新疆维吾尔自治区2009年初中毕业生学业考试(WORD版 有答案)
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数学试题卷考生须知:1.本试卷分为试题卷和答题卷两部分. 2.试题卷共4页,满分150分.考试时间120分钟.3.答题卷共4页,所有答案必须写在答题卷上,写在试题卷上的无效....................... 4.答题前,考生应先在答题卷密封区内认真填写准考证号、姓名、考场号、座位号、地(州、市、师)、县(市、区、团场)和学校. 5.答题时可以使用科学计算器.......... 一、精心选择(本大题共10小题,每小题5分,共50分.每小题所给四个选项中,只有一个是正确的.)1.下列运算正确的是( ) A .246a a a = B .257()x x =C .23y y y ÷=D .22330ab a b -=2.若x y ==,则xy 的值是( )A.B.C .m n +D .m n -3.如图,将三角尺的直角顶点放在直尺的一边上,130250∠=∠=°,°,则3∠的度数等于( ) A .50° B .30° C .20° D .15°4.如图,是由一些相同的小正方体搭成的几何体的三视图,搭成这个几何体的小正方体个数是( ) A .2个 B .3个 C .4个 D .6个5.一副扑克牌,去掉大小王,从中任抽一张,恰好抽到的牌是6的概率是( ) A .154B .113C .152D .146.下列各组图中,图形甲变成图形乙,既能用平移,又能用旋转的是( )新疆维吾尔自治区新疆生产建设兵团 2009年初中毕业生学业考试甲乙甲乙A .B .C .D.甲乙甲乙123 (第3题)主视图左视图主视图7.要反映乌鲁木齐市一天内气温的变化情况宜采用()A.条形统计图B.扇形统计图C.频数分布直方图D.折线统计图8.如图,小正方形的边长均为1,则下列图中的三角形(阴影部分)与ABC△相似的是()9.如图,直线(0)y kx b k=+<与x轴交于点(30),,关于x的不等式0kx b+>的解集是()A.3x<B.3x>C.0x>D.0x<10.如图,直角坐标系中,两条抛物线有相同的对称轴,下列关系不正确...的是()A.h m=B.k n=C.k n>D.00h k>>,二、合理填空(本大题共4小题,每小题5分,共20分)11.若梯形的下底长为x,上底长为下底长的13,高为y,面积为60,则y与x的函数关系是____________.(不考虑x的取值范围)12.某商品的进价为x元,售价为120元,则该商品的利润率可表示为__________.13.如图,在平面直角坐标系中,已知一圆弧过小正方形网格的格点A B C,,,已知A点.B.C.D.AB CC(第14题)(第9题)21()4y x h k=-+n+(第10题)14.如图,60ACB ∠=°,半径为1cm 的O ⊙切BC 于点C ,若将O ⊙在CB 上向右滚动,则当滚动到O ⊙与CA 也相切时,圆心O 移动的水平距离是__________cm . 三、准确解答(本大题共10小题,共80分) 15.(6分)解方程:2(3)4(3)0x x x -+-=.16.(6分)解不等式组:331213(1)8x x x x-⎧+>+⎪⎨⎪---⎩,≤并在数轴上把解集表示出来.17.(6分)下列是两种股票在2009年某周的交易日收盘价格(单位:元),分别计算它们18.(6分)如图,E F ,是四边形ABCD 的对角线AC 上两点, AF CE DF BE DF BE ==,,∥. 求证:(1)AFD CEB △≌△. (2)四边形ABCD 是平行四边形.ABD E F (第18题) C19.(8分)如图,已知菱形ABCD 的边长为1.5cm ,B C ,两点在扇形AEF 的 EF 上,求 BC的长度及扇形ABC 的面积.20.(10分)甲、乙两同学学习计算机打字,甲打一篇3000字的文章与乙打一篇2400字的文章所用的时间相同.已知甲每分钟比乙每分钟多打12个字,问甲、乙两人每分钟各打多少个字?李明同学是这样解答的:设甲同学打印一篇3 000字的文章需要x 分钟, 根据题意,得3000240012x x-= (1) 解得:50x =.经检验50x =是原方程的解. (2)答:甲同学每分钟打字50个,乙同学每分钟打字38个. (3) (1)请从(1)、(2)、(3)三个步骤说明李明同学的解答过程是否正确,若有不正确的步骤改正过来.(2)请你用直接设未知数列方程的方法解决这个问题. 21.(8分)2008年国际金融危机使我国的电子产品出口受到严重影响,在这个情况下有两个电子仪器厂仍然保持着良好的增长势头.(1) 下面的两幅统计图,反映了一厂、二厂各类人员数量及工业产值情况,根据统计图填空:①一厂、二厂的技术员占厂内总人数的百分比分别是________和__________;(结果精确到1%)(第19题) BC DAE F一厂 二厂年②一厂、二厂2008年的产值比2007年的产值分别增长了_______万元和_______万元.(2)下面是一厂、二厂在2008年的销售产品数量占当年产品总数量的百分率统计表,根据(3)仅从以上情况分析,你认为哪个厂生产经营得好?为什么?22.(8分)如图是用硬纸板做成的四个全等的直角三角形,两直角边长分别是a b,,斜边长为c和一个边长为c的正方形,请你将它们拼成一个能证明勾股定理的图形.(1)画出拼成的这个图形的示意图.(2)证明勾股定理.cbac bacbacbacc (第22题)23.(10分)(1)用配方法把二次函数243y x x =-+变成2()y x h k =-+的形成. (2)在直角坐标系中画出243y x x =-+的图象.(3)若1122()()A x y B x y ,,,是函数243y x x =-+图象上的两点,且121x x <<,请比较12y y ,的大小关系.(直接写结果)(4)把方程2432x x -+=的根在函数243y x x =-+的图象上表示出来.24.(12分)某公交公司的公共汽车和出租车每天从乌鲁木齐市出发往返于乌鲁木齐市和石河子市两地,出租车比公共汽车多往返一趟,如图表示出租车距乌鲁木齐市的路程y (单位:千米)与所用时间x (单位:小时)的函数图象.已知公共汽车比出租车晚1小时出发,到达石河子市后休息2小时,然后按原路原速返回,结果比出租车最后一次返回乌鲁木齐早1小时.(1)请在图中画出公共汽车距乌鲁木齐市的路程y (千米)与所用时间x (小时)的函数图象.(2)求两车在途中相遇的次数(直接写出答案)(3)求两车最后一次相遇时,距乌鲁木齐市的路程.(第24题)数学试卷参考答案及评分标准(满分150分)说明:本参考答案供阅卷教师评卷时使用.阅卷中,考生如有其它解法,只要正确、合理,均可得相应分值.一、精心选择(本大题共10小题,每小题5分,共50分)二、合理填空(本大题共4小题,每小题5分,共20分) 11.90y x=12.120120100%x x x x --⎛⎫⨯ ⎪⎝⎭或 13.(10)-, 14三、准确解答(本大题共10小题,共80分) 15.(6分)解法一:2(3)4(3)0x x x -+-=(3)(34)0x x x --+=(3)(53)0x x --= ········································································································· (3分) 30x -=或530x -=12335x x ==, ··············································································································· (6分) 解法二:22694120x x x x -++-=251890xx -+= ··········································································································· (2分) x = ························································································· (4分)181210±=12335x x ==, ··············································································································· (6分) (其它解法可参照给分) 16.(6分)解:解不等式(1)得1x < ········································································ (2分) 解不等式(2)得2x -≥ ······························································································· (3分)········································ (4分) 新疆维吾尔自治区新疆生产建设兵团 2009年初中毕业生学业考试0 1 x所以不等式组的解集为21x -<≤. ············································································ (6分) 17.(6分)解:1(11.6211.5111.9411.1711.01)11.455x =++++=甲 18.50x =乙 ····················································································································· (2分) 2222221(11.6211.45)(11.5111.45)(11.9411.45)(11.1711.45)(11.0111.45)5S ⎡⎤=-+-+-+-+-⎣⎦甲222221(0.170.060.490.280.44)5=++++ 10.54465=⨯ 0.10892= 0.11≈20S =乙 ···························································································································· (4分)甲的极差0.93= ············································································································· (5分) 乙的极差0= ··················································································································· (6分) 18.(6分)证明:(1)DF BE ∥, DFE BEF ∴∠=∠. 180AFD DFE ∠+∠= °,180CEB BEF ∠+∠=°,AFD CEB ∴∠=∠.又AF CE DF BE == ,,AFD CEB ∴△≌△(SAS). ····························· (3分) (2)由(1)知AFD CEB △≌△, DAC BCA AD BC ∴∠=∠=,, AD BC ∴∥.∴四边形ABCD 是平行四边形(一组对边平行且相等的四边形是平行四边形) ······ (6分) 19.(8分)解: 四边形ABCD 是菱形且边长为1.5,1.5AB BC ∴==.又B C 、两点在扇形AEF 的 EF上, 1.5AB BC AC ∴===,ABC ∴△是等边三角形.60BAC ∴∠=°. ············································· (2分) BC 的长60π1.5π1802== (cm ) ···················································································· (5分) 211π31.5π(cm )2228ABC S lR === 扇形·········································································· (8分) 20.(10分)解:(1)李明同学的解答过程中第③步不正确 ······································· (3分)应为:甲每分钟打字300030006050x ==(个) 乙每分钟打字601248-=(个)A B D EFB C DA EF答:甲每分钟打字为60个,乙每分钟打字为48个. ················································· (5分) 解:(2)设乙每分钟打字x 个,则甲每分钟打字(12)x +个, 根据题意得:3000240012x x=+ ························································································ (8分) 解得48x =.经检验48x =是原方程的解.甲每分钟打字12481260x +=+=(个)答:甲每分钟打字为60个,乙每分钟打字为48个. ··············································· (10分) 21.(8分)解:(1)①18%,8% ·················································································· (2分) ②1 500,1 000. ············································································································ (4分) (2)如图72AOB ∠=°.················································ (6分)(3)一厂生产经营得好,因为从题目给出的信息可以发现人少产值高. ················· (8分) (回答合理即可给分) 22.(8分)方法一解:(1)······················································· (3分) (2)证明: 大正方形的面积表示为2()a b + ···························································· (4分) 大正方形的面积也可表示为2142c ab +⨯····································································· (5分) 221()42a b c ab ∴+=+⨯,22222a b ab c ab ++=+,国外50%外地30%本地20%A BO a b c c c cb bb aaa222a b c ∴+=.即直角三角形两直角边的平方和等于斜边的平方. ····················································· (8分) 方法二解:(1)································ (3分) (2)证明: 大正方形的面积表示为:2c , ······························································ (4分) 又可以表示为:214()2ab b a ⨯+- ··············································································· (5分) 2214()2c ab b a ∴=⨯+-, 22222c ab b ab a =+-+, 222c a b ∴=+.即直角三角形两直角边的平方和等于斜边的平方. ····················································· (8分) (其它证法,可参照给分)23.(10分)解:(1)243y x x =-+2(44)34x x =-++-2(2)1x =--. ·············································································································· (3分)(2)对称轴2x =,顶点坐标(21)-,·········································· (6分) ab c(3)12y y > ·················································································································· (8分)(4)如图点C D ,的横坐标34x x ,. ······································································· (10分)24.(12分)解:(1)如图 ···························································································· (3分)······ (5分)(3)如图,设直线AB 的解析式为11y k x b =+,图象过(40)(6150)A B ,,,, 1111406150.k b k b +=⎧∴⎨+=⎩, 1175300.k b =⎧∴⎨=-⎩, 75300y x =-.① ········································································································ (7分) 设直线CD 的解析式为22y k x b =+,图象过(70)(5150)C D ,,,, 2222705150.k b k b +=⎧∴⎨+=⎩, 2275525.k b =-⎧∴⎨=⎩, ∴75525y x =-+.② ·································································································· (7分) 解由①、②组成的方程组得 5.5112.5.x y =⎧⎨=⎩, ∴最后一次相遇时距离乌鲁木齐市的距离为112.5千米. ········································· (12分)。