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2013年全国初中数学联赛初赛 2

发布时间:2014-01-30 11:46:19  

2013年全国初中数学联赛初试解答

一、选择题(本大题满分42分,每小题7分)

1

1、已知-1<x<0,则-x2,x,的大小关系是( )

xA:-x2<x<

1 x

1

B.<-x2<x

x

C.x<-x2<

1 x1

D.<x<-x2

x

解:由-1<x<0,得x+1>0,从而x+x2=x(x+1)<0,所以x<-x2, 11-x2(1-x)(1+x)1又-x==<0,所以<x.故选D. xxxx

2、如图,正方形ABCD,点P是对角线AC上一点,连接BP, 过P作PQ⊥BP,

PQ交CD于Q,若AP

=CQ=2,则正方形ABCD的面积为

A.6+

B.16 C.12+ D.32

解:如图,过P分别作PE、PF、PG垂直于AB、CD、AD,垂足分别为E、F、G.易证Rt△EPB≌Rt△FQP≌

Rt△FDP,所以FQ=FD=EP

此正方形ABCD

的边长为

2+(2+2=12+ 故选C.

3、若实数a,b满足b2+a-2b+2=0,则a的取值范围是( ) A: a≤-1 B:a≥-1 C:a≤1 D:a≥1 解:将原式看作为关于b的一元二次方程,则其判别式 △=(-2)-4(a-2)≥0解得. a≤-1

或:原式化为a=-b+2b-2=-(b-1)-1≤-1 故选A.

4、如图,在四边形ABCD

中,∠B=135°,∠C=120°

,,BC=3-A

. C

.,CD=6,则

AD边的长为( )

2

2

2

B.

D.

解:过A和D

点向BC作垂线,垂足为M和N, 那么

CN=3,DN= MN=BM+BC+CN=6,

所以AD=MN

+(DN-AM)=48,所以AD=. 故选B.

113

5、方程+=的正整数解(x,y)的组数是( )

xy7A.0 解:化简为

2

222

D.5

B.1 C.3

x?y3

= 设:x+y=3a,xy=7a,且a≠0,则x,y可以看成一元二次方程xy7

m-3am+7a=0的两根,而方程的判别式小于0,所以m没有实数解 故答案选A.

xyzx2y2z2

6、已知实数x,y,z满足的值是( ) ++=1,则++

y+zz+xx+yy+zz+xx+y

A.-1

B.0 C.1

1

D.2

解:显然x+y+z≠0,否则

由已知得(xyz++=-3 y+zz+xx+yxyz++)(x+y+z)=x+y+z y+zz+xx+y

x2y2z2

即+x++y++z=x+y+z y+zz+xx+y

x2y2z2

所以++=0. y+zz+xx+y

故选B.

二、填空题(本大题满分28分,每小题7分)

x表示x的正约数个数,则③×④÷⑥等于 . 1、x是正整数,○

解:③=2,④=3,⑥=4,所以③×④÷⑥=33.故填. 22

2、草原上的一片青草,到处长得一样密一样快,70头牛在24天内可以吃完这片青草,30头牛在60天内可以吃完这片青草,则20头牛吃完这片青草需要的天数是 . 解:设草原上原有草量为a,每天长出量为b,并设20头牛在x天内可以吃完这片青草. 因为一头牛一天的吃草量相等,根据题意可得方程组

a?24ba?60ba?bx== 70?2430?6020x

a?24ba?60b由=得a=480b. 70?2430?60

a?60ba?bx3480+x代入=中,得b=b, 1020x30?6020x

解得x=96.故填96.

3、如图,在平行四边形ABCD中,M、N分别是BC、DC

的中点,AM=4,AN=3,且角MAN=60°,则AB的长

是 .

解:延长AM交DC的延长线于F,则△AMB≌△FMC.

则CF=AB,则NF=3AB,过N作NH垂直AF于H, 2

313,

=2213则AH=AN=

,NH=,故HF=2?422

NF=214. 7.所以AB=NF=33

故填14. 3

4、小明将1,2,3,…,n这n个数输入电脑求其平均值,当他认为输完时,电脑上只显示输入(n-1)个数,且平均值为30.75,假设这(n-1)个数输入无误,则漏输入的一个数是 .

2

n(n?1)?m123n2?n?2mn2?n?2n?2?2?2m2解:设输漏的一是m,则====2(n?1)2(n?1)4n?1

n(n?1)?2(n?1)?2(1?m)nm?1123=-+1= 2(n?1)2n?14

n-2(m?1)1191192(m?1)2(m?1)=,n=+。因为1<m<n,所以0<<2, n?122n?1n?1

1191192(m?1)123<+<,即59.5<n<61.5,所以n=60(舍去,m不是正整数)或 n22n?12

=61,当n=61,m=46。

三、(本大题满分20分)

解方程x2-|2x-1|-2=0.

解:当x≥1

时,原方程可化为x2-(2x-1)-

2=0, 2解得x1=

1+x2=1-

1又因为x2=1-,故应舍去. ············································································ 10分 2

当x<1时,原方程可化为x2-(-2x+1)-2=0, 2

解得x3=-3,x4=1

1又因为x4=1>,故应舍去. 2

所以原方程的解为x=1+·································································· 20分 和x=-3. ·

四、(本大题满分25分)

如图,圆内接四边形ABCD中,CB=CD,

22求证:CA-CB=AB×AD; 证明:连结BD、AC交于点E,则

∠BAE=∠CAD,∠ABE=∠ACD

所以△ABE∽△ACD, ······························································································· 5分 所以ABAC, =AEAD

所以AB×AD=AC×AE. ··························································································· 10分 又∠CBE=∠CAB,∠BCE=∠ACB

所以△CBE∽△CAB, ································································································ 15分 所以CBCA, =CECB

2所以,CB=CA×CE ···································································································· 20分

所以CB+AB×AD=CA×CE+AE×AC

CB+AB×AD=AC×(CE+AE) =AC 222

3

所以.CA-CB=AB×AD ························································································· 25分

五、(本大题满分25分)

已知二次函数y=ax2+bx+c和一次函数y=-bx,其中a、b、c满足a>b>c,a+b+c=0.(a,b,c?R).

(1)求证:两函数的图象有两个不同的交点A、B;

(2)过(1)中的两点A、B分别作x轴的垂线,垂足为A1、B1.求线段A1B1的长的取值范围.

ì?y=ax2+bx+c?(1)证明:由í消去y得ax2+2bx+c=0, ???y=-bx22

D=4b2-4ac=4(-a-c)2-4ac=4(a2+ac+c2)=4[(a+c232················ 5分 )+c]. ·24

∵a+b+c=0,a>b>c,

∴a>0,c<0. 3∴c2>0, 4

∴D>0,即两函数的图象有两个不同的交点. ························································· 10分

(2)解:设方程ax2+2bx+c=0的两根为x1和x2,

则x1+x2=-2b,x1x2=c. ·························································································· 15分 aa

|A1B1|2=(x1-x2)2=(x1+x2)2-4x1x2

2b24c4b2-4ac4(-a-c)2-4ac=(-)-==aaa2a2

ccc13······················································· 20分 =4[()2++1]=4[(+)2+]. ·aaa24

∵a+b+c=0,a>b>c,∴a>0,c<0

c1∴a>-a-c>c,解得-2<<-. a2

∴3<|A1B1|2<

12|A

1B1|<······························································· 25分

4

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