2006年广西省玉林市、防城港市初中毕业升学考试数学（非课改卷）亲爱的同学，展示才华的时候到了，相信自己，细心解答，遇到数字运算尽可能使用计算器，定会获得理想的成绩．祝你成功！一、填空题：本大题共10小题，每小题2分，共20分．请将答案直接写在题中的横线上． 1．计算：(2)(1)-⨯-=． 2．写出23a 的一个同类项：．3．已知数据：06625，，，，，那么这组数据的众数是．4．若1003x y +=，2x y -=，则代数式22x y -的值是 ．5．如图1，火焰的光线穿过小孔O ，在竖直的屏幕上形成倒立的实像，像的高度为 1.5cm ，48cm OA =，16cm OC =，则火焰的高度是cm ． 6．已知一元二次方程240x x a ++=两根的和等于这两根的积，则a =．7．如图2，有反比例函数1y x =，1y x=-的图象和一个圆，则S =阴影．8．商店里把塑料凳整齐地叠放在一起，据图3的信息，当有10张塑料凳整齐地叠放在一起时的高度是 cm ．9．某厂前年缴税30万元，今年缴税36.3万元，如果该厂缴税的年平均增长率为x ，那么可列方程为 ．10．如图4，AB 为O 的直径，AB 经过弦CD 的中点E ，150BOC ∠=，则ABD ∠=．图1图2图4二、选择题：本大题共8小题，每小题3分，共24分．在每小题给出的四个选项中，只有 一项是符合题意的，请将你认为正确答案的序号填在题后括号内．11．截至2006年4月15日3时44分，我国神舟六号飞船轨道舱已环绕地球2920圈，用科学记数法表示这个数是（ ） Ａ．42.9210⨯圈 Ｂ．32.9210⨯圈Ｃ．229.210⨯圈Ｄ．40.29210⨯圈12．计算：111x x x +--，正确的结果是（ ） Ａ．1- Ｂ．0 Ｃ．2Ｄ．113．不等式组230.52x x x >-+⎧⎨<⎩，的解集是（ ）Ａ．1x > Ｂ．4x <Ｃ．1x >或4x < Ｄ． 14x <<14．如图5，下列条件不能判定直线a b ∥的是（ ）Ａ．12∠=∠ Ｂ．13∠=∠Ｃ．14180∠+∠=Ｄ．24180∠+∠=15．丽丽买了一张30元的租碟卡，每租一张碟后剩下的余额如表6表示，若丽丽租碟25张，则卡中还剩下（ ） Ａ．5元 Ｂ．10元 Ｃ．20元 Ｄ．14元16．正比例函数(1)y a x =+的图象经过第二、四象限，若a 同时满足方程22(12)0x a x a +-+=，则此方程的根的情况是（ ）Ａ．有两个不相等的实数根Ｂ．有两个相等的实数根 Ｃ．没有实数根Ｄ．不能确定17．如图7，四边形PAOB 是扇形OMN 的内接矩形，顶点P 在MN上，且不与M N ，重合，当P 点在MN 上移动时，矩形PAOB 的形状、大小随之变化，则AB 的长度（ ） Ａ．变大 Ｂ．变小Ｃ．不变 Ｄ．不能确定表6 ＢＮＰＯ图7314ac图5b218．如图8，1O 与2O 相交于A B ，两点，直线PQ 与1O 相切于点P ，与2O 相切于点Q ，AB 的延长线交PQ 于C ，连结PA ，PB ．下列结论：①PC CQ =；②P B B Q>；③PBC APC ∠=∠．其中错误．．的结论有（ ） Ａ．3个 Ｂ．2个Ｃ．1个 Ｄ．0个三八为解答题，满分共76分．解答应写出文字说明，证明过程或演算步骤． 三、本大题共2小题，满分共16分． 19．（本小题满分8分）计算：08(1-． 20．（本小题满分8分） 解方程：651(1)x x x x +=++．四、本大题共2小题，满分共16分． 21．（本小题满分8分）某科技馆座落在山坡M 处，从山脚A 处到科技馆的路线如图9所示．已知A 处海拔高度 为103.4m ，斜坡AB 的坡角为30，40m AB =，斜坡BM 的坡角为18，60m BM =，那么科技馆M 处的海拔高度是多少？（精确到0.1m ）（参考数据：sin180.309= cos180.951= tan180.324= cot18 3.08=）图83018ＢＭ图922．（本小题满分8分）如图10，在ABC △和ABD △中，现给出如下三个论断：①AD BC =；②C D ∠=∠； ③12∠=∠．请选择其中两个论断为条件，另一个论断为结论，构造一个命题． （1）写出所有的真命题（写成“⎫⇒⎬⎭”形式，用序号表示）： ． （2）请选择一个真命题加以证明． 你选择的真命题是：⎫⇒⎬⎭．证明：五、本大题共1小题，满分10分． 23．（本小题满分10分）某制衣厂近四年来关于销售额与总成本的统计图，如图11所示． （1）请你在图12中画出四年利润（利润=销售额-总成本）的统计直方图（要求标出数字）； （2）根据图11，图12分别写出一条你发现的信息；（3）若从2004年到2006年这两年间的利润年平均增长率相同，请你预测2006年的利润是多少万元？图11 2002 2003 2004 2005 年份 利润（万元） 图1221ＡＣＤＢ图1024．（本小题满分10分）为鼓励居民节约用水和保护水资源，A 市城区从2006年3月1日起，对居民生活用水采取按月按户实行阶梯式计量水价收费，其收费标准是：第一阶梯水价为1.28元／3m ；第二阶梯水价为1.92元／3m ．（1）每户人口为4人（含4人）以内的，月用水量332m ≤执行第一阶梯水价，月用水量332m >的部分．．执行第二阶梯水价．如果某户人口4人，3月份用水量330m ，那么应交水费元；4月份用水量335m ，那么应交水费元．（2）每户核定人数超过4人的，月用水量≤（38m ⨯核定人数）执行第一阶梯水阶，月用水量>（38m ⨯核定人数）的部分．．执行第二阶梯水价，若小江家人口有5人，设月用水量3m x ，应交水费y 元． ①请你写出y 与x 的函数关系式；②若小江家某月交水费60.8元，则该月用水量是多少3m ？七、本大题共1小题，满分12分． 25．（本小题满分12分）如图13，已知AB 是O 的直径，弦CD AB ⊥于E ，F 是CE 上的一点，且FC FA =，延长AF 交O 于G ，连结CG ． （1）试判断ACG △的形状（按边分类），并证明你的结论； （2）若O 的半径为5，2OE =，求CF CD 之值．Ａ图1326．（本小题满分12分）在矩形ABCD 中，4AB =，2BC =，以A 为坐标原点，AB 所在的直线为x 轴，建立直角坐标系．然后将矩形ABCD 绕点A 逆时针旋转，使点B 落在y 轴的E 点上，则C 和D 点依次落在第二象限的F 点上和x 轴的G 点上（如图14）． （1）求经过BE G ，，三点的二次函数解析式；（2）设直线EF 与（1）的二次函数图象相交于另一点H ，试求四边形EGBH 的周长． （3）设P 为（1）的二次函数图象上的一点，BP EG ∥，求P 点的坐标．2006年广西省玉林市、防城港市初中毕业升学考试数学试题（非课改）参考答案及评分标准一、填空题：（每小题2分，共20分） 1．22．答案不唯一，如22a -， 3．6 4．20065．4.5 6．4-7．2π8．509．230(1)36.3x +=10．15二、选择题：（每小题3分，共24分） 11．Ｂ 12．Ｄ 13．Ｄ 14．Ｃ 15．Ｂ16．Ａ 17．Ｃ 18．Ｃ三、19．解：原式8=- ····················································································· 6分 8=． ········································································································ 8分 20．解：65x x =+． ···································································································· 2分 55x =． ·········································································································· 4分1x =． ············································································································· 6分检验：当1x =时，120x +=≠，(1)1(11)20x x +=⨯+=≠．∴原方程的解为1x =． ·················································································· 8分 四、21．解：过B 向水平线AC 作垂线BC ，垂足为C ，过M 向水平线BD 作垂线MD ，垂足为D （如右图），则 ··································· 2分 11402022BC AB ==⨯=． ···························4分 sin18MD BM =600.309=⨯18.54=． ········································································································ 6分 ∴科技馆M 处的海拔高度是：103.42018.54141.94141.9(m)++=≈． ······ 8分22．解：（1）真命题是：⎫⇒⎬⎭①②③，⎫⇒⎬⎭②①③ ······························································ 4分 （2）选择命题一：⎫⇒⎬⎭①②③证明：在ABC △和BAD △中，AD BC =∵，12∠=∠，AB BA =， ABC BAD ∴△≌△． ······································································· 7分C D ∠=∠∴． ···················································································· 8分选择命题二：⎫⇒⎬⎭②①③ 3018ＡＢＭ DC证明：在ABC △和BAD △中，C D ∠=∠∵，21∠=∠，AB BA =， ABC BAD ∴△≌△． ······································································· 7分 AD BC =∴． ····················································································· 8分 五、23解：（1）正确画出统计直方图，并标出数字． ························································ 4分（2）答案不唯一．每写出一条正确的信息给1分． ········································ 6分 （3）2004年到2005年的增长率120100100%20%100-=⨯=． ····················· 8分预测2006年的利润为：120(120%)144⨯+=（万元）． ·················· 10分 六、24．解：（1）38.4，46.72 ··························································································· 4分 （2）①当040x ≤≤时， 1.28y x =； ························································· 5分当40x >时，40 1.28(40) 1.92y x =⨯+-⨯1.9225.6x =-． ······················································ 7分② 1.2851.260.840⨯=<∵，可见用水量超过340m ．∴当60.8y =时，1.9225.660.8x -=． ········································· 8分 解得45x =． ··················································································· 9分∴小江家该月用水量为345m ． ······················································· 10分七、25．（1）解：ACG △是等腰三角形．证明如下：CD AB ∵⊥，AD AC =∴． ······················· ······································ 1分G ACD ∠=∠∴． ··········································· ······································ 2分 FC FA =∵，ACD CAG ∠=∠∴． ············· ······································ 3分 G CAG ∠=∠∴．ACG ∴△是等腰三角形． ······················· 4分（2）解：连结AD ，BC ． ······································ 5分由（1）知AC AD =，AC AD =∴． D ACD ∠=∠∴． ···································· 6分D G CAG ∠=∠=∠∴． 又ACF DCA ∠=∠，2002 2003 2004 2005 年份150 100 50ＡACF DCA ∴△∽△． ·········································································· 7分::AC CD CF AC =∴，即2AC CF CD =． ····································· 8分CD AB ∵⊥，2(52)(52)21CE AE EB ==-+=∴． ·············································· 9分 2222(52)2130AC AE CE =+=-+=∴． ······································ 11分 30CF CD =∴． ················································································· 12分 八、26．（1）解：由题意可知，4AE AB ==，2AG AD BC ===． ························· 1分(40)B ，∴，(04)E ，，(20)G -，． ·························································· 2分设经过B E G ，，三点的二次函数解析式是(2)(4)y a x x =+-．把(04)E ，代入之，求得12a =-．·························································· 3分 ∴所求的二次函数解析式是：211(2)(4)422y x x x x =-+-=-++．··············································· 4分（2）解：由题意可知，四边形AEFG 为矩形．F HG B ∴∥，且6GB =． ······································································· 5分∵直线4y =与二次函数图象的交点H 的坐标为(24)H ，， 2EH =∴． ································································································ 6分G ∵与B E ，与H 关于抛物线的对称轴对称，BH EG ===∴ ····························································· 7分 ∴四边形EGBH 的周长262=++⨯8=+ ····························································································· 8分 （3）解法1：设BP 交y 轴于M ．B P E G ∵∥，::AB AG AM AE =∴， 即4:2:4AM =．8AM =∴，于是(08)M -，． ···················· 9分设直线BM 的解析式为y kx b =+．把(40)B ，，(08)M -，代入之，得408.k b b +=⎧⎨=-⎩，解得28.k b =⎧⎨=-⎩，28y x =-∴． ·························································································· 10分联合一次，二次函数解析式组成方程组2281 4.2y x y x x =-⎧⎪⎨=-++⎪⎩，解得620x y =-⎧⎨=-⎩，或40.x y =⎧⎨=⎩，（此组数为B 点坐标）∴所求的P 点坐标为(620)P -，． ···························································· 12分解法2：过P 作PN x ⊥轴于N ．由BP EG ∥，得EGB PBN ∠=∠． 设所求P 点的横坐标为(0)a a <，则纵坐标为214(0)2a a a -++<． ····· 9分 tan PN PBN NB ∠=∵，4tan 22AE EGB AG ∠===， 2PN AENB AG==∴． ······················································································ 10分 4NB NA AB a =+=-∴，22114422PN a a a a ⎛⎫=--++=-- ⎪⎝⎭， 214224a a a--=-∴．解之，得6a =-或4a =． ·········································································· 11分 经检验可知，6a =-是原方程的根；4a =是原方程的增根，故应舍去．当6a =-时，22114(6)642022a a -++=-⨯--+=．∴所求的P 点坐标为(620)P -，．································································ 12分。