广西玉林市博白县2015届高三下学期返校调研考试数学(理)试题 Word版含答案
广西玉林市数学高三理数第二次模拟考试试卷
广西玉林市数学高三理数第二次模拟考试试卷姓名:________班级:________成绩:________一、 单选题 (共 12 题;共 24 分)1. (2 分) 在复平面内,复数 z= (i 是虚数单位)对应的点位于( ) A . 第一象限 B . 第二象限 C . 第三象限 D . 第四象限2. (2 分) (2017 高二下·黄冈期末) 已知集合 A={﹣1, 组成的集合是( )},B={x|mx﹣1=0},若 A∩B=B,则所有实数 mA . {0,﹣1,2}B.{,0,1}C . {﹣1,2}D . {﹣1,0, } 3. (2 分) (2016 高一下·石门期末) 设{an}是公差为正数的等差数列,若 a1+a2+a3=15,a1a2a3=80,则 a11+a12+a13=( ) A . 120 B . 105 C . 90 D . 75 4. (2 分) (2015 九上·沂水期末) 执行如图所示的程序框图,输出的 S 值为( )第 1 页 共 13 页A.B. C. D. 5. (2 分) 由 a,b,c,d,e 这 5 个字母排成一排,a,b 都不与 c 相邻的排法个数为 A . 36 B . 32 C . 28 D . 246. (2 分) 若实数 x,y 满足不等式组 A.6 B.4 C . -2 D . -6则 2x+4y 的最小值是( )7. (2 分) (2017 高三上·湖南月考) 已知椭圆的离心率为 ,双曲线第 2 页 共 13 页() 的离心率为 ,抛物线的离心率为 ,,是( )A.B.C.D.,,则之间的大小关系8. (2 分) (2020·辽宁模拟) 已知椭圆,直线上存在一点 满足A.()的右焦点为,上顶点为,则椭圆的离心率取值范围为( )B.C.D.9. (2 分) (2018 高一下·宜昌期末) 三棱锥中,为 的等边三角形,则该三棱锥外接球的表面积为( )A. B. C.第 3 页 共 13 页且,是边长D.10. (2 分) (2019 高一下·蛟河月考) 为了得到函数 象( )的图象,可以将函数的图A . 向右平移 个单位长度 B . 向右平移 个单位长度 C . 向左平移 个单位长度 D . 向左平移 个单位长度 11.(2 分)(2016 高二下·松原开学考) 已知双曲线的中心为原点,F(3,0)是双曲线的﹣个焦点, 是双曲线的一条渐近线,则双曲线的标准方程为( )A. B. C. D.12. (2 分) (2019 高二下·鹤岗月考) 已知函数,,若存在,使得,则 的取值范围是( )A.B.C.D.二、 填空题 (共 4 题;共 5 分)第 4 页 共 13 页13. (1 分) (2016 高二上·汕头期中) 在△ABC 中,∠C= ,∠B= ,AC=2,M 为 AB 中点,将△ACM 沿 CM 折起,使 A,B 之间的距离为 2 ,则三棱锥 M﹣ABC 的外接球的表面积为________.14. (1 分) (2017·潮州模拟) 在梯形 ABCD 中,AD∥BC, •相交于点 E,⊥,则 •=________=0,||=2,||=4,AC 与 BD15. (1 分) (2015 高三上·石景山期末)的二项展开式中 x 项的系数为________.(用数字作答)16. (2 分) (2019 高三上·汕头期末) 分形几何学是一门以不规则几何形态为研究对象的几何学.分形的外 表结构极为复杂,但其内部却是有规律可寻的.一个数学意义上分形的生成是基于一个不断迭代的方程式,即一种 基于递归的反馈系统.下面我们用分形的方法来得到一系列图形,如图 1,线段 的长度为 ,在线段 上取两个点 , ,使得,以为一边在线段 的上方做一个正六边形,然后去掉线段,得到图 2 中的图形;对图 2 中的最上方的线段 作相同的操作,得到图 3 中的图形;依此类推,我们就得到了以下一系列图形:记第 个图形(图 1 为第 1 个图形)中的所有线段长的和为 ,则(1)________;(2) 如果对,恒成立,那么线段 的长度 的取值范围是________.三、 解答题 (共 7 题;共 55 分)17. (10 分) (2018·东北三省模拟) 已知,且.的内角 , , 的对边分别为 , , ,若(1) 求 的大小;第 5 页 共 13 页(2) 求面积的最大值.18. (5 分) (2016·肇庆模拟) 某市一次全市高中男生身高统计调查数据显示:全市 100 000 名男生的身高 服从正态分布 N(168,16).现从某学校高三年级男生中随机抽取 50 名测量身高,测量发现被测学生身高全部介于 160cm 和 184cm 之间,将测量结果按如下方式分成 6 组:第一组[160,164],第二组[164,168],…,第 6 组[180, 184],如图是按上述分组方法得到的频率分布直方图.(Ⅰ)试评估该校高三年级男生在全市高中男生中的平均身高状况;(Ⅱ)求这 50 名男生身高在 172cm 以上(含 172cm)的人数;(Ⅲ)在这 50 名男生身高在 172cm 以上(含 172cm)的人中任意抽取 2 人,该 2 人中身高排名(从高到低)在 全市前 130 名的人数记为 ξ,求 ξ 的数学期望.参考数据:若 ξ﹣N(μ,σ2),则 p(μ﹣σ<ξ≤μ+σ)=0.6826,p(μ﹣2σ<ξ≤μ+2σ)=0.9544, p(μ﹣3σ<ξ≤μ+3σ)=0.9974.19. (5 分) (2017·山西模拟) 已知如图所示的几何体中,四边形 ABCD 是边长为 2 的菱形,面 PBC⊥面 A BCD,点 E 是 AD 的中点,PQ∥面 ABCD 且点 Q 在面 ABCD 上的射影 Q′落在 AB 的延长线上,若 PQ=1,PB=• =0,=2,且( )(Ⅰ)求证面 PBC⊥面 PBE第 6 页 共 13 页(Ⅱ)求平面 PBQ 与平面 PAD 所成钝二面角的正切值.20. (5 分) (2018·南阳模拟) 已知抛物线交曲线 于两点,交圆于的焦点为,过点 且斜率为 的直线两点(两点相邻).(Ⅰ)若,当时,求 的取值范围;(Ⅱ)过两点分别作曲线 的切线,两切线交于点 ,求与面积之积的最小值.21. (10 分) (2017·扶沟模拟) 已知函数 f(x)=(x+a)ln(x+a),g(x)=﹣+ax.(1)函数 h(x)=f(ex﹣a)+g'(ex),x∈[﹣1,1],求函数 h(x)的最小值;(2)对任意 x∈[2,+∞),都有 f(x﹣a﹣1)﹣g(x)≤0 成立,求 a 的范围.22. (10 分) (2015 高三上·贵阳期末) 选修 4﹣4:坐标系与参数方程.极坐标系与直角坐标系 xoy 有相同的长度单位,以原点为极点,以 x 轴正半轴为极轴,已知曲线 C1 的极坐标方程为 ρ=4cosθ,曲线 C2 的参数方程为(t 为参数,0≤α<π),射线 θ=φ,θ=φ+ ,θ=φ﹣ 与曲线 C1 交于(不包括极点 O)三点 A、B、C.(1) 求证:|OB|+|OC|= |OA|;(2) 当 φ= 时,B,C 两点在曲线 C2 上,求 m 与 α 的值. 23. (10 分) (2018 高二下·衡阳期末) 已知函数 (1) 求实数 的值;的最小值为 .(2) 若,且,求证:.第 7 页 共 13 页一、 单选题 (共 12 题;共 24 分)1-1、 2-1、 3-1、 4-1、 5-1、 6-1、 7-1、 8-1、 9-1、 10-1、 11-1、 12-1、二、 填空题 (共 4 题;共 5 分)13-1、 14-1、 15-1、参考答案第 8 页 共 13 页16-1、 16-2、三、 解答题 (共 7 题;共 55 分)17-1、17-2、18-1、第 9 页 共 13 页第 10 页 共 13 页20-1、21-1、21-2、22-1、22-2、23-1、23-2、。
广西南宁、玉林、柳州、桂林2015届高三第一次适应性检测数学(理)试题
⼴西南宁、⽟林、柳州、桂林2015届⾼三第⼀次适应性检测数学(理)试题2015年⾼中毕业班第⼀次适应性检测数学试卷(理科)第I 卷⼀.选择题:本⼤题共12⼩题,每⼩题5分共60分。
1.已知全集U =R ,集合A ={x |x 2+3x -10>0},B ={x |-2≤x ≤5},则(?U A )∩B 等于(A ){x |=5<x ≤3}(B ){x |-2<x ≤5}(C ){x |-2≤x ≤2}(D ){x |-5≤x ≤5}2.设复数z 满⾜z ?(1-i )=2,则复数z 的模|z |等于(A ) 2(B )2(C ) 5(D )43.设等⽐数列{a n }的前n 项和为S n ,若a 1=1,a 4=-8,则S 5等于(A )-11(B )11(C )31(D )-314.下列函数中,既是偶函数,⼜是在区间(0,+∞)上单调递减的函数是(A )y =ln x(B )y =x 2(C )y =cos x(D )y =2-|x |5.(1-x )5的展开式中,x 2的系数(A )-5(B )5(C )-10(D )106.已知x ,y 满⾜?x -2y +2≥0x ≤4y ≥-2 ,则⽬标函数z =x +y 的最⼤值为(A )4(B )5(C )6(D )77.如图所⽰的程序框图中输出的a 的结果为(A )2(B )-2(C )12(D )-128.已知底⾯为正⽅形的四棱锥,其⼀条侧棱垂直于底⾯,那么该四锥的三视图可以是下列各图中的DCB A府视图府视图正视图正视图正视图正视图正视图侧视图府视图府视图侧视图正视图9.已知函数f (x )=sin(x +π6),其中x ∈[-π3,a ],若f (x )的值域是[-12,1],则a 的取值范围是(A )(-,π3](B )[π3,π2](C )[π2,2π3](D )[π3,π]10.甲和⼄等五名志愿者被随机地分到A 、B 、C 、D四个不同的岗位服务,每个岗位⾄少有⼀名志愿者,则甲和⼄不在同⼀岗位服务的概率为 (A )910(B )110(C )14(D )4862511.双曲线x 2a 2-y 2b 2=1(a >0,b >0)与抛物线y 2=2px (p >0)相交于A 、B 两点,直线AB 恰好过它们的公共焦点F ,则双曲线C 的离⼼率为(A ) 2(B )1+ 2(C )2 2 (D )2+ 212.定义域为[a ,b ]的函数y =f (x )的图象的两个端点为A 、B ,M (x ,y )是f (x )图象上任意⼀点,其中x =λa +(1-λ)b ,向量?→ON =λ?→OA +(1-λ)?→OB ,其中O 为坐标原点,若不等式|?→MN |≤k 恒成⽴,则称函数f (x )在[a ,b ]上“k 阶线性近似”,若函数y =x +1x 在[1,2]上“k 阶线性近似”,则实数k 的取值范围为 (A )[0,+∞)(B )[1,+∞)(C )[32-2,+∞)(D )[32+2,+∞)⼆.填空题:本⼤题共4⼩题,每⼩题5分,共20分。
广西玉林市博白县高三历史下学期返校调研考试试题(含解析)
广西玉林市博白县2015届高三下学期返校调研考试历史试题时量:90分钟总分:100分一、选择题:每小题的四个选项中,只有一个选项符合题目要求,每小题2分,共48分。
将答案代号填入答题卷答题栏内。
1.根据“家”字的构成,最早使用这一词表示居住处所的是A.皇帝 B.贵族 C.平民 D.奴隶【答案】C【解析】考点:汉字“家”的理解。
皇帝居所称“宫”,贵族的称“府”,“家”字本义为穴下有猪,即猪舍。
最初肯定适合平民。
奴隶没有私有财产,不适用此义。
答案为C。
2.“汉代门阀的力量,无论在经济上、社会上、政治上都充分地表现出他们的优势。
……他们凭藉祖先的余荫,不但垄断官吏选举之权,凡州郡掌管中正的官吏,都非由他们中择人任用不可……”这反映的社会深层次问题是A.九品中正制的发展 B.土地兼并的加剧C.中央集权的加强 D.门阀世家的壮大【答案】B【解析】考点:古代选官制度。
本题属区分度题。
解题关键是“社会深层次问题”。
材料是A 和D的表现,不是“深层次问题”,材料与C无关,一般认定是土地问题才是“社会深层次问题”。
答案为B。
3.董仲舒提出:“天生民性,有善质而未能善,于是为之立王以善之,此天意也。
”这一观点在当时的主要影响是A.确立了儒学独尊的地位 B.发展了“天人感应”的理念C.使汉王朝开始重视文化教育 D.为君主受命于天提供理论保障【答案】D【解析】考点:董仲舒的思想。
材料的含义是“君主受命于天”,答案为D。
董仲舒是创立了“天人感应”,不是“发展”。
答案为D4.《旧唐书·食货上》载:“武德七年,始定律令……所授之田,十分之二为世业,八为口分。
世业之田,身死则承户者便授之。
口分,则收入官,更以给人。
”材料表明唐代均田制A.根本上解决了土地兼并问题 B.目的是维护土地国有制C.有利于土地与劳动者的结合 D.有效提高了土地的产量【答案】C【解析】考点:唐代均田制。
本题为材料选择题,材料介绍了唐代的均田制。
广西玉林市博白县高三数学下学期返校调研考试试题 理
广西玉林市博白县2015届高三下学期返校调研考试数学(理)试题(考试时量:120分钟 满分150分) 一:单选题:(本大题共10小题,每小题5分,共50分.在每小题给出的四个选项中,只有一项是符合题目要求的。
) 1.设复数11i z =+,22i ()z x x R =+∈,若12R z z ⋅∈,则x =A .1-B .2-C .1D .22.下列命题中的假命题是A .021>∈∀-xR x , B .212),0x x x>∞+∈∀ , ( C .4001.1,x x x R x x <>∈∃时,恒有 当D .R ∈∃α,使函数 αx y =的图像关于y 轴对称对于定义在实数集R 上的狄利克雷函数⎩⎨⎧∈∉=),(),(Q x Q x x D 10)(,则下列说法中正确的是 A .)(x D 的值域是[]01, B .)(x D 的最小正周期是1C .)(xD 是奇函数 D .)(x D 是偶函数4.若一个底面是正三角形的三棱柱的正视图如图所示,其顶点都在一个球面上,则该球的表面积为A.163πB. 43πC.193πD. 1219π5.已知0>t ,若8)22(0=-⎰tdx x ,则t =A.1B.2-C.2-或4D. 46.已知随机变量X 服从正态分布(3,1)N ,且(15)0.6826P X ≤≤=,则(5)P X >= A .0.1588B .0.1587C .0.1586D .0.15857.已知函数()π()sin (0,0,)2f x A x A ωϕωϕ=+>><的 部分图象如图所示,则ϕ=A .π6-B .6πC .π3-D .π38.在“学雷锋,我是志愿者”活动中,有6名志愿者要分配到3个不同的社区参加服务,每个社区分配2名志愿者,其中甲、乙两人分到同一社区,则不同的分配方案共有 A. 12种 B. 18种 C. 36种 D. 54种9.在平面直角坐标系中,O 为坐标原点,点(2,0)A ,将向量OA u u u r 绕点O 按逆时针方向旋转3π后得向量OB u u u r,若向量a r 满足1a OA OB --=r u u u r u u u r ,则ar 的最大值是A.1- B. C .3 D10.已知函数()y f x =是定义域为R 的偶函数. 当0x ≥时,25(02)16()1()1(2)2x x x f x x ⎧≤≤⎪⎪=⎨⎪+>⎪⎩ 若关于x 的方程2[()]()0f x af x b ++=,a b R ∈、有且仅有6个不同实数根,则实数a 的取值范围是A .59(,)24--B .9(,1)4-- C. 599(,)(,1)244----U D .5(,1)2--二.填空题:(本题共5小题,每小题5分,共25分) 11.已知函数()ln f x x=,则()1f x >的解集为12.春节期间,某销售公司每天销售某种取暖商品的销售额y (单位:万元)与当天的平 均气温x (单位:℃)有关.现收集了春节期间这个销售公司4天的x 与y 的数据列于下 表:平均气温(℃) 2- 3- 5- 6-销售额(万元)20 23 27 30根据以上数据,用线性回归的方法,求得y 与x 之间的线性回归方程y b x a ∧∧∧=+的系数125b ∧=-,则a ∧= .13.在等比数列{}n a 中,11a =,且14a 、22a 、3a 成等差数列,则通项公式n a =14.某程序框图如右图所示,该程序运行后,输出的x 值为31,则a 等于_____ 15.定义函数{}{}()f x x x =⋅,其中{}x 表示不小于x 的最小整数,如{}1.52=,{}2.52-=-.当(]0,x n ∈,*n N ∈时,函数()f x 的值域为n A ,记集合n A 中元素的个数为n a ,则12111n a a a +++=L ________ 三:解答题:(本大题共6小题,共75分。
高三物理月考试题及答案-广西玉林市博白县2015届高三下学期返校调研考试
广西玉林市博白县2015届高三下学期返校调研考试物理试题一、选择题:本题共12小题,每小题给出的四个选项中,1-9题只有一个选项正确,10、11、12有多个选项正确;全部选对得4分,选不全的得2分,有选错或不答的得0分。
(共48分)1.意大利科学家伽利略在研究物体变速运动规律时,做了著名的“斜面实验”,他测量了铜球在较小倾角斜面上的运动情况,发现铜球做的是匀变速直线运动,且铜球加速度随斜面倾角的增大而增大,于是他对大倾角情况进行了合理的外推,由此得出的结论是A .物体都具有保持原来运动状态的属性,即惯性B .力是维持物体运动的原因C .力是产生加速度的原因D .自由落体运动是一种匀变速直线运动2.如图,在灭火抢险的过程中,消防队员有时要借助消防车上的梯子爬到高处进行救人或灭火作业.为了节省救援时间,人沿梯子匀加速向上运动的同时消防车匀速后退,则关于消防队员的运动,下列说法中正确的是A .消防队员做匀加速直线运动B .消防队员做匀变速曲线运动C .消防队员做变加速曲线运动D .消防队员水平方向的速度保持不变3.质量为m 的小球由轻绳a 和b 系于一轻质木架上的A 点和C 点,且b a L L ,如图所示.当轻杆绕轴BC 以角速度ω匀速转动时,小球在水平面内作匀速圆周运动,绳a 在竖直方向、绳b 在水平方向.当小球运动在图示位置时,绳b 被烧断的同时杆也停止转动,则A .小球仍在水平面内作匀速圆周运动B .在绳被烧断瞬间,a 绳中张力突然增大C .在绳被烧断瞬间,小球所受的合外力突然变小D .若角速度ω较大,小球可以在竖直平面ABC 内作圆周运动4.如图所示,ABCD 为竖直平面内正方形的四个顶点,AD 水平,分别从A 点和D 点以速度1v 、2v 各平抛一个小球,两小球均能经过AC 上的E 点,且从D 点抛出的小球经过E 时的速度方向与AC 垂直,不计空气阻力。
则下列正确的是A .两小球的到达E 点所用时间不等B .两小球的速度变化不相同C .两小球的初速度大小关系为:D .若1v 、2v 取合适的值,则E 可以是AC 的中点5.一质量为m 的物体,只受重力和另一恒力F 作用,在竖直平面内以初速度0v 沿直线从O 点运动到A 点,已知l OA =,与竖直方向的夹角为θ,如图所示,则下列说法正确的是A .当θsin mg F =时,物体作加速度为θsin g 的加速运动B .若θtan mg F =,则恒力F 一定做正功C .当θcos mg F =时,物体重力势能的减少量等于其动能的增加D .物体可能作减速运动6.地球赤道地面上有一物体随地球的自转而做圆周运动,所受的向心力为F 1,向心加速度为a 1,线速度为v 1,角速度为ω1;绕地球表面附近做圆周运动的人造卫星(高度忽略)所受的向心力为F 2 ,向心加速度为a 2,线速度为v 2,角速度为ω2;地球同步卫星所受的向心力为F 3 ,向心加速度为a 3,线速度为v 3 ,角速度为ω3;地球表面重力加速度为g ,第一宇宙速度为v ,假设三者质量相等,则下列结论正确的是A .F 1=F 2>F 3B .a 1=a 2=g >a 3C .v 1=v 2=v >v 3D .ω1=ω3<ω27.光滑水平面上有一边长为L 的正方形区域处在电场强度为E 的匀强电场中,电场方向与正方形一边平行。
高三英语月考试题及答案-玉林市博白县2015届高三下学期返校调研考试
广西玉林市博白县2015届高三下学期返校调研考试英语试题本试卷分为四个部分,包括听力、语言知识运用、阅读理解和书面表达。
考试结束后,将本试卷和答题卡一并交回。
Part I Listening Comprehension (30 marks)Section A (22.5 marks)Directions: In this section you’ll hear six conversations between two speakers. For each conversation, there are several questions and each question is followed by 3 choices marked A, B and C. Listen carefully and then choose the best answer for each question.You will hear each conversation TWICE.Example;When will the magazine probably arrive?A. Wednesday.B. Thursday.C. Friday.The answer is B.Conversation 11. What is the possible relationship between the speakers?A. Professor and student.B. Waiter and customer.C. Doctor and patient.2. What’s wrong with the woman?A. She has a backache.B. She has a sore throat.C. She has a fever. Conversation 23. What will the woman do this weekend?A. See a movie.B. Return books.C. Buy a machine.4. What can we know about the woman from the conversation?A. She will try to use a machine.B. She will go to the library before 8:00 am.C. She is not interested in movies.Conversation 35. Where does the man want to go?A. To the White House.B. To some museums.C. To the Capital Building.6. What does the man think of the White House?A. It’s the most famous.B. It’s historic.C. It’s not interesting. Conversation 47. How long does it take to send a package to the USA by air?A. About 10 days.B. Three days.C. Four days.8. Which way does the woman finally choose to send the package?A. By air.B. By express mail.C. By sea.9. How much does the woman pay for the package?A. 325 yuan.B. 352 yuan.C. 253 yuan. Conversation 510. What did the man’s neighbors do last night?A. They had a party.B. They had an argument.C. They watched TV.11. What does the man ask the woman to do?A. Clean the room.B. Change the bed sheets.C. Get fresh towels.12. What is the man going to do?A. Close the window.B. See his friends.C. Enjoy some fresh air. Conversation 613. What foreign languages can the woman speak?A. English and Spanish.B. French and Spanish.C. Russian and French.14. What su bject is the woman’s best one?A. French.B. Writing.C. Spanish.15. How is the woman’s Spanish?A. She can’t speak it at all.B. It is as good as her mother tongue.C. Her written Spanish is not very good.SECTION BDirections: In this section, you will hear a short passage. Listen carefully and then fill in the numbered blanks with the information you’ve heard. Fill in each blank with NO MORETHAN .You will hear the short passage TWICE.Part П Language Knowledge (45 marks)Section A (15 marks)Directions: Beneath each of the following sentences there are four choices marked A, B, C and D. Choose the one answer that best completes the sentence.Example:The wild flowers looked like a soft orange blanket the desert.A. coveringB. coveredC. coverD. to coverThe answer is A.21. The main street is lined with small stands and shops _____ sell almost anything you canimagine.A. whatB. whereC. whoD. which22. —Wow, you have a really good voice. I_____ you were good at singing.—Thank you.A. haven't knownB. hadn't knownC. didn't knowD. don't know23. The scientist spent years alone in the forest, ______ by eating wild fruit and small animals.A. survivingB. survivedC. to surviveD. having survived24. Mom, don’t you think it's OK to play once in a while it doesn’t affect my studies?A. even ifB. unlessC. so thatD. as long as25. —Why not find a new job?—Why should I? I _____ as much, but I like what I'm doing.A. didn't earnB. don't earnC. hadn't earnedD. haven’t earned26. Anyone be in a rough life time, whether he is “Bai Fumei” or “Gao Fushuai.”A. mustB. willC. canD. should27. We believe the time and hard work _______ in completing such an important project are worthwhile.A. involvedB. involvingC. to involveD. to be involved28. When word came _______ Zhang Bichen became the winner of the “V oice of China 2014”,her fans, greatly excited, stood up cheering for her.A. whatB. thatC. whichD. where29. —Look, the paint is starting to come off ______ the wall is damp.—Well, we’d better have the wall repainted some day.A. whenB. onceC. ifD. where30. Time to us all is limited. So in the days _______, we must work out a practical plan for ourstudy and keep to it strictly.A. followingB. to followC. followedD. being followed31. If Susan ______ for ten more minutes at the party last night, she might have met Johnny Depp,the famous Hollywood star.A. stayedB. would stayC. had stayedD. would havestayed32. —Will you go to Mary’s wedding party tonight?—I’m afraid not. I _______ my brother’s.A. will attendB. have attendedC. am attendingD. will be attending33. Playing a role in the hit show “Daddy, Where Are We Going?”, Kimi, together with the otherfour kids, _______ popular with many teenagers now.A. isB. areC. wasD. were34. It is not doing the things we like, but liking the things we have to do ______ makes life happy.A. whichB. thatC. whatD. who35. ______yourself with positive people and you will keep focused on what you can do instead ofwhat you can’t.A. SurroundingB. SurroundedC. SurroundD. Having surroundedSection B (18 marks)Directions: For each blank in the following passage there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.My family suffered a lot three years ago when my father died in a car accident. It 36 my mother, two younger brothers and me alone.At that time, I was in a senior high school. As the eldest son, I had no choice but to 37 of school and work in a factory. Life went on without any wonder. I dared not ask for more and just thought about 38 my two brothers. However, it wasn’t easy, for I couldn’t39 their tuition (学费) even if I worked from day to night without rest. I also had to look after my sick mother. I wanted to go back to school, but it seemed to be an impractical idea, since I needed to work to support my family.A thread of 40 appeared during those gloomy (阴暗的)days. It was a rainy dusk when I went outside into the rain and walked in the street. Suddenly the rain stopped! I raised my head, and found that “the sky” was in fact a dark blue umbrella. Then I heard a deep voice say, “Why not 41 without an umbrella?” It was a one-legged man, “If you run, you won’t get so wet.”His words 42 me deeply. Without my father’s protection, was I only a slave to fate﹙命运﹚?While walking with him in the rain, I knew that his dream was once 43 by an accident. He was glad that he didn’t lose 44 and still “ran” on the road of life. Facing this guy, I had no pity, but admiration.Inspired by his 45 , I went to a city in the south and became an insurance representative. After two years’“running”, I got somewhere and my family situation became better 46 . I went back to school and eventually succeeded in being admitted to a university.Everything is so simple: to run without an umbrella! When you run out of the 47season of your life, there will be a bright sky ahead of you.36. A. left B. carried C. had D. forced37. A. take out B. drop out C. make out D. jump out38. A. bringing in B. taking away C. bringing up D. taking over39. A. receive B. realize C. manage D. afford40. A. chance B. need C. hope D. money41. A. drive B. walk C. travel D. run42. A. shocked B. puzzled C. discouraged D. hurt43. A. burnt B. accepted C. ruined D. discovered44. A. face B. heart C. sight D. control45. A. images B. achievements C. signs D. remarks46. A. simply B. gradually C. normally D. immediately47. A. rainy B. cloudy C. sunny D. windy Section C (12 marks)Directions: Complete the following passage by filling in each blank with one word that best fits the context.Being able to learn to relax can be very important, especially if you have a stressful life. When you build up too much tension without being able to control or release it, over time, 48 can develop some serious health problems.If you want to learn 49 to relax, one simple way is to remind yourself that most of the things that we worry about never even happen. It is good to think ahead to make sure that future problems don’t show up, 50 many people spend a lot of time worrying about things that are totally out of their control. Learn to relax by only focusing on things you 51 control. Relaxing doesn’t always mean sitting back and 52 doing anything. If you have things to do, you can do them in a relaxed manner.You can still relax and do all the things you need to do. Just 53 you have demands and responsibilities, it doesn’t mean you have to do everything in 54 stressed manner. The key is to simplify your life. Just relax by focusing 55 the moment.Part Ш Reading Comprehension (30 marks)Directions: Read the following three passages. Each passage is followed by several questions or unfinished statements. For each of them there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage.AIn the fall of 1985, I was a bright-eyed girl heading off to Howard University, aiming at a legal career and dreaming of sitting on a Supreme Court bench somewhere. Twenty-one years later I am still a bright-eyed dreamer and one with quite a different tale to tell.My grandma, an amazing woman, graduated from college at the age of 65. She was the first in our family to reach that goal. But one year after I started college, she developed cancer. I made the choice to withdraw from college to care for her. It meant that school and my personal dream would have to wait.Then I got married with another dream: building my family with a combination of adopted and biological children. In 1999, we adopted our first son. To lay eyes on him was fantastic and very emotional. A year later came our second adopted boy. Then followed son No. 3. In 2003, I gave birth to another boy.You can imagine how fully occupied I became, raising four boys under the age of 8. Our home was a complete zoo---a joyous zoo. Not surprising, I never did make it back to college full-time. But I never gave up on the dream either. I had only one choice: to find a way. That meant taking as few as one class each semester.The hardest part was feeling guilty about the time I spent away from the boys. They often wanted me to stay home with them. There certainly were times I wanted to quit, but I knew I should set an example for them to follow through the rest of their lives.In 2007, I graduated from the University of North Carolina. It took me over 21 years to get my college degree!I am not special, just single-minded. It always struck me that when you’re looking at a big challenge from the outside it looks huge, but when you’re in the middle of it, it just seems normal. Everything you want won’t arrive in your life on one day. It’s a process. Remember: little steps add up to big dreams.56. When the author went to Howard University, her dream was to be __________.A. a writerB. a teacherC. a judgeD. a doctor57. Why did the author quit school in her second year of college?A. She wanted to study by herself.B. She fell in love and got married.C. She suffered from a serious illness.D. She decided to look after her grandma.58. What can we learn about the author from Paragraphs 4 and 5?A. She was busy yet happy with her family life.B. She ignored her guilty feeling for her sons.C. She wanted to remain a full-time housewife.D. She was too confused to make a correct choice.59. What does the author mostly want to tell us in the last paragraph?A. Failure is the mother of success.B. Little by little, one goes far.C. Every coin has two sides.D. Well begun, half done.60. Which of the following can best describe the author?A. Caring and determined.B. Honest and responsible.C. Ambitious and sensitive.D. Innocent and single-minded.BBig Brothers Big Sisters is based on the simplicity and power of friendship. It is a program which provides friendship and fun by matching vulnerable young people (ages 7-17) with a volunteer adult who can be both a role model and a supportive friend.V olunteer tutors come from all walks of life—married, single, with or without children. Big Brothers and Big Sisters are not replacement parents or social workers. They are tutors: someone to trust, to have fun with, to talk and go to when needed.A Big Sister and Little Sister will generally spend between one and four hours together three or four times each month for at least twelve months. They enjoy simple activities such as a picnic at a park, cooking, playing sport or going to a football match. These activities improve the friendship and help the young person develop positive self-respect, confidence and life direction.Big Brothers Big Sisters organizations exist throughout the world. It is the largest and most well-known provider of tutor services internationally and has been operating for 25 years.Emily and Sarah have been matched since March 2008. Emily is a 10-year-old girl who has experienced some difficulties being accepted by her schoolmates at school. "I was pretty sure there was something wrong with me.”Emily’s mum came across Big Brothers Big Sisters and thought it would be of benefit to Emily by "providing different feedback(反馈)about herself other than just relying on schoolmates to measure her self-worth. ”Sarah wanted to get involved in a volunteer program. "I paid close attention to it and found out how to be a part of it. I thought it would be fun for me to get involved in making time to do something because sometimes it’s all work and no play. ”Big Brothers Big Sisters has been of great benefit and enjoyment to both Emily and Sarah. They love and look forward to their time together and the partnership has certainly helped Emily be more comfortable in being the wonderful, happy and unique girl she is!61. What is the aim of Big Brothers Big Sisters?A. To offer students public services.B. To provide partnership and fun for young people.C. To organize sport activities for young people.D. To help students improve their grades62. A volunteer is usually expected to work within a year for at least .A.24 hoursB. 72 hoursC.48 hoursD. 36 hours63. According to Emily’s mother, this program may provide Emily with .A. a new way to assess herselfB. advice from her teachersC. a new way to judge her schoolmatesD. more comments from her schoolmates64. Why did Sarah want to get involved in the program?A. She used to be a volunteer.B. She felt a bit bored with her life.C. She needed a part-time job.D. She wanted to get a challenging job.65. According to the passage, ‘vulnerable young people’ underlined in Paragraph1 are probablythose who are .A. popular at schoolB. rather weak physicallyC. easily hurt emotionallyD. confident in themselvesCIn a world with limited land, water and other natural resources, the harm from the traditional business model is on the rise. Actually, the past decade has seen more and more forests disappearing and the globe becoming increasingly warm. People now realize that this unhealthy situation must be changed, and that we must be able to develop in sustainable ways. That means growth with low carbon or development of sustainable products. In other words, we should keep the healthy while using its supply of natural resources.Today, sustainable development is a popular trend in many countries. According to a recent study, the global market for low-carbon energy will become three times bigger over the next decade. China, for example, has set its mind on leading that market, hoping to seize chances in the new round of the global energy resolution. It is now trying hard to make full use of wind and solar energy, and is spending a huge amount of money making electric cars and high-speed trains. In addition, we are also seeing great growth in the global markets for sustainable products such as palm oil(棕榈油), which is produced without cutting down valuable rainforest. In recent years the markets for sustainable products have grown more than 50%.Governments can fully develop the potential of these new markets. First, they can set high targets for reducing carbon emissions(排放) and targets for saving and reusing energy. Besides, stronger arrangement of public resources like forests can also help to speed up the development. Finally, governments can avoid the huge public expenses that are taking us in the wrong direction, and redirecting some of those expenses can accelerate the change from the traditional model to a sustainable one.The major challenge of this century is to find ways to meet the needs of a growing population within the limits of this single planet. That is no small task, but it offers abundant new chances for sustainable product industries.66. The traditional business model is harmful because of all the following EXCEPT that .A. it makes the world warmerB. it consumes natural resourcesC. it brings severe damage to futureD. it makes growth hard to continue67. What can we infer from Paragraph2?A. China lacks wind and solar energy.B. China is the leader of the low-carbon market.C. High-speed trains are a low-carbon development.D. Palm oil is made at the cost of valuable forests.68. To fully develop the low-carbon markets governments can______.A. cut public expensesB. forbid carbon emissionsC. develop public resourcesD. encourage energy conservation69. We can learn from the last paragraph that businesses have many chances to _______.A. develop sustainable productsB. explore new natural resourcesC. make full use of natural resourcesD. deal with the major challenge70. What is the main purpose of the passage?A. To introduce a new business modelB. To compare two business modelsC. To predict a change of the global marketsD. To advocate sustainable development Part IV Writing (45 marks)Section A (10 marks)Directions: Read the following passage. Fill in the numbered blanks by using the information from the passage.Write NO MORE THAN THREE WORDS for each answer.Romanticism (浪漫主义) was a literary, artistic and intellectual movement that originated in Europe towards the end of the 18th century. And it reached its peak from 1800 to 1840.Romanticism urged people to look at nature and surroundings from a scientific point of view. It was also a rebellion against scientific rationalization (合理化) of nature. It permitted a person's imagination and freedom in art. The concept of romanticism was present in all the major art forms, like literature, the visual arts and music.Friedrich Schlegel, a German philosopher, writer and critic, used the term “Romantic” for the first time to name a new school of literature which arose in opposition to “Classicism”. Though the concept of Romanticism was identified much earlier, critics believe that Romanticism in English literature dates from the Lyrical Ballads. The romantics were very interested in mystery, ambition and adventure.In the visual arts, the term Romanticism refers to a trend that appeared in the 19th century, which was characterized by opposition to the classical forms and its rules. Romanticism in thevisual arts focused more on the spiritual and emotional representation of nostalgia(怀旧). Romantics used objects, like wild trees, moonlight, and so on, to convey their ideas and concepts.In music, Romanticism was characterized by the freedom of forms and an emphasis on the emotions. It was German composers who used romanticism widely and developed this concept. Many famous composers worked in smaller forms of music that had flexible structure; for instance, ballads, solo piano music, and so on.Romanticism influenced the literature and arts of the 18th and 19th century. The popularity of this movement declined gradually with passing time, but it has a significant place in the history of literature and art development.Section B (10 marks)Directions: Read the following passa0ge. Answer the questions according to the information given in the passage .He seems an unlikely hero, especially one that would save a kid from the jaws of a wild cougar (美洲豹) .Shen Huigang is just now getting recognition for his bravery in fighting off a cougar on Vancouver Island, Canada, during a family outing on Aug .30.Shen, also known as Ian, was then an exchange student at Kwantlen Polytechnic University enjoying the afternoon on a beach near Ucluelet, a small town on the edge of the Pacific Ocean.With him was a friend, Myles Hagar, and Hagar’s two grandchildren.Silently and suddenly a cougar appeared out of nowhere.By the time the two adults spotted the cat, believed to be young but still weighing 30 to 35 kilograms, it already had the head of 18-month old Julien in its mouth. Instinctively﹙本能地﹚, the young man gestured as if he were ready for a fight, and tried to scare the beast off with the bagin his hands.On hearing the noise Shen made, the animal dropped the kid and Hagar grabbed his grandson from the cougar’s jaws. Shen and Hagar gradually chased the animal back into the woods.“We also moved slowly to our vehicle, as we waved our fists and bags, pretending we wanted to fight with it,” Shen said. “The vehicle wasn’t far away but it felt like it took us a century to travel the short journey.”“Any hesitation, at any moment, even a second delay, would have resulted in certain death for Julien. The cougar was just about to break his neck and carry him away to be eaten in the forest.” Hagar said. Julien has since made a full recovery.Parks Canada spokeswoman Arlene Armstrong told the National Post newspaper of Canada in an interview in August. “The two men acted properly by maintaining eye contact with the big cat and aggressively scaring it off.”81. Why is Shen Huigang getting recognition on Vancouver Island?(No more than 9 words )82. What happened to Julien by the time the two adults saw the young cougar?(No more than 9 words )83. Why did the cougar give up eating the kid? (No more than 7 words )84. How can you act properly when you are fighting with a wild cougar? (No more than 11 words )Section C (25 marks)Directions: Write an English composition according to the instructions given below in Chinese.假设你是红星中学高三(1)班的学生李华,请根据以下四幅图的先后顺序,为校刊“英语园地”写一篇短文,讲述上周你参加学校英文歌曲演唱比赛的过程以及你的感受。
2015届广西学业水平考试数学模拟考及详细参考答案
2015届广西普通高中数学学业水平考试模拟考一、选择题(本大题共20小题,每小题3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.设集合}2,1{},3,2,1{==N M ,则N M 等于A .}2,1{B .}3,1{C .}3,2{D .}3,2,,1{ 2.函数)2lg()(-=x x f 的定义域是A .),2[+∞B .),2(+∞C .),3(+∞D .),3[+∞ 3.0410角的终边落在A .第一象限B .第二象限C .第三象限D .第四象限 4.抛掷一枚骰子,得到偶数点的概率是A .61 B .41 C .31 D .215.在等差数列}{n a 中,11=a ,公差2=d ,则8a 等于 A .13 B .14 C .15 D .16 6.下列函数中,在区间),0(+∞内单调递减的是 A .2x y = B .xy 1=C .x y 2=D .x y 2log = 7.直线0=-y x 与02=-+y x 的交点坐标是A .)1,1(B .)1,1(--C .)1,1(-D .)1,1(-8.命题甲“sin 0x >”,命题乙“0x >”,那么甲是乙的( ) (A )充分而不必要条件 ( B )必要而不充分条件 (C )充要条件 (D )既不充分又不必要条件(第16题图)正(主)视图侧(左)视图俯视图9.圆0622=-+x y x 的圆心坐标和半径分别是A .9),0,3(B .3),0,3(C .9),0,3(-D .3),0,3(- 10.313tanπ的值是 A .33- B .3- C .33 D .311.在ABC ∆中,角C B A ,,的对边分别是c b a ,,,已知0120,2,1===C b a ,则c 等于 A .2 B .5 C .7 D .4 12.在等比数列}{n a 中,44=a ,则62a a ⋅等于 A .32 B .16 C .8 D .4 13.将函数)3sin(2π+=x y 的图象上所有点的横坐标缩短到原来的21(纵坐标不变),所得图象对应的表达式为A .321sin(2π+=x yB .)621sin(2π+=x yC .32sin(2π+=x y D .)322sin(2π+=x y 14.在ABC ∆中,角C B A ,,的对边分别是c b a ,,,若B c b si n 2=,则C sin 等于 A .1 B .23 C .22 D .21 15.曲线x x x y 223-+=在1-=x 处的切线斜率是( )(A) 1 (B) -1 (C) 16.如图是一个空间几何体的三视图,则这个几何体侧面展开图的面积是 A .4π B .2πC .πD .π2甲 乙85 0 1 2 3 2 2 8 8 95 2 3 5 第25题图17.不等式组⎪⎩⎪⎨⎧≥-+≤≤0111y x y x 表示的平面区域面积是A .21B .41C .1D .218.容量为100的样本数据被分为6组,如下表第3组的频率是 A .15.0 B .16.0 C .18.0 D .20.0 19.若c b a >>,则下列不等式中正确的是A .bc ac >B .c b b a ->-C .c b c a ->-D .b c a >+ 20.如图所示的程序框图,其输出的结果是 A .11 B .12 C .131 D .132 二、填空题(本大题共4小题,每小题3分,共12分)21.已知函数⎩⎨⎧<≥=0,0,)(2x x x x x f ,则=)3(f ____________.22.过点)1,0(且与直线02=-y x 垂直的直线方程的一般式是____________. 23.等差数列}{n a 的前n 项和为n S .已知36=a ,则=11S ___________.24、甲、乙两名篮球运动员在六场比赛中得分的茎叶图如图所示,记甲的平均分为a ,乙的平均分为b ,则=-a b ___.2015届学业水平考试模拟考(二)数学科答题卡MCV ABD第27题图一、选择题(本大题共20小题,每小题3分,共60分)二、填空题(本大题共4小题,每小题3分,共12分)21 22 23 24 三、解答题(本大题共4小题,共28分,解答应写出文字说明,证明过程或演算步骤)25.(本题满分6分)已知抛物线的焦点和双曲线224520x y -=的一个焦点重合,求抛物线的标准方程.26.(本小题满分6分)已知向量a =)3,sin 1(x +,b =)3,1(.设函数=)(x f b a ⋅,求)(x f 的最大值及单调递增区间. 27.(本小题满分8分)已知:如图,在四棱锥ABCD V -中,底面ABCD 是 平行四边形,M 为侧棱VC 的中点.求证://VA 平面BDM 28.(本小题满分8分)已知函数)(5)1(23)(2R k k x k x x f ∈++-+=在区间)2,0(内有零点,求k 的取值范围.参 考 答 案一、选择题(本大题共20小题,每小题3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)二、填空题(本大题共5小题,每小题3分,共15分)21 9 22 x+2y-2=0 23 33 24 0.5三、解答题(本大题共3小题,共25分,解答应写出文字说明,证明过程或演算步骤)25、抛物线的标准方程为:x y 122±=26函数f(x)=a·b=1+sinx+3= sinx+4,所以最大值是5.增区间是[2kπ-π/2,2kπ+π/2],k ∈Z.27、连结AC 交BD 于O 点,连结OM.底面ABCD 是平行四边形,所以O 为AC 中点,又因M 为侧棱VC 的中点,所以VA ∥OM,因为OM ⊂平面BDM,VA ⊄平面BDM, 所以//VA 平面BDM .28、f(x)=3x2+2(k -1)x +k +5在区间(0,2)内有零点, 等价于方程3x2+2(k -1)x +k +5=0在(0,2)内有实数根,则:(1)判别式△=4(k -1)2-12(k +5)=0时,得:k=7或者k=-2,此时方程的根分别是: k=7时,根是:x1=x2=-2;k=-2时,根是:x1=x2=1. 因为方程在(0,2)内有实数根,所以k=-2.(k=7舍去) (2)若判别式大于0,则:k>7或k<-2.此时:①若两根都在(0,2)内,则:对称轴x=-(k-1)/3在(0,2)内、f(0)>0、f(2)>0,即⎪⎪⎪⎩⎪⎪⎪⎨⎧>++-+=>+=<--<>+--=05)1(412)2(05)0(23100)5(12)1(42k k f k f k k k △解得:⎪⎪⎪⎩⎪⎪⎪⎨⎧>><<<>513-5-15-2-7k k k k k 或得:2-513-<<k . ②若在(0,2)内存在一个根,则:f(0)×f(2)<0,得:-5<k<-13/5. (3)当f(2)=0时,即12+4(k-1)+k+5=0,k=-13/5.此时f(0)=k+5=12/5>0,所以k=-13/5符合题意.当f(0)=k+5=0时,k=-5,此时f(2)= 12+4(k-1)+k+5=-12<0,不符合题意,舍去.得:k=513-.综上可得:-5<k ≤-2.。
2015年高考全国卷2理科数学试题及答案解析(word精校版)
2015年高考全国卷2理科数学试题及答案解析(word 精校版)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,考生务必先将自己的姓名、准考证号码填写在答题卡上。
2.回答第I 卷时,选出每小题的答案后,用铅笔把答题卡上对应题目的答案标号涂黑,如需改动,用橡皮擦干净后,再选涂其他答案标号。
写在本试卷上无效。
3.回答第II 卷时,将答案写在答题卡上,写在本试卷上无效。
4.考试结束后,将本试卷和答题卡一并交回。
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A={-2,-1,0,1,2},B={x|(X-1)(x+2)<0},则A∩B=( )(A ){--1,0} (B ){0,1} (C ){-1,0,1} (D ){,0,,1,2}【答案】A【解析】由已知得{}21B x x =-<<,故{}1,0AB =-,故选A(2)若a 为实数且(2+ai )(a-2i )=-4i,则a=( ) (A )-1 (B )0 (C )1 (D )2 【答案】B(3)根据下面给出的2004年至2013年我国二氧化硫排放量(单位:万吨)柱形图。
以下结论不正确的是( )(A ) 逐年比较,2008年减少二氧化硫排放量的效果最显著 (B ) 2007年我国治理二氧化硫排放显现(C ) 2006年以来我国二氧化硫年排放量呈减少趋势 (D ) 2006年以来我国二氧化硫年排放量与年份正相关 【答案】D【解析】由柱形图得,从2006年以来,我国二氧化硫排放量呈下降趋势,故年排放量与年份负相关.(4)等比数列{a n }满足a 1=3,135a a a ++ =21,则357a a a ++= ( )(A )21 (B )42 (C )63 (D )84 【答案】B(5)设函数211log (2),1,()2,1,x x x f x x -+-<⎧=⎨≥⎩,2(2)(log 12)f f -+=( )(A )3 (B )6 (C )9 (D )12 【答案】C【解析】由已知得2(2)1log 43f -=+=,又2log 121>,所以22log 121log 62(log 12)226f -===,故2(2)(log 12)9f f -+=.(6)一个正方体被一个平面截去一部分后,剩余部分的三视图如右图,则截去部分体积与剩余部分体积的比值为(A )81 (B )71 (C )61 (D )51 【答案】D【解析】由三视图得,在正方体1111ABCD A B C D -中,截去四面体111A A B D -,如图所示,,设正方体棱长为a ,则11133111326A A B D V a a -=⨯=,故剩余几何体体积为3331566a a a -=,所以截去部分体积与剩余部分体积的比值为51.CBADD 1C 1B 1A 1(7)过三点A (1,3),B (4,2),C (1,-7)的圆交于y 轴于M 、N 两点,则MN =(A )26 (B )8(C )46 (D )10 【答案】C(8)右边程序抗土的算法思路源于我国古代数学名著《九章算术》中的“更相减损术”。
广西壮族自治区玉林市博白县中学高三数学理月考试题含解析
广西壮族自治区玉林市博白县中学高三数学理月考试题含解析一、选择题:本大题共10小题,每小题5分,共50分。
在每小题给出的四个选项中,只有是一个符合题目要求的1. 已知向量=(2,4),=(﹣1,1),则2﹣=( )A.(5,7)B.(5,9)C.(3,7)D.(3,9)参考答案:A【考点】平面向量的坐标运算.【专题】平面向量及应用.【分析】直接利用平面向量的数乘及坐标减法运算得答案.【解答】解:由=(2,4),=(﹣1,1),得:2﹣=2(2,4)﹣(﹣1,1)=(4,8)﹣(﹣1,1)=(5,7).故选:A.【点评】本题考查平面向量的数乘及坐标减法运算,是基础的计算题.2. 一个几何体的三视图如图所示,且其侧(左)视图是一个等边三角形,则这个几何体的体积为( )A.B.C.2D.参考答案:B考点:由三视图求面积、体积.专题:空间位置关系与距离.分析:此几何体是底面积是S==1的三棱锥,与底面是边长为2的正方形的四棱锥构成的组合体,它们的顶点相同,底面共面,高为,即可得出.解答:解:此几何体是底面积是S==1的三棱锥,与底面是边长为2的正方形的四棱锥构成的组合体,它们的顶点相同,底面共面,高为,∴V==.点评:本题考查了三棱锥与四棱锥的三视图、体积计算公式,属于基础题.3. 图中阴影部分的面积S是h的函数(0≤h≤H),则该函数的大致图像是()A B C D参考答案:B略4. 已知双曲线﹣=1(a>0,b>0)的一条渐近线平行于直线l:y=2x+10,双曲线的一个焦点在直线l上,则双曲线的方程为()A.﹣=1B.﹣=1C.﹣=1D.﹣=1参考答案:A【考点】双曲线的标准方程.【分析】先求出焦点坐标,利用双曲线﹣=1(a>0,b>0)的一条渐近线平行于直线l:y=2x+10,可得=2,结合c2=a2+b2,求出a,b,即可求出双曲线的方程.【解答】解:∵双曲线的一个焦点在直线l上,令y=0,可得x=﹣5,即焦点坐标为(﹣5,0),∴c=5,∵双曲线﹣=1(a>0,b>0)的一条渐近线平行于直线l:y=2x+10,∴=2,∵c2=a2+b2,∴a2=5,b2=20,∴双曲线的方程为﹣=1.故选:A.5. 已知函数f(x)=|log3(x+1)|,实数m,n满足一1<m<n,且f(m)=f(n).若f(x)在[m2,n]上的最大值为2,则(A) -6 (B) -8 (C) -9 (D) -12参考答案:C6. 已知集合,则(A){一2) (B){3) (C)(-2,3} (D)参考答案:B7. 已知集合A={x||x﹣1|<1},B={x|1﹣≥0},则A∩B=()A.{x|1≤x<2} B.{x|0<x<2} C.{x|0<x≤1} D.{x|0<x<1}参考答案:A【分析】求出A,B中不等式的解集,找出A与B的交集即可.【解答】解:由|x﹣1|<1,即﹣1<x﹣1<1,即0<x<2,即A={x|0<x<2},由1﹣≥0,即≥0,解得x≥1或x<0,即B={x|x≥1或x<0}则A∩B={x|1≤x<2},故选:A【点评】此题考查了交集及其运算,熟练掌握交集的定义是解本题的关键.8. 过抛物线的焦点的直线交抛物线于两点,点是原点,若,则的面积为()A. B.C. D.参考答案:C9. 平行四边形ABCD中,AB=3,AD=2,∠BAD=600,若,且DB⊥AE,则λ的值为A.3B.4C.5D.6参考答案:D10. 的值为A. —B.C.D.参考答案:B略二、填空题:本大题共7小题,每小题4分,共28分11. 已知,函数,若实数满足,则的大小关系为.参考答案:12. 执行如图所示的程序框图,若输出的b的值为31,则图中判断框内①处应填的整数为.参考答案:4【考点】程序框图.【分析】根据框图的流程依次计算程序运行的结果,直到输出的b的值为31,确定跳出循环的a值,从而确定判断框的条件.【解答】解:由程序框图知:第一次循环b=2+1=3,a=2;第二次循环b=2×3+1=7,a=3;第三次循环b=2×7+1=15,a=4;第四次循环b=2×15+1=31,a=5.∵输出的b的值为31,∴跳出循环的a值为5,∴判断框内的条件是a≤4,故答案为:4.13. 设△ABC的内角A,B,C的对边分别为a,b,c,且a=1,b=2,,则sinB=________参考答案:14. 定义:如果函数在定义域内给定区间上存在,满足,则称函数是上的“平均值函数”,是它的一个均值点,如是上的平均值函数,0就是它的均值点.现有函数是上的平均值函数,则实数的取值范围是.参考答案:略15. 函数的定义域是.参考答案:(0,1]【考点】函数的定义域及其求法;对数函数的定义域.【专题】计算题.【分析】令被开方数大于等于0,然后利用对数函数的单调性及真数大于0求出x的范围,写出集合区间形式即为函数的定义域.【解答】解:∴0<x≤1∴函数的定义域为(0,1]故答案为:(0,1]【点评】求解析式已知的函数的定义域应该考虑:开偶次方根的被开方数大于等于0;对数函数的真数大于0底数大于0小于1;分母非0.16.如果复数的实部和虚部相等,则实数等于。
广西玉林市博白县2015届高三下学期返校调研考试语文试卷 Word版含答案
资料概述与简介 广西玉林市博白县2015届高三下学期返校调研考试语文试题 本试题卷共七道大题,22道小题,共10页。
时量150分钟。
满分150分。
语言文字运用(12分,每小题3分) 1.下列中字形和加点的字的读音全都正确的一项是 陈氏每天都在衣不蔽体、食不腹、儿啼女哭中苦度。
xuàn yào)性消费色彩。
C.为阻止污水进一步泄漏,东京电力公司向碎(cuì)石层内注入了俗称“水玻璃”的硅酸钠和调整硬度的其他药剂。
D.对金钱的癖(pì)好使他行为怪诞,全身充满着铜臭味,他总是找借口克扣仆人的佣金,为了节省柴火经常吃夹生饭。
2.下面语段横线处应填入的词语,最恰当的一组是 长期以来,人们把图书馆当成“知识宝库”,当成“知识殿堂”,似乎对图书馆有加;然而,“宝库”和“殿堂”虽好,但它们离普通百姓很远,甚至很遥远,普通百姓往往。
一些人从不进图书馆,恐怕同这种心态不无关系,图书馆“宝库”“殿堂”,“知识公园”。
A.崇敬 望而生畏 不仅是 而且是 B.尊崇 望而生畏 不应是 而应是 C.崇敬 敬而远之 不应是 而应是 D.尊崇 敬而远之 不仅是 而且是 3.下面这首宋词《定风波》中的横线处,填入最贴切的句子是 江水沉沉帆影过,游鱼到晚透寒波。
渡口双双飞白鸟,烟袅,芦花深处隐渔歌。
扁舟短棹归兰浦,人去,萧萧竹径透青莎。
深夜无风新雨歇,凉月,?。
? A.几番离恨无意满渠沟B.岸草欣欣无恨色? C.露迎珠颗入圆荷D.数帆带雨烟中落? 4.一位记者采访某著名作家之子。
下面是开场白,共四句话,其中不得体的一处是 ①我们都知道您父亲是一位蜚声文坛的作家,作品广为流传,读者甚众,影响极广。
②我上中学的时候就拜读过他老人家的诸多大作,至今还能背诵其中的段落。
③您是他老人家的爱子,耳濡目染间,对令尊的作品一定有独到见解。
④今天,您能于百忙中有幸受访,我们深表感谢。
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广西玉林市博白县2015届高三下学期返校调研考试化学试卷 Word版含答案.pdf
广西玉林市博白县2015届高三下学期返校调研考试化学试题 本卷包括试卷和答卷,二个大题,共18个小题,满分100分,考试时间90分钟。
考试结束后,请考生将试卷自行保存,交答卷。
注意事项: 1.在进行答题前,请考生务必将自己的姓名、考号、座位号和答案用钢笔或圆珠笔填写到答卷上的相应栏目内,并按答题要求进行作答,答案不得答在试卷上。
2.可能用到的数据: O─16 P─31 Cl─35.5 一、选择题(本题包括14小题,每小题3分,共42分,每小题只有一个选项符合题意,请将符合题意的选项填在答题卷上的相应栏目内。
) .化学与生活息息相关.下列说法不正确的是 A.可用淀粉KI试纸和食醋检验真假碘盐 B.用饱和氯化铵溶液可以清洗金属表面的锈迹 C.次氯酸钠溶液是生活中常用的消毒剂 D.经常食用含明矾的食品能中和过多的胃酸 .已知:Al(OH)3的电离方程式为:AlO2-+H++H2OAl(OH)3Al3++3OH- 电解熔融AlCl3不能得到金属铝,AlCl3溶于水的电离方程式为:AlCl3=Al3++3Cl- PbSO4难溶于水,易溶于醋酸钠溶液,反应的化学方程式为: PbSO4+2CH3COONa=Na2SO4+(CH3COO)2Pb 则下列关于Al(OH)3、AlCl3和(CH3COO)2Pb的说法中正确的是 A.均为强电解质B.均为弱电解质 C.均为离子化合物D.均为共价化合物 .对H2O的电离平衡不产生影响的微粒是 .下列选项中哪一种可与陶瓷、普通玻璃、水泥归为同种类型的材料 A.铝合金B.高温结构氮化硅陶瓷 C.有机玻璃D.砖瓦 .关于下图中各装置的叙述不正确的是 A.装置①是中和滴定法测定硫酸的物质的量浓度 B.装置中手捂烧瓶(橡胶管已被弹簧夹夹紧),发现导管中有液柱上升并保持稳定,则说明装置不漏气 C.装置中X若为四氯化碳,可用于吸收氨气和氯化氢,并防止倒吸 D.装置可用于检验溴乙烷发生消去反应得到的气体中含有乙烯(假定每个装置中吸收完全) .埋在地下的铸铁输油管道,在下列各种情况下,被腐蚀速率最慢的是 A.在潮湿疏松的碱性土壤中 B.在含铁元素较多的酸性土壤中 C.在干燥致密不透气的土壤中 D.在含碳粒较多,潮湿透气的中性土壤中 .下列图示与对应的叙述相符的是: 图1 图2 图3 图4 A.图1表示常温下,稀释HA、HB两种酸的稀溶液时,溶液pH随加水量的变化,则NaA溶液的pH小于同浓度的NaB溶液的pH。
广西壮族自治区玉林市博白县沙河中学高三数学理联考试卷含解析
广西壮族自治区玉林市博白县沙河中学高三数学理联考试卷含解析一、选择题:本大题共10小题,每小题5分,共50分。
在每小题给出的四个选项中,只有是一个符合题目要求的1. 已知抛物线y2=2px(p>0)的焦点F恰好是双曲线-=1(a>0,b>0)的右焦点,且两曲线的交点连线过点F,则该双曲线的离心率为A.B.C.1+D.1+参考答案:C2. 在含有30个个体的总体中,抽取一个容量为5的样本,则个体a被抽到的概率为A.B. C. D.参考答案:B略3. 设是两条不同的直线,是两个不同的平面,则能得出的是()A.,, B.,,C.,, D.,,参考答案:C4. 点O为△ABC内一点,且满足,设△OBC与△ABC的面积分别为S1、S2,则=()A.B.C.D.参考答案:B【考点】向量的线性运算性质及几何意义.【专题】计算题;转化思想;综合法;平面向量及应用.【分析】延长OC到D,使OD=4OC,延长CO交AB与E,由已知得O为△DABC重心,E为AB中点,推导出S△AEC=S△BEC,S△BOE=2S△BOC,由此能求出结果.【解答】解:延长OC到D,使OD=4OC,延长CO交AB与E,∵O为△ABC内一点,且满足,∴=,∴O为△DABC重心,E为AB中点,∴OD:OE=2:1,∴OC:OE=1:2,∴CE:OE=3:2,∴S△AEC=S△BEC,S△BOE=2S△BOC,∵△OBC与△ABC的面积分别为S1、S2,∴=.故选:B.【点评】本题考查两个三角形面积比值的求法,是中档题,解题时要认真审题,注意向量、三角形重心等知识的合理运用.5. 函数的定义域是,若对于任意的正数,函数都是其定义域上的增函数,则函数的图象可能是参考答案:A6. 设变量x,y满足约束条件则z=|x-3y|的最大值为( )(A)8 (B)4 (C)2 (D)参考答案:A7. 已知点在圆上,则函数的最小正周期和最小值分别为()A.,B. ,C. ,D. ,参考答案:B略8. 函数的值域是()A.R B.(-∞,0) C.(-∞,1) D.(0,+∞)参考答案:D9. 已知函数为常数)的图象过点的反函数,则的值域为()A. [2,5]B. [1,5]C. [1,10] D. [2,10]参考答案:答案:B10. 集合A={x||x|≤4,x∈R},B={x|(x+5)(x-a)≤0},则“A B”是“a>4”的A.充分不必要条件 B.必要不充分条件 C.充要条件 D.既不充分也不必要条件参考答案:B二、填空题:本大题共7小题,每小题4分,共28分11. 如右图,它满足:(1)第行首尾两数均为;(2)表中的递推关系类似杨辉三角,则第行()第2个数是.参考答案:.设第行()第2个数为,则.从而通过累加可知,又=2,所以可知.12. 函数,其中,若动直线与函数的图像有三个不同的交点,它们的横坐标分别为,则是否存在最大值?若存在,在横线处填写其最大值;若不存在,直接填写“不存在”______________.参考答案:1 略13. 一个几何体的三视图如图所示,则该几何体的体积为___________.参考答案:略14. 不等式,对恒成立的实数的取值范围参考答案:略15. 若圆C 的半径为1,圆心在第一象限,且与直线和轴都相切,则该圆的标准方程是________.参考答案:试题分析:依据条件确定圆心纵坐标为1,又已知半径是1,通过与直线4x-3y=0相切,圆心到直线的距离等于半径求出圆心横坐标,写出圆的标准方程.∵圆C 的半径为1,圆心在第一象限,且与直线4x-3y=0和x 轴都相切,∴半径是1,圆心的纵坐标也是1,设圆心坐标(a ,1),∴该圆的标准方程是;考点:圆的标准方程,圆的切线方程 16. 已知实数、、满足,,则的最大值为为_______.参考答案:17. 已知,,且,,则与的夹角为 .参考答案:略三、 解答题:本大题共5小题,共72分。
广西玉林市博白县高三数学下学期返校调研考试试题 文
广西玉林市博白县2015届高三下学期返校调研考试数学(文)试题(考试时量:120分钟 满分150分) 一:单选题:(本大题共10小题,每小题5分,共50分.在每小题给出的四个选项中,只有一项是符合题目要求的。
) 1.已知集合{}{}()12,1R A x x B x x A C B =-≤≤=<I ,则=A .{}1x x > B .{}1x x ≥C .{}2x x 1<≤ D . {}2x x 1≤≤2.如果复数()2bib R i -∈的实部和虚部互为相反数,那么b 等于A .2B . 2-C .2-D .2 3.某产品在某零售摊位的零售价x (单位:元)与每天的销售量y (单位:个)的统计资料如下表所示:由右表可得回归直线方程ˆˆˆy bx a =+中的ˆ4b =-,据此模型预测零售价为15元时,每天的销售量为A .51个B .50个C .49个D .48个 4.已知命题p :对任意的x R ∈,有ln 1x >,则p ⌝是 A .存在0x R ∈,有0ln 1x ≤B .对任意的x R ∈,有ln 1x <C .存在0x R∈,有0ln 1x <D .对任意的x R ∈,有ln 1x ≤5.某程序框图如图所示,则该程序运行后输出的k 值是 A .8 B .7C .6D .56.已知2)(,6,1=-⋅==a b a b a ,则向量a 与b 的夹角为x 16 17 18 19 y50344131A.2πB .3πC .4πD . 6π7.如图是一个几何体的三视图,则此三视图所描述几何体的表面积为A .(1243)π+B .28πC .(2043)π+D .20π8.将函数sin 2y x =的图象向右平移π8个单位后,所得图象的一条对称轴方程是A .π8x =B .π8x =-C .π4x =D .π4x =-9.若函数()()()01x x f x ka a a a -=->≠-∞+∞且在,上既是奇函数又是增函数,则()()log a g x x k =-的图象是A B C D10.定义()()()f x g x h x <<对任意x D ∈恒成立,称()g x 在区间D 上被()()f x h x 、所夹. 若ln y x =在()0,+∞被ay x =-和(1)y a x =-所夹,则实数a 的取值范围A .2(0,)eB .11(,)e e e -C .12(,)e e e - D .2(,1)e二:填空题:(本题共5小题,每小题5分,共25分。
最新广西玉林市博白县届高三5月高考模拟理科数学试题及答案优秀名师资料
广西玉林市博白县2016届高三5月高考模拟理科数学试题及答案2016年高考模拟考试试题数学(理科)本试卷分第?卷(选择题)和第?卷(非选择题)两部分.满分150分,考试时间120分钟.第?卷(选择题共60分)一、选择题:(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).2UR,1. 已知全集,集合AxxBxx,,,,,,10,11,则ðAB:,,,,,,,U( )。
A. B. C. xx,,,,21xx,,,10xx,,,11,,,,,,D. xx01,,,,12,iizz,,2. 已知为虚数单位,若复数的共轭复数为,则z2,i( )。
zz,,25,11 A. B. C. 925,D. 93. 下列有关命题的说法正确的是( )。
22x,1x,1x,1x,1 A.命题“若,则”的否命题为“若,则”2x,,1xx,,,560B.“”是“”的必要不充分条件2,,xR,,xRxx,,,10C.命题“,使得”的否定是“,均有2xx,,,10” xy,D.命题“若,则sinsinxy,”的逆否命题为真命题a,bab4. 已知??=2,??=4,,以为邻边的平行四边形的面积43ab为,则和的夹角为( )A.30?B.60?C.120?D.60?或120?yx,,,,3,yx,5. 已知可行域是则下列目标函数中, ,,xy,,4,,能够在点取得最小值是( )。
(3,1)1 A. B. C. zxy,,,zxy,,2zxy,,,22D. zxy,,26.某几何体的三视图如图所示,且该几何体的体积是2,则正视图的面积等于( )。
93 A.2 B. C. 223D.1的图象是( )。
7. 函数fxx()ln(),,xA B CD8. 某学校组织的数学竞赛中,学生的竞赛成绩X服从正态分412,布即,,则的最PXaPXb(120),(80100),,,,,XN (100,),ab小值为( )。
广西玉林市博白县2015届高三地理下学期返校调研考试试题
XXXX市博白县2015届高三下学期返校调研考试地理试题时量:90分钟满分:100分一、选择题:1—25题,每题只有一个选项符合题意,每题2分,共50分。
水浇地是指除水田、菜地外,有水保证和灌溉设施,在一般年景能正常灌溉的耕地。
望天田是指无灌溉工程设施,主要依靠天然降雨种植水生作物的耕地。
下图示意我国部分省区耕地利用结构。
读图回答1~3题。
1.据图推测,A最有可能是A.XX B.XX C.XX D.XX2.有关解决我国“望天田”问题的措施,不合理的是A.发展灌溉排水系统B.进行规模化人工降雨C.实施高效的旱作农业D.调整耕作区的种植结构3.关于图中各省区耕地结构的说法正确的是A.XX因为降水少,旱地的面积较大B.在各省区中XX的水浇地面积最大C.XX气候湿润,水田面积居各省区中之首D.XX的耕地结构主要与降水有关蒙古包是蒙古等民族的传统民居。
下图示意锡林郭勒(约116°E,44°N)草原上的蒙古包及某河的景观图。
读图回答4~5题。
4.蒙古包选址原则是“春洼、夏岗、秋平、冬阳”。
下列关于选址原因的叙述错误的是A.春季洼地可避风沙B.夏季高地透风凉爽C.秋季平地温差最小D.冬季阳坡温暖避风5.关于图中河流水系主要特征的说法正确的是A.流程长,流域面积广B.落差大,多峡谷C.河网密集,呈放射状D.河道弯曲,支流少某开发商分别在(约40°N)和XX(约28°N)开发了两个楼盘。
两地各有朝向和楼高相同的户型结构如图所示。
读图回答6~8题。
6.冬季卧室采光条件最好的是A.北卧室B.XX南卧室C.南卧室D.XX北卧室7.若XX楼盘每层2.8米共20层,南北楼距为28米。
甲图②栋10层住户一年中不能直接获得太阳照射的时间最接近A.11个月B.9个月C.7个月D.2个月8.在上级部门对两地小区楼盘规划设计图(甲图)审批时,几乎相同的设计在却没有通过,被要求修改。
最终修改的方案可能是A.扩大南北楼的楼间距B.缩小南北楼的楼间距C.降低南侧楼房的高度D.降低北侧楼房的高度不同类型的商业中心,其服务对象和服务X围不同。
广西玉林市博白县高三数学下学期返校调研考试试题 文
广西玉林市博白县2015届高三下学期返校调研考试数学(文)试题(考试时量:120分钟 满分150分) 一:单选题:(本大题共10小题,每小题5分,共50分.在每小题给出的四个选项中,只有一项是符合题目要求的。
) 1.已知集合{}{}()12,1R A x x B x x AC B =-≤≤=<,则=A .{}1x x > B .{}1x x ≥C .{}2x x 1<≤ D . {}2x x 1≤≤2.如果复数()2bib R i -∈的实部和虚部互为相反数,那么b 等于AB .C .2-D .2 3.某产品在某零售摊位的零售价x (单位:元)与每天的销售量y (单位:个)的统计资料如下表所示:由右表可得回归直线方程ˆˆˆy bx a =+中的ˆ4b =-,据此模型预测零售价为15元时,每天的销售量为A .51个B .50个C .49个D .48个 4.已知命题p :对任意的x R ∈,有ln 1x >,则p ⌝是 A .存在0x R ∈,有0ln 1x ≤B .对任意的x R ∈,有ln 1x <C .存在0x R ∈,有0ln 1x <D .对任意的x R ∈,有ln 1x ≤5.某程序框图如图所示,则该程序运行后输出的k 值是 A .8 B .7C .6D .562)(,6=-⋅a b a ,则向量a 与b 的夹角为A .2πB .3πC .4πD . 6π7.如图是一个几何体的三视图,则此三视图所描述几何体的表面积为A.(12π+ B .28π C.(20π+ D .20π8.将函数sin 2y x =的图象向右平移π8个单位后,所得图象的一条对称轴方程是A .π8x =B .π8x =-C .π4x =D .π4x =-9.若函数()()()01x x f x ka a a a -=->≠-∞+∞且在,上既是奇函数又是增函数,则()()log a g x x k =-的图象是A B C D10.定义()()()f x g x h x <<对任意x D ∈恒成立,称()g x 在区间D 上被()()f x h x 、所夹. 若ln y x =在()0,+∞被ay x =-和(1)y a x =-所夹,则实数a 的取值范围A .2(0,)eB .11(,)e e e -C .12(,)e e e - D .2(,1)e二:填空题:(本题共5小题,每小题5分,共25分。
2017-2018届广西玉林市博白县高三模拟试题(博白统测)理科数学试题及答案
2017-2018届⼴西⽟林市博⽩县⾼三模拟试题(博⽩统测)理科数学试题及答案参考公式:如果事件A B ,互斥,那么球的表⾯积公式()()()P A B P A P B +=+24πS R =如果事件A B ,相互独⽴,那么其中R 表⽰球的半径()()()P A B P A P B =球的体积公式如果事件A 在⼀次试验中发⽣的概率是p ,那么 34π3V R =n 次独⽴重复试验中事件A 恰好发⽣k 次的概率其中R 表⽰球的半径()(1)(012)k k n k n n P k C p p k n -=-=,,,…,⼀、选择题1、已知复数1Z i =+,则221Z Z Z --等于A .2B .2-C .2iD .2i - 2、“ p 或q 是假命题”是“⾮p 为真命题”的A .充分不必要条件B .必要不充分条件C .充要条件D .既不充分也不必要条件 3、已知3sin cos()65παα+-=,则cos()3πA. BC .45-D .454、已知4a = ,3b = ,(23)(2)61a b a b -?+=,则a 与b 的夹⾓θ为A .30B .45C .60D .1205、已知ABC ?的顶点B 、C 在椭圆2221()x y a a+=>1上,顶点A 是椭圆的⼀个焦点,且椭圆的另⼀个焦点在BC 边上,ABC ?的周长为则该椭圆的离⼼率为 ABC.D .236、正四棱柱1111ABCD A B C D -中,12AA AB =,则异⾯直线1A B 与1AD 所成⾓的余弦值为A .15B .25D .457、已知函数()23g x x =-,23(())1x f g x x =-则1()2f =A .2-B .12C .15-D .308、等⽐数列{}n a 的前n 项和为n S ,若102S =,3014S =,则40S =A .80B .30C .26D .16 9、如果⼀条直线与⼀个平⾯平⾏,那么称此直线与平⾯构成⼀个“平⾏线⾯组”,在⼀个长⽅体中,由两个顶点确定的直线与含有四个顶点的平⾯构成的“平⾏线⾯组”的个数是 A .60B .48C .36D .2410、有⼀块直⾓三⾓板ABC ,30A ∠= ,90C ∠= ,BC 边在桌⾯上,当三⾓板和桌⾯成45 时,AB 边与桌⾯所成⾓的正弦值为AB1B1A1D 1C CDA .12B . CD 11、已知两点(10)A ,,(0)B b ,,若抛物线24y x =上存在点C 使ABC ?为等边三⾓形,则b =B .5或-13 C .4 D . 4或2-12、已知函数2()2f x x x =-,则满⾜条件()()0()()0f x f y f x f y +≤??-≥?的点(,)x y 所形成区域的⾯积为A .πB .32π C .2π D .4π第Ⅱ卷注意事项:1.答题前,考⽣先在答题卡上⽤直径0.5毫⽶⿊⾊墨⽔签字笔将⾃⼰的姓名、准考证号填写清楚,然后贴好条形码.请认真核准条形码上的准考证号、姓名和科⽬.2.第Ⅱ卷共2页,请⽤直径0.5毫⽶⿊⾊墨⽔签字笔在答题卡上各题的答题区域内作答,在试题卷上作答⽆效.3.本卷共10题,共90分.⼆、填空题:本⼤题共4⼩题,每⼩题5分,共20分.把答案填在横线上.(注意:在试题卷上作答⽆效)132x ≤-的解集为14、ABC ?中,已知(4,)A b ,(4,0)B -,(4,0)C ,D 为BC 上⼀点,且AD平分BAC ∠,则AD 所在的直线⽅程为. 15、64(1(1+展开式中的常数项为. 16、已知正四棱锥S ABCD -中,AB =2,则当该棱锥外接球体积最⼩时,它的⾼等于.三、解答题:本⼤题共6⼩题,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10分)(注意:在试题卷上作答⽆效)在ABC ?中,内⾓A B C 、、的对边分别为a b c 、、,且tan 21+tan A cB b=(2)若a =,求ABC ?⾯积的最⼤值.18、(本⼩题满分12分)(注意:在试题卷上作答⽆效)某化妆品⽣产公司计划在郑州的“五⼀社区”举⾏为期三天的“健康使⽤化妆品知识讲座”。
2015年广西高考数学试题及答案发布
2015年广西文科数学试题及答案
理科数学
2015年广西理科数学试题及答案
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2015年广西高考数学试题及答案发布
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广西玉林市博白县2015届高三下学期返校调研考试数学(理)试题(考试时量:120分钟 满分150分)一:单选题:(本大题共10小题,每小题5分,共50分.在每小题给出的四个选项中,只有一项是符合题目要求的。
) 1.设复数11i z =+,22i ()z x x R =+∈,若12R z z ⋅∈,则x = A .1- B .2- C .1D .22.下列命题中的假命题是A .021>∈∀-x R x ,B .212),0x x x>∞+∈∀ , (C .4001.1,x x x R x x <>∈∃时,恒有 当D .R ∈∃α,使函数 αx y =的图像关于y 轴对称3.对于定义在实数集R 上的狄利克雷函数⎩⎨⎧∈∉=),(),(Q x Q x x D 10)(,则下列说法中正确的是A .)(x D 的值域是[]01,B .)(x D 的最小正周期是1C .)(xD 是奇函数 D .)(x D 是偶函数4.若一个底面是正三角形的三棱柱的正视图如图所示,其顶点都在一个球面上,则该球的表面积为A.163πB. 43πC.193π D.1219π5.已知0>t ,若8)22(0=-⎰tdx x ,则t =A.1B.2-C.2-或4D. 46.已知随机变量X 服从正态分布(3,1)N ,且(15)0.6826P X ≤≤=,则(5)P X >= A .0.1588B .0.1587C .0.1586D .0.15857.已知函数()π()sin (0,0,)2f x A x A ωϕωϕ=+>><的部分图象如图所示,则ϕ=A .π6- B .6πC .π3-D .π38.在“学雷锋,我是志愿者”活动中,有6名志愿者要分配到3个不同的社区参加服务,每个社区分配2名志愿者,其中甲、乙两人分到同一社区,则不同的分配方案共有A. 12种B. 18种C. 36种D. 54种9.在平面直角坐标系中,O 为坐标原点,点(2,0)A ,将向量OA 绕点O 按逆时针方向旋转3π后得向量OB ,若向量a 满足1a OA OB --=,则a 的最大值是 A .1-B .C .3D10.已知函数()y f x =是定义域为R 的偶函数. 当0x ≥时,25(02)16()1()1(2)2x x x f x x ⎧≤≤⎪⎪=⎨⎪+>⎪⎩ 若关于x 的方程2[()]()0f x af x b ++=,a b R ∈、有且仅有6个不同实数根,则实数a 的取值范围是A .59(,)24-- B .9(,1)4-- C. 599(,)(,1)244---- D .5(,1)2--二.填空题:(本题共5小题,每小题5分,共25分) 11.已知函数()ln f x x =,则()1f x >的解集为12.春节期间,某销售公司每天销售某种取暖商品的销售额y (单位:万元)与当天的平 均气温x (单位:℃)有关.现收集了春节期间这个销售公司4天的x 与y 的数据列于下 表:y b x a ∧∧∧=+的系数125b ∧=-,则a ∧= .13.在等比数列{}n a 中,11a =,且14a 、22a 、3a 成等差数列,则通项公式n a =14.某程序框图如右图所示,该程序运行后,输出的x 值为31,则a 等于_____15.定义函数{}{}()f x x x =⋅,其中{}x 表示不小于x 的最小整数,如{}1.52=,{}2.52-=-.当(]0,x n ∈,*n N ∈时,函数()f x 的值域为n A ,记集合n A 中元素的个数为n a ,则12111na a a +++=________ 三:解答题:(本大题共6小题,共75分。
解答应写出文字学明、证明过程或演算步骤) 16.(本小题满分12分)某重点大学自主招生考试过程依次为自荐材料审查、笔试、面试共三轮考核。
规定:只能通过前一轮考核才能进入下一轮的考核,否则将被淘汰;三轮考核都通过才算通过该高校的自主招生考试。
学生甲三轮考试通过的概率分别为23、34、45,且各轮考核通过与否相互独立。
(Ⅰ)求甲通过该高校自主招生考试的概率;(Ⅱ)若学生甲每通过一轮考核,则家长奖励人民币1000元作为大学学习的教育基金。
记学生甲得到教育基金的金额为X ,求X 的分布列和数学期望。
17.(本小题满分12分)在ABC ∆中,内角A B C 、、所对的边分别是a b c 、、.已知3a =,cos A =2B A π=+. (Ⅰ)求b 的值; (Ⅱ)求ABC ∆的面积 18.(本小题满分12分)如图,三棱柱111ABC A B C -中,112AB AC AA BC ====,1160AAC ∠=︒,平面1ABC ⊥平面11AAC C ,1AC 与1AC 相交于点D .(Ⅰ) 求证:BD ⊥平面11AAC C ; (Ⅱ) 求二面角1C AB C --的余弦值.19.(本小题满分13分)某城市旅游资源丰富,经调查,在过去的一个月内(以30天计),第t 天的旅游人数()t f (万人)近似地满足()tt f 14+=,而人均消费()t g (元)近似地满足()25125--=t t g .(Ⅰ)求该城市的旅游日收益()W t (万元)与时间t (130t ≤≤,t N +∈)的函数关系式; (Ⅱ)求该城市旅游日收益的最小值. 20.(本小题满分13分)若数列{}n a 的前n 项和为n S ,对任意正整数n 都有612n n S a =-. (Ⅰ)求数列{}n a 的通项公式; (Ⅱ)令()11112241(1)log log n n n n n b a a -++=-⋅⋅,求数列{}n b 的前n 项和nT 21.(本小题满分13分)已知函数()2ln 1f x x a x =--,函数()1F x a =--(Ⅰ)如果()f x 在[]3,5上是单调递增函数,求实数a 的取值范围;(Ⅱ)当2,0a x =>且1x ≠时,比较()1f x x -与()F x 的大小.数学(理科)答案(考试时量:120分钟 满分150分)参考答案一:单选题:(本大题共10小题,每小题5分,共50分.在每小题给出的四个选项中,只有一项是符合11. ()(),,e e -∞-+∞ 12.77513. 12n - 14. 3 15.21nn + 三:解答题:(本大题共6小题,共75分。
解答应写出文字学明、证明过程或演算步骤) 16.(本小题满分12分)解析:(Ⅰ)设“学生甲通过该高校自主招生考试”为事件A ,则P(A)=23423455⨯⨯= 所以学生甲通过该高校自主招生考试的概率为254分 (Ⅱ)X 的可能取值为0元,1000元,2000元,3000元 5分21(0)133P X ==-=, 231(1000)(1)346P X ==⨯-=,2341(2000)(1)34510P X ==⨯⨯-=2342(3000)3455P X ==⨯⨯= 9分所以,X 的分布列为数学期望为11124700()0100020003000361053E X =⨯+⨯+⨯+⨯=12分 17.(本小题满分12分)解析:(Ⅰ)因0A π<<,故sin A ===. 2分因2B A π=+,故sin sin cos 2B A A π⎛⎫=+== ⎪⎝⎭. 4分CC 1B 1AA 1BDH由正弦定理sin sin a bA B=,得sin sin a B b A ===. 6分 (Ⅱ)cos cos sin 2B A A π⎛⎫=+=-= ⎪⎝⎭. 8分 ()()sin sin sin C A B AB π=-+=+⎡⎤⎣⎦ sincos cos sin A BA B =+13⎛=+= ⎝. 10分 则ABC ∆的面积为111sin 3223ab C =⨯⨯=分 18.(本小题满分12分)解:(Ⅰ)依题意,侧面11AAC C 是菱形,D 是1AC 的中点,因为1BA BC =,所以1BD AC ⊥,又平面1ABC ⊥平面11AAC C ,且BD ⊂平面1ABC , 平面1ABC 平面111AAC C AC =所以BD ⊥平面11AAC C . 4分 (Ⅱ)方法一:由(Ⅰ)知BD ⊥平面11AAC C ,CD ⊂面11AAC C ,所以CD BD ⊥,又1CD AC ⊥,1AC BD D =,所以CD ⊥平面1ABC ,过D 作DH AB ⊥,垂足为H ,连结CH ,则CH AB ⊥,所以DHC ∠为二面角1C AB C --的平面角. 7分在Rt DAB ∆中,1,2ADBD AB ===,所以AD DB DH AB ⋅==,CH ==分所以cos DH DHC CH ∠==, 即二面角1C AB C --分 方法二:以D 为原点,建立空间直角坐标系D xyz -如图所示, 5分由已知可得112,1,AC AD BD A DDC BC ======故()()(()()10,0,0,1,0,0,,1,0,0,D A B C C -,B则()(1,0,3,0,3,AB BC =-=, 7分 设平面ABC 的一个法向量是(),,x y z =n ,则00AB BC ⎧⋅=⎪⎨⋅=⎪⎩n n ,即00x ⎧-=⎪=,解得x y z ⎧=⎪⎨=⎪⎩令1z =,得)=n 9分显然()DC =是平面1ABC 的一个法向量, 10分 所以cos ,5DC DC DC⋅<>===n n n , 即二面角1C AB C --分19.(本小题满分13分)解:(Ⅰ)()()()()2512514--⎪⎭⎫ ⎝⎛+==t t t g t f t W()()⎪⎪⎩⎪⎪⎨⎧≤<-+≤≤++=⎪⎪⎩⎪⎪⎨⎧≤<-⎪⎭⎫ ⎝⎛+≤≤+⎪⎭⎫ ⎝⎛+=3025,4150599251,10044013025,15014251,10014t t t t t t t t t t t t 6分(Ⅱ)①当t ∈[1,25]时,W (t )=401+4t +100t ≥401+24t ·100t=441(当且仅当tt 1004=时取等号) 所以,当5=t 时,W (t )取得最小值441. 8分 ②当t ∈(25,30]时,因为W (t )=t t4150599-+单调递减, 所以t =30时,W (t )有最小值()44148430>=W , 11分 综上,t ∈[1,30]时,旅游日收益W (t )的最小值为441万元. 13分 20.(本小题满分13分)解析:(Ⅰ)由11612S a =-,得11612a a =-,解得118a =. 2分 由612n n S a =- ……①,当2n ≥时,有11612n n S a --=- ……②, 3分 ①-②得:114n n a a -=, 4分 ∴数列{}n a 是首项118a =,公比14q =的等比数列 5分12111111842n n n n a a q-+-⎛⎫⎛⎫∴==⨯= ⎪ ⎪⎝⎭⎝⎭, 6分(Ⅱ)由(Ⅰ)知2111221log log 212n n a n +⎛⎫==+ ⎪⎝⎭. 7分()()11111224141(1)(1)log log (21)(23)n n n n n n n b a a n n --+++=-⋅=-⋅⋅+⋅+ 所以111(1)2123n n b n n -⎛⎫=-⋅+⎪++⎝⎭ 9分 当n 为偶数时,11111111355*********n T n n n n ⎛⎫⎛⎫⎛⎫⎛⎫=+-++++-+⎪ ⎪ ⎪ ⎪-+++⎝⎭⎝⎭⎝⎭⎝⎭11323n =-+ 11分 当n 为奇数时,11111111355*********n T n n n n ⎛⎫⎛⎫⎛⎫⎛⎫=+-++-+++⎪ ⎪ ⎪ ⎪-+++⎝⎭⎝⎭⎝⎭⎝⎭11323n =++ 所以1132311,323n n n T n n ,为奇数为偶数⎧+⎪⎪+=⎨⎪-⎪+⎩13分21.(本小题满分13分)解:(Ⅰ)∵2()ln 1f x x a x =--在[3,5]上是单调递增函数, ∴()20af x x x'=-≥在[3,5]上恒成立, 2分 ∴22a x ≤在[3,5]上的最小值为18, ∴18a ≤∴所求的a 的取值范围为(,18]-¥. 6分(Ⅱ)当2a =时,2()2ln 111f x x x x x --=--,0x >且1x ≠,()11F x a =--=,0x ≥. ∴当2a =,0x >且1x ≠时,()()1f x F x x -=- 8分设2()2ln 2h x x x x =--+-,则()h x 的定义域为0x >,2()21h x xx '=--+=. ∴当01x <<时,()0h x '<,此时,()h x 单调递减; 当1x >时,()0h x '>,此时,()h x 单调递增.∴当0x >且1x ≠时,()(1)0h x h >=. 10分。