北京市中国人民大学附属中学2013届高考冲刺数学(理)试卷(九)

人大附中2012届高考适应性练习数学试题(理科)2012.5.27一、选择题1．已知集合2{N 4}A x x =∈<，2{R 230}B x x x =∈--<，则A ∩B ＝( )A ．{－1,0,1}B ．{0,1}C ．{21<<-x x }D ．{32<<-x x }2．已知复数z 满足z ·(1－i )＝2，其中i 为虚数单位，则z ＝( ) A ．1＋i B ．1－i C ．－1＋i D ．－1－i 3．一个几何体的三视图如下，其中主视图和俯视图都是边长为2的正方形，则该几何体的体积是( ) A ．4 B ．8 C ．34D ．38 4．已知向量b a ,满足1=+==b a b a ，则向量b a ,夹角的余弦值为( ) A ．21 B ．－21 C ．23 D ．－23 5．已知数列}{n a 是等差数列，4,843==a a ，则前n 项和n S 中最大的是( ) A ．3S B ．4S 或5S C ．5S 或6SD ．6S6．已知双曲线)0,0(12222>>=-b a by a x 的渐近线方程为x y 2±=，则其离心率为( )A ．5B ．25C ．5或3D ．5或25 7．已知y x ,满足222(1)x y x y y a x -≥⎧⎪+≤⎨⎪≥-⎩，且y x z +=能取到最小值，则实数a 的取值范围是( )A ．1-<aB ．2≥aC ．21<≤-aD ．1-<a 或2≥a8．已知函数①21)(x x f =，②π()sin 2xf x =，③1ln 21)(+=x x f ，则以下四个命题对已知的三个函数都能成立的是( ) 命题p ：)1(+x f 是偶函数；命题q ：)1(+x f 在(0,1)上是增函数； 命题r ：)(x f 恒过定点(1,1)； 命题s ：21)21(>f ． A ．命题p 、qB ．命题q 、rC ．命题r 、sD ．命题s 、p二、填空题9．5)1(xx -的二项展开式中x 项的系数为__________________________．10．已知直线2)1(:++=x k y l ，圆⎩⎨⎧=+=θθsin 21cos 2:y x C ，则圆心C 的坐标是________；若直线l 与圆C 有公共点，则实数k 的取值范围是________．11．如图，已知PAB 是⊙O 的割线，点C 是PB 的中点，且PA ＝AC ，PT 是⊙O 的切线，TC 交⊙O 于点D ，TC ＝8，CD ＝7，则PT 的长为________________________________．12．如图所示程序框图运行的结果是____________________．13．一艘轮船在江中向正东方向航行，在点P 处观测到灯塔A ,B 在一直线上，并与航线成30°角，轮船沿航线前进1000米到达C 处，此时观测到灯塔A 在北偏西45°方向，灯塔B 在北偏东15°方向，则此时轮船到灯C 之间的距离CB 为_______米． 14．若)(x f 是定义在R 上的奇函数，且对0≥∀x ，总存在正数T ，使得T x f T x f =-+)()(成立，则称)(x f 具有“性质P ”．已知函数)(x g 具有“性质P ”，且)(x g 在[0，T ]上的解析式为)(x g ＝2x ，则(1)常数T ＝__________________________；(2)若当]3,3[T T x -∈时，函数kx x g y -=)(恰有9个零点，则k ＝______． 三、解答题15．(本小题满分13分)已知函数34sin 324cos 4sin2)(2+-=xx x x f ． (Ⅰ)求函数)(x f 的最大值，并写出相应的x 取值集合；(Ⅱ)令f (α＋3π)＝510，且α(0,π)∈，求tan2α的值．16．(本小题满分14分)如图所示，在四棱锥P －ABCD 中，四边形ABCD 为菱形，△PAD 为等边三角形，平面PAD ⊥平面ABCD ，且∠DAB ＝60°，AB ＝2，E 为AD 的中点． (Ⅰ)求证：AD ⊥PB ；(Ⅱ)求二面角A －PD －C 的余弦值；(Ⅲ)在棱PB 上是否存在点F ，使EF ∥平面PDC ？并说明理由．17．(本小题满分13分)如图，某工厂2011年生产的A ,B ,C ,D 四种型号的产品产量用条形图表示，现用分层抽样的方法从中抽取50件样品参加今年五月份的一个展销会． (Ⅰ)问A ，B ，C ，D 型号的产品各抽取了多少件？(Ⅱ)从50件样品中随机抽取2件，求这2件产品恰好是不同型号的产品的概率； (Ⅲ)在50件样品中，从A ，C 两种型号的产品中随机抽取3件，其中A 种型号的产品有X 件，求随机变量X 的分布列和数学期望E (X )． 18．(本小题满分13分)已知函数)1ln(1221)(2+++-=x x mx x f ． (Ⅰ)当23-=m 时，求函数)(x f 的极值点； (Ⅱ)当1≤m 时，曲线)(:x f y C =在点P (0,1)处的切线l 与C 有且只有一个公共点，求实数m 的范围．19．(本小题满分14分)已知椭圆)0(1:22221>>=+b a by a x C 经过点)23,1(M ，且其右焦点与抛物线x y C 4:22=的焦点F 重合．(Ⅰ)求椭圆1C 的方程；(Ⅱ)直线l 经过点F 与椭圆1C 相交于A 、B 两点，与抛物线2C 相交于C 、D 两点．求CDAB 的最大值．20．(本小题满分13分)已知集合S ＝{1,2,3,...，2011,2012}，设A 是S 的至少含有两个元素的子集，对于A 中的任意两个不同的元素)(,y x y x >，若y x -都不能．．．整除y x +，则称集合A 是S 的“好子集”．(Ⅰ)分别判断数集P ＝{2,4,6,8}与Q ＝{1,4,7}是否是集合S 的“好子集”，并说明理由；(Ⅱ)求集合S 的“好子集”A 所含元素个数的最大值；(Ⅲ)设m A A A A ,...,,,321是集合S 的m 个“好子集”，且两两互不包含，记集合i A 的元素个数为),...,2,1(m i k i =，求证：1!(2012)!2012!miii k k =-≤∑中国人民大学附属中学高三模拟考试数学试题(理科)参考答案一、选择题：BAAB BACC 二、填空题： 9．－5；10．(1,0),(－∞，0]； 11．74； 12．10； 13．2500； 14．1，462- 三、解答题：15．(本小题满分13分)解：(Ⅰ)因为34sin 324cos 4sin2)(2+-=xx x x f )sin 21(32sin2x x-+= 2分)3π2sin(22cos 32sin+=+=x x x 4分 因为．1)3π2sin(1≤+≤-x所以．)(x f 的最大值为2 6分相应值的集合为π{|4π(Z)}3x x k k =+∈ 7分 (Ⅱ)由(Ⅰ)知，2cos 22π21sin 23π)3π(21sin 2)3π(xx x x f =+=++=+．5102cos2=α，所以10102cos =α． 5412cos 2cos 2-=-=αα 10分 又因为)π,0(∈α 所以53cos 1sin 2=-=αα． 43cos sin tan -==ααα． 724tan 1tan 22tan 2-=-=ααα 13分 16．(本小题满分14分)(Ⅰ)证明：连结EB ，在△AEB 中，AE ＝1，AB ＝2，∠EAB ＝60°，∴BE 2＝AE 2＋AB 2－2AE ·AB ·cos60°＝1＋4－2＝3． ∵AE 2＋BE 2＝AB 2， ∴AD ⊥EB 2分∵△PAD 为等边三角形，E 为AB 的中点， ∴AD ⊥PE ． 又EB ∩PE ＝E ． ∴AD ⊥平面PEB ． ∴AD ⊥PB ． 4分 (Ⅱ)∵平面PAD ⊥平面ABCD ，平面PAD ∩平面ABCD ＝AD，且PE ⊥AD ，∴PE ⊥平面ABCD ， ∴PE ⊥EB ．以点E 为坐标原点，EA ，EB ，EP 为x ，y ，z 轴的正半轴建立空间直角坐标系，如图，则A (1,0,0),B (0,3,0)P (0,0,3),D (－1,0,0),)0,3,1(-==AB DC ．设平面PCD 的一个法向量为),,(z y x n =，则00n DC n DP ⎧⋅=⎪⎨⋅=⎪⎩，即(,,)(0(,,0x y z x y z ⎧-=⎪⎨=⎪⎩，∴00x x ⎧-=⎪⎨+=⎪⎩ 令z ＝－1，则3=x ，y ＝1，故)1,1,3(-=n ． 平面PAD 的一个法向量为)0,3,0(=．所以cos 53EB n EB n EB n⋅〈⋅〉===⋅．又二面角A －PD －C 为钝角， ∴二面角A －PD －C 的余弦值为55-． (Ⅲ)假设棱PB 上存在点F ，使EF ∥平面PDC ，设F (0,m ,n )，λ=，则：)3,3,0()3,,0(-=-λn m ，∴λλ33,3-==n m ， ∴)33,3,0(λλ-=． ∵EF ∥平面PDC ，∴⊥，即)33,3,0(λλ-·)1,1,3(-＝0． ∴21,0333==+-λλλ， 故当点F 为PB 的中点时，EF ∥平面PDC ．17．(本小题满分13分)解：(Ⅰ)从条形图上可知，共生产产品有50＋100＋150＋200＝500(件)，样品比为10150050=，所以，A ,B ,C ,D 四种型号的产品分别取,20200101,10100101=⨯=⨯ 15150101,550101=⨯=⨯， 即样本中应抽取A 产品10件，B 产品20件，C 产品5件，D 产品15件 3分(Ⅱ)从50件产品中任取2件共有C250＝1225种方法，2件恰为同一产品的方法数为C210＋C220＋C25＋C215＝350种．所以2件恰好为不同型号的产品的概率为7512253501=-7分 (Ⅲ)依题意，X 的可能取值为0,1,2,3， 8分则P (X ＝0)＝4551031535=C C P (X ＝1)＝45510031525110=CC C P (X ＝2)＝45522531515210=CC C P (X ＝3)＝455120315310=C C 故X 的分布列为12分 所以EX ＝2455120345522524551001455100=⨯+⨯+⨯+⨯13分 18．(本小题满分13分)解：(Ⅰ)当23-=m 时，)1ln(1243)(2+++--=x x x x f ， 其定义域为(－1,＋∞)， 1分)1(2273)('2+++-=x x x x f 令,0)('=x f 即02732=++x x 2分 解得2,3121-=-=x x ， 3分经检验，函数)(x f 的极值点为31-4分 (Ⅱ)由题设知，点P (0,1)在曲线C 上且1)0('-=f ，切线l 的方程为1+-=x y ，5分于是方程：)1ln(122112+++-=+-x x mx x 即方程0)1ln(212=++-x x mx 在(－1,＋∞)上有且只有一个实数根； 设)1ln(21)(2++-=x x mx x g ，即求m 的取值使得函数)(x g 在(－1,＋∞)上有且只有一个零点 7分因为0)0(=g ，)1(1)1()('2->+-+=x x xm mx x g ， 所以分以下几种情形讨论： ①当m ＝0时，1)('+-=x xx g ，)(x g 在(－1,0)上递增，在(0,＋∞)上递减， 所以，)(x g 在(－1,＋∞)上有且只有一个零点0 8分②当m ＝1时，01)('2≥+=x x x g ，)(x g 在(－1,＋∞)上递增， 所以，)(x g 在(－1,＋∞)上有且只有一个零点0 9分③当m ＜0时，由01)1()('2=+-+=x xm mx x g 得01=x 或112-=m x ，且111-<-m． 当x 变化时，)('x g 和)(x g 变化如下表：故)(x g 在(－1,＋∞)上有且只有一个零点0 10分④当0＜m ＜1时，由01)1()('2=+-+=x xm mx x g 得01=x 或112-=m x ，且011>-m． 当x 变化时，)('x g 和)(x g 变化如下表：因为，0)1ln()(,0)0()(>+==<mm g g m g ， 所以，必存在)2,1(0mm m x -∈，使0)(0=x g ． 故)(x g 在(－1,＋∞)存在两个零点0和0x ，不符合题意 12分综上，m 的取值范围为(－∞,0]∪{1} 13分19．(本小题满分14分)解：(Ⅰ)解法1：由抛物线方程，得焦点F (1,0)，∴c ＝1 1分故1222==-c b a ① 2分 又椭圆C 1经过点M (1,23)， ∴149122=+b a ② 3分 由①②消去2a 并整理，得：099424=--b b ，解得32=b ，或432-=b (舍去)， 4分 从而42=a ，故椭圆的方程为13422=+y x 5分 解法2：由抛物线方程，得焦点F (1,0)，∴c ＝1．∴4)23()11()23()11(22222=+-+++=a ， ∴3,422==b a ，故椭圆的方程为13422=+y x ． (Ⅱ)①当直线l 垂直于x 轴时，则)2,1(),2,1(),23,1(),23,1(--D C B A ， ∴43=CD AB6分 ②当直线l 与x 轴不垂直，设其斜率为k (k ≠0)，则直线l 的方程为)1(-=x k y 由22(1)143y k x x y =-⎧⎪⎨+=⎪⎩得01248)43(2222=-+-+k x k x k 显然01>∆，∴该方程有两个不等的实数根，设),(),,(2211y x B y x A2221222143)3(4,438k k x x k k x x +-=⋅+=+……8分 所以，2122122212214)(1)()(x x x x k y y x x AB -++=-+-=＝2222222243)1(1243)3(16)438(1kk k k k k k ++=+--+⋅+……10分 由2(1)4y k x y x=-⎧⎨=⎩得0)42(2222=++-k x k x k 显然02>∆，∴该方程有两个不等的实数根，设),(),,(4433y x D y x C∵k ≠0，∴24342kx x +=+． 由抛物线的定义，得22243)1(4442kk k x x CD +=+=++= 12分∴22222221213333344(1)3444AB k k k CD k k k k +=⋅==<++++（）. 综上，当直线l 垂直于x 轴时，ABCD 取得最大值43. 14分 20．(本小题满分13分)解：(Ⅰ)由于4－2＝2整除4＋2＝6，所以集合P 不是集合S 的“好子集”；由于4－1＝3不能整除4＋1＝5，7－1＝6不能整除7＋1＝8，7－4＝3不能整除7＋4＝11，所以集合Q 是集合S 的“好子集” 3分(Ⅱ)设集合)...}(,...,,,{21321n n a a a a a a a A <<<=是集合S 的一个“好子集”．令：)1,...,2,1(1-==-+n i b a a i i i ．由于A 是S “好子集”，所以)1,...,2,1(1-=≠n i b i ，从而)1,...,2,1(2-=≥n i b i ． 若存在某个2=i b ，则此时1+i a 与i a 同奇偶，从而2能整除i i a a ++1与A 是S “好子集”矛盾，故：)1,...,2,1(3-=≥n i b i 5分于是：)1(3...1211-≥+++=--n b b b a a n n从而：201112012)1(31=-≤-≤-a a n n 所以：671≤n另一方面：取}2011,2008,...,7,4,1{=A ，此时集合A 有671个元素，且是集合S 的一个“好子集”，故集合S 的“好子集”A 所含元素个数的最大值为671 8分(Ⅲ)将S 的2012个元素1,2,3,...,2012作全排列，其不同的排法总数为2012!个；另一方面：将子集i A 的i k 个元素排在前i k 个位置，子集i A 在S 中的补集的元素排在后2012－i k 个位置，即排成),...,,,,...,,(20122121i i k k y y y x x x -，这样的排列共有i k !(2012－i k )!个，它们全包含在全排列2012!中． 10分以下还要说明：以集合S 的m 个“好子集”m A A A A ,...,,,321中的元素排在前ik个位置，以它们对应的在S 中的补集的元素排在后2012－i k 个位置的各个排列中，在题设条件下，没有两个是相同的．不妨设：f i k k ≤，由条件f i A A ,互不包含，所以f i A A ⊄，所以排列),...,,,,...,,(2012'2'1''2'1'i i k k y y y x x x -中的前面i k 个元素不可能与排列),...,,,,...,,(20122121i i k k y y y x x x -中的前面i k 个元素完全相同，否则就有f i A A ⊆而与条件矛盾．故：!2012)!2012(!1≤-∑=i i mi k k 13分。

中国人民大学附属中学高考冲刺卷数 学(理) 试 卷（七）第Ⅰ卷（选择题 共40分）一、选择题：本大题共8小题，每小题5分，共40分. 在每小题列出的四个选项中，选出符合题目要求的一项.1.已知集合{0,1}A =，{1,0,3}B a =-+，且A B ⊆，则a 等于（A ）1 （B ）0 （C ）2- （D ）3- 2.已知i 是虚数单位，则复数23z i+2i 3i =+所对应的点落在（A ）第一象限 （B ）第二象限 （C ）第三象限 （D ）第四象限 3.在ABC ∆中，“0AB BC ⋅>”是“ABC ∆为钝角三角形”的（A ）充分不必要条件 （B ）必要不充分条件 （C ）充要条件 （D ）既不充分又不必要条件4.已知六棱锥P ABCDEF -的底面是正六边形，PA ⊥平面ABC .则下列结论不正确．．．的是 （A ）//CD 平面PAF （B ）DF ⊥平面PAF （C ）//CF 平面PAB （D ）CF ⊥平面PAD 5.双曲线22221x y a b-=的渐近线与圆22(2)1x y +-=相切，则双曲线离心率为（A ）2 （B ）3 （C ）2 （D ）3 6.函数sin()(0)y x ϕϕ=π+>的部分图象如图所示，设P 是图象的最高点，,A B 是图象与x 轴的交点，则tan APB ∠=（A ）10 （B ）8 （C ）87 （D ）47第4题图 第6题图 7．已知数列{}n a 的通项公式为13n a n =-，那么满足119102k k k a a a +++++=的整数k（A ）有3个 （B ）有2个 （C ）有1个 （D ）不存在 8．设点(1,0)A ，(2,1)B ，如果直线1ax by +=与线段AB 有一个公共点，那么22a b + （A ）最小值为15 （B ）最小值为5 （C ）最大值为15 （D ）最大值为5第Ⅱ卷（非选择题 共110分）二、填空题：本大题共6小题，每小题5分，共30分. 9．在ABC ∆中，若2B A =，:1:3a b =，则A =_____.10.在521()x x+的展开式中，2x 的系数是_____.OAB PD C•xA BPyO11．如图，AB 是圆O 的直径，P 在AB 的延长线上，PD切圆O 于点C .已知圆O 半径为3，2OP =，则 PC =______；ACD ∠的大小为______.12.在极坐标系中，点(2,)2A π关于直线:cos 1l ρθ=的对称点的一个极坐标为_____.13．定义某种运算⊗，a b ⊗的运算原理如右图所示.设()(0)(2)f x x x x =⊗-⊗.则(2)f =______；()f x 在区间[2,2]-上的最小值为______.14.数列{}n a 满足11a =，11n n n a a n λ+-=+，其中λ∈R ， 12n =，，．①当0λ=时，20a =_____；②若存在正整数m ，当n m >时总有0n a <，则λ的取值范围是_____.三、解答题：本大题共6小题，共80分. 解答应写出必要的文字说明、证明过程或演算步骤.15.（本小题满分13分）已知函数cos 2()sin()4xf x x π=+.（Ⅰ）求函数()f x 的定义域；（Ⅱ）若4()3f x =，求sin 2x 的值.16.（本小题满分13分）如图，已知菱形ABCD 的边长为6，60BAD ∠=,ACBD O =.将菱形ABCD 沿对角线AC 折起，使32BD =，得到三棱锥B ACD -.（Ⅰ）若点M 是棱BC 的中点，求证://OM 平面ABD ； （Ⅱ）求二面角A BD O --的余弦值；（Ⅲ）设点N 是线段BD 上一个动点，试确定N 点的位置，使得42CN =，并证明你的结论.Ma b ≥开始 输入,a b否结束S b= S a =输出S 是17.（本小题满分13分）甲班有2名男乒乓球选手和3名女乒乓球选手，乙班有3名男乒乓球选手和1名女乒乓球选手，学校计划从甲乙两班各选2名选手参加体育交流活动.（Ⅰ）求选出的4名选手均为男选手的概率.（Ⅱ）记X 为选出的4名选手中女选手的人数，求X 的分布列和期望.18.（本小题满分14分）已知函数()(1)e (0)xa f x x x=->，其中e 为自然对数的底数.（Ⅰ）当2a =时，求曲线()y f x =在(1,(1))f 处的切线与坐标轴围成的面积； （Ⅱ）若函数()f x 存在一个极大值点和一个极小值点，且极大值与极小值的积为5e ，求a 的值.19.（本小题满分14分）已知椭圆2222:1x y M a b +=(0)a b >>，且椭圆上一点与椭圆的两个焦点构成的三角形周长为246+．（Ⅰ）求椭圆M 的方程；（Ⅱ）设直线l 与椭圆M 交于,A B 两点，且以AB 为直径的圆过椭圆的右顶点C ， 求ABC ∆面积的最大值．20.（本小题满分13分）若m A A A ,,,21 为集合2}(,,2,1{≥=n n A 且)n ∈*N 的子集，且满足两个条件：①12m A A A A =；②对任意的A y x ⊆},{，至少存在一个},,3,2,1{m i ∈，使}{},{x y x A i = 或}{y . 则称集合组m A A A ,,,21 具有性质P . 如图，作n 行m 列数表，定义数表中的第k 行第l 列的数为⎩⎨⎧∉∈=)(0)(1l l kl A k A k a .（Ⅰ）当4n =时，判断下列两个集合组是否具有性质P ，如果是请画出所对应的表格，如果不是请说明理由；集合组1：123{1,3},{2,3},{4}A A A ===； 集合组2：123{2,3,4},{2,3},{1,4}A A A ===. （Ⅱ）当7n =时，若集合组123,,A A A 具有性质P ，请先画出所对应的7行3列的一个数表，再依此表格分别写出集合123,,A A A ；（Ⅲ）当100n =时，集合组12,,,t A A A 是具有性质P 且所含集合个数最小的集合组，求t 的值及12||||||t A A A ++的最小值.（其中||i A 表示集合i A 所含元素的个数）中国人民大学附属中学高考冲刺卷 数学(理)试卷（七）参考答案一、选择题：本大题共8小题，每小题5分，共40分.二、填空题：本大题共6小题，每小题5分，共30分.9. 30 10. 5 11.1；75 12.)4π（或其它等价写法） 13.2-；6- 14.120；(21,2),k k k -∈*N . 注：11、13、14题第一问2分，第二问3分.三、解答题：本大题共6小题，共80分.若考生的解法与本解答不同，正确者可参照评分标准给分.15.（本小题满分13分） 解：（Ⅰ）由题意，sin()04x π+≠， ………………2分 所以()4x k k π+≠π∈Z ， ………………3分 所以()4x k k π≠π-∈Z ，………………4分函数()f x 的定义域为{xx ≠,4k k ππ-∈Z }. ………………5分（Ⅱ）cos 2cos 2()sin()sin cos cos sin444x xf x x x x ==πππ++ ………………7分 2sin cos x x x=+ ………………8分 22sin )sin )sin cos x x x x x x -==-+. ………………10分因为4()3f x =，所以cos sin 3x x -=. ………………11分所以，2sin 21(cos sin )x x x =-- ………………12分81199=-= . ………………13分16.（本小题满分13分）（Ⅰ）证明：因为点O 是菱形ABCD 的对角线的交点，所以O 是AC 的中点.又点M 是棱BC 的中点，所以OM 是ABC ∆的中位线，//OM AB . ………………1分 因为OM ⊄平面ABD ,AB ⊂平面ABD ,所以//OM 平面ABD . ………………3分 （Ⅱ）解：由题意，3OB OD ==，因为BD =所以90BOD ∠=，OB OD ⊥. ………………4分 又因为菱形ABCD ，所以OB AC ⊥，OD AC ⊥. 建立空间直角坐标系O xyz -，如图所示.(0,3,0),A D (0,0,3)B .所以(33,0,3),(33,3,0),AB AD =-=-………………6分设平面ABD 的法向量为n =(,,)x y z ，则有0,0AB AD ⎧⋅=⎪⎨⋅=⎪⎩n n 即：30,30z y ⎧-+=⎪⎨-+=⎪⎩令1x =，则y z ==n =(1. ………………7分 因为,AC OB AC OD ⊥⊥,所以AC ⊥平面BOD . 平面BOD 的法向量与AC 平行，所以平面BOD 的法向量为0(1,0,0)=n . ………………8分000cos ,⋅〈〉===n n n n n n 因为二面角A BD O --是锐角， 所以二面角A BD O --的余弦值为7. ……………9分 （Ⅲ）解：因为N 是线段BD 上一个动点，设111(,,)N x y z ，BN BD λ=，则111(,,3)(0,3,3)x y z λ-=-，所以1110,3,33x y z λλ===-， ……………10分 则(0,3,33)N λλ-，(33,3,33)CN λλ=-，由CN ==，即29920λλ-+=，…………11分解得13λ=或23λ=， ……………12分 所以N 点的坐标为(0,2,1)或(0,1,2). ……………13分（也可以答是线段BD 的三等分点，2BN ND =或2BN ND =） 17.（本小题满分13分）解：（Ⅰ）事件A 表示“选出的4名选手均为男选手”.由题意知232254()C P A C C = ………………3分11110220=⨯=. ………………5分 （Ⅱ）X 的可能取值为0,1,2,3. ………………6分23225431(0)10620C P X C C ====⨯， ………………7分11212333225423337(1)10620C C C C P X C C +⨯⨯+====⨯， ………………9分 21332254333(3)10620C C P X C C ⨯====⨯， ………………10分 (2)1(0)(1)(3)P X P X P X P X ==-=-=-=920=. ………………11分X………………12分179317()01232020202010E X =⨯+⨯+⨯+⨯=. ………………13分18、（本小题满分14分）解：（Ⅰ）22()e xx ax a f x x -+'=， ………………3分 当2a =时，2222()e xx x f x x -+'=， 12122(1)e e 1f -+'=⨯=，(1)e f =-， 所以曲线()y f x =在(1,(1))f 处的切线方程为e 2e y x =-， ………………5分 切线与x 轴、y 轴的交点坐标分别为(2,0)，(0,2e)-， ………………6分所以，所求面积为122e 2e 2⨯⨯-=. ………………7分（Ⅱ）因为函数()f x 存在一个极大值点和一个极小值点，所以，方程20x ax a -+=在(0,)+∞内存在两个不等实根， ………………8分则240,0.a a a ⎧∆=->⎨>⎩………………9分所以4a >. (10)分设12,x x 为函数()f x 的极大值点和极小值点，则12x x a +=，12x x a =， ………………11分 因为，512()()e f x f x =，所以，1251212e e e x x x a x a x x --⨯=， ………………12分 即1225121212()e e x x x x a x x a x x +-++=，225e e a a a a a -+=，5e e a =， 解得，5a =，此时()f x 有两个极值点，所以5a =. ………………14分19.（本小题满分14分）解：（Ⅰ）因为椭圆M 上一点和它的两个焦点构成的三角形周长为246+，所以24622+=+c a ， ……………1分又椭圆的离心率为，即3c a =，所以3c a =， ………………2分所以3a =，c =………………4分所以1b =，椭圆M 的方程为1922=+y x . ………………5分 （Ⅱ）方法一：不妨设BC 的方程(3),(0)y n x n =->，则AC 的方程为)3(1--=x ny .由22(3),19y n x x y =-⎧⎪⎨+=⎪⎩得0196)91(2222=-+-+n x n x n ， ………………6分 设),(11y x A ，),(22y x B ，因为222819391n x n -=+，所以19327222+-=n n x ， ………………7分同理可得2219327nn x +-=， ………………8分 所以1961||22++=n n BC ，222961||nn n n AC ++=， ………………10分 964)1()1(2||||212+++==∆n n n n AC BC S ABC ， ………………12分 设21≥+=n n t ，则22236464899t S t t t ==≤++， ………………13分当且仅当38=t 时取等号，所以ABC ∆面积的最大值为83. ………………14分方法二：不妨设直线AB 的方程x ky m =+.由22,1,9x ky m x y =+⎧⎪⎨+=⎪⎩ 消去x 得222(9)290k y kmy m +++-=， ………………6分 设),(11y x A ，),(22y x B ，则有12229km y y k +=-+，212299m y y k -=+. ① ………………7分因为以AB 为直径的圆过点C ，所以 0CA CB ⋅=. 由 1122(3,),(3,)CA x y CB x y =-=-，得 1212(3)(3)0x x y y --+=. ………………8分 将1122,x ky m x ky m =+=+代入上式，得 221212(1)(3)()(3)0k y y k m y y m ++-++-=.将 ① 代入上式，解得125m =或3m =（舍）. ………………10分 所以125m =（此时直线AB 经过定点12(,0)5D ，与椭圆有两个交点），所以121||||2ABC S DC y y ∆=-12==……………12分 设211,099t t k =<≤+，则ABC S ∆=所以当251(0,]2889t =∈时，ABC S ∆取得最大值83. ……………14分20.（本小题满分13分）（Ⅰ）解：集合组1具有性质P . ………………1分所对应的数表为： (3)分集合组2不具有性质P . ………………4分因为存在{{2,3}1,2,3,4}⊆， 有123{2,3}{2,3},{2,3}{2,3},{2,3}A A A ===∅，与对任意的A y x ⊆},{，都至少存在一个{1,2,3}i ∈，有}{},{x y x A i = 或}{y 矛盾，所以集合组123{2,3,4},{2,3},{1,4}A A A ===不具有性质P . ………………5分（Ⅱ）1 1 1 1 1 11 1 10 0 0 0 0 0 00 00 1 1 00 0 0 1 1 0 0 1……………7分123{3,4,5,7},{2,4,6,7},{1,5,6,7}A A A ===. ………………8分 （注：表格中的7行可以交换得到不同的表格，它们所对应的集合组也不同） （Ⅲ）设12,,,t A A A 所对应的数表为数表M ，因为集合组12,,,t A A A 为具有性质P 的集合组，所以集合组12,,,t A A A 满足条件①和②， 由条件①：12t A A A A =， 可得对任意x A ∈，都存在{1,2,3,,}i t ∈有i A x ∈， 所以1=xi a ，即第x 行不全为0，所以由条件①可知数表M 中任意一行不全为0. (9)分由条件②知，对任意的A y x ⊆},{，都至少存在一个{1,2,3,,}i t ∈，使}{},{x y x A i = 或}{y ，所以yi xi a a ,一定是一个1一个0，即第x 行与第y 行的第i 列的两个数一定不同.所以由条件②可得数表M 中任意两行不完全相同. ………………10分因为由0,1所构成的t 元有序数组共有2t个，去掉全是0的t 元有序数组，共有21t-个，又因数表M 中任意两行都不完全相同，所以10021t≤-， 所以7t ≥.又7t =时，由0,1所构成的7元有序数组共有128个，去掉全是0的数组，共127个，选择其中的100个数组构造100行7列数表，则数表对应的集合组满足条件①②，即具有性质P .所以7t =. ………………12分 因为12||||||t A A A +++等于表格中数字1的个数， 所以，要使12||||||t A A A +++取得最小值，只需使表中1的个数尽可能少，而7t =时，在数表M 中，1的个数为1的行最多7行；1的个数为2的行最多2721C =行； 1的个数为3的行最多3735C =行；1的个数为4的行最多4735C =行；因为上述共有98行，所以还有2行各有5个1，所以此时表格中最少有722133543552304+⨯+⨯+⨯+⨯=个1. 所以12||||||t A A A +++的最小值为304. ………………14分。

中国人民大学附属中学2013届高三高考冲刺卷（四）（英语）第二部分：知识运用（共两节，45分）第一节：单项填空（共15小题；每小题1分，共15分）21．As Susan stared at ______ piles of plastic in her recycling bin, ______ strong sense of responsibility made her stick to her recycling work．A．不填；不填 B．不填; a C．the ; a D．the; the22．Tom's mother kept telling him that he should work harder, but ______didn't help．A．he B．which C．it D．one23．Put yourself in situations where you're forced to communicate in English, ______ you'll see more progress over time．A．until B．or C．but D．and24．The United Nations are evaluating the damage caused in Haiti earthquake, while in Chile the government itself ______ the lead role．A．is taking B．was taking C．took D．had taken25．What's your opinion of Mr．Li's request that we ______ spend half an hour reading English aloud every morning?A．would B．should C．must D．could26．The actor still remembers the excitement in his class when a female classmate _____ for a key role in a Zhang Yimou film．A．is chosen B．s chosen C．chooses D．chose27．It was the President himself______ opened the door．A．who B．when C．which D．where28．Head coach Li Yan bursts into tears of joy when she sees Zhou Yang first _____ the finish line at the women's 1500m short track speed skating final．A．to cross B．having crossed C．cross D．crossed29．The Mekong River Commission has found no evidence ______ the dams on the upper reaches have an influence on the water flow downstream．A．which B．that C．where D．what30．in four different countries, Jessie is a typical third culture kid, the term of which refers to children who spend a period of time in one or more cultures．A．Bring up B．Having brought up C．Bringing up D．Brought up31．all his courage, he invites Celine．to get off the train with him．A．To gather B．Gathered C．Gathering D．Being gathered32．I don't expect children to be rude, nor ______ them to be disobeyed．A．did I expect B．do I expect C．I expect D．I expected33．In the past few months, the central bank governor and others ______ measures to prevent prices from rising too fast．A．called for B．are calling for C．have called for D．will call for34．the whole process for the studies of cancer took more than a year, the technology is moving so fast that it will soon take just weeks．A．How B．Because C．If D．While35．Beijing took steps to limit the kinds of high-risk of borrowing money from the banks that can create the high price of housing, ______ happened in the United States．a．as B．that C．how D．while第二节：完形填空（共20小题；每小题1．5分，共30分）One afternoon I toured an art museum while waiting for my husband to finish a business meeting．I was looking forward to a quiet 36 of the splendid artwork．A young 37 viewing the paintings ahead of me 38 nonstop between themselves．I watched them a moment and decided the lady was doing all the talking．I admired the man's 39 for putting up with her 40 stream of words．41 by their noise, I moved on．I met them several times as I moved 42 the various rooms of art．Each time I heard her continuous flow of words, I moved away 43 .I was standing at the counter of the museum gift shop making a 44 when the couple approached the45 ．Before they left, the man 46 into his pocket and pulled out a white object．He 47 it into a long stick and then 48 his way into the coatroom to get his wife's jacket．"He's a 49 man, " the clerk at the counter said．" Most of us would give up if we were blinded at such a young age．During his recovery, he made a promise his life wouldn't change．So, as before, he and his wife come in 50 there is a new art show．""But what does he get out of the art?" I asked．"He can't see．""Can't see! You're 51 ．He sees a lot．More than you and I do," the clerk said．"His wife 52 each painting so he can see it in his head．"I learned something about patience, 53 and love that day．I saw the patience of a young wife describing paintings to a person without 54 and the courage of a'' husband who would not 55 blindness to change his life．And I saw the love shared by two people as I watched this couple walk away, their arms intertwined．36．A．view B．touch C．wander D．stare37．A．lady B．couple C．man D．clerk38．A．yelled B．argued C．screamed D．chatted39．A．attempt B．independence C．patience D．wisdom40．A．constant B．vivid C．casual D．vague41．A．Adopted B．Adapted C．Disturbed D．Conducted42．A．from B．to C．towards D．through43．A．anxiously B．quickly C．urgently D．sensibly44．A．comment B．purchase C．decision D．profit45．A．exit B．entrance C．front D．queue46．A．plugged B．reached C．held D．bent47．A．lengthened B．made C．brought D．broadened48．A．led B．found C．tapped D．forced49．A．generous B．rough C．smart D．brave50．A．wherever B．whatever C．whenever D．whichever51．A．unique B．silly C．equal D．wrong52．A．decorates B．draws C．shoves D．describes53．A．kindness B．pride C．courage D．enthusiasm54．A．sight B．support C．expectation D．confidence55．A．get B．allow C．hope D．cause第三部分：阅读理解（共两节，40 分）第一节：（共15小题；每小题2分，共30分）AShopping in the United States changes a lot．About ninety years ago most people shopped in small stores that were owned by one person or a family．Women went from the bakery to the butcher's to the grocer and on to the fruit and vegetable seller in order to get their food for the week．Then about sixty years ago, supermarkets were born．In a supermarket, people could get all the different kinds of food they needed without going to different stores．The next big change in shopping in the United States was the shopping mall．A shopping mall is a group of stores under one roof．Because malls allowed people to shop without worrying about the weather, they soon became very popular．The mall became a place for people to socialize in addition to shopping．If you walk through a mall, you will see older people sitting, chatting and drinking coffee．Malls are places for teenagers to hang out．Many teens will often just "go to the mall" and spend time with their friends．The recent change in American shopping was the superstore．Large chain stores such as Wal-Mart, Office Depot and Toys "R" Us have been built all across the United States．Because they are so large, they can buy goods at a great discount and sell them much cheaper than smaller stores．Sometimes, when they are built near small towns, many of the small town stores have to close．They just cannot compete with their giant neighbors．And now, online shopping is becoming more and more popular all over the States．People are too busy to go to the physical stores, so they go shopping over the Internet．Online shopping has lots of advantages．For instance, online stores are usually available 24 hours a day．Searching or browsing online shops can be faster than browsing the physical stores．While, online shopping also has its disadvantages．People are at higher risk of being cheatedon the part of the merchant than in a physical store．And privacy of personal information may be let out．56．Which of the following shows the right order of shopping development in the United States?①small stores ②superstores ③shopping malls ④shopping online ⑤supermarketsA．①②③④⑤B．①⑤②③④C．①⑤③②④D．②③④①⑤57．Which is the place for people to spend time with others according to the passage?A．Shopping malls．B．Small stores．C．Supermarkets．D．Superstores．58．Why can the superstores sell products at much lower prices?A．Because they are built near small towns．B．Because they are across the United States．C．Because they sell all kinds of products people need．D．Because they can buy goods at a reduction in the price．59．What's the disadvantage of online shopping according to the passage?A．Wasting time．B．Leaking personal information．C．Fixed prices．D．No chance to do physical activitiesBV olunteers, as an essential part of a successful world exposition, are a major channelfor the public to participate in, serve and share the world exposition and a means toshowcase the image of the host country and city. The following information is about thevolunteer for the World Exposition 2010 Shanghai China．I．Basic Requirements for V olunteers● Be willing to participate in voluntary services of Expo 2010;● Age limit: Expo Site volunteers must be born before April 30,1992 and Expo City Voluntary Service Station volunteers beforeApril 30, 1994;● Obey the laws and regulations of the PRC;● Be able to participate in training and relevant activities before the opening of Expo 2010;● Possess necessary knowledge and skills needed by the position;● Be in good heal th to meet the requirements of corresponding voluntary positions．II．Further Information for V olunteers● SourceResidents of Chinese mainland, Hong Kong, Macau and Taiwan, as well as overseas Chinese, and foreigners can all apply to be the volunteers．●Signup methodsApplicants may log in onto the official websites for online signup．They may also consult or connect with the Expo V olunteer Stations．● TimeMay 1 - December 31, 2009Ⅲ．V olunteer TrainingV olunteer training includes general training, special training and position training．General training is carried out through internet, while special training and position training are provided through classroom lectures and field practice．IV．V olunteer Types● Expo Site volunteers refer to those offerin g voluntary services to visitors and the Organizer in the Expo Site, mainly including information, visitor flow management, reception, translation and interpretation, assistance for the disabled, and assistance in media service, event and conference organization and．volunteer management．● Information booth volunteers are stationed in the Expo's information booths at key transportation centers, commercial outlets, tourist attractions, restaurants, hotels and cultural event places outside the Expo Site．They offer services including information, translation, interpretation and even first aid．60．If you were born in April 1993, where can you be a volunteer?A．In the Expo City．B．In the host country．C．In the Expo Site．D．In Chinese mainland 61．Which of the training will be done on the Internet?A．Position training．B．General training．C．Classroom training．D．Special training．62．Which of the following service is offered by information booth volunteers?A．Visitor flow management．B．Helping the disabledC．Assistance in media service． D．Emergency First aid．CWith alarming regularity, we read about oil tankers having accidents near land and the terrible consequences of the oil spills （泄露）on people, nature, and the environment．Millions of dollars have been used in developing special chemicals to help dismiss the spills and to clean up the animals, beaches, and land spoiled by the oil．Unfortunately, when many of these chemicals are used, more damage is caused to the environment, especially to lives in the sea．Of all of today's environmental disasters, an oil spill may actually be one of the least serious．Although oil is poisonous, it is a natural material．In the end, it breaks down naturally．There are, of course, long-term effects, but it is usually more serious in the short term．Nature by itself works better than chemical materials, but when there is a spill we demand that governments act immediately with as much hi-tech knowledge as possible．In 1967 the tanker Torrey Canyon sank off the Scilly Isles near the coast of England and spilled 120,000 tones of oil into the ocean．If you go there today, you will find it hard to see any sign that it ever happened．Governments seem to accept the risk of transporting millions of tons of oil by ship every day so that we can fill up our cars and drive around and cause even more environmental damage．Interestingly, the biggest companies in the world produce cars, and the next biggest supply the gasoline to make them run IWe should be thinking more about reducing our dependency on oil．Governments should be encouraging research into new technologies, such as cars run by solar power （太阳能）, electricity, hydrogen, and soon．Much of this research has, in the past, been held back by the oil, gas, and coal．If the world's millions of cars were 10% more efficient （高效的）—and the industry could easily produce cars at least twice as efficient ?we would need many fewer tankers crossing the oceans each year．If this happened, the risks of oil spills would be reduced, and the air we breathe would be cleaner and fresher, too．63．What is the passage mainly talking about?A．Oil spills pollution．B．What oil pollution is．C．Oil tanker accidents. D．How to reduce oil pollution. 64．How does the author support the idea that oil spills are not as serious as people believe?A．By giving a description．B．By making an argument．C．By giving an example．D．By drawing a diagram．65．What does the underlined word "risk" in Paragraph 5 refer to?A．Transportation depending more on oil．B．Poisonous oil breaking down naturally．C．Millions of tons of oil spilling into the sea．D．More environmental damage being caused．66．Which suggestion, is made for reducing oil tank accidents according to the passage?A．We should build safer tankers in the near future．B．We should develop new technologies to cut oil use．C．Tankers should not be allowed to sail near the coastlines．D．Countries should build more oil pipelines under the sea．DRecently there was a major discovery in the scientific research—the mapping of all DNA in a human gene （基因）is complete．Couple of years ago, this seems an impossible task for scientist to accomplish．All this progress in science leads us to believe that the day, when the human being will be cloned, is not far away．Human cloning has always been a topic of argument, in terms of morality or religion．Taking a look at why cloning might be beneficial, among many cases, it is arguable that parents who are known to be at risk of passing a genetic limitation to a child could make use of cloning．If the clone was free of genetic limitations．then the other clone would be as well．The latter could foe inserted in the woman and allowed to ripen to term．Moreover, cloning would enable women, who can't get pregnant, to have children of their own．Cloning humans would also mean that organs could be cloned, so it would be a source of perfect transfer organs．This, surely would be greatly beneficial to millions of unfortunate people around the world that are expected to lose their lives due to failure of single （or more）organ （s）．It is also arguable that a ban on cloning may be unlawful and would rob people of the right to reproduce and limit the freedom of scientists．Arguments against cloning are also on a perfectly practical side．Primarily, I believe that cloning would step in the normal "cycle" of life．There would be a large number of same genes, which reduce the chances of improvement, and, in turn, development—the fundamental reason how living things naturally adapt to the ever-changing environment．Life processes failing to do so might result in untimely disappearance．Furthermore, cloning would make the uniqueness that each one of us possesses disappear．Thus, leading to creation of genetically engineered groups of people for specific purposes and, chances are, that those individuals would beregarded as "objects" rather than people in the society．Scientists haven't 100 per cent．guaranteed that the first cloned humans will be normal．Thus this could result in introduction of additional limitations in the human "gene-pool"．Regarding such arguable topics in "black or white" approach seems very innocent to me personally．We should rather try to look at all "shades" of it．I believe that cloning is only legal if its purpose is for cloning organs; not humans．Then we could regard this as for "saving life" instead of "creating life"．I believe cloning humans is morally and socially unacceptable．67．Which of the following is true according to the passage?A．Genetic limitation will be beneficial for some women．B．A large number of genes will prevent us from developing．C．Prohibition of cloning might limit the freedom of scientists．D．First cloned humans might be normal according to scientists．68．What's the author's opinion on cloning?A．Cloning should be entirely banned．B．Cloning should be used in creating life．C．Cloning will take away the right to reproduce．D．Cloning is acceptable if it is used for cloning organs．69．Where can you read this article?A．In a story book．B．In a magazine．C．In a science fiction．D．In a brochure．70．Which of the following shows thestructure of the passage?第二节（共5小题；每小题2分，共10分）On a typical hot August day in Xianyou County, Fujian Province, Zeng Demei, a retired worker in his seventies, hurries down a busy street．In his hand is a black leather bag．Zeng opens his bag, taking out two forms．71 Each of the forms contains detailed information of a student．On his arrival two hours later a woman greets him and leads him to her office where another man is waiting．They are the two village officials．They inspect the forms handed to them by Zeng and immediately recognize the girls．72 ? "It’s a pity but it doesn't matter．" says Zeng, who wastes no time in deciding to look for the remaining child, Su Qiuju．After half an hour, they stop outside a small house made of mud brick．A middle-aged man and a girl in a faded pink dress greet them．Su Qiuju is eight years old．She was forced to drop out of school after both her parents died．She is now living with her uncle who cannot afford his own children's education．However, the year of education Su Qiuju did complete was a successful one．73 ．When they are about to leave, Zeng says, "I must find a supporter for this girl to sponsor her education．" Zeng has made it his retirement task to help children complete their schooling．Back in 1999, Zeng took part in a campaign started by the local women's organization to help students from poor families．74 His task had begun and since then he has spent his time persuading his friends and neighbors and others to donate money．"To me, children's education is the most important．75 I have to find sufficient funding before the school opens in September．"When asked how long he will keep up his vital work as the community's guardian angel, he has a simple reply．" Not until my eyes can't see, and my feet can't move．"A．They were having problems with their schoolworkB．These are for the two girls he's going to visit this morningC．They live in a small village not very far, though only one of the girls is still living at homeD．She displayed a talent for handwriting, writing her three-character name neatly and beautifullyE．The thought of students dropping out of school bothers me so much that I can't get to sleep at nightF．Of course, some people question why I would want to give up my retirement to go to so much trouble G．He was so overcome by the tough situation of many poor children ?that he donated all his money to help out a girl第四部分：书面表达（共两节，35分）第一节：情景作文（20分）假设你是红星中学高三（1）班的班长李华，你们全班同学在“五一节”放假期间开展了“体验一天低碳生活”的活动。

北京市海淀区中国人民大学附属中学2024-2025学年九年级上学期开学考试数学试题一、单选题1．在当地时间7月27日结束的巴黎奥运会10米气步枪混合团体比赛中，中国选手黄雨婷/盛李豪夺得本届奥运会首枚金牌，右图是巴黎奥运会射击项目图标，这个图案的对称轴条数为（ ）A ．6B ．4C ．2D ．12．如图，直线a b P ，直线l 与a b ，分别交于点A B ，，过点A 作AC b ⊥于点C ．若155∠=°，则2∠的大小为（ ）A ．35︒B ．45︒C ．55︒D ．125︒3．a ，b 在数轴上的对应点的位置如图所示，则正确的结论是（ ）A ．2a >-B ．a b <C ．0ab >D ．a b <- 4．2024年第33届巴黎奥运会是史上第一届男女比例完全平衡的奥运会，参赛的男女运动员分别为5250，5250名，本届奥运会的运动员总数用科学记数法表示为（ ） A ．35.2510⨯ B ．45.2510⨯ C ．.41510⨯ D ．41.0510⨯ 5．把抛物线23y x =向左平移2个单位长度，再向上平移5个单位长度，得到的抛物线的解析式为（ ）A ．2)3(25y x =+-B ．23(5)2y x =++C ．23(2)5y x =-+D ．23(2)5y x =++6．如图，在点M N P Q ，，，中，一次函数2(0)y kx k =->的图象可能经过的点是（ ）A ．点MB ．点NC ．点PD ．点Q7．当1x =时，下列各式中有意义的是（ ）A ．31x -BC ．()012x -D ．212x x x-÷ 8．在马拉松、公路自行车等耐力运动的训练或比赛中，为合理分配体能，运动员通常会记录“配速”，即每行进1km 所用的时间（单位：min ）．小宇参加5km 的公路自行车骑行训练，他骑行的“配速”情况如图所示，下列说法①第1km 所用的时间最长；②第5km 的平均速度最大；③前3km 的平均速度大于最后2km 的平均速度；所有正确说法的序号是（ ）A ．①②B ．①③C ．②③D ．①②③二、填空题9．计算：1112-⎛⎫+= ⎪⎝⎭．10．一个正五边形的外角和为︒．11．分解因式：3a 2﹣12=．12．某工厂加工了一批共360个工件，质检员小字从中随机抽取了12个工件检测了它们的质量（单位：g ），得到的数据如下：31.02 30.97 31.05 30.99 31.02 31.0530.98 31.02 30.97 31.01 30.96 31.01当一个工件的质量x （单位：g ）满足：30.9731.03x ≤≤时，评定该工件为一等品，根据以上数据，估计这一批工件中一等品的个数是．13．如图，在3×3的正方形网格中，每个小正方形边长为1，点A ，B ，C 均为格点，以点A 为圆心，AB 长为半径作弧，交格线于点D ，则CD 的长为．14．某乡镇下设有六个村庄，村庄之间有道路相通，如图所示，图中的黑线即代表村庄间连通的道路，道路上标志的数字为该道路的长度（单位：千米），小宇要为该乡镇设计自来水管道线路，为了铺设及检修方便，所有的自来水管道均要沿着村庄间的道路铺设，且要求六个村庄都能通过管道相连．请回答：所铺设自来水管道总长度的最小值为千米．三、解答题15．如图，在ABC V 中，90ACB AD AB ∠=︒⊥，且AD AB =，点E 在AC 上，且AE BC =，连接DE ．求证：DE AC ⊥．16．已知关于x 的一元二次方程20x mx n ++=．(1)当2,5m n ==-时，求方程的根；(2)当2m n =+时，求证：方程有两个不相等的实数根．17．在平面直角坐标系xOy 中，抛物线2y x bx c =-++经过点()()3,0,0,3A B -．(1)求此抛物线的解析式；(2)在坐标系中画出这条抛物线（不用列表）；(3)过点(),0P n 作x 轴的垂线，分别交抛物线于点M ，交直线AB 于点N ，记点M 的纵坐标为M y ，点N 的纵坐标为N y ，若M N y y <，结合图象，直接写出n 的取值范围．18．小宇要对一幅书法作品进行装裱，装裱后如图所示，上、下空白处分别称为天头和地头，左、右空白处统称为边，已知原作品的长为60cm ，宽为24cm ，在装裱后左右两边的边宽相等，天头长与地头长也相等，且均为一边宽的5倍，如果在装裱后，原作品的面积恰好是装裱后作品总面积的920，那么装裱后左右两边的边宽分别是多少？19．如图1是一个轨道的示意图，其中四边形ABCD 为菱形，边长2m,120AB ABC =∠=︒，对角线AC 与BD 交于点O ，在此菱形的四条边及对角线上均装有轨道，同时在点B 处安装了一台观测仪．小宇操作机器人以1m/min 的速度沿轨道匀速运动，机器人从点B 出发，依照设定的顺序分别经过O ，C ，D 三点各一次并最终到达点A ．记机器人运动的时间为min x ，机器人到观测仪的距离为m y ，机器人在轨道中转弯所用时间忽略不计．在机器人运动结束后，小宇发现观测仪出现故障，只得到了部分观测结果．经整理后，观测仪中所记录的y 与x 的函数关系的部分对应值如表1所示，其部分函数图象如图2所示．表1根据上述信息回答：(1)机器人的运动路线是：B→______→______→______→A（请选填“O”，“C”，“D”）；(2)补全图2中的函数图象；(3)a=______，b=______．20．巴黎奥运会男子50米步枪三姿决赛于当地时间8月1日上午结束，中国运动员刘宇坤不负众望，最终夺冠，小宇观看了比赛的直播，并记录和分析了比赛数据，得到如下信息：a．决赛共有8名选手参加，先后进行跪姿、卧姿、立姿三种姿势的射击，具体规则为：b．基础射击结束后8名选手的三种姿势平均成绩如下表所示c．决赛结束后，最终获得前三名的选手恰好是基础射击中立姿平均成绩排名前三的选手，且他们最终的排名顺序与他们跪姿的排名顺序一致．这三人单发淘汰赛的成绩如下表所示d ．中国选手刘宇坤在决赛中全部15发立姿射击的总环数为152.1环．根据上述信息回答：(1)从基础射击的平均成绩来看，在这三种姿势中，平均成绩最好的姿势是______，选手之间成绩差异最大的姿势是______；（两空均选填“跑姿”，“卧姿”或“立姿”）(2)在基础射击中，这8名选手立姿平均成绩的中位数为______；(3)在决赛中最终获得前三名的选手分别是：第一名______，第二名______，第三名______；（三空均从~A H 中选填）(4)m 的值为______．21．有这样一个问题：如图1，在矩形ABCD 中，3,30AB ADB =∠=︒，点E ，F 在对角线BD 上，满足BE BF <，点M ，N 分别在线段AD ，BC 上，连接EM ，FM ，EN ，FN ，设EF α=，当a 取何值时，存在M 、N ，使得四边形EMFN 是正方形？小宇为了解决这个问题，进行了如下探究，请补充完整：假设符合题意的正方形存在，(1)画出示意图．．．，如图2，由于四边形EMFN 是正方形，那么它一定是平行四边形，由平行四边形的性质①______（填依据），可知,EO OF MO ON ==，结合ABCD 是矩形，可得BON DOM ≌△△，于是BO DO =，因此，四边形EMFN 的对角线交点恰好是BD 的中点，如图3所示．(2)在图3的基础上，由于EMFN 是正方形，那么它还同时是菱形和矩形．于是由菱形的性质②______（填依据），可得MN EF ⊥于O ，于是MN 垂直平分BD ；又由矩形的性质可得OM ON OE OF ===，这样就能够确定点E ，F ，M ，N 的位置了．(3)根据（1）（2）的分析，在图4中作出正方形EMFN （尺规作图，保留作图痕迹）；(4)结合上述的探索，小宇发现符合题意的正方形EMFN 是唯一的，此时a 的值为______； 解决问题后，小宇又有了进一步的思考：(5)若将原问题改为：当a 取何值时，存在M ，N ，使得四边形EMFN 为矩形？请参照上面的思考，直接写出a 的最小值．22．如图，在ABC V 中，AB BC =，D ，E 分别是AB ，AC 的中点，AF DE ∥，EF AD ∥．(1)求证：四边形ADEF 是菱形；(2)连接CF ，若10,12AB AC ==，求CF 的长．23．在平面直角坐标系xOy 中，直线l ：y kx =与直线y x k =-+交于点A ，直线y x k =-+与x 轴交于点B ．(1)求点B 的坐标（用含k 的代数式表示）；(2)横、纵坐标都是整数的点叫做整点．将AOB V 内（不含边界）的整点个数记为m ， ①当4k =时，结合函数图象，直接写出m 的值；②若1m =，直接写出k 的取值范围．24．在平面直角坐标系xOy 中，已知二次函数()20y ax bx c a =++≠的图象经过不重合的三点()()()1,0,1,,,2A B m C n -，其对称轴为直线x t =．(1)若1,0==m n ，则a ______0（填“>”或“<”）；(2)若2,1m t ==-，求此时二次函数的解析式；(3)当0a >时，对于某个n ，若存在12m <<，使得10t -<<成立，结合图象，直接写出n 的取值范围．25．如图1，四边形ABCD 是平行四边形，AC 为对角线，45ACB ∠=︒，过点D 作AC 的垂线，分别交直线AC BC ，于E F ，，连接AF ．(1)设BAC α∠=，求BAF ∠的度数（用含α的式子表示）；(2)过点B 作AF 的垂线，分别交直线AC AF ，于点M N ，，①依题意补全图形；②用等式表示AM BF DE ，，的数量关系，并证明．26．在平面直角坐标系xOy 中，对于相交的直线1l ，2l 和图形W ，给出如下定义：如果在图形W 上存在两个不重合的点M ，N ，使得点M 到直线1l 的距离与点N 到直线2l 的距离相等，则称图形W 是直线1l ，2l 的“相合图形”．如图1，直线1l ，2l 交于点P ，三角形W 是直线1l ，2l 的“相合图形”(1)已知点()()()1,2,22A B m m ≠，，线段AB 上任一点到x 轴的距离为______，若线段AB 是x 轴，y 轴的“相合图形”，写出一个m 的值为______；(2)点C ，D 在直线4y x =+上，点C 在点D 左侧且CD =CD 是直线1x =，x 轴的“相合图形”，直接写出点C 的横坐标c x 的取值范围；(3)直线22y x =-与x 轴，y 轴分别交于E ，F 两点，边长为2的正方形Γ的四条边分别与两坐标轴垂直，其中心T 在直线142y x =+上，若在线段EF 上存在点(),m n ，使得正方形Γ是直线x m y n ==，的“相合图形”，直接写出点T 的横坐标t 的取值范围．。