山西省太原市2014年高三年级模拟试题
山西太原市2014年高三年级模拟考试(一)英语试卷(带解析)
山西太原市2014年高三年级模拟考试(一)英语试卷(带解析)Everybody knows how to learn. Learning is a natural thing. It begins the we are born. Our first teachers are our f amilies. At home we learn to talk and to and feed ourselves. We learn these and other skills by .Then we go to school. A teacher tells us to learn and how to learn. Many teachers teach us, and we pass many tests and exams. Then people say we are .Are we really educated? Let's the real meaning of learning. Knowing facts doesn't being able to solve problems. Solving problems requires creativity, not just a good . Some people who don’t know many can also be good at solving problems.Henr Ford is a good . He left school at the age of 15. Later, when his company couldn’ t build cars enough, he solved the problem. He of the assembly line. Today the answer seems . Yet, just think of the many university graduates who solve any problems.What does a good teacher do? Does he students factct remember? Well, yes, we must sometimes remember facts. But a good teacher how to find answers. He brings us to the stream of knowledge so we can think for ourselves. When we are , we know where to go.True learning combines intake with output. We take information our brains. Then we use it. Think of a ; it stores a lot of information, but it can’t think. It only obeys commands. A person who only remembers facts hasn’t really learned. Learning takes only when a person can use what he knows. 1.A. month B. minute C. time D. day2.A .wear B. put on C. have on D. dress3.A. asking B. listening C. following D. drilling 4.A. what B. when C. that D. who5.A. controlled B. educated C. suffered D. passed6.A. pick up B. turn away C. set out D.think about7.A. mean B. say C. suggest D. show 8.A. word B. thing C. memory D. condition9.A. people B. facts C. techniques D. ways 10.A. learner B. teacher C. example D. driver11.A. new B. fast C. beautiful D. cheap12.A. complained B. heard C. talked D. thought 13.A. ordinary B. strange C. simple D.special14.A. never B. almost C. seldom D. ever15.A. make B. understand C. master D. give 16.A. knows B. shows C. orders D. encourages17.A. thirsty B. tired C. bored D. free 18.A. of B. for C. about D. into 19.A. radio B. computer C. record D. machine20.A. measures B. notes C. place D. time【答案】1.B2.D3.C4.A5.B6.D7.A8.C9.B10.C11.B12.D13.C14.A15.D16.B17.A18.D19.B20.C【解析】试题分析:文章介绍真正的学习不是记忆事实而是使用学到的知识解决问题。
2014年山西省太原市高考一模数学试卷(文科)【解析版】
2014年山西省太原市高考数学一模试卷(文科)一、选择题:本题共12小题,每小题5分,共60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.1.(5分)已知集合M={x|﹣1<x<1},N={x|3x>1},则M∩N=()A.∅B.{x|x>0}C.{x|x<1}D.{x|0<x<1} 2.(5分)复数等于()A.﹣i B.i C.﹣i D.i3.(5分)已知函数f(x)=sin(2x+α)在x=时有极大值,则α的一个可能值是()A.B.﹣C.D.﹣4.(5分)下列命题中,真命题是()A.B.∀x∈(3,+∞),x2>2x+1C.∃x∈R,x2+x=﹣1D.5.(5分)在五个数字1,2,3,4,5中,若随机取出三个数字,则剩下两个数字的和是奇数的概率是()A.0.3B.0.4C.0.5D.0.66.(5分)已知数列{a n},若点(n,a n)(n∈N*)在经过点(8,4)的定直线l 上,则数列{a n}的前15项和S15=()A.12B.32C.60D.1207.(5分)如图是一算法的程序框图,若输出结果为S=720,则在判断框中应填入的条件是()A.k≤6?B.k≤7?C.k≤8?D.k≤9?8.(5分)如图是一个几何体的正(主)视图和侧(左)视图,其俯视图是面积为的矩形,则该几何体的表面积是()A.8B.C.16D.9.(5分)设M(x0,y0)为抛物线C:x2=8y上一点,F为抛物线C的焦点,以F为圆心、|FM|为半径的圆和抛物线C的准线相交,则y0的取值范围是()A.(0,2)B.[0,2]C.(2,+∞)D.[2,+∞)10.(5分)过双曲线﹣=1(a>0,b>0)的左焦点F(﹣c,0)(c>0)作圆x2+y2=的切线,切点为E,延长FE交双曲线右支于点P,若=(+),则双曲线的离心率为()A.B.C.D.11.(5分)已知球的直径SC=4,A,B是该球球面上的两点,AB=,∠ASC =∠BSC=30°,则棱锥S﹣ABC的体积为()A.3B.2C.D.112.(5分)已知方程=k在(0,+∞)上有两个不同的解α,β(α<β),则下面结论正确的是()A.sinα=αcosβB.sinα=﹣αcosβC.cosα=βsinβD.sinβ=﹣βsinα二、填空题:本大题共4小题,每小题5分,共20分.13.(5分)已知sinα+cosα=,则cos4α=.14.(5分)已知P是直线3x+4y+8=0上的动点,C是圆x2+y2﹣2x﹣2y+1=0的圆心,那么|PC|的最小值是.15.(5分)已知点P(x,y)满足条件(k为常数),若z=x+3y的最大值为8,则k=.16.(5分)在数列{a n}中,已知a1=1,a n+2=,a100=a96,则a9+a10=.三、解答题:解答应写出文字说明,证明过程或演算步骤.17.(12分)已知△ABC中,三个内角A,B,C的对边分别为a,b,c,若△ABC 的外接圆的半径为,且a sin A﹣c sin C=(a﹣b)sin B.(1)求∠C;(2)求△ABC的面积S的最大值.18.(12分)某园艺师培育了两种珍稀树苗A与B,株数分别为8与12,现将这20株树苗的高度编写成如图所示茎叶图(单位:cm).若树高在175cm以上(包括175cm)定义为“生长良好”,树高在175cm以下(不包括175cm)定义为“非生长良好”,且只有“B生长良好”的才可以出售.(1)对于这20株树苗,如果用分层抽样的方法从“生长良好”和“非生长良好”中共抽取5株,再从这5株中任选2株,那么至少有一株“生长良好”的概率是多少?(2)若从所有“生长良好”中选2株,求所选中的树苗都能出售的概率.19.(12分)如图,在斜三棱柱ABC﹣A1B1C1中,点O是A1C1的中点,AO⊥平面A1B1C1.已知∠BCA=90°,AA1=AC=BC=2.(1)求证:AB1⊥A l C;(2)求点C到平面AA1B1的距离.20.(12分)已知函数f(x)=(2﹣a)x﹣2(1+lnx)+a.(1)当a=1时,求f(x)的单调区间;(2)若函数f(x)在区间(0,)无零点,求a的最小值.21.(12分)已知中心在原点O,左右焦点分别为F1,F2的椭圆的离心率为,焦距为2,A,B是椭圆上两点.(1)若直线AB与以原点为圆心的圆相切,且OA⊥OB,求此圆的方程;(2)动点P满足:=+3,直线OA与OB的斜率的乘积为﹣,求动点P的轨迹方程.选修4一1:几何证明选讲22.(10分)如图,已知P A与圆O相切于点A,经过点O的割线PBC交圆O 于点B,C,∠APC的平分线分别交AB,AC于点D,E.(Ⅰ)证明:∠ADE=∠AED;(Ⅱ)若AC=AP,求的值.选修4-4:坐标系与参数方程23.在平面直角坐标系中,曲线C1的参数方程为(a>b>0,φ为参数),且曲线C1上的点M(2,)对应的参数φ=.且以O为极点,X 轴的正半轴为极轴建立极坐标系,曲线C2是圆心在极轴上且经过极点的圆,射线θ=与曲线C2交于点D(,).(1)求曲线C1的普通方程,C2的极坐标方程;(2)若A(ρ1,θ),B(ρ2,θ+)是曲线C1上的两点,求+的值.选修4-5:不等式选讲24.选修4﹣5:不等式选讲已知函数f(x)=|2x+1|﹣|x﹣3|.(Ⅰ)解不等式f(x)≤4;(Ⅱ)若存在x使得f(x)+a≤0成立,求实数a的取值范围.2014年山西省太原市高考数学一模试卷(文科)参考答案与试题解析一、选择题:本题共12小题,每小题5分,共60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.1.(5分)已知集合M={x|﹣1<x<1},N={x|3x>1},则M∩N=()A.∅B.{x|x>0}C.{x|x<1}D.{x|0<x<1}【解答】解:由N中的不等式变形得:3x>1=30,得到x>0,∴N={x|x>0},∵M={x|﹣1<x<1},∴M∩N={x|0<x<1}.故选:D.2.(5分)复数等于()A.﹣i B.i C.﹣i D.i【解答】解:==.故选:D.3.(5分)已知函数f(x)=sin(2x+α)在x=时有极大值,则α的一个可能值是()A.B.﹣C.D.﹣【解答】解:原函数f(x)=sin(2x+α)在x=时有极大值,即2×=2kπ+,α=2kπ+,k∈Z,当k=0时,α=.故选:A.4.(5分)下列命题中,真命题是()A.B.∀x∈(3,+∞),x2>2x+1C.∃x∈R,x2+x=﹣1D.【解答】解:B项是正确的.∀x∈(3,+∞),x2﹣(2x+1)=(x﹣1)2﹣2>2>0,由于对∀x∈R,sin x+cos x ≤,故A错误,方程x2+x+1=0无实根,故C项错误;对于∀x∈(,π)tan x<0<sin x,故D错误.故选:B.5.(5分)在五个数字1,2,3,4,5中,若随机取出三个数字,则剩下两个数字的和是奇数的概率是()A.0.3B.0.4C.0.5D.0.6【解答】解:根据题意,从5个数字中选3个,共有C53=10种情况,满足条件的是剩下两个数字的和是奇数,即取出的三个数为两奇一偶;有C32C21=6种结果,故剩下两个数字的和是奇数的概率是P=.故选:D.6.(5分)已知数列{a n},若点(n,a n)(n∈N*)在经过点(8,4)的定直线l 上,则数列{a n}的前15项和S15=()A.12B.32C.60D.120【解答】解:由题意可得a8=4∵点(n,a n)(n∈N*)在经过点(8,4)的定直线l上=k(为常数)∴a n可写为关于n的一次函数即可设a n=kn+m,则a n﹣a n﹣1∴{a n}为等差数列由等差数列的性质可知,a1+a15=2a8=8∴=15a8=60故选:C.7.(5分)如图是一算法的程序框图,若输出结果为S=720,则在判断框中应填入的条件是()A.k≤6?B.k≤7?C.k≤8?D.k≤9?【解答】解:根据程序框图,运行结构如下:S K第一次循环10 9第二次循环90 8第三次循环720 7此时退出循环,故应填K≤7?故选:B.8.(5分)如图是一个几何体的正(主)视图和侧(左)视图,其俯视图是面积为的矩形,则该几何体的表面积是()A.8B.C.16D.【解答】解:此几何体是一个三棱柱,且其高为=4,由于其底面是一个等腰直角三角形,直角边长为2,所以其面积为×2×2=2,又此三棱柱的高为4,故其侧面积为,(2+2+2)×4=16+8,表面积为:2×2+16+8=20+8.故选:B.9.(5分)设M(x0,y0)为抛物线C:x2=8y上一点,F为抛物线C的焦点,以F为圆心、|FM|为半径的圆和抛物线C的准线相交,则y0的取值范围是()A.(0,2)B.[0,2]C.(2,+∞)D.[2,+∞)【解答】解:由条件|FM|>4,由抛物线的定义|FM|=y0+2>4,所以y0>2故选:C.10.(5分)过双曲线﹣=1(a>0,b>0)的左焦点F(﹣c,0)(c>0)作圆x2+y2=的切线,切点为E,延长FE交双曲线右支于点P,若=(+),则双曲线的离心率为()A.B.C.D.【解答】解:∵|OF|=c,|OE|=,∴|EF|=,∵,∴|PF|=2,|PF'|=a,∵|PF|﹣|PF′|=2a,∴2﹣a=2a,∴,故选:C.11.(5分)已知球的直径SC=4,A,B是该球球面上的两点,AB=,∠ASC =∠BSC=30°,则棱锥S﹣ABC的体积为()A.3B.2C.D.1【解答】解:设球心为点O,作AB中点D,连接OD,CD.因为线段SC是球的直径,所以它也是大圆的直径,则易得:∠SAC=∠SBC=90°所以在Rt△SAC中,SC=4,∠ASC=30°得:AC=2,SA=2又在Rt△SBC中,SC=4,∠BSC=30°得:BC=2,SB=2则:SA=SB,AC=BC因为点D是AB的中点所以在等腰三角形ASB中,SD⊥AB且SD===在等腰三角形CAB中,CD⊥AB且CD===又SD交CD于点D所以:AB⊥平面SCD即:棱锥S﹣ABC的体积:V=AB ,•S△SCD因为:SD=,CD=,SC=4 所以由余弦定理得:cos∠SDC=(SD2+CD2﹣SC2)=(+﹣16)==则:sin∠SDC==由三角形面积公式得△SCD的面积S=SD•CD•sin∠SDC==3==所以:棱锥S﹣ABC的体积:V=AB•S△SCD故选:C.12.(5分)已知方程=k在(0,+∞)上有两个不同的解α,β(α<β),则下面结论正确的是()A.sinα=αcosβB.sinα=﹣αcosβC.cosα=βsinβD.sinβ=﹣βsinα【解答】解:∵有两个根,∴函数y=|sin x|和函数y=kx在(0,+∞)上有两个交点,x>0且k>0,画出两个函数的图象,如图(1)图1函数y=|sin x|和函数y=kx在(0,π)上有一个交点A(α,sinα),在(π,2π)上有一个切点B(β,﹣sinβ)时满足题意,α,β是方程的根.当x∈(π,2π)时,f(x)=|sin x|=﹣sin x,f′(x)=﹣cos x,∴在B处的切线为y+sinβ=f′(β)(x﹣β),将x=0,y=0代入方程,得sinβ=﹣β×(﹣cosβ),∴=cosβ,∵O,AB三点共线,∴,∴=﹣cosβ,∴sinα=﹣αcosβ.故选:B.二、填空题:本大题共4小题,每小题5分,共20分.13.(5分)已知sinα+cosα=,则cos4α=﹣.【解答】解:∵sinα+cosα=,平方可得1+sin2α=,∴sin2α=﹣.∴cos4α=1﹣2sin22α=1﹣2×=﹣,故答案为:﹣.14.(5分)已知P是直线3x+4y+8=0上的动点,C是圆x2+y2﹣2x﹣2y+1=0的圆心,那么|PC|的最小值是3.【解答】解:由圆方程化为变形方程得:(x﹣1)2+(y﹣1)2=1,∴圆心C(1,1),∵圆心C到直线3x+4y+8=0的距离d==3,∴|PC|的最小值为3.故答案为:315.(5分)已知点P(x,y)满足条件(k为常数),若z=x+3y的最大值为8,则k=﹣6.【解答】解:画出可行域将z=x+3y变形为y=,画出直线平移至点A时,纵截距最大,z最大,联立方程得,代入,∴k=﹣6.故答案为﹣616.(5分)在数列{a n}中,已知a1=1,a n+2=,a100=a96,则a9+a10=.【解答】解:∵a1=1,a n+2=,∴a3=,a5==,a7==,a9==,∵a n+2=,a100=a96,∴a100=a96==,∴a962+a96﹣1=0,∴a96=,∴a94=,∴a10=,∴a9+a10=+=.故答案为:.三、解答题:解答应写出文字说明,证明过程或演算步骤.17.(12分)已知△ABC中,三个内角A,B,C的对边分别为a,b,c,若△ABC 的外接圆的半径为,且a sin A﹣c sin C=(a﹣b)sin B.(1)求∠C;(2)求△ABC的面积S的最大值.【解答】解:(1)已知等式a sin A﹣c sin C=(a﹣b)sin B,利用正弦定理化简得:a2﹣c2=ab﹣b2,∴a2+b2﹣c2=ab,∴cos C==,∵C为三角形内角,∴C=;(2)∵sin C=sin=,==2R,即a=2R sin A,b=2R sin B,∴S=ab sin C=•2R sin A•2R sin B=2sin A sin B,∵A+B=π﹣C=,即B=﹣A,代入上式得:S=ab sin C=2sin A sin B=2sin A sin(﹣A)=2sin A(cos A+sin A)=(sin2A﹣cos2A+)=sin(2A﹣)+≤+=,则当2A﹣=,即A=时,S max=.18.(12分)某园艺师培育了两种珍稀树苗A与B,株数分别为8与12,现将这20株树苗的高度编写成如图所示茎叶图(单位:cm).若树高在175cm以上(包括175cm)定义为“生长良好”,树高在175cm以下(不包括175cm)定义为“非生长良好”,且只有“B生长良好”的才可以出售.(1)对于这20株树苗,如果用分层抽样的方法从“生长良好”和“非生长良好”中共抽取5株,再从这5株中任选2株,那么至少有一株“生长良好”的概率是多少?(2)若从所有“生长良好”中选2株,求所选中的树苗都能出售的概率.【解答】解:(1)根据茎叶图,可知“生长良好”有8株,“非生长良好”的有12株,用分层抽样的方法抽取,每株被抽取的概率是=,从“生长良好”中共抽取株,“非生长良好”的有株.设“生长良好”的两株为1,2.“非生长良好”的3株为a,b,c.则所有的基本事件有:(1,2),(1,a),(1,b),(1,c),(2,a),(2,b),(2,c),(a,b),(a,c),(b,c)共有10种,至少有一株“生长良好”的事件有7个∴至少有一株“生长良好”的概率是.(2)依题意,一共有8株生长良好,其中A有5株,B有3株,所有可能的基本事件共有个,树苗都能出售的事件包含的基本事件为个,∴所求概率为.19.(12分)如图,在斜三棱柱ABC﹣A1B1C1中,点O是A1C1的中点,AO⊥平面A1B1C1.已知∠BCA=90°,AA1=AC=BC=2.(1)求证:AB1⊥A l C;(2)求点C到平面AA1B1的距离.【解答】(1)证明:∵AO⊥平面A1B1C1,∴AO⊥,又∵A1C1⊥B1C1,且A1C1∩AO=0,∴B1C1⊥平面A1C1CA,∴A1C⊥B1C1,又∵AA1=AC,∴四边形A1C1CA为菱形,∴A1C⊥AC1,且B1C1∩AC1=C1,∴A1C⊥平面AB1C1,∴AB1⊥A1C.(2)∵CC1∥平面AA1B1,∴点C到平面AA1B2的距离与点C1到平面AA1B1的距离相等,设C1到平面AA1B1的距离为d,∵=,∴,又∵在△AA 1B1中,,,∴d=.20.(12分)已知函数f(x)=(2﹣a)x﹣2(1+lnx)+a.(1)当a=1时,求f(x)的单调区间;(2)若函数f(x)在区间(0,)无零点,求a的最小值.【解答】解:(1)当a=1时,f(x)=x﹣1﹣2lnx,则f′(x)=1﹣,定义域x∈(0,+∞),由f′(x)>0,得x>2;由f′(x)<0,得0<x<2,故f(x)的单调减区间为(0,2),单调增区间为(2,+∞).(2)f(x)=(2﹣a)(x﹣1)﹣2lnx,令m(x)=(2﹣a)(x﹣1),x>0;h(x)=2lnx,x>0,则f(x)=m(x)﹣h(x),①当a<2时,m(x)在(0,)上为增函数,h(x)在(0,)上为增函数.结合图象可知,若f(x)在(0,)无零点,则m()≥h().即(2﹣a)×(﹣1)≥2ln,∴a≥2﹣4ln2.∴2﹣4ln2≤a<2.②当a≥2时,在(0,)上m(x)≥0,h(x)<0.∴f(x)>0,∴f(x)在(0,)上无零点,由①②得a≥2﹣4ln2,∴a min=2﹣4ln2.21.(12分)已知中心在原点O,左右焦点分别为F1,F2的椭圆的离心率为,焦距为2,A,B是椭圆上两点.(1)若直线AB与以原点为圆心的圆相切,且OA⊥OB,求此圆的方程;(2)动点P满足:=+3,直线OA与OB的斜率的乘积为﹣,求动点P的轨迹方程.【解答】解:(1)设椭圆方程为,由,解得:.∴椭圆方程为.①设直线AB为:y=kx+m,A(x1,y1),B(x2,y2),代入椭圆方程得:(1+3k2)x2+6kmx+3(m2﹣1)=0.∴,∵OA⊥OB,∴,即x1x2+y1y2=x1x2+(kx1+m)(kx2+m)==,即4m2﹣3k2﹣3=0.∵直线AB与以原点为圆心的圆相切,∴圆的半径,则.∴圆的方程为;②当直线AB的斜率不存在时,直线AB的方程为,满足上述方程.综上,所求圆的方程为:.(2)设P(x,y),又A(x1,y1),B(x2,y2),由:=+3,得,又直线OA与OB的斜率的乘积为﹣,∴,即x1x2+3y1y2=0.∵A,B在椭圆上,∴.联立,消去x1,x2,y1,y2,得x2+3y2=30.当OA斜率不存在时,即x 1=0,得y1=±1,y2=0,.此时.同理OB斜率不存在时,.∴动点P的轨迹方程为x2+3y2=30().选修4一1:几何证明选讲22.(10分)如图,已知P A与圆O相切于点A,经过点O的割线PBC交圆O 于点B,C,∠APC的平分线分别交AB,AC于点D,E.(Ⅰ)证明:∠ADE=∠AED;(Ⅱ)若AC=AP,求的值.【解答】解:(Ⅰ)∵P A是切线,AB是弦,∴∠BAP=∠C.又∵∠APD=∠CPE,∴∠BAP+∠APD=∠C+∠CPE.∵∠ADE=∠BAP+∠APD,∠AED=∠C+∠CPE,∴∠ADE=∠AED.…(5分)(Ⅱ)由(Ⅰ)知∠BAP=∠C,∵∠APC=∠BP A,∵AC=AP,∴∠APC=∠C∴∠APC=∠C=∠BAP.由三角形内角和定理可知,∠APC+∠C+∠CAP=180°.∵BC是圆O的直径,∴∠BAC=90°.∴∠APC+∠C+∠BAP=180°﹣90°=90°.∴.在Rt△ABC中,,即,∴.∵在△APC与△BP A中∠BAP=∠C,∠APB=∠CP A,∴△APC∽△BP A.∴.∴.…(10分)选修4-4:坐标系与参数方程23.在平面直角坐标系中,曲线C1的参数方程为(a>b>0,φ为参数),且曲线C1上的点M(2,)对应的参数φ=.且以O为极点,X 轴的正半轴为极轴建立极坐标系,曲线C2是圆心在极轴上且经过极点的圆,射线θ=与曲线C2交于点D(,).(1)求曲线C1的普通方程,C2的极坐标方程;(2)若A(ρ1,θ),B(ρ2,θ+)是曲线C1上的两点,求+的值.(1)由曲线C1上的点M(2,)对应的参数φ=可得:,【解答】解:解得,∴曲线C1的普通方程为.设圆C2的半径为R,由于射线θ=与曲线C2交于点D(,).可得,解得R=1.∴圆C2的极坐标方程为ρ=2cosθ.(2)曲线C1的极坐标方程为:,化为,∵A(ρ1,θ),B(ρ2,θ+)是曲线C1上的两点,∴+==+==.选修4-5:不等式选讲24.选修4﹣5:不等式选讲已知函数f(x)=|2x+1|﹣|x﹣3|.(Ⅰ)解不等式f(x)≤4;(Ⅱ)若存在x使得f(x)+a≤0成立,求实数a的取值范围.【解答】解:(Ⅰ)f(x)=|2x+1|﹣|x﹣3|=,如图,它与y=4的交点为(﹣8,4)和(2,4).故不等式f(x)≤4的解集为[﹣8,2].(Ⅱ)由f(x)的图象知,x =﹣时,f(x )有最小值﹣,存在x使得f(x)+a≤0成立,等价于﹣a ≥﹣,a ≤,故实数a 的取值范围为(﹣∞,].第21页(共21页)。
太原市2014年高三年级模拟试题(二)
太原市2014年高三年级模拟试题(二)文科综合能力测试第Ⅰ卷 (选择题共140分)本卷共35小题,每小题4分,共140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。
24.图6为中国秦汉至唐初城乡结构示意图,对该图的描述符合史实的是A.乡村自给自足状态被打破B.城乡之间交流日益频繁C.城乡之间道路四通八达D.城市政治中心功能显著25.明清之际的学者董含说:“昔士大夫以清望为重,乡里富人,羞与为伍,有攀附者必峻绝之。
今人崇尚财货,贿拥资厚者,反屈体降志,或订忘形之交,或结婚姻之雅。
”这说明当时A.重农抑商政策在民间已走向破产B.进步知识分子大力提倡工商皆本C.传统儒学提倡的义利观遭到摒弃D.商品经济发展冲击了传统价值观26.顾炎武说:“君子之为学,以明道也,以救世也。
徒以诗文而已,所谓雕虫篆刻,亦何益哉!”该言论旨在A.维护儒学独尊地位 B.反对文人参加科举C.大力提倡求实学风 D.强调加强自身修养27.公元前5世纪的古希腊学者安提丰认为:“一个人应该以他最有利的方式对待法律。
在众目睽睽之下,他应非常尊重法律。
但当可以自主而又无人在场的情况下,他可以随本性驱动行事。
”这种主张A.反映出古希腊法律是习惯法 B.体现了智者学派的思想特征C.说明了古希腊实行民主政体 D.表明知识分子倡导以人为本28.“自从阶级对立产生以来,正是人的恶劣的情欲──贪欲和权势欲成了历史发展的杠杆。
”可以印证这一观点的史实是A.地理大发现 B.哥白尼提出“日心说”C.美国制定1787年宪法 D.清朝闭关锁国29.郑观应在《商务叹》中写道:“轮船招商开平矿,创自商人尽商股。
……总办商董举自官,不依商律由商举。
……名为保商实剥商,官督商办势如虎;华商因此不及人,为丛驱爵成怨府。
”郑观应感叹的是A.股份经营弊端丛生 B.政府大力干预经济C.民族工业发展艰难 D.外商挤压民族工业30.1892年,康有为撰写《孔子改制考》,认为未来社会是实行共和制的大同社会;1913年7月他又发表文章说:“夫共和之立国……今所模欧师美者……但模仿其政俗之末,而失其道德之本,此其政俗所以在欧、美成功之效,而在我为败坏之由。
百强重点名校高考备考_2014年高考(665)山西太原五中高三年级5月模拟考试(完美整理版)
2014年高考(665)山西太原五中高三年级5月模拟考试高中语文模拟试题0530 1050:山西省太原市第五中学2014届高三第二学期5月月考语文试题第Ⅰ卷阅读题甲必考题一、现代文阅读(9分,每小题3分)阅读下面文字,完成1~3题。
习家池的修禊习俗修禊,是中国传统文化中的精粹。
东汉应劭《风俗通义·祀典》说:“禊者,洁也。
谨按《周礼》男巫掌望祀,旁招以茅,女巫掌岁时以祓除衅浴。
”但从《诗经·溱洧》来看,修禊由纯祭祀活动逐步演变为带有节日性的活动。
三月桃花盛开,百花馥郁,河水荡漾,水盈碧泛,阳光明媚的时候,郑国的青年男女到溱洧两水上沐浴,以洁身除秽邪之气,临别互赠花草。
《韩诗》载:“郑俗,三月上巳,之溱、洧之上,招魂续魄。
秉兰草,拂不祥。
”这应该就是修禊的前身。
但是到了两汉,修禊的形式发生了变化。
其神秘繁复的色彩减少了很多,祭祀只是象征性的,在水曲隈处喝酒吟诗成主要内容,成为一种官民同乐的高雅有趣的节日活动。
到了宋代其内容已经是根据喜好随意增减了。
文人雅士多曲水流觞,临河赋诗歌咏。
一般民众则游春踏青,赏览春光。
史上最著名的一次修禊集会是东晋永和九年(公元353年),王羲之父子、谢安、孙绰等共41人在山阴(今绍兴)兰亭举行的。
贤士名流分列两岸,参差坐于茂林修竹中,曲水赋诗,纵酒狂欢。
王羲之作《兰亭集序》,成为天下第一行书。
但影响巨大波及全国的修禊活动是清康乾年间扬州瘦西湖畔的三次“红桥修禊”,主持者皆为名士,参加者近万,规模空前,成为中国诗歌史上的盛举。
其方式是在洗濯后,列坐水畔,随水流羽杯,举觞吃酒,吟诗作词。
襄阳也是一个有修禊传统的地方,而习家池就是主要的修禊地点。
《湖广通志·襄阳府》载:“县东十里,有白马泉,晋习凿齿居焉,因名习家池。
”宋祝穆撰《方舆胜览·襄阳府》载:“每年三月三日,刺史禊饮于此。
”六朝时的《荆楚岁时记》云:“三月三日,四民并出水滨,为流杯曲水之饮,取黍曲菜汁和蜜为饵,以厌时气。
2014年山西省太原市高三理科二模数学试卷
2014年山西省太原市高三理科二模数学试卷一、选择题(共12小题;共60分)1. 已知集合,且,则集合与集合的关系是A. B. C. D.2. 若复数满足(为虚数单位),则在复平面内对应的点在A. 第一象限B. 第二象限C. 第三象限D. 第四象限3. 双曲线的顶点到渐近线的距离为A. B. C. D.4. 一个几何体的三视图如图所示,则该几何体的体积(单位:)为A. B. C. D.5. 设函数,将的图象向右平移个单位长度后,所得的图象与原图象重合,则的最小值等于A. B. C. D.6. 某程序框图如图所示,该程序运行后输出的为A. B. C. D.7. 设,满足约束条件则的取值范围是A. B. C. D.8. 设的内角,,所对的边分别为,,,若,,则的取值范围是A. B. C. D.9. 已知椭圆的左焦点,与过原点的直线相交于,两点,连接,,若,,,则的离心率为A. B. C. D.10. 已知一个四面体的每个面都是有两条边长为,一条边长为的三角形,则该四面体的外接球的表面积为A. B. C. D.11. 已知定义在上的函数满足:,且,,则方程在区间上的所有实数根之和为A. B. C. D.12. 已知函数,数列中,,数列中,,,则下列说法正确的是A. 是递增数列且,是递减数列且B. 是递增数列且,是递增数列且C. 是递增数列且,是递减数列且D. 是递减数列且,是递增数列且二、填空题(共4小题;共20分)13. 二项式的展开式中的常数项为.14. 某校从名教师中选派名教师同时去个边远地区支教,每地人,其中甲和乙不同去,甲和丙只能同去或同不去,则不同的选派方案共种.15. 三角形中,过中线的中点作直线分别与边和交于,两点,若,,则的最小值是.16. 定义在上的函数满足:,当时,,则不等式的解集为.三、解答题(共8小题;共104分)17. 已知各项均为正数的数列满足,且,其中.(1)求数列的通项公式;(2)设数列满足是否存在正整数,,,使得,,成等比数列?若存在,求出所有的,的值,若不存在,请说明理由.18. 中国航母“辽宁舰”是中国第一艘航母,“辽宁”号以台蒸汽轮机为动力,为保证航母的动力安全性,科学家对蒸汽轮机进行了余项技术改进,增加了某项新技术,该项新技术要进入试用阶段前必须对其中的三项不同指标甲、乙、丙进行通过量化检测.假如该项新技术的指标甲、乙、丙独立通过检测合格的概率分别为,,.指标甲、乙、丙合格分别记为分,分,分;若某项指标不合格,则该项指标记分,各项指标检测结果互不影响.(1)求该项技术量化得分不低于分的概率;(2)记该项新技术的三个指标中被检测合格的指标个数为随机变量,求的分布列与数学期望.19. 三棱锥,底面为边长为的正三角形,平面平面,,为上一点,,为底面三角形中心.(1)求证 面;(2)求证:;(3)设为中点,求二面角的余弦值.20. 已知抛物线的焦点为,直线与抛物线交于不同的两点,,直线与抛物线交于不同的两点,.(1)当过时,在上取不同于的点,使得,求点的轨迹方程;(2)若与相交于点,且倾斜角互补时,,求实数的值.21. 设函数(为常数).(1)若函数在区间上是单调递增函数,求实数的取值范围;(2)若函数有两个极值点,,且,求证:.22. 如图,是直角三角形,.以为直径的圆交于点,点是边的中点.连接交圆于点.(1)求证:,,,四点共圆;(2)求证:.23. 已知在直角坐标系中,直线的参数方程为(为参数),在极坐标系(与直角坐标系取相同的长度单位,且以原点为极点,以轴正半轴为极轴)中,曲线的极坐标方程为.(1)求直线普通方程和曲线的直角坐标方程;(2)设点是曲线上的一个动点,求它到直线的距离的取值范围.24. 设.(1)若不等式的解集为.求的值;(2)若,,求的取值范围.答案第一部分1. C 【解析】根据题意,对集合分类讨论可得:①时,或,或;②时,无论取何值,都有;③时,或,或.综上知.则有.2. A 【解析】因为,所以,所以所以在复平面内对应的点为,即为第一象限内的点.3. B 【解析】双曲线的顶点坐标,渐近线方程为,所以所求的距离为.4. A 【解析】由三视图,该组合体上部是一三棱锥,下部是一圆柱由图中数据知圆柱,三棱锥垂直于底面的侧面是边长为的等边三角形,且边长是,故其高即为三棱锥的高,高为故棱锥高为,由于棱锥底面为一等腰直角三角形,且斜边长为,故两直角边的长度都是,底面三角形的面积是,,故棱锥故该几何体的体积是.5. C【解析】将向右平移个单位长度得,则,所以.又,所以,当时,有最小值.6. A 【解析】由程序框图知:第一次循环,;第二次循环,;第三次循环,;第四次循环,,,所以值的周期为,因为跳出循环的值为,所以输出.7. D 【解析】作出不等式组对应的平面区域如图:其中,,,设,则的几何意义为平面区域内的点到定点的斜率,由图象知的斜率最小,的斜率最大,则的斜率,的斜率为,即,则,,即的取值范围是.8. D 【解析】因为,,由余弦定理可得,所以,因为,所以,所以,解得,当且仅当时取等号,又由可得,故的取值范围为:.9. B 【解析】如图所示,在中,由余弦定理可得,所以,化为,解得.设为椭圆的右焦点,连接,.根据对称性可得四边形是矩形.所以,.所以,,解得,.所以.10. C【解析】设长方体的长宽高分别是,,,其四个顶点就构成一个四面体满足每个面的边长为,,,则,,,则,即长方体的外接球直径,故外接球的表积.11. C 【解析】因为,且,所以,又,则,所以,当,时,上述两个函数都是关于对称,由图象可得:方程在区间上的实根有个,满足,满足,,满足,,满足,,所以方程在区间上的所有实根之和为.12. D【解析】根据题意,得,;所以,,所以是递减的数列,且;,;所以,,所以,所以是递增数列.第二部分13.【解析】二项式的展开式中的通项公式为,令,求得,故展开式中的常数项为.14.【解析】分两步,第一步,先选三名老师,又分两类.第一类,甲去,则丙一定去,乙一定不去,有种不同选法.第二类,甲不去,则丙一定不去,乙可能去也可能不去,有种不同选法.所以不同的选法有种第二步,三名老师去个边远地区支教,有,根据分步计数原理得不同的选派方案共有,.15.【解析】由题意得:.所以.同理,;因为与共线,所以存在实数,使;所以;所以所以所以所以的最小值是.16.【解析】因为定义在上的函数满足:,两边对求导,得,所以,令,则,因为当时,,所以,所以,即,又,直线过原点,所以,所以,都有,令,则,即是上的单调减函数,且,所以不等式,即,即,所以.所以原不等式的解集为.第三部分17. (1)因为,所以,因为,所以有,即,所以数列是公比为的等比数列,由得,解得.从而数列的通项公式为.(2),若,,成等比数列,则,,,使得,,成等比数列.18. (1)该项新技术的三项不同指标甲、乙、丙独立通过检测合格分别为事件,,,则事件“得分不低于分”表示为.所以与为互斥事件,且,,为彼此独立,所以(2)该项新技术的三个指标中被检测合格的指标个数的取值为,,,,因为,,随机变量的分布列为所以.19. (1)连接交于点,连接,如图,因为为正三角形的中心,所以,且为中点.又,所以,因为平面,平面,所以 面.(2)因为,且为中点,所以,又平面平面,所以平面,由(Ⅰ)知,,所以平面,所以.连接,如图,则,又,所以平面,所以.(3)由(Ⅰ)(Ⅱ)知,,,两两互相垂直,且为中点,所以分别以,,所在直线为,,轴,建立空间直角坐标系,如图,则,,,,,,所以,.设平面的法向量为,则令,则.由(Ⅱ)知平面,所以为平面的法向量,所以,由图可知,二面角的余弦值为.20. (1)设,,,因不同于,知不在线段上,设:,代入,得,则,,设,,,在轴的投影分别是,,,,则,,由于,异号,不在线段上,则与同号,所以,即,所以,而,所以,所以点的轨迹方程为.(2)设,,,,,:,:,代入,得,,则,,,,则,,而又在上也在上,所以则,所以,所以式可化为所以.21. (1)根据题意知:在上恒成立.即在区间上恒成立.因为在区间上的最大值为,所以;经检验:当时,,.所以的取值范围是.(2)在区间上有两个不相等的实数根,即方程在区间上有两个不相等的实数根.记,则有解得.所以,,,.所以.令,.,记.所以,,.在使得.当,;当时,.而在单调递减,在单调递增,因为,,所以当,,所以在单调递减,即.22. (1)如图1,连接,,则.因为是的中点,所以.又,,所以,所以,所以,,,四点共圆.(2)延长交圆于点.如图2,由(1)知为圆的切线,所以所以,所以.23. (1)由,化为直角坐标方程:,即曲线的方程为,由直线的参数方程为(为参数)消去,得直线的方程是:.(2)曲线的标准方程为,圆心,半径为.所以圆心到直线的距离为:.所以点到直线的距离的取值范围是.24. (1),其图象如下:当时,.当时,;当时,.所以,.(2)不等式,即.因为的最小值为,所以问题等价于.解得,或.故的取值范围是.。
太原市2014年高三年级模拟试题(一)
太原市2014年高三年级模拟试题(一)二、选择题:本题包括8小题,每小题给出的四个选项中,14~18小题只有一个选项正确,19~21小题有多个选项正确,全部选对得6分,选对但不全得3分,有选错的得0分。
14.如图所示,一只小鸟沿着较粗的均匀树枝从右向左缓慢爬行,在小鸟从A运动到B的过程中A.树枝对小鸟的作用力先减小后增大B.树枝对小鸟的摩擦力先减小后增大C.树枝对小鸟的弹力先减小后增大D.树枝对小鸟的弹力保持不变15.CTMD(中国战区导弹防御体系)是一种战术型导弹防御系统,可以拦截各类型的短程及中程超音速导弹。
在某次演习中,检测系统测得关闭发动机的导弹在距地面高为H处,其速度为v且恰好水平,反应灵敏的地面拦截系统同时以初速度V竖直向上发射一颗炮弹成功拦截。
已知发射时炮弹与导弹的水平距离为s,不计空气阻力,则A.V=H/sv B.V=[KF(]H/s[KF)]vC.V=s/Hv D.V=v16.如图是原、副线圈都有中心抽头(匝数一半处)的理想变压器。
原线圈通过单刀双掷开关S1与电流表连接,副线圈通过另一单刀双掷开关S2与定值电阻R0相连接,通过S1、S2可以改变原、副线圈的匝数。
现在原线圈上加一电压为U的正弦交流电,当S1接a,S2接c时,电流表的示数为I,下列说法正确的是A.当S1接a,S2接d时,电流为2IB.当S1接a,S2接d时,电流为I/2C.当S1接b,S2接c时,电流为4ID.当S1接b,S2接d时,电流为I/217.如图1所示,正电荷Q均匀分布在半径为r的金属球面上。
以圆心O为坐标原点,向右建立x轴,选取无穷远处电势为零,用E表示沿x轴上各点电场强度的大小、φ表示沿x轴上各点电势的高低。
下列E - x和φ - x图象大致正确的是18.将小球以10 m/s的初速从地面竖直向上抛出,取地面为零势能面,小球在上升过程中的动能Ek、重力势能EP与上升高度h间的关系分别如图中两直线所示。
取g = 10 m/s2,下列说法正确的是A.小球的质量为0.2 kgB.小球受到的阻力(不包括重力)大小为0.20 NC.小球动能与重力势能相等时的高度为 20/13mD.小球上升到2 m时,动能与重力势能之差为0.5 J19.如图所示,带等量异种电荷的平行金属板a、b处于匀强磁场中,磁场方向垂直纸面向里。
山西省太原市重点中学2014届下学期高三年级4月月考理综试卷 有答案
山西省太原市重点中学2014届下学期高三年级4月月考理综试卷,有答案第Ⅰ卷(选择题共126分)本试卷共21小题,每小题6分,共126分。
合题目要求的。
以下数据可供解题时参考:相对原子质量(原子量):H 1 C 12 N 14 O16 Na 23 S 32 Cl 35.5 Cr 52 Fe 56一、选择题:本大题共13小题,每小题6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。
1.下列哪一项试剂在两个实验中作用相同A.酒精在“观察植物细胞有丝分裂”和“检测生物组织的脂肪”中的作用B.硫酸铜在“检测生物组织中的还原糖”和“检测生物组织的蛋白质”中的作用C.盐酸在“观察植物细胞有丝分裂”和“低温诱导植物染色体数目的变化”中的作用D.蒸馏水在“提取纯净的动物细胞膜”和“探究生长素类似物促进插条生根的最适浓度”中的作用2.酵母菌的线粒体在饥饿和光照等条件下损伤后发生特异性的“自噬”现象。
这是由于损伤后的线粒体产生一种外膜蛋白,导致高尔基体片层结构包裹线粒体形成“自噬体”,与溶解体结合形成“自噬溶酶体”(如下图所示),下列说法不正确的是A.若线粒体均遭“损伤”酵母菌将无法产生ATPB.内容物降解后形成的产物,可以为细胞提供营养C.线粒体产生的“外膜蛋白”是一种特异性受体D.“自噬溶酶体”的形成依赖生物膜的流动性3.肉毒杆菌是一种生长在缺氧环境下的细菌,在罐头食品及密封腌渍食物中生存能力极强,其繁殖过程中会分泌一种蛋白质类的肉毒毒素,其危害是阻断乙酰胆碱的分泌。
致残率、病死率极高。
据此分析下列有关说法正确的是A.肉毒杆菌是异养生物,生命活动所需能量大部分来自线粒体B.肉毒毒素进入人体内环境能刺激人体产生抗毒素并发生特异性反应,属于细胞免疫C.肉毒毒素阻止乙酰胆碱的释放,能达到麻痹肌肉的效果,可用于美容除皱D.肉毒杆菌细胞中没有染色体,其增殖方式为无丝分裂4.乳糖酶催化乳糖水解。
有两项与此相关的实验,其实验条件均设置为最适条件,实验结果如下:以下分析正确的是A.实验一如果继续增加酶浓度,相对反应速率不再加大B.实验二增加酶浓度时,相对反应速率将增加C.两个实验中都可以通过斐林试剂检验反应速率D.两个实验中若提高反应温度条件5℃,相对反应速率都将增大5. G418是一种抗生素,可通过影响核糖体的功能而阻断蛋白质合成,而neo基因的表达产物可使G418失效。
太原市2014年高三模拟地理试题(一)解析版纯word重绘图
第I卷本卷共35个小题,每小题4分,共140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。
2014年2月15日,山西省卫计委宣布“单独二孩”政策有望于6月前后正式实施。
所谓“单独二孩”政策,是指“夫妻一方为独生子女,并且只有一个子女的,可以生育第二个孩子”。
据悉,目前已有浙江、江西、安徽、天津、北京等省份正式实施“单独二孩”政策。
据此,完成1~2题。
1.下列已实施“单独二孩”政策的省份中,人口生育率最大的是A.北京B.天津C.江西D.浙江【答案】C【解析】经济发展水平相对较低的省市,人口生育率较高,在所给的四个省份中江西省经济发展水平最低,所以其人口生育率最大。
【知识点】人口增长模式及人口问题2.实施“单独二孩”政策,短期内可能A.改变我国人口增长的模式B.影响劳动力人口的职业构成C.加大适龄人口的就业压力D.减轻家庭及社会的养老负担【答案】B【解析】注意题干中的关键词“短期内”,实施“单独二孩”政策后人口出生率会略有升高,但是不至于改变人口增长模式,所以A错误;C、D两项都不是短期内会发生的变化;只有B项在短期内可能发生,因为实施“单独二孩”政策后,怀孕妇女增多,一些适龄妇女将不能参加劳动,所以劳动人口中妇女人口将减少。
【知识点】人口增长模式及人口问题产品生命周期依次可划分为创新、成熟和标准化三个阶段。
当有不同类型的国家净出口该产品时,其生命周期进入下一个阶段。
图1示意某产品生命周期,图中①、②、③分别表示三种不同类型的国家。
读图1,完成3~4题。
图13.该产品处于标准化阶段的时期是A.T4—T6B.T5—T6C.T3—T5D.T1—T3【答案】B【解析】要解答此题目需要从已知材料中提取有效信息,如:“产品生命周期依次可划分为创新、成熟和标准化三个阶段。
当有不同类型的国家净出口该产品时,其生命周期进入下一个阶段”,这样据图可以看出,T1—T3为创新阶段,T3—T5为成熟阶段,T5—T6为标准化阶段。
太原市高三上期中14年试卷
太原市2014-2015学年高三年级测评英语试卷第Ⅰ卷听力测试(共30分)听力略第二部分阅读理解(共两节,满分60分)第一节(共15小题;每小题3分,满分45分)阅读下列短文,从每题所给的四个选项(A,B,C和D)中,选出最佳选项,答案写在答题卡上。
ATeenage VolunteeringTeenage volunteering is a fantastic way to make new friends, see the world and participate in unique community service programs. Join GoEco’s great p rograms around the world. Teenage volunt eering abroad is also a great way to cultivate the foreign language you are studying in school, mee t incredible people and of course do your part to give back to less-fortunate communities abroad. Tiger, Lion and Wildlife ParkVolunteer with rare species of tigers and lions in this beautiful region of South Africa. Learn about conservation of Africa’s large cats and have a great time with an international team of up to 32 volunteers.Duration: 6 to 9 WeeksStarts from $1,580Akha Hilltribe Aid and ExperienceVolunteer in Thailand assisting with the education and development of the Akha hill tribes. Ex perience unique new cultures and traditions while providing much needed aid to the local communit ies.Duration: 2 to 6 WeeksStarts from $990Giant Panda CenterVolunteers are welcome to assist in an exciting and rewarding program with giant pandas in C hina. Make a real impact in saving this endangered species.Duration: 1 to 6 WeeksStarts from $840Sea Turtle ConservationExperience Mediterranean life while volunteering to protect the fragile baby sea turtles along G reece’s beaches. Make a difference in sea turtle conservation by monitoring the sand for these maje stic creatures as well as reaching out to educate the community on the subject.Duration: 2 to 12 WeeksStarts from $1,69021. Teenage volunteering with GoEco is mentioned as being helpful EXCEPT in ______.A. making new friendsB. developing leadershipC. improving a foreign languageD. traveling around the world22. Which project is the best for those who are interested in Asian tribes?A. Tiger, Lion and Wildlife ParkB. Akha Hilltribe Aid and ExperienceC. Giant Panda CentreD. Sea Turtle Conservation23. Which of the volunteering projects will last the longest?A. Sea Turtle ConservationB. Tiger, Lion and Wildlife ParkC. Giant Panda CenterBAdam Braun established the organization--Pencils of Promise in 2008. Its goa is to make sure all children have a chance for education. Six years later, the non-profit group is building a new sc hool somewhere every 90 hours. It has helped more than 22,000 children in Africa, Asia, and Lati n America.It all started when Braun was a college student. He was visiting India when a boy stopped hi m on street. The boy asked for money, Mr Braun asked the boy what he would want if he could have anything in the world. “I thought the answer was going to be a ‘house’, or ‘a car’ or a ‘boa t’. His answer was ‘a pencil’. So I give him my pencil and he just lit up with joy. I realized he never had been to school before, and that was the reality for 57 million children around the worl d”. Braun said.Mr. Braun started working in finance after he graduated from college.But he never forgot the boy and the problem he represented. “We live in a world in which every single child can have access to quality education, because we hav everything necessary already. We have the capability of educating every child. So I became immensely committed to helping create that world.” Braun said.Mr. Braun raised money for his project by using social media. He paid for building the first P encils of Promise School in Laos five years ago. Since then, his group has helped pay for more 2 00 schools in rural areas of Laos, Nicaragua, Guatemala, Ghana.24. Which is NOT true about the organization Pencils of Promise?A. It was set up in 2008.B. It’s a non-profit organization.C. It has benefited more than 22,000 children around the world.D. It helps build a new school every 90 days.25. What happened when Adam Braun was visiting India?A. He went to teach at a local college.B. He bought a car to travel around the country.C. He gave his pencil to a boy on the street.D. A boy stopped him to ask for a house.26. How did Mr. Braun’s experience in India affect him?A. He decided to help bring education to all children.B. He decided to establish a factory to produce pencils.C. He decided to finish his college education.D. He decided to take a job in public media.27. What is the passage mainly about?A. The schools Pencils of Promise has set up.B. The benefits provided for 57 million children.C. The success Braun’s organization has gained.CAmericans love Christmas trees. The National Christmas Tree Association says more than 35 million of them were bought last year in the United States.In large American cities, many people buy trees that have already been cut. But many people drive to a tree farm and cut them own tree. Some buy one from a catalogue or online, and the tree is then delivered to their home. It estimated that 10.9 million artificial trees were sold in 2012, comp ared to 24.5 million natural trees. The average cost of a natural tree was $40, and about $70 for a n artificial one.Most Christmas trees are now grown on farms. The trees take six to ten years to grow. In so me states, fewer than half of the planted trees survived the weather conditions to become Christmas trees. To make sure there is always a supply, farmers usually plant one to three new saplings for every tree they cut down.The Christmas Tree Association says real trees are better for the environment. That is because as Christmas trees grow, they collect carbon dioxide and other gases while supplying fresh oxyge n.The trees also protect water supplies.Christmas trees are grown on soil that does not support other crops. Some people throw away their natural tree when the holiday season ends. But many people recycle them. There are many ways to do this. Christmas trees can be cut up to be used as fertilizer. They can also be cut and used to prevent the wearing down of soil. Christmas trees helped many communities to prevent the sand from blowing away.Artificial trees, which are mostly made of plastics, can be reused for years. Families use them for an average of six to nine years before throwing them away. But these trees can remain in a l andfill(垃圾填埋场)for hundreds of years.28. What can be inferred from the passage?A. The lower pric e may contribute to people’s preference for natural trees.B. In 2012 more natural Christmas trees were sold than artificial ones.C. Farmers plant more new Christmas trees because of their increasing prices.D. On average an artificial Christmas tree is used for six to nine years.29. Which of the following is TRUE about the real Christmas trees?A. Half of the young trees can become Christmas trees.B. It takes six to ten months for Christmas trees to grow.C. Farmers plant one new tree for each tree they cut down.D. Christmas trees are not grown together with other crops.30. What might be the author’s attitude toward the use of artificial Christmas trees?A. FavorableB. NegativeC. CasualD. Optimistic31. Which is NOT the way real Christmas trees are recycled according to the passage?A. They can be cut up to be used as fertilizer.B. They can be cut and used to prevent the wearing down of soil.C. They can be used to prevent the sand from blowing away.D. They can be cut up and used as firewood.解析:文章大意:真正的圣诞树和人工圣诞树的区别及利弊;28. A 推理题,根据原文第二段The average cost of a natural treewas $40, and about $70 for an artificial one. B.C选项文章未提及;D选项文章最后一段第二句原句Families use them for anaverage of six to nine years before throwing them away. 不属于推理题。
太原市2014年高三年级模拟试题(一)
太原市2014年高三年级模拟试题(一)英语试卷(考试时间:下午2:30—4:30)注意事项:1.本试题分第I卷(选择题)和第II卷(非选择题)两部分。
第I卷1至12页,第II卷13至14页。
考试结束后,将本试题和答题卡一并交回。
2.答卷前,考生务必将自己的姓名、准考证号填写在试题和答题卡上。
3.回答第I卷时,选出每小题答案后,用2B铅笔把答题卡上对应题目的答案标号涂黑。
如需改动,用橡皮擦干净后,再选涂其他答案标号。
不能答在本试题上,否则无效。
4.回答第II卷时,须用0.5毫米黑色字迹的签字笔将答案写在答题卡上相对应的答题区域内,写在本试题上无效。
5.第I卷共三部分。
听力满分30分,不计入总分,考试成绩录取时提供给高校作参考。
第I卷第一部分听力(共两节,满分30分)做题时,先将答案标在试题上。
录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。
第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1.What are they talking about?A.Giving tips.B.Reducing the price.C.Ordering a dinner2.What time should Jenny come here?A.At 1:50pm.B.At 2:20pm.C.At 2:30pm.3.Where are the two speakers?A.At home.B.At school.C.In a library.4.How much did they pay for the repair of the bike?A.25 yuan.B.50 yuan.C.100 yuan.5.Why will Tom be invited to the party?A.Because the woman likes him.B.Because he is the man’s good friend.C.Because the man’s mother wants him to come.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。
山西省太原市2014届高三模拟考试(一)数学(文)试题
1 , a100 a96 ,则 a9 a10 an 1证明过程或演算步骤. 17. (本小题满分 12 分) 已 知 △ ABC 中 , 三 个 内 角 A , B , C 的 对 边 分 别 为 a, b, c , 若 △ ABC 的 外 接 圆 的 半 径 为 2 , 且
2 2 14 .已知 P 是直线 3x 4y 8 0 上的动点, C 是圆 x y 2x 2 y 1 0 的圆心,那么 |PC| 的最小值
是
. .
x≥0 15.已知点 P ( x, y ) 满足条件y≤x (k 为常数),若 z x 3 y 的最大值为 8,则实数 k= 2 x + y + k ≤ 0
20. (本小题满分 12 分) 已知函数 f ( x) (2 a) x 2(1 ln x) a . (I)当 a =1 时,求 f ( x ) 的单调区间; (Ⅱ)若函数 f ( x ) 在区间(0, 21. (本小题满分 12 分) 已知中心在原点 O,左右焦点分别为 F1,F2 的椭圆的离心率为 6 ,焦距为 2 2 ,A,B 是椭圆上两点. 3 1 ,求动点 P 的轨迹方程. 3 1 )无零点,求 a 的最小值. 2
(I)若直线 AB 与以原点为圆心的圆相切,且 OA⊥OB,求此圆的方程; (Ⅱ)动点 P 满足: OP = OA +3 OB ,直线 OA 与 OB 的斜率的乘积为→ → →
9.设 M ( x0 , y0 ) 为抛物线 C: x 8 y 上一点,F 为抛物线 C 的焦点,以 F 为圆心、|FM|为半径的圆和抛物线 C
2
的准线相交,则 y0 的取值范围是
山西省太原市高三语文模拟考试试题(太原三模)新人教版
太原市2014年高三年级模拟试题(三)语文试卷注意事项:1,本试卷分第Ⅰ卷(阅读题)和第Ⅱ卷(表达题)两部分,第Ⅰ卷1至8页,第Ⅱ卷9至10页。
2.答题前,考生务必将自己的姓名、准考证号填写在本试题与答题卡相应的位置。
3.全部答案在答题卡上完成,答在本试题上无效。
4.考试结束后,将本试卷和答题卡一并交回。
第Ⅰ卷阅读题甲必考题一、现代文阅读(9分,每小题3分)阅读下面的文字,完成1~3题。
过去30年,中国创造的经济奇迹如今正与令人咋舌的老龄化现象相伴而来。
过快的老龄化速度正成为中国经济增长的最大挑战。
目前,中国有13.5亿人口,并在继续保持增长,但中国人口总数预计到2030年左右将开始缓慢减少。
要认识到老龄化对中国的影响,可以先看看劳动人口与65岁以上的老年人口占中国总人口的比例。
1979年中国开始市场化改革时,劳动人口与老年人口的比例约为7:1,目前这一比例约为5.5:1。
到了2035年,劳动人口与老年人口的比例将骤降至2.5:1。
此外,劳动人口的年龄分布也很重要。
过去的研究已经证明,30岁到40岁的工作人口最有生产力和创新能力,而这正是所谓的“中国人口红利”的基础。
1979年改革以后,劳动人口的增长速度一直快于人口老化的速度,但从2015年开始,这一趋势将发生逆转。
到2035年将有1.5个年纪较大的工人(50到64岁)对一个年轻工人(15到29岁)。
这一趋势在低龄人口中也是如此,中国小学新生从1995年的2500万下降到了2008年的1670万。
这表明,中国正面临着与韩国、澳大利亚和西欧其他国家一样的老龄化处境。
中国的人口老龄化主要是因为人均寿命的显著增加和生育率的急速下降。
1980年中国的人均寿命不到65岁,目前是75岁;1980年每个妇女平均生育2.63个孩子,2011年这个数字只有1.5。
在中国的发达城市上海,生育率只有0.6,这很可能是全世界所有大城市中最低的。
这表明,中国新兴的中产阶层与他们的西方同伴一样,比起家庭更看重生活方式和职业发展。
【VIP专享】太原市2014年高三年级模拟试题(三)
太原市2014年高三年级模拟试题(三)文科综合能力测试第Ⅰ卷 (选择题 共140分)本卷共35小题,每小题4分,共140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。
24. “《礼》者,法之大分,类之纲纪也……隆礼至法则国有常,尚贤使能则民知方……”“君人者隆礼尊贤而王,重法爱民而霸……”材料反映了A .荀子的“礼法并施”B .孔子的“克己复礼” C.墨子的“尚贤使能” D .韩非的“以法治国 ”25.《新唐书·狄仁杰传》记载:“后(武则天)欲以武三思为太子……二人(狄仁杰与王方庆)同辞对曰:‘……姑侄与母子孰亲?陛下立庐陵王(武则天儿子李显),则千秋万岁后常享宗庙;三思立,庙不袱(祭祀)姑。
’后感悟,遂立庐陵王。
”材料主要反映了A .唐朝政治斗争激烈B .宗法制影响深远C .相权对皇权的制衡D .武则天善于纳谏26.按照古制,士大夫祭于庙,庶人只能祭于寝,但明嘉靖之后,商人建祠之风盛行,主要集中于商贸发达的地区。
这反映了明朝A .社会等级制度已被打破B .重农抑商政策遭到破坏C .商人地位超过士人阶层D .经济发展冲击传统观念27.顾炎武曾想革命排满,但深感社会没有力量。
后在山西见一裴姓村落,其祖先在唐代几任宰相,直至明末,还是几百几千家聚族而居。
顾由此认为,社会要封建才有力量,纵使政府垮台,依然可以反抗强敌。
材料表明顾炎武主张A .重新封邦建国B .废除君主专制 C.加强地方自治 D .发动反清复明28.《十二铜表法》中判处死刑的罪有诽谤、行贿、伪证、盗窃农产品、纵火等,但在罗马城内,被判处死刑的市民可以从任何一级起向百人会议上诉,只有独裁者的宣判不得上诉。
材料主要表明《十二铜表法》A .用刑轻罪重罚B .注重司法程序 C.维护私有财产 D .适用范围广泛29.16世纪宗教改革时,有人观察发现:“天主教徒愿意睡得香,新教教徒愿意吃得好。
”以下对材料的解读正确的是A .天主教可遏制物欲横流B .新教徒沉迷于物质享受C .追求世俗幸福推动改革D .基督教享乐思想根深蒂固30.今年又逢甲午年,在韩国,一般称中日甲午战争为“清日战争”,并定义为“东学农民运动引发的清日两国间的纷争,是无视朝鲜主权的动乱”。
山西省太原市2014年高三年级模拟语文试题
山西省太原市2014年高三年级模拟语文试题(二)第I卷阅读题甲必考题一、现代文阅读(9分,每小题3分)阅读下面的文字,完成1~3题。
春秋战国上下五百余载,是中国历史上最充满活力的黄金时代,是“礼崩乐坏,瓦釜雷鸣;高岸为谷,深谷为陵”的剧烈变化时代。
依今人之说,那是个大毁灭、大创造、大沉沦、大崛起,从而在社会整体上大转型的时代,这使得那个时代的人无论是政治家、思想家,还是军事家、教育家,是侠,是士,其生命状态无一不是饱满昂扬奋进向上的,充溢着一种不可遏止的进取精神和非凡的创造力。
那是个讲究谋略的阳谋时代,所以智慧丛生色彩斑斓;那是个本色人生的时代,所以仕学争鸣侠隐飘逸,摇唇鼓舌皆成风流;那是个实力竞争的时代,所以强国富民为本,虚伪的文过饰非的理论无法泛滥;那是个深刻思索、创造思想、成就学问、铸造精神的时代,是中华文化的原生代,所以出现了学术思想百家争鸣的灿烂辉煌的景象。
在我们耳熟能详的中国伟人中,有一半多的伟人属于那个伟大的时代,政治、经济、哲学、文学艺术、科学、神秘文化……几乎所有基本领域,都在那个时代开山立宗并创造了我们民族精神的最高经典,不仅成为我们中华民族文化的源头,而且当之无愧地进入了人类文化的殿堂。
春秋战国时代是我国历史上政治变革最为活跃的时代,五霸迭兴,三家分晋,田氏代齐,七雄兴衰,此起彼伏。
在那个波澜壮阔的时代里,教育从戎与祀的礼制中挣脱出来,孔子私学,稷下学宫,最终实现“以法为教”“以吏为师”的学吏教育制度;文学形成了中国古典文学史上第一个黄金时代。
文艺作品林林总总,诗歌、辞赋、小说、散文皆为后世之滥觞,这时期有《楚辞》的铺陈抒情、浪漫想象;有《论语》《孟子》的雍容和顺、灵活善譬;有《庄子》的想象丰富、奇气袭人;有《韩非子》的锋利峭刻、研箴析理。
艺术更见洋洋大观,青铜器绚烂多彩,金玉精琢叹为观止,铭文风韵为篆刻艺术之典范;宋音楚舞,边磬编钟,宫殿廓城,髹漆彩绘无一不在世界艺术史上熠熠生辉,光彩夺目。
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山西省太原市2014年高三年级模拟试题(一)英语第二部分:阅读理解(共两节,满分60分)第一节:(共15小题,每小题3分,满分45分)阅读下列短文,从每题所给的四个选项(A 、B、C和D)中,选出最佳选项,并在答题卡上将该项涂黑。
Some people have travelled to Canada while others may have just heard of it. As is known to all, the Canadian red and white maple leaf flag is officially called The National Flag of Canada. The Canadian flag shows a stylized red maple leaf with 11 points on a white background, with red borders down each side. The Canadian flag is twice as long as its width. The white square containing the red maple leaf is the same width as the flag. Canada is a very large country, too. It is the second largest country in the world. By contrast, it has a very small population. There are only about 29 million people there. Most Canadians are of British or French origin, and French is an official language as well English. About 45 % of the people are of British origin, that is, they or their parents or grandparents, etc, come from Britain. Nearly 30 % are of French origin. Most of theFrench-Canadians live in the province of Quebec. Over the years people have come to live in Canada from many countries in the world. They are mostly from European countries and also from China, as well as other Asian countries.However, Canada was not an empty country when the Europeans began to arrive.Canadian-Indian lived along the coast, by the rivers and lakes and in forests. Today there are only 350,000 Canadian-Indians in the whole country, with their own language. In the far north live the Intuits.There are only 27, 000 Canadian-Intuits. Their life is hard in such a harsh climate.21. What is the population of Quebec?A. More than 29, 000.000B. About 30% of the total population.C. Over 45% of the total populationD. Less than 30% of the French-Canadians.22. Which of the following stands for Canada?23. Which of the following is TRUE according to this passage?A. The Canadian flag I twice as wide as its length.B. Most Indians are now forced to live along the coast.C. Nowadays Inuit still have difficult living conditions.D. Nobody existed when the Europeans began to arrive in Canada.BThe greatest recent social changes have been in the lives of women. During the twentieth century there has been a remarkable shortening of the proportion of a woman's life spent in caring for the children. A woman marrying at the end of the nineteenth century would probably havebeen in her middle twenties, and would be likely to have seven or eight children, four or five of whom lived till they were five years old. By the time the youngest was fifteen, the mother would have been in her early fifties and would expect to live a further twenty years, during which health made it unusual for her to get paid work. Today women marry younger and have fewer children. Usually a woman's youngest child will be fifteen when she is forty-five years and can be expected to live another thirty-five years and is likely to take paid work until retirement at sixty. Even while she has to take care of children, her work is lightened by modern living conditions.This important change in women's life-patterns has only recently begun to have its full effect on women's economic position. Even a few years ago most girls took a full-time job after they left school. However, when they married, they usually left work at once and never returned to it. Today the school- leaving age is sixteen, many girls stay at school after that age, and though women usually marry older, more married women stay at work at least until shortly before their first child is born. Many more afterwards return to full-or-part-time work. Such changes have led to a new relationship in marriage, with the husband accepting a greater share of the duties and satisfactions of family life, and with both husband and wife sharing more equally in providing the money, and running the home, according to the abilities and interests of each of them.24. At what age did most women marry around the 1890 according to the passage?A. At about twenty-fiveB. In their earl fiftiesC. At the age of fifteenD. At any age from fifteen to forty-five25. What happened to an ordinary family in about 1900?A. The youngest child could live to fifteen.B. Four of five children died after they were five.C. Seven or eight children lived to be more than five.D. Some children died when they were very young.26. When she was over fifty, a late nineteenth-century mother ______.A. was usually expected to die fairly soonB. would expect to work until she diedC. would be healthy enough to take up paid jobsD. was less like to find a job even if she wanted to27. What do we know from the passage?A. Husbands and wives share equal responsibilities at home.B. More and more women are looked down upon by husbands.C. Today women prefer to get married rather than get jobs.D. A husband plays a greater part in looking after the children.CThe adder(蝰蛇)is the only poisonous snake native to Britain. Adders have the most highly developed poison injecting mechanism of all snakes, but they are not aggressive animals. Adders will only use their poison as a last means of defense, usually if caught or trodden on. No one has died from adder bites in Britain for over 20 years. By far the most common snake in Britain is the adder. In Scotland, in fact, there are no other snakes at all. The adder is also the only British snake with a poisonous bite. It can be found almost anywhere, but it prefers sunny hillsides and rough open country, including highlands. In Ireland,there are no snakes at all.Most people regard snake bites as a fatal misfortune, but not all bites are serious, and very few are fatal. Sometimes attempts at emergency treatment turn out to be more dangerous than the bite itself, with amateurs heroically, but mistakenly, trying do-it-yourself surgery and other unnecessary measures.All snakes have small teeth, so it follows the tall snakes can bite, but only the bite of the adder presents any danger. British snakes are shy animals and are far more frightened of you than you could possibly be of them. The adder will attack only if it feels threatened, as can happen if you take it by surprise and step on it accidentally, or if you try to catch it or pick it up, which it dislikes intensely. If it hears you coming, it will normally get out of the way as quickly as it can, but adders cannot move very rapidly and may attack before moving if you are very close.The effect of a bite varies considerably. It depends upon several things, one of which is the body-weight of the person bitten. The bigger the person, the less harmful the bite is likely to be, which is why children suffer far more seriously from snake bites in Britain ,and thought these bites can make some people very ill, and there are probably just as many cases of bites having little or no effect, as there are of serious illness.28. What does the underlined phrase “trodden on” probably mean?A. Found outB. Heard fromC. Stepped on.D. Robbed of.29. Where are adders to be found?A. In fertile fields in Scotland.B. On wild land throughout Britain.C. In many parts of Britain and Ireland.D. Everywhere in Britain except Scotland30. What should you do if you are with someone who is bitten by an adder?A. Catch the biting adder at once.B. Don’t worry about the bitten person.C. Don’t try to treat the bite by yourself.D. Operate on the person as soon as possible.31. What’s the best title of the passage?A. Poisonous SnakesB. Adders in BritainC. Take Care of Sake BitesD. Protect Endangered AddersDMany children first learn the value of money by receiving an allowance (pocket money). The purpose is to let children learn from experiences at an age when financial mistakes are not very costly.The amount of money that parents give to their children to spend as they wish differs from family to family. Timing is another consideration. Some children get a weekly allowance. Others get a monthly allowance.In any case, parents should make clear what, if anything, the child is expected to pay for with the money. At first, young children may spend all of their allowance soon after they receive it. If they do this, they will learn the hard way that spending must be done within a budget (预算). Parents are usually advised not to offer more money until the next allowance. The object is to show young people that a budget demands a choice between spending and saving. Older children may be responsible enough to save money for larger costs, like clothing or electronics.Many people who have written on the subject of allowances say it is not a good idea to pay your child for work around the home. These jobs are a normal part of family life. Paying children to do extra work around the house, however, can be useful. It can even provide an understanding of how a business works.Allowances give children a chance to experience the three things they can do with money. Theycan share it in the form of gifts or giving to a good cause. They can spend it by buying things they want. Or they can save it.Saving helps children understand that costly goals require sacrifice. You have to cut costs and plan for the future. Requiring children to save part of their allowances can also open the door to future saving and investing. Many banks offer services to help children and teenagers learn about personal finance. A saving account is an excellent way to learn about the power of compound interest.Compounding works by paying interest. So, for example, one dollar invested at two percent interest for two years will earn two cents in the first year. The second year, the money will earn two percent of one dollar and two cents, and so on. That may not seem like a lot, but over time it adds up.32. Parents give their children allowances in order to _____.A. show off their wealth of familyB. let them make more moneyC. learn the value of moneyD. help children manage family finance33. What is the second paragraph mainly about?A. The time to give allowances.B. The way to spend allowances.C. the amount of allowances given to children.D. Considerations taken to give allowances.34. Why are parents advised to offer young children allowances strictly as planned?A. To help children learn to make a budget.B. To have children save money for larger costs.C. To let children make fewer mistakes.D. To give them more allowances next time.35. What does the author intend to tell the readers?A. The methods of limiting allowances.B. The advice on a better family budget.C. Teaching children to use allowances wisely.D. Offering allowances to children properly.第二节(共5小题;每小题3分,满分15分)根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。