2011江津中考数学试题及答案

2011江津中考数学试题及答案题号一二三总分得分一、填空题(本大题共6小题，每小题3分，共18分)1．|31|=．2．函数y=2x的自变量取值范围是．3．观察下列各式：12×2=12+2，23×3=23+3，34×4=34+4，45×5=45+5…想一想，什么样的两数之积等于这两数之和？设n表示正整数，用关于n的等式表示这个规律为．4．如果反比例函数y=xk的图象经过点P（-3，1）那么k=．5．如果一个角的补角是120°，那么这个角的余角是．6．如图：AB∥CD，直线EF分别交AB、CD于E、F，EG平分∠BEF，若∠1=72°,则∠1=72°,则∠2=．二、选择题（本大题共8个小题，每小题4分，满分32分）7．下列计算正确的是（）A．（-4x2）(2x2+3x-1)=-8x4-12x2-4xB．(x+y)(x2+y2)=x3+y3C．(-4a-1)(4a-1)=1-16a2D．(x-2y)2=x2-2xy+4y28．把x2-1+2xy+y2的分解因式的结果是（）A．（x+1）(x-1)+y(2x+y)B．(x+y+1)(x-y-1)C．(x-y+1)(x-y-1)D．(x+y+1)(x+y-1)9．已知关于x的方程x2-2x+k=0有实数根，则k的取值范围是（）A．k＜1B．k≤1C．k≤-1D．k≥110．某电视台举办的通俗歌曲比赛上，六位评委给1号选手的评分如下：909691969594这组数据的众数和中位数分别是（）A．94.5，95B．95，95C．96，94.5D．2，9611．面积为2的△ABC，一边长为x，这边上的高为y，则y与x的变化规律用图象表示大致是（）得分评卷人得分评卷人ABCD12．有如下结论（1）有两边及一角对应相等的两个三角形全等；（2）菱形既是轴对称图形又是中心对称图形；（3）对角线相等的四边形是矩形；（4）平分弦的直径垂直于弦，并且平分弦所对的两条弧；（5）两圆的公切线最多有4条，其中正确结论的个数为（）A．1个B．2个C．3个D．4个13．已知：如图梯形ABCD中，AD∥BC，AB=DC，AC与BD相交于点O，那么图中全等三角形共有（）对.A．1对B．2对C．3对D．4对14．如图四边形ABCD内接于⊙O，若∠BOD=100°，则∠DAB=（）A．50°B．80°C．100°D．130°三、解答题（本大题共7个小题，共70分）15．（本小题6分）计算：18+1212-481得分评卷人16．（本小题7分）解方程：3212x-8x2+12=017．（6分）某中学团委到位于A市南偏东60°方向50海里的B基地慰问驻车，然后乘船前往位于B基地正西方向的C哨所看望值班战士，C哨所位于A市的南偏西43°方向，求C到A的距离（精确到1海里，以下数据供选用sin43°≈0.68,cos43°≈0.73）得分评卷人得分评卷人18．（4分）平原上有A、B、C、D四个村庄，为解决当地缺水问题，政府准备投资修建一个蓄水池，不考虑其它因素，请你画图确定蓄水池H点的位置，使它与四个村庄的距离之和最小.（不作证明）19．（本小题5分）阅读材料，解答问题：如图，在锐角△ABC中，BC=a,CA=b,AB=c,△ABC的外接圆的半径为R，则Aasin=Bbsin=Ccsin=2R证明：连结CO并延长交⊙O的直径，∴∠DBC=90°,在Rt△DBC中，∵sinD=DCBC=Ra2,∴sinA=sinD=Ra2∴Aasin=2R同理Bbsin=2R,Ccsin=2R前面的阅读材料中略去Bbsin=2R,Ccsin=2R的证明过程，请你把BCcsin=2R的证明过程补写出来.得分评卷人得分评卷人20．（12分）某图书馆开展两种方式的租书业务，一种是使用会员卡，另一种是使用租书卡，使用这两种卡租书，租书金额y（元）与租书时间x(天)之间的关系如下图所示.（1）分别求出用租书卡和会员卡租书的金额y（元）与租书时间x（天）之间的函数关系式.（2）两种租书方式每天租书的收费分别是多少元？（x≤100）21．（8分）某校师生去外地参加夏令营活动，车站提出两种车票价格的优惠方案供学校选择.第一种方案是教师按原价付款，学生按原价的78%付款；第二种方案是师生都按原价的80%付款；该校有5名教师参加这项活动，试根据夏令营的学生人数选择购票付款的最佳方案？得分评卷人得分评卷人22．（8分）初三（几何）课本中有这样的一道习题，若⊙O1与⊙O2外切于点A，BC是两圆的一条外公切线，B、C为切点，则AB⊥AC（1）若⊙O1和⊙O2外离，BC为两圆的外公切线，B，C为切点，连心线O1O2分别交⊙O1，⊙O2于M，N，设BM与CN的延长线交于A，试问AB与AC是否垂直？证明你的结论.（2）若⊙O1与⊙O2相交，AB与AC垂直吗？得分评卷人23．（14分）函数y=-43x-12的图象分别交x轴，y轴于A,C两点，（1）求出A、C两点的坐标.（2）在x轴上找出点B，使△ACB~△AOC，若抛物线经过A、B、C三点，求出抛物线的解析式.（3）在（2）的条件下，设动点P、Q分别从A、B两点同时出发，以相同的速度沿AC、BA向C、A运动，连结PQ，设AP=m，是否存在m值，使以A、P、Q为顶点的三角形与△ABC相似，若存在，求出所有的m值；若不存在，请说明理由.得分评卷人参考答案一、1．312．x≥23．nn1(n+1)=nn1+(n+1)4．-35．30°6．54°二、7．C8．D9．B10．C11．C12．B13．C14．D三、15．解：原式=32+3-22············································4分=3·····························································6分16．解：设2x2-3=y,则原方程变形为y1-4y=0·····························1分整理得1-4y2=0解这个方程得，y1=21,y2=-21·················································3分当y1=21时，2x2-3=21，解得：x=±27··································4分当y2=-21时，2x2-3=-21，解得：x=±25································5分经检验知，它们都是原方程的根.所以原方程的根是：x=±27，x=±25·································7分17．解：过A作AD⊥BC于D，············································1分在Rt△ADB中，∠DAB=60°,AB=50海里，AD=ABcos60°=25(海里)·······················································································3分在Rt△ADC中，cos43°=AC25，AC=73.025≈34（海里）···········5分答：C到A的距离约是34海里.18．作法：连结AC与BD交点为H点，则H点即为所求的点.······4分19．连结BO并延长交⊙O于E，连结EA，则∠ACB=∠E············1分∵BE是⊙O的直径，∴∠BAE=90°·····································································2分在Rt△ABE中，sinE=BEAB=Rc2·············································3分∴sin∠ACB=sinE=Rc2··························································4分∴ACBcsin=2R·························································································5分20．解：（1）租书卡：设y=kx································································1分观察图象知，当x=100时，y=50，∴100k=50，解得k=21∴y=21x···········································3分用会员卡时，设y=kx+b·························································4分∵(0,20),(100,50)在直线y=kx+b上，∴201003501002020bkbkbb，解得·······························6分∴y=1003x+20（2）用租书卡的方式租书，每天租书的收费为50÷100=0.5（元）··9分用会员卡的方式租书，每天租书的收费为（50-20）÷100=0.3（元）························································································11分答：略.··············································································12分21．解：设参加夏令营活动的学生x人，每张车票的原价为a元，按第一种方案购票应付款y1元，按第二种方案购票应付款y2元，依题意得：···············1分y1=5a+a×78%·xy2=(x+5)·a·80%······································4分（1）当y2＞y1时，（x+5）·a·80%＞5a+a×78%·x解得x＞50··5分（2）当y2=y1时,（x+5）·a·80%=5a+a×78%·x解得x=50········6分（3）当y2＜y1时，（x+5）·a·80%＜5a+a×78%x解得x＜50·······7分答：当学生多于50人时，按第一种方案，当学生等于50人时，两种方案都可以，当学生少于50人时，按第二种方案.···············································8分22．解：（1）猜想：AB⊥AC··················································1分证明：分别连结O1B、O2C，··················································2分则O1B⊥BCO2C⊥BC，从而O1B∥O2C··································4分∴∠BO1O2+∠CO2O=180°·····················································5分∵∠ABC+∠ACB=21∠BO1O2+21∠CO2O1=90°·························6分∴∠BAC=90°即AB⊥AC······················································7分（2）亦有AB⊥AC，证明与（1）类似，略.·······························8分23．（1）A（-16，0）C（0，-12）··········································2分（2）过C作CB⊥AC，交x轴于点B，显然，点B为所求，·········3分则OC2=OA×OB此时OB=9，可求得B（9，