南京2018年中考模拟试卷一
数学
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
....上)
...相应位置
1.计算│-5+3│的结果是
A.-8B.8C.-2D.2
2.计算(-xy2)3的结果是
A.-x3y6 B.x3y6 C.x4y5 D.-x4y5
3.中国是严重缺水的国家之一.若每人每天浪费的水量为0.4 L,那么8 000 000人每天浪费的水量用科学记数法表示为
A.3.2×108 L B.3.2×107 L C.3.2×106 L D.3.2×105 L
4.如果m=27,那么m的取值范围是
A.3<m<4B.4<m<5C.5<m<6D.6<m<7
5.在平面直角坐标系中,点A的坐标是(1,3),将点A绕原点O顺时针旋转90°得到点A′,则点A′的坐标是
A.(-3,1)B.(3,-1)C.(-1,3)D.(1,-3)
6.如图,⊙O1与⊙O2的半径均为5,⊙O1的两条弦长分别为6和8,⊙O2的两条弦长均为7,则图中阴影部分面积的大小关系为
A.S1>S2B.S1<S2C.S1=S2D.无法确定
(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. 9的平方根是 ▲ .
8.若式子x +3在实数范围内有意义,则x 的取值范围是 ▲ . 9.计算(8 -
1
2
)×2的结果是 ▲ . 10.分解因式3a 2-6a +3的结果是 ▲ .
11.为了解居民用水情况,小明在某小区随机抽查了20户家庭的月用水量,结果如下表:
则这20户家庭的月用水量的众数是 ▲ m 3,中位数是 ▲ m 3. 12.已知方程x 2-x -3=0的两根是x 1、x 2,则x 1+x 2= ▲ , x 1x 2= ▲ .
13.函数y = k 1
x
与y =k 2 x (k 1、k 2均是不为0的常数,)的图像交于A 、B 两点,若点A 的坐
标是(2,3),则点B 的坐标是 ▲ .
14.如图,在△ABC 中,AC =BC ,把△ABC 沿AC 翻折,点B 落在点D 处,连接BD , 若∠CBD =16°,则∠BAC = ▲ °.
15.如图,在⊙O 的内接五边形ABCDE 中,∠B +∠E =210°,则∠CAD = ▲ °.
16. 如图,在四边形ABCD 中,AD ∥BC (BC >AD ),∠D =90°,∠
ABE =45°,BC =CD , 若AE =5,CE =2,则BC 的长度为 ▲ .
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、证明过程或演算步骤)
17.(6分)解不等式组???3x +2 >x ,2(x +1)≥4x -1.
B
(第14题)
(第15题)
E
D
C
B
A
O
A
B
C
D
E (第16题)
18.(7分)先化简,再求值: ????1- 1 a -1 ÷ a 2
-4
a -1. 其中a =-3.
19.(7分)某厂为支援灾区人民,要在规定时间内加工1500顶帐篷.在加工了300顶帐篷后,
厂家把工作效率提高到原来的1.5倍,结果提前4天完成任务,求该厂原来每天加工多少顶帐篷?
20.(8分)城南中学九年级共有12个班,每班48名学生,学校对该年级学生数学学科学业
水平测试成绩进行了抽样分析,请按要求回答下列问题: 【收集数据】
(1)要从九年级学生中抽取一个48人的样本,你认为以下抽样方法中最合理的是 ▲ .①随机抽取一个班级的48名学生;②在九年级学生中随机抽取48名女学生; ③在九年级12个班中每班各随机抽取4名学生. 【整理数据】
(2)将抽取的48名学生的成绩进行分组,绘制成绩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:
①表中m 的值为 ▲ ;
② B 类部分的圆心角度数为 ▲ °;
③估计C 、D 类学生大约一共有 ▲ 名.
九年级学生数学成绩频数分布表
【分析数据】
(3)教育主管部们为了解学校学生成绩情况,将同层次的城南、城北两所中学的抽样数
请你评价这两所学校学生数学学业水平测试的成绩,提出一个解释来支持你的观点.
九年级学生数学成绩分布扇形统计图
数据来源:学业水平考试数学成绩抽样
A 类 50%
B 类 25%
C 类
D 类
(第23题)
21.(8分)甲、乙、丙三人到某商场购物,他们同时在该商场的地下车库等电梯,三人都任意从1至3层的某一层出电梯.
(1)求甲、乙两人从同一层楼出电梯的概率;
(2)甲、乙、丙三人从同一层楼出电梯的概率为 ▲ .
22.(7分)如图,在△ABC 中,AD 是BC 边上的中线,点E 是AD 的中点,过点A 作
AF ∥BC 交BE 的延长线于F ,连接CF . (1)求证:△AEF ≌△DEB ;
(2)若∠BAC =90°,求证:四边形ADCF 是菱形.
23.(8分)如图,在建筑物AB 上,挂着35 m 长的宣传条幅AE ,从另一建筑物CD 的顶部D 处看条幅顶端A 处,仰角为45°,看条幅底端E 处,俯角为37°.求两建筑物间的距离BC . (参考数据:sin37°≈0.6,cos37°≈0.8, tan37°≈0.75)
24.(8分)已知二次函数y =ax 2+bx +c 中,函数y 与自变量x 的部分对应值如下表:
x … -1 0 1 2 3 … y
…
8
3
-1
…
(1)当ax 2+bx +c =3时,则 x = ▲ ; (2)求该二次函数的表达式;
(3)将该函数的图像向上(下)平移,使图像与直线y =3只有一个公共点,直接写出平移后的函数表达式.
F
E
D
C
B
A (第22题)
25.(8分)如图,在半径为3的⊙O中,AB是直径,AC是弦,且AC=4 2 .过
点O作直径DE⊥AC,垂足为点P,过点B的直线交AC的延长线和DE的延长线于点F、G.
(1)求线段AP、CB的长;
(2)若OG=9,求证:FG是⊙O的切线.
26.(10分)如图①,点A表示小明家,点B表示学校.小明妈妈骑车带着小明去学校,到达C处时发现数学书没带,于是妈妈立即骑车原路回家拿书后再追赶小明,同时小明步行去学校,到达学校后等待妈妈.假设拿书时间忽略不计,小明和妈妈在整个运动过程中分别保持匀速.妈妈从C处出发x分钟时离C处的距离为y1米,小明离C处的距离为y2米,如图②,折线O-D-E-F表示y1与x的函数图像;折线O-G-F表示y2与x的函数图像.
(1)小明的速度为▲m/min,图②中a的值为▲.
(2)设妈妈从C处出发x分钟时妈妈与小明之间的距离为y米.
①写出小明妈妈在骑车由C处返回到A处的过程中,y与x的函数表达式及x的取值范围;
②在图③中画出整个过程中y与x的函数图像.(要求标出关键点的坐标)
①
③x/min
O ②
D
(第25题)
27.(11分)如图,矩形ABCD 中,AB =4,BC =m (m >1),点E 是AD 边上一定点,
且AE =1.
(1)当m =3时,AB 上存在点F ,使△AEF 与△BCF 相似,求AF 的长度.
(2)如图②,当m =3.5时.用直尺和圆规在AB 上作出所有使△AEF 与△BCF 相似的点F .(不写作法,保留作图痕迹) (3)对于每一个确定的m 的值,AB 上存在几个点F ,使得△AEF 与△BCF 相似?
② ③
A
B
C
D E A
B
C
D
E F
①
①
2018年中考模拟试卷一 数学试卷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(本大题共6小题,每小题2分,共12分)
二、填空题(本大题共10小题,每小题2分,共20分)
7.±3
8.x ≥-3 9.3
10.3(a -1)2 11.5;5.5
12.1;-3 13.(-2, -3); 14.37 15.30 16.6 三、解答题(本大题共11小题,共88分) 17.(本题6分)
解:解①,得x >-1. ············································································ 2分 解②,得x ≤3
2. ····················································································· 4分
∴不等式组的解集为-1<x ≤3
2. ······························································· 6分
18.(本题7分)
解:? ???
?a -1 a -1 - 1 a -1 ·
a -1(a +2)( a -2)
3分 =a -2 a -1 ·a -1
(a +2)( a -2)
4分
=
1
a +2
. ···························································································· 5分 当a =-3时,原式=-1 ········································································ 7分 19.(本题7分)
解:设原来每天加工x 顶帐篷,根据题意得
1500 x =300x +1200
1.5x +4 ··················································································· 4分 解得x =100. ··························································································· 6分 经检验:x =100是原方程的解.
答:原来每天加工100顶帐篷. ···································································· 7分 20.(本题8分)
解:(1)③. ······················································································ 2分 (2)① 1
6
. ········································································· 3分
②90 ···························································································· 4分 ③144 ·························································································· 6分 (3)本题答案不惟一.城南中学成绩好,因为虽然平均数相同,但城南中学成绩的方
差小,说明成绩波动小;或城北中学成绩好,因为虽然平均数相同,但城北中学成绩中A 、B 类的频率和大,说明优秀学生多. ································ 8分
21.(本题8分)
解:(1)甲、乙两人出电梯的可能结果共有9种,即(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3),每种结果出现的可能性相等.甲、乙两人从同一层楼出电梯(记为事件A )的结果有3种,所以P (A )=1
3 ........ .......6分
(2)1
9
. ·························································································· 8分
22.(本题7分)证明:(1)∵E 是AD 的中点,∴AE =DE , ································ 1分
∵AF ∥BC ,∴∠AFE =∠DBE , ··············································· 2分 ∵∠AEF =∠DEB ,∴△AEF ≌△DEB ; ····································· 3分
(2)∵△AEF ≌△DEB ,∴AF =DB , ················································· 5分
∵AD 是BC 边上的中线,∴DC =DB , ∴AF =DC ,∵AF ∥DC ,
∴四边形ADCF 是平行四边形, ················································ 6分 ∵∠BAC =90°,AD 是BC 边上的中线,
∴AD =DC ,∴□ADCF 是菱形. ·············································· 7分
23.(本题8分)
解:过点D 作DF AB 交AB 于点F ,
由已知,BC =DF ................. ........ ........ ........ ........ ........ .............1分 在Rt △ADF 中,∠ADF =45°,则AF =DF .... ............................3分 在Rt △DFE 中,∠EDF =37°,则EF =DF ·tan37°.... ..............5分 又因为AF +EF =AE 所以DF +DF ·tan37°=35
解得DF =BC =20(m ) ...................... ........ ........ ....................7分 答:两建筑物间的距离BC 为20m ............ ........ ........ ............8分 24.(本题8分)
解:(1)0或4.……………………………………………………………………………2分 (2)设y =a (x -2)2-1 ······································································· 3分
∵过点(0,3),
∴3=a (0-2)2-1 ······································································· 4分 ∴a =1 ······················································································ 5分 ∴y = (x -2)2-1= x 2-4x +3 ························································· 6分 (3)y = (x -2)2+3 ············································································· 8分
25.(本题8分)
解:(1)∵DE 是⊙O 的直径,且DE ⊥AC ,
∴AP =PC =1
2
AC …………………………………1分
∵AC =4 2 ,∴AP =2 2 …………………………………2分 又∵OA =3,∴OP =1 又AB 是⊙O 的直径, ∴O 为AB 的中点,
∴OP =1
2 BC ,∴BC =2OP =2. …………………………………4分
(2)∵
OG OA =93=3,OB OP =93=13, ∴ OG OA =OB OP
…………………………………5分
∠BOG =∠POA ,
∴△BOG ∽△POA , …………………………………6分
∴∠GBO =∠OP A =90° …………………………………7分 又∵点B 在⊙O 上,
∴FG 是⊙O 的切线. …………………………………8分
26.(本题10分)
解:(1)60;33. ···················································································· 4分 (2)①小明妈妈的速度为4800
24=200 m/min , ·············································· 5分
∵小明妈妈在骑车由C 回到A 的过程中,小明与妈妈相向而行,小明的速度为60 m/min , ∴y =260x , …………………………………7分
x 的取值范围是0≤x ≤12. …………………………………8分 ②
································································ 10分
27.(本题11分)
解:(1)当∠AEF =∠BFC 时,
要使△AEF ∽△BFC ,需AE BF =AF BC ,即14-AF =AF
3
,
解得AF =1或3;.…………………………………………………2分 当∠AEF =∠BCF 时,
要使△AEF ∽△BCF ,需AE BC =AF BF ,即1 3=AF
4-AF ,
解得AF =1;
综上所述AF =1或3.…………………………………………………4分
(2)
…………………………………7分
提示:延长DA,作点E关于AB的对称点E′,连结CE′,交AB于点F1;
连结CE,以CE为直径作圆交AB于点F2、F3.
(3)当1<m<4且m≠3时,有3个;…………………………………8分当m=3时,有2个;…………………………………………………9分
当m=4时,有2个;………………………………………………10分
当m>4时,有1个.………………………………………………11分