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    2015年初中毕业生毕业升学考试

    数学试卷

    考试时间:120分钟   试卷满分:150

    注意事项:

    1本试卷分第一部分(客观题)和第二部分(主观题)两部分。答卷前,考生务必将自己的姓名、准考证号填写在答题卡上。

    2回答第一部分时,选出每小题答案后,用铅笔把答题卡上对应题目的答案标号涂黑。如需改动,用橡皮擦干净后,再选涂其它答案标号写在本试卷上无效。

    3回答第二部分(主观题)时,将答案写在答题卡对应的区域内,写在本试卷上或答题卡指定的区域外无效。

    4.考试结束后,将本试卷和答题卡一并交回。

    分(客观题)

    一、选择题(下列各题的备选答案中,只有一个是正确的,每小题3分,共30分)

    1.下列计算正确的是

    A   B    C     D

    2如右图,是由若干个相同的小立方体搭成的几何体的俯视图和左视图,则小立方体的个数有可能

    A56         B57

    C456     D567

    3.函数中自变量的取值范围是

    A x3    B         Cx3  Dx3

    4ABCD中,对角线ACBD交于点ODAC=42ºCBD=23º,COD

    A61º       B63º          C65º           D67º   

    5云南鲁甸发生地震后,某社区开展献爱心活动,社区党员积极向灾区捐款,如图是该社区部分党员捐款情况的条形统计图,那么本次捐款钱数的众数和中位数分别是

    A100100  B100200  C200100  D200200

    6若关于的分式方程有增根的值是

    A      B        C          D

    7将弧长为2πcm圆心角为120º的扇形围成一个圆锥的侧面则这个圆锥的高及侧面积分别是

    A B C   D

    8如图,ABECDE是以点E为位似中心的位似图形,已知A34,点C22,点D31则点D的对应点B的坐标是

    A.(42          B.(41      C.(52         D.(51

                    

    9如图,在平面直角坐标系中A(31),以点O为直角顶点作等腰直角三角形AOB双曲线第一象限内的图象经过B,设直线AB的解析式为,当的取值范围是

    A  B   C   D

    10如图,点PAOB内任意一点,OP=5cm,点M和点N分别是射线OA和射线OB上的动点,PMN周长的最小值是5cm,则AOB的度数是

    A       B            C        D        

    分(主观题)

    二、填空题(每小题3分,共24分)

    11.分解因式:=         

    12过度包装既浪费资源又污染环境.据测算,如果全国每年减少十分之一的包装纸用量,那么能减少3 120 000吨二氧化碳的排放量.把数据3 120 000用科学记数法表示为         

    13不等式组的所有正整数解的和为         

    14圆内接正六边形的边心距为,则这个正六边形的面积为          cm2

    15.如图,正方形内的阴影部分是由四个直角边长都是13的直角三角形组成的,假设可以在正方形内部随意取点,那么这个点取在阴影部分的概率为         

    16某服装店购进单价为15元童装若干件,销售一段时间后发现:当销售价为25元时平均每天能售出8件,而当销售价每降低2元,平均每天能多售出4件.当每件的定价为          元时,该服装店平均每天的销售利润最大.

    17定义:只有一组对角是直角的四边形叫做损矩形,连接它的两个非直角顶点的线段叫做这个损矩形的直径,即损矩形外接圆的直径

    如图,ABC中,ABC=90º,以AC为一边向形外作菱形ACEF,点D是菱形ACEF对角线的交点,连接BD,若DBC=60ºACB=15ºBD=,则菱形ACEF的面积为         

    18如图,边长为n的正方形OABC的边OAOC分别在x轴和y轴的正半轴上,A1A2A3An-1OAn等分点,B1B2B3Bn-1CBn等分点,连接A1B1A2B2A3B3An-1Bn-1,分别交)于点C1C2C3Cn-1,当时,则n=         

    三、解答题(19小题10分,20小题10分,共20分)

    19先化简,再求值:.其中满足一元二次方程

    组别

    雾霾天气的主要成因

    百分比

    A

    工业污染

    45

    B

    汽车尾气排放

    C

    炉烟气排放

    15

    D

    其他(滥砍滥伐等)

    20雾霾天气严重影响市民的生活质量.在今年寒假期间,某校八年一班的综合实践小组同学对“雾霾天气的主要成因”随机调查了所在城市部分市民,并对调查结果进行了整理,绘制了如下不完整的统计

    图表,观察分析并回答下列问题.

    本次被调查的市民共有多少人?

    分别补全条形统计图和扇形统计图,                           

    并计算图2中区域B所对应的扇形圆

    心角的度数.

    若该市有100万人口,请估计持有

    AB两组主要成因的市民有多少人?

    四、解答题(21小题12分,22小题12分,共24分)

    21.某化妆品专卖店,为了吸引顾客,在母亲节当天举办了甲、乙两种品牌化妆品有奖酬宾活动,凡购物满88元,均可得到一次摇奖的机会.已知在摇奖机内装有2个红球和2个白球,除颜色外其它都相同,摇奖者必须从摇奖机中一次连续摇出两个球,根据球的颜色决定送礼金券的多少(如下表):

    甲种品牌

    化妆品

    两红

    一红一白

    两白

    礼金卷(元)

    6

    12

    6

    乙种品牌

    化妆品

    两红

    一红一白

    两白

    礼金卷(元)

    12

    6

    12

    1)请你用列表法(或画树状图法)求一次连续摇出一红一白两球的概率;

    2)如果一个顾客当天在本店购物满88元,若只考虑获得最多的礼品卷,请你帮助分析选择购买哪种品牌的化妆品?并说明理由.

    22如图,我南海某海域A处有一艘捕鱼船在作业时突遇特大风浪,船长马上向我国渔政搜救中心发出求救信号,此时一艘渔政船正巡航到捕鱼船正西方向的B处,该渔政船收到渔政求救中心指令后前去救援,但两船之间有大片暗礁,无法直线到达,于是决定马上调整方向,先向北偏东60 º方向以每小时30海里的速度航行半小时到达C处,同时捕鱼船低速航行到A点的正北1.5海里D处,渔政船航行到点C处时测得点D在南偏东53 º方向上.

    (1)CD两点的距离;

    (2)渔政船决定再次调整航向前去救援,若两船航速不变,并且在点E处相会合,求ECD的正弦值.    (参考数据:)

    五、解答题(23小题12分,24小题12分,共24分)

    23.如图,点PO 外一点,PAO于点AABO的直径,连接OP,过点BBCOPO于点C,连接ACOP于点D

    (1)求证:PCO的切线;

    (2)PD=cmAC=8cm求图中阴影部分的面积;

    (3)(2)的条件下,若点E的中点,连接CE,求CE的长

       

    24.某粮油超市平时每天都将一定数量的某些品种的粮食进行包装以便出售,已知每天包装大黄米的质量是包装江米质量的倍,且每天包装大黄米和江米的质量之和为45千克.

    (1)求平均每天包装大黄米和江米的质量各是多少千克?

    (2)为迎接今年620日的端午节,该超市决定在节日前20天增加每天包装大黄米和江米的质量,二者的包装质量与天数的变化情况如图所示,节日后又恢复到原来每天的包装质量.分别求出在这20天内每天包装大黄米和江米的质量随天数变化的函数关系式,并写出自变量的取值范围.

    (3)假设该超市每天都会将当天包装后的大黄米和江米全部出售,已知大黄米成本价为每千克7.9元,江米成本价为每千克9.5元,二者包装费用平均每千克均为0.5元,大黄米售价为每千克10元,江米售价为每千克12元,那么在这20天中有哪几天销售大黄米和江米的利润之和大于120?     [总利润=售价额-成本-包装费用]


    六、解答题(本题满分14分)

    25.【问题探究】

    1)如图1,锐角ABC中,分别以ABAC为边向外作等腰ABE和等腰ACD,使AE=ABAD=ACBAE=∠CAD,连接BDCE,试猜想BDCE的大小关系,并说明理由.

    【深入探究】

    2)如图2,四边形ABCD中,AB=7cmBC=3cm,∠ABC=ACD=ADC=45º,求BD的长.

    3)如图3,在(2)的条件下,当ACD在线段AC的左侧时,求BD的长.


    七、解答题(本题满分14分)

    26如图1,一条抛物线与轴交于AB两点(点A在点B的左侧),与轴交于点C,且当x=1x=3时,的值相等.直线与抛物线有两个交点,其中一个交点的横坐标是6,另一个交点是这条抛物线的顶点M

    (1)求这条抛物线的表达式.

    (2)动点P从原点O出发,在线段OB上以每秒1个单位长度的速度向点B运动,同时动点Q从点B出发,在线段BC上以每秒2个单位长度的速度向点C运动,当一个点到达终点时,另一个点立即停止运动,设运动时间为秒.

    ①若使BPQ为直角三角形,请求出所有符合条件的值;

    为何值时,四边形ACQ P的面积有最小值,最小值是多少?

    (3)如图2,当动点P运动到OB的中点时,过点PPD轴,交抛物线于点D,连接ODOMMDODM,将OPD沿轴向左平移个单位长度(),将平移后的三角形与ODM重叠部分的面积记为,求的函数关系式.



    2015年初中毕业生毕业升学考试

    数学试卷参考答案及评分标准

    说明:

    1此答案仅供参考,阅卷之前请做答案。

    2.如果考生的解法与本解法不同,可参照本评分标准制定相应评分细则。

    3.为阅卷方便,本解答中的推算步骤写得较为详细,但允许考生在解答过程中,合理省略非关键性的推算步骤。

    4.解答右端所注分数,表示考生正确做到这一步应得的累加分数。

    一、选择题(每小题3分,共30分)

    1C    2D   3D    4C   5B   6A    7B    8C    9D    10B

    二、填空题(每小题3分,共24分)

    11     12     136             14

    15                  1622            17        1875

    三、解答题19小题10分,20小题10分,共20分)

    19解:

    =······························································································· 2

    =··························································································· 3

    ==··············································································································· 5

    化简方程  得: 

    ··················································································································· 6

    解得·········································································································· 8

    因为当时原式无意义,所以舍去; ··············································································· 9

    时,原式=··························································································· 10

    20.解:(1)()

    本次被调查的市民共有200 ····················································································· 2(2)补全统计图如图所示:········································································································································ 6


    由题意可得

    2中区域B所对应的扇形圆心角为

    ················································································································· 8

    (3)题意得

    ()

    估计持有AB两种主要成因的市民有75万人. ·························································· 10

    四、解答题(21小题12分,22小题12分,共24分)

    21.(1)解法一

    用列表列出所有可能出现的结果

    第二个

    第一个

    1

    2

    1

    2

    1

    (红1,红2

    (红1,白1

    (红1,白2

    2

    (红2,红1

    (红2,白1

    (红2,白2

    1

    (白1,红1

    (白1,红2

    (白1,白2

    2

    (白2,红1

    (白2,红2

    (白2,白1

    ··········································································································································· 3

    由表格可知所有可能出现的结果共有12且每种结果出现的可能性相同其中恰好连续摇出

    一红一白的结果有8所以P(一红一白)=······························································ 5

    答:一次连续摇出一红一白两球的概率为········································································ 6

    解法二:

    用树状图列出所有可能的结果

    ········································································································ 3

    由树状图可知所有可能出现的结果共有12且每种结果出现的可能性相同其中恰好连续摇出

    一红一白的结果有8所以P(一红一白)= ··························································· 5

    答:一次连续摇出一红一白两球的概率为········································································ 6

    2)若顾客在本店购物满88元,我认为该顾客应选择购买甲品牌的化妆品 ······················ 7

    理由如下:

    (1)P(两红)= P(两白)= ······································································ 9

    若购买甲品牌化妆品则获得礼品卷为+12×+6×=10() ···································· 10

    若购买乙品牌化妆品则获得礼品卷为12×+6×+12×=8()····································· 11

    因为108所以顾客应选择购买甲品牌的化妆品 ···························································· 12

    22.解:(1)如图,过点CCGAB于点GDFCG于点F················································ 1

    则在RtCBG中,由题意知CBG=30°

    CG=BC==7.5 ···························································································· 2

    ∵∠DAG=90°

    四边形ADFG是矩形,

    GF= AD=1.5

    CF= CGGF=7.51.5=6·································· 4

    RtCDF中,CFD=90º

    ∵∠DCF =53°

    cosDCF=

    (海里) ·························································································· 7

    答:CD两点距离为10海里 ······························································································· 8

    2)如图,设渔政船调整方向后t小时能与捕渔船相会合,

    由题意知CE=30tDE=1.5×2×t=3t EDC=53° ······························································· 9

    过点EEHCD于点H EHD=CHE=90º

    sinEDH=

    EH=EDsin53°=····························································································· 11

    RtEHC中,sinECD=

    答:sinECD=············································································································· 12

    五、解答题(23小题12分,24小题12分,共24分)

    23证明如图,连接OC

    PAOA

    ∴∠PAO=90º ····················································································································· 1

    OPBC

    ∴∠AOP=∠OBCCOP=∠OCB

    OC=OB

    ∴∠OBC=∠OCB

    ∴∠AOP=∠COP ··············································································································· 3

      OA=OCOP=OP

      ∴△PAO≌△PCO  (SAS)

    ∴∠PAO=∠PCO=90 º

    OCO的半径,

      ∴PCO的切线. ·············································································································· 5解法一:

    由(1)得PAPC都为圆的切线,

    PA=PCOP平分APCADO=∠PAO=90 º

    ∴∠PAD+DAO=∠DAO+AOD

    ∴∠PAD =∠AOD

    ∴△ADO∽△PDA ············································································································· 6

    AC=8 PD=

    AD=AC=4OD=3AO=5····························································································· 7

    由题意知OD为△ABC的中位线,

    BC=2OD=6AB=10 ······································································································· 8

    S=SOSACB=

    答:阴影部分的面积为··················································································· 9

    解法二:

    ABO的直径,OPBC

    ∴∠PDC=∠ACB=90º

    ∵∠PCO=90 º

    ∴∠PCD+∠ACO=∠ACO+∠OCB=90 º

    PCD=∠OCB

    ∵∠OBC =∠OCB

    ∴∠PCD=∠OBC

    ∴△PDC∽△ACB ······································ 6

    AC=8 PD=

    AD=DC=4PC=··················································································· 7

    CB=6AB=10 ················································································································ 8

    S=SO-SACB=

    答:阴影部分的面积为··················································································· 9

    3)如图,连接AEBE过点BBMCE于点M························································ 10

    ∴∠CMB=∠EMB=AEB=90º

    E的中点,

    ∴∠ECB=∠CBM=ABE=45ºCM=MB =BE=ABcos45º=······························· 11

    EM=

    CE=CM+EM=

    答:CE的长为cm ····································································································· 12

    24.解:(1)解法一:

    设平均每天包装大黄米和江米的质量分别为a千克和b千克, ··············································· 1

    ,解得····························································································· 3

    答:平均每天包装大黄米和江米的质量分别为25千克20千克 ········································ 4

    解法二:

    设平均每天包装大黄米的质量为a千克,那么包装江米的质量(45a)千克, ····················· 1

        解得 a=25,所以4525=20 ···························································· 3

    答:平均每天包装大黄米和江米的质量分别为25千克20千克 ········································ 4

    2)观察图象,可设平均每天包装大黄米的质量与天数的关系式为平均每天包装江米的质量与天数的关系式为

    时,

    由题意的图象过点

    则可列方程组为,解得

    ······················································································································· 5

    由题意的图象过点

    则可列方程组为,解得,

    ···················································································································· 6

    时,

    由题意的图象过点

    则可列方程组为,解得

    ··················································································································· 7

    由题意的图象过点

    则可列方程组为,解得

    ················································································································ 8

      ···························· 9  (3)设第天销售的总利润为元,

    时,

    由题意

    x的取值范围为

    由题意知································································································ 10

    时,

    由题意

    x的取值范围为

    由题意知················································································································· 11

    答:由可知在第111213141516天中销售大黄米和江米的总利润大于120元.     12

    六、解答题(本题满分14分)

    251)答:BD =CE ··········································································································· 1

    理由:∵∠BAE=∠CAD

    ∴∠BAE+∠BAC=∠CAD+∠BAC,即EAC=∠BAD··························································· 2

    AE=ABAC=AD

    ∴△EAC≌△BAD  (SAS)

    BD=CE ··························································································································· 4

    2)解:如图1,在ABC的外部,以点A为直角顶点作等腰直角三角形BAE,使BAE=90ºAE=AB,连接EAEBEC ····································································································································· 5

    ∴∠BAE=

    ∴∠BAE+∠BAC=∠CAD+∠BAC

    EAC=∠BAD

    ∴△EAC≌△BAD  (SAS) ························· 7

    BD=CE

    AE=AB=7

    AEC=∠AEB=45º

    ∵∠ABC=45º

    ∴∠ABC+∠ABE=45º+45º=90º ···························································································· 8

    EC==

    答:BD长是cm ········································································································ 9

    (3)如图2,在线段AC的右侧过点AAEABA,交BC的延长线于点E ···················· 10

    ∴∠BAE=90º

    ∵∠ABC=45º

    ∴∠E=∠ABC=45º

    AE=AB=7····················································································· 11

    ∵∠ACD=∠ADC=45 º

    ∴∠BAE= ∠DAC=90º

    ∴∠BAEBAC=∠DACBAC

    EAC=∠BAD

    ∴△EAC≌△BAD  (SAS)

    BD=CE ····································· 13

    BC=3

    BD=CE=(cm)

    答:BD长是()cm ································································································· 14

    七、解答题(本题满分14分)

    26.解:(1) 时,的值相等,抛物线的对称轴为直线,把分别代入中,得顶点,另一个交点坐标为(66) ····································································· 2

    则可设抛物线的表达式为,将(66)代入其中,解得

    抛物线的表达式为,即··············································· 3

     (2)如图1,当时, 解得 由题意知,A(20)B(40)

    所以OA=2OB=4;当时,,所以点C(0,-3)OC=3,由勾股定理知BC=5

    OP=1×t=tBQ=···································································································· 4

    ①∵∠PBQ是锐角,

    ∴有PQB=90ºBPQ=90º两种情况:

    PQB=90º时, 可得PQB∽△COB

    ······························································· 5

    BPQ=90º时, 可得BPQ∽△BOC

     ∴

    ····························································· 6

    由题意知

    时,以BPQ为顶点的三角形是直角三角形. ······································ 7

    ②如图1,过点QQGABG ∴△BGQ∽△BOC

    ····································································································· 8

    S四边形ACQP=SABC- SBPQ==

    ==

    0 ∴四边形ACQP的面积有最小值, 又∵满足

    时,四边形ACQP的面积最小,最小值是 ····················································· 10

    (3)如图2,由OB=4OP=2 代入中,得,所以D23),直线CDx轴,设直线OD的解析式为,则,所以,因为P1O1D1是由POD 沿x向左平移m个单位得到的,

    所以P12m0),D1(2m3)E2m).··········································· 11

    设直线OM的解析式为,则,所以

    时,作FH轴于点H,由题意O1(m0),又O1D1OD,所以直线O1D1的解析式为

    联立方程组,解得

    所以,所以FH=

    S四边形OFD1E=SOO1D1D-SOO1F-SDD1E

    =

    ==········································································· 13

    如图3,当时,设D1P1OM于点F,直线OM的解析式为,所以,所以

    ∴SOEF===

    综上所述,

    ························································································································ 14

     
     



      
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