2008年资阳市高中阶段学校招生统一考试中考数学试卷及解析

资阳市2008年高中阶段学校招生统一考试数 学全卷分为第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分．第Ⅰ卷1至2页,第Ⅱ卷3至8页．全卷满分120分,考试时间共120分钟．答题前,请考生务必在答题卡上正确填涂自己的姓名、考号和考试科目,并将试卷密封线内的项目填写清楚；考试结束,将试卷和答题卡一并交回．第Ⅰ卷(选择题 共30分)注意事项:每小题选出的答案不能答在试卷上,须用铅笔在答题卡上把对应题目．．．．的答案标号涂黑．如需改动,用橡皮擦擦净后,再选涂其它答案．一、选择题:(本大题共10个小题,每小题3分,共30分)在每小题给出的四个选项中,只有一个选项符合题意．1．4的平方根是 A ．4B ．2C ．－2D ．2或－22．如图1,在数轴上表示到原点的距离为3个单位的点有 A ．D 点B ．A 点C ．A 点和D 点D ．B 点和C 点3．下列运算正确的是 A ．(ab )5＝ab 5B ．a 8÷a 2＝a 6C ．(a 2)3＝a 5D ．(a －b )2＝a 2－b 24．如图2,CA ⊥BE 于A ,AD ⊥BF 于D ,下列说法正确的是 A ．α的余角只有∠BB ．α的邻补角是∠DACC ．∠ACF 是α的余角D ．α与∠ACF 互补5．下列说法正确的是A ．频数是表示所有对象出现的次数B ．频率是表示每个对象出现的次数C ．所有频率之和等于1D ．频数和频率都不能够反映每个对象出现的频繁程度6．2008年5月5日,奥运火炬手携带着象征“和平、友谊、进步”的奥运圣火火种,离开海拔5200米的“珠峰大本营”,向山顶攀登．他们在海拔每上升100米,气温就下降0.6°C 的低温和缺氧的情况下,图2图1于5月8日9时17分,成功登上海拔8844.43米的地球最高点．而此时“珠峰大本营”的温度为－4°C,峰顶的温度为(结果保留整数)A．－26°C B．－22°C C．－18°C D．22°C7．已知a、b、c分别是三角形的三边,则方程(a + b)x2 + 2cx + (a + b)＝0的根的情况是A．没有实数根B．可能有且只有一个实数根C．有两个相等的实数根D．有两个不相等的实数根8．已知矩形ABCD的边AB＝15,BC＝20,以点B为圆心作圆,使A、C、D三点至少有一点在⊙B 内,且至少有一点在⊙B外,则⊙B的半径r的取值范围是A．r＞15 B．15＜r＜20 C．15＜r＜25 D．20＜r＜259．在平面直角坐标系中,如果抛物线y＝2x2不动,而把x轴、y轴分别向上、向右平移2个单位,那么在新坐标系下抛物线的解析式是A．y＝2(x－2)2 + 2 B．y＝2(x + 2)2－2C．y＝2(x－2)2－2 D．y＝2(x + 2)2 + 210．如图3,已知Rt△ABC≌Rt△DEC,∠E＝30°,D为AB的中点,AC＝1,若△DEC绕点D顺时针旋转,使ED、CD分别与Rt△ABC的直角边BC相交于M、N,则当△DMN为等边三角形时,AM的值为A．3B．23C．3D．1图3资阳市2008年高中阶段学校招生统一考试数 学第Ⅱ卷(非选择题 共90分)题号 二 三总 分总分人1718192021222324得分注意事项:本卷共6页,用黑色或蓝色钢笔或圆珠笔直接答在试卷上．请注意准确理解题意、明确题目要求,规范地表达、工整地书写解题过程或结果.二、填空题:(本大题共6个小题,每小题3分,共18分)把答案直接填在题中横线上．11．如图4,□ABCD 中,对角线AC 、BD 交于点O ,请你写出其中的一对全等三角形_________________．12．计算:cot60°－2－2 + 20080＋233＝__________． 13．若A (1x ,1y )、B (2x ,2y )在函数12y x的图象上,则当1x 、2x 满足_______________时,1y ＞2y . 14．如图5,校园内有一块梯形草坪ABCD ,草坪边缘本有道路通过甲、乙、丙路口,可是有少数同学为了走捷径,在草坪内走了一条直“路”EF ,假设走1步路的跨度为0.5米,结果他们仅仅为了少走________步路,就踩伤了绿化我们校园的小草(“路”宽忽略不计).15．资阳市某学校初中2008级有四个绿化小组,在植树节这天种下柏树的颗数如下:10,10,x ,8,若这组数据的众数和平均数相等,那么它们的中位数是________颗．16．如图6,在地面上有一个钟,钟面的12个粗线段刻度是整点时时针(短针)所指的位置．根据图中时针与分针(长针)所指的位置,该钟面所显示的时刻是______时_______分．图4图5图6三、解答题:(本大题共8个小题,共72分)解答应写出必要的文字说明、证明过程或演算步骤．17．(本小题满分7分)先化简,再求值:(21 2x x-－2144x x-+)÷222x x-,其中x＝1．18．(本小题满分7分)如图7,在△ABC中,∠A、∠B的平分线交于点D,DE∥AC交BC于点E,DF∥BC交AC于点F．(1)点D是△ABC的________心；(2)求证:四边形DECF为菱形．图719．(本小题满分8分)惊闻5月12日四川汶川发生强烈地震后,某地民政局迅速地组织了30吨食物和13吨衣物的救灾物资,准备于当晚用甲、乙两种型号的货车将它们快速地运往灾区.已知甲型货车每辆可装食物5吨和衣物1吨,乙型货车每辆可装食物3吨和衣物2吨,但由于时间仓促,只招募到9名长途驾驶员志愿者.(1) 3名驾驶员开甲种货车,6名驾驶员开乙种货车,这样能否将救灾物资一次性地运往灾区？(2)要使救灾物资一次性地运往灾区,共有哪几种运货方案？20．(本小题满分9分)大双、小双的妈妈申购到一张北京奥运会的门票,兄弟俩决定分别用标有数字且除数字以外没有其它任何区别的小球,各自设计一种游戏确定谁去.大双:A袋中放着分别标有数字1、2、3的三个小球,B袋中放着分别标有数字4、5的两个小球,且都已各自搅匀,小双蒙上眼睛从两个口袋中各取出1个小球,若两个小球上的数字之积为偶数,则大双得到门票；若积为奇数,则小双得到门票.小双:口袋中放着分别标有数字1、2、3的三个小球,且已搅匀,大双、小双各蒙上眼睛有放回．．．地摸1次,大双摸到偶数就记2分,摸到奇数记0分；小双摸到奇数就记1分,摸到偶数记0分,积分多的就得到门票(若积分相同,则重复第二次)．(1)大双设计的游戏方案对双方是否公平？请你运用列表或树状图说明理由；(2)小双设计的游戏方案对双方是否公平？不必说理．21．(本小题满分9分)若一次函数y ＝2x －1和反比例函数y ＝2kx的图象都经过点(1,1)． (1)求反比例函数的解析式；(2)已知点A 在第三象限,且同时在两个函数的图象上,求点A 的坐标；(3)利用(2)的结果,若点B 的坐标为(2,0),且以点A 、O 、B 、P 为顶点的四边形是平行四边形,请你直接写出点P 的坐标．·22．(本小题满分10分)如图8,小唐同学正在操场上放风筝,风筝从A 处起飞,几分钟后便飞达C 处,此时,在AQ 延长线上B 处的小宋同学,发现自己的位置与风筝和旗杆PQ 的顶点P 在同一直线上.(1)已知旗杆高为10米,若在B 处测得旗杆顶点P 的仰角为30°,A 处测得点P 的仰角为45°,试求A 、B 之间的距离；(2)此时,在A 处背向旗杆又测得风筝的仰角为75°,若绳子在空中视为一条线段,求绳子AC 约为多少？(结果可保留根号)23．(本小题满分10分)阅读下列材料,按要求解答问题:图8如图9－1,在ΔABC中,∠A＝2∠B,且∠A＝60°．小明通过以下计算:由题意,∠B＝30°,∠C＝90°,c ＝2b,a＝3b,得a2－b2＝(3b)2－b2＝2b2＝b·c．即a2－b2＝bc．于是,小明猜测:对于任意的ΔABC,当∠A＝2∠B时,关系式a2－b2＝bc都成立．(1)如图9－2,请你用以上小明的方法,对等腰直角三角形进行验证,判断小明的猜测是否正确,并写出验证过程；(2)如图9－3,你认为小明的猜想是否正确,若认为正确,请你证明；否则,请说明理由；(3)若一个三角形的三边长恰为三个连续偶数,且∠A＝2∠B,请直接写出这个三角形三边的长,不必说明理由．图9-1图9-2图9-324．(本小题满分12分)如图10,已知点A的坐标是(－1,0),点B的坐标是(9,0),以AB为直径作⊙O′,交y 轴的负半轴于点C,连接AC、BC,过A、B、C三点作抛物线．(1)求抛物线的解析式；(2)点E 是AC延长线上一点,∠BCE 的平分线CD 交⊙O ′于点D ,连结BD ,求直线BD 的解析式； (3)在(2)的条件下,抛物线上是否存在点P ,使得∠PDB ＝∠CBD ?如果存在,请求出点P 的坐标；如果不存在,请说明理由．资阳市2008年高中阶段学校招生统一考试数学试题参考答案及评分意见说 明:1. 解答题中各步骤所标记分数为考生解答到这一步应得分数的累计分数；2. 参考答案中的解法只是该题解法中的一种或几种,如果考生的解法和参考答案所给解法不同,请参照本答案中的标准给分；3. 评卷时要坚持每题评阅到底,当考生的解答在某一步出现错误、影响了后继部分时,如果该步以后的解答未改变问题的内容和难度,可视影响程度决定后面部分的给分,但不得超过后继部分应给分数的一半；如果这一步后面的解答有较严重的错误,就不给分；若是几个相对独立的得分点,其中一处错误不影响其他得分点的得分；4. 给分和扣分都以1分为基本单位；5. 正式阅卷前应进行试评,在试评中须认真研究参考答案和评分意见,不能随意拔高或降低给分标准,统一标准后须对全部试评的试卷予以复查,以免阅卷前后期评分标准宽严不同.一、选择题:(每小题3分,共10个小题,满分30分) 1-5. DCBDC ；6-10. AACBB.二、填空题:(每小题3分,共6个小题,满分18分)11．答案不唯一,ΔAOB ≌ΔCOD 、ΔAOD ≌ΔCOB 、ΔADB ≌ΔCBD 、ΔABC ≌ΔCDA 之一均可；12．343＋(或34＋3)；13．答案不唯一,x 1<x 2<0,或 0<x 1<x 2,或210x x <<或122,3x x ==-等之一均可； 14． 4； 15．10 ； 16．9,12；三、解答题:(共9个小题,满分72分)17．原式=[1(2)x x -–21(2)x -]×(2)2x x - ··················································3分图10图7=1(2)x x -×(2)2x x -–21(2)x -×(2)2x x - =12–2(2)x x -·····················································································4分 =22(2)x x --–2(2)x x - =12x - ·····························································································5分 当x =1时,原式=121- ·······················································································6分= 1 ··································································································7分 说明:以上步骤可合理省略 . 18．(1) 内. ······················································································2分 (2) 证法一:连接CD , ···········································································3分 ∵ DE ∥AC ,DF ∥BC ,∴ 四边形DECF 为平行四边形, ····························································4分 又∵ 点D 是△ABC 的内心,∴ CD 平分∠ACB ,即∠FCD ＝∠ECD , ·····················································5分 又∠FDC ＝∠ECD ,∴ ∠FCD ＝∠FDC ∴ FC ＝FD , ·······················································································6分 ∴ □DECF 为菱形． ···········································································7分 证法二:过D 分别作DG ⊥AB 于G ,DH ⊥BC 于H ,DI ⊥AC 于I ． ······························3分 ∵AD 、BD 分别平分∠CAB 、∠ABC , ∴DI =DG , DG =DH ．∴DH =DI ． ·······················································································4分∵DE ∥AC ,DF ∥BC ,∴四边形DECF 为平行四边形, ······························································5分 ∴S □DECF =CE ·DH =CF ·DI , ∴CE =CF ． ·······················································································6分 ∴□DECF 为菱形． ···········································································7分19．(1) ∵3×5+6×3=33＞30,3×1+6×2=15＞13, ···········································1分∴3名驾驶员开甲种货车,6名驾驶员开乙种货车,这样能将救灾物资一次性地运到灾区． ······································································································2分 (2) 设安排甲种货车x 辆,则安排乙种货车(9–x )辆, ·····································3分由题意得:53(9)30,2(9)13.x x x x +-≥⎧⎨+-≥⎩ ·································································5分解得:1.5≤x ≤5 ···················································································6分注意到x 为正整数,∴x =2,3,4,5 ·······························································7分∴安排甲、乙两种货车方案共有下表4种:方案方案一方案二方案三方案四甲种货车 2 3 4 5乙种货车7 6 5 4 ······································································································8分说明:若分别用“1、8”,“2、7”等方案去尝试,得出正确结果,有过程．．．也给全分. 20．(1) 大双的设计游戏方案不公平．····················································1分可能出现的所有结果列表如下:1 2 34 4 8 125 5 10 15或列树状图如下:0··············································4分∴P(大双得到门票)= P(积为偶数)=46=23,P(小双得到门票)= P(积为奇数)=13, ·························································6分∵23≠13,∴大双的设计方案不公平． ······················································7分(2) 小双的设计方案不公平．································································9分参考:可能出现的所有结果列树状图如下:A袋积B袋21．(1) ∵反比例函数y =2k x 的图象经过点(1,1), ∴1=2k ·····························································································1分 解得k =2, ··························································································2分∴反比例函数的解析式为y =1x． ···························································3分 (2) 解方程组211.y x y x =-⎧⎪⎨=⎪⎩，得11x y =⎧⎨=⎩，；122.x y ⎧=-⎪⎨⎪=-⎩， ··············································5分 ∵点A 在第三象限,且同时在两个函数图象上,∴A (12-,–2)． ···················································································6分 (3) P 1(32,–2),P 2(52-,–2),P 3(52,2)．(每个点各1分) ·····································9分 22. (1) 在Rt △BPQ 中,PQ =10米,∠B =30°,则BQ =cot30°×PQ ＝103, ···························································· 2分 又在Rt △APQ 中,∠P AB =45°,则AQ =cot45°×PQ =10,即:AB =(103+10)(米)； ··············································· 5分(2) 过A 作AE ⊥BC 于E ,在Rt △ABE 中,∠B =30°,AB =103+10,∴ AE =sin30°×AB =12(103+10)=53+5, ··········································· 7分 ∵∠CAD =75°,∠B =30°,∴ ∠C =45°, ················································································· 8分 在Rt △CAE 中,sin45°＝AE AC, ∴AC =2(53+5)=(56+52)(米) ··············································· 10分23. (1) 由题意,得∠A =90°,c =b ,a =2b ,∴a 2–b 2=(2b )2–b 2=b 2=bc ． ·········································· 3分(2) 小明的猜想是正确的． ··········································· 4分理由如下:如图3,延长BA 至点D ,使AD =AC =b ,连结CD ,·············································································· 5分则ΔACD 为等腰三角形．∴∠BAC =2∠ACD ,又∠BAC =2∠B ,∴∠B =∠ACD =∠D ,∴ΔCBD为等腰三角形,即CD =CB =a , ················································ 6分又∠D ＝∠D ,∴ΔACD ∽ΔCBD , ······································ 7分图9-3图8∴AD CDCD BD=．即b aa b c=+．∴a2=b2＋bc．∴a2–b2= bc······· 8分(3) a=12,b=8,c=10． ··················································10分24．(1) ∵以AB为直径作⊙O′,交y轴的负半轴于点C,∴∠OCA+∠OCB=90°,又∵∠OCB+∠OBC=90°,∴∠OCA=∠OBC,又∵∠AOC= ∠COB=90°,∴ΔAOC∽ ΔCOB, ···············································································1分∴OA OC OC OB=．又∵A(–1,0),B(9,0),∴19OCOC=,解得OC=3(负值舍去)．∴C(0,–3), ······································································································3分设抛物线解析式为y=a(x+1)(x–9),∴–3=a(0+1)(0–9),解得a=1 3 ,∴二次函数的解析式为y=13(x+1)(x–9),即y=13x2–83x–3．····························4分(2) ∵AB为O′的直径,且A(–1,0),B(9,0),∴OO′=4,O′(4,0), ·················································································5分∵点E是AC延长线上一点,∠BCE的平分线CD交⊙O′于点D,∴∠BCD=12∠BCE=12×90°=45°,连结O′D交BC于点M,则∠BO′D=2∠BCD=2×45°=90°,OO′=4,O′D=12AB=5．∴D(4,–5)．·······················································································6分∴设直线BD的解析式为y=kx+b(k≠0)∴90,4 5.k bk b+=⎧⎨+=-⎩·················································· 7分解得1,9. kb=⎧⎨=-⎩∴直线BD的解析式为y=x–9. ································ 8分(3) 假设在抛物线上存在点P,使得∠PDB=∠CBD,解法一:设射线DP交⊙O′于点Q,则BQ CD=．分两种情况(如答案图1所示):①∵O′(4,0),D(4,–5),B(9,0),C(0,–3)．∴把点C、D绕点O′逆时针旋转90°,使点D与点B重合,则点C与点Q1重合,因此,点Q1(7,–4)符合BQ CD=,图10答案图1。