福建省三明一中2011届高三第三次月考数学(理)
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2011年三明市区高三毕业班三校联考试卷文科数学(考试时间:120分钟;试卷满分:150分)参考公式:柱体体积公式V Sh = 锥体体积公式13V Sh =(其中S 为底面面积,h 为高)球体表面积、体积公式:2344,3S R V Rππ==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,只有一项是符合题目要求的.请在答题卷相应题目的答题区域内作答1.已知全集{1,2,3,4,5,6},{1,2,5},{4,5,6},UU A C B A===集合则B=A .{1,2}B .{5}C .{1,2,3}D .{3,4,6}2、已知复数2121,21,3z z i z bi z若-=-=是实数,则实数b 的值为A .6-B .0C .61 D .6 3、已知命题p :01,2>+-∈∀x x R x ,则:p ⌝A . 01,2≤+-∈∃x x R xB .01,2≤+-∈∀x xR x C .01,2>+-∈∃x x R xD .01,2≥+-∈∀x xR x4、已知锐角△ABC 的面积为33,BC =4,CA =3, 则角C 的大小为A. 75° B 。
60° C. 45° D. 30°5、已知直线l 经过坐标原点,且与圆22430x y x +-+=相切,切点在第四象限,则直线l 的 方程为A .3y x =-B .3y x =C .33y x =-D .33y x =6、若函数b ax x f +=)(的零点为2,那么函数ax bx x g -=2)(的零点是A .0,2B .0,21 C .0,21- D .2,217、如果执行右面的程序框图,输入6,4n m ==,那么输出的p 等于A .720B .360C .240D .1208、半圆的直径AB =4, O 为圆心,C 是半圆上不同于A 、B 的任意一点,若P 为半径OC 的中点,则PC PB PA •+)(的值是A. -2 B . -1 C . 2 D 。
等差数列、等比数列的概念及求和题组一一、选择题1.(浙江省杭州二中2011届高三11月月考试题文)已知数列{}n a 中,111,34(*2)n n a a a n N n -==+∈≥且,则数列{}n a 通项公式n a 为 ( )A .13n -B .138n +-C .32n -D .3n答案 C.2. (甘肃省甘谷三中2011届高三第三次检测试题)已知等差数列}{n a 的前n 项和为n S ,若45818,a a S =-=则( )A .18 B. 36 C. 54 D. 72 答案 D.3. (福建省安溪梧桐中学2011届高三第三次阶段考试理)已知公差不为0的等差数列{}n a 满足134,,a a a 成等比数列,n n S 为{a }的前n 项和,则3253S S S S --的值为( )A .2B .3C .15D .4答案 A.4.(福建省三明一中2011届高三上学期第三次月考理)数列{}n a 是公差不为0的等差数列,且137,,a a a 为等比数列{}n b 的连续三项,则数列{}n b 的公比为( )A .B .4C .2D .12答案 C.5 . (福建省四地六校2011届高三上学期第三次联考试题理)已知数列{a n }的通项公式为2245n a n n =-+ 则{a n }的最大项是( ) A .a 1B .a 2C .a 3D .a 4答案 B. 6.(浙江省杭州二中2011届高三11月月考试题文)等比数列{}n a 中,372,8,a a == 则5a =( )A .4±B .4C .6D .4-答案 B.7.(福建省厦门外国语学校2011届高三11月月考理) 已知等差数列{}n a 的公差为2-,且245,,a a a 成等比数列,则2a 等于( )A .-4B .-6 cC .-8D .8 答案 D.8.(浙江省温州市啸秋中学2010学年第一学期高三会考模拟试卷)已知数列{a n }的前n 项和S n =312n a n +=+,则A .201 B .241 C .281 D .321答案 A.9. (广东省华附、中山附中2011届高三11月月考理)已知等差数列{}n a 的前n 项和为n S ,且2510,55S S ==,则过点(,)n P n a 和2(2,)n Q n a ++(n ÎN *)的直线的斜率是A .4B .3C .2D .1答案 A.10. (甘肃省甘谷三中2011届高三第三次检测试题)设{}n a 是公差为正数的等差数列,若12315a a a ++=,12380a a a =,则111213a a a ++=( )A .120B .105C .90D .75答案 B.11.(北京四中2011届高三上学期开学测试理科试题)已知等差数列的前项和为,若,且A 、B 、C 三点共线(该直线不过 原点),则=( )A .100 B. 101 C. 200 D. 201 答案 A.12. (贵州省遵义四中2011届高三第四次月考理)在等差数列{}n a 中,351024a a a ++=,则此数列的前13项的和等于( )A .13B .26C .8D .16答案 A.13. (河南省郑州市四十七中2011届高三第三次月考文)在等比数列{}n a 中,已知13118a a a =,那么28a a =(A )3 (B )4 (C )12 (D )16 答案 B.14.(黑龙江大庆实验中学2011届高三上学期期中考试理)若一个等差数列前3项的和为34,最后3项的和为146,且所有项的和为390,则这个数列有( ).13A 项.12B 项 .11C 项 .10D 项答案 A.15.(浙江省杭州二中2011届高三11月月考试题文)已知等差数列{}n a 的前n 项和为n S ,且3711315a a a ++=,则13S =( )A . 104B . 78C . 52D . 39答案 D.16.(福建省厦门双十中学2011届高三12月月考题理)如果数列103*,8,,)}({a a a a a N n m R a a n m n m n n 那么且满足对任意=⋅=∈∈+等于( ) A .256B .510C .512D . 1024答案 D.17.(北京龙门育才学校2011届高三上学期第三次月考)(理科)已知数列{}n a 满足1133,2,+-==n na a a n则n a n的最小值为 ( )A .10B .10.5C .9D .8答案 B.18.(重庆市重庆八中2011届高三第四次月考理)等差数列{}n a 满足:296a a a +=,则9S = ( )A .2-B .0C .1D .2答案 B.19.(重庆市南开中学高2011级高三1月月考理)在数列{}n a 中,*111001,,(),n n a a a n n N a +=-=∈则的值为( )A .55050B .5051C .4950D .4951答案 D.20.(浙江省诸暨中学2011届高三12月月考试题文)在等差数列{a n }中,a 1+3a 8+a 15=120,则2a 6-a 4的值为A .24B .22C .20D .-8 答案 A.21.(浙江省温州市啸秋中学2010学年第一学期高三会考模拟试卷)若{a n }为等差数列,且a 2+a 5+a 8=39,则a 1+a 2+…+a 9的值为A .117B .114C .111D .108 答案 A.22. (甘肃省甘谷三中2011届高三第三次检测试题)已知a b c d ,,,成等比数列,且曲线223y x x =-+的顶点是()b c ,,则a d 等于( )A.3 B.2 C.1 D.2-答案 B.23.(浙江省温州市啸秋中学2010学年第一学期高三会考模拟试卷)数列{}n a 满足⎪⎪⎩⎪⎪⎨⎧<≤-<≤=+)121(12)210(21n n n n n a a a a a若761=a ,则=8aA .76 B .75 C .73 D .71答案 B. 二、填空题24.(浙江省杭州二中2011届高三11月月考试题文)已知等差数列{}n a 的前n 项和为n S ,且13140,0,S S ><若10t t a a +<则t = . 答案:7.25.(福建省厦门外国语学校2011届高三11月月考理)已知等比数列{}n a 各项均为正数,前n 项和为n S ,若22a =,1516a a =.则5S =▲▲. 答案 31. 三、简答题26.(浙江省温州市啸秋中学2010学年第一学期高三会考模拟试卷) 已知{}n a 为等比数列,且364736,18.a a a a +=+=(1)若12n a =,求n ;(2)设数列{}n a 的前n 项和为n S ,求8S .答案 解:设11n n a a q-=,由题意,解之得112812a q =⎧⎪⎨=⎪⎩,进而11128()2n n a -=⋅ (1)由111128()22n n a -=⋅=,解得9.n = ………3分(2)1(1)1256[1()]12nnn a q S q-==--881256[1()]255.2S ∴=-= ………3分27.(浙江省诸暨中学2011届高三12月月考试题文)(本小题满分14分)已知数列{}n a 是公比为d )1(≠d 的等比数列,且231,,a a a 成等差数列. (Ⅰ) 求d 的值;(Ⅱ) 设数列{}n b 是以2为首项,d 为公差的等差数列,其前n 项和为n S ,试比较n S 与n b 的大小.答案 (Ⅰ) 解:012,2,221121213=--∴+=∴+=d d d a a d a a a a21,1-=∴≠d d(Ⅱ) 解:,25221)1(2+-=⎪⎭⎫ ⎝⎛-⋅-+=n n b n ,492)(21nn b b n S n n +-=+=4)10)(1()252(492---=+--+-=-∴n n n nn b S n n;101n n b S n n ===∴时,或 ;,92n n b S n >≤≤时n n b S ,n <≥时11.28.(重庆市南开中学高2011级高三1月月考理)(13分)已知数列{}n a 是公比大于1的等比数列,n S 为数列{}n a 的前n 项和,37S =,且1233,3,4a a a ++成等差数列。
随机抽样及方法一、考点解读:(1)理解随机抽样的必要性和重要性;(2)会用简单随机抽样方法从总体中抽取样本;了解分层抽样和系统抽样方法。
二、知识清单:四、真题剖析:【例1】(2010湖北,6题,5分)将参加夏令营的600名学生编号为:001,002,……600,采用系统抽样方法抽取一个容量为50的样本,且随机抽得的号码为003.这600名学生分住在三个营区,从001到300在第Ⅰ营区,从301到495住在第Ⅱ营区,从496到600在第Ⅲ营区,三个营区被抽中的人数一次为()A.26, 16, 8, B.25,17,8 C.25,16,9 D.24,17,9【答案】B【解析】依题意可知,在随机抽样中,首次抽到003号,以后每隔12个号抽到一个人,则分别是003、015、027、039……构成以3为首项,12为公差的等差数列,故可分别求出在001到300中有25人,在301至495号中共有17人,则496到600中有8人,所以B正确【丢分陷阱】:本题主要考查系统抽样方法.并能转化等差数列的思想【例2】(2010安徽,14题,5分)某地有居民100 000户,其中普通家庭99 000户,高收入家庭1 000户.从普通家庭中以简单随机抽样方式抽取990户,从高收入家庭中以简单随机抽样方式抽取l00户进行调查,发现共有120户家庭拥有3套或3套以上住房,其中普通家庭50户,高收人家庭70户.依据这些数据并结合所掌握的统计知识,你认为该地拥有3套或3套以上住房的家庭所占比例的合理估计是________ .【答案】5.7%【解析】该地拥有3套或3套以上住房的家庭可以估计有:户,所以所占比例的合理估计是÷=.5700100000 5.7%【丢分陷阱】:利用题目给出的比例采用分层抽样模型,【方法总结】:本题分层抽样问题,首先根据拥有3套或3套以上住房的家庭所占的比例,得出100 000户,居民中拥有3套或3套以上住房的户数,它除以100 000得到的值,为该地拥有3套或3套以上住房的家庭所占比例的合理估计. 【例3】(2010北京,题11,5分)(11)从某小学随机抽取100名同学,将他们的身高(单位:厘米)数据绘制成频率分布直方图(如图)。
2011届高三年级第三次月考数学试卷一、选择题(10×5=50分) 1、0sin(330)-的值为( ) A .12B .-12CD .2、若34sin ,cos 55θθ==-,则2θ所在象限是( ) A .一B .二C .三D .四3、如图中的图象所表示的函数的解析式为( )A .3|1|(02)2y x x =-≤≤B .33|1|(02)22y x x =--≤≤C .3|1|(02)2y x x =--≤≤D .1|1|(02)y x x =--≤≤4、函数()y f x =图象如图所示,则函数12log ()y f x = 图象大致是( )5、函数32()ln 2f x xπ=-的零点一定位于区间( ) A .(1,2)B .(2,3)C .(3,4) D .(4,5)6、直线1ln()y x y x a =+=+与曲线相切,则a 的值为( ) A .1B .2C .-1D .-27、已知1sin 2sin ,'2y x x y =+则是( ) A .仅有最小值的奇函数 B .既有最大值又有最小值的偶函数 C .仅有最大值的偶函数D .既不是奇函数也不是偶函数8、函数3()1f x ax x =++有极值的充要条件是( ) A .0a >B .0a ≥C .0a <D .0a ≤AB C D9、函数32()6f x ax ax b =-+在[-1,2]上最大值为3,最小值为-29(a>0),则( ) A .a=2,b=-29B .a-3, b=2C .a=2, b=3D .以上都不对10、函数21()ln 22f x x ax x =--存在单调递减区间,则a 的取值范围是( ) A .(1,)-+∞ B .[0,1) C .(-1,0]D .(,)-∞+∞二、填空题(6×4=24分)11、设230.311331log ,log ,(),,,2a b c a b c ===则大小关系为 。
直线和圆题组一一、选择题1.(北京龙门育才学校2011届高三上学期第三次月考)直线x-y+1=0与圆(x+1)2+y 2=1的位置关系是( ) A .相切 B .直线过圆心 C .直线不过圆心但与圆相交 D .相离 答案 B.2.(北京五中2011届高三上学期期中考试试题理)若过定点)0,1(-M 且斜率为k 的直线与圆05422=-++y x x 在第一象限内的部分有交点,则k 的取值范围是( ))(A 50<<k )(B 05<<-k )(C 130<<k )(D 50<<k答案 A.3、(福建省三明一中2011届高三上学期第三次月考理)两圆042222=-+++a ax y x 和0414222=+--+b by y x 恰有三条公切线,若R b R a ∈∈,,且0≠ab ,则2211b a +的最小值为 ( )A .91B .94C .1D .3答案 C.3.(福建省厦门双十中学2011届高三12月月考题理)已知点P 是曲线C:321y x x =++上的一点,过点P 与此曲线相切的直线l 平行于直线23y x =-,则切线l 的方程是( ) A .12+=x y B .y=121+-xC .2y x =D .21y x =+或2y x =答案 A.4. (福建省厦门双十中学2011届高三12月月考题理)设斜率为1的直线l 与椭圆124:22=+y x C 相交于不同的两点A 、B ,则使||AB 为整数的直线l 共有( ) A .4条 B .5条 C .6条 D .7条 答案 C.5.(福建省厦门外国语学校2011届高三11月月考理) 已知圆22670x y x +--=与抛物线22(0)y px p =>的准线相切,则p = ( ▲ )A 、1B 、2C 、3D 、4答案 B.6.(甘肃省天水一中2011届高三上学期第三次月考试题理)过点M(1,5)-作圆22(1)(2)4x y -+-=的切线,则切线方程为( ) A .1x =-B .512550x y +-=C .1512550x x y =-+-=或D .15550x x y =-+-=或12答案 C.7.(甘肃省天水一中2011届高三上学期第三次月考试题理)已知圆222410x y x y ++-+=关于直线220ax by -+=41(0,0),a b a b>>+对称则的最小值是( )A .4B .6C .8D .9答案 D.8.(广东省惠州三中2011届高三上学期第三次考试理)已知直线x y a +=与圆224x y +=交于A 、B 两点,O 是坐标原点,向量OA 、OB满足||||OA OB OA OB +=-,则实数a 的值是( )(A )2 (B )2- (C 或 (D )2或2- 答案 D.9. (广东省清远市清城区2011届高三第一次模拟考试理)曲线321y x x x =-=-在处的切线方程为( A .20x y -+= B .20x y +-= C . 20x y ++= D .20x y --=答案 C.10.(贵州省遵义四中2011届高三第四次月考理)若直线02=+-c y x 按向量)1,1(-=a 平移后与圆522=+y x 相切,则c 的值为( )A .8或-2B .6或-4C .4或-6D .2或-8邪恶少女漫画/wuyiniao/ 奀莒哂答案 A.11.(黑龙江大庆实验中学2011届高三上学期期中考试理) 若直线y x =是曲线322y x x ax =-+的切线,则a =( ).1A .2B .1C - .1D 或2 答案 D.邪恶少女漫画/wuyiniao/ 奀莒哂12.(黑龙江哈九中2011届高三12月月考理)“3=a ”是“直线012=--y ax ”与“直线046=+-c y x 平行”的 ( )A .充分不必要条件 C .必要不充分条件D .充要条件D .既不充分也不必要条件答案 B.13.(湖北省南漳县一中2010年高三第四次月考文)已知α∥β,a ⊂α,B ∈β,则在β内过点B 的所有直线中A .不一定存在与a 平行的直线B .只有两条与a 平行的直线C .存在无数条与a 平行的直线D .存在唯一一条与a 平行的直线 答案 D.14.(重庆市南开中学2011届高三12月月考文)已知圆C 与直线040x y x y -=--=及都相切,圆心在直线0x y +=上,则圆C 的方程为( )A .22(1)(1)2x y ++-=B .22(1)(1)2x y -++=C .22(1)(1)2x y -+-=D .22(1)(1)2x y +++=答案 B. 二、填空题14.(湖北省南漳县一中2010年高三第四次月考文)已知两点(4,9)(2,3)P Q --,,则直线PQ 与y 轴的交点分有向线段PQ的比为 .答案 2.15. (福建省厦门外国语学校2011届高三11月月考理)已知椭圆的中心为坐标原点O ,焦点在x 轴上,斜率为1且过椭圆右焦点的直线交椭圆于A 、B 两点,)1,3(-=+与共线,求椭圆的离心率▲▲.答案 36=e . 16.(甘肃省天水一中2011届高三上学期第三次月考试题理)设直线30ax y -+=与圆22(1)(2)4x y -+-=相交于A 、B 两点,且弦AB 的长为a = 答案 0.17. (广东省中山市桂山中学2011届高三第二次模拟考试文) 在极坐标中,圆4cos ρθ=的圆心C 到直线sin()4πρθ+=的距离为 .18.(河南省郑州市四十七中2011届高三第三次月考文)如下图,直线PC 与圆O 相切于点C ,割线PAB 经过圆心O ,弦CD ⊥AB 于点E , 4PC =,8PB =,则CE = .答案12519.(黑龙江省哈尔滨市第162中学2011届高三第三次模拟理)已知函数()x f 的图象关于直线2=x 和4=x 都对称,且当10≤≤x 时,()x x f =.求()5.19f =_____________。
三明一中2010~2011上学期学段考试卷高三(理)科数学 (总分150分,时间:120分钟)(注意:请将所有题目的解答都写到“答题卷”上)一、选择题(本题10小题,每小题5分,共50分。
每小题只有一个选项符合题意,请将正确答案填入答题卷中。
)1.设z =1+i (i 是虚数单位),则 错误! + z 2 = ( )A .-1-iB .-1+iC .1-iD .1+i2.设A ={x |x 2-2x -3>0},B ={x |x 2+ax +b ≤0},若 A ∪B =R ,A ∩B =(3,4],则 a +b 等于 ( ) A .7B .-1C .1D .-73.设()f x 为定义在R 上的奇函数,当0x ≥时,()22xf x x b =++ (b 为常数),则(1)f -=( ) (A) 3 (B ) 1 (C )-3 (D )1-4。
如图,在底面ABCD 为平行四边形的四棱柱ABCD -A 1B 1C 1D 1中,M 是AC 与BD 的交点,若AB =a ,11A D =b ,11A A =c ,则下列向量中与1B M 相等的向量是 ( )A .-错误!a +错误!b +cB 。
错误! a +错误!b +cC.错误! a -错误!b +c D .-错误!a -错误!b +c5.已知{}na 为等差数列,135105aa a ++=,24699a a a ++=.以n S 表示{}n a 的前n项和,则使得nS 达到最大值的n 是 ( ) (A )21 (B )20 (C )19 (D )186.已知直线12:(3)(4)10,:2(3)230,l k x k y lk x y -+-+=--+=与平行,则k 的值是( )(A ) 1或3 (B)1或5 (C)3或5 (D)1或27.设,x y 满足约束条件360,20,0,0,x y x y x y --⎧⎪-+⎨⎪⎩若目标函数(0,z ax by a b =+>>0)的最大值为12,则32ab +的最小值为 ( ) (A )256 (B ) 83(C ) 113(D ) 48。
2011年三明市区高三毕业班三校联考试卷英语试题第一卷(共115分)I. 听力理解(共二节,满分30分)II.单项填空(共15小题;每小题1分,满分15分)21. ---I'm so sorry that I broke your vase.---Oh, really? ________.A. All right.B. It doesn't matterC. It's right with meD. I don't care about it22. Jack is ____ honest and hard-working boy and he must be _____ useful man in the future.A. a; aB. an; anC. an; aD. a; an23. ____ , the boys were shouting and singing.A. Happy and excitedB. Happily and excitedC. Happily and excitedlyD. Happy and excitedly24.Barack Obama delivered a speech to 500 local youths during his visit to China, many of ____ from Fudan University and Tongji University.A. whomB. whichC. whoD. them25. He______ good ideas for the product promotionA. came upB. came outC. came up withD. thought over 26.Exercise One should be done in class as an example, while the rest _____ as homework.A. is to be finishedB. are to finishC. are to be finishedD. is to finish27. Mr. Li suggested that we ______over the texts carefully before the coming exam.A. wentB. need goC. goD. must go28. Lucy has got into a situation in her study ______ she finds it hard to go further.A. thatB. whereC. whichD. at which29. ---Sorry, the tickets have already been sold out.--- Really? Maybe I ________a little earlier.A. should comeB. should have comeC. could have comeD. must have come30.. –--Excuse me, but would you mind if I smoke here? —__________.A.Y es, I like the smell of the cigarette.B. No, you can’t do that.C.Of course not. It’s not allowed here.D. I’d rather you didn’t, actually.31. The fog was so heavy this morning that drivers could hardly ________ the things just tenmeters away from them.A. figure outB. make outC. look outD. rule out32. Not only I but also Lilei and Han Meimei ________tired of so much homework everyday.A. amB. isC. areD. was33. _______ many times, he finally managed to settle the problem.A. TriedB. TryingC. Having been triedD. Having tried34. Only by introducing scientific management, I suppose, _____our company to a bright futureand make more profits in the next years.A. can we bringB. we have broughtC. we can bringD. have webrought35. ________ didn’t attend the lecture yesterday won’t be given full marks.A. Any oneB. WhoeverC. WhoD. The person III.完形填空(共20小题; 每小题1.5分, 满分30分)Recently, my husband and I had the opportunity to do something good for two people who were complete strangers to us. It made us feel so good to be able to do it that I thought I should 36 . We were traveling down to Mexico for Thanksgiving week with our family. 37 we were sitting in the 38 first two seats in the first row of first class waiting for the plane to take off, I 39 heard one of the flight attendants(空乘) telling 40 that there was a couple sitting in the 41 of the plane who had just been married the previous day and were 42 off on their honeymoon. They had 43 their flight eight months in 44 but, owing to some schedule changes, had not been able to get 45 together, and no one else on the plane was 46 to move and the bride was in 47 .I turned to my husband and told him what I heard. We 48 agreed and I called the flight attendant 49 to tell her that we’d be happy to give up our seats to this couple. The flight attendant seemed amazed and said ―Really? Are you sure?‖ We said ―Absolutely !‖So, we 50 to the back of the plane in separate seats. The flight attendants were extremely 51 and took good care of us even though we were no longer sitting in first 52 , and we both made friends with the people sitting around us, who, as it 53 didn’t know why the young woman was crying or that they were newly married, and had a great flight. I had to go back to the front of the plane to 54 up a forgotten item at one point during the flight and 55 that the couple were sitting very close together, happily enjoying champagne(香槟酒). It really made my day and Thanksgiving week get off to such a wonderful start !36. A. share B. show C. say D. prove37. A. Since B. After C. Before D. Though38. A. quite B. same C. very D. just39. A. carelessly B. accidentally C. happily D. sadly40. A. others B. us C. them D. everyone41. A. front B. middle C. head D. back42. A. getting B. heading C. taking D. hurrying43. A. fixed B. booked C. checked D. got44. A. air B. time C. church D. advance45. A. seats B. champagne C. food D. tickets46. A. interested B. willing C. eager D. friendly47. A. comfort B. surprise C. tears D. fear48. A. immediately B. differently C. finally D. unbelievably49. A. up B. off C. over D. out50. A. left B. looked C. returned D. moved51. A. beautiful B. careful C. thoughtful D. grateful52. A. class B. flight C. seat D. plane53. A. went on B. turned out C. got along D. took off54. A. pick B. give C. clean D. look55. A. told B. informed C. observed D. heardIV.阅读理解(共20小题;每小题2分,满分40分)AIt is time for students to sell such things as chocolate bars and greeting cards to raise money for their school, class or club .It is inevitable that they will knock on your door and you will easily hand over your cash for overpriced items that you really do not want .That is okay, though, because there are many reasons why children should be allowed to raise money for their schools and clubs .Fundraising (募集资金) is a great way to help children learn social skills .It is not easy to go up to a complete stranger and ask him for his money .They have to nicely ask for help, show the interested buyer what they have to offer and explain how it will help them in school .If someone refuses to buy an item, that child has to take the failure in stride (不特别费力地), and that is a learning lesson as well .Students can learn how to deal with money by fundraising .Of course, it might seem safer for us to take charge of our children’s earnings from their fundraising before it is turned into the school .However, by making them keep track of it, count it, and make sure everyone pays the right amount, they are learning an important lesson .Dealing with more Fundraising helps improve their schools .It is the children’s school .They have to learn there and grow there .Why not let them help in making it a better place?Fundraising allows for more life experiences for the child .The raised money is used towards things like parties, trips, or for the music club to go to see a Broadway play .The children receive the rewards for their hard work at raising the money .Without fundraising, these field trips and special school memories would be missed .In a word, fundraising helps children a lot in many ways .56.The underlined word ―inevitable‖ in the first paragraph means ―_______‖.A .unlikelyB .improperC .unavoidableD .unrealistic57.One of the important indications that children are grown up is that __________.A .children learn to care for othersB .children can deal with moneyC .children like to make upD .children make a date with friends of the other sex58.The author thinks that fundraising _______.A .adds to the family ’s burdenB .wastes the learning timeC .builds up the children’s bodiesD .helps to develop the children’s character59.Which of the following shows the structure of the passage?CP: Central point P: Point Sp: Sub-point (次要点) C: Conclusion P1 P2 P3 P4 P5Sp2 Sp1 P1 P2 P3CP P1 P3 CP Sp2 Sp1 Sp4Sp3 P1 P2 P3 P4CPBLipitorAbout LipitorLipitor is a prescription medicine. Along with diet and exercise, it lowers ―bad‖cholesterol(胆固醇) in your blood. It can also raise ―good‖ cholesterol.Lipitor can lower the risk of heart attack in patients with several common risk factors, including family history of early heart disease, high blood pressure, age and smoking.Who is Lipitor for?Who can take Lipitor:☆People who cannot lower their cholesterol enough with diet and exercise.☆Adults and children over 10.Who should NOT take Lipitor?☆Women who are pregnant, or may become pregnant, Lipitor may harm your unborn baby.☆Women who are breast-feeding. Lipitor can pass into your breast milk and may harm your baby.☆People with liver problems.Possible side effects of LipitorSerious side effects in a small number of people:☆Muscle problems that can lead to kidney problems, including kidney failure.☆Liver problems. Y our doctor may do blood tests to check your liver before you start Lipitor and while you are taking it.Call your doctor right away if you have:☆Unexplained muscle pain or weakness, especially you have a fever or feel very fired.☆Swelling of the face, lips, tongues and/or throat that may cause difficulty in breathing or swallowing.☆Stomach pain.Some common side effects of Lipitor are:☆Muscle pain.☆Upset stomach.☆Changes in some blood test.How to take LipitorDo:☆ Take Lipitor as prescribed by your doctor.☆Try to eat heart-healthy foods while taking Lipitor.☆Take Lipitor at any time of day, with or without food.☆If you miss a dose(一剂), take it as soon as you remember. But if it has been more than 12 hours since you missed a dose, wait. Take the next dose at your next time.Don’t:☆Do not change or stop your dose before talking to your doctor.☆Do not start new medicine before talking to your doctor.60. What is the major function of Lipitor?A. To help to cure liver problems.B. To control blood pressure.C. To help to deal with muscle problems.D. To lower ―bad‖ cholesterol.61. According to the passage, taking Lipitor is beneficial to _________.A. breast-feeding womenB. children suffering from stomachC. adults having heart troubleD. teenagers with muscle problems62. If it has been over 12 hours since you missed a dose, what should you do?A. Reduce the amount of your next dose.B. Eat more when taking your next dose.C. Have a dose as soon as you remember missing it.D. Just take the next dose at your regular time.63. Which of the following is the common side effect of taking Lipitor?A. Throat swelling.B. Upset stomach.C. Kidney failure.D. Muscle weakness.C―Any time! Any where! Decades ago there was no such thing‖ –―Communication‖.Then, September 7th 1987, the global system for mobile communication or GSM was born. And international agreements that laid out the standards, regulations and practices gave rise to a global mobile phone industry.To be honest, the world’s first mobiles were not so attractive and the range of effectiveness wasn’t very good. But they became a must-have among those wealthy people who could afford that. However, by advantage of GSM which has many different elements to it, we can all enjoy the ability to go around the world in 217 countries, land in that country and know that a phone would work.There are other cell phone systems using different technology in the world. The majority of the United States and parts of South America have been using something called CDMA which is very rare in Europe. In some Asian countries like China, GSM and CDMA both exist at the same time. But the GSM Association claims 85% of the global mobile phone market. They estimated there are now about 2.5 billion different users who make more than 7 trillion minutes of calls everyday, and that’s not all.20 years later, the mobile phone is so much more than just a phone. Y ou can use it to send text messages, take pictures, show video, even surf the internet.―The phone itself is involved from just being a communication too l, to be a tool for round-the-clock connectivity, you can not live without it even in a minute.‖ Mobile consultant Nick Lane also points out with so many customized styles and features, your mobile phone will become a symbol of you.Where will the global mobile phone industry be in another 20 years? Certainly, there will be more connections than better coverage. As for where else technology will take us, one can only imagine.64. According the passage, we can know that GSM is ____________.A. a global-used mobile phoneB. a global mobile phone industryC. a global mobile phone associationD. a global mobile communication system65. From this passage, we can infer that ____________.A. as soon as the first mobiles appeared, they became popular and many people have oneB. with a GSM mobile phone, you can make a phone call in most parts of the USAC. there are only two cell phone systems in the whole world: GSM and CDMAD. most of the mobile phone users in the world now are using the GSM mobile phone66. B y pointing out ―your mobile phone will become a symbol of you‖, what does Nick Lanewant to tell us about the mobile phone in the future?A. Y ou can have a mobile phone with the unique look and functions as you like.B. Others can find us without any difficulties if we carry our mobile phone.C. The mobile phone can be used as a permit when you enter some places.D. The mobile phone we carry can show others how wealthy we are.67. What does the writer feel about the GSM development in the following 20 years?A. Satisfied.B. Worried.C. Confident.D. Confused.DBeldon and Canfield are two seashore towns, not far apart.Both towns have many hotels, and in summer the hotels are full of holiday-makers and other tourists.Last August there was a fire at the Seabreeze Hotel in Beldon.The next day, this news appeared on page two of the town’s newspaper.The Beldon Post:FIRE AT SEABREEZELate last night firemen hurried to the Seabreeze Hotel and quickly put out a small fire in a bedroom.The hotel manager said that a cigarette started the fire. We say again to all our visitors:―Please don’t smoke cigarettes in bed.‖ This was Beldon’s first hotel fire for five years.The Canfield Times gave the news in these words on page one:ANOTHER BELDON HOTEL CATCHES FIRELast night Beldon firemen arrived just too late to save clothing, bedclothes and some furniture at the Seabreeze Hotel.An angry holiday-maker said, ―An electric lamp probably started the fire.The bedroom lamps are very old at some of these hotels.When I put my bed side light on, I heard a funny noise from the lamp.‖ We are glad to tell our readers that this sort of adventure does not happen in Canfield.What are the facts, then? It is never easy to find out the exact truth about an accident.There was a fire at the Seabreeze Hotel last August: that is one fact.Do we know anything else? Y es, we know that firemen went to the hotel.Now what do you think of the rest of the ―news‖ ?68.Which of the following best gives the main idea of this text?A. It was not easy to find out exact truth from newspapers.B.Beldon and Canfield are both good places for tourists in summer.C.A fire broke out at night in Seabreeze Hotel last summer.D. Two newspapers gave reports on the same matter.69.Which of the following are probably facts?a.The fire broke out in a bedroom at the hotel.b.A cigarette started the fire.c.An old lamp started the fire.d.The fire broke out at night.e.There has never been a fire in Canfield.A.b and cB. a and dC. c and eD.a and c70.The Canfield Times used the headline like this in order to make its readers think_____________.A.hotels in Beldon often catch fireB.hotels in Beldon don’t often catch fireC. this was the second fire at the Seabreeze HotelD. Beldon was a good place except that hotels there are not quite safe71.The Canfield newspaper gave a report just the opposite to the Beldon Post by saying that_______.A.the bedroom lamps were very old at the Seabreeze HotelB.the bedroom lights made funny noise when the fire took placeC.such accidents never happened in Canfield for the past 5 yearsD.the firemen failed to save clothing, bedclothes and other thingsEIn the old days, children were familiar with birth and death as part of life. Now this is perhaps the first generation of American youngsters who have never been close by during of the birth a baby and have never experienced the death of a family member.Nowadays when people grow old, we often send them to nursing homes. When they get sick, we send them to a hospital, where children are forbidden to visit terminally (晚期的)in patients—even when those patients are their parents. This deprives(剥夺)the dying patient offamily members during the last few days of his life and it deprives the children of an experience of death, which is an important learning experience.Some of my colleagues and I once interviewed and followed about 500 terminally in order to find out what they could teach us and how we could be of more benefit, not just to them but to the members of their families as well. We were most impressed by the fact that even those patients who were not told of their serious illness were quite aware of its potential(潜在的)outcome.It is important for family members, and doctors and nurses to understand these patients’ communication in order to truly understand their needs, fears and fantasies(幻想). Most of our patients welcomed another human being with whom they could talk openly, honestly, and frankly about their trouble. Many of them shared with us their great need to be informed, to be kept up - to - date on their medical condition and to be told when the end was near. We found out that patients who had been dealt with openly and frankly were better able to cope with the coming of death and finally to reach a true stage of acceptance before death.72. The elders of today's Americans _______ .A. are often absent when a family member is born or dyingB. are unfamiliar with birth and deathC. usually see the birth or death of a family memberD. have often experienced the fear of death as part of life73. Children in America are deprived of the chance to________.A. visit a patient at hospitalB. visit their family membersC. learn how to face deathD. look after the patients74. The need of a dying patient for people to accompany(陪伴)him shows________.A. his wish for communication with other peopleB. his fear of deathC. his unwillingness to dieD. he feels very upset about his condition75. It may be concluded from the passage that________.A. dying patients should be truthfully informed of their conditionB. dying patients are afraid of being told of the coming of deathC. most patients are unable to accept death until it can’t be avoidedD. most doctors and nurses understand what dying patients need2011年三明市区高三毕业班三校联考试卷英语试题参考答案单选:21--25 BCADC 26--30 CCBBD 31--35 BCDAB完形填空: 35-40. ACCBA 41-45 DBBDA 46-50 BCACD51-55 DABAC阅读理解: A篇 56--59 :CBDAB 篇 60--63:DCDBC 篇 64--67:DDACD 篇 68--71:ABCDE 篇 72--75:CCAA短文填词:76.while 77.rabbit 78.up 79. himself 80.threw81.dreaming 82.finally 83.dead 84 by 85.rewards书面表达:【参考范文】In the picture, an old couple are waiting eagerly for their children and grandchildren to come back home for a reunion meal on Mid-Autumn Day. They felt lonely without their children around on such a special day. Actually, with the growing of the aging people in our country, we have to face the challenge of caring for the old.One the one hand, our government should take effective measures to provide enough money for their medicine and health care so that they can support themselves and live a happy life free from anxiety. And it’s necessary to encourage the elders to take part in social activities, during which they can have a good time with others. On the other hand, we young people should never ignore our aging parents however busy we are. For example, visit our parents regularly and keep them company or keep in touch with them by telephone. In short, much can be done to help the elders overcome loneliness.。
三明一中10~11学年上学期高三月考(二)理科数学(总分150分,时间:120分钟)(注意:请将所有题目的解答都写到“答题卷”上)一、选择题(本题10小题,每小题5分,共50分。
每小题只有一个选项符合题意,请将正确答案填入答题卷中。
)1、已知m1+i=1-n i,其中m、n是实数,i是虚数单位,则m+n i=( )A.1+2i B.1-2i C.2+i D.2-i2、若函数y=f(x)的定义域是[0,2],则函数g(x)=的定义域是( )A。
[0,1] B.[0,1) C。
[0,1)∪(1,4] D.(0,1)3、若△ABC的周长等于20,面积是103,A=60°,则BC边的长是( )A.5 B.6 C.7 D.84、在平行四边形ABCD中,AC与BD交于点O,E是线段OD的中点,AE的延长线与CD交于点F.若AC =a ,BD =b ,则AF = ( )A.错误!a +错误!b B 。
错误!a +错误!b C.错误!a +错误!b D.错误!a +错误!b5、设等差数列{}na 的前n 项和为nS ,若111a=-,466a a +=-,则当n S 取最小值时,n 等于 ( )A .6B .7C .8D .96、已知向量(1,2)=a ,(2,3)=-b .若向量c 满足()//+c a b ,()⊥+c a b ,则c =( )A .77(,)93B .77(,)39-- C .77(,)39D .77(,)93--7、把函数y =sin(ωx +φ)(ω>0,|φ|<错误!)的图象向左平移错误!个单位长度,所得的曲线的一部分图象如图所示,则ω、φ的值分别是 ( )A .1,3π B .1,—3π C .2,3π D .2,-3π8、设a >0,b >0。
若错误!是3a 与3b 的等比中项,则错误!+错误!的最小值为 ( )A .8B .4C .1 D.错误!9、某加工厂用某原料由甲车间加工出A产品,由乙车间加工出B产品.甲车间加工一箱原料需耗费工时10小时可加工出7千克A 产品,每千克A产品获利40元,乙车间加工一箱原料需耗费工时6小时可加工出4千克B产品,每千克B产品获利50元.甲、乙两车间每天共能完成至多70箱原料的加工,每天甲、乙两车间耗费工时总和不得超过480小时,甲、乙两车间每天总获利最大的生产计划为( )(A)甲车间加工原料10箱,乙车间加工原料60箱(B)甲车间加工原料15箱,乙车间加工原料55箱(C)甲车间加工原料18箱,乙车间加工原料50箱(D)甲车间加工原料40箱,乙车间加工原料30箱10、某商场中秋前30天月饼销售总量f(t)与时间t(0〈t≤30)的关系大致满足f(t)=t2+10t+16,则该商场前t天平均售出(如前10天的平均售出为错误!)的月饼最少为( )A.18 B.27 C.20 D.16二、填空题(本题5小题,每小题4分,共20分)11、平面向量a与b的夹角为060,(2,0)a=,1b=则2+=。
2011年三明市区高三毕业班“三校联考”试卷(理科)数学 (总分150分,时间:120分钟)(注意:请将所有题目的解答都写到“答题卷"上)一、选择题(本题10小题,每小题5分,共50分。
每小题只有一个选项符合题意,请将正确答案填入答题卷中。
)1.已知实数集R ,集合{||2|2}M x x =+<, N=3{|1}1x x <+,则M ∩(∁R N) =( )A .{|40}x x -<<B .{|10}x x -<≤C .{|10}x x -≤<D .{|0,2}x x x <>或 2.已知命题p:21,04x R x x ∀∈-+≥ ,则命题p 的否定p⌝是( ) A 。
21,04x R x x ∃∈-+< B 。
21,04x R xx ∀∈-+≤ C 。
21,04x R x x ∀∈-+< D 。
2,x R x x ∃∈-+3.=a 是复数),(R b a bi a ∈+为纯虚数的( )A .充分不必要条件B .必要不充分条件C .充要条件D .既不充分又不必要条件4.如图是将二进制数11111(2)化为十进制数的一个程序框图,判断框内应填入的条件是( )A .i ≤5B .i ≤4C .i >5D .i >45。
已知某几何体的三视图如下,则该几何体的表面积是( )A. 24 B 。
36+62正视图侧视图俯视图4 43C 。
36D 。
36+6。
如图所示,墙上挂有边长为a 的正方形木板,分都是以正方形的顶点为圆心,半径为2a 的圆孤,某人向此板投镖,中木板,且击中木板上每个点的可能性都一样,则它击中阴影部分的概率是 ( )A .4π B . 1—4π C .1-8π D .与a 的取值有关7.已知βα,是三次函数bx ax xx f 22131)(23++=的两个极值点,且)2,1(),1,0(∈∈βα,则12--a b 的取值范围是 ( )A)1,41( B)1,21( C)41,21(- D)21,21(- 8.已知点F 为抛物线y 2 = -8x 的焦点,O 为原点,点P 是抛物线准线上一动点,点A 在抛物线上,且|AF |=4,则|PA|+|PO|的最小值为 ( )A 。
2011三角函数的概念、同角三角函数的关系和诱导公式(专练1)一、选择题1、(安徽省百校论坛2011届高三第三次联合考试理)已知3cos()||,tan 22ππϕϕϕ-=<且则等于( )A .BCD 2、(浙江省金丽衢十二校2011届高三第一次联考文) 函数()sin sin(60)f x x x =++ 的最大值是( )A B .2C .2D .13、(山东省莱阳市2011届高三上学期期末数学模拟6理)已知)2,2(,31sin ππθθ-∈-=,则)23sin()sin(θππθ--的值是( ) A 、922 B 、922- C 、91- D 、914、(湖南省嘉禾一中2011届高三上学期1月高考押题卷)在区间[1,1]-上随机取一个数,cos 2xx π的值介于0到12之间的概率为( )A .13B .2πC .12D .235、(湖北省补习学校2011届高三联合体大联考试题理) 已知cos()0,cos()0,2πθθπ+<->下列不等式中必成立的是( )A.tancot22θθ> B.sincos22θθ> C.tancot22θθ< D.sincos22θθ<6、(河南省鹿邑县五校2011届高三12月联考理)函数()3sin 23f x x π⎛⎫=-⎪⎝⎭的图像为C,如下结论中正确的是( )A .图像C 关于直线6x π=对称; B .图像C 关于点,06π⎛⎫⎪⎝⎭对称; C .函数()f x 在区间5,1212ππ⎛⎫- ⎪⎝⎭内是增函数; D .由3sin 2y x =的图像向右平移3π个单位长度可以得到图像C 。
7、(河南省辉县市第一高级中学2011届高三12月月考理)若cos 2sin αα+=则tan α=( )A.1-B.2C.1D.-28、(北京四中2011届高三上学期开学测试理科试题) 已知53sin ,,2=⎪⎭⎫⎝⎛∈αππα,则⎪⎭⎫ ⎝⎛+4tan πα等于( )A .7B .7-C .71 D .71- 9、(福建省三明一中2011届高三上学期第三次月考理) 已知函数)(sin cos )(R x x x x f ∈=,给出下列四个命题:①若;),()(2121x x x f x f -=-=则 ②)(x f 的最小正周期是π2; ③)(x f 在区间]4,4[ππ-上是增函数; ④)(x f 的图象关于直线43π=x 对称; ⑤当⎥⎦⎤⎢⎣⎡-∈3,6ππx 时,)(x f 的值域为.43,43⎥⎦⎤⎢⎣⎡-其中正确的命题为( ) A .①②④ B .③④⑤ C .②③ D .③④10、(浙江省温州市啸秋中学2010学年第一学期高三会考模拟试卷)函数()sin cos f x x x =⋅的最小值是( ) A .1- B .12-C .12D .1 11、(浙江省嵊州二中2011届高三12月月考试题文) 函数()2cos sin cos y x x x =+的最大值为( )(A )2 (B 1(C(D 112、(山东省日照市2011届高三第一次调研考试文)已知4sin ,sin cos 0,5θθθ=<则θ2sin 的值为( ) (A)2524-(B)2512- (C)54- (D)2524 13、(福建省四地六校2011届高三上学期第三次联考试题理) 已知22ππθ-<<,且s i n c o s ,a θθ+=其中()0,1a ∈,则关于tan θ的值,在以下四个答案中,可能正确的是 ( )A .3-B .3 或13C .13-D .3-或13- 14、(甘肃省甘谷三中2011届高三第三次检测试题)tan 690°的值为( )A.D.15、(甘肃省甘谷三中2011届高三第三次检测试题)若sin([0,])2θθπ=∈,则tan θ=( )A. 4-B. 4C. 0D. 0或4-选择题参考答案:1—5:D 、A 、B 、D 、A ; 6—10:C 、B 、C 、D 、B ; 11—15:B 、A 、C 、A 、D ;二、填空题16、(重庆市重庆八中2011届高三第四次月考文)在ABC ∆中,如果sin :A sin :B sin C =5:6:8,则此三角形最大角的余弦值是 .17、(重庆市南开中学高2011级高三1月月考文)若3(0,),cos(),sin 5θππθθ∈+==则 。
福建省三明一中2010—2011学年度高三上学期第三次考试数学试题(理)(总分150分,时间:120分钟)一、选择题(本题10小题,每小题5分,共50分。
每小题只有一个选项符合题意,请将正确答案填入答题卷中。
) 1.设a ∈R ,若i )i (2-a (i 为虚数单位)为正实数,则a = ( )A .2B .1C .0D .1-2.若全集U=R ,集合A={1|2||≥+x x },B={021|≤-+x x x },则C U (A ∩B )为( ) A .{x |1-<x 或2>x } B .{x |1-<x 或2≥x } C .{x |1-≤x 或2>x } D .{x |1-≤x 或2≥x }3.下列有关命题的说法正确的是( )A .命题“若1,12==x x 则”的否命题为:“若1,12≠=x x 则”B .“x=-1”是“0652=--x x ”的必要不充分条件C .命题“01,2<++∈∃x x R x 使得”的否定是:“01,2<++∈∀x x R x 均有”D .命题“若y x y x sin sin ,==则”的逆否命题为真命题4.展开式的第6项系数最大,则其常数项为( ) A . 120 B . 252 C . 210 D . 455.程序框图如图:如果上述程序运行的结果S=1320,那么判断框中应填入( )A .K<10?B .K ≤10?C .K<11?D .K ≤11?6.设三条不同的直线a b c 、、,两个不同的平面αβ、,,b αα⊂⊄c 。
则下列命题不成立的是( )A .若//,αβα⊥c ,则β⊥cB .“若b β⊥,则αβ⊥”的逆命题C .若a 是c 在α的射影,b a ⊥则b ⊥cD .“若//b c ,则//c α”的逆否命题7.数列{}n a 是公差不为0的等差数列,且137,,a a a 为等比数列{}n b 的连续三项,则数列{}n b的公比为( )A .2B .4C .2D .128.两圆042222=-+++a ax y x 和0414222=+--+b by y x 恰有三条公切线,若R b R a ∈∈,,且0≠ab ,则2211ba +的最小值为 ( )A .91B .94 C .1 D .39.设,x y 满足约束条件04312x y x x y ≥⎧⎪≥⎨⎪+≤⎩,则231x y x +++取值范围是( ).A [1,5] .B [2,6] .C [3,10] .D [3,11]10.已知函数)(sin cos )(R x x x x f ∈=,给出下列四个命题: ①若;),()(2121x x x f x f -=-=则 ②)(x f 的最小正周期是π2; ③)(x f 在区间]4,4[ππ-上是增函数;④)(x f 的图象关于直线43π=x 对称;⑤当⎥⎦⎤⎢⎣⎡-∈3,6ππx 时,)(x f 的值域为.43,43⎥⎦⎤⎢⎣⎡-其中正确的命题为( )A .①②④B .③④⑤C .②③D .③④二、填空题(本题5小题,每小题4分,共20分)11.已知非零向量a ϖ.b ϖ,满足a ϖ⊥b ϖ,且a ϖ+2b ϖ与a ϖ-2b ϖ的夹角为1200,则||||b a ϖϖ等于12.双曲线221169x y -=上一点P 到右焦点的距离是实轴两端点到右焦点距离的等差中项,则P 点到左焦点的距离为 .13.甲、乙等五名志愿者被分配到上海世博会中国馆.英国馆.澳大利亚馆.俄罗斯馆四个不同的岗位服务,每个岗位至少一名志愿者,则甲.乙两人各自独立承担一个岗位工作的分法共有 种。
(用数字做答)14.某几何体的三视图,其中正视图是腰长为2的等腰三角形,侧视图是半径为1的半圆,.15.已知一系列函数有如下性质:函数1y x x=+在(0,1]上是减函数,在[1,)+∞上是 增函数;函数2y x x=+在上是减函数,在)+∞上 是增函数;函数3y x x=+在上是减函数,在)+∞上是增函数; ………………利用上述所提供的信息解决问题:若函数3(0)my x x x=+>的值域是[6,)+∞,则实数m 的值是___________.三、解答题(共6题,80分),解答应写出文字说明,证明过程或演算步骤. 16.(本题满分13分)A 、B 是直线)0(1)3cos(2cos 2)(02>-++==ωπωωx xx f y 与函数图像的两个相邻交点,且.2||π=AB(I )求ω的值;(II )在锐角ABC ∆中,a ,b ,c 分别是角A ,B ,C 的对边,若ABC c A f ∆=-=,3,23)( 的面积为33,求a 的值.17.(本题满分13分)某学校数学兴趣小组有10名学生,其中有4名女同学;英语兴趣小组有5名学生,其中有3名女学生,现采用分层抽样方法(层内采用不放回简单随机抽样)从数学兴趣小组.英语兴趣小组中共抽取3名学生参加科技节活动。
(1)求从数学兴趣小组.英语兴趣小组各抽取的人数; (2)求从数学兴趣小组抽取的学生中恰有1名女学生的概率;(3)记ξ表示抽取的3名学生中男学生数,求ξ的分布列及数学期望。
18.(本题满分13分)已知三次函数)(x f 的导函数ax x x f 33)(2-=',b f =)0(,a .b 为实数。
(1)若曲线=y )(x f 在点(1+a ,)1(+a f )处切线的斜率为12,求a 的值; (2)若)(x f 在区间[-1,1]上的最小值.最大值分别为-2.1,且21<<a ,求函数)(x f 的解析式。
19.(本题满分13分)如图,在六面体ABCDEFG 中,平面ABC ∥平面DEFG ,AD ⊥平面DEFG ,AC AB ⊥,DG ED ⊥,EF ∥DG .且2====DG DE AD AB ,1==EF AC . (Ⅰ)求证: BF ∥平面ACGD ; (Ⅱ)求二面角F CG D --的余弦值; (Ⅲ) 求五面体ABCDEFG 的体积.20.(本题满分14分)已知点P 是⊙O :229x y +=上的任意一点,过P 作PD 垂直x 轴于D ,动点Q 满足23DQ DP =u u u r u u u r。
(1)求动点Q 的轨迹方程;(2)已知点(1,1)E ,在动点Q 的轨迹上是否存在两个不重合的两点M .N ,使1()2OE OM ON =+u u u r u u u u r u u u r(O 是坐标原点),若存在,求出直线MN 的方程,若不存在,请说明理由。
21.(本题满分14分)已知数列{}n a 中,()112,202,n n a a a n n n N -=--=≥∈. (1)写出23a a 、的值(只写结果)并求出数列{}n a 的通项公式; (2)设12321111n n n n nb a a a a +++=+++⋅⋅⋅+,若对任意的正整数n ,当[]1,1m ∈-时,不等式2126n t mt b -+>恒成立,求实数t 的取值范围。
参考答案(总分150分,时间:120分钟)一.选择题(本题10小题,每小题5分,共50分。
)1.B 2.B 3.D 4.C 5.A 6.B 7.C 8.C 9.D 10. D 二.填空题(本题5小题,每小题4分,共20分) 11.33212.13 13.72 14.)3(2+=πS 15.2. 三.解答题(共6题,80分) 16.(本题满分13分)17.(本题满分13分)解(1)抽取数学小组的人数为2人;英语小组的人数为1人;………………………2分(2)2101416C C C p ⋅==158; ………………………5分(3)2421032(0)525C p C ξ===g ,11642103(1)5C C p C ξ⋅==+g 24210228575C C =g , 262103(2)5C p C ξ==+g 1164210231575C C C ⋅=g ,2621022(3)515C p C ξ===g 。
∴ξ的分布列为:153752751250⨯+⨯+⨯+⨯=ξE 5=. ………………………13分18.(本题满分13分)解:(1)由导数的几何意义)1(+'a f =12 ∴ 12)1(3)1(32=+-+a a a ∴ 93=a ∴ 3=a ………………3分 (2)∵ ax x x f 33)(2-=',b f =)0(∴ b ax x x f +-=2323)( ……5分 由 0)(3)(=-='a x x x f 得01=x ,a x =2 ∵ ∈x [-1,1],21<<a∴ 当∈x [-1,0]时,0)(>'x f ,)(x f 递增;当∈x (0,1)时,0)(<'x f ,)(x f 递减。
…8分∴ )(x f 在区间[-1,1]上的最大值为)0(f ∵ b f =)0(,∴ b =1 …………10分 ∵ a a f 2321231)1(-=+-=,a a f 231231)1(-=+--=- ∴ )1()1(f f <-∴ )1(-f 是函数)(x f 的最小值, ∴ 223-=-a ∴ 34=a ∴ )(x f =1223+-x x ……13分 19.(本题满分13分) 解法一 向量法A B C D E GF M N AC由已知,AD .DE .DG 两两垂直,建立如图的坐标系, 则A (0,0,2), B (2,0,2),C (0,1,2),E (2,0,0), G (0,2,0),F (2,1,0)(Ⅰ)(2,1,0)(2,0,2)(0,1,2)BF =-=-u u u r, (0,2,0)(0,1,2)(0,1,2)CG =-=-u u u r∴BF CG =u u u r u u u r,所以BF ∥CG .又BF ⊄平面ACGD ,故 BF//平面ACGD …4分(Ⅱ)(0,2,0)(2,1,0)(2,1,0)FG =-=-u u u r,设平面BCGF 的法向量为1(,,)n x y z =u r ,则112020n CG y z n FG x y ⎧⋅=-=⎪⎨⋅=-+=⎪⎩u r u u u r u r u u u r,令2y =,则1(1,2,1)n =u r , 而平面ADGC 的法向量2(1,0,0)ni ==u u r r∴121212cos ,||||n n n n n n ⋅<>=⋅u r u u ru r u u r u r u u r =故二面角D-CG-F 的余弦值为69分 (Ⅲ)设DG 的中点为M ,连接AM .FM ,则V =ADM-BEF ABC-MFG V V 三棱柱三棱柱+=ADM MFG DE S AD S ⨯+⨯△△=1122122122⨯⨯⨯+⨯⨯⨯=4.……………13分 解法二设DG 的中点为M ,连接AM .FM ,则由已知条件易证四边形DEFM 是平行四边形,所以MF//DE ,且MF =DE 又∵AB//DE ,且AB =DE∴MF//AB ,且MF =AB ∴四边形ABMF 是平行四边形,即BF//AM , 又BF ⊄平面ACGD 故 BF//平面ACGD ……………4分(利用面面平行的性质定理证明,可参照给分)(Ⅱ)由已知AD ⊥面DEFG ∴DE ⊥AD ,DE ⊥DG 即DE ⊥面ADGC ,∵MF//DE ,且MF =DE , ∴MF ⊥面ADGC在平面ADGC 中,过M 作MN ⊥GC ,垂足为N ,连接NF ,则[来源:Zxxk .com] 显然∠MNF 是所求二面角的平面角.∵在四边形ADGC 中,AD ⊥AC ,AD ⊥DG ,AC=DM =MG =1∴CD CG ==∴MN =5=在直角三角形MNF 中,MF =2,MN 5=∴tan MNF ∠=MFMN,cos MNF ∠=6故二面角D-CG-F…………9分 (Ⅲ)ABC-DEFG V 多面体=ADM-BEF ABC-MFG V V 三棱柱三棱柱+=ADM MFG DE S AD S ⨯+⨯△△=1122122122⨯⨯⨯+⨯⨯⨯=4.……………13分20.(本题满分14分) 解:(1)设()00(,),,P x y Q x y ,依题意,则点D 的坐标为0(,0)D x ……1分∴00(,),(0,)DQ x x y DP y =-=u u u r u u u r………………………2分又 23DQ DP =u u u r u u u r∴000002332x x x x y y y y -==⎧⎧⎪⎪⎨⎨==⎪⎪⎩⎩即 ………………………4分∵ P 在⊙O 上,故22009x y += ∴ 22194x y += ………………………5分∴ 点Q 的轨迹方程为22194x y += ………………………6分(2)假设椭圆22194x y +=上存在两个不重合的两点()1122(,),,M x y N x y 满足1()2OE OM ON =+u u u r u u u u r u u u r,则(1,1)E 是线段MN 的中点,且有12121212122212x x x x y y y y +⎧=⎪+=⎧⎪⎨⎨++=⎩⎪=⎪⎩即…9分又 ()1122(,),,M x y N x y 在椭圆22194x y +=上∴ 22112222194194x y x y ⎧+=⎪⎪⎨⎪+=⎪⎩ 两式相减,得()()()()12121212094x x x x y y y y -+-++=……12分∴ 121249MN y y k x x -==-- ∴ 直线MN 的方程为 49130x y +-=∴ 椭圆上存在点M .N 满足1()2OE OM ON =+u u u r u u u u r u u u r,此时直线MN 的方程为49130x y +-= …………………………14分 21.(本题满分14分)解:(1)∵ ()112,202,n n a a a n n n N -=--=≥∈ ∴ 236,12a a == ……………2分当2n ≥时,()11232212,21,,23,22n n n n a a n a a n a a a a ----=-=-⋅⋅⋅-=⨯-=⨯, ∴ ()12132n a a n n -=⎡+-+⋅⋅⋅++⎤⎣⎦,∴()()()121321212n n n a n n n n +=⎡+-+⋅⋅⋅+++⎤==+⎣⎦ …………………5分当1n =时,()11112a =⨯+=也满足上式, ∴数列{}n a 的通项公式为()1n a n n =+…6分 (2)()()()()()1221111111223221n n n n b a a a n n n n n n ++=++⋅⋅⋅+=++⋅⋅⋅++++++ ()()()()()1111111223221n n n n n n =-+-+⋅⋅⋅+-+++++()()21111121231(2)3n n n n n n n=-==++++++ …………………8分 令()()121f x x x x =+≥,则()212f x x'=-,当()1,0x f x '≥>时恒成立∴ ()f x 在[)1,x ∈+∞上是增函数, 故当1x =时,()()13f x f ==min 即当1n =时, 1(6n b =)max ……………11分 要使对任意的正整数n ,当[]1,1m ∈-时,七彩教育网 免费提供Word 版教学资源 七彩教育网 全国最新初中、高中试卷、课件、教案等教学资源免费下载 不等式2126n t mt b -+>恒成立, 则须使2max 112()66n t mt b -+>=, 即[]220,1,1t mt m ->∀∈-对恒成立,∴ 2220,2220t t t t t t ⎧->><-⎨+>⎩解得,或 ∴ 实数t 的取值范围为()(),22,-∞-⋃+∞…14分 另解: 111111111223121221231n n b b n n n n n n n n +⎛⎫-=--+=+-+ ⎪++++++++⎝⎭ 2233340252253n n n n n n ++=-<++++ ∴ 数列{}n a 是单调递减数列, ∴11(6n b b ==)max。