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2024一2025學年度第一學期期末質量監測試卷
九年級數學
本試卷共6頁，23小題，滿分120分。考試用時120分鐘。
注意事項：1.答卷前，考生務必用黑色字跡的鋼筆或簽字筆將自己的學校、姓名和準考證號填寫在答
題卡上。將條形碼粘貼在答題卡“條形碼粘貼處”
2.作答選擇題時，選出每小題答案后，用2B鉛筆把答題卡上對應題目選項的答案信息點涂
黑：如需改動，用橡皮擦干凈后，再選涂其他答案。
3.非選擇題必須用黑色字跡的鋼筆或簽字筆作答，答案必須寫在答題卡各題目指定區域內
相應位置上；如需改動，先劃掉原來的答案，然后再寫上新的答案；不準使用鉛筆和涂
改液。不按以上要求作答的答案無效。
4.考生必須保持答題卡的整潔?？荚嚱Y束后，將試卷和答題卡一并交回。
一、選擇題：本大題共10小題，每小題3分，共30分.在每小題列出的四個選項中，只
有一項是符合題目要求的
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.漢語是中華民族智慧的結晶，成語又是漢語中的精華，是中華文化的一大瑰寶，具有
極強的表現力.下列成語描述的事件屬于隨機事件的是(
A.旭日東升
B.畫餅充饑
C.守株待兔
D.竹籃打水
3.未來將是一個可以預見的AI時代，下列是世界著名人工智能品牌公司的圖標，其中是
中心對稱圖形但不是軸對稱圖形的是()
4.在平面直角坐標系中，點(5，一6)關于原點對稱的點的坐標是（
A.(5,6)
B.(-5,-6)
C.(5,-6)
D.(-5,6)
5.將拋物線y=3x2+2向左平移2個單位長度，再向下平移3個單位長度，得到的拋物線
的解析式為(
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.若關于x的方程x2一6x一m=0沒有實數根，則實數m的取值范圍是(
A.m>9
B.m>-9
C.m<9
D.m<-9
九年級數學試卷第1頁（共6頁）
7.如題7圖，教室內地面有個傾斜的畚箕，箕面AB與水平地面的夾角∠CAB為61°，
小明將它扶起（將畚箕繞點A順時針旋轉）后平放在地面，箕面AB繞點A旋轉的度數
為()
A.119°
B.120°
C.61°
D.121°
8.正多邊形的一部分如題8圖所示，若∠ACB=18°，則該正多邊形的邊數為()
A.7
B.8
C.9
D.10
題7圖
題8圖
題9圖
9.如題9圖，將一把兩邊都帶有刻度的直尺放在半圓形紙片上，使其一邊經過圓心O,另
一邊所在直線與半圓相交于點D、E,量出半徑OC=5cm,弦DE=8cm,則直尺的寬
度為(
A.1cm
B.2cm
C.3cm
D.4cm
10.在同一坐標系中，一次函數y=ar十2與二次函數y=x2+a的圖象可能是(
二、填空題：本大題共5小題，每小題3分，共15分，請將下列各題的正確答案填寫在
答題卡相應的位置上.
11.拋物線y=(x一2)2-1的頂點坐標是
12.點A(1,y),B(2,y2)都在二次函數yx2十1的圖象上，則y1y2.(填“>”、“=”
或“<”)
13.若m是一元二次方程x2-2x一1=0的一個實數根，則代數式m2-2m十2024=
九年級數學試卷第2頁（共6頁）2024—2025 學年度第一學期期末質量監測試卷
九年級數學參考答案及評分標準
說明：除選擇題外，提供的答案不一定是唯一答案，其它合理答案可酌情給分。
一、選擇題：本大題共 10小題，每小題 3分，共 30分。
題號 1 2 3 4 5 6 7 8 9 10
答案 D C A D B D A D C B
二、填空題：本大題共 5小題，每題 3分，共 15分。
題號 11 12 13 14 15
答案 (2，－1) ＜ 2025 3π 5
三、解答題（一）：本大題共 3小題，每小題 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
則 x＋2＝±4， ·········································································6分
x＝±4－2
∴x1＝－6，x2＝2········································································7分
17．（1）作圖如下：
········································ 4分
（2）4m················································································································7分
18．（1 k）根據題意，設 y＝ ， ··························································· 1分
x
把 x＝4，y＝3代入，得 k＝4×3＝12，···································· 3分
y x 12∴ 關于 的函數解析式為 ＝ ． ·········································4分

2 12 12（ ）把 y＝2代入 ＝ ，得 2＝ ，···············································5分

解得 x＝6，·········································································6分
∴ 小孔到蠟燭的距離為 6cm．··············································· 7分
四、解答題（二）：本大題共 3小題，每小題 9分，共 27分．
九年級數學參考答案及評分標準 第 1 頁 (共 6 頁)
19 1 1．（ ） .····································································································3分
3
（2）畫樹狀圖如下：
·····································6分
由樹狀圖可知，共有 9種可能的結果：(A，A)，(A，B)，(A，C)，(B，A)，(B，
B)，(B，C)，(C，A)，(C，B)，(C，C)；兩人恰好選擇同一項目的結果有 3種，
························································································ 8分
3 1
∴兩人恰好選擇同一項目的概率 P＝ ＝ ································ 9分
9 3
20．（1）證明：如 20題答案圖，連接 OD，則 OD＝OB，
答案題 20圖
∴∠ODB＝∠B，································································· 1分
∵AB＝AC，
∴∠C＝∠B，
∴∠ODB＝∠C，
∴OD//AC， ·······································································2分
∵DE⊥AC于點 E，
∴∠ODE＝∠CED＝90°， ·················································· 3分
∵OD是⊙O的半徑，DE⊥OD，
∴DE是⊙O的切線；··························································· 4分
（2）連接 AD，
∵AB是⊙O的直徑，
∴∠ADB＝90°，
∴AD⊥BC，······································································ 5分
∵AB＝AC，
∴BD＝CD，
∵∠B＝∠C＝30°，OD＝OA，
∴∠AOD＝2∠B＝60°，
∴△AOD是等邊三角形， ·····················································6分
九年級數學參考答案及評分標準 第 2 頁 (共 6 頁)
∴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．任務 1：20；40·········································································· 2分
任務 2：設收納盒高為 xcm，依題意得：·········································3分
1 (100－2x)(40－2x)＝350，··················································· 6分
2
∴x1＝15，x2＝55（舍去）····················································· 7分
∴收納盒長、寬、高分別為 35cm、10cm、15cm，······················8分
∵10cm＜15cm，
∴玩具機械狗不能放入該收納盒．···········································9分
五、解答題（二）：本大題共 2小題，第 22題 13分，第 23題 14分，共 27分．
22．（1）∵AE＝2，BE＝4，∠AEB＝90°，
∴AB＝ AE2＋BE2＝ 22＋42＝2 5，·······································1分
∵四邊形 ABCD是正方形，
∴BC＝AB＝2 5，∠ABC＝90°················································2分
∴AC＝ 2AB＝2 10，························································· 3分
由旋轉的性質得：AB'＝AB＝2 5,
∴CB'＝AC－AB'＝2 10－2 5； ············································ 4分
（2）①四邊形 AEFE'是正方形，理由如下：····································· 5分
由旋轉的性質得：
AE'＝AE，∠EAE'＝α＝90°，∠AE'D＝∠AEB＝90°····················· 6分
∵∠AEF＝180°－90°＝90°
∴四邊形 AEFE'是矩形，······················································· 7分
又∵AE'＝AE
∴四邊形 AEFE'是正方形；···················································· 8分
②過點 C作 CG⊥BE于點 G，如答案 22題圖所示：
九年級數學參考答案及評分標準 第 3 頁 (共 6 頁)
答案題 22圖
則∠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）把點 A(－1，0) y 1 x2 3代入拋物線 ＝－ ＋ x＋c中，
2 2
1 3
則－ － ＋c＝0，······························································ 1分
2 2
解得：c＝2， ·····································································2分
1 3
故拋物線的解析式為：y＝－ x2＋ x＋2． ······························ 3分
2 2
（2）∵⊙D經過 B，C兩點，則 DB＝DC， ···································· 4分
設 D(0，y)，則 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．
故點 D坐標為(0，－3)． ······················································7分
（3）證明：在點 P 運動過程中，存在能夠使得∠PBC＝45°的點 P，理由如下：
九年級數學參考答案及評分標準 第 4 頁 (共 6 頁)
························································································· 8分
①設當 P點在 x軸上方拋物線上時，設∠PBC＝45°，如答案 23－4圖所示，
作 PS⊥BC于 S，SM⊥x軸，PN⊥SM于 N，
答案題 23-4圖
∴∠PSB＝∠N＝∠SMB＝90°，∠PBC＝∠SPB＝45°，
∴∠PSN＝∠SBM＝90°－∠MSB，PS＝SB，
∴△SNP≌△BMS， ······························································ 9分
∴SM＝PN，BM＝NS，
設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
設 S s 1，－ s＋2 ，
2
∴SM＝PN 1＝－ s＋2，BM＝NS＝4－s，
2
1 3
∴點 P坐標為 s＋2，6－ s ，
2 2
P 1把點 s＋2，6 3 1 3－ s 代入拋物線 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，
∴點 P不存在；································································ 11分
②設當 點在 下方拋物線上時，如答案 23－5圖所示，
九年級數學參考答案及評分標準 第 5 頁 (共 6 頁)
答案題 23-5圖
作∠PBC＝45°，CR⊥PB于 ，過 作 RT⊥y軸，過 B作 BQ⊥TR于點 Q，
∴∠CTR＝∠CRB＝∠Q＝90°，∠PBC＝∠RCB＝45°，
∴∠CRT＝∠RBQ＝90°－∠BRQ，CR＝BR，
∴△CTR≌△RQB， ···························································· 12分
∴TR＝BQ，CT＝RQ，
設 TR＝BQ＝a，CT＝RQ＝b，
a＋b＝4 a＝1
則 ，解得： ，
b－a＝2 b＝3
∴ R點坐標為(1，－1)．
1 4
則由待定系數法可得直線lBR：y＝ x－ ，······························ 13分3 3
y 1 3 5＝－ x2＋ x＋2 x＝－
聯立 2 2 3
y 1 x 4
，解得： 17，
＝ － y＝－
3 3 9
P 5 17即點 坐標為 － ，－ ．
3 9
5 17
綜上所述，點 P坐標為 － ，－ ．···································· 14分
3 9
九年級數學參考答案及評分標準 第 6 頁 (共 6 頁)

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