2019年哈尔滨南岗中学高三下学期周测三
2019年黑龙江省哈尔滨三中高考数学三模试卷(文科)(有答案解析)
2019年黑龙江省哈尔滨三中高考数学三模试卷（文科）题号一二三总分得分一、选择题（本大题共12小题，共60.0分）1.若复数z=a+i（a∈R）的模为，则a=（）A. 1B. ±1C. 2D. ±22.设命题：∀x∈R，x2-3x+2≤0，则￢p为（）A. ∃x0∈R，x02-3x0+2≤0B. ∀x∈R，x2-3x+2＞0C. ∃x0∈R，x02-3x0+2＞0D. ∀x∈R，x2-3x+2≥03.已知集合A={x|＜0}，B={x|y=），则A∩B=（）A. （-1，2）B. [-1，2）C. [-1，2]D. [-2，2]4.已知函数（f（x）＝A sin（ωx+φ）（A＞0，ω＞0，|φ|＜）的部分图象如图所示，则f（x）的解析式可以为（）A. y＝2sin（2x+）B. y＝2sin（x+）C. y＝2sin（2x﹣）D. y＝2sin（x﹣）5.过抛物线y2=4x的焦点作一条倾斜角为的直线，与抛物线交于A，B两点，则|AB|=（）A. 4B. 6C. 8D. 166.函数y=4x+2x+1+3（x∈R）的值域为（）A. [2，+∞）B. （3，+∞）C. （，+∞）D. [9，+∞）7.若直线y=kx+1与圆x2+y2=1相交于P，Q两点，且△POQ为等边三角形（其中O为原点），则k的值为（）A. 或-B.C. 或-D.8.已知某个几何体的三视图如图（主视图中的弧线是半圆），根据图中标出的尺寸（单位：cm），可得这个几何体的表面积是（）cm2A. 20+2πB. 20+3πC. 24+2πD. 24+3π9.在边长为2的正方形ABCD内任取一点P，使得∠APB≤的概率为（）A. 1-B.C.D. 1-10.阅读右面的程序框图，如果输入的实数x的取值范围是（-∞，1]∪[2，+∞），那么输出的函数值f（x）取值范围是（）A. [0，2]B. [，2]C. [，4]D. [，2]∪{4}11.已知函数f（x）=a sin x+b cos x，且f（）是它的最大值（其中a，b为常数，且m≠0），给出下列命题：①函数f（x-）为奇函数②函数f（x）的图象关于x=对称；③函数f（-）是函数的最小值④函数f（x）的图象在y轴右侧与直线y=的交点按横坐标从小至大依次记为P1，P2，P3，P4…则|P2P4|=2π．其中正确的个数为（）A. 1B. 2C. 3D. 412.函数f（x）=，若存在实数m，使得方程f（x）=m有三个相异实根，则实数a的范围是（）A. [，+∞）B. [0，]C. （-∞，2]D. [，2）二、填空题（本大题共4小题，共20.0分）13.已知向量=（1，-2），=（t，3），若∥，则t=______14.等比数列{a n}中，a1=1，a3•a5=64，则a2019=______15.设变量x，y满足约束条件：，则目标函数z=3x-2y的最小值为______．16.数学家华罗庚曾说：“数缺形时少直，形少数时难入微”．事实上，很多代数问题可以转化为几何问题加以解法，例如，与相关的代数问题，可以转化为点A（x，y）与点B（a，b）之间距离的几何问题．结合上述观点，可得方程=4的解为______三、解答题（本大题共7小题，共82.0分）17.已知等差效列{a n}的前n项和为S n，且a1=1，S2+a2=4．（1）求数列{a n}的通项公式；（2）设数列b n=2n+1•a n求{b n}的前项和T n．18.棉花的优质率是以其纤维长度来衡量的，纤维越长的棉龙品质越高．棉花的品质分类标准为纤维长度小于等于28mm的为粗绒棉，纤维长度在（25，33]为细绒棉，纤维长度大于33mm的为长绒棉，其中纤维长度在38mm以上的棉花又名“军海1号”，某采购商从新疆某一棉花基地抽测了100根棉花的纤维长度，得到数据如下图频率分有表所示纤维长度（mm）≤25（25，33]（33，38]＞38根数2384020（）若将频率作为概率，根据以上数据，能否认为该基地的这批棉花符合“长绒棉占全部棉花的50%以上的要求（2）用样本估计总体，若这批棉共有10000kg，基地提出了两种销售方案给采购商参考．方案一：不分等级卖出，每千克按13.5元计算．方案二：对10000kg棉花先分等级再销售，分级后不同等级的棉花售价如表纤维长度（mm）≤25（25，33]（33，38]＞38根数281525从采购商的角度，请你帮他决策一下该用哪个方案．（3）用分层抽样的方法从长绒棉中抽取6根棉花，再从6根棉花中取两根进行检验，求抽到的两根棉花只有一根是“军海1号”的概率．19.如图，在五棱锥P-ABCDE中，AB∥DE，BC∥AE，AE⊥平面PDE，AB=AE=PD=2DE=2BC=4，∠PDE=60°．（1）证明：PE⊥CD；（2）过点D作平行于平面PAE的截面，与直线AB，PB，PC分别交于F，G，H，求夹在该截面与平面PAE之间的几何体体积．20.已知函数f（x）=x-1--ln x．（1）若a=0，求f（x）在x=1处的切线方程（2）若函数f（x）存在两个极值点x1和x2，求证：f（x1x2）+≥2ln2-1．21.已知定点P（2，0），圆M：x2+y2+4x-60=0，过点P的直线l₁交圆M于R，S两点，过点P作直线l2∥MS交直线MR于Q点（1）求Q点的轨迹方程E（2）若A，B，C，D是曲线E上不重合的四个点，且AC与BD交于点（-2，0），•=0，求||+||的取值范围22.在平面直角坐标系xOy中，曲线C1的参数方程为（t为参数，0≤α＜π），以坐标原点O为极点，x轴正半籼为极轴；建立极坐标系，曲线C2的极坐标方程为．（1）求曲线C2的直角坐标方程；（2）若曲线C1与C2交于A，B两点，且|AB|=2，求α的值．23.设函数f（x）=|x-a|，如果不等式f（x）≤1的解集为{x|0≤x≤2}．（1）求a的值；（2）当x∈（0，1），证明：．-------- 答案与解析 --------1.答案：B解析：解：∵z=a+i（a∈R）的模为，∴，解得a=±1．故选：B．直接利用复数模的计算公式列式求解．本题考查复数代数形式的乘除运算，考查复数模的求法，是基础题．2.答案：C解析：解：命题为全称命题，命题：∀x∈R，x2-3x+2≤0，则￢p为∃x0∈R，x02-3x0+2＞0，故选：C．根据含有量词的命题的否定即可得到结论．本题主要考查含有量词的命题的否定，比较基础．3.答案：B解析：解：集合A={x|＜0}=（-2，2），∵B={x|y=），∴-x2+x+2≥0，解得-1≤x≤2，即B=[-1，2]，∴A∩B=[-1，2），故选：B．化简集合A、B，根据交集的定义写出A∩B．本题考查了集合的化简与运算问题，是基础题．4.答案：A解析：解：由函数f（x）=A sin（ωx+φ）的部分图象知，A=2，T=-=，解得T=π.∴ω==2；又ωx+φ=2×+φ=，解得φ=.∴f（x）=2sin（2x+）．故选：A．由函数f（x）的部分图象求得A、T、ω和φ的值，即可写出f（x）．本题考查了三角函数的图象与性质的应用问题，是基础题.5.答案：D解析：解：抛物线的焦点坐标为F（1，0），p=2，过焦点的直线的斜率k=tan=，则直线方程为y=（x-1），代入y2=4x得（x-1）2=4x，整理得x2-14x+1=0，设A，B的坐标分别为（x1，y1），（x2，y2），则x1+x2=14，则|AB|=x1+x2+p=14+2=16，故选：D．求出焦点坐标和直线方程，结合过焦点直线方程，利用设而不求的思想进行求解即可．本题主要考查直线和抛物线的应用，联立方程组，利用设而不求思想，结合抛物线的弦长公式进行计算是解决本题的关键．6.答案：B解析：解：令t=2x（t＞0），∴函数y=4x+2x+1+3（x∈R）化为f（t）=t2+2t+3=（t+1）2+2（t＞0），∴f（t）＞3．即函数y=4x+2x+1+3（x∈R）的值域为（3，+∞）．故选：B．令t=2x（t＞0），把原函数转化为关于t的一元二次函数求解．本题考查利用换元法及配方法求函数的值域，是基础题．7.答案：C解析：【分析】本题考查直线和圆的位置关系，是基础题．由已知可得，圆心（0，0）到直线的距离d=，结合点到直线的距离公式可求k．【解答】解：∵y=kx+1与圆x2+y2=1过点（0，1），设P（0，1），∵△POQ为等边三角形，边长为1，∴圆心（0，0）到直线的距离d=，解可得，k=，故选C．8.答案：B解析：解：三视图复原几何体是一个组合体，上部是横卧的圆柱的一半，底面是一个半圆，其中半径为1，高为2的半圆柱；下部是正方体，棱长为：2，半圆柱的侧面积为π×1×2+π×12=3π，正方体部分的侧面积为2×2×5=20，所以组合体的表面积为20+3π（cm2）．故选：B．三视图复原几何体是一个组合体，上部是圆柱的一半，下部是正方体，根据三视图的数据，求出几何体的表面积．本题考查由三视图求组合体的表面积，考查空间想象能力，计算能力，是基础题．9.答案：A解析：解：如图正方形的边长为2，图中白色区域是以AB为直径的半圆，当P落在半圆内时，∠APB＞；当P落在半圆上时，∠APB=；当P落在半圆外时，∠APB＜．故使∠APB＜的概率P==1-．故选：A．由题意画出图形，再由测度比是面积比得答案．本题考查几何概型概率的求法，明确P点的位置是解答该题的关键，是基础题．10.答案：D解析：解：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：该程序的作用是计算分段函数f（x）=的函数值．可得当-∞＜x＜-2时，f（x）=2；当-2≤x≤1时，f（x）∈[，2]；当x=2时，f（x）=4；当x＞2时，f（x）=2；综上，可得输入的实数x的取值范围是（-∞，1]∪[2，+∞）时，输出的函数值f（x）取值范围是[，2]∪{4}．故选：D．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：该程序的作用是计算分段函数f（x）=的函数值，由已知分类讨论即可求解．本题考查的知识点是选择结构，其中根据函数的流程图判断出程序的功能是解答本题的关键，属于基本知识的考查．11.答案：C解析：解：由于函数f（x）=a sin x+b cos x=sin（x+∅），且f（）是它的最大值，∴+∅=2kπ+，k∈Z，∴∅=2kπ+，∴tan∅==1．∴f（x）=|a|sin（x+）．对于①，由于f（x-）=|a|sin x．是奇函数，故①正确；对于②，由于当x=时，f（x）=|a|，故函数f（x）的图象不关于x=对称，故②不正确；对于③，由于f（-）=|a|sin（-+）=-|a|，为函数f（x）的最小值，故③正确；对于④，函数f（x）的图象即把函数y=|a|sin x的图象向左平移个单位得到的，故|P2P4|等于一个周期2π，故④正确．故选：C．由题意可得f（x）=sin（x+∅），对于①，由于f（x-）=|a|sin x．是奇函数，可判断①；对于②，由于x=时，f（x）=|a|，可判断②；对于③，由f（-）=|a|sin（-+）=-|a|，是函数f（x）的最小值，可判断③；对于④，由题意可得，|P2P4|等于一个周期2π，可判断④．本题考查两角和正弦公式，正弦函数的最值，对称性，奇偶性，函数图象的变换，得到f（x）=|a|sin（x+）是解题的关键，属于中档题．12.答案：D解析：解：当-2≤x≤0时，f（x）=2x3+3x2+1．∴f′（x）=6x2+6x=6x（x+1）令f′（x）=0⇒x=0或x=-1；令f′（x）＞0⇒-2＜x＜-1；令f′（x）＜0⇒-1＜x＜0；且最大值为f（-1）=-2+3+1=2；f（-2）=-16+12+1=-3；f（0）=1；当0≤x≤2时，f′（x）=ae x，则若a＜0时，可得f′（x）＜0恒成立，即f（x）在（0，2）上单调递减且最大值为f（0）＜0，不存在有三个相异实根，故不成立舍掉；同理，当a=0时也不存在舍掉；即实数a必须大于0；故当a＞0时，f′（x）＞0恒成立，即f（x）在（0，2）上单调递增，若想f（x）=m有三个相异实根，必须满足⇒．故选：D．分情况讨论，通过函数的单调性求出满足条件的方程的充要条件，列出不等式求解即可得答案．本题考查了函数与方程的综合应用，直线与抛物线的关系的应用，属于中档题．13.答案：-解析：解：向量=（1，-2），=（t，3），若∥，则3×1-（-2）×t=0，解得t=-．故答案为：-．根据平面向量的共线定理，列方程求出t的值．本题考查了平面向量的共线定理应用问题，是基础题．14.答案：解析：解：依题意，数列{a n}是等比数列，设其公比为q，则a3•a5=64=，即a6=64=26，所以q=2或q=-2，所以a2019==22018，故答案为：22018．数列{a n}是等比数列，设其公比为q，则a3•a5=64=，即a6=64=26，所以q=2或q=-2，代入即可．本题考查了等比数列的通项公式，等比数列的性质，属于基础题．15.答案：-1解析：解：画出不等式组表示的平面区域，如图阴影所示；由图形知，当目标函数z=3x-2y过点A时，z取得最小值；由，求得A（1，2），所以z的最小值为z min=3×1-2×2=-1．故答案为：-1．画出不等式组表示的平面区域，结合图形找出最优解，计算目标函数的最小值．本题考查了不等式组表示平面区域的应用问题，也考查了数形结合思想，是中档题．16.答案：解析：解：由=4，得，其几何意义为平面内动点（x，2）与两定点（-3，0），（3，0）距离差的绝对值为4．平面内动点与两定点（-3，0），（3，0）距离差的绝对值为4的点的轨迹为．联立，解得x=．故答案为：．由=4，得，其几何意义为平面内动点（x，2）与两定点（-3，0），（3，0）距离差的绝对值为4．求出平面内动点与两定点（-3，0），（3，0）距离差的绝对值为4的点的轨迹方程，取y=2求得x 值即可．本题考查曲线与方程的求法，考查数学转化思想方法，是中档题．17.答案：解：（1）等差效列{a n}的公差设为d，且a1=1，S2+a2=4，可得1+1+d+1+d=4，解得d=，则a n=1+（n-1）=；（2）b n=2n+1•a n=（n+1）•2n，前n项和T n=2•2+3•4+4•8+…+（n+1）•2n，2T n=2•4+3•8+4•16+…+（n+1）•2n+1，相减可得-T n=4+4+8+16+…+2n-（n+1）•2n+1=2+-（n+1）•2n+1，化为T n=n•2n+1．解析：（1）等差效列{a n}的公差设为d，运用等差数列的通项公式，解方程可得d，进而得到所求通项公式；（2）求得b n=2n+1•a n=（n+1）•2n，运用数列的错位相减法求和，结合等比数列的求和公式，化简可得所求和．本题考查等差数列的通项公式，等比数列的求和公式，以及数列的错位相减法求和，考查方程思想和运算能力，属于中档题．18.答案：解：（1）将频率作为概率，根据以上数据，长绒棉占全部棉花的比例为P==60%，∴该基地的这批棉花符合“长绒棉占全部棉花的50%以上的要求“．（2）方案一：13.5×10000=135000．方案二：2×200+8×3800+15×4000+25×2000=140800．∴从采购商的角度，该用方案一．（3）用分层抽样的方法从长绒棉中抽取6根棉花，其中“军海1号”抽取到：6×=2，再从6根棉花中取两根进行检验，基本事件总数n==15，抽到的两根棉花只有一根是“军海1号”包含的基本事件个数m==8，∴抽到的两根棉花只有一根是“军海1号”的概率p=．解析：（1）将频率作为概率，能求出该基地的这批棉花符合“长绒棉占全部棉花的50%以上的要求“．（2）方案一：13.5×10000=135000．方案二：2×200+8×3800+15×4000+25×2000=140800．从采购商的角度，该用方案一．（3）用分层抽样的方法从长绒棉中抽取6根棉花，其中“军海1号”抽取到2根，再从6根棉花中取两根进行检验，利用古典概型能求出抽到的两根棉花只有一根是“军海1号”的概率．本题考查概率的求法，考查古典概型、排列组合等基础知识，考查运算求解能力，是基础题．19.答案：（1）证明：∵DE=2，PD=4，∠PDE=60°，∴PE==2，∴PE⊥DE．∵AE⊥平面PDE，PE⊂平面PDE，∴AE⊥PE，又AE∩DE=E，AE⊂平面ABCDE，DE⊂平面ABCDE，∴PE⊥平面ABCDE，又CD⊂平面ABCDE，∴PE⊥CD．（2）解：∵平面PAE∥平面DFGH，∴DF∥AE，PA∥GF，又BC∥AE，AB∥DE，∵AE⊥平面PDE，DE⊂平面PDE，∴AE⊥DE，∴四边形AEDF是矩形，∴V P-AEDF=S矩形AEDF•PE==．∵AP∥GF，∴P到平面DFGH的距离等于A到平面DFGH的距离，由（1）可知PE⊥平面ABCDE，故而PE⊥AF，又AF⊥AE，AE∩PE=E，∴AF⊥平面PAE，∴AF⊥平面DFGH，∵BC∥AE，DF∥AE，∴BC∥DF，又BC⊄平面DFGH，DF⊂平面DFGH，∴BC∥平面DFGH，又BC⊂平面PBC，平面PBC∩平面DFGH=GH，∴BC∥GH，∵AF=DE=AB，故F为AB的中点，∴G为PB的中点，∴H是PC的中点，∴GH=BC=1，又梯形DFGH的高为PE=，∴V P-DFGH=V A-DFGH=•AF=×（1+4）××2=．∴夹在该截面与平面PAE之间的几何体体积V=V P-AEDF+V P-DFGH=7．解析：（1）根据AE⊥DE，PE⊥DE可得PE⊥平面ABCDE，于是PE⊥CD；（2）求出梯形DFGH的面积，分别计算棱锥P-AEDF和棱锥P-DFGH的体积．本题考查了线面垂直的判定与性质，考查棱锥的体积计算，属于中档题．20.答案：解：（1）函数f（x）=x-1--ln x．若a=0，f（x）=x-1-ln x，f′（x）=1-，f（1）=0，f′（1）=0，f（x）在x=1处的切线方程为y=0，（2）证明：函数f（x）=x-1--ln x．f′（x）=，因为函数f（x）存在两个极值点x1和x2，所以f′（x）=0，x1=1，x2=，a∈（0，）∪（，1），f（x1x2）+=-2-ln，令t=，t∈（0，1）∪（1，+∞），h（t）=4t-ln t-2，h′（t）=4-=0，t=，所以y=h（t）在（0，）单调递减，在（，1），（1，+∞）单调递增；所以h（t）最小值为h（）=2ln2-1；即h（t）≥2ln2-1；即f（x1x2）+≥2ln2-1．解析：（1）将a=0代入函数，求函数的导数和函数的切点的坐标，利用点斜式可求f（x）在x=1处的切线方程；（2）函数f（x）存在两个极值点x1和x2，求证：f（x1x2）+≥2ln2-1．即证明f（x1x2）+=-2-ln≥2ln2-1，令t=，t∈（0，1）∪（1，+∞），转换成新函数h（t）=4t-ln t-2≥2ln2-1，即求函数h（t）的最小值大于等于2ln2-1即可；本题考查了导数的综合应用，属于中档题．21.答案：解：（1）如图，可得QP=QR，所以QM+QP=QM+QR=MR=8＞MP=4，所以Q点的轨迹是以M，P点为焦点的椭圆，其中a=4，c=2，所以b2=12，故点Q的轨迹方程为；（2）由（1）可知左焦点（-2，0），且AC⊥BD，①当直线AC、BD中有一条直线的斜率不存在时，||+||=6+8=14；②当直线AC的斜率为k，k≠0，其方程为：y=k（x+2），联立，得（3+4k2）x2+16k2x+16k2-48=0，设A（x1，y1），C（x2，y2），则x1+x2=-，x1x2=，所以==，同理可得：=，所以||+||=，令1+k2=t（t＞1），||+||==∈[，14），综上，||+||的取值范围是[，14]．解析：（1）根据题意画出图象，可得QM+QP＞MP，即可知Q点的轨迹是以M，P点为焦点的椭圆；（2）由条件可判断出AC、BD过椭圆左焦点，分别讨论AC、BD斜率存在与不存在的情况，表示出||+||，即可求出取值范围．本题考查点的轨迹方程，利用数形结合判断出轨迹为椭圆是关键，属于中档题．22.答案：解：（1）曲线C2的极坐标方程为．利用三角函数的展开式，转换为直角坐标方程为（x-1）2+（y-1）2=4，（2）曲线C1的参数方程为（t为参数，0≤α＜π），转换为直角坐标方程为y+1=k（x-1），（k=tanα），所以圆心（1，1）到直线l的距离d=，所以，解得k=，所以．解析：（1）直接利用转换关系，把极坐标方程转换为直角坐标方程．（2）利用勾股定理和点到直线的距离公式的应用求出直线的斜率，进一步求出直线的倾斜角．本题考查的知识要点：参数方程极坐标方程和直角坐标方程之间的转换，点到直线的距离公式的应用，勾股定理的应用，主要考查学生的运算能力和转换能力，属于基础题型．23.答案：解：（1）∵f（x）≤1的解集为{x|0≤x≤2}，∴0和2为方程|x-a|=1的两实根，∴|a|=1且|2-a|=1，∴a=1，∴a的值为1；（2）证明：当x∈（0，1）时，=====4，当且仅当即x=时取等号，∴．解析：（1）由f（x）≤1的解集为{x|0≤x≤2}，可知0和2为方程|x-a|=1的两实根，将0和2代入方程|x-a|=1中可求出a的值；（2）由题意可得=，利用基本不等式可得的最小值，从而证明≥4．本题考查了不等式的解集与方程根之间的关系，基本不等式和利用综合法证明不等式，考查了方程思想和转化思想，属中档题．。
2019届黑龙江省哈尔滨市三中高三三模理综物理试题
A 错误
B．从图示位置转出磁场的过程中， 穿过线圈的磁通量减小， 根据楞次定律可知， 线圈中产生顺时针方向的
感应电流， B 错误
C．线圈中产生的感应电动势的最大值满足
Em NBS ，带入数据可得感应电动势的最大值为
C 正确
3BL2 ， 4
D．外力做 功转化为线圈的焦耳热，所以外力的功率等于线圈的热功率，可得
1
期质量变为原来的 ， C 错误
4
D．半衰期不会随着温度而改变， D 错误
所以经过两个半衰
2. 初速度为零的某质点的加速度随时间变化的规律如图所示，则下列说法正确的是（
）
A. 质点做匀加速直线运动 B. 第 1s 内与第 3s 内质点动量变化量之比为 1： 3 C. 第 1s 内与第 3s 内质点动量变化率之比为 1： 3 D. 第 1s 内与第 3s 内质点通过的位移之比为 1： 3 【答案】 D
之比为 1：1， C 错误
D．前 3s 物体先做初速度为 0 的匀加速直线，再做匀速直线，接着做匀加速直线，将中间的匀速去掉，可
知第 1s 与第 3s 组成一个连续的匀加速直线运动，且初速度为
0，通过计算可知第 1s 内与第 3s 内质点通
过的位移之比为 1：3， D 正确
3. 如图所示， 长为 L=5m的水平传送带以 v0=6m/s 逆时针匀速转动， 质量 m=1kg 的物块以水平初速度 v=4m/s
滑上传送带，物块与传送带间的动摩擦因数为
0.2 ，不考虑传送带轮轴摩擦等能量损失，重力加速度
g=10m/s 2，则物块从滑上到离开传送带的过程中，下列说法正确的是（
）
A. 物块离开传送带时的速度大小为 B. 摩擦力对物块一直做负功
精选黑龙江省哈尔滨市2019年高三模拟试题(三)英语试卷(有答案)
2019年哈尔滨市高考模拟试题（三）英语第二部分阅读理解（共两节, 满分40分）第一节（共15小题；每小题2分，满分30分）阅读下列短文，从每题所给的A、B、C和D四个选项中，选出最佳选项。
AGuide dogs enjoy their work very much, and they get a lot of satisfaction from a job well done, but there is no room for typical dog fun during the work day. Games, treats and praise distract the dog from helping its handler navigate (指引方向) the course. Ev en when the handler doesn’t need assistance, a guide dog on the job is trained to ignore distractions and keep still. This is because a guide dog must be able to come to handler’s workplace or be in public places without creating a disturbance(干扰).When you see a guide dog on the job, it is extremely important that you recognize that it is at work. Petting or talking to the dog spreads its concentration, which weakens the handler’s ability to get around in his or her surroundings. People are very impressed with guide dogs and so we have a natural inclination（倾向）to praise them, but the best thing you can do to help a guide dog is to leave it alone so that it can pay attention to its surroundings and maintain its focus on its handler. Guiding is very complicated, and it requires a dog’s undivided attention.When a guide dog gets home at the end of the day, however, it will play and soak up praise just like an ordinary pet. Guide dogs make the difference between work and play based on their lead harness（牵狗用的皮带）：When the harness is on, they must stay completely focused---- when it comes off, it’s play time. Guide dogs work very hard every day, but they lead extremely happy lives, full of lots of attention and stimulation.Working as a guide dog requires good physical and mental shape, so guide dogs typically retire just before they enter old age. Retirement is usually at age 8 or 10; but some work for a little longer, and some guide dogs retire earlier if they’re having trouble with the work.21. What do you think is the typical job of a guide dog?A.Help the policemen to search for the murderer.B.Help the blind or weak-sighted people get around .C.Help the doctor to save the injured people.D.Help the owner to watch their house.22. What does the writer want to tell the readers through the passage?A.Guide dogs are so helpful that we should make friends with them.B.Guide dogs are different from other dogs and they have some special talents.C.Guide dogs’ life style and job situation.D.Guide dogs’ trained way by their handlers.23. What does the underlined word “distract” mean in the passage?A.to stop the dog paying attention to what he is doingB.to make the dog think very carefully about what he is doingC.to interrupt the dog so that he cannot do the work wellD.to stop the dog from doing something24. What should you do when meeting with a guide dog at its job?A. Praise it.B. Feed it.C. Play with it.D. Leave it alone.BGriffith Observatory (天文台) is a national leader in public astronomy, and one of the most popular attractions in Los Angeles. It is located on the southern slope (山坡) of Mount Hollywood in Griffith Park at 1,134 feet above sea level.Visitors may drive to the Observatory and park in its parking lot or on nearby roads. No reservation (预定) is required to visit. Parking is limited, and the busiest times are weekend. Buses, taxis, and carpools are welcome. LADOT provides weekend public bus service from the Sunset/Vermont Metro Red Line station.Griffith Observatory is open six days a week. Admission and parking are free.Hours of OperationTuesday---Friday 12:00 noon---10:00 p.m.Saturday---Sunday 10:00 a.m.---10:00 p.m.Monday ClosedThanksgiving Day and Christmas Day ClosedSamuel Oschin PlanetariumThe Samuel Oschin Planetarium theater offers 8 to 10 live, half-hour presentations each day. There are usually four different shows from which to choose.Shows are presented every 60 to 90 minutes. Check the website, information desks, or the box office for each day’s show times. Late seating is not permitted.Samuel Oschin Planetarium tickets must be purchased at the Observatory and are only available on the day of the show. The ticket prices for shows are：Adults (13-59 years old) $7.00Children (5-12 years old) $3.00Seniors (60 years and older) $5.00Students $5.00Children under 5 years will be admitted only to the first show each day.Hearing assist devices are available upon request.Public TelescopesFree public telescopes are available each evening the Observatory is open and skies are clear. The Zeiss telescope on the roof is generally open by 7:00 p.m.. All observing must be completed by 9:45 p.m..25. What can we learn about Griffith Observatory according to the passage?A. It is open all the year around.B. It is mostly visited at weekends.C. It becomes famous because of Hollywood.D. It is the most attractive place in Los Angeles.26. Which of the following best describes the Samuel Oschin Planetarium theater?A. Hearing assist devices are provided to seniors.B. The shows there generally last 60 to 90 minutes.C. Visitors are required to be seated before the show starts.D. The tickets for its shows can be bought through the website.27. To watch the show, a young couple with a 7-year-old son should pay________.A. $13.00B. $15.00C. $17.00D. $20.0028. Which of the following is charged?A. Parking.B. Telescopes.C. Hearing assist devices.D. LADOT bus service.CWhen we hear of dangerous diseases, cancer and heart disease are often what first cometo mind. Recently, a virus called Ebola has reappeared in Guinea, killing 62 people so far. It leads to high fever, bleeding and so on. Ebola can kill 90% of those infected (感染), especially in underdeveloped societies like those in Africa.Ebola is named after the Ebola River, where it was first discovered in 1976. There are five different types of the Ebola virus, each named after where they first happened: Sudan, Ivory Coast, Reston, Bundibugyo, and Zaire. The deadliest of the five, Zaire, was responsible for the 2012 outbreak, and is believed to be attacking Guinea.Ebola is naturally found in fruit bats, which pass on the virus to other animals by biting or sucking on their blood. Humans who are suffering from the Ebola infection might have touched the bodily fluids (体液) of the infected animals. Once infected, a human becomes a carrier of the deadly virus.Since we live in an interconnected world, where the situation in one country can affect us all, the effects of Ebola are huge: damaging trade relations, affecting foreign visitors, and weakening entire countries. Already the Guinea virus is spreading fast with a few cases spotted in Conakry—the capital city of Guinea, far away from the origin of the virus. It is feared that the disease may have already reached neighboring Liberia and Sierra Leone, too.Unfortunately, there are no disease-specific treatments for Ebola. Healthcare workers only supplythe infected people with water to keep them in good condition. Since there have been many cases of nurses catching the disease from patients, they are forced to wear strict protective clothes, and in some cases, not even allowed to get close to the infected. The fact that there is no cure for the Ebola virus is what makes the outbreak a challenging one to control.29.From the text we know that ________.A. Ebola is naturally found in fruitB. Ebola is a highly infectious virusC. Ebola is a newly discovered diseaseD. Ebola is recently controlled in Guinea30. What is Paragraph 4 mainly about?A. The causes of Ebola.B. The origin of Ebola.C. The treatment for Ebola.D. The influences of Ebola.31. In Ebola infected areas, healthcare workers _______.A. face the danger of catching virusB. take measures to protect the nursesC. give water and medicine to the patientsD. do whatever they can to save the infected32.The text probably comes from ________.A. an English newspaperB. a tourist guideC. a safety brochureD. a medical reportDThere is an English saying : “ Laughter is the best medicine.” Until recently, few people took the saying very seriously. Now however, doctors have begun to study laughter and the effects it has on the human body. They have found facts that laughter really can improve people’s health.Tests were carried out to study the effects of laughter on the body. People watched funny films while doctors checked their heart rate, blood pressure, breathing and muscles. It was found that laughter has similar effects to physical exercise. It increases blood pressure, the heart rate and the rate of breathing; it also works several groups of muscles in the face, the stomach, and even the feet. If laughter exercises the body, it must be beneficial.Other tests have shown that laughter appears to be able to reducing the effect of pain on the body. In one experiment doctors produce pain on the body. In one experiment doctors produce pain in groups of students who listened to different radio programs. The group which tolerated(忍受) the pain for the longest time was the groups which listened to a funny program. The reason why laughter can reduce pain seems to be that it helps to produce endorphins in the brain. These are natural chemicals which diminish both stress and pain.There is also some fact to suggest that laughter helps the body’s immune system, that is, the systemwhich fights infection. In an experiment, one group of students watched a funny video while another group of students served as the control group ---- in other words, a group with which to compare the first group. Doctors checked the blood of the students in both groups and found that the people in the group that watch the video had an increase in the activity of their white blood cells, that is, the cells which fight infection.As a result of these discoveries, some doctors and psychiatrists in the United States now hold laughter clinics, in which they t ry to improve their patients’ condition by encouraging them to laugh.They have found that even if their patients do not really feel like laughing, making them smile is enough to produce beneficial effects similar to those caused by laughter.33. It can be learnt from the passage that laughter can_______.A. make people feel youngerB. change people’s habitsC. improve peoples healthD. make people love their lives34. Smiling can produce____.A. more effects than laughterB. the same effects as laughterC. less effects on the human bodyD. no effects on the human body35. The main idea of this passage is that ______.A. there are several ways of studying the benefits of laughterB. laughter and physical exercise have the same effects on human bodyC. the doctors should learn how to make people laughD. tests show that laughter can produce beneficial effects on human body第二节（共5小题；每小题2分，满分10分）根据短文内容，从短文后的选项中选出能填入空白处的最佳选项。
黑龙江省哈尔滨市2019届高三模拟试卷(三)英语试卷【含答案及解析】
黑龙江省哈尔滨市2019届高三模拟试卷（三）英语试卷【含答案及解析】姓名___________ 班级____________ 分数__________一、阅读理解1. Guide dogs enjoy their work very much, and they get a lot of satisfaction from a job well done, but there is no room for typical dog fun during the work day. Games, treats and praise distract the dog from helping its handler navigate ( 指引方向 ) the course. Even when the handler doesn’t need assistance, a guide dog on the job is trained to ignore distractions and keep still. This is because a guide dog must be able to come to handler’s workplace or be in public places without creating a disturbance( 干扰 ).When you see a guide dog on the job, it is extremely important that you recognize that it is at work. Petting or talking to the dog spreads its concentration, which weakens the handler’s ability to get around in his or her surroundings. People are very impressed with guide dogs and so we have a natural inclination （倾向） to praise them, but the best thing you can do to help a guide dog is to leave it alone so that it can pay attention to its surroundings and maintain its focus on its handler. Guiding is veryc omplicated, and it requires a dog’s undivided attention.When a guide dog gets home at the end of the day, however, it will play and soak up praise just like an ordinary pet. Guide dogs make the difference between work and play based on their lead harness （牵狗用的皮带）： When the harness is on, they must stay completely focused---- when it comes off, it’s play time. Guide dogs work very hard every day, but they lead extremely happy lives, full of lots of attention and stimulation.Working as a guide dog requires good physical and mental shape, so guide dogs typically retire just before they enter old age. Retirement is usually at age 8 or 10; but some work for a little longer, and some guide dogs retire earlier if they’re having trouble with the work.1. What do you think is the typical job of a guide dog?A. Help the policemen to search for the murderer.B. Help the blind or weak-sighted people get around .C. Help the doctor to save the injured people.D. Help the owner to watch their house.2. What does the writer want to tell the readers through the passage?A. Guide dogs are so helpful that we should make friends with them.B. Guide dogs are different from other dogs and they have some special talents.C. Guide dogs’ life style and job situati on.D. Guide dogs’ trained way by their handlers.3. What does the underlined word “distract” mean in the passage?A. to stop the dog paying attention to what he is doingB. to make the dog think very carefully about what he is doingC. to interrupt the dog so that he cannot do the work wellD. to stop the dog from doing something4. What should you do when meeting with a guide dog at its job?A. Praise it.B. Feed it.C. Play with it.D. Leave it alone.2. Griffith Observatory ( 天文台 ) is a national leader in public astronomy, and one of the most popular attractions in Los Angeles. It is located on the southern slope ( 山坡 ) of Mount Hollywood in Griffith Park at 1,134 feetabove sea level.Visitors may drive to the Observatory and park in its parking lot or onnearby roads. No reservation ( 预定 ) is required to visit. Parking is limited, and the busiest times are weekend. Buses, taxis, and carpools are welcome. LADOT provides weekend public bus service from the Sunset/Vermont Metro RedLine station.Griffith Observatory is open six days a week. Admission and parking are free. Hours of OperationTuesday---Friday 12:00 noon---10:00 p.m.Saturday---Sunday 10:00 a.m.---10:00 p.m.Monday ClosedThanksgiving Day and Christmas Day ClosedSamuel Oschin PlanetariumThe Samuel Oschin Planetarium theater offers 8 to 10 live, half-hour presentations each day. There are usually four different shows from which to choose.Shows are presented every 60 to 90 minutes. Check the website, informationdesks, or the box office for each day’s show times. Late seating is not permitted.Samuel Oschin Planetarium tickets must be purchased at the Observatory andare only available on the day of the show. The ticket prices for shows are：Adults (13-59 years old) $7.00Children (5-12 years old) $3.00Seniors (60 years and older) $5.00Students $5.00Children under 5 years will be admitted only to the first show each day.Hearing assist devices are available upon request.Public TelescopesFree public telescopes are available each evening the Observatory is open and skies are clear. The Zeiss telescope on the roof is generally open by 7:00p.m.. All observing must be completed by 9:45 p.m..1. What can we learn abou t Griffith Observatory according to the passage?A. It is open all the year around.B. It is mostly visited at weekends.C. It becomes famous because of Hollywood.D. It is the most attractive place in Los Angeles.2. Which of the following best describes the Samuel Oschin Planetarium theater?A. Hearing assist devices are provided to seniors.B. The shows there g enerally last 60 to 90 minutes.C. Visitors are required to be seated before the show starts.D. The tickets for its shows can be bought through the website.3. To watch the show, a young couple with a 7-year-old son should pay________.A. $13.00B. $15.00C. $17.00D. $20.004. Which of the following is charged?A. Parking.B. Telescopes.C. Hearing assist devices.D. LADOT bus service.3. When we hear of dangerous diseases, cancer and heart disease are often what first cometo mind. Recently, a virus called Ebola has reappeared in Guinea, killing 62 people so far. It leads to high fever, bleeding and so on. Ebola can kill 90%of those infected ( 感染 ), especially in underdeveloped societies like thosein Africa.Ebola is named after the Ebola River, where it was first discovered in 1976.There are five different types of the Ebola virus, each named after where they first happened: Sudan, Ivory Coast, Reston, Bundibugyo, and Zaire. The deadliest of the five, Zaire, was responsible for the 2012 outbreak, and is believed to be attacking Guinea.Ebola is naturally found in fruit bats, which pass on the virus to other animals by biting or sucking on their blood. Humans who are suffering from the Ebola infection might have touched the bodily fluids ( 体液 ) of the infected animals. Once infected, a human becomes a carrier of the deadly virus.Since we live in an interconnected world, where the situation in one country can affect us all, the effects of Ebola are huge: damaging trade relations, affecting foreign visitors, and weakening entire countries. Already the Guinea virus is spreading fast with a few cases spotted in Conakry—the capital city of Guinea, far away from the origin of the virus. It is feared that the disease may have already reached neighboring Liberia and Sierra Leone, too.Unfortunately, there are no disease-specific treatments for Ebola. Healthcare workers only supply the infected people with water to keep them in good condition. Since there have been many cases of nurses catching the disease from patients, they are forced to wear strict protective clothes, and in some cases, not even allowed to get close to the infected. The fact that there is no cure for the Ebola virus is what makes the outbreak a challenging one to control.1. From the text we know that ________.A. Ebola is naturally found in fruitB. Ebola is a highly infectious virusC. Ebola is a newly discovered diseaseD. Ebola is recently controlled in Guinea2. What is Paragraph 4 mainly about?A. The causes of Ebola.B. The origin of Ebola.C. The treatment for Ebola.D. The influences of Ebola.3. In Ebola infected areas, healthcare workers _______.A. face the danger of catching virusB. take measures to protect the nursesC. give water and medicine to the patientsD. do whatever they can to save the infected4. The text probably comes from ________.A. an English newspaperB. a tourist guideC. a safety brochureD. a medical report4. There is an English saying : “ Laughter is the best medicine.” Until recently, few people took the saying very seriously. Now however, doctors have begun to study laughter and the effects it has on the human body. They have found facts that laughter really can improve people’s health.[Tests were carried out to study the effects of laughter on the body. People watched funny films while doctors checked their heart rate, blood pressure, breathing and muscles. It was found that laughter has similar effects to physical exercise. It increases blood pressure, the heart rate and the rate of breathing; it also works several groups of muscles in the face, the stomach, and even the feet. If laughter exercises the body, it must be beneficial.Other tests have shown that laughter appears to be able to reducing theeffect of pain on the body. In one experiment doctors produce pain on the body. In one experiment doctors produce pain in groups of students who listened to different radio programs. The group which tolerated( 忍受 ) the pain for the longest time was the groups which listened to a funny program. The reason why laughter can reduce pain seems to be that it helps to produce endorphins inthe brain. These are natural chemicals which diminish both stress and pain.There is also some fact to suggest that laughter help s the body’s immune system, that is, the system which fights infection. In an experiment, onegroup of students watched a funny video while another group of students served as the control group ---- in other words, a group with which to compare thefirst group. Doctors checked the blood of the students in both groups andfound that the people in the group that watch the video had an increase in the activity of their white blood cells, that is, the cells which fight infection. As a result of these discoveries, some doctors and psychiatrists in theUnited States now hold laughter clinics, in which they try to improve their patients’ condition by encouraging them to laugh.They have found that even if their patients do not really feel like laughing, making them smile is enough to produce beneficial effects similar to those caused by laughter.1. It can be learnt from the passage that laughter can_______.A. make people feel youngerB. change people’s habitsC. improve peoples healthD. make people love their lives2. Smiling can produce____.A. more effects than laughterB. the same effects as laughterC. less effects on the human bodyD. no effects on the human body3. The main idea of this passage is that ______.A. there are several ways of studying the benefits of laughterB. laughter and physical exercise have the same effects on human bodyC. the doctors should learn how to make people laughD. tests show that laughter can produce beneficial effects on human body二、七选五5. Thi nk for a moment about the teachers you’ve had at junior or senior high school. Which one did you like best? And why? Were the teachers you liked best also the ones who were the best teachers, in your opinion? 1. And then begin to read further.Some very common answers to this question are that teachers need to lovetheir students, that they need to have expert knowledge of their subjects and that they should devote themselves completely to their work. All of these ideas are, of course, true to a certain extent. 2.It’s impossible for anyone to love everyone he knows, and teachers deal with a very large number of students over the years. On the other hand, teachers should certainly be able to make their students feel that they’re interested in them as people. 3. A deep knowledge of the subject is especially important. 4. That’s to say, a teacher needs to be trained in the skills of teaching. These skills include how to control a class. Finally, teachers have to devote a lot of time and energy to their work, of course. However, because they’re also models that their students must follow, it’s important that they should be well-balanced people with interests outside their school work—families, friends, hobbies, etc. 5.A.They’re perhaps a little too simp le.B.Students ask too much from teachers.C.Well- qualified teachers should be educated and capable.D.Consider for a minute the qualities that make a teacher outstanding.E.A teacher who only lives for work is likely to become narrow-minded.F.Equally important is the ability to pass that knowledge on to the students effectively.G.They also pay attention to the development of both their brains and their characters.三、完形填空6. About 10 years ago I was having my annual party and my niece came to see me. As she ______ around the room, she noted that my employees seemed happy. Then, I asked her how she thought I did that. “I’m sure you treat them______ ,”she replied. “That’s half of it,” I said. “Do you know ______ the other half is?”She didn’t have the answer. So what’s the answer? I ______ the unhappy people. People laughed at this point. I wish I were ______ . I’m not. I have learned that as a manager you cannot make everyone happy. Good ______requires training, communicating and patience.Don’t ______ me wrong. This doesn’t happen a lot. There’s no joy in the______ of firing someone. And it’s not always the employee’s ______ —there are many bad bosses out there. And not all employees ______ your company. I don’t have a Ph.D., an M.B.A, or even an economics degree. What I do have is a ______ company. Now I know some people argue that business is_______ making money, and not everyone has to be happy. When you _____ a company, you have the right to ______ yourself with the people you choose.I have a good day today. Not ______ I’ve got a big order or great _______ reports. I have wonderful people working for me. They care. They arecommitted( 承担 ). ______ , they understand the whole customer-staff-company triangle, where all of the legs ________ each other. When you have the right people, business is much ________ . I know because I have ________ it.1. A. showed B. looked C. turned D. came2. A. nicely B. firmly C. strictly D. seriously3. A. which B. that C. what D. who4. A. trick B. hire________ C. dislike D. fire5. A. crying B. joking C. cheating D. regretting6. A. assessment B. agreement C. management D. employment7. A. bring B. think C. treat D. get8. A. way B. chance C. act D. form9. A. decision B. failure C. fortune D. fault10. A. suit B. believe C. choose D. understand11. A. funny B. large C. happy D. busy12. A. with B. beyond C. into D. about13. A. leave B. join C. own D. share14. A. relax B. help C. amaze D. surround15. A. until B. because C. after D. that16. A. physical B. written C. technical D. financial17. A. Besides B. However C. Otherwise D. Therefore18. A. support B. fight C. hold D. hurt19. A. busier B. easier C. safer D. higher20. A. made B. forgotten C. got D. put四、短文填空7. What kind of music do you like 1. (good)? My 2. (favour) is pop music. It really took off in the 1950s, when the electric guitar 3. (invent). Whe n someone had the bright idea of putting a microphone inside the guitar, a whole new range 4. sounds became possible. The origins of pop music lie in jazz—a type of music which became 5. (increase) popular among young people. Thefirst real pop superstar is Elvis Presley, 6. king of rock music. He was 7. (follow) in the 1960s and 70s by the rock supergroups. And I was still alittle child when the Beatles had the first hit of 68 8. (their). It was amazing! Suddenly, pop music came to life and everyone wanted to be part of it. After the Beatles came the heavy rock: the Rolling Stones, Led Zeppelin, and Queen, and then rock music 69 9. (spread) across the world. Young people 10. live in the countries as far apart as Canada and Japan have been influenced by American and British pop music for over 40 years now.五、短文改错8. 假定英语课上老师要求同桌之间交换修改作文，请你修改你同桌写的以下作文。
2019届黑龙江省哈尔滨市第三中学校高三第三次模拟数学(理)试题附答案
2019届黑龙江省哈尔滨市第三中学校高三第三次模拟数学（理）试题一、单选题1．设命题p ：x ∀∈R ，2320x x -+≤，则p ⌝为（ ）A ．0x ∃∈R ，200320x x -+≤ B ．x ∀∈R ，320x x -+> C ．0x ∃∈R ，200320x x -+>D ．x ∀∈R ，320x x -+≥【答案】C【解析】根据含有量词的命题的否定即可得到结论． 【详解】命题为全称命题,命题p ：x ∀∈R ，2320x x -+≤，则￢p 为0x ∃∈R ，200320x x -+>，故选：C . 【点睛】本题考查命题的否定，对于全称命题的否定，先否定量词，再否定结论即可，属于基础题. 2．若复数31ai-2，则实数a =（ ） A ．2 B ．2-C ．±1D ．2±【答案】D【解析】利用复数代数形式的乘除运算化简，再由复数模的计算公式列式求解． 【详解】 ∵()()()3111+1122a i a a a ai i i i i -===--+-， 2222222a a ⎛⎫⎛⎫+-== ⎪ ⎪⎝⎭⎝⎭， 即a =2±. 故选：D . 【点睛】本题考查复数的运算及向量的模的计算，属于基础题.3．已知集合202xA x Zx ⎧⎫+=∈≥⎨⎬-⎩⎭，{}22B x Z y x x =∈=-++，则A B I 中的元素个数为（ ） A ．2 B ．3C ．4D ．5【答案】B【解析】可以求出集合A ，B ，然后进行交集的运算求出A ∩B ，从而得出A ∩B 中的元素个数． 【详解】由题意A ={x ∈Z |−2≤x <2}={−2,−1,0,1},B ={x ∈Z ∣−x 2+x +2≥0}={x ∈Z |−1≤x ≤2}={−1,0,1,2}， ∴A ∩B ={−1,0,1}， ∴A ∩B 元素个数为3. 故选：B . 【点睛】本题考查交集及其运算，利用分式不等式及函数定义域的求法求解，注意题目中的条件，属于基础题. 4．已知随机变量X 服从二项分布(),B n p .若()2E X =，()43D X =，则p =（ ） A ．34B ．23C ．13D ．14【答案】C【解析】由随机变量X 服从二项分布B （n ，p ），结合期望及方差的公式运算即可得解． 【详解】由随机变量X 服从二项分布B (n ,p ).又E (X )=2, ()43D X =， 所以np =2，np (1−p )= 43，解得：p =13，故选：C . 【点睛】本题考查二项分布与n 次独立重复试验的模型，运用二项分布的期望及方差的公式运算即可求解，属于基础题. 5．为培养学生的综合素养，某校在高二年级开设了A ，B ，C ，D 四门选修课，并规定每位学生必须从这四门课中选修一门，则甲、乙、丙3名同学所选课程互不相同的概率为（ ） A ．38B ．827C ．116D ．481【答案】A【解析】基本事件总数n =34，甲、乙、丙3名同学所选课程互不相同包含的基本事件个数m =34A ，由此能求出甲、乙、丙3名同学所选课程互不相同的概率． 【详解】为培养学生的综合素养，某校在高二年级开设了A ，B ，C ，D 四门选修课， 并规定每位学生必须从这四门课中选修一门， 基本事件总数n =34，甲、乙、丙3名同学所选课程互不相同包含的基本事件个数m =34A ，则甲、乙、丙3名同学所选课程互不相同的概率为343348n A m p ===.故选：A . 【点睛】本题考查古典概型及其概率计算公式，属于简单题.6．《张丘建算经》是我国北魏时期大数学家张丘建所著，约成书于公元466~485年间.其中记载了这样一道题：今有女子善织，日自倍，五日织五尺，问日织几何？译文是：今有一妇女善于织布，每天织的布都是前一天的2倍，已知她5天里共织布5尺.问这位妇女每天织布多少？在该问题中，若设此女子第n 天织布n a 尺，则135a a a ++=（ ） A ．7 B ．10531C ．73D ．3531【答案】B【解析】推导出数列{a n }是首项为a 1，公比为2的等比数列，由则()51512512a S -==-，解得1531a =，由此能求出a 1+a 3+a 5． 【详解】设此女子第n 天织布a n 尺，则数列{a n }是首项为a 1，公比为2的等比数列， 则()51512512a S -==-，解得1531a =， ∴()2413511151052214163131a a a a a a ++=++=++=， 故选：B ． 【点睛】本题考查等比数列的前n 项和及通项公式的运用，根据题意列出方程计算即可求解，属于基础题. 7．已知函数()()cos f x A x ωϕ=+（0A >，0>ω，2πϕ≤）的部分图象如图所示，现将函数()f x 图象上的所有点的横坐标缩短为原来的12倍，得到函数()g x 的图象，则函数()g x 的解析式为（ ）A ．()2cos 24g x x π⎛⎫=-⎪⎝⎭B ．()2cos 8g x x π⎛⎫=-⎪⎝⎭C ．()2cos 42g x x π⎛⎫=-⎪⎝⎭ D ．()2cos 44g x x π⎛⎫=- ⎪⎝⎭【答案】D【解析】由函数f （x ）的图象求得A 、T 、ω和φ的值，写出f （x ）的解析式，再根据图象变换法则写出函数g （x ）的解析式． 【详解】由函数f (x )=Acos (ωx +φ)的图象知，最大值为2， A =2,周期为73288T πππ⎛⎫=⨯-= ⎪⎝⎭， ∴22Tπω==； 由3282ππϕ⨯+=,解得4πϕ=-，∴()2cos 24f x x π⎛⎫=-⎪⎝⎭； 将函数f (x )图象上的所有点的横坐标缩短为原来的12倍， 得到函数g (x )的图象,则函数()2cos 44g x x π⎛⎫=-⎪⎝⎭.故选：D . 【点睛】本题考查函数y =Asin （ωx +φ）的图象变换，由y =Asin （ωx +φ）的部分图象确定其解析式，A 、ω和φ的值分别由最值、周期、定点来确定求解，属于中等题.8．已知1a =r ，22=b r 0a b ⋅=r r ，=+r r r c a b ，1-=ur r d c ，则u r d 的取值范围是（ ）A ．[]0,3B ．[]2,3C ．[]2,4D ．[]3,4【答案】C【解析】由题意对=+r r r c a b 进行平方可计算||3c =r ，设c r ，d r 的夹角为α，由1d c -=r r 得2||8cos 6||d d α+=r r ，由1cos 1α-≤≤可得d 的取值范围.【详解】∵1a =r ，22=b r 0a b ⋅=r r ，=+r r r c a b ，1-=ur r d c ， 2222()2189c a b a b a b ∴=+=++⋅=+=r r rr r r r ，||3c ∴=r ，设c r ，d r 的夹角为α，由1d c -=r r 得22||2||1d c d c -⋅+=r r r r，∴2||6||cos 91d d α-+=r r，2||8cos 6||d d α+∴=r r ， 设||d x =r，由1cos 1α-≤≤得28116x x+-≤≤，解得242x x ≤≤⎧⎨≥-⎩或244x x ≤≤⎧⎨≤-⎩，24x ∴≤≤，综上所述，故选：C . 【点睛】本题考查平面向量数量积的性质及其运算，涉及知识点有向量的模、向量的垂直关系、向量夹角等，属于中等题.9．函数()2ln 4x f x x =-的图象大致为（ ）A ．B ．C ．D ．【答案】C【解析】利用函数的单调性排除A ，B ，通过函数的最值排除D ，推出结果即可． 【详解】根据函数()2ln 4x f x x =-，可得()()212022x x f x x x x-'=-=>，由f ′(x )>0,得x 2,即函数f (x )在(2,+∞)上单调递增， 由f ′(x )<0得0<x 2,即函数f (x )在2)上单调递减， 可以排除A 、B ，当x 2时,函数f (x )有最小值,f (x )min =f 2)=11ln 2022->， 于是对任意的x ∈(0,+∞),有f (x )>0，故排除D ， 故选：C . 【点睛】本题考查利用导数研究函数的单调性、函数的图象，常用的方法为排除法，可依据函数的单调性、奇偶性、定义域、值域、零点、特殊值等进行排除，属于中等题.10．已双曲线22134x y -=的一条渐近线与椭圆C ：22221x y a b+=（0a b >>）在第一象限的交点为P ，1F ，2F 为椭圆C 的左、右焦点，若1260F PF ∠=︒，则椭圆C 的离心率为（ ）A ．33B ．32C ．13D ．23【答案】A【解析】求得双曲线的渐近线方程，联立椭圆方程，求得P 的坐标，设|PF 1|=m ，|PF 2|=n ，运用椭圆的定义和三角形的余弦定理和面积公式可得12231602PF F S mnsin b =︒=V ,又122212234PF F PS c y b a =⋅⋅=+V ，结合a ，b ，c 的关系和离心率公式可得所求值. 【详解】设双曲线22134x y -=的一条渐近线方程为3y x =， 代入椭圆方程可得2222,3344P b a b a⎛⎫⎪⎪ ⎪+ ⎝⎭+⎪， 设|PF 1|=m ，|PF 2|=n ，可得m +n =2a ， 由余弦定理可得（2c ）2=m 2+n 2-2mncos 60°， 化为（m +n ）2-2mn -mn =4c 2，即为mn =243b则12231602PF F S mnsin b =︒=V ,又122212234PF F P S c y b a =⋅⋅=+V , 可得222433b b a=+，结合b 2=a 2-c 2， 化为7a 4-22c 2a 2+3c 4=0， 可得a 2=3c 2或c 2=7a 2（舍去）， 则e =3=c a , 故选：A ． 【点睛】本题考查椭圆的性质，双曲线的性质，圆锥曲线的综合应用，涉及到知识点多且计算量大，属于较难题. 11．图中的网格是边长为1的正方形，一个三棱锥的三视图如图所示，则该三棱锥的外接球体积为（ ）A ．323π B ．1256π C ．5003π D ．642π 【答案】B【解析】画出几何体的直观图，判断几何体的形状，然后求解几何体的外接球的半径，然后求解体积． 【详解】由题意可知几何体是长方体的一部分，直观图如图：长方体的底面的边长为：23， 所以三棱锥的外接球就是长方体的外接球， 224+3=5,外接球的半径为：52， 则该三棱锥的外接球体积为：345321256ππ⎛⎫= ⎪⎝⎭⨯.故选：B . 【点睛】本题考查三视图求几何体外接球体积，解题的关键是找出球心的位置和球的半径，而三棱锥通常利用补形法求解，转化为长方体外接球的问题，从而得到快速解决，本题属于中等题.12．已知n ∈+N ，直线y ax b =+与曲线()()ln 2f x x n =--相切，设ab 的最大值为n c ，数列{}n c 的前n 项和为n S ，则（ ）A ．存在0n N +∈，00n c <B ．{}n c 为等差数列C ．对于n ∈+N ，11n S e <- D ．321e S e+=【答案】C【解析】设直线y =ax +b 与曲线f （x ）=lnx -（n -2）相切于点（x 0，y 0）．根据()()1,0f x x x'=>，则()()000012=,f x a ax b lnx n x '=+=--，可得：b =−lna −n +1，ab =−alna +a (1−n ).令g (a )=−alna +a (1−n ).n ∈N +，a >0．利用导数研究其单调性极值即可得出． 【详解】设直线y =ax +b 与曲线f (x )=lnx −(n −2)相切于点()00,x y .()()1,0f x x x'=>.则()()000012=,f x a ax b lnx n x '=+=--. 可得：b =−lna −n +1. ∴ab =−alna +a (1−n ).令g (a )=−alna +a (1−n ).n ∈N +，a >0. g ′(a )=−lna −n ，可得n a e -=时,函数g (a )取得极大值即最大值.()n n n g e e c --==,∴数列{}n c 为等比数列，且0nn c e -=>，∴数列{}n c 的前n 项和1111111111n n n e e e S e e e--==<---⎛⎫ ⎪⎝⎭.23331111e e e S e e -++==-. 可知A ，B ，D 错误. 因此只有C 正确. 故选：C . 【点睛】本题考查导数的应用及数列的求和，利用导数研究曲线上某点切线方程，将原问题进一步转化为数列的性质与求和问题，是导数与数列的综合问题，考查综合分析和转化思想，属于较难题.二、填空题13．61x x ⎫⎪⎭的展开式的常数项为____________． 【答案】15【解析】试题分析：由题意得61xx ⎛⎫-⎪⎝⎭的展开式中的通项为63213(1)rr rrT C x-+=-，令632r-=，解得2r=，所以展开式的常数项为15．【考点】二项式定理．14．设变量x，y满足约束条件：3123x yx yx y+≥⎧⎪-≥-⎨⎪-≤⎩，则目标函数32z x y=-的最小值为______.【答案】1-【解析】根据不等式组画可行域，根据目标函数32z x y=-即3122y x z=-12z-表示斜率为32的直线的纵截距，数形结合确定最值点，求出最值点代入目标函数求解即可，【详解】画出可行域，如图：32z x y=-即3122y x z=-，12z-表示斜率为32的直线的纵截距，由图可知，当直线32z x y=-过A点时，12z-取最大，即z取最小值，由31x yx y+=⎧⎨-=-⎩解得(1,2)A，∴32z x y=-的最小值为31221⨯-⨯=-.故答案为：1-.【点睛】本题考查线性规划问题，一般思路是根据不等式组画可行域，根据目标函数确定最值点，求出最值点代入目标函数求解即可，属于基础题.15．过点()()32,5,2A B ---，，且圆心在直线3240x y -+=上的圆的半径为__________． 10【解析】根据弦的垂直平分线经过圆心,结合圆心所在直线方程,即可求得圆心坐标.由两点间距离公式,即可得半径. 【详解】因为圆经过点()()32,5,2A B ---，则直线AB 的斜率为()()()22235k --==---所以与直线AB 垂直的方程斜率为1'2k =-点()()32,5,2A B ---，的中点坐标为()4,0M - 所以由点斜式可得直线AB 垂直平分线的方程为()142y x =-+,化简可得240x y ++= 而弦的垂直平分线经过圆心,且圆心在直线3240x y -+=上,设圆心(),O a b所以圆心满足2403240a b a b ++=⎧⎨-+=⎩解得21a b =-⎧⎨=-⎩所以圆心坐标为()2,1O -- 则圆的半径为()()22322110r OA ==-+++=故答案为: 10【点睛】本题考查了直线垂直时的斜率关系,直线与直线交点的求法,直线与圆的位置关系,圆的半径的求法,属于基础题. 16．已知正项数列{}n a 的前n 项和为n S ，且对任意的*n N ∈满足()()2411n n S a +=+，则36111kk kk k k aa a a =++-=-______.23 【解析】对任意的n ∈N 满足4（S n +1）=（a n +1）2，n ≥2时，4（S n -1+1）=(a n -1+1)2，相减可得：()()1120n n n n a a a a --+--=，根据a n >0，可得a n -a n -1=2，利用等差数列的通项公式可得a n ．再利用裂项求和方法即可得出结果． 【详解】∵()()2411n n S a +=+，∴当1n =时，()()211411a a +=+，∵0n a >，得13a =，当2n ≥时，()()211411n n S a --+=+，∴()()()()2211414111n n n n S S a a --+-+=+-+，∴2211422n n n n n a a a a a --=-+-，∴()()1120n n n n a a a a --+--=， ∴0n a >，∴10n n a a ->+，∴120n n a a ---=，∴12n n a a --=，所以数列{}n a 是以3为首项，2为公差的等差数列， 3(1)221n a n n =+-⨯=+.11(21)23(23)21k kkk k k a a a a k k k k ++=-++-++2123(2123)kk k k k =+⋅++-+111(1)22123k k k ⎛⎫=-⋅-++， ∴3611kk kk k kaa a a =++-1235577375⎛=- ⎝ 12375⎛=-+ ⎝315= 23. 【点睛】本题考查数列的求和、数列递推式，解本题的关键在于裂项的方法，属于中等题.三、解答题17．近些年学区房的出现折射出现行教育体制方面的弊端造成了教育资源的分配不均衡.为此某市出台了政策：自2019年1月1日起，在该市新登记并取得房屋不动产权证书的住房用于申请入学的将不再对应一所学校，实施多校划片.有关部门调查了该市某名校对应学区内建筑面积不同的户型，得到了以下数据：（1）试建立房屋价格y 关于房屋建筑面积的x 的线性回归方程；（2）若某人计划消费不超过100万元购置学区房，根据你得到的回归方程估计此人选房时建筑面积最大为多少？（保留到小数点后一位数字）参考公式：()()()121niii nii x x y y bx x ==--=-∑∑$，a y bx =-$$【答案】（1） 1.317y x =+；（2）面积最大为63.8【解析】（1）由题意计算平均数和回归系数，即可写出回归方程； （2）由题意令1.3x +17≤100，求出x 的取值范围即可． 【详解】 （1）()180+100+120+140=1104x =， ()1120+150+170+200=1604y =， 由公式得12001001001200 1.3900100100900b+++==+++$， $160 1.311017ay bx =-=-⨯=$， 故回归直线方程为 1.317y x =+. （2）由题意得：令1.317100x +≤，解得63.8x ≤，所以估计此人选房时建筑面积最大为63.8平方米. 【点睛】本题考查线性回归方程及应用，根据题目条件将数据代入公式求解即可，考查计算能力，属于基础题. 18．在ABC ∆中，a ，b ，c 分别是角A ，B ，C 所对的边，且满足条件()2tan 1cos 23sin 22A Bb A a += （1）求角B 的大小；（2）若3b =，求22a c -的最大值. 【答案】（1）3B π=；（2）最大值3【解析】（1）由正弦定理，二倍角公式，同角三角函数基本关系式化简已知等式可得3tan2B =B 的值． （2）由正弦定理得23a A =，23cC =，然后代入a 2-c 2，再进行化简整理可得a 2-c 2的最大值． 【详解】（1）由正弦定理得()2sin tan1cos 23sin 22A BB A A += 再由二倍角公式得21cos 2cos 2A A +=，sin 2sin cos 22A A A =，及sin2tan 2cos 2AA A =， 代入可得2sin 232B B =， 进而有3tan2B =3B π=； （2）由正弦定理得23a A =，3c C =，故()222212sin sin a c A C -=-1cos 21cos 21222A C --⎛⎫=- ⎪⎝⎭()6cos2cos2C A =- 222212sinsin22A C A C+-= ()63sin A C =-∴当()sin 1A C -=，2A C π-=即当712A π=，12C π=时取得最大值3【点睛】本题考查正弦定理，二倍角公式，和差化积公式的应用，一定要熟练掌握并灵活应用，特别是二倍角公式的各种变化形式要熟记于心，一般难度不大，但是综合性较强，属于中等题.19．已知正三棱柱111ABC A B C -中，所有棱长都是3，点D ，E 分别是线段1BB 和AC 上的点，1BD =.（1）试确定点E 的位置，使得//BE 平面1ADC ，并证明； （2）若直线1C E 与平面11BCC B 39，求二面角1A BE C --的余弦值的大小. 【答案】（1）E 为AC 三等分点，且3AC AE =，证明见解析；（2）30【解析】（1）取E 为AC 的三等分点，且AC =3AE ，过E 作EK ∥CC 1，且113EK CC BD ==，得到四边形BEKD 为平行四边形，有BE ∥KD ，由线面平行的判定可得BE ∥平面ADC 1；（2）设AC 中点为M ，设A 1C 1的中点为P ，分别以MA ，MB ，MP 所在直线为x ，y ，z 轴建立空间直角坐标系．由直线1C E 与平面11BCC B 39，可得E 点坐标为1,0,02E ⎛⎫ ⎪⎝⎭，然后分别求出平面ABE 与平面BEC 1的一个法向量，由两法向量所成角的余弦值可得二面角A -BE -C 1的余弦值． 【详解】（1）取E 为AC 三等分点，且3AC AE =，过E 作1EK CC ∥， 则113EK CC BD ==，所以BEKD 为平行四边形， 所以BE KD ∥，又1BE ADC ⊄，1DK ADC ⊂， 所以//BE 平面1ADC ，证毕；（2）设AC 中点为M ，设11A C 中点为P ，分别以MA ，MB ，MP 为x ，y ，z 建立空间直角坐标系，则A （32，0，0），C （32-，0，0），B （0，332，0），1C （32-，0，3）， 3332CB ⎛⎫= ⎪ ⎪⎝⎭u u u r ，()10,0,3CC =uuu r ，设平面11BCC B 的一个法向量为(),,m x y z =u r，由13330230m CB x y mCC z ⎧⋅=+=⎪⎨⎪⋅==⎩u u vu u u v u u u u v v ，取3x = 可得()3,1,0m =-u r，设E 点坐标为(),0,0E a ，13,0,32C E a ⎛⎫=+- ⎪⎝⎭u u u u r ，由直线1C E 与平面11BCC B 39， 123332cos ,32923913a m C E a --==⎛⎫++ ⎪⎝⎭u r u u u u r 解得12a =， 可得E 点坐标为1,0,02E ⎛⎫ ⎪⎝⎭， 即3AC AE =，易求平面ABE 法向量()10,0,1n =u r， 设平面1BEC 法向量()2111,,n x y z =u u r，133,,022EB ⎛⎫=- ⎪ ⎪⎝⎭u u u r ，()12,0,3C E =-u u u u r ，由211211113302230n EB x y n C E x z ⎧⋅=-=⎪⎨⎪⋅=-=⎩u u v u u u v u u v u u u u v，取12z =， 可得232n ⎛⎫= ⎪ ⎪⎝⎭u u r ，12121230cos ,10n n n n n n ⋅==⋅u r u u ru r u u r u r u u r ，又因为二面角1A BE C --为钝角， 所以所求余弦值为30. 【点睛】本题考查直线与平面平行的证明，二面角的平面角及求法，立体几何中的探索问题等知识，意在考查学生的转化能力和计算求解能力，对于二面角的求法，通常需要建立适当的空间直角坐标系，求出直线的方向向量和平面的法向量，即可求解，属于常考题型.20．已知动圆M 与直线1y =-相切，且与圆N ：()2221x y +-=外切 （1）求动圆圆心M 的轨迹C 的方程；（2）点O 为坐标原点，过曲线C 外且不在y 轴上的点P 作曲线C 的两条切线，切点分别记为A ，B ，当直线OP 与AB 的斜率之积为1-时，求证：直线AB 过定点.【答案】（1）28x y =；（2）见解析【解析】（1）直接利用直线与圆的位置关系式，圆和圆的位置关系式的应用求出结果． （2）利用直线与曲线的相切和一元二次方程根和系数关系式的应用求出结果． 【详解】（1）设动圆圆心M （x ，y ），由于圆M 与直线y =-1相切，且与圆N ：()2221x y +-=外切．利用圆心到直线的距离和圆的半径和圆心距之间的关系式，可知C 的轨迹方程为：28x y = （2）设直线AB ：y kx m =+，()11,A x y ，()22,B x y ，因为28x y =，4x y '=，所以两条切线的斜率分别为14x ，24x ，则直线AP 的方程是()211184x x y x x -=-，直线BP 的方程是()222284x xy x x -=-.两个方程联立得P 点坐标为1212,28x x x x +⎛⎫⎪⎝⎭， 121212132P AB OP P y y y x x k k x x x -⋅=⋅==--Q ，1232x x ∴=-，由28y kx mx y =+⎧⎪⎨=⎪⎩联立得：208x kx m --= 12832x x m ∴=-=-，4m ∴=故直线AB 过定点()0,4. 【点睛】本题考查轨迹方程、直线过定点问题，求轨迹方程，可根据题目关系及圆锥曲线的几何性质直接求解即可，而对于直线过定点问题，一般根据题目给定条件，利用直线与曲线的关系和一元二次方程根和系数关系式，找出定点即可，属于中等题.21．已知定义在()0,∞+上的函数()()2ln 11ax f x x x x=--++.（1）讨论()f x 的单调区间（2）当223ln ,ln 443e e a ⎛⎫∈ ⎪⎝⎭时，存在0M >，使得对任意()0,x M ∈均有()0f x <，求实数M 的最大值.【答案】（1）见解析；（2）1【解析】（1）先求导，再分类讨论，根据导数和函数单调性的关系即可求出单调区间， （2）根据（1）的结论可得故存在021,21a x a -⎛⎫∈⎪-⎝⎭，使得()00f x =，且当()00,x x ∈时()0f x <恒成立，由()00f x =可得()00020011ln 1x x a x x x ++=-+，再构造函数()()211ln 1x x g x x x x++=-+（0x >），利用导数求出函数的最值即可． 【详解】（1）()()()()21211a x a x f x x ---⎡⎤⎣⎦'=+， ①12a ≤时，()0f x '>，()f x 在()0,∞+上单调递增； ②112a <<时，令()0f x '>得211a x a ->-，故增区间为21,1a a -⎛⎫+∞⎪-⎝⎭， 令()0f x '>得2101a x a -<<-，故减区间为210,1a a -⎛⎫ ⎪-⎝⎭； ③1a ≥时，()0f x '<，则()f x 在()0,∞+上单调递减.（2）易知2231ln ,ln ,14432e e ⎛⎫⎛⎫⊂ ⎪ ⎪⎝⎭⎝⎭，由（1）知：()f x 在210,1a a -⎛⎫ ⎪-⎝⎭上单调递减，在21,1a a -⎛⎫+∞⎪-⎝⎭上单调递增， 则()21001a f f a -⎛⎫<=⎪-⎝⎭， 又()244322ln 32ln ln 303343e f a =-->-⨯-=，故存在021,21a x a -⎛⎫∈⎪-⎝⎭，使得()00f x =， 且当()00,x x ∈时()0f x <恒成立， 故0M x ≤.由()00f x =可得()00020011ln 1x x a x x x ++=-+， 设()()211ln 1x x g x x x x++=-+（0x >）， 则()()()32ln 12x x x g x x ++-'=，令()()()2ln 12h x x x x =++-（0x >）， 则()()2ln 121x h x x x +'=++-+， ()()201xh x x ''=>+，则()h x '在()0,∞+上单调递增，故()()00h x h ''>=，则()h x 在()0,∞+上单调递增，故()()00h x h >=， 则()0g x '>，()g x 在()0,∞+上单调递增， 又()0a g x =，()21ln 4e g =，()332ln 43e g =，故()()()012g g x g <<，则012x <<，又0M x ≤，故1M ≤，即M 的最大值为1. 【点睛】本题考查了函数的单调性，以及不等式恒成立问题，考查导数的应用以及分类讨论思想，转化思想，是一道综合题．不等式恒成立问题常见方法：① 分离参数；② 数形结合；③ 讨论最值；④ 讨论参数，排除不合题意的参数范围，筛选出符合题意的参数范围.22．已知函数()2f x x =-，()21g x x a =-+. （1）当1a =时，解不等式()()f x g x ≤；（2）若()()f x g x ≤恒成立，求实数a 的取值范围.【答案】（1）{0x x ≤或2}3x ≥；（2）3+2⎡⎫∞⎪⎢⎣⎭，【解析】（1）令h （x ）=|x -2|-|2x -1|，由a =1时，不等式f （x ）≤g （x ），可得h （x ）≤1，然后解绝对值不等式即可； （2）由（1）可得h （x ）的最大值，然后根据f （x ）≤g （x ）恒成立，可知a ≥h （x ）max ，从而得到a 的取值范围． 【详解】（1）令()1,2122133,2211,2x x h x x x x x x x ⎧⎪--≥⎪⎪=---=-+<<⎨⎪⎪+≤⎪⎩，由a =1时，不等式f （x ）≤g （x ）， 可得()1h x ≤，∴112x x --≤⎧⎨≥⎩或331122x x -+≤⎧⎪⎨<<⎪⎩或11,12x x +≤⎧⎪⎨≤⎪⎩∴x ≥2或223x <≤或x ≤0，解集为：{0x x ≤或2}3x ≥；（2）由（1）可知：只需()max a h x ≥， 当12x =时，()max 1322h x h ⎛⎫== ⎪⎝⎭， ∴a 的取值范围为3+2⎡⎫∞⎪⎢⎣⎭，． 【点睛】本题考查绝对值不等式的解法、不等式恒成立问题，解绝对值不等式可以用分类讨论思想，利用“零点分段法”求解，解不等式恒成立问题，利用函数的图像及最值求解,属于中等题.。
黑龙江省哈尔滨市2019届高三下学期第三次模拟考试化学试题word版有答案
哈尔滨市第六中学校2019届第三次模拟考试理科综合能力测试本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分，其中第Ⅰ卷第33－38题为选考题，其它题为必考题。
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7．下列说法不正确的是（）A．若升温，水溶液中盐类的水解平衡、弱酸的电离平衡、沉淀溶解平衡不一定正向移动B．将在空气中灼烧呈黑色的铜丝趁热插入盛有乙醇的试管中，铜丝变为红色；反复数次，试管中的液体出现刺激性气味，表明乙醇已被铜氧化为乙醛C．我国古代文献《本草经集注》记载区分硝石( KNO3)与朴硝(Na2SO4)：“以火烧之，紫青烟起，乃真硝石也”，是应用了焰色反应知识D．铝制餐具不宜用来蒸煮或长时间存放咸的食品8．某有机物的分子式是C3H62。
以下说法正确的是 ( )A．不能发生氧化反应 B．至少有三种同分异构体可以水解得到甲酸或甲醇 C．至少有三种同分异构体能与Na0H溶液反应 D．分子中含有碳碳双键或碳氧双键9．下列各项实验操作与现象及结论或解释合理的是（）A．常温下，0．1 m ol·L一的碳酸钠溶液中含有的阴离子数大于0．1NAB．标准状况下，33．6 ml氯气通人足量水中发生反应，转移电子数为1．5×10-3NAC．常温常压下，28 g由C3H6和C4H8组成的混合气体中含共用电子对数目为6NAD．同温下，pH=1体积为1 L的硫酸溶液所含氢离子数与pH=13体积为1 L的氢氧化钠溶液所含氢氧根离子数均为0．1NA11．短周期元素X、Y、Z、W、Q在元素周期表中的相对位置如图所示。
2019年黑龙江哈尔滨南岗区哈尔滨市第三中学高三三模英语试卷-学生用卷
2019年黑龙江哈尔滨南岗区哈尔滨市第三中学高三三模英语试卷-学生用卷一、阅读理解（每小题2分，共20小题，计40分）1、【来源】 2019年黑龙江哈尔滨南岗区哈尔滨市第三中学高三三模（A篇）第21~23题6分(每题2分)◆Go to NYU Precollege Summer ProgramMake the most of your summer vacation with this six-week academic summer camp for high school juniors and seniors. Earn college credits, explore New York City（NYC）, and experience academic and student life at New York University（NYU）. Scholarships are available.◆ParticipantsNYU Precollege summer camp offers rising high school juniors and seniors the opportunity to experience academic and student life at New York University. Teens take college-level courses for academic credits which may be applied to a future degree. Residential（住校的）and non-residential choices are available.◆What to doParticipants take credit-bearing（学分累计的）courses with current college students. Courses are available in over 36 subjects. Students can explore the archeological（考古学的）history of New York City, visit major New York sports equipment, learn the business of sports management, begin to master the Chinese language, practice the art of news reporting, or dig into the service industry while going behind the scenes at Manhattan hotels and restaurants.Additionally, students have the opportunity to participate in free noncredit seminars taught by NYU's distinguished writing experts. The seminars offer students an opportunity to improve writing skills and focus on college-level writing.◆HarvestOutside the classroom, NYU Precollege offers seminars regarding the college admissions and application process, as well as a variety of on-and-off-campus social events and activities. NYU Precollege summer camp allows outstanding teens to get a taste of college while still having time to work or volunteer outside of school, and enjoy the summer vacation.(1) Which is true about NYU Precollege Summer Program?A. It is only for non-residential students.B. It is only intended for high school seniors.C. Its participants should volunteer at NYU.D. It lasts for about one and a half months.(2) What can students do in the seminars?A. Practice oral Chinese with the experts.B. Learn to be more skilled at writing.C. Find out how a post office operates.D. Explore the history of American education.(3) Why does the author write this passage?A. To introduce a summer program.B. To recommend college courses.C. To encourage students to go to college.D. To provide guidance on choosing a college.2、【来源】 2019年黑龙江哈尔滨南岗区哈尔滨市第三中学高三三模（B篇）第24~27题8分(每题2分)Wilbur and Orville Wright were the two youngest sons of Bishop Milton Wright. Wilbur was born on a farm near Millville and Orville was born in August in the house the family would come to occupy for more than 40 years in Dayton.Orville would discover in later years that he and Wilbur's first interests in flight began when they were children, with powered rubber-band-helicopter toys that their father would bring home for them from trips around the country. Orville and Wilbur would build duplicates（复制品） of these toys and bounce them off the ceiling. They were just crazy about these things for the reasons that they did not yet understand.Both boys loved to solve problems that seemed difficult to others. The more difficult the problem was, the more they saw it as a challenge. Exploring the unknown was such a joy for them that Orville once said, "I can remember when Wilbur and I could hardly wait for the morning to come to get at something that interested us. That's happiness."Putting their gift to work, the boys entered the printing business soon after high school. Orville designed and then built a printing press（印刷机） for this enterprise. At first, they did contract printing for other people, and soon one of their customers during the early 1890s was Father Milton's church. They also tried the newspaper business, and although their two newspapers, the West Side News and TheEvening Item were considered to be of high quality, Dayton already had daily newspapers, so the Wrights soon returned to contract printing.Branching out into other areas, they also opened a bicycle business. Bicycling was the fastest growing sport in the country at that time, and their repair shop soon led them into building bicycles of their own design, often with tools of their own construction, such as the lathe（车床） driven by a one-cylinder gasoline engine built by Orville. The bicycle shop would provide the livelihood （生计） that would allow them to carry out their experiments on airplanes.(1) What may inspire Wilbur and Orville Wright to be keen on flight in childhood?A. Their experiences of taking a flight.B. Their wishes to travel by plane.C. The toys their father brought for them.D. Their interest in duplicating toys.(2) What kind of children were the Wright brothers?A. Enthusiastic and curious.B. Quiet and modest.C. Naughty but clever.D. Stubborn but careful.(3) What business did Wilbur and Orville enter shortly after high school?A. The toy business.B. The printing business,C. The lathe business.D. The bicycle business.(4) What will possibly be discussed following this text?A. Who guided them a lot on their way to success.B. What trouble they would meet in their business.C. Why they closed their bicycle shop suddenly.D. How the two brothers began to invent planes.3、【来源】 2019年黑龙江哈尔滨南岗区哈尔滨市第三中学高三三模（C篇）第28~31题8分(每题2分)Hacking isn't just for computers and smartphones. According to a study published last week in the journal Science, scientists have found a way to hack a plant's genes in order to make it use sunlight more quickly. Someday, the results could increase the amount of food produced around the world.Photosynthesis （光合作用） is the word used to describe how plants use sunlight, water, and carbon dioxide to make their own food. Scientists who conducted the new study say this is a very slow process. Plants use less than 1 percent of the energy available to them. But by hacking a plant's genes, the scientists were able to increase the amount of leaf growth on plants between 14 and 20 percent."Specifically, scientists hacked the plant's protective system. Normally, this system is activated when a plant gets too much sunlight, " said scientist Krishna Niyogi, co-author of the study. When the plant senses the light, it gets rid of extra energy and creates more leaves. When the plant is in shade, the protective system is turned off, but the process is slow.Stephen Long is the lead author of the study. He compared a plant's protective system to light-adjusting glasses. When a person wears the glasses outside during the day, the lenses （镜片） darken and lighten depending on how sunny it is. Plants do the same thing, he said. But in plants the adjustment can take anywhere from 10 minutes to an hour. This makes it hard for plants to get the right amount of sunlight needed to create food.The new study sped up the process. By changing the plant's genes, the protective system turned on and off more quickly than normal. As a result, leaf growth on the plants scientists used in the study increased. Leaf growth in two plants increased by 20 percent, while leaf growth on a third plant increased by 14 percent. Scientists conducted the study on tobacco plants. But they think the genetic changes will produce the same results in corn and rice.(1) What will happen to the plants when scientists change plants' genes?A. They will become weaker and weaker.B. They will grow fast to produce more food.C. They will have more beautiful flowers.D. They will depend less on sunlight.(2) Why does the author mention light-adjusting glasses?A. To warn people of the hot and bright sun.B. To explain the principle of plants' protective system.C. To help leaves of the plants absorb sunlight.D. To serve the science experiment smoothly.(3) In the future, the genetic changes will be used to.A. improve tobacco plants' qualityB. increase tobacco plants' leaves and rootsC. improve the flowers' living timeD. increase the production of corn and rice(4) Which of the following can be the theme of this text?A. Hacking plants for more productionB. Leaf growing more quicklyC. Double production of tobacco plantsD. Photosynthesis and production4、【来源】 2019年黑龙江哈尔滨南岗区哈尔滨市第三中学高三三模（D篇）第32~35题8分(每题2分)PHILADELPHIA—The thumb of a Terracotta Warrior was stolen by a member of the public in Philadelphia, where the statue is on display. The cultural relics authority of Northwest China's Shaanxi province will send conservationists（保护工作者）to repair a Terracotta Warrior that had its thumb stolen by a member of the public while on display at a Philadelphia museum.US departments that are involved should guarantee the safety of the relics on display, and parties concerned will be held legally accountable, the provincial cultural heritage department said on Sunday.The statue of a cavalryman（骑兵）, which dates back to at least 209 BC, is one of 10 Chinese Terracotta Warrior statues now on display at the Franklin Institute in Pennsylvania.Since 2009, Wang's team has restored more than 100 of the Terracotta Warriors. According to the Stolen （Lost） Cultural Relics Information Publishing Platform of China, which went online late last year, up to 60 percent of cases of theft have occurred in museums and historical relics institutions. However, most cases occurred in the 1980s and 1990s. As security technology developed in recent years, it has become more difficult to steal a cultural relic from a museum in China, according to the publishing platform.Chen Shiqu, deputy head of the Criminal Investigation Department at the Ministry of Public Security, said during a previous interview that the prices of cultural relics have risen in recent years, leading to more thefts by using upgraded technology."Some criminals are equipped with high-tech tools that can make a tunnel very quickly to steal and transport the cultural relics, " he said. "China will continue to complete the mechanism for fighting against and preventing relics thefts. We will crack down on（打击）the whole market chain, including the trading of the relics and organized thefts. All involved offenders will face criminal liability.”Meanwhile, US authorities said Michael Rohana, 24, from the US state of Delaware, was arrested in connection with the theft and has been released on $15, 000 bail（保释金）, on the condition that he surrender his passport, consent to drug testing and refrain from leaving the country before trial. Museum staff members noticed the missing thumb on Jan 8, and a special agent from the FBI's Art Crime Team tracked down Rohana days later.(1) How is Michael Rohana now?A. He has left America so far.B. He hasn't been caught.C. He hasn't got a passport.D. He has been set free for the moment.(2) What do Chen Shiqu's words mean in Para. 5?A. The rising prices of cultural relics.B. The development of security technology.C. The reason for more thefts of cultural relics.D. Thefts of cultural relics by using upgraded technology.(3) Which of the following words can replace the underlined word "liability" in Para. 6?A. Process.B. Responsibility.C. Influence.D. Activity.(4) Which can serve as the best title for the news report?A. Experts to repair warrior's thumbB. Thumb of Terracotta Warrior was brokenC. US departments are looking into theft of warrior's thumbD. Chinese Terracotta Warrior statues on display in Pennsylvania5、【来源】 2019年黑龙江哈尔滨南岗区哈尔滨市第三中学高三三模第36~40题10分(每题2分)根据短文内容，从短文后的选项中选出能填人空白处的最佳选项。
哈尔滨市高中2018-2019学年高三下学期第三次月考试卷数学
哈尔滨市高中2018-2019学年高三下学期第三次月考试卷数学一、选择题1． 在△ABC 中，角A ，B ，C 所对的边分别是a ，b ，c，若﹣+1=0，则角B 的度数是（ ）A ．60°B ．120°C ．150°D ．60°或120°2． 已知曲线C 1：y=e x 上一点A （x 1，y 1），曲线C 2：y=1+ln （x ﹣m ）（m ＞0）上一点B （x 2，y 2），当y 1=y 2时，对于任意x 1，x 2，都有|AB|≥e 恒成立，则m 的最小值为（ ） A ．1 B．C ．e ﹣1D ．e+13． 若复数满足71i i z+=（为虚数单位），则复数的虚部为（ ） A ．1 B ．1- C ． D ．i -4． 若向量（1，0，x ）与向量（2，1，2）的夹角的余弦值为，则x 为（ ）A ．0B ．1C ．﹣1D ．25． 函数f （x ）=ax 3+bx 2+cx+d 的图象如图所示，则下列结论成立的是（ ）A ．a ＞0，b ＜0，c ＞0，d ＞0B ．a ＞0，b ＜0，c ＜0，d ＞0C ．a ＜0，b ＜0，c ＜0，d ＞0D ．a ＞0，b ＞0，c ＞0，d ＜06． 设x ，y ∈R ，且x+y=4，则5x +5y 的最小值是（ ）A ．9B ．25C ．162D ．507． 1F ，2F 分别为双曲线22221x y a b-=（a ，0b >）的左、右焦点，点P 在双曲线上，满足120PF PF ⋅=，若12PF F ∆，则该双曲线的离心率为（ ）C. 1D. 1【命题意图】本题考查双曲线的几何性质，直角三角形内切圆半径与外接圆半径的计算等基础知识，意在考查基本运算能力及推理能力．8． 已知α∈（0，π），且sin α+cos α=，则tan α=（ ） A．B．C．D．9． 函数y=（x 2﹣5x+6）的单调减区间为（ ）A．（，+∞） B ．（3，+∞）C ．（﹣∞，） D ．（﹣∞，2）班级_______________ 座号______ 姓名_______________ 分数__________________________________________________________________________________________________________________10．（）0﹣（1﹣0.5﹣2）÷的值为（ ）A ．﹣B ．C ．D ．11．下列命题正确的是（ ）A ．已知实数,a b ，则“a b >”是“22a b >”的必要不充分条件B ．“存在0x R ∈，使得2010x -<”的否定是“对任意x R ∈，均有210x ->”C ．函数131()()2xf x x =-的零点在区间11(,)32内D ．设,m n 是两条直线，,αβ是空间中两个平面，若,m n αβ⊂⊂，m n ⊥则αβ⊥12．已知在R 上可导的函数f （x ）的图象如图所示，则不等式f （x ）•f ′（x ）＜0的解集为（ ）A ．（﹣2，0）B ．（﹣∞，﹣2）∪（﹣1，0）C ．（﹣∞，﹣2）∪（0，+∞）D ．（﹣2，﹣1）∪（0，+∞）二、填空题13．命题“若a ＞0，b ＞0，则ab ＞0”的逆否命题是 （填“真命题”或“假命题”．）14．集合A={x|﹣1＜x ＜3}，B={x|x ＜1}，则A ∩B= ．15．若“x ＜a ”是“x 2﹣2x ﹣3≥0”的充分不必要条件，则a 的取值范围为 ．16．如果椭圆+=1弦被点A （1，1）平分，那么这条弦所在的直线方程是 ．17．函数()y f x =图象上不同两点()()1122,,,A x y B x y 处的切线的斜率分别是A B k k ，，规定(),A B k k A B ABϕ-=（AB 为线段AB 的长度）叫做曲线()y f x =在点A 与点B 之间的“弯曲度”，给出以下命题：①函数321y x x =-+图象上两点A 与B 的横坐标分别为1和2，则(),A B ϕ> ②存在这样的函数，图象上任意两点之间的“弯曲度”为常数； ③设点A,B 是抛物线21y x =+上不同的两点，则(),2A B ϕ≤；④设曲线xy e =（e 是自然对数的底数）上不同两点()()112212,,,,1A x y B x y x x -=且，若(),1t A B ϕ⋅<恒成立，则实数t 的取值范围是(),1-∞.其中真命题的序号为________.（将所有真命题的序号都填上）18．若x，y满足线性约束条件，则z=2x+4y的最大值为．三、解答题19．某中学为了普及法律知识，举行了一次法律知识竞赛活动．下面的茎叶图记录了男生、女生各10名学生在该次竞赛活动中的成绩（单位：分）．已知男、女生成绩的平均值相同．（1）求的值；（2）从成绩高于86分的学生中任意抽取3名学生，求恰有2名学生是女生的概率．20．设函数．（1）若x=1是f（x）的极大值点，求a的取值范围．（2）当a=0，b=﹣1时，函数F（x）=f（x）﹣λx2有唯一零点，求正数λ的值．21．已知二阶矩阵M有特征值λ1=4及属于特征值4的一个特征向量=并有特征值λ2=﹣1及属于特征值﹣1的一个特征向量=，=（Ⅰ）求矩阵M；（Ⅱ）求M5．22．求曲线y=x 3的过（1，1）的切线方程．23．（本小题满分12分）某单位共有10名员工，他们某年的收入如下表：员工编号 1 2 3 4 5 6 7 8 9 10 年薪（万元）33．5455．56．577．5850（1）求该单位员工当年年薪的平均值和中位数；（2）从该单位中任取2人，此2人中年薪收入高于5万的人数记为ξ，求ξ的分布列和期望；（3）已知员工年薪收入与工作年限成正相关关系，某员工工作第一年至第四年的年薪分别为3万元、5.4万元、6.5万元、2.7万元，预测该员工第五年的年薪为多少？附：线性回归方程a x b yˆˆˆ+=中系数计算公式分别为： 121()()()niii nii x x y y b x x ==--=-∑∑，x b y aˆˆ-=，其中x 、y 为样本均值．24．设常数λ＞0，a ＞0，函数f （x ）=﹣alnx ．（1）当a=λ时，若f （x ）最小值为0，求λ的值；（2）对任意给定的正实数λ，a ，证明：存在实数x 0，当x ＞x 0时，f （x ）＞0．25．在长方体ABCD﹣A1B1C1D1中，AB=BC=1，AA1=2，E为BB1中点．（Ⅰ）证明：AC⊥D1E；（Ⅱ）求DE与平面AD1E所成角的正弦值；（Ⅲ）在棱AD上是否存在一点P，使得BP∥平面AD1E？若存在，求DP的长；若不存在，说明理由．26．已知函数f（x）=x3﹣x2+cx+d有极值．（Ⅰ）求c的取值范围；（Ⅱ）若f（x）在x=2处取得极值，且当x＜0时，f（x）＜d2+2d恒成立，求d的取值范围．哈尔滨市高中2018-2019学年高三下学期第三次月考试卷数学（参考答案）一、选择题1．【答案】A【解析】解：根据正弦定理有：=，代入已知等式得：﹣+1=0，即﹣1=，整理得：2sinAcosB﹣cosBsinC=sinBcosC，即2sinAcosB=sinBcosC+cosBsinC=sin（B+C），又∵A+B+C=180°，∴sin（B+C）=sinA，可得2sinAcosB=sinA，∵sinA≠0，∴2cosB=1，即cosB=，则B=60°．故选：A．【点评】此题考查了同角三角函数基本关系的运用，熟练掌握基本关系是解本题的关键．2．【答案】C【解析】解：当y1=y2时，对于任意x1，x2，都有|AB|≥e恒成立，可得：=1+ln（x2﹣m），x2﹣x1≥e，∴0＜1+ln（x2﹣m）≤，∴．∵lnx≤x﹣1（x≥1），考虑x2﹣m≥1时．∴1+ln（x2﹣m）≤x2﹣m，令x2﹣m≤，化为m≥x﹣e x﹣e，x＞m+．令f（x）=x﹣e x﹣e，则f′（x）=1﹣e x﹣e，可得x=e时，f（x）取得最大值．∴m≥e﹣1．故选：C．3．【答案】A【解析】试题分析：42731,1i i i i i==-∴==-,因为复数满足71iiz+=,所以()1,1i ii i z iz+=-∴=-，所以复数的虚部为,故选A.考点：1、复数的基本概念；2、复数代数形式的乘除运算.4．【答案】A【解析】解：由题意=，∴1+x=，解得x=0故选A【点评】本题考查空间向量的夹角与距离求解公式，考查根据公式建立方程求解未知数，是向量中的基本题型，此类题直接考查公式的记忆与对概念的理解，正确利用概念与公式解题是此类题的特点．5． 【答案】A【解析】解：f （0）=d ＞0，排除D ， 当x →+∞时，y →+∞，∴a ＞0，排除C ，函数的导数f ′（x ）=3ax 2+2bx+c ，则f ′（x ）=0有两个不同的正实根，则x 1+x 2=﹣＞0且x 1x 2=＞0，（a ＞0），∴b ＜0，c ＞0，方法2：f ′（x ）=3ax 2+2bx+c ，由图象知当当x ＜x 1时函数递增，当x 1＜x ＜x 2时函数递减，则f ′（x ）对应的图象开口向上，则a ＞0，且x 1+x 2=﹣＞0且x 1x 2=＞0，（a ＞0），∴b ＜0，c ＞0，故选：A6． 【答案】D【解析】解：∵5x ＞0，5y＞0，又x+y=4，∴5x +5y ≥2=2=2=50．故选D ．【点评】本题考查基本不等式，关键在于在应用基本不等式时灵活应用指数运算的性质，属于基础题．7． 【答案】D【解析】∵120PF PF ⋅=，∴12PF PF ⊥，即12PF F ∆为直角三角形，∴222212124PF PF F F c +==，12||2PF PF a -=，则222221212122()4()PF PF PF PF PF PF c a ⋅=+--=-， 2222121212()()484PF PF PF PF PF PF c a +=-+⋅=-.所以12PF F ∆内切圆半径12122PF PF F F r c +-==，外接圆半径R c =.12c c =，整理，得2()4ca=+1e =，故选D. 8． 【答案】D【解析】解：将sin α+cos α=①两边平方得：（sin α+cos α）2=1+2sin αcos α=，即2sin αcos α=﹣＜0，∵0＜α＜π，∴＜α＜π，∴sin α﹣cos α＞0，∴（sin α﹣cos α）2=1﹣2sin αcos α=，即sin α﹣cos α=②，联立①②解得：sin α=，cos α=﹣，则tan α=﹣．故选：D ．9． 【答案】B【解析】解：令t=x 2﹣5x+6=（x ﹣2）（x ﹣3）＞0，可得 x ＜2，或 x ＞3，故函数y=（x 2﹣5x+6）的定义域为（﹣∞，2）∪（3，+∞）．本题即求函数t 在定义域（﹣∞，2）∪（3，+∞）上的增区间．结合二次函数的性质可得，函数t 在（﹣∞，2）∪（3，+∞）上的增区间为 （3，+∞）， 故选B ．10．【答案】D【解析】解：原式=1﹣（1﹣）÷=1﹣（1﹣）÷=1﹣（1﹣4）×=1﹣（﹣3）×=1+=． 故选：D ．【点评】本题考查了根式与分数指数幂的运算问题，解题时应细心计算，是易错题．11．【答案】C 【解析】考点：1.不等式性质；2.命题的否定；3.异面垂直；4.零点；5.充要条件．【方法点睛】本题主要考查不等式性质，命题的否定，异面垂直，零点，充要条件.充要条件的判定一般有①定义法:先分清条件和结论（分清哪个是条件,哪个是结论），然后找推导关系（判断,p q q p ⇒⇒的真假），最后下结论（根据推导关系及定义下结论）. ②等价转化法:条件和结论带有否定性词语的命题,常转化为其逆否命题来判断.12．【答案】B【解析】解：由f （x ）图象单调性可得f ′（x ）在（﹣∞，﹣1）∪（0，+∞）大于0， 在（﹣1，0）上小于0，∴f （x ）f ′（x ）＜0的解集为（﹣∞，﹣2）∪（﹣1，0）． 故选B ．二、填空题13．【答案】 真命题【解析】解：若a ＞0，b ＞0，则ab ＞0成立，即原命题为真命题，则命题的逆否命题也为真命题，故答案为：真命题．【点评】本题主要考查命题的真假判断，根据逆否命题的真假性相同是解决本题的关键．14．【答案】 {x|﹣1＜x ＜1} ．【解析】解：∵A={x|﹣1＜x ＜3}，B={x|x ＜1}， ∴A ∩B={x|﹣1＜x ＜1}， 故答案为：{x|﹣1＜x ＜1}【点评】本题主要考查集合的基本运算，比较基础．15．【答案】 a ≤﹣1 ．【解析】解：由x 2﹣2x ﹣3≥0得x ≥3或x ≤﹣1，若“x ＜a ”是“x 2﹣2x ﹣3≥0”的充分不必要条件，则a ≤﹣1， 故答案为：a ≤﹣1．【点评】本题主要考查充分条件和必要条件的应用，根据条件求出不等式的等价是解决本题的关键．16．【答案】 x+4y ﹣5=0 ．【解析】解：设这条弦与椭圆+=1交于P （x 1，y 1），Q （x 2，y 2），由中点坐标公式知x 1+x 2=2，y 1+y 2=2，把P （x 1，y 1），Q （x 2，y 2）代入x 2+4y 2=36，得， ①﹣②，得2（x 1﹣x 2）+8（y 1﹣y 2）=0，∴k==﹣，∴这条弦所在的直线的方程y ﹣1=﹣（x ﹣1），即为x+4y ﹣5=0，由（1，1）在椭圆内，则所求直线方程为x+4y ﹣5=0．故答案为：x+4y ﹣5=0．【点评】本题考查椭圆的方程的运用，运用点差法和中点坐标和直线的斜率公式是解题的关键．17．【答案】②③ 【解析】试题分析：①错：(1,1),(2,5),|||7,A B A B AB k k -=(,)A B ϕ∴=<②对：如1y =；③对；(,)2A B ϕ==≤；④错；1212(,)x x x x A B ϕ==，1211,(,)A B ϕ==>因为1(,)t A B ϕ<恒成立，故1t ≤.故答案为②③.111] 考点：1、利用导数求曲线的切线斜率；2、两点间的距离公式、最值问题、不等式恒成立问题.【方法点晴】本题通过新定义“弯曲度”对多个命题真假的判断考查利用导数求曲线的切线斜率、两点间的距离公式、最值问题、不等式恒成立问题以及及数学化归思想，属于难题.该题型往往出现在在填空题最后两题，综合性较强，同学们往往因为某一点知识掌握不牢就导致本题“全盘皆输”，解答这类问题首先不能慌乱更不能因贪快而审题不清，其次先从最有把握的命题入手，最后集中力量攻坚最不好理解的命题. 18．【答案】 38 ．【解析】解：作出不等式组对应的平面区域如图： 由z=2x+4y 得y=﹣x+，平移直线y=﹣x+，由图象可知当直线y=﹣x+经过点A 时，直线y=﹣x+的截距最大，此时z 最大，由，解得，即A （3，8），此时z=2×3+4×8=6+32=32， 故答案为：38三、解答题19．【答案】（１） 7a =；（２） 310P =． 【解析】试题分析： （１）由平均值相等很容易求得的值；（２）成绩高于86分的学生共五人，写出基本事件共10个，可得恰有两名为女生的基本事件的个数，则其比值为所求．其中恰有2名学生是女生的结果是(96,93,87)，(96,91,87)，(96,90,87)共3种情况． 所以从成绩高于86分的学生中抽取了3名学生恰有2名是女生的概率310P =．1考点：平均数；古典概型．【易错点睛】古典概型的两种破题方法：（1）树状图是进行列举的一种常用方法，适合于有顺序的问题及较复杂问题中基本事件数的探求．另外在确定基本事件时，),(y x 可以看成是有序的，如()1,2与()2,1不同；有时也可以看成是无序的，如)1,2)(2,1(相同．（2）含有“至多”、“至少”等类型的概率问题，从正面突破比较困难或者比较繁琐时，考虑其反面，即对立事件，应用)(1)(A P A P -=求解较好． 20．【答案】【解析】解：（Ⅰ）f （x ）的定义域为（0，+∞），，由f'（1）=0，得b=1﹣a ．∴．…①若a ≥0，由f'（x ）=0，得x=1．当0＜x ＜1时，f'（x ）＞0，此时f （x ）单调递增； 当x ＞1时，f'（x ）＜0，此时f （x ）单调递减． 所以x=1是f （x ）的极大值点．…②若a ＜0，由f'（x ）=0，得x=1，或x=．因为x=1是f （x ）的极大值点，所以＞1，解得﹣1＜a ＜0．综合①②：a 的取值范围是a ＞﹣1．…（Ⅱ）因为函数F （x ）=f （x ）﹣λx 2有唯一零点， 即λx 2﹣lnx ﹣x=0有唯一实数解， 设g （x ）=λx 2﹣lnx ﹣x ，则．令g'（x ）=0，2λx 2﹣x ﹣1=0．因为λ＞0，所以△=1+8λ＞0， 方程有两异号根设为x 1＜0，x 2＞0． 因为x ＞0，所以x 1应舍去．当x ∈（0，x 2）时，g'（x ）＜0，g （x ）在（0，x 2）上单调递减； 当x ∈（x 2，+∞）时，g'（x ）＞0，g （x ）在（x 2，+∞）单调递增． 当x=x 2时，g'（x 2）=0，g （x ）取最小值g （x 2）．… 因为g （x ）=0有唯一解，所以g （x 2）=0，则即因为λ＞0，所以2lnx 2+x 2﹣1=0（*） 设函数h （x ）=2lnx+x ﹣1，因为当x ＞0时， h （x ）是增函数，所以h （x ）=0至多有一解． 因为h （1）=0，所以方程（*）的解为x 2=1， 代入方程组解得λ=1．…【点评】本题考查函数的单调性、极值、零点等知识点的应用，解题时要认真审题，仔细解答，注意合理地进行等价转化．21．【答案】【解析】解：（Ⅰ）设M=则=4=，∴①又=（﹣1）=，∴②由①②可得a=1，b=2，c=3，d=2，∴M=；（Ⅱ）易知=0•+（﹣1），∴M5=（﹣1）6=．【点评】本题考查矩阵的运算法则，考查学生的计算能力，比较基础．22．【答案】【解析】解：y=x 3的导数y ′=3x 2， ①若（1，1）为切点，k=3•12=3， ∴切线l ：y ﹣1=3（x ﹣1）即3x ﹣y ﹣2=0； ②若（1，1）不是切点， 设切点P （m ，m 3），k=3m 2=，即2m 2﹣m ﹣1=0，则m=1（舍）或﹣ ∴切线l ：y ﹣1=（x ﹣1）即3x ﹣4y+1=0． 故切线方程为：3x ﹣y ﹣2=0或3x ﹣4y+1=0．【点评】本题主要考查导数的几何意义、利用导数研究曲线上某点处的切线方程等基础知识，注意在某点处和过某点的切线，考查运算求解能力．属于中档题和易错题．23．【答案】【解析】（1）平均值为10万元，中位数为6万元． （2）年薪高于5万的有6人，低于或等于5万的有4人；ξ取值为0，1，2．152)0(21024===C C P ξ，158)1(2101614===C C C P ξ，31)2(21026===C C P ξ， ∴ξ的分布列为ξ 0 1 2∴()012151535E ξ=⨯+⨯+⨯=．（3）设)4,3,2,1(,=i y x i i 分别表示工作年限及相应年薪，则5,5.2==y x ，21()2.250.250.25 2.255nii x x =-=+++=∑，41()() 1.5(2)(0.5)(0.8)0.50.6 1.5 2.27iii x x y y =--=-⨯-+-⨯-+⨯+⨯=∑，121()()7 1.45()niii nii x x y y b x x ==--===-∑∑，ˆˆ5 1.42.5 1.5a y b x =-=-⨯=，由线性回归方程为 1.4 1.5y x =+．可预测该员工年后的年薪收入为8.5万元．24．【答案】【解析】（1）解：当a=λ时，函数f （x ）=﹣alnx=﹣（x ＞0）．f ′（x ）=﹣=，∵λ＞0，x ＞0，∴4x 2+9λx+3λ2＞0，4x （λ+x ）2＞0．∴当x ＞λ时，f ′（x ）＞0，此时函数f （x ）单调递增；当0＜x ＜λ时，f ′（x ）＜0，此时函数f （x ）单调递减． ∴当x=λ时，函数f （x ）取得极小值，即最小值， ∴f （（λ）==0，解得λ=．（2）证明：函数f （x ）=﹣alnx=﹣alnx=x ﹣﹣alnx ＞x ﹣λ﹣alnx ．令u （x ）=x ﹣λ﹣alnx ． u ′（x ）=1﹣=，可知：当x ＞a 时，u ′（x ）＞0，函数u （x ）单调递增，x →+∞，u （x ）→+∞．一定存在x 0＞0，使得当x ＞x 0时，u （x 0）＞0，∴存在实数x 0，当x ＞x 0时，f （x ）＞u （x ）＞u （x 0）＞0．【点评】本题考查了利用导数研究函数的单调性极值与最值、分类讨论方法、恒成立问题的等价转化方法，考查了推理能力与计算能力，属于难题．25．【答案】【解析】（Ⅰ）证明：连接BD∵ABCD ﹣A 1B 1C 1D 1是长方体，∴D 1D ⊥平面ABCD ， 又AC ⊂平面ABCD ，∴D 1D ⊥AC …1分 在长方形ABCD 中，AB=BC ，∴BD ⊥AC …2分又BD∩D1D=D，∴AC⊥平面BB1D1D，…3分而D1E⊂平面BB1D1D，∴AC⊥D1E…4分（Ⅱ）解：如图建立空间直角坐标系Dxyz，则A（1，0，0），D1（0，0，2），E（1，1，1），B（1，1，0），∴…5分设平面AD1E的法向量为，则，即令z=1，则…7分∴…8分∴DE与平面AD1E所成角的正弦值为…9分（Ⅲ）解：假设在棱AD上存在一点P，使得BP∥平面AD1E．设P的坐标为（t，0，0）（0≤t≤1），则∵BP∥平面AD1E∴，即，∴2（t﹣1）+1=0，解得，…12分∴在棱AD上存在一点P，使得BP∥平面AD1E，此时DP的长．…13分．26．【答案】【解析】解（Ⅰ）∵f（x）=x3﹣x2+cx+d，∴f′（x）=x2﹣x+c，要使f（x）有极值，则方程f′（x）=x2﹣x+c=0有两个实数解，从而△=1﹣4c＞0，∴c＜．（Ⅱ）∵f（x）在x=2处取得极值，∴f′（2）=4﹣2+c=0，∴c=﹣2．∴f（x）=x3﹣x2﹣2x+d，∵f′（x）=x2﹣x﹣2=（x﹣2）（x+1），∴当x∈（﹣∞，﹣1]时，f′（x）＞0，函数单调递增，当x∈（﹣1，2]时，f′（x）＜0，函数单调递减．∴x＜0时，f（x）在x=﹣1处取得最大值，∵x＜0时，f（x）＜恒成立，∴＜，即（d+7）（d﹣1）＞0，∴d＜﹣7或d＞1，即d的取值范围是（﹣∞，﹣7）∪（1，+∞）．【点评】本题考查的知识点是函数在某点取得极值的条件，导数在最大值，最小值问题中的应用，其中根据已知中函数的解析式，求出函数的导函数的解析式，是解答本题的关键．。
2019年黑龙江省哈尔滨三中高考数学三模试卷(文科)
2019年黑龙江省哈尔滨三中高考数学三模试卷（文科）一、选择题：木题共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. 若复数z＝a+i(a∈R)的模为√2，则a＝（）A.1B.±1C.2D.±2【答案】B【考点】复数的模复数的运算【解析】直接利用复数模的计算公式列式求解．【解答】∵z＝a+i(a∈R)的模为√2，∴√a2+1=2，解得a＝±1．2. 设命题：∀x∈R，x2−3x+2≤0，则￢p为（）A.∃x0∈R，x02−3x0+2≤0B.∀x∈R，x2−3x+2>0C.∃x0∈R，x02−3x0+2>0D.∀x∈R，x2−3x+2≥0【答案】C【考点】命题的否定【解析】根据含有量词的命题的否定即可得到结论．【解答】命题为全称命题，命题：∀x∈R，x2−3x+2≤0，则￢p为∃x0∈R，x02−3x0+2> 0，<0}，B＝{x|y=√−x2+x+2)，则A∩B＝（）3. 已知集合A＝{x|x+2x−2A.(−1, 2)B.[−1, 2)C.[−1, 2]D.[−2, 2]【答案】B【考点】交集及其运算【解析】化简集合A、B，根据交集的定义写出A∩B．【解答】<0}＝(−2, 2)，集合A＝{x|x+2x−2∵B＝{x|y=√−x2+x+2)，∴−x2+x+2≥0，解得−1≤x≤2，即B＝[−1, 2]，∴A∩B＝[−1, 2)，4. 已知函数(f(x)＝Asin(ωx+φ)(A>0, ω>0, |φ|<π2)的部分图象如图所示，则f(x)的解析式可以为（）A.y＝2sin(2x+π3) B.y＝2sin(x+π3)C.y＝2sin(2x−π3) D.y＝2sin(x−π3)【答案】A【考点】由y=Asin（ωx+φ）的部分图象确定其解析式【解析】由函数f(x)的部分图象求得A、T、ω和φ的值，即可写出f(x)．【解答】由函数f(x)＝Asin(ωx+φ)的部分图象知，A＝2，34T=5π6−π12=3π4，解得T＝π，∴ω=2πT=2；又ωx+φ＝2×π12+φ=π2，解得φ=π3，∴f(x)＝2sin(2x+π3)．5. 过抛物线y2＝4x的焦点作一条倾斜角为π6的直线，与抛物线交于A，B两点，则|AB|＝（）A.4B.6C.8D.16【答案】D【考点】抛物线的性质【解析】求出焦点坐标和直线方程，结合过焦点直线方程，利用设而不求的思想进行求解即可．【解答】抛物线的焦点坐标为F(1, 0)，p＝2，过焦点的直线的斜率k＝tanπ6=√33，则直线方程为y=√33(x−1)，代入y2＝4x得13(x−1)2＝4x，整理得x 2−14x +1＝0，设A ，B 的坐标分别为(x 1, y 1)，(x 2, y 2)， 则x 1+x 2＝14，则|AB|＝x 1+x 2+p ＝14+2＝16，6. 函数y ＝4x +2x+1+3(x ∈R)的值域为（ ）A.[2, +∞)B.(3, +∞)C.(133, +∞)D.[9, +∞)【答案】 B【考点】函数的值域及其求法 【解析】令t ＝2x (t >0)，把原函数转化为关于t 的一元二次函数求解． 【解答】令t ＝2x (t >0)，∴ 函数y ＝4x +2x+1+3(x ∈R)化为f(t)＝t 2+2t +3＝(t +1)2+2(t >0)， ∴ f(t)>3．即函数y ＝4x +2x+1+3(x ∈R)的值域为(3, +∞)．7. 若直线y ＝kx +1与圆x 2+y 2＝1相交于P ，Q 两点，且△POQ 为等边三角形（其中O 为原点），则k 的值为（ ） A.√3或−√3 B.√3 C.√33或−√33D.√33【答案】 C【考点】直线与圆相交的性质 【解析】由已知可得，圆心(0, 0)到直线的距离d =√32，结合点到直线的距离公式可求k ．【解答】∵ y ＝kx +1与圆x 2+y 2＝1过点(0, 1)，设P(0, 1)， ∵ △POQ 为等边三角形，边长为1， ∴ 圆心(0, 0)到直线的距离d =√1+k2=√32， 解可得，k =±√33，8. 已知某个几何体的三视图如图（主视图中的弧线是半圆），根据图中标出的尺寸（单位：cm ），可得这个几何体的表面积是( )cm 2A.20+2πB.20+3πC.24+2πD.24+3π【答案】B【考点】由三视图求体积【解析】三视图复原几何体是一个组合体，上部是圆柱的一半，下部是正方体，根据三视图的数据，求出几何体的表面积．【解答】三视图复原几何体是一个组合体，上部是横卧的圆柱的一半，底面是一个半圆，其中半径为1，高为2的半圆柱；下部是正方体，棱长为：2，半圆柱的侧面积为π×1×2+π×12＝3π，正方体部分的侧面积为2×2×5＝20，所以组合体的表面积为20+3π(cm2)．9. 在边长为2的正方形ABCD内任取一点P，使得∠APB≤π2的概率为（）A.1−π8B.π8C.π4D.1−π4【答案】A【考点】几何概型计算（与长度、角度、面积、体积有关的几何概型）【解析】由题意画出图形，再由测度比是面积比得答案．【解答】如图正方形的边长为2，图中白色区域是以AB为直径的半圆，当P落在半圆内时，∠APB>π2；当P落在半圆上时，∠APB=π2；当P落在半圆外时，∠APB<π2．故使∠APB<π2的概率P=S−SS=1−π8．10. 阅读右面的程序框图，如果输入的实数x的取值范围是(−∞, 1]∪[2, +∞)，那么输出的函数值f(x)取值范围是（）A.[0, 2]B.[14, 2] C.[14, 4] D.[14, 2]∪{4}【答案】 D【考点】 程序框图 【解析】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：该程序的作用是计算分段函数f(x)={2x −2≤x ≤22x <−2x >2 的函数值，由已知分类讨论即可求解． 【解答】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： 该程序的作用是计算分段函数f(x)={2x −2≤x ≤22x <−2x >2 的函数值． 可得当−∞<x <−2时，f(x)＝2； 当−2≤x ≤1时，f(x)∈[14, 2]；当x ＝2时，f(x)＝4； 当x >2时，f(x)＝2；综上，可得输入的实数x 的取值范围是(−∞, 1]∪[2, +∞)时， 输出的函数值f(x)取值范围是[14, 2]∪{4}．11. 已知函数f(x)＝asinx +bcosx ，且f(π4)是它的最大值（其中a ，b 为常数，且m ≠0），给出下列命题： ①函数f(x −π4)为奇函数②函数f(x)的图象关于x =π2对称； ③函数f(−3π4)是函数的最小值④函数f(x)的图象在y 轴右侧与直线y =a2的交点按横坐标从小至大依次记为P 1，P 2，P 3，P 4…则|P 2P 4|＝2π．其中正确的个数为（ ） A.1 B.2 C.3D.4【答案】 C【考点】命题的真假判断与应用 【解析】由题意可得f(x)=√a 2+b 2sin(x +⌀)，对于①，由于f(x −π4)=√2|a|sinx ．是奇函数，可判断①；对于②，由于x =π2时，f(x)＝|a|，可判断②； 对于③，由f(−3π4)=√2|a|sin(−3π4+π4)=−√2|a|，是函数f(x)的最小值，可判断③；对于④，由题意可得，|P 2P 4|等于一个周期2π，可判断④． 【解答】由于函数f(x)＝asinx +bcosx =√a 2+b 2sin(x +⌀)，且f(π4)是它的最大值， ∴ π4+⌀＝2kπ+π2，k ∈Z ，∴ ⌀＝2kπ+π4，∴ tan⌀=b a =1．∴ f(x)=√2|a|sin(x +π4)．对于①，由于f(x −π4)=√2|a|sinx ．是奇函数，故①正确；对于②，由于当x =π2时，f(x)＝|a|，故函数f(x)的图象不关于x =π2对称，故②不正确；对于③，由于f(−3π4)=√2|a|sin(−3π4+π4)=−√2|a|，为函数f(x)的最小值，故③正确；对于④，函数f(x)的图象即把函数 y =√2|a|sinx 的图象向左平移π4个单位得到的， 故|P 2P 4|等于一个周期2π，故④正确．12. 函数f(x)={2x 3+3x 2+1,−2≤x ≤0ae x ,0≤x ≤2 ，若存在实数m ，使得方程f(x)＝m 有三个相异实根，则实数a 的范围是（ ） A.[1e 2, +∞)B.[0, 1e 2]C.(−∞, 2]D.[1e 2, 2)【答案】D【考点】函数的零点与方程根的关系 【解析】分情况讨论，通过函数的单调性求出满足条件的方程的充要条件，列出不等式求解即可得答案． 【解答】当−2≤x ≤0时，f(x)＝2x 3+3x 2+1．∴ f′(x)＝6x 2+6x ＝6x(x +1) 令f′(x)＝0⇒x ＝0或x ＝−1；令f′(x)>0⇒−2<x <−1；令f′(x)<0⇒−1<x <0；且最大值为f(−1)＝−2+3+1＝2； f(−2)＝−16+12+1＝−3；f(0)＝1；当0≤x ≤2时，f′(x)＝ae x ，则若a <0时，可得f′(x)<0恒成立，即f(x)在(0, 2)上单调递减且最大值为f(0)<0，不存在有三个相异实根，故不成立舍掉； 同理，当a ＝0时也不存在舍掉；即实数a 必须大于0；故当a >0时，f′(x)>0恒成立，即f(x)在(0, 2)上单调递增，若想f(x)＝m 有三个相异实根，必须满足{f(0)=a <2f(2)=ae 2≥1⇒1e 2≤a <2． 二、填空题：本颸共4小颼，每小颼5分，共20分已知向量a →=(1, −2)，b →=(t, 3)，若a → // b →，则t ＝________ 【答案】−32【考点】平面向量共线（平行）的坐标表示 【解析】根据平面向量的共线定理，列方程求出t 的值． 【解答】向量a →=(1, −2)，b →=(t, 3)， 若a → // b →，则3×1−(−2)×t ＝0， 解得t =−32．等比数列{a n }中，a 1＝1，a 3⋅a 5＝64，则a 2019＝________ 【答案】【考点】等比数列的通项公式 【解析】数列{a n }是等比数列，设其公比为q ，则a 3⋅a 5＝64=a 12⋅q 2+4，即a 6＝64＝26，所以q ＝2或q ＝−2，代入即可． 【解答】依题意，数列{a n }是等比数列，设其公比为q ， 则a 3⋅a 5＝64=a 12⋅q 2+4，即a 6＝64＝26， 所以q ＝2或q ＝−2，所以a 2019=a 1⋅q 2018=22018，设变量x ，y 满足约束条件：{x +y ≥3x −y ≥−12x −y ≤3 ，则目标函数z ＝3x −2y 的最小值为________．【答案】 −1【考点】 简单线性规划 【解析】画出不等式组表示的平面区域，结合图形找出最优解，计算目标函数的最小值． 【解答】画出不等式组{x +y ≥3x −y ≥−12x −y ≤3 表示的平面区域，如图阴影所示；由图形知，当目标函数z ＝3x −2y 过点A 时，z 取得最小值； 由{x +y =3x −y =−1，求得A(1, 2)， 所以z 的最小值为z min ＝3×1−2×2＝−1．数学家华罗庚曾说：“数缺形时少直，形少数时难入微”．事实上，很多代数问题可以转化为几何问题加以解法，例如，与√(x −a)2+(y −b)2相关的代数问题，可以转化为点A(x, y)与点B(a, b)之间距离的几何问题．结合上述观点，可得方程|√x 2+6x +13−√x 2−6x +13|=4的解为________ 【答案】 ±6√55【考点】 曲线与方程 【解析】 由|√x 2+6x +13−√x 2−6x +13|=4，得|√(x +3)2+4−√(x −3)2+4|=4，其几何意义为平面内动点(x, 2)与两定点(−3, 0)，(3, 0)距离差的绝对值为4．求出平面内动点与两定点(−3, 0)，(3, 0)距离差的绝对值为4的点的轨迹方程，取y ＝2求得x 值即可． 【解答】由|√x 2+6x +13−√x 2−6x +13|=4，得 |√(x +3)2+4−√(x −3)2+4|=4，其几何意义为平面内动点(x, 2)与两定点(−3, 0)，(3, 0)距离差的绝对值为4． 平面内动点与两定点(−3, 0)，(3, 0)距离差的绝对值为4的点的轨迹为x 24−y 25=1．联立{y =2x 24−y 25=1，解得x =±6√55．三、解答题：共70分.解答应写出文字说明、证明过程或演算步骤，第17～1题为必者题，每个试题考生都必须作答.第22、21题为选考题，考生根据要求作答已知等差效列{a n }的前n 项和为S n ，且a 1＝1，S 2+a 2＝4． （1）求数列{a n }的通项公式；（2）设数列b n ＝2n+1⋅a n 求{b n }的前项和T n ． 【答案】等差效列{a n }的公差设为d ，且a 1＝1，S 2+a 2＝4， 可得1+1+d +1+d ＝4，解得d =12， 则a n ＝1+12(n −1)=n+12；b n ＝2n+1⋅a n ＝(n +1)⋅2n ，前n 项和T n ＝2⋅2+3⋅4+4⋅8+...+(n +1)⋅2n ， 2T n ＝2⋅4+3⋅8+4⋅16+...+(n +1)⋅2n+1，相减可得−T n ＝4+4+8+16+...+2n −(n +1)⋅2n+1 ＝2+2(1−2n )1−2−(n +1)⋅2n+1，化为T n ＝n ⋅2n+1． 【考点】等差数列的通项公式 数列的求和 【解析】（1）等差效列{a n }的公差设为d ，运用等差数列的通项公式，解方程可得d ，进而得到所求通项公式；（2）求得b n ＝2n+1⋅a n ＝(n +1)⋅2n ，运用数列的错位相减法求和，结合等比数列的求和公式，化简可得所求和． 【解答】等差效列{a n }的公差设为d ，且a 1＝1，S 2+a 2＝4， 可得1+1+d +1+d ＝4，解得d =12， 则a n ＝1+12(n −1)=n+12；b n ＝2n+1⋅a n ＝(n +1)⋅2n ，前n 项和T n ＝2⋅2+3⋅4+4⋅8+...+(n +1)⋅2n ， 2T n ＝2⋅4+3⋅8+4⋅16+...+(n +1)⋅2n+1，相减可得−T n ＝4+4+8+16+...+2n −(n +1)⋅2n+1 ＝2+2(1−2n )1−2−(n +1)⋅2n+1，化为T n ＝n ⋅2n+1．棉花的优质率是以其纤维长度来衡量的，纤维越长的棉龙品质越高．棉花的品质分类标准为纤维长度小于等于28mm 的为粗绒棉，纤维长度在(25, 33]为细绒棉，纤维长度大于33mm 的为长绒棉，其中纤维长度在38mm 以上的棉花又名“军海1号”，某采购商从新疆某一棉花基地抽测了100根棉花的纤维长度，得到数据如下图频率分有表所示（1）若将频率作为概率，根据以上数据，能否认为该基地的这批棉花符合“长绒棉占全部棉花的50%以上的要求（2）用样本估计总体，若这批棉共有10000kg，基地提出了两种销售方案给采购商参考．方案一：不分等级卖出，每千克按13.5元计算．方案二：对10000kg棉花先分等级再销售，分级后不同等级的棉花售价如表从采购商的角度，请你帮他决策一下该用哪个方案．（3）用分层抽样的方法从长绒棉中抽取6根棉花，再从6根棉花中取两根进行检验，求抽到的两根棉花只有一根是“军海1号”的概率．【答案】将频率作为概率，根据以上数据，长绒棉占全部棉花的比例为P=60100=60%，∴该基地的这批棉花符合“长绒棉占全部棉花的50%以上的要求“．方案一：13.5×10000＝135000．方案二：2×200+8×3800+15×4000+25×2000＝140800．∴从采购商的角度，该用方案一．用分层抽样的方法从长绒棉中抽取6根棉花，其中“军海1号”抽取到：6×2020+40=2，再从6根棉花中取两根进行检验，基本事件总数n=C62=15，抽到的两根棉花只有一根是“军海1号”包含的基本事件个数m=C41C21=8，∴抽到的两根棉花只有一根是“军海1号”的概率p=mn =815．【考点】分层抽样方法古典概型及其概率计算公式【解析】（1）将频率作为概率，能求出该基地的这批棉花符合“长绒棉占全部棉花的50%以上的要求“．（2）方案一：13.5×10000＝135000．方案二：2×200+8×3800+15×4000+ 25×2000＝140800．从采购商的角度，该用方案一．（3）用分层抽样的方法从长绒棉中抽取6根棉花，其中“军海1号”抽取到2根，再从6根棉花中取两根进行检验，利用古典概型能求出抽到的两根棉花只有一根是“军海1号”的概率．【解答】将频率作为概率，根据以上数据，长绒棉占全部棉花的比例为P=60100=60%，∴该基地的这批棉花符合“长绒棉占全部棉花的50%以上的要求“．方案一：13.5×10000＝135000．方案二：2×200+8×3800+15×4000+25×2000＝140800．∴从采购商的角度，该用方案一．用分层抽样的方法从长绒棉中抽取6根棉花，其中“军海1号”抽取到：6×2020+40=2，再从6根棉花中取两根进行检验，基本事件总数n=C62=15，抽到的两根棉花只有一根是“军海1号”包含的基本事件个数m=C41C21=8，∴抽到的两根棉花只有一根是“军海1号”的概率p=mn =815．如图，在五棱锥P−ABCDE中，AB // DE，BC // AE，AE⊥平面PDE，AB＝AE＝PD ＝2DE＝2BC＝4，∠PDE＝60∘．（1）证明：PE⊥CD；（2）过点D作平行于平面PAE的截面，与直线AB，PB，PC分别交于F，G，H，求夹在该截面与平面PAE之间的几何体体积．【答案】证明：∵DE＝2，PD＝4，∠PDE＝60∘，∴PE=√4+16−2⋅2⋅4⋅cos60=2√3，∴PE⊥DE．∵AE⊥平面PDE，PE⊂平面PDE，∴AE⊥PE，又AE∩DE＝E，AE⊂平面ABCDE，DE⊂平面ABCDE，∴PE⊥平面ABCDE，又CD⊂平面ABCDE，∴PE⊥CD．∵平面PAE // 平面DFGH，∴DF // AE，PA // GF，又BC // AE，AB // DE，∵AE⊥平面PDE，DE⊂平面PDE，∴AE⊥DE，∴四边形AEDF是矩形，∴V P−AEDF=13S矩形AEDF⋅PE=13×2×4×2√3=16√33．∵AP // GF，∴P到平面DFGH的距离等于A到平面DFGH的距离，由（1）可知PE⊥平面ABCDE，故而PE⊥AF，又AF⊥AE，AE∩PE＝E，∴AF⊥平面PAE，∴AF⊥平面DFGH，∵BC // AE，DF // AE，∴BC // DF，又BC平面DFGH，DF⊂平面DFGH，∴BC // 平面DFGH，又BC⊂平面PBC，平面PBC∩平面DFGH＝GH，∴BC // GH，∵AF＝DE=12AB，故F为AB的中点，∴G为PB的中点，∴H是PC的中点，∴GH=12BC＝1，又梯形DFGH的高为12PE=√3，∴V P−DFGH＝VA−DFGH =13S DFGH⋅AF=13×12×(1+4)×√3×2=5√33．∴夹在该截面与平面PAE之间的几何体体积V＝V P−AEDF+V P−DFGH＝7√3．【考点】柱体、锥体、台体的体积计算空间中直线与直线之间的位置关系【解析】（1）根据AE⊥DE，PE⊥DE可得PE⊥平面ABCDE，于是PE⊥CD；（2）求出梯形DFGH的面积，分别计算棱锥P−AEDF和棱锥P−DFGH的体积．【解答】证明：∵DE＝2，PD＝4，∠PDE＝60∘，∴PE=√4+16−2⋅2⋅4⋅cos60=2√3，∴PE⊥DE．∵AE⊥平面PDE，PE⊂平面PDE，∴AE⊥PE，又AE∩DE＝E，AE⊂平面ABCDE，DE⊂平面ABCDE，∴PE⊥平面ABCDE，又CD⊂平面ABCDE，∴PE⊥CD．∵平面PAE // 平面DFGH，∴DF // AE，PA // GF，又BC // AE，AB // DE，∵AE⊥平面PDE，DE⊂平面PDE，∴AE⊥DE，∴四边形AEDF是矩形，∴V P−AEDF=13S矩形AEDF⋅PE=13×2×4×2√3=16√33．∵AP // GF，∴P到平面DFGH的距离等于A到平面DFGH的距离，由（1）可知PE⊥平面ABCDE，故而PE⊥AF，又AF⊥AE，AE∩PE＝E，∴AF⊥平面PAE，∴AF⊥平面DFGH，∵BC // AE，DF // AE，∴BC // DF，又BC平面DFGH，DF⊂平面DFGH，∴BC // 平面DFGH，又BC⊂平面PBC，平面PBC∩平面DFGH＝GH，∴BC // GH，∵AF＝DE=12AB，故F为AB的中点，∴G为PB的中点，∴H是PC的中点，∴GH=12BC＝1，又梯形DFGH的高为12PE=√3，∴V P−DFGH＝VA−DFGH =13S DFGH⋅AF=13×12×(1+4)×√3×2=5√33．∴夹在该截面与平面PAE之间的几何体体积V＝V P−AEDF+V P−DFGH＝7√3．已知函数f(x)＝x−1−a(x−1)2x−lnx．（1）若a＝0，求f(x)在x＝1处的切线方程（2）若函数f(x)存在两个极值点x1和x2，求证：f(x1x2)+4a21−a≥2ln2−1．【答案】函数f(x)＝x−1−a(x−1)2x−lnx．若a＝0，f(x)＝x−1−lnx，f′(x)＝1−1x，f(1)＝0，f′(1)＝0，f(x)在x＝1处的切线方程为y＝0，证明：函数f(x)＝x−1−a(x−1)2x−lnx．f′(x)=(1−a)x2−x+ax2，因为函数f(x)存在两个极值点x 1和x 2，所以f′(x)＝0，x 1＝1，x 2=a1−a ，a ∈(0, 12)∪(12, 1)， f(x 1x 2)+4a 21−a=4a 1−a−2−lna 1−a，令t =a1−a ，t ∈(0, 1)∪(1, +∞)，ℎ(t)＝4t −lnt −2， ℎ′(t)＝4−1t =0，t =14，所以y ＝ℎ(t)在(0, 14)单调递减，在(14, 1)，(1, +∞)单调递增； 所以ℎ(t)最小值为ℎ(14)＝2ln2−1；即ℎ(t)≥2ln2−1； 即f(x 1x 2)+4a 21−a≥2ln2−1．【考点】利用导数研究函数的极值利用导数研究曲线上某点切线方程 【解析】（1）将a ＝0代入函数，求函数的导数和函数的切点的坐标，利用点斜式可求f(x)在x ＝1处的切线方程；（2）函数f(x)存在两个极值点x 1和x 2，求证：f(x 1x 2)+4a 21−a≥2ln2−1．即证明f(x 1x 2)+4a 21−a=4a 1−a−2−lna 1−a≥2ln2−1，令t =a1−a ，t ∈(0, 1)∪(1, +∞)，转换成新函数ℎ(t)＝4t −lnt −2≥2ln2−1，即求函数ℎ(t)的最小值大于等于2ln2−1即可； 【解答】函数f(x)＝x −1−a(x−1)2x−lnx ．若a ＝0，f(x)＝x −1−lnx ，f′(x)＝1−1x ， f(1)＝0，f′(1)＝0，f(x)在x ＝1处的切线方程为y ＝0， 证明：函数f(x)＝x −1−a(x−1)2x−lnx ．f′(x)=(1−a)x 2−x+ax 2，因为函数f(x)存在两个极值点x 1和x 2，所以f′(x)＝0，x 1＝1，x 2=a1−a ，a ∈(0, 12)∪(12, 1)， f(x 1x 2)+4a 21−a =4a1−a −2−ln a 1−a ，令t =a1−a ，t ∈(0, 1)∪(1, +∞)，ℎ(t)＝4t −lnt −2， ℎ′(t)＝4−1t =0，t =14，所以y ＝ℎ(t)在(0, 14)单调递减，在(14, 1)，(1, +∞)单调递增；所以ℎ(t)最小值为ℎ(14)＝2ln2−1；即ℎ(t)≥2ln2−1； 即f(x 1x 2)+4a 21−a≥2ln2−1．已知定点P(2, 0)，圆M:x 2+y 2+4x −60＝0，过点P 的直线l ₁交圆M 于R ，S 两点，过点P 作直线l 2 // MS 交直线MR 于Q 点 （1）求Q 点的轨迹方程E（2）若A ，B ，C ，D 是曲线E 上不重合的四个点，且AC 与BD 交于点(−2, 0)，AC →⋅BD →=0，求|AC →|+|BD →|的取值范围【答案】如图，可得QP ＝QR ，所以QM +QP ＝QM +QR ＝MR ＝8>MP ＝4，所以Q 点的轨迹是以M ，P 点为焦点的椭圆，其中a ＝4，c ＝2，所以b 2＝12， 故点Q 的轨迹方程为x 216+y 212=1；由（1）可知左焦点(−2, 0)，且AC ⊥BD ，①当直线AC 、BD 中有一条直线的斜率不存在时，|AC →|+|BD →|＝6+8＝14； ②当直线AC 的斜率为k ，k ≠0，其方程为：y ＝k(x +2)， 联立{y =k(x +2)x 216+y 212=1，得(3+4k 2)x 2+16k 2x +16k 2−48＝0，设A(x 1, y 1)，C(x 2, y 2)， 则x 1+x 2=−16k 23+4k 2，x 1x 2=16k 2−483+4k 2， 所以|AC →|=√1+k 2|x 1−x 2|=24(1+k 2)3+4k 2，同理可得：|BD →|=24(1+k 2)4+3k ，所以|AC →|+|BD →|=168(1+k 2)2(3+4k 2)(4+3k 2)，令1+k 2＝t(t >1)，|AC →|+|BD →|=168t 2(4t−1)(3t+1)=16812+1t −1t 2∈[967, 14)， 综上，|AC →|+|BD →|的取值范围是[967, 14]．【考点】轨迹方程 【解析】（1）根据题意画出图象，可得QM +QP >MP ，即可知Q 点的轨迹是以M ，P 点为焦点的椭圆；（2）由条件可判断出AC 、BD 过椭圆左焦点，分别讨论AC 、BD 斜率存在与不存在的情况，表示出|AC →|+|BD →|，即可求出取值范围．【解答】如图，可得QP ＝QR ，所以QM +QP ＝QM +QR ＝MR ＝8>MP ＝4，所以Q 点的轨迹是以M ，P 点为焦点的椭圆，其中a ＝4，c ＝2，所以b 2＝12， 故点Q 的轨迹方程为x 216+y 212=1； 由（1）可知左焦点(−2, 0)，且AC ⊥BD ，①当直线AC 、BD 中有一条直线的斜率不存在时，|AC →|+|BD →|＝6+8＝14； ②当直线AC 的斜率为k ，k ≠0，其方程为：y ＝k(x +2)， 联立{y =k(x +2)x 216+y 212=1，得(3+4k 2)x 2+16k 2x +16k 2−48＝0，设A(x 1, y 1)，C(x 2, y 2)， 则x 1+x 2=−16k 23+4k2，x 1x 2=16k 2−483+4k 2， 所以|AC →|=√1+k 2|x 1−x 2|=24(1+k 2)3+4k 2，同理可得：|BD →|=24(1+k 2)4+3k 2，所以|AC →|+|BD →|=168(1+k 2)2(3+4k 2)(4+3k 2)，令1+k 2＝t(t >1)，|AC →|+|BD →|=168t 2(4t−1)(3t+1)=16812+1t −1t 2∈[967, 14)，综上，|AC →|+|BD →|的取值范围是[967, 14]．（二）选考题：共10分，请考生在第22、23题中任选一题作答.如果多做，则按所做的第一题记分[选修4-4坐标系与参数方程]在平面直角坐标系xOy 中，曲线C 1的参数方程为{x =1+tcosαy =−1+tsinα (t 为参数，0≤α<π)，以坐标原点O 为极点，x 轴正半籼为极轴；建立极坐标系，曲线C 2的极坐标方程为ρ2=2√2ρsin(θ+π4)+2．（1）求曲线C 2的直角坐标方程；（2）若曲线C 1与C 2交于A ，B 两点，且|AB|＝2√3，求α的值． 【答案】曲线C 2的极坐标方程为ρ2=2√2ρsin(θ+π4)+2．利用三角函数的展开式，转换为直角坐标方程为(x −1)2+(y −1)2＝4，曲线C 1的参数方程为{x =1+tcosαy =−1+tsinα (t 为参数，0≤α<π)，转换为直角坐标方程为y +1＝k(x −1)，(k ＝tanα)，所以圆心(1, 1)到直线l 的距离d =√r 2−(12|AB|)2=1，所以√1+k 2=1，解得k =±√3，所以α=π32π3．【考点】参数方程与普通方程的互化 圆的极坐标方程 【解析】（1）直接利用转换关系，把极坐标方程转换为直角坐标方程．（2）利用勾股定理和点到直线的距离公式的应用求出直线的斜率，进一步求出直线的倾斜角． 【解答】曲线C 2的极坐标方程为ρ2=2√2ρsin(θ+π4)+2．利用三角函数的展开式，转换为直角坐标方程为(x −1)2+(y −1)2＝4，曲线C 1的参数方程为{x =1+tcosαy =−1+tsinα (t 为参数，0≤α<π)，转换为直角坐标方程为y +1＝k(x −1)，(k ＝tanα)，所以圆心(1, 1)到直线l 的距离d =√r 2−(12|AB|)2=1，所以√1+k 2=1，解得k =±√3，所以α=π32π3．[选修4-5：不等式选讲]设函数f(x)＝|x −a|，如果不等式f(x)≤1的解集为{x|0≤x ≤2}． （1）求a 的值；（2）当x ∈(0, 1)，证明：1f(x)+1f(x+1)≥4． 【答案】∵ f(x)≤1的解集为{x|0≤x ≤2}， ∴ 0和2为方程|x −a|＝1的两实根， ∴ |a|＝1且|2−a|＝1，∴ a ＝1， ∴ a 的值为1；证明：当x ∈(0, 1)时，1f(x)+1f(x+1) =11−x +1x =(11−x +1x)(1−x +x)=2+x 1−x+1−x x=2=2√x 1−x⋅1−x x=4，当且仅当x1−x =1−xx即x =12时取等号，∴ 1f(x)+1f(x+1)≥4．【考点】绝对值不等式的解法与证明 【解析】（1）由f(x)≤1的解集为{x|0≤x ≤2}，可知0和2为方程|x −a|＝1的两实根，将0和2代入方程|x −a|＝1中可求出a 的值；（2）由题意可得1f(x)+1f(x+1)=(11−x +1x )(1−x +x)，利用基本不等式可得1f(x)+1f(x+1)的最小值，从而证明1f(x)+1f(x+1)≥4． 【解答】∵ f(x)≤1的解集为{x|0≤x ≤2}， ∴ 0和2为方程|x −a|＝1的两实根， ∴ |a|＝1且|2−a|＝1，∴ a ＝1， ∴ a 的值为1；证明：当x ∈(0, 1)时，1f(x)+1f(x+1) =11−x +1x =(11−x +1x)(1−x +x) =2+x 1−x+1−x x=2=2√x 1−x⋅1−x x=4，当且仅当x1−x =1−xx即x =12时取等号，∴ 1f(x)+1f(x+1)≥4．。
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2019年哈尔滨南岗中学高三下学期周测三ACities are likely to be affected by overheating, thanks to something called the urban heat island effect. Cities tend to be short of trees, which provide shade, and they are covered with black pavement, which absorbs heat from the sun. Think of how it feels to wear a dark shirt versus a white shirt on a sunny day. A black shirt absorbs light, heating you up. But a white shirt reflects light, keeping you cool.The average temperature in a city of a million or more people can be more than 5 degrees F hotter than surrounding areas. That extra 5 degrees can turn a hot day from uncomfortable to deadly. As temperatures rise, cities will be an especially dangerous place to be during a heat wave. To protect public health, city officials are going to make the city cooler.As part of that effort, Los Angeles is coating its roads in CoolSeal, a gray paint that keeps streets and parking lots 10 degrees cooler than black asphalt（沥青）. It will help Angelinos save money during the summer, when air conditioning sends power bills soaring. And it will save lives by lowering temperatures and improving air quality. Hot weather worsens air pollution by turning car exhaust into smog, which can make life miserable for people with asthma（哮喘）and other breathing problems.Of course, LA will have to do more than paint over a few streets to cool off the city. Angelinos will also need to plant more trees and apply white paint to rooftops — a t least those not already covered in solar panels. While LA is a pioneer of reflective streets, other cities, like New York, are already experimenting with reflective roofs or, like Melbourne, lowering the temperature by planting trees. LA is hardly alone in its effort to stay cool.“This is an urgent challenge, and it’s much bigger than one person,” said Mayor Garcetti in a recent statement. “Climate change is a fact of life that people in Los Angeles and cities around the world live with every day.”1. Which of the following contributes to the urban heat island effect?A. Planting more trees in the streets.B. Covering the streets with white paint.C. Applying reflective paint to rooftops.D. Furnishing every house with air conditioners.2. What should the city officials emphasize during hot summer?A. How to make the city cooler.B. Where to wear a white shirt.C. Why to coat the roads with black paint.D. When to cut off the electricity supply.3. What benefit can people gain from the use of CoolSeal?A. Increasing the indoor activities.B. Promoting the sale of air conditioners.C. Making life easier and more comfortable.D. Reducing the number of cold days.4. What’s the main idea of the passage?A. L.A. adopted many approaches to fight against the heat.B. L.A. calls on people to fight against global warming.C. More and more countries begin to plant trees in the streets.D. L.A. is painting its streets white to keep the city cool.5. Which is the same meaning with the underlined word “miserable” in the last paragraph?A. painfulB. exhaustedC. nativeD. depressedBI still remember my first time going to the Chinese company’s canteen. When I stepped onto the elevator, each one was full of people. 6 However, there was no sign of fire and people were laughing and talking. I followed the flow of the crowd to the doors of the canteen. It occurred to me that in China everyone takes lunch at the same time. 7 The cafeteria followed this schedule, only being open a few hours per day.When I stood in the doorway, I realized I had to act fast or I would be left behind. 8 Everyone lines up and takes a tray but instead of silverware, there are chopsticks and soup spoons.As I was pushed along the line, I saw dishes I didn’t recognize and as people cut in line and picked up dishes left and right. 9 I ended up going with a pork and cucumber dish and a tofu salad.I have been at my company for over a year and a half. 10 I have learned what dishes I like, mostly by trial and error, and a lot of the workers know me and say hello.I am still getting used to eating lunch at “lunch time”, but I think the schedule definitely helps keep me on track during the day. I wish there was something like this at offices in my hometown.A. I thought I had missed a fire drill.B. There was no emergency, it was just lunchtime.C. I noticed things at lunchtime were a bit different.D. I felt that I needed to make some quick decisions.E. The canteen reminded me of my school lunch room.F. And now I am glad to say I am more confident in the canteen.G. I was not very familiar with the local food or restaurants in China.Secord部分：完形填空（共20小题；每小题2分，满分40分）When I first came to Rochester from Austin, Texas — a place where fall is practically non-existent — I was excited for a real northern fall. But, I wasn’t 11 it to feel like Christmas in October.Regardless of this 12 , and the adjustment period that I’m not entirely sure I’m 13 , I am thankful to be somewhere that has a 14 fall. By “real” I 15 a distinctive season, a(n) 16 in the weather and a visual changing of the world around me. The leaves 17 colors, it rains more often, the air is almost always 18 and dry, and a light breeze 19 me to walk a bit faster to get inside. Yes, it is more difficult to be 20outside in a Rochester fall —where it isn’t just “less hot” but 21 , fairly cold. But, I would argue that time spent outside in this season could be more 22 than in the summer.Everyone 23 that fall is about change —I won’t argue against that. Things do 24 in the fall — the weather is 25 becoming worse and worse, the dying leaves become warmly colored, and 26 you know it, we’ll be back to having only eight hours of su nlight a day.This oncoming change is why I think it is so important to 27 fall. Fall is not just about the 28 of change, but also appreciating what will soon be 29 . Go 30 this fall. Enjoy the leaves not because they are turning beautiful deep shades of orange and red, but because there will soon be no leaves in those branches at all.11. A. interested in B. exposed to C. ashamed of D. prepared for12. A. accident B. possibility C. shock D. motivation13. A. beyond B. against C. for D. after14. A. rare B. real C. hard D. strange15. A. take in B. pick up C. call back D. refer to16. A. shift B. conclusion C. improvement D. experience17. A. spoil B. keep C. turn D. avoid18. A. cold B. warm C. hot D. cozy19. A. prevents B. expects C. urges D. suspects20. A. grateful B. creative C. considerate D. comfortable21. A. luckily B. actually C. helpfully D. instantly22. A. terrible B. casual C. unpleasant D. valuable23. A. questions B. emphasizes C. recalls D. analyzes24. A. relax B. change C. harvest D. rest25. A. steadily B. hardly C. helpfully D. narrowly26. A. if B. since C. as D. before27. A. control B. describe C. appreciate D. hide28. A. process B. disadvantages C. fear D. result29. A. disappointing B. puzzling C. gone D. broken30. A. inside B. forward C. backward D. outside第三部分：语法填空(每小题2分，满分20分)China has always been famous for being a “S tate of Etiquettes（礼仪）.” According to historical documents, as early as 2,600 years ago, this nation has already established a thorough set 31 dining etiquettes.A famous 19th century Russian writer, Anton Chekhov, once invited a Chinese man 32 (have) a drink in a bar. Chekhov said, “Before drinking from his cup, he held33 with his hands and presented to me and the bar owner and bar tenders, 34 (say) ‘qing(please).’ This is the custom of China. They are not like us to finish it in one drink, 35 prefer to drink by taking a small amount at a time. With every sip（一小口）, he 36 (eat) some food. Afterwards he handed me some Chinese 37 (coin) to show gratitude. This is a rather interestingly polite nationality…” This was t he most valuable opinion of a Chinese person 38 was given by a foreigner two centuries ago. Chinese traditional dinner procedures used to be long and dealt 39 (serious) with; the 40 (important) the occasion, the more complex the procedures were.答题卡1.[A][B][C][D]2.[A][B][C][D]3.[A][B][C][D]4.[A][B][C][D]5.[A][B][C][D]6.[A][B][C][D] [E][F][G]7.[A][B][C][D] [E][F][G]8.[A][B][C][D] [E][F][G]9.[A][B][C][D] [E][F][G] 10.[A][B][C][D] [E][F][G]Secord部分: 完形填空（共每小题2分，满分40分）11.[A][B][C][D] 12.[A][B][C][D] 13.[A][B][C][D] 14.[A][B][C][D] 15.[A][B][C][D]16.[A][B][C][D] 17.[A][B][C][D] 18.[A][B][C][D] 19.[A][B][C][D] 20.[A][B][C][D]21.[A][B][C][D] 22.[A][B][C][D] 23.[A][B][C][D] 24.[A][B][C][D] 25.[A][B][C][D]26.[A][B][C][D] 27.[A][B][C][D] 28.[A][B][C][D] 29.[A][B][C][D] 30.[A][B][C][D]第三部分：语法填空或词汇填空（共10个小题，每小题2分，满分20分）31.__________; 32. __________; 33. ___________; 34.___________; 35. _____________; 36. ___________; 37._____________; 38. ____________; 39. ___________; 40._______________.短文改错（共10个小题，每小题2分，满分20分）It is essential to take an useful summer vacation after the exam. There are many kinds of activity for us to take part. First, we can visit our relatives and friends or spend a happily time with our family. Second, seeing a film, listening to music or hang out with our friends will also be excited. Third, we should take more exercise to keep healthy. What’s more, we’d better to review what we had learned at school. Finally, it is necessary of us to do some housework, that can help our parents a lot. Above all, I hope all of us can have a pleasant vacation.阅读1-5 DACDA 6-10 ABEDF完型11—15 DCABD 16-20ACACD 21—25 BDBBA 26-30DCACD语法填空31. of 32. to have 33. it 34. saying 35. but36. ate 37. coins 38. that 39. seriously 40. more important短文改错1. an改为a2. activity改为activities3. take part后加in4. happily改为happy5. hang改为hanging6. excited改为exciting7. 删掉we’d better后的to8. had改为have9. of改为for10. that改为which。