江苏省南京市建邺区2018中考二模数学试题(含答案)

1、2018 年中考第一次模拟调研九年级数学学科注意事项：1本试卷共 6 页全卷满分 120 分考试时间为 120 分钟考生答题全部答在答题卡上，答在本试卷上无效2请认真核对监考教师在答题卡上所粘贴条形码的姓名、考试证号是否与本人相符合，再将自己的姓名、考试证号用 0.5 毫米黑色墨水签字笔填写在答题卡及本试卷上3答选择题必须用 2B 铅笔将答题卡上对应的答案标号涂黑如需改动，请用橡皮擦干净后，再选涂其他答案答非选择题必须用 0.5 毫米黑色墨水签字笔写在答题卡上的指定

2、位置，在其他位置答题一律无效4作图必须用 2B 铅笔作答，并请加黑加粗，描写清楚一、选择题（本大题共 6 小题，每小题 2 分，共 12 分在每小题所给出的四个选项中，恰有一项是符合题目要求的，请将正确选项前的字母代号填涂在答题卡相应位置上）1下列计算结果为负数的是（）A(3)(4)B(3)(4)C(3) (4)4D(3)2计算 a6×(a2)3÷a4 的结果是（）Aa3Ba7Ca8Da93若锐角三角函数 tan55°a，则 a

3、0;的范围是（）A0a1B1a2C2a3D3a44下列各数中，相反数、绝对值、平方根、立方根都等于其本身的是（）A0B1C0 和 1D1 和1图                    图5把球放在长方体纸盒内，球的一部分露出盒外，其截面如图所示，已知 EF=CD=4 cm，则球的半径长是（）A2 cmB2.5 cmC3 

4、;cmD4 cm6如图，是一个每条棱长均相等的三棱锥，图是它的主视图、左视图与俯视图若边 AB 的长度为 a，则在这三种视图的所有线段中，长度为 a 的线段条数是（）A12 条B9 条C6 条D5 条EFDAAB主视图左视图OBC俯视图（第 5 题）（第 6 题）二、填空题（本大题共 10 小题，每小题 2 分，共 20 分不需写出解答过程，请把答案直接填写在答题卡相应位置上）7函数 y

5、 1x中，自变量 x 的取值范围是8分解因式 a3a 的结果是9若关于 x 的一元二次方程 x2kx20 有一个根是 1，则另一个根是10辽宁号是中国人民解放军海军第一艘可以搭载固定翼飞机的航空母舰，其满载排水量为67 500 吨用科学记数法表示 67 500 是13如图，四边形 ABCD 是O 的内接四边形，若O 的半径为 3 cm，A110°，则劣弧BD的长为11一组数据&

6、#160;1、2、3、4、5 的方差为 S12，另一组数据 6、7、8、9、10 的方差为 S22，那么 S12S22（填“”、“”或“”）k012在同一平面直角坐标系中，反比例函数y1x（k 为常数，k）的图像与一次函数 y2 xa（a 为常数，a0）的图像相交于 A、B 两点若点 A 的坐标为(m，n)，则点 B 的坐标为cm14如图，点 F、G 在正五边形 ABCDE 的边上，BF、CG&#

7、160;交于点 H，若 CFDG，则BHG°ADABEBOHGCCFD（第 13 题）（第 14 题）16如图，以 AB 为直径的半圆沿弦 BC 折叠后，AB 与 相交于点 D若CD  BD，则B      °1CB15如图，正八边形 ABCDEFGH 的边长为 a，I、J、K、L 分别是各自所在边的中点，且四边形 I

8、JKL 是正方形，则正方形 IJKL 的边长为（用含 a 的代数式表示）3AHILBGADD      E三、解答题（本大题共 11 小题，共 88 分请在答题卡指定区域内作答，解答时应写出文字说明、证明过17（6 分）计算：æèa2 öø÷æèa öøïî21    

9、，CFOJKCB（第 15 题）（第 16 题）程或演算步骤）11aaìï2x0，18（7 分）解不等式组í5x12x1并把它的解集在数轴上表示出来3-3 -2 -10123（第 18 题）19（7 分）如图，四边形 ABCD 是平行四边形，线段 EF 分别交 AD、AC、BC 于点 E、O、F，EFAC，AOCO（1）求证：四边形 AFCE 是平行四边形；（2）在本题三个已知

10、条件中，去掉一个条件， 1）的结论依然成立，这个条件是（直接写出这个条件的序号）AEDOBFC（第 19 题）20（8 分）某天,一蔬菜经营户用 180 元钱从蔬菜批发市场批了西红柿和豆角共 40 千克到菜市场去卖,西红柿和豆角这天的批发价与零售价如下表所示：品名批发价(单位:元/千克)零售价(单位:元/千克)问：他当天卖完这些西红柿和豆角能赚多少钱?西红柿3.65.4豆角4.67.521（8 分）超市水果货架上有四个苹果，重量分别是 100 g、110 g、120 

11、;g 和 125 g（1）小明妈妈从货架上随机取下一个苹果恰是最重的苹果的概率是；（2）小明妈妈从货架上随机取下两个苹果它们总重量超过 232 g 的概率是多少？22（8 分）河西中学九年级共有 9 个班，300 名学生，学校要对该年级学生数学学科学业水平测试成绩进行抽样分析，请按要求回答下列问题：收集数据（1）若从所有成绩中抽取一个容量为 36 的样本，以下抽样方法中最合理的是 在九年级学生中随机抽取 36 名学生的成绩；按男、女各随机抽取&#

12、160;18 名学生的成绩；按班级在每个班各随机抽取 4 名学生的成绩整理数据（2）将抽取的 36 名学生的成绩进行分组，绘制频数分布表和成绩分布扇形统计图如下请根据图表中数据填空：C 类和 D 类部分的圆心角度数分别为°、°；估计九年级 A、B 类学生一共有名成绩（单位：分） 频数频率九年级学生数学成绩分布扇形统计图B 类A 类（80100）181225%B 类（6079）C 类（4059）961416A 类50%

13、D 类（039）3112数据来源：学业水平考试数学成绩抽样（第 22 题）分析数据（3）教育主管部门为了解学校教学情况，将河西、复兴两所中学的抽样数据进行对比，得下表：学校平均数（分）  极差（分）  方差A、B 类的频率和河西中学复兴中学717152804324970.750.82你认为哪所学校本次测试成绩较好，请说明理由23（8 分）下图是投影仪安装截面图教室高 EF3.5 m，投影仪 A 发出的光线夹角BAC30°，投影屏幕高 BC1.2&

14、#160;m固定投影仪的吊臂 AD0.5 m，且 ADDE，ADEF，ACB45°求屏幕下边沿离地面的高度 CF（结果精确到 0.1 m）（参考数据：tan15°0.27，tan30°0.58）DA（第 23 题）EBCF24（9 分）一辆货车从甲地出发以每小时 80 km 的速度匀速驶往乙地，一段时间后，一辆轿车从乙地出发沿同一条路匀速驶往甲地货车行驶 2.5 h 后，在距乙地 160 km&#

15、160;处与轿车相遇图中线段 AB 表示货车离乙地的距离 y1 km 与货车行驶时间 x h 的函数关系（1）求 y1 与 x 之间的函数表达式；（2）若两车同时到达各自目的地，在同一坐标系中画出轿车离乙地的距离y2 与 x 的图像，求该图像与x 轴交点坐标并解释其实际意义ykmA160O2.5B xh（第 24 题）25（8 分）某超市欲购进一种今年新上市的产品，购进价为 20

16、0;元/件，该超市进行了试销售，得知该产品每天的销售量 t（件）与每件销售价 x（元/件）之间有如下关系：t3x90（1）请写出该超市销售这种产品每天的销售利润 y（元）与 x 之间的函数表达式；（2）当 x 为多少元时，销售利润最大？最大利润是多少？26（9 分）ABC 中，ACB90°，AC:BC4:3，O 是 BC 上一点，O 交 AB 于点 D，交 BC 延长线于点 E连接 ED

17、，交 AC 于点 G，且 AGAD（1）求证：AB 与O 相切；（2）设O 与 AC 的延长线交于点 F，连接 EF，若 EFAB，且 EF5，求 BD 的长EAGCODB（第 26 题）F（P27 10 分）图是一张AOB45°的纸片折叠后的图形， 、Q 分别是边 OA、OB 上的点，且 OP2 cm将AOB 沿 P

18、Q 折叠，点 O 落在纸片所在平面内的 C 处（1）当 PCQB 时，OQcm；在 OB 上找一点 Q，使 PCQB（尺规作图，保留作图痕迹）；（2）当折叠后重叠部分为等腰三角形时，求 OQ 的长BBBQCOP        AO        P     &

19、#160; A备用图 1O        P        A备用图 2（第 27 题）2018年中考第一次模拟调研数学参考答案及评分标准说明：本评分标准每题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，参照本评分标准的精神给分一、选择题（每小题 2 分，共计 12 分）题号答案1A2C3B4A5B6D12（n，m）  

20、;  13               14108°        15    a         16. 18°a      &

21、#160;aa     a21a    (a1)(a1)a1二、填空题（每小题 2 分，共计 20 分）7x18a(a+1)(a-1)9-2106.75×1041172+ 232三、解答题（本大题共 10 小题，共计 88 分）17（本题 6 分）a22a1a21解：原式 ÷a22a1a·(a1) 2a·a1 ·

22、83;·················································

23、83;······································ 6 分18（本题 7 分）解：解不等式，得 x2 · &

24、#183;·················································&

25、#183;························· 2 分解不等式，得 x 1 ·················

26、83;·················································

27、83;····· 4 分所以，不等式组的解集是1x2········································

28、;················ 5 分画图-1    012    · ····················&#

29、183;·········································· 7 分19（本题 7 分）解：（1）四边形&#

30、160;ABCD 是平行四边形AECFDACBCA· ···········································

31、···························· 1 分在AOE 和COF 中ìïDAC=ACBí AO=COïîAOE=COFAOECOF（ASA） ····&

32、#183;·················································&

33、#183;············ 3 分AECF四边形 AFCE 是平行四边形· ····························

34、····························· 5 分（2）··················&

35、#183;·················································&

36、#183;························ 7 分   解：（1）···················

37、··················································

38、···························· 2 分    因此，总重量超过 232g 的概率是· ·········

39、········································· 8 分   tan  BAPtan1

40、5°0.27 ···············································&

41、#183;··················· 5 分20（本题 8 分）解：设批发了西红柿 x 千克，豆角 y 千克ì x + y = 40由题意得： í·······

42、3;············································· 3 分î3.6 

43、;x + 4.6 y = 180ì x = 4解得： í6 分î y = 36(5.4  3.6)× 4(7.5  4.6)× 36  111.6（元） ··············

44、3;························· 7 分答：卖完这些西红柿和豆角能赚 111.6 元 ················

45、···································· 8 分21（本题 8 分）14（2）共有 6 种等可能出现的结果，分别为···

46、;··················································

47、;· 3 分（100，110）；（100，120）；（100，125）；（110，120）；（110，125）；（120，125）····································

48、3;······················ 6 分总重量超过 232g 的结果有 2 种，即（110，125），（120，125） ··············&#

49、183;···· 7 分1322（本题 8 分）解：（1）······································

50、83;·················································

51、83;··· 2 分（2） 60°，30°········································

52、;··········································· 4 分 225···

53、··················································

54、······························ 6 分（3）两所学校都可以选择只要理由正确皆可得分··············

55、······························· 8 分选择河西中学，理由是平均分相同，河西中学极差和方差较小，河西中学成绩更稳定选择复兴中学，理由是平均分相同，复兴中学 A，B 类频率和高，复兴中学高分人数更多23（本题 8

56、 分）解：过点 A 作 APEF，垂足为 PADDE，ADE90°ADEF，DEP90°APEF，APEAPC90°,ADEDEPAPE90°四边形 ADEP 为矩形EPAD0.5m ························

57、3;·················································

58、3;·········· 2 分APC90°，ACB45°CAP45°ACB，BAPCAPCAB45°30°15°APCP·························

59、;··················································

60、;···················· 4 分在 APB 中BPAPBP0.27AP0.27CP,BCCPBPCP0.27CP0.73CP1.2mCP1.64m················&#

61、183;·················································&#

62、183;························ 7 分CFEFEPCP3.50.51.641.361.4m··················

63、3;··················· 8 分24（本题 9 分）解：（1）由条件可得 k1801 分设 y180x+b1，过点（2.5，160），可得方程 16080×2.5+b1解得 b1360 ········

64、··················································

65、··························· 3 分y1 80x+360··················&

66、#183;·················································&

67、#183;··········· 4 分（2）当 y1 0 时，可得 x4.5轿车和货车同时到达，终点坐标为（4.5，360）设 y2 k2 x+b2 ，过点（2.5，160）和（4.5，360）解得 k2 100，b2 90y2 100x90 图像如下图·······

68、················································· 

69、7 分与 x 轴交点坐标为(0.9，0) ···········································&

70、#183;························ 8 分说明轿车比货车晚出发 0.9h ··················

71、83;··············································· 9 分2

72、5（本题 8 分）解：（1）表达式为 y(3x+90)(x20)化简为 y3x²+150x1800·····································

73、3;··························· 4 分（2）把表达式化为顶点式 y3(x25)² +75·············

74、3;···································· 6 分当 x25 时，y 有最大值 75答：当售价为 25 元时，有最大利润

75、60;75 元 · ·············································

76、··· 8 分26（本题 9 分）（1）证明：连结 ODACB90°，OED+EGC90° ·································&

77、#183;································ 1 分O，ODOE，ODEOEDAGAD，ADGAGD ··········

78、;··················································

79、;···················· 3 分AGDEGCOED+EGCADG+ODEADO90°ODAB ····················

80、3;·················································

81、3;················· 4 分OD 为半径AB 是O 的切线································