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2024－2025学年第二学期初中毕业班适应性训练
数 学
本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，完卷时间120分钟，满分150分。
注意事项：
1．答题前，考生务必在试题卷、答题卡规定位置填写本人准考证号、姓名等信息．考生要认真核对答题卡上粘贴的条形码的“准考证号、姓名”与考生本人准考证号、姓名是否一致。
2．选择题每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其他答案标号．非选择题答案用0.5毫米黑色墨水签字笔在答题卡上相应位置书写作答，在试题卷上答题无效。
3．作图可先使用2B铅笔画出，确定后必须用0.5毫米黑色墨水签字笔描黑。
4．结束时，考生必须将答题卡交回。
第Ⅰ卷
一、选择题：本题共10小题，每小题4分，共40分。在每小题给出的四个选项中，只有一个选项是符合题目要求的。
1．的倒数是
A． B． C．2 D．-2
2．德化白瓷是“海丝”最具国际影响力的文化名片之一，下图是德化白瓷的直口
杯，它的主视图是
A． B． C． D．
3．作为新兴力量，DeepSeek承载着打破国外技术垄断、为中国AI开拓新局的使命，在全球AI竞技场上崭露头角，助力中国迈向AI强国之列．数据显示，DeepSeek发布20天后，其日活跃用户已达22 150 000人，将数据22 150 000用科学记数法表示为
A．2.215 B．22.15 C．2.215 D．0.2215
4．下列图形中，既是轴对称图形，又是中心对称图形的是
A． B． C． D．
5．估计的值在
A. 4和5之间 B. 5和6之间 C. 6和7之间 D. 20和40之间
6．学校举行“强国有我，筑梦未来”演讲比赛，小明统计了7位评委对某参赛选手的评分并制成如下表格．如果去掉一个最高分和一个最低分，那么下表中的数据一定
不会发生变化的是
众数 中位数 平均数 方差
8.6 8.4 8.5 0.25
A．8.6 B．8.4 C．8.5 D．0.25
7．如果一个正多边形的每一个内角是150°，那么这个正多边形的边数是
A. 16 B. 12 C. 8 D. 6
8．小月同学在手工课上用扇形卡纸制作的简易圆锥形漏斗如图所示，若漏斗的底面圆的直径为6cm，高为4cm，则扇形卡纸的面积是
A．7.5πcm2 B．9πcm2 C．cm2 D．30πcm2
9．幻方起源于中国，是我国古代数学的杰作之一，是一种将数字填在正方形格子中，使每一横行、每一竖列以及两条斜对角线上的数字和都相等的填数游戏．图①是一个幻方，图②是一个未完成的幻方，根据图②可列出的方程组是
A． B．
C． D．
10．如图，在平面直角坐标系中，矩形ABCD的顶点A，C分
别在x轴，y轴的正半轴上，点D（2，3），AD=5，若反
比例函数y=（k＞0，x＞0）的图象经过点B，则k的
值为
A． B．8 C．10 D．
第Ⅱ卷
二、填空题：本题共6小题，每小题4分，共24分。
11．=__________．
12．如图，A、B两点被一个池塘隔开，M、N分别是AC、BC的
中点， 测量MN的长度为150米，那么AB的长度为______米．
春节期间，小明和小亮分别从三部影片《哪吒之魔童降世》《唐探1900》
《封神第二部：战火西岐》中随机选择一部观看，则他们选择的影片相同的
概率为__________．
14．如图，直线y=kx+b与x轴的正半轴相交于点A，与直线
y=3x相交于点B(m,6)，则关于x的不等式kx+b>3x的解
集是__________．
15．已知，则的值为__________．
16．如图是凸透镜成像示意图，CD是蜡烛AB通过
凸透镜MN所成的虚像．已知蜡烛的高AB为6cm，
蜡烛AB离凸透镜MN的水平距离OB为10cm，
该凸透镜的焦距OF为15cm，AE∥OF，则像CD
的高为__________cm．
三、解答题：本题共9小题，共86分。解答应写出文字说明、证明过程或演算步骤。
17．(8分)
解不等式组并写出它的整数解．
18．(8分)
如图，在四边形ABCD中，AD∥BC，点E在对角线BD上，BE=AD且
∠BEC=∠A，求证：AD+DE=BC．
19．(8分)
先化简，再求值：，其中.
20．(8分)
某校气象兴趣小组的同学们想预估一下A市今年6月份日平均气温状况．他们收集了A市近四年6月份每天的日平均气温，从中随机抽取了60天的日平均气温，并绘制成如下统计图：
根据以上信息，回答下列问题：
（1）求这60天的日平均气温的平均数；
（2）若日平均气温在18 ℃～22 ℃的范围内(包含18 ℃和22 ℃)为“舒适温度”．请预估A市今年6月份日平均气温为“舒适温度”的天数．
21．(8分)
如图，已知二次函数y=+bx+c的图象与x轴交于A，B两点，与y轴交于
点C，其中B（3，0），C（0，3）.
（1）求该二次函数的顶点坐标；
（2）若P是二次函数图象上的一点，且点P在第一象限，△PBC的面积是△ABC面积的一半，求点P的坐标.
22．(10分)
如图，在三角形纸片ABC 中，AB>AC．
（1）将三角形纸片折叠，使点C落在AB边上，且折痕经过点A.请用尺规作图作出折痕,并找出AB边上与点C重合的点E.（不写作法，保留作图痕迹）
（2）在（1）的条件下，求证：∠C >∠B．
23．(10分)
已知整数a，b，m，n满足a+b=mn．
（1）求证：为非负数；
（2）若n为偶数，判断am-bm是否可以为奇数，说明你的理由．
24．(12分)
综合实践：
根据以下素材，解决问题．
如何确定拱桥形状？
问题 背景 河面上有一座拱桥，对它的形状，同学们各抒己见．有同学说拱桥的形状是抛物线，也有同学说是圆弧．为确定拱桥的形状，九年级综合实践小组开展了一次探究活动．
素材 1 在正常水位时，小组成员对拱形内水面宽度和拱顶离水面的距离进行了测量并绘制了下图．测得拱形内水面宽AB为40米，拱顶离水面的距离CD为10米．
素材 2 大雨过后，水位上涨．小组成员再对拱形内水面宽度和拱顶离水面的距离进行了两次测量．发现当拱形内水面宽为36米时，水位（相对正常水位）上涨2.05米；当拱形内水面宽为32米时，水位（相对正常水位）上涨3.85米．
素材 3 如何检验探究过程中提出的假设是否符合实际情况呢？ 定义：离差平方和是实际观测值与预测值之间差的平方和，反映了基于假设算得的预测值与实际观测值之间的差异．离差平方和越小，说明预测值与实际观测值之间的误差越小，提出的假设与实际情况更为接近
解决问题
假设 1 小组成员首先假设拱桥形状是抛物线．根据素材1建立如图所示的直角坐标系，求该抛物线的解析式．
假设 2 小组成员又提出拱桥形状可能是圆弧．请根据素材1求出该圆弧的半径．
分析判断 基于假设1和假设2，请分别计算拱形内水面宽36米和32米时水位上涨的预测值，直接填入下表（数据保留两位小数），并结合素材3分别求出两种假设下数据的离差平方和，判断拱桥更接近哪一种形状．（参考数据：） 水面宽36米水面宽32米水位上涨的实际观测值（）2.053.85假设1的预测值（）3.6假设2的预测值（）2.3
25．(14分)
如图1，AB和CD是⊙O的直径，且AB=4，点P是BA延长线上的一点．连接PC交⊙O于点E（点E在线段PC上，且不与点P、点C重合）．
（1）若，求证：PC=PO；
（2）如果PA=2．
①如图2，若= ，求证：E是PC的中点；
②若△BDE为等腰三角形，求线段PE的长．
数学试题 第6页（共6页）2024－2025 学年第二学期初中毕业班适应性训练
数学参考答案
一、选择题：本题共 10 小题，每小题 4 分，共 40 分。
1.D 2. A 3. C 4. D 5. B 6. B 7. B 8. C 9. A 10. D
二、填空题：本题共 6 小题，每小题 4 分，共 24 分。
11.π 12. 300 13. 1 14. x>-2 15. 8 16. 18
3
三、解答题：本题共 9 小题，共 86 分。
17.(8 分)解：解不等式①得 x＜2································································· 2 分
解不等式②得 x 2 ·········································································4分
所以不等式组的解集是－2 ≤ x＜2······················································· 6 分
所以它的整数解是－2，－1，0，1······················································· 8分
18.(8 分)解： AD∥BC，
ADB EBC ······························································· 2 分
在△ADB和△EBC中，
A BEC

AD BE

ADB EBC
△ADB≌△EBC ··························································· 5分
∴ = ·········································································· 6分
∴ + = ···································································7分
∵AD=BE
∴ + = ·································································· 8分
19 (8 ) ( 2a 1 1) a
2 1
. 分 先化简，再求值： ，其中 a 2 1 .
a a
2a 1 2
解： ( 1) a 1
a a
(2a 1 a ) (a 1)(a 1)= ·······································································2 分
a a a
1
a 1 a
= ················································································ 4 分
a (a 1)(a 1)
1
= ·····································································································6 分
a 1
1 1 2
当 a 2 1时，原式 ·············································· 8分
2 1 1 2 2
1
20.(8分)解：（1）x (17 5 18 12 19 13 20 9 21 6 22 4 23 6 24 5)
60
=20····························································································· 3 分
所以这 60天的日平均气温 20 ℃·································································· 4 分
12 13 9 6 4
（2）∵ 30 22
60
∴预估 A市今年 6月份日平均气温为“舒适温度”的天数有 22 天.······················8 分
21.(8 分)解：（1）将 B（3，0），C（0，3）代入 y x2 bx c
9 3b c 0

c 3
b 2 ·······························································································1分
解得
c 3
所以二次函数的表达式为 y x2 2x 3 （x-1）2 4；
顶点坐标为（1,4）······························································ 3分
（2）∵ y x2 2x 3 （x-1）2 4的对称轴为直线 x=1，B（3，0），C(0,3)，
∴A(-1,0),OC=3,
∴AB=3-(-1)=4,
∴S 1 ABC 2 ×AB×OC=6·············································································4 分
1 S
∴S PBC= 2 ABC△ =3
设 BC为 y=kx+3，
∵y=kx+3 过 B（3，0），
∴3k+3=0，解得 k=-1，
∴ y x 3 ····························································································5分
2
设 P（m，-m2+2m+3）,所以过 P 作 PD⊥x 轴交 BC 点 D（m，-m+3）,
∴PD=-m2+3m···························································································· 6 分
1
∵S△PBC= ×PD×OB=32
∴1
2 ×(-m
2+3m)×3=3
∴m2-3m+2=0···························································································· 7 分
解得 m=1 或 m=2
∴P 的坐标为（1，4）或（2，3）·································································8 分
22.(10 分)（1）解：如图，
作出∠BAC 的角平分线 AD 交 BC 于点 D·························································· 2 分
∴线段 AD即为所求作的折痕．····································································3分
在 AB 上截取出 E 点，使 AE=AC
则点 E为所求的折叠后与点 C 重合的点 ·························································4 分
（其他作法相应给分）
（2）证明：∵AD平分∠BAC，
EAD CAD ·······················································································5分
又∵AE=AC，AD=AD，
△EAD≌△CAD(SAS) ·············································································7分
∴∠AED=∠C··························································································· 8 分
AED B ···························································································9分
C B ·····························································································10 分
23.(10 分)
（1）证明：∵ a b mn，
a2 b2 2mnab a2 b2 2(a b)b ·······················································2分
a2 2ab b2 (a b)2 0 ·····································································4分
3
a 2 b2 2abmn为非负数 ······································································5分
（2）不可以，理由如下：
a，b,m,n为整数，n为偶数,
mn为偶数 ··························································································· 6分
a b mn,
a b为偶数 ·························································································7分
a,b同为偶数或同为奇数，
a b为偶数 ·····································································8分
am bm (a b)m为偶数 ····································································· 9分
am bm不可能为奇数 ········································································ 10 分
24(12 分)
解：假设 1：如图所示，建立平面直角坐标系，
∵测得拱形内水面宽 AB 为 40 米，拱顶离水面的距离 CD 为 10 米．
∴B(20,0)，C（0，10）
设抛物线解析式为 = 2+10，将点 B(20,0)代入得，400 + 10 = 0
1
解得： =
40
1
∴抛物线解析式为 y = 2 + 10 ····························································3 分
40
假设 2：如图所示，设圆心 O，连接 OB，过作 OD⊥AB 于 D，交⊙O 于 C.
在Rt△OBD中，OB2 DB2 OD2，设半径OB r，则 OD=r-10.
∴ 2 = 202 + ( 10)2
解得 = 25
∴该圆弧的半径为 25 米 ···········································································6 分
4
分析判断：
对 于 抛 物 线 ， 当 拱 形 内 水 面 宽 为 36 1米 时 ， 将 x=18 代 入 y = 2 + 10 ， 得
40
y = 1 × 182 + 10=1.9
40
∴水位（相对正常水位）上涨 1.9 米.····························································· 7 分
对圆弧，当拱形内水面宽为 32 米时，设 = 32， EF ,OC交于点 G，
则 = 25 10 = 15, = 1 × = 16
2
在Rt△OEG中，OE 2 GE 2 OG2
设DG x，则252 = 162 + ( + 15)2
解得： = 4.2（负值舍去）·········································································9分
填表如下，
水面宽 36 米 水面宽 32 米
水位上涨的实际观测值（m） 2.05 3.85
假设 1 的预测值（m） 1.9 3.6
假设 2 的预测值（m） 2.3 4.2
根据离差平方和的定义，对于假设 1，离差平方和为:
2 2
（2.05 1.9） +（3.85 3.6） = 0.085 ·················································· 10 分
对于假设 2，离差平方和为:
2 2
（2.05 2.3） +（3.85 4.2） = 0.185 ·················································11 分
∵0.085 < 0.185
∴拱桥的形状更接近抛物线········································································12 分
25.(14分)（1）解：连接OE，
∵CO2 CE CP，
5
OC PC
∴ ，
CE OC
又 C C，
∴△POC∽△OEC，
∴∠POC=∠OEC ················································································· 2 分
EO CO，
OEC OCE，
∴∠POC=∠OCE，
∴PC=PO ·····················································································4 分
（2）解：①方法一：过 P作PH OD于 H，连接 DP、DE,
∵直径 AB=4，∴半径 OA=2，
∴ PO 4，
∵ cos BOC cos POH OH 1
PO 4
1
∴OH= PO=1 ·············································································5 分
4
∴HD=OD-OH=1=OH
∵ PH OD
∴PD=PO=4······························································································· 6 分
∵CD=4
∴PD=CD ·················································································· 7 分
CD是直径，
DEC 90 ··············································································· 8分
DE PC，
PE CE ················································································ 9分
①方法二：过点 C 作 CF⊥OB 于 F，连接 AE、BC，
1
∵COS∠COB= = OC=2
4
1
∴OF = ·························································································5分
2
6
9
∴PF=PA+AO+OF =
2
在Rt△ OCF 中，CF= 0 2 2= 15
2
在Rt△ PCF 中，CP= 2 + 2=2 6 ························································ 6 分
∵四边形 ABCE 内接于⊙O，
∴∠PEA=∠PBC,
又∵∠P=∠P
∴ΔPEA∽ΔPBC
= ∴ ····································································································7 分

即 PE×2 6=2×6
解得 PE= 6······························································································ 8 分
∴EC=PC-PE=2 6- 6= 6 =PE
∴E 是 PC 的中点 ·································································· 9 分
（2）②△BDE 为等腰三角形，有三种情况：
（i）当 ED BE时，连接 EO， EA，BC.
又 OB OD,OE OE ，
△OEB≌△OED
OBE ODE
AE E C
AOE COE
AOE 1 AOC
2
又 ABC 1 AOC
2
AOE ABC
OE BC
PE PO 4 2

PC PB 6 3
PC 3 PE
2
∵四边形 ABCE 内接于⊙O，
7
∴∠PEA=∠PBC,
又∵∠P=∠P
∴ΔPEA∽ΔPBC
= ∴

3
即 PE 2 2 6
2
PE 2 2 (取正) ···········································································11 分
（ii）当DE DB时，连接 EA，OE，BC，
又 OE OB
OD垂直平分BE ,垂足为G
OGB 90
AB为直径
AEB 90 OGB
AE OC
A是OP的中点
E是PC的中点,即PC=2PE
由（i）得ΔPEA∽ΔPBC
= ∴

即 2PE 2 2 6
PE （6 取正） ············································································ 13 分
（iii）当 BD BE时，如图，
BD BE
B D B E
OP DE
又∵CD 是直径，
∴∠CED=90°
PC DE
OP PC ,这与OP与PC相交于P点矛盾
BD BE不可能，即BD BE
综上述，PE的长为2 2或 6.
8
···········································································································14 分
9

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