2018届高三年级华附、省实、深中、广雅四校联考
广东省华附+深中+省实+广雅四校联考2023-2024学年高三上学期1月期末数学试题

华附、省实、广雅、深中2024届高三四校联考数学命题学校:广东实验中学 定稿人:杨晋鹏 张淑华本试卷分选择题和非选择题两部分,共4页,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己的姓名、考号填写在答题卷上。
2.选择题每小题选出答案后,用2B 铅笔把答题卷上对应题目的答案标号涂黑;如需改动,用橡皮擦干净后,再选涂其它答案;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,答案必须写在答题卷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,将答题卷收回。
一.单项选择题(本大题共8小题,每小题5分,共40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.已知全集U =R ,集合A ,B 满足A ⊆(A⋂B),则下列关系一定正确的是( )A. A =BB. B ⊆AC. (∁U A)∩B =⌀D. A⋂(∁U B)=⌀2.已知复数z 满足i z i −=+1)1(,则z 2024=( )A. iB. −1C. 1D. −i3.直线x +2y +3=0关于直线y =−x 对称的直线方程是( )A. x +2y −3=0B. 2x +y −3=0C. x −2y −3=0D. 2x +3y +3=04.已知向量a 在b 方向上的投影向量的模为2,向量b 在a 方向上的投影向量的模为1,且)32)b a b a −⊥+((,则向量a 与向量b 的夹角为( )A .6πB .4πC .3πD .43π 5.若椭圆Γ1:x 2a 2+y 2b2=1(a >b >0)的离心率为12,则双曲线Γ2:y 2b2−x 2a 2=1的离心率为( )A.321B.27 C. √ 3 D. √ 56. 在平直的铁轨上停着一辆高铁列车,列车与铁轨上表面接触的车轮半径为R ,且某个车轮上的点P 刚好与铁轨的上表面接触,若该列车行驶了距离S ,则此时P 到铁轨上表面的距离为( ) A .)cos 1(RS R +B .)cos 1(R S R −C .R S R sin 2D .RS R sin7.若1ln )1)1=−=−b c e c a((则a ,b ,c 的大小关系为( ) A . c ≤a <bB . c <a <bC .c <b <aD .b <a ≤c8.数列}{n a 的前n 项和n S ,且1112881−−−++=n n n n a n a a a ,),2(+∈≥N n n ,若11=a ,则 A .3252024<<S B .2522024<<S C .2232024<<S D . 2312024<<S 二.多项选择题(本题共4小题,每小题5分,共20分.在每小题给出的四个选项中,有多个选项是符合题目要求的,全部选对的得5分,选对但不全的得2分,有选错的得0分) 9.下列结论正确的是( )A. 若a >b,c >d ,则ac 2>bd 2B. 若ac 2>bc 2,则a >bC. “ab >1”是“a >1,b >1”成立的充分不必要条件D. 若a >b >1,则)1(log log 1+<+b b a a10. 已知圆C 1:122=+y x ,圆C 2:222)4()3(r y x =++−)(0>r ,P 、Q 分别是圆C 1与圆C 2上的点,则( )A .若圆C 1与圆C 2无公共点,则0<r <4B .当r =5时,两圆公共弦所在直线方程为0186=−−y xC .当r =2时,则PQ 斜率的最大值为−724D .当r =3时,过P 点作圆C 2两条切线,切点分别为A ,B ,则APB ∠不可能等于 π211.已知函数f(x)=x 3−3x 2,满足f (x )=kx +b 有三个不同的实数根x 1,x 2,x 3,则( ) A. 若k =0,则实数b 的取值范围是−4<b <0B. 过y 轴正半轴上任意一点仅有一条与函数 y =f (x )−1 相切的直线C. x 1x 2+x 2x 3+x 1x 3=kD.若 x 1,x 2,x 3成等差数列,则k +b =−212.已知正四面体O −ABC 的棱长为3,下列说法正确的是( )A. 若点P 满足OP ⃗⃗⃗⃗⃗ =x OA ⃗⃗⃗⃗⃗ +y OB ⃗⃗⃗⃗⃗⃗ +z OC ⃗⃗⃗⃗⃗ ,且x +y +z =1,则|OP ⃗⃗⃗⃗⃗ |的最小值为6B. 在正四面体O −ABC 的内部有一个可以任意转动的正四面体,则此四面体体积可能为√ 210C. 若正四面体O −ABC 的四个顶点分别在四个互相平行的平面内,且每相邻平行平面间的距离均相等,则此距离为3√ 1010D.点Q 在△ABC 所在平面内且|QO|=2|QA|,则Q 点轨迹的长度为2√ 303π三、填空题(本大题共4小题,共20.0分)13.已知双曲线1422=−y x ,则此双曲线的渐近线方程为 .14.已知等差数列{a n }的前n 项和为S n (n ∈N ∗),a 4=4,a 7=10,则S n 的最小值为 . 15.已知函数)3(sin )(2πω−=x x f (ω>0)的最小正周期为2π,且f (x )在[0,m]上单调递减,在[2m,5π3]上单调递增,则实数m 的取值范围是 .16. 在同一平面直角坐标系中,M ,N 分别是函数34)(2−+−−=x x x f 和函数x axe ax x g −=)ln()( 图象上的动点,若对任意a >0,有|MN |≥m 恒成立,则实数m 的最大值为______________. 四、解答题(本大题共6小题,共70.0分。
华附、省实、执信、广雅“四大名校”对比帖!

华附、省实、执信、广雅“四大名校”对比帖!华附、省实、执信、广雅“四大名校” 对比帖 !华南师范大学附属中学——稳坐第一把交椅学校简介华南师范大学附属中学是一所由广东省教育厅和华南师范大学双重领导,办学成绩卓著的学校,也是广东省首批国家级示范性高中。
华南师大附中前身可追溯至1888年建立的广州格致书院。
师资情况学校有广东省杰出教师4人,特级教师17人,高级教师近百人,获得省以上表彰的教师达200多人次。
正高级教师2人,华附教师中享受政府特殊津贴的2人,荣获“全国模范、优秀教师”称号10人,广东省“南粤杰出、优秀教师”等称号的25人。
有6位教师被华南师大聘为教育硕士导师,有8位教师先后担任广东省、广州市中学学科教研会会长、副会长。
高考成绩高考成绩多年来一直名列省市的前茅,其中考取重点大学的占85%左右,广东高考状元和单科第一名相当大的一部分是华附学生。
2011年重本率87.80%,其中理科类重本率89.54%,文科类重本率81.4%。
36位同学获得北京大学、清华大学、上海交通大学等著名高校预录资格。
历年录取分数线广东省实验中学——近年异军突起,先声夺人学校简介广东省实验中学(以下简称“省实”)是直属广东省领导的省重点学校,是广州市六所最著名重点学校之一,首批国家级示范性高中之一。
师资情况现有教职工292人,其中特级教师9人,高级教师75人,具有有硕士研究生学历的21人。
近五年来有20多位老师获得省级以上荣誉称号。
高考成绩2011文理科本科率均达到99.5%。
三个班的同学均100%上重点大学录取分数线。
创造了广东省一所学校有3个班100%上重点大学录取分数线新记录。
2010文科重本率80% 理科重本率76% 全校本科率99.11%2009重本率约70%,本科率约93%,其中总分状元2人,总分650分以上79人。
2008文科645分以上6人,约占全广州市1/3;理科650分以上19人,约占全广州市1/4。
广东省四校(华附、省实、广雅、深中)2023-2024学年高二下学期期末联考数学试题(原卷版)

华附、省实、广雅、深中2022级高二下学期四校联考数学命题学校:广东实验中学 命题人:高二数学备课组本试卷共4页,19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己的校名、姓名、考号、座位号等相关信息填写在答题卡指定区域内,并用2B 铅笔填涂相关信息.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.再选涂其它答案;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.一、选择题:本题共8小题,每小题5分,共40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()i 11z +=(i 为虚数单位),则z z −=( )A. 2−B. 2i −C. 2D. 2i 2. 已知等比数列{}n a 中,1241,9a a a ==,则7a =( ) A 3 B. 3或-3 C. 27 D. 27或-27 3. 已知圆22:2O x y +=与抛物线2:2(0)C x py p =>的准线相切,则p 的值为( )A.B. C. 4 D. 24. 如图所示,在正方形铁皮上剪下一个扇形和一个直径为2的圆,使之恰好围成一个圆锥,则圆锥的高为( )A.B.C.D..5. 某校高二年级下学期期中考试数学试卷满分为150分,90分以上(含90分)为及格.阅卷结果显示,全年级800名学生的数学成绩近似服从正态分布,试卷的难度系数(=平均分/150)为0.49,标准差为22,则该次数学考试及格的人数约为( )附:若()2,X Nµσ ,记()()p k P k X k µσµσ=−≤≤+,则()()0.750.547,10.683p p ≈≈. A. 127人B. 181人C. 254人D. 362人 6. 已知双曲线2213y x −=的左、右焦点分别为12,F F ,直线y x =与双曲线的右支交于点P ,则12PF PF ⋅= ( )A. 1−B. 0C. 1D. 27. 现有一组数据0,1,2,3,4,5,若将这组数据随机删去两个数,则剩下数据平均数小于3的概率为( ) A. 23B. 1115C. 45D. 1315 8. 若函数()()21e 12x g x x b x =−+−存在单调递减区间,则实数b 的取值范围是( ) A. [0,)+∞ B. ()0,∞+ C. (],0−∞ D. (),0∞−二、多选题:本题共36分,共18分.在每小题给出的四个选项中,有多项符合题目要求,全部选对的得6分,部分选对的得部分分.9. 若“2x k <−或x k >”是“23x −<<”必要不充分条件,则实数k 的值可以是( )A. 3B. 3−C. 5D. 5−10. 下列关于成对数据统计的表述中,正确的是( )A. 成对样本数据的经验回归直线一定经过点(),x yB. 依据小概率事件0.1α=的2χ独立性检验对零假设0H 进行检验,根据22×列联表中的数据计算发现20.10.837 2.706x χ≈<=,由()2 2.7060.1P χ≥=可推断0H 不成立,即认为X 和Y 不独立,该推断犯错误的概率不超过0.1C. 在残差图中,残差点的分布随解释变量增大呈现扩散的趋势,说明残差的方差不是一个常数,不满足一元线性回归模型对随机误差的假设D. 决定系数2R 越大,表示残差平方和越大,即模型的拟合效果越差11. 如图,心形曲线22:()1L x y x +−=与y 轴交于,A B 两点,点P 是L 上的一个动点,则( ) 的的A.点和()1,1−均在L 上B. 点PC. OP 的最大值与最小值之和为3D. PA PB +≤三、填空题:本题共3小题,每小题5分,共15分.12. 6(21)x y +−的展开式中,所有项的系数和为__________.13. 如图,正八面体ABCDEF 的12条棱长相等,则二面角E AB F −−的余弦值为__________.14. 数列{}n a 的前n 项和为n S ,且111,22n n a a a n +=−=,则满足2024n S >的最小正整数n 为__________.四、解答题:本题共5小题,共77分.解答应写出文字说明、证明过程或演算步骤.15. 已知ABC 的内角,,A B C 的对边分别为,,a b c ,且sin sin sin A B C b c a b+=+−.(1)求A ;(2)如图,若点D 是BC 边上一点,且,2AB AD BD CD ⊥=,求ADB ∠.的16. 如图,四棱锥P ABCD −的侧面PCD 为正三角形,底面ABCD 为梯形,//AB CD ,平面PCD ⊥平面ABCD ,已知44CD AB ==,13PM MD = .(1)证明:AM //平面PBC ;(2)若,AC AD PA ==,求直线AM 与平面PAB 所成角的正弦值. 17. 一个袋子中有30个大小相同的球,其中有10个红球、20个白球,从中随机有放回地逐次摸球作为样本,摸到红球或者第5次摸球之后停止.用X 表示停止时摸球的次数.(1)求X 的分布列和期望;(2)用样本中红球的比例估计总体中红球的比例,求误差的绝对值不超过0.1的概率.18. 已知椭圆2222:1(0)x y E a b a b+=>>的长轴长为()()1,2,0,2,02M N −. (1)求椭圆E 的方程;(2)过()4,0P 作一条斜率存在且不为0的直线l 交E 于,A B 两点.(i )证明:直线AM 和直线BM 斜率均存在且互为相反数;(ii )若直线AM 与直线BN 交于点Q ,求Q 的轨迹方程.19. 拟合(Fittiong )和插值(Imorterpolation )都是利用已知的离散数据点来构造一个能够反映数据变化规律的近似函数,并以此预测或估计未知数据的方法.拟合方法在整体上寻求最好地逼近数据,适用于给定数据可能包含误差的情况,比如线性回归就是一种拟合方法;而插值方法要求近似函数经过所有的已知数据点.适用于需要高精度模型的场景,实际应用中常用多项式函数来逼近原函数,我们称之为移项式插值.例如,为了得到1cos 2的近似值,我们对函数()πcos 2f x x =进行多项式插值.设一次函数()1L x ax b =+满足()()()()11001110L f L f == == ,可得()f x 在[]0,1上的一次插值多项式()11L x x =−+,由此可计算出1cos 2的的“近似值”11111cos 10.6822πππf L =≈=−≈,显然这个“近似值”与真实值的误差较大.为了减小插值估计的误差,除了要求插值函数与原函数在给定节点处的函数值相等,还可要求在部分节点处的导数值也相等,甚至要求高阶导数也相等.满足这种要求的插值多项式称为埃尔米特(Hermite )插值多项式.已知函数()πcos 2f x x = 在[]0,1上的二次埃尔米特插值多项式()2H x ax bx c ++满足()()()()()()001100H f H f H f =′=′ = (1)求()H x ,并证明当[]0,1x ∈时,()()f x H x ;(2)若当[]0,1x ∈时,()()2f x H x x λ− ,求实数λ的取值范围; (3)利用()H x 计算1cos 2的近似值,并证明其误差不超过140. (参考数据:2110.318,0.101ππ≈≈;结果精确到0.001)。
广东省华附、省实、广雅、深中四校2025届高考冲刺押题(最后一卷)语文试卷含解析

广东省华附、省实、广雅、深中四校2025届高考冲刺押题(最后一卷)语文试卷注意事项1.考生要认真填写考场号和座位序号。
2.试题所有答案必须填涂或书写在答题卡上,在试卷上作答无效。
第一部分必须用2B 铅笔作答;第二部分必须用黑色字迹的签字笔作答。
3.考试结束后,考生须将试卷和答题卡放在桌面上,待监考员收回。
1、阅读材料,完成下面小题。
禅对诗的渗透,可以从两方面看:一方面是以禅入诗,另一方面是以禅喻诗。
……对于诗,禅家的工具并不是都有用,诗家只是挑自己需要的。
这可以化简些,说只是“诗作”和“作诗”两个方面有某种性通之处。
诗所写是创造的意境,用来满足诸多渴想而难于在现实中获得的欲求。
意境各式各样,其中一类是淡远,或说有出世间意,与禅境相似,而没有禅那祥决绝。
分别还是在于情的性质以及有无。
陶渊明“采菊东篱下,悠然见南山”是意境超凡,可是其中还蕴涵着对于闲适生活的淡淡的甚至不淡淡的爱。
云峰志璿禅师“不知谁住原西寺,每日钟声送夕阳"也是意境超凡,但心境是随缘兼看破,不是恋恋。
有淡淡的爱是“似禅境”,随缘而不恋恋是“真禅境"。
似非真、可是无妨虽不能之而心向往之,以求越靠近,超凡的味道越浓。
求味道浓,有若有意若无意的,如王维的有些诗作就是。
时代靠后,有意的成分增加,终于形成诗论,如严羽的《沧浪诗话》就是。
再说“妙悟”,也是诗家和禅家有相通之处。
“池塘生春草”,好,何以好,如何好,不可说。
“师姑元是女人作",所悟之境为何,也是不可说。
不可说,作,只好相信文章本天成,妙手偶得之;欣赏,只好付之偶然,期待忽而相视而笑,莫逆于心。
总之,都是尽日觅不得,有时还自来的事,与学习数学,加减乘除,循序渐进,所知明确,功到必成的情况迥然不同。
无明确的法,而仍要作,仍要欣赏,只好乞援于无法之法,就是所谓“妙"。
超凡,两家有类似的希求,妙悟,两家有同样的甘苦,于是,说是物以类聚也好,说是同病相怜也好,诗和禅就携起手来。
广东省深中、华附、省实、广雅四校联考2023-2024学年高三最后一卷生物试卷含解析

广东省深中、华附、省实、广雅四校联考2023-2024学年高三最后一卷生物试卷注意事项:1.答题前,考生先将自己的姓名、准考证号码填写清楚,将条形码准确粘贴在条形码区域内。
2.答题时请按要求用笔。
3.请按照题号顺序在答题卡各题目的答题区域内作答,超出答题区域书写的答案无效;在草稿纸、试卷上答题无效。
4.作图可先使用铅笔画出,确定后必须用黑色字迹的签字笔描黑。
5.保持卡面清洁,不要折暴、不要弄破、弄皱,不准使用涂改液、修正带、刮纸刀。
一、选择题:(共6小题,每小题6分,共36分。
每小题只有一个选项符合题目要求)1.下列关于细胞内蛋白质与核酸的叙述,错误的是()A.有的蛋白质具有催化功能,有的核酸也具有催化功能B.含有蛋白质和核酸的结构,有的可作为遗传物质的载体,有的具有合成蛋白质的功能C.蛋白质和核酸的合成都需要搬运各自组成单位的工具D.分化程度不同的细胞内核酸和蛋白质的种类均有差异2.下列动物细胞的生命活动中,不是由单层膜的细胞器完成的是()A.合成磷脂B.对蛋白质进行分拣C.发出纺锤丝形成纺锤体D.消化细胞自身产生的碎渣3.下列选项中不符合含量关系“c=a+b,且a>b”的是()A.a非必需氨基酸种类、b必需氨基酸种类、c人体蛋白质的氨基酸种类B.a线粒体的内膜面积、b线粒体的外膜面积、c线粒体膜面积C.a各细胞器的膜面积、b细胞核的膜面积、c生物膜系统的膜面积D.a叶肉细胞的自由水、b叶肉细胞的结合水、c叶肉细胞总含水量4.阿托品是一种常见的麻醉药物。
某实验小组将离体的神经一肌肉接头处置于生理盐水中,并滴加阿托品,用针刺神经纤维后肌肉不收缩;再滴加乙酰胆碱酯酶抑制剂后,阿托品的麻醉作用降低甚至解除(突触间隙中的乙酰胆碱酯酶能水解乙酰胆碱)。
据此判断,阿托品抑制突触处的兴奋传递的机制可能是A.破坏突触后膜上的神经递质受体B.阻止突触前膜释放神经递质C.竞争性地和乙酰胆碱的受体结合D.阻断突触后膜上的钠离子通道5.细胞增殖与细胞凋亡都影响到人体内细胞的数量,下列有关叙述错误的是()A.两者都受到严格的遗传机制控制B.两者进行的速率都不受外界影响C.两者的顺利进行都需要有关酶的催化D.两者均能对多细胞生物体的生长发育起作用6.下列说法正确的是()A.环境容纳量是指一定空间中所能维持的种群最大数量B.调查土壤小动物类群丰富度可以用标志重捕法进行采集和调查C.演替过程中草本植物逐渐取代苔藓,主要原因是苔藓寿命较短D.生态系统的能量流动包含能量的输入、传递、转化和散失二、综合题:本大题共4小题7.(9分)酵母菌是单细胞真菌,在自然界中分布广泛,在人类的生活和生产中应用也很多。
华附、省实、广雅、深中四校2014届高三上学期期末联考数学理

5π12-π32Oy x2014届高三上学期期末华附、省实、广雅、深中四校联考理科数学命题学校:深圳中学本试卷分选择题和非选择题两部分,共4页,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己的校名、姓名、考号、座位号等相关信息填写在答题卡指定区域内.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,再选涂其它答案;不能答在试卷上.第一部分选择题(共40分)一、选择题:本大题共8小题,每小题5分,满分40分.在每小题给出的四个选项中,只有一项符合要求.1.若集合{}21,A m =,{}2,4B =,则“2m =”是“{}4A B = ”的A. 充要条件B. 既不充分也不必要条件C. 必要不充分条件D. 充分不必要条件2. 若0.52a =,πlog 3b =,22πlog sin5c =,则 A .b c a >> B . b a c >> C .a b c >> D .c a b >> 3.函数()sin()(0,0)f x A x A ωθω=+>>的部分图象如图所示,则()f x =A .π2sin(2)6x - B. π2sin(2)3x -C. π2sin(4)3x +D. π2sin(4)6x +4.已知圆22:1O x y +=及以下3个函数:①3()f x x =; ②()tan f x x =;③()sin .f x x x =其中图像能等分圆C 面积的函数有A .3个 B. 2个 C. 1 个 D. 0个5. 1231()x x-展开式中的常数项为 A .220 B .220- C .1320 D .1320-6.执行如图所示的程序框图,输出的S 值为A .2- B. 1-C. 0D. 17. 已知数列{}n a 满足:11,7a =对于任意的n *∈N ,17(1),2n n n a a a +=-则14131314a a -=A .27- B. 27 C. 37- D. 378.点O 是平面α内的定点,点(A 与点O 不同)的“对偶点”A '是指:点A '在射线OA 上且1OA OA '⋅=厘米2.若平面α内不同四点,,,P Q R S 在某不过点O 的直线l 上,则它们相应的“对偶点”,,,P Q R S ''''在 A .一个过点O 的圆上 B .一个不过点O 的圆上 C .一条过点O 的直线上 D .一条不过点O 的直线上S=S -T 结束输出ST=T+S开始T ≥0是否T=0,S =1第二部分非选择题(110分)二、填空题:本大题共6小题,每小题5分,共30分.9. 某学校高一、高二、高三年级的学生人数之比为334::,现用分层抽样的方法从该校高中三个年级的学生中抽取容量为50的样本,则应从高二年级抽取名学生.10. 若向量(1,2),(4,)BA CA x ==,且BA 与CA 的夹角为0,︒则BC = .11. 某几何体的三视图如图所示,其中正(主)视图与侧(左)视图的边界均为直角三角形,俯视图的边界为直角梯形,则该几何体的体积为 . 12. 已知直线:l x p =过抛物线2:4C y x =的焦点,直线l 与抛物线C 围成的平面区域的面积为,S 则p =______ ,S = .13. 已知函数1,01()12,12x x x f x x +≤<⎧⎪=⎨-≥⎪⎩,若0a b >≥,且()()f a f b =,则()bf a 的取值范围是 .选做题(请考生在以下两小题中任选一题做答,若两小题都做,则按第14题记分).14.(几何证明选讲选做题) 如图,过点C 作ABC 的外接圆O 的切线交BA 的延长线 于点D .若3CD =, 2AB AC ==,则BC = . 15.(坐标系与参数方程选做题) 在极坐标系O ρθ(0,02π)ρθ≥≤<中,点(2,)2A π关于直线:cos 1l ρθ=的对称点的极坐标为 .三、解答题:本大题共6小题,共80分,解答应写出必要的文字说明、证明过程或演算步骤. 16.(本小题满分12分)在ABC 中,三个内角,,A B C 所对的边分别为,a ,.b c已知2223()2b c a bc +-=,2B A =. (1) 求tan A ; (2) 设ππ(2sin(),1),(sin(),1),44m B n B =-=+-求m n ⋅的值.DBCOA_ 俯视图_ 侧视图_ 正视图 _2 _2_4 _2H PGFED CBA17.(本小题满分12分)盒子中装有四张大小形状均相同的卡片,卡片上分别标有数i,i,2,2,--其中i 是虚数单位.称“从盒中随机抽取一张,记下卡片上的数后并放回”为一次试验(设每次试验的结果互不影响). (1)求事件A “在一次试验中,得到的数为虚数”的概率()P A 与事件B “在四次试验中,至少有两次得到虚数” 的概率()P B ;(2)在两次试验中,记两次得到的数分别为,a b ,求随机变量a b ξ=⋅的分布列与数学期望.E ξ18.(本小题满分14分)如图,四边形ABCD 是正方形,EA ⊥平面ABCD ,EA PD ,2AD PD EA ==,F ,G , H 分别为PB ,EB ,PC 的中点. (1)求证:FG 平面PED ; (2)求平面FGH 与平面PBC 所成锐二面角的大小.已知数列{}n a 的前n 项和为,n S 记11()2(2),.n n n n f n a S n S a n *++=-+∈N (1)若数列{}n a 是首项与公差均为1的等差数列, 求(2014)f ; (2)若121,2,a a ==且数列{}{}212,n n a a -均是公比为4的等比数列,求证:对任意正整数n ,()0.f n ≥在平面直角坐标系中,已知点(2,2)F 及直线:20l x y +-=,曲线1C 是满足下列两个条件的动点(,)P x y 的轨迹:①2,PF d =其中d 是P 到直线l 的距离;②00.225x y x y >⎧⎪>⎨⎪+<⎩(1) 求曲线1C 的方程;(2) 若存在直线m 与曲线1C 、椭圆22222:1(0)x y C a b a b+=>>均相切于同一点,求椭圆2C 离心率e 的取值范围.已知函数22()en nxx x af x --=,其中,,N R n a *∈∈e 是自然对数的底数. (1)求函数12()()()g x f x f x =-的零点;(2)若对任意,N n *∈()n f x 均有两个极值点,一个在区间(1,4)内,另一个在区间[]1,4外,求a 的取值范围;(3)已知,,,N k m k m *∈<且函数()k f x 在R 上是单调函数,探究函数()m f x 的单调性.2014届高三上学期期末华附、省实、广雅、深中四校联考参考答案与评分标准理科数学说明:一、本解答给出了一种或几种解法供参考,如果考生的解法与本解答不同,可根据试题的主要考查内容比照评分标准制订相应的评分细则.二、对计算题当考生的解答在某一步出现错误时,如果后续部分的解答未改变该题的内容和难度,可视影响的程度决定给分,但不得超过该部分正确解答应得分数的一半;如果后续部分的解答有较严重的错误,就不再给分.三、解答右端所注分数,表示考生正确做到这一步应得的累加分数. 四、只给整数分数,选择题和填空题不给中间分数.一、选择题:本大题共8小题,每小题5分,满分40分.在每小题给出的四个选项中,只有一项符合要求.1.【解析】{}244 2.A B m m =⇔=⇔=±2. 【解析】0.50221,>= πππ0log 1log 3log π1,=<<=222πlog sin log 10.5<= 3.【解析】由图知()f x 在5π12x =时取到最大值2,且最小正周期T 满足 35ππ+.4123T =故2,A =32π3π,2,4ωω⨯==5π2sin(2)212θ⨯+=,5πsin()1,6θ+= 5πππ2π,2π,623k k k θθ+=+=-∈Z .所以π2sin ()(2).3x f x -= 或由5(π)212f =逐个检验知π2sin ()(2).3x f x -=4.【解析】圆O 关于原点O 对称. 函数3y x =与函数tan y x =是定义域上的奇函数,其图像关于原点对称, 能等分圆O 面积;而sin y x x =是R 上的偶函数,其图像关于y 轴对称,且当01x <≤时sin 0,x x >不能等分圆O 面积5. 【解析】1231()x x -展开式中的通项为4121231121231()(1)(0,1,2,,12).k k k k k kk T C x C x k x--+=-=-= 1k T +为常数项的充要条件是9.k =常数项91012220.T C =-=-6.【解析】0,11,01,1T S T S T S ==⇒==⇒==-0,11,0.T S T S ⇒==-⇒=-=7. 【解析】11,7a =234716373467613,,,.277727772777a a a =⨯⨯==⨯⨯==⨯⨯=由数学归纳法可证明:当n 为大于1的奇数时, 67n a =;当n 为正偶数时, 3.7n a =故14131314a a -=3.78.【解析】过O 作与直线l 垂直的直线,m 以O 为原点,直线m 为x 轴,单位为1厘米,建立平面直角平面坐标系. 设直线1:(0)l x a a=≠,01(,)P y a 是直线l 上任意一点,它的“对偶点”为(,)P x y ',则存在0,λ>使得OP OP λ'= ,即01,x y y a λλ==,又01x OP OP OP OP y y a''⋅=⋅=+= ,消去λ,得220x y ax +-=.故,,,P Q R S ''''在过点O 的圆22:0x y ax +-=上.题号 1 2 3 4 5 6 7 8 答案 DC B B B CD ABA (2,π2)Of (x )12b a12321yxO二、填空题:本大题共6小题,每小题5分,共30分.9. 15 10. (3,6)-- 11. 8 12. 81,.3 13. 3,24⎡⎫⎪⎢⎣⎭ 14. 23 15. (22,).4π9. 【解析】根据分层抽样的方法步骤,按照一定比例抽取,样本容量为50,那么根据题意得:从高三一共可以抽取人数为:1510350=⨯. 10. 【解析】由BA 与CA 的夹角为0,︒知8x =,(3,6).BC BA AC BA CA =+=-=--11. 【解析】由三视图可知该几何体是一个四棱锥,根据“正侧等高,正俯等长,侧俯等宽”的规则,其体积为11(24)428.32V =⨯+⨯⨯= 12. 【解析】抛物线2:4C y x =的焦点为(1,0)F ,知1p =. 31120282(2)4.33S x dx x ==⨯=⎰ 13. 【解析】如图,()f x 在[)0,1,[)1,+∞上均单调递增, 由0a b >≥及()()f a f b =知11.2a b ≥>≥()()(1)b f a bf b b b ⋅==+的取值范围113(1),(11),2.224⎡⎫⎡⎫++=⎪⎪⎢⎢⎣⎭⎣⎭14. 【解析】由2()CD DA DB DA DA AB =⨯=⨯+知2230D A D A +-=,解得1, 3.DA DB ==由DAC DCB 得AC CD BC BD =,即2 3.AC BDBC CD⨯== 15. 【解析】如图,在极坐标系(0,02π)O ρθρθ≥≤<中,设(2,)2A π关于 直线:cos 1l ρθ=的对称点为(,),B ρθ则2OA AB ==,且.OA AB ⊥ 从而π22,,4OB AOB =∠=即πππ22,.244ρθ==-= 三、解答题:本大题共6小题,共80分,解答应写出必要的文字说明、证明过程或演算步骤.16.解: (1)2223()2,b c a bc +-=2221cos .23b c a A bc +-∴== …………………………………………2分0π,A <<22sin 1cos ,3A A ∴=-=…………………………………………… 4分 sin tan 2.cos AA A== ………………………………………………………6分 (2)(解法一)ππ(2sin(),1),(sin(),1),44m B n B =-=+-ππ2sin()sin()144m n B B ∴⋅=-+-……………………… 7分222(cos sin )(cos sin )122B B B B =⨯-⨯+-22cos sin 1B B =-- ………………………………………… 9分 22sin .B =- ……………………………………………… 10分2B A = ,22sin sin 22sin cos ,3B A A A ∴===16.9m n ⋅=-………12分(2)(解法二)ππ(2sin(),1),(sin(),1),44m B n B =-=+-ππ2s i n ()s i n ()144m n B B ∴⋅=-+-……………………… 7分 πππ2cos ()sin()1244B B ⎡⎤=--+-⎢⎥⎣⎦ππ2cos()sin()144B B =++-πsin(2)12B =+-cos 21B =- ………………………………………………………9分 22sin .B =- ……………………………………………………… 10分2B A = ,22sin sin 22sin cos ,3B A A A ∴===16.9m n ⋅=-………12分(2)(解法三) 2B A = ,22sin sin 22sin cos ,3B A A A ∴===21cos cos 212sin .3B A A ==-=- ………………………9分π24(2sin(),1)(2(cos sin ),1)(,1),43m B B B +∴=-=-=-……10分π242(sin(),1)((sin cos ),1)(,1).426n B B B -=+-=+-=-…11分4242161.369m n +-∴⋅=-⨯-=-………………………12分17.解:(1)21()42P A ==, ……………………………………………………………2分004113441111511()1()1()()()()1.22221616P B P B C C ⎡⎤=-=-+=-=⎢⎥⎣⎦………… 5分(2),,a b ξ的可能取值如下左表所示:i - i 2- 2 i - 1 1 2 2i 1 1 2 22- 2 2 4 4 2 2 24 4……………………………………………………………6分由表可知:418141(1),(2),(4).164162164P =P =P =ξξξ====== ………………9分 所以随机变量X 的分布列为(如上右表) …………………………………… 10分 所以1119()124.4244E =ξ⨯+⨯+⨯= ………………………………………………12分ξ 1 2 4P 14 12 14abξz yx H PGF ED C BAQP HFE D C 18. (1)证明:F ,G 分别为PB ,BE 的中点,FG ∴ PE . …………………………………1分又FG ⊄平面PED ,PE ⊂平面PED , …………………………………3分 FG ∴ 平面PED . ……………………………………………………………5分(2)解:EA ⊥ 平面ABCD ,EA PD ,PD ∴⊥平面.ABCD,AD CD ⊂ 平面,ABCD PD AD ∴⊥,PD CD ⊥. 四边形ABCD 是正方形,AD CD ∴⊥.以D 为原点,分别以直线,,DA DC DP 为x 轴, y 轴,z 轴建立如图所示的空间直角坐标系,设1.EA = ……………………………………7分2AD PD EA == ,D ∴()0,0,0,P ()0,0,2,A ()2,0,0,C ()0,2,0,B ()2,2,0,(2,0,1)E ,(2,2,2)PB =- ,(0,2,2)PC =-.F ,G ,H 分别为PB ,EB ,PC 的中点,F ∴()1,1,1,G 1(2,1,)2,H (0,1,1),1(1,0,)2GF =- ,1(2,0,).2GH =- …… ………8分(解法一)设1111(,,)x y z =n 为平面FGH 的一个法向量,则110GF GH ⎧⋅=⎪⎨⋅=⎪⎩n n , 即11111021202x z x z ⎧-+=⎪⎪⎨⎪-+=⎪⎩,令11y =,得1(0,1,0)=n . …… …………………10分设2222(,,)x y z =n 为平面PBC 的一个法向量,则220PB PC ⎧⋅=⎪⎨⋅=⎪⎩n n , 即222222220220x y z y z +-=⎧⎨-=⎩,令21z =,得2(0,1,1)=n . …… …………………12分 所以12cos ,n n =1212⋅⋅n n n n =22. ……………………………………………13分所以平面FGH 与平面PBC 所成锐二面角的大小为π4(或45︒). …………14分(解法二) (0,1,1)(2,0,0)0DH BC ⋅=⋅-= ,(0,1,1)(0,2,2)0DH PC ⋅=⋅-=,DH ∴是平面PBC 一个法向量. …… ……………… …………………10分(0,2,0)(1,0,0)0DC FH ⋅=⋅-= ,1(0,2,0)(1,0,)02DC FG ⋅=⋅-= ,DC ∴是平面平面FGH 一个法向量. …… ……………… …………………12分22cos ,,222DH DC DH DC DH DC ⋅===⋅ ……… … …………………13分 ∴平面FGH 与平面PBC 所成锐二面角的大小为π4(或45︒). … …………14分(解法三) 延长AE 到,Q 使得,AE EQ =连,.PQ BQ2PD EA AQ == ,EA PD ,∴四边形ADPQ 是平行四边形,.PQ AD四边形ABCD 是正方形,,.BC AD PQ BC ∴F ,H 分别为PB ,PC 的中点,,.FH BC FH PQ ∴FH ⊄平面PED ,PQ ⊂平面PED , FH ∴ 平面PED . ………7分 ,,FH FG F FH FG =⊂ 平面,ADPQ ∴平面FGH 平面.ADPQ ………9分 故平面FGH 与平面PBC 所成锐二面角与二面角D PQ C --相等. … …10分 ,PQ CD PQ PD ⊥⊥ ,,,PD CD D PD DC =⊂ 平面,PDC PQ ∴⊥平面.PDC PC ⊂ 平面,,PDC PQ PC ∴⊥DPC ∠是二面角D PQ C --的平面角. …12分 ,,45.AD PD AD PD DPC =⊥∴∠=︒ … …………13分∴平面FGH 与平面PBC 所成锐二面角的大小为π4(或45︒). … …………14分19. 解:(1) 数列{}n a 是首项与公差均为1的等差数列, …………………………1分∴,N n *∀∈1(1),1,.2n n n n n a n a n S ++==+=………………………………3分 11()2(2)n n n n f n a S n S a ++=-+(1)(1)2(1)2(1)22n n n n n n n ++⎡⎤=+⨯-⨯++⎢⎥⎣⎦22(1)(1)0.n n n n =+-+= ……………………………5分故(2014)0.f = ………………………………………………………6分(2)由题意,n *∀∈N 12221142,n n n a ---=⨯= ………………………………………7分1212242.n n n a --=⨯= ……………………………………8分故12.n n a -= …………………………………………………9分,n *∀∈N 1122,21,12nnn n n a S +-===--11()2(2)n n n n f n a S n S a ++=-+ 1112(21)(222)2(232)2.n n n n n n n n n +++=---+=--+ ……………………10分 (证法一)当1n =时,(1)0f =; ……………………………11分当2n ≥时,[]1124(11)41(1)4n n n n +-=⨯+≥+-=, ……………………………12分1()2(232)22(432)22(2)220.n n n n f n n n n n n n n n +=--+≥--+=-+≥> …………………………………………………………………………………………13分故对任意正整数n ,()0.f n > ………………………………………………………14分 (证法二),n *∀∈N (1)()f n f n +-1212(235)222(232)2n n n n n n n n +++⎡⎤⎡⎤=--++---+⎣⎦⎣⎦2122(235)(232)2n n n n n ++⎡⎤=-----+⎣⎦2(6238) 2.n n n =⨯--+ ……………………………11分012(11)1n n n n C C n =+≥+=+ , ,(1)()2(6638)22(32)2220N n n n n f n f n n n n *∴∀∈+-≥+--+=-+≥+>,数列{}()f n 是递增数列. ………………………………………………………12分 2(1)2(232)20,f =--+= ……………………… …………………………13分 ,()0.N n f n *∴∀∈≥ ……………………………………………………………………14分20. 解:(1)2222(2)(2)22()4PF x y x y x y =-+-=+-++,22x y d +-=,2分由①2,PF d =得:222222()4222()2x y x y x y xy x y +-++=++-++,即 1.xy = ……………………………………………………………4分将1xy =代入②得:1150,0,2x x x x >>+<,解得: 12.2x << 所以曲线1C 的方程为:1y x =1(2).2x << ………………………………6分 (2)(解法一)由题意,直线m 与曲线1C 相切,设切点为1(,)M t t , 12.2t <<则直线m 的方程为2111()()()y x t x t x t t x t'-=⨯-=--=,即212.y x t t =-+……7分将212y x t t=-+代入椭圆2C 的方程222222b x a y a b +=,并整理得:242222222()4(4)0.b t a x a tx a b t t +-+-=由题意,直线m 与椭圆2C 相切于点1(,)M t t,则4222422222242224164()(4)4(4)0a t a b t a b t t a b t a t b t ∆=-+-=-+=,即22424.a b t t += ……………………………………………………………9分又222211,t a b t+= 即242222.b t a a b t += 联解得:22222,2.b a t t == ………10分由12,2t <<及22a b >得1 2.t << 故2222411a b e a t -==-, ……………………………………………………12分 得2150,16e <<又01,e <<故150.4e << 所以椭圆2C 离心率e 的取值范围是15(0,).4 ………………………………14分(2)(解法二)设直线m 与曲线111:(2)2C y x x =<<、椭圆22222:1(0)x y C a b a b+=>> 均相切于同一点1(,),M t t 则222211.t a b t += …………………………………………………7分由1y x =知21y x'=-;由22221(0)x y y a b +=>知221x y b a =-,222222222.211x b bx b x a y a y x x a a a-'==-=--- 故2224221,.1b t a b t t a t-=-= …………………………………………………9分联解222222411t a b ta b t ⎧+=⎪⎨⎪=⎩,得22222,2.b a t t == ……………………………………………10分 由12,2t <<及22a b >得1 2.t << 故2222411a b e a t -==-, ……………………………………………………12分 得2150,16e <<又01,e <<故150.4e << 所以椭圆2C 离心率e 的取值范围是15(0,).4………………………………14分21. 解:(1)222122222(2)(e 1)()()()e e ex x x xx x a x x a x x a g x f x f x -------=-=-=, 44a ∆=+ ① 当1a <-时,0,∆<函数()g x 有1个零点:10.x = ………………………1分 ② 当1a =-时,0,∆=函数()g x 有2个零点:120, 1.x x == ……………………2分 ③ 当0a =时,0,∆>函数()g x 有两个零点:120, 2.x x == ……………………3分 ④ 当1,0a a >-≠时,0,∆>函数()g x 有三个零点:1230,11,1 1.x x a x a ==-+=++ …………………………………………4分(2)222(22)e (2)e 2(1)2().e e nx nx n nx nxx n x x a nx n x a n f x -----+++⋅-'== …………5分 设2()2(1)2n g x nx n x a n =-+++⋅-,()n g x 的图像是开口向下的抛物线.由题意对任意,N n *∈()0n g x =有两个不等实数根12,x x , 且()[]121,4,1,4.x x ∈∉则对任意,N n *∈(1)(4)0n n g g <,即6(1)(8)0n a n a n ⎡⎤⋅+⋅⋅--<⎢⎥⎣⎦, ………………7分又任意,N n *∈68n -关于n 递增,681n->-, 故min 61(8),186 2.a a n-<<--<<-=所以a 的取值范围是()1,2.- ……………………………………………9分(3)由(2)知, 存在,R x ∈22(1)2()0ek kxkx k x a k f x -+++⋅-'=<,又函数()k f x 在R 上是单调函数,故函数()k f x 在R 上是单调减函数, ……………………………………10分从而2224(1)4(2)4(1)0,k k k ka k a k ∆=++-=++≤即21(1).a k≤-+ ……11分所以2222222214()4(1)41(1).m k m m m a m m k k -⎡⎤∆=++≤+-+=⎢⎥⎣⎦ 由,,,N k m k m *∈<知0.m ∆< …………………………………………13分 即对任意,R x ∈22(1)2()0ek kxkx k x a k f x -+++⋅-'=< 故函数()m f x 在R 上是减函数. …………………………………………14分。
广东省华附、广雅、省实、深中2015届高三上学期期末四校联考英语试题含答案

2015届高三上学期华附、省实、广雅、深中四校联考英语命题学校:华师附中命题人:高三英语备课组Ⅰ语言知识及应用(共两节, 满分45分)第一节完形填空(共15小题;每小题2分,满分30分)阅读下面短文,掌握其大意,然后从1~15各题所给的A、B、C和D项中,选出最佳选项,并在答题卡上将该项涂黑。
The day I was given the news was a cold Saturday. My mom sat me on the floor and opened her mouth to give me the most 1 words I'd ever heard: ―Grandma Billie has been diagnosed with cancer.‖ Tears filled my eyes immediately. Mom continued, ―Grandma is old and her body is too 2 . There’s nothing they can do to treat her.‖ I knew this wasn't my mom’s fault but I was so 3 . I was almost yelling at my mom, ―So, they’re just going to let her 4 ?‖We had to just wait for her to die. I couldn't even 5 the thought. We went to the hospital right away. The doctor said, ―Billie has a tumor in her 6 . She won't be able to 7 because it hurts her too much. One option is a feeding tube, but there are times when it will be uncomfortable.‖ At the end, the doctor sighed, ―With a feeding tube, she would probably have three 8 at most.‖My whole family on my mom’s side was really 9 and they were all looking to God to make this easy. I couldn't handle it. If God was so wonderful and so 10 then why did he allow all these awful things to happen to good people? Why was he letting my grandma die?I came and visited every day after 11 practice. Grandma would tell me the 12 stories over and over again, but every time I was just as surprised and excited to hear them. She was getting 13 really fast and every time I came to see her it broke my heart more.3 weeks later, after the softball practice, I was getting ready to see grandma when mom appeared in the locker room. She opened her mouth but at first no words came out. She 14 again, ―You can’t go to see grandma today because she just died.‖ Silence.I couldn’t 15 anyone else in the room. Mom sat next to me and we both held each other and cried.1. A. unforgettable B. regretful C. painful D. indescribable2. A. weak B. small C. pale D. thin3. A. sad B. mad C. desperate D. anxious4. A. pass B. suffer C. go D. die5. A. bear B. hate C. take D. dismiss6. A. skin B. breast C. throat D. lung7. A. feed B. eat C. breathe D. digest8. A. years B. months C. days D. hours9. A. ambitious B. religious C. mysterious D. cautious10. A. delightful B. peaceful C. pitiful D. powerful11. A. softball B. school C. social D. music12. A. short B. strange C. dull D. same13. A. happier B. older C. better D. worse14. A. tried B. said C. cried D. opened15. A. see B. sense C. hear D. feel第二节语法填空(共10小题;每小题1. 5分,满分15分)阅读下面短文,按照句子结构的语法性和上下文连贯的要求,在空格处填入一个适当的词或使用括号中词语的正确形式填空,并将答案填写在答题卡标号为16~25的相应位置上。
广东省华附、省实、广雅、深中2022届高三上学期四校联考数学试题-答案

广东省华附、省实、广雅、深中2022届高三上学期四校联考数学参考答案一、单项选择题:三、填空题:本大题共4小题,每小题5分,共20分.13.π4 14.4 15.827 16.(n 2−2n +3)⋅2n+1−6一、单选题1.设集合A ={y ∣y =√1−x},B ={−1,0,1},则A ∩B =( ) A .{1} B .{0,1} C .{−1,0} D .{−1,0,1}【答案】B2.已知复数i(,)z a b a b R =+∈,且3(1i )2i z +=+,则a b +=( ) A .12 B .32C .1D .2【答案】D3.已知命题p:∃x ,y ∈R ,sin(x +y)=sinx +siny ;命题q:∀x ,y ∈R ,sinx ⋅siny ⩽1,则下列命题中为真命题的是( ) A .p ∧q B .¬p ∧q C .p ∧(¬q) D .¬(p ∨q)【答案】A【解析】当x =0,y =π2时,sin(x +y)=sinx +siny 成立所以命题p 为真命题,则¬p 是假命题; 因为∀x ,y ∈R ,所以sinx ≤1,siny ⩽1,则sinx ⋅siny ⩽1,故命题q 为真命题,则¬q 是假命题; 所以p ∧q 是真命题,¬p ∧q 是假命题, p ∧(¬q)是假命题,¬(p ∨q)是假命题,故选:A 4.声强级L (单位:dB )与声强I 的函数关系式为:L =10lg(I 10−12),若女高音的声强级是75dB ,普通女性的声强级为45dB ,则女高音声强是普通女性声强的( ) A .10倍 B .100倍 C .1000倍 D .10000倍【答案】C【解析】设女高音声强为I 1,普通女性声强为I 2,则101g (I 110−12)=75,所以I 110−12=107.5①,10lg(I210−12)=45,所以I 210−12=104.5②,则①÷②得:I1I 2=1000,故女高音声强是普通女性声强的1000倍.故选:C5.等差数列{a n }中,a 1=1,公差为d(d ∈Z),a 3+λa 9+a 15=17,λ∈(−1,0),则公差d 的值为( ) A .1 B .0 C .−1 D .−2【答案】A【解析】a 3+λa 9+a 15=a 1+2d +λ(a 1+8d )+a 1+14d =(2+λ)a 1+(16+8λ)d =17,整理得:(8d +1)λ=15−16d ,由于d ∈Z ,所以8d +1≠0,故λ=15−16d 8d+1,则λ=15−16d 8d+1∈(−1,0),若8d +1>0,解得:1516<d <2,由于d ∈Z ,所以d =1;若8d +1<0,解得:{d >2d <1516,此时无解,综上:公差d 的值为1,故选:A6.函数())1f x x =+,定义域为R 的函数()g x 满足()()2g x g x −+=,若函数()y f x =与()y g x =图象的交点为()()()112266,,,,,,x y x y x y ,则61()i i i x y =+=∑( )A .0B .6C .12D .24【答案】B【解析】由g(−x)+g(x)=2得y =g(x)的图象关于(0,1)对称,同时函数f(x)=ln(√x 2+1−x)+1定义域也为R ,且f(−x)+f(x)=ln(√x 2+1+x)+1+ln(√x 2+1−x)+1 =ln [(√x 2+1+x)(√x 2+1−x)]+2=ln (x 2+1−x 2)+2=2 即f(−x)+f(x)=2,故也关于(0,1)对称,则函数f(x)=ln(√x 2+1−x)+1与y =g(x)图象的交点关于(0,1)对称, 则不妨设关于点(0,1)对称的坐标为(x 1,y 1),(x 6,y 6),则x 1+x 62=0,y 1+y 62=1,则x 1+x 6=0,y 1+y 6=2,同理可得:x 2+x 5=0,y 2+y 5=2,x 3+x 4=0,y 3+y 4=2, 即∑(x i +y i )6i=1=3×(0+2)=6,故选:B .7.在足球比赛中,球员在对方球门前的不同的位置起脚射门对球门的威胁是不同的,出球点对球门的张角越大,射门的命中率就越高.如图为室内5人制足球场示意图,设球场(矩形)长BC 大约为40米,宽AB 大约为20米,球门长PQ 大约为4米.在某场比赛中有一位球员欲在边线BC 上某点M 处射门(假设球贴地直线运行),为使得张角∠PMQ 最大,则BM 大约为( )(精确到1米)A .8米B .9米C .10米D .11米【答案】C【解析】由题意知,PB =8,QB =12,设∠PMB =α,∠QMB =β,BM =x ,则tanα=8x ,tanβ=12x,所以tan∠PMQ =tan (β−α)=12x −8x 1+12x ⋅8x=4x x 2+96=4x+96x≤2√x⋅x=2√6,当且仅当x =96x,即x =√96时取等号,又因为√96≈10,所以BM 大约为10米. 故选:C. 8.倾斜角为π3的直线经过双曲线C:x 2a 2−y 2b 2=1(a >0,b >0)的右焦点F ,与双曲线C 的右支交于A ,B 两点,且AF⃑⃑⃑⃑⃑ =λFB ⃑⃑⃑⃑⃑ (λ⩾5),则双曲线C 的离心率的取值范围是( ) A .[43,+∞)B .(1,43]C .(1,2)D .[43,2)【答案】D【解析】设l 为双曲线的右准线,过A 、B 作AD ,BE 垂直于l ,D ,E 为垂足, 过A 作AG ⊥EB 于G , 根据双曲线的第二定义,得|AD|=|AF |e,|BE|=|BF|e,∴|AF|=e|AD|,|BF|=e|BE|, ∵ AF⃑⃑⃑⃑⃑ =λFB ⃑⃑⃑⃑⃑ (λ⩾5), ∴|AF⃑⃑⃑⃑⃑ |=λ|FB ⃑⃑⃑⃑⃑ |, ∴|AD|=λ|BE|,∴|AB|=|AF|+|BF|=(1+λ)e|BE|, ∴|BG|=λ|AD|−|BE|=(λ−1)|BE|, ∴cos∠ABG =|BG||AB|=(λ−1)|BE|(λ+1)e|BE|,∴ecos∠ABG =λ−1λ+1,∴e =2(λ−1)λ+1=2−4λ+1,∵λ⩾5,则λ+1⩾6,可得0<4λ+1⩽23, ∴43≤2−4λ+1<2,∴ 43⩽e <2,即离心率的取值范围是[43,2). 故选:D .二、多选题9.已知α,β是两个不同的平面,l 是一条直线,则下列命题中正确的是( ) A .若α || β,l || β,则l || α B .若l ⊥α,l ⊥β,则α || β C .若l ⊥α,l ||β,则α⊥β D .若α⊥β,l || β,则l ⊥α【答案】BC【解析】对于A ,若α//β,l//β,则l//α或l ⊂α,故A 不正确;对于B ,若l ⊥α,l ⊥β,则α//β,故B 正确;对于C ,若l ⊥α,l//β,过l 的平面γ与β相交,设交线为m , ∵l//β,l ⊂γ,β∩γ=m ,则l//m ,∵l ⊥α,则m ⊥α,∵m ⊂β,故α⊥β,故C 正确; 对于D ,若α⊥β,l//β,则l 与α不一定垂直,故D 不正确; 10.设x >0,x,y ∈R ,则( ) A .“x >y ”⇒“x >|y|”B .“x <y ”⇒“x <|y|”C .“x ≥|y|”⇒“x +y ≥|x +y|”D .“x >y ”⇒“x +|y|≥|x +y|”【答案】BCD【解析】A :当x =1,y =−2时,x >|y|不成立,故错误; B :由y >x >0,则x <|y|成立,故正确;C :x >0且x ≥|y|,即x ≥y ≥−x ,则x +y ≥0,故x +y =|x +y|恒成立,故正确;D :当x >0>y 时,x +|y|>|x +y|,当x >y ≥0时,x +|y|=|x +y|,故正确;11.已知抛物线C:y 2=4x ,圆F:(x −1)2+y 2=14(F 为圆心),点P 在抛物线C 上,点Q 在圆F 上,点A(−1,0),则下列结论中正确的是( )A .|PQ|的最小值是12 B .|PF||PA|的最小值是√22C .当∠PAQ 最大时,|AQ |=√152D .当∠PAQ 最小时,|AQ |=√152【答案】ABC【解析】A. |PQ|的最小值是|PF|的最小值减去圆的半径,又|PF|的最小值是1,所以|PQ|的最小值是1-12=12,故正确;B. 设P (4t 2,4t ),则|PF |2=(4t 2−1)2+(4t )2=16t 4+8t 2+1, |PA |2=(4t 2+1)2+(4t )2=16t 4+24t 2+1, 所以|PF |2|PA|2=16t 4+8t 2+116t 4+24t 2+1=1−16t 216t 4+24t 2+1=1−1616t 2+24+1t2≥1−2√16t 2⋅1t 2+24=12, 当且仅当16t 2=1t 2,即t =±12时,等号成立,所以|PF||PA|的最小值是√22,故正确;C.如图所示:当∠PAQ 最大时,直线AQ 与圆相切,则|AQ |=√22−14=√152,故正确;D.当∠PAQ 最小时为0∘,即P ,A ,Q 共线,则|AQ |∈[32,52],故错误; 12.设函数f(x)={−x 2−2x,x ⩽0|lnx |,x >0,则下列命题中正确的是( )A .若方程f(x)=a 有四个不同的实根x 1,x 2,x 3,x 4,则x 1⋅x 2⋅x 3⋅x 4的取值范围是(0,1)B .若方程f(x)=a 有四个不同的实根x 1,x 2,x 3,x 4,则x 1+x 2+x 3+x 4的取值范围是(0,+∞)C .若方程f(x)=ax 有四个不同的实根,则a 的取值范围是(0,1e )D .方程f 2(x)−(a +1a )f(x)+1=0的不同实根的个数只能是1,2,3,6【答案】AD【解析】对于A :作出f(x)的图像如下:若方程f(x)=a 有四个不同的实根x 1,x 2,x 3,x 4,则0<a <1,不妨设x 1<x 2<x 3<x 4,则x 1,x 2是方程−x 2−2x −a =0的两个不等的实数根,x 3,x 4是方程|lnx|=a的两个不等的实数根,所以x 1x 2=a ,−lnx 3=lnx 4,所以lnx 4+lnx 3=0,所以x 3x 4=1, 所以x 1x 2x 3x 4=a ∈(0,1),故A 正确;对于B :由上可知,x 1+x 2=−2,−lnx 3=lnx 4=a ,且0<a <1, 所以x 3x 4=1,所以x 3∈(1e ,1),x 4∈(1,e ),所以x 3+x 4=1x 4+x 4∈(2,1+1e),所以x 1+x 2+x 3+x 4∈(0,1+1e),故B 错误;对于C :方程f(x)=ax 的实数根的个数,即可函数y =f(x)与y =ax 的交点个数,因为y =ax 恒过坐标原点,当a =0时,有3个交点,当a <0时最多2个交点,所以a >0, 当y =ax 与y =lnx(x >1)相切时,设切点为(x 0,lnx 0), 即y ′=1x ,所以y ′|x=x 0=1x 0=lnx 0x 0,解得x 0=e ,所以y ′|x=x 0=1e ,所以a =1e,所以当y =ax 与y =lnx(x >1)相切时, 即a =1e时,此时有4个交点, 若f(x)=ax 有4个实数根,即有4个交点,当a >1e时由图可知只有3个交点,当0<a <1e时,令g (x )=lnx −ax ,x ∈(1,+∞),则g ′(x )=1x−a =1−ax x,则当1<x <1a 时g ′(x )>0,即g (x )单调递增,当x >1a 时g ′(x )<0,即g (x )单调递减,所以当x =1a 时,函数取得极大值即最大值,g (x )max =g (1a )=−lna −1>0,又g (1)=−a <0及对数函数与一次函数的增长趋势可知,当x 无限大时g (x )<0,即g (x )在(1,1a )和(1a ,+∞)内各有一个零点,即f(x)=ax 有5个实数根,故C 错误; 对于D :f 2(x)−(a +1a )f(x)+1=0,所以[f(x)−a][f(x)−1a]=0,所以f(x)=a或f(x)=1a,由图可知,当m>1时,f(x)=m的交点个数为2,当m=1,0时,f(x)=m的交点个数为3,当0<m<1时,f(x)=m的交点个数为4,当m<0时,f(x)=m的交点个数为1,所以若a>1时,则1a∈(0,1),交点的个数为2+4=6个,若a=1时,则1a=1,交点的个数为3个,若0<a<1,则1a>1,交点有4+2=6个,若a<0且a≠−1时,则1a <0且a≠1a,交点有1+1=2个,若a=−1=1a,交点有1个,综上所述,交点可能由1,2,3,6个,即方程不同实数根1,2,3,6,故D正确;三、填空题13.已知向量a⃗,b⃑⃗满足a=(4,0),b⃑⃗=(m,1),|a|=a⃗⋅b⃑⃗,则a⃗与b⃑⃗的夹角为___________.【答案】π4【解析】由题意,向量a=(4,0),b⃑⃗=(m,1),因为|a⃗|=a⃗⋅b⃑⃗,可得4m+0×1=4,解得m=1,即b⃑⃗=(1,1),可得|b⃑⃗|=√2,所以cos⟨a ,b⃑⟩=a⃑ ⋅b⃑|a⃑ |⋅|b⃑|=4×√2=√22,又因为⟨a ,b⃑⟩∈[0,π],所以⟨a ,b⃑⟩=π4.故答案为:π4.14.曲线y=lnx−2x 在x=1处的切线的倾斜角为α,则sinα+cosαsinα−2cosα=___________.【答案】4【解析】f′(x)=1x +2x2,tanα=f′(1)=3,sinα+cosαsinα−2cosα=tanα+1tanα−2=3+13−2=4.15.沙漏是古代的一种计时装置,它由两个形状完全相同的容器和一个狭窄的连接管道组成,开始时细沙全部在上部容器中,利用细沙全部流到下部容器所需要的时间进行计时.如图,某沙漏由上、下两个圆锥组成,这两个圆锥的底面直径和高分别相等,细沙全部在上部时,其高度为圆锥高度(h)的23(细管长度忽略不计).假设细沙全部漏入下部后,恰好堆成一个盖住沙漏底部的圆锥形沙堆.这个沙堆的高与圆锥的高h 的比值为______. 【答案】827【解析】设沙漏上下两个圆锥的底面半径为r ,高为h , 左侧倒圆锥形沙堆的体积V 1=13π(2r 3)22ℎ3=881πr 2ℎ,右侧圆锥形沙堆的体积V 2=13πr 2ℎ′, 由V 1=V 2得ℎ′=827ℎ. 故答案为:827.16.已知数列{a n }满足a 1=2,n 2⋅a n+1=2(n +1)2⋅a n ,S n 为数列{a n }的前n 项和,则S n =___________. 【答案】(n 2−2n +3)⋅2n+1−6【解析】由n 2⋅a n+1=2(n +1)2⋅a n 得a n+1(n+1)2=2×a n n 2,又a 112=2,所以数列{ann 2}是等比数列,公比为2, 所以ann 2=2×2n−1=2n ,即a n =n 2⋅2n .S n =1×2+22×22+32×23+⋯+n 2×2n ,(1)(1)×2得2S n =1×22+22×23+⋯+(n −1)2×2n +n 2×2n+1,(2)(1)-(2)得:−S n =1×2+3×22+5×23+⋯+(2n −1)×2n −n 2×2n+1,(3) (3)×2得:−2S n =1×22+3×23+5×24+⋯+(2n −3)×2n +(2n −1)×2n+1−n 2×2n+2,(4) (3)-(4)得:S n =2+2×22+2×23+⋯+2×2n −(2n −1)×2n+1+n 2×2n+1 =2+8(1−2n−1)1−2−(2n −1)×2n+1+n 2×2n+1=(n 2−2n +3)×2n+1−6.故答案为:(n 2−2n +3)⋅2n+1−6.四、解答题17.已知等差数列{}n a 和正项等比数列{}n b 满足14a =,12b =,212n n n b b b ++=+,332a b =+. (1)求{}n a 和{}n b 的通项公式;(2)对于集合A 、B ,定义集合{A B x x A −=∈且}x B ∉,设数列{}n a 和{}n b 中的所有项分别构成集合A 、B ,将集合A B −的所有元素按从小到大依次排列构成一个新数列{}n c ,求数列{}n c 的前30项和30S .【答案】(1)a n =3n +1,b n =2n ;(2)S 30=1632【解析】(1)设等差数列{a n }公差为d ,等比数列{b n }的公比为q (q >0), ∵b n+2=b n+1+2b n ,∴q 2=q +2,解得q =2或q =−1<0(舍去). 又b 1=2,所以b n =2×2n−1=2n . 所以a 3=b 3+2=10,d =a 3−a 13−1=10−42=3,所以,a n =a 3+(n −3)d =10+3(n −3)=3n +1. (2)∵a 30=91,a 33=100,又b 6=64<121<b 7=128, 所以S 30中要去掉数列{b n }的项最多6项,数列{b n }的前6项分别为2、4、8、16、32、64,其中4、16、64三项是数列{a n }和数列{b n }的公共项,所以{c n }前30项由{a n }的前33项去掉{b n }的b 2=4,b 4=16,b 6=64这3项构成. S 30=(a 1+a 2+⋯+a 33)−(b 2+b 4+b 6)=33×(4+100)2−(4+16+64)=1632.18.已知四边形ABCD ,A ,B ,C ,D 四点共圆,AB =5,BC =2,cos∠ABC =−45.(1)若sin∠ACD =√55,求AD 的长;(2)求四边形ABCD 周长的最大值. 【答案】(1)5;(2)15√2+7【解析】(1)在△ABC 中,由余弦定理得 AC 2=AB 2+BC 2−2AB ⋅BC ⋅cos∠ABC=52+22−2×5×2×(−45)=45,得AC =3√5. 因为cos∠ABC =−45, 0<∠ABC <π,所以sin∠ABC =35. 因为A,B,C,D 四点共圆,所以∠ABC 与角∠ADC 互补, 所以sin∠ADC =35,cos∠ADC =45,在△ACD ,由正弦定理得:ADsin∠ACD =ACsin∠ADC , 所以AD =AC⋅sin∠ACD sin∠ADC=3√5×√5535=5.(2)因为四边形ABCD 的周长为DC +DA +BC +BA =DC +DA +7, 在△ACD 中,由余弦定理得:AC 2=DA 2+DC 2−2DA ⋅DC ⋅cos∠ADC , 即45=DA 2+DC 2−85DA ⋅DC =(DA +DC)2−185DA ⋅DC≥(DA +DC)2−185(DA+DC 2)2=110(DA +DC)2∴(DA +DC)2≤450, ∴DA +DC ≤15√2, 当且仅当DA =DC =15√22时,(DA +DC)max =15√2,所以四边形ABCD 周长的最大值为15√2+7.19.移动支付(支付宝及微信支付)已经渐渐成为人们购物消费的一种支付方式,为调查市民使用移动支付的年龄结构,随机对100位市民做问卷调查得到2×2列联表如下:(1)按年龄35岁以下(含35岁)是否使用移动支付的人群中采用分层抽样的方式抽取10人做进一步的问卷调查,从这10人随机中选出3人颁发参与奖励,设使用移动支付的人数为X ,求X 的分布列及期望.(2)用这100位市民使用移动支付的频率代替全市市民使用移动支付的概率,从全市随机中选出10人,则使用移动支付的人数最有可能为多少? 【答案】(1)分布列见解析,125;(2)6【解析】(1)根据分层抽样知使用移动支付的人数为8人,不使用移动支付的有2人,则X 的可能值为1,2,3, P(X =1)=C 81C 22C 103=115,P(X =2)=C 82C 21C 103=715,P(X =3)=C 83C 20C 103=715,分布列为E(X)=1×115+2×715+3×715=125.(2)从全市随机选出10人,设使用移动支付的人数为Y ,则Y~B(10,35),且P(Y =k)=C 10k(35)k (25)10−k (k ∈N,0≤k ≤10). 由{C 10k (35)k (25)10−k≥C 10k−1(35)k−1(25)11−kC 10k (35)k (25)10−k ≥C 10k+1(35)k+1(25)9−k , 解得285≤k ≤335,因为k ∈N ∗,所以k =6,故使用移动支付的人数最有可能为6.20.如图所示的几何体中,△ABE ,△BCE ,△DCE 都是等腰直角三角形,AB =AE =DE =DC ,且平面ABE ⊥平面BCE ,平面DCE ⊥平面BCE .(1)求证:AD ||平面BCE ; (2)求二面角B −AD −E 的余弦值. 【答案】(1)证明见解析;(2)13【解析】(1)证明:分别取EB,EC 的中点O,H ,连接AO,DH,OH ,设AB =AE =DE =DC =1,则EB =EC =√2, ∵AB =AE,BO =OE,∴AO ⊥BE ,又平面ABE ⊥平面BCE ,平面ABE ∩平面BCE =BE,AO ⊂平面ABE , ∴AO ⊥平面BCE , 同理可证DH ⊥平面BCE , ∴AO//DH ,又因为AO =DH =√22,所以四边形AOHD 是平行四边形,∴AD//OH , 又∵AD ⊄平面BCE,OH ⊂平面BCE , ∴AD//平面BCE ;(2)如图,取BC 的中点为F ,则OF ⊥BE ,以点O 为坐标原点,OB,OF,OA 所在的直线分别为x 轴,y 轴,z 轴,建立空间直角坐标系,则A (0,0,√22),B (√22,0,0),D (−√22,√22,√22),E (−√22,0,0),则BA⃑⃑⃑⃑⃑ =(−√22,0,√22),BD ⃑⃑⃑⃑⃑⃑ =(−√2,√22,√22), 设平面ABD 的一个法向量为m ⃑⃑⃗=(x,y,z ), 则{−√22x +√22z =0−√2x +√22y +√22z =0⇒{−x +z =0−2x +y +z =0,令x =1,得平面ABD 的一个法向量为m ⃑⃑⃗=(1,1,1). 则AE⃑⃑⃑⃑⃑ =(−√22,0,−√22),DE ⃑⃑⃑⃑⃑ =(0,−√22,−√22), 设平面ADE 的一个法向量为n ⃑⃗=(a,b,c ), 则{−√22a −√22c =0−√22b −√22c =0⇒{a +c =0b +c =0,令a =1,得平面ADE 的一个法向量为n⃑⃗=(1,1,−1), 设二面角B −AD −E 的大小为θ,则|cos θ|=|m ⃑⃑⃑⃗⋅n ⃑⃗||m⃑⃑⃑⃗||n ⃑⃗|=√3×√3=13,观察可知θ为锐角,所以二面角B −AD −E 的余弦值为13.21.已知椭圆C :x 2a 2+y 2b 2=1(a >b >0)的右焦点F (c,0)在直线√3x +y −2√3=0上,且离心率为12.(1)求椭圆C 的方程;(2)设A (−a,0),B (a,0),过点A 的直线与椭圆C 交于另一点P (异于点B ),与直线x =a 交于一点M ,∠PFB 的角平分线与直线x =a 交于点N ,是否存在常数λ,使得BN⃑⃑⃑⃑⃑⃑ =λBM ⃑⃑⃑⃑⃑⃑ ?若存在,求出λ的值;若不存在,请说明理由.【答案】(1)x216+y212=1;(2)存在,λ=12,理由见解析【解析】(1)因为右焦点F(c,0)在直线√3x+y−2√3=0上,所以√3c−2√3=0,∴c=2∵e=ca =2a=12,∴a=4,∴b2=16−4=12.所以椭圆C的方程为x216+y212=1.(2)存在,λ=12,理由如下:因为A(−4,0),B(4,0),F(2,0),设M(4,y1),N(4,y2),P(x0,y0). 显然y1y2>0.可设直线AP的方程为x=my−4(m≠0),因为点M在这条直线上,则my1=8,m=8y1.联立{x=my−43x2+4y2=48,得(3m2+4)y2−24my=0的两根为y0和0,∴y0=24m3m2+4,∴x0=my0−4=12m2−163m2+4.∵k PF=y0x0−2=24m3m2+412m2−163m2+4−2=4mm2−4=8y116−y12,k NF=y22.设∠BFN=θ,则∠PFB=2θ,∴tan2θ=2tanθ1−tan2θ=y21−(y22)2=4y24−y22=4mm2−4.∴8y116−y12=4y24−y22,∴(2y2−y1)(y1y2+8)=0,因为y1y2>0,所以2y2−y1=0,∴y2=12y1.故存在常数λ=12,使得BN⃑⃑⃑⃑⃑⃑ =λBM⃑⃑⃑⃑⃑⃑ .22.已知函数f(x)=2x−2(a+2)√x+alnx(a∈R).(1)讨论函数f(x)的单调性;(2)若函数g(x)=f(x2)−2(a+1)lnx的图象与x轴交于两点A(x1,0),B(x2,0),且0<x1<x2,设x0=λx1+μx2,其中常数λ、μ满足条件λ+μ=1,μ⩾λ>0,g′(x)为函数g(x)的导函数,试判断g′(x0)的正负,并说明理由.【答案】(1)答案见解析;(2)g′(x0)>0,理由见解析【解析】(1)由题意,函数f(x)=2x−2(a+2)√x+alnx(a∈R)的定义域为(0,+∞),可得f′(x)=2√x +ax=(√x−1)(2√x−a)x,(x>0),①当a≤0时,可得2√x−a>0,当x∈(0,1)时,f′(x)<0,当x∈(1,+∞)时,f′(x)>0,所以f(x)在(0,1)单调递减,在(1,+∞)单调递增;②当a=2时,可得f′(x)=2(√x−1)2x≥0在(0,+∞)上恒成立,所以函数f(x)在(0,+∞)上单调递增;③当0<a <2时,当x ∈(0,a 24)时,f ′(x )>0;当x ∈(a 24,1)时,f ′(x )<0; 当x ∈(1+∞)时,f ′(x )>0,所以f (x )在(a 24,1)递减,在(0,a 24),(1+∞)递增; ④当a >2时,当x ∈(0,1)时,f ′(x )<0;当x ∈(1,a 24)时,f ′(x )>0; 当x ∈(a 24,+∞)时,f ′(x )>0,所以f (x )在(1,a 24)递减,在(0,1),(a 24,+∞)递增. 综上,当a ≤0时,f (x )在(0,1)递减,在(1,+∞)递增;当a =2时,f (x )在(0,+∞)上单调递增:当0<a <2时,f (x )在(a 24,1)递减,在(0,a 24),(1,+∞)递增: 当a >2时,f (x )在(1,a 24)递减,在(0,1),(a 24,+∞)递增. (2)因为g (x )=f (x 2)−2(a +1)ln x =2[x 2−(a +2)x −ln x ],(x >0), 可得g ′(x )=2[2x −(a +2)−1x ],则g ′(x 0)=2[2x 0−(a +2)−1x 0], 因为函数y =g (x )的图象与x 轴交于两点A (x 1,0),B (x 2,0),且0<x 1<x 2.所以{x 12−lnx 1=(a +2)x 1x 22−lnx 2=(a +2)x 2,两式相减得(x 12−x 22)−(ln x 1−ln x 2)=(a +2)(x 1−x 2), 因为x 1−x 2≠0,所以a +2=(x 1+x 2)−lnx 1−ln x 2x1−x 2, 所以g ′(x 1+x 22)=2[(x 1+x 2)−(a +2)−2x 1+x 2]=2[ln x 1−ln x 2x 1−x 2−2x 1+x 2] =2x 1−x 2[(ln x 1−ln x 2)−2(x 1−x 2)x 1+x 2]=2x 1−x 2[ln x 1x 2−2(x 1x 1−1)x 1x 2+1],令t =x 1x 2,因为0<x 1<x 2,可得0<t <1,令u (t )=ln t −2(t−1)t+1,可得u ′(t )=1t −4(t+1)2=(t−1)2t(t+1)2, 又由0<t <1,所以u ′(t )>0,所以u (t )在(0,1)上是増函数,则u (t )<u (1)=0, 所以ln x 1x 2−2(x 1x 2−1)x 1x 2+1<0, 又因为2x 1−x 2<0,所以g ′(x 1+x 22)>0, 因为x 0=λx 1+μx 2,λ+μ=1,μ≥λ>0,所以μ≥12≥λ>0,所以x 0−x 1+x 22=(1−μ)x 1+μx 2−x 1+x 22=(μ−12)(x 2−x 1)≥0,所以x 0≥x 1+x 22, 因为g ′(x )在(0,+∞)上单调遌增,所以g ′(x 0)≥g ′(x 1+x 22)>0,所以g ′(x 0)>0.。
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2018届高三年级华附、省实、深中、广雅四校联考英语本试卷分选择题和非选择题两部分,共10页,满分120分,考试用时120分钟。
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第Ⅰ卷选择题(共70分)第一部分阅读理解(共两节,满分40分)第一节阅读下列短文,从每题所给的四个选项(A、B、C和D)中,选出最佳选项,并在答题卡上将该项涂黑。
(共15小题;每小题2分,满分30分)AFamily structure has changed dramatically over the last 50 years. The "Leave it to Beaver" family is no longer the standard, and several variations on family have been created.Nuclear FamilyThe nuclear family is the traditional type of family structure. This family type consists of two parents and children. The nuclear family was long held in esteem by society as being the ideal in which to raise children. Children in nuclear families receive strength and stability from the two- parent structure and generally have more opportunities due to the financial ease of two adults. According to . Census data, almost 70 percent of children live in a nuclear family unit.Single Parent FamilyThe single parent family consists of one parent raising one or more children on his own. The single parent family is the biggest change society has seen in terms of the changes in family structures. One in four children is born to a single mother. Single parent families are generally close and find ways to work together to solve problems, such as dividing up household chores. When only one parent is at home, it may be a struggle to find childcare, as there is only one parent working, which limits income and opportunities in many cases.Extended FamilyThe extended family structure consists of two or more adults who are related, either by blood or marriage, living in the same home. This family includes many relatives living together and working toward common goals, such as raising the children and keeping up with the household duties. Many extended families include cousins, aunts or uncles and grandparents living together. This type of family structure may form due to financial difficulties or because older relatives are unable to care for themselves alone. Extended families are becoming increasingly common all over the world.Childless FamilyWhile most people think of family as including children, there are couples who either cannot or choose not to have children. The childless family is sometimes the "forgotten family," as it does not meet the traditional standards set by society. Childless families consist of a husband and wife living and working together. Many childless families take on the responsibility of pet ownership or have extensive contact with their nieces and nephews as a substitute for having their own children. 1.In what aspect do single parent families usually work togetherA.eldercare B.finance C.housework D.tutoring 2.Which of the following belongs to the type of extended familyA.Jack and Betty, who have been married for over 15 years, have two children named Daniel and Chirstine.B.Jose and Maria live with their 3 children. They also live with their parents, an uncle and an aunt.C.Two years ago Darrel and Tanya got married. Before they got married Darrel had three sons and Tanya had a daughter. Now they all live together.D.Harry and his wife Jane have been married for six years. They have one daughter named Melissa.3.What can be inferred from the textA.Nuclear families are most common in America.B.The "Leave it to Beaver" family used to be a family type.C.Childless families tend to live with their nieces and nephews.D.Extended families form to take care of the old.BWhen I was five or six years old, I remember watching TV and I would see these commercials and I was watching other children suffer in other parts of the world and you know the commercials were like, "you can give 25 cents, save a child's life." And I would think to myself like, I wonderhow many 25 cents I could save up to save all the kids in Africa. And I would say to myself, "When I grow up, when I can get rich, I'll save kids all over the world." I just didn't know I would be in the position to do that by the time I was a teenager.At 17 I started my career here in America, and by the age of 18, I started my first charity organization. I went on to team up with other organizations in the following years and met, helped, and even lost some of the most beautiful souls, such as six-year-old Jasmina Anema who passed away in 2010 from leukemia. Her story inspired thousands to volunteer as donors through DKMS. Fast forward to 2012 and then my grandmother, the late Clara Brathwaite, she lost her battle with cancer, which is the very reason and the driving force behind the Clara Lionel Foundation. At CLF, our mission is to impact as many lives as possible, but it starts with just one. Just one.All you need to do is help one person, expecting nothing in return. To me, that is a humanitarian People make it seem too hard, man. The truth is, and what I want the little girl watching those commercials to know, is you don't have to be rich to help somebody. You don't have to be famous. You don't even have to be college-educated.But it starts with your neighbor, the person sitting next to you in class, the kid down the block in your neighborhood, you just do whatever you can to help in any way that you can.4.We know from the passage that ________.A.the speaker thought it hard to save 25 centsB.the speaker liked watching commercials on TVC.the speaker was confused by the TV commercialsD.the speaker helped others earlier than she had thought5.What's the correct order of the events that happeneda.The speaker built up her career in America.b.The speaker wondered when and how she could help kids in Africa.c.The speaker's grandmother passed away.d.Jasmina Anema died from leukemia.e.The speaker set up her first charity organization.f.Clara Lionel Foundation was founded.A.a, b, d, e, c, f B.b, a, e, d, c, f C.b, a, d, c, e, f D.b, d, c, f, a, e 6.Who does "the little girl" in Paragraph 3 probably refer toA.Anyone who wants to help. B.The speaker herself.C.The audience present. D.The speaker's daughter.7.What's the best title for the textA.Life Isn't Always Tough B.It's Easy to Be a HumanitarianC.You don't Have to Be Rich to Help D.Be Always Ready to Take ChallengeCA 12-year-old with end stage cancer, the child's parents had recently moved her from the hospital to her home in the suburbs of Los Angeles. Some days later the girl's breath quickened, and her father phoned the family's hospice nurse.The nurse knew the visit would require more than four hours of her time: a two-hour drive in each direction, plus her time with the girl. "Why don't we connect over FaceTime" she asked. The father agreed, and they connected.The nurse asked the father to move his daughter gently to her side, then to her back, to lift the child's shirt. T he nurse would ask: what do you see, what concerns you, and the father would explain. In this fashion the pair examined the girl—the nurse on her computer, the father his iPad. Together they decided that the nurse's presence was not necessary, that the child had more time.Later, the father reported feeling comforted by the nurse. He appreciated her availability, the fact that she could see what he saw, and their ability to discuss it in real time.Telemedicine has become a trend in America. Some hospitals have gone so far as to specially design telemedicine clinics. The room is staged like an office but with better lighting. There's a nice desk for the clinician to sit behind, a computer situated stage left, and books in the background. A physician taking a video call from home might wear a pair of headphones, equipped with a mic—to ensure whatever the patient says isn't broadcast to anyone off-camera.However, there is something more than the technology bit. According to David, the head of the telemedicine pilot the girl's family had been part of, there are some people who are great in person and you put them on camera they're a dead fish. "Some physicians are camera shy, and thus usually seem ignorant of patients' feelings. For others, the physical isolation can actually help them be more empathetic." He adds."My experience is that, once you get past some initial hurdles, you can maintain an intimate, immediate connection with patients that in some cases may be more useful than even in-person interactions." David says.8.In the case mentioned by the writer, we learn that ________.A.a physician was arranged after the nurse knew it would take a long timeB.the nurse successfully diagnosed the girl with an Ipad over FaceTimeC.the girl's father showed satisfaction and thought highly of the serviceD.the 12-year-old girl was dying and should be sent to hospital in no time9.What do we know about the rooms for telemedicineA.Better lights are provided for a comfortable environment.B.Headphones with mics are equipped to protect privacy of patients.C.Professional books are in place for immediate consultation.D.The room with a computer is situated in the doctor's home.10.What challenges may clinicians face in telemedicineA.They are afraid of being isolated from patients.B.They could not examine the patients carefully.C.They are required to deal with in-person interactions.D.They may not feel quite themselves in front of a camera.11.What does the underlined word "empathetic" in paragraph 6 probably mean A.understanding B.independent C.indifferent D.confidentDEvery country may have to consider what, if anything, to do about "global warming". We should understand that the oft-repeated claim that nearly all scientists demand that something dramatic be done to stop global warming is not true.Perhaps the most inconvenient fact is the lack of global warming for well over 10 years now. This is known to the warming establishment, as one can see from the 2009 "Climategate" email of climate scientist Kevin Trenberth: "The fact is that we can't account for the lack of warming at the moment and it is an irony that we can't." But the warming is only missing if one believes computer models where so-called feedbacks involving water vapor and clouds greatly amplify the small effect of CO2. The fact is that CO2 is not a pollutant and it is a key component of the biosphere(生物圈)'s life cycle. Plants get better growth with more of it and part of the increase of agricultural yields in the past century certainly came from additional CO2 in the atmosphere.Although the number of the scientists who are publicly opposed to the claim is growing, many young scientists secretly say that while they also have serious doubts about the global-warming message, they are afraid to speak up for fear of not being promoted or worse.Why is there so much passion about global warming There are several reasons, but a good place to start is the old question "Cuibono", or the modern update, "Follow the money". Alarmism over climate is of great benefit to many, providing government funding for academic research, and thus those people who benefit from this fiercely defended their dogma and the privileges it brought them.Every country should support rational measures to protect and improve our environment, but it makes no sense at all to back expensive programs that turn away resources from real needs and are based on alarming but shaky claims of "incontrovertible" evidence.12.According to the passage, global warming ________.A.hasn't happened in the latest 10 yearsB.has become an urgent problem these yearsC.has worsened a little bit in the recent yearsD.has never happened since measures were taken13.Which of the following statements about CO2 does the writer probably agree with A.CO2 greatly affects the global warming.B.CO2 makes crops more productive.C.CO2 involves water vapor and clouds.D.CO2 is the most important part of the biosphere.14.Why is there so much passion for global warmingA.Because it is a good chance for young scientists to be promoted.B.Because the scientists want to have an insight of the problem.C.Because the government values the problem and tries to solve it.D.Because some people can greatly benefit from the research about it.15.What's the writer's attitude toward the programs to protect and improve our environment A.To support all the programs as long as they are beneficial.B.To support cheap programs instead of expensive ones.C.To support the programs that are necessary and reliable.D.To support those programs that use less money and resource.第二节根据短文内容,从短文后的选项中选出能填入空白处的最佳选项,并在答题卡上将该项涂黑。