普陀區(qū)2020學年度第二學期初三質量調研數學試卷及答案(二模)

普陀區(qū)學年第二學期初質量調研數學試卷（間100分鐘滿分150分）考注：．試共25題．卷分分考時100分鐘．題，生必答要在題規(guī)的置作，草稿、試上題律效．第、大外其各如特說，必在題的相位上出明計的要驟一選題本題題每分滿分[下列各的個項，且有個項正確，擇確的號填在題的應置下計算中，正確的是（A）

aa；（）2a

a3；（C）

a；（D(2a

)

a

．.下單項式中，以與

y

合并同類項的是（A）

；（B）

3

2

；（）

y；（）2

y

．.方的是（A）x

；（B）

；（C）x

；（D

．．知兩組數據：

、x

、x

、x

、x

和x

、x

、

、

、

，下列有關這兩組數據的說法中，正確的是（A）平均數相等；B）中位數相；（C）眾數相；（）差相等.已eq\o\ac(△,在)eq\o\ac(△,)

eq\o\ac(△,和)eq\o\ac(△,)

中，AB

，AC

，下列條件中，不一定能得到△ABC≌（A）BC（）

；（C）C

；（D）—1—

如1，在平面直角坐標系中eq\o\ac(△,，)eq\o\ac(△,)

的頂點、均y軸，在

軸上，eq\o\ac(△,將)eq\o\ac(△,)ABC繞著頂點旋轉后，點C對應C軸，的對應點比例函數

y

x

在第一象限的圖像上．如果

的坐標分別是(0,、(，么點

的坐標是（A）(3,2)；（C）(2,3)；

（B(,4)（）(4,)

；．二填題本題12題，題分，滿分.因分：a

▲

．.已f(x)

x

，那么f(3)＝

▲

．.不式組

的解集是

▲

．10已正比例函數ykxk

是常數，k的函數值的增大而減小，那的取值范圍是

▲

．.如果于的程x

m有兩個相等的實根，那m

的值等于

▲

．12拋線2ax（a

）的對稱軸是直線

▲

．13為喚起公眾的節(jié)水意識，從1993年，聯合國將每年的月日定為“世界水日.某居委會表彰了社區(qū)內戶節(jié)約用水的家庭月這戶庭節(jié)約用水的情況如下表所示，那么5月這100戶庭節(jié)水量的平均數是每戶節(jié)水量（單位：噸）

▲

噸

7.2節(jié)水戶戶數

14小已有兩根長度分別是和的細竹簽里四根長度分別是4cm7cm、的細竹簽，小明從盒子里意抽取一根細竹簽，恰能與已有的兩根細竹簽首尾順次聯結組成三角形的概率等于

▲

.15如2，兩條平行l(wèi)1

、l

2

分別經過正五邊形ABCDE

的頂點B、C

．如果1=20么=

▲

.—2—

3316如3，已eq\o\ac(△,知)eq\o\ac(△,)ABC

中，、E分為邊AB、

的中點，點F在DE的長線上，EF，

，AF

，那么向量用量、表示是

▲

．17已等腰三角形ABC中，A為心為半徑長作

A以為心

為半徑作B如果A與內，那eq\o\ac(△,么)eq\o\ac(△,)ABC

的面積等于

▲

．18如4形ABCDAB

為邊

的中點

在AE

上F

作MN

，分別交邊、DC

于點、

．聯結

，如△

是以CN

為底邊的等腰三角形，那么FC

▲

．三解題本題題滿分）19題滿分10分）計算：

32720題滿分10分）解方程

x24xxx

．21題滿分10分，第1)小題滿分5分第2)題滿分5）在平面直角坐標系

xOy

如圖5線分與x

軸軸于AB，一個正比例函數的圖像與這直線交于C

，C

的橫坐標是．（1求正比例函數的解析式；（2將正比例函數的圖像向上或向下平移，交直線yx點，平移所得函數圖像的截距

，如果交點D始落在線段上，求b

的取值范圍—3—

22題滿分10分，第1)小題滿分5分第2)題滿分5）如圖6-1扇戶打開后可以用窗鉤將其固定鉤一個端點A固在窗戶底邊O上轉軸底

之間的距離的一個端點B可窗框邊上的滑F

上移動，滑槽OF

的長度為，AB、

、AO

構成一個三角形．當窗鉤端點B與

之間的距離是7

cm位置時（如圖6-2），窗戶打開AOB的數為（1求窗鉤AB的長度精確到；（2需要將窗戶打開的角AOB

的度數調整到45

時求此時窗鉤端點B與

之間的距離精確到．（參考數據sin37，cos37，，）23題滿分12分，第1)小題滿分6分第2)題滿分6）已知如圖7□ABCD中點

E

、

F

分別在邊

邊

的延長線上邊AEFD是菱形，菱形的對角線AF分別交DE、求證1）四邊形為矩形；FQ．

于點Q，

．—4—

24題滿分12分，第1)小題滿分4分第2)題滿分4，(3)小題滿分）在平面直角坐標系

xOy

如拋物線y

12

與軸于點A

、軸于點，點D是第四象限內拋物線上的一個動點，直線AD與線交于點E．（1、c的值和直線BC的達式；（2設CAD點E的標；（3設點D的坐標為d，用含的數式表示△與DCE的積比．—5—

25本滿分分，第小題滿分4分第2)小題滿分，第3)小題滿分4分在梯形中，AD∥BC，BC，AD，CD，

35

(如圖9)M是邊上個動（與點、C重為心，CM為徑作圓，M與線、射線分相交于點E、（1設

18CE，證：四邊形AMCD是行四邊形；5（2聯結EM，設FMBEMC，CE的；（3以D為圓心DA為徑作圓，D與M的共弦恰好經過梯形的一個頂點，求此時M的徑長．備用圖備用圖2—6—

普區(qū)2020學度二期三量研學卷參答及分明一選題本大題共6，每題4分滿分24分．(C)；

．；

．(D)；

．(D)

．；

．(A)．二填題本大題共題，每題，滿分分）．

．

；

．x＜；．k＜0；

11；

12

；．；

1．；4

1592．b；

．7

；

18

5

．三解題（大題共題其中第19---22題每題分第23每題12分第題分滿分分）19解：原=

3

····························································8分=

3

．······················································································（）20解：

2x

24x

．······························································（）去分母，得

x

x

x．···········································2分化簡，得

x21．······························································（）解得x7．····································································）12經檢驗：是方的根，x增，舍去．·····························（1分所以，原方程的根是x．······························································（1）—7—

121解）直線y經點C，2∴把

x

1代入yx得y2

．···············································（2分設所求正比例函數的解析式為kx(k．·········································（）∵正例函數的像經過點C∴把

x

，y

代入，得k

．···········································（）∴正例函數的解析式為

32

．······················································（）1（2由直線x別與2

軸、y軸交于點、B，可得·····································································（2分設平移所得函數的解析式為

32

．因為交點D始終落在線段上，所以由平移所得函數

32

的像經過線段AB的點B，得b．（）由平移所得函數y

32

的像經過線段AB的點，得···（）∴b的值范圍為2

．····························································（）22解）據題意，可知AOB37OA20cm，OBcm．過點作AH⊥OF垂足為點．······················································（）在eq\o\ac(△,Rt)eq\o\ac(△,)OAH中，∵sin

AH

，∴AH20sin37．······································）同理可得．···································································（1分由OB，BH(cm)·······························································（1分在eq\o\ac(△,Rt)eq\o\ac(△,)ABH中AB

AH

12

(cm)．·······················）答：窗鉤的長度約等于．（2根據題意，可知AOBOA，ABcm．過點作AG⊥OF，足為點．在eq\o\ac(△,Rt)eq\o\ac(△,)中AG

2220=10222

．···················（2）在eq\o\ac(△,Rt)eq\o\ac(△,)，由勾股定理得BG(cm)．·········································（）—8—

因為OFcm，所以OB14．·····························（2分）答：此時窗鉤端點B點O之的距離約等于．23證明）

AFBF，PFEF

．·············································（）∵AFBEFP∴AFB∽．···········································（）∴．··········································································（）∵四形AEFD是形，∴AFDE．···············································（）即∴90··············································································（1分∵四形ABCD是平行四邊形，∴四形是形．·······················（1分（2證法一：∵四形是形，∴ADBC，DCB90·······（）∵QCF．又∵PQD，DPQ∽．······································（）∴

DP．·················································································（）FCFQ∵四形AEFD是形，∴ADEF，PE．·······························1分∴．BECF．······························································（）∴

PEDQ．BEFQ即BEFQPE．········································································（）證法二：聯結．∵四形AEFD是形，∴AP，DFQPFE．·······················（）∵FP．∴PFE．∴PBF．··········································································（1分∵四形ABCD是矩形，∴90·········································（）∴DQP．∴DEB．·········································································（）∴FQD．·······································································（）∴

FQDQ．BEPE—9—

即BEFQPE．········································································（）124解）拋物線yx2

與x交于點拋物線的表達式是y

12

··················································（）即：拋物線的表達式是

12

2．因此，b······································································（）可得點C的坐標為可設直線BC的表達式為ykx．因為直線BC過點B，得，解得．···································（）因此，直線BC的達式為····················································（1分（2由45

，∵45ABC．····················································（1分又∵ACB為共角，eq\o\ac(△,∴)eq\o\ac(△,)ACE∽△BCA．··········································（）得

CB∵，CB2，∴CE

2

．·············································（1分由點線段上可設點E的標為由兩點距離公式

(

[66)]2

，得m2=

．解得．8∴點坐標為,3

．·································································（）SAE（3eq\o\ac(△,∵)eq\o\ac(△,)ACE與同，∴ACE．··································（）SED過點作AMAB直于點D作足為點與線相交于點N．可得∥DN．∴

AEAMDE

．························································（1分1由題意得AM，DNd2

d，··················································（1分可得

AEDE2d

．·······································································（）—10—

CHrrCHrr即△ACE與的積比等于

162d

．．解）過點M作，足為點．································（1分由垂徑定理可得CH

9．························································（1分在eq\o\ac(△,Rt)eq\o\ac(△,)中由C

，可得．得CM．···············（）CM∵AD，ADCM．∵∥BC，四邊形AMCD是行四邊形．·····································（）（2設M的半徑長為r．3在eq\o\ac(△,Rt)eq\o\ac(△,)中CHr．可r．··········