题目列表(包括答案和解析)

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5.(08贵州贵阳)25.(本题满分12分)(本题暂无答案)

某宾馆客房部有60个房间供游客居住,当每个房间的定价为每天200元时,房间可以住满.当每个房间每天的定价每增加10元时,就会有一个房间空闲.对有游客入住的房间,宾馆需对每个房间每天支出20元的各种费用.

设每个房间每天的定价增加元.求:

(1)房间每天的入住量(间)关于(元)的函数关系式.(3分)

(2)该宾馆每天的房间收费(元)关于(元)的函数关系式.(3分)

(3)该宾馆客房部每天的利润(元)关于(元)的函数关系式;当每个房间的定价为每天多少元时,有最大值?最大值是多少?(6分)

试题详情

4.(08广东深圳)22.如图9,在平面直角坐标系中,二次函数的图象的顶点为D点,与y轴交于C点,与x轴交于A、B两点, A点在原点的左侧,B点的坐标为(3,0),

OB=OC ,tan∠ACO=

(1)求这个二次函数的表达式.

(2)经过C、D两点的直线,与x轴交于点E,在该抛物线上是否存在这样的点F,使以点A、C、E、F为顶点的四边形为平行四边形?若存在,请求出点F的坐标;若不存在,请说明理由.

(3)若平行于x轴的直线与该抛物线交于M、N两点,且以MN为直径的圆与x轴相切,求该圆半径的长度.

(4)如图10,若点G(2,y)是该抛物线上一点,点P是直线AG下方的抛物线上一动点,当点P运动到什么位置时,△APG的面积最大?求出此时P点的坐标和△APG的最大面积.

(08广东深圳22题解析)22.(1)方法一:由已知得:C(0,-3),A(-1,0) …1分

将A、B、C三点的坐标代入得       ……………………2分

解得:                     ……………………3分

所以这个二次函数的表达式为:      ……………………3分

方法二:由已知得:C(0,-3),A(-1,0)      ………………………1分

设该表达式为:            ……………………2分

将C点的坐标代入得:               ……………………3分

所以这个二次函数的表达式为:      ……………………3分

(注:表达式的最终结果用三种形式中的任一种都不扣分)

(2)方法一:存在,F点的坐标为(2,-3)       ……………………4分

理由:易得D(1,-4),所以直线CD的解析式为:

∴E点的坐标为(-3,0)                ……………………4分

由A、C、E、F四点的坐标得:AE=CF=2,AE∥CF

∴以A、C、E、F为顶点的四边形为平行四边形

∴存在点F,坐标为(2,-3)              ……………………5分

方法二:易得D(1,-4),所以直线CD的解析式为:

∴E点的坐标为(-3,0)                ………………………4分

∵以A、C、E、F为顶点的四边形为平行四边形

∴F点的坐标为(2,-3)或(―2,―3)或(-4,3) 

代入抛物线的表达式检验,只有(2,-3)符合

∴存在点F,坐标为(2,-3)              ………………………5分

(3)如图,①当直线MN在x轴上方时,设圆的半径为R(R>0),则N(R+1,R),

代入抛物线的表达式,解得  …………6分

②当直线MN在x轴下方时,设圆的半径为r(r>0),

则N(r+1,-r),

代入抛物线的表达式,解得  ………7分

∴圆的半径为.  ……………7分

(4)过点P作y轴的平行线与AG交于点Q,

易得G(2,-3),直线AG为.……………8分

设P(x),则Q(x,-x-1),PQ

        ……………………9分

时,△APG的面积最大

此时P点的坐标为.    ……………………10分

试题详情

3.(08广东广州)25、(2008广州)(14分)如图11,在梯形ABCD中,AD∥BC,AB=AD=DC=2cm,BC=4cm,在等腰△PQR中,∠QPR=120°,底边QR=6cm,点B、C、Q、R在同一直线l上,且C、Q两点重合,如果等腰△PQR以1cm/秒的速度沿直线l箭头所示方向匀速运动,t秒时梯形ABCD与等腰△PQR重合部分的面积记为S平方厘米

(1)当t=4时,求S的值

(2)当,求S与t的函数关系式,并求出S的最大值

图11
 

(08广东广州25题解析)25.(1)t=4时,Q与B重合,P与D重合,

重合部分是

试题详情

2.(08甘肃白银等9市)28.(12分)如图20,在平面直角坐标系中,四边形OABC是矩形,点B的坐标为(4,3).平行于对角线AC的直线m从原点O出发,沿x轴正方向以每秒1个单位长度的速度运动,设直线m与矩形OABC的两边分别交于点M、N,直线m运动的时间为t(秒).

(1) 点A的坐标是__________,点C的坐标是__________;

 (2) 当t=    秒或    秒时,MN=AC;

(3) 设△OMN的面积为S,求S与t的函数关系式;

(4) 探求(3)中得到的函数S有没有最大值?若有,求出最大值;若没有,要说明理由.

(08甘肃白银等9市28题解析)28. 本小题满分12分

解:(1)(4,0),(0,3);  ·················································································· 2分

(2) 2,6;  ·········································································································  4分

(3) 当0<t≤4时,OM=t.

由△OMN∽△OAC,得

ON=,S=.  ····································  6分

当4<t<8时,

如图,∵ OD=t,∴ AD= t-4.

方法一:

由△DAM∽△AOC,可得AM=,∴ BM=6-. ···························  7分

由△BMN∽△BAC,可得BN==8-t,∴ CN=t-4. ··································  8分

S=矩形OABC的面积-Rt△OAM的面积- Rt△MBN的面积- Rt△NCO的面积

=12--(8-t)(6-)-

=.  ··························································································· 10分

方法二:

易知四边形ADNC是平行四边形,∴ CN=AD=t-4,BN=8-t.·································· 7分

由△BMN∽△BAC,可得BM==6-,∴ AM=.······ 8分

以下同方法一.

 (4) 有最大值.

方法一:

当0<t≤4时,

∵ 抛物线S=的开口向上,在对称轴t=0的右边, S随t的增大而增大,

∴ 当t=4时,S可取到最大值=6; ················ 11分

当4<t<8时,

∵ 抛物线S=的开口向下,它的顶点是(4,6),∴ S<6.

综上,当t=4时,S有最大值6. ·······································································  12分

方法二:

∵ S=

∴ 当0<t<8时,画出S与t的函数关系图像,如图所示. ······························  11分

显然,当t=4时,S有最大值6.  ···································································  12分

说明:只有当第(3)问解答正确时,第(4)问只回答“有最大值”无其它步骤,可给1分;否则,不给分.

试题详情

1.(08福建莆田)26.(14分)如图:抛物线经过A(-3,0)、B(0,4)、C(4,0)三点.

  (1) 求抛物线的解析式.

  (2)已知AD = AB(D在线段AC上),有一动点P从点A沿线段AC以每秒1个单位长度的速度移动;同时另一个动点Q以某一速度从点B沿线段BC移动,经过t 秒的移动,线段PQ被BD垂直平分,求t的值;

 (3)在(2)的情况下,抛物线的对称轴上是否存在一点M,使MQ+MC的值最小?若存在,请求出点M的坐标;若不存在,请说明理由。

(注:抛物线的对称轴为)

   

(08福建莆田26题解析)26(1)解法一:设抛物线的解析式为y = a (x +3 )(x - 4)

   因为B(0,4)在抛物线上,所以4 = a ( 0 + 3 ) ( 0 - 4 )解得a= -1/3

   所以抛物线解析式为

解法二:设抛物线的解析式为

依题意得:c=4且  解得

 所以  所求的抛物线的解析式为

(2)连接DQ,在Rt△AOB中,

所以AD=AB= 5,AC=AD+CD=3 + 4 = 7,CD = AC - AD = 7 – 5 = 2

因为BD垂直平分PQ,所以PD=QD,PQ⊥BD,所以∠PDB=∠QDB

因为AD=AB,所以∠ABD=∠ADB,∠ABD=∠QDB,所以DQ∥AB

所以∠CQD=∠CBA。∠CDQ=∠CAB,所以△CDQ∽ △CAB

  即

所以AP=AD – DP = AD – DQ=5 –=  

所以t的值是

(3)答对称轴上存在一点M,使MQ+MC的值最小

理由:因为抛物线的对称轴为

所以A(- 3,0),C(4,0)两点关于直线对称

连接AQ交直线于点M,则MQ+MC的值最小

过点Q作QE⊥x轴,于E,所以∠QED=∠BOA=900

   DQ∥AB,∠ BAO=∠QDE,  △DQE ∽△ABO

  即

所以QE=,DE=,所以OE = OD + DE=2+=,所以Q()

设直线AQ的解析式为

  由此得

所以直线AQ的解析式为  联立

由此得  所以M

则:在对称轴上存在点M,使MQ+MC的值最小。

试题详情

96.(08广东佛山25题)25.我们所学的几何知识可以理解为对“构图”的研究:根据给定的(或构造的)几何图形提出相关的概念和问题(或者根据问题构造图形),并加以研究.

例如:在平面上根据两条直线的各种构图,可以提出“两条直线平行”、“两条直线相交”的概念;若增加第三条直线,则可以提出并研究“两条直线平行的判定和性质”等问题(包括研究的思想和方法).

请你用上面的思想和方法对下面关于圆的问题进行研究:

(1) 如图1,在圆O所在平面上,放置一条直线(和圆O分别交于点AB),根据这个图形可以提出的概念或问题有哪些(直接写出两个即可)?

(2) 如图2,在圆O所在平面上,请你放置与圆O都相交且不同时经过圆心两条直线(与圆O分别交于点AB与圆O分别交于点CD).

请你根据所构造的图形提出一个结论,并证明之.

(3) 如图3,其中AB是圆O的直径,AC是弦,D的中点,弦DEAB于点F. 请找出点C和点E重合的条件,并说明理由.

 

(08广东佛山25题解答)解:(1) 弦(图中线段AB)、弧(图中的ACB弧)、弓形、求弓形的面积(因为是封闭图形)等.  (写对一个给1分,写对两个给2分)

(2) 情形1  如图21,AB为弦,CD为垂直于弦AB的直径.  …………………………3分

结论:(垂径定理的结论之一).  …………………………………………………………4分

证明:略(对照课本的证明过程给分).  …………………………………………………7分

情形2  如图22,AB为弦,CD为弦,且ABCD在圆内相交于点P.

结论:.

证明:略.

情形3 (图略)AB为弦,CD为弦,且在圆外相交于点P.

结论:.

证明:略.

情形4  如图23,AB为弦,CD为弦,且ABCD.

结论:  =   .

证明:略.

(上面四种情形中做一个即可,图1分,结论1分,证明3分;

其它正确的情形参照给分;若提出的是错误的结论,则需证明结论是错误的)

(3) 若点C和点E重合,

则由圆的对称性,知点C和点D关于直径AB对称. …………………………………8分

,则.………………………………9分

D是   的中点,所以

.………………………………………………………10分

解得.……………………………………………………………11分

(若求得等也可,评分可参照上面的标准;也可以先直觉猜测点BC是圆的十二等分点,然后说明)

 

试题详情

95.(08山东聊城25题)25.(本题满分12分)如图,把一张长10cm,宽8cm的矩形硬纸板的四周各剪去一个同样大小的正方形,再折合成一个无盖的长方体盒子(纸板的厚度忽略不计).

 

(1)要使长方体盒子的底面积为48cm2,那么剪去的正方形的边长为多少?

(2)你感到折合而成的长方体盒子的侧面积会不会有更大的情况?如果有,请你求出最大值和此时剪去的正方形的边长;如果没有,请你说明理由;

(3)如果把矩形硬纸板的四周分别剪去2个同样大小的正方形和2个同样形状、同样大小的矩形,然后折合成一个有盖的长方体盒子,是否有侧面积最大的情况;如果有,请你求出最大值和此时剪去的正方形的边长;如果没有,请你说明理由.

(08山东聊城25题解答)(本题满分12分)

解:(1)设正方形的边长为cm,则

.························································································ 1分

解得(不合题意,舍去),

剪去的正方形的边长为1cm.·············································································· 3分

(注:通过观察、验证直接写出正确结果给3分)

(2)有侧面积最大的情况.

设正方形的边长为cm,盒子的侧面积为cm2

的函数关系式为:

.······························································································· 5分

改写为

时,

即当剪去的正方形的边长为2.25cm时,长方体盒子的侧面积最大为40.5cm2.········ 7分

(3)有侧面积最大的情况.

设正方形的边长为cm,盒子的侧面积为cm2

若按图1所示的方法剪折,则的函数关系式为:

时,.····························· 9分

若按图2所示的方法剪折,则的函数关系式为:

时,.················································································· 11分

比较以上两种剪折方法可以看出,按图2所示的方法剪折得到的盒子侧面积最大,即当剪去的正方形的边长为cm时,折成的有盖长方体盒子的侧面积最大,最大面积为cm2

说明:解答题各小题只给了一种解答及评分说明,其他解法只要步骤合理,解答正确,均应给出相应分数.

试题详情

94.(08广东梅州23题)23.本题满分11分.

如图11所示,在梯形ABCD中,已知ABCDADDBAD=DC=CBAB=4.以AB所在直线为轴,过D且垂直于AB的直线为轴建立平面直角坐标系.

(1)求∠DAB的度数及ADC三点的坐标;

(2)求过ADC三点的抛物线的解析式及其对称轴L

(3)若P是抛物线的对称轴L上的点,那么使PDB为等腰三角形的点P有几个?(不必求点P的坐标,只需说明理由)

(08广东梅州23题解答)解: (1) DCABAD=DC=CB

 ∠CDB=∠CBD=∠DBA,··············································································· 0.5分

   ∠DAB=∠CBADAB=2∠DBA, ············ 1分

DAB+∠DBA=90DAB=60, ·········· 1.5分

  ∠DBA=30AB=4, DC=AD=2,  ········· 2分

RtAODOA=1,OD=,··························· 2.5分

A(-1,0),D(0, ),C(2, ).  · 4分

(2)根据抛物线和等腰梯形的对称性知,满足条件的抛物线必过点A(-1,0),B(3,0),

故可设所求为  = (+1)( -3) ······························································ 6分

将点D(0, )的坐标代入上式得, =

所求抛物线的解析式为  =   ·········································· 7分

其对称轴L为直线=1.······················································································ 8分

(3) PDB为等腰三角形,有以下三种情况:

①因直线LDB不平行,DB的垂直平分线与L仅有一个交点P1P1D=P1B

P1DB为等腰三角形;  ················································································· 9分

②因为以D为圆心,DB为半径的圆与直线L有两个交点P2P3DB=DP2DB=DP3P2DBP3DB为等腰三角形;

③与②同理,L上也有两个点P4P5,使得 BD=BP4BD=BP5.  ···················· 10分

由于以上各点互不重合,所以在直线L上,使PDB为等腰三角形的点P有5个.

试题详情

93.(08福建南平26题)26.(14分)

(1)如图1,图2,图3,在中,分别以为边,向外作正三角形,正四边形,正五边形,相交于点

①如图1,求证:

②探究:如图1,    

如图2,    

如图3,    

(2)如图4,已知:是以为边向外所作正边形的一组邻边;是以为边向外所作正边形的一组邻边.的延长相交于点

①猜想:如图4,      (用含的式子表示);

②根据图4证明你的猜想.

(08福建南平26题解答)(1)①证法一:均为等边三角形,

························································································ 2分

··············································· 3分

························································ 4分

.··················································· 5分

证法二:均为等边三角形,

························································································ 2分

························································································ 3分

可由绕着点按顺时针方向旋转得到··································· 4分

.··························································································· 5分

.········································································ 8分(每空1分)

(2)①········································································································ 10分

②证法一:依题意,知都是正边形的内角,

,即.····························· 11分

.·························································································· 12分

······ 13分

········································ 14分

证法二:同上可证  .··························································· 12分

,如图,延长

································ 13分

················· 14分

证法三:同上可证  .··························································· 12分

························································ 13分

········································································ 14分

证法四:同上可证  .··························································· 12分

.如图,连接

.···································· 13分

······························· 14分

注意:此题还有其它证法,可相应评分.

试题详情

92.24.(本小题满分12分)

如图10,已知点A的坐标是(-1,0),点B的坐标是(9,0),以AB为直径作⊙O′,交y轴的负半轴于点C,连接AC、BC,过A、B、C三点作抛物线.

(1)求抛物线的解析式;

(2)点E是AC延长线上一点,∠BCE的平分线CD交⊙O′于点D,连结BD,求直线BD的解析式;

(3)在(2)的条件下,抛物线上是否存在点P,使得∠PDB=∠CBD?如果存在,请求出点P的坐标;如果不存在,请说明理由.

(24题解答)(1) ∵以AB为直径作⊙O′,交y轴的负半轴于点C,

∴∠OCA+∠OCB=90°,

又∵∠OCB+∠OBC=90°,

∴∠OCA=∠OBC,

又∵∠AOC= ∠COB=90°,

∴ΔAOC∽ ΔCOB,························································································ 1分

又∵A(–1,0),B(9,0),

,解得OC=3(负值舍去).

∴C(0,–3),

······················································································································ 3分

设抛物线解析式为y=a(x+1)(x–9),

∴–3=a(0+1)(0–9),解得a=

∴二次函数的解析式为y=(x+1)(x–9),即y=x2x–3.···························· 4分

(2) ∵AB为O′的直径,且A(–1,0),B(9,0),

∴OO′=4,O′(4,0),····················································································· 5分

∵点E是AC延长线上一点,∠BCE的平分线CD交⊙O′于点D,

∴∠BCD=∠BCE=×90°=45°,

连结O′D交BC于点M,则∠BO′D=2∠BCD=2×45°=90°,OO′=4,O′D=AB=5.

∴D(4,–5).································································································· 6分

∴设直线BD的解析式为y=kx+b(k≠0)

··························································· 7分

解得

∴直线BD的解析式为y=x–9.····································· 8分

(3) 假设在抛物线上存在点P,使得∠PDB=∠CBD,

解法一:设射线DP交⊙O′于点Q,则

分两种情况(如答案图1所示):

①∵O′(4,0),D(4,–5),B(9,0),C(0,–3).

∴把点C、D绕点O′逆时针旋转90°,使点D与点B重合,则点C与点Q1重合,

因此,点Q1(7,–4)符合

∵D(4,–5),Q1(7,–4),

∴用待定系数法可求出直线DQ1解析式为y=x–.··································· 9分

解方程组

∴点P1坐标为(),[坐标为()不符合题意,舍去].

······················································································································ 10分

②∵Q1(7,–4),

∴点Q1关于x轴对称的点的坐标为Q2(7,4)也符合

∵D(4,–5),Q2(7,4).

∴用待定系数法可求出直线DQ2解析式为y=3x–17.······································ 11分

解方程组

∴点P2坐标为(14,25),[坐标为(3,–8)不符合题意,舍去].

······················································································································ 12分

∴符合条件的点P有两个:P1(),P2(14,25).

解法二:分两种情况(如答案图2所示):

①当DP1∥CB时,能使∠PDB=∠CBD.

∵B(9,0),C(0,–3).

∴用待定系数法可求出直线BC解析式为y=x–3.

又∵DP1∥CB,∴设直线DP1的解析式为y=x+n.

把D(4,–5)代入可求n= –

∴直线DP1解析式为y=x–.························· 9分

解方程组

∴点P1坐标为(),[坐标为()不符合题意,舍去].

······················································································································ 10分

②在线段O′B上取一点N,使BN=DM时,得ΔNBD≌ΔMDB(SAS),∴∠NDB=∠CBD.

由①知,直线BC解析式为y=x–3.

取x=4,得y= –,∴M(4,–),∴O′N=O′M=,∴N(,0),

又∵D(4,–5),

∴直线DN解析式为y=3x–17.······································································ 11分

解方程组

∴点P2坐标为(14,25),[坐标为(3,–8)不符合题意,舍去].

······················································································································ 12分

∴符合条件的点P有两个:P1(),P2(14,25).

解法三:分两种情况(如答案图3所示):

①求点P1坐标同解法二.··············································································· 10分

②过C点作BD的平行线,交圆O′于G,

此时,∠GDB=∠GCB=∠CBD.

由(2)题知直线BD的解析式为y=x–9,

又∵ C(0,–3)

∴可求得CG的解析式为y=x–3,

设G(m,m–3),作GH⊥x轴交与x轴与H,

连结O′G,在Rt△O′GH中,利用勾股定理可得,m=7,

由D(4,–5)与G(7,4)可得,

DG的解析式为,··········································································· 11分

解方程组

∴点P2坐标为(14,25),[坐标为(3,–8)不符合题意,舍去].························ 12分

∴符合条件的点P有两个:P1(),P2(14,25).

说明:本题解法较多,如有不同的正确解法,请按此步骤给分.

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