2009年永州市初中毕业学业水平考试及答案

永州市2009年初中毕业学业水平考试试卷 数 学 (试题卷) 温馨提示: 1.考生作答时,选择题和非选择题均须作答在答题卡上,在本试卷上作答无效．考生在答题卡上按答题卡中注意事项的要求答题． 2.考试结束后,将本试卷和答题卡一并交回． ３.本试卷包括试题卷和答题卡．满分120分,考试时量120分钟．本试卷共三道大题,25个小题．如缺页,考生须声明． 一、填空题(本大题共8个小题,每小题3分,共24分．请将答案填在答题卡的答案栏内) 1．?2009的相反数是 ． c d 2．函数y?4?2x的自变量x的取值范围是 ． 1 2 （第3题） 3 4 a 3．如图,直线a、b分别被直线c、d所截,如果?1??2, 那么?3??4? 度． 114．已知a?b，则?a?c ?b?c(填＞、＜或＝)． 22b 5．在平面直角坐标系中,点A的坐标为(4，的坐标5),则点A关于x轴的对称点A′为 ． 6．今年“五·一”期间,小亮一家三口人决定去旅游,小亮的理想景点为朝阳公园和浯溪公园,爸爸的理想景点为柳子庙,妈妈的理想景点为阳明山,他们把B 四个景点写在四张相同的卡片上,采用抽签的办法来确定一个旅游景点,那么,抽到小亮的理想景点的概率为 ． 227．若实数a满足a?2a?3，则3a?6a?8的值为 ． 60° 8．某校初三(一)班课外活动小组为了测得学校旗杆的高度,A 他们在离旗杆6米的A处,用高为1.5米的仪器测得旗杆顶部 B处的仰角为60°,如图所示,则旗杆的高度为 D 米．(已知3≈1.732，结果精确到0.1米) C E （第8题） 二、选择题(本大题共8个小题,每小题3分,共24分．每小题只有一个正确选项,请将正确选项的代号填涂在答题卡的答案栏内．) 9．永州市高度重视科技创新工作,全市科技投入从“十一五”初期的3.01亿元,增加到2008年的6.48亿元．请将6.48亿用科学记数法(保留两个有效数字)记为( ) A．6.48?10 8B．6.4?10 C．6.5?10 88D．6.5?10 910．如图,在长方体的数学课本上放有一个圆柱体,则它的主视图为( ) A. （第13题） C. 11．下列计算中,正确的是( ) A．x?x?x 2552B. D. 235 B．x·x?x D．(xy)?xy 32365236C．(?x)?(?x)?0 12．若点M(?3，4)在反比例函数y?例函数图象上的是( ) k(k?0，k是常数)的图象上,则下列点中也在此反比xA．(3，?4) B．(4，3) C．(3，4) D．(?3，?4) 13．下列命题是真命题的是( ) A． 对角线相等且互相垂直的四边形是菱形 B． 平移不改变图形的形状和大小 C． 对角线互相垂直的梯形是等腰梯形 D． 相等的弦所对的弧相等 14．为了了解某校2009年初三学生体育测试成绩,从中随机抽取了50名学生的体育测试成绩如下表: 成绩 (分) 人数 15 1 18 4 19 3 20 4 21 2 22 3 23 2 24 8 25 5 26 5 27 4 28 4 29 3 30 2 则这50名学生的体育测试成绩的众数、中位数分别为( ) A．24,24 B．8,24 C．24,23.5 D．4,23.5 15．用长4米的铝材制成一个矩形窗框,使它的面积为意列出关于x的方程为( ) 242若设它的一边长为x米,根据题米，252424 B．2x(2?x)? 25252424C．x(4?2x)? D．x(2?x)? 252516．右图是蜘蛛结网过程示意图,一只蜘蛛先以O为起点结六条线OA、OB、OC、 OD、OE、OF后,再从线OA上某点开始按逆时针方向D 依次在OA、 OB、OC、OD、OE、OF、…上结网,若将各线上的结点依次记为1、OA、OB、A．x(4?x)?2、3、4、5、6、7、8、…,那么第200个结点在( ) A．线OA上 B．线OB上 C．线OC上 D．线OF上 三、解答题(本题9个小题,共72分,解答题要求写出C 9 3 10 4 O 6 5 11 E 2 1 8 14 B 7 13 A 12 F （第16题） 证明步骤或解答过程) ?2?5?1??1?17．(本小题6分)计算:4cos30°?3?2?? ??．?2??°-27???3???18．(本小题6分)先化简,再求值． x?1?x2?1?x?2，其中x??2009． ???2x?1x?1x?x??19．(本小题6分)如图所示是一块破损的正八边形窗户玻璃的图形,请你利用对称或其它有关知识补全图形．(用尺规作图,不写作法,保留作图痕迹) F E M D C N B A （第19题） 20．(本小题8分)为了了解我市某县参加2008年初中毕业会考的6000名考生的数学成绩,从中抽查了200名学生的数学成绩(成绩为整数,满分120分)进行统计分析,并根据抽查结果绘制了如下的统计表和扇形统计图: 成绩(分) 人数 59.5分以下 59.5~69.5 28 44 69.5~79.5 46 79.5~89.5 89.5~99.5 99.5以上 32 请根据以上图表信息回答下列问题: (1)请将以上统计表和扇形统计图补充完整； (2)若规定60分以下(不含60分)为“不合格”,60分以上(含60分)为“合格”,80分以上(含80分)为“优秀”,试求该样本的合格率、优秀率； (3)在(2)的规定下,请用上述样本的有关信息估计该县本次毕业会考中数学成绩优秀的人数和不合格的人数． 79.5~89.5 89.5~99.5 99.5以上 14% 11% 16% 22% 69.5~79.5 59.5~69.5 59.5以下 （第20题） 21．(本小题8分)如图,平行四边形ABCD，E、F两点在对角线BD上,且BE?DF，连结求证:四边形AECF是平行四边形． AE，EC，CF，FA． A D F E B C （第21题） 22．(本小题8分)某工厂为了扩大生产规模,计划购买5台A、B两种型号的设备,总资金不超过28万元,且要求新购买的设备的日总产量不低于24万件,两种型号设备的价格和日产量如下表．为了节约资金,问应选择何种购买方案？ 价格(万元/台) 日产量(万件/台) 23．(本小题10分)如图,在平面直角坐标系内,O为原点,点A的坐标为(?3，，0)经过A、O两点作半径为A 6 6 B 5 4 5的⊙C，交y轴的负半轴于点B． 2 (1)求B点的坐标； (2)过B点作⊙C的切线交x轴于点D，求直线BD的解析式． y A O C B （第23题） 24．(本小题10分)问题探究: (1)如图①所示是一个半径为D x 3,高为4的圆柱体和它的侧面展开图,AB是圆柱的一条母线,2π一只蚂蚁从A点出发沿圆柱的侧面爬行一周到达B点,求蚂蚁爬行的最短路程．(探究思路:将圆柱的侧面沿母线AB剪开,它的侧面展开图如图①中的矩形ABB′则蚂蚁爬行的最短A′，路程即为线段AB′的长) 2(2)如图②所示是一个底面半径为,母线长为4的圆锥和它的侧面展开图,PA是它的一条母3线,一只蚂蚁从A点出发沿圆锥的侧面爬行一周后回到A点,求蚂蚁爬行的最短路程． (3)如图③所示,在②的条件下,一只蚂蚁从A点出发沿圆锥的侧面爬行一周到达母线PA上的一点,求蚂蚁爬行的最短路程． B B′ A? A? A A 图① 图② （第24题） P P A 图③ 0)(0，?3)，25．(本小题10分)如图,在平面直角坐标系中,点A、C的坐标分别为(?1，、点B在x轴上．已知某二次函数的图象经过A、B、C三点,且它的对称轴为直线x?1点P为，直线BC下方的二次函数图象上的一个动点(点P与B、C不重合),过点P作y轴的平行线 交BC于点F． (1)求该二次函数的解析式； (2)若设点P的横坐标为m，用含m的代数式表示线段PF的长． (3)求△PBC面积的最大值,并求此时点P的坐标． y A O C x=1 F B x P （第25题） 永州市2009年初中毕业学业水平考试 数学参考答案及评分标准 一、填空题(每小题3分,共24分) 1．2009 2．x≤2 ３．180 ４．＜ 5．(4，?5) 6．1 7．1 8．11.9 2二、选择题(每小题3分,共24分) 9．C 10．D 11．C 12．A 13．B 14．A 15．D 16．B 三、解答题 17． 解:4cos30°-3?2?(5?11)°?27?(?)?2 23=4?31 ···································································· 3分 ?2?3?1?33?22?1?????3???=23?2?3?1?33?9·················································································· 5分 =8 ··················································································································· 6分 x?1?x2?1?x?2??218．解: ? ?x?1x?1?x?xx?1?x2?1?x?2??=? ·················································································· 1分 ?x?1x?1?x(x?1)x2?1x(x?1)·2= ························································································ 3分 (x?1)(x?1)x?1x··············································································································· 4分 x?1?20092009E F 当x??2009时,原式= ······························ 6分 ??2009?12010M 19． 连结AE、BF相交于点O ·········································· 2分 G D 分别作C、D两点关于O点的对称点G、H ·························· 4分 O 连结AH、HG、GF ························································ 6分 =H (其它作法参照评分标准进行评分,如利用轴对称作图,利用正八边形 C 的性质作图) N 20． 解:(1)28 22 14% 23% ············································ 4分 B A (2)合格率:1-14%=86% ······················································· 5分 （第19题） 优秀率:14%+11%+16%=41% ················································ 6分 (3)优秀人数:41%×6000=2460 ················································································· 7分 不合格人数:14%×6000=840 ··················································································· 8分 21．证明:连结AC交BD于点O ······························· 2分 四边形ABCD为平行四边形 ···································· 4分 ?OA?OC，OB?OD ···································· 6分 B BE?DF，?OE?OF ·?四边形AECF为平行四边形 ····························· 8分 A E O （第21题） F C D 22．解:设购买A型设备为x台,则购买B型设备为(5?x)台,依题意得: ···························· 1分 ?6x?5(5?x)≤28 ···························································································· 4分 ??6x?4(5?x)≥24解得2≤x≤3 ··································································································· 6分 x为整数,?x?2或x?3． 当x?2时,购买设备的总资金为6×2+5×3=27(万元) 当x?3时,购买设备的总资金为6×3+5×2=28(万元) ?应购买A型设备2台,B型设备3台． ··································································· 8分 23． 解:(1)?AOB?90° y ············································ 1分 ?AB是直径,且AB?5 ·D A 在Rt△AOB中,由勾股定理可得 O x ···························· 3分 BO?AB?AO?5?3?4 ············································· 4分 ?4) ·?B点的坐标为(0，(2)2222C B BD是⊙C的切线,CB是⊙C的半径 （第23题） 即 ?BD?AB，?ABD?90°??DAB??ADB?90° 又?BDO??OBD?90° ···························································································· 5分 ??DAB??DBO ·?AOB??BOD?90° ··························································································· 6分 ?△ABO∽△BDO ·OAOBOB24216???OD??? OBODOA33?16?0? ·························································································· 7分 ?D的坐标为?，3??设直线BD的解析式为y?kx?b(k?0，k、b为常数) ?16?k?b?0则有?3 ······························································································· 8分 ??b??43?k????········································································································ 9分 4 ·??b??4 ?直线BD的解析式为y?3········································································· 10分 x?4 ·4324．解:(1)易知BB′·········································································· 1分 ?2π??3 ·2π2······································································ 3分 AB′=AB2?BB′?42?32?5 ·即蚂蚁爬行的最短路程为5． ················································································· 4分 (2)连结AA则AA的长为蚂蚁爬行的最短路程,设r1为圆锥底面半径,r2为侧面展开图(扇′，′形)的半径,则r1?即2?π?nπr22由题意得:2πr1= ，r2?4，31802n····················································································· 5分 ??π?4 ·3180?n?60 是等边三角形 ························································································ 6分 ?△PAA′?最短路程为AA ·················································································· 7分 ′?PA?4．(3)如图③所示是圆锥的侧面展开图,过A作AC?PA于点C，则线段AC的长就是蚂蚁爬′行的最短路程． ··································································································· 8分 ?AC?PAsin?APA??4?sin60°?4?3?23 ················································· 9分 2?蚂蚁爬行的最短距离为23． ··············································································· 10分 P P 60° C B′B A? A? A? A 图① A 图② 图③ A 25．解:(1)设二次函数的解析式为y?ax?bx?c(a?0，a、b、c为常数),由抛物线的对2?a?b?c?0?称性知B点坐标为(3，，·················································· 1分 0)依题意得:?9a?3b?c?0 ·??c??3 ?3a??3??23?解得:?b?? ························································· 2分 3?????c??3?所求二次函数的解析式为y?y A O C x=1 F B x 3223x?x?3 ··········· 3分 33P （第25题） (2)P点的横坐标为m，?P点的纵坐标为3223m?m?3 ·································· 4分 33设直线BC的解析式为y?kx?b(k?0，k、b是常数)，依题意,得???3k?b?0 ??b??3?3?k???3 ?b??3?故直线BC的解析式为y?3x?3 ····································································· 5分 3??3?点F的坐标为?m，m?3? ??3???PF??32m?3m(0?m?3) ········································································ 6分 3(3)△PBC的面积S?S△CPF?S△BPF?1PF·BO 22?1?323?3?93m?3m?m???=??? ??3??2?2?328?????当m?933时,△PBC的最大面积为 ································································ 8分 82把m?3223533m?m?3得y??代入y? 3342?353??点P的坐标为?，················································································ 10分 ?2?4?? ·??