东北三校2008年部分毕业班第三次摸底考试

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2008届高三年级第三次质量检测文科数学试题及答案

2008届高三年级第三次质量检测文科数学试题及答案

2008届高三年级第三次质量检测文科数学试题本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分, 第Ⅰ卷为1-10题,共50分,第Ⅱ卷为11-21题,共100分.全卷共计150分。

考试时间为120分钟. 注意事项:参考公式:如果事件A 、B 互斥,那么()()()P A B P A P B +=+如果事件A 、B 相互独立,那么()()()P A B P A P B ⋅=⋅球的表面积公式 24πS R = 球的体积公式 34π3V R =其中R 表示球的半径 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,满分50分.在每小题给出的四个选项中,只有一项是符合题目要求的.1.M={}4|2<x x ,N={}032|2<--x x x ,则集合M N=( ). A.{2|-<x x } B.{3|>x x } C.{21|<<-x x } D. {32|<<x x }2. 复数2(2)(1)12i i i+--的值是( ).A .2 B. 2- C. 2i D. 2i - 3. 已知||3a =,||5b =,12a b =,则向量在向量上的投影为( ).A 12BC D4. 方程2sin 2sin 0x x a --=x R ∈在上有解,则a 的取值范围是( ).A .[)+∞-,1B .),1(+∞-C .]3,1[-D .[)3,1-5.“12m =”是“直线(2)310m x my +++=与直线(2)(2)30m x m y -++-=相互垂直”的( ) A 充分必要条件 B 充分而不必要条件C 必要而不充分条件D 既不充分也不必要条件6. 等差数列{}n a 中,n S 是前n 项和,且k S S S S ==783,,则k 的值为( ).A.4 B 11 C.2 D 127. 为了得到函数)62sin(π-=x y 的图象,可以将函数x y 2cos =的图象( ).A.向右平移6π个单位 B.向右平移3π个单位 C.向左平移6π个单位 D.向左平移3π个单位8.若椭圆2215x y m +=的离心率e =,则m 的值为( ). A.13或2539. 在棱长为2的正方体1111ABCD A BC D -中,点E ,F 分别是棱AB ,BC 的中点,则点1C 到平面1B EF 的距离是( ).A.332 B.322 C.32D.3410.10.定义A D D C C B B A ****,,,的运算分别对应下图中的(1)、(2)、(3)、(4),那么下图中的(A )、(B )所对应的运算结果可能是(1) (2) (3) (4) (A ) (B ) A.D A D B **, B.C A D B **, C.D A C B **, D.D A D C **,第Ⅱ部分(非选择题,共100分)二、填空题:本大题共5小题,每小题5分,满分20分,其中14,15题是选做题,考生只能选做一题,,若两题全都做的,只计算前一题的得分.11. 函数()212log 2y x x =-的单调递减区间是 .12.甲、乙两人独立的解决一个问题,甲能解决这个问题的概率为0.6,乙能解决这个问题的概率为0.7,那么甲乙两人中至少有一人解决这个问题的概率是 .13.设x 、y 满足条件310x y y x y +≤⎧⎪≤-⎨⎪≥⎩,则22(1)z x y =++的最小值 . 14.(坐标系与参数方程选做题)自极点O 向直线l 做垂线,垂足为(2,)3H π,则直线l 的极坐标方程是 . 15.(几何证明选讲选做题)已知圆的直径13AB =,C 为圆上一点,过C 作CD AB ⊥于D (AD BD >),若6CD =,则AD 的长为 .三、解答题:本大题共6小题,满分80分.解答须写出文字说明、证明过程或演算步骤. 16.(本小题满分12分) 在ABC △中,1tan 4A =,3tan 5B = (Ⅰ)求角C的大小;(Ⅱ)若AB ,求BC 边的长A B 1BC 117.(本小题满分13分)如图,在直三棱柱111ABC A B C -中, 3AC =, 4BC =, 5AB =, 14AA =, 点D 是AB 的中点.(1)求证:1AC BC ⊥; (2)求证:1AC ∥平面1CDB .18.(本小题满分13分)设数列{}n a 的前n 项和为n S ,点(,)()nS n n N n*∈均在函数32y x =-的图像上. (Ⅰ)求数列{}n a 的通项公式;(Ⅱ)设13+=n n n a a b ,n T 是数列{}n b 的前n 项和,求使得20n m T <对所有n N *∈都成立的最小正整数m .19.(本小题满分14分)已知圆C 过点(0,)A a (0)a >, 且在x 轴上截得的弦MN 的长为2a .(1) 求圆C 的圆心的轨迹方程; (2) 若45MAN ∠=, 求圆C 的方程.20.(本小题满分14分)已知函数2()(0,0)f x ax bx c a bc =++>≠,()0,,()()0.f x x F x f x x >⎧=⎨-<⎩ (Ⅰ)若函数)(x f 的最小值是(1)0f -=,且(0)1f =,求(2)(2)F F +-的值; (Ⅱ)在(Ⅰ)的条件下,k x x f +>)(在区间[3,1]--恒成立,试求k 的取值范围;(Ⅲ)令()2g x a x b =+,若(1)0g =,又()f x 的图象在x 轴上截得的弦的长度为m ,且02m <≤,试确定c b -的符号.21.(本小题满分14分)已知函数2221()()1ax a f x x x -+=∈+R ,其中a ∈R . (Ⅰ)当1a =时,求曲线()y f x =在点(2(2))f ,处的切线方程; (Ⅱ)当0a ≠时,求函数()f x 的单调区间与极值.A B 1BC 08届高三第三次质量检测文科数学参考答案:二、填空题:本大题共5小题,每小题5分,满分20分,其中14,15题是选做题,考生只能选做一题,,若两题全都做的,只计算前一题的得分.11.(2,+∞) 12.0.88 13. 4 14.cos()23πρθ-= 15. 9三、解答题:本大题共6小题,满分80分.解答须写出文字说明、证明过程或演算步骤. 16.(本小题满分12分)解:(Ⅰ)∵ π()C A B =-+, ………………1分∴ 1345tan tan()113145C A B +=-+=-=--………………4分 又 ∵ 0πC <<, ∴ 3π4C = …………………5分(Ⅱ)由22sin 1tan cos 4sin cos 1A A A A A ⎧==⎪⎨⎪+=⎩,,且π02A ⎛⎫∈ ⎪⎝⎭,,…………………7分得sin A =…………………………9分 由正弦定理sin sin AB BC C A =, 得sin 2sin ABC AB C==……………………12分 17.(本小题满分13分)证明: (1) ∵ 三棱柱111ABC A B C -为直三棱柱, ∴ 1C C ⊥平面ABC , ∴1C C AC ⊥, ∵ 3AC =, 4BC =, 5AB =, ∴ 222AC BC AB +=,∴ AC BC ⊥, 又 1CC BC C ⋂=, ∴ AC ⊥平面11CC B B ,∴ 1AC BC ⊥ ……………………………………7分(2) 令1BC 与1CB 的交点为E , 连结DE .∵ D 是AB 的中点, E 为1BC 的中点, ∴ DE ∥1AC . 又 ∵1AC ⊄平面1CDB , DE ⊂平面1CDB , ∴1AC ∥平面1CDB . ………………………13分 18.(本小题满分13分) 解: (1) 由题意得32nS n n=- , 即 232n S n n =-,…………………1分 当2n ≥时 , 22132[3(1)2(1)]65n n n a S S n n n n n -=-=-----=-,…………4分 当1n =时, 111615a S ===⨯-, ………………5分 ∴ 165()n n n a S S n n N *-=-=-∈, ……………………6分 (2) 由(1)得133111()(65)(61)26561n n n b a a n n n n +===--+-+,…………………8分 ∴ 111111[(1)()()]277136561n T n n =-+-++--+ 11(1)261n =-+ . ……………………11分 因此,使得11(1)()26120m n N n *-<∈+成立的m 必须且只需满足1220m≤, 即10m ≥,故满足要求的的最小正整数10m =………………13分19.(本小题满分14分)解: (1)设圆C 的圆心为,)(y C x ,依题意圆的半径 r =……………… 2分∵ 圆C 在x 轴上截得的弦MN 的长为2a . ∴ 222||y a r +=故 2222()||x y a y a +-=+ ………………………… 4分 ∴ 22x ay =∴ 圆C 的圆心的轨迹方程为22x ay = ………………… 6分 (2) ∵ 45MAN ∠= , ∴ 90MCN ∠= ……………………… 9分令圆C 的圆心为00(,)x y , 则有2002x ay = (00y ≥) ,…………… 10分又 ∵ 01||2y MN a == …………………… 11分∴ 0x = ……………………… 12分∴ r == ……………………… 13分∴ 圆C 的方程为 222()()2x y a a +-= …………………… 14分 21.(本小题满分14分)解:(Ⅰ)由已知.12,0,1-=-=+-=abc b a c 且 解得1a =,2b =, …………………2分∴ 2()(1)f x x =+ , ∴ 22(1),(0)()(1),(0),x x F x x x ⎧+>⎪=⎨-+<⎪⎩ …………4分∴ 22(2)(2)(21)[(21)]8F F +-=++--+=. ……………………5分(Ⅱ)在(Ⅰ)条件下,k x x f +>)(在区间[3,1]--恒成立,即210x x k ++->在区间[3,1]--恒成立,从而12++<x x k 在区间[3,1]--上恒成立,…………………8分 令函数2()1p x x x =++,则函数2()1p x x x =++在区间[3,1]--上是减函数,且其最小值min ()(1)1p x p =-=, ∴ k 的取值范围为(,1)-∞…………………………10分(Ⅲ)由(1)0g =,得20a b +=,∵ 0a > ∴20b a =-<,………………11分 设方程0)(=x f 的两根为21,x x ,则122bx x a+=-=,12c x x a =,∴12||m x x =-=∵ 02m <≤, ∴ 01<≤, ∴01c a ≤<, ∵ 0a >且0bc ≠, ∴ 0c >,∴ 0c b ->……………14分 21.(本小题满分14分)解: (Ⅰ)解:当1a =时,22()1x f x x =+,4(2)5f =,……………1分又22222222(1)422()(1)(1)x x x f x x x +--'==++,则6(2)25f '=-.…………………3分 所以,曲线()y f x =在点(2(2))f ,处的切线方程为46(2)525y x -=--, 即625320x y +-=.……………4分(Ⅱ)解:2222222(1)2(21)2()(1)()(1)(1)a x x ax a x a ax f x x x +--+--+'==++.…………6分 由于0a ≠,以下分两种情况讨论.(1)当0a >时,令()0f x '=,得到11x a=-,2x a =,当x 变化时,()()f x f x ',的变化情况如下表:所以()f x 在区间a ⎛⎫--⎪⎝⎭,∞,()a +,∞内为减函数,在区间a a ⎛⎫- ⎪⎝⎭,内为增函数 故函数()f x 在点11x a =-处取得极小值1f a ⎛⎫- ⎪⎝⎭,且21f a a ⎛⎫-=- ⎪⎝⎭,函数()f x 在点2x a =处取得极大值()f a ,且()1f a =.…………………10分(2)当0a <时,令()0f x '=,得到121x a x a==-,,当x 变化时,()()f x f x ',的变化情况如下表:所以()f x 在区间()a -,∞,a ⎛⎫- ⎪⎝⎭,+∞内为增函数,在区间a a ⎛⎫- ⎪⎝⎭,内为减函数. 函数()f x 在1x a =处取得极大值()f a ,且()1f a =. 函数()f x 在21x a=-处取得极小值1f a ⎛⎫- ⎪⎝⎭,且21f a a ⎛⎫-=- ⎪⎝⎭.………………14分。

08年哈尔滨市第三中学第三次高考模拟考试

08年哈尔滨市第三中学第三次高考模拟考试

2008年哈尔滨市第三中学第三次高考模拟考试理科综合试卷(生物)本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分,共31小题,共10页,总分300分,考试时问150分钟。

考试结束后将本试题卷和答题卡一并交回。

注意事项:1.答题前,考生先将自己的姓名、准考汪号、考场号、座位号填写在试题指定的位置上,将条形码准确粘贴在条形码区域内。

2.选择题选出答案后,必须使用2B铅笔将答题卡上对应题目的答案标号涂黑,如需改动,用橡皮擦干净后,再选涂其他答案,不能答在试题卷上。

3.非选择题使用0.5毫米黑色字迹的签字笔在答题卡上书写,字体工整、笔迹清楚。

4.非选择题必须按照题号顺序在答题卡上各题目的答题区域内作答,超出答题区域或在其它题的答题区域内书写的答案无效;在草稿纸、试题卷上答题无效。

5.作图可先用铅笔画出,确定后必须用黑色字迹的签字笔描黑。

6.保持卡面清洁,不要折叠、不要弄破、弄皱,不准使用涂改液、修正带、刮纸刀。

以下数据可供解题时参考:相对原子质量(原子量):H—1 C—12 O—16 Cl—35.5 Fe—56第Ⅰ卷选择题:共21个小题,每小题6分,共126分。

一、选择题(本题共13小题。

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.菠菜根的分生区细胞不断分裂使根向远处生长,在分裂过程中不会发生的是A.高尔基体参与细胞壁的形成 B.染色体形态较固定、数目较清晰C.基因的表达 D.同源染色体的分开2.下列关于特异性免疫的叙述,不正确的是A.细胞免疫感应阶段中,T细胞释放淋巴因子能促进B细胞增殖分化为效应B细胞。

B.风湿性心脏病、系统性红斑狼疮等疾病是免疫系统攻击自身的正常组织和器官而引发的疾病C.注射减毒或者灭活的乙肝病毒、注射破伤风抗毒素分别属于比较常见的免疫预防和免疫治疗的措施D.在特异性免疫反应阶段中都能产生记忆细胞,并在二次免疫中特异性识别同种抗原3.下列有关生物工程的叙述,正确的是A.基因工程中目的基因与运载体的结合只是通过碱基互补配对就可以实现B.细胞的全能性是植物组织培养和动物细胞培养的理论基础C.单克隆抗体的制备是典型的动物细胞融合技术和动物细胞培养技术的综合应用D.发酵工程产谷氨酸时,要不断调整PH值,若PH值过大,则代谢产物是乳酸或琥珀酸4.下图1为某种遗传病系谱图,下列说法错误的是A.该病不可能的遗传方式是X染色体隐性遗传病B.若该疾病为常染色体遗传病,则系谱中所有的显性个体均为杂合子C.该疾病最有可能为X染色体显性遗传病,则系谱中所有患病女性均为杂合子D.如果该图中患病男子与另外一个患病女子婚配能够生下正常男孩,则再生一个男孩不患病的几率为1/45.下列关于动物生命活动调节的叙述,正确的是A.反射弧包括感受器,传人神经,中枢神经,传出神经,效应器五个部分B.兴奋在细胞间的传递具有定向性,化学递质需要穿过突触前膜、突触间隙和突触后膜,然后才能引起下一个神经元的兴奋C.冷觉感受器感知寒冷刺激时,正常机体能够依靠下丘脑分泌的某种激素直接增加产热以维持体温平衡D.刺激某一完整反射弧的感受器或传入神经,可使效应器产生相同的反30.(22分)下图19中A、B为人体内两种不同组织处的毛细胞血管,①②③示某些化学物质。

2007-2008学年度东北师大附中上学期高三第三次摸底考试

2007-2008学年度东北师大附中上学期高三第三次摸底考试

2007-2008学年度东北师大附中上学期高三第三次摸底考试历史试题本试卷分为第I 卷(选择题)和第n 卷(非选择题)两部分。

共100分,考试时间90分钟。

注意事项:1•第I 卷的答案用铅笔涂写在答题卡上,第n 卷的答案或答题过程均写在答题纸内的 指定处,写在试题卷上的无效。

2•答题前,考生务必将自己的“班级” 、“学号”和“姓名”写在答题卡和答题纸上。

3•考试结束,只交答题卡和答题纸。

第I 卷本卷共24小题,每小题2分,共48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 •春秋战国是古代社会转型的一个重要历史时期,以下相关描述中最能反映这一历史时期( )② 社会形态上由奴隶制向封建制过渡 ④思想文化上,由百家争鸣到独尊儒术 C .①③ D .②④( )② 认为世界的本源为道” ④认为 理”是万物的本源 B •范缜、朱熹、老子、庄子 D •王充、庄子、老子、朱熹 3•下图为1979年甘肃省出土的东汉时期古希腊文铅饼(外国货币)两大发展趋势的是①地方管理上由分封制向郡县制发展 ③国家形势上由割据纷争走向大统一 A .①②B .②③2.下列思想依次是谁提出的①认为万物由元气构成 ③认为世界是我”的主观产物 A •王充、老子、庄子、朱熹 C .王充、朱熹、庄子、老子,由此我们可以获取得的历史信息是( )A •这是张骞出使西域时使用的货币B •丝绸之路开通后,西方铸币技术东传C.长安城各地商人来来往往,商业繁荣D •丝绸之路开通后,有外国商人来华贸易4•中国古代科技文化灿烂辉煌,下列科技著作与称谓不符的是()A • 《天工开物》——中国17 世纪的工艺百科全书B •《本草纲目》——东方医药巨典C •《九章算术》一一当时世界上最先进的理论数学著作D .《梦溪笔谈》一一中国科学史的里程碑5 •下列内容不属于魏晋南北朝社会经济特点的是()A •江南经济开发,中原相对停止,甚至倒退B •士族庄园经济、寺院经济占有重要地位C •商品经济发展水平比较低D •各民族经济交流加强6•史学家白寿彝指出:“隋炀帝开运河,适应了新形势的需要。

第三次三校联考试题答案

第三次三校联考试题答案

08—09学年度第三次三校联考英语试题参考答案2009年04月16日一、听力〔20小题,每一小题1分, 共20分〕1-5:ACABA6-10:ABACB11-15:AACAC16-20:ABCBC二、单项选择〔15小题,第小题1分,15分〕21-25:DCBCB26-30: AAABD31-35:CABBC三、完形填空〔20小题,每一小题2分,共40分〕36-40 DACBC41-45 CBADD46-50 BBCDA51-55 DBCBA四、阅读理解(20小题,第小题3分,共60分)56-60:ACCDC 61-65:AACBA 66-70:ADDBC 71-75:DACCD五、短文改错: (10小题, 每小1分, 共10分)76. very 前加a 77. and改为but 78. being改为be 79. for改为by 80.√81. card改为cards 82. quarreled改为quarrels 83.fixing改为fixed 84. good改为well 85. 去掉to六、书面表达:〔25分〕一、评分原如此1.此题总分为25分,按5个档次给分。

2.评分时,先根据文章的内容和语言初步确定其所属档次,然后以该档次的要求来衡量,确定或调整档次,最后给分。

3.词数少于80和多于120的,从总分中减去2分。

4.评分时,应注意的主要内容为:内容要点、应用词汇和语法结构的数量和准确性与上下文的连贯性。

5.拼写与标点符号是语言准确性的一个方面,评分时,应视其对交际的影响程度予以考虑英、美拼写与词汇用法均可承受。

6.如书写较差,以至影响交际,将分数降低一个档次。

What We High School Students Learn fromthe Global Financial CrisisThe global financial crisis broke out in 2008, which caused serious economic slide and increased job loss.What can we learn from it?First , we should put all our heart into study to broaden our knowledge in various fields. Second, it is required that we should develop our abilities in many ways and improve our personal qualities to achieve our goals in this highly competitive society. Next, learning to cooperate with others is another thing we must do, because sense of teamwork contributes to success more quickly. Last but not least, as young generation in China, w e’d better set our long-term plans to meet what society needs.At present I think only by working hard can our ability be fully embodied. I’ll spar e no effort to study hard to achieve my goal.。

2008届高三年级第三次统一考试 语文试卷

2008届高三年级第三次统一考试 语文试卷

2008届高三年级第三次统一考试语文试卷2008.3本试卷分为第I卷(选择题)和第Ⅱ卷(非选择题)两部分。

第I卷1至5页,第Ⅱ卷6至16页。

满分150分,考试时间150分钟。

笫I卷(选择题共30分)注意事项:1.答第I卷前,考生务必将自己的姓名、准考证号、考试科目、试卷类型用铅笔填涂在答题卡上。

2.每小题选出答案后,用铅笔把答题卡上对应题目的答案标号涂黑。

如要改动,用橡皮擦干净后,再选涂其它答案标号。

不能答在试卷上。

3.考试结束,将本试卷和答题卡一并交回。

一、(12分,每小题3分)1.下列词语中加点的字,每对读音都不相同的一组是A.拗.口/执拗.重.复/安土重.迁省.悟/反躬自省.B.露.脸/雨露.塞.外/茅塞.顿开咀.嚼/含英咀.华C.稽.首/稽.查屏.障/敛声屏.气气度./审时度.势D.徘徊./低徊.纤.细/纤.毫毕见和.诗/和.衷共济2.依次填入下面横线处的词语,最恰当的一组是①国家发展改革委宣布从1月15日起启动临时价格机制,主要涉及粮食、食用植物油、肉类及其制品、牛奶、鸡蛋、液化石油气等。

②翻翻日历数着回家过年的日子,张丹丹和男友着要在年三十这天从北京赶回哈尔滨,带父母品尝一顿西餐厅里的年夜饭。

③然而,在时尚的潮流中,新年的春联、中秋的月饼、街市上掠过的大红唐装,还是让我们依稀看到传统的倔强身影。

A.干预合计偶尔B.干涉核计偶尔C.干预核计偶然D.干涉合计偶然3.下列各句中,加点的成语使用恰当的一句是A.这种草药能治疗高血压病,对人体没有任何副作用,它的原理多见于各种医学书刊,临床使用又屡试不爽,你还有什么可怀疑的?B.名著续写难似乎是世界级难题,除《飘》的续集《斯佳丽》赢得赞誉外,像高鹗续书之于《红楼梦》,果戈理付之一炬的《死魂灵(二)》之于原著,都难逃画虎类犬的命运。

C.央视一套黄金时段播出的电视剧《闯关东》引起强烈反响,人们评头品足,称赞该剧将近年来花样男子的娇柔一扫而光,重新定位了中国男人应有的气质。

吉林省东北师大附中2007—2008学年上学期高三第三次摸底考试

吉林省东北师大附中2007—2008学年上学期高三第三次摸底考试

吉林省东北师大附中2007—2008学年上学期高三第三次摸底考试语文试题本试卷分为第Ⅰ卷和第Ⅱ卷两部分。

满分150分,考试时间为150分钟。

注意事项:1.答题前,考生务必将自己的“姓名”、“班级”、和“考号”写在答题纸和作文纸上;2.请同学们认真阅读试题要求,并把答案分别落实在答题卡和答题纸上;3.考试结束,分交答题卡、答题纸和作文纸。

第Ⅰ卷一、(12分,每小题3分)1.下列词语中加点的字,每对的读音都不相同的一组是()A.落照/落色场院/场馆累赘/累累失误跑马卖解/苏三起解B.中听/中肯宿仇/宿将一幢/灯影幢幢安步当车/螳臂当车C.簇拥/箭镞巡视/驯服秕谷/如丧考妣缀玉连珠/苦学不辍D.剽掠/慓悍花圃/浦口裨将/俾众周知每况愈下/坚贞不渝2.下列词语中,错别字最多的一项是()A.震憾侯车室飞扬跋扈欲加之罪,何患无词B.蕴籍座右铭金碧辉煌螳螂捕蝉,黄鹊在后C.膺品大姆指沧海一粟华而不实,脆而不坚D.飘渺捅娄子喧然大波管中窥豹,可见一班3.下列句子标点符号的使用正确的一项是A.周杰伦暌违3年的个人巡演“2007世界巡回演唱会”,照惯例从台北起跑。

他穿着新定制的武士装,霸气十足地从舞台中缓缓升起,一连热唱“黄金甲”、“无双”两首中国风快歌。

B.国家统计局网站发布了10月份全国工业品出厂价格,数据显示:工业品出厂价格同比上涨3.2%,原材料、燃料、动力购进价格上涨4.5%。

C.增加清明、端午、中秋这三个传统节日为法定节假日是没有问题的;把除夕纳入假期也是没有问题的;但是要取消五一黄金周却在某些人的讨论中是有问题的。

D.十七大报告指出,“改革开放是决定当代中国命运的关键抉择,是发展中国特色社会主义、实现中华民族伟大复兴的必由之路。

”近日,记者分别就相关问题采访了有关专家。

4.下列各句中,没有语病的一句是()A.字写得好坏,作文写得有文采,是由一个人的艺术修养、文学功底的高低所决定的。

B.国安队虽然失去了联赛冠军,但是主帅李章洙认为,国安队还是以超过鲁能队6分的战绩稳居第二名,这意味着国安获得了亚冠联赛资格。

2008年哈尔滨师大附中高三第三次模拟考试理

2008年哈尔滨师大附中高三第三次模拟考试理

2008年哈尔滨师大附中高三第三次模拟考试数学试题(理科)本试卷分选择题和非选择两部分,共22题,共150分。

考试结束后,将本试卷和答题卡一并交回。

注意事项:1.答题前,考生将自己的姓名、准考证号码填写清楚,将条形码准确粘贴在条形码区域内。

2.选择题必须使用2B 铅笔填涂;非选择题必须使用0.5毫米黑色字迹的签字笔书写,字体工整、笔迹清楚。

3.请按照题号顺序在各题目的答题区域内作答,超出答题区域书写的答案无效;在草稿纸、试题卷上答题无效。

4.作图可先使用铅笔画出,确定必在用黑色迹的签字笔描黑。

5.保持卡面清洁,不要折叠,不要弄破、弄破,不准使用涂改液、修正带、刮纸刀。

一、选择题:(每小题5分,满分60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.复数2)11(i-的值为 ( )A .2B .—2C .i 2D .—i 2 2.函数||sin )(x x f =的一个单调递减区间为( )A .)4,4(ππ-B .)4,43(ππ--C .)2,0(π D .)0,2(π-3.集合B A x x x B x x A 则},06|{},3|12||{2≤-+=>+== ( )A .]2,1[]2,3[ --B .),1()2,3(+∞--C .[)(]2,12,3 --D .(](]2,13, -∞-4.直线)3,2(02:-==+-m y x l 按向量平移后得到的直线5)1()2(221=++-y x l 与圆相切,则m 的值为( )A .9或—1B .5或—5C .—7或7D .3或135.已知向量m b n m a 其中),sin ,(cos ),,(θθ==、n 、2|,|4||.λθ<⋅=∈b a b a R 则当若恒成立时,实数λ的取值范围是( )A .22-<>λλ或B .22-<>λλ或C .22<<-λD .22<<-λ6.已知相交直线l 、m 都在平面α内,并且都不在平面β内,若p :l 、m 中至少有一条与平面β相交;q :平面q p 是则相交与,βα的( )A .充分不必要条件B .必要不充分条件C .充要条件D .既不充分也不必要条件7.已知随机变量ηηξηξD E B 和则若),6.0,10(~,8=+分别为 ( )A .6和2.4B .2和2.4C .2和5.6D .6和6.68.定义在R 上的奇函数[),1,0,2)(时当的周期函数是周期为∈x x f ,12)(-=xx f 则)83(log 21f 的值为( )A .21-B .25-C .—5D .—69.设nx )21(+展开式的各项系数的和为n a ,各二项式系数的和为nn nn n n b a a b b +-++∞→11lim ,则=( )A .31-B .31 C .—1 D .010.已知函数)3(),1(),4(,,sin )(ππf f f R x x x x f -∈=则的大小关系为 ( )A .)4()1()3(ππ->>f f fB .)1()4()3(f f f >->ππC .)3()1()4(ππf f f >>-D .)4()3()1(ππ->>f f f11.已知点F 1、F 2为双曲线)0,0(12222>>=-b a by a x 的左、右焦点,P 为双曲线右支上一点,点P 到右准线的距离为d ,若|PF 1|,|PF 2|,d 依次等差数列,此双曲线离心率的取值范围是( )A .(]3,1B .[)+∞+,32C .]32,32[+-D .(]32,1+12.已知S A S ⊆=},2008,,3,2,1{ 且A 中有三个元素,若A 中的元素可构成等差数列,则这样的集合A 共有( )A .32008C 个B .32004A 个 C .221004A 个 D .222004C 个第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.已知数列n n n a a n n 项和最大时则其前的通项公式为,29}{-=的值为 。

哈尔滨市第三中学高三第三次高考模拟(理)

哈尔滨市第三中学高三第三次高考模拟(理)

2008年哈尔滨市第三中学高三第三次高考模拟数学试题(理科)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,在每小题给出的四个选项中,只有一个是符合题目要求的)1.设复数t z R t i t t t z 则对应的点在第三象限,),()1()352(2∈++-+=的取值范围为( )A .13-<<-tB .211<<-t C .213<<-t D .211-<<-t2.下列四个函数中,最小正周期为ππ125,=x 且图象关于直线对称的函数是 ( )A .)32sin(π+=x yB .)32sin(π-=x yC .)32sin(π-=x yD .)32sin(π+=x y3.函数a x x x a x x x x x f 则处连续在点,1)1)(1(log )1(132)(22=⎪⎩⎪⎨⎧≥+<--+=的值是 ( )A .—3B .6C .4D .24.数列s a a n 且对于任意的正整数中,6,}{1=、20085,a a a a t t t s 则都有+=+等于( )A .2008B .2006C .12042D .120485.若对于任意的a ax x R x 则实数恒成立不等式,4||,≥∈的取值范围是 ( )A .41-<a B .41||≤a C .41||<a D .41-≥a 6.已知三个不同的平面n m ,,,和两条不重合的直线γβα,有下列4个命题: ①n m n m //,,//则=⋂βαα ②βαβα⊥⊂⊥则,,//,n n m m ③γαβγβα//,,则⊥⊥ ④γαγβα⊥⊥=⋂则,,m mA .2B .3C .4D .17.体育课下课后,老师要求体育委员把5个相同的篮球、3个相同的排球、2个相同的橄榄球排成一排放好,则不同的放法有( )A .420种B .1260种C .5040种D .2520种8.函数a a f f x ex x x f x 则满足,2)()1(,)0()01)(sin()(12=+⎪⎩⎪⎨⎧≥<<-=-π所有可能的值为 ( )A .221-或 B .22-C .1D .221或9.过椭圆l F b a by a x 的直线作倾斜角为的右焦点θ)0(12222>>=+交椭圆于A 、B 两点,P 为右准线上一点,使2π=∠APB 的点P( )A .有1个B .有2个C .有无数个D .不存在10.在平面直角坐标系中,已知△ABC 三个顶点的坐标分别为A (2,1),B (—1,—1),C(1,3),点P 在直线BC 上运动,动点Q 满足PC PB PA PQ ++=,则点Q 的轨迹方程为( ) A .032=--y x B .032=--y xC .042=-+y xD .032=-+y x11.已知球的半径为2,球面上的三个大圆所在平面两两垂直,则以三个大圆的交点为顶点的八面体的体积为( )A .316 B .332 C .364 D .3212.已知}491|{4)(232)(2≤≤=+=++=x x M x x x q px x x f 是定义在集合和ϕ上的函数,对任意的),()(),()(,,000x x x f x f M x M x ϕϕ≥≥∈∈使得存在常数且)()(00x x f ϕ=,则函数)(x f 在集合M 上的最大值为( )A .4B .833C .6D .8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,将答案填在答题卡上)13.已知等式14142210324)1()12(x a x a x a a x x x ++++=+-- 成立,则1442a a a +++ = .14.已知随机变量)5.0(,78.0)5.2(),,5.1(2≤=≤ξξσξP P N 则服从正态分布= . 15.设)(),(x g x f 分别是定义在R 上的奇函数和偶函数,当0])()([,0>'<x g x f x 时,且0)()(,0)2(<=x g x f g 则不等式的解集是 . 16.已知q p m m x x q x p ⌝⌝>≤-+-≤--是而),0(012:,2|321:|22的必要不充分条件,则实数m 的取值范围为 .三、解答题(本大题共6个小题,共70分,解答应写出文字说明,证明过程或演算步骤) 17.(本小题满分10分)已知二次函数)()(2R t t tx x x f ∈+-=,同时满足:①不等式0)(≤x f 的解集有且只有一个元素;②若}{.)()(,01212n a x f x f x x 设数列成立总有不等式<<<的前n 项和).(n f S n =(I )求数列}{n a 的通项公式; (II )求).1111(lim 2534231+∞→++++n n n a a a a a a a a 18.(本小题满分12分)向量.)()(),21,12cos 2(),2,(cos ),1,sin 3(2p n m p n m ⋅+=-==-=x f x x x 设函数 (I )将函数x y x f 2sin ,)(=可得到函数平移的图象按向量d 的图象,求模最小的向量d ;(II )设△ABC 的三边a ,b ,c 依次成等比数列,且边b 所对的角为x ,求x 的取值范围及函数)(x f 的最大值和最小值.19.(本小题满分12分)某班艺体特长生中,每人至少具备艺术特长、体育特长两项中的一项,已知具备艺术特长的有2人,具有体育特长的有5人.现从中任选2人,设ξ为选出的2人中既具备艺术特长又具备体育特长的人数,且.7.0)0(=>ξP (I )求该班艺体特长生的人数;(II )求ξ的概率分布,并计算ξ的数学期望Eξ.(用数字作答)20.(本小题满分12分)在如图所示的几何体中,EA ⊥平面ABC ,DB ⊥平面ABC ,AC ⊥BC ,且BC=BD=a AC a AE 2,23==,AM=2MB. (I )求证:CM ⊥EM ;(II )求直线CD 与平面MCE 所成角的大小.21.(本小题满分12分)已知两个动点A 、B 和一个定点)0(2),(200>=p px y y x M 均在抛物线上(A 、B 与M 不重合).设F 为抛物线的焦点,Q 为对称轴上一点,若0)21(=⋅+,且|||,||,|成等差数列. (I )求的坐标; (II )若25||,3||==,A 、B 两点在抛物线)0(22>=p px y 准线上的射影分别为A 1、B 1,求四边形ABB 1A 1面积的取值范围.22.(本小题满分12分) 函数).0(1ln )(>+=a x a x f(I )当)11(1)(:,0xa x f x -≥->求证时;(II )在区间a x x f e 求实数恒成立上,)(),1(>的取值范围; (III )当).)(11(2)1()3()2(:,21*N n n n n f f f a ∈+-+>++++=求证时。

08年高中理科综合毕业班第三次摸底考试

08年高中理科综合毕业班第三次摸底考试

2008年高中理科综合毕业班第三次摸底考试理科综合第I卷(选择题共126分)以下数据可供解题时参考:相对原子质量(原子量):H:1 C:12 N:14 O:16一、选择题(本题包括13小题。

每小题只有一个选项符合题意。

每小题6分,总计78分)1.下列对原核细胞与真核细胞的有关叙述正确的是()A.原核细胞没有叶绿体,不能进行光合作用B.原核细胞较真核细胞简单,细胞内只具有一种核酸C.真核细胞具有染色体;原核细胞具有染色质D.真核细胞才可以进行减数分裂或有丝分裂2.下列对动物生命活动调节的有关叙述正确的是()A.性行为是以性激素为生理基础的行为,因而动物在完成性行为时主要靠性激素调节B.在正常情况下,内环境的各项理化性质经常处于变动之中,但都保持在适宜的范围内C.已获得免疫的机体,再次受到相同抗原刺激即发生过敏反应D.在血糖调节过程中,甲状腺激素与胰岛素之间表现拮抗作用3.根据现代生物技术原理,为解决如下四个问题:①快速培养名贵花卉②克服远缘植物有性杂交不亲和的障碍③提高农作物的抗病能力④干扰素的工业化生产。

依次可以4.豌豆的种皮灰色(G)对白色(g)为显性,现有基因型分别为GG和gg的两个亲本杂交产生F1,然后再连续自交,根据基因的分离定律,豌豆种皮灰色:白色=3:l的分离比出现于()A.F l代每株植株所结种子B.F1代所有植株所结种子C.F2代每株植株所结种子D.F2代所有植株所结种子5.乙肝疫苗是将乙肝表面蛋白抗原基因在酵母菌中克隆和表达得到的乙肝表面抗原,实际上是中空的病毒颗粒。

请判断以下选项不正确的是()A.乙肝属于获得性免疫缺陷病B.乙肝疫苗需要低温保存C.乙肝病毒内部的核酸储存着病毒的全部遗传信息,控制着病毒的一切性状D.感染乙肝病毒后通常需要体液免疫和细胞免疫共同发挥作用6.美国和俄罗斯科学家将大量48Ca离子加速去轰击人造元素249Cf,从而制造出3颗新原子。

每颗新原子的原子核包含118个质子和179个中子。

2007-2008学年度东北师大附中上学期高三第三次摸底考试

2007-2008学年度东北师大附中上学期高三第三次摸底考试

2007-2008学年度东北师大附中上学期高三第三次摸底考试历史试题本试卷分为第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分。

共100分,考试时间90分钟。

注意事项:1.第Ⅰ卷的答案用铅笔涂写在答题卡上,第Ⅱ卷的答案或答题过程均写在答题纸内的指定处,写在试题卷上的无效。

2.答题前,考生务必将自己的“班级”、“学号”和“姓名”写在答题卡和答题纸上。

3.考试结束,只交答题卡和答题纸。

第Ⅰ卷本卷共24小题,每小题2分,共48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.春秋战国是古代社会转型的一个重要历史时期,以下相关描述中最能反映这一历史时期两大发展趋势的是()①地方管理上由分封制向郡县制发展②社会形态上由奴隶制向封建制过渡③国家形势上由割据纷争走向大统一④思想文化上,由百家争鸣到独尊儒术A.①②B.②③C.①③D.②④2.下列思想依次是谁提出的()①认为万物由元气构成②认为世界的本源为“道”③认为世界是“我”的主观产物④认为“理”是万物的本源A.王充、老子、庄子、朱熹B.范缜、朱熹、老子、庄子C.王充、朱熹、庄子、老子D.王充、庄子、老子、朱熹3.下图为1979年甘肃省出土的东汉时期古希腊文铅饼(外国货币),由此我们可以获取得的历史信息是()A.这是张骞出使西域时使用的货币B.丝绸之路开通后,西方铸币技术东传C.长安城各地商人来来往往,商业繁荣D.丝绸之路开通后,有外国商人来华贸易4.中国古代科技文化灿烂辉煌,下列科技著作与称谓不符的是()A.《天工开物》——中国17世纪的工艺百科全书B.《本草纲目》——东方医药巨典C.《九章算术》——当时世界上最先进的理论数学著作D.《梦溪笔谈》——中国科学史的里程碑5.下列内容不属于魏晋南北朝社会经济特点的是()A.江南经济开发,中原相对停止,甚至倒退B.士族庄园经济、寺院经济占有重要地位C.商品经济发展水平比较低D.各民族经济交流加强6.史学家白寿彝指出:“隋炀帝开运河,适应了新形势的需要。

东北三校2008年部分毕业班第三次摸底考试

东北三校2008年部分毕业班第三次摸底考试

东北三校2008年部分毕业班第三次摸底考试数 学第I 卷(选择题 共60分)一、选择题:本大题共12个小题,每小题5分,共60分,在每个小题的四个选项中,只有一个是符合题目要求的. 1.已知集合=<+-=≥=N M x x x N x x M 则},013|{},1|||{ ( )A .}31|{<≤x xB .}11|{≤<-x xC .}3|{>x xD .}1|{-<x x 2.函数)1(21≥=-x y x的反函数是( )A .)1)(1(log 21≥+=x x yB .)1(1log 21≥+=x x yC .)10)(1(log 21≤<+=x x yD .)10(1log 21≤<+=x x y3.由数字0,1,2,3,5组成的没有重复数字的三位奇数的个数为 ( )A .60B .48C .36D .274.某商店两个进价不同的商品都卖了80元,其中一个盈利60%,另一个亏本20%,问在这次买卖中,这家商店 ( ) A .不赚不赔 B .赚了10元 C .赔了10元 D .赚了50元 5.过点A (2,-1),圆心在直线x y 2-=上,且与直线1=+y x 相切的圆的方程为( ) A .2)56()53(22=++-y x B .2)56()53(22=-++y xC .2)2()1(22=-++y xD .2)2()1(22=++-y x6.在半径为35的球面上有A 、B 、C 三点,AB=6,BC=8,CA=10,则球心到平面ABC 的距离为( )A .5B .24C .25D .107.函数)1(2log 2>-=a x x y a 在其定义域内( )A .有最大值,无最小值B .有最小值,无最大值C .有最大值,也有最小值D .既无最大值,也无最小值 8.对于直线l 、m 和平面α、β,α⊥β的一个充分条件是 ( ) A .βα⊥⊥m l m l ,//, B .αβα⊥=⊥m l m l ,,C .αβ⊂⊥l m m l ,,//D .βα⊥⊥m l m l ,,//9.已知函数),(,)1(4)1()1(12)(2+∞-∞⎩⎨⎧->+--≤+-=在x a x a x ax ax x f 内是减函数,则实数a 的取值范围是( )A .)1,(-∞B .)0,(-∞C .),1(+∞D .(0,1)10.将周期为π的函数)0(c o s c o s s i n 2s i n 22>-+=ωωωωωx x x x y 的图象按)1,8(π-=a 平移后,所得函数图象的解析式为( )A .12sin 2+=x yB .1)44sin(2-+=πx yC .1)82sin(2+-=πx yD .x y 2cos 21-=11.设x 、y 满足条件41,033042022--=⎪⎩⎪⎨⎧≤--≥+-≥-+x y u y x y x y x 则的取值范围是 ( )A .],31[)1,(+∞--∞ B .]31,41[-C .]31,1[-D .]43,51[12.如右图,已知在梯形ABCD 中,|AB|=2|CD|,点E 分有向线段所成的比为52,双曲线过C 、D 、E 三点,且以AB 为 焦点,则此双曲线的离心率为 ( ) A .2 B .3C .2D .3选择题答题栏第Ⅱ卷(非选择题 共90分)二、填空题:本大题共4个小题,每小题4分,共16分. 13.10)212(xx x -的展开式中,常数项为 .(用数字作答)14.已知=-==|32|,3,1||,2||b a b a b a 则的夹角为与π.15.已知数列=∈++==*+n n n n a N n n a a a a 则且满足),(12,1}{11 .16.在△ABC 中,内角A 、B 、C 所对的边分别为a 、b 、c ,给出下列结论:①若A >B >C ,则C B A sin sin sin >>; ②若C B A c b a cos cos cos ,>>>>则;③必存在A 、B 、C ,使C B A C B A tan tan tan tan tan tan ++<成立; ④若︒===25,20,40B b a ,则△ABC 必有两解.其中,真命题的编号为 .(写出所有真命题的编号) 三、解答题:本大题共6个小题,共74分. 解答应写出文字说明,证明过程或演算步骤. 17.(本小题满分12分)已知)sin(,44,434,1411)43sin(,71)4cos(βαπβππαπβπαπ+<<-<<=+=-求的值.甲、乙两位同学站在罚球线进行投篮,已知每次投篮,甲投中的概率为21,乙投中的概率为32,若甲、乙两位同学各投篮4次,求: (1)甲恰好投中3次的概率; (2)乙至少投中1次的概率;(3)乙恰好比甲多投中2次的概率. 19.(本小题满分12分)已知在等比数列}{n a 中,n S 是其前n 项的和,若12,,++m m m S S S 成等差数列,则m a ,12,++m m a a 成等差数列.(1)写出这个命题的逆命题:(2)判断逆命题是否为真,并给出证明.如图,在直三棱住ABC —A 1B 1C 1中,AC=AB=4,241=BB ,2=BM ,︒=∠90BAC , 点N 在CA 1上,且131NA CN =. (1)求证:MN//平面A 1B 1C 1; (2)求点A 1到平面AMC 的距离; (3)求二面角C —A 1M —B 的大小. 21.(本小题满分12分)已知定义在R 上的函数)(x f 满足:)()()(y f x f y x f +=+,当.0)(,0<<x f x 时 (1)求证:)(x f 是奇函数;(2)解关于x 的不等式:m m m f x m f x f mx f 且,0).((2)()(2)(22>->-为常数).已知椭圆的中心在原点O ,短轴长为22,其焦点F (c ,0)(c >0)对应的准线l 与x轴交于A 点,|OF|=2|FA|,过A 的直线与椭圆交于P 、Q 两点. (1)求椭圆的方程;(2)若0=⋅,求直线PQ 的方程;(3)设)1(>=λλ,过点P 且平行于准线l 的直线与椭圆相交于另一点M. 求证F 、M 、Q 三点共线.数学(理)参考答案一、选择题:每小题5分,共60分. 1—12ADDBD CBCDA CB二、填空题:每小题4分,共16分. 13.840 14.13 15.n 2 16.①④ 三、解答题:本大题共6个小题,共74分.解答应写出文字说明,证明过程或演算步骤. 17.解:042434<-<-∴<<απππαπ…………1分又734)71(1)4(cos 1)4sin(71)4cos(22-=--=---=-∴=-απαπαπ (4)分πβπππβπ<+<∴<<-43244又1435)1411(1)43(sin 1)43cos(,1411)43sin(22-=--=+--=+∴=+βπβπαπ ……7分+-+=--+=++∴)4cos()43cos()]4()43cos[()](2cos[απβπαπβπβαπ23)4sin()43sin(-=-+απβπ ………………11分 23)](2cos[)sin(=++-=+∴βαπβα……………………12分18.解:(1)甲恰好投中3次的概率为4121)21(334=⨯⨯C ;……………………3分 (2)∵乙投篮4次都未投中的概率为181)31(4=, ∴乙至少投中1次的概率为8180)31(14=-;………………6分(3)设乙恰好比甲多投中2次为事件A ,乙恰好投中2次而甲恰好投中0次为事件B 1,乙恰好投中3次而甲恰好投中1次为事件B 2,乙恰好投中4次而甲恰好投中2次为事件B 3,则A=B 1+B 2+B 3 .……………………………………8分∴乙恰好比甲多投中2次的概率 P (A )=P (B 1)+P (B 2)+P (B 3) (9)分2244443143344042224)21()32()21(2131)32()21()31()32(⋅⋅⋅+⋅⋅⋅⋅⋅+⋅⋅⋅⋅=C C C C C C.16231)21(2=⋅………………………………12分19.解:(1)逆命题:在等比数列}{n a 中,n S 是其前n 项的和,若12,,++m m m a a a 成等差数列,则12,,++m m m S S S 成等差数列. ………………………………2分 (2)当q=1时,逆命题为假;当q ≠1时,逆命题为真.证明:设}{n a 的公比为q ,若12,,++m m m a a a 成等差数列,则有122+++=m m m a a a …3分012,0,0,22111111=--≠≠+=∴-+q q q a q a q a q a m m m 解得,211-==q q 或…5分当q=1时,11111)1(,)2(,a m S a m S ma S m m m +=+==++1212,,2++++∴+≠∴m m m m m m S S S S S S 不成等差数列.………………7分当])21(1[34211])21(1[22,2121212+++--=+--=-=m m m a a S q 时 …………8分211])21()2(1)21(41[211])21(1[211])21(1[2211111+-⋅--+-⋅-=+--++--=+++++m m m m m m a a a S S ])21(1[34211])21(22[2121++--=+-⋅-=m m a a1212,,,2++++∴+=∴m m m m m m S S S S S S 成等差数列. ………………11分综上得,当q=1时,逆命题为假;当q ≠1时,逆命题为真. …………12分 20.法一:(1)证明:在C 1A 1上取点D ,使11141A C D C =,连结ND 、B 1D , 11111////,4131BB A C ND CA NA CN ∴==………………1分又M B BB CC ND 111234343====,∴四边形B 1MND 为平行四边形, ∴MN//B 1D …………………………………………………………2分又⊄MN 平面A 1B 1C 1,B 1D ⊂平面A 1B 1C 1 ∴MN//平面A 1B 1C 1 …………4分 (2)∵三棱柱ABC —A 1B 1C 1为直三棱柱 AC A A ⊥∴1又︒=∠90BAC , ⊥∴AC 平面A 1ABB 1在平面A 1ABB 1中过A 1作A 1H ⊥AM ,又AC ⊥A 1H ,∴A 1H 为点A 1到平面AMC 的距离 ……………………………………………………6分在23,24,4,22111=+====∆MB AB AM BB A A AB MA A 中,316,2821211111==⋅=⋅=∴∆H A AB A A AM H A S M AA , 即A 1到平面AMC 3161=⋅m AA ……………………8分 (3)在平面A 1ABB 1中过A 作AE ⊥A 1M ,交A 1M 于点E ,连结CE ,则AE 为CE 在平面A 1ABB 1内的射影,ACE M A CE ∠∴⊥∴,1为二面角C —A 1M —B 的平面角 ……………………10分在MA A 1∆中,28,341212111==+=∴∆M AA S MB B A M A 2161=⋅∴AE M A417tan ,171716==∠∴=∴AE AC AEC AE ………………11分 ,417arctan=∠∴AEC 即二面角C —A 1M —B 的大小为417arctan …………12分 法二:(1)∵三棱柱ABC —A 1B 1C 1为直三棱柱,A 1A ⊥AC ,A 1A ⊥AB ,又∠BAC=90°∴可以点A 为坐标原点,以AC 、AB 、AA 1所在直线分别为x 轴、y 轴、z 轴建立空间直角坐标系,由已知得A (0,0,0)、B 1(0,4,24)、C 1(4,0,24)、 C (4,0,0)、A 1(0,0,24)、M (0,4,2)、N (3,0,2),在C 1A 1上取点D ,使)24,0,3(,41111D A C D C 则=B B 11)0,4,3(),0,4,3(=∴-=-=∴ ………………2分⊄∴MN DB MN 又1//平面A 1B 1C 1,B 1D ⊂平面A 1B 1C 1 ∴MN//平面A 1B 1C 1…4分(2))2,4,0(=、)2,4,4(-=,设平面AMC 的法向量为)1,,(b a =, 则00=⋅=⋅CM m AM m ⎪⎩⎪⎨⎧=++-=+∴0244024b a b ,解得42,0-==b a ……6分)24,0,0(),1,42,0(1=-=∴AA 又 ∴A 1到平面AMC 的距离为316………………8分 (3))24,0,4(1-=CA 、)2,4,4(-=CM ,设平面A 1MC 的法向量为)1,,(d c =,则0,01=⋅=⋅CA)1,423,2(.423,202440244=∴==⎪⎩⎪⎨⎧=++-=+-∴n d c d c c 解得又⊥平面A 1ABB 1,∴所求得二面角的大小为><,,………………11分 而),0,0,4(= 33334446624||||,cos =⨯=⋅>=<∴n AC ∴二面角C —A 1M —B 的大小为.33334arccos………………12分 21.解:(1).0)0(),0()0()0(,0),()()(=+===+=+f f f f y x y f x f y x f 即得令令)()(),()()0(,,0x f x f x f x f f x y y x -=-∴-+=-==+得即 )(x f ∴是奇函数…3分(2)设x 1、x 2∈R ,且,0,2121<-<x x x x 则由已知得0)(21<-x x f .0)()()()()(212121<-=-+=-∴x x f x f x f x f x f)()()(21x f x f x f 即<∴在R 上是增函数. ……………………6分又)2()(2).2()()()(2x f x f m f m f m f m f ==+=同理 ………………7分 )2()()2()()(2)()(2)(2222x f x m f m f mx f m f x m f x f mx f +>+⇔->- 02)2(22)2()2(222222>++-⇔+>+⇔+>+⇔m x m mx x x m m mx x x m f m mx f 0))(2(02)2(,02>--∴>++-∴>m x m x x mm x m …………9分 当2,2><m m m即时,不等式的解集为}2|{m x m x x ><或; 当20,2<<>m m m 即时,不等式的解集为}.2|{m x m x x ><或 …………12分 22.解:(1)由题意,设椭圆的方程为:)2(12222>=+a y ax 由已知得⎪⎩⎪⎨⎧-==-)(22222c c a c c a ,解得∴==,2,6c a 椭圆的方程为:.12622=+y x …………3分 (2)由(1)可得,准线l 的方程为:)0,3(.3A x ∴=设直线PQ 的方程为:)3(-=x k y 由方程组062718)13()3(126222222=-+-+⎪⎩⎪⎨⎧-==+k x k x k x k y y x 解得…………5分 依题意),(),,(.3636,0)32(1221112y x Q y x P k k 设得<<->-=∆,则 13182221+=+k k x x ① 136272221+-=k k x x ② ………………7分 由直线PQ 的方程得)3(),3(2211-=-=x k y x k y ,]9)(3[)3)(3(2121221221++-=--=∴x x x x k x x k y y ③ 又0,02121=+∴=⋅y y x x ④ ………………9分由①②③④得)36,36(55,152-∈±=∴=k k ,直线PQ 的方程为 035035=-+=--y x y x 或 ………………10分(3)证明:).,3(),,3(2211y x y x -=-= 由题意可得方程组.215,1126126)3(32222221212121λλλλλ-=∴>⎪⎪⎪⎩⎪⎪⎪⎨⎧=+=+=-=-x y x y x y y x x 又 ………………12分 ),21(),1)3((),2(),,(),0,2(1121111y y x y x y x M F --=-+-=--=∴-λλ ),21(2y λλλ--= 而y y x λλλ-=∴-=-=),,21(),2(222 //∴,又∵直线FM 、直线FQ 有公区点F 故F 、M 、Q 三点共线. ……………………14分。

东北师大附中

东北师大附中

东北师大附中2007—2008学年上学期高三第三次摸底考试物 理 试 题考试时间:90分钟 试卷满分:100分本试卷分为第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分,全卷共有19道题. 注意事项:1.第Ⅰ卷的答案用2B 铅笔涂写在答题卡上,第Ⅱ卷的答案或解答过程均写在答题纸 内的指定处,写在试卷上无效!2.答题前,考生务必将自己的“班级”、“考号”、“姓名”写在答题卡和答题纸上.3.考试结束,只交答题卡和答题纸.第Ⅰ卷(选择题 共48分)一、本大题共12道题,每题4分,共48分.每题给出的四个选项中至少有一个选项是正确的,全部选对的得4分,选对但不全的得2分,有选错或不答的得0分.1.如图所示,示波管是示波器的核心部件,它由电子枪、偏转电极和荧光屏组成。

如果在荧光屏上P 点出现亮斑,那么示波管中的 ( )A .极板X 应带正电B .极板X '应带正电C .极板Y 应带正电.D .极板Y '应带正电2.如图为一列沿x 轴正方向传播的简谐横波在0=t 时的波形,当R 处质点在0=t 时的振动状态传到S 处时,关于P 、R 处的质点,下列说法正确的是 ( ) A .P 处的质点位于波峰.B .P 处的质点速度达到最大.C .R 处的质点位于波峰.D .R 处的质点速度达到最大.3.右图是某地区1、7月份气温与气压的 对照表,7月份与1月份相比较,从 平均气温和平均大气压的角度来说正 确的是 ( )A .空气分子无规则热运动的激烈情况几乎相同.B .空气分子无规则热运动减弱了.vC .单位时间内空气分子对单位面积的地面撞击次数增多了.D .单位时间内空气分子对单位面积的地面撞击次数减少了.4.由于科学研究的需要,常常将质子)(11H 和α粒子)(42He 等带电粒子贮存在圆环状空腔中,圆环状空腔置于一个与圆环平面垂直的匀强磁场中,磁感应强度为B ,如果质子和α粒 子在空腔中做圆周运动的轨迹相同(如图中虚线所示),比较质子和α粒子在圆环状空腔中运动的动能H E 和αE ,周期H T 和αT 的大小,下面选项正确的是( ) A .ααT T E E H H ≠≠, B .ααT T E E H H ==, C .ααT T E E H H =≠,D .ααT TE E H H ≠=,5.单匝闭合线框在匀强磁场中,绕垂直于磁场方向的转轴匀速转动,在转动过程中,穿过线框的最大磁通量为m ϕ,线框中的最大感应电动势为m E ,下列说法中正确的是( ) A .在穿过线框的磁通量为2m ϕ的时刻,线框中的感应电动势为2m EB .在穿过线框的磁通量为2mϕ的时刻,线框中的感应电动势为2m EC .线框每转动一周,线框中的感应电流方向改变一次.D .线框转动的角速度为mmE ϕ6.下面是2007年10月26日新华社记者田兆运和王玉山的题为“嫦娥一号成功实施第二次变轨,11分钟冲上7万公里”报道中的一段描述:“月球是地球唯一的天然卫星,它以3683公里/小时的速度绕地球运行,绕地球一周的公转周期为27.3个地球日,月球在绕地球公转的同时进行自转”. “在地球上只能看见月球的正面,而永远看不见月球的背面”,月球的背面给人类蒙上了一层十分神秘的色彩,通过对月球运动的分析,说明人们在地球上看不到月球背面的原因是 ( ) A .月球的自转周期与地球的自转周期相同. B .月球的自转周期与地球的公转周期相同. C .月球的公转周期与地球的自转周期相同. D .月球的公转周期与月球的自转周期相同.7.我国发射的“嫦娥一号”探月卫星简化后的路线示意图如图所示,卫星由地面发射后经过发射轨道进入停 泊轨道,然后在停泊轨道经过调速后进入地月转移轨 道,再次调速后进入工作轨道,卫星开始对月球进行 探测。

东北三校实验班三摸

东北三校实验班三摸

东北三校实验班三摸学习资料收录大全东北三校2008年部分毕业班第三次摸底考试英语题号一二三总分得分第I卷(选择题共95分)第一部分英语知识运用 (共3节满分50分)第一节:语音知识 (共5小题每小题1分,满分5分)从A、B、C、D中找出其划线部分与所给单词的划线部分读音相同的选项1. grade A. temperature B. classmate C. necklace D. fortunate2.south A. courage B. soup C. southern D. trousers 3. smooth A. feather B. tooth C. thief D. warmth 4. official A. concert B. century C. coast D. ocean 5. surprise A. performance B. further C. work D. nurse 第二节:语法和词汇知识(共15小题,每小题1分,共15分)6. The children burst .A. out laughB. into laughC. out laughingD. into laughing 7. Heheard nothing this matter.A. concernedB. conceptionC. concerningD. revealing 8. I didn‘tsee her in the meeting room this morning. She have spoken at the meeting.A. mustn‘tB. shouldn‘tC. needn‘tD.couldn‘t9. I can see, there is only one possible way to keep away from the danger.A. As long asB. As far asC. Just asD. Even if 10.It is for us to remember on such a short time.A. a too long listB. too a long listC. a long too listD. too longa list 11.It was because of bad weather the foot match had to be put off.A. soB. so thatC. whenD. that 12. the people present, hecouldn‘t carry on his speech.A. DisappointedB. To the disappointment ofC. DisappointingD. To disappoint13.The mot her didn‘t know who for the broken glass.A. to blameB. to be blamedC. blamedD. is to be blamed 14. Mybrother‘s pale face suggested that he ill, and my parents suggestedthat he______ a medical examination.A. be; should haveB. was; haveC. should be; hadD. was; has 15.----May I look at the menu for a little while?----Of course, sir.A. don‘t worryB. it doesn‘t matterC. enjoy yourselfD. take your time16. Walk about 100 meters the bridge, and you will find the schoolon the right. A. beyond B. beside C. before D. over 17.New reports say peace talks between the two countries with no agreement reachedA. have broken downB. have broken outC. have broken inD. have broken up 18. In a library books are usually by subjects.学习资料收录大全A. classifiedB. definedC. separatedD. governed 19. About in 1979, he went to high school and began to study law by himself.A. a, theB. /, /C. /, theD. /, a 20. If only he quietly as the doctor instructed, he would not suffer so much now.A. liesB. layC. had lainD. should lie 第三节:完型填空(共20小题,每小题1.5分,满分30分)阅读下列短文,从短文后所给的各题的四个选项(A.B.C.D )中选出最佳选项Scott and his companions were terribly disappointed. When they gotto the South Pole, theyfound the Norwegians(挪威人) had 21 them in the race to be the first ever to reach it. After 22 the British flag at the Pole, they took a photograph of themselves 23 they started the 950– mile journey back.The journey was unexpectedly 24 , and the joy and excitement about the Pole had gone out of them. The sun hardly 25 . The snow storms always made it impossible to sight the stones they had 26 to mark their way home. To make things 27 , Evans, whom they hadall thought of 28 the strongest of the five, fell badly into a deep hole in the ice. Having 29 along for several days, he suddenly fell down and died.The four who were 30 pushed on at the best speed they could 31 . Captain Oates hadbeen suffering for some time from his 32 feet; at night his feet swelled ( 肿胀 ) so large thathe could 33 put his boot on the next morning, and he walked bravely although he was in great 34 . He knew his slowness was making it less likely that 35 could save themselves. He askedthem to leave him behind in his sleeping – bag, but they refused, and helped him 36 a fewmore miles, until it was time to put up the 37 for another night.The following morning, 38 the other three were still in their sleeping –bags, he said, ―Iam just going outside and may be 39 some time .‖ He was never seen again. He had walked out 40 into the snow storm, hoping that his death would help his companions. 21(A(hit B(fought C(won D(beaten 22(A(growing B(putting C(planting D(laying23(A(after B(until C(while D(before24(A(safe B(fast C(short D(slow25(A(rose B(set C(appeared D(disappeared26(A(taken up B(cut up C(set up D(picked up27(A(easier B(better C(bitter D(worse28(A(to B(upon C(as D(in29(A(battled B(struggled C(speeded D(waited30(A(left B(lost C(defeated D(saved31(A(manage B(try C(employ D(find32(A(ached B(frozen C(harden D(harmed33(A(hardly B(never C(seldom D(nearly34(A(pain B(fear C(trouble D(danger35(A(all others B(some others C(others D(the others36(A(away B(with C(off D(on37(A(bed B(tent C(blanket D(sleeping , bag38(A(while B(since C(for D(once39(A(missed B(separated C(passed D(gone学习资料收录大全40(A(patiently B(lonely C(alone D(worriedly第二部分阅读理解(共25题,第一节每小题2分,第二节每小题1分)AEnglish author, Richard Savage, was once living in London in great poverty. In order to earn a little money, he had written the story of his life, but not many copies of the book had been sold in the shop, and Savage was living from hand to mouth. As a result of his lack of food, he becamevery ill, but after a time, because of the skill of the doctor who had looked after him, he got well again. After a week or two, the doctor sent a bill to Savage for his visits, but poor Savage hadn‘t any money and couldn‘t pay it. The doctor waited for another month and sent the bill again. But still no money came. After several weeks he sent it to him again asking for h is money. In the end he came to Savage‘s house and asked him for payment, saying to Savage, ―You know you owe yourlife to me and I expected some gratitude from you.‖―I agree,‖ said Savage, ―that I owe my life to you, and to prove to you that I am notu ngrateful for your work, I will give my life to you.‖ With these words he handed to him two copies entitled, ?The Life of Richard Saves‘.41. The best title for this text should be ________.A. A Life for a LifeB. A Skilled DoctorC. A Poor English WriterD. The Life of Richard Savage42. By saying ―Savage was living from hand to mouth‖, the author means that ________.A. Savage was very poor and illB. Savage was too poor to live onC. Savage had clothes to wear and food to eatD. Savage had no money to buy clothes and food43. The writer wrote this story just to tell us ________.A. a funny storyB. an unhappy storyC. a miserable storyD. an ungrateful story44. What ending can we infer from the story?A. Savage paid the bill with his life.B. Savage didn‘t owe the doctor his life.C. The doctor was grateful for the books.D. The doctor had to accept the two books.BA fireman rescued a 76-year-old woman from a burning apartmentbuilding by tossing (抛)her off a fire escape. The woman, Mrs. Mary Togers, landed safely in the arms of another fireman waiting on the roof of the next building.Fireman Edward Lane had risked his life to reach the woman, who was trapped in her top-floor apartment in the five-storey building. It was full of thick smoke. The fire was burning out of control as Lane carried the almost unconscious woman onto a fire escape. As flames threatened them, Lane tried without success to first climb to the roof and then to go below the height of the flames. He could not do so while carrying the dead weight of the helpless 113-pound woman.Then he saw another fireman, Mike Mays, motioning (做手势) to him with his arms held out.Mays was on the roof of the next building, which was level with the fire escape. A space of only four feet separated the two buildings, but there was a fifty-foot drop to the ground. Lane tried to pass Mrs. Rogers over to Mays, but it was too wide and Mays could not reach the old lady.学习资料收录大全Then the two men decided that the only chance to save the woman was to try something very dangerous. Lane reached back and threw Mrs. Rogers across the space between the buildings. For a moment the woman was not supported above the ground. Then Mays caught her by the neck and shoulders, and pulled her safely onto the roof. Mrs. Rogers was uninjured. Lane then climbed to the roof of the burning building, went down another fire escape and back to fight the fire. 45. Lane could not pass over Mrs. Rogers to Mays because ________.A. she was too heavyB. the flames from the fire were too closeC. Mrs. Rogers kept slipping from his handsD. the space between the buildings was too wide46. Lane decided to toss Mrs. Rogers over to the next building because ________.A. he was strongB. Mrs. Rogers told him toC. he was getting breathlessD. it was the only way to send her to safety47. One thing Edward Lane did NOT do was ________.A. risking his own lifeB. carrying Mrs. Rogers down the fire escapeC. taking a chance with the life of Mrs. RogersD. getting Mike Mays to help rescue Mrs. Rogers48. After Edward Lane saved Mrs. Rogers, he ________.A. stopped to restB. helped carry her downstairsC. went back to fight the fireD. helped her come back to lifeCIn New York, there is now a school with no daily lessons, no class teachers, not even a school building. The ―school‖ is a project. It is called ―City-as-School‖ and the name means just that, the city itself is the place where 350 students, between the age of 15 and 18, learn their lessons.Students choose areas of work which interest them and then they help to do that work. For instance, one girl spends her week in the offices of a Congresswoman — an elected official —helping the public with problems such as pensions(养老金),housing, etc. Then she goes to helpin a theatre for a day and she spends one day a week taking first-year courses at college.City-as-School is 30 years old. The education system in New York accepts it now, as an alternative to final years at school. But can it replace ordinary lessons? Well, students have to pass maths and science exams before they enter the ―school‖. These subjects are not easy to provide for in ― City-as-School‖. Teachers monitor the progress of the students. 80 to 85% of the students goto college (or university, as it is called in Britain) after their time at ―City-as-School‖. The successrate is high. And the students are enthusiastic about their―school‖. They like the responsibility of their work, and the sense of purpose it gives them.When the students leave ―City-as-School‖, they don‘t have a normal academic education; butthey do know a lot about different kinds of work in the city!49. ―City-as-School‖ is special because it is ________.A. in New York, one of the most famous cities in the worldB. not a common school we usually see and knowC. a school having special students学习资料收录大全D. free of charge for the citizens50. Which of the following is NOT included in the permission of entering the ―school‖?A. Between 15 and 18 years old.B. Passing maths and science exams.C. Finishing first-year courses at college.D. Students being in their final years at school.51. What do the teachers at ―City-as-School‖ do?A. To monitor the progress of the students.B. To give lessons to the students.C. To evaluate the students‘ performance.D. To protect the students.52. What is the author‘s attitude to ―City-as-School‖?A. Agreeable.B. Indifferent.C. Neutral (中立的).D. Disagreeable.DBill Clinton was born on Aug.19, 1946. Three months before his birth, his father had died when driving home to his pregnant wife, Virginia, he went off a high way, was thrown from the car and drowned in a river.When Bill was 4, his mother remarried Roger Clinton. And there were always troubles: a sometimes violent , alcoholic stepfather and a half-brother. Only one year after the marriage, the drunken stepfather fireda shotgun at the ceiling to keep his bride and stepson from leaving thehouse. Virginia was very much frightened. So Roger Clinton beat Virginia from time to time. But teenager Clinton played a role of protector of his mother bravely. The stepfather never laid another band on Virginia.In high school, he was very good at Latin and maths. He also played saxophone in the hand. At age 16, as a member of a youth group, Clinton met President John F. Kennedy at the White House, it led him to the life of public service. Once he set his mind to do something, he did not give up. He was elected governor of Arkansas at the age of 32.Clinton has said he ran for president to make the country a better place for people like Chelsea, his daughter. He did win. At the age of 46, he became the third youngest president in the nation's history.53.When this passage was published, Clinton was_________.A. governor of ArkansasB. a famous professorC. President of U.S.A.D. President of a university54.Clinton's own father died_________.A. before Clinton was bornB. after Clinton was bornC. from drinking too much brandyD. when Clinton's mother was giving birth55.Clinton protected his mother by_________.A. fighting against his stepfatherB. beating his stepfatherC. having long talks with his stepfatherD. the means we don't know56.The word ―it‖ in ―It led him to the life of public service‖ refers to_________.学习资料收录大全A. Clinton's high school educationB. becoming a member of a youth groupC. Clinton's visiting President KennedyD. doing public serviceEReflecting is a special kind of thinking. In the first place, it's a both active and controlled. When ideas pass aimlessly through your head, that is not reflecting. When someone tells you a story and it suddenly brings to mind something that happened to you, that is not reflecting either. Reflecting means focusing your attention. It means weighing, considering, and choosing. Suppose you're going home, and when you get there, you turn the knob, the door opens and you step in. Getting into your home does not require reflection. But now suppose that when youturn the knob, the door does not open. To get into the house, some reflecting is in order. You have to think about what you are going to do. You have to imagine possibilities and consider choices.The second way that reflecting is different from some other kinds of thinking is that it's persistent. It requires continuous effort. Suppose you're still trying to get through your front door. You check yourpocket for the key. You walk around the house looking for an open window. You go to a phone to call a family member who has a key. Such behavioris proof of persistent reflective thinking. And if someone asks you what you are doing, you may say that you are trying to figure out how to get into your house. But suppose, instead, you go to a nearby record store and look through the new records. If someone asks you what you are doing and you say that you are trying to figure out how to get into your house, that will not make sense. You are only reflecting as long as you stickto the problem or task.The third way that reflecting is different from some other kinds of thinking is that it's careful. It aims at making sense. This doesn't mean that reflecting cannot be imaginative. A great deal of reflection could go into writing a science-fiction story about people who can move through solid objects. The ability to walk through walls could make sense in a science-fiction story. But it wouldn't make much sense in trying to get through your locked front door. Such imagining would be a kind of thinking, but it would not be reflection.57. According to the passage, reflecting is _______.A. not a kind of thinking but an actionB. giving all your attention to figuring out somethingC. bringing to mind something that happened to youD. letting ideas pass quickly through your head58. If you don't take your key with you from work, which of the following does not showreflecting?A. You go through the open window.B. You telephone your family member.C. You go to the nearby record store.D. You ask the policeman for help. 59. The underlined word "persistent" probably means _____.A. existingB. continuingC. tryingD. imagining60. Which of the following is the difference between thinking and reflecting?A. Reflecting takes more time than thinking.B. Reflecting is more important than thinking.C. Reflecting is a kind of more imaginative thinking.D. Reflecting is a kind of deeper thinking.学习资料收录大全第二节补全对话根据对话内容从对话后的选项中选出最能填入空白处的最佳选项-------What do you think I ought to see first in London? I‘m told one ought to see the BritishMuseum. Do you think I shall have time for that? ------- 61 But if I were you, I should leave that for some other day. You could spend a whole day there. It‘s much too big to be seen in an hour or so.------I suppose it is. 62------That‘s not a bad idea. You could spend a coup le of hours there comfortably, or even a wholeafternoon, watching the wild animals and all those birds. You could have tea there too.------I‘ll do that, there. How do I get there?------ 63 Where are we now? Oh, there‘s that building. I think your best way from hereis to take Baker Street.------ 64------Oh, no, a quarter of an hour or so, but, if you‘re in a hurry, why not take a taxi?------I think I will. 65 Taxi!A. Let me see.B. Well, you might.C. What time is it now?D. Is it much of a walk?E. Ah, here‘s one coming.F. What about going to the zoo?G. Must I stay in London for long?第?卷(非选择题共55分)第三部分:写作(共三节满分55分)第一节:单词拼写(共10小题,每小题1分,满分10分)66. Write your name and address in the b spaces at the top of the page. 67. He wants to help you but he is u to .68. She r her mother in the way she moves her hands when she talks.69. Do you remember the old saying, ―No pains ,no g ‖?70. Can you explain it again ? I didn‘t make s of it.71. The room is twenty in .(长度)72. What is the government (政策)on education?73. The prisoners became (绝望的)in their attempts to escape. 74. Extra lesson on Sundays put more (压力)on us students who are heavily burdened75.What are the (消费)of moving house?第二节:短文改错(共10小题,每小题1.5分,满分15分)此题要求改正所给短文中的错误。

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东北三校2008年部分毕业班第三次摸底考试数 学第I 卷(选择题 共60分)一、选择题:本大题共12个小题,每小题5分,共60分,在每个小题的四个选项中,只有一个是符合题目要求的. 1.已知集合=<+-=≥=N M x x x N x x M 则},013|{},1|||{ ( )A .}31|{<≤x xB .}11|{≤<-x xC .}3|{>x xD .}1|{-<x x 2.函数)1(21≥=-x y x 的反函数是( )A .)1)(1(log 21≥+=x x yB .)1(1log 21≥+=x x yC .)10)(1(log 21≤<+=x x yD .)10(1log 21≤<+=x x y3.由数字0,1,2,3,5组成的没有重复数字的三位奇数的个数为 ( )A .60B .48C .36D .274.某商店两个进价不同的商品都卖了80元,其中一个盈利60%,另一个亏本20%,问在这次买卖中,这家商店 ( ) A .不赚不赔 B .赚了10元 C .赔了10元 D .赚了50元 5.过点A (2,-1),圆心在直线x y 2-=上,且与直线1=+y x 相切的圆的方程为( ) A .2)56()53(22=++-y x B .2)56()53(22=-++y xC .2)2()1(22=-++y xD .2)2()1(22=++-y x6.在半径为35的球面上有A 、B 、C 三点,AB=6,BC=8,CA=10,则球心到平面ABC 的距离为( )A .5B .24C .25D .107.函数)1(2log 2>-=a x x y a 在其定义域内( )A .有最大值,无最小值B .有最小值,无最大值C .有最大值,也有最小值D .既无最大值,也无最小值 8.对于直线l 、m 和平面α、β,α⊥β的一个充分条件是 ( ) A .βα⊥⊥m l m l ,//, B .αβα⊥=⊥m l m l ,,C .αβ⊂⊥l m m l ,,//D .βα⊥⊥m l m l ,,//9.已知函数),(,)1(4)1()1(12)(2+∞-∞⎩⎨⎧->+--≤+-=在x a x a x ax ax x f 内是减函数,则实数a 的取值范围是( )A .)1,(-∞B .)0,(-∞C .),1(+∞D .(0,1)10.将周期为π的函数)0(c o s c o s s i n 2s i n 22>-+=ωωωωωx x x x y 的图象按)1,8(π-=平移后,所得函数图象的解析式为( )A .12sin 2+=x yB .1)44sin(2-+=πx yC .1)82sin(2+-=πx y D .x y 2cos 21-=11.设x 、y 满足条件41,033042022--=⎪⎩⎪⎨⎧≤--≥+-≥-+x y u y x y x y x 则的取值范围是 ( )A .],31[)1,(+∞--∞ B .]31,41[-C .]31,1[-D .]43,51[12.如右图,已知在梯形ABCD 中,|AB|=2|CD|,点E 分有向线段所成的比为52,双曲线过C 、D 、E 三点,且以AB 为 焦点,则此双曲线的离心率为 ( ) A .2 B .3C .2D .3选择题答题栏第Ⅱ卷(非选择题 共90分)二、填空题:本大题共4个小题,每小题4分,共16分. 13.10)212(xx x -的展开式中,常数项为 .(用数字作答)14.已知=-==|32|,3,1||,2||b a b a b a 则的夹角为与π.15.已知数列=∈++==*+n n n n a N n n a a a a 则且满足),(12,1}{11 . 16.在△ABC 中,内角A 、B 、C 所对的边分别为a 、b 、c ,给出下列结论:①若A >B >C ,则C B A sin sin sin >>; ②若C B A c b a cos cos cos ,>>>>则;③必存在A 、B 、C ,使C B A C B A tan tan tan tan tan tan ++<成立; ④若︒===25,20,40B b a ,则△ABC 必有两解.其中,真命题的编号为 .(写出所有真命题的编号) 三、解答题:本大题共6个小题,共74分. 解答应写出文字说明,证明过程或演算步骤. 17.(本小题满分12分)已知)sin(,44,434,1411)43sin(,71)4cos(βαπβππαπβπαπ+<<-<<=+=-求的值.甲、乙两位同学站在罚球线进行投篮,已知每次投篮,甲投中的概率为21,乙投中的概率为32,若甲、乙两位同学各投篮4次,求: (1)甲恰好投中3次的概率; (2)乙至少投中1次的概率;(3)乙恰好比甲多投中2次的概率. 19.(本小题满分12分)已知在等比数列}{n a 中,n S 是其前n 项的和,若12,,++m m m S S S 成等差数列,则m a ,12,++m m a a 成等差数列.(1)写出这个命题的逆命题:(2)判断逆命题是否为真,并给出证明.如图,在直三棱住ABC —A 1B 1C 1中,AC=AB=4,241=BB ,2=BM ,︒=∠90BAC , 点N 在CA 1上,且131NA CN =. (1)求证:MN//平面A 1B 1C 1; (2)求点A 1到平面AMC 的距离; (3)求二面角C —A 1M —B 的大小. 21.(本小题满分12分)已知定义在R 上的函数)(x f 满足:)()()(y f x f y x f +=+,当.0)(,0<<x f x 时 (1)求证:)(x f 是奇函数;(2)解关于x 的不等式:m m m f x m f x f mx f 且,0).((2)()(2)(22>->-为常数).已知椭圆的中心在原点O ,短轴长为22,其焦点F (c ,0)(c >0)对应的准线l 与x轴交于A 点,|OF|=2|FA|,过A 的直线与椭圆交于P 、Q 两点. (1)求椭圆的方程;(2)若0=⋅,求直线PQ 的方程;(3)设)1(>=λλ,过点P 且平行于准线l 的直线与椭圆相交于另一点M. 求证F 、M 、Q 三点共线.数学(理)参考答案一、选择题:每小题5分,共60分. 1—12ADDBD CBCDA CB二、填空题:每小题4分,共16分. 13.840 14.13 15.n 2 16.①④ 三、解答题:本大题共6个小题,共74分.解答应写出文字说明,证明过程或演算步骤. 17.解:042434<-<-∴<<απππαπ…………1分又734)71(1)4(cos 1)4sin(71)4cos(22-=--=---=-∴=-απαπαπ ………4分πβπππβπ<+<∴<<-43244又1435)1411(1)43(sin 1)43cos(,1411)43sin(22-=--=+--=+∴=+βπβπαπ (7)分+-+=--+=++∴)4cos()43cos()]4()43cos[()](2cos[απβπαπβπβαπ23)4sin()43sin(-=-+απβπ ………………11分 23)](2cos[)sin(=++-=+∴βαπβα……………………12分 18.解:(1)甲恰好投中3次的概率为4121)21(334=⨯⨯C ;……………………3分 (2)∵乙投篮4次都未投中的概率为181)31(4=, ∴乙至少投中1次的概率为8180)31(14=-;………………6分(3)设乙恰好比甲多投中2次为事件A ,乙恰好投中2次而甲恰好投中0次为事件B 1,乙恰好投中3次而甲恰好投中1次为事件B 2,乙恰好投中4次而甲恰好投中2次为事件B 3,则A=B 1+B 2+B 3 .……………………………………8分∴乙恰好比甲多投中2次的概率 P (A )=P (B 1)+P (B 2)+P (B 3) (9)分2244443143344042224)21()32()21(2131)32()21()31()32(⋅⋅⋅+⋅⋅⋅⋅⋅+⋅⋅⋅⋅=C C C C C C.16231)21(2=⋅………………………………12分 19.解:(1)逆命题:在等比数列}{n a 中,n S 是其前n 项的和,若12,,++m m m a a a 成等差数列,则12,,++m m m S S S 成等差数列. ………………………………2分 (2)当q=1时,逆命题为假;当q ≠1时,逆命题为真.证明:设}{n a 的公比为q ,若12,,++m m m a a a 成等差数列,则有122+++=m m m a a a …3分012,0,0,22111111=--≠≠+=∴-+q q q a q a q a q a m m m 解得,211-==q q 或…5分当q=1时,11111)1(,)2(,a m S a m S ma S m m m +=+==++1212,,2++++∴+≠∴m m m m m m S S S S S S 不成等差数列.………………7分当])21(1[34211])21(1[22,2121212+++--=+--=-=m m m a a S q 时 …………8分211])21()2(1)21(41[211])21(1[211])21(1[2211111+-⋅--+-⋅-=+--++--=+++++m m m m m m a a a S S ])21(1[34211])21(22[2121++--=+-⋅-=m m a a1212,,,2++++∴+=∴m m m m m m S S S S S S 成等差数列. ………………11分综上得,当q=1时,逆命题为假;当q ≠1时,逆命题为真. …………12分 20.法一:(1)证明:在C 1A 1上取点D ,使11141A C D C =,连结ND 、B 1D , 11111////,4131BB A C ND CA NA CN ∴==………………1分又M B BB CC ND 111234343====,∴四边形B 1MND 为平行四边形, ∴MN//B 1D …………………………………………………………2分又⊄MN 平面A 1B 1C 1,B 1D ⊂平面A 1B 1C 1 ∴MN//平面A 1B 1C 1 …………4分 (2)∵三棱柱ABC —A 1B 1C 1为直三棱柱 AC A A ⊥∴1又︒=∠90BAC , ⊥∴AC 平面A 1ABB 1在平面A 1ABB 1中过A 1作A 1H ⊥AM ,又AC ⊥A 1H ,∴A 1H 为点A 1到平面AMC 的距离 ……………………………………………………6分在23,24,4,22111=+====∆MB AB AM BB A A AB MA A 中,316,2821211111==⋅=⋅=∴∆H A AB A A AM H A S M AA , 即A 1到平面AMC 316||1=m ……………………8分 (3)在平面A 1ABB 1中过A 作AE ⊥A 1M ,交A 1M 于点E ,连结CE ,则AE 为CE 在平面A 1ABB 1内的射影,ACE M A CE ∠∴⊥∴,1为二面角C —A 1M —B 的平面角 ……………………10分在MA A 1∆中,28,341212111==+=∴∆M AA S MB B A M A 2161=⋅∴AE M A417tan ,171716==∠∴=∴AE AC AEC AE ………………11分 ,417arctan=∠∴AEC 即二面角C —A 1M —B 的大小为417arctan…………12分 法二:(1)∵三棱柱ABC —A 1B 1C 1为直三棱柱,A 1A ⊥AC ,A 1A ⊥AB ,又∠BAC=90°∴可以点A 为坐标原点,以AC 、AB 、AA 1所在直线分别为x 轴、y 轴、z 轴建立空间直角坐标系,由已知得A (0,0,0)、B 1(0,4,24)、C 1(4,0,24)、 C (4,0,0)、A 1(0,0,24)、M (0,4,2)、N (3,0,2),在C 1A 1上取点D ,使)24,0,3(,41111D A C D C 则=B B 11)0,4,3(),0,4,3(=∴-=-=∴ ………………2分⊄∴MN DB MN 又1//平面A 1B 1C 1,B 1D ⊂平面A 1B 1C 1 ∴MN//平面A 1B 1C 1…4分(2))2,4,0(=、)2,4,4(-=,设平面AMC 的法向量为)1,,(b a =, 则00=⋅=⋅CM m AM m ⎪⎩⎪⎨⎧=++-=+∴0244024b a b ,解得42,0-==b a ……6分)24,0,0(),1,42,0(1=-=∴AA 又 ∴A 1到平面AMC 的距离为316………………8分 (3))24,0,4(1-=CA 、)2,4,4(-=CM ,设平面A 1MC 的法向量为)1,,(d c =,则0,01=⋅=⋅CA)1,423,2(.423,202440244=∴==⎪⎩⎪⎨⎧=++-=+-∴d c d c c 解得又⊥平面A 1ABB 1,∴所求得二面角的大小为><,,………………11分 而),0,0,4(= 33334446624||||,cos =⨯=⋅>=<∴n AC ∴二面角C —A 1M —B 的大小为.33334arccos………………12分 21.解:(1).0)0(),0()0()0(,0),()()(=+===+=+f f f f y x y f x f y x f 即得令令)()(),()()0(,,0x f x f x f x f f x y y x -=-∴-+=-==+得即 )(x f ∴是奇函数…3分(2)设x 1、x 2∈R ,且,0,2121<-<x x x x 则由已知得0)(21<-x x f .0)()()()()(212121<-=-+=-∴x x f x f x f x f x f)()()(21x f x f x f 即<∴在R 上是增函数. ……………………6分又)2()(2).2()()()(2x f x f m f m f m f m f ==+=同理 ………………7分 )2()()2()()(2)()(2)(2222x f x m f m f mx f m f x m f x f mx f +>+⇔->- 02)2(22)2()2(222222>++-⇔+>+⇔+>+⇔m x m mx x x m m mx x x m f m mx f 0))(2(02)2(,02>--∴>++-∴>m x m x x mm x m …………9分 当2,2><m m m即时,不等式的解集为}2|{m x m x x ><或; 当20,2<<>m m m 即时,不等式的解集为}.2|{m x m x x ><或 …………12分 22.解:(1)由题意,设椭圆的方程为:)2(12222>=+a y ax 由已知得⎪⎩⎪⎨⎧-==-)(22222c c a c c a ,解得∴==,2,6c a 椭圆的方程为:.12622=+y x …………3分 (2)由(1)可得,准线l 的方程为:)0,3(.3A x ∴=设直线PQ 的方程为:)3(-=x k y 由方程组062718)13()3(126222222=-+-+⎪⎩⎪⎨⎧-==+k x k x k x k y y x 解得…………5分 依题意),(),,(.3636,0)32(1221112y x Q y x P k k 设得<<->-=∆,则 13182221+=+k k x x ① 136272221+-=k k x x ② ………………7分 由直线PQ 的方程得)3(),3(2211-=-=x k y x k y ,]9)(3[)3)(3(2121221221++-=--=∴x x x x k x x k y y ③ 又0,02121=+∴=⋅y y x x ④ ………………9分由①②③④得)36,36(55,152-∈±=∴=k k ,直线PQ 的方程为 035035=-+=--y x y x 或 ………………10分(3)证明:).,3(),,3(2211y x y x -=-= 由题意可得方程组.215,1126126)3(32222221212121λλλλλ-=∴>⎪⎪⎪⎩⎪⎪⎪⎨⎧=+=+=-=-x y x y x y y x x 又 ………………12分 ),21(),1)3((),2(),,(),0,2(1121111y y x y x FM y x M F --=-+-=--=∴-λλ ),21(2y λλλ--= 而FQ FM y y x FQ λλλ-=∴-=-=),,21(),2(222 //∴,又∵直线FM 、直线FQ 有公区点F 故F 、M 、Q 三点共线. ……………………14分。

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