题目列表(包括答案和解析)
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)
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
∴圆的半径为或. ……………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
当时,△APG的面积最大
此时P点的坐标为,. ……………………10分
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
|
(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=
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.
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
则 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。
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A、B),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的两条直线和(与圆O分别交于点A、B,与圆O分别交于点C、D).
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是的中点,弦DE⊥AB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AB与CD在圆内相交于点P.
结论:.
证明:略.
情形3 (图略)AB为弦,CD为弦,且与在圆外相交于点P.
结论:.
证明:略.
情形4 如图23,AB为弦,CD为弦,且AB∥CD.
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
设,则,.………………………………9分
又D是 的中点,所以,
即.………………………………………………………10分
解得.……………………………………………………………11分
(若求得或等也可,评分可参照上面的标准;也可以先直觉猜测点B、C是圆的十二等分点,然后说明)
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
即.
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,
则与的函数关系式为:
.
即.······························································································· 5分
改写为.
当时,.
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,则与的函数关系式为:
.
即.
当时,.····························· 9分
若按图2所示的方法剪折,则与的函数关系式为:
.
即.
当时,.················································································· 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.
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AB∥CD, AD⊥DB,AD=DC=CB,AB=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A、D、C三点的坐标;
(2)求过A、D、C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) DC∥AB,AD=DC=CB,
∠CDB=∠CBD=∠DBA,··············································································· 0.5分
∠DAB=∠CBA, ∠DAB=2∠DBA, ············ 1分
∠DAB+∠DBA=90, ∠DAB=60, ·········· 1.5分
∠DBA=30,AB=4, DC=AD=2, ········· 2分
RtAOD,OA=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L与DB不平行,DB的垂直平分线与L仅有一个交点P1,P1D=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P2、P3,DB=DP2,DB=DP3, P2DB, P3DB为等腰三角形;
③与②同理,L上也有两个点P4、P5,使得 BD=BP4,BD=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.
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:与均为等边三角形,
,························································································ 2分
且··············································· 3分
,
即························································ 4分
.··················································· 5分
证法二:与均为等边三角形,
,························································································ 2分
且························································································ 3分
可由绕着点按顺时针方向旋转得到··································· 4分
.··························································································· 5分
②,,.········································································ 8分(每空1分)
(2)①········································································································ 10分
②证法一:依题意,知和都是正边形的内角,,,
,即.····························· 11分
.·························································································· 12分
,,······ 13分
,
········································ 14分
证法二:同上可证 .··························································· 12分
,如图,延长交于,
,
································ 13分
················· 14分
证法三:同上可证 .··························································· 12分
.
,
························································ 13分
即········································································ 14分
证法四:同上可证 .··························································· 12分
.如图,连接,
.···································· 13分
即······························· 14分
注意:此题还有其它证法,可相应评分.
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=x2–x–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
解方程组得
∴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
解方程组得
∴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
解方程组得
∴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
解方程组得
∴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
解方程组得
∴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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