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泉州市 2026届初中毕业班模拟考试
数学试卷评分参考
评分说明:
1.本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容，比照
评分参考制定相应的评分细则。
2.对计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响
的程度决定后继部分的给分，但不得超过该部分正确解答应给分数的一半;如果后继部分的解答有较严重
的错误，就不再给分。
3.解答右端所注分数，表示考生正确做到这一步应得的累加分数。选择题和填空题不给中间分。
一、选择题（每小题 4 分，共 40 分）
1.A 2.C 3.B 4.A 5.B 6.B 7.C 8.D 9.C 10.B
二、填空题（每小题 4 分，共 24 分）
11. x 3 12. x x 2 13. 2 14. 1 15. 2 16. 5
3 3
三、解答题（共 86 分）
17.（8分）
解：原式 2 2 2 1 1 ·································································································· 6分
3 2 . ·············································································································· 8分
(其它解法，请参照以上评分参考)
18.（8分）
证明：∵四边形 ABCD为矩形，
∴ A D 90 ，··········································································································· 2分
在△AEF 和△DCE中，
A D，

AEF DCE，

EF CE.
∴△AEF ≌△DCE，········································································································ 6分
∴ AF DE．···················································································································8分
(其它解法，请参照以上评分参考)
19.（8分）
x 1 1 x 2 1
解：原式 2 ·····························································································2分x 1 x 4x 4
x 2 (x 1)(x 1)
2 ······························································································ 5分x 1 (x 2)
x 1
··············································································································· 6分
x 2
数学学科评分参考 第 1 页 共 7 页
当 x 3时，
3 1
原式 ················································································································7分
3 2
4．················································································································· 8分
(其它解法，请参照以上评分参考)
20.（8分）
解：（1） 93， 90；··········································································································· 4分
（2） S 2 1B [(90 93)
2 (95 93)2 (90 93)2 (95 93)2 (95 93)2 ] 6 ；··································· 6分
5
（3）因为 x 2 2A 93， xB 93， SA 26， SB 6，
所以 x x ， S 2 2A B A SB ，······································································································7分
所以 B款机器人的清扫效率更稳定．···················································································· 8分
(其它解法，请参照以上评分参考)
21．（8分）
解：（1）如图 1，点 E为所求作的点；··················································································· 4分
A A
E
E
O D O
B B D
C C
图 1 图 2
（2）如图 2，
∵ AB∥CD，CE∥AD，
∴四边形 ADCE 为平行四边形．··························································································· 6分
∵ D是 AC中点，
∴ AD C D，
∴ AD CD，··················································································································· 7分
∴ ADCE 为菱形．·········································································································· 8分
(其它解法，请参照以上评分参考)
22.（10分）
解：（1）设体积V 与温度 t的函数表达式为V kt b(k 0) ，根据题意，
10k b 1000.3，
得 ···········································································································2分
60k b 1002.3，
k 0.04，
解得 ·················································································································4分
b 999.9.
所以体积V 与温度 t的函数表达式为V 0.04t 999.9．·····························································5分
数学学科评分参考 第 2 页 共 7 页
当温度为 200 C 时合金球的体积为V 0.04 200 999.9 1007.9(cm 3) ．······································ 6分
（2）将 t 200代入V 0.0 397t 999.95，得
V 0.0 397 200 999.95 1007.89，···················································································· 7分
所以1007.9 1007.89 0.01，1 007.89 1006 1.89，
因为 0.01 1.89，·············································································································· 9分
所以小明的结果更接近最佳表达式．·····················································································10分
(其它解法，请参照以上评分参考)
23.（10分）
解：（1）因为二次函数的图象经过点 B 2 m,5 ，C 2 m,5 ，
所以二次函数 y x2 (a b)x bc的图象的对称轴为直线 x 2 ，············································· 2分
a b
所以 2 ，
2
所以 a b 2 2．··········································································································· 3分
因为二次函数 y x2 (a b)x bc的图象经过点 A 1，2 ，
所以12 (a b) bc 2 ，································································································· 4分
所以 bc 3 2 1．············································································································ 5分
（2）解法一：
不存在．理由如下：·········································································································· 6分
假设存在正整数 c，使得 a为正整数，
由（1）得， a b 2 2，则 b a 2 2 ，
所以 bc a 2 2 c ac 2 2c .
又因为 bc 3 2 1，
所以 ac 2 2c 3 2 1，···································································································7分
所以 ac 1 2 3 2c ，
因为 c为正整数，
所以 3 2c 0，
ac 1
所以 2 ，·············································································································8分
3 2c
又因为 a， c为正整数，
ac 1
所以 为有理数，······································································································· 9分
3 2c
所以 2 为有理数，
这与 2 为无理数矛盾，····································································································· 10分
数学学科评分参考 第 3 页 共 7 页
所以假设不成立，即不存在正整数 c，使得 a为正整数．
解法二：
不存在．理由如下：·········································································································· 6分
假设存在正整数 c，使得 a为正整数，
由（1）得， a b 2 2，则 b a 2 2 ，又 bc 3 2 1，
所以 a 2 2 c 3 2 1，································································································· 7分
因为 c为正整数，
所以 c 0，
3 2 1
所以 a 2 2 ，·····································································································8分
c
因为 c为正整数，
所以 c≥1，
3 2 1
所以 ≤3 2 1，······································································································9分
c
3 2 1
所以 2 2≤3 2 1 2 2= 2 1 1，
c
即 a 1，
所以 a不为正整数，
这与 a为正整数矛盾，······································································································· 10分
所以假设不成立，即不存在正整数 c，使得 a为正整数．
(其它解法，请参照以上评分参考)
24.（12分）
解：（1）① ADE ，② CDE，③ 360 .··············································································· 3分
（2） k1 k2 k3 是定值，··································································································· 4分
理由如下：
∵△PAF∽△PQR，△EDR∽△PQR，
PF ER
∴ k1， k3，·······································································································5分PR PR
∵△BQC∽△PQR，且 BC EF ，
BC EF
∴ k2 ················································································································6分PR PR
k k k PF EF ER PR∴ 1 2 3 1，PR PR PR PR
数学学科评分参考 第 4 页 共 7 页
即 k1 k2 k3 1为定值.······································································································ 7分
S 1 S S
（3）由（2）知 1 k 21
2 2
， 2 k2 ，
3 k3 ，····································································8分S 9 S S
1 S4 S (S S S ) S当 k1 时，
1 2 3 1 S1 S 2 3 8 k 22 k 23 .············································· 9分3 S S S S S 9
∵ k1 k2 k3 1，
∴ k3 1 k
2
1 k2 k3 2
，
S 8 2 24 k 2
2 1 2 2
∴ k

2
k ≤ ，·································································· 10 分
S 9 2 3 2 2 3 3 3
k 2∵ 2 k3 1 k1 ，且 k 0， k 0，3 2 3
2
∴ 0 k2 ，·················································································································· 11分3
k 1 S 2∴当 42 时， 取得最大值为 .······················································································ 12分3 S 3
(其它解法，请参照以上评分参考)
25.（14分）
（1）证明：∵△ABC是等边三角形，
∴ ACB 60 ，···············································································································1 分
由旋转可知 EAF EAC CAF 60 ，············································································ 2分
又∵ ACB AFB CAF ，
∴ AFB EAC；··········································································································· 3分
（2）如图 1，延长 AE至点 K，使得 EK EA，连接 BK ， DK ， FK ， B K
又∵点 E为 BD中点，
∴四边形 BADK 为平行四边形，·························································································· 4分
E
∴ BK∥DA，
∴ KBA BAC 180 .
C
∵△ABC是等边三角形， A D
∴ BA CA， ACB BAC 60 ， 图 1 F
∴ KBA 180 BAC 120 ，
∵ FCA 180 ACB 120 ，
∴ KBA FCA .··············································································································5分
由（1）得， BAC EAF 60 ，
数学学科评分参考 第 5 页 共 7 页
∴ BAC KAC EAF KAC ，即 BAK CAF ，
∴△KAB≌△FAC，··········································································································6分
∴ KA FA，
∴△KAF 为等腰三角形.
又∵ KAF 60 ，
∴△KAF 为等边三角形.···································································································· 7分
又∵ EK EA，
∴ AE⊥EF；··················································································································· 8分
（3）结论①正确，理由如下：
由（2）得， AE⊥EF ，
在Rt△AEF 中， EF AF sin EAF 3 AF ，······································································ 9分
2
如图 2，设 DK 与 BF 相交于点 P． B K
由（2）得，△KAB≌△FAC，
∴ BK CF . P
E
∵四边形 BADK 为平行四边形， H
∴ AB∥KD，
M
∴ PDC BAC 60 . A CD
∵ DCP 60 ，
图 2 F
∴ DPC 180 PDC DCP 60 ，
∴ PDC DCP DPC ，
∴△PDC为等边三角形.····································································································· 10分
∵ DH PC ，
∴ PH CH .···················································································································· 11分
∵ BK∥AD，
∴ BKP PDC 60 ，
又∵ BPK DPC 60 ，
∴ BPK BKP，
∴ BK BP，···················································································································12分
∴CF BP，
∴ BH FH ，即 DH 是线段 BF的垂直平分线，
∴ BM FM .··················································································································· 13分
∵ BM EM FM EM EF ，
∴ BM EM 3 AF ，
2
数学学科评分参考 第 6 页 共 7 页
即 2(BM EM ) 3AF .····································································································· 14分
结论③正确，理由如下：
如图 3，延长 AC至点Q，使得QC AD，连接MQ，则 DQ DC QC DC AD AC .
设 DK 与 BF 相交于点 P．
由（2）证得，△KAB≌△FAC，
∴ BK CF .
∵四边形 BADK 为平行四边形，
∴ AB∥KD，
∴ PDC BAC 60 .
∵ DCP 60 ，
∴ DPC 180 PDC DCP 60 ，
∴ PDC DCP DPC ，
∴△PDC为等边三角形.····································································································· 9分
又 DH PC ，
∴ MDK MDQ .
∵ DK AB AC ，
∴ DK DQ .···················································································································· 10分
∵MD MD，
∴△MDK≌△MDQ，
∴MK MQ .····················································································································11分
∵ AE EK ， AE EF B， K
∴MA MK ， P
∴MA MQ . E H
过点M 作MN AC 于点 N，则 NA NQ ，
∴ NA AD NQ QC ，即 DN CN ， M
A D N C Q
∴MN 垂直平分CD，
∴MD
F
MC ， 图 3
∴ MCD MDC .··········································································································· 12分
由 DH BC ， DCH 60 可得， MDC 30 ，
∴ MCD 30 ，·············································································································· 13分
∴ HCM ACB MCD 30 MCD ，
即CM 平分 ACB .············································································································ 14分
(其它解法，请参照以上评分参考)
数学学科评分参考 第 7 页 共 7 页保密★启用前
泉州市 2026 届初中毕业班模拟考试
2026.05
初 三 数 学
（本卷共25题；满分：150分；考试时间：120分钟）
友情提示：所有答案必须填写到答题卡相应的位置上。
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，只有一项
是符合要求的。
1． 下列各数中，负数是
A． 1 B． 0 C．1 D． 2
2． 据报道， 2026年春节假期，泉州市文旅市场供需两旺，累计接待游客 1 618.39万人次，比
去年同期增长 59.98%．数据 16 183 900用科学记数法表示为
A． 0.161 839 108 B． 0.161 839 109
C．1.618 39 107 D．1.618 39 108
3． 德化瓷烧制技艺是福建德化地方传统手工技艺，被列入第一批国家级非物质文化遗产名
录．如图，是德化陶瓷茶杯，关于它的三视图，下列说法正确的是
A．主视图与俯视图相同 B．主视图与左视图相同
C．左视图与俯视图相同 D．三视图都相同
4． 下列运算正确的是
A． 2a2 3a2 5a2 B． a2 a4 a8
C． a6
3
a3 a2 D． 2a2 6a6
5． 如图，△ABC中，借助直角三角板作 AB边上的高，将三角板按如图所示摆放，
其中点 A， B， E在同一直线上，点 E，C，D在同一直线上， A 25 ，
E 90 ，则 ACD的大小为
A．110 B．115 C．120 D．125
6． 某校开展“向海图强，我是先锋”红领巾讲解员大赛，评分设置“主题内容”“语言表达”
“仪态台风”三项，依次按 5:3: 2的比例计算综合得分，某选手三项得分（百分制）依次为
94分，90分，92分，则该选手综合得分为
A. 92分 B. 92.4分 C. 92.8分 D. 94分
7． 如图， O 是△ABC的外接圆， OAC 23 ，则 B的大小为
A． 46° B． 60 C． 67° D． 77
初三数学试题 第 1 页 （共 6 页）
8． 我国自主研制的全超导托卡马克核聚变实验装置 EAST 实现了上亿度 1066秒稳态长脉冲
高约束模等离子体运行，刷新世界纪录．下表是该装置实现稳态长脉冲高约束模运行时间
的突破历程：
年份 2012 2016 2017 2023 2025
运行时间（秒） 30 60 101 403 1066
若 2023年至 2025年运行时间的年平均增长率设为 x，则符合题意的方程为
A． 403 1 2x 1066 B． 403 1 x2 1066
C 2 2．1066 1 x 403 D． 403 1 x 1066
9．已知 a＞b，下列说法不一定正确的是
A．若 b＞c，则 a＞c B．若 c＞d，则 a c＞b d
C．若 c＞d，则 ac＞bd D．若 c＞d，则 a d＞b c
10．已知二次函数 y ax2 bx c a 0 的图象经过点 A 4，k ， B 2，k 两点，若关于 x的方
2
程 a x 3 b x 3 c k 1有两个不相等的实数根m，n m n ，则下列结论正确的是
A． n -m = 4 B． n +m = 4 C． n -m = 6 D． n +m = 6
二、填空题：本题共 6 小题，每小题 4 分，共 24 分。
1
11．若分式 有意义，则 x的取值范围是________．
x 3
12．因式分解： x2-2x= ________．
13．在Rt△ABC 中， C 90 ， AC 3， BC 2，则 tan A的值为________．
14．某班从“均衡饮食”“体育锻炼”“心理健康”三个健康主题中随机选两个开展班会，则恰
好选中“均衡饮食”与“心理健康”的概率为________．
15．如图，在平面直角坐标系 xOy 中，点 A在第一象限，点 B在 x轴的正
k
半轴上， AO AB，反比例函数 y k 0 的图象经过△AOB的重
x
心G，若△AOB的面积为 6，则 k的值为________．
16．如图，正方形 ABCD的对角线相交于点O，点 E是OA的中点，点 F，G
是 BD上两个动点（ F 在G 的下方），且满足 AE FG，若正方形边
长为 2，则 AG EF的最小值为________．
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）
计算： 8 2 1 30．
初三数学试题 第 2 页 （共 6 页）
18．（8分）
如图，点 E， F 分别在矩形 ABCD的边 AD， AB上， AEF DCE， EF CE．求证：
AF DE．
19．（8分）
1 x2 4x 4
先化简，再求值： 1 ，其中 x 3．
x 1 x2 1
20.（8分）
某科技小组对 A，B两款智能扫地机器人进行清扫效率测试，在相同测试环境下，各清扫 5
次，测得每分钟清扫面积（单位：平方分米）如下表：
款式 第 1次 第 2次 第 3次 第 4次 第 5次 平均数
A 95 90 95 85 100 a
B b 95 90 95 95 93
（1）表格中 a ，b ；
（2）请计算 B款机器人每分钟清扫面积的方差；
（3）若 A款机器人每分钟清扫面积的方差为 26，根据两款机器人每分钟清扫面积的平均数
与方差，判断哪款机器人的清扫效率更稳定，并说明理由．
21．（8分）
如图，四边形 ABCD内接于 O， AB∥CD．
（1）在 AB上求作点 E，使得CE∥AD；（要求：尺规作图，不写作法，保留作图痕迹）
（2）在（1）的条件下，D是 AC的中点，求证：四边形 ADCE 是菱形．
A
O D
B
C
初三数学试题 第 3 页 （共 6 页）
22．（10分）
某研究小组为探究一种固态合金球的体积随温度变化的规律，测得不同温度下合金球的体
积数据如下表（已知该合金的熔点为1000℃）：
温度 t / C 40 20 10 0 10 20 40 60
体积V / cm3 998.3 999.2 999.6 1 000 1000.3 1000.7 1001.6 1002.3
（1）小明认为V 与 t 之间近似地符合一次函数关系，他选取其中两组数据 10，1000.3 和
60，1002.3 ，请帮他求出函数的表达式，并算出温度为 200℃时合金球的体积；
（2）小华选取其它数据算出温度为 200℃时，合金球的体积为1006cm3．研究小组认为这
批数据可能分布在某条直线附近，于是他们利用某 AI（人工智能）平台对全部数据进行分析，
得到一次函数的最佳表达式为V 0.0 397t 999.95．小明和小华计算 200℃时合金球的体积结果，
哪位同学的结果更接近最佳表达式？请说明理由.
23．（10分）
二次函数 y x2 a b x bc的图象经过点 A 1， 2 ， B 2 m，5 ，C 2 m，5 ．
（1）求 bc的值；
（2）是否存在正整数 c，使得 a为正整数？若存在，请求出所有符合条件的 c的值；若不存
在，请说明理由．
初三数学试题 第 4 页 （共 6 页）
24．（12分）
综合与实践主题：废料再利用，瓷砖的密铺与优化设计
【项目情境】某工地在铺设地面过程中，产生了一批规格相同的三角形瓷砖废料．为了废料
利用，工人师傅希望从这些三角形瓷砖中，切割出对边分别平行且相等的六边形（称为“平行六
边形”）瓷砖，并用于地面铺设．现需解决两个问题：仅用这种平行六边形能否铺满地面；在一
定条件下，为了充分利用三角形瓷砖，如何切割才能使得平行六边形的面积最大．
【活动一：密铺可行性探究】猜想：仅用规格相同的“平行六边形”可以铺满地面，铺设效
果如图 1所示． 如图 2，平行六边形 ABCDEF中，AF∥CD，AB∥DE ，BC∥EF，AF CD，
AB DE， BC EF．
试说明：平行六边形 ABCDEF可以铺满地面．
证明：连接 AD，
∵ AF∥CD， AB∥DE ，
∴ FAD ADC， BAD ① ，
∴ BAF ② ，
同理， B E， C F，
∵六边形的内角和为 720 ，
∴ 2 BAF B C 720 ，
∴ BAF B C ③ ．
即在每个顶点周围放置三个规格相同的平行六边形，恰好组成一个周角，所以可以铺满地面．
【活动二：废料图形性质探究】按图 3的方式，从三角形瓷砖中切割出一个平行六边形后，
会产生三个新的小三角形：△PAF，△BQC，△EDR ，它们都与△PQR相似，记它们与△PQR
的相似比分别为 k1， k2， k3 ，探究 k1， k2， k3 的数量关系．
【活动三：面积最大化探究】在一定条件下，为了充分利用
三角形瓷砖，如何切割才能使得平行六边形的面积最大？记
△PAF，△BQC，△EDR ，六边形 ABCDEF，△PQR的面积分
别为 S
S
1， S2 ， S3， S4 ， S，探究 4 的最大值．S
阅读以上材料，并回答下列问题：
（1）补全活动一证明过程①②③所缺的内容；
（2）活动二探究中， k1 k2 k3 是否定值，若是，请说明理由；若不是，请举一个反例说明．
1 S
（3）活动三探究中，当 k 41 时，求 的最大值．3 S
初三数学试题 第 5 页 （共 6 页）
25．（14分）
如图 1，△ABC是等边三角形，D为 AC边上不与 A，C重合的一点，点 E为 BD中点，连
接 AE，将射线 AE绕点 A顺时针旋转 60 交 BC的延长线于点 F ．
（1）求证： AFB EAC；
（2）求证： AE EF ；
（3）过点 D作DH BC 于点 H ，交 EF 于点M ，连接 AM ， BM ，CM ，如图 2．已知
下列三个结论中，至少有一个是正确的，请你选择其中正确的一个结论，并证明．
结论：① 2 BM EM 3AF ；
② AM BM ；
③CM 平分 ACB．
注：如果选择的结论不正确，那么第（3）问得 0分；如果选择多个正确的结论分别解答，
按第一个解答计分．
B B
E E
H
M
A D C A D C
图1 F 图2 F
初三数学试题 第 6 页 （共 6 页）

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