数学(文)卷·2014届湖南省益阳市箴言中学高三上学期第五次模拟考试试题(2014.01)
湖南省益阳市箴言中学2014届高三上学期第五次模拟考试试题 化学 Word版含答案.pdf

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2014年湖南省益阳市箴言中学高考数学一模试卷(理科)

2014年湖南省益阳市箴言中学高考数学一模试卷(理科)学校:___________姓名:___________班级:___________考号:___________一、选择题(本大题共8小题,共40.0分)1.设i为虚数单位,则等于()A.iB.-iC.1+iD.1-i【答案】A【解析】解:===i答案为:i.故选A.分子和分母同乘以分母的共轭复数,分母上进行复数的乘法运算,分母上进行复数的乘法运算,得到最简形式,约分得到结果.本题考查复数的除法运算,分子和分母同乘以分母的共轭复数,把复数整理成整式形式,这是经常出现的一个复数题目,在解题时注意运算,一定能得分.2.已知集合M={x||x-7|<9},N={x|y=},且M,N都是全集U的子集,则如图所示韦恩图中阴影部分表示的集合()A.{x|-3<x<-2}B.{x|-3≤x≤-2}C.{x|x≥16}D.{x|x>16}【答案】B【解析】解:M={x||x-7|<9}={x|-2<x<16},则C U M═{x|x≥16或x≤-2},又N={x|y=}={x|-3≤x≤3},∴C U M∩N={x|-3≤x≤-2}.故选B.先化简M集合,再求得其补集,再与N集合求交集即可本小题主要考查V enn图表达集合的关系及运算、V enn图的应用等基础知识,考查数形结合思想.属于基础题.3.按照如图的程序框图执行,若输出结果为15,则M处条件为()A.k≥16B.k<8C.k<16 D.k≥8【答案】A【解析】解:程序运行过程中,各变量的值如下表示:S k是否继续循环循环前01/第一圈12是第二圈34是第三圈78是第四圈1516否故退出循环的条件应为k≥16故选A分析程序中各变量、各语句的作用,再根据流程图所示的顺序,可知:该程序的作用是累加k值到S并输出S.算法是新课程中的新增加的内容,也必然是新高考中的一个热点,应高度重视.程序填空也是重要的考试题型,这种题考试的重点有:①分支的条件②循环的条件③变量的赋值④变量的输出.其中前两点考试的概率更大.此种题型的易忽略点是:不能准确理解流程图的含义而导致错误.4.给出下列命题中①向量,满足||=||=|-|,则与+的夹角为30°;②•>0,是,的夹角为锐角的充要条件;③将函数y=|x-1|的图象按向量=(-1,0)平移,得到的图象对应的函数表达式为y=|x|;④若,则△ABC为等腰三角形;以上命题正确的个数是()A.4个B.1个C.3个D.2个【答案】D【解析】解:对于①,取特值零向量时,命题错误,若前提为非零向量由向量加减法的平行四边形法则与夹角的概念正确.对②•>0时,,的夹角为锐角或零角,不一定是锐角,故充分性不成立.对于③,注意按向量平移的意义,就是图象向左移1个单位,故结论正确.对于④;由于向量的数量积满足分配律运算,故结论正确,故选D.对于①,当,中有一个为0时,结论不成立.对②•>0时,,的夹角为锐角或零角.按向量平移的意义③正确.由向量的数量积满足分配律运算,以及=|AB|2,故④正确.本题考查两个向量的加减混合运算及其几何意义,用两个向量的数量积表示两个向量的夹角.5.已知某几何体的正视图和侧视图均是边长为1的正方形,则这个几何体的体积不可能是()A. B. C.1 D.【答案】D【解析】解:∵几何体的正视图和侧视图均是边长为1的正方形,故它必是一个柱体当它的底面是一个以1为两直角边的直角梯形时其面积为,故排除A;当它的底面是一个以1为直径的圆时其面积为,故排除B;当它的底面是一个以1为边长的正方形时其面积为1,故排除C;由于正视图和侧视图均是边长为1的正方形,故俯视图的面积最大为1×1=1即几何体的体积最大为1而>1,故这个几何体的体积不可能是故选D由已知中几何体的正视图和侧视图均是边长为1的正方形,可得俯视图的面积最大为1×1=1,即几何体的体积最大为1,分析四个答案,可得结论.本题考查的知识点是由三视图求体积,其中分析出几何体的底面面积,即俯视图最大面积是1是解答的关键.6.有下列四种说法:①命题:“∃x0∈R,使得x2-x>0”的否定是“∀x∈R,都有x2-x≤0”;②已知随机变量x服从正态分布N(1,σ2),P(x≤4)=0.79,则P(x≤-2)=0.21;③函数f(x)=2sinxcosx-1,(x∈R)图象关于直线对称,且在区间,上是增函数;④设实数x,y∈[0,1],则满足:x2+y2<1的概率为.其中错误的个数是()A.0B.1C.2D.3【答案】A【解析】解:由含有一个量词的命题的否定得①显然正确;由②可得μ=1(即平均数为1),P(x≥4)=1-0.79=0.21,又图象对称轴为x=μ=1,所以P(x≤-2)=P(x≥4)=0.21,故②正确;③由于函数f(x)=2sinxcosx-1,(x∈R),即f(x)=sin(2x)-1的对称轴为x=,单调增区间为[kπ-,kπ+](k∈Z),所以f(x)的图象关于直线对称,且在区间,上是增函数,故③正确;④这是几何概率,区域D:边长为1的正方形,区域d:第一象限内的个单位圆,测度:面积,所以满足:x2+y2<1的概率是,故④正确.故选:A①由全称命题与存在性命题的否定推断;②运用正态分布的特点可计算求得;③运用正弦函数的对称轴方程和单调增区间判断;④由几何概率知识可得.本题考查简易逻辑的基础知识:命题的否定,以及正弦函数的对称轴和单调区间,几何概率问题,意在考查学生的综合运用知识的能力,是一道基础题.7.函数 , , > ,若方程f (x )=x +a 恰有两个不等的实根,则a 的取值范围为( )A.(-∞,0)B.[0,1)C.(-∞,1)D.[0,+∞) 【答案】 C【解析】解:由函数, , >,可得f (x )的图象和函数y =x +a 有两个不同的交点, 如图所示:故有a <1, 故选C .由题意可得f (x )的图象和函数y =x +a 有两个不同的交点,结合图象,求出a 的取值范围.本题考查根的存在性及根的个数判断,以及函数与方程的思想、数形结合的数学思想,解答关键是运用数形结合的思想,属于中档题.8.已知f (x )为R 上的可导函数,且对∀x ∈R ,均有f (x )>f ′(x ),则有( ) A.e 2013f (-2013)<f (0),f (2013)<e 2013f (0) B.e 2013f (-2013)<f (0),f (2013)>e 2013f (0) C.e 2013f (-2013)>f (0),f (2013)<e 2013f (0) D.e 2013f (-2013)>f (0),f (2013)>e 2013f (0) 【答案】 C【解析】 解:令,则 ′′,因为f (x )>f '(x ),所以g ′(x )<0,所以函数g (x )为R 上的减函数, 所以g (-2013)>g (0), 即>,所以e 2013f (-2013)>f (0),<,所以f (2013)<e 2013f (0).故选C .根据题目给出的条件:“f (x )为R 上的可导函数,且对∀x ∈R ,均有f (x )>f '(x )”,结合给出的四个选项,设想寻找一个辅助函数g (x )=,这样有以e 为底数的幂出现,求出函数g (x )的导函数,由已知得该导函数小于0,得出函数g (x )为减函数,利用函数的单调性即可得到结论.本题考查了导数的运算,由题目给出的条件结合选项去分析函数解析式,属逆向思维,属中档题.二、填空题(本大题共8小题,共40.0分)9.如图,割线PBC经过圆心O,PB=OB=1,PB绕点O逆时针旋120°到OD,连PD交圆O于点E,则PE= ______ .【答案】【解析】解:由余弦定理得,PD2=OD2+OP2-2OD•OP cos120°=1+4-2×1×2×(-)=7,所以PD=.根据割线定理PE•PD=PB•PC得,PE=1×3,所以PE=.故答案为.先由余弦定理求出PD,再根据割线定理即可求出PE,问题解决.已知三角形两边与夹角时,一定要想到余弦定理的运用,之后做题的思路也许会豁然开朗.10.(坐标系与参数方程选做题)若直线l的极坐标方程为,曲线C:ρ=1上的点到直线l的距离为d,则d的最大值为______ .【答案】【解析】解:直线的直角坐标方程为x+y-6=0,曲线C的方程为x2+y2=1,为圆;d的最大值为圆心到直线的距离加半径,即为故答案为:.求出直线的直角坐标方程,圆的直角坐标方程,通过圆心到直线的距离求出d的最大值.本题考查极坐标方程与直角坐标方程的互化,直线与圆的位置关系,考查计算能力.11.若不等式|a-1|≥x+2y,对满足x2+y2=5的一切实数x,y恒成立,则实数a的取值范围是______ .【答案】a≥6或a≤-4【解析】解:∵x2+y2=5,∴令x=,,∴x+2y==,令tanφ=2,∴x+2y=5sin(θ+φ)≤5,∵不等式|a-1|≥x+2y,对满足x2+y2=5的一切实数x,y恒成立,∴|a-1|≥5,则a-1≤-5或a-1≥5,解得:a≤-4或a≥6.故答案为:a≥6或a≤-4.由题意令x=,,由三角函数的化积公式求出x+2y的最大值,再由不等式|a-1|≥x+2y,对满足x2+y2=5的一切实数x,y恒成立,可得|a-1|大于等于x+2y 的最大值,求解绝对值得不等式得a的取值范围.本题考查了函数恒成立问题,考查了圆的参数方程,体现了数学转化思想方法,是中档题.12.设,则二项式的展开式中,x2项的系数为______ .【答案】60【解析】解:=-6cosx=-6cos+6cos0=6T r+1=x6-r(-2)r x-r=(-2)r x6-2r,当6-2r=2即r=2T3=(-2)2x2=60x2,故答案为:60先利用定积分的运算法则求出n的值,然后研究的展开式通项公式,使x的指数为2,可求出x2的项,得到所求.本题主要考查了定积分的应用,以及二项式定理的应用,同时考查了运算求解的能力,属于中档题.13.设实数x,y满足条件:①x≥0,y≥0;②3x-y-6≤0;③x-y+2≥0,目标函数z=ax+by (a>0,b>0)的最大值为12,则的最小值是______ .【答案】【解析】解:不等式表示的平面区域如图所示阴影部分,由ax+by=z(a>0,b>0),得,则的斜率k=<,平移直线得,由图象可知当直线得经过点B时,直线的截距最大,此时z最大,由,解得,即直线x-y+2=0与直线3x-y-6=0的交点为B(4,6),此时目标函数z=ax+by(a>0,b>0)取得最大12,即4a+6b=12,∴2a+3b=6,即,∴=()()=,当且仅当即a=b时取等号,∴的最小值是,故答案为:已知2a+3b=6,求的最小值,可以作出不等式的平面区域,先用乘积进而用基本不等式解答.本题综合地考查了线性规划问题和由基本不等式求函数的最值问题.要求能准确地画出不等式表示的平面区域,并且能够求得目标函数的最值.利用数形结合是解决本题的关键.14.设数列{a n},{b n}都是正项等比数列,S n,T n分别为数列{lga n}与{lgb n}的前n项和,且,则= ______ .【答案】【解析】解:设正项等比数列{a n}的公比为q,设正项等比数列{b n}的公比为p,则数列{lga n}是等差数列,公差为lgq,{lgb n}是等差数列,公差为lgp.故S n=n•lga1+,同理可得T n=n•lgb1+.又=,∴====,故答案为.设{a n}的公比为q,{b n}的公比为p,则数列{lga n}是等差数列,公差为lgq,{lgb n}是等差数列,公差为lgp.求出S n和T n,由于=,根据===,运算求得结果.本题主要考查等比数列的定义和性质,等比数列的通项公式,对数的运算性质以及等差数列的前n项和公式的应用,属于中档题.15.已知点F为抛物线y2=-8x的焦点,O为原点,点P是抛物线准线上一动点,点A在抛物线上,且|AF|=4,则|PA|+|PO|的最小值为______ .【答案】2【解析】解:∵|AF|=4,由抛物线的定义得,∴A到准线的距离为4,即A点的横坐标为-2,又点A在抛物线上,∴从而点A的坐标A(-2,4);坐标原点关于准线的对称点的坐标为B(4,0)则|PA|+|PO|的最小值为:|AB|=故答案为:2.利用抛物线的定义由|AF|=4得到A到准线的距离为4,即可求出点A的坐标,根据:“|PA|+|PO|”相当于在准线上找一点,使得它到两个定点的距离之和最小,最后利用平面几何的方法即可求出距离之和的最小值.此题考查学生灵活运用抛物线的简单性质解决最小值问题,灵活运用点到点的距离、对称性化简求值,是一道中档题.16.设集合M={1,2,3,4,5,6},对于a i,b i∈M,记且a i<b i,由所有e i组成的集合设为:A={e1,e2,…,e k},则k的值为______ ;设集合B=′′,,对任意e i∈A,e′j∈B,则e i+e′∈M的概率为______ .【答案】11;【解析】解:由题意知,a i,b i∈M,a i<b i,∵首先考虑M中的二元子集有{1,2},{1,3},…,{5,6},共15个,即为C62=15个.又a i<b i,满足的二元子集有:{1,2},{2,4},{3,6},这时,{1,3},{2,6},这时,{2,3},{4,6},这时,共7个二元子集.故集合A中的元素个数为k=15-7+3=11.列举A={,,,,,,,,,,}B={2,3,4,5,6,,,,,,},,,,,共6对.∴所求概率为:.故答案为:11;.由题意知本题是一个古典概型,a i,b i∈M,a i<b i,首先考虑M中的二元子集有C62=15个,通过列举得到集合A中共有11个元素,列举A和B集合,满足条件的共有6种结果,根据古典概型概率公式得到结果.本题考查古典概型,是一个通过列举来解决的概率问题,从这个题目上体会列举法的优越性和局限性.是一个基础题.三、解答题(本大题共6小题,共75.0分)17.在△ABC中,角A,B,C的对边分别是a、b、c,已知向量=(cos A,cos B),=(a,2c-b),且∥.(Ⅰ)求角A的大小;(Ⅱ)若a=4,求△ABC面积的最大值.【答案】解:(I)∵向量=(cos A,cos B),=(a,2c-b),且∥,∴acos B-(2c-b)cos A=0,利用正弦定理化简得:sin A cos B-(2sin C-sin B)cos A=0,∴sin A cos B+cos A sin B-2sin C cos A=0,即sin(A+B)=sin C=2sin C cos A,∵sin C≠0,∴cos A=,又0<A<π,则A=;(II)由余弦定理a2=b2+c2-2bccos A,得:16=b2+c2-bc≥bc,即bc≤16,当且仅当b=c=4时,上式取等号,∴S△ABC=bcsin A≤4,则△ABC面积的最大值为4.【解析】(I)由两向量的坐标及两向量平行,利用平面向量的数量积运算法则列出关系式,再利用正弦定理化简,整理后再利用两角和与差的正弦函数公式化简,根据sin C不为0,求出cos A的值,由A为三角形的内角,利用特殊角的三角函数值即可求出A的度数;(II)由a与cos A的值,利用余弦定理列出关系式,整理后利用基本不等式求出bc的最大值,再由bc的最大值与sin A的值即可得到三角形ABC面积的最大值.此题考查了正弦、余弦定理,三角形的面积公式,基本不等式的运用,以及平面向量的数量积运算法则,熟练掌握定理及公式是解本题的关键.18.2012年“双节”期间,高速公路车辆较多.某调查公司在一服务区从七座以下小型汽车中按进服务区的先后每间隔50辆就抽取一辆的抽样方法抽取40名驾驶员进行询问调查,将他们在某段高速公路的车速(km/h)分成六段:[60,65)[65,70)[70,75)[75,80),[80,85)[85,90),得到如图的频率分布直方图.问:(1)某调查公司在采样中,用到的是什么抽样方法?(2)求这40辆小型车辆车速的众数和中位数的估计值.(3)若从车速在(60,70)的车辆中任抽取2辆,求抽出的2辆车中速车在[65,70)的车辆数ξ的分布列及其均值(即数学期望).【答案】解:(1)由题意知这个抽样是按进服务区的先后每间隔50辆就抽取一辆的抽样方法抽取40名驾驶员进行询问调查,是一个具有相同间隔的抽样,并且总体的个数比较多,这是一个系统抽样.故调查公司在采样中,用到的是系统抽样,(2分)(2)众数的估计值为最高的矩形的中点,即众数的估计值等于77.5(4分)设图中虚线所对应的车速为x,则中位数的估计值为:0.01×5+0.02×5+0.04×5+0.06×(x-75)=0.5,解得x=77.5,即中位数的估计值为77.5(6分)(3)从图中可知,车速在[60,65)的车辆数为:m1=0.01×5×40=2(辆),(7分)车速在[65,70)的车辆数为:m2=0.02×5×40=4(辆)(8分)∴ξ=0,1,2,P(ξ=0)=,P(ξ=1)=,P(ξ=2)=,ξ的分布列为:(11分)数学期望Eξ=0×+1×+2×=.(12分)【解析】(1)这个抽样是按进服务区的先后每间隔50辆就抽取一辆的抽样方法抽取40名驾驶员进行询问调查,是一个具有相同间隔的抽样,并且总体的个数比较多,这是一个系统抽样;(2)选出直方图中最高的矩形求出其底边的中点即为众数;求出从左边开始小矩形的面积和为0.5对应的横轴的左边即为中位数;利用各个小矩形的面积乘以对应矩形的底边的中点的和为数据的平均数.(3)从车速在(60,70)的车辆中任抽取2辆,根据题意抽出的2辆车中速车在(65,70)的车辆数ξ可能为0、1、2,求出相应的概率,即可求得分布列和期望.解决频率分布直方图的有关特征数问题,利用众数是最高矩形的底边中点;中位数是左右两边的矩形的面积相等的底边的值;平均数等于各个小矩形的面积乘以对应的矩形的底边中点的和.此题把统计和概率结合在一起,比较新颖,也是高考的方向,应引起重视.19.如图,在三棱锥P-ABC中,PA=PB=AB=2,BC=3,∠ABC=90°,平面PAB⊥平面ABC,D、E分别为AB、AC中点.(1)求证:DE∥平面PBC;(2)求证:AB⊥PE;(3)求二面角A-PB-E的大小.【答案】解:(Ⅰ)∵D、E分别为AB、AC中点,∴DE∥BC.∵DE⊄平面PBC,BC⊂平面PBC,∴DE∥平面PBC.…(4分)(Ⅱ)连接PD,∵PA=PB,D为AB中点,∴PD⊥AB.….(5分)∵DE∥BC,BC⊥AB,∴DE⊥AB…(6分)又∵PD∩DE=D,PD,DE⊂平面PDE∴AB⊥平面PDE…(8分)∵PE⊂平面PDE,∴AB⊥PE…(9分)(Ⅲ)∵AB⊥平面PDE,DE⊥AB…(10分)如图,以D为原点建立空间直角坐标系,由PA=PB=AB=2,BC=3,则B(1,0,0),P(0,0,),E(0,,0),∴=(1,0,),=(0,,).设平面PBE的法向量,,,∴令得,,…(11分)∵DE⊥平面PAB,∴平面PAB的法向量为,,.…(12分)设二面角的A-PB-E大小为θ,由图知,<,>,所以θ=60°,即二面角的A-PB-E大小为60°…(14分)【解析】(Ⅰ)由三角形中位线定理可得DE∥BC,进而由线面平行的判定定理得到DE∥平面PBC(II)连接PD,由等腰三角形三线合一,可得PD⊥AB,由DE∥BC,BC⊥AB可得DE⊥AB,进而由线面垂直的判定定理得到AB⊥平面PDE,再由线面垂直的性质得到AB⊥PE;(Ⅲ)以D为原点建立空间直角坐标系,分别求出平面PBE的法向量和平面PAB的法向量,代入向量夹角公式,可得二面角A-PB-E的大小.本题考查的知识点是二面角的平面角及求法,直线与平面平行的判定,直线与平面垂直的判定,熟练掌握空间直线与平面位置关系的判定,性质是解答(I)和(II)的关键,而(III)的关键是建立空间坐标系,将空间角问题转化为向量夹角问题.20.如图,某隧道设计为双向四车道,车道总宽20m,要求通行车辆限高5m,隧道全长2.5km,隧道的两侧是与地面垂直的墙,高度为3米,隧道上部拱线近似地看成半个椭圆.(1)若最大拱高h为6m,则隧道设计的拱宽l是多少?(2)若要使隧道上方半椭圆部分的土方工程量最小,则应如何设计拱高h和拱宽l?(已知:椭圆+=1的面积公式为S=πab,柱体体积为底面积乘以高.)(3)为了使隧道内部美观,要求在拱线上找两个点M、N,使它们所在位置的高度恰好是限高5m,现以M、N以及椭圆的左、右顶点为支点,用合金钢板把隧道拱线部分连接封闭,形成一个梯形,若l=30m,梯形两腰所在侧面单位面积的钢板造价是梯形顶部单位面积钢板造价的倍,试确定M、N的位置以及h的值,使总造价最少.【答案】解:(1)如图建立直角坐标系,则点P(10,2),椭圆方程为+=1.将b=h-3=3与点P坐标代入椭圆方程,得a=,此时l=2a=,因此隧道的拱宽约为m.(5分)(2)要使隧道上方半椭圆部分的土方工程量最小,由柱体的体积公式可知:只需半椭圆的面积最小即可.由椭圆方程+=1,得+=1.因为+≥,即ab≥40,(8分)所以半椭圆面积S=≥.(10分)当S取最小值时,有==,得a=10,b=.此时l=2a=20,h=b+3=+3(12分)故当拱高为(+3)m、拱宽为20m时,隧道上方半椭圆部分的土方工程量最小(13分)(3)根据题意设M(x,2),N(-x,2),则10≤x<15设=2[x+],(10≤x<15)(11分)则′令f'(x)=0,⇒x2-30x+221=0⇒x=13(x=17舍去),且10≤x<13时,f′(x)<0,13<x<15时,f′(x)>0,∴x=13时,f(x)取最小值,此时M(13,2),N(-13,2),代入椭圆方程得∴(13分)【解析】(1)先建立直角坐标系,找到对应椭圆方程再把b=h-3=3与点P坐标代入椭圆方程,即可求出隧道设计的拱宽l是多少;(2)转化为求半椭圆的面积最小值问题,对椭圆方程用基本不等式即可求出对应的半椭圆面积以及满足要求的拱高h和拱宽l.(3)先求出总造价的表达式,再利用导函数研究其最值即可.本题是对椭圆方程在实际生活中应用的考查.涉及到了函数的最值问题.一般在研究函数的最值问题时,通常借助于导函数.21.已知椭圆C的中心在原点,离心率等于,它的一个短轴端点点恰好是抛物线x2=8y的焦点.(1)求椭圆C的方程;(2)已知P(2,3)、Q(2,-3)是椭圆上的两点,A,B是椭圆上位于直线PQ两侧的动点,①若直线AB的斜率为,求四边形APBQ面积的最大值;②当A、B运动时,满足∠APQ=∠BPQ,试问直线AB的斜率是否为定值,请说明理由.【答案】解:(Ⅰ)设C方程为>>,则.由,,得a=4∴椭圆C的方程为.…(4分)(Ⅱ)①解:设A(x1,y1),B(x2,y2),直线AB的方程为,代入,得x2+tx+t2-12=0由△>0,解得-4<t<4…(6分)由韦达定理得x1+x2=-t,x1x2=t2-12.∴==.由此可得:四边形APBQ的面积∴当t=0,.…(8分)②解:当∠APQ=∠BPQ,则PA、PB的斜率之和为0,设直线PA的斜率为k则PB的斜率为-k,直线PA的直线方程为y-3=k(x-2)由(1)代入(2)整理得(3+4k2)x2+8(3-2k)kx+4(3-2k)2-48=0∴…(10分)同理直线PB的直线方程为y-3=-k(x-2),可得∴,…(12分)所以AB的斜率为定值.…(14分)【解析】(Ⅰ)根据椭圆C的一个顶点恰好是抛物线的焦点,离心率等于.由此列式解出出a,b的值,即可得到椭圆C的方程.(Ⅱ)①设A(x1,y1),B(x2,y2),直线AB的方程为,将直线的方程代入椭圆的方程,消去y得到关于x的一元二次方程,再结合根系数的关系利用弦长公式即可求得四边形APBQ的面积,从而解决问题.②设直线PA的斜率为k,则PB的斜率为-k,PA的直线方程为y-3=k(x-2)将直线的方程代入椭圆的方程,消去y得到关于x的一元二次方程,再结合根系数的关系利用弦长公式即可求得x1+2,同理PB的直线方程为y-3=-k(x-2),可得x2+2,从而得出AB的斜率为定值.本题考查的知识点是椭圆的标准方程,直线与圆锥曲线的综合问题,其中根据已知条件计算出椭圆的标准方程是解答本题的关键.22.已知函数g(x)=x2-(2a+1)x+alnx(Ⅰ)当a=1时,求函数g(x)的极值;(Ⅱ)求函数g(x)在区间[1,e]上的最小值;(Ⅲ)在(Ⅰ)的条件下,设f(x)=g(x)+4x-x2-2lnx,证明:>(n≥2).参考数据:ln2≈0.6931.【答案】解:(Ⅰ)∵a=1,可得g(x)=x2-3x+lnx,(x>0)∴g′(x)=2x-3+==,令g′(x)=0,x1=,x2=1,g′(x)>0,即x>1或x<,g(x)为增函数,g′(x)<0,即<x<1,g(x)为减函数,g(x)在x=出取极大值,g(x)极大值=g()=--ln2,g(x)在x=1出取极小值,g(x)极小值=g(1)=-2,(Ⅱ)g′(x)=2x-(2a+1)+==当a≤1时,x∈[1,e],g′(x)≥0,g(x)单调递增,g(x)min=g(1)=-2a,当1<a<e时,x∈[1,a]时,g′(x)<0,g(x)单调递减,x∈[a,e]时,g′(x)>0,g(x)单调递增,∴g(x)min=g(a)=-a2-a+alna,当a≥e时,x∈[1,e],g′(x)<0,g(x)单调递减,g(x)min=g(e)=e2-(2a+1)e+a,∴g(x)的最小值为:g(a)=<<(III)依题意可得,f(x)=g(x)+4x-x2-lnx=x-lnx(x>0)∴k-f(k)=lnk,令h(x)=lnx-(x2-1),∵x∈[2,+∞)时,h′(0)=<0,∴h(x)≤h(2)=ln2-<0,即lnx<(x2-1),∴>=2(-)∴==++…+>2(1-+-+…+-+-)=2(1+--)=,(n≥2);【解析】(Ⅰ)把a=1代入g(x)然后对g(x)进行求导,令g′(x)=0,可得极值点,从而求出函数g(x)的极值;(Ⅱ)已知函数g(x)=x2-(2a+1)x+alnx,对其进行求导,利用导数研究单调性,对a值进行讨论,从而求解;(III)根据题意,f(x)=g(x)+4x-x2-2lnx=x-lnx(x>0),利用不等式lnx<(x2-1),对要证明的不等式左边进行放缩,来进行证明;本题考查导数知识的运用,考查函数的极值问题,考查不等式的证明,用好导数是关键,本题考查的知识点比较全面,是一道难题,计算量也很大,同学们要认真做好笔记;。
2014届高三数学一模文科试卷(附答案)

2014届高三数学一模文科试卷(附答案)箴言中学2013年高三第一次学月考试(时量120分钟满分 150分)一、选择题:本大题共9小题,每小题5分,共45分,每小题只有一项符合题目要求. 1.已知全集,集合,,则 =__________. A. {1,2,4} B. {2,3,4} C. {0,2,4} D . {0,2,3,4} 2.复数为虚数单位)在复平面内所对应的点在__________. A.第一象限 B.第二象限C.第三象限 D.第四象限 3.设 , 则“ ”是“ ”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 4.四名同学根据各自的样本数据研究变量之间的相关关系,并求得回归直线方程,分别得到以下四个结论:① y与x负相关且;② y与x负相关且;③ y与x正相关且;④ y与x正相关且 . 其中一定不正确的结论的序号是__________. A.①② B.②③ C.③④ D.①④ 5.下列函数中,既是偶函数又在区间上单调递减的是__________. A. B. C. D. 6.已知向量,,若,则=__________. A. B. C. D. 7.已知点在圆外, 则直线与圆的位置关系是_______. A.相切 B.相交 C.相离 D.不确定 8.若 ,则的取值范围是__________. A. B. C. D. 9.形如的函数因其函数图象类似于汉字中的�遄郑�故生动地称为“�搴�数”。
则当时的“�搴�数”与函数的交点个数为__________. A.2 B.3 C.4 D.5 二、填空题:本大题共6小题,每小题5分,共30分. 10.直线(为参数)的倾斜角为__________. 11.某学员在一次射击测试中射靶10次,命中环数如下:7,8,7,9,5,4,9,10,7,4, 则命中环数的方差为 . (注:方差,其中为的平均数) 12. 某几何体的三视图如图1所示,它的体积为__________. 13. 阅读图2的程序框图, 该程序运行后输出的的值为 __. 14. 设F1,F2是椭圆C:的两个焦点,若在C上存在一点P, 使PF1⊥PF2,且∠PF1F2=30°,则C的离心率为_____________. 15.已知函数的定义域为,部分对应值如下表,的导函数,的图象如图所示.�1 0 2 4 5 1 2 0 2 1 (1)的极小值为_______;(2)若函数有4个零点,则实数的取值范围为_________.箴言中学2013年高三第一次学月考试文科数学答题卷一、选择题:本大题共9小题,每小题5分,共45分,序号 1 2 3 4 5 6 7 8 9 答案二、填空题:本大题共6小题,每小题5分,共30分. 10.____________11.____________ 12..____________ 13.____________14.____________ 15.____________ _____________ 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12分) 若函数在R上的最大值为5. (1)求实数m的值; (2)求的单调递减区间。
数学(理)卷·2014届湖南省益阳市箴言中学高三上学期第五次模拟考试试题(2014.01)

19、(本大题满分 12 分)某投资公司拟投资开发某项新产品,市场评估能获 10--1 000 万元
第3页共9页
的投资收益,现公司准备制定一个对科研课题组的奖励方案:奖金 y(单位:万元)随投资收 益 x(单位:万元)的增加而增加,且奖金不低于 1 万元,同时奖金不超过投资收益的 20%, (Ⅰ)设奖励方案的函数模拟为 f(x),试用数学语言表述公司对奖励方案的函数模型 f(x) 的基本要求; (Ⅱ)下面是公司预设的两个奖励方案的函数模型: (1) y= +2;(2)y=4lgx-3. 试分别分析这两个函数模型是否符合公司要求?
2 sin B cos A = 3 sin A cos C + 3 sin C cos A = 3 sin( A + C)
则 2sinBcosA= 3 sinB……………………………………………………4 分
所以 cosA= 3 于是A= π ……………………………………………6 分
2
6
(2)由(1)知A= π ,又AC=BC,所以 C= 2π ………………7 分
故 x 的取值范围是 (3, 4] ………………………………12 分
17. 解:(1)由 m • n = 0 (2b − 3c) cos A = 3a cos C ……………………………………………1 分
所以 (2 sin B − 3 sin C) cos A = 3 sin A cos C ………………………2 分
设 g(x) = 4 lg x − 2 − x ,则 g ′(x) = 4 lg e − 1
5
x5
当 x≥10 时, g ′(x) = 4 lg e − 1 ≤ 2 lg e − 1 = lg e2 − 1 < 0
湖南省益阳市箴言中学2014届高三上第五次模拟考试试题

2014届高三第五次模拟考试英语试题(命题:刘劲审题:熊世平宋杰2014/1/8)本卷共四个部分,时量120分钟,满分150分PART ONE LISTENING COMPREHENSION (30marks)SECTION ADirections:In this section, you’ll 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.Conversation 11. What’s the girl doing?A. She is having a hot teaB. She is having a mealC. She is playing with her mother2. What is the girl given at last?A. A knifeB. A spoonC. A cupConversation 23. Why does the man want to live in an apartment?A. He doesn’t want to be aloneB. He wants his own spaceC. He can’t stand his friends4. What will the man do this weekend?A. Learn how to cookB. Meet his friendsC. Visit the woman Conversation 35. How much should the man pay for the fruit?A. 18 yuanB. 12 yuanC. 15 yuan6. What will the man do next?A. Go homeB. Buy more food and drinkC. Eat some breadConversation 47. What do they have for breakfast today?A. Carrot juice and breadB. Apple juice and fried eggsC. Carrot juice and fried eggs8. Which subject doesn’t John have today?A. MathB. MusicC. History9. What can we learn from the conversation?A. John’s mother will pack the schoolbag for him.B. They are eating in the room upstairs.C. John doesn’t like apple juice and bread at all.Conversation 510. Where is the man going to spend his holidays?A. In AustraliaB. In AmericaC. In China11. How will the man travel?A. By seaB. By trainC. By air12. What is the weather like in the place where the woman will spend her summer holidays?A. It’s coldB. It’s rainyC. It’s warm26 –Did you see the CEO in his office? -- Yes, he ______ by the journalist from BBC.A. has been interviewedB. is being interviewedC. had been interviewedD. was being interviewed27. –Jack failed the driving test again.-- The question is ______ we can help him avoid making the same mistake.A. whyB. howC. whatD. that28.I really can’t think of ______ told me you were not coming back till tomorrow.A. it was who thatB. who was it thatC. that who was itD. who it was that 29.The boy stood there astonished, with his eyes ______ on the strange moving object in the sky.A. to fixB. fixingC. fixedD. having fixed30.--What’s the matter with Philips?-- Something must have happened to him. He left home three hours ago, but ______ at the party.A. didn’t arriveB. isn’t arrivingC. hadn’t arrivedD. hasn’t arrived31.After Queen Elizabeth officially opened the 2012 Summer Olympics, London become the first city ______ three Olympic Games in history.A. to have hostedB. hostingC. hostedD. having hosted32.If ever the occasion ______ when I want advice, you’re the first person I’ll come to.A. will ariseB. has arisenC. is arisingD. arises33. My tutor always teaches us “A man ______ fail many times, but he isn’t a failure until be began to blame somebody else.”A. shallB. mustC. wouldD. can34. ______ time at Hunan Normal University, I’m very impressed with the quality of the teachers and students.A. Having spentB. Beijing spentC. To spendD. spending35. A large amount of money ______ spent on the Hope Project and a good many schools ______ set up.A, are; has been B. is; have been C. are; have been D. is; has beenSection 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.It was the end of my junior year, the weekend before a party. I was __36__ around that night with some friends, and a bottle got passed around the car. Everyone was taking swigs (大口喝). Then the bottle got to me. I took a swig. That was the biggest __37__ of my life – I had to drive home later that night.About a mile from my house, a policeman turned his light bar on and I pulled over and waited for him to come to my door. “Can I see your __38__ and registration please, Miss?” He took them and went back to his car for a while. When he returned, he asked if I had __39__.“Yes sir.” I said. He asked me to step out of my car.All of a sudden I felt cold metal on my wrists. He __40__ me right then and there. As I was sitting in the back seat of his car, all I could think was, “God, I’m in so much __41__ my parents are going to kill me.”The cop drove me to a town twenty minutes from my home. Being arrested was so __42__.The next week I got a lawyer and I went to court two months after that. I got $ 700 in fines, forty hours of community __43__, and had to attend a MADD (Mothers Against Drunk Driving) meeting. I’ll be on probation (缓刑) for a year, and I have to __44__ driver’s education again. And I __45__ my license for a whole year!This has been a huge learning experience for me. It opened my eyes to how easy it is to make a __46__ decision. I want everyone reading this to know that it’s not right to drink and drive, and it __47__ isn’t worth losing your license.36.A.running B.driving C.riding D. walking 37.A.problem B.mistake C.change D.challenge 38.A.license B.card C.identity D.passport 39.A.eaten B.drunk C.practiced D. quarreled 40.A.forgave B.scolded C.arrested D.fined 41.A.mess B.trouble C.wonder D.confusion 42.A.common B.different C.scary D.boring 43.A.communication B.activity C.service D.survey 44.A.bring B.give C.take D.make 45.A.held B.got C.kept D.lost 46.A.happy B.quick C.difficult D.stupid 47.A.gradually B.necessarily C.possibly D.definitelySection C(12marks)Directions:Complete the following passage by filling in each blank with one word that best fits the context.Many people think the search for cleaner energy leads only to renewable resources like sun, wind and water. But it __48__ leads to a fossil fuel. Natural gas is considered __49__ cleanest of the fossil fuels, the fuel created by plant and animal remains over millions __50__ years. Burning it releases fewer pollutants __51__ oil or coal. The gas is mainly methane (甲烷). It produces half the carbon dioxide of other fossil fuels. So __52__ may help cut the production of carbon gases linked to climate change.Russia is first in __53__ are called "proved reserves" of natural gas. The United States is sixth. Over the years, big oil __54__ gas companies recovered much of the easily reached supplies of gas in America. They drilled straight down into formations where gas collects. __55___these supplies were used up, big drillers looked for similar formations in other countries.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 and D.Choose the one that fits best according to the information given in the passage.AFamily traditions were important in our house, and none was more appreciated than the perfect Christmas tree.“Dad, can we watch when you trim (修剪) the tree?”My eldest son, Dan, nine, and his seven-year-old brother John, asked.“I won’t be cutting this year,” my husband Bob said. “Dan, you and John are old enough tomeasure things. Do it all by yourselves. Think you boys can handle it?”Dan and John seemed to grow six inches in their chairs at the thought of such an amazing responsibility. “We can handle it.” Dan promised. “We won’t let your down.”A few years before Christmas, Dan and John rushed in after school. They gathered the tools they’d need and brought them out to the yard, where the tree waited. I was cooking when I heard the happy sounds as the boys carried the tree into the living room. Then I heard the sound that every mother knows is trouble: dead silence. I hurried out to them. The tree was cut too short. John crossed his arms tight across his chest. His eyes filled with angry tears.I felt worried. The tree was central to our holiday. I didn’t want the boys to feel ashamed every time they looked at it. I couldn’t lower the ceiling, and I couldn’t raise the floor either. There was no way to undo the damage done. Suddenly, a thought came to my mind, which turned the problem into the solution.”“We can’t make the tree taller,” I said. “But we can put it on a higher position.”Dan nodded his head sideways. “We could put it on the coffee table, It just might work! Let’s try it!”When Bob got home and looked at the big tree on top of the coffee table, Dan and John held their breath.“What a good idea!” he declared. “Why didn’t I ever think of such a thing?”John broke into a grin. Dan’s chest swelled with pride.56. Who trimmed the Christmas tree this year?A. Dan and JohnB. The writerC. BobD. The whole family57. The underlined part “grow six inches (Para 4) implies the brothers felt _____,A. scaredB. embarrassedC. nervousD. proud58. What happened after the brothers moved the Christmas tree into the living room?A. They got angry with each other. .B. They found the tree was cut short.C. They rushed to schoolD. They began to decorate the tree.59. How could the short tree be turned into a perfect one?A. By making the tree taller.B. By lowering the ceiling.C. By placing it on a coffee table.D. By raising the floor.60. What Bob said in the last but one paragraph showed_________.A. he was too stupid to think of the ideaB. he was a little disappointedC. he should not have given them the taskD. he appreciated what the brothers had doneBDebate: Can stricter gun control laws decrease the rate of crimes involving guns?Yes, because with fewer weapons, you get fewer crimes.Take a look at the crime rates of countries like the United Kingdom. Their crime rate is very low, compared to ours here in the United States. Just by simply putting stricter laws in place on who can own them, it will greatly reduce the crimes involving guns.Posted by: 5h4yGloryPast PracticeThere was an assault weapon(攻击性武器) ban back in 1994 under Bill Clinton. Shootings and crime did not go down for the 10 years when this law was in effect. Criminals do not follow rules. The law only affects those who try to obey the law to arm them to protect themselves.Posted by: jb1988Wake up, you Americans!It's hardly surprising that America has one of the highest gun related murder rates of all countries! And of course guns don't kill people, but the more people own guns, the easier it is to get access to one, the higher the chance of people using it to kill somebody is !In Germany, we have really strict gun laws and only a few people own guns, and obviously we don't need them. There's no reason that a normal citizen would need a gun! Guns are made for one reason only: killing.Posted by: cstephie13More gun control laws, fewer shootingsOn the same day as the Sandy Hook tragedy, a man in China entered a primary school with a knife and injured 22 students on December, 14. However, nobody was killed. Compare that to the 26 students and teachers killed at SandyHook in Connecticut with a powerful gun. We must leam something from this tragedy.Posted by: Jasonx986Gun violence goes up with more gun controlCriminals will take advantage of the situation. They will do whatever they want and take whatever they want and kill whoever gets in their way. They will get guns illegally. Crime rates go up as a result of more gun control, because criminals know you don't have one, and they will kill you as they like. Banning guns doesn't make any sense !Posted by: eebnflow61. According to the first post, in the United Kingdom ___________.A. the crime rate is higher than America'sB. the crime rate is rising sharplyC. there is no gun control lawsD. buying guns is very hard62. We can infer that jb1988___________.A. is against gun control lawsB. was working under Bill ClintonC. is in favor of gun control lawsD. has a gun63. Which of the following does cstephie13 probably agree with?A. The more people own guns, the more people will be killed by them.B. The USA has one of the highest murder rates in the world.C. People need guns to protect themselves in time of danger.D. Guns are made to protect rather than kill.64. What do we know about the Sandy Hook tragedy?A. 22 students and teachers were killed by a powerful gun.B. 22 students were injured without any deaths.C. The tragedy took place on December 14 in Connecticut.D. 26 primary students were killed by a knife.65. Which of the following thinks that guns should not be controlled?A. Jasonx986B. eebnflowC. 5h4yGloryD. cstephie13CWith the average temperature for January standing at -50C, it is no wonder the Russian village of Oymyakon is the coldest permanently inhabited settlement in the world. Known as the “Pole of Cold”, the coldest ever temperature recorded in Oymyakon was -71.2C. This is the lowest recorded temperature for any permanently inhabited location on Earth and the lowest temperature recorded in the Northern Hemisphere(北半球).Ironically, Oymyakon actually means “non-freezing water” due to a nearby hot spring. Most homes in Oymyakon still burn coal and wood for heat and enjoy few modern conveniences.Nothing grows there so people eat reindeer(驯鹿) meat and horsemeat. A single shop provides the town's bare necessities and the locals work as reindeer-breeders, hunters and ice-fisherman.There are few modern conveniences in the village -- with many buildings still having outdoor toilets -- and most people still burn coal and wood for heat. When coal deliveries are irregular, the power station starts burning wood. If the power ceases, the town shuts down in about five hours, and the pipes freeze and crack.Daily problems that come with living in Oymyakon include pen ink freezing, glasses freezing to people's faces and batteries losing power. Locals are said to leave their cars running all day for fear of not being able to restart them. Even if there was coverage for mobile phone reception, the phones themselves would not work in such cold conditions.Another problem caused by the frozen temperatures is burying dead bodies, which can take anything up to three days. The earth must first have thawed(融解) sufficiently in order to dig it, so a bonfire is lit for a couple of hours. Hot coals are then pushed to the side and a hole couple of inches deep is dug. The process is repeated for several days until the hole is deep enough to bury the coffin.Travel companies offer tourists the opportunity to visit the village and sample life in the freezing conditions.66. Oymyakon gets its name because ______.A. it is extremely cold thereB. its water is non-freezingC. there is a hot spring nearbyD. it is known as the “Pole of Cold”67. From the passage we can infer that people in Oymyakon _______.A. seldom have fresh vegetables and fruitsB. prefer meat to any other kind of foodC. can buy anything in the shopD. might row a boat to catch fish68. Why do the locals have their cars running all day?A. They have little awareness of savingB. They want to keep warm in the car.C. They have abundant oil resources.D. They are afraid they can’t get the car started.69. Mobile phones would not work in Oymyakon because ______.A. batteries can’t be charged in such conditionsB. the extreme coldness makes the batteries lose powerC. the butteries are of poor qualityD. there was no coverage for phone reception.70. Which of the following is the first step when burying a dead body?A. Hot coals have to be pushed to the side.B. A bonfire has to be lit first.C. The coffin must be put in place.D. A hole couple of inches deep is dug.PART FOUR WRITING(45’)SECTION A(10’)Directions: Read the following passage. Complete the diagram by using the information for the passage. Write NO MORE THAN 3 WORDS for each answer.Who would have thought that just breathing could solve a lot of our health problems? Don’t we all breathe, all day, every day? It’s automatic, and we don’t even have to think about it.There’s a big difference between regular breathing and deep breathing. Regular breathing comes from the lungs, using the chest muscles. It provides oxygen to the heart which in turn makes sure the oxygen gets to all the cells in the body. There’s not an organ in the body that can operate without oxygen.Deep breathing involves learning to slow the breathing and use the diaphragm, the muscle located beneath the lungs, and not just the chest muscles. Now sit up straight and take another deep breath using the diaphragm. Your chest will rise and you will feel the diaphragm move upwards. Exhale(呼出)slowly, preferably through pursed (噘起)lips. That is an effective deep breathing.Deep breathing allows the body to take in more oxygen and release more carbon dioxide. This leads to many health benefits: a lowering of blood pressure, slowing of heart rate, and relaxation of the muscles. It calms the mind, helping to reduce sleeplessness. An increase of energy and reduction of fatigue with reduced anxiety and stress are desired effects, too.Eastern cultures have practiced deep breathing for centuries. They know it is essential to keep the mind and body in a positive relationship. The practices of yoga, qigong, and t’ai chi include deep breathing in their relaxation techniques. Could this be the reason why Asian women suffer fewer menopausal(更年期)symptoms?If we all did deep breathing exercises, even as little as a few minutes a day, we could improve our mental outlook and most likely see an improvement in our physical health as well.Title: Deep BreathingⅠ. 71. _______________●Coming from the lungs by using the chest muscles●72. _______________ to the heart and sent to all the cells.Ⅱ. Deep breathing●73. _______________the breathing and use the diaphragm●More oxygen74. ________________ and more carbon dioxide releasedⅢ. 75. _______________ of deep breathing●Blood pressure lowered and 76. _______________●Muscles relaxed and mind calmed●77. _______________, fatigue, anxiety and stress reduced●78. _______________ increasedⅣ. 79. _______________ of deep breathing●Yoga, qigong and t’ai chiⅤ. SummaryBreathing 80. _______________ every day a few minutes could improve our mental outlook and physical healthSection B (10 marks)Directions:Read the following passage. Answer the questions according to the information given in the passage.I have a friend named John Roberts who owns a horse ranch (牧马场) in San Ysidro. The last time I was there he shared a story with me.“It is a story about a young man who was the son of an itinerant(流动的) horse trainer. When he was a senior,he was asked to write a paper about what he wanted to be and do when he grew up.”“That night he wrote a seven-page paper describing his goal of someday owning a horse ranch. He wrote about his dream in great detail and he even drew a diagram of a 200-acre ranch,showing the location of all the buildings,the stables and the track.”“The next day he handed it in to his teacher. Two days later he received his paper back. On the front page was a large red F with a note that read, ‘See me after class.’”“The boy went to see the teacher after class and asked,‘Why did I receive an F?’”“The teac her said, ‘This is an unrealistic dream for a young boy like you. You have no money. You come from an itinerant family. You have no resources. Owning a horse ranch requires a lot of money. You have to buy the land. You have to pay for the original breeding stock and later you’ll have to pay large stud fees. There‘s no way you could ever do it.’Then the teacher added,‘If you will rewrite this paper with a more realistic goal, I will reconsider your grade.’”“The boy went home and asked his father what he should do. His father said, ‘Look, son,you have to make up your own mind on this. However, I think it is a very important decision for you.’”“Finally,after sitting with it for a week, the boy turned in the same paper, making no changes at all. He stated,‘You can keep the F and I’ll keep my dream.‘”John continued, “I tell you this story because you are sitting in the middle of my 200-acre horse ranch. I still have that school paper framed over the fireplace.” He added, “If I had changed my paper at that tim e, maybe we would not have the chance to be here talking about the dreams.”81. What was the boy’s goal described in his seven-page paper? (No more than 6 words) (2 marks)________________________________________________________________________ 82. Why di d the teacher give him an F”? (No more than 14 words) (2 marks)________________________________________________________________________ 83. How did John’s father respond to his son’s puzzlement? (No more than10 words) (3marks)________________________________________________________________________ 84. What can we learn from the article? (No more than 8 words.) (3 marks)________________________________________________________________________Section C (25 marks)Directions: Write an English composition according to the instructions given below in Chinese.假定你是即将毕业的高三学生李华,受学校英文报的邀请,写一篇英语短文,谈谈你高中学习和生活的收获。
2014年湖南省益阳市箴言中学高考数学五模试卷(理科)

2014年湖南省益阳市箴言中学高考数学五模试卷(理科)学校:___________姓名:___________班级:___________考号:___________一、选择题(本大题共8小题,共40.0分)1.集合M={4,5,-3m},N={-9,3},若M∩N≠∅,则实数m的值为()A.3或-1B.3C.3或-3D.-1【答案】A【解析】解:由M∩N≠∅,可知-3m=-9,或-3m=3,解得m=3或-1,故选A.利用M∩N≠∅,列出关系式,直接求出m的值即可.本题考查集合的基本运算,集合的交集的应用,考查计算能力.2.关于复数,下列说法中正确的是()A.在复平面内复数z对应的点在第一象限B.复数z的共轭复数C.若复数z1=z+b(b∈R)为纯虚数,则b=1D.设a,b为复数z的实部和虚部,则点(a,b)在以原点为圆心,半径为1的圆上【答案】C【解析】解:=.对应的点的坐标为(-1,1)位于第二象限,∴A错误.,∴B错误.z1=z+b=b-1+i,若复数z1=z+b(b∈R)为纯虚数,则b=1成立.∴C正确.∵|z|=,∴点(a,b)在以原点为圆心,半径为的圆上,∴D错误.故选:C.化简复数z,然后分别进行判断即可.本题主要考查复数的计算和化简,利用复数的四则运算法则和复数的几何意义是解决本题的关键,比较基础.3.“2a>2b”是“log2a>log2b”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【答案】B【解析】解:2a>2b⇒a>b,当a<0或b<0时,不能得到log2a>log2b,反之由log2a>log2b即:a>b>0可得2a>2b成立.故选B.分别解出2a>2b,log2a>log2b中a,b的关系,然后根据a,b的范围,确定充分条件,还是必要条件.本题考查对数函数的单调性与特殊点,必要条件、充分条件与充要条件的判断,是基础题.4.阅读如图所示的程序框图,输出结果s的值为()A. B. C. D.【答案】C【解析】解:由题意,该程序按如下步骤运行经过第一次循环得到s=cos,n=2,;经过第二次循环得到s=cos cos,n=3;经过第三次循环得到s=cos cos cos,n=4;经过第四次循环得到s=cos cos cos cos,n=5此时不满足n≥4,输出最后的s因此,输出结果s=cos cos cos cos=×=×=×=×=故选:C由程序中的变量、各语句的作用,结合流程图所给的顺序,可知该程序经过四次循环,得到当n=5时不满足n≥4,输出最后的s=cos cos cos cos,再用三角恒等变换进行化简整理,即可得到本题答案.题给出程序框图,求最后输出的s值,着重考查了程序框图的理解、用三角恒等变换求三角函数值等知识,属于基础题.5.已知、是非零向量且满足,,则△ABC的形状是()A.等腰三角形B.直角三角形C.等边三角形D.等腰直角三角形【答案】C【解析】解:∵、是非零向量且满足,,∴,∴,∴, BAC=60°.∴△ABC是等边三角形,故选:C.由、是非零向量且满足,,利用向量垂直与数量积的关系可得,进而得到,即可得出.本题考查了向量垂直与数量积的关系、等边三角形的判定方法,属于基础题.6.将函数y=sin2x的图象向左平移个单位,再向上平移1个单位,所得图象的函数解析式是()A.y=cos2xB.y=2cos 2xC.D.y=2sin2x【答案】B【解析】解:令y=f(x)=sin2x,则f(x+)=sin2(x+)=cos2x,再将f(x+)的图象向上平移1个单位,所得图象的函数解析式是y=cos2x+1=2cos2x,故选:B.利用函数y=A sin(ωx+φ)的图象变换规律及三角函数间的关系式即可得到答案.本题考查函数y=A sin(ωx+φ)的图象变换,考查升幂公式的应用,属于中档题.7.已知空间不共面的四点A,B,C,D,则到这四点距离相等的平面有()个.A.4B.6C.7D.5【答案】C【解析】解:一个点在平面一侧,另三个点在另一侧,这样满足条件的平面有四个,都是中截面如图:二个点在平面一侧,另两个点在另一侧,这样满足条件的平面有三个如图:故到这四点距离相等的平面有7个故选:C四个点在平面同侧不可能存在与空间不共面四点距离相等的平面,那么可分为一个点在平面一侧,另三个点在另一侧,中截面满足条件,这样的情形有4个,还有一类是二个点在平面一侧,另两个点在另一侧,这样满足条件的平面有三个,即可求出所有满足条件的平面.本小题主要考查平面的基本性质及推论、确定平面的条件、空间距离等基础知识,考查空间想象力、化归与转化思想.属于基础题.8.设<,g(x)是二次函数,若f(g(x))的值域是[0,+∞),则g(x)的值域是()A.(-∞,-1]∪[1,+∞)B.(-∞,-1]∪[0,+∞)C.[0,+∞)D.[1,+∞)【答案】C【解析】解:∵,,<,∴当|x|≥1时,f(x)≥1;当0≤x<1时,0≤f(x)<1.∴x≥0或x≤-1时,f(x)≥0,∵f[g(x)]的值域为[0,+∞),∴g(x)≥0或g(x)≤-1.∵g(x)是二次函数,∴g(x)的值域应该是g(x)≥0或g(x)≤-1中的一个,若g(x)≤-1,则f(g(x))的值域是[1,+∞),不符合题意,若g(x)≥0,则f(g(x))的值域是[0,+∞),故g(x)≥0,即g(x)的值域为[0,+∞).故选C.由,,<,知当|x|≥1时,f(x)≥1.当0≤x<1时,0≤f(x)<1,得到x≥0或x≤-1时,f(x)≥0,由此根据f[g(x)]的值域为[0,+∞),能求出g(x)的值域.本题考查二次函数的性质和应用,是基础题,解题时要认真审题,仔细解答,注意分段函数的性质和应用.二、填空题(本大题共7小题,共35.0分)9.(坐标系与参数方程选做题)已知直线:为参数,:(s为参数),若l1∥l2,则k= ______ ;若l1⊥l2,则k= ______ .【答案】4;-1【解析】解:直线l1的方程即kx+2y-k-4=0,直线l2的方程即2x+y-1=0.若l1∥l2,则-2=,k=4.若l1⊥l2,则-2•=-1,k=-1.故答案为:4;-1.先把直线的方程化为普通方程,再利用两直线平行,斜率相等,两直线垂直,斜率之积等于-1,分别求出k值.本题考查两直线平行、垂直的性质,两直线平行,斜率相等,两直线垂直,斜率之积等于-1.10.某几何体的三视图如图所示,则其表面积为______ .【答案】+2.【解析】解:由三视图知几何体为半个圆锥,圆锥的底面圆半径为1,高为2,∴圆锥的母线长为,∴几何体的表面积S=×π×12+×π×1×+×2×2=.故答案是+2.由三视图知几何体为半个圆锥,根据三视图的数据求底面面积与高,代入棱锥的体积公式计算.本题考查了由三视图求几何体的表面积,考查了圆锥的侧面积公式,解题的关键是由三视图判断几何体的形状及三视图的数据所对应的几何量.11.若函数f(a)=(2+sinx)dx,则f()= ______ .【答案】π+1【解析】解:===π+1.故答案为π+1.利用微积分基本定理即可得出.熟练掌握微积分基本定理是解题的关键.12.已知一元二次不等式f(x)<0的解集为{<<,则f(2x)>0的解集为______ .【答案】{x|x<-1或x>1}【解析】解:∵一元二次不等式f(x)<0的解集为{<<,∴一元二次不等式f(x)>0的解集为{x|x>或<},由f(2x)>0,得2x>2或2x<,解得x>1或x<-1,即f(2x)>0的解集为{x|x<-1或x>1}.故答案为:{x|x<-1或x>1}.根据一元二次不等式解集得到f(x)>0的解集,然后解指数不等式即可.本题主要考查一元二次不等式的解法和性质,以及指数不等式的解法,考查学生运算能力.13.已知实数x,y满足,则的最大值为______ .【答案】【解析】解:满足的可行域如图所示,三个顶点的坐标分别为(1,),(1,2),(2,1),∴的值分别为,2,令=t,则t∈[,2]∴===t-∴该函数在[,2]上单调递增,∴t=2时,的最大值为.故答案为:.作出满足的可行域,可得的范围,化简目标函数,利用其单调性,可求最大值.本题考查线性规划知识的运用,考查函数的单调性,确定可行域是关键.14.已知数列{a n}的通项公式为a n=-n+p,数列{b n}的通项公式为b n=2n-5.设c n=,,>,若在数列{c n}中,c8>c n(n∈N*,n≠8),则实数p的取值范围是______ .【答案】(12,17)【解析】解:当a n≤b n时,c n=a n,当a n>b n时,c n=b n,∴c n是a n,b n中的较小者,因为a n=-n+p,所以{a n}是递减数列;因为b n=2n-5,所以{b n}是递增数列,因为c8>c n(n≠8),所以c8是c n的最大者,则n=1,2,3,…7,8时,c n递增,n=8,9,10,…时,c n递减,因此,n=1,2,3,…7时,2n-5<-n+p总成立,当n=7时,27-5<-7+p,∴p>11,n=9,10,11,…时,2n-5>-n+p总成立,当n=9时,29-5>-9+p,成立,∴p<25,而c8=a8或c8=b8,若a8≤b8,即23≥p-8,所以p≤16,则c8=a8=p-8,∴p-8>b7=27-5,∴p>12,故12<p≤16,若a8>b8,即p-8>28-5,所以p>16,∴c8=b8=23,那么c8>c9=a9,即8>p-9,∴p<17,故16<p<17,综上,12<p<17.故答案为:(12,17).由c n表达式知c n是a n,b n中的较小者,易判断{a n}是递减数列,{b n}是递增数列,由c8>c n(n≠8)知c8是c n的最大者,从而可知n=1,2,3,…7,8时,c n递增,n=8,9,10,…时,c n递减,进而可知a n与b n的大小关系,且c8=a8或c8=b8,分两种情况讨论,当c8=a8时,a8>b7,当c8=b8时,b8>a9,分别解出p的范围,再取并集即可;本题考查等差数列、等比数列的综合、数列的函数特性,考查分类讨论思想,考查学生分析问题解决问题的能力,考查学生逻辑推理能力,难度较大.15.对于集合N={1,2,3,…,n}的每一个非空子集,定义一个“交替和”如下:按照递减的次序重新排列该子集,然后从最大数开始交替地减、加后继的数.例如集合{1,2,4,6,9}的交替和是9-6+4-2+1=6,集合{5}的交替和为5.当集合N中的n=2时,集合N={1,2}的所有非空子集为{1},{2},{1,2},则它的“交替和”的总和S2=1+2+(2-1)=4,请你尝试对n=3、n=4的情况,计算它的“交替和”的总和S3、S4,并根据其结果猜测集合N={1,2,3,…,n}的每一个非空子集的“交替和”的总和S n= ______ .【答案】n•2n-1【解析】解:S1=1S2=4当n=3时S3=1+2+3+(2-1)+(3-1)+(3-2)+(3-2+1)=12S4=1+2+3+4+(2-1)+(3-1)+(4-1)+(3-2)+(4-2)+(4-3)+(3-2+1)+(4-2+1)+(4-3+1)+(4-3+2)+(4-3+2-1)=32∴根据前4项猜测集合N={1,2,3,…,n}的每一个非空子集的“交替和”的总和S n=n•2n-1故答案为:n•2n-1根据“交替和”的定义:按照递减的次序重新排列该子集,然后从最大数开始交替地减、加后继的数可求出“交替和”的总和S3、S4,并根据其结果猜测集合N={1,2,3,…,n}的每一个非空子集的“交替和”的总和S n即可.本题主要考查了数列的应用,同时考查了归纳推理的能力,属于中档题.三、解答题(本大题共6小题,共75.0分)16.函数y=f(x)对于x>0有意义,且满足条件f(2)=1,f(xy)=f(x)+f(y),f (x)是增函数.(1)证明:f(1)=0;(2)若f(x)+f(x-3)≥2成立,求x的取值范围.【答案】解:(1)在f(xy)=f(x)+f(y)中,令x=2,y=1,则f(2×1)=f(2)+f(1),又由f(2)=1,则f(1)=0;(2)令x=2,y=2,则f(2×2)=f(4)=f(2)+f(2)=2,所以f(x)+f(x-3)=f(x2-3x)≥f(4),又f(x)为增函数所以>>,综上,x≥4.【解析】(1)令x=2,y=1,并代入f(xy)=f(x)+f(y),即可求出f(1)的值;(2)令x=2,y=2,代入求得f(4),结合题意可将f(x)+f(x-3)≥2转化为f(x2-3x)≥f(4),结合函数的单调性与函数的定义域,可得>>,解可得x的值.本题考查抽象函数的应用,解(2)的关键是根据题意,分析出f(4)=2,进而用f(4)替换2,其次要注意函数的定义域.17.设三角形ABC的内角为A,B,C所对的边长分别为a,b,c,,,,,且.(Ⅰ)求角A的大小;(Ⅱ)若AC=BC,且BC边上的中线AM的长为,求△ABC的面积.【答案】解:(Ⅰ)∵=(cos A,cos C),=(c-2b,a),⊥,∴•=0,即(c-2b)cos A+acos C=0,利用正弦定理化简得:(sin C-2sin B)cos A+sin A cos C=0,即2sin B cos A=(sin A cos C+cos A sin C)=sin(A+C)=sin B,∴cos A=,则A=;(Ⅱ)由第一问知A=,又AC=BC,∴C=,设AC=x,则MC=x,AM=,在△AMC中,由余弦定理得AC2+MC2-2AC•MC cos C=AM2,即x2+(x)2-2x•cos=()2,解得:x=2,则S△ABC=×2×2×sin=.【解析】(Ⅰ)由两向量的坐标,根据两向量垂直时数量积为0列出关系式,再利用正弦定理化简,求出cos A的值,即可确定出角A的大小;(Ⅱ)由A的度数求出C的度数,设AC=x,表示出MC,再由AM的长,在三角形AMC 中,利用余弦定理列出关于x的方程,求出方程的解得到x的值,利用三角形面积公式即可求出三角形ABC面积.此题考查了正弦、余弦定理,平面向量的数量积运算,以及三角形的面积公式,熟练掌握定理及公式是解本题的关键.18.在如图所示的空间几何体中,平面ACD⊥平面ABC,△ACD与△ACB是边长为2的等边三角形,BE=2,BE和平面ABC所成的角为60°,且点E在平面ABC上的射影落在 ABC的平分线上.(Ⅰ)求证:DE∥平面ABC;(Ⅱ)求二面角E-BC-A的余弦值.【答案】(本小题满分12分)解:(Ⅰ)由题意知,△ABC,△ACD都是边长为2的等边三角形,取AC中点O,连接BO,DO,则BO⊥AC,DO⊥AC,…(2分)又∵平面ACD⊥平面ABC,∴DO⊥平面ABC,作EF⊥平面ABC,那么EF∥DO,根据题意,点F落在BO上,∵BE和平面ABC所成的角为60°,∴ EBF=60°,∵BE=2,∴,…(4分)∴四边形DEFO是平行四边形,∴DE∥OF,∵DE不包含于平面ABC,OF⊂平面ABC,∴DE∥平面ABC.…(6分)(Ⅱ)解法一:作FG⊥BC,垂足为G,连接EG,∵EF⊥平面ABC,∴EF⊥BC,又EF∩FG=F,∴BC⊥平面EFG,∴EG⊥BC,∴ EGF就是二面角E-BC-A的平面角.…(9分)R t△EFG中,°,,.∴.即二面角E-BC-A的余弦值为.…(12分)解法二:建立如图所示的空间直角坐标系O-xyz,B(0,,0),C(-1,0,0),E(0,,),∴=(-1,-,0),=(0,-1,),平面ABC的一个法向量为,,设平面BCE的一个法向量为,,则,∴,∴,,.…(9分)所以<,>,又由图知,所求二面角的平面角是锐角,二面角E-BC-A的余弦值为.…(12分)【解析】(Ⅰ)取AC中点O,连接BO,DO,由题设条件推导出DO⊥平面ABC,作EF⊥平面ABC,由已知条件推导出 EBF=60°,由此能证明DE∥平面ABC.(Ⅱ)法一:作FG⊥BC,垂足为G,连接EG,能推导出 EGF就是二面角E-BC-A的平面角,由此能求出二面角E-BC-A的余弦值.法二:以OA为x轴,以OB为y轴,以OD为z轴,建立空间直角坐标系O-xyz,利用向量法能求出二面角E-BC-A的余弦值.本题考查直线与平面平行的证明,考查二面角的余弦值的求法,解题时要注意空间思维能力的培养,注意向量法的合理运用.19.某投资公司投资开发某项新产品,市场评估能获得10~1000万元的投资收益.现公司准备制定一个对科研课题组的奖励方案:奖金y(单位:万元)随投资收益x(单位:万元)的增加而增加,且奖金不低于1万元,同时不超过投资收益的20%.(1)设奖励方案的函数模型为f(x),试用数学语言表述公司对奖励方案的函数模型f (x)的基本要求.(2)下面是公司预设的两个奖励方案的函数模型:①f(x)=+2;②f(x)=4lgx-2.试分别分析这两个函数模型是否符合公司要求.【答案】解:(Ⅰ)由题意知,公司对奖励方案的函数模型f(x)的基本要求是:当x∈[10,1000]时,①f(x)是增函数;②f(x)≥1恒成立;③恒成立,(Ⅱ)(i)对于函数模型,当x∈[10,1000]时,f(x)是单调递增函数,则f(x)≥1显然恒成立,若函数在[10,1000]上恒成立,即29x≥300恒成立,又∵(29x)min=290,∴不恒成立,综上所述,函数模型满足基本要求①②,但是不满足③,故函数模型不符合公司要求;(ii)对于函数模型f(x)=4lgx-2,当x∈[10,1000]时,f(x)是单调递增函数,则f(x)min=f(10)=4lg10-2=2>1,∴f(x)≥1恒成立,令,则,∵当x≥10时,<,∴g(x)在[10,1000]上是减函数,∴g(x)≤g(10)=4lg10-2-2=0,即,∴,∴恒成立,综上所述,函数模型f(x)=4lgx-2满足基本条件①②③,故函数模型f(x)=4lgx-2符合公司要求.【解析】(Ⅰ)根据题意可知奖励方案描述的是函数的单调性和最值,从而运用数学语言描述出即可;(Ⅱ)分别对两个函数模型研究它们的单调性和恒成立问题,判断是否符合(1)中的基本要求即可.本题主要考查函数模型的选择与应用.解决实际问题通常有四个步骤:(1)阅读理解,认真审题;(2)引进数学符号,建立数学模型;(3)利用数学的方法,得到数学结果;(4)转译成具体问题作出解答,其中关键是建立数学模型.本题的解题关键是理解题意,将题意转化为数学问题进行求解.属于中档题.20.设数列{a n}的前n项和为S n,已知(n∈N*).(Ⅰ)求数列{a n}的通项公式;(Ⅱ)求证:当x>0时,>;(Ⅲ)令,数列{c n}的前2n项和为T2n.利用(2)的结论证明:当n∈N*且n≥2时,<.【答案】(Ⅰ)解:由,得(n≥2)…(2分)两式相减,得,即(n≥2)于是,所以数列是公差为1的等差数列…..….(3分)又,所以a1=4.所以,故.….(5分)(Ⅱ)证明:令>,则>,(7分)∴g(x)在(0,+∞)时单调递增,g(x)>g(0)=0,即当x>0时,>….(9分)(Ⅲ)证明:因为,所以当n≥2时,==.…(11分)下面证<令,由(2)可得>,所以>,>,…,>以上n个式相加,即有>∴<…(14分)【解析】(Ⅰ)由已知条件推导出(n≥2),,由此能求出数列{a n}的通项公式.(Ⅱ)令>,利用导数求出g(x)>g(0)=0,从而能证明当x>0时,>.(Ⅲ)由,知当n≥2时,=,由此能证明当n∈N*且n≥2时,<.本题考查数列的通项公式的求法,考查不等式的证明,解题时要认真审题,注意导数性质和分组求和法的合理运用.21.已知a∈R,函数,g(x)=(lnx-1)e x+x(其中e为自然对数的底数).(1)求函数f(x)在区间(0,e]上的最小值;(2)是否存在实数x0∈(0,e],使曲线y=g(x)在点x=x0处的切线与y轴垂直?若存在,求出x0的值;若不存在,请说明理由.【答案】解:(1)∵,∴令f'(x)=0,得x=a.①若a≤0,则f'(x)>0,f(x)在区间(0,e]上单调递增,此时函数f(x)无最小值.②若0<a<e,当x∈(0,a)时,f'(x)<0,函数f(x)在区间(0,a)上单调递减,当x∈(a,e]时,f'(x)>0,函数f(x)在区间(a,e]上单调递增,所以当x=a时,函数f(x)取得最小值lna③若a≥e,则f'(x)≤0,函数f(x)在区间(0,e]上单调递减,所以当x=e时,函数f(x)取得最小值..综上可知,当a≤0时,函数f(x)在区间(0,e]上无最小值;当0<a<e时,函数f(x)在区间(0,e]上的最小值为lna;当a≥e时,函数f(x)在区间(0,e]上的最小值为.(2)∵g(x)=(lnx-1)e x+x,x∈(0,e],∴g'(x)=(lnx-1)e x+(lnx-1)(e x)+1=.由(1)可知,当a=1时,.此时f(x)在区间(0,e]上的最小值为ln1=0,即.(10分)当x0∈(0,e],>,,∴>.曲线y=g(x)在点x=x0处的切线与y轴垂直等价于方程g'(x0)=0有实数解.(13分)而g'(x0)>0,即方程g'(x0)=0无实数解.、故不存在x0∈(0,e],使曲线y=g(x)在点x=x0处的切线与y轴垂直.【解析】(1)讨论满足f (x)=0的点附近的导数的符号的变化情况,来确定极值,将f(x)的各极值与其端点的函数值比较,其中最小的一个就是最小值;(2)将曲线y=g(x)在点x=x0处的切线与y轴垂直转化成方程g'(x0)=0有实数解,只需研究导函数的最小值即可.本题主要考查了利用导数求闭区间上函数的最值,以及利用导数研究曲线上某点切线方程,属于中档题.。
2014年湖南省益阳市箴言中学高考数学三模试卷(文科)

2014年湖南省益阳市箴言中学高考数学三模试卷(文科)学校:___________姓名:___________班级:___________考号:___________一、选择题(本大题共9小题,共45.0分)1.已知全集U={0,1,2,3,4},集合A={1,2,3},B={2,4},则(∁U A)∪B为()A.{1,2,4}B.{2,3,4}C.{0,2,3,4}D.{0,2,4}【答案】D【解析】解:∵∁U A={0,4},∴(∁U A)∪B={0,2,4};故选D.由题意,集合∁U A={0,4},从而求得(∁U A)∪B={0,2,4}.本题考查了集合的运算,属于基础题.2.“x<-1”是“x<0”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【答案】A【解析】解:当x<-1时,x<0成立,当x=-时,满足x<0,但x<-1不成立,即“x<-1”是“x<0”的充分不必要条件,故选:A.利用充分条件和必要条件的定义进行判断.本题主要考查充分条件和必要条件的判断,利用不等式的之间的关系是解决本题的关键,比较基础.3.若(a-2i)i=b-i,其中a,b∈R,i是虚数单位,则a+b=()A.1B.-1C.2D.-2【答案】A【解析】解:因为(a-2i)i=b-i所以2+ai=b-i,可得b=2a=-1所以a+b=1.故选A.先化简方程,利用复数相等的条件,求出a、b的值,可求a+b本题考查复数相等的条件,是基础题.4.如图是一个空间几何体的三视图,则该几何体的侧面积为()A. B. C.8 D.12【答案】C【解析】解:由三视图可知:该几何体是正四棱锥,其中底边长=2,侧面的斜高为2,由此得该几何体的侧面积=4×=8.故选C.由三视图可知:该几何体是正四棱锥,其中底边长2,侧面的斜高为2,据此可计算出侧面积.本题考查由三视图计算原几何体的侧面积,正确恢复原几何体是解决问题的关键.5.函数y=sinxcosxcos2x的最小正周期为()A. B. C.π D.2π【答案】A【解析】解:∵y=sinxcosxcos2x==∴最小正周期为T==.故选:A.求三角函数的最小正周期,首先要把函数化成正弦型函数的标准形式,即化成y=A sin (ωx+φ)+B的形式,然后利用T=求出周期.本题是求三角函数周期的基本题型,解答本题的关键是化成正弦型函数的标准形式.6.若双曲线-=1(a>0)的离心率为2,则a=()A.2B.C.D.1【答案】B【解析】解:=2,(a>0),∴a=.故选B.由题意知=2,(a>0),由此可以求出a的值.本题考查双曲线的离心率,比较简单.会利用公式就能求出实数a.7.方程x2-4x+4=lnx的解的个数有()A.1个B.2个C.3个D.4个【答案】B【解析】解:方程x2-4x+4=lnx的解的个数,即为函数y=x2-4x+4与y=lnx的图象交点的个数,在同一坐标系中画出两个函数的图象如下图所示:由图可得,两个函数的图象共有2个交点,故方程x2-4x+4=lnx的解的个数有2个,故选:B方程x2-4x+4=lnx的解的个数,即为函数y=x2-4x+4与y=lnx的图象交点的个数,在同一坐标系中画出两个函数的图象,可得答案.本题主要考查方程根的个数的判断,构造函数,利用方程和函数之间的关系,即可得到结论.8.若a=20.5,b=logπ3,c=log2sin,则()A.a>b>cB.b>a>cC.c>a>bD.b>c>a【答案】A解:<<,由指对函数的图象可知:a>1,0<b<1,c<0,故选A利用估值法知a大于1,b在0与1之间,c小于0.估值法是比较大小的常用方法,属基本题.9.已知函数f(x)(x∈R)满足f(1)=1,且f′(x)<,则不等式f(lg2x)<+的解集为()A.(0,)B.(0,)∪(10.+∞)C.(,10)D.(10,+∞)【答案】B【解析】解:令lg2x=t,(t>0),则不等式<即为不等式<,令,则′′<,所以F(t)=f(t)-在(0,+∞)内单调递减,又,所以<的解集为(1,+∞),由>,得<<或>,所以不等式<的解集为,,∞.故选:B.令lg2x=t,(t>0),则<,令,则′′<,由此利用导数性质能求出不等式<的解集.本题考查不等式的解法,是中档题,解题时要认真审题,注意导数性质的合理运用.二、填空题(本大题共6小题,共30.0分)10.2log510+log50.25= ______ .【答案】2【解析】解:∵2log510+log50.25=log5100+log50.25=log525=2故答案为:2.根据对数运算法则nlog a b=log a b n和log a M+log a N=log a(MN)进行求解可直接得到答案.本题主要考查对数的运算法则,解题的关键是对对数运算法则的熟练程度,属于基础题.11.若f(x)=3x+a•3-x是奇函数,则a= ______ .【答案】-1【解析】解:∵f(x)=3x+a•3-x是奇函数,∴f(x)+f(-x)=0∴3x+a•3-x+3-x+a•3x=0∴(1+a)3x+(1+a)3-x=0∴(1+a)(3x+3-x)=0∴1+a=0∴a=-1故答案为-1由函数是奇函数的性质建立方程求参数的值即可.由奇函数的定义可得f(x)+f(-x)=0,将函数解析式代入,利用恒成立的关系求参数本题考查函数奇偶性的性质,解题的关键是利用函数是奇函数这一条件得到方程f(x)+f(-x)=0根据其恒成立的性质求出参数a的值.12.已知变量x,y满足约束条件,则目标函数z=y+2x的最大值为______ .【答案】13【解析】解:作出可行域如图,由z=y+2x知,y=-2x+z,所以动直线y=-2x+z的纵截距取得最大值时,目标函数取得最大值.结合可行域可知当动直线经过点A(5,3)时,目标函数去的最大值13.故答案为:13.先根据约束条件画出可行域,再利用几何意义求最值,只需求出直线z=y+2x过点A(5,3)时,z最大值即可.本题主要考查了简单的线性规划,以及利用几何意义求最值,属于基础题.13.运行如图的程序框图,输出的结果是______【答案】510【解析】解:此问题相当于以下问题:已知数列{a n}的通项,求其前8项和S8.∵=510.故答案为510.此问题相当于以下问题:已知数列{a n}的通项,求其前8项和S8.利用等比数列的前n项和公式即可得出.把问题正确等价转化和熟练掌握等比数列的定义及其前n项和公式是解题的关键.14.已知向量,满足||=4,||=3,且(2-3)•(2+)=61,则与的夹角为______ .【答案】【解析】解:设与的夹角为θ,由题意可得(2-3)•(2+)=4-4-3=61,代入数据可得4×42-4×4×3×cosθ-3×32=61,解得cosθ=-,∴θ=故答案为:设与的夹角为θ,由数量积的运算代入已知式子计算可得.本题考查平面向量的数量积与夹角,属基础题.15.设棱长为1的正方体为图形C1,以C1各个面的中心为顶点的正八面体为图形C2,以C2各个面的中心为顶点的正方体为图形C3,以C3各个面的中心为顶点的正八面体为图形C4,…,以此类推.设正多面体C n(n∈N+)的棱长为a n(各棱长相等的多面体称为正多面体),则:(1)a1=1,a2= ______ ;(2)当n为奇数时,a n= ______ .【答案】;【解析】解:(1)正方体C1各面中心为顶点的凸多面体C2为正八面体,它的中截面(垂直平分相对顶点连线的界面)是正方形,该正方形对角线长等于正方体的棱长,所以它的棱长a2==;(2)以C2各个面的中心为顶点的正方体为图形C3是正方体,正方体C3面对角线长等于C2棱长的,(正三角形中心到对边的距离等于高的),因此对角线为=,所以a3==,以上方式类推,得,=,…,当n为奇数时,a n=,故答案为:(1);(2).(1)根据条件先求出a2,(2)根据条件依次求出a3,a4,a5,然后利用归纳推理得到:n为奇数时,a n的表达式.本题主要考查等比数列得通项公式,以及归纳推理的应用,可以从中找到规律,分奇数项、偶数项讨论,可以求a n通项公式.三、解答题(本大题共6小题,共75.0分)16.已知函数f(x)=cos(x-),x∈R.(1)求f()的值;(2)若cosθ=,θ∈(0,),求f(2θ-).【答案】解:(1)∵f(x)=cos(x-),∴f()=cos=×=1;(2)∵cosθ=∈(,),θ∈(0,),∴θ∈(,),2θ∈(,),∴sinθ==,∴sin2θ=2sinθcosθ=,cos2θ=-=-;∴f(2θ-)=cos(2θ--)=cos(2θ-)=(cos2θcos+sin2θsin)=(×(-)+×)=.【解析】(1)易求f()=cos的值;(2)依题意,可求得sinθ,利用二倍角的正弦公式与两角差的余弦即可求得f(2θ-).本题考查两角和与差的余弦函数,求得sinθ及sin2θ的值是关键,着重考查运算求解能力,属于中档题.17.在△ABC中,内角A,B,C所对的边分别是a,b,c.已知cos2A+6sin2=4.(Ⅰ)求角A的度数;(Ⅱ)若a=,b+c=3,求△ABC的面积S.【答案】解:(1)∵==1+cos A.∴4=cos2A+6sin2=2cos2A-1+3(1+cos A),化为2cos2A+3cos A-2=0,又|cos A|≤1,解得cos A=,解得A∈(0,π).∴.(II)由余弦定理可得:a2=b2+c2-2bccos A,∴=,∴3=32-2bc-bc,化为bc=2.∴S===.【解析】(1)利用倍角公式、诱导公式、余弦函数的单调性即可得出;(2)利用余弦定理、三角形的面积计算公式即可得出.本题考查了倍角公式、诱导公式、余弦函数的单调性、余弦定理、三角形的面积计算公式,考查了计算能力与推理能力,属于中档题.18.如图,在四棱锥P-ABCD中,底面ABCD是边长为4的正方形,PA⊥平面ABCD,E为PB中点,PB=4.(I)求证:PD∥面ACE.(Ⅱ)求三棱锥E-ABC的体积.【答案】(I)证明:连接BD,交AC于F,连接EF.∵四边形ABCD为正方形∴F为BD的中点∵E为PB的中点,∴EF∥PD又∵PD⊄面ACE,EF⊂面ACE,∴PD∥平面ACE…(5分)(Ⅱ)解:取AB中点为G,连接EG∵E为AB的中点∵PA⊥平面ABCD,∴EG⊥平面ABCD,在R t△PAB中,PB=4,AB=4,则PA=4,EG=2…(10分)∴…(12分)【解析】(I)连接BD,交AC于F,连接EF,证明EF∥PD,利用线面平行的判定定理,可得结论;(II)取AB中点为G,连接EG,证明EG⊥平面ABCD,即可求三棱锥E-ABC的体积.本题考查线面平行,考查三棱锥体积的计算,考查学生分析解决问题的能力,属于中档题.19.某村庄拟修建一个无盖的圆柱形蓄水池(不计厚度).设该蓄水池的底面半径为r米,高为h米,体积为V立方米.假设建造成本仅与表面积有关,侧面积的建造成本为100元/平方米,底面的建造成本为160元/平方米,该蓄水池的总建造成本为12000π元(π为圆周率).(Ⅰ)将V表示成r的函数V(r),并求该函数的定义域;(Ⅱ)讨论函数V(r)的单调性,并确定r和h为何值时该蓄水池的体积最大.【答案】解:(Ⅰ)∵蓄水池的侧面积的建造成本为200•πrh元,底面积成本为160πr2元,∴蓄水池的总建造成本为200•πrh+160πr2元即200•πrh+160πr2=12000π∴h=(300-4r2)∴V(r)=πr2h=πr2•(300-4r2)=(300r-4r3)又由r>0,h>0可得0<r<5故函数V(r)的定义域为(0,5)(Ⅱ)由(Ⅰ)中V(r)=(300r-4r3),(0<r<5)可得V′(r)=(300-12r2),(0<r<5)∵令V′(r)=(300-12r2)=0,则r=5∴当r∈(0,5)时,V′(r)>0,函数V(r)为增函数当r∈(5,5)时,V′(r)<0,函数V(r)为减函数且当r=5,h=8时该蓄水池的体积最大【解析】(I)由已知中侧面积和底面积的单位建造成本,结合圆柱体的侧面积及底面积公式,根据该蓄水池的总建造成本为12000π元,构造方程整理后,可将V表示成r的函数,(Ⅱ)根据(I)中函数的定义值及解析式,利用导数法,可确定函数的单调性,根据单调性,可得函数的最大值点.本题考查的知识点是函数模型的应用,其中(Ⅰ)的关键是根据已知,求出函数的解析式及定义域,(Ⅱ)的关键是利用导数分析出函数的单调性及最值点.20.设数列{a n}的前n项和为S n,若对于一切n∈N+,=t(t为非零常数),则称数列{a n}为“和谐数列”,t为“和谐比”.(1)设数列{b n}是首项为1,公差为2的等差数列,证明:数列{b n}为“和谐数列”,并求出“和谐比”;(2)设正项等比数列{c n}的首项为c1,公比为q(q≠1),若数列{lgc n}为“和谐数列”,试探究c1与q之间的关系,并说明理由.【答案】(1)证明:b n=1+(n-1)×2=2n-1,s n=n+(n-1)×2=n2,s2n=4n2,∴t==即数列{b n}为“和谐数列”;(2)∵c n=c1q n-1,lgc n-lgc n-1=lgq,∴{lgc n}为等差数列,∴,∴lgc1=lgq,且4t=1,∴q=c12,t=.【解析】(1)应用等差数列的通项和求和公式,即可得到t;(2)应用等比数列的通项和等差数列的通项和求和公式,即可得到t、q和c1的关系.本题考查新定义及应用,考查等差数列和等比数列的通项和求和公式,解题的关键是对新定义的理解,属于中档题.21.已知函数f(x)=(ax2+bx+c)e x在x=1处取得极小值,其图象过点A(0,1),且在点A处切线的斜率为-1.(Ⅰ)求f(x)的解析式;(Ⅱ)设函数g(x)的定义域D,若存在区间[m,n]⊆D,使得g(x)在[m,n]上的值域也是[m,n],则称区间[m,n]为函数g(x)的“保值区间”.证明:当x>1时,函数f(x)不存在“保值区间”.【答案】解:(Ⅰ)∵f(x)=(ax2+bx+c)e x,∴f′(x)=[ax2+(2a-b)x+(b+c)]e x,(2分)∵函数f(x)=(ax2+bx+c)e x在x=1处取得极小值,其图象过点A(0,1),且在点A处切线的斜率为-1.∴′,即,解得′所以f(x)的解析式为f(x)=(x2-2x+1)e x.(4分)(Ⅱ)由(Ⅰ)得f'(x)=(x2-1)e x,假设当x>1时,f(x)存在“保值区间”[m,n](n>m>1)因为当x>1时,f'(x)=(x2-1)e x>0,所以f(x)在区间(1,+∞)上是增函数∴即于是问题转化为(x-1)2e x-x=0有两个大于1的不等实根.(6分)现在考查函数h(x)=(x-1)2e x-x(x≥1),h′(x)=(x2-1)e x-1令φ(x)=(x2-1)e x-1,∴φ′(x)=(x2+2x-1)e x,当x>1时,φ′(x)>0∴φ(x)在(1,+∞)上是增函数,即h′(x)在(1,+∞)上是增函数∴h′(1)=-1<0,,h′(2)=3e2-1>0∴存在唯一x0∈(1,2),使得h′(x0)=0(10分)当x变化时,h′(x),h(x)的变化情况如下表:所以,h(x)在(1,x0)上单调递减,在(x0,+∞)上单调递增.∴h(x0)<h(1)=-1<0∵h(2)=e2-2>0∴当x>1时,h(x)的图象与x轴有且只有一个交点即方程(x-1)2e x-x=0有且只有一个大于1的根,与假设矛盾故当x>1时,f(x)不存在“保值区间”.(13分)【解析】(Ⅰ)求导函数,根据函数f(x)=(ax2+bx+c)e x在x=1处取得极小值,其图象过点A(0,1),且在点A处切线的斜率为-1,建立方程组,从而可得f(x)的解析式为f(x)=(x2-2x+1)e x.(Ⅱ)由(Ⅰ)得f'(x)=(x2-1)e x,假设当x>1时,f(x)存在“保值区间”[m,n](n>m>1),进而问题转化为(x-1)2e x-x=0有两个大于1的不等实根,构造新函数h(x)=(x-1)2e x-x(x≥1),可判断存在唯一x0∈(1,2),使得h′(x0)=0,h(x)在(1,x0)上单调递减,在(x0,+∞)上单调递增,从而可得当x>1时,h(x)的图象与x轴有且只有一个交点,即方程(x-1)2e x-x=0有且只有一个大于1的根,与假设矛盾,故可得证.本题以函数的性质为载体,考查导数知识的运用,考查函数的解析式,考查新定义,同时考查反证法思想的运用,综合性强.高中数学试卷第11页,共11页。
2014年湖南省益阳市箴言中学高三理科一模数学试卷

2014年湖南省益阳市箴言中学高三理科一模数学试卷一、选择题(共8小题;共40分)1. 设i为虚数单位,则1+i1−i等于 A. iB. −iC. 1+iD. 1−i2. 设全集U=R,集合A=x x2−1<0,B=x x x−2≥0,则A∩∁U B= ______A. x0<x<2B. x0<x<1C. x0≤x<1D. x−1<x<03. 按照如图的程序框图执行,若输出结果为15,则M处条件为 A. k≥16B. k<8C. k<16D. k≥84. 已知a,b是非零向量,则a=b是 a+b与 a−b垂直的______ 条件.A. 充分不必要B. 必要不充分C. 既不充分也不必要D. 充要5. 已知某几何体的正视图和侧视图均是边长为1的正方形,则这个几何体的体积不可能是 A. 12B. π4C. 1D. π36. 有下列四种说法:①命题:“∂x0∈R,使得x2−x>0”的否定是“∀x∈R,都有x2−x≤0”;②已知随机变量x服从正态分布N1,σ2,P x≤4=0.79,则P x≤−2=0.21;③函数f x=2sin x cos x−1,x∈R图象关于直线x=3π4对称,且在区间 −π4,π4上是增函数;④设实数x,y∈0,1,则满足:x2+y2<1的概率为π4.其中错误的个数是 A. 0B. 1C. 2D. 37. 已知函数f x=2−x−1,x≤0,f x−1,x>0.若方程f x=x+a有且只有两个不相等的实数根,则实数a的取值范围是______A. −∞,1B. −∞,1C. 0,1D. 0,+∞8. 已知 f x 为 R 上的可导函数,且对 ∀x ∈R ,均有 f x >fʹ x ,则有 A. e 2013f −2013 <f 0 ,f 2013 <e 2013f 0 B. e 2013f −2013 <f 0 ,f 2013 >e 2013f 0 C. e 2013f −2013 >f 0 ,f 2013 <e 2013f 0D. e 2013f −2013 >f 0 ,f 2013 >e 2013f 0二、填空题(共8小题;共40分)9. 如图,割线 PBC 经过圆心 O ,PB =OB =1,PB 绕点 O 逆时针旋 120∘ 到 OD ,连 PD 交圆 O 于点 E ,则 PE = ______.10. 若直线 l 的极坐标方程为 ρcos θ−π4 =3 2,曲线 C :ρ=1 上的点到直线 l 的距离为 d ,则 d的最大值为______.11. 若不等式 a −1 ≥x +2y ,对满足 x 2+y 2=5 的一切实数 x ,y 恒成立,则实数 a 的取值范围是______.12. 设 n =∫0π6sin x d x ,则二项式 x −2x n的展开式中,x 2 项的系数为______.13. 设实数 x ,y 满足条件:①x ≥0,y ≥0;②3x −y −6≤0;③x −y +2≥0,目标函数z =ax +by a >0,b >0 的最大值为 12,则 2a+3b 的最小值是______.14. 设数列 a n , b n 都是正项等比数列,S n ,T n 分别为数列 lg a n 与 lg b n 的前 n 项和,且S n T n=n2n +1,则 log b 5a 5= ______.15. 已知点 F 为抛物线 y 2=−8x 的焦点,O 为原点,点 P 是抛物线准线上一动点,点 A 在抛物线上,且 AF =4,则 PA + PO 的最小值为 ______.16. 设集合 M = 1,2,3,4,5,6 , 对于 a i ,b i ∈M ,记 e i =a ib i且 a i <b i ,由所有 e i 组成的集合设为:A = e 1,e 2,⋯,e k ,则 k 的值为______;设集合B = eʹi eʹi =1e i,e i ∈A ,对任意 e i ∈A ,eʹj ∈B ,则 e i +eʹj ∈M 的概率为______.三、解答题(共6小题;共78分)17. 在 △ABC 中,角 A ,B ,C 的对边分别是 a ,b ,c ,已知向量 m = cos A ,cos B ,n = a ,2c −b ,且 m ∥n .(1)求角 A 的大小; (2)若 a =4,求 △ABC 面积的最大值.18. 2012 年“双节”期间,高速公路车辆较多.某调查公司在一服务区从七座以下小型汽车中按进服务区的先后每间隔 50 辆就抽取一辆的抽样方法抽取 40 名驾驶员进行询问调查,将他们在某段高速公路的车速(km/h)分成六段:60,65,65,70,70,75,75,80,80,85,85,90,得到如图的频率分布直方图.问:(1)某调查公司在采样中,用到的是什么抽样方法?(2)求这40辆小型车辆车速的众数和中位数的估计值.(3)若从车速在60,70的车辆中任抽取2辆,求抽出的2辆车中速车在65,70的车辆数ξ的分布列及其均值(即数学期望).19. 如图,在三棱锥P−ABC中,PA=PB=AB=2,BC=3,∠ABC=90∘,平面PAB⊥平面ABC,D,E分别为AB,AC中点.(1)求证:DE∥平面PBC;(2)求证:AB⊥PE;(3)求二面角A−PB−E的大小.20. 如图,某隧道设计为双向四车道,车道总宽20 m,要求通行车辆限高5 m,隧道全长2.5 km,隧道的两侧是与地面垂直的墙,高度为3 m,隧道上部拱线近似地看成半个椭圆.(1)若最大拱高 为6 m,则隧道设计的拱宽l是多少?(2)若要使隧道上方半椭圆部分的土方工程量最小,则应如何设计拱高 和拱宽l?(已知:椭圆x2a2+y2b2=1的面积公式为S=πab,柱体体积为底面积乘以高.)(3)为了使隧道内部美观,要求在拱线上找两个点 M ,N ,使它们所在位置的高度恰好是限高5 m ,现以 M ,N 以及椭圆的左、右顶点为支点,用合金钢板把隧道拱线部分连接封闭,形成一个梯形,若 l =30 m ,梯形两腰所在侧面单位面积的钢板造价是梯形顶部单位面积钢板造价的 2 倍,试确定 M ,N 的位置以及 的值,使总造价最少.21. 已知椭圆 C 的中心在原点,离心率等于 12,它的一个短轴端点恰好是抛物线 x 2=8 3y 的焦点.(1)求椭圆 C 的方程;(2)已知 P 2,3 ,Q 2,−3 是椭圆上的两点,A ,B 是椭圆上位于直线 PQ 两侧的动点,①若直线 AB 的斜率为 12,求四边形 APBQ 面积的最大值;②当 A ,B 运动时,满足 ∠APQ =∠BPQ ,试问直线 AB 的斜率是否为定值,请说明理由.22. 已知函数 g x =x 2− 2a +1 x +a ln x(1)当 a =1 时,求函数 g x 的极值; (2)求函数 g x 在区间 1,e 上的最小值;(3)在(Ⅰ)的条件下,设 f x =g x +4x −x 2−2ln x ,证明: 1k−f kn k =2>3n 2−n−2n n +1n ≥2 .答案第一部分1. A2. B3. A4. B5. D6. A7. A8. C第二部分9. 37710. 32+111. a≥6或a≤−412. 6013. 25614. 91915. 216. 11;6121第三部分17. (1)因为m=cos A,cos B,n=a,2c−b,且m∥n,所以a cos B−2c−b cos A=0,利用正弦定理化简得:sin A cos B−2sin C−sin B cos A=0,所以sin A cos B+cos A sin B−2sin C cos A=0,即sin A+B=sin C=2sin C cos A,因为sin C≠0,所以cos A=1,2.又0<A<π,则A=π3(2)由余弦定理a2=b2+c2−2bc cos A,得:16=b2+c2−bc≥bc,即bc≤16,当且仅当b=c=4时,上式取等号,bc sin A≤43,所以S△ABC=12则△ABC面积的最大值为418. (1)由题意知这个抽样是按进服务区的先后每间隔50辆就抽取一辆的抽样方法抽取40名驾驶员进行询问调查,是一个具有相同间隔的抽样,并且总体的个数比较多,这是一个系统抽样.故调查公司在采样中,用到的是系统抽样.(2)众数的估计值为最高的矩形的中点,即众数的估计值等于77.5.设图中虚线所对应的车速为x,则中位数的估计值为:0.01×5+0.02×5+0.04×5+0.06×x−75=0.5,解得x=77.5,即中位数的估计值为77.5.(3)从图中可知,车速在60,65的车辆数为:m1=0.01×5×40=2(辆),车速在65,70的车辆数为:m2=0.02×5×40=4(辆).所以ξ=0,1,2,Pξ=0=C22C40C62=115,Pξ=1=C21C41C62=815,Pξ=2=C20C42C62=615,ξ的分布列为:ξ012P115815615数学期望Eξ=0×115+1×815+2×615=43.19. (1)∵D,E分别为AB,AC中点,∴DE∥BC.∵DE⊄平面PBC,BC⊂平面PBC,∴DE∥平面PBC.(2)连接PD,∵PA=PB,∴PD⊥AB.∵DE∥BC,BC⊥AB,∴DE⊥AB.又∵PD∩DE=D,∴AB⊥平面PDE.∵PE⊂平面PDE∴AB⊥PE.(3)∵平面PAB⊥平面ABC,平面PAB∩平面ABC=AB,PD⊥AB,∴PD⊥平面ABC.如图,以D为原点建立空间直角坐标系B1,0,0,P 0,0,3,E0,32,0,PB=1,0,−3,PE=0,32,−3.PBE的法向量n1=x,y,z,∴x−3z=0,32y−3z=0,令z=3得n1=3,2,3.∵DE⊥平面PAB,∴平面PAB的法向量为n2=0,1,0.设二面角的A−PB−E大小为∠BAC=90∘,由图知cosθ=cos n1,n2=n1 ⋅n2n1⋅ n2=12,所以θ=60∘,即二面角的A−PB−E大小为60∘.20. (1)如图建立直角坐标系,P10,2,椭圆方程为x2a +y2b=1.将b= −3=3与点P的坐标代入椭圆方程,得a=65,此时l=2a=125,因此隧道的拱宽约为12 5 m.(2)要使隧道上方半椭圆部分的土方工程量最小,由柱体的体积公式可知:只需半椭圆的面积最小即可.由椭圆方程x 2a +y2b=1得102a+22b=1.因为102a2+22b2≥2×10×2ab,即ab≥40,所以半椭圆面积S=πab2≥40π2=20π.当S取最小值时,有102a2=22b2=12,得a=102,b=22.此时l=2a=20 =b+3=2+3,故当拱高为22+3m,拱宽为20 2 m时,隧道上方半椭圆部分的土方工程量最小.(3)根据题意,设M x,2,N−x,2,则10≤x<15,设f x=2x+22 x=2 x+2x2−30x+29910≤x<15.则fʹx= x 2−30x+229+2x2−30x+229.令fʹx=0,⇒x2−30x+221=0⇒x=13(x=17舍去),且10≤x<13时,fʹx<0,13<x<15时,fʹx>0,所以x=13时,f x取最小值,此时M13,2,N−13,2,代入椭圆方程得b=251414,所以 =3+b=3+251414.21. (1)设C方程为x2a2+y2b2=1a>b>0,则b=23.由ca =12,a2=c2+b2,得a=4.所以椭圆C的方程为x 216+y212=1.(2)①设A x1,y1,B x2,y2,直线AB的方程为y=12x+t,代入x216+y212=1,得x2+tx+t2−12=0.由Δ>0,解得−4<t<4.由韦达定理得x1+x2=−t,x1x2=t2−12.所以x1−x2= x1+x22−4x1x2=22=48−3t2.由此可得:四边形APBQ的面积S=12×6×x1−x2=348−3t2所以当t=0,S max=123.②当∠APQ=∠BPQ,则PA,PB的斜率之和为0,设直线PA的斜率为k,则PB的斜率为−k,直线PA的直线方程为y−3=k x−2.由y−3=k x−2, ⋯⋯1 x216+y212=1, ⋯⋯21代入2整理得3+4k2x2+83−2k kx+43−2k2−48=0,所以x1+2=82k−3k3+4k.同理直线PB的直线方程为y−3=−k x−2,可得x2+2=−8k−2k−33+4k =8k2k+33+4k.所以x1+x2=16k2−123+4k2,x1−x2=−48k3+4k2,k AB=y1−y2x1−x2=k x1−2+3+k x2−2−312=k x1+x2−4kx1−x2=12,所以AB的斜率为定值12.22. (1)因为a=1,可得g x=x2−3x+ln x,x>0所以gʹx=2x−3+1x =2x2−3x+1x=2x−1x−1x,令gʹx=0,x1=12,x2=1,gʹx>0,即x>1或x<12,g x为增函数,gʹx<0,即12<x<1,g x为减函数,g x在x=12出取极大值,g x极大值=g12=−54−ln2,g x在x=1出取极小值,g x极小值=g1=−2.(2)gʹx=2x−2a+1+ax =2x2−2a+1x+ax=2x−1x−ax.当a≤1时,x∈1,e,gʹx≥0,g x单调递增,g x min=g1=−2a,当1<a<e时,x∈1,a时,gʹx<0,g x单调递减,x∈a,e时,gʹx>0,g x单调递增,所以g x min=g a=−a2−a+a ln a,当a≥e时,x∈1,e,gʹx<0,g x单调递减,g x min=g e=e2−2a+1e+a,所以g x的最小值为:g a=−2a,a≤1−a2−a+a ln a,1<a<e e2−2a+1e+a,a≥e.(3)依题意可得,f x=g x+4x−x2−ln x=x−ln x x>0所以k−f k=ln k,令 x=ln x−14x2−1,因为x∈2,+∞时, ʹ0=2−x 22x2<0.所以 x≤ 2=ln2−34<0,即ln x<14x2−1,所以1ln x >4x−1x+1=21x−1−1x+1所以1nk=2=1 nk=2=1+1+⋯+1>21−1+1−1+⋯+1−1+1−1=21+12−1n−1n+1=3n2−n−2n n+1n≥2.。
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箴言中学 2014 届高三第五次模拟考试
文科数学试题卷
时量 120 分钟 满分 150 分
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×
12k 18k 2 +
(1)求函数 f (x) 在[t, 2t] (t > 0) 上的单调区间;
(2)若函数 h(x) = f (x) + g(x) 有两个不同的极值点 x1, x2 ( x1 < x2 ) , 且 x2 − x1 < ln 2 ,求实数 a 的取值范围.
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五模数(文)参考答案
题1
2
3
4
5
6
7
(1)求椭圆 C1 的方程;
(2)过点
S
(0,
−
1 3
)
的动直线
l
交椭圆
C1
于
A、B
两点,试问:在直角坐标平面上
是否存在一个定点 T,使得以 AB 为直径的圆恒过定点 T?若存在,求出 T
的坐标;若不存在,请说明理由。
21. (本小题 13 分) 已知函数 f ( x) = x ln x , g(x) = −x2 + ax − 2 .
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e= c = 2 ⇒a= a2
2
所以椭圆 C1
:
x2 2
+
y2
=1
…………………………5 分
(Ⅱ)当直线 l 与 x 轴平行时,以 AB 为直径的圆方程为 x2 + ( y + 1)2 = ( 4)2 33
当直线 l 与 y 轴重合时,以 AB 为直径的圆方程为 x2 + y2 = 1
∆ABC 的面积为 3 ,求 a 的值. 2
17. (本小题 12 分) M 公司从某大学招收毕业生,经过综合测试,录用了 14 名男生和 6 名 女生,这 20 名毕业生的测试成绩如茎叶图所示(单位:分),公司规定:成绩在 180 分以 上者到“甲部门”工作;180 分以下者到“乙部门”工作。
(I)求男生成绩的中位数及女生成绩的平均值; (II)如果用分层抽样的方法从“甲部门”人选和“乙部门”人选中共选取 5 人,
(2)由(1)可知 C1D ⊥ 平面A1ABB1 ,所以 AC1 与平面 A1ABB1 所成的角为 ∠C1 AD ,
在
RT ∆C1AD中,由 sin
∠C1 AD
=
C1D AC1
=
10
,
10
∴ A1A = 2
2
∴ VA1 − AC1D
=
VC1
−
A1
AD
=
1 3
S∆
A1
AD
.C1D
=
2 ………………12 分 3
∴T = 2π = π . (2)由 f ( A) = 4 ,∴ f ( A) = 2sin(2A + π ) + 3 = 4 ,∴sin(2A + π ) = 1 .
2
6
62
又Q A为∆ABC 的内角,∴ π < 2A + π < 13 π ,∴2 A + π = 5 π ,
6
66
66
∴
A
=
π 3
A. i > 9
B. i ≥ 7
C. i ≥ 9
D. i > 5
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8.在区间[0,π ]上随机取一个数 x ,则事件“ sin x + cos x ≥ 6 ”发生的概率为_________. 2
A. 1 B. 2 C. 1 D. 1 432 3
9. 定义方程 f (x) = f ′(x) 的实数根 x0 叫做函数的“新驻点”,若函数 g(x) = sin x (0 < x < π ) ,
e
e
(1)当 0 < t ≤ 1 时,函数 f(x)在[t, 2t] 上递减;(2)当 1 < t < 1 时,函数 f(x)在
2e
2e e
t,
1 e
上递减,在
1 e
,
2t
上递增;(3)当
t
≥
1 e
时,函数
f(x)在 [t ,
2t
]
上上单调递增。
(II)y=f(x)+g(x)=xlnx﹣x2+ax﹣2,则 y′=lnx﹣2x+1+a 题意即为 y′=lnx﹣2x+1+a=0 有两个不同的实根 x1,x2(x1<x2), 即 a=﹣lnx+2x﹣1 有两个不同的实根 x1,x2(x1<x2), 等价于直线 y=a 与函数 G(x)=﹣lnx+2x﹣1 的图象有两个不同的交点
8
9
号
答C
A
D
B
B
A
C
D
B
案
10.{4} 11. [-1,6] 12. 14π 13. 40 14. 2 15. ⑴4015;⑵ an = n × 2n−1
16.解:(1) f (x) = 3 sin 2x + 2 cos2 x + 2 = 3 sin 2x + cos 2x + 3 = 2sin(2x + π ) + 3 6