2014年高考第一次模拟考试题

东北三省三校2014年高三第一次联合模拟考试参考答案第一部分：1. C2. A3. B4. B5. A6. C7. B8. A9. C 10. B 11. C 12. A 13. A 14. B 15. C 16. C 17. A 18. B 19. A 20. C第二部分：21. A 22. B 23. A 24. C 25. D 26. A 27. C28. B 29. D 30. B 31. C 32. A 33. C 34. A35. C 36. G 37. C 38. A 39. D 40. F第三部分：41. B 42. C 43. A 44. D 45. B 46. B 47. C48. D 49. B 50. A 51. C 52. D 53. B 54. A55. C 56. B 57. A 58. C 59. D 60. A61. a 62. at 63. screaming 64. Impatient 65. bothered66. pieces 67. until / before 68. have finished 69. there 70. It/ That/This第四部分：第一节：Of all my teachers, Mr. Smith is the one who impresses me most. Because he is 54,Though/Althoughhe looks very young at his age. And he’s one of the most popular teachers in ourforschool. Compared with other teacher, Mr. Smith pays more attention to his way ofteachersteaching. He tries various ways to make his class lively and interestingly. In hisinterestingopinion, we would not only know “what”, but also understand “why”. So, instead ofshouldgiving us answers immediately, he encourages us to think by themselves wheneverourselveshe puts forward questions, With his help, we /\learned how to analyze and settlehavedown problems. What a wonderful world of “what” he leads us to! He is such a去掉whylearning person that we all admire him very much.learned第二节：One Possible Version:Dear Editor,We took part in a voluntary activity organized by our school last Sunday. We gathered at the Culture Square, passing leaflets to passers-by from which they can learn what the earthquake is and how to protect themselves against the earthquake. The leaflets drew the attention of the public to the damage that an earthquake can cause and told them to take action in a right and efficient way. Some made a public speech, telling people about the best place they can hide when an earthquake occurs.We have learned much from the voluntary activity, which offers us an opportunity to contribute to society. We told the public about the situation because we thought everyone should be concerned about each other and take responsibility for promoting a public cause.Best wishes!Yours,Li HuaDear Editor,We took part in a voluntary activity organized by our school last Sunday. I am writing the letter for the purpose of reporting to you the related information concerning our voluntary activity.To improve students’ ability to protect ourselves and enrich after-class activities, our school organized a school-wide voluntary activity on Sunday last week at the Culture Square. This content of voluntary activity is concerned with advertisement of common sense of earthquake and quakeproof knowledge. And meanwhile, we volunteers distributed/ handed out leaflets to passers-by and made speeches on what the earthquake was and how to protect themselves against the earthquake. By means of participating in this activity, we have learned much and realized that we should devote ourselves to making contributions to society and take responsibility for promoting a public cause.My appreciation to you for your adopting my report is beyond words. I look forward to a favorable reply at your earliest convenience.Best wishes!Yours,Li Hua听力部分录音稿1. W: John, why don’t you watch NBA games on TV?M: Oh, I’m studying for a Chinese test tomorrow.2. M: Where is Adam？He was supposed to meet us here half an hour ago, but he hasn’t come yet.W: I’ve just phoned and we’ve had a quarrel. He was busy with the contract and didn’t notice the time.3. M: May I help you?W: Yes, I’d like to apply for a loan. We need the loan to build an addition to our house.M: All right. I’ll give you an application form. You can fill it out here, or take it home if you like.4. M: Hello, Kate. Do you know how many students will go to the mountain?W: Our monitor said that the whole class will go except Tom. He has something urgent to do.That is, 20 students.M: Oh, 18 students have been there by now.5. W: Chris, did you know about Gemma Hudson? Have you heard that she’s leaving?M: Really? How do you know that?W: She told me. She’s leaving next month.M: Actually, I’m not surprised. I didn’t think she was very happy working here.6. W: What is the tall building over there?M: It’s the library, which houses over half a million books. It also has computer labs and quiet study spaces.W: And the building next to it?M: That’s the computer center. The IT facilities are really advanced there. Some of its classrooms are open 24 hours a day.W: Is there a sports center on campus?M: Of course. The sport facilities are excellent. A stadium is on the northeast side of the campus.A 50-metre swimming pool is just beside it. There is a football match this afternoon. Let’sgo there and have a look.7. W: So, do you like watching the World Cup?M: Um, no. I can’t stand it.W: Mmm, I hate it too. But I want to watch the World Cup Final this weekend. But our TV won’t work.M: I can’t play, that’s my problem. If I’m not very good at a sport, I don’t like watchin g it.W: So, what do you like, then?M: I play golf.W: Oh, right. My husband plays too. I’m not interested in it, though. I go horse riding.M: I’d love to have a go at riding if I have time. I bet it’s good fun.8. M: So, Sally, we should probably start thinking about our winter vacation plans.W: Yeah, I think we should fly down to Florida for a couple of weeks.M: Uh, I’d really prefer not to. Those holiday travel packages cost way too much.W: Well, how about skiing up at Sky Valley Resort for a week or so?M: I’d rather not. We did it last year. Remember?W: And we had a really good time. Do you have any suggestions?M: Umm… you know, I never have any good ideas.W: OK. Here’s something different. How about going to the New Year’s Eve party in New York City? We’ve always wanted to do that.M: But I don’t really like big parties. They’re always crowded.W: You’re driving me crazy, Tom! All right. Let’s just stay home for two weeks!M: No way! That sounds too boring! Can’t you think of anything else?9. W: Andy, are you busy at the moment?M: Oh, hi. Well, actually I’m expecting an important call from the office any minute but go ahead.W: I’m sorry, but it’s really important. It’s about the car. It was broken into last night…M: Oh, no!W: I want to know if you saw or heard anything.M: Did I see anything? Where?W: In the road. Did you see anybody looking suspicious?M: No, I don’t think I looked out of the window.W: Can you remember if you heard anything? They can’t have done it completely silently. I came back late from a party and I parked it at about 11:30, and then I had a shower and went to bed.M: No, I’m sorry. What have they taken?W: Well, this morni ng I found the driver’s window smashed. There’s glass everywhere, and the CD player has gone.M: How annoying! Is anything else missing?W: Well, it could have been worse. Luckily, there wasn’t much in the car anyway.10. M: In today’s lecture we’ll begin with Doxey’s model for the socio-cultural impact of tourism.He identifies four stages.He identifies the first stage as happiness. Because initially the tourists are regarded as a new and interesting thing.As tourist development begins to increase, some sections of the local population become involved with tourists while others don’t, and it is increasingly the case that commercial rather than social factors are influencing relationships between tourists and the host community. People are less interested in the tourists for their own sake. This is identified as the second stage by Doxey.If development continues to increase, the hosts get annoyed with the tourists, and this is the third stage.Well, development of the tourist area may start to get out of control, which is going to make life difficult for local people. So the policy makers, the government, the local authorities, and so on, provide more roads, more car parks and so on to try to help cope with the coming of tourists. But the lives of the local people are made increasingly difficult and in the final stage of the model, the hosts show unkindness to the tourists.。

绝密★启用前 试卷类型:A2014年第一次高考模拟考试 数学试卷（理科）第一部分 选择题（共40分）一、选择题：（本大题共8小题，每小题5分，共40分，在每小题给出的四个选项中，只有一项是符合题目要求的）1、若集合A={x -2＜x ＜1}，B={x 0＜x ＜2}, 则集合A ∩B=（ ）A. {x -1＜x ＜1}B. {x -2＜x ＜1}C. {x -2＜x ＜2}D. {x 0＜x ＜1} 2、在复平面内，复数)21(i i z +=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 3、设条件:0p a >；条件2:0q a a +≥，那么p 是q 的（ ） 条件A ．充分非必要B ．必要非充分C ．充分且必要D ．非充分非必要4、设}{n a 是等差数列,若,13,372==a a 则数列}{n a 前8项和为（ ）A ．128 B.80 C.64 D.565、顶点在原点，准线与x轴垂直，且经过点（1的抛物线方程是（ ）A ．22y x =-B ．22y x =C ．22x y = D. 22x y =- 6、某程序框图如图所示，现输入如下四个函数， 则可以输出的函数是（ ）A ．2()f x x = B ．1()f x x=C ．()xf x e = D ．()sin f x x =7、已知函数x x f lg )(=－sinx ，则()f x 在（0,＋∞）上 的零点个数为（ ）A ．2B ．3C ． 4D ． 无数个 8、 定义域为[],a b 的函数()y f x =的图象的两个端点为A, B, M ()(),x y x 是f 图象上任意 一点，其中()()()[0,11,1]x a b ON OA OB λλλλλ=+-=+-∈ 向量，若不等式MN k ≤恒成立，则称函数()[],f x a b 在上“k 阶线性近似”. 若函数[]112y x x=-在，上“k 阶线性近似”，则实数k 的取值范围为( ) A.[)0+∞，B.[)1+∞，C.32⎡⎫+∞⎪⎢⎣⎭D.32⎡⎫+∞⎪⎢⎣⎭第二部分 非选择题（共110分）二、填空题（本题共6小题，第14、15题任选一道作答，多选的按14小题给分，共30分） (一)必做题(9～13题)9、已知，，若，则10、右图是一个几何体的三视图，根据图中数据，可得该几何体 的表面积．．．是___________ 11、61(2)2x x-的展开式的常数项是12、已知函数2y x =与(0)y kx k =>的图象所围成的阴影部分 （如图所示）的面积为43，则k =_______. 13、在平面直角坐标系上，设不等式组00(4)x y y n x >⎧⎪>⎨⎪≤--⎩所表示的平面区域为n D ，记n D 内的整点（即横坐标和纵坐标均为整数的点）的个数为 ()n a n N *∈. 则1a ＝ ，经推理可得到n a ＝ ．14、（坐标系与参数方程选做题）已知直线l 的参数方程为：2,14x t y t=⎧⎨=+⎩（t 为参数），圆C 的极坐标方程为2cos ρθ=，则圆C 的圆心到直线l 的距离为 .三、解答题（本大题共6小题，共80分，解答须写出文字说明、证明过程或演算步骤，）16、（本小题满分12分）设锐角三角形ABC 的内角A,B,C 的对边分别为c b a ,,,且A b a sin 2=。

2014年邢台市普通高考模拟考试文科数学参考答案及评分标准(Ⅱ)解法一：由(Ⅰ)知60B =，又23b =222cos6012a c ac ∴+-= 2()312a c ac ∴+-= ……………8分22()1233()2a c a c ac +∴+-=≤，当且仅当a c =时取”＝” ,2()48a c ∴+≤a c ∴+≤……………………………………………………………………………10分又,,a b c 是三角形的三边,a c +>,∴(a b +∈ …………………12分解法二：由(Ⅰ)知60B =，又23b =．由正弦定理得4sin sin sin a c b A C B ==== ……………8分 4sin ,4sin ,a A c C ∴==又120,120A C C A +=∴=-314sin 4sin(120)6sin 23cos (sin cos )43sin(30)2a c A A A A A A A ∴+=+-=+=+=+………………………………10分10120sin(30)12A A <<∴<+≤ a c ∴<+≤………………………………………………………………………12分18. 解:（Ⅰ）证明：连结PC ,交DE 与N ,连结MN ，PAC ∆中，,M N 分别为两腰,PA PC 的中点 , ∴//MN AC . ……………2分因为MN ⊂面MDE ,又AC ⊄面MDE ，所以//AC 平面MDE . ……………………………………………………………4分（Ⅱ）解：由四边形PDCE 为矩形，知,DC PD ⊥又平面⊥PDCE 平面ABCD ，,ABD PD 平面⊥∴ ……………6分三棱锥ABD P -的体积为6221161213131=⨯⨯⨯=⨯⨯⨯=⨯=∆-PD AD AB PD S V ABD D AB P . ………8分由已知,DC AD ⊥又平面⊥PDCE 平面ABCD ，,PDCE AD 平面⊥∴,//CD AB 四棱锥的体积为322122313131=⨯⨯⨯=⨯⨯=⨯==--AD PD CD AD S V V PDCE PDCE A PDCE B 矩形.…………………10分46B PDCEP ABD V V --==，或者143P ABD B PDCE V V --==, N所以原几何体被平面PBD 所分成的两部分的体积比144或. …………………12分 19.（Ⅰ）解：成绩在[)14,13的人数为500.042⨯=，设为,x y ;成绩在[)18,17的人数为500.063⨯=，设为C B A ,,; ……………2分 [)14,13,∈n m 时有xy 一种情况．[)18,17,∈n m 时有BC AC AB ,,三种情况;n m ,分别在[)[)18,1714,13和时有A,,,,,x xB xC yA yB yC 六种情况……………4分 所以3(2)5P m n ->=……………………………………………………6分……………8分 （Ⅱ）2250(241268)8.33332183020K ⨯⨯-⨯=≈⨯⨯⨯ ………………………………………………10分由于2 6.635K >，故有99%的把握认为“体育达标与性别有关” ……………… 12分20.解：（Ⅰ）当0 ()e x a f x a '>=-时，，…………………………………………2分 由e 0x a ->， 得单调增区间为()ln ,a +∞；由e 0x a -<，得单调减区间为(,ln )a -∞ ， …………………4分由上可知min ()(ln )ln 1f x f a a a a ==-- ……………………………………………6分（II ）若()0f x ≥对x ∀∈R 恒成立，即min ()0f x ≥，由（I ）知问题可转化为ln 10a a a --≥对0a >恒成立 ．令()ln 1(0)g a a a a a =--> ， ()ln g a a '=-， ……………8分()g a 在(0,1)上单调递增，在(1,)+∞上单调递减，∴max ()(1)0g a g ==． …………………………………………………………………10分即ln 10a a a --≤ ， ∴ln 10a a a --= ．由()g a 图象与x 轴有唯一公共点知，所求a 的值为1． …………………………… 12分21. 解：(Ⅰ)由已知，2b =，又36=e ， 即3642=-a a ，解得32=a ， ……………2分 所以椭圆C 的方程为112422=+y x . ………………………………………………4分(Ⅱ)假设存在点)0,(0x Q 满足题设条件.当x MN ⊥轴时，由椭圆的对称性可知恒有NQP MQP ∠=∠，即0x ∈R ； ………6分 当MN 与 轴不垂直时，设MN 所在直线的方程为(1)y k x =-，代入椭圆方程化简得： 0122)3(2222=-+-+k x k x k ， 设),(),,(2211y x N y x M ，则312,3222212221+-=+=+k k x x k k x x , 022011x x y x x y k k NQ MQ -+-=+121020(1)(1)k x k x x x x x --=+-- 1202101020(1)()(1)()()()k x x x k x x x x x x x --+--=--， …………………………………8分 ∵ 120210120120(1)()(1)()2(1)()2x x x x x x x x x x x x --+--=-+++ 02202223)1(23)12(2x k k x k k +++-+-= ， 若NQP MQP ∠=∠， 则0=+NQ MQ k k ， ……………10分即[]023)1(23)12(2022022=+++-+-x k k x k k k ， 整理得0(4)0k x -=，x∵k ∈R ，∴04x =.Q 的坐标为)0,4(Q .综上,在x 轴上存在定点)0,4(Q ，使得NQP MQP ∠=∠. ………………………12分22.解：（Ⅰ） 由PA 是圆O 的切线，因此PAD ∠=ACD ∠，……………2分在等腰OCD ∆中，OD OC =，可得ACD CDE ∠=∠， …………4分所以PAD CDE ∠=∠. ……………… 5分（Ⅱ） EC PBD PEC ∆∆连结，由与相似可知， PB BD PE CE=，……………7分 由切割线定理可知，2PA PB PC =⋅，则2PA PB PC =，……………9分 又EC AD =，可得 2PA BD PC PE AD=⋅. ……10分 23. 解: （Ⅰ）曲线C 的普通方程为221324x y += ……………3分 直线l 的参数方程为8cos ()2sin x t y t ααα=+⎧⎨=+⎩为参数 ………………………………5分（Ⅱ）将l 的参数方程为代入曲线C 的方程得：()228+cos 8(2sin )32t t αα++= ……………7分 1212264||||[8,64]17sin PM PM t t α∴⋅==∈+ ………………………………………10分24.解: （Ⅰ）当1a =时，不等式为3135x x -++≤ ……………1分当13x ≤时，不等式即1335x x -++≤，12x ≥- 1123x ∴-≤≤ ……………3分当13x >时，不等式即3135x x -++≤，34x ≤ 1334x ∴<≤ 综上，不等式的解集为1324x x ⎧⎫-≤≤⎨⎬⎩⎭………………………………………5分（Ⅱ）1(3)23()3131(3)43a x x f x x ax a x x ⎧++≥⎪⎪=-++=⎨⎪-+<⎪⎩……………6分 当3a <-时，()f x 单调递减，无最小值； ……………7分当33a -≤≤时，()f x 在区间1,3⎛⎫-∞ ⎪⎝⎭上单调递减，在1,3⎛⎫+∞ ⎪⎝⎭上单调递增， 13x ∴= 处取得最小值； ……………8分当3a >时，()f x 单调递增，无最小值； ……………9分综上， 33a -≤≤ …………………………………………………………………10分数学23题更正及评分标准理科第17题(Ⅱ)方法2理科第17题(Ⅱ)方法217（Ⅱ）在ABC ∆中,由余弦定理得令t a c =+,则c t a =-22()()1242a a t a t a -+--= ………8分 即227104480a ta t -+-=,因为a 有解且0a > 所以22121210028(448)000t t a a a a ⎧∆=--≥⎪+>⎨⎪>⎩………10分 212112t ∴<≤t ∴<≤因此a c +的最大值为………12分。

第6题图俯视图2014年高考数学模拟考试试卷第Ⅰ卷(选择题，共50分)一、选择题：本大题共10小题，每小题5分，共50分.在每小题给出的四个选项中，只 有一项是符合题目要求的.1.已知集合{}ln A x y x =|=，集合{}2,1,1,2B =--，则A B =A.(1,2)B.{}1,2C.{}1,2--D.(0,)+∞2.若(4i)i i a b +=+其中,a b ∈R ，i 是虚数单位，则a b - = A.3B.5C.3-D.5-3.设0.32a =，20.3b =，2log (0.3)(1)x c x x =+>，则,,a b c 的大小关系是 A.a b c << B.b a c << C.c b a << D.b c a <<4.不等式2311x x +≥-的解集是 A.[4,)-+∞ B.(4,)-+∞ C.[4,1)- D.(,4](1,)-∞-+∞5.“1a =”是“函数22cos sin y ax ax =-的最小正周期为π”的 A.充分不必要条件 B.必要不充分条件C.充要条件D.既非充分条件也不是必要条件6.一个四棱锥的三视图如图所示，其左视图是等边三角形， 该四棱锥的体积等于B.C.D.7.袋中有4个形状大小一样的球，编号分别为1,2,3,4，从中任取2个球，则这2个球的编 号之和为偶数的概率为 A.16 B.23 C.12D.13 8.已知等比数列}{n a 满足：354321=++++a a a a a ，122524232221=++++a a a a a ，则54321a a a a a +-+-的值是A.2B.9C.4D.149.设函数3()f x x =+sin x ，若02θπ≤≤时, (cos )(1)0f m f m θ+->恒成立，则实数 m 的取值范围是A.(0,1)B.(,0)-∞C.1(,)2-∞ D.(,1)-∞D CBA10.当n *∈N 且2n ≥时，24112225n p q -++++=+(其中p 、q 为非负整数，且05q ≤≤，则q 的值为A.0B.1C.3D.与n 有关第Ⅱ卷 (非选择题，共90分)二、填空题：本大题共5小题，每小题5分，共20分.将正确答案填在答题卷上对应题号 的横线上.11.若下框图所给的程序运行结果为20S =，那么判断框中应填入的关于k 的条件是 .12.函数()37ln f x x x =-+的零点位于区间(,1)()n n n +∈N ，则n = . 13.已知锐角三角形的边长分别为2、4、x ，试求x 的取值范围 . 14.对于函数321()(2)3f x x ax a x b=-+-+，若()f x 有六个不同的单调区间，则a 的取值范围为 .15.（文科做②；理科从①②两小题中任意选作一题） ①（坐标系与参数方程选做题）在极坐标系中，直线π()6θρ=∈R 截圆π2cos()6ρθ=- 的弦长是 .②(不等式选做题)关于x 的不等式|||1|1x a x ---≤在R 上恒成立（a 为常数），则实数a 的取值范围是 .三、解答题：本大题共6小题，满分75分.解答应写出文字说明，证明过程或演算步骤 16.(本大题满分12分)在ABC ∆中，已知45ABC ∠=，AB =，D 是BC 边上的一点，5,3AD DC ==，求AC 的长.17. (本大题满分12分)A 、B 两个口袋，A 袋中有6张卡片，其中1张写0，2张写1，3张写有2；B 袋中7张卡片，其中4张写有0，1张写有1，2张写有2，从A 袋中取1张卡片，B 袋中取2张卡片，共3张卡片， 求： （1）取出的3张卡片都写0的概率；（2）取出的3张卡片数字之积是4的概率； （3）取出的3张卡片数字之积的数字期望.18.(本大题满分12分)如图，已知AB ⊥平面ACD ，DE ⊥平面ACD ，△ACD 为等边三角形，2AD DE AB ==，F 为CD 的中点． （1）求证：//AF 平面BCE ； （2）求证：平面BCE ⊥平面CDE ；（3）求直线BF 和平面BCE 所成角的正弦值．19.(本大题满分12分)设数列{}n a 的前n 项和为n S ，且(1)n n S a λλ=+-，其中λ是不等于1-和0的常数. （1）证明：数列{}n a 是等比数列；（2）设数列{}n a 的公比()q f λ=，数列{}n b 满足111,()3n n b b f b -==（n *∈N ，且2n ≥），求数列1n b ⎧⎫⎨⎬⎩⎭的前n 项和为n T .20.(本大题满分13分)已知函数()sin f x ax b x =+，当3x π=时，()f x取得极小值3π-. （1）求,a b 的值；（2）设直线:()l y g x =，曲线:()S y f x =.若直线l 与曲线S 同时满足下列两个条件： ①直线l 与曲线S 相切且至少有两个切点；②对任意x ∈R 都有()()g x f x ≥.则称直线l 为曲线S 的“上夹线”.试证明：直线:2l y x =+为曲线:sin S y ax b x =+“上夹线”.A BCDEF21.(本大题满分14分)一直线过抛物线22(0)y px p =>的焦点F ，且交抛物线于,A B 两点，C 为抛物线准线的一点 （1）求证：ACB ∠不可能是钝角；（2）是否存在这样的点C ，使得ABC ∆为正三角形？若存在，请求出点C 的坐标；若不存在，请说明理由.参考答案一、选择题：1～5. BBBDA ； 6～10. ADCDA. 二、填空题：11.8k >； 12.2; 13.1t ≤<； 14.(1,2)； 15. ①2；②[]0,2. 三、解答题：16.解：在ABD ∆中，由正弦定理得sin 22sin 5AB B ADB AD ∠∠===∴3ADB π∠=或23π，①若3ADB π∠=，则23ADC π∠=，ADC ∆中，由余弦定理得222cos 49AC AD DC AD DC ADC =+-⋅∠=2 ∴7AC =，②若23ADB π∠=，则3ADC π∠=，ADC ∆中，由余弦定理得222cos 19,AC AD DC AD DC ADC =+-⋅∠=2∴AC =17.（文科）（1）每颗骰子出现的点数都有6种情况，∴基本事件总数为3666=⨯个.记“点),(y x P 在直线1-=x y 上”为事件A ，A 有5个基本事件：)}5,6(),4,5(),3,4(),2,3(),1,2{(=A ， .365)(=∴A P（2）记“点),(y x P 满足x y 42<”为事件B ，则事件B 有17个基本事件: 当1=x 时，;1=y 当2=x 时，2,1=y ；当3=x 时，3,2,1=y ；当4=x 时，;3,2,1=y 当5=x 时，4,3,2,1=y ；当6=x 时，4,3,2,1=y ．.3617)(=∴B P （理科）解：（1）设事件A 表示：“取出的3张卡片都写0”F HG EMDCBA2427C 11()6C 21P A =⋅=（2）设事件B 表示：“取出的3张卡片数字之积是4”2112122277C C C 234()6C 6C 63P B =⋅+⋅=（3）设取出的3张卡片数字之积为随机变量ξ，则ξ可取0，2，4，82327C 1537(0)(1)66C 42P ξ==+⋅-=； 111227C C 22(2)6C 63P ξ==⋅= 11121222C C C 234(4)6C 6C 63P ξ==⋅+⋅=； 222C 31(8)6C 42P ξ==⋅= 24863634263E ξ=⋅+⋅+⋅=18.解（1） 证法一：取CE 的中点G ，连FG BG 、．∵F 为CD 的中点，∴//GF DE 且12GF DE =． ∵AB ⊥平面ACD ，DE ⊥平面ACD ， ∴//AB DE ，∴//GF AB ． 又12AB DE =，∴GF AB =． ∴四边形GFAB 为平行四边形，则//AF BG ． ∵AF ⊄平面BCE ，BG ⊂平面BCE ， ∴//AF 平面BCE ．证法二：取DE 的中点M ，连AM FM 、． ∵F 为CD 的中点，∴//FM CE ．∵AB ⊥平面ACD ，DE ⊥平面ACD ，∴//DE AB ． 又12AB DE ME ==， ∴四边形ABEM 为平行四边形，则//AM BE ． ∵FM AM ⊄、平面BCE ，CE BE ⊂、平面BCE ， ∴//FM 平面BCE ，//AM 平面BCE ． 又FMAM M =，∴平面//AFM 平面BCE ．∵AF ⊂平面AFM ， ∴//AF 平面BCE ．（2）证：∵ACD ∆为等边三角形，F 为CD 的中点，∴AF CD ⊥．∵DE ⊥平面ACD ，AF ⊂平面ACD ，∴DE AF ⊥． 又CDDE D =，故AF ⊥平面CDE ．∵//BG AF ，∴BG ⊥平面CDE ． ∵BG ⊂平面BCE ， ∴平面BCE ⊥平面CDE ．（3）平面CDE 内，过F 作FH CE ⊥于H ，连BH ∵平面BCE ⊥平面CDE ，∴FH ⊥平面BCE ∴FBH ∠为BF 和平面BCE 所成的角设22AD DE AB a ===，则2sin 45FH CF==2BF a ==，Rt FHB ∆中，sin FH FBH BF ∠==∴直线BF 和平面BCF 19.（1）证明：∵(1)n n S a λλ=+-∴11(1)(2)n n S a n λλ--=+-≥∴1n n n a a a λλ-=-+，即1(1)n n a a λλ-+= 又1λ≠-且0λ≠，∴11n n a a λλ-=+ 又11a =，∴数列{}n a 是以1为首项，1λλ+为公比的等比数列.（2）解：由（1）知：()1q f λλλ==+∴111()(2)1n n n n b b f b n b ---==≥+故有1111111n n n n b b b b ---+==+，∴1111(2)n n n b b --=≥ ∴数列1n b ⎧⎫⎨⎬⎩⎭是以3为首项，1为公差的等差数列，∴2(1)53()22n n n n nT n n *-+=+=∈N20.解：（1）∵()sin f x ax b x =+，∴()cos f x a b x '=+而由已知得：10233a b a ⎧+=⎪⎪⎨ππ⎪⋅+=⎪⎩∴1,2a b ==-此时()2sin f x x x =-，∴()12cos f x x '=-，当(0,)3x π∈时，()0f x '<，当(,)32x ππ∈时，()0f x '>∴当3x π=时，()f x取得极小值3π即1,2a b ==-符合题意（2）由()12cos 1f x x '=-=，得cos 0x =当2x π=-时，cos 0x =，此时1222y x π=+=-+，22sin 22y x x π=-=-+12y y =，∴(,2)22ππ--+是直线l 与曲线S 的切点当2x 3π=时，cos 0x =，此时1222y x 3π=+=+，22sin 22y x x 3π=-=+ 12y y =，∴(,2)223π3π+也是直线l 与曲线S 的切点∴直线l 与曲线S 相切且至少有两个切点对任意x ∈R ，()()(2)(2sin )22sin 0g x f x x x x x -=+--=+≥即()()g x f x ≥，因此直线:2l y x =+为曲线:2sin S y x x =-“上夹线”21.解：设1122(,),(,),(,)2pA x yB x yC m -，直线AB 方程为2p x ty =+由222p x ty y px ⎧=+⎪⎨⎪=⎩得：2220y pty p --=，则212122,y y pt y y p +==-∴2212122,4p x x pt p x x +=+=（1）11(,)2p CA x y m =+-，22(,)2pCB x y m =+- ∴2()0CA CB pt m ⋅=-≥∴,CA CB <>不可能为钝角，故ACB ∠不可能是钝角（2）假设存在点C ，使得ABC ∆为正三角形 由（1）得：线段AB 的中点为2(,)2pM pt pt +①若直线AB 的斜率不存在，这时0t =，(,),(,)22p p A p B p -，点C 的坐标只可能是(,)2pp -，由CM AB =，得：2p p =，矛盾，于是直线AB 的斜率必存在 ②由CM AB ⊥，得：1CM AB k k ⋅=-，即21122pt m p p t pt -⋅=-++∴32m pt pt =+，∴3(,2)2pC pt pt -+2(CM p t =+22(1)AB p t =+由CM =，得：t =，∴(,)2p C -±故存在点(,)2pC -±，使得ABC ∆为正三角形。