若从中随机取出2个球,则所有取出的2个球颜色不同的概率等于 A .310 B .25 C .35 D .123、根据如下图所示的框图,对大于2的正数N ,输出的数列的通项公式是A .2n a n =B .2(1)n a n =-C .2n n a =D .12n n a -=4、某几何体的三视图如上图所示,且该几何体的体积是3,则正视图中的x 的值 A .2 B .3 C .32 D .925、设:{|lg(1)},:{|21}xp x x y x q x x -∈=-∈<,则p 是q 的A .充分且不必要条件B .必要且不充分条件C .充要条件D .既不充分也不必要条件 6、在ABC ∆中,0120,2,3,,ABC BA BC D E ∠===是线段AC 的三等分点,则BD BE ⋅的值为 A .659 B .119 C .419 D .139-7、将函数()2sin(2)4f x x π=+的图象向右平移(0)ϕϕ>个单位,再讲图象上没一点的横坐标缩短到原来的12(纵坐标不变),所得图象关于直线4x π=对称,则ϕ的最小值为A .18πB .14πC .38πD .12π 8、已知函数()2log ,02sin(),2104x x f x x x π⎧<<⎪=⎨≤≤⎪⎩,若存在实数1234,,,x x x x 满足1234()()()()f x f x f x f x ===,且1234x x x x <<<,则3412(1)(1)x x x x --的取值范围是A .(9,21)B .(20,32)C .(8,24)D .(15,25)第Ⅱ卷二、填空题:本大题共4小题,每小题5分,共20分,把答案填在答题卷的横线上.. 9、设i 为虚数单位,复数z 满足3(2)z i i -=,则复数z 的虚部为 10、()21ln 2f x x x =-+在1[,]e e上的最大值是 11、已知函数()12cos (0),()2,()0f x wx wx w f x f x +>=-=,且12x x -的最小值 等于π,则w = 12、已知直线:l y =,点(,)P x y 是圆22(2)1x y -+=上的动点,则点P 到直线l 的距离的最小值为13、如图,12,F F 是双曲线22221(0,0)x y a b a b-=>>的左右焦点,过1F 的直线与双曲线的左右两支分别交于点,A B ,若2ABF ∆为 等边三角形,则双曲线的离心率为 14、已知下列命题: ①函数()f x =有最小值2;②“2450x x --=”的一个必要不充分条件是“5x =”;③命题:,tan 1p x R x ∃∈=;命题2:,10q x R x x ∀∈-+>,则命题“()p q ∧⌝”是假命题; ④函数()3231f x x x =-+在点(2,(2))f 处的切线方程为3y =-.其中正确命题的序号是三、解答题:本大题共6小题,满分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13分)在ABC ∆中,角,,A B C 的对边分别为,,a b c ,且3,sin 2sin a b C A ===. (1)求c 的值; (2)求sin(2)4A π-的值.18、(本小题满分13分)某人准备投资1200万元办一所中学,为了考虑社会效益和经济效益,对该地区教育市场进行调查,得出一组数据,列表如下(以班级为单位): 市场调查表:根据物价部门的有关文件,初中是义务教育阶段,收费标准适当控制,预计除书本费、办公费外,初中每人每年可以取600元,高中每人每年可收取1500元,因生源和环境等条件限制,办学规模以20至30个班为宜(含20个与30个),教师实行聘任制,初、高中的教育周期均为三年,设初中编制为x 个班,高中编制为y 个班,请你合理安排招生计划,使年利润最大.17、(本小题满分13分)如图,在四棱锥P ABCD -中,底面ABCD 是边长为2的正方形,侧面PAD ⊥底面ABCD ,且,,2PA PD AD E F ==分别为,PC BD 的中点. (1)求证:平面//EF 平面PAD ; (2)求证:平面PAB ⊥平面PDC ;18、(本小题满分13分)已知数列{}n a 满足1221,3,32(,2)n n a a a a n N n ++===-∈≥.(1)证明:数列{}1n n a a +-是等比数列,并求出{}n a 的通项公式;(2)设数列{}n b 满足242log (1)n n b a =+,证明:对一切正整数n ,有2221211111112n b b b +++<--- .19、(本小题满分14分)已知椭圆2222:1(0)x y Ca b a b+=>>. (1)求椭圆C 的方程; (2)设与圆223:4O x y +=相切的直线l 交椭圆C 于,A B 两点,求OAB ∆面积的最大值, 及取得最大值时直线l 的方程.20、(本小题满分14分 已知函数()3212()32a f x x x x a R =-+-∈. (1)当3a =时,求函数()f x 的单调区间;(2)若对于任意[1,)x ∈+∞都有()2(1)f x a '<-成立,求实数a 的取值范围; (3)过过点1(0,)3-可作函数()y f x =图象的三条不同切线,求实数a 的取值范围.高三数学(文)(1705)一、选择题(每小题5分,共40分)二、填空题(每小题5分,共30分)9.52-10.21- 11.12121 13.7 14.③④ 三、解答题(本大题共6小题,共80分) (15)(本小题满分13分) (Ⅰ) 根据正弦定理,, (2)因为,所以. (5)(Ⅱ)根据余弦定理,得 , (8)于是 ,从而,, (11). (13)(16)(本小题满分13分)设初中编制为 个班,高中编制为 个班,则依题意有 (4)又设年利润为 万元,那么,即 (7)在直角坐标系中作出不等式组所表示的可行域,如图所示. (10)问题转化为在如图所示的阴影部分中,求直线在轴上的截距的最大值.显然图中的点是符合题意的最优解.解方程组得即. (11)所以.故学校规模以初中个班、高中个班年利润最大 (13)(17)(本小题满分13分)(Ⅰ)连接,为正方形,为中点,为中点.所以在中,,且,所以. (4)(Ⅱ)因为,为正方形,,所以. (6)所以, (7)又,所以是等腰直角三角形,且即 (9),且所以又,所以. (13)(18)(本小题满分13分)(Ⅰ)因为, 所以,因为,,所以, (3)所以数列是以为首项, 为公比的等比数列,则所以 (7)(Ⅱ)nn 2)112(log 224=+-= (9)则 (13)(19)(本小题满分14分)(Ⅰ)由题意可得:2212133a bc a⎧+=⎪⎪⎨⎪=⎪⎩ ..........................2 22223,1,13x a b y ==∴+= (4)(Ⅱ)①当k不存在时,x y =∴=,1324OAB S ∆∴== (5)②当k 存在时,设直线为y kx m =+,()()1122,,,,A x y B x y222221,(13)63303x y k x km m y kx m ⎧+=⎪+++-=⎨⎪=+⎩....................8 212122263313,13km m x x x x k k --+==++ (9)2243(1)d r m k =⇒=+ (10)||AB ===2=≤ (12)当且仅当2219,k k =即k =时等号成立..........................13 11222OAB S AB r ∆∴=⨯≤⨯=∴OAB ∆面积的最大值为,此时直线方程1y x =±. (14)(20)(本小题满分14分) (Ⅰ)当时,,得. (1)因为232-+-=x x x f )(’=)1(2---x x )( , 所以当时,,函数单调递增; 当或时,,函数单调递减.所以函数的单调递增区间为,单调递减区间为............4 (Ⅱ)方法1:由x x a x x f 2231)(23-+-=,得.因为对于任意都有成立,即对于任意都有成立, 即对于任意都有成立,令 ,要使对任意 都有成立,必须满足 或即 或所以实数 的取值范围为 . (9)方法2:由x x a x x f 2231)(23-+-=,得 ,因为对于任意都有成立,所以问题转化为,对于任意 都有.因为 ,其图象开口向下,对称轴为.①当 时,即时,在上单调递减,所以 ,由 ,得,此时 .②当 时,即时,在上单调递增,在上单调递减,所以 ,由,得,此时.综上①②可得,实数 的取值范围为 . (9)(Ⅲ)设点是函数图象上的切点,则过点 的切线的斜率为 ,所以过点 的切线方程为 .因为点 在切线上,所以即.若过点可作函数图象的三条不同切线,则方程有三个不同的实数解.令,则函数与轴有三个不同的交点.令,解得或.因为,,所以必须,即.所以实数的取值范围为. (14)。
高三生物答案(2012、05)
一、选择题:每小题6分,共36分。
1. B
2.C
3.D
4.A
5.D
6.C
二、非选择题:3道题,共44分。
7.(12分)
(1)体液免疫和细胞免疫
(2)摄取和处理 淋巴因子 记忆细胞的增殖、分化(3)T细、B细胞、记忆细胞和吞噬细胞(答出3个即给分)(4)①→③→⑤以及①→⑥→④
8.(16分)
(1)RNA (四种)脱氧核苷酸
(2)F DNA连接酶
(3)感受态细胞抗生素A
(4)显微注射转基因小鼠间进行杂交
9.(16分)
(1)还原糖是酶促反应的产物,60℃是相关酶的最适温度(2)土豆不含淀粉酶
(3)②还原糖与斐林试剂反应生成砖红色沉淀
第二步:质量分数为3%的淀粉溶液
第三步:斐林试剂50-65℃水浴保温2min
实验结论:未出现砖红色沉淀出现砖红色沉淀。
英语试题本试卷分为第I卷(选择题)和第II卷(非选择题)两部分,共130分,考试用时100分钟。
第I卷(共95分)第一部分:英语知识运用(共两节,满分45分)第一节:单项填空(共15小题:每小题1分,满分15分)从A、B、C、D四个选项中,选出可以填入空白处的最佳选项。
例:Stand over there____youll be able to see it better.A.or B.and C.but D.while答案是B.1.-Hi, Betty, are you free at present? I have to ask you for a favor.-_____.With pleasure.A.Sorry, I am busy B.Go aheadC.Help yourself D.Ask, please2.We'd better take umbrellas-I'm sure it____when we amve in London; it's always wet there at this time of year.A.will rain B.is raining C.will be raining D.would rain3.If you are feeling so tired, perhaps a little sleep would____.A.act B.help C.serve D.last4.I must be getting fat -I can____do my trousers up.A.fairly B.hardly C.nearly D.seldom5.-Oh, we have finished our task and can have a good rest.- ________.Our manager took another order yesterday.A.Yes, it will B.I don't take much of itC.It's possible D. It's hard to say6.- Father, you promised!-Well,____.But it was you who didn't keep your word first.A.so was I B.so did I C.so I was D.so I did7.The Chinese government has made the Mid-Autumn Festival a legal holiday,____ people will have a day off.A.when B.which C.where D.what8.-Long time no see.Haven't you graduated from college?-Yes.I____business management for 4 years in Beijing.A.study B.have studied C.studied D.am studying9.She is quite____to office work.You had better offer her some suggestions when necessary.A.familiar B.fresh C.similar D.sensitive10.Many of the creatures in Rowling's world are not real, and much of____happens is strange. A.which B.that C.what D.it11.-Next week I will go to a job interview.Will you give me some suggestions?-Smiling is a great way to make yourself .A. stand out B.turn out C.work out D.pick out12.Be careful when you cross the road in rush hours so as to avoid by a car.A. being knocked down B. knocking downC. to be knocked down D. to knock down13.The aim of the government is to make every citizen better off,____, to help them livea happy life.A. as a result B.for a start C. for one thing D.in other words14.-Did you visit the famous cultural relics last month?-No, we____ it, but we spent too much time shopping.A. could have visited B.must have visitedC. can't have visited D.shouldn't have visited15.____I feel pity on earthquake victims in Sichuan, I can’t really do very much to help them but contribute my pocket money.A. As long as B.When C.While D.As笫二节:完型填空(共20小题;每小题1.5分,满分30分)阅读下面短文,掌握其大意,然后从16-35各题所给的A、B、C、D四个选项中,选出最佳选项。
2015年天津红桥区高三二模英语试卷-学生用卷一、单项选择1、【来源】 2015年天津红桥区高三二模第1题—I hear they will get married next week.—____.A. Good luckB. Oh, really?C. Congratulations!D. Yes, sir2、【来源】 2015年天津红桥区高三二模第2题2017~2018学年江苏无锡高二上学期期末第21题2017~2018学年江苏无锡高二上学期期末第11题1分The population of Africa is so rapidly as to cause concern of the whole world.A. expandingB. stretchingC. extendingD. spreading3、【来源】 2015年天津红桥区高三二模第3题—I’m green at everything.—Don’t worry. ____you get older, you will get more experience.A. AsB. TillC. UnlessD. Since4、【来源】 2015年天津红桥区高三二模第4题—Do you regret up your opinion that every child should take part in service learning?—No. On the contrary, I think it is very necessary.A. bringingB. to bringC. to be broughtD. being brought5、【来源】 2015年天津红桥区高三二模第5题It is _____ ill health that she is now not so active in the movement.A. instead ofB. in place ofC. thanks toD. in face of6、【来源】 2015年天津红桥区高三二模第6题Those who are not _____ with the progress they have made will work harder.A. proudB. afraidC. popularD. content7、【来源】 2015年天津红桥区高三二模第7题2017~2018学年北京东城区北京市广渠门中学高二上学期期中第26题1分Allen has decided that he will be an engineer by the time he ____ twenty.A. had beenB. is going to beC. will beD. is8、【来源】 2015年天津红桥区高三二模第8题—Do you let your kids travel alone at night?—_____!A. Forget itB. Go aheadC. Absolutely notD. At any time9、【来源】 2015年天津红桥区高三二模第9题Our bodies are strengthened by taking exercise._____, our minds are developed by learning.A. ProbablyB. SimilarlyC. LikelyD. Generally10、【来源】 2015年天津红桥区高三二模第10题—I missed the first part of the film. It was really a pity.—You_____ home half an hour earlier.A. should goB. must have goneC. should leaveD. should have left11、【来源】 2015年天津红桥区高三二模第11题Do tell me the ways you think of _____ the problem as soon as possible.A. to solveB. solvingC. solveD. being solved12、【来源】 2015年天津红桥区高三二模第12题At school, some students are active_____ some are shy, yet they can be good friends with one another.A. althoughB. whileC. soD. as13、【来源】 2015年天津红桥区高三二模第13题Well, we meant to give you a delightful surprise. But just now I made a slip of the tongue and ____ our secret.A. threw awayB. put awayC. gave awayD. carried away14、【来源】 2015年天津红桥区高三二模第14题He'll never forget his stay in the park he found his cell phone which had gone missing days before.A. thatB. whereC. whyD. which15、【来源】 2015年天津红桥区高三二模第15题—Could you please tell us about some good ways to learn English?— My suggestion is that reading often and writing regularly______.A. are neededB. is neededC. be neededD. should need二、完形填空16、【来源】 2015年天津红桥区高三二模第16题Each year her husband sent her valentine roses, tied with pretty bows, and a card. The cardworld1say, "I love you even more than last year on this day."The year he2, the roses were delivered to her door as usual. The card said, "Be my valentine. My love for you will always3with every passing year." She thought this was the4time that the roses would appear.She5he ordered roses6without knowing that he would7. He always liked to do things ahead. Then, if he got busy, everything would work out8. She put9in a very special vase, and then placed the vase beside the portrait of his smiling face. She would sit for hours, staring at his picture and the roses.A year went by,10it was tough to live without him. Then, on Valentine's Day, the doorbell rang, and she saw roses sitting11her door. She brought the roses in, and then looked at them in surprise. Later she12the flower-shop and asked them to explain13valentine roses were sent to her.The owner replied, "I know your husband passed away a year ago. The flowersyou14today were paid for two years ago. Your husband always plannedahead. Here is15of many years. You'll get them every year till… and he also wrote a special little16for you…he did this years ago. He said, if you find out that I'm no longer here, the card should be sent to my wife the following year."She thanked him and bung up the phone, her tears17her face. She found a card among the roses and read it slowly, "Hello, my love. I know it's been a year since I was gone. I hope it's not too hard for you to overcome. I know you must be18and the pain is real. The19we shared made everything so beautiful in life. I loved you more than20can say. Please try not to be too sad. That is why the roses will be sent to you for years."A. sometimesB. alwaysC. occasionallyD. rarelyA. diedB. returnedC. disappearedD. retiredA. comeB. plantC. growD. leaveA. firstB. secondC. lastD. everyA. guessesB. rememberedC. decidedD. informedA. in returnB. in addition toC. in realityD. in advanceA. live throughB. pass awayC. break downD. leave behindA. freeB. earlyC. quietD. fineA. oneB. someC. themD. itA. butB. andC. thenD. thereforeA. byB. outC. insideD. uponA. visitedB. wentC. cameD. calledA. whenB. whetherC. whyD. howA. sentB. receivedC. possessedD. acceptedA. an orderB. an excuseC. a proposalD. a programA. letterB. articleC. noteD. cardA. coming intoB. running downC. going outD. getting awayA. crazyB. cautiousC. lonelyD. embarrassedA. marriageB. lifeC. loveD. familyA. flowersB. rosesC. cardsD. words三、阅读理解17、【来源】 2015年天津红桥区高三二模(A篇)第17~21题Besides entertainment and beautiful lanterns, another important part of the Lantern Festival, or Yuanxiao Festival is eating small dumpling balls made of glutinous rice flour. We call these balls Yuanxiao or Tangyunan. Obviously, they get the name from the festival itself. It is said that the custom of eating Yuanxiao originated during the Eastern Jin Dynasty in the fourth century, then became popular during the Tang and Song periods.The fillings inside the dumplings or Yuanxiao are either sweet or salty. Sweet fillings are made of sugar, Walnuts, sesame(芝麻), osmanthus flowers(桂花), rose petals, sweetened tangerine(橘子)peel, bean paste, or jujube paste(枣子酱). A single ingredient or any combination can be used as the filling. The salty variety is filled with minced meat(肉末儿), vegetables or a mixture.The way to make Yuanxiao also varies between northern and southern China. The usual method followed in southern provinces is to shape the dough of rice flour in to balls, make a hole, insert the filling, then close the hole and smooth out the dumpling by rolling it between your hands. In North China, sweet or nutmeat stuffing is the usual ingredient. The fillings are pressed into hardened cores, dipped lightly in water and rolled in a flat basket containing day glutinous rice flour. A layer of the flour sticks to the filling, which is then again dipped in water and rolled a second time in the rice flour. And so it goes, like rolling a snowball, until the dumpling is the desired size.The custom of eating Yuanxiao dumplings remains. This tradition encourages both old and new stores to promote their Yuanxiao products. They all try best to improve the taste and quality of the dumpling is to attract more customers.(1) Which of the following is NOT true?A. The custom of eating Yuanxiao dates back to the fourth century.B. Sugar, rose petals and minced meat are all fillings of Yuanxiao.C. Sweet Yuanxiao are usually made in southern China.D. People in northern China usually make Yuanxiao by rolling like a snowball.(2) Which country does the Lantern Festival come from?A. ChinaB. South KoreaC. Jin DynastyD. Tang and Song periods.(3) Choose the right order of making Tangyuan in southern provinces.1) make a hole and insert the filling2) roll the dough between your hands3) shape the dough of rice flour into balls4) close the hole5) prepare some tice flourA. 5 3 2 1 4B. 3 5 1 2 4C. 3 1 4 2 5D. 5 3 1 4 2(4) What is the best title of this passage?A. The Lantern FestivalB. The ingredients of YuanxiaoC. The difference between Yuanxiao and TangyuanD. China's Traditional Food—Yuanxiao(5) Why do we eat Yuanxiao today?A. Because it's a traditionB. Because it's a tasty foodC. Because we love our countryD. Because this food can make money18、【来源】 2015年天津红桥区高三二模(B篇)第18题"I get no respect in this house."That's the phrase I use most in my home, when I'm again and again ignored by my teenage daughters."I would never have spoken to my parents like that—I'd have been thrown out of the house, " I say.They speak just as disrespectfully to my husband, in a manner that my dad would never have tolerated. The furthest I'd go by way of challenging him was raising my eyes as he told me off.The disturbing thing is, according to other parents, that this is quite common.Lack of respect isn't only within families, it is universal. Only last week, while discussing problem teens with my friend, we both commented upon how our children feel far more relaxed with adults than we did. If I went to a friend's house I would always call their parents Mr. and Mrs., I would never havesaid, ” Hi Jean, what are we having for tea? ”Nowadays, children use your first name from a young age. I can't think of one occasion when my daughters' friends have called me Mrs. Mead. It has always been "Helen'. Even "Hi" , which is commonly used, is too casual. It's fine if you're living in a teenage surfing community, but in child-adult relationships, "hello" is far more polite.More and more Britons are becoming annoyed by this lack of respect from others. The survey of 1, 000 adults by the Internet search engine Ask Jeeves found that the over-60s in particular were most annoyed with using their first names.All this part of the disrespectful child problem. Youngsters grow up with too familiar an approach to adults. I'm not saying they should be made to stand to attention, but I think they need to be aware of the age difference and our greater life experience."What do you know? " I hear from my daughters. "A lot more than you." I say. But it falls on deaf ears.We are far too soft, we need to reclaim our authority. But I have an awful feeling it may be too late.(1) The author doesn't like the way.A. her daughters laugh at herB. her daughters talk to herC. her husband talked to herD. her father speak to her(2) The underlined phrase "told me off" in Para. 4 probably means "".A. drove me awayB. said hard words to meC. spoke ill of meD. made me fell bored(3) The author is worried because.A. she doesn't like to be called by her first nameB. young people are not taught be ambitiousC. young people show no respect for older peopleD. he daughters are very relaxed in front of her(4) According to the survey,.A. more and more English people enjoy the casual way of greetingsB. people over 60s don't like it when strangers call them by their first namesC. it's quite common for English people to say "Hi" to adultsD. English teenagers always use insulting words on their parents(5) The author feels about the improvement of the situation.A. hopelessB. hopefulC. doubtfulD. certain19、【来源】 2015年天津红桥区高三二模(C篇)第19~23题Most American schools have boys and girls in the same classroom. But a public high school in the small, rural town of Boonsboro, Maryland, offers a special program called the Academy. Top students in the ninth and tenth grades can attend single-sex classes for math, science, English and social studies.The aim is to help teenagers keep their mind on their work by keeping males and females apart. What they really want to do is take that top group of kids and take them to the very highest level they can achieve there, so that they're prepared for college.Students are invited to enter this program after they finish middle school, and placement in the Academy is voluntary. However, they need high marks and test scores and strong teacher recommendations. They also need to be involved in activities.Almost 350 students have taken part in the Academy. It began as an experiment in 2010. Michael Bair has been at Boonsboro High for twenty years and directs the Academy. These novels deal with the arrogance(傲慢) of man and the pride of man which finally leads to man's downfall. One of those books, for example, is The Call of the Wild by Jack London. This classic story tells of a dog stolen from his home and sold to work as a sled dog in Canada's Klondike Gold Rush. Morgan Fleet, a senior atBoonsboro High, likes being in the academy. She says boys and girls act differently when they are together in a classroom. She believe this difference seems to prevent her from developing herself because she's more focused on thought like "Oh, I wish they would shut up. Oh, what do they think of me? " instead of focusing on "What's the homework? What's going on in this class? What's the lesson? " .There is one activity that girls and boys in the Academy to do together. At the end of the day, many of them stay and help tutor students who are having trouble in school.(1) What is the purpose of the Academy program?A. To offer a special program in high school.B. To help students concentrate on study.C. To strengthen sex education for teenagers.D. To start learning college courses earlier.(2) To enter the Academy program, students are NOT required to.A. do well in their studyB. be recommended by teachersC. take part in some activitiesD. have worked as volunteers(3) The Call of the Wild is given as an example to show that male students.A. should learn more about literatureB. should have a sense of protecting the wildlifeC. are reading more manly novelsD. are interested in stories on animals(4) Which is likely to be Morgan Fleet's opinion?A. A girl cares about the boys' opinions of her.B. Single-sex classes should be stopped.C. Girls are more focused when studying with boys.D. Boys and girls act the same in separate classrooms.(5) The passage is intended to.A. introduce the single-sex class projectB. explain the differences between boys and girlsC. persuade students to enter the Academy programD. urge students to work hard in their study20、【来源】 2015年天津红桥区高三二模(D篇)第20~24题When I was growing up, 16 was always a magical age, a symbol of maturity, responsibility and of course more independence and freedom. I sat through the hours of Driver's Ed classes eager to get out on the road. I couldn't wait to get my driver's license.But it's a different story for today's teens. In January, the U.S. teens got their license at age 16. Twenty years before, that number was almost 45 percent.There are numbers of reasons for the fall-off. The growing responsibilities like paying for insurance and high gas prices discourage teens from getting behind the wheel. Plus, many teens today are so busy with homework, endless hours of activities and part-time jobs that finding the time for Driver's Ed classes may be more difficult than ever.In addition, many states have raised the driving age, or restricted when teens can drive and who they can have in the car. Parents may also be making their own personal restrictions until they feel their teens are responsible enough to drive safely.Driving is part of the American culture, but it's not the central focus like it was 27 years ago. They have so many other things to do now. one of the more interesting factors delaying teens driving might be the change of their social life, today, teens need to look no further than Facebook or other social networking sites to connect with their friends, there is simply less need, maybe less desire, to be able to grab the keys and go.Michelle Wel got her license as a senior in high school because her digital social life made it easy not to drive. "If I couldn't get a ride to see my friends who live a town over, " the 19-year-old said, " I could talk on IM or Skype."Research has shown that these online relationships can lead to higher quality friendships so it isn't necessarily a bad thing. However, it's important to find a balance. If old face-to-face friendships are good, why not drive to find them?(1) We can infer from the first paragraph that.A. Driver's Ed classes allowed teens to know what maturity wasB. getting a driving license at 16 was a must for American teensC. 16 was considered an age when one could get his driver’s licenseD. teens could d rive on the road without taking Driver’s Ed classes(2) What does the passage mainly talk about?A. Desire to drive on the road declines among American teens.B. Getting a license costs much more than ever before.C. Social networking sites are changing the life of American teens.D. American teens are becoming more responsible than ever.(3) Which is NOT a fact delaying American teens getting their driving licenses?A. The cost of insurance and gas.B. Too many school course and activitiesC. Restrictions from families and staresD. Pursuit of more freedom and independence(4) Michaels Wei's example is used to explain.A. why American teens are crazy about digital social lifeB. what social networking websites are bridging to American teensC. to what degree the Internet is affecting the American car cultureD. what the Internet does to help teens to get a driving license(5) The last paragraph is reminding the readers that.A. the Internet has a bad effect on the teen’s social lifeB. teens should keep a balance in choosing their lifestyleC. actual contacts can be replaced by talking on lineD. face-to-face friendship is always the best choice四、任务型阅读21、【来源】 2015年天津红桥区高三二模第21~25题We all know eating out can be a lot of fun. There's the experience of having someone wait on your every need, the social aspect of being in a restaurant with friends and, of course, the great benefit of the lack of in-home cleanup.We all also know eating out can be expensive, and I'm pretty sure college isn't the most financially successful time in a normal person's life. That's why eating at home for a majority of meals really is the only TAL#NBSP option for many college students.Eating at home doesn't have to be dull, boring, or difficult. It can and should be an energetic, creative experience that transforms the way you look at food. When you cook for yourself, you know exactly how your pasta(意大利面) is going to turn out. You know where the greens in your salad come from. You know the olive oil you're using really comes from olives that are pressed for the first time. Use eating at home as an opportunity to learn and develop your kitchen skills. I promise you they will remain useful and rewarding for the rest of your life.Another important concept is to keep in mind that when it comes to shopping for food—less quantity, more quality. Buy base ingredients that are fresh, as local as possible and beautiful, and anything you cook will improve greatly. I promise you will be just as satisfied by eating a little bit less of something with a lot more quality.I am not saying eating out is a bad thing. But I insist there are many more places that are not worth your cash. Put a little more effort into cooking for yourself at home and you will gain experience and appreciation for the world of food.(1) List at least two benefits of eating out according to the text. (No more than 15 words)(2) What does the underlined word "option" most probably mean? (1 word)(3) What's Paragraph 3 mainly about? (No more than 7 words)(4) What advice is given concerning shopping for food in Paragraph 4? (No more than 4 words)(5) Do you like eating at home? State your own reasons. (No more than 20 words)五、书面表达22、【来源】 2015年天津红桥区高三二模第22题请以The Most Unforgettable Person I Ever knew 为题目写一篇英语短文,参加你校英语教研组举办的英语短文比赛:(1)我生活中最难以忘怀的人士……;(2)为什么他(或她)一直在我心中;(3)结论。
天津市红桥区2016届高三数学下学期第二次模拟考试试题理(扫描版)高三数学(理)2016、05一、选择题:每小题5分,共40分二、填空题:每小题5分,共30分.三、解答题:共6小题,共80分. (15)(本小题满分13分)已知21()2cos 2f x x x =+- (Ⅰ)求函数()f x 的最小正周期及在区间0,2π⎡⎤⎢⎥⎣⎦的最大值;(Ⅱ)若01()3f x =.0,612x π5π⎡⎤∈⎢⎥⎣⎦,求0sin 2x 的值 解 (Ⅰ)∵f (x )=32sin 2x +1+cos 2x 2-12=32sin 2x +12cos 2x =sin ⎝ ⎛⎭⎪⎫2x +π6(余弦二倍角1分两角和公式1分,特殊角函数值1分)-----------------------------------------------------------------------3分∴()sin(2)6f x x π=+,∴最小正周期为π,--------------------------------------------------------------------------------------5分∵x ∈⎣⎢⎡⎦⎥⎤0,π2,故π0,6x ⎡⎫∈⎪⎢⎣⎭()f x 单调增,ππ,62x ⎡⎤∈⎢⎥⎣⎦单调减∴sin ⎝⎛⎭⎪⎫2x +π6∈⎣⎢⎡⎦⎥⎤-12,1, 所以f (x )在区间⎣⎢⎡⎦⎥⎤0,π2的最大值是1.---------------------------(单调性1分,最大值1分)---7分(Ⅱ)∵()sin(2)6f x x π=+,01()3f x =,∴01sin(2)63x π+=,又0,612x π5π⎡⎤∈⎢⎥⎣⎦所以0262x ππ⎡⎤+∈,π⎢⎥⎣⎦,故0cos(2)6x π+=---------------(判断区间1分,结论1分)--9分所以0000sin 2sin(2)sin(2)cos cos(2)sin 666666x x x x ππππππ=+-=+-+1132⎛=⨯ ⎝⎭--------------(变形,公式、特殊角、结论各1分)-------13分(16)(本小题满分13分)甲、乙两队参加听歌猜歌名游戏,每队3人.随机播放一首歌曲,参赛者开始抢答,每人只有一次抢答机会,答对者为本队赢得一分,答错得零分.假设甲队中每人答对的概率均为23,乙队中3人答对的概率分别为23,13,12,且各人回答正确与否相互之间没有影响.(Ⅰ) 若比赛前随机从两队的6个选手中抽取两名选手进行示范,求抽到的两名选手在同一个队的概率;(Ⅱ)用ξ表示甲队的总得分,求随机变量ξ的分布列和数学期望; (Ⅲ)求两队得分之和大于4的概率.解:(Ⅰ) 6个选手中抽取两名选手共有26651521C ⨯==⨯种结果, 抽到的两名选手在同一个队包括同在甲队或乙队,共有:2326C =种结果, 用A 表示事件:“从两队的6个选手中抽取两名选手,求抽到的两名选手在同一个队”62()155P A ==.-----------------------------(公式2分,结论1分)-----------3分故从两队的6个选手中抽取两名选手进行示范,抽到的两名选手在同一个队的概率为25.(Ⅱ)解法一:由题意知,ξ的可能取值为0,1,2,3,且P (ξ=0)=C 03×⎝ ⎛⎭⎪⎫1-233=127,P (ξ=1)=C 13×23×⎝ ⎛⎭⎪⎫1-232=29,P (ξ=2)=C 23×⎝ ⎛⎭⎪⎫232×⎝ ⎛⎭⎪⎫1-23=49,P (ξ=3)=C 33×⎝ ⎛⎭⎪⎫233=827.所以ξ的分布列为-----------------------7分ξ的数学期望E (ξ)=0×127+1×29+2×49+3×827=2.----------------------------9分解法二:根据题设可知ξ~B ⎝ ⎛⎭⎪⎫3,23, 因此ξ的分布列为P (ξ=k )=C k3×⎝ ⎛⎭⎪⎫23k ×⎝ ⎛⎭⎪⎫1-233-k=C k3×2k 33,k =0,1,2,3.因为ξ~B ⎝ ⎛⎭⎪⎫3,23,所以E (ξ)=3×23=2. (Ⅲ)用B 表示事件:两队得分之和大于4包括:两队得分之和为5,两队得分之和为6,用1A 表示事件:两队得分之和为5,包括甲队3分乙队2分和乙队3分甲队2分。
2015红桥二模语文高三语文答案一、1.B 2.C 3.D 4.A 5.D二、6. C(亲情背景下的社会互助和应急保障只能“小范围”“小社会”,笼统地表述为“促进社会救助,实现社会的应急保障”则会让人觉得不可思议。
)7. A(作者于文中所言“‘八婆’特征”是缘于私人场合之下且带有亲情色彩的“叽叽喳喳地说个不停”,与通常意义的“八婆”有着本质的区别,作者以此表明沟通交流的重要,并非认为“八婆”值得效仿。
)8. D(由原文可知,作者不赞同由政府承担全部的社会保障义务,但现代社会的普遍观点是“把社会保障的责任全部放在政府身上”,两者并不一致。
只不过“政府要承担全部社会保障责任”与“现代社会不主张政府的权力太大”两者又形成了矛盾。
)三、9.C 10.D 11.B 12.C四、13.(1)这之后纯粹、古朴、隐隐约约、旷远之声音才出现,然而从中能有所领悟、喜欢听的人,十个人中没有一两个。
【3分】(2)柯孝廉于是和韫修各自弹奏一曲,音律高低交错,不能分辨出是两个人弹的。
【2分】(3)然则,没有琴弦弹琴,已经足以改变人的性情,何况是像成连、伯牙那类妙手呢?【3分】14.(1)不能。
“似觉”含有不确定之意。
诗人多年来仕途奔波,宦海沉浮,似乎觉得谙练世情,但又感到政治风云变幻难测。
“似觉”一词准确地写出了诗人对前途把握不定的心理态度,能很好地呼应下句中的“空更”一词。
“犹觉”则没有这样的表达效果。
【2分】(2)①“云容山意商量雪”描绘出乌云堆积山头,酝酿着一场大雪的景象,衬托出诗人游宦在外的艰辛和飘泊羁旅的寂苦,也暗含了作者对家乡的思念之情;【3分】②“柳眼桃腮领略春”形象地描绘出春回大地、桃红柳绿的景象,表达了他在长期贬谪之后,终于盼来了重新起用的喜悦心情。
【1分】(3)尾联展开想象,采用推己及人(对写)的写作技巧,不说自己如何想念家人,而说儿女们在“围炉”时念叨自己这个远行人,写他们对自己的思念,从而把作者真切思乡之情表达得委婉、含蓄、生动。
【写法1分,分析1分,效果2分】15.【5分,每正确默写一题得1分】(1)足以极视听之娱(2)夫天地者万物之逆旅也(3)丛菊两开他日泪(4)故国不堪回首月明中(5)一蓑烟雨任平生(6)满地黄花堆积五、16. (1)真实自然,拉近与描写对象的距离,增强文章的感染力。
(2)突破时空限制,便于直接抒发对麦子的深情。
【4分,每点2分】17. (1)承接上文,丰富了文章的内容;(2)暗示了以往生活的艰难;(3)突出了对麦子的感情由来已久。
【3分。
每点1分】18. (1)以小麦的生长成熟过程为线索,条理清晰;(2)在小麦生长过程的纵向描写中穿插了对母亲、哥哥的叙述,形散神聚。
【4分,每点2分】20. (1)文章的开篇和结尾,反复使用的相同语句,体现了作者对作为生命给养的麦子的真挚感情,充满感恩之情;【2分】(2)在作者看来,自己仿佛也是母亲、哥哥辛勤种植的麦子,在他们的精心呵护下,自己才得以茁壮成长。
【4分,合理探究,分析中肯,逻辑清晰。
】六、21.略。
22. ⑤⑥②④③①(⑤是“情义”总说句,再根据前文提示,先“情”后“义”的顺序排定。
)【2分,错一个,全题不得分。
】23. “恭敬莅临”不得体,应为“拨冗莅临”;“敬祈惠存”不恰当,应为“敬祈惠允”。
【4分,每正确修改一处,得2分。
】七、24.作文评分标准【备注】:1.无标题扣2分。
2.不足800字的,每少50字扣1分。
3.三个错别字扣1分,不重复扣分,最多扣3分。
参考译文:丁未年夏天,我再次寄住在姑苏的桃花坞中,唐伯虎的旧房就在那里。
我当时闲居失意,又被阴雨天气所困。
邻居张韫修善于弹琴,于是我就到(张韫修)那里去拜访。
他为我弹奏一曲《洞庭秋月》。
刚一弹奏,就感觉波涛声在屋梁间远处传来,忽然变为了汹涌澎湃、鱼龙出没的声音,即使不一定身在君山,但是寒烟升起、树木凋落,“美丽的公主下落到北岸”的景象,好像看见得一清二楚,而且还听到了饰物清脆的响声。
啊,多奇妙呀,技艺达到了这种地步!张韫修说:“我弹琴有很多年了。
开始的时候,喜好弹奏声调细微悦耳的旋律,大概听过这种旋律并且喜欢的人曾超过了百人。
接下来才明白这样做不正确,于是冥思苦想,废寝忘食,为了这件事用心良苦,仿佛模模糊糊,似有似无,然后纯古旷远之音才出现,然而以之为德,喜欢听的人,十个人中没有一两个。
知道这样音律的人,难道只有柯孝廉吗?”过了没多久,柯孝廉拜访我,童子背着两个丝袋子,一个袋子里装有能发出金石之声的琴,素培道士在琴的背面刻有铭文;另一个袋子里就是柯孝廉自己写的诗。
秋风萧瑟,虫子的叫声凄怆,柯孝廉于是和韫修各自弹奏一曲,音律高低交错,不能分辨出是两个人弹的。
读他的诗,水大流急,山峰巍峨,都是纯古高远的声音,不是细微靡靡之音,诗和琴也是相通的。
弹琴的技巧高超呀,探究它的精微,足以用来感动植物、鬼神,我不知道现在的人所弹奏的是否吻合古人所弹奏的?然则,没有琴弦弹琴,已经足以改变人的性情,何况是像成连、伯牙那类妙手呢?我听说吴中两个洞庭的面积都比楚地要小,但是山谷风景优美,它的最高峰是缥缈峰,乘小船、穿木鞋,连宿两晚就可到达。
柯孝廉带两个袋子,登上缥缈峰顶弹琴,应当会有灵威老人出来听。
而石公林屋的奇特,枫叶橘子的美丽,都可以作为写作的素材。
宋琬不敏捷,只能慎重地拄着杖跟随在他的后面。
等级 类别 一等 (20-17分) 二等 (16-12分) 三等 (11-7分) 四等 (6-0分)内 容 切合题意 中心突出 内容充实 感情真挚 符合题意 中心明确 内容较充实 感情真实 基本符合题意 中心基本明确 内容单薄 感情基本真实 偏离题意中心不明或立意不当 内容空洞感情虚假或思想不健康 表 达 符合文体要求 结构严谨 语言流畅 字体工整 大体符合文体要求结构完整 语言通顺 字体较工整 基本符合文体要求 结构基本完整 语言基本通顺 字迹清楚 不符合文体要求 结构混乱语言不通顺,语病多 字迹潦草难辨 特 征深刻 丰富 有文采 有创意较深刻 较丰富 较有文采 较有创意个别段落较深刻 个别细节例子较好 个别语句较生动 略显个性无明显特征2015红桥二模英语高三英语答案单项选择:1-5BAAAC 6-10DDCBD 11-15ABCBC完型填空:16-20BACCA 21-25DBDCB 26-30ADCBA 31-35DBCCD阅读理解:36-40CADDA 41-45BBCBC 46-50BDCAA 51-55CADCB阅读表达:56. ① The waiters can wait on you.② You can stay with your friends.③ You don’t have to do the clean-up.57. Choice./Selection./Alternative.58. The advantages/benefits of eating at home.59. Less quantity, more quality.60. (略)评分建议:能基于篇章内容,紧扣主题,表述合理正确。
书面表达:61. One possible version:The Most Unforgettable Person I Ever knewIn my life I have met many people who are really worth my recalling. But perhaps the most unforgettable person I ever knew is my Chinese language teacher.What frequently brings back memories of my school teacher is his special qualities. First of all, I was attracted by his lively wit. I remember we always long for his class with great eagerness because his lecture were humorously delivered, never failing to provoke loud laughs. Second, he gave us the greatest gift a teacher can offer --- an awakening of a passion for learning. He not only led us to an appreciation of the beauty and perfection of Chinese and literature, but also aroused our great interest in exploring something deeper in this field. Finally, I was deeply impressed by the respect he showed for us, for he treated us like friends rather than students.Although it is nearly 10 years since I attended his last class, he is the talk of our old classmates, and I know part of him has already stayed in my heart.2015红桥二模数学文高三数学(文)答案一、选择题:每小题5分,共40分. 题号 1 2 3 4 5 6 7 8 答案CDABDCBC二、填空题:每小题5分,共30分.题号 91011121314答案1+i436138314三、解答题:共6小题,共80分. (15)(本小题满分13分)甲、乙两位学生参加数学竞赛培训.现分别从他们在培训期间参加的若干次预赛成绩中随机抽取8次.得到甲、乙两位学生成绩的茎叶图.(Ⅰ)现要从中选派一人参加数学竞赛,对预赛成绩的平均值和方差进行分析,你认为哪位学生的成绩更稳定?请说明理由;(Ⅱ)若将频率视为概率,求乙同学在一次数学竞赛中成绩高于84分的概率;(Ⅲ)在甲同学的8次预赛成绩中,从不小于80分的成绩中随机抽取2个成绩,列出所有结果,并求抽出的2个成绩均大于85分的概率. 解:(Ⅰ)派甲参加比较合适,理由如下:85)35124889290480270(81=++++++++⨯+⨯+⨯=甲x ,1(70180490353535)858x =⨯+⨯+⨯+++++=乙, …………………………… 3分222222)8585()8583()8580()8579()8578(81-+-+-+---=甲S])8595()8592()8590(222-+-+-+=35.5, 222222)8585()8583()8580()8580()8575[(81-+-+-+-+-=乙S])8595()8592()8590(222-+-+-+=41, ……………………………… 6分22,乙甲乙甲S S x x <= , ∴甲的成绩比较稳定. …………………………………………………………… 7分 (Ⅱ)D4182p ==. …………………………………………………………… 9分 (Ⅲ)从不小于80分的成绩中随机抽取2个成绩,所有结果为(8182),,(8184),,(8188),,(8193),,(8195),,(8284),,(8288),,(8293),,(8295),,(8488),,(8493),,(8495),,(8893),,(8895),,(9395),,共15个. ………………………………………………… 11分其中,满足2个成绩均大于85分的有(8893),,(8895),,(9395),,共3个, 所以,所求概率为31155=. …………………………………………………13分 (16)(本小题满分13分)设函数()23sin 22sin 1f x x x =+-.(Ⅰ)求函数()f x 的最大值和最小正周期;(Ⅱ)已知ABC △中,角A B C ,,的对边分别为a b c ,,,若()2f C =,3CA CB ⋅=,112a b +=,求边c . 解:(Ⅰ)2()3sin 22sin 1f x x x =+-3sin 2cos2x x =-π2sin(2)6x =-,所以函数()f x 的最大值为2,最小正周期πT =. …………………………… 6 分(Ⅱ)由()2f C =,得π3C =. 由3CA CB ⋅=,得6ab =. 又112a b +=,得22273()24a b a b ab +=+-=, 由余弦定理:2222cos c a b ab C =+-7349644=-=. 所以72c =. …………………………………………………………………13分(17)(本小题满分13分)如图,已知PA ⊥平面ABC ,PQ ⊥平面QBC ,且2PA PQ AB AC BC =====. (Ⅰ)求证:BC ⊥平面PAQ ; (Ⅱ)求二面角P BC Q --的正弦值.解:(Ⅰ)证明:∵PA ⊥平面ABC ,PQ ⊥平面QBC ,BC ⊂平面ABC ,BC ⊂平面QBC ,∴PA ⊥BC ,PQ ⊥BC ,又PA PQ P =,∴BC ⊥平面PAQ . ………………………………………………4分(Ⅱ)∵PA ⊥平面ABC ,90PAC PAB ∠=∠=, 已知2PA AB AC BC ====,∴22PB PC ==, ∵PQ ⊥平面QBC ,∴90PQB PQC ∠=∠=,又∵2,22PQ PQ PB PC ====,∴PQB PQC ∆≅∆ ∴2BQ CQ ==, 取BC 的中点D ,连结QD PD ,,则QD BC PD BC ⊥⊥,, ∴PDQ ∠为二面角P BC Q --的平面角, ∵37AD PD ==,,∴227sin 77PQ PDQ PD ∠===. 故二面角P BC Q --的正弦值为277. …………………………………13 分(18)(本小题满分13分)已知数列{}n a 的前n 项和为n S ,且满足1a a =,1n n a a λ+=(0λ≠),其中*n ∈N . (Ⅰ)当2a λ==时,求5a 和5S ;(Ⅱ)已知*m n t ∈N ,,,若m S ,n S ,t S 成等差数列,求证:对任意的自然数k ,m k a +,n k a +,t k a +成等差数列;(Ⅲ)在(Ⅰ)的条件下,设数列{}n b 满足(21)2nn nna b n =+⋅,是否存在正整数r s ,(1)r s <<,使得1r s b b b ,,成等比数列?若存在,求出所有的r s ,的值;若不存在,请说明理由.解:(Ⅰ)当2a λ==时,数列{}n a 为12a =,公比为2的等比数列,所以55232a ==,552(12)6212S -==-.--------------------------------3分(Ⅱ)当0a =时,0n a =,0n S =,所以m k a +,n k a +,t k a +成等差数列. 当0a ≠时,{}n a 为1a a =,公比为λ的等比数列,所以1n n a a λ-=⋅.-----4分1λ=时,{}n a 为常数列,所以m k a +,n k a +,t k a +成等差数列;1λ≠时,(1)1n n a S λλ-=-.因为m S ,n S ,t S 成等差数列,所以2m t n S S S +=,得(1)(1)2(1)m t n a a a λλλ-+-=-,即2m t n λλλ+=⋅.-------------------7分所以(m k a ++)t k a +1112()2m k t k n k n k a a a a λλλ+-+-+-+-=⋅+⋅-⋅1(2)0k m t n a λλλλ-=⋅+-⋅=, 所以m k a ++t k a +=2n k a +,即m k a +,n k a +,t k a +成等差数列.----------------9分 (Ⅲ)当2a λ==时,n n n n na b 2)12(⋅+==12+n n,若1r s b b b ,,成等比数列, 则21()()21321r s r s =++,即2244163r s r r s =+++,可得223241r r s r -++=. 所以22410r r -++>,解得:661122r -<<+. 又*r ∈N ,且1r >,所以2r =,此时12s =.故当且仅当2r =,12s =使得1r s b b b ,,成等比数列. …………………………13分 (19)(本小题满分14分)如图椭圆22:143x y C +=的右顶点是A ,上下两个顶点分别为B D ,,四边形OANB 是矩形(O 为原点),点E M ,分别为线段OA AN ,的中点. (Ⅰ)证明:直线DE 与直线BM 的交点在椭圆C 上;(Ⅱ) 已知圆O :221x y +=,直线l :1mx ny +=,当点()P m n ,在椭圆C 上运动时,求直线l 被圆O 所截得的弦长的取值范围.解:(Ⅰ)由题意,得(20)A ,,(03)B ,,(03)D -,,(10)E ,,3(2)2M ,, 所以直线DE 的方程33y x =-,直线BM 的方程为334y x =-+, ………… 2分 由33334y x y x ⎧=-⎪⎨=-+⎪⎩,, 得85335x y ⎧=⎪⎪⎨⎪=⎪⎩,. 所以直线DE 与直线BM 的交点坐标为833()55,, ………………………………… 4分 因为22833()()55143+=,所以点833()55,在椭圆22:143x y C +=上. ……………… 6分y xN MEDBAO(Ⅱ)∵点()P m n ,在椭圆C 上运动, ∴ 22143m n+=,22334n m =- ①.圆心到直线的距离221d m n=+.-----------------------8分∵直线l 被圆O 所截得的弦长22121L m n =-+,----------------------------10分将①代入,得222112121134L m n m =-=-++. ∵2m 0≤≤4,∴21344m +3≤≤,----------------------------------------12分所以2633L ≤≤. 故直线l 被圆O 所截得的弦长的取值范围26[3]3,. ……………………… 14分(20)(本小题满分14分)已知函数()mf x mx x=-,()2ln g x x =. (Ⅰ)当2m =时,若直线l 过点(04)-,且与曲线()y f x =相切,求直线l 的线方程; (Ⅱ)当1m =时,判断方程()g()f x x =在区间()1,+∞上有无实根;(Ⅲ)若(]1x e ∈,时,不等式()()2f x g x -<恒成立,求实数m 的取值范围. 解:(Ⅰ)令切点为00()x y ,,当2m =时,()22f x x x =-,()22'2f x x=+, ∴ ()0202'2k f x x ==+,切线l 的方程为0020022(2)(2)()y x x x x x --=+-,又直线l 过点(04)-,,01x ∴=, ∴切线方程为44y x =-. ……………………………………………………… 5分(Ⅱ)1m =时,令()()()12ln h x f x g x x x x=-=--, ()222112'()10x h x x x x -=+-=≥,()h x ∴在()0,+∞上为增函数,又(1)0h =,所以()()f x g x =在()1+∞,内无实数根. ………………………… 10分 (Ⅲ)2ln 2mmx x x--<恒成立, 即()2122ln m x x x x -<+恒成立, 又210x ->,则当(]1,x e ∈时,222ln 1x x xm x +<-恒成立,令,只需m 小于()G x 的最小值,()()2222(ln ln 2)'1x x x G x x-++=-, ………………………………………… 11分1x e <≤,ln 0x ∴> ,∴当(]1x e ∈,时()'0G x <,()G x ∴在(]1,e 上单调递减,()G x ∴在(]1,e 的最小值为()241eG e e =-, 则m 的取值范围是241e e ⎛⎫-∞ ⎪-⎝⎭,. ……………………………………14分2015红桥二模 数学理高三数学(理)答案一、选择题:每小题5分,共40分 题号 1 2 3 4 5 6 7 8 答案CDABCCBA二、填空题:每小题5分,共30分.题号 91011121314答案2i436221583288三、解答题:共6小题,共80分. (15)(本小题满分13分)甲、乙两位学生参加数学竞赛培训.现分别从他们在培训期间参加的若干次预赛成绩中随机抽取8次.得到甲、乙两位学生成绩的茎叶图.(Ⅰ)现要从中选派一人参加数学竞赛,对预赛成绩的平均值和方差进行分析,你认为哪位学生的成绩更稳定?请说明理由;(Ⅱ)若将频率视为概率,对学生甲在今后的三次数学竞赛成绩进行预测,记这三次成绩中高于80分的次数为ξ,求ξ的分布列及数学期望E ξ. 解:(Ⅰ)派甲参加比较合适,理由如下:85)35124889290480270(81=++++++++⨯+⨯+⨯=甲x ,85)53535390480170(81=+++++⨯+⨯+⨯=乙x , ………………………… 3分222222)8585()8583()8580()8579()8578(81-+-+-+---=甲S])8595()8592()8590(222-+-+-+=35.5,222222)8585()8583()8580()8580()8575[(81-+-+-+-+-=乙S])8595()8592()8590(222-+-+-+=41, ………………………… 6分22,乙甲乙甲S S x x <= ,∴甲的成绩比较稳定. ……………………………… 7分(Ⅱ)记“甲同学在一次数学竞赛中成绩高于80分”为事件A , 则4386)(==A P . ……………………………………………………………… 9分随机变量ξ的可能取值为0,1,2,3, 且ξ服从B (43,3),,)431()43()(3313k C k P --⋅==∴ξk=0,1,2,3. ξ的分布列为:……………………………………………………… 11分49642736427264916410=⨯+⨯+⨯+⨯=∴ξE(或49433=⨯==np E ξ). ………13分(16)(本小题满分13分)己知函数π()sin()(00||)2f x A x A ωφωφ=+>><,,的部分图象如图所示. (Ⅰ)求ω的值和函数()f x 的解析式;(Ⅱ)已知ABC △中,角A B C ,,的对边分别为a b c ,,,若()12f C =,边72c =,ABC △的面积为332,求a b +的值. 解:(Ⅰ)由图象知1A =,5ππ4()π126T =-=,故,2π2Tω==,将点π(1)6,代入π()sin(2)(||)2f x x φφ=+<,得π6φ=,故()f x 的解析式为π()sin(2)6f x x =+. ……………………………6 分(Ⅱ)由()12f C =,得π3C =, ξ123P641649 6427 6427NDCBAMzyxMABCDN已知72c =,ABC ∆的面积为332,则22331πsin 22349π2cos 43ab a b ab ⎧=⎪⎪⎨⎪=+-⎪⎩,226494ab a b ab =⎧⎪⇒⎨+=+⎪⎩ 所以2212111242a b a b ab +=++==. ………………………… 13分 (17)(本小题满分13分)如图,四边形ABCD 是边长为1的正方形,MD ABCD ⊥平面,NB ABCD ⊥平面,且1MD NB == .(Ⅰ)求证:AN ⊥CM ;(Ⅱ)求二面角A MN C --的余弦值;(Ⅲ)若E 为AM 中点,在线段AB 上是否存在点S ,使得//ES 平面CMN ?若存在,求线段ES 的长;若不存在,请说明理由. 解:证明:如图,以D 为坐标原点,建立空间直角坐标D xyz -,依题意,得(0,0,0)(1,0,0),(1,1,0),(0,1,0),(0,0,1),(1,1,1)D A B C M N .--------------1分(Ⅰ)(0,1,1)MC =-,(0,1,1)AN =,所以0MC AN ⋅=,故AN ⊥CM . ……………… 4分(Ⅱ)设平面AMN 的法向量(),,a b c =1n ,(1,1,0)MN =,00MN AN ⎧⋅=⎪⎨⋅=⎪⎩11n n 00a b b c +=⎧⇒⎨+=⎩,得一个(1,1,1)=-1n . 设平面CMN 的法向量()2,,e f g =n ,(1,0,1)CN =,2200MN CN ⎧⋅=⎪⎨⋅=⎪⎩n n 00e f e g +=⎧⇒⎨+=⎩,得一个2(1,1,1)=--n . 故12||||3==n n ,21111⋅=+-=1n n , 所以12121cos ||||3θ⋅==⋅n n n n ,所以二面角A MN C --的余弦值为13. ……………………………………9 分(Ⅲ)假设在线段AB 上存在点S ,使得//ES 平面CMN .y N MB因为11(,0,)22E ,设(1,,0)S y ,又11,,22ES y ⎛⎫=- ⎪⎝⎭,由(Ⅱ)知平面CMN 的法向量2(1,1,1)=--n ,若//ES 平面CMN ,则ES ⊥2n ,即11,,(1,1,1)022y ⎛⎫-⋅--= ⎪⎝⎭,解得1y =.故当点S 与点B 重合时,使得//ES 平面CMN ,此时22116()1()222ES =++-=. ………………………………………13 分(18)(本小题满分13分)如图椭圆22:143x y C +=的右顶点是A ,上下两个顶点分别为B D ,,四边形OANB 是矩形(O 为原点),点E M ,分别为线段OA AN ,的中点. (Ⅰ)证明:直线DE 与直线BM 的交点在椭圆C 上;(Ⅱ) 已知圆O :221x y +=,直线l :1mx ny +=,当点()P m n ,在椭圆C 上运动时,求直线l 被圆O 所截得的弦长的取值范围.解:(Ⅰ)由题意,得(20)A ,,(03)B ,,(03)D -,,(10)E ,,3(2)2M ,, 所以直线DE 的方程33y x =-,直线BM 的方程为334y x =-+, ………… 2分 由33334y x y x ⎧=-⎪⎨=-+⎪⎩,, 得85335x y ⎧=⎪⎪⎨⎪=⎪⎩,. 所以直线DE 与直线BM 的交点坐标为833()55,, ………………………………… 4分 因为22833()()55143+=,所以点833()55,在椭圆22:143x y C +=上. ……………… 6分(Ⅱ)∵点()P m n ,在椭圆C 上运动, ∴22143m n +=,22334n m =- ①.圆心到直线的距离221d m n=+.∵直线l 被圆O 所截得的弦长22121L m n =-+,将①代入,得222112121134L m n m =-=-++. ∵2m 0≤≤4,∴21344m +3≤≤,所以2633L ≤≤. 故直线l 被圆O 所截得的弦长的取值范围26[3]3,. ……………………… 13分 (19)(本小题满分14分)已知数列{}n a 的前n 项和为n S ,且满足1a a =()a ∈R ,10n n a a λ+-=(0λ≠),其中*n ∈N .(Ⅰ)当2a λ==时,求5a 和5S ;(Ⅱ)已知*m n t ∈N ,,,若m S ,n S ,t S 成等差数列,求证:对任意的自然数k ,m k a +,n k a +,t k a +成等差数列;(Ⅲ)当2a λ==时,若1n n nsa a a λ+-=,s ∈R ,用数学归纳法证明:4s -≥时,对一切n *∈N ,2n a ≥成立.解:(Ⅰ)当2a λ==时,数列{}n a 为12a =,公比为2的等比数列,所以55232a ==,552(12)6212S -==-.-----------------------3分(Ⅱ)当0a =时,0n a =,所以m k a +,n k a +,t k a +成等差数列.当0a ≠时,{}n a 为1a a =,公比为λ的等比数列,所以1n n a a λ-=⋅,-------------4分1λ=时,{}n a 为常数列,m k a +,n k a +,t k a +成等差数列;1λ≠时,(1)1n n a S λλ-=-.因为m S ,n S ,t S 成等差数列,所以2m t n S S S +=,得(1)(1)2(1)m t n a a a λλλ-+-=-,即2m t n λλλ+=⋅.------------------------7分 所以(m k a ++)t k a +1112()2m k t k n k n k a a a a λλλ+-+-+-+-=⋅+⋅-⋅1(2)0k m t n a λλλλ-=⋅+-⋅=, 所以m k a ++t k a +=2n k a +,即m k a +,n k a +,t k a +成等差数列.---------------------9分 (Ⅲ)当2a λ==时,12n n nsa a a +=+. 下面用数学归纳法证明:4s -≥时,对一切n *∈N ,2n a ≥成立. 证明:(1)显然1n =时,结论成立;(2)假设(1)n k k =≥时结论成立,即2k a ≥.-------------------------------------------------10分当1n k =+时,12k k ksa a a +=+.---------------------------------------------------------------------11分考察函数()2sf x x x =+,[2)x ∈+∞,, ① 若40s -≤≤,由2'()20sf x x =->,知()f x 在区间[2)+∞,上单调递增. 由假设得12k k k s a a a +=+42s≥+2≥.② 若0s >,对[2,)x ∈+∞总有()242sf x x x=+>>, 则由假设得122k k ksa a a +=+>. 所以,1n k =+时,结论成立,综上可知:当4s ≥-时,对一切n *∈N ,2n a ≥成立. ………………14分 (20)(本小题满分14分)已知函数()mf x mx x=-,()2ln g x x =. (Ⅰ)当2m =时,若直线l 过点(04)-,且与曲线()y f x =相切,求直线l 的线方程; (Ⅱ)当1m =时,判断方程()g()f x x =在区间()1,+∞上有无实根;(Ⅲ)若(]1x e ∈,时,不等式()()2f x g x -<恒成立,求实数m 的取值范围. 解:(Ⅰ)令切点为00()x y ,,当2m =时,()22f x x x =-,()22'2f x x=+,∴ ()0202'2k f x x ==+,切线l 的方程为0020022(2)(2)()y x x x x x --=+-, 又直线l 过点(04)-,,01x ∴=, ∴切线方程为44y x =-. ……………………………………………………… 5分(Ⅱ)1m =时,令()()()12ln h x f x g x x x x=-=--, ()222112'()10x h x x x x -=+-=≥,()h x ∴在()0,+∞上为增函数, 又(1)0h =,所以()f xg x =在()1+∞,内无实数根. ………………………… 10分 (Ⅲ)2ln 2mmx x x--<恒成立, 即()2122ln m x x x x -<+恒成立, 又210x ->,则当(]1,x e ∈时,222ln 1x x xm x +<-恒成立,令,只需m 小于()G x 的最小值,()()2222(ln ln 2)'1x x x G x x-++=-, ………………………………………… 11分1x e <≤,ln 0x ∴> ,∴当(]1x e ∈,时()'0G x <,()G x ∴在(]1,e 上单调递减,()G x ∴在(]1,e 的最小值为()241eG e e =-, 则m 的取值范围是241e e ⎛⎫-∞ ⎪-⎝⎭,. ……………………………………14分2015红桥二模 政治。