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北京市西城区2010初三数学抽样测试题及答案

发布时间:2013-12-08 15:45:52  

初三数学试卷

一、选择题(本题共32分,每小题4分)

1.若两圆的半径分别是4 cm和5cm,圆心距为9 cm,则这两圆的位置关系是( ).

A.内切 B.相交 C.外切 D.外离

2.关于x的一元二次方程(a?1)x2?x?a2?1?0有一个根为0,则a的值等于( ).

A.-1 B.0 C.1 D.1或-1

3.抛物线y?(x?1)(x?3)的对称轴是直线( ).

A. x??1 B. x?1 C.x??3 D.x?3

4.如图,在平面直角坐标系中,以P (4,6)为位似中心,

把△ABC缩小得到△DEF,若变换后,点A、B的对应点分别

为点D、E,则点C的对应点F的坐标应为( ).

A. (4,2) B. (4,4) C. (4,5) D. (5,4)

5.某汽车销售公司2007年盈利1500万元, 2009年盈利2160万元,且从2007年到2009年,每年盈利的年增长率相同.设每年盈利的年增长率为x,根据题意,下面所列方程正确的是( ).

(1?x)2?2160 B.1500x?1500x2?2160 A.1500

2(1?x)?1500(1?x)2?2160 C.1500x?2160 D. 1500

6.如图,在Rt△ABC中,∠ACB=90°,M为AB边的中点, 将Rt△ABC绕点M旋转,使点A与点C重合得到△CED,连结MD. 若∠B=25°,则∠BMD等于( ).

A. 50° B.80° C.90° D.100°

7.如图,PA、PB分别与⊙O相切,切点分别为A、B,PA =3,

∠P=60°,若AC为⊙O的直径,则图中阴影部分的面积为( ). A.?

2

D. ? 8.对于实数c、d,我们可用min{ c,d }表示c、d两数中较小的数,如min{3,?1}=?1.若关于x的函数y = min{2x2,a(x?t)2}的图象关于直线x?3对称,则a、t的值可能是

A.3,6 B.2,?6

C.2,6 D.?2,6

初三数学 第 1 页 共 16 页

二、填空题(本题共16分,每小题4分)

9.如图,在△ABC中,DE∥BC分别交AB、AC于点D、E,若DE=1,

BC=3,那么△ADE与△ABC面积的比为.

10.如图,AB为⊙O的直径,弦CD⊥AB,E为AD上一点,

若∠BOC=70°,则∠BED的度数为 °.

D(0,3),11.如图,平面直角坐标系xOy中,

正方形ABCD的顶点B,D的坐标分别为B(0,?1),

A点在第二象限.则A点的坐标为,以B点为顶点,经过A,C两点的抛物线 的解析式为 .

12.如图,平面直角坐标系xOy中,M点的坐标为(3,0),⊙M的半径为2,过M点的直

线与⊙M的交点分别为A,B,则△AOB的面积的最大值为 ,此时A,B两点 所在直线与x轴的夹角等于 °.

三、解答题(本题共30分,每小题5分)

13.计算:2sin45??sin60??cos30??tan260?.

214.已知关于x的方程x?3x?3m?0. 4

(1)如果此方程有两个不相等的实数根,求m的取值范围;

(2)在(1)中,若m为符合条件的最大整数,求此时方程的根.

初三数学 第 2 页 共 16 页

15.已知二次函数y = x2 +4x +3.

(1)用配方法将y = x2 +4x +3化成y = a (x -h) 2 + k(2(3)写出当x为何值时,y>0.

16.已知:如图,在Rt△ABC中, ∠C=90°, D、E分别为

AB、 AC边上的点,且AD?3AE,连结DE.若AC=3,AB=5

5

猜想DE与AB有怎样的位置关系?并证明你的结论.

17.已知:如图,⊙O的半径为5,AB为直径,CD为弦, CD⊥AB于E,

(1)若AE=2,求CD的长.

(2)若弦MN∥CD,交直径AB于F点,且CD=6,求EF的长度.

初三数学 第 3 页 共 16 页

18.列方程解应用题 为了鼓励居民节约用电,某地区规定: 如果每户居民一个月的用电量不超过a

度时,每度电按0.40元交费;如果每户居民一个月的用电量超出a度时,则该户居民的电费将使用二级电费计费方式,即其中有a度仍按每度电0.40元交费,超出a度部分则按每度电a元交费.下表是该地区一户居民10月份、11月份的用电情况.根据表中的数据,求在该150

地区规定的电费计费方式中,a度用电量为多少?

四、解答题(本题共20分,第19题6分,第20题4分,第21题4分,第22题6分)

19.已知:抛物线C1 :y?ax2?bx?c经过点A(-1,0)、

B (3,0)、C(0,-3).

(1)求抛物线C1的解析式;

(2)将抛物线C1向左平移几个单位长度,可使所得的抛物

线C2经过坐标原点,并写出C2的解析式;

(3)把抛物线C1绕点A(-1,O)旋转180,写出所得抛物

线C3顶点D的坐标.

初三数学 第 4 页 共 16 页

o

20.已知:如图,一座商场大楼的顶部竖直立有一个矩形广告牌,小红

同学在地面上选择了在一条直线上的三点A(A为楼底)、D、E,她在D处测得广告牌顶端C的仰角为60°,在E两处测得商场大楼楼顶B 的仰角为45°,DE=5米.已知,广告牌的高度BC=2.35米,求这座商场大楼的高度AB(取1.73,2取1.41,小红的身高不计,结果保留整数).

21.如图所示,AC为⊙O的直径且PA⊥AC,BC是⊙O的一条弦,

DBDC2直线PB交直线AC于点D,. DPDO3

(1)求证:直线PB是⊙O的切线;

(2)求cos∠BCA的值.

初三数学 第 5 页 共 16 页

22.阅读下列材料:

对于李老师所提出的问题,请给出你认为正确的解答(写出BD的取值范围,并在 备用图中画出对应的图形,不写作法,保留作图痕迹).

五、解答题(本题共22分,第23题7分,第24题8分,第25题7分)

223.已知关于x的一元二次方程ax?2bx?c?0(a?0)①. (1)若方程①有一个正实根c,且2ac?b?0.求b的取值范围;

2(2)当a=1 时,方程①与关于x的方程4x?4bx?c?0②有一个相同的非零实根,求

8b2?c

8b2?c 的值.

初三数学 第 6 页 共 16 页

24.已知:如图,AB是⊙O的直径,C是⊙O上一点, 过C点的切线与AB的延长线交于

点D,CE∥AB交⊙O于点E ,连结AC、BC、AE.

(1)求证:①∠DCB=∠CAB;

②CD?CE?CB?CA;

(2)作CG⊥AB于点G.若tan?CAB?

的值(用含k的式子表示).

初三数学 第 7 页 共 16 页 1EC(k>1),求kGB

25.已知:抛物线y?x2?(m?1)x?m与x轴交于点A(x1,0)、B

(x2,0)(A在B的左侧),与y轴交于点C.

(1)若m>1,△ABC的面积为6,求抛物线的解析式;

(2)点D在x轴下方,是(1)中的抛物线上的一个动点,且

在该抛物线对称轴的左侧,作DE∥x轴与抛物线交于另

一点E,作DF⊥x轴于F,作EG⊥x轴于点G,求矩形

DEGF周长的最大值;

(3)若m<0,以AB为一边在x轴上方做菱形ABMN(∠NAB为锐角), P是AB边的

中点,Q是对角线AM上一点,若cos?NAB?

面积最大时,求点A的坐标. 4,QB?PQ?6 ,当菱形ABMN的5

初三数学 第 8 页 共 16 页

北京市西城区2009——2010学年度第一学期期末 初三数学试卷答案及评分参考 2010.1

一、选择题(本题共32分,每小题4分)

二、填空题(本题共16分,每小题4分)

三、解答题(本题共30分,每小题5分)

13.解:2sin45??sin60??cos30??tan260?.

=2?

23

························································································· 4分 ???()2. ·

222

=2?3. ······················································································································ 5分

14.(1)解:a?1,b?3,c?

3

m. 4

3m

?9?3m. ······················································· 1分 4

??b2?4ac?32?4?1?

∵ 该方程有两个不相等的实数根, ∴ 9?3m?0. ································································································· 2分 解得 m?3.

∴ m的取值范围是m?3. ·············································································· 3分

(2)解:∵m?3,

∴ 符合条件的最大整数是 m?2. ································································ 4分

3

?0, 2

3

?3?32?4?1?

?3?32

x??解得 .

22

2

此时方程为 x?3x?

∴方程的根为 x1?

?3??3?3

,x2?. ············································ 5分 22

初三数学 第 9 页 共 16 页

15.解:(1)y?x2?4x?3

?x2?4x?4?1

································································································· 2分 ?(x?2)2?1.

(2)列表:

图象见图1.··························································· 4分 (3)x<-3或x>-1. ········································ 5分

16.DE与AB ························ 1分

证明:∵AC=3,AB=5,AD?

3

AE, 5

图1

A

ACAB

?. ··········································· 2分 ADAE

∵ ∠A =∠A , ·········································· 3分 ∴ △ADE ∽△ACB . ······························ 4分 E∵∠C=90°, B∴∠ADE =∠C=90°. 图2 ∴DE⊥AB. ·················································································

························ 5分

17.(1)解:连接CO.

∵AB为直径,CD为弦,CD⊥AB于E, ∴?OEC?90?,CD?2CE. ………1分 ∵OA?5,AE?2,

∴OE?3,OC?5. ……………………2分 在Rt△OCE中,

CE??4.

∴CD?8

.

∴CD的长为8. ………………………………3分 (2)解: EF=1或7

初三数学 第 10 页 共 16 页

18.解: 因为 80×0.4=32,100×0.4=40<42,

所以 80?a?100. ·································································································· 1分

a······························································· 3分 ?42. ·150

去分母,得 60a?(100?a)a?42?150. 由题意得 0.4a?(100?a)

整理,得 a2?160a?6300?0.

解得 a1?90,a2?70. ···················································································· 4分 因为 a?80,

所以 a2?70不合题意,舍去.

所以 a?90.

答:在该地区规定的电费计费方式中,a度用电量为90度. ······································ 5分

四、解答题(本题共20分,第19题6分,第20题4分,第21题4分,第22题6分)

19.解:(1)∵y?ax2?bx?c经过点A (-1,0) 、B (3,0)、 C (0,-3) . ?a?b?c?0,∴ ? ································· 2分 ?9a?3b?c?0,

?c??3.?

?a?1,

解得 ??b??2,

?c??3.?∴ 所求抛物线C1的解析式为:y

?x2?2x?3图5 ························································· 3分 (2)抛物线C1向左平移3个单位长度,可使得到的抛物线C2经过坐标原点 ···· 4分

所求抛物线C2的解析式为:y?x(x?4)?x2?4x. ···································· 5分 (3)D点的坐标为(-3,4). ·······················································

························· 6分

20.解:设AB为x米.

依题意,在Rt△ABE中, ∠BEA=45°,

∴ AE=AB=x.

∴ AD =AE-DE=x-5,AC = BC+ AB =2.35+x. ············· 2在Rt△ADC中, ∠CDA=60°,

∴ AC=AD?tan?CDA=AD.

∴ x+2.35=3( x-5). ···························································································· 3分 ∴ (3-1 )x=2.35+5.

解得 x?5?2.35

?1x .

∴ x≈15.

答:商场大楼的高度AB约为15米. ·············· 4分

初三数学 第 11 页 共 16 页

21.(1)证明:连接OB、OP ………………………………………………………(1分)

DBDC2??DPDO3 且∠D=∠D ∵

∴ △BDC∽△PDO∴ ∠DBC=∠DPO

∴ BC∥OP∴ ∠BCO=∠POA ∠CBO=∠BOP

∵ OB=OC∴ ∠OCB=∠CBO

∴ ∠BOP=∠POA又∵ OB=OA OP=OP

∴ △BOP≌△AOP∴ ∠PBO=∠PAO

又∵ PA⊥AC∴ ∠PBO=90°

∴ 直线PB是⊙O的切线 …………………………………(3分)

(2)由(1)知∠BCO=∠POA 设PB?a,则BD?2a

又∵ PA?PB?a∴ AD?22a

又∵ BC∥OP∴

∴ DC?CA?DC?2 CO1?22a?2a∴ OA?2a∴ OP?a 222

∴ cos∠BCA=cos∠POA= ……………………………………………(5分) 3

22.解:BD=msin?或BD≥m.(各1分)

见图7、图8;(各1分)

五、解答题(本题共22分,第23题7分,第图8 23.解:(1)∵ c为方程的一个正实根(c?0),

2∴ ac?2bc?c?0. ······················································································· 1分

∵c?0,

∴ ac?2b?1?0,即ac??2b?1.································································ 2分 ∵ 2ac?b?0,

∴ 2(?2b?1)?b?0.

2解得 b??. ································································································· 3分 3

又ac?0(由a?0,c?0).

∴ ?2b?1?0.

1解得 b??. 2

21∴ ??b??. ····························································································· 4分 32

2(2)当a?1时,此时方程①为 x?2bx?c?0.

初三数学 第 12 页 共 16 页

设方程①与方程②的相同实根为m,

2∴ m?2bm?c?0③

2∴ 4m?4bm?c?0④

2④-③得 3m?2bm?0.

整理,得 m(3m?2b)?0.

∵m≠0,

∴3m?2b?0.

2b解得 m??. ································································································· 5分 3

2b222把m??代入方程③得 (?b)?2b(?b)?c?0. 333

8b2

?c?0,即8b2?9c. ∴?9

8b2?c42?. ·当8b?9c时,2···········································································7分 8b?c5

24.(1)证明:①如图10, 解法一:作直径CF,连结BF.

∴ ∠CBF=90°, ········································ 1分

A 则 ∠CAB=∠F =90°-∠1. ∵ CD切⊙O于C,

∴ OC⊥CD , ············································ 2分 则 ∠BCD =90°-∠1. 图10

∴ ∠BCD =∠CAB . ················································

解法二:如图11, 连结OC.

∵ AB是直径, A ∴ ∠ACB=90°. ········································ 1分

则∠2 =90°-∠OCB.

∵ CD切⊙O于C,

图11 ∴ OC⊥CD . ································································································2分

则 ∠BCD =90°-∠OCB.

∴ ∠BCD =∠2.

∵ OA=OC,

∴ ∠2 =∠CAB .

∴ ∠BCD =∠CAB . ··················································································· 3分 ② ∵ EC∥AB ,∠BCD =∠3,

∴ ∠4 =∠3=∠BCD. ····················································································· 4分 ∵ ∠CBD+∠ABC=180°,

∵ ∠AEC+∠ABC=180°,

初三数学 第 13 页 共 16 页

∴ ∠CBD=∠AEC . ······················································································ 5分 ∴ △ACE∽△DCB .

∴ CACD. ?CECB

∴ CD?CE?CB?CA. ····················································································6分

(2)连结EB,交CG于点H,

∵ CG⊥AB于点G, ∠ACB=90°.

∴ ∠3=∠BCG.

∵ ∠3 =∠4.

∴ AE = ∴ ∠3=∠EBG .

∴ ∠BCG=∠EBG .

∵ tan?CAB?1(k>1), k

GH1?. GBk∴ 在Rt△HGB中,tan?HBG?

在Rt△BCG中,tan?BCG?BG1?. CGk

设HG =a,则BG= ka,CG= k2a. CH=CG-HG=( k2-1)a.

∵ EC∥AB ,

∴ △ECH∽△BGH .

ECCH(k2?1)a???k2?1. ·∴ ··········································································· 8分 GBHGa

解法二: 如图10-2,作直径FC,连结FB、EF,则∠CEF=90°.

∵CG⊥AB于点G,

在Rt△ACG中,tan?CAB?CG1? AGk

11a,CF=AB=AG+BF=(k?)a. kk

设CG =a,则AG= ka,BG?∵ EC∥AB , ∠CEF=90°, ∴直径AB⊥EF.

∴EF=2CG= a.

EC=CF2?EF2?(k?)2a2?(2a)2)=( k?

1(k?)EC?1. ∴?1BGk2?1

k1kA 1)a. k图10-2

初三数学 第 14 页 共 16 页

解法三:如图11-2,作EP⊥AB于点P,

在Rt△ACG中,tan?CAB?

CG1

? AGk1

a, k

1

a. k

图11-2

A

设CG =a,则AG= ka,BG?

可证△AEP≌△BCG,则有AP=BG?EC=AG-AP=(k?

1

)a. k

1(k?)EC?1. ∴?

1BGk2?1k

25.解:∵ 抛物线与x轴交于点A(x1,0)、B(x2,0),

∴ x1、x2是关于x的方程x2?(m?1)x?m?0的解.

解方程,得x?1或x?m. ······························································································· 1分

(1)∵ A在B 的左侧,m>1,

∴x1?1 ,x2?m. ································································································· 2分 ∴ AB=m-1.

抛物线与y轴交于C(0,m)点. ∴ OC=m.

11 △ABC的面积S=AB?OC=(m?1)m?6. 22 解得 m1?4,m2??3(不合题意,舍去).

∴ 抛物线解析式为y?x2?5x?4. ······················· 3分

(2)∵ 点D在(1)中的抛物线上,

5

∴ 设D(t, t?5t?4)(1?t?).

2

2

x

2

∴ F(t,0),DF=?t2?5t?4. 又抛物线对称轴是直线x?

图12

5

,DE与抛物线对称轴交点记为R(如图12), 2

5

∴ DR=?t,DE=5?2t.

2

设矩形DEGF的周长为L,则 L=2(DF+DE).

∴ L =2(?t2?5t?4+5?2t)

=?2t2?6t?2

313

=?2(t?)2?. ······························································································ 4分

22

初三数学 第 15 页 共 16 页

∵ 1?t?5, 2

3时,L有最大值. 2

313当t?时,L最大=. 22

13∴ 矩形周长的最大值为. ···················································································· 5分 2∴ 当且仅当t?

(3)∵ A在B 的左侧,m<0,

∴x1?m ,x2?1.

∴ AB=1-m.

如图13,作NH⊥AB于H,连结QN.

在Rt△AHN中, cos?NAB?AH4?. AN5图13 设AH=4k(k>0), 则AN=5k,NH=3k.

∴ AP=

PN=PH2?HN2=11553AB=AN=k,PH=AH -AP=4k?k=k,222223k. 2

∵ 菱形ABMN是轴对称图形,

∴ QN=QB.

∴ PQ+QN = PQ+QB=6.

∵ PQ+QN≥PN(当且仅当P、Q、N三点共线时,等号成立).

k, 2

4解得 k≤. ············································································································· 6分 5

∵ S菱形ABMN=AB·NH=15 k2≤48. ∴ 6?∴ 当菱形面积取得最大值48时,k=

此时AB=5k=1-m =4.

解得 m=1-4. 4. 5

∴ A点的坐标为(1-4,0). ············································································· 7分

初三数学 第 16 页 共 16 页

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