2020学年奉贤区质量调研九年级数学(202104)(完卷时间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）1．计算23a a的结果是（▲）（A）5a；（B）25a；（C）6a；（D）26a．2．在下列各式中，二次根式b的有理化因式是（▲）（Ab；（Bb；（C；（D3．某校对进校学生进行体温检测，在某一时段测得6名学生的体温分别为36.8℃，36.9℃，36.5℃，36.6℃，36.9℃，36.5℃，那么这6名学生体温的平均数与中位数分别是（▲）（A）36.7℃，36.7℃；（B）36.6℃，36.8℃；（C）36.8℃，36.7℃；（D）36.7℃，36.8℃．4．下列函数中，函数值y随自变量x的值增大而减小的是（▲）（A）2=yx ；（B）2yx；（C ）2y x；（D ）2y x．5．如图，在梯形ABCD中，AB∥DC，对角线AC、BD交于点O.下列条件中，不一定能判断梯形ABCD是等腰梯形的是（▲）（A）AD=BC；（B）∠ABC=∠BAD；（C）AB=2DC；（D）∠OAB=∠OBA．6．如图，在Rt△ABC中，∠C=90°，BC=18，AC=24，点O在边AB上，且BO=2OA．以点O为圆心，r为半径作圆，如果⊙O与Rt△ABC的边有3个公共点，那么下列各值中，半径r不可以取的是（▲）（A）6；（B）10；（C）15；（D）16．二、填空题（本大题共12题，每题4分，满分48分）7．9的平方根是▲．A BCDO第5题图AC第6题图8．函数1xy x =-的定义域是 ▲ ． 9．如果抛物线2y ax bx c =++在对称轴左侧呈上升趋势，那么a 的取值范围是 ▲ ．10．如果一元二次方程230x px -+=有两个相等的实数根，那么p 的值是 ▲ ． 11．将π，23，0，-1这5个数分别写在5张相同的卡片上，字面朝下随意放在桌上，任取一张，取到无理数的概率为 ▲ .12．某小区一天收集各类垃圾共2.4吨，绘制成各类垃圾收集量的扇形图，其中湿垃圾在扇形图中对应的圆心角为135°，那么该小区这一天湿垃圾共收集了 ▲ 吨． 13．某品牌汽车公司大力推进技术革新，新款汽车油耗从每百公里8升下降到每百公里6.8升，那么该汽车油耗的下降率为 ▲ ．14． 如图△ABC 中，点D 在BC 上，且CD =2BD .设AB a =，AC b =，那么AD = ▲ （结果用a 、b 表示）．15．已知传送带和水平面所成斜坡的坡度i =1：3，如果物体在传送带上经过的路程是30米，那么该物体上升的高度是 ▲ 米（结果保留根号）．16．如图，⊙O 的半径为6，如果弦AB 是⊙O 内接正方形的一边，弦AC 是⊙O 内接正十二边形的一边，那么弦BC 的长为 ▲ ．17. 我们把反比例函数图像上到原点距离相等的点叫做反比例函数图像上的等距点.如果点A （2,4）与第一象限内的点B 是某反比例函数图像上的等距点，那么点A 、B 之间的距离是 ▲ ．18．如图，在△ABC 中，AD 是BC 边上的中线，∠ADC=60°，BC=3AD ．将△ABD 沿直线AD 翻折，点B 落在平面上的B'处，联结AB'交BC 于点E ，那么CEBE的值为 ▲ ．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：2422331x x xx xx x ，其中3x 错误!未找到引用源。

． 第14题图CBD第16题图第18题图A BCD20．（本题满分10分）解不等式组：132221132x x x x ，并把解集在数轴上表示出来．21．（本题满分10分，每小题满分5分）如图，已知，在Rt △ABC 中，∠C =90°，AB=4，BC=2，点D 是AC 的中点，联结BD 并延长至点E ，使∠E=∠BAC . （1）求sin ∠ABE 的值； （2）求点E 到直线BC 的距离.22．（本题满分10分，每小题满分5分）为了预防“诺如病毒”，某校对专用教室采取“药熏”消毒.从开始消毒到结束，室内含药量y （毫克/立方米）与时间x （分）这两个变量之间的关系如图中折线OA -AB 所示. （1）求20分钟至60分钟时间段之间的含药量y 与时间x 的函数解析式（不要求写定义域）； （2）开始消毒后，消毒人员在某一时刻对该专用教室的含药量进行第一次检测，时隔半小时进行了第二次跟踪检测，发现室内含药量比第一次检测时的含药量下降了2毫克/立方米，求第一次检测时的含药量.23．（本题满分12分，每小题满分6分）如图，已知，在平行四边形ABCD 中，E 为射线CB 上一点，联结DE 交对角线AC 于点F ，∠ADE =∠BAC . （1）求证：CF CA CB CE ⋅=⋅；（2）如果AC=DE ，求证：四边形ABCD 是菱形.ABCDEF第20题图-112546第21题图A BCED第22题图分)2024．（本题满分12分，第(1)小题满分3分，第(2)小题满分4分，第(3)小题满分5分）如图，在平面直角坐标系xOy 中，已知B (0，2)，31,2C ⎛⎫- ⎪⎝⎭，点A 在x 轴正半轴上，且OA =2OB ．抛物线20yax bx a 经过点A 、C ．（1）求这条抛物线的表达式；（2）将抛物线先向右平移m 个单位，再向上平移1个单位，此时点C 恰好落在直线AB 上的点C'处，求m 的值；（3）设点B 关于原抛物线对称轴的对称点为B'，联结AC ，如果点F 在直线AB'上，∠ACF =∠BAO ，求点F 的坐标．25．（本题满分14分，第(1)小题满分4分，第(2)小题满分5分，第(3)小题满分5分）如图，已知扇形AOB 的半径OA =4，∠AOB =90°，点C 、D 分别在半径OA 、OB 上（点 C 不与点A 重合），联结CD ．点P 是弧AB 上一点，PC=PD ． （1）当cot ∠ODC 34=，以CD 为半径的圆D 与圆O 相切时，求CD 的长； （2）当点D 与点B 重合，点P 为弧AB 的中点时，求∠OCD 的度数； （3）如果OC =2，且四边形ODPC 是梯形，求ΔΔPCDOCDS S 的值．C第24题图备用图AB 备用图ABP第25题图DAB O奉贤区2020学年度九年级数学质量调研参考答案及评分说明（202104）一、选择题：（本大题共6题，每题4分，满分24分）1．D ； 2．B ； 3．A ； 4．D ； 5．C ； 6．C . 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，其中19-22题每题10分，23、24题每题12分，25题14分，满分78分）19．解原式=1234313131x x x x x x x x x x x ····································· （3分）=2242631xx x x xx x ······························································································· （2分）=2331x xx x ·············································································································· （1分） =3311x x x x x x ． ····························································································· （1分）当3x 时，原式333231． ···································································· （3分）20．解：由①得2x ································································································ （3分） 由②得5x ·································································································· （3分）所以原不等式组得解集为25x ·································································· （2分） 作图正确 ················································································································ （2分）21.（1）解：过点D 作DHAB ，垂足为H ．在Rt △ABC 中，∠C =90°，AB=4，BC=2，∴AC =，∠BAC =30° ∵点D 是AC 的中点，∴3AD CD···································································· （1分）7. 3±；8. 1x ；9. 0a <； 10. ± 11．25； 12. 0.9； 13．15%； 14．21+33a b ； 15．； 16．17．18．37．在Rt △BCD 中，∠C =90°，3AD ，∠BAC =30°，∴DH=2························ （1分） 在Rt △ABC 中，∠C =90°，3CD，BC=2，∴BD····································· （1分）在Rt △BDH 中，sin ∠2114········································································ （2分） （2）过点E 作EGBC ，垂足为G ．∵∠E=∠BAC ，∠ABE=∠DBA ，∴ΔABD ∽ΔABE ················································ （1分） ∴AB BD BE AB ，474BE ，∴1677BE ································································· （2分） ∵,DC BC EG BG ，∴DC ∥EG ，∴EGBEDC BD，得77·················· （1分） ∴1637EG ，∴点E 到直线BC·············································· （1分）22.（1）设直线OA 的解析式(0)y kx k ，把（15，6）代入得解析式得25k，解析式为25y x ·········································· （1分） 当x =20时，y =8，∴A （20，8） ················································································ （1分） 设直线AB 的解析式(0)y kxb k，由它经过点A 、B ，得208600k b kb， 解得1512kb·············································· （2分） ∴直线AB 的解析式为1125yx ． ······································································ （1分） （2）设第一次检测时间为a 分钟，则第一次检测时的含药量为25a 毫克/立方米，（1分）第二次检测时间为b 分钟，则第二次检测时的含药量为1125b 毫克/立方米.（1分） 由题意得，302112255b a a b ，解得4031303ab·········································· （2分）∴2240165533a ····························································································· （1分）答：第一次检测时的含药量为163.23．证明：（1）∵四边形ABCD是平行四边形，∴AD∥CE，∴ADE E ···········（2分）∵∠ADE=∠BAC，∴∠E=∠BAC ··········································································（1分）∵∠ACB=∠ECF，∠E=∠BAC，∴△ACB∽△ECF············································（1分）∴CF CBCE CA，∴CF CA CB CE⋅=⋅ ······································································（2分）（2）∵AD∥CE，∴CF EFAC DE·················································································（1分）∵AC=DE，∴CF=EF···························································································（1分）∴∠E=∠FCE········································································································（1分）又∵∠E=∠BAC，∴∠BAC=∠FCE···································································（1分）∴AB=BC···············································································································（1分）又∵四边形ABCD是平行四边形，∴四边形ABCD是菱形 ···························（1分）24．解：（1）由题意，抛物线2yax bx 经过点A (4，0)，31,2C ⎛⎫- ⎪⎝⎭得164032a bab ，解得122a b ··················································································· （2分） 抛物线的表达式是2122y x x =-. ·············································································· （1分） （2）设直线AB 的解析式(0)y kx b k ，由它经过点A 、B ，得402k bb ， 解得122kb∴直线AB 的解析式为122yx ··········································································· （1分） ∵将抛物线先向右平移m 个单位，再向上平移1个单位，设C'11,2m ········· （2分）将C'11,2m 代入122y x ， 解得m =4． ··················································· （1分） （3）∵11tan tan 22OACBAO ，，∴OAC BAO ······································ （1分） ∵点B 关于原抛物线对称轴的对称点为B'，∴B'（4，2），∴直线AB'为x =4 ········································································· （1分） 当点F 在直线x =4上，且∠ACF =∠BAO 时，（i ）过点C 作x 轴平行线交直线x =4于点1F ，此时点1F 的坐标为34,2··········· （1分） （ii ）作2ACF BAO ，射线2CF 交x 轴于点D设D （n ，0），∵2ACF BAOCAO ，∴DCDA∴29414nn ，解得178n ，∴D （178，0） ································· （1分） ∴直线CD 的解析式为41736y x，当4x ，52y ,∴252,2F ············· （1分）25.解：（1）在Rt △ABC 中，∠AOB =90°，∵cot ∠ODC 34， 设3,4OD k OCk ，∴5CD k ················································································· （1分） ∵以CD 为半径的圆D 与圆O 相切，∴OB CD OC ············································ （1分） 即435k k ，解得12k ······························································································ （1分） ∴52CD························································································································· （1分） （2）联结OP 、AP .∵P 为弧AB 的中点，∴45o AOP BOP ，PA PB ········································· （1分） ∵ OA OP OB ，可得67.5o OAP OPAOBP又∵PC PB ，PA PB ，∴PA PC ，可得67.5o OAPPCA ···················· （1分） ∴45o APC（或22.5o OPC ）············································································· （1分） 可得ΔPCD 是等腰直角三角形∴45o PCB ······························································ （1分） ∴67.5o OCD ·············································································································· （1分） （3）联结OP （i ）当CP ∥OD 时过点P 作PQ ⊥OB ，垂足为Q ． ∵ 2,4CO OP ，∴23PCPD ······································································· （1分）在Rt △PDQ 中，∠PQD =90°，PQ=2，23PD ， ∴22DQ ，2322OD··················································································· （1分）∴ΔΔ23362322PCDOCDS PC S OD ············································································ （1分）（ii ）当CO ∥PD 时过点P 作PK ⊥OA ，垂足为K由题意得四边形OKPD 是矩形，设KO PD x ,则2,KC x PCx在Rt △PCK 中，222KP PC KC ,则44KP OD x 在Rt△OPD 中，222OD PD OP ,则2224x∴解得262x,262PD ················································································· （1分）。