重庆市2009年初中毕业暨高中招生考试数 学 试 卷(本卷共五个大题,满分150分,考试时间120分钟)参考公式:抛物线2y ax bx c =++(0a ≠)的顶点坐标为2424b ac b a a ⎛⎫-- ⎪⎝⎭,,对称轴公式为2bx a=-. 一、选择题:(本大题10个小题,每小题4分,共40分)在每个小题的下面,都给出了代号为A 、B 、C 、D 的四个答案,其中只有一个是正确的,请将正确答案的代号填在题后的括号中.1.5-的相反数是( ) A .5B .5-C .15D .15-2.计算322x x ÷的结果是( ) A .xB .2xC .52xD .62x3.函数13y x =+的自变量x 的取值范围是( ) A .3x >- B .3x <- C .3x ≠- D .3x -≥4.如图,直线AB CD 、相交于点E ,DF AB ∥.若100AEC ∠=°, 则D ∠等于( )A .70°B .80°C .90°D .100° 5.下列调查中,适宜采用全面调查(普查)方式的是( )A .调查一批新型节能灯泡的使用寿命B .调查长江流域的水污染情况C .调查重庆市初中学生的视力情况D .为保证“神舟7号”的成功发射,对其零部件进行检查6.如图,O ⊙是ABC △的外接圆,AB 是直径.若80BOC ∠=°,则A ∠等于( )A .60°B .50°C .40°D .30°7.由四个大小相同的正方体组成的几何体如图所示,那么它的左视图是( )A .B .C .D .8.观察下列图形,则第n 个图形中三角形的个数是( )C AEB FD4题图 (1)第2个第3个6题图A .22n +B .44n +C .44n -D .4n9.如图,在矩形ABCD 中,2AB =,1BC =,动点P 从点B 出发, 沿路线B C D →→作匀速运动,那么ABP △的面积S 与点P 运动 的路程x 之间的函数图象大致是( )10.如图,在等腰Rt ABC △中,908C AC ∠==°,,F 是AB 边上的中点,点D 、E 分别在AC 、BC 边上运动,且保持AD CE =.连接DE 、DF 、EF .在此运动变化的过程中,下列结论:①DFE △是等腰直角三角形; ②四边形CDFE 不可能为正方形, ③DE 长度的最小值为4;④四边形CDFE 的面积保持不变;⑤△CDE 面积的最大值为8.其中正确的结论是( ) A .①②③ B .①④⑤ C .①③④ D .③④⑤ 二、填空题:(本大题6个小题,每小题4分,共24分)在每小题中,请将答案直接填在题后的横线上.11.据重庆市统计局公布的数据,今年一季度全市实现国民生产总值约为7840000万元.那么7840000万元用科学记数法表示为 万元.12.分式方程1211x x =+-的解为 . 13.已知ABC △与DEF △相似且面积比为4∶25,则ABC △与DEF △的相似比为 .14.已知1O ⊙的半径为3cm ,2O ⊙的半径为4cm ,两圆的圆心距12O O 为7cm ,则1O ⊙与2O ⊙的位置关系是 .15.在平面直角坐标系xOy 中,直线3y x =-+与两坐标轴围成一个AOB △.现将背面完全相同,正面分别标有数1、2、3、12、13的5张卡片洗匀后,背面朝上,从中任取一张,将该卡片上的数作为点P 的横坐标,将该数的倒数作为点P 的纵坐标,则点P 落在AOB△内的概率为 .16.某公司销售A 、B 、C 三种产品,在去年的销售中,高新产品C 的销售金额占总销售金额的40%.由于受国际金融危机的影响,今年A 、B 两种产品的销售金额都将比去年减少20%,因而高新产品C 是今年销售的重点.若要使今年的总销售金额与去年持平,那么今年高新产品C 的销售金额应比去年增加 %. 三、解答题:(本大题4个小题,每小题6分,共24分)解答时每小题必须给出必要的演算过程或推理步骤.A .B .C .D .D C P BAC E B A FD10题图17.计算:1021|2|(π(1)3-⎛⎫-+⨯-- ⎪⎝⎭.18.解不等式组:303(1)21x x x +>⎧⎨--⎩,①≤.②19.作图,请你在下图中作出一个以线段AB 为一边的等边ABC △.(要求:用尺规作图,并写出已知、求作,保留作图痕迹,不写作法和结论)已知: 求作:20.为了建设“森林重庆”,绿化环境,某中学七年级一班同学都积极参加了植树活动,今年4月该班同学的植树情况的部分统计如下图所示:(1)请你根据以上统计图中的信息,填写下表:该班人数 植树株数的中位数 植树株数的众数(2)请你将该条形统计图补充完整.A B19题图(株) 20题图植树2株的人数占32%四、解答题:(本大题4个小题,每小题10分,共40分)解答时每小题必须给出必要的演算过程或推理步骤.21.先化简,再求值:22121124x x x x ++⎛⎫-÷ ⎪+-⎝⎭,其中3x =-.22.已知:如图,在平面直角坐标系xOy 中,直线AB 分别与x y 、轴交于点B 、A ,与反比例函数的图象分别交于点C 、D ,CE x ⊥轴于点E ,1tan 422ABO OB OE ∠===,,.(1)求该反比例函数的解析式; (2)求直线AB 的解析式.23.有一个可自由转动的转盘,被分成了4个相同的扇形,分别标有数1、2、3、4(如图所示),另有一个不透明的口袋装有分别标有数0、1、3的三个小球(除数不同外,其余都相同),小亮转动一次转盘,停止后指针指向某一扇形,扇形内的数是小亮的幸运数,小红任意摸出一个小球,小球上的数是小红的吉祥数,然后计算这两个数的积. (1)请你用画树状图或列表的方法,求这两个数的积为0的概率;(2)小亮与小红做游戏,规则是:若这两个数的积为奇数,小亮赢;否则,小红赢.你认为该游戏公平吗?为什么?如果不公平,请你修改该游戏规则,使游戏公平.24.已知:如图,在直角梯形ABCD 中,AD ∥BC ,∠ABC =90°,DE ⊥AC 于点F ,交BC 于点G ,交AB 的延长线于点E ,且AE AC =. (1)求证:BG FG =;(2)若2AD DC ==,求AB 的长.DC EB GA24题图F x 23题图五、解答题:(本大题2个小题,第25小题10分,第26小题12分,共22分)解答时每小题必须给出必要的演算过程或推理步骤.25.某电视机生产厂家去年销往农村的某品牌电视机每台的售价y (元)与月份x 之间满足函数关系502600y x =-+,去年的月销售量p (万台)与月份x 之间成一次函数关系,其中两个月的销售情况如下表:月份 1月 5月 销售量 3.9万台 4.3万台(1)求该品牌电视机在去年哪个月销往农村的销售金额最大?最大是多少?(2)由于受国际金融危机的影响,今年1、2月份该品牌电视机销往农村的售价都比去年12月份下降了%m ,且每月的销售量都比去年12月份下降了1.5m%.国家实施“家电下乡”政策,即对农村家庭购买新的家电产品,国家按该产品售价的13%给予财政补贴.受此政策的影响,今年3至5月份,该厂家销往农村的这种电视机在保持今年2月份的售价不变的情况下,平均每月的销售量比今年2月份增加了1.5万台.若今年3至5月份国家对这种电视机的销售共给予了财政补贴936万元,求m 的值(保留一位小数).5.8315.9166.0836.164)26.已知:如图,在平面直角坐标系xOy 中,矩形OABC 的边OA 在y 轴的正半轴上,OC 在x 轴的正半轴上,OA =2,OC =3.过原点O 作∠AOC 的平分线交AB 于点D ,连接DC ,过点D 作DE ⊥DC ,交OA 于点E .(1)求过点E 、D 、C 的抛物线的解析式;(2)将∠EDC 绕点D 按顺时针方向旋转后,角的一边与y 轴的正半轴交于点F ,另一边与线段OC 交于点G .如果DF 与(1)中的抛物线交于另一点M ,点M 的横坐标为65,那么EF =2GO 是否成立?若成立,请给予证明;若不成立,请说明理由;(3)对于(2)中的点G ,在位于第一象限内的该抛物线上是否存在点Q ,使得直线GQ 与AB 的交点P 与点C 、G 构成的△PCG 是等腰三角形?若存在,请求出点Q 的坐标;若不存在,请说明理由.26题图x重庆市2009年初中毕业暨高中招生考试数学试题参考答案及评分意见一、选择题1.A 2.B 3.C 4.B 5.D 6.C 7.A 8.D 9.B 10.B 二、填空题11.67.8410⨯ 12.3x =- 13.2:5 14.外切 15.3516.30 三、解答题17.解:原式23131=+⨯-+ ···································································· (5分) 3=. ·················································································· (6分) 18.解:由①,得3x >-. ········································································ (2分)由②,得2x ≤. ········································································· (4分) 所以,原不等式组的解集为32x -<≤. ·········································· (6分)19.解:已知:线段AB . ········································································· (1分) 求作:等边ABC △. ··············································································· (2分) 作图如下:(注:每段弧各1分,连接线段AC BC 、各1分)······························································ (6分)20.解:(1)填表如下:该班人数 植树株数的中位数 植树株树的众数 50 3 2············································· (4分)(2)补图如下:························ (6分)四、解答题:21.解:原式221(1)2(2)(2)x x x x x +-+=÷++- ······················································ (4分)A BC(株)21(2)(2)2(1)x x x x x ++-=++ ·············································································· (6分) 21x x -=+. ······························································································· (8分) 当3x =-时,原式325312--==-+.····························································· (10分) 22.解:(1)42OB OE ==,,246BE ∴=+=. CE x ⊥轴于点E .1tan 2CE ABO BE ∴∠==,3CE ∴=. ························································· (1分) ∴点C 的坐标为()23C -,. ······································································· (2分) 设反比例函数的解析式为(0)my m x =≠. 将点C 的坐标代入,得32m=-, ································································· (3分)6m ∴=-. ···························································································· (4分) ∴该反比例函数的解析式为6y x=-. ·························································· (5分)(2)4OB =,(40)B ∴,.····································································· (6分) 1tan 2OA ABO OB ∠==, 2OA ∴=,(02)A ∴,. ············································································ (7分) 设直线AB 的解析式为(0)y kx b k =+≠.将点A B 、的坐标分别代入,得240.b k b =⎧⎨+=⎩,·················································· (8分)解得122.k b ⎧=-⎪⎨⎪=⎩, ························································································· (9分) ∴直线AB 的解析式为122y x =-+. ······················································· (10分) 23.解:(1)画树状图如下:···················· (4分)0 1 30 2 60 3 90 4 120 1 3 0 1 3 0 1 3 0 1 3 2 3 4 1 幸运数 吉祥数 积或列表如下: 幸运数积吉祥数1 2 3 4 0 0 0 0 0 1 1 2 3 4 3 3 6 9 12············································································································· (4分) 由图(表)知,所有等可能的结果有12种,其中积为0的有4种, 所以,积为0的概率为41123P ==. ···························································· (6分) (2)不公平. ························································································· (7分) 因为由图(表)知,积为奇数的有4种,积为偶数的有8种.所以,积为奇数的概率为141123P ==, ························································ (8分) 积为偶数的概率为282123P ==. ································································ (9分) 因为1233≠,所以,该游戏不公平.游戏规则可修改为:若这两个数的积为0,则小亮赢;积为奇数,则小红赢. ································ (10分) (只要正确即可) 24.(1)证明:90ABC DE AC ∠=°,⊥于点F ,ABC AFE ∴∠=∠. ································ (1分) AC AE EAF CAB =∠=∠,, ABC AFE ∴△≌△ ································· (2分) AB AF ∴=. ········································· (3分) 连接AG , ·············································· (4分)AG AG AB AF ==,,Rt Rt ABG AFG ∴△≌△. ······················ (5分) BG FG ∴=. ········································· (6分) (2)解:AD DC DF AC =,⊥,1122AF AC AE ∴==. ··········································································· (7分) 30E ∴∠=°.30FAD E ∴∠=∠=°, ············································································ (8分) AF ∴= ························································································· (9分) AB AF ∴== ··············································································· (10分)五、解答题:25.解:(1)设p 与x 的函数关系为(0)p kx b k =+≠,根据题意,得D CE B G AF3.954.3.k b k b +=⎧⎨+=⎩,························································································· (1分) 解得0.13.8.k b =⎧⎨=⎩,所以,0.1 3.8p x =+. ·························································· (2分)设月销售金额为w 万元,则(0.1 3.8)(502600)w py x x ==+-+. ···················· (3分) 化简,得25709800w x x =-++,所以,25(7)10125w x =--+.当7x =时,w 取得最大值,最大值为10125.答:该品牌电视机在去年7月份销往农村的销售金额最大,最大是10125万元. ··· (4分) (2)去年12月份每台的售价为501226002000-⨯+=(元), 去年12月份的销售量为0.112 3.85⨯+=(万台), ········································ (5分) 根据题意,得2000(1%)[5(1 1.5%) 1.5]13%3936m m -⨯-+⨯⨯=. ················· (8分) 令%m t =,原方程可化为27.514 5.30t t -+=.1415t ∴==. 10.528t ∴≈,2 1.339t ≈(舍去)答:m 的值约为52.8. ············································································ (10分) 26.解:(1)由已知,得(30)C ,,(22)D ,,90ADE CDB BCD ∠=-∠=∠°,1tan 2tan 212AE AD ADE BCD ∴=∠=⨯∠=⨯=. ∴(01)E ,. ····························································································· (1分) 设过点E D C 、、的抛物线的解析式为2(0)y ax bx c a =++≠. 将点E 的坐标代入,得1c =.将1c =和点D C 、的坐标分别代入,得42129310.a b a b ++=⎧⎨++=⎩,······················································································ (2分) 解这个方程组,得56136a b ⎧=-⎪⎪⎨⎪=⎪⎩故抛物线的解析式为2513166y x x =-++. ··················································· (3分) (2)2EF GO =成立. ············································································ (4分) 点M 在该抛物线上,且它的横坐标为65, ∴点M 的纵坐标为125. ··········································································· (5分) 设DM 的解析式为1(0)y kx b k =+≠, 将点D M 、的坐标分别代入,得1122612.55k b k b +=⎧⎪⎨+=⎪⎩, 解得1123k b ⎧=-⎪⎨⎪=⎩,. ∴DM 的解析式为132y x =-+. ······························································ (6分) ∴(03)F ,,2EF =. ·············································································· (7分) 过点D 作DK OC ⊥于点K , 则DA DK =.90ADK FDG ∠=∠=°, FDA GDK ∴∠=∠.又90FAD GKD ∠=∠=°, DAF DKG ∴△≌△. 1KG AF ∴==.1GO ∴=. ···························································································· (8分) 2EF GO ∴=.(3)点P 在AB 上,(10)G ,,(30)C ,,则设(12)P ,.∴222(1)2PG t =-+,222(3)2PC t =-+,2GC =.①若PG PC =,则2222(1)2(3)2t t -+=-+, 解得2t =.∴(22)P ,,此时点Q 与点P 重合.∴(22)Q ,. ···························································································· (9分) ②若PG GC =,则22(1)22t 2-+=, 解得 1t =,(12)P ∴,,此时GP x ⊥轴.GP 与该抛物线在第一象限内的交点Q 的横坐标为1,x。