【天道原创】2013年SAT官方每日一题1月-2月汇总之数学选择题

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2013年全国各地高考数学试题及解答分类汇编大全(17 计数原理、二项式定理)

2013年全国各地高考数学试题及解答分类汇编大全（17计数原理、二项式定理）一、选择题：1．(2013江西理) ⎝⎛⎭⎫x 2－2x 35展开式中的常数项为( ) A ．80 B ．－80 C ．40 D ．－40 答案 C解析 T r ＋1＝C r 5(x 2)5－r ⎝⎛⎭⎫－2x 3r ＝C r 5(－2)r x 10－5r ，令10－5r ＝0得r ＝2.∴常数项为T 3＝C 25(－2)2＝40.2．(2013辽宁理) 使得()3nx n N n+⎛∈ ⎝的展开式中含有常数项的最小的为A ．4B ．5C ．6D ．7【答案】B 【解析】通项52(3)3n r r n rrr n rnnC x C x---=，常数项满足条件52n r =，所以2r =时5n =最小3．(2013全国大纲文) (x ＋2)8的展开式中x 6的系数是( )．A ．28B ．56C ．112D ．224 答案：C解析：T 2＋1＝28C x 8－2·22＝112x 6.故选C ．4．(2013全国大纲理) (1＋x )8(1＋y )4的展开式中x 2y 2的系数是( )．A ．56B ．84C ．112D ．168 答案：D解析：因为(1＋x )8的展开式中x 2的系数为28C ，(1＋y )4的展开式中y 2的系数为24C ，所以x 2y 2的系数为2284C C 168=.故选D.5．(2013全国新课标Ⅱ理)已知(1＋ax )(1＋x )5的展开式中x 2的系数为5，则a 等于( ) A ．－4 B ．－3 C ．－2 D ．－1 答案 D解析 (1＋ax )(1＋x )5中含x 2的项为：(C 25＋C 15a )x 2，即C 25＋C 15a ＝5，a ＝－1.6、(2013全国新课标Ⅰ理) 设m 为正整数，2()m x y +展开式的二项式系数的最大值为a ，21()m x y ++展开式的二项式系数的最大值为b ，若13a =7b ，则m ＝ ( )A 、5B 、6C 、7D 、8【命题意图】本题主要考查二项式系数最大值及组合数公式，考查方程思想，是容易题. 【解析】由题知a =2mm C ，b =121m m C ++，∴132mm C =7121m m C ++，即13(2)!!!m m m ⨯=7(21)!(1)!!m m m ⨯++，解得m =6，故选B.7．(2013山东理) 用0，1，…，9十个数字，可以组成有重复数字的三位数的个数为（A ）243 （B ）252 （C ）261 （D ）279 【答案】B【解析】有重复数字的三位数个数为91010900⨯⨯=。

2013年高考数学理知识与能力测试题

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2013年SAT真题

2013年SAT真题精选（部分）1、Jason was truly -------, for he squandered a great deal of money with no thought for the future.(A) prescient(B) infallible(C) reticent(D) improvident(E) sedulous2、The 1990`s were ------- years for the organization: the staff seas happy, customers weresatisfied, and profits were excellent.(A) halcyon(B) notorious(C) somnolent(D) capricious(E) expedient3、It is her supremely skillful use of sophisticated laboratory instruments that makes Veronica the------- research technician that she is.(A) susceptible(B) consummate(C) visionary(D) vitriolic(E) doctrinaire4、The critic noted that the ------- tone that characterizes much of the writer`s work stands in starkcontrast to his gentle disposition.(A) benign(B) somber(C) stoic(D) conciliatory(E) strident5、African American poet Lucille Clifton writes in a notably ------- style, achieving great impactin a few unadorned words.(A) incantatory(B) economical(C) disaffected(D) unstinting(E) evenhanded6、The scientist ------- the value of an interdisciplinary approach to environmental studies, arguingsuch an approach was of ------- importance in promoting environmental literacy among students.(A) invoked .. marginal(B) touted .. paramount(C) ignored .. unprecedented(D) disparaged .. unparalleled(E) extolled .. questionable7、Jessica was ------- by Jon`s angry outburst: she literally did not know what to say think, or do.(A) disenchanted(B) peeved(C) assuaged(D) beguiled(E) nonplussed8、The slogan "What goes up must come down" was so universally accepted by economists that itwas considered(A) a conjecture(B) an axiom(C) a fad(D) a testimonial(E) an argument9、Sally, thoroughly convinced of her own importance, often acts without --------: she feels noguilt for example, about appropriating her brother`s possessions.(A) compunction(B) gratification(C) aplomb(D) indignation(E) inducement10、Tchaikovsky’s Nutcracker leaves an ------ impression on audiences: children especiallyremember the dazzling costumes and stirring music.(A) amorphous(B) indelible(C) ineffable(D) innocuous(E) inscrutable。

【精校】2013年普通高等学校招生全国统一考试(北京卷)数学理

2013年普通高等学校招生全国统一考试（北京卷）数学理一、选择题共8小题，每小题5分，共40分.在每小题列出的四个选项中，选出符合题目要求的一项.1.(5分)已知集合A={-1，0，1}，B={x|-1≤x＜1}，则A∩B=( )A. {0}B. {-1，0}C. {0，1}D. {-1，0，1}解析：∵A={-1，0，1}，B={x|-1≤x＜1}，∴A∩B={-1，0}.答案：B2.(5分)在复平面内，复数(2-i)2对应的点位于( )A. 第一象限B. 第二象限C. 第三象限D. 第四象限解析：复数(2-i)2=4-4i+i2=3-4i，复数对应的点(3，-4)，所以在复平面内，复数(2-i)2对应的点位于第四象限.答案：D.3.(5分)“φ=π”是“曲线y=sin(2x+φ)过坐标原点”的( )A. 充分而不必要条件B. 必要而不充分条件C. 充分必要条件D. 既不充分也不必要条件解析：φ=π时，曲线y=sin(2x+φ)=-sin2x，过坐标原点.但是，曲线y=sin(2x+φ)过坐标原点，即O(0，0)在图象上，将(0，0)代入解析式整理即得sinφ=0，φ=kπ，k∈Z，不一定有φ=π.故“φ=π”是“曲线y=sin(2x+φ)过坐标原点”的充分而不必要条件.答案：A.4.(5分)执行如图所示的程序框图，输出的S值为( )A. 1B.C.D.解析：框图首先给变量i和S赋值0和1.执行，i=0+1=1；判断1≥2不成立，执行，i=1+1=2；判断2≥2成立，算法结束，跳出循环，输出S的值为.答案：C.5.(5分)函数f(x)的图象向右平移1个单位长度，所得图象与曲线y=e x关于y轴对称，则f(x)=( )A. e x+1B. e x-1C. e-x+1D. e-x-1解析：函数y=e x的图象关于y轴对称的图象的函数解析式为y=e-x，而函数f(x)的图象向右平移1个单位长度，所得图象与曲线y=e x的图象关于y轴对称，所以函数f(x)的解析式为y=e-(x+1)=e-x-1.即f(x)=e-x-1.答案：D.6.(5分)若双曲线的离心率为，则其渐近线方程为( )A. y=±2xB.C.D.解析：由双曲线的离心率，可知c=a，又a2+b2=c2，所以b=a，所以双曲线的渐近线方程为：y==±x.答案：B.7.(5分)直线l过抛物线C：x2=4y的焦点且与y轴垂直，则l与C所围成的图形的面积等于( )A.B. 2C.D.解析：抛物线x2=4y的焦点坐标为(0，1)，∵直线l过抛物线C：x2=4y的焦点且与y轴垂直，∴直线l的方程为y=1，由，可得交点的横坐标分别为-2，2.∴直线l与抛物线围成的封闭图形面积为=( x-)|=. 答案：C.8.(5分)设关于x，y的不等式组表示的平面区域内存在点P(x0，y0)，满足x0-2y0=2，求得m的取值范围是( )A.B.C.D.解析：先根据约束条件画出可行域，要使可行域存在，必有m＜-2m+1，要求可行域包含直线y=x-1上的点，只要边界点(-m，1-2m)，在直线y=x-1的上方，且(-m，m)在直线y=x-1的下方，故得不等式组，解之得：m＜-.答案：C.二、填空题共6小题，每小题5分，共30分.9.(5分)在极坐标系中，点(2，)到直线ρsinθ=2的距离等于.解析：在极坐标系中，点化为直角坐标为( ，1)，直线ρsinθ=2化为直角坐标方程为y=2，(，1)，到y=2的距离1，即为点到直线ρsinθ=2的距离1，答案：1.10.(5分)若等比数列{a n}满足a2+a4=20，a3+a5=40，则公比q= ；前n项和S n= . 解析：设等比数列{a n}的公比为q，∵a2+a4=20，a3+a5=40，∴，解得.∴==2n+1-2.答案：2，2n+1-2.11.(5分)如图，AB为圆O的直径，PA为圆O的切线，PB与圆O相交于D，若PA=3，PD：DB=9：16，则PD= ，AB= .解析：由PD：DB=9：16，可设PD=9x，DB=16x.∵PA为圆O的切线，∴PA2=PD·PB，∴32=9x·(9x+16x)，化为，∴.∴PD=9x=，PB=25x=5.∵AB为圆O的直径，PA为圆O的切线，∴AB⊥PA.∴==4. 答案：，4.12.(5分)将序号分别为1，2，3，4，5的5张参观券全部分给4人，每人至少1张，如果分给同一人的2张参观券连号，那么不同的分法种数是.解析：5张参观券全部分给4人，分给同一人的2张参观券连号，方法数为：1和2，2和3，3和4，4和5，四种连号，其它号码各为一组，分给4人，共有4×=96种. 答案：96.13.(5分)向量，，在正方形网格中的位置如图所示，若，则= .解析：以向量、的公共点为坐标原点，建立如图直角坐标系，可得=(-1，1)，=(6，2)，=(-1，-3)，∵，∴，解之得λ=-2且μ=-因此，==4答案：414.(5分)如图，在棱长为2的正方体ABCD-A1B1C1D1中，E为BC的中点，点P在线段D1E 上，点P到直线CC1的距离的最小值为 .解析：如图所示，取B1C1的中点F，连接EF，ED1，∴CC1∥EF，又EF⊂平面D1EF，CC1⊄平面D1EF，∴CC1∥平面D1EF.∴直线C1C上任一点到平面D1EF的距离是两条异面直线D1E与CC1的距离.过点C1作C1M⊥D1F，∵平面D1EF⊥平面A1B1C1D1.∴C1M⊥平面D1EF.过点M作MP∥EF交D1E于点P，则MP∥C1C. 取C1N=MP，连接PN，则四边形MPNC1是矩形.可得NP⊥平面D1EF，在Rt△D1C1F中，C1M·D1F=D1C1·C1F，得=.∴点P到直线CC1的距离的最小值为.答案：三、解答题共6小题，共50分.解答应写出文字说明，演算步骤15.(13分)在△ABC中，a=3，b=2，∠B=2∠A.(Ⅰ)求cosA的值；(Ⅱ)求c的值.解析：(Ⅰ)由条件利用正弦定理和二倍角公式求得cosA的值.(Ⅱ)由条件利用余弦定理，解方程求得c的值.答案：(Ⅰ)由条件在△ABC中，a=3，，∠B=2∠A，利用正弦定理可得，即=.解得cosA=. (Ⅱ)由余弦定理可得 a2=b2+c2-2bc·cosA，即 9=+c2-2×2×c×，即 c2-8c+15=0.解方程求得 c=5，或 c=3.当c=3时，此时a=c=3，根据∠B=2∠A，可得B=90°，A=C=45°，△ABC是等腰直角三角形，但此时不满足a2+c2=b2，故舍去.综上，c=5.16.(13分)如图是某市3月1日至14日的空气质量指数趋势图，空气质量指数小于100表示空气质量优良，空气质量指数大于200表示空气重度污染.某人随机选择3月1日至3月13日中的某一天到达该市，并停留2天.(Ⅰ)求此人到达当日空气重度污染的概率；(Ⅱ)设X是此人停留期间空气质量优良的天数，求X的分布列与数学期望；(Ⅲ)由图判断从哪天开始连续三天的空气质量指数方差最大？(结论不要求证明)解析：(Ⅰ)由题意此人随机选择某一天到达该城市且停留2天，因此他必须在3月1日至13日的某一天到达该城市，由图可以看出期间有2天属于重度污染，据此即可得到所求概率；(Ⅱ)由题意可知X所有可能取值为0，1，2.由图可以看出在3月1日至14日属于优良天气的共有7天.①当此人在3月4号，5号，8号，9号，10号这5天的某一天到达该城市时，停留的2天都不是优良天气；②当此人在3月3号，6号，7号，11号，这4天的某一天到达该城市时，停留的2天1不是优良天气1天是优良天气；③当此人在3月1号，2号，12号，13号，这4天的某一天到达该城市时，停留的2天都是优良天气根据以上分析即可得出P(X=0)，P(X=1)，p(x=2)及分布列与数学期望.(Ⅲ)由图判断从3月5天开始连续三天的空气质量指数波动最大，因此方差最大.答案：(Ⅰ)设“此人到达当日空气重度污染”为事件A.因为此人随机选择某一天到达该城市且停留2天，因此他必须在3月1日至13日的某一天到达该城市，由图可以看出期间有2天属于重度污染，故P(A)=.(Ⅱ)由题意可知X所有可能取值为0，1，2.由图可以看出在3月1日至14日属于优良天气的共有7天.①当此人在3月4号，5号，8号，9号，10号这5天的某一天到达该城市时，停留的2天都不是优良天气，故P(X=0)=；②当此人在3月3号，6号，7号，11号，这4天的某一天到达该城市时，停留的2天中的1天不是优良天气1天是优良天气，故P(X=1)=；③当此人在3月1号，2号，12号，13号，这4天的某一天到达该城市时，停留的2天都是优良天气，故P(X=2)=.故X的分布列如下.∴E(X)==.(Ⅲ)由图判断从3月5日开始连续三天的空气质量指数波动最大，因此方差最大.17.(14分)如图，在三棱柱ABC-A1B1C1中，AA1C1C是边长为4的正方形.平面ABC⊥平面AA1C1C，AB=3，BC=5.(Ⅰ)求证：AA1⊥平面ABC；(Ⅱ)求证二面角A1-BC1-B1的余弦值；(Ⅲ)证明：在线段BC1上存在点D，使得AD⊥A1B，并求的值.解析：(I)利用AA1C1C是正方形，可得AA1⊥AC，再利用面面垂直的性质即可证明；(II)利用勾股定理的逆定理可得AB⊥AC.通过建立空间直角坐标系，利用两个平面的法向量的夹角即可得到二面角；(III)设点D的竖坐标为t，(0＜t＜4)，在平面BCC1B1中作DE⊥BC于E，可得D，利用向量垂直于数量积得关系即可得出.答案：(I)∵AA1C1C是正方形，∴AA1⊥AC.又∵平面ABC⊥平面AA1C1C，平面ABC∩平面AA1C1C=AC，∴AA1⊥平面ABC.(II)由AC=4，BC=5，AB=3.∴AC2+AB2=BC2，∴AB⊥AC.建立如图所示的空间直角坐标系，则A1(0，0，4)，B(0，3，0)，B1(0，3，4)，C1(4，0，4)，∴，，.设平面A1BC1的法向量为，平面B1BC1的法向量为=(x2，y2，z2). 则，令y1=4，解得x1=0，z1=3，∴.，令x2=3，解得y2=4，z2=0，∴.===.∴二面角A1-BC1-B1的余弦值为.(III)设点D的竖坐标为t，(0＜t＜4)，在平面BCC1B1中作DE⊥BC于E，可得D，∴=，=(0，3，-4)，∵，∴，∴，解得t=.∴.18.(13分)设l为曲线C：y=在点(1，0)处的切线.(Ⅰ)求l的方程；(Ⅱ)证明：除切点(1，0)之外，曲线C在直线l的下方.解析：(Ⅰ)求出切点处切线斜率，代入代入点斜式方程，可以求解；(Ⅱ)利用导数分析函数的单调性，进而分析出函数图象的形状，可得结论.答案：(Ⅰ)∵∴∴l的斜率k=y′|x=1=1∴l的方程为y=x-1，(Ⅱ)令f(x)=x(x-1)-lnx，(x＞0)，曲线C在直线l的下方，即f(x)=x(x-1)-lnx＞0，则f′(x)=2x-1-=，∴f(x)在(0，1)上单调递减，在(1，+∞)上单调递增，又f(1)=0，∴x∈(0，1)时，f(x)＞0，即＜x-1，x∈(1，+∞)时，f(x)＞0，即＜x-1，即除切点(1，0)之外，曲线C在直线l的下方.19.(14分)已知A，B，C是椭圆W：上的三个点，O是坐标原点.(Ⅰ)当点B是W的右顶点，且四边形OABC为菱形时，求此菱形的面积；(Ⅱ)当点B不是W的顶点时，判断四边形OABC是否可能为菱形，并说明理由.解析：(I)根据B的坐标为(2，0)且AC是OB的垂直平分线，结合椭圆方程算出A、C两点的坐标，从而得到线段AC的长等于.再结合OB的长为2并利用菱形的面积公式，即可算出此时菱形OABC的面积；(II)若四边形OABC为菱形，根据|OA|=|OC|与椭圆的方程联解，算出A、C的横坐标满足=r2-1，从而得到A、C的横坐标相等或互为相反数.再分两种情况加以讨论，即可得到当点B不是W的顶点时，四边形OABC不可能为菱形.答案：(I)∵四边形OABC为菱形，B是椭圆的右顶点(2，0)，∴直线AC是BO的垂直平分线，可得AC方程为x=1，设A(1，t)，得，解之得t=(舍负)，∴A的坐标为(1，)，同理可得C的坐标为(1，-)，因此，|AC|=，可得菱形OABC的面积为S=|AC|·|B0|=；(II)∵四边形OABC为菱形，∴|OA|=|OC|，设|OA|=|OC|=r(r＞1)，得A、C两点是圆x2+y2=r2，与椭圆的公共点，解之得=r2-1，设A、C两点横坐标分别为x1、x2，可得A、C两点的横坐标满足x1=x2=·，或x1=·且x2=-·，①当x1=x2=·时，可得若四边形OABC为菱形，则B点必定是右顶点(2，0)；②若x1=·且x2=-·，则x1+x2=0，可得AC的中点必定是原点O，因此A、O、C共线，可得不存在满足条件的菱形OABC 综上所述，可得当点B不是W的顶点时，四边形OABC不可能为菱形.20.(13分)已知{a n}是由非负整数组成的无穷数列，该数列前n项的最大值记为A n，第n 项之后各项a n+1，a n+2…的最小值记为B n，d n=A n-B n.(Ⅰ)若{a n}为2，1，4，3，2，1，4，3…，是一个周期为4的数列(即对任意n∈N*，a n+4=a n)，写出d1，d2，d3，d4的值；(Ⅱ)设d是非负整数，证明：d n=-d(n=1，2，3…)的充分必要条件为{a n}是公差为d的等差数列；(Ⅲ)证明：若a1=2，d n=1(n=1，2，3，…)，则{a n}的项只能是1或者2，且有无穷多项为1.解析：(Ⅰ)根据条件以及d n=A n-B n 的定义，直接求得d1，d2，d3，d4的值.(Ⅱ)设d是非负整数，若{a n}是公差为d的等差数列，则a n=a1+(n-1)d，从而证得d n=A n-B n=-d，(n=1，2，3，4…).若d n=A n-B n=-d，(n=1，2，3，4…).可得{a n}是一个不减的数列，求得d n=A n-B n=-d，即 a n+1-a n=d，即{a n}是公差为d的等差数列，命题得证.(Ⅲ)若a1=2，d n=1(n=1，2，3，…)，则{a n}的项不能等于零，再用反证法得到{a n}的项不能超过2，从而证得命题.答案：(Ⅰ)若{a n}为2，1，4，3，2，1，4，3…，是一个周期为4的数列，∴d1=A1-B1=2-1=1，d2=A2-B2=2-1=1，d3=A3-B3=4-1=3，d4=A4-B4=4-1=3.(Ⅱ)充分性：设d是非负整数，若{a n}是公差为d的等差数列，则a n=a1+(n-1)d，∴A n=a n=a1+(n-1)d，B n=a n+1=a1+nd，∴d n=A n-B n=-d，(n=1，2，3，4…).必要性：若 d n=A n-B n=-d，(n=1，2，3，4…).假设a k是第一个使a k-a k-1＜0的项，则d k=A k-B k=a k-1-B k≥a k-1-a k＞0，这与d n=-d≤0相矛盾，故{a n}是一个不减的数列.∴d n=A n-B n=a n-a n+1=-d，即 a n+1-a n=d，故{a n}是公差为d的等差数列.(Ⅲ)证明：若a1=2，d n=1(n=1，2，3，…)，首先，{a n}的项不能等于零，否则d1=2-0=2，矛盾.而且还能得到{a n}的项不能超过2，用反证法证明如下：假设{a n}的项中，有超过2的，设a m是第一个大于2的项，由于{a n}的项中一定有1，否则与d1=1矛盾.当n≥m时，a n≥2，否则与d m=1矛盾.因此，存在最大的i在2到m-1之间，使a i=1，此时，d i=A i-B i=2-B i≤2-2=0，矛盾. 综上，{a n}的项不能超过2，故{a n}的项只能是1或者2.下面用反证法证明{a n}的项中，有无穷多项为1.若a k是最后一个1，则a k是后边的各项的最小值都等于2，故d k=A k-B k=2-2=0，矛盾，故{a n}的项中，有无穷多项为1.综上可得，{a n}的项只能是1或者2，且有无穷多项为1.考试高分秘诀是什么？试试这四个方法，特别是中考和高考生谁都想在考试中取得优异的成绩，但要想取得优异的成绩，除了要掌握好相关的知识定理和方法技巧之外，更要学会一些考试技巧。

2013年全国高考理科数学试题分类汇编10：排列、组合及二项式定理

2013年全国高考理科数学试题分类汇编10：排列、组合及二项式定理一、选择题1 ．（2013年普通高等学校招生统一考试新课标Ⅱ卷数学（理））已知5)1)(1(x ax ++的展开式中2x 的系数为5,则=a（）A ．4-B ．3-C ．2-D ．1-【答案】D2 ．（2013年普通高等学校招生统一考试山东数学（理））用0,1,,9十个数字,可以组成有重复数字的三位数的个数为（）A ．243B ．252C ．261D ．279【答案】B3 ．（2013年高考新课标1（理））设m 为正整数,2()m x y +展开式的二项式系数的最大值为a ,21()m x y ++展开式的二项式系数的最大值为b ,若137a b =,则m =（）A ．5B ．6C ．7D ．8【答案】B4 ．（2013年普通高等学校招生统一考试大纲版数学（理））()()8411+x y +的展开式中22x y 的系数是（）A ．56B ．84C ．112D ．168【答案】D5 ．（2013年普通高等学校招生统一考试福建数学（理））满足{},1,0,1,2a b ∈-,且关于x 的方程220ax x b ++=有实数解的有序数对(,)a b 的个数为（）A ．14B ．13C ．12D ．10【答案】B6 ．（2013年上海市春季高考）10(1)x +的二项展开式中的一项是（）A ．45xB ．290xC ．3120xD ．4252x【答案】C7 ．（2013年普通高等学校招生统一考试辽宁数学（理））使得()3nx n N n+⎛∈ ⎝的展开式中含有常数项的最小的为（）A ．4B ．5C ．6D ．7【答案】B8 ．（2013年高考四川卷（理））从1,3,5,7,9这五个数中,每次取出两个不同的数分别为,a b ,共可得到lg lg a b -的不同值的个数是（）A ．9B ．10C ．18D ．20【答案】C9 ．（2013年高考陕西卷（理））设函数61,00.,()x x f x x x ⎧⎛⎫-<⎪ ⎪=⎝≥⎭⎨⎪⎩ , 则当x>0时, [()]f f x 表达式的展开式中常数项为（）A ．-20B ．20C ．-15D ．15【答案】A10．（2013年高考江西卷（理））(x 2-32x )5展开式中的常数项为（）A ．80B ．-80C ．40D ．-40【答案】C 二、填空题11．（2013年上海市春季高考）36的所有正约数之和可按如下方法得到:因为2236=23⨯,所以36的所有正约数之和为22222222(133)(22323)(22323)(122)133)91++++⨯+⨯++⨯+⨯=++++=（参照上述方法,可求得2000的所有正约数之和为________________________【答案】483612．（2013年高考四川卷（理））二项式5()x y +的展开式中,含23xy 的项的系数是_________.(用数字作答)【答案】1013．（2013年上海市春季高考）从4名男同学和6名女同学中随机选取3人参加某社团活动,选出的3人中男女同学都有的概率为________(结果用数值表示).【答案】4514．（2013年普通高等学校招生统一考试浙江数学（理））将F E D C B A ,,,,,六个字母排成一排,且B A ,均在C 的同侧,则不同的排法共有________种(用数字作答)【答案】48015．（2013年普通高等学校招生统一考试重庆数学（理））从3名骨科.4名脑外科和5名内科医生中选派5人组成一个抗震救灾医疗小组,则骨科.脑外科和内科医生都至少有1人的选派方法种数是___________(用数字作答) 【答案】59016．（2013年普通高等学校招生统一考试天津数学（理））6x ⎛⎝的二项展开式中的常数项为______.【答案】1517．（2013年普通高等学校招生统一考试浙江数学（理））设二项式53)1(xx -的展开式中常数项为A ,则=A ________. 【答案】10-18．（2013年高考上海卷（理））设常数a R ∈,若52a x x ⎛⎫+ ⎪⎝⎭的二项展开式中7x 项的系数为10-,则______a =【答案】2a=-19．（2013年高考北京卷（理））将序号分别为1,2,3,4,5的5张参观券全部分给4人,每人至少1张,如果分给同一人的2张参观券连号,那么不同的分法种数是_________.【答案】9620．（2013年普通高等学校招生统一考试安徽数学（理））若8x⎛⎝的展开式中4x的系数为7,则实数a=______.【答案】2121．（2013年普通高等学校招生统一考试大纲版数学（理））6个人排成一行,其中甲、乙两人不相邻的不同排法共有____________种.(用数字作答).【答案】480。

2013年高考真题解析分类汇编(理科数学)10：排列、组合及二项式定理2013年高考真题解析分类汇编

2013高考试题解析分类汇编（理数）10：排列、组合及二项式定理一、选择题1 ．（2013年普通高等学校招生统一考试新课标Ⅱ卷数学（理）（纯WORD 版含答案））已知5)1)(1(x ax ++的展开式中2x 的系数为5,则=a（）A ．4-B ．3-C ．2-D ．1-D已知（1+ax ）（1+x ）5的展开式中x 2的系数为+a •=5，解得a=﹣1，故选D ．2 ．（2013年普通高等学校招生统一考试山东数学（理）试题（含答案））用0,1,,9十个数字,可以组成有重复数字的三位数的个数为（）A ．243 B ．252 C ．261 D ．279B有重复数字的三位数个数为91010900⨯⨯=。

没有重复数字的三位数有1299648C A =,所以有重复数字的三位数的个数为900648=252-，选B.仁为太傅谢安的孙子试卷试题等到平定京邑后化学教案高祖进驻石头城化学教案景仁与百官同去拜见高祖化学教案高祖注视着他3 ．（2013年高考新课标1（理））设m 为正整数,2()m x y +展开式的二项式系数的最大值为a ,21()m x y ++展开式的二项式系数的最大值为b ,若137a b =,则m =（）A ．5 B ．6 C ．7 D ．8 B因为m 为正整数，由（x+y ）2m 展开式的二项式系数的最大值为a ，以及二项式系数的性质可得a=，同理，由（x+y ）2m+1展开式的二项式系数的最大值为b ，可得 b=．再由13a=7b ，可得13=7，即 13×=7×，即 13=7×，即 13（m+1）=7（2m+1）．解得m=6，故选B ．4 ．（2013年普通高等学校招生统一考试大纲版数学（理）WORD 版含答案（已校对））()()8411+x y +的展开式中22x y 的系数是（）A ．56B ．84C ．112D ．168D（x+1）3的展开式的通项为T r+1=C 3r x r 令r=2得到展开式中x 2的系数是C 32=3，（1+y ）4的展开式的通项为T r+1=C 4r y r 令r=2得到展开式中y 2的系数是C 42=6，（1+x ）3（1+y ）4的展开式中x 2y 2的系数是：3×6=18，故选D ．5 ．（2013年普通高等学校招生统一考试福建数学（理）试题（纯WORD 版））满足{},1,0,1,2a b ∈-,且关于x 的方程220ax x b ++=有实数解的有序数对(,)a b 的个数为（）A ．14 B ．13C ．12D ．10B方程220ax x b ++=有实数解，分析讨论①当0a =时，很显然为垂直于x 轴的直线方程，有解．此时b 可以取4个值．故有4种有序数对②当0a ≠时，需要440ab ∆=-≥，即1ab ≤．显然有3个实数对不满足题意，分别为（1,2），（2,1），（2,2）．(,)a b 共有4*4=16中实数对，故答案应为16-3=13．6 ．（2013年普通高等学校招生统一考试辽宁数学（理）试题（WORD 版））使得()13nx n N n x x +⎛⎫+∈ ⎪⎝⎭的展开式中含有常数项的最小的为（）A ．4 B ．5 C ．6 D ．7B展开式的通项公式为5211(3)()3k n kn kkk n kk nnT C x C xx x---+==。

2013年普通高等学校招生全国统一考试理数(大纲卷) (含答案)

绝密★启用前2013年普通高等学校招生全国统一考试数学（理科）一、选择题：本大题共12小题，每小题5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）设集合{}{}{}1,2,3,4,5,|,,,A B M x x a b a A b B ====+∈∈则M 中元素的个数为（A ）3 （B ）4 （C ）5 （D ）6（2）()3=（A ）8- （B ）8 （C ）8i - （D ）8i （3）已知向量()()()()1,1,2,2,,=m n m n m n λλλ=+=++⊥-若则（A ）4- （B ）-3 （C ）2- （D ）-1 （4）已知函数()()()-1,021f x f x -的定义域为，则函数的定义域为（A ）()1,1- （B ）11,2⎛⎫- ⎪⎝⎭ （C ）()-1,0 （D ）1,12⎛⎫ ⎪⎝⎭（5）函数()()1=log 10f x x x ⎛⎫+> ⎪⎝⎭的反函数()1=f x - （A ）()1021x x >- （B ）()1021xx ≠- （C ）()21x x R -∈ （D ）()210xx -> （6）已知数列{}n a 满足{}12430,,103n n n a a a a ++==-则的前项和等于（A ）()-10-61-3 （B ）()-1011-39（C ）()-1031-3 （D ）()-1031+3（7）()()342211+x y x y +的展开式中的系数是（A ）56 （B ）84 （C ）112 （D ）168（8）椭圆22122:1,,46x y C A A P C PA +=的左、右顶点分别为点在上且直线斜率的取值范围是[]12,1,PA --那么直线斜率的取值范围是（A ）1324⎡⎤⎢⎥⎣⎦，（B ）3384⎡⎤⎢⎥⎣⎦，（C ）112⎡⎤⎢⎥⎣⎦，（D ）314⎡⎤⎢⎥⎣⎦，（9）若函数()211=,2f x x ax a x ⎛⎫++∞ ⎪⎝⎭在是增函数，则的取值范围是（A ）[]-1，0 （B ）[]-∞1，（C ）[]0，3 （D ）[]3∞，+ （10）已知正四棱锥1111112,ABCD A B C D AA AB CD BDC -=中，则与平面所成角的正弦值等于（A ）23 （B）3 （C）3 （D ）13（11）已知抛物线()2:82,2,C C y x M k C =-与点过的焦点，且斜率为的直线与交于,0,A B MA MB k ==u u u r u u u rg 两点,若则（A ）12（B）2 （C（D ）2（12）已知函数()=cos sin 2,f x x x 下列结论中正确的是（A ）()(),0y f x π=的图像关于中心对称（B ）()2y f x x π==的图像关于对称（C ）()f x （D ）()f x 既是奇函数，又是周期函数二、填空题：本大题共4小题，每小题5分.（13）已知1sin ,cot 3a a a =-=是第三象限角，则 .（14）6个人排成一行，其中甲、乙两人不相邻的不同排法共有种.（用数字作答）（15）记不等式组0,34,34,x x y x y ≥⎧⎪+≥⎨⎪+≤⎩所表示的平面区域为.D 若直线()1y a x D a =+与有公共点，则的取值范围是 .（16）已知圆O 和圆K 是球O 的大圆和小圆，其公共弦长等于球O 的半径，3602OK O K =o ，且圆与圆所在的平面所成角为，则球O 的表面积等于 . 三、解答题：解答应写出文字说明、证明过程或演算步骤.17．（本小题满分10分）等差数列{}n a 的前n 项和为232124.=,,,n S S a S S S 已知且成等比数列，求{}n a 的通项式.18．（本小题满分12分）设()(),,,,,.ABC A B C a b c a b c a b c ac ∆++-+=的内角的对边分别为（I ）求;B（II ）若31sin sin , C.4A C -=求19．（本小题满分12分）如图，四棱锥902,P ABCD ABC BAD BC AD PAB PAD -∠=∠==∆∆o中，，与都是等边三角形.（I ）证明：;PB CD ⊥（II ）求二面角.A PD C --的大小20．（本小题满分12分）甲、乙、丙三人进行羽毛球练习赛，其中两人比赛，另一人当裁判，每局比赛结束时，负的一方在下一局当裁判，设各局中双方获胜的概率均为1,2各局比赛的结果都相互独立，第1局甲当裁判.（I ）求第4局甲当裁判的概率；（II ）X 表示前4局中乙当裁判的次数，求X 的数学期望.21．（本小题满分12分）已知双曲线()221222:10,0x y C a b F F a b -=>>的左、右焦点分别为，，离心率为3，直线2 6.y C =与的两个交点间的距离为（I ）求,;a b ；（II ）2F l C A B 设过的直线与的左、右两支分别相交于、两点，且11,AF BF -证明：22.AF AB BF 、、成等比数列22．（本小题满分12分）已知函数()()()1=ln 1.1x x f x x xλ++-+ （I ）若()0,0,x f x λ≥≤时求的最小值;；（II ）设数列{}211111,ln 2.234n n n n a a a a n n=+++⋅⋅⋅+-+>的通项证明：2013年普通高等学校统一考试试题大纲全国理科答案一．二、填空题13. 1[,4]216. 16π三、解答题17. 设{}n a 的公差为d.由232S a =得2223a a =，故20a =或23a =.由124,,S S S ，成等比数列得2214=S S S .又12S a d =-，222S a d =-，4242S a d =+，故2222(2)()(42)a d a d a d -=-+.若20a =，则222d d =-，所以0d =，此时0n S =，不合题意；若23a =，则2(6)(3)(122)d d d -=-+，解得0d =或2d =. 因此{}n a 的通项公式为3n a =，或21n a n =-.18. （Ⅰ），因为()()a b c a b c ac ++-+=，所以222a cb ac +-=-.由余弦定理得2221cos 22a cb B ac +-==-，因此0120B =.（Ⅱ）由（Ⅰ）知060A C +=，所以 cos()cos cos sin sin A C A C A C -=+cos cos sin sin 2sin sin A C A C A C =-+cos()2sin sin A C A C =++13122-=+⨯ 3=，故030A C -=或030A C -=-，因此015C =或045C =.19. （Ⅰ）证明：取BC 的中点E ，连结DE ，则ABED 为正方形. 过P 作PO ⊥平面ABCD ，垂足为O.连结OA ，OB,OD,OE.由PAB ∆和PAD ∆都是等边三角形知PA=PB=PD ，所以OA=OB=OD ，即点O 为正方形ABED 对角线的交点，故OE BD ⊥，从而PB OE ⊥.因为O 是BD 的中点， E 是BC 的中点，所以OE//CD.因此PB CD ⊥. （Ⅱ）解法一：由（Ⅰ）知CD PB ⊥，CD PO ⊥，PB PO P =I . 故CD ⊥平面PBD.又PD ⊂平面PBD ，所以CD PD ⊥. 取PD 的中点F ，PC 的中点G ，连结FG ，则FG//CD ，FG//PD.连结AF ，由APD ∆为等边三角形可得AF ⊥PD.所以， AFG ∠为二面角A-PD-C 的平面角. ……8分连结AG ，EG ，则EG//PB. 又PB ⊥AE ，所以EG ⊥AE.设AB=2，则22AE =，112EG PB ==，故223AG AE EG =+=.在AFG ∆中，122FG CD ==，3AF =，3AG =，所以2226cos 23FG AF AG AFG FG AF +-∠==-⨯⨯.因此二面角A-PD-C 的大小为6arccos π-. 解法二：由（Ⅰ）知，OE,OB,OP 两两垂直.以O 为坐标原点， OE uuu r的方向为x 轴的正方向建立如图所示的空间直角坐标系O-xyz. 设||2AB =u u u r，则(2,0,0)A -，(0,2,0)D -，(22,2,0)C -，(0,0,2)P .(22,2,2)PC =u u u r ，(0,2,2)PD =-u u u r. 2,0,2)AP =u u u r ，(2,2,0)AD =u u u r.设平面PCD 的法向量为1(,,)n x y z =u r，则 1(,,)(22,2,2)0n PC x y z •=•--=u r u u u r， 1(,,)(0,2,2)0n PD x y z •=•--=u r u u u r，可得20x y z --=，0y z +=.取1y =-，得0,1x z ==，故1(0,1,1)n =-u r. 设平面PAD 的法向量为2(,,)n m p q =u u r，则 2(,,)(2,02)=0n AP m p q •=•u u r u u u r，，2(,,)n AD m p q •=•u u r u u u r，可得0,0m p m p +=-=.取m=1，得1,1p q ==-，故2(1,1,1)n =-u u r.于是121212cos ,3||||n n n n n n •<>=u r u u ru r u u r u r u u r .由于， 12,n n <>u r u u r等于二面角A-PD-C 的平面角，所以二面角A-PD-C的大小为arccos3π-. 20. （Ⅰ）记1A 表示事件， “第2局结果为甲胜”， 2A 表示事件“第3局甲参加比赛时，结果为甲负”， A 表示事件“第4局甲当裁判”.则12=A A A •.12121()=P()()()4P A A A P A P A •==. （Ⅱ）X 的可能取值为0,1,2.记3A 表示事件“第3局乙和丙比赛时，结果为乙胜丙”，1B 表示事件“第1局结果为乙胜丙”，2B 表示事件“第2局乙和甲比赛时，结果为乙胜甲”，3B 表示事件“第3局乙参加比赛时，结果为乙负”.则1231231(0)()()()()8P X P B B A P B P B P A ==••==13131(2)()()=4P X P B B P B P B ==•=（），115(1)1-(0)(2)1848P X P X P X ===-==--=，9()0(0)1(=1)+2(2)8E X P X P X P X =•=+••==.21.（Ⅰ）由题设知3ca=，即2229a b a +=，故228b a =.所以C 的方程为22288x y a -=.将y=2代入上式，求得x =由题设知，=21a =.所以1,a b ==（Ⅱ）由（Ⅰ）知，1(3,0)F -，2(3,0)F ，C 的方程为2288x y -=. ①由题意可设l 的方程为(3)y k x =-， ||k <，代入①并化简得2222(8)6980k x k x k --++=.设11(,)A x y ，22(,)B x y ，则11x ≤-，21x ≥，212268k x x k +=-，2122988k x x k +•=-.于是11||(31)AF x ===-+，12||31BF x ===+由11||||AF BF =得12(31)31x x -+=+，即1223x x +=-. 故226283k k =--，解得245k =，从而12199x x •=-.由于21||13AF x ===-，22||31BF x ===-.故2212||||||23()4AB AF BF x x =-=-+=，221212||||3()9-116AF BF x x x x •=+-=.因而222|||||AB|AF BF •=，所以2||AF 、||AB 、2||BF 成等比数列.22. （Ⅰ）由已知(0)0f =， 2'2(12)()(1)x x f x x λλ--=+，'(0)0f =.若12λ<，则当02(12)xλ<<-时，'()0f x>，所以()0f x>.若12λ≥，则当0x>时，'()0f x<，所以当0x>时，()0f x<.综上，λ的最小值是12.（Ⅱ）证明：令12λ=.由（Ⅰ）知，当0x>时，()0f x<，即(2)ln(1) 22x xxx+>++.取1xk=，则211ln()2(1)k kk k k++>+.于是212111()422(1)nn nk na an k k-=-+=++∑21212(1)nk nkk k-=+=+∑211lnnk nkk-=+>∑ln2lnn n=-ln2=.所以21ln24n na an-+>.。

【天道原创】SAT Essay真题汇总2013版

SAT Essay 真题汇总2013版天道教育SAT写作组编目录O FFICIAL G UIDE 官方指南题目 (4)C OLLEGE B OARD样题 (7)O NLINE C OURSE 在线练习试题 (8)历年真题(05.03-12.12) (10)2005.03 (10)2005.05 (11)2005.06 (13)2005.10 (14)2005.11 (15)2005.12 (16)2006.01 (18)2006.05 (19)2006.06 (20)2006.10 (22)2006.11 (23)2006.12 (25)2007.01 (26)2007.03 (27)2007.05 (29)2007.06 (30)2007.10 (31)2007.11 (33)2007.12 (34)2008.01 (36)2008.03 (37)2008.05 (38)2008.06 (39)2008.10 (40)2008.11 (42)2008.12 (43)2009.01 (45)2009.03 (46)2009.05 (47)2009.06 (49)2009.10 (50)2009.11 (51)2009.12 (53)2010.01 (54)2010.03 (56)2010.05 (57)2010.06 (58)2010.10 (59)2010.11 (61)2010.12 (62)2011.01 (64)2011.03 (65)2011.05 (66)2011.06 (67)2011.10 (69)2011.11 (70)2011.12 (71)2012.1 (73)2012.3 (74)2012.5 (75)2012.6 (77)2012.10 (78)2012.11 (79)2012.12 (81)Official Guide 官方指南题目Practice Test 11.Think carefully about the issue presented in the following excerpt and the assignment below To change is to risk something, making us feel insecure. Not to change is a bigger risk, though we seldom feel that way. There is no choice but to change. People, however, cannot be motivated to change from the outside. All of our motivation comes from within. Assignment:What motivates people to change?Practice Test 22.Think carefully about the issue presented in the following excerpt and the assignment below Technology promises to make our lives easier, freeing up time for leisure pursuits. But the rapid pace of technological innovation and the split second processing capabilities of computers that can work virtually nonstop have made all of us feel rushed. We have adopted the relentless pace of the very machines that were supposed to simplify our lives, with the result that, whether at work or play, people do not feel like their lives have changed for the better.Assignment: Do changes that make our lives easier not necessarily make them better?Practice Test 33.Think carefully about the issue presented in the following excerpt and the assignment belowA mistakenly cynical view of human behavior holds that people are primarily driven by selfish motives：the desire for wealth, for money, or for fame. Yet history gives us many examples of individuals who have sacrificed their own welfare for a cause or a principle that they regarded as more important than their lives. Conscience—that powerful inner voice that tells us what is right and what is wrong—can be a more compelling force than money, power, or fame.Assignment: Is conscience a more powerful motivator than money, fame, or power?Practice Test 44.Think carefully about the issue presented in the following excerpt and the assignment below The old saying, ―be careful what you wish for‖, may be an appropriate warning. The drive to achieve a particular goal can dangerously narrow one’s perspective and encourage the fantasy that success in one endeavor will solve all of life’s difficulties. In fact, success can sometimes have unexpected consequences. Those who propel themselves toward the achievement of one goal often find that their lives are worse once ―success‖ is achieved than they were before. Assignment:Can success be disastrous?Practice Test 55.Think carefully about the issue presented in the following excerpt and the assignment belowA better understanding of other people contributes to the development of moral virtues. We shall be both kinder and fairer in our treatment of others if we understand them better. Understanding ourselves and understanding others are connected, since as human beings we all have things in common.Assignment:Do we need other people in order to understand ourselves?Practice Test 66.Think carefully about the issue presented in the following excerpt and the assignment below There is, of course, no legitimate branch of science that enables us to predict the future accurately. Yet the degree of change in the world is so overwhelming and so promising that the future, I believe, is far brighter than anyone has contemplated since the end of the Second World War.Assignment:Is the world changing for the better?Practice Test 77.Think carefully about the issue presented in the following excerpt and the assignment below ―T ough challenges reveal our strengths and weaknesses.‖This statement is certainly true; adversity helps us discover who we are. Hardships can often lead us to examine who we are and to question what is important in life. In fact, people who have experienced seriously adverse events frequently report that they were positively changed by their negative experiences. Assignment:Do you think that ease does not challenge us and that we need adversity to help us discover who we are?Practice Test 88.Think carefully about the issue presented in the following excerpt and the assignment below Traditionally the term ―heroism‖ has been applied to those who have braved physical danger to defend a cause or to protect others. But one of the most feared dangers people face is that of disapproval by their family, peers, or community. Sometimes acting courageously requires someone to speak out at the risk of such rejection. We should consider those who do so true heroes.Assignment: S hould heroes be defined as people who say what they think when we ourselves lack the courage to say it?Official Guide Sample Test 19.Think carefully about the issue presented in the following excerpt and the assignment below Some people believe that there is only one foolproof plan, perfect solution, or correct interpretation. But nothing is ever that simple. For better or worse, for every so-called final answer there is another way of seeing things. There is always a ―however‖.Assignment: Is there always another explanation or another point of view?Official Guide Sample Test 210.Think carefully about the issue presented in the following excerpt and the assignment below Honesty is important, of course, but deception can actually make it easier for people to get along. In a recent study, for example, one out of every four of the lies told by participants was told solely for the benefit of another person. In fact, most lies are harmless social untruths in which people pretend to like someone or something more than they actually do (―Your muffins are the best!‖)Adapted from Allison Kornet, ―The Truth About Lying‖Assignment: Is deception ever justified?College Board样题Test 111.Think carefully about the issue presented in the following excerpt and the assignment belowA sense of happiness and fulfillment, not personal gain, is the best motivation and reward for one’s achievements. Expecting a reward of wealth or recognition for achieving a goal can lead to disappointment and frustration. If we want to be happy in what we do in life, we should not seek achievement for the sake of winning wealth and fame. The personal satisfaction of a job well done is its own reward.Assignment: Are people motivated to achieve by personal satisfaction rather than by money or fame?Test 212.Think carefully about the issue presented in the following excerpt and the assignment below Many persons believe that to move up the ladder of success and achievement, they must forget the past, repress it, and relinquish it. But others have just the opposite view. They see old memories as a chance to reckon with the past and integrate past and present.Adapted from Sara Lawrence-Lightfoot, I’ve Known Rivers: Lives of Loss and Liberation Assignment: Do memories hinder or help people in their effort to learn from the past and succeed in the present?（2005年6月真题）Test 313.Think carefully about the issue presented in the following excerpt and the assignment below Given the importance of human creativity, one would think it should have a high priority among our concerns. But if we look at the reality, we see a different picture. Basic scientific research is minimized in favor of immediate practical applications. The arts are increasingly seen as dispensable luxuries. Yet as competition heats up around the globe, exactly the opposite strategy is needed.Assignment: Is creativity needed more than ever in the world today?Test 414.Think carefully about the issue presented in the following excerpt and the assignment below Nowadays nothing is private: our culture has become too confessional and self-expressive. People think that to hide one’s thoughts or feelings is to pretend not to have those thoughtsor feelings. They assume that honesty requires one to express every inclination and impulse.Adapted from J. David V elleman, “The Genesis of Shame”Assignment: Should people make more of an effort to keep some things private? （2005年10月真题）Online Course 在线练习试题Online Course Test 115.Think carefully about the issue presented in the following excerpt and the assignment below. We often hear that we can learn much about someone or something just by casual observation. We are not required to look beneath the surface or to question how something seems. In fact, we are urged to trust our impressions, often our first impressions, of how a person or a situation seems to be. Yet appearances can be misleading. What ―seems‖ isn’t always what is. Assignment: Is the way something seems to be not always the same as it actually is?Online Course Test 216.Think carefully about the issue presented in the following excerpt and the assignment below. For a variety of reasons, people often make choices that have negative results. Later, they regret these choices, finding out too late that bad choices can be costly. On the other hand, decisions that seem completely reasonable when they are made may also be the cause of later disappointment and suffering. What looks like a wonderful idea at one time can later seem like the worst decision that could have been made. Good choices, too, can be costly. Assignment: Are bad choices and good choices equally likely to have negative consequences?Online Course Test 317.Think carefully about the issue presented in the following excerpt and the assignment below. The people we call heroes do not usually start out as unusual. Often they are ordinary people subject to ordinary human weaknesses—fear, doubt, and self-interest. In fact, they live ordinary lives until they distinguish themselves by having to deal with an injustice or a difficult situation. Only then, when they must respond in thought and in action to an extraordinary challenge, do people begin to know their strengths and weaknesses. Assignment: Do people learn who they are only when they are forced into action?Online Course Test 418.Think carefully about the issue presented in the following excerpt and the assignment below. People's lives are the result of the choices they make—or fail to make. The path one takes in life is not arbitrary. Choices and their consequences determine the course of every person's life. All people, whatever their circumstances, make the choices on which their lives depend.Assignment: Are people's lives the result of the choices they make?Online Course Test 519.Think carefully about the issue presented in the following excerpt and the assignment below. Many people believe that ―closed doors make us creative.‖ These people argue that obstacles and restrictions are necessary, for without them we would never be forced to come up with new solutions. But ―closed doors,‖ either in the form of specific obstacles or a lack of opportunities, often prevent people from reaching their full creative potential.Assignment: Do closed doors make us creative?Online Course Test 620.Think carefully about the issue presented in the following excerpt and the assignment below. People who like to think of themselves as tough-minded and realistic tend to take it for granted that human nature is ―selfish‖ and that life is a struggle in which only the fittestmay survive. According to this view, the basic law by which people must live is the law of the jungle. The ―fittest‖ are those people who ca n bring to the struggle superior force, superior cunning, and superior ruthlessness.Assignment: Do people have to be highly competitive in order to succeed?历年真题(05.03-12.12)2005.0321.Think carefully about the issue presented in the following excerpt and the assignment below. We must seriously question the idea of majority rule. The majority grinned and jeered when Columbus said the world was round. The majority threw him into a dungeon for his discoveries. Where is the logic in the notion that the opinion held by a majority of people should have the power to influence our decisions?Adapted from James A. Reed, "Majority Rule"Assignment:Is the opinion of the majority—in government or in anyother circumstances—a poor guide?22.Think carefully about the issue presented in the following excerpt and the assignment below. Even scientists know that absolute objectivity has yet to be attained. It's the same for absolute truth. But, as many news reporters have observed, the idea of objectivity as a guiding principle is too valuable to be abandoned. Without it, the pursuit of knowledge is hopelessly lost.Adapted from “Focusing Our Values,” Nieman ReportsAssignment: Are people better at making observations, discoveries, and decisions if they remain neutral and impartial?23.Think carefully about the issue presented in the following excerpt and the assignment below. If you think that what you do is your own business, you are wrong. In this world your conduct affects not only you but the conduct of other people as well. If you behave in a way that is considered unacceptable and other people copy your behavior, you are responsible for the consequences.Adapted from Margaret Banning, “Letter to Susan”Assignment: Is a person responsible, through the example he or she sets, for the behavior of other people?2005.0524.Think carefully about the issue presented in the following excerpt and the assignment below.I t's easy to see why—aside from the income it provides—having a job is so desirable in our culture. Work works for us. It structures our time and imposes a rhythm on our lives. It gets us organized into various kinds of communities and social groups. And perhaps most important, work tells us what to do every day.Adapted from Joanne B. Ciulla, The Working LifeAssignment: Do people depend on work—whether it is a job,schoolwork, or volunteer work—to determine what their dailyactivities and interactions with others should be?25.Think carefully about the issue presented in the following excerpt and the assignment below. There is no progress unless someone comes up with a new way of looking at things, of trying things that have never been done or thought of before. We cannot move forward by looking backward to old customs and past experience. There can be no advancement or improvement unless there are people who look forward in pursuit of the new and untried.Assignment: Does progress depend on people with new ideas rather than on people whose ideas are based on the current way of doing things?26.Think carefully about the issue presented in the following excerpt and the assignment below. We are afraid that our cause is unjust, or that it is unclear, or that it is too insignificant to justify the horrors of a confrontation with Authority. We will endure almost any inconvenience before undertaking head-on, I'm-here-to-tell-you complaint.Assignment: Are people afraid to speak out against authority, whether the authority is an individual, a group, or a government?27.Think carefully about the issue presented in the following excerpt and the assignment below. Alone we can afford to be wholly whatever we are and to feel whatever we feel absolutely. With others we are busy wondering what does my companion see or think of this, and what do I think of it? The original impact of our feelings gets lost or reduced.Adapted from May Sarton, The Rewards of Living a Solitary LifeAssignment: Does worrying too much about other people’s opinions prevent us from seeing things clearly?2005.0628.Think carefully about the issue presented in the following excerpt and the assignment below.I cannot comprehend those who emphasize or recognize only what is useful. I am concerned that learning for learning's sake is no longer considered desirable, that everything we do and think must be directed toward the solution of a practical problem. More and more we seem to try to teach how to make a good living and not how to live a good life.Assignment: Do people put too much emphasis on learning practical skills?29.Think carefully about the issue presented in the following excerpt and the assignment below. Most of our schools are not facing up to their responsibilities. We must begin to ask ourselves whether educators should help students address the critical moral choices and social issues ofour time. Schools have responsibilities beyond training people for jobs and getting students into college.——Svi ShapiroAssignment: Should schools help students understand moral choices and social issues?30.Think carefully about the issue presented in the following excerpt and the assignment below. The media not only transmit information and culture, they also decide what information is important. In that way, they help to shape culture and values.——Alison BernsteinAssignment: Do newspapers, magazines, television, radio, movies,the Internet, and other media determine what isimportant to most people?2005.1031.Think carefully about the issue presented in the following excerpt and the assignment below.1. Success in life is largely a matter of luck. It has little correlation with merit. And in all fields of life there have always been people of great merit who did not succeed. Karl Popper, Popper Selections2. As Colin Powell said, ―there are no secrets to success. Don’t waste time looking for them. Success is the result of preparation, hard work, and learning from failure.‖ Adapted from Barry Farber, ―Selling Points‖Assignment:Is success in life earned or do people succeed because they are lucky?32.Think carefully about the issue presented in the following excerpt and the assignment below.1. Celebrities have the power to attract ―communities‖ of like-minded followers; they provide an identity that people can connect to and call their own. Celebrities are trusted; they stand for certain ideas and values to which followers can express allegiance.Adapted from William Greider, Who Will Tell the People?2. Admiration for celebrities is often accompanied by contempt for ―average‖ people. As we focus on the famous, other people become less important to us. The world becomes populated with a few ―some bodies‖ and an excess of ―near-nobodies.‖Adapted from Norman Solomon and Jeff Cohen, Wizards of Media OzAssignment:Is society’s admiration for famous people beneficial or harmful?33.Think carefully about the issue presented in the following excerpt and the assignment below.This is a time for shallowness. Seriousness is so rare these days that wetend to make all kinds of allowances for those who only seem to possessit. In this way, shallow ideas are not recognized for what they are, and they are increasingly mistaken for deep thoughts.Adapted from Margaret Talbot, ―The Perfectionist‖Assignment: Do we live in a time when people do not engage in serious thinking?2005.1134.Think carefully about the issue presented in the following excerpt and the assignment below. Beauty is not a quality in people or in objects themselves. It exists in the mind that perceives those objects, and each mind perceives beauty differently. To seek real beauty, in some absolute sense, is pointless. Where one person sees beauty, another may even see the opposite. For this reason, we all ought to accept our own perceptions of who or what is beautiful, and not be influenced by the perceptions of others.Adapted from David Hume, ―Of the Standard of Taste‖Assignment:Should our perceptions of beauty be influenced by the perceptions of beauty of other people?35.Think carefully about the issue presented in the following excerpt and the assignment below. There are situations where flattery is mandatory: The bride is always beautiful. If we look at someone's artwork, we are obliged to say something complimentary to the artist. If we visit someone with a new baby, we are required to say the infant is cute. In such situations, to say nothing is interpreted as rudeness. We complement each other because we understand that flattery makes life run smoothly.Assignment:Is praising others, even if the praise is excessive or undeserved, a necessary part of life?36.Think carefully about the issue presented in the following excerptand the assignment below.Conflict is not necessarily bad, and it does not necessarily indicate a failed interaction. It is a signal, a message that says, "Things aren't working around here. We've got to do something different." Thus, conflict can be a catalyst-a motivating force-encouraging people to interact and communicate in ways that are more satisfying. Conflict can actually benefit people by pushing them to make necessary changes.Adapted from Beverly Potter, From Conflict to CooperationAssignment: Is conflict helpful?37.Think carefully about the issue presented in the following excerpt and the assignment below. What explains our increasing obsession with money and the things it can buy? It seems as though the acquisition of money is gradually replacing real measures of success, such as integrity, honesty, skill, and hard work.Adapted from Alan Durning, "Limiting Consumption: Toward a Sustainable Culture" Assignment: Has the acquisition of money and possessions replaced more meaningful ways of measuring our achievements?2005.1238.Think carefully about the issue presented in the following excerpt and the assignment below. We like to think that if someone has "the right stuff," he or she will naturally rise to the top. But it isn't true. In that same way that acting talent doesn't guarantee stardom, the capacity for leadership doesn't guarantee that one will run a corporation or a government. In fact, at least in our time, genuine achievement is not highly valued, and those who are skilled at achieving greatness are not necessarily those who are ready to lead.Adapted from Warren Bennis, On Becoming a LeadåAssignment: Are leaders necessarily people who are most capable of leadership?39.Think carefully about the issue presented in the following excerpt and the assignment below. How valuable is history for our generation? On the surface this question is not as easy as it once might have been, for there is a widespread belief that history may no longer be relevant to modern life. We live, after all, in an age that appears very different from the world that came before us. Adapted f rom Stephen Vaughn, ―History: Is It Relevant?‖Assignment: Is knowledge of the past no longer useful for us today?40.Think carefully about the issue presented in the following excerpt and the assignment below. The free expression of thoughts and opinions is one of humanity's most precious rights. Every citizen must be able to speak, write, and publish freely, provided that he or she is held accountable for the abuse of this liberty in cases determined by the law.Adapted from Thomas Paine, Rights of ManAssignment: Is it necessary to limit or put restrictions on freedom of thought and expression?41.Think carefully about the issue presented in the following excerpt and the assignment below. Progress is likely to slow down once science and technology have met our basic human needs. New developments in science and technology will not continue to produce more societal benefits. In fact, the promise that science and technology will continue to benefit us is increasingly doubtful when so many individuals find their lives changing in ways they cannot control and in directions they do not desire.Adapted from Daniel Sarewitz, ―Social Change and Science Policy‖Assignment: Do the benefits of scientific and technological developments come at the cost of undesirable changes to people's lives?2006.0142.Think carefully about the issue presented in the following excerpt and the assignment below.A colleague of the great scientist James Watson remarked that Watson was always ―lounging around, arguing about pr oblems instead of doing experiments.‖ He concluded that ―there is more than one way of doing good science.‖ It was Watson’s form of idleness, the scientist went on to say, that allowed him to solve ―the greatest of all biological problems: the discovery of the structure of DNA.‖ It is a point worth remembering in a society overly concerned with efficiency. Adapted from John C. Polanyi, “understanding Discovery”Assignment: Do people accomplish more when they are allowed to do things in their own way?43.Think carefully about the issue presented in the following excerpt and the assignment below.I do not feel terrible about my mistakes, though I grieve the pain they have sometimes caused others. Our lives are ―experiments with truth,‖ and in an experim ent negative results are lat least as important as successes. I have no idea how I would have learned the truth about myself andmy calling without the mistakes I have made.Adapted from Parker Palmer, Let Your Life SpeakAssignment:Is it necessary to make mistakes, even when doing so has negative consequences for other people?44.Think carefully about the issue presented in the following excerpt and assignment belowAn actor, when his cue came, was unable to move onto the stage. He said, ―I can’t get i n, the chair is in the way.‖ And the producer said, ―use the difficulty. If it’s a drama, pick the chair up and smash it. If it’s a comedy fall over it.‖ From this experience the actor concluded that in any situation in life that is negative, there is something positive you can do with it.Adapted from Lawrence Eisenberg, “Caine Scrutiny”Assignment: Can any obstacle or disadvantage be turned into something good?45.Think carefully about the issue presented in the following excerpt and assignment below Every important discovery results from patience, perseverance, and concentration--sometimes continuing for months or years--on one specific subject. A person who wants to discover a new truth must remain absorbed by that one subject, must pay no attention to any thought that is unrelated to the problem.Adapted from Santiago Ramon Cajal, Advice for a Young InvestigatorAssignment: Are all important discoveries the result of focusing on one subject?2006.0546.Think carefully about the issue presented in the following excerpt and assignment below Some people claim that each individual is solely responsible for what happens to him or her. But the claim that we ought to take absolute responsibility for the kinds of people we are and the kinds of lives we lead suggests that we have complete control over our lives. We do not. The circumstances of our lives can make it more or less impossible to make certain kinds of choices. Adapted fro m Gordon D. Marino, “I think You Should Be Responsible; Me, I’m not so Sure”Assignment: Are we free to make our own decisions or are we limited in the choices we can make?47.Think carefully about the issue presented in the following excerpt and assignment below Certainly anyone who insists on condemning all lies should think about what would happen if we could reliably tell when our family, friends, colleagues, and government leaders were deceiving us. It is tempting to think that the world would become a better place without the hand, perhaps there is such a thing as too much honesty.Adapted from Allison Kornet, “The Truth About Lying”。

2013年普通高等学校招生全国统一考试(答案)

2013年普通高等学校招生全国统一考试(课标全国卷Ⅱ)一、选择题1.C 由题意得M∩N={-2,-1,0}.选C.2.C 21+i =2(1-i)2=|1-i|=2.选C.3.B 由约束条件得可行域(如图),当直线2x-3y-z=0过点A(3,4)时,z min=2×3-3×4=-6.故选B.4.B 由正弦定理bsin B =csin C及已知条件得c=22.又sin A=sin(B+C)=12×22+32×22=2+64,从而S△ABC=12bcsin A=12×2×22×2+64=3+1.故选B.5.D 在Rt△PF2F1中,令|PF2|=1,因为∠PF1F2=30°,所以|PF1|=2,|F1F2|=3.所以e=2c2a =|F1F2||PF1|+|P F2|=33.故选D.6.A cos2 α+π4=1+cos2α+π22=1-sin2α2=16.选A.评析本题考查了三角函数的化简求值,考查了降幂公式、诱导公式的应用.7.B 由框图知循环情况为:T=1,S=1,k=2;T=12,S=1+12,k=3;T=12×3,S=1+12+12×3,k=4;T=12×3×4,S=1+12+12×3+12×3×4,k=5>4,故输出S.选B.8.D∵3<2<3,1<2<5,3>2,∴log33<log32<log33,log51<log52<log55,log23>log22,∴12<a<1,0<b<12,c>1,∴c>a>b.故选D.9.A 在空间直角坐标系中,易知O(0,0,0),A(1,0,1),B(1,1,0),C(0,1,1)恰为单位正方体的四个顶点.因此该几何体以zOx平面为投影面所得的正视图为A.评析本题考查了三视图和直观图,考查了空间想象能力.把几何体补成正方体是求解的关键.10.C 设直线AB 与抛物线的准线x=-1交于点C.分别过A,B 作AA 1垂直准线于A 1,BB 1垂直准线于B 1.由抛物线的定义可设|BF|=|BB 1|=t,|AF|=|AA 1|=3t.由三角形的相似得|BC ||AB |=|BC |4t=12,∴|BC|=2t,∴∠B 1CB=π6,∴直线的倾斜角α=π3或23π.又F(1,0),∴直线AB 的方程为y= 3(x-1)或y=- 3(x-1).故选C.11.C 由三次函数的值域为R 知, f(x)=0必有解,A 项正确;因为f(x)=x 3+ax 2+bx+c 的图象可由曲线y=x 3平移得到,所以y=f(x)的图象是中心对称图形,B 项正确;若y=f(x)有极值点,则其导数y=f '(x)必有2个零点,设为x 1,x 2(x 1<x 2),则有f '(x)=3x 2+2ax+b=3(x-x 1)(x-x 2),所以f(x)在(-∞,x 1)上递增,在(x 1,x 2)上递减,在(x 2,+∞)上递增,则x 2为极小值点,所以C 项错误,D 项正确.选C.评析本题考查了三次函数的图象和性质,考查了利用导数研究函数的单调性和极值.掌握基本初等函数的图象和性质是解题关键.12.D 由2x(x-a)<1得a>x-12,令f(x)=x-12,即a>f(x)有解,则a>f(x)min ,又y=f(x)在(0,+∞)上递增,所以f(x)>f(0)=-1,所以a>-1,选D.评析本题考查了函数的值域与最值的求法,考查了分离参变量的方法,熟悉基本初等函数的单调性是解题关键. 二、填空题 13.答案 0.2解析任取两个不同的数的情况有(1,2),(1,3),(1,4),(1,5),(2,3),(2,4),(2,5),(3,4),(3,5),(4,5),共10种,其中和为5的有2种,所以所求概率为210=0.2. 14.答案 2解析解法一:AE·BD= AD+12AB·(AD-AB)=AD2-12AB2+0=22-12×22=2.解法二:以A为原点建立平面直角坐标系(如图).则A(0,0),B(2,0),C(2,2),D(0,2),E(1,2).∴AE=(1,2),BD=(-2,2).从而AE·BD=(1,2)·(-2,2)=1×(-2)+2×2=2.评析本题考查了向量的基本运算.向量的运算可以利用运算法则也可以利用坐标运算.15.答案24π解析设底面中心为E,则|AE|=12|AC|=62,∵体积V=13×|AB|2×|OE|=|OE|=322,∴|OA|2=|AE|2+|OE|2=6.从而以|OA|为半径的球的表面积S=4π·|OA|2=24π.评析本题考查了正四棱锥和球,考查了表面积和体积,考查了空间想象能力和运算求解能力.计算错误是失分的主要原因.16.答案56π解析令y=f(x)=cos(2x+φ),将其图象向右平移π2个单位后得f x-π2=cos2 x-π2+φ =cos(2x+φ-π)=sin(2x+φ-π)+π2=sin2x+φ-π2的图象,因为其与y=sin2x+π3的图象重合,所以φ-π2=π3+2kπ(k∈Z),所以φ=2kπ+56π(k∈Z),又-π≤φ<π,所以φ=56π.三、解答题17.解析(Ⅰ)设{a n}的公差为d.由题意得,a112=a1a13,即(a1+10d)2=a1(a1+12d).于是d(2a1+25d)=0.又a1=25,所以d=0(舍去)或d=-2.故a n=-2n+27.(Ⅱ)令S n=a1+a4+a7+…+a3n-2.由(Ⅰ)知a3n-2=-6n+31,故{a3n-2}是首项为25,公差为-6的等差数列.从而S n=n2(a1+a3n-2)=n2(-6n+56)=-3n2+28n.18.解析(Ⅰ)证明:连结AC 1交A1C于点F,则F为AC1中点.又D是AB中点,连结DF,则BC1∥DF.因为DF⊂平面A1CD,BC1⊄平面A1CD,所以BC1∥平面A1CD.(Ⅱ)因为ABC-A1B1C1是直三棱柱,所以AA1⊥CD.由于AC=CB,D为AB的中点,所以CD⊥AB.又AA1∩AB=A,于是CD⊥平面ABB1A1.由AA1=AC=CB=2,AB=22得∠ACB=90°,CD=,A1D=,DE=1E=3,故A1D2+DE2=A1E2,即DE⊥A1D.所以VC-A1DE =13×12×6×3×2=1.评析本题考查了三棱柱的性质,考查了直线与平面平行的判定和体积的计算,考查了空间想象能力和运算求解能力.正确地选择方法和规范化解题至关重要.19.解析(Ⅰ)当X∈[100,130)时,T=500X-300(130-X)=800X-39 000.当X∈[130,150]时,T=500×130=65 000.所以T=800X-39000, 100≤X<130, 65000,130≤X≤150.(Ⅱ)由(Ⅰ)知利润T不少于57 000元当且仅当120≤X≤150.由直方图知需求量X∈[120,150]的频率为0.7,所以下一个销售季度内的利润T 不少于57 000元的概率的估计值为0.7.20.解析 (Ⅰ)设P(x,y),圆P 的半径为r. 由题设得y 2+2=r 2,x 2+3=r 2.从而y 2+2=x 2+3. 故P 点的轨迹方程为y 2-x 2=1. (Ⅱ)设P(x 0,y 0),由已知得00 2= 22. 又P 在双曲线y 2-x 2=1上,从而得|x 0-y 0|=1,y 02-x 02=1.由 x 0-y 0=1,y 02-x 02=1得 x 0=0,y 0=-1.此时,圆P 的半径r= 3.由 x 0-y 0=-1,y 02-x 02=1得 x 0=0,y 0=1.此时,圆P 的半径r= 3.故圆P 的方程为x 2+(y-1)2=3或x 2+(y+1)2=3. 21.解析 (Ⅰ)f(x)的定义域为(-∞,+∞), f '(x)=-e -xx(x-2).①当x∈(-∞,0)或x∈(2,+∞)时, f '(x)<0; 当x∈(0,2)时, f '(x)>0.所以f(x)在(-∞,0),(2,+∞)上单调递减,在(0,2)上单调递增.故当x=0时, f(x)取得极小值,极小值为f(0)=0;当x=2时, f(x)取得极大值,极大值为f(2)=4e -2.(Ⅱ)设切点为(t, f(t)),则l 的方程为 y=f '(t)(x-t)+f(t). 所以l 在x 轴上的截距为 m(t)=t-f (t )f '(t )=t+t t -2=t-2+2t -2+3.由已知和①得t∈(-∞,0)∪(2,+∞).令h(x)=x+2x (x≠0),则当x∈(0,+∞)时,h(x)的取值范围为[2 x∈(-∞,-2)时,h(x)的取值范围是(-∞,-3).所以当t∈(-∞,0)∪(2,+∞)时,m(t)的取值范围是(-∞,0)∪[2 +3,+∞). 综上,l 在x 轴上的截距的取值范围是(-∞,0)∪[2 2+3,+∞).评析本题考查了导数的应用,均值定理求最值,考查了综合解题的能力,正确地求导是解题的关键.22.解析(Ⅰ)证明:因为CD为△ABC外接圆的切线,所以∠DCB=∠A,由题设知BCFA =DCEA,故△CDB∽△AEF,所以∠DBC=∠EFA.因为B,E,F,C四点共圆,所以∠CFE=∠DBC,故∠EFA=∠CFE=90°.所以∠CBA=90°,因此CA是△ABC外接圆的直径.(Ⅱ)连结CE,因为∠CBE=90°,所以过B,E,F,C四点的圆的直径为CE,由DB=BE,有CE=DC,又BC2=DB·BA=2DB2,所以CA2=4DB2+BC2=6DB2.而DC2=DB·DA=3DB2,故过B,E,F,C四点的圆的面积与△ABC外接圆面积的比值为12.23.解析(Ⅰ)依题意有P(2cos α,2sin α),Q(2cos 2α,2sin 2α),因此M(cos α+cos 2α,sin α+sin 2α).M的轨迹的参数方程为x=cosα+cos2α,y=sinα+sin2α(α为参数,0<α<2π).(Ⅱ)M点到坐标原点的距离d= x2+y2=α<2π). 当α=π时,d=0,故M的轨迹过坐标原点.24.证明(Ⅰ)由a2+b2≥2ab,b2+c2≥2bc,c2+a2≥2ca得a2+b2+c2≥ab+bc+ca.由题设得(a+b+c)2=1,即a2+b2+c2+2ab+2bc+2ca=1.所以3(ab+bc+ca)≤1,即ab+bc+ca≤13.(Ⅱ)因为a 2b +b≥2a,b2c+c≥2b,c2a+a≥2c,故a 2b +b2c+c2a+(a+b+c)≥2(a+b+c),即a 2b +b2c+c2a≥a+b+c.所以a 2b +b2c+c2a≥1.。

2013年高考数学倒计时20天正能量第1辑金题强化卷05 理 (解析版)

2013年普通高等学校招生全国统一考试金题强化卷数学理（5）第I 卷一．选择题：本大题共10小题，每小题5分，共50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. 【某某省某某市2011—2012学年度高三第三次模拟测试】设{|23},{|}A x x B x x a =<<=<，若A ⊆B 则a 的取值X 围是 A ．a ≥3B ．a ≥2C ．a ≤2D ．a ≤32. 【某某市2013届高中毕业班第一次诊断性检测】函数()41log xf x x-=的定义域是 (A){}01x x <<(B){}01x x ≤< (C){}01x x <≤(D){}01x x ≤≤3. 【2012某某示X 高中高第二学期高三联考】下列函数中，在(1, 1)内有零点且单调递增的是( ) A ．12log yx B ．21xyC ．212yx D ． 3y x4. 【华中师大一附中2013届高考适应性考试】已知函数sin (0)y ax b a =+>的图象如图所示，则函数log ()a y x b =+的图象可能是( )．5. 【某某省某某市2012年普通高等学校招生适应性训练】设,a b R ，则“0,0a b ”是“2a bab ”的A ．充分条件但不是必要条件B ．必要条件但不是充分条件C ．充分必要条件D ．既不充分条件也不必要条件6.【某某省某某市2013届高三模拟考试】试题如果数列1a ，21a a ，32a a ，…，1n n a a -，…是首项为1，公比为2-的等比数列，则5a 等于（）A ．32B ．64C ．32-D ．64-7.【某某市2012届高中毕业班第二次诊断性检测】设直线l ：()110mx m y +--=（m 为常数），圆()22:1+4C x y -=，则 (A) 当m 变化时，直线L 恒过定点(-1，1)(B) 直线L 与圆C 有可能无公共点(C) 若圆C 上存在关于直线L 对称的两点，则必有m=0(D) 若直线l 与圆C 有两个不同交点M 、N ，则线段MN 的长的最小值为238. 【2012年某某市高中毕业班第一次模拟考试】已知实数x ，y 满足2003x y x y x +-≤⎧⎪-≤⎨⎪≥⎩则4z x y =+的最大值为A. 9B. 17C. 5D. 15 【答案】B【解析】画出符合约束条件的可行域如右图，当直线144zy x =-+过点A(-3，-3)时，z 的值为15，当直线144zy x =-+过点B(-3，5)时，z 的值为17，∴Z 的最大值为17，故选B 。

2013全国卷一卷二数学试卷及答案

2013全国卷一数学满分:班级：_________ 姓名：_________ 考号：_________一、单选题（共12小题）1．已知集合A=｛1，2，3，4｝，，则A∩B=（）A．｛1,4｝B．｛2，3｝C．{9，16}D．｛1,2}2．（）A．B．C．D．3．从1，2，3，4中任取2个不同的数，则取出的2个数之差的绝对值为2的概率是（）A．B．C．D．4．已知双曲线的离心率为，则的渐近线方程为（）A．B．C．D．5．已知命题，；命题，，则下列命题中为真命题的是（）A ．B ．C ．D ．6．设首项为，公比为的等比数列的前项和为，则（）A．B．C．D．7．执行右面的程序框图，如果输入的t∈[－1，3]，则输出的s属于（）A．[－3B．[－5, 4 ], 2 ]C ．[－4 , 3 ]D ．[－2 , 5 ]8．为坐标原点，为抛物线的焦点，为上一点，若，则的面积为（）A．B．C．D．9．函数在的图像大致为（）A ．B ．C ．D ．10．已知锐角的内角的对边分别为，，，，则（）A．B．C．D．11．某几何函数的三视图如图所示，则该几何的体积为（）A．B．C．D．12．已知函数，若，则的取值范围是（）A．B．C．D．二、填空题（共4小题）13.设满足约束条件，则的最大值为______。

14.已知是球的直径上一点，，平面，为垂足，截球所得截面的面积为，则球的表面积为_______。

15.设当时，函数取得最大值，则______.16.已知两个单位向量的夹角为，，若，则_____。

三、解答题（共8小题）17.已知等差数列的前项和满足，。

（Ⅰ）求的通项公式；（Ⅱ）求数列的前项和。

18.为了比较两种治疗失眠症的药（分别成为A药，B药）的疗效，随机地选取20位患者服用A药，20位患者服用B药，这40位患者服用一段时间后，记录他们日平均增加的睡眠时间（单位：h）实验的观测结果如下：服用A药的20位患者日平均增加的睡眠时间：服用B药的20位患者日平均增加的睡眠时间：（1）分别计算两组数据的平均数，从计算结果来看，哪种药的效果好？（2）完成茎叶图，从茎叶图来看，哪种药疗效更好？19.如图，三棱柱中，，，。

2013年普通高等学校招生全国统一考试数学理试题(广东卷,含答案)

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∵ A’ O⊥平面 BCDE,CD 平面 BCDE, ∴CD⊥ A’O
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(完整版)sat数学考试试题(可编辑修改word版)

SAT 数学真题精选1. If 2 x + 3 = 9, what is the value of 4 x – 3 ?(A) 5 (B) 9 (C) 15 (D) 18 (E) 212. If 4(t + u) + 3 = 19, then t + u = ?(A) 3 (B) 4 (C) 5 (D) 6 (E) 73.In the xy-coordinate (坐标) plane above, the line contains the points (0,0) and (1,2). If line M (not shown) contains the point (0,0) and is perpendicular （垂直）to L, what is an equation of M?(A) y = -1/2 x(B) y = -1/2 x + 1(C)y = - x(D)y = - x + 2(E)y = -2x4.If K is divisible by 2,3, and 15, which of the following is also divisible by these numbers?(A) K + 5 (B) K + 15 (C) K + 20 (D) K + 30 (E) K + 455.There are 8 sections of seats in an auditorium. Each section contains at least 150 seats but not more than 200 seats. Which of the following could be the number of seats in this auditorium?(A) 800 (B) 1,000 (C) 1,100 (D) 1,300 (E) 1,7006.If rsuv = 1 and rsum = 0, which of the following must be true?(A) r < 1 (B) s < 1 (C) u= 2 (D) r = 0 (E) m = 07.The least integer of a set of consecutive integers (连续整数) is –126. if the sum of these integers is 127, how many integers are in this set?(A) 126 (B) 127 (C) 252 (D) 253 (E) 2548.A special lottery is to be held to select the student who will live in the only deluxe room in a dormitory. There are 200 seniors, 300 juniors, and 400 sophomores who applied. Each senior’s name is placed in the lottery 3 times; each junior’s name, 2 time; and each sophomore’s name, 1 times. If a student’s name is chosen at random from the names in the lottery, what is the probability that a senior’s name will be chosen?（A）1/8 (B) 2/9 (C) 2/7 (D) 3/8 (E) 1/2Question #1: 50% of US college students live on campus. Out of all students living on campus, 40% are graduate students. What percentage of US students are graduate students living on campus?(A) 90% (B) 5% (C) 40% (D) 20% (E) 25% Question #2: In the figure below, MN is parallel with BC and AM/AB = 2/3. What is the ratio between the area of triangle AMN and the area of triangle ABC?(A) 5/9 (B) 2/3 (C) 4/9 (D) 1/2 (E) 2/9Question #3: If a2 + 3 is divisible by 7, which of the following values can be a?(A)7 (B)8 (C)9 (D)11 (E)4Question #4: What is the value of b, if x = 2 is a solution of equation x2 - b · x + 1 = 0?(A)1/2 (B)-1/2 (C)5/2 (D)-5/2 (E)2Question #5: Which value of x satisfies the inequality | 2x | < x + 1 ?(A)-1/2 (B)1/2 (C)1 (D)-1 (E)2Question #6: If integers m > 2 and n > 2, how many (m, n) pairs satisfy the inequality m n < 100?(A)2 (B)3 (C)4 (D)5 (E)7Question #7: The US deer population increase is 50% every 20 years. How may times larger will the deer population be in 60 years ?(A)2.275 (B)3.250 (C)2.250 (D)3.375 (E)2.500 Question #8: Find the value of x if x + y = 13 and x - y = 5.(A)2 (B)3 (C)6 (D)9 (E)4Question #9:The number of medals won at a track and field championship is shown in the table above. What is the percentage of bronze medals won by UK out of all medals won by the 2 teams?(A)20% (B)6.66% (C)26.6% (D)33.3% (E)10%Question #10: The edges of a cube are each 4 inches long. What is the surface area, in square inches, of this cube?(A)66 (B)60 (C)76 (D)96 (E)65Question #1: The sum of the two solutions of the quadratic equation f(x) = 0 is equal to 1 and the product of the solutions is equal to -20. What are the solutions of the equation f(x) = 16 - x ?(a) x1 = 3 and x2 = -3 (b) x1 = 6 and x2 = -6(c) x1 = 5 and x2 = -4 (d) x1 = -5 and x2 = 4(e) x1 = 6 and x2 = 0Question #2: In the (x, y) coordinate plane, three lines have the equations:l1: y = ax + 1l2: y = bx + 2l3: y = cx + 3Which of the following may be values of a, b and c, if line l3 is perpendicular to both lines l1 and l2?(a) a = -2, b = -2, c = .5 (b) a = -2, b = -2, c = 2(c) a = -2, b = -2, c = -2 (d) a = -2, b = 2, c = .5(e) a = 2, b = -2, c = 2Question #3: The management team of a company has 250 men and 125 women. If 200 of the managers have a master degree, and 100 of the managers with the master degree are women, how many of the managers are men without a master degree? (a) 125 (b) 150 (c) 175 (d) 200 (e) 225 Question #4: In the figure below, the area of square ABCD is equal to the sum of the areas of triangles ABE and DCE. If AB = 6, then CE =(a) 5 (b) 6 (c) 2 (d) 3 (e) 4Question #5:If α and β are the angles of the right triangle shown in the figure above, then sin2α + sin2β is equal to:(a) cos(β)(b) sin(β)(c) 1 (d) cos2(β)(e) -1 Question #6: The average of numbers (a + 9) and (a - 1) is equal to b, where a and b are integers. The product of the same two integers is equal to (b - 1)2. What is the value of a?(a) a = 9 (b) a = 1 (c) a = 0 (d) a = 5 (e) a = 11Question #1: If f(x) = x and g(x) = √x, x≥ 0, what are the solutions of f(x) = g(x)? (A) x = 1 (B)x1 = 1, x2 = -1(C)x1 = 1, x2 = 0 (D)x = 0(E)x = -1Question #2: What is the length of the arc AB in the figure below, if O is the center of the circle and triangle OAB is equilateral? The radius of the circle is 9(a) π(b) 2 ·π(c) 3 ·π(d) 4 ·π(e) π/2 Question #3: What is the probability that someone that throws 2 dice gets a 5 and a 6? Each dice has sides numbered from 1 to 6.(a)1/2 (b)1/6 (c)1/12 (d)1/18 (e)1/36 Question #4: A cyclist bikes from town A to town B and back to town A in 3 hours. He bikes from A to B at a speed of 15 miles/hour while his return speed is 10 miles/hour. What is the distance between the 2 towns?(a)11 miles (b)18 miles (c)15 miles (d)12 miles (e)10 miles Question #5: The volume of a cube-shaped glass C1 of edge a is equal to half the volume of a cylinder-shaped glass C2. The radius of C2 is equal to the edge of C1. What is the height of C2?(a)2·a/π(b)a / π(c)a / (2·π)(d)a / π(e)a + πQuestion #6: How many integers x are there such that 2x < 100, and at the same time the number 2x + 2 is an integer divisible by both 3 and 2?(a)1 (b)2 (c) 3 (d) 4 (e)5Question #7: sin(x)cos(x)(1 + tan2(x)) =(a)tan(x) + 1 (b)cos(x)(c)sin(x) (d)tan(x)(e)sin(x) + cos(x)Question #8: If 5xy = 210, and x and y are positive integers, each of the following could be the value of x + y except:(a)13 (b) 17 (c) 23 (d)15 (e)43Question #9: The average of the integers 24, 6, 12, x and y is 11. What is the value of the sum x + y?(a)11 (b)17 (c)13 (d)15 (e) 9Question #10: The inequality |2x - 1| > 5 must be true in which one of the following cases?I. x < -5 II. x > 7 III. x > 01.Three unit circles are arranged so that each touches the other two. Find the radiiof the two circles which touch all three.2.Find all real numbers x such that x + 1 = |x + 3| - |x - 1|.3.(1) Given x = (1 + 1/n)n, y = (1 + 1/n)n+1, show that x y = y x.(2) Show that 12 - 22 + 32 - 42 + ... + (-1)n+1n2 = (-1)n+1(1 + 2 + ... + n).4.All coefficients of the polynomial p(x) are non-negative and none exceed p(0). Ifp(x) has degree n, show that the coefficient of x n+1 in p(x)2 is at most p(1)2/2.5.What is the maximum possible value for the sum of the absolute values of thedifferences between each pair of n non-negative real numbers which do not exceed 1?6.AB is a diameter of a circle. X is a point on the circle other than the midpoint ofthe arc AB. BX meets the tangent at A at P, and AX meets the tangent at B at Q.Show that the line PQ, the tangent at X and the line AB are concurrent.7.Four points on a circle divide it into four arcs. The four midpoints form aquadrilateral. Show that its diagonals are perpendicular.8.Find the smallest positive integer b for which 7 + 7b + 7b2 is a fourth power.9.Show that there are no positive integers m, n such that 4m(m+1) = n(n+1).10.ABCD is a convex quadrilateral with area 1. The lines AD, BC meet at X. Themidpoints of the diagonals AC and BD are Y and Z. Find the area of the triangle XYZ.11.A square has tens digit 7. What is the units digit?12.Find all ordered triples (x, y, z) of real numbers which satisfy the following systemof equations:xy = z - x - yxz = y - x - zyz = x - y - z第11 页共11 页。

2013年普通高等学校招生全国统一考试数学理试题(大纲版)

2013年普通高等学校招生全国统一考试数学（理科）一、选择题：本大题共12小题，每小题5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）设集合{}{}{}1,2,3,4,5,|,,,A B M x x a b a A b B ====+∈∈则M 中元素的个数为（A ）3 （B ）4 （C ）5 （D ）6（2）()3=（A ）8- （B ）8 （C ）8i - （D ）8i （3）已知向量()()()()1,1,2,2,,=m n m n m n λλλ=+=++⊥-若则（A ）4- （B ）-3 （C ）2- （D ）-1 （4）已知函数()()()-1,021f x f x -的定义域为，则函数的定义域为（A ）()1,1- （B ）11,2⎛⎫- ⎪⎝⎭ （C ）()-1,0 （D ）1,12⎛⎫ ⎪⎝⎭（5）函数()()1=log 10f x x x ⎛⎫+> ⎪⎝⎭的反函数()1=f x - （A ）()1021x x >- （B ）()1021xx ≠- （C ）()21x x R -∈ （D ）()210xx -> （6）已知数列{}n a 满足{}12430,,103n n n a a a a ++==-则的前项和等于（A ）()-10-61-3 （B ）()-1011-39（C ）()-1031-3 （D ）()-1031+3 （7）()()342211+x y x y +的展开式中的系数是（A ）56 （B ）84 （C ）112 （D ）168（8）椭圆22122:1,,46x y C A A P C PA +=的左、右顶点分别为点在上且直线斜率的取值范围是[]12,1,PA --那么直线斜率的取值范围是（A ）1324⎡⎤⎢⎥⎣⎦，（B ）3384⎡⎤⎢⎥⎣⎦，（C ）112⎡⎤⎢⎥⎣⎦，（D ）314⎡⎤⎢⎥⎣⎦，（9）若函数()211=,2f x x ax a x ⎛⎫++∞ ⎪⎝⎭在是增函数，则的取值范围是（A ）[]-1，0 （B ）[]-∞1，（C ）[]0，3 （D ）[]3∞，+ （10）已知正四棱锥1111112,ABCD A B C D AA AB CD BDC -=中，则与平面所成角的正弦值等于（A ）23（B）3 （C）3 （D ）13（11）已知抛物线()2:82,2,C C y x M k C =-与点过的焦点，且斜率为的直线与交于,0,A B MA MB k ==两点,若则（A ）12（B）2 （C（D ）2（12）已知函数()=cos sin 2,f x x x 下列结论中正确的是（A ）()(),0y f x π=的图像关于中心对称（B ）()2y f x x π==的图像关于对称（C ）()f x （D ）()f x 既是奇函数，又是周期函数二、填空题：本大题共4小题，每小题5分.（13）已知1sin ,cot 3a a a =-=是第三象限角，则 .（14）6个人排成一行，其中甲、乙两人不相邻的不同排法共有种.（用数字作答）（15）记不等式组0,34,34,x x y x y ≥⎧⎪+≥⎨⎪+≤⎩所表示的平面区域为.D 若直线()1y a x D a =+与有公共点，则的取值范围是 . （16）已知圆O 和圆K 是球O 的大圆和小圆，其公共弦长等于球O 的半径，3602OK O K =，且圆与圆所在的平面所成角为，则球O 的表面积等于 . 三、解答题：解答应写出文字说明、证明过程或演算步骤. 17．（本小题满分10分）等差数列{}n a 的前n 项和为232124.=,,,n S S a S S S 已知且成等比数列，求{}n a 的通项式. 18．（本小题满分12分）设()(),,,,,.ABC A B C a b c a b c a b c ac ∆++-+=的内角的对边分别为（I ）求;B（II）若sin sin C.A C =求19．（本小题满分12分）如图，四棱锥902,P ABCD ABC BAD BC AD PAB PAD -∠=∠==∆∆中，，与都是等边三角形.（I ）证明：;PB CD ⊥（II ）求二面角.A PD C --的大小 20．（本小题满分12分）甲、乙、丙三人进行羽毛球练习赛，其中两人比赛，另一人当裁判，每局比赛结束时，负的一方在下一局当裁判，设各局中双方获胜的概率均为1,2各局比赛的结果都相互独立，第1局甲当裁判.（I ）求第4局甲当裁判的概率；（II ）X 表示前4局中乙当裁判的次数，求X 的数学期望. 21．（本小题满分12分）已知双曲线()221222:10,0x y C a b F F a b-=>>的左、右焦点分别为，，离心率为3，直线2y C =与（I ）求,;a b ；（II ）2F l C A B 设过的直线与的左、右两支分别相交于、两点，且11,AF BF -证明：22.AF AB BF 、、成等比数列22．（本小题满分12分）已知函数()()()1=ln 1.1x x f x x xλ++-+（I ）若()0,0,x f x λ≥≤时求的最小值;；（II ）设数列{}211111,ln 2.234n n n n a a a a n n=+++⋅⋅⋅+-+>的通项证明：。

2013年普通高等学校招生全国统一考试数学理试题(新课标Ⅱ卷)

2013年普通高等学校招生全国统一考试( 新课标Ⅱ)数学(理科) 第Ⅰ卷一、选择题：本大题共12小题。

每小题5分，共60分。

在每个小题给出的四个选项中，只有一项是符合题目要求的。

1．已知集合{}{}2|(1)4,,1,0,1,2,3M x x x R N =-<∈=-，则=N M（A ）{}2,1,0 （B ）{}2,1,0,1-（C ）{}3,2,0,1- （D ）{}3,2,1,0 2．设复数z 满足(1)2i z i -=，则=z（A ）i +-1（B ）i --1 （C ）i +1（D ）i -13．等比数列{}n a 的前n 项和为n S ，已知12310a a S +=，95=a ，则=1a（A ）31 （B ）31-（C ）91 （D ）91-4．已知n m ,为异面直线，⊥m 平面α，⊥n 平面β。

直线l 满足,,,l m l n l l αβ⊥⊥⊄⊄，则（A ）βα//，且α//l（B ）βα⊥，且β⊥l（C ）α与β相交，且交线垂直于l（D ）α与β相交，且交线平行于l5．已知5)1)(1(x ax ++的展开式中2x 的系数为5，则=a （A ）4-（B ）3-（C ）2-（D ）1-6．执行右面的程序框图，如果输入的10N =，那么输出的S =（A ）1111+2310+++…… （B ）1111+2310+++……！！！（C ）1111+2311+++…… （D ）1111+2311+++……！！！7．一个四面体的顶点在空间直角坐标系O xyz -中的坐标分别是(1,0,1),(1,1,0),(0,1,1),(0,0,0)，画该四面体三视图中的正视图时，以zOx 平面为投影面，则得到正视图可以为(A) (B) (C) (D)8．设357log 6,log 10,log 14a b c ===，则（A ）c b a >> （B ）b c a >> （C ）a c b >> （D ）a b c >>9．已知0a >，,x y 满足约束条件13(3)x x y y a x ≥⎧⎪+≤⎨⎪≥-⎩，若2z x y =+的最小值为1，则a =（A ）14（B ）12（C ）1（D ）210．已知函数32()f x x ax bx c =+++，下列结论中错误的是（A ）0x ∃∈R ，0()0f x =（B ）函数()y f x =的图像是中心对称图形（C ）若0x 是()f x 的极小值点，则()f x 在区间0(,)x -∞上单调递减（D ）若0x 是()f x 的极值点，则0'()0f x =11．设抛物线2:2(0)C y px p =>的焦点为F ，点M 在C 上，5MF =，若以MF 为直径的圆过点)2,0(，则C 的方程为（A ）24y x =或28y x = （B ）22y x =或28y x =（C ）24y x =或216y x = （D ）22y x =或216y x =12．已知点(1,0),(1,0),(0,1)A B C -，直线(0)y ax b a =+>将△ABC 分割为面积相等的两部分，则b 的取值范围是（A ）(0,1) (B) 1(1)2( C) 1(1]3 (D) 11[,)32 第Ⅱ卷本卷包括必考题和选考题，每个试题考生都必修作答。

2013年高考数学倒计时20天正能量第1辑金题强化卷07 理 (解析版)

2013年普通高等学校招生全国统一考试金题强化卷数学理（7）第I 卷一．选择题：本大题共10小题，每小题5分，共50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.【某某省某某市2012届高三第二次模拟】已知集合{},3M m =-,{}22730,N x x x x =++<∈Z ,如果MN ≠∅,则m 等于A ．1-B ．2-C ．2-或1-D ．32-2.【改编题】复数201321i i -（i 为虚数单位）的虚部是（）A ．15iB ．15 C ． 15i -D ．15-3. 【2012届某某某某高三二模考试】设函数)(x f 为定义在R 上的奇函数，当0x ≥时，()22x f x x b =++（b 为常数），则(1)f -=（A ）52-（B ）1-（C ）3-（D ）3 4. 【原创题】若(,)2παπ∈，且3cos2sin()4παα=-，则sin 2α的值为（）A.118B. 118-C.1718 D.1718- 【答案】D.【解析】cos 2sin(2)sin[2()]24ππααα=-=-2sin()cos()44ππαα=--代入3cos2sin()4παα=-得：1cos()46πα-=，展开2sin cos 6αα+= 平方上式得：172sin cos sin 218ααα==-. 5.【济钢高中2012届高三5月份高考冲刺题】下列结论错误的．．．是（） A ．命题“若p ，则q ”与命题“若,q ⌝则p ⌝”互为逆否命题; B ．命题:[0,1],1xp x e ∀∈≥，命题2:,10,q x R x x ∃∈++<则p q ∨为真; C ．“若22,am bm <则a b <”的逆命题为真命题; D ．若q p ∨为假命题，则p 、q 均为假命题．6. 【“华安、连城、永安、漳平一中，龙海二中，泉港一中”六校联考2012-2013学年上学期第三次月考】函数[]2()2,55f x x x x =--∈-，,定义域内任取一点0x ，使0()0f x ≤的概率是( )101.A 32.B 103.C 54.D7.【济钢高中2012届高三5月份高考冲刺题】如右图所示是某一容器的三视图，现向容器中匀速注水，容器中水面的高度h 随时间t 变化的可能图象是（）A ．B ．C ．D ．8.【原创改编题】已知点P （x ，y ）满足条件20,250,0,x y x y y a --≤⎧⎪+-≥⎨⎪-≤⎩点A （2，1），且||cos OP AOP ⋅∠的最大值为25，则a 的值是A ．-2B ．lC ．1D ．29.【某某省某某市2012届高三第二次模拟】函数()295y x =--的图象上存在不同的三点到原点的距离构成等比数列，则以下不可能成为该等比数列的公比的数是 A ．34B ．2C ．3D ．510.【某某省莱芜市2012届高三4月高考模拟试题】定义域为[a,b]的函数()y f x =图像的两个端点为A 、B ，M （x ,y ）是()f x 图象上任意一点，其中(1)[,]=+-∈x a b a b λλ,已知向量(1)ON OA OB λλ=+-，若不等式||MN k ≤恒成立，则称函数()[,]f x a b 在上“k阶线性近似”。