九年级上期末测试题

人教版九年级数学上册期末测试题（附参考答案）满分120分考试时间120分钟一、选择题：本大题共10个小题，每小题3分，共30分。

每小题只有一个选项符合题目要求。

1.方程x2＋4x＋3＝0的两个根为( )A．x1＝1，x2＝3B．x1＝－1，x2＝3C．x1＝1，x2＝－3D．x1＝－1，x2＝－32.一个口袋里装有4个白球，5个黑球，除颜色外，其余如材料、大小、质量等完全相同，随意从中抽出一个球，抽到白球的概率是( )A．49B．59C．14D．193.将抛物线y＝x2向右平移3个单位长度，再向上平移4个单位长度，得到的抛物线是( )A．y＝(x－3)2＋4 B．y＝(x＋3)2＋4C．y＝(x＋3)2－4 D．y＝(x－3)2－44.如图，在方格纸中，将Rt△AOB绕点B按顺时针方向旋转90°后得到Rt△A′O′B，则下列四个图形中正确的是( )A BC D5.如图，AB切⊙O于点B，连接OA交⊙O于点C，BD∥OA交⊙O于点D，连接CD.若∠OCD＝25°，则∠A的度数为( )A．25°B．35°C．40°D．45°6.若关于x的一元二次方程x2－8x＋m＝0的两根为x1，x2，且x1＝3x2，则m的值为( )A．4 B．8C．12 D．167.如图，抛物线y＝ax2＋bx＋c(a≠0)与x轴交于点A，B，与y轴交于点C，对称轴为直线x＝－1.若点A的坐标为(－4，0)，则下列结论正确的是( )A．2a＋b＝0B．4a－2b＋c>0C．x＝2是关于x的一元二次方程ax2＋bx＋c＝0(a≠0)的一个根D．点(x1，y1)，(x2，y2)在抛物线上，当x1>x2>－1时，y1<y2<08.图1是一把扇形纸扇，图2是其完全打开后的示意图，外侧两竹条OA和OB 的夹角为150°，OA的长为30 cm，贴纸部分的宽AC为18 cm，则CD⏜的长为( )A．5π cm B．10π cmC．20π cm D．25π cm9.如图，⊙O与正五边形ABCDE的两边AE，CD相切于A，C两点，则∠AOC的度数是( )A．144°B．130°C．129°D．108°10.在如图所示的运算程序中，若开始输入x的值为48，我们发现第一次输出的结果为24，第二次输出的结果为12……则第2 023次输出的结果为( )A．6 B．3C．622 021D．322 022二、填空题：本题共6个小题，每小题3分，共18分。

（第4题）九年级数学上册期末复习综合测试题（含答案）一、选择题（本大题共6小题，每小题2分，共12分．） 1．一元二次方程 x 2＝x 的根是（ ）A ．x 1＝0，x 2＝1B ．x 1＝0，x 2＝－1C ．x 1＝x 2＝0D ．x 1＝x 2＝12．一个不透明布袋中有2个红球，3个白球，这些球除颜色外无其他差别，摇匀后从中随机摸出一个小球，该小球是红色的概率为（ ）A ．12B ．23C ．15D ．253．若一组数据 2，3，4，5，x 的方差比另一组数据 5，6，7，8，9 的方差大，则 x 的值可能是（ ） A ．1B ．4C ．6D ．84．如图，OA 、OB 是⊙O 的半径，C 是⊙O 上一点．若∠OAC ＝16°，∠OBC ＝54°，则 ∠AOB 的度数是（ ）A ．70°B ．72°C ．74°D ．76°5．若关于x 的一元二次方程ax 2＋k ＝0的一个根为2，则二次函数y ＝a (x ＋1)2＋k 与x 轴的交点坐标为（ ） A ．(－3，0)、(1，0) B ．(－2，0)、(2，0) C ．(－1，0)、(1，0)D ．(－1，0)、(3，0)6．如图，在Rt △ABC ，∠ACB ＝90°，AC ＝4，BC ＝3，点D ，E 分别在AB ，AC 上，连接DE ，将△ADE 沿DE 翻折，使点A 的对应点F 落在BC 的延长线上，若FD 平分∠EFB ，则AD 的长为（ ） A ． 157B ．207C ．258D ．259二、填空题（本大题共10小题，每小题2分，共20分．） 7（第12题）l 1 l 2l 3A BCEFD （第11题）8．若a b ＝43，则a －b b＝ ．9．设x 1、x 2是方程x 2＋mx －m ＋3＝0的两个根，则x 1＋x 2－x 1x 2＝ ．10．把抛物线y ＝－x 2向左平移2个单位，然后向上平移3个单位，则平移后该抛物线相应的函数表达式为 ．11．如图，l 1∥l 2∥l 3，若AD ＝1，BE ＝3，CF ＝6，则ABBC的值为 ．12．如图，点A 、B 、C 在⊙O 上，⊙O 的半径为3，∠AOC ＝的长为 ． 13．已知关于x 的函数y ＝x 2＋2mx ＋1，若x ＞1时，y 随x 的增大而增大，则m 的取值范围是 ．14．如图，弦AB 是⊙O 的内接正六边形的一边，弦AC 是⊙O 的内接正方形的一边，若 BC ＝2＋23，则⊙O 的半径为 ．15．如图，正方形ABCD 的边长是4，点E 在DC 上，点F 在AC 上，∠BFE ＝90°，若 CE ＝116．如图，在矩形ABCD 中，AB ＝2，AD ＝4，点E 、F 分别为AD 、CD 边上的点，且EF 的长为2，点G 为EF 的中点，点P 为BC 上一动点，则P A ＋PG 的最小值为 ． 三、解答题（本大题共11小题，共88分．请在答题卡指定区域．．．．．．．内作答，解答时应写出文字说明、证明过程或演算步骤）17．（8分）解方程：（1）x 2－4x －5＝0； （2）x 2－4＝2x (x －2)．18．（8分）甲乙两人在相同条件下完成了5次射击训练，两人的成绩（单位：环）如下（1）甲射击成绩的中位数为 环，乙射击成绩的众数为 环；（2）计算两人射击成绩的方差；（3）根据训练成绩，你认为选派哪一名队员参赛更好，为什么？19．（8分）某校开展秋季运动会，需运动员代表进行发言，从甲、乙、丙、丁四名运动员中随机抽取．（1）若随机抽取1名，甲被抽中的概率为 ； （2）若随机抽取2名，求甲在其中的概率．20．（7分）如图，在△ABC 中，点D 、E 分别在AB 、AC 上，且∠BCE ＋∠BDE ＝180°． （1）求证：△ADE ∽△ACB ；（2）连接BE 、CD ，求证：△AEB ∽△ADC ．21．（8分）如图是二次函数y ＝－x 2＋bx ＋c 的图像． （1）求该二次函数的关系式及顶点坐标； （2）当y ＞0时 x 的取值范围是 ；（3）当m ＜x ＜m ＋4时，－5＜y ≤4，则m 的值为 ．22．（7分）在Rt △ABC ，∠BAC ＝90°，AB ＝AC ，D 、E、F 分别为BC 、AB 、AC 边上的点，且∠EDF ＝45°．（1）求证：△EBD ∽△DCF ；（2）当D 是BC 的中点时，连接EF ，若CF ＝5，DF ＝4，则EF 的长为 ．23．（8分）某超市销售一种商品，成本为每千克50元．当每千克售价60元时，每天的销售量为60千克，经市场调查，当每千克售价增加1元，每天的销售量减少2千克． （1）为保证某天获得750元的销售利润，则该天的销售单价应定为多少？ （2）当销售单价定为多少时，才能使当天的销售利润最大？最大利润是多少？24．（8分）如图，AB 为⊙O 的直径，弦CD ⊥AB 于点P ，连接BC ，过点D 作DE ⊥CD ，交⊙O 于点E ，连接AE ，F 是DE 延长线上一点，且∠BCD ＝∠F AE ． （1）求证：AF 是⊙O 的切线；（2）若AF ＝2，EF ＝1，求⊙O 的半径．25．（8分）已知二次函数y ＝(x －2)(x －m )（m 为常数）． （1）求证：不论m 为何值，该函数的图像与x 轴总有公共点；（2）若M （－1，0）， N （3，0），该函数图像与线段MN 只有1个公共点，直接写出 m 的取值范围；（3）若点A （－1，a ），B （1，b ），C （3，c ）在该函数的图像上，当abc ＜0时，结合函数图像，直接写出m 的取值范围.26．（8分）如图，四边形ABCD 内接于⊙O ，AB ＝AC ，BD ⊥AC ，垂足为E ． （1）求证：∠BAC ＝2∠DAC ； （2）若AB ＝10，CD ＝5，求BC 的长．27．（10分）定义：圆心在三角形的一边上，与另一边相切，且经过三角形一个顶点（非切点）的圆，称为这个三角形圆心所在边上的“伴随圆”．（1） 如图①，在△ABC 中，∠C ＝90°，AB ＝5，AC ＝3，则BC 边上的伴随圆的半径为 ． （2）如图②，△ABC 中，AB ＝AC ＝5，BC ＝6，直接写出它的所有伴随圆的半径． （3）如图③，△ABC 中，∠ACB ＝90°，点E 在边AB 上，AE ＝2BE ，D 为AC 的中点，且∠CED ＝90°．①求证：△CED 的外接圆是△ABC 的AC 边上的伴随圆； ②DE的值为 ．参考答案说明：本评分标准每题给出了一种或几种解法供参考，如果考生的解法与本解答不同，参照本评分标准的精神给分．一、选择题（本大题共6小题，每小题2分，共12分）二、填空题（本大题共10小题，每小题2分，共20分）7．9 8．13 9．－3 10．y ＝－(x ＋2)2＋3 11．2312．2π 13．m ≥－1 14． 2 2 15．322 16．4 2 －1三、解答题（本大题共11小题，共88分） 17．（8分）（1）解：x 2－4x －5＝0 x 2－4x ＋4＝5＋4(x －2)2＝9 ········································································································ 1分x －2＝±3 ········································································································ 2分 ∴ x 1＝5，x 2＝－1． ··························································································· 4分 （2）解：x 2－4＝2x (x －2) x 2－4＝2x 2－4xx 2－4x ＋4＝0 ··································································································· 5分 (x －2)2＝0 ········································································································ 6分 ∴ x 1＝x 2＝2． ··································································································· 8分 18．（8分）（1）7；8 ········································································································ 2分 （2）s 2甲＝(7－8)2＋(7－8) 2＋(10－8)2＋(9－8)2＋(7－8)25＝1.6环2． ······························ 4分s 2乙＝(8－8)2＋(8－8) 2＋ (7－8)2＋(8－8)2＋(9－8)25＝0.4环2． ······································ 6分（3）选择乙．因为甲乙两人平均数相同均为8，说明两人实力相当，但s 2乙＜s 2甲，乙的成绩更加稳定，所以选乙． ······················································································· 8分19．（8分）（1）14． ·········································································································· 2分（2）解：随机抽取两名运动员，共有6种等可能性结果：（甲，乙）、（甲，丙）、（甲，丁）、（乙，丙）、（乙，丁）、（丙，丁）．其中满足“有甲运动员”(记为事件A )的结果只有3种，所以P (A )＝12． ·································································································· 8分20．（7分）（1）证明：∵ ∠BCE ＋∠BDE ＝180°， ∠EDA ＋∠BDE ＝180°，∴ ∠EDA ＝∠BCE ． ·························································································· 1分 又 ∠A ＝∠A ， ································································································· 2分 ∴ △ADE ∽△ACB ． ·························································································· 3分 （2）∵ △ADE ∽△ACB ， ∴ AD AC ＝AE AB， ·········································· 4分 ∴AD AE ＝ACAB， ······································· 5分 又 ∠A ＝∠A ， ········································ 6分 ∴ △AEB ∽△ADC ． ································· 7分21．（8分）（1）将(0，3)、 (3，0)代入，得⎩⎨⎧3＝c ，0＝－9＋3b ＋c································································································· 1分解得⎩⎨⎧c ＝3，b ＝2····································································································· 2分∴ y ＝－x 2＋2x ＋3 ····························································································· 3分 ∴ 顶点坐标为(1，4) ························································································ 4分 （2）－1＜x ＜3. ······························································································ 6分 （3）－2或0 ···································································································· 8分 22．（7分）（1）解：∵∠BAC ＝90°，AB ＝AC ，∴ ∠B ＝∠C ＝45°． ··························································································· 1分 ∴ 在△BDE 中，∠BED ＋∠BDE ＝180°－∠B ＝135°， ∵ ∠EDF ＝45°，∴ ∠BDE ＋∠CDF ＝135°，∴ ∠BED ＝∠CDF ． ·························································································· 3分 ∵ ∠B ＝∠C ，∴ △EBD ∽△DCF ． ·························································································· 5分 （2 ········································································································ 7分23．（8分）（1）解：设每千克的销售价增加x 元，根据题意，得(60＋x －50) (60－2x )＝750 ··················································································· 2分 ∴ x 1＝5，x 2＝15． ····························································································· 3分 60＋5＝65或60＋15＝75 ···················································································· 4分 答：销售单价为65或75元时获得利润750元． （2）解：每千克的销售价增加x 元，利润为w 元．w ＝(60＋x －50) (60－2x ) ···················································································· 6分 ＝－2(x －10)2＋800 ···························································································· 7分 ∵ a ＝－2＜0，∴ 当x ＝10时，w 有最大值800． ········································································ 8分 60＋10＝70答：当销售单价为70元时获得最大利润，为800元． 24．（8分） （1）连接BD ．∵ AB 为⊙O 的直径，CD ⊥AB ，∴ ⌒BC ＝ ⌒BD ， ························································· 1分 ∴ ∠BDC ＝∠BCD ．∵ 四边形ABDE 为⊙O 的内接四边形，∴ ∠BDE ＋∠BAE ＝180°，即∠BDC ＋∠CDF ＋∠BAE ····· 2分∵ DE ⊥CD ， ∴ ∠CDF ＝90°， ∴ ∠BDC ＋∠BAE ＝90°．∵ ∠BCD ＝∠F AE ， ·························································································· 3分 ∴ ∠BAE ＋∠F AE ＝90°，即∠F AB ＝90°， ∴ AF ⊥AB ． 又 点A 在⊙O 上，∴ AF 与⊙O 相切． ·························································································· 4分 （2）过点O 作OG ⊥DF 垂足为G ． ∵ ∠F AB ＝∠D ＝∠APD ＝90°， ∴ 四边形APDF 是矩形， ∴ ∠F ＝90°．∵ ∠F AB ＝∠F ＝∠OGF ＝90°， ∴ 四边形AOGF 是矩形，∴ AF ＝OG ，AO ＝GF ． ···················································· 5分 设OE ＝OA ＝r ，则GE ＝r －1．在Rt △OGE 中，由勾股定理得OG 2＋GE 2＝OE 2， ···················································· 6分 即4＋(r －1)2＝r 2， ···························································································· 7分 解得r ＝5 2 ． ····································································································· 8分25．（8分）（1）令y ＝0，即(x －2)(x －m )＝0 ········································································· 1分 ∴ x 1＝2，x 2＝m ． ····························································································· 2分 当m ＝2时，x 1＝x 2，方程有两个相等的实数根； 当m ≠2时，x 1≠x 2，方程有两个不等的实数根． ∴ 不论m 为何值，方程总有实数根；∴ 不论m 为何值，该函数的图像与x 轴总有公共点． ·············································· 3分 （2）m ＝2或m ＞3或m ＜－1． ··········································································· 6分 （3）－1＜m ＜1或m ＞3． ·················································································· 8分 26．（8分）。

2023-2024学年度上学期九年级期末调研测试英语试卷第Ⅰ卷一、单项选择（本题共10分，每小题1分）选择最佳答案。

1. —Boys and girls, let’s make ________ decision to have all around development from now on.—Only then will we have a better future.A. aB. anC. /2. —Our parents will take us to Disney as a special ________ next summer vacation.—Sounds like fun!A. treatB. shameC. energy3. — What do you think of your English teacher?— He is always ________ us. What a good teacher!A. absent fromB. tired ofC. patient with4. Learning to work with your classmates can help you find pleasure in what you do, even when the job ________ is hard.A. ourselvesB. itselfC. themselves5. Though they didn’t carry out their plan, they succeeded in ________ their minds.A. regrettingB. improvingC. stealing6. The population problem is very serious in this area. We must come up with good ways to ________ it.A. argue withB. deal withC. agree with7. —________ almost three years since we met for the first time.—Yeah, how time flies! We need to value the time we spent with each other.A. It wasB. It’s beenC. It will be8. —Who is the woman with Ann?—I’m not sure. It ________ be her mother.A. mayB. mustC. can’t9. Kids can own their favorite hobbies at holidays ________ they are given too much homework.A. ifB. unlessC. until10. —Excuse me, Mr Zhang. Could you please tell me ________ ?—Next Wednesday. Work hard and get ready for it.A. where I can pass the examB. if we will have an examC. when we will have the exam二、完形填空（本题共10分，每小题1分）I made a promise to myself a couple of months ago. I had put some money in my wallet with the purpose of ____11____ to the first person I met who needed it. Time passed, and as my daily life got busy, I was unable to keep my word. ____12____ I opened my wallet and saw the money, I thought of that, and I would feel a little disappointed for doing nothing. The voice inside told me that I should do something soon.Today, when I was walking to work on a busy street, I passed a ____13____ old lady with her hand out on the corner. Seeing her aged face and eyes that failed to see, I stopped ____14____. A sense of sympathy (同情) washed over me, and I remembered the money that had been lying in my wallet all these five ____15____.I took out the money from the wallet and bent down (弯腰) to her level. ____16____, I took her hand, placed the notes into it, and closed her fingers over the notes ____17____ she could feel the money and understand me.After I got up, I slowly moved away. I felt like I had achieved something. I had kept my word to myself. The lady nodded her head a little bit. I felt ____18____ and moved on.After that, I thought about why such a simple act had taken so long. This small act was a huge step in facing and walking ____19____ my own fears around giving. What a gift it is to have these small but meaningful____20____!根据短文内容选择最佳答案。

人教版九年级上册数学期末测试卷一、单选题(共15题，共计45分)1、用20cm长的绳子围成一个矩形，如果这个矩形的一边长为xcm，面积是Scm2，则S与x的函数关系式为（）A.S＝x（20﹣x）B.S＝x（20﹣2x）C.S＝x（10﹣x）D.S＝2x（10﹣x）2、若抛物线与轴的交点为，则下列说法不正确的是（）A.抛物线开口向上B.抛物线的对称轴是C.当时, 的最大值为D.抛物线与轴的交点为和3、如图，△ABC内接于⊙O，∠ABC=71°，∠CAB=53°，点D在AC弧上，则∠ADB的大小为A.46°B.53°C.56°D.71°4、已知⊙O的面积为4π，则其内接正三角形的面积为（）A. B. C.3 D.35、在下列四个图形中，是中心对称图形的是（）A. B. C. D.6、下列关于方程x2+x﹣1=0的说法中正确的是（）A.该方程有两个相等的实数根B.该方程有两个不相等的实数根，且它们互为相反数C.该方程有一根为D.该方程有一根恰为黄金比例7、在以下图形中，是中心对称图形的是（）A.等边三角形B.等腰梯形C.平行四边形D.正五边形8、将抛物线向上平移3个单位长度，再向右平移5个单位长度，所得到的拋物线为（）A. B. C. D.9、如图，把一个量角器放在∠BAC的上面，请你根据量角器的读数判断∠BAC的度数是( )A.30°B.60°C.15°D.20°10、方程x2+6x﹣5=0的左边配成完全平方后所得方程为（）A.（x+3）2=14B.（x﹣3）2=14C.（x+3）2=4D.（x﹣3）2=411、已知二次函数的图象如图所示，则这个二次函数的表达式为（）A.y=x 2﹣2x+3B.y=x 2﹣2x﹣3C.y=x 2+2x﹣3D.y=x 2+2x+312、下列方程是一元二次方程的是（）A. +x 2=0B.3x 2﹣2xy=0C.x 2+x﹣1=0D.ax 2﹣bx=013、在平面直角坐标系中，将抛物线y=3x2先向右平移1个单位，再向上平移2个单位，得到的抛物线的解析式是（）A.y=3（x+1）2+2B.y=3（x+1）2﹣2C.y=3（x﹣1）2+2 D.y=3（x﹣1）2﹣214、在一个不透明的袋子中装有4个除颜色外完全相同的小球，其中黄球1个，红球1个，白球2个，“从中任意摸出2个球，它们的颜色相同”这一事件是（）A.必然事件B.不可能事件C.随机事件D.确定事件15、如图，抛物线y＝ax2+bx+1的顶点在直线y＝kx+1上，对称轴为直线x＝1，有以下四个结论：①ab＜0，②b＜，③a＝﹣k，④当0＜x＜1时，ax+b ＞k，其中正确的结论是（）A.①②③B.①③④C.①②④D.②③④二、填空题(共10题，共计30分)16、一个不透明的袋子中装有8个大小、形状、都一样的小球，其中有3个红球与5个黄球，从这8个球中任取一个球是红球的概率是：________.17、如图，抛物线（）与轴交于点，与轴交于，两点，其中点的坐标为，抛物线的对称轴交轴于点，，并与抛物线的对称轴交于点．现有下列结论：① ；② ；③ ；④ ．其中所有正确结论的序号是________．18、一个不透明的口袋里有10个黑球和若干个黄球，它们除颜色外其余都相同，从口袋中随机摸出一球记下其颜色，再把它放回口袋中摇匀，重复上述过程，共实验200次，其中有120次摸到黄球，由此估计袋中的黄球有________个．19、欧阳修在《卖油翁》中写道：“（翁）乃取一葫芦置于地，以钱覆其扣，徐以杓酌油沥之，自钱孔入，而钱不湿，因曰：我亦无他，唯手熟尔．”可见技能通过反复苦练而达到熟能生巧．若铜钱是直径为4cm的圆，中间有边长为1cm的正方形孔，你随机向铜钱上滴一滴油，则油（油滴的大小忽略不计）正好落入孔中的概率为________．（结果保留π）20、扬州园林中有许多花窗，图案中蕴含着对称之美,现从中选取如图的四种窗格图案,其中既是中心对称图形又是轴对称图形的有________个.21、已知关于x的一元二次方程的一个根是2，则另一个根是________.22、某超市销售某种玩具，进货价为20元．根据市场调查：在一段时间内，销售单价是30元时，销售量是400件，而销售单价每上涨1元，就会少售出10件玩具，超市要完成不少于300件的销售任务，又要获得最大利润，则销售单价应定为________ 元．23、如图，AB是⊙O的一条弦，点C是⊙O上一动点，且∠ACB=30°，点E、F 分别是AC、BC的中点，直线EF与⊙O交于G、H两点．若⊙O的半径为7，则GE+FH的最大值为________．24、如图，方格纸上一圆经过（2，6）、（﹣2，2）、（2，﹣2）、（6，2）四点，则该圆圆心的坐标为________．25、如图，AB是⊙O的直径，∠C=20°，则∠BOC的度数是________ ．三、解答题(共5题，共计25分)26、解方程组：．27、如图为桥洞的形状，其正视图是由和矩形ABCD构成．O点为所在⊙O的圆心，点O又恰好在AB为水面处．若桥洞跨度CD为8米，拱高（OE⊥弦CD于点F ）EF为2米．求所在⊙O的半径DO．28、如图，⊙O的半径为17cm，弦AB∥CD，AB=30cm，CD=16cm，圆心O位于AB，CD的上方，求AB和CD的距离．29、把一副扑g牌中的三张黑桃牌（它们的正面数字分别为3、4、5）洗匀后正面朝下放在桌面上．小王和小李玩摸牌游戏，游戏规则如下：先由小王随机抽取一张牌，记下牌面数字后放回，洗匀后正面朝下，再由小李随机抽取一张牌，记下牌面数字．当两张牌的牌面数字相同时，小王赢；当两张牌的牌面数字不同时，小李赢．现请你分析游戏规则对双方是否公平，并说明理由．30、在-2，-1，0，1，2这五个数中任取两数m，n，用列表或画树状图的方法求二次函数的顶点在坐标轴上的概率.参考答案一、单选题(共15题，共计45分)1、C2、C3、C4、D5、D6、D7、C8、D9、C10、A11、B12、C13、C14、C15、B二、填空题(共10题，共计30分)17、18、19、20、21、22、23、24、25、三、解答题(共5题，共计25分)26、28、29、。