2018_2019学年广东广州番禺区广州大学附属中学大学城校区高二上学期期中英语试卷

广东省第二师范学院番禺附属中学2018-2019学年高二物理下学期期中试题一.选择题（共16题，54分。

1-10为单选题，每题3分；11-16为不定项，每题4分，选不全得2分，选错得零分）1．关于电磁感应现象的有关说法中，正确的是（）A．只要穿过闭合电路的磁通量不为零，闭合电路中就一定有感应电流产生B．穿过闭合电路的磁通量变化越快，闭合电路中感应电动势越大C．穿过闭合电路的磁通量越大，闭合电路中的感应电动势越大D．穿过闭合电路的磁通量减少，则闭合电路中感应电流就减小2．如图所示，绕在铁芯上的线圈与电源、滑动变阻器和电键组成闭合回路，在铁芯的右端套有一个表面绝缘的铜环A，下列各种情况中铜环A中没有感应电流的是（）A．线圈中通以恒定的电流B．通电时，使变阻器的滑片P作匀速移动C．通电时，使变阻器的滑片P作加速移动D．将电键突然断开的瞬间3．如图所示，两水平平行金属导轨间接有电阻R，置于匀强磁场中，导轨上垂直搁置两根金属棒ab、cd．当用外力F拉动ab棒向右运动的过程中，cd棒将会（）A．向右运动 B．向左运动 C．保持静止 D．向上跳起4．如图所示，有界匀强磁场垂直纸面向里，一闭合导线框abcd从高处自由下落，运动一段时间后进入磁场，下落过程线框始终保持竖直，对线框进入磁场过程的分析正确的是（）A．感应电流沿顺时针方向B．a端电势高于b端C．可能匀加速进入D．感应电流的功率可能大于重力的功率5．由交变电流瞬时表达式i＝10sin500t(A)可知，从开始计时起，第一次出现电流峰值所需要的时间是( )A．2 ms B．1 ms C．6.28 ms D．3.14 ms6．理想变压器的原、副线圈匝数比为10 ∶1，下列说法中正确的是( )A．穿过原、副线圈每一匝的磁通量之比是10 ∶1B．穿过原、副线圈每一匝的磁通量之比是1 ∶10C．正常工作时，原、副线圈的输入、输出电压之比为10 ∶1D．正常工作时，原、副线圈的输入、输出电压之比为1 ∶107．古时有“守株待兔”的寓言，设兔子的头部受到大小等于自身重力的撞击力时即可致死，并设兔子与树桩的作用时间为0.2 s，则被撞死的兔子奔跑的速度可能为（g取10 m/s2）（）．①1 m/s ②1.5 m/s ③2 m/s ④2.5 m/sA．③B．③④ C．②③④ D．①②③④8．在氢原子光谱中，可见光区域中有4条，其颜色为一条红色，一条蓝色，两条紫色，它们分别是从n＝3、4、5、6能级向n＝2能级跃迁时产生的，则以下判断正确的是( ) A．红色光谱线是氢原子从n＝6能级到n＝2能级跃迁时产生的B．紫色光谱线是氢原子从n＝6和n＝5能级向n＝2能级跃迁时产生的C．若从n＝6能级跃迁到n＝1能级将产生红外线D．若从n＝6能级跃迁到n＝2能级所辐射的光子不能使某金属产生光电效应，则从n ＝6能级向n＝3能级跃迁时辐射的光子将可能使该金属产生光电效应9．氦原子被电离出一个核外电子，形成类氢结构的氦离子，已知基态的氦离子能量为E1＝－54.4 eV，氦离子的能级示意图如图所示，在具有下列能量的光子或者电子中，不能被基态氦离子吸收而发生跃迁的是( )A．42.8 eV(光子) B．43.2 eV(电子)C．41.0 eV(电子) D．54.4 eV(光子)10.用导线把验电器与锌板相连接，当用紫外线照射锌板时，发生的现象是（）A有电子从锌板飞出 B有光子从锌板飞出C锌板带负电 D验电器内的金属箔带负电11-16为不定项11．如图所示，匀强磁场垂直于金属导轨平面向里，导体棒ab与导轨接触良好，当导体棒ab在金属导轨上做下述哪种运动时，能使闭合金属线圈c向右摆动（）A．向右加速运动 B．向右减速运动 C．向左加速运动 D．向左减速运动12．如图所示，带铁芯的电感线圈的电阻与电阻器R的阻值相同，A1和A2是两个完全相同的电流表，则下列说法中正确的是（）A．闭合S瞬间，电流表A1示数小于A2示数B ．闭合S 瞬间，电流表A 1示数等于A 2示数C ．断开S 瞬间，电流表A 1示数大于A 2示数D ．断开S 瞬间，电流表A 1示数等于A 2示数13．现在低碳环保问题已引起广大人民群众的普遍关注，深圳市民林先生自主研发出一种磁悬浮风力发电机，其发电原理可以简化为：一个矩形线圈绕在其平面内并且垂直于匀强磁场的轴做匀速转动而产生感应电动势，产生的感应电动势图象如图所示，并且接在原、副线圈匝数比为n 1n 2＝15的理想变压器原线圈两端，则( )A ．变压器副线圈两端电压的有效值为U 2＝110 VB ．感应电动势的瞬时值表达式为e ＝222sin10πt (V)C ．t ＝0.005 s 时穿过线圈平面的磁通量最大D ．t ＝0.005 s 时线圈平面和磁场方向平行14．在匀强磁场中，一矩形金属线框绕与磁感线垂直的转轴匀速转动，如图乙所示，产生的交变电动势随时间变化规律的图象如图甲所示，已知发电机线圈内阻为1.0 Ω，外接一只电阻为9.0 Ω的灯泡，则( )A ．电压表V 的示数为20 VB ．电路中的电流方向每秒改变10次C ．灯泡实际消耗的功率为36 WD ．电动势随时间变化的瞬时表达式为e ＝20cos5πt (V)15．如图所示，质量分别为m 和2m 的A 、B 两个木块用轻弹簧相连，放在光滑水平面上，A 靠紧竖直墙．用水平力F 将B 向左压，使弹簧被压缩一定长度，静止后弹簧储存的弹性势能为E ．这时突然撤去F ，关于A 、B 和弹簧组成的系统，下列说法中正确的是（ ）．A ．撤去F 后，系统的动量守恒，机械能守恒B ．撤去F 后，在A 离开竖直墙前，系统的动量不守恒，机械能守恒C ．撤去F 后，在A 离开竖直墙后，弹簧的弹性势能的最大值为ED ．撤去F 后，在A 离开竖直墙后，弹簧的弹性势能的最大值为E /316．在光滑水平面上动能为E 0、动量大小为p 0的小钢球1与静止小钢球2发生碰撞，碰撞前后球1的运动方向相反，将碰撞后球1的动能和动量大小分别记为E 1、p 1，球2的动能和动量大小分别记为E 2、p 2，则必有( )A ．E 1<E 0B ．p 2>p 0C ．E 2>E 0D ．p 1>p 0二、实验题(共10分)17．气垫导轨是常用的一种实验仪器，它是利用气泵使带孔的导轨与滑块之间形成气垫，使滑块悬浮在导轨上，滑块在导轨上的运动可视为没有摩擦，我们可以用带竖直挡板C和D 的气垫导轨和滑块A和B验证动量守恒定律，实验装置如图所示，采用的实验步骤如下：a．松开手的同时，记录滑块A、B运动时间的计时器开始工作，当A、B滑块分别碰到C、D挡板时计时器结束计时，分别记下A、B到达C、D的运动时间t1和t2．b．在A、B间水平放入一个轻弹簧，用手压住A、B使弹簧压缩，放置在气垫导轨上，并让它静止在某个位置．c．给导轨送气，调整气垫导轨，使导轨处于水平．d．用刻度尺测出A的左端至C板的距离L1，B的右端至D板的距离L2．（1）实验步骤的正确顺序是_________________．（2）实验中还需要的测量仪器是_____________．还需要测量的物理量是____________________．（3）利用上述测量的实验数据，验证动量守恒定律的表达式是____________________．（4）实验中弹簧与A、B滑块之一是粘连好还是不粘连好？__________好．三、解答题(本题共3小题，共36分)18.如图所示为一个冲击摆，它可以用来测量高速运动的子弹的速率。

2017-2018学年上学期期中三校联考
高二（文）数学
第Ⅰ卷（选择题部分，共60分）
一、选择题：（本大题共12小题，每小题5分，在每小题给出的四个选项中，只有一项是符
合题目要求的）
1. 已知集合，，则（）．
A. B. C. D. 或
【答案】A
【解析】∵集合
∴集合
∴
故选C
2. 已知命题甲：，命题乙：或，则命题甲是命题乙的（）．
A. 充分不必要条件
B. 必要不充分条件
C. 充要条件
D. 既不充分也不必要
【答案】D
【解析】∵“若或，则”为假命题
∴它的等假命题“若，则且”为假命题；
∵“若，则或”为假命题
∴其等价命题“若且，则”为假命题
∴命题甲：，是命题乙：且的既不充分也不必要条件．
故选D
3. 某校举行演讲比赛，位评委给选手打出的分数如茎叶图所示，统计员在去掉一个最高分
和一个最低分后，算得平均分为，复核员在复核时，发现有一个数字（茎叶图中的）无法看清，若统计员计算无误，则数字应该是（）．
A. B. C. D.
【答案】D
【解析】记分员在去掉一个最高分和一个最低分后，余下的个数字的平均数是，
，，故选 D.
4. 若实数，满足，且，则下列四个数中最大的是（）．
A. B. C. D.
【答案】C
【解析】∵且，
∴，
∴，
∴，即，
又，
∴，
∴最大的一个数为．
故选C
5. 如图所示，四个相同的直角三角形与中间的小正方形拼成一个边长为的大正方形，若直角三角形中较小的锐角，现在向该正方形区域内随机地投掷一枚飞镖，则飞镖落在小正方形内的概率是（）．
A. B. C. D.。

2018学年第二学期广东二师附中中段测试高二级试题数学（理）本试卷共4页，22小题， 满分150分。

考试用时120分钟。

注意事项：1.本试卷分第Ⅰ卷（选择题）和第Ⅱ卷（非选择题）两部分。

答卷前，考生务必将自己的姓名和考生号、试室号、座位号填写在答题卡上，并用2B 铅笔在答题卡的相应位置填涂考生号。

2．作答第Ⅰ卷时，选出每小题答案后，用2B 铅笔把答题卡上对应题目选项的答案信息点涂黑；如需改动，用橡皮擦干净后，再选涂其他答案。

写在本试卷上无效。

3．第Ⅱ卷必须用黑色字迹的钢笔或签字笔作答，答案必须写在答题卡各题目指定区域内的相应位置上；如需改动，先划掉原来的答案，然后再写上新的答案；不准使用铅笔和涂改液。

不按以上要求作答无效。

4．考生必须保持答题卡的整洁。

考试结束后，将答题卡一并交回，试卷自己保管好。

第Ⅰ卷一、选择题：本题共12小题，每小题5分, 在每小题给出的四个选项中，只有一项是符合题目要求的. 1.若集合{}04A x x =<<，{}42B x x =-<≤，则A B =A.()0 4，B.(]4 2-，C.(]0 2，D.()4 4-，2.若复数z 满足1i 1iz -=-，则z =A.1B.C.2D.3．已知向量(1,2),(2,1),(1,)a b c λ==-=，若()a b c +⊥，则λ的值为A ．3-B ．13-C ．13D ．34．若3sin(2)25πα-= ，则44sin cos αα-的值为 A ．45 B ．35 C ．45-D ．35-5. 函数()f x 在[0,)+∞单调递减，且为偶函数．若(12)f =-，则满足3()1x f -≥-的x 的取值范围是A ．[1,5]B ．[1,3]C ．[3,5]D ．[2,2]-B6．某工厂为提高生产效率，开展技术创新活动，提出了完成某项生产任务的两种新的生产方 式．为比较两种生产方式的效率，选取40名 工人，将他们随机分成两组，每组20人， 第一组工人用第一种生产方式，第二组工人 用第二种生产方式．根据工人完成生产任务 的工作时间（单位：min ）绘制了如右茎叶图： 则下列结论中表述不正确．．．的是 A. 第一种生产方式的工人中，有75%的工人完成生产任务所需要的时间至少80分钟 B. 第二种生产方式比第一种生产方式效率更高 C. 这40名工人完成任务所需时间的中位数为80D. 无论哪种生产方式的工人完成生产任务平均所需要的时间都是80分钟.7. 如图，网格纸上虚线小正方形的边长为1，实线画出的是某几何体 的三视图，则该几何体的体积为 A ．643B ．52C ．1533D ．568．某班星期五上午安排5节课，若数学2节，语文、物理、化学各1节， 且物理、化学不相邻，2节数学相邻，则星期五上午不同课程安排种数为 A ．6 B ．12 C ．24 D ．489. 过双曲线22221(0,0)x y a b a b -=>>两焦点且与x 轴垂直的直线与双曲线的四个交点组成一个正方形，则该双曲线的离心率为 A1BC ．32D ．210. 右图为中国古代刘徽的《九章算术注》中研究“勾股容方”问题的图形，图中△ABC 为直角三角形，四边形DEFC 为它的内接正方形，记正方 形为区域Ⅰ，图中阴影部分为区域Ⅱ，在△ABC 上任取一点，此点取 自区域Ⅰ、Ⅱ的概率分别记为1p 、2p ，则A ．12p p =B ．12p p <C ．12p p ≤D ．12p p ≥11．已知△ABC 中，AB=AC=3，sin 2sin ABC A ∠= ，延长AB 到D 使得BD=AB ，连结CD ，则CD 的长为A ．BCD．12.在平面直角坐标系xOy 中，圆C 经过点(0，1)，(0，3)，且与x 轴正半轴相切，若圆C 上存在点M ，使得直线OM 与直线y kx =(0k >)关于y 轴对称，则k 的最小值为A.C.第Ⅱ卷二、填空题：本题共4小题，每小题5分，共20分．13．命题“对2[1,1],310x x x ∀∈-+->”的否定是 _______；14.已知变量y x ,满足⎪⎩⎪⎨⎧≥≥+-≤-003302y y x y x ，则y x z +=的最小值为 ；15．在曲线()sin cos f x x x =-，(,)22x ππ∈-的所有切线中，斜率为1的切线方程为 ；16．已知圆锥的顶点为S ，底面圆周上的两点A 、B 满足SAB ∆为等边三角形，且面积为轴截面的面积为8，则圆锥的表面积为 .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.17(本小题满分12分) 在ABC ∆中，角A B C ，，的对边分别是a b c ，，.已知sin sin 03b C c B π⎛⎫--= ⎪⎝⎭.(Ⅰ)求角C 的值；(Ⅱ)若4a c ==，，求ABC ∆的面积.18(本小题满分12分) 设数列｛n a ｝的前n 项和为n s ，已知344n n s a =-，*n N ∈． （1)求数列｛n a ｝的通项公式； (2）令2211log log n n n b a a +=，求数列｛n b ｝的前n 项和Tn.19.(本小题满分12分)面BCGF ，如图，三棱台ABC EFG -的底面是正三角形，平面ABC ⊥平2CB GF =，BF CF =.(Ⅰ)求证：AB CG ⊥；(Ⅱ)若BC CF =，求直线AE 与平面BEG 所成角的正弦值.20.（本小题满分12分）已知点P 在椭圆C :22221(0)x y a b a b+=>>上，椭圆C 的焦距为2. （1）求椭圆C 的方程；（2）斜率为定值k 的直线l 与椭圆C 交于A 、B 两点，且满足22||||OA OB +的值为常数，（其中O 为坐标原点）（i ）求k 的值以及这个常数；（ii ）写出一般性结论（不用证明）：斜率为定值k 的直线l 与椭圆22221(0)x y a b a b+=>>交于A 、B 两点，且满足22||||OA OB +的值为常数，则k 的值以及这个常数是多少？21.（本小题满分12分）设函数1()ln f x ax x b x=-++()a b R ∈、， （1）讨论()f x 的单调性；（2）若函数()f x 有两个零点1x 、2x ，求证：121222x x ax x ++>．22（本小题满分10分)已知()32f x x =+.(Ⅰ)求()1f x ≤的解集；(Ⅱ)若()2f x a x ≥恒成立，求实数a 的最大值.2018学年第二学期广东二师附中中段测试高二级试题数学（理）参考答案一、选择题解析: 5. 法一：因函数()f x 在[0,)+∞单调递减，且为偶函数，则函数()f x 在(,0)-∞单调递增，由()(22)1f f =-=-，则23215x x -≤-≤⇒≤≤.故选A.法二：由3()1x f -≥-得()2)3(f x f ≥-或3()(2)x f f ≥--，即303532x x x -≥⎧⇒≤≤⎨-≤⎩或301332x x x -<⎧⇒≤<⎨-≥-⎩，综合得15x ≤≤. 7.由三视图知该几何体为一长方体与一直三棱柱的组合体，其体积为2143414562⨯+⨯⨯⨯=. 8. 第一步：将两节数学捆在一起与语文先进行排列有22A 种排法，第二步：将物理、化学在第一步排后的3个空隙中选两个插进去有23A 种方法，根据乘法原理得不同课程安排种数为222312=A A .9.将x c =代入双曲线得4222b b y y a a =⇒=±，则222b c ac c a a =⇒=-11e e⇒-=，解得e =.10. 法一：设△ABC 两直角边的长分别为,a b ，其内接正方形的边长为x ，由x b x a b -=得abx a b=+， 则122()ab p a b =+，222122211()()ab a b p p a b a b +=-=-=++22()aba b ≥+（当且仅当a b =时取等号）． 法二（特殊法）：设1,2,BC AC ==CD x =，则23x =，故12445,1999p p ==-=，从而排除A 、D ，当△ABC 为等腰直角三角形时12p p =，排除B ，故选C ． 11. 由sin 2sin ABC A ∠=结合正弦定理得1322BC AC ==，在等腰三角形ABC 中，311c o s 434ABC ∠=⨯=，从而1co s 4D B C ∠=-，由余弦定理得：2222cos CD BD BC BD BC DBC =+-⋅⋅∠272=，故2CD =.12.（略）二、填空题解析：15.设切点为00(,)x y ，则由000'()cos sin 1f x x x =+=且0(,)22x ∈-，得00x =，01y =-，故所求的切线方程为10x y --=（或1y x =-）.16. 设圆锥母线长为l ,由SAB ∆为等边三角形，且面积为24l =⇒=，又设圆锥底面半径为r ，高为h ，则由轴截面的面积为8得8rh =，又2216r h +=，解得r =（或设轴截面顶角为S ，则由21sin 82l S =得90S =︒，可得圆锥底面直径2r =，）故 2=1)S rl r πππ+=表.三、解答题17.(本小题满分12分) 解: (Ⅰ)∵sin sin 03b C c B π⎛⎫--= ⎪⎝⎭，∴1sin sin sin sin 022B C C C B ⎛⎫--= ⎪ ⎪⎝⎭，∴1sin 02C C +=，∴sin 03C π⎛⎫+= ⎪⎝⎭.∵()0C π∈，，∴23C π=. …………………………6分(Ⅱ)∵2222cos ca b ab C =+-，∴24120b b +-=，∵0b >，∴2b =，∴11sin 2422S ab C ==⨯⨯=. …………………………12分18(本小题满分12分)．解：（1）∵344n n s a =-， ①∴ 当n ≥2时，11344n n S a --=-．② ………………………………………2分 由①-②得1344n n n a a a -=-，即14n n a a -=(n ≥2)． ………………………3分当n =1时，得11344a a =-，即14a =．∴ 数列{a n }是首项为4，公比为4的等比数列．……………………………5分 ∴ 数列{a n }的通项公式为4n n a =． …………………………………………6分 （2）∵ 2211log log n n n b a a +=⋅=1221log 4log 4n n +⋅=1111()2(22)41n n n n =-⋅++． …………………………………8分∴ 数列{b n }的前n 项和123n n T b b b b =+++⋅⋅⋅+11111111[(1)()()()]4223341n n =-+-+-+⋅⋅⋅+-+ 11(1)414(1)n n n =-=++． ………………………12分19.(本小题满分12分)解：(Ⅰ)取BC 的中点为D ，连结DF .由ABC EFG -是三棱台得，平面ABC ∥平面EFG ，从而//BC FG . ∵2CB GF =，∴//CD GF =， ∴四边形CDFG 为平行四边形，∴//CG DF . ∵BF CF =，D 为BC 的中点， ∴DF BC ⊥，∴CG BC ⊥.∵平面ABC ⊥平面BCGF ，且交线为BC ，CG ⊂平面BCGF ， ∴CG ⊥平面ABC ，而AB ⊂平面ABC ，∴CG AB ⊥. ………………………5分 (Ⅱ)连结AD .由ABC ∆是正三角形，且D 为中点得，AD BC ⊥. 由(Ⅰ)知，CG ⊥平面ABC ，//CG DF ，∴DF AD DF BC ⊥⊥，，∴DB DF DA ，，两两垂直.以DB DF DA ，，分别为x y z ，，轴，建立如图所示的空间直角坐标系D xyz -. 设2BC =，则A(0 0 ，，E(12-)，B (1，0，0)，G (-1，0)，∴12AE ⎛=- ⎝⎭，()2 0BG =-，32BE ⎛=- ⎝⎭. 设平面BEG 的一个法向量为()n x y z =，，.由00BG n BE n ⎧⋅=⎪⎨⋅=⎪⎩可得，20 302x x ⎧-=⎪⎨-=⎪⎩，.令x =21y z ==-，，∴()3 2 1n =-，，.设AE 与平面BEG 所成角为θ，则6sin cos AE n AE n AE nθ⋅=<>==⋅，.………12分 20. (本小题满分12分) 解：（1）由点P 在椭圆上得223112a b +=，2c =2，-----------------------------1分 2222322b a a b ∴+=，c =1，又222a b c =+， 222232(1)2(1)b b b b ∴++=+，422320b b ∴--=，解得22b =，得23a =，∴椭圆C 的方程为22132x y +=；----------------------------------------------4分 （2）（i ）设直线l 的方程为y kx t =+，联立22132x y +=， 得222(32)6360k x ktx t +++-=，∴2121222636(1)(2)3232ktt x x x x k k -+=-=++----------------------5分又22112(1)3x y =-，22222(1)3x y =-， 2222221122||||()()OA OB x y x y +=+++22121()43x x =++212121[()2]43x x x x =+-+ 22221636[()2]433232kt t k k -=-⨯+++ 222221(1812)362443(32)k t k k -++=⨯++--------------------------------8分 要使22||||OA OB +为常数，只需218120k -=，得223k =，---------------9分∴22||||OA OB +212424453(22)+=⨯+=+，∴k ==，这个常数为5；-------------------------------10分（ii ）b k a=±，这个常数为22a b +．----------------------------------------12分 21. 解：(本小题满分12分)（1）222111'()(0)ax x f x a x x x x --=--=>，----------------------------------1分 设2()1(0)g x ax x x =-->，①当0a ≤时，()0g x <，'()0f x <；----------------------------------------------2分 ②当0a >时，由()0g x =得12x a +=或102x a-=<，记12x a+=0x =则20()1()(0)g x ax x a x x x x =--=->，∵0x >∴当0(0,)x x ∈时，()0g x <，'()0f x <，当0(,)x x ∈+∞时，()0g x >，'()0f x >，--------------------------------------4分 ∴当0a ≤时，()f x 在(0,)+∞上单调递减； 当0a >时，()f x在1(0,2a+上单调递减，在1(,)2a ++∞上单调递增．-----5分 （2）不妨设12x x <，由已知得1()0f x =，2()0f x =， 即1111ln ax x b x =--，2221ln ax x b x =--，---------------------------------------6分 两式相减得21212111()ln ln ()a x x x x x x -=---， ∴212121ln ln 1x x a x x x x -=+-，----------------------------------------------7分要证121222x x ax x ++>， 即要证2112122121ln ln 122()x x x x x x x x x x -++>+-，只需证21121221ln ln 2x x x x x x x x -+>⋅⋅-，只需证222121212ln x x xx x x ->，即要证2121212ln x x x x x x ->，---------------------------9分设21x t x =，则1t >，只需证12ln t t t->，-------------------------------------------10分 设1()2ln (1)h t t t t t=-->，只需证()0h t >，222221221(1)'()10t t t h t t t t t -+-=+-==>， ()h t ∴在(1,)+∞上单调递增，()(1)0h t h ∴>=，得证．---------------12分22. (本小题满分10分)解：(Ⅰ)由()1f x ≤得，|32|1x +≤，所以，1321x -≤+≤，解得113x -≤≤-， 所以，()1f x ≤的解集为113⎡⎤--⎢⎥⎣⎦，. …………………………5分 (Ⅱ)()2f x a x ≥恒成立，即232+≥x a x 恒成立.当0x =时，a R ∈；当0x ≠时，23223+≤=+x a x x x.因为23x x +≥(当且仅当23x x =，即x =)，所以a ≤a的最大值是…………………………10分。

2018-2019学年广东省广州市华南师大附中高二（上）期中物理试卷试题数：17.满分：1001.（单选题.3分）两带电小球A、B.A固定不动.质量为m的B只在库仑力作用下由静止开始运动．已知初始时A、B间的距离为d.B的加速度为a；经过一段时间后.B的加速度变为a4、速度变为v.此时A、B间的距离为（）A. d2 .此过程中电势能增加12mv2B.2d.此过程中电势能增加12mv2C.2d.此过程中电势能减少12mv2D. d2 .此过程中电势能减少12mv22.（单选题.3分）一带负电的粒子只在电场力作用下沿x轴正向运动.其电势能E p随位移x变化的关系如图所示.其中0-x2段是关于直线x=x1对称的曲线.x2-x3段是直线.则下列说法正确的是（）A.x1处电场强度最小.但不为零B.x2～x3段电场强度大小方向均不变.为一定值C.粒子在0～x2段做匀变速运动.x2～x3段做匀速直线运动D.在0、x1、x2、x3处电势φ0、φ1、φ2、φ3的关系为φ3＞φ2=φ0＞φ13.（单选题.3分）有一条横截面积为S的铜导线.通过的电流为I．已知铜的密度为ρ.摩尔质量为M.阿伏加德罗常数为N A.电子的电量为e．若认为导线中每个铜原子贡献一个自由电子.则铜导线中自由电子定向移动的速率可表示为（）A. IMρSN A eB. IρMSN A eC. IMSρN A eD. IMρSN A e4.（单选题.3分）关于电源的电动势.下列说法中正确的是（）A.电源电动势的大小.等于非静电力在电源内部把正电荷从负极送到正极所做的功的大小B.电动势公式E= Wq 中W与电压U= Wq中的W是一样的.都是电场力做的功C.电源不接入电路时.电源两极间的电压大小等于电动势D.电源的电动势越大.电源所能提供的电能就越多5.（单选题.3分）如图所示电路中.当滑动变阻器R3的滑片P向上端a滑动时.电流表A1、A2及电压表V的示数的变化情况是（）A.电流表A1增大.A2减小.电压表V减小B.电流表A1减小.A2增大.电压表V减小C.电流表A1增大.A2增大.电压表V增大D.电流表A1减小.A2减小.电压表V增大6.（单选题.3分）两根同种材料制成的电阻丝a和b.a的长度和横截面的直径均为b的两倍.要使两电阻丝接入电路后消耗的电功率相等.加在它们两端的电压之比U a：U b为（）A.1：1B.2：1C. √2：1D.1：√27.（多选题.4分）下列说法正确的是（）A.汤姆生用油滴实验测定了电子的电荷量B.库仑通过扭秤实验确认了点电荷间的作用力与电荷量、距离之间的关系C.库仑用实验测出了静电力常量的值为1.60×10-10N•m2/C2D.法拉第采用了电场线的方法形象地描述电场8.（多选题.4分）如图所示.虚线a、b、c代表电场中一簇等势线.相邻等势线之间的电势差相等.实线为一带电质点（重力不计）仅在电场力作用下通过该区域时的运动轨迹.P、Q是这条轨迹上的两点.据此可知（）A.a、b、c三条等势线中.a的电势最高B.电场中Q点处的电场强度比P点处小C.该带电质点在P点处的速率比在Q点处大D.该带电质点在P点具有的电势能比在Q点具有的电势能大9.（多选题.4分）等量异种点电荷的连线和其中垂线如图所示.现将一个带负电的检验电荷先从图中a点沿直线移到b点.再从b点沿直线移到c点．下列说法正确的是（）A.从a点到b点.电势逐渐升高B.从a点到b点.检验电荷受电场力先增大后减小C.从a点到c点.检验电荷所受电场力的方向始终不变D.从a点到c点.检验电荷的电势能先不变后减小10.（多选题.4分）如图所示.一电子沿Ox轴射入电场.在电场中的运动轨迹为OCD.已知OA= AB .电子过C、D两点时竖直方向的分速度为v Cy和v Dy；电子在OC段和OD动能变化量分别为ΔE K1和ΔE K2.则（）A.v cy：v Dy=1：2B.v cy：v Dy=1：4C.ΔE k1：ΔE k2=1：3D.ΔE k1：ΔE k2=1：411.（多选题.4分）一个带正电的微粒放在电场中.场强的大小和方向随时间变化的规律如图所示．带电微粒只在电场力的作用下.由静止开始运动.则下列说法中正确的是（）A.微粒在0～1 s内的加速度与1～2 s内的加速度相同B.微粒将沿着一条直线运动C.微粒做往复运动D.微粒在第1 s内的位移与第3 s内的位移相同12.（多选题.4分）如图所示.电容器极板间有一可移动的电介质板.介质与被测物体相连.接入电路后.通过极板上物理量的变化可确定被测物体的位置.则以下说法正确的是（）A.若电容器极板间的电压不变.x变大.电容器极板上带电量减小B.若电容器极板电量不变.x变小.电容器极板间电压变小C.若电容器极板间的电压不变.x变大.有电流流向电容器的正极板D.若电容器极板间的电压不变.x变大.有电流流向电容器的负极板13.（问答题.0分）在“描绘小灯泡的伏安特性曲线”的实验中.需测量一个标有3V.1.5W”灯泡两端的电压和通过灯泡的电流.现有如下器材：直流电源（电动势3.0V.内阻不计）开关导线若干电流表A1（量程3A.内阻约0.1Ω）电压表V1（量程3V.内阻约3kΩ）电流表A2（量程0.6A.内阻约5Ω）电压表V2（量程15V.内阻约200kΩ）滑动变阻器R1（阻值0～10Ω.额定电流1A）滑动变阻器R2（阻值0～1kΩ.额定电流300mA）（1）请在甲图方框内完成电路图.并在乙图中将相应的实验电路图连接好．（2）实验中电压表应选用 ___ .电流表应选用 ___ .滑动变阻器应选用 ___ ．（用对应序号的字母表示）（3）图丙是用上述方法测定某电路元件的伏安特性曲线数据.由图线可知该元件的电阻随着电压的增大而 ___ （选填“变大”、“变小”或“不变”）.在电压为U1时.电阻值为 ___ ．（4）某同学在实验时的电压表和电流表如图丁所示.其中电压表接0～3V量程.电流表接0～0.6A量程.则电压表的读数为 ___ .电流表的读数为 ___ ．（5）由于实验用的电流表损坏.所以需将一量程为I g3=600μA的表头改装成量程为0～0.6A的电流表.已知其内阻为R=999Ω.则需要 ___ （填“串联”或“并联”）一个值为 ___ 的定值电阻．14.（问答题.0分）同学们要测量某一均匀新材料制成的圆柱体的电阻率ρ。

广东省第二师范学院番禺附属中学2018-2019学年高二英语上学期期中试题本试卷分四部分，共10页，满分150分。

考试时间120分钟。

注意事项：1．答卷前，考生务必在答题卡上用黑色字迹的钢笔或签字笔填写自己的测试证号和姓名；填写考区考场室号、座位号，再用2B铅笔把对应的两个号码涂黑。

2．选择题每小题选出答案后，用2B铅笔把答题卡上对应题目的答案标号涂黑；如需改动，用橡皮擦干净后，再选涂其他答案；不能答在试卷上。

3．非选择题必须用黑色字迹的钢笔或签字笔作答。

答案必须写在答题卡各题目指定区域内的相应位置上；如需改动，先划掉原来的答案，然后再写上新的答案；不准使用铅笔、圆珠笔和涂改液。

不按以上要求作答的答案无效。

4．考生必须保持答题卡的整洁，考试结束时，将答题卷和答题卡一并交回。

I 听力 (共两节，满分20分)第一节听力理解 (4段共10小题，每小题1.5分，满分15分) 每段播放两遍。

各段后有几个小题，各段播放前每小题有5秒钟的阅题时间。

请根据各段播放内容及其相关小题，在5秒钟内从题中所给的A、B、C项中，选出最佳选项，并在答题卡上将该项涂黑。

听下面一段对话，回答第1~2题。

1. Who are the speakers waiting for?A. May.B. May’s parents.C. May’s teachers.2. What drinks will the speakers order?A. Lemon tea, green tea and coffee.B. Green tea, coffee and orange juice.C. Lemon tea, green tea and orange juice.听下面一段对话，回答第3~5题。

3. What was the man doing before he spoke with the woman?A. Talking on the phone.B. Reading the newspaper.C. Walking to the university.4. How much is the man’s rent now each week?A. $40.B. $80.C. $120.5. What does the man NOT like about the new room?A. The location.B. The noise.C. The size.听下面一段对话，回答第6~8题。

密★启用前试卷类型:A2018-2019学年度第二学期二师附中高二期中教学质量检测文科数学★祝考试顺利★ 注意事项：1、答题前，请先将自己的姓名、准考证号用0.5毫米黑色签字笔填写在试题卷和答题卡上的相应位置，并将准考证号条形码粘贴在答题卡上的指定位置。

用2B 铅笔将答题卡上试卷类型A 后的方框涂黑。

2、选择题的作答：每个小题选出答案后，用2B 铅笔把答题卡上对应题目的答案标号涂黑。

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4、选考题的作答：先把所选题目的题号在答题卡上指定的位置用2B 铅笔涂黑。

答案用0.5毫米黑色签字笔写在答题卡上对应的答题区域内，写在试题卷、草稿纸和答题卡上的非选修题答题区域的答案一律无效。

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一、选择题：本题共12小题，每小题5分，共60分。

在每小题给出的四个选项中，只有一项是符合题目要求的。

1. 已知集合{|12}A x x =-≤≤，{1,2,3}B =，则A B =（A ）{1} （B ）{2}（C ）{1,2}（D ）{1,2,3}2.已知a 为实数，若复数()()1a i i +-为纯虚数，则a = A.1- B.12-C. 1D.2 3.已知3sin()5πθ +=，则sin(2)2πθ -= A.45 B.725- C. 725D.354.已知等差数列{}n a 的前n 项和为n S ，且23415a a a ++=，713a =，则5S =（ ）A) 28 (B) 25(C) 20(D) 185、设x ，y 满足约束条件030426x y x y ≤≤⎧⎪≤≤⎨⎪+≥⎩，则3z x y =+的最大值为（A ）7 （B ）9 （C ）13 （D ）156、如图所示程序框图，若判断框内为“3i ≤”，则输出S =（ ） A ．2 B ．6 C. 10 D ．347、设x ∈R ，则“20x -≥”是“11x -≤”的（ ） A ．充分而不必要条件 B ．必要而不充分条件 C ．充要条件 D ．既不充分也不必要条件8、如图，某几何体的三视图是三个全等的等腰直角三角形，且直角边长都等于1，则该几何体的外接球的体积为（ ） （A) 12π(B)2 (C) 3π (D) 43π 9、刘徽是我因魏晋时期的数学家，在其撰写的《九章算术注》中首创“割圆术”，所谓“割圆术”，是用圆内接正多边形的面积去无限逼近圆面积并以此求取圆周率的方法，如图所示，圆内接正十二边形的中心为圆心O ，圆O 的半径为2，现随机向圆O 内段放a 粒豆子，其中有b 粒豆子落在正十二边形内(,,a b N b a *∈<)，则圆固率的近似值为A.b a B.a b C.3a b D.3ba10、已知1F 和2F 分别是双曲线()222210,0x y a b a b-=>>的两个焦点，A 和B 是以O 为圆心，以1OF 为半径的圆与该双曲线左支的两个交点，且2F AB 是等边三角形，则该双曲线的离心率为 ( )(A)(B) 1(C)1 (D) 211.函数()2sin f x x x x =+的图象大致为12．已知函数()1,0()ln ,0kx x f x x x ->⎧⎪=⎨--<⎪⎩，若函数()f x 的图象上关于原点对称的点有2对，则实数k 的取值范围是（ ）(A) (,0)-? (B) 1(0,)2(C) (0,)+? (D) (0,1) 二、填空题：本题共4小题，每小题5分，共20分。

2019-2020学年广州大学附属中学大学城校区高三英语期中考试试卷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AThe COVID -19 pandemic has affected all aspects of life, including the way we travel. But for those who are looking to expand their horizons while still staying safe, the following three travel trends in 2021 may provide inspirations. Let’s take a look.StaycationWith many travel restrictions during the pandemic, people preferred traveling to nearby places in 2020. This trend continues in 2021. According to search data, 62 percent of people are interested in taking a vacation within driving distance of home. People who live in large cities want to get back in touch with nature. Travelers are looking for places different from their everyday accommodations, for example, farm stays, villas and cottages.Pod travelWhile 2020 saw a rise in solo travel and isolated adventures, 2021 shows that people want to be more connected. “Pod travel”, or gathering in isolated spaces with loved ones, is growing in popularity. 85 percent of survey respondents favor traveling with family or friends, and over half of the trips searched include three or more people. Pod travel is here to stay for those who want to safely be together while reducing risks associated with socializing with others.Remote working and travelingMany people worked and learned from home in 2020 because of the pandemic. Remote working blurs the line between working and traveling. There was a 128 percent increase in the mention of phrases such as “relocation”, “relocate”, “remote work” and “trying a new neighborhood”. People are actively booking longer stays (e. g. two plus week trips) in small to mid—size cities with access to immersive natural surroundings and wide—open spaces.1.What can we learn about Staycation?A.Travelling to the countryside.B.Taking an isolated adventure.CHaving holidays in nearby places. D.Staying indoors all by oneself.2.What’s special about Pod travel?A.Traveling alone.B.Traveling far away.C.Traveling while working.D.Traveling with loved ones.3.Where might we find the text in a magazine?A.Medicine.cation.C.TourismD.Career.BHappiness is not a warm phone, according to anew study exploring the link between young life satisfaction and screen time. The study was led by professor of psychology Jean M. Twenge at San Diego State University (SDSU).To research this link, Twenge, along with colleagues Gabrielle Martin at SDSU and W. Keith Campbell at the University of Georgia, dealt with data from the Monitoring the Future (MtF) study, a nationally representative survey of more than a million U. S. 8th-, 10th-, and 12th-graders. The survey asked students questions about how often they spent time on their Phones, tablets and computers, as well as questions about their face-to-face social interactions and their overall happiness.On average found that teens who spent more time in front of screen devices — playing computer games, using social media, texting and video chatting — were less happy than those who invested more time in non-screen activities like sports, reading newspapers and magazines, and face-to-face social interactions."The key to digital media use and happiness is limited use," Twenge said. "Aim to spend no more than two hours a day on digital media, and try to increase the amount of time you spend seeing friends face-to-face and exercising — two activities reliably linked to greater happiness."Looking at historical trends from the same age groups since the 1990s, it's easy to find that the increase of screen devices over time happened at the same time as a general drop-off in reported happiness inU. S.teens. Specifically, young peopled life satisfaction and happiness declined sharply after 2012. That's the year when the percentage of Americans who owned a smartphone rose above 50 percent. By far the largest change in teens' lives between 2012 and 2016 was the increase in the amount of time they spent on digital media, and the following decline in in-person social activities and sleep.4. Which method did Twenge's team use for the study?A. Calculating students' happiness.B. Asking students certain questions.C. Analyzing data from a survey.D. Doing experiments on screen time.5. How does the author develop the finding of the study in paragraph 3?A. By making a comparison.B. By giving an example.C. By making an argument.D. By introducing a concept.6. What is the purpose of the last paragraph?A. To draw a conclusion from the study.B. To offer some advice to the readers.C. To prove social activities' importance.D. To support the researchers' finding.7. Which of the following can be the best title for the text?A. Quitting Phones Equals HappinessB. Screen Time Should Be BannedC. Teens' Lives Have Changed SharplyD. Screen-addicted Teens Are UnhappierCWhen I was a boy, there was but one permanent ambition among my comrades in our village on the west bank of Mississippi River. That was, to be a steamboat man. We had temporary ambitions of other sorts, but they were only temporary.My father was a justice of the peace, and I supposed he possessed the power of life and death over all men and could hang anybody that offended him. This was distinction enough for me as a general thing；butthe desire to be a steamboat man kept intruding, nevertheless. One of our boys in town, who went away and was not heard of for-a long time, turned up as apprentice engineer on a steamboat. This thing shook the bottom out of all my Sunday—school teachings. That boy was notoriously worldly, and I was just the opposite. There was nothing generous about this fellow in his greatness. He would always manage to have a rusty nail to scrub while his boat stopped at our town, and he would sit on the inside guard and scrub it, where we could all see him. And wherever his boat was laid up he would come home and show off in the town in his blackest and greasiest clothes, so that nobody could help remembering that he was a steamboat man; and he used all sorts of steamboat technical terms in his talk, as if he were so used to them that he forgot common people could not understand them.This creature's career could produce but one result, and it was speedily followed. Boy after boy managed to get on the river. Despite many choices, pilot was the grandest position of all. The pilot, even in those days of trivial wages, had a princely salary—from 150—250 dollars a month, and no board payment.But our parents would not let us and our worry was the next year would find us hunting for jobs with low pay again. So by and by I ran away. I said I never would come home again till I was a pilot and could come in glory.8. Why does the writer mention his father's job in Paragraph 2?A. To show that his father was in power.B. To show that his father is cruel.C. To emphasize the job he prefers.D. To emphasize his love for his father.9. Which of the following can best conclude the writer's attitude toward the boy?A. He thought the boy was material but pitiful.B. He thought the boy was annoying but still envied him.C. He thought the boy was shallow but knowledgeable.D. He thought the boy was disrespectful but still liked him.10. Which of the following statements is Not True?A. The boy talked in a way to make others feel jealous.B. The boy's experience made other boys follow suit.C. The pilot's salary was ly high but without meals covered.D. The writer was ambitious to make his childhood dream come true.11. What rhetorical method does the underlined sentence have?A. Simile.B. Personification.C. Parallelism.D. Irony.DSport is not only physically challenging, but can also be mentally challenging. Criticism from coaches, parents and other teammates, as well as pressure to win can create too much anxiety or stress for young athletes. Stress canbe physical, emotional, or psychological and research has indicated that it can lead to burnout. Burnout has been described as dropping or quitting of an activity that was at one time enjoyable.The early years of development are critical years for learning about oneself. The sport setting is one where valuable experiences can take place. Young athletes can, for example, learn how to cooperate with others, make friends, and gain other social skills that will be used throughout their lives. Coaches and parents should be aware, at all times, that their feedback to youngsters can greatly affect their children. Youngsters may take their parents and coaches’ criticisms to heart and find faults in themselves.Coaches and parents should also pay attention that youth sport participation does not become work for children. The outcome of the game should not be more important than the process of learning the sport and other life lessons. In today’s youth sport setting young athletes may be worrying more about who will win instead of enjoying themselves and the sport. Following a game many parents and coaches focus on the outcome and find faults with youngsters’ performances. Positive support should be provided regardless of the outcome. Research indicates that positive support motivates and has a greater effect on learning than criticism. Again, criticism can create high levels of stress, which can lead to burnout.12. An effective way to prevent the burnout of young athletes is_____________.A. to make sport less competitiveB. to increase their sense of successC. to reduce their mental stressD. to make sport more challenging13. According to the passage, sport is positive for young people in that_____________.A. it can help them learn more about societyB. it enables them to find faults in themselvesC. it can provide them with valuable experiencesD. it teaches them how to set realistic goals for themselves14. Many coaches and parents are in the habit of criticizing young athletes_____________.A. believing that criticism does good to their early developmentB. without realizing criticism may destroy their self­confidenceC. in order to make them remember life’s lessonsD. so as to put more pressure on them15. According to the passage parents and coaches should_____________.A. pay more attention to letting children enjoy sportB. help children to win every gameC. train children to cope with stressD. enable children to understand the positive aspect of sport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。

广东省第二师范学院番禺附属中学 2018-2019学年高二数学下学期期中试题理本试卷共4页，22小题， 满分150分。

考试用时120分钟。

注意事项：1.本试卷分第I 卷（选择题）和第H 卷（非选择题）两部分。

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写在本试卷 上无效。

3 •第n 卷必须用黑色字迹的钢笔或签字笔作答，答案必须写在答题卡各题目指定区域内的相应位置上；如需改动，先划掉原来的答案，然后再写上新的答 案；不准使用铅笔和涂改液。

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第I 卷一、选择题：本题共 12小题，每小题5分，在每小题给出的四个选项中，只有一项是符合 题目要求的.1. 若集合 A = 'x 0 v x v 4> , B ={x —4 <x ^2>，贝y A" B =A. 0, 4B.-4,2 1 C. 0, 21 D. -4, 4.彳12. 若复数Z 满足Z —I =1 —「，则z =iA. 1B. .3C.2D.3•已知向量 a =(1,2),b =(2, -1),c =(1, ■)，若(a • b) _ c ，则’的值为44，贝U sin ：- - cos ：-的值为A.A • -3B.D. 3D.2一个正方形，则该双曲线的离心率为D. 25.函数f(x)在［0, •：：)单调递减，且为偶函数•若f(2)-_1，则满足f(x_3)__1的x的取值范围是A . [1,5]B • [1,3]C • [3,5] D. [_2,2]A. 第一种生产方式的工人中，有 75%的工人完成生产任务所需要的时间至少 80分钟B. 第二种生产方式比第一种生产方式效率更高C. 这40名工人完成任务所需时间的中位数为80D. 无论哪种生产方式的工人完成生产任务平均所需要的时间都是 80分钟.且物理、化学不相邻，2节数学相邻，则星期五上午不同课程安排种数为 A. 6 B . 12 C . 24 D . 482 29.过双曲线 笃 与=1(a 0,b 0)两焦点且与a b 6.某工厂为提高生产效率，开展技术创新活动， 提出了完成某项生产任务的两种新的生产方 式.为比较两种生产方式的效率，选取 40名 工人，将他们随机分成两组，每组20人，第一组工人用第一种生产方式，第二组工人 用第二种生产方式•根据工人完成生产任务 的工作时间(单位： min )绘制了如右茎叶图： 则下列结论第聃生产施 | 第二种生产方式8 6 5 5 6 H 9 9 7 6 2 1 {)1223 4 5 6 ^8 9 8 7 7 6 5 4J32 i 14 4 52 110 0 9 fl7.如图，网格纸上虚线小正方形的边长为1,实线画出的是某几何体的三视图，则该几何体的体积为A.64 3B . 52C . 53-D . 563&某班星期五上午安排5节课，若数学2节，语文、物理、化学各1节,x轴垂直的直线与双曲线的四个交点组成ADE10. 右图为中国古代刘徽的《九章算术注》中研究“勾股容方”问题的图形，图中△ ABC为直角三角形，四边形DEFC为它的内接正方形，记正方形为区域I,图中阴影部分为区域□，在厶ABC上任取一点，此点取自区域I、n的概率分别记为p1、p2，贝yA. P i 二P2 B . P i ：：P2 C . P i _ P2 D . P i _ P2ii .已知△ ABC中，则CD的长为AB=AC=3 sin NABC =2sin A，延长AB到D使得BD=AB 连结CDA. 3 32B .班C•乂 D. 362 212.在平面直角坐标系xOy中，圆C经过点（0 , i）,（0 , 3），且与X轴正半轴相切，若圆C上存在点M，使得直线0M与直线y =心（k 0）关于y轴对称，则k的最小值为A.^13 B. 3 C. 2 3 D.4 3第n卷二、填空题：本题共4小题，每小题5分，共20分.13. 命题“对于x€[_i,i],x2+3x_i >0 ”的否定是______________ ；2x - y 乞014. 已知变量x, y满足x-3y，3_0，贝U z = x，y的最小值为___________ ；|沙0i5 .在曲线f（x）二si nx-cosx, x （一了?）的所有切线中，斜率为i的切线方程i6•已知圆锥的顶点为S,底面圆周上的两点A、B满足SAB为等边三角形，且面积为4 - 3，又知圆锥轴截面的面积为8,则圆锥的表面积为 ________ . _____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i7（本小题满分i2分）在厶ABC中，角A, B, C的对边分别是a, b, c.已知 bsin C —二—csin B=0I 3丿 (I )求角C 的值；(n )若 a = 4, c =2, 7，求；_ ABC 的面积.12分)设数列｛ a n ｝的前n 项和为S n ，已知3S n =4a n -4 , n ・N(1)求数列｛ a n ｝的通项公式;19. (本小题满分12分)如图，三棱台 ABC-EFG 的底面是正三角形，平面 ABC_平 面 BCGF ,CB =2GF , BF 二 CF .(I )求证：AB _ CG ；(n )若BC 二CF ,求直线AE 与平面BEG 所成角的正弦值.20. (本小题满分12 分)2 2C : ^2 ^7 = 1(a b 0)上，椭圆C 的焦距为2. a b(1)求椭圆C 的方程;(2)斜率为定值k 的直线l 与椭圆C 交于A B 两点，且满足|OA|2 • |OB|2的值为常18(本小题满分(2)令b n1log 2allog 2a n 1，求数列｛b n ｝的前n 项和Tn.6已知点P (-歹，1)在椭圆数，(其中O为坐标原点)(i )求k的值以及这个常数；(ii )写出一般性结论(不用证明)：斜率为定值k的直线I与椭圆2 2XV 2 22 2 =1(a b 0)交于A B两点，且满足|0A| | OB |的值为常数，贝U k的值以及a b这个常数是多少？21. (本小题满分12分)1 设函数f (x) =ax -In x b (a、b R),x(1)讨论f (x)的单调性；(2)若函数f (x)有两个零点x1、x2，求证：x i x2 2 2ax1x2.22 (本小题满分10分)已知f x;=3x 2 .(I )求f x乞1的解集；(D )若f x2 _a x恒成立，求实数a的最大值.则p-i2ab 2(a b)P 2 =1 _ P 1 =12 22ab a b2 — 2(a b)(a b)2ab (a b)(当且仅当a 二b 时取等号)•2018学年第二学期广东二师附中中段测试高二级试题数学(理)参考答案、选择题解析：5.法一：因函数f(x)在［0,v)单调递减，且为偶函数，则函数 f(x)在(-二，0)单调递增，由f(2) = f (_2) =-1，则-2乞x-3冬2= 1乞x 乞5.故选A.fx-3>0法二：由 f (x -3) _ -1 得 f (x -3) _ f (2)或 f (x -3) _ f(-2)，即=• 3 乞 x ^5(x -3兰 2x 一3 ：： 0或：仁x ：：： 3，综合得1 _ x _ 5 •x -3 一 -27. 由三视图知该几何体为一长方体与一直三棱柱的组合体，其体积为21 42 34 1 4 =56.228. 第一步：将两节数学捆在一起与语文先进行排列有 A 2种排法，第二步：将物理、化学在第一步排后的3个空隙中选两个插进去有A 3种方法，根据乘法原理得不同课程安排种数为2 2 A 2A 3=12.x h — x 10.法一：设厶ABC 两直角边的长分别为 a,b ，其内接正方形的边长为x ，由得a bab x =a +b2b 4.b 2 nt b 2 2 21 ,小y 2 二y - 贝U cac = c - a 二e 1，解aa ae9.将x = c 代入双曲线得2 4 4 5法二（特殊法）：设BC 八AC =2, CD =x ，则x = 3，故P -9，P^^r 9，从而排除A D,当厶ABC 为等腰直角三角形时 p^ p 2，排除B,故选C.1311.由sin._ ABC =2si nA 结合正弦定理得 BC AC，在等腰三角形ABC 中,223 111 cos._ABC， 从而c o ._ DB C，由余弦定理得4 3 4412.(略)n n解析：15.设切点为(X o ,y °)，则由 f'(X o ) =COSX 0• SIn X o =1 且x^ (,)，得 x °= 0，2 2y= -1，故所求的切线方程为 x-y-1^0 （或y^x-1）16.设圆锥母线长为I ,由SAB 为等边三角形，且面积为4、一3得 3|2 =4■■一3= I = 4，4又设圆锥底面半径为r ，高为h ，则由轴截面的面积为 8得rh =8，又r 2 h 2 =16，解得 r =2-2，（或设轴截面顶角为S ,则由ll 2s In S=8得S=90，可得圆锥底面直径22r =4V2 ,）故 S 表 = url +^r 2 =8（72 +1）兀.三、解答题17.（本小题满分12分）2 2 2CD =BD BC -2BD BC cos DBC27，故CD3、62（n \解：（I ）•/ bsin C csin B = 0 ,I 3丿—cosC -sin C sin B = 0 2 丿si nC 2 ••• sin C 二 9I 3丿sinB 】sinC2T C °,二，••. C = —3 ................................ 6 分2 2 2 2(n )••• c 二 a b - 2abcosC , • b 4b - 12 = 0 ,•/ b 0 ,• b = 2 , •- SabsinC2 42卞?3................................... 12 分2 2 218（本小题满分12分）•解：（1 ）••• 3s n =4a n -4 , ① •当n 》2时，3S n 」=4a n 丄一4 •②............................. 2 分由①-②得3a n =4a n -4a n 」，即a . =4a n 」（n 》2） . .............................. 3 分当 n =1 时，得 3a^4a^4，即 a 1=4.•数列｛a n ｝是首项为4,公比为4的等比数列. (5)分数列｛a n ｝的通项公式为a n=4n .1 1 1 1 = ( )•2n (2n2) 4 n n 1数列｛b n ｝的前n 项和T n=b 1b> b 3亠b n=丄［（1 一1）（丄一1）（丄一丄）4：川■（丄 4 2 2 3 3 4 n=丄(1—)n. .........................................12 分4 n 1 4(n 1)19.（本小题满分12分）解：（I ）取BC 的中点为D ，连结DF .由ABC -EFG 是三棱台得，平面 ABC //平面 EFG ，从而BC 〃 FG . •/ CB =2GF , • CD //GF ,•四边形CDFG 为平行四边形，二 CG//DF . ••• BF 二CF , D 为 BC 的中点， • DF _BC , • CG _BC .(21： b n :log 2a n 1 _____ =ga n 11 log2 4n log 2 4n+)]•••平面ABC _平面BCGF，且交线为BC , CG 平面BCGF , • CG丄平面ABC，而AB二平面ABC ,。

2019-2020学年广州大学附属中学大学城校区高三英语上学期期中考试试卷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ARome can be pricey for travelers, which is why many choose to stay in a hostel (旅社). The hostels in Rome offer a bed in a dorm room for around $25 anight, and for that, you’ll often get to stay in a central location (位置) with security and comfort.Yellow HostelIf I had to make just one recommendation for where to stay in Rome, it would be Yellow Hostel. It’s one of the best-rated hostels in the city, and for good reason. It’s affordable, and it’s got a fun atmosphere without being too noisy. As an added bonus, it’s close to the main train station.Hostel Alessandro PalaceIf you love social hostels, this is the best hostel for you in Rome. Hostel Alessandro Palace is fun. Staff members hold plenty of bar events for guests like free shots, bar crawls and karaoke. There’s also an area on the rooftop for hanging out with other travelers during the summer.Youth Station HostelIf you’re looking for cleanliness and a modern hostel, look no further than Youth Station. It offers beautiful furnishings and beds. There are plenty of other benefits, too; it doesn’t charge city tax; it has both air conditioning and a heater for the rooms; it also has free Wi-Fi in every room.Hotel and Hostel Des ArtistesHotel and Hostel Des Artistes is located just a 10-minute walk from the central city station and it’s close to all of the city’s main attractions. The staff is friendly and helpful, providing you with a map of the city when you arrive, and offering advice if you require some. However, you need to pay 2 euros a day for Wi-Fi.1.What is probably the major concern of travelers who choose to stay in a hostel?fort.B.Security.C.Price.D.Location.2.Which hotel best suits people who enjoy an active social life?A.Yellow Hostel.B.Hostel Alessandro Palace.C.Youth Station Hostel.D.Hotel and Hostel Des Artistes.3.What is the disadvantage of Hotel and Hostel Des Artistes?A.It gets noisy at night.B.Its staff is too talkative.C.It charges for Wi-Fi.D.It’s inconveniently located.BIf you've ever had a dog, you know just howdeep a connection you can develop with “man's best friend”. But a dog's life is much shorter than humans, about 12 to 15 years long, which means every dog owner has to go through the heart­breaking moment when their loving pet passes away.Why not make a clone of that dog then? This is the solution offered by a South Korean company, Sooam Biotech Research Foundation. The company has already successfully cloned at least 400 dogs, mostly for US customers, ever since it pioneered the technique in 2005. Now, Sooam Biotech has introduced its business toUKdog owners as well, offering them dogs that look just like their lost ones.To clone a dog, researchers first need to take a skin cell from a living dog or one that has just died. Meanwhile,another dog is selected to supply an egg. Researchers then replace the DNA in the egg with that from the skin cell and implant the egg into the womb (子宫) of a female dog. The egg grows into a puppy over the following two months. The whole process takes less than a day, but it comes at a shockingly high price — around ￡63,000.But if you can't afford it now, you can also save the cell in a laboratory andaccess it at a later date.However, magical as cloning might sound, there is no guarantee that the cloned dog will be a perfect copy of the original one. Just like identical twins of humans, they share the exactly same DNA but there will still be small differences between them. “The spots on a Dalmatian (斑点狗) clone will be different, for example” Insung Hwang, head of Sooam Biotech, told The Guardian.Dog owners will also have to accept the fact that personality is not “cloneable”. Apart from genes, personality is also determined by upbringing and environment, which are both random elements that cloning technologies simply cannot overcome, Professor Tom Kirkwood atNewcastle University,UK, told The Telegraph.Perhaps bringing our dogs back by cloning is not the best way to remember them after all.Kirkwood, a dog owner himself, pointed out, “An important aspect of our relationship with them is coming to terms with the pain of letting go.”4. What service does Sooam Biotech Research Foundation offer?A. Making copies of pet dogs.B. Giving pet dogs identical twinsC. Helping dogs give birth to more puppies.D.Helping dog owners love their dogs more.5. Which order is correct in the dog cloning process?a. An egg is taken from another dog.b. A skin cell is taken from the pet dog.c. The egg grows into a puppy in two months.d. The egg is placed in the womb of a female dog.e. The DNA in the egg is replaced by the DNA from the skin cell.A.a→d→b→e→c.B. a→e→b→d→cC. b→a→d→e→c.D. b→a→e→d→c.6. What can we learn about dog cloning from the passage?A. It has not been put into practice until recently.B. It is very popular among US andUKpet owners.C. It might not give the owners an exactlysame dog.D. It is very expensive and usually takes half a year to complete.7. What doesKirkwoodthink of dog cloning?A. He disagrees with it.B. He supports it.C. He is curious about it.D. He thinks it unbelievable.CA satellite is about to demonstrate a new way of capturing space junk with magnets for the first time. With the frequency of space launches dramatically increasing in recent years, the potential for a disastrous collision above Earth is continually growing. Now, Japanese orbital clean-up company Astroscale is testing a potential solution.The firm's End-of-Life Services by Astroscale demonstration mission is scheduled to lift off on 20 March aboard a Russian Soyuz rocket. It consists of two spacecraft: a smal “client” satellite and a larger “servicer” satellite, or “chaser”. The smaller satellite is equipped with a magnetic (磁力的) plate which allows the chaser todock withit.The two stacked spacecraft will perform three tests once in orbit, each of which will involve the servicer satellite releasing and then recapturing the client satellite. The first test will be the simplest, with the client satellite drifting a short distance away and then being recaptured. In the second test, the servicer satellite will setthe client satellite tumbling before catching up with it and matching its motion to grab it.Finally, if those two tests go well, the chaser will live up to its name by letting the client satellite float a few hundred metres away before finding it and attaching to it. All of these tests will be performed autonomously, with little to no human input once they are set in motion.“These kinds of demonstrations have never been done before in space - they are very different to, say, an astronaut controlling a robotic arm on the International Space Station,” says Jason Forshaw at AstroscaleUK.“This is more of an autonomous mission.” At the end of the tests, both spacecraft will burn up in Earth's atmosphere.If companies wanted to use this capability, they would have to attach a magnetic plate to their satellites so they could be captured later. Because of the growing space garbage problem, many countries now require firms to have a way to bring back their satellites once they run out of fuel or fail, so this could be a fairly simple likely plan, Forshaw says. Right now, each chaser can only nab one satellite, but Astroscale is working on a version that could drag three or four out of orbit at once.8. Which of the following can replace the underlined word "dock with" in Paragraph 2?A. Deal with.B. Keep up with.C. Join together.D. Crash.9. Why many countries now require firms to have a way to bring back their satellites?A. Because of the growing space waste problem.B. Because the frequency of space launches are dramatically increasing.C. Because they can earn large profits from it.D. Because Astroscale has found a new method of capturing the space garbage.10. What will Astroscale do to solve the space junk problem?A. An astronaut controls a robotic arm on the International Space to capture the “client” satellite.B. Through a magnetic plate remotely controlled by humans on the ground to catch the “client” satellite.C. Finding the "client "satellite and attaching to it with a magnetic plate automatically.D. Tumbling to match the motion of “client satellite the drag three or four satellites out of its orbit into atmosphere.11. What can we infer from the passage?A. People will bum the space junk up in Earth's atmosphere in the future.B. Japan andRussiawill conduct space debris cleanup experiment together.C. These kinds of demonstrations have never been done before.D. The demonstration mission will be divided into three phases.DThe man who invented the World Wide Web a few decades ago is calling for major changes to make it better for humans. In an open letter published on Tuesday, Berners-Lee said that the web was used by half the world's population.Berners-Lee said the web had clearly created great opportunities for humans to progress and had made life easier for millions of people. Actually, it also has offered opportunities to groups traditionally not heard a new voice in society. However, he added that the web had also provided new ways for cheats to commit crimes (犯罪).“Against the background of news stories about how the web is misused, it's understandable that many people feel afraid and unsure if the web is really a force for good,” he wrote.Berners-Lee created a group called the World Wide Web Foundation. He islooking for help from governments, companies and people to become more involved in shaping the web to do more good for humans. His actual plan is called the “Contract (合同) for the Web”.Under this contract, governments are called on to take steps to makesure all people can connect to the Internet and that personal privacy is respected. Businesses are asked to keep the Internet prices low so more people can use the web. In addition, companies should respect privacy and develop technologies that aim to put people first.The plan also calls on people to create materials for the web and work with others to make sure that is rich, quality information for everyone. Besides, people should seek to “build strong communities that respect personal speech and human equality.” “The path to make the Internet better is the responsibility of everyone who uses it,” Bermers-Lee added, “Making big changes will not be easy, but will be very well worth it in the end.”12. What does Berners-Lee think of the World Wide Web?A. It is his greatest regret.B. It stops the progress of humans.C. It needs improving.D. It does more harm than good.13. What's wrong with the web according to Berners-Lee?A. It is misused for bad purposes.B. It is misunderstood by all people.C. It blocks out a new voice in society.D. It is expensive for half the world’s population.14. What are governments called on to do under the “Contract for the Web”?A. Put technology first.B. Create materials for the web.C. Popularize the Internet.D. Make the Internet free of charge.15. What should people do with the Internet in Berners-Lee's opinion?A. Be responsible for it.B. Absolutely reject it.C. Completely rely on it.D. Be unconcerned about it.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学年广州大学附属中学大学城校区高三英语上学期期中考试试卷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ABook reading is certainly one of the most absorbing habits. For young adults who love to read, finding some good books to read is very essential. Writing a book review can help you to improve your language and writing skills.The Book ThiefListed onThe New York Times Children’s Best Seller List for over 100 weeks, The Book Thief by Markus Zusak is the story of a young girl in the Nazi camps set during World War II. So, if you love history and wish to learn how the life was during Adolf Hitler’s time, read this historic book.The Diary of Young GirlEven Anne Frank can not have imagined that her personal diary written during World War II would become such a popular book. It’s a must read that describes the situation of a family in the evils of wars through the eyes of a teenager.Animal FarmAnimal Farm is one of the most popular books by George Orwell. It is just a reflection of the Stalin and World War II period that has been so creatively presented in this book. It is an interesting example of how literature can be used to present conditions common in the society.Adventures of Huckleberry FinnMark Twain’s Adventures of Huckleberry Finn is one of the great American novels in history, and is certainly a great pick for young adults. Young Huck Finn and his mischief along with the color1 ful description of people around theMississippi Rivermake this novel a great book to read.1.Which book describes the author’s own experiences according to this passage?A.The Book ThiefB.The Diary of Young GirlC.Animal FarmD.Adventures of Huckleberry Finn2.What do the first three books have in common?A.All of them are about wars.B.All of them are about farms.C.All of them are intended for history lovers.D.All of them were written during World War II.3.The purpose of this passage is to _________.A.instruct youngsters how to improve skillsB.tell youngsters some wonderful reading habitsC.introduce several good books to youngstersD.give youngsters advice on writing a book reviewBThose who are concerned that robots are taking over the world can rest easy—for now. Though the androids have proved useful at performing ordinary tasks, they are not ready for the greatest time. At least that appears to be the case atJapan’s Henn-na Hotel chain where over half of the robot staff are being replaced by humans.The first location of the unique hotel opened in July 2015 was atNagasaki’s Huis Ten Bosch Theme Park. The hotel’s owner, Hideo Sawada, promised the hotel to be managed primarily by robots. Guests were greeted and checked-in by a dinosaur robot, while a cute android called Churi, placed inside each room, provided information about attractions. Not surprisingly, the lodging, recognized in 2016 as the world’s first robot-staffed hotel by Guinness World Records, drew in curious visitors from all around the world.But as the years have passed, the hotel’s main draw is becoming less novel and more unsatisfactory. Also as the robots are “aging”, they are costing more to repair. Among the 283 androids being replaced are the chain’s two dinosaur receptionists. In addition to scaring young guests, they are also unable to photocopy guests’ passports, forcing human employees to step in each time. Also out are the cute Churi robots, which annoyed guests by interrupting their conversations. For example, one guest told The Wall Street Journal that Churi mistook his snoring for a command and kept asking him to repeat his request all night.Sawada told The Wall Street Journal, “When you actually use robots you realize there are places where they aren’t needed—or just annoy people.” While Sawada may be cutting back on his use of androids, the recently-opened Smart LYZ Hotel and the Fly Zoo Hotel inChina, are run entirely by robots, with not a human in sight. Whether the employees have more competence than those “hired” by the Henn-na Hotel chain remains to be seen.4. What makesJapan’s Henn-na Hotel unique?A. Its robot employees.B. Its advanced equipment.C. Its convenient location.D. Its successful management.5. What is the author’s purpose with the example in paragraph 3?A. To entertain readers.B. To prove Churi’s drawback.C. To introduce Churi’s functions.D. To persuade people not to book the hotel.6. What does the owner ofJapan’s Henn-na Hotel think of his robot staff now?A. Attractive.B. Costly.C. Pioneering.D. Disappointing.7. What is the best title for the text?A. Robots Are Taking Over the World.B. The Boom of Robots-staffed Hotel.C. Robot Staff Are Fired For No Competence.D. The First Robots-staffed Hotel Won Guinness World Record.CJules Verne was born on 8 February1828 inthe French city of Nantes. From an early age, he had a fascination with exploration and discovery. When he was six, his teacher, Madame Sambin, told him stories about her husband, who disappeared while traveling the world on a ship 30 years before. She told her class that he was like Robinson Crusoe, a fictional castaway who lived on a desert island. Verne would later write stories about similar characters.In 1847, Verne was sent by his family to study law at a university in Paris, but he preferred to write novels, poems and plays. After graduating, he realized he wanted to write adventure stories based on science and technology.Thishad never been done before, but Verne was sure that it would be a success. His first story, Five Weeks in a Balloon, was published in September 1862. His career lasted for more than 40 years, during which time he wrote more than 60 gripping stories.To begin with, Verne wrote positive and optimistic books. Many of these were to be his most popular creations. Although some included fantastical elements, they were usually based in scientific fact, making them believable. These happy stories weren’t to last. As he got older, Verne became less confident in the idea that science and technology were always good for the planet. His books started to include more scientists who used technology for their own-sometimes evil - purposes. Verne died on 24 March 1905, but new books continued to be published until 1919. These stories were based on ideas Verne had written about while he was still alive, but featured new characters and plots created by his son, Michel.In the 20th century, his books were translated into more than 140 languages and several successful filmversions were released. His creations have been recognized as an inspiration for many scientists and inventors. Many of the futuristic ideas from his most popular books have since come true.8. What is the purpose of Paragraph 1?A To show Verne’s discovery.B. To offer the background information of Verne.C. To tell of the adventure of Sambin’s husband.D. To explain how Verne began his writing career.9. What does the underlined word “This” refer to in Paragraph 2:A. Writing novels, poems and plays.B. Studying law.C. Graduating from university.D. Wring adventure stories.10. What can be learned about Verne’s late works?A. They covered happy elements.B. They were unbelievable stories.C. They revealed Verne’s doubt on science.D. They were partly written by Berne’s son.11. Which of the following best describes Verne?A. Talented and productive.B. Popular and caring.C. Optimistic and generous.D. Friendly and honest.DWatching what you eat can be easier said than done, but a recent study shows it might not just be about what's on your plate — it could be about how quickly it disappears.Japanese researchers followed 1,083 adultsfor five years, splitting them into three categories based on how quickly they ate: slow, normal, and fast. They also answered a questionnaire at the beginning of the study, sharing their diet, physical activity, and medical history. In the beginning, none of the volunteers had metabolic syndrome (新陈代谢综合征) - meaning at least three risk factors — which can lead to health problems like heart conditions and diabetes.When the participants reported back five years later 84 had been diagnosed (诊断) with metabolic syndrome — and their eating speed was a major predictor, according to the results in the journal Circulation. The fast eaters were 89 percent more likely to have metabolic syndrome than slow and normal eaters. Just 2.3 percent of slow eaters received the diagnosis, compared to 11.6 percent of fast eaters. But that's not all. Fast eaters also saw more weight gain, larger waistlines, and higher blood sugar levels than slow eaters.The researchers saygobblingmakes it easier not to take notice of fullness before your body has a chance to signal you to stop. “So when people eat fast they are more likely to overeat,” said Takayuki Yamaji, MD, study author and cardiologist at Hiroshima University in Japan in a statement.Previous research backs up the weight benefits of slow eating, too. One study of New Zealand women found fast eaters have higher body-mass indexes (指数), and a Chinese study found that both healthy and fat men ate less when told to chew 40 times instead of 15 times before swallowing. Initial research even suggests chewing your food longer could bum more calories - up to about 1,000 extra every month.12. What are the participants divided by?A. Medical history.B. Health condition.C. Physical activity.D. Eating speed.13. Which may be the result of the study?A. Fast eaters are 4 times more likely to have metabolic syndrome.B. Normal and slow eaters don’t have metabolic illness.C. 89% of fast eaters have higher blood pressure.D. Slow caters are healthier than fast eaters.14. What does the underlined word “gobbling” in Paragraph 4 best mean?A. Tasting slowly.B. Digesting quickly.C. Eating greedily.D. Cooking carefully.15. What does the last paragraph tell us?A. The importance of eating speed.B. The advantage of eating slowly.C. The result of a Chinese study.D. Fast eating and overeating.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。

广东省第二师范学院番禺附属中学2018-2019学年高二数学上学期期末考试试题 理本试卷共4页，全卷三大题22小题，满分150分，考试用时120分钟。

注意事项:1.答卷前,考生务必用黑色字迹的钢笔或签字笔将自已的姓名、考试科目、班级和考生号等信息填写在答题卡上，并用2B 铅笔将考号在答题卡相关的区域内涂黑。

2.选择题每小题选出答案后，用2B 铅笔把答题卡对应的答案符号涂黑；如需改动,用橡皮擦干净后,再选涂其它答案，答案不能答在试卷上。

3．非选择题必须用黑色字迹的钢笔或签字笔作答，答案必须写在答题卡各题目指定区域内的相应位置上；如需改动，先划掉原来的答案，然后再写上新的答案；不准使用铅笔和涂改液。

不按以上要求作答的答案无效。

4.考生必须保持答题卡的整洁,考试结束后,将答题卡答卷交给监考老师。

一、选择题：本题共12小题，每小题5分，共60分．在每小题给出的四个选项中，只有一项是符合题目要求的．(1)集合，，则= （ ）}{220A x x x =--≤{}10B x x =-<A B ⋂（A ） (B) (C) （D ） }{1x x ≥{}1x x <-}{11x x -≤<{}21x x -≤<(2)在中，已知，那么一定是（ ）ABC ∆cos cos a B b A =ABC ∆(A)等腰三角形 (B)直角三角形 (C)等腰三角形或直角三角形 (D)等腰直角三角形 (3)已知在上是奇函数,且满足,当时,，()f x R ()()4f x f x +=()0,2x ∈()22f x x =则（ ） （A ） （B ） （C ） （D ）()7f =29898-2-(4)要得到函数sin 43y x π⎛⎫=-⎪⎝⎭的图象，只需要将函数sin 4y x =的图象（ ） （A ）向右平移12π个单位 （B ）向左平移12π个单位（C ）向右平移3π个单位 （D ）向左平移3π个单位(5)已知双曲线（）的渐近线方程为, 则双曲线的：C 22221y x a b -=0,0>>b a x y 21±=C离心率为（ ） (A) (B) (C) (D)626525(6)在上随机取一个实数，能使函数,在上有零点的概率[]4,3-m ()22f x x =+R 为（ ） (A)(B) (C) （D ） 27374757(7)下列命题中，真命题的是（ ） (A) (B) (C)的充要条件是0,0x x R e∃∈≤2,2x x R x ∀∈>0a b +=1ab=-（D ）若，且，则中至少有一个大于1 ,x y R ∈2x y +>,x y(8)《九章算术》中，将底面是直角三角形的直三棱柱称之为“堑堵”，将底面为矩形，一棱垂直于底面的四棱锥称之为“阳马”，某“堑堵”与某“阳马”组合而成的几何体的三视图如图所示，已知该几何体的体积为，则图中（ ）635=x32.2.3.1.D C B A (９)阅读如下程序框图，运行相应的程序，则程序运行后输出的结果为（ ）(A) 7 (B) 9 (C) 10 (D) 11 (１０)甲、乙、丙、丁四位同学一起去向老师询问成语竞赛的成绩。

广东第二师范番禺附属中学2018-2019学年高一上学期期中考试数学试题第Ⅰ卷（共60分）一、选择题：本小题共12题，每小题5分，在每小题给出的四个选项中，只有一项是符合题目要求的. 1. 已知集合，，则等于（ ）A.B.C.D.2. 下列各函数中与函数()f x x =为同一函数的是（ ）A ．2x y x=B. 2y =C. yD.ln e xy =3. 已知函数23,0,()log ,0,x x f x x x ⎧≤=⎨>⎩则12f f ⎡⎤⎛⎫ ⎪⎢⎥⎝⎭⎣⎦的值是（ ） A.1B.13C.1- D. 3-4.下列是函数为奇函数，且在(,)-∞∞上为增函数的是（ ） A.3()f x x = B.()||1f x x =+ C.()21xf x =-D.()lg f x x =5. 函数的零点位于区间( ) A.B.C.D.6. 已知不等式的解集为,则的值为（ ） A. -10 B. -14C.14D.107. 已知，则的大小关系是（ ）A. B. C.D.8. 设函数，，则的最小值和最大值分别为（ ）{}1,2,5M ={|2}N x x =≤M N ⋂{}1{}5{}1,2{}2,5220ax bx ++>1123x x ⎧⎫-<<⎨⎬⎩⎭a b +01a <<22log 2a a a ，，22log 2a a a <<222log a a a <<22log 2aa a <<222log a a a <<()2263f x x x =-+[]1,4x ∈()f xA. -1，11B. -1，3C.，4D.，119. 函数2()ln f x x x =的图象大致是（ ）AB CD10. 某书店对购书者实行优惠，规定：①如一次购书不超过100元，则不予折扣；②如一次购书超过100元但不超过300元的，按九折付款；③如一次购书超过300元的，其中300元按第②条给予优惠，超过300元的部分则按八折付款．某人两次去购书，分别付款88元与243元，如他一次去购买同样的书，则应付款( ) A ．358元B ．326．4元C ．316．4元D ．286．4元11. 已知偶函数()f x 在区间[)0,+∞上是增函数，则(1)f -与2(22)f a a -+的大小关系是（ ）A ．2(1)(22)f f a a --≤+B ．2(1)(22)f f a a --≥+C ．2(1)(22)f f a a --<+D ．2(1)(22)f f a a -->+ 12. 对实数与，定义新运算“”：． 设函数，．若函数的零点恰有两个，则实数的取值范围是（ ）A ．B ．C ．D ．第Ⅱ卷（共90分）二、填空题：本小题共4题，每小题5分.13. 已知幂函数()f x x α=的图象过点(，则函数()f x 的解析式是 . 14. 若函数()f x 与函数()3xg x =互为反函数，则(3)f 的值是________.32-32-⎩⎨⎧-≤-=⊗1＞,1,b a b b a a b a )()2()(22x x x x f -⊗-=R ∈x 311,,44⎛⎫⎡⎫--+∞ ⎪⎪⎢⎝⎭⎣⎭15. 若对任意实数x ∈R ，不等式2230x mx m ++-≥恒成立，则实数m 的取值范围是____. 16已知函数1(0xy a a -=>且1)a ≠的图象恒过点A , 若点A 满足方程()100mx ny mn +-=>上, 则12m n+的最小值为________. 三、解答题：解答题应写出文字说明、证明过程或演算步骤. 17.化简求值：（本小题满分10分）（1）；（2）.18.（本小题满分12分）已知集合{}36A x x =<<，．（Ⅰ）求，；（Ⅱ）已知集合，若，试求实数的取值范围．13134210.064160.258-⎛⎫--++ ⎪⎝⎭3log 22311lg25lg2log 9log 223⎛⎫++-⨯ ⎪⎝⎭{|29}B x x =<<A B ⋂()()R R C A C B ⋃{|21}C x a x a =<<-B C B ⋃=a19.（本小题满分12分） 已知函数是定义在上的偶函数，当时，. （1）画出图象； （2）求出的解析式；（3）观察图象，若函数()y f x =与函数y m =的图象有四个交点时，求的取值范围.20.(本小题满分12分) 已知a ∈R , 函数()()11,0,11,0.x f x xa x x ⎧->⎪=⎨⎪-+≤⎩(1) 求()1f 的值;(2)证明: 函数()f x 在()0,+∞上单调递增; (3)求函数()f x 的零点.()f x R 0x ≥()24f x x x=-()f x ()f x m21.(本小题满分12分) “足寒伤心，民寒伤国”，精准扶贫是巩固温饱成果、加快脱贫致富、实现中华民族伟大“中国梦”的重要保障．某地政府在对石山区乡镇企业实施精准扶贫的工作中，准备投入资金将当地农产品进行二次加工后进行推广促销，预计该批产品销售量Q 万件（生产量与销售量相等）与推广促销费x 万元之间的函数关系为24x Q +=（其中推广促销费不能超过3万元）．已知加工此批农产品还要投入成本14Q Q ⎛⎫+⎪⎝⎭万元（不包含推广促销费用），若加工后的每件成品的销售价格定为204Q ⎛⎫+⎪⎝⎭元/件． （1）试将该批产品的利润y 万元表示为推广促销费x 万元的函数； （利润=销售额-成本-推广促销费）（2）当推广促销费投入多少万元时，此批产品的利润最大？最大利润为多少？22、(本小题满分12分) 设(R ). (1)若，当2x <时，求的单调递增区间；(2)若，写出的单调区间；(3)若存在，使得方程有三个不相等的实数解，求的取值范围.()2f x x x a x =-+a ∈2a =()f x 2a >()f x []2,4a ∈-()()f x tf a =t【参考答案】一、选择题二、填空题13. ()12f x x=或()f x=14. 1 15. [2，6] 16.3+三、解答题17.解：（1）. （2）lg25+lg2+-23log9log2⨯=lg5+lg2+-()232log3log2⨯⨯=1+-2=.18.解：（Ⅰ）∵集合，，∴，∴；（Ⅱ）∵，且，∴，当时，，计算得出；当时，计算得出；综上，实数的取值范围是或.19.解：(1)1313421510.064160.251810822-⎛⎫--++=-++=⎪⎝⎭123log213⎛⎫⎪⎝⎭3log23-1212-{|36}A x x=<<{}|29B x x=<<{|36}A B x x⋂=<<()()()R R RC A C B C A B⋃=⋂{|36}x x x=≤≥或{|21}C x a x a=<<-B C B⋃=C B⊆C=∅21a a≥-1a≤C≠∅25a≤≤a1a≤25a≤≤（2）当x <0时-x >0,,为偶函数,, .（3）最小值为,由（1）问图像可知若函数()y f x =与函数y m =的图象有四个交点时,.20. (1)解: 当0x >时, ()11f x x =-,∴()11101f =-=. (2)证明:在()0,+∞上任取两个实数12,x x ,且12x x <, 则()()12121111f x f x x x ⎛⎫⎛⎫-=--- ⎪ ⎪⎝⎭⎝⎭2111x x =-1212x x x x -=. ∵120x x <<,∴12120,0x x x x -<>. ∴12120x x x x -<, 即()()120f x f x -<.∴()()12f x f x <. ∴函数()f x 在()0,+∞上单调递增. (3) (ⅰ)当0x >时, 令()0f x =, 即110x-=, 解得10x =>. ∴1x =是函数()f x 的一个零点.（ⅱ）当0x ≤时, 令()0f x =, 即()110a x -+=.(※) 当1a >时, 由(※)得101x a =<-,∴11x a=-是函数()f x 的一个零点; 当1a =时, 方程(※)无解; 当1a <时, 由(※)得101x a=>-,(不合题意,舍去). 综上所述, 当1a >时, 函数()f x 的零点是1和11a-; 当1a ≤时, 函数()f x 的零点是1. 21.解：（1）20144y Q Q x Q Q ⎛⎫⎛⎫=+⋅-+- ⎪ ⎪⎝⎭⎝⎭ 420x Q =--()1620,032x x x =--≤≤+.（2）由（1）得()161620222221422y x x x x ⎡⎤=--=-++≤=⎢⎥++⎣⎦， 当且仅当1622x x =++且03x ≤≤，即2x =时等号成立. 当2x =时，max 14y =. 答：当推广促销费投入2万元时，利润最大为14万元.22.解：（1）当，当2x <时，()24f x x x =-+, 对称轴：()42221b x a =-=-=⨯-, 0a <抛物线开口向下.的单调递增区间是(],2-∞.（2）,,,当时，，在为增函数,当时，，即,在为增函数，在为减函数, 则的单调增区间为和，单调减区间. （3）由（2）可知，当时，为增函数，方程不可能有三个不相等实数根,当时，由（2）得,,即在有解, 由在上为增函数,当时，的最大值为,则.2a =∴()f x ()()()222,{2,x a x x af x x a x x a-++<=+-≥2a >022a a a ∴<-<<+x a ≥22a a ->∴()f x (),a +∞x a <22022a a a +--=<22a a+<∴()f x 2,2a +⎛⎫-∞ ⎪⎝⎭2,2a a +⎛⎫ ⎪⎝⎭()f x 2,2a +⎛⎫-∞ ⎪⎝⎭(),a +∞2,2a a +⎛⎫ ⎪⎝⎭22a -≤≤()f x 24a <≤()()22a f a tf a f +⎛⎫<< ⎪⎝⎭()22224a a at +<<()2218a t a+<<(]2,4()22118822a a aa +=++(]2,4∴4a =()228a a+98918t <<。