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平谷区 2024 年一模试卷评分标准
初 三 数 学
2025 年 4 月
一、选择题（本题共 16 分，每小题 2 分）
题号 1 2 3 4 5 6 7 8
答案 D B B C C A A D
二、填空题（本题共 16 分，每小题 2 分）
题号 9 10 11 12 13 14 15 16
答案 x 5 a(x +1（) x 1） x =1 50° ＞ 96 5 6;4
2
三、解答题（共 68 分，第 17—19 题，每题 5 分，第 20-21 题，6 分，第 22 题，5 分，第 23 题，6 分，第 24—25 题，
每题 5 分，第 26 题 6 分；第 27—28 题，每题 7 分）解答应写出文字说明、演算步骤或证明过程.
0
3tan 30 ( 2025) + 2 12
17.解：
3
=3 1+ 2 2 3
3 ····················································································4
=1- 3 ········································································································5
（2 x 1）＞x 3

x 14
＜ 2x
18.解不等式组： 3
x 1
解①得 ····························································································································· 2
x 2
解②得 ······························································································································ 4
1 x 2 ·································································································5
19.先化简，再求值：
（2 x 2y）+ 2x
x 2 2xy + y2
2x 4y + 2x
=
（x-y）2 ························································································································· 2
（4 x y）
=
（x y）2 ································································································································· 3
4
=
x y ······································································································································· 4
x y = 2,
4
原式 = = 2
2 ························································································································· 5
20．解：
（1）∵AE∥BF，AF∥BE
∴四边形 AFBE 是平行四边形 ···································································································· 1
∵矩形 ABCD
∴∠D=90°
∴∠DAE+∠AED=90°
∵∠DAE=∠BEC
∴∠BEC+∠AED=90° ················································································································· 2
∴∠AEB=90°
∴四边形 AFBE 是矩形 ················································································································ 3
（2）∵∠DAE=30°，∠D=90°，DE=1
∴AE=2
∵矩形 ABCD
∴∠BAD=90°，
∴∠BAE=60°
∵∠AEB=90°
∴∠ABE=30° ······························································································································ 5
∴AB=2AE=4
∵矩形 AFBE
∴EF=AB=4 ····································································································6
21. 解：设明明第一天领取 x 分钟，妹妹第一天领取了 y 分钟. ·································2
x =（1+50%）y

x =（1+50%）
2y 15 ·············································································································· 4
x = 30

y = 20 解得: ················································································································· 5
答：
明明第一天领取 30 分钟，妹妹第一天领取了 20 分钟 ·······································6
（方法不唯一，其他方法依步骤给分）
22.（1）∵函数 y=2x 过点（1，m）
∴m=2 ··········································································································1
∵函数 y＝kx＋3(k≠0)过点（1，2）∴k=-1 ·············································································· 2
y = x + 3 ······························································································································· 3
（2） a 2. ··············································································································· 5
23.（1）证明：连接 BD
∵AB 是直径
∴∠ADB=90° ································································································1
∵AB=BC，∠ADB=90°
∴AD=DC ······································································································································· 2
∵O 是 AB 的中点
∴OD∥BC
······················································································································································ 3
C
（2）过点 D 作 DH⊥AB 于点 H
D
AC = 4 5
AD = 2 5
A BH O
∵AB=BC=10
由勾股定理得：BD = 4 5
················································································4
∵∠ADB=∠DHA=90°
E
AD BD=AB DH
2 5 4 5=10 DH
∴DH=4
∴HO=3 ········································································································5
∵BE 是⊙O 的切线
∴∠ABE=90°
∵DH⊥AB
∴△DHO∽△EBO
3 5
=
4 BE
20
解得：BE =
3 ······························································································6
24.解（1)补全函数图象 ·············································································································· 2
脂肪氧化率(g/min)
0.7
0.6
0.5
0.4
0.3
0.2
0.1
O 45 50 55 60 65 70 75 80 85 90运动强度（%vo2max）
(2)①0.55（0.54-0.58 均可） ····································································································· 3
52 x 78（下限 51-53 之间，上限 77-79 之间均可）
② ·····································4
③ 8 ············································································································5
.
25.(1) m=__88____，n=__87___；
················································································· 2
(2) 补全频数分布直方图；
······················································································································································ 3
（3）乙；甲 ································································································································ 5
26.（1）当 a=1 时，
抛物线 y = x
2 2x +C
抛物线的对称轴为 x=1 ··············································································································· 1
点A（ 1，y），B (m，y1 2 )
因为 y1 = y2
所以点 A 与点 B 关于 x=1 对称
∴m=3
··········································································································································· 2
（2）∵抛物线的对称轴为 x=a，······························································································ 3
情况 1：当 a＞0 时
A( a，y1 )一定位于对称轴的左侧
x=a
A（-a，y ） A，1 (3a，y1）
4
2
a 2 4

3a 4 解得
a
此时，有 3 ············································································4
情况 2：情况 1：当 a＜0 时
A( a，y1 )一定位于对称轴的右侧
x=a
a 2解 得 a 2
此时，有
-2 a<0
····························A··，··(·3·a·，···y··）···························A··（··-·a·，···y··）····································5 1 1
综上，
4
a 或-2 a<0 2
3 4
······················································································································································ 6
27.(1)
∵将线段 BC 绕着点 C 逆时针旋转β°得到线段 CD
∴CB=CD，∠BCD=β
180
∠CBD=
2 ..........................................1
180
+ =180
2
2 =180 E
..........................................2
(2)
C
1
结论：EG = BD
2 .........................................3
F G
依题意补全图形；.........................................4
证明： 取 BD 的中点 K，连接 CK、FK、EF. D
A B K
3
=
2
∴∠ABC=135°，∠BCD=90°
∵K 为 BD 中点
∴CK⊥BD
∵AE⊥DC
∴∠AEC=∠AKC=90°
∵点 F 是 AC 中点
∴FK=AF=FC=EF
∵BC=CD，∠BCD=90°
∴∠D=45°
∵∠AED=90°
∴∠EAD=45°
∴∠EFK=∠EFC+∠CFK
=2∠EAC+2∠CAD
=2∠EAD
=90°
∵AB=BC，F 为 AC 中点
∴BF⊥AC
∠BFG=90°
∴∠BFK=∠EFG
∵FB=FG
∴△EFG≌△KFB..........................................6
∴EG=BK
1
EG = BD
2 .....................................7
28．解：（1）A1B1，A3B3； ········································································································ 2
1 1
（0， 1+ 3）或（0， 1 3）
（2） 4 4 ············································································· 4
（3）2; ······································································································································· 5
2
2 1
1
x
–3 –2 –1 O 1 2 3
x
–3 –2 –1 O 1 2 3
–1
–1
–2
–2
–3 –3
3+ 13 3 + 39 1 13 3 39
（ ， ）或（ ， ）
8 8 8 8 ··························································· 7

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