2014年普通高等学校招生全国统一考试湖南卷

2014湖南省高考语文试卷答案解析2014年普通高等学校招生全国统一考试（湖南卷）语文本试题卷共七道大题，22小题，共8页。

时量150分钟。

满分150分。

一、语言的运用家风是一个家族世代相传沿袭下来的体现家族成员精神风貌、道德品质、审美格调和整体气质的家族文化风格。

一个家族之链上某一个任务出类拔（）、深（）众望而为家族其他成员所宗仰追慕。

其懿行（）言便成为家风之源，再经过家族子孙代代接力式的（）守祖训，流风余韵，绵延不绝，就形成了一个家族鲜明的家风。

1、下列汉字依次填入语段中括号内，字音和字形全部正确的一组是A、萃孚fóu佳恪géB、粹负fú佳恪kèC、粹负fù嘉恪géD、萃孚fú佳恪kè2、将下列各句中没有语病的一句填入语段中画横线处，选项是A、家风是一个影响力和美誉度都好的家庭必备的要素，也是一个家庭最为宝贵的精神财富。

B、家风即便是一个家庭最为宝贵的精神财富，也是一个有影响力有美誉度的家族必备的要素。

C、家风是一个有影响力有美誉度的家庭必备的要素，也是一个家庭最为宝贵的精神财富。

D、家风是最为宝贵的一个家族的精神财富，也是一个有影响力有美誉度的家庭必备的要素。

中国美术馆和台湾长流美术馆共同举办的“江山万里————张大千艺术展”今日与观众见面。

走进中国美术馆五层展厅，迎面的青绿山水画《谷口人家》前站满了凝神观看的人。

这青绿泼彩渲染的画面，令人叹为观止。

在展厅右侧，一副《江山万里图》静静地面对着来来往往的人流。

“”开阔的画面及其意境，向人们展示着笔墨点染间的艺术动力，也仿佛在诉说着画家眷恋故土的情怀。

3、下列选项中的“人家”与语段中加点词“人家”意义相同的一项是A、人家在何许？云外一声鸡B、小女子已许配了人家C、你走了，教人家怎么办呢D、诗书门第，勤俭人家二、文言文阅读。

阅读下面的文言文，完成5-9题。

雪屋记（明杜琼）吴有儒曰徐孟祥氏，读书绩文，志行高洁，家光福山中。

绝密★启用前2014年普通高等学校招生全国统一考试（湖南卷）语文本试卷满分150分，考试时间150分钟。

注意事项：1. 答题前，考生务必将自己的姓名、准考证号写在答题卡和本试题卷的封面上，并认真核对答题卡条形码上的姓名、准考证号和科目。

2. 选择题和非选择题均须在答题卡上作答，在本试题卷和草稿纸上作答无效。

考生在答题卡上按如下要求答题：（1）选择题部分请按题号用2B 铅笔填涂方框，修改时用橡皮擦干净，不留痕迹； （2）非选择题部分请按题号用0.5毫米黑色墨水签字笔书写，否则作答无效； （3）请勿折叠答题卡。

保持字体工整、笔迹清晰、卡面清洁。

3. 本试题卷共8页。

如缺页，考生须及时报告监考老师，否则后果自负。

4. 考试结束后，将本试题卷和答题卡一并交回。

一、语言文字运用（12分。

每小题3分）家风是一个家族世代相传沿袭下来的体现家族成员精神风貌、道德品质、审美格调和整体气质的家族文化风格。

一个家族之链上某一个人物出类拔（ ）、深（ ）众望而为家族其他成员所宗仰追慕。

其懿行（ ）言便成为家风之源，再经过家族子孙代代接力式的（ ）守祖训，流风余韵，绵延不绝，就形成了一个家族鲜明的家风。

____________1. 下列汉字依次填入语段中括号内，字音和字形全都正确的一组是（ ）A. 萃 孚f óu 佳 恪g éB. 粹 负f ú 佳 恪k èC. 粹 负f ù 嘉 恪g éD. 萃 孚f ú 嘉 恪k è2. 将下列各句中没有语病的一句填入语段中画横线处，选项是（ ）A. 家风是一个影响力和美誉度都好的家族必备的要素，也是一个家族最为宝贵的精神财富。

B. 家风即便是一个家族最为宝贵的精神财富，也是一个有影响力有美誉度的家族必备的要素。

C. 家风是一个有影响力有美誉度的家族必备的要素，也是一个家族最为宝贵的精神财富。

D. 家风是最为宝贵的一个家族的精神财富，也是一个有影响力有美誉度的家族必备的要素。

2014年普通高等学校招生全国统一考试(湖南卷)数学(文史类)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.(2014湖南,文1)设命题p:∀x∈R,x2+1>0,则p为()A.∃x0∈R,x02+1>0B.∃x0∈R,x02+1≤0C.∃x0∈R,x02+1<0D.∀x∈R,x2+1≤0答案:B解析:因为全称命题的否定为特称命题,所以p为∃x0∈R,x02+1≤0.故选B.2.(2014湖南,文2)已知集合A={x|x>2},B={x|1<x<3},则A∩B=()A.{x|x>2}B.{x|x>1}C.{x|2<x<3}D.{x|1<x<3}答案:C解析:由交集的概念,结合数轴(数轴略)可得A∩B={x|2<x<3}.故选C.3.(2014湖南,文3)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为p1,p2,p3,则()A.p1=p2<p3B.p2=p3<p1C.p1=p3<p2D.p1=p2=p3答案:D解析:由随机抽样的原则可知简单随机抽样、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p1=p2=p3,故选D.4.(2014湖南,文4)下列函数中,既是偶函数又在区间(-∞,0)上单调递增的是()A.f(x)=1x2B.f(x)=x2+1C.f(x)=x3D.f(x)=2-x答案:A解析:由偶函数的定义知,A,B为偶函数.A选项,f'(x)=-2x3在(-∞,0)恒大于0;B选项,f'(x)=2x在(-∞,0)恒小于0.故选A.5.(2014湖南,文5)在区间[-2,3]上随机选取一个数X,则X≤1的概率为()A.45B.35C.25D.15答案:B解析:由几何概型的概率公式可得P(X≤1)=3,故选B.6.(2014湖南,文6)若圆C1:x2+y2=1与圆C2:x2+y2-6x-8y+m=0外切,则m=()A.21B.19C.9D.-11答案:C解析:易知圆C1的圆心坐标为(0,0),半径r1=1.将圆C2化为标准方程(x-3)2+(y-4)2=25-m(m<25),得圆C2的圆心坐标为(3,4),半径r2=25-m(m<25).由两圆相外切得|C1C2|=r1+r2=1+25-m=5,解方程得m=9.故选C.7.(2014湖南,文7)执行如图所示的程序框图.如果输入的t∈[-2,2],则输出的S属于()A .[-6,-2]B .[-5,-1]C .[-4,5]D .[-3,6]答案:D解析:当t ∈[-2,0)时,执行以下程序:t=2t 2+1∈(1,9],S=t-3∈(-2,6];当t ∈[0,2]时,执行S=t-3∈[-3,-1],因此S ∈(-2,6]∪[-3,-1]=[-3,6].故选D .8.(2014湖南,文8)一块石材表示的几何体的三视图如图所示,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 .4答案:B 解析:由三视图可得原石材为如右图所示的直三棱柱A 1B 1C 1-ABC ,且AB=8,BC=6,BB 1=12.若要得到半径最大的球,则此球与平面A 1B 1BA ,BCC 1B 1,ACC 1A 1相切,故此时球的半径与△ABC 内切圆半径相等,故半径r=6+8-102=2.故选B .9.(2014湖南,文9)若0<x 1<x 2<1,则( ) A .e x 2−e x 1>ln x 2-ln x 1 B .e x 2−e x 1<ln x 2-ln x 1 C .x 2e x 1>x 1e x 2 D .x 2e x 1<x 1e x 2答案:C解析:设f (x )=e x -ln x ,则f'(x )=x ·e x -1.当x>0且x 趋近于0时,x ·e x -1<0;当x=1时,x ·e x -1>0,因此在(0,1)上必然存在x 1≠x 2,使得f (x 1)=f (x 2),因此A,B 不正确;设g (x )=e x x,当0<x<1时,g'(x )=(x -1)e xx 2<0,所以g (x )在(0,1)上为减函数.所以g (x 1)>g (x 2),即e x 1x 1>e x 2x 2,所以x 2e x 1>x 1e x 2.故选C .10.(2014湖南,文10)在平面直角坐标系中,O 为原点,A (-1,0),B (0, 3),C (3,0),动点D 满足|CD |=1,则|OA +OB +OD |的取值范围是( ) A .[4,6] B .[ -1, +1] C .[2 3,2 7] D .[ 7-1, 7+1]答案:D解析:设动点D 的坐标为(x ,y ),则由|CD |=1得(x-3)2+y 2=1,所以D 点的轨迹是以(3,0)为圆心,1为半径的圆.又OA +OB +OD =(x-1,y+ ),所以|OA +OB +OD |= (x -1)2+(y + 3)2,故|OA +OB +OD |的最大值为(3,0)与(1,- )两点间的距离加1,即 1,最小值为(3,0)与(1,- )两点间的距离减1,即 1.故选D . 二、填空题:本大题共5小题,每小题5分,共25分. 11.(2014湖南,文11)复数3+i i 2(i 为虚数单位)的实部等于 .答案:-3解析:由题意可得3+i i2=3+i-1=-3-i,故复数的实部为-3. 12.(2014湖南,文12)在平面直角坐标系中,曲线C : x =2+ 2t ,y =1+ 2t(t 为参数)的普通方程为 . 答案:x-y-1=0解析:两式相减得,x-y=2-1,即x-y-1=0.13.(2014湖南,文13)若变量x ,y 满足约束条件 y ≤x ,x +y ≤4,y ≥1,则z=2x+y 的最大值为 .答案:7解析:不等式组表示的平面区域如图阴影部分所示,作直线l 0:2x+y=0并平移,当直线经过点A (3,1)时,在y 轴上的截距最大,此时z 取得最大值,且最大值为7. 14.(2014湖南,文14)平面上一机器人在行进中始终保持与点F (1,0)的距离和到直线x=-1的距离相等.若机器人接触不到过点P (-1,0)且斜率为k 的直线,则k 的取值范围是 . 答案:(-∞,-1)∪(1,+∞)解析:由题意知,机器人行进的路线为抛物线y 2=4x.由题意知过点P 的直线为y=kx+k (k ≠0),要使机器人接触不到过点P 的直线,则直线与抛物线无公共点,联立方程得k4y 2-y+k=0,即Δ=1-k 2<0,解得k>1或k<-1. 15.(2014湖南,文15)若f (x )=ln(e 3x +1)+ax 是偶函数,则a= . 答案:-3解析:由题意得f (-x )=ln(e -3x +1)-ax=ln 1+e 3xe3x -ax=ln(1+e 3x )-ln e 3x -ax=ln(e 3x +1)-(3+a )x ,而f (x )为偶函数,因此f (-x )=f (x ),即ax=-(3+a )x ,所以a=-3.三、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. 16.(本小题满分12分)(2014湖南,文16)已知数列{a n }的前n 项和S n =n 2+n2,n ∈N *. (1)求数列{a n }的通项公式;(2)设b n =2a n +(-1)n a n ,求数列{b n }的前2n 项和.分析:在第(1)问中,通过S n 可求出a n ,在求解过程中要注意分n=1和n ≥2两种情况进行讨论;在第(2)问中,充分利用第(1)问的结论得到b n =2n +(-1)n n ,然后利用分组求和法分别计算(21+22+…+22n )和(-1+2-3+…+2n ),最后相加得到{b n }的前2n 项和. 解:(1)当n=1时,a 1=S 1=1;当n ≥2时,a n =S n -S n-1=n 2+n −(n -1)2+(n -1)=n.故数列{a n }的通项公式为a n =n.(2)由(1)知,b n =2n +(-1)n n.记数列{b n }的前2n 项和为T 2n ,则T 2n =(21+22+…+22n )+(-1+2-3+4-…+2n ). 记A=21+22+ (22),B=-1+2-3+4-…+2n ,则A=2(1-22n )1-2=22n+1-2,B=(-1+2)+(-3+4)+…+[-(2n-1)+2n ]=n.故数列{b n }的前2n 项和T 2n =A+B=22n+1+n-2.17.(本小题满分12分)(2014湖南,文17)某企业有甲、乙两个研发小组,为了比较他们的研发水平,现随机抽取这两个小组往年研发新产品的结果如下:(a ,b ),(a ,b ),(a ,b ),(a ,b ),(a ,b ),(a ,b ),(a ,b ),(a ,b ),(a ,b ),(a ,b ),(a ,b ),(a ,b ),(a ,b ),(a ,b ),(a ,b ) 其中a ,a 分别表示甲组研发成功和失败;b ,b 分别表示乙组研发成功和失败.(1)若某组成功研发一种新产品,则给该组记1分,否则记0分.试计算甲、乙两组研发新产品的成绩的平均数和方差,并比较甲、乙两组的研发水平;(2)若该企业安排甲、乙两组各自研发一种新产品,试估计恰有一组研发成功的概率.分析:在第(1)问中,通过已知条件可分别写出甲、乙两组的成绩,然后利用平均数公式分别计算甲、乙两组的平均成绩,再结合方差公式得到甲、乙两组的方差,进而比较甲、乙两组的研发水平;在第(2)问中,充分利用古典概型的概率公式,转化为计算基本事件的个数,从而求得概率. 解:(1)甲组研发新产品的成绩为1,1,1,0,0,1,1,1,0,1,0,1,1,0,1,其平均数为x 甲=1015=23; 方差为s 甲2=115 1-23 2×10+ 0-232×5 =29.乙组研发新产品的成绩为1,0,1,1,0,1,1,0,1,0,0,1,0,1,1, 其平均数为x 乙=9=3; 方差为s 乙2=115 1-352×9+ 0-352×6 =625. 因为x 甲>x 乙,s 甲2<s 乙2,所以甲组的研发水平优于乙组. (2)记E={恰有一组研发成功}.在所抽得的15个结果中,恰有一组研发成功的结果是(a ,b ),(a ,b ),(a ,b ),(a ,b ),(a ,b ),(a ,b ),(a ,b ),共7个.故事件E 发生的频率为715.将频率视为概率,即得所求概率为P (E )=7.18.(本小题满分12分)(2014湖南,文18)如图,已知二面角α-MN-β的大小为60°,菱形ABCD 在面β内,A ,B 两点在棱MN 上,∠BAD=60°,E 是AB 的中点,DO ⊥面α,垂足为O. (1)证明:AB ⊥平面ODE ;(2)求异面直线BC 与OD 所成角的余弦值.分析:在第(1)问中,可利用线面垂直的判定定理证明,由DO ⊥平面α可得到DO ⊥AB ,然后利用△ABD 为正三角形得到DE ⊥AB ,最后根据线面垂直的判定定理得出所证结论;在第(2)问中,充分利用第(1)问的结论AB ⊥平面ODE ,从而得到二面角α-MN-β的平面角,达到立几化平几的目的,即转化为求∠ADO 的余弦,然后利用解直角三角形的方法求出余弦值.解:(1)如图a,因为DO ⊥α,AB ⊂α,所以DO ⊥AB.图a连接BD ,由题设知,△ABD 是正三角形. 又E 是AB 的中点,所以DE ⊥AB. 而DO ∩DE=D ,故AB ⊥平面ODE.(2)因为BC ∥AD ,所以BC 与OD 所成的角等于AD 与OD 所成的角,即∠ADO 是BC 与OD 所成的角. 由(1)知,AB ⊥平面ODE ,所以AB ⊥OE.又DE ⊥AB ,于是∠DEO 是二面角α-MN-β的平面角,从而∠DEO=60°. 不妨设AB=2,则AD=2.易知DE= 3. 在Rt △DOE 中,DO=DE ·sin 60°=3. 连接AO ,在Rt △AOD 中,cos ∠ADO=DO=32=3.故异面直线BC 与OD 所成角的余弦值为34.19.(本小题满分13分)(2014湖南,文19)如图,在平面四边形ABCD 中,DA ⊥AB ,DE=1,EC= 7,EA=2,∠ADC=2π3,∠BEC=π3. (1)求sin ∠CED 的值; (2)求BE 的长.分析:在第(1)问中,通过已知条件,借助余弦定理得到CD 的长,然后在△CDE 中,利用正弦定理得到∠CED 的正弦值;在第(2)问中,利用∠CED 的正弦值求得其余弦值,然后利用角之间的关系表示出∠AEB ,进而表示出∠AEB 的余弦值,最后在Rt △EAB 中利用边角关系,求得BE 的长. 解:如题图,设∠CED=α.(1)在△CDE 中,由余弦定理,得EC 2=CD 2+DE 2-2CD ·DE ·cos ∠EDC. 于是由题设知,7=CD 2+1+CD ,即CD 2+CD-6=0. 解得CD=2(CD=-3舍去). 在△CDE 中,由正弦定理,得EC sin ∠EDC=CDsin α. 于是,sin α=CD ·sin 2π3EC =2× 327=217,即sin ∠CED= 21.(2)由题设知,0<α<π3,于是由(1)知,cos α= 1-sin 2α= 1-2149=2 77. 而∠AEB=2π-α,所以cos ∠AEB=cos 2π-α =cos 2πcos α+sin 2πsin α=-1cos α+ 3sin α=-1×2 7+ 3×21=7.在Rt △EAB 中,cos ∠AEB=EA =2,故BE=2= 714=4 7.20.(本小题满分13分)(2014湖南,文20)如图,O 为坐标原点,双曲线C 1:x 2a 12−y 2b 12=1(a 1>0,b 1>0)和椭圆C 2:y 2a 22+x 2b 22=1(a 2>b 2>0)均过点P2 3,1 ,且以C 1的两个顶点和C 2的两个焦点为顶点的四边形是面积为2的正方形.(1)求C 1,C 2的方程;(2)是否存在直线l ,使得l 与C 1交于A ,B 两点,与C 2只有一个公共点,且|OA +OB |=|AB |?证明你的结论.分析:在第(1)问中,利用已知条件结合图形以及双曲线、椭圆中a ,b ,c 的几何意义,列出关于a 1,b 1,a 2,b 2的方程,得到它们的值,从而求出双曲线C 1、椭圆C 2的方程;在第(2)问中,首先对直线l 的斜率进行分类讨论,当斜率k 不存在时易得A ,B 两点的坐标,进而判断满足题设条件的直线l 不存在;当斜率k 存在时,可先设出l 的方程,然后代入曲线方程,利用根与系数的关系并结合向量的运算,依此判断满足题设条件的直线l 不存在. 解:(1)设C 2的焦距为2c 2,由题意知,2c 2=2,2a 1=2.从而a 1=1,c 2=1.因为点P 2 33,1 在双曲线x 2-y 2b 12=1上,所以2 332−1b 12=1.故b 12=3. 由椭圆的定义知2a 2= 2 332+(1-1)+ 2 332+(1+1)=2 3.于是a 2= 3,b 22=a 22−c 22=2.故C 1,C 2的方程分别为x 2-y 23=1,y 23+x 22=1. (2)不存在符合题设条件的直线.①若直线l 垂直于x 轴,因为l 与C 2只有一个公共点,所以直线l 的方程为x= 或x=- .当x= 2时,易知A ( 2, 3),B ( 2,- 3), 所以|OA +OB |=2 2,|AB |=2 3. 此时,|OA+OB |≠|AB |. 当x=- 2时,同理可知,|OA +OB |≠|AB |.②若直线l 不垂直于x 轴,设l 的方程为y=kx+m. 由 y =kx +m ,x 2-y 2=1得(3-k 2)x 2-2kmx-m 2-3=0. 当l 与C 1相交于A ,B 两点时,设A (x 1,y 1),B (x 2,y 2),则x 1,x 2是上述方程的两个实根,从而x 1+x 2=2km 3-k2,x 1x 2=m 2+3k 2-3.于是y 1y 2=k 2x 1x 2+km (x 1+x 2)+m 2=3k 2-3m 2k 2-3.由 y =kx +m ,y 2+x 2=1得(2k 2+3)x 2+4kmx+2m 2-6=0. 因为直线l 与C 2只有一个公共点,所以上述方程的判别式Δ=16k 2m 2-8(2k 2+3)(m 2-3)=0. 化简,得2k 2=m 2-3,因此OA·OB =x 1x 2+y 1y 2=m 2+3k 2-3+3k 2-3m 2k 2-3=-k 2-3k 2-3≠0,于是OA 2+OB 2+2OA ·OB ≠OA 2+OB 2-2OA ·OB , 即|OA +OB 2|≠|OA −OB 2|,故|OA +OB |≠|AB |. 综合①,②可知,不存在符合题设条件的直线.21.(本小题满分13分)(2014湖南,文21)已知函数f (x )=x cos x-sin x+1(x>0).(1)求f (x )的单调区间;(2)记x i 为f (x )的从小到大的第i (i ∈N *)个零点,证明:对一切n ∈N *,有1x 12+1x 22+…+1n 2<2.分析:在第(1)问中,通过已知条件,借助导数,转化为判断导数在(0,+∞)上的符号,进而得出函数的单调区间;在第(2)问中,充分利用第(1)问的结论,得到f (x )在(n π,(n+1)π)上存在零点,从而得出n π<x n+1<(n+1)π,然后分n=1,n=2,n ≥3三种情况讨论112+122+…+1n 2的值与2的大小关系,即可得证. 解:(1)f'(x )=cos x-x sin x-cos x=-x sin x.令f'(x )=0,得x=k π(k ∈N *).当x ∈(2k π,(2k+1)π)(k ∈N )时,sin x>0,此时f'(x )<0; 当x ∈((2k+1)π,(2k+2)π)(k ∈N )时,sin x<0,此时f'(x )>0.故f (x )的单调递减区间为(2k π,(2k+1)π)(k ∈N ),单调递增区间为((2k+1)π,(2k+2)π)(k ∈N ). (2)由(1)知,f (x )在区间(0,π)上单调递减. 又f π=0,故x 1=π.当n ∈N *时,因为f (n π)f ((n+1)π)=[(-1)n n π+1][(-1)n+1(n+1)π+1]<0,且函数f (x )的图象是连续不断的,所以f (x )在区间(n π,(n+1)π)内至少存在一个零点.又f (x )在区间(n π,(n+1)π)上是单调的,故n π<x n+1<(n+1)π.因此,当n=1时,1x 12=4π2<23; 当n=2时,1x 12+1x 22<1π2(4+1)<23;当n ≥3时,1x 12+1x 22+…+1x n 2<1π2 4+1+122+…+1(n -1)2 <1π2 5+11×2+…+1(n -2)(n -1) <12 5+ 1-1 + 1-1 +…+ 1n -2-1n -1 =1π2 6-1n -1<6π2<23. 综上所述,对一切n ∈N *,1x 12+1x 22+…+1x n 2<23.。

2014年湖南高考语文试题及答案以下是查字典语文小编给大家整理编辑的2014年湖南高考语文试题及答案,一起来看看吧！ 2014年普通高等学校招生全国统一考试（湖南本试题卷共七道大题，22小题，共8页。

时量150 分钟。

满分150分亠、语言文字运用（12分。

每小题3分）家风是一个家族世代相传沿袭下来的体现家族成员精神风貌、道德品质、审美格调和整体气质的家族文化风格。

一个家族之链上某一个人物出类拔（）、深（）众望而为家族其他成员所宗仰追慕，其懿行（）言便成为家风之源，再经过家族子孙代代接力式的（）守祖训，流风余韵，绵延不绝，就形成了一个家族鲜明的家风。

1.下列汉字依次填入语段中括号内，字音和字形全部正确的一组是萃孚f&oacute;u 佳恪g&eacute; B. 粹负A.f&uacute; 佳恪k&egrave;C.粹负f&ugrave; 嘉恪g&eacute;D. 萃孚f&uacute; 嘉恪k&egrave;【答案】D【解析】出类拔萃：拔：超出；类：同类；萃：原为草丛生的样子，引申指同类丛聚。

后以出类拔萃形容卓越出众，不同一般。

萃字从草不从米，据义定形。

深孚众望：使大家信服，符合大家的期望。

孚：使人信服、信任、相信。

读f&uacute;,褒义词。

深负众望：指辜负了大家的期望。

负：辜负，读f&ugrave;，贬义词。

懿行嘉言：嘉，美好的意思，不能写作佳。

常指有益的言论和高尚的行为。

2009年湖南卷字音题曾考过嘉言懿行(y&igrave;) 。

恪守：谨慎而恭敬遵守。

恪读k&egrave;，形声不能套读半边。

试题分析：本题属于一题多考，既考字音、字形，又考成语运用。

题目新颖，含金量极高。

考点：识记现代汉语普通话常用字的字音。

能力层级为识记A。

考点：识记并正确书写现代常用规范汉字。

能力层级为识记A。

考点：正确使用词语（包括熟语）。

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)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 ＋iz ＝i(i 为虚数单位)的复数z ＝( )A.12＋12iB.12－12i C ．－12＋12i D ．－12－12i2．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为p 1、p 2、p 3，则( )A ．p 1＝p 2<p 3B ．p 2＝p 3<p 1C ．p 1＝p 3<p 2D ．p 1＝p 2＝p 33．已知f (x )，g (x )分别是定义在R 上的偶函数和奇函数，且f (x )－g (x )＝x 3＋x 2＋1，则f (1)＋g (1)＝( )A ．－3B ．－1C ．1D ．34.⎝⎛⎭⎫12x －2y 5的展开式中x 2y 3的系数是( ) A ．－20 B ．－5 C ．5 D ．205．已知命题p ：若x >y ，则－x <－y ：命题q ：若x >y ，则x 2>y 2，在命题①p ∧q ；②p ∨q ；③p ∧(q )；④(p )∨q 中，真命题是( )A ．①③B ．①④C ．②③D ．②④6．执行如图所示的程序框图，如果输入的t ∈[－2,2]，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体的三视图如图所示，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.p ＋q 2B.(p ＋1)(q ＋1)－12C.pqD.(p ＋1)(q ＋1)－19．已知函数f (x )＝sin(x －φ)，且＝0，则函数f(x)的图象的一条对称轴是( )A ．x ＝5π6B ．x ＝7π12C ．x ＝π3D ．x ＝π610．已知函数f (x )＝x 2＋e x －12(x <0)与g (x )＝x 2＋ln(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.⎝⎛⎭⎫－∞，1e B ．(－∞，e) C.⎝⎛⎭⎫－1e ，e D.⎝⎛⎭⎫－e ，1e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π4的直线l 与曲线C ：(a 为参数)交于A ，B 两点，且|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，已知AB ，BC 是⊙O 的两条弦，AO ⊥BC ，AB ＝3，BC ＝22，则⊙O 的半径等于________．13．若关于x 的不等式|ax －2|<3的解集为⎩⎨⎧⎭⎬⎫x －53<x <13，则a ＝________.(二)必做题(14－16题)14．若变量x ，y 满足约束条件且z ＝2x ＋y 的最小值为－6，则k ＝________.15.如图，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y 2＝2px (p >0)经过C ，F 两点，则ba＝________.16．在平面直角坐标系中，O 为原点，A (－1,0)，B (0，3)，C (3，0)，动点D 满足||＝1，则||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．(1)求至少有一种新产品研发成功的概率；(2)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，在平面四边形ABCD 中，AD ＝1，CD ＝2，AC ＝7(1)求cos ∠CAD 的值； (2)若cos ∠BAD ＝－714，sin ∠CBA ＝216，求BC 的长．19．(本小题满分12分)如图，四棱柱ABCD -A 1B 1C 1D 1的所有棱长都相等，AC ∩BD ＝O ，A 1C 1∩B 1D 1＝O 1，四边形ACC 1A 1和四边形BDD 1B 1均为矩形．(1)证明：O 1O ⊥底面ABCD ；(2)若∠CBA ＝60°，求二面角C 1-OB 1-D 的余弦值．20．(本小题满分13分)已知数列{a n }满足a 1＝1，|a n ＋1－a n |＝p n ，n ∈N *.(1)若{a n }是递增数列，且a 1,2a 2,3a 3成等差数列，求p 的值；(2)若p ＝12，且{a 2n －1}是递增数列，{a 2n }是递减数列，求数列{a n }的通项公式．21．(本小题满分13分)如图，O 为坐标原点，椭圆C 1：x 2a 2＋y 2b 2＝1(a >b >0)的左、右焦点分别为F 1，F 2，离心率为e 1：双曲线C 2：x 2a 2－y 2b 2＝1的左、右焦点分别为F 3，F 4，离心率为e 2.已知e 1e 2＝32，且|F 2F 4|＝3－1.(1)求C 1、C 2的的方程；(2)过F 1作C 1的不垂直于y 轴的弦AB ，M 为AB 的中点，当直线OM 与C 2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a >0，函数f (x )＝ln(1＋ax )－2xx ＋2.(1)讨论f (x )在区间(0，＋∞)上的单调性；(2)若f (x )存在两个极值点x 1、x 2，且f (x 1)＋f (x 2)>0，求a 的取值范围．2014年普通高等答案一、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．解析：选B 去掉分母，得z ＋i ＝z i ，所以(1－i)z ＝－i ，解得z ＝－i 1－i ＝12－12i ，选B.2．解析：选D 根据抽样方法的概念可知，简单随机抽样、系统抽样和分层抽样三种抽样方法，每个个体被抽到的概率都是nN，故p 1＝p 2＝p 3，故选D.3．解析：选C 用“－x ”代替“x ”，得f (－x )－g (－x )＝(－x )3＋(－x )2＋1，化简得f (x )＋g (x )＝－x 3＋x 2＋1，令x ＝1，得f (1)＋g (1)＝1，故选C.4.解析：选A 由二项展开式的通项可得，第四项T 4＝C 35⎝⎛⎭⎫12x 2(－2y )3＝－20x 2y 3，故x 2y 3的系数为－20，选A.5．解析：选C 由不等式的性质可知，命题p 是真命题，命题q 为假命题，故①p ∧q 为假命题，②p ∨q 为真命题，③q 为真命题，则p ∧(q )为真命题，④p 为假命题，则(p )∨q 为假命题，所以选C.6．解析：选D 由程序框图可得S ＝，其值域为(－2,6]∪[－3，－1]＝[－3,6]，故选D.7．解析：选B 该几何体为直三棱柱，底面是边长分别为6,8,10的直角三角形，侧棱长为12，故能得到的最大球的半径等于底面直角三角形内切圆的半径，其半径为r ＝2×12×6×86＋8＋10＝2，故选B.8．解析：选D 设年平均增长率为x ，原生产总值为a ，则(1＋p )(1＋q )a ＝a (1＋x )2，解得x ＝(1＋p )(1＋q )－1，故选D.9．解析：选A 由定积分sin (x －φ)d x ＝－cos (x －φ)＝12cos φ－32sin φ＋cos φ＝0，得tan φ＝3，所以φ＝π3＋k π(k ∈Z )，所以f (x )＝sin ⎝⎛⎭⎫x －π3－k π(k ∈Z )，由正弦函数的性质知y ＝sin ⎝⎛⎭⎫x －π3－k π与y ＝sin ⎝⎛⎭⎫x －π3的图象的对称轴相同，令x －π3＝k π＋π2，则x ＝k π＋5π6(k ∈Z )，所以函数f (x )的图象的对称轴为x ＝k π＋56π(k ∈Z )，当k ＝0，得x ＝5π6，选A.10．解析：选B 由题意可得，当x >0时，y ＝f (－x )与y ＝g (x )的图象有交点，即g (x )＝f (－x )有正解，即x 2＋ln(x ＋a )＝(－x )2＋e －x －12有正解，即e －x －ln(x ＋a )－12＝0有正解，令F (x )＝e －x －ln(x ＋a )－12，则F ′(x )＝－e －x －1x ＋a <0，故函数F (x )＝e －x －ln(x ＋a )－12在(0，＋∞)上是单调递减的，要使方程g (x )＝f (－x )有正解，则存在正数x 使得F (x )≥0，即e －x －ln(x＋a )－12≥0，所以a ≤e e －x －12－x ，又y ＝e e －x －12－x 在(0，＋∞)上单调递减，所以a <e e －0－12－0＝e 12，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(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．解析：曲线C 的方程为(x －2)2＋(y －1)2＝1，圆心C (2，1)，半径r ＝1，又弦长|AB |＝2，故AB 为圆C 的直径，即直线l 过圆心C (2,1)，又直线l 的斜率k ＝1，所以直线l 的方程为x －y ＝1，极坐标方程为ρ(cos θ－sin θ)＝1.答案：ρ(cos θ－sin θ)＝112．解析：设AO ，BC 的交点为D ，由已知可得D 为BC 的中点，则在直角三角形ABD 中，AD ＝AB 2－BD 2＝1，设圆的半径为r ，延长AO 交圆O 于点E ，由圆的相交弦定理可知BD ·CD ＝AD ·DE ，即(2)2＝2r －1，解得r ＝32.答案：3213．解析：由不等式的解集可知－53，13为不等式对应的方程|ax －2|＝3的根，即，解得a ＝－3.答案：－3(二)必做题(14－16题)14．解析：画出可行域(图略)，由题意可知不等式组表示的区域为一三角形，平移参照直线2x ＋y ＝0，可知在点(k ，k )处z ＝2x ＋y 取得最小值，故z min ＝2k ＋k ＝－6，解得k ＝－2.答案：－215.解析：由正方形的定义可知BC ＝CD ，结合抛物线的定义得点D 为抛物线的焦点，所以|AD |＝p ＝a ，D ⎝⎛⎭⎫p 2，0，F ⎝⎛⎭⎫p 2＋b ，b ，将点F 的坐标代入抛物线的方程得b 2＝2p ⎝⎛⎭⎫p2＋b ＝a 2＋2ab ，变形得⎝⎛⎭⎫b a 2－2b a －1＝0，解得b a ＝1＋2或b a ＝1－2(舍去)，所以ba＝1＋ 2. 答案：1＋ 216．解析：设D (x ，y )，由||＝1，得(x －3)2＋y 2＝1，向量＝(x －1，y ＋3)，故||＝(x －1)2＋(y ＋3)2的最大值为圆(x －3)2＋y 2＝1上的动点到点(1，－3)距离的最大值，其最大值为圆(x －3)2＋y 2＝1的圆心(3,0)到点(1，－3)的距离加上圆的半径，即(3－1)2＋(0＋3)2＋1＝1＋7.答案：1＋7三、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．解析：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P (E )＝23，P (E )＝13，P (F )＝35，P (F )＝25.且事件E 与F 、E 与F 、E 与F 、E 与F 都相互独立． (1)记H ＝{至少有一种新产品研发成功}，则H ＝E F ，于是 P (H )＝P (E )P (F )＝13×25＝215，故所求的概率为P (H )＝1－P (H )＝1－215＝1315.(2)设企业可获利润为X (万元)，则X 的可能取值为0,100,120,220. 因P (X ＝0)＝P (E F )＝13×25＝215，P (X ＝100)＝P (E F )＝13×35＝315，P (X ＝120)＝P (E F )＝23×25＝415，P (X ＝220)＝P (EF )＝23×35＝615.故所求的分布列为数学期望为E (X )＝0×215＋100×315＋120×415＋220×615＝300＋480＋1 32015＝2 10015＝140.18．解析：(1)如题图，在△ADC 中，由余弦定理，得cos ∠CAD ＝AC 2＋AD 2－CD 22AC ·AD .故由题设知，cos ∠CAD ＝7＋1－427＝277.(2)如题图，设∠BAC ＝α，则α＝∠BAD －∠CAD . 因为cos ∠CAD ＝277，cos ∠BAD ＝－714，所以sin ∠CAD ＝1－cos 2∠CAD ＝1－⎝⎛⎭⎫2772＝217，sin ∠BAD ＝1－cos 2∠BAD ＝1－⎝⎛⎭⎫－7142＝32114.于是sin α＝sin(∠BAD －∠CAD )＝sin ∠BAD cos ∠CAD －cos ∠BAD sin ∠CAD ＝32114×277－⎝⎛⎭⎫－714×217 ＝32. 在△ABC 中，由正弦定理，BC sin α＝ACsin ∠CBA. 故BC ＝AC ·sin αsin ∠CBA＝7×32216＝3.19．解析：(1)如图(a )，因为四边形ACC 1A 1为矩形，所以CC 1⊥AC ，同理DD 1⊥BD ，因为CC 1∥DD 1，所以CC 1⊥BD ，而AC ∩BD ＝O ，因此CC 1⊥底面ABCD .由题设知，O 1O ∥C 1C ，故O 1O ⊥底面ABCD .(2)法一：如图(a)，过O 1作O 1H ⊥OB 1于H ，连接HC 1.由(1)知，O 1O ⊥底面ABCD ，所以O 1O ⊥底面A 1B 1C 1D 1，于是O 1O ⊥A 1C 1. 又因为四棱柱ABCD -A 1B 1C 1D 1的所有棱长都相等，所以四边形A 1B 1C 1D 1是菱形， 因此A 1C 1⊥B 1D 1，从而A 1C 1⊥平面BDD 1B 1，所以A 1C 1⊥OB 1，于是OB 1⊥平面O 1HC 1， 进而OB 1⊥C 1H ，故∠C 1HO 1是二面角C 1-OB 1-D 的平面角． 不妨设AB ＝2.因为∠CBA ＝60°，所以OB ＝3，OC ＝1，OB 1＝7. 在Rt △OO 1B 1中，易知O 1H ＝OO 1·O 1B 1OB 1＝237，而O 1C 1＝1，于是C 1H ＝O 1C 21＋O 1H 2＝1＋127＝197.故cos ∠C 1HO 1＝O 1HC 1H ＝237197＝25719.即二面角C 1-OB 1-D 的余弦值为25719.法二：因为四棱柱ABCD -A 1B 1C 1D 1的所有棱长都相等，所以四边形ABCD 是菱形，因此AC ⊥BD .又O 1O ⊥底面ABCD ，从而OB ，OC ，OO 1两两垂直．如图(b)，以O 为坐标原点，OB ，OC ，OO 1所在直线分别为x 轴、y 轴、z 轴，建立空间直角坐标系O -xyz .不妨设AB ＝2，因为∠CBA ＝60°，所以OB ＝3，OC ＝1.于是相关各点的坐标为：O (0,0,0)，B 1(3，0,2)，C 1(0,1,2)．易知，n 1＝(0,1,0)是平面BDD 1B 1的一个法向量．设n 2＝(x ，y ，z )是平面OB 1C 1的一个法向量，则即⎩⎨⎧3x ＋2z ＝0，y ＋2z ＝0.取z ＝－3，则x ＝2，y ＝23，所以n 2＝(2,23，－3)，设二面角C 1-OB 1-D 的大小为θ，易知θ是锐角，于是cos θ＝|cos 〈n 1，n 2〉|＝⎪⎪⎪⎪n 1·n 2|n 1|·|n 2|＝2319＝25719.故二面角C 1-OB 1-D 的余弦值为25719.20．解析：(1)因为{a n }是递增数列，所以a n ＋1－a n ＝|a n ＋1－a n |＝p n . 而a 1＝1，因此a 2＝p ＋1，a 3＝p 2＋p ＋1.又a 1,2a 2,3a 3成等差数列，所以4a 2＝a 1＋3a 3，因而3p 2－p ＝0， 解得p ＝13或p ＝0.当p ＝0时，a n ＋1＝a n ，这与{a n }是递增数列矛盾，故p ＝13.(2)由于{a 2n －1}是递增数列，因而a 2n ＋1－a 2n －1>0，于是(a 2n ＋1－a 2n )＋(a 2n －a 2n －1)>0. ① 但122n <122n －1，所以|a 2n ＋1－a 2n |<|a 2n －a 2n －1|. ② 由①，②知，a 2n －a 2n －1>0，因此a 2n －a 2n －1＝⎝⎛⎭⎫122n －1＝(－1)2n22n －1. ③因为{a 2n }是递减数列，同理可得，a 2n ＋1－a 2n <0，故a 2n ＋1－a 2n ＝－⎝⎛⎭⎫122n ＝(－1)2n ＋122n. ④由③，④即知，a n ＋1－a n ＝(－1)n ＋12n.于是a n ＝a 1＋(a 2－a 1)＋(a 3－a 2)＋…＋(a n －a n －1) ＝1＋12－122＋…＋(－1)n 2n －1＝1＋12·1－⎝⎛⎭⎫－12n －11＋12＝43＋13·(－1)n2n －1. 故数列{a n }的通项公式为a n ＝43＋13·(－1)n2n －1.21．解析：(1)因为e 1e 2＝32，所以a 2－b 2a ·a 2＋b 2a ＝32，即a 4－b 4＝34a 4，因此a 2＝2b 2，从而F 2(b,0)，F 4(3b,0)．于是3b －b ＝|F 2F 4|＝3－1，所以b ＝1，a 2＝2，故C 1，C 2的方程分别为x 22＋y 2＝1，x 22－y 2＝1.(2)因AB 不垂直于y 轴，且过点F 1(－1,0)，故可设直线AB 的方程为x ＝my －1. 由⎩⎪⎨⎪⎧x ＝my －1，x 22＋y 2＝1得(m 2＋2)y 2－2my －1＝0. 易知此方程的判别式大于0.设A (x 1，y 1)，B (x 2，y 2)，则y 1，y 2是上述方程的两个实根，所以y 1＋y 2＝2mm 2＋2，y 1y 2＝－1m 2＋2.因此x 1＋x 2＝m (y 1＋y 2)－2＝－4m 2＋2，于是AB 的中点为M ⎝ ⎛⎭⎪⎫－2m 2＋2，m m 2＋2，故直线PQ 的斜率为－m 2，PQ 的方程为y ＝－m2x ，即mx ＋2y ＝0.由⎩⎨⎧y ＝－m 2x ，x22－y 2＝1得(2－m 2)x 2＝4，所以2－m 2>0，且x 2＝42－m 2，y 2＝m 22－m 2，从而|PQ |＝2x 2＋y 2＝2m 2＋42－m 2. 设点A 到直线PQ 的距离为d ，则点B 到直线PQ 的距离也为d ，所以2d ＝|mx 1＋2y 1|＋|mx 2＋2y 2|m 2＋4. 因为点A ，B 在直线mx ＋2y ＝0的异侧，所以(mx 1＋2y 1)(mx 2＋2y 2)<0，于是|mx 1＋2y 1|＋|mx 2＋2y 2|＝|mx 1＋2y 1－mx 2－2y 2|， 从而2d ＝(m 2＋2)|y 1－y 2|m 2＋4. 又因为|y 1－y 2|＝(y 1＋y 2)2－4y 1y 2＝22·1＋m 2m 2＋2，所以2d ＝22·1＋m 2m 2＋4. 故四边形APBQ 的面积S ＝12|PQ |·2d ＝22·1＋m 22－m 2＝22·－1＋32－m 2. 而0<2－m 2≤2，故当m ＝0时，S 取得最小值2.综上所述，四边形APBQ 面积的最小值为2. 22．解析：(1)f ′(x )＝a 1＋ax －2(x ＋2)－2x (x ＋2)2＝ax 2＋4(a －1)(1＋ax )(x ＋2)2. (*) 当a ≥1时，f ′(x )>0.此时，f (x )在区间(0，＋∞)上单调递增． 当0<a <1时，由f ′(x )＝0得x 1＝21－a a ⎝ ⎛⎭⎪⎫x 2＝－21－a a 舍去. 当x ∈(0，x 1)时，f ′(x )<0；当x ∈(x 1，＋∞)时，f ′(x )>0.故f (x )在区间(0，x 1)上单调递减，在区间(x 1，＋∞)上单调递增．综上所述，当a ≥1时，f (x )在区间(0，＋∞)上单调递增；当0<a <1时，f (x )在区间⎝ ⎛⎭⎪⎫0，21－a a 上单调递减，在区间⎝ ⎛⎭⎪⎫21－a a ，＋∞上单调递增． (2)由(*)式知，当a ≥1时，f ′(x )≥0，此时f (x )不存在极值点．因而要使得f (x )有两个极值点，必有0<a <1.又f (x )的极值点只可能是x 1＝21－a a 和x 2＝－21－a a ，且由f (x )的定义可知，x >－1a 且x ≠－2，所以－21－a a >－1a ，－21－a a ≠－2，解得a ≠12.此时，由(*)式易知，x 1，x 2分别是f (x )的极小值点和极大值点．而f (x 1)＋f (x 2)＝ln(1＋ax 1)－2x 1x 1＋2＋ln(1＋ax 2)－2x 2x 2＋2＝ln[1＋a (x 1＋x 2)＋a 2x 1x 2]－4x 1x 2＋4(x 1＋x 2)x 1x 2＋2(x 1＋x 2)＋4＝ln(2a －1)2－4(a －1)2a －1＝ln(2a －1)2＋22a －1－2. 令2a －1＝x ，由0<a <1且a ≠12知 当0<a <12时，－1<x <0；当12<a <1时，0<x <1. 记g (x )＝ln x 2＋2x－2. (ⅰ)当－1<x <0时，g (x )＝2ln(－x )＋2x －2，所以g ′(x )＝2x －2x 2＝2x －2x 2<0.因此，g (x )在区间(－1,0)上单调递减，从而，g (x )<g (－1)＝－4<0.故当0<a <12时，f (x 1)＋f (x 2)<0.(ⅱ)当0<x <1时，g (x )＝2ln x ＋2x －2，所以g ′(x )＝2x －2x 2＝2x －2x 2<0. 因此，g (x )在区间(0,1)上单调递减，从而g (x )>g (1)＝0.故当12<a <1时，f (x 1)＋f (x 2)>0. 综上所述，满足条件的a 的取值范围为⎝⎛⎭⎫12，1.。

2014·湖南卷(课标语文)一、语言文字运用(12分。

每小题3分)家风是一个家族世代相传沿袭下来的体现家族成员精神风貌、道德品质、审美格调和整体气质的家族文化风格。

一个家族之链上某一个人物出类拔()、深()众望而为家族其他成员所宗仰追慕，其懿行()言便成为家风之源，再经过家族子孙代代接力式的()守祖训，流风余韵，绵延不绝，就形成了一个家族鲜明的家风。

____________________ 1．[2014·湖南卷] 下列汉字依次填入语段中括号内，字音和字形全都正确的一组是()A．萃孚fóu佳恪géB．粹负fú佳恪kèC．粹负fù嘉恪géD．萃孚fú嘉恪kè1．D[解析] 本题考查“识记现代汉语普通话常用字的字音”和“识记并正确书写现代常用规范汉字”，并杂有成语的辨析。

出类拔萃：形容超出同类。

萃，草丛生的样子，引申为聚集，指成群的人或物。

懿行嘉言：有教育意义的好言语和好行为。

嘉，美好。

“恪”读“kè”。

“孚”和“负”需要认真辨析：“孚”意指“使……信服”，“负”意为“辜负”。

根据语境应该选“深孚众望”，即“使大家信服，符合大家的期望”。

2．[2014·湖南卷] 将下列各句中没有．．语病的一句填入语段中画横线处，选项是() A．家风是一个影响力和美誉度都好的家族必备的要素，也是一个家族最为宝贵的精神财富。

B．家风即便是一个家族最为宝贵的精神财富，也是一个有影响力有美誉度的家族必备的要素。

C．家风是一个有影响力有美誉度的家族必备的要素，也是一个家族最为宝贵的精神财富。

D．家风是最为宝贵的一个家族的精神财富，也是一个有影响力有美誉度的家族必备的要素。

2．C[解析] 本题考查辨析病句。

A项，因语序不当而有歧义，“影响力和美誉度都好”既可以理解成“家族”的定语，也可以理解为“要素”的定语。

2014年普通高等学校招生全国统一考试（湖南卷)英语本试题卷分四个部分，共12页。

时量120分钟。

满分150分。

Part I Listening Comprehension (30 marks)Section A (22.5 marks)Directions: In this section，you will hear six conversations between two speakers. For each conversation, there are several questions and each question is followed by three choices marked A, B and C. Listen carefully and then choose the best answer for each question.You will hear each conversation TWICE.Example:When will the magazine probably arrive?A. Wednesday.B. Thursday.C. Friday.The answer is B.Conversation 11. What will the woman do first?A. Take a shower.B. Go camping.C. Set up a time.2. When will the man probably call the woman?A. Thursday.B. Friday.C. Sunday.Conversation 23. What is the man going to do?A. Have a coffee break.B. See a doctor.C. Buy a pet.4. What happened to the man?A. He fell ill.B. He lost his dog.C. He slept badly.Conversation 35. What is the woman?A. A bus driver.B. A waitress.C. A tour guide.6. What does the man want to get?A. Some gifts.B. A menu.C. A bus schedule.Conversation 47. What did the man do yesterday?A. He saw a movie.B. He watched TV.C. He visited some friends.8. What time will the speakers probably meet this Saturday evening?A. At 6:30.B. At 7:00.C. At 7:30.9. Which of the following will the man buy?A. Some drinks.B. A birthday cake.C. Concert tickets.Conversation 510. What is the woman doing now?A. She is serving a customer.B. She is conducting an interview.C. She is doing some recording.11 .When does the man go to the nursing home?A. Tuesdays.B. Thursdays.C. Sundays.12. Where will the man probably be working next Monday?A. At the airport nearby.B. In the studio next doorC. At the store downtown.Conversation 613. Why does the woman call the man?A. The oven doesn‟t work.B. The heater won‟t start.C. The plug is broken.14. Who will handle the problem first tomorrow evening?A. The woman.B. The man.C. A worker.15. Who is the woman speaking to?A. Her husband.B. Her house owner.C. Her boss.Section B (7.5 marks)Directions: In this section, you will hear a short passage. Listen carefully and then Jill in the numbered blanks with the information you have heard. Fill in each blank with NO MORE THAN THREE WORDS.Part II Language Knowledge (45 marks)Section A (15 marks)Directions: For each of the following unfinished sentences there are four choices marked A, B, C and IX Choose the one that best completes the sentence.The wild flowers looked like a soft orange blanket the desert,A. coveringB. coveredC. coverD. to coverThe answer is A.21.Children，when by their parents, are allowed to enter the stadium.A. to be accompaniedB. to accompanyC. accompanyingD. accompanied22.If Mr. Dewey _____ present, he would have offered any possible assistance to the people there.A. wereB. had beenC. should beD. was23. ____your own needs and styles of communication is as important as learning to convey your affection and emotions.A. UnderstandingB. To be understoodC. Being understoodD. Having understood24. As John Lennon once said，life is_____ happens to you while you are busy making other plans.A. whichB. thatC. whatD. where25. —I‟ve prepared all kinds of food for the picnic.—Do you mean we_____ bring anything with us?A. can‟tB. mustn‟tC. shan‟tD. needn't26；You will never gain success you are fully devoted to your work.A. whenB. becauseC. afterD. unless27. There is no greater pleasure than lying on my back in the middle of the grassland, _____at the night sky.A. to stareB. staringC. stared D, having stared28. Since the time humankind started gardening, we _____to make our environment more beautiful.A. tryB. have been tryingC. are tryingD. will try29. Only when you can find peace in your heart _____good relationships with others.A. will you keepB. you will keepC. you keptD. did you keep30. what you‟re doing today important, because you‟re trading a day of your life for itA. MakeB. To makeC. MakingD. Made31.1 am looking forward to the day my daughter can read this book and know my feelings for her.A. asB. whyC. whenD. where32. All we need a small piece of land where we can plant various kinds of fruit trees throughout thegrowing seasons of the year.A. areB. wasC. isD. were33. It‟s not doing the things we like, but liking the things we have to do makes life happy.A. thatB. whichC. whatD. who34. Whenever you , a present, you should think about it from the receiver‟s point of view.A. boughtB. have boughtC. will buyD. buy35. ourselves from the physical and mental tensions，we each need deep thought and inner quietness.A. Having freedB. FreedC. To freeD. FreeingSection B (18 marks)Directions: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the contextThe summer before I went off to college, Mom stood me in her usual spot behind the ironing board (烫衣板)and said, “Pay attention: I‟m going to teach you to iron.”Mom clearly explained her 36 for this lesson. I was going to be 37 and needed to learn this vital skill. Also, I would be meeting new people, and properly ironed clothes would help me make a good 38 . “Learn to iron a sh irt,” com Mom said, “and you can iron anything.”But ironing shirts was not 39 work. It didn‟t make use of long muscles we used to throw a baseball，and it wasn‟t a40 operation like ice-skating. Ironing was like driving a car on a street that has a stop sign every 10 feet, Moreover, an iron produced steam and it carried an element of 41 .If you touched the wrong part of it, you‟d get burnt. If you forgot to turn it off when you 42 ,you might burn down the house.As for technique, Mom 43 me to begin with the flat spaces outward, always pushing the iron forward into wrinkled (有褶皱的）parts. Collars had to be done right. Mom said they were close to your face, where everyone would 44 them.Over the years, I‟ve learned to iron shirts skillfully, which gives me a sense of 45 Whatever failures I suffer in my life, an ironed shirt tells me I am good at something. 46 ,through ironing I‟ve learned the method for solving even the most troublesome problems. “47 wrinkles one at a time,” as Mom might have said, “and before long everything will get ironed out.”36. A. reasons B. rules C. emotions D. methods37.A. helpful B. confident C. powerful D. independent38. A. conclusion B. suggestion C. impression D. observation39. A. useful B. easy C. special D. suitable40. A. direct B. single C. smooth D. strange41. A. doubt B. pressure C. surprise D. danger42. A. went away B. fell down C. jumped off D. looked up43. A. taught B. chose C. forced D. sent44. A. touch B. design C. see D. admire45. A. honesty B. freedom C. justice D. pride46. A. Instead B. Besides C. Otherwise D. However47. A. Make up B. Deal with C Ask for D. Rely onSection C (12 marks)Directions: Complete the following passage by filling in each blank with one word that best fits the contextWe can choose our friends, but usually we cannot choose our neighbors. However, to get a happy home life, we have to get along with 48 as well as possible.An important quality in a neighbor is consideration for 49 . People should not do things 50 will disturb their neighbors unnecessarily. For example, television sets need not be played at full volume (音量）51 loud pop music should not be played very late at night. By avoiding things likely to upset your neighbors, you can enjoy 52 friendly relationship with them.An equally important quality is tolerance. Neighbors should do all they can to avoid disturbing other people，53 there are times when some level of disturbance is unavoidable. 54 neighbors want to get along well with each other, they have to show their tolerance. In this way, everyone will live 55 peace.Section CPart Ⅲ Reading Comprehension (30 marks)Directions: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage.AWant to improve your writing skills? New Writing South is directing the way!·Towner Writer Squad (班组）for kids aged 13-17Led by comedy and TV writer, Marian Kilpatrick, Towner Writer Squad will meet once a month at the contemporary art museum for 11 months, starting 12 October, 2014.The FREE squad sessions will include introductions to a wide range of writing styles, from poetry to play writing and lyrics (抒情诗）to flash fiction, to support the development of young writers.Application & SelectionIf you would like to apply to be part of the Towner Writer Squad, please send a sample piece of your writing (about 500 words), responding to the title …LUNCH,‟ with your name, age, address and e-mail address to:debo@.Once all applications are in, com you will be invited to an open selection event on 17 September，4-5pm, at the gallery of Towner. This will be an informal opportunity to meet the Squad Leader, Squad Associate and other young people.You will also have a c hance to get to know the fantastic gallery space and get a taste of what‟s to come.Deadline for applications: 8 September, 2014For further information go to: /towner or or Any questions 一feel free to send your e-mail to Towner Writer Squad Associate: whame@ ﹡﹡﹡﹡﹡﹡﹡﹡﹡·Beginner Writing Project for kids aged 10-13Due to popular demand, a writing project will be started for eager beginners.Start time: 6 September, 2014Meet every other Saturday，2-4pm, at the Towner Study Centre.Study and write at your own pace 一you do not have to rush 一as you have a year to go through the project. Practise under the guidance of some experienced writers and teachers who can help you with basic writing skills. Most importantly, build confidence and have fun while writing!No previous experience or special background is required. Many others have been successful this way. If they can do it, why can‟t you?Fee: £179For more information go to: or 56. Towner Writer Squad will be started _______.A. t o train comedy and TV writersB. to explore the fantastic gallery spaceC. t o introduce a contemporary art museumD. to promote the development of young writers57. To join the Writer Squad，each applicant should first _______.A. provide a piece of their writingB. meet the Writer Squad LeaderC. offer their family informationD. complete an application form58. Applications for the Writer Squad should be e-mailed no later than _______.A. 6 September, 2014B. 8 September, 2014C. 17 September, 2014D. 12 October, 201459. What is most important for the beginners?A. Practising as much as possible.B. Gaining confidence and having fun.C. Studying and writing at their own pace.D. Learning skills from writers and teachers.60. More information about Beginner Writing Project can be found at _______.A. /townerB. C. D. BIn the mid-1950s, I was a somewhat bored early-adolescent male student who believed that doing any more than necessary was wasted effort. One day, this approach threw me into embarrassmentIn Mrs. Totten‟s eighth-grade math class at Central Avenue School in Anderson, Indiana, we were learning to add and subtract decimals (小数).Our teacher typically assigned daily homework, which would be recited in class the following day. On most days, our grades were based on our oral answer to homework questions.Mrs. Totten usually walked up and down the rows of desks requesting answers from student after student in the order the questions had appeared on our homework sheets. She would start either at the front or the back of the classroom and work toward the other end.Since I was seated near the middle of about 35 students, it was easy to figure out which questions I might have to answer. This particular time, I had completed my usual two or three problems according to my calculations.What I failed to expect was that several students were absent, which threw off my estimate. As Mrs. Totten made her way from the beginning of the class，I desperately tried to determine which math problem I would get. I tried to work it out before she got to me, but I had brain freeze and couldn‟t function.When Mrs. Totten reached my desk，s he asked what answer I‟d got for problem No. 14. “I…I didn‟t get anything,” I answered，and my face felt warm.“Correct,” she said.It turned out that the correct answer was zero.What did I learn that day? First, always do all your homework. Second, in r eal life it isn‟t always what you say but how you say it that matters. Third，I would never make it as a mathematician.If I could choose one school day that taught me the most, it would be that one.61. What does the underlined part in Paragraph 1 indicate?A. It is wise to value one‟s time.B. It is important to make an effortC. It is right to stick to one‟s belief.D. It is enough to do the necessary.62. Usually, Mrs. Totten asked her students to _______.A. recite their homework togetherB. grade their homework themselvesC. answer their homework questions orallyD. check the answers to their homework questions63. The author could work out which questions to answer since the teacher always _______.A. asked questions in a regular wayB. walked up and down when asking questionsC. chose two or three questions for the studentsD. requested her students to finish their usual questions64. The author failed to get the questions he had expected because _______.A. the class didn‟t begin as usualB. several students didn‟t come to schoolC. he didn‟t try hard to make his estimateD. Mrs. Totten didn‟t start from the back of the class65，Which of the following can be the best title for the passage?A. An Unforgettable TeacherB. A Future MathematicianC. An Effective ApproachD. A Valuable LessonCThe behaviour of a building‟s users may be at least as important as its design when it comes to energy use, according to new research from the UK Energy Research Centre (UKERC). The UK promises to reduce its carbon emissions (排放）by 80 percent by 2050, part of which will be achieved by all new homes being zero-carbon by 2016. But this report shows that sustainable building design on its own — though extremely important- is not enough to achieve such reductions: the behaviour of the people using the building has to change too.The study suggests that the ways that people use and live in their homes have been largely ignored by existing efforts to improve energy efficiency (效率)，which instead focus on architectural and technological developments.…Technology is going to assist but it is not going to do everything, explains Katy Janda, a UKERC senior researcher，…consumption patterns of building users can defeat the most careful design. ‟In other words，old habits die hard, even in the best-designed eco-home.Another part of the problem is information. Households and bill-payers don‟t have the knowledge they need to change their energy-use habits. Without specific information，it‟s hard to estimate the costs and benefits of making different choices. Feedback (反馈）facilities, like smart meters and energy monitors，could help bridge this information gap by helping people see how changing their behaviour directly affects their energy use; some studies have shown that households can achieve up to 15 percent energy savings using smart meters.Social science research has added a further dimension (方面），suggesting that individuals‟ behaviour in the home can be personal and cannot be predicted 一whether people throw open their windows rather than turn down the thermostat (恒温器) , for example.Janda argues that education is the key. She calls for a focused programme to teach people about buildings and their own behaviour in them.66. As to energy use, the new research from UKERC stresses the importance of________.A. zero-carbon homesB. the behaviour of building usersC. sustainable building designD. the reduction of carbon emissions67. The underlined word “which” in Paragraph 2 refers to”________.”A. the waysB. their homesC. developmentsD. existing efforts68. What are Katy Janda‟s words mainly about?A. The importance of changing building users, habits.B. The necessity of making a careful building design.C. The variety of consumption patterns of building users.D. The role of technology in improving energy efficiency.69. The information gap in energy use _______.A. can be bridged by feedback facilitiesB. affects the study on energy monitorsC. brings about problems for smart metersD. will be caused by building users‟ old habits70. What does the dimension added by social science research suggest?A. The social science research is to be furthered.B. The education programme is under discussion.C. The behaviour of building users is unpredictable.D. The behaviour preference of building users is similar.Part IV Writing (45 marks)Section A (10 marks)Directions: Read the following passage. Fill in the numbered blanks by using the information from the passage. Write NO MORE THAN THREE WORDS for each answer.Many of us invest valuable time，energy and money planning our vacations. We do this because we know for sure that going on vacations must be good for us. Research proves this feeling without a doubt. Vacations help us perform better at work, improve our sleep quality and cushion us against depression.Yet, despite these benefits, many of us return home with a feeling that our last vacation was OK - but not great. In order to change this, some mistakes should be avoided. A classic one for vacation planners is attempting to maximize value for money by planning trips that have too many components (组成部分). Perhaps you‟re planning a trip to Europe, seven cities in 10 days，and you realize it will cost only a little more to add two more destinations to the list Sounds fine in theory, but hopping from one place to the next hardly gives an opportunity to experience what psychologists call mindfulness - time to take in our new surroundings, time to be present and absorb our travel experiences. Another mistake is that we worry too much about strategic issues such as how to find a good flight deal，how to get from A to B，or which destinations to add or subtract from our journey. These issues may seem important, but our psychological state of mind is far more important.Actually, vacation happiness is based on the following top rules. First, choose your travel companions wisely, because nothing contributes more significantly to a trip than the right companions. Second，don‟t spend your vacation time in a place where everything is too expensive so as to maintain a positive mood. Third, shop wisely, for meaningful experiences provide more long-term happiness than physical possessions.Section B (10 marks)Directions: Read the following passage. Answer the questions according to the information given in the passage.Kids and PondsYears ago there was a group of kids who would hang around at some local ponds in the woods near their houses in Warwick, Rhode Island. In summer they caught frogs and fish. When winter arrived they couldn‟t wait to go skating. Time passed, and the ponds became the only open space for the kids to enjoy themselves in that neighborhood.One day. a thirteen-year-old boy from this group of kids read in the local newspaper that a developer wanted to fill in the ponds and build over a hundred small houses called condominiums. So the boy went door to door and gathered more than two hundred signatures (签名）to stop the development A group of citizens met and decided to support him.At the meeting of the town planning board (委员会)，the boy was quite nervous at first and spoke very softly. But when he saw the faces of his friends and neighbors in the crowd and thought about what was happening to their favorite ponds，his voice grew louder. He told the town officials that they should speak for the citizens. He also insisted that they should leave enough space for children. A few days later，the developer stopped his plan. Nine years later, when that teen was a senior in college, he was informed that the developer was back with his proposal to build condominiums. Now twenty-two years old, he was studying wetlands ecology. He again appeared before the town planning board. This time as an expert witness, he used environmental protection laws to explain restrictions on development in and around wetlands and the knowledge of wetlands ecology to help improve the development. Finally some condominiums were built, but less than half the number the developer wanted. The ponds where those kids used to hang around were protected by a strip of natural land，and are still there today.81. What did the kids like to do at the local ponds in winter?(No more than 6 words) (2 marks)_______________________________________________________________________________82. How did the boy win the citizens‟ support?(No more than 10 words) (2 marks)_______________________________________________________________________________83. What did the boy tell the town officials?(No more than 16 words) (3 marks)_______________________________________________________________________________84. What helped the boy to protect the ponds successfully nine years later?(No more than 12 words) (3 marks)_______________________________________________________________________________Section C (25 marks)Directions: Write an English composition according to the instructions given below.学校正在组织科技创新大赛，你想为日常生活中某件物品（如钢笔、书包、鞋子……)设计添加新功能来参赛。

绝密★启用前2014年普通高等学校招生全国统一考试(湖南卷)语文本试卷共22题，共150分。

考试结束后，将本试卷和答题卡一并交回。

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

2．答题时请按要求用笔。

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

4．作图可先使用铅笔画出，确定后必须用黑色字迹的签字笔描黑。

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

一．语言文字运用(12分。

每小题3分)家风是一个家族世代相传沿袭下来的体现家族成员精神风貌．道德品质．审美格调和整体气质的家族文化风格。

一个家族之链上某一个任务出类拔（）、深（）众望而为家族其他成员所宗仰追慕。

其懿行（）言便成为家风之源，再经过家族子孙代代接力式的（）守祖训，流风余韵，绵延不绝，就形成了一个家族鲜明的家风。

1．下列汉字依次填入语段中括号内，字音和字形全部正确的一组是A．萃孚fóu 佳恪géB．粹负fú佳恪kèC．粹负fù嘉恪géD．萃孚fú佳恪kè2．将下列各句中没有．．语病的一句填入语段中画横线处，选项是A．家风是一个影响力和美誉度都好的家庭必备的要素，也是一个家庭最为宝贵的精神财富。

B．家风即便是一个家庭最为宝贵的精神财富，也是一个有影响里有美誉度的家族必备的要素。

C．家风是一个有影响力有美誉度的家庭必备的要素，也是一个家庭最为宝贵的精神财富。

D．家风是最为宝贵的一个家族的精神财富，也是一个有影响力有美誉度的家庭必备的要素。

中国美术馆和台湾长流美术馆共同举办的“江山万里——张大千艺术展”今日与观众见面。

走进中国美术馆五》前站满了凝神观看的人。

这青绿泼彩渲染的画面，墨色浑厚华滋，层次分层展厅，迎面的青绿山水画《谷口人家．．明；章法疏密相间，错落有致；勾皴笔法遒劲雄健，开阖有度，令人叹为观止。

2014年普通高等学校招生全国统一考试(湖南卷)数学(文史类)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.(2014湖南,文1)设命题p:∀x∈R,x2+1>0,则 p为()A.∃x0∈R,x02+1>0B.∃x0∈R,x02+1≤0C.∃x0∈R,x02+1<0D.∀x∈R,x2+1≤0答案:B解析:因为全称命题的否定为特称命题,所以 p为∃x0∈R,x02+1≤0.故选B.2.(2014湖南,文2)已知集合A={x|x>2},B={x|1<x<3},则A∩B=()A.{x|x>2}B.{x|x>1}C.{x|2<x<3}D.{x|1<x<3}答案:C解析:由交集的概念,结合数轴(数轴略)可得A∩B={x|2<x<3}.故选C.3.(2014湖南,文3)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为p1,p2,p3,则()A.p1=p2<p3B.p2=p3<p1C.p1=p3<p2D.p1=p2=p3答案:D解析:由随机抽样的原则可知简单随机抽样、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p1=p2=p3,故选D.4.(2014湖南,文4)下列函数中,既是偶函数又在区间(-∞,0)上单调递增的是()A.f(x)=1x2B.f(x)=x2+1C.f(x)=x3D.f(x)=2-x答案:A解析:由偶函数的定义知,A,B为偶函数.A选项,f'(x)=-2x3在(-∞,0)恒大于0;B选项,f'(x)=2x在(-∞,0)恒小于0.故选A.5.(2014湖南,文5)在区间[-2,3]上随机选取一个数X,则X≤1的概率为()A.45B.35C.25D.15答案:B解析:由几何概型的概率公式可得P(X≤1)=35,故选B.6.(2014湖南,文6)若圆C1:x2+y2=1与圆C2:x2+y2-6x-8y+m=0外切,则m=()A.21B.19C.9D.-11答案:C解析:易知圆C1的圆心坐标为(0,0),半径r1=1.将圆C2化为标准方程(x-3)2+(y-4)2=25-m(m<25),得圆C2的圆心坐标为(3,4),半径r2=√25-m(m<25).由两圆相外切得|C1C2|=r1+r2=1+√25-m=5,解方程得m=9.故选C.7.(2014湖南,文7)执行如图所示的程序框图.如果输入的t∈[-2,2],则输出的S属于()A.[-6,-2]B.[-5,-1]C .[-4,5]D .[-3,6]答案:D解析:当t ∈[-2,0)时,执行以下程序:t=2t 2+1∈(1,9],S=t-3∈(-2,6];当t ∈[0,2]时,执行S=t-3∈[-3,-1],因此S ∈(-2,6]∪[-3,-1]=[-3,6].故选D .8.(2014湖南,文8)一块石材表示的几何体的三视图如图所示,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 .4答案:B 解析:由三视图可得原石材为如右图所示的直三棱柱A 1B 1C 1-ABC ,且AB=8,BC=6,BB 1=12.若要得到半径最大的球,则此球与平面A 1B 1BA ,BCC 1B 1,ACC 1A 1相切,故此时球的半径与△ABC 内切圆半径相等,故半径r=6+8-102=2.故选B .9.(2014湖南,文9)若0<x 1<x 2<1,则( ) A .e x 2−e x 1>ln x 2-ln x 1 B .e x 2−e x 1<ln x 2-ln x 1 C .x 2e x 1>x 1e x 2 D .x 2e x 1<x 1e x 2答案:C解析:设f (x )=e x -ln x ,则f'(x )=x ·e x -1x.当x>0且x 趋近于0时,x ·e x -1<0;当x=1时,x ·e x -1>0,因此在(0,1)上必然存在x 1≠x 2,使得f (x 1)=f (x 2),因此A,B 不正确;设g (x )=e x x,当0<x<1时,g'(x )=(x -1)e xx 2<0,所以g (x )在(0,1)上为减函数.所以g (x 1)>g (x 2),即e x 1x 1>e x 2x 2,所以x 2e x 1>x 1e x 2.故选C .10.(2014湖南,文10)在平面直角坐标系中,O 为原点,A (-1,0),B (0,√3),C (3,0),动点D 满足|CD ⃗⃗⃗⃗⃗ |=1,则|OA ⃗⃗⃗⃗⃗ +OB ⃗⃗⃗⃗⃗ +OD ⃗⃗⃗⃗⃗⃗ |的取值范围是( ) A .[4,6] B .[√19-1,√19+1] C .[2√3,2√7] D .[√7-1,√7+1]答案:D解析:设动点D 的坐标为(x ,y ),则由|CD⃗⃗⃗⃗⃗ |=1得(x-3)2+y 2=1,所以D 点的轨迹是以(3,0)为圆心,1为半径的圆.又OA ⃗⃗⃗⃗⃗ +OB ⃗⃗⃗⃗⃗ +OD ⃗⃗⃗⃗⃗⃗ =(x-1,y+√3),所以|OA ⃗⃗⃗⃗⃗ +OB ⃗⃗⃗⃗⃗ +OD ⃗⃗⃗⃗⃗⃗ |=√(x -1)2+(y +√3)2,故|OA ⃗⃗⃗⃗⃗ +OB ⃗⃗⃗⃗⃗ +OD ⃗⃗⃗⃗⃗⃗ |的最大值为(3,0)与(1,-√3)两点间的距离加1,即√7+1,最小值为(3,0)与(1,-√3)两点间的距离减1,即√7-1.故选D .二、填空题:本大题共5小题,每小题5分,共25分. 11.(2014湖南,文11)复数3+i i 2(i 为虚数单位)的实部等于 .答案:-3 解析:由题意可得3+i i2=3+i-1=-3-i,故复数的实部为-3.12.(2014湖南,文12)在平面直角坐标系中,曲线C :{x =2+√22t ,y =1+√22t(t 为参数)的普通方程为 .答案:x-y-1=0解析:两式相减得,x-y=2-1,即x-y-1=0.13.(2014湖南,文13)若变量x ,y 满足约束条件{y ≤x ,x +y ≤4,y ≥1,则z=2x+y 的最大值为 .答案:7解析:不等式组表示的平面区域如图阴影部分所示,作直线l 0:2x+y=0并平移,当直线经过点A (3,1)时,在y 轴上的截距最大,此时z 取得最大值,且最大值为7. 14.(2014湖南,文14)平面上一机器人在行进中始终保持与点F (1,0)的距离和到直线x=-1的距离相等.若机器人接触不到过点P (-1,0)且斜率为k 的直线,则k 的取值范围是 . 答案:(-∞,-1)∪(1,+∞)解析:由题意知,机器人行进的路线为抛物线y 2=4x.由题意知过点P 的直线为y=kx+k (k ≠0),要使机器人接触不到过点P 的直线,则直线与抛物线无公共点,联立方程得k4y 2-y+k=0,即Δ=1-k 2<0,解得k>1或k<-1. 15.(2014湖南,文15)若f (x )=ln(e 3x +1)+ax 是偶函数,则a= . 答案:-32解析:由题意得f (-x )=ln(e -3x +1)-ax=ln 1+e 3xe 3x-ax=ln(1+e 3x )-ln e 3x -ax=ln(e 3x +1)-(3+a )x ,而f (x )为偶函数,因此f (-x )=f (x ),即ax=-(3+a )x ,所以a=-32.三、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. 16.(本小题满分12分)(2014湖南,文16)已知数列{a n }的前n 项和S n =n 2+n2,n ∈N *. (1)求数列{a n }的通项公式;(2)设b n =2a n +(-1)n a n ,求数列{b n }的前2n 项和.分析:在第(1)问中,通过S n 可求出a n ,在求解过程中要注意分n=1和n ≥2两种情况进行讨论;在第(2)问中,充分利用第(1)问的结论得到b n =2n +(-1)n n ,然后利用分组求和法分别计算(21+22+…+22n )和(-1+2-3+…+2n ),最后相加得到{b n }的前2n 项和. 解:(1)当n=1时,a 1=S 1=1;当n ≥2时,a n =S n -S n-1=n 2+n 2−(n -1)2+(n -1)2=n.故数列{a n }的通项公式为a n =n.(2)由(1)知,b n =2n +(-1)n n.记数列{b n }的前2n 项和为T 2n ,则T 2n =(21+22+…+22n )+(-1+2-3+4-…+2n ). 记A=21+22+ (22),B=-1+2-3+4-…+2n ,则A=2(1-22n )1-2=22n+1-2,B=(-1+2)+(-3+4)+…+[-(2n-1)+2n ]=n.故数列{b n }的前2n 项和T 2n =A+B=22n+1+n-2.17.(本小题满分12分)(2014湖南,文17)某企业有甲、乙两个研发小组,为了比较他们的研发水平,现随机抽取这两个小组往年研发新产品的结果如下:(a ,b ),(a ,b ),(a ,b ),(a ,b ),(a,b ),(a ,b ),(a ,b ),(a ,b ),(a ,b ),(a ,b ),(a,b ),(a ,b ),(a ,b ),(a ,b ),(a ,b ) 其中a ,a 分别表示甲组研发成功和失败;b ,b 分别表示乙组研发成功和失败.(1)若某组成功研发一种新产品,则给该组记1分,否则记0分.试计算甲、乙两组研发新产品的成绩的平均数和方差,并比较甲、乙两组的研发水平;(2)若该企业安排甲、乙两组各自研发一种新产品,试估计恰有一组研发成功的概率.分析:在第(1)问中,通过已知条件可分别写出甲、乙两组的成绩,然后利用平均数公式分别计算甲、乙两组的平均成绩,再结合方差公式得到甲、乙两组的方差,进而比较甲、乙两组的研发水平;在第(2)问中,充分利用古典概型的概率公式,转化为计算基本事件的个数,从而求得概率. 解:(1)甲组研发新产品的成绩为1,1,1,0,0,1,1,1,0,1,0,1,1,0,1,其平均数为x 甲=1015=23; 方差为s 甲2=115[(1-23)2×10+(0-23)2×5]=29.乙组研发新产品的成绩为1,0,1,1,0,1,1,0,1,0,0,1,0,1,1, 其平均数为x 乙=915=35; 方差为s 乙2=115[(1-35)2×9+(0-35)2×6]=625. 因为x 甲>x 乙,s 甲2<s 乙2,所以甲组的研发水平优于乙组. (2)记E={恰有一组研发成功}.在所抽得的15个结果中,恰有一组研发成功的结果是(a ,b ),(a ,b ),(a ,b ),(a ,b ),(a ,b ),(a ,b ),(a ,b ),共7个.故事件E 发生的频率为715.将频率视为概率,即得所求概率为P (E )=715.18.(本小题满分12分)(2014湖南,文18)如图,已知二面角α-MN-β的大小为60°,菱形ABCD 在面β内,A ,B 两点在棱MN 上,∠BAD=60°,E 是AB 的中点,DO ⊥面α,垂足为O. (1)证明:AB ⊥平面ODE ;(2)求异面直线BC 与OD 所成角的余弦值.分析:在第(1)问中,可利用线面垂直的判定定理证明,由DO ⊥平面α可得到DO ⊥AB ,然后利用△ABD 为正三角形得到DE ⊥AB ,最后根据线面垂直的判定定理得出所证结论;在第(2)问中,充分利用第(1)问的结论AB ⊥平面ODE ,从而得到二面角α-MN-β的平面角,达到立几化平几的目的,即转化为求∠ADO 的余弦,然后利用解直角三角形的方法求出余弦值.解:(1)如图a,因为DO ⊥α,AB ⊂α,所以DO ⊥AB.图a连接BD ,由题设知,△ABD 是正三角形. 又E 是AB 的中点,所以DE ⊥AB. 而DO ∩DE=D ,故AB ⊥平面ODE.(2)因为BC ∥AD ,所以BC 与OD 所成的角等于AD 与OD 所成的角,即∠ADO 是BC 与OD 所成的角. 由(1)知,AB ⊥平面ODE ,所以AB ⊥OE.又DE ⊥AB ,于是∠DEO 是二面角α-MN-β的平面角,从而∠DEO=60°. 不妨设AB=2,则AD=2.易知DE=√3. 在Rt △DOE 中,DO=DE ·sin 60°=32. 连接AO ,在Rt △AOD 中,cos ∠ADO=DOAD=322=34.故异面直线BC 与OD 所成角的余弦值为34.19.(本小题满分13分)(2014湖南,文19)如图,在平面四边形ABCD 中,DA ⊥AB ,DE=1,EC=√7,EA=2,∠ADC=2π3,∠BEC=π3.(1)求sin ∠CED 的值; (2)求BE 的长.分析:在第(1)问中,通过已知条件,借助余弦定理得到CD 的长,然后在△CDE 中,利用正弦定理得到∠CED 的正弦值;在第(2)问中,利用∠CED 的正弦值求得其余弦值,然后利用角之间的关系表示出∠AEB ,进而表示出∠AEB 的余弦值,最后在Rt △EAB 中利用边角关系,求得BE 的长.解:如题图,设∠CED=α.(1)在△CDE 中,由余弦定理,得EC 2=CD 2+DE 2-2CD ·DE ·cos ∠EDC.于是由题设知,7=CD 2+1+CD ,即CD 2+CD-6=0. 解得CD=2(CD=-3舍去). 在△CDE 中,由正弦定理,得EC sin∠EDC=CDsinα. 于是,sin α=CD ·sin 2π3EC =2×√32√7=√217,即sin ∠CED=√217.(2)由题设知,0<α<π3,于是由(1)知,cos α=√1-sin 2α=√1-2149=2√77. 而∠AEB=2π3-α,所以cos ∠AEB=cos (2π3-α)=cos 2π3cos α+sin 2π3sin α=-12cos α+√32sin α=-12×2√77+√32×√217=√714.在Rt △EAB 中,cos ∠AEB=EA BE =2BE ,故BE=2cos∠AEB=2√714=4√7.20.(本小题满分13分)(2014湖南,文20)如图,O 为坐标原点,双曲线C 1:x 2a 12−y 2b 12=1(a 1>0,b 1>0)和椭圆C 2:y 2a 22+x 2b 22=1(a 2>b 2>0)均过点P (2√33,1),且以C 1的两个顶点和C 2的两个焦点为顶点的四边形是面积为2的正方形.(1)求C 1,C 2的方程;(2)是否存在直线l ,使得l 与C 1交于A ,B 两点,与C 2只有一个公共点,且|OA⃗⃗⃗⃗⃗ +OB ⃗⃗⃗⃗⃗ |=|AB ⃗⃗⃗⃗⃗ |?证明你的结论.分析:在第(1)问中,利用已知条件结合图形以及双曲线、椭圆中a ,b ,c 的几何意义,列出关于a 1,b 1,a 2,b 2的方程,得到它们的值,从而求出双曲线C 1、椭圆C 2的方程;在第(2)问中,首先对直线l 的斜率进行分类讨论,当斜率k 不存在时易得A ,B 两点的坐标,进而判断满足题设条件的直线l 不存在;当斜率k 存在时,可先设出l 的方程,然后代入曲线方程,利用根与系数的关系并结合向量的运算,依此判断满足题设条件的直线l 不存在. 解:(1)设C 2的焦距为2c 2,由题意知,2c 2=2,2a 1=2.从而a 1=1,c 2=1.因为点P (2√33,1)在双曲线x 2-y 2b 12=1上,所以(2√33)2−1b 12=1.故b 12=3.由椭圆的定义知2a 2=√(2√33)2+(1-1)+√(2√33)2+(1+1)=2√3.于是a 2=√3,b 22=a 22−c 22=2.故C 1,C 2的方程分别为x 2-y 23=1,y 23+x 22=1. (2)不存在符合题设条件的直线.①若直线l 垂直于x 轴,因为l 与C 2只有一个公共点,所以直线l 的方程为x=√2或x=-√2. 当x=√2时,易知A (√2,√3),B (√2,-√3), 所以|OA ⃗⃗⃗⃗⃗ +OB ⃗⃗⃗⃗⃗ |=2√2,|AB ⃗⃗⃗⃗⃗ |=2√3. 此时,|OA ⃗⃗⃗⃗⃗ +OB ⃗⃗⃗⃗⃗ |≠|AB ⃗⃗⃗⃗⃗ |.当x=-√2时,同理可知,|OA ⃗⃗⃗⃗⃗ +OB ⃗⃗⃗⃗⃗ |≠|AB ⃗⃗⃗⃗⃗ |. ②若直线l 不垂直于x 轴,设l 的方程为y=kx+m. 由{y =kx +m ,x 2-y 23=1得(3-k 2)x 2-2kmx-m 2-3=0.当l 与C 1相交于A ,B 两点时,设A (x 1,y 1),B (x 2,y 2),则x 1,x 2是上述方程的两个实根,从而x 1+x 2=2km 3-k2,x 1x 2=m 2+3k 2-3.于是y 1y 2=k 2x 1x 2+km (x 1+x 2)+m 2=3k 2-3m 2k 2-3.由{y =kx +m ,y 23+x 22=1得(2k 2+3)x 2+4kmx+2m 2-6=0.因为直线l 与C 2只有一个公共点,所以上述方程的判别式Δ=16k 2m 2-8(2k 2+3)(m 2-3)=0. 化简,得2k 2=m 2-3,因此OA⃗⃗⃗⃗⃗ ·OB ⃗⃗⃗⃗⃗ =x 1x 2+y 1y 2=m 2+3k 2-3+3k 2-3m 2k 2-3=-k 2-3k 2-3≠0,于是OA⃗⃗⃗⃗⃗ 2+OB ⃗⃗⃗⃗⃗ 2+2OA ⃗⃗⃗⃗⃗ ·OB ⃗⃗⃗⃗⃗ ≠OA ⃗⃗⃗⃗⃗ 2+OB ⃗⃗⃗⃗⃗ 2-2OA ⃗⃗⃗⃗⃗ ·OB ⃗⃗⃗⃗⃗ , 即|OA ⃗⃗⃗⃗⃗ +OB ⃗⃗⃗⃗⃗ 2|≠|OA ⃗⃗⃗⃗⃗ −OB ⃗⃗⃗⃗⃗ 2|,故|OA ⃗⃗⃗⃗⃗ +OB ⃗⃗⃗⃗⃗ |≠|AB ⃗⃗⃗⃗⃗ |.综合①,②可知,不存在符合题设条件的直线.21.(本小题满分13分)(2014湖南,文21)已知函数f (x )=x cos x-sin x+1(x>0). (1)求f (x )的单调区间;(2)记x i 为f (x )的从小到大的第i (i ∈N *)个零点,证明:对一切n ∈N *,有1x 12+1x 22+…+1x n 2<23.分析:在第(1)问中,通过已知条件,借助导数,转化为判断导数在(0,+∞)上的符号,进而得出函数的单调区间;在第(2)问中,充分利用第(1)问的结论,得到f (x )在(n π,(n+1)π)上存在零点,从而得出n π<x n+1<(n+1)π,然后分n=1,n=2,n ≥3三种情况讨论1x 12+1x 22+…+1x n 2的值与23的大小关系,即可得证. 解:(1)f'(x )=cos x-x sin x-cos x=-x sin x.令f'(x )=0,得x=k π(k ∈N *).当x ∈(2k π,(2k+1)π)(k ∈N )时,sin x>0,此时f'(x )<0;当x ∈((2k+1)π,(2k+2)π)(k ∈N )时,sin x<0,此时f'(x )>0.故f (x )的单调递减区间为(2k π,(2k+1)π)(k ∈N ),单调递增区间为((2k+1)π,(2k+2)π)(k ∈N ). (2)由(1)知,f (x )在区间(0,π)上单调递减. 又f (π2)=0,故x 1=π2.当n ∈N *时,因为f (n π)f ((n+1)π)=[(-1)n n π+1][(-1)n+1(n+1)π+1]<0,且函数f (x )的图象是连续不断的,所以f (x )在区间(n π,(n+1)π)内至少存在一个零点.又f (x )在区间(n π,(n+1)π)上是单调的,故n π<x n+1<(n+1)π.因此,当n=1时,1x 12=4π2<23; 当n=2时,1x 12+1x 22<1π2(4+1)<23;当n ≥3时,1x 12+1x 22+…+1x n 2<1π2[4+1+122+…+1(n -1)2] <1π2[5+11×2+…+1(n -2)(n -1)] <1π2[5+(1-12)+(12-13)+…+(1n -2-1n -1)] =1π2(6-1n -1)<6π2<23. 综上所述,对一切n ∈N *,1x 12+1x 22+…+1x n 2<23.。

2014年普通高等学校招生全国统一考试（湖南卷）答案语文1．D 2．C 3．A 4．B 5．A 6．C 7．D 8．B9．(1)（房屋）已经建成，天上正好下起了雪，于是用“雪屋”给它命名。

（2）与徐孟祥交往的士大夫作诗来歌咏它，叫我给它作（一篇）记。

（3）在树上居住会跌落，在洞穴里居住会生病。

10．（1）重章叠句（2）以“桃之夭夭”起兴，通过铺垫和渲染，热烈而真挚地表达了对新娘的赞美和祝福。

以桃设比，通过对桃花、桃实、桃叶的描写，在赞美新娘美丽贤淑的同时，从不同的角度祝福新娘婚后夫妻和睦、子孙繁衍、家庭兴旺；联想巧妙、形象鲜明、意趣盎然。

11．（1）卒相与欢。

（2）将有事于西畴。

或棹孤舟。

（3）皓腕凝霜雪。

还乡须断肠。

12．专业化教育，命意13．C 14．D15．①以人们日常见面最常用的问候语开篇，亲切自然。

②通过对这种问候语不同态度的对比，引出粮食话题，突出其重要性。

16．①机器切断了人与粮食之间的联系，对粮食的处理简单粗暴、毫无情感，颠覆了粮食、吃粮的人与吃本身。

②化肥、激素和农药的出现，改写了季节、雨水，改写了生命的密码，食物因多而贱，人们对食物不再怀有敬意。

17．①“养猪送猪”的细节，综合运用了多种描写手法，如为猪搔痒梳毛、垫麻袋的动作描写，“于心不忍”的心理描写，呼唤“猪娃子”的语言描写，以及猪肥膘颤动晃荡的形象描写等，细腻逼真，生动形象。

②细节具体而形象地呈现出人们参与生产的完整过程及与粮食的密切联系，表现了爷爷奶奶对猪的深厚感情，与后文大机器时代人们对粮食的态度形成对比。

18．①文章充满温情地叙写了农业时代粮食与人的血肉联系，并以之与大机器时代粮食生产、消费方式进行对比，表明了粮食是我们生命的源头和全部，表达了应珍爱粮食、尊重自然，敬畏生命的思想感情。

②可以从珍爱粮食、尊重自然、敬畏生命等角度来联系现实谈自己的看法。

19．答题要求：所写文字必须紧扣阿必“不肯睡觉”，能体现人物天真、可爱等特点，并符合生活逻辑。

2014年普通高等学校招生全国统一考试(湖南卷)一.选择题.1.【答案】B 【解析】由题可得()111122z i i i z i zi z i i z i z i +-=⇒+=⇒-=-⇒==--,故选B. 【考点定位】复数2.【答案】D【解析】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.【考点定位】抽样调查3.【答案】C【解析】分别令1x =和1x =-可得()()113f g -=且()()111f g ---=()()111f g ⇒+=,则()()()()()()1131211111f g f f g g -==⎧⎧⎪⎪⇒⎨⎨+==-⎪⎪⎩⎩()()111f g ⇒+=,故选C. 【考点定位】奇偶性4.【答案】A【解析】第1n +项展开式为()55122nn n C x y -⎛⎫- ⎪⎝⎭, 则2n =时, ()()2532351121022022n n nC x y x y x y -⎛⎫⎛⎫-=-=- ⎪ ⎪⎝⎭⎝⎭,故选A. 【考点定位】二项式定理5.【答案】C【解析】当x y >时,两边乘以1-可得x y -<-,所以命题p 为真命题,当1,2x y ==-时,因为22x y <,所以命题q 为假命题,所以②③为真命题,故选C.【考点定位】命题真假 逻辑连接词6.【答案】D【解析】当[)2,0t ∈-时,运行程序如下,(](]2211,9,32,6t t S t =+∈=-∈-,当[]0,2t ∈时 ,则(][][]2,63,13,6S ∈---=-,故选D.【考点定位】程序框图 二次函数7.【答案】B【解析】由图可得该几何体为三棱柱,所以最大球的半径为正视图直角三角形内切圆的半径r ,则862r r r -+-⇒=,故选B.【考点定位】三视图 内切圆 球8.【答案】D【解析】设两年的平均增长率为x ,则有()()()2111x p q +=++1x ⇒=,故选D.【考点定位】实际应用题9.【答案】A【解析】函数()f x 的对称轴为2x k πϕπ-=+2x k πϕπ⇒=++, 因为()2302sin 0cos cos 03x dx ππϕϕϕ⎛⎫-=⇒--+= ⎪⎝⎭⎰sin 03πϕ⎛⎫⇒-= ⎪⎝⎭, 则56x π=是其中一条对称轴,故选A. 【考点定位】三角函数图像 辅助角公式10.【答案】B【解析】由题可得存在()0,0x ∈-∞满足()()0220001ln 2xx e x x a +-=-+-+ ()001ln 2x e x a ⇒--+-0=,当0x 取决于负无穷小时,()001ln 2x e x a --+-趋近于-∞,因为函数()1ln 2x y e x a =--+-在定义域内是单调递增的,所以ln a a <⇒<,故选B.【考点定位】指对数函数 方程二.填空题. 11.【答案】sin 42πρθ⎛⎫-=- ⎪⎝⎭ 【解析】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 =,所以圆心在直线l 上,故1y x=-sin cos 1sin 42πρθρθρθ⎛⎫⇒=-⇒-=- ⎪⎝⎭,故填sin 42πρθ⎛⎫-=- ⎪⎝⎭.【考点定位】极坐标 参数方程12.【答案】32【解析】设线段AO 交BC 于点D 延长AO 交圆与另外一点E ,则BD DC ==由三角形ABD 的勾股定理可得1AD =,由双割线定理可得2BD DC AD DE DE =⇒=,则直径332AE r =⇒=,故填32. 【考点定位】勾股定理 双割线定理13.【答案】3- 【解析】由题可得52331233a a ⎧--=⎪⎪⎨⎪-=⎪⎩3a ⇒=-,故填3-. 【考点定位】绝对值不等式14.【答案】2-【解析】求出约束条件中三条直线的交点为()(),,4,k k k k -(),2,2,且,4y x x y ≤+≤的可行域如图,所以2k ≤,则当(),k k 为最优解时,362k k =-⇒=-,当()4,k k -为最优解时,()24614k k k -+=-⇒=,故填2-.【考点定位】线性规划15.1【解析】由题可得,,,22a a C a F b b ⎛⎫⎛⎫-+ ⎪ ⎪⎝⎭⎝⎭,则2222a pa a b p b ⎧=⎪⎨⎛⎫=+ ⎪⎪⎝⎭⎩1a b ⇒=,故填1.【考点定位】抛物线16.【答案】【解析】动点D 的轨迹为以C 为圆心的单位圆,则设为()[)()3cos ,sin 0,2θθθπ+∈,则(3OA OB OD ++=cos θθ的最大值为2,++的最大值为=,故填所以OA OB OD【考点定位】参数方程圆三角函数。

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姓名准考证号祝你考试顺利!绝密★启用前2014年普通高等学校招生全国统一考试（湖南4）理 科 数 学一、选择题1．若集合{}{}1,2,3,4,5,(,)|,,A B x y x A y A x y A ==∈∈-∈，则B 中所含元素的个数为 A ．3 B ．6 C ．8 D ．10 2．若,a b R ∈，则“2a b =”是“复数12a bii+-为纯虚数”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3．一个几何体的三视图形状都相同、大小均相等，那么这个几何体不可以是 A ．球 B ．三棱锥 C ．正方体 D ．圆柱 4．将函数()sin f x x ω=（其中0ω>）的图象向右平移4π个单位长度，所得图象经过点3(,0)4π，则ω的最小值是 A ．13 B ．1 C ．53 D ．25．已知2222360,20x y z a x y z a ++-=+++-=，则实数a 的取值范围为A ．[1,4]B ．(,1][4,)-∞⋃+∞C ．(1,4)D ．(,1)(4,)-∞⋃+∞ 6．两人进行乒乓球比赛,先赢三局着获胜,决出胜负为止,则所有可能出现的情形(各人输赢局次的不同视为不同情形)共有( )A. 10种B.15种C. 20种D. 30种7．下图是用模拟方法估计圆周率π的程序框图,P 表示估计结果,则图中空白框内应填入( )A. 1000N P =B. 41000N P =C. 1000M P =D. 41000MP =8．在数列{}n a 中，已知1222,7,n a a a +==等于*1()n n a a n N +⋅∈的个位数字，则2013a 的值为 A ．8 B ．6 C ．4 D ．2 9．下列函数中，在(0,)2π上有零点的函数是A ．()sin f x x x =-B ．2()sin f x x x π=- C ．2()sin f x x x =-D ．22()sin f x x x π=-10．如图，P 为椭圆221259x y +=上第一象限内的任意一点，过椭圆的右顶点A ，上顶点B 分别作y 轴，x 轴的平行线，它们相交于点C ，过P 引BC 、AC 的平行线交AC 于N ，交BC 于M ，交AB 于D 、E ，记矩形PMCN 的面积为1S ，三角形PDE 的面积为2S ，则12:S S =A ．1B ．2C ．12D ．与点P 的坐标有关 二、填空题11．设向量(1,2),(1,1),(2,)a m b m c m ==+= ，若()a c b +⊥ ，则||a =。

语文试卷 第1页（共14页）语文试卷 第2页（共8页）绝密★启用前2014年普通高等学校招生全国统一考试（湖南卷）语文本试卷满分150分，考试时间150分钟。

注意事项：1. 答题前，考生务必将自己的姓名、准考证号写在答题卡和本试题卷的封面上，并认真核对答题卡条形码上的姓名、准考证号和科目。

2. 选择题和非选择题均须在答题卡上作答，在本试题卷和草稿纸上作答无效。

考生在答题卡上按如下要求答题：（1）选择题部分请按题号用2B 铅笔填涂方框，修改时用橡皮擦干净，不留痕迹； （2）非选择题部分请按题号用0.5毫米黑色墨水签字笔书写，否则作答无效； （3）请勿折叠答题卡。

保持字体工整、笔迹清晰、卡面清洁。

3. 本试题卷共8页。

如缺页，考生须及时报告监考老师，否则后果自负。

4. 考试结束后，将本试题卷和答题卡一并交回。

一、语言文字运用（12分。

每小题3分）家风是一个家族世代相传沿袭下来的体现家族成员精神风貌、道德品质、审美格调和整体气质的家族文化风格。

一个家族之链上某一个人物出类拔（ ）、深（ ）众望而为家族其他成员所宗仰追慕。

其懿行（ ）言便成为家风之源，再经过家族子孙代代接力式的（ ）守祖训，流风余韵，绵延不绝，就形成了一个家族鲜明的家风。

____________1. 下列汉字依次填入语段中括号内，字音和字形全都正确的一组是（ ）A. 萃 孚f óu 佳 恪g éB. 粹 负f ú 佳 恪k èC. 粹 负f ù 嘉 恪g éD. 萃 孚f ú 嘉 恪k è2. 将下列各句中没有语病的一句填入语段中画横线处，选项是（ ）A. 家风是一个影响力和美誉度都好的家族必备的要素，也是一个家族最为宝贵的精神财富。

B. 家风即便是一个家族最为宝贵的精神财富，也是一个有影响力有美誉度的家族必备的要素。

C. 家风是一个有影响力有美誉度的家族必备的要素，也是一个家族最为宝贵的精神财富。

2014年普通高等学校招生全国统一考试（湖南卷)英语本试题卷分四个部分，共12页。

时量120分钟。

满分150分。

Part I Listening Comprehension (30 marks)Section A (22.5 marks)Directions: In this section，you will hear six conversations between two speakers. For each conversation, there are several questions and each question is followed by three choices marked A, B and C. Listen carefully and then choose the best answer for each question.You will hear each conversation TWICE.Example:When will the magazine probably arrive?A. Wednesday.B. Thursday.C. Friday.The answer is B.Conversation 11. What will the woman do first?A. Take a shower.B. Go camping.C. Set up a time.2. When will the man probably call the woman?A. Thursday.B. Friday.C. Sunday.Conversation 23. What is the man going to do?A, Have a coffee break. B. See a doctor. C. Buy a pet.4. What happened to the man?A. He fell ill.B. He lost his dog.C. He slept badly.Conversation 35. What is the woman?A. A bus driver.B. A waitress.C. A tour guide.6. What does the man want to get?A. Some gifts.B. A menu.C. A bus schedule.Conversation 47. What did the man do yesterday?A. He saw a movie.B. He watched TV.C. He visited some friends.英语试题第1页（共12页)8. What time will the speakers probably meet this Saturday evening?A. At 6:30.B. At 7:00.C. At 7:30.9. Which of the following will the man buy?A. Some drinks.B. A birthday cake. G. Concert tickets.Conversation 510. What is the woman doing now?A. She is serving a customer.B. She is conducting an interview.C. She is doing some recording.11 .When does the man go to the nursing home?A. Tuesdays.B. Thursdays.C. Sundays.12. Where will the man probably be working next Monday?A. At the airport nearby.B. In the studio next door,C. At the store downtown.Conversation 613. Why does the woman call the man?A. The oven doesn‟t work.B. The heater won‟t start.C. The plug is broken.14. Who will handle the problem first tomorrow evening?A. The woman.B. The man. C A worker.15. Who is the woman speaking to?A. Her husband.B. Her house owner.C. Her boss.Section B (7.5 marks)Directions: In this section, you will hear a short passage. Listen carefully and then Jill in thenumbered blanks with the information you have heard. Fill in each blank with NO MORE THAN THREE WORDS.You will hear the short passage TWICE.Part II Language Knowledge (45 marks)Section A (15 marks)Directions: For each of the following unfinished sentences there are four choices marked A, B, C and IX Choose the one that best completes the sentence.ZXXKExample:The wild flowers looked like a soft orange blanket the desert,A. coveringB. coveredC. coverD. to coverThe answer is A.21.Children，when b y their parents, are allowed to enter the stadium.A. to be accompaniedB. to accompanyC. accompanyingD. accompanied22.If Mr. Dewey _____ present, he would have offered any possible assistance tothe people there.A. wereB. had been G. should be D. was23.____your own needs and styles of communication is as important as learning toconvey your affection and emotions.A. UnderstandingB. To be understood€. Being understood D. Having understood24. As John Lennon once said，life is_____ happens to you while you are busy making other plans.A. whichB. thatC. whatD. where25. —I‟ve prepared all kinds of food for the picnic.—Do you mean we_____ bring anything with us?A. can‟tB. mustn‟tC. shan‟tD. needn't26. You will never gain success you are fully devoted to your work*A. whenB. becauseC. afterD. unless27. There is no greater pleasure than lying on my back in the middle of the grassland, _____at the night sky.A. to stareB. staringC. stared D, having stared28. Since the time humankind started gardening, we _____to make our environment more beautiful.A. tryB. have been tryingC. are tryingD. will try29. Only when you can find peace in your heart _____good relationships with others.A. will you keepB. you will keepC. you keptD. did you keep30. what you‟re doing today important, because you‟re trading a day of your lifefor itA. MakeB. To makeC. MakingD. Made31.I am looking forward to the day my daughter can read this book and know my feelings for her.A. asB. whyC. whenD. where32. All we need a small piece of land where we can plant various kinds of fruittrees throughout the growing seasons of the year.A. areB. wasC. isD. were33. It‟s not doing the things we like, but liking the things we have to do makes life happy.A. thatB. whichC. whatD. who34. Whenever you , a present, you should think about it from the receiver‟s point of view.A. boughtB. have boughtC. will buyD. buy35. ourselves from the physical and mental tensions，we each need deep thoughtand inner quietness.A. Having freedB. FreedC. To freeD. FreeingSection B (18 marks)Directions: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context ZXXKThe summer before I went off to college, Mom stood me in her usual spot behind the ironing board (烫衣板)and said, “Pay attention: I‟m going to teach you to iron.”Mom clearly explained her 36 for this lesson. I was going to be 37 and needed to learn this vital skill. Also, I would be meeting new people, and properly ironed clothes would help me make a good 38 .“Learn to iron a shirt,” zxxkcom Mom said, “and you can iron anything.”But ironing shirts was not 39 work. It didn‟t make use of long muscles we used to throw a baseball，and it wasn‟t a40 operation like ice-skating. Ironing was like driving a car on a street that has a stop sign every 10 feet, Moreover,an iron produced steam and it carried an element of41 .If you touched the wrong part of it, you‟d get burnt. If you forgot to turn it off when you42 ,you might bum down the house.As for technique, Mom 43 me to begin with the flat spaces outward, always pushing the iron forward into wrinkled (有褶皱的）parts. Collars had to be done right. Mom said they were close to your face, where everyone would 44 them.Over the years, I‟ve learned to iron shirts skillfully, which gives me a sense of 45 Whatever failures I suffer in my life, an ironed shirt tells me I am good at something. 46 ,through ironing I‟ve learned the method for solving even the most troublesome problems. “47 wrinkles one at a time,” as Mom might have said, “and before long everything will get ironed out.”36. A. reasons B. rules C. emotions D.methods37.A. helpful B. confident C. powerful D. independent38. A. conclusion B. suggestion C. impression D. observation39. A. useful B. easy C. special D. suitable40. A. direct B. single C. smooth D. strange41. A. doubt B. pressure C. surprise D. danger42. A. went away B. fell down C. jumped off D. looked up43. A. taught B. chose C. forced D. sent44. A. touch B. design C. see D. admire45. A. honesty B. freedom C. justice D. pride46. A. Instead B. Besides C.Otherwise D. However47. A. Make up B. Deal with C Ask for D. Rely onSection C (12 marks)Directions: Complete the following passage by filling in each blank with one word that best fits the contextWe can choose our friends, but usually we cannot choose our neighbors. However, to get a happy home life, we have to get along with 48 as well as possible.An important quality in a neighbor is consideration for 49 . People should not do things 50 will disturb their neighbors unnecessarily. For example, television sets need not be played at full volume (音量）51 loud pop music should not be played very late at night. By avoiding things likely to upset your neighbors, you can enjoy 52 friendly relationship with them.An equally important quality is tolerance. Neighbors should do all they can to avoid disturbing other people，53 there are times when some level of disturbance is unavoidable. 54 neighbors want to get along well with each other, they have to show their tolerance. In this way, everyone will live 55 peace.Part ⅢReading Comprehension (30 marks)Directions: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C andD. Choose the one that fits best according to the information given in the passage.AWant to improve your writing skills? New Writing South is directing the way!·Towner Writer Squad (班组）for kids aged 13-17Led by comedy and TV writer, Marian Kilpatrick, Towner Writer Squad will meet once a month at the contemporary art museum for 11 months, starting 12 October, 2014.The FREE squad sessions will include introductions to a wide range of writing styles, from poetry to play writing and lyrics (抒情诗）to flash fiction, to support the development of young writers. Application & SelectionIf you would like to apply to be part of the Towner Writer Squad, please send a sample piece ofyour writing (about 500 words), responding to the title…LUNCH,‟with your name, age, address and e-mail address to: debo@.Once all applications are in,zxxk com you will be invited to an open selection event on 17 September，4-5pm, at the gallery of Towner. This will be an informal opportunity to meet the Squad Leader, Squad Associate and other young people.You will also have a chance to get to know the fantastic gallery space and get a taste of what‟s to come.Deadline for applications: 8 September, 2014For further information go to: /towner or or Any questions 一feel free to send your e-mail to Towner Writer Squad Associate:whame@﹡﹡﹡﹡﹡﹡﹡﹡﹡Beginner Writing Project for kids aged 10-13Due to popular demand, a writing project will be started for eager beginners.Start time: 6 September, 2014Meet every other Saturday，2-4pm, at the Towner Study Centre.Study and write at your own pace 一you do not have to rush 一as you have a year to go through the project. Practise under the guidance of some experienced writers and teachers who can help you with basic writing skills. Most importantly, build confidence and have fun while writing! No previous experience or special background is required. Many others have been successful this way. If they can do it, why can‟t you?Fee: £179For more information go to: or 56. Towner Writer Squad will be started _______.A. to train comedy and TV writersB. to explore the fantastic gallery spaceC. to introduce a contemporary art museumD. to promote the development of young writers57. To join the Writer Squad，each applicant should first _______.A. provide a piece of their writingB. meet the Writer Squad LeaderC. offer their family informationD. complete an application form58. Applications for the Writer Squad should be e-mailed no later than _______.A. 6 September, 2014B. 8 September, 2014C. 17 September, 2014D. 12 October, 201459. What is most important for the beginners?A. Practising as much as possible.B. Gaining confidence and having fun.C. Studying and writing at their own pace.D. Learning skills from writers and teachers.60. More information about Beginner Writing Project can be found at _______.A. /townerB. C. D. BIn the mid-1950s, I was a somewhat bored early-adolescent male student who believed that doing any more than necessary was wasted effort. One day, this approach threw me into embarrassment In Mrs. Totten‟s eighth-grade math class at Central Avenue School in Anderson, Indiana, we were learning to add and subtract decimals (小数).Our teacher typically assigned daily homework, which would be recited in class the following day. On most days, our grades were based on our oral answer to homework questions.Mrs. Totten usually walked up and down the rows of desks requesting answers from student after student in the order the questions had apeared on our homework sheets. She would start either at the zxxk front or the back of the classroom and work toward the other end.Since I was seated near the middle of about 35 students, it was easy to figure out which questions I might have to answer. This particular time, I had completed my usual two or three problemsaccording to my calculations.What I failed to expect was that several students were absent, which threw off my estimate. As Mrs. Totten made her way from the beginning of the class，I desperately tried to determine which math problem I would get. I tried to work it out before she got to me, but I had brain freeze and couldn‟t function.When Mrs. Totten reached my desk，she asked what answer I‟d got for problem No. 14. “I (I)didn‟t get anything,” I answered，and my face felt warm.“Correct,” she said.It turned out that the correct answer was zero.What did I learn that day? First, always do all your homework. Second, in real life it isn‟t always what you say but how you say it that matters. Third，I would never make it as a mathematician. If I could choose one school day that taught me the most, it would be that one.61. What does the underlined part in Paragraph 1 indicate?A. It is wise to value one‟s time.B. It is important to make an effortC. It is right to stick to one‟s belief.D. It is enough to do the necessary.62. Usually, Mrs. Totten asked her students to _______.A. recite their homework togetherB. grade their homework themselvesC. answer their homework questions orallyD. check the answers to their homework questions63. The author could work out which questions to answer since the teacher always _______.A. asked questions in a regular wayB. walked up and down when asking questionsC. chose two or three questions for the studentsD. requested her students to finish their usual questions64. The author failed to get the questions he had expected because _______.A. the class didn‟t begin as usualB. several students didn‟t come to schoolC. he didn‟t try hard to make his estimateD. Mrs. Totten didn‟t start from the back of the class65，Which of the following can be the best title for the passage?A. An Unforgettable TeacherB. A Future MathematicianC. An Effective ApproachD. A Valuable Lesson62. Usually, Mrs. A. recite theirCThe behaviour of a building‟s users may be at least as important as its design when it comes to energy use, according to new research from the UK Energy Research Centre (UKERC). The UK promises to reduce its carbon emissions (排放）by 80 percent by 2050, part of which will be achieved by all new homes being zero-carbon by 2016. But this report shows that sustainable building design on its own — though extremely important- is not enough to achieve such reductions: the behaviour of the people using the building has to change too.The study suggests that the ways that people use and live in their homes have been largely ignored by existing efforts to improve energy efficiency (效率)，which instead focus on architectural and technological developments.…Technology is going to assist but it is not going to do everything,‟explains Katy Janda, a UKERC senior researcher，…consumption patterns of building users can defeat the most careful design. ‟In other words，old habits die hard, even in the best-designed eco-home.Another part of the problem is information. Households and bill-payers don‟t have the knowledge they need to change their energy-use habits. Without specific information，it‟s hard to estimate the costs and benefits of making different choices. Feedback (反馈）facilities, like smart meters and energy monitors，could help bridge this information gap by helping people see how changing their behaviour directly affects their energy use; some studies have shown that households can achieve up to 15 percent energy savings using smart meters.Social science research has added a further dimension (方面），suggesting thatindividuals‟behaviour in the home can be personal and cannot be predicted 一whether people throw open their windows rather than turn down the thermostat (恒温器) , for example.Janda argues that education is the key. She calls for a focused programme to teach people about buildings and their own behaviour in them.66. As to energy use, the new research from UKERC stresses the importance of________.A. zero-carbon homesB. the behaviour of building usersC. sustainable building designD. the reduction of carbon emissions67. The underlined word “which” in Paragraph 2 refers to”________.”A. the waysB. their homesC. developmentsD. existing efforts68. What are Katy Janda‟s words mainly about?A. The importance of changing building users, habits.B. The necessity of making a careful building design.C. The variety of consumption patterns of building users.D. The role of technology in improving energy efficiency. ZXXK69. The information gap in energy use _______.A. can be bridged by feedback facilitiesB. affects the study on energy monitorsC. brings about problems for smart metersD. will be caused by building users‟ old habits70. What does the dimension added by social science research suggest?A. The social science research is to be furthered.B. The education programme is under discussion.C. The behaviour of building users is unpredictable.D. The behaviour preference of building users is similar.Part IV Writing (45 marks)Section A (10 marks)Directions: Read the following passage. Fill in the numbered blanks by using the information from the passage.Write NO MORE THAN THREE WORDS for each answer.Many of us invest valuable time,energy and money planning our vacations. We do this because we know for sure that going on vacations must be good for us. Research proves this feeling without adoubt. Vacations help us perform better at work, improve our sleep quality and cushion us against depression.Yet, despite these benefits, many of us return home with a feeling that our last vacation was OK - but not great. In order to change this, some mistakes should be avoided. A classic one for vacation planners is attempting to maximize value for money by planning trips that have too many components (组成部分. Perhaps you‟re planning a trip to Europe, seven cities in 10 days，and you realize it will cost only a little more to add two more destinations to the list Sounds fine in theory, but hopping from one place to the next hardly gives an opportunity to experience what psychologists call mindfulness - time to take in our new surroundings, time to be present and absorb our travel experiences. Another mistake is that we worry too much about strategic issues such as how to find a good flight deal，how to get from A to B,or which destinations to add or subtract from our journey. These issues may seem important, but our psychological state of mind is far more important.Actually, vacation happiness is based on the following top rules. First, choose your travel companions wisely, because nothing contributes more significantly to a trip than the right companions. Second，don‟t spend your vacation time in a place where everything is too expensive so as to maintain a positive mood. Third, shop wisely, for meaningful experiences provide more long-term happiness than physical possessions.Section B (10 marks)Directions: Read the following passage. Answer the questions according to theinformation given in the passage.Kids and PondsYears ago there was a group of kids who would hang around at some local ponds in the woods near their houses in Warwick, Rhode Island. In summer they caught frogs and fish. When winter arrived they couldn‟t wait to go skating. Time passed, and the ponds became the only open space for the kids to enjoy themselves in that neighborhood.One day. a thirteen-year-old boy from this group of kids read in the local newspaper that a developer wanted to fill in the ponds and build over a hundred small houses called condominiums. So the boy went door to door and gathered more than two hundred signatures (签名）to stop the development A group of citizens met and decided to support him.At the meeting of the town planning board (委员会)，the boy was quite nervous at first and spoke very softly. But when he saw the faces of his friends and neighbors in the crowd and thought about what was happening to their favorite ponds，his voice grew louder. He told the town officials that they should speak for the citizens. He also insisted that they should leave enough space for children. A few days later，the developer stopped his plan.Nine years later, when that teen was a senior in college, he was informed that the developer was back with his proposal to build condominiums. Now twenty-two years old, he was studying wetlands ecology. He again appeared before the town planning board. This time as an expert witness, he used environmental protection laws to explain restrictions on development in and around wetlands and the knowledge of wetlands ecology to help improve the development. Finally some condominiums were built, but less than half the number the developer wanted. The ponds where those kids used to hang around were protected by a strip of natural land，and are still there today.81. What did the kids like to do at the local ponds in winter?(No more than 6 words) (2 marks)Section C (25 marks)Directions: Write an English composition according to the instructions given below. 学校正在组织科技创新大赛，你想为日常生活中某件物品（如钢笔、书包、鞋子……)设计添加新功能来参赛。

2014年普通高等学校招生全国统一考试(湖南卷)一.选择题. 1.【答案】B 【解析】由题可得()111122z i i i z i zi z i i z i z i +-=⇒+=⇒-=-⇒==--,故选B. 【考点定位】复数2.【答案】D【解析】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. 【考点定位】抽样调查3.【答案】C【解析】分别令1x =和1x =-可得()()113f g -=且()()111f g ---=()()111f g ⇒+=,则()()()()()()1131211111f g f f g g -==⎧⎧⎪⎪⇒⎨⎨+==-⎪⎪⎩⎩()()111f g ⇒+=,故选C.【考点定位】奇偶性4.【答案】A【解析】第1n +项展开式为()55122nn n C x y -⎛⎫- ⎪⎝⎭, 则2n =时, ()()2532351121022022nn n C x y x y x y -⎛⎫⎛⎫-=-=- ⎪ ⎪⎝⎭⎝⎭,故选A.【考点定位】二项式定理5.【答案】C【解析】当x y >时,两边乘以1-可得x y -<-,所以命题p 为真命题,当1,2x y ==-时,因为22x y <,所以命题q 为假命题,所以②③为真命题,故选C. 【考点定位】命题真假 逻辑连接词6.【答案】D【解析】当[)2,0t ∈-时,运行程序如下,(](]2211,9,32,6t t S t =+∈=-∈-,当[]0,2t ∈时,[]33,1S t =-∈--,则(][][]2,63,13,6S ∈---=-,故选D.【考点定位】程序框图 二次函数7.【答案】B【解析】由图可得该几何体为三棱柱,所以最大球的半径为正视图直角三角形内切圆的半径r ,则862r r r -+-==,故选B.【考点定位】三视图 内切圆 球8.【答案】D【解析】设两年的平均增长率为x ,则有()()()2111x p q +=++1x ⇒=,故选D.【考点定位】实际应用题9.【答案】A【解析】函数()f x 的对称轴为2x k πϕπ-=+2x k πϕπ⇒=++,因为()232sin 0cos cos 03x dx ππϕϕϕ⎛⎫-=⇒--+= ⎪⎝⎭⎰sin 03πϕ⎛⎫⇒-= ⎪⎝⎭, 所以23k πϕπ=+或423k ππ+,则56x π=是其中一条对称轴,故选A. 【考点定位】三角函数图像 辅助角公式10.【答案】B【解析】由题可得存在()0,0x ∈-∞满足()()0220001ln 2xx e x x a +-=-+-+ ()001ln 2x e x a ⇒--+-0=,当0x 取决于负无穷小时,()001ln 2x e x a --+-趋近于-∞,因为函数()1ln 2x y e x a =--+-在定义域内是单调递增的,所以()01ln 002e a-+->ln a a ⇒<⇒<故选B.【考点定位】指对数函数 方程二.填空题.11.【答案】sin 42πρθ⎛⎫-=- ⎪⎝⎭ 【解析】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 =,所以圆心在直线l上,故1y x=-sin cos 1sin 42πρθρθρθ⎛⎫⇒=-⇒-=- ⎪⎝⎭,故填sin 42πρθ⎛⎫-=- ⎪⎝⎭.【考点定位】极坐标 参数方程12.【答案】32【解析】设线段AO 交BC 于点D 延长AO 交圆与另外一点E ,则BD DC ==由三角形ABD 的勾股定理可得1AD =,由双割线定理可得2BD DC AD DE DE =⇒=,则直径332AE r =⇒=,故填32.【考点定位】勾股定理 双割线定理13.【答案】3-【解析】由题可得52331233a a ⎧--=⎪⎪⎨⎪-=⎪⎩3a ⇒=-,故填3-.【考点定位】绝对值不等式14.【答案】2-【解析】求出约束条件中三条直线的交点为()(),,4,k k k k -(),2,2,且,4y x x y ≤+≤的可行域如图,所以2k ≤,则当(),k k 为最优解时,362k k =-⇒=-,当()4,k k -为最优解时,()24614k k k -+=-⇒=, 因为2k ≤,所以2k =-,故填2-.【考点定位】线性规划15.1【解析】由题可得,,,22a a C a F b b ⎛⎫⎛⎫-+ ⎪ ⎪⎝⎭⎝⎭,则2222a paa b p b ⎧=⎪⎨⎛⎫=+ ⎪⎪⎝⎭⎩1a b ⇒=,1. 【考点定位】抛物线16.【答案】【解析】动点D 的轨迹为以C 为圆心的单位圆,则设为()[)()3c o s,s i n 0,2θθθπ+∈,则(3OA OB OD ++==,因为cos sin θθ+的最大值为2,所以OA OB OD ++的最大值为=,故填【考点定位】参数方程 圆 三角函数17.某企业甲,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,乙两组的研发是相互独立的. (1)求至少有一种新产品研发成功的概率;(2)若新产品A 研发成功,预计企业可获得120万元,若新产品B 研发成功,预计企业可获得利润100万元,求该企业可获得利润的分布列和数学期望. 17.【答案】(1)1315(2)详见解析 【解析】(1)解:设至少有一组研发成功的事件为事件A 且事件B 为事件A 的对立事件,则事件B 为一种新产品都没有成功,因为甲,乙成功的概率分别为23,35, 则()2312211353515P B ⎛⎫⎛⎫=-⨯-=⨯= ⎪ ⎪⎝⎭⎝⎭,再根据对立事件概率之间的公式可得()()13115P A P B =-=,所以至少一种产品研发成功的概率为1315. (2)由题可得设该企业可获得利润为ξ,则ξ的取值有0,1200+,1000+,120100+,即0,120,100,220ξ=,由独立试验的概率计算公式可得:()2320113515P ξ⎛⎫⎛⎫==-⨯-= ⎪ ⎪⎝⎭⎝⎭;()23412013515P ξ⎛⎫==⨯-= ⎪⎝⎭;()2311001355P ξ⎛⎫==-⨯= ⎪⎝⎭;()232220355P ξ==⨯=;所以ξ的分布列如下:则数学期望0120100220151555E ξ=⨯+⨯+⨯+⨯322088130=++=.【考点定位】分布列 期望 独立试验的概率18.如图5,在平面四边形ABCD 中,1,2,AD CD AC ===(1)求cos CAD ∠的值;(2)若cos BAD ∠=sin CBA ∠=求BC 的长.18.【答案】(1) cos CAD ∠=(2)67【解析】解:(1)由DAC ∆关于CAD ∠的余弦定理可得222cos 2AD AC DC CAD AD AC +-∠===所以cos CAD ∠=.(2)因为BAD ∠为四边形内角,所以s i n 0BAD ∠>且sin 0CAD ∠>,则由正余弦的关系可得s i n BAD ∠==且sin CAD ∠==,再有正弦的和差角公式可得()sin sin sin cos sin cos BAC BAD CAD BAD CAD CAD BAD ∠=∠-∠=∠∠-∠∠⎛= ⎝⎭==再由ABC ∆的正弦定理可得 sin sin AC BCCBA BAC =∠∠6BC ⇒=⎝⎭67=.【考点定位】正余弦定理 正余弦之间的关系与和差角公式19.如图6,四棱柱1111ABCD A B C D -的所有棱长都相等,11111,AC BD O AC B D O ==,四边形11ACC A 和四边形11BDD B 为矩形. (1)证明:1O O ⊥底面ABCD ;(2)若060CBA ∠=,求二面角11C OB D --的余弦值.19.【答案】(1) 详见解析 (2) 19【解析】(1)证明:四棱柱1111ABCD A B C D -的所有棱长都相等∴四边形ABCD 和四边形1111A B C D 均为菱形11111,ACBD O AC B D O ==∴1,O O 分别为11,BD B D 中点四边形11ACC A 和四边形11BDD B 为矩形∴1//OO 11//CC BB 且11,CC AC BB BD ⊥⊥ 11,OO BD OO AC ∴⊥⊥又AC BD O =且,AC BD ⊆底面ABCD1OO ∴⊥底面ABCD .(2)过1O 作1B O 的垂线交1B O 于点E ,连接11,EO EC .不妨设四棱柱1111ABCD A B C D -的边长为2a .1OO ⊥底面ABCD 且底面ABCD //面1111A B C D 1OO ∴⊥面1111A B C D又11O C ⊆面1111A B C D111O C OO ∴⊥四边形1111A B C D 为菱形1111O C O B ∴⊥又111O C OO ⊥且1111OO O C O =,111,O O O B ⊆面1OB D11O C ∴⊥面1OB D又1B O ⊆面1OB D111B O O C ∴⊥又11B O O E ⊥且1111O C O E O =,111,O C O E ⊆面11O EC1B O ∴⊥面11O EC∴11O EC ∠为二面角11C OB D --的平面角,则1111cos O EO EC EC ∠=060CBA ∠=且四边形ABCD 为菱形11O C a ∴=,11,B O=112,OO a B O ===,则111111111221sin 37O O O E B OO B O B O aa B O a=∠=== 再由11O EC ∆的勾股定理可得1EC===, 则1111cos O E O EC EC ∠=19a==,所以二面角11C OB D --. 【考点定位】线面垂直 二面角20.已知数列{}n a 满足111,nn n a a a p +=-=,*n N ∈.(1)若{}n a 为递增数列,且123,2,3a a a 成等差数列,求P 的值; (2)若12p =,且{}21n a -是递增数列,{}2n a 是递减数列,求数列n a 的通项公式. 20.【答案】(1)13p = (2) 1141,33241,332n n n n a n --⎧-⎪⎪=⎨⎪+⎪⎩为奇数为偶数【解析】解:(1)因为数列{}n a 为递增数列,所以10n n a a +-≥,则11n nn n n n a a p a a p ++-=⇒-=,分别令1,2n =可得22132,a a p a a p-=-=2231,1a p a p p ⇒=+=++,因为123,2,3a a a 成等差数列,所以21343a a a =+()()224113130p p p p p ⇒+=+++⇒-=13p ⇒=或0,当0p =时,数列n a 为常数数列不符合数列{}n a 是递增数列,所以13p =.(2)由题可得122122212121111,222n n n n n n n n n a a a a a a +-++-+-=⇒-=-=,因为{}21n a -是递增数列且{}2n a 是递减数列,所以2121n n a a +->且222n n a a +<,则有22221221222121n n n n n n n n a a a a a a a a +-++-+-<-⎧⇒-<-⎨<⎩,因为(2)由题可得122122212121111,222n n n n n n n n n a a a a a a +-++-+-=⇒-=-=,因为{}21n a -是递增数列且{}2n a 是递减数列,所以21210n n a a+-->且2220n n a a +-<()2220n n a a +⇒-->,两不等式相加可得()21212220n n n n a a a a +-+--->2212221n n n n a a a a -++⇒->-,又因为2212112n n n a a ---=22212112n n n a a +++>-=,所以2210n n a a -->,即2212112n n n a a ---=,同理可得2322212n n n n a a a a +++->-且2322212n n n n a a a a +++-<-,所以212212n n n a a +-=-,则当2n m =()*m N ∈时,21324322123211111,,,,2222m m m a a a a a a a a ---=-=--=-=,这21m -个等式相加可得2113212422111111222222m m m a a --⎛⎫⎛⎫-=+++-+++⎪ ⎪⎝⎭⎝⎭212222111111111224224113321144m m m -----=-=+--22141332m m a -⇒=+. 当21n m =+时, 2132432122321111,,,,2222m m m a a a a a a a a +-=-=--=-=-,这2m 个等式相加可得2111321242111111222222m m m a a +-⎛⎫⎛⎫-=+++-+++ ⎪ ⎪⎝⎭⎝⎭2122211111111224224113321144m m m---=-=--- 21241332m m a +=-,当0m =时,11a =符合,故212241332m m a --=- 综上1141,33241,332n n n n a n --⎧-⎪⎪=⎨⎪+⎪⎩为奇数为偶数.【考点定位】叠加法 等差数列 等比数列21.如图7,O 为坐标原点,椭圆1:C ()222210x y a b a b +=>>的左右焦点分别为12,F F ,离心率为1e ;双曲线2:C 22221x y a b-=的左右焦点分别为34,F F ,离心率为2e ,已知12e e =,且241F F =.(1)求12,C C 的方程;(2)过1F 点的不垂直于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.21.【答案】(1) 2212x y += 2212x y -= (2)4 【解析】解:(1)由题可得12e e ==,且12F F =,因为12e e =,且24F F =,所以22212b a+=且1a ⇒=且1,b a ==所以椭圆1C 方程为2212x y +=,双曲线2C 的方程为2212x y -=. (2)由(1)可得()21,0F -,因为直线AB 不垂直于y 轴,所以设直线AB 的方程为1x ny =-,联立直线与椭圆方程可得()222210n y ny +--=,则222A B n y y n +=+,则22mny n =+,因为(),M M M x y 在直线AB 上,所以2222122M n x n n -=-=++,因为AB 为焦点弦,所以根据焦点弦弦长公式可得21222M AB e x n =+=++)2212n n +=+,则直线PQ 的方程为2M M y ny x y x x =⇒=-,联立直线PQ 与双曲线可得22202n x x ⎛⎫---= ⎪⎝⎭2284x n ⇒=-,22224n y n =-则24022n n ->⇒-<<,所以,P Q 的坐标为,⎛ ⎝,则点,P Q 到直线AB 的距离为221224n n n d n +-=,222224n nn d n --=,因为点,Q P 在直线AB 的两端所以()2221222224n n nn d dn ++-+==+,则四边形APBQ 面积()1212S AB d d =+= =,因为2440n ≥->,所以当242n n =⇒=±时, 四边形APBQ 面积取得最小值为4.【考点定位】弦长 双曲线 椭圆 最值22.已知常数0a >,函数()()2ln 12xf x ax x =+-+. (1)讨论()f x 在区间()0,+∞上的单调性;(2)若()f x 存在两个极值点12,x x ,且()()120f x f x +>,求a 的取值范围. 【答案】(1)详见解析【解析】解:(1)对函数()f x 求导可得()()24'12a f x ax x =-++()()()()2224112a x ax ax x +-+=++()()()224112ax a ax x --=++,因为()()2120ax x ++>,所以当10a -≤时,即1a ≥时,()'0f x ≥恒成立,则函数()f x 在()0,+∞单调递增,当1a ≤时, ()'0f x x =⇒=,则函数()f x 在区间⎛ ⎝⎭单调递减,在⎫⎪+∞⎪⎝⎭单调递增的.(2) 解:(1)对函数()f x 求导可得()()24'12a f x ax x =-++()()()()2224112a x ax ax x +-+=++()()()224112ax a ax x --=++,因为()()2120ax x ++>,所以当10a -≤时,即1a≥时,()'0f x ≥恒成立,则函数()f x 在()0,+∞单调递增,当1a <时, ()'0f x x =⇒=,则函数()f x 在区间⎛ ⎝⎭单调递减,在⎫⎪+∞⎪⎝⎭单调递增的.(2)函数()f x的定义域为1,a⎛⎫-+∞⎪⎝⎭,由(1)可得当01a<<时,()'0f x x=⇒=,则1a>-⇒12a≠,则()f x的两个极值点,()()12ln1ln1f x f x⎡⎡+=++-+⎣⎣()ln141a a=--+⎡⎤⎣⎦,因为112a<<或12a<<,则12<<,则设t=12t⎛⎫<<⎪⎝⎭,则()()()212ln144f x f x t t+=-+,设函数()()2ln144g x x x=-+12t⎛⎫<<⎪⎝⎭, 后续有待更新!!!【考点定位】导数含参二次不等式对数。