江阴市华士实验中学2016九年级下学期期中数学试卷及答案

初三年级数学学科期中考试试卷 命题人：丁建峰 审核人：陈艳 一、选择题（本大题共 10 小题，每小题 3 分，共 30分。在每小题所给出的四个选项中，只有一 项是正确的，） 1.﹣3的绝对值是 （ ） A．﹣3 B．3 C．－ 1 3 D．1 3 2.二次根式 x−1中字母 x的取值范围是 （ ） A．x＜1 B．x≤1 C．x＞1 D．x≥1 3.未来三年，国家将投入 8450亿元用于缓解群众“看病难、看病贵”的问题．将 8450亿元用科学 记数法表示为 （ ） A．0.845×104亿元 B．8.45×103亿元 C．8.45×104亿元 D．84.5×102亿元 4.方程 2x﹣1=3的解是 （ ） A．x=2 B．x=0.5 C．x=1 D．x= −1 5.在同一平面直角坐标系中，函数 y=mx+m与 y=m x （m≠0）的图象可能是 （ ） A． B． C． D． 6.下列命题： ①平行四边形的对边相等； ②正方形既是轴对称图形，又是中心对称图形； ③对角线相等的四边形是矩形； ④一条对角线平分一组对角的平行四边形是菱形． 其中真命题的个数是 （ ） A．1 B．2 C．3 D．4 7.如图，已知△ABC的三个顶点均在格点上，则 cosA的值为 （ ） A． 1 3 3 B． 1 5 5 C．2 5 5 D． 2 3 3 8.如图，一个多边形纸片按图示的剪法剪去一个内角后，得到一个内角和为 2340°的新多边形， 则原多边形的边数为 （ ） A．13 B．14 C．15 D．16 第 7题 第 8题 第 9题 9.过正方体中有公共顶点的三条棱的中点切出一个平面，形成如图几何体，其正确展开图为（ ） A． B． C． D． 10.已知一次函数y=2x−4的图像与 x 轴、y轴分别相交于点 A、B，点 P在该函数图像上， P到 x轴、y轴的距离分别为 d1、d2，若 d1+d2=m，当 m为何值时，符合条件点 P有且只有两个（ ） (A)m＞2 (B) 2＜m＜4 (C) m≥4 (D) 0＜m＜4 二、填空题（本大题共 8 小题，每小题 2 分，共 16 分。） 11.分解因式：x2y﹣y= ． 12.方程 4x−12 x−2 =3的解是 x= ． 13.将一次函数 y=3x+1的图象沿 y轴向上平移 2个单位后，得到的图象对应的函数与 x轴的交点 坐标为 ． 14. 如图，菱形中，对角线 AC、BD交于点 O，E为 AD边中点，菱形 ABCD的周长为 28，则 OE的长等于 ． 第 14题 第 15题 第 16题 第 17题 15.如图，在平面直角坐标系中，直线 y =－x＋2与反比例函数 y=1 x 的图象有唯一公共点. 若直线 y=−x＋b与反比例函数 y=1 x 的图象有 2个公共点，求 b的取值范围是 ； 16.如图，在矩形 ABCD中，AB=4，AD=2，以点 A为圆心，AB长为半径画圆 弧交边 DC于点 E，则弧 BE的长度为 ． 21*04*4 17.设△ABC的面积为 9，如图将边 BC、AC分别 3等份，BE1、AD1相交于点 O， 则△AOB的面积为 ． 18.如右图，四边形 ABCD是以 AC所在直线为对称轴的轴对称图形，∠B=90°， ∠BAD=40°，AC=3，点 E，F分别为线段 AB、AD上的动点（不含端点），则 EF+CF长度的最小值为 ． 三、解答题（本大题共 10 小题，共 84 分。解答时应写出文字说明、证明过程 或演算步骤） 19.计算：⑴计算： ﹣|﹣3|﹣（﹣π）0+2015； ⑵     2x 5 x 1 x 2    20.⑴解方程： x2﹣4x﹣5=0 ⑵解不等式组： 21. 如图，在△ABC中，CD是 AB边上的中线，E是 CD的中点，过点 C作 AB的平行线交 AE 的延长线于点 F，连接 BF． (1) 求证：CF＝AD； (2) 若 CA＝CB，∠ACB＝90°，试判断四边形 CDBF的形状，并说明理由． 22. 如图，AB为⊙O的切线，切点为 B，连接 AO，AO与⊙O交于点 C，BD为⊙O的直径，连 接 CD．若点 C为 AO的中点，⑴求∠A的度数；⑵若⊙O的半径为 2，求图中阴影部分的面积. [来源:Z.Com] 23. 初中生在数学运算中使用计算器的现象越来越普遍，某校一兴趣小组随机抽查了本校若干名 学生使用计算器的情况．以下是根据抽查结果绘制出的不完整的条形统计图和扇形统计图： 请根据上述统计图提供的信息，完成下列问 题： （1）这次抽查的样本容量是 ； （2）请补全上述条形统计图和扇形统计图； （3）若从这次接受调查的学生中，随机抽查 一名学生恰好是“不常用”计算器的百分比是 多少？ A B D E C F 24. 有三张正面分别写有数字 0，1，2的卡片，它们背面完全相同，现将这三张卡片背面朝上洗 匀后随机抽取一张，以其正面数字作为 a的值，然后将其放回，再从三张卡片种随机抽取一张， 以其正面的数字作为 b的值， ⑴求点（a，b）在第一象限的概率；（请画“树状图”或者“列表”等方式给出分析过程） ⑵在点（a，b）所有可能中，任取两个点，它们之间的距离为 5的概率是 ； 25.如图，在平面直角坐标系中，点 A（10，0），以 OA 为直径在第一象限内作半圆，B 为半圆上一 点，连接 AB 并延长至 C，使 BC=AB，过 C 作 CD⊥ x轴于点 D，交线段 OB 于点 E，已知 CD=8，抛 物线经过点 O、E、A 三点。 （1）∠OBA= 。 （2）求抛物线的函数表达式。 （3）若 P 为抛物线上位于 AE 部分上的一个动点，以 P、O、A、E为顶点的四边形的面积记为 S， 求点 P 在什么位置时？ 面积 S 的最大值是多少？ 26.某商店销售 10台 A型和 20台 B型电脑的利润为 4000元，销售 20台 A型和 10台 B型电脑 的利润为 3500元． ⑴求每台 A型电脑和 B型电脑的销售利润； ⑵该商店计划一次购进两种型号的电脑共 100台，其中 B型电脑的进货量不超过 A型电脑的 2 倍。设购进 A型电脑 x台，这 100台电脑的销售总利润为 y元. ①求 y与 x的关系式； ②该商店购进 A型、B型各多少台，才能使销售利润最大？ ⑶实际进货时，厂家对 A型电脑出厂价下调 m（0＜m＜100）元，且限定商店最多购进 A 型电脑 70 台.若商店保持两种电脑的售价不变，请你根据以上信息及（2）中的条件，设计出使这 100 台电脑销售总利润最大的进货方案. 27. 已知点 A（3，4），点 B为直线 x=−1上的动点，设 B（－1，y）， （1）如图①，若△ABO是等腰三角形且 AO=AB时，求点 B的坐标； （2）如图②，若点 C（x，0）且－1＜x＜3，BC⊥AC垂足为点 C； ①当 x=0 时，求 tan∠BAC 的值； ②若 AB 与 y轴正半轴的所夹锐角为α，当点 C 在什么位置时？tanα的值最大? 图① 图② [来源:Z.Com] 28.如图，等边△ABC边长为 6，点 P、Q是 AC、BC边上的点，P从 C向 A点以每秒 1个单位 运动，同时 Q从 B向 C以每秒 2个单位运动，若运动时间为 t秒（0＜t＜3） x=-1 x=-1 ⑴如图①，当 t=2时，求证 AQ=BP； ⑵如图②，当 t为何值时，△CPQ的面积为 3； ⑶如图③，将△CPQ沿直线 PQ翻折至△C′PQ， ①点 C′ 落在△ABC内部（不含△ABC的边上），确定 t的取值范围 ； ②在①的条件下，若 D、E为边 AB边上的三等分点，在整个运动过程中，若直线 CC′与 AB的交 点在线段 DE上，总共有多少秒？ 图① 图② 图③ 2016.4初三数学期中考试答案： 选择题：1~5 BDBAA 6~10 CCBBA 填空题：11、y(x-1)(x+1) 12、6 13、（-1,0） 14、3.5 15、 b＞2或 b＜-2 16、 2π/3 17、 1.8 18、3 3/2 计算题： 19、⑴2016 ·················································································································· （4分） ⑵2x2−1 ·························································································································· （4分） 20、⑴x1=5，x2=-1 ··············································································································（4分） ⑵-5＜x＜-2······················································································································ （4分） 21、⑴∵AB∥CF ∴∠EAD=∠EFC, ∠ADE=∠FCE,·························································································· （1分） ∵E是 CD的中点 ∴DE=CE···························································································································（2分） ∴△ADE≌FCE ∴AD=CF···························································································································（3分） ∵CD是 AB边上的中线 ∴AD=BD ∴BD=CF···························································································································（4分） (2)由（1）知 BD=CF 又∵BD∥CF ∴四边形 CDBF是平行四边形································································································（6分） ∵CA=CB,AD=BD ∴∠CDB=90°,CD=BD=AD···································································································· （7分） ∴四边形 CDBF 是正方形.·····································································································（8分） 22、⑴连接 BC ∵AB为⊙O的切线，切点为 B ∴∠OBA=90°·····················································································································（1分） ∵点 C为 AO的中点 ∴AC=OC=BC·····················································································································（2分） ∵OB=CO ∴OB=OC=BC即△OBC是等边三角形·····················································································（3分） ∴∠BOC=60° ∴∠A=30°························································································································· （4分） ⑵由⑴可知∠BOC=60°，则∠DOC=120°···················································································· （5分） S 扇形= 4π 3 ·····························································································································（6分） S△ODC = 3···························································································································（7分） S 阴影= 4π 3 − 3···················································································································· （8分） 23、⑴160 ························································································································· （2分） ⑵略 图中一个空 1分············································································································（5分） ⑶25%······························································································································（7分） 24、⑴ 4 9 图 3分+共 9种等可能情况 1分+结论 1分·············································································（5分） ⑵ 2 9 ·································································································································· （7分） 25、（1）90.························································································································ （2分） （2）如答图 1，连接 OC， [来源:学。科。网Z。X。X。K] ∵由（1）知 OB⊥AC，又 AB=BC， ∴OB是的垂直平分线. ∴OC=OA=10. （3分） 在 Rt△OCD中，OC=10，CD=8，∴OD=6. ∴C(6，8)，B(8，4). （4分） ∴OB所在直线的函数关系为 y=1 2 x. 又 E点的横坐标为 6，∴E点纵坐标为 3，即 E(6，3)． ∵抛物线过 O(0，0)，E(6，3) ，A(10，0)， （5分） ∴设此抛物线的函数关系式为 y=ax（x-10）， 把 E点坐标代入得 3=a·6（6-10），解得 a=− 1 8 .[来源:学*科*网 Z*X*X*K] ∴此抛物线的:函数关系式为 y=− 1 8 x（x-10），即 y= − 1 8 x2+5 4 x． （6分） （3）∵E(6，3) ，A(10，0) ∴AE y= − 3 4 x+9 2 ······································································（7分） PQ∥y轴 设 P(a，− 1 8 a2+5 4 a) Q(a，－ 3 4 a＋15 2 ) PQ=− 1 8 a2+5 4 a+3 4 a-15 2 =− 1 8 a2+2a -15 2 =− 1 8 (a2-16a+64)+8−15 2 =− 1 8 (a-8) 2+1 2 当 a=8时，PQmax=1 2 ············································································································（8分） S=S△OAE+S△AEP=15+ 1 2 PQ·(a-6)+1 2 PQ(10-a)=15+2PQ==− 1 4 (a-8) 2+16····················································（9分） S＝16，点 P（8，2）··········································································································（10分） 26、解：⑴每台 A型电脑销售利润为 100元，每台 B型电脑的销售利润为 150元····························（2分） ⑵①y=-50x+15000················································································································（4分） ②商店购进 34台 A型电脑和 66台 B型电脑的销售利润最大．···················································· （6分） ⑶据题意得，y=（100+m）x+150（100-x），即 y=（m-50）x+15000，············································（7分） ①当 0＜m＜50时，y随 x的增大而减小， ∴当 x=34时，y取最大值，··································································································（8分） 即商店购进 34台 A型电脑和 66台 B型电脑的销售利润最大． ②m=50时，m-50=0，y=15000，····························································································（9分） 即商店购进 A型电脑数量满足 331 3 ≤x≤70的整数时，均获得最大利润； ③当 50＜m＜100时，m-50＞0，y随 x的增大而增大， ∴当 x=70时，y取得最大值．······························································································（10分） 即商店购进 70台 A型电脑和 30台 B型电脑的销售利润最大． 27、解：⑴如图，在 Rt△ABE 中 （4-y） 2 +4 2 =5 2 ； y=1 或 7 B（-1，1） 或者 B（-1，7） ·······························································································（2分） ⑵① 1 4 易证△AOF≌△OBG ·········································································································（4分） BO:AO=OG:AF=1：4 ········································································································（5分） tan∠BAC(或者 tan∠BAO)= 1 4 ······························································································（6分） ②由平行可知：∠ABH=α 在 Rt△ABE 中 tanα= 4 BH ···························································（7分 ） ∵ tanα随 BH的增大而减小 ∴当 BH最小时 tanα有最大值；即 BG最大时，tanα有最大值。·········（8分） 易证△ACF≌△CBG 得 BG/CF=CG/AF y/x-3=x+1/4 y=-1 4 （x+1）（3-x） y=-1 4 （x-1） 2+1··································································································································· （9分） 当 x=1时，ymax=1 当 C(1，0)时，tanα有最大值 4 3 ································································· （10分） 28、⑴略 t=2 时 CP=CQ 1 分 全等 1 分 ···············································································（2分） ⑵ 3t（3-t）·1 2 = 3 ····································································································· （3分） t1=1 t2=2············································································································· （4分） ⑶①1.5＜x＜2.4··············································································································· （6分） ②如图过点 C，作 FG∥AB ∵FG∥AB ∴FC′:AD=CC′:CD=C′G:BD ∵D 是 AB 上的三等分点， ∴BD=2AD ∴C ′G=2FC′ 即 C′是 FG 上的三等分点·······················································································（7分） 易证△CFG 是等边三角形 易证：△C′FP∽△QGC′ CP:CQ= C′P:C′Q=△C′FP 的周长：△QGC′的周长=t：（6-2t）；··························（8分） ∵C′是 FG 上的三等分点 ∴ △C′FP 的周长 △QGC′的周长 = t 6-2t = 4 5 t= 24 13 ···························································································（9分） 同理 t 6-2t = 5 4 t= 15 7 时间为 15 7 - − 24 13 = 27 91 ··············································································（10分） 不用注册，免费下载！