广东省深圳明德外语实验高三数学上学期期终考试题理苏教版【会员独享】

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广东省深圳市明德外语实验学校2024-2025学年九年级上学期9月月考数学试题

广东省深圳市明德外语实验学校2024-2025学年九年级上学期9月月考数学试题一、单选题1．方程2430x x ++=的两个根为（）A ．121,3x x ==B ．121,3x x =-=C ．121,3x x ==-D ．121,3x x =-=- 2．已知关于x 的一元二次方程2610x x k -++=的两个实数根为1x ，2x ，且221224x x +=，则k 的值为（）A ．5B ．6C ．7D ．83．下列四种说法：①矩形的两条对角线相等且互相垂直；②菱形的对角线相等且互相平分；③有两边相等的平行四边形是菱形；④有一组邻边相等的菱形是正方形．其中正确的有（） A ．0个 B ．1个 C ．2个 D ．3个4．根据下列表格的对应值：由此可判断方程212150x x +-=必有一个根满足（）A ．1 1.1x <<B ．1.1 1.2x <<C ．1.2 1.3x <<D ． 1.3x >5．若关于x 的一元二次方程()2500ax bx a ++=≠的一个解是=1x -，则2017a b -+的值是（）A ．2016B ．2018C ．2020D ．20226．如图，▱ABCD 的对角线AC 、BD 相交于点O ，那么下列条件中，能判断▱ABCD 是菱形的为（）A．AO＝CO B．AO＝BO C．∠AOB＝∠BOC D．∠BAD＝∠ABC 7．如图，在矩形ABCD中，对角线AC、BD交于点O，自点A作AE⊥BD于点E，且BE：ED=1：3，过点O作OF⊥AD于点F，若OF=3cm，则BD的长为（）cm．A．6 B．9 C．12 D．158．如图，在菱形ABCD中，菱形的边长为5，对角线AC的长为8，延长AB至E，BF平分CBE，则ACGV的面积为（）A．20B．C．12D．249．如图，在菱形ABCD中，P是对角线AC上一动点，过点P作PE⊥BC于点E，PF⊥AB 于点F．若菱形ABCD的周长为20，面积为24，则PE＋PF的值为()A．4 B．245C．6 D．48510．如图，在矩形ABCD中，AB＝2，AD＝1，E为AB的中点，F为EC上一动点，P为DF 中点，连接PB，则PB的最小值是（）A ．2B ．4C ．D ．2二、填空题11．若关于x 的方程2(1)210k x x +--=有实数根，则k 的取值范围是．12．如图，三个边长均为2的正方形重叠在一起，1O 、2O 是其中两个正方形的中心，则阴影部分的面积是．13．已知：如图所示，E 是正方形ABCD 边BC 延长线一点，若EC AC =，AE 交CD 于F ，则AFC ∠=度．14．如图，在菱形ABCD 中，AC ＝24，BD ＝10，AC 、BD 相交于点O ，若CE //BD ，BE //AC ，连接OE ，则OE 的长是．15．如图，菱形ABCD 中，∠ABC ＝60°，AB ＝2，E 、F 分别是边BC 和对角线BD 上的动点，且BE ＝DF ，则AE +AF 的最小值为．三、解答题16．解方程：(1)22950x x --=(2)244x x x -=-17．阅读下面的例题：分解因式：221x x +-．解：令2210x x +-=得到一个关于x 的一元二次方程，121a b c ===-Q ，，，1x ∴===-解得11x =-21x =-()()(((212211111x x x x x x x x x x ⎡⎤⎡⎤∴+-=--=----=++⎣⎦⎣⎦．这种因式分解的方法叫求根法，请你利用这种方法完成下面问题：(1)已知代数式22x x k --对应的方程解为5-和7，则代数式22x x k --分解后为；(2)将代数式231x x --分解因式．18．如图，在矩形ABCD 的BC 边上取一点E ，连接AE ，使得AE ＝EC ，在AD 边上取一点F ，使得DF ＝BE ，连接CF ．过点D 作DG ⊥AE 于G ．（1）求证：四边形AECF 是菱形；（2）若AB ＝4，BE ＝3，求DG 的长．19．某农场要建一个饲养场（矩形ABCD ）两面靠现有墙（AD 位置的墙最大可用长度为27米，AB 位置的墙最大可用长度为15米），另两边用木栏围成，中间也用木栏隔开，分成两个场地及一处通道，并在如图所示的三处各留1米宽的门（不用木栏）．建成后木栏总长45米．设饲养场（矩形ABCD ）的一边AB 长为x 米．（1）饲养场另一边BC=____米（用含x 的代数式表示）．（2）若饲养场的面积为180平方米，求x 的值．20．如图，已知在菱形ABCD 中，对角线AC 与BD 交于点O ，延长DC 到点E ，使C E C D =，延长BC 到点F ，使CF BC =，顺次连接点B ，E ，F ，D ，且1BD =，AC =(1)求菱形ABCD 的面积；(2)求证：四边形BEFD 是矩形；(3)求四边形BEFD 的周长及面积．21．数学课上，师生们以“利用正方形和矩形纸片折叠特殊角”为主题开展数学活动．(1)操作判断小明利用正方形纸片进行折叠，过程如下：步骤①：如图1，对折正方形纸片ABCD ，使AD 与BC 重合，得到折痕EF ，把纸片展平；步骤②：连接AF ，BF ．可以判定ABF △的形状是：．（直接写出结论）小华利用矩形纸片进行折叠，过程如下：如图2，先类似小明的步骤①，得到折痕EF 后把纸片展平；在BC 上选一点P ，沿AP 折叠AB ，使点B 恰好落在折痕EF 上的一点M 处，连接AM ．小华得出的结论是：30BAP PAM MAD ∠=∠=∠=︒．请你帮助小华说明理由．(2)迁移探究小明受小华的启发，继续利用正方形纸片进行探究，过程如下：如图3，第一步与步骤①一样；然后连接AF ，将AD 沿AF 折叠，使点D 落在正方形内的一点M 处，连接FM 并延长交BC 于点P ，连接AP ，可以得到：PAF ∠= ︒（直接写出结论）；同时，若正方形的边长是4，可以求出BP 的长，请你完成求解过程．(3)拓展应用如图4，在矩形ABCD 中，6AB =，8BC =．点P 为BC 上的一点（不与B 点重合，可以与C 点重合），将ABP V 沿着AP 折叠，点B 的对应点为M 落在矩形的内部，连接MA ，MD ，当△MAD 为等腰三角形时，可求得BP 的长为．（直接写出结论） 22．如图1，在正方形ABCD 和正方形BEFG 中，点,,A B E 在同一条直线上，P 是线段DF 的中点，连接PG ，PC ．(1)探究PG 与PC 的位置关系及PG PC的值（写出结论，不需要证明）； (2)如图2，将原问题中的正方形ABCD 和正方形BEFG 换成菱形ABCD 和菱形BEFG ，且60ABC BEF ∠=∠=度．探究PG 与PC 的位置关系及PG PC的值，写出你的猜想并加以证明； (3)如图3，将图2中的菱形BEFG 绕点B 顺时针旋转，使菱形BEFG 的边BG 恰好与菱形ABCD 的边AB 在同一条直线上，问题（2）中的其他条件不变．你在（2）中得到的两个结论是否发生变化？写出你的猜想并加以证明．。

高三数学理每周一练第11周

明德外语实验学校2010-2011学年第一学期期中考试高三数学姓名时间120分钟满分150分一、选择题：本大题共10小题，每小题5分，满分50分，在每小题给出的四个选项中，只有一项是符合题目要求的。

1．函数x y sin 2=的定义域为A ，值域为B ，则A B ⋂=( ) A ．A B ．B C ．]1,1[- D ．A 2 2．集合{}2,4,6M =的真子集的个数为（）A ．6B ．7C ．8D ．93．不等式2320x x -+<的解集是（）A.{}21x x x <->-或B.{}12x x x <>或 C.{}12x x << D ．{}21x x -<<-4. 在△ABC 中，“B A >”是“B A sin sin >”的（）A.充分而不必要条件B.必要而不充分条件C.充分必要条件D.既不充分也不必要条件 5．已知向量)1,1(=a ,),2(n b =,若b a b a ⋅=+||，则n = ( )A.3-B.1-C.1D.36．已知向量a 、b 满足1,4,a b ==，且2a b ⋅= ，则a 与b 的夹角为（）A ．6π B ．4π C ．3π D ．2π 7. 已知3(,),sin ,25παπα∈=则tan()4πα+等于（）A ．17B ．7C ．17- D ．7-8.定义运算a ⊕b ⎩⎨⎧=b a )()(b a b a >≤，则函数1)(=x f ⊕x2的大致图象是（）9．函数()22log 1log 1x f x x -=+，若()()1221f x f x +=（其中1x 、2x 均大于２），则()12f x x 的最小值为（） A ．35 B ．23 C ．45 D二、填空题：本大题共5小题，每小题4分，满分20分。

11.设R a ∈，若i i a 2)(-（i 为虚数单位）为正实数，则=a 12.已知)sin()(ϕω+=x A x f ,0)(,)(==βαf A f ,||βα-的最小值3π,则正数=ω 13.不等式142x x -<-+的解集是．14.在极坐标系中，点()1,0到直线()cos sin 2ρθθ+=的距离为． 15.如图2所示，AB 与CD 是O 圆的直径，AB ⊥CD ，P 是AB 延长线上一点，连PC 交O 圆于点E ，连DE 交AB 于点F ，若42==BP AB ，则=PF ．三、解答题：本大题共6小题，满分80分．解答须写出文字说明、证明过程和演算步骤.16．（本小题满分12分）在△ABC 中，角,,A B C 所对的边分别为,,a b c ，已知2a =，3c =，1cos 4B =．（1）求b 的值；（2）求sinC 的值． 17．（本小题满分12分）已知函数()1cos 2cos 2xf x x x a +=++（a 为常数）．（1）求函数()x f 的最小正周期，并指出其单调减区间；（2）若函数()x f 在⎥⎦⎤⎢⎣⎡20π，　上的最大值是2,试求实数a 的值.C 图218. (本小题满分12分)在海岸A 处，发现北偏东045方向，距A 处)13(-n mile 的B 处有一艘走私船，在A 处北偏西075的方向，距离A 处2n mile 的C 处的缉私船奉命以的速度310n h mile /追截走私船。

深圳明德外语实验学校数学高三上期末(培优专题)

一、选择题1．设,x y 满足约束条件 202300x y x y x y --≤⎧⎪-+≥⎨⎪+≤⎩，则46y x ++的取值范围是A ．3[3,]7- B ．[3,1]- C ．[4,1]-D ．(,3][1,)-∞-⋃+∞2．在ABC ∆中，2AC =,BC =135ACB ∠=，过C 作CD AB ⊥交AB 于D ，则CD =( ) ABCD3．若n S 是等差数列{}n a 的前n 项和，其首项10a >，991000a a +>，991000a a ⋅< ，则使0n S >成立的最大自然数n 是（） A ．198B ．199C ．200D ．2014．已知函数f （x ）＝x 2﹣2x +k ，若对于任意的实数x 1，x 2，x 3，x 4∈[1，2]时，f （x 1）+f （x 2）+f （x 3）＞f （x 4）恒成立，则实数k 的取值范围为（） A ．（23，+∞） B ．（32，+∞） C ．（﹣∞，23） D ．（﹣∞，32） 5．设等比数列{}n a 的前n 项和为n S ，若633S S =，则96S S =（） A ．2B ．73C ．83D ．36．设数列{}n a 是以2为首项，1为公差的等差数列，{}n b 是以1为首项，2为公比的等比数列，则1210b b b a a a ++⋯+=（） A ．1033B ．1034C ．2057D ．20587．在ABC ∆中，,,a b c 是角,,A B C 的对边，2a b =，3cos 5A =，则sinB =（） A ．25B ．35C ．45 D ．858．在等差数列{a n }中，a 1＞0，a 10·a 11＜0，若此数列的前10项和S 10＝36，前18项的和S 18＝12，则数列{|a n |}的前18项和T 18的值是（） A ．24B ．48C ．60D ．849．已知等比数列{}n a 的各项均为正数，前n 项和为n S ，若26442,S 6a S a =-=，则5a = A ．4B ．10C ．16D ．3210．设2z x y =+，其中,x y 满足2000x y x y y k +≥⎧⎪-≤⎨⎪≤≤⎩，若z 的最小值是12-，则z 的最大值为（） A ．9-B ．12C ．12-D ．911．已知01x <<，01y <<，则）AB ．CD ．12．已知点(),M a b 与点()0,1N -在直线3450x y -+=的两侧，给出以下结论：①3450a b -+>；②当0a >时，+a b 有最小值，无最大值；③221a b +>；④当0a >且1a ≠时，11b a +-的取值范围是93,,44⎛⎫⎛⎫-∞-⋃+∞ ⎪ ⎪⎝⎭⎝⎭，正确的个数是（） A ．1B ．2C ．3D ．413．设数列{}n a 的前n 项和为n S ，若2，n S ，3n a 成等差数列，则5S 的值是（） A ．243-B ．242-C ．162-D ．24314．一个递增的等差数列{}n a ，前三项的和12312a a a ++=，且234,,1a a a +成等比数列，则数列{}n a 的公差为 ( ) A ．2±B ．3C ．2D ．115．已知函数1()2xf x ⎛⎫= ⎪⎝⎭，则不等式()24(3)f a f a ->的解集为( )A ．(4,1)-B ．(1,4)-C ．(1,4)D ．(0,4)二、填空题16．若首项为1a ，公比为q （1q ≠）的等比数列{}n a 满足21123lim()2n n a q a a →∞-=+，则1a 的取值范围是________.17．《九章算术》“竹九节”问题：现有一根9节的竹子，自上而下各节的容积成等差数列，上面4节的容积共3升，下面3节的容积共4升，则第5节的容积为升； 18．已知向量()()1,,,2a x b x y ==-，其中0x >，若a 与b 共线，则yx的最小值为__________．19．已知变量,x y 满足约束条件2{41y x y x y ≤+≥-≤，则3z x y =+的最大值为____________．20．若实数,x y 满足约束条件200220x y x y x y +≥⎧⎪-≤⎨⎪-+≥⎩，则3z x y =-的最小值等于_____．21．已知数列{}n a 满足51()1,62,6n n a n n a a n -⎧-+<⎪=⎨⎪≥⎩，若对任意*n N ∈都有1n n a a +>，则实数a 的取值范围是_________.22．等差数列{}n a 前9项的和等于前4项的和.若141,0k a a a =+=，则k = . 23．设(3()lg f x x x =+，则对任意实数,a b ，“0a b +≥”是“()()0f a f b +≥”的_________条件．（填“充分不必要”.“必要不充分”.“充要”.“既不充分又不必要”之一）24．已知n S 是数列{}n a 的前n 项和，122n n S a +=-，若212a =，则5S =__________． 25．已知()()0f x kx k =>，若正数a 、b 满足()()()()f a f b f a f b +=，且4a b f f k k ⎛⎫⎛⎫+ ⎪ ⎪⎝⎭⎝⎭的最小值为1，则实数k 的值为______．三、解答题26．在ABC ∆中，,,a b c 分别是角,,A B C 所对的边，且2sin 3tan c B a A =.（1）求222b c a+的值；（2）若2a =，求ABC ∆面积的最大值.27．在数列{}n a 中, 已知11a =，且数列{}n a 的前n 项和n S 满足1434n n S S +-=, n *∈N . （1）证明数列{}n a 是等比数列；（2）设数列{}n na 的前n 项和为n T ，若不等式3()1604nn aT n+⋅-<对任意的n *∈N 恒成立, 求实数a 的取值范围.28．记等差数列{}n a 的前n 项和为n S ,已知2446,10a a S +==．（Ⅰ）求数列{}n a 的通项公式;（Ⅱ）令2n n n b a =⋅*()n N ∈,求数列{}n b 的前n 项和n T ．29．已知数列{}n a 的前n 项和为n S ，且4133n n S a =-. （1）求{}n a 的通项公式；（2）若1n b n =+，求数列{}n n a b 的前n 项和n T .30．记n S 为数列{}n a 的前n 项和.已知0n a >，2634n n n S a a =+-.（1）求{}n a的通项公式；（2）设2211n nnn na aba a+++=，求数列{}n b的前n项和n T.【参考答案】2016-2017年度第*次考试试卷参考答案**科目模拟测试一、选择题1．B 2．A 3．A 4．B 5．B 6．A 7．A 8．C 9．C 10．B 11．B 12．B 13．B 14．C 15．B二、填空题16．【解析】【分析】由题意可得且即且化简可得由不等式的性质可得的取值范围【详解】解：故有且化简可得且即故答案为：【点睛】本题考查数列极限以及不等式的性质属于中档题17．【解析】试题分析：由题意可知解得所以考点：等差数列通项公式18．【解析】【分析】根据两个向量平行的充要条件写出向量的坐标之间的关系之后得出利用基本不等式求得其最小值得到结果【详解】∵其中且与共线∴即∴当且仅当即时取等号∴的最小值为【点睛】该题考查的是有关向量共线19．11【解析】试题分析：由题意得作出不等式组所表示的可行域如图所示由得平移直线则由图象可知当直线经过点时直线的截距最大此时有最大值由解得此时考点：简单的线性规划20．【解析】【分析】先画出可行域改写目标函数然后求出最小值【详解】依题意可行域为如图所示的阴影部分的三角形区域目标函数化为：则的最小值即为动直线在轴上的截距的最大值通过平移可知在点处动直线在轴上的截距最21．【解析】【分析】由题若对于任意的都有可得解出即可得出【详解】∵若对任意都有∴∴解得故答案为【点睛】本题考查了数列与函数的单调性不等式的解法考查了推理能力与计算能力属于中档题22．10【解析】【分析】根据等差数列的前n项和公式可得结合等差数列的性质即可求得k 的值【详解】因为且所以由等差数列性质可知因为所以则根据等差数列性质可知可得【点睛】本题考查了等差数列的前n项和公式等差数23．充要【解析】所以为奇函数又为单调递增函数所以即是的充要条件点睛：充分必要条件的三种判断方法1定义法：直接判断若则若则的真假并注意和图示相结合例如⇒为真则是的充分条件2等价法：利用⇒与非⇒非⇒与非⇒非24．【解析】【分析】由题意首先求得然后结合递推关系求解即可【详解】由题意可知：且：整理可得：由于故【点睛】本题主要考查递推关系的应用前n项和与通项公式的关系等知识意在考查学生的转化能力和计算求解能力25．9【解析】【分析】由求出满足的关系然后利用基本不等式求出的最小值再由最小值为1可得【详解】∵∴即∴当且仅当时等号成立∴故答案为：9【点睛】本题考查基本不等式求最值解题时需用凑配法凑出基本不等式所需的三、解答题26．27．28．29．30．2016-2017年度第*次考试试卷参考解析【参考解析】**科目模拟测试一、选择题1．B解析：B【解析】【分析】【详解】先作可行域，而46yx++表示两点P（x,y）与A（-6,-4）连线的斜率，所以46yx++的取值范围是[,][3,1]AD ACk k=-，选B.点睛：线性规划问题，首先明确可行域对应的是封闭区域还是开放区域、分界线是实线还是虚线，其次确定目标函数的几何意义，是求直线的截距、两点间距离的平方、直线的斜率、还是点到直线的距离等等，最后结合图形确定目标函数最值取法、值域范围.2．A【解析】【分析】先由余弦定理得到AB 边的长度，再由等面积法可得到结果. 【详解】根据余弦定理得到2222AC BC AB AC BC +-=⨯⨯将2AC =,BC =,代入等式得到AB=再由等面积法得到1122225CD CD ⨯=⨯⨯⇒=故答案为A. 【点睛】这个题目考查了解三角形的应用问题，涉及正余弦定理，面积公式的应用，在解与三角形有关的问题时，正弦定理、余弦定理是两个主要依据.解三角形时，有时可用正弦定理，有时也可用余弦定理，应注意用哪一个定理更方便、简捷一般来说 ,当条件中同时出现ab 及2b 、2a 时，往往用余弦定理，而题设中如果边和正弦、余弦函数交叉出现时，往往运用正弦定理将边化为正弦函数再结合和、差、倍角的正余弦公式进行解答.3．A解析：A 【解析】【分析】先根据10a >，991000a a +>，991000a a ⋅<判断出991000,0a a ><；然后再根据等差数列前n 项和公式和等差中项的性质，即可求出结果．【详解】∵991000a a ⋅<， ∴99a 和100a 异号； ∵1991000,0a a a >+>，991000,0a a ∴><，有等差数列的性质可知，等差数列{}n a 的公差0d <，当99,*n n N ≤∈时，0n a >；当100,*n n N ≥∈时，0n a <；又()()119899100198198198022a a a a S +⨯+⨯==> ，()119919910019919902a a S a+⨯==<，由等差数列的前n 项和的性质可知，使前n 项和0n S >成立的最大自然数n 是198. 故选：A ．【点睛】本题主要考查了等差数列的性质．考查了学生的推理能力和运算能力．4．B【解析】【分析】由于2()(1)1f x x k =-+-，分析对称轴，得到min max ()1,()f x k f x k =-=，转化f（x 1）+f （x 2）+f （x 3）＞f （x 4）恒成立，为1min 2min 3min 4max ()()()()f x f x f x f x ++>，即得解. 【详解】由于2()(1)1f x x k =-+- ，当[1,2]x ∈，min max ()(1)1,()(2)f x f k f x f k ==-==，对于任意的实数x 1，x 2，x 3，x 4∈[1，2]时，f （x 1）+f （x 2）+f （x 3）＞f （x 4）恒成立，即：1min 2min 3min 4max ()()()()f x f x f x f x ++> 即：33(1)2k k k ->∴> 故选：B 【点睛】本题考查了二次函数的恒成立问题，考查了学生转化划归，综合分析，数学运算的能力，属于中档题.5．B解析：B 【解析】【分析】首先由等比数列前n 项和公式列方程，并解得3q ，然后再次利用等比数列前n 项和公式，则求得答案．【详解】设公比为q ，则616363313(1)1113(1)11a q S q q q a q S qq---===+=---， ∴32q =，∴93962611271123S q S q --===--．故选：B ．【点睛】本题考查等比数列前n 项和公式，考查函数与方程思想、转化与化归思想，考查逻辑推理能力、运算求解能力，求解时也可以利用连续等长片断的和序列仍然成等比数列，进行求解.6．A【解析】【分析】【详解】首先根据数列{a n }是以2为首项，1为公差的等差数列，{b n }是以1为首项，2为公比的等比数列，求出等差数列和等比数列的通项公式，然后根据a b1+a b2+…+a b10=1+2+23+25+…+29+10进行求和．解：∵数列{a n }是以2为首项，1为公差的等差数列， ∴a n =2+（n-1）×1=n+1， ∵{b n }是以1为首项，2为公比的等比数列， ∴b n =1×2n-1，依题意有：a b1+a b2+…+a b10=1+2+22+23+25+…+29+10=1033，故选A ．7．A解析：A 【解析】试题分析：由3cos 5A =得，又2a b =，由正弦定理可得sin B =.考点：同角关系式、正弦定理．8．C解析：C 【解析】试题分析：∵11011101100000a a a d a a ⋅∴＞，＜，＜，＞，＜， ∴18110111810181060T a a a a S S S =+⋯+--⋯-=--=（），选C . 考点：1.等差数列的求和；2.数列的性质.9．C解析：C 【解析】由64S S -=6546a a a +=得，()22460,60q q a q q +-=+-=，解得2q，从而3522=28=16a a =⋅⨯，故选C.10．B解析：B 【解析】【分析】作出不等式对应的可行域，当目标函数过点A 时，z 取最小值，即min 12z =-，可求得k 的值，当目标函数过点B 时，z 取最大值，即可求出答案. 【详解】作出不等式对应的可行域，如下图阴影部分，目标函数可化为2y x z =-+，联立20x y y k +=⎧⎨=⎩，可得()2,A k k -，当目标函数过点A 时，z 取最小值，则()2212k k ⨯-+=-，解得4k =，联立0x y y k -=⎧⎨=⎩，可得(),B k k ，即()4,4B ，当目标函数过点B 时，z 取最大值，max 24412z =⨯+=.故选：B.【点睛】本题考查线性规划，考查学生的计算求解能力，利用数形结合方法是解决本题的关键，属于基础题.11．B解析：B 【解析】【分析】2222++≥x y x y 222+x y ()2212+-x y ，()2212-+x y ()()22112-+-x y 边分别相加求解。

深圳外国语学校数学高三上期末经典练习(含答案解析)

一、选择题1．等差数列{}n a 中，已知70a >，390a a +<，则{}n a 的前n 项和n S 的最小值为（） A ．4SB ．5SC ．6SD ．7S2．已知数列{}n a 的前n 项和为n S ，且1142n n a -⎛⎫=+- ⎪⎝⎭，若对任意*N n ∈，都有()143n p S n ≤-≤成立，则实数p 的取值范围是（）A ．()2,3B ．[]2,3C ．92,2⎡⎤⎢⎥⎣⎦D ．92,2⎡⎫⎪⎢⎣⎭3．已知数列{}n a 的前n 项和2n S n =，()1nn n b a =-则数列{}n b 的前n 项和n T 满足（） A ．()1nn T n =-⨯ B ．n T n = C ．n T n =-D ．,2,.n n n T n n ⎧=⎨-⎩为偶数，为奇数4．程大位《算法统宗》里有诗云“九百九十六斤棉，赠分八子做盘缠.次第每人多十七，要将第八数来言.务要分明依次弟，孝和休惹外人传.”意为：996斤棉花，分别赠送给8个子女做旅费，从第一个开始，以后每人依次多17斤，直到第八个孩子为止.分配时一定要等级分明，使孝顺子女的美德外传，则第八个孩子分得斤数为（） A ．65B ．184C ．183D ．1765．已知在ΔABC 中，a ，b ，c 分别为角A ，B ，C 的对边，A 为最小角，且a =√3，b =2，cosA =58，则ΔABC 的面积等于（）A ．7√316B ．√3916C ．√394D ．7√346．“干支纪年法”是中国历法上自古以来就一直使用的纪年方法，干支是天干和地支的总称，把干支顺序相配正好六十为一周，周而复始，循环记录，这就是俗称的“干支表”甲、乙、丙、丁、戊、己、庚、辛、癸等十个符号叫天干，子、丑、寅、卯、辰、巳、午、未、申、酉、戌、亥等十二个符号叫地支，如公元1984年农历为甲子年，公元1985年农历为乙丑年，公元1986年农历为丙寅年，则公元2047年农历为 A ．乙丑年B ．丙寅年C ．丁卯年D ．戊辰年7．已知函数223log ,0(){1,0x x f x x x x +>=--≤，则不等式()5f x ≤的解集为 ( )A ．[]1,1-B ．[]2,4-C ．(](),20,4-∞-⋃D ．(][],20,4-∞-⋃ 8．我国的《洛书》中记载着世界上最古老的一个幻方：将1，2，...，9填入33⨯的方格内，使三行、三列、两对角线的三个数之和都等于15 (如图）.一般地，将连续的正整数1，2，3，…，2n 填入n n ⨯的方格内，使得每行、每列、每条对角线上的数的和相等，这个正方形就叫做n 阶幻方.记n 阶幻方的一条对角线上数的和为n N (如：在3阶幻方中，315N =)，则10N =（）A ．1020B ．1010C ．510D ．5059．若直线2y x =上存在点(,)x y 满足30,230,,x y x y x m +-≤⎧⎪--≥⎨⎪≥⎩则实数m 的最大值为A ．2-B ．1-C ．1D ．310．设实数,x y 满足242210x y x y x -≤⎧⎪+≤⎨⎪-≥⎩，则1y x +的最大值是（）A ．-1B ．12 C ．1 D ．3211．在ABC ∆中，角,,A B C 的对边分别为a ，b ，c ．若ABC ∆为锐角三角形，且满足sin (12cos )2sin cos cos sin B C A C A C +=+，则下列等式成立的是（）A ．2a b =B ．2b a =C ．2A B =D ．2B A =12．在等差数列{a n }中，a 1＞0，a 10·a 11＜0，若此数列的前10项和S 10＝36，前18项的和S 18＝12，则数列{|a n |}的前18项和T 18的值是（） A ．24B ．48C ．60D ．8413．已知等比数列{}n a 的各项均为正数，前n 项和为n S ，若26442,S 6a S a =-=，则5a =A ．4B ．10C ．16D ．3214．已知数列{}n a 的前n 项和为n S ，点(,3)n n S +*()n N ∈在函数32xy =⨯的图象上，等比数列{}n b 满足1n n n b b a ++=*()n N ∈，其前n 项和为n T ，则下列结论正确的是（） A ．2n n S T =B ．21n n T b =+C ．n n T a >D ．1n n T b +<15．在直角梯形ABCD 中，//AB CD ，90ABC ∠=，22AB BC CD ==，则cos DAC ∠=（）A 25B 5C 310D ．1010二、填空题16．已知变数,x y 满足约束条件340{210,380x y x y x y -+≥+-≥+-≤目标函数(0)z x ay a =+≥仅在点(2,2)处取得最大值，则a 的取值范围为_____________．17．数列{}21n-的前n 项1,3,7..21n-组成集合{}()*1,3,7,21nn A n N=-∈，从集合nA中任取()1,2,3?··n k k =个数，其所有可能的k 个数的乘积的和为（若只取一个数，规定乘积为此数本身），记12n n S T T T =++⋅⋅⋅+，例如当1n =时，{}1111,1,1===A T S ；当2n =时，{}21221,2,13,13,13137A T T S ==+=⨯=++⨯=，试写出n S =___18．在平面直角坐标系中，设点()0,0O，(A ，点(),P x y的坐标满足200y x y -≤+≥⎨⎪≥⎪⎩，则OA 在OP 上的投影的取值范围是__________ 19．已知数列{}n a 中，45n a n =-+，等比数列{}n b 的公比q 满足1(2)n n q a a n -=-≥，且12b a =，则12n b b b +++=__________．20．在数列{}n a 中，“()n 12n a n N*n 1n 1n 1=++⋯+∈+++，又nn n 11b a a +=，则数列{}n b 的前n 项和n S 为______．21．若关于 x 的不等式 ()2221x ax -< 的解集中的整数恰有 3 个，则实数 a 的取值范围是________________．22．若正项数列{}n a 满足11n n a a +-<,则称数列{}n a 为D 型数列,以下4个正项数列{}n a 满足的递推关系分别为:①2211n n a a +-= ②1111n na a ③121nn n a a a +=+ ④2121n n a a +-=，则D 型数列{}n a 的序号为_______.23．设等比数列{}n a 满足a 1 + a 2 = –1, a 1 – a 3 = –3，则a 4 = ___________． 24．在数列{}n a 中，11a =，且{}n a 是公比为13的等比数列．设13521T n n a a a a -=++++，则lim n n T →∞=__________．(*n ∈N ) 25．数列{}n a 满足：1a a =（a R ∈且为常数），()()()*13343n n n n n a a a n N a a +⎧->⎪=∈⎨-≤⎪⎩，当100a =时，则数列{}n a 的前100项的和100S 为________. 三、解答题26．在条件①()(sin sin )()sin a b A B c b C +-=-，②sin cos()6a Bb A π=+，③sinsin 2B Cb a B +=中任选一个，补充到下面问题中，并给出问题解答.在ABC ∆中，角,,A B C 的对边分别为,,a b c ，6b c +=，a =， . 求ABC ∆的面积.27．在数列{}n a 中, 已知11a =，且数列{}n a 的前n 项和n S 满足1434n n S S +-=, n *∈N . （1）证明数列{}n a 是等比数列；（2）设数列{}n na 的前n 项和为n T ，若不等式3()1604nn aT n+⋅-<对任意的n *∈N 恒成立, 求实数a 的取值范围.28．在ABC △中，内角A ，B ，C 所对的边分别为a ，b ，c.已知sin cos 6b A a B π⎛⎫=- ⎪⎝⎭. （1）求角B 的大小；（2）设a =2，c =3，求b 和()sin 2A B -的值.29．在ABC ∆sin cos C c A =．（Ⅰ）求角A 的大小；（Ⅱ）若ABC S ∆，2b c +=+a 的值．30．已知在公比为q 的等比数列{}n a 中，416a =，()34222a a a +=+. （1）若1q >，求数列{}n a 的通项公式；（2）当1q <时，若等差数列{}n b 满足31b a =，512b a a =+，123n n S b b b b =+++⋅⋅⋅+，求数列1n S ⎧⎫⎨⎬⎩⎭的前n 项的和.【参考答案】2016-2017年度第*次考试试卷参考答案**科目模拟测试一、选择题 1．C2．B3．A4．B5．C6．C7．B8．D9．B10．D11．A12．C13．C14．D15．C二、填空题16．【解析】【分析】【详解】试题分析：由题意知满足条件的线性区域如图所示：点而目标函数仅在点处取得最大值所以考点：线性规划最值问题17．【解析】【分析】通过计算出并找出的共同表示形式进而利用归纳推理即可猜想结论【详解】当时则由猜想：故答案为：【点睛】本题考查元素与集合关系的判断以及数列前项和的归纳猜想属于中档题18．【解析】【分析】根据不等式组画出可行域可知；根据向量投影公式可知所求投影为利用的范围可求得的范围代入求得所求的结果【详解】由不等式组可得可行域如下图阴影部分所示：由题意可知：在上的投影为：本题正确结19．【解析】【分析】【详解】所以所以故答案为20．【解析】【分析】运用等差数列的求和公式可得可得由数列的裂项相消求和化简可得所求和【详解】解：则可得数列的前n项和故答案为【点睛】本题考查数列的前项和首先运用数列的裂项法对项进行分解然后重新组合最终达21．【解析】试题分析：关于x的不等式（2x－1）2<ax2等价于其中且有故有不等式的解集为所以解集中一定含有123可得所以解得考点：含参数的一元二次方程的解法22．①②③④【解析】【分析】根据D型数列的定义逐个判断正项数列是否满足即可【详解】对①因为且正项数列故故所以成立对②故成立对③成立对④故成立综上①②③④均正确故答案为：①②③④【点睛】本题主要考查了新定23．-8【解析】设等比数列的公比为很明显结合等比数列的通项公式和题意可得方程组：由可得：代入①可得由等比数列的通项公式可得【名师点睛】等比数列基本量的求解是等比数列中的一类基本问题解决这类问题的关键在于24．【解析】【分析】构造新数列计算前n项和计算极限即可【详解】构造新数列该数列首项为1公比为则而故【点睛】本道题考查了极限计算方法和等比数列前n项和属于中等难度的题目25．【解析】【分析】直接利用分组法和分类讨论思想求出数列的和【详解】数列满足：（且为常数）当时则所以（常数）故所以数列的前项为首项为公差为的等差数列从项开始由于所以奇数项为偶数项为所以故答案为：【点睛】三、解答题26．27．28．29．30．2016-2017年度第*次考试试卷参考解析【参考解析】**科目模拟测试一、选择题1．C解析：C【解析】【分析】先通过数列性质判断60a <，再通过数列的正负判断n S 的最小值. 【详解】∵等差数列{}n a 中，390a a +<，∴39620a a a +=<，即60a <.又70a >，∴{}n a 的前n 项和n S 的最小值为6S . 故答案选C 【点睛】本题考查了数列和的最小值，将n S 的最小值转化为{}n a 的正负关系是解题的关键.2．B解析：B 【解析】11111444222n n S -⎛⎫⎛⎫⎛⎫=+-++-+⋅⋅⋅++- ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭11221244133212nnn n ⎛⎫-- ⎪⎛⎫⎝⎭=+=+-⋅- ⎪⎛⎫⎝⎭-- ⎪⎝⎭()143n p S n ≤-≤即22113332n p ⎛⎫⎛⎫≤-⋅-≤ ⎪ ⎪ ⎪⎝⎭⎝⎭对任意*n N ∈都成立，当1n =时，13p ≤≤ 当2n =时，26p ≤≤当3n =时，443p ≤≤ 归纳得：23p ≤≤故选B点睛：根据已知条件运用分组求和法不难计算出数列{}n a 的前n 项和为n S ，为求p 的取值范围则根据n 为奇数和n 为偶数两种情况进行分类讨论，求得最后的结果3．A解析：A 【解析】【分析】先根据2n S n =,求出数列{}n a 的通项公式,然后利用错位相减法求出{}n b 的前n 项和n T .【详解】解:∵2n S n =,∴当1n =时,111a S ==;当2n ≥时,()221121n n n a S S n n n -=-=--=-, 又当1n =时,11a =符合上式,∴21n a n =-, ∴()()()1121nnn n b a n =-=--,∴()()()()()123113151121nn T n =⨯-+⨯-+⨯-+⋅⋅⋅+--①,∴()()()()()2341113151121n n T n +-=⨯-+⨯-+⨯-+⋅⋅⋅+--②,①-②,得()()()()()()23412121111211n n n T n +⎡⎤=-+⨯-+-+-+⋅⋅⋅+---⨯-⎣⎦()()()()()()211111122112111n n n n n -+⎡⎤---⎣⎦=-+⨯--⨯-=---,∴()1nn T n =-,∴数列{}n b 的前n 项和()1nn T n =-.故选:A . 【点睛】本题考查了根据数列的前n 项和求通项公式和错位相减法求数列的前n 项和,考查了计算能力,属中档题.4．B解析：B 【解析】分析：将原问题转化为等差数列的问题，然后结合等差数列相关公式整理计算即可求得最终结果.详解：由题意可得，8个孩子所得的棉花构成公差为17的等差数列，且前8项和为996，设首项为1a ，结合等差数列前n 项和公式有：811878828179962S a d a ⨯=+=+⨯=，解得：165a =，则81765717184a a d =+=+⨯=. 即第八个孩子分得斤数为184. 本题选择B 选项.点睛：本题主要考查等差数列前n 项和公式，等差数列的应用，等差数列的通项公式等知识，意在考查学生的转化能力和计算求解能力.5．C解析：C 【解析】【分析】根据同角三角函数求出sinA ；利用余弦定理构造关于c 的方程解出c ，再根据三角形面积公式求得结果. 【详解】cosA =58⇒sinA =√1−cos 2A =√398由余弦定理得：a 2=c 2+b 2−2bccosA ，即3=c 2+4−5c 2解得：c =12或c =2∵A 为最小角 ∴c >a ∴c =2∴S ΔABC =12bcsinA =12×2×2×√398=√394本题正确选项：C 【点睛】本题考查余弦定理解三角形、三角形面积公式的应用、同角三角函数关系，关键是能够利用余弦定理构造关于边角关系的方程，从而求得边长.6．C解析：C 【解析】记公元1984年为第一年，公元2047年为第64年，即天干循环了十次，第四个为“丁”，地支循环了五次，第四个为“卯”,所以公元2047年农历为丁卯年. 故选C.7．B解析：B 【解析】分析：根据分段函数，分别解不等式，再求出并集即可．详解：由于()223log ,01,0x x f x x x x +>⎧=⎨--≤⎩，当x ＞0时，3+log 2x≤5，即log 2x≤2=log 24，解得0＜x≤4，当x≤0时，x 2﹣x ﹣1≤5，即（x ﹣3）（x+2）≤0，解得﹣2≤x≤0， ∴不等式f （x ）≤5的解集为[﹣2，4]，故选B ．点睛：本题考查了分段函数以及不等式的解法和集合的运算，分段函数的值域是将各段的值域并到一起，分段函数的定义域是将各段的定义域并到一起，分段函数的最值，先取每段的最值，再将两段的最值进行比较，最终取两者较大或者较小的.8．D解析：D 【解析】n 阶幻方共有2n 个数，其和为()222112...,2n n n n ++++=阶幻方共有n 行，∴每行的和为()()2221122n n n n n++=，即()()2210110101,50522n n n N N+⨯+=∴==，故选D.9．B解析：B 【解析】【分析】首先画出可行域，然后结合交点坐标平移直线即可确定实数m 的最大值. 【详解】不等式组表示的平面区域如下图所示，由2230y x x y =⎧⎨--=⎩，得：12x y =-⎧⎨=-⎩，即C 点坐标为（－1，－2），平移直线x ＝m ，移到C 点或C 点的左边时，直线2y x =上存在点(,)x y 在平面区域内，所以，m ≤－1，即实数m 的最大值为－1.【点睛】本题主要考查线性规划及其应用，属于中等题.10．D解析：D 【解析】【分析】由约束条件确定可行域，由1y x+的几何意义，即可行域内的动点与定点P （0，-1）连线的斜率求得答案．【详解】由约束条件242210x y x y x -≤⎧⎪+≤⎨⎪-≥⎩，作出可行域如图，联立10220x x y -=⎧⎨+-=⎩，解得A （112，），1y x+的几何意义为可行域内的动点与定点P （0，-1）连线的斜率，由图可知，113212PAk +==最大．故答案为32．【点睛】本题考查简单的线性规划，考查了数形结合的解题思想方法，属于中档题型．11．A解析：A 【解析】sin()2sin cos 2sin cos cos sin A C B C A C A C ++=+所以2sin cos sin cos 2sin sin 2B C A C B A b a =⇒=⇒=，选A.【名师点睛】本题较为容易，关键是要利用两角和差的三角函数公式进行恒等变形. 首先用两角和的正弦公式转化为含有A ，B ，C 的式子，用正弦定理将角转化为边，得到2a b =.解答三角形中的问题时，三角形内角和定理是经常用到的一个隐含条件，不容忽视. 12．C 解析：C 【解析】试题分析：∵11011101100000a a a d a a ⋅∴＞，＜，＜，＞，＜， ∴18110111810181060T a a a a S S S =+⋯+--⋯-=--=（），选C . 考点：1.等差数列的求和；2.数列的性质.13．C【解析】由64S S -=6546a a a +=得，()22460,60q q a q q +-=+-=，解得2q，从而3522=28=16a a =⋅⨯，故选C.14．D解析：D 【解析】【分析】【详解】由题意可得：332,323n nn n S S +=⨯=⨯- ，由等比数列前n 项和的特点可得数列{}n a 是首项为3，公比为2的等比数列，数列的通项公式：132n n a -=⨯ ，设11n nb b q -= ，则：111132n n n b q b q --+=⨯ ，解得：11,2b q == ，数列{}n b 的通项公式12n nb -= ，由等比数列求和公式有：21nn T =- ，考查所给的选项：13,21,,n n n n n n n n S T T b T a T b +==-<< .本题选择D 选项.15．C解析：C 【解析】【分析】设1BC CD ==，计算出ACD ∆的三条边长，然后利用余弦定理计算出cos DAC ∠．【详解】如下图所示，不妨设1BC CD ==，则2AB =，过点D 作DE AB ⊥，垂足为点D ，易知四边形BCDE 是正方形，则1BE CD ==，1AE AB BE ∴=-=，在Rt ADE ∆中，222AD AE DE =+=，同理可得225AC AB BC =+=，在ACD ∆中，由余弦定理得2222521310cos 210252AC AD CD DAC AC AD +-+-∠===⋅⨯⨯，故选C ．本题考查余弦定理求角，在利用余弦定理求角时，首先应将三角形的边长求出来，结合余弦定理来求角，考查计算能力，属于中等题．二、填空题16．【解析】【分析】【详解】试题分析：由题意知满足条件的线性区域如图所示：点而目标函数仅在点处取得最大值所以考点：线性规划最值问题解析：1(,)3+∞【解析】【分析】【详解】试题分析：由题意知满足条件的线性区域如图所示：，点(22)A ，，而目标函数(0)z x ay a =+≥仅在点(2,2)处取得最大值，所以1133AB k a a ->=-∴> 考点：线性规划、最值问题.17．【解析】【分析】通过计算出并找出的共同表示形式进而利用归纳推理即可猜想结论【详解】当时则由猜想：故答案为：【点睛】本题考查元素与集合关系的判断以及数列前项和的归纳猜想属于中档题解析：1()221n n +-【解析】【分析】通过计算出3S ，并找出1S 、2S 、3S 的共同表示形式，进而利用归纳推理即可猜想结论．【详解】当3n =时,{}31,3,7A =，则113711T =++=,213173731T =⨯+⨯+⨯=,313721T =⨯⨯=,∴312311312163S T T T =++=++=,由1212112121S ⨯==-=-，2332272121S ⨯==-=-，34623632121S ⨯==-=-，⋯猜想：(1)221n n n S +=-.故答案为：1()221n n +-.【点睛】本题考查元素与集合关系的判断以及数列前n 项和的归纳猜想,属于中档题.18．【解析】【分析】根据不等式组画出可行域可知；根据向量投影公式可知所求投影为利用的范围可求得的范围代入求得所求的结果【详解】由不等式组可得可行域如下图阴影部分所示：由题意可知：在上的投影为：本题正确结解析：[]3,3-【解析】【分析】根据不等式组画出可行域，可知5,66AOP ππ⎡⎤∠∈⎢⎥⎣⎦；根据向量投影公式可知所求投影为cos OA AOP ∠，利用AOP ∠的范围可求得cos AOP ∠的范围，代入求得所求的结果.【详解】由不等式组可得可行域如下图阴影部分所示：由题意可知：6AOB π∠=，56AOC π∠=OA 在OP 上的投影为：cos 9323OA AOP AOP AOP ∠=+∠=∠AOB AOP AOC ∠≤∠≤∠ 5,66AOP ππ⎡⎤∴∠∈⎢⎥⎣⎦33cos ,22AOP ⎡∴∠∈-⎢⎣⎦[]cos 3,3OA AOP ∴∠∈-本题正确结果：[]3,3- 【点睛】本题考查线性规划中的求解取值范围类问题，涉及到平面向量投影公式的应用；关键是能够根据可行域确定向量夹角的取值范围，从而利用三角函数知识来求解.19．【解析】【分析】【详解】所以所以故答案为解析：41n -【解析】【分析】【详解】()()145[415]4n n q a a n n -=-=-+---+=-，124253b a ==-⨯+=-，所以()11134n n n b b q --=⋅=-⋅-，()113434n n n b --=-⋅-=⋅，所以211214334343434114n n n n b b b --++⋯+=+⋅+⋅+⋯+⋅=⋅=--，故答案为41n -.20．【解析】【分析】运用等差数列的求和公式可得可得由数列的裂项相消求和化简可得所求和【详解】解：则可得数列的前n 项和故答案为【点睛】本题考查数列的前项和首先运用数列的裂项法对项进行分解然后重新组合最终达解析：4nn 1+ 【解析】【分析】运用等差数列的求和公式可得()n 11na n n 1n 122=⋅+=+，可得()n n n 11411b 4a a n n 1n n 1+⎛⎫===- ⎪++⎝⎭，由数列的裂项相消求和，化简可得所求和．【详解】解：()n 12n 11na n n 1n 1n 1n 1n 122=++⋯+=⋅+=++++，则()n n n 11411b 4a a n n 1n n 1+⎛⎫===- ⎪++⎝⎭，可得数列{}n b 的前n 项和n 1111111S 4122334n n 1⎛⎫=-+-+-+⋯+- ⎪+⎝⎭14n 41n 1n 1⎛⎫=-=⎪++⎝⎭．故答案为4nn 1+．【点睛】本题考查数列的前n 项和，首先运用数列的裂项法对项进行分解，然后重新组合，最终达到求和目的，考查化简整理的运算能力，属于基础题．21．【解析】试题分析：关于x 的不等式（2x －1）2<ax2等价于其中且有故有不等式的解集为所以解集中一定含有123可得所以解得考点：含参数的一元二次方程的解法解析：2549,916⎡⎤⎢⎥⎣⎦【解析】试题分析：关于x 的不等式（2x －1）2<ax 2等价于2(4)410a x x -+-+<，其中40a ∆=>且有40a ->，故有04a <<，不等式的解集为1122x a a <<+-，所以111422a <<+解集中一定含有1,2,3，可得，所以53{74a a ≥≤，解得2549916a ≤≤. 考点：含参数的一元二次方程的解法.22．①②③④【解析】【分析】根据D 型数列的定义逐个判断正项数列是否满足即可【详解】对①因为且正项数列故故所以成立对②故成立对③成立对④故成立综上①②③④均正确故答案为：①②③④【点睛】本题主要考查了新定解析：①②③④ 【解析】【分析】根据D 型数列的定义,逐个判断正项数列{}n a 是否满足11n n a a +-<即可. 【详解】对①,因为2211n n a a +-=,且正项数列{}n a .故()222211211n n n n n a a a a a +=+<++=+,故11n n a a +<+.所以11n n a a +-<成立. 对②,1111111111n n n nn nn a a a a a a a ,故22101111n n n n nn n n n n n a a a a a a a a a a a +--=---++==<<+成立. 对③, 112221101111n nn n n n n n n n a a a a a a a a a a ++⎛⎫=⇒-=-=-<< ⎪+++⎝⎭成立对④, ()2222112121211n n n n n n n a a a a a a a ++-=⇒=+<++=+.故11n n a a +<+,11n n a a +-<成立. 综上, ①②③④均正确.故答案为：①②③④ 【点睛】本题主要考查了新定义的问题,需要根据递推公式证明11n n a a +-<.属于中等题型.23．-8【解析】设等比数列的公比为很明显结合等比数列的通项公式和题意可得方程组：由可得：代入①可得由等比数列的通项公式可得【名师点睛】等比数列基本量的求解是等比数列中的一类基本问题解决这类问题的关键在于解析：-8 【解析】设等比数列{}n a 的公比为q ，很明显1q ≠-，结合等比数列的通项公式和题意可得方程组：()()12121311113a a a q a a a q ⎧+=+=-⎪⎨-=-=-⎪⎩，①，②，由②①可得：2q =-，代入①可得11a =，由等比数列的通项公式可得3418a a q ==-.【名师点睛】等比数列基本量的求解是等比数列中的一类基本问题，解决这类问题的关键在于熟练掌握等比数列的有关公式并能灵活运用，尤其需要注意的是，在使用等比数列的前n 项和公式时，应该要分类讨论，有时还应善于运用整体代换思想简化运算过程.24．【解析】【分析】构造新数列计算前n 项和计算极限即可【详解】构造新数列该数列首项为1公比为则而故【点睛】本道题考查了极限计算方法和等比数列前n 项和属于中等难度的题目解析：9lim 8n n T →∞=【解析】【分析】构造新数列{}21n a -，计算前n 项和，计算极限，即可。

广东省深圳明德外语实验高三数学上学期期终考试题理苏教版

明德外语实验学校高三上期末考试数学试卷（理科）（时间：120分钟满分：150分）一、选择题：本大题共8小题，每小题5分，满分40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. 定义集合运算：A ※B ＝｛t|t ＝xy ，x ∈A ，y ∈B}，设A ＝｛1，2｝，B ＝｛0，2｝，则集合A ※B 的所有元素之和为（）A. 6B. 3C. 2D. 02. 点P (4，－2)与圆x 2＋y 2＝4上任一点连线的中点轨迹方程是( )A ． (x －2)2＋(y ＋1)2＝4B ．(x －2)2＋(y ＋1)2＝1C ． (x ＋4)2＋(y －2)2＝4D ．(x ＋2)2＋(y －1)2＝1 3. 函数xx g x x f -=+=122)(log 1)(与在同一直角坐标系下的图象大致是（）A B C D4. 已知等差数列{}n a 中，10795=-+a a a ，记n n a a a S +++= 21，则13S 的值为（） A. 130 B. 260C. 156D. 1685. 下列结论错误．．的是（） A ．命题：“若20232==+-x x x ，则”的逆否命题为:“若2≠x ，则0232≠+-x x ”.B. 命题:“存在x 为实数，02>-x x ”的否定是“任意x 是实数，02≤-x x ”.C. “22bc ac >”是“b a >”的充分不必要条件.D. 若p 且q 为假命题，则p 、q 均为假命题.6. 如右图，在一个长为π，宽为2的矩形OABC 内，曲线()sin 0y x x π=≤≤与x 轴围成如图所示的阴影部分，向矩形OABC 内随机投一点（该点落在矩形OABC 内任何一点是等可能的），则所投的点落在阴影部分的概率是（） A ．4π B . 3πC.2πD.1π7. 方程(1＋4k )x －(2－3k )y ＋2－14k ＝0所确定的直线必经过点( )A ．(2,2)B ．(－2,2)C ．(－6,2)D ．(3，－6)8. 用数学归纳法证明：12＋22＋…＋n 2＋…＋22＋12＝n 2n 2＋13，第二步证明由“k 到k ＋1”时，左边应加( )A ．k 2B ．(k ＋1)2C ．k 2＋(k ＋1)2＋k 2D ．(k ＋1)2＋k 2，二、填空题：本大题共7小题，考生作答6小题，每小题5分，满分30分. （一）必做题（9～13题） 9. 已知03πθ<<，且3cos()35πθ-=，则sin θ= ． 10. 已知A 、B 、C 是圆O ：221x y +=上三点且OA OB OC +=，则AB OA ⋅= ． 11．曲线ln(21)y x =-上的点到直线230x y -+=的最短距离是． 12. 若不等式413x x a a++-≥+对任意的实数x 恒成立，则实数a 的取值范围是13.已知)(x f 是偶函数，在),0[+∞上是增函数，若(1)(2)f ax f x +≤+（1||≥a ）在]1,21[∈x 上恒成立，则实数a 的取值范围为．（二）选做题（14～15题，考生只能从中选做一题） 14、(坐标系与参数方程选做题)在平面直角坐标系xOy 中，已知直线l 的参数方程为21,()42x t t R y t =-⎧∈⎨=-⎩参数，以直角坐标原点为极点，x 轴的正半轴为极轴建立相应的极坐标系。

高考模拟练习—广东省深圳市外国语高级中学2022届高三上学期期末数学试题(含答案解析)

广东省深圳市福田区2022届高三上学期数学试题试卷副标题注意事项：1．答题前填写好自己的姓名、班级、考号等信息 2．请将答案正确填写在答题卡上第I 卷（选择题）请点击修改第I 卷的文字说明一、单选题 1．若全集{}1,2,3,4,5,6U =，{}1,3,4M =，{}2,3,4N =，则集合()UM N 等于（）A ．5,6B ．{}1,5,6C ．{}2,5,6D ．{}1,2,5,62．有下列四个命题：①x R ∀∈10>；②x N ∀∈，20x >；③x N ∃∈，[3x ∈-，1)-；④∃∈x Q ，22x =．其中真命题的个数为（）A ．1B ．2C ．3D ．43．已知i 是虚数单位，则复数22(1i)z =-的共轭复数为（） A ．2iB ．-2iC ．iD ．-i4．不等式()()2230x x -->的解集是（） A ．()3,2,2⎛⎫-∞⋃+∞ ⎪⎝⎭ B ．RC ．3,22⎛⎫⎪⎝⎭D ．∅5．已知3log 0.5a =，0.53b =，0.50.3c =则（） A ．a b c <<B ．c a b <<C ．b c a <<D ．a c b <<6．已知函数2()e 2e x f x x x =-，若曲线()y f x =在1x =处的切线与直线230x ay -+=垂直，则=a （） A ．2e -B ．2e-C ．e 2D ．2e7．已知a ，b 为正实数，直线2y x a =-与曲线()ln y x b =+相切，则12a b+的最小值是（） A ．6B ．C ．8D ．8．已知函数()y f x =是定义在R 上的偶函数，且()()2f x f x -=，当01x ≤≤时，()f x x =，设函数()()5log g x f x x =-，则()g x 的零点的个数为（）A ．6B ．7C ．8D ．9二、多选题 9．若,,a b m R ∈，则下列说法正确的是（）A ．命题“0,21x x ∀<<”的否定为：“000,21xx ∃≥≥”B ．235log 27log 25log 818⋅⋅=C ．若a ＋2b ＝2，则396a b +≥D ．“幂函数2m y x -=在(0,)+∞上单调递增”的充要条件是“指数函数(2)x y m =-单调递增” 10．下列函数中满足“对任意x 1，x 2∈（0，＋∞），都有1212()()f x f x x x --＞0”的是（）A ．f （x ）＝－2xB ．f （x ）＝－3x ＋1C ．f （x ）＝x 2＋4x ＋3D ．f （x ）＝x －1x11．已知函数()f x 为偶函数，且()()22f x f x +=--，则下列结论一定正确的是（） A ．()f x 的图象关于点(2,0)-中心对称 B ．()f x 是周期为4的周期函数 C ．()f x 的图象关于直线2x =-轴对称D ．(4)f x +为偶函数12．已知函数()222e xx x f x +-=，则下列结论正确的是（）A ．函数()f x 有极小值也有最小值B ．函数()f x 存在两个不同的零点C ．当260ek -<<时，()f x k =恰有三个不相等的实根 D ．当[]0,x t ∈时，()f x 的最大值为26e ，则t 的最小值为2 第II 卷（非选择题）请点击修改第II 卷的文字说明三、填空题13．若向量()3,a m=，()2,1b=-，且a b⊥，则实数m的值为______.14．已知复数z满足()2i43iz+=+（i为虚数单位），则z的虚部为__________．15．已知不等式x2＋2x＋a2－3＞0的解集为R，则a的取值范围是_____________.16．已知平面向量PA，PB满足PA＝PB＝1，PA·PB＝－12，若BC＝1，则AC的最大值为______.四、解答题17．若不等式20ax bx c++≥的解集是{}123x x-≤≤，求不等式20cx bx a++<的解集．18．已知函数21()2ln2f x x x x=--（1）求函数()f x在1x=处的切线方程；（2）求函数()f x在[1,4]上的最小值．19．已知)(f x为二次函数，满足)(03f=，)()(121f x f x x+-=-（1）求函数)(f x的解析式（2）函数)(12xg x⎛⎫=⎪⎭⎝，求函数)()(g f x的值域20．已知函数()321f x x x x=+-+.(1)求函数()f x的单调区间和极值；(2)若函数()y f x=的图象与直线y a=仅有一个公共点，求实数a的取值范围. 21．已知函数()()3xx e xf a=-+.（1）当1a=时，求()f x的最小值；（2）若()f x有两个零点，求实数a的取值范围.22．已知函数221()e2xf x a x ax=--，a∈R.（1）当1a=时，求函数2()()g x f x x=+的单调区间；（2）当44e1a<<-，时，函数()f x有两个极值点1x，2x（12x x<），证明：212x x->.参考答案：1．D 【解析】【分析】利用集合的交集运算，计算M N ⋂，再利用集合的补集运算即得解【详解】由题意，{3,4}MN =再由{}1,2,3,4,5,6U = 可得(){}1,2,5,6UM N =故选：D 【点睛】本题考查了集合的交集、补集运算，考查了学生概念理解，综合分析能力，属于基础题 2．A 【解析】【分析】逐一判断全称量词命题或存在量词命题的真假即可判断作答. 【详解】对于①，x R ∀∈010>，①是真命题；对于②，因0x =时，x ∈N ，20x =，②是假命题；对于③，因x N ∀∈，0x ≥，即[3,1)x ∉--，③是假命题；对于④，因当且仅当x x =22x =Q ，且Q ，④是假命题，所以真命题的序号是①，共1个．故选：A 3．D 【解析】【分析】根据复数的除法运算求得z ，进而可求得其共轭复数. 【详解】2221=i (1i)2i iz -=-=-=，其共轭复数为i -. 故选：D. 4．A 【解析】【分析】直接利用解一元二次不等式的解法求解即可. 【详解】解：由()()2230x x -->，得32x <或2x >. 所以不等式()()2230x x -->的解集为()3,2,2⎛⎫-∞⋃+∞ ⎪⎝⎭，故选：A. 5．D 【解析】【分析】利用指数函数和对数函数的性质判断,,a b c 与中间量0，1的关系，从而可得答案【详解】因为3log y x =在(0,)+∞上递增，且0.51<，所以33log 0.5log 10<=，即0a <，因为3x y =在R 上递增，且0.50>，所以00.5133>=，即1b >，因为0.3x y =在R 上递减，且0.50>，所以0.5000.30.31<<=，即01c <<，所以a c b <<，故选：D 6．A 【解析】【分析】求得()'1f ，根据两条直线相互垂直求得a .【详解】()()'22e 2e x f x x x =+⋅-，()'13e 2e e f =-=，由于曲线()y f x =在1x =处的切线与直线230x ay -+=垂直所以2e 12e a a⋅=-⇒=-.故选：A 7．C 【解析】【分析】设切点为（m ，n ），求出曲线对应函数的导数，可得切线的斜率，代入切点坐标，解方程可得n ＝0，进而得到2a +b ＝1，再由乘1法和基本不等式，即可得到所求最小值．【详解】设切点为（m ，n ）， y ＝ln （x +b ）的导数为1y x b'=+，由题意可得1m b+=1，又n ＝m ﹣2a ，n ＝ln （m +b ），解得n ＝0，m ＝2a ，即有2a +b ＝1，因为a 、b 为正实数，所以12124=()2248b a a b a b a b a b +++=+++≥+=，当且仅当122a b ==时取等号，故12a b+的最小值为8．故选：C ． 8．C 【解析】【分析】由题设知()g x 的零点可转化为()f x 与5log x 的交点问题，而()[0,1]f x ∈且周期为2，关于y………外…………○…………装…………○…………订…………○…………线…………○…………学校:___________姓名：___________班级：___________考号：___________………内…………○…………装…………○…………订…………○…………线…………○…………轴对称的函数；5log x 且关于y 轴对称，当55x -≤≤时有5log (,1]x ∈-∞，画出(0,)+∞的草图即可确定交点个数，利用对称性确定总交点数. 【详解】由题意知：()f x 关于1x =对称，而()g x 的零点即为()5=log f x x 的根，又∵()f x 在R 上的偶函数，知：()[0,1]f x ∈且周期为2，关于y 轴对称的函数，而55x -≤≤时5log (,1]x ∈-∞且关于y 轴对称 ∴()f x 与5log x 在(0,)+∞的图象如下，∴共有4个交点，由偶函数的对称性知：在(,0)-∞上也有4个交点，所以共8个交点. 故选：C. 【点睛】关键点点睛：将函数零点转化为两个函数的交点问题，应用数形结合的方法，由函数的周期性、奇偶对称性判断交点的个数. 9．BC 【解析】【分析】根据全称命题的否定、对数运算的性质、基本不等式及等价法判断充分必要条件，即可知各选项的正误. 【详解】A ：由全称命题的否定知：命题“0,21x x ∀<<”的否定为“000,21xx ∃<≥”，故错误；B ：23523525log 27log 25log 818log 3log 5log 218log 5log 218⋅⋅=⋅⋅⋅=⋅⋅=，故正确；C ：由2(1)a b =-，则939969a b bb ++≥==，当且仅当1a =，12b =时等号成立，故正确；D ：幂函数2m y x -=在(0,)+∞上单调递増，则20m ->即2m >，而指数函数(2)x y m =-单调递增，则21m ->即3m >，故“幂函数2m y x -=在(0,)+∞上单调递増”的必要不充分条件是“指数函数(2)x y m =-单调递增”，故错误；故选：BC 10．ACD 【解析】【分析】先由题意判断f （x ）为（0，＋∞）上的增函数.再对四个选项一一验证: 对于A ：利用反比例函数的单调性直接判断; 对于B ：利用一次函数的单调性直接判断; 对于C ：利用二次函数的单调性直接判断; 对于D ：先判断出1y x =和21y x=-在（0，＋∞）上的单调性,即可判断【详解】因为“对任意x 1，x 2∈（0，＋∞），都有1212()()f x f x x x --＞0”所以不妨设0< x 1<x 2,都有12()()f x x <, 所以f （x ）为（0，＋∞）上的增函数.对于A ：f （x ）＝－2x 在（0，＋∞）上为增函数,故A 正确;对于B ：f （x ）＝－3x ＋1在（0，＋∞）上为减函数,故B 错误;对于C ：f （x ）＝x 2＋4x ＋3对称轴为x =-2,开口向上,所以在（0，＋∞）上为增函数,故C 正确;对于D ：f （x ）＝x －1x ,因为1y x =在（0，＋∞）上为增函数, 21y x =-在（0，＋∞）上为增函数,所以f （x ）＝x －1x在（0，＋∞）上为增函数, 故D 正确;故选:ACD 11．AD【解析】【分析】由()2()2f x f x +=--，可知()f x 的图象关于点()2,0中心对称；结合函数()f x 为偶函数可得()f x 是周期为8以及关于直线4x =轴对称，结合周期，对称中心和对称轴可判断出()4f x +为偶函数【详解】因为()2()2f x f x +=--，所以()f x 的图象关于点()2,0中心对称，又因为函数()f x 为偶函数，所以()f x 是周期为8的周期函数，且它的图象关于点(2,0)-中心对称和关于直线4x =轴对称，所以()4f x +为偶函数. 故选：AD. 12．ABD 【解析】【分析】求出导函数()'f x ，由()'f x 确定函数的单调性、极值、函数的变化趋势，然后逐项分析即可．【详解】解：由222()x x x f x e +-=，得()222(22)(22)4()x x x x x e x x e x f x e e +-+--+'==，令()0f x '=，则2x =-或2x =，当2x <-或2x >时，()0f x '<；当22x -<<时，()0f x '> ，所以()f x 在(,2)-∞-和(2,+)∞上单调递减，在(2,2)-上单调递增，所以()f x 有极小值()2244222f e e ---==--，有极大值()224+4262f e e-==，当x →-∞时，()f x →+∞，当x →+∞时，()0f x →，故函数的图象如图，………外…………○…………装…………○…………订…………○…………线…………○…………学校:___________姓名：___________班级：___________考号：___________………内…………○…………装…………○…………订…………○…………线…………○…………数形结合可知：函数()f x 有极小值也有最小值，故A 正确；因为函数()f x 与x 轴有两个交点，故函数()f x 存在两个不同的零点，故B 正确；当260e k -<<时，()f x k =恰有三个不相等的实根等价于直线y k =与函数()f x 有三个不同的交点，故260e k <<，故C 错误；当[]0,x t ∈时，()f x 的最大值为26e ，则2t ≥，故D 正确. 所以选项ABD 正确，故选：ABD 【点睛】本题考查用导数确定函数的单调性、极值，确定方程根的个数问题．解题思路是求得导函数，然后求出导函数的零点，确定导函数的正负得函数的单调性，可得极值、最值，确定函数的变化趋势，可确定方程()f x k =的解的个数． 13．6 【解析】【分析】由a b ⊥可得0a b ⋅=，从而可求出实数m 的值【详解】因为()3,a m =，()2,1b =-，且a b ⊥，所以60a b m ，得6m =，故答案为：6 14．1- 【解析】【分析】由复数的模长和运算法则化简，由复数的基本概念可得虚部. 【详解】∵复数z 满足()2i 43i z +=+, ∴()2i 5z +==, ∴()()()52i 5105i 2i 2i 2i 2i 5z --====-++-, ∴z 的虚部为1-. 故答案为：1-. 15．()(),22,-∞-+∞【解析】要使不等式x 2＋2x ＋a 2－3＞0的解集为R ，只需x 2＋2x ＋a 2－3=0的判别式小于零即可. 【详解】因为不等式x 2＋2x ＋a 2－3＞0的解集为R ，则244(3)0a ∆=--<，解得2a >，或2a <-因此，实数a 的取值范围是()(),22,-∞-+∞.故答案为：()(),22,-∞-+∞.【点睛】本题考查利用一元二次不等式在实数集上恒成立求参数的取值范围问题，如果二次项系数为………外…………○…………装…………○…………订…………○…………线…………○…………学校:___________姓名：___________班级：___________考号：___________………内…………○…………装…………○…………订…………○…………线…………○…………参数，要对参数分等于零和不等于零两种情况讨论，结合二次项系数的符号和判别式的符号来进行求解，考查分析问题和解决问题的能力，属于中等题. 16．31+##13+ 【解析】【分析】以P 为原点，P A 为x 轴建立坐标系，求出C 的轨迹即可求解. 【详解】如图，以P 为原点，P A 为x 轴建立坐标系.∵PA ＝PB ＝1，PA ·PB ＝－12，∴∠APB ＝120°， ∵BC ＝1，故C 在以B 为圆心，1为半径的圆B 上， ()0,0P ，1,0A ，132⎛- ⎝⎭B ，∴AC 的最大值为：22131113122AB ⎛⎫⎛⎫+=++= ⎪ ⎪ ⎪⎝⎭⎝⎭. 31. 17．{132x x ⎫-<<⎬⎭.【解析】【分析】由题可得13-，2为方程20ax bx c ++=的两个根，得出,,a b c 关系即可求解.【详解】由20ax bx c ++≥的解集为{}123x x -≤≤，知0a <，且13-，2为方程20ax bx c ++=的两个根，∴53b a -=，23c a =-，∴53b a =-，23c a =-．∴不等式20cx bx a ++<变为225033a x a x a ⎛⎫⎛⎫-+-+< ⎪ ⎪⎝⎭⎝⎭，即22530ax ax a +->，又0a <，∴22530x x +-<，解得132x -<<， ∴所求不等式的解集为{132x x ⎫-<<⎬⎭．故答案为：{132x x ⎫-<<⎬⎭.18．（1）4230x y +-=；（2）min ()2ln 2f x =-. 【解析】【分析】（1）根据导数的几何意义求解即可；（2）求导分析()f x 的单调性，再求区间内的最小值即可【详解】（1）11(1)2ln1122f =--=- ∴切点为11,2⎛⎫- ⎪⎝⎭，2()1f x x x =--'(1)1212f =--'∴=-∴切线方程为：12(1)2y x +=-- 故函数()f x 在1x =处的切线方程4230x y +-=（2）2(2)(1)()1(0)x f x x x x x x'-+=--=>令()0f x '=2x ∴=或1x =-（舍）min ()(2)2ln 2f x f ∴==-19．（1）)(223f x x x =-+；（2）10,4⎛⎤ ⎥⎝⎦【解析】【分析】（1）设)(2f x ax bx c =++()0a ≠，利用)(03f =可得c 的值，由)()(1f x f x +-21x =-，利用对应系数相等列方程可得a ，b 的值，进而可得)(f x 的解析式；（2）)()(22312x x g f x -+⎛⎫= ⎪⎝⎭由12ty ⎛⎫= ⎪⎝⎭和223t x x =-+复合而成，求出223t x x =-+的范围，再由指数函数的单调性即可求解. 【详解】（1）设)(2f x ax bx c =++()0a ≠，因为)(03f c ==，)(23f x ax bx =++由)()(121f x f x x +-=-可得：()()22113321a x b x ax bx x ++++---=-，整理可得：221ax a b x ++=-，所以221a a b =⎧⎨+=-⎩，可得12a b =⎧⎨=-⎩，所以)(223f x x x =-+；（2）由)(12xg x ⎛⎫=⎪ ⎭⎝，可得)()(22312x x g f x -+⎛⎫= ⎪⎝⎭，因为)()(22312x x g f x -+⎛⎫= ⎪⎝⎭是由12ty ⎛⎫= ⎪⎝⎭和223t x x =-+复合而成，因为()2223122t x x x =-+=-+≥，即[)2,t ∈+∞，12t y ⎛⎫= ⎪⎝⎭在R 上单调递减，所以2111224t y ⎛⎫⎛⎫=≤= ⎪ ⎪⎝⎭⎝⎭，………外…………○…………装…………○…………订…………○…………线…………○…………学校:___________姓名：___________班级：___________考号：___________………内…………○…………装…………○…………订…………○…………线…………○…………又因为102t y ⎛⎫=> ⎪⎝⎭，所以110,24ty ⎛⎫⎛⎤=∈ ⎪ ⎥⎝⎭⎝⎦，所以函数)()(g f x 的值域为10,4⎛⎤⎥⎝⎦.20．(1)增区间是：(),1-∞-，1,3⎛⎫+∞ ⎪⎝⎭，减区间是：11,3⎛⎫- ⎪⎝⎭，极大值(1)2f -=，极小值2227；(2){|2a a >或22}27a <. 【解析】【分析】(1)利用导数研究单调性即可求解；(2)根据(1)中极值和单调性作出f (x )近似图像即可求解. (1)2()321(31)(1)f x x x x x '=+-=-+，易得当13x >或1x <-时，()0f x '>，当113x -<<时，()0f x '<，∴函数的增区间是：(),1-∞-，1,3⎛⎫+∞ ⎪⎝⎭，减区间是：11,3⎛⎫- ⎪⎝⎭，当1x =-时函数取得极大值(1)2f -=，当13x =时，函数取得极小值2227；(2)画出函数图像，由图形知，当2a >或2227a <时，y a =与()y f x =只有一个交点．故a 的范围{|2a a >或22}27a <．21．（1）2-；（2）21,e ⎛⎫+∞ ⎪⎝⎭.【解析】【分析】（1）利用导数可确定()f x 的单调性，由此得到()()min 0f x f =；（2）当0a ≤时，可知()f x 单调递增，不符合题意；当0a >时，可得()f x 单调性，得到()()min ln f x f a =，将问题转化为()ln 0f a <，解不等式可求得结果.【详解】（1）当1a =时，()3xf x e x =--，则()f x 的定义域为(),-∞+∞，且()1xf x e '=-，∴当(),0x ∈-∞时，()0f x '<；当()0,x ∈+∞时，()0f x '>；()f x ∴在(),0-∞上单调递减，在()0,∞+上单调递增， ()f x ∴的最小值为()02f =-.（2）由题意知：()f x 定义域为(),-∞+∞，()xf x e a '=-；①当0a ≤时，()0xf x e a '=->恒成立，()f x ∴在(),-∞+∞上单调递增，不符合题意；②当0a >时，令()0f x '=，解得：ln x a =，∴当(),ln x a ∈-∞时，()0f x '<，()f x 单调递减；当()ln ,x a ∈+∞时，()0f x '>，()f x 单调递增；即当0a >时，()f x 有极小值也是最小值为()()ln 2ln f a a a =-+. 又当x →-∞时，()f x →+∞；当x →+∞时，()f x →+∞；∴要使()f x 有两个零点，只需()ln 0f a <即可，则2ln 0a +>，解得：21a e >；综上所述：若()f x 有两个零点，则a 的取值范围为21,e ⎛⎫+∞ ⎪⎝⎭.22．（1）减区间为(),0-∞，增区间为()0,∞+；（2）具体见解析.【解析】【分析】（1）对函数求导，根据导函数和原函数的关系得出单调区间；（2）先求出导函数()'f x ，设()()h x f x '=，进而通过()h x '的符号得出()'f x 的单调区间，再通过特值法和放缩法判断出()'f x 零点的位置，进而得到()'f x 的符号，从而得出原函数的单调区间和极值点，最后再通过放缩法证明问题. 【详解】（1）R x ,∈21()e 2xg x x =-，2()e 1x g x '=-，0x >时，()0g x '>，0x <时，()0g x '<，则函数()g x 在(),0-∞单调递减，在()0,∞+单调递增.（2）R x ,∈2()e 2x f x a x a '=--，令()()h x f x '=，∵440e 1a <<-，则()22e 2xh x a '=-在R 上单调递增，∴11,ln 2x a ⎛⎫∈-∞ ⎪⎝⎭时，()0h x '<，()'f x 单调递减，11ln ,2x a ⎛⎫∈+∞ ⎪⎝⎭时，()0h x '>，()'f x 单调递增，∴()'f x 在11ln 2x a=处取得极小值，且(0)=0f '.令()()e 1xx x ϕ=-+，()e 1x x ϕ'=-，则()0,x ∈+∞时，()0x ϕ'>，()x ϕ单调递增，∴()()00ϕϕ>=x ，∴x >0时，()0e 1xx x ϕ>⇒>+，则2e +12x x>，于是x >0时，2222e +11e 124xx x x x ⎛⎫>>+⇒>+ ⎪⎝⎭.∴()()()222()e 2112221x f x a x a x x ax x x ax '=->+--=-=--，∴2x a >时，()0f x '>，于是2112ln ,2x a a ⎛⎫∃∈ ⎪⎝⎭（x 2唯一），使得2()0f x '=. ∴(),0x ∈-∞时，()0f x '>，()f x 单调递增，()20,x x ∈时，()0f x '<，()f x 单调递减，()2,x x ∈+∞时，()0f x '>，()f x 单调递增.则函数()f x 在10x =处取得极小值，在2x x =处取得极大值. 又∵440e 1a <<-，∴()()444414(2)e 4e 440e 1e 1f a a a '--=--=-<⋅-=-，∴222,x a ⎛⎫∈ ⎪⎝⎭，∴212x x ->. 【点睛】本题第（2）问有难度，看似是双变量的问题实际上是单变量问题，在探讨()'f x 的零点时首先要想到特值，本题含指数函数可以尝试验证x =0是否是零点；在判断第二个零点时用到了放缩法，因此我们需要对课本上的常见放缩不等式进行总结和归纳，比如常见的e 1,1ln x x x x ≥+-≥等等.。

广东省深圳市明德外语实验学校2021-2022学年高一数学理期末试卷含解析

广东省深圳市明德外语实验学校2021-2022学年高一数学理期末试卷含解析一、选择题：本大题共10小题，每小题5分，共50分。

在每小题给出的四个选项中，只有是一个符合题目要求的1. 在平面直角坐标系中，角的顶点与原点重合，始边与x的非负半轴重合，终边过点，则（）A. B. C. D.参考答案：A【分析】由三角函数定义得到cosα,然后由诱导公式即可得到答案.【详解】角的终边过点，则，则，故选：A【点睛】本题考查三角函数定义和诱导公式的应用，属于基础题.2. 已知，则（）A. －1B.C. 1D.参考答案：D【分析】由已知求得，再利用诱导公式及同角三角函数基本关系式化弦为切即可求解。

【详解】由，得，即，则．故选D．【点睛】本题主要考查三角函数的化简求值，诱导公式与同角三角函数基本关系式的应用。

3. 已知y = f ( x ) 是定义在R上的单调函数，则（）（A）函数x = f – 1 ( y ) 与y = f ( x )的图象关于直线y = x对称（B）函数f ( – x ) 与f ( x )的图象关于原点对称（C）f – 1 ( x )和f ( x )的单调性相反（D）函数f ( x + 1 ) 和f – 1 ( x ) – 1的图象关于直线y = x对称参考答案：D4. 已知函数的定义域为，若其值域也为，则称区间为的保值区间．若的保值区间是，则的值为（）A．1 B． C．D．参考答案：A5. 数列{a n}满足，且对任意的都有，则数列的前100项的和为（）．A. B. C. D.参考答案：B【分析】先利用累加法求出，再利用裂项相消法求解.【详解】∵，∴，又，∴∴，∴数列的前100项的和为：．故选:B．【点睛】本题主要考查数列通项的求法，考查裂项相消求和，意在考查学生对这些知识的理解掌握水平和分析推理能力.6. （5分）已知sin（π+α）=，α为第三象限角，则tanα=（）A．B．﹣C．D．﹣参考答案：A考点：同角三角函数基本关系的运用．专题：三角函数的求值．分析：已知等式利用诱导公式化简求出sinα的值，根据α为第三象限角，利用同角三角函数间基本关系求出cosα的值，即可确定出tanα的值．解答：∵sin（π+α）=﹣sinα=，即sinα=﹣，α为第三象限角，∴cosα=﹣=﹣，则tanα==，故选：A．点评：此题考查了同角三角函数基本关系的运用，熟练掌握基本关系是解本题的关键．7. 若=,且，则=( )A．- B.- C. D.参考答案：C8. 已知集合A={x|x2+x﹣2＜0}，B={x|x＞0}，则集合A∩B等于（）A．{x|x＞﹣2} B．{x|0＜x＜1} C．{x|x＜1} D．{x|﹣2＜x＜1}参考答案：B【考点】交集及其运算．【专题】集合．【分析】求出A中不等式的解集确定出A，找出A与B的交集即可．【解答】解：由A中不等式变形得：（x﹣1）（x+2）＜0，解得：﹣2＜x＜1，即A={x|﹣2＜x＜1}，∵B={x|x＞0}，∴A∩B={x|0＜x＜1}，故选：B．【点评】此题考查了交集及其运算，熟练掌握交集的定义是解本题关键．9. 下列说法不正确的是（）A. 四边相等的四边形是菱形；B．空间中，一组对边平行且相等的四边形是一定是平行四边形；C. 两两相交的且不共点的三条直线确定一个平面；D. 两组对边平行的四边形是平行四边形参考答案：A10. 已知是定义在R上不恒为0的函数，且对任意，有成立，，令，则有( )A. {a n}为等差数列B. {a n}为等比数列C. {b n}为等差数列D. {b n}为等比数列参考答案：C令,得到得到，.,说明为等差数列，故C正确，根据选项，排除A，D.∵.显然既不是等差也不是等比数列。

2019-2020学年明德实验学校高三英语期末试题及答案解析

2019-2020学年明德实验学校高三英语期末试题及答案解析第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AA Lifelong Devotion to Keeping People FedYuan Longping is a Chinese agricultural scientist and educator, known for developing the hybrid rice varieties.Yuan graduated from the Southwest Agriculture Institute in 1953 andbegan his teaching career at an agriculture school.In the 1960s, when a serious food shortage sweptChina, Yuan decided to devote himself to studying how to increase the yields of rice. He then began a lifelong connection with rice.Yuan succeeded in growing the world’s first high-yielding hybrid rice varieties in 1973, which could reach a yield of over 500 kg per mu (about 0.067 hectares), rising from the previous yield of only 300 kg per mu. For the next four decades, he continued to work on research and development of hybrid rice, achieving increasingly higher outputs. In 2020, hybrid rice developed by Yuan’s team achieved 1,500 kilograms per mu in two growing seasons, a new world record.InChina, where rice is the main food for the majority of the 1.4 billion people, the planting area of hybrid rice has reached 16 million hectares, or 57 % of the total planting area of rice, helping feed an extra 80 million people a year.Hybrid rice has also been grown in over 40 countries, including theU.S.,Brazil,India,Vietnam, thePhilippinesandMadagascar. The total planting area of the hybrid rice has reached 8 million hectares overseas.Even after a great success, Yuan never held himself back from making new breakthroughs. In 2017, his team started to grow seawater rice inQingdao. The rice was designed to grow in saline-alkaline land and survive even after being completely in seawater. His team planned to develop a type of seawater rice that could be planted in 6.67 million hectares of saline-alkaline land acrossChinato boost the country’s rice harvest by about 20 %. In 2018, Yuan’s team was invited to plant the saline-alkaline tolerant rice in experimental fields inDubai, which achieved huge success. In June 2020, his team started to grow seawater rice on a farm at an altitude of 2,800 meters in northwestChina’sQinghaiProvince. The experiment succeeded.Yuan had two dreams — to “enjoy the cool under the rice crops taller than men” and that hybrid rice could be grown all over the world to help solve the global food shortage.1. What made Yuan Longping decide to study rice?A. A serious food shortage.B. Agriculture development.C. His interest in the rice experiment.D. His wish to plant the tallest rice in the world.2. From the passage, we know that Yuan Longping ________.A. developed a variety of hybrid riceB. worked as a scientist after graduationC. started to grow seawater rice inDubaiin 2017D. grew the first high-yielding hybrid rice varieties in 19533. We can infer from the passage that Yuan Longping’s most outstanding qualities are________.A. modest and outgoingB. honest and creativeC. generous and optimisticD. responsible and devotedBWho is a genius? This question has greatly interested humankind for centuries.Let's state clearly: Einstein was a genius. His face is almost the international symbol for genius. But we want to go beyond one man and explore the nature of genius itself. Why is it that some people are so much more intelligent or creative than the rest of us? And who are they?In the sciences and arts, those praised as geniuses were most often white men, of European origin. Perhaps this is not a surprise. It's said that history is written by the victors, and those victors set the standards for admission to the genius club. When contributions were made by geniuses outside the club—women, or people of a different color1 or belief—they were unacknowledged and rejected by others.A study recently published bySciencefound that as young as age six, girls are less likely than boys to say that members of their gender(性别)are “really, really smart.” Even worse, the study found that girls act on that belief: Around age six they start to avoid activities said to be for children who are “really, really smart.” Can our planet afford to have any great thinkers become discouraged and give up? It doesn't take a genius to know the answer: ly not.Here's the good news. In a wired world with constant global communication, we're all positioned to see flashes of genius wherever they appear. And the more we look, the more we will see that social factors(因素)like gender, race, and class do not determine the appearance of genius. As a writer says, future geniuses come from those with “intelligence, creativity, perseverance(毅力), and simple good fortune, who are able to change the world.”4. Whatdoes the author think of victors' standards for joining the genius club?A. They're unfair.B. They're conservative.C. They're objective.D. They're strict.5. What can we infer about girls from the study inScience?A. They think themselves smart.B. They look up to great thinkers.C. They see gender differences earlier than boys.D. They are likely to be influenced by social beliefs6. Why are more geniuses known to the public?A. Improved global communication.B. Less discrimination against women.C.Acceptance of victors' concepts.D. Changes in people's social positions.7. What is the best title for the text?A. Geniuses Think AlikeB. Genius Takes Many FormsC. Genius and IntelligenceD. Genius and LuckCAge has never been a problem for 16-year-old Thessalonika Arzu-Embry. After all, she’s already got her master’s degree.The North Chicago-area teen started homeschooling at the age of 4. She began having an influence on others soon after. When she was 6 years old, she was an inspirational speaker at an organization called Tabitha House Community Service, which is for people who were forced to leave their homes because of earthquake, flood and other natural disasters.At the age of 11, she graduated from high school and then earned her bachelor’s degree in psychology in 2013. She completed those classes online as she was traveling for church events and leadership meetings.She doesn’t stop there, though. The teen plans to focus on aviation psychology (航空心理学) for her further study, a decision inspired by her father who is a pilot. She grew up around airplanes and took fights all the time. Her goal is to use it to help pilots deal with problems that could have deadly results once the plane takes off — a topic that has been in the news lately. For her, it’s a mix of two of her interests.In her free time, Thessalonika enjoys playing tennis, swimming and being active in her youth group at church.She also has three self-published books, which are on her site. Jump the Education Barrier is written to help students finish college, and in the future aims to help business owners with trends. Her third book The Genius Race is designed to help people to be talents in various areas of life.8. Which of the following is TRUE about Thessalonika?A. When she was 6 years old, she started homeschooling.B. She gained her master’s degree at the age of 11.C. She majored in science and technology.D. In 2013 she got her bachelor’s degree through completing courses online.9. What is her next plan according to the passage?A. Major in aviation psychology.B. Deliver inspiring speeches for church events and leadership meetings.C. Be active in her youth group at church.D. Write another book to help people to be talents.10. Why does she write the book Jump the Education Barrier?A. It aims to help people to be geniuses.B. It is intended to give students a hand to complete college.C. It is designed to arouse people’s awareness of psychology.D. The author hope to share her own experience with others.11. Which ofthe following can be the best title for the text?A. The Story of ThessalonikaB. To be a talentC. Three Published BooksD. HomeschoolingDWhen you walk on a sandy beach, it takes more energy than striding down a sidewalk — because the weight of your body pushes into the sand. Turns out, the same thing is true for vehicles driving on roads. The weight of the vehicles creates a very shallow indentation (凹陷) in the pavement (路面) — and it makes it such that it’s continuously driving up a very shallow hill.Jeremy Gregory, a sustainability scientist at M.I.T. and histeam modeled how much energy could be saved — and green-house gases avoided — by simply stiffening (硬化) the nation’s roads and highways. And they found that stiffening 10 percent of the nation’s roads every year could prevent 440 megatons of carbon dioxide equivalent emissions over the next five decades — enough to offset half a percent of projected transportationsector emissions over that time period. To put those emissions savings into context — that amount is equivalent to how much CO2 you’d spare the planet by keeping a billion barrels of oil in the ground — or by growing seven billion trees — for a decade.The results are in the Transportation Research Record.As for how to stiffen roads? Gregory says you could mix small amounts of synthetic fibers orcarbon nanotubes into paving materials. Or you could pave with cement-based concrete, which is stiffer than asphalt (沥青).This system could also be a way to shave carbon emissions without some of the usual hurdles. Usually, when it comes to reducing emissions in the transportation sector, you’re talking about changing policies related to vehicles and also driver behavior, which involves millions and millions of people — as opposed to changing the way we design and maintain our pavements. That’s just on the order of thousands of people who are working in transportation agencies. And when it comes to retrofitting (翻新) our streets and highways —those agencies are where the rubber meets the road.12. Why does the author mention “walk on a sandy beach” in paragraph 1?A. To present a fact.B. To make a contrast.C. To explain a rule.D. To share an experience.13. What suggestion does the author give to reduce CO2 emissions?A. Hardening the road.B. Keeping oil in the ground.C. Growing trees for decades.D. Improving the transportation.14. What is the advantage of this suggestion?A. Gaining more support.B. Consuming less money.C. Involving more people.D. Facing fewer usual obstacles.15. What does the underlined part mean in the last paragraph?A.Those agencies are likely to make more rules.B. Those agencies will change some related policies.C. Those agenciesmight put more rubber tires on the roads.D. Those agencies will play a key role in making this happen.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

广东省深圳市明德外语实验学校高三数学文模拟试题含解析

广东省深圳市明德外语实验学校高三数学文模拟试题含解析一、选择题：本大题共10小题，每小题5分，共50分。

在每小题给出的四个选项中，只有是一个符合题目要求的1. 若如下框图所给的程序运行结果为，那么判断框中应填入的关于的条件是A．B．C．D．参考答案：D略2. 已知函数为奇函数,且当时,,则( )A B 0 C 1 D 2参考答案：A3.集合,若时，，则运算可能是（）A.加法B.除法C.减法D.乘法参考答案：答案：D4. 如图，三棱柱ABC-A1 B1C1的侧面A1ABB1⊥BC，且A1C与底面成45°角，AB=BC=2，则该棱柱体积的最小值为 A． B． C．4 D．3参考答案：C5. 若集合，则（）A．B． C. D．参考答案：B6. 某四面体的三视图如图所示，该四面体四个面的面积中，最大的是A．8 B． C．10 D．参考答案：C本题考查了三视图的相关知识，难度中等.由三视图可知，该四面体可以描述为：面,,且,从而可以计算并比较得面的面积最大，为10，故应选C.7. 不等式|x﹣5|+|x+3|≥10的解集是（）A．[﹣5，7] B．[﹣4，6] C．（﹣∞，﹣5]∪[7，+∞）D．（﹣∞，﹣4]∪[6，+∞）参考答案：D考点：绝对值不等式的解法．专题：集合．分析：解法一：利用特值法我们可以用排除法解答本题，分别取x=0，x=﹣4根据满足条件的答案可能正确，不满足条件的答案一定错误，易得到答案．解法二：我们利用零点分段法，我们分类讨论三种情况下不等式的解，最后将三种情况下x的取值范围并起来，即可得到答案．解：法一：当x=0时，|x﹣5|+|x+3|=8≥10不成立可排除A，B当x=﹣4时，|x﹣5|+|x+3|=10≥10成立可排除C故选D法二：当x＜﹣3时不等式|x﹣5|+|x+3|≥10可化为：﹣（x﹣5）﹣（x+3）≥10解得：x≤﹣4当﹣3≤x≤5时不等式|x﹣5|+|x+3|≥10可化为：﹣（x﹣5）+（x+3）=8≥10恒不成立当x＞5时不等式|x﹣5|+|x+3|≥10可化为：（x﹣5）+（x+3）≥10解得：x≥6故不等式|x﹣5|+|x+3|≥10解集为：（﹣∞，﹣4]∪[6，+∞）故选D【点评】本题考查的知识点是绝对值不等式的解法，其中利用零点分段法进行分类讨论，将绝对值不等式转化为整式不等式是解答本题的关键．8. 已知点A（0，1），B（﹣2，3）C（﹣1，2），D（1，5），则向量在方向上的投影为（）A．B．﹣C．D．﹣参考答案：D【考点】平面向量数量积的运算．【专题】平面向量及应用．【分析】先求出，，根据投影的定义，在方向的投影为，所以根据两向量夹角的余弦公式表示出，然后根据向量的坐标求向量长度及数量积即可．【解答】解：∵；∴在方向上的投影为==．故选D．【点评】考查由点的坐标求向量的坐标，一个向量在另一个向量方向上的投影的定义，向量夹角的余弦的计算公式，数量积的坐标运算．9. 已知直线a和平面，那么a//的一个充分条件是A．存在一条直线b，a//b且bB．存在一条直线b，a b且bC．存在一个平面，a∥且//D．存在一个平面，//且//参考答案：10. 已知集合, 集合, 则A． B． C． D．参考答案：D二、填空题:本大题共7小题,每小题4分,共28分11. 若函数图象的两条相邻的对称轴之间的距离为，且该函数图象关于点成中心对称，，则▲.参考答案：12. 已知实数满足，则的取值范围是参考答案：13. 正数a ，b ，c 满足，则的取值范围是______.参考答案：【分析】构造空间向量，，利用得到结论.【详解】令z=，则，又，记，，则，又，∴，即.【点睛】本题考查了三维向量坐标的运算，考查了的应用，考查了分析问题、转化问题的能力，属于发散思维的综合性问题. 14. 圆的圆心到直线的距离是_____.参考答案：圆的标准方程为，圆心为，半径为1，圆心到直线的距离为，答案为1. 15. “”是“”的( )条件A ．充分不必要B ．必要不充分C ．充分必要D ．既不充分也不必要参考答案：B 略16. 在二项式的展开式中，含项的系数为（结果用数值表示）.参考答案：7017. 用数字1，2，3，4，5，6，7，8，9组成没有重复数字，且至多有一个数字是偶数的四位数，这样的四位数一共有___________个.（用数字作答）参考答案：1080三、解答题：本大题共5小题，共72分。

广东省深圳市明德外语实验学校高二数学理期末试题含解析

广东省深圳市明德外语实验学校高二数学理期末试题含解析一、选择题：本大题共10小题，每小题5分，共50分。

在每小题给出的四个选项中，只有是一个符合题目要求的1. 直线3x+4y﹣13=0与圆（x﹣2）2+（y﹣3）2=1的位置关系是（）A．相离B．相交C．相切D．无法判定参考答案：C【考点】直线与圆的位置关系．【专题】计算题．【分析】由圆的方程找出圆心坐标和圆的半径r，然后利用点到直线的距离公式求出圆心到已知直线的距离d，发现d=r，故直线与圆相切．【解答】解：由圆的方程得到：圆心坐标为（2，3），半径r=1，所以圆心到直线3x+4y﹣13=0的距离d==1=r，则直线与圆的位置关系为相切．故选C【点评】此题考查了直线与圆的位置关系，以及点到直线的距离公式．其中直线与圆的位置关系的判定方法为：当0≤d＜r时，直线与圆相交；当d=r时，直线与圆相切；当d＞r时，直线与圆相离．2. 若集合，，则=（）A． B．C． D．参考答案：C3. 如表是某厂节能降耗技术改造后，在生产甲产品过程中记录的产量x（吨）与相应的生产能耗y （吨）的几组对应数据：若根据如表提供的数据，用最小二乘法可求得对的回归直线方程是，则表中的值为（）A．4 B．4.5 C. 3 D．3.5参考答案：A4. 若，则A. 8B. 7C. 6D. 4参考答案：A【分析】根据排列数，组合数的公式，求得，即可求解，得到答案．【详解】由题意，根据排列数、组合数的公式，可得，即，解得，故选A．【点睛】本题主要考查了排列数，组合数的应用，其中解答中熟记排列数，组合数的计算公式，准确化简、运算是解答的关键，着重考查了运算与求解能力，属于基础题．5. 双曲线的虚轴长是实轴长的2倍，则双曲线的渐近线方程为A．B．C．D ．参考答案：A6. 已知{a n}为等差数列，，若{b n}为等比数列，，则{b n}的类似结论是（）A． B．C． D．参考答案：D略7. 下列各式中，最小值等于２的是A． B．Ｃ．Ｄ．参考答案：D8. 已知圆锥的高为8，底面圆的直径为12，则此圆锥的侧面积是( )A．24πB．30πC．48πD．60π参考答案：D【考点】旋转体（圆柱、圆锥、圆台）．【专题】计算题；转化思想；综合法；空间位置关系与距离．【分析】圆锥的侧面积是一个扇形，根据扇形公式计算即可．【解答】解：底面圆的直径为12，则半径为6，∵圆锥的高为8，根据勾股定理可知：圆锥的母线长为10．根据周长公式可知：圆锥的底面周长=12π，∴扇形面积=10×12π÷2=60π．故选：D．【点评】本题主要考查了圆锥的侧面积的计算方法．解题的关键是熟记圆锥的侧面展开扇形的面积计算方法．9. 下表是降耗技术改造后生产甲产品过程中记录的产量x（吨）与相应的生产能耗y（吨标准煤）的几组对应数据，根据表中提供的数据，求出y关于x的线性回归方程为=0.7x+0.35，那么表中m值为（）参考答案：D【考点】BK ：线性回归方程．【分析】根据表格中所给的数据，求出这组数据的横标和纵标的平均值，表示出这组数据的样本中心点，根据样本中心点在线性回归直线上，代入得到关于m 的方程，解方程即可．【解答】解：∵根据所给的表格可以求出==4.5， ==∵这组数据的样本中心点在线性回归直线上，∴=0.7×4.5+0.35，∴m=3，故选：D．10. 下列函数中，最小值为2的函数为A． B．C． D．参考答案：D二、填空题:本大题共7小题,每小题4分,共28分11. 若不等式在上的解集是空集，则的取值范围是．参考答案：略12. 若关于x的不等式x2﹣ax﹣a≤﹣3的解集不是空集，则实数a的取值范围是．参考答案：{a|a≤﹣6，或a≥2}【考点】二次函数的性质．【分析】不等式x2﹣ax﹣a≤﹣3的解集不是空集，即b2﹣4ac≥0即可，从而求出a的取值范围．【解答】解：∵不等式x2﹣ax﹣a≤﹣3，∴x2﹣ax﹣a+3≤0；∴a2﹣4（﹣a+3）≥0，即a2+4a﹣12≥0；解得a≤﹣6，或a≥2，此时原不等式的解集不是空集，∴a的取值范围是{a|a≤﹣6，或a≥2}；故答案为：{a|a≤﹣6，或a≥2}．13. 方程（为参数）的曲线的焦距为．参考答案：略14. 现有5位同学报名参加两个课外活动小组,每位同学限报其中的一个小组,则不同的报名方法共有________种.参考答案：32 .15. 某程序框图如图所示，则输出的??????????????????????? .参考答案：2616. P为椭圆上一点，F1、F2为左右焦点，若∠F1PF2=60°，则△F1PF2的面积为．参考答案：【考点】椭圆的简单性质．【分析】先利用椭圆定义求出|PF1|+|PF2|和|F1F2|的值，因为知道焦点三角形的顶角，利用余弦定理求出|PF1||PF2|的值，再代入三角形的面积公式即可．【解答】解：由椭圆方程可知，a=5，b=3，∴c=4∵P点在椭圆上，F1、F2为椭圆的左右焦点，∴|PF1|+|PF2|=2a=10，|F1F2|=2c=8在△PF1F2中，cos∠F1PF2=====cos60°=∴72﹣4|PF 1||PF 2|=2|PF 1||PF 2|，∴|PF 1||PF 2|=12 又∵在△F 1PF 2中， =|PF 1||PF 2|sin∠F 1PF 2∴=×12sin60°=3故答案为317. 已知则．参考答案：－1/9略三、解答题：本大题共5小题，共72分。

2019-2020学年明德实验学校高三英语上学期期末考试试卷及参考答案

2019-2020学年明德实验学校高三英语上学期期末考试试卷及参考答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AWhat are you waiting for? A new series of movies shown this year can’t be missed. Have you enjoyed them already?Frozen IIFrozen was the highest grossing (票房) animated film ever. In Frozen II Elsa, Anna, Olaf and Krist left off Arendelle to seek thesource of Elsa’s icy magic. Although the millions of children who loved the first film are older now, they might give it a reception.Last ChristmasA festive romantic comedy, Emilia Clarke stars in Last Christmas as Bridget Jonesy , a shop assistant, whose life in London is a mess, and Henry Golding as the eligible bachelor(黄金单身汉)who tidies it up. The film’s director, Paul Feig, and co-writer, Emma Thompson, promise that the film is worth expecting.A Beautiful Day in the NeighborhoodTom Hanks stars in A Beautiful Day in the Neighborhood as the only American celebrity(名人) more famous than he is. As the host of Mister Rogers’ Neighborhood for more than 30 years, Fred Rogers is a legend of pre-school children’s television, which appeals to a large audience.Charlie’s AngelsHollywoodaction movies starring women are rare. But have you seen a movie starred, written and directed by women, too? Charlie’s Angels is one of the first. A reboot (翻拍) of the 1970s TV series, not to mention the two films from 2000 and 2003, the new version is directed by Elizabeth Banks. She also plays Bosley, one of the female detectives who are employed by Charles Townsend to go on global adventures.1. Which moviebecame the most popular cartoon film this year?A. Frozen II.B. Last Christmas.C. A Beautiful Day in the Neighborhood.D. Charlie’s Angels.2. Which character works on a TV station?A. Krist.B. Emilia Clarke.C. Fred Rogers.D. Bosley.3. Which action movie was directed and starred by a woman?A. Frozen II.B. Last Christmas.C. A Beautiful Day in the Neighborhood.D. Charlie’s Angels.BWhen I was 13, I lost my sight. Since then, I had learned to get about with a walking stick, but had to stay at home because my parents thought I would get lost or robbed, even get hit by a car.I, however, believed I could regain my way if I lost it. A neighbor told me that a public library was offering a free course designed for the blind. That's an important opportunity for me to kill two birds with one stone: I could practice my getting — about skills on my way to learning practical technology. My parentssettled forit.But how would I plan my course? I knew that the blind singer Ray Charles, get around without a walking stick by counting steps. But I couldn't seem to do that the way he had. I developed the power of my imagination, catching the layout（布局）of places I visited and taking note of landmarks in my mind. Every time I visited a place, the mental map I'd drawn would turn up and helped me with the direction. But that doesn't mean I didn't lose my way in the process of acquiring this skill. I'd have to swallow（吞下）my pride to ask kind strangersfor help.On those days I lost my way, I'd go to bed feeling down. But my desires to beat blindness and further my education were usually enough to get me out of bed the next day and try again. Today, I'm a published reporter and audio producer.Yes, I've lost my way at times and found it again. And when people ask me,"Aren't you afraid to be out on your own?” the answer to me is clear：I'd rather risk and find happiness than stick to safety and be painful.Now, impressed by my progress, my father told my mother, "Our boy can see!".4. What does the underlined phrases “settled for" in the second paragraph mean?A. Talked about.B. Stuckto.C. Agreed to.D. Cared about.5. How did the author go around on his own after losing his sight?A.He created pictures of places in his mind.B. He drew a map on the paper to help him.C. He was always asking strangers for directions.D.He threw away the walking stick and counted steps.6. Which of the following can best describe the author?A. Determined and adventurous.B. Patient and intelligent.C. Warm-hearted and positive.D. Adventurous and outgoing.7. How did the author's parents feel about his progress?A.Concerned.B. Surprised.C. Confident.D. Proud.CYou’re in a crowd of people who are all asking for the same thing. How do you make your voice heard above the rest? Be different. Don’t shout. Lisa, 25, was waiting to board a plane flying fromLondontoAustriafor Christmaswhen the flight was cancelled.“There were about a hundred of us unable to leave,” she says. “Everyone else was shouting at the airport staff. Instead of joining in, I walked up to the man behind the ticket desk very quietly and said, ‘This must be so awful for you! I don’t know how you deal with these situations—it’s not even your fault. I could never handle it as well as you are.’ Without my even asking, he found me a seat on another airline with an upgrade to first class. He was happy to do a favor forsomeone who was appreciative instead of unfriendliness.”Flattery (恭维) is an essential element of the sweet-talk strategy. “It’s human psychology that stroking a person’s ego (自我) with a few well-directed praises makes them want to prove you right,” says apsychologist. “Tell someone they’re pretty and they’ll instantly fix their hair; praise their sense of humor and they’ll tell a joke.”You need help and there’s ly no reason that the person will want to lend a hand. Allison, 26. a lawyer, realized she’d made a huge mistake on a batch of documents. “The only way I could fix the problem was to get the help of a colleague who I knew didn’t like me,” she said.Allison then went to the woman’s office and explained her problem. “As I was saying to the boss the other day you’re the only person who would know how to handle a situation like this, what would you suggest I do?” “Feeling pumped up (鼓励), she set about helping me and we finished the job on time, and she was happy to help.” Allison said.8. Whatwould have happened at the airport according to paragraph 1?A. The departure hall was filled with noise.B. Someone screamed just lo be different.C. The passengers waited on board patiently.D. The airport stuff were rude to the passengers.9. Why did the man put Lisa on another airline?A. He admired Lisa’s beauty.B. He appreciated her attitude.C. He was ready to help others.D. He was blamed for the cancellation.10. What is the third paragraph mainly about?A. The potential benefits of ego.B. The strategy to start small talk.C. The great importance of flattery.D. The value of humor in daily life.11. What can we learn about Allison’s colleague?A. She was a popular lawyer.B. She was always ready to help others.C. She always got praise from Allison.D. She did a great favor for Allison eventually.DWho is a genius? This question has greatly interested humankind for centuries.Let's state clearly: Einstein was a genius. His face is almost the international symbol for genius. But we want to go beyond one man and explore the nature of genius itself. Why is it that some people are so much more intelligent or creative than the rest of us? And who are they?In the sciences and arts, those praised as geniuses were most often white men, of European origin. Perhaps this is not a surprise. It's said that history is written by the victors, and those victors set the standards for admission to the genius club. When contributions were made by geniuses outside the club—women, or people of a different color1 or belief—they were unacknowledged and rejected by others.A study recently published bySciencefound that as young as age six, girls are less likely than boys to say that members of their gender(性别)are “really, really smart.” Even worse, the study found that girls act on that belief: Around age six they start to avoid activities said to be for children who are “really, really smart.” Can our planet afford to have any great thinkers become discouraged and give up? It doesn't take a genius to know the answer: ly not.Here's the good news. In a wired world with constant global communication, we're all positioned to see flashes of genius wherever they appear. And the more we look, the more we will see that social factors(因素)like gender, race, and class do not determine the appearance of genius. As a writer says, future geniuses come from those with “intelligence, creativity, perseverance(毅力), and simple good fortune, who are able to change the world.”12. Whatdoes the author think of victors' standards for joining the genius club?A. They're unfair.B. They're conservative.C. They're objective.D. They're strict.13. What can we infer about girls from the study inScience?A. They think themselves smart.B. They look up to great thinkers.C. They see gender differences earlier than boys.D. They are likely to be influenced by social beliefs14. Why are more geniuses known to the public?A. Improved global communication.B. Less discrimination against women.C.Acceptance of victors' concepts.D. Changes in people's social positions.15. What is the best title for the text?A. Geniuses Think AlikeB. Genius Takes Many FormsC. Genius and IntelligenceD. Genius and Luck第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

2021年明德实验学校高三英语上学期期末考试试卷及答案解析

2021年明德实验学校高三英语上学期期末考试试卷及答案解析第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项AEver wonder why there are so many people polluting the earth? Ever say to yourself：Hey, I wish that I could do more to help the environment? Have you ever thought about trying to help the earth but never really did it? Well, here are some pretty easyand skillful ideas for that green - earth desire inside you.●Turn off your computer. By leaving it on all day you are creating more CO2than a regular passenger would, driving to and from work in one day.● Ride your bike or carpool (合用汽车).Obviously, youare creating less CO2which will help the ozone(臭氧).● Make a garden. Even simply grow some plants in your kitchen, which will help produce more oxygen while eating up some of that evil CO2.● Buy local groceries. It creates less impact on the environment. Besides, you're supporting your local farmers.● Recycle. You had to see this coming. But you have no idea how much you are helping the environment by simply reusing a water bottle instead of buying a huge pack at the store.● Don't run the water while brushing. It saves you money and helps the water resources.● Open the curtains. Natural light is much prettier and it will keep the energy usage down.● Rechargeable batteries. You have no idea how much it takes to get rid of batteries. Do yourself a favor. Save some money and some energy.1. If you don't want to create more CO2, you may_______.A. turn off your computer or open the curtainsB. turn off your computer or ride your bike or carpoolC. make a garden or open the curtainsD. use rechargeable batteries or make a garden2. Which of the following can best describe the function of the first paragraph?A. Main body.B. Argument.C. Lead - in.D. Conclusion.3. The main idea of the passage is about________.A. the importance of environmental protectionB. some ways about how to prevent pollutionC. some suggestions about how to save energyD. some suggestions about environmental protectionBSummer heat can be dangerous, and heat leads to tragedy far toooften. According to kidsandcars, org, an average of 37 young children per year die of car heat in the US, when they are accidentally left in a hot vehicle.For Bishop Curry, a fifth grader from Mckinney, Texas, one such incident hit close to home. A six-month-old baby from his neighborhood died after hours in a hot car. After hearing about her death, Curry decided that something needed to be done. Young Curry, who turned 11 this year, has always had a knack for inventing things, and he drew up a sketch (草图) of a device he called “Oasis.”The device would attach to carseats and watch the temperature inside the car. If it reached a certain temperature in the car, and the device sensed a child in the carseat, it would begin to circulate cool air. Curry alsodesigns the device using GPS and Wi-Fi technology, which would alarm the child’s parents and, if there was no response from them, the police.Curry’s father believes that the invention has potential. “The cool thing about Bishop’s thinking is none of this technology is new,” he said. “We feel like the way he’s thinking and combining all these technologies will get to production faster.” His father even introduced the device to Toyota, where he works as an engineer. The company was so impressed that they sent Curry and his father to a car safety conference in Michigan.In January, Curry’s father launched a campaign for the invention. They hope to raise money to finalize the patent, build models, and find a manufacturer. Their goal was $20,000, but so many people believed in Oasis’ potential that they have raised more than twice that — over $46,000.Curry’s father remembers the first time he saw his son’s sketch. “I was so proud of him for thinking of a solution,” he said. “We always just complain about things and rarely offer solutions.”4. What inspired Curry to invent Oasis?A. His narrow escape from death after being locked in a car.B. His knowledge of many children’s death because of car heat.C. The death of his neighbor’s baby after being left in a hot car.D. The injury of 37 children in his school in a car accident.5. What would Oasis do if it was hot in a car with a child?A. It would inform the parents or even the police.B. It would pump out the hot air in the car.C. It would sound the alarm attached to the car.D. It would get the window open to save the child.6. What does Curry’s father think is cool about Curry’s invention?A. It used some of the most advanced technology.B. It simply combined technologies that existed.C. It could accelerate production of new technology.D. It is the most advanced among similar products.7. Why did Curry’s father start a campaign to raise money?A. To conduct experiments to test the invention.B. To get other children devoted to inventions.C. To support a charity of medical aid for children.D. To get the patent and bring it to production.CNot long after the first fitness magazine was published, a list probably followed soon after, ranking the best fitness equipment. This tradition has continued, with the implied message: usethisand exercise willbe yours.And that's part of the problem, says Dr. Lieberman, a professor of Harvard University. There isn't one “best” anything to achieve fitness. Besides, people understand exercise is good for them. Knowledge about exercise still doesn't motivate.Before you can answer why, it helps to look at history. Before the Industrial Revolution, people fetched water and walked up stairs because they had to. But then technology made life and work easier. Exercise has become something that people have to carve out time for. “It's a fundamental instinct to avoid physical activity when it's neither necessary nor rewarding,” he says.It would seem like being healthy would qualify as necessary, but a doctor's prescription to exercise “can make it like taking cod liver oil,” Lieberman says. “Sometimes it works, but more often than not, it doesn't. And it's still coming across as an order, and “not having a heart attack in five years is not an immediate reward,” says Dr. Beth Frates, assistant professor at Harvard Medical School.People might not want to exercise because it's never been enjoyable. Most of us probably have memories of gym class, not being picked for a team, or being in a fitness center that's filled with in-shape people. The majoritydon't feel excited. They feel that exercise isn't for them, but it can be. Coaching people in an empowering and motivating way can work much better than ordering someone to exercise. It starts with an expanded definition of what counts as exercise, and an injection of what's rarely used to describe exercise, but is certainly allowed: namely, fun.8. What does the underlined “this” in paragraph 1 refer to?A. The magazine.B. The tradition.C. The equipment.D. The message.9. What can we infer about technology?A. It improves life quality.B. It saves people's time.C. It drives social progress.D. It makes exercise less likely.10. Why does the author mention “cod liver oil” in paragraph 4?A. To attach importance to health.B. To present a doctor's prescription.C. To explain exercise is considered inessential.D. To introduce the latest medical application.11. What is the main idea of the last paragraph?A. Exercise should be made more joyful.B. It's more fun to work out with others.C. We may encounter bad workout experiences.D. Orders work well to motivate people to exercise.DBabies who frequently communicate with their caregivers using eye contact and vocalisations(发声)at the age of one are more likely to develop greater languages skills by the time they reachtwo,according to new research.In the study, researchers looked at 11-and 12-month-od babies' vocalisations. gestures and gaze behaviours ,and at how their caregivers responded to them.To measure he interactions ,the researchers videoed infants(婴儿)and caregiver at home,and asked them to play as usual.They took those recordings back to the universityThe scientists then used statistical models to find that the best predictor of vocabulary at 24 months was when infants were seen to use vocalsatioms while looking at their caregiver's face when they were about a yearold.The benefits were even greater when these interactions were followed by responses from the caregiver.The statistics showed that at 19 months,children had an average of about 100 words.Those who exhibited the beneficial interactive behaviour earlier in life were seen to have an average of about 30 extra words."The message of this paper is thatitis the result of a joint effort; noticing what your child is attending to and talking to them about it will support their language development." said McGillion, a co-author of the work."The joy of this message is that that can happen in any context... across any part of your day.It's not something that requires special equipment or even lots of time.I can happen when you're doing the laundry,for example—when you're taking out the socks, you can talk about socks...in the park, in the car, at mealtimes,at bathtimes.This finding can be used in any context,"added McGillion."This is a developmental snapshot in the first year of life, but children are constantly growing and changing and so are their behaviours. It would be interesting to look at these sorts of behaviours again as children progress through the second year of life to see what's happening there,"said Donnellan,the lead author on the study.12. How did the researchers get the findings?A. By interacting with babies.B. By asking babies to vocalize.C. By analyzing relevant recordings.D. By referring to the previous statistics.13. What does he underlined word "it"in Paragraph 5 mean?A. Infants' eye contact.B. Infants' larger vocabulary.C. The response from caregivers.D. The best predictor of vocabulary.14. What did McGilion say about infants' interactive behaviour?A. I's easy to perform.B. It's complex to understand.C. It's difficult to copy.D. It's interesting to video.15. What might further studies be on?A. Children's academic progress.B. Children's growing environment.C. Children's potential physical development.D. Children's behaviours across more age ranges.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

2020-2021学年明德实验学校高三英语上学期期末试题及答案

2020-2021学年明德实验学校高三英语上学期期末试题及答案第一部分阅读（共两节，满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项ALocated in the beautiful Sichuan Basin, Chongqing is a magical 8D city. The natural history and cultural scenery of the area provide children with learning opportunities because they can enjoy the many wonders of this area.Fengjie Tiankeng Ground JointTiankeng Diqiao Scenic Area is located in the southern mountainous area of Fengjie County. The Tiankeng pit is 666 meters deep and is currently the deepest tiankeng in the world. The scenic spot is divided into ten areas including Xiaozhai Tiankeng, Tianjingxia Ground, Labyrinth River, and Longqiao River. There are many and weird karst cave shafts, and countless legends haunt them.Youyang Peach GardenYouyang Taohuayuan Scenic Area is a national forest park, a national 5A-level scenic spot, and a national outdoor sports training base. Located in the hinterland of Wuling Mountain. The Fuxi Cave in the scenic spot is about 3,000 meters long, with winding corridors, deep underground rivers, and color1 ful stalactites. The landscape is beautiful.Jinyun Mountain National Nature ReserveJinyun Mountain is located in Beibei District of Chongqing City, about 45 kilometers away from the Central District of Chongqing City. The nine peaks of Jinyun Mountain stand upright and rise from the ground. The ancient trees on the mountain are towering, the green bamboos form the forest, the environment is quiet, and the scenery is beautiful, so it is called "Little Emei". Among them, Yujian Peak is the highest, 1050 meters above sea level; Lion Peak is the most precipitous and spectacular, and the other peaks are also unique.Chongqing People's SquareChongqing's Great Hall of the People, one of the landmarks of Chongqing, gives people the deepest impression than its magnificent appearance resembling the Temple of Heaven. It also uses the traditional method of central axis symmetry, with colonnade-style double wings and a tower ending, plus a large green glazed roof, large red pillars, white railings, double-eave bucket arches, and painted carved beams.1.How deep is the Tiankeng Ground Joint?A.666mB.3,000mC.45kmD.1050m2.Which of the following rocks can you see in Youyang Peach Garden?A.LimestoneB.StalactiteC.MarbleD.Quartzite3.Which attraction is closest to downtown Chongqing?A.Fengjie Tiankeng Ground JointB.Jinyun Mountain National Nature ReserveC.Chongqing People's SquareD.Youyang Peach GardenBHave you ever done something that was really dangerous just because you thought it was safe？Maybe you did a dangerous trick on your bicycle or skateboard because you were wearing a helmet and thought you couldn’t get hurt. The psychology(心理) of this sort of behavior is called the Peltzman Effect, named after Sam Peltzman, professor of economics at the University of Chicago. Peltzman believes that those moments when people think they are the safest are the times when they act most dangerously.Peltzman said that people drove more dangerously when they wore seat belts(安全带) . Driving a large four-wheel drive vehicle has a similar effect on drivers’ behavior. Because drivers of large vehicles sit up higher and can see better, they feel they can make better judgments when they drive. They are better protected in accidents,so they act more dangerously. This makes driving morehazardousto other drivers.The Peltzman Effect isn’t just limited to driving. In 1972, the American Food and Drug Administration (FDA) passed a law requiring child safety caps on most medicine bottles. The safety caps were designed to prevent children from accidentally taking the medicine, especially painkillers such as aspirin. Requiring safety caps sounded like a great idea, but there was an unexpected side effect. Because the safety caps are so hardto take off, some people leave them off altogether．Worse, some parents leave the bottles where kids can reach them because they feel that it is safe because of the cap. A study on the Peltzman Effect showed that more than 3,500 children have been harmedby aspirin because of the safety caps.The Peltzman Effect describes how we’re likely to take more risks and act more dangerously when we feel safest. What’s more, the effects of these behaviors can be quite different from what we expect.4. What is the Peltzman Effect？A. People behave less safely when they feel safe．B. People feel safest when they are under protection．C. Something that seems dangerous turns out to be safe．D. People who act dangerously are likely to be together．5. What does the underlined word“hazardous”in Paragraph 2 mean？A. Interesting．B. Expensive．C. Dangerous．D. Important．6. Medicine bottles with safety caps ________．A. are required throughout the worldB. meet the demands of the Peltzman EffectC. sell well in the worldD. are not completely safe7. What would be the best title for the text？A. Unsafe Safety MeasuresB. Types of Decision MakingC. People’s Fear of Taking RisksD. Different Behaviors of People in DangerCAn anti-obesity program for Australian girls didn’t lead to any improvements in their diet, physical activities or body weight a year later, according to a new report.Findings from the school-based intervention (介入), which involved exercise sessions and nutrition workshops for lower-income girls, are the latest disappointment in a lot of research attempting tohead offadult obesity and the disease risks that come with it.Especially during the middle-and high-school years, girls’ physical activity reduces obviously, according to lead researcher David Lubans, from theUniversityofNewcastleinNew South Wales,Australia. He said, “In the future we need to make the programs more interesting and exciting and present information in a way that is meaningful to adolescent girl.”Lubans and his workmates conducted their study in 12 schools in low-income areas ofNew South Wales. At the start of the study, girls in both groups weighed an average of close to 130pounds, with about four in ten considered overweight. Over the next year, adolescents in the intervention group were given pedometers (计步器) to encourage walking and running and invited to nutrition workshops and regular exercise sessions during theschoolday and at lunchtime. Participation in some of those activities were less than ideal. For example, the girls went to only one-quarter of lunchtime exercise sessions, and less than one in ten completed at-home physical activity or nutrition challenges, the researchers reported. At the end of the year, girls in both groups had gained a similar amount of weight and there was no difference in their average body fat.Preventive medicine researcher Robert Klesges said that although some anti-obesity programs have helped adults lose weight, the teen population has always been a source of failure for researchers. “The common belief is: nothing works,” he said. “And we have got to get beyond that.”“We need to think outside the box,” said Klesges, who wasn’t involved in the new study. “That could include learning from what has worked in adult studies, such as giving meal replacement drinks or prepared foods to teens who have trouble making changes to their diet. Or, it could mean using a “step-care” method — rather than researchers or their doctor telling them to keep doing the same thing.” Klesges said.8. The underlined words “head off” in Paragraph 2 can best be replaced by “________”.A. damageB. defendC. preventD. affect9. The methods used in the program to stop obesity don’t include ________.A. walking and runningB. inviting them to nutrition workshopsC. joining exercise sessions regularlyD. giving meal replacement drinks10. The main reason for the failure of the anti-obesity program is probably that ________.A. the participants didn’t take an active part in itB. the program was not interesting and exciting to participantsC. the participants didn’t get extra nutrition or exercise helpD. the program didn’t pay attention to healthy exercise11. What can be inferred from the last paragraph?A. As researchers, it is important to have creative research methods.B. Researchers need to give meals or prepare foods to participants.C. Teen girls have no difficulty in making changes to their diet.D. Some ant-obesity programs have not helped adults lose weight.DIn Australia, plenty of wild things can bite or sting(刺伤) you. Strangely enough, one of them is a tree. Nowscientists have figured out what makes the tree’s sting so bad.The rainforests of eastern Australia are home to a stinging tree known as Dendrocnide. Many people callit the gympie-gympie tree—a name given to the tree by native Australians. It’s covered with sharp, needle-like hairs that carry poison. If you touch a gympie-gympie tree, you won’t forget it anytime soon. The pain can stay with you for hours, days or weeks. In some cases, it’s been reported to stay for months.Scientists have long looked for the source of this powerful sting. Now researchers at the University of Queensland have discovered what makes this stinging plant so painful. After carefully studying different kinds of gympie-gympie trees, the scientists were able to separate out different chemicals that the trees produce. This allowed them to identify a group of chemicals that they believed was responsible for the pain.The researchers created artificial versions of these chemicals, which they call “gympietides”. Sure enough, when the scientists injected mice with gympietides, the mice licked(舔) at the places where they’d been injected, indicating that they hurt in those places. When the scientists studied the way gympietides were built, they found that they formed a knot-like shape. The shape makes the chemicals very stable, which helps explain why the pain stays so long.The knot-like shape of the gympietides was similar to the shape of poisons produced by poisonous spiders and cone snails. The scientists were surprised to see three very different kinds of life all using similar poisons. Spiders and cone snails carry poisons because they catch food by stinging other creatures. It’s not clear how stinging helps the gympie-gympie tree.Though the tree’s sting may stop some animals from eating it, it doesn’t stop all animals. Beetles and pademelons (small s of the kangaroo) are able to eat the plant without trouble.12. Why is a touch on the stinging tree unforgettable?A. Because it has so unusual an appearance.B. Because it is extremely rare in existence.C. Because touching it creates a quite strange feeling.D. Because the pain caused by it doesn’t go away quickly.13. What do scientists fail to find out about the stinging tree?A. How it produces poisons.B. What poisons it produces.C. How it benefits from the sting.D. The consequences of its sting.14. What does the text imply about the stinging tree?A. It produces the same poisons as spiders.B. Poisonous as it is, it also has natural enemies.C. Animals are wise enough to stay away from it.D. Only one chemical in it causes pain to thetoucher.15. What’s the best title for the text?A. Scientists Discover Stinging Tree's SecretB. Caution: Stinging Tree Can Bite and Poison YouC. Scientists Discover a Strange Species in AustraliaD. Effective Ways to Avoid Being Hurt by Stinging Tree第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

2020-2021深圳明德外语实验学校高中必修三数学上期中试卷带答案

【详解】
由题意可知，首次抽到003号，以后每隔12个号抽到一个人，则抽到的号构成以3为首项，12为公差的等差数列，记为，其中，公差，则第个号 .
令，即，所以第一营区抽人；
令，即，所以第二营区抽人；
三个营区共抽人，所以第三营区抽人.
故选： .
【点睛】
本题考查系统抽样，属于基础题.
①平均来说一队比二队防守技术好；②二队比一队防守技术水平更稳定；③一队防守有时表现很差，有时表现又非常好；④二队很少不失球.
A．1个B．2个C．3个D．4个
3．已知变量之间满足线性相关关系，且之间的相关数据如下表所示：
x
1
2
3
4
y
0.1
m
3.1
4
则实数（）
A．0.8B．0.6C．1.6D．1.8
，得到频率分布直方图如图所示，若不低于140分的人数为110.① ；② ；③100分以下的人数为60；④分数在区间的人数占大半．则说法正确的是（）
A．①②B．①③C．②③D．②④
6．已知某地区中小学生人数和近视情况分别如图1和如图2所示，为了了解该地区中小学生的近视形成原因，用分层抽样的方法抽取的学生进行调查，则样本容量和抽取的高中生近视人数分别为（）
所以说法正确的是①③.
故选B.
【点睛】
本题主要考查了频率分布直方图的应用，其中解答熟记频率分布直方图的性质，以及在频率分布直方图中，各小长方形的面积表示相应各组的频率，所有小长方形的面积的和等于1，着重考查了分析问题和解答问题的能力，属于基础题.
6．B
解析：B
【解析】
【分析】
【详解】
试题分析：由题意知，样本容量为，其中高中生人数为，

广东省深圳外国语学校2024-2025学年高三上学期第二次月考数学试题

广东省深圳外国语学校2024-2025学年高三上学期第二次月考数学试题一、单选题1．已知集合{A x y ==，{}21x B y y ==+，则A B =I （）A ．(]1,2B ．(]0,1C ．[]1,2D ．[]0,22．已知命题:1,1p x x ∀>>，则命题p 的否定为（） A ．1,1x x ∃>≤ B ．1,1x x ∃≤≤ C ．1,1x x ∀><D ．1,1x x ∀≤>3．设函数()()3x x af x -=在区间30,2⎛⎫ ⎪⎝⎭上单调递减，则实数a 的取值范围是（）A ．(),1∞--B ．)[3,0-C ．(]0,1D ．[)3,+∞4．函数()1cos ex x x f x -=的图象大致为（）A ．B ．C ．D ．5．设正实数a 、b 、c 满足2240a ab b c -+-=，则当cab 取得最小值时，236a b c+-的最大值为（） A ．1B ．2C ．3D ．46．已知函数()f x 的定义域为(),e x y f x =+R 是偶函数，()3e xy f x =-是奇函数，则()ln3f 的值为（） A ．73B ．3C ．103D ．1137．根据公式3sin33sin 4sin ααα=-，sin10︒的值所在的区间是（）A ．11,76⎛⎫ ⎪⎝⎭B ．11,65⎛⎫ ⎪⎝⎭C ．11,54⎛⎫ ⎪⎝⎭D ．11,43⎛⎫ ⎪⎝⎭8．已知函数()()()22241,f x mx m x g x mx =--+=，若对于任意的实数(),x f x 与()g x 至少有一个为正数，则实数m 的取值范围是（） A ．()0,2B ．()0,8C ．[)2,8D ．(),0-∞二、多选题9．下列说法正确的是（）A ．若函数()f x 定义域为[]1,3，则函数()21f x +的定义域为[]0,1B ．若定义域为R 的函数()f x 值域为[]1,5，则函数()21f x +的值域为[]0,2C ．函数15xy ⎛⎫= ⎪⎝⎭与5log y x =-的图象关于直线y x =对称D ．a b >成立的一个必要条件是1a b ->10．若log 1a b >，则下列不等式一定成立的是（）A ．a b <B ．1ab a b +>+C ．11a b a b->- D ．11a b a b+<+11．已知定义在R 上的偶函数()f x 和奇函数()g x 满足()()21f x g x ++-=，则（）A ．()f x 的图象关于点()2,1对称B ．()f x 是以8为周期的周期函数C ．()()8g x g x +=D ．20241(42)2025k f k =-=∑三、填空题12．已知函数()cos 2f x x =，则0ππ()()66lim x f x f x∆→+∆-=∆. 13．已知函数223,2()(06log ,2a x x x f x a x x ⎧-++≤=>⎨+>⎩且1)a ≠，若函数()f x 的值域是(],4∞-，则实数a 的取值范围是14．若()e 1xa xb ≥++对一切x ∈R 恒成立，则()1a b +的最大值为.四、解答题15．设函数()32.f x x ax bx c =+++（I ）求曲线().y f x =在点()()0,0f 处的切线方程；（II ）设4a b ==，若函数()f x 有三个不同零点，求c 的取值范围 16．记ABC V 的角,,A B C 的对边分别为,,a b c ，已知sin sin sin A B Cb c a b-=++.(1)求A ；(2)若点D 是BC 边上一点，且,2AB AD CD BD ⊥=，求sin ADB ∠的值. 17．如图，四棱锥P ABCD -中，底面ABCD 是边长为2的菱形，π3ABC ∠=，已知E 为棱AD 的中点，P 在底面的投影H 为线段EC 的中点，M 是棱PC 上一点.(1)若2CM MP =，求证：//PE 平面MBD ；(2)若,PB EM PC EC ⊥=，确定点M 的位置，并求二面角B EM C --的余弦值. 18．已知函数()()()2ln 1cos 2g x x x =--+--.(1)函数()f x 与()g x 的图像关于1x =-对称，求()f x 的解析式； (2)()1f x ax -≤在定义域内恒成立，求a 的值； (3)求证：2111ln 42nk n f k =+⎛⎫-< ⎪⎝⎭∑，*N n ∈. 19．设自然数3n ≥，若由n 个不同的正整数1a ，2a ，…，n a 构成的集合{}12,,,n S a a a =L 满足：对集合S 的任何两个不同的非空子集A 、B ，A 中所有元素之和与B 中所有元素之和均不相等，则称集合S 具有性质P ．(1)试分别判断在集合{}11,2,3,4S =与{}21,2,4,8S =是否具有性质P ，不必说明理由；(2)已知集合{}12,,,n S a a a =L 具有性质P ．①记121k i k i a a a a ==+++∑L ，求证：对于任意正整数k n ≤，都有121kki i a =≥-∑；②令12i i i d a -=-，1kk i i D d ==∑，求证：0k D ≥；(3)在（2）的条件下，求12111na a a +++L 的最大值．。

2022年广东省深圳市明德实验学校高三数学理期末试卷含解析

2022年广东省深圳市明德实验学校高三数学理期末试卷含解析一、选择题：本大题共10小题，每小题5分，共50分。

在每小题给出的四个选项中，只有是一个符合题目要求的1. 设双曲线的左、右焦点分别为F1，F2.以F1为圆心，| F1F2|为半径的圆与双曲线在第一、二象限内依次交于AB两点.若|F1B|=3|F2A|,则该双曲线的离心率是（）A. B. C. D. 2参考答案：C如图所示，根据已知可得，，又，所以，即，又因为，所以,所以。

2. 若点O和点F分别为椭圆的中心和左焦点，点P为椭圆上的任意一点，则的最大值为（）A. 2B. 3C. 6D. 8参考答案：C3. 已知为坐标原点，直线与圆分别交于两点．若，则实数的值为（）A．1 B．C．D．参考答案：D略4. 已知定义在R上的奇函数，满足，且在区间[0，2]上是增函数，则(A) (B)(C) (D)参考答案：D略5. 函数的图象向右平移个单位后关于原点对称，则函数在上的最大值为（）A. B. C. D.参考答案：B【分析】由条件根据函数的图象变换规律，正弦函数的图象的对称性可得，，由此根据求得的值，得到函数解析式即可求最值．【详解】函数的图象向右平移个单位后，得到函数的图象，再根据所得图象关于原点对称，可得，，∵，∴，，由题意，得，∴，∴函数在区间的最大值为，故选B．【点睛】本题主要考查函数的图象变换规律，正弦函数的图象的对称性，考查了正弦函数最值的求法，解题的关键是熟练掌握正弦函数的性质，能根据正弦函数的性质求最值，属于基础题．6. 若等比数列{a n}的前n项和为S n， =（）A．3 B．7 C．10 D．15参考答案：D【考点】8G：等比数列的性质．【分析】根据等比数列的性质可知：可设其中公比为q，根据=3求出q4，再代入进行求解．【解答】解：∵据=3，（q≠1），若q=1可得据=2≠3，故q≠1，∴==3，化简得1﹣q8=3（1﹣q4），可得q8﹣3q4+2=0，解得q4=1或2，q≠1，解得q4=2，===15．故选：D．7. 命题“对任意的”的否定是（）A．不存在 B．存在C．存在 D．对任意的参考答案：C8. 具有线性相关关系的变量x，y ，满足一组数据如右表所示.若与的回归直线方程为，则m的值是A. 4B.C. 5D. 6参考答案：A9. 设、是两个不同的平面，是一条直线，以下命题：①若，，则；②若，，则；③若，，则；④若，，则；其中正确命题的个数是（）A. B. C. D.参考答案：B略10. （5分）已知{a n}为各项都是正数的等比数列，若a4?a8=4，则a5?a6?a7=（）A． 4 B． 8 C． 16 D． 64参考答案：B【考点】：等比数列的通项公式．【专题】：等差数列与等比数列．【分析】：由等比数列的性质可得a 6=2，而a 5?a 6?a 7=a 63，代值计算可得．解：∵{a n }为各项都是正数的等比数列且a 4?a 8=4， ∴由等比数列的性质可得a 62=a 4?a 8=4，∴a 6=2，再由等比数列的性质可得a 5?a 6?a 7=a 63=8，故选：B ．【点评】：本题考查等比数列的性质，属基础题．二、填空题:本大题共7小题,每小题4分,共28分11. 函数y=的f （x+1）单调递减区间是．参考答案：（﹣∞，0]考点：复合函数的单调性．专题：函数的性质及应用．分析：根据复合函数单调性之间的关系即可得到结论．解答：解：函数y==，则函数y==，的单调递减区间为（﹣∞，1]，即函数f （x ）的单调递减区间为（﹣∞，1]，将函数f （x ）向左平移1个单位得到f （x+1]，此时函数f （x+1）单调递减区间为（﹣∞，0]，故答案为：（﹣∞，0]点评：本题主要考查复合函数单调性的判断，根据复合函数之间的关系是解决本题的关键．12. （几何证明选讲选做题）如图4，是圆外一点，过引圆的两条割线、，，，则____________．参考答案：213. 已知单位向量与向量的夹角为，则________.参考答案： 114. 给出以下四个命题：①已知命题；命题则命题是真命题；②过点且在轴和轴上的截距相等的直线方程是； ③函数在定义域内有且只有一个零点；④若直线和直线垂直，则角其中正确命题的序号为．(把你认为正确的命题序号都填上) 参考答案：①③15. 已知，且，则sinα=．参考答案：解答：解：∵α∈（0，），β∈（﹣，0），∴α﹣β∈（0，π），又cos （α﹣β）=，sinβ=﹣，∴sin （α﹣β）==，cosβ==，则sinα=sin[（α﹣β）+β]=sin （α﹣β）cosβ+cos （α﹣β）sinβ=×+×（﹣）=．故答案为：16. 设，若是的充分不必要条件，则实数的取值范围为 .参考答案：试题分析：，解，解得；由，得，得，由于是的充分不必要条件，，解得，又由于，，故答案为考点：1、绝对值不等式的解法；2、充分条件必要条件的应用17. 观察下列等式：13+23=32，13+23+33=62，13+23+33+43=102，…，根据上述规律，第五个等式为 ▲ ．参考答案：13 + 23 + 33 + 43 +53 + 63 = 212 略三、解答题：本大题共5小题，共72分。

江苏省明德实验学校2022年高三英语第一学期期末学业质量监测模拟试题含解析

2022-2023高三上英语期末模拟试卷考生须知：1．全卷分选择题和非选择题两部分，全部在答题纸上作答。

选择题必须用2B铅笔填涂；非选择题的答案必须用黑色字迹的钢笔或答字笔写在“答题纸”相应位置上。

2．请用黑色字迹的钢笔或答字笔在“答题纸”上先填写姓名和准考证号。

3．保持卡面清洁，不要折叠，不要弄破、弄皱，在草稿纸、试题卷上答题无效。

第一部分（共20小题，每小题1.5分，满分30分）1．After having some alcohol, drivers tend to speed when passing streets with less people.A．pull up B．put up C．rise up D．pick up2．We live in this society now ________ literally someone is always helping.A．when B．whereC．that D．what3．He used to study in a village school, ______was rebuilt two years ago.A．which B．thatC．where D．whose4．Some drunken drivers think that they may be lucky to________a fine, which may cost their own lives.A．get through B．get along withC．get down to D．get away with5．—Our team has just narrowly won the game. I feel so relieved now.—__________!A．Cheer up B．You betC．What a pity D．Well done6．The goal, which they are unlikely to live to see , is to “cure, prevent or manage all diseases” in the next 80 or so years.A．accomplish B．accomplishedC．accomplishing D．being accomplished7．—Thank God I passed the interview yesterday. I was sweating heavily.—Me too. I ________ when I was sitting outside waiting.A．looked down my nose B．let my hair downC．had butterflies in my stomach D．chanced my arm8．Smoking ______ in the kitchen in my house.A．doesn’t allow B．didn’t allowC．isn’t allowed D．won’t allow9．I hope that we will be able to make it through the tough times and back to thebusiness of working together ________ our common goals.A．on behalf of B．in honor of C．on top of D．in search of10．The president stood by a window inside the room, ________ I entered, looking over the square.A．where B．into whichC．which D．that11．Mr. Wilson is a man of patience and kindness, and his good temper never ______ him.A．fails B．disappointsC．controls D．worries12．You should set a goal and see ________ you can achieve it in the coming exam. A．which B．whatC．whether D．when13．I don’t mind her criticizing me,but___is how she does it that I object to.A．it B．thatC．this D．which14．I’m not quite sure how to get there, ---------- I’d better _____ a map.A．watch B．look up C．consult D．read15．Children exposed to air pollution are more to suffering from different diseases.A．possible B．probable C．likely D．certainly16．Taught in a more professional way, you _________ the swimming skill much better. A．might master B．would have masteredC．can have mastered D．could master17．—Do you really plan to drop out of the football team?—________ It’s time for me to concentrate on my study.A．I’m just kidding. B．Definitely not.C．I mean it D．What a pity!18．The company needs to _____ its outdated image to promote its newly-released product.A．lay out B．call upC．shake off D．give away19．Life teaches us not to regret over yesterday, for it ________ and is beyond our control.A．passed B．will passC．has passed D．had passed20．One of our rules is that every student _______ wear school uniform while at school. A．might B．couldC．shall D．will第二部分阅读理解（满分40分）阅读下列短文，从每题所给的A、B、C、D四个选项中，选出最佳选项。

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1. 定义集合运算：A ※B ＝｛t|t ＝xy ，x ∈A ，y ∈B}，设A ＝｛1，2｝，B ＝｛0，2｝，则集合A ※B 的所有元素之和为（）A. 6B. 3C. 2D. 02. 点P (4，－2)与圆x 2＋y 2＝4上任一点连线的中点轨迹方程是( )A ． (x －2)2＋(y ＋1)2＝4B ．(x －2)2＋(y ＋1)2＝1C ． (x ＋4)2＋(y －2)2＝4D ．(x ＋2)2＋(y －1)2＝1 3. 函数x x g x x f -=+=122)(log 1)(与在同一直角坐标系下的图象大致是（）A B C D4. 已知等差数列{}n a 中，10795=-+a a a ，记n n a a a S +++= 21，则13S 的值为（） A. 130 B. 260C. 156D. 1685. 下列结论错误．．的是（） A ．命题：“若20232==+-x x x ，则”的逆否命题为:“若2≠x ，则0232≠+-x x ”.B. 命题:“存在x 为实数，02>-x x ”的否定是“任意x 是实数，02≤-x x ”.C. “22bc ac >”是“b a >”的充分不必要条件.D. 若p 且q 为假命题，则p 、q 均为假命题.6. 如右图，在一个长为π，宽为2的矩形OABC 内，曲线()sin 0y x x π=≤≤与x 轴围成如图所示的阴影部分，向矩形OABC 内随机投一点（该点落在矩形OABC 内任何一点是等可能的），则所投的点落在阴影部分的概率是（） A ．4πB .3πC.2πD.1π7. 方程(1＋4k )x －(2－3k )y ＋2－14k ＝0所确定的直线必经过点( ) A ．(2,2) B ．(－2,2) C ．(－6,2) D ．(3，－6)8. 用数学归纳法证明：12＋22＋…＋n 2＋…＋22＋12＝n n 2＋3，第二步证明由“k 到k ＋1”时，左边应加( )A ．k 2B ．(k ＋1)2C ．k 2＋(k ＋1)2＋k 2D ．(k ＋1)2＋k 2，二、填空题：本大题共7小题，考生作答6小题，每小题5分，满分30分. （一）必做题（9～13题） 9. 已知03πθ<<，且3cos()35πθ-=，则sin θ= ． 10. 已知A 、B 、C 是圆O ：221x y +=上三点且OA OB OC +=，则AB OA ⋅= ． 11．曲线ln(21)y x =-上的点到直线230x y -+=的最短距离是． 12. 若不等式413x x a a++-≥+对任意的实数x 恒成立，则实数a 的取值范围是13.已知)(x f 是偶函数，在),0[+∞上是增函数，若(1)(2)f ax f x +≤+（1||≥a ）在]1,21[∈x 上恒成立，则实数a 的取值范围为．（二）选做题（14～15题，考生只能从中选做一题） 14、(坐标系与参数方程选做题)在平面直角坐标系xOy 中，已知直线l 的参数方程为21,()42x t t R y t =-⎧∈⎨=-⎩参数，以直角坐标原点为极点，x 轴的正半轴为极轴建立相应的极坐标系。

在此极坐标系中，若圆C 的极坐标方程为2cos ρθ=，则圆心C 到直线l 的距离为。

15、(几何证明选讲选做题)如图,已知PA 是⊙O 的切线，A 是切点，直线PO 交⊙O 于B 、C 两点，D 是OC 的中点，连结AD 并延长交⊙O 于点E ，若PA=APB=30︒，则AE=三、解答题：本大题共6小题，满分80分，解答须写出文字说明、证明过程和演算步骤。

16． (12分)函数2()lg(2)f x x x =--的定义域为集合A ，函数()g x =的定义域为集合B . （1）求A B 和AB ；（2）若{}AC p x x C ⊆<+=,04|，求实数p 的取值范围.17. (12分) 已知函数2()(sin cos )+cos2f x x x x =+.（1）求函数()f x 的最小正周期；（2）当0,2x π⎡⎤∈⎢⎥⎣⎦时，求函数()f x 的最大值，并写出x 相应的取值.18. （本小题满分14分）如图11－4－1，在平面直角坐标系xOy 中，已知圆C 1：(x ＋3)2＋(y－1)2＝4和圆C 2：(x －4)2＋(y －5)2＝4.(1)若直线l 过点A (4,0)，且被圆C 1截得的弦长为2 3，求直线l 的方程；(2)设P 为平面上的点，满足：存在过点P 的无穷多对互相垂直的直线l 1和l 2，它们分别与圆C 1和圆C 2相交，且直线l 1被圆C 1截得的弦长与直线l 2被圆C 2截得的弦长相等，试求所有满足条件的点P 的坐标．19. 已知)(x f 是二次函数，)(x f '是它的导函数，且对任意的R ∈x ，2)1()(x x f x f ++='恒成立．（1）求)(x f 的解析表达式；（2）设0>t ，曲线C ：)(x f y =在点))(,(t f t P 处的切线为l ，l 与坐标轴围成的三角形面积为)(t S ．求)(t S 的最小值．20. （本小题满分14分）已知数列{n a }满足1(,2),(1,)n n x a y a x y +=-=⊥且，23+a 是2a 与4a 的等差中项. （1）求数列{n a }的通项公式n a ；（2）若满足12132log n nb a =+，12n n S b b b =++⋅⋅⋅+，求n S 的最大值.21．（本小题满分14分）已知函数211()24f x x x =-+，()f x '为函数()f x 的导函数．（1）若数列{}n a 满足：11a =，1()()n n a f a f n +''=+（n N *∈），求数列{}n a 的通项n a ；（2）若数列{}n b 满足：1b b =，12()n n b f b +=（n N *∈）．①当12b =时，数列{}n b 是否为等差数列？若是，请求出数列{}n b 的通项n b ；若不是，请说明理由；②当112b <<时，求证：11221ni ib b =<-∑．．明德外语实验学校高三上期末考试数学参考答案（理科）（时间：120分钟满分：150分）一、选择题：本大题共8小题，每小题5分，满分40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。

二、填空题：本大题共7小题，每小题5分，满分30分。

10.3 2 -(){},02-∞ 13、[4,1][1,2]--14、三、解答题：本大题共6小题，满分80分，解答须写出文字说明、证明过程和演算步骤。

第17题（本小题满分12分）解：（1）因为2()(sin cos)cos2f x x x x=++22sin2sin cos cos cos2x x x x x=+++1sin2cos2x x=++)4xπ+所以：22Tππ==，即函数()f x的最小正周期为π（2）因为02x π≤≤，得52444x πππ≤+≤，所以有sin(2)14x π≤+≤1)4x π-≤+≤01)14x π≤+≤所以，函数()f x的最大值为1此时，因为52444x πππ≤+≤，所以，242x ππ+=，即8x π=.18．风向标教师用书 247页附答案19. 解：（1）设c bx ax x f ++=2)(（其中0≠a ），则b ax x f +=2)('， …………2分c b a x b a ax c x b x a x f +++++=++++=+)2()1()1()1(22．由已知，得22(1)(2)ax b a x a b x a b c +=++++++，∴⎪⎩⎪⎨⎧=++=+=+b c b a a b a a 2201，解之，得1-=a ，0=b ，1=c ， ∴1)(2+-=x x f ． ………………5分（2）由（1）得，)1,(2t t P -，切线l 的斜率t t f k 2)('-==，∴切线l 的方程为)(2)1(2t x t t y --=--，即122++-=t tx y ． ………………7分从而l 与x 轴的交点为)0,21(2tt A +，l 与y 轴的交点为)1,0(2+t B ， ∴tt t S 4)1()(22+=（其中0>t ）． ………………9分∴224)13)(13)(1()('t t t t t S -++=． ………………11分当330<<t 时，0)('<t S ，)(t S 是减函数；当33>t 时，0)('>t S ，)(t S 是增函数． (13)分∴93433)]([min =⎪⎪⎭⎫⎝⎛=S t S ． ………………14分说明：本题主要考查二次函数的概念、导数的应用等知识，以及运算求解能力． 20.21．（本小题满分14分）解：（1）1()22f x x '=-， 111(2)(2)22122n n n a a n a n +∴=-+-=+-，即12(1)12(21)n n a n a n ++++=++11a =， ∴数列{21}n a n ++是首项为4，公比为2的等比数列．12142n n a n -∴++=⋅，即1221n n a n +=--．（2）① 12()n n b f b +=2122n n b b =-+，2112()2n n n b b b +∴-=-． ∴当112b =时，212b =．假设12k b =，则k k b b =+1．由数学归纳法，得出数列{}n b 为常数数列，是等差数列，其通项为12n b =． ②21122n n n b b b +=-+， 2112()2n n n b b b +∴-=-．∴当1112b <<时，2112b b >>．假设12k b >，则 112k k b b +>>．由数学归纳法，得出数列12n b >(1,2,3,)n =．又1112()22n n n b b b +-=-，11122111n n n b b b +∴=---，即11122111n n n b b b +=---． ∴11ni i b =∑11112211()ni i i b b =+=---∑11112211n b b +=---． 112n b +>，111211221ni ib b b =∴<=--∑．。