福建省龙海二中2014届高三第二学期期初高考模拟考试试卷数学(文) Word版含答案

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福建省龙海市第二中学2014届高考历史第二学期期初模拟考试试卷

福建省龙海市第二中学2014届高考历史第二学期期初模拟考试试卷

2014届龙海二中高三第二学期期初高考模拟考试卷(考试时间:90分钟总分:100分)一、选择题(每题2分,29题共58分)1.《左传》记载,晋国赵鞅率军与郑国交战于铁,临战发表誓词:“克敌者,上大夫受县,下大夫受郡,士田十万。

”这反映出当时晋国A.分封制度受到冲击B.郡县制度全面推行C.宗法等级秩序变动D.国家政权趋向松散2.周代文献中中国共有五种不同含义,《史记·齐大公世家》载:“是时周室微,唯齐楚秦晋为强。

晋初与会,献公死,国内乱。

秦穆公辟远,不与中国会盟。

楚成王初收荆蛮有之,夷狄自置。

唯独齐为中国会盟,而齐桓公能宣其德,故诸侯宾会。

”材料“中国”指的是?A.内陆地区B.周天子的直辖区C.少数民族地区D.华夏族的诸侯国3.宋朝形成了“中书主民,枢密院主兵,三司主财,各不相知”的局面。

这反映出A.宰相职权范围扩大B.专制皇权达到顶峰C.君权对相权的制约D.中央对地方控制加强4.《古代行政区划史》曰:元朝大政委于中书省,今冀、晋、鲁、内蒙及河南的黄河以北地区称“腹里”,由中书省直辖。

中书省也称都省,为全国行政中枢。

总领各行省,又兼辖腹里。

以下观点符合上述引文的是A.元朝时黄河以北地区,归中央政府直辖B.元朝中书省集行政、监察、军事大权于一身C.元朝中书省掌管全国最高行政事务D.元朝中书省职权与唐朝中书省大体相当5.科举制是我国古代的选官制度。

下列各项中,确切反映唐朝科举制情况的是A.其形式都是以官举士B.采用八股考试方式C.分进士等科目考试D.要求应试士人熟读四书五经6.洪秀全在定都天京后颁行的《天父诗》中写道:“只有媳错无爷错,只有婶错无哥错,只有人错无天错,只有臣错无主错。

”可见,太平天国政权具有A.革命性B.进步性C.封建性D.空想性7.恩格斯说:“没有哪一次极大的历史灾难不是以历史的进步为补偿的”。

抗日战争对中国历史进程最重要的补偿是A.加速了近代民主革命的进程B.实现了中华民族的独立C.贏得了世界各国的普遍尊重D.壮大了社会主义的力量8.古代雅典是世界民主政治的源头,但也存在明显的缺陷。

【数学】福建省龙海市第二中学高三下学期第二次模拟考试(理)

【数学】福建省龙海市第二中学高三下学期第二次模拟考试(理)

福建省龙海市第二中学 高三下学期第二次模拟考试(满分150分, 考试时间120分钟)本试卷分为第I 卷(选择题)和第II 卷(非选择题)两部分,满分150分,考试时间120分钟.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,只有一项是符合题意.)1.复数z 满足1)43(=-⋅i z (i 是虚数单位),则|z|= ( )A .251B .255C .51D .55 2. 已知函数122+--=x x y 的定义域为集合,集合{}Z n n x x B ∈-==,12,则为( )A.{}3,1 B. {}3,1,1,3--C. {}3,1,1-D.{}1,1,3--3. 《九章算术》是我国古代的数学名著,书中有如下问题:“今有五人分五钱,令上二人所得与下三人等。

问各得几何。

”其意思为“已知甲、乙、丙、丁、戊五人分5钱,甲、乙两人所得与丙、丁、戊三人所得相同,且甲、乙、丙、丁、戊所得依次成等差数列。

问五人各得多少钱?”(“钱”是古代的一种重量单位).这个问题中,甲所得为_ __钱。

A.54 B. 43 C.32 D.534.已知非零向量b a ,的夹角为30,且,3,1==b a 则=-b a 2( )A.32-B.1C.2D.2 5. 已知点x ,y 满足约束条件⎩⎪⎨⎪⎧x +y -2≥0x -2y +4≥0x -2≤0,则z =3x +y 的最大值与最小值之差为( )A .5B .6C .7D .86.执行如图所示的程序框图,若输入7,6x y ==,则输出的有序数对为( ) A .(11,12)B .(12,13)C .(13,14)D .(13,12)A AB ⋂7.函数x e x y -=22在]2,2[-的图像大致为( )A .B .C .D .8.如图为某几何体的三视图,则该几何体的表面积为( ) A .10+ B .10+ C .6+2+D .6++9.过抛物线)0(22>=p px y 的焦点F 作斜率为3的直线,与抛物线在第一象限内交于点A ,若4=AF ,则=p ( )A.4B.2C.1D.310.已知)),0(,0,0(),cos()(πϕωϕω∈>>-=A x A x f ,)(x f 的导函数...)(x f '的部分图象如图所示,则下列对)(x f 的说法正确的是( )A.最大值为2且关于点)0,2(π-中心对称B.最小值为2-且在]23,2[ππ上单调递减 C.最大值为4且关于直线2π-=x 对称D.最小值为4-且在]23,0[π上的值域为]4,0[11.已知双曲线()2222:10,0x y C a b a b-=>>的右顶点为A , 以A 为圆心的圆与双曲线C 的某一条渐近线交于两点,P Q .若60PAQ ∠=,且3OQ OP =(其中O 为原点),则双曲线C 的离心率为( )A .2B .7CD .12.已知R λ∈,函数1,0,()lg ,0,x x f x x x ⎧+<=⎨>⎩2()414g x x x λ=-++,若关于x 的方程(())f g x λ=有6个解,则λ的取值范围为 ( )A .2(0,)3B .12(,)23C .21(,)52D .2(0,)5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.已知随机变量2(1,2)N ξ,且(13)0.7P ξ-≤≤=,则(1)P ξ≤-=__________.14. 在231()nx x -的展开式中含有常数项,则正整数n 的最小值是 . 15.在四面体ABCD 中,6,4,5AB CD AC BD AD BC ======,则四面体ABCD 的外接球的表面积等于 .16. 设函数()21f x x =,()222()f x x x =-, 记1021|()()||()()|k k k k k S f a f a f a f a =-+-+9998|()()|k k f a f a +-,其中99i ia =,(0,1,2,,99i =),1,2k =,则=-219S S ____.三、解答题(本大题共8小题,共70分.解答应写出文字说明、证明过程或演算步骤.) 17.(本小题满分12 分)在ABC △中,角C ,B ,A 的对边分别为c ,b ,a ,且,,a b c 成等比数列,5sin 13B =. (Ⅰ)求11tan tan A C+的值; (Ⅱ)若12BA BC ⋅=,求a c +的值.18. (本小题满分12 分)如图,四棱锥P -ABCD 的底面是直角梯形,AD BC ,90ADC ∠=︒,2AD BC = ,PA ⊥平面ABCD .(Ⅰ)设E 为线段PA 的中点,求证:BE //平面PCD ; (Ⅱ)若PA AD DC ==,求平面PAB 与平面PCD 所成二面角的余弦值.19.(本题满分12分)为了增强高考与高中学习的关联度,考生总成绩由统一高考的语文、数学、外语3个科目成绩和高中学业水平考试3个科目成绩组成.保持统一高考的语文、数学、外语科目不变,分值不变,不分文理科,外语科目提供两次考试机会.计入总成绩的高中学业水平考试科目,由考生根据报考高校要求和自身特长,在思想政治、历史、地理、物理、化学、生物、信息技术七科目中自主选择三科.(1)某高校某专业要求选考科目物理,考生若要报考该校该专业,则有多少种选考科目的选择;(2)甲、乙、丙三名同学都选择了物理、化学、历史组合,各学科成绩达到二级的概率都是0.8,且三人约定如果达到二级不参加第二次考试,达不到二级参加第二次考试,如果设甲、乙、丙参加第二次考试的总次数为X ,求X 的分布列和数学期望.20.(本题满分12分)设P 为椭圆22221x y a b+=()0a b >>上任一点,F 1,F 2为椭圆的焦点,|PF 1|+|PF 2|=4,离心率为23. (Ⅰ)求椭圆的标准方程;(Ⅱ)直线l :()0y kx m m =+≠与椭圆交于P 、Q 两点,试问参数k 和m 满足什么条件时,直线OP ,PQ ,OQ 的斜率依次成等比数列;(III)求OPQ ∆面积的取值范围.21.(本题满分12分)已知函数ax a x x f ln )()(-=,21()()1g x x a x a=-++(R ∈a ,1a >). (Ⅰ)若函数)(x f 在a x =处的切线l 斜率为2,求l 的方程;(Ⅱ)是否存在实数a ,使得当1(,)x a a∈时, ()()f x g x >恒成立.若存在,求a 的值;若不存在,说明理由.请考生在第(22),(23)二题中任选一题作答,如果多做,则按所做的第一题记分.作答时用2B 铅笔在答题卡上把所选题目对应的题号涂黑. 22. (本小题满分10分)选修4-4:坐标系与参数方程 在直角坐标系xOy 中,曲线C 的参数方程为2cos 4sin x y θθ=⎧⎨=⎩(θ为参数),直线l 的参数方程为1cos 2sin x t y t αα=+⎧⎨=+⎩(t 为参数).(1)求C 和l 的直角坐标方程;(2)若曲线C 截直线l 所得线段的中点坐标为(1,2),求l 的斜率.23.(本小题满分10分)选修4-5:不等式选讲 已知函数()2+=x x f .(Ⅰ)解不等式()14+->x x f ;(Ⅱ)已知()0,02>>=+b a b a ,求证:()ba x f x 1425+≤--.参考答案一、选择题1—5:CBBBC 6—10:ADCBD 11—12:AD二、13. 0.15 14 . 5 15.2774.2S R ππ== 16. 121911S S -= 解析:16. 解析:当1k =时,()21f x x =在[0,)+∞单调递增,所以1011199()()()f a f a f a <<<所以1021|()()||()()|k k k k k S f a f a f a f a =-+-+9998|()()|k k f a f a +-11101211199198()()()()()()f a f a f a f a f a f a =-+-++-19910()()f a f a =-11(1)(0)1f f =-=当2k =时,()222()f x x x =-在1[0,]2单调递增,在1[,)2+∞单调递减 所以2021249250251299()()()()()()f a f a f a f a f a f a <<<=>>>所以1021|()()||()()|k k k k k S f a f a f a f a =-+-+9998|()()|k k f a f a +- 21202221249248()()()()()()f a f a f a f a f a f a =-+-++-++250251251252298299()()()()()()f a f a f a f a f a f a -+-++-24920250299()()()()f a f a f a f a =-+- 2492()f a =所以1224912()S S f a -=-222249509998100114999999⨯-⨯=-⨯==所以111=17.(本小题满分12分)解:(Ⅰ)因为,,a b c 成等比数列,所以2b ac =----------------------------1分 由正弦定理可得2sin sin sin A C B =-----------------------------2分 所以11cos cos tan tan sin sin A CA C A C+=+-----------------------------3分 sin cos cos sin sin sin C A C AA C+=sin()sin sin A C A C+=--------------------------------4分sin sin sin BA C=--------------------------------5分113sin 5B ==--------------------------------6分 (Ⅱ)由12BA BC ⋅=得cos 12ac B =知cos 0B >----------------------7分 由5sin 13B =得12cos 13B =---------------------------------------8分 所以21213cos b ac B===---------------------------------------9分 由余弦定理得2222cos b a c ac B =+-得22()22cos b a c ac ac B =+----------------------------------10分 即21213()213(1)13a c =+-⨯⨯+--------------------------------11分 解得37a c +=----------------------------------------------12分 18. (本题满分12分) (Ⅰ)证明:设线段的中点为, 连接,. 在△中,为中位线,故. 又平面,平面,所以平面.在底面直角梯形中, ,且,故四边形为平行四边形, 即.又平面,平面,所以平面.又因为平面,平面,且,所以平面平面.又平面,所以有平面. ……………6分(Ⅱ)如右图所示,以为坐标原点,的方向为轴正方向,建立空间直角坐标系. 设,则,,,.,,,,设是平面的法向量,则,,可取,同理,设是平面的法向量,则,可取,从而. …12分19. (本题满分12分)(1)考生要报考该校该专业,除选择物理外,还需从其他六门学科中任选两科,故共有2615C =种不同选择.(2)因为甲乙丙三名同学每一学科达到二级的概率都相同且相互独立,所以参加第二次考试的总次数X 服从二项分布()9,0.2B ,所以分布列为所以X 的数序期望()90.2 1.8E X =⨯=.20.(本题满分12分)(Ⅰ)2a =4,a =2,c =ae =3,b =1,所以椭圆方程:1422=+y x ......3分 (Ⅱ)设点()11,y x P ,()22,y x Q ,则由⎪⎩⎪⎨⎧=++=1422y x m kx y ,消y ,得()044814222=-+++m kmx x k ,因为直线与椭圆交于不同的两点,所以0)14)(1(16642222>+--=∆k m m k ,.....5分解得2214m k >+,由韦达定理得,148221+-=+k kmx x ,14442221+-=k m x x ....6分 由题意知,OQ OP k k k •=2,即212212122122121221212)()(x x m x x x x km k x x m x x km x x k x x y y k +++=+++==, 所以0)(2122121=++x x m x x x x km ,即412=k ,所以202<<m ................9分(III)设点O 到直线PQ 的距离为d ,则21km d +=,()()221221y y x x PQ -+-==21k +241k +∆=22224k14114+-++m k k ,.....10分所以2221m m d PQ S OPQ -==∆,则()2222m m S OPQ -=∆,...11分 所以()(]1,02222∈-=∆m m SOPQ,所以OPQ ∆面积的取值范围是(]1,0....................12分21.(本小题满分12分) 解:(Ⅰ)因为()ln()1af x ax x'=-+,()2f a '=,……………………………2分 所以2ln 2a =,解得a e =或a e =-(舍去). ………………………………………3分 因为()()ln f x x e ex =-,所以()0f e =,切点为(),0e , 所以l 的方程为22y x e =-.………………………5分(Ⅱ)由()()f x g x >得,21()ln ()1x a ax x a x a ->-++,1()ln ()()x a ax x a x a->--,又1(,)x a a∈,所以1ln ax x a<-,1ln 0ax x a -+<.…………………………2分令1()ln h x ax x a =-+(1(,)x a a ∈),则11()1xh x x x-'=-=,所以,当11x a<<时,()0h x '>,()h x 单调递增;当1x a <<时,()0h x '<,()h x 单调递减, 所以当1x =时,函数()h x 取得最大值()11ln 1h a a=+-.…………………………9分 故只需1ln 10a a +-<(*). 令1()ln 1x x x ϕ=+-(1x >),则22111()x x x x xϕ-'=-=,所以当1x >时,()0x ϕ'>,()g x 单调递增,所以()()10x ϕϕ>=.…………11分 故不等式(*)无解.综上述,不存在实数a ,使得当1(,)x a a∈时, ()()f x g x >恒成立. …………12分 请考生在第(22),(23),二题中任选一题作答,如果多做,则按所做的第一题记分.作答时用2B 铅笔在答题卡上把所选题目对应的题号涂黑. 22. (本小题满分10分)选修4-4:坐标系与参数方程解:(1)曲线C 的直角坐标方程为:221416x y += 当cos 0α≠时,l 的直角坐标方程为:tan 2tan y x αα=⋅+-, 当cos 0α=时,l 的直角坐标方程为:1x =…………5分 (2)将l 的参数方程代入C 的直角坐标方程,得22(13cos )4(2cos sin )80t t ααα+++-=①因为曲线C 截直线l 所得线段中点(1,2)在C 内,所以①有两解1t ,2t ,则120t t += 又1224(2cos sin )13cos t t ααα++=+故2cos sin 0αα+= 于是直线l 的斜率tan 2k α==-.…………10分23.(本小题满分10分)选修4-5:不等式选讲解:(Ⅰ)()14+->x x f ,即为412>+++x x , 该不等式等价于如下不等式组:1)274122-<⇒⎩⎨⎧>-----<x x x x ,2)∅<⇒⎩⎨⎧>--+-<≤-x x x x 41212, 3)214121>⇒⎩⎨⎧>+++-≥x x x x , 所以原不等式的解集为⎭⎬⎫⎩⎨⎧>-<2127x x x 或…………5分 (Ⅱ)由()2922525≤+--=--x x x f x , 而()()29452141421142114=+≥⎪⎭⎫ ⎝⎛+++=⎪⎭⎫ ⎝⎛++=+b a a b b a b a b a , 所以()b a x f x 1425+≤--.…………10分。

福建省四地六校2014届高三上学期第二次月考数学文试题 Word版含答案

福建省四地六校2014届高三上学期第二次月考数学文试题 Word版含答案

“华安、连城、永安、漳平一中、龙海二中、泉港一中”六校联考2013-2014学年上学期第二次月考高三(文科)数学试卷(考试时间:120分钟 总分:150分)一、选择题(本题共12个小题,每小题5分,共60分。

在每小题给出的四个选项中,只有一个项是符合要求的)1. 已知集合{}{}2540,1,2,3,4,M x Z x x N =∈-+<=则MN = ( )A .{}1,2,3B .{}2,3,4C .{}2,3D .{}1,2,4 2. 已知命题:,cos 1p x R x ∀∈≤,则 ( )A .:,cos 1p x R ⌝∀∈>B .:,cos 1p x R ⌝∀∈≥C .:,cos 1p x R ⌝∃∈≥D .:,cos 1p x R ⌝∃∈>3. 已知复数i(2i)z =-(其中i 为虚数单位),则复数z 在复平面内对应的点在( ) A .第一象限 B .第二象限 C .第三象限 D .第四象限4. 在各项都为正数的等比数列}{n a 中,首项为3,前3项和为21,则3a 等于( ) A .15 B .12 C .9 D .65. 下列函数中既是偶函数,又在区间0+∞(,)上单调递增的函数是( )A .3y x =B .21y x =-+C . ||1y x =+ (D )2x y = 6. 已知向量p ()2,3=-,q (),6x =,且//p q ,则+p q 的值为 ( )A B .5C D .137. 若,且,则( )8. 已知0a b >>,则下列不等式中总成立的是 ( ) A.11a b b a +>+ B.11a b a b +>+ C. 11b b a a +>+ D.11b a b a->-CA9. 要得到函数sin(2)6y x π=+的图象,只需将函数sin 2y x =的图象( )A .向左平移12π个单位长度 B .向右平移12π个单位长度C .向左平移6π个单位长度D .向右平移6π个单位长度 10. 如图,已知=,=,=3,用,表示,则等于( ) ++C +. +A .1()f x x x =- B .e ()x f x x=C .21()1f x x=- D .ln ()x f x x =12. 已知定义域为R 的函数()f x 满足(4)3f =-,且对任意x R ∈总有()3f x '<, 则不等式()315f x x <-的解集为( )A.(),4-∞B.(),4-∞-C.()(),44,-∞-+∞D.()4,+∞第Ⅱ卷 (非选择题 共90 分)二、填空题(本大题共4小题,每小题4分,共16分)13. 等差数列{a n }的前n 项和为S n ,若a 2+a 7+a 12=30,则S 13的值是14. 已知实数,x y 满足约束条件⎪⎩⎪⎨⎧≤≤-≥++0005y y x y x ,则24z x y =+的最小值是 .15. 已知集合(){}2()lg 23A x f x x x ==--,{}2,2x B y y a x ==-≤.若A B A =,则a 的取值范围是 .16. 将全体正整数排成一个三角形数阵:按照以上排列的规律,第n 行(n ≥3)从左向右的第3个数为三、解答题(本大题共6小题,共74分。

福建省四地六校2014届高三高考模拟试题 数学文 含答案

福建省四地六校2014届高三高考模拟试题 数学文 含答案

“四地六校”数学(文科)2014年高考模拟试题(考试时间:120分钟 总分:150分)命题人:龙海二中 姚跃宣 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,满分60分.在每小题给出的四个选项中,只有一项是符合题目要求的.1、{}{}则,设集合,2202<<-=<-=x x B x x x A ( )A B A A =⋃、 R B A B =⋃、 A B A C =⋂、 ∅=⋂B A D 、2、已知为虚数单位,a 为实数,复数(2i)(1+i)z a =-在复平面内对应的点为M ,则“2=a ”是“点M 在坐标轴上”的 ( ) A .充分而不必要条件 B .必要而不充分条件 C .充要条件 D .既不充分也不必要条件3、设向量b a ,是同一平面内所有向量的一组基底,若)2(||)(b a b a-+λ则实数λ的值为( )A 、2B 、—2C 、21D 、21- 4、若一个圆台的的正视图如图所示,则其侧面积等于( )A .6B .6πC .35πD .65π5、如图所示,执行右边的程序框图,输出的y=( )A 、 3B 、 5C 、 7D 、96、设l 、m 是两条不同的直线,α是一个平面,则下列命题正确的是( )A . 若l m ⊥,m α⊂,则l α⊥B . 若l α⊥,l m //,则m α⊥C . 若l α//,m α⊂,则l m //D . 若l α//,m α//,则l m // 7、若x ,y 满足约束条件{3131≤+≤≤-≤y x x y ,则2x-y 的最小值为( )A 、-6B 、 —4C 、 -3D 、—18、已知直线22x y +=与x 轴,y 轴分别交于,A B 两点,若动点(,)P a b 在线段AB 上,则ab 的最大值为( )A 、B 、2C 、3D 、9、{}===k a S S S n a knn,则,,若项和为的前的等差数列设公差不为00218( ) A 、14 B 、 15 C 、 16 D 、2110、已知定义在R 上的函数f(x)满足==-++)2014(,3)12012(2)(f x x x f x f 则( )A 、0B 、 2010C 、 —2010D 、201411、设双曲线22221(0,0)x y a b a b-=>>的离心率为2e =,右焦点为(,0)f c ,方程20ax bx c --=的两个实根分别为1x 和2x ,则点P (1x ,2x )( )A 。

福建省龙海市第二中学2014-2015学年高二下学期期末考试数学(文)试卷 Word版含答案

福建省龙海市第二中学2014-2015学年高二下学期期末考试数学(文)试卷 Word版含答案

龙海二中2014-2015学年第二学期期末考试 高二数学(文科)试卷 (考试时间:120分钟总分:150分) 本试卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题).本试卷共6页,满分150分.考试时间120分钟. 第Ⅰ卷(选择题共60分) 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,只有一项是符合题目要求的. 1.设全集U=M∪N={1,2,3,4,5},M∩N={2,4},则N= ( )A.{1,2,3}B.{1,3,5}C.{1,4,5}D.{2,3,4} 2.若a∈R,则“a=1”是“|a|=1”的 ( )A.充分而不必要条件B.必要而不充分条件C.充要条件D.既不充分又不必要条件 3.若函数f(x)=(m-2)x2+(m2-1)x+1是偶函数,则在区间(-∞,0]上,f(x)是( ) A.增函数 B.减函数 C.常数函数 D.可能是增函数,也可能是常数函数 4.已知函数f(x)=ax+logax(a>0且a≠1)在[1,2]上的最大值与最小值之和为loga2+6,则a的值为() A. B. C.2 D.4 5.函数f(x)=ex+x-2的零点所在的一个区间是 ( )A.(-2,-1)B.(-1,0)C.(0,1)D.(1,2) 6.如图是张大爷晨练时所走的离家距离(y)与行走时间(x)之间的函数关系的图象,若用黑点表示张大爷家的位置,则张大爷散步行走的路线可能是() 7.函数f(x)=sin(ωx+φ)(ω>0,|φ|<)的最小正周期为π,且其图象向左平移个单位后,得到的函数为奇函数,则f(x)的图象 ( )A.关于点对称B.关于直线对称C.关于点对称D.关于直线对称 8.已知2弧度的圆心角所对的弦长为4,那么这个圆心角所对的弧长为 ( )A.4B.sin 2 C. D.4sin 1 9.函数y=sin(-2x)的单调增区间是()A.[kπ- π,kπ+ π](k∈Z)B.[kπ+ π,kπ+ π](k∈Z)C.[kπ- π,kπ+π](k∈Z)D.[kπ+ π,kπ+ π](k∈Z) 10.已知函数f(x)的部分图象如图所示,A为最高点,则f(x)的解析式可能为() A. B. C. D. 11.在△ABC中,内角A,B,C的对边分别是a,b,c,若,则A=( )A.30°B.60°C.120°D.150° 12.某班设计了一个八边形的班徽(如图),它由腰长为1,顶角为α的四个等腰三角形及其底边构成的正方形所组成,该八边形的面积为( )A.2sin α-2cos α+2B.sin α-cos α+3C.3sin α-cos α+1D.2sin α-cos α+1 第Ⅱ卷(非选择题共90分) 二、填空题:本大题共4小题,每小题4分,共16分.将答案填在答题卡的相应位置 13.已知f(x+1)=4x+3,则f(x)= . 14.已知△ABC中,tan A=,则cos A=. 15. 已知命题p:m<1,命题q:函数是减函数,若p与q一真一假,则实数m的取值范围是 . 16.函数f(x)=Asin(ωx+φ)(A>0,ω>0,|φ|<)的图象关于直线x=对称,它的最小正周期为π,则y=f(x)的图象上离坐标原点O最近的对称中心是 . 三、解答题:本大题共6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) 已知幂函数f(x)的图象过点,函数g(x)是偶函数且当x∈[0,+∞)时,. (1)求f(x),g(x)的解析式. (2)解不等式f (x)0)的最小正周期为π. (1)求ω的值; (2)求函数f(x)在区间[0,π]上的取值范围. 21.(本小题满分12分) 已知向量a=(cos 2α,sin α),向量b=(1,2sin α-1),α∈,a·b=. (1) 求sin α的值 (2)求的值. 22(本小题满分分) x3+ax2+bx,且f′(-1)=0. (1)试用含a的代数式表示b; (2)求f(x)的单调区间; 龙海二中2014-2015学年第二学期期末考试 高二数学(文科)答案 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,只有一项是符合题目要求的. 1.设全集U=M∪N={1,2,3,4,5},M∩N={2,4},则N= ( )A.{1,2,3}B.{1,3,5}C.{1,4, 5}D.{2,3,4} 解析:画出韦恩图,可知N={1,3,5}. 答案:B 2.若a∈R,则“a=1”是“|a|=1”的 ( )A.充分而不必要条件B.必要而不充分条件C.充要条件D.既不充分又不必要条件 解析:当a=1时,|a|=1成立;反过来,若|a|=1时,a=±1,即a=1不一定成立. 答案:A 3.若函数f(x)=(m-2)x2+(m2-1)x+1是偶函数,则在区间(-∞,0]上,f(x)是( ) A.增函数 B.减函数 C.常数函数 D.可能是增函数,也可能是常数函数 解析:易知m2-1=0,即m=±1.将m=±1代入函数中, 得m-2<0,所以f(x)在(-∞,0]上为增函数. 答案:A 4.已知函数f(x)=ax+logax(a>0且a≠1)在[1,2]上的最大值与最小值之和为loga2+6,则a的值为() A. B. C.2 D.4 解析:因为f(x)在[1,2]上单调,故f(1)+f(2)=a+a2+loga2=loga2+6,则a+a2=6,又a>0,故a=2. 答案:C 5.函数f(x)=ex+x-2的零点所在的一个区间是 ( )A.(-2,-1)B.(-1,0)C.(0,1)D.(1,2) 解析:因为f(-1)=e-1-1-2<0,f(0)=e0+0-2=-1<0,f(1)=e1+1-2=e-1>0,所以函数f(x)=ex+x-2的零点所在的一个区间是(0,1),故选C. 答案:C 6.如图是张大爷晨练时所走的离家距离(y)与行走时间(x)之间的函数关系的图象,若用黑点表示张大爷家的位置,则张大爷散步行走的路线可能是() 解析:由函数关系的图象知所走路线为圆弧,故选D. 答案:D 7.函数f(x)=sin(ωx+φ)(ω>0,|φ|<)的最小正周期为π,且其图象向左平移个单位后,得到的函数为奇函数,则f(x)的图象 ( )A.关于点对称B.关于直线对称C.关于点对称D.关于直线对称 解析:因为,所以ω=2.所以f(x)=sin(2x+φ),图象向左平移个单位后得到.由它为奇函数可得2×0+φ+=kπ(k∈Z),所以φ=kπ-.因为|φ|<,所以φ=-.所以.由(k∈Z)得对称中心为,所以A、C错误.由(k∈Z)得对称轴为(k∈Z),所以B正确. 答案:B 8.已知2弧度的圆心角所对的弦长为4,那么这个圆心角所对的弧长为 ( )A.4B.sin 2 C. D.4sin 1 解析:如图所示,O为BC中点,则AO⊥BC,在△AOB中,∠BAO=1 rad,BO=2, 设半径为R,所以sin 1=.所以R=,所以弧长l=2×R=2×=. 答案:C 9.函数y=sin(-2x)的单调增区间是()A.[kπ- π,kπ+ π](k∈Z)B.[kπ+ π,kπ+ π](k∈Z)C.[kπ- π,kπ+π](k∈Z)D.[kπ+ π,kπ+ π](k∈Z) 解析:y=sin(-2x)=-sin(2x-). 令2kπ+≤2x-≤2kπ+π, 解得kπ+π≤x≤kπ+π(k∈Z). 所以单调增区间为[kπ+ π,kπ+ π](k∈Z). 答案:D 10.已知函数f(x)的部分图象如图所示,则f(x)的解析式可能为() A. B. C. D. 【解析】设函数f(x)=Asin(ωx+),由函数的最大值为2知A=2,又由函数图象知该函数的周期T=4×=4π,所以ω=.将点(0,1)代入得=,所以f(x)=【答案】A 11.在△ABC中,内角A,B,C的对边分别是a,b,c,若,则A=( )A.30°B.60°C.120°D.150° 解析:由可得, 由余弦定理得,所以A=30°. 答案:A 12.某班设计了一个八边形的班徽(如图),它由腰长为1,顶角为α的四个等腰三角形及其底边构成的正方形所组成,该八边形的面积为( )A.2sin α-2cos α+2B.sin α-cos α+3C.3sin α-cos α+1D.2sin α-cos α+1 解析:四个等腰三角形面积之和为4××1×1×sin α=2sin α,再由余弦定理可得正方形的边长为,所以总面积为2sin α+=2sin α-2cos α+2. 答案:A 第Ⅱ卷(非选择题共90分) 二、填空题:本大题共4小题,每小题4分,共16分.将答案填在答题卡的相应位置 13.已知f(x+1)=4x+3,则f(x)= . 解析:因为f(x+1)=4x+3=4(x+1)-1,所以f(x)=4x-1. 答案:4x-1 14.已知△ABC中,tan A=,则cos A=. 解析:由tan A=知A为钝角,cos A<0,所以cos A=. 答案: 15. 已知命题p:m<1,命题q:函数是减函数,若p与q一真一假,则实数m的取值范围是 . 解析:p:m<1,q:m<2. 因为p与q一真一假,所以p真q假或p假q真. 所以 答案: 1≤m<2 16.函数f(x)=Asin(ωx+φ)(A>0,ω>0,|φ|<)的图象关于直线x=对称,它的最小正周期为π,则y=f(x)的图象上离坐标原点O最近的对称中心是 . 解析:由题意知.由得故. 令f(x)=0可得,即. 故f(x)的图象上离原点O最近的对称中心是. 答案: 三、解答题:本大题共6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) 已知幂函数f(x)的图象过点,函数g(x)是偶函数且当x∈[0,+∞)时,. (1)求f(x),g(x)的解析式. (2)解不等式f(x)<g(x).(提示:结合图形,可直接写出答案) 解:(1)设f(x)=,因为其图象过点, 故,即,所以α=3, 故.....................................................4分 令x∈(-∞,0),则-x∈(0,+∞), 所以. 因为g(x)是偶函数,故g(-x)=g(x), 所以,x∈(-∞,0), 所以 故 (x∈R)............................................8分 (2)在同一坐标系下作出f(x)=与g(x)的图象如图所示,由图象可知f(x)1时,1-2a<-1. 当x变化时,f′(x)与f(x)的变化情况如下表: 由此得,函数f(x)的单调增区间为(-∞,1-2a)和(-1,+∞),单调减区间为(1-2a,-1). ②当a=1时,1-2a=-1.此时,f′(x)≥0恒成立,且仅在x=-1处f′(x)=0, 故函数f(x)的单调增区间为R. ③当a-1,同理可得函数f(x)的单调增区间为(-∞,-1)和(1-2a,+∞),单调减区间为(-1,1-2a). 综上:当a>1时,函数f(x)的单调递增区间为(-∞,1-2a)和(-1,+∞),单调减区间为(1-2a,-1); 当a=1时,函数f(x)的单调增区间为R; 当a<1时,函数f(x)的单调增区间为(-∞,-1)和(1-2a,+∞),单调减区间为(-1,1-2a)…14分 龙海二中2014-2015学年第二学期期末考试 高二数学(文科)试题答题卷 (考试时间:120分钟总分:150分) 选择题(本大题共12小题,每小题5分,共60分) 题号 1 2 3 4 5 6 7 8 9 10 11 12 答案二、填空题(本大题共4小题,每小题4分,共16分) 13、 14、 15、 16、 三、解答题:本大题共6小题,共74分.解答应写出文字说明,证明过程或演算步骤. ………………………………..密………………………………………………封…………………………………线……………………………………… 21、(12分) 22、(14分) 20、(12分) 19、(12分) 18、(12分) 17、(12分)。

福建省龙海市第二中学2014届高考数学第二学期期初模拟考试试卷 理

福建省龙海市第二中学2014届高考数学第二学期期初模拟考试试卷 理

2014届龙海二中高三第一学期期初数学科高考模拟卷(考试时间:120分钟 总分:150分) 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。

每小题给出的四个选项中,只有一项是符合题目要求的。

1.双曲线2214x y -=的一个焦点坐标是( )A.( B .(2,0)- C. D .(1,0)2.已知函数()f x 是定义在R 上的奇函数,当0x >时,()2,(3)xf x f =-则的值是( )A .18B .18-C .8D .-83.已知为虚数单位,a 为实数,复数2(1)z i ai =⋅+在复平面内对应的点为M ,则“0a >”是“点M 在第二象限”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 4.已知具有线性相关的两个变量,x y 之间的一组数据如下:且回归方程是0.95,6,y x a x y =+=则当时的预测值为( ) A .8.4 B .8.3 C .8.2 D .8.15.用若干个棱长为1cm 的小正方体叠成一个几何体,图1为其正视图,图2为其俯视图,若这个几何体的体积为7cm3,则其侧视图为 ( )6.已知集合231{|(6)(10)0,},()()nM m m m m N x n M x =--≤∈-∈若的二项展开式中存在常数项,则n 等于( ) A .7 B .8C .9D .107.在区间[0,]π上随机取一个数x ,则事件“sin cos x x +≥”发生的概率为( )A .14B .13C .12D .238.已知不等式组1010330x y x y x y -+≥⎧⎪+-≥⎨⎪--≤⎩表示的平面区域为D ,若直线1y kx =+将区域D 分成面积相等的两部分,则实数k 的值是( )A .15B .14C .13 D .129.若双曲线)0(12222>>=-b a b y a x 的左右焦点分别为1F 、2F ,线段21F F 被抛物线22y bx = 的焦点分成5:7的两段,则此双曲线的离心率为( )A .98 BC.D.10.记集合31212323{1,2,3,4,5,6},{|,,,}101010a a a A M m m a a a A ===++∈,将M 中的元素按从小到大排列,则第70个是( ) A .0.264 B .0.265 C .0.431 D .0.432第Ⅱ卷(非选择题,共100分)二、填空题:本大题共5小题,每小题4分,共20分,把答案填写在答题卡的相应位置。

福建省龙海市高三下学期期初考试数学试卷(文)含答案

福建省龙海市高三下学期期初考试数学试卷(文)含答案

高三第二学期期初考试数学(文科)试卷(考试时间:120分钟 总分:150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,只有一项是符合题目要求的。

请将正确的答案填涂在答题卡上。

)1.已知集合{123}A =,,,2{|9}B x x =<,则A B =(A ){210123}--,,,,, (B ){21012}--,,,, (C ){123},, (D ){12},2.设复数z 满足i 3i z +=-,则z =(A )12i -+ (B )12i - (C )32i + (D )32i - 3.函数=sin()y A x ωϕ+的部分图像如图所示,则(A )2sin(2)3y x π=- (B )2sin(2)6y x π=-(C )2sin(2+)6y x π= (D )2sin(2+)3y x π=4.体积为8的正方体的顶点都在同一球面上,则该球的表面积为 (A )12π (B )323π(C )8π (D )4π 5.已知双曲线2222:1x y C a b-=(0,0a b >>)的离心率为2,则C 的渐近线方程为(A )(B ) (C ) (D )6. 已知函数1()1x f x x +=-的图像在点()2,(2)f 处的切线与直线10ax y ++=平行,则 实数a =(A )2 (B )12 (C )12-(D )2-7.等比数列的前项和为,若,,则( )(A ) 18 (B ) 10 (C ) -14 (D ) -22 8.函数的部分图像大致为( )y =2y x =±y =y =(A ) (B ) (C) (D )9.已知函数在单调递增,则的最大值是( )(A )(B )(C )(D )10. 若实数x ,y 满足不等式组()()125002x y x y x --+-≥⎧⎪⎨≤≤⎪⎩,,则2z x y =-的取值范围是(A )[]5,5- (B )[]5,1- (C )[]1,3 (D )[]5,3- 11.在边长为1的正方形中,动点在以点为圆心且与相切的圆上.若,则的最大值是( )(A )3 (B) (C ) (D ) 4 12.已知函数,对于任意,,恒成立,则的取值范围是( ) (A )(B )(C ) (D )二、填空题(本大题共4小题,每小题5分,共20分) 13.已知132a =,则()2log 2a =. 14.已知向量,,若,则__________.15.设函数()113,1,,1,x e x f x x x -⎧<⎪=⎨⎪≥⎩则使得()2f x ≤成立的x 的取值范围是________.16.如图,为测量山高MN ,选择A 和另一座山的山顶C 点.从A 点测得 M 点的仰角60MAN ∠=︒,C 点的仰角45CAB ∠=︒以及75MAC ∠=︒;从C 点测得60MCA ∠=︒.已知山高100BC m =,则山高MN =三、解答题:(本大题共6小题,共70分,解答应写出文字说明,证明过程或演算步骤)17.(本小题12分)已知a ,b ,c 分别为ABC ∆三个内角A ,B ,C 的对边,sin cos c C c A =-. (Ⅰ)求A ;(Ⅱ)若a =2,ABC ∆b ,c .18.(本小题12分)(17)(本小题满分12分)设正项数列{}n a 的前n 项和为{}n S ,且11a =,当2n ≥时,n a =(Ⅰ)求数列{}n a 的通项公式; (Ⅱ)设数列{}n b 满足1121212222n nn n nb b b b a --+++=,求{}n b 的前n 项和n T .19.(本小题12分)如图,四棱锥P ABCD -中,底面ABCD 是直角梯形,90ABC ∠=︒,222BC AD AB ===,,PB PC PD ⊥=(Ⅰ)求证:平面PBC ⊥平面ABCD ;(Ⅱ)若PC PB =,求点D 到平面PAB 的距离.20.(本小题满分12分)已知动圆C 过定点(1,0)F ,且与定直线1x =-相切. (1)求动圆圆心C 的轨迹E 的方程;(2)过点()2,0M -的任一条直线l 与轨迹E 交于不同的两点,P Q ,试探究在x 轴上是否存在定点N(异于点M ),使得QNM PNM π∠+∠=?若存在,求点N 的坐标;若不存在,说明理由. 21. (本小题满分12分)已知函数.(1)求曲线在点处的切线方程;(2)函数与函数的图像总有两个交点,设这两个交点的横坐标分别为,.(ⅰ)求的取值范围; (ⅱ)求证:.选考题:请考生在第22、23两题中任选一题作答。

2014年福建省高考数学试卷及解析(文科)

2014年福建省高考数学试卷及解析(文科)

2014年福建省高考数学试卷(文科)一、选择题:本大题共12小题,每小题5分,共60分1、(5分)若集合P={x|2≤x<4},Q={x|x≥3},则P∩Q等于()A、{x|3≤x<4}B、{x|3<x<4}C、{x|2≤x<3}D、{x|2≤x≤3}2、(5分)复数(3+2i)i等于()A、﹣2﹣3iB、﹣2+3iC、2﹣3iD、2+3i3、(5分)以边长为1的正方形的一边所在所在直线为旋转轴,将该正方形旋转一周所得圆柱的侧面积等于()A、2πB、πC、2D、14、(5分)阅读如图所示的程序框图,运行相应的程序,输出的n的值为()A、1B、2C、3D、45、(5分)命题“∀x∈[0,+∞),x3+x≥0”的否定是()A、∀x∈(﹣∞,0),x3+x<0B、∀x∈(﹣∞,0),x3+x≥0C、∃x0∈[0,+∞),x03+x0<0D、∃x0∈[0,+∞),x03+x0≥06、(5分)已知直线l过圆x2+(y﹣3)2=4的圆心,且与直线x+y+1=0垂直,则l的方程是()A、x+y﹣2=0B、x﹣y+2=0C、x+y﹣3=0D、x﹣y+3=07、(5分)将函数y=sinx的图象向左平移个单位,得到函数y=f(x)的函数图象,则下列说法正确的是()A、y=f(x)是奇函数B、y=f(x)的周期为πC、y=f(x)的图象关于直线x=对称D、y=f(x)的图象关于点(﹣,0)对称8、(5分)若函数y=log a x(a>0,且a≠1)的图象如图所示,则下列函数正确的是()A、B、C、D、9、(5分)要制作一个容积为4m3,高为1m的无盖长方体容器,已知该容器的底面造价是每平方米20元,侧面造价是每平方米10元,则该容器的最低总造价是()A、80元B、120元C、160元D、240元10、(5分)设M为平行四边形ABCD对角线的交点,O为平行四边形ABCD所在平面内任意一点,则等于()A、B、2 C、3 D、411、(5分)已知圆C:(x﹣a)2+(y﹣b)2=1,设平面区域Ω=,若圆心C∈Ω,且圆C与x轴相切,则a2+b2的最大值为()A、49B、37C、29D、512、(5分)在平面直角坐标系中,两点P1(x1,y1),P2(x2,y2)间的“L﹣距离”定义为|P1P2|=|x1﹣x2|+|y1﹣y2|、则平面内与x轴上两个不同的定点F1,F2的“L ﹣距离”之和等于定值(大于|F1F2|)的点的轨迹可以是()A、B、C、D、二、填空题:本大题共4小题,每小题4分,共16分13、(4分)如图,在边长为1的正方形中随机撒1000粒豆子,有180粒落到阴影部分,据此估计阴影部分的面积为、14、(4分)在△ABC中,A=60°,AC=2,BC=,则AB等于、15、(4分)函数f(x)=的零点个数是、16、(4分)已知集合{a,b,c}={0,1,2},且下列三个关系:①•a≠2;②‚b=2;③ƒc≠0有且只有一个正确,则100a+10b+c等于、三、解答题:本大题共6小题,共74分.17、(12分)在等比数列{a n}中,a2=3,a5=81、(Ⅰ)求a n;(Ⅱ)设b n=log3a n,求数列{b n}的前n项和S n、18、(12分)已知函数f(x)=2cosx(sinx+cosx)、(Ⅰ)求f()的值;(Ⅱ)求函数f(x)的最小正周期及单调递增区间、19、(12分)如图,三棱锥A﹣BCD中,AB⊥平面BCD,CD⊥BD、(Ⅰ)求证:CD⊥平面ABD;(Ⅱ)若AB=BD=CD=1,M为AD中点,求三棱锥A﹣MBC的体积、20、(12分)根据世行2013年新标准,人均GDP低于1035美元为低收入国家;人均GDP为1035﹣4085美元为中等偏下收入国家;人均GDP为4085﹣12616美元为中等偏上收入国家;人均GDP不低于12616美元为高收入国家、某城市有5个行政区,各区人口占该城市人口比例及人均GDP如下表:行政区区人口占城市人口比例区人均GDP(单位:美元)A25%8000B30%4000C15%6000D10%3000E20%10000(Ⅰ)判断该城市人均GDP是否达到中等偏上收入国家标准;(Ⅱ)现从该城市5个行政区中随机抽取2个,求抽到的2个行政区人均GDP 都达到中等偏上收入国家标准的概率、21、(12分)已知曲线Γ上的点到点F(0,1)的距离比它到直线y=﹣3的距离小2、(Ⅰ)求曲线Γ的方程;(Ⅱ)曲线Γ在点P处的切线l与x轴交于点A、直线y=3分别与直线l及y轴交于点M,N,以MN为直径作圆C,过点A作圆C的切线,切点为B,试探究:当点P在曲线Γ上运动(点P与原点不重合)时,线段AB的长度是否发生变化?证明你的结论、22、(14分)已知函数f(x)=e x﹣ax(a为常数)的图象与y轴交于点A,曲线y=f(x)在点A处的切线斜率为﹣1、(1)求a的值及函数f(x)的极值;(2)证明:当x>0时,x2<e x;(3)证明:对任意给定的正数c,总存在x0,使得当x∈(x0,+∞)时,恒有x <ce x、参考答案与试题解析一、选择题:本大题共12小题,每小题5分,共60分1、(5分)若集合P={x|2≤x<4},Q={x|x≥3},则P∩Q等于()A、{x|3≤x<4}B、{x|3<x<4}C、{x|2≤x<3}D、{x|2≤x≤3}分析:由于两集合已是最简,直接求它们的交集即可选出正确答案解答:解:∵P={x|2≤x<4},Q={x|x≥3},∴P∩Q={x|3≤x<4}、故选:A、点评:本题考查交集的运算,理解好交集的定义是解题的关键2、(5分)复数(3+2i)i等于()A、﹣2﹣3iB、﹣2+3iC、2﹣3iD、2+3i分析:直接由复数代数形式的乘法运算化简求值、解答:解:(3+2i)i=3i+2i2=﹣2+3i、故选:B、点评:本题考查了复数代数形式的乘法运算,是基础的计算题、3、(5分)以边长为1的正方形的一边所在所在直线为旋转轴,将该正方形旋转一周所得圆柱的侧面积等于()A、2πB、πC、2D、1分析:边长为1的正方形,绕其一边所在直线旋转一周,得到的几何体为圆柱,从而可求圆柱的侧面积、解答:解:边长为1的正方形,绕其一边所在直线旋转一周,得到的几何体为圆柱,则所得几何体的侧面积为:1×2π×1=2π,故选:A、点评:本题是基础题,考查旋转体的侧面积的求法,考查计算能力、4、(5分)阅读如图所示的程序框图,运行相应的程序,输出的n的值为()A、1B、2C、3D、4分析:根据框图的流程模拟运行程序,直到不满足条件2n>n2,跳出循环,确定输出的n值、解答:解:由程序框图知:第一次循环n=1,21>1;第二次循环n=2,22=4、不满足条件2n>n2,跳出循环,输出n=2、故选:B、点评:本题考查了当型循环结构的程序框图,根据框图的流程模拟运行程序是解答此类问题的常用方法、5、(5分)命题“∀x∈[0,+∞),x3+x≥0”的否定是()A、∀x∈(﹣∞,0),x3+x<0B、∀x∈(﹣∞,0),x3+x≥0C、∃x0∈[0,+∞),x03+x0<0D、∃x0∈[0,+∞),x03+x0≥0分析:全称命题的否定是一个特称命题,按此规则写出其否定即可得出正确选项、解答:解:∵命题“∀x∈[0,+∞),x3+x≥0”是一个全称命题、∴其否定命题为:∃x0∈[0,+∞),x03+x0<0故选:C、点评:本题考查全称命题的否定,掌握此类命题的否定的规则是解答的关键、6、(5分)已知直线l过圆x2+(y﹣3)2=4的圆心,且与直线x+y+1=0垂直,则l的方程是()A、x+y﹣2=0B、x﹣y+2=0C、x+y﹣3=0D、x﹣y+3=0分析:由题意可得所求直线l经过点(0,3),斜率为1,再利用点斜式求直线l 的方程、解答:解:由题意可得所求直线l经过点(0,3),斜率为1,故l的方程是y﹣3=x﹣0,即x﹣y+3=0,故选:D、点评:本题主要考查用点斜式求直线的方程,两条直线垂直的性质,属于基础题、7、(5分)将函数y=sinx的图象向左平移个单位,得到函数y=f(x)的函数图象,则下列说法正确的是()A、y=f(x)是奇函数B、y=f(x)的周期为πC、y=f(x)的图象关于直线x=对称D、y=f(x)的图象关于点(﹣,0)对称分析:利用函数图象的平移法则得到函数y=f(x)的图象对应的解析式为f(x)=cosx,则可排除选项A,B,再由cos=cos(﹣)=0即可得到正确选项、解答:解:将函数y=sinx的图象向左平移个单位,得y=sin(x+)=cosx、即f(x)=cosx、∴f(x)是周期为2π的偶函数,选项A,B错误;∵cos=cos(﹣)=0,∴y=f(x)的图象关于点(﹣,0)、(,0)成中心对称、故选:D、点评:本题考查函数图象的平移,考查了余弦函数的性质,属基础题、8、(5分)若函数y=log a x(a>0,且a≠1)的图象如图所示,则下列函数正确的是()A、B、C、D、分析:根据对数函数的图象所过的特殊点求出a的值,再研究四个选项中函数与图象是否对应即可得出正确选项、解答:解:由对数函数的图象知,此函数图象过点(3,1),故有y=log a3=1,解得a=3,对于A,由于y=a﹣x是一个减函数故图象与函数不对应,A错;对于B,由于幂函数y=x a是一个增函数,且是一个奇函数,图象过原点,且关于原点对称,图象与函数的性质对应,故B正确;对于C,由于a=3,所以y=(﹣x)a是一个减函数,图象与函数的性质不对应,C错;对于D,由于y=log a(﹣x)与y=log a x的图象关于y轴对称,所给的图象不满足这一特征,故D错、故选:B、点评:本题考查函数的性质与函数图象的对应,熟练掌握各类函数的性质是快速准确解答此类题的关键、9、(5分)要制作一个容积为4m3,高为1m的无盖长方体容器,已知该容器的底面造价是每平方米20元,侧面造价是每平方米10元,则该容器的最低总造价是()A、80元B、120元C、160元D、240元分析:设池底长和宽分别为a,b,成本为y,建立函数关系式,然后利用基本不等式求出最值即可求出所求、解答:解:设池底长和宽分别为a,b,成本为y,则∵长方形容器的容器为4m3,高为1m,∴底面面积S=ab=4,y=20S+10[2(a+b)]=20(a+b)+80,∵a+b≥2=4,∴当a=b=2时,y取最小值160,即该容器的最低总造价是160元,故选:C、点评:本题以棱柱的体积为载体,考查了基本不等式,难度不大,属于基础题,由实际问题向数学问题转化是关键、10、(5分)设M为平行四边形ABCD对角线的交点,O为平行四边形ABCD所在平面内任意一点,则等于()A、B、2 C、3 D、4分析:虑用特殊值法去做,因为O为任意一点,不妨把O看成是特殊点,再代入计算,结果满足哪一个选项,就选哪一个、解答:解:∵O为任意一点,不妨把A点看成O点,则=,∵M是平行四边形ABCD的对角线的交点,∴=2=4故选:D、点评:本题考查了平面向量的加法,做题时应掌握规律,认真解答、11、(5分)已知圆C:(x﹣a)2+(y﹣b)2=1,设平面区域Ω=,若圆心C∈Ω,且圆C与x轴相切,则a2+b2的最大值为()A、49B、37C、29D、5分析:作出不等式组对应的平面区域,利用圆C与x轴相切,得到b=1为定值,此时利用数形结合确定a的取值即可得到结论、解答:解:作出不等式组对应的平面区域如图:圆心为(a,b),半径为1∵圆心C∈Ω,且圆C与x轴相切,∴b=1,则a2+b2=a2+1,∴要使a2+b2的取得最大值,则只需a最大即可,由图象可知当圆心C位于B点时,a取值最大,由,解得,即B(6,1),∴当a=6,b=1时,a2+b2=36+1=37,即最大值为37,故选:B、点评:本题主要考查线性规划的应用,利用数形结合是解决线性规划题目的常用方法、12、(5分)在平面直角坐标系中,两点P1(x1,y1),P2(x2,y2)间的“L﹣距离”定义为|P1P2|=|x1﹣x2|+|y1﹣y2|、则平面内与x轴上两个不同的定点F1,F2的“L ﹣距离”之和等于定值(大于|F1F2|)的点的轨迹可以是()A、B、C、D、分析:设出F1,F2的坐标,在设出动点M的坐标,由新定义列式后分类讨论去绝对值,然后结合选项得答案、解答:解:设F1(﹣c,0),F2(c,0),再设动点M(x,y),动点到定点F1,F2的“L﹣距离”之和等于m(m>2c>0),由题意可得:|x+c|+|y|+|x﹣c|+|y|=m,即|x+c|+|x﹣c|+2|y|=m、当x<﹣c,y≥0时,方程化为2x﹣2y+m=0;当x<﹣c,y<0时,方程化为2x+2y+m=0;当﹣c≤x<c,y≥0时,方程化为y=;当﹣c≤x<c,y<0时,方程化为y=c﹣;当x≥c,y≥0时,方程化为2x+2y﹣m=0;当x≥c,y<0时,方程化为2x﹣2y﹣m=0、结合题目中给出的四个选项可知,选项A中的图象符合要求、故选:A、点评:本题考查轨迹方程的求法,考查了分类讨论的数学思想方法,解答的关键是正确分类,是中档题、二、填空题:本大题共4小题,每小题4分,共16分13、(4分)如图,在边长为1的正方形中随机撒1000粒豆子,有180粒落到阴影部分,据此估计阴影部分的面积为0.18、分析:根据几何槪型的概率意义,即可得到结论、解答:解:正方形的面积S=1,设阴影部分的面积为S,∵随机撒1000粒豆子,有180粒落到阴影部分,∴几何槪型的概率公式进行估计得,即S=0.18,故答案为:0.18、点评:本题主要考查几何槪型的概率的计算,利用豆子之间的关系建立比例关系是解决本题的关键,比较基础、14、(4分)在△ABC中,A=60°,AC=2,BC=,则AB等于1、分析:利用余弦定理列出关系式,将AC,BC,以及cosA的值代入即可求出AB 的长、解答:解:∵在△ABC中,A=60°,AC=b=2,BC=a=,∴由余弦定理得:a2=b2+c2﹣2bccosA,即3=4+c2﹣2c,解得:c=1,则AB=c=1,故答案为:1点评:此题考查了余弦定理,以及特殊角的三角函数值,熟练掌握定理是解本题的关键、15、(4分)函数f(x)=的零点个数是2、分析:根据函数零点的定义,直接解方程即可得到结论、解答:解:当x≤0时,由f(x)=0得x2﹣2=0,解得x=或x=(舍去),当x>0时,由f(x)=0得2x﹣6+lnx=0,即lnx=6﹣2x,作出函数y=lnx和y=6﹣2x在同一坐标系图象,由图象可知此时两个函数只有1个交点,故x>0时,函数有1个零点、故函数f(x)的零点个数为2,故答案为:2点评:本题主要考查函数零点个数的判断,对于比较好求的函数,直接解方程f (x)=0即可,对于比较复杂的函数,由利用数形结合进行求解、16、(4分)已知集合{a,b,c}={0,1,2},且下列三个关系:①•a≠2;②‚b=2;③ƒc≠0有且只有一个正确,则100a+10b+c等于201、分析:根据集合相等的条件,列出a、b、c所有的取值情况,再判断是否符合条件,求出a、b、c的值后代入式子求值、解答:解:由{a,b,c}={0,1,2}得,a、b、c的取值有以下情况:当a=0时,b=1、c=2或b=2、c=1,此时不满足题意;当a=1时,b=0、c=2或b=2、c=0,此时不满足题意;当a=2时,b=1、c=0,此时不满足题意;当a=2时,b=0、c=1,此时满足题意;综上得,a=2、b=0、c=1,代入100a+10b+c=201,故答案为:201、点评:本题考查了集合相等的条件的应用,以及分类讨论思想,注意列举时按一定的顺序列举,做到不重不漏、三、解答题:本大题共6小题,共74分.17、(12分)在等比数列{a n}中,a2=3,a5=81、(Ⅰ)求a n;(Ⅱ)设b n=log3a n,求数列{b n}的前n项和S n、分析:(Ⅰ)设出等比数列的首项和公比,由已知列式求解首项和公比,则其通项公式可求;(Ⅱ)把(Ⅰ)中求得的a n代入b n=log3a n,得到数列{b n}的通项公式,由此得到数列{b n}是以0为首项,以1为公差的等差数列,由等差数列的前n项和公式得答案、解答:解:(Ⅰ)设等比数列{a n}的公比为q,由a2=3,a5=81,得,解得、∴;(Ⅱ)∵,b n=log3a n,∴、则数列{b n}的首项为b1=0,由b n﹣b n﹣1=n﹣1﹣(n﹣2)=1(n≥2),可知数列{b n}是以1为公差的等差数列、∴、点评:本题考查等比数列的通项公式,考查了等差数列的前n项和公式,是基础的计算题、18、(12分)已知函数f(x)=2cosx(sinx+cosx)、(Ⅰ)求f()的值;(Ⅱ)求函数f(x)的最小正周期及单调递增区间、分析:(Ⅰ)利用三角恒等变换化简函数的解析式为f(x)=sin(2x+)+1,从而求得f()的值、(Ⅱ)根据函数f(x)=sin(2x+)+1,求得它的最小正周期、令2kπ﹣≤2x+≤2kπ+,k∈Z,求得x的范围,可得函数的单调递增区间、解答:解:(Ⅰ)∵函数f(x)=2cosx(sinx+cosx)=sin2x+1+cos2x=sin(2x+)+1,∴f()=sin(+)+1=sin+1=+1=2、(Ⅱ)∵函数f(x)=sin(2x+)+1,故它的最小正周期为=π、令2kπ﹣≤2x+≤2kπ+,k∈Z,求得kπ﹣≤x≤kπ+,故函数的单调递增区间为[kπ﹣,kπ+],k∈Z、点评:本题主要考查三角函数的恒等变换,三角函数的周期性和单调性,属于中档题、19、(12分)如图,三棱锥A﹣BCD中,AB⊥平面BCD,CD⊥BD、(Ⅰ)求证:CD⊥平面ABD;(Ⅱ)若AB=BD=CD=1,M为AD中点,求三棱锥A﹣MBC的体积、分析:(Ⅰ)证明:CD⊥平面ABD,只需证明AB⊥CD;=V C﹣ABM=S△ABM•CD,即可求出三棱锥A﹣MBC的(Ⅱ)利用转换底面,V A﹣MBC体积、解答:(Ⅰ)证明:∵AB⊥平面BCD,CD⊂平面BCD,∴AB⊥CD,∵CD⊥BD,AB∩BD=B,∴CD ⊥平面ABD ;(Ⅱ)解:∵AB ⊥平面BCD ,BD ⊂平面BCD , ∴AB ⊥BD 、 ∵AB=BD=1, ∴S △ABD =, ∵M 为AD 中点, ∴S △ABM =S △ABD =, ∵CD ⊥平面ABD ,∴V A ﹣MBC =V C ﹣ABM =S △ABM •CD=、点评:本题考查线面垂直,考查三棱锥A ﹣MBC 的体积,正确运用线面垂直的判定定理是关键、20、(12分)根据世行2013年新标准,人均GDP 低于1035美元为低收入国家;人均GDP 为1035﹣4085美元为中等偏下收入国家;人均GDP 为4085﹣12616美元为中等偏上收入国家;人均GDP 不低于12616美元为高收入国家、某城市有5个行政区,各区人口占该城市人口比例及人均GDP 如下表: 行政区 区人口占城市人口比例区人均GDP (单位:美元)A 25% 8000B 30% 4000C 15% 6000D 10% 3000 E20%10000(Ⅰ)判断该城市人均GDP 是否达到中等偏上收入国家标准;(Ⅱ)现从该城市5个行政区中随机抽取2个,求抽到的2个行政区人均GDP 都达到中等偏上收入国家标准的概率、分析:(Ⅰ)利用所给数据,计算该城市人均GDP ,即可得出结论; (Ⅱ)利用古典概型概率公式,即可得出结论、解答:解:(Ⅰ)设该城市人口总数为a ,则该城市人均GDP 为=6400∴该城市人均GDP达到中等偏上收入国家标准;(Ⅱ)从该城市5个行政区中随机抽取2个,共有=10种情况,GDP都达到中等偏上收入国家标准的区域有A,C,E,抽到的2个行政区人均GDP都达到中等偏上收入国家标准,共有=3种情况,∴抽到的2个行政区人均GDP都达到中等偏上收入国家标准的概率、点评:本题考查概率与统计等基础知识,考查数据处理能力、运算求解能力、应用意识,考查必然、或然思想、21、(12分)已知曲线Γ上的点到点F(0,1)的距离比它到直线y=﹣3的距离小2、(Ⅰ)求曲线Γ的方程;(Ⅱ)曲线Γ在点P处的切线l与x轴交于点A、直线y=3分别与直线l及y轴交于点M,N,以MN为直径作圆C,过点A作圆C的切线,切点为B,试探究:当点P在曲线Γ上运动(点P与原点不重合)时,线段AB的长度是否发生变化?证明你的结论、分析:(Ⅰ)设S(x,y)曲线Γ上的任意一点,利用抛物线的定义,判断S满足配额我想的定义,即可求曲线Γ的方程;(Ⅱ)通过抛物线方程利用函数的导数求出切线方程,求出A、M的坐标,N的坐标,以MN为直径作圆C,求出圆心坐标,半径是常数,即可证明当点P在曲线Γ上运动(点P与原点不重合)时,线段AB的长度不变、解答:解:(Ⅰ)设S(x,y)曲线Γ上的任意一点,由题意可得:点S到F(0,1)的距离与它到直线y=﹣1的距离相等,曲线Γ是以F为焦点直线y=﹣1为准线的抛物线,∴曲线Γ的方程为:x2=4y、(Ⅱ)当点P在曲线Γ上运动(点P与原点不重合)时,线段AB的长度不变,证明如下:由(Ⅰ)可知抛物线的方程为y=,设P(x0,y0)(x0≠0)则y0=,由y得切线l的斜率k==∴切线l的方程为:,即、由得,由得,又N(0,3),所以圆心C(),半径r==∴点P在曲线Γ上运动(点P与原点不重合)时,线段AB的长度不变、点评:本题考查轨迹方程的求法,直线与抛物线的位置关系的应用,圆的方程函数的导数等指数的应用,难度较大、22、(14分)已知函数f(x)=e x﹣ax(a为常数)的图象与y轴交于点A,曲线y=f(x)在点A处的切线斜率为﹣1、(1)求a的值及函数f(x)的极值;(2)证明:当x>0时,x2<e x;(3)证明:对任意给定的正数c,总存在x0,使得当x∈(x0,+∞)时,恒有x <ce x、分析:(1)利用导数的几何意义求得a,再利用导数法求得函数的极值;(2)构造函数g(x)=e x﹣x2,利用导数求得函数的最小值,即可得出结论;(3)利用(2)的结论,令x0=,则e x>x2>x,即x<ce x、即得结论成立、解答:解:(1)由f(x)=e x﹣ax得f′(x)=e x﹣a、又f′(0)=1﹣a=﹣1,∴a=2,∴f(x)=e x﹣2x,f′(x)=e x﹣2、由f′(x)=0得x=ln2,当x<ln2时,f′(x)<0,f(x)单调递减;当x>ln2时,f′(x)>0,f(x)单调递增;∴当x=ln2时,f(x)有极小值为f(ln2)=e ln2﹣2ln2=2﹣ln4、f(x)无极大值、(2)令g(x)=e x﹣x2,则g′(x)=e x﹣2x,由(1)得,g′(x)=f(x)≥f(ln2)=e ln2﹣2ln2=2﹣ln4>0,即g′(x)>0,∴当x>0时,g(x)>g(0)>0,即x2<e x;(3)对任意给定的正数c,总存在x0=>0、当x∈(x0,+∞)时,由(2)得e x>x2>x,即x<ce x∴对任意给定的正数c,总存在x0,使得当x∈(x0,+∞)时,恒有x<ce x。

数学_2014年福建省某校高考数学模拟试卷(文科)(含答案)

数学_2014年福建省某校高考数学模拟试卷(文科)(含答案)

2014年福建省某校高考数学模拟试卷(文科)一、选择题:(本大题共12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. 计算i(1−i)2的值等于( ) A 4 B 2 C −2i D 4i2. 已知集合M ={x|y =ln(1−x)},集合N ={x|x 2−x =0},则M ∩N =( ) A {0, 1} B {0} C {1} D ⌀3. 若a ,b 为非零实数,且a <b ,则下列命题成立的是( ) A 1a >1b B a 2<b 2 C a 2b <ab 2 D a 3<b 34. 已知角α的终边与单位圆x 2+y 2=1交于P(12, y 0),则cos2α=( )A −12B 1C 12D −√325. 为了调查任教班级的作业完成的情况,将班级里的52名学生随机编号,用系统抽样的方法抽取一个容量为4的样本,已知5号、31号、44号同学在样本中,那么样本中还有一位同学的编号应该是( )A 13B 17C 18D 216. 若命题p:∃x ∈R ,x >lnx −2,命题q:∀x ∈R ,2x >1,那么( )A 命题“p 或q”为假B 命题“p 且q“为真C 命题,“¬p 或q”为假D 命题“p 且¬q“为假7. 设P 是△ABC 所在平面内的一点,BC →+BA →=2BP →,则( )A PA →+PB →=0→B PC →+PA →=0→C PB →+PC →=0→D PA →+PB →+PC →=0→8. 已知流程图如图所示,该程序运行后,为使输出的b 值为16,则循环体的判断框内①处应填( )A 2B 3C 4D 5 9. 函数f(x)=cos(πx)x 2的图象大致是( )A B C D10. 已知曲线C 1:x 23+y 2=1和C 2:x 2−y 2=1,且曲线C l 的焦点分别为F 1、F 2,点M 是C 1和C 2的一个交点,则△MF 1F 2的形状是( )A 锐角三角形B 直角三角形C 钝角三角形D 都有可能11.如图是某几何体的三视图,其中正视图、左视图均为正方形,俯视图是腰长为2的等腰三角腰形,则该几何体的体积是( ) A 83 B 83√2 C 43D 4 12. 已知函数f(x)={1−|x −1|(x ≤2)−14x 2+2x −3(x >2),如在区间(1, +∞)上存在n(n ≥1)个不同的数x 1,x 2,x 3,…,x n 使得比值f(x 1)x 1=f(x 2)x 2=...f(x n )x n成立,则n 的取值集合是( )A {1, 2, 3, 4}B {1, 2, 3}C {2, 3}D {2, 3, 4}二、填空题:(本大题共4小题,每小题4分,共16分,把答案填在相应横线上.) 13. 若等差数列{a n }的前n 项和为S n ,且S 3=15,a 1=2,则a 4=________. 14. 若不等式x 2−2ax −b 2+4≤0恰有一解,则ab 的最大值为________.15. 设点(a, b)是区域{x +y −4≤0x >0,y >0内的随机点,函数f(x)=ax 2−4bx +1在区间[1, +∞)上是增函数的概率为________.16. 设[t]表示不超过实数t 的最大整数,则在坐标平面xoy 上,满足[x]24+[y]29=1的点P(x, y)所形成的图形的面积为________.三、解答题:(本大题共6小题,共74分,解答瘦写出文字说明、证明过程或演算步骤.) 17. 为了解某校学生的视力情况,现采用随机抽样的方式从该校的A ,B 两班中各抽5名学生进行视力检测.检测的数据如下:A 班5名学生的视力检测结果:4.3,5.1,4.6,4.1,4.9.B 班5名学生的视力检测结果:5.1,4.9,4.0,4.0,4.5.(1)分别计算两组数据的平均数,从计算结果看,哪个班的学生视力较好? (2)由数据判断哪个班的5名学生视力方差较大?(结论不要求证明) (3)根据数据推断A 班全班40名学生中有几名学生的视力大于4.6?18. 如图,在平行四边形么BCD 中,∠DAB =60∘,AD =4,AB =2,将△CBD 沿BD 折起到△EBD 的位置. (1)求证:BD ⊥平面CDE ;(2)当∠CDE 取何值时,三棱锥E −ABD 的体积取最大值?并求此时三棱锥E −ABD 的侧面积.19. 某同学用“五点法”画函数f(x)=Asin(ωx +φ)+B(A >0, ω>0, |φ|<π2)在某一个周期内的图象时,列表并填入的部分数据如下表:(1)请求出上表中的x l ,x 2,x 3,并直接写出函数f(x)的解析式.(2)将f(x)的图象沿x 轴向右平移23个单位得到函数g(x),若函数g(x)在x ∈[0, m](其中m ∈(2, 4))上的值域为[−√3, √3],且此时其图象的最高点和最低点分别为P ,Q ,求OQ →与QP →夹角θ的大小.20.如图,设椭圆C:x 2a 2+y 2b 2=1(a >b >0)的左右焦点为F 1,F 2,上顶点为A ,点B ,F 2关于F 1对称,且AB ⊥AF 2 (1)求椭圆C 的离心率;(2)已知P 是过A ,B ,F 2三点的圆上的点,若△AF 1F 2的面积为√3,求点P 到直线l:x −√3y −3=0距离的最大值.21. 数列{a n }的前n 项和为S n ,a 1=1,a n+1=2S n +1,等差数列{b n }满足b 3=3,b 5=9, (1)分别求数列{a n },{b n }的通项公式;(2)若对任意的n ∈N ∗,(S n +12)⋅k ≥b n 恒成立,求实数k 的取值范围. 22. 已知函数f(x)=(x −e)(lnx −1)(e 为自然对数的底数).(I)求曲线y=f(x)在x=1处的切线方程;(II)若m是f(x)的一个极值点,且点A(x1, f(x1)),B(x2, f(x2))满足条件:(1−lnx1)(1−lnx2)=−1.①求m的值;②若点P(m, f(m)),判断A,B,P三点是否可以构成直角三角形?请说明理由.2014年福建省某校高考数学模拟试卷(文科)答案1. B2. B3. D4. A5. C6. C7. B8. B9. A10. B11. A12. B13. 1114. 215. 1316. 2(4.3+5.1+4.6+4.1+4.9)=4.6,17. 解:(1)A班5名学生的视力平均数为x A¯=15B班5名学生的视力平均数为x B¯=1(5.1+4.9+4.0+4.0+4.5)=4.5,.5从数据结果来看A班学生的视力较好.(2)B班5名学生视力的方差较大.(3)在A班抽取的5名学生中,视力大于4.6的有2名,.所以这5名学生视力大于4.6的频率为25=16名,所以全班40名学生中视力大于4.6的大约有40×25则根据数据可推断A班有16名学生视力大于4.6.18. (1)证明:在△ABD中,∵ ∠DAB=60∘,AB=2,AD=4,∴ BD =√4+16−2×2×4×12=2√3,∴ AB 2+BD 2=AD 2, ∴ AB ⊥DE ∵ AB // CD ,∴ BD ⊥CD ,BD ⊥DE ,又∵ CD ∩DE =D ,∴ BD ⊥平面CDE …(2)解:设E 点到平面ABCD 距离为ℎ,则ℎ≤ED =2. 由(1)知BD ⊥DE 当DE ⊥CD 时,∵ BD ∩CD =D ,∴ ED ⊥平面ABCD ,∴ 当∠CDE =90∘时,ℎ=ED =2,三棱锥E −ABD 的体积取最大值. 此时ED ⊥平面ABCD ,∴ ED ⊥AD 、ED ⊥BD 在Rt △DBE 中,∵ DB =2√3,DE =DC =AB =2, ∴ S △BDE =12DB ⋅DE =2√3,在Rt △ADE 中,S △ADE =12AD ⋅DE =4,∵ AB ⊥BD ,BD ⊥DE ,BD ∩DE =D ,∴ AB ⊥平面BDE ,∴ AB ⊥BE . ∵ BE =BC =AD =4, ∴ S △ABE =12AB ⋅BE =4,综上,∠CDE =90∘时,三棱锥E −ABD 体积取最大值,此时侧面积S =8+2√3.… 19. 解:(1)由题意可得ω⋅13+φ=π2,ω⋅73+φ=3π2,∴ ω=π2,φ=π3,再结合表格中的数据,可得函数f(x)=√3sin(π2x +π3).再根据π2x 1+π3=0,π2x 2+π3=π,π2x 3+π3=2π,求得x l =−23,x 2=43,x 3,=103;(2)将f(x)的图象沿x z 轴向右平移23个单位, 得到函数g(x)=√3sin[π2(x −2π3)+π3]=√3sin π2x 的图象,若函数g(x)在x ∈[0, m](其中m ∈(2, 4))上的值域为[−√3, √3], 且此时其图象的最高点和最低点分别为: P(1, √3)、Q(3, −√3),∴ OQ →=(3, −√3)、QP →=(−2, 2√3). 设OQ →与QP →夹角θ的大小为θ,则cosθ=OQ →⋅QP →|OQ →|⋅|QP →|=√12⋅√4+12=−√32, ∴ θ=5π6.20. 解:(1)由题意,B(−3c,0),AB =√9c 2+b 2,AF 2=a ,BF 2=4c…由AB ⊥AF 2及勾股定理可知AB 2+AF 22=BF 22,即9c 2+b 2+a 2=16c 2… 因为b 2=a 2−c 2,所以a 2=4c 2,解得e =c a=12…(2)由(1)可知△AF 1F 2是边长为a 的正三角形,所以S =√34a 2=√3解得a =2,c =1,b =√3…由AB ⊥AF 2可知直角三角形ABF 2的外接圆以F 1(−1, 0)为圆心,半径r =2 即点P 在圆(x +1)2+y 2=4上,…因为圆心F 1到直线l :x −√3y −3=0的距离为d =|1+3|2=2=r…故该圆与直线l 相切,所以点P 到直线l 的最大距离为2r =4… 21. 解:(1)由a n+1=2S n +1①, 得a n =2S n−1+1②,①−②得a n+1−a n =2(S n −S n−1), ∴ a n+1=3a n (n ≥2)又a 2=3,a 1=1也满足上式, ∴ a n =3n−1;∵ b 5−b 3=2d =6, ∴ d =3,∴ b n =3+(n −3)×3=3n −6; (2)S n =a 1(1−q n )1−q =1−3n 1−3=3n −12,∴ (3n −12+12)k ≥3n −6对n ∈N ∗恒成立,∴ k ≥6n−123n对n ∈N ∗恒成立,令c n =3n−63n,c n −c n−1=3n−63n−3n−93n−1=−2n+73n−1,当n ≤3时,c n >c n−1,当n ≥4时,c n <c n−1, (c n )max =c 3=19,所以实数k 的取值范围是k ≥29.22. 解:(I)f′(x)=lnx −ex ,f ′(1)=−e ,又f(1)=e −1, ∴ 曲线y =f(x)在x =1处的切线方程为y −(e −1)=−e(x −1), 即ex +y −2e +1=0.(II )①对于f′(x)=lnx −ex ,定义域为(0, +∞).当0<x <e 时,lnx <1,−e x <−1,∴ f′(x)=lnx −ex <0; 当x =e 时,f ′(x)=1−1=0;当x >e 时,lnx >1,−ex>−1,∴ f′(x)=lnx −ex>0∴ f(x)存在唯一的极值点e ,∴ m =e ,则点P 为(e, 0)②若x 1=e ,则(1−lnx 1)(1−lnx 2)=0,与条件(1−lnx 1)(1−lnx 2)=−1不符, 从而得x 1≠e .同理可得x 2≠e .若x 1=x 2,则(1−lnx 1)(1−lnx 2)=(1−lnx 1)2≥0, 与条件(1−lnx 1)(1−lnx 2)=−1不符,从而得x 1≠x 2. 由上可得点A ,B ,P 两两不重合. PA →⋅PB →=(x 1−e,f(x 1))⋅(x 2−e,f(x 2))=(x 1−e)(x 2−e)+(x 1−e)(x 2−e)(lnx 1−1)(lnx 2−1) =(x 1−e)(x 2−e)(lnx 1lnx 2−lnx 1x 2+2)=0 从而PA ⊥PB ,点A ,B ,P 可构成直角三角形.。

20132014学年第二学期龙海市高考模拟卷

20132014学年第二学期龙海市高考模拟卷

2013-2014学年第二学期龙海市高考模拟卷(考试时间:120分钟总分:150分)命题人:龙海二中陈群贤第Ⅰ卷(选择题共115分)第一部分听力(共两节, 满分30分)略第二部分英语知识运用第一节单项填空(共15小题;每小题1分,满分15分)21. — Tom, you failed the vocabulary test again.— _____? I had been working hard on it.A.How come B.So what C.Why not D.What for22. Young couples in China rushed to register their marriage on Jan 3rd, 2014, because the pronunciation of the number “201413” in Chinese _____ “love you all my life”.A. stands outB. stands forC. looks outD. looks for23. More than 400 people waited in line to buy lunch at a branch of Qing Feng Steamed Bun Shop in Beijing, _____ President Xi Jinping queued up, ordered and paid for a simple lunch a day earlier.A. whichB. whereC. whatD. that24. “Where Are We Going, Dad?”, a popular TV show _____ by Hunan Satellite TV, are popular with both parents and children.A. producedB. having been producedC. having producedD. producing25. It is beyond my understanding that many adults ______ be so crazy about Harry Potter series.A.shall B.can C.must D.should26. The change to the family planning policy, allowing some couples to have a second child, will be performed in some _____ in the first quarter of 2014.A. religionsB. regionsC. reservesD. preserves27. It makes little difference to me _____ we go or stay.A.whether B.where C.that D.how28. People can accept the fact that house prices tend to rise year by year but at a(an) _______ rate.A. modestB. sharpC. regularD. amazing29. _____ that Mike was sure to be admitted into the art school.A. So successful was his performanceB. So successful his performance wasC. So his performance was successfulD. So was his successful performance30. LeBron James was announced as The Associated Press’ 2013 male athlete of the year, _____ the 3rd basketball player after Michael Jordan and Larry Bird to win the award.A. to becomeB. becomeC. becomingD. having become31. I am blue. I met a classmate in my new school, but when I stopped and smiled, she _____ me and walked on.A. missedB. refusedC. deniedD. ignored32. Japanese Prime Minister Shinzo Abe visited the Yasukuni shrine (靖国神社) _____ strong opposition from neighboring countries.A. in terms ofB. due toC. apart fromD. in spite of33. Mum is coming. What present _____ for your birthday?A. you expect she has gotB. you expect has she gotC. do you expect she has gotD. do you expect has she got34. — Sorry, I overslept. I forgot to set the alarm.— Your clock _____. Perhaps you should buy a new one.A. never workedB. never worksC. had never workedD. is never working35. Without the help of her coach, Li Na, a 31-year-old tennis player, _____ the 2014 Australian Open.A. would winB. would have wonC. wouldn’t winD. wouldn’t have won第二节完形填空(共20 小题;每小题1.5分,满分30分)During my studying and living abroad in Britain, I was 36 of my mother's Chinese English, for which she was often treated 37 . People in department stores, at banks, and at restaurants did not give her good 38 , or take her seriously. 39 they pretended not to understand her, or even acted 40 they did not hear her.My mother has long 41 the limitations of her English as well. When I was fifteen, she 42 have me call people on the phone to 43 I was she. I was forced to ask for information or even to yell at people who had been rude to her. One time I had to call her stockbroker (股票经纪人). I said in an adolescent voice that was not very 44 , "This is Mrs. Tan." And my mother was standing beside me, 45 loudly, "Why he don't send me check, already two week late." And then, in perfect English I said, "I'm getting rather 46 . You agreed to send the check two weeks ago, but it 47 . "Then s he talked more loudly. "What h e want? I come to New York tell him front of his boss. "And so I 48 the stockbroker again, "I can't 49 any more excuse. If I don't receive the check immediately, I am going to have to speak to your manager when I am in London next week. "The next week we 50 in London. While I was sitting there 51 , my mother, the real Mrs. Tan, was shouting to his boss in her broken English.When I was a teenager, my mother's 52 English embarrassed me. But now, I see it differently. To me, my mother's English is perfectly clear, perfectly natural. It is my mother tongue. Her language, as I hear it, is vivid, direct, and full of observation and 53 . It was the language that helped 54 the way I saw things, expressed ideas, and made 55 of the world.36. A. proud B. ashamed C. afraid D. convinced37. A. anxiously B. instantly C. carefully D. unfairly38. A. checks B. attitudes C. expressions D. service39. A. otherwise B. therefore C. instead D. however40. A. even though B. as if C. now that D. if only41. A. recognized B. realized C. specialized D. organized42. A. could B. would C. might D. should43. A. pretend B. protect C. prevent D. prefer44. A. awkward B. convincing C. perfect D. useful45. A. whispering B. shouting C. talking D. crying46. A. pleased B. astonished C. interested D. concerned47. A. isn’t to arrive B. doesn’t arrive C. d idn’t arrive D. hasn’t arrived48. A. came to B. shouted to C. turned to D. referred to49. A. receive B. hold C. tolerate D. notice50. A. set off B. ended up C. settled down D. called on51. A. scared B. excited C. bored D. embarrassed52. A. broken B. perfect C. excellent D. great53. A. wisdom B. importance C. use D. value54. A. set B. choose C. shape D. build55. A. use B. fun C. sense D. comment第三部分阅读理解(共20小题;每小题2分,满分40分)AI am a mother of three girls. Getting married and having children was my childhood goal in life. By the time I had my third daughter, my middle daughter was not quite two and my oldest was three and a half. I loved them so much, yet there were some days I referred to as “too much of a good thing”.As any mother of young children knows, when the children are quiet, they’re either sleeping or in trouble. This morning, I knew that the only sleeping child at this time of day could be the baby. I was wondering whether I should enjoy the silence or ruin my peace --- and the girls’ fun.Finally, I thought I’d better go and find out what those two were doing. I could hear a noise. It was coming from the bathroom. As I got closer, I could see that the door was closed. I said to myself, “Oh no, please not the toilet again!”Sure enough, when I opened the door I saw Candace and Charity standing around the toilet, dipping their toothbrushes in --- and brushing their teeth! Not only that, but the toilet was unflushed (没冲洗的)!I hardly had any words to speak. As I snatched the toothbrushes out of their hands and flushed the toilet I shouted,“What are you doing?”Candace, my oldest daughter, answered, “We’re brushing our teeth!” They both smiled at me.Hoping this was the only time they had ever done this shocking thing, I asked, “You never did this before, did you?”Candace answered, “Oh, we do it all the time, mommy. We do it with yours and Daddy’s toothbrush, too!”The moral of the story: What you don’t know won’t hurt you; but when you do find out, it just might kill you!56. What’s the writer’s general impression of young children?A. They are polite and helpful.B. They are lovely but troublesome.C. They are clever and full of imagination.D. They are careless and make silly mistakes.57. The underlined word “moral” is closed in meaning to “_____”.A. lessonB. endingC. secretD. place58. The writer must feel worst when _____.A. she heard a noise coming from the bathroom.B. she couldn’t find her two daughters in the morning.C. she found the toilet in the bathroom had not been flushed.D. she learned her daughters had been using her toothbrush.BWhile all cultures share the same basic emotions, the body language used throughout different cultures of the world vary enormously. What can mean one thing in one country can often mean something completely different in another.For example, in North America and Europe people tend to prefer direct eye contact. But in some Asian countries longer eye contact is considered rude. So when communicating with people, always be aware of different cultural customs that may exist.A V sign in the US could mean victory, but in England, it stands for a rude challenge, which has the same meaning as showing the middle finger in the US.The OK gesture in America and England is given to mean everything is good or well. But in Latin America is looked on as a rude sign.The thumbs up sign in America and most of Europe means that something is good or well done, but it is considered rude in many Asian countries.Putting your feet on the table is generally not thought to be rude in America and England. However, in Thailand it is really rude.Telling someone to come to you by curling your index finger is acceptable in America and England, but this gesture signifies death in Singapore.Raising your hand up means stop in America and England. In some Asian countries this gesture is used when asking for permission to speak.In most westernized countries it is considered normal for two men to shake hands. In some Asian countries it is quite normal for men to kiss each other, while in most westernized countries men kissing in public would be viewed as homosexual behavior.If you would like more information on different cultural gestures all over the world, visit our website often. And we do have lots of information that interests you.59. What’s the best title for this passage?A. How gestures can vary in different cultures?B. What gestures can be acceptable in western countries?C. What can we do when traveling?D. How to be friendly to others?60. In Singapore when someone died, people there usually ________.A. put thumbs upB. curl their index fingersC. curl their middle fingersD. raise their hands up61. To what main clue is the passage written when the writer tries telling readers different cultural gestures?A. Asian countriesB. Latin AmericaC. America and EnglandD. Southern Asia62. The passage may come from ___________.A. a radio programB. a TV channelC. a magazine columnD. the InternetCShould parents ever hit their children?Research suggests many of us are likely to respond “no”, and public support for spanking(打屁股)has been fallingover the years.But surveys also show that 75 percent to nearly 90 percent of parents admit to spanking their child at least once.I was raised in a zero-tolerance home for disrespect, and my parents often turned to physical punishment.And, no, I don' t feel I was damaged by it.Nothing is more annoying than watching ill-mannered behavior from children.But there is data to suggest that a return to old-school spanking isn't the answer.Two years ago, Newsweek reported that it had found data suggesting that teens whose parents used physical punishment were more likely to become aggressive.Murray Straus, professor at the University of New Hampshire in America, has studied the topic of children and spanking for decades.He said that children who were physically punished have lower IQs than their peers.It may be that children with lower IQs were more likely to get spanked, but the punishment may have been counterproductive(反作用的)to their mental development, as well.Some researchers make the argument that occasional open-handed smacks(用巴掌打)on the bottom are not only harmless but can have some benefit.Last year, Marjorie Gunnoe, a psychologist at Calvin College, studied teens who have never spanked.There are a greater number of children growing up without ever having been physically punished.Gunnoe’s research suggests they don' t turn out any better than those who were sometimes spank.There are some parents who simply cannot control their tempers(脾气).But I still believe that the best parents are the ones who are able to offer fair and firm discipline without ever turning to physical punishment.63.According to the first three paragraphs, the author was probably hit by her parents when .A.they were dissatisfied with her grade B.she showed no respect for the elderC.they cannot control their temper D.their discipline turns out to be not strict enough 64.According to Murray Straus, children who are physically punished____ _.A.are less aggressive toward others when they get older B.have slower physical developmentC.benefit from occasional spanking D.may develop lower IQs than their peer 65.Which of the following statements is TRUE according to the article?A.40 percent of children grow up without ever being spankedB.Children who suffer less physical punishment are better studentsC.Occasional open-handed spanking on the bottom are mentally harmfulD.Researchers disagree over whether smacking is mentally harmful to children66.The author seems to agree that _ _____.A.parents should determine whether a child needs to be smacked or notB.children who have been spanked tend to behave better than those who haven'tC.good parents discipline their children in a fair and reasonable wayD.physical punishment should be the last resort(手段) of any parentDA German taxi-driver, Franz Bussman, recently found his brother who was thought to have been killed twenty years ago.While on a walking tour with his wife, he stopped to talk to a workman. After they had gone on, Mrs. Bussman said that the workman was closely like her husband and even suggested that he might be his brother. Franz laughed at the idea, pointing out that his brother had been killed in action during the war. Though Mrs. Busman knew this story quite well, she thought there was a chance in a million that she might be right.A few days later, she sent a boy to the workman to ask him if his name was Hans Bussman. Needless to say, the man’s name was Hans Bussman. And he really was Franz’s long-lost brother.When the brothers were reunited, Hans explained how it was that he was still alive.After having been wondered towards the end of the war, he had been sent to hospital and was separated from his unit(部队). The hospital had been bombed and Hans had made his way back into Western Germany on foot. Meanwhile, his unit was lost and all records of him had been destroyed. Hans returned to his home, but the house had been bombed up. Guessing that his family had all been killed during an air-raid(空袭), Hans settled down in a village fifty miles away where he had remained ever since.67. Which of the following can be used as the best title of the passage?A. Living Not FarB. Coming Back to LifeC. A Chance in a MillionD. Back after the War68. Walking along the street, _______________.k*s*5uA. Mr. Bussman recognized his brother at the first sight.B. Mr. Bussman happened to meet a workman and talked to him.C. Mrs. Bussman thought of the workman as her long-lost brother.D. Mr. and Mrs. Bussman talked to the workman for he was like his brother.69. How to understand the sentence “There was a chance in a million that she might be right.”?A. It was impossible for her to be right.B. She had no chance to meet his brother any more.C. There were many chances for her to meet his brother again.D. There was a little possibility of what she suggested, though little.70. Which of the following orders is RIGHT according to the passage?a. He walked back to Western Germany.b. He was wounded when the war was coming to the end.c. The hospital was destroyed by bombs.d. He came back to his family house.e. He was then sent to hospital.f. His unit of Germany didn’t exist any longer.A. b-a-e-d-f-cB. b-e-c-a-f-dC. b-e-a-c-d-fD. c-b-f-d-a-eEThe Nez Perce Indians are a tribe that lived in the Pacific Northwest ofthe United States. At the time of the Lewis and Clark expedition(远征队),which was one of the first journeys by Americans from the Atlantic coast tothe Pacific coast and back again, the Nez Perce territory(领土) covered about17 million acres, covering parts of Washington, Oregon, and Idaho. But thatwas a brief sweet history.In September 1805, when Lewis and Clark came off the Rockies on their westward journey, the entire exploring party was hungry and ill—too weak to defend themselves. Had the Nez Perce chosen to attack them, they could have put an end to the Lewis and Clark expedition there on the banks of Clearwater River. Instead the Nez Perce welcomed the white Americans and looked after them until theymade a full recovery.Thus began a long friendship between the Nez Perce and white Americans. But white men’s greed for land and go ld finally broke the friendship.In 1855 Governor Isaac Stevens of Washington Territory invited the Nez Perce to a peace conference. He said there were a great many white people in the country, and many more would come. But the chief of the tribe at that time, Old Joseph, replied, “Take away your paper. I will not touch it with my hand.”Things were quiet for a while after that, but not for long. When Old Joseph died, the chiefship was passed onto Young Joseph. In the late 1870s, government officials came to order the Nez Perce to leave the Wallowa Valley and then began hard battles between the Nez Perce and the white soldiers. After the tribe had fought thirteen battles and moved 1,600 miles towards Canada in an attempt to retreat (撤退) north, Young Joseph, gave in to the United States Army. Here was his famous statement, “Hear me my chiefs. I am tired; my heart is sick and sad. From where the sun now stands I will fight no more forever.”In 1885, Chief Joseph was sent along with many of his band to the Colville Reservation in Washington where Joseph continued to lead his band for another 25 years, at times coming into conflict with the leaders of 11 other tribes living on the reservation.71. What is TRUE of the Lewis and Clark expedition?A. They were well cared for by the Nez Perce.B. They ended their expedition on the banks of Clearwater River.C. They started the first battle with the Nez Perce.D. They were the first Americans to travel from the Atlantic to the Pacific.72. What ended the friendship between the Nez Perce and the Whites?A. Old Joseph’s proud manner.B. The breakdown of the peace talk.C. A growing number of white men in the land.D. W hite men’s increasing demand for land and gold.73. Young Joseph gave in at last because ________.A. he grew olderB. he was terribly illC. he hated the warD. he lost other chiefs’ support74. Which map most probably shows the path the Nez Perce took in the late 1870s?A BCD75. The passage might be followed by a paragraph about ________.A. the customs and traditions of the Nez Perce IndiansB. the last years of Chief Joseph in the Colville ReservationC. lasting fights between the Nez Perce and the whitesD. constant conflicts between the Nez Perce and other tribes第二卷(非选择题共35分)第四部分写作(共两节,满分35分)第一节短文填词(共10小题;每小题1分,满分10分)阅读下面短文,根据以下提示:1)汉语提示,2)首字母提示,3)语境提示,在每个空格内填入一个适当的英语单词,所填单词要求意义准确,拼写正确,并将该词完整地写在答题卡中相应的横线上。

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2014届龙海二中高三第二学期期初数学(文科)高考模拟考试卷(考试时间:120分钟 总分:150分)命题人:龙海二中 姚跃宣一、选择题(每小题5分,共60分)1.设全集{1,3,5,7}U =,集合{3,5}A =,{1,3,7}B =,则u A (C B)⋂=u ( ) A .{5} B .{3,5} C .{1,5,7} D .∅ 2.命题:x R ∀∈,220x x -+≥的否定是( )A .x R ∃∈,220x x -+≥ B .x R ∃∈,220x x -+< C .x R ∀∈,220x x -+> D .x R ∀∈,220x x -+< 3.已知复数1Z i =+,则2Z=( ) A .2i - B .2i C .1i - D .1i +4.如图,在一个边长为2的正方形中随机撒入200粒豆子,恰有120粒落在阴影区域内,则该阴影部分的面积约为( ) A .35 B .125C .65D .185 5.已知等比数列{}n a 的公比为正数,且23952a a a ⋅=,21a =,则1a =( )A .12 B .2 C .2 D .226.直线3230x y +-=截圆224x y +=得到的劣弧所对圆心角等于( )A .6π B .4π C .3πD .2π7.已知m ,n 是平面α外的两条直线,且//m n ,则“//m α”是“//n α”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 8.平面向量a 与b 的夹角为60,(2,0)a =,||1b =,则|2|a b +=( ) A .3 B .23 C .4 D .129.在某五场篮球比赛中,甲、乙两名运动员得分的茎叶图如右。

下列说法正确的是( )A .在这五场比赛中,甲的平均得分比乙好,且甲比乙稳定B .在这五场比赛中,甲的平均得分比乙好,但乙比甲稳定C .在这五场比赛中,乙的平均得分比甲好,且乙比甲稳定D .在这五场比赛中,乙的平均得分比甲好,但甲比乙稳定 10.某程序框图如图所示,该程序运行后输出的k 的值是( )A .4B .5C .6D .7 11.将函数)46sin(π+=x y 的图像上各点向右平移8π个单位,则得到新函数的解析式为( )A .cos 6y x =-B .cos6y x =C .)856sin(π+=x y D .)86sin(π+=x y 12.已知函数()f x 是R 上的偶函数,且(1)(1)f x f x -=+,当[0,1]x ∈时,2()f x x =,则函数7()log y f x x =-的零点个数为( )A .3B .4C .5D .6 二、填空题(每小题4分,共16分)13.如图,一个空间几何体的正视图和侧视图都是边长为1的正方形,俯视图是直径为1的圆,那么该几何体的侧面积为 。

14.若实数x 、y 满足2045x y x y +-≥⎧⎪≤⎨⎪≤⎩,则z x y =+的最大值为 。

15.若椭圆2214x y m +=的离心率等于32,则m = 。

16.观察下列两等式的规律,请写出一个(包含下面两命题)一般性的命题: ① 4330sin 30sin 30sin 30sin 022=⋅++; ② 4320sin 40sin 20sin 40sin 000202=⋅++。

三、解答题(共74分)17.(本题满分12分)已知ABC ∆的内角A 、B 、C 所对的边分别为,,a b c ,且2a =,3cos 5B =。

(1) 若4b =,求cos 2A 的值。

(2)若ABC ∆的面积4S =,求,b c 的值。

18.(本题满分12分)等比数列{n a }的前n 项和为n s ,已知1S ,3S ,2S 成等差数列正 侧俯(1) 求{n a }的公比q ;(2)若1a -3a =3,求n s 。

19.(本题满分12分)为了了解某市开展群众体育活动的情况,拟采用分层抽样的方法从,,A B C 三个区中抽取7个工厂进行调查,已知,,A B C 区中分别有18,27,18个工厂。

(1)求从,,A B C 区中应分别抽取的工厂个数;(2)若从抽得的7个工厂中随机地抽取2个进行调查结果的对比,请计算这2个工厂中至少有一个来自A 区的概率。

20.(本小题满分12分)如图,在直角梯形ABCD 中,,,,,24//90===︒=∠CD AD AB AB CD ADC 将ADC ∆沿AC 折起,使平面ADC ⊥平面ABC ,得到几何体D -ABC 。

(Ⅰ)求证:BC ⊥平面ACD ; (Ⅱ)求几何体D -ABC 的体积21.(本小题满分12分)已知椭圆C 的焦点在x 轴上,它的一个顶点恰好是抛物线y =41x 2的焦点,离心率为552。

(Ⅰ)求椭圆C 的方程;(Ⅱ)过椭圆C 的右焦点作直线l 交椭圆C 于A 、B 两点,交y 轴于点M ,若MA =λ1AF ,MB =2λBF ,求证:λ1+2λ为定值22.(本小题满分14分)已知函数23)(ax x f =图象上斜率为3的两条切线间的距离为,5102函数.33)()(2+-=abxx f x g (Ⅰ)若函数)(,1)(x g x x g 求处有极值在=的解析式; (Ⅱ)若函数]1,1[)(-在区间x g 上为增函数,且]1,1[)(42-≥+-在区间x g mb b 上都成立,求实数m 的取值范围..AB CMD DC ABM参考答案一、选择题(每小题5分,共60分)二、填空题(每小题4分,共16分)13.π 14.9 15.1或16 16.43)60sin(sin )60(sin sin 0022=-⋅+-+θθθθ 三、解答题(共74分) 17.∴217cos 212sin 25A A =-=18.(Ⅰ)依题意有)(2)(2111111q a q a a q a a a ++=++由于 01≠a ,故022=+q q ,又0≠q ,从而21-=q …… 6分 (Ⅱ)由已知可得321211=--)(a a ,故41=a从而))(()())((n nn211382112114--=----=S …………………………12分 19.(1)解: 工厂总数为18+27+18=63,样本容量与总体中的个体数比为71639= ,所以从A,B,C 三个区中应分别抽取的工厂个数为2,3,2.(2)设1A ,2A 为在A 区中抽得的2个工厂,1B ,2B ,3B 为在B 区中抽得的3个工厂,1C ,2C 为在C 区中抽得的2个工厂,这7个工厂中随机的抽取2个,全部的可能结果有21种,随机的抽取的2个工厂至少有一个来自A 区的结果有 11种。

所以所求的概率为1121。

题号 1 2 3 4 5 6 7 8 9 10 11 12 答案 A B C B D C C B C AA D20.证明:(Ⅰ)由于AC =BC =22,从而,222AB BC AC =+故AC ⊥BC ,………3分 又平面ADC ⊥平面ABC ,平面ADC 平面ABC =AC ,BC ⊂平面ABC,从而BC ⊥平面ACD 。

………………………………………6分 (Ⅱ)由(1)知,BC 为三棱锥B -ACD 的高,BC =22,…………8分,2=∆ACD S (10)所以=-ABC D V 3242223131=⨯⨯==-Sh V ACD B .………………………12分 21、解:(Ⅰ)设椭圆C 的方程为22a x +22by =1(a>b>0),则由题意知b =1,∴222ab a -=552,即211a -=552∴a 2=5 ………3分 ∴椭圆的方程为52x +y 2=1 ……………………………4分 (Ⅱ)设A 、B 、M 的坐标分别为(x 1,y 1),(x 2,y 2),(0,y 0),易知点F 的坐标为(2,0) MA =λ1AF ∴(x 1,y 1—y 0)=λ1(2—x 1,-y 1)……………………6分 ∴ x 1=1112λλ+ , y 1=101λ+y…………………7分将A (x 1,y 1)坐标代入椭圆方程得51⎪⎪⎭⎫ ⎝⎛+1112λλ2+⎪⎪⎭⎫ ⎝⎛+101λy 2=1…………8分整理得:λ12+10λ1+5—5 y 02=0 同理由MB =2λBF得:λ22+10λ2+5—5 y 02=0 ………………………………10分∴λ1、λ2是方程x 2+10x +5—5 y 02=0的两根,得λ1+2λ=-10………12分22、解:(Ⅰ)223)(x a x f ⋅=' 3322=⋅∴x a解得a x ±=, 即切点坐标为),(),,(a a a a -- ……………………………2分∴切线方程为)(3),(3a x a y a x a y +=+-=-或023023=+-=--a y x a y x 或 …………………………4分5102)1(3|22|22=-+--∴a a ,解得,1±=a 3)(x x f =∴ 33)(3+-=∴bx x x g ………………6分(Ⅱ)1)(,33)(2=-='x x g b x x g 在 处有极值,,0)1(='∴g即03132=-⨯b , 解得1=b ………………………8分]1,1[)(-在函数x g 上递增,033)(2≥-='∴b x x g 在区间[-1,1]上恒成立,,0≤∴b ………………………10分又]1,1[)(42-≥+-在区间x g mb b 上恒成立,),1(42g mb b ≥+-∴ …………………………12分即b mb b 3442-≥+-,]0,(3-∞∈+≥∴b b m 在上恒成立,3≥∴m ),3[+∞∴的取值范围是m ……………………………14分2014届龙海二中高三第二学期期初数学(文科)考试卷(考试时间:120分钟 总分:150分)命题人:龙海二中 姚跃宣一、选择题:本大题共12小题,每小题5分,共60分二、填空题:本大题共4小题,每小题4分,共16分13、 14、 15、 16、三、解答题:本大题共6小题,共80分.解答应写出文字说明,证明过程或演算步骤。

题号 1 2 3 4 56789101112答案17、(本小题满分12分)18、(本小题满分12分)19、(本小题满分12分)20、(本小题满分12分)21、(本小题满分12分)..A BCMD DCA BM22、(本小题满分14分)。

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