2019-2020年黑龙江省哈尔滨市质检一：哈尔滨市2019届高三第一次质量检测数学(理)试题-附答案精品

2019届黑龙江省哈尔滨市第三中学校高三上学期第一次调研考试语文试题★祝考试顺利★注意事项：1、考试范围：高考考查范围。

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一、现代文阅读（36分）（一）论述类文本阅读（9分，每小题3分）阅读下面的文字，完成1～3题。

商品经济以货币为价值尺度和流通媒介。

帝制国家控制商业的手段，除了超经济强制性的法规——如界定商人社会地位、户籍及垄断个别生产与交换部门等之外，最重要的手段是税收政策、货币政策和金融体制。

中国帝制时代大多数时期，政府控制货币的制作、发行、各币种比价。

政府对货币的一般控制，包括统一货币和控制货币供给量，对于维系市场秩序关系甚巨。

这是帝制体系与商业长期共生的基础之一。

但货币既被政府权力牵系，更由经济本身左右。

即使集权专制的政府，也不能完全控制货币运行。

一般说来，商品经济愈发达，货币运行控制的复杂性愈高；市场开放性愈强，政府对货币的控制力愈弱。

黑龙江省哈尔滨市重点中学2019届高三第一次模拟考试试题语文Word版含解析一、现代文阅读（一）阅读下面的文字，完成下列小题。

欧阳修与“君子之勇”孙明在宋代的政治史、文学史、学术史和思想史上，欧阳修都是一个承上启下的人。

他以参与庆历改革而将“以姑息为安”的政治推向王安石变法；以提倡古文而上接韩柳，下开一代文风；以“庆历正学”而从唐代经学注疏传统通往王安石的新学和朱熹的理学集大成。

这些发展史统统绕不过他，后来者却也指摘他。

受他提携的王安石，当政后阻挠神宗召他回朝参与变法，说他不知经，不识义理。

朱熹也批评他经学造诣不高，并且不能躬行实践。

上世纪六十年代是检讨中国文化的年代，刘子健先生站在历史家的立场，做宏观趋势的评判：“从欧阳的经历，看到经学兴而求致用，古文兴而议论更犀利，士大夫权力提高而反不稳定。

总之，儒家思想虽已部分实现，而官僚政治的纠纷，反因而愈变愈坏，至于不可收拾！”可是当推崇宋学的陈寅恪要托一位先贤来表明心迹时，只找到了欧阳修。

他曾将自己的议论比于曾国藩和张之洞，到“奄奄垂死，将就木矣”的时候，这两个人却还当不得他一生的旨趣。

正是欧阳修高扬儒家君子的理想，“贬斥势利，尊崇气节”，形塑、引出了宋代的优秀文化，也正是在这个意义上，史家的“空文”乃是有益于“治道学术”的。

生逢一个活泼时代，欧阳修活得波澜跌宕而又丰富多姿。

王安石、朱熹用经学不深批评他，却都不能不承认欧阳修的文章好，欧阳本心也未必把文学看得低于经学。

他也作艳词，广为歌妓传唱，这是道学家攻击他不重实践的原因之一，今天说起来却让人更觉得欧阳的自由与可爱。

欧阳修喜欢酒，他不仅以“醉翁”自况，同游诸君也多是爱酒的人，连称赞诗僧，也是在“酒友”的语境中豪阔道出的：“曼卿隐于酒，秘演隐于浮屠，皆奇男子也。

然喜为歌诗以自娱。

当其极饮大醉，歌吟笑呼，以适天下之乐，何其壮也！”美文、美人、美酒，若止步于人生快意，欧阳修也就是个烂漫文人，在如此快意中，他对衰老的惶惑更见生命经验的厚度。

2019-2020学年黑龙江省哈尔滨三中高三（上）第一次调研物理试卷一、单选题（本大题共7小题，共28.0分）1.在电磁学发展过程中，许多科学家做出了贡献．下列说法正确的是()A. 奥斯特经过多年研究，发现了电磁感应现象B. 美国物理学家劳伦斯发明了回旋加速器，能在实验室中产生大量的高能粒子C. 欧姆发现大多数金属温度降到某一值时，都会出现电阻突然降为零的超导现象D. 库仑通过实验精确测定了元电荷e的数值2.文艺复兴时期，达·芬奇曾提出如下正确的原理：如果力F在时间t内使静止的质量为m的物体移动一段距离s，那么()A. 相同的力在相同的时间内使质量是一半的物体移动相同的距离B. 相同的力在相同的时间内使质量是两倍的物体移动相同的距离C. 相同的力在两倍的时间内使质量是两倍的物体移动相同的距离D. 一半的力在相同的时间内使质量是一半的物体移动相同的距离3.如图为A、B两物体沿直线运动的v−t图象，根据图象可以判断()A. 开始时B物体在前，A物体在后B. 4s末两物体可能相遇C. 4s后两物体间距离不断增大D. A物体的加速度小于B物体的加速度4.将一个半球体置于水平地面上，半球的中央有一光滑小孔，上端有一光滑的小滑轮，柔软光滑的轻绳绕过滑轮，两端分别系有质量为m1、m2的物体(两物体均可看成质点，m2悬于空中)时，整个装置处于静止状态，如图所示。

已知此时m1与半球的球心O的连线与水平线成53°角(sin53°=0.8,cos53°=0.6)，m1与半球面的动摩擦因数为0.5，并假设m1所受到的最大静摩擦力可认为等于滑动摩擦力。

则在整个装置处于静止的前提下，下列说法正确的是()A. 当m1m2的比值不同时，地面对半球体的摩擦力也不同B. 当m1m2=53时，半球体对m1的摩擦力为零C. 当1≤m1m2<53时，半球体对m1的摩擦力的方向垂直于图中的虚线向上D. 当53<m1m2≤5时，半球体对m1的摩擦力的方向垂直于图中的虚线向下5.一个作为匀加速直线运动的物体，通过A点时速度为1m/s，通过B点时速度为3m/s，则通过AB的中点时的速度和这一过程中间时刻的速度分别为()A. √5m/s，2m/sB. 2m/s，√5m/sC. √52m/s，2m/s D. 2m/s，√52m/s6.如图所示，重力为20N的物体在水平面上向左运动，同时受到大小为6N、方向向右的水平力F的作用，物体与地面间的动摩擦因数为0.1，则物体所受的摩擦力的大小和方向为()A. 2N向右B. 2N向左C. 6N向右D. 8N向左7.一个球在墙角处挨着竖直墙壁停放在水平地板上，在给出的球的受力示意图中，正确的是()A. B. C. D.二、多选题（本大题共5小题，共20.0分）8.粗糙的水平地面上一物体在水平拉力作用下做直线运动，水平拉力F及运动速度v随时间变化的图象如图甲和图乙所示。

2019-2020学年黑龙江省哈尔滨三中高三（上）第一次验收数学试卷2（8月份）一、选择题（本大题共12小题，共60.0分）1.集合A={0,2}，B={x∈N|x<3}，则A∩B=()A. {2}B. {0,2}C. (0,2]D. [0,2]2.已知sin(π+α)=−√63，α为第二象限角，则cosα−sinα=()A. √3+√63B. −√3+√63C. √3−√63D. √6−√333.角α的终边经过点，则cos2α+sin2α=()A. 1−2√65B. 1+2√65C. 1−√65D. 1+3√654.已知函数f(x)=2sin(x−π6)，则下列结论中正确的是()A. y=f(x)的一个周期为πB. y=f(x)的图象关于点(π6,2)对称C. y=f(x)的图象关于直线x=π6对称D. y=f(x)在区间(π6,2π3)上单调递增5.已知sinα+3cosα2cosα−sinα=2，则sin2α+sinαcosα+1等于()A. 115B. 25C. 85D. 756.已知f(x)=x3+3x，a=20.3，b=0.32，c=log20.3，则()A. f(a)<f(b)<f(c)B. f(b)<f(c)<f(a)C. f(c)<f(b)<f(a)D. f(c)<f(a)<f(c)7.函数f(x)=log32(|x|−1)的大致图象是()A. B.C. D.8.当−π2≤x≤π2时，函数f(x)=2sin(x+π3)有()A. 最大值1，最小值−1B. 最大值1，最小值−12 C. 最大值2，最小值−2D. 最大值2，最小值−19. 已知命题p ：∃x 0∈[3,+∞),x 0+4x 0≤4，命题q ：∀x ∈R ，4x −2×2x ≥−1，则以下命题为真命题的是( )A. p ∧qB. p ∨(￢q)C. (￢p)∧qD. (￢p)∧(￢q)10. 已知函数f(x)=2sinωx(ω>0)在[−π4,π6]上为增函数，则ω的最大值为( )A. 3B. 2C. 32D. 5411. 对任意x ∈R ，函数y =f(x)的导数都存在，若f(x)+f ′(x)>0恒成立，且a >0，则下列说法正确的是( )A. f(a)<f(0)B. f(a)>f(0)C. e a ⋅f(a)<f(0)D. e a ⋅f(a)>f(0)12. 已知函数f(x)=2a −x 2(1e ⩽x ⩽e,e 为自然对数的底数)与的图象上存在关于x 轴对称的点，则实数a 的最小值为( )A. 12B. 12e −1 C. 12e +1D. e 22−1二、填空题（本大题共4小题，共20.0分）13. 函数y =log a (x −3)+1( a >0，a ≠1)的图象恒过定点坐标______ ． 14. 函数f (x )=cos x2(sin x2−√3cos x2)的最小正周期为___________． 15. 若sin(π4−θ)=13，θ∈(0,π2)，则cos2θ=______．16. 函数f(x)=e x −ax −2恰有一个零点，则实数a 的取值范围是______ ． 三、解答题（本大题共6小题，共70.0分）17. 在极坐标系中，已知点(4,π4)，直线为ρsin(θ+π4)=1．(1)求点(4,π4)的直角坐标系下的坐标与直线的普通方程； (2)求点(4,π4)到直线ρsin(θ+π4)=1的距离．18. 已知cosα=17，cos(α−β)=1314，且0<β<α<π2．(1)求tan2α的值；(2)求β的值．19.设函数f(x)=sin(ωx−π6)+cos(π−ωx)，其中0<ω<3，f(π6)=0．(1)求函数f(x)的最小正周期及单调增区间；(2)将函数f(x)的图象上各点的横坐标变为原来的2倍(纵坐标不变)，再将得到的图象向左平移π4个单位，得到函数g(x)的图象，求g(x)在(−π4,3π4)上的值域．20.设f(x)=|x+2|+|x−3|．(1)求不等式f(x)≥6的解集；(2)若不等式f(x)>m2−4m恒成立，求实数m的取值范围．21.已知函数f(x)=|x2−4|+a|x−2|，x∈[−3,3].若f(x)的最大值是0，则实数a的取值范围是．22.设函数f(x)=ln(x+1)+a(x2−x)，其中a∈R．(1)当a=0时，求证：f(x)<x，对任意的x∈(0,+∞)成立；(2)讨论函数f(x)极值点的个数，并说明理由；(3)若∀x>0，f(x)≥0成立，求a的取值范围．-------- 答案与解析 --------1.答案：B解析：解：集合A ={0,2}，B ={x ∈N|x <3}={0,1，2}， 则A ∩B ={0,2}． 故选：B ．根据交集的定义写出A ∩B ．本题考查了交集的定义与计算问题，是基础题．2.答案：B解析： 【分析】本题考查了诱导公式和同角三角函数的基本关系，属于基础题．由诱导公式得sinα=√63，再由同角三角函数的基本关系得出cosα，即可得出结果．【解答】解：由sin(π+α)=−√63，α为第二象限角，则sinα=√63，所以cosα=−√1−sin 2α=−(√63)=−√33，所以cosα−sinα=−√33−√63．故选B ．3.答案：A解析： 【分析】本题考查了任意角的三角函数，诱导公式，二倍角公式，属于基础题．先求出角α的终边经过点，也就可求出sinα，cosα，根据二倍角公式可以求出cos2α+sin2α的值． 【解答】 解：，∴角α的终边经过点(√32,−√22)，∴r =√(√32)2+(−√22)2=√52，∴sinα=−√105，cosα=√155， ∴cos2α+sin2α=cos 2α−sin 2α+2sinαcosα =(√155)2−(−√105)2+2×(−√105)×√155=1−2√65． 故选A ．4.答案：D解析：解：函数f(x)=2sin(x −π6)，其最小正周期为2π，A 错误； 令x −π6=kπ，k ∈Z ， 得x =π6+kπ，k ∈Z ，x =π6时，f(π6)=2sin(π6−π6)=0， ∴f(x)的图象关于点(π6,0)对称，且不关于直线x =π6对称，∴B 、C 都错误； x ∈(π6,2π3)时，x −π6∈(0,π2)，∴f(x)是单调递增函数，D 正确． 故选：D ．根据正弦型函数的图象与性质，对选项中的命题判断真假性即可． 本题考查了正弦型函数的图象与性质的应用问题，是基础题．5.答案：D解析：解：∵sinα+3cosα2cosα−sinα=2，∴tanα=13，∴sin 2α+sinαcosα+1=sin 2α+sinαcosαsin 2α+cos 2α+1=tan 2α+tanαtan α+1+1=75，故选：D ．由已知求得tanα，结合平方关系把sin 2α+sinαcosα+1化弦为切求解．本题考查三角函数的化简求值，考查了同角三角函数基本关系式的应用，是基础题．6.答案：C解析：本题考查复合函数的单调性、指数函数及其性质、对数函数及其性质的知识点，属于基础题．根据函数f(x)=x3+3x在R上单调递增且得到答案．【解答】解：函数f(x)=x3+3x在R上单调递增，又，所以a>b>c，则有，故选C．7.答案：B解析：解：函数f(x)=log32(|x|−1)，可知函数f(x)是偶函数，排除C，D；定义域满足：|x|−1>0，可得x<−1或x>1．当x>1时，y=log32(x−1)是递增函数，排除A；故选：B．利用奇偶性结合单调性即可选出答案．本题考查了函数图象变换，是基础题．8.答案：D解析：【分析】本题考查正弦函数的性质，根据x的范围得出x+π3的范围，根据正弦函数的单调性得出f(x)的最值．【解答】解：因为−π2≤x≤π2，所以−π6≤x+π3≤5π6，所以−12≤sin(x+π3)≤1，所以−1≤f(x)≤2．9.答案：C【分析】本题考查利用函数的单调性求最值，考查复合命题的真假判断，属于中档题．由函数的单调性求得x+4x的范围判断p，由配方法说明q为真命题，再由复合命题的真假判断得答案．【解答】解：当x∈[3,+∞)时，x+4x 单调递增，(x+4x)min=3+43>4，故命题p为假命题；∵4x−2×2x+1=(2x−1)2≥0，∴∀x∈R，4x−2×2x≥−1，故命题q为真命题．∴p∧q为假命题；p∨(￢q)为假命题；(￢p)∧q为真命题；(￢p)∧(￢q)为假命题．故选：C．10.答案：B解析：【分析】本题主要考查正弦型函数的单调性，属于基础题．由题意利用正弦型函数的单调性，求出ω的最大值．【解答】解：∵函数f(x)=2sinωx(ω>0)在[−π4,π6]上为增函数，∴−π4⋅ω≥−π2，且π6⋅ω≤π2，求得ω≤2，又ω>0，所以0<ω≤2，则ω的最大值为2，故选B．11.答案：D解析：【分析】本题考查函数单调性的应用和构造函数，利用导数研究函数的单调性，由单调性比较大小，属于一般题．构造函数g(x)=e x f(x)，通过求导结合题意可得函数g(x)的单调性即可求解．解：设g(x)=e x f(x)，则g′(x)=e x [f(x)+f′(x)]， 因为 f(x)+f ′(x)>0， 所以g(x)在R 上为增函数，因为a >0，所以g(a)=e a f(a)>g(0)=e 0f(0)=f(0)， 故选D ．12.答案：A解析： 【分析】本题考查了构造函数法求方程的解及参数范围；关键是将已知转化为方程2a −x 2=−2lnx ⇔−a =lnx −x 22在[1e ,e]上有解．转化为函数图象交点问题，一般题型．【解答】解：由已知，得到方程2a −x 2=−2lnx ⇔−a =lnx −x 22在[1e ,e]上有解．设ℎ(x)=lnx −x 22，求导得：ℎ′(x)=1x −x =(1−x)(1+x)x，∵1e ≤x ≤e ，∴ℎ′(x)=0在x =1有唯一的极值点， ∵ℎ(1e )=−1−12e 2，ℎ(e)=1−e 22，ℎ(x)极大值=ℎ(1)=−12，且知ℎ(e)<ℎ(1e )，故方程−a =lnx −x 22在[1e ,e]上有解等价于1−e 22≤−a ≤−12．所以a ≥12，从而a 的最小值为12． 故选A ．13.答案：(4,1)解析：解：∵log a 1=0，∴当x −3=1，即x =4时，y =1，则函数y =log a (x −3)+1的图象恒过定点(4,1)． 故答案为：(4,1)．由log a 1=0得x −3=1，求出x 的值以及y 的值，即求出定点的坐标 本题考查对数函数的性质和特殊点，主要利用log a 1=0，属于基础题14.答案：2π解析：本题考查三角函数的周期，解题的关键是利用二倍角公式以及两角和差公式将函数式化简.利用二倍角公式以及两角和差公式将函数式化简为正弦型函数可得结果．【解答】解：f(x)=cos x2(sin x2−√3cos x2)=sinx2cosx2−√3cos2x2=12sinx−√3×1+cosx2=12sinx−√32cosx−√32=sin(x−π3)−√32，∴函数的最小正周期为2π．故答案为2π．15.答案：4√29解析：【分析】本题主要考查同角三角函数的基本关系，诱导公式、二倍角公式的应用，属于基础题．利用同角三角函数的基本关系求得cos(π4−θ)的值，再利用诱导公式、二倍角公式求得cos2θ=sin(π2−2θ)的值．【解答】解：若sin(π4−θ)=13，θ∈(0,π2)，∴π4−θ∈(−π4,π4)，∴cos(π4−θ)=√1−sin2(π4−θ)=2√23，则cos2θ=sin(π2−2θ)=2sin(π4−θ)cos(π4−θ)=4√29．故答案为：4√29．16.答案：(−∞,0]解析：【分析】本题主要考查了函数的零点的个数的判断，主要采用了转化为判断函数的图象的交点的个数，注意体会数形结合思想与转化思想在解题中的应用，属于中档题．由函数f(x)=e x−ax−2恰有一个零点，可得e x=ax+2只有一个根，可得函数y=e x与函数y= ax+2的图象只有一个交点，结合函数的图象可求a的取值范围．【解答】解：由函数f(x)=e x−ax−2恰有一个零点，可得e x=ax+2只有一个根，从而可得函数y=e x与函数y=ax+2的图象只有一个交点，结合函数的图象可得，a>0时不符合条件，a≤0时符合条件．故a≤0．故答案为：(−∞,0]．17.答案：解：(1)点(4,π4)化成直角坐标为(2√2,2√2)，直线ρsin(θ+π4)=1，化成直角坐标方程为√22x+√22y=1，即x+y−√2=0．(2)由题意可知，点(4,π4)到直线ρsin(θ+π4)=1的距离，就是点(2√2,2√2)到直线x+y−√2=0的距离，由距离公式可得为d=√2+2√2−√2|√2=3．解析：(1)利用极坐标与直角坐标互化的方法，可得结论；(2)利用点到直线的距离公式，可得结论．本题考查极坐标与直角坐标互化的方法、点到直线的距离公式，比较基础．18.答案：解：(1)由cosα=17，0<α<π2，得sinα=√1−cos 2α=√1−(17)2=4√37， ∴tanα=sinαcosα=4√37×71=4√3， 于是tan2α=2tanα1−tan 2α=√31−(4√3)2， =−8√347； (2)由0<β<α<π2，得0<α−β<π2，又∵cos(α−β)=1314，∴sin(α−β)=√1−cos 2(α−β)=√1−(1314)2=3√314， 由β=α−(α−β)，得cosβ=cos[α−(α−β)]=cosαcos(α−β)+sinαsin(α−β)=17×1314+4√37×3√314=12，∴β=π3．解析：本试题主要是考查了两角和差的三角函数变换的运用，以及构造角的思想求解角的综合运用．(1)依题意可求得tanα，再利用二倍角的正切即可求得tan2α的值；(2)由cos(α−β)=1314，0<β<α<π2可求得sin(α−β)，由于β=α−(α−β)，利用两角差的余弦即可求得cosβ．19.答案：解：(1)∵函数f(x)=sin(ωx −π6)+cos(π−ωx)=√32sinωx −12cosωx −cosωx =√32sinωx −32cosωx =√3sin(ωx −π3)，其中0<ω<3． ∵f(π6)=0=√3sin(πω6−π3)，∴ωπ6−π3=kπ，k ∈Z ，∴ω=2，f(x)=√3sin(2x −π3). 最小正周期， 令2kπ−π2≤2x −π3≤2kπ+π2，求得kπ−π12≤x ≤kπ+5π12，故函数f(x)的增区间为[kπ−π12,kπ+5π12]，k ∈Z ；(2)将函数f(x)的图象上各点的横坐标变为原来的2倍(纵坐标不变)，可得y =√3sin(x −π3)的图象；再将得到的图象向左平移π4个单位，得到函数g(x)=√3sin(x +π4−π3)=√3sin(x −π12)的图象， 在(−π4,3π4)上，x −π12∈(−π3,2π3)， 故当x −π12=π2时，函数g(x)取得最大值为√3，当x −π12=−π3时，函数g(x)=−32，故g(x)的值域为(−32,√3].解析：本题主要考查三角恒等变换，正弦函数的周期性和单调性，函数y =Asin(ωx +φ)的图象变换规律，正弦函数的定义域和值域，属于中档题．(1)利用三角恒等变换化简f(x)的解析式，再根据正弦函数的周期性和单调性，得出结论；(2)利用函数y =Asin(ωx +φ)的图象变换规律求得g(x)的解析式，再利用正弦函数的定义域和值域，求得g(x)在(−π4,3π4)上的值域．20.答案：解：(1)f(x)≥6可化为：|x +2|+|x −3|≥6，①当x <−2时，−x −2−x +3≥6，解得x ≤−52；②当−2≤x ≤3时，x +2−x +3≥6，不成立；③当x >3时，x +2+x −3≥6，解得x ≥72，综上所述，f(x)≥6的解集为{x|x ≤−52或x ≥72};(2)∵|x +2|+|x −3|≥|(x +2)−(x −3)|=5，即f(x)min =5，又不等式f(x)>m 2−4m 恒成立等价于f(x)min >m 2−4m ，∴5>m 2−4m ，解得−1<m <5，∴实数m 的取值范围是(−1,5)．解析：本题考查了绝对值不等式的解法，考查不等式恒成立问题，属于中档题．(1)通过讨论x 的范围，去掉绝对值，从而解出不等式的解集；(2)不等式f(x)>m 2−4m 恒成立等价于f(x)min >m 2−4m ，即5>m 2−4m ，即可求实数m 的取值范围．21.答案：(−∞,−5]解析：【分析】本题主要考查函数的最值以及函数与方程的综合运用，属于一般题．解析：解：由题有函数f(x)=|x 2−4|+a|x −2|=|x −2|(|x +2|+a )≤0 ，x ∈[−3,3]．当x =2时，f (x )=0恒成立；当x ≠2时，a ≤(−|x +2|)min =−5 ，x ∈[−3,3]则a 的取值为a ∈(−∞,−5]故答案为(−∞,−5]．22.答案：解：(1)a =0时，f(x)=ln(x +1)，定义域是(−1,+∞)，令g(x)=ln(x +1)−x ，g′(x)=1x+1−1=−x x+1<0，∴g(x)在(0,+∞)递减，∴g(x)<g(0)=0，故f(x)<x ，对任意的x ∈(0,+∞)成立；(2)函数f(x)=ln(x +1)+a(x2−x)，其中a ∈R ，x ∈(−1,+∞)．f′(x)=2ax 2+ax−a+1x+1，令g(x)=2ax 2+ax −a +1．(i)当a =0时，g(x)=1，此时f′(x)>0，函数f(x)在(−1,+∞)上单调递增，无极值点． (ii)当a >0时，△=a 2−8a(1−a)=a(9a −8)．①当0<a ≤89时，△≤0，g(x)≥0，f′(x)≥0，函数f(x)在(−1,+∞)上单调递增，无极值点． ②当a >89时，△>0，设方程2ax 2+ax −a +1=0的两个实数根分别为x 1，x 2，x 1<x 2． ∵x 1+x 2=−12，∴x 1<−14，x 2>−14．由g(−1)>0，可得−1<x 1<−14，∴当x ∈(−1,x 1)时，g(x)>0，f′(x)>0，函数f(x)单调递增；当x ∈(x 1,x 2)时，g(x)<0，f′(x)<0，函数f(x)单调递减；当x ∈(x 2,+∞)时，g(x)>0，f′(x)>0，函数f(x)单调递增．因此函数f(x)有两个极值点．(iii)当a <0时，△>0.由g(−1)=1>0，可得x1<−1<x2．∴当x ∈(−1,x 2)时，g(x)>0，f′(x)>0，函数f(x)单调递增；当x ∈(x 2,+∞)时，g(x)<0，f′(x)<0，函数f(x)单调递减．因此函数f(x)有一个极值点．综上所述：当a <0时，函数f(x)有一个极值点；当0≤a ≤89时，函数f(x)无极值点；当a >89时，函数f(x)有两个极值点．(3)由(2)可知：①当0≤a≤8时，函数f(x)在(0,+∞)上单调递增．9∵f(0)=0，∴x∈(0,+∞)时，f(x)>0，符合题意．<a≤1时，由g(0)≥0，可得x2≤0，函数f(x)在(0,+∞)上单调递增．②当89又f(0)=0，∴x∈(0,+∞)时，f(x)>0，符合题意．③当1<a时，由g(0)<0，可得x2>0，∴x∈(0,x2)时，函数f(x)单调递减．又f(0)=0，∴x∈(0,x2)时，f(x)<0，不符合题意，舍去；>0．④当a<0时，设ℎ(x)=x−ln(x+1)，x∈(0,+∞)，ℎ′(x)=xx+1∴ℎ(x)在(0,+∞)上单调递增．因此x∈(0,+∞)时，ℎ(x)>ℎ(0)=0，即ln(x+1)<x，可得：f(x)<x+a(x2−x)=ax2+(1−a)x，当x>1−1时，aax2+(1−a)x<0，此时f(x)<0，不合题意，舍去．综上所述，a的取值范围为[0,1]．解析：(1)求出f(x)的表达式，令g(x)=ln(x+1)−x，根据函数的单调性求出g(x)<g(0)=0，从而证出结论；(2)求出f(x)的导数，令g(x)=2ax2+ax−a+1，通过讨论a的范围，判断函数的单调性，从而求出函数的极值的个数；(3)通过讨论a的范围，结合函数的单调性求出满足题意的a的范围即可．本题考查了函数的单调性、最值、极值问题，考查导数的应用以及分类讨论思想，是一道综合题．。

哈尔滨第三中学2019—2020第一学期高三第一次调研化学相对原子质量：H-1 N-14 O-16 F-19 S-32 Fe-56 Cu-64 I-127一、选择题（本题包括20个小题，每小题只有一个选项符合题意，1-10题每小题2分，11-20题每小题3分，共50分）1. 化学与生产、生活、社会密切相关，下列说法正确的是A.“雾霾天气”、“温室效应”、“光化学烟雾”的形成都与氮的氧化物无关B.高铁酸钾是一种新型高效、多功能的水处理剂，既能杀菌消毒又能净水C.石灰石、铁粉、硅胶是食品包装中常用的干燥剂D.纯碱是焙制糕点所用的发酵粉的主要成分之一，也可用纯碱除去物品表面的油污2. 材料与国民经济建设、国防建设和人民生活密切相关。

下列关于材料的说法中正确的是A.太阳能电池板中的二氧化硅可用作光导纤维B.古代的陶瓷、砖瓦、现代的有机玻璃、水泥都是硅酸盐产品C.高纯度的硅可用于制造计算机芯片D.石英玻璃耐强酸、强碱，可用来制造化学仪器3. 下列说法正确的是A.可用酒精清洗试管中附着的硫粉B.钙元素的焰色反应是绿色C.2mol Na2O2与足量水反应转移4mol电子D.在钢中加入一些稀土元素可以增强钢的耐磨性和耐热性4. 在给定的条件下，下列选项所示的物质间转化均能一步实现的是A.B.C.D.5. 下列说法不正确的是A.氨气易液化，液氨气化时吸热，因此液氨常用作制冷剂B.氮的固定是指将游离态的氮转变为氮的化合物C.漂白粉和漂粉精的有效成分都是Ca(ClO)2D.SO2和CO2都是造成酸雨的主要气体6.化学在生活中有着广泛的应用，下列对应关系错误的是氢氟酸在玻璃器皿上刻蚀标记①实验室收集氨气采用图1所示装置②实验室做氯气与钠的反应实验时采用图2所示装置③实验室中用玻璃棒分别蘸取浓盐酸和浓氨水做氨气与酸生成铵盐的实验④实验室中采用图3所示装置进行铜与稀硝酸的反应A.②③④B.①②③C.①②④D.①③④8. 用于“点豆腐”的氯化镁结晶水合物化学式是MgCl2•nH2O，取10.15g此结晶水合物溶于水，加入足量的硝酸银溶液，得到氯化银沉淀14.35g 。

2019届黑龙江省哈尔滨市高三下学期第一次模拟考试英语试卷【含答案及解析】姓名___________ 班级____________ 分数__________一、阅读理解1. With fuel costs rising and airlines finding more fees to impose （强加，征税）on travelers every day, airfare isn’t getting any cheaper. Since youcan’t drive to all your dream destinations, flying is the only way to go sometimes and, undeniably, the fastest. Luckily, there are plenty of ways to find the most affordable fares and also avoid paying as many extra charges as possible when you plan ahead.Get the best fare. Airlines put out their fare sales on Tuesday morning, making this day the best day to book a flight for less.Fly during the least popular times. Tuesday, Wednesday and Saturday are the slowest days to fly, which means cheaper deals than the rest of the week. You can also find reduced rates on early morning flights, since many people don’t like to get up before the sun to get to the airport. Earlier boarding timescan also considerably cut down your chances of getting bumped on an overbooked flight or delayed because of other delayed flights or mechanical issues.Choose your seat later. Some airlines charge you to pick your seat when you book online, adding even more to the bottom line of your ticket cost. If you show up early on your travel day, you can still get suitable seats. Some ofthe best seats get held back until flight day, unless others are willing topay extra for them ahead of time, so you still have the chance at one of those. Fly on holidays. You already know that summer is the most expensive time to fly, and even though most other times are more affordable, the dayssurrounding holidays can be crazy. Save big if you’re willing to travel on major holidays, such as Thanksgiving and Christmas.Don’t wait until t he last-minute to book. Many travelers don’t know that there’s a sweet spot for booking and getting the best price on your tickets. Book too early or too late and you could end up paying more than you need to.The best time to book is between three months and six weeks from when you want to travel.1. The passage is intended to _______.A. inform us how to book a cheaper ticketB. tell us how to choose the seat and time for a flightC. persuade us into travelling by airD. inform us of some best ways to save money on a flight2. Which of the following statements is not true?A. Airline travelers are supposed to pay more for the rising fuel costsB. Early morning flight will help avoid some bumps and delays.C. You can only book your seat online before or on your flight day.D. Booking too early may cost you more money on your tickets.3. Which is the best time to book the flight for October 2nd?A. May 15thB. August 15thC. June 15thD. September 15th2. I’m seventeen. I had worked as a box boy at a supermarket in Los Angeles. People came to the counter and you put things in their bags for them and carried things to their cars. It was hard work.While working, you wear a plate with your name on it. I once met someone I knew years ago. I remembered his name and said, "Mr. Castle, how are you?" We talked about this and that. As he left, he said, "It was nice talking to you, Brett." I felt great, he remembered me. Then I looked down at my name plate. Oh, no. He didn’t remembe r me at all. He just read the name plate. I wish I had put "Irving" down on my name plate. If he’d have said, "Oh yes, Irving, how could I forget you?" I’d have been ready for him. There’s nothing personal here.The manager and everyone else who were a step above the box boys often shouted orders. One of these was: you couldn’t accept tips. Okay, I’m outside and I put the bags in the car. For a lot of people, the natural reaction is to take a quarter and give it to me. I’d say, "I’m sorry, I can’t." They’d get angry. When you give someone a tip, you’re sort of being polite. You take a quarter and you put it in their hand and you expect them to say, "Oh, thanks a lot." When you say, "I’m sorry, I can’t." they feel a little put down . They say, "No one will know." And they put it in your pocket. You say, "I really can’t."It gets to a point where you almost have to hurt a person physically to prevent him from tipping you. It was not in agreement with the store’s belief in being friendly. Accepting tips was a friendly thing and made the customerfeel good. I just couldn’t understand the strangeness of some people’s ideas. One lady actually put it in my pocket, got in the car, and drove away. I would have had to throw the quarter at her or eaten it or something.I had decided that one year was enough. Some people needed the job to stay alive and fed. I guess I had the means and could afford to hate it and give it up.1. What can be the best title for this text?A. The Art of Taking TipsB. Why I Gave up My JobC. How Hard Life Is for Box BoysD. Getting along with Customers2. From the second paragraph, we can infer that ________.A. the writer didn’t like the impersonal part of his jobB. Mr. Castle mistook Irving for BrettC. with a name plate, people can easily start talkingD. Irving was the writer’s real name3. The box boy refused to accept tips because ________.A. customers only gave small tipsB. he didn’t want to fight with the customersC. the store didn’t allow the box boys to take tipsD. some customers had strange ideas about tipping4. The underlined phrase "put down" in the third paragraph probably means________.A. misunderstoodB. defeatedC. hatefulD. hurt3. In colleges around the country, most students are also workers.The reality of college can be pretty different from the images presented in movies and television. Instead of the students who wake up late, party all the time, and study only before exams, many colleges are full of students with pressing schedules of not just classes and activities, but real jobs, too.This isn’t a temporary phenomenon. The share of working students has been on the rise since the 1970s, and one-fifth of students work all year round. About one-quarter of those who work while attending school have both a full-course load and a full-time job. The arrangement can help pay for tuition ( 学费 )and living costs, obviously. And there’s value in it beyond the direct cause: such jobs can also be critical for developing important professional andsocial skills that make it easier to land a job after graduation. With many employers looking for students with already-developed skill sets, on-the-jobtraining while in college can be the best way to ensure a job later on.But it’s not all upside. Even full-time work may not completely cover the cost of tuition and living expenses. The study notes that if a student worked a full-time job at the federal minimum wage, they would earn just over $15,000 each year, certainly not enough to pay for tuition, room, and board at many colleges without some serious financial aid. That means that though they’re sacrificing time away from the classroom, many working students will still graduate with at least some debt. And working full time can reduce the chance that students will graduate at all, by cutting into the time available for studying and attending classes.There is little reward for attending but not finishing college. Students who wind up leaving school because of difficulty in managing work and class are likely to find themselves stuck in some of the same jobs they might have gotten if they hadn’t gone at all. The difficulty of working too much while in school can create a cycle that pushes students further into debt without receiving any of the financial or career benefits.1. According to the passage, the reality of college students is that ______.A. they throw parties a lotB. they stay up late every nightC. they pay no attention to examsD. they work besides attending classes2. What is the indirect cause of an increasing number of working students?A. The lack of summer jobs for young adults.B. The need of developing social networks.C. The chance of finding a job after graduation.D. The expenses of high tuition and living costs.3. We can learn from the passage that ______.A. dropping out of college may not help students get career benefitsB. students can cover all their college expenses through workingC. students receive a huge reward for managing work and classD. working students are more likely to finish college4. What is the best title for the passage?A. The Difficulties of Landing a JobB. The Struggle of Work-School BalanceC. The Reward of Working While StudyingD. The Images of Working College Students4. A survey has shown that what you do on a plane can be determined b y which nationality is listed on your passport.According to the results of an international passenger survey, Australiansare the biggest drinkers on board with 36 percent choosing to down the hatch, compared to 35 percent of Americans and 33 percent of Brits.The Airline Passenger Experience Association(APEX) spoke to around 1,500 people, aged 18 and older, who have travelled by plane at least once during the last three months and were living in the United States, United Kingdom, Germany, Japan, China, Singapore, Australia and Brazil.The results found Chinese travelers are most likely to nod off once the seat-belt sign switches off. They are also the first to take out their credit card for some in-flight shopping and the biggest fans of gaming. Americans on the other hand like to use their time in the air more productively—when not drinking—choosing to work while flying at 35,000 feet.Meanwhile, Brits and Germans are the best at making chat with random strangers—spending 50 percent more time than any other nationality. Comparatively, Brazilians conduct their conversations online via email, messaging apps or social media.Despite plane food having a bad reputation, seven out of ten interviewees said they were happy to eat up on the selection of in-flight snacks and meals. In-flight magazines were also popular with four out of five passengers.The international flyers did however express their desire for better in-flight entertainment. “The industry has greatly improved the comfort, entertainment and on board service, and passengers are accepting those improvements” said Russell Lemieux, APEX executive director. “At the same time, passengers are demanding more from their air travel experiences which will drive more improvements touching all aspects of the journey. ” he added.1. What can you probably see in the flight according to the passage?A. Brazilians choose to drink.B. Americans do in-flight shopping.C. Germans chat to kill the time.D. The Chinese switch off the seat-belt sign.2. When on board the plane, ______.A. passengers from one nation have little in commonB. most passengers like to read in-flight magazinesC. more than half of the passengers don’t enjoy plane foodD. most people tend to use in-flight time to have a good sleep.3. What can we learn from the last paragraph?A. Flyers care little about entertainment.B. Flyers are not satisfied with the improvements.C. Flyers are expecting better flight experiences.D. Flyers have more and more demands from airlines.4. What’s the purpose of the passage?A. To entertain readers with interesting stories.B. To encourage people to behave well in public.C. To criticize impolite behaviors on the plane.D. To inform readers of the results of a survey.二、七选五5.You must have written your research paper, your personal essay, your book review-----whatever your school class requires. You think you have provided good information in the needed number of words. 1.But is it really done? Many teachers and professional writers believe that writing is revision. 2.Revision of writing is a necessary skill for students. The classroom is agood place to practice patience, concentration and listening. There arerewards with spending time with your thoughts and really taking time to compose your ideas in an orderly and reasonable way. You should put away your paper after you have written a first version, or draft. Wait several hours, maybe overnight, before working on it more. 3. Not only are you refreshed,but you’re looking at things through different eyes. That’s what revision literally means—to see again through different eyes.Following a four-step process may help you with your paper. The first step in the process is invention. It includes forming many questions about your subject. It is called “question-storming”. 4. Then comes the revision period. Take your time to read what you’ve written, to think about it, and maybe tore-shape it based on what you see now, as a kind of new person looking at it with a reader’s eyeglasses rather than a writer’s. The fourth step is called “publication”. 5. In a sense, anytime you turn it over to another person, that’s publica tion.Probably, the process takes away some of the tension of writing. And worry about the quality of your writing often disappears when you share that writing.A. Perfect writing is not possible.B. In the second step, you draft and compose a paper.C. And you feel good because your work is finished.D. This is just like returning to a job after a vacation.E. This does not mean your writing is professional publication.F. What is most important is getting your thoughts and ideas on paper.G. In other words, writing well means making needed changes and rewriting.三、完形填空6. I was going through my son Matthew’s backpack when I saw an envelope inthe bottom of it. Immediately, I knew it was a “thank you” card from one of his ________ . Totally not necessary since my Christmas gifts to them are my way of saying, “Thank you”. I ________ I read it quickly. And then I stopped.I ________ the card and read it again. One word caught my attention. “Ilove working with our Matthew.” One word. Our. That one word ________ the meaning of the sentence for me. If she had written “I love working with Matthew”, I would know that she loves working with my son. ________ byadding that one word, “our”, it meant “I love working with this boy who________ here, is accepted here and we all take responsibility for caring for.”I ________ knew this, of course, seeing a blog I wrote previously, butit’s always good to be _______ . In that blog post I mentioned ten reasonswhy his ________ is the right place for him. Since that blog we have had his IEP (Individualized Education Program) meeting, where I was ________ of that feeling again. In that meeting, someone ________ “Everyone loves Matthew. We all love Matthew”. And it was genuine and __________ . A s we went around the room and the staff ________ us on information about Matthew, it was apparent that it went way beyond sharing what he is doing ________ and behaviorally. Each person had a unique little ________ to tell about Matthew. Stories that show that they really know who Matthew is and that they ________ him.In fact just today I had written a note in his communication book that it was ________ Matthew to see new snow and not be able to play in it. Later in the day I got an email and a picture of Matthew ________ with snow in a big container inside the school.As I was reflecting on this, I realized that as a family we are reallylucky ________ school isn’t the only place where they think of him as “our Matthew”. It ________ to other parts of our lives as well — our friends,our family, our neighborhood, and our church.1. A. teachers B. friends C. classmates D. doctors2. A. realize B. admit C. imagine D. predict3. A. gave away B. tore up C. put away D. opened up4. A. simplified B. changed C. determined D. created5. A. Or B. And C. So D. But6. A. stays B. lives C. belongs D. remains7. A. already B. also C. even D. still8. A. reached B. accepted C. adored D. reminded9. A. school B. book C. home D. room10. A. informed B. convinced C. suspected D. cured11. A. commented B. insisted C. guaranteed D. recalled12. A. formal B. casual C. sincere D. severe13. A. advised B. judged C. updated D. congratulated14. A. accurately B. academically C. steadily D. securely15. A. secret B. lie C. joke D. story16. A. understand B. greet C. envy D. embarrass17. A. killing B. influencing C. calming D. inspiring18. A. meeting B. playing C. fighting D. dealing19. A. until B. unless C. though D. because20. A. flies B. extends C. applies D. switches四、短文填空7.When you take a walk in any of the cities in the West, you often see a lot of people 1. (walk) with dogs. It is still true that the dog is the most useful and faithful animal in the world. But the reasons why people keep a dog 2. (change). In the old days, people used to train dogs to protect themselves against the attacks from 3. beasts. And later they came 4. ( realize) that the dog was not only useful 5. willing to obey its master. For example, when people used dogs for hunting, the dogs would not eat what they caught without 6. (permit). But now people in the city need not protect themselves against attacks of animals. Why do they keep dogs, then? Some people keep dogs to protect themselves from robbery, but the 7. (important) reason is for companionship. For a child, 8. dog is his best friend when he has no friends to play with; for a young couple, a dog is 9. child when they have no children; for old couples, a dog is also their child when their real children have grown up. So the main reason why people keep dogs has changed 10. protection to friendship.五、短文改错8. 假定英语课上英语老师要求同桌之间互相修改作文，请你修改你同桌的以下作文。

黑龙江省哈尔滨市2019届高三第一次模拟考试理科数学哈师大附中、东北师大附中、辽宁省实验中学三省名校高三联考一、选择题：本大题共12个小题，每小题5分，共60分.在每小题给出的四个选项中，只有一项是符合题目要求的. 1.复数11212i i +++（其中i 为虚数单位）的虚部为（ ） A ．35 B ．35i C ．35- D ．35i -2.若集合{|12}A x x =<<，{|,}B x x b b R =>∈，则A B ⊆的一个充分不必要条件是（ ） A ．2b ≥ B ．12b <≤ C ．1b ≤ D ．1b <3.已知某7个数的平均数为4，方差为2，现加入一个新数据4，此时这8个数的平均数为x ，方差为2s ，则（ ）A ．4x =，22s <B ．4x =，22s >C ．4x >，22s <D ．4x >，22s >4.已知椭圆C ：22221(0)x y a b a b +=>>，若长轴长为6，且两焦点恰好将长轴三等分，则此椭圆的标准方程为（ ）A ．2213632x y += B ．22198x y += C ．22195x y += D ．2211612x y += 5.已知正项等比数列{}n a 满足31a =，5a 与432a 的等差中项为12，则1a 的值为（ ） A ．4 B ．2 C ．12 D ．146.已知变量x ，y 满足约束条件40221x y x y --≤⎧⎪-≤<⎨⎪≤⎩，若2z x y =-，则z 的取值范围是（ ）A ．[5,6)-B ．[5,6]-C ．(2,9)D ．[5,9]- 7.七巧板是一种古老的中国传统智力游戏，被誉为“东方魔板”.如图，这是一个 用七巧板拼成的正方形，其中1号板与2号板为两个全等的等腰直角三角形，3 号板与5号板为两个全等的等腰直角三角形，7号板为一个等腰直角三角形，4号板为一个正方形，6号板为一个平行四边形.现从这个正方形内任取一点，则此点取 自阴影部分的概率是（ ） A ．18 B ．14 C ．316 D ．388.已知函数()sin()f x x ωϕ=+)x ωϕ+0,2πωϕ⎛⎫>< ⎪⎝⎭的最小正周期为π，且()3f x f x π⎛⎫-=⎪⎝⎭，则（ ）A ．()f x 在0,2π⎛⎫⎪⎝⎭上单调递减 B ．()f x 在2,63ππ⎛⎫⎪⎝⎭上单调递增 C ．()f x 在0,2π⎛⎫⎪⎝⎭上单调递增 D ．()f x 在2,63ππ⎛⎫⎪⎝⎭上单调递减 9.某程序框图如图所示，该程序运行后输出M ，N 的值分别为（ ）A ．13，21B ．34，55C ．21，13D ．55，34 10.设函数212()log (1)f x x =+112x++，则使得()(21)f x f x ≤-成立的x 的取值范围是（ ） A ．(,1]-∞ B ．[1,)+∞ C ．1,13⎡⎤⎢⎥⎣⎦D ．[)1,1,3⎛⎤-∞+∞ ⎥⎝⎦11.设1F ，2F 分别为双曲线22221(0,0)x y a b a b-=>>的左、右焦点，过1F 作一条渐近线的垂线，垂足为M ，延长1F M 与双曲线的右支相交于点N ，若13MN F M =，则此双曲线的离心率为（ ）A B ．53 C ．43 D12.设1x ，2x 分别是函数()xf x x a -=-和()log 1a g x x x =-的零点（其中1a >），则124x x +的取值范围是（ ）A ．[4,)+∞B ．(4,)+∞C ．[5,)+∞D ．(5,)+∞ 二、填空题：本题共4小题，每小题5分，共20分.13.已知向量(1,1)a =，(2,)b x =，若a b +与3a b -平行，则实数x 的值是 ．14.某几何体的三视图如图所示，其中主视图的轮廓是底边为 高为1的等腰三角形，俯视图的轮廓为菱形，左视图是个半圆. 则该几何体的体积为 ．15.512a x x x x ⎛⎫⎛⎫-- ⎪⎪⎝⎭⎝⎭的展开式中各项系数的和为2，则该展开式中含4x 项的系数为 ．16.如图所示，将平面直角坐标系中的格点（横、纵坐标均为整数的点）按如下规则标上标签：原点处标数字0，记为0a ；点(1,0)处标数字1，记为1a ； 点(1,1)-处标数字0，记为2a ；点(0,1)-处标数字-1，记为3a ； 点(1,1)--处标数字-2，记为4a ；点(1,0)-处标数字-1，记为5a ； 点(1,1)-处标数字0，记为6a ；点(0,1)处标数字1，记为7a ； …以此类推，格点坐标为(,)i j 的点处所标的数字为i j +（i ，j 均为整数），记12n n S a a a =++⋅⋅⋅+，则2018S = ．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17～21题为必考题，每个试题考生都必须作答.每22、23题为选考题，考生根据要求作答. （一）必考题：共60分.17.在ABC ∆中，内角A ，B ，C 所对的边分别为a ，b ，c ，且cos cos 2b A a B c -=. （1）证明：tan 3tan B A =-；（2）若222b c a +=+，且ABC ∆a .18.如图1，在高为6的等腰梯形ABCD 中，//AB CD ，且6CD =，12AB =，将它沿对称轴1OO 折起，使平面1ADO O ⊥平面1BCOO .如图2，点P 为BC 中点，点E 在线段AB 上（不同于A ，B 两点），连接OE 并延长至点Q ，使//AQ OB .（1）证明：OD ⊥平面PAQ ；（2）若2BE AE =，求二面角C BQ A --的余弦值.19.2019年2月22日上午，山东省省委、省政府在济南召开山东省全面展开新旧动能转换重大工程动员大会，会议动员各方力量，迅速全面 展开新旧动能转换重大工程.某企业响应号召，对现有设备进行改造， 为了分析设备改造前后的效果，现从设备改造前后生产的大量产品中 各抽取了200件产品作为样本，检测一项质量指标值，若该项质量指 标值落在[20,40)内的产品视为合格品，否则为不合格品.图3是设备 改造前的样本的频率分布直方图，表1是设备改造后的样本的频数分 布表.表1：设备改造后样本的频数分布表（1）完成下面的22⨯列联表，并判断是否有99%的把握认为该企业生产的这种产品的质量指标值与设备改造有关；（2）根据图3和表1提供的数据，试从产品合格率的角度对改造前后设备的优劣进行比较；（3）企业将不合格品全部销毁后，根据客户需求对合格品．．．进行等级细分，质量指标值落在[25,30)内的定为一等品，每件售价240元；质量指标值落在[20,25)或[30,35)内的定为二等品，每件售价180元；其它的合格品定为三等品，每件售价120元.根据表1的数据，用该组样本中一等品、二等品、三等品各自在合格品中的频率．．．．．．．．代替从所有产品中抽到一件相应等级产品的概率.现有一名顾客随机购买两件产品，设其支付的费用为X （单位：元），求X 的分布列和数学期望. 附：22()()()()()n ad bc K a b c d a c b d -=++++ 20.在平面直角坐标系xOy 中，抛物线1C ：24x y =，直线l 与抛物线1C 交于A ，B 两点.（1）若直线OA ，OB 的斜率之积为14-，证明：直线l 过定点； （2）若线段AB 的中点M 在曲线2C：214(4y x x =--<<上，求AB 的最大值.21.已知函数2()ln (21)f x a x x a x =-+-()a R ∈有两个不同的零点. （1）求a 的取值范围；（2）设1x ，2x 是()f x 的两个零点，证明：122x x a +>.（二）选考题：共10分.请考生在22、23题中任选一题作答，如果多做，则按所做的第一题记分.22.[选修4-4：坐标系与参数方程]在直角坐标系xOy 中，过点(1,2)P 的直线l的参数方程为1122x t y ⎧=+⎪⎪⎨⎪=⎪⎩（t 为参数）.以原点O 为极点，x 轴正半轴为极轴建立极坐标系，曲线C 的极坐标方程为4sin ρθ=. （1）求直线l 的普通方程和曲线C 的直角坐标方程；（2）若直线l 与曲线C 相交于M ，N 两点，求11PM PN+的值.23.[选修4-5：不等式选讲] 已知函数()222f x x x =--+. （1）求不等式()6f x ≥的解集；（2）当x R ∈时，()f x x a ≥-+恒成立，求实数a 的取值范围.理科数学参考答案 一、选择题1-5: CDABA 6-10: ACDBC 11、12：BD 二、填空题15. -48 16. -249 三、解答题 17.【解析】 （1）根据正弦定理，由已知得：sin cos cos sin B A B A -2sin 2sin()C A B ==+， 展开得：sin cos cos sin B A B A -2(sin cos cos sin )B A B A =+， 整理得：sin cos 3cos sin B A B A =-，所以，tan 3tan B A =-.（2）由已知得：222b c a +-=，∴222cos 2b c a A bc +-=22bc ==，由0A π<<，得：6A π=，tan 3A =，∴tan B =由0B π<<，得：23B π=，所以6C π=，a c =，由12sin 23S ac π=212==2a =. 18.【解析】（1）【解法一（几何法）】取1OO 的中点为F ，连接AF ，PF ；∴//PF OB ， ∵//AQ OB ，∴//PF AQ ，∴P 、F 、A 、Q 四点共面， 又由图1可知1OB OO ⊥， ∵平面1ADO O ⊥平面1BCOO ， 且平面1ADO O平面11BCO O OO =，∴OB ⊥平面1ADOO ， ∴PF ⊥平面1ADOO ， 又∵OD ⊂平面1ADOO ， ∴PF OD ⊥.在直角梯形1ADOO 中，1AO OO =，1OF O D =，1AOF OO D ∠=∠，∴1AOF OO D ∆≅∆，∴1FAO DOO ∠=∠，∴190FAO AOD DOO AOD ∠+∠=∠+∠=， ∴AF OD ⊥. ∵AFPF F =，且AF ⊂平面PAQ ，PF ⊂平面PAQ ，∴OD ⊥平面PAQ .（1）【解法二（向量法）】由题设知OA ，OB ，1OO 两两垂直，所以以O 为坐标原点，OA ，OB ，1OO 所在直线分别为x 轴、y 轴、z 轴，建立如图所示的空间直角坐标系，设AQ 的长度为m ，则相关各点的坐标为(0,0,0)O ，(6,0,0)A ，(0,6,0)B ，(0,3,6)C ，(3,0,6)D ，(6,,0)Q m . ∵点P 为BC 中点，∴9(0,,3)2P ，∴(3,0,6)OD =，(0,,0)AQ m =，9(6,,3)2PQ m =--， ∵0OD AQ ⋅=，0OD PQ ⋅=，∴OD AQ ⊥，OD PQ ⊥，且AQ 与PQ 不共线， ∴OD ⊥平面PAQ .（2）∵2BE AE =，//AQ OB ，∴132AQ OB ==， 则(6,3,0)Q ，∴(6,3,0)QB =-，(0,3,6)BC =-.设平面CBQ 的法向量为1(,,)n x y z =，∵1100n QB n BC ⎧⋅=⎪⎨⋅=⎪⎩，∴630360x y y z -+=⎧⎨-+=⎩，令1z =，则2y =，1x =，则1(1,2,1)n =，又显然，平面ABQ 的法向量为2(0,0,1)n =，设二面角C BQ A --的平面角为θ，由图可知，θ为锐角，则12126cos 6n n n n θ⋅==⋅. 19.【解析】（1）根据图3和表1得到22⨯列联表：将22⨯列联表中的数据代入公式计算得：22()()()()()n ad bc K a b c d a c b d -=++++2400(172828192)20020036436⨯⨯-⨯=⨯⨯⨯12.210≈. ∵12.210 6.635>，∴有99%的把握认为该企业生产的这种产品的质量指标值与设备改造有关. （2）根据图3和表1可知，设备改造前产品为合格品的概率约为1724320050=，设备改造后产品为合格品的概率约为1922420025=；显然设备改造后产品合格率更高，因此，设备改造后性能更优. （3）由表1知：一等品的频率为12，即从所有产品中随机抽到一件一等品的概率为12； 二等品的频率为13，即从所有产品中随机抽到一件二等品的概率为13；三等品的频率为16，即从所有产品中随机抽到一件三等品的概率为16.由已知得：随机变量X 的取值为：240，300，360，420，480.240P X =（）1116636=⨯=，300P X =（）12111369C =⨯⨯=，360P X =（）1211115263318C =⨯⨯+⨯=，420P X =（）12111233C =⨯⨯=，480P X =（）111224=⨯=.∴随机变量X 的分布列为：∴240300369EX =⨯+⨯（）3604204804001834+⨯+⨯+⨯=. 20.【解析】设()11,A x y ，()22,B x y ，（1）由题意可知直线l 的斜率存在，设直线l 的方程为y kx m =+，由24x y y kx m⎧=⎨=+⎩，得：2440x kx m --=， ()2160k m ∆=+>，124x x k +=，124x x m =-，1212OA OBy y k k x x ⋅⋅=⋅2212121144x x x x ⋅=⋅12164x x m ⋅==-， 由已知：14OA OB k k ⋅=-，所以1m =， ∴直线l 的方程为1y kx =+，所以直线l 过定点(0,1).（2）设()00,M x y ，则12022x x x k +==，2002y kxm k m =+=+， 将()00,Mx y 带入2C ：214(4y x x =--<<得：22124(2)4k m k +=-，∴243m k =-.∵0x -<<2k -<k <<又∵()216k m ∆=+22216(43)32(2)0k k k =+-=->，∴k <<故k 的取值范围是：(k ∈.AB ==243m k =-代入得：AB =22≤=当且仅当2212k k +=-，即k =所以AB 的最大值为21.【解析】 （1）【解法一】函数()f x 的定义域为：(0,)+∞.'()221a f x x a x =-+-(21)()x a x x+-=， ①当0a ≤时，易得'()0f x <，则()f x 在(0,)+∞上单调递增， 则()f x 至多只有一个零点，不符合题意，舍去. ②当0a >时，令'()0f x =得：x a =，则∴max ()()f x f x =极大()(ln 1)f a a a a ==+-. 设()ln 1g x x x =+-，∵1'()10g x x=+>，则()g x 在(0,)+∞上单调递增. 又∵(1)0g =，∴1x <时，()0g x <；1x >时，()0g x >. 因此：（i ）当01a <≤时，max ()()0f x a g a =⋅≤，则()f x 无零点， 不符合题意，舍去.（ii ）当1a >时，max ()()0f x a g a =⋅>，∵12()(1)f a e e =-2110e e --<，∴()f x 在区间1(,)a e上有一个零点， ∵(31)ln(31)f a a a -=-2(31)(21)(31)a a a --+--[ln(31)(31)]a a a =---， 设()ln h x x x =-，(1)x >，∵1'()10h x x=-<， ∴()h x 在(1,)+∞上单调递减，则(31)(2)ln 220h a h -<=-<， ∴(31)(31)0f a a h a -=⋅-<，∴()f x 在区间(,31)a a -上有一个零点，那么，()f x 恰有两个零点. 综上所述，当()f x 有两个不同零点时，a 的取值范围是(1,)+∞. （1）【解法二】函数的定义域为：(0,)+∞.'()221a f x x a x =-+-(21)()x a x x+-=， ①当0a ≤时，易得'()0f x <，则()f x 在(0,)+∞上单调递增， 则()f x 至多只有一个零点，不符合题意，舍去. ②当0a >时，令'()0f x =得：x a =，则增∴max ()()f x f x =极大()(ln 1)f a a a a ==+-.∴要使函数()f x 有两个零点，则必有()(ln 1)0f a a a a =+->，即ln 10a a +->， 设()ln 1g a a a =+-，∵1'()10g a a=+>，则()g a 在(0,)+∞上单调递增， 又∵(1)0g =，∴1a >； 当1a >时： ∵12()(1)f a e e =-2110e e--<， ∴()f x 在区间1(,)a e上有一个零点； 设()ln h x x x =-，∵11'()1x h x x x-=-=，∴()h x 在(0,1)上单调递增，在(1,)+∞上单调递减， ∴()(1)10h x h ≤=-<，∴ln x x <，∴2()ln (21)f x a x x a x =-+-22(21)3ax x a x ax x x ≤-+-=--23(3)ax x x a x ≤-=-， 则(4)0f a <，∴()f x 在区间(,4)a a 上有一个零点， 那么，此时()f x 恰有两个零点.综上所述，当()f x 有两个不同零点时，a 的取值范围是(1,)+∞. （2）【证法一】由（1）可知，∵()f x 有两个不同零点，∴1a >，且当(0,)x a ∈时，()f x 是增函数； 当(,)x a ∈+∞时，()f x 是减函数； 不妨设：12x x <，则：120x a x <<<； 设()()(2)F x f x f a x =--，(0,2)x a ∈， 则：'()'()'(2)F x f x f a x =--2(21)2a ax a x a x=-+-+-2(2)(21)a x a --+- 22()22(2)a a x a x a x x a x -=+-=--. 当(0,)x a ∈时，'()0F x >，∴()F x 单调递增，又∵()0F a =， ∴()0F x <，∴()(2)f x f a x <-， ∵1(0,)x a ∈，∴11()(2)f x f a x <-， ∵12()()f x f x =，∴21()(2)f x f a x <-，∵2(,)x a ∈+∞，12(,)a x a -∈+∞，()f x 在(,)a +∞上单调递减， ∴212x a x >-，∴122x x a +>. （2）【证法二】由（1）可知，∵()f x 有两个不同零点，∴1a >，且当(0,)x a ∈时，()f x 是增函数； 当(,)x a ∈+∞时，()f x 是减函数；不妨设：12x x <，则：120x a x <<<； 设()()()F x f a x f a x =+--，(0,)x a ∈， 则'()'()'()F x f a x f a x =++-2()(21)a a a x a a x a x=-++-++-2()(21)a x a --+- 222()()a a x a x a x a x a x =+-=+-+-. 当(0,)x a ∈时，'()0F x >，∴()F x 单调递增， 又∵(0)0F =，∴()0F x >，∴()()f a x f a x +>-， ∵1(0,)a x a -∈，∴12()()f x f x =11(())(())f a a x f a a x =--<+-1(2)f a x =-， ∵2(,)x a ∈+∞，12(,)a x a -∈+∞，()f x 在(,)a +∞上单调递减， ∴212x a x >-，∴122x x a +>. 22.【解析】（1）由已知得：11222x t y ⎧-=⎪⎪⎨⎪-=⎪⎩，消去t得21)y x -=-，20y -+=， 即：l20y -+=.曲线C ：4sin ρθ=得，24sin ρρθ=，即224x y y +=，整理得22(2)4x y +-=， 即：C ：22(2)4x y +-=.（2）把直线l的参数方程1122x t y ⎧=+⎪⎪⎨⎪=+⎪⎩（t 为参数）代入曲线C 的直角坐标方程中得：221(1))42t ++=，即230t t +-=，设M ，N 两点对应的参数分别为1t ，2t ，则121213t t t t +=-⎧⎨⋅=-⎩，∴11PM PN +1212PM PN t t PM PN t t ++==⋅⋅1212t t t t -==⋅3=. 23.【解析】（1）当2x ≤-时，()4f x x =-+，∴()646f x x ≥⇒-+≥2x ⇒≤-，故2x ≤-； 当21x -<<时，()3f x x =-，∴()636f x x ≥⇒-≥2x ⇒≤-，故x φ∈； 当1x ≥时，()4f x x =-，∴()646f x x ≥⇒-≥10x ⇒≥，故10x ≥； 综上可知：()6f x ≥的解集为(,2][10,)-∞+∞.（2）由（1）知：4,2()3,214,1x x f x x x x x -+≤-⎧⎪=--<<⎨⎪-≥⎩，【解法一】如图所示：作出函数()f x 的图象，由图象知，当1x =时，13a -+≤-，解得：2a ≤-， ∴实数a 的取值范围为(,2]-∞-. 【解法二】当2x ≤-时，4x x a -+≥-+恒成立，∴4a ≤， 当21x -<<时，3x x a -≥-+恒成立，∴2a ≤-， 当1x ≥时，4x x a -≥-+恒成立，∴2a ≤-，-∞-. 综上，实数a的取值范围为(,2]。

黑龙江省哈尔滨市第三中学2019-2020学年高三下学期第一次调研语文试题(word无答案)一、现代文阅读(★★) 1 . 阅读下面的文字，完成下面小题。

我国古代的全民文化水准低，识文断字的是少数，老百姓的文学创作和享受几乎与文字无缘，而文学史又只能由文字来写，能写的又只有所谓君子和士大夫这一类人。

所以，留传下来的古代文学作品和古代文学史其主流就是贵族文学史，也就是所谓高雅文学的文学史。

这并不意味着我国古代就没有劳动人民的文学，或曰之为俗文学。

由于古代劳动人民口口相传的民谚、诗歌、史诗之类的客观存在和一些贵族文人自觉不自觉的保护、保存、培育和发展，我们还是能把握出其明显的脉络的。

应该说，我国古代俗文学的第一个高潮在“诗经”时代。

东周之前的三代，虽有战争，但对经济和社会生活的破坏程度比后来“礼崩乐坏”的战国、汉魏晋等朝代要小得多，中央集权专制较之后的时代为弱，意识形态上也比“焚书坑儒”和“独尊儒术”以后要宽松得多。

人民在相对稳定的社会环境下男耕女织，夫唱妇随，偶有战争，也不至于引起较大范围的社会动乱，且人民也能以正确和平和的心态待之。

而这一切,都反映在那个时代的民歌集大成之著作——《诗经·国风》中。

《国风》是我国古代俗文学第一个高潮的标志。

读“国风”，我们既可以欣赏到先民们那发自内心的愉悦、痛快淋漓的愤怒，又可以由此对当时的社会状况作一个广泛而比较细致的了解。

那是一个真正意义上的不自觉而然的全民诗歌运动,是后来人为发起的如唐代的“新乐府运动”所不可比的。

《诗经》中的“雅、颂”，属于高雅文学,但其文学表现手法和音乐形制同“国风”没有很大的区别和对立，“风”和“雅”“颂”三而合一作为诗经“六义”中的三义一并称之，这实在是阶级矛盾在一定程度上调和甚至是相融在文学上曲折的反映。

汉时,乐府的民间采诗活动在目的上和诗的规模、内容以及语言的丰富性上,与《诗经·国风》不可同日而语。

以后，随着语言文白的分野,城市的发展，作为广大农民参与创作和流传的、以口头文学形式出现并被记录下来的俗文学便再也没有缓过劲来。