上海市2016青浦区初三数学一模试卷(含答案)

静安区2015学年第一学期期末教学质量调研

九年级数学试卷 2016.1

（完成时间：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．

12的相反数是

（A）2； （B）2； （C）2．下列方程中，有实数解的是

（A）x2x10； （B）x21x； （C）3．化简(x11)1的结果是 （A）

22； （D）． 221x1x； （D）01．

x2xx2xxx； （B）； （C）x1； （D）1x． 1xx14．如果点A（2，m）在抛物线yx2上，将此抛物线向右平移3个单位后，点A同时平移到点A’ ，那么A’坐标为

（A）（2，1）； （B）（2，7） （C）（5，4）； （D）（－1，4）． 5．在Rt△ABC中，∠C＝90°，CD是高，如果AD＝m，∠A=, 那么BC的长为 （A）mtancos； （B）mcotcos； （C）

mtanmtan； （D）．

sincos6．如图，在△ABC与△ADE中，∠BAC=∠D，要使△ABC与△ADE相似，还需满足下列条件中的 （A）（C）

ACBCACAB； （B）； ADDEADAEACABACBC； （D）． ADDEADAED A E C B （第6题图） 二、填空题：（本大题共12题，每题4分，满分48分） 7．计算：(2a2)3 ▲ ． 8．函数f(x)x3的定义域为 ▲ ． x29．方程x5x1的根为 ▲ ．

10．如果函数y(m3)x1m的图像经过第二、三、四象限，那么常数m的取值范围

为 ▲ ．

11．二次函数yx26x1的图像的顶点坐标是 ▲ ．

12．如果抛物线yax22ax5与y轴交于点A，那么点A关于此抛物线对称轴的对称点坐

标是 ▲ ．

13．如图，已知D、E分别是△ABC的边AB和AC上的点，DE//BC，BE与

CD相交于点F，如果AE=1，CE=2，那么EF∶BF等于 ▲ .

D F C

（第13题图）

A E 114．在Rt△ABC中，∠C＝90°，点G是重心，如果sinA，BC＝2，

3B

那么GC的长等于 ▲ .

15．已知在梯形ABCD中，AD//BC，BC=2AD，设ABa，BCb，那么CD ▲ ．（用向量a、

b的式子表示）；

16．在△ABC中，点D、E分别在边AB、AC上，∠AED＝∠B，AB＝6，BC＝5，AC＝4，

如果四边形DBCE的周长为

25，那么AD的长等于 ▲ . 2B A D 17．如图，在□ABCD中，AE⊥BC，垂足为E，如果AB=5， 4BC=8，sinB．那么tanCDE ▲ ．

5E （第17题图）

C 18. 将□ABCD（如图）绕点A旋转后，点D落在边AB上的点D’，点C落到C’，且点C’、B、C在一直线上，如果AB=13，AD=3，那么∠A的余弦值为 ▲ ．

19．（本题满分10分）

D A

C B

（第18题图）

三、解答题：（本大题共7题，满分78分）

x2x6x26x9化简：，并求当x32时的值． 22x4x2x

20．（本题满分10分）

用配方法解方程：2x23x30．

21．（本题满分10分, 其中第（1）小题6分，第（2）小题4分））

1

4，第一象限内的点B在这个反比 x与反比例函数的图像交于点A（3，a）

31例函数图像上，OB与x轴正半轴的夹角为，且tan．

3y （1）求点B的坐标； 如图，直线y（2）求△OAB的面积．

22．（本题满分10分）

A B O （第21题图）

x 如图，从地面上的点A看一山坡上的电线杆PQ，测得杆顶端点P的仰角是26.6°，向前 走30米到达B点，测得杆顶端点P和杆底端点Q的仰角分别是45°和33.7°．求该电线杆PQ的高度（结果精确到1米）．

(备用数据：sin26.60.45,cos26.60.89,tan26.60.50,cot26.62.00，

sin33.70.55,cos33.70.83,tan33.70.67,cot33.71.50．)

23．（本题满分12分，其中每小题6分）

已知：如图，在△ABC中，点D、E分别在边BC、AB上，BD=AD=AC，AD与CE相交 于点F，AE2EFEC．

（1） 求证：∠ADC=∠DCE+∠EAF；

（2） 求证：AF·AD=AB·EF． 24．（本题满分12分，其中每小题6分）

如图，直线y直线yB

E

F D

（第23题图）

C A A P Q

B

（第22题图）

1x1与x轴、y轴分别相交于点A、B，二次函数的图像与y轴相交于点C，与21x1相交于点A、D，CD//x轴，∠CDA=∠OCA． 2（1） 求点C的坐标；

C y D

（2） 求这个二次函数的解析式．

（第24题图）

25．（本题满分14分，其中第（1）小题4分，第（2）、（3）小题各5分）

4，点E在对角线AC上，且CE＝5AD，BE的延长线与射线AD、射线CD分别相交于点F、G．设AD=x，△AEF的面积为y．

（1）求证：∠DCA＝∠EBC；

（2）如图，当点G在线段CD上时，求y关于x的函数解析式，并写出它的定义域； （3）如果△DFG是直角三角形，求△AEF的面积．

A D F

已知：在梯形ABCD中，AD//BC，AC＝BC＝10，cosACB

G

E B

（第25题图）

C

静安区2015学年第一学期期末教学质量调研 九年级数学试卷

参考答案及评分说明2016.1

一、选择题：

1．D； 2．D； 3．A； 4．C； 5．C； 6．C． 二、填空题：

7．8a6； 8．x2； 9．x4； 10．1m3； 11．（3, -8）； 12．（2, 5）； 13．

1115； 14．2； 15．ab； 16．2； 17．； 18．． 32132三、解答题：

(x2)(x3)(x3)219．解：原式＝ ············································································（4分）

(x2)(x2)x(x2)＝

(x2)(x3)x(x2) ················································································（1分）

(x2)(x2)(x3)2

＝

x． ········································································································（2分） x3132当x20．解：x23时，原式＝

3113． ·································（3分）

2331333····································································································（1分） x0, ·

2233············································································································（1分） x2x,·

223339······················································································（2分） x2x()2, ·

24216333, ··········································································································（2分） (x)2416333, ··········································································································（2分） x44333333,x2． ··············································································（2分） x12424

21．解：（1）∵直线y∴a4， x与反比例函数的图像交于点A（3，a）

34． ······························································（1分） 3=4，∴点的坐标A（3，4）

3k设反比例函数解析式为y， ·············································································（1分）

xk12∴4,k12，∴反比例函数解析式为y． ···········································（1分）

3x过点B作BH⊥x轴，垂足为H， 由tan∴mBH1·······················（1分） ，设BH=m，则OB=3m，∴B（3m，m） ·

OB312，m2（负值舍去）， ······································································（1分） 3m∴点B的坐标为（6，2）． ······················································································（1分）

（1） ···································· 过点A

作AE⊥x轴，垂足为E，

SOABSOAES梯形AEHBSOBH ············································································（1分）

111···············································（1分） AEOE(AEBH)EHOHBH ·

222111=34(42)3629． ······················································（2分） 222=

22．解：延长PQ交直线AB于点H，由题意得．

由题意，得PH⊥AB，AB=30，∠PAH=26 .6°，∠PBH=45°，∠QBH=33.7°， 在Rt△QBH中，cotQBHBH····················（2分） 1.50，设QH=x，BH=1.5x， ·

QH在Rt△PBH中，∵∠PBH=45°，∴PH= BH=1.5x， ···············································（2分） 在Rt△PAH中，cotPAHAH··································（2分） 2.00，AH=2PH=3x， ·

PH∵AH–BH=AB，∴3x1.5x30，x20． ·························································（2分） ∴PQ=PH–QH=1.5xx0.5x10． ·····································································（1分） 答：该电线杆PQ的高度为10米． ·················································································（1分）

23．证明：（1）∵AE2EFEC，∴

AEEC． ··························································（1分） EFAE又∵∠AEF=∠CEA，∴△AEF∽△CEA． ·······················································（2分） ∴∠EAF=∠ECA， ····························································································（1分） ∵AD=AC，∴∠ADC=∠ACD， ·······································································（1分） ∵∠ACD =∠DCE+∠ECA=∠DCE+∠EAF． ·····················································（1分）

（2）∵△AEF∽△CEA，∴∠AEC=∠ACB． ·······························································（1分）

∵DA=DB，∴∠EAF=∠B． ················································································（1分） ∴△EAF∽△CBA． ·····························································································（1分）

AFEF． ···································································································（1分） BAACAFEF∵AC=AD，∴． ················································································（1分） BAAD∴

∴AFADABEF． ····················································································（1分）

24．解：（1）∵直线y1x1与x轴、y轴分别相交于点A、B， 2∴A（–2，0）、B（0，1）．∴OA=2，OB=1． ······················································（2分） ∵CD//x轴，∴∠OAB=∠CDA，∵∠CDA=∠OCA，∴∠OAB=∠OCA． ··············（1分） ∴tan∠OAB=tan∠OCA， ·························································································（1分）

OBOA12，∴， ··················································································（1分） 2OCOAOC∴OC4，∴点C的坐标为（0，4）． ································································（1分）

CDBC（2）∵CD//x轴，∴．··················································································（1分） AOBO∴

∵BC=OC–OB=4–1=3,∴

CD3． ·························（1分） ，∴CD=6，∴点D（6，4）

21设二次函数的解析式为yax2bx4， ····························································（1分）

1a,04a2b4,4 · ………………（1分） ········································（1分） 3436a6b4,b.2∴这个二次函数的解析式是yx2

25．解：（1）∵AD∥BC，∴∠DAC=∠ECB． ········································································（1分）

又∵AD=CE，AC=CB，∴△DAC≌△ECB． ······························································（2分） ∴∠DCA＝∠EBC． ···································································································（1分） （2）过点E作EH⊥BC，垂足为H．AE=AC–CE=10x．

143·················································（1分） x4． ·

2CHCEcosACBSCBE43···························································（1分） x，∴EH=x． ·

55113·······························································（1分） BCEH10x3x． ·

225SAEFAE2()， ·············································（1分） SCEBCE∵AF//BC．∴△AEF∽△CEB，∴

y(10x)23x260x300∴，∴y． ······················································（1分） 3xxx2定义域为0x555． ·····················································································（1分） （3）由于∠DFC=∠EBC<∠ABC, 所以∠DFC不可能为直角．

（i）当∠DGF=90°时，∠EGC=90°，由∠GCE=∠GBC，可得△GCE∽△GBC．

∴tanGBCCGCEx····································································（1分） ． ·

GBCB103xEH3x5在Rt△EHB中， tanGBC． ···························（1分）

4BH504x10x5x3x，解得x0(舍去),或x5． 10504x352605300∴SAEF····························································（1分） 15． ·

5∴

（ii）当∠GDF=90°时，∠BCG=90°，由△GCE∽△GBC，

可得∠GEC=90°，∠CEB=90°， ·····································································（1分）

33，SAEF． ············································（1分） 223综上所述，如果△DFG是直角三角形，△AEF的面积为15或．

2可得BE=6，CE=8，AE=2，EF=