【南平中考数学试题及答案】2008

二00八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．一、选择题（本大题共9小题，每小题4分，共36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．计算：3-=（ ） A ．3B ．3-C ．13D ．13-2．计算：53x x ÷=（ ） A ．2xB ．53xC ．8xD ．13．有一个质地均匀且六个面上分别刻有1到6的点数的正方体骰子，掷一次骰子，向上的一面的点数为2的概率是（ ） A ．0B ．12C ．16D ．14．下列图形中，既是轴对称图形，又是中心对称图形的是（ ）A ．等腰梯形B ．平行四边形C ．正三角形D ．矩形 5．小丽家下个月的开支预算如图所示．如果用于教育的支出是150元， 则她家下个月的总支出为（ ） A ．625元 B ．652元 C ．750元 D ．800元 6．如图1所示的几何体的主视图．．．是（ ）7．已知ABC DEF △∽△，相似比为3，且ABC △的周长为18，则D E F △的周长为（ ） A ．2 B ．3 C ．6 D ．548．有一人患了流感，经过两轮传染后共有100人患了流感，那么每轮传染中平均一个人传染的人数为（ ） A ．8人 B ．9人 C ．10人 D ．11人9．如图，正比例函数(0)y kx k =>与反比例函数4y x=的图 象相交于A C ，两点，过点A 作x 轴的垂线交x 轴于点B ， 连接BC ，则ABC △的面积等于（ ）A ．2B ．4C ．6D ．8二、填空题（本大题共9小题，每小题3分，共27分．请将答案填入答题卡．．．的相应位置） 10= ． 11．当x = 时，分式12x -没有意义． 12．如图，奥运五环标志里，包含了圆与圆的位置关系中的 外离．．和 ． 13．为了解一批节能灯的使用寿命，宜采用 的方式进行调查．（填：“全面调查”或“抽样调查”） 14．“明天会下雨”是 事件．（填“必然”或“不可能” 或“可能”）15．如图，菱形ABCD 中，O 是对角线AC BD ，的交点， 5cm AB =，4cm AO =，则BD = cm ． 16．因式分解：322a a a ++= ．17．如图，ABC △中，AB AC >，D E ，两点分别在边AC AB ，上， 且DE 与BC 不平行．请填上一个．．你认为合适的条件： ， 使ADE ABC △∽△．（不再添加其他的字母和线段；只填一个条件，多填不给分！） 18．定义：a 是不为1的有理数，我们把11a-称为a 的差倒数．．．． 如：2的差倒数是1112=--，1-的差倒数是111(1)2=--．已知113a =-，2a 是1a 的差倒数，3a 是2a 的差倒数，4a 是3a 的差倒数，……，依此类推，则2009a = ． 三、解答题（本大题共8小题，共87分．请在答题卡．．．的相应位置作答） 19．（8分）先化简，再求值：()()(2)a b a b b b +-+-，其中1a =-，1b =．20．（8分）解不等式组：23432x x x x +<+⎧⎪⎨->⎪⎩①②21．（9分）如图，线段AB 经过圆心O ，交O 于点A C ，，点D 在O 上，连接AD BD ，，30A B ∠=∠= ．BD 是O 的切线吗？请说明理由．22．（10分）某商场家电销售部有营业员20名，为了调动营业员的积极性，决定实行目标管理，即确定一个月的销售额目标，根据目标完成情况对营业员进行适当的奖惩．为此，商场统计了这20名营业员在某月的销售额，数据如下：（单位：万元） 25 26 21 17 28 26 20 25 26 30 20 21 20 26 30 25 21 19 28 26 （1）请根据以上信息完成下表：（2上述数据中，众数是 万元，中位数是 万元，平均数是 万元； （3）如果将众数作为月销售额目标，能否让至少一半的营业员都能达到目标？请说明理由． 23．（12分）“母亲节”到了，九年级（1）班班委发起慰问烈属王大妈的活动，决定在“母亲节”期间全班同学利用课余时间去卖鲜花筹集慰问金．已知同学们从花店按每支1.2元买进鲜花，并按每支3元卖出．（1）求同学们卖出鲜花的销售额y （元）与销售量x （支）之间的函数关系式；（2）若从花店购买鲜花的同时，还总共用去40元购买包装材料，求所筹集的慰问金w （元）与销售量x （支）之间的函数关系式；若要筹集不少于500元的慰问金，则至少要卖出鲜花多少支？（慰问金=销售额－成本）24．（12分）2008年初，我国南方部分省区发生了雪灾，造成通讯受阴．如图，现有某处山坡上一座发射塔被冰雪从C 处压折，塔尖恰好落在坡面上的点B 处，在B 处测得点C 的仰角为38，塔基A 的俯角为21，又测得斜坡上点A 到点B 的坡面距离AB 为15米，求折．断前．．发射塔的高．（精确到0.1米）25．（14分）如图，平面直角坐标系中有一矩形纸片OABC ，O 为原点，点A C ，分别在x轴，y 轴上，点B 坐标为(（其中0m >），在BC 边上选取适当的点E 和点F ，将OCE △沿OE 翻折，得到OGE △；再将ABF △沿AF 翻折，恰好使点B 与点G 重合，得到AGF △，且90OGA ∠=．（1）求m 的值；（2）求过点O G A ，，的抛物线的解析式和对称轴； （3）在抛物线的对称轴．．．上是否存在点P ，使得OPG △是等腰三角形？若不存在，请说明理由；若存在，直接答出．．．．所有满足条件的点P 的坐标（不要求写出求解过程）． 【提示：抛物线2(0)y ax bx c a =++≠的对称轴是2b x a =-，顶点坐标是2424b ac b aa ⎛⎫-- ⎪⎝⎭，】26．（14分）（1）如图1，图2，图3，在ABC △中，分别以AB AC ，为边，向ABC △外作正三角形，正四边形，正五边形，BE CD ，相交于点O ．①如图1，求证：ABE ADC △≌△；②探究：如图1，BOC ∠=；如图2，BOC ∠=；如图3，BOC ∠=．（2）如图4，已知：AB AD ，是以AB 为边向ABC △外所作正n 边形的一组邻边；AC AE ，是以AC 为边向ABC △外所作正n 边形的一组邻边．BE CD ，的延长相交于点O ．①猜想：如图4，BOC ∠=（用含n 的式子表示）；②根据图4证明你的猜想．2008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9小题，每小题4分，共36分）1．A ； 2．A ； 3．C ； 4．D ； 5．C ； 6．B ； 7．C ； 8．B ； 9．B ． 二、填空题（本大题共9小题，每小题3分，共27分） 10．4； 11．2； 12．相交； 13．抽样调查； 14．可能； 15．6； 16．2(1)a a +；17．1B ∠=∠或2C ∠=∠或AE ADAC AB=； 18．34． 三、解答题（本大题共8小题，共87分）19．解：原式2222a b b b =-+- ························································································· 4分22a b =- ································································································································ 6分当1a =-，1b =时，原式2(1)21=--=- ········································································ 8分 20．解：由①得，243x x -<- ··························································································· 1分 1x < ········································································································································ 3分 由②得，32x x -> ················································································································ 4分 3x -> ····································································································································· 5分 3x <- ····································································································································· 6分∴不等式组的解集为3x <- ·································································································· 8分 21．答：BD 是O 的切线． ······························································································· 2分 理由1：连接OD ，OA OD = ，30ADO A ∴∠=∠=·················································· 4分30A B ∠=∠= ，180()120BDA A B ∴∠=-∠+∠= ················································ 7分 90BDO BDA ADO ∴∠=∠-∠= 即OD BD ⊥BD ∴是O 的切线． ····························································· 9分 理由2：连接OD ，OA OD = ，30ADO A ∴∠=∠= ······························································ 4分 60BOD ADO A ∴∠=∠+∠= ·············································· 7分 30B ∠= ，180()90BDO BOD B ∴∠=-∠+∠= ，即OD BD ⊥BD ∴是O 的切线． ··········································································································· 9分理由3：连接OD ，OA OD = ，30ADO A ∴∠=∠=·················································· 4分 在BD 的延长线上取一点E ，30A B ∠=∠=60ADE A B ∴∠=∠+∠= ·································································································· 7分 90EDO ADO ADE ∴∠=∠+∠= ，即OD BD ⊥BD ∴是O 的切线． ··········································································································· 9分理由4：连接OD ，OA OD = ，30ADO A ∴∠=∠=·················································· 4分 连接CD ，则90ADC ∠=··································································································· 5分60ODC ADC ADO ∴∠=∠-∠= ····················································································· 6分 OD OC = ，60OCD ∴∠=30B ∠= ，30BDC OCD B ∴∠=∠-∠= ··································································· 7分 90ODB ODC BDC ∴∠=∠+∠= ，即OD BD ⊥BD ∴是O 的切线． ··········································································································· 9分22．解：（1）3，5，2，2（每空1分） ················································································ 4分 （2）26，25，24（每空1分） ······························································································ 7分 （3）不能 ································································································································ 8分 因为此时众数26万元>中位数25万元 ·············································································· 10分 （或：因为从统计表中可知20名营业员中，只有9名达到或超过目标，不到半数） 23．解：（1）3y x = ·············································································································· 4分 （2）3 1.240w x x =--········································································································ 7分 1.840x =- ····························································································································· 8分 ∴所筹集的慰问金w （元）与销售量x （支）之间的函数关系式为 1.840w x =- 解法一：当500w ≥时，1.840500x -≥ ········································································ 10分 解得300x ≥∴若要筹集不少于500元的慰问金，至少要售出鲜花300支 ············································ 12分 解法二：由1.840500x -=，解得300x = ······································································ 11分 1.840w x =- 中1.80>w ∴随x 的增大而增大，∴若要筹集不少于500元的慰问金，至少要售出鲜花300支 ············································ 12分 24．解：作BD AC ⊥于D ，由已知得：38CBD ∠=，21ABD ∠= ，15AB =米在Rt ADB △中，sin ADABD AB∠=，sin 15sin 21 5.38AD AB ABD ∴=∠=⨯ ≈ ······················· 3分 cos BDABD AB∠=， cos 15cos 2114.00BD AB ABD ∴=∠=⨯ ≈ ···················· 5分在Rt BDC △中，tan CDCBD BD∠=tan 14.00tan3810.94CD BD CBD ∴=∠⨯ ≈≈ ··························································· 8分 cos BD CBD BC ∠=，14.0017.77cos cos38BD BC CBD ∴=∠≈≈ ····································· 10分5.3810.9417.7734.09AD CD BC ∴++++=≈ ··························································· 11分 34.1≈答：折断前发射塔的高约为34.1米． ················································································· 12分 注意：按以下方法进行近似计算视为正确，请相应评分． ①若到最后再进行近似计算结果为：15cos2115sin 2115cos 21tan 3834.1cos38AD CD BC ++=⨯+⨯+≈； ②若解题过程中所有三角函数值均先精确到0.01，则近似计算的结果为： 5.4010.8817.6633.9433.9AD CD BC ++++=≈≈．25．（1）解法一：(B ，由题意可知AG AB =OG OC ==OA m = ················································ 2分90OGA ∠= ，222OG AG OA ∴+= ·············································································· 3分 222m ∴+=．又0m > ，2m ∴= ·················································································· 4分解法二：(B ，由题意可知AG AB =OG OC ==OA m = ················································ 2分90OGA ∠= ，45GOA GAO ∴∠=∠= ········································································ 3分2cos cos 45OG m OA GOA ∴====∠················································································ 4分 （2）解法一：过G 作直线GH x ⊥轴于H ，则1OH =，1HG =，故(11)G ，． ······················································································ 5分。