2013江苏
高考
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数学卷
2013江苏高考数学
,1、函数的最小正周期为 y,3sin(2x,)4
22、设(i为虚数单位),则复数z的模为 z,(2,i)
22xy3、双曲线的两条渐近线方程为 ,,1169
4、集合共有 个子集. ,,,1,0,1
5、右图是一个算法的
流程
快递问题件怎么处理流程河南自建厂房流程下载关于规范招聘需求审批流程制作流程表下载邮件下载流程设计
图,则输出的n的值是
6、抽样统计甲、乙两位射击运动员的5次训练成绩(单位:环),结果如下: 运动员 第1次 第2次 第3次 第4次 第5次
甲 87 91 90 89 93
乙 89 90 91 88 92 则成绩较为稳定(方差较小)的那位运动员成绩的方差为
7、现有某类病毒记作其中正整数可以任意选取,则m,n都取到奇XYm,n(m,7,n,9)mn
数的概率为
8、如图,在三棱柱中,D,E,F分别是AB,AC,的中点,设三棱锥AAABC,ABC1111
F,ADE的体积为、三棱柱的体积为,则= VVABC,ABCV:V1212111
C1
B1 A1
F C
E D A B
29、抛物线在x=1处的切线与两坐标轴围成三角形区域为D(包括三角形内部和边y,x
界)。若点P(x,y)是区域D内的任意一点则x+2y的取值范围是
21,ABCAD,ABBE,BCD,EAB,BC10.设分别是的边上的点,,,若23
,,,,,,(为实数),则的值为_______ DE,,AB,,AC121212
2x,0f(x)f(x),x11.已知是定义在R上的奇函数,当时,,则不等式的f(x),x,4x
statistical data is not difficult to see that external transport road transport accounts for the absolute dominance of passenger traffic over the proportion of over 97%, cargo accounted for above 81%, and rail and air transport development up until 2012, borne freight rate is low, there is enough potential in the future. At this stage, single railway lines, and dominated by freight, low level of passenger services, passenger transport is not sufficiently competitive. CAI BA, Yibin airport capacity insufficient air passenger
解集区间表示为_____________
22xy12.在平面直角坐标系中,椭圆C的
标准
excel标准偏差excel标准偏差函数exl标准差函数国标检验抽样标准表免费下载红头文件格式标准下载
方程,右焦点为,,,1(a,0,b,0)xOyF22ab
ll右准线为,短轴的一个端点为,设原点到直线的距离为,到的距离为,ddBFBF12若,则椭圆C的离心率_______ d,6d21
113.在平面直角坐标系中,设定点,P是函数y,图像上一动点,若xOyA(a,a)(x,0)x
点之间的最短距离为,则满足条件的实数的所有值为_____ 22aP,A
114.在正项等比数列a,,a,a,3中,,则满足的最大{a}a,a,...,a,aa...a567n12n12n2
正整数的值为_____ n
15.已知 a,(cos,,sin,),b,(cos,,sin,),0,,,,,,
(1)若求证: a,b|a,b|,2
(2)设,若,求的值 a,b,c,,,c,(0,1)
SABSBCS,ABCAB,BCAS,AB16.如图,在三棱锥中,平面?平面,,,过作A
GAF,SB,垂足为,点,分别棱中点 SA,SCFE
EFGABC证明(1)平面?平面
BCSA(2)?
S
G E
F A C
B statistical data is not difficult to see that external transport road transport accounts for the absolute dominance of passenger traffic over the proportion of over 97%, cargo accounted for above 81%, and rail and air transport development up until 2012, borne freight rate is low, there is enough potential in the future. At this stage, single railway lines, and dominated by freight, low level of passenger services, passenger transport is not sufficiently competitive. CAI BA, Yibin airport capacity insufficient air passenger
17.图,在平面直角坐标系中,点,直线,设圆C的半径为1,圆xOyA(0,3)l:y,2x,4
心在直线上 l
(1)若圆心C也在直线上,过点A作圆C的切线,求切线方程 y,x,1
(2)若圆C存在点M,使MA=2MO,求圆心C的横坐标的取值范围 a
l y
A
O x
18如图,游客从某旅游景区的景点A处下山至C处有两种路径。一种是从A沿直线不行到C,另一种是先从A沿索道乘缆车到B,然后从B沿直线步行到C
现有甲、乙两位游客从A处下山,甲沿AC匀速步行,速度为50m/min,在甲出发2min后,乙从A乘缆车到B,在B处停留1min后,再从B匀速步行到C,假设缆车匀速直线运动的
123cosA,,cosC,速度为130m/min,山路AC长为1260m,经测量, 135(1)求索道AB长
(2)问乙出发多久后,乙在缆车上与甲的距离最近,
(3)为使两位游客在C处互相等待的时间不超过3分钟,乙步行的速度应该控制在什么范围内,
statistical data is not difficult to see that external transport road transport accounts for the absolute dominance of passenger traffic over the proportion of over 97%, cargo accounted for above 81%, and rail and air transport development up until 2012, borne freight rate is low, there is enough potential in the future. At this stage, single railway lines, and dominated by freight, low level of passenger services, passenger transport is not sufficiently competitive. CAI BA, Yibin airport capacity insufficient air passenger
nSnd19.设是首项为,公差为的等差数列,为其前项和,记,b,{a}San(d,0)nnn2n,c
*,其中为实数 n,Nc
2*c,0(1)若,且成等比数列,证明: b,b,bS,nS(k,n,N)124nkk
c,0(2)若是等差数列,证明: {b}n
x20.设函数其中a为实数 f(x),lnx,ax,g(x),e,ax,
(1)若在上是单调减函数,且在有最小值,求a的取值范围 f(x)(1,,,)g(x)(1,,,)
(2)若在上是单调增函数,试求的零点个数,并证明你的结论 g(x)(,1,,,)f(x)
statistical data is not difficult to see that external transport road transport accounts for the absolute dominance of passenger traffic over the proportion of over 97%, cargo accounted for above 81%, and rail and air transport development up until 2012, borne freight rate is low, there is enough potential in the future. At this stage, single railway lines, and dominated by freight, low level of passenger services, passenger transport is not sufficiently competitive. CAI BA, Yibin airport capacity insufficient air passenger