2023 年初中学业质量检查数学试题
参考答案及评分标准
说明：
（一）考生的正确解法与“参考答案”不同时，可参照“参考答案及评分标准”的精神进行评分．
（二）如解答的某一步出现错误，这一错误没有改变后续部分的考查目的，可酌情给分，但原则上不超过后
面应得的分数的二分之一；如属严重的概念性错误，就不给分．
（三）以下解答各行右端所注分数表示正确做完该步应得的累计分数．
一、选择题（每小题 4分，共 40分）
1.B 2.D 3.A 4. C 5.A 6. C 7.D 8.C 9.B 10.D
二、填空题（每小题 4分，共 24分）
11. 如 1(答案不唯一) 12. 0, 1 1 13. 不可能 14.4 15. 16. x3 x2 2 x1
三、解答题（共 86分）
17.（8分）
解：原式 2 3 2 3 7 ································································································· 6分
9 3 3 ············································································································ 8分
(其它解法，请参照以上评分标准)
18.（8分）
a 2 a 2 a 2 a
解：原式 2 ···················································································· 2分 a 1 a 1 a 1
a 2 a 2 a 1 a
2 ······················································································ 3分 a 1 a 2 a 1
a 2 a
········································································································· 4分
a 1 a 1
2
.·················································································································· 5分
a 1
当 a 3 1 2时，原式 ················································································ 6分
3 1 1
2 2 3
.················································································ 8分
3 3
(其它解法，请参照以上评分标准)
19.（8分）
D
证明：∵ ACB ABC，
E C
∴ AB AC，·········································································· 2分
∵ AB∥CE ， A B
(第 19题图)
∴ BAD ACE，·············································································································4分
2023 年初中学业质量检查数学试题参考答案 第 1 页 （共 6 页）
AB AC ,
在△ABD与△CAE中， BAD ACE , ·················································································5分
AD CE
∴△ABD≌△CAE S.A.S ，································································································ 6分
∴ AE BD .······················································································································· 8分
(其它解法，请参照以上评分标准)
20．（8分）
证明：∵四边形 ABCD C是矩形， D F P
∴ ADC DAB 90 ， AD∥ BC .······································ 2分 E
∵ DAC 30 Q，
A B
∴ ACD 90 30 60 .····················································3分 (第 20题图)
由旋转的性质可得 DAC FAE 30 ， ACD E 60 ，·····················································4分
∴ DAE 30 30 60 ，·································································································· 5分
∵ AD∥ BC，
∴ PQE DAE 60 ，····································································································· 6分
∴ QPE 180 60 60 60 ，·························································································· 7分
∴△PQE 是等边三角形.······································································································· 8分
(其它解法，请参照以上评分标准)
21.（8分）
解：(1)80 40% 200 (人)；··································································································3分
受调查学生喜欢运动项目条形统计图
(2)补全图形如图所示，
人数(单位：名)
30
360 54 ， 80
200
∴扇形统计图中喜欢跳绳的扇形圆心角的度数为 54 . 50
40
30
A B C D 项目
(3)法一：
画出树状图如下： 小张 A B C D
小王 A B C D A B C D A B C D A B C D
······································································································································· 6分
2023 年初中学业质量检查数学试题参考答案 第 2 页 （共 6 页）
共有 16 4 1种等可能结果，其中同时选中相同项目的共有 4种，故 P(项目相同) .
16 4
······································································································································· 8分
法二：
列表如下：
A B C D
A A, A A, B A, C A, D
B B, A B, B B, C B, D
C C, A C, B C, C C, D
D D, A D, B D, C D, D
······································································································································· 6分
4 1
共有 16种等可能结果，其中同时选中相同项目的共有 4种，故 P(项目相同) .
16 4
······································································································································· 8分
(其它解法，请参照以上评分标准)
22.（10分）
解：(1)由表中数据猜想： y是 x的反比例函数.·············································································· 1分
k k
设 y k 0 ，将 x 225， y 40，代入 y ，得 k 225 40 9000，
x x
9000
∴故所求函数关系式为 y .································································································4分
x
(2)由题意得 x 130 y 4500 ，································································································· 6分
9000 9000
把 y 代入，得 x 130 4500，··············································································· 7分
x x
解得 x 260，························································································································ 8分
经检验， x 260是原方程的根，且符合题意.····································· 9分
∴若直销店计划每天的销售利润为 4500元，则其售价应定为 260元．·····10分 A
(其它解法，请参照以上评分标准) O
23.（10分） G
(1)解：如图 1，⊙O是所求作的圆.·············································3分
B M C
(2)延长GO交⊙O于点 F ，连接OB、 BF ，则 F BEG 180 ， (第 23题图 1)
又∵ GEM BEG 180 ， GEM 45 ，
F A
∴ F GEM 45 ， O
∴ BOG 90 ， D
∵⊙O与 AM 相切于点G， G
∴ AG OG， AGD 90
B E M C
2023 年初中学业质量检查数学试题参考答案 第 3 页 （共 6 页(第）23题图 2)
∴ AGD BOG ，
∴OB∥ AG，························································································································4分
∴△BOD∽△AGD，
BO OD BO AG
∴ ，即 ，····································································································5分
AG GD OD GD
∵G是△ABC的重心，
∴ AG 2GM ，······················································································································ 6分
又∵ DG GM 1，
∴ AG 2，··························································································································· 7分
BO 2
∴ 2， BO 2OD，·····································································································8分
OD 1
设半径OB OG r ，则OD r 1，
∴ r 2 r 1 ，解得 r 2，······································································································ 9分
S 90 2
2 1
∴ 2 2 2 .·························································································· 10分弓形 360 2
(其它解法，请参照以上评分标准)
24.（13分）
(1)证明：
∵CD与CE 都是⊙O的切线，
∴CD CE ，即△CDE是等腰三角形.·············································A············································3分
(2)如图 1，作 AM 直线 DE M于点M ，作 BN 直线DE 于点 N，
D
则 AMD BNE 90 .
由(1)证得CD CE ，
F O
∴ CDE CED， G C
又∵ CDE ADM ， CED BEN ，
E
∴ ADM BEN , B N
∴△AMD∽△BNE ,··············································································(·第···2·4·题···图··1·)······················ 5分
AM AD
∴ ，
BN BE
∵ AD与 AF 都是⊙O的切线，
∴ AD AF ，同理可证 BF BE，····························································································· 6分
∵ AM DE， BN DE， FG DE，
∴ AM ∥ FG∥ BN ，
MG AF AD
∴ ，
GN BF BE
AM MG
∴ ，
BN GN
又∵ AMD BNE 90 ，
∴△AMG∽△BNG ,················································································································7分
∴ AGM BGN ，
∵ FG DE ,
∴ FGM FGN 90 ,
∴ FGM AGM FGN BGN ，
∴ AGF BGF ，即 FG 平分 AGB .························································································8分
(3) AT BF ，理由如下：
2023 年初中学业质量检查数学试题参考答案 第 4 页 （共 6 页）
过点Q作⊙O的切线HK ，分别交 AC、BC于点 H 、K，连接OA、OB、
OK、OH . A
∵ AB与 HK 都是⊙O的切线，∴ AB FQ，HK FQ， D
∴ AB∥ HK ， AFO OQH 90 ，∴ BAH AHK 180 H，
T
∵ AB与 AC都是⊙O的切线，
F O Q C
∴OA平分 BAC 1，即 OAH BAH OHA 1，同理可证 AHK ，
2 2
E K
B
∴ OAH OHA 1 BAH AHK 1 180 90 ，
2 2 (第 24题图 2)
∴ AOH 180 OAH OHA 90 ，
∴ HOQ AOF 90 ，
∵ FAO AOF 90 ，∴ HOQ FAO ，
∴△AFO∽△OQH ，··············································································································9分
FO AF
∴ ，又∵ FO OQ，∴ FO2 AF QH ，
QH OQ
FO BF
同理可证得△BFO∽△OQK ，∴ ，即 FO2 BF QK ，
QK OQ
∴ AF QH BF AF QK QK ，即 ，························································································ 11分
BF QH
HQ CQ CQ QK
∵ AB∥ HK ，∴△CHQ∽△CAT ，∴ ，同理可证 ，
AT CT CT TB
HQ QK QK TB
∴ ， ，
AT TB HQ AT
AF TB
∴ ，························································································································12分
BF AT
BF AT BF AT
，即 ，
AF BF AT TB AB AB
∴ AT BF .··························································································································· 13分
(其它解法，请参照以上评分标准)
25.（13分）
解：(1)由题意，得 y x n 2 2 4n 7，
令 x 0，则 y n2 3，
∴点 F 的坐标为 0, n2 3 .····································································································· 2分
(2) 2抛物线 y x n 2 4n 7的对称轴为直线 x 2 n，顶点 E的坐标为 2 n, 4n 7 ，
∵当 x 2时， y随 x的增大而减小，
∴ 2 n 2，解得 n 0，········································································································· 3分
∵点 F 在 y轴的正半轴，
∴ n2 3 0，解得 3 n 3，···························································································· 4分
2023 年初中学业质量检查数学试题参考答案 第 5 页 （共 6 页）
∴ 0 n 3 .此时 2 n 0，由点 A的纵坐标为 t 3，得 4n 7 3，解得 n 1；······························ 5分
综上， n的取值范围是1 n 3 .······························································································· 7分
(3)∵ BC∥ AE， y
∴△AQE ∽△CQB， E
EQ AQ C Q
∴ ，
BQ CQ T B A
如图，过点C作CT AB于点T，则 CTB EBA 90 ， O x
∵CT ∥QB， D
AQ AB
∴ ，
CQ TB
EQ AB
∴ .···········································································································(·第··2·5·题··图··)······
BQ TB
由题意得抛物线的顶点为 E h, k ，设 AB m m 0 ，则 A m h, m2 k ， B h, m2 k ，
由 A m h, m2 k 与 E h, k ，可求得直线 AE的解析式为 y mx mh k，···································9分
故经过点 B h, m2 k 且与直线 AE平行的直线CD的解析式为 y mx mh k m 2 m x h m k ，
············································································································································10分
2 y x h k , 2
联立得 2 ，化简并整理 x h m x h m 0得，·········································11分
y m x h m k
x h 1 5解得 m ，
2
C D 1 5∵ 在 的左侧，∴ xC h m，······················································································ 12分2
EQ AB xA xB xA xB m 5 1∴ .···························································· 13分
BQ TB xB xT xB xC h h 1 5
2
m
2
(其它解法，请参照以上评分标准)
2023 年初中学业质量检查数学试题参考答案 第 6 页 （共 6 页）2023年初中学业质量综合测试
数学试题
(试卷满分：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
顶是符合要求的。
1.到等于
A.3
B.-3
C.±3
D.3
2.据报道，2022年我国机械工业累讨实现营业收入28900000000000元，将
28900000000000用科学记数法表示为
A.2.89×10
B.28.9×1012
C.0.289×104
D,2.89×1013
3.如图是由圆弧及正方形组成的几何图形，则对该几何图形描述正确的是
A.既是轴对称图形，又是中心对称图形
B.不是轴对称图形，而是中心对称图形
C,是轴对称图形，而不是中心对称图形
D.不是轴对称图形，也不是中心对称图形
(第3题图)
4.对于非零实数4，·下列运算正确的是
A.(a)-a
B.(a2）=a
C.a.a3=a
D.a3.a2=a
5.如图是某零件抽象而成的几何体，则该几何体的左视图是
主坝方向
《第5题图)
6.某校欲组建一支33人的鼓号队，需要从初一年段挑选部分身高大小比较均匀的同学参
加，设初一全年段同学身高的平均数和方差分别为x、52,则入选该鼓号队的同学身高
平均数和方差分别为无、，则下列结论一定成立的是
A.xB.x>
C.s2>2
D.22023年初中学业质兰综合测试数学试题第1页共6页
7.我国古代数学家梅毂成在其数学著作《增删算法统宗》有题如下：“群羊一百四十，剪
毛不弹劳，群中有母有羊羔，先剪二羊比较：·大羊剪毛斤二，一十二两羔毛，百五十
斤是根苗，子母各该多少？”其大意是：今有一群羊140只，母羊与羊羔都可以剪毛.首
先剪两只羊的毛后知道：每只大羊可剪毛18两，每只羊羔可剪毛12两.现在总共剪得
羊毛150斤（注：1斤=16两）.试问大羊与羊羔各有多少？”若设大羊x只，羊羔y只，
则依题意可列方程组为
A.
x+y=150,
B.
x+y=140,
12x+18y=140×16
112x+18y=150×16
C.
|x+y=150,
D
x+y=140,
18x+12y=140×16
18x+12y=150×16
8.图，在Rt△ABC中，∠ACB=90°，AC=8,BC=6,点G是△ABC的重心，则CG
等于
A哥
8
B.
3
c.
3
D.4
(第8题网)
9.若点A(m,n-3)、B(m+1,-2n)都在一次函数y=-kx+b的x轴下方的图象上，则n的
值可以是
A.3
B.2
C.1
D.0
10.如图，在菱形ABCD中，线段EF经过对角线AC的中点O,
D
交AD于点E,EF⊥BC,垂足为F,将四边形EFCD沿
D
EF翻折得到四边形EFCD',若AD=5,AC=8,
则sin∠AOC的值为
B.
4
(第10思圈
c号
D.
24
25
二、填空题：本题共6小题，每小题4分，共24分。
11.若双曲线y=在第一、三象限，则k可以是
(写出…个k的值即可).
12.直线y=2x-1与y轴的父点坐标是
13.在一个不透明的袋子中装有3个黑球和2个白球，小球除了颜色不同外，其余均完全相
同，现随机从袋子中摸出一个小球，摸到的球恰好是红球，则该事件是
事
件.（填“随机”、“必然”或“不可能”）.
2023年初中学业质量综合测试数学试题第2页共6页
