2019年福建省龙岩市中考数学试题及答案(word版)

2019年龙岩市初中毕业、升学考试

数 学 试 题

（满分：150分 考试时间：120分钟）

考室座位号 注意：

请把所有答案填涂或书写到答题卡上！请不要错位、越界答题！ 在本试题上答题无效.

一、选择题（本大题共10小题，每小题4分，共40分．每小题的四个选项中，只有一项

是符合题目要求） 1．计算：5+(-2)= A．3

B．3

C．7

D．7

2．右图是由四个相同的小正方体组合而成的立体图形，它的俯视图是

A B C D 正面

3．下列计算正确的是 （第2题图）

326A．a+a=a2 B．a2?a3a6 C．(-a)=-a D．a7?a5a2 4．下列图形，既是中心对称图形，又是轴对称图形的是 A．等边三角形 B．平行四边形

C．正五边形

D．正六边形

5．在九年级某次体育测试中，某班参加仰卧起坐测试的一组女生（每组8人）成绩如下（单位：次/分）：45、44、45、42、45、46、48、45，则这组数据的平均数、众数分别为 A．44、45

B．45、45

C．44、46

D．45、46 6．如图，A、B、P是半径为2的⊙O上的三点，∠APB＝45°， 则弦AB的长为

A．2 B．2 C．22

D．4

（第6题图） 7．若我们把十位上的数字比个位和百位上的数字都大的三位数称为凸数，如：786，465．则由1，2，3这三个数字构成的，数字不重复的三位数是“凸数”的概率是 5A．1 B．1 C．2 D．

36328．若二次函数y=ax2+bx+c（a¹0）的图象如图所示，则下列选项正确的是

A．a>0

B．c>0 C．ac>0 D．bc<0

数学试卷

（第8题图） （第9题图） （第10题图）

9．如图，边长分别为4和8的两个正方形ABCD和CEFG并排放在一起，连结BD并延长交EG于点T，交FG于点P，则GT＝ A．2

B．22 C．2

D．1

10．如图，在平面直角坐标系xoy中，A(0，2)，B(0，6)，动点C在直线y＝x上．若以A、B、C三点为顶点的三角形是等腰三角形，则点C的个数是 A．2

B．3

C．4

D．5

二、填空题（本大题共7小题，每小题3分，共21分） 11．分解因式a22a＝______________．

12．已知x＝3是方程x2-6x+k=0的一个根，则k=______． 13．已知|a-2|+b-3=0，则ab＝____________. 14．如图，PA是⊙O的切线，A为切点，B是⊙O上一点，

BC⊥AP于点C，且OB=BP=6，则BC=_____________．

（第14题图） 15．如图，AB∥CD，BC与AD相交于点M，N是射线CD上的一点．

若∠B＝65°，∠MDN＝135°，则∠AMB＝_________． 16．下列说法：

①对顶角相等；

②打开电视机，“正在播放《新闻联播》”是必然事件；

（第15题图） 1③若某次摸奖活动中奖的概率是，则摸5次一定会中奖；

5④想了解端午节期间某市场粽子的质量情况，适合的调查方式是抽样调查；

⑤若甲组数据的方差s2＝0.01，乙组数据的方差s2＝0.05，则乙组数据比甲组数据更稳定．

其中正确的说法是________________．（写出所有正确说法的序号） 17．对于任意非零实数a、b，定义运算“Å”，使下列式子成立：

321213，2?1，(-2)?5，5?(2)=-，…，则a?b___________．

101022三、解答题（本大题共8小题，共89分）

（背面还有试题） 18．(本题满分10分)

1?2-（1）计算：38(3)0(1)2013|23|; （2）解方程：

4x=+1． 2x+12x+1x31，其中x=2． 缸22x-34x-92x+319．(本题满分8分)先化简，再求值：

20．(本题满分10分)如图，四边形ABCD是平行四边形，

（第20题图） 数学试卷

E、F是对角线AC上的两点，∠1＝∠2． （1）求证：AE=CF；

（2）求证：四边形EBFD是平行四边形． 21．(本题满分10分)某市在2019年义务教育质量监测过程中，为了解学生的家庭教育情况，

就八年级学生平时主要和谁在一起生活进行了抽样调查.下面是根据这次调查情况制作的不完整的频数分布表和扇形统计图．

频数分布表

代码 和谁一起生活 A B C D 父母 爷爷奶奶 外公外婆 其它 合计 频数 4200 660 600 b 6000 频率 0.7 a 0.1 0.09 1 （第21题图） 请根据上述信息，回答下列问题:

（1）a_______________，b_______________; （2）在扇形统计图中，和外公外婆一起生活的学生所对应扇形圆心角的度数是________; （3）若该市八年级学生共有3万人，估计不与父母一起生活的学生有_______________人.

22．(本题满分12分)如图①，在矩形纸片ABCD中，AB=3+1，AD=3．

（1）如图②，将矩形纸片向上方翻折，使点D恰好落在AB边上的D¢处，压平折痕交CD于点E，则折痕AE的长为_______________； （2）如图③，再将四边形BCED¢沿D¢CED?,BⅱC交E向左翻折，压平后得四边形BⅱAE于点F，则四边形BⅱFED的面积为_______________；

（3）如图④，将图②中的DAED¢绕点E顺时针旋转a角，得DAⅱED?，使得EA¢恰好经过顶点B，求弧DD的长.（结果保留）

图① 图② 图③ 图④

（第22题图）

数学试卷

23．(本题满分12分)某公司欲租赁甲、乙两种设备，用来生产A产品80件、B产品100

件．已知甲种设备每天租赁费为400元，每天满负荷可生产A产品12件和B产品10件；乙种设备每天租赁费为300元，每天满负荷可生产A产品7件和B产品10件． （1）若在租赁期间甲、乙两种设备每天均满负荷生产，则需租赁甲、乙两种设备各多少天恰好完成生产任务？

（2）若甲种设备最多只能租赁5天，乙种设备最多只能租赁７天，该公司为确保完成生产任务，决定租赁这两种设备合计10天（两种设备的租赁天数均为整数），问该公司共有哪几种租赁方案可供选择？所需租赁费最少是多少？

24．(本题满分13分)如图，将边长为４的等边三角形AOB放置于平面直角坐标系xoy中，

kF是AB边上的动点（不与端点A、B重合），过点F的反比例函数y=(k>0,x>0)与

xOA边交于点E，过点F作FC^x轴于点C，连结EF、OF． （1）若SDOCF=3，求反比例函数的解析式;

（2）在（1）的条件下，试判断以点E为圆心，EA长 为半径的圆与y轴的位置关系，并说明理由; （3）AB边上是否存在点F，使得EF^AE？

（第24题图） 25．(本题满分14分)如图，四边形ABCD是菱形，对角线AC与BD交于点O，且AC=80，

动点M、N分别以每秒１个单位的速度从点A、D同时出发，分别沿ABD=60．

和D®A运动，当点N到达点A时，M、N同时停止运动．设运动时间为t秒． （1）求菱形ABCD的周长；

（2）记DDMN的面积为S, 求S关于t的解析式，并求S的最大值；

（3）当t=30秒时，在线段OD的垂直平分线上是否存在点P，使得∠DPO＝∠DON？若存在，这样的点P有几个？并求出点P到线段OD的距离；若不存在，请说明理由．

（第25题图） 若存在，请求出BF:FA的值；若不存在，请说明理由． OD数学试卷

2019年龙岩市初中毕业、升学考试 参 考 答 案 及 评 分 标 准

数 学

说明：评分最小单位为1分，若学生解答与本参考答案不同，参照给分． 一、选择题（本大题共10题，每题4分，共40分） 1 2 3 4 5 6 7 8 9 题号 答案 A C D D B C A C B 10 B 二、填空题（本大题共7题，每题3分，共21分．注：答案不正确、不完整均不给分） 11．a(a2)

12．9 13．8

14．3

15．70 16．①④

a2b2 17．．

ab三、解答题（本大题共8题，共89分）

18．(10分，第（1）小题5分，第（2）小题5分)

（1）解：原式=21(1)23 ·················································································· 4分 = 23 ······································································································ 5分 （2）解：方程两边同乘（2x+1），得 4=x+2x+1 ································································· 2分 3=3x

x=1 ······················································································································ 3分 检验：把x=1代入2x+1=3≠0 ········································································· 4分 ∴原分式方程的解为x=1. ················································································ 5分 19．（8分）解：原式=

x(2x3)(2x3)1 ······················································ 4分 2x332x3x= ······································································································· 6分 32当x=2时，原式=. ······················································································· 8分

320．(10分)

(1)证明：（法一）如图：∵四边形ABCD是平行四边形

∴AD=BC，AD∥BC，∠3=∠4 ························· 1分

∵∠1=∠3+∠5, ∠2=∠4+∠6 ············································································ 2分 ∠1=∠2

∴∠5=∠6 ··········································································································· 3分

数学试卷

∴△ADE≌△CBF ····························································································· 5分 ∴AE=CF ··········································································································· 6分 （法二）如图：连接BD交AC于点O ····················· 1分 在平行四边形ABCD中

OA=OC，OB=OD ················································ 2分 ∵∠1=∠2,∠7=∠8

∴△BOF≌△DOE ································································································ 4分 ∴OE=OF ············································································································· 5分 ∴OA－OE=OC－OF

即AE=CF.·············································································································· 6分

(2) )证明：（法一）∵∠1=∠2,

∴DE∥BF ······································································································· 7分

∵△ADE≌△CBF

∴DE=BF ········································································································· 9分 ∴四边形EBFD是平行四边形. ·································································· 10分

（法二）∵OE=OF，OB=OD ······················································································· 9分

∴四边形EBFD是平行四边形. ································································ 10分

其他证法，请参照标准给分.

21．（10分，第（1）小题4分，第（2）小题3分，第（3）小题3分） （1） 0.11 ， 540 ； （注：每空2分） （2）36；

（3）9000．

22．（12分，每小题4分）

（1）6 ···························································································································· 4分 （2）31 ························································································································ 8分 2（3）∵∠C＝90，BC=3，EC＝1

BC=3 CE∴∠BEC＝60 ···················································································································· 9分 由翻折可知：∠DEA＝45 ······························································································· 10分 ∴AEA75＝DED ···························································································· 11分

∴tan∠BEC＝∴l 755323 ······················································································· 12分 36012数学试卷

23．（12分，第（1）小题5分，第（2）小题7分）

解：（1）设需租赁甲、乙两种设备分别为x、y天． ······················································ 1分

x7y8012则依题意得  ···································································· 3分

10x1y0100x2解得  ·································································································· 4分

y8答：需租赁甲种设备2天、乙种设备8天. ···································································· 5分 （2）设租赁甲种设备a天、乙种设备(10－a)天，总费用为元． ····························· 6分

依题意得

a510a712a7(10a)8010a10(10a)100∴3≤a≤5． ∵a为整数，

∴a＝3、4、5． ············································································································ 8分 方法一：

∴共有三种方案．

方案（1）甲3天、乙7天，总费用400×3+300×7＝3300; ·························· 9分 方案（2）甲4天、乙6天，总费用400×4+300×6＝3400; ························ 10分 方案（3）甲5天、乙5天，总费用400×5+300×5＝3500. ······················ 11分 ∵3300<3400<3500 ∴方案（1）最省，最省费用为3300元． ······· 12分

方法二：

则＝400a+300(10－a)＝100a+3000 ························································· 10分 ∵100＞0，

∴随a的增大而增大．

∴当a＝3时，最小＝3300． ······································································· 11分 答：共有3种租赁方案：①甲3天、乙7天；②甲4天、乙6天；③甲5天、乙5天.最少租赁费用3300元． ····································································· 12分

方法三：能用穷举法把各种方案枚举出来，并得出三种符合条件的方案，求出最省费用的，参照标准酌情给分．

24.（1）设F（x，y），(x＞0，y＞0) ．

则OC=x, CF=y ·················································································································· 1分

1∴SOCFxy3． ··································································································· 2分

2数学试卷

∴xy=23．

∴k=23．····················································································································· 3分 23(x＞0) ． ··································································· 4分 x（2）该圆与y轴相离． ·································································································· 5分

理由：过点E作EH⊥x轴，垂足为H，过点E作EG⊥y轴，垂足为G． 在△AOB中，OA=AB=4，∠AOB=∠ABO=∠A=60．

∴反比例函数解析式为y=设OH=m,则tanAOB∴EH=3m，OE=2m． ∴E坐标为（m, ∵E在反比例y=∴3m=EH3． OH3m）． ··············· 6分 23图像上， x23． m∴m1=2， m2=-2(舍去)．

∴OE=22，EA=422，EG=2 ······································································· 7分 ∵422＜2，

∴EA＜EG．

∴以E为圆心，EA垂为半径的圆与y轴相离．················································· 8分

（3） 存在． ···························································································································· 9分 方法一：假设存在点F，使AE⊥FE．过点F作FC⊥OB于点 C，过E点作EH⊥OB于点H．

设BF= x.

∵△AOB是等边三角形，

∴AB=OA=OB=4，∠AOB=∠ABO=∠A=60．

1∴BC=FB·cos∠FBC＝x

2FC=FB·sin∠FBC=3x 21∴AF=4－x，OC=OB －BC=4－x

2∵AE⊥FE

1∴AE=AF·cos∠A=2－x

21∴OE=O A－AE=x+2

21∴OH=OE·cos∠AOB=x1，

4数学试卷

EH=OE·sin∠AOB=3x3 43311∴E(x1,····························································· 11分 x3)，F(4－x，x)

4242k∵E、F都在双曲线y=的图象上，

x3311∴（x1）（x3）＝（4－x）x

42424解得 x1=4，x2=． ····························································································· 12分

5当BF=4时，AF=0，当BF=

BF不存在，舍去． AF416BF1时，AF=，··········································································· 13分 ． ·55AF4方法二：假设存在点F，使AE⊥FE．过E点作EH⊥OB于 H.

∵△AOB是等边三角形，设E(m,

3m)，则OE=2m, AE=4－2m．

∴AB=OA=AB=4，∠AOB=∠ABO=∠A=60．

AE1， AF2∴AF=2AE=8－4m，FB=4m－4． ∵CosA∴FC=FB·sin∠FBC=23m－23， BC=FB·cos∠FBC=2m－2． ∴OC=6－2m

∴F(6－2m, 23m－23)． ···················································································· 11分 ∵E、F都在双曲线y=

k上， x∴m·3m=(6－2m)(23m－23) 化简得：5m2-16m+12=0

6解得： m1=2，m2=． ···························································································· 12分

5当m＝2时，AF=8－4m=0，BF=4，F与B重合，不合题意，舍去．

616164当m＝时，AF=8－4m＝BF=4－=．

55，55 ∴BF:FA1:4． ····································································································· 13分

1125. (1)在菱形ABCD中， OAAC40,ODBD3022∵AC⊥BD

∴AD=302402=50.

∴菱形ABCD的周长为200. ····························· 4分 (2) 过点M作MP⊥AD，垂足为点P.

数学试卷

①当0＜t≤40 ∵SinOADMPOD3 AMAD53∴MP=t

51∴SDNMP

232t ························································································································· 6分 10②当40MPAO∵SinADO= MDAD =

4∴MP=(70t)

5∴SDMN1DNMP 222·························································································· 8分 t228t(t35)2490

5532t,0t4010∴S

22(t35)490,40t505当0＜t≤40时，S随t的增大而增大，当t＝40时，最大值为480. 当40＜t≤50时，S随t的增大而减小，当t＝40时，最大值为480. 综上所述，S的最大值为480. ····························································································· 9分 （3）存在2个点P，使得∠DPO=∠DON. ········································································ 10分 方法一：过点N作NF⊥OD于点F，

40120则NFNDSinODA3024505，DF=NDCosODA30309018.505

NF242 ···································································· 11分 OF12作NOD的平分线交NF于点G，过点G作GH⊥ON于点H.

1111∴SONFOFNFSOGNSOFGOFFGONGH(OFON)FG 2222∴OF=12，∴tanNODOFNF122424∴FG=OFON1212515

24GF152∴tanGOFOF12 15设OD中垂线与OD的交点为K，由对称性可知：

11∴DPKDPODONFOG ·································································· 12分

22数学试卷

∴tanDPK∴PK=DK152PKPK15

15(51) ··········································································································· 13分 2根据菱形的对称性可知，在线段OD的下方存在与点P关于OD轴对称的点P'.

15(51). ·························································· 14分 2方法二：如图，作ON的垂直平分线，交EF于点I，连结OI，IN.

过点N作NG⊥OD，NH⊥EF,垂足分别为G，H. 当t=30时，DN=OD=30，易知△DNG∽△DAO，

∴存在两个点P到OD的距离都是∴即

DNNGDG． DAAOOD30NGDG． 504030∴NG=24,DG=18． ······································································································· 10分 ∵EF垂直平分OD，

∴OE= ED＝15，EG=NH=3． ······················································································ 11分 设OI=R，EI=x,则

在Rt△OEI中，有R2=152+x2 ① 在Rt△NIH中，有R2=32+(24-x)2 ② 15x2由①、②可得：

155R215155． ···························································································· 13分 2根据对称性可得，在BD下方还存在一个点P'也满足条件．

∴PE=PI+IE=15(51)∴存在两个点P，到OD的距离都是. ····················································· 14分

2(注：只求出一个点P并计算正确的扣1分.)

（第25题图）