初中数学中考复习 广西玉林市玉州区2019届九年级数学第一次模拟试题
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这是一份初中数学中考复习 广西玉林市玉州区2019届九年级数学第一次模拟试题,共16页。试卷主要包含了 本试卷满分120分, 非选择是,用直径0, 使分式值为零的的值为, 下列计算中,正确的是等内容,欢迎下载使用。
广西玉林市玉州区2019届九年级数学第一次模拟试题注意事项:1. 本试卷满分120分。考试时间为120分钟。2. 选择题每小题选出答案后,用2B铅笔把答题卡上对应题目的选项标号涂黑。3. 非选择是,用直径0.5毫米黑色墨水签字笔在答题卡上各题的答题区区域内作答,答在式试卷上无效。考试结束,将本试卷和答题卡一并交回。一、选择题:本大题共12小题,每小题3分,共36分。在每小题给出的四个选项中,只有一项是符合题目要求的,把正确答案的标号填(涂)在答题卡内相应的位置上。1. 的倒数等于( )A. B. C. D. 2. 下列四个图形中,既是轴对称图形又是中心对称图形的是( ) A B C D3. 国家开发银行2018年有力有序落实“一带一路”2500亿元专项贷款,落实“十三五”规划,全面提升国际发展质量。其中2500亿用科学记数法表示为( )A. B. C. D. 4. 使分式值为零的的值为( )A. B. C. D. 5. 从一副扑克牌中随机抽出一张牌,得到方块或者的概率是( )A. B. C. D. 6. 如图所示,将绕点按顺时针旋转一定角度得到,点的对应点恰好落在边上,若,,则的长为( ) 第6题图A. B. C. D. 7. 一组2.3.4.3.3的众数、中位数、方差分别是( )A. B. C. D. 8. 下列计算中,正确的是( )A. B. C. D. 9. 如图,在中,,,,则下列三角函数表示正确的是( )第9题图A. B. C. D. 10. 中国“一带一路”战略沿线国家和地区带来很大的经济效益,沿线某地区居民2017年人均收入为美元,预计2019年人均收入将达到美元,设2017年到2019年该地区居民年人均收入平均增长率为,可列方程为( )A. B. C. D. 11. 如图,将沿弦MN折叠,圆弧恰好经过圆心,点劣弧上一点,则的度数为( )A. B. C. D. 12. 已知二次函数(h为常数),在自变量的值满足的情况下,与其对应的函数值的最大值为0,则的值为( )A. 和 B. 和 C. 和 D. 和二、填空题:本大题共6小题,每小题3分,共18分,把答案填在答题卡的横线上。13. 使二次根式在实数范围内有意义的的取值范围是 ;14. 分解因式: ;15. 如果,那么 ;16. 如果点,在抛物线上,那么的值为 ;17. 如图,母线长为的圆锥的侧面展开图是一个扇形,若圆锥的底面圆半径是,则展开图扇形的圆心角底数为 ;第17题图18. 如图,正方形的边长为10,点在上,,过M作,分别交、于、两点,若、分别为、的中点,则的长为 第18题图三、解答题:本大题共8小题,满分66分,解答过程写在答题卡上,解答写出文字说明,证明过程或演算步骤。19.(6分)计算:20.(6分)化简,并从中选择一个合适的数求代数式值。21.(6分)如图,三个顶点的坐标分别为.(1)请画出向左平移个单位长度后得到的;(2)请画出关于原点对称的;(3)请轴上求作一点,使的周长最小,请画出,并直接写出的坐标.22.(8分)某班为了解学生每周进行体育锻炼的时间情况,对全班名学生进行调查,按每周进行体育锻炼的时间(单位:小时),将学生分成五类:类,类,类,类,类.绘制成尚不完整的条形统计图如图. 根据以上信息,解答下列问题:(1)类学生有 人,补全条形统计图;(2)类学生人数占被调查总人数的 %;(3)从该班每周进行体育锻炼时间在的学生中任选人人,求这人每周进行体育锻炼时间都在中的概率.23.(8分)如图,已知是的直径,是切线,连接交于点,且为中点。(1)求证:;(2)若的直径长为8,①求弧的长;②求阴影部分的面积.24.(10分)蔬菜基地种植了娃娃菜和油菜两种蔬菜共亩,设种植娃娃菜亩,总收益为万元,有关数据见下表: 成本(单位:万元/亩)销售额(单位:万元/亩)娃娃菜2.43油菜22.5(1)求关于的函数关系式(收益 = 销售额 – 成本);(2)若计划投入的总成本不超过万元,要使获得的总收益最大,基地应种植娃娃菜和油菜各多少亩?(3)已知娃娃菜每亩地需要化肥kg,油菜每亩地需要化肥kg,根据(2)中的种植亩数,基地计划运送所需全部化肥,为了提高效率,实际每次运送化肥的总量是原计划的倍,结果运送完全部化肥的次数比原计划少次,求基地原计划每次运送多少化肥.25.(10分)如图1,在正方形中,,分别为,的中点,连接,,交点为. 若正方形的边长为.图1(1)求证:;(2)将沿对折,得到(如图2),延长交的延长线于点,求的长;图2(3)将绕点逆时针方向旋转,使边正好落在上,得到(如图3),若和相交于点,求四边形面积.图326.(12分)已知:如图,直线与轴负半轴交于点,与轴正半轴交于点,线段的长是方程的一个根,请解答下列问题:(1)求点的坐标;(2)双曲线与直线交于点,且,求的值;(3)在(2)的条件下,点在线段上,,直线轴,垂足为,点在直线上,在直线上的坐标平面内是否存在点,使以点、、、为顶点的四边形是矩形?若存在,请求出点的坐标;若不存在,请说明理由。2019年玉州区第一次中考模拟试题数 学 参考答案一、ACC ACA BBD BCA二、13. 14. 15. 16. 17. 18. 18题解答过程:过作于点,过作于点,于点∵四边形是正方形∴,∵∴四边形是矩形∴∴∵是正方形的对角线∴m]∴∵分别为的中点又∵∴∴∴在中, 三、19.解:原式 ····················································· 4分 ····················································· 5分 ····························································· 6分20.解:···························································· 2分······················································· 3分······································································· 4分∵,且,且∴,且,且∴在中,的值只能取∴原式······························································· 6分21.(1)如图所示,为所求作的三角形········································· 2分(2)如图所示,所求作的三角形 ············································ 4分(3)如图所示,为所求作的三角形,点坐标为························ 6分22. 解:(1),补全图形如下: ··················································· 2分(2) ········································································· 4分(3)∵每周进行体育锻炼时间在内的两人记为甲、乙,在内的人记为,从中任选两人有:甲乙、甲、甲、甲、乙、乙、乙、这种可能结果,其中人每周进行体育锻炼时间都在中的有这种结果∴这人每周进行体育锻炼时间都在中的概率为······························ 8分23.(1)证明:连接∵是的直径∴∴∵为中点∴垂直平分∴········································································ 3分(2)①连接]∵是的切线,是的直径∴∴∵∴∴∵的直径长为∴的半径长为∴弧的长··································································· 6分②∵∴∴························································ 8分24.解:(1)由题意得;························· 3分(2)由题意知,解得······································· 5分对于,∵,∴随的增大而增大,∴当时,所获总收益最大,此时.答:基地应种植娃娃菜亩,种植油菜亩;············································· 6分(3)设原计划每次运送化肥,实际每次运送,需要运送的化肥总量是,由题意可得························································· 8分解得.经检验,是原分式方程的解.答:基地原计划每次运送化肥··················································· 10分25.解:(1)证明:如图1,∵分别是正方形边的中点,∴,在和中,,∴,······················································· 2分,又∵,∴,∴,∴.······································································ 3分 图1(2)解:如图2,根据题意得,∵,∴,∴,∴,∵,····················································· 5分在中,设,∴,∴,∴.························································· 6分图2(3)解:∵正方形边长为,∵,∴,∵,∴,····································································· 8分∴∴ ∴∴,∴,∴四边形的面积是.························································ 10分26.解:(1)解方程得或.∵线段的长是方程的一个根,∵的长是正数∴,∴.将代入,得,∴,∴.············································································ 3分(2)在中,,∴.图1如图①,过点作轴于点,则,∴····································································· 5分∴ 即 解得,∴,∴.∵双曲线()经过点,∴······································································ 7分(3)存在①当为以点为顶点的矩形的一边时,过点作轴于点,作交直线于点,如图②所示,∴,图2∴··································································· 8分∴ ∴,∴,∴.∵,∴设直线的函数表达式为,把代入,得,解得,∴直线的函数表达式为当时,,∴,∴.(注:也可以用三角形相似求解∴如图3图3 ∵∴点的坐标为;(点的平移)················································· 9分当为以点为顶点的矩形的一边时,同理得出满足条件的另一点的坐标为;··········································································· 11分②当为以点为顶点的矩形的对角线时,点在直线的下方,不符合题意。∴满足条件的的坐标为或;······································ 12分
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