普陀区2017学年度第二学期初三质量调研数 学 试 卷(时间:100分钟,满分:150分)考生注意:1.本试卷含三个大题,共25题.答题时,考生务必按答题要求在答题纸规定的位置上作答,在草稿纸、本试卷上答题一律无效.2.除第一、二大题外,其余各题如无特别说明,都必须在答题纸的相应位置上写出证明或计算的主要步骤.一、选择题:(本大题共6题,每题4分,满分24分)[下列各题的四个选项中,有且只有一个选项是正确的,选择正确项的代号并填涂在答题纸的相应位置上]1. 下列计算中,错误的是 ············································································· (▲) (A )120180=; (B )422=-;(C )2421=; (D )3131=-.2.下列二次根式中,最简二次根式是 ······························································ (▲) (A )a 9; (B )35a ; (C )22b a +; (D )21+a . 3.如果关于x 的方程022=++c x x 没有实数根,那么c 在2、1、0、3-中取值是 ··· (▲) (A )2; (B )1; (C )0; (D )3-. 4.如图1,已知直线CD AB //,点E 、F 分别在AB 、CD 上,CFE ∠:EFB ∠3=:4,如果40B ∠=,那么BEF ∠=········································································· (▲) (A )20; (B )40; (C )60; (D )80.5. 自1993年起,联合国将每年的3月22日定为“世界水日”,宗旨是唤起公众的节水意识,加强水资源保护.某校在开展“节约每一滴水”的活动中,从初三年级随机选出20名学生统计出各自家庭一个月的节约用水量,有关数据整理如下表.ABCDFE图1图 2节约用水量(单位:吨) 1 1.2 1.4 2 2.5 家庭数46532这组数据的中位数和众数分别是 ···································································· (▲) (A )1.2,1.2; (B )1.4,1.2; (C )1.3,1.4; (D )1.3,1.2. 6. 如图2,已知两个全等的直角三角形纸片的直角边分别为a 、b )(b a ≠,将这两个三角形的一组等边重合,拼合成一个无重叠的几何图形,其中轴对称图形有 ···················· (▲) (A )3个; (B )4个; (C )5个; (D )6个.二、填空题:(本大题共12题,每题4分,满分48分)7.计算:xy x 3122⋅= ▲ .8.方程32x x =+的根是 ▲ .9.大型纪录片《厉害了,我的国》上映25天,累计票房约为402700000元,成为中国纪录电影票房冠军.402700000用科学记数法表示是 ▲ .10.用换元法解方程312122=+-+x x x x 时,如果设y xx =+21,那么原方程化成以y 为“元”的方程是 ▲ .11.已知正比例函数的图像经过点M (2-,1)、),(11y x A 、),(22y x B ,如果21x x <,那么1y ▲ 2y .(填“>”、“=”、“<”)12.已知二次函数的图像开口向上,且经过原点,试写出一个符合上述条件的二次函数的解析式: ▲ .(只需写出一个)13.如果一个多边形的内角和是720,那么这个多边形的边有 ▲ 条.14.如果将“概率”的英文单词 probability 中的11个字母分别写在11张相同的卡片上,字面朝下随意放在桌子上,任取一张,那么取到字母b 的概率是 ▲ .15.2018年春节期间,反季游成为出境游的热门,中国游客青睐的目的地仍主要集中在温暖的东南亚地区.据调查发现2018年春节期间出境游约有700万人, 游客目的地分布情况的扇形图如图3所示,从中可知出境游东南亚 地区的游客约有 ▲ 万人.A 东南亚欧美澳新16%港澳台 15%韩日11%其他13%y xO ABC图 6ABCDEF图 4BCDOA 图516. 如图4,在梯形ABCD 中,BC AD //,AD BC 3=,点E 、F 分别是边AB 、CD 的中点.设a AD =,b DC =,那么向量EC 用向量、表示是 ▲ .17. 如图5,矩形ABCD 中,如果以AB 为直径的⊙O 沿着BC 滚动一周,点B 恰好与点C 重合,那么ABBC的值等于 ▲ .(结果保留两位小数)18. 如图6,在平面直角坐标系xOy 中,△ABC 的顶点A 、C 在坐标轴上,点B 的坐标是(2,2).将△ABC 沿x 轴向左平移得到△111A B C ,点1B 落在函数6y x=-的图像上.如果此时四边形11AA C C 的面积等于552,那么点1C 的坐标是 ▲ . 一、选择题:(本大题共6题,每题4分,满分24分)1.(B); 2.(C); 3.(A); 4.(C); 5.(D); 6.(B). 二、填空题:(本大题共12题,每题4分,满分48分)三、解答题:(本大题共7题,满分78分)19.(本题满分10分)先化简,再求值:42442222---++÷+x x x x x x x ,其中22x -. 19.解:原式()()22+22(2)22x x x x x x x -=-+-+ ················································· (3分) 7.323x y ; 8. 3x =; 9. 810027.4⨯ ; 10. 32=-yy ; 11.>;12. 2y x =等;13.6; 14.112; 15.315; 16.b a212+;17.3.14;18.(5-,211).ABCDE 图7122x x x =-++ ····································································· (2分) 12x x -=+. ··············································································· (1分) 当22x =时,原式221222--=-+······················································· (1分)22=···························································· (1分) 2322-=. ························································· (2分)20.(本题满分10分)求不等式组()7153,31>34x x x x ⎧++⎪⎨--⎪⎩≥的整数解.20.解:由①得,2x ≥-. ·········································································· (3分)由②得,x <3. ··········································································· (3分) ∴原不等式组的解集是2<3x -≤. ················································· (2分) 所以,原不等式组的整数解是2-、1-、0、1、2. ··························· (2分)21.(本题满分10分)如图7,在Rt △ABC 中,90C ∠=,点D 在边BC 上,DE ⊥AB ,点E 为垂足,7AB =,45DAB ∠=,3tan 4B =. (1)求DE 的长;(2)求CDA ∠的余弦值. 21.解:(1)∵DE ⊥AB ,∴︒=∠90DEA又∵45DAB ∠=,∴AE DE =. ······················································ (1分) 在Rt △DEB 中,︒=∠90DEB ,43tan =B ,∴43=BE DE . ······················· (1分) 设x DE 3=,那么x AE 3=,x BE 4=.∵7AB =,∴743=+x x ,解得1=x . ··············································· (2分)∴3=DE . ··················································································· (1分) (2) 在Rt △ADE 中,由勾股定理,得23=AD .····································· (1分)同理得5=BD . ············································································· (1分) 在Rt △ABC 中,由43tan =B ,可得54cos =B .∴528=BC . ················ (1分) ∴53=CD . ·················································································· (1分)∴102cos ==∠AD CD CDA . ····························································· (1分)即CDA ∠的余弦值为210.22.(本题满分10分)小张同学尝试运用课堂上学到的方法,自主研究函数21y x =的图像与性质.下面是小张同学在研究过程中遇到的几个问题,现由你来完成: (1)函数21y x =的定义域是 ▲ ; (2)下表列出了y 与x 的几组对应值:x… 2-32- m34- 12- 1234 1 32 2… y…144911694 416914914…表中m 的值是 ▲ ;(3)如图8,在平面直角坐标系xOy 中,描出以表中各组对应值为坐标的点,试由描出的点画出该函数的图像; (4)结合函数21y x=的图像,写出这个 函数的性质: ▲ .(只需写一个) 22.解:(1)0x ≠的实数; ················································································· (2分) (2)1-; ······························································································ (2分) (3)图(略); ·························································································· (4分) (4)图像关于y 轴对称;图8图像在x 轴的上方;在对称轴的左侧函数值y 随着x 的增大而增大,在对称轴的右侧函数值y 随着x 的增大而减小;函数图像无限接近于两坐标轴,但永远不会和坐标轴相交等. ···················· (2分) 23.(本题满分12分)已知:如图9,梯形ABCD 中,AD ∥BC ,DE ∥AB ,DE 与对角线AC 交于点F ,FG ∥AD ,且FG EF =.(1)求证:四边形ABED 是菱形; (2)联结AE ,又知AC ⊥ED ,求证:212AE EF ED =.23.证明:(1)∵ AD ∥BC ,DE ∥AB ,∴四边形ABED 是平行四边形. ···················· (2分)∵FG ∥AD ,∴FG CFAD CA=. ······························································ (1分) 同理EF CFAB CA =. ··········································································· (1分) 得FG AD=EF AB∵FG EF =,∴AD AB =. ······························································· (1分) ∴四边形ABED 是菱形. ··································································· (1分) (2)联结BD ,与AE 交于点H .∵四边形ABED 是菱形,∴12EH AE =,BD ⊥AE .···························· (2分) 得90DHE ∠= .同理90AFE ∠=.∴DHE AFE ∠∠=. ········································································ (1分) 又∵AED ∠是公共角,∴△DHE ∽△AFE . ······································· (1分)∴EH DEEF AE =. ··············································································· (1分) ∴212AE EF ED =. ······································································· (1分) A B CDE F G图924.(本题满分12分)如图10,在平面直角坐标系xOy 中,直线3y kx =+与x 轴、y 轴分别相交于点A 、B ,并与抛物线21742y x bx =-++的对称轴交于点()2,2C ,抛物线的顶点是点D .(1)求k 和b 的值;(2)点G 是y 轴上一点,且以点B 、C 、G 为顶点的三角形与△BCD 相似,求点G 的坐标;(3)在抛物线上是否存在点E :它关于直线AB 的对称点F 恰好在y 轴上.如果存在,直接写出点E 的坐标,如果不存在,试说明理由.24.解:(1) 由直线3y kx =+经过点()2,2C ,可得12k =-. ····································· (1分) 由抛物线21742y x bx =-++的对称轴是直线2x =,可得1b =. ················· (1分) (2)∵直线132y x =-+与x 轴、y 轴分别相交于点A 、B ,∴点A 的坐标是()6,0,点B 的坐标是()0,3. ········································ (2分)∵抛物线的顶点是点D ,∴点D 的坐标是92,2⎛⎫ ⎪⎝⎭. ·································· (1分) ∵点G 是y 轴上一点,∴设点G 的坐标是()0,m .∵△BCG 与△BCD 相似,又由题意知,GBC BCD ∠=∠,∴△BCG 与△BCD 相似有两种可能情况: ··········································· (1分) ①如果BG BC CB CD =,那么552=,解得1m =,∴点G 的坐标是()0,1. ··· (1分) 图10xy1 1O②如果BG BC CD CB =,那么35552m -=,解得12m =,∴点G 的坐标是10,2⎛⎫ ⎪⎝⎭. (1分) 综上所述,符合要求的点G 有两个,其坐标分别是()0,1和10,2⎛⎫ ⎪⎝⎭ .(3)点E 的坐标是91,4⎛⎫- ⎪⎝⎭或92,2⎛⎫ ⎪⎝⎭. ···················································· (2分+2分)25.(本题满分14分)已知P 是O ⊙的直径BA 延长线上的一个动点,P ∠的另一边交O ⊙于点C 、D ,两点位于AB 的上方,AB =6,OP m =,1sin 3P =,如图11所示.另一个半径为6的1O ⊙经过点C 、D ,圆心距1OO n =. (1)当6m =时,求线段CD 的长;(2)设圆心1O 在直线AB 上方,试用n 的代数式表示m ;(3)△1POO 在点P 的运动过程中,是否能成为以1OO 为腰的等腰三角形,如果能,试求出此时n 的值;如果不能,请说明理由.25.解:(1)过点O 作OH ⊥CD ,垂足为点H ,联结OC .在Rt △POH 中,∵1sin 3P =,6PO =,∴2OH =. ······························ (1分) ∵AB =6,∴3OC =. ······································································ (1分) 由勾股定理得 5CH =. ·································································· (1分)OAB备用图PDOABC图11∵OH ⊥DC ,∴2CD CH == ················································· (1分)(2)在Rt △POH 中,∵1sin 3P =, PO m =,∴3mOH =. ·························· (1分)在Rt △OCH 中,2293m CH ⎛⎫- ⎪⎝⎭=. ···················································· (1分)在Rt △1O CH 中,22363m CH n ⎛⎫-- ⎪⎝⎭=. ············································· (1分)可得 2236933m m n ⎛⎫⎛⎫--- ⎪ ⎪⎝⎭⎝⎭=,解得23812n m n -=. ······························ (2分)(3)△1POO 成为等腰三角形可分以下几种情况:● 当圆心1O 、O 在弦CD 异侧时①1OP OO =,即m n =,由23812n n n-=解得9n =. ······························ (1分)即圆心距等于O ⊙、1O ⊙的半径的和,就有O ⊙、1O ⊙外切不合题意舍去.(1分)②11O P OO =n =,解得23m n =,即23n 23812n n -=,解得n ······························· (1分) ● 当圆心1O 、O 在弦CD 同侧时,同理可得 28132n m n-=.∵1POO ∠是钝角,∴只能是m n =,即28132n n n -=,解得n . ········ (2分)综上所述,n。