2012年湖北省荆门市中考真题及答案
荆门市二O一二年初中毕业生学业及升学考试试卷
数学
注意事项:
1.答题前,考生务必将自己的姓名、准考证号填写在答题卡上,并将准考证号条形码
粘贴在答题卡指定位置.
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3.填空题和解答题用0.5毫米的黑色墨水签字笔答在答题卡上每题对应的答题区域内.答在试题卷上无效.
4.考试结束,请将本试题卷和答题卡一并上交.
一、选择题(本大题12个小题,每小题只有唯一正确答案,每小题3分,共36分)
1.下列实数中,无理数是( )
A.-5 B.π C
D.|-2|
2
2.用配方法解关于x的一元二次方程x2-2x-3=0,配方后的方程可以是( ) A.(x-1)2=4 B.(x+1)2=4 C.(x-1)2=16 D.(x+1)2=16 3.已知:直线l1∥l2,一块含30°角的直角三角板如图所示放置,∠1=25°,则∠2等于( ) A.30° B.35° C.40° D.45° 4
|x-y-3|互为相反数,则x+y的值为( ) A.3 B.9 C.12 D.27
5.对于一组统计数据:2,3,6,9,3,7,下列说法错误的是( ) ..
第3题图
l1 l
2
A.众数是3 B.中位数是6 C.平均数是5 D.极差是7 6.已知点M(1-2m,m-1)关于x轴的对称点在第一象限,则m的取值范围在数轴上表示...正确的是( )
7.下列4×4的正方形网格中,小正方形的边长均为1,三角形的顶点都在格点上,则与△ABC相似的三角形所在的网格图形是( ) C
A
A. B. C. D.
8.如图,点A是反比例函数y=2(x>0)的图象上任意一点,AB∥x轴交反
x
A. B. C. D.
第8题图
比例函数y=-3 的图象于点B,以AB为边作□ABCD,其中C、D在x轴上,则S□ABCD
为( )
A.2 B.3 C.4 D.5
9.如图,△ABC是等边三角形,P是∠ABC的平分线BD上一点,PE⊥AB于点E,线段BP的垂直平分线交BC于点F,垂足为点Q.若BF=2,则PE的长为( )
A.2 B.
C
D.3
D
第9题图
10.如图,已知正方形ABCD的对角线长为
ABCD沿直线EF折叠,则图中阴影部分的周长为( )
A.
B.
C.8 D.6 11.已知:多项式x2-kx+1是一个完全平方式,则反比例函数y=k1的
x
第10题图
解析式为( )
A.y=1 B.y=-3 C.y=1或y=-3 D.y=2或y=-2
12.已知:顺次连结矩形各边的中点,得到一个菱形,如图①;再顺次连结菱形各边的中点,得到一个新的矩形,如图②;然后顺次连结新的矩形各边的中点,得到一个新的菱形,如图③;如此反复操作下去,则第2012个图形中直角三角形的个数有( ) A.8048个 B.4024个 C.2012个 D.1066个
二、填空题(本大题共5个小题,每小题3分,共15分)
图① 图② 图③
(-2)-2-
-2)0=__▲__. 14.如图,在直角坐标系中,四边形OABC是直角梯形,BC∥OA,⊙P分别与OA、OC、BC相切于点E、D、B,与AB交于点F.已知A(2,0),B(1,2),则tan∠FDE=.
13
第15题图
图(1) 图(2)
第17题图
15.如图是一个上下底密封纸盒的三视图,请你根据图中数据,计算这个密封纸盒的表面积为__▲__cm2.(结果可保留根号)
16.新定义:[a,b]为一次函数y=ax+b(a≠0,a,b为实数)的“关联数”.若“关联数”[1,m-2]的一次函数是正比例函数,则关于x的方程1+1=1的解为__▲__.
x1m
17.如图(1)所示,E为矩形ABCD的边AD上一点,动点P、Q同时从点B出发,点P沿折
线BE—ED—DC运动到点C时停止,点Q沿BC运动到点C时停止,它们运动的速度都是1cm/秒.设P、Q同发t秒时,△BPQ的面积为ycm2.已知y与t的函数关系图象如图(2)(曲线OM为抛物线的一部分),则下列结论:AD=BE=5;cos∠ABE=3;当0<t≤5
5
时,y=2t2;当t=29秒时,△ABE∽△QBP;其中正确的结论是__▲__(填序号).
54
三、解答题(本大题共7个小题,共69分)
18.(本题满分8分)先化简,后求值:
(1a21)(a3),其中a
1. a3a1
19.(本题满分9分)如图,Rt△ABC中,∠C=90°,将△ABC沿AB向下翻折后,再绕点A按顺时针方向旋转α度(α<∠BAC),得到Rt△ADE,其中斜边AE交BC于点F,直角边DE分别交AB、BC于点G、H. (1)请根据题意用实线补全图形; (2)求证:△AFB≌△AGE.
B
第19题图
20.(本题满分10分)“端午节”是我国的传统佳节,民间历来有吃“粽子”的习俗.我市某食品厂为了解市民对去年销量较好的肉馅粽、豆沙馅粽、红枣馅粽、蛋黄馅粽(以下分别用A、B、C、D表示)这四种不同口味粽子的喜爱情况,在节前对某居民区市民进行了抽样调查,并将调查情况绘制成如下两幅统计图(尚不完整).
请根据以上信息回答:
(1)本次参加抽样调查的居民有多少人? (2)将两幅不完整的图补充完整;
(3)若居民区有8000人,请估计爱吃D粽的人数;
(4)若有外型完全相同的A、B、C、D粽各一个,煮熟后,小王吃了两个.用列表或画树状图的方法,求他第二个吃到的恰好是C粽的概率.
21.(本题满分10分)如图所示为圆柱形大型储油罐固定在U型槽上的横截面图.已知图中ABCD为等腰梯形(AB∥DC),支点A与B相距8m,罐底最低点到地面CD距离为1m.设油罐横截面圆心为O,半径为5m,∠D=56°,求:U型槽的横截面(阴影部分)的面积.(参考数据:sin53°≈0.8,tan56°≈1.5,π≈3,结果保留整数)
第21题图
22.(本题满分10分)荆门市是著名的“鱼米之乡”.某水产经销商在荆门市长湖养殖场批发购进草鱼和乌鱼(俗称黑鱼)共75千克,且乌鱼的进货量大于40千克.已知草鱼的批发单价为8元/千克,乌鱼的批发单价与进货量的函数关系如图所示.
(1)请直接写出批发购进乌鱼所需总金额y(元)与进货量x(千克)之间的函数关系式;
(2)若经销商将购进的这批鱼当日零售,草鱼和乌鱼分别可卖出89%、95%,要使总零售量不低于进货量的93%,问该经销商应怎样安排进货,才能使进货费用最低?最低费用是多少?
)
第22题图
23.(本题满分10)已知:y关于x的函数y=(k-1)x2-2kx+k+2的图象与x轴有交点. (1)求k的取值范围;
(2)若x1,x2是函数图象与x轴两个交点的横坐标,且满足(k-1)x12+2kx2+k+2=4x1x2. ①求k的值;②当k≤x≤k+2时,请结合函数图象确定y的最大值和最大值.
24.(本题满分12分)如图甲,四边形OABC的边OA、OC分别在x轴、y轴的正半轴上,顶点在B点的抛物线交x轴于点A、D,交y轴于点E,连结AB、AE、BE.已知tan∠CBE=1,A(3,0),D(-1,0),E(0,3).
3
(1)求抛物线的解析式及顶点B的坐标; (2)求证:CB是△ABE外接圆的切线;
(3)试探究坐标轴上是否存在一点P,使以D、E、P为顶点的三角形与△ABE相似,若存在,直接写出点P的坐标;若不存在,请说明理由; ....(4)设△AOE沿x轴正方向平移t个单位长度(0<t≤3)时,△AOE与△ABE重叠部分的面积为s,求s与t之间的函数关系式,并指出t的取值范围. 图甲
图乙(备用图)
荆门市二O一二年初中毕业生学业及升学考试
数学试题参考答案及评分标准
一、选择题(每选对一题得3分,共36分)
1.B 2.A 3.B 4.D 5.B 6.A 7.B 8.D 9.C 10.C
11.C 12.B
二、填空题(每填对一题得3分,共15分)
13.-1 14.1 15.(
360 ) 16.x=3 17.①③④
2
18.解:原式=1a3=2.„„„„„„„„„„„„„„„„„„„„„„5分
a1
当a
+1
.„„„„„„„„„„„„„„„„„„8分
19.解:(1)画图,如图1;„„„„„„„„„„„„„„„„„„„„„„„„„4分 (2)由题意得:△ABC≌△AED.„„„„„„„„„„„„„„„„„„„„„„„5分
a1
∴AB=AE,∠ABC=∠E
6分 在△AFB和△AGE中,
ABCE,B ABAE,
,
图1
∴△AFB≌△AGE(ASA).„„„„„„„„„„„„„„„„„„„„„„„„„„9分 20.解:(1)60÷10%=600(人).
答:本次参加抽样调查的居民有600人.„„„„„„„„„„„„„„„„„„„„2分 (2)如图2;„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„5分
图2
(3)8000×40%=3200(人).
答:该居民区有8000人,估计爱吃D粽的人有3200人.„„„„„„„„„„„„7分 (4)如图3;
开始
A D B C D A B D
图3
(列表方法略,参照给分).„„„„„„„„„„„„„„„„„„„„„„„„„„8分 P(C粽)=3=1.
124
答:他第二个吃到的恰好是C粽的概率是1.„„„„„„„„„„„„„„„„„10分
4
21.解:如图4,连结AO、BO.过点A作AE⊥DC于点EN,
ON交⊙O于点M,交AB于点F.则OF⊥AB. ∵OA=OB=5m,AB=8m, ∴AF=BF=1AB=4(m),∠AOB=2∠AOF.3分 2在Rt△AOF中,sin∠AOF=AF=0.8=sin53°.
图4
∴∠AOF=53°,则∠AOB=106°.„„„„„„„„„„„„„„„„„„„„„„„5分
∵OF3(m),由题意得:MN=1m,
∴FN=OM-OF+MN=3(m).„„„„„„„„„„„„„„„„„„„„„„„„6分 ∵四边形ABCD是等腰梯形,AE⊥DC,FN⊥AB, ∴AE=FN=3m,DC=AB+2DE.
在Rt△ADE中,tan56°=AE=3,∴DE=2m,DC=12m.„„„„„„„„„„„7分
∴S阴=S梯形ABCD-(S扇OAB-S△OAB)=1(8+12)×3-(106π×52-1×8×3)=20(m2).
23602
答:U型槽的横截面积约为20m2.„„„„„„„„„„„„„„„„„„„„„„10分
26x (20≤x≤40),
22.解:(1)y=„„„„„„„„„„„„„„„„„„„„„„„4分
24x (x40).
(2)设该经销商购进乌鱼x千克,则购进草鱼(75-x)千克,所需进货费用为w元.
x40,
由题意得:
89%(75x)95%x≥93%75.
解得x≥50.„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„6分
由题意得w=8(75-x)+24x=16x+600.„„„„„„„„„„„„„„„„„„„„8分 ∵16>0,∴w的值随x的增大而增大.
∴当x=50时,75-x=25,W最小=1400(元).
答:该经销商应购进草鱼25千克,乌鱼50千克,才能使进货费用最低,最低费用为1400元.„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„10分 23.解:(1)当k=1时,函数为一次函数y=-2x+3,其图象与x轴有一个交点.„„1分 当k≠1时,函数为二次函数,其图象与x轴有一个或两个交点, 令y=0得(k-1)x2-2kx+k+2=0.
△=(-2k)2-4(k-1)(k+2)≥0,解得k≤2.即k≤2且k=1.„„„„„„„„„„„2分
综上所述,k的取值范围是k≤2
3分 (2)①∵x1≠x2,由(1)知k<2且k=1.
由题意得(k-
1)x12+(k+2)=2kx1.(*)4分
将(*)代入(k-1)x12+2kx2+k+2=4x1x2中得: 2k(x1+x2)=4x1x25分 2kk2又∵x1+x2=,x1x2=, k1k1
图5 ∴2k·2k=4·k2.„„„„„„„„„„„„„„„„„„„„„„„„„„6分
解得:k1=-1,k2=2(不合题意,舍去).
∴所求k值为-1.„„„„„„„„„„„„„„„„„„„„„„„„„„„„„7分 ②如图5,∵k1=-1,y=-2x2+2x+1=-2(x-1)2+3.
22
且-1≤x≤1.„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„8分 由图象知:当x=-1时,y最小=-3;当x=1时,y最大=3.„„„„„„„„„„9分
22
∴y的最大值为3,最小值为-3.„„„„„„„„„„„„„„„„„„„„„„10分
2
24.(1)解:由题意,设抛物线解析式为y=a(x-3)(x+1). 将E(0,3)代入上式,解得:a=-1. ∴y=-x2+2x+3.
则点B(1,4).„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„„2分 (2)如图6,证明:过点B作BM⊥y于点M,则M(0,4)
在Rt△AOE中,OA=OE=3,
∴∠1=∠2=45°,AE
. 在Rt△EMB中,EM=OM-OE=1=BM,
∴∠MEB=∠MBE=45°,BE
. ∴∠BEA=180°-∠1-∠MEB=90°.
∴AB是△ABE外接圆的直径.„„„„„„„„„„„„„„„„„„„„„„„„3分 在Rt△ABE中,tan∠BAE=BE=1=tan∠CBE,
AE3
∴∠BAE=∠CBE.
在Rt△ABE中,∠BAE+∠3=90°,∴∠CBE+∠3=90°. ∴∠CBA=90°,即CB⊥AB.
∴CB是△ABE外接圆的切线.„„„„„„„„„„„„„„„„„„„„„„„„5分 (3)P1(0,0),P2(9,0),P3(0,-1).„„„„„„„„„„„„„„„„„„„„„8分
3
(4)解:设直线AB的解析式为y=kx+b.
k2,3kb0,
将A(3,0),B(1,4)代入,得解得
b6.kb4.
∴y=-2x+6.
过点E作射线EF∥x轴交AB于点F,当y=3时,得x=3,∴F(3,3).„„„„9分
22
情况一:如图7,当0<t≤3时,设△AOE平移到△SNM的位置,MS交AB于点H,MN
2
交AE于点G.
则ON=AS=t,过点H作LK⊥x轴于点K,交EF于点L. 由△AHS ∽△FHM,得ASHK.即
t3t2
FMHL
HK
.解得HK=2t.
3HK
∴S阴=S△MNS -S△GNA-S△HAS=1×3×3-1(3-t)2-1t·2t=-3t2+3t.„„„„11分
图7
图8
情况二:如图8,当3<t≤3时,设△AOE平移到△PQR的位置,PQ交AB于点I,交AE
2
于点V.由△IQA∽△IPF,得
AQIQ3tIQ
.即.解得IQ=2(3-t).
FPIPt∴S阴=S△IQA-S△VQA=1×(3-t)×2(3-t)-1(3-t)2=1(3-t)2=1t2-3t+9.
22222
332
t3t (0t≤),22综上所述:s=„„„„„„„„„„„„„„„„„„„„12分 1t23t9 (3t≤3).222