北师大版数学九年级上册期末模拟试卷04（含答案）

这是一份北师大版数学九年级上册期末模拟试卷04（含答案），共13页。试卷主要包含了选择题，填空题，解答题等内容，欢迎下载使用。
北师大版数学九年级上册期末模拟试卷一、选择题1．点（一1，一）所在的象限为  A．第一象限    B．第二象限    c．第三象限    D．第四象限2．反比例函数y＝的图象生经过点（1，－2），则k的值为  A．－1    B．－2    C．1    D．23．若y＝ kx－4的函数值y随x的增大而减小，则k的值可能是下列的  A．－4    B．0    C．1    D．34．在平面直角坐标系中，函数y＝ －x＋1的图象经过  A．第一，二，三象眼    B.第二，三，四象限  C．第一，二，四象限    D．第一，三，四象限5．如图，AB是⊙O的直径，点C在⊙O上，若∠B＝50°，则∠A的度数为A．80°    B．60°    C．50°    D．40°6．如图，点A(t，3)在第一象限，OA与x轴所夹的锐角为α，tanα＝A．1    B．1.5    C．27．抛物线y＝－3x2－x＋4与坐标轴的交点的个数是A．3    B．2    C．1    D．08．在同一平面直角坐标系中，函数y＝mx＋m与y＝－ (m≠0)的图象可能是    9．如图，点A是反比例函数y＝(x＞0）的图象上任意一点，AB//x轴，交反比例函数y＝－的  图象于点B，以AB为边作ABCD，其中C、D在x轴上，则SABCD为A. 2           B. 3          C. 4             D. 5         10．如图，在平面直角坐标系中，⊙O的半径为1，则直线y＝x一与⊙O的位置关系是   A．相离    B.相切     C．相交    D．以上三种情况都有可能11．竖直向上发射的小球的高度h(m)关于运动时间t(s)的函数表达式为h＝at2＋bt，其图象如图  所示，若小球在发射后第2秒与第6秒时的高度相等，则下列时刻中小球的高度最高的是  A．第3秒    B.第3.9秒    C．第4.5秒    D．第6.5秒       12．如图，将抛物线y＝(x—1)2的图象位于直线y＝4以上的部分向下翻折，得到新的图像，若直线y＝－x＋m与新图象有四个交点，则m的取值范围为A.＜m＜3     B.＜m＜7      C.＜m＜7    D.＜m＜3二、填空题13．直线y＝kx＋b经过点(0，0)和(1，2)，则它的解析式为_____________14．如图，A、B、C是⊙O上的点，若∠AOB＝70°，则∠ACB的度数为__________      15．如图，己知点A(O，1)，B（O，－1），以点A为圆心，AB为半径作圆,交x轴的正半轴于点C．则∠BAC等于____________度．16．如图，在平面直角坐标系中，抛物线y＝x2经过平移得到抛物线y＝x2－2x,其对称轴与两段抛物线弧所围成的阴影部分的面积为______________17．如图，已知点A、C在反比例函数y＝(a＞0)的图象上，点B、D在反比例函数y＝(b＜0)的图象上，AB∥CD∥x轴，AB，CD在x轴的两侧，AB＝3，CD＝2，AB与CD的距离为5，则a－b的值是________________      18．如图所示，⊙O的面积为1，点P为⊙O上一点，令记号【n,m】表示半径OP从如图所示的位置开始以点O为中心连续旋转n次后，半径OP扫过的面积．旋转的规则为：第1次旋转m度；第2次从第1次停止的位置向相同的方向再次旋转度：第3次从第2次停止的位置向相同的方向再次旋转度；第4次从第3次停止的位置向相同的方向再次旋转度……依此类推．例如【2，90】＝，则【2017, 180】＝_______________三、解答题19.(1)计算sin245°＋cos30°•tan60°   (2)在直角三角形ABC中，已知∠C＝90°，∠A＝60°，BC＝3，求AC.     20．如图，⊙O的直径CD＝10，AB是⊙O的弦，AB⊥CD，垂足为M, OM∶OC＝3∶5.求AB的长度．   21．如图，点(3，m)为直线AB上的点.求该点的坐标．   22．如图，在⊙O中，AB，CD是直径，BE是切线，连结AD，BC，BD.    (1)求证：△ABD≌△CDB;    (2)若∠DBE＝37°，求∠ADC的度数．   23．某体育用品店购进一批单价为40元的球服，如果按单价60元销售，那么一个月内可售出240套，根据销售经验，提高销售单价会导致销售量的减少，即销售单价每提高5元，销售量相应减少20套.求当销售单价为多少元时，才能在一个月内获得最大利润？最大利润是多少？       24.如图所示，某数学活动小组要测量小河对岸大树BC的高度，他们在斜坡上D处测得大树顶端B的仰角是30°，朝大树方向下坡走6米到达坡底A处，在A处测得大树顶端B的仰角是48°，若坡角∠FAE＝30°，求大树的高度．（结果保留整数，参考数据：sin48°≈0.74，cos48°≈0.67, tan48°≈l.ll, ≈1.73)          25.如图，矩形OABC的顶点A、C分别在x轴、y轴的正半轴上，点D为对角线OB的中点，点E(4，n)在边AB上，反比例函数y＝(k≠0)在第一象限内的图象经过点D、E，且tan∠BOA＝．(1)求边AB的长；(2)求反比例函数的解析式和n的值；(3)若反比例函数的图象与矩形的边BC交于点F，将矩形折叠，使点D与点F重合，折痕分别与x、y轴正半轴交于H、G，求线段OG的长  26．如图，抛物线y＝(x2＋3x一4）与x轴交于A、B两点，与y轴交于点C．(1)求点A、点C的坐标，(2)求点D到AC的距离。(3)看点P为抛物线上一点，以2为半径作⊙P，当⊙P与直线AC相切时，求点P的横坐标．   27．(1)如图l，Rt△ABD和Rt△ABC的斜边为AB，直角顶点D、C在AB的同侧，求证：A、B、C、D四个点在同一个圆上．(2)如图2，△ABC为锐角三角形，AD⊥BC于点D，CF⊥AB于点F，AD与CF交于点G，连结BG并延长交AC于点E，作点D关于AB的对称点P，连结PF.求证：点P、F、E三点在一条直线上．(3)如图3，△ABC中，∠A＝30°，AB＝AC＝2，点D、E、F分别为BC、CA、AB边上任意一点，△DEF的周长有最小值，请你直接写出这个最小值．                     参考答案一、选择题：题号123456789101112答案CBACDCAADBBD二、填空题：13. y=2x14. 3515. 6016.417. 618. 或三、解答题：19.(1) 解：=  1分=  2分························································=2 ························································3分  (2)解：∵∠B=90°－∠A=90°－60°=30°·························1分tanB=······················································2分∴AC=3·tanB=3tan30°=3×=．··································3分20. 解：连接OB，··············································1分∵⊙O的直径CD＝10，∴OC＝5，····················································2分又∵OM︰OC＝3︰5，∴OM＝3，····················································3分∵AB⊥CD，且CD为⊙O的直径，∴△BOM是直角三角形，且AB＝2BM；································4分在Rt△BOM中，OB＝5，OM＝3，∴BM＝，·····················································5分∴AB＝2BM＝8·················································6分21. 解：设直线AB的解析式为由图象可知，直线AB过点(－1，2)和(－2，0)·························1分∴ ························································2分(1)－(2)得k=2，把k=2代入(1)得2=－2+b，∴b=4··································3分∴ ∴直线AB的解析式为y=2x+4······································4分当x=3时，y=2×3+4=10·········································5分∴该点坐标为（3，10）········································6分22.（1）证明：∵AB、CD为⊙O直径∴ ∠ADB＝∠CBD＝90°，·······································1分又∵∠A＝∠C，AB＝CD，∴△ABD≌△CDB（AAS）．········································3分（2）∵BE与⊙O相切于B，∴AB⊥BE，·················································· 4分又∵∠ADB为直角，∴∠A和∠DBE都是∠ABD的余角，································· 5分∴∠A＝∠DBE＝37°，··········································6分∵OA＝OD，∴∠ADC＝∠A＝37°．··········································7分23.解：设销售单价为x元，一个月内获得的利润为ｗ元，根据题意，得······1分ｗ=(x－40)(240－×20)·········································4分=（x－40）（－4x+480）=－4x2+640x－19200=－4（x－80）2+6400·········································5分所以抛物线顶点坐标为（80，6400）抛物线的对称轴为直线x=80， ∵a=－10＜0，∴当x=80时，ｗ的最大值为6400．································· 6分∴当销售单价为80元时，才能在一个月内获得最大利润，最大利润是6400元····························································7分24.解：如图，过点D作DM⊥EC于点M，DN⊥BC于点N， 设BC=h. ············2分在Rt△DMA中，∵AD=6，∠DAE=30°，∴DM=3，AM= ，···············································3分则CN=3，BN=h－3；·············································4分在Rt△BDN中，∵∠BDN=30°，∴DN= ；·····················································5分在Rt△ABC中，∵∠BAC=48°，∴AC=.··········································6分∵AM+AC=DN，·················································7分∴+=，解之得h≈13.故大树的高度为13米.···········································8分    25.解：(1)∵在Rt△BOA中，点E(4，n)在直角边AB上，∴OA=4，·····················································1分∴AB=OA×tan∠BOA=2.··········································2分(2)∵点D为OB的中点，点B（4，2），∴点D（2，1），又∵点D在的图象上，∴k=2，∴，························································3分又∵点E在图象上，∴4n=2，∴n=.·······················································4分(3)设点F（a，2），∴2a=2，∴CF=a=1 ，··················································5分连结FG，设OG=t，则OG=FG=t ，CG=2－t，·········································6分在Rt△CGF中，GF2=CF2＋CG2 ，···································7分∴t2=（2－t)2＋12 ，解得t =，∴OG=t=．····················································8分26.解：⑴∵当x=0时，y=－，∴C(0，－)，·················································1分∵当y=0时，，得，，∴A(－4，0), B(1，0)··········································2分⑵∵A(－4，0), C(0，－)，∴AO=4, CO=，在Rt△AOC中，∵tan∠OAC==，∴∠OAC=30°，···············································3分作OD⊥AC于D，∴OD= AOsin∠OAC=2.···········································4分⑶∵A(－4，0), C(0，－)，∴可解得直线AC的解析式为，·····································5分当⊙P与直线AC相切时，点P到直线AC的距离为2，若点P在直线AC的上方，由(2)可知，点P在过点O且平行于直线AC的直线上，此时，直线OP的表达式为：， ····································6分∴，解得或，·····················································7分若点P在直线AC的下方，可得点P在直线上，·············································8分∴，∴解得，∴点P的横坐标为或或－2.·········································9分     27.解：  (1) 取AB的中点O，连结OD，OC，···························1分∵Rt△ABD和Rt△ABC的斜边为AB，∴OD=，OC=，·················································2分∴OA=OB=OC=OD,∴A、B、C、D四个点在同一个圆上.·································3分(2)如图，连结DF，·············································4分∵点D、P关于AB对称，∴∠1=∠2，··················································5分∵AD⊥BC于点D，CF⊥AB于点F，∴∠2+∠3=90°，∠4+∠BCE=90°，BE⊥AC，点A、C、D、F四点共圆，∴点B、F、E、C四点共圆，∠3=∠4，·······························6分∴∠2=∠BCE，∠BFE+∠BCE=180°，∴∠2+∠BFE=180° ，··········································7分∴∠1+∠BFE=180°，∴点P、F、E三点在一条直线上.···································8分(3).························································9分