广东省六校(广州二中深圳实验珠海一中中山纪念)2018届高三下学期第三次联考数学(文)试题Word版含答案
广东省六校 2018届高三下学期第三次联考数学(理)试题(教师版)
2018届广东省六校第三次联考理科数学一、选择题：本题共12小题，每小题5分，共60分．在每小题给出的四个选项中，只有一项是符合题目要求的．1.已知集合(){}22,|,,2M x y x y xy =+=为实数且，(){},|,,2N x y x y x y =+=为实数且，则M N ⋂的元素个数为（ ） A. 0 B. 1C. 2D. 3【答案】B 【解析】 【分析】：集合M 与集合N 表示的集合都是点集，所以可以把两个方程联立，通过求方程的判别式来判定交点的个数． 【详解】：联立方程组2222x y x y ⎧+=⎨+=⎩ 所以2210x x -+=判别式0∆= ，所以M N ⋂ 的解集只有一个． 故选B【点睛】：本题考查了两个集合的交点个数问题，主要注意两个集合都为点集，所以交集的个数也就是两个方程的解的个数，因此可以通过方程思想来解，属于简单题．2.设等差数列{}n a 的前n 项和为n S ，若39S =，530S =，则789a a a ++= A. 63 B. 45C. 36D. 27【答案】A 【解析】由题意3239S a ==，23a =，53530S a ==，36a =，∴32633d a a =-=-=，12330a a d =-=-=，7898133(7)3(073)63a a a a a d ++==+=⨯+⨯=，故选A ．3.若变量,x y 满足约束条件0210430y x y x y ≤⎧⎪--≥⎨⎪--≤⎩，则35z x y =+的取值范围是A. [)3,+∞ B. []8,3-C. (],9-∞D. []8,9-【答案】D 【解析】画出不等式组表示的可行域（如图阴影部分所示）．由35z x y =+得355z y x =-+，平移直线355zy x =-+，结合图形可得，当直线经过可行域内的点A 时，直线在y 轴上的截距最大，此时z 取得最大值，由题意得点A 的坐标为（3,0），∴max 339z =⨯=．当直线经过可行域内的点B 时，直线在y 轴上的截距最小，此时z 取得最小值，由210430x y x y --=⎧⎨--=⎩，解得11x y =-⎧⎨=-⎩，故点B 的坐标为(1,1)--，∴min 3(1)5(1)8z =⨯-+⨯-=-．综上可得89z -≤≤，故35z x y =+的取值范围是[8,9]-．选D ． 4.函数1ln sin 1ln xy x x-=⋅+的图象大致为( )A. B. C. D.【答案】A 【解析】 设1ln ()sin 1ln xf x x x -=⋅+，由1ln 0x +≠得1x e ≠±，则函数的定义域为1111(,)(,)(,)e e e e-∞-⋃-⋃+∞．∵1ln 1ln ()sin()sin ()1ln 1ln x x f x x x f x xx----=⋅-=-⋅=-+-+，∴函数()f x 为奇函数，排除D ． 又11e>，且(1)sin1>0f =，故可排除B ． 211e e<，且2222211ln11(2)11()sin sin 3sin 01121ln e f x e e e e---=⋅=⋅=-⋅<-+，故可排除C ．选A ． 5.设函数())f x ϕ=+，其中常数ϕ满足0πϕ-<<．若函数()()()g x f x f x '=+（其中()f x '是函数()f x 的导数）是偶函数，则ϕ等于A. 3π-B. 56π-C. 6π-D. 23π-【答案】A 【解析】由题意得()()()))g x f x f x ϕϕ=+'=++)3πϕ=++，∵函数()g x 为偶函数， ∴,3k k Z πϕπ+=∈．又0πϕ-<<， ∴3πϕ=-．选A ．6.执行如图的程序框图，如果输入的,,a b k 分别为1,2,3，输出的158M =，那么判断框中应填入的条件为（ ）A. n k <B. n k ≥C. 1n k <+D. 1n k ≥+【答案】C 【解析】分析：直接按照程序运行即可找到答案. 详解：依次执行程序框图中的程序，可得：①1331,2,,2222M a b n =+====，满足条件，继续运行； ②28382,,,33323M a b n =+====，满足条件，继续运行；③3315815,,,428838M a b n =+====，不满足条件，停止运行，输出158．故判断框内应填n ＜4，即n ＜k+1． 故选C ．点睛：本题主要考查程序框图和判断框条件，属于基础题,直接按照程序运行，一般都可以找到答案.7.已知02012(1)(2)(2)(2)(2)n n n i b i b i b i b i -+=-++-++-+++-+ （2n ≥，i 为虚数单位），又数列{}n a 满足：当1n =时， 12a =-；当2n ≥，n a 为22(2)b i -+的虚部．若数列2n a ⎧⎫-⎨⎬⎩⎭的前n 项和为n S ，则2018S =A.20172018B.20182017C.40352018D.40332017【答案】C 【解析】由题意得1(1)[1(2)](2),0,1,,n n r rr n i i T C i r n +-+=+-+=-+=，∴当2n ≥时，22(1)2n n n b C -==， 又 ()222222(34)34b i b i b b i -+=-=-， 故当2n ≥时，242(1)n a b n n =-=--， ∴当2n ≥时，221112(1)(1)1n a n n n n n n--===-----． ∴201811111140351(1)()()1(1)2232017201820182018S =+-+-++-=+-=．选C ． 8.如图，在同一个平面内，三个单位向量,,OA OB OC 满足条件：OA 与OC 的夹角为α，且t a n 7α=，OB与OC 与的夹角为45°.若(),OCmOA nOB m n R =+∈，则m n +的值为( )A. 3 C. D.2【答案】B 【解析】建立如图所示的平面直角坐标系，由tan 7α=知α为锐角，且sin 1010αα==，故3cos(45)5α+︒=-， 4sin(45)5α+︒=．∴点B,C的坐标为34(,),(551010-，∴342(,),(,551010OB OC =-=． 又OC mOA nOB =+， ∴27234(,)(,)(1,0)101055m n=-+， ∴354510mn n ⎧-+=⎪⎪⎨⎪=⎪⎩，解得8m n ⎧=⎪⎪⎨⎪=⎪⎩，∴882m n +=+=．选B ． 9.四面体S ABC -中，三组对棱的长分别相等，依次为5，4，x ，则x的取值范围是 A. B. (3,9)C.D. (2,9)【答案】C 【解析】【详解】由于四面体的三组对棱分别相等，故可构造在长方体内的三棱锥P ABC -(如图所示)，其中5,4,PA BC PC AB PB AC x ======．设长方体的三条棱长分别为,,a b c ，则有22222222516a b x a c c b ⎧+=⎪+=⎨⎪+=⎩①②③．（1）由②-③得22a b 9-=，又222a b x +=， ∴22290b x =->，解得3x >．（2）由②+③得222241a b c ++=，又222a b x +=，∴222410c x =->，解得x <综上可得3x <<x 的取值范围是．选C ．点睛：由于长方体的特殊性，因此解题时构造长方体中的四面体是解答本题的关键，借助几何模型使得解题过程顺利完成，这也是解答立体几何问题的常用方法．10.从2个不同的红球、2个不同的黄球、2个不同的蓝球共六个球中任取2个，放入红、黄、蓝色的三个袋子中，每个袋子至多放入一个球，且球色与袋色不同，那么不同的放法有( ) A. 42种 B. 36种 C. 72种 D. 46种【答案】A 【解析】 分以下几种情况：①取出的两球同色，有3种可能，取出球后则只能将两球放在不同色的袋子中，则共有22A 种不同的方法，故不同的放法有2236A =种．②取出的两球不同色时，有一红一黄、一红一蓝、一黄一蓝3种取法，由于球不同，所以取球的方法数为1122312C C =种；取球后将两球放在袋子中的方法数有2213A +=种，所以不同的放法有12336⨯=种．综上可得不同的放法有42种．选A ．11.已知点F 为双曲线()2222:1,0x y E a b a b -=>的右焦点，直线(0)y kx k =>与E 交于,M N 两点，若MF NF ⊥，设MNF β∠=，且126ππβ⎡⎤∈⎢⎥⎣⎦，，则该双曲线的离心率的取值范围是（ ）A.B. 1⎤⎦C. ⎡⎣D. 1⎡⎤⎣⎦【答案】D 【解析】如图，设双曲线的左焦点为F '，连,MF NF ''．由于MF NF⊥，所以四边形F NFM '为矩形，故2MN FF c '==．在Rt NFM ∆中，2cos ,2sin FN c FM c ββ==，由双曲线的定义可得22cos 2sin a NF NF NF FM c c ββ=-='-=-cos()4πβ=+，∴1cos()4c e a πβ==+． ∵126ππβ≤≤，∴53412πππβ≤+≤，cos()42πβ≤+≤，1e ≤．即双曲线的离心率的取值范围是1]．选D ． 点睛：求双曲线的离心率时，将提供的双曲线的几何关系转化为关于双曲线基本量,,a b c 的方程或不等式，利用222b c a ＝－和e=ca转化为关于e 的方程或不等式，通过解方程或不等式求得离心率的值或取值范围． 12.已知()()1122,,A x y B x y 、是函数()ln x f x x =与()2kg x x =图象的两个不同的交点，则()12f x x +的取值范围是( ) A. 2ln ,2e e ⎛⎫+∞⎪⎝⎭B. 21ln ,2e e e ⎛⎫⎪⎝⎭ C. 10e ⎛⎫ ⎪⎝⎭，D. 2ln ,02e e ⎛⎫⎪⎝⎭ 【答案】D 【解析】 由2ln x kx x=得ln x x k =，设()ln (0)h x x x x =>，则()1ln h x x =+'，∴当10x e <<时函数单调递减，当1x e >时函数单调递增，故min 1()h x e=-．由题意得12,x x （令12x x <）是函数()y h x =图象与直线y k =的两个交点的横坐标，即12()()h x h x =，结合图象可得12101x x e <<<<． 设21()()()()x h x h x x e e ϕ=-->，则22()()()(1ln )[1ln()]0x h x h x x x e eϕ=+-=+++-''>'，∴()x ϕ在1(,)e+∞上单调递增，∴1121()()()()0x h h e e e e ϕϕ>=--=， ∴21()()()h x h x x e e >->．∴222()()h x h x e >-，∴122()()h x h x e >-∵21x e >，故221x e e -<，且()h x 在1(0,)e 上单调递减，∴122x x e >-，即122x x e +>．由ln ()(0)x f x x x =>，得21ln ()x f x x-'=，故()f x 在(0,)e 上单调递增． ∴122e 2()()ln 2f x x f e e+>=．设ln ()(01)1xt x x x =<<-，可得函数()t x 在(0,1)上单调递减， ∴12()()t x t x >，即1212ln ln 11x x x x >--， 又1122ln ln x x x x =，∴121212ln 1ln 1x x x x x x -=>-， ∴222211x x x x -<-，即2121()(1)0x x x x -+-<，∴211x x +<， ∴12()0f x x +<． 综上可得12e 2ln ()02f x x e <+<，即所求范围为e 2(ln ,0)2e．选D ．二、填空题：本题共4小题，每小题5分，共20分．13.已知()f x 是定义在R 上的奇函数，则311[(2)]f x dx x-+=⎰_____；【答案】ln3 , 【解析】33311111311[(2)](2)()ln 1f x dx f x dx dx f t dt x x x --+=-+=+⎰⎰⎰⎰ln3=． 14.已知函数()sin cos f x a x b x =-，若ππ44f x f x ⎛⎫⎛⎫-=+ ⎪ ⎪⎝⎭⎝⎭，则函数13ax b y ++=的图象恒过定点_____． 【答案】(1,3) 【解析】 由题意4x π=是()f x图象的一条对称轴，∴())4f a b π=-=0a b +=，因此在13ax b y ++=中令1a =，则133a b y ++==，即过定点(1,3)．15.已知几何体的三视图如图所示，其中俯视图为一正方形，则该几何体的表面积为__________．【答案】2 【解析】由三四图可得，该几何体为如图所示的三棱锥1D ABC -．∵正方体的棱长为2，∴11AC CD D A ===∴111221122,222ABC ABD BCD ACD S S S S ∆∆∆∆=⨯===⨯⨯=== ∴该几何体的表面积为2+． 答案：2+16.若函数()f x 的图象上存在不同的两点11(,)A x y ，22(,)B x y ，其中1122,,,x y x y 使得1212x x y y +的最大值为0，则称函数()f x 是“柯西函数”．给出下列函数： ①()ln (03)f x x x =<<； ②1()(0)f x x x x=+>； ③()f x =④()f x =其中是“柯西函数”的为 ___．（填上所有正确答案的序号） 【答案】① ④ 【解析】设()()1122,,,OA x y OB x y ==，由向量的数量积的可得||||||OA OB OA OB ⋅≤⋅，当且仅当向量OA OB ，共线（,,O A B 三点共线）时等号成立．故1212122x x y y y x y +⋅+的最大值为0时，当且仅当,,O A B 三点共线时成立．所以函数()f x 是“柯西函数”等价于函数()f x 的图象上存在不同的两点,A B ，使得,,O A B 三点共线． 对于①，函数()ln (03)f x x x =<<图象上不存在满足题意的点； 对于②，函数()1(0)f x x x x=+>图象上存在满足题意的点；对于③，函数()f x =图象上存在满足题意的点；对于④，函数()f x =故函数① ④是“柯西函数”． 答案：① ④点睛：（1）本题属于新定义问题，读懂题意是解题的关键，因此在解题时得到“柯西函数”即为图象上存在两点A,B ，使得O,A,B 三点共线是至关重要的，也是解题的突破口． （2）数形结合是解答本题的工具，借助于图形可使得解答过程变得直观形象．三、解答题：共70分．解答应写出文字说明、证明过程或演算步骤．第1721题为必考题，每个试题考生都必须作答．第22、23题为选考题，考生根据要求作答． （一）必考题：共60分.17.设数列{}n a 的前n 项和为n S ，数列{}n S 的前n 项和为n T ，满足22,n n T S n n N *=-∈.(Ⅰ)求123,,a a a 的值； (Ⅱ)求数列{}n a 的通项公式.【答案】(Ⅰ)11a =，24a =，310a =；(Ⅱ)1322n n a -=⨯-.【解析】 试题分析：(Ⅰ)在22n n T S n =-中，分别令1,2,3n =可得到123,,S S S ，然后可得到123,,a a a 的值．(Ⅱ)先由1n n T T --得到221(*)n n S a n n N =-+∈，再由1n n S S --可得122(2)n n a a n -=+≥，故可得122(2)(2)n n a a n -+=+≥，因此得到数列{2}n a +为等比数列，由此可求得数列{}n a 的通项公式．试题解析： （Ⅰ）∵，，∴；∵，∴；∵，∴．（Ⅱ）∵… ①，∴…②，∴①-②得，，又也满足上式，∴…③，∴…④，③-④得，∴．又，∴数列是首项为3，公比为的等比数列．∴，∴．点睛：数列的通项a n与前n项和S n的关系是11,1,2nn nS naS S n-=⎧=⎨-≥⎩．在应用此结论解题时要注意：若当n＝1时，a1若适合1n nS S--，则n＝1的情况可并入n≥2时的通项an；当n＝1时，a1若不适合1n nS S--，则用分段函数的形式表示．18.某小店每天以每份5元的价格从食品厂购进若干份食品，然后以每份10元的价格出售.如果当天卖不完，剩下的食品还可以每份1元的价格退回食品厂处理.(Ⅰ)若小店一天购进16份，求当天的利润y （单位：元）关于当天需求量n （单位：份，n N ∈）的函数解析式；(Ⅱ)小店记录了100天这种食品的日需求量（单位：份），整理得下表：以100天记录的各需求量的频率作为各需求量发生的概率.(i)小店一天购进16份这种食品，X 表示当天的利润（单位：元），求X 的分布列及数学期望； (ii)以小店当天利润的期望值为决策依据，你认为一天应购进食品16份还是17份？ 【答案】(Ⅰ)()964,1680,16n n y n N n -<⎧=∈⎨≥⎩；(Ⅱ)(i)答案见解析；(ii)17份.【解析】 试题分析：(Ⅰ) 分16n ≥和16n <两种情况分别求得利润，写成分段的形式即可得到所求．(Ⅱ)(i) 由题意知X 的所有可能的取值为62，71，80，分别求出相应的概率可得分布列和期望； (ii)由题意得小店一天购进17份食品时，利润Y 的所有可能取值为58，67,76,85，分别求得概率后可得Y 的分布列和期望，比较()()E X E Y 和的大小可得选择的结论．试题解析： （Ⅰ）当日需求量时，利润，当日需求量时，利润，所以关于的函数解析式为．（Ⅱ）（i ）由题意知的所有可能的取值为62，71，80， 并且，，．∴的分布列为：∴元．（ii ）若小店一天购进17份食品，表示当天的利润(单位：元)，那么的分布列为∴的数学期望为元．由以上的计算结果可以看出，即购进17份食品时的平均利润大于购进16份时的平均利润． ∴所以小店应选择一天购进17份．19.如图，在四棱锥P ABCD -中，ABCD 是平行四边形，1AB BC ==，120BAD ∠=，PB PC ==2PA =，E ，F 分别是AD ，PD 的中点．（Ⅰ）证明：平面EFC ⊥平面PBC ； （Ⅱ）求二面角A BC P --的余弦值．【答案】（1）见解析（2） 【解析】 试题分析：（Ⅰ）运用几何法和坐标法两种方法进行证明可得结论．（Ⅱ）运用几何法和坐标法两种方法求解，利用坐标法求解时，在得到两平面法向量夹角余弦值的基础上，通过图形判断出二面角的大小，最后才能得到结论．试题解析：解法一：（Ⅰ）取中点，连，∵，∴，∵是平行四边形，，，∴，∴是等边三角形，∴，∵，∴平面，∴.∵分别是的中点，∴∥，∥，∴，，∵，∴平面，∵平面，∴平面平面.（Ⅱ）由（Ⅰ）知，，∴是二面角的平面角., ，，在中，根据余弦定理得，∴二面角的余弦值为．解法二：（Ⅰ）∵是平行四边形，，，∴，∴是等边三角形，∵是的中点，∴，∵∥，∴.以为坐标原点，建立如图所示的空间直角坐标系.则，，，，，设，由，，可得，，，∴，∵是的中点，∴，∵0⋅=，CB CF∴，∵，，∴平面，∵平面，∴平面平面.（Ⅱ）由（Ⅰ）知，，．设是平面的法向量，由031022n CB y nCP x y z ⎧⋅==⎪⎨⋅=-++=⎪⎩，得0y zx =⎧⎪⎨=⎪⎩， 令2x =，则(2,0,3)n =． 又是平面的法向量，∴3cos ,77m n m n m n ⋅===-⋅，由图形知二面角A BC P --钝角，∴二面角的余弦值为. 20.已知椭圆2222:1(0)x y C a b a b +=>>1A 、2A分别为椭圆C 的左、右顶点，点(2,1)P -满足121PA PA ⋅=． （Ⅰ）求椭圆C 的方程；（Ⅱ）设直线l 经过点P 且与C 交于不同的两点M 、N ，试问：在x 轴上是否存在点Q ，使得直线 QM 与直线QN 的斜率的和为定值？若存在，请求出点Q 的坐标及定值；若不存在，请说明理由．【答案】（1）2214x y += （2）(2,0)Q ，定值为1.【解析】 试题分析：（Ⅰ）由121PA PA ⋅=可得2a =，再根据离心率求得c =21b =，故可得椭圆的方程．（Ⅱ）由题意可得直线l 的斜率存在，设出直线方程后与椭圆方程联立消元后得到一元二次方程，求出直线 QM 与直线QN 的斜率，结合根与系数的关系可得QM QN k k +222(48)24(2)8(2)t k tt k t k t -+=-+-+，根据此式的特点可得当2t =时，QM QN k k +为定值．试题解析： （Ⅰ）依题意得、，，∴1=，解得．∵，∴，∴，故椭圆的方程为．（Ⅱ）假设存在满足条件的点.当直线与轴垂直时，它与椭圆只有一个交点，不满足题意. 因此直线的斜率存在，设直线的方程为，由消去整理得，设、，则，，∵222(48)24(2)8(2)t k tt k t k t -+=-+-+，∴要使对任意实数，为定值，则只有，此时．故在轴上存在点，使得直线与直线的斜率的和为定值．点睛：解决解析几何中定值问题的常用方法（1）从特殊入手，求出定值，再证明这个值与变量无关．（2）直接对所给要证明为定值的解析式进行推理、计算，并在计算推理的过程中消去变量得到常数，从而证明得到定值，这是解答类似问题的常用方法．21.已知函数2()(1)e 2xa f x x x =--，其中R a ∈． （Ⅰ）函数()f x 的图象能否与x 轴相切？若能，求出实数a ，若不能，请说明理由；（Ⅱ）求最大的整数a ，使得对任意12R,(0,)x x ∈∈+∞，不等式12122()()2f x x f x x x +-->- 恒成立．【答案】（1）不能（2）3 【解析】 试题分析：（Ⅰ）假设函数()f x 的图象能与x 轴相切．设切点为(,0)t ，根据导数的几何意义得到关于t 的方程，然后判断此方程是否有解即可得到结论．（Ⅱ）将不等式变形为()()()()12121212f x x x x f x x x x +++>-+-,设()()g x f x x =+，则问题等价于()()1212g x x g x x +>-对任意()12,0,x R x ∈∈+∞恒成立，故只需函数()()212xa g x x e x x =--+在R 上单调递增，因此()10x g x xe ax =-+≥'在R 上恒成立即可，由(1)10g e a -+'=≥可得1a e ≤+，即为()0g x '≥成立的必要条件，然后再证3a =时，310x xe x -+≥即可得到结论．试题解析：（Ⅰ）∵()()21e 2xa f x x x =--，∴．假设函数的图象与轴相切于点，则有， 即．显然，将代入方程中可得． ∵，∴方程无解．故无论a 取何值，函数的图象都不能与轴相切． （Ⅱ）由题意可得原不等式可化为， 故不等式在R 上恒成立．设，则上式等价于， 要使对任意恒成立， 只需函数在上单调递增，∴在上恒成立．则，解得，∴在上恒成立的必要条件是：． 下面证明：当时，恒成立．设，则，当时，，()h x 单调递减；当时，，()h x 单调递增．∴，即．则当时，，；当时，，． ∴恒成立．所以实数的最大整数值为3．点睛： （1）解决探索性问题时，可先假设结论成立，然后在此基础上进行推理，若得到矛盾，则假设不成立；若得不到矛盾，则假设成立．（2）解答本题的关键是构造函数()g x ，将问题转化为函数()g x 单调递增的问题处理，然后转化为()0g x '≥恒成立，可求得实数a 的值．（二）选考题：共10分. 请考生在第22、23题中任选一题作答. 如果多做，则按所做的第一题计分.22.已知直线l 的参数方程为cos sin x m t y t αα=+⎧⎨=⎩（t 为参数，0απ≤<），以坐标原点为极点，以x 轴正半轴为极轴建立极坐标系，曲线C 的极坐标方程为4cos ρθ=，射线,444πππθφφθφ⎛⎫=-<<=+ ⎪⎝⎭，4πθφ=-分别与曲线C 交于、、A B C 三点（不包括极点O ）. (Ⅰ)求证：OB OC OA +=； (Ⅱ)当12πφ=时，若B C 、两点在直线l 上，求m 与α的值.【答案】(Ⅰ)证明见解析；(Ⅱ)22,3m πα==. 【解析】试题分析： (Ⅰ)由曲线C 的极坐标方程可得点A B C 、、的极径，即得到,,OA OB OC ，计算后即可证得结论正确．(Ⅱ)根据12πφ=可求得点B,C 的极坐标，转化为直角坐标后可得直线BC 的直角坐标方程，结合方程可得m 与α的值．试题解析： （Ⅰ）证明：依题意，，，，则．（Ⅱ）当时，两点的极坐标分别为，， 故两点的直角坐标为，. 所以经过点的直线方程为，又直线经过点，倾斜角为， 故，．23.已知函数()222f x x a x a =+-+-.(Ⅰ)若()13f <，求实数a 的取值范围；(Ⅱ)若不等式()2f x ≥恒成立，求实数a 的取值范围. 【答案】(Ⅰ)2433⎛⎫- ⎪⎝⎭，；(Ⅱ)26,,55⎛⎫⎛⎫-∞-⋃+∞ ⎪ ⎪⎝⎭⎝⎭. 【解析】试题分析：（Ⅰ）由()13f <可得123a a +-<，根据分类讨论法解不等式组即可．（Ⅱ）根据绝对值的几何意义求得()f x 的最小值为(1)2a f -，由(1)22a f -≥可得实数a 的取值范围．试题解析： （Ⅰ）由可得，， ①当时，不等式化为，解得，∴； ② 当时，不等式化为，解得，∴； ③ 当时，不等式化为，解得，∴. 综上实数的取值范围是． （Ⅱ）由及绝对值的几何意义可得，当时，取得最小值．f x≥恒成立，∵不等式()2∴，即，解得或．∴实数的取值范围是.。
广东省六校2018届高三下学期第三次联考文综地理试题【解析】
广东省六校（广州二中，深圳实验，珠海一中，中山纪念，东莞中学，惠州一中）2018届高三下学期第三次联考文科综合地理试题第I卷（选择题）一、选择题海洋中上下层水温度的差异，蕴藏着一定的能量，叫做海水温差能，或称海洋热能。
利用海水表层（热源）和深层（冷源）之间的温度差发电的电站，叫海水温差发电站，可以连续性输出电力且伴生淡水。
1930年在法国首次试验成功，但当时发出的电能还不如耗去的电力多。
近年来世界各国海水温差发电的研究取得了实质性进展。
据此完成下面小题。
1．根据海水温差发电原理，下列海域最有利于海水温差发电的海域是A.、美国东海岸B.寒暖流交汇处C.地中海沿岸D.南纬20度到北纬20度的海洋洋面2．下列有关海水温差发电的说法，正确的是A.发电成本低B.海洋污染大C.发电量稳定D.能源总量小3．从深海抽取的冷海水营养成分丰富且无菌，有多种用途，深海水的利用不包括A.产制淡水B.冷冻、空调C.养殖、制药D.提炼锰结核黄土高原河流的径流和泥沙主要来源于几次大的暴雨过程。
延河流域分别在1977年7月和2013年7月发生了两次极端降水事件，而洪水过程及水沙特征表现差异较大。
读图和表，回答下面小题。
4．关于延河流域泥沙特征的描述正确的是A.1977年流域泥沙颗粒较大，泥沙颗粒越大数量越多B.2013年小粒径泥沙比重明显增加，泥沙颗粒明显变细C.2013年流域泥沙颗粒较小，0.01mm以下粒径泥沙最多D.2013较1977年，各粒径泥沙数量均有明显增加5．导致延河流域泥沙粒径变化的最可能原因是A.降雨量减少，河流径流量减少，侵蚀作用减弱B.流域地形平坦，流速缓慢，沉积作用强C.大量修筑淤地坝起到显著的拦沙效果D.修建梯田，破坏破面，水土流失加剧下图为南京市区某处路段的人工隧道示意图，该隧道是在开挖山体后又建造的，并未采用传统山体基部掏空的方法。
隧道长100米，顶高10米，隧道两侧是小山丘的相对高度为20米，隧道顶部与山体表面都有茂密的植被覆盖。
广东省六校(珠海一中,东莞中学,惠州一中)2018届高三下学期第三次联考文科综合历史试题+Word版含解析
广东省六校（广州二中，深圳实验，珠海一中，中山纪念，东莞中学，惠州一中）2018届高三下学期第三次联考文科综合历史试题1. 《左传》记载，公元前516年，齐景公问政于晏婴。
晏婴献策回答：“唯礼可以已之”，若行礼制则“民不迁，农不移，工贾不变，士不滥，官不滔，大夫不收公利”。
这表明先秦的“礼”是A. 治国安邦的重要措施B. 以人为本的民本思想C. 维护宗法分封的工具D. 强化等级秩序的手段【答案】A【解析】材料“若行礼制则民不迁，农不移，工贾不变，士不滥，官不滔”反映了礼制有利于稳定社会、治国安邦，故A正确；材料无法体现以人为本的民本思想，故B错误；材料未涉及维护宗法分封的问题，故C错误；材料也没有反映强化等级秩序，故D错误。
故选A。
点睛：本题解题的关键是紧扣材料关键信息“行礼制则民不迁，农不移，工贾不变，士不滥，官不滔”，学生应该结合所学知识从礼制与治国安邦的关系出发，即可排除无关选项，得出正确答案。
2. 下表反映了秦汉两朝不同时期对地方监察官的设立情况据此能够认定的历史事实是A. 秦汉时丞相负有地方监察职责B. 西汉时地方监察官的职权广泛C. 秦汉时对地方官吏的管理加强D. 秦汉时监察官官职高于地方官【答案】C【解析】表中四段材料并不能说明秦汉时丞相负有地方监察职责，故A错误；“遣御史监三辅郡……刺史，掌奉诏条察州”也不能说明西汉时地方监察官的职权广泛，故B错误；四段材料都反映了秦汉时派官对地方监察，说明对地方官吏的管理加强，故C正确；材料也不能表明秦汉时监察官官职高于地方官，故D错误。
故选C。
点睛：本题解题的关键是仔细观察表格内容，注意结合所学知识找出四段材料的共同信息，排除表述过于片面的或史实错误的选项，即可从“地方官吏的管理加强”的角度得出正确答案。
3. 唐宋很多皇帝各种宗教都提倡。
如唐玄宗亲自对《孝经》、《金刚经》和《道德经》三部书进行注释，并将注本颁行天下；南宋孝宗的《三教论》说：“以佛修心，以道养生，以儒治世”。
广东省六校2018届高三英语下学期第三次联考试题
广东省六校2018届高三英语下学期第三次联考试题试卷共10页，卷面满分120分，折算成135分计入总分。
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第一部分阅读理解（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，从短文后每题所给的A、B、C和D四个选项中，选出最佳选项，并在答题卡上将该项涂黑。
AQuestioning the existence of aliens is something that scientists have done for decades. In fact, most people do believe that aliens exist in some fashion. The main necessities for life are water and some form of energy source. Not surprisingly, there are some planets, exoplanets and moons that fit the bill. Here are several best chances at finding life in the universe.◆TRAPPIST-1TRAPPIST-1 is a planetary system a few dozen light-years away, whose discovery was announced in early 2017. This system consists of seven Earth-like exoplanets orbiting an "ultra-cool" star, and it is one of our shots at finding possible life beyond our own solar system.◆TitanTitan is the largest moon of Saturn, the sixth planet from our Sun. This moon could potentially harbor life but possibly not in the sense that we think. Titan does not exactly fit the description of being in a typical habitable zone.Titan has water, and it has liquid. It just doesn't have liquid water. The water on this moon is completely solid because of the extremely cold temperature.◆EuropaEuropa is one of Jupiter's moons here in our own solar system. It is another candidate due to its potential to hold liquid water. Europa is thought to have all the necessities for life including water, energy sources, and the right chemical build-up.◆MarsThe Red Planet, the fourth from the Sun, is probably one of the most talked-about potential candidates for extraterrestrial(地球外的) life and even for human colonization. Despite some different voices, finding extraterrestrial life on Mars really is a serious possibility.We know by now that we won't find little green men or any intelligent form of life that we understand. However, there is evidence that there was and may still be microscopic life on the small red planet.1. The common point of these celestial bodies is that _______.A. solid water exists on them respectivelyB. life might exist on each of themC. they all contain liquid waterD. each of them has living things on it2. Which celestial body could have life different from human imagination?A. TRAPPIST-1.B. Titan.C. Europa.D. Mars.3. Which of the following statements is TRUE?A. Europa is the most talked-about potential candidate for extraterrestrial life.B. TRAPPIST-1 is the only body that contains life beyond the solar system.C. All the scientists do believe that aliens exist in some manner.D. There is proof that life existed on the celestial body Mars.BAbout a month after I joined Facebook, I got a call from Lori Goler, a highlyregarded senior director of marketing at eBay. She made it clear this was a business call. “I want to apply to work with you at Facebook,” she said. “Instead of recommending myself, I want to ask you: What is your biggest problem, and how can I solve it?”My jaw hit the floor. I had hired thousands of people over the previous decade and no one had ever said anything remotely like that. People usually focus on finding the right role for themselves, with the implication that their skills will help the company. Lori put Facebook’s needs front and center. It was a killer approach. I responded, “Recruiting is my biggest problem. And, yes, you can solve it.”Lori never dreamed she would work in recruiting, but she jumped in. She even agreed to trade earnings for acquiring new skills in a new field. Lori did a great job running recruiting and within months was promoted to her current job, leading People@Facebook.The most common metaphor for careers is a ladder, but this concept no longer applies to most workers. As of 2010, the average American had eleven jobs from the ages of eighteen to forty-six alone. Lori often quotes Pattie Sellers, who came up with a much better metaphor: “Careers are a jungle gym, not a ladder.”As Lori describes it, there’s only one way to get to the top of a ladder, but there are many ways to get to the top of a jungle gym. The jungle gym model benefits everyone, but especially women who might be starting careers, switching careers, getting blocked by external barriers, or reentering the workforce after taking time off. The ability to create a unique path with occasional dips, detours (弯路), and even dead ends presents great views of many people, not just those at the top. On a ladder, most climbers are stuck staring at the butt of the person above.4. Why did Lori make the call?A. She helped Facebook to solve the biggest problem.B. She wanted to make a business deal with Facebook.C. She tried to ask for a pay rise in Facebook.D. She wanted to become an employee in Facebook.5. What impressed “I” by Lori?A. Lori was good at running recruiting.B. Lori attached great importance to Facebook’s needs.C. Lori jumped in Facebook with no adequate experience.D. Lori was skilled in marketing at eBay.6. What can we infer from the passage?A. Now all people don’t tend to climb the ladder.B. None on the ladder can enjoy the great views.C. Jungle gyms offer limited exploration for employees.D. A pregnant woman, jobless, benefits little from the jungle gyms.7. What is the best title of the passage?A. It’s a Jungle Gym, Not a Ladder.B. Facebook’s Biggest Problem.C. Applying for a Job in Facebook.D. A Jungle Gym is Better than a Ladder.CCareers in zoology are extremely varied and unique, and can provide incredible learning and work opportunities for anyone devoted to animal studies and welfare. Zoology, or the study of animals, is a wide field with many specialties, including research, conservation, veterinary (兽医的) medicine, and the care of animals. For people with a love of animals and some training, dreams of careers in zoology may come true.Some careers in zoology focus on research and scientific studies. These careers may allow those with a good theoretical science background to develop and run studies that improve human understanding of the animal world. Research in zoology can help create safer and more effective products for animals or can teach humans more about animal behavior in order to aid in conservation, breeding programs, and habitat preservation.Careers in zoology that focus on conservation attach great importance to the continued survival and increased protection of animal species. Conservationists may work with political groups or governments to help make laws to protect and preserveanimals, or may work in the field gathering information on potential threats to the health of global ecosystems. Some conservationists work in educational fields, trying to improve human efforts to save animals from extinction.Animals are subject to illness and injury, and some careers in zoology help to create a safe, stable animal population. Veterinary medicine is an important specialty field, and may take several years of intensive training to qualify as a certified veterinarian. While many veterinarians focus on the small-animal practice of domestic pets, vets in rural areas often work with large farm animals and more adventurous veterinarians may work with exotic species in zoos and wildlife preserves. .Humans love to observe animals, and modern-day zoos and wildlife preserves help meet that interest while providing facilities to assist with conservation programs. Zoology careers in zoos can range from overseeing breeding programs, to creating the proper diet for a deer, to cleaning the tiger’s cage. Many volunteer and entry-level jobs are available for people that love animals.8. What is the purpose of researching on animals in zoology?A. To help governments make laws.B. To improve researchers’ scientific knowledge.C. To create safe and effective products for humans.D. To educate people to protect animals.9. What can careers in zoology that focus on veterinary medicine help?A. Better understand animals.B. P rotect animals’ health.C. C lean the animals’ cages.D. Increase the number of animals.10. Which of the following shows the structure of the whole passage?(P1=paragraph 1; P2= paragraph 2; P3= paragraph 3; P4= paragraph 1; P5= paragraph 5;)11. What does the passage mainly present?A. Volunteers’ passion for zoology.B. A new idea of careers in zoology.C. Different types of careers in zoology.D. Effective measures of animal protection.DNEW YORK－A massive winter storm paralyzed much of the US East Coast on Thursday and Friday, dumping as much as 46 centimeters of snow from the Carolinas to Maine. It also caused flooding on the streets of Boston due to swelling storm tides, forced the cancellation of nearly 5,000 flights and closed businesses, offices and schools.Some meteorologists (气象学家) classified the storm as a "bomb cyclone" for its sharp drop in atmospheric pressure would be followed immediately by a blast of cold air that could break records in more than two dozen cities and bring wind chills as low as -40℃ during the weekend.From Baltimore, Maryland, to Caribou in Maine, efforts were underway to clear roadways of ice and snow as wind chill temperatures were to plunge during the day, reaching -40℃ in some parts after sundown, according to the National Weather Service.Utility companies across the East worked to repair downed power lines early on Friday as about 21,000 customers remained without electricity, down from almost 80,000 the day before, and issued warnings that temperatures may become dangerously low."If the temperature in your home begins to fall, we recommend taking shelter elsewhere until service can be restored. You can find warming centers by contacting local authorities," National Grid power company, which serves Massachusetts, said on Twitter.Airlines canceled 4,000 flights on Thursday and hundreds more on Friday, according to , an online tracking service.Mayor Bill de Blasio said it could feel like -30℃ on Friday and Saturday nights with the wind chill. "This is a serious, serious storm, and may be the most severein 2018," he said at a news conference.New York Governor Andrew Cuomo declared a state of emergency for the southern part of the state, while New Jersey Governor Chris Christie declared a state of emergency for several counties.12.Which of the following is NOT the consequence of the massive winter storm?A.Breakdown of the US coastB.Flooding in Boston.C.Cancellation of nearly 5,000 flights.D.Closure in business, offices, and schools.13.Why did some meteorologists classify this winter storm as “bomb cyclone”?A.Because it will cause changeable atmospheric pressure.B.Because the scale of this storm could break records in lots of cities.C.Because of the deadly and destructive effects of this storm.D.Because of the extremely cold air and the wind chills in this storm.14.Which of the following best explains the underlined word “plunge” in P3?A. decline steadilyB. drop dramaticallyC. fluctuate violentlyD. fall moderately15.About the rescue work, which of the following statements is INCORRECT?A.People were making great efforts to rid roadways of ice and snow.B.About 101 thousand people were left without electricity on Thursday.C.Residents can stay in warming centers until heating service is restored.D.Aviation service hasn't been brought back to normal yet until Friday.第二节 (共5小题；每小题2分，满分10分)根据短文内容，从短文后的选项中选出能填入空白处的最佳选项, 并在答题卡上将该项涂黑。
广东省六校2018届高三下学期第三次联考理科综合物理试题(精品解析版)
广东省六校（广州二中，深圳实验，珠海一中，中山纪念，东莞中学，惠州一中）2018届高三下学期第三次联考理科综合物理试题二、选择题1. 伽利略对自由落体运动的研究，是科学实验和逻辑思维的完美结合，如图所示，可大致表示其实验和思维的过程，对这一过程的分析，下列正确的是A. 伽利略认为自由落体运动的速度是均匀变化的，这是他用实验直接进行了验证的B. 其中丁图是实验现象，甲图是经过合理外推得到的结论C. 运用甲图实验，可“冲淡”重力的作用，更方便进行实验测量D. 运用丁图实验，可“放大”重力的作用，从而使实验现象更明显【答案】C2. 如图所示，水平细杆上套一环A，环A与球B间用一轻质绳相连，质量分别为m A、m B，由于球B受到水平风力作用，环A与球B一起向右匀速运动。
已知细绳与竖直方向的夹角为θ，则下列说法正确的是A. 环A受到的摩擦力大小为m B gtanθB. 风力增大时，轻质绳对球B的拉力保持不变C. 杆对环A的支持力随着风力的增大而增大D. 环A与水平细杆间的动摩擦因数为【答案】A【解析】以整体为研究对象，分析受力如图1所示，根据平衡条件得知：杆对环A的支持力N=（m A+m B）g，f=F；所以杆对环A的支持力保持不变．故C错误．以B球为研究对象，分析受力如图2所示，由平衡条件得到：轻质绳对球B的拉力T=，风力F=m B gtanθ，风力F增大时，θ增大，cosθ减小，T增大．故B错误．环A受到的摩擦力大小为f=F=m B gtanθ，选项A正确；由图1得到f=F，环A与水平细杆间的动摩擦因数为．故D错误．故选A.点睛：本题是两个物体的平衡问题，要灵活选择研究对象．几个物体的速度相同时，可以采用整体法和隔离法相结合的方法研究．3. 如图为某一物理量y随另一物理量x变化的函数图象，关于该图象与坐标轴所围面积（图中阴影部分）的物理意义，下列说法正确的是A. 若图象表示加速度随时间的变化，则面积等于质点在相应时间内的位移B. 若图象表示力随位置的变化，则面积等于该力在相应位移内所做的功C. 若图象表示电容器充电电流随时间的变化，则面积等于相应时间内电容器储存的电能D. 若图象表示电势随位置的变化，则面积等于电场在x0位置处的电场强度【答案】B【解析】若图象表示加速度随时间的变化，则根据∆v=at可知面积等于质点在相应时间内的速度变化量，选项A错误；若图象表示力随位置的变化．则根据W=Fx可知面积等于该力在相应位移内所做的功，选项B正确；若图象表示电容器充电电流随时间的变化，则根据q=It可知面积等于相应时间内电容器储存的电量，选项C错误；若图象表示电势随位置的变化，则图像在x0位置的切线的斜率表示电场在x0位置处的电场强度，面积无任何物理意义，选项D错误；故选B.4. 由于行星自转的影响，行星表面的重力加速度会随纬度的变化而有所不同。
广东省珠海一中等六校高三第三次联考数学理试题Word版含答案
2018届广东省六校第三次联考理科数学 第Ⅰ卷（共60分）一、选择题：本大题共12个小题,每小题5分,共60分.在每小题给出的四个选项中，只有一项是符合题目要求的.1.已知集合y x y x M ,|),{(=为实数，且}222=+y x ，y x y x N ,|),{(=为实数，且}2=+y x ，则N M 的元素个数为( )A ．0B ．1C ．2D ．32.设等差数列{}n a 的前n 项和为n S ，若30953==S S ，,则=++987a a a ( ) A ．63 B ．45 C ．36 D ．273.若变量y x ,满足约束条件⎪⎩⎪⎨⎧≤--≥--≤0340120y x y x y ，则y x z 53+=的取值范围是( )A ．[)∞+，3 B ．[]3,8- C ．(]9,∞- D ．[]9,8- 4.函数x x x y sin ||ln 1||ln 1⋅+-=的部分图象大致为( )A ．B ．C. D ．5.设函数()()ϕ+=x x f 3cos ，其中常数ϕ满足0<ϕ<π-.若函数)(')()(x f x f x g +=（其中)('x f 是函数)(x f 的导数）是偶函数，则ϕ等于( )A ．3π-B ．π-65 C. 6π- D ．32π- 6.执行下面的程序框图，如果输入的k b a ,,分别为1，2，3，输出的815=M ,那么，判断框中应填入的条件为( )A ．k n <B ．k n ≥ C.1+<k n D ．1+≤k n7.已知()()()()()nn ni b i b i b i b i +-+++-++-++-=+-2222122100 i n ,2≥（为虚数单位），又数列{}n a 满足：当1=n 时，21-=a ；当2≥n ，n a 为()222i b +-的虚部，若数列⎭⎬⎫⎩⎨⎧-n a 2的前n 项和为n S ，则=2018S ( ) A ．20182017 B ．20172018 C.20184035 D ．201740338.如图，在同一个平面内，三个单位向量OC OB OA ,,满足条件：与的夹角为α，且7tan =α，OB 与OC 与的夹角为45°.若()R n m OB n OA m OC ∈+=,，则n m +的值为( )A ．3B ．223 C.23 D ．229.四面体ABC S -中，三组对棱的长分别相等，依次为x ，，45，则x 的取值范围是( ) A ．()412， B ．()93， C. ()413， D ．()92，10.从2个不同的红球、2个不同的黄球、2个不同的篮球共六个球中任取2个，放入红、黄、蓝色的三个袋子中，每个袋子至多放入一个球，且球色与袋色不同，那么不同的放法有( ) A ．42种 B ．36种 C.72种 D ．46种11.已知点F 为双曲线()0,1:2222>=-b a by a x E 的右焦点，直线)0(>=k kx y 与E 交于NM ,两点，若NF MF ⊥,设β=∠MNF ,且⎥⎦⎤⎢⎣⎡ππ∈β612，，则该双曲线的离心率的取值范围是( ) A ．[]62,2+ B ．[]13,2+ C. []62,2+ D ．[]13,2+12.已知()()2211,,y x B y x A 、是函数()x x x f ln =与()2xkx g =图象的两个不同的交点，则()21x x f +的取值范围是( )A ．⎪⎭⎫⎝⎛+∞,2ln 2e e B ．⎪⎭⎫ ⎝⎛e e e 1,2ln 2 C.⎪⎭⎫ ⎝⎛e 10， D ．⎪⎭⎫⎝⎛0,2ln 2e e 第Ⅱ卷（共90分）二、填空题（每题5分，满分20分，将答案填在答题纸上） 13.已知函数)(x f y =是定义在R 上的奇函数，则()⎰=⎥⎦⎤⎢⎣⎡+-3112dx x x f ． 14.已知函数()x b x a x f cos sin -=，若⎪⎭⎫⎝⎛+π=⎪⎭⎫ ⎝⎛-πx f x f 44，则函数13++=b ax y 恒过定点 ．15.已知几何体的三视图如图所示，其中俯视图为一正方形，则该几何体的表面积为 ．16.若函数()x f 的图象上存在不同的两点()()2211,,,y x B y x A ，其中2211,,,y x y x 使得222221212121y x y x y y x x +⋅+-+的最大值为0，则称函数()x f 是“柯西函数”.给出下列函数：①()()30ln <<=x x x f ； ②()()01>+=x xx x f ； ③()822+=x x f ； ④()822-=x x f .其中是“柯西函数”的为 （填上所有正确答案的序号）．三、解答题 （本大题共6小题，共70分.解答应写出文字说明、证明过程或演算步骤.） 17. 设数列{}n a 的前n 项和为n S ，数列{}n S 的前n 项和为n T ，满足*∈-=N n n S T n n ,22. (Ⅰ)求321,,a a a 的值； (Ⅱ)求数列{}n a 的通项公式.18.某小店每天以每份5元的价格从食品厂购进若干份食品，然后以每份10元的价格出售.如果当天卖不完，剩下的食品还可以每份1元的价格退回食品厂处理.(Ⅰ)若小店一天购进16份，求当天的利润y （单位：元）关于当天需求量n （单位：份，N n ∈）的函数解析式；(Ⅱ)小店记录了100天这种食品的日需求量（单位：份），整理得下表： 日需求量n 14 15 16 17 18 19 20 频数10201616151310以100天记录的各需求量的频率作为各需求量发生的概率.(i)小店一天购进16份这种食品，X 表示当天的利润（单位：元），求X 的分布列及数学期望；(ii)以小店当天利润的期望值为决策依据，你认为一天应购进食品16份还是17份？19如图，在四棱锥ABCD P -中，ABCD 是平行四边形，︒=∠==120,1BAD BC AB ，2==PC PB ,F E PA ,,2=分别是PD AD ,的中点.(Ⅰ)证明：平面⊥EFC 平面PBC ； (Ⅱ)求二面角P BC A --的余弦值.20.已知椭圆()01:2222>>=+b a b y a x C 的离心率为23，21A A 、分别为椭圆C 的左、右顶点点()1,2-P 满足121=⋅PA . (Ⅰ)求椭圆C 的方程；(Ⅱ)设直线l 经过点P 且与C 交于不同的两点N M 、，试问：在x 轴上是否存在点Q ，使得QM 与直线QN 的斜率的和为定值？若存在，请求出点Q 的坐标及定值；若不存在，请说明理由.21.已知函数()()221x a e x x f x--=，其中R a ∈. (Ⅰ)函数()x f 的图象能否与x 轴相切？若能，求出实数a ，若不能，请说明理由； (Ⅱ)求最大的整数a ，使得对任意()+∞∈∈,0,21x R x ，不等式()()221212x x x f x x f ->--+恒成立.请考生在22、23两题中任选一题作答，如果多做，则按所做的第一题记分. 22.选修4-4：坐标系与参数方程 已知直线l 的参数方程为⎩⎨⎧α=α+=sin cos t y t m x （t 为参数，π<α≤0），以坐标原点为极点，以x轴正半轴为极轴建立极坐标系，曲线C 的极坐标方程为θ=ρcos 4，射线4,44π+ϕ=θ⎪⎭⎫ ⎝⎛π<ϕ<π-ϕ=θ，4π-ϕ=θ分别与曲线C 交于C B A 、、三点（不包括极点O ）.(Ⅰ)求证：OA OC OB 2=+；(Ⅱ)当12π=ϕ时，若C B 、两点在直线l 上，求m 与α的值.23.选修4-5：不等式选讲已知函数()a x a x x f 222-+-+=. (Ⅰ)若()31<f ，求实数a 的取值范围；(Ⅱ)若不等式()2≥x f 恒成立，求实数a 的取值范围.2018 届广东省六校第三次联考理科数学参考答案一、选择题1-5: BADAA 6-10: CCBCA 11、12：DD 二、填空题13.3ln 14.()31， 15. 23224++ 16.① ④ 三、解答题17.解：(Ⅰ)∵12111-==S T S ,111a S ==，∴11=a . ∵422221-==+S T S S ,∴42=a . ∵9233321-==++S T S S S ,∴103=a .(Ⅱ)∵ 22n S T n n -=①，()21112--=--x S T n n …②，∴①-②得，()2122≥+-=n n a S n n ，∵112211+⨯-=a S ， ∴()1122≥+-=n n a S n n …③，32211+-=--n a S n n …④， ③-④得，()2221≥+=-n a a n n ， )2(221+=+-n n a a .∵321=+a ，∴{}2+n a 是首项3公比2的等比数列，1232-⨯=+n n a ， 故2231-⨯=-n n a .18.解：(Ⅰ)当日需求量16≥n 时，利润80=y ， 当日需求量16<n 时，利润649)16(45-=--=n n n y ，所以y 关于n 的函数解析式为()N n n n n y ∈⎩⎨⎧≥<-=16,8016,649.(Ⅱ)(i)X 可能的取值为62，71，80，并且()()2.071,1.062====X P X P ,()7.080==X P .X 的分布列为：X 62 71 80 P0.10.20.7X 的数学期望为()4.767.0802.0711.062=⨯+⨯+⨯=X E 元.(ii)若小店一天购进17份食品，Y 表示当天的利润（单位：元），那么Y 的分布列为Y 58 67 76 85 P0.10.20.160.54Y 的数学期望为()26.7754.08516.0762.0671.058=⨯+⨯+⨯+⨯=Y E 元.由以上的计算结果可以看出，()()Y E X E <,即购进 17 份食品时的平均利润大于购进 16份时的平均利润．所以，小店应选择一天购进 17 份. 19.解法一：(Ⅰ)取BC 中点G ，连AC AG PG ,,,∵PC PB =,∴BC PG ⊥, ∵ABCD 是平行四边形，1==BC AB ,120=∠BAD ,∴60=∠ABC ,∴ABC ∆是等边三角形，∴BC AG ⊥,∵G PG AG = ,∴⊥BC 平面PAG ,∴PA BC ⊥. ∵F E ,分别是PD AD , 的中点,∴PA EF //,AG EC //, ∴EF BC ⊥,EC BC ⊥,∵E EC EF = ,∴⊥BC 平面EFC , ∵⊂BC 平面PBC ,∴平面⊥EFC 平面PBC . (Ⅱ)由(Ⅰ)知BC AG BC PG ⊥⊥,, ∴PGA ∠是二面角P BC A --的平面角. ∵2,23,27412===-=PA AG PG , 在PAG ∆中，根据余弦定理得,7212cos 222=⋅-+=∠AG PG PA AG PG PGA , ∴二面角P BC A --的余弦值为721-. 解法二：(Ⅰ)∵ABCD 是平行四边形，1==BC AB ,120=∠BAD ，∴60=∠ADC ,∴ADC ∆是等边三角形，∵E 是AD 的中点， ∴AD CE ⊥,∵BC AD //, ∴BC CE ⊥.分别以,的方向为x 轴、y 轴的正方向，C 为坐标原点， 如图建立空间直角坐标系. 则()⎪⎪⎭⎫ ⎝⎛⎪⎪⎭⎫⎝⎛0,21,23,0,0,23,0,0,0A E C ,⎪⎪⎭⎫⎝⎛-0,21,23D ,设()z y x P ,,2==4=,解得1,21,23==-=z y x ， ∴可得⎪⎪⎭⎫ ⎝⎛-1,21,23P ,∵F 是PD 的中点，∴⎪⎭⎫ ⎝⎛21,0,0F ,∵0=∙,∴CF CB ⊥,∵BC CE ⊥,C CF CE = ,∴⊥BC 平面EFC ,∵⊂BC 平面PBC ,∴平面⊥EFC 平面PBC .(Ⅱ)由(Ⅰ)知，()0,1,0=CB ,⎪⎪⎭⎫ ⎝⎛-=1,21,23,设z y x ,,=是平面PBC 的法向量，则⎪⎩⎪⎨⎧⊥⊥，∴⎪⎩⎪⎨⎧=++-=∙==∙021230z y x n CP y ， 令2-=x ，则)3,0,2(--=，又)1,0,0(=是平面ABC 的法向量，∴721,cos -=<， ∴二面角P BC A --的余弦值为721-. 注：直接设点()z F ，，00，或者说⊥CF 平面ABCD ,AD PA ⊥，酌情扣分. 20.解：(Ⅰ)依题意，()0,1a A -、()0,2a A ，()12-，P ，∴()22151,2)1,2a a a PA PA -=-⋅--=⋅（, 由121=⋅PA ,0>a ,得2=a ，∵23==a c e ， ∴1,3222=-==c a b c ，故椭圆C 的方程为1422=+y x . (Ⅱ)假设存在满足条件的点()0,t Q .当直线l 与x 轴垂直时， 它与椭圆只有一个交点，不满足题意.因此直线l 的斜率k 存在，设)2(1:-=+x k y l ，由⎪⎩⎪⎨⎧=+-=+14)2(122y x x k y ，消y 得 ()()01616816412222=+++-+k k x k kx k ，设()()2211,,y x N y x M 、，则22212221411616,41816kkk x x k k k x x ++=++=+， ∵()()()()()()t x t x t x k kx t x k kx tx yt x y k k QN QM -----+---=-+-=+21122122111212 ()()()()()()()2222212121212824284122122tk t k t t k t t x x t x x tk x x kt k x kx +-+-+-=++-+++++-=， ∴要使对任意实数Q N Q M k k k +,为定值，则只有2=t ，此时，1=+Q N Q M k k . 故在x 轴上存在点()0,2Q ,使得直线QM 与直线QN 的斜率的和为定值1.21.解：(Ⅰ)由于ax xe x f x -=)('.假设函数()x f 的图象与x 轴相切于点()0,t ，则有⎩⎨⎧==0)('0)(t f t f ，即()⎪⎩⎪⎨⎧=-=--0'02'12at te t a e t . 显然0',0>=≠a e t 代入方程()02'12=--t a e t 中得，0222=+-t t . ∵04<-=∆，∴无解．故无论a 取何值，函数()x f 的图象都不能与x 轴相切. (Ⅱ)依题意，()()()()21212121x x x x x x f x x f +-->--+()()()()21212121x x x x f x x x x f -+->+++⇔恒成立.设()x x f x g +=)(，则上式等价于()()2121x x g x x g ->+，要使()()2121x x g x x g ->+对任意()+∞∈∈,0,21x R x 恒成立，即使()()x x a e x x g x +--=221在R 上单调递增， ∴01)('≥+-=ax xe x g x 在R 上恒成立.则1,01)1('+≤≥+-=e a a e g ，∴0)('≥x g 在R 上成立的必要条件是：1+≤e a .下面证明：当3=a 时，013≥+-x xe x 恒成立.设()1--=x e x h x ，则1)('-=xe x h ，当0<x 时，0)('<x h ，当0>x 时，0)('>x h ， ∴0)0()(min ==h x h ，即1,+≥∈∀x e R x x .那么，当0≥x 时，()011213,222≥-=+-≥+-+≥x x x x xe x x xe x x ； 当0<x 时，0)13(13,1>+-=+-<xe x x xe e x x x ，∴013≥+-x xe x 恒成立. 因此，a 的最大整数值为 3.22.解：(Ⅰ)证明：依题意，ϕ=cos 4OA ，⎪⎭⎫ ⎝⎛π-ϕ=⎪⎭⎫ ⎝⎛π+ϕ=4cos 4,4cos 4OC OB ， 则OA OC OB 2cos 244cos 44cos 4=ϕ=⎪⎭⎫ ⎝⎛π-ϕ+⎪⎭⎫ ⎝⎛π+ϕ=+.(Ⅱ)当12π=ϕ时，C B 、两点的极坐标分别为⎪⎭⎫ ⎝⎛π-⎪⎭⎫ ⎝⎛π63232，，，， 化直角坐标为()()3331-，，，C B .经过点C B 、的直线方程为()23--=x y ，又直线l 经过点()0,m ，倾斜角为α，故32,2π=α=m . 23.解：(Ⅰ)∵()31<f ，∴321<-+a a ， ①当0≤a 时，得32,3)21(-><-+-a a a ，∴032≤<-a ； ②当210<<a 时，得2,3)21(-><-+a a a ，∴210<<a ； ③当21≥a 时，得34,3)21(<<--a a a ，∴3421<≤a . 综上所述，实数a 的取值范围是⎪⎭⎫⎝⎛-3432，. (Ⅱ)∵()a x a x x f 2122-+-+=，根据绝对值的几何意义知，当21a x -=时，()x f 的值最小, ∴221≥⎪⎭⎫ ⎝⎛-a f ，即2251>-a ， 解得56>a 或52-<a .∴实数a 的取值范围是⎪⎭⎫ ⎝⎛+∞⎪⎭⎫ ⎝⎛-∞-,5652, .。
广东省六校(广州二中,深圳实验等)2018届高三下学期第三次联考语文试题
广东六校联盟2018届高三第三次联考语文阅读下面的文字，完成下列小题。
人们的生活处于不停的变动之中，文化也随之而不断变化和发展。
不同文化的交流与碰撞带来文化的融合与冲突，历史学家汤因比发现这是人类文明兴衰的一个重要机制。
这一过程的结果是，文化自信随着社会生活的变迁而变化。
在生产力发展的驱动下，现代资本主义生产方式深刻改变了西方世界的经济社会结构，进而引发了文化的剧烈变革。
资本主义国家由于不断扩大市场的内在需求而推动了全球化进程，带来不同文化之间的激烈碰撞。
马克思、恩格斯观察到:“资产阶级，由于一切生产工具的迅速改进，由于交通的极其便利，把一切民族甚至最野蛮的民族都卷到文明中来了。
它的商品的低廉价格，是它用来摧毁一切万里长城、征服野蛮人最顽强的仇外心理的重炮。
它迫使一切民族在自己那里推行所谓的文明，即变成资产者。
”这种社会生产方式的变革推动了世界各国的文化震荡，引发了普遍的文化危机，人们在新的世界格局下如何重建自己的文化自信成为普遍问题。
文化认同危机冲击、瓦解了传统的文化自信，引发了人们对既有文化的反省性认识，在文化批判中形成了文化自信的发展机制。
一方面，作为社会意识的文化弥散于人们的日常生活和社会心理中，具有天然的传承性和保守性，由此也形成了走向僵化的可能性。
另一方面，文化也具有一种自我发展的潜能，作为一种能动的因素，它通过新思想的引入而吹响变革的号角，从而成为克服僵化机制进而维系社会系统活力的积极力量。
文化批判意味着以批判性的立场认识和对待自己，不是盲目地肯定或否定，而是在理性地反思与省察之上客观地予以认识和对待，它的对立面不是对象本身而是拒绝理性的思想方法，克服这种思想方法正是启蒙理性的要求。
康德将启蒙理解为脱离人加之于自己的“不成熟状态”。
所以，批判精神乃是实现精神成长的真实表现。
对于一个民族而言，这意味着文化的成熟，表现出该文化的理性自觉和现代意蕴。
可见，人类文化随着社会生活的变迁而发展，现代文化更是在现代性各种因素的驱动下处于不停的变动之中，由此形成了文化自信生成与发展的辩证逻辑。
广东省六校2018届高三下学期第三次联考数学(理)试题(教师版)
D. 8,9
由 z 3x 5y 得 y
3z x ，平移直线 y
55
3z x ，结合图形可得，当直线经过可行域内的点
55
A时，
直线在 y 轴上的截距最大，此时 z 取得最大值，由题意得点 A 的坐标为（ 3,0 ），∴ zmax 3 3 9 ．当直线
x 2y 1 0
x1
经过可行域内的点 B 时，直线在 y 轴上的截距最小， 此时 z 取得最小值， 由
an
2
1
11
．
2n( n 1) n(n 1) n 1 n
∴ S2018 1 (1 1 ) ( 1 1 ) 2 23
(1
1 ) 1 (1
1)
4035
．选 C．
2017 2018
2018 2018
8. 如图，在同一个平面内， 三个单位向量 OA, OB, OC 满足条件： OA 与 OC 的夹角为 ，且 tan
方程的解的个数，因此可以通过方程思想来解，属于简单题．
2. 设等差数列 { an} 的前 n 项和为 Sn ，若 S3 9 ， S5 30 ，则 a7 a8 a9
A. 63
【答案】 A 【解析】
B. 45
C. 36
D. 27
由题意 S3 3a2 9 ，a2 3 ，S5 5a3 30 ，a3 6 ，∴ d a3 a2 6 3 3 ，a1 a2 d 3 3 0 ，
A. 0
B. 1
C. 2
D. 3
【答案】 B
【解析】
【分析】
：集合 M 与集合 N 表示的集合都是点集，所以可以把两个方程联立，通过求方程的判别式来判定交点的个
数．
【详解】：
联立方程组
广东省六校2018届高三下学期第三次联考英语试题学校__
广东省六校（广州二中，深圳实验，珠海一中，中山纪念，东莞中学，惠州一中）2018届高三下学期第三次联考英语试题学校:___________姓名：___________班级：___________考号：___________一、阅读选择Questioning the existence of aliens is something that scientists have done for decades. In fact, most people do believe that aliens exist in some fashion. The main necessities for life are water and some form of energy source. Not surprisingly, there are some planets, exoplanets and moons that fit the bill. Here are several best chances at finding life in the universe.◆TRAPPIST-1TRAPPIST-1 is a planetary system a few dozen light-years away, whose discovery was announced in early 2017. This system consists of seven Earth-like exoplanets orbiting an "ultra-cool" star, and it is one of our shots at finding possible life beyond our own solar system.◆TitanTitan is the largest moon of Saturn, the sixth planet from our Sun. This moon could potentially harbor life but possibly not in the sense that we think. Titan does not exactly fit the description of being in a typical habitable zone.Titan has water, and it has liquid. It just doesn't have liquid water. The water on this moon is completely solid because of the extremely cold temperature.◆EuropaEuropa is one of Jupiter's moons here in our own solar system. It is another candidate due to its potential to hold liquid water. Europa is thought to have all the necessities for life including water, energy sources, and the right chemical build-up.◆MarsThe Red Planet, the fourth from the Sun, is probably one of the most talked-about potential candidates for extraterrestrial(地球外的) life and even for human colonization. Despite some different voices, finding extraterrestrial life on Mars really is a serious possibility.We know by now that we won't find little green men or any intelligent form of life that we understand. However, there is evidence that there was and may still be microscopic life on the small red planet.1．The common point of these celestial bodies is that _______.A．solid water exists on them respectivelyB．life might exist on each of themC．they all contain liquid waterD．each of them has living things on it2．Which celestial body could have life different from human imagination? A．TRAPPIST-1. B．Titan.C．Europa. D．Mars.3．Which of the following statements is TRUE?A．Europa is the most talked-about potential candidate for extraterrestrial life. B．TRAPPIST-1 is the only body that contains life beyond the solar system.C．All the scientists do believe that aliens exist in some manner.D．There is proof that life existed on the celestial body Mars.About a month after I joined Facebook, I got a call from Lori Goler, a highly regarded senior dir ector of marketing at eBay. She made it clear this was a business call. “I want to apply to work with you at Facebook,” she said. “Instead of recommending myself, I want to ask you: What is your biggest problem, and how can I solve it?”My jaw hit the floor. I had hired thousands of people over the previous decade and no one had ever said anything remotely like that. People usually focus on finding the right role for themselves, with the implication that their skills will help the company. Lori put Face book’s needs front and center. It was a killer approach. I responded, “Recruiting is my biggest problem. And, yes, you can solve it.”Lori never dreamed she would work in recruiting, but she jumped in. She even agreed to trade earnings for acquiring new skills in a new field. Lori did a great job running recruiting and within months was promoted to her current job, leading People@Facebook.The most common metaphor for careers is a ladder, but this concept no longer applies to most workers. As of 2010, the average American had eleven jobs from the ages of eighteen to forty-six alone. Lori often quotes Pattie Sellers, who came up with a much better metaphor: “Careers are a jungle gym, not a ladder.”As Lori describes it, there’s only one way to get to the top of a ladder, but there are many ways to get to the top of a jungle gym. The jungle gym model benefits everyone, but especiallywomen who might be starting careers, switching careers, getting blocked by external barriers, or reentering the workforce after taking time off. The ability to create a unique path with occasional dips, detours (弯路), and even dead ends presents great views of many people, not just those at the top. On a ladder, most climbers are stuck staring at the butt of the person above. 4．Why did Lori make the call?A．She helped Facebook to solve the biggest problem.B．She wanted to make a business deal with Facebook.C．She tried to ask for a pay rise in Facebook.D．She wanted to become an employee in Facebook.5．What impressed “I” b y Lori?A．Lori was good at running recruiting.B．Lori attached great importance to Facebook’s needs.C．Lori jumped in Facebook with no adequate experience.D．Lori was skilled in marketing at eBay.6．What can we infer from the passage?A．Now all people don’t tend to climb the ladder.B．None on the ladder can enjoy the great views.C．Jungle gyms offer limited exploration for employees.D．A pregnant woman, jobless, benefits little from the jungle gyms.7．What is the best title of the passage?A．It’s a Jungle Gym, Not a Ladder.B．Facebook’s Biggest Problem.C．Applying for a Job in Facebook.D．A Jungle Gym is Better than a Ladder.Careers in zoology are extremely varied and unique, and can provide incredible learning and work opportunities for anyone devoted to animal studies and welfare. Zoology, or the study of animals, is a wide field with many specialties, including research, conservation, veterinary (兽医的) medicine, and the care of animals. For people with a love of animals and some training, dreams of careers in zoology may come true.Some careers in zoology focus on research and scientific studies. These careers may allow those with a good theoretical science background to develop and run studies that improvehuman understanding of the animal world. Research in zoology can help create safer and more effective products for animals or can teach humans more about animal behavior in order to aid in conservation, breeding programs, and habitat preservation.Careers in zoology that focus on conservation attach great importance to the continued survival and increased protection of animal species. Conservationists may work with political groups or governments to help make laws to protect and preserve animals, or may work in the field gathering information on potential threats to the health of global ecosystems. Some conservationists work in educational fields, trying to improve human efforts to save animals from extinction.Animals are subject to illness and injury, and some careers in zoology help to create a safe, stable animal population. Veterinary medicine is an important specialty field, and may take several years of intensive training to qualify as a certified veterinarian. While many veterinarians focus on the small-animal practice of domestic pets, vets in rural areas often work with large farm animals and more adventurous veterinarians may work with exotic species in zoos and wildlife preserves. .Humans love to observe animals, and modern-day zoos and wildlife preserves help meet that interest while providing facilities to assist with conservation programs. Zoology careers in zoos can range from overseeing breeding programs, to creating the proper diet for a deer, to cleaning the tiger’s cage. Many volunteer and entry-level jobs are available for people that love animals.8．What is the purpose of researching on animals in zoology?A．To help governments make laws.B．To improve researchers’ scientific knowledge.C．To create safe and effective products for humans.D．To educate people to protect animals.9．What can careers in zoology that focus on veterinary medicine help?A．Better understand animals. B．Protect animals’ health.C．Clean the animals’ cages.D．Increase the number of animals. 10．Which of the following shows the structure of the whole passage?(P1=paragraph 1; P2= paragraph 2; P3= paragraph 3; P4= paragraph 1; P5= paragraph 5;)A．B．C．D．11．What does the passage mainly present?A．V olunteers’ passion for zoology.B．A new idea of careers in zoology.C．Different types of careers in zoology.D．Effective measures of animal protection.NEW YORK－A massive winter storm paralyzed much of the US East Coast on Thursday and Friday, dumping as much as 46 centimeters of snow from the Carolinas to Maine. It also caused flooding on the streets of Boston due to swelling storm tides, forced the cancellation of nearly 5,000 flights and closed businesses, offices and schools.Some meteorologists (气象学家) classified the storm as a "bomb cyclone" for its sharp drop in atmospheric pressure would be followed immediately by a blast of cold air that could break records in more than two dozen cities and bring wind chills as low as -40℃ during the weekend.From Baltimore, Maryland, to Caribou in Maine, efforts were underway to clear roadways of ice and snow as wind chill temperatures were to plunge during the day, reaching -40℃ in some parts after sundown, according to the National Weather Service.Utility companies across the East worked to repair downed power lines early on Friday as about 21,000 customers remained without electricity, down from almost 80,000 the day before, and issued warnings that temperatures may become dangerously low."If the temperature in your home begins to fall, we recommend taking shelter elsewhere until service can be restored. You can find warming centers by contacting local authorities," National Grid power company, which serves Massachusetts, said on Twitter.Airlines canceled 4,000 flights on Thursday and hundreds more on Friday, according to , an online tracking service.Mayor Bill de Blasio said it could feel like -30℃ on Friday and Saturday nights with thewind chill. "This is a serious, serious storm, and may be the most severe in 2018," he said at a news conference.New York Governor Andrew Cuomo declared a state of emergency for the southern part of the state, while New Jersey Governor Chris Christie declared a state of emergency for several counties.12．Which of the following is NOT the consequence of the massive winter storm? A．Breakdown of the US coastB．Flooding in Boston.C．Cancellation of nearly 5,000 flights.D．Closure in business, offices, and schools.13．Why did some meteorologists classify this winter storm as “bomb cyclone”? A．Because it will cause changeable atmospheric pressure.B．Because the scale of this storm could break records in lots of cities.C．Because of the deadly and destructive effects of this storm.D．Because of the extremely cold air and the wind chills in this storm.14．Which of the following best explains the underlined word “plunge” in P3?A．decline steadily B．drop dramaticallyC．fluctuate violently D．fall moderately15．About the rescue work, which of the following statements is INCORRECT? A．People were making great efforts to rid roadways of ice and snow.B．About 101 thousand people were left without electricity on Thursday.C．Residents can stay in warming centers until heating service is restored.D．Aviation service hasn't been brought back to normal yet until Friday.二、七选五"A question brought me to the point of ending my life when I was fifty years old. My question was the simplest one that lies in every person. It is the question without which life is impossible: Why do I live? 16．Is there anything in my life that will not be destroyed by my death?"These are the words of the famous Russian writer Leo Tolstoy. Many people ask these difficult questions. And they struggle to find meaning in their life. Tolstoy spent his whole life trying to answer difficult questions like these. 17．In the 1850s, Leo Tolstoy wrote his first stories. He wrote about his experiences in the military. He also told stories about when he was a child. 18．He was finally successful. He earned respect from many wealthy and intelligent men, who talked a lot about faith and the meaning of life. But soon he found they were proud and they made very bad moral choices.19．He opened a school for the children of people who worked on his land. They were very poor. He wanted to help them because he thought they were more honest than the wealthy people he knew.Tolstoy learned many things from his workers. He respected how they worked hard to provide for their families. 20．So in 1862, Leo Tolstoy married a young woman named Sonya Behrs. The next 15 years were the best years of Tolstoy's life. It was during this time that he wrote his most famous books-- War and Peace and Anna Karenina. They communicate what he thought was the answer to all his questions-- humans were supposed to live a simple life and take care of their families.A．These works were published and Tolstoy became a well-known writer.B．How can I realize my dream if I have one?C．Why do I wish for anything, or do anything?D．He gained a lot of inspirations which had great influence on his following actions.E.His search for answers influenced his writing.F.So in the 1860s, Tolstoy tried a different way to find the meaning of life.G.He began to believe that marriage and family would give his life meaning.三、完形填空I come from one of those families where you have to yell at the dinner table to get ina word. Everyone has a strong 21 , and talks at the same time, and no one has a 22 leading to heated arguments. We often talk or even debate with each other on different topics.23 a family like mine has made me more 24 about the world around me, making me tend to question anything anyone tells me. But it has also made me realize that I’m not a good listener. And when I say “listening”, I’m not 25 to the nodding-your-head-and- 26 -answering-Uh-huh-or-Ooh-I-see variety. I mean the kind of listening where you find yourself deeply 27 with the person you’re speaking with, when his story becomes so 28 that your world becomes less about you and more about him. No, I was never very good at that.I spent summer in South Africa two years ago. I worked for a good non-profit 29 called Noah, which works 30 on behalf of children affected by AIDS. But 31 you asked me what I really did in South Africa, I’d tell you one thing: I listened, and I listened. Sometimes I 32 , but mostly listened.And had I not spent two months 33 , I might have missed the 34 moment when a quiet little girl at one of Noah’s community centers, orphaned（孤儿）at the age of three, whispered after a long 35 , “I love you.”36 that summer, I knew how to hear. I could sit down with anyone and hear their37 and nod and respond at the 38 time—but most of the time I was 39 about the next words out of my own mouth. Ever since my summer in South Africa, I have noticed that it’s in those moments when my mouth is closed and my 40 is wide open that I’ve learned the most about other people, and perh aps about myself. 21．A．qualification B．influence C．opinion D．assumption 22．A．commitment B．problem C．schedule D．request 23．A．Belonging to B．Believing in C．Bringing up D．Struggling for 24．A．anxious B．curious C．nervous D．adventurous 25．A．objecting B．appealing C．turning D．referring 26．A．rudely B．loudly C．politely D．gratefully 27．A．identifying B．quarreling C．debating D．competing 28．A．vivid B．magical C．mind-numbing D．time-consuming 29．A．school B．organization C．factory D．church 30．A．effortlessly B．timelessly C．aimlessly D．tirelessly 31．A．unless B．because C．although D．if 32．A．applauded B．spoke C．wept D．complained 33．A．studying B．traveling C．listening D．working 34．A．touching B．frustrating C．astonishing D．fascinating 35．A．delay B．course C．journey D．silence 36．A．Before B．After C．Except D．Since 37．A．needs B．stories C．comments D．cases 38．A．valuable B．free C．right D．same 39．A．talking B．arguing C．learning D．thinking 40．A．sympathy B．spirit C．mind D．family四、用单词的适当形式完成短文A lot of films have tried to describe the afterlife and our memories of family members, 41．few have done as well as Coco, Disney Pixar’s 42．(late) masterpiece animation, which hit the big screen on Nov. 24, 2017.43．(inspire) by the Mexican holiday of Dia de los Muertos — Day of the Dead — the film’s production team created a young boy, Miguel, who wants his family to understand his love of music. 44．that year’s Dia de los Muertos, 45．accident takes him to the Land of the Dead. In this land, there are friendly skeletons who can cross a bridge made of flower petals to visit their living family — that is, as long as their family still puts their photos on the family shrine (神龛). Those spirits who 46．(forget) by their family will disappear completely. So it’s in this magical world 47．Miguel gets to meet and discover the truth about hisgreat-great-grandpa.In an era 48．young people are so 49．(easy) attracted by celebrities, Coco reveals the emptiness of such flattery, teaching kids to preserve and respect the memory of their elders while 50．(remind) them that the source of true creativity is so often personal.五、短文改错51．假定英语课上老师要求同桌之间交换修改作文，请你修改你同桌写的以下作文。
广东省六校2018届高三下学期第三次联考理科综合化学试题
广东省六校（广州二中，深圳实验，珠海一中，中山纪念，东莞中学，惠州一中）2018届高三下学期第三次联考理科综合化学试题1.化学与人类的生活、生产息息相关，下列说法不正确的是( )A. “地沟油”禁止食用，但可以用来制肥皂或燃油B. BaCO3在医学上可用作“钡餐”C. 光导纤维的主要成分是SiO2，太阳能电池使用的材料是单质硅D. 臭氧、醋酸、双氧水都能杀菌消毒2.对于下图所示实验，下列实验现象预测或操作正确的是( )A. 实验甲：匀速逐滴滴加盐酸时，试管中没气泡产生和有气泡产生的时间段相同B. 实验乙：充分振荡后静置，下层溶液为橙红色，上层为无色C. 实验丙：由MgCl2•6H2O制备无水MgCl2D. 装置丁：酸性KMnO4溶液中有气泡出现，且溶液颜色会逐渐变浅乃至褪去3. 都属于多环烃类，下列有关它们的说法错误的是( )A. 这三种有机物的二氯代物同分异构体数目相同B. 盘烯能使酸性高锰酸钾溶液褪色C. 棱晶烷和盘烯互为同分异构体D. 等质量的这三种有机物完全燃烧，耗氧量相同4. 锂空气电池放电时的工作原理如图所示。
下列叙述正确的是( )A. 放电时Li＋由B极向A极移动B. 电池放电时总反应方程式为4Li＋O2 ＋2H2O===4LiOHC. 电解液a、b之间可采用阴离子交换膜D. 电解液a可能为LiCl水溶液5. 如表所示的五种元素中，W、X、Y、Z为短周期元素，这四种元素的原子最外层电子数之和为22。
下列说法不正确的是( )A. X、Y能形成不止一种氢化物，而Z只能形成一种B. W和Z形成的共价化合物中，所有原子最外层均满足8电子稳定结构C. W和T都具有良好的半导体性能D. X和Y最多可以形成5种化合物6. 下列陈述I、Ⅱ均正确且有因果关系的是( )陈述I 陈述ⅡA CO2能与水反应生成碳酸CO2属于酸性氧化物B AlCl3属于共价化合物AlCl3属于非电解质C 漂白粉中的Ca(ClO)2会与空气中CO2、H2O反应漂白粉应密封保存D H2O2和SO2均能使酸性高锰酸钾溶液褪色H2O2有还原性， SO2有漂白性7. 室温下，将0.10 mol·L－1盐酸滴入20.00 mL 0.10mol·L－1氨水中，溶液中pH和pOH随加入盐酸体积变化曲线如图所示。
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绝密★启用前2018届广东省六校第三次联考文科数学注意事项：1．答题前，考生务必将自己的姓名、准考证号、座位号、学校、班级等考生信息填写在答题卡上。
2．作答选择题时，选出每个小题答案后，用2B 铅笔把答题卡上对应题目的答案信息点涂黑。
如需改动，用橡皮擦干净后，再选涂其它答案，写在本试卷上无效。
3．非选择题必须用黑色字迹签字笔作答，答案必须写在答题卡各题指定的位置上。
4．考试结束后，将本试卷和答题卡一并交回。
一、选择题：本题共12小题，每小题5分，共60分. 在每题给出的四个选项中，只有一项符合题目要求. 1.函数()ln(1)f x x =++的定义域为（ ） A ．(2,)+∞ B ．(1,2)(2,)-+∞ C ．(1,2)- D ．(]1,2- 2.如果复数ibi212+-(其中i 为虚数单位，b 为实数)的实部和虚部互为相反数，那么b 等于( ) A ．6- B ．32 C ．32- D ．23.高考结束后，同学聚会上，某同学从《爱你一万年》，《非你莫属》，《两只老虎》，《单身情歌》四首歌中选出两首歌进行表演，则《爱你一万年》未选取的概率为（ ）A ．13B ．12 C ．23 D ．564.圆22(2)4x y -+=关于直线y x =对称的圆的方程是（ ）A ．22((1)4x y +-=B ．22((4x y +=C ．22(2)4x y +-=D ．22(1)(4x y -+=5．某几何体的三视图如图所示，且该几何体的体积是3，则正视图中的x 的值是( )A ．2 B.29 C. 23D ．3 6.已知sin()3cos()sin()2πθπθθ++-=-，则2sin cos cos θθθ+=（ ）A ．15B ．25C ．35D7.实数x 、y 满足000x y x y c ≤⎧⎪≤⎨⎪+-≥⎩，且x y -的最大值不小于1，则实数c 的取值范围是( )A ．1c ≤-B ．1c ≥- C．c ≤．c ≥8.函数x x x f cos )(=的导函数)(x f '在区间],[ππ-上的图像大致是（ ）A. B. C. D. 9.三棱锥ABC P -中，ABC PA 平面⊥且2=PA ，ABC ∆是边长为3的等边三角形，则该三棱锥外接球的表面积为( )A.34πB ．π4C ．π8D ．π20 10.自主招生联盟成行于2009年清华大学等五校联考，主要包括“北约”联盟，“华约”联盟，“卓越”联盟和“京派”联盟．在调查某高中学校高三学生自主招生报考的情况，得到如下结果：①报考“北约”联盟的学生，都没报考“华约”联盟②报考“华约”联盟的学生，也报考了“京派”联盟③报考“卓越”联盟的学生，都没报考“京派”联盟④不报考“卓越”联盟的学生，就报考“华约”联盟根据上述调查结果，下列结论错误的是（ ）A ．没有同时报考“华约” 和“卓越”联盟的学生B ．报考“华约”和“京派”联盟的考生一样多C ．报考“北约” 联盟的考生也报考了“卓越”联盟D ．报考“京派” 联盟的考生也报考了“北约”联盟 11.设120172016,log log a b c ===，则,,a b c 的大小关系为（ ）A ．a b c >>B ．a c b >> C. b a c >> D ．c b a >>12. 已知双曲线E ： 22x a﹣22y b =1（0,0>>b a ），点F 为E 的左焦点，点P 为E 上位于第一象限内的点，P 关于原点的对称点为Q ，且满足FQ 3PF =，若b =OP ，则E 的离心率为（ ）A.二．填空题：本题共4小题，每小题5分。
13.若向量,2,()a b a b a b a ==-⊥满足，则向量a 与b 的夹角等于 ．14.执行如图所示的程序框图，则输出S 的结果为 ．15.已知函数)(x f y =在点))2(,2(f 处的切线方程为12-=x y ，则函数)()(2x f x x g +=在点))2(,2(g 处的切线方程为________．16．已知平面四边形ABCD 为凸四边形（凸四边形即任取平面四边形一边所在直线， 其余各边均在此直线的同侧），且2=AB ，4=BC ，5=CD ，3=DA ， 则平面四边形ABCD 面积的最大值为________．三．解答题：共70分，解答应写出文字说明，证明过程或演算步骤。
（一）必考题：共60分。
17.（本小题满分12分）已知数列{}n a 的前n 项和为n S ，且满足22n n S n -=.(n N *∈)DC 1A 1B 1CBA（1）求数列{}n a 的通项公式；（2）设22,(21)2,(2)(1)(1)n a n n n n k b n k a a +⎧=-⎪=⎨=⎪--⎩（k N *∈），求数列{}n b 的前n 2项和n T 2.18.（本小题满分12分）如图,在三棱柱111ABC A B C -中，侧棱1AA ⊥底面ABC ,,AB BC D ⊥为AC 的中点,12A A AB ==,3BC =.(1)求证：1//AB 平面1BC D ；(2) 求四棱锥11B AAC D -的体积.19.（本小题满分12分）随着社会的发展，终身学习成为必要，工人知识要更新，学习培训必不可少，现某工厂有工人1000名，其中250名工人参加过短期培训（称为A 类工人），另外750名工人参加过长期培训（称为B 类工人），从该工厂的工人中共抽查了100名工人，调查他们的生产能力（此处生产能力指一天加工的零件数）得到A 类工人生产能力的茎叶图（左图），B 类工人生产能力的频率分布直方图（右图）.（1）问A 类、B 类工人各抽查了多少工人，并求出直方图中的x ;（2）求A 类工人生产能力的中位数，并估计B 类工人生产能力的平均数（同一组中的数据用该组区间的中点值作代表）;（3) 若规定生产能力在[130，150]内为能力优秀，由以上统计数据在答题卡上完成下面的2⨯2列联表，并判断是否可以在犯错误概率不超过0.1%的前提下,认为生产能力与培训时间长短有关.能力与培训时间列联表参考数据：参考公式：22(),()()()()n ad bc K a b c d a c b d -=++++ 其中d c b a n +++=. 20.（本小题满分12分）已知动点M 到定点)0,1(F 的距离比M 到定直线2-=x 的距离小1.（1）求点M 的轨迹C 的方程；（2）过点F 任意作互相垂直的两条直线21l l 和，分别交曲线C 于点B A ,和N K ,．设线段AB ，KN 的中点分别为Q P ,，求证：直线PQ 恒过一个定点.21.（本小题满分12分）已知函数()()1ln 122+-++-=x x a x x x f (其中R a ∈,且a 为常数) ． （1）若对于任意的()+∞∈,1x ，都有()0>x f 成立，求a 的取值范围；（2）在（Ⅰ）的条件下，若方程()01=++a x f 在(]2,0∈x 上有且只有一个实根，求a 的取值范围．（二）选考题：共10分。
请考生在第22、23题中任选一题作答。
如果多做，则按所做的第一题计分。
22.（本题满分10分）[选修4-4：坐标系与参数方程]在直角坐标系xOy 中，曲线1C 的参数方程为⎪⎩⎪⎨⎧+-=-=ty t x 542532（t 为参数）．以坐标原点为极点，以x 轴正半轴为极轴建立极坐标系，曲线2C 的极坐标方程为cos tan ρθθ=．（1）求曲线1C 的普通方程和曲线2C 的直角坐标方程；（2）若1C 与2C 交于A B ,两点，点P 的极坐标为π4⎛⎫- ⎪⎝⎭，求11||||PA PB +的值． 23.（本题满分10分）[选修4-5：不等式选讲]设函数()221(0)f x x a x a =-++>，()2g x x =+． （Ⅰ）当1a =时，求不等式()()f x g x ≤的解集； （Ⅱ）若()()f x g x ≥恒成立，求实数a 的取值范围．2018届广东省六校第三次联考文科数学参考答案与评分标准一．选择题：本大题共12小题，每小题5分。
二．填空题：本大题共4小题，每小题5分。
13．；14．30；15．；16．；16. 解：设AC=,在中由余弦定理有同理，在中，由余弦定理有：，即①，又平面四边形面积为，即②. ①②平方相加得，当时，取最大值.三、解答题（本大题共6小题，共70分.解答应写出文字说明、证明过程或演算步骤.）17. (本小题满分12分)解：（1）当时，…2分（），…………………………………3分当时，由得，…………………………………4分显然当时上式也适合，∴…………………………………5分（2）∵…………………………………6分∴…………………………………7分…………………9分…………………………………11分…………………………………12分18. (本小题满分12分)解：（1）证明:连接,设与相交于点,连接,∵四边形是平行四边形, ∴点为的中点.∵为的中点，∴为△的中位线,∴. ……………………… 2分∵平面,平面,∴平面. ……………………………………4分(2)解法1: ∵平面,平面,∴平面平面，且平面平面.作，垂足为，则平面，…………… 6分∵，，在Rt△中，，，…8分∴四棱锥的体积…… 10分.∴四棱锥的体积为. …… 12分解法2: ∵平面,平面,∴.∵,∴.∵,∴平面. …… 6分取的中点,连接,则,∴平面.三棱柱的体积为, …… 8分则,.…… 10分而,∴. ∴.∴四棱锥的体积为. …… 12分19(本小题满分12分）解：（1）由茎叶图知A类工人中抽查人数为25名, .......................................1分∴B类工人中应抽查100-25=75（名）. ......................................................2分由频率分布直方图得 (0.008+0.02+0.048+x)´10=1，得x=0.024. (3)分（2）由茎叶图知A类工人生产能力的中位数为122 ………………………………4分由（1）及频率分布直方图，估计B类工人生产能力的平均数为115´0.008´10+125´0.020´10+135´0.048´10+145´0.024´10=133.8 (6)分（3…………9分由上表得＞10.828 ……11分因此,可以在犯错误概率不超过0.1%的前提下,认为生产能力与培训时间长短有关.………12分20.(本小题满分12分)解：(1)由题意可知：动点到定点的距离等于到定直线的距离，根据抛物线的定义可知，点的轨迹是抛物线。