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2006年上海市中考数学试题和答案

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2006 年上海市初中毕业生统一学业考试 数学试卷 (满分 150 分,考试时间 100 分钟) 题号 一 二 三 四 总分 17 18 19 20 21 22 23 24 25 得分 考生注意: 1.本卷含四大题,共 25 题; 2.除第一、二大题外,其余各题如无特别说明,都必须写出证明或计算的主要步骤. 一.填空题:(本大题共 12 题,满分 36 分) 【只要求直接写出结果,每个空格填对得 3 分,否则得零分】 1.计算: 4  __________. 2.计算: 12 xx__________. 3.不等式 60x 的解集是__________. 4.分解因式: 2x xy__________. 5.函数 1 3y x  的定义域是__________. 6.方程 2 1 1x 的根是__________. 7.方程 2 3 4 0xx   的两个实数根为 1x , 2x ,则 12xx __________. 8. 用 换 元 法 解 方 程 2 2 21221 xx xx  时,如果设 2 21 xy x  , 那 么 原 方 程 可 化 为 __________. 9.某型号汽油的数量与相应金额的关系如图 1 所示,那么这种汽油的单价是每升__________ 元. 10.已知在 ABC△ 和 1 1 1A B C△ 中, 11AB A B , 1AA∠ ∠ ,要使 1 1 1ABC A B C△ ≌△ , 还需添加一个条件,这个条件可以是__________. 11.已知圆O 的半径为1,点 P 到圆心O 的距离为 2 ,过点 P 引圆O 的切线,那么切线长 是__________. 12.在中国的园林建筑中,很多建筑图形具有对称性.图 2 是一个破损花窗的图形,请把它 补画成中心对称图形. 金额(单位:元) 509 0 100 数量(单位:升) 图 1 图 2 二.选择题:(本大题共 4 题,满分 16 分) 【下列各题的四个结论中,有且只有一个结论是正确的,把正确结论的代号写在题后的圆 括号内,选对得 4 分;不选、错选或者多选得零分】 13.在下列方程中,有实数根的是( ) A. 2 3 1 0xx   B. 4 1 1x    C. 2 2 3 0xx   D. 1 11 x xx 14.二次函数  213yx    图象的顶点坐标是( ) A. 13 , B. 13, C. 13, D. 13, 15.在 ABC△ 中, AD 是 BC 边上的中线,G 是重心.如果 6AG  ,那么线段 DG 的长 为( ) A. 2 B.3 C.6 D.12 16.在下列命题中,真命题是( ) A.两条对角线相等的四边形是矩形 B.两条对角线互相垂直的四边形是菱形 C.两条对角线互相平分的四边形是平行四边形 D.两条对角线互相垂直且相等的四边形是正方形 三.(本大题共 5 题,满分 48 分) 17.(本题满分 9 分) 先化简,再求值: 2111 x xx  ,其中 2x  . 18.(本题满分 9 分) 解方程组: 2 30 10 xy xy        , . 19.(本题满分 10 分,每小题满分各 5 分) 已知:如图 3,在 ABC△ 中,AD 是边 BC 上的高,E 为边 AC 的中点, 14BC  , 12AD  , 4sin 5B  .求(1)线段 DC 的长;(2) tg EDC∠ 的值. 20.(本题满分 10 分,第(1)小题满分 3 分,第(2)小题满分 4 分,第(3)小题满分 3 分) 某市在中心城区范围内,选取重点示范路口进行交通文明状况满意度调查,将调查结果的满 意度分为:不满意、一般、较满意、满意和非常满意,依次以红、橙、黄、蓝、绿五色标识.今 年五月发布的调查结果中,橙色与黄色标识路口数之和占被调查路口总数的15% .结合未 画完整的图 4 中所示信息,回答下列问题: (1)此次被调查的路口总数是__________; (2)将图 4 中绿色标识部分补画完整,并标上相应的路口数; (3)此次被调查路口的满意度能否作为该市所有路口交通文明状况满意度的一个随机样 本? 答:____________________. 21.(本题满分 10 分) 本市新建的滴水湖是圆形人工湖.为测量该湖的半径,小杰和小丽沿湖边选取 A ,B ,C 三 根木柱,使得 A , B 之间的距离与 A ,C 之间的距离相等,并测得 BC 长为 240 米, A 到 BC 的距离为5 米,如图 5 所示.请你帮他们求出滴水湖的半径. A E C D B 图 3 40 30 20 10 0 0 1 8 41 红 橙 黄 蓝 绿 路口数 标识 图 4 B A C 图 5 四.(本大题共 4 题,满分 50 分) 22.(本题满分 12 分,第(1)小题满分 5 分,第(2)小题满分 7 分) 如图 6,在直角坐标系中,O 为原点.点 A 在第一象限,它的纵坐标是横坐标的 3 倍,反比 例函数 12y x 的图象经过点 A . (1)求点 A 的坐标; (2)如果经过点 A 的一次函数图象与 y 轴的正半轴交于点 B ,且OB AB ,求这个一次 函数的解析式. 23.(本题满分 12 分,每小题满分各 6 分) 已知:如图 7,在梯形 ABCD中, AD BC∥ , AB DC .点 E , F ,G 分别在边 AB , BC ,CD 上, AE GF GC. (1)求证:四边形 AEFG 是平行四边形; (2)当 2FGC EFB∠ ∠ 时,求证:四边形 AEFG 是矩形. 24.(本题满分 12 分,第(1)小题满分 5 分,第(2)小题满分 3 分,第(3)小题满分 4 分) 如图 8,在直角坐标系中,O 为原点.点 A 在 x 轴的正半轴上,点 B 在 y 轴的正半轴上, tg 2OAB ∠ .二次函数 2 2y x mx   的图象经过点 A , B ,顶点为 D . (1)求这个二次函数的解析式; y A x O 图 6 B E A D G C 图 7 F (2)将 OAB△ 绕点 A 顺时针旋转90 后,点 B 落到点C 的位置.将上述二次函数图象沿 y 轴向上或向下平移后经过点C .请直接写出点C 的坐标和平移后所得图象的函数解析式; (3)设(2)中平移后所得二次函数图象与 y 轴的交点为 1B ,顶点为 1D .点 P 在平移后的 二次函数图象上,且满足 1PBB△ 的面积是 1PDD△ 面积的 2 倍,求点 P 的坐标. 25.(本题满分 14 分,第(1)小题满分 4 分,第(2)小题满分 7 分,第(3)小题满分 3 分) 已知点 P 在线段 AB 上,点O 在线段 AB 延长线上.以点O 为圆心,OP 为半径作圆,点C 是圆O 上的一点. (1)如图 9,如果 2AP PB , PB BO .求证: CAO BCO△ ∽△ ; (2)如果 AP m ( m 是常数,且 1m  ), 1BP  ,OP 是OA ,OB 的比例中项.当 点C 在圆O 上运动时,求 :AC BC 的值(结果用含 m 的式子表示); (3)在(2)的条件下,讨论以 BC 为半径的圆 B 和以CA 为半径的圆C 的位置关系,并写 出相应 m 的取值范围. y B A x O 图 8 C A P B O 图 9 2006 年上海市初中毕业生统一学业考试 数学试卷答案要点与评分标准 说明: 1.解答只列出试题的一种或几种解法.如果考生的解法与所列解法不同,可参照解答中评 分标准相应评分. 2.第一大题只要求直接写出结果,每个空格填对得 3 分,否则得零分;第二大题每题选对 得 4 分,不选、错选或者多选得零分;17 题至 25 题中右端所注的分数,表示考生正确 做对这一步应得分数.评分时,给分或扣分均以 1 分为单位. 答案要点与评分标准 一.填空题:(本大题共 12 题,满分 36 分) 1. 2 ; 2. 3 x ; 3. 6x  ; 4.  x x y ; 5. 3x≠ ; 6.1; 7. 4 ; 8. 2 2 1 0yy   (或 1 2y y); 9.5.09; 10. 1BB∠ ∠ (或 1CC∠ ∠ ,或 11AC AC ); 11. 3 ; 12.答案见图 1. 二.选择题:(本大题共 4 题,满分 16 分) 13.A; 14.B; 15.B; 16.C. 三.(本大题共 5 题,满分 48 分) 17.解:原式 211xx xx  ····································································· (2 分)   111 xxx xx  ·························································· (2 分)    1 11 xx x x x   ···························································· (1 分) 1 1x  , ············································································ (2 分) 当 2x  时,原式 1 21 21     . ············································· (2 分) 18.解:消去 y 得 2 20xx   , ····························································· (3 分) 图 1 得 1 2x  , 2 1x  , ··································································· (3 分) 由 1 2x  ,得 1 5y  , ······························································ (1 分) 由 2 1x  ,得 2 2y  , ······························································· (1 分) 原方程组的解是 1 1 2 5 x y    , ; 2 2 1 2 x y    , . ············································ (1 分) 19.解:(1)在 Rt BDA△ 中, 90BDA ∠ , 12AD  , 4sin 5 ADB AB, ·· (1 分) 15AB. ················································································· (1 分) 2 2 2 215 12 9BD AB AD     . ·········································· (2 分) 14 9 5DC BC BD     . ······················································ (1 分) (2)[方法一]过点 E 作 EF DC⊥ ,垂足为 F , EF AD ∥ . ········· (1 分) AE EC , 15 22DF DC   , 1 62EF AD. ························· (2 分) 在 Rt EFD△ 中, 90EFD ∠ , 12tg 5 EFEDC DF∠ . ················· (2 分) [方法二]在 Rt ADC△ 中, 90ADC ∠ , 12tg 5 ADC DC. ············ (2 分) DE 是斜边 AC 上的中线, 1 2DE AC EC   . ····························· (1 分) EDC C∠ ∠ . ········································································· (1 分) 12tg tg 5EDC C  ∠ . ······························································· (1 分) 20.(1)60 ; ······················································································ (3 分) (2)图略(条形图正确,得 2 分;标出数字 10,得 2 分); ······················ (4 分) (3)不能.······················································································· (3 分) 21.解:设圆心为点O ,连结OB ,OA ,OA 交线段 BC 于点 D . ·················· (1 分) AB AC , AB AC . OA BC ⊥ ,且 1 1202BD DC BC   . ································································································ (1 分) 由题意, 5DA  . ······································································ (1 分) 在 Rt BDO△ 中, 2 2 2OB OD BD, ··········································· (2 分) 设OB x 米, ············································································ (1 分) 则  2225 120xx   , ······························································ (2 分) 1442.5x . ·········································································· (1 分) 答:滴水湖的半径为1442.5米. ···················································· (1 分) 四.(本大题共 4 题,满分 50 分) 22.解:(1)由题意,设点 A 的坐标为 3aa, , 0a  . ······························· (1 分) 点 A 在反比例函数 12y x 的图象上,得 123a a , ···························· (1 分) 解得 1 2a  , 2 2a  , ·································································· (1 分) 经检验 1 2a  , 2 2a  是原方程的根,但 2 2a  不符合题意,舍去. ···· (1 分) 点 A 的坐标为 26, . ·································································· (1 分) (2)由题意,设点 B 的坐标为 0 m, . ··········································· (1 分) 0m ,  2 262mm    . ··················································· (2 分) 解得 10 3m  ,经检验 10 3m  是原方程的根,点 B 的坐标为 100 3   , . ·· (1 分) 设一次函数的解析式为 10 3y kx, ················································· (1 分) 由于这个一次函数图象过点  26A , , 1062 3k   ,得 4 3k  . ············ (1 分) 所求一次函数的解析式为 4 10 33yx. ·········································· (1 分) 23.证明:(1) 在梯形 ABCD中, AB DC , BC∠ ∠ . ·················· (2 分) GF GC , C GFC∠ ∠ . ····················································· (1 分) B GFC∠ ∠ , AB GF ∥ ,即 AE GF∥ . ································ (1 分) AE GF ,四边形 AEFG 是平行四边形. ··································· (2 分) (2)过点G 作GH FC⊥ ,垂足为 H . ············································ (1 分) GF GC , 1 2FGH FGC∠ ∠ . ············································· (1 分) 2FGC EFB∠ ∠ , FGH EFB∠ ∠ .····································· (1 分) 90FGH GFH∠ ∠ , 90EFB GFH  ∠ ∠ . ······················ (1 分) 90EFG∠ . ·········································································· (1 分) 四边形 AEFG 是平行四边形,四边形 AEFG 是矩形. ···················· (1 分) 24.解:(1)由题意,点 B 的坐标为 02, , ·············································· (1 分) 2OB, tg 2OAB ∠ ,即 2OB OA  . 1OA.点 A 的坐标为 10, . ··················································· (2 分) 又 二次函数 2 2y x mx   的图象过点 A , 20 1 2m    . 解得 3m  , ··············································································· (1 分) 所求二次函数的解析式为 2 32y x x   . ······································ (1 分) (2)由题意,可得点C 的坐标为 31, , ············································ (2 分) 所求二次函数解析式为 2 31y x x   . ············································· (1 分) (3)由(2),经过平移后所得图象是原二次函数图象向下平移1个单位后所得的图 象,那么对称轴直线 3 2x  不变,且 111BB DD. ····································· (1 分) 点 P 在平移后所得二次函数图象上,设点 P 的坐标为 2 31x x x, . 在 1PBB△ 和 1PDD△ 中, 11 2PBB PDDSS△ △ , 边 1BB 上的高是边 1DD 上的高的 2 倍. ①当点 P 在对称轴的右侧时, 32 2xx ,得 3x  ,点 P 的坐标为 31, ; ②当点 P 在对称轴的左侧,同时在 y 轴的右侧时, 32 2xx ,得 1x  , 点 P 的坐标为 11, ; ③当点 P 在 y 轴的左侧时, 0x  ,又 32 2xx   ,得 30x (舍去), 所求点 P 的坐标为 31, 或 11, . ················································ (3 分) 25.(1)证明: 2AP PB PB BO PO    , 2AO PO . 2AO PO PO BO   . ········································································ (2 分) PO CO , ··············································································· (1 分) AO CO CO BO. COA BOC∠ ∠ , CAO BCO△ ∽△ .················· (1 分) (2)解:设OP x ,则 1OB x,OA x m , OP 是OA ,OB 的比例中项,   2 1x x x m    , ·································································· (1 分) 得 1 mx m  ,即 1 mOP m  . ························································· (1 分) 1 1OB m . ············································································· (1 分) OP 是OA ,OB 的比例中项,即 OA OP OP OB , OP OC , OA OC OC OB. ·························································· (1 分) 设圆O 与线段 AB 的延长线相交于点Q ,当点C 与点 P ,点Q 不重合时, AOC COB∠ ∠ , CAO BCO△ ∽△ . ······································ (1 分) AC OC BC OB. ············································································· (1 分) AC OC OP mBC OB OB    ;当点C 与点 P 或点Q 重合时,可得 AC mBC  , 当点C 在圆O 上运动时, :AC BC m ; ········································ (1 分) (3)解:由(2)得, AC BC ,且    11AC BC m BC m    ,  1AC BC m BC   ,圆 B 和圆C 的圆心距 d BC , 显然  1BC m BC ,圆 B 和圆C 的位置关系只可能相交、内切或内含. 当圆 B 与圆C 相交时,   11m BC BC m BC    ,得02m, 1m  , 12m   ; ·································································· (1 分) 当圆 B 与圆C 内切时, 1m BC BC,得 2m  ; ·························· (1 分) 当圆 B 与圆C 内含时,  1BC m BC ,得 2m  . (1 分)