2009年湖南永州市初中毕业学业考试

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 ． 2．函数y =x 的取值范围是 ． 3．如图，直线a b 、分别被直线c d 、所截，如果12∠=∠， 那么34∠+∠= 度．4．已知a b >，则12a c -+ 12b c -+（填＞、＜或＝）．5．在平面直角坐标系中，点A 的坐标为(45)，，则点A 关于x 轴的对称点A ′的坐标为 ．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 ．7．若实数a 满足223a a -=，则2368a a --的值为 ．8．某校初三（一）班课外活动小组为了测得学校旗杆的高度，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 米． 1.732，结果精确到0.1米）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 ．86.4810⨯B ．86.410⨯C ．86.510⨯D ．96.510⨯10．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（ ）（第8题）ACD EB 60° abc d 1 23 4（第3题）11．下列计算中，正确的是（ ） A ．235x x x +=B ．236x x x =·C ．2552()()0x x -+-=D ．32365()x y x y =12．若点(34)M -，在反比例函数(0)ky k k x=≠，是常数的图象上，则下列点中也在此反比例函数图象上的是（ ）A ．(34)-，B ．(43)，C ．(34)，D ．(34)--， 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则这50名学生的体育测试成绩的众数、中位数分别为（ ）A ．24，24B ．8，24C ．24，23.5D ．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22425米，若设它的一边长为x 米，根据题意列出关于x 的方程为（ ）A ．24(4)25x x -=B ．242(2)25x x -=C ．24(42)25x x -= D ．24(2)25x x -=16．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依次在OA 、OB OC 、、OD 、OE 、OF 、OA OB 、、…上结网，若将各线上的结点依次记为1、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A .B .D .C . （第13题） （第16题）三、解答题（本题9个小题，共72分，解答题要求写出证明步骤或解答过程）17．（本小题6分）计算：2114cos30223-⎛⎫⎛⎫+- ⎪ ⎪ ⎪⎝⎭⎝⎭°．18．（本小题6分）先化简，再求值．2221111xx x x x x x ++⎛⎫+÷⎪+-+⎝⎭，其中2009x =-． 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）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：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．E F MN A B CD （第19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AE EC CF FA ，，，．求证：四边形AECF 是平行四边形．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？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(30)-，，经过A O 、两点作半径为52的C ⊙，交y 轴的负半轴于点B ． 99.5以上59.5以下 （第20题）CDA B E F（第21题）（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．24．（本小题10分）问题探究： （1）如图①所示是一个半径为32π，高为4的圆柱体和它的侧面展开图，AB 是圆柱的一条母线，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 A ′′，则蚂蚁爬行的最短路程即为线段AB ′的长）（2）如图②所示是一个底面半径为23，母线长为4的圆锥和它的侧面展开图，PA 是它的一条母线，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．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(10)(0-，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1x =，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（第23题） BA A 'B ′ 图① A ' PA图② P A 图③（第24题）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．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 2．2x ≤ ３．180 ４．＜ 5．(45)-，6．127．1 8．11.9 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：214cos302()3-+-°=(21421213⨯-+-⎛⎫- ⎪⎝⎭····················································································· 3分=219- ······································································································· 5分 =8 ·················································································································································· 6分18．解： 2221111x x x x x x x++⎛⎫+÷ ⎪+-+⎝⎭ =221111(1)xx x x x x x ++⎛⎫+÷ ⎪+-+⎝⎭········································································································ 1分（第25题）=221(1)(1)(1)1x x x x x x +++-+· ··············································································································· 3分 =1x x - ··········································································································································· 4分 当2009x =-时，原式=20092009200912010-=-- ··································· 6分19． 连结AE BF 、相交于点O ····················································· 2分分别作C D 、两点关于O 点的对称点G H 、 ································· 4分 连结AH HG GF 、、 ······································································· 6分（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 的性质作图）20． 解：（1）28 22 14% 23% ················································· 4分（2）合格率：1-14%=86% ······························································ 5分优秀率：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 OA OC OB OD ∴==， ·············································· 4分BE DF OE OF =∴= ， ············································· 6分 ∴四边形AECF 为平行四边形 ···································· 8分 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65(5)2864(5)24x x x x +-⎧⎨+-⎩≤≥ ···················································································································· 4分 解得23x ≤≤····························································································································· 6分 x 为整数，23x x ∴==或．当2x =时，购买设备的总资金为6×2+5×3=27（万元） 当3x =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）90AOB ∠= ° AB ∴是直径，且5AB = ······················································ 1分在AOB Rt △中，由勾股定理可得4BO ····································· 3分 B ∴点的坐标为(04)-， ························································· 4分（2）BD 是C ⊙的切线，CB 是C ⊙的半径 BD AB ∴⊥，即90ABD ∠=° 90DAB ADB ∴∠+∠=° 又90BDO OBD ∠+∠= °（第21题）CDA BE FO （第23题）（第19题）E F MNAB C DG H ODAB DBO ∴∠=∠ ······················································································································ 5分 90AOB BOD ∠=∠= ° ABO BDO ∴△∽△ ···················································································································· 6分2241633OA OB OB OD OB OD OA ∴=∴=== D ∴的坐标为1603⎛⎫ ⎪⎝⎭，·················································································································· 7分 设直线BD 的解析式为(0)y kx b k k b =+≠，、为常数则有16034k b b ⎧+=⎪⎨⎪=-⎩ ························································································································ 8分344k b ⎧=⎪∴⎨⎪=-⎩ ····································································································································· 9分 ∴直线BD 的解析式为344y x =- ····························································································· 10分 24．解：（1）易知32π32πBB =⨯=′ ························································································· 1分5AB ==′= ·························································································· 3分 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1r 为圆锥底面半径，2r 为侧面展开图（扇形）的半径，则12243r r ==，，由题意得：21π2πr =180n r 即22ππ43180n⨯⨯=⨯⨯ ············································································································· 5分 60n ∴= PAA ∴△′是等边三角形 ··············································································································· 6分∴最短路程为4AA PA ==′． ······································································································· 7分 （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sin 4sin 604AC PA APA '∴=∠=⨯== °······························································ 9分 ∴蚂蚁爬行的最短距离为·································································································· 10分BB ′P C60°A 'P25．解：（1）设二次函数的解析式为2(0)y ax bx c a a b c =++≠，、、为常数，由抛物线的对称性知B 点坐标为(30)，，依题意得：0930a b c a b c c ⎧-+=⎪++=⎨⎪=⎩······················································· 1分解得：3a b c ⎧=⎪⎪⎪⎪=⎨⎪⎪⎪=⎪⎩······································································ 2分∴所求二次函数的解析式为233y x x =--·············· 3分 （2）P 点的横坐标为m ，P ∴2-····································· 4分 设直线BC 的解析式为(0)y kx b k k b =+≠，、是常数，依题意，得30k b b +=⎧⎪⎨=⎪⎩k b ⎧=⎪∴⎨⎪=⎩故直线BC的解析式为y x =······················································································· 5分 ∴点F的坐标为m ⎛ ⎝2(03)PF m m ∴=+<< ··························································································· 6分（第25题）。