【数学】山东省烟台二中2014届高三10月月考(理)2

2014年山东省烟台市高考数学二模试卷（文科）学校:___________姓名：___________班级：___________考号：___________一、选择题(本大题共10小题，共50.0分)1.设复数z=1+（其中i为虚数单位），则z+3的虚部为（）A.4iB.4C.-4iD.-4【答案】B【解析】解：∵z=1+，∴．则=．即的虚部为：4．故选：B．由复数z求出z的共轭复数，然后代入z+3化简求值即可得到答案．本题考查了复数代数形式的除法运算，考查了共轭复数的求法，是基础的计算题．2.（0≤a≤2）的最大值为（）A.0B.C.D.【答案】C【解析】解；=显然当a=时取最大值，最大值为，故选：C．直接利用配方法求出函数的最值．本题属于求表达式的最值问题，利用配方法求最值是众多方法之一，本题是一道基础题．3.下列有关命题的说法正确的是（）A.命题“若x2=1，则x=1”的否命题为：“若x2=1，则x≠1”B.“x=-1”是“x2-5x-6=0”的必要不充分条件C.命题“∃x∈R，使得x2+x+1＜0”的否定是：“∀x∈R，均有x2+x+1＜0”D.命题“若x=y，则sinx=siny”的逆否命题为真命题【答案】D【解析】解：对于A：命题“若x2=1，则x=1”的否命题为：“若x2=1，则x≠1”．因为否命题应为“若x2≠1，则x≠1”，故错误．对于B：“x=-1”是“x2-5x-6=0”的必要不充分条件．因为x=-1⇒x2-5x-6=0，应为充分条件，故错误．对于C：命题“∃x∈R，使得x2+x+1＜0”的否定是：“∀x∈R，均有x2+x+1＜0”．因为命题的否定应为∀x∈R，均有x2+x+1≥0．故错误．由排除法得到D正确．故答案选择D．对于A：因为否命题是条件和结果都做否定，即“若x2≠1，则x≠1”，故错误．对于B：因为x=-1⇒x2-5x-6=0，应为充分条件，故错误．对于C：因为命题的否定形式只否定结果，应为∀x∈R，均有x2+x+1≥0．故错误．由排除法即可得到答案．此题主要考查命题的否定形式，以及必要条件、充分条件与充要条件的判断，对于命题的否命题和否定形式要注意区分，是易错点．4.已知α∈（，π），sin（α+）=，则sinα=（）A. B. C.或 D.【答案】B【解析】解：∵α∈（，π），sin（α+）=，∴α+∈（，π），∴cos（α+）=-，∴sinα=sin[（α+）-]=sin（α+）cos-cos（α+）sin=+=，故选：B．根据角的范围利用同角三角函数的基本关系求出cos（α+）的值，再根据sinα=sin[（α+）-]，利用两角差的正弦公式计算求得结果．本题主要考查两角和差的正弦公式，同角三角函数的基本关系，属于中档题．5.已知向量=（x-1，2），=（4，y），若⊥，则9x+3y的最小值为（）A.2B.C.6D.9【答案】C【解析】解：∵⊥，∴（x-1，2）•（4，y）=0，化为4（x-1）+2y=0，即2x+y=2．∴9x+3y≥===6，当且仅当2x=y=1时取等号．故选C．由于⊥⇔=0，即可得出x，y的关系，再利用基本不等式即可得出9x+3y的最小值．本题考查了⊥⇔=0、基本不等式的性质，属于基础题．6.若双曲线C：4x2-y2=λ（λ＞0）与抛物线y2=4x的准线交于A，B两点，且|AB|=2，则λ的值是（）A.1B.2C.4D.13【答案】A【解析】解：抛物线y2=4x的准线方程为x=1，代入双曲线C：4x2-y2=λ，可得y=±，∵|AB|=2，∴2=2，∴λ=1．故选：A．求出抛物线y2=4x的准线方程为x=1，代入双曲线，求出A，B两点的纵坐标，利用|AB|=2，即可求出λ的值．本题考查抛物线、双曲线的性质，考查学生的计算能力，比较基础．根据上表可得回归方程，据此模型预报当x为5时，y的值为（）A.6.9 B.7.1 C.7.04 D.7.2【答案】B【解析】解：由题意，==2.5，==4.5∵回归方程=1.04x+，∴4.5=1.04×2.5+，∴=1.9∴=1.04x+1.9，∴当x=5时，=1.04×5+1.9=7.1故选：B．确定样本中心点，利用回归方程=1.04x+，求出，即可求得回归方程，从而可预报x为5时，y的值．本题考查回归方程，考查学生的计算能力，属于基础题．8.已知函数g（x）是R上的奇函数，且当x＜0时g（x）=-ln（1-x），函数f（2-x2）＞f（x），则实数x的取值范围是（）＞，若A.（-2，1）B. ，，，C.（-1，2）D.，，，【答案】A【解析】解：∵奇函数g（x）满足当x＜0时，g（x）=-ln（1-x），∴当x＞0时，g（-x）=-ln（1+x）=-g（x），得当x＞0时，g（x）=-g（-x）=ln（1+x）∴f（x）的表达式为＞，∵y=x3是（- ，0）上的增函数，y=ln（1+x）是（0，+ ）上的增函数，∴f（x）在其定义域上是增函数，由此可得：f（2-x2）＞f（x）等价于2-x2＞x，解之得-2＜x＜1故选A根据奇函数g（x）当x＜0时g（x）=-ln（1-x），可得当x＞0时，g（x）=ln（1+x）．结合f（x）表达式可得f（x）在其定义域上是增函数，得f（2-x2）＞f（x）等价于2-x2＞x，解之即得本题答案．本题给出分段函数，要我们解关于x的不等式，着重考查了基本初等函数的单调性和函数的奇偶性等知识，属于中档题．9.已知空间几何体的三视图如图所示，则该几何体的体积是（）A. B. C.4 D.8【答案】B【解析】解：由三视图可知：该几何体是由底面边长为2的正方形，高为2的四棱锥．因此该几何体的体积==．故选B．由三视图可知：该几何体是由底面边长为2的正方形，高为2的四棱锥．据此可求出该几何体的体积．本题考查了由三视图求原几何体的体积，正确恢复原几何体是解决问题的关键．10.设a，b，c为实数，f（x）=（x+a）（x2+bx+c），g（x）=（ax+1）（cx2+bx+1）．记集合S=|x|f（x）=0，x∈R|，T=|x|g（x）=0，x∈R|，若card S，card T分别为集合元素S，T的元素个数，则下列结论不可能的是（）A.card S=1，card T=0B.card S=1，card T=1C.card S=2，card T=2D.card S=2，card T=3【答案】D【解析】解：∵f（x）=（x+a）（x2+bx+c），当f（x）=0时至少有一个根x=-a当b2-4c=0时，f（x）=0还有一根只要b≠2a，f（x）=0就有2个根；当b=2a，f（x）=0是一个根当b2-4c＜0时，f（x）=0只有一个根；当b2-4c＞0时，f（x）=0只有二个根或三个根当a=b=c=0时card S=1，card T=0当a＞0，b=0，c＞0时，card S=1且card T=1当a=c=1，b=-2时，有card S=2且card T=2故选D．根据函数f（x）的解析可知f（x）=0时至少有一个根x=-a，然后讨论△=b2-4c可得根的个数，从而得到g（x）=0的根的个数，即可得到正确选项．本题主要考查了方程根的个数，同时考查了元素与集合的关系，分类讨论是解题的关键，属于基础题．二、填空题(本大题共5小题，共25.0分)11.执行如图所示的程序框图，若输入A的值为2，则输出P的值为______ ．【答案】4【解析】解：当P=1时，S=1+；当P=2时，S=1++；当P=3时，S=1+++；当P=4时，S=1++++=；不满足S≤2，退出循环．则输出P的值为4故答案为：4．由已知中的程序框图及已知中输入2，可得：进入循环的条件为S≤2，即P=1，2，3，4，模拟程序的运行结果，即可得到输出的P值．本题考查的知识点是程序框图，在写程序的运行结果时，我们常使用模拟循环的变法，但程序的循环体中变量比较多时，要用表格法对数据进行管理．12.如图，目标函数z=ax-y的可行域为四边形OACB（含边界）．若点C（3，2）是该目标函数取最小值时的最优解，则a的取值范围是______ ．【答案】【解析】解：由可行域可知，直线AC的斜率K AC==-2直线BC的斜率K BC==-，当直线z=ax-y的斜率介于AC与BC之间时，C是该目标函数z=ax-y的最优解，所以a∈[-2，-]故答案为：-2根据约束条件对应的可行域，利用几何意义求最值，z=ax-y表示直线在y轴上的截距的相反数，结合图象可求a的范围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的方法反求参数的范围，属于基础题．13.在圆x2+y2-2x-6y=0内，过点E（0，1）的最长弦和最短弦分别是AC和BD，则四边形ABCD的面积为______ ．【答案】10【解析】解：圆x2+y2-2x-6y=0即（x-1）2+（y-3）2=10表示以M（1，3）为圆心，以为半径的圆．由圆的弦的性质可得，最长的弦即圆的直径，AC的长为2．∵点E（0，1），∴ME==．弦长BD最短时，弦BD和ME垂直，且经过点E，此时，BD=2=2=2．故四边形ABCD的面积为=10，故答案为10．根据圆的标准方程求出圆心M的坐标和半径，最长的弦即圆的直径，故AC的长为2，最短的弦BD和ME垂直，且经过点E，由弦长公式求出BD的值，再由ABCD的面积为求出结果．本题主要考查直线和圆的位置关系，两点间的距离公式，弦长公式的应用，体现了数形结合的数学思想，属于中档题．14.一艘海轮从A处出发，以每小时20海里的速度沿南偏东40°方向直线航行．30分钟后到达B处．在C处有一座灯塔，海轮在A处观察灯塔，其方向是南偏东70°，在B处观察灯塔，其方向是北偏东65°，那么B、C两点间的距离是______ ．【答案】海里【解析】解：如图，由已知可得，∠BAC=30°，∠ABC=105°，AB=10，从而∠ACB=45°．°=海里．在△ABC中，由正弦定理可得BC=°故答案为：海里．先根据题意画出图象确定∠BAC、∠ABC的值，进而可得到∠ACB的值，最后根据正弦定理可得到BC的值．本题主要考查正弦定理的应用，考查对基础知识的掌握程度，属于中档题．15.已知函数f（x）的定义域[-1，5]，部分对应值如表，f（x）的导函数y=f′（x）的图象如图所示，下列关于函数f（x）的命题：①函数f（x）的值域为[1，2]；②函数f（x）在[0，2]上是减函数；③当1＜a＜2时，函数y=f（x）-a最多有4个零点；④如果当x∈[-1，t]时，f（x）的最大值是2，那么t的最大值为4．其中正确命题的序号是______ （写出所有正确命题的序号）【答案】①②③【解析】解：①由图象得：f（0），f（4）是极大值，而f（2）是极小值，f（-1），f（5）是端点值，∴最大值在f（0），f（4），f（-1）中取，最小值在f（2），f（5）中取；结合表格得：①正确．②由图象得：在[0，2]上，f′（x）＜0，∴f（x）是减函数，故②正确．③画出函数y=f（x）-a的草图，可以发现，当a=1.5时，有三个零点，当a=2时有两个零点，当1.5＜a＜2时，有4个零点，故③正确．④由图象得函数f（x）的定义域[-1，5]，f（x）的最大值是2，t的最大值是5．故答案为：①②③．通过函数的图象，再结合表格可直接读出．本题考察了函数的单调性，极值，导数的应用，以及读图的能力．三、解答题(本大题共6小题，共75.0分)16.某数学兴趣小组有男女生各5名．以下茎叶图记录了该小组同学在一次数学测试中的成绩（单位：分）．已知男生数据的中位数为125，女生数据的平均数为126.8．（1）求x，y的值；（2）现从成绩高于125分的同学中随机抽取两名同学，求抽取的两名同学恰好为一男一女的概率．【答案】解：（1）男生成绩为119，122，120+x，134，137，其中位数为125，故x=5．…（3分）女生成绩为119，125，120+y，128，134，平均数为126.8=，解之得y=8…（6分）（2）设成绩高于125的男生分别为a1、a2，记a1=134，a2=137，设成绩高于125的女生分别为b1、b2、b3，记b1=128，b2=128，b3=134，从高于12（5分）同学中取两人的所有取法：（a1，a2），（a1，b1），（a1，b2），（a1，b3），（a2，b1），（a2，b2），（a2，b3），（b1，b2），（b1，b3），（b2，b3）共10种，…（8分）其中恰好为一男一女的取法：（a1，b1），（a1，b2），（a1，b3），（a2，b1），（a2，b2），（a2，b3）共6种，…（10分）∵故抽取的两名同学恰好为一男一女的概率为．…（12分）【解析】（1）由已知中男生数据的中位数为125，可知120+x=125，由女生数据的平均数为126.8，可知126.8=，解方程可得x，y的值；（2）分别计算从成绩高于125分的同学中随机抽取两名同学的取法种数，和抽取的两名同学恰好为一男一女的取法种数，代入古典概型概率公式，可得答案．此题考查了古典概型概率计算公式，茎叶图，掌握古典概型概率公式：概率=所求情况数与总情况数之比是解题的关键．17.设函数f（x）=sin（2ωx+）+2sin2ωx（ω＞0），其图象的两个相邻对称中心的距离为．（1）求函数f（x）的解析式；（2）若△ABC的内角为A，B，C，所对的边分别为a，b，c（其中b＜c），且f（A）=2，a=，△ABC面积为，求b，c的值．【答案】解：（1）==…（3分）由题意知T=π，∴，ω=1，∴函数的解析式为：…（6分）（2）由f（A）=2，得，0＜A＜π，∴，∴即bc=6，…（8分）又a2=b2+c2-2bccos A，将，代入得b2+c2=13，…（10分）又b＜c解得…（12分）【解析】（1）通过两角和与差的三角函数化简函数的表达式为一个角的一个三角函数的形式，图象的两个相邻对称中心的距离为．求出函数的周期，然后求函数f（x）的解析式；（2）利用解析式通过f（A）=2，求出A，通过a=，△ABC面积为，以及余弦定理即可求b，c的值．本题考查两角和与差的三角函数，函数的解析式的求法，余弦定理的应用，三角形的面积的求法，考查计算能力．18.如图，四边形PCBM是直角梯形，∠PCB=90°，PM∥BC，PM=1，BC=2．又AC=1，∠ACB=120°，AB⊥PC，直线AM与直线PC所成的角为60°．（Ⅰ）求证：PC⊥AC；（Ⅱ）求三棱锥V B-MAC的体积．【答案】（I）证明：∵PC⊥BC，PC⊥AB，BC∩AB=B，∴PC⊥平面ABC，∵AC⊂平面ABC，∴PC⊥AC．（II）解：∵PC⊥平面ABC，PC⊂平面PCBM，∴平面PCBM⊥平面ABC，如图，在平面ABC中过A作AD垂直于BC的延长线与D，则AD⊥平面PCBM，则AD为三棱锥A-MBC的高，∵∠ACB=120°，∴∠ACD=60°，在直角三角形ADC中，AD=AC sin60°=1×．又S△BMC=S四边形PCBM-S△MPC=（PM+BC）•PC-PM•PC=（1+2）×1-×1×1=1∴V B-MAC=V A-MBC==∴三棱锥B-MAC的体积为．【解析】（Ⅰ）利用线面垂直的判定定理，证明PC⊥平面ABC，然后证明PC⊥AC；（Ⅱ）由PC⊥平面ABC，根据面面垂直的判定可得面ABC⊥面PVBM，再由两面垂直的性质定理可得三棱锥A-MBC的高，解直角三角形求出三棱锥A-MBC的高，则体积可求．本题主要考查了直线与平面、平面与平面垂直的判定和性质，考查三棱锥B-MAC的体积的计算，考查考查空间想象能力、运算能力和推理论证能力，属于中档题．19.在数列{a n}中，已知a1=，，b n+2=3a n（n∈N*）．（1）求数列{a n}、{b n}的通项公式；（2）设数列{c n}满足c n=a n•b n，求{c n}的前n项和S n．【答案】解：（1）∵a1=，，∴数列{a n}是公比为的等比数列，∴，又，故b n=3n-2（n∈N*）．（2）由（1）知，，，∴，，∴，于是．两式相减，得=．∴【解析】（1）由条件建立方程组即可求出数列{a n}、{b n}的通项公式；（2）根据错位相减法即可求{c n}的前n项和S n．本题主要考查等差数列和等比数列的通项公式的计算，以及利用错位相减法进行求和的内容，考查学生的计算能力．20.已知向量=（x，y），=（1，0），且（+）•（-）=0．（1）求点Q（x，y）的轨迹C的方程；（2）设曲线C与直线y=kx+m相交于不同的两点M、N，又点A（0，-1），当|AM|=|AN|时，求实数m的取值范围．【答案】解：（1）由题意向量=（x，y），=（1，0），且（+）•（-）=0，∴，化简得，∴Q点的轨迹C的方程为．…（4分）（2）由得（3k2+1）x2+6mkx+3（m2-1）=0，由于直线与椭圆有两个不同的交点，∴△＞0，即m2＜3k2+1．①…（6分）（i）当k≠0时，设弦MN的中点为P（x P，y P），x M、x N分别为点M、N的横坐标，则，从而，，…（8分）又|AM|=|AN|，∴AP⊥MN．则，即2m=3k2+1，②将②代入①得2m＞m2，解得0＜m＜2，由②得＞，解得＞，故所求的m的取值范围是（，2）．…（10分）（ii）当k=0时，|AM|=|AN|，∴AP⊥MN，m2＜3k2+1，解得-1＜m＜1．…（12分）综上，当k≠0时，m的取值范围是（，2），当k=0时，m的取值范围是（-1，1）．…（13分）【解析】（1）利用向量的数量积公式，结合（+）•（-）=0，即可求点Q（x，y）的轨迹C的方程；（2）直线方程代入椭圆方程，分类讨论，设弦MN的中点为P，利用|AM|=|AN|，AP⊥MN，即可求出实数m的取值范围．本题考查轨迹方程，考查直线与椭圆的位置关系，考查小时分析解决问题的能力，属于中档题．21.已知函数f（x）=x2+ax-lnx，a∈R．（Ⅰ）若a=0时，求曲线y=f（x）在点（1，f（1））处的切线方程；（Ⅱ）若函数f（x）在[1，2]上是减函数，求实数a的取值范围；（Ⅲ）令g（x）=f（x）-x2，是否存在实数a，当x∈（0，e]（e是自然常数）时，函数g（x）的最小值是3，若存在，求出a的值；若不存在，说明理由．【答案】解：（I）a=0时，曲线y=f（x）=x2-lnx，∴f′（x）=2x-，∴g′（1）=1，又f（1）=1曲线y=f（x）在点（1，f（1））处的切线方程x-y=0．（II）′在[1，2]上恒成立，令h（x）=2x2+ax-1，有得，得（II）假设存在实数a，使g（x）=ax-lnx（x∈（0，e]）有最小值3，′=①当a≤0时，g（x）在（0，e]上单调递减，g（x）min=g（e）=ae-1=3，（舍去），②当＜＜时，g（x）在，上单调递减，在，上单调递增∴，a=e2，满足条件．③当时，g（x）在（0，e]上单调递减，g（x）min=g（e）=ae-1=3，（舍去），综上，存在实数a=e2，使得当x∈（0，e]时g（x）有最小值3．【解析】（I）欲求在点（1，f（1））处的切线方程，只须求出其斜率的值即可，故先利用导数求出在x=1处的导函数值，再结合导数的几何意义即可求出切线的斜率．从而问题解决．（II）先对函数f（x）进行求导，根据函数f（x）在[1，2]上是减函数可得到其导函数在[1，2]上小于等于0应该恒成立，再结合二次函数的性质可求得a的范围．（III）先假设存在，然后对函数g（x）进行求导，再对a的值分情况讨论函数g（x）在（0，e]上的单调性和最小值取得，可知当a=e2能够保证当x∈（0，e]时g（x）有最小值3．本题主要考查导数的运算和函数的单调性与其导函数的正负之间的关系，当导函数大于0时原函数单调递增，当导函数小于0时原函数单调递减．。

高三英语阶段检测题第一节：单项填空1. With the successful launch of Tiangong-l, which was _____ breakthrough in space field,China's space dream took a step closer to _____ reality.A a;/ B. the a C. a; the D the;/2. Believe it or not, no bread eaten by us is as sweet as _______ earned by our own labor.A one B. that C. such D. what3.1 am surprised that a company with the good _______ would produce such poor quality goods.A enjoyment B. appreciation C. entertainment D. reputation4.If you can't________ your shyness, you’ll never pass the interview.A take offB .get overC .break off D. give away5. -Is your father a teacher of English in the No.l Middle School?-No! But he _________ English there for ten years.A. has taughtB. has been teachingC. taughtD. had taught6. He devoted all the money he had _________a school.A. set upB. to set upC. to setting upD. setting up7. I feel sure that ______ qualification, ability and experience, our headmaster is abundantly equal to theposition we have in mind.A. in case ofB. in terms ofC. in the opinion ofD.in the course of8. With the nuclear crisis worsening in Iran, the world's attention is fixed again on ______is called the MiddleEastA. whichB. thatC. whatD. it9.一Are you satisfied with the result of the exam?一Not at all. I can't have _______.A.a worse oneB. a better oneC. the worst oneD. the best one10. The real-name policy for train tickets has been applied to all trains since January l, 2012 inChina,________ will make it easy for people to get tickets.A. whichB. whenC. whatD. where11. The government _____ air quality in urban areas from level one to five: excellent, fairly good,slightly polluted, poor and dangerous.A arranges B. classifies C. distributes D. divides12. -What has made him in such a low mood recently?—_________ by his father for not passing the examA Being criticized B. Criticized C.Having criticized D. To criticize13. Only when _______ be possible to sign the papers.A does the lawyer come with it B. the lawyer comes will itC. has the lawyer come it willD. the lawyer comes it will14. Find ways to praise your children often, _________ you'II find they will open their hearts toyou.A till B. or C. and D. but15. —No photos, please!—______________.A. Don't mention itB.I suppose notC. Never mindD. I'm terribly sorry16. Three foreign films will be on this month. One is made in Korea, _______ two are made in England.A. restB. anotherC. otherD. the other17. Teachers _______ be sensitive to the development level of each student so as to help them better.A. need toB. mayC. are able toD. can18. Most students think they should have _______ at school if there were no examinations.A. the happiest timeB. a more happier timeC. much happiest timeD. a much happier time19. The manager listened to the customers’ complaints attentively with great patience, _______ to miss any point.A. not tryingB. trying notC. to try notD. not to try20. Rain and high winds today _______ to take the place of yesterday’s mild conditions.A. expectB. are expectedC. are expectingD. has expected21. The adoption of orphans and physically challenged children has been a subject of public debate _______ a firein an unregistered orphanage last month.A. according toB. in case ofC. ever sinceD. such as22. In fact, I think it’s very much nicer without the naughty boy, if you don’t mind me _______ so.A. sayB. to sayC. sayingD. to saying23. The number of the dead pigs _______ out of the Huangpu River in Shanghai’s Songjiang District had risen to5,916 by March 12.A. fishedB. to fishC. to be fishedD. fishing24. A 7-year-old boy received an operation on Tuesday successfully _______ he has many other health problems.A. sinceB. howeverC. thoughD. therefore25. China’s plan _______ its installed nuclear power capacity by 20 percent this year shows that the country isdeveloping new energy in an efficient way.A. raisedB. to raiseC. raisingD. has raised26. It was announced that only when the terrible disease was under control _______ to return to their homes.A. the residents would decideB. would the residents decideC. would the residents be decidedD. the residents would be decided27. Some abstract modern paintings were ______ my understanding; I simply couldn’t figure out what thepainters really wanted to tell us.A. withinB. behindC. aroundD. beyond28. — What did you have for breakfast during your visit to Britain?— ______ but bread and butter. That was the only food they served day in and day out.A. NothingB. EverythingC. AnythingD. Something29. The document film ______ next week aims at introducing local cultures to the world.A. releasedB. being releasedC. to be releasedD. having been released30. The school authority had a discussion regarding ______ they should set up more scholarships to meet theincreasing demands.A. whetherB. whatC. thatD. whom31. Something has to be done to stop the rivers from being further polluted, ?A. doesn’t itB. hasn’t itC. does itD. has it32. A warning signal suddenly appeared on the screen of my computer, ______ the computer was being attackedby a certain virus.A. indicatedB. indicatingC. to indicateD. having indicated33. On top of the hill ______, whose style can be traced to the ages of the Roman Empire.A. there standing a churchB. does a church standC. a church standsD. stands a church34 —Do you have the time? I’ve got something to tell you.— Ok, ______ you make your story short.A. now thatB. even ifC. so long asD. in case35. IP addresses are the unique sequence of numbers ______ to each computer, websites or otherInternet-connected devices.A. to assignB. assigningC. assignedD. being assigned第二部分完形填空(每小题1分；满分20分)We should show respect to everybody, especially our elders because they are ahead of us — in age, in wisdom and maturity, in experience and education. Our 36 have done a lot for us, directly or indirectly and most of us 37 everything to their kindness and love.When we 38 them respect, whether it is by bowing to them, or 39 them with a smile, or offering them any help they need, it is one way of 40 our own love and gratitude to them. 41 , elders have also been through all the years you are 42 and know a little more about the world than you do.It is 43 that you do not agree with the belief of your elders, but this is nothing new. All younger generations have always 44 with their elders and it is these differences that bring changes in human 45 . However much you disagree with them, give them credit for their 46 .With changing times and 47 influences, youngsters no longer know what is interpreted as disrespect to elders. Youngsters should 48 express their views and if there are arguments, they should not 49 their voices.If there is no space on sofas or chairs, children will immediately 50 their places, and sit on the carpet. In buses and trains, youngsters are 51 to give up their places to older people. This is not a 52 of who has more rights. It is simply that those who are younger have the strength to bear 53 , or tolerate unpleasantness, so it is natural to show consideration to those who are older and perhaps at a 54 disadvantage.When you do simple things as a mark of respect, elders become 55 that youngsters care for them, and they respond with affection and kindness.36. A. youngsters B. elders C. parents D. juniors37. A. devote B. owe C. pay D. contribute38. A. show B. explain C. exhibit D. point39. A. greeting B. receiving C. declaring D. showing40. A. expressing B. describing C. sending D. suggesting41. A. However B. Therefore C. Besides D. Though42 A. experiencing with B. going through C. suffering from D. worrying out43. A. true B. likely C. reasonable D. strange44. A . quarreled B. dealt C. lived D. disagreed45. A. community B. organization C. society D. public46. A. experience B. reality C. emotion D. information47. A. cultural B. economical C. political D. psychological48. A. quietly B. slightly C. silently D. coldly49. A. rise B. raise C. support D. force50. A. give away B. get rid of C. give up D. send out51. A. expected B. forced C. needed D. reminded52. A. doubt B. question C. wonder D. challenge53. A. suffering B. upset C. trouble D. discomfort54. A. serious B. light C. heavy D. slight55. A. aware B. alive C. knowing D. sensible第三部分阅读理解ABenin is one of the smallest African states. It lies in West Africa on the Gulf (海湾) of Guinea, to the south of Burkina Faso and Niger, between Togo on the west and Nigeria on the east.Benin used to be called Dahomey and was controlled and ruled by France from 1893 to 1960, when it became independent. In 1963 the army general Soglo overthrew (推翻) the first president Maga. Soglo set up an army government and called himself head of state in 1965, but was overthrown and replaced by a civilian (非军人) government in 1967. In December 1969 Benin had another change of power with the army again taking over. In May 1970, Maga and two other men set up a new government, with each of them acting as president in turn for two years. However, half a year after Maga turned over power to the second man Ahomadegbe, the three-man government was overthrown by the army once more and General Kerekou became president. In November 1975 Kerekou changed the name of the nation from Dahomey to Benin, Benin being the name of a 17th century kingdom covering the same place. Kerekou also announced that Benin would be a People’s Republic based on Marxism-Leninism.56. Which of the following maps shows rightly the positions of Benin and its neighbouring countries?Bn=Benin;Tg=Togo;Nr=Niger;BF=Burkina Faso;Na=Nigeria;GG=Gulf of Guinea57. For how long was Benin an independent state before itbecame a People’s Republic?A. 30 years.B. 25 years.C. 20 years.D. 15 years.58. Choose the right order in which the following people ruled in Benin. (Ah=Ahomadegbe; Ke=Kerekou;Ma=Maga; So=Soglo)A. Ma, So, Ma, Ke, AhB. So, Ma, Ah, Ma, KeC. Ma, So, Ma, Ah, KeD. So, Ma, Ke, Ma, Ah59. When and how did Benin get its two names — Benin and Dahomey?A. Benin was its oldest name. The name Dahomey was used later, but has been replaced by Benin again.B. Dahomey was its oldest name, but it has been replaced by Benin.C. Dahomey was its oldest name. The name Benin was used later, but has been replaced by Dahomey again.D. Benin was its oldest name, but it has been replaced by Dahomey.(B)Winter begins in the north on December 22nd. People and animals have been doing what they always do to prepare for the colder months. Squirrels(松鼠), for example, have been busy gathering nuts from trees. Well, scientists have been busy gathering information about what the squirrels do with the food they collect.They examined differences between red squirrels and gray squirrels in the American state of Indiana. Thescientists wanted to know how these differences could affect the growth of black walnut (黑胡桃) trees. The black walnut is the nut of choice for both kinds of squirrels. The black walnut tree is also a central part of some hardwood forests.Rob Swihart of Purdue University did the study with Jake Goheen, a former Purdue student now at the University of New Mexico. The two researchers estimate that several times as many walnuts grow when gathered by gray squirrels as compared to red squirrels. Gray squirrels and red squirrels do not store nuts and seeds in the same way. Gray squirrels bury nuts one at a time in a number of places. But they seldom remember where they buried every nut. So some nuts remain in the ground. Conditions are right for them to develop and grow the following spring. Red squirrels, however, store large groups of nuts above ground. Professor Swihart calls “death traps for seeds”.Gray squirrels are native to Indiana. But Professor Swihart says their numbers began to decrease as more forests were cut for agriculture. Red squirrels began to spread through the state during the past century.The researchers say red squirrels are native to forests that stay green all year, unlike walnut trees. They say the cleaning of forest land for agriculture has helped red squirrels invade Indiana. Jake Goheen calls them a sign of an environmental problem more than a cause.60. The study done by Rob Swihart and Jake Goheen is to ________.A. find out the living conditions for squirrelsB. learn squirrels’ influence on black walnut treesC. do something to get rid of squirrelsD. save the forests in the American state of Indiana61 The difference between gray squirrels and red squirrels mainly lies in ________.A. the way they gather the walnutB. the time they have winter sleepC. the place they have winter sleepD. the place they store the walnuts62. When Professor Swihart says “death traps for seeds”, he actually means that ________.A. red squirrels eat more nuts than gray squirrelsB. gray squirrels and red squirrels will have severe fightsC. nuts above the ground will not develop into plantsD. seeds can be traps for other animals in the forest63. According to the passage, which of the following is true?A. The black walnut is equally attractive to both gray and red squirrels.B. Gray squirrels do more harm to the forest than red squirrels.C. Red squirrels and gray squirrels have helped the spread of walnut trees.D. The cleaning of forest land benefits red squirrels directly.(C)Television is a relatively stable advertising medium. In many ways, the television ads today are almost the same to those two decades ago. Most television ads still feature actors, still run 30 or 60 seconds, and still show a product. However, the different medium of the Internet causes unique challenges to advertisers, forcing them to adapt their practices and techniques.In the early days of Internet marketing, online advertisers used banner (框式广告) and pop-up ads (弹出式广告) to attract customers. These techniques reached large audiences, led to many sales leads, and came at a low cost. However, a small number of Internet users began to consider these advertising techniques annoying. Yet because marketing strategies relying heavily on banners and pop-ups produced results, companies invested growing amounts of money into purchasing these ad types. As consumers became more complicated, frustration with these online advertising techniques grew. Independent programmers began to develop tools that blocked banner and pop-up ads.A major development in online marketing came with the introduction of pay-per-click ads. Unlike banner or pop-up ads, which originally required companies to pay every time a website visitor saw an ad, pay-per-click ads allowed companies to pay only when an interested potential customer clicked on an ad. More importantly, however, these ads are not affected by the pop-up and banner blockers. As a result of these advantages and theincredible growth in the use of search engines, which provide excellent places for pay-per-click advertising, a great number of companies began turning to pay-per-click marketing. However, as with the banner and pop-up ads, pay-per-click ads came with their shortcomings. When companies began pouring billions of dollars into this emerging medium, online advertising specialists started to notice the presence of what would later be called click fraud (欺诈): representatives of a company with no interest in the product advertised by a competitor click on the competitor’s ads simply to increase the marketing cost of the competitor. Click fraud grew so rapidly that marketers sought to diversify(摆脱) their online positions away from pay-per-click marketing through new mediums.Although pay-per-click advertising remains a common and effective advertising tool, marketers adapted yet again to the changing elements of the Internet by adopting new techniques such as pay-per-performance advertising. As the pace of the Internet’s evolution increases, it seems all the more likely that advertising successfully on the Internet will require a strategy that avoids constancy (持续性) and welcomes change.64. What is the main idea of the passage?A. The pace of the Internet’s evolution is increasing and will only increase in the future.B. Internet advertising fails to reach Internet users, causing ads to be blocked.C. The Internet has experienced dramatic changes in short periods of time.D. Rapid development of the Internet calls for new advertising strategies and mediums.65. As an advertising medium, the television and the Internet mainly differ in ________.A. the type of individual each medium reachesB. whether the medium is interactiveC. the pace at which the medium developsD. the cost of advertising with each medium66. According to the passage, which of the following is a typical click fraud?A. Using software to block competitors’ advertisements.B. Clicking on the pay-per-click ads of competitors.C. Clicking on the banner advertisements of opponent companies.D. Using search engine to attack the pages of competitors.67. What does the author imply about the future of pay-per-performance advertising?A. It will eventually become less popular just like other forms of Internet advertising.B. It will not face shortcomings due to its differing approach to online marketing.C. Internet users will develop free software to block its effectiveness.D. Although it improves on pay-per-click advertising, it still suffers from click fraud.(D)Humans have sewn by hand for thousands of years. It was said that the first thread was made from animal muscle and sinew. And the earliest needles were made from bones. Since those early days, many people have been involved in the process of developing a machine that could do the same thing more quickly and with greater efficiency.Charles Wiesenthal, who was born in Germany, designed and received a patent on a double-pointed needle that eliminated the need to turn the needle around with each stitch(缝合) in England in 1755. Other inventors of that time tried to develop a functional sewing machine, but each design had at least one serious imperfection.Frenchman Barthelemy Thimonnier finally engineered a machine that really worked. However, he was nearly killed by a group of angry tailors when they burned down his garment factory. They feared that they would lose their jobs to the machine.American inventor Elias Howe, born on July 9, 1819, was awarded a patent for a method of sewing that used thread from two different sources. Howe’s machine had a needle with an eye at the point, and it used the two threads to make a special stitch called a lockstitch. However, Howe faced difficulty in finding buyers for his machines in America. In frustration, he traveled to England to try to sell his invention there. When he finally returned home, he found that dozens of manufacturers were adapting his discovery for use in their own sewing machines.Isaac Singer, another American inventor, was also a manufacturer who made improvements to the design of sewing machines. He invented an up-and-down-motion mechanism that replaced the side-to-side machines. He also developed a foot treadle(脚踏板) to power his machine. This improvement left the sewer’s hands free. Undoubtedly, it was a huge improvement of the hand-cranked machine of the past. Soon the Singer sewing machine achieved more fame than the others for it was more practical, it could be adapted to home use and it could be bought on hire-purchase. The Singer sewing machine became the first home appliance, and the Singer company became one of the first American multinationals.However, Singer used the same method to create a lockstitch that Howe had already patented. As a result, Howe accused him of patent infringement（侵犯）. Of course, Elias Howe won the court case, and Singer was ordered to pay Howe royalties（版税）. In the end, Howe became a millionaire, not by manufacturing the sewing machine, but by receiving royalty payments for his invention.68. Barthelemy Thimonni er’s garment factory was burned down because _____________.A. people did not know how to put out the fireB. Elias Howe thought Thimonnier had stolen his inventionC. the sewing machines was couldn’t work finallyD. workers who feared the loss of their jobs to a machine set fire69. Which of the following is NOT TRUE according to the passage?A. Singer is an American inventor and manufacturer.B. The Singer sewing company became more practical.C. The foot treadle helped to make the sewer’s hands fr ee.D. Singer made improvements to the design of sewing machines.70. Why did the court force Isaac Singer to pay Elisa Howe a lifetime of royalties?A. Because the judge was against Singer for his surly attitude.B. Because Howe had already patented the lockstitch used by Singer.C. Because Singer had borrowed money from Howe and never repaid it.D. Because Singer and Howe had both invented the same machine.71. Which of the following would be the best title for this passage?A. A Stitch in Time Saves NineB. The Case between Howe and SingerC. Patent Laws on the Sewing MachineD. The Early History of the Sewing MachineENew technology links the world as never before. Our planet has shrunk. It’s now a “global village” where countries are only seconds away by fax or phone or satellite link. And, of course, our ability to benefit from this high-tech communications equipment is greatly increased by foreign language skills.Deeply involved with this new technology is a kind of modern businesspeople who have a growing respect for the economic value of doing business abroad. In modern markets, success overseas often helps support domestic business efforts.Overseas assignments (指派) are becoming increasingly important to advancement within executive (行政) ra nks. The executive stationed in another country no longer need fear being “out of sight and out of mind.” He or she can be sure that the overseas effort is central to the company’s plan for success, and that promotions often follow or accompany an assignment abroad. If an employee can succeed in a difficult assignment overseas, superiors will have greater confidence in his or her ability to manage back in the United States where cross-cultural considerations and foreign language issues are becoming more and more common.Thanks to a variety of relatively inexpensive communications devices (装置) with business applications, even small businesses in the United States are able to get into international markets.English is still the international language of business. But there is an ever-growing need for people who can speak another language. A second language isn’t generally required to get a job in business, but having language skills gives a candidate the edge when other qualifications appear to be equal.The employee posted abroad who speaks the country’s language has an opportunity to fast-forward certain negotiations, and can have the cultural insight (洞察力) to know when it is better to move more slowly. The employee at the home office who can communicate well with foreign customers over the telephone or by fax machine is an obvious asset (有价值的人或物) to the firm.72. With the increased use of high-tech communications equipment, businesspeople ________.A. are eager to work overseasB. have to get familiar with modern technologyC. are gaining more economic benefits from domestic operationsD. are attaching more importance to their overseas business73. In this passage, “out of sight and out of mind” (Line 3, Para. 3) probably means ________.A. leaving all care and worry behindB. being unable to think properly for lack of insightC. being totally out of touch with business at homeD. missing opportunities for promotion when abroad74. According to the passage, what is an important consideration of international corporations in employingpeople today?A. Ability to speak the customer’s language.B. Connections with businesses overseas.C. Technical know-how.D. Business experience.75. The advantage of employees having foreign language skills is that they can ________.A. fast-forward their proposals to headquartersB. better control the whole negotiation processC. easily make friends with businesspeople abroadD. easily find new approaches to meet market needsCThroughout this long, tense election, everyone has focused on the presidential candidates and how they’ll change America. Rightly so, but se lfishly, I’m more fascinated by Michelle Obama and what she might be able to do, not just for this country, but for me as an African-American woman. As the potential First Lady, she would have the world’s attention. And that means that for the first time p eople will have a chance to get up close and personal with the type of African-American woman they so rarely see.Usually, the lives of black women go largely unexamined. The prevailing theory seems to be that we’re all hot-tempered single mothers who ca n’t keep a man. Even in the world of make-believe, black women still can’t escape the stereotype of being eye-rolling, oversexed females raised by our never-married, alcoholic (酗酒的) mothers.These images have helped define the way all women are viewed, including Michelle Obama. Before she ever gets the chance to commit to a cause, charity or foundation as First Lady, her most urgent and perhaps most complicated duty may be simple to be herself.It won’t be easy. Because few mainstream publications have done in-depth features on regular African-American women, little is known about who we are, what we think and what we face on a regular basis. For better or worse, Michelle will represent us all.Just as she will have her critics, she will also have millions of fans who usually have little interest in the First Lady. Many African-American blogs have written about what they’d like to see Michelle bring to the White House—mainly showing the world that a black woman can support her man and raise a strong black family. Michelle will have to work to please everyone—an impossible task. But for many African-American women like me, just a little of her poise (沉着), confidence and intelligence will go a long way in changing an image that’s been around for far too long.76. Why does Michelle Obama hold a strong fascination for the author?A) She serves as a role model for African women.B) She possesses many admirable qualities becoming a First Lady.C) She will present to the world a new image of African-American women.D) She will pay closer attention to the interests of African-American women.77. What is the common stereotype of African-American women according to the author?A) They are victims of violence. B) They are of an inferior violence.C) They use quite a lot of body language. D) They live on charity and social welfare.78. What do many African-Americans write about in their blogs?A) Whether Michelle can live up to the high expectations of her fans.B) How Michelle should behave as a public figure.C) How proud they are to have a black woman in the White House.D) What Michelle should do as wife and mother in the White House.79. What does the author say about Michelle Obama as a First Lady?A) However many fans she has, she should remain modest,B) She shouldn’t disappoint the African-American community.C) However hard she tries, she can’t expect to please everybody.D) She will give priority to African-American women’s concerns.80. What do many African-American women hope Michelle Obama will do?A) Help change the prevailing view about black women.B) Help her husband in the task of changing America.C) Outshine previous First Lady.D) Fully display her fine qualities.。

本试卷共4页.分第I 卷（选择题）和第II 卷（非选择题）两部分.共150分.考试时间120分钟.第I 卷（选择题，共60分）注意事项：1.答第I 卷前，考生必将自己的姓名、准考证号、考试科目用铅笔涂写在答题卡上.2.每题选出答案后，用2B 铅笔把答题卡对应题目的答案标号涂黑.如需改动.用橡皮擦干净后，再改涂其它答案标号.一、选择题：本题共12小题，每小题5分，共60分.在每小题给出的四个选项中，只有一项是符合题目要求的.1.若集合{}{}223,1,M x x N y y x x R =-<<==+∈，则集合M N ⋂= A.()2,-+∞ B.()2,3- C.[)1,3 D.R2.已知函数(),0,1ln ,0,x e x f x f f e x x ⎧<⎡⎤⎛⎫==⎨ ⎪⎢⎥>⎝⎭⎣⎦⎩则 A.1e - B.e - C.e D.1e3.已知弧度数为2的圆心角所对的弦长也是2，则这个圆心角所对的弧长是A.2B.2sin1C.2sin1D.sin24.下列命题中，真命题是A.存在,0x x R e ∈≤B.1,11a b ab >>>是的充分条件C.任意2,2x x R x ∈>D.0a b +=的充要条件是1a b=- 5.已知角θ的顶点在坐标原点，始边与x 轴正半轴重合，终边在直线20x y -=上，则()()3sin cos 2sin sin 2πθπθπθπθ⎛⎫++- ⎪⎝⎭⎛⎫--- ⎪⎝⎭= A.2- B.2 C.0 D.236.若,,,a b c R a b ∈>且，则下列不等式一定成立的是A.a c b c +≥-B.()20a b c -≥C.ac bc >D.20c a b>- 7.若命题“2000230x R mx m ∃∈++-<，使得x ”为假命题，则实数m 的取值范围是A.[]2,6B.[]6,2--C.()2,6D.()6,2--8.已知函数()()sin ,,1,113f x x x f f f ππ⎛⎫⎛⎫=-- ⎪ ⎪⎝⎭⎝⎭则的大小关系为 A.()1311f f f ππ⎛⎫⎛⎫->-> ⎪ ⎪⎝⎭⎝⎭B.()1311f f f ππ⎛⎫⎛⎫->-> ⎪ ⎪⎝⎭⎝⎭C.()1113f f f ππ⎛⎫⎛⎫>->- ⎪ ⎪⎝⎭⎝⎭D.()1311f f f ππ⎛⎫⎛⎫->>- ⎪ ⎪⎝⎭⎝⎭ 9.已知函数()f x 满足：()()()14412x x f x x x f x ⎛⎫≥=<=+ ⎪⎝⎭，则；当时f ， ()22log 3f +=则 A.38 B.18 C.112 D.12411.如果函数()y f x =图像上任意一点的坐标(),x y 都满足方程()111g x y gx gy +=+，那么正确的选项是A.()()14y f x xy =+∞≥是区间，上的减函数，且B.()()14y f x xy =+∞≤是区间，上的增函数，且C.()()14y f x xy =+∞≤是区间，上的减函数，且D.()()14y f x xy =+∞≥是区间，上的增函数，且12.设定义在R 上的函数()f x 是最小正周期为2π的偶函数，()()f x f x '是的导函数，当[]()()()0,010,022x f x x x x f x ππππ⎛⎫'∈<<∈≠-< ⎪⎝⎭时，；当且时，.则方程 ()[]cos 2,2f x x ππ=-在上的根的个数为A.2B.5C.8D.4第II 卷（非选择题，共90分）注意事项:1.将第II 卷答案用0.5mm 的黑色签字笔答在答题纸的相应位置上。

山东省数学高考模拟试题精编二【说明】本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，满分150分．考试时间120分钟．请将第Ⅰ卷的答案填入答题栏内,第Ⅱ卷可在各题后直接作答.一、选择题(本大题共12小题,每小题5分，共60分．在每小题给出的四个选项中，只有一项是符合题目要求的．）1．设A＝｛1,4，2x}，B＝{1，x2}，若B⊆A，则x＝( )A．0 B．－2C．0或－2 D．0或±22．命题“若x＞1，则x＞0”的否命题是( )A．若x＞1，则x≤0 B．若x≤1，则x＞0C．若x≤1，则x≤0 D．若x＜1,则x＜03．若复数z＝2－i,则z＋错误!＝（)A．2－i B．2＋iC．4＋2i D．6＋3i4．（理)已知双曲线错误!－错误!＝1的一个焦点与抛物线y2＝4x的焦点重合,且双曲线的离心率等于错误!，则该双曲线的方程为（）A．5x2－错误!y2＝1 B。

错误!－错误!＝1C.y25－错误!＝1 D．5x2－错误!y2＝1(文)已知双曲线错误!－错误!＝1（a＞0，b＞0)的离心率为错误!，则双曲线的渐近线方程为（）A．y＝±错误!x B．y＝±错误!xC．y＝±2x D．y＝±错误!x5．设函数f（x）＝sin x＋cos x，把f（x)的图象按向量a＝（m,0)（m ＞0）平移后的图象恰好为函数y＝－f′(x)的图象，则m的最小值为( ）A。

错误! B.错误!C.错误!D。

错误!6．（理)已知错误!n的展开式的各项系数和为32，则展开式中x4的系数为（)A．5 B．40C．20 D．10(文)采用系统抽样方法从960人中抽取32人做问卷调查，为此将他们随机编号为1,2，……，960，分组后在第一组采用简单随机抽样的方法抽到的号码为9.抽到的32人中，编号落入区间［1,450］的人做问卷A，编号落入区间［451，750]的人做问卷B，其余的人做问卷C。

2014山东高考数学模拟题（理科）本试卷分第I 卷和第II 卷两部分，共7页。

满分150分。

考试用时120分钟。

考试结束后，将本试卷和答题卡一并交回。

注意事项：1．答卷前，考生务必用0.5毫米黑色墨水签字笔将自己的姓名、座号、准考证号、县区和科类填写在答题卡和试卷规定的位置上。

2．第I 卷每小题选出答案后，用2B 铅笔把答题卡上对应题目的答案标号涂黑；如需改动，用橡皮擦干净后，再选涂其他答案标号。

3．第II 卷必须用0.5毫米黑色签字笔作答，答案必须写在答题卡各题目指定区域内的位置，不能写在试卷上；如需改动，先划掉原来的答案，然后再写上新的答案；不能使用涂改液、胶带纸、修正带。

不按以上要求作答的答案无效。

4．填空题请直接填写答案，解答题应写出文字说明、证明过程或演算步骤。

第Ⅰ卷（选择题 共50分）一、选择题：本大题共10小题，每小题5分，共50分．在每小题给出的四个选项中，只有一项是符合题目要求的。

1.已知1m m R i∈-，复数在复平面内对应的点在直线0x y -=上，则实数m 的值是A.1-B.0C.1D.22．设集合}1,0,1{-=M ，},{2a a N =则使M ∩N ＝N 成立的a 的值是A ．1B ．0C ．－1D ．1或－13.下列函数中，既是偶函数，又是在区间（0，+∞）上单调递减的函数为（ ）A.Y=ln x1 B.Y=3x C.Y=x2 D.Y=COSX4.已知直线l ⊥平面α,直线m ⊂平面β,给出下列命题:①α∥β⇒l ⊥m ;②α⊥β⇒l ∥m ;③l ∥m ⇒α⊥β;④l ⊥m ⇒α∥β.其中正确的命题是( ).A .①②③B .②③④C .②④D .①③5. 已知4sin()5απ+=-，且α是第二象限的角，那么tan()4πα+等于 A ．17- B ．17C ．7D ．7-6．设变量,x y 满足约束条件22022010x y x y x y --≤⎧⎪-+≥⎨⎪+-≥⎩，则11y s x +=+的取值范围是（ ）A ．3[1,]2 B ．1[,1]2 C ．[1,2] D ．1[,2]27.在y=2x ,y=log 2x,y=x 2,y=cos2x,这四个函数中，当0<x 1<x 2<1时，使)2(21x x f +>2)()(f 21x x f +恒成立的函数的个数是 （ ）A. 0B. 1C. 2D. 38.．二项式(1x－x 3x )n 展开式中含有x 4项，则n 的可能取值是（ ）A ．5B ．6C ．3D ．7 9. 从5位男数学教师和4位女数学教师中选出3位教师派到3个班担任班主任（每班1位班主任），要求这3位班主任中男女教师都有，则不同的选派方案共有（ ） A ．210 B ．420C ．630D ．84010.函数f(x)满足条件f(x)=-f(6-x)和f(x)=f(2-x).若f(a)=f(2014),a [5,9),且f(x)在[5,9)上单调，则a 的值是 （ ）A. 6B.5C. 7D. 8 二．填空题（每小题5分，共25分）11．下图是某高三学生进入高中三年来的数学考试成绩茎叶图，第1次到12次的考试成绩依次记为1212A A A ，，…，．图2-2是统计茎叶图中成绩在一定范围内考试次数的一个算法流程图．那么算法流程图输出的结果是 ． 12.曲线y ＝x ，y ＝2－x ，y ＝－13x 所围成图形的面积是__________。

山东省潍坊一中2014届高三数学10月阶段性检测试题理（含解析）新人教B版第Ⅰ卷（共60分）一、选择题：本大题共12个小题,每小题5分,共60分.在每小题给出的四个选项中，只有一项是符合题目要求的.1.若集合2{|23},{|1,},M x x N y y x x=-<<==+∈R则集合M N=（）A．（-2，+∞）B．（-2，3）C．[)1,3D．R2.已知函数e,0,()ln,0,x xf xx x⎧<=⎨>⎩则1[()]ef f=( )A．-1eB．e-C．e D．1e3.已知弧度数为2的圆心角所对的弦长也是2，则这个圆心角所对的弧长是（）A．2 B．2sin1C．2sin1D．sin24.下列命题中，真命题是（ ） A ．存在,e 0x x ∈≤RB ．1,1a b >>是1ab >的充分条件C ．任意2,2x x x ∈>RD ．0a b +=的充要条件是1ab=-5.已知角θ的顶点在坐标原点，始边与x 轴正半轴重合，终边在直线20x y -=上，则3πsin()cos(π-)2πsin()sin(π-)2θθθθ++=--（ ） A ．-2B ．2C ．0D ．236.若,,a b c ∈R ，且a b >，则下列不等式一定成立的是( ) A ．a c b c +≥- B ．2()0a b c -≥C ．ac bc >D ．20c a b>-7.若命题“0,x ∃∈R 使得200230x mx m ++-<”为假命题，则实数m 的取值范围是（ ）A ．[2，6]B ．[-6，-2]C ．（2，6）D ．（-6，-2）8.已知函数()sin ,f x x x =则π()11f ，(1)f -，π()3f -的大小关系为( )A ．ππ()(1)()311f f f ->->B ．ππ(1)()()311f f f ->->C ．ππ()(1)()113f f f >->-D ．ππ()()(-1)311f f f ->>9.已知函数()f x 满足：4x ≥，则1()()2x f x =；当4x <时，()(1),f x f x =+则2(2log 3)f +=( )A ．38B ．18C ．112D ．12410.如图所示为函数π()2sin()(0,0)2f x x ωϕωϕ=+>≤≤的部分图像，其中A ，B 两点之间的距离为5，那么(1)f -=( ) A ．-1 B ．3- C ．3D ．111.如果函数()y f x =图像上任意一点的坐标(,)x y 都满足方程lg()lg lg x y x y +=+，那么正确的选项是（ ）A ．()y f x =是区间(1,)+∞上的减函数，且4xy ≥B ．()y f x =是区间(1,)+∞上的增函数，且4xy ≤C ．()y f x =是区间(1,)+∞上的减函数，且4xy ≤D ．()y f x =是区间(1,)+∞上的增函数，且4xy ≥12.设定义在R上的函数()f x是最小正周期为2π的偶函数，'()()f x f x是的导函数，当[]0,πx∈时，0()1f x<<；当(0,π)x∈且π2x≠时，π()'()02x f x-<，则方程()cosf x x=在[]2π,2π-上的根的个数为( )A． 2 B．5 C．8 D．4第Ⅱ卷（共90分）二、填空题（每题4分，满分16分，将答案填在答题纸上）13.已知3sin5α=，且α为第二象限角，则tanα的值为 .14.曲线2y x =，y x =所围成的封闭图形的面积为 .15.若函数()log a f x x =(其中a 为常数且0,1a a >≠)，满足23()()f f a a >，则1(1)1f x ->的解集是.111x a<<-.考点：1.不等式求解；2.对数的单调性；3.函数的定义域.16.设,x y 满足约束条件.32020,0,0x y x y x y --≤⎧⎪-≥⎨⎪≥≥⎩若目标函数(0,0)z ax by a b =+>>的最大值为1，则a bab+的最小值为 .三、解答题 （本大题共6小题，共74分.解答应写出文字说明、证明过程或演算步骤.） 17.（本小题满分12分）设命题p ：函数2()lg()16af x ax x =-+的定义域为R ；命题q ：39x x a -<对一切的实数x 恒成立，如果命题“p 且q ”为假命题，求实数a 的取值范围.18.（本小题满分12分）设函数()sin 3cos f a αα=+，其中，角α的顶点与坐标原点重合，始边与x 轴非负半轴重合，终边经过点(,)P x y ，且0απ≤≤. （1）若P 点的坐标为（3,1），求()f α的值；（2）若点(,)P x y 为平面区域11x y y x y +≥⎧⎪≥⎨⎪≤⎩上的一个动点，试确定角α的取值范围，并求函数()f α的值域.19.（本小题满分12分）已知一企业生产某产品的年固定成本为10万元，每生产千件需另投入2.7万元，设该企业年内共生产此种产品x 千件，并且全部销售完，每千件的销售收入为()f x 万元，且22110.8(0<10)30()1081000(10)3x x f x x xx ⎧-≤⎪⎪=⎨⎪->⎪⎩（1）写出年利润P （万元）关于年产品x （千件）的函数解析式； （2）年产量为多少千件时，该企业生产此产品所获年利润最大？（注：年利润=年销售收入-年总成本）20.（本小题满分12分）若()sin(2)6f x x πω=-的图象关于直线3x π=对称，其中15(,).22ω∈-（1）求()f x 的解析式； （2）将()y f x =的图象向左平移3π个单位，再将得到的图象的横坐标变为原来的2倍（纵坐标不变）后得到()y g x =的图象；若函数π()(,3π)2y g x x =∈的图象与y a =的图象有三个交点且交点的横坐标成等比数列，求a 的值.21.（本小题满分12分）定议在R 上的单调函数()f x 满足3(2)2f =，且对任意,x y R ∈都有()()().f x y f x f y +=+（1）求证：()f x 为奇函数； （2）若(3)(392)0x x xf k f ⋅+--<对任意x ∈R 恒成立，求实数k 的取值范围.22.（本小题满分14分）已知函数2()ln f x a x bx =-图象上一点(2,(2))P f 处的切线方程为32ln 22y x =-++.（1）求,a b 的值；（2）若方程()0f x m +=在1,e e ⎡⎤⎢⎥⎣⎦内有两个不等实根，求m 的取值范围（其中e 为自然对数的底数）；（3）令()()g x f x kx =-，若()g x 的图象与x 轴交于12(,0),(,0)A x B x （其中12x x <），AB 的中点为0(,0)C x ，求证：()g x 在0x 处的导数0()0.g x '≠12120ln 1x x x x x ∴=-。

高三物理检测试题一、选择题（5分×10=50分，对而不全得3分）1、关于物体的运动，可能发生的是( )A．加速度大小逐渐减小，速度也逐渐减小B．加速度方向不变，而速度方向改变C．加速度和速度都在变化，加速度最大时，速度最小D．加速度为零时，速度的变化率最大2有关滑动摩擦力的下列说法中，正确的是 ( )A．有压力一定有滑动摩擦力B．有滑动摩擦力一定有压力C．滑动摩擦力总是与接触面上的压力垂直D．只有运动物体才受滑动摩擦力3、亚丁湾索马里海域六艘海盗快艇试图靠近中国海军护航编队保护的商船，中国特战队员发射爆震弹成功将其驱离。

假如其中一艘海盗快艇在海面上运动的v－t图象如图所示，设运动过程中海盗快艇所受阻力不变．则下列说法正确的是 ( )A．海盗快艇在0～66 s内从静止出发做加速度增大的加速直线运动B．海盗快艇在66 s末开始调头逃离C．海盗快艇在96 s末开始调头逃离D．海盗快艇在96 s～116 s内做匀减速直线运动4、如图所示，一重为10 N的球固定在支杆AB的上端，今用一段绳子水平拉球，使杆发生弯曲，已知绳的拉力为7.5 N，则AB杆对球的作用力 ( )A．大小为7.5 NB．大小为10 NC．方向与水平方向成53°角斜向右下方D．方向与水平方向成53°角斜向左上方5、如图所示，质量分别为m1、m2的两个物体通过轻弹簧连接，在力F的作用下一起沿水平方向做匀速直线运动(m1在地面上，m2在空中)，力F与水平方向成θ角．则m1所受支持力F N和摩擦力F f正确的是 ( )A．F N＝m1g＋m2g－F sin θB．F N＝m1g＋m2g－F cos θC．F f＝F cos θD．F f＝F sin θ6、如图所示，有一质量不计的杆AO，长为R，可绕A自由转动．用绳在O点悬挂一个重为G的物体，另一根绳一端系在O点，另一端系在以O 点为圆心的圆弧形墙壁上的C 点．当点C 由图示位置逐渐向上沿圆弧CB 移动过程中(保持OA 与地面夹角θ不变)，OC 绳所受拉力的大小变化情况是 ( ) A ．逐渐减小 B ．逐渐增大C ．先减小后增大D ．先增大后减小7、一轻杆一端固定质量为m 的小球，以另一端O 为圆心，使小球在竖直面内做半径为R 的圆周运动，如图所示，则下列说法正确的是 ( ) A ．小球过最高点时，杆所受到的弹力可以等于零 B ．小球过最高点的最小速度是gRC ．小球过最高点时，杆对球的作用力一定随速度增大而增大D ．小球过最高点时，杆对球的作用力一定随速度增大而减小8、如图所示，物体A 叠放在物体B 上，B 置于光滑水平面上，A 、B 质量分别为m A ＝6 kg ，m B ＝2 kg ，A 、B 之间的动摩擦因数μ＝0.2，开始时F ＝10 N ，此后逐渐增加，在增大到45 N 的过程中，则 ( ) A ．当拉力F <12 N 时，物体均保持静止状态B ．两物体开始没有相对运动，当拉力超过12 N 时，开始相对运动C ．两物体从受力开始就有相对运动D ．两物体始终没有相对运动9、质量为m 的探月航天器在接近月球表面的轨道上飞行，其运动视为匀速圆周运动。

章丘四中2014届高三理科数学阶段检测试题一、选择题（每题5分,共60分）1．设集合{}6,5,4,3,2,1=A ,{}8,7,6,5,4=B ,满足A S ⊆且φ≠B S 的集合S 为A.57B. 56C. 49D. 82. 已知,a b 是实 数，则“0a >且0b >”是“0a b +>且0ab >”的 A ．充分而不必要条件 B ．必要而不充分条件 C ．充分必要条件D ．既不充分也不必要条件3. 命题“存在0x ∈R ，02x ≤0”的否定是 A.不存在0x ∈R, 02x >0 B. 存在0x ∈R, 02x ≥0 C. 对任意的x ∈R, 2x ≤0 D. 对任意的x ∈R, 2x >04. 命题“若一个数是负数，则它的平方是正数”的逆命题是A ．“若一个数是负数，则它的平方不是正数”B ．“若一个数的平方是正数，则它是负数”C ．“若一个数不是负数，则它的平方不是正数”D ．“若一个数的平方不是正数，则它不是负数”5. 已知函数()cos (0)f x x x ωωω=+>，()y f x =的图像与直线2y =的两个相邻交点的距离等于π，则()f x 的单调递增区间是 A. 5[,],1212k k k Z ππππ-+∈ B. 511[,],1212k k k Z ππππ++∈ C. [,],36k k k Z ππππ-+∈ D. 2[,],63k k k Z ππππ++∈ 6.设()()sin f x x ωϕ=+，其中0ω>，则()f x 是偶函数的充要条件是A.()01f =B.()00f =C.()'01f =D.()'00f =7. 设函数2()()f x g x x =+，曲线()y g x =在点(1,(1))g 处的切线方程为21y x =+，则曲线()y f x =在点(1,(1))f 处切线的斜率为A ．4B ．14-C ．2D ．12- 8. △ABC 平面内一点O 满足OA OB OB OC OC OA ==，则O 为△ABCA.内心B.外心C.重心D.垂心9.设函数()cos (0)f x x ωω=>，将()y f x =的图像向右平移3π个单位长度后，所 得的图像与原图像重合，则ω的最小值等于A ．31 B ．3 C ．6 D ．9 10. 已知奇函数()x f y =在区间(－∞，＋∞)上是单调减函数．,,R αβγ∈， 且0,0,0αββγγα+>+>+>，则()()()f f f αβγ++的值A. 大于0B. 小于0C. 大于等于0D. 小于等于011.函数()23x f x x =+的零点所在的一个区间是A ．（-2，-1）B.（-1,0） C.（0,1） D.（1,2）12.已知定义在R 上的奇函数)(x f ，满足(4)()f x f x -=-,且在区间[0,2]上是增函数A.(25)(11)(80)f f f -<<B. (80)(11)(25)f f f <<-C. (11)(80)(25)f f f <<-D. (25)(80)(11)f f f -<<二、填空题：（每题4分，共16分） 13. =⎰dx x ee 11 .14. 若函数2()1x a f x x +=+在1x =处取极值，则a =_____________. 15. 函数3()log f x x =在区间[],a b 上的值域为[]0,1则b a -的最大值为________.16.直线1y =与曲线2y x x a =-+有四个交点，则a 的取值范围是 .三、解答题：（共6题.17至20题每题12分，21、22题每题13分，共74分）17. 在ABC △中，5cos 13B =-，4cos 5C =． （Ⅰ）求sin A 的值；（Ⅱ）设ABC △的面积332ABC S =△，求BC 的长．18. 已知函数2()sin sin()2f x x x x πωωω=++（0ω>）的最小正周期为π．（Ⅰ）求ω的值； （Ⅱ）求函数()f x 在区间2π03⎡⎤⎢⎥⎣⎦，上的取值范围．19. 用长为18 cm 的钢条围成一个长方体形状的框架，要求长方体的长与宽之比为2：1，问该长方体的长、宽、高各为多少时，其体积最大？最大体积是多少？20. 设函数329()62f x x x x a =-+-． （1）对于任意实数x ，()f x m '≥恒成立，求m 的最大值；（2）若方程()0f x =有且仅有一个实根，求a 的取值范围．21、设函数ax x x a x f +-=22ln )(，0>a（Ⅰ）求)(x f 的单调区间；（Ⅱ）求所有实数a ，使2)(1e x f e ≤≤-对],1[e x ∈恒成立．22. 设函数()f x 满足122()()41f x f x x x-=-+，数列{}n a 和{}n b 满足下列条件：1111,2(),,23n n n n n n n a a a f n b a a c a n ++=-==-=++．(1)求()f x 的解析式；(2)证明{}n c 成等比数列，并求{}n b 的通项公式n b ；。

2014年山东省济宁市高考数学二模试卷（理科）学校:___________姓名：___________班级：___________考号：___________一、选择题(本大题共10小题，共50.0分)1.复数z为纯虚数，若（2-i）z=a+i（i为虚数单位），则实数a的值为（）A.-B.2C.-2D.【答案】D【解析】解：由（2-i）z=a+i，得：，∵z为纯虚数，∴，解得：a=．故选：D．把等式两边同时乘以，然后利用复数代数形式的除法运算化简，由实部等于0且虚部不等于0求解实数a的值．本题考查复数代数形式的除法运算，考查了复数的基本概念，是基础题．2.已知集合A={x∈R||x-1|≤2}，B={x∈R|x2≤4}，则A∩B=（）A.（-1，2）B.[-1，2]C.（0，2]D.[-2，3]【答案】B【解析】解：由A中的不等式解得：-2≤x-1≤2，即-1≤x≤3，即A=[-1，3]，由B中的不等式解得：-2≤x≤2，即B=[-2，2]，则A∩B=[-1，2]．故选：B．求出A与B中不等式的解集确定出A与B，找出A与B的交集即可．此题考查了交集及其运算，熟练掌握交集的定义是解本题的关键．x、y之间的一组数据如下表：，则当x=6时，y的预测值为（）A.8.46B.6.8C.6.3D.5.76【答案】C【解析】解：∵==2，==4.5，2+3.6=4.5，解得：=0.45，∴=0.45x+3.6，当x=6时，=6.3，故选：C．根据已知中的数据，求出数据样本中心点的坐标，代入求出回归直线方程，进而将x=6代入可得答案．本题考查线性回归方程的求法和应用，是一个中档题，这种题目解题的关键是求出回归直线方程，数字的运算不要出错．4.设变量x、y满足约束条件：，则目标函数z=5x+3y的最大值为（）A.18B.17C.27D.【答案】C【解析】解：不等式组对应的平面区域如图：由z=5x+3y得y=-，平移直线y=-，则由图象可知当直线y=-经过点B时直线y=-的截距最大，此时z最大，由，解得，即B（3，4），此时M=z=5×3+3×4=27，故选：C．作出不等式组对应的平面区域，根据z的几何意义，利用数形结合即可得到最大值．本题主要考查线性规划的应用，根据z的几何意义，利用数形结合是解决本题的关键．5.已知函数f（x）=cos（2x+φ）的图象沿x轴向左平移个单位后，得到函数g（x）的图象，则“φ=-”是“g（x）为偶函数”的（）A.充分不必要条件B.必要不充分条件C.充要条件D.既不充分又不必要条件【答案】A【解析】解：函数f（x）=cos（2x+φ）的图象沿x轴向左平移个单位后，得到函数g（x）的图象，∴g（x）=cos（2x++φ），当φ=-时，g（x）=cos2x是偶函数，但是g（x）为偶函数，φ=kπ-，k∈Z．∴“φ=-”是“g（x）为偶函数”的充分不必要条件．故选：A．求出平移后的函数的解析式，然后判断函数的奇偶性，即可得到结果．本题考查函数的图象变换，函数的解析式的求法以及函数的奇偶性的应用，充要条件的判断，基本知识的考查．6.若某空间几何体的三视图如图所示，则该几何体的体积是（）A.16B.32C.48D.144【答案】C【解析】解：由三视图知：几何体为四棱锥，且四棱锥的一条侧棱与底面垂直，如图：其中BC=2，AD=6，AB=6，SA⊥平面ABCD，SA=6，∴几何体的体积V=××6×6=48．故选：C．几何体为四棱锥，结合直观图判断相关几何量的数据，把数据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体积，根据三视图判断几何体的形状及数据所对应的几何量是解答本题的关键．7.函数f（x）=1-x+lgx的图象大致是（）A. B. C. D.【答案】A【解析】解：定义域为（0，+∞），=，∴当x∈（0，lge），时f （x）＜0，f（x）单调递减，当x∈（lge，+∞）时，f （x）＞0，f（x）单调递增，当x=lge时，f（x）取得极大值也是最大值，f（lge）=1-lge+lg（ge）=＞0，∴f（x）的图象为A．故选；A．利用函数的单调性，和极大值，就能判断函数的图象．考查函数的单调性，极值和最值．属于基础题．8.向量=（1，2），=（1，-λ），在区间[-5，5]上随机取一个数λ，使向量2+与-的夹角为锐角的概率为（）A. B. C. D.【答案】D【解析】解：∵=（1，2），=（1，-λ），∴2+=（3，4-λ），-=（0，2+λ），若2+与-的夹角为锐角，则（2+）•（-）＞0，即（4-λ）（2+λ）＞0，解得-2＜λ＜4，则向量2+与-的夹角为锐角的概率为=，故选：D根据向量数量积的应用，求出向量2+与-的夹角为锐角的等价条件，利用几何概型的概率公式即可得到结论．本题主要考查几何概型的概率的计算，利用向量数量积的应用求出向量2+与-的夹角为锐角的等价条件，是解决本题的关键．9.已知双曲线-=1（a＞0，b＞0）的右焦点与抛物线y2=12x的焦点重合，且双曲线的一条渐近线被圆（x-3）2+y2=8截得的弦长为4，则此双曲线的渐近线方程为（）A.y=±2x B.y=±x C.y=±x D.y=±2x【答案】B解：抛物线y2=12x的焦点坐标为（3，0），∵双曲线-=1（a＞0，b＞0）的右焦点与抛物线y2=12x的焦点重合，∴c=3，∵双曲线的一条渐近线被圆（x-3）2+y2=8截得的弦长为4，∴圆心到渐近线的距离为2，设渐近线方程为bx+ay=0，则=2，∴b=2，∴a=，∴双曲线的渐近线方程为y=±x．故选：B．求出抛物线的焦点坐标，可得c=3，利用双曲线的一条渐近线被圆（x-3）2+y2=8截得的弦长为4，可得圆心到渐近线的距离为2，从而可求a，b，即可求出双曲线的渐近线方程．本题考查双曲线的渐近线方程，考查直线与圆的位置关系，考查学生的计算能力，属于中档题．10.已知函数y=f（x）的定义域为（-π，π），且函数y=f（x-1）的图象关于直线x=1对称，当x∈（0，π）时，f（x）=-f （）sinx-πlnx（其中f （x）是f（x）的导函数）．若a=f（π0.2），b=f（logπ3），c=f（log9），则a，b，c的大小关系式（）A.b＞a＞cB.a＞b＞cC.c＞b＞aD.b＞c＞a【答案】A【解析】解：由x∈（0，π）时，f（x）=-f （）sinx-πlnx，∴f（x）=-f （）cosx-，∴=-2，∴f（x）=2sinx-πlnx，∴当x∈（0，π）时，f （x）=2cosx-，≤x＜π，2cosx＜0；0＜x＜，2cosx＜2，＞2，则有f （x）＜0．则f（x）在x∈（0，π）上为减函数．又函数y=f（x-1）的图象关于直线x=-1对称，则函数y=f（x）为偶函数，∵log9＜-3而1＜π0.3＜2，0＜logπ3＜1．∴f（logπ3＞f（π0.2）＞f（log9）∴b＞a＞c．由题意可知函数为偶函数，把给出的函数解析式求导后求出的值，代入导函数解析式判断导函数的符号，得到原函数的单调性，由单调性得答案．本题考查了函数的单调性与导函数之间的关系，考查了函数的奇偶性的性质，解答的关键在于判断函数在（0，π）上的单调性，是中档题．二、填空题(本大题共5小题，共25.0分)11.设随机变量X服从正态分布N（1，σ2），若P（1＜X＜2）=p，则P（X＜0）= ______ ．【答案】-p【解析】解：随机变量ξ服从正态分布N（1，σ2），∴曲线关于x=1对称，∴P（X＜0）=P（X＞2）=-P（1＜X＜2）=-p．故答案为：-p．随机变量ξ服从正态分布N（1，σ2），得到曲线关于x=1对称，根据曲线的对称性得到小于0的概率和大于2的概率是相等的，根据概率的性质得到结果．本题主要考查正态分布曲线的特点及曲线所表示的意义、函数图象对称性的应用等基础知识，属于基础题．12.阅读如图所示的程序图，运行相应的程序，输出的结果s= ______【答案】16【解析】解：由已知中的程序框图：当n=1时，S=1，a=3，满足继续循环的条件，n=2；当n=2时，S=4，a=5，满足继续循环的条件，n=3；当n=3时，S=9，a=7，满足继续循环的条件，n=4；当n=4时，S=16，a=9，不满足继续循环的条件故输出的s值为16，故答案为：16由已知中的程序框图可知：该程序的功能是利用循环结构计算并输出变量s的值，模拟程序的运行过程，可得答案；本题考查的知识点是程序框图，当循环的次数不多，或有规律时，常采用模拟循环的方法解答．13.若函数y=e-x在点（0，1）处的切线为l，则由曲线y=e-x，直线x=1，切线l所围成封闭图形的面积为______ ．【答案】-【解析】解：∵y=e-x，∴y y=-e-x，则在（0，1）处的切线斜率k=-1，则切线方程为y-1=-（x-0）=-x，即y=-x+1，则阴影部分的面积S==--=-e-x|=1-=-，故答案为：-利用导数的几何意义，求出切线方程，利用积分的几何意义，即可求出封闭区域的面积．本题主要考查导数的几何意义以及积分的几何意义，要求熟练掌握函数的导数公式和积分公式．14.设x，y∈R，a＞1，b＞1，若a x=b y=3，a+b=2，则+的最大值为______ ．【答案】3【解析】解：∵a＞1，b＞1，a x=b y=3，∴xlga=ylgb=lg3，∴====3，当且仅当a=b=3时取等号．∴+的最大值为3．故答案为：3．利用对数的换底公式、对数的运算法则、基本不等式的性质即可得出．本题考查了对数的换底公式、对数的运算法则、基本不等式的性质，属于基础题．15.对于三次函数f（x）=ax3+bx2+cx+d（a≠0），定义f″（x）是y=f（x）的导函数y=f （x）的导函数，若方程f″（x）=0有实数解x0，则称点（x0，f（x0））为函数y=f（x）的“拐点”．可以证明，任何三次函数都有“拐点”，任何三次函数都有对称中心，且“拐点”就是对称中心．请你根据这一结论判断下列命题：①存在有两个及两个以上对称中心的三次函数；②函数f（x）=x3-3x2-3x+5的对称中心也是函数y=tan x的一个对称中心；③存在三次函数h（x）方程h （x）=0有实数解x0，且点（x0，h（x0））为函数y=h （x）的对称中心；④若函数g（x）=x3-x2-，则g（）+g（）+g（）+…+g（）=-1006.5其中正确命题的序号为______ （把所有正确命题的序号都填上）．【答案】②③④【解析】解：任何三次函数都有且只有一个对称中心，故①不正确；∵f（x）=x3-3x2-3x+5，∴f （x）=3x2-6x-3，∴f″（x）=6x-6，令f″（x）=6x-6=0，解得x=1，f（1）=0，∴f（x）=x3-3x2-3x+5的对称中心是（1，0），当x=1时，（，0）是函数y=tan x的一个对称中心，故②正确，∵任何三次函数都有对称中心，且“拐点”就是对称中心，∴存在三次函数f （x）=0有实数解x0，点（x0，f（x0））为y=f（x）的对称中心，故③正确．∵g（x）=x3-x2-，∴g （x）=x2-x，g''（x）=2x-1，令g''（x）=2x-1=0，解得x=，g（）==，∴函数g（x）=x3-x2-的对称中心是（，）∴g（x）+g（1-x）=-1，∴g（）+g（）+g（）+…+g（）=-1006.5，故④正确．所以正确命题的序号为②③④故答案为：②③④．①③利用三次函数对称中心的定义和性质进行判断；②根据新定义求出对称中心，而y=tan x的对称中心是（，），继而判断；④求得函数g（x）=x3-x2-的对称中心（，），g（x）+g（1-x）=-1，继而求出值．本小题考查新定义，考查函数与导数等知识，考查化归与转化的数学思想方法，考查计算能力，属于中档题．三、解答题(本大题共6小题，共75.0分)16.已知向量=（-，2cosx），=（cos2x+sin2x，cosx），记函数f（x）=•．（Ⅰ）求f（x）的最小正周期及单调减区间；（Ⅱ）记△ABC的内角A、B、C的对边长分别为a、b、c，若f（）=1，b=3，c=2，求sin A的值．【答案】解：（I）f（x）=•=-（cos2x+sin2x）+2cos2x=-（cos2x+sin2x）+cos2x+1=cos2x-sin2x+1=cos（2x+）+1∴f（x）的最小正周期为π令2kπ＜2x+＜2kπ+π，k∈z，解得kπ-＜x＜kπ+，k∈z∴f（x）的单调减区间为（kπ-，kπ+），k∈z（II）由f（）=1，得cos（B+）+1=1．即cos（B+）=0，又B是三角形的内角，故B=由正弦定理得得sin C=，又b＞c，故C是锐角∴cos C==∴sin A=sin（B+C）=sin B cos C+cos B sin C=【解析】（I）先求出f（x）的解析式，再由周期公式及复合三角函数的性质求单调区间；（II）由f（）=1求出B，再由正弦定理求出sin C，再由sin A=sin（B+C）结合和角公式即可求出sin A的值．本题考查正弦定理的应用以及三角恒等变换公式，三角函数的周期公式及单调区间的求法，综合性较强，属于高考中常见的题型17.袋子里有完全相同的3只红球和4只黑球，今从袋子里随机取球．（1）若有放回地取3次，每次取一个球，求取出1个红球2个黑球的概率；（2）若无放回地取3次，每次取一个球，若取出每只红球得2分，取出每只黑球得1分，求得分ξ的分布列和数学期望．【答案】解：（1）从袋子里有放回地取3次球，相当于做了3次独立重复试验，每次试验取出红球的概率为，取出黑球的概率为，设事件A=“取出1个红球2个黑球”，则P（A）==；（2）ξ的取值有四个：3、4、5、6，P（ξ=3）==，P（ξ=4）==，P（ξ=5）==，P（ξ=6）==．分布列为：…（10分）从而得分ξ的数学期望Eξ=3×+4×+5×+6×=．【解析】（1）确定每次试验取出红球、黑球的概率，利用独立重复试验的概率公式，即可求取出1个红球2个黑球的概率；（2）确定ξ的取值，求出相应的概率，可得分布列与数学期望．本题考查概率的求解，考查离散型随机变量的分布列与数学期望，正确求概率是关键．18.如图1，在R t△ABC中，∠ACB=30°，∠ABC=90°，D为AC中点，AE⊥BD于E，延长AE交BC于F，将△ABD沿BD折起，使平面ABD⊥平面BCD，如图2所示．（1）求证：AE⊥平面BCD；（2）求二面角A-DC-B的余弦值；（3）已知点M在线段AF上，且EM∥平面ADC，求的值．【答案】（1）证明：∵平面ABD⊥平面BCD，交线为BD，又在△ABD中，AE⊥BD于E，AE⊂平面ABD，∴AE⊥平面BCD．（2）解：由（1）知AE⊥平面BCD，∴AE⊥EF，由题意知EF⊥BD，又AE⊥BD，如图，以E为坐标原点，分别以EF、ED、EA所在直线为x轴，y轴，z轴，建立空间直角坐标系E-xyz，设AB=BD=DC=AD=2，则BE=ED=1，∴AE=，BC=2，BF=，则E（0，0，0），D（0，1，0），B（0，-1，0），A（0，0，），F（，，），C（，，），，，，，，，由AE⊥平面BCD知平面BCD的一个法向量为，，，设平面ADC的一个法向量，，，则，取x=1，得，，，∴cos＜，＞=-∴二面角A-DC-B的余弦值为．（3）设，其中λ∈[0，1]，∵，，，∴，，，∴，，，由，得，解得，，∴在线段AF上存在点M使EM∥平面ADC，且．【解析】（1）由平面ABD⊥平面BCD，交线为BD，AE⊥BD于E，能证明AE⊥平面BCD．（2）以E为坐标原点，分别以EF、ED、EA所在直线为x轴，y轴，z轴，建立空间直角坐标系E-xyz，利用向量法能求出二面角A-DC-B的余弦值．（3）设，利用向量法能求出在线段AF上存在点M使EM∥平面ADC，且．本题考查直线与平面垂直的证明，考查二面角的余弦值的求法，考查满足条件的点是否存在的判断与求法，解题时要认真审题，注意向量法的合理运用．19.已知数列{b n}满足S n+b n=，其中S n为数列{b n}的前n项和．（1）求证：数列{b n-}是等比数列，并求数列{b n}的通项公式；（2）如果对任意n∈N*，不等式≥2n-7恒成立，求实数k的取值范围．【答案】解：（1）∵S n+b n=，∴S n+1+b n+1=，两式相减得b n+1=b n+，即b n+1-=（b n-），∵S1+b1=，即b1=，∴数列{b n-}是首项为b1-=3，公比q=的等比数列，∴b n-=3×，即b n=3×+．则数列{b n}的通项公式b n=3×+；（2）∵b n=3×+；∴S n=3×（1++…+）+=+=6（1-）+；∵不等式≥2n-7，化简得k，设c n=，则c n+1-c n==，当n≥5时，c n+1≤c n，c n为单调递减数列，当1≤n＜5时，c n+1＞c n，c n为单调递增数列，＜，∴当n=5时，c n取得最大值，即要使不等式≥2n-7恒成立，则实数k的取值范围是k≥．【解析】（1）求证：数列{b n-}是等比数列，并求数列{b n}的通项公式；（2）求出S n的表达式，将不等式恒成立，转化为最值问题即可得到结论．本题主要考查等差数列的判断，以及不等式恒成立的证明，综合考查学生的运算性质．20.已知函数f（x）=+lnx（a∈R）．（1）求f（x）的最小值；（2）当a=2时，求证：ln（n+1）+2＞nln（2e）（n∈N*）．【答案】解：（1）∵f （x）=，x＞0，①a≤0时，f （x）≥0，f（x）在（0，+∞）上递增，∴f（x）无最值，②a＞0时，令f （x）＞0，解得：x＞a，令f x）＜0，解得：0＜x＜a，∴f（x）在（0，a）递减，在（a，+∞）递增，∴f（x）min=f（a）=lna+1，综上，a≤0时，f（x）无最值，a＞0时，f（x）min=f（a）=lna+1，（2）a=2时，由（1）得f（x）≥ln2+1，即+lnx≥ln2+1，从而lnx≥ln2+1-=ln（2e）-（*），∴分别令x=，…，代入（*）得下列n个不等式，ln＞ln（2e）-=ln（2e）-2×，ln＞ln（2e）-=ln（2e）-2×，…，ln＞ln（2e）-（2×），将所述n个不等式相加得：ln++ln+…+ln＞nln（2e）-2（+2×+…+2×），∴ln（n+1）＞nln（2e）-2（++…+），即ln（n+1）+2＞nln（2e）．【解析】（1）先求出f （x）=，x＞0，再讨论①a≤0时，②a＞0时的情况，从而求出函数的最小值；（2）a=2时，由（1）得f（x）≥ln2+1，从而lnx≥ln2+1-=ln（2e）-（*），分别令x=，…，代入（*）得下列n个不等式，得ln++ln+…+ln＞nln（2e）-2（+2×+…+2×），进而证明ln（n+1）+2＞nln（2e）．本题考察了函数的单调性，函数的最值问题，导数的应用，不等式的证明，是一道综合题．21.如图所示的曲线C由曲线C1：+=1（a＞b＞0，y≥0）和曲线C2：x2+y2=a2（y＜0）组成，已知曲线C1过点（，），离心率为，点A，B分别为曲线C与x轴、y轴的一个交点．（1）求曲线C1和C2的方程；（2）若点Q是曲线C2上的任意一点，求△QAB面积的最大值及点Q的坐标；（3）若点F为曲线C1的右焦点，直线l；y=kx+m与曲线C1相切于点M，且与直线x=交于点N，过点P做MN，垂足为H，求证|FH|2=|MH|+|HN|．【答案】（1）解：由已知得，①又e=，∴，即a2=4b2，②由①②得a2=4，b2=1，∴曲线C1的方程为=1．（y≥0）．曲线C2的方程为x2+y2=4（y＜0）．（2）解：由（1）知A（-2，0），B（0，1），∴AB所在直线为x-2y+2=0，由题意知当曲线C2在点Q上的切线与直线AB平行时，△QAB面积最大，设此时切线方程为x-2y+t=0，t＜0，由直线与圆相切得：，∴t=-2或t=2（舍）此时△QAB的高为：h==2+，（S△QAB）max===，由，得x=，y=-，∴Q（，），∴△QAB面积的最大值为，此时点Q坐标为（，）．（3）证明：由题意得F（，0），N（，），设M（x0，y0），由，得（1+4k2）x2+8kmx+4m2-4=0，又直线l与曲线C1相切于M，∴△=（8km）2-4（1+4k2）（4m2-4）=0，即m2=4k2+1，，，∴M（-，），∴，，，，=（-）×+==，又m2=4k2+1，∴=0，∴=0，∴△MFN为直角三角形，在R t△MFN中，FH⊥MN，∴|FH|2=|MH|+|HN|．【解析】（1）由已知得，e=由此能求出曲线C1的方程和曲线C2的方程．（2）由（1）知AB所在直线为x-2y+2=0，由题意知当曲线C2在点Q上的切线与直线AB平行时，△QAB面积最大，由此能求出△QAB面积的最大值及点Q的坐标．（3）设M（x0，y0），由，得（1+4k2）x2+8kmx+4m2-4=0，由直线l与曲线C1相切于M，得m2=4k2+1，由此利用向量知识结合已知条件能证明|FH|2=|MH|+|HN|．本题考查曲线方程的求法，考查三角形面积的最大值的相应的点的坐标的求法，考查等式的证明，解题时要认真审题，注意函数与方程思想的合理运用．。

山东省烟台市高三统一质量检测考试数学（文科）试卷第Ⅰ卷（共50分）一、选择题：本大题共10个小题,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. 设集合}032|{2<--=x x x M ，2{|log 0}N x x =<，则N M 等于（ ） A ．)0,1(-B ．)1,1(-C ．)1,0(D ．)3,1(2.若复数z 的实部为1，且||2z =，则复数z 的虚部是（ ）A ．．． D3.若命题:p α∃∈R ，cos()cos παα-=；命题:q ∀R ∈x ，012>+x . 则下面结论正确的是（ ）A.p 是假命题B.q ⌝是真命题 C.p ∧q 是假命题 D.p ∨q 是真命题 【答案】D 【解析】试题分析：由cos()cos παα-=得，cos cos ,cos 0,,2k k Z πααααπ-===+∈，所以，p 是真命题；又012>+x 恒成立，所以，q 是真命题；因此，p ∨q 是真命题，故选D ．考点：简单逻辑联结词，存在性命题，全称命题.4.若函数()21,1ln ,1x x f x x x ⎧+≤=⎨>⎩， 则((e))f f =（其中e 为自然对数的底数） （ ）A ．0B ．1C ．2 D.2ln(e 1)+5.若一个三棱锥的三视图如图所示，其中三个视图都是直角三角形，则在该三棱锥的四个面中，直角三角形的个数为（ ）A ．1B ．2C ．3D ．4 【答案】D 【解析】试题分析：观察三视图可知，该三棱锥底面BCD 是直角三角形，CD BC ⊥,侧面,ABC ABD 是直角三角形；由CD BC ⊥，CD AB ⊥,知CD ABC ⊥平面，CD AD ⊥，侧面ACD 也是直角三角形， 故选D ..考点：三视图，几何体的结构特征.6.在等差数列}{n a 中，12012a =- ，其前n 项和为n S ，若2012102002201210S S -=，则2014S 的值等于 （ ） A.2011 B. -2012 C.2014 D. -20137.如图是某班50位学生期中考试数学成绩的频率分布直方图，其中成绩分组区间是：[)4050，，[)5060，，[)6070，，[)7080，，[)8090，，[]90100，，则图中x 的值等于 （ ）A ．0.12B ．0.18．0.012 D ．0.018【答案】D 【解析】试题分析：由频率分布直方图，(0.0060.0060.010.0540.006)101x +++++⨯=， 所以，0.018x =，故选D ． 考点：频率分布直方图8.函数x x y sin =在[]ππ,-上的图象是（ ）【答案】A 【解析】试题分析：函数x x y sin =是偶函数，所以，其图象关于y 轴对称，排除D ； 由x π=时，0y =，排除C ； 由 2x π=时，2y π=，排除B ；选A .考点：函数的奇偶性，函数的图象.9.若函数()2sin()(214)84f x x x ππ=+-<<的图象与x 轴交于点A ，过点A 的直线l 与函数的图象交于B 、C 两点，则=⋅+)(（其中O 为坐标原点） （ ）A ．32-B ．32C ．72-D ．7210.对任意实数,m n ，定义运算m n am bn cmn *=++，其中c b a ,,为常数，等号右边的运算是通常意义的加、乘运算．现已知12=4*，23=6*，且有一个非零实数t ，使得对任意实数x ，都有x t x *=，则t =（ ） A ．4 B ．5 C ．6 D ．7第Ⅱ卷（共100分）二、填空题（每题5分，满分25分，将答案填在答题纸上）11.若直线10ax by -+=平分圆22:241C x y x y ++-+0=的周长,则ab 的取值范围是 .12.若某程序框图如右图所示，则该程序运行后输出的i 值为 .13.已知变量y x ,满足约束条件⎪⎩⎪⎨⎧≥+-≥≤+-042042k y x y y x ，且目标函数y x z +=3的最小值为1-，则实常数=k .【答案】9 【解析】试题分析：画出可行域及直线03=+y x ，如图所示.平移直线03=+y x 可知，当其经过直线2y =与直线40x y k -+=的交点(8,2)A k -时，y x z +=3的最小值为1-，所以3(8)21,9k k -+=-=故答案为9．考点：简单线性规划的应用14.对大于或等于2的正整数的幂运算有如下分解方式：3122+= 53132++= 753142+++= … 5323+= 119733++= 1917151343+++= …根据上述分解规律，若115312++++= m ，3p 的分解中最小的正整数是21，则=+p m .∵3p 的分解中最小的数是21， ∴33p 5p 5==，，m p 6511∴+=+=,故答案为11.考点：归纳推理，等差数列的求和.15.已知抛物线24y x =的准线与双曲线22221x y a b-=的两条渐近线分别交于A ，B两点，且||AB =，则双曲线的离心率e 为.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）全国第十二届全国人民代表大会第二次会议和政协第十二届全国委员会第二次会议，2014年3月在北京开幕．期间为了了解国企员工的工资收入状况，从108名相关人员中用分层抽样方法抽取若干人组成调研小组，有关数据见下表：（单位：人）（1）求x ，y ；（2）若从中层、高管抽取的人员中选2人，求这二人都来自中层的概率．选中的2人都来自中层的事件包含的基本事件有：12(,)b b ，13(,)b b ，23(,)b b 共3种. 由古典概型概率的计算公式即得.17.（本小题满分12分）已知函数27()sin 22sin 1()6f x x x x π⎛⎫=--+∈ ⎪⎝⎭R ， （1）求函数()f x 的周期及单调递增区间；（2）在ABC ∆中，三内角A ，B ，C 的对边分别为c b a ,,，已知函数()f x 的图象经过点1,,,,2A b a c ⎛⎫ ⎪⎝⎭成等差数列，且9AB AC ⋅=，求a 的值. 【答案】（1）最小正周期：22T ππ==， 递增区间为：[,]()36k k k Z ππππ-+∈;（2）a =【解析】试题分析：首先应用和差倍半的三角函数公式，化简得到()f x sin(2)6x π=+（1）最小正周期：22T ππ==，利用“复合函数的单调性”，求得()f x 的单调递增区间[,]()36k k k Z ππππ-+∈; （2）由1()sin(2)62f A A π=+=及0<A<π可得3A π=, 根据,,b a c 成等差数列，得2a b c =+， 根据1cos 9,2AB AC bc A bc ⋅=== 得18bc =，应用余弦定理即得所求. 试题解析：2711()sin(2)2sin 1cos 22cos 2cos 2262222f x x x x x x x x π=--+=-++=+ sin(2)6x π=+ ………………………………………………3分18.（本题满分12分）如图1，在直角梯形ABCD 中，//AD BC ，90,ADC BA BC ∠==.把BAC ∆沿AC 折起到PAC ∆的位置，使得P 点在平面ADC 上的正投影O 恰好落在线段AC 上，如图2所示，点E F 、分别为棱PC CD 、的中点.（1）求证：平面//OEF 平面APD ；（2）求证：CD ⊥平面POF ；（3）若3,4,5AD CD AB ===，求四棱锥E CFO -的体积.【答案】（1）证明见解析；（2）证明见解析.（3【解析】试题分析：（1）因为点P 在平面ADC 上的正投影O 恰好落在线段AC 上，所以PO ⊥平面ABC ，PO ⊥AC ；由AB BC =，知O 是AC 中点，得到//OE PA ， //OE PAD 平面；同理//OF PAD 平面；根据,OE OF O OE OF OEF =⊂、平面，得到平面//OEF 平面PDA .所以平面//OEF 平面PDA …………………5分19.（本小题满分12分）已知数列{}n a 的前n 项和为n S ，且22n n S a +=，数列{}n b 满足11b =，且12n n b b +=+.（1）求数列{}n a ,{}n b 的通项公式；（2）设1(1)1(1)22n nn n n c a b --+-=-，求数列{}n c 的前2n 项和2n T ．∴{}n a 是等比数列，公比为2，首项12a =， ∴2n n a =. ………3分由12n n b b +=+，得{}n b 是等差数列，公差为2. ……………………4分 又首项11=b ，∴ 21n b n =-. ……………………………6分（2）2(21)n n c n ⎧=⎨--⎩ 为偶数为奇数n n ……………………8分3212222[37(41)]n n T n -=+++-+++- ……………10分 2122223n n n +-=--． ……………………………12分 考点：等差数列、等比数列的通项公式及其求和公式，“分组求和法”.20.（本小题满分13分）已知函数+1()ln +1a f x x ax x=+-. （1）当1a =时，求曲线()y f x =在点(2,(2))f 处的切线方程；（2）当102a -≤≤时，讨论()f x 的单调性. 【答案】（1）ln20x y -+=（2）当0a =时，在(0,1),()f x 单调递减，在'(1,)()0f x +∞>,()f x 单调递增； 当12a =-时,在(0,)+∞单调递减 当102a -<<时，在1(0,1),(,)a a +-+∞()f x 单调递减，()f x 在1(1,)a a -单调递增； 【解析】试题分析：（1）利用切点处的导函数值是切线的斜率，应用直线方程的点斜式即得；（2）2'222111(1)(1)()a ax x a ax a x f x a x x x x ++--++-=+-==， ……6分 当0a =时，'21()x f x x-=，此时，在'(0,1)()0f x <,()f x 单调递减， 在'(1,)()0f x +∞>,()f x 单调递增； …………………………… 8分 当102a -≤<时，'21()(1)()a a x x a f x x++-=， 当11a a +-=即12a =-时2'2(1)()02x f x x-=-≤在(0,)+∞恒成立， 所以()f x 在(0,)+∞单调递减； …………………………………10分 当102a -<<时，11a a +->，此时在'1(0,1),(,)()0a f x a+-+∞<,()f x 单调递减，()f x 在'1(1,)()0a f x a->单调递增； ………………………………12分 综上所述：当0a =时，()f x 在(0,1)单调递减，()f x 在(1,)+∞单调递增； 当102a -<<时， ()f x 在1(0,1),(,)a a -+∞单调递减，()f x 在1(1,)a a-单调递增； 当12a =-时()f x 在(0,)+∞单调递减. ……………………………………13分 考点：应用导数研究函数的单调性，导数的几何意义，直线方程的点斜式.21.(本题满分l4分) 已知椭圆)0(1:2222>>=+b a by a x C 经过点(1,2P ，且两焦点与短轴的两个端点的连线构成一正方形.（1）求椭圆C 的方程；（2）直线l 与椭圆C 交于A ，B 两点，若线段AB 的垂直平分线经过点1(0,)2-，求AOB ∆ （O 为原点）面积的最大值.试题解析：（1）∵椭圆)0(1:2222>>=+b a by a x C 的两焦点与短轴的两个端点的连线构成正方形，∴a =， ∴222212x y b b+=， …………2分又∵椭圆经过点P ，代入可得21b =， ∴故所求椭圆方程为22 1.2x y += …………4分12|||AB x x =-=所以1=||||2AOB S AB d ∆考虑到2212k t +=且02t <<化简得到AOB S ∆…………………13分 因为02t <<，所以当1t =时，即k =时，AOB S ∆综上，AOB ∆面积的最大值为. …………………14分 考点：椭圆的几何性质，直线与圆锥曲线的位置关系，三角形面积公式.。