发布时间 : 星期六 文章上海市普陀区2019年初三二模数学试卷(含答案)更新完毕开始阅读060684dfba68a98271fe910ef12d2af90342a881
普陀区2018学年第二学期初三质量调研
数 学 试 卷
(时间:100分钟,满分:150分)
考生注意:
1.本试卷含三个大题,共25题.答题时,考生务必按答题要求在答题纸规定的位置上作答,在草稿纸、本试卷上答题一律无效.
2.除第一、二大题外,其余各题如无特别说明,都必须在答题纸的相应位置上写出证明或 计算的主要步骤.
一、选择题:(本大题共6题,每题4分,满分24分)
[下列各题的四个选项中,有且只有一个选项是正确的,选择正确项的代号并填涂在答题纸的相应位置上]
1.下列计算中,正确的是 ········································································································ (▲) (A)(a2)3?a5; (B)a2?a3?a6; (C)2a?3a?6a2; (D)2a?3a?5a2. 2.如图1,直线l1l2?1?30??2?50??3?20?80?90?100?3.14列函数中,如果x?0,yl1
1 的值随x的值增大而增大,那么这个函数是 ······································································· (▲) (A)y??2x; (B)y?2; 2 x图1
3 l2
(C)y??x?1; (D)y?x2?1.
5.如果一组数据3、4、5、6、x、8的众数是4,那么这组数据的中位数是 ················· (▲) (A)4; (B); (C)5; (D).
6.如图2,ABCD的对角线AC、BD交于点O,顺次联结ABCD各边中点得到的一个新的四边形,如果添加下列四个条件中的一个条件:①AC⊥BD;②C△ABO?C△CBO;③?DAO??CBO;④?DAO??BAO,可以使这个新的四边形成为矩形,那么这样的条件个数是 ··························································································································· (▲)
A D
(A)1个; (B)2个;
O (C)3个; (D)4个.
C B
图2 二、填空题:(本大题共12题,每题4分,满分48分)
7.分解因式:a2?2a= ▲ . 8.函数y?1的定义域是 ▲ . 3x?1?2x?1?0,9.不等式组?的解集是 ▲ .
x?3≤4x?10.月球离地球近地点的距离为363300千米,数据363300用科学记数法表示是 ▲ . 11.如果a?2、b??1,那么代数式2a?b的值等于 ▲ .
12.如果关于x的方程x2?3x?m?2?0有两个相等的实数根,那么m的值等于 ▲ . 13.抛物线y?ax2?2ax?5的对称轴是直线 ▲ .
14.张老师对本校参加体育兴趣小组的情况进行调查,图3-1和图3-2是收集数据后绘制的两幅不完整统计图.已知参加体育兴趣小组的学生共有80名,其中每名学生只参加一个兴趣小组.根据图中提供的信息,可知参加排球兴趣小组的人数占参加体育兴趣小组总人数的百分数是 ▲ . 人数
24
排球 足球 图3-1
A 2米
篮球45% 足球
篮球兴趣小组
图3-2
排球
B
图4
C 15.如图4,传送带AB和地面BC所成斜坡的坡度为1:3,如果它把物体从地面送到离地面2米高的地方,那么物体所经过的路程是 ▲ 米.(结果保留根号)
uuurruuurruuur16.如图5,AD、BE是△ABC的中线,交于点O,设OB?a,OD?b,那么向量AB用
向量a、b表示是 ▲ .
B
D 图5
A
A
E O
C
B
图6
E D
图7
C
17.如图6,一个大正方形被平均分成9个小正方形,其中有2个小正方形已经被涂上阴影,
在剩余的7个白色小正方形中任选一个涂上阴影,使图中涂上阴影的三个小正方形组成轴对称图形,这个事件的概率是 ▲ .
18.如图7,AD是△ABC的中线,点E在边AB上,且DE⊥AD,将△BDE绕着点D旋转,使得点B与点C重合,点E落在点F处,联结AF交BC于点G,如果
AE5?,那么BE2GF的值等于 ▲ . AB
三、解答题:(本大题共7题,满分78分) 19.(本题满分10分)
?1?计算:2sin60??2?27????(?1)2019.
?2?
20.(本题满分10分)
解方程:
21.(本题满分10分)
12?34x2??1. 2x?9x?3 如y图8/,已y (万元吨) 知点D、E分别在△ABC的边AB和AC上,
DEBC 6 A DE12A D E ?ADEABCBC?9cotB??AEDyxyBxABCDAD?BCEAD BC33D B(D) E B D ?ACE??BCDECABAB2?ED?BCxOy?C EAACB B 图10 212A C y??x?4m(m?0)xyABCABAB?2BCmx?bx?10ACC PB Cy??备用图33图8 A A C O C O 1 4x(吨) 0 20 28 ?AB?5cos?60 ?S△PAB?2S△OBCP?BACACB?90OACACOAO⊙O⊙OABDCx 5A O 1 EC2?ED?EAABCDCD⊙COA?xDBxDAB⊙CABEADAE?yyxxO⊙CABx二、填空题:(本大题共12题,每题4分,满分48分) 图12
图13
图9 7. a(a?2);
图11
8. x?1; 39. ?1≤x?12.
1; 210. 3.633?105; 13.x?1;
11.5; 14.25%; 17.
17; 415.210; 18.
rr16.a?2b;
三、解答题
5; 710. 63(本大题共7题,其中第19---22题每题10分,第23、24题每题12分,第25题14分,满分78分) 19.解:原式=2?3······································································· (6分) ?2?33?8?(?1)
2=2?3?33?8?1 ···················································································· (2分) =23?5. ···································································································· (2分)
20.解:去分母得,4x?2(x?3)?(x2?9). ···································································· (3分)
整理得,x2?2x?3?0. ····················································································· (3分) 解得 x?1,x??3. ······················································································· (2分) 经检验,x??3是增根,舍去. ········································································ (1分) 所以,原方程的解是x?1. ··············································································· (1分)