2017年高考模拟试卷(6)

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2017英语高考模拟卷6

2017英语高考模拟卷6

2016年普通高等学校全国统一考试英语时间:120分钟满分150分注意事项:1. 本试卷分第I卷(选择题)和第II卷(非选择题)两部分。

2 . 考生务必将自己的姓名、考生号填写在答题卡上。

3. 作答时,将答案写在答题卡上。

写在本试卷上无效。

4. 考试结束后•将本试卷和答题卡一并交回。

第I卷第一部分听力(共两节,满分30分)做题时,先将答案标在试卷上。

录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节(共5个小题;每小题1.5分,满分7.5分)听下面5段对话。

每段对话后有一个小题,从每题所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。

每段对话仅读一遍。

例: How much is the shirt?A. £ 19.15.B. £ 9.18.C.£ 9.15.答案是C o1. What will Lucy do at 11:30 tomorrow?A. Go out for lunch.B. See her den tist.C. Visit a friend.2. What is the weather like now?A. It 'unny.B. It 'ainy.C. It 'clsudy.3. Why does the man talk to Dr. Simpso n?A. To make an apology.B. To ask for help.C. To discuss his studies.4. How will the woma n get back from the railway statio n?A. By bus.B.By car.C.Tak ing a bicycle5. What does Jenny decide to do first?A. Look for a job.B. Go on a trip. D. Get an assista nt.第二节(共15小题:每小题1.5分,满分22.5分)听下面5段对话或独白。

浙江省杭州市2017届高考模拟命题比赛数学试卷6.doc

浙江省杭州市2017届高考模拟命题比赛数学试卷6.doc

浙江省杭州市2017届高考模拟命题比赛数学试卷6一、选择题:(本大题共10小题,每小题4分,共40分.在每小题给出的四个选项中,只有一个是符合题目要求的.)1.已知集合{}2230M x x x =∈+-≤R ,{}50N x x =∈-≤≤R ,则()C M N ⋃=R ( ) A.()53--, B. ](53--, C. )53--⎡⎣, D.]()5301--⋃⎡⎣,, 2.已知复数z 满足()13i 1i z +=+,则z =( ) A .22B .2-C .2D . 2 3.已知,a b 是实数,则“0a >且0b >”是“0a b +>且0ab >”的( ) A.充分而不必要条件B.必要而不充分条件C.充分必要条件D.既不充分也不必要条件4.若tan 3θ=,则22sin 3sin cos θθθ-=( )A.110B.37C.910D.135.已知直线1y x =+与曲线ln()y x a =+相切,则a 的值为( ) A 2B .1C .-1D . 2 6.若实数,x y 满足10,0,0,x y x y x -+≥⎧⎪+≥⎨⎪≤⎩则2ex y+的最小值是( ) A.1B.12eC. eD.2e7.已知(),B n p ξ:,且5E ξ=,3D ξ=,则p 等于( )A.13B.35 C.25 D.238.在ABC ∆中,已知10AB =u u u r ,边AB 上的高为3,则当AC BC u u u r u u u r g 最小时,AC BC +=u u u r u u u r( )A. B.10C.313D1039.已知双曲线122:22=-x y C ,直线l 过点()0,2A ,斜率为k ,当10<<k 时,双曲线的上支上有且仅有一点B 到直线l 的距离为2,则k =( )A.1B.5C.510.给定函数()2,f x x ax b =++设,p q 是满足1p q +=的实数,若对于任意的实数,x y 均有:()()()pf x q x f px qx +≥+,则( ) A.0q p ≤≤B.0p q ≤≤C.0p q ≤≤D.0q p ≤≤二、填空题:(本大题共7小题,第11-14题每题6分,第15-17每题4分,共36分.)11.抛物线24y x =的焦点坐标是________,若直线10ax y -+=经过抛物线焦点,则实数a = .12.在ABC ∆中, 3B π∠=,三边长,,a b c 成等差数列,且6ac =,则ABC S ∆=____,b 的值是_____________.13.已知四棱锥P ABCD -的三视图如下图所示,E 是侧棱PC 上的动点.四棱锥P ABCD -的体积位__________________,异面直线与所成角为_____________.14. 已知数列{}n a 的首项1122,,1,2,31n n n a a a n a +===+….则3a = 数列n n a ⎧⎫⎨⎬⎩⎭的前n 项和n S ,则n S =____________.15.在一次晚会上,9位舞星共上演n 个“三人舞”节目,若在这些节目中,任二人都曾合作过一次,且仅合作一次,则n =__________.16.若曲线22120C x y x +-=:与曲线()20C y mx m --=:x 有两个不同的公共点,则m的取值所组成的集合是_________.17.设二次函数()()20f x ax bx c a b c a =++∈≠R ,,,满足条件:(1)当x ∈R 时,()()42f x f x -=-,且();f x x ≥ABCD P E(2)当()0,2x ∈时,()21;2x f x +⎛⎫≤ ⎪⎝⎭(3)()f x 在R 上的最小值为0.若存在,t ∈R 只要[]1,x m ∈(1m >),就有()f x t x +≤.则m 的最大值为_________. 三、解答题:(本大题共5小题,共74分.解答应写出文字说明,证明过程或演算步骤.) 18.(本题满分14分)在ABC △中,内角,,A B C 所对的边长分别是,,a b c . (1)若()sin sin sin 2A B C C +-=,试判断ABC △的形状. (2)若2,3a A π==,且ABC △的面积3=S ,求,b c 的值;19.(本题满分15分)如图所示,直角梯形ACDE 与等腰直角ABC ∆所在平面互相垂直,F 为BC 的中点,90BAC ACD ∠=∠=o,//AE CD ,22DC AC AE ===. (Ⅰ)求证:AF ∥平面BDE ; (Ⅱ)求二面角B DE C --的余弦值.20.(本题满分15分)已知函数2()f x ax ax =+和()g x x a =-.其中a ∈R .0a ≠且.(1)若函数()f x 与的()g x 图像的一个公共点恰好在x 轴上,求a 的值; (2)若p 和q 是方程()()0f x g x -=的两根,且满足10p q a<<<,证明:当()0,x p ∈时,()()g x f x p a <<-.21.设曲线()2212:1x C y a a+=为正常数与()222C y x m =+:在x 轴上方仅有一个公共点P .(1)求实数m 的取值范围;()a 用表示(2)O 为原点,若1C 与x 轴的负半轴交于点A ,当102a <<时,试求AOP V 的面积的最大值. ()a 用表示22.给定正整数n 和正数M .对于满足条件2211n a a M ++≤的所有等差数列123,,,a a a …,1221=n n n S a a a +++++…+,(1)求证:2251S Mn ⎛⎫≤ ⎪+⎝⎭(2)求1221=n n n S a a a +++++…+的最大值.参考答案一、选择题1-5 CACCD 6-10 ACBBC二、填空题11. ()1,0,1- 12.2 13. 23,90︒14. 89,22n n + 15. 12 16.33⎧⎫⎪⎪-⎨⎬⎪⎪⎩⎭17.9 三、解答题18.解:(1)因为()sin sin sin cos sin B C B C B C -=-①()(),sin sin ()=sin() =sin cos cos sin A B C A B C B C B c B C=π-+=π-+++=sin cos cos sin B c B C +② sin 22sin cos C C C =③将①②③代入sin sin()sin 2A B C C +-= 化简可得:sin sin B C =因为在ABC ∆中,所以B C =,ABC ∆为等腰三角形.(2)因为在ABC ∆中,1,sin 32A S bc A π===所以4bc = ④又因为2221cos 22b c a A bc +-==,且2a =,⑤ 由④⑤解得2,2b c ==19.解:(Ⅰ)取BD 的中点P ,连结,EP FP ,则1//2PF CD , 又因为1//2EA CD ,所以//EA PF ,所以四边形AFPE 是平行四边形, 所以//AF EP ,又因为EP ⊂面,BDE AF ⊄平面BDE , 所以//AF BDE 面(Ⅱ)以CA CD 、所在直线分别作为x 轴,z 轴,以过C 点和AB 平行的直线作为y 轴,建立如图所示坐标系. 由22DC AC AE ===可得:()()()2,0,0,2,2,0,2,0,1,A B E ()0,0,2D则(0,2,0),(0,2,1),AB BE ==-u u u r u u u r (2,2,2)BD =--u u u r.因为面ACDE ⊥面ABC ,面ACDE I 面,ABC AC AB AC =⊥, 所以AB ⊥面.ACDE所以(0,2,0)AB =u u u r是面CDE 的一个法向量.设面BDE 的一个法向量(),,n x y z =r,则BE n ⊥r u u u r ,BD n ⊥r u u u r .所以00,BE BD ⎧=⎪⎨=⎪⎩u u u r g u u u rg n n 即202220,y z x y z -+=⎧⎨--+=⎩ 整理,得200.y z x y z -=⎧⎨+-=⎩令1y =,则2,1,z x ==所以()1,1,2n =r是面CDE 的一个法向量.故2226cos ,||||2112AB AB AB 〈〉===⨯++u u u ru u u r g u u u r n n n . 图形可知:二面角B DE C --的平面角π(0,)2θ∈,所以其余弦值为66. 20.解:(1)设函数()g x 图像与x 轴的交点坐标为(a ,0),又因为点(a ,0)也在函数()f x 的图像上, 所以320a a +=. 而0a ≠,所以1a =-.(2)由题意可知()()()()f x g x a x p x q -=--. 因为10x p q a<<<<,所以()()0a x p x q -->, 所以当()0,x p ∈时,()()0,f x g x ->即()()f x g x >.又()()()()()()(1)f x p a a x p x q x a p a x p ax aq --=--+---=--+,0,110,x p ax aq aq -<-+>->且所以()()f x p a --<0,所以()f x p a <-,综上可知,()()g x f x p a <<-.21.解:联立方程组()22221,2,x y a y x m ⎧+=⎪⎨⎪=+⎩消去y 得2222220x a x a m a ++-= ① 故()222222f x x a x a m a =++-,问题(1)转化为方程①在(),x a a ∈-上有唯一解或等根.只需讨论以下三种情况:1︒0=∆得212a m +=,此时2P x a =-,当且仅当2a a a -<-<,即01a <<时适合; 2︒ ()()0f a f a -<g ,当且仅当a m a -<<;3︒ ()0f a -=得m a =,此时22P x a a =-,当且仅当22a a a a -<-<,即01a <<时适合;()0f a =得m a =-,此时22P x a a =--,当且仅当22a a a a -<-<,无解,从而m a ≠-.综上可知,当01a <<时,212a m +=或a m a-<≤;当1a ≥时,a m a -<<.(2)AOP ∆的面积12P S ay =. 因为102a <<,故当a m a -<≤时,20a a <-+<,由唯一性得2P x a =-+m a =时,P x 取得最小,此时0P y >,从而P y =取得最大,此时P y =,从而S =当212a m +=时,2P x a =-,P y =12S =下面比较与12令12=13a =.故当103a <<时,12≤max 12S =当1132a <<时,12>max S =22. 解:(1) 设公差为d ,1n a a +=,则()()12211112n n n S a a a n a n n d+++=++=+++…+ 故1,21S a nd n +=+又()()22221122411=4310210n M a a a nd aa nd a nd +≥+=-+⎛⎫++- ⎪⎝⎭24,101S n ⎛⎫≥ ⎪+⎝⎭(2)因为 24,101S M n ⎛⎫≥ ⎪+⎝⎭所以1S n ≤+且当a =d =()()12=1=12n S n n n =+++ 由于此时43a nd =,所以22114410101104n S a a M M n +⎛⎫+=== ⎪+⎝⎭g 所以,S的最大值为12n +。

【高考模拟】2017年普通高等学校招生全国统一考试模拟数学(文)试题(六)含答案

【高考模拟】2017年普通高等学校招生全国统一考试模拟数学(文)试题(六)含答案

核 心 八 模2017年普通高等学校招生全国统一考试模拟试题数学(文科)(六) 第Ⅰ卷(选择题 共60分)一、选择题:本大题共12小题,每小题5分,共60分.在每个小题给出的四个选项中,有且只有一项符合题目要求.1.已知集合()(){}{}|210,|11A x x x B x Z x =-+<=∈-≤≤,则A B = A. {}1,0- B. {}0,1 C. {}1,0,1- D. {}1,2- 2.方程26130x x ++=的一个根是A. 32i -+B. 32i +C.23i -+D.23i +3.已知()f x 是定义在R 上的偶函数,且在区间(),0-∞上单调递增,若实数a 满足()(12a f f ->,则实数a 的取值范围是A. 1,2⎛⎫-∞ ⎪⎝⎭B. 13,,22⎛⎫⎛⎫-∞+∞ ⎪⎪⎝⎭⎝⎭ C. 13,22⎛⎫ ⎪⎝⎭ D.3,2⎛⎫+∞ ⎪⎝⎭4.某路口人行横道的信号灯为红灯和绿灯交替出现,红灯持续时间为40秒,若一名行人来到路口遇到红灯,则至少需要等待15秒才出现绿灯的概率为A.710B. 58C. 38D. 3105.执行如图所示的程序框图,若输出的86s =,则判断框内的正整数的值为A.7B. 6,7C. 6,7,8D.8,96.向量,a b 满足a b += ,且()0a b a -⋅= ,则,a b的夹角的余弦值为A. 0B.13 C. 12 D. 27.已知等差数列{}n a 中,n S 为其前n 项和,若()244n S an n a a R =++-∈,记数列1n S ⎧⎫⎨⎬⎩⎭的前项和为n T ,则10T =A.18 B. 14 C. 940 D.5228.已知,,a b c 均为正数,且()()2a c b c ++=,则23a b c ++的最小值是9.某几何体的三视图如下图所示,且该几何体的体积为x 的值为2310.已知A,B 是球O 的球面上两点,90AOB ∠= ,C 为该球面上的动点,若三棱锥O ABC -的体积的最大值为36,则球O 的表面积为A. 36πB. 64πC. 144πD.256π11.已知点12,F F 是双曲线()2222:10,0x y C a b a b-=>>的左、右焦点,O 为坐标原点,点P在双曲线C 的右支上,且满足12122,3F F OP PF PF =≥,则双曲线C 的离心率的取值范围为A. ()1,+∞B. ⎫+∞⎪⎪⎣⎭C. ⎛ ⎝⎦D.51,2⎛⎤⎥⎝⎦ 12.已知关于x 的方程11202kx x ---=+有三个不相等的实根,则实数k 的取值范围是A. 1⎛⎫+∞ ⎪ ⎪⎝⎭B. 1⎛-+ ⎝⎭C.()0,1,02⎛--∞ ⎝⎭ D. ()1,022⎛-+-∞ ⎝⎭第Ⅱ卷(非选择题 共90分)二、填空题:本大题共4小题,每小题5分,共20分.13.已知,x y 满足约束条件20220220x y x y x y +-≤⎧⎪--≤⎨⎪-+≥⎩,若20x y k ++≥恒成立,则实数k 的取值范围为 .14.某事业单位公开招聘一名职员,从笔试成绩合格的6(编号为1—6)名应试者中通过面试选聘一名.甲、乙、丙、丁四人对入选者进行预测.甲:不可能是6号;乙:不是4号就是5号;丙:是1,2,3号中的一名;丁:不可能是1,2,3号.已知四人中只有一人预测正确,则入选者应为 .15.已知点A,F 分别是椭圆()2222:10x y C a b a b+=>>的上顶点和左焦点,若AF 与圆22:4O x y +=相切于点T ,且点T 是线段AF 靠近点A 的三等分点,则椭圆C 的标准方程为 .16.若数列{}n a 满足2133431n n a a a a a a a a +->->->>-> ,则称数列{}n a 为“差递减”数列.若数列{}n a 是“差递减”数列,且其通项n a 与其前n 项和()n S n N *∈满足()2321n n S a n N λ*=+-∈,则实数λ的取值范围为 .三、解答题:本大题共6小题,共70分.解答应写出必要的文字说明或推理、验算过程. 17.(本题满分12分)在ABC ∆中,角,,A B C 的对边分别为,,a b c ,且a c >,已知12,cos , 3.3BA BC B b ⋅=== ,求:(1)a 和c 的值; (2)()cos B C -的值.18.(本题满分12分)从某企业生茶的某种产品中抽取100件,测量这些产品的一项质量指标值,由测量结果得如下频数分布表:(1)在答题卡上作出这些数据的频率分布直方图;(2)估计这种产品质量指标值的平均值及方差(同一组数据用该组区间的中点值代表); (3)根据以上抽样的数据,能否认为该企业生产的这种产品符合“质量指标值不低于95的产品至少占全部产品的80%”的规定?19.(本题满分12分)如图所示,在四棱锥P ABCD -中,底面ABCD为菱形,E 为AC 与BD 的交点,PA ⊥平面ABCD ,M 为PA 的中点,N 为BC 的中点.(1)证明:直线//MN 平面PCD ;(2)若点Q 为PC 的中点,120,1BAD PA AB ∠=== ,求三棱锥A QCD -的体积.20.(本题满分12分)已知曲线C 上的任意一点到点()0,1F 的距离减去它到x 轴的距离的差都是1. (1)求曲线C 的方程;(2)设直线()0y kx m m =+>与曲线C 在x 轴及x 轴上方部分交于A,B 两点,若对于任意的k R ∈都有0FA FB ⋅<,求m 的取值范围.21.(本题满分12分) 已知函数()()21130.a f x ax a a x-=++->(1)当1a =时,求函数()y f x =在点()()2,2f 处的切线方程;(2)若不等式()()1ln f x a x ≥-在[)1,x ∈+∞时恒成立,求实数a 的取值范围.请考生在第22、23两题中任选一题作答,如果两题都做,则按照所做的第一题给分;作答时,请用2B 铅笔将答题卡上相应的题号涂黑。

2017届高考数学模拟试卷(六)含答案

2017届高考数学模拟试卷(六)含答案

2017届⾼考数学模拟试卷(六)含答案江苏省2017届⾼考数学模拟试卷(六)⾼三数学试卷(⽂科)第Ⅰ卷(共60分)⼀、填空题:本⼤题共14个⼩题,每⼩题5分,共70分.请把答案直接填在答题卡相应位置上.1.已知集合{}2|20A x x x =-=,{}0,1,2B =,则A B = .2.若31zi i=+-,i 是虚数单位,则复数z 的虚部为. 3.函数22()log (6)f x x =-的定义域为. 4.已知函数()sin()5f x kx π=+的最⼩正周期是3π,则正数k 的值为.5.已知幂函数()y f x =的图象经过点1(4,)2,则1()4f 的值为.6.“三个数a ,b ,c 成等⽐数列”是“2b ac =”的条件.(填“充分不必要、充要、必要不充分、既不充分也不必要”)7.已知53cos()25πα+=,02πα-<<,则sin 2α的值是. 8.已知函数()f x 是奇函数,当0x <时,2()3sin 2xf x x a π=-,且(3)6f =,则a = .9.若等差数列{}n a 的前5项和525S =,且43a =,则7a = . 10.若直线y x b =+是曲线ln y x x =的⼀条切线,则实数b = . 11.函数3sin(2)4y x π=+的图象向左平移?(02)个单位后,所得函数图象关于原点成中⼼对称,则?= .12.数列{}n a 定义如下:11a =,23a =,122(1)22n n n n a na a n n +++=-++,1,2,n =….若201642017m a >+,则正整数m 的最⼩值为. 13.已知点O 为△ABC 内⼀点,且230OA OB OC ++=,则△AOB ,△AOC ,△BOC 的⾯积之⽐等于.14.定义在R 上的奇函数()f x ,当0x ≥时,2,[0,1),()11|3|,[1,),xx f x x x x -?∈?=+??--∈+∞?则函数1()()F x f x π=-的所有零点之和为.⼆、解答题(本⼤题共6⼩题,共90分.解答应写出⽂字说明、证明过程或演算步骤.)15.在△ABC 中,a ,b ,c 分别为内⾓A ,B ,C 所对的边,且满⾜a b c <<,2sin b a B =.(1)求A 的⼤⼩;(2)若2a =,23b =,求△ABC 的⾯积.16.已知函数()|1|f x x =-,2()65g x x x =-+-(x R ∈).(1)若()()g x f x ≥,求x 的取值范围;(2)求()g x ()f x -的最⼤值.17.已知锐⾓△ABC 中的三个内⾓分别为A ,B ,C .(1)设BC CA CA AB ?=?,判断△ABC 的形状;(2)设向量(2sin,3)s C =-,2(cos 2,2cos 1)2C t C =-,且//s t ,若1sin 3A =,求sin()3B π-的值.18.某地拟建⼀座长为640⽶的⼤桥AB ,假设桥墩等距离分布,经设计部门测算,两端桥墩A ,B 造价为100万元,当相邻两个桥墩的距离为x ⽶时(其中64100x <<).中间每个桥墩的平均造价为x 万元,桥⾯每1⽶长的平均造价为(2)640x x +万元.(1)试将桥的总造价表⽰为x 的函数()f x ;(2)为使桥的总造价最低,试问这座⼤桥中间(两端桥墩A ,B 除外)应建多少个桥墩?19.已知各项都为正数的等⽐数列{}n a 的前n 项和为n S ,数列{}n b 的通项公式,1,n n n b n n ?=?+?为偶数为奇数(*n N ∈),若351S b =+,4b 是2a 和4a 的等⽐中项.(1)求数列{}n a 的通项公式;(2)求数列{}n n a b ?的前n 项和n T .20.已知函数1()1ln a f x x x=-+(a 为实数).(1)当1a =时,求函数()f x 的图象在点11(,())22f 处的切线⽅程;(2)设函数2()32h a a a λ=-(其中λ为常数),若函数()f x 在区间(0,2)上不存在极值,且存在a 满⾜1()8h a λ≥+,求λ的取值范围;(3)已知*n N ∈,求证:11111ln(1)12345n n+<++++++….江苏省2017届⾼考数学模拟试卷(六)⾼三数学试卷(⽂科)⼀、填空题 1.{}0,2 2.2- 3.(,6)(6,)-∞-+∞ 4.6 5.2 6.充分不必要7.241258.5 9.3- 10.1- 11.38π 12.8069 13.3:2:1 14.112π- ⼆、解答题15.解:(1)2sin b a B =,∴sin 2sin sin B A B =,∵sin 0B >,∴1sin 2A =,由于a b c <<,所以A 为锐⾓,∴6A π(2)由余弦定理2222cos a b c bc A =+-,∴234122232c c =+-, 2680c c -+=,2c =或4c =,由于a b c <<,4c =,所以1sin 232S bc A ==.当1x <时,()1f x x =-,由()()g x f x ≥,得2651x x x -+-≥-,整理得(1)(6)0x x --≤,所以[]1,6x ∈,由1,16x x综上x 的取值范围是[]1,4.(2)由(1)知,()()g x f x -的最⼤值必在[]1,4上取到,所以22599()()65(1)()244g x f x x x x x -=-+---=--+≤,所以当52x =时,()()g x f x -取到最⼤值为94. 17.解:(1)因为BC CA CA AB ?=?,所以()0CA BC AB ?-=,⼜0AB BC CA ++=,∴()CA AB BC =-+,所以()()0AB BC BC AB -+?-=,所以220AB BC -=,所以22||||AB BC =,即||||AB BC =,故△ABC 为等腰三⾓形.(2)∵//s t ,∴22sin (2cos 1)22CC C -=,∴sin 22C C =,即tan 2C = ∵C 为锐⾓,∴2(0,)C π∈,∴223C π=,∴3C π=,∴23A B π=-,∴2sin()sin ()333B B πππ??-=--sin()3A π=-,⼜1sin 3A =,且A 为锐⾓,∴cos A =sin()sin()sin cos cos sin 3333B A A A ππππ-=-=-=. 18.解:(1)由桥的总长为640⽶,相邻两个桥墩的距离为x ⽶,知中间共有640(1)x-个桥墩.于是桥的总造价640()640(2(1)f x x=+-100+.即3112226408080()138033f x x x x -?=+-+3112225120080138033x x x -=+-+(64100x <<).(2)由(1)可求13122236404040'()233f x x x x --?=--,整理得3221'()(98064080)6f x x x x -=--?.由'()0f x =,解得180x =,26409x =-(舍去),⼜当(64,80)x ∈时,'()0f x <;当(80,100)x ∈时,'()0f x >,所以当80x =,桥的总造价最低,此时桥墩数为6401780-=个. 19.解:(1)∵数列{}n b 的通项公式,1,n n n b n n ?=?+?为偶数为奇数(*n N ∈),∴56b =,44b =.设各项都为正数的等⽐数列{}n a 的公⽐为q ,0q >,∵3517S b =+=,∴21117a a q a q ++=,①∵4b 是2a 和4a 的等⽐中项,∴224316a a a ==,解得2314a a q ==,②由①②得23440q q --=,解得2q =或23q =-(舍去),∴11a =,12n n a -=.(2)当n 为偶数时,0(11)2n T =+?[]2342122(31)242(51)2(1)122n n n n --+?++?+?++?++-+?+?…0231022(22232422)(222)n n n --=+?+?+?++?++++……,设023*********n n H n -=+?+?+?++?…,③则2312 2 2232(1)22n n n H n n -=+?+?++-?+?…,④③-④,得0231222222n nn H n --=+++++-? (1212)n-=-2n n -?(1)21n n =-?-,∴(1)21n n H n =-?+,∴21422(1)21()21433nnn n T n n -=-?++=-?+-.当n 为奇数,且3n ≥时,11(1)2n n n T T n --=++?1115222()2(1)2(2)23333n n n n n n ---=-?+++?=-?+,经检验,12T =符合上式.∴122(2)2,3322()2,33n n n n n T n n -?-?+??=??-?+??为奇数,为偶数.20.解:(1)当1a =时,11()1ln f x x x =-+,211'()f x x x=-,则1()4222f =-=,1()12ln 2ln 212f =-+=-,∴函数()f x 的图象在点11(,())22f 处的切线⽅程为:1(ln 21)2()2y x --=-,即2ln 220x y -+-=.(2)221'()a a xf x x x x-=-=,由'()0f x =,解得x a =,由于函数()f x 在区间(0,2)上不存在极值,所以0a ≤或2a ≥,由于存在a 满⾜1()8h a λ≥+,所以max 1()8h a λ≥+,对于函数2()32h a a a λ=-,对称轴34a λ=,①当304λ≤或324λ≥,即0λ≤或83λ≥时,2max 39()()48h a h λλ==,由max 1()8h a λ≥+,即29188λλ≥+,结合0λ≤或83λ≥可得:19λ≤-或83λ≥;②当3014λ<≤,即403λ<≤时,max ()(0)0h a h ==,由max 1()8h a λ≥+,即108λ≥+,结合403λ<≤可知:λ不存在;③当3124λ<<,即4833λ<<时,max ()(2)68h a h λ==-;由max 1()8h a λ≥+,即1688λλ-≥+,结合4833λ<<可知:13883λ≤<,综上可知,λ的取值范围是113(,][,)98-∞-+∞.(3)证明:当1a =时,21'()xf x x-=,当()0,1x ∈时,'()0f x >,()f x 单调递增;当(1,)x ∈+∞时,'()0f x <,()f x 单调递减,∴11()1ln f x x x =-+在1x =处取得最⼤值(1)0f =,即()f x 111ln x x =-+(1)0f ≤=,∴11ln x x x -≤,令1n x n =+,则11ln n n n +<,即1ln(1)ln n n n+-<,∴ln(1)ln(1)ln1n n +=+-[][]111ln(1)ln ln ln(1)(ln 2ln1)11n n n n n n =+-+--++-<++++……,故1111ln(1)1234n n+<+++++….。

2017年普通高等学校招生全国统一考试仿真卷 理科数学(六)教师版

2017年普通高等学校招生全国统一考试仿真卷 理科数学(六)教师版

绝密 ★ 启用前2017年普通高等学校招生全国统一考试仿真卷理科数学(六)本试题卷共2页,23题(含选考题)。

全卷满分150分。

考试用时120分钟。

★祝考试顺利★注意事项:1、答题前,先将自己的姓名、准考证号填写在试题卷和答题卡上,并将准考证号条形码粘贴在答题卡上的指定位置。

用2B 铅笔将答题卡上试卷类型A 后的方框涂黑。

2、选择题的作答:每小题选出答案后,用2B 铅笔把答题卡上对应题目的答案标号涂黑。

写在试题卷、草稿纸和答题卡上的非答题区域均无效。

3、填空题和解答题的作答:用签字笔直接答在答题卡上对应的答题区域内。

写在试题卷、草稿纸和答题卡上的非答题区域均无效。

4、选考题的作答:先把所选题目的题号在答题卡上指定的位置用2B 铅笔涂黑。

答案写在答题卡上对应的答题区域内,写在试题卷、草稿纸和答题卡上的非答题区域均无效。

5、考试结束后,请将本试题卷和答题卡一并上交。

第Ⅰ卷一、选择题:本大题共12小题,每小题5分,在每小题给出的四个选项中,只有一项是符合题目要求的。

1.[2017云师附中]复数3ii 1z =-,则其共轭复数z 在复平面内对应的点位于( ) A .第一象限 B .第二象限C .第三象限D .第四象限【答案】C 【解析】∵i 1i i 12z --+==-,其共轭复数为1i 2z --=,对应点为11()22--,在第三象限,故选C . 2.[2017重庆联考]已知某品种的幼苗每株成活率为p ,则栽种3株这种幼苗恰好成活2株的概率为( ) A .2p B .()21p p -C .223C pD .()2231C p p -【答案】D【解析】由题设可知32n k ==,,则所求事件的概率()()22311n kkkn C p p C p p --=-,应选答案D .3.[2017枣庄模拟]若集合{}{}22|22,|log A x x B x y x =∈-<<==Z ,则A B =I ( )A .{}1,1-B .{}1,0,1-C .{}1D .{}0,1【答案】A【解析】因为{}{}22|22{1,0,1},|log {|0}A x x B x y x x x =∈-<<=-===≠Z ,所以A B I ={1,1}-,故选A .4.[2017怀仁一中]某几何体的三视图如图所示,图中的四边形都是边长为2的正方形,两条虚线互相垂直,则该几何体的体积是( )A .203B .163C .π86-D .π83-【答案】A【解析】该几何体为一个正方体去掉一个倒四棱锥,其中正方体棱长为2,倒四棱锥顶点为正方体中心,底面为正方体上底面,因此体积是3212021233-⨯⨯=,选A . 5.[2017重庆模拟]我国古代数学算经十书之一的《九章算术》有一衰分问题:今有北乡八千一百人,西乡七千四百八十八人,南乡六千九百一十二人,凡三乡,发役三百人,则北乡遣( ) A .104人 B .108人C .112人D .120人【答案】B【解析】由题设可知这是一个分层抽样的问题,其中北乡可抽取的人数为:8100810030030010881007488691222500⨯=⨯=++,应选答案B .6.[2017长沙一中]如图是用模拟方法估计圆周率π值的程序框图,P 表示估计结果,则图中空白框内应填入( )班级 姓名 准考证号 考场号 座位号此卷只装订不密封A .2000MP =B .42000MP =C .2000NP =D .42000NP =【答案】B【解析】由题意得以及程序框图可知,用模拟方法估计圆周率的程序框图,M 是圆周内的点的次数,当i 大于2000时,圆周内的点的次数为4M ,总试验次数为2000,所以要求的概率42000M ,所以空白框内应填入的表达式是42000MP =,故选B . 7.[2017重庆一中],x y 满足约束条件40240240x y x y x y +-⎧⎪--⎨⎪-+⎩≤≤≥,若z ax y =-取得最大值的最优解不唯一...,则实数a 的值为( ) A .1- B .2C .12D .2或1-【答案】C【解析】作出不等式组对应的平面区域如图:(阴影部分).由z ax y =-得z ax y -=,即直线的截距最小,z 最大.若0=a ,此时z y -=,此时,目标函数只在B 处取得最大值,不满足条件,若0>a ,目标函数z ax y -=的斜率0>=a k ,要使z ax y =-取得最大值的最优解不唯一,则直线z ax y -=与直线042=--y x 平行,此时21=a ,若0<a ,不满足,故选C .8.[2017郑州一中]函数()12cos 12x x f x x ⎛⎫-= ⎪+⎝⎭的图象大致为( ) A . B .C .D .【答案】C【解析】∵()()()1221cos cos 1221x x x x f x x x f x --⎛⎫---=-==- ⎪++⎝⎭,所以()f x 为奇函数,排除选项A ,B .又π02x ⎛⎫∈ ⎪⎝⎭,时,()0f x <,图像在x 轴下方,故本题正确答案为C .9.[2017南白中学]已知在三棱锥P ABC -中,1PA PB BC ===,AB =AB BC ⊥,平面PAB ⊥平面ABC ,若三棱锥的顶点在同一个球面上,则该球的表面积为( )AB .3πCD .2π【答案】B【解析】由题意得,AC为截面圆的直径,且AC =ABC 的距离为d ,设球的半径为R ,因为1PA PB BC ===,AB =PA PB ⊥,因为平面PAB ⊥平面ABC ,所以点P 到平面ABC的距离为,由勾股定理可得222221())2R d d =+=+,解得的取值范围是( ) A .[]06, B .[]26-,C .[]02,D .[]22-,【答案】B【解析】以三角形的外接圆圆心为原点建立平面直角坐标系,设()(()202cos 2sin A B P θθ-,,,,. 则()()22cos 2sin 12cos2sin PA PB θθθθ=--=--u u u r u u u r,,, ()()())22cos 12cos 2sin 2sin 22cos PA PB θθθθθθ⋅=---+-=--u u u r u u u r[]π24sin 266θ⎛⎫=-+∈- ⎪⎝⎭,,故选B .11.[2017正定中学]函数()f x 在定义域R 内可导,若()()2f x f x =-,且当()1x ∈-∞,时,()()10x f x -'<,设()0a f =,12b f ⎛⎫=⎪⎝⎭,()3c f =,则( ) A .a b c << B .c a b <<C .c b a <<D .b c a <<【答案】B【解析】由()()2f x f x=-可知,()f x 的图象关于1x =对称,根据题意又知()1x ∈-∞,时,()0f x '>,此时()f x 为增函数,()1x ∈+∞,时,()0f x '<,()f x 为减函数,所以()()()13102f f f f ⎛⎫=- ⎪⎝⎭<<,即c a b <<,故选B12.[2017重庆一模]设a b c ∈R ,,且0c ≠,若上表中的对数值恰有两个是错误的,则a 的值为( ) A .2lg21B .13lg 214C .13lg 27D .6lg7【答案】B【解析】由题设可知lg3lg2lg6lg51lg14a b c a b c a c =+=-=+=-+=,,,,()lg83b a c a -=-,都是正确的,所以lg3lg142a -=,即13lg 214a =,应选答案B .第Ⅱ卷本卷包括必考题和选考题两部分。

2017届安徽省普通高中高考模拟卷(六)数学(理科)试卷

2017届安徽省普通高中高考模拟卷(六)数学(理科)试卷

2017届安徽省普通高中高考模拟卷(六)数学(理科)试卷本试卷分第一部分(必考部分)和第二部分(选考部分)两部分。

满分150分,考试时间120分钟。

必考部分(共140分)一、选择题:本题共12小题,每小题5分。

在每小题给出的四个选项中,只有一项是符合题目要求的。

1. 已知集合{}R 12,1,0,1,2,|02x A B x x -⎧⎫=--=≥⎨⎬+⎩⎭ð,则A B = ( ) A .{}1,0,1- B .{}1,0- C .{}2,1,0-- D .{}0,1,2 2. 在复平面内,复数ii+1的对应点位于 A.第一象限 B.第二象限 C.第三象限 D.第四象限 3.阅读程序框图,当输入x 的值为2时,运行相应程序,则输出x 的值为( )A .5B .11C .23D . 47 4.下列命题中真命题的个数是( ) ①若q p ∧是假命题,则,p q 都是假命题;②命题“01,23≤+-∈∀x x R x ”的否定是“32000,10x R x x ∃∈-+>”; ③若,11:,1:<≤xq x p 则p ⌝是q 的充分不必要条件. A .0B .1C .2D .35. 已知数列{}n a 为等差数列,且满足1590a a +=.若(1)mx -展开式中2x 项的系数等于数列{}n a 的第三项,则m 的值为( )A .6B .8C .9D .106.设ABC ∆的内角,,A B C 所对边的长分别为,,a b c .若sin 2sinB A =,4,3c C π==,则ABC ∆的面积为( )A .83B .163C .D .7.若291(4)()x x x-+的展开式中3x 的系数为( ) A .36 B .-144 C.60 D .-608.过圆锥顶点的平面截去圆锥一部分,所得几何体的三视图如图所示,则原圆推的体积为( )A .1B .23π C. 43π D .83π 9.已知2220182018201720172ln ,2ln ,2017201720162016a b ⎛⎫⎛⎫=-=- ⎪ ⎪⎝⎭⎝⎭2201620162ln 20152015c ⎛⎫=- ⎪⎝⎭,则( ) A .a b c >> B .a c b >> C .c a b >> D .c b a >>10.已知函数f (x )=2sin (ωx+φ)(ω>0,0<φ<π)的图象上相邻两个最高点的距离为π.若将函数f (x)的图象向左平移个单位长度后,所得图象关于y 轴对称.则函数f (x )的解析式为( ) A .f (x )=2sin (x+) B .f (x )=2sin (x+)C .f (x )=2sin (2x+) D .f (x )=2sin (2x+)11.点A ,B ,C ,D 在同一个球的球面上,,∠ABC=90°,若四面体ABCD 体积的最大值为3,则这个球的表面积为( )A .2πB .4πC . 8πD .16π12.已知函数()1,0,,0,x e m x f x ax b x ⎧+-≥=⎨+<⎩其中1m <-,对于任意1x R ∈且10x ≠,均存在唯一实数2x ,使得()()21f x f x =,且12x x ≠,若()()f x f m =有4个不相等的实数根,则a 的取值范围是( )二、填空题:本题共4小题,每小题5分.13. 已知向量,a b 满足()4,3,3a b =-=- ,若向量,a b 的夹角为23π,则 23a b += __________.14. 已知椭圆()22122:10x y C a b a b +=>>与椭圆()22222:10y x C a b a b+=>>相交于,,,A B C D 四点,若椭圆1C的一个焦点为()F ,且四边形ABCD 的面积为163,则椭圆1C 的离心率为 __________. 15. 已知实数,x y 满足240300x y x y y -+≥⎧⎪+-≤⎨⎪≥⎩,若66ax y +≥-恒成立,则实数的取值范围为_________.16. 向如图所示的边长为2的正方形区域内任投一点,则该点落入阴影部分的概率为.三、解答题:解答应写出文字说明、证明过程或演算步骤. 17.(本小题满分10分)某厂生产的产品在出厂前都要做质量检测,每件一等品都能通过检测,每件二等品通过检测的概率为12.现有10件产品,其中6件是一等品,4件是二等品. (Ⅰ)随机选取3件产品,设至少有一件通过检测为事件A ,求事件A 的概率; (Ⅱ)随机选取3件产品,其中一等品的件数记为X ,求X 的分布列及数学期望EX .18.(本小题满分12分)如图,已知菱形ABCD 与直角梯形ABEF 所在的平面互相垂直,其中BE AF ∥ ,AB AF ⊥,122AB BE AF ===,3CBA π∠=,P 为DF 的中点. (Ⅰ)求证:PE ∥平面ABCD ; (Ⅱ)求二面角D EF A --的余弦值;(Ⅲ)设G 为线段AD 上一点,AG AD λ=, 若直线FG 与平面ABEF 求AG 的长.19. (本小题满分12分)如图,将数字1,2,3,…,2n(n≥3)全部填入一个2行n列的表格中,每格填一个数字。

2017年高考模拟试卷(6)

2017年高考模拟试卷(6)

开始结束输出S n ←1, S ←0S < 100 n ←n + 1S ←S + 2n NY (第5题)2017年高考模拟试卷(6)第Ⅰ卷(必做题,共160分)一、填空题:本大题共14小题,每小题5分,共70分 . 1. 设集合A = {1,x },B = {2,3,4},若A ∩B ={4},则 x = ▲ .2. 若复数z 1=2+i ,z 1·-z 2=5,则z 2= ▲ .3. 从数6,7,8,9,10,11六个数中,任取两个不同的数, 则两个数互质的概率是 ▲ .4.已知一组数据x 1,x 2,…,x 100的方差是2,则数据 3x 1,3x 2,…,3x 100 的标准差为 ▲ .5.执行右边的程序框图,则输出的S 的值为 ▲ .6.设正四棱柱ABCD —A 1B 1C 1D 1的底面ABCD 是单位正方形,其表面积14,则AA 1= ▲ . 7.不等式组⎩⎨⎧y ≤x +2y ≥x0≤y ≤4x ≥0表示的平面区域的面积为S ,则S 的值为 ▲ .8.函数y =sin(ωx +π4)(ω>0)的图象在[0,1]上恰有三个最高点,则ω的取值范围是 ▲ .9.若两个非零向量a ,b 的夹角为60°,且(a +2b )⊥(a -2b ),则向量a +b 与a -b 的夹角的余弦值是 ▲ .10.已知函数f (x )=e x -1-tx ,∃x 0∈R ,f (x 0)≤0,则实数t 的取值范围 ▲ .11.已知数列{a n }是一个等差数列,首项a 1>0,公差d ≠0,且a 2、a 5、a 9依次成比数列,则 使a 1+a 2+…+a n >100a 1的最小正整数k 的值是 ▲ .12.抛物线y 2=2px (p >0)和双曲线x 2a 2-y 2b2=1(a >0,b >0)有一个相同的焦点F 2(2,0),而双曲线的另一个焦点F 1,抛物线和双曲线交于点B 、C ,若△BCF 1是直角三角形,则双曲线的离心率是 ▲ .13.△ABC 中,∠A 、∠B 、∠C 的对边分别为a 、b 、c ,若a 2cos A =b 3cos B =c 6cos C,则cos A cos B cos C = ▲ .14.已知函数f (x )=2x 3+7x 2+6x x 2+4x +3,x ∈[0,4],则f (x )最大值是 ▲ .AA 1B 1 CD 1 B C 1D MO 115.(本小题满分14分)已知α∈(0,π),且sin(α+π3)=6-24.(1)求sin(α-π4)的值;(2)求cos(2α-π3)的值.16.(本小题满分14分)如图,四棱柱ABCD -A 1B 1C 1D 1中,底面ABCD 是菱形,M 是AB的中点,O 1是A 1C 1与B 1D 1的交点. (1)求证:O 1M ∥平面BB 1C 1C ;(2)若平面AA 1C 1C ⊥平面ABCD ,求证:四边形BB 1D 1D 是矩形.17.(本小题满分14分)如图所示,一根绳穿过两个定滑轮,且两端分别挂有3(N)、2(N)的重物.现在两个滑轮之间的绳上挂一个重量为m (N)的重物,恰好使系统处于平衡状态. (1)若∠AOB =120°,求m 的值; (2)求m 的取值范围.18. 椭圆C :x 2+y 2=1的左、右顶点分别为A 、B ,F 为椭圆C 的右焦点,在椭圆C 上任取异ABO 3Nm (N)2N于A 、B 的点P ,直线P A 、PB 分别与直线x =3交于点M ,N ,直线MB 与椭圆C 交于点Q . (1)求FM →·FN →的值;(2)证明:A 、Q 、N 三点共线.19.(本小题满分16分)已知数列{}n a 满足123n n a a n ++=-,n *∈N .(1)若数列{}n a 为等差数列,求1a ;(2)设1(0)a a a =>,2n n *∀∈N ≥,,不等式22113n n n n a a a a ++++≥成立,求实数a 的最小值.20.(本小题满分16分)已知二次函数f (x )=ax 2+bx +1,g (x )=a 2x 2+bx +1.(1)若f (x )≥g (x )对任意实数x 恒成立,求实数a 的取值范围;(2)若函数f (x )有两个不同零点x 1,x 2;函数g (x )有两个不同零点x 3,x 4. (i)若x 3<x 1<x 4,试比较x 2,x 3,x 4的大小关系; (ii)若x 1=x 3<x 2,m 、n 、p ∈1(,)x -∞,()()()()()()f m f n f pg n g p g m '''==,求证m =n =p .第Ⅱ卷(附加题,共40分)21.[选做题]本题包括A 、B 、C 、D 四小题,每小题10分;请选定其中两题,并在相应的答题区域.................内作答.... A .(选修4-1:几何证明选讲)如图,AB 是半圆的直径,C 是半圆上一点,D 是弧AC 的中点,DE ⊥AB 于E ,AC 与DE 交于M ,求证:AM =DM . AEBCDMB .(选修4-2:矩阵与变换)已知二阶矩阵M 属于特征值3的一个特征向量为a =⎣⎡⎦⎤11,并且矩阵M 对应的变换将点(-1,2)变成点(9,15),求出矩阵M..C .(选修4-4:坐标系与参数方程)已知圆C 的极坐标方程是4cos ρθ=,以极点为平面直角坐标系的原点,极轴为x 轴的正半轴,建立平面直角坐标系,直线l 的参数方程是2222x t m y t ⎧=+⎪⎪⎨⎪=⎪⎩(t 是参数).若直线l 与圆C 相切,求实数m 的值.D .(选修4-5:不等式选讲)设函数()|1||1|f x x x =-++,若不等式|||2|||()a b a b a f x +--≤⋅对任意,a b R ∈且0a ≠恒成立,求实数x 的范围.【必做题】第22题、第23题,每题10分,共计20分.请在答题卡指定区域内作答..........,解答时应写出文字说明、证明过程或演算步骤.22.(本小题满分10分) 如图,在四棱锥O -ABCD 中,底面ABCD 是边长为1的菱形,∠ABC =45°,OA ⊥底面ABCD ,OA =2,M 为OA 的中点. (1)求异面直线AB 与MD 所成角的大小; (2)求平面OAB 与平面OCD 所成锐二面角的余弦值.23.设a 0<a 1<a 2<…<a n (i ∈N *,i =1,2,…,n ),以[b ,c ]表示正整数b ,c 的最小公倍数.求证:1[a 0,a 1]+1[a 1,a 2]+…+1[a n -1,a n ]≤1-12n .2017年高考模拟试卷(5)参考答案M D O A B C1.{1,2,3,6}. 2.1i +. 3. 391. 4. 18. 5.29.6.充分不必要. 7.4. 8.76. 9.10.10.已知函数()sin(2)3f x x π=+(0x <π≤),且1()()3f f αβ==(βα≠),则=+βα ▲ .10.76π.由0x <π≤,知2333x ππ7π+≤≤,因为31()()32f f αβ==<,所以()()3π222332αβππ+++=⨯, 所以76αβπ+=.11.(1,2]. f (f (x ))=⎩⎪⎨⎪⎧x 2-2x ,x <0,2-x 2,0≤x <1,x 4-2x 2,x ≥1.作出函数f (f (x ))的图像可知,当1<k ≤2时,函数y=f (f (x ))-k 有3个不同的零点.12.12.由2AB AC AO +=可得OB OC +=0,即BO OC =,所以圆心在BC 上,且AB AC ⊥. 注意到||||=2AB AO =,所以ππ,,4,2336B C BC AC ====,所以12CA CB ⋅=.13.212-.由()a a b c bc ++=,得1b c b c a a a a ++=⋅,设,b c x y a a ==,则1x y xy ++=, 1a b c x y =++,因为21()2x y x y xy +++=≤,所以222x y ++≥,所以a b c +的最大值为212-.14.设a 为实数,记函数f (x )=ax -ax 3(x ∈[12,1])的图象为C .如果任何斜率不小于1的直线与C 都至多有一个公共点,则a 的取值范围是 ▲ .14.1,42⎡⎤-⎢⎥⎣⎦.由任何斜率不小于1的直线与C 都至多有一个公共点,也即x ∈[12,1]时,曲线()y f x =上任意两点连线的斜率都小于1,所以()1f x '≤在x ∈[12,1]上恒成立.由2()31f x a ax '=-≤,即2310ax a -+≥,设()31g t at a =-+,1,14t ⎡⎤∈⎢⎥⎣⎦,只需1()04g ≥,且(1)0g ≥,所以142a -≤≤.15.解:(1)由正弦定理知,b sin A =a sin B =2,①又a cos B =1, ②①,②两式平方相加,得(a sin B )2+(a cos B )2=3, 因为sin 2B +cos 2B =1, 所以a =3(负值已舍);(2)由(1)中①,②两式相除,得sin B cos B=2,即tan B =2,因为A -B =π4,所以tan A =tan(B +π4)=tan B +tanπ41-tan B tanπ4 =1+21-2=-3-22.(14分)16.证:(1)方法1:取线段PD 的中点M ,连结FM 、AM .因为F 为PC 的中点,所以FM ∥CD ,且FM =12CD .因为四边形ABCD 为矩形,E 为AB 的中点,所以EA ∥CD ,且EA =12CD .所以FM ∥EA ,且FM =EA .所以四边形AEFM 为平行四边形.所以EF ∥AM . 又AM ⊂平面P AD ,EF ⊄平面P AD ,所以EF ∥平面P AD .方法2:连结CE 并延长交DA 的延长线于N ,连结PN .因为四边形ABCD 为矩形,所以AD ∥BC , 所以∠BCE =∠ANE ,∠CBE =∠NAE . 又AE =EB ,所以△CEB ≌△NEA . 所以CE =NE .又F 为PC 的中点,所以EF ∥NP . 又NP ⊂平面P AD ,EF ⊄平面P AD , 所以EF ∥平面P AD .方法3:取CD 的中点Q ,连结FQ 、EQ .在矩形ABCD 中,E 为AB 的中点,所以AE =DQ ,且AE ∥DQ .所以四边形AEQD 为平行四边形, 所以EQ ∥AD .又AD ⊂平面P AD ,EQ ⊄平面P AD , 所以EQ ∥平面P AD .(2分)因为Q 、F 分别为CD 、CP 的中点,又PD ⊂平面P AD ,FQ ⊄平面P AD ,所以FQ ∥平面P AD .又FQ 、EQ ⊂平面EQF ,FQ ∩EQ =Q ,所以平面EQF ∥平面P AD .(5分) 因为EF ⊂平面EQF ,所以EF ∥平面P AD . (2) 设AC 、DE 相交于G .在矩形ABCD 中,因为AB =2BC ,E 为AB 的中点,所以DA AE =CDDA= 2.又∠DAE =∠CDA ,所以△DAE ∽△CDA , 所以∠ADE =∠DCA .又∠ADE +∠CDE =∠ADC =90°, 所以∠DCA +∠CDE =90°. 由△DGC 的内角和为180°,得∠DGC =90°. 即DE ⊥AC .因为点P 在平面ABCD 内的正投影O 在直线AC 上,所以PO ⊥平面ABCD . 因为DE ⊂平面ABCD ,所以PO ⊥DE . 因为PO ∩AC =O ,PO 、AC ⊂平面P AC , 所以DE ⊥平面P AC ,又DE ⊂平面PDE ,所以平面P AC ⊥平面PDE .17.解:(1)设n *()n ∈N 年内所建安置房面积之和首次不低于3 000万m 2, 依题意,每年新建安置房面积是以200为首项,50为公差的等差数列, 从而n 年内所建安置房面积之和为(1)200502n n n -⎡⎤+⨯⎢⎥⎣⎦m 2, 则(1)200502n n n -+⨯≥3 000,整理得,271200n n +-≥, 解得8 (15)n n -≤≥舍去.答:8年内所建安置房面积之和首次不低于3 000万m 2.(2)依题意,每年新建住房面积是以500为首项,1.1为公比的等比数列, 设第m 年所建安置房面积占当年新建住房面积的比为()p m , 则1120050(1)3()500(10.1)10 1.1m m m m p m --+-+==⋅+⨯, 由()(1)p m p m =+得,13410 1.110 1.1m mm m -++=⨯⨯,解得7m =.答:第7年和第8年,所建安置房面积占当年新建住房面积的比保持不变. ·····14分 18.解:(1)分别过点A 、B 作直线l 的垂线,垂足为11,B A , 由题意得11BB AA =,由点到直线距离公式得112a AA BB ==,因为圆A 以1AF 为半径,所以半径为c ,被直线l 截得的弦长为222()2a c -,圆B 以OB 为半径,∴半径为a ,被直线l 截得的弦长为222()2a a -.因为直线l :33y x =-被圆A 和圆B 截得的弦长之比为156,所以()222222241563324a c c a a a --==,解得a c 34=(a >c >0). 因为c e a=,所以所求的离心率为34,(2)存在点P ,使得过点P 有无数条直线被圆A 和圆B 截得的弦长之比为34,设点0(,0)P x ,由题意可得直线方程为0()y k x x =-, 直线截圆A 所得的弦长为2202|(7)|21k x c k ⎛⎫--- ⎪+⎝⎭, 直线截圆B 所得的弦长为2202|(7)|21k x a k ⎛⎫-- ⎪+⎝⎭,[][]22022220220202(7)2(1)(7)134(1)(7)(7)21k x c k c k x k k a k x k x a k ⎛⎫+- ⎪+-++⎝⎭==+--⎛⎫-- ⎪+⎝⎭,化简得22222220016(7)9(7)(1)(169)k x k x k c a +--=+-(*),由(1)离心率为34,得22169c a =,即方程(*)为0)1)(49(002=++x x k ,解得10-=x 或490-=x , 即存在2个点)0,1(-和)0,49(-;当10-=x 时,22||61||81k c kk a k ⎧<+⎪⎨<+⎪⎩,解得7157151515k -<<,当490-=x 时,22||421||561k c kk a k⎧<+⎪⎨<+⎪⎩,解得7772121k -<<,即有无数条直线;故存在2个点P ,使得过点P 有无数条直线被圆A 和圆B 截得的弦长之比为34.19.解:(1)∵()()e ,0x f x x k x '=->.(i )当0k ≤时,()0恒成立'>f x ,∴()f x 的递增区间是0+(,)∞,无递减区间;无极值.(ii )当0>k 时,由()0'>f x 得,>x k ;由()0'<f x 得,0<<x k ;∴()f x 的递减区间是(0,)k ,递増区间是(,+)∞k ,()f x 的极小值为()e k f k =-,无极大值. (2)①由()4f x x <,可得(1)e 40x x k x ---<,因为e 0x >,所以41e x x x k --<,即41ex x k x >--对任意[1,2]x ∈恒成立, 记4()1ex xg x x =--,则4(1)e 4(1)()1e e x x x x x g x -+-'=-=, 因为[1,2]x ∈,所以()0g x '>,即()g x 在[1,2]x ∈上单调递增,故2max228e 8()(2)1e e g x g -==-=.所以实数k 的取值范围为22e 8(,)e-+∞.②由已知1212()()()f x f x x x =≠,结合(1)可知, 0k >,()f x 在(,)-∞k 上单调递减,在(,+)∞k 上单调递增,又(1)0+=f k ,1<+x k 时,()0<f x .不妨设121<<<+x k x k ,此时2x k >,12->k x k ,故要证122+<x x k ,只要证122k x x ->,只要证12(2)()f k x f x ->, 因12()()f x f x =,即证11(2)()f k x f x ->.设()(2)()h x f k x f x =--2(1)(1)()k xxx k x k x k -+-=---<e e e ,2()e ()()e e k xxx k h x x k -'=--22()()k x x x k --=e e e , ∴当<x k 时,()0h x '<,()h x 在(,)-∞k 上单调递减,∴(,)x k ∈-∞时,()()0k k h x h k >=-+=e e , 故当<x k 时,(2)()->f k x f x ,即11(2)()->f k x f x 成立,∴122+<x x k .20.解:(1)111312A B d ===,,;222413A B d ===,,;333716A B d ===,,. …………………………………………………………………3分(2)① 当1n =时,11(1)1a a λλ-=-+,所以11a =;当2n ≥时,由21(1)33n n S a n λλ-=-++,则1121(1)(1)33n n S a n λλ---=-+-+, 两式相减得12(1)3n n n a a a λλλ--=-++,即123n n a a λ-=+, 所以11122233(1)3(1)n n n n b a a b λλλλλ---⎡⎤=++=+==⎢⎥--⎣⎦.……………………………6分 因为112313(1)3(1)b a λλλ-=+=--,所以当13λ≠时,数列{}n b 满足1n n bb λ-=(2n ≥),即数列{}n b 是以313(1)λλ--为首项,λ为公比的等比数列;当13λ=时,数列{}n b 不是等比数列. …………………………………………………8分② 由①知,当13λ≠时,13123(1)3(1)n n a λλλλ--=⋅---;当13λ=时,23(1)n a λ=--.……………………………………………………………10分又{}{}1212max min i i i i n d a a a a a a ++=-,,,,,,, {}{}112123max min i i i i n d a a a a a a ++++=-,,,,,,.由于{}{}1223min min i i n i i n a a a a a a ++++,,,≤,,,,所以由1i i d d +>可得,{}{}12121max max i i a a a a a a +<,,,,,,.所以{}1211max i i a a a a ++=,,,对任意的正整数1232i n =-,,,,恒成立, 即数列{}n a 的前1n -项单调递增是题设成立的必要条件,易知13λ≠.………………12分因为1i i i d a a +=-,112i i i d a a +++=-,所以1212i i i i i d d a a a +++-=+-1231(12)3(1)i λλλλλ--=⋅+--1231(1)3(1)i λλλλ--=⋅--.当1λ>时,由1n n a a +>,得3103(1)λλ->-,解得1λ>, 此时10i i d d +-≥,不符合1i i d d +>,舍去;当01λ<<,由1n n a a +>,得3103(1)λλ-<-,解得113λ<<,此时10i i d d +-<,符合1i i d d +>.综上所述,λ的取值范围是()113,. ……………………………………………………16分第II 卷(附加题,共40分)21A .证:因为P A 是圆O 在点A 处的切线,所以∠P AB =∠ACB .因为PD ∥AC ,所以∠EDB =∠ACB , 所以∠P AE =∠P AB =∠ACB =∠BDE .又∠PEA =∠BED ,故△P AE ∽△BDE . …………………… 10分21B .解:设点(x 0,y 0)为曲线|x |+|y |=1上的任意一点,在矩阵10103M ⎛⎫⎪= ⎪ ⎪⎝⎭对应的变换作用下得到的点为(,)x y '',则0010103x x y y ⎛⎫'⎡⎤⎡⎤ ⎪=⎢⎥⎢⎥ ⎪' ⎪⎣⎦⎣⎦⎝⎭,所以003x x y y ='⎧⎨='⎩ ……5分 所以曲线|x |+|y |=1在矩阵10103M ⎛⎫⎪= ⎪ ⎪⎝⎭对应的变换作用下得到的曲线为|x |+3|y |=1, 所围成的图形为菱形,其面积为1222233⨯⨯= .……10分21C .解:(1)将(2,3)M 及对应的参数3πϕ=代入cos ,(0,sin x a a b y b ϕϕϕ=⎧>>⎨=⎩为参数),得2cos 33sin 3a b ππ⎧=⎪⎪⎨⎪=⎪⎩,所以42a b =⎧⎨=⎩,所以曲线1C 的普通方程为221164x y +=. ……4分 (2)曲线1C 的极坐标方程为2222cos sin 1164ρθρθ+=,将12(,),(,)2A B πρθρθ+代入得222211cos sin 1164ρθρθ+=,222222sin cos 1164ρθρθ+=,所以221211516ρρ+=. ……10分21D .解:因为a >0,b >0,a +b =1,所以(2a +1)+(2b +2)=5,从而(12a +1+2b +1 )[(2a +1)+(2b +2)]=1+4+2b +22a +1+4(2a +1)2b +2≥5+22b +22a +1×4(2a +1)2b +2=9. …………………… 6分 所以12a +1+2b +1≥95.当且仅当2b +22a +1=4(2a +1)2b +2,且a +b =1,即a =13,b =23 时,12a +1+2b +1取得最小值95. …………………… 10分22.解:因为在直三棱柱111ABC A B C -中,AB AC ⊥,所以分别以AB 、AC 、1AA 所在的直线为x 轴、y 轴、z 轴,建立空间直角坐标系, 则111(0,0,0),(2,0,0),(0,4,0),(0,0,3),(2,0,3),(0,4,3)A B C A B C ,因为D 是BC 的中点,所以(1,2,0)D ,……………………………………………………2分 (1)因为111(0,4,0),(1,2,3)AC A D ==-,设平面11A C D 的法向量1111(,,)n x y z =, 则1111100n AC n A D ⎧⋅=⎪⎨⋅=⎪⎩,即111140230y x y z =⎧⎨+-=⎩,取111301x y z =⎧⎪=⎨⎪=⎩,所以平面11A C D 的法向量1(3,0,1)n =,而1(1,2,3)DB =-, 所以111111335cos ,35n DB n DB n DB ⋅<>==⋅, 所以直线1DB 与平面11A C D 所成角的正弦值为33535;…………………………………5分 (2)11(2,0,0)A B =,1(1,2,3)DB =-,设平面11B A D 的法向量2222(,,)n x y z =, 则2112100n A B n DB ⎧⋅=⎪⎨⋅=⎪⎩,即222220230x x y z =⎧⎨-+=⎩,取22232x y z =⎧⎪=⎨⎪=⎩,平面11B A D 的法向量2(0,3,2)n =,所以121212130cos ,65n n n n n n ⋅<>==⋅, 二面角111B A D C --的大小的余弦值13065.……………………………………………10分 23.(1)证明:0)(121)2()(31222222>-=+-++=-b a b a b ab a B A (2)证明:11,1B A n ==;,)2(,11,311nn n n n b a B b a b a n A n +=--+=≥++令,,y b a x b a =-=+且0,>y x ,于是,)2(],)()[()1(21)2()2(1111111n n n n n n n n x B y x y x y n y y x y x n A =--++=--++=+++++ 因为y x C y x C y x C y x y x nn n n n n n n 11323111112)22(])()[(+-++++≥++=--+ ,所以n n n n n n n n B x x y x C y n A ===⋅+≥++)2(22)1(21111.。

浙江省杭州市萧山区2017年高考模拟命题比赛数学试卷6

浙江省杭州市萧山区2017年高考模拟命题比赛数学试卷6

2017年高考模拟试卷数学卷注意事项:1.答题前,考生务必将自己的姓名、准考证号用黑色字亦的签字笔或钢笔镇写在答题纸规定的位置上。

2.每小题选出后,用2B 铅笔把答题纸上对应题目的答案标号涂黑,如需改动,用橡皮擦干净后,再选涂其它答案标号。

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参考公式:如果事件A , B 互斥, 那么 棱柱的体积公式P (A +B )=P (A )+P (B )V =Sh如果事件A , B 相互独立, 那么 其中S 表示棱柱的底面积, h 表示棱柱的高 P (A ·B )=P (A )·P (B )棱锥的体积公式如果事件A 在一次试验中发生的概率是p , 那么n V =31Sh次独立重复试验中事件A 恰好发生k 次的概率 其中S 表示棱锥的底面积, h 表示棱锥的高 P n (k )=C kn p k (1-p )n -k (k = 0,1,2,…, n ) 球的表面积公式棱台的体积公式S = 4πR 2)2211(31S S S S h V ++=球的体积公式其中S 1, S 2分别表示棱台的上.下底面积, h 表示棱台 V =34πR 3的高 其中R 表示球的半径选择题部分(共40分)一、选择题。

(本大题共10小题,每小题4分,共40分.在每小题给出的四个选项中,只有一项是符合题目要求的.) 1、(原创)设全集}1|{},03|{,2-<=>--==x x B x x x A R U ,则图中阴影部分表示的集合为 ( ) A .}0|{>x xB .}13|{-<<-x xC .}03|{<<-x xD .}1|{-<x x2、(改编)若R a ∈,则2=a 是复数i a a z )2(42++-=是纯虚数的 ( ) A .充分非必要条件 B .必要非充分条件 C .充要条件 D .既不充分也不必要条件 3、(改编)如图是一个几何体的三视图,则该几何体的表面积为 ( )A. π++36B. π343218++ C. π++3218 D. π+32 4、(改编)在数列{}n a 中,21=a ⎩⎨⎧+=+为偶数)为奇数)n a n a a n n n (2(21则=6a ( )A.11B.17C.22D.235、(原创)定义在R 上的奇函数()f x 满足:当0>x 时,x x f x2017log 2017)(+=,则在R 上方程()0f x =的实根个数为A .1B .2C .3D .46、(改编)在1,2,3,4,5这五个数中,任取两个不同的数记作,a b ,则满足2()f x x ax b =-+有两个零点的概率是( ). A.52 B.209 C.109 D.217、(名校联盟卷改编)已知定义在),(20π上的函数)(x f ,其导函数为)(x f ',若对任意的)2,0(π∈x 恒有0tan )()(<'-x x f x f 成立,则 A 、)3(2)4(3ππf f <B 、1sin )6(2)1(πf f <C 、)4()6(2ππf f >D 、)3()6(3ππf f >8、(名校联盟卷改编)已知双曲线12222=+b y a x ,圆222a y x =+,过双曲线第一象限内任意一点),(00y x P 作圆C 的两条切线,其切点分别为A 、B ,若AB 与x 轴、y 轴分别交于M 、N 两点,且3||||2222=-ON a OM b ,则双曲线的离心率为 A. 3 B. 2 C.3 D. 29、(五校联盟卷原题)三棱锥BCD A -的底面是正三角形,侧棱相等且两两垂直,点P 是该棱锥表面(包括棱)上一点,且P 到四个顶点的距离有且只有两个不同的值,则这样的点P 的个数有 A. 5 B. 6 C. 8 D. 1110、(五校联盟卷原题),P Q 是两个定点,点M 为平面内的动点,且MP MQλ=(0λ>且1λ≠),点M 的轨迹围成的平面区域的面积为S ,设()S f λ=(0λ>且1λ≠)则以下判断正确的是( )A .)(λf 在)1,0(上是增函数,在),(∞+1上是减函数B .)(λf 在)1,0(上是减函数,在),(∞+1上是减函数C .)(λf 在)1,0(上是增函数,在),(∞+1上是增函数D .)(λf 在)1,0(上是减函数,在),(∞+1上是增函数 非选择题部分(共110分)二、填空题:本大题共7小题,多空题每小题6分,单空题每小题4分,共36分。

枣阳市2017届高三第六次模拟考试数学试卷(理)有答案

枣阳市2017届高三第六次模拟考试数学试卷(理)有答案

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浙江省杭州市2017届高考模拟命题比赛数学试卷6.doc

浙江省杭州市2017届高考模拟命题比赛数学试卷6.doc

浙江省杭州市2017届高考模拟命题比赛数学试卷6一、选择题:(本大题共10小题,每小题4分,共40分.在每小题给出的四个选项中,只有一个是符合题目要求的.)1.已知集合{}2230M x x x =∈+-≤R ,{}50N x x =∈-≤≤R ,则()C M N ⋃=R ( ) A.()53--, B. ](53--, C. )53--⎡⎣, D.]()5301--⋃⎡⎣,, 2.已知复数z 满足()13i 1i z +=+,则z =( ) A .22B .2-C .2D . 2 3.已知,a b 是实数,则“0a >且0b >”是“0a b +>且0ab >”的( ) A.充分而不必要条件B.必要而不充分条件C.充分必要条件D.既不充分也不必要条件4.若tan 3θ=,则22sin 3sin cos θθθ-=( )A.110B.37C.910D.135.已知直线1y x =+与曲线ln()y x a =+相切,则a 的值为( ) A 2B .1C .-1D . 2 6.若实数,x y 满足10,0,0,x y x y x -+≥⎧⎪+≥⎨⎪≤⎩则2ex y+的最小值是( ) A.1B.12eC. eD.2e7.已知(),B n p ξ:,且5E ξ=,3D ξ=,则p 等于( )A.13B.35 C.25 D.238.在ABC ∆中,已知10AB =u u u r ,边AB 上的高为3,则当AC BC u u u r u u u r g 最小时,AC BC +=u u u r u u u r( )A. B.10C.313D1039.已知双曲线122:22=-x y C ,直线l 过点()0,2A ,斜率为k ,当10<<k 时,双曲线的上支上有且仅有一点B 到直线l 的距离为2,则k =( )A.1B.5C.510.给定函数()2,f x x ax b =++设,p q 是满足1p q +=的实数,若对于任意的实数,x y 均有:()()()pf x q x f px qx +≥+,则( ) A.0q p ≤≤B.0p q ≤≤C.0p q ≤≤D.0q p ≤≤二、填空题:(本大题共7小题,第11-14题每题6分,第15-17每题4分,共36分.)11.抛物线24y x =的焦点坐标是________,若直线10ax y -+=经过抛物线焦点,则实数a = .12.在ABC ∆中, 3B π∠=,三边长,,a b c 成等差数列,且6ac =,则ABC S ∆=____,b 的值是_____________.13.已知四棱锥P ABCD -的三视图如下图所示,E 是侧棱PC 上的动点.四棱锥P ABCD -的体积位__________________,异面直线与所成角为_____________.14. 已知数列{}n a 的首项1122,,1,2,31n n n a a a n a +===+….则3a = 数列n n a ⎧⎫⎨⎬⎩⎭的前n 项和n S ,则n S =____________.15.在一次晚会上,9位舞星共上演n 个“三人舞”节目,若在这些节目中,任二人都曾合作过一次,且仅合作一次,则n =__________.16.若曲线22120C x y x +-=:与曲线()20C y mx m --=:x 有两个不同的公共点,则m的取值所组成的集合是_________.17.设二次函数()()20f x ax bx c a b c a =++∈≠R ,,,满足条件:(1)当x ∈R 时,()()42f x f x -=-,且();f x x ≥ABCD P E(2)当()0,2x ∈时,()21;2x f x +⎛⎫≤ ⎪⎝⎭(3)()f x 在R 上的最小值为0.若存在,t ∈R 只要[]1,x m ∈(1m >),就有()f x t x +≤.则m 的最大值为_________. 三、解答题:(本大题共5小题,共74分.解答应写出文字说明,证明过程或演算步骤.) 18.(本题满分14分)在ABC △中,内角,,A B C 所对的边长分别是,,a b c . (1)若()sin sin sin 2A B C C +-=,试判断ABC △的形状. (2)若2,3a A π==,且ABC △的面积3=S ,求,b c 的值;19.(本题满分15分)如图所示,直角梯形ACDE 与等腰直角ABC ∆所在平面互相垂直,F 为BC 的中点,90BAC ACD ∠=∠=o,//AE CD ,22DC AC AE ===. (Ⅰ)求证:AF ∥平面BDE ; (Ⅱ)求二面角B DE C --的余弦值.20.(本题满分15分)已知函数2()f x ax ax =+和()g x x a =-.其中a ∈R .0a ≠且.(1)若函数()f x 与的()g x 图像的一个公共点恰好在x 轴上,求a 的值; (2)若p 和q 是方程()()0f x g x -=的两根,且满足10p q a<<<,证明:当()0,x p ∈时,()()g x f x p a <<-.21.设曲线()2212:1x C y a a+=为正常数与()222C y x m =+:在x 轴上方仅有一个公共点P .(1)求实数m 的取值范围;()a 用表示(2)O 为原点,若1C 与x 轴的负半轴交于点A ,当102a <<时,试求AOP V 的面积的最大值. ()a 用表示22.给定正整数n 和正数M .对于满足条件2211n a a M ++≤的所有等差数列123,,,a a a …,1221=n n n S a a a +++++…+,(1)求证:2251S Mn ⎛⎫≤ ⎪+⎝⎭(2)求1221=n n n S a a a +++++…+的最大值.参考答案一、选择题1-5 CACCD 6-10 ACBBC二、填空题11. ()1,0,1- 12.2 13. 23,90︒14. 89,22n n + 15. 12 16.33⎧⎫⎪⎪-⎨⎬⎪⎪⎩⎭17.9 三、解答题18.解:(1)因为()sin sin sin cos sin B C B C B C -=-①()(),sin sin ()=sin() =sin cos cos sin A B C A B C B C B c B C=π-+=π-+++=sin cos cos sin B c B C +② sin 22sin cos C C C =③将①②③代入sin sin()sin 2A B C C +-= 化简可得:sin sin B C =因为在ABC ∆中,所以B C =,ABC ∆为等腰三角形.(2)因为在ABC ∆中,1,sin 32A S bc A π===所以4bc = ④又因为2221cos 22b c a A bc +-==,且2a =,⑤ 由④⑤解得2,2b c ==19.解:(Ⅰ)取BD 的中点P ,连结,EP FP ,则1//2PF CD , 又因为1//2EA CD ,所以//EA PF ,所以四边形AFPE 是平行四边形, 所以//AF EP ,又因为EP ⊂面,BDE AF ⊄平面BDE , 所以//AF BDE 面(Ⅱ)以CA CD 、所在直线分别作为x 轴,z 轴,以过C 点和AB 平行的直线作为y 轴,建立如图所示坐标系. 由22DC AC AE ===可得:()()()2,0,0,2,2,0,2,0,1,A B E ()0,0,2D则(0,2,0),(0,2,1),AB BE ==-u u u r u u u r (2,2,2)BD =--u u u r.因为面ACDE ⊥面ABC ,面ACDE I 面,ABC AC AB AC =⊥, 所以AB ⊥面.ACDE所以(0,2,0)AB =u u u r是面CDE 的一个法向量.设面BDE 的一个法向量(),,n x y z =r,则BE n ⊥r u u u r ,BD n ⊥r u u u r .所以00,BE BD ⎧=⎪⎨=⎪⎩u u u r g u u u rg n n 即202220,y z x y z -+=⎧⎨--+=⎩ 整理,得200.y z x y z -=⎧⎨+-=⎩令1y =,则2,1,z x ==所以()1,1,2n =r是面CDE 的一个法向量.故2226cos ,||||2112AB AB AB 〈〉===⨯++u u u ru u u r g u u u r n n n . 图形可知:二面角B DE C --的平面角π(0,)2θ∈,所以其余弦值为66. 20.解:(1)设函数()g x 图像与x 轴的交点坐标为(a ,0),又因为点(a ,0)也在函数()f x 的图像上, 所以320a a +=. 而0a ≠,所以1a =-.(2)由题意可知()()()()f x g x a x p x q -=--. 因为10x p q a<<<<,所以()()0a x p x q -->, 所以当()0,x p ∈时,()()0,f x g x ->即()()f x g x >.又()()()()()()(1)f x p a a x p x q x a p a x p ax aq --=--+---=--+,0,110,x p ax aq aq -<-+>->且所以()()f x p a --<0,所以()f x p a <-,综上可知,()()g x f x p a <<-.21.解:联立方程组()22221,2,x y a y x m ⎧+=⎪⎨⎪=+⎩消去y 得2222220x a x a m a ++-= ① 故()222222f x x a x a m a =++-,问题(1)转化为方程①在(),x a a ∈-上有唯一解或等根.只需讨论以下三种情况:1︒0=∆得212a m +=,此时2P x a =-,当且仅当2a a a -<-<,即01a <<时适合; 2︒ ()()0f a f a -<g ,当且仅当a m a -<<;3︒ ()0f a -=得m a =,此时22P x a a =-,当且仅当22a a a a -<-<,即01a <<时适合;()0f a =得m a =-,此时22P x a a =--,当且仅当22a a a a -<-<,无解,从而m a ≠-.综上可知,当01a <<时,212a m +=或a m a-<≤;当1a ≥时,a m a -<<.(2)AOP ∆的面积12P S ay =. 因为102a <<,故当a m a -<≤时,20a a <-+<,由唯一性得2P x a =-+m a =时,P x 取得最小,此时0P y >,从而P y =取得最大,此时P y =,从而S =当212a m +=时,2P x a =-,P y =12S =下面比较与12令12=13a =.故当103a <<时,12≤max 12S =当1132a <<时,12>max S =22. 解:(1) 设公差为d ,1n a a +=,则()()12211112n n n S a a a n a n n d+++=++=+++…+ 故1,21S a nd n +=+又()()22221122411=4310210n M a a a nd aa nd a nd +≥+=-+⎛⎫++- ⎪⎝⎭24,101S n ⎛⎫≥ ⎪+⎝⎭(2)因为 24,101S M n ⎛⎫≥ ⎪+⎝⎭所以1S n ≤+且当a =d =()()12=1=12n S n n n =+++ 由于此时43a nd =,所以22114410101104n S a a M M n +⎛⎫+=== ⎪+⎝⎭g 所以,S的最大值为12n +。

2017年高考模拟命题比赛英语试卷6 含答案

2017年高考模拟命题比赛英语试卷6 含答案

2017年高考模拟试卷英语卷考生须知:本试卷共8页,满分150分,考试时间120分钟。

第Ⅰ卷第一部分:听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话,每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题,每段对话仅读一遍。

1.How will the woman pay?A.By chequeB.By credit cardC.In cash2.Why can’t the woman speak to William Grogan now?A.He is having a meeting.B.He has gone out for lunch.C.He is calling someone else.3.Where does the conversation take place?A.In a restaurantB.In a food storeC.In the man’s house4.What does the man mean?A.The woman should buy a new car.B.It’s a cheaper to repair the old car.C.He’d like to exchange cars with the woman.5.What are the speakers mainly talking about?A.A courseB.A teacherC.A paper第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白,每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。

2017年普通高等学校招生全国统一考试模拟(六)英语试题含答案

2017年普通高等学校招生全国统一考试模拟(六)英语试题含答案

第Ⅰ卷第一部分听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。

每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

听完每段对话后,你都有10秒钟的时间回答有关小题和阅读下一小题。

每段对话仅读一遍。

1.Where does this conversation probably tae place?A.At home.B.In a shop.C.At the doctor's.2.What will the woman probably do?A.Buy boos.B.Borrow boos.C.Read in the library.3.How long has the woman been woring inside?A.Five days.B.One wee.C.Three days.4.What do we now about Charlene?A.She made an ecuse.B.She always had a good reason.C.She was willing to tae the advice.5.Why does the woman refuse to lend the man her bag?A.She doesn't have a bag but a baset.B.She has to use it to carry boos.C.She has many things in it.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或对白,每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置,听每段对话或独白前,你将有时间阅读各个小题。

每小题5秒钟;听完后,各小题给出5秒钟的作答时间。

每段对话或独白读两遍。

听第6段材料,回答第6、7题。

2017年高考英语阅读理解模拟题6

2017年高考英语阅读理解模拟题6

阅读下列材料,从每题所给的四个选项(A、B、C和D)中,选出最佳选项。

ACosmo Books Ltd., 14, Woodman Road,Hertford Estate, Two Bridges,Rickmansworth, West Sussex.Middx.25th FebruaryDear sir,Just over six months ago, I saw an advertisement in the Morning Mail for a set of the complete works of William Shakespeare. Your company, Cosmo books Ltd., offered this set ( eight books of plays and two books of poetry) at what was claimed to be a ‘remarkable’ price: fifteen pounds and fifty pence, including postage and packing. I had wanted a set of Shakespeare’s plays and poems for some time, and these books, in red imitation leather, looked particularly attractive; so I sent for them.Two weeks later, the books arrived, together with a set of the complete works of Charles Dickens which I had not ordered. So I returned the Dickens books to you, with a cheque for fifteen pounds and fifty pence for the works of Shakespeare. Two more weeks passed. Then there arrived on my door step a second set of the works of Shakespeare, the same set of novels by Dickens and a six book set of the plays of Moliere, in French. Since I do not read French, these were of no use to me at all. However, I could not afford to post all these books back to you, so I wrote to you at the end of August of last year, instructing you to come and collect all the books that I did not want, and asking you not to send any other books until further notice.You did not reply to that letter. Instead you sent me a bill for forty two pounds , and a set of the plays of Schiller, in German. Since then, a new set of books has arrived every two weeks, the works of Goethe, the poems of Milton, the plays of Strindberg; I hardly know what I have. The books are still all in their boxes, in the garage, and my car has to stand in the rain outside.I have no room for any more books, and even if I read from now until the Last Judgement, I should not finish reading all the books that you have sent me.Please send no more books, send no more bills, send no more angry letters demanding payment. Just send one large lorry and take all the books away, leaving me only with the one set of the complete works of Shakespeare for which I have paid.Yours faithfully,SIMON WALKER1. Simon Walker wrote the letter to ________A. complain about sending him books he had not ordered.B. urge Cosmo Books Ltd. to take away the books he had not ordered.C. laugh at Cosmo Books Ltd..D. advise readers not to order books from Cosmo Books Ltd..2. The advertisement that Mr. Walker saw in the Morning Mail was for ____A. unlimited number of Cosmo Books.B. a set of 10 books of the works of Shakespeare.C. a book containing all the plays and poems of Shakespeare.D. fifteen pounds and fifty pence.3. Mr. walker answered the advertisement because ________A. he wanted a set of Shakespeare’s works, and this set was cheap, and looked attractive.B. he claimed that the books were being offered at a remarkable price.C. he had ordered the set and had been waiting for them to come for some time.D. the set he already had was not particularly attractive.4. Cosmo Books have _______A. sent bills for books that they have not sent.B. continued to send books that Mr. Walker did not order.C. still not sent Mr. Walker the books that he ordered.D. made a gift to Mr. Walker of several sets of books.5. The tone of the letter is that of _______A. bitternessB. respectC. annoyanceD. humorBIt is pretty much a one-way street. While it may be common for university researchers to try their luck in the commercial world, there is very little traffic in the opposite direction. Pay has always been the biggest deterrent, as people with families often feel they cannot afford the drop in salary when moving to a university job. For some industrial scientists, however, the attractions of academia (学术界) outweigh any financial considerations.Helen Lee took a 70% cut in salary when she moved from a senior post in Abbott Laboratories to a medical department at the University of Cambridge. Her main reason for returning to academia mid-career was to take advantage of the greater freedom to choose research questions. Some areas of inquiry have few prospects of a commercial return, and Lee’s is one of them.The impact of a salary cut is probably less severe for a scientist in the early stages of a career. Guy Grant, now a research associate at the Unilever Centre for Molecular Informatics at the University of Cambridge, spent two years working for a pharmaceutical (制药的) company before returning to university as a post-doctoral researcher. He took a 30% salary cut but felt it worthwhile for the greater intellectual(知识的)opportunities.Higher up the ladder, where a pay cut is usually more significant, the demand for scientists with a wealth of experience in industry is forcing universities to make the transition (转换) to academia more attractive, according to Lee. Industrial scientists tend to receive training that academics do not, such as how to build a multidisciplinary team, manage budgets and negotiate contracts. They are also well placed to bring something extra to the teaching side of an academic role that will help students get a job when they graduate, says Lee, perhaps experience in manufacturing practice or product development. “Only a small number of undergraduates will continue in an academic career. So someone leaving university who already has the skills needed to work in an industrial lab has far more potential in the job market than someone who has spent all their time on a narrow research project.”6.By “a one-way street” (Line 1, Para. 1), the author means ________.A. university researchers know little about the commercial worldB. there is little exchange between industry and academiaC. few industrial scientists would quit to work in a universityD. few university professors are willing to do industrial research7. The word “deterrent” (Line 2, Para. 1) most probably refers to something that ________.A. keeps someone from taking actionB. helps to move the trafficC. attracts people’s attentionD. brings someone a financial burden8. What was Helen Lee’s major consideration when she changed her job in the middle of her career?A. Flexible work hours.B. Her research interests.C. Her preference for the lifestyle on campus.D. Prospects of academic accomplishments.9. Guy Grant chose to work as a researcher at Cambridge in order to ________.A .do financially more rewarding workB .raise his status in the academic worldC. enrich his experience in medical researchD. exploit better intellectual opportunities10. What contribution can industrial scientists make when they come to teach in a university?A. Increase its graduates’ competitiveness in the job market.B. Develop its students’ potential in research.C. Help it to obtain financial support from industry.D. Gear its research towards practical applications.CBeing sociable looks like a good way to add years to your life. Relationships with family, friends, neighbours, even pets, will all do the trick, but the biggest longevity (长寿) boost seems to come from marriage or an equivalent relationship. The effect was first noted in 1858 by William Farr, who wrote that widows (寡妇)and widowers were at a much higher risk of dying than their married peers. Studies since then suggest that marriage could add as much as seven years to a man’s life and two to a woman’s. Th e effect holds for all causes of death, whether illness, accident or self-harm.Even if the odds are stacked against you, marriage can more than compensate. Linda Waite of the University of Chicago has found that a married older man with heart disease can expect to live nearly four years longer than an unmarried man with a healthy heart. Likewise, a married man who smokes more than a pack a day is likely to live as long as a divorced man who doesn’t smoke. There’s a flip side, however, as partners are more likely to become ill or die in the couple of years following their spouse’s death, and caring for a spouse with mental disorder can leave you with some of the same severe problems. Even so, the odds favour marriage. In a 30-year study of more than 10,000 people, Nicholas Christakis of Harvard Medical School describes how all kinds of social networks have similar effects.So how does it work? The effects are complex, affected by socio-economic factors, health-service provision, emotional support and other more physiological (生理的) mechanisms. For example, social contact can boost development of the brain and immune system, leading to better health and less chance of depression later in life. People in supportive relationships may handle stress better. Then there are the psychological benefitsof a supportive partner.A life partner, children and good friends are all recommended if you aim to live to 100. The ultimate social network is still being mapped out, but Christakis says: “People are interconnected, so their health is interconnected.”11. William Farr’s study and other studies show that _________.A .social life provides an effective cure for illnessB. being sociable helps improve one’s quality of lifeC. women benefit more than men from marriageD. marriage contributes a great deal to longevity12. Linda Waite’s studies support the idea that _________.A. older men should quit smoking to stay healthyB. marriage can help make up for ill healthC. the married are happier than the unmarriedD. unmarried people are likely to suffer in later life13. It can be inferred from the context that the “flip side” (Line 4, Para. 2) refers to _________.A. the disadvantages of being marriedB. the emotional problems arising from marriageC. the responsibility of taking care of one’s familyD. the consequence of a broken marriage14. What does the author say about social networks?A. They have effects similar to those of a marriage.B. They help develop people’s community spirit.C. They provide timely support for those in need.D. They help relieve people of their life’s burdens.15. What can be inferred from the last paragraph?A. It’s important that we develop a social network when young.B. To stay healthy, one should have a proper social network.C. Getting a divorce means risking a reduced life span.D. We should share our social networks with each other.DA few years ago I asked my children’s governess, Julia Vassilyevna, to come into my study.“ Sit down, Julia Vassilyevna,” I said.“Let’s settle our accounts. Although you most likely need some money, you stand on ceremony and won’t ask for it yourself. Now then, we agree on thirty rubles a month…”“ Forty.”“ No, thirty. I made a note of it. I always pay the governess thirty. Now then, you’ve been here two months, so…”“ Two months and five days.”“Exactly two months. I made a specific note of it. That means you have sixty rubles coming to you. Subtract nine Sundays… you know you didn’t work with Kolya on Sundays, you only took walks. And three holidays…”Julia Vassilyevna flushed a deep red and picked at the flounce of her dress, but--- not a word.“ Three holidays, therefore take off twelve rubles. Four days Kolya was sick and there were no lessons, as you were occupied only with Vanya. Three days you had a toothache and my wife gave you permission not to work after lunch. Twelve and seven---nineteen. Subtract…that leaves…hmm…forty-one rubles. Correct?”Julia Vassilyena’s left eye reddened and filled with moisture. Her chin trembled; she coughed nervously and blew her nose, but---not a word.“ Around New Year’s you broke a teacup and saucer: take off two rubles. The cup cost more, it was an heirloom, but---let it go. When didn’t I take a loss? Then, due to your neglect, Kolya climbed a tree and tore his jacket: take off ten. Also due to your heedlessness the maid stole Vanya’s shoes. You ought to watch everything! You get paid for it. So, that means five more rubles off. The tenth of January I gave you ten rubles…”“ You didn’t ” whispered Julia Vassilyevna.“ But I made a note of it.”“ Well…all right.”“ Take twenty-seven from forty-one ---that leaves fourteen.”Both eyes filled with tears. Perspiration appeared on the thin, pretty little nose. Poor girl!“ Only once was I given any money,” she said in a trembling voice, “ and that was by your wife. Three rubles, nothing more.”“ Really? You see now, and I didn’t make a note of it! Take three from fourteen…leaves eleven. Here’s your money, my dear. Three , three, three, one and one. Here it is!”I handed her eleven rubles. She took them and with trembling fingers stuffed them into her pocket.“ Merci,” she whispered.I jumped up and started pacing the room. I was overcome with anger.“ For what, this ---‘merci’?” I asked.“ For the money.”“But you know I’ve cheated you, God’s sake---robbed you! I have actually stolen from you! Why this ‘merci’?”“ In my other places they didn’t give me anything at all.”“ They didn’t give you anything? No wonder! I played a little joke on you, a cruel lesson, just to teach you …I’m going to give you the entire eighty rubles! Here they are in an envelope all ready for you…Is it really possible to be so spineless? Why don’t you protest? Why be silent? Is it possible in this world to be without teeth and claws--- to be such a nincompoop?”She smiled crookedly and I read in her expression: “ It is possible.”I asked her pardon for the cruel lesson and , to her great surprise, gave her the eighty rubles. She murmured her litter “merci” several times and went out. I looked after her and thought: “How easy it is to crush the weak in this world!”16. When the employer called Julia Vassilyevna in to talk with him, what he really wanted to do was _______A. to settle their accountsB. to criticize her for neglecting her duties as a governess.C. to play a joke on her so as to amuse himself.D. to teach her not to be so spineless.17. The employer made all those criticisms about Julia Vassiliyevna’s work because __________A. she had neglected her duties.B. he wanted to pay her as little money as possible.C. he wanted to make her realize that she was being cheated and protest against it.D. he wanted to make her feel miserable.18. Julia Vassilyevna accepted everything her employer said because ________A. she had in fact neglected her duties.B. she was a very dumb girl.C. she thought it was of no use to protest to her employer.D. she loved the children she taught.19. The employer became very angry when Julia Vassilyevna said “ Merci” because ________A. she didn’t say “Merci” loudly.B. she didn’t protest to him.C. he thought that by simply saying “ Merci” she wasn’t polite enough.D. he didn’t like the way in which she stuffed the money into per pocket and expressed her thanks.20. The text is about _______A. how a governess was cheated by her employer.B. how an employer tried to teach the governess a cruel lesson.C. how a governess was fired by her employer for being a nincompoop.D. how an employer punished the governess for not having done a good job.BBABC CABDA DBAAB DCCBB。

2017届江苏高考模拟信息卷六解析

2017届江苏高考模拟信息卷六解析

•B
• 5.下列诗句与所描写景点对应错误的一项是, • A: 大漠孤烟直,长河落日圆——金沙江 B: 接天莲叶无穷碧,映日荷花别样红——西湖 C: 两岸青山相对出,孤帆一片日边来——长江 D: 岱宗何崔嵬,群山无与比——泰山 参考答案: A
• “大漠孤烟直,长河落日圆”出自王维《使至塞上》, 其中的长河指黄河。
• “接天莲叶无穷碧,映日荷花别样红”出自杨万里的 《晓出静慈寺送林子方》; • “两岸青山相对出,孤帆一片日边来”出自李白的《望 天门山》; • “岱宗何崔嵬,群山无与比”出自贾鲁的登泰山。

严郢,字叔敖,华州华阴人。及进士第,补太常协律 郎,守东都太庙。
• 吕諲镇江陵,表为判官。方士申泰芝以术得幸肃宗,遨 游湖、衡间,以妖幻诡众,奸赃钜万,潭州刺史庞承鼎 按治。帝不信, 召还泰芝,下承鼎江陵狱。郢具言泰 芝左道,帝遣中人与諲杂讯有状,帝不为然。
• 宰相杨炎请屯田丰州,发关辅民凿陵阳渠,郢习朔边病 利,即奏止之。渠卒不成,弃之。炎恶异己,阴讽御史 张著劾郢匿发民浚渠,使怨归上。系金吾。长安中日数 千人遮建福门讼郢冤,帝微知之,削兼御史中丞。人知 郢得原,皆迎拜。会秋旱,郢请蠲租税,炎令度支御史 按覆,以不实,罢为大理卿。
• 炎之罢,卢杞引郢为御史大夫,共谋炎罪。即逮捕河中 观察使赵惠伯下狱,楚掠惨棘,锻成其罪,卒逐炎崖州, 惠伯费州。天下以郢挟宰相报仇为不直。然杞用 郢败 炎,内忌郢才,因按蔡廷玉事,杀御史郑詹,出郢为费 州刺史。道逢柩殡,问之,或曰:“赵惠伯之殡。”郢 内惭,忽忽岁余卒。
2017届江苏高考模拟信息卷六
•1.1.在下面一段话的空缺处依次填入词语,最恰当的一组是(3 分) •写作,要有语言创新意识。常用的汉字只有两三千个,但是如 何 ▲ 应用却没有止境。有创新意识的作者,才有可能让笔下的 文字散发出无穷的魅力。在我们读过的作品中,有些词句的妙用让 人 ▲ ,有些语言的精美令读者 ▲ ,成功的表达无不体现了 作者的创新精神。 • A.糅合 拍手称快 刻骨铭心 B.糅合 拍案叫绝 耿耿于怀 • C.组合 拍案叫绝 刻骨铭心 D.组合 拍手称快 耿耿于怀

2017高考仿真卷+英语(六)含解析

2017高考仿真卷+英语(六)含解析

2017高考仿真卷·英语(六)第一部分听力(略)第二部分阅读理解(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下面短文,从每题所给的A、B、C和D四个选项中,选出最佳选项。

ADear Interested Pre-Applicant,On January 4,2017,The Housing Authority of Coo County (HACC) will begin accepting pre-applications for one (1) bedroom standard and (1) bedroom Uniform Federal Accessibility Standard units (UFAS) at the following Affordable Housing Community.North Suburban Housing215 W.Miner in Arlington Heights,IL & 9238 Gross Point Road in Soie,ILNoteUniform Federal Accessibility Standards unitsThese apartments are all one bedroom,one bathroom units that include accessible features such as lower cabinets in the itchen as well as grab bars and a shower seat in the bathroom.These units also have wider doorways for easier access for wheelchairs and walers and accessible controls for heating and air conditioning as well as the shower set at a height that is easier for a wheelchair user to reach.To qualify for the wait list opening▶The head of household or co-head/spouse must be eligible for one of following preferencesa person 62 years of age or older OR a person 18 years of age or older with a disability;▶The annual income for a one-person household cannot eceed(超过) $32,340 and the annual income for a two-person household cannot eceed $36,960;▶Household will be required to pay rent equal up to 30% of their annual income.Please note that the HACC will only accept the first 300 completed pre-applications.Beginning January 4,2017,Wait List Pre-Applications will be available at the following locations during noted business hoursHACC Wait List & Leasing Department175 West Jacson Blvd.Suite 350 Chicago,IL 60604,Fa(312) 542-4752North Suburban Housing Offices215 W.Miner In Arlington Heights,IL 60005 OR 9238 Gross Point Road in Soie,IL 60077Business Hours For All OfficesMonday to Friday,830 am to 430 pm,Closed WednesdaysImportantIf you or anyone in your household is a person with a disability and requires a specific accommodation or sees assistance with the completion of the pre-application,please contact the housing authority via any of the methods listed above,or call (312) 542-4786.Please contact our office with any further questions at (312) 542-4788.Sincerely,Wait List and Leasing Department21.What does the underlined word “eligible”probably mean?A.Qualified.B.Affordable.C.Responsible.D.Prepared.22.According to the letter,you can’t apply for the housing if .A.you are over 62 years oldB.you earn less than $32,340 a yearC.you are under 18 years oldD.you are a disabled middle-aged man23.If you have any further questions about the pre-application,you can .A.write to HACCB.call (312) 542-4788C.Fa (312) 542-4752D.visit HACC personally24.What can we learn from the letter?A.The HACC will accept all the pre-applications.B.The pre-application must be completed by yourself.C.Wait List Pre-Applications will be available at any time.D.The apartments have special designs for disabled people.B“Worrying is lie a rocing chair.It gives you something to do,but doesn’t get you anywhere.”This saying has been flowing through my mind these past few wees.I recently had shouldersurgery,and I am being challenged by new situations in my job.Even though I now challenge is good,I have been worrying here and there.Will the surgery go well?Will I feel relief from the discomfort I have been eperiencing in that shoulder?How will I fit physical therapy into my weely schedule?Have I prepared my students well for their fall term eams?How will I get all the fall term grades and comments completed by the deadline with other wor responsibilities and my son’s birthday party and our eldest daughter coming home from school for Thansgiving brea?The fact is,everything will get done.There are ways to squeee in wor and appointments.What is important is the attitude you have throughout.Worry and aniety can feel uncomfortable.Such emotions can trouble us,and as the above saying implies,will eep us stuc in place.If we recognie that these emotions are factoring into a current situation,there is always the opportunity to redirect the energy that goes into worry and aniety towards faith and action.Faith that you can handle the issue or tas at hand and taing action with that faith supporting you as you face challenges.Fleeing away from worry and aniety is not easy,nor is it without its own strains,but it is possible.How satisfying it feels to complete something you were worried and anious about.I will get through this tas and healing with faith and action,and with the support and understanding of my family.But,perhaps I should do what a dear friend just told me she did,which was to find a copy of my grandfather Norman Vincent Peale’s landmar boo,The Power of Positive Thining,and reread it.This will be part of my action plan,an action plan that will support my faith and attitude,thus reducing my worry and aniety.25.Which of the following can replace the underlined word “therapy”in Paragraph 2?A.eaminationB.painC.treatmentD.change26.What does the author want to tell us by listing so many worries and tass?A.Teachers are easily fed up with their job.B.Everybody has to answer for their own fault.C.We sometimes worried unnecessarily in daily life.D.Some people are always addicted to unhappiness.27.What does the author intend us to do in Paragraph 3?A.Focus our energy on faith and action.B.Realie the harm of worry and aniety.C.Thin twice before taing action.D.Ignore the troubles in life.28.What will be the author’s unique way to improve her mood?A.Escape from worry immediately. Bplete some tass.C.Get support from her family.D.Read again her grandfather’s boo.CBEIJING—Family planning officers will handle the illegal birth case of film director hang Yimou by following legal procedures,a government spoesman said on Tuesday,December 10,2013.When ased to comment on hang’s case at a press conference,Yao Hongwen,spoesman for the National Health and Family Planning Commission (NHFPC),said “Every citien is equal before the law and nobody is allowed to have privileges(特权) to give birth to more children than allowed.”Yao’s comment came after media stories claimed that hang had remarried and fathered a total of seven childrenone with his former wife,three with his present wife,Chen Ting,and another three from two lovers.“The NHFPC’s attitude towards citiens’birth violations(违法) is consistent and clear.Every Chinese citien should obey and safeguard the seriousness of laws and regulations.”NHFPC spoesman Yao Hongwen said.China’s family planning policy,introduced in the late 1970s to slow down the country’s growing population,limits most urban couples to one child and most rural couples to two.The policy was recently relaed,allowing couples to have two children if one of them is an only child.Those who brea the policy face severe fines.The amount of the fine is generally several times the yearly annual income of the two spouses(配偶).Some media calculations put the fine that hang may face at as much as 160 million yuan (over $26 million).hang’s birth case has caused criticism online about famous people who have more children than the policy allows.In an open letter on December 1,hang admitted that the couple has two sons and daughter and they are willing to accept investigation and punishment according to China’s laws and regulations.He apologied to the public.However,the letter denied that hang had lovers and was father to as many as seven children from various mothers.hang,one of the “fifth generation”of Chinese filmmaers,made his first directorial wor in 1987 with “Red Sorghum”and has won a number of award worldwide.He was chief director of the opening and closing ceremonies of the Beijing Olympic Games.29.The purpose of the news report is to .A.claim that hang Yimou remarried and fathered a total of seven childrenrm us that hang’s illegal birth case will be dealt with by the governmentC.convince us that hang is a famous director and his case will be pardonedD.confirm that hang’s illegal birth case has resulted in lots of criticism online30.According to Yao Hongwen,which of the following statements is true?A.Every Chinese citien should obey and safeguard laws and regulations.B.Few people have the privileges to give birth to more children than allowed.C.China’s family planning policy has never been relaed since introduced.D.hang will be fined as much as 160 million yuan for his violations.31.Judging from the recent relaed policy,a couple .A.can have as many as three children if they live in a rural areaB.will definitely be fined if they have two children nowadaysC.can give birth to two children if one of them is an only childD.can only give birth to two children if they live in an urban areaDChildren grow quicly,especially their feet,but many families cannot buy new shoes each time a child needs them.As a result,about 300 million children around the world go barefoot.Those children ris picing up diseases and parasites from the soil.An American man may have a solution for those barefoot ids—a shoe that grows with the child.enton Lee is the creator of The Shoe That Grows.He eplains how the shoe wors and how it is able to last so long.“It grows in three placesthe front;it can also grow on the side with Velcro;and on the bac with our bucle.It can last up to five years.The bottom is compressed rubber lie tire rubber.The top is just high-quality leather.”The shoe comes in two siessmall that fits children aged 4 to 9 and large for children aged 9 to 14.Lee says he got the idea for the shoe while woring as a volunteer in an orphanage in enya.“...and there was a little girl in a white dress waling net to me.And I just remember looing down and seeing how small her shoes were.They were so small that she had to cut open the front of her shoes to let her toes stic out.And I just remember thining,wouldn’t it be nice if there were a pair of shoes that could grow with her feet?”However,Lee says it was not easy to turn his idea into a reality.“I ept writing down I should do this because of this or because of that.”Finally,in 2009 Lee founded a non-profit organiation called Because International.In the office of Because International,Lee eeps a pair of his own shoes to help him remember his promise.“I told myself I would not get new shoes until the idea was done.So,those are the shoes that I wore for a little over five years.”32.Which of the following is the characteristic of the newly invented shoes?A.They are more epensive.B.They are not so comfortable.C.They can grow with a child.D.They come in different sies.33.Who is this ind of newly invented shoes intended for?A.All the teenagers.B.Children aged 4 to 14.C.Middle-aged women in Africa.D.Senior citiens with low income.34.What does Para.4 mainly tal about?A.How the shoes wor.B.How Lee got his inspiration.C.How to mae the new shoes.D.Where to order this ind of shoes.35.Why does enton Lee eep a pair of his own shoes in the office?A.To remind him of his promise.B.To copy the design of them.C.To get a change conveniently.D.To use them as an eample.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,选项中有两项为多余选项。

【核心八模】2017年普通高等学校招生全国统一考试模拟(六)理科综合试题(PDF版)

【核心八模】2017年普通高等学校招生全国统一考试模拟(六)理科综合试题(PDF版)

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浙江省杭州市萧山区2017年高考模拟命题比赛数学试卷6

浙江省杭州市萧山区2017年高考模拟命题比赛数学试卷6

2017年高考模拟试卷数学卷注意事项:1.答题前,考生务必将自己的姓名、准考证号用黑色字亦的签字笔或钢笔镇写在答题纸规定的位置上。

2.每小题选出后,用2B 铅笔把答题纸上对应题目的答案标号涂黑,如需改动,用橡皮擦干净后,再选涂其它答案标号。

不能答在试题卷上。

参考公式:如果事件A , B 互斥, 那么 棱柱的体积公式P (A +B )=P (A )+P (B )V =Sh如果事件A , B 相互独立, 那么 其中S 表示棱柱的底面积, h 表示棱柱的高 P (A ·B )=P (A )·P (B )棱锥的体积公式如果事件A 在一次试验中发生的概率是p , 那么n V =31Sh次独立重复试验中事件A 恰好发生k 次的概率 其中S 表示棱锥的底面积, h 表示棱锥的高 P n (k )=C kn p k (1-p )n -k (k = 0,1,2,…, n ) 球的表面积公式棱台的体积公式S = 4πR 2)2211(31S S S S h V ++=球的体积公式其中S 1, S 2分别表示棱台的上.下底面积, h 表示棱台 V =34πR 3的高 其中R 表示球的半径选择题部分(共40分)一、选择题。

(本大题共10小题,每小题4分,共40分.在每小题给出的四个选项中,只有一项是符合题目要求的.) 1、(原创)设全集}1|{},03|{,2-<=>--==x x B x x x A R U ,则图中阴影部分表示的集合为 ( ) A .}0|{>x xB .}13|{-<<-x xC .}03|{<<-x xD .}1|{-<x x2、(改编)若R a ∈,则2=a 是复数i a a z )2(42++-=是纯虚数的 ( ) A .充分非必要条件 B .必要非充分条件 C .充要条件 D .既不充分也不必要条件 3、(改编)如图是一个几何体的三视图,则该几何体的表面积为( )3A. π++36B. π343218++ C. π++3218 D. π+32 4、(改编)在数列{}n a 中,21=a ⎩⎨⎧+=+为偶数)为奇数)n a n a a n n n (2(21则=6a ( )A.11B.17C.22D.235、(原创)定义在R 上的奇函数()f x 满足:当0>x 时,x x f x2017log 2017)(+=,则在R 上方程()0f x =的实根个数为A .1B .2C .3D .46、(改编)在1,2,3,4,5这五个数中,任取两个不同的数记作,a b ,则满足2()f x x ax b =-+有两个零点的概率是( ). A.52 B.209 C.109 D.217、(名校联盟卷改编)已知定义在),(20π上的函数)(x f ,其导函数为)(x f ',若对任意的)2,0(π∈x 恒有0tan )()(<'-x x f x f 成立,则 A 、)3(2)4(3ππf f <B 、1sin )6(2)1(πf f <C 、)4()6(2ππf f >D 、)3()6(3ππf f >8、(名校联盟卷改编)已知双曲线12222=+b y a x ,圆222a y x =+,过双曲线第一象限内任意一点),(00y x P 作圆C 的两条切线,其切点分别为A 、B ,若AB 与x 轴、y 轴分别交于M 、N 两点,且3||||2222=-ON a OM b ,则双曲线的离心率为 A. 3 B. 2 C.3 D. 29、(五校联盟卷原题)三棱锥BCD A -的底面是正三角形,侧棱相等且两两垂直,点P 是该棱锥表面(包括棱)上一点,且P 到四个顶点的距离有且只有两个不同的值,则这样的点P 的个数有 A. 5 B. 6 C. 8 D. 1110、(五校联盟卷原题),P Q 是两个定点,点M 为平面内的动点,且MP MQλ=(0λ>且1λ≠),点M 的轨迹围成的平面区域的面积为S ,设()S f λ=(0λ>且1λ≠)则以下判断正确的是( )A .)(λf 在)1,0(上是增函数,在),(∞+1上是减函数B .)(λf 在)1,0(上是减函数,在),(∞+1上是减函数C .)(λf 在)1,0(上是增函数,在),(∞+1上是增函数D .)(λf 在)1,0(上是减函数,在),(∞+1上是增函数 非选择题部分(共110分)二、填空题:本大题共7小题,多空题每小题6分,单空题每小题4分,共36分。

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开始结束输出S n ←1, S ←0S < 100 n ←n + 1S ←S + 2n NY (第5题)2017年高考模拟试卷(6)南通市数学学科基地命题第Ⅰ卷(必做题,共160分)一、填空题:本大题共14小题,每小题5分,共70分 . 1. 设集合A = {1,x },B = {2,3,4},若A ∩B ={4},则 x = ▲ .2. 若复数z 1=2+i ,z 1·-z 2=5,则z 2= ▲ .3. 从数6,7,8,9,10,11六个数中,任取两个不同的数, 则两个数互质的概率是 ▲ .4.已知一组数据x 1,x 2,…,x 100的方差是2,则数据 3x 1,3x 2,…,3x 100 的标准差为 ▲ .5.执行右边的程序框图,则输出的S 的值为 ▲ .6.设正四棱柱ABCD —A 1B 1C 1D 1的底面ABCD 是单位正方形,其表面积14,则AA 1= ▲ . 7.不等式组⎩⎨⎧y ≤x +2y ≥x0≤y ≤4x ≥0表示的平面区域的面积为S ,则S 的值为 ▲ .8.函数y =sin(ωx +π4)(ω>0)的图象在[0,1]上恰有三个最高点,则ω的取值范围是 ▲ .9.若两个非零向量a ,b 的夹角为60°,且(a +2b )⊥(a -2b ),则向量a +b 与a -b 的夹角的余弦值是 ▲ .10.已知函数f (x )=e x -1-tx ,∃x 0∈R ,f (x 0)≤0,则实数t 的取值范围 ▲ .11.已知数列{a n }是一个等差数列,首项a 1>0,公差d ≠0,且a 2、a 5、a 9依次成比数列,则 使a 1+a 2+…+a n >100a 1的最小正整数k 的值是 ▲ .12.抛物线y 2=2px (p >0)和双曲线x 2a 2-y 2b2=1(a >0,b >0)有一个相同的焦点F 2(2,0),而双曲线的另一个焦点F 1,抛物线和双曲线交于点B 、C ,若△BCF 1是直角三角形,则双曲线的离心率是 ▲ .13.△ABC 中,∠A 、∠B 、∠C 的对边分别为a 、b 、c ,若a 2cos A =b 3cos B =c 6cos C,则cos A cos B cos C = ▲ .14.已知函数f (x )=2x 3+7x 2+6x x 2+4x +3,x ∈[0,4],则f (x )最大值是 ▲ .第 2页,共 13页AA 1B 1 CD 1 B C 1D MO 1二、解答题:本大题共6小题,共90分.15.(本小题满分14分)已知α∈(0,π),且sin(α+π3)=6-24.(1)求sin(α-π4)的值;(2)求cos(2α-π3)的值.16.(本小题满分14分)如图,四棱柱ABCD -A 1B 1C 1D 1中,底面ABCD 是菱形,M 是AB的中点,O 1是A 1C 1与B 1D 1的交点. (1)求证:O 1M ∥平面BB 1C 1C ;(2)若平面AA 1C 1C ⊥平面ABCD ,求证:四边形BB 1D 1D 是矩形.17.(本小题满分14分)如图所示,一根绳穿过两个定滑轮,且两端分别挂有3(N)、2(N)的重物.现在两个滑轮之间的绳上挂一个重量为m (N)的重物,恰好使系统处于平衡状态. (1)若∠AOB =120°,求m 的值; (2)求m 的取值范围.ABO 3Nm (N)2N18. 椭圆C :x 24+y 23=1的左、右顶点分别为A 、B ,F 为椭圆C 的右焦点,在椭圆C 上任取异于A 、B 的点P ,直线P A 、PB 分别与直线x =3交于点M ,N ,直线MB 与椭圆C 交于点Q . (1)求FM →·FN →的值;(2)证明:A 、Q 、N 三点共线.19.(本小题满分16分)已知数列{}n a 满足123n n a a n ++=-,n *∈N .(1)若数列{}n a 为等差数列,求1a ;(2)设1(0)a a a =>,2n n *∀∈N ≥,,不等式22113n n n n a a a a ++++≥成立,求实数a 的最小值.20.(本小题满分16分)已知二次函数f (x )=ax 2+bx +1,g (x )=a 2x 2+bx +1.(1)若f (x )≥g (x )对任意实数x 恒成立,求实数a 的取值范围;(2)若函数f (x )有两个不同零点x 1,x 2;函数g (x )有两个不同零点x 3,x 4. (i)若x 3<x 1<x 4,试比较x 2,x 3,x 4的大小关系; (ii)若x 1=x 3<x 2,m 、n 、p ∈1(,)x -∞,()()()()()()f m f n f pg n g p g m '''==,求证m =n =p .第Ⅱ卷(附加题,共40分)21.[选做题]本题包括A 、B 、C 、D 四小题,每小题10分;请选定其中两题,并在相应的答题区域.................内作答.... A .(选修4-1:几何证明选讲)如图,AB 是半圆的直径,C 是半圆上一点,D 是弧AC 的中点,DE ⊥AB 于E ,AC 与DE 交于M ,求证:AM =DM .A EBCD M第 4页,共 13页B .(选修4-2:矩阵与变换)已知二阶矩阵M 属于特征值3的一个特征向量为a =⎣⎡⎦⎤11,并且矩阵M 对应的变换将点(-1,2)变成点(9,15),求出矩阵M..C .(选修4-4:坐标系与参数方程)已知圆C 的极坐标方程是4cos ρθ=,以极点为平面直角坐标系的原点,极轴为x 轴的正半轴,建立平面直角坐标系,直线l 的参数方程是2222x t m y t ⎧=+⎪⎪⎨⎪=⎪⎩(t 是参数).若直线l 与圆C 相切,求实数m 的值.D .(选修4-5:不等式选讲)设函数()|1||1|f x x x =-++,若不等式|||2|||()a b a b a f x +--≤⋅对任意,a b R ∈且0a ≠恒成立,求实数x 的范围.【必做题】第22题、第23题,每题10分,共计20分.请在答题卡指定区域内作答..........,解答时应写出文字说明、证明过程或演算步骤.22.(本小题满分10分) 如图,在四棱锥O -ABCD 中,底面ABCD 是边长为1的菱形,∠ABC =45°,OA ⊥底面ABCD ,OA =2,M 为OA 的中点. (1)求异面直线AB 与MD 所成角的大小; (2)求平面OAB 与平面OCD 所成锐二面角的余弦值.23.设a 0<a 1<a 2<…<a n (i ∈N *,i =1,2,…,n ),以[b ,c ]表示正整数b ,c 的最小公倍数.求证:1[a 0,a 1]+1[a 1,a 2]+…+1[a n -1,a n ]≤1-12n .2017年高考模拟试卷(5)参考答案一、填空题M D O A B C1.{1,2,3,6}. 2.1i +. 3. 391. 4. 18. 5.29.6.充分不必要. 7.4. 8.76. 9.10.10.已知函数()sin(2)3f x x π=+(0x <π≤),且1()()3f f αβ==(βα≠),则=+βα ▲ .10.76π.由0x <π≤,知2333x ππ7π+≤≤,因为31()()32f f αβ==<,所以()()3π222332αβππ+++=⨯, 所以76αβπ+=.11.(1,2]. f (f (x ))=⎩⎪⎨⎪⎧x 2-2x ,x <0,2-x 2,0≤x <1,x 4-2x 2,x ≥1.作出函数f (f (x ))的图像可知,当1<k ≤2时,函数y=f (f (x ))-k 有3个不同的零点.12.12.由2AB AC AO +=可得OB OC +=0,即BO OC =,所以圆心在BC 上,且AB AC ⊥. 注意到||||=2AB AO =,所以ππ,,4,2336B C BC AC ====,所以12CA CB ⋅=.13.212-.由()a a b c bc ++=,得1b c b c a a a a ++=⋅,设,b c x y a a==,则1x y xy ++=,1a b c x y =++,因为21()2x y x y xy +++=≤,所以222x y ++≥,所以a b c+的最大值为212-. 14.设a 为实数,记函数f (x )=ax -ax 3(x ∈[12,1])的图象为C .如果任何斜率不小于1的直线与C 都至多有一个公共点,则a 的取值范围是 ▲ .14.1,42⎡⎤-⎢⎥⎣⎦.由任何斜率不小于1的直线与C 都至多有一个公共点,也即x ∈[12,1]时,曲线()y f x =上任意两点连线的斜率都小于1,所以()1f x '≤在x ∈[12,1]上恒成立.由2()31f x a ax '=-≤,即2310ax a -+≥,设()31g t at a =-+,1,14t ⎡⎤∈⎢⎥⎣⎦,只需1()04g ≥,且(1)0g ≥,所以142a -≤≤.二、解答题第 6页,共 13页15.解:(1)由正弦定理知,b sin A =a sin B =2,①又a cos B =1, ②①,②两式平方相加,得(a sin B )2+(a cos B )2=3, 因为sin 2B +cos 2B =1, 所以a =3(负值已舍);(2)由(1)中①,②两式相除,得sin B cos B=2,即tan B =2,因为A -B =π4,所以tan A =tan(B +π4)=tan B +tanπ41-tan B tanπ4 =1+21-2=-3-22.(14分)16.证:(1)方法1:取线段PD 的中点M ,连结FM 、AM .因为F 为PC 的中点,所以FM ∥CD ,且FM =12CD .因为四边形ABCD 为矩形,E 为AB 的中点,所以EA ∥CD ,且EA =12CD .所以FM ∥EA ,且FM =EA .所以四边形AEFM 为平行四边形.所以EF ∥AM . 又AM ⊂平面P AD ,EF ⊄平面P AD , 所以EF ∥平面P AD .方法2:连结CE 并延长交DA 的延长线于N ,连结PN .因为四边形ABCD 为矩形,所以AD ∥BC , 所以∠BCE =∠ANE ,∠CBE =∠NAE . 又AE =EB ,所以△CEB ≌△NEA . 所以CE =NE .又F 为PC 的中点,所以EF ∥NP . 又NP ⊂平面P AD ,EF ⊄平面P AD , 所以EF ∥平面P AD .方法3:取CD 的中点Q ,连结FQ 、EQ . 在矩形ABCD 中,E 为AB 的中点,所以AE =DQ ,且AE ∥DQ .所以四边形AEQD 为平行四边形, 所以EQ ∥AD .又AD ⊂平面P AD ,EQ ⊄平面P AD , 所以EQ ∥平面P AD .(2分)因为Q 、F 分别为CD 、CP 的中点, 所以FQ ∥PD .又PD ⊂平面P AD ,FQ ⊄平面P AD ,所以FQ ∥平面P AD .又FQ 、EQ ⊂平面EQF ,FQ ∩EQ =Q ,所以平面EQF ∥平面P AD .(5分) 因为EF ⊂平面EQF ,所以EF ∥平面P AD . (2) 设AC 、DE 相交于G .在矩形ABCD 中,因为AB =2BC ,E 为AB 的中点,所以DA AE =CDDA= 2.又∠DAE =∠CDA ,所以△DAE ∽△CDA , 所以∠ADE =∠DCA .又∠ADE +∠CDE =∠ADC =90°, 所以∠DCA +∠CDE =90°. 由△DGC 的内角和为180°,得∠DGC =90°. 即DE ⊥AC .因为点P 在平面ABCD 内的正投影O 在直线AC 上,所以PO ⊥平面ABCD . 因为DE ⊂平面ABCD ,所以PO ⊥DE . 因为PO ∩AC =O ,PO 、AC ⊂平面P AC , 所以DE ⊥平面P AC ,又DE ⊂平面PDE ,所以平面P AC ⊥平面PDE .17.解:(1)设n *()n ∈N 年内所建安置房面积之和首次不低于3 000万m 2, 依题意,每年新建安置房面积是以200为首项,50为公差的等差数列, 从而n 年内所建安置房面积之和为(1)200502n n n -⎡⎤+⨯⎢⎥⎣⎦m 2,则(1)200502n n n -+⨯≥3 000,整理得,271200n n +-≥, 解得8 (15)n n -≤≥舍去.答:8年内所建安置房面积之和首次不低于3 000万m 2.(2)依题意,每年新建住房面积是以500为首项,1.1为公比的等比数列, 设第m 年所建安置房面积占当年新建住房面积的比为()p m , 则1120050(1)3()500(10.1)10 1.1m m m m p m --+-+==⋅+⨯, 由()(1)p m p m =+得,13410 1.110 1.1m mm m -++=⨯⨯,解得7m =. 答:第7年和第8年,所建安置房面积占当年新建住房面积的比保持不变. ·····14分 18.解:(1)分别过点A 、B 作直线l 的垂线,垂足为11,B A , 由题意得11BB AA =,由点到直线距离公式得112a AA BB ==,因为圆A 以1AF 为半径,所以半径为c ,被直线l 截得的弦长为222()2a c -,第 8页,共 13页圆B 以OB 为半径,∴半径为a ,被直线l 截得的弦长为222()2a a -.因为直线l :33y x =-被圆A 和圆B 截得的弦长之比为156,所以()222222241563324a c c a a a --==,解得a c 34=(a >c >0). 因为c e a =,所以所求的离心率为34,(2)存在点P ,使得过点P 有无数条直线被圆A 和圆B 截得的弦长之比为34,设点0(,0)P x ,由题意可得直线方程为0()y k x x =-, 直线截圆A 所得的弦长为2202|(7)|21k x c k ⎛⎫--- ⎪+⎝⎭, 直线截圆B 所得的弦长为2202|(7)|21k x a k ⎛⎫-- ⎪+⎝⎭,[][]22022220220202(7)2(1)(7)134(1)(7)(7)21k x c k c k x k k a k x k x a k ⎛⎫+- ⎪+-++⎝⎭==+--⎛⎫-- ⎪+⎝⎭,化简得22222220016(7)9(7)(1)(169)k x k x k c a +--=+-(*),由(1)离心率为34,得22169c a =,即方程(*)为0)1)(49(002=++x x k ,解得10-=x 或490-=x , 即存在2个点)0,1(-和)0,49(-;当10-=x 时,22||61||81k c kk a k ⎧<+⎪⎨<+⎪⎩,解得7157151515k -<<,当490-=x 时,22||421||561k c k k a k⎧<+⎪⎨<+⎪⎩,解得7772121k -<<,即有无数条直线;故存在2个点P ,使得过点P 有无数条直线被圆A 和圆B 截得的弦长之比为34.19.解:(1)∵()()e ,0x f x x k x '=->.(i )当0k ≤时,()0恒成立'>f x ,∴()f x 的递增区间是0+(,)∞,无递减区间;无极值. (ii )当0>k 时,由()0'>f x 得,>x k ;由()0'<f x 得,0<<x k ;∴()f x 的递减区间是(0,)k ,递増区间是(,+)∞k ,()f x 的极小值为()e k f k =-,无极大值. (2)①由()4f x x <,可得(1)e 40x x k x ---<,因为e 0x >,所以41e x x x k --<,即41ex x k x >--对任意[1,2]x ∈恒成立, 记4()1ex xg x x =--,则4(1)e 4(1)()1e e x x x x x g x -+-'=-=, 因为[1,2]x ∈,所以()0g x '>,即()g x 在[1,2]x ∈上单调递增,故2max228e 8()(2)1e e g x g -==-=.所以实数k 的取值范围为22e 8(,)e-+∞.②由已知1212()()()f x f x x x =≠,结合(1)可知,0k >,()f x 在(,)-∞k 上单调递减,在(,+)∞k 上单调递增,又(1)0+=f k ,1<+x k 时,()0<f x .不妨设121<<<+x k x k ,此时2x k >,12->k x k ,故要证122+<x x k ,只要证122k x x ->,只要证12(2)()f k x f x ->, 因12()()f x f x =,即证11(2)()f k x f x ->.设()(2)()h x f k x f x =--2(1)(1)()kx xx k x k x k -+-=---<e e e , 2()e ()()e e k xxx k h x x k -'=--22()()k x x x k --=e e e , ∴当<x k 时,()0h x '<,()h x 在(,)-∞k 上单调递减,∴(,)x k ∈-∞时,()()0k k h x h k >=-+=e e ,故当<x k 时,(2)()->f k x f x ,即11(2)()->f k x f x 成立,∴122+<x x k .20.解:(1)111312A B d ===,,;222413A B d ===,,;333716A B d ===,,. …………………………………………………………………3分(2)① 当1n =时,11(1)1a a λλ-=-+,所以11a =;当2n ≥时,由21(1)33n n S a n λλ-=-++,则1121(1)(1)33n n S a n λλ---=-+-+, 两式相减得12(1)3n n n a a a λλλ--=-++,即123n n a a λ-=+, 所以11122233(1)3(1)n n n n b a a b λλλλλ---⎡⎤=++=+==⎢⎥--⎣⎦.……………………………6分因为112313(1)3(1)b a λλλ-=+=--,第 10页,共 13页所以当13λ≠时,数列{}n b 满足1n n bb λ-=(2n ≥),即数列{}n b 是以313(1)λλ--为首项,λ为公比的等比数列;当13λ=时,数列{}n b 不是等比数列. …………………………………………………8分② 由①知,当13λ≠时,13123(1)3(1)n n a λλλλ--=⋅---;当13λ=时,23(1)n a λ=--.……………………………………………………………10分又{}{}1212max min i i i i n d a a a a a a ++=-,,,,,,, {}{}112123max min i i i i n d a a a a a a ++++=-,,,,,,.由于{}{}1223min min i i n i i n a a a a a a ++++,,,≤,,,,所以由1i i d d +>可得,{}{}12121max max i i a a a a a a +<,,,,,,.所以{}1211max i i a a a a ++=,,,对任意的正整数1232i n =-,,,,恒成立,即数列{}n a 的前1n -项单调递增是题设成立的必要条件,易知13λ≠.………………12分因为1i i i d a a +=-,112i i i d a a +++=-,所以1212i i i i i d d a a a +++-=+-1231(12)3(1)i λλλλλ--=⋅+--1231(1)3(1)i λλλλ--=⋅--.当1λ>时,由1n n a a +>,得3103(1)λλ->-,解得1λ>, 此时10i i d d +-≥,不符合1i i d d +>,舍去;当01λ<<,由1n n a a +>,得3103(1)λλ-<-,解得113λ<<,此时10i i d d +-<,符合1i i d d +>.综上所述,λ的取值范围是()113,. ……………………………………………………16分第II 卷(附加题,共40分)21A .证:因为P A 是圆O 在点A 处的切线,所以∠P AB =∠ACB .因为PD ∥AC ,所以∠EDB =∠ACB , 所以∠P AE =∠P AB =∠ACB =∠BDE .又∠PEA =∠BED ,故△P AE ∽△BDE . …………………… 10分21B .解:设点(x 0,y 0)为曲线|x |+|y |=1上的任意一点,在矩阵10103M ⎛⎫⎪= ⎪ ⎪⎝⎭对应的变换作用下得到的点为(,)x y '',则0010103xx y y ⎛⎫'⎡⎤⎡⎤ ⎪=⎢⎥⎢⎥ ⎪' ⎪⎣⎦⎣⎦⎝⎭,所以003x x y y ='⎧⎨='⎩ ……5分 所以曲线|x |+|y |=1在矩阵10103M ⎛⎫⎪= ⎪ ⎪⎝⎭对应的变换作用下得到的曲线为|x |+3|y |=1, 所围成的图形为菱形,其面积为1222233⨯⨯= .……10分21C .解:(1)将(2,3)M 及对应的参数3πϕ=代入cos ,(0,sin x a a b y b ϕϕϕ=⎧>>⎨=⎩为参数),得2cos 33sin3a b ππ⎧=⎪⎪⎨⎪=⎪⎩,所以42a b =⎧⎨=⎩,所以曲线1C 的普通方程为221164x y +=. ……4分 (2)曲线1C 的极坐标方程为2222cos sin 1164ρθρθ+=,将12(,),(,)2A B πρθρθ+代入得222211cos sin 1164ρθρθ+=,222222sin cos 1164ρθρθ+=,所以221211516ρρ+=. ……10分21D .解:因为a >0,b >0,a +b =1,所以(2a +1)+(2b +2)=5,从而(12a +1+2b +1 )[(2a +1)+(2b +2)]=1+4+2b +22a +1+4(2a +1)2b +2≥5+22b +22a +1×4(2a +1)2b +2=9. …………………… 6分 所以12a +1+2b +1≥95.当且仅当2b +22a +1=4(2a +1)2b +2,且a +b =1,即a =13,b =23 时,12a +1+2b +1取得最小值95. …………………… 10分22.解:因为在直三棱柱111ABC A B C -中,AB AC ⊥,第 12页,共 13页所以分别以AB 、AC 、1AA 所在的直线为x 轴、y 轴、z 轴,建立空间直角坐标系, 则111(0,0,0),(2,0,0),(0,4,0),(0,0,3),(2,0,3),(0,4,3)A B C A B C ,因为D 是BC 的中点,所以(1,2,0)D ,……………………………………………………2分 (1)因为111(0,4,0),(1,2,3)A C A D ==-,设平面11A C D 的法向量1111(,,)n x y z =,则1111100n A C n A D ⎧⋅=⎪⎨⋅=⎪⎩,即111140230y x y z =⎧⎨+-=⎩,取111301x y z =⎧⎪=⎨⎪=⎩,所以平面11A C D 的法向量1(3,0,1)n =,而1(1,2,3)DB =-, 所以111111335cos ,35n DB n DB n DB ⋅<>==⋅, 所以直线1DB 与平面11A C D 所成角的正弦值为33535;…………………………………5分 (2)11(2,0,0)A B =,1(1,2,3)DB =-,设平面11B A D 的法向量2222(,,)n x y z =,则2112100n A B n DB ⎧⋅=⎪⎨⋅=⎪⎩,即222220230x x y z =⎧⎨-+=⎩,取222032x y z =⎧⎪=⎨⎪=⎩,平面11B A D 的法向量2(0,3,2)n =,所以121212130cos ,65n n n n n n ⋅<>==⋅, 二面角111B A D C --的大小的余弦值13065.……………………………………………10分 23.(1)证明:0)(121)2()(31222222>-=+-++=-b a b a b ab a B A (2)证明:11,1B A n ==;,)2(,11,311nn n n n b a B b a b a n A n +=--+=≥++令,,y b a x b a =-=+且0,>y x ,于是,)2(],)()[()1(21)2()2(1111111n n n n n n n n x B y x y x y n y y x y x n A =--++=--++=+++++ 因为y x C y x C y x C y x y x nn n n n n n n 11323111112)22(])()[(+-++++≥++=--+ ,所以n n n n nn n n B x x y x C y n A ===⋅+≥++)2(22)1(21111.。

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