2012年四川省眉山市中考真题(word版含答案)

眉山市2012年初中学业水平暨高中阶段教育学校招生考试数 学 试 卷注意事项：1．本试卷分为A 卷和B 卷，A 卷分第Ⅰ卷和第Ⅱ卷两部分，第Ⅰ卷共12个小题，共36分，第1页至第2页；第Ⅱ卷共12个小题，共64分，第3页至第6页；B 卷共2个小题，共20分，第7页至第8页，全卷满分120分，考试时间120分钟．2．答第Ⅰ卷前，考生务必将姓名、准考证号、考试科目用铅笔涂写在答题卡上相应的位置，并请将密封线内的内容填写清楚．第Ⅰ卷不能答在试卷上，第Ⅱ卷和B 卷答在试卷上．3．不允许使用计算器进行运算，凡无精确度要求的题目，结果均保留准确值，解答题A 卷（共100分） 第Ⅰ卷（选择题 共36分）一、选择题：本大题共12个小题，每小题3分，共36分．在每个小题给出的四个选项中，只有一项是正确的，请把正确选项的字母填涂在答题卡上相应的位置． 1．若x =5，则x 的值是（ ）（A ）5 （B ）5- （C ）5± （D ）152．下列运算正确的是（ ）（A ）5510a a a += （B ）339a a a ∙= （C ）339(3)9a a = （D ）1239a a a ÷=3．函数y =x 的取值范围是（ ）（A ）2x > （B ）2x ≥ （C ）2x ≤ （D ）2x <4．某种微粒子，测得它的质量为0.00006746克，这个质量用科学记数法表示（保留三个有效数字）应为（ ）（A ）56.7510-⨯克 （B ）56.7410-⨯克 （C ）66.7410-⨯克 （D ）66.7510-⨯克5．若关于x 的一元二次方程220x x m -+=中有两个不相等的实数根，则m 的取值范围是（ ）（A ）1m < （B ）1m <- （C ）1m > （D ）1m >- 6．下列命题中，真命题是（ ）（A ）有两条对角线相等的四边形是等腰梯形（B ）两条对角线互相垂直且平分的四边形是正方形 （C ）等边三角形既是轴对称图形又是中心对称图形 （D ）有一个角是60°的等腰三角形是等边三角形 7．如图，在ABC △中，90ACB ∠=°，20A ∠=°，若将ABC △沿CD 折叠，使B 点落在AC 边上的E 处，则ADE ∠的度数是（ ）（A ）30° （B ）40° （C ）50° （D ）55°8．一组数据为235734、、、、、，对于这组数据，下列说法错误的是（ ） （A ）平均数是4 （B ）极差是5 （C ）众数是3 （D ）中位数是69．用一些大小相同的小正方体组成的几何体的左视图和俯视图如图所示，则组成这个几何体的小正方体的块数，最多．．可能是（ ） （A ）17 （B ）18 （C ）19 （D ）2010．若m n 、是一元二次方程2520x x --=的两个实数根，则m n m n +-的值是（ ） （A ）7- （B ）7 （C ）3 （D ）3-11．圆锥底面圆的半径为1cm ，母线长为6cm ，则圆锥侧面展开图的圆心角是（ ）（A ）30° （B ）60° （C ）90° （D ）120°12．已知：如图，在直角坐标系中，有菱形OABC ，A 点的坐标为(100)，，对角线OB 、AC相交于D 点，双曲线(0)ky x x =>经过D 点，交BC 的延长线于E 点，且160OB AC ∙=，有下列四个结论①双曲线的解析式为20(0)y x x=> ②E 点的坐标是(48)，③4sin 5COA ∠=④AC OB += ）（A ）1个 （B ）2个 （C ）3个 （D ）4个第Ⅱ卷 （非选择题 共64分）二、填空题：本大题共6个小题，每小题3分，共18分．将正确答案直接填在题中横线上．13．因式分解：22ax ax a -+=___________．14．如图，ABCD Y 中，5AB =，3AD =，AE 平分DAB ∠交BC 的延长线于F 点，则CF =___________．15．已知，如图，PA PB 、与O ⊙相切于A 点、B 点，1OA =，PA =部分的面积是___________（结果保留π）．16．某学校有80名学生，参加音乐、美术、体育三个课外小组（每人只参加一项），这80人中若有40%的人参加体育小组，35%的人参加美术小组，则参加音乐小组的有___________人．17．直线(3)2y a x b =-+-在直角坐标系中的图象如图所示，化简：2b a b --=___________．18．在ABC △中，5AB =，3AC =，AD 是BC 边上的中线，则AD 的取值范围是___________．三、计算题：本大题共2个小题，每小题6分，共12分．19．计算：214cos30()1)2-+--°20．解方程：11322x x x-+=--四、本大题共2个小题，每小题8分，共16分．21．如图，图中的小方格都是边长为1的正方形，ABC △的顶点坐标分别为(30)A -，，(12)B --，，(22)C -，．（1）请在图中画出ABC △绕B 点顺时针旋转180°后的图形；（2）请直接写出以A B C 、、为顶点的平行四边形的第四个顶点D 的坐标．22．如图，在与河对岸平行的南岸边有A B D 、、三点，A B D 、、三点在同一直线上，在A 点处测得河对岸C 点在北偏东60°方向；从A 点沿河边前进200米到达B 点，这时测得C 点在北偏东30°方向，求河宽CD ．五、本大题共2个小题，每小题9分，共18分．23．有质地均匀的A B C D 、、、四张卡片，上面对应的图形分别是圆、正方形、正三角形、平行四边形，将这四张卡片放入不透明的盒子中摇匀，从中随机抽出一张（不放回），再随机抽出第二张．（1）如果要求抽出的两张卡片上的图形，既有圆又有三角形，请你用列表或画树状图的方法，求出出现这种情况的概率；（2）因为四张卡片上有两张上的图形，既是中心对称图形，又是轴对称图形，所以小明和小东约定做一个游戏，规则是：如果抽出的两个图形，既是中心对称图形又是轴对称图形，则小明赢；否则，小东赢．问这个游戏公平吗？为什么？如果不公平，请你设计一个公平的游戏规则．24．青神竹编，工艺精美，受到人们的喜爱，有一客商到青神采购A B 、两种竹编工艺品回去销售，其进价和回去的售价如右表所示，若该客商计划采购A B 、两种竹编工艺品共60件，所需总费用为y 元，其中A 型工艺品x 件．（1）请写出y 与x 之间的函数关系式；（不求出x 的取值范围） （2）若该客商采购的B 型工艺品不少于14件，且所获总利润要求不低于2500元，那么他有几种采购方案？写出每种采购方案，并求出最大利润．B 卷（共20分）一、本大题共1个小题，共9分25．已知：如图，四边形ABCD 是正方形，BD 是对角线，BE 平分DBC ∠交DC 于E 点，交DF 于M ，F 是BC 延长线上一点，且CE CF =． （1）求证：BM DF ⊥；（2）若正方形ABCD 的边长为2，求ME MB ∙．二、本大题共1个小题，共11分．26．已知：如图，直线33y x =+与x 轴交于C 点，与y 轴交于A 点，B 点在x 轴上，OAB △是等腰直角三角形．（1）求过A B C 、、三点的抛物线的解析式；（2）若直线CD AB ∥交抛物线于D 点，求D 点的坐标；（3）若P 点是抛物线上的动点，且在第一象限，那么PAB △是否有最大面积？若有，求出此时P 点的坐标和PAB △的最大面积；若没有，请说明理由．眉山市2012年初中学业水平暨高中阶段教育学校招生考试数学参考答案及评分意见一、选择题：（每小题3分，共36分）1.C2.D3.B4.A5.A6.D7.C8.D9.C 10.B 11.B 12.C 二、填空题：（每小题3分，共18分）13.()21a x - 14. 2 15.π316. 20 17. 1 18. 14AD << 三、计算题：(每小题6分，共12分)19.解：原式()4412=⨯+- ······································································· 3分41=-··································································································· 5分 5= ······························································································································ 6分20.解：原式11322x x x -+=--- ()()1321x x +-=--······························································································· 3分 1361x x +-=-+24x = 2x = ··························································································································· 5分 经检验：2x =是方程的增根，舍去∴原方程无解 ··············································································································· 6分 四、解答题：(每小题8分，共16分)21.（1）如图所示，A BC △′′为所求作图形 ····························································· 2分（2）()100D ， ············································································································· 4分 ()224D --， ················································································································ 6分 ()344D -， ···················································································································· 8分 22.解：由题意可知 在Rt BCD △中6030CBD BCD ==∠°，∠°设BD x =，则2BC x CD ==， ···················· 3分在Rt ACD △中，3060CAD ACD ==∠°，∠° ∵tan CDCAD AD=∠=解得100x = ················································································································ 7分∴CD =答：河宽CD为. ·························································································· 8分 五、解答题：(每小题9分，共18分) 23.解：（1）树状图如下：··········································································································· 4分或列表法如右表：()21126P ==既有圆又有三角形 ···························································································· 5分 （2）∵()21126P ==既是中心对称图形又是轴对称图形 ∴游戏不公平可以改为：抽出的两个图形中，如果有圆，小明赢；如果没有圆，则小东赢 ········ 9分 24.解：（1）()1508060y x x =+-704800x =+ ············································································································· 3分（2）由题意得：()()()601420015010080602500x x x -⎧⎪⎨-+--⎪⎩≥≥ ····························· 5分解之得：130463x ≤≤ ∵x 为正整数∴44x =或45或46 ··································································································· 6分 ∴有如下三种方案：方案一：购买A 型工艺品44件，B 型工艺品16件； 总利润为：445016202520⨯+⨯=（元）方案二：购买A 型工艺品45件，B 型工艺品15件； 总利润为：455015202550⨯+⨯=（元）方案三：购买A 型工艺品46件，B 型工艺品14件； 总利润为：465014202580⨯+⨯=（元）综上所述第三种方案所获利润最大，最大利润为2580元 ········································ 9分B 卷一、 (1个小题共9分)25.（1）证明：∵正方形ABCD∴BC DC =又∵90CE CF BCE DCF ===，∠∠°∴()SAS BCE DCF △≌△ ························· 2分∴12=∠∠∴132490+=+=∠∠∠∠° ∴BM DF ⊥ ·············································································································· 4分 （2）解：∵BM 平分DBF ∠，根据（1），可知： 67.5BDM F ==∠∠° ∴BD BF = ∴DM FM = ············································································································· 5分 ∵正方形ABCD 的边长为2∴2BD BF CE CF ==== 在Rt DCF △中()22224216DF DC CF =+=+=-∴2242DF DM ⎛⎫==- ⎪⎝⎭·················································································· 7分 ∵21DBM ==∠∠DME BMD =∠∠∴DME BMD △∽△ ······························································································· 8分∴DM MEMB DM= 即2DM ME MB =∴4ME MB =-······························································································· 9分 二、 (1个小题共11分)26.解：（1）∵A C 、分别是直线33y x =+与y 轴和x 轴的交点 ∴()()0310A C -，，，又∵OAB △是等腰直角三角形∴()30B ， ···················································································································· 1分 设过A B C 、、三点的抛物线解析式为：2y ax bx c =++∴30930c a b c a b c =⎧⎪-+=⎨⎪++=⎩解得：123a b c =-⎧⎪=⎨⎪=⎩∴223y x x =-++ ···································································································· 3分 （2）设直线AB 解析式为：y kx b =+ ∵()()0330A B ，，，∴330b k b =⎧⎨+=⎩ 解得13k b =-⎧⎨=⎩∴直线AB ：3y x =-+ ···························································································· 4分 ∵CD AB ∥∴设直线CD 的解析式为：1y x b =-+ ∵直线CD 过点()10C -， ∴11b =-∴直线CD ：1y x =-- ····························································································· 5分 又∵直线1y x =--与抛物线223y x x =-++相交于D 点∴2123y x y x x =--⎧⎨=-++⎩ 解得：1110x y =-⎧⎨=⎩（与C 点重合，舍去）或2245x y =⎧⎨=-⎩∴()45D -， ················································································································ 7分 （3）PAB △有最大面积. ·························································································· 8分 ∴P 点在抛物线上∴设()223P x x x -++，，过P 作PE x ⊥轴于E 点 PAB PEB AOB AOEP S S S S =+-△△△梯形 ()()()2232332333222x x x x x x -++--++⨯=+-23922x x =-+ ············································································································· 9分∴当32x =时，278PAB S =△最大 ··············································································· 10分当32x =，215234y x x =-++=∴31524P ⎛⎫ ⎪⎝⎭， ············································································································· 11分。