上海市华师大一附中等八校2012届高三2月联合调研考试卷(数学文)
2024-2025学年上海华二附中高三上学期数学月考试卷及答案(2024.09)
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1华二附中2024学年第一学期高三年级数学月考2024.09一、填空题(本大题共有12题,满分54分,第1-6题每题4分,第7-12题每题5分) 1.已知i 为虚数单位,复数12iz i+=,则z 的实部为________. 2.若函数()133x xf x a =⋅+为偶函数,则实a =________. 3.若事件A 、B 发生的概率分别为1()2P A =,2()3P B =,且相互独立,则()P A B =________.4.已知集合(){}2|log 1A y y x ==−,{}3|27B x x =≤,则A B =________.5.设{}n a 是等比数列,且13a =,2318a a +=,则n a =________.6.现有一球形气球,在吹气球时,气球的体积V 与直径d 的关系式为36d V π=,当2d =时,气球体积的瞬时变化率为________. 7.已知随机变量X 的分布为123111236⎛⎫⎪ ⎪ ⎪⎝⎭,且3Y aX =+,若[]2E Y =−,则实数a =________. 8.记函数()()()cos 0,0f x x =ω+ϕω><ϕ<π的最小正周期为T ,若()f T =,9x π=为()f x 的零点,则ω的最小值为________.9.若6(0)b ⎛> ⎝的展开式中含x 项的系数为60,则2a b +的最小值为________.10.顶点为S 的圆锥的母线长为60cm ,底面半径为25cm ,A ,B 是底面圆周上的两点,O 为底面中心,且35AOB π∠=,则在圆锥侧面上由点A 到点B 的最短路线长为____cm .(精确到0.1cm )11.已知△ABC 中,22AB BC ==,AB 边上的高与AC 边上的中线相等,则tan B =2________.12.给定公差为d 的无穷等差数列{}n a ,若存在无穷数列{}n b 满足: ①对任意正整数n ,都有1n n b a −≤②在21b b −,32b b −,…,20252024b b −中至少有1012个为正数,则d 的取值范围是________. 二、单选题(本大题共4小题,共18.0分.在每小题列出的选项中,选出符合题目的一项) 13.“1a b +>”是“33a b >”的( ) A .充分不必要条件 B .必要不充分条件C .充要条件D .既不充分也不必要条件14.如果两种证券在一段时间内收益数据的相关系数为正数,那么表明( ) A .两种证券的收益之间存在完全同向的联动关系,即同时涨或同时跌 B .两种证券的收益之间存在完全反向的联动关系,即涨或跌是相反的 C .两种证券的收益有同向变动的倾向 D .两种证券的收益有反向变动的倾向15.设0k >,若向量a 、b 、c 满足::1::3a b c k =,且2()b a c b −=−,则满足条件的k 的取值可以是( )A .1B .2C .3D .416.设1A ,1B ,1C ,1D 分别是四棱锥P ABCD −侧棱PA ,PB ,PC ,PD 上的点.给出以下两个命题,①若ABCD 是平行四边形,但不是菱形,则1111A B C D 可能是菱形;②若ABCD 不是平行四边形,则1111A B C D 可能是平行四边形.( ) A .①真②真 B .①真②假 C .①假②真 D .①假②假三、解答题(本大题共5小题,共78.0分.)17.(本小题14.0分)如图,在圆柱中,底面直径AB等于母线AD,点E在底面的圆周⊥,F是垂足.(1)求证:AF DB⊥;(2)若圆柱与三棱锥D ABE−的体积的比等于3π,求直线DE与平面ABD所成角的大小.3418.(本小题14.0分)李先生是一名上班旋,为了比较上下班的通勤时间,记录了20天个工作日内,家里到单位的上班时间以及同路线返程的下班时间(单位:分钟),如下茎叶图显示两类时间的共40个记录:(1)求出这40个通勤记录的中们数M ,并完成下列22⨯列联表:(2)根据列联表中的数据,请问上下班的通勤时间是否有显著差异?并说明理由. 附:()()()()()22n ad bc a b c d a c b d −χ=++++,()2 3.8410.05P χ≥≈.519.(本小题14.0分)如图,某城市小区有一个矩形休闲广场,20AB =米,广场的一角是半径为16米的扇形BCE 绿化区域,为了使小区居民能够更好的在广场休闲放松,现决定在广场上安置两排休闲椅,其中一排是穿越广场的双人靠背直排椅MN (宽度不计),点M 在线段AD 上,并且与曲线CE 相切;另一排为单人弧形椅沿曲线CN (宽度不计)摆放,已知双人靠背直排椅的造价每米为2a 元,单人弧形椅的造价每米为a 元,记锐角NBE ∠=θ,总造价为W 元。
文科2012年上海市虹口区高三年级二模数学(含答案)
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2012年上海各区高三数学二模真题系列卷——虹口区数学(文科)2012年上海市虹口区高三年级二模试卷——数学(文科)2012年4月一、填空题(每小题4分,满分56分)1、已知集合{}0)2)(5(<-+=x x x M ,{}51££=x x N ,则=ÇN M . 2、设i z -=1(i 为虚数单位),则=+22z z. 3、若非零向量a 、b ,满足b a =,且0)2(=×+b b a ,则a 与b 的夹角大小为的夹角大小为. 4、已知椭圆15222=+ty tx 的焦点为)6,0(±,则实数=t. 5、若等比数列{}n a 满足n n naa 91=×+,则公比=q .6、一平面截一球得到直径为2的圆面,球心到这平面的距离为3,则该球的体积是,则该球的体积是 .7、如果nxx )1(+展开式中,第4项与第6项的系数相等,则该展开式中,常数项的值是项的系数相等,则该展开式中,常数项的值是 . 8、在同一平面直角坐标系中,在同一平面直角坐标系中,函数函数)(x g y =的图像与x y 3=的图像关于直线x y =对称,而函数)(x f y =的图像与)(x g y =的图像关于y 轴对称,若1)(-=a f ,则a 的值是的值是9、在约束条件:ïïîïïíì³³£-+£-+00062062y x y x y x 下,目标函数y x z -=2的最大值为的最大值为 . 10、执行如图所示的程序框图,若输入A 的值为2,则输出的P 值是值是.11、从{1,2,3,4,5,6}中随机选一个数a ,从{1,2,3}中随机选一个数b ,则b a <的概率等于的概率等于 . 12、在ABC D 中,边2=BC ,3=AB ,则角C 的取值范围是的取值范围是. 13、函数ïîïíì<-³+=0404)(22x xx x x x x f ,则不等式5)(->x f 的解集是的解集是. 14、R b a Î,,b a >且1=ab ,则ba b a -+22的最小值等于的最小值等于 . 二、选择题(每小题5分,满分20分)15、命题A :若函数)(x f y =是幂函数,则函数)(x f y =的图像不经过第四象限.的图像不经过第四象限.那么命题那么命题A 的逆命题、的逆命题、否命题、否命题、否 是 输出p 输入A结束结束开始开始 1=p 1=S A S £1+=p p p S S 1+=正前方PDCBAOB 1D 1C 1A 1D CBADCBA主视图俯视图左视图俯视图主视图左视图左视图主视图俯视图左视图俯视图主视图16、右图是底面为正方形的四棱锥,其中棱PA 垂直于底面,它的三视图正确的是(垂直于底面,它的三视图正确的是( )17、P 为双曲线11222=-y x 上一点,1F 、2F 分别是左、右焦点,若2:3:21=PF PF ,则21F PF D 的面积是( ).A 36 .B 312 .C 12.D 24 18、等差数列{}n a 中,如果存在正整数k 和l (l k ¹),使得前k 项和l k S k =,前l 项和klS l =,则(,则( ) .A 4>+l k S B 4=+l k S .C 4<+l k S.D l k S +与4的大小关系不确定的大小关系不确定三、解答题(满分74分)分)19、(本题满分12分)在长方体1111D C B A ABCD -中,6==BC AB ,用过1A ,B ,1C 三点的平面截去长方体的一个角后,留下几何体111D C A ABCD -的体积为120. (1)求棱1AA 的长;的长;(2)若O 为11C A 的中点,求异面直线BO 与11D A 所成角的大小.所成角的大小.20、(本题满分12分)已知n m x f ×=)(,其中)1,cos 2(x m =,)2sin 3,cos (x x n =)(R x Î.(1)求)(x f 的最小正周期及单调递增区间;的最小正周期及单调递增区间;(2)在ABC D 中,a 、b 、c 分别是角A 、B 、C 的对边,若2)(=A f ,1=b ,ABC D 面积为233,求:边a 的长及ABC D 的外接圆半径R .21、(本题满分14分)已知:函数b ax ax x g ++-=12)(22)1,0(<¹b a ,在区间]3,2[上有最大值4,最小值1,设函数xx g x f )()(=.(1)求a 、b 的值及函数)(x f 的解析式;的解析式;(2)若不等式02)2(³×-xx k f 在]1,1[-Îx 时恒成立,求实数k 的取值范围;的取值范围;B 1B 2B n B n+1A n+1A nA 2A1Oyx22、(本题满分18分)已知:曲线C 上任意一点到点)0,1(F 的距离与到直线1-=x 的距离相等.的距离相等.(1)求曲线C 的方程;的方程;(2)过点)0,1(F 作直线交曲线C 于M ,N 两点,若MN 长为316,求直线MN 的方程;的方程;(3)设O 为坐标原点,如果直线)1(-=x k y 交曲线C 于A 、B 两点,是否存在实数k ,使得0=×OB OA ?若存在,求出k 的值;若不存在,说明理由.的值;若不存在,说明理由.23、(本题满分18分)如图,平面直角坐标系中,射线x y =(0³x )和0=y (0³x )上分别依次有点1A 、2A ,……,n A ,……,和点1B ,2B ,……,n B ……,其中)1,1(1A ,)0,1(1B ,)0,2(2B .且21+=-n n OA OA ,n n n n B B B B 1121-+=4,3,2(=n ……). (1)用n 表示n OA 及点n A 的坐标;的坐标;(2)用n 表示1+n n B B 及点n B 的坐标;的坐标;(3)写出四边形n n n n B B A A 11++的面积关于n 的表达式)(n S ,并求)(n S 的最大值.2012年虹口区高三年级二模数学试卷(文科)参考答案和评分标准说明:1、本解答仅列出试题的一种解法,如果考生的解法与所列解答不同,可参考解答中的评分精神进行评分.2、评阅试卷,应坚持每题评阅到底,不要因为考生的解答中出现错误而中断对该题的评阅,当考生的解答在某一步出现错误,影响了后继部分,但该步以后的解答未改变这一题的内容和难度时,可视影响程度决定后面部分的给分,这时原则上不应超过后面部分应给分数之半,如果有较严重的概念性错误,就不给分.一、填空题(每小题4分,满分56分)分)1、{}21<£x x ; 2、i -1; 3、°120; 4、2,3; 5、3; 6、31040p ; 7、70; 8、 31- ;9、6; 10、4;11、61; 12、]3,0(p ; 13、),1(¥+-; 14、22二、选择题(每小题5分,满分20分)分)15、C ; 16、B ; 17、C ; 18、A ; 三、解答题(满分74分)分) 19、(12分)(1)设h AA =1,12062131622=×××-×=h h V\41==h AA …………6分(2) 11//D A BC ,\OBC Ð是所求异面直线所成的角…………8分 在OBC D 中,34)23(422=+==OC OB ,6=BC ,\34343arccos=ÐOBC …………12分 20、(12分)(1)1)62sin(22sin 3cos 2)(2++=+=px x x x f …………2分p =T ………………3分单调递增区间]6,3[pp pp +-k k )(Z k Î……………4分(2)21)62sin(2)(=++=pA A f ,由21)62sin(=+pA ,得3p=A …………6分2333sin 121=´´´pc ,\6=c …………8分3sin 31sin 2p==Aa R ,\393=R …………12分21、(14分)(1)b ax ax x g ++-=12)(22,由题意得:,由题意得:°1ïîïíì=++==+=>413)3(11)2(0b a g b g a 得îíì==01b a , 或 °2 ïîïíì=++==+=<113)3(41)2(0b a g b g a 得îíì>=-=131b a (舍去)(舍去) \1=a ,0=b …………6分12)(2+-=x x x g ,21)(-+=xx x f …………7分(2)不等式02)2(³×-xx k f ,即xx x k 22212׳-+,\1)21(2)21(2+×-£x x k ……10分设]2,21[21Î=xt ,\2)1(-£t k , 0)1(min2=-t ,\0£k …………14分22、(18分)(1)x y 422=…………4分(2)当直线MN 的斜率不存在时,不合题意。
2012年(全国卷II)(含答案)高考文科数学
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上海市华师大一附中等八校2012届高三2月联合调研考试卷(数学理)
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2012届高三联合调研考试数学试卷(理科)(本试卷满分150分,测试时间120分钟)参加学校:华师大一附中、曹杨二中、市西、市三女子、控江、格致、市北、(育才、晋元高中)一、填空题(本大题共56分,每小题4分)1.计算:1i2i-=+____________ (其中i 为虚数单位). 2.已知向量(5,3)a =-,(2,)b x =,若向量a 、b 互相平行,则x =____________. 3.已知向量a 与b 的夹角为3π,||1a =,||2b =,若b a λ-与a 垂直,则实数λ=_________. 4.在二项式8(ax -的展开式中,若含2x 项的系数为70,则实数a =_____________. 5.已知θ是第二象限角,若4sin 5θ=,则tan()24θπ-的值为_______________. 6.若21316log 1a a M a -+=-,[4,17]a ∈,则M 的取值范围是_________________.7.关于x 的方程组(1)21y q x y qx =-+⎧⎪⎨=-⎪⎩有唯一的一组实数解,则实数q 的值为_____________. 8.把编号为1、2、3、4、5的5位运动员排在编号为1、2、3、4、5的5条跑道中,若有且只有两位运动员的编号与其所在跑道编号相同,则不同的排法种数共有___________种. 9.过点1(,1)2M 的直线l 与圆C :22(1)4x y -+=交于A 、B 两点,C 为圆心,当ACB ∠最小时,直线l 的方程为_________________.10.在平面直角坐标系xOy 中,函数()()1f x k x =-(1k >)的图像与x 轴交于点A ,它的反函数()1y fx -=的图像与y 轴交于点B ,并且这两个函数的图像交于点P .若四边形OAPB 的面积是3,则k =___________.11.已知Z k ∈,向量(,1)AB k =,(2,4)AC =,若||10AB ≤,则ABC ∆为直角三角形的概率是_______________.12.已知ABC ∆中,2AC =,1BC =,23ACB π∠=,D 为AB 上的点,若2AD DB =,则CDB ∠=____________(结果用反三角表示).13.设直线:l 220x y ++=关于原点对称的直线为l ',若l '与椭圆2214y x +=的交点为A ,B 两点,点P 是椭圆上的动点,则使PAB ∆的面积为12的点P 的个数为_____________.14.如图所示的程序框图中, ,函数int()x 表示不超过x 的最大整数,则由框图给出的计算结果是____________.二、选择题(本大题满分20分,每小题5分)15.若函数21y a x =⋅,22x y c =⋅,33y b x =⋅,则由表中数据确定()f x 、()g x 、()h x 依次对应 ( ). (A) 1y 、2y 、3y (B) 2y 、1y 、3y (C ) 3y 、2y 、1y (D) 1y 、3y 、2y 16.在证券交易过程中,常用到两种曲线,即时价格曲线()y f x =及平均价格曲线()y g x = (如(2)3f =是指开始买卖后二个小时的即时价格为3元;(2)3g =表示二个小时内的平均价格为3元),在下图给出的四个图像中实线表示()y f x =,虚线表示()y g x =其中可能正确的是 ( ).(A ) (B ) (C ) (D )17.一个四棱锥和一个三棱锥恰好可以拼接成一个三棱柱.这个四棱锥的底面为正方形, 且底面边长与各侧棱长相等,这个三棱锥的底面边长与各侧棱长也都相等.设四棱锥、 三棱锥、三棱柱的高分别为1h ,2h ,h ,则12::h h h = ( ).2:22218.函数()y f x =的定义域为R ,若对于任意的正数a ,函数()()()g x f x a f x =+-都是其定义域上的增函数,则函数()y f x =的图像可能是 ( ).(A ) (B) (C) (D)三、解答题(本大题满分74分)19.(本题满分12分)第1小题满分6分,第2小题满分6分. 已知函数()()22sin cos 2cos 2f x x x x =++-.(1)求函数()f x 的最小正周期; (2 )当3,44x ππ⎡⎤∈⎢⎥⎣⎦时,求函数()f x 的最大值,最小值.20.(本题满分12分)第1小题满分6分,第2小题满分6分.在一个棱长为2+的正方体的八个顶角上分别截去一个三棱锥,使截掉棱锥后的多面体有六个面为正八边形,八个面为正三角形(如图所示),(1)求异面直线AB 与GH 所成角的大小; (2)求此多面体的体积(结果用最简根式表示).21.(本题满分12分)第1小题满分5分,第2小题满分7分.已知O 为坐标原点,点(2,1),(1,2)A B ,对于k N *∈有向量k OP kOB OA =+, (1)试问点k P 是否在同一条直线上,若是,求出该直线的方程;若不是,请说明理由;(2)是否在存在k N *∈使k P 在圆22(2)5x y +-=上或其内部,若存在求出k ,若不存在说明理由.22.(本题满分19分)第1小题满分5分,第2小题满分5分,第3小题满分9分. 已知函数()y f x =的图像(如图所示)过点(0,2)、(1.5,2)和点(2,0),且函数图像关于点(2,0)对称;直线1x =和3x =及0y =是它的渐近线.现要求根据给出的函数图像研究函数1()()g x f x =的相关性质与图像, (1)写出函数()y g x =的定义域、值域及单调递增区间;(2)作函数()y g x =的大致图像(要充分反映由图像及条件给出的信息);(3)试写出()y f x =的一个解析式,并简述选择这个式子的理由(按给出理由的完整性及表达式的合理、简洁程度分层给分).23.(本题满分19分)第1小题满分5分,第2小题满分8分,第3小题满分6分. 由下面四个图形中的点数分别给出了四个数列的前四项,将每个图形的层数增加可得到这四个数列的后继项.按图中多边形的边数依次称这些数列为“三角形数列”、“四边形数列”,将构图边数增加到n 可得到“n 边形数列”,记它的第r 项为(,)P n r ,1,3,6,10 1,4,9,16 1,5,12,22 1,6,15,28 (1) 求使得(3,)36P r >的最小r 的取值; (2) 试推导(,)P n r 关于n 、r 的解析式;( 3) 是否存在这样的“n 边形数列”,它的任意连续两项的和均为完全平方数,若存在,指出所有满足条件的数列并证明你的结论;若不存在,请说明理由.参考答案: 1.13i 55- 2. 65- 3. 1 4. 1± 5. 136. 3[2log 2,2]--- (或3[log 18,2]--等7.12或 18.209.2430x y -+= (或11()(1)022x y ---=等)10.32 11. 41912. 13.2 14.1 15.D 16.C 17.B 18.D19. 解: (1)()sin 2cos 224f x x x x π⎛⎫=+=+ ⎪⎝⎭. 4分∴()f x 的最小正周期为π. 6分(2).337,,244444x x πππππ⎡⎤∈∴≤+≤⎢⎥⎣⎦, 8分,1sin 242x π⎛⎫∴-≤+≤⎪⎝⎭ 10分∴()1f x ≤≤. 12分∴当3,44x ππ⎡⎤∈⎢⎥⎣⎦时,函数()f x 的最大值为1,最小值.20. 解: (1) 易知//FE AB ,//GH EC ,所以FEC ∠就是异面直线AB 与GH 所成的余角). 3分经计算得: 351)48FEC ππ∠=-=(也可以直接用4522.567.5+=做)所以异面直线AB 与GH 所成的角的大小为38π1),arc . 6分(2,则由题意得:2x x +=,所以, 9分 设多面体的体积为V ,则311(2832V =+-⨯⨯=563+ 12分21.解:(1)点k P 在同一条直线上,直线方程为23y x =-. 2分证明如下:设点(,)k k k P x y ,则(,)(1,2)(2,1)k k x y k =+ 即2,21,k kx k y k =+⎧⎨=+⎩所以23k k y x =-.所以,点k P 在直线23y x =-上. 5分 (文科)按证明情况酌情给分(2)由圆22(2)5x y +-=的圆心(0,2)到直线23y x =-=可知直线与圆相切, 所以直线与圆及内部最多只有一个公共点 10分而切点的坐标为:(2,1),此时0k =不满足题意,所以不存在k N *∈满足题意. 12分22.解: (1) 定义域为:{|1,2,3,}x x x x x R ≠≠≠∈ 2分 值域为: (,0)(0,)-∞⋃+∞ 3分 函数的单调递增区间为: (1,2)和(2,3) 5分 (2)图像要求能反映出零点((1,0)和(3,0),渐近线2x =,过定点,单调性正确. 5分(3) 结论可能各异如:3(2)()|1||3|x f x x x -=--,2222(3)()22(1)xx x f x x x x -⎧>⎪-⎪=⎨-⎪<⎪-⎩211()2tan()132233x x f x x x x x π-⎧<⎪-⎪⎪=-<<⎨⎪-⎪>⎪-⎩,等层次一:函数图像能满足题意, 但没有说明理由 4分 层次二: 函数图像能满足题意,能简述理由(渐近线、定点等部分内容) 6分层次三: 函数图像能满足题意,能说明过定点、渐近线、单调性及对称性 9分解: (1)(1)(3,)2r r P r +=, 3分 由题意得(1)362r r +>,所以,最小的9r =. 5分(2)设n 边形数列所对应的图形中第r 层的点数为r a ,则12(,)r P n r a a a =++⋅⋅⋅+ 从图中可以得出:后一层的点在2n -条边上增加了一点,两条边上的点数不变, 所以12r r a a n +-=-,11a =所以{}r a 是首项为1公差为2n -的等差数列, 所以(,)[2(1)(2)]2r P n r r n =+--.(或(2)(1)2n r r r --+等) 13分 (3)2(,1)(,)(2)21P n r P n r n r r ++=-++ 16分 显然3n =满足题意, 17分 而结论要对于任意的正整数r 都成立,则2(2)21n r r -++的判别式必须为零, 所以,44(2)0n --=,3n = 19分所以,满足题意的数列为“三角形数列”.(文科)(2)为第50项,(3)同理科(2).。
上海市华师大一附中等八校高三2月联合调研考试卷英语
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上海2012届高三下学期(八)校联合调研考试英语试卷(八校联考)2012.2.16 参加学校:华师大一附中、曹杨二中、市西、市三女子、控江、格致、市北、(育才、晋元高中)第I卷(共105 分)II. Grammar and VocabularySection A: GrammarDirections: Beneath each of the following sentences there are four choices marked A, B, C and D. Choose the one answer that best completes the sentence.25. Keeping a foreign pet is dangerous, as the owner might be infected ________ the diseases brought along with the animal.A. inB. withC. byD. for26. All the ladies wear beautiful clothes for the celebration. Some are dressed in red, _______ in purple.A. otherB. anotherC. othersD. the other27. According to some theories ________ from psychoanalysis, life is supposedly easier; and more pleasant when anxiety is overcome.A. obtainB. obtainingC. to obtainD. obtained28. To understand the complex situation completely requires more thought than ________ so far.A. gaveB. has givenC. has been givenD. is being given29. To show our respect, we usually take our gloves off _______ we are to shake hands with.A. whoeverB. wheneverC. whicheverD. wherever30. You _______ an A on your term paper, but you quoted a little bit more from others' papers.A. should receiveB. could have receivedC. would receiveD. must have received31. Eric, a middle-aged fashion designer, is very happy because the clothes be designed have never been ________.A. most popularB. the most popularC. less popularD. more popular32. We are strongly against the company's decision _______ it will fire half of its staff in the following years.A. whichB. thatC. whatD. when33. Nancy, ________ for about half a year to apply for a job as an airline hostess, finally took a position at a shopping center.A. struggledB. having struggledC. strugglingD. to struggle34. Don’t start reading a book ________ you find that it’s one you can read with ease and understanding.A. whenB. ifC. unlessD. though35. I can’t believe that anyone else in the school can swim as fast as you, ________?A. can IB. can't theyC. can't youD. can they36. One of the things that made me long ________ back in prison was that I had little opportunity thinking and reflection.A. beB. beingC. to beD. to being37. When you have a job interview, it's ________ you're dressed that sets the tone of it.A. howeverB. howC. whatD. whatever38. The new lecture course differs from the old one, ________ the students aren't required to attend lectures.A. which thatB. thatC. whichD. in which39. Fortunately, what the witness reported to the police on the phone led to ___________.A. the suspect's being capturedB. capture the suspectC. the suspect having capturedD. be captured by the suspect40. _______ they may have, the CEO of the corporation calls on all the staff to gather their courage to strive it.A. What a serious problemB. However a serious problemC. How serious a problemD. Whatever a serious problemSection B: VocabularyDirections: Complete the passage by choosing the proper words in the box. Each word can onlyRealizing your dreams is not an easy ___41___.Last fall I decided to write a new book for my publisher. Writing a book is a ___42___ goal, which got off to a terrific start last October. The writing is flowing well. Then I got sick. In fact, I got ill that I needed surgery and the ___43___ was long and exhausting. I did not work from the first week of November until the second week of January. By then I was nervous about meeting the April deadline for submitting the new manuscript to my publisher.Worried, I asked my author friends for some help, and they gave me this key piece of advice, “Let’s start writing!" they all said. So I did. It was not an immediate ___44___ t o my depression, but after a few weeks of ___45___, I got back to normal. Several people in my circle of supporters helped me make some good connections and I got the book ___46___, and to the publisher on time. It was an exciting goal for me to reach, so I took my family to Hawaii as my ___47___. Sometimes you ___48___ your own dreams because of self-doubt, fear, or external complications. You can think of many different excuses to ___49___ those dreams aside, but, if you go after your dreams, your world will become more exciting and you will begin to live a more passionate and meaningful life.So, now, take a minute to write down three goals you want to accomplish this year.III. Reading ComprehensionSection ADirections: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.In 1867 the United States faced the task of rebuilding after the destruction of the Civil War, so it looked westward for the raw materials needed to fuel industrial growth. Geological surveys and mapping journeys were set forth to explore this 50 territory. These groups, in turn, hired mapmakers, scientists, cooks, drivers, and doctors. They also hired painters and photographers as part of the teams. Painters needed few supplies, making it relatively easy for them to travel in the wilderness, ____51____ photographers were not so lucky; they had to transport a fully stocked darkroom on the ___52____.Until the late 1870s, most photographers used the ___53___ wet-collodion process. The first step was to wash a clean sheet of glass with a sticky mixture of collodion and chemicals, (collodion or "gun-cotton" was a recent medical discovery used to cover wounds because the viscous (粘性的) solution turned into a protective film when dry.) After it was washed, the plate went into another bath that the picture was ___54___ getting darker; Finally, the glass negative (底片) was washed clean with fresh water. ___55___ a photograph from the negative had to wait until the photographer went back to the studio. The ___56___ of the negative depended on the size of the camera. Some negatives could be as large as 20 by 24 inches.Imagine the ___57____ of taking photographs in the 1860s and 1870s in the remote western wilderness! Photographers went over Rocky Mountains and through rushing rivers. They were ___58____ in the terrible desert heat, with cameras, sheets of glass, and vats of chemicals. Bad weather, equipment failures, and accidents were frequent problems. They persevered, but success in creating a negative did not ___59___ the production of a photograph; plates still had to be ___60___ transported back to the studio before the image could be printed on paper. A photographer could carry 120 pounds of many miles to ____61____ a magnificent view only to have the easily broken plate ___62___ in transportation.___63___, once photographers were successful, the results were superb and much admired. Photographs were put on exhibition, and people bought albums filled with pictures by Timothy O'Sullivan, Carleton Watkins, and William Henry Jackson. Jackson's photographs of Yellowstone's natural wonders, along with the paintings of fellow Thomas Moran, even helped ___64___ Congress to preserve thousands of acres of this land in 1872 as the nation's first national park.Section BDirections: Read the following four passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.(A)I once had my Chinese MBA students brainstorming on “two-hour business plans”. I separated them into six groups and gave them an example: a restaurant chain. The more original their idea, the better, I said. Finally, five of the six groups presented plans for restaurant chains. The sixth proposed a catering service. Though I admitted the time limit had been difficult, I expressed my disappointment.My students were middle managers, financial analysts and financiers from state owned enterprises and global companies. They were not without talent or opinions, but they had been shaped by an educational system that rarely stressed or rewarded critical thinking or inventiveness. The scene I just described came in different forms during my two years’ teaching at the school. Papers were often copied from the Web and the Harvard Business Review. Case study debates were written up and just memorized. Students frequently said that copying is a superior business strategy, better than inventing and creating.In China, every product you can imagine has been made and sold. But so few well developed marketing and management minds have been raised that it will be a long time before most people in the world can name a Chinese brand.With this problem in mind, partnerships with institutions like Yale and MIT have been established. And then there’s the “thousand talent scheme”: this new government program is intended toimprove technological modernization by attracting top foreign trained scientists to the mainland with big money. But there are worries about China’s research environment. It’s hardly known for producing independent thinking and openness, and even big salary offers may not be attractive enough to overcome this.At last, for China, becoming a major world creator is not just about setting up partnerships with top Western universities. Nor is it about gathering a group of well-educated people and telling them t o think creatively. It’s about establishing a rich learning environment for young minds. It’s not that simple.65. Why does the author feel disappointed at his students?A. Because there is one group presenting a catering service.B. Because the six groups made projects for restaurant chains.C. Because all the students copied a case for the difficult topic.D. Because the students’ ideas were lacking in creativeness.66. We can infer from the passage that ________.A. China can make and sell any product all over the worldB. high pay may not solve the problem of China’s research environmentC. cooperation with institutions has been set up to make a Chinese brandD. the new government program are aimed at encouraging imagination67. Which is the best title of the passage?A. Look for a New Way of Learning.B. Reward Creative Thinking.C. How to Become a Creator.D. Establish a technical Environment.(B)Every summer, no matter how urgent work schedule is, I take off one day exclusively for my son. We call it dad-son day. This year our third stop was the amusement panic, where we discoveredthat he was tall enough to ride one of the fastest roller coasters (过山车)in the world. We experienced through face-stretching turns and circles for ninety seconds. Then, as we stepped off the ride, in a calm voice, he remarked that it was not as exciting as other rides he’d been on. As I listened, I began to sense something seriously out of balance.Throughout the season, I noticed similar events all around me. Parents found it hard to find new stimulations for cold kids. Surrounded by ever-greater stimulation, their young feces were looking disappointed and boredFacing their children's complaints of "nothing to do", parents were spending large numbers of dollars for various forms of entertainment In many cases the money seemed to do little more than buy relief from the terrible complaint of their bored children. This set me thinking the obvious question: "How can it be so hard for kids to find something to do when there's been such a range of stimulating entertainment available to them?"What really worries me is the strength of the stimulation. I watch my little daughter's &ce as she absorbs bloody special effects in movies.Why do children facing such excitement seem starved for more? Thai was, I realized, the point I discovered during my own adolescence that what creates excitement is not going fast, but goingfaster. Excitement has less to do with speed than changes in speed.I am concerned about the increasing effect of years at these levels of feverish activity. It is no mystery to me why many teenagers appear uninterested and burned out, with a "been there, done that" air of indifference toward much of life. As increasing numbers of frie nds’ children are advised to take medicine to deal with inattentiveness at school or anti-depressants (抗抑郁药)to help with the loss of interest and joy in their lives—I question the role of kids’ boredom in some of the diagnoses (处方).My own work is focused on the chemical imbalances and biological factors related to behavioral and emotional disorders. These are complex problems. Yet I’ve been reflecting more and more on how lie pace of life and the strength of stimulation may be contributing to the rising rates of psychological problems among children and adolescents in our society.68. The reason why the author felt surprised in the amusement park was that ________.A. his son was not as excited by the roller coasters ride as expectedB. his songs enjoyed turns and circles with his face stretchedC. his son appeared upset but calm while riding the roller coastersD. his son could keep his balance so well on the fast moving roller coasters69. According to the author, children will probably feel excited ________.A. if their parents allow them to ride roller coasters very oftenB. since parents spend money on the same form of entertainmentC. after they take anti-depressants according to the diagnosesD. if they are often exposed to more stimulating entertainment70. From his own experience, the author came to the conclusion that children seem to expect _______.A. a much wider variety of sports facilitiesB. activities that require complicated skillsC. the change of the forms of recreationD. more challenging physical exercise71. In order to relieve children's boredom, the author would probably suggest ________.A. adjusting the pace of life and strength of stimulationB. promoting the practice of dad-son daysC. consulting a specialist in child psychologyD. balancing school work with after school activities(C)Cultural rules determine every aspect of food consumption. Who eats together defines social units. For example, in some societies, the nuclear family is the unit that regularly eats together. The anthropologist Mary Douglas has pointed out that, for the English, the kind of meal and the kind of food that is served relate to the kinds of social links between people who are eating together. She distinguishes between regular meals, Sunday meals when relatives may come, and cocktail parties for relatives and friends. The food served symbolizes the occasion and reflects who is present. For example, only snacks are served at a cocktail party. It would be inappropriate to serve a steak or hamburgers. The distinctions among cocktails, regular meals, and specialdinners mark the social boundaries between those guests who are invited for drinks, those who are invited to dinner, and those who come to a family meal. In this example, the type of food symbolizes the category of guest and with whom it is eaten.In some New Guinea societies, the nuclear family is not the unit that eats together. The men take their meals in a men's house, separately from their wives and children. Women prepare and eat their food in their own houses and take the husband's portion to the men's house. The women eat with their children in their own houses. This pattern is also widespread among Near Eastern societies.Eating is a metaphor that is sometimes used to signify marriage. In many New Guinea societies, like that of the Lese on the island of New Ireland in the Pacific and that of the Trobriand Islanders, marriage is symbolized by the couple's eating together for the first time. Eating symbolizes their new status as a married couple. In U.S. society, it is just the reverse. A couple may go out to dinner on a first date.Other cultural rules have to do with taboos against eating certain things. In some societies, members of a family group, arc not allowed to eat the animal or bird that is their ancestor. Since they believe themselves to be children of that ancestor, it would be like eating that ancestor or eating themselves.There is also an association between food prohibitions and rank, which is found in its most extreme form in the caste (social class) system of India. A caste system consists of ranked groups, each with a different economic specialization. In India, there is an association between caste and the idea of pollution. Members of highly ranked groups can be polluted by coming into contact with the bodily secretions, particularly saliva(唾液),of individuals of lower-ranked castes. Because of the fear of pollution, Brahmans and other high-ranked individuals will not share food with, not eat from the same plate as, not even accept food from an individual or from a low-ranking class.72. According to the passage, who will NOT eat together?A. The English during regular meals.B. Americans on their first date.C. Men and women in Near Eastern societies.D. Newly-married people on the island of New Ireland.73. In Paragraph 4, the underlined word "taboos" means _____________.A. favorsB. prohibitionsC. hatredD. gossips74. According to the passage, eating together indicates all the following EXCEPT .A. the type of foodB. social relations.C. marital statusD. family ties.75. What is the main idea of the passage?A. Different kinds of food in western countries.B. Relations between food and social units.C. Symbolic meanings of different kinds of food.D. Food consumption in different cultures.Section CDirections: Read the following text and choose the most suitable heading from the list A-F forA. The description of using amateur records to encourage the public.B. The description of old records kept by amateur naturalists.C. Concerns over amateur data for lacking objectivity and precision.D. The necessity of encouraging amateur collection.E. How people react to their involvement in data collection.F. The application of amateur records to phonology.76._______________Tim Sparks slides a small leather-bound notebook out of an envelope. The book's yellowing pages contain beekeeping notes made between 1941 and 1969 by the late Walter Coates of Kilworth, Leicestershire. He adds it to his growing pile of local journals, birdwatchers' lists and gardening diaries. "We're uncovering about one major new record each month," he says, "I still get surprised." Around two centuries before Coates, Robert Marsham, a landowner from Norfolk in east of England, began recording the life cycles of plants and animals on his estate. Successive Marshams continued recording these notes for 211 years.77._______________Today, such records are being put to uses that their authors couldn't possibly have expected. These data sets, and others like them, are proving valuable to ecologists interested in the timing of biological events, or phonology. By combining the records with climate data, researchers can reveal how, for example, changes in temperature affect the arrival of spring, allowing ecologists to make improved predictions about the impact of climate change.78._______________But not all professionals are happy to use amateur data. "A lot of scientists won't touch them, they say they're too full of problems," says Root. Because different observers can have different ideas of what forms, for example, an open snowdrop. "The biggest concern with ad hoc (临时的) observations is how carefully and systematically they w ere taken,” says Mark Schwarts of the University of Wisconsin, Milwaukee, who studies the interactions between plants and climate. "We need to know pretty precisely what a person's been observing—if they just say ‘I noted when the leaves came out’, it might not be that useful.” Measuring the onset of autumn can be particularly problematic because deciding when leaves change color is a more subjective process than noting when they appear.79._______________Overall, most phrenologists arc positive about the contribution that amateurs can make. "They get the raw power of science: careful observation of the natural world," says Sagarin. Others suggest that the right statistics can iron out some of the problems with amateur data. Togetherwith colleagues at Wageoingen University in the Netherlands, environmental scientist Arnold van Vliet is developing statistical techniques to account for the uncertainty in amateur phonological data. Besides, the data are cheap to collect, and can provide breadth in space, time and range of species," It’s very difficult to collect data on a large geographical scale without enlisting an army of observers, says Root.80._______________Phonology also helps to drive home messages about climate change. “Because the public under stand these records, they accept them,” says Sparks. It can also illustrate potentially unpleasant consequences, he adds, such as the finding that more rat infestations are reported to local councils in warmer years. And getting people involved is great for public relations. "People are excited to think that the data they have been collecting as a hobby can be used for something scientific—it empowers them” says Root.Section DDirections: Read the passage carefully. Then answer the questions or complete the statements in the fewest possible wordsThe automobile may not be closely associated with modem American culture, but it has occupied a central role in America's economic and social history.No one can deny the status of Henry Ford in car history. When the first Ford Model T rolled off the assembly line in 1908, businessman Henry Ford transformed the car from a luxury attachment for the rich to an automobile product for the middle classes. The rush of cars into the community forced all levels of government to build new and better roads. Better roads fed the demand for larger, faster, more stylish vehicles, and a host of companies rushed to meet that demand.If there was a first Golden Age of automobile, it may well have been the 1950s. It was an age of prosperity. Large, regular paychecks encouraged the public display of wealth through costly items such as new cars. Americans, moreover, needed those cars as they moved away from the cities into the suburbs, where such things as stores, jobs, and schools were seldom within walking distance. Cars became essential if people were to get to work or to the grocery store.As Ac 1950s slipped into 1960s, it became apparent that these fashionable wheels were gas-consuming road cruisers, dangerous in an accident, and often full of faults. Under pressure from a variety of groins, the federal government required that newer models provide greater fuel efficiency and cleaner emissions. Seatbelts became standard equipment as well Rising fad prices in the 1970s, coupled with concern for the environment, made the smaller cars produced by foreign companies for European and Asian markets very popular.The 1980s and 1990s saw an upswing in the popularity of big cars. New models including minivans and sport utility vehicles have become main products in auto dealers' showrooms. Traffic jams on the roads have become part of the American way of life. In 1911 a horse could travel through rush-hour traffic in Los Angeles at 11 miles per hour. In 2000 a car covering the same territory at the same time of day moved at about 4 miles per hour. But perhaps that is not important. When a car is equipped with a telephone and television set, a computer, and global positioning satellite connections, it can feel just like home.(Note: Answer the questions or complete the statements in NO MORE THAN TEN WORDS.)81. made Henry Ford well-known in car history.82. That the car industry prospered in the 1950s in America resulted from ________.83. Why were smaller cars once popular in the 1970s?84. In today's society, how can a car make people ignore traffic jams?第II卷(共45分)I. TranslationDirections: Translate the following sentences into English, using the words given in the brackets.1. 实验结果比预想的要满意得多。
2012年上海市高考数学试卷(文科)答案与解析
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2012年上海市高考数学试卷(文科)参考答案与试题解析一、填空题(本大题共有14题,满分56分)1.(4分)(2012•上海)计算:=1﹣2i(i为虚数单位).考点: 复数代数形式的乘除运算.专题: 计算题.分析:由题意,可对复数代数式分子与分母都乘以1﹣i,再由进行计算即可得到答案解答:解:故答案为1﹣2i点评:本题考查复数代数形式的乘除运算,解题的关键是分子分母都乘以分母的共轭,复数的四则运算是复数考查的重要内容,要熟练掌握2.(4分)(2012•上海)若集合A={x|2x﹣1>0},B={x||x|<1},则A∩B=(,1).考点:交集及其运算.专题:计算题.分析:由题意,可先化简两个集合A,B,再求两个集合的交集得到答案解答:解:由题意A={x|2x﹣1>0}={x|x>},B={x|﹣1<x<1},∴A∩B=(,1)故答案为(,1)点评:本题考查交的运算,是集合中的基本题型,解题的关键是熟练掌握交集的定义3.(4分)(2012•上海)函数的最小正周期是π.考点: 二阶矩阵;三角函数中的恒等变换应用;三角函数的周期性及其求法.专题:计算题.分析:先根据二阶行列式的公式求出函数的解析式,然后利用二倍角公式进行化简,最后根据正弦函数的周期公式进行求解即可.解答:解:=sinxcosx+2=sin2x+2∴T==π∴函数的最小正周期是π故答案为:π点评:本题主要考查了二阶行列式,以及三角函数的化简和周期的求解,同时考查了运算求解能力,属于基础题.4.(4分)(2012•上海)若是直线l的一个方向向量,则l的倾斜角的大小为arctan(结果用反三角函数值表示)考点: 平面向量坐标表示的应用.专题: 计算题.分析:根据直线的方向向量的坐标一般为(1,k)可得直线的斜率,根据tanα=k,最后利用反三角可求出倾斜角.解答:解:∵是直线l的一个方向向量∴直线l的斜率为即tanα=则l的倾斜角的大小为arctan故答案为:arctan点评:本题主要考查了直线的方向向量,解题的关键是直线的方向向量的坐标一般为(1,k),同时考了反三角的应用,属于基础题.5.(4分)(2012•上海)一个高为2的圆柱,底面周长为2π,该圆柱的表面积为6π.考点:旋转体(圆柱、圆锥、圆台).专题:计算题.分析:求出圆柱的底面半径,然后直接求出圆柱的表面积即可.解答:解:因为一个高为2的圆柱,底面周长为2π,所以它的底面半径为:1,所以圆柱的表面积为S=2S底+S侧=2×12×π+2π×2=6π.故答案为:6π.点评:本题考查旋转体的表面积的求法,考查计算能力.6.(4分)(2012•上海)方程4x﹣2x+1﹣3=0的解是x=log23.考点: 有理数指数幂的运算性质.专题:计算题.分析:根据指数幂的运算性质可将方程4x﹣2x+1﹣3=0变形为(2x)2﹣2×2x﹣3=0然后将2x 看做整体解关于2x的一元二次方程即可.解答:解:∵4x﹣2x+1﹣3=0∴(2x)2﹣2×2x﹣3=0∴(2x﹣3)(2x+1)=0∵2x>0∴2x﹣3=0∴x=log23故答案为x=log23点评:本题主要考差了利用指数幂的运算性质解有关指数类型的方程.解题的关键是要将方程4x﹣2x+1﹣3=0等价变形为(2x)2﹣2×2x﹣3=0然后将2x看做整体再利用因式分解解关于2x的一元二次方程.7.(4分)(2012•上海)有一列正方体,棱长组成以1为首项、为公比的等比数列,体积分别记为V1,V2,…,V n,…,则(V1+V2+…+V n)═.考点: 数列的极限;棱柱、棱锥、棱台的体积.专题:计算题.分析:由题意可得,正方体的体积=是以1为首项,以为公比的等比数,由等不数列的求和公式可求解答:解:由题意可得,正方体的棱长满足的通项记为a n则∴=是以1为首项,以为公比的等比数列则(V1+V2+…+v n)==故答案为:点评:本题主要考查了等比数列的求和公式及数列极限的求解,属于基础试题8.(4分)(2012•上海)在的二项式展开式中,常数项等于﹣20.考点:二项式定理的应用.专题:计算题.分析:研究常数项只需研究二项式的展开式的通项,使得x的指数为0,得到相应的r,从而可求出常数项.解答:解:展开式的通项为T r+1=x6﹣r(﹣)r=(﹣1)r x6﹣2r令6﹣2r=0可得r=3 常数项为(﹣1)3=﹣20故答案为:﹣20点评:本题主要考查了二项式定理的应用,解题的关键是写出展开式的通项公式,同时考查了计算能力,属于基础题.9.(4分)(2012•上海)已知y=f(x)是奇函数,若g(x)=f(x)+2且g(1)=1,则g(﹣1)=3.考点: 函数奇偶性的性质;函数的值.专题:计算题.分析:由题意y=f(x)是奇函数,g(x)=f(x)+2得到g(x)+g(﹣x)=f(x)+2+f(﹣x)+2=4,再令x=1即可得到1+g(﹣1)=4,从而解出答案解答:解:由题意y=f(x)是奇函数,g(x)=f(x)+2∴g(x)+g(﹣x)=f(x)+2+f(﹣x)+2=4又g(1)=1∴1+g(﹣1)=4,解得g(﹣1)=3故答案为:3点评:本题考查函数奇偶性的性质,解题的关键是利用性质得到恒成立的等式,再利用所得的恒等式通过赋值求函数值10.(4分)(2012•上海)满足约束条件|x|+2|y|≤2的目标函数z=y﹣x的最小值是﹣2.考点: 简单线性规划.分析:作出约束条件对应的平面区域,由z=y﹣x可得y=x+z,则z为直线在y轴上的截距,解决越小,z越小,结合图形可求解答:解:作出约束条件对应的平面区域,如图所示由于z=y﹣x可得y=x+z,则z为直线在y轴上的截距,截距越小,z越小结合图形可知,当直线y=x+z过C时z最小,由可得C(2,0),此时Z=﹣2最小故答案为:﹣2点评:借助于平面区域特性,用几何方法处理代数问题,体现了数形结合思想、化归思想.线性规划中的最优解,通常是利用平移直线法确定.11.(4分)(2012•上海)三位同学参加跳高、跳远、铅球项目的比赛,若每人只选择一个项目,则有且仅有两人选择的项目相同的概率是(结果用最简分数表示)考点: 古典概型及其概率计算公式;列举法计算基本事件数及事件发生的概率.专题:概率与统计.分析:先求出三个同学选择的所求种数,然后求出有且仅有两人选择的项目完全相同的种数,最后利用古典概型及其概率计算公式进行求解即可.解答:解:每个同学都有三种选择:跳高与跳远;跳高与铅球;跳远与铅球三个同学共有3×3×3=27种有且仅有两人选择的项目完全相同有××=18种其中表示3个同学中选2个同学选择的项目,表示从三种组合中选一个,表示剩下的一个同学有2种选择故有且仅有两人选择的项目完全相同的概率是=故答案为:点评:本题主要考查了古典概型及其概率计算公式,解题的关键求出有且仅有两人选择的项目完全相同的个数,属于基础题.12.(4分)(2012•上海)在矩形ABCD中,边AB、AD的长分别为2、1,若M、N分别是边BC、CD上的点,且满足,则的取值范围是[1,4].考点:平面向量数量积的运算.专题: 计算题.分析:先以所在的直线为x轴,以所在的直线为x轴,建立坐标系,写出要用的点的坐标,根据两个点的位置得到坐标之间的关系,表示出两个向量的数量积,根据动点的位置得到自变量的取值范围,做出函数的范围,即要求得数量积的范围.解答:解:以所在的直线为x轴,以所在的直线为x轴,建立坐标系如图,∵AB=2,AD=1,∴A(0,0),B(2,0),C(2,1),D(0,1),设M(2,b),N(x,1),∵,∴b=∴,=(2,),∴=,∴1,即1≤≤4故答案为:[1,4]点评:本题主要考查平面向量的基本运算,概念,平面向量的数量积的运算,本题解题的关键是表示出两个向量的坐标形式,利用函数的最值求出数量积的范围,本题是一个中档题目.13.(4分)(2012•上海)已知函数y=f(x)的图象是折线段ABC,其中A(0,0)、、C(1,0),函数y=xf(x)(0≤x≤1)的图象与x轴围成的图形的面积为.考点: 分段函数的解析式求法及其图象的作法.专题:计算题;压轴题.分析:先利用一次函数的解析式的求法,求得分段函数f(x)的函数解析式,进而求得函数y=xf(x)(0≤x≤1)的函数解析式,最后利用定积分的几何意义和微积分基本定理计算所求面积即可解答:解:依题意,当0≤x≤时,f(x)=2x,当<x≤1时,f(x)=﹣2x+2∴f(x)=∴y=xf(x)=y=xf(x)(0≤x≤1)的图象与x轴围成的图形的面积为S=+=x3+(﹣+x2)=+=故答案为:点评:本题主要考查了分段函数解析式的求法,定积分的几何意义,利用微积分基本定理和运算性质计算定积分的方法,属基础题14.(4分)(2012•上海)已知,各项均为正数的数列{a n}满足a1=1,a n+2=f(a n),若a2010=a2012,则a20+a11的值是.考点: 数列与函数的综合.专题:综合题;压轴题.分析:根据,各项均为正数的数列{a n}满足a1=1,a n+2=f(a n),可确定a1=1,,,a7=,,,利用a2010=a2012,可得a2010=(负值舍去),依次往前推得到a20=,由此可得结论.解答:解:∵,各项均为正数的数列{a n}满足a1=1,a n+2=f(a n),∴a1=1,,,a7=,,∵a2010=a2012,∴∴a2010=(负值舍去),由a2010=得a2008=…依次往前推得到a20=∴a20+a11=故答案为:点评:本题主要考查数列的概念、组成和性质、同时考查函数的概念.理解条件a n+2=f(a n),是解决问题的关键,本题综合性强,运算量较大,属于中高档试题.二、选择题(本大题共有4题,满分20分)15.(5分)(2012•上海)若i是关于x的实系数方程x2+bx+c=0的一个复数根,则() A.b=2,c=3 B.b=2,c=﹣1 C.b=﹣2,c=﹣1 D.b=﹣2,c=3考点:复数代数形式的混合运算;复数相等的充要条件.专题:计算题.分析:由题意,将根代入实系数方程x2+bx+c=0整理后根据得数相等的充要条件得到关于实数a,b的方程组,解方程得出a,b的值即可选出正确选项解答:解:由题意1+i是关于x的实系数方程x2+bx+c=0∴1+2i﹣2+b+bi+c=0,即∴,解得b=﹣2,c=3故选D点评:本题考查复数相等的充要条件,解题的关键是熟练掌握复数相等的充要条件,能根据它得到关于实数的方程,本题考查了转化的思想,属于基本计算题16.(5分)(2012•上海)对于常数m、n,“mn>0”是“方程mx2+ny2=1的曲线是椭圆"的()A.充分不必要条件B.必要不充分条件C.充分必要条件D.既不充分也不必要条件考点:必要条件、充分条件与充要条件的判断.专题:常规题型.分析:先根据mn>0看能否得出方程mx2+ny2=1的曲线是椭圆;这里可以利用举出特值的方法来验证,再看方程mx2+ny2=1的曲线是椭圆,根据椭圆的方程的定义,可以得出mn >0,即可得到结论.解答:解:当mn>0时,方程mx2+ny2=1的曲线不一定是椭圆,例如:当m=n=1时,方程mx2+ny2=1的曲线不是椭圆而是圆;或者是m,n都是负数,曲线表示的也不是椭圆;故前者不是后者的充分条件;当方程mx2+ny2=1的曲线是椭圆时,应有m,n都大于0,且两个量不相等,得到mn >0;由上可得:“mn>0”是“方程mx2+ny2=1的曲线是椭圆”的必要不充分条件.故选B.点评:本题主要考查充分必要条件,考查椭圆的方程,注意对于椭圆的方程中,系数要满足大于0且不相等,本题是一个基础题.17.(5分)(2012•上海)在△ABC中,若sin2A+sin2B<sin2C,则△ABC的形状是()A.钝角三角形B.直角三角形C.锐角三角形D.不能确定考点: 三角形的形状判断.专题:三角函数的图像与性质.分析:利用正弦定理将sin2A+sin2B<sin2C,转化为a2+b2<c2,再结合余弦定理作出判断即可.解答:解:∵在△ABC中,sin2A+sin2B<sin2C,由正弦定理===2R得,a2+b2<c2,又由余弦定理得:cosC=<0,0<C<π,∴<C<π.故△ABC为钝角三角形.故选A.点评:本题考查三角形的形状判断,着重考查正弦定理与余弦定理的应用,属于基础题.18.(5分)(2012•上海)若(n∈N*),则在S1,S2,…,S100中,正数的个数是()A.16 B.72 C.86 D.100考点:数列与三角函数的综合.专题: 计算题;综合题;压轴题.分析:由于sin>0,sin>0,…sin>0,sin=0,sin<0,…sin<0,sin=0,可得到S1>0,…S13>0,而S14=0,从而可得到周期性的规律,从而得到答案.解答:解:∵sin>0,sin>0,…sin>0,sin=0,sin<0,…sin<0,sin=0,∴S1=sin>0,S2=sin+sin>0,…,S8=sin+sin+…sin+sin+sin=sin+…+sin+sin>0,…,S12>0,而S13=sin+sin+…+sin+sin+sin+sin+…+sin=0,S14=S13+sin=0+0=0,又S15=S14+sin=0+sin=S1>0,S16=S2>0,…S27=S13=0,S28=S14=0,∴S14n﹣1=0,S14n=0(n∈N*),在1,2,…100中,能被14整除的共7项,∴在S1,S2,…,S100中,为0的项共有14项,其余项都为正数.故在S1,S2,…,S100中,正数的个数是86.故选C.点评:本题考查数列与三角函数的综合,通过分析sin的符号,找出S1,S2,…,S100中,S14n=0,S14n=0是关键,也是难点,考查学生分析运算能力与冷静坚持的态度,属于难题.﹣1三、解答题(本大题共有5题,满分74分)19.(12分)(2012•上海)如图,在三棱锥P﹣ABC中,PA⊥底面ABC,D是PC的中点,已知∠BAC=,AB=2,,PA=2,求:(1)三棱锥P﹣ABC的体积;(2)异面直线BC与AD所成的角的大小(结果用反三角函数值表示)考点:异面直线及其所成的角;棱柱、棱锥、棱台的体积.专题:常规题型;综合题.分析:(1)首先根据三角形面积公式,算出直角三角形ABC的面积:S△ABC=,然后根据PA⊥底面ABC,结合锥体体积公式,得到三棱锥P﹣ABC的体积;(2)取BP中点E,连接AE、DE,在△PBC中,根据中位线定理得到DE∥BC,所以∠ADE(或其补角)是异面直线BC、AD所成的角.然后在△ADE中,利用余弦定理得到cos∠ADE=,所以∠ADE=arccos是锐角,因此,异面直线BC与AD所成的角的大小arccos.解答:解:(1)∵∠BAC=,AB=2,,∴S△ABC=×2×=又∵PA⊥底面ABC,PA=2∴三棱锥P﹣ABC的体积为:V=×S△ABC×PA=;(2)取BP中点E,连接AE、DE,∵△PBC中,D、E分别为PC、PB中点∴DE∥BC,所以∠ADE(或其补角)是异面直线BC、AD所成的角.∵在△ADE中,DE=2,AE=,AD=2∴cos∠ADE==,可得∠ADE=arccos(锐角)因此,异面直线BC与AD所成的角的大小arccos.点评:本题给出一个特殊的三棱锥,以求体积和异面直线所成角为载体,考查了棱柱、棱锥、棱台的体积和异面直线及其所成的角等知识点,属于基础题.20.(14分)(2012•上海)已知f(x)=lg(x+1)(1)若0<f(1﹣2x)﹣f(x)<1,求x的取值范围;(2)若g(x)是以2为周期的偶函数,且当0≤x≤1时,g(x)=f(x),求函数y=g(x)(x∈[1,2])的反函数.考点:函数的周期性;反函数;对数函数图象与性质的综合应用.专题:计算题.分析:(1)应用对数函数结合对数的运算法则进行求解即可;(2)结合函数的奇偶性和反函数知识进行求解.解答:解:(1)f(1﹣2x)﹣f(x)=lg(1﹣2x+1)﹣lg(x+1)=lg(2﹣2x)﹣lg(x+1),要使函数有意义,则由解得:﹣1<x<1.由0<lg(2﹣2x)﹣lg(x+1)=lg<1得:1<<10,∵x+1>0,∴x+1<2﹣2x<10x+10,∴.由,得:.(2)当x∈[1,2]时,2﹣x∈[0,1],∴y=g(x)=g(x﹣2)=g(2﹣x)=f(2﹣x)=lg(3﹣x),由单调性可知y∈[0,lg2],又∵x=3﹣10y,∴所求反函数是y=3﹣10x,x∈[0,lg2].点评:本题考查对数的运算以及反函数与原函数的定义域和值域相反等知识,属于易错题.21.(14分)(2012•上海)海事救援船对一艘失事船进行定位:以失事船的当前位置为原点,以正北方向为y轴正方向建立平面直角坐标系(以1海里为单位长度),则救援船恰好在失事船正南方向12海里A处,如图,现假设:①失事船的移动路径可视为抛物线;②定位后救援船即刻沿直线匀速前往救援;③救援船出发t小时后,失事船所在位置的横坐标为7t(1)当t=0.5时,写出失事船所在位置P的纵坐标,若此时两船恰好会合,求救援船速度的大小和方向.(2)问救援船的时速至少是多少海里才能追上失事船?考点:圆锥曲线的综合.专题:应用题.分析:(1)t=0.5时,确定P的横坐标,代入抛物线方程中,可得P的纵坐标,利用|AP|=,即可确定救援船速度的大小和方向;(2)设救援船的时速为v海里,经过t小时追上失事船,此时位置为(7t,12t2),从而可得vt=,整理得,利用基本不等式,即可得到结论.解答:解:(1)t=0.5时,P的横坐标x P=7t=,代入抛物线方程中,得P的纵坐标y P=3.…2分由|AP|=,得救援船速度的大小为海里/时.…4分由tan∠OAP=,得∠OAP=arctan,故救援船速度的方向为北偏东arctan 弧度.…6分(2)设救援船的时速为v海里,经过t小时追上失事船,此时位置为(7t,12t2).由vt=,整理得.…10分因为,当且仅当t=1时等号成立,所以v2≥144×2+337=252,即v≥25.因此,救援船的时速至少是25海里才能追上失事船.…14分点评:本题主要考查函数模型的选择与运用.选择恰当的函数模型是解决此类问题的关键,属于中档题.22.(16分)(2012•上海)在平面直角坐标系xOy中,已知双曲线C:2x2﹣y2=1.(1)设F是C的左焦点,M是C右支上一点,若,求点M的坐标;(2)过C的左焦点作C的两条渐近线的平行线,求这两组平行线围成的平行四边形的面积; (3)设斜率为k()的直线l交C于P、Q两点,若l与圆x2+y2=1相切,求证:OP⊥OQ.考点: 直线与圆锥曲线的综合问题;直线与圆的位置关系;双曲线的简单性质.专题:计算题;综合题;压轴题;转化思想.分析:(1)求出双曲线的左焦点F的坐标,设M(x,y),利用|MF|2=(x+)2+y2,求出x 的范围,推出M的坐标.(2)求出双曲线的渐近线方程,求出直线与另一条渐近线的交点,然后求出平行四边形的面积.(3)设直线PQ的方程为y=kx+b,通过直线PQ与已知圆相切,得到b2=k2+1,通过求解=0.证明PO⊥OQ.解答:解:(1)双曲线C1:的左焦点F(﹣),设M(x,y),则|MF|2=(x+)2+y2,由M点是右支上的一点,可知x≥,所以|MF|==2,得x=,所以M().(2)左焦点F(﹣),渐近线方程为:y=±x.过F与渐近线y=x平行的直线方程为y=(x+),即y=,所以,解得.所以所求平行四边形的面积为S=.(3)设直线PQ的方程为y=kx+b,因直线PQ与已知圆相切,故,即b2=k2+1…①,由,得(2﹣k2)x2﹣2bkx﹣b2﹣1=0,设P(x1,y1),Q(x2,y2),则,又y1y2=(kx1+b)(kx2+b).所以=x1x2+y1y2=(1+k2)x1x2+kb(x1+x2)+b2==.由①式可知,故PO⊥OQ.点评:本题考查直线与圆锥曲线的综合问题,圆锥曲线的综合,向量的数量积的应用,设而不求的解题方法,点到直线的距离的应用,考查分析问题解决问题的能力,考查计算能力.23.(18分)(2012•上海)对于项数为m的有穷数列{a n},记b k=max{a1,a2,…,a k}(k=1,2,…,m),即b k为a1,a2,…,a k中的最大值,并称数列{b n}是{a n}的控制数列,如1,3,2,5,5的控制数列是1,3,3,5,5.(1)若各项均为正整数的数列{a n}的控制数列为2,3,4,5,5,写出所有的{a n}.(2)设{b n}是{a n}的控制数列,满足a k+b m﹣k+1=C(C为常数,k=1,2,…,m),求证:b k=a k (k=1,2,…,m).(3)设m=100,常数a∈(,1),a n=a n2﹣n,{b n}是{a n}的控制数列,求(b1﹣a1)+(b2﹣a2)+…+(b100﹣a100).考点: 数列的应用.专题:综合题;压轴题;点列、递归数列与数学归纳法.分析:(1)根据题意,可得数列{a n}为:2,3,4,5,1;2,3,4,5,2;2,3,4,5,3;2,3,4,5,4,;2,3,4,5,5;(2)依题意可得b k+1≥b k,又a k+b m﹣k+1=C,a k+1+b m﹣k=C,从而可得a k+1﹣a k=b m﹣k+1﹣b m﹣k≥0,整理即证得结论;(3)根据,可发现,a4k﹣3=a(4k﹣3)2+(4k﹣3),a4k=a(4k﹣2)2+(4k﹣2),a4k﹣1=a(4k﹣1)2﹣(4k﹣1),a4k=a(4k)2﹣4k,通过比较﹣2大小,可得a4k﹣2>a4k﹣1,a4k>a4k﹣2,而a4k+1>a4k,a4k﹣1﹣a4k﹣2=(a﹣1)(8k﹣3),从而可求得(b1﹣a1)+(b2﹣a2)+…+(b100﹣a100)=(a2﹣a3)+(a6﹣a7)+…+(a98﹣a99)=(a4k﹣2﹣a4k﹣1)=2525(1﹣a).解答:解:(1)数列{a n}为:2,3,4,5,1;2,3,4,5,2;2,3,4,5,3;2,3,4,5,4,;2,3,4,5,5;…4分(2)∵b k=max{a1,a2,…,a k},b k+1=max{a1,a2,…,a k+1},∴b k+1≥b k…6分∵a k+b m﹣k+1=C,a k+1+b m﹣k=C,∴a k+1﹣a k=b m﹣k+1﹣b m﹣k≥0,即a k+1≥a k,…8分∴b k=a k…10分(3)对k=1,2,…25,a4k﹣3=a(4k﹣3)2+(4k﹣3),a4k﹣2=a(4k﹣2)2+(4k﹣2),a4k﹣1=a(4k﹣1)2﹣(4k﹣1),a4k=a(4k)2﹣4k,…12分比较大小,可得a4k﹣2>a4k﹣1,∵<a<1,∴a4k﹣1﹣a4k﹣2=(a﹣1)(8k﹣3)<0,即a4k﹣2>a4k﹣1;a4k﹣a4k﹣2=2(2a﹣1)(4k﹣1)>0,即a4k>a4k﹣2,又a4k+1>a4k,从而b4k﹣3=a4k﹣3,b4k﹣2=a4k﹣2,b4k﹣1=a4k﹣2,b4k=a4k,…15分∴(b1﹣a1)+(b2﹣a2)+…+(b100﹣a100)=(a2﹣a3)+(a6﹣a7)+…+(a98﹣a99)=(a4k﹣2﹣a4k﹣1)=(1﹣a)(8k﹣3)=2525(1﹣a)…18分点评:本题考查数列的应用,着重考查分析,对抽象概念的理解与综合应用的能力,对(3)观察,分析寻找规律是难点,是难题.。
精品解析上海市华师大一附中等八校2012届高三2月联合调研考试(物理)
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上海市华师大一附中等八校2012届高三2月联合调研考试卷物理参加学校:华师大一附中、曹杨二中、市西、市三女子、控江、格致、市北、育才、晋元高中本试卷分为第Ⅰ卷(第1页~4页)和第Ⅱ卷(第4页~8页)两部分,全卷共8页。
满分150分,考试时间120分钟。
本卷g 均取10m/s 2。
第Ⅰ卷(共56分)考生注意:1.答第Ⅰ卷前,考生务必在答题卡和答题纸上用蓝色或黑色的钢笔或圆珠笔清楚填写学校、班级、学号、姓名。
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答案需要更改时,必须将原选项用橡皮擦去,重新选择。
答案不能涂写在试卷上,涂写在试卷上一律不给分。
一.单项选择题(每小题2分,共16分。
每小题只有一个正确选项。
答案涂写在答题卡上。
)1.用比值法定义物理量是物理学中一种重要的思想方法,下列物理量的表达式不属于用比值法定义的是( )A .加速度m F a =B .功率t W P =C .速度S v t =D .电阻IU R = 1答案:A 解析:根据牛顿第二定律得出的加速度m F a =,是加速度的决定式,不属于用比值法定义的加速度定义式。
2.水流在推动水轮机的过程中做了3×108J 的功,这句话应理解为( )A .水流在推动水轮机前具有3×108J 的能量B .水流在推动水轮机的过程中具有3×108J 的能量C .水流在推动水轮机后具有3×108J 的能量D.水流在推动水轮机的过程中能量减少了3×108J2答案:D解析:根据功能关系,水流在推动水轮机的过程中做了3×108J的功,水流在推动水轮机的过程中能量减少了3×108J,选项D正确。
3.了解物理规律的发现过程,学会像科学家那样观察和思考,往往比掌握知识本身更重要。
以下符合事实的是()A.伽利略通过“理想实验”得出“力是维持物体运动的原因”B.牛顿建立了万有引力定律,并测出了万有引力常量C.奥斯特发现了电流的磁效应,使人们突破了对电与磁认识的局限性D.楞次发现了电磁感应现象,使人们对电与磁内在联系的认识更加完善4.下列关于电磁波的说法正确的是()A.电磁波在真空和介质中传播的速度相同B.变化的磁场能够在空间产生电场C.电磁波的波长、波速、周期的关系为v =λ•TD.电磁波既可能是横波,也可能是纵波4.答案:B解析:电磁波在真空和介质中传播的速度不相同,选项A错误;变化的磁场能够在空间产生电场,选项B正确;电磁波的波长、波速、周期的关系为v =λ/T,选项C错误;电磁波只可能是横波,选项D错误。
八省八校(T8联考)第二次联考语文试题
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2022届高三第二次T8联考语文试题命题学校:华中师大一附中 考试时间:2022年3月21日上午8:00−10:30试卷满分150分 考试用时150分钟 注意事项:1.答卷前,考生务必将自己的姓名、准考证号填写在答题卡上。
2.回答选择题时,选出每小题答案后,用铅笔把答题卡对应题目的答案标号涂黑。
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写在本试卷上无效。
3.考试结束后,将本试卷和答题卡一并交回。
一、现代文阅读(35分)(一)现代文阅读Ⅰ(本题共5小题,19分)阅读下面的文字,完成1~5题。
材料一:《千里江山图》是北宋宫廷画家王希孟创作的绢本设色画,现收藏于北京故宫博物院,是中国的十大传世名画之一。
画作在设色上以青绿为主色调,在施色时注重手法的变化,色彩张弛有度,或浑厚,或轻盈,使画面层次分明,色如宝石光彩照人,画中的内容让人观后赞叹,不失雄阔之境界与恢宏之气势。
舞蹈诗剧《只此青绿》中由演员饰演的展卷人(文物工作者)对《千里江山图》潜心钻研后“走”入了画家希孟的内心世界,目睹希孟绘制画作,让观众追随展卷人的视角,徜徉在富有传奇色彩的中国传统美学意趣之中。
舞台被巧妙设计成多层同心圆,还原了中国传统画卷的“展卷”过程。
圆环相扣,亦是时空相交。
当卷轴在舞台上被缓缓打开,左手拉开了“过去”的时间线,右手推动着未来的新图景,那转动不息的舞台仿若镌刻民族记忆的年轮,引领观者踏入时光的循环里。
另外,“圆”是中国传统文化的一个重要精神原型,是揭示中国艺术生命秘密的不可忽视的因素。
在剧中,除了舞台上的圆装置,还有视频画面中贯穿全场的明月是圆的,织卷人手中的竹盘是圆的,制墨人的石磨是圆的……这些偶然,都让剧中“圆”的寓意愈发朦胧而神秘,也越发自然。
在舞台装置的启动下,让人有一种“看一次‘青广东实验中学 东北育才中学 石家庄二中 华中师大一附中西南大学附中 南京师大附中 湖南师大附中 福州一中 八校绿’,启一日中国情”的观后直感,这种精神内涵的承载和历久弥新的文化传统远远大于剧作故事本身,让我们牢牢记住了中华民族的价值观念及以和谐为美的社会伦理思想。
上海市十校2012届高三第二次联考数学(文)试题及答案
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上海市十校2011—2012学年度高三第二学期考试数 学 试 题(文)一、填空题(本大题满分56分)本大题共有14题,考生应在答题纸相应编号的空格内直接填写结果,每个空格填对得4分,否则一律得零分. 1.若函数()f x 的反函数为13()log f x x -=,则()f x = .2.若复数(1)(3)bi i +-是纯虚数(i 是虚数单位,b 为实数),则=b _________.3.已知不等式20 1 x a x+≤的解集为[]1,b -,则实数b a +的值为_________.4.已知线性方程组的增广矩阵为116 02a ⎛⎫⎪⎝⎭,若该线性方程组解为42⎛⎫ ⎪⎝⎭,则实数a =___. 5.若向量a 、b 满足||1,||2a b == ,且a 与b 的夹角为3π,则()a a b ⋅+ =________.6.已知圆锥的母线长为5,侧面积为π15,则此圆锥的体积为__________. 7.在ABC ∆中,已知sin :sin :sin 2:3:4A B C =,则cos C = ..8.在约束条件⎪⎩⎪⎨⎧≥+-≥+≤02,0,3y x y x x 下,则目标函数y x z 2-=的最小值是________.9.若实数m 、∈n {2-,1-, 1,2,3},且n m ≠,则方程122=+ny m x 表示焦点在y 轴上的双曲线的概率是__________. 10..已知1()31xf x a =--是奇函数, 则()f x 的值域为 . 11.数列{}n a 中,12a =,对于任意*n N ∈,都有14n n a a +=+,n S 是{}n a 的前n 项和,则lim1nn nna S →∞=+_______.12.已知双曲线22221x y a b-=的两个焦点分别为12F F 、,该双曲线与抛物线28y x =有一个公共的焦点1F ,且两曲线的一个交点为P ,1||5F P =,则12F PF ∠的大小为_ (结果用反三角函数表示)13.毛毛的计算器中的“开根号”键最近“感冒”了,输出的结果千奇百怪.细心的毛毛在复习资料上发现有一个真命题:已知对于任意正数x x ≠,x 和31x x ++之 间;并且31x x ++比x.则输出的y=_______.(结果用 分数表示)14、下图展示了一个区间(0,k )(k 是一个给定的正实数)到实数集R 的对应过程:区间(0,k )中的实数m 对应线段AB 上的点M ,如图1;将线段AB 弯成半圆弧,圆心为H , 如图2;再将这个半圆置于直角坐标系中,使得圆心H 坐标 为(0,1),直径AB 平行x 轴,如图3;在图形变化过程中, 图1中线段AM 的长度对应于图3中的圆弧AM 的长度,直线HM 与直线1y =-相交与点N (,1)n -.,则与实数m 对应的实数就是n ,记作()n f m =.给出下列命题:(1)()64kf =;(2)函数()n f m =是奇函数;(3)()n f m =是定义域上的单调递增函数;(4)()n f m =的图象关于点(,0)2k 对称;(5)方程()2f m =的解是34m k =. 其中正确命题序号为_______.二、选择题(本大题满分20分)本大题共有4题,每题有且只有一个正确答案.考生必须在答题纸的相应编号上,将代表答案的小方格涂黑,选对得5分,否则一律得零分.15.“2a =”是“直线210x ay +-=与直线220ax y +-=平行”的( )(A )充要条件 (B )充分不必要条件 (C )必要不充分条件 (D )既不充分也不必要条件16.某林场有树苗30000棵,其中松树苗4000棵.为调查树苗的生长情况,采用分层抽样的方法抽取一个容量为150的样本,则样本中松树苗的数量为 ( ) (A ) 20; (B ) 15; (C ) 25; (D )30;M Am 图图图17.若M 为ABC ∆所在平面内一点,且满足()()02=-+⋅-,则∆ABC 的形状为( )(A )正三角形 (B )直角三角形(C )等腰三角形(D )等腰直角三角形18.在平面直角坐标系中,设点(,)P x y ,定义[]||||OP x y =+,其中O 为坐标原点.对于下列结论: (1)符合[]1OP =的点P 的轨迹围成的图形的面积为2;(2)设点P220y +-=上任意一点,则min []1OP =;(3)设点P 是直线: 1 ()y kx k R =+∈上任意一点,则“使得[]OP 最小的点P 有无数个”的充要条件是“1k =±”; (4)设点P 是圆221x y +=上任意一点,则max []OP =其中正确的结论序号为 ( ) (A ) (1)、(2) 、(3) (B )(1)、(3)、(4) (C ) (2)、(3)、(4) (D )(1)、(2)、(4)三、解答题(本大题满分74分)本大题共有5题,解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤. 19.(本题满分12分)本题共有3个小题,第(1)小题满分4分,第(2)小题满分4分,第(3)小题满分4分. 某高速公路收费站入口处的安全标识墩如图1所示,墩的上半部分是正四棱锥P -EFGH,下半部分是长方体ABCD -EFGH .图2、图3分别是该标识墩的主视图和俯视图. (1)请画出该安全标识墩的侧(左)视图; (2)求该安全标识墩的体积; (3)证明:直线BD ⊥平面PEG .20.(本题满分14分)本题共有2个小题,第1小题满分7分,第2小题满分7分.mm 图2 图3已知函数21()cos cos ()2f x x x x x R =-+∈. (1)求函数()f x 的最小正周期;(2)求函数()f x 在区间[0,]2π上的函数值的取值范围.21.(本题满分14分)本题有2个小题,第1小题满分7分,第2小题满分7分. 已知数列}{n a 的前n 项和)1(23-=n n a S ,n ∈*N . (1)求}{n a 的通项公式;(2)若对于任意的n ∈*N ,有14+≥⋅n a k n 成立,求实数k 的取值范围.22.(本题满分16分)本题共有3个小题,第1小题满分4分,第2小题满分6分,第3小题满分6分.已知过点(1,0)A -的动直线l 与圆22:(3)4C x y +-=相交于,P Q 两点,M 是PQ 中点,l 与直线:360m x y ++=相交于N . (1)求证:当l 与m 垂直时,l 必过圆心C ;(2)当PQ =l 的方程;(3)探索AM AN ∙是否与直线l 的倾斜角有关?若无关,请求出其值;若有关,请说明理由.23.(本题满分18分)本题共有3个小题,第1小题满分4分,第2小题满分6分,第3小题满分8分. 已知函数2()3,()2f x mx g x x x m =+=++,设函数()G x =()()1f x g x --。
上海名校2012年三模冲刺试卷集(华附二中文科)
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浦东新区2012年高三综合练习数学(文科)试卷2012.05注意:1.答卷前,考生务必在答题纸上将学校、班级、姓名、考号填写清楚. 2.本试卷共有23道试题,满分150分,考试时间120分钟.一、填空题(本大题满分56分)本大题共有14题,考生应在答题纸编号的空格内直接填写结果,每个空格填对得4分,否则一律得零分.1.函数21y x =-的单________.2. 已知(2,3),(4,),//,a b x a b x ==且则=______.3. 已知,x y R ∈,i 为虚数单位,且(2)1x i y i --=-+,则x y +=_____.4.已知3cos 5x =,,02x π⎛⎫∈- ⎪⎝⎭, 则2sin cos 1sin x x x =_____5. 已知01x <<,则(1)x x -的最大值是_______.6.方程lg(2)lg(3)lg12x x -+-=的解是_________.7.数列{}n a 的前n 项和为n S ,若点(,)n n S (*n N ∈)在函数2log (1)y x =+的反函数的图像上,则n a =________.8.在5张卡片上分别写上数字1,2,3,4,5,然后把它们混合,再任意排成一行,组成5位数,则得到能被5整除的5位数的概率为______。
9. 若复数z a bi =+(i 为虚数单位)满足1a ≤且1b ≤,则z 在复平面内所对应的图形的面积为__.10.若直线340x y m ++=与曲线 222440x y x y +-++=没有公共点,则实数m 的取值范围是____________.11.若一个底面为正三角形、侧棱与底面垂直的棱柱的三视图如下图所示,则这个棱柱的表面积为___________12. 已知函数22()(4)2f x x b a x a b =+--+-是偶函数,则函数图像与y 轴交点的纵坐标的最大值是______.13. 定义一个对应法则f :()()()/,,,0,0P m n Pm n m n →≥≥.现有点()/1,3A 与()/3,1B 点,点/M 是线段//A B 上一动点,按定义的对应法则f :/M M →。
上海市八校联考2024学年高三第二学期期末质量抽测数学试题
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上海市八校联考2024学年高三第二学期期末质量抽测数学试题请考生注意:1.请用2B 铅笔将选择题答案涂填在答题纸相应位置上,请用0.5毫米及以上黑色字迹的钢笔或签字笔将主观题的答案写在答题纸相应的答题区内。
写在试题卷、草稿纸上均无效。
2.答题前,认真阅读答题纸上的《注意事项》,按规定答题。
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.已知集合{}2|230A x x x =--<,集合{|10}B x x =-≥,则()A B ⋂=R( ).A .(,1)[3,)-∞+∞B .(,1][3,)-∞+∞C .(,1)(3,)-∞+∞D .(1,3)2.如图,在三棱锥S ABC -中,SA ⊥平面ABC ,AB BC ⊥,现从该三棱锥的4个表面中任选2个,则选取的2个表面互相垂直的概率为( )A .12B .14C .13D .233.设x ∈R ,则“|1|2x -< “是“2x x <”的( ) A .充分而不必要条件 B .必要而不充分条件 C .充要条件D .既不充分也不必条件4.已知我市某居民小区户主人数和户主对户型结构的满意率分别如图和如图所示,为了解该小区户主对户型结构的满意程度,用分层抽样的方法抽取30%的户主进行调查,则样本容量和抽取的户主对四居室满意的人数分别为A .240,18B .200,20C .240,20D .200,185.已知R 为实数集,{}2|10A x x =-≤,1|1B x x ⎧⎫=≥⎨⎬⎩⎭,则()A B =R( )A .{|10}x x -<≤B .{|01}x x <≤C .{|10}x x -≤≤D .{|101}x x x -≤≤=或6.波罗尼斯(古希腊数学家,的公元前262-190年)的著作《圆锥曲线论》是古代世界光辉的科学成果,它将圆锥曲线的性质网罗殆尽,几乎使后人没有插足的余地.他证明过这样一个命题:平面内与两定点距离的比为常数k (k >0,且k≠1)的点的轨迹是圆,后人将这个圆称为阿波罗尼斯圆.现有椭圆2222x y a b+=1(a >b >0),A ,B 为椭圆的长轴端点,C ,D 为椭圆的短轴端点,动点M 满足MA MB=2,△MAB 面积的最大值为8,△MCD 面积的最小值为1,则椭圆的离心率为( ) A .23B .33C .22D .327.已知函数()()()2cos 0,0f x x ωϕωϕπ=+><≤的图象如图所示,则下列说法错误的是( )A .函数()f x 在1711,1212ππ⎡⎤--⎢⎥⎣⎦上单调递减B .函数()f x 在3,2ππ⎡⎤⎢⎥⎣⎦上单调递增 C .函数()f x 的对称中心是(),026k k Z ππ⎛⎫-∈ ⎪⎝⎭ D .函数()f x 的对称轴是()5212k x k Z ππ=-∈ 8.已知抛物线2:4C y x =和点(2,0)D ,直线2x ty =-与抛物线C 交于不同两点A ,B ,直线BD 与抛物线C 交于另一点E .给出以下判断:①以BE 为直径的圆与抛物线准线相离;②直线OB 与直线OE 的斜率乘积为2-;③设过点A ,B ,E 的圆的圆心坐标为(,)a b ,半径为r ,则224a r -=. 其中,所有正确判断的序号是( ) A .①②B .①③C .②③D .①②③9.甲、乙、丙、丁四人通过抓阄的方式选出一人周末值班(抓到“值”字的人值班).抓完阄后,甲说:“我没抓到.”乙说:“丙抓到了.”丙说:“丁抓到了”丁说:“我没抓到."已知他们四人中只有一人说了真话,根据他们的说法,可以断定值班的人是( ) A .甲B .乙C .丙D .丁10.设x ,y 满足约束条件21210x y x y x y +≤⎧⎪+≥-⎨⎪-≤⎩,若32z x y =-+的最大值为n ,则12n x x ⎛⎫- ⎪⎝⎭的展开式中2x 项的系数为( ) A .60 B .80C .90D .12011.函数的图象可能是下面的图象( )A .B .C .D .12.如图,在等腰梯形ABCD 中,//AB DC ,222AB DC AD ===,60DAB ∠=︒,E 为AB 的中点,将ADE ∆与BEC ∆分别沿ED 、EC 向上折起,使A 、B 重合为点F ,则三棱锥F DCE -的外接球的体积是( )A .68B .64C .32π D .23π 二、填空题:本题共4小题,每小题5分,共20分。
上海市闵行区2012年高考二模数学(文)试题
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上海市闵行区2012届高三4月教学质量调研(二模)数 学 试 卷(文科)本试卷共有23道题,共4页.满分150分,考试时间120分钟.一. 填空题(本大题满分56分)本大题共有14题,考生应在答题纸上相应编号的空格 内直接填写结果,每个空格填对得4分,否则一律得零分. 1.不等式128x ≤≤的解是 . 2.计算23lim(2)n nn n →∞+++=+ .3.在等差数列{}n a 中,33a =,45a =,则13a = .4.已知复数z =(i为虚数单位),则z z ⋅= . 5.已知两条直线1l :230ax y --=,2l :0164=-+y x .若1l 的一个法向量恰为2l的一个方向向量,则=a .6.函数2cos cos y x x x =+的最小值为 .7.二项式41)x的展开式的各项系数的和为p ,所有二项式系数的和为q ,则:p q 的值为 .8.如右图,若输入的 5.54a b c =-==-,,则执行该程序框图所得的结果是 .9.已知大小、形状、颜色完全相同的n (*n ∈N )个乒乓球中有5个是次品,从中随机抽取5个加以检验,若至少抽到3个次品的概率是(01)P P <<,则至多抽到2个次品的概率是(用含P 的式子表示) .10.已知实数x y ,满足33010x x y x y ≤⎧⎪+-≥⎨⎪-+≥⎩,则22x y +的最小值是 .11.设P 为双曲线2213x y -=虚轴的一个端点,Q 为双曲线上的一个动点,则PQ 的最小值为 .12.已知曲线C :922=+y x )0,0(≥≥y x 与直线4x y +=相交于点1122()()A x y B x y ,,,,则1221x y x y +的值为 .13.问题“求不等式345x x x +≤的解”有如下的思路:不等式345x x x +≤可变为34()()155x x +≤,考察函数34()()()55x x f x =+可知,函数()f x 在R 上单调递减,且(2)1f =,∴原不等式的解是2x ≥.仿照此解法可得到不等式:33(23)(23)x x x x -+>+-的解是 .14.若1)(+=x xx f ,)()(1x f x f =,()[]()*1()2n n f x f f x n n -=≥∈N ,,则()()++21f f …()()()()1220122012111f f f f +++++ = .二. 选择题(本大题满分20分)本大题共有4题,每题只有一个正确答案.考生应在答题纸的相应编号上,将代表答案的小方格涂黑,选对得5分,否则一律得零分. 15.已知向量a b 、都是非零向量,“||||a b a b ⋅=⋅”是“//a b ”的 [答]( )(A )充分非必要条件. (B) 必要非充分条件.(C )充要条件. (D )既非充分也非必要条件. 16.将sin 2y x =的图像向右平移6π个单位,即得到()y f x =的图像,则[答]( ) (A) ()sin(2)6f x x π=-. (B) ()sin(2)6f x x π=+.(C) ()sin(2)3f x x π=-. (D) ()sin(2)3f x x π=+.17.如图几何体由前向后方向的正投影面是平面EFGH ,则该几何体的主视图是 [答]( )18.方程||||1169y y x x +=-的曲线即为函数)(x f y =的图像,对于函数)(x f y =,有如下结论:①)(x f 在R 上单调递减;②函数()4()3F x f x x =+不存在零点;③函数)(x f y =的值域是R ;④若函数()g x 和)(x f 的图像关于原点对称,则()y g x =由方程||||1169y y x x +=确定.其中所有正确的命题序号是 [答]( ) (A) ①③. (B) ①④. (C) ①③④. (D) ①②③.三. 解答题(本大题满分74分)本大题共有5题,解答下列各题必须在答题纸相应编号的规定区域内写出必要的步骤.19.(本题满分12分)已知p :(1)4z x i =-+ (其中x ∈R ,i 是虚数单位)的模不大于5,和3223100x q x x -<:,若利用p q 、构造一个命题“若p ,则q ”,试判断该命题及其逆命题的真假,并说明理由.20.(本题满分14分)本题共有2个小题,每小题满分各7分.如图,在四棱锥P A B C D -中,底面A B C D 是矩形,PA ⊥平面A B C D ,22PA AD AB ===,E 是PB 的中点.(1)求三棱锥P ABC -的体积;E FGH(C ) (B ) (A ) (D )E DBCA P(2)求异面直线EC 和AD 所成的角(结果用反三角函数值表示).21.(本题满分14分)本题共有2个小题,每小题满分各7分.如图,两铁路线垂直相交于站A ,若已知AB =100千米,甲火车从A 站出发,沿AC 方向以50千米/小时的速度行驶,同时乙火车从B 站出发,沿BA 方向以v 千米/小时的速度行驶,至A 站即停止前行(甲车仍继续行驶)(两车的车长忽略不计). (1)求甲、乙两车的最近距离(用含v 的式子表示); (2)若甲、乙两车开始行驶到甲,乙两车相距最近时所用时间为0t 小时,问v 为何值时0t 最大?22.(本题满分16分)本题共有3个小题,第(1)小题满分4分,第(2)小题满分6分,第(3)小题满分6分.已知椭圆22142x y +=的两焦点分别为12F F 、,P 是椭圆在第一象限内的一点,并满足121PF PF ⋅=,过P 作倾斜角互补的两条直线PA PB 、分别交椭圆于A B 、两点. (1)求P 点坐标;(2)当直线PA经过点(时,求直线AB 的方程;(3)求证直线AB 的斜率为定值.23.(本题满分18分)本题共有3个小题,第1小题满分4分,第2小题满分6分,第3小题满分8分.如图,在y 轴的正半轴上依次有点12n A A A 、、、、,其中点1(0,1)A 、2(0,10)A ,且||3||11+-=n n n n A A A A ),4,3,2( =n ,在射线)0(≥=x x y 上依次有点12n B B B 、、、、,点1B 的坐标为(3,3),且22||||1+=-n n OB OB ),4,3,2( =n . (1)求||1+n n A A (用含n 的式子表示);(2)求点n A 、n B 的坐标(用含n 的式子表示);(3)设四边形11n n n n A B B A ++面积为n S ,问{}n S 中是否存在两项n S ,k S (1,)n k n k <<∈N 、,使得1S ,n S ,k S 成等差数列?若存在,求出所有这样的两项,若不存在,请说明理由.参考答案与评分标准说明:A B CA1.本解答仅列出试题的一种或两种或三种解法,如果考生的解法与所列解答不同,可参考解答中的评分精神进行评分.2.评阅试卷,应坚持每题评阅到底,不要因为考生的解答中出现错误而中断对该题的评阅,当考生的解答在某一步出现错误,影响了后继部分,但该步以后的解答未改变这一题的内容和难度时,可视影响程度决定后面部分的给分,这时原则上不应超过后面部分应给分数之半,如果有较严重的概念性错误,就不给分.一、(第1题至第14题) 1.[]0,3; 2.12; 3.23; 4.13;5.3; 6.12-; 7.文16,理4; 8(或b );9.文1P -,理30;10.文92;11.文212.9;13.文3x <-,理1x <-或3x >; 14.2012.二、(第15题至第18题) 15.A ; 16. C ; 17.D ; 18.D . 三、(第19题至第23题) 19.解:由p 得22(1)42524x x -+≤⇒-≤≤, (4分)由q 得3223100x x x -<2230x x ⇒--≤13x ⇒-≤≤, (8分)由[24][ 1 3]--,,Ý,即p q ⇒,但q p ⇒,∴命题“若p 则q ”是假命题(10分) 而其逆命题“若q 则p ”是真命题. (12分) 20. [解](文) (1) 依题意,PA ⊥平面ABCD ,底面ABCD 是矩形,高2PA =,2BC AD ==,1AB = (2分)∴12112ABC S =⋅⋅=△ (4分) 故121233P ABC V -=⨯⨯=. (7分) (2)∵//BC AD ,所以ECB ∠或其补角为异面直线EC 和AD 所成的角θ,(2分)又∵PA ⊥平面ABCD ,∴PA BC ⊥,又BC AB ⊥,∴BC PAB ⊥面,∴BC PB ⊥,于是在Rt CEB ∆中,2BC =,12BE PB ===, (4分)tan BE BC θ===, (6分)∴异面直线EC 和AD所成的角是(或. (7分) (理)(1) 解法一:分别以AB AD AP 、、为x 轴、y 轴、z 轴建立空间直角坐标系,依题意,42AD AB ==,,则各点坐标分别是 (0 0 0)A ,,,(2 0 0)B ,,,(2 4 0)C ,,,(0 4 0)D ,,,(0 0 2)P ,,,∴(1 0 1)E ,,,(1 2 1)F ,,,(1 4 1)EC =-,,, 又∵AB ⊥平面PAD ,∴平面PAD 的法向量为(2,0,0)n AB ==, (2分) 设直线EC 与平面PAD 所成的角为α,则EDB CAP2sin 18||||EC n EC n α⋅===⋅, (6分) ∴直线EC 与平面PAD 所成的角为arcsin(7分)解法二:∵PA ⊥平面ABCD ,∴C D P A ⊥,又C D A D ⊥,∴CD ⊥平面PAD ,取PA 中点G ,CD 中点H ,联结EG GH GD 、、,则E G A B C////且1=12EG AB =,EGHC∴是平行四边形,∴HGD∠即为直线EC 与平面PAD 所成的角. (2分)在Rt GAD ∆中,GD =,在Rt GHD ∆中,tanHD HGD GD ∠===,(6分) ∴直线EC 与平面PAD 所成的角为. (7分)(2)解法一:由(1)解法一的建系得,(1 21)AF =,,,(0 4 0)AD =,,,设平面AFD 的法向量为(,,)n x y z =,点P 到平面AFD 的距离为d ,由0AF n ⋅=,0AD n ⋅=得20x y z ++=且40y =,取1x =得(1,0,1)n =-,∴22AP n d n⋅===(2分) 又6AF FD ==,∴2AFD S ==△(4分)∴1433P AFD V -=⨯=. (7分) 解法二:易证PE 即为三棱锥PAFD -底面上的高,且PE = (2分) 底面AFD △边AD 上的高等于AE ,且AE =AFD S =△(4分) 1144323P AFD V -=⨯⨯=. (7分)解法三:依题意,//EF 平面PAD ,∴P AFD F PAD E PAD D PAE V V V V ----===(4分) 11114224322123D PAE V PA AB AD -=⨯⨯⨯⨯⨯=⨯⨯⨯=. (7分)21. [解](1)设两车距离为d ,则22222100(100)(50)(2500)20010000(0)dvt t v t vt t v=-+=+-+≤≤ (3分)210010002500v v v <<+,∴当21002500v t v =+时,min d =千米; (7分)(2)当两车相距最近时,02100100125002500v t v v v==≤++, (3分)此时50v =千米/小时. (5分)即当车速50v =千米/小时,两车相距最近所用时间0t 最大,最大值是1小时.(7分)22. [解](1)由题可得1(F ,2F ,设)0,0(),(00000>>y x y x P 则100(,)PF x y =-,200(2,)PF x y =-,∴22120021PF PF x y ⋅=+-=,(1分)∵点),(00y x P 在曲线上,则220012x y +=,(2分)解得点P 的坐标为. (4分) (2)当直线PA 经过点(时,则PA 的斜率为1-,因两条直线PA PB 、的倾斜角互F E DBCAPH G补,故PB 的斜率为1,由222131)20142y x x x y x -=-+⎧⎪⇒-++=⎨+=⎪⎩得,12x x ==即A x =,故A y =(2分)同理得B x =By =(4分)∴直线AB的方程为23y x =- (6分)(3) 依题意,直线PA PB 、的斜率必存在,不妨设BP 的方程为:1(0)y k x k -=>.由221(142y k x y x -=⎧⎪⎨+=⎪⎩ 得222(21)41)420k x k x k +--+--=,(2分)设),(B B y x B ,则241)21B k x k -+=+,22421B k x k -=+,同理22421A k x k +=+,则2821A B kx x k -=+,同理2(21A B A By y k x x k -=-+-=+.(4分) 所以:AB的斜率2A B AB A B y y k x x -==-为定值. (6分) 23. [解](1)9110||,31||||2111=-==-+A A A A A A n n n n 且 , (2分) 311211)31()31(9)31(||||---+===∴n n n n n A A A A (4分)(2)由(1)的结论可得12231||||||n n A A A A A A -+++4412711931()()3223n n --=++++=- (2分) n A 点∴的坐标42911(0,())223n --, (3分)1||||n n OB OB --=2,3,n =)且1||OB ={||}n OB ∴是以23为首项,22为公差的等差数列 (5分)||((2n OB n n ∴=-=+n B 的坐标为(21,21)n n ++.(6分) (3)(文)连接1+n n B A ,设四边形11n n n n A B B A ++的面积为n S ,则111n n n n nn n A A B B B A SS S +++∆∆=+341112911[()](23)[()232223n n n --=⋅++⋅-32923n n-=+ (2分) 由1S ,n S ,k S (1,)n k n k <<∈N 、成等差数列,332929292()(9)()23223n k n k--+=+++即123()36k n n k =⋅-,①(4分)∵111120333n n n n n n +++--=<,∴3n n ⎧⎫⎨⎬⎩⎭是单调递减数列. 当3n ≥时,139n n ≤,①式右边小于0,矛盾, (6分)当2n =时,得23k k -=,易知3k =是唯一解,∴1S ,2S ,3S 成等差数列. 即当3n ≥时,{}n S 中不存在1S ,n S ,k S 三项成等差数列.综上所述,在数列{}n S 中,有且仅有1S ,2S ,3S 成等差数列. (8分) (理)连接1+n n B A ,设四边形11n n n n A B B A ++的面积为n S ,则111n n n n n n n A A B B B A S S S +++∆∆=+341112911[()](23)[()232223n n n --=⋅++⋅-32923n n -=+ (2分) 不妨设 (1 )m n k S S S m n k m n k ≤<<∈N ,,,、、成等差数列, 又12120,3n n n nS S +---=<,1n n S S <+即}{n S ∴是单调递减数列.n S ∴是等差中项,即2n m k S S S =+,∴3332929292()()()232323n m k n m k ---+=+++,即2333n m k n m k=+1)当1m =,2n =时,得23k k -=,3k =是唯一解,∴1S ,2S ,3S 成等差数列(4分)2)当1m =,3n ≥时,即123()36k n n k =⋅-,① ∵111120333n n n n n n +++--=<,∴3n n ⎧⎫⎨⎬⎩⎭是单调递减数列.当3n ≥时,139n n ≤,①式右边小于0,矛盾, (6分)3)当2m ≥时,2n m k S S S =+不可能成立. ∵111120333n n n n n n +++--=<,∴数列{}3n n 是递减数列,当2m ≥时,32(1)m m ≥+,由2m n k ≤<<(m n k ∈N 、、)知,1n m ≥+ ∴112(1)323333m m m n m m m n +++=≥≥(当且仅当23m n ==,时等号成立) ∴2333m k n m k n+>对任意2m n k ≤<<(m n k ∈N 、、)恒成立, 即当2m ≥时,{}n S 中不存在不同的三项恰好成等差数列.综上所述,在数列{}n S 中,有且仅有123S S S ,,成等差数列. (8分)。
【恒心】【好卷速递】上海市华师大一附中等八校2012届高三2月联合调研考试卷(英语)
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上海2012届高三下学期(八)校联合调研考试英语试卷(八校联考)2012.2.16 参加学校:华师大一附中、曹杨二中、市西、市三女子、控江、格致、市北、(育才、晋元高中)第I卷(共105 分)II. Grammar and VocabularySection A: GrammarDirections: Beneath each of the following sentences there are four choices marked A, B, C and D. Choose the one answer that best completes the sentence.25. Keeping a foreign pet is dangerous, as the owner might be infected ________ the diseases brought along with the animal.A. inB. withC. byD. for26. All the ladies wear beautiful clothes for the celebration. Some are dressed in red, _______ in purple.A. otherB. anotherC. othersD. the other27. According to some theories ________ from psychoanalysis, life is supposedly easier; and more pleasant when anxiety is overcome.A. obtainB. obtainingC. to obtainD. obtained28. To understand the complex situation completely requires more thought than ________ so far.A. gaveB. has givenC. has been givenD. is being given29. To show our respect, we usually take our gloves off _______ we are to shake hands with.A. whoeverB. wheneverC. whicheverD. wherever30. You _______ an A on your term paper, but you quoted a little bit more from others' papers.A. should receiveB. could have receivedC. would receiveD. must have received31. Eric, a middle-aged fashion designer, is very happy because the clothes be designed have never been ________.A. most popularB. the most popularC. less popularD. more popular32. We are strongly against the company's decision _______ it will fire half of its staff in the following years.A. whichB. thatC. whatD. when33. Nancy, ________ for about half a year to apply for a job as an airline hostess, finally took a position at a shopping center.A. struggledB. having struggledC. strugglingD. to struggle34. Don‟t start reading a book ________ you find that it‟s one you can read with ease and understanding.A. whenB. ifC. unlessD. though35. I can‟t believe that anyone else in the school can swim as fast as you, ________?A. can IB. can't theyC. can't youD. can they36. One of the things that made me long ________ back in prison was that I had little opportunity thinking and reflection.A. beB. beingC. to beD. to being37. When you have a job interview, it's ________ you're dressed that sets the tone of it.A. howeverB. howC. whatD. whatever38. The new lecture course differs from the old one, ________ the students aren't required to attend lectures.A. which thatB. thatC. whichD. in which39. Fortunately, what the witness reported to the police on the phone led to ___________.A. the suspect's being capturedB. capture the suspectC. the suspect having capturedD. be captured by the suspect40. _______ they may have, the CEO of the corporation calls on all the staff to gather their courage to strive it.A. What a serious problemB. However a serious problemC. How serious a problemD. Whatever a serious problemSection B: VocabularyDirections: Complete the passage by choosing the proper words in the box. Each word can only beRealizing your dreams is not an easy ___41___.Last fall I decided to write a new book for my publisher. Writing a book is a ___42___ goal, which got off to a terrific start last October. The writing is flowing well. Then I got sick. In fact, I got ill that I needed surgery and the ___43___ was long and exhausting. I did not work from the first week of November until the second week of January. By then I was nervous about meeting the April deadline for submitting the new manuscript to my publisher.Worried, I asked my author friends for some help, and they gave me this key piece of advice, “Let‟s start writing!" they all said. So I did. It was not an immediate ___44___ t o my depression, but after a few weeks of ___45___, I got back to normal. Several people in my circle of supporters helped me make some good connections and I got the book ___46___, and to the publisher on time. It was an exciting goal for me to reach, so I took my family to Hawaii as my ___47___. Sometimes you ___48___ your own dreams because of self-doubt, fear, or external complications. You can think of many different excuses to ___49___ those dreams aside, but, if you go after your dreams, your world will become more exciting and you will begin to live a more passionate and meaningful life.So, now, take a minute to write down three goals you want to accomplish this year.III. Reading ComprehensionSection ADirections: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.In 1867 the United States faced the task of rebuilding after the destruction of the Civil War, so it looked westward for the raw materials needed to fuel industrial growth. Geological surveys and mapping journeys were set forth to explore this 50 territory. These groups, in turn, hired mapmakers, scientists, cooks, drivers, and doctors. They also hired painters and photographers as part of the teams. Painters needed few supplies, making it relatively easy for them to travel in the wilderness, ____51____ photographers were not so lucky; they had to transport a fully stocked darkroom on the ___52____.Until the late 1870s, most photographers used the ___53___ wet-collodion process. The first step was to wash a clean sheet of glass with a sticky mixture of collodion and chemicals, (collodion or "gun-cotton" was a recent medical discovery used to cover wounds because the viscous (粘性的) solution turned into a protective film when dry.) After it was washed, the plate went into another bath that the picture was ___54___ getting darker; Finally, the glass negative (底片) was washed clean with fresh water. ___55___ a photograph from the negative had to wait until the photographer went back to the studio. The ___56___ of the negative depended on the size of the camera. Some negatives could be as large as 20 by 24 inches.Imagine the ___57____ of taking photographs in the 1860s and 1870s in the remote western wilderness! Photographers went over Rocky Mountains and through rushing rivers. They were ___58____ in the terrible desert heat, with cameras, sheets of glass, and vats of chemicals. Bad weather, equipment failures, and accidents were frequent problems. They persevered, but success in creating a negative did not ___59___ the production of a photograph; plates still had to be ___60___ transported back to the studio before the image could be printed on paper. A photographer could carry 120 pounds of many miles to ____61____ a magnificent view only to have the easily broken plate ___62___ in transportation.___63___, once photographers were successful, the results were superb and much admired. Photographs were put on exhibition, and people bought albums filled with pictures by Timothy O'Sullivan, Carleton Watkins, and William Henry Jackson. Jackson's photographs of Yellowstone's natural wonders, along with the paintings of fellow Thomas Moran, even helped ___64___ Congress to preserve thousands of acres of this land in 1872 as the nation's first national park.Section BDirections: Read the following four passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.(A)I once had my Chinese MBA students brainstorming on “two-hour business plans”. I separated them into six groups and gave them an example: a restaurant chain. The more original their idea, the better, I said. Finally, five of the six groups presented plans for restaurant chains. The sixth proposed a catering service. Though I admitted the time limit had been difficult, I expressed my disappointment.My students were middle managers, financial analysts and financiers from state owned enterprises and global companies. They were not without talent or opinions, but they had been shaped by an educational system that rarely stressed or rewarded critical thinking or inventiveness. The scene I just described came in different forms during my two years‟ teaching at the school. Papers were often copied from the Web and the Harvard Business Review. Case study debates were written up and just memorized. Students frequently said that copying is a superior business strategy, better than inventing and creating.In China, every product you can imagine has been made and sold. But so few well developed marketing and management minds have been raised that it will be a long time before most people in the world can name a Chinese brand.With this problem in mind, partnerships with institutions like Yale and MIT have been established. And then there‟s the “thousand talent scheme”: this new government program is intended toimprove technological modernization by attracting top foreign trained scientists to the mainland with big money. But there are worries about China‟s research environment. It‟s hardly known for producing independent thinking and openness, and even big salary offers may not be attractive enough to overcome this.At last, for China, becoming a major world creator is not just about setting up partnerships with top Western universities. Nor is it about gathering a group of well-educated people and telling them t o think creatively. It‟s about establishing a rich learning environment for young minds. It‟s not that simple.65. Why does the author feel disappointed at his students?A. Because there is one group presenting a catering service.B. Because the six groups made projects for restaurant chains.C. Because all the students copied a case for the difficult topic.D. Because the students‟ ideas were lacking in creativeness.66. We can infer from the passage that ________.A. China can make and sell any product all over the worldB. high pay may not solve the problem of China’s research environmentC. cooperation with institutions has been set up to make a Chinese brandD. the new government program are aimed at encouraging imagination67. Which is the best title of the passage?A. Look for a New Way of Learning.B. Reward Creative Thinking.C. How to Become a Creator.D. Establish a technical Environment.(B)Every summer, no matter how urgent work schedule is, I take off one day exclusively for my son. We call it dad-son day. This year our third stop was the amusement panic, where we discoveredthat he was tall enough to ride one of the fastest roller coasters (过山车)in the world. We experienced through face-stretching turns and circles for ninety seconds. Then, as we stepped off the ride, in a calm voice, he remarked that it was not as exciting as other rides he‟d been on. As I listened, I began to sense something seriously out of balance.Throughout the season, I noticed similar events all around me. Parents found it hard to find new stimulations for cold kids. Surrounded by ever-greater stimulation, their young feces were looking disappointed and boredFacing their children's complaints of "nothing to do", parents were spending large numbers of dollars for various forms of entertainment In many cases the money seemed to do little more than buy relief from the terrible complaint of their bored children. This set me thinking the obvious question: "How can it be so hard for kids to find something to do when there's been such a range of stimulating entertainment available to them?"What really worries me is the strength of the stimulation. I watch my little daughter's &ce as she absorbs bloody special effects in movies.Why do children facing such excitement seem starved for more? Thai was, I realized, the point I discovered during my own adolescence that what creates excitement is not going fast, but goingfaster. Excitement has less to do with speed than changes in speed.I am concerned about the increasing effect of years at these levels of feverish activity. It is no mystery to me why many teenagers appear uninterested and burned out, with a "been there, done that" air of indifference toward much of life. As increasing numbers of frie nds‟ children are advised to take medicine to deal with inattentiveness at school or anti-depressants (抗抑郁药)to help with the loss of interest and joy in their lives—I question the role of kids‟ boredom in some of the diagnoses (处方).My own work is focused on the chemical imbalances and biological factors related to behavioral and emotional disorders. These are complex problems. Yet I‟ve been reflecting more and more on how lie pace of life and the strength of stimulation may be contributing to the rising rates of psychological problems among children and adolescents in our society.68. The reason why the author felt surprised in the amusement park was that ________.A. his son was not as excited by the roller coasters ride as expectedB. his songs enjoyed turns and circles with his face stretchedC. his son appeared upset but calm while riding the roller coastersD. his son could keep his balance so well on the fast moving roller coasters69. According to the author, children will probably feel excited ________.A. if their parents allow them to ride roller coasters very oftenB. since parents spend money on the same form of entertainmentC. after they take anti-depressants according to the diagnosesD. if they are often exposed to more stimulating entertainment70. From his own experience, the author came to the conclusion that children seem to expect _______.A. a much wider variety of sports facilitiesB. activities that require complicated skillsC. the change of the forms of recreationD. more challenging physical exercise71. In order to relieve children's boredom, the author would probably suggest ________.A. adjusting the pace of life and strength of stimulationB. promoting the practice of dad-son daysC. consulting a specialist in child psychologyD. balancing school work with after school activities(C)Cultural rules determine every aspect of food consumption. Who eats together defines social units. For example, in some societies, the nuclear family is the unit that regularly eats together. The anthropologist Mary Douglas has pointed out that, for the English, the kind of meal and the kind of food that is served relate to the kinds of social links between people who are eating together. She distinguishes between regular meals, Sunday meals when relatives may come, and cocktail parties for relatives and friends. The food served symbolizes the occasion and reflects who is present. For example, only snacks are served at a cocktail party. It would be inappropriate to serve a steak or hamburgers. The distinctions among cocktails, regular meals, and special dinners markthe social boundaries between those guests who are invited for drinks, those who are invited to dinner, and those who come to a family meal. In this example, the type of food symbolizes the category of guest and with whom it is eaten.In some New Guinea societies, the nuclear family is not the unit that eats together. The men take their meals in a men's house, separately from their wives and children. Women prepare and eat their food in their own houses and take the husband's portion to the men's house. The women eat with their children in their own houses. This pattern is also widespread among Near Eastern societies.Eating is a metaphor that is sometimes used to signify marriage. In many New Guinea societies, like that of the Lese on the island of New Ireland in the Pacific and that of the Trobriand Islanders, marriage is symbolized by the couple's eating together for the first time. Eating symbolizes their new status as a married couple. In U.S. society, it is just the reverse. A couple may go out to dinner on a first date.Other cultural rules have to do with taboos against eating certain things. In some societies, members of a family group, arc not allowed to eat the animal or bird that is their ancestor. Since they believe themselves to be children of that ancestor, it would be like eating that ancestor or eating themselves.There is also an association between food prohibitions and rank, which is found in its most extreme form in the caste (social class) system of India. A caste system consists of ranked groups, each with a different economic specialization. In India, there is an association between caste and the idea of pollution. Members of highly ranked groups can be polluted by coming into contact with the bodily secretions, particularly saliva(唾液),of individuals of lower-ranked castes. Because of the fear of pollution, Brahmans and other high-ranked individuals will not share food with, not eat from the same plate as, not even accept food from an individual or from a low-ranking class.72. According to the passage, who will NOT eat together?A. The English during regular meals.B. Americans on their first date.C. Men and women in Near Eastern societies.D. Newly-married people on the island of New Ireland.73. In Paragraph 4, the underlined word "taboos" means _____________.A. favorsB. prohibitionsC. hatredD. gossips74. According to the passage, eating together indicates all the following EXCEPT .A. the type of foodB. social relations.C. marital statusD. family ties.75. What is the main idea of the passage?A. Different kinds of food in western countries.B. Relations between food and social units.C. Symbolic meanings of different kinds of food.D. Food consumption in different cultures.找家教,可以找柯南东升,可以关注824135830空间,更多精彩请加821435830Section CDirections: Read the following text and choose the most suitable heading from the list A-F for76._______________Tim Sparks slides a small leather-bound notebook out of an envelope. The book's yellowing pages contain beekeeping notes made between 1941 and 1969 by the late Walter Coates of Kilworth, Leicestershire. He adds it to his growing pile of local journals, birdwatchers' lists and gardening diaries. "We're uncovering about one major new record each month," he says, "I still get surprised." Around two centuries before Coates, Robert Marsham, a landowner from Norfolk in east of England, began recording the life cycles of plants and animals on his estate. Successive Marshams continued recording these notes for 211 years.77._______________Today, such records are being put to uses that their authors couldn't possibly have expected. These data sets, and others like them, are proving valuable to ecologists interested in the timing of biological events, or phonology. By combining the records with climate data, researchers can reveal how, for example, changes in temperature affect the arrival of spring, allowing ecologists to make improved predictions about the impact of climate change.78._______________But not all professionals are happy to use amateur data. "A lot of scientists won't touch them, they say they're too full of problems," says Root. Because different observers can have different ideas of what forms, for example, an open snowdrop. "The biggest concern with ad hoc (临时的) observations is how carefully and systematically they were taken,” says Mar k Schwarts of the University of Wisconsin, Milwaukee, who studies the interactions between plants and climate. "We need to know pretty precisely what a person's been observing—if they just say …I noted when the leaves came out‟, it might not be that useful.” Measuring the onset of autumn can be particularly problematic because deciding when leaves change color is a more subjective process than noting when they appear.79._______________Overall, most phrenologists arc positive about the contribution that amateurs can make. "They get the raw power of science: careful observation of the natural world," says Sagarin. Others suggest that the right statistics can iron out some of the problems with amateur data. Together withcolleagues at Wageoingen University in the Netherlands, environmental scientist Arnold van Vliet is developing statistical techniques to account for the uncertainty in amateur phonological data. Besides, the data are cheap to collect, and can provide breadth in space, time and range of spe cies," It‟s very difficult to collect data on a large geographical scale without enlisting an army of observers, says Root.80._______________Phonology also helps to drive home messages about climate change. “Because the public understand these records, they accept them,” says Sparks. It can also illustrate potentially unpleasant consequences, he adds, such as the finding that more rat infestations are reported to local councils in warmer years. And getting people involved is great for public relations. "People are excited to think that the data they have been collecting as a hobby can be used for something scientific—it empowers them” says Root.Section DDirections: Read the passage carefully. Then answer the questions or complete the statements in the fewest possible wordsThe automobile may not be closely associated with modem American culture, but it has occupied a central role in America's economic and social history.No one can deny the status of Henry Ford in car history. When the first Ford Model T rolled off the assembly line in 1908, businessman Henry Ford transformed the car from a luxury attachment for the rich to an automobile product for the middle classes. The rush of cars into the community forced all levels of government to build new and better roads. Better roads fed the demand for larger, faster, more stylish vehicles, and a host of companies rushed to meet that demand.If there was a first Golden Age of automobile, it may well have been the 1950s. It was an age of prosperity. Large, regular paychecks encouraged the public display of wealth through costly items such as new cars. Americans, moreover, needed those cars as they moved away from the cities into the suburbs, where such things as stores, jobs, and schools were seldom within walking distance. Cars became essential if people were to get to work or to the grocery store.As Ac 1950s slipped into 1960s, it became apparent that these fashionable wheels were gas-consuming road cruisers, dangerous in an accident, and often full of faults. Under pressure from a variety of groins, the federal government required that newer models provide greater fuel efficiency and cleaner emissions. Seatbelts became standard equipment as well Rising fad prices in the 1970s, coupled with concern for the environment, made the smaller cars produced by foreign companies for European and Asian markets very popular.The 1980s and 1990s saw an upswing in the popularity of big cars. New models including minivans and sport utility vehicles have become main products in auto dealers' showrooms. Traffic jams on the roads have become part of the American way of life. In 1911 a horse could travel through rush-hour traffic in Los Angeles at 11 miles per hour. In 2000 a car covering the same territory at the same time of day moved at about 4 miles per hour. But perhaps that is not important. When a car is equipped with a telephone and television set, a computer, and global positioning satellite connections, it can feel just like home.(Note: Answer the questions or complete the statements in NO MORE THAN TEN WORDS.) 81. made Henry Ford well-known in car history.82. That the car industry prospered in the 1950s in America resulted from ________.83. Why were smaller cars once popular in the 1970s?84. In today's society, how can a car make people ignore traffic jams?找家教,可以找柯南东升,可以关注824135830空间,更多精彩请加821435830第II卷(共45分)I. TranslationDirections: Translate the following sentences into English, using the words given in the brackets.1. 实验结果比预想的要满意得多。
上海华师大一附中2012届高三第二学期开学检测数学试题
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华师大一附中2012届高三数学第二学期开学检测试题一、填空题(本大题满分56分)本大题共有14题,考生应在答题纸相应编号的空格内直接填写结果,每个空格填对得4分,否则一律得零分。
1.若集合{}{}1,2,3,4,2A B x N x ==∈≤,则____________A B =。
2.若复数1111i iz mi i+-=+-+(i 为虚数单位)为实数,则实数____________m =。
3.已知0a >,且函数212sin ()y ax =-的最小正周期为π,则_________a =。
4.在二项式452,)1(x xx 含的展开式中-的项的系数是__________。
5.(理)若直线:l y kx =与曲线{2cos :sin x C y θθ=+=(参数∈θR )有唯一的公共点,则实数k = 。
(文)若直线:l y kx =与圆22:(2)1C x y -+=有唯一的公共点,则实数k = 。
6.已知函数1()2f x x=+的反函数为1()f x -,若1()0f x ->,则x 的取值范围为_______。
7.某程序框图如图所示, 该程序运行后输出的k 的值是__________。
8.若数列{}n a 满足21211,1,22n n a a a a +=-==,则()12lim ________n n a a a →∞+++=。
9.如图,在ABC △中,3AB =,2AC =,D 是边BC 的中点,则________AD BC ⋅=。
10.高三⑴班共有56人,学号依次为1,2,3,,56,现用系统抽样的办法抽取一个容量为4的样本,已知学号为6,34,48的同学在样本中,那么还有一个同学的学号应为(第7题)CAB_______________。
11.(理)已知数列{}n a 的通项公式为3n n a =,集合*{|,99,}i A y y a i i N ==≤∈,*{|41,}B y y m m N ==+∈。
上海市十三校2012届高三12月第一次联考数学(文)试卷
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2012年高三年级十三校第一次联考数学(文科)试卷考试时间:120分钟 满分:150分一、填空题(本大题满分56分)本大题共有14题,每题4分.1. 已知*n N ∈,则1lim32n n n →∞+=- . 2. 如图,U 是全集,A U B U ⊆⊆,,用集合运算符号表示图中阴影部分的集合是 .3. 函数1()sin 2cos 22f x x x =-+的最小正周期是.4. 若2i +是方程20( )x bx c b c R ++=∈、的根,其中i 是 虚数单位,则b c += . 5. 若函数12()log a f x x -=在(0 )+∞,上单调递减, 则实数a 的取值范围是 .6. 图中是一个算法流程图,则输出的 正整数n 的值是 .7. 设函数212() 0()2log (2) 0x x f x x x ⎧⎪-≤=⎨+>⎪⎩的反函数 为1()y f x -=,若1()4f a -=,则实数a 的值是 .8. 如图,在ABC ∆中,90 6 BAC AB D ∠==,,在斜 边BC 上,且2CD DB =,则AB AD ⋅的值为 . 9. 对于任意的实数k ,如果关于x 的方程()f x k =最多有2个不同的实数解,则|()|f x m =(m 为实常数)的不同的实数解的个数最多为 .10. 已知01a <<,则函数|||log |x a y a x =-的零点的个数为 .11. 已知等差数列{}n a 的公差4d =,且711a =,若112k k a a ++>,则正整数k 的最小值 为 .12. 设不等式21log (0 1)a x x a a -<>≠且,的解集为M ,若(1 2)M ⊆,,则实数a 的取值范围是 . 13. 已知函数()2arctan x f x x =+,数列{}n a 满足*111()()()402312n n na a f a f n N a +==∈,-,则2012()f a = .14. 设 a b c ,,是平面内互不平行的三个向量,x R ∈,有下列命题: ①方程20(0)ax bx c a ++=≠不可能有两个不同的实数解;②方程20(0)ax bx c a ++=≠有实数解的充要条件是240b a c -⋅≥;③方程22220a x a bx b +⋅+=有唯一的实数解b x a=-; ④方程22220a x a bx b +⋅+=没有实数解.其中真命题有 .(写出所有真命题的序号)二、选择题(本大题满分20分)本大题共有4题,每小题5分.(第2题图)DA B C(第8题图)15. 满足不等式3121x x -≥+的实数x 的取值范围是 ( ) A.( 4]-∞-, B.1[4 ]2--,C.1( 4]( )2-∞--+∞,,D.1[4 )2--,16. 设角 αβ、是锐角,则“4παβ+=”是“(1tan )(1tan )2αβ++=”成立的 ( )A.充分非必要条件B.必要非充分条件C.充要条件D.既非充分也非必要条件17. 对于复数 a b c d 、、、,若集合{ }S a b c d =,,,具有性质:“对任意 x y S ∈,,都有xy S∈”,则当2211a b c b=⎧⎪=⎨=⎪⎩时,b c++的值是( )A.1B.1-C.iD.i -18. 某个QQ 群中有n 名同学在玩一个数字哈哈镜游戏,这些同学依次编号为1 2 3 n ,,,,.在哈哈镜中,每个同学看到的像用数对( )()p q p q <,表示,规则如下:若编号为k 的同学看到像为( )p q ,,则编号为1k +的同学看到像为( )q r ,,且*( )q p k p q r N -=∈,,.已知编号为1的同学看到的像为(5 6),.请根据以上规律,编号为3和n 的同学看到的像分别是 ( )A.(7 10) (4 24)n n ++,;,B.22838(10 13) ( )22n n n n ++++,;,C.222545(10 13) ( )22n n n n ++++,;, D.221010(8 11) ( )22n n n n -+++,;, 三、解答题(本大题共5小题,满分74分)19. (本题满分12分)已知矩阵||5||10x x +⎛⎫⎪+ ⎝的某个列向量的模不大于行列式2143的值,求实数x 的取值范围.20. (本题满分14分,第1小题满分7分,第2小题满分7分)为了研究某种癌细胞的繁殖规律和一种新型抗癌药物的作用,将癌细胞注入一只小白鼠体内进行实验,经检测,癌细胞的繁殖规律与天数的关系如下表.已知这种癌细胞在小白鼠体内的个数超过810时小白鼠将会死亡,注射这种抗癌药物可杀死其体内癌细胞的98%.天数t 1 2 3 4 5 6 7 …癌细胞个数N1 2 4 8 16 32 64 … (1)要使小白鼠在实验中不死亡,第一次最迟应在第几天注射该种药物?(精确到1天) (2)若在第10天,第20天,第30天,……给小白鼠注射这种药物,问第38天小白鼠是否仍然存活?请说明理由.21. (本题满分14分,第1小题满分7分,第2小题满分7分)已知函数())(0)3f x x πωω=+>.(1)若()(0)2y f x πθθ=+<<是最小正周期为π的偶函数,求ω和θ的值;(2)若()(3)g x f x =在(0 )3π,上是增函数,求ω的最大值;并求此时()f x 在[0 ]π,上的取值范围.22. (本题满分16分,第1小题满分4分,第2小题满分6分,第3小题满分6分)设等比数列}{n a 的前n 项和为n S ,已知*122()n n a S n N +=+∈.(1)求数列}{n a 的通项公式;(2)在n a 与*1()n a n N +∈之间插入n 个1,构成如下的新数列:1234 1 1 1 1 1 1 a a a a ,,,,,,,,,,,求这个数列的前2012项的和;(3)在n a 与1n a +之间插入n 个数,使这2n +个数组成公差为n d 的等差数列(如:在1a 与2a 之间插入1个数构成第一个等差数列,其公差为1d ;在2a 与3a 之间插入2个数构成第二个等差数列,其公差为2d ,…以此类推),设第n 个等差数列的和是n A . 是否存在一个关于n的多项式()g n ,使得()n n A g n d =对任意*n N ∈恒成立?若存在,求出这个多项式;若不存在,请说明理由.23. (本题满分18分,第1小题满分5分,第2小题满分5分,第3小题满分8分)已知函数22()(1)(1)x b f x x D a x=-+-∈,,其中0a b <<.(1)当(0 )D =+∞,时,设xba x t +=,()()f x g t =,求()y g t =的解析式及定义域; (2)当(0 )D =+∞,,12ab ==,时,求)(x f 的最小值;(3)设0k >,当22(1)a k b k ==+,时,1()9f x ≤≤对任意[ ]x a b ∈,恒成立,求k 的取值范围.2012年高三年级十三校第一次联考数学(文科)答题纸题号15 1617 18 答案三.解答题考试时间:120分钟 满分:150分一、填空题(本大题满分56分)本大题共有14题,每题4分.24. 已知*n N ∈,则1lim32n n n →∞+=- .1325. 如图,U 是全集,A U B U ⊆⊆,,用集合运算符号 表示图中阴影部分的集合是 .U A B ð26. 函数1()sin 2cos 22f x x x =-+的最小正周期是 .π27. 若2i +是方程20( )x bx c b c R ++=∈、的根,其中i 是虚数单位,则b c += .1 28. 若函数12()log a f x x -=在(0 )+∞,上单调递减,则实数a 的取值范围是 .102a << 29. 图中是一个算法流程图,则输出的 正整数n 的值是 .1130. 设函数212() 0()2log (2) 0x x f x x x ⎧⎪-≤=⎨+>⎪⎩的反函数 为1()y f x -=,若1()4f a -=,则实数a 的值是 .2log 631. 如图,在ABC ∆中,90 6 BAC AB D ∠==,,在斜 边BC 上,且2CD DB =,则AB AD ⋅的值为 .2432. 对于任意的实数k ,如果关于x 的方程()f x k =最多有2个不同的实数解,则|()|f x m =(m 为实常数)的不同的实数解的个数最多为 .433. 已知01a <<,则函数|||log |x a y a x =-的零点的个数为 .234. 已知等差数列{}n a 的公差4d =,且711a =,若112k k a a ++>,则正整数k 的最小值 为 . 635. 设不等式21log (0 1)a x x a a -<>≠且,的解集为M ,若(1 2)M ⊆,,则实数a 的取值范围是 . (1 36. 已知函数()2arctan x f x x =+,数列{}n a 满足*111 ()()()402312n n n a a f a f n N a +==∈,-,则2012()f a = .24π+37. 设 a b c ,,是平面内互不平行的三个向量,x R ∈,有下列命题: ①方程20(0)ax bx c a ++=≠不可能有两个不同的实数解;②方程20(0)ax bx c a ++=≠有实数解的充要条件是240b a c -⋅≥;③方程22220a x a bx b +⋅+=有唯一的实数解b x a=-; (第2题图)DA B C(第8题图)④方程22220a x a bx b +⋅+=没有实数解.其中真命题有 .(写出所有真命题的序号) ①④二、选择题(本大题满分20分)本大题共有4题,每小题5分. 38. 满足不等式3121x x -≥+的实数x 的取值范围是 ( D ) A.( 4]-∞-, B.1[4 ]2--,C.1( 4]( )2-∞--+∞,,D.1[4 )2--, 39. 设角 αβ、是锐角,则“4παβ+=”是“(1tan )(1tan )2αβ++=”成立的 ( C)A.充分非必要条件B.必要非充分条件C.充要条件D.既非充分也非必要条件40. 对于复数 a b c d 、、、,若集合{ }S a b c d =,,,具有性质:“对任意 x y S ∈,,都有xy S∈”,则当2211a b c b=⎧⎪=⎨=⎪⎩时,b c++的值是( B )A.1B.1-C.iD.i -41. 某个QQ 群中有n 名同学在玩一个数字哈哈镜游戏,这些同学依次编号为1 2 3 n ,,,,.在哈哈镜中,每个同学看到的像用数对( )()p q p q <,表示. 规则如下:若编号为k 的同学看到像为( )p q ,,则编号为1k +的同学看到像为( )q r ,,且*( )q p k p q r N -=∈,,.已知编号为1的同学看到的像为(5 6),.请根据以上规律,编号为3和n 的同学看到的像分别是 ( D )A.(7 10) (4 24)n n ++,;,B.22838(10 13) ( )22n n n n ++++,;,C.222545(10 13) ( )22n n n n ++++,;, D.221010(8 11) ( )22n n n n -+++,;, 三、解答题(本大题共5题,满分74分)42. (本题满分12分)已知矩阵||5||10x x +⎛⎫⎪+ ⎝的某个列向量的模不大于行列式2143的值,求实数x 的取值范围. 解:依题意,21243=, (4)分显然列向量||5||10x x a +⎛⎫⎪+= ⎪ ⎪⎝⎭的模不大于2,即||52||1x x +≤+,…………………………………8分 解得3x ≥,或3x ≤-∴满足条件的实数x 的取值范围是( 3][3 )-∞-+∞,,…………………………………12分43. (本题满分14分,第1小题满分7分,第2小题满分7分)为了研究某种癌细胞的繁殖规律和一种新型抗癌药物的作用,将癌细胞注入一只小白鼠体内进行实验,经检测,癌细胞的繁殖规律与天数的关系如下表.已知这种癌细胞在小白鼠体内的个数超过810时小白鼠将会死亡,注射这种抗癌药物可杀死其体内癌细胞的98%.天数t 1 2 3 4 5 6 7 …癌细胞个数N1 2 4 8 16 32 64 … (1)要使小白鼠在实验中不死亡,第一次最迟应在第几天注射该种药物?(精确到1天) (2)若在第10天,第20天,第30天,……给小白鼠注射这种药物,问第38天小白鼠是否仍然存活?请说明理由.解:(1)依题意,18210t -≤……………………………………………………………………2分 ∴82log 10127.58t ≤+≈……………………………………………………………………5分 即第一次最迟应在第27天注射该种药物. …………………………………………………7分(2)设第n 次注射药物后小白鼠体内的这种癌细胞个数为n a ,则912(198%)a =-,且1012(198%)n n a a +=-,∴1012(198%)n nn a -=-…………………10分于是1031332(198%)a ⨯-=-,即第3次注射后小白鼠体内的这种癌细胞个数为3232100,…12分到第38天小白鼠体内的这种癌细胞个数为32878322 1.11010100⨯≈⨯<……………………14分∴第38天小白鼠仍然存活.(注:列举法求解的也行,请按步骤评分) 44. (本题满分14分,第1小题满分7分,第2小题满分7分)已知())(0)3f x x πωω=+>.(1)若()(0)2y f x πθθ=+<<是最小正周期为π的偶函数,求ω和θ的值; (2)若()(3)g x f x =在(0 )3π,上是增函数,求ω的最大值;并求此时()f x 在[0 ]π,上的取值范围.解:(1)∵())3f x x πθωωθ+=++ (1)分又()y f x θ=+是最小正周期为π的偶函数,∴2ππω=,即2ω=, (3)分 且232k ππθπ+=+,即()212k k Z ππθ=+∈ …………………………………………………6分 注意到02πθ<<,∴2 12πωθ==,为所求; (7)分(2)因为()f x 在(0 )3π,上是增函数,∴53023212()12326332k k k Z k k ππωππππωωπ⎧⎧⨯+≥-≤⎪⎪⇒∈⎨⎨≤+⨯+≤+⎪⎪⎩⎩,…………………………………………9分∵0ω>,∴1206k +>,∴151212k -<<,于是0k =,∴106ω<≤,即ω的最大值为61, (12)分此时,()3sin()23x g x π=+,∴510sin()1()3236223x x x g x πππππ≤≤⇒≤+≤⇒≤+≤⇒∈……………………14分45. (本题满分16分,第1小题满分4分,第2小题满分6分,第3小题满分6分)设等比数列}{n a 的前n 项和为n S ,已知*122()n n a S n N +=+∈.(1)求数列}{n a 的通项公式;(2)在n a 与*1()n a n N +∈之间插入n 个1,构成如下的新数列:1234 1 1 1 1 1 1 a a a a ,,,,,,,,,,,求这个数列的前2012项的和;(3)在n a 与1n a +之间插入n 个数,使这2n +个数组成公差为n d 的等差数列(如:在1a 与2a 之间插入1个数构成第一个等差数列,其公差为1d ;在2a 与3a 之间插入2个数构成第二个等差数列,其公差为2d ,…以此类推),设第n 个等差数列的和是n A . 是否存在一个关于n 的多项式()g n ,使得()n n A g n d =对任意*n N ∈恒成立?若存在,求出这个多项式;若不存在,请说明理由. 解:(1)设11n n a a q -=,由)(22*1N n S a n n ∈+=+知,112111222()2a q a a q a a q =+⎧⎨=++⎩, (2)分 解得{123a q ==, ∴123n na -=⨯ (4)分(2)依题意,到n a 为止新的数列共有(1)12342n n n ++++++=项,…………………6分令(1)20122n n +=,得62.9n =≈, 即到62a 为止新的数列共有62(621)12346219532++++++==项…………………8分 故该数列的前2012项的和为626212622(13)1261(20121953)19503194913a a a ⨯-++++++++-=+=+-(或626212622(13)(201262)19503194913a a a ⨯-++++-=+=+-)………………10分(3)依题意,1123234311n n n n d n n --⨯-⨯⨯==++;11(2323)(2)4(2)32n n n n n A n --⨯+⨯+==+⨯ 要使()n n A g n d =,则11434(2)3()1n n n g n n --⨯+⨯=⨯+, (14)分∴2()(2)(1)32g n n n n n =+⨯+=++,即存在2()32g n n n =++满足条件. ………16分46. (本题满分18分,第1小题满分5分,第2小题满分5分,第3小题满分8分)已知函数22()(1)(1)x b f x x D a x=-+-∈,,其中0a b <<.(1)当(0 )D =+∞,时,设xba x t +=,()()f x g t =,求()y g t =的解析式及定义域; (2)当(0 )D =+∞,,12ab ==,时,求)(x f 的最小值;(3)设0k >,当22(1)a k b k ==+,时,1()9f x ≤≤对任意[ ]x a b ∈,恒成立,求k 的取值范围.解:(1)设(0)x b t x a x =+>,则t ≥ab x =时取等号,………………2分此时222222()(1)(1)(1)1(1)1x b x b b b f x t a x a x a a=-+-=+--+=--+,………………4分即22()(1)1b g t t a =--+,其定义域为)+∞………………………………………5分 (2)由(1)知,当12a b ==,时,2()(1)3(g t t t =--≥……………………………7分函数2()(1)3g t t =--在)+∞上单调递增,∴2min ()1)36f x g ==-=-10分(3) 设2222(1)([ (1)])x k t x k k k x+=+∈+,,则2(1)k t k +≥,当且仅当(1)x k k =+时取等号,显然22(1)[ (1)]k k k k +∈+,且当2x k =和2(1)x k =+时,都有22(1)1k t k +=+………………………………………13分此时2222222(1)2(1)()()(1)[1](1)1k k x f x g t t k x k ++==-+-=--+, 其中222(1)(1)[ 1]k k t k k ++∈+,………………………………………………………14分 函数2222(1)()(1)1k g t t k +=--+在222(1)(1)[ 1]k k k k +++,上单调递增, ∴22min 222(1)2(1)2(1)2()[][1]1k k k f x g k k k k +++==--+= 222222max 2222(1)(1)2(1)(1)()[][]1[1]k k k k f x g k k k k++++==-+=-…………………………16分 又1()9f x ≤≤对任意22[ (1)]x k k ∈+,恒成立, ∴222221(1)[1]9k k k⎧≥⎪⎨+-≤⎪⎩,即113k k k ≤≤⎧⎪⎨≥≤-⎪⎩或,注意到0k >,∴1k ≤≤. …………………………………………………18分。
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2012届高三联合调研考试数学试卷(文科)(本试卷满分150分,测试时间120分钟)参加学校:华师大一附中、曹杨二中、市西、市三女子、控江、格致、市北、(育才、晋元高中)一、填空题(本大题共56分,每小题4分)1.计算:1i2i-=+____________ (其中i 为虚数单位). 2.已知向量(5,3)a =- ,(2,)b x =,若向量a 、b 互相平行,则x =____________.3.已知向量a 与b 的夹角为3π,||1a =,||2b = ,若b a λ- 与a 垂直,则实数λ=_________.4.在二项式81()ax x-的展开式中,常数项为70,则实数a =_____________. 5.已知θ是第三象限角,若3cos 5θ=-,则cos sin 1θθ-的值为_______________. 6.若133log (|5|9)4M a =-+,[4,17]a ∈,则M 的取值范围是_________________.7.关于x 的方程组(1)21y q x y qx =-+⎧⎪⎨=-⎪⎩有唯一的一组实数解,则实数q 的值为_____________. 8.把编号为1、2、3、4、5的5位运动员排在编号为1、2、3、4、5的5条跑道中,若有且只有两位运动员的编号与其所在跑道编号相同,则不同的排法种数共有___________种.9.过点1(,1)2M 的直线l 与圆C :22(1)4x y -+=交于A 、B 两点,C 为圆心,当ACB ∠最小时,直线l 的方程为_________________.10.在平面直角坐标系xOy 中,函数()()1f x k x =-(1k >)的图像与x 轴交于点A ,它的反函数()1y fx -=的图像与y 轴交于点B ,并且这两个函数的图像交于点P .若四边形OAPB 的面积是3,则k =___________.11.已知Z k ∈,向量(,1)AB k = ,(2,4)AC = ,若||10AB ≤,则ABC ∆为直角三角形的概率是_______________.12.已知ABO ∆中, (1OA =- , (1,0)OB =,D 为AB 上的点,若2AD DB = ,则ODB ∠=____________(结果用反三角表示).13.设直线:l 220x y ++=关于原点对称的直线为l ',若l '与椭圆2214y x +=的交点为A ,B 两点,点P 是椭圆上的动点,则使PAB ∆的面积为12的点P 的个数为_____________.14.如图所示的程序框图中, ,函数int()x 表示不超过x 的最大整数,则由框图给出的计算结果是____________.二、选择题(本大题满分20分,每小题5分)15.若函数21y a x =⋅,22x y c =⋅,33y b x =⋅,则由表中数据确定()f x 、()g x 、()h x 依次对应 ( ).(A) 1y 、2y 、3y (B) 2y 、1y 、3y (C ) 3y 、2y 、1y (D) 1y 、3y 、2y16.在证券交易过程中,常用到两种曲线,即时价格曲线()y f x =及平均价格曲线()y g x = (如(2)3f =是指开始买卖后二个小时的即时价格为3元;(2)3g =表示二个小时内的平均价格为3元),在下图给出的四个图像中实线表示()y f x =,虚线表示()y g x =其中可能正确的是 ( ).(A ) (B ) (C ) (D )17. 正四面体ABCD 的表面积为S ,其中四个面的中心分别是E 、F 、G 、H .设四面体EFGH 的表面积为T ,则TS等于 ( ).(A) 49 (B) 19 (C)14 (D)1318.函数()y f x =的定义域为R ,若对于任意的正数a ,函数()()()g x f x a f x =+-都是其定义域上的增函数,则函数()y f x =的图像可能是 ( ).(A ) (B) (C) (D)三、解答题(本大题满分74分)19.(本题满分12分)第1小题满分6分,第2小题满分6分.已知函数()()22sin cos 2cos 2f x x x x =++-.(1)求函数()f x 的最小正周期; (2 )当3,44x ππ⎡⎤∈⎢⎥⎣⎦时,求函数()f x 的最大值,最小值.20.(本题满分12分)第1小题满分6分,第2小题满分6分.在一个棱长为2+锥,使截掉棱锥后的多面体有六个面为正八边形,八个面为正三角形(如图所示),(1)求异面直线AB 与GH 所成角的大小; (2)求此多面体的体积(结果用最简根式表示).21.(本题满分12分)第1小题满分5分,第2小题满分7分.已知O 为坐标原点,点(2,1),(1,2)A B ,对于k N *∈有向量k OP kOB OA =+,(1)试证明k P 都在同一条直线23y x =-上;(2)是否在存在k N *∈使k P 在圆22(2)5x y +-=上或其内部,若存在求出k ,若不存在说明理由.22.(本题满分19分)第1小题满分5分,第2小题满分5分,第3小题满分9分. 已知函数()y f x =的图像(如图所示)过点(0,2)、(1.5,2)和点(2,0),且函数图像关于点(2,0)对称;直线1x =和3x =及0y =是它的渐近线.现要求根据给出的函数图像研究函数1()()g x f x =的相关性质与图像, (1)写出函数()y g x =的定义域、值域及单调递增区间;(2)作函数()y g x =的大致图像(要充分反映由图像及条件给出的信息);(3)试写出()y f x =的一个解析式,并简述选择这个式子的理由(按给出理由的完整性及表达式的合理、简洁程度分层给分).23.(本题满分19分)第1小题满分5分,第2小题满分8分,第3小题满分6分. 由下面四个图形中的点数分别给出了四个数列的前四项,将每个图形的层数增加可得到这四个数列的后继项.按图中多边形的边数依次称这些数列为“三角形数列”、“四边形数列” ,将构图边数增加到n 可得到“n 边形数列”,记它的第r 项为(,)P n r ,1,3,6,10 1,4,9,16 1,5,12,22 1,6,15,28 (1)求使得(3,)36P r >的最小r 的取值;(2)3725是否为“五边形数列”中的项,若是,为第几项;若不是,说明理由; ( 3) 试推导(,)P n r 关于n 、r 的解析式.参考答案: 1.13i 55- 2. 65- 3. 1 4. 1± 5. 136. 3[2log 2,2]--- (或3[log 18,2]--等7.12或 18.209.2430x y -+= (或11()(1)022x y ---=等)10.32 11. 419 12. 13.2 14.1 15.D 16.C 17.B 18.D19. 解: (1)()sin 2cos 224f x x x x π⎛⎫=+=+ ⎪⎝⎭. 4分∴()f x 的最小正周期为π. 6分(2).337,,244444x x πππππ⎡⎤∈∴≤+≤⎢⎥⎣⎦, 8分,1sin 242x π⎛⎫∴-≤+≤⎪⎝⎭ 10分∴()1f x ≤≤. 12分∴当3,44x ππ⎡⎤∈⎢⎥⎣⎦时,函数()f x 的最大值为1,最小值.20. 解: (1) 易知//FE AB ,//GH EC ,所以FEC ∠就是异面直线AB 与GH 所成的余角). 3分经计算得: 351)48FEC ππ∠=-= (也可以直接用4522.567.5+=做)所以异面直线AB 与GH 所成的角的大小为38π1),arc . 6分(2,则由题意得:2x x +=所以, 9分设多面体的体积为V ,则311(2832V =-⨯⨯=563+ 12分21.解:(1)点k P 在同一条直线上,直线方程为23y x =-. 2分证明如下:设点(,)k k k P x y ,则(,)(1,2)(2,1)k k x y k =+即2,21,k kx k y k =+⎧⎨=+⎩所以23k k y x =-.所以,点k P 在直线23y x =-上. 5分 (文科)按证明情况酌情给分(2)由圆22(2)5x y +-=的圆心(0,2)到直线23y x =-=可知直线与圆相切, 所以直线与圆及内部最多只有一个公共点 10分而切点的坐标为:(2,1),此时0k =不满足题意,所以不存在k N *∈满足题意. 12分22.解: (1) 定义域为:{|1,2,3,}x x x x x R ≠≠≠∈ 2分 值域为: (,0)(0,)-∞⋃+∞ 3分 函数的单调递增区间为: (1,2)和(2,3) 5分 (2)图像要求能反映出零点((1,0)和(3,0),渐近线2x =,过定点,单调性正确. 5分(3) 结论可能各异如:3(2)()|1||3|x f x x x -=--,2222(3)()22(1)xx x f x x x x -⎧>⎪-⎪=⎨-⎪<⎪-⎩211()2tan()132233x x f x x x x x π-⎧<⎪-⎪⎪=-<<⎨⎪-⎪>⎪-⎩,等层次一:函数图像能满足题意, 但没有说明理由 4分 层次二: 函数图像能满足题意,能简述理由(渐近线、定点等部分内容) 6分层次三: 函数图像能满足题意,能说明过定点、渐近线、单调性及对称性 9分解: (1)(1)(3,)2r r P r +=, 3分 由题意得(1)362r r +>, 所以,最小的9r =. 5分(2)设n 边形数列所对应的图形中第r 层的点数为r a ,则12(,)r P n r a a a =++⋅⋅⋅+ 从图中可以得出:后一层的点在2n -条边上增加了一点,两条边上的点数不变, 所以12r r a a n +-=-,11a =所以{}r a 是首项为1公差为2n -的等差数列, 所以(,)[2(1)(2)]2r P n r r n =+--.(或(2)(1)2n r r r --+等) 13分 (3)2(,1)(,)(2)21P n r P n r n r r ++=-++ 16分 显然3n =满足题意, 17分 而结论要对于任意的正整数r 都成立,则2(2)21n r r -++的判别式必须为零, 所以,44(2)0n --=,3n = 19分所以,满足题意的数列为“三角形数列”.(文科)(2)为第50项,(3)同理科(2).。