湛江中考数学试题及答案_中考数学试题及答案

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2011年湛江中考数学试题及答案

湛江市2011 年初中毕业生学业考试 数 学 试 卷 说明: 1.本试卷满分150 分,考试时间90 分钟. 2.本试卷共6 页,共5 大题. 3.答题前,请认真阅读答题卡上的“注意事项”,然后按要求将答案写在答题卡相应的位 置上. 4.请考生保持答题卡的整洁,考试结束,将试卷和答题卡一并交回. 注意:在答题卡上作图必须用黑色字迹的钢笔或签字笔.

一、选择题:本大题10 个小题,其中1~5 每小题3 分,6~10 每小题4 分,共35 分.在每 小题给出的四个选项中,只有一项是符号题目要求的)1.下列四个数中,在 1  和2 之间的数是()A.0 B. 2  C. 3  D.3 2.下列各式中,与 2(1)x  相等的是()A. 2 1 x  B. 2 2 1 x x   C. 2 2 1 x x   D. 2 x 3.湛江是个美丽的海滨城市,三面环海,海岸线长达1556000 米,数据1556000 用科学记 数法表示为()A. 7 1.5 5 6 1 0  B. 8 0.1 5 5 6 1 0  C. 5 1 5.5 6 1 0  D. 6 1.5 5 6 1 0  4.在右图的几何体中,它的左视图是()5.沃尔玛商场为了了解本商场的服务质量,随机调查了本商场的100 名顾客,调查的结果 如图所示,根据图中给出的信息,这 100 名顾客中对该商场的服务质量表示不满意的有()A.6 人 B.11 人 C.39 人 D.44 人 第4 题图 A. B. C. D. A 44% B 39% C 11% D A:很满意 B:满意 C:说不清 D:不满意 第5 题图 A B C D E 第6 题图

6.如图,在等边 A B C △ 中,D E、分别是A B A C、的中点,3 D E ,则 A B C △ 的周长 是()A.6 B.9 C.18 D.24 7.如图,在平面直角坐标系中,菱形O A C B 的顶 点O 在原点,点C 的坐标为(4 0),点B 的纵坐标 是 1 ,则顶点A 的坐标是()A.(2 1),B.(1 2),C.(1 2),D.(2 1),8.根据右图所示程序计算函数值,若输入的x 的值为 5 2,则输出的 函数值为()A. 3 2 B. 2 5 C. 4 2 5 D. 2 5 4 9.下列说法中: ①4 的算术平方根是±2;② 2 与 8  是同类二次根式; ③点(2 3)P ,关于原点对称的点的坐标是(2 3) ,; ④抛物线 2 1(3)1 2 y x     的顶点坐标是(3 1),. 其中正确的是()A.①②④ B.①③ C.②④ D.②③④ 10.如图,小林从P 点向西直走12 米后,向左转,转动的角度为,再走12 米,如此重复,小林共 走了108 米回到点P,则()A.3 0 ° B.4 0 ° C.8 0 ° D.不存在二、填空题:本大题共10 个小题,其中11~15 每小题3 分,16~20 每小题4 分,共35 分. 11. 2  的相反数是 . 12.要使分式 1 3 x  有意义,则x 的取值范围是 . 13.如图,已知 1 5 5 A B C D   ‖,°,则 2  = . 14.分解因式: 2 2 m n   . 15.已知在一个样本中,40 个数据分别落在4 个组内,第一、二、四组数据个数分别为5、12、8,则第三组的频数为 . O A B C y x 第7 题图 P   第10 题图 输入x 值 1 y x  (1 0)x   ≤ 1 y x (2 4)x ≤ ≤ 2 y x (0 2)x  ≤ 输出y 值 第8 题图 A B C D 1 2 第13 题图 16.如图,A B 是 O ⊙ 的直径,C D E、、是 O ⊙ 上的点,则 1 2     °. 17.一件衬衣标价是 132 元,若以 9 折降价出售,仍可获 利10%,则这件衬衣的进价是 元. 18.如图,1 2 O O ⊙、⊙ 的直径分别为2cm和4cm,现将 1 O ⊙ 向 2 O ⊙ 平移,当 1 2 O O = cm时,1 O ⊙ 与 2 O ⊙ 相切. 19.已知 2 2 2 2 3 3 2 2 3 3 3 3 8 8      ,2 4 4 4 4 1 5 1 5   ,�6�7�6�7,若 2 8 8 a a b b   (a、b 为正整数)则a b   . 20.如图,在梯形A B C D 中,9 0 5 1 1 A B C D A B C D A B       ‖,°,,点M N、分别为A B C D、的中点,则线段M N  .

三、解答题:本大题共2 小题,每小题8 分,共16 分.21.如图,一只蚂蚁从点A 沿数轴向右直爬2 个单位到达点B,点A 表示 2 ,设点B 所 表示的数为m.(1)求m 的值;(2)求 0 1(6)m m    的值. A B C D E O 1 2 第16 题图 第18 题图 O 1 O 2 A B C D N M 第20 题图 1 2 0-1-2 A B 第21 题图 22.如图,点O A B、、的坐标分别为(0 0)(3 0)(3 2),、,、,将 OAB △ 绕点O 按逆时针方向 旋转9 0 °得到 O A B  

△ .(1)画出旋转后的 O A B   △,并求点B  的坐标;(2)求在旋转过程中,点A 所经过的路径A A  的长度.(结果保留π)

四、解答题:本大题共4 小题,每小题10 分,共40 分.23.某语文老师为了了解中考普通话考试的成绩情况,从所任教的九年级(1)、(2)两班各 随机抽取了10 名学生的得分,如图所示:(1)利用图中的信息,补全下表: 班级 平均数(分)中位数(分)众数(分)九(1)班 16 16 九(2)班 16(2)若把16 分以上(含16 分)记为“优秀”,两班各有60 名学生,请估计两班各有多少 名学生成绩优秀. B y x A O 第22 题图 0 1 2 3 4 5 6 7 8 9 10 2 20 18 16 14 12 10 6 4 8 成绩(分)编号 九(1)班 0 1 2 3 4 5 6 7 8 9 10 2 20 18 16 14 12 10 6 4 8 成绩(分)编号 九(2)班 第23 题图 24.如图,某军港有一雷达站P,军舰M 停泊在雷达站P 的南偏东6 0 °方向 36 海里处,另一艘军舰N 位于军舰M 的正西方向,与雷达站P 相距1 8 2 海里.求:(1)军舰N 在雷达站P 的什么方向?(2)两军舰M N、的距离.(结果保留根号)25.六张大小、质地均相同的卡片上分别标有1、2、3、4、5、6,现将标有数字的一面朝 下扣在桌面上,从中随机抽取一张(放回洗匀),再随机抽取第二张.(1)用列表法或树状图表示出前后两次抽得的卡片上所标数字的所有可能结果;(2)记前后两次抽得的数字分别为m、n,若把

m、n 分别作为点A 的横坐标和纵坐标,求 点()A m n,在函数 1 2 y x  的图象上的概率. 26.如图,A B 是 O ⊙ 的切线,切点为B A O,交 O ⊙ 于点C,过点C 作D C O A ,交A B 于 点D.(1)求证: C D O B D O    ;(2)若 3 0 A O   °,⊙ 的半径为4,求阴影部分的面积.(结果保留π)第24 题图 N M P 北 O A B C D 第26 题图

五、解答题:本大题共2 小题,每小题12 分,共24 分.27.某公司为了开发新产品,用A、B 两种原料各360 千克、290 千克,试制甲、乙两种 新型产品共50 件,下表是试验每件 .. 新产品所需原料的相关数据: A(单位:千克)B(单位:千克)甲 9 3 乙 4 10(1)设生产甲种产品x 件,根据题意列出不等式组,求出x 的取值范围;(2)若甲种产品每件成本为70 元,乙种产品每件成本为90 元,设两种产品的成本总额为 y 元,写出成本总额y(元)与甲种产品件数x(件)之间的函数关系式;当甲、乙两 种产品各生产多少件时,产品的成本总额最少?并求出最少的成本总额. 28.已知矩形纸片O A B C 的长为4,宽为3,以长O A 所在的直线为x 轴,O 为坐标原点建 立平面直角坐标系;点P 是O A 边上的动点(与点O A、不重合),现将 P O C △ 沿P C 翻折 得到 P E C △,再在A B 边上选取适当的点D,将 P A D △ 沿P D 翻折,得到 P F D △,使得 直线P E P F、重合.(1)若点E 落

在B C 边上,如图①,求点P C D、、的坐标,并求过此三点的抛物线的函 数关系式;(2)若点E 落在矩形纸片O A B C 的内部,如图②,设O P x A D y  ,当x 为何值时,y 取得最大值?(3)在(1)的情况下,过点P C D、、三点的抛物线上是否存在点Q,使 P D Q △ 是以P D 为直角边的直角三角形?若不存在,说明理由;若存在,求出点Q 的坐标 原 料 含 量 产 品 C y E B F D A P x O 图① A B D F E C O P x y 图② 第28 题图 湛江市2011 年初中毕业生学业考试 数学试卷参考答案与评分说明

一、选择题:本大题共10 小题,其中1~5 小题每题3 分,6~10 小题每题4 分,共35 分. 题号 1 2 3 4 5 6 7 8 9 10 答案 A B D B A C D B C B

二、填空题:本大题共10 小题,其中11~15 每小题3 分,16~20 每小题4 分,共35 分. 11.2 12. 3 x  13.1 2 5 ° 14.()()m n m n   15.15 16.90 17.108 18.1 或3 19.71 20.3

三、解答题:本大题共2 小题,每小题8 分,共16 分. 21.解:(1)由题意可得 2 2 m   ······························································································ 2 分(2)把m 的值代入得: 0 0 1(6)2 2 1(2 2 6)m m          ··································· 3 分 = 0 1 2(8 2)   ·····································································

··········································· 4 分 = 2 1 1   ································································································································ 7 分 = 2 ··········································································································································· 8 分 22.解:(1)如图 O A B   △ 为所示,点B  的坐标为(2 3),; ·····················································································4 分(2)O A B △ 绕点O 逆时针旋转9 0 °后得 O A B   △,点A 所经过的路径A A  是圆心角为9 0 °,半径为3 的扇形O A A 的弧长,所以 1 3(2 π 3)π 4 2 l     . ··················································································· 7 分 即点A 所经过的路径A A  的长度为 3 π 2 . ·············· 8 分

四、解答题:本大题共4 小题,每小题10 分,共40 分. 23.解:(1)班级 平均数(分)中位数(分)众数(分)九(1)班 16 16 16 九(2)班 16 16 14 ······························································································································································ 6 分(2)7 6 0 4 2 1 0  (名),6 6 0 3 6 1 0  (名).  九(1)班有42 名学生成绩

优秀,九(2)班有36 名学生成绩优秀. ··································· 10 分 第22 题图 B y x A O A  B  24.解:过点P 作P Q M N ,交M N 的延长线于点Q. ······················································· 1 分(1)在R t P Q M △ 中,由 6 0 M P Q   °,得 3 0 P M Q   ° 又 3 6 P M  1 1 3 6 1 8 2 2 P Q P M     (海里)······················································································· 3 分 在R t P Q N △ 中,1 8 2 c o s 2 1 8 2 P Q Q P N P N    ,4 5 Q P N    ° 即军舰N 到雷达站P 的东南方向(或南偏东4 5 °)······························································ 5 分(2)由(1)知R t P Q N △ 为等腰直角三角形,1 8 P Q N Q   (海里)····················· 7 分 在R t P Q M △ 中,t a n 1 8 t a n 6 0 1 8 3 M Q P Q Q P M     · · °(海里)1 8 3 1 8 M N M Q N Q     (海里)··············································································· 9 分 答:两军舰的距离为   1 8 3 1 8  海里. ··············································································· 10 分 25 解:(1)列表: 1 2 3 4 5 6 1(1,1)(2,1)(3,1)(4,1)(5,1)(6,1)2(1,2)(2,2)(3,2)(4,2)(5,2)(6,2)3(1,3)(2,3)(3,3)(4,3)(5,3)(6,3)4(1,4)(2,4)(3,4)(4,4)(5,4)(6,4)5(1,5)(2,5)(3,5)(4,5)(5,5)(6,5)6(1,6)(2,6)(3,6)(4,6)(5,6)(6,6)······························································································································································ 4 分 由表可看出,前后两次抽得的卡片上所标数字的所有可能结果有36 种. ································· 5 分 或画树状图: 第 一 第 二 次 次 第24 题图 N M P 北 Q 1 1 2 3 4 5 6 2 1 2 3 4 5 6 3 1 2 3 4 5 6 4 1 2 3 4 5 6 5 1 2 3 4 5 6 6 1 2 3 4 5 6 第一次: 第二次: 从树状图可以看出,所有可能出现的结果有36 种,即: ····························································· 3 分(1,1)、(1、2)、(1、3)、(1、4)、(1、5)、(1、6)、(2,1)、(2、2)、(2、3)、(2、4)、(2、5)、(2、6)(3,1)、(3、2)、(3、3)、(3、4)、(3、5)、(3、6)(4,1)、(4、2)、(4、3)、(4、4)、(4、5)、(4、6)(5,1)、(5、2)、(5、3)、(5、4)、(5、5)、(5、6)(6,1)、(6、2)、(6、3)、(6、4)、(6、5)、(6、6)······························································· 5 分(2)有4 个点(2,6)、(3,4)、(4,3)、(6,2)在函数 1 2 y x  的图象上 ···························· 8 分  所求概率 4 1 3 6 9 P   ····················································································································· 10 分 26.解:(1)A B 切 O ⊙ 于点B ∴O B A B ,即 9 0 B   ° ················································································································· 1 分 又 9 0 D C O A O C D    ,° ······································································································ 2 分 在R t C O D △ 与R t B O D △ 中 O D O D O B O C  ,R t R t()C O D B O D H L  △ ≌

△ ····································································································· 3 分 C D O B D O     . ·························································································································· 4 分(2)在R t A B O △ 中,3 0 4 A O B    °,8 O A   8 4 4 A C O A O C       ······································································· 5 分 在R t A C D △ 中,t a n C D A A C   又 3 0 4

A A C    °,4 3 t a n 3 0 3 C D A C    · ° ······································································· 7 分 1 4 3 1 6 3 2 2 4 2 3 3 O C D O C D B S S        △ 四 边

形 ········································································ 8 分 又 3 0 6 0 A B O C      °,°. 2 6 0 π 4 8 π 3 6 0 3 O B C S    扇

形 · . ········································································································ 9 分 1 6 3 8 π 3 3 O C D B O B C S S S      阴 影 四 边 形 扇 形 . ······································································· 10 分

五、解答题:本大题共2 小题,每小题12 分,共24 分. 27.解:(1)依题意列不等式组得 9 4(5 0)3 6 0 3 1 0(5 0)2 9 0 x x x x        ≤ ≤ ·················································· 3 分 O A B C D 第26 题图 由不等式①得 3 2 x ≤ ······················································································································· 4 分 由不等式②得 3 0 x ≥ ······················································································································· 5 分 x  的取值范围为3 0 3 2 x ≤

≤ ····································································································· 6 分(2)7 0 9 0(5 0)y x x    ········································································································ 8 分 化简得 2 0 4 5 0 0 y x    2 0 0 y   ,随x 的增大而减小. ························································································· 9 分 而3 0 3 2 x ≤ ≤  当 3 2 x ,5 0 1 8 x   时,2 0 3 2 4 5 0 0 3 8 6 0 y      最 小 值(元)························· 11 分 答:当甲种产品生产32 件,乙种18 件时,甲、乙两种产品的成本总额最少,最少的成本总 额为3860 元. ···························································································································· 12 分 28.解:(1)由题意知,P O C P A D △、△ 均为等腰直角三角形,可得(3 0)(0 3)(4 1)P C D,、,、,··································································································· 2 分 设过此三点的抛物线为 2(0)y a x b x c a    ,则 3 9 3 0 1 6 4 1 c a b c a b c             1 2 5 2 3 a b c                过P C D、、三点的抛物线的函数关系式为 2 1 5 3 2 2 y x x 

  ········································ 4 分(2)由已知P C 平分 O P E P D ,平分 A P F ,且P E P F、重合,则 9 0 C P D   ° C y E B F D A P x O 图① A B D F E C O P x y 图② 第28 题图 9 0 O P C A P D      °,又 9 0 A P D A D P     ° O P C A D P     . R t R t P O C D A P  △ ∽ △ . O P O C A D A P  ,即 3 4 x y x   ··································································································· 6 分 2 2 1 1 4 1 4(4)(2)(0 4)3 3 3 3 3 y x x x x x x             当 2 x  时,y 有最大值 4 3 . ······························································································ 8 分(3)假设存在,分两种情况讨论: ①当 9 0 D P Q   °时,由题意可知 9 0 D P C   °,且点C 在抛物线上,故点C 与点Q 重合,所求的点Q 为(0,3)·················································································································· 9 分 ②当 9 0 D P Q   °时,过点D 作平行于P C 的直线D Q,假设直线D Q 交抛物线于另一点 Q,点(3 0)0 3 P C,、(,), 直线P C 的方程为 3 y x   ,将直线P C 向上平移2 个单位 与直线D Q 重合, 直线D Q 的方程为 5 y

x    ································································· 10 分 由 2 5 1 5 3 2 2 y x y x x            得 1 6 x y       或 4 1 x y      又点(4 1)(1 6)D Q  ,,. 故该抛物线上存在两点(0 3)(1 6)Q ,、,满足条件. ································································ 12 分 说明:以上各题如有其他解(证)法,请酌情给分. y x A B E C Q O P D F(Q)第28 题图

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