2012大连初中毕业升学考试数学试卷(word版)

专题02 中考折叠问题的归类解析【专题综述】折叠问题在近年来各地的中考试卷中频频出现，解决这一类问题主要抓住两点：折叠前后重合的角相等，重合的边也相等．【方法解读】一、折叠与平行例1：如图，在四边形ABCD中，∠A＝100°，∠C＝70°.将△BMN沿MN翻折，得△FMN，若MF∥AD，FN∥DC，则∠B＝___．【来源】2013-2014学年江苏省宜兴市和桥学区七年级下学期期中考试数学试卷（带解析）【答案】95°在△BMN中，∠B=180°-（∠BMN+∠BNM）=180°-（50°+35°）=180°-85°=95°．考点：1．平行线的性质；2．三角形内角和定理；3．翻折变换（折叠问题）．【解读】根据两直线平行，同位角相等求出∠BMF，∠BNF，再根据翻折的性质求出∠BMN和∠BNM，然后利用三角形的内角和定理列式计算即可得解．【举一反三】如图，将平行四边形ABCD沿对角线BD进行折叠，折叠后点C落在点F处，DF交AB于点E．（1）求证：EDB EBD∠=∠；（2）判断AF与BD是否平行，并说明理由．【来源】2015中考真题分项汇编第1期专题4 图形的变换【答案】【解析】试题解析：（1）由折叠可知：∠CDB =∠EDB∵四边形ABCD是平行四边形∴DC∥AB∴∠CDB =∠EBD∴∠EDB=∠EBD(2) ∵∠EDB=∠EBD∴DE=BE由折叠可知：DC=DF∵四边形ABCD是平行四边形∴DC=AB∴AE=EF∴∠EAF=∠EFA△BED中, ∠EDB+∠EBD+∠DEB=180°即2∠EDB+∠DEB=180°同理△AEF中，2∠EFA+∠AEF=180°∵∠DEB=∠AEF∴∠EDB= ∠EFA∴AF∥BD考点：折叠变换，平行四边形的性质，等腰三角形的性质与判定，三角形的内角和二、折叠与全等例2：如图，在□ABCD中，点E，F分别在边DC，AB上，DE=BF，把平行四边形沿直线EF折叠，使得点B，C分别落在点B′，C′处，线段EC′与线段AF交于点G，连接DG，B′G。

2012年高考新课标卷数学（文科数学、理科数学）试卷真题及参考答案(河南、河北、黑龙江、吉林、宁夏、山西、内蒙古、新疆、云南)绝密*启用前2012年普通高等学校招生全国统一考试（新课标卷）文科数学注息事项:1.本试卷分第Ⅰ卷（选择题）和第Ⅱ卷(非选择题)两部分。

答卷前，考生务必将自己的姓名、准考证号填写在本试卷和答题卡相应位置上。

2.问答第Ⅰ卷时。

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写在本试卷上无效.3.回答第Ⅱ卷时。

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第Ⅰ卷一、选择题：本大题共12小题，每小题5分，在每小题给同的四个选项中，只有一项是符合题目要求的。

1、已知集合A={x |x 2－x －2<0}，B={x |－1<x <1}，则（A ）A ⊂≠B （B ）B ⊂≠A （C ）A=B （D ）A ∩B=∅（2）复数z ＝－3+i 2+i 的共轭复数是（A ）2+i （B ）2－i （C ）－1+i （D ）－1－i3、在一组样本数据（x 1，y 1），（x 2，y 2），…，（x n ，y n ）（n ≥2，x 1,x 2,…,x n 不全相等）的散点图中，若所有样本点（x i ，y i ）(i =1,2,…,n )都在直线y =12x +1上，则这组样本数据的样本相关系数为（A ）－1 （B ）0 （C ）12 （D ）1（4）设F 1、F 2是椭圆E ：x 2a 2＋y 2b 2＝1(a >b >0)的左、右焦点，P 为直线x =3a 2上一点，△F 1PF 2是底角为30°的等腰三角形，则E 的离心率为（ ）（A ）12 （B ）23 （C ）34 （D ）455、已知正三角形ABC 的顶点A(1,1)，B(1,3)，顶点C 在第一象限，若点（x ，y ）在△ABC 内部，则z =－x+y 的取值范围是（A ）(1－3，2) （B ）(0，2) （C ）(3－1，2) （D ）(0，1+3)（6）如果执行右边的程序框图，输入正整数N(N ≥2)和实数a 1,a 2,…,a N ，输出A,B ，则（A ）A+B 为a 1,a 2,…,a N 的和（B ）A ＋B 2为a 1,a 2,…,a N 的算术平均数（C ）A 和B 分别是a 1,a 2,…,a N 中最大的数和最小的数（D ）A 和B 分别是a 1,a 2,…,a N 中最小的数和最大的数（7）如图，网格纸上小正方形的边长为1，粗线画出的是某几何体的三视图，则此几何体的体积为（A）6（B）9（C）12（D）18(8)平面α截球O 的球面所得圆的半径为1，球心O 到平面α的距离为2，则此球的体积为（A ）6π （B ）43π （C ）46π （D ）63π（9）已知ω>0，0<φ<π，直线x =π4和x =5π4是函数f (x )=sin(ωx +φ)图像的两条相邻的对称轴，则φ=（A ）π4 （B ）π3 （C ）π2 （D ）3π4（10）等轴双曲线C 的中心在原点，焦点在x 轴上，C 与抛物线y 2=16x 的准线交于A ，B 两点，|AB|=43，则C 的实轴长为（A ） 2 （B ）2 2 （C ）4 （D ）8(11)当0<x ≤12时，4x <log a x ，则a 的取值范围是（A ）(0，22) （B ）(22，1) （C ）(1，2) （D ）(2，2)（12）数列{a n }满足a n +1＋(－1)n a n ＝2n －1，则{a n }的前60项和为（A ）3690 （B ）3660 （C ）1845 （D ）1830第Ⅱ卷本卷包括必考题和选考题两部分。

2012年福建省南平市初中毕业、升学考试数 学 试 题（满分：150分；考试时间：120分钟）★友情提示：① 所有答案都必须填在答题卡相应的位置上，答在本试卷上一律无效；② 可以携带使用科学计算器，并注意运用计算器进行估算和探究； ③ 未注明精确度、保留有效数字等的计算问题不得采取近似计算．★参考公式：抛物线c bx ax y ++=2（a ≠0）的对称轴是，a bx 2-=顶点坐标是⎪⎪⎭⎫ ⎝⎛--a b ac ab 4422， 一、选择题（本大题共10小题，每小题4分，共40分．每小题只有一个正确的选项，请在答题卡．．．的相应位置填涂） 1．－3的相反数是 A ．31B ．31-C ．3D ．－32．计算：210÷＝A ．5B ．5C ．25 D ．210 3．若要对一射击运动员最近5次训练成绩进行统计分析，判断他的训练成绩是否稳定，则需要知道他这5次训练成绩的 A ．中位数 B ．平均数 C ．众数 D ．方差 4．正多边形的一个外角等于30°，则这个多边形的边数为 A ．6 B ．9 C ．12 D ．15 5．下列计算正确．．的是A . 523a a a =+B . a a a =÷45 C . 44a a a =⋅D ．632)(ab ab =6．为验证“掷一个质地均匀的骰子，向上的一面点数为偶数的概率是0.5”，下列模拟实验中，不．科学．．的是 A ．袋中装有1个红球1个绿球，它们除颜色外都相同，计算随机摸出红球的频率 B ．用计算器随机地取不大于10的正整数，计算取得奇数的频率 C ．随机掷一枚质地均匀的硬币，计算正面朝上的频率D ．如图，将一个可以自由转动的转盘分成甲、乙、丙3个相同的扇形，转动转盘任其自由停止，计算指针指向甲的频率7．一个三角形的周长是36，则以这个三角形各边中点为顶点的三角形的周长．．是A ． 6B ． 12C ． 18D ． 368．已知反比例函数xy 1=的图象上有两点A (1，m )、B (2，n )， 则m 与n 的大小关系为 A ．m ＞n B ．m ＜n C ．m ＝n D ．不能确定9．如图所示，水平放置的长方体的底面是长为4和宽为2的矩形， 它的主视图的面积为12，则长方体的体积．．等于 A ．16 B ．24C ．32D ．4810．如图，正方形纸片ABCD 的边长为3，点E 、F 分别在边BC 、CD 上，将AB 、AD 分别沿AE 、AF 折叠，点B 、D 恰好都落在点G 处，已知BE ＝1，则EF 的长为A ．23B ．25C ．49D ．3二、填空题（本大题共8小题，每小题3分，共24分．请将答案填入答题卡．．．的相应位置）11．计算：38＝ ．12．样本数据2，8，3，5，6的极差是 ．13．分解因式：2422+-x x = ．14．如图，△ABC 为⊙O 的内接三角形，AB 为⊙O 的直径，点D •在⊙O 上，∠ADC ＝68°，则∠BAC ＝ °． 15．将直线x y 2=向上平移1个单位长度后得到的直线 是 ．16．如图，在山坡AB 上种树，已知∠C ＝90°，∠A ＝28°，AC ＝6米，则相邻两树的坡面距离AB ≈ 米．（精确到0.1米） 17．某校举行A 、B 两项趣味比赛，甲、乙两名学生各自随机选择参加其中的一项，则他们恰好参加同一项比赛的概率 是 ． 18．设)x 表示大于．．x 的最小整数，如[)3＝4，[)2.1-＝－1，则下列结论中正确．．的是 ．（填写所有正确结论的序号） ① [)00=； ② )[)f x x x =-的最小值是0；③ )[)f x x x =-的最大值是1； ④ 存在实数x ，使)[)f x x x =-＝0.5成立.（第16题图）BCAB（第14题图）（第9题图）EGF B C AD （第10题图）（第21题图）ABCD（第22题图）90童装童车儿童玩具类 别儿童玩具%25%童车 %童装抽查件数三、解答题（本大题共8小题，共86分．请在答题卡．．．的相应位置作答） 19．（1）（7分）计算：01220124π313--+⎪⎭⎫ ⎝⎛⨯--）（．（2）（7分）解不等式组：⎩⎨⎧+-.823712x x x ，20．（8分）解分式方程：03632=+-+-x xx x ．21．（8分）如图，已知四边形ABCD 是平行四边形，若点E 、F 分别在边BC 、AD 上，连接AE 、CF ．请再从下列三．个．备选条件．．．．中，选择添加．．一个恰．．．当．的条件，使四边形AECF 是平行四边形，并予以证明． 备选条件：AE ＝CF ，BE ＝DF ，∠AEB ＝∠CFD ． 我选择添加的条件是： ． （注意：请根据所选择的条件在答题卡相应试题的图中，画出符合要求的示意图，并加以证明）22．（10分）“六·一”前夕，质检部门从某超市经销的儿童玩具、童车和童装中共抽查了300件儿童用品．以下是根据抽查结果绘制出的不完整的统计表和扇形图：请根据上述统计表和扇形图提供的信息，完成下列问题： （1）分别补全上述统计表和扇形图；（2）已知所抽查的儿童玩具、童车、童装的合格率分别为90%、88%、80%，若从该超市的这三类儿童用品中随机购买一件，请估计能买到合格品的概率是多少？23．（10分）如图，直线l 与⊙O 交于C 、D 两点，且与半径OA 垂直，垂足为H ，已知OD＝2，∠O ＝60°． （1）求CD 的长；（2）在OD 的延长线上取一点B ，连接AB 、AD ，若AD ＝BD ，求证：AB 是⊙O 的切线．①② ＜ ＜ （第23题图）A24．（10分）某乡镇决定对小学和初中学生按照每生每天3元的标准进行营养补助，其中家庭困难寄宿生的补助标准为：小学生每生每天4元，初中生每生每天5元．已知该乡镇现有小学和初中学生共1 000人，且小学、初中均有2%的学生为家庭困难寄宿生． 设该乡镇现有小学生x 人． （1）用含x 的代数式表示：该乡镇小学生每天共需营养补助费是 元； 该乡镇初中生每天共需营养补助费是 元； （2）设该乡镇小学和初中学生每天共需营养补助费为y 元，求y 与x 之间的函数关系式； （3）若该乡镇小学和初中学生每天共需营养补助费为3029元，问小学生、初中生分别有多少人？25．（12分）在平面直角坐标系中，矩形OABC 如图所示放置，点A 在x 轴上，点B 的坐标为（m ，1）（m ＞0）．将此矩形绕点O 逆时针旋转90°，得到矩形C B A O '''． （1）写出点A 、A '、C '的坐标；（2）设过点A 、A '、C '的抛物线解析式为c bx ax y ++=2，求此抛物线的解析式；（c b a 、、可用含m 的式子表示）（3）试探究：当m 的值改变时，点B 关于点O 的对称点D是否可能落在（2）中的抛物线上？若能，请求出此时m 的值．26．（14分）如图，在△ABC 中，点D 、E 分别在边BC 、AC 上，连接AD 、DE ，且∠1＝∠B ＝∠C ． （1）由题设条件，请写出三个正确．．．．结论；（要求：不再添加其它字母和辅助线，找结论过程中添加的字母或辅助线不能出现在结论中，不必证明） 答：结论一： ；结论二： ； 结论三： ． （2）若∠B ＝45°，BC ＝2，当点D 在BC 上运动时（点D 不与点B 、C 重合）， ① 求CE 的最大值；② 若△ADE 是等腰三角形，求此时BD 的长．BCE A（第26题图）B C（备用图）A1 （第25题图）（注意：在第（2）小题求解过程中，若有运用（1）中得出的结论，须加以证明）2012年福建省南平市初中毕业、升学考试数学试题参考答案及评分说明说明：（1） 解答右端所注分数，表示考生正确作完该步应得的累计分数，全卷满分150分． （2） 对于解答题，评卷时要坚持每题评阅到底，勿因考生解答中出现错误而中断本题的评阅．当考生的解答在某一步出现错误时，如果后续部分的解答未改变该题的考试要求，可酌情给分，但原则上不超过后面应得分数的一半，如果有较严重的错误，就不给分．（3） 如果考生的解法与本参考答案不同，可参照本参考答案的评分标准相应评分． （4） 评分只给整数分．一、选择题（本大题共10小题，每小题4分，共40分）1．C ； 2．A ； 3．D ； 4．C ； 5．B ； 6．D ； 7．C ； 8．A ； 9．B ； 10．B ． 二、填空题（本大题共8小题，每小题3分，共24分） 11．2； 12．6； 13．2)1(2-x ； 14．22；15．x y 2=＋1；16．6.8；17．21（或0.5）； 18．③ ④． 三、解答题（本大题共8小题，共86分） 19．（1）解：原式＝1)4(39---⨯π ············································································· 4分 ＝1427--+π ···················································································· 6分 ＝π-30 ······························································································· 7分 （2）由 ① 得 2x ＜7＋1 ························································································· 2分 x ＜4 ································································································ 3分 由 ② 得 823<-x x ······················································································ 5分 x ＜8 ································································································ 6分 ∴不等式组的解集为 x ＜4 ··············································································· 7分20．解法一：原方程化为0)6()3)(3(2=-++-x x x x ··················································· 4分∴06922=-+-x x x ························································································ 6分解得 x ＝23··········································································································· 7分 经检验，x ＝23是原分式方程的解．∴原方程的解是x ＝23························································································· 8分解法二：原方程化为0)6()3(3)3(2=-++-+x x x x x ·········································· 4分（以下与解法一相同）21．情形一：选择添加的条件是BE =DF ······················· 2分证法一：∵四边形ABCD 是平行四边形∴AD =BC ，AD ∥BC ············································ 4分 ∵BE =DF ，∴AD －DF =BC －BE 即 AF =CE ·························· 6分 ∴四边形AECF 是平行四边形 ································· 8分 证法二：∵四边形ABCD 是平行四边形， ∴AB =CD ，∠B =∠D 又∵BE =DF ， ∴△ABE ≌△CDF∴AE =CF ······················································································································· 4分 又∵AD =BC ，∴AD －DF =BC －BE 即AF =CE ··············································································· 6分 ∴四边形AECF 是平行四边形 ····················································································· 8分情形二：选择添加的条件是∠AEB =∠CFD ································································· 2分 证法一：∵四边形ABCD 是平行四边形，∴AD ∥BC ···················································································································· 4分 ∴∠AEB =∠EAF ············································································································ 5分 又∵∠AEB =∠CFD ，∴∠EAF =∠CFD ············································································································ 6分 ∴AE ∥CF ······················································································································ 7分 又∵AF ∥EC ，∴四边形AECF 是平行四边形 ····················································································· 8分 证法二：∵∠AEB =∠CFD ，∴180°－∠AEB =180°－∠CFD ，即∠AEC =∠CF A ······················································ 4分 又∵四边形ABCD 是平行四边形， ∴AB =CD ，∠B =∠D ∴△ABE ≌△CDF∴∠BAE =∠DCF ········································································································· 6分 又∵∠BAD =∠DCB ，∴∠BAD －∠BAE =∠DCB －∠DCF ，即∠EAF =∠FCE ·········································· 7分 ∴四边形AECF 是平行四边形 ················································································ 8分AB CDEF22．解：（1）（补全统计图表每空2分，共8分） ································································· 8分 （2）85.0300%80135%8875%9090=⨯+⨯+⨯答：从该超市这三大类儿童用品中随机购买一件能买到合格品的概率是0.85 ·························································································································· 10分90抽查件数童装童车儿童玩具类 别儿童玩具% 25%童车%童装 75 13545 3023．（1）解法一：∵O A ⊥CD ，∴∠OHD =90° ································································ 1分在Rt △OHD 中，∠O =60°，OD =2，∴O ∠sin sin HDDOHOD∠= ····································· 2分 ∴HD OD DOH =⋅∠O ∠sin ····································3分=360sin 2=︒ ··········································4分 ∴2CD HD == ····································5分 解法二：∵OA ⊥CD ，∴∠OHD =90° ··········1分 在Rt △OHD 中，∵∠O =60°，∴∠ODH =30° ···························2分 ∴OH =12OD ＝1 ···········································3分 ∴3122222=-=-=OH OD HD ························································ 4分 ∴2CD HD == ···························································································· 5分 （2）证法一：∵OA =OD ，∠O =60°，∴△AOD 为等边三角形∴∠OAD =∠ODA =60° ·························································································· 6分 ∵AD =BD ，∴∠DAB =∠B ·················································································· 7分 ∵∠ODA =∠DAB +∠B ，∴∠DAB =12∠ODA =12×60°=30° ········································································ 8分 ∴∠OAB =∠OAD +∠DAB =60°+30°=90° ······························································ 9分 即OA ⊥AB ，∴AB 是⊙O 的切线 ······································································ 10分 证法二：∵OA =OD ，∠O =60°，∴△AOD 为等边三角形∴OD = AD ············································································································ 6分 又∵AD =BD ，∴OD =BD ····················································································· 7分 ∵DH ⊥OA ，∴OH =AH ，∠OHD =90° ································································ 8分 ∴HD ∥AB ，∴∠OAB =∠OHD =90° ··································································· 9分 即OA ⊥AB ，∴AB 是⊙O 的切线 ······································································ 10分 证法三：∵OA =OD ，∠O =60°，∴△AOD 为等边三角形∴∠O =∠DAO ······································································································· 6分 ∵AD =BD ，∴∠B =∠DAB ·················································································· 7分 ∵∠OAB +∠O +∠B =180°，∴∠OAB +∠DAO +∠DAB =180° ········································································ 8分 ∴2∠OAB =180°，∴∠OAB =90° ······································································· 9分 即OA ⊥AB ，∴AB 是⊙O 的切线 ··································································· 10分。

2012年玉林市防城港市初中毕业暨升学考试数 学（全卷共三大题，共4页，满分120分，考试时间120分钟）注意事项：1．请将答案填写在答题卡上，在试卷上作答无效．．．．．．．．．考试结束，将本试卷和答题卡一并交回．2．选择题每小题选出答案后，用2B 铅笔把答题卡上对应题目的选项标号涂黑．3．非选择题，用直径0.5毫米黑色墨水签字笔在答题卡上各题的答题区域内作答．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，把正确答案的标号填在答题卡内相应的位置上．1．计算：22=（ ）（A ）1 （B ）2 （C ）4 （D ）82．如图，a b ∥，c 与a ，b 都相交，150∠=°，则2∠=（ ）（A ）40° （B ）50° （C ）100° （D ）130°3．计算：=（ ）（A ）3 （B （C ） （D ）4．下列基本几何体中，三视图都是相同图形的是（ ）圆柱 三棱柱 球 长方体（A ） （B ） （C ） （D ）5．正六边形的每个内角都是（ ）（A ）60° （B ）80° （C ）100° （D ）120°6．市农科所收集统计了甲、乙两种甜玉米各10块试验田的亩产量后，得到其方差分别是20002S =甲．、2001S =乙．，则（ ） （A ）甲比乙的亩产量稳定（B ）乙比甲的亩产量稳定（C ）甲、乙的亩产量的稳定性相同（D ）无法确定哪一品种的亩产量更稳定7．一次函数1y mx m =+-的图象过点(02)，，且y 随x 的增大而增大，则m =（ ） （A ）1- （B ）3 （C ）1 （D ）1-或38．如图，在菱形ABCD 中，对角线AC ，BD 相交于点O ，且AC BD ≠，则图中全等三角形有（ ）（A ）4对 （B ）6对 （C ）8对 （D ）10对9．如图，Rt ABC △的内切圆O ⊙与两直角边AB ，BC 分别相切于点D ，E ，过劣弧DE（不包括端点D ，E ）上任一点P 作O ⊙的切线MN 与AB ，BC 分别交于点M ，N ，若O ⊙的半径为r ，则Rt MBN △的周长为（ ）（A ）r （B ）32r （C ）2r （D ）52r 10．如图，正方形ABCD 的两边BC ，AB 分别在平面直角坐标系的x 、y 轴的正半轴上，正方形A B C D ''''与正方形ABCD 是以AC 的中点O '为中心的位似图形，已知AC =若点A '的坐标为(12)，，则正方形A B C D ''''与正方形ABCD 的相似比是（ ）（A ）16 （B ）13 （C ）12 （D ）2311．二次函数2(0)y ax bx c a =++≠的图象如图所示，其对称轴为1x =，有如下结论：①1c > ②20a b += ③24b ac <④若方程20ax bx c ++=的两根为1x ，2x ，则122x x +=．则正确的结论是（ ）（A ）①② （B ）①③ （C ）②④ （D ）③④12．一个盒子里有完全相同的三个小球，球上分别标上数字1-、1、2．随机摸出一个小球（不放回）其数字记为p ，再随机摸出另一个小球其数字记为q ，则满足关于x 的方程20x px q ++=有实数根的概率是（ ）（A ）12 （B ）13 （C ）23 （D ）56第Ⅱ卷（非选择题共84分）二、填空题：本大题共6小题，每小题3分，共18分，把答案填在答题卡的横线上．13．既不是正数也不是负数的数是___________．14．某种原子直径21210-⨯．纳米，把这个数化为小数是___________纳米．15．在平面直角坐标系中，一青蛙从点(10)A-，处向右跳2个单位长度，再向上跳2个单位长度到点A'处，则点A'的坐标为___________．16．如图，矩形OABC内接于扇形MON，当C N C O=时，NMB∠的度数是___________．17．如图，两块相同的三角板完全重合在一起，30A =∠，10AC=，把上面一块绕直角顶点B逆时针旋转到A BC''△的位置，点C'在AC上，A C''与AB相交于点D，则C D'=___________．18．二次函数29(2)4y x=--+的图象与x轴围成的封闭区域内（包括边界），横、纵坐标都是整数的点有___________个（提示：必要时可利用下面的备用图画出图象来分析）．三、解答题：本大题共8小题，满分共66分．解答过程写在答题卡上，解答应写出文字说明、证明过程或演算步骤．19．（6分）计算：2(2)4(1)a a-+-．20．（6分）求不等式组11121122xx⎧-⎪⎪⎨⎪-⎪⎩≥≤的整数解．21．（6分）已知等腰ABC △的顶角36A =∠（如图）．（1）作底角ABC ∠的平分线BD ，交AC 于点D （用尺规作图，不写作法，但保留作图痕迹，然后用墨水笔加黑）；（2）通过计算说明ABD △和BDC △都是等腰三角形．22．（8分）某奶品生产企业，2010年对铁锌牛奶、酸牛奶、纯牛奶三个品种的生产情况进行了统计，绘制了图1、2的统计图，请根据图中信息解答下列问题：（1）酸牛奶生产了多少万吨？把图1补充完整；酸牛奶在图2中所对应的圆心角是多少度？（2）由于市场不断需求，据统计，2011年酸牛奶的生产量比2010年增长20%，按照这样的增长速度，请你估算2012年酸牛奶的生产量是多少万吨？23．（8分）如图，已知点O 为Rt ABC △斜边AC 上一点，以点O 为圆心，OA 长为半径的O ⊙与BC 相切于点E ，与AC 相交于点D ，连接AE ．（1）求证：AE 平分CAB ∠；（2）探求图中∠1与C ∠的数量关系，并求当AE EC =时tan C 的值．24．（10分）一工地计划租用甲、乙两辆车清理淤泥，从运输量来估算；若租两车合运，10天可以完成任务；若单独租用乙车完成任务则比单独租用甲车完成任务多用15天．（1）甲、乙两车单独完成任务分别需要多少天？（2）已知两车合运共需租金65 000元，甲车每天的租金比乙车每天的租金多1 500元．试问：租甲乙两车、单独租甲车、单独租乙车这三种方案中，哪一种租金最少？请说明理由．25．（10分）如图，在平面直角坐标系xOy 中，梯形AOBC 的边OB 在x 轴的正半轴上，AC OB ∥，BC OB ⊥，过点的双曲线k y x=的一支在第一象限交梯形对角线OC 于点D ，交边BC 于点E ．（1）填空：双曲线的另一支在第___________象限，k 的取值范围是___________；（2）若点C 的坐标为(22)，，当点E 在什么位置时，阴影部分面积S 最小？（3）若12OD OC =，2OAC S =△，求双曲线的解析式．26．（12分）如图，在平面直角坐标系xOy 中，矩形AOCD 的顶点A 的坐标是(04)，，现有两动点P ， Q ，点P 从点O 出发沿线段OC （不包括端点O ，C ）以每秒2个单位长度的速度匀速向点C 运动，点Q 从点C 出发沿线段CD （不包括端点C ，D ）以每秒1个单位长度的速度匀速向点D 运动．点P ，Q 同时出发，同时停止．设运动时间为t （秒），当2t =（秒）时，PQ =（1）求点D 的坐标，并直接写出t 的取值范围；（2）连接AQ 并延长交x 轴于点E ，把AE 沿AD 翻折交CD 延长线于点F ，连接EF ，则AEF △的面积S 是否随t 的变化而变化？若变化，求出S 与t 的函数关系式；若不变，求出S 的值．（3）在（2）的条件下，t 为何值时，四边形APQF 是梯形？2012年玉林市防城港市初中毕业暨升学考试数学参考答案及评分标准一、选择题：本大题共12小题，每小题3分，共36分．1. C2. B3. C4. C5. D6. A7. B 8. C 9. C 10. B 11. C 12. A二、填空题：本大题共6小题，每小题3分，共18分．13．0 14．0.012 15．(12)， 16．30° 17．52 18．7 三、解答题：本大题共8小题，满分共66分．19．解：原式24444a a a =-++- ····················································································4分 2a = ··················································································································6分 或原式[]2(1)14(1)a a =--+- ······················································································1分2(1)2(1)1a a =-+-+ ·························································································3分 2(11)a =-+ ··········································································································5分 2a = ·······················································································································6分 20．解：解①122x ≥， 4x ∴≥······················································································2分 解②132x ≤， 6x ∴≤ ·····················································································4分 46x ∴≤≤ ···········································································································5分 ∴不等式组的整数解为4，5， 6 ········································································· 6分 21．解：（1）作图如右． ·······································································································2分（2）1803636722A C ABC -∠=∠=∠==°°°，° 72362ABD DBC ∴∠=∠==°° ··································· 3分 72BDC A ABD ∴∠=∠+∠=°····································· 4分A ABD ABD ∴∠=∠，△是等腰三角形； ·····························································5分C BDC BCD ∠=∠，△是等腰三角形． ···························································6分22．解：（1）设总量为x 万吨，则12050%x= 240x ∴= ··············································1分 ∴酸牛奶为：2404012080--=（万吨） ·················································2分 补全直方图 ·······································································································3分 它在图2所对应的圆心角是：80360120240⨯=°°．·····································5分 （2）2012年酸牛奶的生产量为：280(120%)+ ····················································7分=115.2（万吨） ···················································8分23．证明：（1）连接OE ，BC 是O ⊙切线，则OE BC ⊥，又AB BC ⊥，AB OE ∴∥，32∴∠=∠ ········· 1分又OA OE =，12∴∠=∠ ···································· 2分13∴∠=∠ ····························································· 3分AE ∴平分CAB ∠ ················································· 4分（2）1221EOC ∠=∠+∠=∠········································································5分∴在Rt OEC △中，2190C ∠+∠=° ·······················································6分若AE EC =，则1C ∠=∠，30C ∴∠=° ···············································7分tan tan 30C ∴=°= ··········································································8分 24．解：（1）设甲车单独完成任务需要x 天，则乙车单位完成任务需要（x +15）天 ········1分则：1010115x x +=+ ························································································2分 得10(15)10(15)x x x x ++=+即：251500x x --=(15)(10)0x x ∴-+= 得115x =，210x =-（不合题意，舍去） ········4分经检验知：15x =是原方程的解，1530x ∴+=答：甲、乙单独完成任务分别需要15天、30天． ·······································5分（2）设乙车每天租金a 元，则甲车每天租金(1500)a +元． ·······························6分10(1500)65000a a ∴++=，得2500a =，15004000a += ··················7分方案一：甲、乙两车合作完成任务需租金：65000元方案二：甲车单独完成任务需租金：40001560000⨯=元 ························ 8分 方案三：乙车单独完成任务需租金：25003075000⨯=元 ························9分∴甲车单独完成任务租金最少． ··································································· 10分25．解：（1）三，0k >（每个空1分） ··············································································2分 （2）由(22)C ，，则(2)2k A ，，(2)2k E ， ··································································3分AEC OBE S S S ∴=+△△ 11(2)(2)222222k k k =--+⨯⨯ ····························· 4分 211(42)242k k k =-++ 21(4)28k k =-+ 213(2)82k =-+ ······························································································5分当2k =时，即(21)E ，为BC 中点时，S 最小． ···············································6分（3）方法一：令()C a b ，，由12OD OC =，则11()22D a b ，，而()k A b b ， ·············8分 又11()()222OAC k S a b ab k b =-∙=-=△，4ab k ∴=+ ···························································· 9分114(4)443k ab k k ∴==+∴=，， ∴双曲线解析式为43y x=．································· 10分 方法二：令()D a b ，，由12OD OC =，则(22)C a b ，，而(2)2k A b b， ·············· 8分 又11(2)2(4)2222OAC k S a b ab k b =-∙=-=△，44k ab +∴= ····················9分 44k k ab +∴==，43k =，∴双曲线解析式为43y x =． ··························· 10分26．解：（1）由4OP =，2CQ =，则在Rt PCQ △中，22222216PC PQ CQ =-=-=，4PC ∴=， ·············································· 1分84OC OP PC CD OA ∴=+===，，(84)D ∴， ······································2分04t << ·······································································································3分（2）S 值没变化，理由是：Rt Rt QCE QDA △∽△ ·······································4分 则CE AD CQ DQ=即84CE t t =-，84t CE t ∴=-． ············································ 5分 22(4)QF QD t ==-18()(4)(8)3224t S QF AD CE t t∴=+=-+=- ··········································6分 S ∴值不发生变化，32S =． ······································································ 7分 （3）方法一：若四边形APQF 是梯形，则PQ AF ∥． ···································8分PCQ ADF ∴△∽△，PC AD CQ DF ∴=，又82PC t =-， 4DF QD t ==-，8284t t t -∴=- ·····························································9分 则2224320t t -+=即212160t t -+= ······················································ 10分1212622t ±±∴===±··································· 11分 04t <<∴当6t =-APQF 是梯形． ································· 12分 方法二：若四边形APQF 是梯形，则PQ AF ∥． ·····································8分 90CQP AFD AQD CQE QCP QCE ∠=∠=∠=∠∠=∠=，°CQ CQ =，QCP QCE ∴△≌△ PC CE ∴= 即8824t t t -=- ······································································9分 则2224320t t -+=即212160t t -+= ······················································ 10分6t ∴===±··································· 11分 04t <<∴当6t =-APQF 是梯形． ································· 12分。