湖南省桑植一中皇仓中学2014届高三第一次联考(9月)数学试卷(理科)

2014年普通高等学校招生全国统一考试(湖南卷)数学(理工农医类)本试卷包括选择题、填空题和解答题三部分,共5页,时间120分钟,满分150分.一、选择题:本大题共10小题,每小题5分,共50分.在每小题给出的四个选项中,只有一项是符合题目要求的. 1.满足(z i i i z+=为虚数单位)的复数z =( )A ．1122i + B ．1122i - C ．1122i -+ D ．1122i--2.对一个容量为N 的总体抽取容量为n 的样本,当选取简单随机抽样、系统抽样和分层抽样三种不同方法抽取样本时,总体中每个个体被抽中的概率分别是123,,p p p 则( )A ．123p p p =< B ．231pp p =< C ．132pp p =< D ．123pp p ==3.已知(),()f x g x 分别是定义在R 上的偶函数和奇函数,且32()()1f x g x x x -=++,则(1)(1)f g +=()A ．－3B ．－1C ．1D ．34.51(2)2x y -的展开式中23x y 的系数是( )A ．－20B ．－5C ．5D ．205.已知命题:p 若x y >,则x y -<-,命题:q 若x y >,则22x y >.在命题:①p q ∧②p q ∨③()p q ∧⌝④()p q ⌝∨中,真命题是( ) A ．①③ B ．①④ C ．②③ D ．②④6.执行如图右所示的程序框图,如果输入的[t ∈于( )A. [6,2]-- B ．[5,1]-- C ．[4,5]- D ．[3,6]-7. 打磨,加工成球,则能得到的最大球的半径等于( ) A ．1 B ．2 C ．3 D ．48.某市生产总值连续两年持续增加,第一年的增长率为p ,第二年的增长率为q ,则该市这两年生产总值的年平均增长率为( ) A ．2p q + B ．(1)(1)12p q ++- C D ．19.已知函数()sin(),f x x ϕ=-且230()0f x dx π=⎰,则函数()f x 的图象的一条对称轴是( )正视图侧视图俯视图A ．56x π= B ．712x π= C ．3x π= D ．6x π=10.已知函数21()(0)2x f x x e x =+-<与2()ln()g x x x a =++的图象上存在关于y 轴对称的点,则a 的取值范围是( ) A．(-∞ B．(-∞ C．( D．(二、填空题:本大题共6小题,考生作答5小题,每小题5分,共25分. (一)选做题(请考生在第11、12、13三题中任选两题作答,如果全做,则按前两题记分)11.在平面直角坐标系中,倾斜角为4π的直线l 与曲线2c os ,:(1sin x C y ααα=+⎧⎨=+⎩为参数)交于 A B 、两点,且||2AB =,以坐标原点O为极点,x 轴正半轴为极轴建立极坐标系,则直线l 的极坐标方程是 .12.如图右,已知,AB BC 是O 的两条弦,,AO BC AB BC ⊥=则O 的半径等于.13.若关于x的不等式|2|3ax -<的解集为51{|}33x x -<<,则a = .(二)必做题(14-16题)ABDCO14.若变量,x y 满足约束条件,4,y x x y y k ≤⎧⎪+≤⎨⎪≥⎩,且2z x y =+的最小值为－6,则k = .15.如图右,正方形ABCD 和正方形DEFG 的边长分别为,()a b a b <, 原点O 为AD 的中点,抛物线22(0)ypx p =>经过,C F 两点,则b a =.16.在平面直角坐标系中,O 为原点,(1,0),(3,0)A B C -,动点D 满足||1CD =,则||OA OB OD ++的最大值是.三、解答题:本大题共6小题,共75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)某企业有甲、乙两个研发小组,他们研发新产品成功的概率分别为23和35.现安排甲组研发新产品A ,乙组研发新产品B .设甲、乙两组的研发相互独立.(Ⅰ)求至少有一种新产品研发成功的概率;(Ⅱ)若新产品A 研发成功,预计企业可获利润120万元;若新产品B 研发成功,预计企业可获利润100万元.求该企业可获利润的分布列和数学期望.18.(本小题满分12分)如图右,在平面四边形ABCD 中,1,2,AD CD AC ===(Ⅰ)求cos CAD ∠的值;(Ⅱ)若cos BAD CBA ∠=∠=求BC 的长.19.(本小题满分12分)如图,四棱柱1111ABCD A B C D -的所有棱长都相等,11111,,ACBD O AC B D O ==四边形11ACC A 和四边形11BDD B 均为矩形.(Ⅰ)证明:1O O ⊥底面ABCD ;(Ⅱ)若60CBA ∠=,求二面角11C OB D --的余弦值.A 1B 1C 1D 1O 1ACDBO已知数列{}na 满足*111,||,.n n n aa a p n N +=-=∈(Ⅰ)若{}na 是递增数列,且123,2,3a a a 成等差数列,求p 的值; (Ⅱ)若12p =,且21{}n a-是递增数列,2{}n a 是递减数列,求数列{}n a 的通项公式.21.(本小题满分13分)如图右,O 为坐标原点,椭圆22122:1(0)x y C a b a b +=>>的左、右焦点分别为12,F F ,离心率为1e ;双曲线22222:x y C a b-1=的左、右焦点分别为34,F F ,离心率为2e .已知12e e =且24|| 1.F F =(Ⅰ)求12,C C 的方程;(Ⅱ)过1F 作1C 的不垂直于y 轴的弦,AB M 为AB 的中点.当直线OM 与2C 交于,P Q 两点时,求四边形APBQ 面积的最小值.已知常数0a >,函数2()ln(1).2xf x ax x =+-+ (Ⅰ)讨论()f x 在区间(0,)+∞上的单调性;(Ⅱ)若()f x 存在两个极值点12,,x x 且12()()0,f x f x +>求a 的取值范围.参考答案一.选择题1【解】选B.由(1)1111(1)(1)222z i i i i i iz i zi i i +---=====-----,即选B.2【解】选D. 根据随机抽样的原理可得简单随机抽样、分层抽样、系统抽样都必须满足每个个体被抽到的概率相等,即123p p p ==,故选D.3【解】选C.由函数奇偶性,联想转 化:32(1)(1)(1)(1)(1)(1)11f g f g +=---=-+-+=.4【解】选A.二项式51(2)2x y -的通项为()5151()2,,2r r rr T C x y r r N -+=-≤5∈,令3r =时,()33223451()2202TC x y x y =-=-,故选A.5【解】选C.显然p 真q 假,6【解】选D. 由程序框图可知①当[)2,0t ∈-时,运行程序如下,(]2211,9,3t t S t =+∈=-∈②当[]0,2t ∈时,则[]33,1S t =-∈--;综上①②可知,(][][]2,63,13,6S ∈---=-故选D.7【解】选B.由三视图可得该几何体为三棱柱(正视图侧视图宽为6的矩形侧面与地面接触).易知不存在球与该三棱 柱的上、下底面及三个侧面同时相切,故最大的球是与其 三个侧面同时相切,所以最大球的半径为上(下)底面直角 三角形内切圆的半径r ,则681022r +-==,故选B.8【解】选D.设两年的年平均增长率为x , 则有()()()2111x p q +=++1x ⇒=,故选D.9【解】选A.由230()0f x dx π=⎰得,23cos()00x πϕ--=,即2cos cos()03πϕϕ--=,可化为3cos 02ϕϕ=,即tan ϕ可得,3k k Z πϕπ=+∈,也所以()sin()sin()3f x x x πϕ=-=±-,经检验可知A 选项符合.10【解】选B.依题意在曲线()g x 取一点(,())(0)x g x x >,则在曲线()f x 上存在一点(,())x f x --与之对应(关于y 轴对称),所以()()f xg x -=在0x >上有解,即221ln()2x x e x x a -+-=++,也即1ln()2x x a e -+=-在x >有解,由于121ln(),2x y x a y e -=+=-分别为(0,)+∞减函数,于是结合图象易知,方程1ln()2x x a e -+=-有解的充要条件为01ln 2a e -<-,即a 选B.二、填空题:本大题共6小题,考生作答5小题,每小题5分,共25分. 11【解】填(c o s s i n)ρθθ-=.依题意曲线C的普通方程为()()22211x y -+-=,设直线l 的方程为y x b =+,因为弦长2AB =,所以圆心()2,1到直线l 的距离0d =,y AB O所以圆心在直线l 上,故1y x =-sinsin )1ρθθθ=-=. 12【解】填32.设AOBC D =,易知BD =ABD ∆中由勾股定理可得1AD =,连接2222232(1)2OB BD OD r r r =+⇔=+-⇒=. 13【解】填-3 .由题可得52331233a a ⎧--=⎪⎨-=⎪⎩3a ⇒=-,故填3-. (二)必做题(14-16题) 14【解】填 -2 .如右图所示,2k <,故当目标函数2y x z =-+过点(,)A k k 时,z 即63k -=,即2k =-.15【解】填1.由条件可知(,),(,)22a a C a Fb b + 上,代入C 点易得p a =,又代入F 点得2220b ab a --=,可化为2()210bb a a --=,得1b a =,又因为1b a >,所以1a=,即为所求.16【解】填1.由||1CD =知,动点(,)D x y 在:(C x 设(1,OA OB OD x y =++=-m ,则22||(1)(x y =-+m 其几何意义为C 上动点(,)D x y 与定点(1,E 如右图所示,由平面几何知,max||||1EC r =+=m .三.解答题A17【解】(Ⅰ)记E ={甲组研发新产品成功},F ={乙组研发新产品成功}.由题设知,E F相互独立,且23(),()35P E P F ==,又记事件 “至少有一种新产品研发成功”为M , 则2313()1()1()1()()13(1)(1)3515P M P M P EF P E P F =-=-=-=---= (6)分(Ⅱ)记该企业可获利润为ξ(万元),则ξ的可能取值有0,100,120,220. 且易知122133(0),(100)35153515P P ξξ==⨯===⨯=;224236(120),(220)35153515P P ξξ==⨯===⨯=; 故所求的分布列为(如右表所示): 且2346()010012022014015151515E ξ=⨯+⨯+⨯+⨯= (12)分18【解】(Ⅰ)如图右,在ADC ∆中,由余弦定理,得2227c o s 2AC AD CD CAD AC AD +-∠===⋅ (5)分(Ⅱ)设BAC α∠=,则BAD CAD α=∠-∠, 因为cos CAD BAD ∠∠=,且,(0,)CAD BAD π∠∠∈,所以sin CAD ∠==同理sin 14BAD ∠=, A CDBACDB于是sin sin()sin cos cos sin BAD CAD BAD CAD BAD CAD α=∠-∠=∠⋅∠-∠⋅∠,(147=-⋅=, (10)分所以在ABC ∆中,由正弦定理有sin 3sin AC BC CBAα===∠, 即为所求.………………12分19【解】(Ⅰ)证明:如图右,因为四边形11ACC A 为矩形,所以1C C A C ⊥,同理1DD BD ⊥, 因为11//CC DD ,所以1CCBD ⊥,而AC BD O =,因此1CC ⊥底面ABCD .由题设知11//OO CC ,故1O O ⊥底面ABCD ;………………6分(Ⅱ)解法1 如图右,由(Ⅰ)知1O O ⊥底面ABCD ,所以1O O ⊥底面1111A B C D ,于是1O O ⊥11AC .又由题设知四边形1111A B C D 是菱形,所以1111ACB D ⊥,而1111B D OO O =,故11AC ⊥平面11BDD B ,于是过点1O 作11O H B O ⊥于H ,连结1HC 则11HC B O ⊥(三垂线定理),故11C HO ∠是二面角11C OB D --的平面角.不妨设2AB =,因为60CBA ∠=,所以11,OB OC OB ===在11Rt OO B ∆中,11111OO O BO H OB ⋅==,而111OC=,于是1C H ==,A 1B 1HC 1D 1O 1ACDBO故11Rt HO C ∆中,有1111cos O H C HOC H ∠=== 即二面角11COB D --的余弦值为解法2 由题设知四边形ABCD 是菱形,所以AC 又(Ⅰ)已证1O O ⊥底面ABCD ,从而1,,OB OC OO 两两垂直,如图右,以O 为原点,1,,OB OC OO 所在直线分 别分x 轴,y 轴,z 轴,建立空间直角坐标系O xyz -. 不妨设2AB =,因为60CBA ∠=,所以1OB OC =,于是相关各点的坐标为11(0,0,0),(0,1,2)O B C ,易知1(0,1,0)=n是平面11BDD B 的一个法向量.设2(,,)x y z =n 是平面11OB C 一个法向量,则21210,0,OB OC ⎧⋅=⎪⎨⋅=⎪⎩n n ,即20,20.z y z +=+=⎪⎩ ,令z =则2,x y ==,故2=n ,设二面角11C OB D --的大小为θ,由图可知θ为锐角,于是121212||cos |cos ,|||||θ⋅=<>===⨯n n n n n n ,故二面角11COB D --的余弦值为分20【解】(Ⅰ)因为{}na 是递增数列,所以11||n n n n n aa a a p ++-=-=,而11a =,因为2231,1ap a p p =+=++,又123,2,3a a a 成等差数列,所以21343aa a =+,因而230p p -=,解得1,3p =或0p =,1当0p =时,1n n aa +=,这与{}n a 是递增数列矛盾.故13p =;………………………………6分(Ⅱ)由于21{}n a -是递增数列,因而21210n n a a +-->,于是212221()()0n n n n a a a a +--+->,……①而22121222111||()||()22n n n n n n a a a a -+--=<-=,……②由①②知,2210n n a a -->,即212211()2n n n a a ---=,……③因为2{}na 是递减数列,同理可得2120n n aa +-<,故22121()2n n n a a +-=-……④由③④即知,11*111(1)()(),2,22n n n nn a a n n N ----=-=--≥∈,所以112211()()()(2)nn n n n aa a a a a a a n ---=-+-++-+≥121111[()()()]1222n n --=--+-++-+1111()141121()()1233212n n ----=--=--+, 又当1n =时,11a=也适合上式,故1*411(),332n n a n N -=--∈.………………………13分21【解】(Ⅰ)因为12e e =所以22221222(1)(1)b b e e a a =-+= 得222ab =,从而24(,0),,0)F b F ,24||1b F F -=,即21,2b a ==,故12,C C 的方程分别为22221,122x x y y +=-=(Ⅱ)由(Ⅰ)易知1(1,0)F -,依题意设:1AB x my =-,1122(,),(,)A x y B x y ,由221,22x my x y =-⎧⎨+=⎩,得22(2)210m y my +--=,显然0>恒成立, 所以12122221,21m y y y y m m -+==++, 故121224()22x x m y y m -+=+-=+,于是AB 的中点222(,)22mM m m -++, 故直线PQ 的斜率为2m k -=,即直线:2m PQ y x =-,即20mx y +=,由22,222m y x x y ⎧=-⎪⎨⎪-=⎩得22(2)4m x -=,即2224(2)2x m m =<-,由双曲线的对称性易PQ =,由M 为AB 的中点,显然,A B 到直线PQ 的距离相等,即d =,所以2d =,又因为,A B 在直线20mx y +=的两侧,故1122(2)(2)0mx y mx y +⋅+<,于是22d ==,又因为12||y y -==即2d ,故四边形APBQ的面积为21||22)2S PQ d m =⋅=≤<, 由2022m <-≤,故当0m =时,S 有最小值2,综上所述,四边形APBQ 面积的最小值为2.………………13分22【解】(Ⅰ)由2222(2)24(1)'()1(2)(1)(2)a x x ax a f x ax x ax x +-+-=-=++++,(0x >)①当1a ≥时,'()0f x >;②当01a <<时,由()0f x '=得,12x x ==-舍去),且由于二次函数24(1)y ax a =+-的图象是开口向上的抛物线,故易知:当10x x <<时,'()0f x <,当1x x >时,'()0f x >,综上所述,当1a ≥时,()f x 在区间(0,)+∞上单调递增; 当01a <<时,()f x在区间上递减,在区间)+∞上递增.……6分 (Ⅱ)由(Ⅰ)知224(1)'()(1)(2)ax a f x ax x +-=++,所以①当1a ≥时,()0f x '≥,此时()f x 不存在极值点. ②当01a <<时,'()0f x =的两根为12x x ==- 依题意12,x x 是()f x 定义域上的两个极值点,故必有221,2x x a=--≠-, 解得12a ≠,结合二次函数24(1)y ax a =+-的图象可知,当101,2a a <<≠时,12,x x 分别是()f x 的极小值、极大值点.且12124(1)0,a x x x x a-+==. 而1212121222()()ln(1)ln(1)22x x f x f x ax ax x x +=+-++-++,212121212121244()ln[1()]2()4x x x x a x x a x x x x x x ++=+++-+++224(1)2ln(21)ln(21)2,2121a a a a a -=--=-+--- 令21(1,0)(0,1)t a =-∈-,则2122()()()ln 2,(11,0)f x f x g t t t t t+==+--<<≠,于是22(1)'()0t g t t -=<,即()g t 在(1,0),(0,1)t ∈-上递减,所以 ①当10t -<<时,()(1)40g t g <-=-<,与12()()()0f x f x g t +=>的题意矛盾,舍去;②当01t <<时,()(1)0g t g >=,符合题意.综上可知,要使12()()0,f x f x +>则必须有21(0,1)t a =-∈,即1(,1)2a ∈为所求.……13分。

2014届高三联考试卷(一)数 学(理科)领航教育数学命题组本卷分第Ⅰ卷(选择题)和第Ⅱ卷(非选择题)两部分㊂总分150分㊂考试时间120分钟㊂第Ⅰ卷(选择题,共40分)一㊁选择题(本大题共8小题,每小题5分,共40分,在每小题给出的四个选项中,只有一项是符合题目要求的)1.已知集合S ={y |y =2x},T ={x |y =l g (x -1)},则S ɘT =( )A.(0,+ɕ) B .[0,+ɕ) C .(1,+ɕ) D.[1,+ɕ)2.已知命题p ʒ∃x ɪR ,x -2>l gx ,命题q ʒ∀x ɪR ,x 2>0,则( )A.命题p ᶱq 是假命题B .命题p ɡq 是真命题C .命题p ᶱ(췍q )是假命题 D.命题p ɡ(췍q )是真命题3.函数y =l o g a (x +3)-1(a >0,且a ʂ1)的图象恒过定点A ,且点A 在直线m x +n y +1=0上(其中m ,n >0),则1m +2n的最小值等于( )A.16B .12C .9 D.84.设a ɪR ,函数f (x )=e x+a ㊃e -x 的导函数是f ᶄ(x ),且f ᶄ(x )是奇函数.若曲线y =f (x )的一条切线的斜率是32,则切点的横坐标为( )A.l n 2B .-l n 2C .l n 22 D.-l n 225.已知定义域为R 的函数f (x )满足f (-x )=-f (x +4),当x >2时,f (x )单调递增,若x 1+x 2<4,且(x 1-2)(x 2-2)<0,则f (x 1)+f (x 2)与0的大小关系是( )A.f (x 1)+f (x 2)>0B .f (x 1)+f (x 2)=0C .f (x 1)+f (x 2)<0 D.f (x 1)+f (x 2)ɤ06.已知函数f (x )=e x -1,g (x )=-x 2+4x -3.若存在实数a ,b ,使得f (a )=g (b ),则b 的取值范围是( )A.[2-2,2+2]B .(2-2,2+2)C .[1,3] D.(1,3)7.若关于x 的方程|x |x +2=k x2有四个不同的实数解,则实数k 的取值范围为( )A.(0,1)B .12,()1C .12,()+ɕ D.(1,+ɕ)8.若对于定义在R 上的函数f (x ),其图象是连续不断的,且存在常数λ(λɪR )使得f (x +λ)+λf (x )=0对任意实数x 都成立,则称f (x )是一个 λ 伴随函数 .有下列关于 λ 伴随函数 的结论:①f (x )=0是常数函数中唯一一个 λ 伴随函数 ;②f (x )=x 不是 λ 伴随函数 ;③f (x )=x 2是 λ 伴随函数 ;④ 12 伴随函数至少有一个零点.其中正确结论是多少个( )A.1B .2C .3D.4第Ⅰ卷(选择题)答题表题号12345678答案第Ⅱ卷(非选择题,共110分)二㊁填空题(本大题共7小题,每小题5分,共35分)9.函数y =l o g a (x 2+2x -3).当x =2时,y <0,则此函数的单调递减区间是.10.设函数f (x )=x 2-4x +3,g (x )=3x-2,集合M ={x ɪR |f (g (x ))>0},N ={x ɪR |g (x )<2},则M ɘN 为 .11.设函数f (x )=(x +1)2+s i n x x 2+1的最大值为M ,最小值为m ,则M +m =.12.用m i n {a ,b ,c }表示a ,b ,c 三个数中的最小值㊂设f (x )=m i n {2x,x +2,10-x }(x ȡ0),则f (x )的最大值为 .13.已知函数f (x )=2x-a , x ɤ0x 2-3a x +a ,x >{,有三个不同的零点,则实数a 的取值范围是.14.某商品在最近100天内的单价f (t )与时间t 的函数关系是f (t )=t 4+22 (0ɤt <40,t ɪN )-t 2+52 (40ɤt ɤ100,t ɪN ìîíïïïï),日销售量g (t)与时间t 的函数关系是g (t )=-t 3+1093(0ɤt ɤ100,t ɪN ).则这种商品的日销售额的最大值为.15.函数f (x )的定义域为D ,若对于任意x 1,x 2ɪD ,当x 1<x 2时,都有f (x 1)ɤf (x 2),则称函数f (x )在D 上为非减函数㊂设函数f (x )为定义在[0,1]上的非减函数,且满足以下三个条件:①f (0)=0;②f (1-x )+f (x )=1,xɪ[0,1];③当x ɪ0,[]14时,f (x )ȡ2x 恒成立㊂则f ()37+f ()59=.三㊁解答题(本大题共6小题,共75分,解答应写出文字说明,证明过程或演算步骤)16.(本小题满分12分)已知集合A ={x |x 2-2x -8ɤ0},B ={x |x 2-(2m -3)x +m (m -3)ɤ0,m ɪR }.(1)若A ɘB =[2,4],求实数m 的值;(2)设全集为R ,若A ⊆C R B ,求实数m 的取值范围.17.(本小题满分12分)已知命题pʒx1和x2是方程x2-m x-2=0的两个实根,不等式a2-5a-3ȡ|x1-x2|对任意实数mɪ[-1,1]恒成立;命题qʒ不等式a x2+2x-1>0有解,若pᶱq为真命题,pɡq为假命题,求a的取值范围.18.(本小题满分12分)设f(x)=3a x2+2b x+c,若a+b+c=0,f(0)>0,f(1)>0,求证:(1)a>0且-2<b a<-1;(2)方程f(x)=0在(0,1)内有两个实根.19.(本小题满分13分)设函数f(x)=a x-(k-1)a-x(a>0且aʂ1)是定义域为R的奇函数.(1)求k的值;(2)若f(1)=32,且g(x)=a2x+a-2x-2m㊃f(x)在[1,+ɕ)上的最小值为-2,求m的值.20.(本小题满分13分)某蔬菜基地2013年2月2日有一批黄瓜进入市场销售,通过市场调查,预测黄瓜的价格f (x)(单位:元/k g)与时间(表示距2月10日的天数,单位:天,xɪ(0,8])的数据如下表:时间x862价格8420(1)根据上表数据,从下列函数中选取一个函数描述黄瓜价格f(x)与上市时间x的变化关系:f(x)=a x+b,f(x)=a x2+b x+c,f(x)=a㊃b x,f(x)=a㊃l o g b x,其中aʂ0;并求出此函数;(2)为了控制黄瓜的价格,不使黄瓜的价格过于偏高,经过市场调研,引入一控制函数h(x)=e x-(12-2m)x+39.(x>0)m称为控制系数.求证:当m>l n2-1时,总有f(x)<h(x).21.(本小题满分13分)已知函数f(x)=12x2-a l n x(a>0).(1)若a=2,求f(x)在(1,f(1))处的切线方程;(2)求f(x)在区间[1,e]上的最小值;(3)若f(x)在区间(1,e)上恰有两个零点,求a的取值范围.2014届数学参考答案(联考试卷一)一㊁选择题:1.C2.D3.D4.A5.C6.B7.D8.B解析:1.S ={y |y >0},T ={x |x >1},ʑS ɘT =(1,+ɕ),选C .2.由图象可知p 真,又q 假故选D .3.a =(-2,-1),ʑ2m +n =1,ʑ1m +2n =1m +2()n ㊃(2m +n )=4+n m +4m nȡ8.选D .4.fᶄ(x )=e x-a ㊃e -x ,又f ᶄ(x )为奇函数,ʑf ᶄ(0)=0,ʑa =1,设切点横坐标为x 0则f ᶄ(x 0)=e x 0-e -x 0=32,即e x 0=2,x 0=l n 2,选A .5.(理)不妨设x 1<x 2则x 1<2,x 2>2,又x 1+x 2<4,ʑ4-x 1>x 2>2,ʑf (4-x 1)>f (x 2),ʑ-f (x 1)>f (x 2),即f (x 1)+f (x 2)<0,故选C .6.f (x )=e x -1>1,ʑg (b )=-b 2+4b -3>-1,ʑ2-2<b <2+2,故选B .7.|x |x +2=k x 2=k |x |2,ʑx =0或1x +2=k |x |,ʑy =1x +2与y =k |x |有不为0的三个交点,ʑk >1,故选D .8.①λ=-1时f (x )可为任一常数函数②f (x )=x 时λx +(x +λ)=0不恒成立③f (x )=x2代入显然不是④λ=12时,f x ()+1=-12f (x ),ʑf ()12=-12f (0),又f (x )图象连续不断,ʑf (x )在0,[]12上至少有一个零点,故选B .二㊁填空题:9.(1,+ɕ) 10.{x |x <1} 11.2 12.6 13.49<a ɤ1 14.808.5 15.1三㊁解答题:16.解:(1)ȵA =[-2,4],B =[m -3,m ],A ɘB =[2,4].(2分)………………………………………………………ʑm -3=2m ȡ{4,ʑm =5.(6分)…………………………………………………………………………………(2)C R B ={x |x <m -3,或x >m },ȵA ⊆B ,ʑm <-2,或m -3>4,ʑm >7或m <-2.(12分)……………17.解:ȵx 1,x 2是方程x 2-mx -2=0的两个实根,ʑx 1+x 2=m x 1x 2{=-2,ʑ|x 1-x 2|=(x 1+x 2)2-4x 1x 2=m 2+8ʑ当m ɪ[-1,1]时,|x 1-x 2|m a x =3.由不等式a 2-5a -3ȡ|x 1-x 2|对任意实数m ɪ[-1,1]恒成立,可得:a 2-5a -3ȡ3ʑa ȡ6或a ɤ-1.ʑ命题p 为真命题时a ȡ6或a ɤ-1,命题p 为假命题时-1<a <6.(5分)………………命题q ʒ不等式a x 2+2x -1>0有解.①当a >0时,显然有解;②当a =0时,2x -1>0有解;③当a <0时,ȵa x 2+2x -1>0有解.ʑΔ=4+4a >0,ʑ-1<a <0.从而命题p :不等式a x 2+2x -1>0有解时a >-1ʑ命题q 是真命题时a >-1,命题q 是假命题时a ɤ-1.(10分)………………………………………………ȵp ᶱq 真,p ɡq 假,ʑp 与q 有且仅有一个为真.(1)当命题p 是真命题且命题q 是假命题时a ɤ-1.(2)当命题p 是假命题且命题q 是真命题时-1<a <6综上所述:a 的取值范围为a <6.(12分)……………………………………………………………………………18.解:(1)ȵf (0)>0,f (1)>0,所以c >0,3a +2b +c >0.由条件a +b +c =0,消去b 得a >c >0;由条件a +b +c =0消去c ,得a +b <0,2a +b >0.故-2<b a<-1.(6分)…………………………………………………(2)抛物线f (x )=3a x 2+2b x +c 的对称轴为x =-b 3a ,由-2<b a <-1得13<-b 3a <23.即对称轴x ɪ13,()23;而ә=4b 2-12a c =4[(-a -c )2-3a c ]=4(a 2+c 2-a c )>0,且f (0)>0,f (1)>0,所以方程f (x )=0在区间(0,1)内有两个不等的实根.(12分)………………………19.解:(1)由题意,对任意x ɪR ,f (-x )=-f (x ),即a -x -(k -1)a x =-a x+(k -1)a -x ,即(k -1)(a x +a -x )-(a x +a -x )=0,(k -2)(a x+a -x )=0,因为x 为任意实数,所以k =2.(4分)………………………………………………………………………(2)由(1)f (x )=a x-a -x ,因为f (1)=32,所以a -1a =32,解得a =2.故f (x )=2x -2-x ,g (x )=22x +2-2x -2m (2x-2-x ),令t =2x -2-x ,则22x +2-2x =t 2+2,由x ɪ[1,+ɕ),得t ɪ32,[)+ɕ,所以g (x )=h (t )=t 2-2m t +2=(t -m )2+2-m 2,t ɪ32,[)+ɕ,当m <32时,h (t )在ɪ32,[)+ɕ上是增函数,则h ()32=-2,94-3m +2=-2,解得m =2512(舍去)当m ȡ32时,则h (m )=-2,2-m 2=-2,解得m =2,或m =-2(舍去).综上,m 的值是2.(13分)…………………………………………………………………………………………20.解:(1)根据表中数据,表述黄瓜价格f (x )与上市时间x 的变化关系的函数决不是单调函数,这与函数f (x )=a x +b ,f (x )=a ㊃b x,f (x )=a ㊃l o g b x ,均具有单调性不符,所以,在a ʂ0的前提下,可选取二次函数f (x )=a x 2+b x +c 进行描述,把表格提供的三对数据代入该解析式得到:64a +8b +c =836a +6b +c =44a +2b +c ìîíïïï=20,解得a =1,b =-12,c =40.所以,黄瓜价格f (x )与上市时间x 的函数关系是f (x )=x 2-12x +40.x ɪ(0,8].(6分)………………(2)设函数g (x )=h (x )-f (x )=e x -x 2+2m x -1,求导,结果见下表.gᶄ(x )=e x -2x +2m ,继续对g ᶄ(x )求导得g ᵡ(x )=e x-2.表格如下:x(0,l n 2)l n 2(l n 2,+ɕ)g ᵡ(x )-0+gᶄ(x )减极小值增由上表可知g ᶄ(x )ȡg ᶄ(l n 2),而gᶄ(l n 2)=e l n 2-2l n 2+2m =2-2l n 2+2m =2(m -l n 2+1),由m >l n 2-1知g ᶄ(l n 2)>0,所以g ᶄ(x )>0,即g (x )在区间(0,+ɕ)上为增函数.于是有g (x )>g (0),而g (0)=e 0-02+2m ˑ0-1=0,故g (x )>0,即当m >l n 2-1且x >0时,e x >x 2-2m x +1.即h (x )>f (x ).(13分)………………………21.解:(1)a =2,f (x )=12x 2-2l n x ,f ᶄ(x )=x -2x ,f ᶄ(1)=-1,f (1)=12,f (x )在(1,f (1))处的切线方程为2x +2y -3=0.(3分)…………………………………………………(2)由f ᶄ(x )=x -a x =x 2-a x,由a >0及定义域为(0,+ɕ),令f ᶄ(x )=0得x =a .①若a ɤ1,即0<a ɤ1,在(1,e )上,f ᶄ(x )>0,f (x )在[1,e ]上单调递增,因此,f (x )在区间[1,e ]的最小值为f (1)=12.②若1<a <e ,即1<a <e 2,在(1,a )上,f ᶄ(x )<0,f (x )单调递减;在(a ,e )上,f ᶄ(x )>0,f (x )单调递增,因此f (x )在区间[1,e ]上的最小值为f (a )=12a (1-l n a ).③若a ȡe ,即a ȡe 2,在(1,e )上,f ᶄ(x )<0,f (x )在[1,e ]上单调递减,因此,f (x )在区间[1,e ]上的最小值为f (e )=12e 2-a .综上,当0<a ɤ1时,f m i n (x )=12;当1<a <e 2时,f mi n x =12a (1-l n a );当a ȡe 2时,f mi n (x )=12e 2-a .(9分)………………………………………………………………………(3)由(2)可知当0<a ɤ1或a ȡe 2时,f (x )在(1,e )上是单调递增或递减函数,不可能存在两个零点.当1<a <e 2时,要使f (x )在区间(1,e)上恰有两个零点,则ʑ12a (1-l n a )<0f (1)=12>0f (e )=12e 2-a >ìîíïïïïïï0,即a >e a <12e {2,此时,e <a <12e 2.所以,a 的取值范围为e ,12e ()2.(13分)………………………………………………………………………。

湖南省桑植一中皇仓中学2014届高三第一次联考（9月）英语试卷（时量：120分钟满分：150分）PART ONE: LISTENING COMPREHENSION（30 points）Section A (22.5 points)Directions: In this section, you will hear 6 conversations between 2 speakers. For each conversation, there are several questions and each question is followed by 3 choices. Listen to the conversations carefully and then answer the questions by marking the corresponding letter（A, B or C）on the question booklet. You will hear each conversation TWICE.Conversation 11. What is the probable relationship between the two speakers?A. Teacher and student.B. Salesgirl and customer.C. Doctor and patient.2. What does the man think will help the woman?A. Some medicine.B. Breathing slowly.C. Doing some tests. Conversation 23. Why does the man request the girl to stop watching TV?A. Because she should finish her homework first.B. Because she should protect her eyes.C. Because they will go out for a tour soon..4. What can we know about the man?A. He is a heavy smoker.B. He likes reading books in bed.C. He always breaks his promise. Conversation 35. What does the man like to do this evening?A. Hold a party.B. Watch TV.C. Go to the cinema.6. When will the company party start?A. At 5:30.B. At 7:30.C. At 8:00.Conversation 47. What are the two speakers mainly talking about?A. People working in shops.B. Goods in various qualities.C. Shopping in different places.8. What is the man?A. A salesman in a small shop.B. A manager of a supermarket.C. A staff of a department store.9. What might the woman think of supermarket staff?A. They‟re unfriendly.B. They‟re very nice.C. They‟re well paid. Conversation 510. Where did the man see the ad for the bike sale?A. On TV.B. In a department store.C. In the newspaper.11. Which bike is the latest model?A. The Curzon.B. The Anderson.C. The Instant.12. What does the man decide to do in the end?A. Buy the cheapest one.B. Go to the shop to have a look first.C. Buy the one recommended by the woman.Conversation 613. What is the purpose of the woman‟s visit?A. Business.B. Pleasure.C. Business and pleasure14. What are in the woman‟s luggage?A. Clothes, a computer and books.B. A CD player, clothes and books.C. Some gifts, books and a CD player.15. What do we know about the woman?A. Her parents are on the same trip.B. She enjoys travelling to different countries.C. She was born in that country.Section B (7.5 points)Directions: In this section, you will hear mini-talk. Listen carefully and then fill in the numbered blanks with t he information you’ve got. Fill in each blank with NO MORE THAN 3 WORDS .You willPAER TWO: LANGUAGE KNOWLEDGE （45 points）Section A（15 points）Directions: Beneath each of the following sentences there are 4 choices marked A, B, C and D. Choose one answer that best completes the sentence.21. ——Do you know the answers to these questions asked by your little brother?——Of course not, but I can‟t let him know .A. themB. oneC. thoseD. that22. With all the novels he was interested in , he left the library and went back home.A. borrowB. borrowingC. to borrowD. borrowed23. Playing QQ‟s “Happy Farm” by planting, watering, fertilizing, spraying, harves ting and selling of virtual vegetables, fruits and flowers, we can find out all the excitement is about.A. thatB. ifC. whatD. when24. I don‟t think David could have done such a stupid thing last night, ?A. did heB. didn‟t heC. do ID. don‟t I25. 31. Chen Yang has brains．In fact，I doubt whether anyone in the class has IQ.A．a high B．a higher C．the higher D．the highest26. If my brother doesn‟t go to the evening party, .A. neither do IB. I will eitherC. either will ID. nor will I27. It is what all people wish for that has broken the law will face justice.A. whoeverB. anyoneC. whoD. everyone28. Mr. Robinson in China for many years, but now he is in Australia.A. has livedB. had livedC. livedD. was living29. The music sounded cheerful and encouraging, all the people present in high spirits.A. makingB. to makeC. madeD. having made30. ——What made the little so upset?——.A. She got separated from her parentsB. Because she had lost her wayC. Being too frightenedD. Not allowed to watch TV31. Nowadays almost everyone prefers to use a smart mobile phone functions are more practical.A. whichB. whereC. whatD.whose32. According to the schedule, the flight No. 232 to Beijing at 10:30. We‟d better set out right now, or we‟ll be late.A. leavesB. leftC. will leaveD. has left33. When we were in the college, we stay in the library, devoting ourselves to books.A. shouldB. wouldC. canD. need34. and we‟ll finish the work on time.A. If we are given two more daysB. Two more daysC. Having two more daysD. Given two more days35. We are always told that only through hard work our goals in our study.A. we will achieveB. we have achievedC. have we achievedD. will we achieve Section B（18 points）Directions: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in blanks with a word or phrase that best fits the context.A man and his girlfriend got married. Everyone could tell that the love they had for each other was true.A few months later, the wife came to the husband with a proposal, “I read in a magazine about how we can 36 our marriage. Each of us will write a list of the things that we find a bit 37 with the other. Then, we can talk about how we can 38 them together and make our lives happier together.”The husband 39 . So each of them went to a(n) 40 room and thought about the things that annoyed them about the other for the rest of the day.The next morning, at the breakfast table, they decided that they would go over their 41 .“ I‟ll start,” offered the wife. She 42 her list. It had many items on it, enough to fill 3 pages. After the wife had read all three pages to her husband, 43 the husband seated, “I don‟t have anything on my list. I think that you are perfect the 44 that you are. I don‟t want you to 45 anything for me.”The wife, touched by his honesty by his depth of his love for her and his 46 of her, turned her head and wept.In life, there are enough times when we are disappointed, depressed and annoyed. We don‟t really have to go 47 for them. We have a wonderful world that is full of beauty, light and promise. Why waste time seeking the bad, disappointing or annoying when we can look around us, and see the wonderful things before us?36. A. enrich B. strengthen C. lengthen D. deepen37. A. annoying B. surprising C. satisfying D. inspiring38. A. fix B. do C. find D. set39. A. smiled B. refused C. agreed D. hesitated40. A. spare B. separate C. empty D. special41. A. dishes B. magazines C. lives D. lists42. A. put away B. handed in C. took out D. wrote down43. A. disappointedly B. unluckily C. bitterly D. quietly44. A. appearance B. way C. situation D. position45. A. change B. accept C. decide D. drop46. A. advice B. warning C. acceptance D. pride47. A. leaving B. waiting C. asking D. lookingSection C (12 points)Directions: Complete the following passage by filling in each blank with one word that best fits the context.Do you feel it difficult to be happy all the time? Now I‟ll give you some tips on 48 to make yourself happy. One way is being unselfish because unselfishness is the key factor required if you want to get along well 49 others. By saying being unselfish I mean you shouldn‟t want everything your own way or demand 50 best share of everything.51 way is to look for good points in other people. You will find most people pleasant to go with and it will surely make you happy. Third, you can‟t expect to be too perfect, 52 don‟t be too unhappy when you make a mistake. Everything will be OK 53 you try to make things right. Finally, it‟s important to remember that while you are no worse 54 others, chances you have may not be much better. 55 this case, the surest way to be happy is to hold optimistic attitude. PART THREE: READING COMPREHENSION（30 points）Directions: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are 4 choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage.AYesterday I went to our local grocery store. I often go to their deli counter, and I understand that it can get a little confused for the workers. So normally none of them ever take the time to smile or seem overly friendly. But yesterday it was completely different. I was pleasantly surprised to be greeted with a very welcoming smile by the young man behind the counter. He never sighed heavily or slum ped back and forth like he didn‟t want to be there. He was all by himself and quite a long line. But not once did he act concerned about it. He just did his job efficiently and acted very kind the entire time.I was so impressed that I approached the manager. I explained to her that I often visit the deli counter and I have never been greeted with such kindness. She agreed with me that he was a wonderful person and she thanked me for sharing my feelings with her.As I was walking away，I could hear her approaching the young man with，“I just got a wonderful compliment(praise) about you.” I couldn't hear everything she was saying，but I knew that she did thank him. I couldn't help but smile!Later I had to pass by the deli counter to get onions. There was no one there，except the diligent young man. He didn't say anything，he just smiled at me. I realized that I hadn't done a huge deed that day，but that small deed made a small difference to someone．I love seeing people smile. I just received my smile cards and I wish I had one with me that day. Maybe I will drop one off at the deli a different day! It's amazing how good I felt after that.So，friends，the next time you are in a grocery store，retail store，restaurant，or anywhere that someone is working hard，letting them know in some way can mean so much. I hope you get a smile out of it like I did!56. The passage is mainly about .A. friendship between the writer and a young manB. the power of a simple complimentC. a grocery storeD. a diligent assistant57. The writer found it was different yesterday because .A. the young man was very busyB. the young man was concerned about so many customersC. the writer was warmly greeted by a young manD. the young man acted very kind all the time58. The writer approached the manager to .A. thank himB. say hello to himC. complain about the serviceD. praise the young man59. The underlined sentence in Paragraph 4 implies that the writer will .A. smile to others at the deli one dayB. give one card to the young man one dayC. drop in at the deli one dayD. go to the deli for a job one day60. What can we learn from the last paragraph?A. Praising others' hard work means nothing.B. Helping others is always rewarding.C. We should never hesitate to praise hard-working persons.D. Not all good deeds deserve praise.BOne is never too old to learn. Life learning (sometimes called un-schooling or self-directed learning) is one of t hose concepts that are almost easier to explain by saying what it isn‟t than what it is. And that‟s probably because our own schooled backgrounds have convinced us that learning happens only in a fine building on certain days, between certain hours, and managed by a specially trained p rofessional.Within that schooling framestudy, no matter how hard teachers try and no matter how good their textbooks, many bright students get bored, many slower students struggle and give up or lose their self-respect, and most of them reach the end of the process unprepared to enter into society. They have memorized a certain body of knowledge long enough to rush back the information on tests, but the y haven‟t really learnt much, at least of the official curriculum.Life learners, on the other hand, know that learning is not difficult, that people learn things quite easily if they‟re not compelled and forced, if they see a need to learn something, and if they are trusted and respected enough to learn it on their own timetable, at their own speed, in their own way—no matter what age and no matter whether we‟re at school or at home.Life learning is independent of time, location or the presence of teacher. It does not require mom or dad to teach, or kids to work in workbooks at the table from 9 to noon. Life learning is learner-driven. It involves living and learning—in and from the real world. It is about exploring, questioning, experimenting, making messes, taking risks without fear of making mistakes, being laughed at and trying again.Furthermore, life learning is about trusting kids to learn what they need to know and about helping them to learn and grow in their own ways. It is about providing positive experiences that enable children to understand the world and their culture and to interact with it.61. It is implied in the text that it is hard to ___________.A. find a specially trained teacherB. carry life learning throughC. learn without going to schoolD. tell the nature of life learning62. According to the author, the schooling framestudy often ____________.A. produces slow students with poor memoriesB. fails to provide enough knowledge about lifeC. ignores some parts of the official curriculumD. gives little care to the quality of teaching materials63. Life learners recognize that learning will not be difficult if they are ____________.A. clear about why to learnB. careful to make a time tableC. able to respect other peopleD. cautious about any mistakes64. According to the author, life learning ___________.A. could be a road full of trials and errorsB. teaches a kid how to avoid being laughed atC. makes a kid independent of his parentsD. could prevent one from running risks65. Through life learning, children ___________.A. will grow without the assistance from parentsB. will be separated from the negative side of societyC. will be driven to learn necessary life knowledgeD. will learn to communicate with the real societyCThe world is filled with smart, talented, educated and gifted people. We meet them every day. A few days ago, my car was not running well. I pulled it into a garage and the young mechanic had it fixed in just a few minutes. He knew what was wrong by simply listening to the engine. I was amazed. The sad truth is, great talent is not enough.I am constantly shocked at how little talented people earn. I heard the other day that less than 5 percent of Americans earn more than $100,000 a year. A business consultant who specializes in the medical trade told me how many doctors and dentists struggled financially. It was this business consultant who gave me the phrase, “They are one skill away from great wealth.”There is an old saying that goe s, “Job means…just over broke（破产）‟.” And unfortunately, I would say that the saying applies to millions of people. Because school does not think financial intelligence is intelligence, and most workers live within their means. They work and they pay the bills. Instead I recommend to young people to seek work for what they will learn, more than what they will earn.When I ask the classes I teach, “How many of you can cook a better hamburger than McDonald‟s?”，almost all the students raise their hands. I the n ask, “So if most of you can cook a bett er hamburger, how come McDonald‟s makes more money than you?” The answer is obvious: McDonald‟s is excellent at business systems. The reason why so many talented people are poor is that they focus on making a better hamburger but know little or nothing about business systems. The world is filled with talented poor people. They focus on perfecting their skills at building a better hamburger rather than the skills of selling and delivering the hamburger.66. The author mentions the mechanic in the first paragraph to show that .A. he is just one of the talented peopleB. he is ready to help othersC. he has a sharp sense of hearingD. he knows little about car repairing67. The underlined part in the third paragraph can be best replaced by“”.A. spend more than they can affordB. do in their own wayC. live in their own circleD. live within what they earn68. Why do talented people earn so little according to the author?A. They don‟t work hard enough.B. They lack financial intelligence.C. They don‟t make full use of their talents.D. They have no specialized skills.69. The success of McDonald‟s lies in its .A. skills at making hamburgersB. good business systemsC. talented workersD. excellent service70. The main purpose of the author is to tell us .A. how young people can find a satisfactory jobB. what schools should teach studentsC. why so many talented people are poorD. how McDonald‟s makes much moneyPART FOUR: WRITING（45 points）Section A (10 points)Directions: Read the following passage. Complete the diagram/ Fill in the numbered blanks by using the in formation in the passage. Write NO MORE THAN 3 WORDS for each answer. Write your answers on your answer sheet.A gap year (also known as year abroad, year out, year off, deferred year, bridging year, time off and time out) is a year during which students take time off and do something other than schooling, such as travel or work. The gap year is most commonly taken after secondary school and before starting university.The practice of taking a gap year developed in the United Kingdom in the 1960s. During a gap year, a student might travel, engage in volunteer work overseas or undertake a working holiday abroad.In 1978, the Prince of Wales and Colonel John Blashford-Snell began what is now known as Raleigh International by launching "Operation Drake," a gap year expedition voyage around the world following Sir Francis Drake's route. The Center for Intermin Programs in the United States was originally founded in 1980 to assist students in making a more effective transition from high school to college.The gap year has grown very popular among students in the UK, Australia, New Zealand and Canada. A trend for gap years is to participate in international education programs that combine language study, homestay, cultural immersion, community service, and independent study.The high pressure increasingly leaves students of high school tired out and longing for refreshment.Also, counselors are coming to bless the gap-year option, and colleges increasingly are offering a delayed enrollment option as more and more “gappers” arrive on campus with enhanced motivation and maturity—all of which prepares well foe their undergraduate years in college.In 2010, gap year travel has increased among school, college and university leavers, as this isSection B (10 points)Directions: Read the following passage. Answer the questions according to the information givenin the passage and required word limit. Write your answers on your answer sheet.In the animal kingdom, weakness can bring about aggression in other animal. This sometimes happens with humans also. But I have found that my weakness brings out the kindness in people. I see it every day when people hold doors for me, pour cream into my coffee, or help me to put on my coat. And I have discovered that it makes them happy.From my wheelchair experience, I see the best in people, bur sometimes I feel sad because those who appear independent miss the kindness I see daily. They don‟t get to see this soft side of others often, we try every way possible to avoid showing our weakness, which includes a lot of pretending. But only when we stop pretending we‟re brave or strong do we allow people to show the kindness that‟s in them.Last month, when I was driving home on a busy highway, I began to feel unwell and drove more slowly than usual. People behind me began to get impatient and angry, with some speeding up alongside me, horning (按喇叭) or even shouting at me. At the moment I decided to do something I had never done in twenty fore years of driving. I put on the car flashlights and drove on at a really low speed.No more angry shouts and no more horns! When I put on my flashlights, I was saying to other drivers, “I have a problem here. I am weak and doing the best I can.” And everyone understood. Several times, I saw drivers who wanted to pass. They couldn‟t get around me because of the stream of passing traffic. But instead of getting impatient and angry, they waited, knowing the driver in front of them was in some way weak.Sometimes situations call for us to act str ong and brave even when we don‟t feel that way. But those are and far between. More often, it would be better if we don‟t pretend we feel strong when we feel weak or pretend that we are brave when we are scared.81. How do people feel when they offer their help?(No more than 2 words)____________________________________________________________________82. What reaction did other drivers have when the author drove very slowly at first? (No more than 5 words)____________________________________________________________________83. Why did other drivers behave differently when the author put on the car flashlights? (No more than 8 words)____________________________________________________________________84. What does the author advise us to do at last? (No more than 8 words)____________________________________________________________________Section C (25 points)Directions: Write an English composition according to the instructions given below .高三阶段学习比较紧张，正确的学习方法尤为重要。

湖南省桑植一中皇仓中学2014届高三第一次联考（9月）物理试卷时量：90分钟 分值：100分一 选择题（每小题4分，共48分。

其中1-10小题每小题只有一个选项是正确的，选对的给4分，有错选和未选的得0分。

11-12小题每小题有多个选项正确，全部选对的给4分，选对但不全的得2分，有错选和未选的得0分） 1.如图所示，质量为m 的等边三棱柱静止在水平放置的斜面上。

已知三棱柱与斜面之间的动摩擦因数为μ，斜面的倾角为o30，则斜面对三棱柱的支持力与摩擦力的大小分别为 A ．23mg 和21mg B ．21mg 和23mgC ．21mg 和21μmg D ．23mg 和23μmg2.用轻弹簧竖直悬挂质量为m 的物体,静止时弹簧伸长量为L.现用该弹簧沿斜面方向拉住质量为2m 的物体,系统静止时弹簧伸长量也为L.斜面倾角为300,如图所示.则物体所受摩擦力 A.等于零B.大小为mg 21,方向沿斜面向下 C.大小为m g 23,方向沿斜面向上 D.大小为mg,方向沿斜面向上3.公园里的喷泉的喷嘴与地面相平且竖直向上，某一喷嘴喷水流量Q=5L/s ，水的出口速度0v =20 m/s,不计空气阻力。

则处于空中的水的体积是（2g 10m /s =）A. 15 LB. 40 LC. 20 LD. 30 L4.如图所示，物体B 叠放在物体A 上，A 、B 的质量均为m ，且上、下表面均与斜面平行，它们以共同速度沿倾角为θ的固定斜面C 匀速下滑，则A ．A ，B 间没有静摩擦力B ．A 受到B 的静摩擦力方向沿斜面向上C ．A 受到斜面的滑动摩擦力大小为2mgsin θD ．A 与B 间的动摩擦因数μ＝tan θ5.质量分别为m 、2m 、3m 的物块A 、B 、C 叠放在光滑的水平地面上,现对B 施加一水平力F,已知AB 间、BC 间最大静摩擦力均为f 0,为保证它们能够一起运动,F 最大值为A.6f0B.4f0C.3f0D.2f06.如图所示，一根轻绳跨过光滑的定滑轮，两端分别系一个质量为m1、m2的物块。

湖南省桑植一中皇仓中学2014届高三第一次联考（9月）历史试卷时量：90分钟满分：100分一、选择题（2×24=48分）1、战国以前，“百姓”是对贵族的总称；战国以后，“百姓”成为民众的通称。

导致这一变化的主要原因是( )A．分封制的加强B．宗法制的衰落C．百家争鸣局面的出现D．井田制的推行2、“（隋朝）明确规定九品以上地方官一律由尚书省所属吏部任免，每年由吏部进行考核……通过考试选拔人才，首先设立秀才、明经等科，一律按才学标准录取……增设进士科，放宽录取标准。

”由此可见，隋朝官吏制度的特点（）A. 官员选拔制度包括九品中正制和科举制B. 官吏任免权和考核权由中央掌握C. 才学是选拔官员的唯一标准D. 中央官员是从地方官员中选拔的3、《十二铜表法》规定：“此时如债务人仍不清偿，又无人为其担保，则债权人得将其押至家中拘留，系以皮带或脚镣，但重量最多为十五磅，愿减轻者听便。

”由此可见《十二铜表法》( )A.对平民没有一点好处B.维护平民利益C.维护贵族的利益D.是对后世影响最广泛的古代法律4、中国古代加强皇权的措施中，常以内侍、贵戚监督和逐渐代替重臣，以近臣演变为重臣，以辅佐皇帝办事的部门发展为正式的国家机关。

以下机构体现这种方式的是（）A．中书门下内阁军机处B．中朝内阁军机处C．中朝中书门下内阁D．尚书台内阁议政王大臣会议5、1742年，辉格党领袖、内阁首相罗伯特·沃尔波，因失去了下院多数的信任而被迫率内阁集体辞职，此后，“内阁失去下院多数信任必须辞职”就成为英国一种不成文的制度。

这一制度有利于( )A．防止国家元首专制独裁B．形成内阁对议会负责的关系C．进一步增强内阁首相的权力D．协调内阁与国王的关系6、有学者指出：“从18世纪70—80年代起，一直到19世纪30年代……英国出现了这样一幅奇特的景象：保守主义的政治与快速发展的经济同行。

”对这一时期英国保守主义政治的特征理解正确的是（）A.英国的责任内阁制开始确立B.工业资产阶级未能充分参与国家政权C.议会开始成为国家权力中心D.选举中的高额财产限制被打破7、从一项统计数据看，（德国）全国食糖消费量从1876年的平均每人6公斤增加到1913年的21.4公斤，棉花消费量从1871年平均每人不到3公斤增加到1913年的7.6公斤，储蓄存款从1870年的约15亿马克增加到1914年的约200亿马克。

桑植一中2011高三数学9月月考理科试卷（时量：120分钟 总分：150分）第I 卷一．选择题：本大题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有一项是符合要求的，请将所选答案填在答题卷中对应位置． 1．设实数R a ∈且i i a ⋅-)(（其中i 是虚数单位）为正实数，则a 的值为 （ ） A ．－1 B ．0 C ．0或－1 D ．1 2．函数()sin()4f x x π=-图像的对称轴．．．方程可以是 （ ） A ．2x π=B ．4x π=C ．2x π=-D ．4x π=-3．已知向量a 、b 满足6,8,a b ==且,a b a b +=-则a b += （ ） A ．10 B ．20 C ．21 D ．304．“a =1”是“直线0x y +=和直线0x ay -=相互垂直”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5．执行下面的程序框图，输出的S 值为 （ ）A ．910B ．718C ．89D ．256．不等式11x a x+>+对于一切非零实数x 均成立，则实数a 的取值范围是 （ ） A. [－1,1] B. (1,1)- C. (－2,2) D.[－2,2]7．直线223(3)(2)4y kx x y =+-+-=与圆相交于M ，N 两点，32≥MN ，则k 的取值范围是（ ）A ．3[,0]4-B ．[)3,0,4⎛⎤-∞-⋃+∞ ⎥⎝⎦ C ．[ D ．2[,0]3-8．定义方程()()f x f x '=的实数根x 0叫做函数()f x 的“新驻点”，如果函数()g x x =，()ln(1)h x x =+，()cos x x ϕ=（()x π∈π2，）的“新驻点”分别为α，β，γ，那么α，β，γ的大小关系是 （ ）A ．γβα<<B ．βγα<<C ．βαγ<<D ．γαβ<< 二．填空题：本大题共7小题，每小题5分，共35分．把答案填在答题卷中对应题号后 的横线上．（一）选做题（9～11题，考生只能从中选做两题）9．（参数方程与极坐标选讲）在极坐标系中,圆C 的极坐标方程为: 22cos 0ρρθ+=,点P 的极坐标为2,2π⎛⎫⎪⎝⎭,过点P 作圆C 的切线,则两条切线夹角的正切值是 10．（几何证明选讲）如图，已知PA 是圆O 的切线，切点为A ， 直线PO 交圆O 于,B C 两点， 2AC =,120PAB ∠=， 则圆O 的面积为 11．（优选法选讲）用0.618法对某一试验进行优选，存优范围 是[]2000,8000，则第二个试点2x 是 （二）必做题（12～16题）12．集合A=｛4，5，7，9｝，B=｛3，4，7，8，9｝，全集U=A B ，集合)(B A C U =13．已知平面区域}1|),{(22≤+=Ωy x y x ，0(,)01x M x y y x y ⎧≥⎫⎧⎪⎪⎪=≥⎨⎨⎬⎪⎪⎪+≤⎩⎩⎭，若在区域Ω上随机投 一点P ，则点P 落在区域M 的概率为 14．设函数()mf x x ax =+的导函数是()'21fx x =+，则()21f x dx -⎰的值等于15．不等式0)6(62<-+-a a x x 的解集中恰有三个整数，实数a 的取值范围为___________16．设1a ，2a ，…，n a 是1，2，…，n 的一个排列，把排在i a 的左边．．且比i a 小．的数的个数称为i a 的顺序数（12i n =，，，）．如在排列6，4，5，3，2，1中，5的顺序数为1，3的顺序数为0．则在由1、2、3、4、5、6、7、8这八个数字构成的全排列中，同时满足8的顺序数为2，7的顺序数为3，5的顺序数为3的不同排列的种数为_________ _第II 卷三．解答题：本大题共6小题，共75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）已知函数)0,0)(sin()(πϕωϕω≤≤>+=x x f 为偶函数，其图像上相邻的两个最高点间的距离为π2 。

试卷满分300分考试用时150分2014年1月注意事项：1.本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分。

答题前，考生务必将自己的姓名和准考证号写在题卷和答题卡上。

2.回答第Ⅰ卷时，选出每小题答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑，如需改动，用橡皮擦干净后，再选涂其他答案标号框。

写在本试卷上无效。

3.回答第Ⅱ卷时，将答案写在答题卡上，考试结束，务必将答题卡上交。

12．“网络秒杀”是现在商家推出的一种营销模式，这样的营销活动可以让一些用户以非常低廉的价格买到自己心仪的东西，因此受到了广大消费者的追捧。

小李最近通过秒杀买到了自己心仪已久的相机。

这一过程反映了“网络秒杀”的出现①丰富了商品交换的形式和手段②方便了消费者购物并减少了现金使用③减少了商品的价值量④促进了商品所有权和使用权的分离A．①②B．③④C．①③D．②④13.下图反映的是甲、乙两种互不关联的商品，当各自价格变动时对其需求量的影响程度。

根据图示，下列推断正确的是A.甲商品价格上涨不会导致消费者对其需求量大幅下降B.乙商品需求弹性大，更适合采取“降价促销”的方式C.如果居民收人不断增长，则更适合扩大甲商品的生产D.如果宏观经济不景气，则乙商品生产受到的冲击较大14.假设2012年单位甲国货币/单位乙国货币购买力之比为1:6。

2013年甲国因为超量发行货币而导致该国通货膨胀20%，同时，乙国的货币的购买力贬值20%。

其他条件不变，从购买力角度看，则两国间的汇率应为A．1：6.25 B．1:6 C．1:5.76 D．1:4.6415．中国制造在国际市场上素以物美价廉著称，但现在，低价鞋给贸易保护主义者以口实，国外反倾销已成为我国鞋类出口的重大障碍，鞋类商品在海外遭抵制的事件年年都有发生，对此，我国企业应考虑A.提高商品价格，反对贸易保护主义B.降低成本，加强管理C.树立国际品牌，增加商品附加值D.积极实施走出去战略16．刚买房就被装修的盯上；到医院孩子还没生，各种奶粉生产厂家就“闻风而动”……近年来各种个人信息遭泄露的事件频频发生，给人们的生活带来严重困扰。

一、选择题：本大题共8小题，每小题5分，共40分.在每小题给出的四个选项中，只有一项是符合题目要求的.1、设全集{}1,2,3,4,5U =,集合{}2,3,4A =,{}2,5B =,则()U B C A =（ ） A.{}5B. {}125， ，C. {}12345， ， ， ，D.∅2、下列命题中正确的是( )A.命题“x R ∀∈，2x x -0≤”的否定是“2,0x R x x ∃∈-≥”B.命题“p q ∧为真”是命题“p q ∨为真”的必要不充分条件C.若“22am bm ≤，则a b ≤”的否命题为真D.若实数,[1,1]x y ∈-，则满足221x y +≥的概率为4π.3、下列四个几何体中，各几何体的三视图有且仅有两个视图相同的是A.①②B.②③C.②④D.①③4、过原点且倾斜角为60︒的直线被圆2240x y y +-=所截得的弦长为5、等比数列{}n a 的前n 项和为n S ,已知12310a a S +=,95=a ,则=1aA.31B.31-C.91D.91- 6、已知0x 是xx f x 1)21()(+=的一个零点，)0,(),,(0201x x x x ∈-∞∈，则A.0)(,0)(21<<x f x fB.0)(,0)(21>>x f x fC.0)(,0)(21<>x f x fD.0)(,0)(21><x f x f7、已知函数()f x 是R 上的偶函数，若对于0≥x ，都有)()2(x f x f =+，且当)2,0[∈x时，)1(log )(2+=x x f ，则f (2013)f (2014)-+的值为 A ．2- B ．1- C ．1 D ．28、设函数1(1)|-1|)=1(=1)x x f x x ⎧≠⎪⎨⎪⎩（，若关于x 的方程2[()]+()+c=0f x bf x 有三个不同的实 数根123,,x x x ，则222123++x x x 等于 A. 13B. 5C. 223c +2c D. 222b +2b二、填空题：本大题共8小题，考生作答7小题，每小题5分，共35分. 9、函数ln(1-x)的定义域为______.10、若)2,0(πα∈,且2cos α+ 1sin(2)22πα+=,则tan α=______.11、在如图所示的正方体1111ABCD A B C D -中,异面直线1A B 与1B C 所成角的大小为_______12、已知向量AB 与AC 的夹角为120°,且3AB =,2AC =,若AP AB AC λ=+,且AP BC ⊥,则实数λ的值为__________.13、抛物线22(0)x py p =>的焦点为F,其准线与双曲线22133x y -=相交于,A B 两点,若ABF ∆为等边三角形,则P =_____________14、已知函数f （x ）=⎩⎨⎧>≤--.1,log 1,1)2(x x ，x x a a 若f （x ）在（-∞，+∞）上单调递增，则实数a 的取值范围为________.15、设n S 为数列{}n a 的前n 项和,1(1),,2n n n n S a n N *=--∈则(1)3a =_____; (2)2399++⋅⋅⋅+=S S S ___________.凤凰中学2014届高三第一次月考理科数学答卷（A 卷）一、选择题（每小题5分，共40分）9、 10、 11、 12、 13、 14、 15、三、解答题：本大题共6小题，共75分。

湖南省桑植一中皇仓中学2014届高三第一次联考（9月）地理试卷考试范围：区域地理、地球地图、地球运动。

一、选择题（25小题，每题2分。

每题中只有一个正确选项，共50分）读某区域经纬网分布示意图，回答１－２题1、下列有关甲乙两地的叙述正确的是：A：甲地区属于亚热带季风气候，水热充足。

B：甲地区是世界上商品谷物农业的主要分布区C：乙地区生产规模大，机械化程度高D：乙地区人口众多，市场需求量大。

2、当出现图示气压中心时，甲地区容易出现的气象灾害是A：沙尘暴B：寒潮C：洪涝D：干旱读下图，回答３－５题3、若此图表示某个国家的人口资源环境状况，则该国最有可能位于：A：非洲B：西欧C：大洋洲D：北美4、此图反映的人口、资源、环境关系的说法，错误的是：A：人类与环境之间相互影响、相互制约B：图中反映的问题，发展中国家比发达国家严重C：只要注意治理，环境会恢复到污染前的原来状况D：有些环境问题可能会影响其它国家甚至全球5、有关图中反映２０世纪后半期人口、资源、污染的变化，解释正确的是A：人口过快增长主要是城市化的结果B：资源递减是因为资源为非可再生资源C：污染加剧与人类片面追求经济增长有关D：人口过快增长是环境污染加剧的唯一原因。

读下面经纬网图，回答6～7题。

6、从A地到B地，若不考虑地形因素，最近的走法是A.一直向东走B、先向正北，再向正南C、先向东北，再向东，最后向东南走D、先向东南，再向东，最后向东北走7、读经纬网图，纬线AB约是DE长度的A、一半B、等长 C 、1.5倍D、2倍右图中阴影区域为夜半球，①③两点的地方时相同且所在的弧线两侧的日期不同，②③两点永远昼夜等长，且两点的经度差90°，④为②③的中点。

据此回答8－9题8、此时北京时间为A、8；00 B：9：00C、20：00 D：0：009、该日，图中④点的正午太阳高度为A、66°34′B、90°C、0°D、23°26′图中大圆表示昼半球范围，N表示北极点，B为北京市，OB距离约为2220Km，据此回答10一12题10、图中A点坐标为A．70°N 64°W B．70°N 116°EC．20°S 64°W D．20°S 116°E11、此刻，下列地点将会迎来日出的是A．麦加B．珀斯C．中山站D开普敦12、这一天，可能出现在A．5月底B．6月初C．7月底 D 8月初下图为我国西南某地区1月平均气温等温线图（比例尺为1/20万，数字单位：0C），据图回答13～14题13、图中所示地区较适宜种植A．甜菜B．天然橡胶C．青稞D．茶树14、甲乙两地的相对高度有可能是A．800m B．1500mC．1800m D．2500m下图为某地区太阳高度的等值线图,ad为经线。

2014湖南高考数学试卷理科一、 选择题：本大题共10个小题，每小题5分，共50分．在每个小题给出的四个选项中，只有一项是符合题目要求的 1． 满足(z ii i z +=为虚数单位）的复数z = A ．1122i + B ．1122i - C ．1122i -+ D ．1122i--2．对一个容量为N 的总体抽取容量为n 的样本，学科网当选取简单随机抽样、zxxk 系统抽样和分层抽样三种不同方法抽取样本时，总体中每个个体被抽中的概率分别是123,,,p p p 则 A ．123p p p =< B ．231p p p =< C ．132p p p =< D ．123p p p == 3．已知(),()f x g x 分别是定义在R 上的偶函数和奇函数，且32()()1,f x g x x x -=++(1)(1)f g +则＝A ．－3B ．－1C ．1D ．3 4．51(2)2x y -的展开式中23x y 的系数是zxxk A ．－20 B ．－5 C ．5 D ．205．已知命题22:,;:,.p x y x y q x y x y >-<->>若则命题若则在命题 ①p q ∧②p q ∨③()p q ∧⌝④()p q ⌝∨中，真命题是 A ．①③ B ．①④ C ．②③ D ．②④6．执行如图1所示的程序框图，如果输入的[2,2]t ∈-，则输出的S 属于 A ．[6,2]--B ．[5,1]--C ．[4,5]-D ．[3,6]-7．一块石材表示的几何何的三视图如图2所示，将该石材切削、打磨，加工成球，则能得到的最大球的半径等于A ．1B ．2C ．3D ．48．某市生产总值连续两年持续增加，第一年的增长率为p ，第二年的增长率为q ，则该市这两年生产总值的年平均增长率为A ．2p q + B ．(1)(1)12p q ++- C D 1 9．已知函数230()sin(),()0,f x x f x dx πϕ=-=⎰且则函数()f x 的图象的一条对称轴是A ．56x π=B ．712x π=C ．3x π=D ．6x π= 10．已知函数zxxk 221()(0)()ln()2x f x x e x g x x x a =+-<=++与的图象上存在关于y 轴对称的点，则a 的取值范围是 A．(-∞ B．(-∞ C．( D．( 二、填空题：本大题共6小题，考生作答5小题，每小题5分，共25分．（一）选做题（请考生在第11，12，13三题中任选两题作答，学科网如果全做，则按前两题记分）11．在平面直角坐标系中，倾斜角为4π的直线l 与曲线2cos :,(1sin x C y ααα=+⎧⎨=+⎩为参数）交于A B ，两点，则AB ｜｜＝2，以坐标原点O 为极点，x 轴正半轴为极轴建立极坐标系，则直线l 的极坐标方程是 12．如图3，已知,AB BC 是O的两条弦，,AO BC AB BC ⊥==则O 的半径等于13．若关于x 的不等式|2|3ax -<的解集为51{|}33x x -<<，则a = （二）必做题（14－16题）14．若变量,x y 满足约束条件4y xx y y k ≤⎧⎪+≤⎨⎪≥⎩，且2z x y =+的最小值为－6，则k =15．如图4，正方形ABCD DEFG 和正方形的边长分别为,()a b a b <，原点O 为AD 的中点，抛物线22(0)y px p =>经过,bC F a=两点，则16．在平面直角坐标系中，O为原点，(1,0),(3,0),A B C -动点D 满足||1,CD OA OB OD =++则｜｜的最大值是 三、解答题：本大题共6小题，共75分．学科网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）某企业有甲、乙两个研发小组，他们研发新产品成功的概率分别为2335和．现安排甲组研发新产品A ，乙组研发新产品B ．设甲、乙两组的研发相互独立．（I ） 求至少有一种新产品研发成功的概率； （II ） 若新产品A 研发成功，预计企业可获利润120万元；若新产品B 研发成功，预计企业可获利润100万元．求该企业可获利润的分布列和数学期望．18． （本小题满分12分）如图5，在平面四边形ABCD 中，127.AD CD AC ＝，＝，＝ （I ） 求cos CAD ∠的值； （II ） 若721cos ,sin ,BAD CBA ∠=-∠=求zxxk BC 的长．19． （本小题满分12分）如图6，四棱柱1111ABCD A BC D -的所有棱长都相等，11111,,ACBD O AC B D O ==四边形1111ACC A BDD B 和四边形均为矩形．（I ） 证明：1;O O ABCD ⊥底面（II ）若1160,CBA C OB D ∠=--求二面角的余弦值．20． （本小题满分13分）已知数列{n a }满足*111,||,.n n n a a a p n N +=-=∈ （I ） 若{n a }是递增数列，且12,3,23a a a 成等差数列，求p 的值； （II ）若12p =，且{21n a -}是递增数列，{2n a }学科网是递减数列，zxxk 求数列{n a }的通项公式．21． （本小题满分13分）如图7，O 为坐标原点，椭圆22122:1(0)x y C a b a b+=>>的左、右焦点分别为12,F F ，离心率为1e ；双曲线22222:1x y C a b -=的左、右焦点分别为34,F F ，离心率为2e ．已知12e e =且24|| 1.F F（I ） 求12,C C 的方程；（II ） 过1F 作1C 的不垂直于y 轴的弦AB 的中点．当直线OM 与2C 交于,P Q 两点时，求四边形APBQ 面积的最小值．22． （本小题满分13分）已知常数20,()ln(1).2xa f x ax x >=+-+函数 （I ） 讨论()f x 在区间(0,)+∞上的单调性；（II ） 若()f x 存在学科网两个极值点12,,x x 且12()()0,f x f x +>求a 的zxxk 取值范围．答案一、选择题1、B2、D3、C4、A5、C6、D7、B8、D9、A 10、B 二、填空题11、(cos sin )1p θθ-=12、3213、3- 14、2-15、116、1三、解答题 17、（本小题满分12份） 解：（I ）记E=｛甲组研发新产品成功｝，F=｛乙组研发新产品成功｝.由题设知2132(),(),(),(),3355P E P E P F P F ====故所求的概率为（Ⅱ）设企业可获利润为X （万元），则X 的可能取值为0，100,120,220.因122(0)()3515P X P EF ===⨯=, 133(100)()3515P X P E F ===⨯= 224(120)()3515P X P EF ===⨯=, 235(220)()3515P X P EF ===⨯=, 故所求的分布为数学期望为2()015E X =⨯+310015⨯+412015⨯+622015⨯=300480132021001401515++==18、（本小题满分12份）解：（I ）如图5，在ADC ∆中，由余弦定理，得222cos .2AC AD CD CAD AC AD+-∠=⋅故由题设知，cos CAD ∠==sin BAD ∠=== 于是sin x =sin()BAD CAD ∠-∠=sin cos cos sin BAD CAD BAD CAD ∠∠-∠∠=(147-在ABC ∆中，由正弦定理，BC=sin 3sin AC aCBA⋅==∠ 19、（本小题满分12份） 解：（I ）如图（a ），因为四边形11ACC A 为矩形，所以1CC AC ⊥.同理1DD BD ⊥。