高中阶段教育学校招生考试数学试卷及答案 (全word)及初中毕业会考数学试卷本试卷分会考卷和加试卷两部分，会考卷1至6页，满分100分；加试卷7至10页，满分60分.全卷满分160分，120分钟完卷. 注意事项：1.答题前，考生务必将密封线内的内容填写清楚，将自己的姓名、准考证号、考试科目等涂写在机读卡上.2.答第Ⅰ卷时，每小题选出答案后，用铅笔把机读卡上对应题目的答案标号涂黑.如需改动，用橡皮擦干净后再选涂其它答案.3.只参加毕业会考的考生只需做会考卷，要参加加升学考试的考生须完成会考卷和加试卷两部分.4.考试结束后，将本试卷和机读卡一并收回.第Ⅰ卷（选择题 共36分）一、选择题（本大题共12小题，每小题3分，共36分.在每小题给出的四个选项中，只有一项是符合题目要求的.） 1.12010-的倒数是 A ．2010- B. 2010 C.12010 D. 12010- 2.截止2010年4月20日23时35分，央视“情系玉树，大爱无疆”赈灾晚会共收到社会各界为玉树捐款2 175 000 000元，用科学记数法表示捐款数应为A ．102.17510⨯元 B. 92.17510⨯元 C. 821.7510⨯元 D. 7217.510⨯元 3.下列图形是正方体的表面展开图的是4．下列事件中为必然事件的是 A ．早晨的太阳一定从东方升起 B.打开数学课本时刚好翻到第60页 Ｃ从一定高度落下的图钉，落地后钉尖朝上. Ｄ.今年14岁的小云一定是初中学生ABCD5．将一副三角板如图放置，使点A 在DE 上，BC DE ∥，则AFC ∠的度数为A.45°B. 50°C. 60°D. 75° 6.函数y x=中，自变量x 的取值范围是 A.1x -≥ B. 1x >-C. 1x -≥且0x ≠D. 1x >-且0x ≠ 7.方程()12x x -=的解是A ．1x =- B. 2x =- C. 1212x x ==-， D.1212x x =-=， 8.某品牌服装折扣店将某件衣服按进价提高50%后标价，再打8折（标价的80%）销售，售价为240元.设这件衣服的进价为x 元，根据题意，下面所列的方程正确的是A ．50%80%240x ⨯=· B.()150%80%240x +⨯=·C.24050%80%x ⨯⨯=D. ()150%24080%x +=⨯·9.学剪五角星：如图，先将一张长方形纸片按图①的虚线对折，得到图②，然后将图②沿虚线折叠得到图③，再将图③沿虚BC 剪下ABC △，展开即可得到一个五角星.如果想得到一个正五角星（如图④），那么在图③中剪下ABC △时，应使ABC ∠的度数为A.126°B. 108°C. 100°D. 90°10.在四张完全相同的卡片上分别印有等边三角形、平行四边形、等腰梯形、圆的图案，现将印有图案的一面朝下，混合后从中一次性随机抽取两张，则抽到的卡片上印有的图案都是轴对称图形的概率为 A ．14 B. 13 C. 12 D. 34① ② ③ ④11.如图，反比例函数()0ky x x=>的图象经过矩形OABC 对角线的交点M ，分别与AB BC 、相交于点.D E 、若四边形ODBE 的面积为6，则k 的值为A ．1 B. 2 C. 3 D. 412.如图，梯形ABCD 中，AD BC ∥，点E 在BC 上，AE BE =，点F 是CD 的中点，且 AF AB ⊥，若 2.746AD AF AB ===，，，则CE 的长为A ． B. 1 C. 2.5 D. 2.3内江市二O 一O 年高中阶段教育学校招生考试及初中毕业会考试卷数学第Ⅱ卷（非选择题 共64分）注意事项：1.第Ⅱ卷共4页，用钢笔或圆珠笔将答案直接答在试卷上.2.答题前将密封线内的项目填写清楚.二、填空题（本大题共4小题，每小题5分，共20分.请将最后答案直接填在题中横线上.） 13.在一次演讲比赛中，某选手的得分情况如下：87、91、91、93、87、89、96、97，这组数据的中位数是_________.14.化简：2111x x x x x+++=--_________. 15.如图，为了测量某棵树的高度，小明用长为2m 的竹竿做测量工具，移动竹竿，使竹竿、树的顶端的影子恰好落在地面的同一点.此时，竹竿与这一点距离相距6m ，与树相距15m ，则树的高度为_________m.16.如图,圆内接四边形ABCD 是由四个全等的等腰梯形组成，AD 是O ⊙的直径，则BEC ∠为___________度.三、解答题（本大题共5小题，共44分） 17．(7分)已知()1012cos 451201013a b c d π-⎛⎫==+=-= ⎪⎝⎭，°，，（1）请化简这四个数；（2）根据化简结果，列式表示这四个数中“有理数的和”与“无理数的积”的差，然后计算结果.18．(9分)如图，ACD △和BCE △都是等腰直角三角形，90ACD BCE AE ∠=∠=°，交CD 于点F BD ，分别交CE AE 、于点.G H 、试猜测线段AE 和BD 的数量和位置关系，并说明理由.19．(9分)学校为了解学生参加体育活动的情况，对学生“平均每天参加体育活动的时间”进行了随机抽样调查，下图是根据调查结果绘制的两幅不完整的统计图.请你根据统计图提供的信息，解答以下问题：（1）“平均每天参加体育活动的时间”“为0.5~1小时”部分的扇形统计图的圆心角为______度；（2）本次一共调查了_________名学生；（3）将条形统计图补充完整；（4）若该校有2000名学生，你估计全校可能有多少名学生平均每天参加体育活动的时间在0.5小时以下.20.（9分）为建设“宜居宜业宜游”山水园林式城市，内江市正在对城区沱江河段进行区域性景观打造. 如图，某施工单位为测得某河段的宽度，测量员先在河对岸边取一点A，再在河这边沿河边取、，在点B处测得点A在北偏东30°方向上，在点C处测得点A在西北方向上，量两点B C得BC长为200米.请你求出该河段的宽度（结果保留根号）.21. （10分）一家蔬菜公司收购到某种绿色蔬菜140吨，准备加工后进行销售，销售后获利的情况如下表所示：销售方式粗加工后销售精加工后销售每吨获利(元) 1000 2000已知该公司的加工能力是：每天能精加工5吨或粗加工15吨，但两种加工不能同时进行.受季节等条件的限制，公司必须在一定时间内将这批蔬菜全部加工后销售完.（1）如果要求12天刚好加工完140吨蔬菜，则公司应安排几天精加工，几天粗加工？（2）如果先进行精加工，然后进行粗加工.①试求出销售利润W元与精加工的蔬菜吨数m之间的函数关系式；②若要求在不超过10天的时间内，将140吨蔬菜全部加工完后进行销售，则加工这批蔬菜最多获得多少利润？此时如何分配加工时间？内江市二O一O年高中阶段教育学校招生考试及初中毕业会考试卷数学加试卷（共60分）二总分总分人题号一5 6 7得分注意事项：加试卷共4页，请将答案直接写在试卷上. 一、选择题（本大题共4小题，每小题6分，共24分.请将最简答案直接填写在题中横线上.）1.已知2510m m --=，则22125m m m-+=___________. 2.下面的方格图案中的正方形顶点叫做格点，图1中以格点为顶点的等腰直角三角形共有4个，图2中以格点为顶点的等腰直角三角形共有___________个，图3中以格点为顶点的等腰直角三角形共有___________个，图4中以格点为顶点的等腰直角三角形共有___________个.3.已知非负数a b c ，，满足条件75a b c a +=-=，，设S a b c =++的最大值为m ，最小值为n ，则m n -的值为___________. 4.如图，在ABC △中，AB AC =，点E F 、分别在AB 和AC 上，CE 与BF 相交于点D ，若AE CF D =，为BF 的中点，AE AF :的值为___________.二、解答题（本大题共3个小题，每小题12分，共36分.解答题必须写出必要的文字说明、证明过程或推演步骤.） 5.（12分） 阅读理解：我们知道，任意两点关于它们所连线段的中点成中心对称，在平面直角坐标系中，任意两点()()1122P x y Q x y ，、，的对称中心的坐标为1212.22x x y y ++⎛⎫⎪⎝⎭，观察应用：（1）如图，在平面直角坐标系中，若点()()120123P P -、，的对称中心是点A ，则点A 的坐 标为_________；（2）另取两点()()1.62.110.B C --，、，有一电子青蛙从点1P 处开始依次关于点A B C 、、作循环对称跳动，即第一次跳到点1P 关于点A 的对称点2P 处，接着跳到点2P 关于点B 的对 称点3P 处，第三次再跳到点3P 关于点C 的对称点4P 处，第四次再跳到点4P 关于点A 的对称点5P 处，…则点38P P 、的坐标分别为_________、_________.拓展延伸：（3）求出点2012P 的坐标，并直接写出在x 轴上与点2012P 、点C 构成等腰三角形的点的坐标.6.（12分）如图，在Rt ABC △中，90C ∠=°，点E 在斜边AB 上，以AE 为直径的O ⊙与BC 相切于点.D（1）求证：AD 平分.BAC ∠ （2）若3 4.AC AE ==，①求AD 的值；②求图中阴影部分的面积.7.（12分）如图，抛物线()2230y mx mx m m =-->与x 轴交于A B 、两点，与y 轴交于C 点.（1）请求出抛物线顶点M 的坐标（用含m 的代数式表示），A B 、两点的坐标； （2）经探究可知，BCM △与ABC △的面积比不变，试求出这个比值；（3）是否存在使BCM △为直角三角形的抛物线？若存在，请求出；如果不存在，请说明 理由.参考答案及评分意见会考卷（共100分）一、选择题（本大题共12小题，每小题3分，共36分.在每小题给出的四个选项中，只有一项是符合题目要求的.）1．A 2．B 3．C 4．A 5．D 6．C 7．D 8．B 9．A 10．C 11．B 12．D 二、填空题（本大题共4小题，每小题5分，共20分.请将最后答案直接填在题中横线上.）13．91 14．1x + 15．7 16．30 三、解答题（本大题共5小题，共44分）17.解：（1）11()33n -==，2cos451212b =+=⨯+°1=，0(2010π)c =- 1=，11d == ······································································································ 4分（2）a c ，为有理数，b d ，为无理数， ··································································· 5分311)a c bd ∴+-=+- ········································································· 6分 =4(21)3--= ·························································································· 7分18.解：猜测 AE BD AE BD =，⊥． ·········································································· 2分理由如下：90ACD BCE ∠=∠= °，ACD DCE BCE DCE ∴∠+∠=∠+∠，即.ACE DCB ∠=∠ ····································· 3分 ACD ∴△和BCE △都是等腰直角三角形. AC CD CE CB ∴==，， ································································································ 4分 ACE DCB ∴△≌△． ····································································································· 5分 AE BD ∴=， ··················································································································· 6分 .CAE CDB ∠=∠ ··········································································································· 7分 90AFC DFH DHF ACD ∠=∠∴∠=∠= ，°． ·························································· 8分 AE BD ∴⊥． ·················································································································· 9分 19.解：（1）54·················································································································· 2分 （2）200 ·························································································································· 4分········································································································································· 7分（3）20005%100⨯=（人） ······································ 9分 20．解：过点A 作AD BC ⊥于点D . ··························· 1分据题意，90306045ABC ACD ∠=-=∠=°°°，°． ···· 2分45CAD ACD CAD ∴∠=∴∠=∠°，， AD CD ∴=，200.BD BC CD AD ∴=-=- ······································ 4分 在Rt ABD △中，tan ADABD BD∠=，tan (200)tan60)AD BD ABD AD AD ∴=∠=-=-··°．····························· 7分AD ∴=300AD ∴==- ··················································································· 9分答：该河段的宽度为（300-.21．解：（1）设应安排x 天进行精加工，y 天进行粗加工， ······································ 1分 根据题意得 12515140.x y x y +=⎧⎨+=⎩，······················································································ 3分解得48.x y =⎧⎨=⎩，答：应安排4天进行精加工，8天进行粗加工. ··························································· 4分 （2）①精加工m 吨，则粗加工（140m -）吨，根据题意得20001000(140)W m m =+-=1000140000m + ································································································ 6分 ② 要求在不超过10天的时间内将所有蔬菜加工完，14010515m m-∴+≤ 解得 5m ≤ ······································································ 8分 05m ∴<≤ 又 在一次函数1000140000W m =+中，10000k =>，W ∴随m 的增大而增大，∴当5m =时，5140000145000.W ⨯+=最大=1000 ············································· 9分 ∴精加工天数为55÷=1，粗加工天数为(1405)159-÷=．∴安排1天进行精加工，9天进行粗加工，可以获得最多利润为145000元. ······· 10分 加试卷（共60分）一、填空题（本大题共4小题，每小题6分，共24分.请将最简答案直接填在题中横线上.）1．28 2．10，28，50 3．7 4二、解答题（本大题共3个小题，每小题12分，共36分.解答题必须写出必要的文字说明、证明过程或推演步骤.）5.解：（1）（1，1） ·········································································································· 2分（2）（5.21.2-，） ··········································································································· 4分（2，3） ·························································································································· 6分 （3）1(01)P ，-→2(23)P ，→3( 5.21.2)P -，→4(3.2 1.2)P -，→5( 1.23.2)P -，→6(21)P -，→7(01)P -，→8(23)P ，… ∴7P 的坐标和1P 的坐标相同，8P 的坐标和2P 的坐标相同，即坐标以6为周期循环.20126÷= 335…2，2012P ∴的坐标与2P 的坐标相同，为2012(23)P ，； ·························································· 8分 在x 轴上与点2012P 、点C 构成等腰三角形的点的坐标为()(20)10)(50)-，0，，，， ······································································ 12分 6.（1）证明：连接OD ，则OA OD =，DAO ODA ∴∠=∠. ··································· 1分BC 是O ⊙的切线， .OD BC ∴⊥ AC BC OD AC ∴ ⊥，∥， ············································ 2分 .CAD ODA ∴∠=∠ DAO CAD AD ∴∠=∠∴，平分.BAC ∠ ························ 4分（2）①连结ED ，AE 为直径，90ADE C ∴∠=∠=°．又由（1）知DAO CAD ADE ACD ∠=∠∴，△∽△， ·················································· 6分AD ACAE AD∴=， ················································································································· 7分 34AC AE == ，， 23412AD AE AC ∴==⨯=·，AD ∴==······································································································· 8分②在Rt ADE △中，cos AD DAE AE ∠=== 30DAE ∴∠=°． ············································································································· 9分120 2.AOD DE ∴∠==°，111222AOD ADE S S AD DE ∴==⨯=△△· ································································ 10分2120π24π.3603AOD S ⨯=扇形= ··························································································· 11分4π3AOD AOD S S S ∴-=-△阴影扇形= ·········································································· 12分7.解：（1）22223(23)(1)4y mx mx m m x x m x m =--=--=-- ，∴抛物线顶点M 的坐标为（1，4-m ） ······································································· 2分 抛物线223(0)y mx mx m m =-->与x 轴交于A B 、两点，∴当0y =时，2230mx mx m --=，20230.m x x >∴--= ，解得1213x x =-=，，A B ∴、两点的坐标为（10-，）、（30，）. ··································································· 4分（2）当0x =时，3y m =-，∴点C 的坐标为(03)m ，-.13(1)366.2ABC S m m m ∴=⨯--⨯-==△ ······························································· 5分过点M 作MD x ⊥轴于点D ，则12OD BD OB OD ==-=，，44.MD m m =-=BCM BDM OBC OCMD S S S S ∴=+-△△△梯形=111()222BD DM OC OM OD OB OC ++-··· =11124(34)133222m m m m ⨯⨯++⨯-⨯⨯=3m. ················································································································ 7分 :1:2.BCM ABC S S ∴=△△ ································································································· 8分 （3）存在使BCM △为直角三角形的抛物线.过点C 作CN DM ⊥于点N ，则C M N △为Rt △，13CN OD DN OC m ====，，.MN DM DN m ∴=-=22221.CM CN MN m ∴=+=+在Rt OBC △中，222299BC OB OC m =+=+， 在Rt BDM △中，2222416.BM BD DM m =+=+①如果BCM △是Rt △，且90BMC ∠=°，那么222CM BM BC +=，即222141699m m m +++=+，。