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2024一2025学年度第一学期期末质量监测试卷
九年级数学
本试卷共6页，23小题，满分120分。考试用时120分钟。
注意事项：1.答卷前，考生务必用黑色字迹的钢笔或签字笔将自己的学校、姓名和准考证号填写在答
题卡上。将条形码粘贴在答题卡“条形码粘贴处”
2.作答选择题时，选出每小题答案后，用2B铅笔把答题卡上对应题目选项的答案信息点涂
黑：如需改动，用橡皮擦干净后，再选涂其他答案。
3.非选择题必须用黑色字迹的钢笔或签字笔作答，答案必须写在答题卡各题目指定区域内
相应位置上；如需改动，先划掉原来的答案，然后再写上新的答案；不准使用铅笔和涂
改液。不按以上要求作答的答案无效。
4.考生必须保持答题卡的整洁。考试结束后，将试卷和答题卡一并交回。
一、选择题：本大题共10小题，每小题3分，共30分.在每小题列出的四个选项中，只
有一项是符合题目要求的
1.下列方程中，是一元二次方程的是(
A.x-1=0
B.x+y=2
c.2+2=1
D.x2-1=0
2.汉语是中华民族智慧的结晶，成语又是汉语中的精华，是中华文化的一大瑰宝，具有
极强的表现力.下列成语描述的事件属于随机事件的是(
A.旭日东升
B.画饼充饥
C.守株待兔
D.竹篮打水
3.未来将是一个可以预见的AI时代，下列是世界著名人工智能品牌公司的图标，其中是
中心对称图形但不是轴对称图形的是()
4.在平面直角坐标系中，点(5，一6)关于原点对称的点的坐标是（
A.(5,6)
B.(-5,-6)
C.(5,-6)
D.(-5,6)
5.将抛物线y=3x2+2向左平移2个单位长度，再向下平移3个单位长度，得到的抛物线
的解析式为(
A.y=3x+2)2+3
B.y=3x+2)2-1
C.y=3(x-2)2+3
D.y=3(x-2)2-1
6.若关于x的方程x2一6x一m=0没有实数根，则实数m的取值范围是(
A.m>9
B.m>-9
C.m<9
D.m<-9
九年级数学试卷第1页（共6页）
7.如题7图，教室内地面有个倾斜的畚箕，箕面AB与水平地面的夹角∠CAB为61°，
小明将它扶起（将畚箕绕点A顺时针旋转）后平放在地面，箕面AB绕点A旋转的度数
为()
A.119°
B.120°
C.61°
D.121°
8.正多边形的一部分如题8图所示，若∠ACB=18°，则该正多边形的边数为()
A.7
B.8
C.9
D.10
题7图
题8图
题9图
9.如题9图，将一把两边都带有刻度的直尺放在半圆形纸片上，使其一边经过圆心O,另
一边所在直线与半圆相交于点D、E,量出半径OC=5cm,弦DE=8cm,则直尺的宽
度为(
A.1cm
B.2cm
C.3cm
D.4cm
10.在同一坐标系中，一次函数y=ar十2与二次函数y=x2+a的图象可能是(
二、填空题：本大题共5小题，每小题3分，共15分，请将下列各题的正确答案填写在
答题卡相应的位置上.
11.抛物线y=(x一2)2-1的顶点坐标是
12.点A(1,y),B(2,y2)都在二次函数yx2十1的图象上，则y1y2.(填“>”、“=”
或“<”)
13.若m是一元二次方程x2-2x一1=0的一个实数根，则代数式m2-2m十2024=
九年级数学试卷第2页（共6页）2024—2025 学年度第一学期期末质量监测试卷
九年级数学参考答案及评分标准
说明：除选择题外，提供的答案不一定是唯一答案，其它合理答案可酌情给分。
一、选择题：本大题共 10小题，每小题 3分，共 30分。
题号 1 2 3 4 5 6 7 8 9 10
答案 D C A D B D A D C B
二、填空题：本大题共 5小题，每题 3分，共 15分。
题号 11 12 13 14 15
答案 (2，－1) ＜ 2025 3π 5
三、解答题（一）：本大题共 3小题，每小题 7分，共 21分．
16．解：x2＋4x＝12，······································································ 1分
x2＋4x＋4＝12＋4，·································································· 3分
(x＋2)2＝16， ··········································································· 5分
则 x＋2＝±4， ·········································································6分
x＝±4－2
∴x1＝－6，x2＝2········································································7分
17．（1）作图如下：
········································ 4分
（2）4m················································································································7分
18．（1 k）根据题意，设 y＝ ， ··························································· 1分
x
把 x＝4，y＝3代入，得 k＝4×3＝12，···································· 3分
y x 12∴ 关于 的函数解析式为 ＝ ． ·········································4分
2 12 12（ ）把 y＝2代入 ＝ ，得 2＝ ，···············································5分
解得 x＝6，·········································································6分
∴ 小孔到蜡烛的距离为 6cm．··············································· 7分
四、解答题（二）：本大题共 3小题，每小题 9分，共 27分．
九年级数学参考答案及评分标准 第 1 页 (共 6 页)
19 1 1．（ ） .····································································································3分
3
（2）画树状图如下：
·····································6分
由树状图可知，共有 9种可能的结果：(A，A)，(A，B)，(A，C)，(B，A)，(B，
B)，(B，C)，(C，A)，(C，B)，(C，C)；两人恰好选择同一项目的结果有 3种，
························································································ 8分
3 1
∴两人恰好选择同一项目的概率 P＝ ＝ ································ 9分
9 3
20．（1）证明：如 20题答案图，连接 OD，则 OD＝OB，
答案题 20图
∴∠ODB＝∠B，································································· 1分
∵AB＝AC，
∴∠C＝∠B，
∴∠ODB＝∠C，
∴OD//AC， ·······································································2分
∵DE⊥AC于点 E，
∴∠ODE＝∠CED＝90°， ·················································· 3分
∵OD是⊙O的半径，DE⊥OD，
∴DE是⊙O的切线；··························································· 4分
（2）连接 AD，
∵AB是⊙O的直径，
∴∠ADB＝90°，
∴AD⊥BC，······································································ 5分
∵AB＝AC，
∴BD＝CD，
∵∠B＝∠C＝30°，OD＝OA，
∴∠AOD＝2∠B＝60°，
∴△AOD是等边三角形， ·····················································6分
九年级数学参考答案及评分标准 第 2 页 (共 6 页)
∴OD＝AD 1＝ AB＝1 ···························································· 7分
2
∵∠ADE＝∠ODE－∠ODA＝90°－60°＝30°，
∴ 1＝ ·············································································8分
2
∴ ＝ 2 2＝ 12 ( 1 )2 3＝ ······································ 9分
2 2
21．任务 1：20；40·········································································· 2分
任务 2：设收纳盒高为 xcm，依题意得：·········································3分
1 (100－2x)(40－2x)＝350，··················································· 6分
2
∴x1＝15，x2＝55（舍去）····················································· 7分
∴收纳盒长、宽、高分别为 35cm、10cm、15cm，······················8分
∵10cm＜15cm，
∴玩具机械狗不能放入该收纳盒．···········································9分
五、解答题（二）：本大题共 2小题，第 22题 13分，第 23题 14分，共 27分．
22．（1）∵AE＝2，BE＝4，∠AEB＝90°，
∴AB＝ AE2＋BE2＝ 22＋42＝2 5，·······································1分
∵四边形 ABCD是正方形，
∴BC＝AB＝2 5，∠ABC＝90°················································2分
∴AC＝ 2AB＝2 10，························································· 3分
由旋转的性质得：AB'＝AB＝2 5,
∴CB'＝AC－AB'＝2 10－2 5； ············································ 4分
（2）①四边形 AEFE'是正方形，理由如下：····································· 5分
由旋转的性质得：
AE'＝AE，∠EAE'＝α＝90°，∠AE'D＝∠AEB＝90°····················· 6分
∵∠AEF＝180°－90°＝90°
∴四边形 AEFE'是矩形，······················································· 7分
又∵AE'＝AE
∴四边形 AEFE'是正方形；···················································· 8分
②过点 C作 CG⊥BE于点 G，如答案 22题图所示：
九年级数学参考答案及评分标准 第 3 页 (共 6 页)
答案题 22图
则∠BGC＝90°＝∠AEB··························································· 9分
∴∠CBG＋∠BCG＝∠CBG＋∠ABE＝90°
∴∠BCG＝∠ABE，······························································ 10分
在△BCG和△ABE中，
∠BGC＝∠AEB
∠BCG＝∠ABE
BC＝AB
∴△BCG≌△ABE AAS ························································· 11分
∴CG＝BE＝4，BG＝AE＝2，···············································12分
∴EG＝BE－BG＝4－2＝2
∴CE＝ CG2＋EG2＝ 42＋22＝2 5······································· 13分
23．（1）把点 A(－1，0) y 1 x2 3代入抛物线 ＝－ ＋ x＋c中，
2 2
1 3
则－ － ＋c＝0，······························································ 1分
2 2
解得：c＝2， ·····································································2分
1 3
故抛物线的解析式为：y＝－ x2＋ x＋2． ······························ 3分
2 2
（2）∵⊙D经过 B，C两点，则 DB＝DC， ···································· 4分
设 D(0，y)，则 CD＝|2－y|， ·················································5分
BD2＝(0－y)2＋42＝16＋y2，CD2＝(2－y)2，
∴16＋y2＝(2－y)2， ···························································· 6分
解得：y＝－3．
故点 D坐标为(0，－3)． ······················································7分
（3）证明：在点 P 运动过程中，存在能够使得∠PBC＝45°的点 P，理由如下：
九年级数学参考答案及评分标准 第 4 页 (共 6 页)
························································································· 8分
①设当 P点在 x轴上方抛物线上时，设∠PBC＝45°，如答案 23－4图所示，
作 PS⊥BC于 S，SM⊥x轴，PN⊥SM于 N，
答案题 23-4图
∴∠PSB＝∠N＝∠SMB＝90°，∠PBC＝∠SPB＝45°，
∴∠PSN＝∠SBM＝90°－∠MSB，PS＝SB，
∴△SNP≌△BMS， ······························································ 9分
∴SM＝PN，BM＝NS，
设lBC：y＝kx＋b，代入 B(4，0)，C(0，2)，可得：
b＝2 k
1
＝－
，解得 2，
4k＋b＝0 b＝2
故l 1BC：y＝－ x＋2，························································· 10分2
设 S s 1，－ s＋2 ，
2
∴SM＝PN 1＝－ s＋2，BM＝NS＝4－s，
2
1 3
∴点 P坐标为 s＋2，6－ s ，
2 2
P 1把点 s＋2，6 3 1 3－ s 代入抛物线 y＝－ x2＋ x＋2中可得
2 2 2 2
3 26－ s 1 1 s 2 3 1＝－ ＋ ＋ s＋2 ＋2，解得s1＝4，s ＝6，2 2 2 2 2 2
∵0＜s＜4，
∴点 P不存在；································································ 11分
②设当 点在 下方抛物线上时，如答案 23－5图所示，
九年级数学参考答案及评分标准 第 5 页 (共 6 页)
答案题 23-5图
作∠PBC＝45°，CR⊥PB于 ，过 作 RT⊥y轴，过 B作 BQ⊥TR于点 Q，
∴∠CTR＝∠CRB＝∠Q＝90°，∠PBC＝∠RCB＝45°，
∴∠CRT＝∠RBQ＝90°－∠BRQ，CR＝BR，
∴△CTR≌△RQB， ···························································· 12分
∴TR＝BQ，CT＝RQ，
设 TR＝BQ＝a，CT＝RQ＝b，
a＋b＝4 a＝1
则 ，解得： ，
b－a＝2 b＝3
∴ R点坐标为(1，－1)．
1 4
则由待定系数法可得直线lBR：y＝ x－ ，······························ 13分3 3
y 1 3 5＝－ x2＋ x＋2 x＝－
联立 2 2 3
y 1 x 4
，解得： 17，
＝ － y＝－
3 3 9
P 5 17即点 坐标为 － ，－ ．
3 9
5 17
综上所述，点 P坐标为 － ，－ ．···································· 14分
3 9
九年级数学参考答案及评分标准 第 6 页 (共 6 页)
