资源简介

机密★启用前
2025年广东省初中学业水平模拟考试（二）
数学
本试卷共4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总点涂黑：
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。在每小题给出的四个选项中，只有一
项是符合题目要求的.
1.一5的倒数是()
A.5
B
D.-0.5
2.截至2025年3月15日，国产动画电影《哪吒之魔童闹海》突破150亿元票房，登顶全球动画电
影票房榜，数据150亿用科学记数法表示为（)
A.15000000000
B.1.5×1010
C.1.5×1011
D.0.15×101
3.如图是由一个长方体和一个圆柱组成的几何体，它的俯视图是()
主视方向
4.
B
C.
D.
4.如图，直线c与直线a、b都相交.若a/1b,∠1=37°，则∠2=()
b
A.143°
B.63°
C.53°
D.37°
5.下列运算正确的是()
A.2ab+3ab=5ab
B.(a2)3=a
C.2·a4=a8
D.a÷a=a3
6.小美和小好同学做“石头、剪刀、布”的游戏，两人同时出相同的手势，这个事件是（)
A.随机事件
B.不可能事件
C.必然事件
D.确定性事件
数学试卷第1页（共4页）
7.在剪纸活动中，小花同学想用一张矩形纸片剪出一个正五边形，其中正五边形的一条边与矩形
的边重合，如题7图所示，则∠a的大小为()
A.54
B.60°
C.70°
D.72°
D
/P
题7图
题9图
题10图
8.下列运算结果为x一1的是()
A.1
B.
2-1x
c.
x+1.1
x2+2x+1
D.
xx+1
÷
x+1
9.已知，二次函数y=ax2十bx十c的图象如题9图所示，则点P(一a,一b)所在的象限是()
A.第一象限
B.第二象限
C.第三象限
D.第四象限
10.如题10图，在四边形ABCD中，AD/BC,BC=5,CD=3.按下列步骤作图：①以点D为
圆心，适当长度为半径画弧，分别交DA,DC于E,F两点；②分别以点E,F为圆心以大丁
2F的长为半径画弧，两弧交于点P:③连接DP并延长交BC于点G.则BG的长是()
A.2
B.3
C.4
D.5
二、填空题：本大题共5小题，每小题3分，共15分.
11.写出一个比2大且比3小的无理数
12.若a2一2a-2=0,则3a2-6a十1=
13.已知点P(4一m,m)在第二象限，则m的取值范围是
14.如图，在平行四边形ABCD中，BD=CD,AE⊥BD于点E,若∠C=67°，则∠BAE=
0
15.“赵爽弦图”巧妙利用面积关系证明了勾股定理.如图所示的“赵爽弦图”是由四个全等直角
三角形和中间的小正方形拼成的一个大正方形.设直角三角形的两条直角边长分别为m,n(m
>n).若小正方形面积为6，(m+n)2=22,则大正方形面积为
m
n
数学试卷第2页（共4页）2025 年广东省初中学业水平模拟考试（二）
数学参考答案及评分标准
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 C B D D A A D B B A
二、填空题（本大题共 5小题，每小题 3分，共 15分．
11． 5(不唯一） 12．7 13．m＞4 14．44 15．14
三、解答题（一）本大题共 3小题，每小题 7分，共 21分.
16 4 3．解：原式＝ × 2 ＋1－2 3＋2························································································ 4分
＝3········································································································································ 7分
17．（1）证明：Δ＝m2－4n······································································································1分
＝m2－4(m－2)···················································································································2分
＝m2－4m＋8·······················································································································3分
＝(m－2)2＋4＞0
∴方程总有两个不相等的实数根······················································································ 4分
（2）解：令 m＝2，则 n＝0
代入得x2＋2x＝0·················································································································5分
解得x1＝0，x2＝－2··········································································································· 7分
18．解：
（1）设出 A，B两款纪念品的进货单价分别为 x元，y元．·················································1分
3x－2y＝120
则 ，·············································································································· 2分
x＋2y＝200
x＝80
解得 ，······················································································································ 3分
y＝60
答：A，B两款纪念品的进货单价分别为 80元和 60元．············································ 4分
（2）设购买 m个 B款纪念品，(70－m)个 A款纪念品，····················································· 5分
根据题意，得 60m＋80(70－m)≤5000，······································································· 6分
解得 m≥30，
答：至少应购买 B款纪念品 30个．················································································7分
四、解答题（二）本大题共 3小题，每小题 9分，共 27分
19．（1）51.2······························································································································· 2分
（2）30，30，补全条形统计图如下：
································································································ 5分
数学参考答案及评分标准 第 1 页 (共 6 页)
（3）画树状图如下：
·························································· 7分
共有 6种等可能情况，其中恰好为一男一女的有 4种；·············································· 8分
∴P 4 2（恰好是一男一女）＝ 6＝ 3．··················································································9分
20．解：
k
（1）∵y＝ 2x（x＜0）过点 P（－2，1），
∴k2＝－2 y
2
，∴ ＝－ x········································································································ 1分
把 Q（－1，m）代入 y 2＝－ x，得 m＝2，∴Q（－1，2）··········································· 2分
－2k1＋b＝1，
把 P（－2，1）、Q（－1，2）代入 y＝k1x＋b，得
－k1＋b＝2，
k1＝1，
解得： ···················································································································3分
b＝3.
∴一次函数的解析式为 y＝x＋3······················································································· 4分
（2）如图，作点 P关于 x轴的对称点 P′，连接 P′Q交 x轴于点 E，
此时 EP＋EQ的值最小······································································································5分
设直线 P′Q的解析式为 y＝ax＋c（a≠0），
－2a＋c＝－1，
把 P′（－2，－1）、Q（－1，2）代入 y＝ax＋c，得
－a＋c＝2，
a＝3，
解得
c＝5.
∴y 5＝3x＋5，∴E（－ 3，0）.·························································································· 6分
设直线 PQ与 x轴的交点为 F，∴F（－3，0），··························································· 7分
∴S OEPQ＝SΔOFQ－SΔEFP··························································································四边形 8分
1 1 5
＝ ×3×2－ ×(－ ＋3)×1
2 2 3
7
＝ ．···································································································································· 9分
3
数学参考答案及评分标准 第 2 页 (共 6 页)
21．证明：
（1）如图，连接 OD，··············································································································· 1分
∵AB＝AC，
∴∠ABC＝∠ACB，
∵OB＝OD，
∴∠OBD＝∠ODB，
∴∠ODB＝∠ACB，
∴AC∥OD，······················································································································· 2分
∴∠DFC＝∠ODF，
∵DE⊥AC，
∴∠ODF＝∠DFC＝90°，
∴OD⊥DE，·······················································································································3分
∵OD是⊙O的半径，
∴DE是⊙O的切线；········································································································4分
（2）∵AC＝AB＝6
∴AO＝OB＝OD＝3，········································································································5分
3
∵OD⊥DE，tanE＝
4
OD 3
∴ ＝
DE 4
∴DE＝4，···························································································································6分
∴OE＝ OD2＋DE2＝ 32＋42＝5
∴AE＝OE－OA＝2，········································································································ 7分
∵AC∥OD，
∴△AEF∽△OED，·········································································································· 8分
AE AF
∴ ＝ ，
OE OD
2 AF
∴ ＝ ，
5 3
∴AF 6＝ 5．···························································································································9分
五、解答题（三）本大题共 2小题，第 22题 13分，第 23题 14分，共 27分）
22．解：
（1）∵四边形 ABCD为矩形，
∴∠ADC＝90°，················································································································ 1分
∵CE＝2CD，
∴DE＝CD，
∴AD垂直平分 CE，
数学参考答案及评分标准 第 3 页 (共 6 页)
∴CF＝EF＝2，·················································································································· 2分
∵DF＝DC，
∴ DCF为等腰直角三角形，
2
∴CD＝DF＝ 2 CF＝ 2，································································································ 3分
∵AD＝ 3DC＝ 6，
∴AF＝AD－DF＝ 6－ 2；····························································································· 4分
（2）①DM和 DN的数量关系为 DM＝DN，理由如下：······················································ 5分
连接 DF，如图：
，
∵将 Rt△EFC绕点 D逆时针旋转，点 C的对应点为 G，使点 F在矩形内部
∴FD⊥EG，DF＝DE＝DG＝CD，△DGF和△DEF为等腰直角三角形，··············· 6分
∴∠DFE＝∠G＝45°，∠FDG＝90°
∴∠FDN＋∠NDG＝90°
∵∠ADC＝90°
∴∠FDN＋∠MDF＝90°
∴∠MDF＝∠NDG·············································································································7分
∠DFE＝∠G
在△MDF和△NDG中 DF＝DG
∠MDF＝∠NDG
∴ MDF≌ NDG(ASA)，
∴DM＝DN；······················································································································ 8分
②在图 1中，由（1）得：DF＝DC
∴AF＝AD－DF＝ 3DC－DC＝( 3－1)DC···································································9分
在图 3中，连接 DF1，
∵在△ADC中，AD＝ 3DC
∴tan∠ACD AD 3DC＝DC＝ DC ＝ 3
∴∠ACD＝60°，∴∠CAD＝30°·····················································································10分
∴AC＝2DC
又∵DF1＝DC
数学参考答案及评分标准 第 4 页 (共 6 页)
∴ DF1C是等边三角形··································································································· 11分
∴F1C＝F1D＝DC＝AF1
∴∠AF1D＝180°－∠DF1C＝180°－60°＝120°
∴∠AF1M＝∠AF1D－∠ 1F1D＝120°－45°＝75°
∴∠AMF1＝180°－∠F1AM－∠AF1M＝180°－30°－75°＝75°
∴∠AMF1＝∠AF1M
∴AM＝AF1＝CD
∴DM＝AD－AM＝ 3CD－CD＝( 3－1)CD·······························································12分
由①可得：DM＝DN，
∴综上，图 3中所有与图 1中的 AF相等的线段为 DM、DN．································ 13分
23．（1）解：∵抛物线 y＝ax2＋bx＋4 a≠0 与 x轴交于 A －4，0 、B 2，0 两点，
16a－4b＋4＝0
∴ ，········································································································· 1分
4a＋2b＋4＝0
a 1＝－
解得： 2，
b＝－1
1
∴抛物线的解析式为 y＝－ 2 x
2－x＋4．·········································································2分
（2）(思路：将等腰三角形的问题转化为直角三角形的问题来解决)
过点 Q作 QM⊥EF于点 M，如图：
则∠QME＝90°，
∵FQ＝EQ，QM⊥EF，
∴EF＝2EM，
∵A －4，0 ，D 0，3 ，
∴OA＝4，OD＝3，
在 Rt△AOD中，由勾股定理得 AD＝5．········································································3分
∵PQ⊥x轴，
∴PQ∥OC，
∴∠QEM＝∠ADO，
∴cos∠QEM＝cos∠ADO，······························································································ 4分
EM OD 3
∴ QE＝ AD＝ 5，
∴EM 3＝ 5QE，EF
6
＝ 5QE，
数学参考答案及评分标准 第 5 页 (共 6 页)
FEQ 16∴△ 的周长＝QE＋EF＋FQ＝ 5 QE，·································································· 5分
∴当 QE最大时，△FEQ的周长最大．
设 Q m 1，－ 22m －m＋4 ，其中－4≤m≤0．
∵A －4，0 ，D 0，3 ，
3
∴直线 AD的解析式为 y＝ 4 x＋3，
∴E m 3， 4m＋3 ，············································································································6分
QE 1m2 m 4 3 m 3 1 7 1 7
2 81
∴ ＝－ 2 － ＋ － 4 ＋ ＝－ 2m
2－ 4m＋1＝－ 2 m＋ 4 ＋ 32，··············7分
1
∵－ 2＜0，
7 81
∴m＝－ 4时，QE有最大值，最大值为 ，32
16 81 81
∴△FEQ周长的最大值为 × ＝ ．······································································ 8分
5 32 10
3 1
－1
（ ）∵抛物线 y＝－ 22 x －x＋4的对称轴为 x＝－ ＝－1．2×(－12)
1
由题知：平移后的抛物线的解析式为 y＝－ 22 x －x＋4±d．
x n y 1设 N＝ ，则 N＝－ 22 n －n＋4±d．
3
又∵直线 AD的解析式为 y＝ 4 x＋3，点 N在 AD上，
∴y 3N＝ 4 n＋3，
1
∴－ 2 n
2－n＋4±d 3＝ 4 n＋3，
d 1 2 7∴ ＝ 2 n ＋ 4 n－1 ，
∵H 1，0 ，A －4，0 ，
∴AH＝1－ －4 ＝5．
当△AHN是等腰三角形时，
2 3 2
①若 AN＝AH，则 n＋4 ＋ 4 n＋3 ＝5
2，
解得n1＝－8（舍去），n2＝0，
∴d 1＝ 2×0
7
＋ 4×0－1 ＝1；
②若 AN＝NH，则 n＋4＝1－n，
3
解得 n＝－ 2＜－1（舍去）；
③若 AH＝NH，
2 3 2
则 n－1 ＋ 24 n＋3 ＝5 ，
n 12解得： 1＝ 5，n2＝－4（舍去），
d 1 12
2 7 12 152
∴ ＝ 2× 5 ＋ 4× 5 －1 ＝ 25．
152
综上，抛物线的平移距离 d的值为 或 1．······························································· 14分
25
备注：第（3）问无需过程，每个答案 3分，共 6分）
数学参考答案及评分标准 第 6 页 (共 6 页)

展开更多......

收起↑