2016年静安区九年级数学一模试卷
静安区2015学年第一学期九年级期末学业质量调研测试
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.
2
1的相反数是
(A )2; (B )2-; (C )22; (D )2
2
-. 2.下列方程中,有实数解的是
(A )012=+-x x ; (B )x x -=-12; (C )012=--x x x ; (D )112=--x
x x
. 3.化简11)1(---x 的结果是 (A )
x x
-1; (B )1
-x x ; (C )1-x ; (D )x -1. 4.如果点A (2,m )在抛物线2x y =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 =α, 那么BC 的长为 (A )ααcos tan ??m ; (B )ααcos cot ??m ; (C )
α
αcos tan ?m ; (D )αα
sin tan ?m .
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 )
AE
AB
AD AC =; (B )DE BC AD AC =
; (C )DE AB AD AC =; (D )AE
BC
AD AC =
. (第6题图)
E
二、填空题:(本大题共12题,每题4分,满分48分) 7.计算:=-32)2(a ▲ . 8.函数2
3
)(+-=
x x x f 的定义域为 ▲ . 9.方程15-=+x x 的根为 ▲ .
10.如果函数m x m y -+-=1)3(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162+-=x x y 的图像的顶点坐标是 ▲ .
12.如果抛物线522+-=ax ax y 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 ▲ .
14.在Rt △ABC 中,∠C =90°,点G 是重心,如果31
sin =A ,BC =2,
那么GC 的长等于 ▲ .
15.已知在梯形ABCD 中,AD //BC ,BC =2AD ,设a AB =,b BC =,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10,那么AD 17.如图,在□ABCD 中,AE ⊥BC ,垂足为E ,如果BC =8,5
4
sin =
B .那么=∠CDE tan ▲ . 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,A
D =3,那么∠A 的余弦值为 ▲ .
(第17题图)
B
A C
E
D (第13题图)
F
(第18题图)
D A
B
C
三、解答题:(本大题共7题,满分78分) 19.(本题满分10分)
化简:x
x x x x x x 29
6462
222-+-÷---,并求当21
3=x 时的值.
20.(本题满分10分)
用配方法解方程:03322=--x x .
21.(本题满分10分, 其中第(1)小题6分,第(2)小题4分))
如图,直线x y 3
4
=与反比例函数的图像交于点A (3,a )
,第一象限内的点B 在这个反比 例函数图像上,OB 与x 轴正半轴的夹角为α,且3
1
tan =α.
(1)求点B 的坐标; (2)求△OAB 的面积. 22.(本题满分10分)
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:00.26.26cot ,50.06.26tan ,89.06.26cos ,45.06.26sin =?=?=?=?,
50.17.33cot ,67.07.33tan ,83.07.33cos ,55.07.33sin =?=?=?=?.)
(第21题图)
A
B
Q
P (第22题图)
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 ,EC EF AE ?=2.
(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12
1
+=x y 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 ,与直线12
1
+=
x y 相交于点A 、D ,CD //x 轴,∠CDA =∠OCA . (1) 求点C 的坐标; (2) 求这个二次函数的解析式.
25.(本题满分14分,其中第(1)小题4
已知:在梯形ABCD 中,AD //BC ,AC =BC =10,5
4
cos =
∠ACB ,点E 在对角线AC 上,且CE =AD ,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
B
F E
(第23题图)
A B C
D F G
E (第25题图)
静安区2015学年第一学期期末学业质量调研 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:
7.68a -; 82x ≠-.; 9.4=x ; 10.13m <<; 11.(3, -8); 12.(2, 5); 13.
31; 14.2; 15.b a 2
1--; 16.4或无解; 17.21; 18.135
. 三、解答题:
19.解:原式= )
2()3()2)(2()3)(2(2
--÷
-+-+x x x x x x x ························································· (4分) =
2
)
3()
2()2)(2()3)(2(--?-+-+x x x x x x x ···························································· (1分) =
3
-x x
. ·············································································· (2分) 当332
1==x
时,原式=
2
3
13
113
33+-
=-=
-. ························ (3分) 20.解:2
33
022x x -
-=, ·································································· (1分) 23322x x -=, ········································································ (1分)
223339()24216x x -+=+, ························································ (2分)
2333()416
x -=, ······································································ (2分)
34x -
=·················································································· (2分)
1x =
2x = ······························································· (2分)
21.解:(1)∵直线x y 3
4
=
与反比例函数的图像交于点A (3,a )
, ∴33
4
?=
a =4,∴点的坐标A (3,4)
. ·············································· (1分) 设反比例函数解析式为x
k
y =, ·························································· (1分)
∴12,34==k k ,∴反比例函数解析式为x
y 12=. ································ (1分)
过点B 作BH ⊥x 轴,垂足为H , 由3
1
tan ==OB BH α,设BH =m ,则OB =m 3,∴B (m 3,m ) ·
··············· (1分) ∴m
m 312
=
,2±=m (负值舍去), ···················································· (1分) ∴点B 的坐标为(6,2). ································································ (1分)
(2)过点A 作AE ⊥x 轴,垂足为E ,
OBH AEHB OAE OAB S S S S ???-+=梯形 ·
························································ (1分) =BH OH EH BH AE OE AE ?-?++?21
)(2121 ·
·································· (1分) ==??+?++??262
1
3)24(2143219. ·········································· (2分)
22.解:延长PQ 交直线AB 于点H ,由题意得.
由题意,得PH ⊥AB ,AB =30,∠P AH =26 .6°,∠PBH =45°,∠Q BH =33.7°, 在Rt △QBH 中,50.1cot ==
∠QH
BH
QBH ,设QH =x ,BH =x 5.1, ············· (2分) 在Rt △PBH 中,∵∠PBH =45°,∴PH = BH =x 5.1, ······························· (2分) 在Rt △P AH 中,00.2cot ==
∠PH
AH
PAH ,AH =2PH =x 3, ·
······················· (2分) ∵AH –BH =AB ,∴305.13=-x x ,20=x . ········································· (2分) ∴PQ =PH –QH =105.05.1==-x x x . ·················································· (1分) 答:该电线杆PQ 的高度为10米. ···························································· (1分)
23.证明:(1)∵EC EF AE ?=2,∴
AE
EC
EF AE =
. ······································· (1分) 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)
∴
AC
EF
BA AF =
. ·········································································· (1分) ∵AC =AD ,∴AD
EF
BA AF =
. ··························································· (1分) ∴EF AB AD AF ?=?. ······························································· (1分)
24.解:(1)∵直线12
1
+=
x y 与x 轴、y 轴分别相交于点A 、B , ∴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) ∴
OC
OA OA OB =
,∴OC 2
21=, ······························································ (1分) ∴4=OC ,∴点C 的坐标为(0,4). ················································ (1分) (2)∵CD //x 轴,∴
BO
BC
AO CD =
. ····························································· (1分) ∵BC =OC –OB=4–1=3,∴1
3
2=CD ,∴CD =6,∴点D (6,4)
. ·············· (1分) 设二次函数的解析式为42++=bx ax y , ············································· (1分)
???++=+-=,46364,4240b a b a ………………(1分) ??
???
=-=.
23,41b a ······························ (1分) ∴这个二次函数的解析式是42
3
41
2++-=x x y . ·
···································· (1分)
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 =x -10.
x ACB CE CH 54cos =∠?=,∴EH =x 5
3
. ·
··········································· (1分) x x EH BC S CBE
35
3
102121=??=?=?.
················································ (1分) ∵AF//BC .∴△AEF ∽△CEB ,∴
2
)(CE
AE S S CEB AEF =?, ······························· (1分) ∴2
2
)10(3x x x y -=,∴x x x y 3006032+-=. ········································ (1分)
定义域为05x <≤. ······························································· (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 .
∴10
tan x
CB CE GB CG GBC ===
∠.
···················································· (1分) 在Rt △EHB 中, x x x
x
BH EH GBC 45035
1053tan -=-==∠. ·················· (1分)
∴
x
x
x 450310-=
,解得),(0舍去=x 或5=x . ∴155
300
560532=+?-?=?AEF S . ·············································· (1分)
(ii )当∠GDF =90°时,∠BCG =90°,
可得∠GEC =90°,∠CEB =90°, ················································· (1分)
可得BE =6,CE =8,AE =2,EF =
23,2
3
=?AEF S . ······························ (1分) 综上所述,如果△DFG 是直角三角形,△AEF 的面积为15或2
3
.