北山中学初2015级初三上期第一月月考数学参考答案

1北山中学初2015级初三上期第一学月月考数学参考答案一、选择题：（本大题12个小题，每小题4分，共48分）在每小题的下面，都给出了代号为A、B、C、D的四个答案，其中只有一个是正确答案，请将正确答案的代号填在答卷的对应位置。题号123456789101112答案ACBCDAADDCBC二、填空题：（本大题6个小题，每小题4分，共24分）在每小题中，请将答案填在答卷的对应位置。13.____3___14._0562xx_15.上，y轴（或x=0）16.2)2(2xy或_242xxy17.____20%_____18.___①③____三、解答题：（本大题共2个小题，每小题7分，共14分）解答时每小题必须给出必要的演算过程或推理步骤.19．解方程：（x-1)2-4(x-1)+4=020．已知二次函数y=-x2+4x-2（1）把它化成顶点式为2)2-(-2xy；·········2分（2）在给出的坐标系中画出函数的图象（略）四、解答题：（本大题共4个小题，每小题10分，共40分）解答时每小题必须给出必要的演算过程或推理步骤.21．关于x的一元二次方程230xxk有两个不相等的实数根．（1）求k的取值范围．（2）请选择一个k的负整数值，并求出方程的根．解：0)21(2x·········4分∴321xx··············7分解法二：原方程化简得x2-6x+9=0·····2分即（x-3）2=0····4分∴321xx··············7分（其他做法也可）解：（1）由题意得△=)(14)3(422kacb······························2分∴9+4k0···········································4分解得49k···········································6分（2）当k=-2时，原方程为0232xx，···································8分解得,21x,12x······································10分当k=-1时（略）列表正确······4分画图正确······7分2ABCD16米草坪1米0.5米2米2.5米22.如图所示，某幼儿园有一道长为16米的墙，计划用32米长的围栏靠墙围成一个矩形的草坪ABCD．(1)围成的矩形草坪ABCD的面积为120平方米时．求该矩形草坪BC边的长．（2）围成的矩形草坪ABCD的面积可以是140平方米吗？为什么？解：(1)设AB长为xm，则BC为（32-2x）m，···············1分由题意得120)232(xx··························3分解得x=6或x=10··································4分当x=6时，32-2x=20＞16，不合题意，舍去当x=10时，32-2x=12＜16，符合题意···················5分答：该矩形草坪BC边的长为12米。·······················6分(2)设AB长为xm，则BC为（32-2x）m，由题意得140)232(xx·······················7分∴070162xx∴△=0247014)16(422acb·········8分∴此方程无实数根·····························9分∴不能围成面积是140平方米的矩形草坪ABCD。··········10分23.如图，小明的父亲在相距2米的两棵树间拴了一根绳子，给他做了个简易秋千，栓绳子的地方离地面都是2.5米，绳子自然下垂呈抛物线状，身高1米的小明距较近的那棵树0.5米时，头部刚好接触到绳子，则绳子最低点距离地面的距离为多少米？解：（1）如图，建立直角坐标系．·······························2分由图可设抛物线的解析式为：caxy2．··············4分把（0.5，1）、（1，2.5）代入得：5.21)5.0(2caca····································6分解得2a，21c··································8分∴绳子所在抛物线的函数关系式为：5.022xy.∵当x=0时，5.05.022xy，····················9分∴绳子最低点距离地面的距离为0.5米.·····················10分（其他建立平面直角坐标系的方法也可）324.某商品的进价为每件50元，售价为每件60元，每个月可卖出200件.如果每件商品的售价上涨1元，则每个月少卖10件（每件售价不能高于72元）.设每件商品的售价上涨x元（x为整数），每个月的销售利润为y元，（1）求y与x的函数关系式，并直接写出x的取值范围；（2）每件商品的售价定为多少元时，每个月可获得最大利润？最大月利润是多少元？解：（1）)10200)(10(xxy·················································4分=-10x2+100x+200·······································································其中0≤≤12；··················································5分（2）2250)5(1020001001022xxxy·······························7分∵满足0≤≤12，·····················································8分∴5x时，2250最大y·····················································9分答：每件商品的售价定为65元时，每个月可获得最大利润，最大月利润是2250元。·······10分五、解答题：（本大题共2个小题，每小题12分，共24分）解答时每小题必须给出必要的演算过程或推理步骤.25.已知ABC△的两边AB、AC的长是关于x的一元二次方程22(23)320xkxkk的两个实数根，第三边BC的长为5．（1）当k为何值时，ABC△是直角三角形；（2）当k为何值时，ABC△是等腰三角形，并求出ABC△的周长．解：（1）△=01)23(14)32(4222kkkacb·························1分∴,11kx,22kx·······················································2分又由直角三角形三边关系得：25)2()1(22kk···①或22)2(25)1(kk···②·····················4分解①得：k=-5或k=2又因为k+1和k+2是直角三角形的边长，故为正数，所以k=2························5分解②得：k=11答：当k=2或k=11时，ABC△是直角三角形。·······················6分（2）∵△ABC的两边AB、AC的长是这个方程的两个实数根，由（1）知，AB≠AC，△ABC第三边BC的长为5，且△ABC是等腰三角形，∴必然有AB=5或AC=5，即x=5是原方程的一个解.························8分将x=5代入方程0)12(22kkxkx，25-5(2k+1)+k2+k=0，解得k=4或k=5．························10分当k=3时，x的另外一个根是4，△ABC的周长为14；························11分当k=4时，x的另外一个根是6，△ABC的周长是16。························12分426.已知抛物线y＝ax2＋2x＋c的图像与x轴交于点A（3，0）和点C，与y轴交于点B（0，3）．（1）求抛物线的解析式；（2）在抛物线的对称轴上找一点D，使得点D到点B、C的距离之和最小，并求出点D的坐标；（3）在第一象限的抛物线上，是否存在一点P，使得△ABP的面积最大？若存在，求出点P的坐标；若不存在，请说明理由．解：（1）∵抛物线cxaxy22的图象经过点A（3，0）和点B（0，3），∴3069cca，······································································2分解得a=-1，c=3，······································································3分∴抛物线的解析式为：322xxy．···················································4分（2）对称轴为x=ab2=1，令322xxy，解得,31x12x，∴C（-1，0）．·······································································5分如图1所示，连接AB，与对称轴x=1的交点即为所求之D点，由于A、C两点关于对称轴对称，则此时DB+DC=DB+DA=AB最小．··································6分设直线AB的解析式为y=kx+b，由A（3，0）、B（0，3）可得：303bbk，解得k=-1，b=3，···················································7分∴直线AB解析式为y=-x+3．当x=1时，y=2，∴D点坐标为（1，2）．················································8分（3）结论：存在．如图2所示，设P（x，y）是第一象限的抛物线上一点，过点P作PN⊥x轴于点N，则ON=x，PN=y，AN=OA-ON=3-x．S△ABP=S梯形PNOB+S△PNA-S△AOB······························································9分=21（OB+PN）ON+21PN×AN-21OA×OB=21（3+y）x+21y（3-x）-21×3×3=23（x+y）-29，······························································10分∵P（x，y）在抛物线上，∴322xxy，代入上式得：S△ABP=23（x+y）-29=827)23(232x，∴当x=23时，S△ABP取得最大值．···················································11分当x=23时，322xxy=415，∴P（23，415）．所以，在第一象限的抛物线上，存在一点P，使得△ABP的面积最大；P点的坐标为（23，415）．···················································12分5