济南市2020届高三第一次模拟考试及答案详解(精校版)

2020年山东省济南市高考数学一模试卷（文科）一、选择题：本大题共10个小题，每小题5分，共50分．在每小题给出的四个选项中，只有一项是符合题目要求的．1．设复数z满足z•（2+i）=10﹣5i（i为虚数单位），则z的共轭复数为（）A．﹣3+4i B．﹣3﹣4i C．3+4i D．3﹣4i2．已知集合M={x|﹣x≤x＜3}，集合N={x|y=}，则M∪N=（）A．M B．N C．{x|﹣1≤x≤2}D．{x|﹣3≤x＜3}3．某校高三（1）班共有48人，学号依次为1，2，3，…，48，现用系统抽样的办法抽取一个容量为6的样本．已知学号为3，11，19，35，43的同学在样本中，那么还有一个同学的学号应为（）A．27 B．26 C．25 D．244．已知直线ax+by=1经过点（1，2），则2a+4b的最小值为（）A．B．2C．4 D．45．设m，n是两条不同的直线，α，β是两个不同的平面，给出下列四个命题：①若m∥n，m⊥β，则n⊥β；②若m∥α，m∥β，则α∥β；③若m∥n，m∥β，则n∥β；④若m⊥α，m⊥β，则α⊥β其中真命题的个数为（）A．1 B．2 C．3 D．46．已知命题p：∃x0∈R，使sinx0=；命题q：∀x∈（0，），x＞sinx，则下列判断正确的是（）A．p为真B．￢q为假C．p∧q为真D．p∨q为假7．函数f（x）=2sin（ωx+φ）（w＞0，|φ|＜）的部分图象如图所示，则f（0）+f（）的值为（）A．2﹣B．2+C．1﹣D．1+8．已知x，y满足约束条件，则z=的范围是（）A．[，2]B．B[﹣，]C．[，]D．[，]9．已知函数f（x）=ax2﹣bx2+x，连续抛掷两颗骰子得到的点数分别是a，b，则函数f （x）在x=1处取得最值的概率是（）A．B．C．D．10．已知抛物线y2=2px（p＞0），△ABC的三个顶点都在抛物线上，O为坐标原点，设△ABC三条边AB，BC，AC的中点分别为M，N，Q，且M，N，Q的纵坐标分别为y1，y2，y3．若直线AB，BC，AC的斜率之和为﹣1，则++的值为（）A．﹣B．﹣C．D．二、填空题：（本题共5小题，每题5分，共25分）11．设ln3=a，ln7=b，则e a+e b=_______．（其中e为自然对数的底数）12．已知向量，，其中||=，||=2，且（﹣）⊥，则向量和的夹角是_______．13．已知过点（2，4）的直线l被圆C：x2+y2﹣2x﹣4y﹣5=0截得的弦长为6，则直线l的方程为_______．14．公元263年左右，我国数学家刘徽发现当圆内接正多边形的边数无限增加时，多边形面积可无限逼近圆的面积，并创立了“割圆术”．利用“割圆术”刘徽得到了圆周率精确到小数点后两位的近似值3.14，这就是著名的“徽率”．如图是利用刘徽的“割圆术”思想设计的一个程序框图，则输出n的值为_______．（参考数据：sin15°=0.2588，sin7.5°=0.1305）15．已知函数f（x）=，g（x）=kx+1，若方程f（x）﹣g（x）=0有两个不同实根，则实数k的取值范围为_______．三、解答题:本大题共6小题，共75分16．近日，济南楼市迎来去库存一系列新政，其中房产税收中的契税和营业税双双下调，对住房市场持续增长和去库存产生积极影响．某房地产公司从两种户型中各拿出9套进行促销活动，其中A户型每套面积100平方米，均价1.1万元/平方米，B户型每套面积80平方米，均价1.2万元/平方米．下表是这18套住宅平方米的销售价格：（单位：万元/平方米）：房号/户型 1 2 3 4 5 6 7 8 9A户型0.98 0.99 1.06 1.17 1.10 1.21 a 1.09 1.14B户型 1.08 1.11 1.12 b 1.26 1.27 1.26 1.25 1.28（I）求a，b的值；（II）张先生想为自己和父母买两套售价小于100万元的房子，求至少有一套面积为100平方米的概率．17．在△ABC中，内角A，B，C的对边分别为a，b，c，已知2ccosA+a=2b（Ⅰ）求角C的值；（Ⅱ）若c=2，且△ABC的面积为，求a，b．18．如图，四棱锥P﹣ABCD的底面为正方形，侧面PAD⊥底面ABCD，PA⊥AD，E，E，H分别为AB，PC，BC的中点（Ⅰ）求证：EF∥平面PAD；（Ⅱ）求证：平面PAH⊥平面DEF．19．已知数列{a n}为公差不为零的等差数列，其前n项和为S n，满足S5﹣2a2=25，且a1，a4，a13恰为等比数列{b n}的前三项（Ⅰ）求数列{a n}，{b n}的通项公式；（Ⅱ）设T n是数列{}的前n项和，是否存在k∈N*，使得等式1﹣2T k=成立，若存在，求出k的值；若不存在，说明理由．20．设椭圆C： +=1（a＞b＞0），定义椭圆C的“相关圆”方程为x2+y2=．若抛物线y2=4x的焦点与椭圆C的一个焦点重合，且椭圆C短轴的一个端点和两个焦点构成直角三角形（Ⅰ）求椭圆C的方程和“相关圆”E的方程；（Ⅱ）过“相关圆”E上任意一点P的直线l：y=kx+m与椭圆交于A，B两点，O为坐标原点，若OA⊥OB，证明原点O到直线AB的距离为定值，并求m的取值范围．21．设函数f（x）=ax2+b（lnx﹣x），g（x）=﹣2+（1﹣b）x，已知曲线y=f（x）在点（1，f（1））处的切线与直线x﹣y+1=0垂直．（Ⅰ）求a的值；（Ⅱ）求函数f（x）的极值点；（Ⅲ）若对于任意b∈（1，+∞），总存在x1，x2∈[1，b]，使得f（x1）﹣f（x2）﹣1＞g（x1）﹣g（x2）+m成立，求实数m的取值范围．2020年山东省济南市高考数学一模试卷（文科）参考答案与试题解析一、选择题：本大题共10个小题，每小题5分，共50分．在每小题给出的四个选项中，只有一项是符合题目要求的．1．设复数z满足z•（2+i）=10﹣5i（i为虚数单位），则z的共轭复数为（）A．﹣3+4i B．﹣3﹣4i C．3+4i D．3﹣4i【考点】复数代数形式的乘除运算．【分析】由z•（2+i）=10﹣5i，得z=，再由复数代数形式的乘除运算化简复数z，则z的共轭复数可求．【解答】解：由z•（2+i）=10﹣5i，得=3﹣4i，则z的共轭复数=3+4i．故选：C．2．已知集合M={x|﹣x≤x＜3}，集合N={x|y=}，则M∪N=（）A．M B．N C．{x|﹣1≤x≤2}D．{x|﹣3≤x＜3}【考点】并集及其运算．【分析】分别求出集合M、N的范围，从而求出其并集即可．【解答】解：集合M={x|﹣x≤x＜3}={x|0≤x＜3}，集合N={x|y=}={x|﹣3≤x≤2}，则M∪N={x|﹣3≤x＜3}，故选：D．3．某校高三（1）班共有48人，学号依次为1，2，3，…，48，现用系统抽样的办法抽取一个容量为6的样本．已知学号为3，11，19，35，43的同学在样本中，那么还有一个同学的学号应为（）A．27 B．26 C．25 D．24【考点】系统抽样方法．【分析】根据系统抽样的特征，从48名学生从中抽取一个容量为6的样本，则系统抽样的分段间隔为8，可求得余下的同学的编号．【解答】解：∵从48名学生从中抽取一个容量为6的样本，∴系统抽样的分段间隔为=8，∵学号为3，11，19，35，43的同学在样本中，∴抽取的另一个同学的学号应为27，故选：A．4．已知直线ax+by=1经过点（1，2），则2a+4b的最小值为（）A．B．2C．4 D．4【考点】基本不等式．【分析】直线ax+by=1经过点（1，2），可得：a+2b=1．再利用基本不等式的性质、指数的运算性质即可得出．【解答】解：∵直线ax+by=1经过点（1，2），∴a+2b=1．则2a+4b≥==2，当且仅当时取等号．故选：B．5．设m，n是两条不同的直线，α，β是两个不同的平面，给出下列四个命题：①若m∥n，m⊥β，则n⊥β；②若m∥α，m∥β，则α∥β；③若m∥n，m∥β，则n∥β；④若m⊥α，m⊥β，则α⊥β其中真命题的个数为（）A．1 B．2 C．3 D．4【考点】命题的真假判断与应用．【分析】①根据线面垂直的性质定理进行判断．②根据线面平行的判定定理进行判断．③根据线面平行的判定定理进行判断．④根据线面垂直和面面垂直的判定定理进行判断．【解答】解：①若m∥n，m⊥β，则n⊥β成立，故①正确；②若m∥α，m∥β，则α∥β不一定成立，有可能相交，故②错误；③若m∥n，m∥β，则n∥β或n⊂β；故③错误，④若m⊥α，m⊥β，则α∥β，故④错误，故正确的是①，故选：A6．已知命题p：∃x0∈R，使sinx0=；命题q：∀x∈（0，），x＞sinx，则下列判断正确的是（）A．p为真B．￢q为假C．p∧q为真D．p∨q为假【考点】复合命题的真假．【分析】分别判断出p，q的真假，从而判断出复合命题的真假即可．【解答】解：∀x∈R，都有sinx≤1，故命题p：∃x0∈R，使sinx0=是假命题；令f（x）=x﹣sinx，f′（x）=1+cosx＞0，y=f（x）在区间（0，）上单调递增，∴f（x）＞f（0）=0，故命题q：∀x∈（0，），x＞sinx是真命题，故B正确，故选：B．7．函数f（x）=2sin（ωx+φ）（w＞0，|φ|＜）的部分图象如图所示，则f（0）+f（）的值为（）A．2﹣B．2+C．1﹣D．1+【考点】正弦函数的图象．【分析】根据函数f（x）的部分图象，求出周期T与ω的值，再计算φ的值，写出f（x）的解析式，从而求出f（0）+f（）的值．【解答】解：根据函数f（x）=2sin（ωx+φ）（w＞0，|φ|＜）的部分图象，得T=﹣（﹣）=，又T==π，∴ω=2；当x=﹣时，函数f（x）取得最小值﹣2，∴2×（﹣）+φ=﹣+2kπ，k∈Z，解得φ=﹣+2kπ，k∈Z，又|φ|＜，∴φ=﹣，∴f（x）=2sin（2x﹣）；∴f（0）+f（）=2sin（﹣）+2sin（2×﹣）=2×（﹣）+2sin=2﹣．故选：A．8．已知x，y满足约束条件，则z=的范围是（）A．[，2]B．B[﹣，]C．[，]D．[，]【考点】简单线性规划．【分析】画出满足条件的平面区域，求出角点的坐标，根据z=的几何意义求出z的范围即可．【解答】解：画出满足条件的平面区域，如图示：，由，解得A（1，2），由，解得B（3，1），而z=的几何意义表示过平面区域内的点与（﹣1，﹣1）的直线的斜率，显然直线AC斜率最大，直线BC斜率最小，K AC==，K BC==，故选：C．9．已知函数f（x）=ax2﹣bx2+x，连续抛掷两颗骰子得到的点数分别是a，b，则函数f （x）在x=1处取得最值的概率是（）A．B．C．D．【考点】利用导数求闭区间上函数的最值．【分析】所有的（a，b）共计6×6=36个，函数f′（x）=ax2﹣bx在x=1处取得最值等价于f″（1）=2a﹣b=0，用列举法求得满足条件的（a，b）有3个，再根据概率公式计算即可．【解答】解：连续抛掷两颗骰子得到的点数分别是a，b，共有36种等可能事件，∵f（x）=ax3﹣bx2+x，∴f′（x）=ax2﹣bx+1，∵函数f′（x）=ax2﹣bx+1在x=1处取得最值，∴f″（x）=2ax﹣b，∴f″（1）=2a﹣b=0，即2a=b，满足的基本事件有（1，2），（2，4），（3，6），共3种，故函数f′（x）在x=1处取得最值的概率为=，故选：C．10．已知抛物线y2=2px（p＞0），△ABC的三个顶点都在抛物线上，O为坐标原点，设△ABC三条边AB，BC，AC的中点分别为M，N，Q，且M，N，Q的纵坐标分别为y1，y2，y3．若直线AB，BC，AC的斜率之和为﹣1，则++的值为（）A．﹣B．﹣C．D．【考点】抛物线的简单性质．【分析】设AB，BC，AC的方程，联立方程组消元，利用根与系数的关系解出y1，y2，y3，根据斜率之和为﹣1化简++即可得出答案．【解答】解：设AB的方程为x=m1y+t1，BC的方程为x=m2y+t2，AC的方程为x=m3y+t3，联立方程组，消元得：y2﹣2pm1y﹣2pt1=0，∴y1=pm1，同理可得：y2=pm2，y3=pm3，∵直线AB，BC，AC的斜率之和为﹣1，∴++=﹣1．∴则++=++=（++）=﹣．故选：B．二、填空题：（本题共5小题，每题5分，共25分）11．设ln3=a，ln7=b，则e a+e b=10．（其中e为自然对数的底数）【考点】对数的运算性质．【分析】使用对数恒等式解出．【解答】解：∵ln3=a，ln7=b，∴e a=3，e b=7，∴e a+e b=10．故答案为10．12．已知向量，，其中||=，||=2，且（﹣）⊥，则向量和的夹角是．【考点】平面向量数量积的运算．【分析】利用向量垂直的数量积为0列出方程；利用向量的平方等于向量模的平方及向量的数量积公式将方程用模与夹角表示求出夹角．【解答】解：设两个向量的夹角为θ，∵||=，||=2，且（﹣）⊥，∴（﹣）•=||2﹣•=||2﹣||•||cosθ=3﹣2cosθ=0，解得cosθ=，∵0≤θ≤π，∴θ=，故答案为：．13．已知过点（2，4）的直线l被圆C：x2+y2﹣2x﹣4y﹣5=0截得的弦长为6，则直线l的方程为x﹣2=0或3x﹣4y+10=0．【考点】直线与圆的位置关系．【分析】设过点（2，4）的直线l的方程为y=k（x﹣2）+4，求出圆C的圆心C（1，2），半径r=，圆心C（1，2）到直线l的距离d，由此能求出直线l的方程；当直线l的斜率不存在时，直线l的方程为x=2也满足条件．由此能求出直线l的方程．【解答】解：设过点（2，4）的直线l的方程为y=k（x﹣2）+4，圆C：x2+y2﹣2x﹣4y﹣5=0的圆心C（1，2），半径r==，圆心C（1，2）到直线l的距离d==，∵过点（2，4）的直线l被圆C：x2+y2﹣2x﹣4y﹣5=0截得的弦长为6，∴由勾股定理得：，即，解得k=，∴直线l的方程为y=（x﹣2）+4，即3x﹣4y+10=0，当直线l的斜率不存在时，直线l的方程为x=2，圆心C（1，2）到直线x=2的距离d=1，满足，故x﹣2=0是直线l的方程．综上，直线l的方程为x﹣2=0或3x﹣4y+10=0．故答案为：x﹣2=0或3x﹣4y+10=0．14．公元263年左右，我国数学家刘徽发现当圆内接正多边形的边数无限增加时，多边形面积可无限逼近圆的面积，并创立了“割圆术”．利用“割圆术”刘徽得到了圆周率精确到小数点后两位的近似值3.14，这就是著名的“徽率”．如图是利用刘徽的“割圆术”思想设计的一个程序框图，则输出n的值为24．（参考数据：sin15°=0.2588，sin7.5°=0.1305）【考点】程序框图．【分析】列出循环过程中S与n的数值，满足判断框的条件即可结束循环．【解答】解：模拟执行程序，可得n=6，S=3sin60°=，不满足条件S≥3.10，n=12，S=6×sin30°=3，不满足条件S≥3.10，n=24，S=12×sin15°=12×0.2588=3.1056，满足条件S≥3.10，退出循环，输出n的值为24．故答案为：24．15．已知函数f（x）=，g（x）=kx+1，若方程f（x）﹣g（x）=0有两个不同实根，则实数k的取值范围为（，1）∪（1，e﹣1]．【考点】根的存在性及根的个数判断；函数的零点与方程根的关系．【分析】方程f（x）﹣kx=1有两个不同实根可化为函数f（x）与函数y=kx+1有两个不同的交点，作函数f（x）与函数y=kx+1的图象，结合函数的图象求解．【解答】解：∵g（x）=kx+1，∴方程f（x）﹣g（x）=0有两个不同实根等价为方程f（x）=g（x）有两个不同实根，即f（x）=kx+1，则等价为函数f（x）与函数y=kx+1有两个不同的交点，当1＜x≤2，则0＜x﹣1≤1，则f（x）=f（x﹣1）=e x﹣1，当2＜x≤3，则1＜x﹣1≤2，则f（x）=f（x﹣1）=e x﹣2，当3＜x≤4，则2＜x﹣1≤3，则f（x）=f（x﹣1）=e x﹣3，…当x＞1时，f（x）=f（x﹣1），周期性变化；函数y=kx+1的图象恒过点（0，1）；作函数f（x）与函数y=kx+1的图象如下，C（0，1），B（2，e），A（1，e）；故k AC=e﹣1，k BC=；在点C处的切线的斜率k=e0=1；结合图象可得，实数k的取值范围为（，1）∪（1，e﹣1]；故答案为：三、解答题:本大题共6小题，共75分16．近日，济南楼市迎来去库存一系列新政，其中房产税收中的契税和营业税双双下调，对住房市场持续增长和去库存产生积极影响．某房地产公司从两种户型中各拿出9套进行促销活动，其中A户型每套面积100平方米，均价1.1万元/平方米，B户型每套面积80平方米，均价1.2万元/平方米．下表是这18套住宅平方米的销售价格：（单位：万元/平方米）：房号/户型 1 2 3 4 5 6 7 8 9A户型0.98 0.99 1.06 1.17 1.10 1.21 a 1.09 1.14B户型 1.08 1.11 1.12 b 1.26 1.27 1.26 1.25 1.28（I）求a，b的值；（II）张先生想为自己和父母买两套售价小于100万元的房子，求至少有一套面积为100平方米的概率．【考点】列举法计算基本事件数及事件发生的概率；分层抽样方法．【分析】（Ⅰ）由已知利用平均数公式能求出a，b．（Ⅱ）A户型小于100万的有2套，B户型小于100万的有4套，先求出买两套售价小于100万的房子所含基本事件总数，再列举法求出事件A=“至少有一套面积为100平方米住房所含基本事件个数，由此能求出至少有一套面积为100平方米的概率．【解答】解：（Ⅰ）由已知得：（0.98+0.99+1.06+1.17+1.10+1.21+a+1.09+1.14）=1.1，解得a=1.16，（1.08+1.11+1.12+b+1.26+1.27+1.26+1.25+1.28）=1.2，解得b=1.17．…（Ⅱ）A户型小于100万的有2套，设为A1，A2，B户型小于100万的有4套，设为B1，B2，B3，B4…买两套售价小于100万的房子所含基本事件总数为=15，…令事件A=“至少有一套面积为100平方米住房”，则A中所含基本事件有{A1，A2}，{A1，B1}，{A1，B2}，{A1，B3}，{A1，B4}，{A2，B1}，{A2，B2}，{A2，B3}，{A2，B4}，共9个…∴P（A）=，∴至少有一套面积为100平方米的概率为．．17．在△ABC中，内角A，B，C的对边分别为a，b，c，已知2ccosA+a=2b（Ⅰ）求角C的值；（Ⅱ）若c=2，且△ABC的面积为，求a，b．【考点】正弦定理；余弦定理．【分析】（Ⅰ）利用两角和的正弦函数公式，正弦定理，三角形内角和定理化简已知等式可得sinA=2sinAcosC，由于sinA≠0，解得，又C是三角形的内角，即可得解C的值．（Ⅱ）利用三角形面积公式可求ab=4，又由余弦定理可解得a+b=4，联立即可解得a，b的值．【解答】（本题满分为12分）解：（Ⅰ）∵2ccosA+a=2b，∴2sinCcosA+sinA=2sinB，…∴2sinCcosA+sinA=2sin（A+C），即2sinCcosA+sinA=2sinAcosC+2cosAsinC，∴sinA=2sinAcosC，∴，又∵C是三角形的内角，∴…（Ⅱ）∵，∴，∴ab=4，…又∵c2=a2+b2﹣2abcosC，∴4=（a+b）2﹣2ab﹣ab，∴a+b=4，∴a=b=2．…18．如图，四棱锥P﹣ABCD的底面为正方形，侧面PAD⊥底面ABCD，PA⊥AD，E，E，H分别为AB，PC，BC的中点（Ⅰ）求证：EF∥平面PAD；（Ⅱ）求证：平面PAH⊥平面DEF．【考点】平面与平面垂直的判定；直线与平面平行的判定．【分析】（Ⅰ）取CD中点N，连接FN，EN，则FN∥PD，EN∥AD，故而平面EFN∥平面PAD，所以EF∥平面PAD；（Ⅱ）由侧面PAD⊥底面ABCD可得PA⊥平面ABCD，故PA⊥DE，由正方形的性质可得DE⊥AH，故DE⊥平面PAH，于是平面PAH⊥平面DEF．【解答】证明：（Ⅰ）取CD中点N，连接FN，EN．∵在△CPD中，F，N为中点，∴FN∥PD．∵正方形ABCD中，E，N为中点，∴EN∥AD，∵EN⊂平面EFN，FN⊂平面EFN，EN∩FN=N，PD⊂平面PAD，AD⊂平面PAD，PD∩AD=D，∴平面EFN∥平面PAD，∵EF⊂平面EFN，∴EF∥平面PAD．（Ⅱ）∵侧面PAD⊥底面ABCD，PA⊥AD，侧面PAD∩底面ABCD=AD，∴PA⊥底面ABCD，∵DE⊂底面ABCD，∴DE⊥PA，∵E，H分别为正方形ABCD边AB，BC中点，∴Rt△ABH≌Rt△ADE，则∠BAH=∠ADE，∴∠BAH+∠AED=90°，则DE⊥AH，∵PA⊂平面PAH，AH⊂平面PAH，PA∩AH=A，∴DE⊥平面PAH，∵DE⊂平面EFD，∴平面PAH⊥平面DEF．19．已知数列{a n}为公差不为零的等差数列，其前n项和为S n，满足S5﹣2a2=25，且a1，a4，a13恰为等比数列{b n}的前三项（Ⅰ）求数列{a n}，{b n}的通项公式；（Ⅱ）设T n是数列{}的前n项和，是否存在k∈N*，使得等式1﹣2T k=成立，若存在，求出k的值；若不存在，说明理由．【考点】数列的求和；数列递推式．【分析】（I）利用等差数列与等比数列的通项公式及其前n项和公式即可得出；（II）利用“裂项求和”与数列的单调性即可得出．【解答】解：（Ⅰ）设等差数列{a n}的公差为d（d≠0），∴，解得a1=3，d=2，∵b1=a1=3，b2=a4=9，∴．（Ⅱ）由（I）可知：a n=3+2（n﹣1）=2n+1．，∴=，∴，单调递减，得，而，所以不存在k∈N*，使得等式成立．20．设椭圆C： +=1（a＞b＞0），定义椭圆C的“相关圆”方程为x2+y2=．若抛物线y2=4x的焦点与椭圆C的一个焦点重合，且椭圆C短轴的一个端点和两个焦点构成直角三角形（Ⅰ）求椭圆C的方程和“相关圆”E的方程；（Ⅱ）过“相关圆”E上任意一点P的直线l：y=kx+m与椭圆交于A，B两点，O为坐标原点，若OA⊥OB，证明原点O到直线AB的距离为定值，并求m的取值范围．【考点】椭圆的简单性质．【分析】（Ⅰ）由抛物线y2=4x的焦点为（1，0）与椭圆C的一个焦点重合，椭圆C短轴的一个端点和其两个焦点构成直角三角形，得到b=c=1，由此能求出椭圆C的方程和“相关圆”E 的方程．（Ⅱ）联立方程组得（1+2k2）x2+4kmx+2m2﹣2=0，由此利用根的判别式、韦达定理、点到直线距离公式，结合已知条件能证明原点O到直线AB的距离为定值，并能求出m的取值范围．【解答】解：（Ⅰ）因为若抛物线y2=4x的焦点为（1，0）与椭圆C的一个焦点重合，所以c=1又因为椭圆C短轴的一个端点和其两个焦点构成直角三角形，所以b=c=1故椭圆C的方程为，“相关圆”E的方程为…证明：（Ⅱ）设A（x1，y1），B（x2，y2）联立方程组得（1+2k2）x2+4kmx+2m2﹣2=0△=16k2m2﹣4（1+2k2）（2m2﹣2）=8（2k2﹣m2+1）＞0，即2k2﹣m2+1＞0…，由条件OA⊥OB得3m2﹣2k2﹣2=0…所以原点O到直线l的距离是由3m2﹣2k2﹣2=0得为定值．…此时要满足△＞0，即2k2﹣m2+1＞0，又，即，所以，即或…21．设函数f（x）=ax2+b（lnx﹣x），g（x）=﹣2+（1﹣b）x，已知曲线y=f（x）在点（1，f（1））处的切线与直线x﹣y+1=0垂直．（Ⅰ）求a的值；（Ⅱ）求函数f（x）的极值点；（Ⅲ）若对于任意b∈（1，+∞），总存在x1，x2∈[1，b]，使得f（x1）﹣f（x2）﹣1＞g（x1）﹣g（x2）+m成立，求实数m的取值范围．【考点】利用导数研究函数的极值；利用导数研究函数的单调性．【分析】（Ⅰ）求出函数的导数，得到f′（1）=2a=﹣1，求出a的值即可；（Ⅱ）求出f（x）的导数，结合二次函数的性质，通过讨论b的范围，确定函数的单调区间，求出函数的极值点即可；（Ⅲ）令F（x）=f（x）﹣g（x），x∈[1，b]，求出F（x）的导数，得到F（x）max﹣F（x）min=F（b）﹣F（1）=blnb﹣b+1，问题转化为即blnb﹣b＞m对任意b∈（1，+∞）成立．构造函数：t（b）=blnb﹣b，b∈[1，+∞），通过讨论函数t（b）的单调性，求出m的范围即可．【解答】解：（Ⅰ），所以k=f'（1）=2a=﹣1，所以…（Ⅱ），其定义域为（0，+∞），，令h（x）=﹣x2﹣bx+b，x∈（0，+∞）△=b2+4b（i）当﹣4≤b≤0时，△=b2+4b≤0，有h（x）≤0，即f'（x）≤0，所以f（x）在区间（0，+∞）上单调递减，故f（x）在区间（0，+∞）无极值点；（ii）当b＜﹣4时，△＞0，令h（x）=0，有，，x2＞x1＞0，当x∈（0，x1）时，h（x）＜0，即f'（x）＜0，得f（x）在（0，x1）上递减；当x∈（x1，x2）时，h（x）＞0，即f'（x）＞0，得f（x）在（x1，x2）上递增；当x∈（x2，+∞）时，h（x）＜0，即f'（x）＜0，得f（x）在（x2，+∞）上递减．此时f（x）有一个极小值点和一个极大值点．（iii）当b＞0时，△＞0，令h（x）=0，有，，当x∈（0，x2）时，h（x）＞0，即f'（x）＞0，得f（x）在（0，x2）上递增；当x∈（x2，+∞）时，h（x）＜0，即f'（x）＜0，得f（x）在（x2，+∞）上递减．此时f（x）唯一的极大值点，无极小值点．综上可知，当b＜﹣4时，函数f（x）有一个极小值点和一个极大值点．当﹣4≤b≤0时，函数f（x）在（0，+∞）上有无极值点；当b＞0时，函数f（x）有唯一的极大值点，无极小值点；…（III）令F（x）=f（x）﹣g（x），x∈[1，b]，则F（x）==blnx﹣x若总存在x1，x2∈[1，b]，使得f（x1）﹣f（x2）﹣1＞g（x1）﹣g（x2）+m成立，即总存在x1，x2∈[1，b]，使得f（x1）﹣g（x1）＞f（x2）﹣g（x2）+m+1成立，即总存在x1，x2∈[1，b]，使得F（x1）﹣F（x2）＞m+1成立，即F（x）max﹣F（x）min＞m+1，因为x∈[1，b]，所以F'（x）≥0，即F（x）在[1，b]上单调递增，所以F（x）max﹣F（x）min=F（b）﹣F（1）=blnb﹣b+1，即blnb﹣b+1＞m+1对任意b∈（1，+∞）成立，即blnb﹣b＞m对任意b∈（1，+∞）成立．构造函数：t（b）=blnb﹣b，b∈[1，+∞），t'（b）=lnb，当b∈[1，+∞）时，t'（b）≥0，∴t（b）在[1，+∞）上单调递增，∴t（b）min=t（1）=﹣1．∴对于任意b∈（1，+∞），∴t（b）＞t（1）=﹣1．所以m≤﹣1…2020年9月12日。

山东省2020年新高考模拟济南一模试题（含答案及解析）英语第一部分阅读（共两节，满分50分）第一节（共15小题;每小题2. 5分，满分37. 5分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项。

AIf you want to learn why everything is usually cheaper, and what items are the best deals at your local or chain dollar or discount store, store employees have a lot of useful information. They tend to know what shoppers want — even before shoppers know themselves.Everyone buys unnecessary thingsWhether you shop at a chain dollar store such as Dollar Tree, Family Dollar or your local 99 cents store, not everything that’s on sale is necessarily worth purchasing. Bryan Waring, a former Dollar Tree employee, says that you’re not alone in buying more than necessary from these stores. “It seems basic, but everyone falls for this trap,” he says. “You go into a store where everything is cheap, and you walk out with things you don't need.” He suggests going into a store — yes, even the dollar store — with a checklist of things you truly need.Everything is cheaper after the holiday seasonPatricia, a seasonal worker at a Dollar Tree, says that the post-Christmas season means even more deals on everything from decorations to sweets. “After Christmas, all the gift wrap paper went to 50 cents, and all Christmas items were half price,” she says. “Even candy bars are 89 cents versus $ 1.”Products are less expensive because of their sizeIn order for dollar stores to keep their prices low, product sizes are usually smaller than normal, according to Cheapism. Dollar stores aren't the only ones guilty of this trick. Cheapism also reports Walmart is guilty of doing the same thing to attract customers.1. What is Bryan’s advice against buying unnecessary things?A. Making a purchase alone.B. Writing a to-buy list ahead.C. Shopping at your local store.D. Buying basic things separately.2. Which of the following is a better time for shopping according to Patricia?A. In the Christmas sales.B. At a particular discount.C. After the Christmas season.D. During some holiday seasons.3. How do stores make their products cheaper? A. By reducing product sizes. B. By lowering product costs.C. By adopting discount strategies.D. By attracting more customers.BIt was just a normal day for Ruth Miller, a 63-year-old woman until everything went horribly wrong. She was walking to her car after shopping when the unthinkable happened.Right as she was unlocking her car, a man quickly came up behind her and tried to wrestle her purse away. She was in shock. Luckily she remembered she had her Safe Personal Alarm (SPA) on her purse, and since she was too scared to scream for help, she quickly reached for the alarm and pulled the pin (保险栓). Immediately her SPA started just screaming. The man didn't know what to do! He froze for a second, and then ran away like a bat out of hell!SPA is a safety device capable of creating a 125db sound that attracts attention and scares away potential attackers. To compare, it's the same volume as a military jet during takeoff.Paul Davidson, the inventor of SPA, knows all too well the type of situation that Ruth found herself in. But that's not the only type of situation that SPA helps protect against. Parents can give it to their kids as an extra means of protection. Teenagers can use it so they can feel safe walking home. Women can know it's there when they have to use the parking lot at night. “My mother, who is 76 years old, carries it around in case she falls and needs to ask people for help. I only wish I'd have thought of it earlier,” said Paul.The police have been recommending SPA since it first hit the market. In fact, since its launch, SPA has been in a state, shifting between in stock to sold out nearly every other week, and it's also got tons of loyal followers worldwide.4. What does the underlined part “the unthinkable” in Paragraph 1 refer to?A. An attempted robbery.B. A wrestling match.C. An angry argument.D. A car accident.5. How did Ruth react to the unexpected situation?A. She fought violently.B. She froze in great fear.C. She cried desperately for help.D. She sounded her safety device.6. Why does Paul mention his mother?A. To imply the elderly need more care.B. To suggest he cares about his mother.C. To show SPA can be widely used.D. To make an advertisement for SPA.7. What can be learned from the text?A. SPA is well received in the market.B. People hesitate to pay for security.1C. SPA was sold out in the first two weeks.D. SPA is not important in life.CLike clockwork, nearly every fourth February includes one extra day. February 29th, otherwise known as Leap Day, isn’t exactly a holiday. Instead, it’s there to keep your calendar consistent with the earth’s rotation (旋转) around the sun.According to History, com, Roman emperor Julius Caesar is the “father” of Leap Year. Until he came along, people used a 355-day calendar, which was 10.25 days shorter than the solar year. Roman officials were supposed add an extra month every now and then to keep the seasons exactly where they should be. But that didn’t work out all that well. When special o ccasions started shifting into different seasons around 45 BCE, Caesar consulted with astronomers and decreed (下令) that the empire should use a 12-month, 365-day calendar, which he named after himself, Caesar’s Julian calendar included a Leap Day every fou r years.Though Leap Day keeps your calendar in line with the earth’s rotation around the sun, it causes a different kind of problem for leapsters. When should these February 29th babies celebrate their birthdays during the other three-quarters of their lives? Some party on February 28th, while others prefer a two-day celebration that spans the last day of February and the first day of March.Leap Day can be a nuisance in the legal system. In 2006, a court in Massachusetts was deciding whether criminal John Melo could be released a day early since his 10-year sentence included a Leap Day. In the case, the judge decided that since the man was sentenced to prison for years, not days. Leap Day didn’t make a bit of difference.Though a few timekeepers have pushe d for calendars that don’t include Leap Day, almost all astronomers and societies agree that Leap Day is the best method to keep the calendar on track.8. Why was Leap Day created?A. To celebrate special occasions.B. To honor Emperor Julius Caesar.C. To keep pace with the solar year.D. To keep track of all the seasons.9. What is the problem with the birthday celebration of February 29th babies?A. It is sometimes delayed.B. It lasts at least two days.C. It has to be held every other year.D. It may take place on different dates.10. What does the underlined word “nuisance” in Paragraph 4 mean?A. Joke.B. Topic.C. Trouble.D. Mistake.11. what is the attitude of most astronomers towards Leap Day?A. Critical.B. Supportive.C. Doubtful.D. Cautious.DScientists have developed a new type of smart bandage (绷带) that can signal the type of bacterial (细菌的) infection it’s protecting, just like a traffic light, as well as release the right type of drugs on demand. The traffic light system works just like this: Green means no bacteria or a low concentration of bacteria, yellow means drug-sensitive (DS) bacteria responsive to standard antibiotics (抗生素) and causes antibiotic release, and red means drug-resistant (DR) bacteria that need extra help to be wiped out.In testing the bandage on mice, the research team was able to successfully treat both DS and DR infections using the new method. However, the common methods of sensing resistance are limited by time, the requirement for professional personnel, and expensive instruments. Moreover, the abuse of antibiotics causes the accelerated process of bacterial resistance.It’s easy to see how a simple bandage and light could overcome some of these limitations. Treatment doesn’t have to wait for a doctor to make a diagnosis, and the bandage can get the right sort of drugs applied at the earliest opportunity. What’s more, the person wearing the bandage gets real-time feedback on what’s happening with the infection, if there’s an infection at all. The researchers say it offers numerous benefits over existing treatments that make use of light, including photodynamic therapy or PDT.We’ve been seeing quite a few upgrades to the traditional bandage in recent years, thanks to advances in science — like the nanofiber mesh that attracts bacteria and draws some of it out, speeding up the healing process. Then there’s the novel bandage for treating burns, which stops bacteria from multiplying and lowers the risk of infection.The more work that a bandage can do while it’s protecting a wound, the better. Efforts to improve bandages continue and now we've got a bandage that not only releases antibiotics, but also tells the patient exactly what’s going on too.12. What is the smart bandage mainly designed to do?A. Avoid the use of antibiotics.B. Clear out harmful bacteria.C. Detect bacterial infections.D. Increase treatment options.13. What is the advantage of the smart bandage?A. It saves much time and cost.B. It removes the risk of infection.C. It prevents the bacterial resistance.D. It improves doctor-patient relationship.14. What can be inferred from the last two paragraphs?A. Traditional bandages are out of use now.B. More smart bandages will be developed.2C.Progress in science calls for more research.D. People are urged to study medical science.15. What does the text focus on?A. A successful test on mice.B. A colour-changing bandage.C. Sensing drug-resistant bacteria.D. Preventing abuse of antibiotics.第二节（共5小题;每小题2. 5分，满分12. 5分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

山东省济南市2020届高三英语第一次模拟考试试题（含解析）本试卷共12页。

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考试用时120分钟。

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录音内容结束后，你将有两分钟的时间将试卷上的答案转涂到答题卡上。

第一节(共5小题；每小题1.5分，合计7.5分)听下面5段对话。

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1.【此处有音频，请去附件查看】Where are probably the speakers?A. At a concert.B. In a restaurant.C. In a cinema.【答案】B【解析】【详解】此题为听力题，解析略。

2.【此处有音频，请去附件查看】When does the conversation take place?A. At 2:45 P.m.B. At 3:00 P.m.C. At 3:15 P.m.【答案】C【解析】【详解】此题为听力题，解析略。

3.【此处有音频，请去附件查看】What is the woman going to do now?A. Look for her keys.B. Go to work by bus.C. Clean up the room.【答案】B【解析】【详解】此题为听力题，解析略。

2020届山东省济南市高考数学一模试卷一、单选题（本大题共12小题，共60.0分）1. 若复数z 满足z =，则z 对应的点位于复平面的( )A. 第一象限B. 第二象限C. 第三象限D. 第四象限 2. 已知集合A ={x||x|≤2,x ∈Z}，B ={x|1x+1≤0,x ∈R}，则A ∩∁R B =( )A. (−1,2]B. [−1,2]C. {−1,0，1，2}D. {0,1，2}3. 已知等比数列{a n }满足a n >0，n =1.2，…，且a 5・a 2n−5=22n (n ≥2).则当n ≥1时，log 2a 1+log 2a 3+⋯+log 2a 2n−1=.A. n(2n −1)B. (n +1)2C. n 2D. (n −1)24. 在等比数列{a n }中，a 5=3，则a 1⋅a 2⋅a 3…a 9=39，若数列{b n }为等差数列，b 5=3，则数列{b n }的类似结论为( )A. b 1b 2…b 9=39B. b 1+b 2+⋯+b 9=39C. b 1b 2…b 9=3×9D. b 1+b 2+⋯+b 9=3×95. 若x ，y 为不等式组{x +y ≥12x −y ≤2y −2≤0表示的平面区域中的一点，且使得mx +y 取得最小值的点(x,y)有无数个，则m =( )A. 1B. 2C. −1D. 1或−26. 我国古代在珠算发明之前多是用算筹为工具来记数、列式和计算的．算筹实际上是一根根相同长度的小木棍，如图，算筹表示数1～9的方法有“纵式”和“横式”两种，规定个位数用纵式，十位数用横式，百位数用纵式，千位数用横式，万位数用纵式，…，以此类推，交替使用纵横两式．例如：627可以表示为“”.如果用算筹表示一个不含“0”且没有重复数字的三位数，这个数至少要用7根小木棍的概率为( )A. 1114B. 1721C. 2021D. 79847. 执行如图所示的程序，若输入的x =3，则输出的所有x 的值的和为( )A. 243B. 363C. 729D. 10928. 下列命题中是假命题的是( ) A.; B. 使得函数是偶函数； C. 使得； D. 是幂函数，且在上递减；9. 已知四棱柱的侧棱长为2，且侧棱垂直于底面，底面是边长为2且有一个内角为60°的菱形，若该四棱柱的俯视图的面积与四棱柱的底面积相等，则该四棱柱左视图面积的最小值是( )A. 4√3B. 2√3C. 2D. √3 10. 已知点，在单位圆上， (为坐标原点)，则的取值范围是 A. B. C. D.11. a ⃗ =(8+12x,x)，b ⃗ =(x +1,2)(其中x >0)，若a ⃗ //b ⃗ ，则x 的值为( )A. 8B. 4C. 2D. 012. 下列命题中是真命题的是( )A. x ∈R ，使得sinxcosx =B. x ∈(−∞,0)，2x >1C. x ∈R ，x 2≥x +1D. x ∈(0,)，tanx >sinx二、单空题（本大题共4小题，共20.0分）13. 设G 为△ABC 的重心，且sinA ⋅GA ⃗⃗⃗⃗⃗ +sinB ⋅GB ⃗⃗⃗⃗⃗ +sinC ⋅GC ⃗⃗⃗⃗⃗ =0⃗ ，则角B 的大小为______．14.将全体正偶数排成一个三角数阵：按照以上排列的规律，第n行(n≥3)从左向右的第3个数为______．+x的最小值为______ ．15.已知x>3，则函数y=1x−316.已知数列{a n}是公比为2的等比数列，且a3，a4+2，a5成等差数列，则数列{a n}的前5项和S5=______．三、解答题（本大题共7小题，共82.0分）17.△ABC的内角A、B、C的对边分别为a、b、c，面积为S，已知3a2−4√3S=3b2+3c2．(1)求A；(2)若a=3，求△ABC周长的取值范围．18.如图，在四棱锥P−ABCD中，PC⊥底面ABCD，底面ABCD是直角梯形，AB⊥AD，AB//CD，AB=2AD=2CD=2，E是PB的中点．(1)求证：平面EAC⊥平面PBC；(2)若二面角P−AC−E的余弦值为，求直线PA与平面EAC所成角的正弦值．(a,b∈R)，若f(x)在点(1,f(1))处的切线斜率为1．19.设函数f(x)=ax+bx(Ⅰ)用a表示b；(Ⅱ)设g(x)=lnx−f(x)，若g(x)≤−1对定义域内的x恒成立，求实数a的取值范围．20.某社区居委会用“百分制”调查该社区居民对社区的治安满意度，现从调查的居民中随机选取14名，将他们的治安满意度分数的数据绘制成如下的茎叶图，若治安满意度不低于89分，则称该居民对社区的治安满意度为“非常好”．(1)若从这14人中任选3人，求至多有2人对社区治安满意度为“非常好”的概率；(2)若从这14人中任选2人，记X表示这2人中对社区治安满意度为“非常好”的人数，求X的分布列及数学期望．21. 已知函数f(x)=a |x|−|x|+2．(1)若x ∈[12,2]时f(x)≥0恒成立，求实数a 的取值范围；(2)当a =−34时，求函数f(x)在x ∈[−2,0)上的最大值．22. 在直角坐标系中，以原点为极点，x 轴的正半轴为极轴建坐标系，已知曲线C ：ρsin 2θ=2acosθ(a >0)，已知过点P(−2,−4)的直线l 的参数方程为{x =−2+t y =−4+t，直线l 与曲线C 分别交于M ，N ．(1)写出曲线C 和直线l 的普通方程；(2)若|PM|，|MN|，|PN|成等比数列，求a 的值．23.已知函数f(x)=|x+2|+|2x−4|．(1)求f(x)<6的解集；(2)若关于x的不等式f(x)≥m2−3m的解集是R，求m的取值范围．【答案与解析】1.答案：B解析：试题分析：对应的点为，位于第二象限，故B正确．考点：复数的运算、复数的几何意义．2.答案：C解析：解：∵集合A={x||x|≤2,x∈z}={−2,−1,0，1，2}，B={x|1≤0,x∈R}={x|x<−1}，x+1∴C R B={x|x≥−1}，∴A∩∁R B={−1,0，1，2}．故选：C．先求出集合B，再求出C R B，由此利用交集定义能求出A∩∁R B.本题考查的知识点是集合的交集，补集运算，集合的包含关系判断及应用，难度不大，属于基础题．3.答案：C解析：解析：，所以a n=2n，又有log2a1+log2a3+⋯+log2a2n−1=log2(a1×a3×…×a2n−1)=log2(2n)n=n2．4.答案：D解析：等差和等比的类比时，在等比中为积在等差中为和，按此规律写出规律即可．本题考查等差和等比数列的类比、考查利用所学知识解决问题的能力．解：因为在等比数列中有a1⋅a9=a2⋅a8=a3⋅a7=a4⋅a6=a52有a1⋅a2⋅…⋅a9=a59，而等差数列中有b1+b9=b2+b8=b3+b7=b4+b6=2b5，故在等差数列{b n}中，类似地，有b1+b2+⋯+b9=9b5=3×9．故选D ．5.答案：D解析：解：作出不等式组{x +y ≥12x −y ≤2y −2≤0对应的平面区域：由题意，z =mx +y 取得最小值的最优解有无数个，最优解应在线段AC 或BC 上取到，故mx +y =0应与直线AC 或BC 平行，∴−m =−1，或−m =2即m =1或m =−2．故选：D ．由题设条件，目标函数z =x +ay ，取得最小值的最优解有无数个知取得最优解必在边界上而不是在顶点上，故目标函数的斜率为正，最小值应在左上方边界AC 上取到，即ax +y =0应与直线AC 或BC 平行，进而计算可得a 值．本题考查线性规划最优解的判定，作出不等式组对应的平面区域，利用目标函数的几何意义，结合数形结合进行求解是解决本题的关键．6.答案：D解析：解：至少要用7根小木棍的对立事件为用5根小木棍和6根小木棍这两种情况， 用5根小木棍为126这一种情况的全排列，用6根小木棍为123，127，163，167这四种情况的全排列，故至少要用7根小木棍的概率为1−5A 33A 93=7984． 故选：D ．利用已知条件，推出对立事件的个数，利用古典概型概率的求法，转化求解即可．本题考查古典概型概率的求法，对立事件的概率的求法，分析题意的解题的关键，中档题． 7.答案：D解析：由已知中的程序算法可知：该程序的功能是利用循环结构计算并输出变量x的值，模拟程序的运行过程，分析循环中各变量值的变化情况，可得答案．本题考查了程序框图的应用问题，解题时应模拟程序框图的运行过程，分析程序的功能，以便得出正确的结论，是基础题．解：模拟程序的运行可得：当x=3时，y是整数，当x=32时，y是整数，依此类推可知当x=3n(n∈N∗)时，y是整数，则由x=3n>1000，得n≥7，所以输出的所有x的值为3，9，27，81，243，729，其和为1092．故选D．8.答案：A解析：答案A当时，，所以该命题是假命题，选A．考点：全称命题与特称命题真假判断．9.答案：B解析：解：由已知四棱柱的侧棱长为2，且侧棱垂直于底面，底面是边长为2且有一个内角为60°的菱形，若该四棱锥的俯视图的面积与四棱柱的底面积相等，说明棱柱是放倒图形，如图：侧视图是菱形，侧视图的面积的最小值为：2×2sin60°=2√3．故选：B．通过题意判断四棱柱放置的形状，画出图形，然后确定侧视图的形状，即可得到结果．本题考查简单几何体的三视图的画法，视图面积是的求法．考查空间想象能力以及计算能力．10.答案：B解析：本题考查三角函数的定义、向量的数量积、三角函数的化简求值，考查计算能力．求出的表达式，结合角的范围，求出其取值范围．解：设，，则，则则，当中一个为240°，另一个为120°时，有最大值为，当中一个为60°，另一个为−60°时，有最小值为，故选B．11.答案：B解析：解：∵a ⃗ //b ⃗ ，且x >0； ∴2(8+12x)−x(x +1)=0； 解得x =4，或x =−4(舍去)． 故选：B ．根据a ⃗ //b ⃗ 即可得出2(8+12x)−x(x +1)=0，再根据x >0，即可解出x 的值． 考查向量坐标的定义，以及向量平行时的坐标关系．12.答案：D解析：当x ∈(0,)时，0<cosx <1，0<sinx <1， ∴>sinx ，即tanx >sinx ．13.答案：π3解析：解：∵G 是△ABC 的重心，∴GA ⃗⃗⃗⃗⃗ +GB ⃗⃗⃗⃗⃗ +GC ⃗⃗⃗⃗⃗ =0⃗ ，则GB ⃗⃗⃗⃗⃗ =−(GA ⃗⃗⃗⃗⃗ +GC ⃗⃗⃗⃗⃗ ) ∵sinA ⋅GA⃗⃗⃗⃗⃗ +sinB ⋅GB ⃗⃗⃗⃗⃗ +sinC ⋅GC ⃗⃗⃗⃗⃗ =0⃗ ， ∴sinA ⋅GA ⃗⃗⃗⃗⃗ −sinB ⋅(GA ⃗⃗⃗⃗⃗ +GC ⃗⃗⃗⃗⃗ )+sinC ⋅GC ⃗⃗⃗⃗⃗ =0⃗ ，即(sinA −sinB)GA ⃗⃗⃗⃗⃗ +(sinC −sinB)GC ⃗⃗⃗⃗⃗ =0⃗ ， 又∵GA ⃗⃗⃗⃗⃗ 与GC⃗⃗⃗⃗⃗ 不共线， ∴sinA −sinB =sinC −sinB =0， ∴sinA =sinB =sinC ． ∴a =b =c ． ∴A =B =C =π3．故答案为：π3．由G 是△ABC 的重心，可得GA⃗⃗⃗⃗⃗ +GB ⃗⃗⃗⃗⃗ +GC ⃗⃗⃗⃗⃗ =0⃗ .又sinA ⋅GA ⃗⃗⃗⃗⃗ +sinB ⋅GB ⃗⃗⃗⃗⃗ +sinC ⋅GC ⃗⃗⃗⃗⃗ =0⃗ ，可得sinA ⋅GA ⃗⃗⃗⃗⃗ −sinB ⋅(GA ⃗⃗⃗⃗⃗ +GC ⃗⃗⃗⃗⃗ )+sinC ⋅GC ⃗⃗⃗⃗⃗ =0⃗ ，即(sinA −sinB)GA ⃗⃗⃗⃗⃗ +(sinC −sinB)GC ⃗⃗⃗⃗⃗ =0⃗ ，由于GA ⃗⃗⃗⃗⃗ 与GC ⃗⃗⃗⃗⃗ 不共线，可得sinA −sinB =sinC −sinB =0，即可得出a =b =c ．本题考查了三角形的重心性质、共面向量定理、正弦定理，考查了推理能力和计算能力，属于难题．14.答案：n2−n+6解析：解：观察三角形数阵知第n行有n个正偶数，则第n行(n≥3)前共有1+2+3+⋯+(n−1)=(n−1)n2个数，所以第n行(n≥3)从左向右的第3个数为2[(n−1)n2+3]=n2−n+6，故答案为：n2−n+6．首先找出三角形数阵的规律，求出前n−1行正偶数的个数，然后由偶数的特点求出第n行第3个偶数．本题考查了归纳推理，难点在于发现其中的规律，考查观察、分析、归纳能力．15.答案：5解析：解：x>3，则函数y=1x−3+x=1x−3+x−3+3≥2√(x−3)⋅1x−3+3=2+3=5，当且仅当x=4时取等号，故函数y=1x−3+x的最小值为5，故答案为：5．根据基本不等式即可求出最小值．本题考查了基本不等式的应用，关键掌握一正二定三相等，属于基础题．16.答案：31解析：解：由题意可得2(a4+2)=a3+a5，即2(8a1+2)=4a1+16a1，解得a1=1．∴S5=1×(1−25)1−2=31．故答案为：31．由已知列式求得等比数列的首项，再由等比数列的前n项和求解S5．本题考查等差数列和等比数列的通项与前n项和，是基础题．17.答案：解：(1)∵S=12bcsinA，∴由已知得：b2+c2−a2=−4√33S=−4√33⋅12bcsinA，∴化简得：b2+c2−a22bc =−√33sinA=cosA，∴tanA=−√3，A∈(0,π)，∴A=2π3．(2)在△ABC中，由正弦定理得：．∴b=2√3sinB，c=2√3sinC=2√3sin(π3−B)，记△ABC周长为y，∴y=a+b+c=2√3sinB+2√3sin(π3−B)+3．化解得：y=2√3sinB+2√3(√32cosB−12sinB)+3=2√3sin(B+π3)+3．∵B∈(0,π3)，∴周长y∈(6,3+2√3]综上所述：△ABC周长的取值范围(6,3+2√3].解析：本题考查了正弦、余弦定理，以及三角形的面积公式，熟练掌握公式及定理是解本题的关键．(1)已知等式利用面积、余弦定理化简，整理后求出A的度数即可；(2)记△ABC周长为y，y=a+b+c=2√3sinB+2√3sin(π3−B)+3=2√3sin(B+π3)+3.根据B∈(0,π3)，可得ABC周长的取值范围．18.答案：(1)见解析(2)解析：(1)∵PC⊥平面ABCD，AC⊂平面ABCD，∴AC⊥PC.∵AB=2，AD=CD=1，∴AC=BC=．∴AC2+BC2=AB2.∴AC⊥BC．又BC∩PC=C，∴AC⊥平面PBC．∵AC ⊂平面EAC ， ∴平面EAC ⊥平面PBC ． (2)如图，以点C 为原点，，，分别为x 轴、y 轴、z 轴正方向，建立空间直角坐标系，则C(0,0，0)，A(1,1，0)，B(1,−1,0)，设P(0,0，a)(a >0)， 则E，=(1,1，0)，=(0,0，a)，=.取m =(1,−1,0)，则m ·=m ·=0，m 为面PAC 的法向量．设n =(x,y ，z)为面EAC 的法向量，则n ·=n ·=0，即取x =a ，y =−a ，z =−2，则n =(a,−a,−2)，依题意，|cos 〈m ，n 〉|===，则a =2.于是n =(2,−2,−2)，=(1,1，−2).设直线PA 与平面EAC 所成角为θ，则sinθ=|cos 〈，n 〉|==，即直线PA 与平面EAC 所成角的正弦值为19.答案：解：(Ⅰ)函数的导数为f′(x)=a −bx 2，因为f(x)在点(1,f(x))处的切线斜率为1， 所以f′(1)=a −b =1，解得b =a −1； (Ⅱ)因为g(x)=lnx −f(x)， 所以g(x)=lnx −f(x)=lnx −(ax +a−1x)=lnx −ax −a−1x，要使g(x)≤−1恒成立，即g(x)max ≤−1． g′(x)=1x −a +a−1x =−ax 2+x+a−1x =−(ax+a−1)(x−1)x ，①当a =0时，g′(x)=x−1x 2，当x ∈(0,1)，g′(x)<0，g(x)单调递减，当x∈(1,+∞)，g′(x)>0，g(x)单调递增，则g(x)min=g(1)=1，不符题意；②当a≠0时，g′(x)=−(ax+a−1)(x−1)x2=−a[x−(−1+1a)](x−1)x2=0⇒x=1,x=−1+1a，(1)若a<0，−1+1a<0，当x∈(0,1)，g′(x)<0，g(x)单调递减；当x∈(1,+∞)，g′(x)>0，g(x)单调递增，则g(x)min=g(1)=1−2a>1>−1，不符题意；(2)若a>0，若0<a≤12，−1+1a>1，当x∈(0,1)，g′(x)<0，g(x)单调递减，这时g(−1+1a )=ln(−1+1a)+2a−1>−1，不符题意；若12<a<1，0<−1+1a<1，x∈(0,−1+1a)，g′(x)<0，g(x)单调递减，这时g(1)=1−2a>1−2=−1，不符题意；若a≥1，−1+1a≤0，x∈(0,1)，g′(x)>0，g(x)单调递增；当x∈(1,+∞)，g′(x)<0，g(x)单调递减，则g(x)max=g(1)=1−2a≤−1，符合题意；综上，得g(x)≤−1恒成立，实数a的取值范围为a≥1．解析：(Ⅰ)由f(x)在点(1,f(1))处的切线斜率为1，则得到f′(1)=1，进而可得结果；(Ⅱ)由于g(x)≤−1恒成立，等价于g(x)max≤−1.利用导数可求得函数的最大值，可验证此时满足要求，从而得到a的范围．本题考查导数的几何意义、利用导数求函数的最值，考查转化思想，本题综合性强，运算量大，对能力要求较高．20.答案：解：(1)由茎叶图得这14人中对社区治安满意度为“非常好”的人数为5，从这14人中任选3人，基本事件总数n=C143=364，至多有2人对社区治安满意度为“非常好”包含的基本事件个数：m=C143−C53=354，∴至多有2人对社区治安满意度为“非常好”的概率p=mn =354364=177182．(2)由题意得X的可能取值为0，1，2，P(X=0)=C92C142=3691，P(X=1)=C91C51C142=4591，P(X=2)=C52C142=1091∴X的分布列为：E(X)=0×3691+1×4591+2×1091=6591．解析：(1)由茎叶图得这14人中对社区治安满意度为“非常好”的人数为5，从这14人中任选3人，基本事件总数n=C143=364，至多有2人对社区治安满意度为“非常好”包含的基本事件个数：m= C143−C53=354，由此能求出至多有2人对社区治安满意度为“非常好”的概率．(2)由题意得X的可能取值为0，1，2，分别求出相应的概率，由此能求出X的分布列和数学期望．本题考查概率的求法，考查离散型随机变量的概率分布、数学期望的求法，考查排列组合、古典概型等基础知识，考查运算求解能力，考查函数与方程思想，是中档题．21.答案：解：(1)由x∈[12,2]时f(x)≥0恒成立，可得ax−x+2≥0恒成立，故a≥x2−2x在x∈[12,2]时恒成立，由二次函数的性质可知，x∈[12,2]，y=x2−2x∈[−1,0]，故a≥0，∴实数a的取值范围[0,+∞)，(2)当a=−34时，x∈[−2,0)，f(x)=34x+x+2，结合对勾函数的性质可知，f(x)在[−2,0)上先增后减，当x=−√32时，函数取得最大值2−√3．解析：(1)由已知进行分离参数后转化为求解二次函数的范围，即可求解a 的范围； (2)把a 的值代入后，然后结合对勾函数的性质可求．本题主要考查了由不等式的恒成立求解参数范围问题，体现了转化思想的应用．22.答案：解：(1)由曲线C ：ρsin 2θ=2acosθ(a >0)，可得ρ2sin 2θ=2aρcosθ，化为y 2=2ax ．由直线l 的参数方程为{x =−2+ty =−4+t ，消去参数t 可得直线l ：y =x −2．(2)联立{y =x −2y 2=2ax，化为x 2−(4+2a)x +4=0， ∵直线l 与抛物线相交于两点，∴△=(4+2a)2−16>0，解得a >0或a <−4.(∗) ∴x 1+x 2=4+2a ，x 1x 2=4．∴|MN|=√(1+1)[(x 1+x 2)2−4x 1x 2]=√2[(4+2a)2−16]=√8a 2+32a ． |PM|=√(x 1+2)2+(y 1+4)2=√2|x 1+2|，|PN|=√2|x 2+2|．∴|PM||PN|=2|(x 1+2)(x 2+2)|=2|x 1x 2+2(x 1+x 2)+4| =2|16+4a|∵|PM|，|MN|，|PN|成等比数列， ∴|MN|2=|PM||PN|，∴(√8a 2+32a)2=2|16+4a|， 化为a(4+a)=|4+a|， ∵a >0或a <−4． 解得a =1． ∴a =1．解析：(1)利用极坐标化为直角坐标方程的公式x =ρcosθ，y =ρsinθ可得曲线C 的方程；消去参数t 即可得到直线l 的方程；(2)把直线的方程代入抛物线的方程得到根与系数的关系，利用两点间的距离公式和等比数列的定义即可得出．本题考查了把极坐标方程化为直角坐标方程、参数方程化为普通方程、直线与抛物线相交问题转化为把直线的方程与抛物线的方程联立得到根与系数的关系、两点间的距离公式和等比数列的定义等基础知识与基本技能方法，属于难题．.23.答案：解：(1)由题设知：当x≥2时，不等式等价与x+2+2x−4<6，即2≤x<83当2>x>−2时，不等式等价与x+2+4−2x<6，即2>x>0.当x≤−2时，不等式等价于−x−2+4−2x<6，x无解．}.综上可得，满足不等式的解是{x|0<x<83(2)由函数f(x)的图象可得f(x)=|x+2|+|2x−4|的最小值为4，则由题意可得m²−3m≤4，解之得，−1≤m≤4．即m的范围为−1≤m≤4．解析：本题考查带绝对值的函数的应用，绝对值不等式的解法，绝对值的几何意义是解题的关键．(1)分当x≥2时、当2>x>−2时，当x≤−2时三种情况，分别去掉绝对值，求得不等式的解集，再取并集，即得所求．(2)求出函数的最小值，然后求解m²−3m≤4，得到实数m的取值范围．。

保密★启用前济南市2020届高三年级统一第一次模拟考试语文试题注意事项:1.答卷前，考生务必将自己的姓名、考号和座位号填写在答题卡上。

2.回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑。

如需改动，用橡皮擦干净后，再选涂其它答案标号。

回答非选择题时，用0.5毫米的黑色签字笔将答案写在答题卡上。

写在本试卷上无效。

3.考试结束后，将答题卡交回。

一、现代文阅读（35分）（一）现代文阅读I（本题共5小题，19分）阅读下面的文字，完成1～5题。

材料一:我们应该把焦点放在人类会话的形式上。

我形象地使用“会话”这个词，并不仅仅指语言，同时也指一切使某个文化中的人民得以交流信息的技巧和技术。

在这样的意义上，整个文化就是一次会话，或者更准确地说，是以不同象征方式展开的多次会话的组合。

信息内容在缺乏媒介的世界里是不能存在的。

我并不是说，火灾战争这些从来没有在这个世界的任何地方发生过，我想说的是，如果没有用来宣传它们的技术，人们就无法了解，无法将这一切纳入自己的日常生活，这些信息就不能作为文化的内容而存在。

马歇尔·麦克卢汉有一句著名的警句“媒介即信息”。

他认为，深入一种文化最有效的途径是了解这种文化用于会话的工具。

语言无愧为一种原始而不可或缺的媒介，它使我们成为人，保持人的特点，事实上还定义了人的含义。

但这并不是说，除了语言之外没有任何其他媒介，人们还能够同样方便地以另一种方式讲述同样的事情。

虽然文化多是语言的产物，但是每一种媒介都会对它进行再创造——从绘画到象形符号，从字母到电视。

和语言一样，每一种媒介都为思考、表达思想和抒发情感的方式提供了新的定位，从而创造出独特的话语符号。

这就是麦克卢汉所说的“媒介即信息”。

摘编自尼尔·波兹曼《娱乐至死》材料二:最深的恐惧是对未知的恐惧，面对新型冠状病毒这一全然陌生的敌人，阅读新闻，成为一个公民在如此灾难面前，保持知情、消解恐慌、积极防护最简单最有效的办法。

2020届济南中学高中部高三英语一模试题及答案解析第一部分阅读（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、D四个选项中选出最佳选项AMany workers have had no choice but to adapt to working from home in recent months since offices shut down due to the COVID-19 pandemic (新冠疫情). And the noisy situation and endless housework may result in a terrible emotion. A new option is waiting foryou. That is WFH: work from a hotel.Hotel FigueroA special program titled Work Perks aims to reposition some of 94-year-old Hotel Figuero’s 268 rooms as day-use offices.According to Managing Director Connie Wang, the set-up launched in June and is a great opportunity to get out of their houses with high-speed Wi-Fi, unlimited printing privileges and free parking. The 350-square-foot rooms sell for $ 129 per day, with an option to extend to an overnight stay for an additional $ 20.The WytheA boutique hotel inBrooklyn. The hotel recently announced a partnership with co-working office space company Industrious through which it is recycling 13 second-story guest rooms to serve as offices for up to four people.Each of the rooms has a small outdoor platform, and dogs are welcome. Pricing starts at $ 200 and goes up to $ 275, depending on how many people use the space.The SawyerThe Sawyer, in Sacramento, California, is offering pool cabanas (更衣室) for use as outdoor offices, complete with fast Wi-Fi, free parking and catered lunch for $ 150 per day.HotelsByDayYannis Moati founded HotelsByDay back in 2015. That company has grown to include more than 1,500 hotels, and has seen a significant increase in the number of inquiries for day-use bookings lately.Moati said the current situation will force hotels to upgrade themselves to stay alive, and he predicted that offering rooms for day-use only is one of the directions they will go.1. How much should one pay for a 24-hour stay in Hotel Figuero?A. $ 129.B. $ 149.C. $ 150.D. $ 200.2. Which hotel allows pets in?A. The Wythe.B. The Sawyer.C. HotelsByDay.D. Hotel Figuero.3. What do we know about Yannis Moati?A. He started a program titledWork Perks.B. He has upgraded at least 1,500 rooms.C. He usually predicts everything correctly.D. He is optimistic about the WFH trend.BThe China International Search and Rescue Team(CISAR) was formed in 2001 and is now made up of several hundred rescue workers and about 20 police dogs. The team brings help and hope to those whose lives are changed by astorm, flood, earthquake, or any other natural disasters.After long and careful training, the team went on its first international rescue tasks in 2003. That year, the Chinese team helped save lives after earthquakes inAlgeriaandIran. It was the first time that a Chinese team had worked outsideChinawhose members won high praise for bravery and skill.Since then, the CISAR has completed many tasks. The list of people to whom help has been given is long. The team treated more than 3,000 people who were wounded in the 2006 earthquake inIndonesia, helped 2,500 wounded people after the earthquake that hitHaitiin 2010, and spent several months giving aid to over 25,000 people suffering from the 2010 floods inPakistan. On April 26, 2015, a group of 62 people from CISAR went toNepalafter the 8.1 magnitude earthquake that happened there.Rescue workers are trained to find people, treat wounds, and hand out food, water, and other supplies. They have to be able to do work that is difficult under conditions which can be very dangerous. After a disaster, there is usually no electricity or water, and there may be diseases and other dangers. Rescue workers get to save lives, but they must also bury the dead. That means they have to be strong in both body and mind.Rescue workers must have big hearts, too. It takes a lot of love and courage to risk one’s own life to save someone else’s. The members of the CISAR have plenty of both and are always ready to go wherever help is needed.4. What is the function of the numbers in Paragraph 3?A. To advertise for the CISAR.B. To add some basic information.C. To praise Recue Workers’ contributions.D. To stress the dangers Rescue Workers face.5. What is the author’s attitude towards Rescue Workers?A. Hopeful.B. Respectful.C. Curious.D. Supportive.6. What are the last two paragraphs mainly about?A. The duty rescue workers must perform.B. The qualities rescue workers must own.C. The difficulties rescue workers must go through.D. The willingness rescue workers should require.7. What may be the best title of the passage?A. China to the RescueB. How to train CISARC. Welcome to CISARD. Rescue on requestCWhen I was 13, I lost my sight. Since then, I had learned to get about with a walking stick, but had to stay at home because my parents thought I would get lost or robbed, even get hit by a car.I, however, believed I could regain my way if I lost it. A neighbor told me that a public library was offering a free course designed for the blind. That's an important opportunity for me to kill two birds with one stone: I could practice my getting — about skills on my way to learning practical technology. My parentssettled forit.But how would I plan my course? I knew that the blind singer Ray Charles, get around without a walking stick by counting steps. But I couldn't seem to do that the way he had. I developed the power of my imagination, catching the layout（布局）of places I visited and taking note of landmarks in my mind. Every time I visited a place, the mental map I'd drawn would turn up and helped me with the direction. But that doesn't mean I didn't lose my way in the process of acquiring this skill. I'd have to swallow（吞下）my pride to ask kind strangersfor help.On those days I lost my way, I'd go to bed feeling down. But my desires to beat blindness and further my education were usually enough to get me out of bed the next day and try again. Today, I'm a published reporter and audio producer.Yes, I've lost my way at times and found it again. And when people ask me,"Aren't you afraid to be out on your own?” the answer to me is clear：I'd rather risk and find happiness than stick to safety and be painful.Now, impressed by my progress, my father told my mother, "Our boy can see!".8. What does the underlined phrases “settled for" in the second paragraph mean?A. Talked about.B. Stuckto.C. Agreed to.D. Cared about.9. How did the author go around on his own after losing his sight?A.He created pictures of places in his mind.B. He drew a map on the paper to help him.C. He was always asking strangers for directions.D.He threw away the walking stick and counted steps.10. Which of the following can best describe the author?A. Determined and adventurous.B. Patient and intelligent.C. Warm-hearted and positive.D. Adventurous and outgoing.11. How did the author's parents feel about his progress?A.Concerned.B. Surprised.C. Confident.D. Proud.DNaomi Cooke was walking with a friend and their dogs through her local park in Burnside, on Tuesday when she heard someone shout to watch out. Cooke turned and hardly had time to react before a flying disc hit her in the face with a "big bang”, leaving her right cheek swollen almost to the size of a golfball.Two men playing disc golf at the course in Jellie Park were about 20 metres fromthe pairwhen one of them threw the disc hard, aiming for a nearby goal.After being hit Cooke immediately went to the emergency department, where two CT scans on her face and cheek found she had escaped any broken bones. "I'm lucky it didn't hit my eye because I think I would have lost it." Cooke said.Cooke often walks her dog at the park and said it was always busy with people playing disc golf, but it was not until after Tuesday that she became concerned about public safety there.There were no signs about the disc golf course in the park, she said, and the area is shared with children and people walking their dogs.“If it had hit one of the kids in the head, it could have killed them.” Cooke did not think she was the only person who had been hit before, and said there would be others who share her concerns.Cooke planned to go to the council, saying it needed to realise how dangerous it was for the space to be shared by everyone and to provide disc golfers with a space where they can play safely. "There should be rules about how it's done, making it safe for everyone.”12. What happened to Cooke on Tuesday?A. She was struck by a golf ball.B. She was hit by a flying disc.C. She was beaten by two men.D. She was frightened by a mad dog.13. What do the underlined words "the pair" in Paragraph 2 refer to?A. Cooke and her friend.B. Cooke and her dog.C. The two disc golfers.D. The two CT scans.14. How did Cooke feel about people playing disc golf in the park?A. Acceptable.B. Shocked.C. Angry.D. Worried.15. Why did Cooke plan to go to the council?A. To get the two men in trouble.B. To call for a ban on disc golf.C. To ask for personal protection.D. To call for safer places for disc golf.第二节（共5小题；每小题2分，满分10分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

2020年山东省济南市高三一模数学试题、单项选择题：本题共8小题，每小题5分，共40分。

3.加强体育锻炼是青少年生活学习中非常重要的组成部分。

某学生做引体向上运动，处于如图所示的平衡状态时，若两只胳膊的夹角为60 ,每只胳膊的拉力大小均为 400N,则该学生的体重（单位：kg ）约为（参考数据：取重力加速度大小为 g = 10m/s ；J3 1.732 ）5 .方舱医院的创设，在抗击新冠肺炎疫情中发挥了不可替代的重要作用。

每名护士从周一到周日轮流安排一个夜班。

若甲的夜班比丙晚一天，丁的夜班比戊晚两天，乙的夜班比庚 早三天，己的夜班在周四，且恰好在乙和丙的正中间，则周五值夜班的护士为6 .已知抛物线y 2=4x 的焦点为F,直线l 过F 且与抛物线交于 A, B 两点，过A 作抛物线准线的垂线，垂足为M,MAF 的角平分线与抛物线的准线交于点P,线段AB 的中点为Q 。

若7 .洛书，古称龟书，是阴阳五行术数之源，被世界公认为组合数学的鼻祖，它是中华民族对人类的伟大贡献之一。

在古代传说中有神龟出于洛水， 其甲壳上有图1: “以五居中，五方白圈皆阳数，四隅黑点为阴数这就是最早的三阶幻方，按照上述说法，将1到9这九个数字，填在如图 2所示的九宫格里，九宫格的中1.已知全集U = R,集合2A={X x>x},则 0A = 0,1B. (0,1)C.,1D.2.设复数z=二1 i（其中 i 为虚数单位），则复数 z 在复平面内对应的点所在的象限为B. C. D.第四象限A. f( x) =x+ tanxB. f(x) = x+ sin2x1 .八 C. f (x) = x —sin 2x21D. f(x) = x -cosx 2某方舱医院医疗小组有七名护士,B.C. 戊D.庚B.4C. 6D. 8AB =8,则 PQ”,y= f （x ）的部分图象如图，则D. 8175 f （x ）的解析式可能是4.已知函数的概率是3 .8 .已知直线y= ax+ b b>0）与曲线y= x 有且只有两个公共点 2x 1+ x 2 =A.1B. 0C. 1D. a二、多项选择题：本题共4小题，每小题5分，共20分。

2020年济南市高三一模考试英语试题及答案解析2020年高三模拟考试英语试题注意事项：1.答题前，考生务必将自己的姓名、座号、考号填写在答题卡和试卷指定位置上。

2.回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑。

如需改动，用橡皮擦干净后，再选涂其他答案标号。

回答非选择题时，将答案写在答题卡上。

写在本试卷上无效。

3.考试结束后，讲本试卷和答题卡并交回第一部分阅读（共两节，满分50分）第一节（共15小题;每小题2. 5分,满分37. 5分）阅读下列短文，从每题所给的A、B、C、D四个选项中选出最佳选项。

AIf you want to learn why everything is usually cheaper, and what items are the best deals at your local or hhain dollar or discount store, store employees have a lot of useful information. They tend to know what shoppers want—even before shoppers know themselves.Everyone buys unnecessary thingsWhether you shop at a chain dollar store such as Dollar Tree 9 Family Dollar or your local 99 cents store, not everything that's on sale is necessarily worth purchasing. Bryan Waring, a former Dollar Tree employee, says that you're not alone in buying more than necessary from these stores. "It seems basic, but everyone falls for this trap," he says. “You go into a store where everything is cheap, and you walk out with things you don't nee d. ” He suggests going into a store—yes, even the dollar store—with a checklist of things you truly need.Everything is cheaper after the holiday seasonPatricia, a seasonal worker at a Dollar Tree, says that the post-Christmas season means even more deals on everything from decorations to sweets. "After Christmas, all the gift wrap paper went to 50 cents, and all Christmas items were half price," she says. “Even candy bars are 89 cents versus $ 1.”Products are less expensive because of their sizeIn order for dollar stores to keep their prices low, product sizes are usually^ smaller than normal, according to CheapisnL Dollar stores aren’t the only ones guilty of this[trick. Cheapism also reports Walmart is guilty of doing the same thing to attract customers.1. What is Bryan's advice against buying unnecessary things?A Making a purchase alone.B. Writing a to-buy list ahead.C. Shopping at your local store.D. Buying basic things separately2. Which o£ the following is a better time for shopping according to Patricia?A. In the Christmas salesB. At a particular discount.C. After the Christmas season.D. During some holiday seasons.3. How do stores make their products cheaper?A. By reducing product sizes.B. By dowering product costs.-C. By adopting discount strategies.D. By attracting more customers.BIt was just a normal day for Ruth Miller, a 63-year-old woman until everything went horribly wrong. She was walking to her car after shopping when the unthinkable happened.Right as she was unlocking her car, a man quickly came up behind her and tried to wrestle her purse away. She was in shock. L uckily she remembered she had her Safe Personal Alarm （SPA） on her purse, and since she was too scared to scream for help, she quickly reached for the alarm and pulled the pin (保险栓). Immediately her SPA started just screaming. The man didn't know what to do! He froze for a second, and then ran away like a bat out of hell!SPA is a safety device capable of creating a 125db sound that attracts attention and scares away potential attackers. To compare, it's the same volume as a military jet during takeoff.Paul Davidson, the inventor of SPA, knows all too well the type of situation that Ruth found herself in. But that's not the only type of situation that SPA helps protect against. Parents can give it to their kids as an extra means of protection. Teenagers can use it so they can feel safe walking home. Women can know it's there when they have to use the parking lot at night. “My mother, who is 76 years old, carries it around in case she falls and needs to ask people for help. I only wish I'd have thought of it earlier,” said Paul.The police have been recommending SPA since it first hit the market. In fact, since its launch SPA has been in a state, shifting between in stock to sold out nearly every other week and it's also got tons of loyal followers worldwide.4.What does the underlined part “the unthinkable” in Paragraph 1 refer to?A. An attempted robbery.B. A wrestling match.C. An angry argument.D. A car accident.5. How did Ruth react to the unexpected situation?A.She fought violently.B.She froze in great fear.C.She cried desperately for help.D.She sounded her safety device.6.Why does Paul mention his mother?A.To imply the elderly need more care.B.To suggest he cares about his mother.C.To show SPA can be widely used.D.To make an advertisement for SPA.7.What can be learned from the text?A.SPA is well received in the market.B.People hesitate to pay for security.C.SPA was sold out in the first two weeks.D.The police always recommend products.CLike clockwork, nearly every fourth February includes one extra day. February 29th, otherwise known as Leap Day, isn't exactly a holiday. Instead, it's there to keep your calendar consistent with the earth's rotation（旋转）around the sun.According to History, com, Roman emperor Julius Caesar is the "father” of Leap Year. Until he came along, people used a 355-day calendar, which was 10. 25 days shorter than the solar year, Roman officials were supposed to add an extra month every now and then to keep the seasons exactly where they should be. But that didn't work out all that well. When special occasions started shifting into different seasons around 45 BCE, Caesar consulted with astronomers and decreed（下令）that the empire should use a 12-month, 365- day calendar, which he named after himself, Caesar’s Julian calendar included a L eap Day every four years.Though Leap Day keeps your calendar in line with the earth's rotation around the sun, it causes a different kind of problem for leapsters. When should these February 29th babies celebrate their birthdays during the other three-quarters of their lives? Some party on February 28th, while others prefer a two-day celebration that spans the last day of February and the first day of March.Leap Day can be a nuisance in the legal system. In 2006, a court in Massachusetts was deckling whether criminal John Melo could be released a day early since his 10-year sentence included a Leap Day. In the case, the judge decided that since the man was sentenced to prison for years, not days, Leap Day didn't make a bit of difference.Though a few timekeepers have pushed for calendars that don’t include L eap Day, almost all astronomers and societies agree that Leap Day is the best method to keep the calendar on track.8. Why was the Leap Day created?A. To create special occasions.B. To honor Emperor Julius Caesar.C. To keep pace with solar year.D. To keep track of all the seasons.9. What is the problem with the birthday celebration of February 29th babies?A. It is sometimes delayed.B. It lasts at least two days.C. It has to be held every other year.D. It may take place on different dates.10. What does the underlined word "nuisance" in Paragraph 4 mean?A. JokeB. Topic.C. TroubleD. Mistake.11. What is the attitude of most astronomers towards Leap Day?A. CriticalB. SupportiveC. DoubtfulD. CautiousDScientists have developed a new type of smart bandage（绷带）that can signal the type of bacterial（细菌的）infection it's protecting, just like a traffic light, as well as release the right type of drugs on demand. The traffic light system works just like this: Green means no bacteria or a low concentration of bacteria, yellow means drug-sensitive （DS）bacteria responsive to standard antibiotics（抗生素）and causes antibiotic release, and red means drug-resistant （DR） bacteria that need extra help to be wiped out.In testing the bandage on mice, the research team was able to successfully treat both DS and DR infections using the new method. However, the common methods of sensing resistance are limited by time, the requirement for professional personnel, and expensive instruments. Moreover, the abuse of antibiotics causes the accelerated process of bacterial resistance.It's easy to see how a simple bandage and light could overcome some of these limitations. Treatment doesn't have to wait for a doctor to make a diagnosis, and the bandage can get the right sort of drugs applied at the earliest opportunity. What's more, the person wearing the bandage gets real-time feedback on what's happening with the infection, if there's an infection at all. The researchers say it offers numerous benefits over existing treatments that make use of light, including photodynamic therapy or PDT.We've been seeing quite a few upgrades to the traditional bandage in recent years, thanks to advances in science—like the nanofibre mesh that attracts bacteria and draws some of it out, speeding up the healing process. Then there's the novel bandage for treating bums, which stops bacteria from multiplying and lowers the risk of infection.The more work that a bandage can do while it's protecting a wound, the better. Efforts to improve bandages continue and now we've got a bandage that not only releases antibiotics, but also tells the patient exactly what'sgoing on too.12.What is the smart bandage mainly designed to do?A.Avoid the use of antibiotics.B.Clear out harmful bacteria.C.Detect bacterial infections.D.Increase treatment options.13.What is the advantage of the smart bandage?A.It saves much time and cost.B.It removes the risk of infection.C.It prevents the bacterial resistance.D.It improves doctor-patient relationship.14.What can be inferred from the last two paragraphs?A.Traditional bandages are out of use now.B.More smart bandages will be developed.C.Progress in science calls for more research.D.People are urged to study medical science.15.What does the text focus on?A.A successful test on mice.B.A colour changing bandageC.Sensing drug-resistant bacteria.D.Preventing abuse of antibiotics.第二节（共5小题海小题2. 5分，满分12.5分）阅读下面短文，从短文后的选项中选出可以填入空白处的最佳选项。

2020年济南市第一中学高三语文模拟试卷及参考答案一、现代文阅读（36分）(一)现代文阅读I(9分)阅读下面的材料，完成下面小题。

材料一：据新华社北京4月3日电（记者叶昊鸣）记者3日从人力资源社会保障部了解到，近日正式向社会发布13个新职业信息，包括人工智能工程技术人员、物联网工程技术人员等。

通过分析可以发现，这13个新职业主要集中在高新技术领域，这与我国当前所处社会发展阶段、产业结构升级发展、科技创新能力提升、信息化广泛使用等方面有着莫大的关联。

近几年，随着我国人工智能、物联网、大数据和云计算的广泛运用，与此相关的高新技术产业成为我国经济新的增长点，对从业人员的需求大幅增长，形成了相对稳定的从业人群。

在这一背景下，以较高的专业技术知识和能力为支撑、从业人员普遍具有较高学历的专业技术类新职业应运而生。

（摘编《新华每日电讯·新闻纵深》2019年4月4日）材料二：我们的时代，科技发展日新月异，云计算、物联网、大数据、人工智能等新技术促进社会进步的速度比以往任何时代都快。

麻省理工学院的教授迈克斯·泰格马克认为最不可能被取代的职业呈现三个条件：其一是需要与人交互的社交情商，其二是涉及创造性、需要聪敏的方法，其三是在不可预料的环境中工作。

所有这一切正是劳动和职业启蒙教育可以帮助我们培养和造就的。

劳动和职业启蒙教育是人与人合作的劳动，在劳动中体现每个人的价值，劳动中明白自己的职业倾向，在劳动中体验与人合作的快乐。

同时，人们追求幸福人生的需要对人的职业生涯规划比以往任何时代都紧迫，而劳动和职业启蒙教育正是建立职业生涯规划的起点，非常重要、不可替代。

大力开展劳动和职业启蒙教育是新时代赋予职业教育的历史使命。

《国家职业教育改革实施方案》指明“职业教育与普通教育是两种不同教育类型，具有同等重要地位”。

职业教育作为一种类型就必然是系统化的，在中小学阶段当然要有这一类教育的基础，那就是劳动和职业启蒙教育。