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5,“计里画方”是中国古代一种按比例尺绘制地图的传统方法.绘图
时先在图上布满方格，然后按方格绘制地图内容.小华按照“计里
太原市2024年初中学业水平模拟考试（二）
画方”的方法，绘制了蒙山大佛旅游区的局部示意图（如右）.若该图
数学
中“开化寺”与“蒙山晓月”两处景点的坐标分别为(-1，-2)、(1，-2)，
则景点“蒙山氧吧”的坐标为
(第5题图)
(考试时间：上午8：30一10：30)
A.(-1,5)
B.(-2,5)
0.(-2,2)
D.(2,-2)
注意事项：
6.如图，同学们将平行于凸透镜主光轴的红光AB和紫光C射人同一个凸透镜，折射光
1.本试卷分第1卷和第Ⅱ卷两部分，全卷共8页，满分120分，考试时间120分钟
线BM,DW交于点O,与主光轴分别交于点F,F2,由此发现凸透镜的焦点略有偏差，
2.签题前，考生务必将自己的姓名、准考证号谟写在本试卷相应的位置
3,答案全部在答题卡上完成，答在本试卷上无效。
若∠ABM=165°，∠CD=160°，则LF,OF的度数为
红光义
B
。0
4.考试结来后，将太试卷和答题卡一并交回
A.165
B.160
紫光c
C.155
D.1459
(第6愿图)
第I卷选择题（共30分）
7,语文课上，同学们以“并州犹是诗故乡一唐代山西诗人群像”为主题展开研习活动.小彬
一、选择题（本大题共10个小题，每小题3分，共30分.在每个小题给出的四个选项中，只有
和小颖计划从王维、柳宗元、白居易、王勃四位唐代山西诗人中任选一位撰写研习报告，则
一项符合题目要求，请选出并在答题卡上将该项涂黑)
他们恰好选择的是同一位诗人的概率是
1.下列各数中最小的是
B.0
D.-4
B时
2.国家安全人人有责，维护国家安全人人可为.今年4月15日是第九个全民国家安全教育
日.下列国家安全图标中，文字上方的部分是中心对称图形的是
c号
王
机宗元
白居场
王动
8.知图，四边形ABCD内接于⊙0，AB=CD,连接0A,0C.若∠BAD=80°，
则LAOC的度数为
核安全
国士安军
生加安全
军事安全
B
C
A.100°
B.160
3.下列运算正确的是
C.120
D.135
A.m2·m=m
B.5m 6n 11mn
(第8题图
9.已知点A(6,m),B(-3,2),C(6,-m)在同一个函数图象上，则这个函数图象可能是
C.(4x2y2+y)÷y=4xy2
4.一个棱柱的侧面展开图如右图所示，则该棱柱底面的形状是
是平
(第4题图)
数学试卷（二）第1页（共8页）
数学试卷（二）第2页（共8页）2024 年太原市初中学业水平模拟考试（二）
数学试题参考答案及评分标准
一、选择题（本大题共 10道小题，每小题 3分，共 30分）
题号 1 2 3 4 5 6 7 8 9 10
选项 D C D B C D A B C B
二、填空题（本大题共 5道小题，每小题 3分，共 15分）
11.2 2 12.甲 13. 4 14.55 15.3 5
三、解答题（本大题共 8道小题，满分 75分）
16.（本题共 2个小题，每小题 5分，共 10分）
3 3 1(1)解：原式= 4 ·························································3分
2
=0 2 ································································ 4分
=2. ··········································································· 5分
（2）①乘法对加法的分配律 ··································································1分
② 2， 2 2 1 x 3.······························································· 3分
③x 3 . ············································································ 5分
2
17.（本题 6分）
证明：∵四边形 ABCD是平行四边形，对角线相交于点 O， O
∴OB=OD. ..........................................................2分
∵BE=DE，
∴OE⊥BD，即∴AC⊥BD.·····································································4分
∵四边形 ABCD是平行四边形.
∴□ABED是菱形.··············································································· 6分
18.（本题 9分）
解：（1）2 79 15%················································································ 3分
（2）合理. ···························································································4分
理由：根据统计表提供的信息，甲、乙两个方案的平均分相同，但是其他统计量不同.
第 1页
从优等率看，乙方案为 15%，甲方案为 10%，乙方案中优等实验样品的数量高于甲方案中优
等实验样品的数量，乙方案优于甲方案；
从中位数看，乙方案的中位数为 79 分，甲方案的为 78 分，说明乙方案试验样品综合评分的
中位数高于甲方案的，乙方案优于甲方案；
从方差看，乙方案样本方差小于甲方案样本方差，说明乙方案样实验样品的表现更为稳定.
综上所述，乙方案优于甲方案.·········································································8分
【说明：每运用一种统计量说明理由的，得 1 分，共 4 分】
（3）100×（10%+15%）=25（个）.
答：第二阶段要分析的试验样品共约 25个.························································9分
19.（本题 8 分）
解：（1）①(a－b)((a+b－1) ② ③a+b=1··················································· 3分
1 2 2023 1 2023 2
（2） + = + ··············································5分
2024 2024 2024 2024
（3）证明：方法 1：
2 m2 n2 mn m2 n2 mn
等式左边= + = 2 2 2 . .........................6分 n n n
2 mn n2 2mn m2 m2 n2 mn
等式右边= + 2 2 2 . ...........7 分 n n n
所以左边=右边，等式成立.································································· 8分
方法 2：
由材料可知，当 a≠b时，等式 a2+b＝a+b2只要满足条件 a+b=1即可成立.
2 +
2
而等式 = + （其中 m，n为任意实数，且 n≠0，n≠2m）

= 与材料中的条件等式结构相同，其中 a， =b. ·························· 6分


所以，该等式只要满足条件 + =1，等式即成立.

·······························································································7分
+ + 而 = =1, 所以，等式成立. ·································· 8分

第 2页
20.（本题 9 分）
解：如图，过点 A作 AH⊥BC交 CB的延长线于点 H，过点 E作 QN⊥EF垂足为点 E，交
AH于点 Q，交 DM于点 N.············································································· 1分
由题意得，四边形 AQEF和四边形 QHMN都是矩形，
∴HQ=MN，AQ=FE=48cm. .................................. 2分 Q N
∠DCM =180°-∠BCD=180°-101°=79°. ....... 3分
在△CDM中，∠CDM+∠DCM =90°，
H
∴∠CDM=90°-79°=11°.
而∠EDN=∠CDE+∠CDM=59°+11°=70°. ······················································4分
在 Rt△CDM中，∠CMD=90°，
tan DM DCM ,
CM
∴DM CM tan79 30 5.14 154.（2 cm）. ···················································· 5分
在 Rt△EDN中，∠END=90°，
cos EDN DN ,
DE
∴DN DE cos70 50 0.34 1（7 cm）. ·······················································6分
∵MN=DM-DN，
∴MN 154.2 17 137.（2 cm）. ······································································ 7分
∴AH=AQ+QH=EF+MN=48+137.2=185.2≈185（cm）. ········································· 8分
答：点 A到地面的距离约为 185cm. ·································································9分
21.（本题 9分）
解：设购进西芹 x箱，购进西红柿（400-x）箱.·················································· 1分
根据题意，得 18 4 x 24 6 400 x 10800.···································· 3分
解，得 x 150. ·································································4分
设销售这两种蔬菜的总利润为 y元.
根据题意，得 y 28 18 4 x 40 24 6 400 x . ······························· 6分
化简，得y 4x 4000.
可知 y是 x的一次函数，且 k = -4＜0，
所以，y随 x的增大而减小，当 x取最小值时，y最大.
第 3页
因为 x 150，且 x为整数，
所以，x最小为 150, 此时 400-x=250.
···········································································································7分
即，当 x=150 时，y取得最大值，y 最大值= -4×150+4000=3400（元）. ...................8 分
答：当购进西芹 150 箱，西红柿 250 箱时，销售这两种蔬菜的利润最大，
最大利润为 3400 元.·········································································· 9分
22.（本题 10 分）
2
解：（1）将 A（-1,0），B（3,0）代入 y x bx c,
1 b c 0,
得， ·······································································1分
9 3b c 0.
b 2,
解，得 ················································································· 2分
c 3.
2
∴抛物线的函数表达式为 y x 2x 3. ······················································· 3分
点 C,D的坐标为 C（0,3）, D（1,0）.······························································5分
（2）连接 OP，过点 P 作 PE⊥x轴于点 E.
∵点 P为抛物线上对称轴右侧的一个动点，点 P横坐标为 m，
∴点 P 2的坐标为（m， m 2m 3)，
E
∵PE⊥x轴于点 E，
∴ PE m2 2m 3,OE m . ··································································· 6分
∵点 C,D的坐标为 C（0,3）, D（1,0），∴OC =3, OD =1.
S PCD S PCO S POD S COD，························································7分
S 1 PCD OC OE
1 1
OD PE OC OD.
2 2 2
S 3m 1 3 3 1 5 PCD

m2 m m2 m. ·······································8分
2 2 2 2 2 2
2
S 1 m 5 25 PCD .········································································ 9分2 2 8
1 m 3 1 0 m 5 25＜ ＜ ， ＜ ，∴当 时,S PCO有最大值为 . ································10分2 2 8
第 4页
23.（本题 14 分）
解：（1）EM=FN.
证明：∵四边形 ABEF是矩形，
2 4
∴∠BAF=∠ABE=∠E=∠F=90°, AF=BE. ....................1分
∵△ABC是等边三角形， 1
3
∴∠1=∠2=60°. ................................................................2分
∴∠BAF-∠2=∠ABE-∠1，
∴∠3=∠4. ······························································································· 3分
∴△BME≌△ANF，
∴EM=FN.···································································································4分
（2）连接 DN.
∵点 N是 EF的中点，∴EF=2FN=2EN.
∵四边形 ADEF是矩形，
∴AD=EF.
∵AD=2DE,
∴2AF=2DE=2FN=2EN.
∴AF=DE=FN=EN.······················································································· 5分
∴∠5=∠6,∠NDE=∠8.
∵∠F=90°，∴∠5=∠6=180
90
45 . ···························································· 6分
2

同理，∠NDE=∠8 45 .
∴∠DNA=180°-∠6-∠8=90°.
∴∠DNC=180°-∠DNA=90°. ·······································································7分
∵AD∥EF，∴∠DAN=∠6，∠ADN=∠8=45°，
∴∠DAN=∠ADN,∴NA=ND.·········································································· 8分
DN
在 Rt△NDC，∠DNC=90°，∴tan∠C= .
CN
DN
∵∠C=60°，∴ tan60°= 3 .
CN
设 CN=k,∴AN=ND= 3 k. ·············································································9分
∵∠DNA=∠7=45°，∠C=∠C,
第 5页
∴△CMN∽△CDA,······················································································10分
MN CN CN k 3 1
∴ . ··············································· 11分
AD CA CN NA k 3k 2
2 3 6 7 3
（3） 或 . ········································································· 14分
2 6
【说明】上述解答题的其他解法，请参照此标准评分.
第 6页

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