广东省广州市第六中学2019-2020年高一下学期黄埔班4月10.11周末数学试题(无答案)

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广东省广州市第六中学高一数学文联考试题含解析

广东省广州市第六中学高一数学文联考试题含解析一、选择题：本大题共10小题，每小题5分，共50分。

在每小题给出的四个选项中，只有是一个符合题目要求的1. （4分）半径R的半圆卷成一个圆锥，则它的体积为（）A．πR3 B．πR3 C．πR3 D．πR3参考答案：A考点：旋转体（圆柱、圆锥、圆台）．专题：计算题．分析：求出扇形的弧长，然后求出圆锥的底面周长，转化为底面半径，求出圆锥的高，然后求出体积．解答：2πr=πR，所以r=，则h=，所以V=故选A点评：本题是基础题，考查圆锥的展开图与圆锥之间的计算关系，圆锥体积的求法，考查计算能力．2. 已知自然对数的底数，在下列区间中，函数的零点所在区间为（）A．B．(1,2) C．(2,e) D．(e,3)参考答案：C函数是单调递增函数，根据零点存在定理得到故零点存在于(2,e)之间。

3. 函数的零点所在的区间为（）A. B. C. D.参考答案：C略4. （5分）在空间直角坐标系中，点B是点A（2，﹣3，5）关于xOy面的对称点，则|AB|=（）A．10 B．C．D．38参考答案：A考点：空间两点间的距离公式．专题：空间位置关系与距离．分析：先求出点P关于坐标平面的对称点，进而即可求出向量的坐标及模．解答：∵点A（2，﹣3，5）关于xoy平面的对称点B（2，﹣3，﹣5），∴=（0，0，﹣10），∴|AB|==10．故选：A．点评：本题考查空间两点的距离公式，对称知识的应用，熟练掌握向量的模的求法是解题的关键．5. 若函数f（x）=x3+x2﹣2x﹣2的一个正数零点附近的函数值用二分法计算，其参考数据如下：那么方程x3+x2﹣2x﹣2=0的一个近似根（精确到0.1）为（）A．1.2 B．1.3 C．1.4 D．1.5参考答案：C【考点】二分法求方程的近似解．【分析】由图中参考数据可得f（1.43750＞0，f（1.40625）＜0，又因为题中要求精确到0.1可得答案．【解答】解：由图中参考数据可得f（1.43750）＞0，f（1.40625）＜0，又因为题中要求精确到0.1，所以近似根为 1.4故选 C．6. 函数的图像为M，则下列结论中正确的是（）A．图像M关于直线对称B．由的图像向左平移得到MC. 图像M关于点对称D．在区间上递增参考答案：C由的图像向左平移得到,f(x)在区间上有增有减，图像M关于点对称.7. 如图，它表示电流在一个周期内的图象，则的解析式为A． B.C. D.参考答案：A 略8. 已知是R上的减函数，则a的取值范围是（）A．（0，1）B．C．D．参考答案：B【考点】函数单调性的性质．【专题】计算题．【分析】由题意可得0＜a＜1，且3a﹣1＜0，（3a﹣1）×1+4a＞a，于是可求得a的取值范围．【解答】解：∵f（x）=是R上的减函数，∴0＜a＜1，①且3a﹣1＜0，②（3a﹣1）×1+4a≥a，③由①②③得：≤a＜．故选B．【点评】本题考查函数单调性的性质，难点在于对“f（x）=是R上的减函数”的理解与应用，易错点在于忽视“（3a﹣1）×1+4a≥a”导致解的范围扩大，考查思维的缜密性，属于中档题．9. 已知全集I=R，M=，N=，则(C M)∩N等于（）A、 B、 C、 D、参考答案：A10. 已知是函数的两个零点,则（）A. B. C. D.参考答案：B二、填空题:本大题共7小题,每小题4分,共28分11. 通过观察所给两等式的规律， ①②请你写出一个（包含上面两命题）一般性的命题：．参考答案：12. 函数，的值域是_____________．参考答案：[0，4] 略13. 已知等比数列的前项和，则.参考答案：略14. 已知等差数列{a n }的公差d 不为0，且a 1，a 3，a 7成等比数列，则= ．参考答案：2【考点】84：等差数列的通项公式；85：等差数列的前n 项和．【分析】由题意可得，解之可得a 1=2d≠0，变形可得答案．【解答】解：由题意可得：，即d （2d ﹣a 1）=0，因为公差d 不为0，故2d ﹣a 1=0，解得a 1=2d≠0，故==2，故答案为：2 15. 给出如下结论：①函数是奇函数；②存在实数，使得；③若是第一象限角且，则；④是函数的一条对称轴方程；⑤函数的图形关于点成中心对称图形.其中正确的结论的序号是．（填序号）参考答案：①④①函数=﹣sin ，是奇函数，正确；②存在实数α，使得sin α+cos α=sin （α+）≤，故错误；③α，β是第一象限角且α＜β．例如：45°＜30°+360°，但tan45°＞tan （30°+360°），即tanα＜tanβ不成立；④是函数，f （）=﹣1，是一条对称轴方程，故正确；⑤函数的图象关于点，f （）=1，不是对称中心，故错误．故答案为：①④．16. ，，，且，求实数的取值范围 . 参考答案：17. 阅读材料：某同学求解的值其过程为：设，则，从而，于是，即，展开得，，，化简，得，解得，，（舍去），即．试完成以下填空：设函数对任意都有成立，则实数的值为．参考答案：4略三、解答题：本大题共5小题，共72分。

2019-2020学年广东省广州市二中、广雅、执信、六中四校联考高一下学期期末数学试卷 (解析版)

2019-2020学年广东省广州市二中、广雅、执信、六中四校联考高一第二学期期末数学试卷一、选择题（共12小题）.1．已知集合A＝{x2﹣3x+2＜0}，B＝{x|}，则（）A．A⊆B B．B⊆A C．A∩∁R B＝∅D．A∩B＝∅2．下列函数中，既是偶函数又在区间（﹣∞，0）上单调递增的是（）A．y＝﹣x﹣2B．y＝ln|x|C．y＝x3﹣3D．y＝﹣e|x|3．已知cos（α+）＝，则＝（）A．B．C．D．4．已知函数f（x）＝sin x，，将f（x）的图象经过下列哪种变换可以与g（x）的图象重合（）A．向左平移个单位，再把各点的横坐标缩短到原来的B．向左平移个单位，再把各点的横坐标缩短到原来的C．向左平移个单位，再把各点的横坐标伸长到原来的2倍D．向左平移个单位，再把各点的横坐标伸长到原来的2倍5．设等差数列{a n}的前n项和为S n，若3a6＝6+a4，则S13的值等于（）A．26B．39C．52D．656．已知函数f（x）＝sin（ωx+φ）（ω＞0）的图象相邻两对称轴间的距离等于，若∀x∈R．f （x）≤，则正数φ的最小值为（）A．B．C．D．7．设a＝2﹣1.5，b＝log23，，则（）A．c＜a＜b B．a＜c＜b C．a＜b＜c D．c＜b＜a8．已知数列{a n}的首项为1，数列{b n}为等比数列且，若b5b6＝2，则a11＝（）A．16B．21C．31D．329．已知正方体ABCD﹣A1B1C1D1的边长为2，M是BB1的中点，点P在正方体内部或表面上，且MP∥平面AB1D1，则动点P的轨迹所形成的区城面积是（）A．B．C．D．10．设O是△ABC的外接圆圆心、且，则∠BOC＝（）A．B．C．D．11．如图，网格纸上正方形小格的边长为1，图中粗线画出的是某几何体的三视图，则该几何体的体积为（）A．B．C．D．12．已知定义域为R的函数f（x）满足f（x）＝3f（x+2），且设f（x）在[2n﹣2，2n）上的最大值为a n（n∈N*），且数列{a n}的前n项和为S n，若S n＜k对任意的正整数n均成立，则实数k的取值范围为（）A．B．C．D．二、填空题（共4小题）.13．已知＝（2，﹣1），＝（1，t），若（2﹣）⊥，则＝．14．已知数列{a n}的前项和为S n，S n＝4n﹣3．则数列{a n}的通项公式为．15．已知三角形ABC中，AB＝AC＝，BC＝2，将它沿高AD翻折，使点B与点C间的距离为．此时四面体A﹣BCD外接球的表面积．16．函数在区间[﹣4，8]上的所有零点之和为．三、解答题：解答应写出文字说明、证明过程成演算步骤17．设函数f（x）＝sinωx（sinωx+cosωx）+cos2ωx最小正周期为π．（1）求ω；（2）求函数f（x）区间上的值域．18．设数列{a n}前n项和为S n且2a1＝a2＝2，等差数列{b n}满足b1＝1，b2+b5＝b8且b2S n+1+b5S n＝b8S n（n≥2，n∈N*）．﹣1（1）求{a n}和{b n}的通项公式；（2）求数列{a n b n}的前n项和T n．19．已知△ABC的外接圆半径为R，a，b，c分别是角A，B，C的对边，b＝2且b sin B﹣a sin A＝2R（sin B﹣sin C）sin C．（1）求角A；（2）若AD是BC边上的中线AD＝，求△ABC的面积．20．如图，已知四棱锥P﹣ABCD的底面ABCD为梯形，AB∥CD，PA＝PD＝，BC＝CD＝1，AB＝2，∠BCD＝，点P在底面ABCD上的投影落在AB边上．（1）若E为PD上一点且PE＝2ED，证明：PB‖平面ACE；（2）求二面角P﹣AD﹣B的余弦值．21．某化工厂从今年一月起若不改善生产环境，按生产现状每月收入为75万元，同时将受到环保部门的处罚，第一个月罚7万元，以后每月增加2万元，如果从今年一月起投资600万元添加回收净化设备（改设备时间不计），一方面可以改善环境，另一方面可以大大降低原料成本，设添加回收净化设并投产后n个月的累计收入为g（n），据测算，当1≤n≤5（n∈N*）时，g（n）＝n2+kn（k是常数），且前4个月的累计收入为416万元，从第6个月开始，每个月的收入都与第5个月相同，同时，该厂不但不受处罚，而且还将得到环保部门的一次性奖励200万元．（1）求添加回收净化设备后前7个月的累计收入；（2）从第几个月起投资开始见效，即投资改造后的纯收入（累计收入连同奖励减去改造设备费）多于不改造的纯收入（累计收入减去罚款）？22．设a为正数，函数f（x）＝ax2+bx+c满足f（0）＝1且f（x）＝f．（1）若f（1）＝1，求f（x）；（2）设g（x）＝log2（x﹣2+2），若对任意实数t，总存在x1，x2∈[t﹣1，t+1]，使得f（x1）﹣f（x2）≥g（x3）﹣g（x4）对所有x3，x4∈都成立，求a的取值范围．参考答案一、选择题（共12小题）.1．已知集合A＝{x2﹣3x+2＜0}，B＝{x|}，则（）A．A⊆B B．B⊆A C．A∩∁R B＝∅D．A∩B＝∅解：∵A＝{x|1＜x＜2}，B＝{x|log8x＞log82}＝{x|x＞2}，∴∁R B＝{x|x≤2}，A∩∁R B≠∅，A∩B＝∅．故选：D．2．下列函数中，既是偶函数又在区间（﹣∞，0）上单调递增的是（）A．y＝﹣x﹣2B．y＝ln|x|C．y＝x3﹣3D．y＝﹣e|x|解：根据题意，依次分析选项：对于A，y＝﹣x﹣2＝﹣，是偶函数，在区间（﹣∞，0）上单调递减，不符合题意；对于B，y＝ln|x|＝，是偶函数，在区间（﹣∞，0）上单调递减，不符合题意；对于C，y＝x3﹣3，不是偶函数，不符合题意；对于D，y＝﹣e|x|＝，既是偶函数又在区间（﹣∞，0）上单调递增，符合题意；故选：D．3．已知cos（α+）＝，则＝（）A．B．C．D．解：因为﹣α＝﹣（α），所以sin（﹣α）＝sin[﹣（α+）]＝cos（α+）＝，故选：A．4．已知函数f（x）＝sin x，，将f（x）的图象经过下列哪种变换可以与g（x）的图象重合（）A．向左平移个单位，再把各点的横坐标缩短到原来的B．向左平移个单位，再把各点的横坐标缩短到原来的C．向左平移个单位，再把各点的横坐标伸长到原来的2倍D．向左平移个单位，再把各点的横坐标伸长到原来的2倍解：将函数f（x）＝sin x的图象向左平移个单位，再将函数上各点的横坐标缩短为原来的，得到的图象．故选：A．5．设等差数列{a n}的前n项和为S n，若3a6＝6+a4，则S13的值等于（）A．26B．39C．52D．65解：等差数列{a n}中，3a6＝6+a4，∴a4+a8+a6＝6+a4，∴a8+a6＝6，S13＝＝＝39．故选：B．6．已知函数f（x）＝sin（ωx+φ）（ω＞0）的图象相邻两对称轴间的距离等于，若∀x∈R．f （x）≤，则正数φ的最小值为（）A．B．C．D．解：∵函数f（x）＝sin（ωx+φ）（ω＞0）的图象相邻两对称轴间的距离等于，∴•＝，∴ω＝4．若∀x∈R，f（x）≤，∴4×+φ＝kπ+，k∈Z．故当k＝1时，正数φ取得最小值为，故选：D．7．设a＝2﹣1.5，b＝log23，，则（）A．c＜a＜b B．a＜c＜b C．a＜b＜c D．c＜b＜a解：∵2﹣1.5＜20＝1，，，∴a＜c＜b．故选：B．8．已知数列{a n}的首项为1，数列{b n}为等比数列且，若b5b6＝2，则a11＝（）A．16B．21C．31D．32解：∵数列{a n}的首项为1，数列{b n}为等比数列且，∴b1•b2•b3…b10＝••…＝＝a11．∵b5b6＝2，∴b1•b2•b3…b10＝＝25，∴a11＝25＝32，故选：D．9．已知正方体ABCD﹣A1B1C1D1的边长为2，M是BB1的中点，点P在正方体内部或表面上，且MP∥平面AB1D1，则动点P的轨迹所形成的区城面积是（）A．B．C．D．解：如图所示，E，F，G，H，N分别为B1C1，C1D1，DD1，DA，AB的中点，则EF∥B1D1∥NH，MN∥B1A∥FG，∴平面MEFGHN∥平面AB1D1，∴动点P的轨迹是六边形MEFGHN及其内部．∵正方体ABCD﹣A1B1C1D1的边长为2，∴EF＝FG＝GH＝HN＝NM＝ME＝，即六边形EFGHNM是边长为的正六边形，则其面积S＝．故选：C．10．设O是△ABC的外接圆圆心、且，则∠BOC＝（）A．B．C．D．解：因为，所以，两边平方得：＝，（设θ＝∠BOC），∵，∴7＝4+4cosθ+1，∴，∵，∴．故选：B．11．如图，网格纸上正方形小格的边长为1，图中粗线画出的是某几何体的三视图，则该几何体的体积为（）A．B．C．D．解：由题意可知几何体的直观图是正四棱柱的一部分，如图A﹣BCD所示，因为AD＝4，AB＝2，所以三棱锥的体积为：﹣＝．故选：D．12．已知定义域为R的函数f（x）满足f（x）＝3f（x+2），且设f（x）在[2n﹣2，2n）上的最大值为a n（n∈N*），且数列{a n}的前n项和为S n，若S n＜k对任意的正整数n均成立，则实数k的取值范围为（）A．B．C．D．解：当x∈[0，2）时，且，可得0≤x＜1时，f（x）的最大值为f（0）＝2；1＜x≤2时，f（x）的最大值为f（）＝，即有0≤x＜2时，f（x）的最大值为；当2≤x＜4时，f（x）＝f（x﹣2）的最大值为；当4≤x＜6时，f（x）＝f（x﹣2）的最大值为；…可得{a n}为首项为，公比为的等比数列，可得S n＝＝（1﹣）＜，由S n＜k对任意的正整数n均成立，可得k≥．即k的最小值为．故选：B．二、填空题：本大题共4小题，每小题5分13．已知＝（2，﹣1），＝（1，t），若（2﹣）⊥，则＝．解：已知＝（2，﹣1），＝（1，t），若（2﹣）⊥，则（2﹣）•＝2﹣•＝2×5﹣（2﹣t）＝0，∴t＝﹣8，则＝＝，故答案为：．14．已知数列{a n}的前项和为S n，S n＝4n﹣3．则数列{a n}的通项公式为a n＝．解：根据题意，数列{a n}的前项和S n＝4n﹣3，①则当n≥2时，S n﹣1＝4n﹣1﹣3，②①﹣②可得：a n＝S n﹣S n﹣1＝3×4n﹣1，（n≥2）当n＝1时，a1＝S1＝4﹣3＝1，不符合a n＝3×4n﹣1，故a n＝；故答案为：a n＝15．已知三角形ABC中，AB＝AC＝，BC＝2，将它沿高AD翻折，使点B与点C间的距离为．此时四面体A﹣BCD外接球的表面积8π．解：因为AB＝AC＝，BC＝2，将它沿高AD翻折，使点B与点C间的距离为．所以可得BD＝CD＝1，AD＝＝＝2，在△BCD中，由余弦定理可得cos∠BCD＝＝＝﹣，所以sin∠BDC＝，设△BCD的外接圆的半径为r，则2r＝＝，所以r＝1；因为BD⊥AD，CD⊥AD，BD∩CD＝D，所以AD⊥面BCD，将此三棱锥放在直棱柱AB1C1﹣DBC中，设其外接球的半径为R，则R2＝r2+（）2＝12+12＝2，所以外接球的表面积S＝4πR2＝4π•2＝8π，故答案为：8π．16．函数在区间[﹣4，8]上的所有零点之和为16．解：由题意函数y＝sin x﹣，x∈[﹣4，8]的零点，即sin x﹣＝0的根；作出函数y＝sin x与y＝，如下图所示：由图可得：两个函数的图象有8个不同的交点，且两两关于点（2，0）对称，故8个点横坐标之和为16，即函函数在区间[﹣4，8]上的所有零点之和为16，故答案为：16．三、解答题：解答应写出文字说明、证明过程成演算步骤17．设函数f（x）＝sinωx（sinωx+cosωx）+cos2ωx最小正周期为π．（1）求ω；（2）求函数f（x）区间上的值域．解：（1）因为f（x）＝sinωx（sinωx+cosωx）+cos2ωx＝sin2ωx+sinωx cosωx+cos2ωx＝+sin2ωx+cos2ωx＝sin（2ωx+）+，所以最小正周期T＝＝π，所以ω＝±1；（2）当ω＝1时，f（x）＝sin（2x+）+，x∈[，]，则2x+∈[，]，令t＝2x+∈[，]，因为t∈[，]，f（t）＝sin t+，单调递减，所以f（t）min＝f（）＝+＝0，f（t）max＝f（）＝•1+＝，而t∈[，），f（t）单调递增，f（t）min＝f（）＞f（），所以函数f（x）区间上的值域为[0，]．当ω＝﹣1时，f（x）＝﹣sin（2x﹣]+，x∈[，]，则2x﹣∈[﹣，]，令t＝2x﹣∈[﹣，]，所以f（t）＝﹣sin t+，当t∈[﹣，]，f（t）单调递减，f（t）min＝f（）＝﹣+＝；f（t）max＝f（﹣）＝﹣•sin（﹣）+＝+＝+＝+＝，t∈（，]，f（t）单调递增，f（t）max＝f（）＝+＝1，所以函数f（x）区间上的值域为[，1]．综上所述ω＝1时函数f（x）区间上的值域[0，]；ω＝﹣1时函数f（x）区间上的值域[，1]．18．设数列{a n}前n项和为S n且2a1＝a2＝2，等差数列{b n}满足b1＝1，b2+b5＝b8且b2S n+1+b5S n＝b8S n（n≥2，n∈N*）．﹣1（1）求{a n}和{b n}的通项公式；（2）求数列{a n b n}的前n项和T n．解：（1）设公差为d的等差数列{b n}满足b1＝1，b2+b5＝b8，则b1+d+b1+4d＝b1+7d，解得d＝，所以．数列{a n}前n项和为S n且2a1＝a2＝2，且b2S n+1+b5S n﹣1＝b8S n，整理得，即（S n+1﹣S n）＝3（S n﹣S n﹣1），n≥2，即S n+1﹣S n＝2（S n﹣S n﹣1），即a n+1＝2a n，n≥2，∵2a1＝a2＝2，∴数列{a n}是以1为首项，2为公比的等比数列，∴a n＝2n﹣1；（2）a n b n＝（n+1）2n﹣1＝（n+1）2n﹣2，∴T n＝2×2﹣1+3×20+4×21+…+（n+1）2n﹣2，①，2T n＝2×20+3×21+4×22+…+（n+1）2n﹣1，②，∴﹣T n＝1+20+21+22+…+2n﹣2﹣（n+1）2n﹣1＝1+﹣（n+1）2n﹣1＝﹣n×2n﹣1，∴T n＝n×2n﹣1．19．已知△ABC的外接圆半径为R，a，b，c分别是角A，B，C的对边，b＝2且b sin B﹣a sin A＝2R（sin B﹣sin C）sin C．（1）求角A；（2）若AD是BC边上的中线AD＝，求△ABC的面积．解：（1）∵由正弦定理＝2R，可得b＝2R sin B，c＝2R sin C，∴由已知可得：b sin B﹣a sin A＝（b﹣c）sin C，∴b2﹣a2＝c（b﹣c）＝bc﹣c2，即b2+c2﹣a2＝bc，∴由余弦定理可得cos A＝＝＝，∵A∈（0，π），∴A＝．（2）∵BC边上的中线AD＝，b＝2，又，两边平方，可得：2＝（++2），∴＝（c2+22+2×c×2×cos），整理可得：c2+2c﹣3＝0，解得c＝1，或﹣3（舍去），∴S△ABC＝bc sin A＝＝．20．如图，已知四棱锥P﹣ABCD的底面ABCD为梯形，AB∥CD，PA＝PD＝，BC＝CD＝1，AB＝2，∠BCD＝，点P在底面ABCD上的投影落在AB边上．（1）若E为PD上一点且PE＝2ED，证明：PB‖平面ACE；（2）求二面角P﹣AD﹣B的余弦值．解：（1）证明：连接AC，BD，设AC∩BD＝F，∵AB∥CD，∴△DFC∽△BFA，又AB＝2DC，∴DF：FB＝1：2，连接FE，∵PE＝2ED，∴DE：EP＝1：2，则EF∥PB．∵EF⊂平面ACE，PB⊄平面ACE，∴PB∥平面ACE；（2）∵点P在底面ABCD上的投影落在AB边上，设O为P在AB上的投影，则PO⊥平面ABCD，取AD中点G，连接PG，OG，∵PA＝PD，∴PG⊥AD，OG⊥AD，则∠PGO为二面角P﹣AD﹣B的平面角．在梯形ABCD中，∵BC＝CD＝1，∠BCD＝，∴三角形BCD为等边三角形，得BD＝1，且∠ABD＝，又AB＝2，∴AD＝．∴∠BAD＝，则OG＝．在△PAD中，由PA＝PD＝，AD＝，得PG＝．在Rt△POG中，可得cos∠PGO＝．∴二面角P﹣AD﹣B的余弦值为．21．某化工厂从今年一月起若不改善生产环境，按生产现状每月收入为75万元，同时将受到环保部门的处罚，第一个月罚7万元，以后每月增加2万元，如果从今年一月起投资600万元添加回收净化设备（改设备时间不计），一方面可以改善环境，另一方面可以大大降低原料成本，设添加回收净化设并投产后n个月的累计收入为g（n），据测算，当1≤n≤5（n∈N*）时，g（n）＝n2+kn（k是常数），且前4个月的累计收入为416万元，从第6个月开始，每个月的收入都与第5个月相同，同时，该厂不但不受处罚，而且还将得到环保部门的一次性奖励200万元．（1）求添加回收净化设备后前7个月的累计收入；（2）从第几个月起投资开始见效，即投资改造后的纯收入（累计收入连同奖励减去改造设备费）多于不改造的纯收入（累计收入减去罚款）？解：（1）据题意g（4）＝42+4k＝416，解得k＝100，∴g（n）＝n2+100n，（n≤5）第5个月的收入为g（5）﹣g（4）＝525﹣416＝109万元，∴g（7）＝g（5）+2×109＝743万元．（2）g（n）＝，即g（n）＝﹒若不投资改造，则前n个月的总罚款7n+×2＝n2+6n，令g（n）﹣600+200＞75n﹣（n2+6n），得：g（n）+n2﹣69n﹣400＞0．显然当n≤5时，上式不成立；当n＞5时，109n﹣20+n2﹣69n﹣400＞0，即n（n+40）＞420，又n∈N*，解得n≥9．故从第9个月投资开始见效．22．设a为正数，函数f（x）＝ax2+bx+c满足f（0）＝1且f（x）＝f．（1）若f（1）＝1，求f（x）；（2）设g（x）＝log2（x﹣2+2），若对任意实数t，总存在x1，x2∈[t﹣1，t+1]，使得f（x1）﹣f（x2）≥g（x3）﹣g（x4）对所有x3，x4∈都成立，求a的取值范围．解：（1）函数f（x）＝ax2+bx+c满足f（0）＝1且f（x）＝f，可得c＝1且﹣＝，即b＝﹣2，又f（1）＝1，可得a+b+c＝1，解得a＝2，则f（x）＝2x2﹣2x+1；（2）g（x）＝log2（x﹣2+2）＝log2[（﹣1）2+1]，当x∈[，4]可得g（x）的最小值为g（1）＝0，最大值为g（4）＝1，g（x3）﹣g（x4）的最大值为1﹣0＝1，所以对任意的实数t，总存在x1，x2∈[t﹣1，t+1]，使得f（x1）﹣f（x2）≥1．设f（x）＝ax2﹣2x+1在[t﹣1，t+1]上最大值为M（t），最小值为m（t），f（x）的对称轴为直线x＝，令h（t）＝M（t）﹣m（t），则对任意的实数t，h（t）≥1．①当≤t﹣1时，f（x）在[t﹣1，t+1]上递增，可得M（t）＝f（t+1），m（t）＝f（t﹣1），则h（t）＝M（t）﹣m（t）＝4at﹣4，此时h（t）≥4a（+1）﹣4＝4a≥1，∴a≥；②当t﹣1＜≤t时，M（t）＝f（t+1），m（t）＝f（）＝1﹣，h（t）＝M（t）﹣m（t）≥f（+1）﹣（1﹣）＝a（+1）2﹣2（1+）+1﹣（1﹣）＝a≥1，∴a≥1．③当t＜＜t+1时，M（t）＝f（t﹣1），m（t）＝f（）＝1﹣，h（t）＝M（t）﹣m（t）≥f（﹣1）﹣（1﹣）＝a（﹣1）2﹣2（﹣1）+1﹣（1﹣）＝a≥1，∴a≥1；④当≥t+1时，f（x）在[t﹣1，t+1]递减，可得M（t）＝f（t﹣1），m（t）＝f（t+1），则h（t）＝M（t）﹣m（t）＝﹣4at+4，此时h（t）≥﹣4a（﹣1）+4＝4a≥1，∴a≥，综上，a的取值范围是[1，+∞）．。

2020年广东省广州市第六中学高一数学理测试题含解析

2020年广东省广州市第六中学高一数学理测试题含解析一、选择题：本大题共10小题，每小题5分，共50分。

在每小题给出的四个选项中，只有是一个符合题目要求的1. 2012年学期末，某学校对100间学生公寓进行综合评比，依考核分数分为A，B，C，D四种等级，其中分数在[60，70）为D等级，有15间；分数在[70，80）为C等级，有40间；分数在[80，90）为B等级，有20间；分数在[90，100）为D等级，有25间．考核评估后，得其频率直方图如图所示，估计这100间学生公寓评估得分的中位数是（）B略2. 已知集合M={1,2,3}，N={1,2,3,4}，定义函数.若点A(1,(1))、B(2,)、C(3,)，ΔABC的外接圆圆心为D，且,则满足条件的函数有（）A． 6个 B． 10个 C． 12个 D． 16个参考答案：C3. 如图，某圆拱桥的水面跨度16m，拱高4m．现有一船宽10m，则该船水面以上的高度不得超过（）A．+6 B．C．－6 D．+6参考答案：C【考点】直线与圆的位置关系．【分析】可得R2=（R﹣4）2+82，解得R=10，由如图得DM=EH=5，OH=OD+DH=6+DH 由OE2=EH2+OH2，得102=52+（6+DH）2，解得DH=5，即可得该船水面以上的高度不得超过5m【解答】解：如图，设圆拱所在圆的圆心为O，依题意得AD=8，OA=R，OD=R﹣4，由OA2=OD2+AD2，即R2=（R﹣4）2+82，解得R=10，如图DM=EH=5，OH=OD+DH=6+DH，由OE2=EH2+OH2，得102=52+（6+DH）2，解得DH=5，∴该船水面以上的高度不得超过5m，故选：C ．4. 在同一坐标系中，函数与的图像之间的关系是（）A ．关于轴对称B ．关于原点对称C ．关于轴对称 D．关于直线对称参考答案：D略5. 已知，则cos2α=（）A．B．C．D．参考答案：B【考点】GT：二倍角的余弦．【分析】直接应用二倍角的余弦公式cos2α=2cos2α﹣1代入求得结果．【解答】解：cos2α=2cos2α﹣1=﹣故选B6. 已知3a=2，那么log38﹣2log36用a表示是( )A．a﹣2 B．5a﹣2 C．3a﹣（1+a）2 D．3a﹣a2参考答案：A【考点】二次函数的性质．【专题】函数的性质及应用．【分析】先表示出a=，结合对数的运算性质，从而得到答案．【解答】解：∵3a=2，∴a=，∴﹣2=3﹣2（+1）=3a﹣2（a+1）=a﹣2，故选：A．【点评】本题考查了对数函数的性质，考查了导数的运算，是一道基础题．7. 函数y=sin（2x+）图象的对称轴方程可能是（）A．x=﹣B．x=﹣C．x=D．x=参考答案：D【考点】函数y=Asin（ωx+φ）的图象变换．【分析】令2x+=求出x的值，然后根据k的不同取值对选项进行验证即可．【解答】解：令2x+=，∴x=（k∈Z）当k=0时为D选项，故选D．8. 在△ABC中，a=3，b=5，sin A=，则sin B=( )A. B. C. D. 1参考答案：B 试题分析：由正弦定理得，故选B．9. 若，且，则下列不等式一定成立的是（）A. B.C. D.参考答案：B【分析】根据不等式性质确定选项.【详解】当时，不成立；因为，所以；当时，不成立；当时，不成立；所以选B.【点睛】本题考查不等式性质，考查基本分析判断能力，属基础题.10. 在△ABC中，内角A，B，C所对的边分别是a，b，c，已知，，则cos A= ( )A．B．C．D．参考答案：A在△ABC中,∵b?c=a,2sinB=3sinC,利用正弦定理可得2b=3c,求得a=2c,b=c.再由余弦定理可得.本题选择A选项.二、填空题:本大题共7小题,每小题4分,共28分11. 不等式的解集是参考答案：12. 将正方形ABCD 沿对角线BD 折成直二面角A －BD－C，有如下四个结论：（1）AC⊥BD；（2）△ACD是等边三角形（3）AB与平面BCD所成的角为60°；（4）AB与CD所成的角为60°。

广东省广州六中高一数学下学期期末学业水平测试试题新人教A版

广州市第六中学第二学期期末学业水平测试高一数学期末考试试卷一、选择题(本大题共10小题，每小题5分，满分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．函数y ＝sin2x ＋cos 2x 是( )A ．周期为π的偶函数B ．周期为π的奇函数C ．周期为2π的增函数D ．周期为2π的减函数2．已知向量a ＝(1,3)，b ＝(3，x)，若a ⊥b ，则实数x 的值为( )A ．9 B. －9 C ．1 D ．－13．已知{an}是等差数列，前n 项和为Sn ，a1＝120，公差d ＝－4，若Sn≤an(n≥2)，则n 的最小值为( )A ．60B ．62C ．70D ．724．设|a|＝5，|b|＝4，a·b＝－10，则a 与b 的夹角为( )A ．30° B．60° C．120° D．150°5．若实数x ，y 满足⎩⎪⎨⎪⎧ x －y ＋1≤0，x>0，则y x 的取值范围是( ) A ．(0,1) B ．(0,1] C ．(1，＋∞) D．[1，＋∞) 6．已知角θ的终边与单位圆交于点P －55，255，则cos(π－θ)的值为( )A ．－255B ．－55 C.55 D.2557．已知数列{an}是等比数列，且an>0，公比q≠1，则a1＋a8与a4＋a5的大小关系是( )A ．a1＋a8>a4＋a5B ．a1＋a8≥a4＋a5C ．a1＋a8<a4＋a5D ．a1＋a8≤a4＋a58．定义在R 上的函数f(x)既是偶函数又是周期函数，若f(x)的最小正周期为π，且当x∈[0，π2]时，f(x)＝sin x ，则f 5π3的值为( ) A ．－12 B. －32 C.32 D.129．若α，β均为锐角，sin α＝255，sin(α＋β)＝35，则cosβ等于( ) A.255 B.2525 C.255或2525 D ．－252510．下列结论中正确的是( )A ．若ac>bc ，则a>bB ．若a8>b8，则a>bC ．若a>b ，c<0，则ac<bcD ．若a<b ，则a>b二、填空题(本大题共5小题，每小题4分，满分20分)11．已知α∈π2，π，且sin α＝35，则tan α的值为____________． 12．若三点A(2,2)，B(a,0)，C(0,4)共线，则a 的值等于_________．13．不等式(x ＋1)2(x －1)<0的解集为__________．14．在△ABC 中，角A ，B ，C 所对的边分别为a ，b ，c ，若c ＝2，b ＝6，B ＝120°，则a ＝________.15．设x ，y 满足约束条件⎩⎪⎨⎪⎧ 3x －y －6≤0x －y ＋2≥0x≥0，y≥0，若目标函数z ＝ax ＋by(a>0，b>0)的最大值为12，则2a ＋3b的最小值为________．三、解答题：本大题共6小题，满分80分．解答须写出文字说明、证明过程和演算步骤．16．(本题满分12分)已知函数f(x)＝Asin(ωx＋φ)A>0，ω>0,0<φ<π2的周期为π，其图象上一个最高点为M π6，2. (1)求f(x)的解析式；(2)当x ∈[0，π4]时，求f(x)的最值及相应x 的值． 17．(本题满分14分)集合D ＝{平面向量}，定义在D 上的映射f ，满足对任意x ∈D ，均有f(x)＝λx(λ∈R 且λ≠0)．(1)若|a|＝|b|，且a 与b 不共线，试证明：[f(a)－f(b)]⊥(a ＋b)；(2)若A(1,2)，B(3,6)，C(4,8)，且f BC→＝AB →，求f(AC →)·AB →.18．(本题满分14分)已知向量a ＝－12，32，OA →＝a －b ，OB →＝a ＋b ，△AOB 是以O 为直角顶点的等腰直角三角形．(1)求向量b ；(2)求△AOB 的面积．19．(本题满分14分)某工厂用两种不同的原料均可生产同一产品，若釆用甲种原料，每吨成本 1 000元，运费500元，可生产产品90千克；若釆用乙种原料，每吨成本1 500元，运费400元，可生产100千克．若每日预算总成本不得超过6 000元，运费不得超过2 000元，问此工厂每日最多可生产多少千克产品？20．(本题满分14分)已知数列{an}的前n 项和为Sn ，设an 是Sn 与2的等差中项，数列{bn}中，b1＝1，bn ＋1＝bn ＋2.(1)求an ，bn ；(2)若数列{bn}的前n 项和为Bn ，比较1B1＋1B2＋…＋1Bn与2的大小；(3)令Tn ＝b1a1＋b2a2＋…＋bn an，是否存在正整数M ，使得Tn<M 对一切正整数n 都成立？若存在，求出M 的最小值；若不存在，请说明理由．广州市第六中学第二学期期末学业水平测试高一数学期末考试参考答案一、选择题(本大题共10小题，每小题5分，满分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)ADBCCCACBC二、填空题(本大题共5小题，每小题4分，满分20分)11．【答案】－34 12．【答案】4 13．【答案】{x|x<1且x≠－1} 14．【答案】 2 15．【答案】256 三、解答题：本大题共6小题，满分80分．解答须写出文字说明、证明过程和演算步骤．16．(本题满分12分)(1)∵周期T ＝π，∴2πω＝π，即ω＝2.又f(x)图象的最高点为M π6，2，∴A ＝2，∴f(x)＝2sin(2x ＋φ)．将点Mπ6，2代入，得sin π3＋φ＝1，∵0<φ<π2，∴φ＝π6，∴f(x)＝2sin 2x ＋π6. (2)∵x ∈[0，π4]，∴2x ＋π6∈[π6，2π3].∴当2x ＋π6＝π6，即x ＝0时，ymin ＝1；当2x ＋π6＝π2，即x ＝π6时，ymax ＝2. 17．(本题满分14分)(1)证明：由题意有[f(a)－f(b)]·(a＋b)＝(λa－λb)(a＋b)＝λ(a2－b2)＝0.∵f(a)－f(b)≠0，a ＋b≠0，∴[f(a)－f(b)]⊥(a ＋b)．(2)AB →＝(2,4)，BC →＝(1,2)，∴f(BC →)＝λ(1,2)＝(2,4)，∴λ＝2.又AC →＝(3,6)，∴f(AC →)·AB →＝2(3,6)·(2,4)＝60.18．(本题满分14分)(1)∵OA ＝OB ，∴a2＝b2，即|a|＝|b|＝1，∴|AB →|＝|OB →－OA →|＝|2b|＝2，∴|a －b|＝|a ＋b|＝2，∴a ⊥b.设b ＝(x ，y)，则⎩⎪⎨⎪⎧ x2＋y2＝1，－12x ＋32y ＝0，解得⎩⎪⎨⎪⎧ x ＝32，y ＝12或⎩⎪⎨⎪⎧ x ＝－32，y ＝－12.∴b ＝32，12或b ＝－32，－12. (2) S△AOB ＝12×(2)2＝1. 19．(本题满分14分)设工厂每日需用甲原料x 吨，乙原料y 吨，可生产产品z 千克．则⎩⎪⎨⎪⎧ 1 000x ＋1 500y≤6 000，500x ＋400y≤2 000，x≥0，y≥0，即⎩⎪⎨⎪⎧ 2x ＋3y≤12，5x ＋4y≤20，x≥0，y≥0，画出可行域，如图所示．目标函数z ＝90x ＋100y(千克)．当直线z ＝90x ＋100y 过直线2x ＋3y ＝12和5x ＋4y ＝20交点A ⎝ ⎛⎭⎪⎫127，207时，z 取得最大值，即zmax ＝90·127＋100·207＝440(千克)．工厂每日最多生产440千克产品．20．(本题满分14分)(1)由题意2an ＝Sn ＋2，∴Sn ＝2an －2，Sn ＋1＝2an ＋1－2，∴an ＋1＝Sn ＋1－Sn ＝2an ＋1－2an ，即an ＋1＝2an ，又2a1＝S1＋2＝a1＋2，∴a1＝2，∴an ＝2n.∵b1＝1，bn ＋1＝bn ＋2，∴bn ＝2n －1.(2)Bn ＝1＋3＋5＋…＋(2n －1)＝n2.1B1＋1B2＋…＋1Bn ＝112＋122＋…＋1n2<1＋11×2＋12×3＋…＋1n －1·n ＝1＋⎝ ⎛⎭⎪⎫1－12＋⎝ ⎛⎭⎪⎫12－13＋…＋⎝ ⎛⎭⎪⎫1n －1－1n ＝2－1n <2. (3)Tn ＝12＋322＋…＋2n －12n ，∴12Tn ＝122＋323＋…＋2n －12n ＋1，两式相减，得12Tn ＝12＋2⎝ ⎛⎭⎪⎫122＋123＋…＋12n －2n －12n ＋1＝12＋2×14⎝ ⎛⎭⎪⎫1－12n －11－12－2n －12n ＋1,∴Tn ＝3－12n －2－2n －12n <3.又T1＝12，Tn 单调递增，∴Tn ∈⎣⎢⎡⎭⎪⎫12，3.∴M 的最小值为3.关于数学名言警句大全1、数学家本质上是个着迷者，不迷就没有数学。

广东省广州六中-学年高一数学下学期期末学业水平测试试题新人教A版

广州市第六中学2012—2013学年第二学期期末学业水平测试高一数学期末考试试卷一、选择题(本大题共10小题，每小题5分，满分50分．在每小题给出嘚四个选项中，只有一项是符合题目要求嘚)1．函数y ＝sin2x ＋cos 2x 是( )A ．周期为π嘚偶函数B ．周期为π嘚奇函数C ．周期为2π嘚增函数D ．周期为2π嘚减函数2．已知向量a ＝(1,3)，b ＝(3，x)，若a ⊥b ，则实数x 嘚值为( ) A ．9 B. －9 C ．1 D ．－13．已知{an}是等差数列，前n 项和为Sn ，a1＝120，公差d ＝－4，若Sn≤an(n≥2)，则n 嘚最小值为( )A ．60B ．62C ．70D ．724．设|a|＝5，|b|＝4，a·b＝－10，则a 与b 嘚夹角为( ) A ．30° B．60° C．120° D．150°5．若实数x ，y 满足⎩⎪⎨⎪⎧x －y ＋1≤0，x>0，则y x 嘚取值范围是( )A ．(0,1)B ．(0,1]C ．(1，＋∞) D．[1，＋∞)6．已知角θ嘚终边与单位圆交于点P －55，255，则cos(π－θ)嘚值为( )A ．－255B ．－55C.55D.2557．已知数列{an}是等比数列，且an>0，公比q≠1，则a1＋a8与a4＋a5嘚大小关系是( )A ．a1＋a8>a4＋a5B ．a1＋a8≥a4＋a5C ．a1＋a8<a4＋a5D ．a1＋a8≤a4＋a58．定义在R 上嘚函数f(x)既是偶函数又是周期函数，若f(x)嘚最小正周期为π，且当x ∈[0，π2]时，f(x)＝sin x ，则f 5π3嘚值为( )A ．－12 B. －32 C.32 D.129．若α，β均为锐角，sin α＝255，sin(α＋β)＝35，则cosβ等于( ) A.255 B.2525 C.255或2525 D ．－252510．下列结论中正确嘚是( )A ．若ac>bc ，则a>bB ．若a8>b8，则a>bC ．若a>b ，c<0，则ac<bcD ．若a<b ，则a>b二、填空题(本大题共5小题，每小题4分，满分20分)11．已知α∈π2，π，且sin α＝35，则tan α嘚值为____________．12．若三点A(2,2)，B(a,0)，C(0,4)共线，则a 嘚值等于_________．13．不等式(x ＋1)2(x －1)<0嘚解集为__________．14．在△ABC 中，角A ，B ，C 所对嘚边分别为a ，b ，c ，若c ＝2，b ＝6，B ＝120°，则a ＝________.15．设x ，y 满足约束条件⎩⎪⎨⎪⎧3x －y －6≤0x －y ＋2≥0x≥0，y≥0，若目标函数z ＝ax ＋by(a>0，b>0)嘚最大值为12，则2a ＋3b嘚最小值为________．三、解答题：本大题共6小题，满分80分．解答须写出文字说明、证明过程和演算步骤． 16．(本题满分12分)已知函数f(x)＝Asin(ωx＋φ)A>0，ω>0,0<φ<π2嘚周期为π，其图象上一个最高点为M π6，2.(1)求f(x)嘚解析式；(2)当x ∈[0，π4]时，求f(x)嘚最值及相应x 嘚值．17．(本题满分14分)集合D ＝{平面向量}，定义在D 上嘚映射f ，满足对任意x ∈D ，均有f(x)＝λx(λ∈R 且λ≠0)． (1)若|a|＝|b|，且a 与b 不共线，试证明：[f(a)－f(b)]⊥(a ＋b)； (2)若A(1,2)，B(3,6)，C(4,8)，且f BC →＝AB →，求f(AC →)·AB →.18．(本题满分14分)已知向量a ＝－12，32，OA →＝a －b ，OB →＝a ＋b ，△AOB 是以O 为直角顶点嘚等腰直角三角形．(1)求向量b ；(2)求△AOB 嘚面积．19．(本题满分14分)某工厂用两种不同嘚原料均可生产同一产品，若釆用甲种原料，每吨成本1 000元，运费500元，可生产产品90千克；若釆用乙种原料，每吨成本1 500元，运费400元，可生产100千克．若每日预算总成本不得超过6 000元，运费不得超过2 000元，问此工厂每日最多可生产多少千克产品？20．(本题满分14分)已知数列{an}嘚前n 项和为Sn ，设an 是Sn 与2嘚等差中项，数列{bn}中，b1＝1，bn ＋1＝bn ＋2.(1)求an ，bn ；(2)若数列{bn}嘚前n 项和为Bn ，比较1B1＋1B2＋…＋1Bn与2嘚大小；(3)令Tn ＝b1a1＋b2a2＋…＋bnan ，是否存在正整数M ，使得Tn<M 对一切正整数n 都成立？若存在，求出M 嘚最小值；若不存在，请说明理由．广州市第六中学2012—2013学年第二学期期末学业水平测试高一数学期末考试参考答案一、选择题(本大题共10小题，每小题5分，满分50分．在每小题给出嘚四个选项中，只有一项是符合题目要求嘚) ADBCCCACBC二、填空题(本大题共5小题，每小题4分，满分20分)11．【答案】－3412．【答案】4 13．【答案】{x|x<1且x≠－1}14．【答案】2 15．【答案】256三、解答题：本大题共6小题，满分80分．解答须写出文字说明、证明过程和演算步骤． 16．(本题满分12分)(1)∵周期T ＝π，∴2πω＝π，即ω＝2.又f(x)图象嘚最高点为M π6，2，∴A ＝2，∴f(x)＝2sin(2x ＋φ)．将点M π6，2代入，得sinπ3＋φ＝1，∵0<φ<π2，∴φ＝π6，∴f(x)＝2sin 2x ＋π6. (2)∵x ∈[0，π4]，∴2x ＋π6∈[π6，2π3].∴当2x ＋π6＝π6，即x ＝0时，ymin ＝1；当2x ＋π6＝π2，即x ＝π6时，ymax ＝2.17．(本题满分14分)(1)证明：由题意有[f(a)－f(b)]·(a＋b)＝(λa－λb)(a＋b)＝λ(a2－b2)＝0.∵f(a)－f(b)≠0，a ＋b≠0，∴[f(a)－f(b)]⊥(a ＋b)．(2)AB →＝(2,4)，BC →＝(1,2)，∴f(BC →)＝λ(1,2)＝(2,4)，∴λ＝2.又AC →＝(3,6)，∴f(AC →)·AB →＝2(3,6)·(2,4)＝60.18．(本题满分14分)(1)∵OA ＝OB ，∴a2＝b2，即|a|＝|b|＝1，∴|AB →|＝|OB →－OA →|＝|2b|＝2，∴|a －b|＝|a ＋b|＝2，∴a ⊥b.设b ＝(x ，y)，则⎩⎪⎨⎪⎧x2＋y2＝1，－12x ＋32y ＝0，解得⎩⎪⎨⎪⎧x ＝32，y ＝12或⎩⎪⎨⎪⎧x ＝－32，y ＝－12.∴b ＝32，12或b ＝－32，－12. (2) S △AOB ＝12×(2)2＝1.19．(本题满分14分)设工厂每日需用甲原料x 吨，乙原料y 吨，可生产产品z 千克．则⎩⎪⎨⎪⎧1 000x ＋1 500y≤6 000，500x ＋400y≤2 000，x≥0，y≥0，即⎩⎪⎨⎪⎧2x ＋3y≤12，5x ＋4y≤20，x≥0，y≥0，画出可行域，如图所示．目标函数z ＝90x ＋100y(千克)．当直线z ＝90x ＋100y 过直线2x ＋3y ＝12和5x ＋4y ＝20交点A ⎝ ⎛⎭⎪⎫127，207时，z 取得最大值，即zmax ＝90·127＋100·207＝440(千克)．工厂每日最多生产440千克产品．20．(本题满分14分)(1)由题意2an ＝Sn ＋2，∴Sn ＝2an －2，Sn ＋1＝2an ＋1－2，∴an ＋1＝Sn ＋1－Sn ＝2an ＋1－2an ，即an ＋1＝2an ，又2a1＝S1＋2＝a1＋2，∴a1＝2，∴an ＝2n.∵b1＝1，bn ＋1＝bn ＋2，∴bn ＝2n －1.(2)Bn ＝1＋3＋5＋…＋(2n －1)＝n2.1B1＋1B2＋…＋1Bn ＝112＋122＋…＋1n2<1＋11×2＋12×3＋…＋1n －1·n ＝1＋⎝ ⎛⎭⎪⎫1－12＋⎝ ⎛⎭⎪⎫12－13＋…＋⎝⎛⎭⎪⎫1n －1－1n ＝2－1n <2. (3)Tn ＝12＋322＋…＋2n －12n ，∴12Tn ＝122＋323＋…＋2n －12n ＋1，两式相减，得12Tn ＝12＋2⎝ ⎛⎭⎪⎫122＋123＋…＋12n －2n －12n ＋1＝12＋2×14⎝⎛⎭⎪⎫1－12n －11－12－2n －12n ＋1,∴Tn ＝3－12n －2－2n －12n <3.又T1＝12，Tn 单调递增，∴Tn ∈⎣⎢⎡⎭⎪⎫12，3.∴M 嘚最小值为3.。

广州市第六中学数学高一下期末测试卷(含答案解析)

一、选择题1．(0分)[ID ：12703]已知ABC ∆是边长为4的等边三角形，P 为平面ABC 内一点，则•()PA PB PC +的最小值是（）A ．6-B ．3-C ．4-D ．2-2．(0分)[ID ：12696]已知函数y=f （x ）定义域是[-2，3]，则y=f （2x-1）的定义域是（） A ．50,2⎡⎤⎢⎥⎣⎦B ．[]1,4-C ．1,22⎡⎤-⎢⎥⎣⎦D ．[]5,5-3．(0分)[ID ：12693](2015新课标全国I 理科)《九章算术》是我国古代内容极为丰富的数学名著，书中有如下问题:“今有委米依垣内角，下周八尺，高五尺.问:积及为米几何?”其意思为:“在屋内墙角处堆放米(如图，米堆为一个圆锥的四分之一)，米堆底部的弧长为8尺，米堆的高为5尺，问米堆的体积和堆放的米各为多少?”已知1斛米的体积约为1.62立方尺，圆周率约为3，估算出堆放的米约有A ．14斛B ．22斛C ．36斛D ．66斛4．(0分)[ID ：12691]已知不等式220ax bx ++>的解集为{}12x x -<<，则不等式220x bx a ++<的解集为（）A ．112x x ⎧⎫-<<⎨⎬⎩⎭B ．112x x x ⎧⎫<->⎨⎬⎩⎭或 C ．{}21x x -<<D ．{}21x x x <->或5．(0分)[ID ：12687]C ∆AB 是边长为2的等边三角形，已知向量a ，b 满足2a AB =，C 2a b A =+，则下列结论正确的是（）A ．1b =B ．a b ⊥C ．1a b ⋅=D ．()4C a b +⊥B6．(0分)[ID ：12685]已知()f x 是定义域为(,)-∞+∞的奇函数，满足(1)(1)f x =f +x -，若(1)2f =，则(1)(2)f +f (3)(2020)f f +++=（）A ．50B ．2C ．0D ．50-7．(0分)[ID ：12664]已知0,0a b >>，并且111,,2a b成等差数列，则4a b +的最小值为（）A ．2B ．4C ．5D ．98．(0分)[ID ：12660]函数()lg ||f x x x =的图象可能是（）A ．B ．C ．D ．9．(0分)[ID ：12659]定义在R 上的奇函数()f x 满足()()2f x f x +=-，且当[]0,1x ∈时，()2cos xf x x =-，则下列结论正确的是（）A ．()20202019201832f f f ⎛⎫⎛⎫<< ⎪ ⎪⎝⎭⎝⎭B ．()20202019201832f f f ⎛⎫⎛⎫<< ⎪ ⎪⎝⎭⎝⎭C ．()20192020201823f f f ⎛⎫⎛⎫<<⎪ ⎪⎝⎭⎝⎭D ．()20192020201823f f f ⎛⎫⎛⎫<<⎪ ⎪⎝⎭⎝⎭10．(0分)[ID ：12650]下列四个正方体图形中，A ，B 为正方体的两个顶点，M ，N ，P 分别为其所在棱的中点，能得出//AB 平面MNP 的图形的序号是（）A ．①③B ．②③C ．①④D ．②④11．(0分)[ID ：12649]若tan()24πα+=，则sin cos sin cos αααα-=+（）A ．12B ．2C ．2-D ．12-12．(0分)[ID ：12646]已知圆()()22:341C x y -+-=和两点(),0A m -，()(),00B m m >，若圆C 上存在点P ，使得90APB ∠=︒，则m 的最大值为（）A ．7B ．6C ．5D ．413．(0分)[ID ：12643]已知0.6log 0.5a =，ln0.5b =，0.50.6c =，则（） A ．a c b >>B ．a b c >>C ．c a b >>D ．c b a >>14．(0分)[ID ：12681]若,αβ均为锐角，25sin α=()3sin 5αβ+=，则cos β=A ．255B ．2525C ．255或2525D ．2525-15．(0分)[ID ：12657]函数()(1)lg(1)35f x x x x =-+--的零点个数为（） A ．3B ．2C ．1D ．0二、填空题16．(0分)[ID ：12809]某工厂生产甲、乙、丙、丁四种不同型号的产品，产量分别为200,400,300,100件，为检验产品的质量，现用分层抽样的方法从以上所有的产品中抽取60件进行检验，则应从丙种型号的产品中抽取________ 件.17．(0分)[ID ：12808]一个空间几何体的三视图及部分数据如图所示，则这个几何体的体积是___________18．(0分)[ID ：12806]设a ，b ，c 分别为ABC ∆内角A ，B ，C 的对边.已知233cos cos a b cB C=，则222a c b ac +-的取值范围为______. 19．(0分)[ID ：12804]已知ABC ，135B ∠=，22,4AB BC ==，求AB AC ⋅=______．20．(0分)[ID ：12776]若x ，y 满足约束条件10,{30,30,x y x y x -+≥+-≥-≤则z=x−2y 的最小值为__________.21．(0分)[ID ：12772]()()()()()1tan11tan 21tan31tan 441tan 45︒︒︒︒︒+++++=__________．22．(0分)[ID ：12771]已知l ，m 是平面α外的两条不同直线．给出下列三个论断： ①l ⊥m ；②m ∥α；③l ⊥α．以其中的两个论断作为条件，余下的一个论断作为结论，写出一个正确的命题：__________．23．(0分)[ID ：12736]函数sin 3y x x =-的图像可由函数2sin y x =的图像至少向右平移________个单位长度得到．24．(0分)[ID ：12754]某三棱锥的三视图如下图所示，正视图、侧视图均为直角三角形，则该三棱锥的四个面中，面积最大的面的面积是．25．(0分)[ID ：12752]已知复数z x yi =+，且23z -=，则yx的最大值为__________．三、解答题26．(0分)[ID ：12928]某高校在2012年的自主招生考试成绩中随机抽取100名中学生的笔试成绩，按成绩分组，得到的频率分布表如表所示．组号分组频数频率第1组 [)160,165 5 0.050 第2组 [)165,170① 0.350第3组 [)170,175 30 ②第4组 [)175,180 20 0.200 第5组[)180,185100.100（1）请先求出频率分布表中,①②位置的相应数据，再完成频率分布直方图；（2）为了能选拔出最优秀的学生，高校决定在笔试成绩高的第3,4,5组中用分层抽样抽取6名学生进入第二轮面试，求第3、4、5组每组各抽取多少名学生进入第二轮面试；（3）在（2）的前提下，学校决定在6名学生中随机抽取2名学生接受A 考官进行面试，求：第4组至少有一名学生被考官A 面试的概率． 27．(0分)[ID ：12913]已知函数()f x =│x +1│–│x –2│. （1）求不等式()f x ≥1的解集；（2）若不等式()f x ≥x 2–x +m 的解集非空，求实数m 的取值范围.28．(0分)[ID ：12880]已知二次函数()f x 满足()(1)2f x f x x -+=-且(0)1f =. （1）求()f x 的解析式；（2）当[1,1]x ∈-时，不等式()2x m f x >+恒成立，求实数m 的取值范围.29．(0分)[ID ：12857]如图所示，为美化环境，拟在四边形ABCD 空地上修建两条道路EA 和ED ，将四边形分成三个区域，种植不同品种的花草，其中点E 在边BC 的三等分点处（靠近B 点），3BC =百米，BC CD ⊥，120ABC ∠=，21EA =百米，60AED ∠=.（1）求ABE △区域的面积；（2）为便于花草种植，现拟过C 点铺设一条水管CH 至道路ED 上，求水管CH 最短时的长．30．(0分)[ID ：12843]设函数2()cos 2sin 3f x x x π⎛⎫=++ ⎪⎝⎭．（1）求函数()f x 的最小正周期．（2）求函数()f x 的单调递减区间；（3）设,,A B C 为ABC 的三个内角，若1cos 3B =，124C f ⎛⎫=- ⎪⎝⎭，且C 为锐角，求sin A ．【参考答案】2016-2017年度第*次考试试卷参考答案**科目模拟测试一、选择题1．A2．C3．B4．A5．D6．C7．D8．D9．C10．C11．D12．B13．A14．B15．B二、填空题16．18【解析】应从丙种型号的产品中抽取件故答案为18点睛:在分层抽样的过程中为了保证每个个体被抽到的可能性是相同的这就要求各层所抽取的个体数与该层所包含的个体数之比等于样本容量与总体的个体数之比即ni17．【解析】【分析】先还原几何体再根据柱体体积公式求解【详解】空间几何体为一个棱柱如图底面为边长为的直角三角形高为的棱柱所以体积为【点睛】本题考查三视图以及柱体体积公式考查基本分析求解能力属基础题18．【解析】【分析】把已知式用正弦定理化边为角由两角和的正弦公式和诱导公式化简可求得即角从而得角的范围注意由余弦定理可得结论【详解】因为所以所以即又所以则因为所以而故故答案为：【点睛】本题考查正弦与余弦19．16【解析】【分析】由正余弦定理可得由平面向量的数量积公式有：得解【详解】由余弦定理可得：所以由正弦定理得：所以所以即故答案为16【点睛】本题考查了余弦定理正弦定理及向量的数量积属简单题20．【解析】【分析】【详解】试题分析：由得记为点；由得记为点；由得记为点分别将ABC的坐标代入得所以的最小值为【考点】简单的线性规划【名师点睛】利用线性规划求最值一般用图解法求解其步骤是：（1）在平面直21．【解析】【分析】根据式子中角度的规律可知变形有由此可以求解【详解】根据式子中角度的规律可知变形有所以故答案为：【点睛】本题主要考查两角和的正切公式的应用以及归纳推理的应用属于中档题22．如果l⊥αm∥α则l⊥m或如果l⊥αl⊥m则m∥α【解析】【分析】将所给论断分别作为条件结论加以分析【详解】将所给论断分别作为条件结论得到如下三个命题：（1）如果l⊥αm∥α则l⊥m正确；（2）如果23．【解析】试题分析：因为所以函数的的图像可由函数的图像至少向右平移个单位长度得到【考点】三角函数图像的平移变换两角差的正弦公式【误区警示】在进行三角函数图像变换时提倡先平移后伸缩但先伸缩后平移也经常出24．【解析】试题分析：该三棱锥底面是边长为2的正三角形面积为有两个侧面是底边为2高为2的直角三角形面积为2另一个侧面是底边为2腰为的等腰三角形面积为所以面积最大的面的面积是考点：三视图25．【解析】【分析】根据复数z的几何意义以及的几何意义由图象得出最大值【详解】复数且复数z的几何意义是复平面内以点为圆心为半径的圆的几何意义是圆上的点与坐标原点连线的斜率由图可知：即的最大值为故答案为：三、解答题26．27．28．29．30．2016-2017年度第*次考试试卷参考解析【参考解析】**科目模拟测试一、选择题 1．A 解析：A 【解析】【分析】建立平面直角坐标系，表示出点的坐标，利用向量坐标运算和平面向量的数量积的运算，求得最小值，即可求解. 【详解】由题意，以BC 中点为坐标原点，建立如图所示的坐标系，则(0,23),(2,0),(2,0)A B C -，设(,)P x y ，则(,23),(2,),(2,)PA x y PB x y PC x y =--=---=--，所以22()(2)(23)(2)2432PA PB PC x x y y x y y •+=-⋅-+-⋅-=-+222[(3)3]x y =+--，所以当0,3x y ==时，()PA PB PC •+取得最小值为2(3)6⨯-=-，故选A.【点睛】本题主要考查了平面向量数量积的应用问题，根据条件建立坐标系，利用坐标法是解答的关键，着重考查了推理与运算能力，属于基础题.2．C解析：C 【解析】∵函数y =f (x )定义域是[−2,3]， ∴由−2⩽2x −1⩽3，解得−12⩽x ⩽2，即函数的定义域为1,22⎡⎤-⎢⎥⎣⎦，本题选择C 选项.3．B解析：B 【解析】试题分析：设圆锥底面半径为r ，则12384r ⨯⨯=，所以163r =，所以米堆的体积为211163()5433⨯⨯⨯⨯=3209，故堆放的米约为3209÷1.62≈22，故选B. 考点：圆锥的性质与圆锥的体积公式4．A解析：A 【解析】【分析】根据一元二次不等式的解集与一元二次方程根的关系，结合韦达定理可构造方程求得,a b ；利用一元二次不等式的解法可求得结果.【详解】220ax bx ++>的解集为{}12x x -<<1∴-和2是方程220ax bx ++=的两根，且0a <1212122ba a⎧-=-+=⎪⎪∴⎨⎪=-⨯=-⎪⎩，解得：11a b =-⎧⎨=⎩ 222210x bx a x x ∴++=+-< 解得：112x -<<，即不等式220x bx a ++<的解集为112x x ⎧⎫-<<⎨⎬⎩⎭故选：A 【点睛】本题考查一元二次不等式的解法、一元二次不等式的解集与一元二次方程根的关系等知识的应用；关键是能够通过一元二次不等式的解集确定一元二次方程的根，进而利用韦达定理构造方程求得变量.5．D解析：D 【解析】试题分析：2,2AB a AC a b ==+,AC AB b ∴=+,b AC AB BC ∴=-=．由题意知12,cos1201212b a b a b ⎛⎫=⋅=⋅=⨯⨯-=- ⎪⎝⎭． ()()2422a b BC AB BC BC AB BC BC∴+⋅=+⋅=⋅+212cos1202222402AB BC ⎛⎫=⋅+=⨯⨯⨯-+= ⎪⎝⎭．()4a b BC ∴+⊥．故D 正确．考点：1向量的加减法；2向量的数量积；3向量垂直．6．C解析：C 【解析】【分析】利用()f x 是定义域为(,)-∞+∞的奇函数可得：()()f x f x -=-且()00f =，结合(1)(1)f x =f +x -可得：函数()f x 的周期为4；再利用赋值法可求得：()20f =，()32f =-，()40f =，问题得解.【详解】因为()f x 是定义域为(,)-∞+∞的奇函数，所以()()f x f x -=-且()00f = 又(1)(1)f x =f +x -所以()()()()()21111f x f x f x f x f x ⎡⎤⎡⎤+=++=-+=-=-⎣⎦⎣⎦ 所以()()()()()4222f x f x f x f x f x ⎡⎤⎡⎤+=++=-+=--=⎣⎦⎣⎦ 所以函数()f x 的周期为4，在(1)(1)f x =f +x -中，令1x =，可得：()()200f f ==在(1)(1)f x =f +x -中，令2x =，可得：()()()3112f f f =-=-=- 在(1)(1)f x =f +x -中，令3x =，可得：()()()4220f f f =-=-= 所以(1)(2)f +f ()()()()2020(3)(2020)12344f f f f f f ⎡⎤+++=⨯+++⎣⎦ 50500=⨯=故选C 【点睛】本题主要考查了奇函数的性质及函数的周期性应用，还考查了赋值法及计算能力、分析能力，属于中档题．7．D解析：D 【解析】∵111,,2a b成等差数列， ()111141445529a b a a b a b a b a b b a b ⎛⎫∴+=∴+=++=+++⋅= ⎪⎝⎭，，当且仅当a =2b 即33,2a b ==时“=“成立，本题选择D 选项.点睛：在应用基本不等式求最值时，要把握不等式成立的三个条件，就是“一正——各项均为正；二定——积或和为定值；三相等——等号能否取得”，若忽略了某个条件，就会出现错误．8．D解析：D 【解析】【分析】分析函数()y f x =的定义域、奇偶性及其在()0,1上的函数值符号，可得出结论. 【详解】函数()lg f x x x =的定义域为{}0x x ≠，定义域关于原点对称，()()lg lg f x x x x x f x -=--=-=-，函数()y f x =为奇函数，排除A 、C 选项；当01x <<时，lg 0x <，此时()lg 0f x x x =<，排除B 选项. 故选：D. 【点睛】本题考查由函数的解析式选择函数图象，一般分析函数的定义域、奇偶性、单调性、零点以及函数值符号，考查推理能力，属于中等题.9．C解析：C 【解析】【分析】根据f （x ）是奇函数，以及f （x+2）=f （-x ）即可得出f （x+4）=f （x ），即得出f （x ）的周期为4，从而可得出f （2018）=f （0），2019122f f ⎛⎫⎛⎫=⎪ ⎪⎝⎭⎝⎭，20207312f f ⎛⎫⎛⎫= ⎪ ⎪⎝⎭⎝⎭然后可根据f （x ）在[0，1]上的解析式可判断f （x ）在[0，1]上单调递增，从而可得出结果. 【详解】∵f（x ）是奇函数；∴f（x+2）=f （-x ）=-f （x ）；∴f（x+4）=-f （x+2）=f （x ）； ∴f（x ）的周期为4；∴f（2018）=f （2+4×504）=f （2）=f （0），2019122f f ⎛⎫⎛⎫= ⎪ ⎪⎝⎭⎝⎭,20207 312f f ⎛⎫⎛⎫= ⎪ ⎪⎝⎭⎝⎭∵x∈[0，1]时，f （x ）=2x -cosx 单调递增；∴f(0)＜12f ⎛⎫ ⎪⎝⎭ ＜712f ⎛⎫ ⎪⎝⎭ ∴()20192020201823f f f ⎛⎫⎛⎫<< ⎪ ⎪⎝⎭⎝⎭,故选C. 【点睛】本题考查奇函数，周期函数的定义，指数函数和余弦函数的单调性，以及增函数的定义，属于中档题.10．C解析：C 【解析】【分析】用面面平行的性质判断①的正确性.利用线面相交来判断②③的正确性，利用线线平行来判断④的正确性. 【详解】对于①，连接AC 如图所示，由于//,//MN AC NP BC ，根据面面平行的性质定理可知平面//MNP 平面ACB ，所以//AB 平面MNP .对于②，连接BC 交MP 于D ，由于N 是AC 的中点，D 不是BC 的中点，所以在平面ABC 内AB 与DN 相交，所以直线AB 与平面MNP 相交.对于③，连接CD ，则//AB CD ，而CD 与PN 相交，即CD 与平面PMN 相交，所以AB 与平面MNP 相交.对于④，连接CD ，则////AB CD NP ，由线面平行的判定定理可知//AB 平面MNP .综上所述，能得出//AB 平面MNP 的图形的序号是①④. 故选：C 【点睛】本小题主要考查线面平行的判定，考查空间想象能力和逻辑推理能力，属于基础题.11．D解析：D 【解析】由tan()24πα+=有tan 112,tan 1tan 3ααα+==-，所以11sin cos tan 1131sin cos tan 1213αααααα---===-+++，选D.点睛：本题主要考查两角和的正切公式以及同角三角函数的基本关系式，属于中档题。

2020年广东省广州市第六中学-学年高一下学期期中考试数学试题 Word版含解析

广州第六中学高一下学期数学期中考试一、选择题（本大题12小题，每小题5分，共60分） 1.已知集合{}210,01x A xB x x x x ⎧⎫+=<=-≤⎨⎬-⎩⎭,则A B ⋂=（）A. {}11x x -≤≤B. {}01x x <≤C. {}01x x ≤≤D.{}01x x ≤<【答案】D 【解析】【分析】先求集合A 和集合B ，然后取交集即可. 【详解】{}10|111x A xx x x ⎧⎫+=<=-<<⎨⎬-⎩⎭,{}{}20|01B x x x x x =-≤=≤≤,则{}|01A B x x ⋂=≤<, 故选：D【点睛】本题考查集合的交集运算，属于简单题.2.设,a b R ∈，若0>-b a ，则下列不等式正确的是（） A. 0b a ->B. 0>+a bC. 033<+b aD.022<-b a【答案】B 【解析】【分析】利用不等式的性质对选项逐个进行判断即可.【详解】00||a a b b ⇒>->…, A 项，||a b b >…，则b-a<0,故A 项错误； B 项，||-a b b >…，则a+b>0,故B 项正确；C 项，||a b >⇒333||a b b ≥->，则33+0a b >,故C 项错误；D 项，a >|b |⇒22a b >，即220a b ->，故D 项错误. 故选：B【点睛】本题考查不等式性质的应用，属于基础题.3.如图，设,A B 两点在河的两岸，一测量者在A 的同侧所在的河岸边选定一点C ，测出AC 的距离为50cm ，45,105ACB CAB ∠=︒∠=︒后，可以计算出,A B 两点的距离为（）252B. m 225C. m 250D. m 350【答案】C 【解析】分析：利用正弦定理求解。

2020-2019学年度广东省广州六中高三第二次月考化学试卷

2020-2019学年度广东省广州六中高三第二次月考化学试卷可能用到的相对原子质量：H1 C12 N14 O16 Na23Mg24Al27P31S32Cl35.5K39Ca40Fe56Cu63.5 Sn119一、选择题（本题包括9小题，每小题3分，共27分。

每小题只有一个．．．．选项符合题意） 1．Se 是人体必需微量元素。

下列关于Se 7834和Se 8034的说法正确的是 ( ) A ．Se 7834和Se 8034互为同素异形体 B ．Se 7834和Se 8034互为同位素 C ．Se 7834和Se 8034分别含有44个和46个质子 D ．Se 7834和Se 8034都含有34个中子2．关于新材料新工艺的研究方兴未艾。

下列有关说法正确的是（） A ．用煤制取二甲醚（CH 3OCH 3）替代石油，可以缓解我国石油的供需矛盾 B ．非晶体铜不导电，Cu 原子呈杂乱无章的非晶态排列，可与盐酸反应放出H 2 C ．纳米级CaCO 3与盐酸反应比块状CaCO 3慢 D ．Mg 与被保护金属形成原电池，Mg 作正极被腐蚀3．2019年8月28日上午，我国首条纤维乙醇生产线——河南天冠集团3000吨级纤维乙醇项目，在镇平开发区开工奠基。

这一项目打破了过去单纯以粮食类原料生产乙醇的历史，使利用秸秆类纤维质原料生产乙醇成为现实。

下列有关说法中不正确的是 A ．用这种纤维乙醇部分替代成品油，有助于缓解日益增长的成品油需求 B ．这一项目不仅使秸秆类废物得到科学利用，而且能为国家节约大量粮食 C ．乙醇能部分替代汽油作车用燃料是因为乙醇与汽油组成元素相同，化学成分相似 D ．掺入10％燃料乙醇的乙醇汽油可以使汽车尾气中一氧化碳排放量下降 4．下列实验方案能达到预定目的的是 0.45/0.30 A ．将SO 2通入石蕊溶液中证明S02有漂白性B ．用绿豆大小的钠和钾分别与冷水反应说明钠、钾金属性的强弱C ．向某未知溶液中加入BaCl 2溶液后再加入稀硝酸，检验是否含SO 42-D ．用干燥的pH 试纸可测得新制氯水的pH=25．某学生设计了一个“黑笔写红字”的趣味实验。

广东省广州市黄埔中学2019-2020学年高一数学文模拟试卷含解析

广东省广州市黄埔中学2019-2020学年高一数学文模拟试卷含解析一、选择题：本大题共10小题，每小题5分，共50分。

在每小题给出的四个选项中，只有是一个符合题目要求的1. ----------------( )A． B． C．D．参考答案：A略2. 若指数函数y=（a﹣2）x在（﹣∞，+∞）上是减函数，那么( )A．2＜a＜3 B．﹣2＜a＜1 C．a＞3 D．0＜a＜1参考答案：A【考点】指数函数的单调性与特殊点．【专题】函数的性质及应用．【分析】利用指数函数y=（a﹣2）x在（﹣∞，+∞）上是减函数，可得0＜a﹣2＜1，解出即可．【解答】解：∵指数函数y=（a﹣2）x在（﹣∞，+∞）上是减函数，∴0＜a﹣2＜1，解得2＜a＜3．故选：A．【点评】本题考查了指数函数的单调性与底数的关系，属于基础题．3. 若函数在区间上的最大值是最小值的2倍，则=（）A． B． C．D．参考答案：C4. 如果，那么角的终边所在的象限是A．第一象限B．第二象限C．第三象限D．第四象限参考答案：B略5. 设f（x）=3x+3x﹣8，现用二分法求方程3x+3x﹣8=0在区间（1，2）内的近似解的，计算得f（1）＜0，f（1.25）＜0，f（1.5）＞0，f（2）＞0，则方程的根落在的区间（）A．（1，1.25）B．（1.25，1.5）C．（1.5，2）D．不能确定参考答案：B【考点】二分法求方程的近似解．【分析】根据函数的零点存在性定理，由f（1）与f（1.5）的值异号得到函数f（x）在区间（1，1.5）内有零点，同理可得函数在区间（1.25，1.5）内有零点，从而得到方程3x+3x﹣8=0的根所在的区间．【解答】解：∵f（1）＜0，f（1.5）＞0，∴在区间（1，1.5）内函数f（x）=3x+3x﹣8存在一个零点，又∵f（1.5）＞0，f（1.25）＜0，∴在区间（1.25，1.5）内函数f（x）=3x+3x﹣8存在一个零点，由此可得方程3x+3x﹣8=0的根落在区间（1.25，1.5）内，故选：B．6.已知函数满足对任意的都成立。

广东省广州市六中珠江中学2020年高一数学理月考试题含解析

广东省广州市六中珠江中学2020年高一数学理月考试题含解析一、选择题：本大题共10小题，每小题5分，共50分。

在每小题给出的四个选项中，只有是一个符合题目要求的1. 函数的定义域为，则实数的取值范围是（）A．B． C． D．参考答案：B2. 等腰直角三角形中，是斜边的中点，若，则=( )A.B． C． D．参考答案：A略3. 设全集，，则（）A．B． C．D．参考答案：B4. 函数f（x）=+lg（1+3x）的定义域是（）A．（﹣∞，﹣）B．（﹣，）∪（，+∞）C．（，+∞）D．（，）∪（，+∞）参考答案：B 【考点】函数的定义域及其求法．【分析】由1﹣2x≠0.1+3x＞0，解不等式即可得到所求定义域．【解答】解：由1﹣2x≠0.1+3x＞0，可得x＞﹣，且x≠，则定义域为（﹣，）∪（，+∞），故选：B．5. 如图，设A、B两点在河的两岸，一测量者在A的同侧，在所在的河岸边选定一点C，测出AC的距离为50m，∠ACB=45°，∠CAB=105°后，就可以计算出A、B两点的距离为（）A．mB．mC．mD．参考答案：A6. 若关于x的方程a x﹣x﹣a=0有两个不同的实数根，则实数a的取值范围为（）A．（0，+∞）B．（0，1）C．（0，2）D．（1，+∞）参考答案：D【考点】函数的零点与方程根的关系．【分析】由a x﹣x﹣a=0得a x=x+a，再画出a＞1和0＜a＜1时的两种函数y=a x，y=x+a的图象，根据图象可直接得出答案．【解答】解：由a x﹣x﹣a=0得a x=x+a，则等价为函数y=a x，的图象与直线y=x+a有两个不同的交点．①a＞1时，此时满足两个函数的图象有两个交点，②0＜a＜1时，此时两个函数只有一个交点，不满足两个函数的图象有两个交点，综上，若关于x的方程a x﹣x﹣a=0（a＞0）有两个解，则实数a的取值范围为（1，+∞）故选：D7. 已知函数在R上单调递减,则实数的取值范围是（）A． B．C．D．参考答案：C略8. （5分）若集合A={x|x＞﹣3}，则（）A．0?A B．{0}∈A C．?∈A D．{0}?A参考答案：D考点：元素与集合关系的判断．专题：集合．分析：由已知，明确A集合中含有元素0，然后注意元素与集合关系的符号表示以及集合与集合的关系表示即可．解答：因为0＞﹣3，所以0∈A，{0}?A；故选D．点评：本题考查了元素与集合的关系以及集合与集合的关系，属于基础题．9. 已知点A（1，3），B（4，﹣1），则与向量同方向的单位向量为（）A．(，) B．(，) C．(，) D．(，)参考答案：A【考点】平行向量与共线向量；95：单位向量．【分析】由条件求得=（3，﹣4），||=5，再根据与向量同方向的单位向量为求得结果．【解答】解：∵已知点A（1，3），B（4，﹣1），∴=（4，﹣1）﹣（1，3）=（3，﹣4），||==5，则与向量同方向的单位向量为=，故选A．10. 设集合（）A．B．C．D．参考答案：B二、填空题:本大题共7小题,每小题4分,共28分11. 设向量若A，B，C三点共线，则k＝_______.参考答案：k=-2或k=11略12. 在R上定义运算※，若存在，使不等式※成立，则实数m 的取值范围为．参考答案：(－3,2)13. 在△ABC 中，a，b ，c分别是角A，B ，C的对边，已知a ，b ，c成等比数列，且，则的值为________.参考答案：【分析】利用成等比数列得到，再利用余弦定理可得，而根据正弦定理和成等比数列有，从而得到所求之值.【详解】∵成等比数列，∴.又∵，∴.在中，由余弦定理，因，∴.由正弦定理得，因为，所以，故.故答案为：.【点睛】在解三角形中，如果题设条件是关于边的二次形式，我们可以利用余弦定理化简该条件，如果题设条件是关于边的齐次式或是关于内角正弦的齐次式，那么我们可以利用正弦定理化简该条件，如果题设条件是边和角的混合关系式，那么我们也可把这种关系式转化为角的关系式或边的关系式.14. 已知a+1，a+2，a+3是钝角三角形的三边，则a的取值范围是参考答案：(0，2)15. 将圆心角为，面积为的扇形，作为圆锥的侧面，则圆锥的体积为__________参考答案：16. =__________。

2019-2020学年广州市第六中学高三英语下学期期末试题及答案解析

2019-2020学年广州市第六中学高三英语下学期期末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ALOI English—English Conversation ClassesTake English conversation classes with native English -speaking teachers from theUnited States,IrelandandAustralia, live, 1 - on - 1.This is by far the best way to improve your real spoken English and listening abilities and get to a level where you can use it in real - life situations. Our teachers mainly carry on conversations about work in different fields, such as business, finance, politics and travel. However, if you like, you can have normal day - to - day conversations about events, weather, etc. If you would like to learn some grammar during conversation, we can do that too!Our English Conversation classes：• Your conversations will always be with a native English speaker.• You can choose the time, day and teacher for each class.• The classes are made for each student individually. They will be at your level, studying what you need and what interests you.Packages10 Classes Package $ 229.90Interested in taking 10, 55 - minute conversation English classes?Schedule a first free class with LOI English and talk to the teacher about it. After that, you can arrange to take our 10 - class conversation English course.20 Classes Package $ 439,80Interested in taking 20, 55 - minute conversation English classes?Schedule a first free class with LOI English and talk to the teacher about it. After that, you can arrange to take our 20 - class conversation English course.1. What do the classes mainly focus on?A. Daily conversations.B. Work - related topics.C. Grammar rules.D. Area - based accents.2. Which is NOT offered to the students?A. Native English - speaking teachers.B. Personalized courses.C. Choice of class time and teacher.D. Offline learning materials.3. What should a student do first if he decides on either of the packages?A. Arrange a free class with LOI English.B. Bargain with LOI English for lower fees.C. Choose an experienced LOI English teacher.D. Make a promise to sit through each class.BThe shade of a single tree can provide welcome relief from the hot summer sun. But when that single tree is part of a small forest, it creates a considerable cooling effect. According to a study published today in the Proceedings of the National Academy of Sciences, trees play a big role in keeping our cities cool.According to the study, the right amount of tree cover can lower summer daytime temperatures by as much as 10 degrees Fahrenheit. And the effect is quite noticeable from neighborhood to neighborhood, even down to the scale of a single city block. “We knew that cities are warmer than the surrounding countryside, but we found that temperatures vary just as much within cities,” says Monica Turner, a professor in the department of Integrative Biology, Wisconsin-Madison University and a co-author of the study.With climate change making extreme heat events more common each summer, city planners are working on how to prepare. Heat waves drive up energy demands and costs and can have big human health impacts. One potentially powerful tool, the study's authors say, are organisms that have been around long before human civilizations could appreciate their leafy benefits. And those trees may be the secret to keeping the places we live livable.Essentially, says Turner, roads, sidewalks and buildings absorb heat from the sun during the day and slowly release that heat at night. Trees, on the other hand, not only shade those surfaces from the sun's rays, they also release water into the air through their leaves, a process that cools things down.To get the maximum benefit of this cooling service, the study found that tree cover must be more than 40 percent. In other words, an aerial picture of a single city block would need to be nearly half-way covered by a leafy green network of branches and leaves.4. What can we infer from Paragraph 2?A. Temperatures in cities mainly depend on their green coverage.B. People living and working in cities must plant trees in summer.C. Cities are warmer than the countryside because they don't have trees.D. An area with more trees can be cooler than the other parts within a city.5. Which of the following problems is caused by extreme heat events?A. Severe damage of city facilities.B. Serious human health problems.C. Residents' growing demands for plants.D. Unnecessary waste of energy resources.6. Why are trees crucial to cities?A. They help shade and cool the cities.B. They make urban scenery beautiful.C. They build up city cooling systems.D. They essentially block and release heat.7. What can city planners conclude from the study?A. They should publish the study in a newspaper.B. They should educate citizens to protect forests.C. They should plant trees on roads and sidewalks.D. They should cover nearly half the city with trees.CYou’re in a crowd of people who are all asking for the same thing. How do you make your voice heard above the rest? Be different. Don’t shout. Lisa, 25, was waiting to board a plane flying fromLondontoAustriafor Christmaswhen the flight was cancelled.“There were about a hundred of us unable to leave,” she says. “Everyone else was shouting at the airport staff. Instead of joining in, I walked up to the man behind the ticket desk very quietly and said, ‘This must be so awful for you! I don’t know how you deal with these situations—it’s not even your fault. I could never handle it as well as you are.’ Without my even asking, he found me a seat on another airline with an upgrade to first class. He was happy to do a favor forsomeone who was appreciative instead of unfriendliness.”Flattery (恭维) is an essential element of the sweet-talk strategy. “It’s human psychology that stroking a person’s ego (自我) with a few well-directed praises makes them want to prove you right,” says apsychologist. “Tell someone they’re pretty and they’ll instantly fix their hair; praise their sense of humor and they’ll tell a joke.”You need help and there’s ly no reason that the person will want to lend a hand. Allison, 26. a lawyer, realized she’d made a huge mistake on a batch of documents. “The only way I could fix the problem was to get the help of a colleague who I knew didn’t like me,” she said.Allison then went to the woman’s office and explained her problem. “As I was saying to the boss the other day you’re the only person who would know how to handle a situation like this, what would you suggest I do?”“Feeling pumped up (鼓励), she set about helping me and we finished the job on time, and she was happy to help.” Allison said.8. Whatwould have happened at the airport according to paragraph 1?A. The departure hall was filled with noise.B. Someone screamed just lo be different.C. The passengers waited on board patiently.D. The airport stuff were rude to the passengers.9. Why did the man put Lisa on another airline?A. He admired Lisa’s beauty.B. He appreciated her attitude.C. He was ready to help others.D. He was blamed for the cancellation.10. What is the third paragraph mainly about?A. The potential benefits of ego.B. The strategy to start small talk.C. The great importance of flattery.D. The value of humor in daily life.11. What can we learn about Allison’s colleague?A. She was a popular lawyer.B. She was always ready to help others.C. She always got praise from Allison.D. She did a great favor for Allison eventually.DA male tiger was rescued in Mishan city, Northeast China's Heilongjiang province, on Friday. A tiger was recently seen on the Siner Russian border（边界），but there is no evidence to suggest this is the same tiger or that it had lost its way from Russia.Wherever it may have come from, cross-border protection of endangered species is a problem. According to Ma Jianzhang, a senior wild animal researcher in China, tigers know no borders, which have been built to stop humans from illegally crossing over. However, these borders also stop the free movement of tigers, thus preventing their breeding（繁殖）and exchange of genes.The Sino-Russian border alsocomes in the way ofprotecting tigers in the region. Data show about 500 to 550 Siberian tigers, or over 90 percent of the species, live in the wild in Russia; 27 live in the wild in China, and 50 inthe border region. Sometimes tigers lose their way across the border and that is hindering（妨碍）efforts to stop illegal hunting and protecting them.Fortunately, both China and Russia have realized this problem and are working toward solving it. As early as 2010, the two countries established a cross-border protection network aimed at strengthening communication on protecting tigers in the northeast. The countries have also held anti-poaching（反偷猎）activities together. In 2019, two national parks from the two countries signed a three-year deal to share research data on tigers, greatly improving the living conditions of Siberian tigers in the wild.However, much more needs to be done to protect the species. It is necessary to combine the tiger protection areas in the two countries. That will require greater efforts from both sides and some difficulties might remain even then. Fortunately, both countries have been taking effective measures to provide the endangered species a better home.12. What's the original purpose of setting up the Sino-Russian borders?A. To ban people from illegal crossings.B. To help protect tigers in the region.C. To protect tigers from losing their way.D. To stop tigers' breeding and exchange of genes.13. What does the underlined phrase “comes in the way of” probably mean?A. keeps from.B. is on the way toC. is aimed atD. keeps up14. What is the focus of Paragraph 4 ?A. Recent studies and researches on tigers.B. Joint efforts by China and Russia to protect tigers.C. The possible effects of establishing a networkD. Reasons for improving the living conditions of tigers.15. What's the author's purpose in writing the text?A. To inform the possible dangers of tiger protection.B. To cover the news about a missing tiger on the border.C. To put forward suggestions on finding the missing tiger.D. To report the problems and measures of cross-border tiger protection.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。

2019-2020学年广州市六中学高三英语下学期期末试题及答案

2019-2020学年广州市六中学高三英语下学期期末试题及答案第一部分阅读（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AThough online ordering is popular in the UK today，handson experience still attracts many buyers' attention.Oxford StreetLondon is a shoppers' paradise，and most visitors head for Oxford Street.Halfway along is Selfridges，one of the world's first department stores where the architecture and window dressing impress you even before you go in.There are at least four other major stores on this street，plus Europe's largest music shop.In December the atmosphere is especially festive because of the Christmaslights which are always lighted by a politician，a great sportsman，or a super star.Kensington High StreetA fashionable area for shopping is Kensington High Street.One of its department stores even has a roof garden—good for relaxing between purchases.There's also an organic food superstore.Borough MarketYou can't visit London without going to a supermarketOne of the oldest and best for food is probably Borough Market on the south bank of the Thames.Get off the underground at London Bridge and wander among fresh fish stalls，bakeries，and fruit and vegetable sellers.You can also buy specialities to take away，like cheeses，pork pies，mustard and chutney.More places for bargainsPortobello Road Market on a Saturday is your best choice for jewelry or collector's items.It's a shortwalk from the underground station at Notting Hill Gate and you never know what you will find.For clothes，Camden Market is the place.It gets a little crowded at the weekends，though.Convent Garden Market offers great experience for the simple pleasure of looking at jewelry and gifts.1.Oxford Street becomes special in December for ________.A.the wonderful window dressingB.the Europe's largest music shopC.the lights lighted by a famous personD.many bargains to choose from2.To enjoy food，you can goto ________.A.Camden MarketB.Borough MarketC.Kensington High StreetD.Portobello Road Market3.Where is this passage probably taken from?AA science book. B.A traveler's journal.C.A news report.D.A trip guide.BChildren's average daily time spent watching television or using mobile device increased from 53 minutes at age 12 months to more than 150 minutes at 3 years, according toan analysis by researchers at the National Institutes of Health (NIH). By age 8, children were more likely to log the highest amount of screen time if they had been in home-based children or were born to first-time mothers.“Our results indicate that screen habits begin early, ”said Edwina Yeung, an investigator in National Institute of Child Health and Human Development (NICHD).“This finding suggests that interventions(干涉) to reduce screen time could have a better chance of success if introduced early.”In the research, mothers of 4, 000 children responded to questions on their kids' media habits when they were 12, 18, 24, 30, and 36 months of age.TheAmericanAcademyof Pediatrics recommends avoiding digital media exposure for children under 18 months of age, introducing children 18 to 24 months of age to screen media slowly, and limiting screen time to an hour a day for children from 2 to 5 years of age. In the current study, researchers found that 87% of the children had screen time exceeding (超过) these recommendations. However, while screen time increased throughout toddlerhood, by age 7 and 8, screen time fell to under 1.5 hours per day. The researchers believe this decrease relates to time consumed by school-related activities.The study authors classified the children into two groups based on how much their average daily screen time increased from age 1 to age 3. The first group, 73% of the total had the lowest increase, from an average of nearly 51 minutes a day to nearly an hour and 47 minutes a day. The second group, 27% of the total, had the highest increase, from nearly 37 minutes of screen time a day to about 4 hours a day. Higher levels of parental education were associated with lower odds of inclusion in the second group.4. Who use mobile device longest according to the NIH's analysis?A. 10-year-old born to first-time mothers.B. 3-year-old children in low income families.C. 8-year-old children in home-based childcare.D. 1-year-old children with parents poorly educated.5. What does the research suggest according to Edwina?A. Parents should stop their children using media.B. Parents should limit the use of digital media themselves.C. Parents should reduce their children's screen time earlier.D. Parents should avoid their children using digital media at infancy (婴儿期).6. Why does children's screen time fall when they age 7 and 8?A. They are studying at school.B. They can control themselves.C. They are tired of using them.D. They are forbidden to use them.7. Which of the following may be the best title?A. Keep Away from MediaB. Screen Habits Begin EarlyC. Urgency of Parental EducationD. Harm of Home-based ChildcareCYour house may have an effect on your figure.Experts say the way you design your home could play a role in whether you pack on the pounds or keep them off. You can make your environment work for you instead of against you. Here are some ways to turn your home into part of your diet plan.Open the curtains and turn up the lights. Dark environments are more likely to encourage overeating, for people are often less selfconscious (难为情) when they’re in poorly lit places—and so more likely to eat lots of food. If your home doesn’t have enough window light, get more lamps and flood the place with brightness.Mind the color1 s. Research suggests warm color1 s fuel our appetites. In one study, people who ate meals in a blue room consumed 33 percent less than those in a yellow or red room. Warm color1 s like yellow make food appear more appetizing, while cold color1 s make us feel less hungry. So when it’s time to repaint, go blue.Don’t forget the clock—or the radio. People who eat slowly tend to consume about 70 fewer calories (卡路里) per meal than those who rush through their meals. Begin keeping track of the time, and try to make dinner last at least 30 minutes. And while you’re at it, actually sit down to eat. If you need some help slowing down, turn on relaxing music. It makesyou less likely to rush through a meal.Downsize the dishes. Big serving bowls and plates can easily make us fat. We eat about 22 percent morewhen using a 12inch plate instead of a 10inch plate. When we choose a large spoon over a smaller one, total intake (摄入) jumps by 14 percent. And we’ll pour about 30 percent more liquid into a short, wide glass than a tall, skinny glass.8. The text is especially helpful for those who care about ____________.A. their home comfortsB. their body shapeC. house buyingD.healthy diets9. A home environment in blue can help people ______________.A. digest food betterB. reduce food intakeC. burn more caloriesD. regain their appetites10. What are people advised to do at mealtimes?A. Eat quickly.B. Play fast music.C. Use smaller spoons.D. Turn down the lights.11. What can be a suitable title for the text?A. Is Your House Making You Fat?B. Waysof Serving DinnerC. Effects of SelfConsciousnessD. Is Your Home Environment Relaxing?DZaki was small for his twelve years, and he was angry being treated as a child. Farid, his older brother, had been looked upon as a man long before he was Zaki’s age. Every day Farid and the other young Bahraini men went out in their wooden boats to dive for oysters (牡蛎). Many times Zaki begged to go along, but Faridalways refused to let him.So every day Zaki would go to the shallow water to practice. His grandfather, a former diver, would watch him and advise him. All morning, Zaki would practice diving beneath the waves. Every afternoon, again and again he would go underwater and hold his breath. With each day’s practice, his diving improved and he could hold his breath a little while longer. Soon Zaki felt as much at home in the water as he did out of it.Zaki rose early one day. He wanted to compete with his brother. They dived beneath the waves. Zaki opened his eyes and found himself looking into his brother’s face. Farid was smiling with confidence. Slowly, the smile was disappearing from Farid’s face. As more seconds passed, a worried look appeared on Farid’s face. Farid was realizing that Zaki could possibly beat him. Looking into Farid’s eyes, Zaki suddenly understood what losing could mean to his brother. Never would the villagers allow him to live it down. He would be laughed at by losing to a little child. Almost without thinking, Zaki kicked his feet and rose to the surface of the water a second beforeFarid’s head appeared beside him.The men around them cheered and patted Farid on the back. Farid, however, put his arm around Zaki’s shoulders. “Today,” Farid announced, “we shall have a new diver among us.” Then quietly, for Zaki’s ears alone, he said “Thank you, my brother.” And Zaki knew that they both had learned that it takes more than strength to makea man.12. What is the second paragraph mainly about?A. Zaki’s grandfather was a good diver.B. Zaki liked staying at home every day.C. Zaki practised hard in the water daily.D. Zaki’s grandfather encouraged him to dive.13. Which of the following best describes Zaki according to paragraph 3?A. Considerate.B. Ambitious.C. Confident.D. Adventurous.14. What can we infer from the last sentence of the text?A. Farid beat his little brother easily.B. Zaki was as strong as his brother.C. Zaki regretted losing the competition.D. Both Farid and Zaki had grown up.15. What is the best title for the text?A. Farid’s PrideB. Zaki’s ChallengeC. Brothers’ CompetitionD. Grandfather’s Advice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。

2019-2020学年广州市六中学高三英语下学期期末试题及参考答案

2019-2020学年广州市六中学高三英语下学期期末试题及参考答案第一部分阅读（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AThe following 4 famous paintings, from Jan van Eyck’s portrait to Pablo Picasso’s masterpiece, have stood the test of time.The Arnolfini PortraitJan van Eyck’s Arnolfini Portrait, an oil painting on wood produced in 1434, is undoubtedly one of the masterpieces in the National Gallery,London. This painting is as visually interesting as it is famed. It is also an informative document on fifteenth-century society, through van Eyck’s heavy use of symbolism-while husbands went out to engage in business, wives concerned themselves with domestic duties.The Starry NightVincent van Gogh painted The Starry Night, oil on canvas, a moderately abstract landscape painting of an expressive night sky over a small hillside village, during his 12-month stay at the mental hospital nearSaint-Remy-de-Provence,Francebetween 1889 and 1890. When theMuseumofModern ArtinNew York Citypurchased the painting from a private collector in 1941, it was not well known, but it has since become one of van Gogh's most famous works.The HarvestersThe Harvesters is an oil painting on wood completed by Pieter Bruegel the Elder in 1565. It depicts the harvest time which most commonly occurred within the months of August and September. Nicolaes Jonghelinck, a merchant banker and art collector fromAntwerp, commissioned this painting. The painting has been at the Metropolitan Museum of Art inNew York Citysince 1919.GuernicaGuernica,a large black-and-white oil painting, was painted by the Cubist Spanish painter, Pablo Picasso in 1937. The title ‘Guernica’ refers to the city that was bombed by Nazi planes during the Spanish Civil War. The painting depicts the horrors of war and as a result, has come to be an anti-war symbol and a reminder of the tragedies of war. Today, the painting is housed at the Museo National Centro de Arte ReinaSofiainMadrid.1. What do the four paintings have in common?A. They are oil paintings.B. They are kept inNew York.C. They show different people’s lives.D. They are created during 10th century to 19th century.2. What do we know about The Starry Night?A. It shows the scenery of a small hillside village.B. Van Gogh produced it when living in his home.C. A private collector gave it away to the museum.D. It was not famous before 1941.3. Whose painting shows the destruction of war?A. Jan van Eyck.B. Vincent van Gogh.C. Pieter Bruegel.D. Pablo Picasso.BThe headmaster of a primary school showed on television to support her idea that parents should “dress appropriately in daywear” when they drop off and pick up their kids from school.Kate Chisholm, head teacher atSkerneParkAcademyin Darlington, theU.K., sent a letter home asking parents to set a better example for their children.“I have noticed there has been an increasing tendency for parents to drop off and pick up their kids from school while still wearing their pajamas (睡衣),” Chisholm wrote.“Could I please ask that when you are sending your children, you take the time to dress appropriately in daywear that is suitable for the weather conditions?”Kate Chisholm wants parents at her school to dress nicer. She appeared on British television station ITV to further explain her decision, saying she had started noticing the pajama trend had been picked up by “30 or 40” parents at school.Despite her determination to make school a nicer place to be, Chisholm admits that she can't demand that parents dress up-such as Karen Routh, 49, who wore pajamas to drop off her 8-ycar-old daughter Holly, because she was running late and didn't feel well.“I imagine there might be some people who keep up wearing pajamas for the next six months to prove a point,” Chisholm said. “I can't force people to get dressed but I will keep sending letters home in the hope that they decide to put on a pair of jeans.”Wearing pajamas in public has also become a hot issue for some schools and States in theU.S.In 2015, aFloridaschool board member insisted on a dress rule for parents who showed up in the school in sleepwear.4. The headmaster asks parents to pay attention to ______.A. the way they dressB. the relations with teachersC. the way they treat their kidsD. the clothes they buy for their kids5. How does Chisholm try to change this situation?A. Asking kids to set examples.B. Keeping them out of school.C. Sending letters to persuade them.D. Forcing them to change by laws.6. Why did Kate Chisholm appear on ITV?A. She wanted to force Karen to dress properly.B. Parents spent less time on their clothes.C. She wanted to explain her decision about the parents' dress.D. She wanted to tell us more and more parents wear pajamas to school.7. It can be inferred from the last paragraph that ______.A. strict laws should be passed to stop pajamasB. aFloridaschool will force parents to wear jeansC. people wearing pajamas in public will be punishedD. more and more people are concerned about dressing properly in publicCSpain's tourism industry is looking to Chinese tourists for its high-endmarket, according to Rafael Cascales, president of the Spain-China Tourism Association (ATEC). “It is the kind of tourism that is not only interested in the sun, beach and the “all-included” culture. They enjoy culture, wine, history and nature, and the new Chinese tourists would also want to spend more money in Spain," said Cascales in a recent interview with Xinhua.“They are younger, more women travel and they are more cosmopolitan (见多识广的).They also travel on their own or in couples or in smaller groups. The old-fashioned large groups of visitors have not disappeared, but this new form of traveling is becoming more important,55he said.Speaking of the consumption pattern of the new kind of Chinese tourists, Cascales said, “The money they spend is distributed better because they will book one flight with one airline, the hotel with another company and the restaurant with another.” In his eyes, “Chinese tourists are very important because they combine two things: there are a large number of them and they spend more money than anyone else — almost four times more thantourists from other countries." They not only travel abroad in the summer months when Spain has to compete with the sun and beaches in countries such as Turkey and Egypt, but also travel in the off-peak seasons of a year, according to Cascales.In 2017, Spain is the second most popular tourist destination in the world, only after France. It attracted about 82 million visitors, 700,000 of them from China, a number which is estimated to rise to about 2.2 million by 2022.“We are ready; we have the infrastructure (基础设施) at every level, especially in hotel capacity. Here those visitors can find what they are looking for, including the luxury items which distinguish them,” Cascales noted.8. What are the features of the new Chinese tourists according to Cascales?A. They are cautious about spending money in Spain.B. They are likely to travel in smaller groups now.C. They are only interested in the sun and beach.D. They are mainly male visitors of middle age.9. What can we learn about the consumption pattern of new Chinese tourists?A. They will reserve flights and hotels with different companies.B. They will spend less money than tourists from other countries.C. They will travel abroad only during the off-peak seasons.D. They will spend the money in different cities.10. What is done to meet the demands of Chinese tourists?A. Local cultures in Spain are promoted.B. Well-furnished hotels are provided.C. Best and expensive items are offered for free.D. More shopping sites are constructed.11. What is the purpose of this text?A. To introduce the tourism industry of Spain.B. To show Spain's desire to attract Chinese tourists.C. To describe the features of Chinese tourists.D. To advertise Spain as a top tourist destination.DElizabeth Spelke, a cognitive psychologist at Harvard, has spent her career testing the world’s most complexlearning system-the mind of a baby. Babies might seem like no match for artificial intelligence (AI). They are terrible at labeling images, hopeless at mining text, and awful at video games. Then again, babies can do things beyond the reach of any AI. By just a few months old, they’ve begun to grasp the foundations of language, such as grammar. They’ve started to understand how to adapt to unfamiliar situations.Yet even experts like Spelke don’t understand precisely how babies or adults learn. Consider one of the most impressive examples of AI, Alpha Zero, a programme that plays board games with superhuman skill. After playing thousands of games against itself at a super speed, and learning from winning positions, Alpha Zero independently discovered several famous chess strategies and even invented new ones. It certainly seems like a machineeclipsinghuman cognitive abilities. But Alpha Zero needs to play millions more games than a person during practice to learn a game. Most importantly, it cannot take what it has learned from the game and apply it to another area.To some AI experts, that calls for a new approach. In a November research paper, Francois Chollet, a well-known AI engineer, argued that it’s misguided to measure machine intelligence just according to its skills at specific tasks. “Humans don’t start out with skills; they start out with a broad ability to acquire new skills,” he says. “What a strong human chess player is demonstrating is not only the ability to play chess, but the potential to fulfill any task of a similar difficulty.”12. Compared to an advanced AI programme, a baby might be better at ________.A playing games B. identifying locations C. labeling pictures D. making adjustments13. What does the underlined word “eclipsing” in Paragraph 3 probably mean?A. Imitating.B. Beating.C. Limiting.D. Promoting.14. According to the text, Francois Chollet may agree that ________.A. AI is good at completing certain assignments.B. AI is likely to gain abilities with less training.C. AI lacks the ability of acquiring specific skills.D. AI performs better than humans in cognitive ability.15. Whichwould be the best title for this passage?A. What is exactly intelligence?B. Why is modern AI advanced?C. Where is human intelligence going?D. How do humans face the challenge of AI?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。