2010年福建省龙岩市数学中考真题(word版含答案)

2010年龙岩市初中毕业、升学考试数学试题（满分：150分 考试时间：120分钟）注意：请把所有答案填涂或书写到答题卡上！请不要错位、越界答题！ 在本试题上答题无效.提示：抛物线2(0)y ax bx c a =++≠的对称轴是2bx a =-,顶点坐标是24,24b ac b a a ⎛⎫-- ⎪⎝⎭一、选择题（本大题共10题，每题4分，共40分.每题的四个选项中，只有一个符合题意，请将正确的选项填涂到答题卡．．．上） 1.-3的绝对值是 A ．-3 B.13- C.3 D.132.下列运算正确的是A ．4482x x x += B.235x x x =· C.824x x x ÷= D.248()x x -=-3.下列事件是不可能事件的是A.掷一次质地均匀的正方体骰子，向上的一面是5点B.在只装有红球和绿球的袋子中摸出一个球，结果是黄球C.经过城市中某一有交通信号灯的路口，遇到绿灯D.通常加热到100℃时，水沸腾4．若关于x 的一元二次方程20x x a -+=的一个根为2，则a 的值是A ．6 B.-6 Ｃ.2 Ｄ.-25．如图所示的几何体是由三个同样大小的正方体搭成的，其左视图是6.如图，若圆锥底面圆的半径为3，则该圆锥侧面展开图扇形的弧长为 A.2π B.4π C.6π D.9π7.从4张分别写有数字-6，-4，0，3的卡片中，任意抽取一张，卡片上的数字是正数的概率是 A ．34 B. 12 C. 13 D. 148.把多项式269x x -+分解因式，所得结果正确的是A ．()23x - B. ()23x + C. ()69x x -+ D. ()()33x x +-9.如图，AB 是O ⊙的直径，CD 是O ⊙的切线，C 为切点，25B ∠=°，则D ∠等于A ．25° B. 40° C. 30° D. 50° 10.对于反比例函数ky x=，当0x >时，y 随x 的增大而增大，则二次函数2y kx kx =+的大致图象是二、填空题（本大题共7题，每题3分，共21分.请将答案填入答题卡．．．相应位置） 11.当x =_________时，分式11x -没有．．意义. 12.去年龙岩市固定资产投资约为43 400 000 000元，用科学记数法表示为_________元. 13.若两圆相外切，圆心距为8，其中一个圆的半径为3，则另一个圆的半径是_________. 14.甲、乙两班举行计算机汉字输入比赛，测得每个学生每分钟输入汉字的个数，并进行统计.两个班的平均数、方差分别为：135135;x x ==乙甲，221510s s ==乙甲，．根据统计结果，_________班的成绩波动较小.15.函数y kx b =+的图象如图所示，当0y <时，x 的取值范围是_________.16.如图，在矩形ABCD 中，对角线AC 、BD 相交于点O ，点E 、F 分别是AO 、AD 的中点，若8AC =，则EF =_________.17.右图是圆心角为30°，半径分别是1、3、5、7、…的扇形组成的图形，阴影部分的面积依次记为1S 、2S 3S 、…，则50S =_________(结果保留π).三、解答题（本大题共8题，共89分）18．（10分）（1）计算：(()0320102tan 452+-°（2）先化简，再求值：()22232a a a a ---，其中a =0.01）19.（8分）解方程：21.133x xx x =-++ 20.（10分）如图，在等腰梯形ABCD 中，AB CD ∥，点E 、F 在AB 上，且AE BF =，连接CE 、.DF 求证：.CE DF =21.（10分）我市某化工厂为响应国家“节能减排”的号召，从2006年开始采取措施，控制二氧化硫的排放.图①、图②分别是该厂2006~2009年二氧化硫排放量（单位：吨）的两幅不完整的统计图.请根据图中信息解答下列问题：（1）该厂2006~2009年二氧化硫的排放总量是___________吨，这四年二氧化硫排放量的中位数是___________吨；（2）把图①的折线图补充完整；（3）图②中2006年二氧化硫的排放量对应扇形的圆心角是___________度，2009年二氧化硫的排放量占这四年排放总量的百分比是___________.22.（12分）在平面直角坐标系中，AOB △的位置如图所示.（1）若11AOB △是AOB △关于原点O 的中心对称图形，则顶点1A 的坐标为（_______、_________）； （2）在网格上画出AOB △关于y 轴对称的图形；（3）在网格上画出将AOB △三个顶点的横、纵坐标均扩大为原来的2倍后的图形，并求出变换后图形的周长等于__________；若把AOB △顶点的横、纵坐标均扩大为原来的n 倍，试猜想变换后图形的周长等于___________.23.（12分）某校为迎接县中学生篮球比赛，计划购买A 、B 两种篮球共20个供学生训练使用.若购买A 种篮球6个，则购买两种篮球共需费用720元；若购买A 种篮球12个，则购买两种篮球共需费用840元.（1）Ａ、B 两种篮球单价各多少元？（2）若购买A 种篮球不少于8个，所需费用总额不超过800元.请你按要求设计出所有的购买方案供学校参考，并分别计算出每种方案购买A 、B 两种篮球的个数及所需费用.24.（13分）如图，抛物线交x 轴于点()20A -，，点()40B ，，交y 轴于点()04C -，． （1）求抛物线的解析式，并写出顶点D 的坐标；（2）若直线y x =-交抛物线于M ，N 两点，交抛物线的对称轴于点E ，连接B C E B E C ，，．试判断EBC △的形状，并加以证明；（3）设P 为直线MN 上的动点，过P 作PF ED ∥交直线MN 下方的抛物线于点F ．问：在直线MN 上是否存在点P ，使得以P E D F 、、、为顶点的四边形是平行四边形？若存在，请求出点P 及相应的点F 的坐标；若不存在，请说明理由.25.（14分）如图①，ABC 绕其直角顶点C 顺时针旋转α角()090α<<°°，得111A B C A C△，交AB 于点D ，11A B 分别交于BC AB 、于点E F 、，连接1.AB（1）求证：1;ADC A DF △∽△（2）若30α=°，求11AB A ∠的度数；（3）如图②，当45α=°时，将11A B C △沿C A →方向平移得22222A B C A C △，交AB 于点,G 22B C 交BC 于点,H 设(20,CC x x =<<ABC △与222A B C △的重叠部分面积为,S 试求S 与x 的函数关系式.2010年龙岩市初中毕业、升学考试参考答案及评分标准数学说明：评分最小单位为1分，若学生解答与本参考答案不同，参照给分.11．1 12. 104.3410⨯ 13. 5 14. 乙 15. 2x > 16. 2 17. 66π三、解答题（本大题共8题，共89分） 18．（10分，第（1）小题5分，第（2）小题5分） （1）解：原式=1+5-2+（-8） ··························································································· 4分=-4 ················································································································ 5分（2）解：原式=222322a a a a --+ ···················································································· 1分=a - ················································································································ 3分当a ==············································································ 4分 ≈-2.65 ·········································································· 5分19．（8分）解：()21131x xx x =-++ ································································································· 1分 方程两边同乘()31x +，得 ····················································································· 2分 ()3312x x x =+- ·································································································· 4分 3332x x x =+-3323x x x -+= ····································································································· 5分23x =32x = ···································································································· 6分 检验：当32x =时，()310x +≠ ········································································· 7分∴32x =是原方程的解 ··············································································· 8分20．（10分） 证明：（法一）如图∵AE BF =∴AE EF BF EF +=+ 即AF BE = ············································································································ 2分∵四边形ABCD 是等腰梯形 ∴AD BC A B =∠=∠， ························································································ 5分 ∴ADF BCE △≌△ ····························································································· 8分 ∴CE DF = ·········································································································· 10分 （法二）如图连接DE CF 、 ························································· 1分 ∵四边形ABCD 是等腰梯形 ∴AD BC A B =∠=∠， ········································· 3分 ∵AE BF =∴ADE BCF △≌△ ·············································· 6分 ∴DE CF = ············································································································ 7分 ∵DC AB ∥∴四边形EFCD 是等腰梯形 ·················································································· 8分 ∴CE DF = ·········································································································· 10分21．（10分，第（1）小题4分，第（2）小题2分，第（3）小题4分）（1）100 25 ················································· 4分（2）正确补全折线图（如右图所示） ··························· 2分 （3）144 10% ································································ 4分 注：第（1）、（3）题每空2分，第（2）题正确画出一段得1分 22．（12分，第（1）小题2分，第（2）小题3分，第（3）小题7分）（1）()134A --， ······················································· 2分 （2）正确画出图形 ····················································· 3分（3）正确画出图形 ····················································· 3分 32 ········································································ 5分 16n ···································································· 7分 注：第（1）题每空1分，第（2）、（3）小题每正确画出一个顶点给1分23．（12分，第（1）小题5分，第（2）小题7分）（1）设A 种篮球每个x 元，B 种篮球每个y 元 ··················································· 1分依题意，得614720128840x y x y +=⎧⎨+=⎩ ········································································ 3分解得5030x y =⎧⎨=⎩ ·································································································· 4分答：A 种篮球每个50元，B 种篮球每个30元 ················································· 5分（2）（法一）设购买A 种篮球m 个，则购买B 种篮球()20m -个 ··················· 1分 依题意，得()5030208008m m m +-⎧⎪⎨⎪⎩≤≥ ····················································· 2分解得810m ≤≤ ····························································································· 3分∵篮球的个数必须为整数∴m 只能取8、9、10 ······················································································ 4分 可分别设计出如下三种方案：方案①：当8m =时，2012,m -=5083012760⨯+⨯=即购买A 种篮球8个，B 种篮球12个，费用共计760元 ············· 5分方案②：当9m =时，2011,m -=5093011780⨯+⨯=即购买A 种篮球9个，B 种篮球11个，费用共计780元 ···························· 6分 方案③：当10m =时，2010,m -=50103010800⨯+⨯=即购买A 种篮球10个，B 种篮球10个，费用共计800元 ·························· 7分（法二）设购买篮球的费用共w 元，A 种篮球购买m 个，依题意，可得总费用w （元）与m （个）之间的函数关系式为 ······························································································· 1分 ()503020w m m =+- ()8m ≥ ··················································································· 2分 ∴20600w m =+∵800w ≤∴20600800m +≤ 10m ≤ ∴810m ≤≤ ························································································································ 3分 注：以下过程同（法一）三种方案写对一种分别得1分 24．（13分，第（1）小题4分，第（2）小题4分，第（3）小题5分）（１）解： （法一）设所求的抛物线解析式()20y ax bx c a =++≠ ··················································· 1分∵点A B C 、、均在此抛物线上.∴42016404a b c a b c c -+=⎧⎪++=⎨⎪=-⎩∴1214a b c ⎧=⎪⎪=-⎨⎪=-⎪⎩······················································································································· 2分 ∴所求的抛物线解析式为2142y x x =-- ··································································· 3分 顶点D 的坐标为912⎛⎫-⎪⎝⎭， ···························································································· 4分 （法二）设所求的抛物线解析式()()24y a x x =+- ················································· 1分 ∵点C 在此抛物线上 ∴()()02044a +-=-12a =··························································································································· 2分 ∴所求的抛物线解析式为()()1242y x x =+-即2142y x x =-- ······································································································ 3分顶点D 的坐标为912⎛⎫-⎪⎝⎭， ···························································································· 4分 注：顶点横、纵坐标错一个不给分（2）EBC △的形状为等腰三角形 ······················································································· 1分 证明：（法一）∵直线MN 的函数解析式为y x =-∴ON 是BOC ∠的平分线 ······························································································ 2分 ∵B C 、两点的坐标分别为（4，0），（0，-4） ∴4CO BO ==∴MN 是BC 的垂直平分线 ··························································································· 3分 ∴CE BE =即ECB △是等腰三角形 ································································································· 4分 （法二）∵直线MN 的函数解析式为y x =-∴ON 是BOC ∠的平分线∴COE BOE ∠=∠ ························································································· 2分 ∵B C 、两点的坐标分别为()()4004-，、， ∴4CO BO == 又∵CE BE =。