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人教版初二下册数学期末试卷及答案

人教版初二下册数学期末试卷及答案TTA standardization office【TTA 5AB- TTAK 08- TTA 2C】附中2010-2011学年度下期期末考试初二数学试题(总分:150分 考试时间:120分钟)一、选择题(每小题4分,共40分)1.在22923 3.1407π-,,,,,这六个实数中,无理数的个数是( ) A .4个B .3个C .2个D .1个2.下列计算正确的是( ) A .532-= B .824+=C .2733=D .(12)(12)1+-=3.已知Rt △ABC 中,90C ∠=︒,BC = 8,4sin 5A =,则AC = ( ) A .6B .8C .10D .3234.如图,小正方形的边长均为1,则下列图中三角形(阴影部分)与△ABC 相似的是( )5.如图,设M 、N 分别为直角梯形ABCD 两腰AD 、CB 的中点,DE ⊥AB 于点E ,将△ADE 沿DE 翻折,M 与N 恰好重合,则AE ∶BE 等于( ) A .2∶1 B .1∶2 C .3∶2D .2∶36.关于x 的方程22(21)10k x k x +-+=有实数根,则下列结论正确的是( ) A .当12k =时方程两根互为相反数 B .当14k ≤时方程有实数根C DMN(第5题图)AB C D(第4题图)(第9题图)(第10题图)(第14题图)C .当1k =±时方程两根互为倒数D .当k = 0时方程的根是x = – 17.已知实数x 、y 满足222222()(1)2x y x y x y +++=+,则的值为( ) A .1 B .2C .– 2或1D .2或 – 18.已知22y x =的图象是抛物线,若抛物线不动,将x 轴、y 轴分别向上、向右平移2个单位,在新坐标系下,所得抛物线解析式为( ) A .22(2)2y x =-+ B .22(2)2y x =+- C .22(2)2y x =--D .22(2)2y x =++9.如图,△OAP 、△ABQ 均是等腰直角三角形,点P 、Q 在函数4(0)y x x=>的图象上,直角顶点A 、B 均在x 轴上,则点B 的坐标为( ) A 1,0) B 1,0) C .(3,0)D 1,0)10 2D .13二、填空题(每小题3分,共30分)11.方程220x x +=的解为_________________.12.已知α为锐角,若1sin cos 3αα==,则_________________. 13.已知221(2)m m y m m x +-=+是反比例函数,则m = _________________.14.在实数范围内定义一种运算“*”,其规则为22a b a b *=-,根据这个规则,方程(2)50x +*=的解为_________________.15.如图是一个二次函数当40x -≤≤的图象,则此时函数y 的取值范围是_________________.16.小亮同学想利用影长测量学校旗杆的高度,如图,他在某一时刻立1m 长的标杆测得其影长为2 m ,同时旗杆的投影一部分在地面上,另一部分在某建筑物的墙(第16题图)上,分别测得其长度为9.6 m 和2 m ,则学校旗杆的高度为_________________m .17.一个三角形两边长为3和4角形的形状为_________________18.已知开口向下的抛物线过A (– 1,0),B (3,0)两点,与y 轴交于C,且BC =_________________.19.如图,□ABCD 中,E 为CD 上一点,DE ∶CE = 2∶3,连结AE 、BE 、BD ,且AE 、BD 交于点F ,则S △DEF ∶S △EBF ∶S △ABF = _________________.20.如图,二次函数2y ax bx c =++的图象经过点(– 1,2)和(1,0),且与y 轴交于负半轴,给出以下四个结论:① abc < 0;② 2a + b > 0;③ a + c = 1;④ a > 1.其中正确结论的序号是_________________.(第19题(第20题三、解答题(共80分) 21.(6分) (1)计算:0111()()|tan3023--+︒(6分) (2) 解方程:21302x x ++=22.(10分) 已知抛物线2y ax bx c =++的图象如图所示.(1) 抛物线的解析式为____________________. (2) 抛物线的顶点坐标为______________,且y 有最________(填“大”或“小”)值.(3) 当x _____________时,y 随x 的增大而减小. (4) 根据图象可知,使不等式20ax bx c ++<成立的x 的取值范围是______________________.23.(8分) 如图,在△ABC 中,430sin 105B C AC ∠=︒==,,,求AB 的长.24.(10分) 如图,反比例函数k y x=的图象与一次函数y1,3),B(n ,– 1)两点.(1)求反比例函数与一次函数的解析式;(2) 根据图象,直接写出使反比例函数的值大于一次函数的值的x 的取值范围.(第23题图)(第24题(第22题25.(10分) 西瓜经营户以2元/千克的价格购进一批小型西瓜,以3元/千克的价格出售,每天可售出200千克.为了促销,该经营户决定降价销售,经调查发现,这种小型西瓜每降价元/千克,每天可多售出40千克,另外,每天的房租等固定成本共24元,该经营户要想每天盈利200元,应将每千克小型西瓜的售价降低多少元?26.27.(10分) 如图,在等腰直角三角形ABC中,904,,点D在线段BC上∠=︒==BAC AB AC运动(不与B、C重合),过D作45∠=︒,交AC于E.ADE(1)求证:△ABD∽△DCE;(2)设BD = x,AE = y,求y与x之间的函数关系式,并写出自变量的取值范围.(第26题图)(10分) 如图,甲、乙两辆大型货车同时从A 地出发驶往P 市.甲车沿一条公路向北偏东53︒方向行驶,直达P 市,其速度为30千米/小时.乙车先沿一条公路向正东方向行驶1小时到达B 地,卸下部分货物(卸货的时间不计),再沿一条北偏东37︒方向的公路驶往P 市,其速度始终为35千米/小时. (3) 求AP 间的距离.(4) 已知在P 市新建的移动通信接收发射塔,其信号覆盖面积只可达P 市周围方圆50千米的区域(包括边缘地带),除此以外,该地区无其他发射塔,问甲、乙两司机至少经过多少小时可以互相正常通话? (5)(3434sin37cos53cos37sin53tan37tan535543︒=︒=︒=︒=︒=︒=,,,)53︒37︒(第27题图)28.(10分) 如图,抛物线2y ax bx c =++与x 轴交于A 、B 两点(点A 在点B 的左侧),与y轴交于点C ,这条抛物线的顶点是M (1,– 4),且过点(4,5). (1) 求这条抛物线的解析式;(2) P 为线段BM 上的一点,过点P 向x 轴引垂线,垂足为Q ,若点P 在线段BM 上运动(点P 不与点B 、M 重合),四边形PQAC 的面积能否等于7?如果能,求出点P 的坐标;如果不能,请说明理由.(3) 设直线m 是抛物线的对称轴,是否存在直线m 上的点N ,使以N 、B 、C 为顶点的三角形是直角三角形?若存在,请求出点N 的坐标;若不存在,请说明理由.附中2010—2011学年度下期期末考试初二数学试题参考答案一、选择题(每小题4分,共40分)题号 1 2 3 4 5 6 78 9 10 选项CCABABA BDC二、填空题(每小题3分,共30分) 11.x 1 = 0,x 2 = – 2 12.2213.– 1 14.x 1 = 3,x 2 = – 715.24y -≤≤16.17.等腰或直角三角形 18.223y x x =-++ 19.4∶10∶25 20.②③④三、解答题(共80分) 21.(1) 解:原式23313|3|=+⨯--····················································· 3分 23123=+- ································································· 5分 431=+········································································ 6分 (2) 解:a = 1,b = 3,c =12O xyAQB PMC(第28题图)∵ 21494172b ac -=-⨯⨯= ∴x =···································································· 5分 ∴12x x ==··················································· 6分 22.(1) 213222y x x =-++(2) (32528,) 大 (3) 32>(4) 14x x <->或(每空2分)23.解:过A 作AD ⊥BC 于D ································································ 1分∵ AD ⊥BC ,4sin 5C =,AC = 10∴ sin 8AD AC C == ··························· 5分 ∵ 30B ∠=︒∴ AB = 2AD = 16 ······························ 8分24.解:(1) ∵ A (1,3),B (n ,– 1)在反比例函数k y x=的图象上∴ 31kk n =⎧⎪⎨-=⎪⎩········································································· 2分 ∴ 33k n ==-, ···································································· 3分 ∵ A (1,3),B (n ,– 1)在一次函数y mx b =+的图象上 ∴313m bm b=+⎧⎪⎨-=-+⎪⎩ ·································································· 4分12m b =⎧⎪⎨=⎪⎩解得 ········································································· 5分 ∴ 反比例函数与一次函数的解析式分别为32y y x x==+, ············· 6分(2) 301x x <-<<或 ·································································· 10分25.解:设每千克小型西瓜的售价降低x 元,由题意 ·································· 1分(32)(20040)242000.1xx --+⨯-= ···················································· 5分(第23题图)∴ x 1 = ,x 2 = ········································································· 9分 答:应将每千克小型西瓜的售价降低元或元. ··································· 10分 26.(1) 证明:∵ 90BAC ∠=︒,AB = AC∴ 45B C ∠=∠=︒ ································································ 1分 ∵ 23180C ∠+∠+∠=︒∴ 23135∠+∠=︒ ················2分 ∵ 1218045ADE ADE ∠+∠+∠=︒∠=︒, ∴ 12135∠+∠=︒ ················3分 ∴ 13∠=∠ ························4分 ∴ △ABD ∽△DCE ············5分(2) 解:∵ AB = AC = 4∴ 42BC =········································································· 6分 ∵ BD = x ,AE = y∴ 424CD x CE y =-=-, ······················································· 7分 ∵ △ABD ∽△DCE , ∴42AB BD xDC CE yx ==--4即4 ·················································· 9分 ∴ 2124(042)4y x x x =-+<< ·············································· 10分27.解:(1) 过P 作PD ⊥AB 延长线于D ·················································· 1分由题意知:AB = 35千米 设PD = x 千米∵ 在Rt △PAD 中,tan37PDAD︒= ∴ 4tan373x AD x ==︒··················· 2分 CADE (第26题图)12353︒37︒(第27题图)D∵在Rt △PBD 中,tan53PD BD ︒=∴ 3tan534PD BD x ==︒ ································································ 3分 ∵ AD – BD = AB 即433534x x -=∴ x = 60,即 AD = 60千米 ····················································· 4分 ∴在Rt △PAD 中,sin37PD AP ︒=∴ 60100sin370.6PD AP ===︒千米···················································· 5分 (2) ∵在Rt △PBD 中,sin53PD BD ︒=∴ 60754sin 533PD PB ===︒千米 ···················································· 6分 设甲、乙两司机分别出发t 1、t 2小时后手机有信号∴ 15010050530303PA t --===小时 ··············································· 7分 25035755060123535357PB AB t +-+-====小时 ································ 8分 ∵ 125351236321721t t ==<== ······················································· 9分 ∴ 甲、乙两司机至少经过127小时可以正常通话. ···················· 10分 28.解:(1) 设2(1)4y a x =--代入(4,5)得a = 1,∴ 223y x x =-- ·············· 2分······················································································ 8分 ③若∠C = Rt ∠,则222NB BC NC =+,∴22418(3)1n n +=+++ ∴ 4n =- ······················································································ 9分综上,存在这样的点N,其坐标为(1或(1或(12),或(1,4-) ······················································································ 10分。

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