2024-2025学年鞍山市铁东区九年级下册3月月考数学试卷 （PDF版，含答案）

2024-2025 学年第二学期九年质量监测九年数学
答案及给分标准
一、选择题（每小题 3 分，共 30 分）
1．D 2．C 3．A 4．B 5．D 6．C 7．A 8．C 9．C 10．B
二、填空题（每小题 3 分，共 15 分）
11 50 169． x 3 12． 2 3 2 13． 14．85 15．
3 12
三、解答题
16、（1） 1 3 3 12 2 0
1 3 2 3 1 ······················································································· 2分
3 ·········································································································5分
2 x 2 x
2 4 1
（ ） 2 x 4x 4 x x
x 2 x 2 x 2 1
2 ···········································································2分 x 2 x x
x 2 1
································································································· 4分
x x
x 1
·······································································································5分
x
17、解：（1）设书架上数学书为 x本，语文书为 y本．·················································· 1分
x y 80
············································································ 2分
0.8x 1.2y 82
x 35
解，得： ···············································································3分
y 45
答：略
（2）设数学书有 a本···················································································4分
1.2 10 0.8a≤82··············································································· 5分
a 87 1≤ ···················································································· 6分
2
∵ a为正整数
a≤87 ······················································································ 7分
答：略······························································································ 8分
18、解：（1）5；2．······························································································· 2分
频数/人
男生
7
女生
6
5
4
3
2
1
0
A B C D 组别
······························································5分
2 5 7（ ） 1200 720（人）········································································7分
20
答：胖瘦程度为“正常”的人数约为 720人．···················································· 8分
19、解：过 E作 EP AB交于 P，延长CD交 EP于Q ·····················································1分
∵ CDE 120
∴ EDQ 60
由题意，得： EQD 90
∴ DEQ 90 EDQ 30 ········································································· 2分
在 Rt△ EDQ中 DE 6
DQ 1 ED 3
2 A
∴CQ CD DQ 5 ························································F···························· 3分
∵ QPB PBC BCQ 90
∴四边形 BCQP为矩形···································E··············Q····P·····························4分
∴QC PB 5
∵ FED 73 DEQ 30 D
M N
∴ FEP FED DEQ 43 ······································C··B······························ 5分
在 Rt△ FEP中 EF 8 图 2
sin FEP FP
EF
FP
∴ 0.68
8
∴ FP 5.44
∴ BF PB FP 5 5.44 10.4 m··································································· 7分
答： BF 长约为10.4米．················································································8分
20、解：设降价 x元，每天的销售利润为 13000元··························································1分
60 30 x 300 50x 13000 ·······································································3分
x2 24x 80 0 ······················································································4分
解，得： x1 4 x2 20 ··············································································· 5分
30 x
100%≥ 50%
30
x≤15·····························································································7分
∴ x 20舍去 x 4
∴ 60 4 56 （元）···················································································· 8分
答：略
21、证明：（1）∵ B C B C
∴ CAF BEF
∵ BEF EBD D
EAC BAE CAF 2 EBD D
∴ BAE EBD D 2 EBD D
∴ BAE EBD ··············································································2分
∵ AB是 O的直径
∴ BEA 90
∴ BAE ABE 90
∴ ABD EBD ABE 90 ····························································3分
∴ AB BD
∵ AB是 O的直径
∴ BD为 O的切线··········································································· 4分
（2）∵ FBD 90 DE EF
BE 1∴ DF EF 1 DF
2 2
∴ BE EF
∴ EBF EFB ·············································································· 5分
∵ AE AE
∴ C FBE
∵ EFB AFC
∴ AFC C
∴ AF AC 1 3 ··········································································· 6分
∵ FBE C AFC EFB
∴△ EBF∽△ ACF
BE BF
∴ ∵ AC 2EF
AC CF
EF BF 1
∴
AC CF 2
1
∴ BF CF 1 ················································································8分
2
∴ AB AF BF 1 3 1 2 3
∴ AO 3 1 ··················································································9分
2
22、证明：（1）∵ AD AF D ABC
∴ D AFD ABC
∴ EBC EFA ·············································································· 1分
∵ AD AF AD BC
∴ AF BC
∵ E E
∴△ AFE≌△CBE
∴ AE CE ······················································································ 3分
（2）延长 AB交DC延长线于 E，过 A作 AG DE 交于G，在DE 上取点 F 使得 AF AD
······························································································································· 4分
∵ AD AF D ABC
∴ D AFD ABC
∴ EBC EFA
∵ AD AF AD BC
∴ AF BC
∵ E E
∴△ AFE≌△CBE
∴ AE CE BE EF
∵ AB AE BE
CF CE EF
∴ AB CF
∴ DF CD CF CD AB ································································ 5分
在 Rt△ AGD中
cosD DG 3
AD 5
3
∴ DG AD
5
∴ DF 6 AD ··················································································· 6分
5
∵ AF AD AG DF
1
∴ DG DF
2
∴CD AB 6 AD
5
CD AB 6∴ BC ·············································································7分
5
（3）延长 EA交CD延长线于 F ，在CD取一点G使得 ED EG ，过 E作 EH CD交于 H
······························································································································· 8分
∵ ED EG
∴ EDG EGD
∵ EAC EDC 180
EAC CAF 180
∴ CAF EGF
∵四边形 ABDE是平行四边形
∴ AB DE AE BD 3 AB∥ ED
∵ AB AC
∴ AC DE
∵ F F
∴△ EFD≌△CAF
∴ AF GF EF CF
∵ AE EF AF
CG CF GF
∴ AE CG 3 ················································································· 9分
∵ ED EG EH DG
∴ DH HG
设 DH HG x
在 Rt△ EDH 中
tan EDH EH 8
DH
∴ EH 8DH 8x
∵ EAC EDC 180
P
∴ FAC FDE 180
∵四边形 AFDP内角和 4 2 180 360
∴ FAP FDP F APD 360
∴ F APD 180
∵ APD CPD 180
∴ F CPD
∵ AB∥ ED
∴ BAC CPD F ······································································ 10分
在 Rt△ EFH 中
tan F tan BAC EH 4
FH 3
8x 4
∴
FH 3
∴ FH 6x
EF EH 2 FH 2 10x
∴CF EF 10x
∵CF FH HG CH 7x 3
∴10x 7x 3
x 1····························································································11分
∴ DH 1 EH 8
在 Rt△ EDH 中
ED DH 2 EH 2 65
∴ AB DE 65 ··············································································12分
23、解：（1）①由题意，得：
h x2 bx 1 x
h x2 b 1 x 1 ·············································································1分
x b b 1 1图象对称轴为：直线 ···············································2分
2a 2 4
2b 2 1
b 1
2
y x2 1∴ x 1··············································································· 3分
2
②成立···························································································· 4分
由函数 y x2 bx 1，可得
b b 4ac b2 4 b2

2a 2 4a 4
b 4 b2
∴函数图象顶点坐标为（ ， ）
2 4
b 4 b2
令 x y
2 4
y 4 b
2 2
1 b∴

4 2
∴ y x2 1
∴函数 y x2 bx 1图象顶点在函数 y x21 1的图象上························· 6分
由函数 h x2 b 1 x 1可得
b b 1 4ac b2 4 b 1 2

2a 2 4a 4
b 1 4 b 1 2
∴函数图象顶点坐标为（ ， ）
2 4
x b 1将 代入 y1 x
2 1
2
b 1 2 4 y 1 b 1
2
1 4 4
∴函数 h x2 b 1 x 1图象顶点也在函数 y1 x2 1的图象上················7分
∴小明同学的猜想成立
b
（2）二次函数 y ax2 bx c图象对称轴为：直线 x
2a
h ax2 bx c ax
h ax2 b a x c
h x b b a b 1奇异函数 图象对称轴为：直线
2a 2a 2a 2
1
∴两条对称轴之间的距离 PE
2
由轴对称，可得
DE PE 1
2
∴ PD PE DE 1
∴抛物线 y '由抛物线 y ax2 bx c 向右平移 1个单位得到
∴ y ' a x 1 2 b x 1 c ··································································· 8分
由题意，得： a x 1 2 b x 1 c ax 2 b a x c
H 3∵ 点横坐标为
2
3 2 3 3 2 3
∴ a 1 b 1 c a b a c
2 2 2 2
∴ a 2b ··························································································· 9分
由轴对称，可得
PD QE
PQ 5 PE 1∵
2 2
在 Rt△QEP中
QE PQ2 PE 2 1 ············································································ 10分
y ax2 bx c P 4ac b
2
二次函数 图象顶点 纵坐标为：
4a
2 4ac b a
2
函数 h ax b a x c图像顶点Q纵坐标为：
4a
①当 a 0时
4ac b a 2
QE 4ac b
2 2ab a2
1
4a 4a 4a
a2 2a 0
a1 0（舍去） a2 2
∴b 1 a 1 ······················································································11分
2
②当 a 0时
2
QE 4ac b
2 4ac b a 2ab a2
1
4a 4a 4a
a2 2a 0
a1 0（舍去） a2 2
b 1∴ a 1 ····················································································12分
2
a 2 a 2
综上所述： 或
b 1

b 1
注：所有题其他做法酌情给分