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初三数学公开课

发布时间:2014-06-06 13:59:15  

思如泉涌 德才兼优

厦门市初三数学质检押题卷

1考试范围:八年级:一次函数,反比例函数;九年级上册:二次根式,一元二次方程,旋转,圆,概率;

2试题结构:与中考题一致;

3注意点:

(1)韦达定理不要求,但是可以用,不扣分;

(2)作函数图象:一:直线直接画两点就可以了。

二:双曲线,只要一支一个点,做草图而已。

三:应用题要画图,要求范围,空心,实心要标清楚,不可多画;

(3)如何写分段函数的解析式,必须写出范围,后面跟解析式,不能用大括号连接;

(4)函数的自变量范围,题目有要求才写,或者后面需要用到范围的才写;

(5)关于辅助线,有名称的辅助线都不能称为辅助线,如三角形的高、中线、中位线、半径等,连接两点不是辅助线,如做平行线才是辅助线,且辅助线最多一条;

(6)关于梯形:作为背景出题,只要求掌握梯形性质。明白等腰梯形是轴对称图形,会建立直角坐标系解决问题,点的坐标等,平移辅助线在梯形中补考;

(7)动点问题:端点会很明确的给出; (8)应用题:式子写对才会给分;

(9)圆内接四边形;

一次函数:

(1)变量与函数,自变量与函数值的概念及函数的图象;

(2)正比例函数:一般地,形如y?kx(k是常数,k?0)的函数;

(3)一次函数:一般地,形如y?kx?b(k,b是常数,k?0)的函数;

(4)用函数观点看方程(组)与不等式;

1.一次函数与一元一次方程; 2.一次函数与一元一次不等式;

3.一次函数与二元一次方程(组);

反比例函数:

k(1)定义:形如y?(k为常数,k≠0)的函数称为反比例函数。 x

(2)性质1:反比例函数的图象属于双曲线。且关于原点对称。

性质2:当k>0时,双曲线的两支分别位于第一、第三象限,在每个象限内,y随x的增大而减小。

性质3:当k<0时,双曲线的两支分别位于第二、第四象限,在每个象限内,y随x的增大而增大。

特别注意。

(1)由于反比例函数中,自变量x为分母,所以,为了使函数有意义,x必须不等于0.所以,反比例函数图象肯定不经过原点。

(2)反比例函数中的k是一个不为零的常数,且k是指代分子中的整个数或式子。

求反比例函数的解析式

k待定系数法:如果反比例函数过某点,那么这点肯定满足反比例函数的基本形式y?,把x、x

y分别带入式子,便可求出k的值。

1

思如泉涌 德才兼优

【1】在平面直角坐标系中,O是坐标原点,点P(m,n)在反比例函数y=k的图象上.若m=k,

x

n=k

-2,则k= ;若m+n,OP=2,且此反比例函k

数y=满足:当x?0时,y随x的增大而减小,则k= ;

x【2】如图,平面直角坐标系中,OB在x轴上,∠ABO=90°,点A的坐标为(2,4),将△ABO绕点A逆时针旋转90°,点O的对应点C恰好落在双曲线y?

【3】在函数

y=

k

x

k

(x>0)上,则k; x

(k<0)的图象上有(1,y1),(﹣1,y2),(﹣2,y3)三个点,则下列各式中

正确( )

A.y1<y2<y3 B.y1<y3<y2 C.y3<y2<y1 D.y2<y3<y1

【4】如图1,已知双曲线y=(x?0)经过矩形OABC的边AB,AB的中点F,E,且四

k

x

边形OEBF的面积为2,反比例函数解析式为 . 【5】直线y=kx+b经过点M(-1,m),B(m,1)(m?1),则必有( )

A.k>0,b>0 B.k>0,b<0 C.k<0,b>0 D.k<0,b<0 【6】A(x1,y1)、B(x2,y2)是一次函数y=kx+2(k>0)图象上不同的两点,若

t=(x1-x2)(y1-y2),则( )

A.t<0 B.t=0 C.t>0 D.t≤0 【7】如图,一次函数y?kx?b的图象与反比例函数y?

(1)根据图象,分别写出A、B的坐标; (2)求出两函数解析式;

(3)根据图象回答:当x为何值时,一次函数的函数值大于反比例函数的函数值;

m

的图象相交于A、B两点. x

2

思如泉涌 德才兼优

【1】下列计算正确的是( )

?

?

2 C. 【2】函数y=x?1 +?2?

6??1的自变量x的取值范围是( ) x?2

A、x≥-1 B、x≤-1 C、x≠2 D、x≥-1且x≠2

【3】关于 x的一元二次方方程x2?2x?m?0 没有实数根,则m 的取值范围是( )

A. m??1 B. m??1 C. m?1 D.m?1

【4】透支一个均匀的正六面体骰子,每个面上依次标有1、2、3、4、5和6,掷得的数是“5”或“6”的概率等于( ) 1111A. B. C. D. 3564【5】如果梯形的中位线的长是6cm,上底长是4cm,那么下底长为( )

A. 2cm B. 4cm C. 6cm D. 8cm

【6】为了绿化校园,某校计划经过两年时间,绿地面积增加21%.设平均每年绿地面积增长率为x,则方程可列为( ).

A 、(1+x)2=21% B、(1+x)+(1+x)2=21% C、(1+x)2 =1+21% D、(1+x)+(1+x)2=1+21%

【7】在平面直角坐标系中,已知点O(0,0),A(2,4).将线段OA沿x轴向左平移2个单 位,记点O、A的对应点分别为点O1、A1,则点O1,A1的坐标分别是( )

A.(0,0),(2,4) B.(0,0),(0,4) C.(2,0),(4,4) D.(-2,0),(0,4)

【8】计算 (2?1)(2?1)= ;

【9】x2?3x?_______?(x?___)2;

【10】已知n?16是整数,则n的最小整数值是________________;

【11】五边形的内角和的度数是 ;

【12】已知一个圆锥的底面半径长为3cm、母线长为6cm,则圆锥的侧面积是 cm2;

【13】如图5,已知∠ABC=90°,AB=πr,BC2r的⊙O从点A出发,沿

动到点C时停止.请你根据题意,在图5上画出圆心..O运动路径的示意图;

圆心O运动的路程是 . πr【14】如图,AB是圆O的弦,AB=2,△AOB的面积是3,

则∠AOB= ;

【15】如图4,点D是等边△ABC内一点,如果△ABD绕点A 逆时针旋转后能与△ACE重合,那么旋转了 度;

E3

B图4

思如泉涌 德才兼优

【16】已知a+b=2,ab=-1,则3a+ab+3b= ;a2+b2= ;

【17】在分别写有整数1到10的10张卡片中,随机抽取1张卡片,则该卡片上的数字恰好是奇数的概率是 ;

【18】

; 【19】解方程:3x2?4x?1?0(配方法);

【20】在平面直角坐标系中,已知点A(-4,1),B(-2,0),C(-3,-1),请在图上画出△ABC,并画出与△ABC关于原点O对称的图形;

【21】如图,已知A、B、C是圆O上的三点,∠ACB=90°,BC=5,AC=12,求圆O直径的长度.

【22】甲袋中有三个红球,分别标有数字1、2、3;乙袋中有三个白球,分别标有数字2、3、4.这些球除颜色和数字外完全相同.小明先从甲袋中随机摸出一个红球,再从乙袋中随机摸出一个白球.请画出树状图,并求摸得的两球数字相同的概率;

A B

4

思如泉涌 德才兼优

【23】在平面直角坐标系中,定义:p(x,y),若满足x+y=﹣xy,则称点p为“和谐点”,如点p(0,0)是一个和谐点.

①若和谐点在双曲线y=上,求这个和谐点.

②求证:直线y=x+m上一定有两个和谐点.

【24】某种服装,平均每天可以销售20件,每件盈利44元,在每件降价幅度不超过10元的情况下,若每件降价1元,则每天可多售出5件,如果每天要盈利1600元,每件应降价多少元?

【25】 已知:⊙O是△ABC的外接圆,AB为⊙O的直径,弦CD交AB于E,∠BCD=∠BAC . (1)求证:AC=AD;

(2)过点C作直线CF,交AB的延长线于点F,若∠BCF=30°,则结论“CF一定是⊙O

4x

图85

思如泉涌 德才兼优

【26】已知关于x的方程x2―2x―2n=0有两个不相等的实数根.

(1)求n的取值范围;

(2)若n<5,且方程的两个实数根都是整数,求n的值;

【27】已知点A(1,c)和点B (3,d )是直线y=k1x+b与双曲线y=k2>0)的交点. x

(1)过点A作AM⊥x轴,垂足为M,连结BM.若AM=BM,求点B的坐标;

(2)设点P在线段AB上,过点P作PE⊥x轴,垂足为E,并交双曲线y=k2k2(k>0)x 2

PN1于点N.当 取最大值时,若PN= ,求此时双曲线的解析式; NE2

6

思如泉涌 德才兼优

k(1)解:∵点A(1,c)和点B (3,d )在双曲线y=(k2>0)上, x

∴ c=k2=3d ·················································································· 1分 ∵ k2>0, ∴ c>0,d>0.

A(1,c)和点B (3,d )都在第一象限.

∴ AM=3d. ··················································································· 2分 过点B作BT⊥AM,垂足为T.

∴ BT=2. ······················································································ 3分 TM=d.

∵ AM=BM, ∴ BM=3d.

在Rt△BTM中,TM 2+BT2=BM2,

∴ d2+4=9d2, ∴ d=

点B(3,2 . ················································································ 4分 2

k(2)解1:∵ 点A(1,c)、B(3,d)是直线y=k1x+b与双曲线y=(k2>0)的交点, x

∴ c=k2,,3d=k2,c=k1+b,d=3k1+b. ································ 5分

14∴ k1=-k2,b=k2. 33

∵ A(1,c)和点B (3,d )都在第一象限,∴ 点P在第一象限.

PEk1x+b∴= NEkx

kb=2+ k2k2

14x2+x. ······································································ 6分 33

∵ 当x=1,3时,

又∵当x=2时, PE1; NEPE4. NE3

PE4∴ 1≤. ·············································································· 7分 NE3

∴ PE≥NE. ··················································································· 8分

PNPE14∴ =-1=-x2+x-1. ··················································· 9分 NENE33

∴ 当x=2时,

PN1. ············································································ 10分 NE3

1由题意,此时PN= 2

3∴ NE ······················································································11分 2

7

思如泉涌 德才兼优

3∴ 点N(2) . ∴ k2=3. 2

3∴ y ························································································ 12分 x

解2:∵ A(1,c)和点B (3,d )都在第一象限,∴ 点P在第一象限.

PEk1x+bk2b∵ ==+x, NEk2k2k2

x

PE当点P与点A、B重合时,=1, NE

即当x=1或3时,PE1. NE

kb9k3b∴ 有 +1, =-1. ······································· 5分 k2k2k2k2

14解得,k1=-k2,b2. 33

PE14∴ =-x2+x. ······································································· 6分 NE33

∵ k2=-3k1,k2>0,∴ k1<0.

k3k∵ PE-NE=k1x+b-k1x-4k1+ x x

x2-4x+3k1 (x-1)(x-3)=k1( )= , ············································ 7分 x x

又∵当1≤x≤3时,

(x-1) (x-3) ≤0,

∴ k1( (x-1)(x-3) ) ≥0. x

∴ PE-NE≥0. ············································································· 8分 ∴ PNPE=-1 NENE

14x2+x-1. ································································· 9分 33

PN1∴ 当x=2的最大值是 ················································· 10分 NE3

1由题意,此时PN= 2

3∴ NE ····················································································11分 2

3∴ 点N(2) . ∴ k2=3. 2

3∴ y ························································································ 12分 x

k 解3:∵ 点A(1,c)、B(3,d)是直线y=k1x+b与双曲线y=(k2>0)的交点, x

8

思如泉涌 德才兼优 ∴ c=k2,,3d=k2,c=k1+b,d=3k1+b. ·································· 5分 k2=3d, k1=-d,b=4d.

3d∴ 直线y=-dx+4d,双曲线y=. x

∵ A(1,c)和点B (3,d )都在第一象限,∴ 点P在第一象限.

3d∴ PN=PE-NE=-dx+4d-x

x2-4x+3d (x-1)(x-3)=-d( ) , ······································ 6分 x x

又∵当1≤x≤3时,(x-1) (x-3) ≤0,

d (x-1)(x-3)∴-≥0. x

∴ PN=PE-NE≥0. ···································································· 7分

3d-dx+4d-xPN∴ = ································································· 8分 NE3dx

14x2+x-1. ································································· 9分 33

PN1∴ 当x=2的最大值是 ················································· 10分 NE3

1由题意,此时PN= 2

3∴ NE ······················································································11分 2

3∴ 点N(2) . 2

3∴ k2=3. ∴ y=. ··············································································· x

4课后练习,2013年厦门市九年级质检测试卷;

9

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