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上海市青浦区2017年中考二模数学试卷(含答案)

上海市青浦区2017年中考二模数学试卷(含答案)
上海市青浦区2017年中考二模数学试卷(含答案)

100.5

80.5

60.5

40.5

图1

青浦区2017学年九年级第二次学业质量调研测试

数学试卷 2018.4

(满分150分,100分钟完成)

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、本调研卷上答题一律无效.

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)

[每小题只有一个正确选项,在答题纸相应题号的选项上用2B 铅笔正确填涂] 1.下列实数中,有理数是( ▲ ) (A ;

(B )2.1;

(C )π;

(D )1

3

5.

2.下列方程有实数根的是( ▲ )

(A )4+2=0x ; (B 1-; (C )2+21=0x x -;(D )1

11

x x x =

--. 3.已知反比例函数1

y x

=

,下列结论正确的是( ▲ ) (A )图像经过点(-1,1);

(B )图像在第一、三象限;

(C )y 随着x 的增大而减小; (D )当1x >时,1y <. 4.用配方法解方程241=0x x -+,配方后所得的方程是( ▲ )

(A )2(2)=3x -; (B )2(+2)=3x ; (C )2(2)=3x --;(D )2(+2)=3x -. 5. “a 是实数,2

0a ≥”这一事件是( ▲ )

(A )不可能事件; (B )不确定事件; (C )随机事件; (D )必然事件. 6. 某校40名学生参加科普知识竞赛(竞赛分数都是整数),

竞赛成绩的频数分布直方图如图1所示,成绩的中位数落在( ▲ )

(A )50.5~60.5分; (B )60.5~70.5分; (C )70.5~80.5分; (D )80.5~90.5分.

二、填空题:(本大题共12题,每题4分,满分48分)

[在答题纸相应题号后的空格内直接填写答案] 7.计算:32()=a a ÷- ▲ . 8.因式分解:24=a a - ▲ . 9.

函数y 的定义域是 ▲ .0

10.不等式组1020.x x +≥??->?

的整数解是 ▲ .

11.关于x 的方程=2(1)ax x a +≠的解是 ▲ . 12.抛物线2(3)+1y x =-的顶点坐标是 ▲ .

13.掷一枚材质均匀的骰子,掷得的点数为合数的概率是 ▲ .

14.如果点1P (2,1y )

、2P (3,2y )在抛物线2+2y x x =-上,那么1y ▲ 2y .(填“>”、 “<”或 “=”)

15.如图2,已知在平行四边形ABCD 中,E 是边AB 的中点,F 在边AD 上,且AF ︰FD=2

︰1,如果AB a =,BC b =,那么EF = ▲ .

16.如图3,如果两个相似多边形任意一组对应顶点P 、P '所在的直线都经过同一点O ,

且有(0)OP k OP k '=?≠,那么我们把这样的两个多边形叫位似多边形,点O 叫做位似中心.已知ABC ?与A B C '''?是关于点O 的位似三角形,3OA OA '=,则ABC ?与

A B C '''?的周长之比是 ▲ .

17.如图4,在△ABC 中,BC=7,AC

=,tan 1C =,点P 为AB 边上一动点(点P

不与点B 重合),以点P 为圆心,PB 为半径画圆,如果点C 在圆外,那么PB 的取值范围是 ▲ .

18.已知,在Rt △ABC 中,∠C =90°,AC =9, BC =12,点D 、E 分别在边AC 、BC 上,且

CD ︰CE =3︰4.将△CDE 绕点D 顺时针旋转,当点C 落在线段DE 上的点F 处时,BF

恰好是∠ABC 的平分线,此时线段CD 的长是 ▲ .

图3

A

B

C

D

E F

图 2

图4

P

O

P'

三、解答题:(本大题共7题,满分78分)[将下列各题的解答过程,做在答题纸的相应位置上]

19.(本题满分10分)

计算:1

01

2

152(3)2

-+

--+().

20.(本题满分10分)

先化简,再求值:2

5+3222x x x x ?

?--÷

?++??

()

,其中x =

21. (本题满分10分,第(1)、(2)小题,每小题5分)

如图5,在Rt △ABC 中,∠C =90°,AC=3,BC =4,∠ABC 的平分线交边AC 于点D ,延长BD 至点E ,且BD=2DE ,联结AE . (1)求线段CD 的长; (2)求△ADE 的面积.

22.(本题满分10分)

如图6,海中有一个小岛A ,该岛四周11海里范围内有暗礁.有一货轮在海面上由西向正东方向航行,到达B 处时它在小岛南偏西60°的方向上,再往正东方向行驶10海里后恰好到达小岛南偏西45°方向上的点C 处.问:如果货轮继续向正东方向航行,是否会有触礁的危险?

(参考数据:

1.41

1.73)

23.(本题满分12分,第(1)、(2)小题,每小题6分)

如图7,在梯形ABCD 中,AD ∥BC ,对角线AC 、BD 交于点M ,点E 在边 BC 上,且

DAE DCB ∠=∠,联结AE ,AE 与BD 交于点F .

M

F

E D C

B

A

A

B C

图6

E

D C B

A

图5

(1)求证:2DM MF MB =?; (2)联结DE ,如果3BF FM =,

求证:四边形ABED 是平行四边形.

24.(本题满分12分,第(1)、(2)、(3)小题,每小题4分)

已知:如图8,在平面直角坐标系xOy 中,抛物线2

3y ax bx =++的图像与x 轴交于点

A (3,0),与y 轴交于点

B ,顶点

C 在直线2x =上,将抛物线沿射线AC 的方向平移,当顶点C 恰好落在y 轴上的点

D 处时,点B 落在点

E 处. (1)求这个抛物线的解析式;

(2)求平移过程中线段BC 所扫过的面积;

(3)已知点F 在x 轴上,点G 在坐标平面内,且以点C 、E 、F 、G 为顶点的四边形是矩形,求点F 的坐标. .

25.(本题满分14分,第(1)小题4分,第(2)小题6分,第(3)小题4分)

如图9-1,已知扇形MON

,∠MON =90,点B 在弧MN 上移动,联结BM ,作OD ⊥BM ,垂足为点D ,C 为线段OD 上一点,且OC =BM ,联结BC 并延长交半径OM 于点A ,设OA = x ,∠COM 的正切值为y . (1)如图9-2,当AB ⊥OM 时,求证:AM =AC ; (2)求y 关于x 的函数关系式,并写出定义域; (3)当△OAC 为等腰三角形时,求x 的值.

O

M

N

D

C

B

A

O

M

N

D

C

B

A

N

M

O

青浦区2017学年九年级第二次学业质量调研测试评分参考

一、选择题:

1.B ; 2.C ; 3.B ; 4.A ; 5.D ; 6.C . 二、填空题:

7.a ; 8.()4-a a ; 9.3≥-x ; 10.101、、-; 11. 2

1

-a ; 12.(3,1); 13.13; 14.>; 15.2132-b a ; 16.1︰3; 17.3508

<

三、解答题:

19.解:原式212-+. ································································ (8分)

=1. ············································································· (2分)

20.解:原式=()

22

452

23--+?++x x x x , ····························································· (5分) =

()()()

2

332

2

3+-+?

++x x x x x , ······················································· (1分)

=

33

-+x x . ·················································································· (1分)

当=

x 2. ············································ (3分) 21.解:(1)过点D 作DH ⊥AB ,垂足为点H . ················································ (1分)

∵BD 平分∠ABC ,∠C =90°, ∴DH = DC =x , ········································································ (1分) 则AD =3-x . ∵∠C =90°,AC=3,BC =4,∴AB =5. ··········································· (1分)

∵sin ∠==

HD BC

BAC AD AB

, ∴

4

35

=-x x , ·

········································································· (1分)

∴43=x . ················································································ (1分) (2)11410

52233

=?=??=ABD

S

AB DH . ·

············································ (1分) ∵BD=2DE , ∴

2=

=ABD ADE

S BD

S

DE

, ································································ (3分) ∴1015

323

=

?=ADE

S

. ·

······························································ (1分) 22.解:过点A 作AH ⊥BC ,垂足为点H . ······················································ (1分)

由题意,得∠BAH =60°,∠CAH =45°,BC =10. ····································· (1分) 设AH =x ,则CH =x . ······································································· (1分) 在Rt △ABH 中,

∵tan ∠=

BH BAH AH ,∴10tan 60+?=x

x

, ········································ (3分)

10=+x

,解得513.65=≈x , ······································· (2分) ∵13.65>11, ················································································ (1分)

∴货轮继续向正东方向航行,不会有触礁的危险. ································· (1分) 答:货轮继续向正东方向航行,不会有触礁的危险.

23.证明:(1)∵AD //BC ,∴∠=∠DAE AEB , ·············································· (1分)

∵∠=∠DCB DAE ,∴∠=∠DCB AEB , ································ (1分) ∴AE //DC , ··········································································· (1分)

=

FM AM

MD MC

. ··································································· (1分) ∵AD //BC ,∴=

AM DM

MC MB

, ··················································· (1分) ∴=FM DM MD MB

, ··································································· (1分) 即2

=?MD MF MB .

(2)设=FM a ,则=3BF a ,=4BM a . ··········································· (1分)

由2

=?MD MF MB ,得2

4=?MD a a ,

∴2=MD a , ········································································ (1分)

∴3==DF BF a . ·································································· (1分) ∵AD //BC ,∴

1==AF DF

EF BF

, ·

··················································· (1分)

∴=AF EF , ········································································· (1分) ∴四边形ABED 是平行四边形. ···················································· (1分)

24.解:(1)∵顶点C 在直线2x =上,∴22=-

=b

x a

,∴4=-b a . ·

··············· (1分) 将A (3,0)代入2

3y ax bx =++,得933=0++a b , ··················· (1分) 解得1=a ,4=-b . ································································ (1分) ∴抛物线的解析式为2

43=-+y x x . ·········································· (1分) (2)过点C 作CM ⊥x 轴,CN ⊥y 轴,垂足分别为M 、N .

∵2

43=-+y x x =()

2

21=--x ,∴C (2,1-). ························· (1分)

∵1==CM MA ,∴∠MAC =45°,∴∠ODA =45°, ∴3==OD OA . ···································································· (1分) ∵抛物线2

43=-+y x x 与y 轴交于点B ,∴B (0,3),

∴6=BD . ········································································ (1分) ∵抛物线在平移的过程中,线段BC 所扫过的面积为平行四边形BCDE 的面

积,

∴1

2262122

==???=?=BCDE

BCD

S

S

BD CN . ························ (1分)

(3)联结CE .

∵四边形BCDE 是平行四边形,∴点O 是对角线CE 与BD 的交点,

即 OE OC ==

(i )当CE 为矩形的一边时,过点C 作1CF CE ⊥,交x 轴于点1F ,

设点1F a (,0)

,在1Rt OCF 中,222

11=OF OC CF +, 即 22(2)5a a =-+,解得 52a =

,∴点15

2

F (,0) ································ (1分) 同理,得点25

2

F (-,0) ······································································ (1分) (ii )当CE 为矩形的对角线时,以点O 为圆心,OC 长为半径画弧分别交x 轴于点

3F 、4F ,可得 34=OF OF OC ==3F )、4F ()

····· (2分)

综上所述:满足条件的点有15

2F (,0),252

F (-,0),3F )),4F ().

25.解:(1)∵OD ⊥BM ,AB ⊥OM ,∴∠ODM =∠BAM =90°. ··························· (1分)

∵∠ABM +∠M =∠DOM +∠M ,∴∠ABM =∠DOM . ······················ (1分) ∵∠OAC =∠BAM ,OC =BM ,

∴△OAC ≌△ABM , ·································································· (1分) ∴AC =AM . ············································································· (1分)

(2)过点D 作DE //AB ,交OM 于点E .··············································· (1分)

∵OB =OM ,OD ⊥BM ,∴BD =DM . ··········································· (1分) ∵DE //AB ,

=

MD ME

DM AE

,∴AE =EM , ∵OM

,∴AE

=)

1

2

x . ·

·············································· (1分) ∵DE //AB ,

∴2==

OA OC DM

OE OD OD , ···························································· (1分) ∴2=

DM OA OD OE

∴=

y

(0<≤x ····················································· (2分)

(3)(i ) 当OA =OC 时, ∵111

222

=

==DM BM OC x , 在Rt △ODM

中,=

=OD ∵=DM y OD

∴1

=

x

.解得2=x

,或2=x (舍).(2分) (ii )当AO =AC 时,则∠AOC =∠ACO ,

∵∠ACO >∠COB ,∠COB =∠AOC ,∴∠ACO >∠AOC , ∴此种情况不存在. ·································································· (1分) (ⅲ)当CO =CA 时,

则∠COA =∠CAO=α,

∵∠CAO >∠M ,∠M =90α?-,∴α>90α?-,∴α>45?, ∴290α∠=>?BOA ,∵90∠≤?BOA ,∴此种情况不存在. ········ (1分)

2017年中考语文模拟试卷及答案

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