南平市2020-2021學年第一學期九年級期末質(zhì)量檢測數(shù)學試卷與答案

．．．．．．市—2021年學期級質(zhì)量數(shù)學題（試間120分鐘滿：150分）友提：所答都須填答卡應的置，答本卷一律效②題要求結(jié)取近值，得采近計算一、選擇題本大題共10小，每小題4分，共40．每小題只有一個正確選項，請在題卡的相應位置填涂）1A

BD

2關(guān)xA

2Bxx

Dxx3在個不透明的袋子中裝有5個球球除顏色外完全相同中黑球紅3個，從中隨機摸出一小球，則摸出的小球是紅色的率是

D.

4A

By

1

Dyx25已知ACOB

C

D九年級數(shù)學試題第1頁共頁）

6.x

60

B

D60+60

2857.eq\o\ac(△,Rt)=90=8cmCrrD.8.126A4B3CD19.x(+1x+1=0k

是常數(shù)，＜1）的根的情況A存在一個k，使得方只有一個實數(shù)根．一有兩個不相等的實數(shù)根

B無實數(shù)根D．一定有個相等的實數(shù)根ABCDACD6ABCAC是ABD3+3C6≤3+3

B3BDD．3二、填空題本大題共題，每空4，共24分．將答案填入答題卡的相位置）11-62-m126結(jié)保留13拋物線y頂點坐標是

14如圖，AB是O的徑，弦CD⊥AB于E，且＝CD＝6，則⊙O的徑______九年級數(shù)學試題第2頁共頁）

第題圖

1111111115在做“擲一枚質(zhì)地均勻的硬幣”試驗時下列說法正確的______①不②當；③多④連6.16設函數(shù)y

1

與

y

的圖象的交點坐為，m+1)(n+1)______三、解答題本大題共9小題，共86分．在答題卡的相應置作答）17題分，每小題4分）解方程）；（2x.（題滿分8分如圖，在直角坐系中，點A，B，C的坐標分別為（將△ABC繞著點C順針旋轉(zhuǎn)90°eq\o\ac(△,A)eq\o\ac(△,)BC,其中點的應點為點

1.（1）請畫出旋轉(zhuǎn)后eq\o\ac(△,A)eq\o\ac(△,)BC；（2）求出在旋轉(zhuǎn)過中點A所走的路程.結(jié)保留

第題圖（題滿分分某校開展科技節(jié)覽活動，設置了編號為1至的六個展區(qū)，小佳計劃隨參觀兩個展區(qū)，且每個展區(qū)選中的機會均等，求4號展區(qū)被選中的概率.九年級數(shù)學試題第3頁共頁）

（題滿分分如圖，ABC內(nèi)于O，AB為徑D為AC上點且OD∥BC求證：ADC為等腰三角形．D

O

21（本題滿分）

第題圖某服裝商店計劃售一種男士襯衫，已知銷售x件這種男士襯衫成本每件（元16價每件（n與的系分別，n120（為整數(shù)）55（）若該商店某日銷售種男士襯衫的利潤為600元，求當日銷售；（）求可獲得的最大日潤22（本題滿分10分在扇形中，

AOC60

，點在上，且BC，E在半徑OB上以，為鄰邊作平行邊形，當點C，，F(xiàn)共時

（）求∠的數(shù)；（）求證CF=.

O

第圖

23（本題滿分10分在平面直角坐標xOy中,

△ABC的點（0(30)反例數(shù)y

(k＞的圖象過AC的點（1）求反比例函數(shù)達式；（2）已知點關(guān)點2)的對稱點F，試判斷點F是在反比例函數(shù)的圖象，并說九年級數(shù)學試題第4頁共頁）

121222121222明理由．24題分）在eq\o\ac(△,Rt)中∠°，DEC是△ABC點C逆時旋轉(zhuǎn)90°所得，其中A點B的應點分別是點，點E，長交于F，連接FC.3（1）若∠=30°，=（2）求證：平分EFA；（3）求證：=2FC

，求FB的；

DFE

第題25（本題分）拋物線Cxax的點A在某條拋物線C上物C向平移(b)個單位后，所得物線頂點仍在拋物線上.2（1）求點A坐標(用含a的代數(shù)式表示)（2）求與b關(guān)系式；（3）拋物線C的點為其對稱軸與的交點為D，點是物線上同于頂點的任意一點，線交物線C于另一點M直線EF交線ly求證：直線MN與x軸互相垂直

12

于點，南平市2020學年一學九年級期質(zhì)量測數(shù)試參考答案及評分說明九年級數(shù)學試題第5頁共頁）

說明：（1）解答右端所注數(shù)為考生正確做完該步應得的累計分數(shù)，全卷滿分50分（2）于解答題，評卷時堅持每題評閱到底，勿因考生解中出現(xiàn)錯誤而中斷本題的評閱．當考生的答在某一步出現(xiàn)錯誤時，如果續(xù)部分的解答未改變該題的考試求，可酌情給分，但則上不超過后面應得分數(shù)的一，如果有較嚴重的錯誤，就不給．（3）若考生的解-與本參考答案不，可參照本參考答案的評分標相應評分．（4）評分只給整數(shù)．選擇題和填空題不給中間分．一、選擇題（本題共小，每小題4分，共40分）1B；．；3．D；．；．；6D；．C8D；．A；10．C．9題的鍵方程可是元一方10題示如圖，由圓的性質(zhì)知，點B在半圓AC上∴當BD過心時的長最大二、填空題（本題共小題，每小題4分共24分）11；12．30

；13．（,

．

；

15①③；1625或2．16題答聯(lián)立

y

1x，化為元二次方程為：

2

.因為+1)(n+1)=

yxmn

，點mn在函數(shù)所以.

1

上，而是程

2

的根，所以

1，得

，從而計算

1

5，5

，所以+1)(n+1)=

mnm

.設意：讓學理兩個數(shù)象交點加對完平公的理，生也以接求,三、解答題（本題共小題，共86分17）：x(x,·······················································································x01

，2

.

·············································································4分（2）解：九年級數(shù)學試題第6頁共頁）

2222122221因為,b=1，=-，······································································1分13

·······················································2分

132

，

············································································3分x

，x22

.···························································

4分（1）解：正確畫出圖.··························3分如圖eq\o\ac(△,，)C就所要畫的圖·················分（）點所走過的路是以點C圓心為半徑的弧CA

，····························分所以

l

π10··························7答：點A所走過的徑長為解：

102

π.················分1

2

4

5

61

（2,1）（3,1）（4,1）

（）（6,1）23

（1,2）（1,3）（2,3）

（3,2）（4,2）（4,3）

（）（6,2）（）（6,3）456

（1,4）（2,4）（3,4）（1,5）（2,5）（3,5）（4,5）（1,6）（2,6）（3,6）（4,6）

（）（6,4）（6,5）（）用枚舉法、樹狀或列表列出所有等可能情況均···································6分共有30個可情況，其中含號展的有10個···································7分所以（4展==·························································8分解：記AC與的點為點∵AB是的徑，∴.······························分

∵OD，

E==90°，·····················分∴半徑OD弦，·······················分

O

∴ADCD，································分∴=，7是等腰角形····················

8分九年級數(shù)學試題第7頁共頁）

661221解）根據(jù)意，列方程：x）6005

，······2分解得：x=30.···························································.3分答：每日銷售量件或件時，日利潤為元················4（2）設日利為元································································5分根據(jù)題意，得y

2

25)

2

625

，·····························6分因為是的次數(shù)，且=-1,所以當時y有大值是··································7分答：可獲得的最日利潤為625元·································8分(1解：∵

ABBC

，且∠AOC=60°，∴∠BOC=20°，∠0°.··················分∵OBOC，

∴∠OBC=80°.···························分∵平行四邊形中OA∥EF，

∴∠∠，·························分∴∠∠OBC-∠BEF=°-40°=40°.4

∵平行四邊形中AFE=AOB=40，∴∠·····································5分

O

第圖

(2)接，6分∵∠AOC°，OAOC∴△COA是等邊三角形∴ACOC∠OCA°，∴∠ACF=∠OCB∠°-°°，∴∠BOC∠ACF．·······························分又∵∠OBC∠CFA，∴△≌△，·····························9分∴OBCF，∴=··········································10分注此的鍵是周與圓角關(guān)系.23解：（1）根據(jù)題意得（4·································································1分因為反比例函數(shù)y

kx

(k＞的圖象過點D，九年級數(shù)學試題第8頁共頁）

2222所以=4·························································································4分則反比例函數(shù)表式為y=

.·································································5分（2）因為點B關(guān)點E對點為,4···········································7分所以當=1時，y=

·····································································9分則為14）在反比例函數(shù)的象

···············································分24解：（1）如圖1由轉(zhuǎn)質(zhì)得，≌DCE，∴∠A∠D=30°，ACDC=23，··········∴eq\o\ac(△,Rt)BC，AB=4，··················分∴BD3.

F

D∵∠ACB=90°∴∠ABC=60°，∴∠=∠60°，····················3分∴∠=90°，∴=（23.··················4分

E第題圖1

（）如圖2過作⊥ABCN⊥DE垂分別為，∴∠=∠DNC=90·························分又∵DC，A∠

6分∴△ACM≌DCN，∴=，·······································7分∴FC平8分注此的鍵是角分線理

第題圖2（）如圖3過作⊥，過點作EG⊥FC，垂足分為點HG，∴∠BHC=EGC∠BHF∠EGF°由（）得∠∠BFC=45，∴BH=FH=

，F(xiàn)GEG=EF.···········分2∵∠BCH+∠90°，∠CEG+ECG=°，九年級數(shù)學試題第9頁共頁）

第題圖

222122222222122222∴∠BCH＝∠CEG.·······························∵EC，∴△≌△，·····························分∴EGBHCG∵=∴

，2即EF+FB=2

·································注此的鍵是“2與45°關(guān)”理解．解：（1）因為

=(x2

，所以頂點A的坐為（,-2

分（2）頂點（a,-

+在拋物線C上令x=a，則拋物線的解析式為：

2

.·········································5分因為將拋物線C向平b(b)單位后，所以所得拋物線點B坐標為（,-a+a·6分因為點B仍在拋物線上，所以))即（ba.

整理得b，·····························

8分又因為＞，所以b0.····························································9分（3）為拋物線C：

①頂點為（

1