2012年上海市中考真题(word版含答案)
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2012年上海市初中毕业统一学业考试·数学(满分:150分 时间:120分钟)一、选择题(本大题共6题,每题4分,满分24分)1.在下列代数式中,次数为3的单项式是( )(A )2xy (B )33x y + (C )3x y (D )3xy2.数据5,7,5,8,6,13,5的中位数是( )(A )5 (B )6 (C )7 (D )83.不等式组2620x x -<⎧⎨->⎩,的解集是( ) (A )3x >- (B )3x <- (C )2x > (D )2x <4.在下列各式中,二次根式 )(A(B(C(D5.在下列图形中,为中心对称图形的是( )(A )等腰梯形 (B )平行四边形(C )正五边形 (D )等腰三角形6.如果两圆的半径长分别为6和2,圆心距为3,那么这两圆的位置关系是( )(A )外离 (B )相切 (C )相交 (D )内含二、填空题(本大题共12题,每题4分,满分48分)7.计算112-=________. 8.因式分解xy x -=________.9.已知正比例函数(0)y kx k =≠,点(23)-,在函数的图象上,则y 随x 的增大而________(增大或减小).102=的根是________.11.如果关于x 的一元二次方程260x x c -+=(c 是常数)没有实数根,那么c 的取值范围是________.12.将抛物线2y x x =+向下平移2个单位,所得新抛物线的表达式是________.13.布袋中装有3个红球和6个白球,它们除颜色外其他都相同,如果从布袋里随机摸出一个球,那么所摸到的球恰好为红球的概率是________.14.某校500名学生参加生命安全知识测试,测试分数均大于或等于60且小于100,分数段的频率分布情况如下表所示(其中每个分数段可包括最小值,不包括最大值),结合下表15.如图,已知梯形ABCD ,AD BC ∥,2BC AD =,如果AD =a ,AB =b ,那么AC =________(用a ,b 表示).16.在ABC △中,点D E ,分别在AB AC ,上,AED B ∠=∠,如果2AE =,ADE △的面积为4,四边形BCED 的面积为5,那么边AB 的长为________. 17.我们把两个三角形的重心之间的距离叫做重心距,在同一平面内有两个边长相等的等边三角形,如果当它们的一边重合时重心距为2,那么当它们的一对角成对顶角时重心距为________.18.如图,在Rt ABC △中,90C ∠=°,1BC =,点D 在AC 上,将ADB △沿直线BD 翻折后,将点A 落在点E 处,如果AD ED ⊥,那么线段DE 的长为________. 三、解答题(本大题共7题,满分78分)19.(本题满分10分)计算:121211)3()22-⨯+-.20.(本题满分10分)解方程:261393x x x x +=+--.中,90ACB ∠=°,D 是边AB 的中点,BE CD ⊥,垂足为点E .已知15AC =,3cos 5A =. (1)求线段CD 的长;(2)求sin DBE ∠的值.22.(本题满分10分,第(1)小题满分5分,第(2)小题满分5分)某工厂生产一种产品,当生产数量至少为10吨,但不超过50吨时,每吨的成本y (万元/吨)与生产数量x (吨)的函数关系式如图所示.(1)求y 关于x 的函数解析式,并写出它的定义域;(2)当生产这种产品的总成本为280万元时,求该产品的生产数量.(注:总成本=每吨的成本×生产数量)23.(本题满分12分,第(1)小题满分5分,第(2)小题满分7分)已知:如图,在菱形ABCD 中,点E F 、分别在边BC CD 、上,BAF DAE ∠=∠,AE 与BD 交于点G .(1)求证:BE DF =;(2)当DF AD FC DF=时,求证:四边形BEFG 是平行四边形.分)如图,在平面直角坐标系中,二次函数26y ax x c =++的图象经过点(40)(10)A B -,、,,与y 轴交于点C ,点D 在线段OC 上,OD t =,点E 在第二象限,90ADE ∠=°,1tan 2DAE EF OD ∠=⊥,,垂足为F . (1)求这个二次函数的解析式;(2)求线段EF OF 、的长(用含t 的代数式表示);(3)当ECA OAC ∠=∠时,求t 的值.25.(本题满分14分,第(1)小题满分3分,第(2)小题满分5分,第(3)小题满分6分)如图,在半径为2的扇形AOB 中,90AOB ∠=°,点C 是弧AB 上的一个动点(不与点A B 、重合),OD BC ⊥,OE AC ⊥,垂足分别为D E 、.(1)当1BC =时,求线段OD 的长;(2)在DOE △中是否存在长度保持不变的边?如果存在,请指出并求其长度;如果不存在,请说明理由;(3)设BD x =,DOE △的面积为y ,求y 关于x 的函数关系式,并写出它的定义域.2012年上海市初中毕业统一学业考试数学参考答案一、1. A2. B3. C4. C5. B6. D二、7.128.(1)x y - 9.减小 10.3x = 11.9c > 12.22y x x =+- 13.13 14.150 15.2a b + 16.3 17.4 181 三、19.解:原式=412-································································· (4分)=21···································································· (8分) =3. ·········································································································· (10分)20.解:(3)63x x x -+=+. ······················································································· (3分) 2430x x -+=. ····································································································· (6分) 11x =或23x =. ······································································································ (9分) 经检验:3x =是方程的增根,1x =是原方程的根. ·········································· (10分) 21.解:(1)在Rt ABC △中,因为15AC =,3cos 5A =.则得153cos 5AC A AB AB ===,解得25AB =,再由直角三角形斜边上的中线等于斜边的一半可得12522CD AB ==. ··································································································································· (4分)(2)由15AC =,25AB =,利用勾股定理可得20BC =, ····································· (6分)又因cos sin A ABC =∠,得3sin 5ABC ∠=.又因CD DB =,于是得ECB ABC ∠=∠,由sin sin ABC ECB ∠=∠,得4cos 5ECB ∠=,又因20BC =,解得16EC =. ····································································································································· (8分) 因252CD =,于是72DE =,252DB =,则7sin 25DE DBE DB ∠==. ··················· (10分) 22.解:(1)因为所求函数的图象是一条直线,故设其函数解析式为y kx b =+,又因点(1010),、(506),在这个函数的图象上,将其直接代入y kx b =+可得101050 6.k b k b +=⎧⎨+=⎩,解得110k =-,11b =, 可得11110y x =-+, ····································································································· (4分) 由函数图象可得x 的取值范围为1050x ≤≤. ··························································· (5分)(2)由题意得1(11)28010x x -+=, ··········································································· (8分) 解得140x =,270x =.因为1050x ≤≤,所以40x =. ······································· (9分) 答:该产品的生产数量为40吨. ·················································································· (10分)23.解:(1)因为四边形ABCD 是菱形,则AD AB =,又因BAF DAE ∠=∠,所以BAE DAF ∠=∠,又由菱形的性质可得ADC ABC ∠=∠,于是(ASA)ABE ADF △≌△,则BE DF =. ·································································· (5分)(2)由AD BC ∥结合DF BE =,得AD AD DG DF DF BE GB FC===,得GF BE ∥; ··································································································································· (9分)另外因为DF BE =,DC BC =,所以DC BC DF BE=,则得BG EF ∥, ················· (11分) 于是由平行四边形的判定可得四边形BEFG 是平行四边形. ····································· (12分) 24.解:(1)把(40)(10)A B -,,,,代入26y ax x c =++解得28a c =-=,. ······· (2分)2268y x x ∴=-++ ········································································································ (3分) (2)9090EFD EDA DEF EDF ∠=∠=∴∠+∠=°,°,90EDF ODA ∠+∠=°,DEF ODA ∴∠=∠,EDF DAO ∴△∽△, ···································································································· (5分) 111.222EF ED ED EF DO t EF t DO DA DA DO ∴=∴=∴==∴=,,,, ···································· (7分) 同理12DF ED OA DA ==,且4OA =,2DF ∴=,2OF t =-. ··································· (8分) (3)如图连接EC AC ,,过点A 作EC 的垂线,交CE 于点G . 1(2)2E t t --,,由90GCA OAC AC AC CGA COA ∠=∠=∠=∠=,,°, (AAS)GAC OCA ∴△≌△. ······································· (9分)4CG ∴=,令2268y x x =-++中0x =,解得(08)C ,,即8AG =,(4AE =EG ∴=222EF CF CE +=,2221()(10)4)2t t +-=, ································ (10分) 解得110t =,26t =, ··································································································· (11分) 其中110t =(不舍题意,舍去),6t ∴=. ································································· (12分)25.解:(1)1122OD BC BD BC ⊥∴==,,OD ∴== ····················································································· (3分) (2)存在,DE 是不变的. ··························································································· (4分)如图①,连接AB 且由勾股定理可得AB = ·················· (6分)由垂径定理可得点D E 、分别是BC 和CA 的中点,则DE 是ABC △的中位线,由此求得12DE AB == ···· (8分)(3)2BD x OB ==,,由勾股定理可得OD =,又123490B O A ∠=∠∠=∠∠=,,°,2345DOE ∴∠+∠=∠=°, ·················· (10分) 如图②,过点D 作DF OE ⊥,在Rt ODF △中,45DOF ∠=°,OD =2DF ∴==2DE =2EF x =,)22OE OF EF x x =+==, ································ (12分)1)2y x ∴=21(44x x =-+<< ··········································································· (14分)。