2017年江西省南昌市一模理科成绩

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江西省南昌市2017届高三一模考试理科综合试题(扫描版)

江西省南昌市2017届高三一模考试理科综合试题(扫描版)

NCS20170607项目第一次模拟测试卷理科综合参考答案及评分标准一—二、选择题生物化学物理 三、非选择题 (一)必做题 22.(1)B (2分) (2)0.48 (3分)23.(1)(1)0.154(0.152~0.155均可) (2分)(2) 22.0 (2分)(3)Ⅰ. 如右图所示 (4分)Ⅱ. 20212()x I R r R I I +=- (2分) 24.(14分)解:(1)设物体在斜面上运动的加速度大小为a 1,方向沿斜面向上,则2121t a vt x += 2分 解得21m/s 25.0=a 1分物块在A 点的速度为m/s 50=+=at v v 2分(2)设物块与接触面间的摩擦因数为μ,物块在水平面上运动时,有2ma mg =μ 1分由图线可知 22m/s 2=a 1 分解得 2.0=μ 1 分物块在斜面上运动时,设受的摩擦力为f ,则1sin cos ma f mg F =+-θθ 2分N f μ= 1分θθsin cos F mg N += 1分解得N 1.10sin cos cos sin 1=++-=θμθθμθma mg mg F 2分25.(18分)解:(1)要使粒子能沿图中虚线O 2O 3进入PQ 、MN 之间的区域,则粒子所受到向上的洛伦兹力与向下的电场力大小相等qE B qv =10 2分解得 m/s 105510⨯==B E v 2 分 (2)粒子在矩形efgh 磁场中作匀速圆周运动,设半径为R ,则 Rv m B qv 200= 解得 m 1.00==qB m v R 2分作出粒子在磁场中轨迹图如图所示。

由几何知识可得l R R 21cos =+θ 解得 060=θ 2分粒子射入点的位置在eh 边上距e 点为m 20360sin 0==R x 1 分 射入的速度方向和eh 边成060角向左下方 1分(3)粒子从O 3以速度v 0进入PQ 、MN 之间的区域,先做匀速直线运动,打到ab 板上,以大小为v 0的速度垂直于磁场方向运动.粒子将以半径R 在垂直于磁场的平面内作匀速圆周运动,转动一周后打到ab 板的下部.由于不计板的厚度,所以质子从第一次打到ab 板到第二次打到ab 板后运动的时间为粒子在磁场运动一周的时间,即一个周期T 。

【江西省南昌市】2017届高三第一次模拟数学(理科)试卷-答案

【江西省南昌市】2017届高三第一次模拟数学(理科)试卷-答案
13.120
14. 3 2
15. (3 2)π
16. 3 10 2
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.【解析】(Ⅰ)设等差数列{an} 的公差为 d ,由 S3 S4 S5 可得 a1 a2 a3 a5 , 即 3a2 a5 ,所以 3(1 d ) 1 4d ,解得 d 2 . ∴ an 1 (n 1) 2 2n 1 .
x3 2 y1(x2 2)
x3 2
y12 (x2 2)2
3(1
x12 4
)( x2

2)2
(x1 2)(x2 2)
即 ( x3 2)2 (x2 2)(x1 2) x1x2 2(x1 x2 ) 4 , x3 2 (x1 2)(x2 2) x1x2 2(x1 x2 ) 4
:
y

y2 (x 2) , x2 2
联立两直线方程得 y1 (x 2) y2 (x 2) (其中 x 为 G 点的横坐标)即证: 3y1 y2 ,
x1 2
x2 2
x1 2 x2 2
即 3k(x1 4) (x2 2) k(x2 4) (x1 2) ,即证 4x1x2 10(x1 x2 ) 16 0
又 f '(0) 4a 2 0 , f '(1) 6a 0 ,所以存在 t (0,1) 使得 f '(t) 0
且当 a 0时 t 1 ,当 a 1 时 t 0 ,所以 t 的取值范围是 (0,1) . 2
又当 x (0,t) , f '(x) 0 ,当 x (t, ) 时, f '(x) 0 ,

【江西省南昌】2017届高三第一次模拟数学年(理科)试题答案

【江西省南昌】2017届高三第一次模拟数学年(理科)试题答案

A.{4}
B.{5}
C. [4, 5]
D.{4, 5}
3.我国古代数学名著《九章算术》有“米谷粒分”题:发仓募粮,所募粒中秕不百三则收之(不超过 3%),
现抽样取米一把,取得 235 粒米中夹秕 n 粒,若这批米合格,则 n 不超过( )
A.6 粒
B.7 粒
C.8 粒
D.9 粒
4.已知 13 23 (6 )2 ,13 23 33 1( 2 )2 ,13 23 33 43 2(0 )2 , 若 13 23 33 43
负者得 0 分,在每一场比赛中,甲胜乙的概率为 3 ,丙胜甲的概率为 3 ,乙胜丙的概率为 p ,且各场比赛
5
4
结果互不影响.若甲获第一名且乙获第三名的概率为 1 . 10
(Ⅰ)求 p 的值;
(Ⅱ)设在该次对抗比赛中,丙得分为 X ,求 X 的分布列和数学期望.
19.如图,四棱锥 P-ABCD 的底面 ABCD 为平行四边形,平面 PAB 平面 ABCD,PB PC ,ABC 45 ,

y2
1
分别交于点 A 、 M 和 A 、 N ,记直线 l1 的斜率为 k1 .
(Ⅰ)求 k k1 的值; (Ⅱ)当 k 变化时,试问直线 MN 是否恒过定点?若恒过定点,求出该定点坐标;若不恒过定点,请说明理 由.
21.已知函数 f (x) eax bx(a 0) 在点 (0, f (0)) 处的切线方程为 y 5x 1,且 f (1) f (1) 12 .
y
1 cos sin

为参数).
(Ⅰ)求曲线 C 的极坐标方程;
(Ⅱ)若曲线
C
向左平移一个单位,再经过伸缩变换

2017届南昌市一模通报会

2017届南昌市一模通报会
2017届“一模”测试成绩通报会
沈军红 2017年3月15日
一、数据通报 二、存在问题 三、复习建议
存在问题 一、个体知识“诊断”与辅导缺失
针对作业、周练、月考、模考对学生个体的知识缺陷进行归因分析,通过讲解典型 题例或实验操作、动作训练等个性化辅导,帮助学生改善和提高。
二、知识教学与能力培养存在割裂
拿到教辅就用,拿到试题就做,不问青红皂白,“眉毛胡子一把抓”。
二、多“高考题”,少“押猜题”, 突出复习的前瞻性 把高考题、各地最新模拟题等归类、分解,理清各类题型的方法、技巧,
达到举一反三的效果。
三、多“回头看”,少“题海战”, 突出复习的基础性 要解放学生,给时间回顾教材、笔记、老师归纳的解题技巧、错题档案、 对教材识记和理解相对薄弱的知识点,不要把所有的时间都用来做题, 完全被老师“牵着鼻子走”!
复习建议
四、多“开小灶”,少“一刀切”, 突出复习的针对性 针对模拟考试中存在问题,学校要分析,要有对策,对边缘生可以 “开小灶”,加强个别辅导。
五、多“减压”,少“施压”, 突出复习的指导性
开展心理讲座、心理咨询、科学迎考放松活动,消除心理焦虑,使 考生注意力集中、思维活跃、记忆流畅。
谢谢!Biblioteka 学生的能力是在教师的教和学生的学的过程中获得的,教师在传授知识的同时,一
定要注重能力的培养。学生的能力不是考试考出来的。
三、课堂教学与命题思路有所相悖
课堂教学的思路:事实—概念---原理---方法---实例 命题思路:问题---背景(条件)---解决问题的路径和方法---问题的解决
复习建议
一、多 “研考”,少“拿来主义”,突出复习的导向性 研究课标、考试说明、考点、高考题,突出重点,提高复习效率。不要不问学情

江西2017届高三摸底考试理科排名

江西2017届高三摸底考试理科排名

姓名准考证号学校总分语文总分数学总分英语总分物理总分黄亚坤2300160011抚州一中704.50122.00150.00133.50110.00王志鹏2300150148临川一中699.00121.00144.00142.00105.00方宏昊2300150171临川一中699.00122.50145.00137.50110.00陈依畅2010061111师大附中697.00132.00140.00148.00100.00刘浩文2300150024临川一中697.00122.00150.00138.00107.00喻润一2300150147临川一中696.00121.00150.00135.00108.00周金龙2300150154临川一中696.00123.00148.00135.00110.00袁嘉明2300150165临川一中696.00123.00146.00142.00108.00范禹2300160004抚州一中695.50114.50148.00141.00107.00周仝希2010060507师大附中695.00137.50135.00143.5098.00何纳博2300150155临川一中694.50122.50149.00139.00107.00李绪雯2010061526师大附中693.50125.00146.00142.5099.00周书捷2010062001师大附中692.50130.00139.00142.5099.00李楚含2210071639莲塘一中692.50121.50143.00140.00102.00邓佳逸2300150175临川一中692.50119.00150.00135.50105.00邹睿2300150032临川一中690.50118.00145.00142.50105.00肖鉴2300150104临川一中690.50121.00144.00138.50107.00杨姝桐2300150151临川一中690.50121.00150.00135.50104.00黄凡2210072513莲塘一中689.50119.00150.00132.50107.00华宇成2300150140临川一中689.50122.00149.00130.50107.00危玮2300150144临川一中689.50119.00142.00140.50105.00闵琦文2210071807莲塘一中689.00132.00131.00138.00105.00张嘉俊2300150001临川一中688.50118.00150.00141.5099.00石如琪2010060332师大附中688.00126.00140.00141.00106.00余曼卿2010062021师大附中686.50116.00147.00138.50106.00郑杰2300150022临川一中686.50109.50150.00138.00110.00齐彦清2300150018临川一中686.00121.50141.00135.50106.00傅琛2010060114师大附中685.50120.00137.00145.50102.00刘菁莹2300150150临川一中685.00115.00150.00132.00105.00程思瑾2300150162临川一中685.00113.50149.00135.50110.00鄢韵2300150170临川一中684.50116.00149.00136.50103.00郑志龙2300150141临川一中684.00115.50150.00139.50104.00刘奕晨2010061321师大附中683.50114.50144.00136.00105.00胡伟霖2210072621莲塘一中683.50116.00133.00146.50104.00郑龙韬2300160014抚州一中683.50115.00150.00137.5099.00张婧琳2010062033师大附中683.00114.00137.00142.00110.00涂俊2210071620莲塘一中683.00121.00146.00135.00101.00李嘉彦2010510108南昌外国语682.00112.00148.00144.0099.00肖继真2300130118吉安一中682.00130.00131.00141.00105.00华天泽2010060526师大附中681.50115.00141.00138.50102.00余懿豪2300150095临川一中681.50119.00150.00129.50104.00高瑞麟2010060620师大附中681.00118.00144.00137.00104.00曾照轩2010062009师大附中681.00115.50143.00138.50105.00易昕仪2300150027临川一中681.00111.00143.00138.00107.00胡凌欣2010510110南昌外国语680.50118.50143.00139.00108.00邓峻恺2010060410师大附中680.00113.00145.00142.00103.00王涌2300130055吉安一中680.00118.00148.00133.00104.00谢君2300150043临川一中680.00125.00140.00135.00100.00陈燕怡2300160083抚州一中680.00118.00149.00142.0093.00陶内奇2210071706莲塘一中679.50121.50137.00141.00102.00郭莹清2300130042吉安一中679.50114.50144.00141.00102.00黄翔2300150085临川一中679.50112.50150.00138.00105.00方博懋2010020422南昌二中679.00120.00141.00138.0097.00李欣雨2010510126南昌外国语679.00116.00136.00143.00109.00王铭璇2300150069临川一中679.00111.50143.00141.50105.00叶婷2300150153临川一中679.00117.00145.00138.00100.00叶泽华2300150169临川一中679.00118.50150.00130.50105.00彭森洋2010060430师大附中678.50116.50137.00138.00110.00刘雍2300160003抚州一中678.50104.00145.00136.50110.00袁正2010020606南昌二中678.00118.50145.00142.5096.00糜晓宇2010062004师大附中678.00110.00140.00137.00108.00熊天钰2010061035师大附中677.50115.00145.00138.50104.00尹康2010062010师大附中677.50119.00149.00136.50102.00陈智颖2010510107南昌外国语677.50110.50136.00142.00104.00杨从君2010060433师大附中677.00113.50138.00137.50102.00郑祺城2010062025师大附中677.00119.50146.00136.50103.00孟文祺2300130052吉安一中677.00125.00142.00132.0098.00詹志2300150143临川一中677.00114.50145.00134.50109.00唐明明2300160007抚州一中677.00114.00135.00143.00104.00胡益宇2210071819莲塘一中676.50112.50145.00138.00105.00石子辉2220090116新建二中676.50120.00150.00134.50103.00张梦帆2300150040临川一中676.50121.50148.00135.0099.00李鸣昊2010021922南昌二中676.00120.50140.00138.5097.00熊宇涵2010510101南昌外国语676.00115.00138.00140.00100.00罗策2300150035临川一中676.00123.00137.00138.00101.00丁明琦2010020710南昌二中675.50123.00139.00133.5099.00聂仪昆2010062003师大附中675.50129.00138.00137.5096.00赖蒽2010510102南昌外国语675.50119.50145.00144.0092.00刘文伟2300130049吉安一中675.50110.00148.00139.50108.00王婕2300160054抚州一中675.50104.00146.00139.50105.00陈伟凡2210071816莲塘一中675.00118.50150.00137.5096.00徐健耀2300150008临川一中675.00114.00147.00137.00102.00罗智宇2010062013师大附中674.50127.00131.00136.50106.00吴天2220090114新建二中674.50127.50132.00140.0098.00宋乐佳2300130054吉安一中674.50117.00144.00138.50104.00余亮华2300150015临川一中674.50122.50144.00130.00108.00章俊龙2300150033临川一中674.50119.50140.00135.00106.00徐嘉俊2300150149临川一中674.50111.00143.00137.5098.00韩志超2300160002抚州一中674.50112.50149.00134.00100.00徐沛翔2010060705师大附中674.00118.50142.00131.50104.00谢禹2010060904师大附中674.00122.50139.00138.5097.00宋法兴2300150013临川一中674.00113.00147.00131.00104.00陈佳宇2300150016临川一中674.00118.50142.00132.50104.00章晋哲2010060529师大附中673.50115.50144.00133.00107.00徐欣阳2010062103师大附中673.50126.00136.00139.5099.00左子萱2300130045吉安一中673.50111.50141.00138.00103.00罗益民2010020711南昌二中673.00119.00143.00140.0099.00莫辛夷2010060622师大附中673.00124.00141.00133.00102.00刘龙祥2010062008师大附中673.00124.00143.00135.0093.00吕乐2210071517莲塘一中673.00119.00146.00141.0090.00徐志成2210071631莲塘一中673.00107.50141.00139.50105.00吴欣诚2300150041临川一中673.00118.00142.00136.00104.00张子衿2010060206师大附中672.50111.00136.00143.50108.00黄昕远2010060233师大附中672.50132.00132.00136.50100.00黄磊2300150002临川一中672.50121.00145.00132.5099.00刘珂2300150166临川一中672.50123.50150.00123.0099.00刘建2300150177临川一中672.50121.00144.0012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15.50138.00129.5097.00黄震2300112110永丰中学642.00119.50117.00126.5098.00程子悦2300120062弋阳一中642.00111.00129.00127.0097.00杨延睿2300130093吉安一中642.00103.00142.00125.00105.00陈思婕2300160042抚州一中642.00102.50143.00111.50105.00王梦欣2300160044抚州一中642.00119.50130.00134.5094.00汤世豪2300160122抚州一中642.0095.50150.00121.50105.00汤嫒嫒2010021316南昌二中641.50104.50149.00139.0091.00魏恒威2210072144莲塘一中641.50104.50139.00127.0098.00涂冲2220090123新建二中641.50103.50135.00127.0096.00胡利杨2240102022进贤第一中学641.50108.00131.00123.50104.00陈文昊2300112147永丰中学641.50120.00134.00124.5098.00李嘉敏2300150006临川一中641.50103.00133.00133.50104.00游孜明2300160035抚州一中641.50113.00129.00132.5093.00饶旺2300160076抚州一中641.50112.50145.00116.00101.00陈昱方2300160094抚州一中641.50110.00130.00130.5097.00张鹏飞2300160099抚州一中641.50113.50127.00133.00100.00任标2300170082丰城九中641.50107.50143.00127.0098.00龚文奕2010021327南昌二中641.00101.50142.00131.5094.00舒惠千2010022013南昌二中641.00117.50135.00133.5097.00李耀2010030123南昌三中641.00102.00142.00127.0096.00朱逸航2010062019师大附中641.00111.00126.00130.00102.00王钰婧2010510137南昌外国语641.00127.00121.00134.0089.00熊世龙2210072139莲塘一中641.00115.00134.00120.0099.00刘洋2210072320莲塘一中641.00109.00146.00123.00100.00胡秋辉2220090225新建二中641.00103.00133.00125.00100.00杨景皓2300130110吉安一中641.00110.00135.00133.00107.00黎健成2010020212南昌二中640.50106.50132.00137.0095.00肖鹏宇2010020329南昌二中640.50115.00134.00130.5093.00傅廷杰2010020408南昌二中640.50109.50137.00126.0091.00朱正午2010020909南昌二中640.50106.50131.00133.0095.00洪宇2010021421南昌二中640.50110.50132.00135.00102.00周旻皓2010061001师大附中640.50112.00128.00132.5092.00陈李蔚2010062101师大附中640.50107.00137.00125.5099.00魏恒达2210071714莲塘一中640.50116.50134.00130.0084.00程鸿铭2210072410莲塘一中640.50118.00125.00126.5096.00林启昌2300112107永丰中学640.50111.00120.00133.5096.00肖若冶2300130078吉安一中640.50107.00128.00130.50100.00邵凯2010020634南昌二中640.00104.00136.00142.0083.00刘元正2010060812师大附中640.00111.00135.00137.00102.00颜苗2010062207师大附中640.00116.50137.00125.5098.00高博韬2010062217师大附中640.00113.50129.00138.5090.00熊迪非2010510234南昌外国语640.00108.00133.00135.0095.00应宗佺2210073016莲塘一中640.00116.00122.00131.0096.00缪珂2220090107新建二中640.00114.00143.00123.0093.00徐佳泰2240101706进贤第一中学640.00112.00127.00126.00100.00樊智鹏2240101721进贤第一中学640.00108.00119.00138.00101.00陈昊2300130097吉安一中640.00112.00125.00138.0091.00颜治文2300130117吉安一中640.00103.00135.00128.00102.00李文峤2300150080临川一中640.00103.50131.00136.5094.00朱济人2010510117南昌外国语639.5095.00140.00135.50105.00朱子璇2010510220南昌外国语639.50112.00135.00133.5092.00赖宇琴2300130082吉安一中639.50108.50136.00131.0094.00龚兴2300130100吉安一中639.50115.00135.00117.50107.00徐洋博2300150164临川一中639.50115.50136.00120.0093.00周大强2300160140抚州一中639.50108.00139.00122.5097.00梅斌涛2300160142抚州一中639.50103.50147.00133.0091.00张佳琪2010021218南昌二中639.00112.50124.00132.5097.00赖雨宣2010030109南昌三中639.00126.00110.00139.0098.00朱莅朗2010060129师大附中639.00111.50131.00133.5096.00胡俊2220090111新建二中639.00109.00130.00126.0099.00李若松2300112124永丰中学639.00110.50129.00132.5092.00李博胜2300130066吉安一中639.00110.00131.00125.0096.00杜渤翔2010040107南昌十中638.50111.50140.00121.0094.00崔峻伟2010060403师大附中638.50108.50142.00134.0091.00徐妙2010060634师大附中638.50120.00118.00140.5091.00李可迪2010061403师大附中638.50123.00120.00125.5097.00邱芊2010062310师大附中638.50111.00133.00130.5098.00万文超2220090134新建二中638.50109.50141.00127.0099.00杨鸿建2240332706进贤第二中学638.50116.00124.00126.5098.00顾文韬2300160020抚州一中638.50111.50140.00128.0090.00付煜峻2010020804南昌二中638.00108.50145.00115.5096.00曾皓阳2010021004南昌二中638.00104.50138.00130.5099.00黄敏璇2010021808南昌二中638.00112.50129.00131.5092.00黄琪兴2300130080吉安一中638.00111.50133.00131.5087.00邹通2300150011临川一中638.00110.00144.00123.0091.00余浩文2300150060临川一中638.00110.00133.00131.00103.00熊海军2300160048抚州一中638.00105.50139.00117.50103.00邓雯敏2010020624南昌二中637.50113.00132.00131.5091.00邓文星2010061504师大附中637.50117.50128.00134.00101.00李超2010062029师大附中637.50114.00137.00121.5090.00王惟嘉2010510111南昌外国语637.50102.50139.00139.0091.00王嘉钰2010510222南昌外国语637.50108.00129.00134.5098.00帅文璇2300150063临川一中637.5098.50142.00138.00101.00罗霄航2300170005丰城九中637.50112.50120.00134.00103.00舒梓铧2010021219南昌二中637.0098.00134.00126.00100.00段依田2010030107南昌三中637.00108.50130.00127.5096.00肖含翘2010062036师大附中637.00108.50139.00125.5098.00顾逸君2010560102南大附中637.00109.50128.00133.5094.00万志伟2210072445莲塘一中637.00106.00131.00133.0092.00姜子呈2220290227新建一中637.00116.00131.00129.0095.00陈冰月2240103730进贤第一中学637.00111.00118.00135.00101.00。

2017年江西省南昌市高考数学一模试卷(理科)

2017年江西省南昌市高考数学一模试卷(理科)

2017年江西省南昌市高考数学一模试卷(理科)一.选择题:共12小题,每小题5分,共60分.在每个小题给出的四个选项中,只有一项是符合题目要求的.1.(5分)已知全集U=R,集合A={x|y=lgx},集合B=,那么A∩(∁U B)=()A.∅B.(0,1]C.(0,1) D.(1,+∞)2.(5分)若复数,其中i为虚数单位,则复数z的虚部是()A.﹣1 B.﹣i C.1 D.i3.(5分)已知α,β为第一象限的两个角,则“α>β”是“sinα>sinβ”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4.(5分)设某中学的高中女生体重y(单位:kg)与身高x(单位:cm)具有线性相关关系,根据一组样本数据(x i,y i)(i=1,2,3,…,n),用最小二乘法近似得到回归直线方程为,则下列结论中不正确的是()A.y与x具有正线性相关关系B.回归直线过样本的中心点C.若该中学某高中女生身高增加1cm,则其体重约增加0.85kgD.若该中学某高中女生身高为160cm,则可断定其体重必为50.29kg5.(5分)若圆锥曲线C:x2+my2=1的离心率为2,则m=()A.B.C.D.6.(5分)执行如图所示的程序框图,输出S的值为()A.log210﹣1 B.2log23﹣1 C.D.67.(5分)已知函数的周期为π,若f(α)=1,则=()A.﹣2 B.﹣1 C.1 D.28.(5分)如图,在平面直角坐标系xoy中,直线y=2x+1与圆x2+y2=4相交于A,B两点,则cos∠AOB=()A.B.C.D.9.(5分)我国古代数学名著《九章算术》中有如下问题:今有甲乙丙三人持钱,甲语乙丙:各将公等所持钱,半以益我,钱成九十(意思是把你们两个手上的钱各分我一半,我手上就有90钱);乙复语甲丙,各将公等所持钱,半以益我,钱成七十;丙复语甲乙:各将公等所持钱,半以益我,钱成五十六,则乙手上有()钱.A.28 B.32 C.56 D.7010.(5分)某空间几何体的三视图如图所示(图中小正方形的边长为1),则这个几何体的体积是()A.B.C.16 D.3211.(5分)抛物线y2=8x的焦点为F,设A(x1,y1),B(x2,y2)是抛物线上的两个动点,若x1+x2+4=|,则∠AFB的最大值为()A.B. C. D.12.(5分)定义在R上的偶函数f(x)满足f(2﹣x)=f(x),且当x∈[1,2]时,f(x)=lnx﹣x+1,若函数g(x)=f(x)+mx有7个零点,则实数m的取值范围为()A.B.C.D.二.填空题:本大题共4小题,每小题5分,共20分.13.(5分)在多项式(1+2x)6(1+y)5的展开式中,xy3项的系数为.14.(5分)已知单位向量的夹角为,,则在上的投影是.15.(5分)如图,直角梯形ABCD中,AD⊥DC,AD∥BC,BC=2CD=2AD=2,若将直角梯形绕BC 边旋转一周,则所得几何体的表面积为.16.(5分)已知x2+y2=4,在这两个实数x,y之间插入三个实数,使这五个数构成等差数列,那么这个等差数列后三项和的最大值为.三.解答题:本大题共5小题,共70分.解答应写出文字说明、证明过程或演算步骤.17.(12分)已知等差数列{a n}的前n项和为S n,且a1=1,S3+S4=S5.(Ⅰ)求数列{a n}的通项公式;(Ⅱ)令b n=(﹣1)n﹣1a n a n+1,求数列{b n}的前2n项和T2n.18.(12分)某中学的环保社团参照国家环境标准制定了该校所在区域空气质量指数与空气质量等级对应关系如下表(假设该区域空气质量指数不会超过300):空气质量指数(0,50](50,100](100,150](150,200](200,250](250,300]空气质量等级1级优2级良3级轻度污染4级中度污染5级重度污染6级严重污染该社团将该校区在2016年100天的空气质量指数监测数据作为样本,绘制的频率分布直方图如图,把该直方图所得频率估计为概率.(Ⅰ)请估算2017年(以365天计算)全年空气质量优良的天数(未满一天按一天计算);(Ⅱ)该校2017年6月7、8、9日将作为高考考场,若这三天中某天出现5级重度污染,需要净化空气费用10000元,出现6级严重污染,需要净化空气费用20000元,记这三天净化空气总费用为X元,求X的分布列及数学期望.19.(12分)如图,四棱锥P﹣ABCD中,平面PAD⊥平面ABCD,底面ABCD为等腰梯形,AB∥CD,AD=DC=BC=2,AB=4,△PAD为正三角形.(Ⅰ)求证:BD⊥平面PAD;(Ⅱ)设AD的中点为E,求平面PEB与平面PDC所成二面角的平面角的余弦值.20.(12分)已知椭圆C:=1(a>b>0)的左、右顶点分别为A1,A2,左、右焦点分别为F1,F2,离心率为,点B(4,0),F2为线段A1B的中点.(Ⅰ)求椭圆C的方程;(Ⅱ)若过点B且斜率不为0的直线l与椭圆C的交于M,N两点,已知直线A1M与A2N相交于点G,试判断点G是否在定直线上?若是,请求出定直线的方程;若不是,请说明理由.21.(12分)已知函数f(x)=(2x﹣4)e x+a(x+2)2(x>0,a∈R,e是自然对数的底).(Ⅰ)若f(x)是(0,+∞)上的单调递增函数,求实数a的取值范围;(Ⅱ)当时,证明:函数f(x)有最小值,并求函数f(x)最小值的取值范围.请考生在第(22)、(23)两题中任选一题作答,如果多做,则按所做的第一题记分.[选修4-4:坐标系与参数方程]22.(10分)在平面直角坐标系xoy中,曲线C1过点P(a,1),其参数方程为(t为参数,a∈R).以O为极点,x轴非负半轴为极轴,建立极坐标系,曲线C2的极坐标方程为ρcos2θ+4cosθ﹣ρ=0.(Ⅰ)求曲线C1的普通方程和曲线C2的直角坐标方程;(Ⅱ)已知曲线C1与曲线C2交于A、B两点,且|PA|=2|PB|,求实数a的值.[选修4-5:不等式选讲]23.已知函数f(x)=|2x﹣a|+|x﹣1|,a∈R.(Ⅰ)若不等式f(x)≤2﹣|x﹣1|有解,求实数a的取值范围;(Ⅱ)当a<2时,函数f(x)的最小值为3,求实数a的值.2017年江西省南昌市高考数学一模试卷(理科)参考答案与试题解析一.选择题:共12小题,每小题5分,共60分.在每个小题给出的四个选项中,只有一项是符合题目要求的.1.(5分)已知全集U=R,集合A={x|y=lgx},集合B=,那么A∩(∁U B)=()A.∅B.(0,1]C.(0,1) D.(1,+∞)【分析】由对数函数的定义域求出A,由函数的值域求出B,由补集和交集的运算求出答案,【解答】解:由题意知,A={x|y=lgx}={x|x>0}=(0,+∞),又,则B={y|y≥1}=[1,+∞),即C U B=(﹣∞,1),所以A∩(C U B)=(0,1),故选:C.【点评】本题考查交、并、补集的混合运算,以及对数函数的定义域,属于基础题.2.(5分)若复数,其中i为虚数单位,则复数z的虚部是()A.﹣1 B.﹣i C.1 D.i【分析】利用复数的运算法则即可得出.【解答】解:,故选:C.【点评】本题考查了复数的运算法则,考查了推理能力与计算能力,属于基础题.3.(5分)已知α,β为第一象限的两个角,则“α>β”是“sinα>sinβ”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【分析】根据三件函数的定义和关系式,结合充分条件和必要条件的定义进行判断.【解答】解:∵角α,β的终边在第一象限,∴当α=+2π,β=,满足α>β,但sinα=sinβ,则sinα>sinβ不成立,即充分性不成立,若当α=,β=+2π,满足sinα>sinβ,但α>β不成立,即必要性不成立,故“α>β”是“sinα>sinβ”的既不必要也不充分条件,故选:D.【点评】本题主要考查充分条件和必要条件的判断,比较基础.4.(5分)设某中学的高中女生体重y(单位:kg)与身高x(单位:cm)具有线性相关关系,根据一组样本数据(x i,y i)(i=1,2,3,…,n),用最小二乘法近似得到回归直线方程为,则下列结论中不正确的是()A.y与x具有正线性相关关系B.回归直线过样本的中心点C.若该中学某高中女生身高增加1cm,则其体重约增加0.85kgD.若该中学某高中女生身高为160cm,则可断定其体重必为50.29kg【分析】根据回归分析与线性回归方程的意义,对选项中的命题进行分析、判断正误即可.【解答】解:由于线性回归方程中x的系数为0.85,因此y与x具有正的线性相关关系,A正确;由线性回归方程必过样本中心点,因此B正确;由线性回归方程中系数的意义知,x每增加1cm,其体重约增加0.85kg,C正确;当某女生的身高为160cm时,其体重估计值是50.29kg,而不是具体值,因此D 错误.故选:D.【点评】本题考查了回归分析与线性回归方程的应用问题,是基础题目.5.(5分)若圆锥曲线C:x2+my2=1的离心率为2,则m=()A.B.C.D.【分析】圆锥曲线C:x2+my2=1方程可化为,利用离心率为2,求出m的值.【解答】解:因为圆锥曲线C:x2+my2=1方程可化为,所以离心率为,故选:C.【点评】本题考查双曲线的离心率,考查方程思想,比较基础.6.(5分)执行如图所示的程序框图,输出S的值为()A.log210﹣1 B.2log23﹣1 C.D.6【分析】由题意,模拟程序的运行过程,依次写出每次循环得到的S,i的值,即可得出跳出循环时输出S的值.【解答】解:模拟程序的运行,可得:由,当i=7时,进入循环,得,当i=8退出循环,输出,故选:B.【点评】本题主要考查了循环结构的程序框图应用问题,是基础题目.7.(5分)已知函数的周期为π,若f(α)=1,则=()A.﹣2 B.﹣1 C.1 D.2【分析】根据函数f(x)的周期求出ω的值,再化简f(α+)并求值.【解答】解:因为函数f(x)=Asin(ωx+φ)的周期为T==π,∴ω=2,∴f(x)=Asin(2x+φ),又f(α)=Asin(2α+φ)=1,∴f(α+)=Asin[2(α+)+φ]=Asin(2α+3π+φ)=﹣Asin(2α+φ)=﹣1.故选:B.【点评】本题考查了三角函数的图象与性质的应用问题,是基础题目.8.(5分)如图,在平面直角坐标系xoy中,直线y=2x+1与圆x2+y2=4相交于A,B两点,则cos∠AOB=()A.B.C.D.【分析】求出圆心到直线y=2x+1的距离,由垂径定理得AB,利用余弦定理,可得结论.【解答】解:因为圆心到直线y=2x+1的距离,由垂径定理得:∴由余弦定理有,故选:D.【点评】本题考查点到直线距离公式的运用,考查垂径定理、余弦定理的运用,属于中档题.9.(5分)我国古代数学名著《九章算术》中有如下问题:今有甲乙丙三人持钱,甲语乙丙:各将公等所持钱,半以益我,钱成九十(意思是把你们两个手上的钱各分我一半,我手上就有90钱);乙复语甲丙,各将公等所持钱,半以益我,钱成七十;丙复语甲乙:各将公等所持钱,半以益我,钱成五十六,则乙手上有()钱.A.28 B.32 C.56 D.70【分析】设甲、乙丙各有x钱,y钱,z钱,列出方程组求得甲有72钱,乙有32钱,丙有4钱.【解答】解:设甲、乙丙各有x钱,y钱,z钱,则,解得x=72,y=32,z=4.∴甲有72钱,乙有32钱,丙有4钱.故选:B.【点评】本题考查函数在生产生活中的实际应用,是基础题,解题时要认真审题,注意函数性质的合理运用.10.(5分)某空间几何体的三视图如图所示(图中小正方形的边长为1),则这个几何体的体积是()A.B.C.16 D.32【分析】回归到正方体中,该几何体是一个底面为等腰直角三角形的三棱锥,即如图中的几何体A﹣BCD,其体积是正方体体积的,即可得出结论.【解答】解:回归到正方体中,该几何体是一个底面为等腰直角三角形的三棱锥,即如图中的几何体A﹣BCD,其体积是正方体体积的,等于,故选:A.【点评】本题考查由三视图求体积,考查学生的计算能力,回归到正方体中,该几何体是一个底面为等腰直角三角形的三棱锥是关键.11.(5分)抛物线y2=8x的焦点为F,设A(x1,y1),B(x2,y2)是抛物线上的两个动点,若x1+x2+4=|,则∠AFB的最大值为()A.B. C. D.【分析】利用余弦定理,结合基本不等式,即可求出∠AFB的最大值.【解答】解:因为,|AF|+|BF|=x1+x2+4,所以.在△AFB中,由余弦定理得:=.又.所以,∴∠AFB的最大值为,故选:D.【点评】本题考查抛物线的定义,考查余弦定理、基本不等式的运用,属于中档题.12.(5分)定义在R上的偶函数f(x)满足f(2﹣x)=f(x),且当x∈[1,2]时,f(x)=lnx﹣x+1,若函数g(x)=f(x)+mx有7个零点,则实数m的取值范围为()A.B.C.D.【分析】确定函数为偶函数则其周期为T=2,函数在x∈[1,2]为减函数,作出函数的图象,得出当x<0时,要使符合题意则,根据偶函数的对称性,当x>0时,要使符合题意则.即可得出结论.【解答】解:因为函数f(2﹣x)=f(x)可得图象关于直线x=1对称,且函数为偶函数则其周期为T=2,又因为,当x∈[1,2]时有f'(x)≤0,则函数在x∈[1,2]为减函数,作出其函数图象如图所示:其中,当x<0时,要使符合题意则根据偶函数的对称性,当x>0时,要使符合题意则.综上所述,实数m的取值范围为,故选:A.【点评】本题考查函数的奇偶性、单调性,考查数形结合的数学思想,难度大.二.填空题:本大题共4小题,每小题5分,共20分.13.(5分)在多项式(1+2x)6(1+y)5的展开式中,xy3项的系数为120.【分析】利用二项式展开式的通项公式即可得出.【解答】解:根据题意(1+2x)6(1+y)5=,∴xy3的系数为=120,故答案为:120.【点评】本题考查了二项式定理的应用,考查了推理能力与计算能力,属于基础题.14.(5分)已知单位向量的夹角为,,则在上的投影是.【分析】根据平面向量投影的定义,利用数量积的运算求出对应的值即可.【解答】解:单位向量的夹角为,,则在上的投影是:||cos<,>==•=(2﹣)•=2﹣•=2﹣1×1×1×cos=.故答案为:.【点评】本题考查了平面向量数量积与投影的计算问题,是基础题目.15.(5分)如图,直角梯形ABCD中,AD⊥DC,AD∥BC,BC=2CD=2AD=2,若将直角梯形绕BC边旋转一周,则所得几何体的表面积为.【分析】由圆锥及圆柱的几何特征可得,该几何体由两个底面相待的圆锥和圆柱组合而成,其中圆柱和圆锥的高均为1,代入圆柱和圆锥的体积公式,即可得到答案.【解答】解:由图中数据可得:,S=π×2×圆柱侧1=2π,.所以几何体的表面积为.故答案为:.【点评】本题考查的知识点是圆柱与圆锥的体积及余弦定理,关键是:(1)熟练掌握圆柱和圆锥的体积公式是关键,(2)将空间问题转化为平面问题是解答立体几何常用的技巧.16.(5分)已知x2+y2=4,在这两个实数x,y之间插入三个实数,使这五个数构成等差数列,那么这个等差数列后三项和的最大值为.【分析】设构成等差数列的五个数分别为x,a,b,c,y,推导出.从而等差数列后三项和为.法一:设x=2cosα,y=2sinα,利用三角函数性质能求出这个等差数列后三项和的最大值.法二:令z=x+3y,则x+3y﹣z=0,当直线x+3y﹣z=0与圆x2+y2=4相切时z将有最大值,由此能求出这个等差数列后三项和的最大值.【解答】解:设构成等差数列的五个数分别为x,a,b,c,y,则x+y=a+c=2b,∴.则等差数列后三项和为=.(另解:由等差数列的性质有x+y=a+c=2b,所以.)方法一:因为x2+y2=4,设x=2cosα,y=2sinα,所以.方法二:令z=x+3y,则x+3y﹣z=0,所以当直线x+3y﹣z=0与圆x2+y2=4相切时z将有最大值,此时,即,∴.故答案为:.【点评】本题考查等差数列的后三项的最大值的求法,是中档题,解题时要认真审题,注意等价转化思想的合理运用.三.解答题:本大题共5小题,共70分.解答应写出文字说明、证明过程或演算步骤.17.(12分)已知等差数列{a n}的前n项和为S n,且a1=1,S3+S4=S5.(Ⅰ)求数列{a n}的通项公式;(Ⅱ)令b n=(﹣1)n﹣1a n a n+1,求数列{b n}的前2n项和T2n.【分析】(Ⅰ)设等差数列{a n}的公差为d,根据题意、等差数列的性质以及通项公式列出方程,求出公差d,由等差数列的通项公式求出a n;(Ⅱ)由(I)化简b n=(﹣1)n﹣1a n a n+1,利用并项求和法和等差数列的前n项和公式求出数列{b n}的前2n项和T2n.【解答】解:(Ⅰ)设等差数列{a n}的公差为d,由S3+S4=S5可得a1+a2+a3=a5,﹣﹣﹣﹣﹣﹣﹣(2分)即3a2=a5,则3(1+d)=1+4d,解得d=2﹣﹣﹣﹣﹣(4分)所以a n=1+(n﹣1)×2=2n﹣1.﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣(6分)(Ⅱ)由(Ⅰ)可得:﹣﹣﹣﹣﹣﹣(7分)所以=4[12﹣22+32﹣42+…+(2n﹣1)2﹣(2n)2]﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣(9分)=﹣4(1+2+3+4+…+2n﹣1+2n)=﹣﹣﹣﹣﹣﹣(12分)【点评】本题考查等差数列的性质、通项公式以及前n项和公式,以及并项求和法求数列的和,考查化简、变形能力.18.(12分)某中学的环保社团参照国家环境标准制定了该校所在区域空气质量指数与空气质量等级对应关系如下表(假设该区域空气质量指数不会超过300):空气质量指数(0,50](50,100](100,150](150,200](200,250](250,300]空气质量等级1级优2级良3级轻度污染4级中度污染5级重度污染6级严重污染该社团将该校区在2016年100天的空气质量指数监测数据作为样本,绘制的频率分布直方图如图,把该直方图所得频率估计为概率.(Ⅰ)请估算2017年(以365天计算)全年空气质量优良的天数(未满一天按一天计算);(Ⅱ)该校2017年6月7、8、9日将作为高考考场,若这三天中某天出现5级重度污染,需要净化空气费用10000元,出现6级严重污染,需要净化空气费用20000元,记这三天净化空气总费用为X元,求X的分布列及数学期望.【分析】(I)利用直方图的性质即可得出.(Ⅱ)由题可知,X的所有可能取值为:0,10000,20000,30000,40000,50000,60000,利用二项分布列的概率与数学期望计算公式即可得出.【解答】解:(Ⅰ)由直方图可估算2017年(以365天计算)全年空气质量优良的天数为:(0.1+0.2)×365=0.3×365=109.5≈110(天).﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣(4分)(Ⅱ)由题可知,4级污染以下的概率P=1﹣0.002×100=.X的所有可能取值为:0,10000,20000,30000,40000,50000,60000,﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣(6分)则:,,,,,,.∴X的分布列为X0100002000030000400005000060000P﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣(10分)=9000(元).﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣﹣(12分)【点评】本题考查了频率分布直方图的性质、二项分布列的概率与数学期望计算公式,考查了推理能力与计算能力,属于中档题.19.(12分)如图,四棱锥P﹣ABCD中,平面PAD⊥平面ABCD,底面ABCD为等腰梯形,AB∥CD,AD=DC=BC=2,AB=4,△PAD为正三角形.(Ⅰ)求证:BD⊥平面PAD;(Ⅱ)设AD的中点为E,求平面PEB与平面PDC所成二面角的平面角的余弦值.【分析】(Ⅰ)在等腰梯形ABCD中,过点D作DE⊥AB于点E,推导出AD⊥BD,由此能证明BD⊥平面PAD.(Ⅱ)以D为坐标原点,DA所在直线为x轴,DB所在直线为y轴,过点D平行于PE所在直线为z轴,建立空间直角坐标系.利用向量法能求出平面PEB与平面PDC所成二面角的余弦值.【解答】证明:(Ⅰ)在等腰梯形ABCD中,过点D作DE⊥AB于点E,如图所示:有∴在△ABD中,有AB2=AD2+BD2,即AD⊥BD又因为平面PAD⊥平面ABCD且交线为AD,∴BD⊥平面PAD.﹣﹣﹣﹣﹣(5分)解:(Ⅱ)由平面PAD⊥平面ABCD,且△PAD为正三角形,E为AD的中点,∴PE⊥AD,得PE⊥平面ABCD.如图所示,以D为坐标原点,DA所在直线为x轴,DB所在直线为y轴,过点D 平行于PE所在直线为z轴,建立空间直角坐标系.由条件AD=DC=BC=2,则AE=DE=1,,.则D(0,0,0),E(1,0,0),,.﹣﹣﹣﹣﹣﹣﹣(6分)在等腰梯形ABCD中,过点C作BD的平行线交AD延长线于点F如图所示:则在Rt△CDF中,有,DF=1,∴.﹣﹣﹣﹣﹣﹣﹣(7分)(另解:可不作辅助线,利用求点C坐标)∴,,设平面PDC的法向量则,取,则y1=1,z1=﹣1,∴面PDC的法向量.﹣﹣﹣﹣﹣﹣﹣(9分)同理有,,设平面PBE的法向量则,取y2=1,则,z2=0,∴面PBE的法向量.﹣﹣(10分)设平面PEB与平面PDC所成二面角的平面角为θ,∴.即平面PEB与平面PDC所成二面角的余弦值为.﹣﹣﹣﹣﹣﹣﹣(12分)【点评】本题考查线面垂直的证明,考查二面角的余弦值的求法,是中档题,解题时要认真审题,注意空间中线线、线面、面面间的位置关系的合理运用.20.(12分)已知椭圆C:=1(a>b>0)的左、右顶点分别为A1,A2,左、右焦点分别为F1,F2,离心率为,点B(4,0),F2为线段A1B的中点.(Ⅰ)求椭圆C的方程;(Ⅱ)若过点B且斜率不为0的直线l与椭圆C的交于M,N两点,已知直线A1M与A2N相交于点G,试判断点G是否在定直线上?若是,请求出定直线的方程;若不是,请说明理由.【分析】(Ⅰ)设点A1(﹣a,0),F2(c,0),由题意得a=4﹣2c,由椭圆的离心率,得a=2c,求出a,b,由此能示出椭圆C的方程.(Ⅱ)法一:根据椭圆的对称性猜测点G是与y轴平行的直线x=x0上.假设当点M为椭圆的上顶点时,直线l的方程为,此时点N,联立直线和直线可得点,猜想点G在直线x=1上,对猜想给予证明,得到点G在定直线上x=1上.法二:设M(x1,y1),N(x2,y2),G(x3,y3),由B,M,N三点共线,得:2x1x2﹣5(x1+x2)+8=0,再由A1,M,G三点共线,A2,N,G三点共线,推导出点G在定直线x=1上.法三:设l的方程为y=k(x﹣4),M(x1,y1),N(x2,y2).由得(3+4k2)x2﹣32k2x+64k2﹣12=0,由此利用根的判别式、韦达定理,结合A1,M,G三点共线,A2,N,G三点共线,推导出点G在定直线x=1上.【解答】解:(Ⅰ)设点A1(﹣a,0),F2(c,0),由题意可知:,即a=4﹣2c①又因为椭圆的离心率,即a=2c②联立方程①②可得:a=2,c=1,则b2=a2﹣c2=3所以椭圆C的方程为.﹣﹣﹣﹣﹣﹣﹣(5分)解:(Ⅱ)解法一:根据椭圆的对称性猜测点G是与y轴平行的直线x=x0上.假设当点M为椭圆的上顶点时,直线l的方程为,此时点N,则联立直线和直线可得点据此猜想点G在直线x=1上,下面对猜想给予证明:﹣﹣﹣﹣﹣﹣﹣(7分)设M(x1,y1),N(x2,y2),联立方程可得:(3+4k2)x2﹣32k2x+64k2﹣12=0,△>0由韦达定理可得,(*)﹣﹣﹣﹣﹣﹣﹣(9分)因为直线,,联立两直线方程得(其中x为G点的横坐标)即证:,即3k(x1﹣4)•(x2﹣2)=﹣k(x2﹣4)•(x1+2),即证4x1x2﹣10(x1+x2)+16=0﹣﹣﹣﹣﹣﹣﹣(11分)将(*)代入上式可得此式明显成立,原命题得证.所以点G在定直线上x=1上.﹣﹣﹣﹣﹣﹣﹣(12分)解法二:设M(x1,y1),N(x2,y2),G(x3,y3),x1,x2,x3两两不等,因为B,M,N三点共线,所以,整理得:2x1x2﹣5(x1+x2)+8=0﹣﹣﹣﹣﹣﹣﹣(8分)又A1,M,G三点共线,有:①又A2,N,G三点共线,有:②,将①与②两式相除得:即,﹣﹣﹣﹣﹣﹣﹣(10分)将2x1x2﹣5(x1+x2)+8=0即代入得:解得x3=4(舍去)或x3=1,所以点G在定直线x=1上.﹣﹣﹣﹣﹣﹣﹣(12分)解法三:由题意知l与x轴不垂直,设l的方程为y=k(x﹣4),M(x1,y1),N (x2,y2).由得(3+4k2)x2﹣32k2x+64k2﹣12=0,△>0.﹣﹣﹣﹣﹣﹣﹣(7分)设M(x1,y1),N(x2,y2),G(x3,y3),x1,x2,x3两两不等,则,,,由A1,M,G三点共线,有:①由A2,N,G三点共线,有:②①与②两式相除得:﹣﹣﹣﹣﹣﹣﹣(10分)解得x3=4(舍去)或x3=1,所以点G在定直线x=1上.﹣﹣﹣﹣﹣﹣﹣(12分)【点评】本题考查椭圆方程的求法,考查点是否在定直线上的判断与证明,是中档题,解题时要认真审题,注意椭圆性质、根的判别式、韦达定理、三点共线等知识点的合理运用.21.(12分)已知函数f(x)=(2x﹣4)e x+a(x+2)2(x>0,a∈R,e是自然对数的底).(Ⅰ)若f(x)是(0,+∞)上的单调递增函数,求实数a的取值范围;(Ⅱ)当时,证明:函数f(x)有最小值,并求函数f(x)最小值的取值范围.【分析】(Ⅰ)求出函数的导数,根据函数的单调性求出a的范围即可;(Ⅱ)根据函数的单调性求出f(x)的最小值,从而求出最小值的范围即可.【解答】解:(Ⅰ)f'(x)=2e x+(2x﹣4)e x+2a(x+2)=(2x﹣2)e x+2a(x+2),依题意:当x>0时,函数f'(x)≥0恒成立,即恒成立,记,则=,所以g(x)在(0,+∞)上单调递减,所以,所以;﹣﹣﹣(6分)(Ⅱ)因为[f'(x)]'=2xe x+2a>0,所以y=f'(x)是(0,+∞)上的增函数,又f'(0)=4a﹣2<0,f'(1)=6a>0,所以存在t∈(0,1)使得f'(t)=0且当a→0时t→1,当时t→0,所以t的取值范围是(0,1).﹣﹣﹣﹣﹣﹣﹣(8分)又当x∈(0,t),f'(x)<0,当x∈(t,+∞)时,f'(x)>0,所以当x=t时,.且有由(Ⅰ)知,在(0,+∞)上单调递减,又,g(1)=0,且,故t∈(0,1),∴,t∈(0,1)﹣﹣﹣﹣﹣﹣﹣(10分)记h(t)=e t(﹣t2+t﹣2),则h'(t)=e t(﹣t2+t﹣2)+e t(﹣2t+1)=e t(﹣t2﹣t ﹣1)<0,所以h(1)<h(t)<h(0),即最小值的取值范围是(﹣2e,﹣2).﹣﹣﹣﹣﹣﹣﹣(12分)【点评】本题考查了函数的单调性、最值问题,考查导数的应用以及转化思想,是一道综合题.请考生在第(22)、(23)两题中任选一题作答,如果多做,则按所做的第一题记分.[选修4-4:坐标系与参数方程]22.(10分)在平面直角坐标系xoy中,曲线C1过点P(a,1),其参数方程为(t为参数,a∈R).以O为极点,x轴非负半轴为极轴,建立极坐标系,曲线C2的极坐标方程为ρcos2θ+4cosθ﹣ρ=0.(Ⅰ)求曲线C1的普通方程和曲线C2的直角坐标方程;(Ⅱ)已知曲线C1与曲线C2交于A、B两点,且|PA|=2|PB|,求实数a的值.【分析】(Ⅰ)利用三种方程的转化方法,求曲线C1的普通方程和曲线C2的直角坐标方程;(Ⅱ)根据参数方程的几何意义可知|PA|=2|t1|,|PB|=2|t2|,利用|PA|=2|PB|,分类讨论,求实数a的值.【解答】解:(Ⅰ)曲线C1参数方程为,∴其普通方程x﹣y﹣a+1=0,﹣﹣﹣﹣﹣﹣﹣(2分)由曲线C2的极坐标方程为ρcos2θ+4cosθ﹣ρ=0,∴ρ2cos2θ+4ρcosθ﹣ρ2=0∴x2+4x﹣x2﹣y2=0,即曲线C2的直角坐标方程y2=4x.﹣﹣﹣﹣﹣﹣﹣(5分)(Ⅱ)设A、B两点所对应参数分别为t1,t2,联解得要有两个不同的交点,则,即a>0,由韦达定理有根据参数方程的几何意义可知|PA|=2|t1|,|PB|=2|t2|,又由|PA|=2|PB|可得2|t1|=2×2|t2|,即t1=2t2或t1=﹣2t2﹣﹣﹣﹣﹣﹣﹣(7分)∴当t1=2t2时,有t1+t2=3t2=,t1t2=2t22=,∴a=>0,符合题意.﹣﹣﹣﹣﹣﹣﹣(8分)当t1=﹣2t2时,有t1+t2=﹣t2=,t1t2=﹣2t22=,∴a=>0,符合题意.﹣﹣﹣﹣﹣﹣﹣(9分)综上所述,实数a的值为或.﹣﹣﹣﹣﹣﹣﹣(10分)【点评】本题考查三种方程的转化,考查参数方程的运用,考查参数的几何意义,考查分类讨论的数学思想,属于中档题.[选修4-5:不等式选讲]23.已知函数f(x)=|2x﹣a|+|x﹣1|,a∈R.(Ⅰ)若不等式f(x)≤2﹣|x﹣1|有解,求实数a的取值范围;(Ⅱ)当a<2时,函数f(x)的最小值为3,求实数a的值.【分析】(Ⅰ)由绝对值的几何意义知,由不等式f(x)≤2﹣|x﹣1|有解,可得,即可求实数a的取值范围;(Ⅱ)当a<2时,(x)在单调递减,在单调递增,利用函数f(x)的最小值为3,求实数a的值.【解答】解:(Ⅰ)由题f(x)≤2﹣|x﹣1|,即为.而由绝对值的几何意义知,﹣﹣﹣﹣﹣﹣﹣(2分)由不等式f(x)≤2﹣|x﹣1|有解,∴,即0≤a≤4.∴实数a的取值范围[0,4].﹣﹣﹣﹣﹣﹣﹣(5分)(Ⅱ)函数f(x)=|2x﹣a|+|x﹣1|的零点为和1,当a<2时知,∴﹣﹣﹣﹣﹣﹣﹣(7分)如图可知f(x)在单调递减,在单调递增,∴,得a=﹣4<2(合题意),即a=﹣4.﹣﹣﹣﹣﹣﹣﹣(10分)【点评】本题考查绝对值的几何意义,考查函数的单调性与最小值,考查数形结合的数学思想,属于中档题.。

2017届江西省南昌市高三第一次模拟考试数学(理)试卷(带解析)

2017届江西省南昌市高三第一次模拟考试数学(理)试卷(带解析)

绝密★启用前2017届江西省南昌市高三第一次模拟考试数学(理)试卷(带解析)注意事项:1.答题前填写好自己的姓名、班级、考号等信息2.请将答案正确填写在答题卡上第I卷(选择题)请点击修改第I卷的文字说明一、选择题1.已知全集U=R,集合A={x|y=lgx},集合B={y|y=√x+1},那么A∩(C U B)=()A. ϕB. (0,1]C. (0,1)D. (1,+∞)2.若复数z=21+i3,其中i为虚数单位,则复数z的虚部是()A. -1B. −iC. 1D. i3.已知α,β均为第一象限的角,那么α>β是sinα>sinβ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4.设某中学的高中女生体重y(单位:kg)与身高x(单位:cm)具有线性相关关系,根据一组样本数据(x i,y i)(i=1,2,3,…,n),用最小二乘法近似得到回归直线方程为y^=0.85x−85.71,则下列结论中不正确的是()A. y与x具有正线性相关关系B. 回归直线过样本的中心点(x̅,y̅)C. 若该中学某高中女生身高增加1cm,则其体重约增加0.85kgD. 若该中学某高中女生身高为160cm,则可断定其体重必为50.29kg.5.若圆锥曲线C:x2+my2=1的离心率为2,则m=()A. −√33B. √33C. −13D. 136.执行如图所示的程序框图,输出S的值为()A. log210−1B. 2log23−1C. 92D. 67.已知函数f(x)=Asin(ωx+φ)(A>0,ω>0,0<φ<π2)的周期为π,若f(α)=1,则f(α+3π2)=()A. -2B. -1C. 1D. 28.如图,在平面直角坐标系xOy中,直线y=2x+1与圆x2+y2=4相交于A,B两点,则cos∠AOB=()A. √510B. −√510C. 910D. −9109.我国古代数学名著《九章算术》中有如下问题:今有甲乙丙三人持钱,甲语乙丙:各将公等所持钱,半以益我,钱成九十(意思是把你们两个手上的钱各分我一半,我手上就有90钱);乙复语甲丙,各将公等所持钱,半以益我,钱成七十;丙复语甲乙:各将公等所持钱,半以益我,钱成五十六,则乙手上有()钱.A. 28B. 32C. 56D. 7010.某空间几何体的三视图如图所示(图中小正方形的边长为1),则这个几何体的体积是()A. 323B. 643C. 16D. 3211.抛物线y2=8x的焦点为F,设A(x1,y1),B(x2,y2)是抛物线上的两个动点,x1+x2+ 4=2√33|AB|,则∠AFB的最大值为()A. π3B. 3π4C. 5π6D. 2π312.定义在R上的偶函数f(x)满足f(2−x)=f(x),且当x∈[1,2]时,f(x)=lnx−x+1,若函数g(x)=f(x)+mx有7个零点,则实数m的取值范围为()A. (1−ln28,1−ln26)∪(ln2−16,ln2−18) B. (ln2−16,ln2−18)C. (1−ln28,1−ln26) D. (1−ln28,ln2−16)第II卷(非选择题)请点击修改第II卷的文字说明二、填空题13.在多项式(1+2x)6(1+y)5的展开式中,xy3项的系数为__________.,a=2e1⃗⃗⃗ −e2⃗⃗⃗ ,则a在e1⃗⃗⃗ 上的投影是__________.14.已知单位向量e1⃗⃗⃗ ,e2⃗⃗⃗ 的夹角为π315.如图,直角梯形ABCD中,AD⊥DC,AD//BC,BC=2CD=2AD=2,若将直角梯形绕BC边旋转一周,则所得几何体的表面积为__________.16.已知x2+y2=4,在这两个实数x,y之间插入三个实数,使这五个数构成等差数列,那么这个等差数列后三项和的最大值为__________.三、解答题17.已知等差数列{a n的前n项和为S n,且a1=1,S3+S4=S5.(1)求数列{a n}的通项公式;(2)令b n=(−1)n−1a n a n+1,求数列{b n}的前2n项和T2n.18.某中学的环保社团参照国家环境标准制定了该校所在区域空气质量指数与空气质量等级对应关系如下表(假设该区域空气质量指数不会超过300)该社团将该校区在2016年100天的空气质量指数监测数据作为样本,绘制的频率分布直方图如下图,把该直方图所得频率估计为概率.(1)请估算2017年(以365天计算)全年空气质量优良的天数(未满一天按一天计算);(2)该校2017年6月7、8、9日将作为高考考场,若这三天中某天出现5级重度污染,需要净化空气费用10000元,出现6级严重污染,需要净化空气费用20000元,记这三天净化空气总费用X元,求X的分布列及数学期望.19.如图,四棱锥P−ABCD中,平面PAD⊥平面ABCD,底面ABCD为等腰梯形,AB//CD,AD=DC=BC=2,AB=4,ΔPAD为正三角形.(1)求证:BD⊥平面PAD;(2)设AD的中点为E,求平面PEB与平面PDC所成二面角的平面角的余弦值.20.已知椭圆C:x2a +y2b=1(a>b>0)的左、右顶点分别为A1,A2,左、右焦点分别为F1,F2,离心率为12,点B(4,0),F2为线段A1B的中点.(1)求椭圆C的方程;(2)若过点B且斜率不为0的直线l与椭圆C的交于M,N两点,已知直线A1M与A2M相交于点G,试判断点G是否在定直线上?若是,请求出定直线的方程;若不是,请说明理由.21.已知函数f(x)=(2x−4)e x+a(x+2)2(x>0,a∈R,e是自然对数的底数). (1)若f(x)是(0,+∞)上的单调递增函数,求实数a的取值范围;(2)当a ∈(0,12)时,证明:函数f(x)有最小值,并求函数f(x)最小值的取值范围.22.选修4-4:坐标系与参数方程在平面直角坐标系xOy 中,曲线C 1过点P(a,1),其参数方程为{x =a +√2ty =1+√2t(t 为参数,a ∈R ),以O 为极点,x 轴非负半轴为极轴,建立极坐标系,曲线C 2的极坐标方程为ρcos 2θ+4cosθ−ρ=0.(1)求曲线C 1的普通方程和曲线C 2的直角坐标方程;(2)已知曲线C 1与曲线C 2交于A,B 两点,且|PA|=2|PB|,求实数a 的值. 23.选修4-5:不等式选讲已知函数f(x)=|2x −a|+|x −1|,a ∈R(1)若不等式f(x)≤2−|x −1|有解,求实数a 的取值范围; (2)当a <2时,函数f(x)的最小值为3,求实数a 的值.参考答案1.C 【解析】A ={x|y =lgx}=(0,+∞),B ={y|y =√x +1}=[1,+∞),所以C U B =(−∞,1),A ∩(C U B)=(0,1),选C. 2.C 【解析】z =21+i3=21−i=1+i ,故复数z 的虚部是1,选C3.D【解析】试题分析:因为α,β角的终边均在第一象限,所以当α=π3+2π,β=π3时,满足α>β,但sinα=sinβ,则sinα>sinβ不成立;若当α=π6,β=π3时,满足sinα>sinβ,当α>β不成立,所以“α>β”是“sinα>sinβ”的既不充分也不必要条件,故选D . 考点:充要条件的判定. 4.D【解析】由回归直线方程定义知:因为斜率大于零,所以y 与x 具有正线性相关关系;回归直线过样本的中心点(x̅,y ̅);身高增加每增加1cm ,则其体重约增加k =0.85kg ;身高为160cm ,则可估计其体重约为0.85×160−85.71=50.29kg ,但不可断定.选D. 5.C【解析】x 2+my 2=1⇒x 2−y 21−m=1,所以√1+(1−m)1=2⇒m =−13,选C.6.B【解析】第一次循环,S =3+log 2√2,i =2;第二次循环,S =3+log 2√2+log 2√32,i =3;以此类推得第七次循环,S =3+log 2√2+log 2√32+⋯+log 2√87=3+log 2√8=92,i =8;结束循环输出log 292=2log 23−1,选B.点睛:算法与流程图的考查,侧重于对流程图循环结构的考查.先明晰算法及流程图的相关概念,包括选择结构、循环结构、伪代码,其次要重视循环起点条件、循环次数、循环终止条件,更要通过循环规律,明确流程图研究的数学问题,是求和还是求项. 7.B【解析】由题意得π=2πω⇒ω=2,Asin(2α+φ)=1,所以f(α+3π2)=Asin(2α+3π+φ)=−Asin(2α+φ)=-1,选B. 8.D【解析】圆心O 到直线y =2x +1距离√5所以cos∠AOB 2=1√52=2√5cos∠AOB =2×(2√5)2−1=−910.选D.9.B【解析】设甲乙丙各有x,y,z 钱,则有x +y2+z2=90,x2+y +z2=70,x2+y2+z =56,解得x =72,y=32,z=4,选B.10.A【解析】几何体为一个三棱锥,高为4,底面为一个等腰直角三角形,直角边长为4,所以体积是13×4×12×42=323,选A.11.D【解析】由抛物线定义得AF=x1+2,BF=x2+2,所以由x1+x2+4=2√33|AB|得AF+BF=2√33|AB|,因此cos∠AFB=|AF|2+|BF|2−|AB|22|AF|⋅|BF|=14|AF|2+14|BF|2−32|AF|⋅|BF|2|AF|⋅|BF|≥1 4×2|AF|⋅|BF|−32|AF|⋅|BF|2|AF|⋅|BF|=−12所以0<∠AFB≤2π3,选D.点睛:1.凡涉及抛物线上的点到焦点距离时,一般运用定义转化为到准线距离处理. 2.若P(x0,y0)为抛物线y2=2px(p>0)上一点,由定义易得|PF|=x0+p2;若过焦点的弦AB AB 的端点坐标为A(x1,y1),B(x2,y2),则弦长为|AB|=x1+x2+p,x1+x2可由根与系数的关系整体求出;若遇到其他标准方程,则焦半径或焦点弦长公式可由数形结合的方法类似地得到.12.A【解析】f(2−x)=f(x)=f(−x)⇒T=2,当x∈[1,2]时,f′(x)=1x−1≤0,作出y=f(x)图形,由图可知直线y=−mx过点A(−6,ln2−1)时有六个交点,过点B(−8,ln2−1)时有八个交点,过点C(6,ln2−1)时有六个交点,过点D(8,ln2−1)时有八个交点,因此要使函数g(x)=f(x)+mx有7个零点,需m∈(1−ln28,1−ln26)∪(ln2−16,ln2−18),选A.点睛:利用函数零点的情况求参数值或取值范围的方法(1)利用零点存在的判定定理构建不等式求解.(2)分离参数后转化为函数的值域(最值)问题求解.(3)转化为两熟悉的函数图象的上、下关系问题,从而构建不等式求解. 13.120;【解析】由题意得xy3项的系数为C61×2×C53=120.14.32;【解析】a在e1⃗⃗⃗ 上的投影是a⃗ ⋅e1⃗⃗⃗⃗|e1⃗⃗⃗⃗ |=((2e1⃗⃗⃗⃗ −e2⃗⃗⃗⃗ )⋅e1⃗⃗⃗⃗|e1⃗⃗⃗⃗ |=2−1×1×cosπ3=32.15.(3+√2)π;【解析】几何体为一个圆锥与圆柱的组合体,表面积为πrl+2πrℎ+πr2=π×1×√2+ 2π×1×1+π×1=√2π+3π点睛:空间几何体表面积的求法(1)以三视图为载体的几何体的表面积问题,关键是分析三视图确定几何体中各元素之间的位置关系及数量.(2)多面体的表面积是各个面的面积之和;组合体的表面积注意衔接部分的处理.(3)旋转体的表面积问题注意其侧面展开图的应用.16.3√102【解析】设这五个数为x,x+d,x+2d,x+3d,x+4d=y,则x2+(x+4d)2=4,令x=2cosθ,x+4d=2sinθ,则d=sinθ−cosθ2,因此后三项和等于3(x+3d)=32(cosθ+3sinθ)≤32√10.17.(Ⅰ)a n=2n−1(Ⅱ)−8n2−4n【解析】试题分析: (Ⅰ)求等差数列通项公式,一般方法为待定系数法,即根据条件列出关于首项与公差的方程组,解出首项与公差再代入通项公式即可,(Ⅱ)涉及符号数列求和,一般方法为分组求和,即按奇偶,项的正负重新组合,利用平方差公式转化为求特殊数列(如等差数列)的和.试题解析: (Ⅰ)设等差数列{a n}的公差为d,由S3+S4=S5可得a1+a2+a3=a5,即3a2=a5,所以3(1+d)=1+4d,解得d=2.∴a n=1+(n−1)×2=2n−1.(Ⅱ)由(Ⅰ)可得:b n=(−1)n−1⋅(2n−1)(2n+1)=(−1)n−1⋅(4n2−1).∴T2n=(4×12−1)−(4×22−1)+(4×32−1)−(4×42−1)+⋯+(−1)2n−1⋅[4×(2n)2−1]=4[12−22+32−42+⋯+(2n−1)2−(2n)2]=−4(1+2+3+4+⋯+2n−1+2n)=−4×2n(2n+1)2=−8n2−4n.点睛:本题采用分组转化法求和,即通过两个一组进行重新组合,将原数列转化为一个等差数列. 分组转化法求和的常见类型有分段型(如a n={n,n为奇数2n,n为偶数)及本题的符号型(如a n=(−1)n n2)18.(Ⅰ)110(Ⅱ)EX=9000【解析】试题分析: (Ⅰ)根据频率分布直方图知小长方形面积为对应区间概率,先计算空气质量优良区间对应的概率,再根据频数等于总数乘以概率得空气质量优良的天数,(Ⅱ)先确定随机变量取法,再分别求对应概率,列表得分布列,最后根据期望公式求数学期望.试题解析: (Ⅰ)由直方图可估算2017年(以365天计算)全年空气质量优良的天数为(0.1+0.2)×365=0.3×365=109.5≈110(天). (Ⅱ)由题可知,X 的所有可能取值为:0,10000,20000,30000,40000,50000,60000, 则:P(X =0)=(45)3=64125,P(X =10000)=C 31×110×(45)2=24125P(X =20000)=C 32×(110)2×(45)+C 31×(110)×(45)2=108500=27125 P(X =30000)=(110)3+C 31×110×C 21×110×45=491000P(X =40000)=C 32×(110)2×110+C 32×(110)2×45=271000P(X =50000)=C 32×(110)2×110=31000P(X =60000)=(110)3=11000.EX =0×64125+10000×48250+20000×27125+30000×491000+40000×271000+50000×31000+60000×11000 =9000(元). 19.(Ⅰ)详见解析(Ⅱ)7√6565【解析】试题分析: (Ⅰ)已知面面垂直,所以利用面面垂直的性质定理证明线面垂直.而要利用面面垂直性质定理,需在其中一平面内寻找或论证与两平面交线的垂线,本题通过计算,利用勾股定理得到线线垂直.(Ⅱ)求二面角的方法,一般为利用空间向量数量积进行求解:先根据条件建立恰当的空间直角坐标系,设立各点坐标,根据方程组解各面法向量,再根据向量数量积求两法向量夹角余弦值,最后根据二面角与法向量夹角之间关系确定二面角余弦值.试题解析:(Ⅰ)在等腰梯形ABCD 中,过点D 作DE ⊥AB 于点E , 如图所示:有AE =1,DE =√3,BD =2√3∴在ΔABD 中,有AB 2=AD 2+BD 2,即AD ⊥BD又因为平面PAD ⊥平面ABCD 且交线为AD ,∴BD ⊥平面PAD .(Ⅱ) 由平面PAD ⊥平面ABCD ,且ΔPAD 为正三角形,E 为AD 的中点, ∴PE ⊥AD ,得PE ⊥平面ABCD .如图所示,以D 为坐标原点,DA 所在直线为x 轴,DB 所在直线为y 轴,过点D 平行于PE 所在直线为z 轴,建立空间直角坐标系.由条件AD =DC =BC =2,则AE =DE =1,PE =√3,BD =2√3. 则D(0,0,0),E(1,0,0),B(0,2√3,0),P(1,0,√3).------- 6分在等腰梯形ABCD 中,过点C 作BD 的平行线交AD 延长线于点F 如图所示:则在RtΔCDF 中,有CF =√3,DF =1,∴C(−1,√3,0).(另解:可不做辅助线,利用AB⃗⃗⃗⃗⃗ =2DC ⃗⃗⃗⃗⃗ 求点C 坐标) ∴CD ⃗⃗⃗⃗⃗ =(1,−√3,0),PD ⃗⃗⃗⃗⃗ =(−1,0,−√3),设平面PDC 的法向量n 1⃗⃗⃗⃗ =(x 1,y 1,z 1)则{n 1⃗⃗⃗⃗ ⋅CD⃗⃗⃗⃗⃗ =x 1−√3y 1=0 n 1⃗⃗⃗⃗ ⋅PD ⃗⃗⃗⃗⃗ =−x 1−√3z 1=0 ,取x 1=√3,则y 1=1,z 1=−1,∴面PDC 的法向量n 1⃗⃗⃗⃗ =(√3,1,−1).同理有PE ⃗⃗⃗⃗⃗ =(0,0,−√3),PB ⃗⃗⃗⃗⃗ =(−1,2√3,−√3),设平面PBE 的法向量n 2⃗⃗⃗⃗ =(x 2,y 2,z 2) 则{n 2⃗⃗⃗⃗ ⋅PE ⃗⃗⃗⃗⃗ =−√3z 2=0 n 2⃗⃗⃗⃗ ⋅PB ⃗⃗⃗⃗⃗ =−x 2+2√3y 2−√3z 2=0 , 取y 2=1,则x 2=2√3,z 2=0,∴面PBE 的法向量n 2⃗⃗⃗⃗ =(2√3,1,0).--10分设平面PEB 与平面PDC 所成二面角的平面角为θ, ∴cosθ=|cos <n 1⃗⃗⃗⃗ ,n 2⃗⃗⃗⃗ >|=√3×2√3+13+1+1×12+1=7√6565.即平面PEB 与平面PDC 所成二面角的余弦值为7√6565. 点睛:利用法向量求解空间线面角的关键在于“四破”:第一,破“建系关”,构建恰当的空间直角坐标系;第二,破“求坐标关”,准确求解相关点的坐标;第三,破“求法向量关”,求出平面的法向量;第四,破“应用公式关”. 20.(Ⅰ)x 24+y 23=1(Ⅱ)详见解析【解析】试题分析: (Ⅰ)求椭圆标准方程,一般方法为待定系数法,即根据条件建立关于a,b,c 的两个独立条件,再与a 2=b 2+c 2联立方程组,解出a,b,c 的值,(Ⅱ)先根据特殊直线或椭圆几何性质确定定直线x =1,再根据条件证明点G 横坐标为1.由题意设M,N 两点坐标,用M,N 两点坐标表示点G 横坐标.根据直线l 方程与椭圆方程联立方程组,利用韦达定理得M,N 两点坐标关系(用直线l 斜率表示),并代入点G 横坐标表达式,化简可得为定值. 试题解析: (Ⅰ)设点A 1(−a,0),F 2(c,0),由题意可知:c =−a+42,即a =4−2c ①又因为椭圆的离心率e =ca =12,即a =2c ② 联立方程①②可得:a =2,c =1,则b 2=a 2−c 2=3 所以椭圆C 的方程为x 24+y 23=1.(Ⅱ)方法一:根据椭圆的对称性猜测点G 是与y 轴平行的直线x =x 0上. 假设当点M 为椭圆的上顶点时,直线l 的方程为√3x +4y −4√3=0,此时点N(85,3√35), 则联立直线l A 1M :√3x −2y +2√3=0和直线l A 2N :3√3x +2y −6√3=0可得点G(1,3√32) 据此猜想点G 在直线x =1上,下面对猜想给予证明:设M(x 1,y 1),N(x 2,y 2),联立方程{y =k(x −4)x 24+y 23=1可得:(3+4k 2)x 2−32k 2x +64k 2−12=0,Δ>0由韦达定理可得x 1+x 2=32k 23+4k 2,x 1x 2=64k 2−123+4k 2 (*)因为直线l A 1M :y =y 1x 1+2(x +2),l A 2N :y =y 2x2−2(x −2),联立两直线方程得y 1x1+2(x +2)=y 2x 2−2(x −2)(其中x 为G 点的横坐标)即证:3y 1x 1+2=−y 2x 2−2,即3k(x 1−4)⋅(x 2−2)=−k(x 2−4)⋅(x 1+2),即证4x 1x 2−10(x 1+x 2)+16=0 将(*)代入上式可得4⋅(64k 2−12)3+4k 2−10×32k 23+4k 2+16=0⇔16k 2−3−20k 2+3+4k 2=0此式明显成立,原命题得证.所以点G 在定直线上x =1上. 方法二:设M(x 1,y 1),N(x 2,y 2),G(x 3,y 3),x 1,x 2,x 3两两不等, 因为B,M,N 三点共线,所以y 1x 1−4=y 2x2−4⇒y 12(x1−4)2=y 22(x2−4)2⇒3(1−x 124)(x1−4)2=3(1−x 224)(x2−4)2,整理得:2x 1x 2−5(x 1+x 2)+8=0又A 1,M,G 三点共线,有:y 3x 3+2=y 1x 1+2 ①又A 2,N,G 三点共线,有:y 3x3−2=y 2x 2−2② 将①与②两式相除得:x 3+2x 3−2=y 2(x 1+2)y 1(x 2−2)⇒(x 3+2x 3−2)2=y 22(x 1+2)2y 12(x 2−2)2=3(1−x 224)(x 1+2)23(1−x 124)(x 2−2)2=(x 2+2)(x 1+2)(x 1−2)(x 2−2) 即(x 3+2x 3−2)2=(x 2+2)(x 1+2)(x 1−2)(x 2−2)=x 1x 2+2(x 1+x 2)+4x 1x 2−2(x 1+x 2)+4,将2x 1x 2−5(x 1+x 2)+8=0即x 1x 2=52(x 1+x 2)−4=0代入得:(x 3+2x 3−2)2=9解得x 3=4(舍去)或x 3=1,所以点G 在定直线x =1上.方法三:显然l 与x 轴不垂直,设的方程为y =k(x −4),M(x 1,y 1),N(x 2,y 2). 由{y =k(x −4)x 24+y 23=1得(3+4k 2)x 2−32k 2x +64k 2−12=0,Δ>0.设M(x 1,y 1),N(x 2,y 2),G(x 3,y 3),x 1,x 2,x 3两两不等, 则x 1+x 2=32k 23+4k2,x 1x 2=64k 2−123+4k 2,|x 1−x 2|=√(x 1+x 2)2−4x 1x 2=12√1−4k 23+4k 2,由A 1,M,G 三点共线,有:y 3x 3+2=y 1x 1+2 ①由A 2,N,G 三点共线,有:y 3x3−2=y 2x 2−2②①与②两式相除得:x 3+2x 3−2=y 2(x 1+2)y 1(x 2−2)=k(x 2−4)(x 1+2)k(x 1−4)(x 2−2)=x 1x 2−(x 1+x 2)+3(x 1−x 2)−8x 1x 2−3(x 1+x 2)+(x 1−x 2)+8=−13 解得x 3=4(舍去)或x 3=1,所以点G 在定直线x =1上.点睛:定点、定值问题通常是通过设参数或取特殊值来确定“定点”是什么、“定值”是多少,或者将该问题涉及的几何式转化为代数式或三角问题,证明该式是恒定的. 定点、定值问题同证明问题类似,在求定点、定值之前已知该值的结果,因此求解时应设参数,运用推理,到最后必定参数统消,定点、定值显现. 21.(Ⅰ)a ≥12(Ⅱ)(−2e,−2)【解析】试题分析: (Ⅰ)先将单调性转化为不等式恒成立:当x >0时,函数f′(x)≥0恒成立,再变量分离转化为对应函数最值:−2a ≤(2x−2)e xx+2的最小值,最后根据导数求函数g(x)=(2x−2)e xx+2最值,(Ⅱ)利用二次求导,确定导函数为单调递增函数,再利用零点存在定理确定导函数有且仅有一个零点,根据导函数符号变化规律得函数在此零点(极小值点)取最小值.最后利用导函数零点表示函数最小值,并根据导函数零点取值范围,利用导数方法确定最小值函数的值域.试题解析: (Ⅰ)f′(x)=2e x +(2x −4)e x +2a(x +2)=(2x −2)e x +2a(x +2), 依题意:当x >0时,函数f′(x)≥0恒成立,即(2x−2)e xx+2≥−2a 恒成立,记g(x)=(2x−2)e xx+2,则g′(x)=2xe x (x+2)−(2x−2)e x(x+2)=(2x 2+2x+2)e x(x+2)>0,所以g(x)在(0,+∞)上单调递增,所以g(x)>g(0)=−1,所以−2a ≤−1,即a ≥12; (Ⅱ)因为[f′(x)]′=2xe x +2a >0,所以y =f′(x)是(0,+∞)上的增函数, 又f′(0)=4a −2<0,f′(1)=6a >0 ,所以存在t ∈(0,1)使得f′(t)=0 且当a →0时t →1,当a →12时t →0,所以t 的取值范围是(0,1). 又当x ∈(0,t),f′(x)<0,当x ∈(t,+∞)时,f′(x)>0,所以当x =t 时,f(x)min =f(t)=(2t −4)e t+a(t +2)2.且有f′(t)=0⇒a =−(t−1)e t t+2∴f(x)min =f(t)=(2t −4)e t −(t −1)(t +2)e t =e t (−t 2+t −2).记ℎ(t)=e t (−t 2+t −2),则ℎ′(t)=e t (−t 2+t −2)+e t (−2t +1)=e t (−t 2−t −1)<0,所以ℎ(1)<ℎ(t)<ℎ(0),即最小值的取值范围是(−2e,−2). 22.(Ⅰ)y 2=4x (Ⅱ)a =136或94.【解析】试题分析: (Ⅰ)根据加减相消法将曲线C 1参数方程化为普通方程,利用ρcosθ=x,ρsinθ=y,ρ2=x 2+y 2将曲线C 2化为直角坐标方程;(Ⅱ)先将直线参数方程转化为{x =a +√2t2y =1+√2t2(t 为参数,a ∈R ),再根据直线参数方程几何意义由|PA|=2|PB|得|t 1|=2|t 2|,最后将直线参数方程代入C 2化为直角坐标方程,利用韦达定理得关于a 的方程,解得a 的值. 试题解析: (Ⅰ)曲线C 1参数方程为{x =a +√2t y =1+√2 ,∴其普通方程x −y −a +1=0,由曲线C 2的极坐标方程为ρcos 2θ+4cosθ−ρ=0,∴ρ2cos 2θ+4ρcosθ−ρ2=0 ∴x 2+4x −x 2−y 2=0,即曲线C 2的直角坐标方程y 2=4x .(Ⅱ)设A 、B 两点所对应参数分别为t 1,t 2,联解{y 2=4xx =a +√2t y =1+√2t 得2t 2−2√2t +1−4a =0要有两个不同的交点,则Δ=(2√2)2−4×2(1−4a)>0,即a >0,由韦达定理有{t 1+t 2=√2 t 1⋅t 2=1−4a 2根据参数方程的几何意义可知|PA|=2|t 1|,|PB|=2|t 2|,又由|PA|=2|PB|可得2|t 1|=2×2|t 2|,即t 1=2t 2或t 1=−2t 2 ∴当t 1=2t 2时,有{t 1+t 2=3t 2=√2 t 1⋅t 2=2t 22=1−4a 2⇒a =136>0,符合题意. 当t 1=−2t 2时,有{t 1+t 2=−t 2=√2 t 1⋅t 2=−2t 22=1−4a2⇒a =94>0,符合题意. 综上所述,实数a 的值为a =136或94.23.(Ⅰ)[0,4](Ⅱ)a =−4.【解析】试题分析: (Ⅰ)先化简不等式f(x)≤2−|x −1|得|x −a2|+|x −1|≤1,再根据绝对值三角不等式得|x −a 2|+|x −1|≥|a 2−1|,最后根据不等式有解得|a2−1|≤1,解不等式得实数a 的取值范围;(Ⅱ)根据绝对值定义将函数f(x)分成三段,结合函数图像可得f(x)min =f(a2),最后根据方程f(a2)=3求实数a 的值.试题解析: (Ⅰ)由题f(x)≤2−|x −1|,即为|x −a2|+|x −1|≤1. 而由绝对值的几何意义知|x −a2|+|x −1|≥|a2−1|,------- 2分由不等式f(x)≤2−|x−1|有解,∴|a2−1|≤1,即0≤a≤4.∴实数a的取值范围[0,4].------- 5分(Ⅱ)函数f(x)=|2x−a|+|x−1|的零点为a2和1,当a<2时知a2<1∴f(x)={−3x+a+1(x<a2)x−a+1(a2≤x≤1)3x−a−1 (x>1)------- 7分如图可知f(x)在(−∞,a2)单调递减,在[a2,+∞)单调递增,∴f(x)min=f(a2)=−a2+1=3,得a=−4<2(合题意),即a=−4.。

江西省南昌市2017届高三一模考试理综化学试题

江西省南昌市2017届高三一模考试理综化学试题

NCS20170607项目第一次模拟测试卷理科综合参考答案及评分标准一—二、选择题化学26.(14分)(1)吸水(1分)(2)Si+2HCl 250℃~260℃SiH2Cl2(2分)(3)defgbcd(3分)(4)P2O5或CaCl2(2分)尾气处理和防止空气中的水进入B中(2分)(5)SiCl4、SiHCl3等(2分)(6)SiH 2Cl2+2Cl2SiCl4+2HCl (或SiH2Cl2+Cl2SiHCl3+HCl)(2分)27.(14分)(1)70℃(1分)温度进一步升高同等幅度,转化率几乎不变,但能耗增加(2分)(2)Mg(OH)2+2CO2 === Mg(HCO3)2(2分)(3)煅烧反应(1分)热解反应(1分)(4)① 0.01(2分)②C(2分)③方案2(1分)加入碳酸钠的产品中易混入更多杂质,不易提纯(2分)28.(15分)(1)3NO(g)+O3(g)===3NO2(g) ΔH=-317.1 kJ·mol-1(2分)(2)否(1分)该反应是焓增、熵减的反应,根据G=ΔH -T·ΔS,G>0 (2分)(3)①HCOO-+H2O HCOOH + OH-(2分)②酸(1分)c (Na+)>c(HCOO-)>c (Cl-)>c(H+)>c(OH-)(2分)(4)①放热(1分)②不变(2分)③0.25(2分)35.(15分)(1)>(1分);硅烷为分子晶体,随相对分子质量增大,分子间作用力增强,熔沸点升高(2分)(2)①CuO中铜的价层电子排布为3d94s0,Cu2O中铜的价层电子排布为3d10,后者处于稳定的全充满状态而前者不是(2分)②Cu+H2O2+4NH3·H2O===Cu(NH3) 42++2OH-+4H2O(2分)③9.0 g·cm-3(2分)(3)1s22s22p63s23p63d104s24p1(或3d104s24p1) (2分)sp3(1分)配位(1分)H3BO3+H2O[B(OH)4]-+H+(2分)36.(15分)(1)(1分)碳碳双键、醛基(2分)(2)KMnO4(H+)在氧化醛基的同时,还可以氧化碳碳双键(2分)(3)(2分)取代反应(1分)(4)6(2分)(2分)(5)(3分)。

2017年江西一本录取分数线:文科533分理科503分

2017年江西一本录取分数线:文科533分理科503分

6⽉23⽇,省教育考试院召开新闻发布会,公布我省2017年普通⾼校招⽣各批次录取⽂化控制分数线。

省委教育⼯委委员、省教育考试院党委书记刘润保主持,省教育考试院党委副书记、院长刘雪平进⾏新闻发布。

2017年江西⼀本录取分数线为:
第⼀批本科⽂史类533分、理⼯类503分;
今年继续实⾏本专科分段填报志愿,⾼职(专科)的分数线待⼋⽉初考⽣填完志愿后再划定。

今年的录取⼯作将于7⽉8⽇⾄8⽉19⽇在江西教育发展⼤厦录取中⼼进⾏,共分六个批次:
1.7⽉8⽇-12⽇,提前批本科录取;
2.7⽉13⽇-15⽇,国家专项本科录取;
3.7⽉16⽇-23⽇,第⼀批本科录取;
4.7⽉24⽇-8⽉8⽇,第⼆批本科录取;
其中:
7⽉24⽇-8⽉1⽇,⽂理类⼆本,三校⽣⽂理类本科,艺术、体育类⼆本A段录取;
8⽉2⽇-8⽉8⽇,艺术、体育类⼆本B段录取,三校⽣艺术类本科录取;
5.8⽉9⽇-12⽇,提前批⾼职(专科)录取;
6.8⽉13⽇-19⽇,⾼职(专科)录取。

2017年南昌城区中考高分段成绩表

2017年南昌城区中考高分段成绩表
育华
胡思贤
附中5班
育华
付艺轩
附中5班
23中
邹静
附中6班
育华
左奇坤
附中6班
育华
邹和泰
附中6班
育华
周仕睿
附中6班
育华
郁林珑
附中6班
育华
杨海玥
附中6班
育华
徐可
附中6班
育华
谢栎
附中6班
育华
吴玥萱
附中6班
育华
吴文涵
附中6班
育华
王玉玲
附中6班
育华
王思诗
附中6班
育华
王麒杰
附中6班
育华
邱静雯
附中6班
育华
欧昱锟
附中6班
初中学校
中考分数
录取高中学校
心远
秦升荣
730
苏二
心远
李岱泽
720
苏二
育华
李思涵
720
滨江竞赛班
凤凰城
陈子丹
720
滨江竞赛班
育华
邱宇萌
719
滨江竞赛班
凤凰城
余楚菁
716
滨江竞赛班
心远
张歆媛
715
附中6班
育华
刘家蔚
715
附中5班
其它
欧阳骏
713
附中6班
育华
徐馨薏
713
附中6班
育华
习逸田
713
附中5班
育华
梅家乐
680
苏二
心远
程力扬
680
红二
育华
万梓航
680
附中7
育华
邓少轩
680

南昌一模考试(市成绩)

南昌一模考试(市成绩)

高三(15)班南昌一模考试成绩表 班主任:夏莹学号姓名语文数学英语物理化学生物总分班次级次学号姓名语文数学英语物理化学生物总分班次级次1546谭志诚86669855555441411211550余翊杰71259420121623841838 1554金蒙82.58262476043376.522531555胡颖慧9333553193523841838 1537余凡超78617957434836632941503夏淑娟81345118361323343851 1562邓家建82597957404536243301549熊艳伟76563123222323144853 1519李佳文76536634605434354201570朝君泽67107138143123144853 1505陈翔70527733444432065281517金瑞清52358613162823046856 1544陈朝曦86715448253431875411551李小勇80295223321122747861 1501周渊92446532384631785461513喻祥5599930171422448866 1524于兰87478925402831695511533陈涛70255926281622448866 1536万美华856577263227312105731566余勇52256922351722050874 1510周正874080403131309115871540万文强57205727164221951877 1516罗莉丹735182394117303126161525李锋833947326921652883 1559邓志武695685224722301136231553徐俊74353924122220653893 1523邓杰744257415432300146261534万佳俊77104427331220354897 1545王登祥793277423525290156681542胡雅筠6827572691620354897 1514周超733586313430289166721518涂余谭643067251520156901 1561夏圆624065563333289166721565陈青4456314433220156901 1509黄云清754588123830288186761528刘平平67304417301119958904 1511黄国强764580373019287196791527熊木生35406424181419559906 1508凌智坚775073331437284206921556黄明星4741442526919260910 1515邹仑643581531239284206921522李焕4515682293018961915 1547万根华595174234037284206921558李鑫燚73322426151618662918 1526蔡思843794132728283237021564叶婷632560881517963925 1504闵学梁584098283321278247181552杨兰冲50154920191216564933 1521熊年年854175262426277257231568谈志勇4635341713265948 1530徐伟剑814481142626272267391548陈霞719129266958 1538缪鹏882742265732272267391507江绍胜755868113621269287521512夏志艳764387162025267297601529丁康694665183136265307691539何招娣754379173318265307691532朱娜腊763961253129261327801531朱斌724659392314253338001520李德冠754757272420250348071557胡双双652786232521247358151560魏敏692585152626246368171506邓龙霞793863173313243378221502余勇692247424516241388291543叶摇摇8980262520240398311535邹万茹79415918222023940835。

【江西省南昌】2017学年高考一模(理科)数学年试题答案

【江西省南昌】2017学年高考一模(理科)数学年试题答案

14.若直线 ax by1 0( a 0, b 0)过曲线 y 1 sinπx(0 x 2)的对称中心,则 1 2 的最小值为 ab
______.
15.已知在三棱锥
P-ABC
中,VP﹣ABC

43 3
,APC

π 4
,BPC

π 3
,PA
AC
,PB

BC
,且平面
A.{x | x 0}
B. {x | x 1}
C.{x | 0 x 1}
D.{x | 0 x 1}源自3.下列选项错误的是( )
A.命题“若 x 1,则 x2 3x 2 0 ”的逆否命题是“若 x2 3x 2 0 ,则 x 1 ”
B.“ x 2 ”是“ x2 3x 2 0 ”的充分不必要条件
B.[ 1 , 3] 44
C.[ 3 , 1] 44
D.[1 , 3] 44
9.《张丘建算经》卷上第 22 题为“今有女善织,日益功疾,初日织五尺,今一月日织九匹三丈.”其意思
为:现有一善于织布的女子,从第 2 天开始,每天比前一天多织相同量的布,第 1 天织了 5 尺布,现在一
月(按 30 天计算)共织 390 尺布,记该女子一月中的第 n 天所织布的尺数为 an ,则 a14 a15 a16 a17 的值
为( )
A.55
B.52
C.39
D.26
10.已知某几何体的三视图如图所示,则该几何体的体积为( )
A. 8π 3
B. 3π
C. 10π 3
D. 6π
11.将函数 f (x) sin2x 的图像向右平移(0 π) 个单位后得到函数 g(x) 的图像.若对满足 2

江西省南昌市2017年3月高三数学研讨会资料-南昌市2017高三一模质量分析 (共37张PPT)

江西省南昌市2017年3月高三数学研讨会资料-南昌市2017高三一模质量分析 (共37张PPT)

市一批重点文科摸底与一模考试情况比较
学校 1 2 3 4 5 6 7 8 9 10 11 12 13 平均分1 92.51 117.85 98.12 56.21 73.91 66.97 99.05 102.04 84.05 98.24 90.12 73.35 62.02 6 1 5 13 9 11 3 2 8 4 7 10 12 7 1 4 13 8 9 6 3 11 2 5 10 12
平均分2 优秀人数1优秀人数2 93.98 37 35 112.90 277 320 100.29 73 83 79.18 25 26 89.84 14 21 79.51 1 2 99.57 110 41 114.02 245 311 90.94 84 50 106.78 166 255 99.26 69 73 87.09 89 133 82.81 18 29
市一批重点理科摸底与一模考试情况比较
学校 1 2 3 4 5 6 7 8 9 10 11 12 13 平均分1 92.14 110.78 95.68 67.98 81.64 71.67 102.11 109.70 90.79 97.25 89.04 79.63 69.36 6 1 5 13 9 11 3 2 7 4 8 10 12 7 2 4 13 9 12 5 1 8 3 6 10 11
高考改革:五种能力与两个意识 数学能力包括运算求解能力、数据处理能力、空间想象能力、 抽象概括能力、推理论证能力、应用意识和创新意识。 1.能力的分类和要求与以前有不同,必然要反映在命题中。 2.特别应注意新增加的“数据处理能力”和“应用意识和创新 意识”。前者与统计有关,后者与应用问题有关。 3.“推理论证能力”有别于先前四大能力之一的“逻辑思维能 力”,逻辑思维能力注重是演绎推理,“合情推理”也应引起 我们的重视,它可以有效地培养学生的创新意识,这正是我们 国家现在大力提倡的。 4.鼓励考生思维活跃, 提倡考生发散思维, 也应该给与特殊方法, 特殊技能一定的地位。

【江西省南昌市】2017年高考一模(理科)数学试卷-答案

【江西省南昌市】2017年高考一模(理科)数学试卷-答案

江西省南昌市2017年高考一模(理科)数学试卷答 案1~5.CCDDC 6~10.BBDBA 11~12.DA 13.120 14.3215.(3π16 17.解:(Ⅰ)设等差数列{}n a 的公差为d , 由345S S S +=可得1235a a a a ++=, 即253a a =,则3114d d +=+(),解得2d = 所以11221n a n n =+⨯=(-)-. (Ⅱ)由(Ⅰ)可得:112(1) (21)(21)(1) (41)n n n b n n n --=--+=--g g所以:22222122(411)(421)(431)(441)(1) [4(2)1]n n T n -=⨯--⨯-+⨯--⨯-++-⨯-L g=22222241234(21)(2)[]n n +++-L --- =4(1234212)n n -++++++L - =22(21)4842n n n n +-⨯=-- 18.解:(Ⅰ)由直方图可估算2017年(以365天计算)全年空气质量优良的天数为: (0.10.2)3650.3365109.5110+⨯=⨯=≈(天). (Ⅱ)由题可知,X 的所有可能取值为:0,10 000,20 000,30 000,40 000,50 000,60 000, 则:3464(0)()5125P X ===,1231424(10000)()105125P X C ==⨯⨯=, 221233141410827(20000)()()C ()()105105500125P X C ==⨯⨯+⨯⨯==, 31132111449(30000)()C C 10101051000P X ==+⨯⨯⨯⨯=,222233111427(40000)()C ()10101051000P X C ==⨯⨯+⨯⨯=,223113(50000)()10101000P X C ==⨯⨯=,311(60000)()101000P X ===.∴X 的分布列为X 0 10 000 20 000 30 000 40 000 50 000 60 000P64125 24125 27125 491 000 271 000 31 000 11 000644827492731()01000020000300004000050000600001252501251000100010001000E X =⨯+⨯+⨯+⨯+⨯+⨯+⨯=9000(元).19.证明:(Ⅰ)在等腰梯形ABCD 中,过点D 作DE AB ⊥于点E ,如图所示:有1,AE DE BD ==∴在ABD △中,有222AB AD BD =+,即AD BD ⊥又因为平面PAD ⊥平面ABCD 且交线为AD ,∴BD PAD ⊥平面.解:(Ⅱ)由平面PAD ⊥平面ABCD ,且PAD △为正三角形,E 为AD 的中点, ∴PE AD ⊥,得PE ⊥平面ABCD .如图所示,以D 为坐标原点,DA 所在直线为x 轴,DB 所在直线为y 轴,过点D 平行于PE 所在直线为z 轴,建立空间直角坐标系.由条件2AD DC BC ===,则1AE DE ==,PEBD =. 则(0,0,0)D ,(1,0,0)E,B,P .在等腰梯形ABCD 中,过点C 作BD 的平行线交AD 延长线于点F 如图所示: 则在t R CDF △中,有CF ,1DF =,∴(C -.(另解:可不作辅助线,利用2AB DC =u u u r u u u r求点C 坐标)∴(1,CD =u u u r,(1,0,PD =-u u u r ,设平面PDC 的法向量1111(,,z )n x y =u u r则111111 0 0n CD x n PD x ⎧==⎪⎨=-=⎪⎩u u r u u u u r g u u r u u u r g,取1x =1111y z ==,-, ∴面PDC的法向量11)n =-u u r.同理有(0,0,PE =u u u r,(PB =-u u u r ,设平面PBE 的法向量2222(,,z )n x y =u u r则222222 0 0n PE n PB x ⎧==⎪⎨=-+=⎪⎩u u r u u u r g u u r u u u r g , 取21y =,则2x =20z =,∴面PBE的法向量2n =u u r.设平面PEB 与平面PDC 所成二面角的平面角为θ,∴12cos |cos(,)|n n θ===u u r u u r即平面PEB 与平面PDC20.解:(Ⅰ)设点12(,0)(,0)-,A a F c ,由题意可知:42a c -+=,即42a c =﹣① 又因为椭圆的离心率1e 2c a ==,即2a c =② 联立方程①②可得:21a c ==,,则2223b a c ==-所以椭圆C 的方程为22143x y +=.解:(Ⅱ)解法一:根据椭圆的对称性猜测点G 是与y 轴平行的直线0x x =上.假设当点M 为椭圆的上顶点时,直线l40y +-=,此时点8(5N ,则联立直线120A M l y -+和直线220A N l y +-=可得点G据此猜想点G 在直线1x =上,下面对猜想给予证明:设11(,)M x y ,22(,)N x y ,联立方程22(4)143y k x x y =-⎧⎪⎨+=⎪⎩可得:2222(34)3264120k x k x k ++=--,0V >由韦达定理可得21223234k x x k +=+,2122641234k x x k-=+(*) 因为直线111:(2)2A M y l y x x =++,222:(2)2A N y l y x x =--, 联立两直线方程得1212(2)(2)22y y x x x x +=-+-(其中x 为G 点的横坐标)即证: 1212322y y x x -=+-, 即22113(4)(2)(4)(2 )k x x k x x -=+g g ﹣--,即证1212410160x x x x ++=-()将(*)代入上式可得22222224 (6412)1032160163203403434k k k k k k k-⨯-+=⇔--++=++g 此式明显成立,原命题得证.所以点G 在定直线上1x =上.解法二:设11,()M x y ,22(,)N x y ,33(,)G x y ,123,,x x x 两两不等, 因为B M N ,,三点共线,所以221222121222221212123(1)3(1)4444(4)(4)(4)(4)x x y y y y x x x x x x --=⇒=⇒=------, 整理得:12122580x x x x ++=-() 又1A M G ,,三点共线,有:313122y yx x =++① 又2A N G ,,三点共线,有:323222y y x x =--②, 将①与②两式相除得:2222212332121212222131********(1)(2)22(2)(2)(2)(2)4()2(2)2(2)(2)(2)3(1)(2)4x x x x y x y x x x x x y x x y x x x x -+++++++=⇒===--------即2321121231212122(2)(2)2()4()2(2)(2)2()4x x x x x x x x x x x x x x ++++++==----++,将12122580x x x x ++=-()即12125()402x x x x =+-=代入得:2332()92x x +=- 解得34x =(舍去)或31x =,所以点G 在定直线1x =上.解法三:由题意知l 与x 轴不垂直,设l 的方程为(4)y k x =-,11(,)M x y ,22(,)N x y .由22(4)143y k x x y =-⎧⎪⎨+=⎪⎩得2222(34)3264120k x k x k ++=--,0V >.设11,()M x y ,22(,)N x y ,33(,)G x y ,两两不等,则21223234k x x k +=+,2122641234k x x k-=+,12||x x -=, 由1A M G ,,三点共线,有:313122y yx x =++① 由2A N G ,,三点共线,有:323222y y x x =--② ①与②两式相除得:32121121212312121212122(2)(4)(2)()3()812(2)(4)(2)3()()83x y x k x x x x x x x x x y x k x x x x x x x x ++-+-++--====-----++++ 解得34x =(舍去)或3x =1,所以点G 在定直线1x =上.21.解:(Ⅰ)()2e (24)e 2(2)(22)e 2(2)x x x f x x a x x a x '=+-++=-++,依题意:当0x >时,函数'()0f x ≥恒成立,即(1)e 2≥x x a x -+恒成立,记(1)e ()2xx g x x -=+,则222e (2)(1)e (1)e '()0(2)(2)<x x x x x x x x g x x x +--++==-++, 所以()g x 在(0,)+∞上单调递减,所以1()(0)2<g x g =,所以12≥a ; (Ⅱ)因为()2[]e 20>x f x x a =+'',所以()y f x '=是(0,)+∞上的增函数, 又(0)420f a '=-<,(1)60>f a '=,所以存在0,1()t ∈使得0()f t '= 且当0a →时1t →,当12a →时0t →,所以t 的取值范围是(0,1). 又当(0,)x t ∈,()0<f x ',当(,)x t ∈+∞时,()0>f x ',所以当x t =时,2min()()(24)e (2)tf x f t t a t ==-++.且有(1)e (t)02tt f a t -'=⇒=-+由(Ⅰ)知(1)e ()2tt a g t t -=-=+,在(0,)+∞上单调递减,又1(0)2g =,(1)0g =,且1(0,)2a ∈,故(0,1)t ∈,∴2min ()()(24)e (1)(2)e e (2)t t t f x f t t t t t t ==---+=-+-,(0,1)t ∈ 记2()e (2)-t h t t t =+-,则22()e (2)e (21)e (1)0----<t t t h t t t t t t '=-+++=-, 所以(1)()(0)h h t h <<,即最小值的取值范围是(2e,2)--.22.解:(Ⅰ)曲线1C参数方程为1x a y ⎧=⎪⎨=+⎪⎩,∴其普通方程10x y a --+=,由曲线2C 的极坐标方程为cos24cos 0-ρθθρ+=,∴22cos24cos 0-ρθρθρ+= ∴22240,x x x y +-=-即曲线2C 的直角坐标方程24y x =.(Ⅱ)设A B 、两点所对应参数分别为12、,t t联解241y xx a y ⎧=⎪⎪=⎨⎪=+⎪⎩得22140t a -+-=要有两个不同的交点,则242(14)0a =-⨯-V >,即0a >,由韦达定理有121214 2t t a t t ⎧+=⎪⎨-=⎪⎩g根据参数方程的几何意义可知1||2||PA t =,2||2||PB t =, 又由||2||PA PB =可得12||2|22|t t =⨯,即122t t =或122t t =-∴当122t t =时,有1223t t t +==21221422a t t t -==,∴1036a =>,符合题意. 当212t t =-时,有122t t t +==-21221422a t t t -=-=,∴904a =>,符合题意.综上所述,实数a 的值为136a =或94.23.解:(Ⅰ)由题()2|1|≤-f x x -,即为|||1|12≤ax x -+-.而由绝对值的几何意义知|||1||1|22≥a ax x -+--,由不等式|(|)21≤--f x x 有解,∴|1|12≤a-,即04≤≤a .∴实数a 的取值范围[0,4].(Ⅱ)函数(||)1|2|--f x x a x =+的零点为2a 和1,当2a <时知12<a,∴31()2()1(1)231(1)<≤≤>a x a x a f x x a x x a x ⎧-++⎪⎪⎪-+⎨⎪--⎪⎪⎩如图可知()f x 在(,)2a -∞单调递减,在[,)2a +∞单调递增, ∴min()()1322a af x f ==-+=,得42a =-<(合题意),即4a =-.江西省南昌市2017年高考一模(理科)数学试卷解析1.解:由题意知,A={x|y=lgx}={x|x>0}=(0,+∞),又,则B={y|y≥1}=[1,+∞),即CUB=(﹣∞,1),所以A∩(CUB)=(0,1),故选C.2.解:,故选:C.3.解:∵角α,β的终边在第一象限,∴当α=+2π,β=,满足α>β,但sinα=sinβ,则sinα>sinβ不成立,即充分性不成立,若当α=,β=+2π,满足sinα>sinβ,但α>β不成立,即必要性不成立,故“α>β”是“sinα>sinβ”的既不必要也不充分条件,故选:D.4.解:由于线性回归方程中x的系数为0.85,因此y与x具有正的线性相关关系,A正确;由线性回归方程必过样本中心点,因此B正确;由线性回归方程中系数的意义知,x每增加1cm,其体重约增加0.85kg,C正确;当某女生的身高为160cm时,其体重估计值是50.29kg,而不是具体值,因此D错误.故选:D.5.解:因为圆锥曲线C:x2+my2=1方程可化为,所以离心率为,故选:C.6.解:模拟程序的运行,可得:由,当i=7时,进入循环,得,当i=8退出循环,输出,故选:B.7.解:因为函数f(x)=Asin(ωx+φ)的周期为T==π,∴ω=2,∴f(x)=Asin(2x+φ),又f(α)=Asin(2α+φ)=1,∴f(α+)=Asin[2(α+)+φ]=Asin(2α+3π+φ)=﹣Asin(2α+φ)=-1.故选:B.8.解:因为圆心到直线y=2x+1的距离,由垂径定理得:∴由余弦定理有,故选D.9.解:设甲、乙丙各有x钱,y钱,z钱,则,解得x=72,y=32,z=4.∴甲有72钱,乙有32钱,丙有4钱.故选:B.10.解:回归到正方体中,该几何体是一个底面为等腰直角三角形的三棱锥,即如图中的几何体A﹣BCD,其体积是正方体体积的,等于,故选A.11.解:因为,|AF|+|BF|=x1+x2+4,所以:.在△AFB中,由余弦定理得:=.又.所以,∴∠AFB的最大值为,故选D.12.解:因为函数f(2﹣x)=f(x)可得图象关于直线x=1对称,且函数为偶函数则其周期为T=2,又因为,当x∈[1,2]时有f'(x)≤0,则函数在x∈[1,2]为减函数,作出其函数图象如图所示:其中,当x<0时,要使符合题意则根据偶函数的对称性,当x>0时,要使符合题意则.综上所述,实数m的取值范围为,故选A.13.解:根据题意(1+2x)6(1+y)5=,∴xy3的系数为=120,故答案为:120.14.解:单位向量的夹角为,,则在上的投影是:||cos<,>==•=(2﹣)•=2﹣•=2﹣1×1×1×cos=.故答案为:32.15.解:由图中数据可得:,S圆柱侧=π×2×1=2π,.所以几何体的表面积为.故答案为:(3π+.16.解:设构成等差数列的五个数分别为x,a,b,c,y,则x+y=a+c=2b,∴.则等差数列后三项和为=.(另解:由等差数列的性质有x+y=a+c=2b,所以.)方法一:因为x2+y2=4,设x=2cosα,y=2sinα,所以.方法二:令z=x+3y,则x+3y﹣z=0,所以当直线x+3y﹣z=0与圆x2+y2=4相切时z将有最大值,此时,即,∴..17.(Ⅰ)设等差数列{an}的公差为d,根据题意、等差数列的性质以及通项公式列出方程,求出公差d,由等差数列的通项公式求出an;(Ⅱ)由(I)化简bn=(﹣1)n﹣1anan+1,利用并项求和法和等差数列的前n项和公式求出数列{bn}的前2n项和T2n.18.(I)利用直方图的性质即可得出.(Ⅱ)由题可知,X的所有可能取值为:0,10000,20000,30000,40000,50000,60000,利用二项分布列的概率与数学期望计算公式即可得出.19.(Ⅰ)在等腰梯形ABCD中,过点D作DE⊥AB于点E,推导出AD⊥BD,由此能证明BD⊥平面PAD.(Ⅱ)以D为坐标原点,DA所在直线为x轴,DB所在直线为y轴,过点D平行于PE所在直线为z轴,建立空间直角坐标系.利用向量法能求出平面PEB与平面PDC所成二面角的余弦值.20.(Ⅰ)设点A1(﹣a,0),F2(c,0),由题意得a=4﹣2c,由椭圆的离心率,得a=2c,求出a,b,由此能示出椭圆C的方程.(Ⅱ)法一:根据椭圆的对称性猜测点G是与y轴平行的直线x=x0上.假设当点M为椭圆的上顶点时,直线l的方程为,此时点N,联立直线和直线可得点,猜想点G在直线x=1上,对猜想给予证明,得到点G在定直线上x=1上.法二:设M(x1,y1),N(x2,y2),G(x3,y3),由B,M,N三点共线,得:2x1x2﹣5(x1+x2)+8=0,再由A1,M,G三点共线,A2,N,G三点共线,推导出点G在定直线x=1上.法三:设l的方程为y=k(x﹣4),M(x1,y1),N(x2,y2).由得(3+4k2)x2﹣32k2x+64k2﹣12=0,由此利用根的判别式、韦达定理,结合A1,M,G三点共线,A2,N,G三点共线,推导出点G 在定直线x=1上.21.(Ⅰ)求出函数的导数,根据函数的单调性求出a的范围即可;(Ⅱ)根据函数的单调性求出f(x)的最小值,从而求出最小值的范围即可.22.(Ⅰ)利用三种方程的转化方法,求曲线C1的普通方程和曲线C2的直角坐标方程;(Ⅱ)根据参数方程的几何意义可知|PA|=2|t1|,|PB|=2|t2|,利用|PA|=2|PB|,分类讨论,求实数a的值.23.(Ⅰ)由绝对值的几何意义知,由不等式f(x)≤2﹣|x﹣1|有解,可得,即可求实数a的取值范围;(Ⅱ)当a<2时,(x)在单调递减,在单调递增,利用函数f(x)的最小值为3,求实数a的值.2017年3月15日。

2017年高三摸底考试应届理科成绩表

2017年高三摸底考试应届理科成绩表

学生人数 总分平均分 语文平均分 语文优秀率 语文良好率 语文及格率 语文低分率 数学平均分 28623 400.41 92.34 0.74 17.51 62.99 2.04 73.77 189 358.23 92.39 0.53 17.99 60.85 2.12 58.21 76 246.71 67.33 0 0 2.67 22.67 41.31 76 262.95 83.87 0 2.63 38.16 3.95 39.21 1033 397.27 92.29 0.1 11.42 62.92 0.58 74.76 958 397.49 92.52 0.42 13.39 64.23 0.52 74.05 1358 498.25 103.02 1.92 45.17 93.43 0.15 90.29 759 299.02 90.29 0 4.48 58.23 0.92 67.08 523 416.17 93.35 0 13.24 67.18 0.77 78.34 507 287.94 76.28 0 0.2 12.43 6.9 48.01 159 221.12 70.54 0 0.63 3.16 14.56 36.74 1109 454.69 98.16 0.9 25.07 82.6 0.09 83.49 499 376.09 85.05 0 5.01 36.27 2.4 70.67 166 219.45 72.19 0 1.86 13.66 17.39 34.3 710 437.11 96.44 0.99 23.45 76.41 1.27 82.02 112 271.39 81.43 0 2.7 27.03 4.5 49.23 617 441.3 96.25 0.16 23.18 74.23 0.16 79.6 215 254.82 77.4 0 3.26 23.26 12.09 40.46 637 400.45 94.49 0.47 18.84 69.39 0.78 73.68 907 432.19 95.26 0.33 17.11 74.28 0.22 77.07 88 303.36 83.91 0 3.45 36.78 1.15 58.2 801 398.22 94.02 0.25 16.88 68.5 0.5 71.04 85 261.6 75.55 0 1.19 9.52 9.52 46.61 451 404.59 90.85 0.22 12.64 59.42 1.77 78.13 41 178.56 66.3 0 0 0 30 26.46 413 380.92 89.73 0 9.2 57.63 2.66 68.42 1050 454.79 95.96 0.77 19.71 75.87 0.29 85.92 677 368.72 85.55 0 1.33 35.01 0.59 65.04 200 260.7 80.94 0 0 25.63 5.53 41.58 183 266.48 66.36 0 0.55 15.47 33.7 47.97 23 326.65 90.96 0 4.35 65.22 0 52.17 1463 449.1 96.83 0.89 24.91 77 0.75 81.39 930 420.99 97.33 0.86 26.02 78.49 0.54 76.75 469 364.39 88.15 0.21 8.15 49.57 1.5 68.1 321 315.27 78.25 0 1.25 15.58 5.61 58.25 106 213.76 69.89 0 0 4.81 19.23 32.33 406 348.84 87.83 0.25 8.96 49 3.73 62.95 1060 434.61 96.25 0.09 18.07 78.05 0.38 80.13 562 370.13 91.22 0 10.5 56.41 0.53 68.36 545 282.44 84.36 0 6.24 36.15 4.22 47.74 541 347.48 89.94 0 9.81 52.78 0.93 66.42 1039 458.31 98.41 0.48 27.81 83.04 0 86 447 396.47 89.63 0 2.68 55.03 1.12 77.38 457 358.68 88.86 0 8.32 53.39 2.84 71.11
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姓名准考证号学校语文数学英语物理化学生物总分省排名刘方麒2010020316南昌二中116.00146.00138.00106.0092.0079.00677.0013方博懋2010020422南昌二中120.50139.00135.00104.0087.0083.00668.5032韩铖霖2010020208南昌二中111.50143.00139.50106.0091.0076.00667.0041万臻哲2010020224南昌二中108.50146.00140.0097.0088.0080.00659.5083丁明琦2010020704南昌二中117.00143.00128.0097.0089.0082.00656.00108徐嘉忆2010020532南昌二中120.00137.00137.50101.0080.0079.00654.50116兰焜耀2010020501南昌二中111.00145.00131.5094.0093.0079.00653.50125陈琛2010020102南昌二中115.00132.00142.0092.0094.0077.00652.00135李雨聪2010020534南昌二中119.50129.00132.5097.0091.0083.00652.00135肖鹏宇2010020330南昌二中122.50129.00134.5094.0088.0083.00651.00145袁正2010020533南昌二中130.00140.00138.5096.0087.0059.00650.50150方雨婵2010020517南昌二中126.00133.00136.0088.0089.0078.00650.00160江哲昊2010021034南昌二中112.50145.00132.5097.0085.0078.00650.00160龙茜雅2010021002南昌二中122.50129.00143.0081.0090.0083.00648.50173何东2010021424南昌二中112.00141.00136.00108.0084.0066.00647.00191汪建之2010020815南昌二中105.00133.00139.50104.0084.0079.00644.50230王江畅2010020631南昌二中106.50141.00131.00100.0089.0077.00644.50230曹婉琳2010020132南昌二中113.00143.00129.5099.0079.0080.00643.50245 12吴可鸣2010020832南昌二中104.50141.00131.00100.0089.0078.00643.50245谢文宇2010021706南昌二中104.50134.00139.50107.0085.0071.00641.00277余凯2010020438南昌二中111.00146.00130.0088.0091.0075.00641.00277卢德铭2010020425南昌二中119.00143.00119.5090.0092.0077.00640.50285晏明熙2010020911南昌二中107.00141.00144.0086.0085.0075.00638.00311吴昊2010020116南昌二中117.00142.00130.0091.0076.0082.00638.00311杨泽俊2010021315南昌二中118.00129.00133.5091.0089.0077.00637.50319张恒煜2010021222南昌二中115.00138.00127.0090.0093.0074.00637.00330 15邬帅2010021732南昌二中111.00144.00132.5088.0084.0076.00635.50345付航2010021211南昌二中109.50133.00138.0083.0091.0080.00634.50363粟烨2010021437南昌二中119.00132.00133.5092.0080.0078.00634.50363邓力友2010020503南昌二中105.50147.00126.0098.0081.0076.00633.50380 14匡浩民2010020816南昌二中120.50129.00130.0085.0086.0083.00633.50380游镇宇2010020238南昌二中110.50127.00133.50106.0083.0073.00633.00392龚文奕2010021223南昌二中128.50128.00133.0082.0085.0076.00632.50399林世颖2010021704南昌二中115.00132.00131.0085.0088.0081.00632.00403朱继伟2010021115南昌二中109.50134.00132.0088.0095.0073.00631.50410肖人杰2010020223南昌二中114.50136.00119.5093.0094.0074.00631.00418廖子源2010020516南昌二中109.50132.00123.00100.0094.0072.00630.50425戴旒熙2010021331南昌二中116.50131.00129.0094.0084.0076.00630.50425 15李萧铭2010021319南昌二中117.00139.00116.5085.0091.0081.00629.50440陈俊达2010020117南昌二中115.00125.00129.0099.0086.0075.00629.00450熊逸钦2010022018南昌二中123.00134.00132.5075.0087.0077.00628.50456 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10何俊2010021335南昌二中109.00139.00132.0084.0083.0075.00622.00574罗益民2010020627南昌二中108.00138.00124.0099.0078.0074.00621.00596吴高华2010021209南昌二中119.50135.00108.0091.0091.0076.00620.50604曾麟岚2010020226南昌二中122.00131.00133.5088.0085.0061.00620.50604蔡顺哲2010021616南昌二中110.00123.00126.00102.0082.0077.00620.00616李韦伊2010020737南昌二中110.00129.00132.0082.0091.0076.00620.00616袁若宸2010020803南昌二中106.50124.00131.00101.0080.0077.00619.50630陈一涛2010021037南昌二中124.00137.00126.5075.0082.0075.00619.50630 13彭莉雯2010020723南昌二中105.00139.00131.5087.0081.0076.00619.50630熊文田2010021423南昌二中121.50110.00131.0089.0088.0079.00618.50651王恬恬2010021527南昌二中123.50130.00126.0095.0075.0069.00618.50651陈雨欣2010021615南昌二中100.00139.00130.0085.0087.0077.00618.00664伍金文2010021330南昌二中115.00129.00128.0086.0084.0076.00618.00664熊海明2010021333南昌二中114.50131.00119.0095.0081.0077.00617.50673黄云开2010020301南昌二中115.00139.00128.5079.0085.0071.00617.50673 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13姜雅思2010020612南昌二中107.5094.00118.5072.0079.0074.00545.002991殷悦2010020401南昌二中113.50104.00130.0080.0071.0046.00544.503002蔡岷2010020624南昌二中102.00135.00125.0061.0078.0043.00544.003017曾毓秀2010020432南昌二中109.00107.00127.0079.0062.0060.00544.003017熊笑靥2010020623南昌二中99.00117.00128.5071.0076.0052.00543.503033何佳昆2010020507南昌二中99.00125.00120.5064.0073.0062.00543.503033江舸2010020710南昌二中114.50111.00127.0064.0071.0056.00543.503033魏坤2010020734南昌二中114.00111.00127.0069.0070.0052.00543.003060刘喆2010022221南昌二中106.50104.00114.5064.0087.0067.00543.003060刘钰萱2010020309南昌二中105.5094.00121.5078.0078.0065.00542.003097万虹辰2010021826南昌二中97.00123.00121.0074.0076.0051.00542.003097龚智诚2010021503南昌二中115.50135.00116.5074.0047.0054.00542.003097周昕宇2010021311南昌二中105.5096.00129.0070.0081.0060.00541.503119陶琢2010022208南昌二中102.00109.00125.5068.0079.0058.00541.503119万宇豪2010020135南昌二中114.00116.00113.0071.0073.0054.00541.003140周捷2010021618南昌二中118.00109.00130.5054.0077.0052.00540.503160方彪2010022238南昌二中108.00116.00104.5075.0073.0064.00540.503160高奕奕2010020903南昌二中101.50103.00129.0066.0071.0070.00540.503160胡志明2010020107南昌二中112.50123.00117.5062.0065.0060.00540.003187熊扬涛2010020811南昌二中99.50121.00110.5073.0074.0062.00540.003187李峥2010022107南昌二中100.50114.00120.5069.0083.0052.00539.003217李雯2010021401南昌二中104.50114.00119.5073.0083.0045.00539.003217黄新昊2010022010南昌二中109.00118.00110.5072.0079.0050.00538.503235朱子骏2010021910南昌二中101.50111.00140.0077.0057.0052.00538.503235潘恒昌2010021414南昌二中103.00102.00128.0063.0074.0068.00538.003258王文军2010020203南昌二中100.50115.00120.5061.0080.0061.00538.003258胡雨晞2010020130南昌二中97.00120.00130.0060.0081.0050.00538.003258王紫玥2010021831南昌二中108.00121.00128.0069.0054.0058.00538.003258吴晶雅2010021902南昌二中106.50121.00125.5057.0068.0060.00538.003258张子阳2010021103南昌二中102.50131.0089.0082.0074.0059.00537.503291万昕妍2010021038南昌二中100.50102.00127.0072.0080.0056.00537.503291方堃2010021620南昌二中105.00133.00113.5059.0068.0059.00537.503291过辰希2010021535南昌二中103.00107.00127.0067.0082.0051.00537.003308肖泽辉2010021720南昌二中103.50109.00100.5082.0085.0057.00537.003308刘宇飞2010022129南昌二中106.00101.00119.0065.0082.0064.00537.003308熊漠淞2010021636南昌二中108.50111.00122.5067.0070.0058.00537.003308彭聪2010021219南昌二中113.50124.00105.5073.0070.0051.00537.003308熊绘乾2010020638南昌二中95.50122.00119.5065.0076.0059.00537.003308张良坤2010020111南昌二中104.00103.00122.5064.0078.0065.00536.503332戴逢镁2010020637南昌二中117.5082.00125.5061.0083.0067.00536.003354张文昊2010020715南昌二中107.50115.00108.5055.0073.0077.00536.003354胡非凡2010020234南昌二中106.50129.00109.5057.0071.0063.00536.003354童吉枥2010022133南昌二中113.00122.00111.0068.0077.0045.00536.003354彭楚轩2010021028南昌二中104.50116.00122.5059.0076.0058.00536.003354彭子昱2010020120南昌二中108.50109.00131.5059.0061.0067.00536.003354骆劲宇2010020716南昌二中103.00103.00116.5076.0075.0062.00535.503378王雨豪2010020807南昌二中97.00116.00127.0071.0075.0049.00535.003399汪志2010021712南昌二中95.50117.00117.5076.0076.0053.00535.003399彭莉媛2010020514南昌二中111.50105.00125.5070.0060.0063.00535.003399徐浩轩2010021305南昌二中99.50101.00128.0081.0067.0058.00534.503416王士越2010020304南昌二中110.00111.00117.0069.0060.0067.00534.003432龙志武2010022106南昌二中101.00112.00122.0050.0085.0064.00534.003432胡琦帆2010021216南昌二中112.50113.00108.5061.0079.0060.00534.003432李籽2010020622南昌二中111.50117.0099.0073.0078.0055.00533.503455刘炜齐2010020109南昌二中101.00113.00127.0066.0068.0058.00533.003475涂力2010020235南昌二中102.50107.00118.0083.0075.0047.00532.503496胡翔宇2010021631南昌二中105.50105.00119.0065.0069.0069.00532.503496官震宇2010021534南昌二中108.0096.00124.0071.0077.0056.00532.003511金一帆2010020336南昌二中109.0086.00135.0070.0074.0058.00532.003511陶姗2010020932南昌二中109.00114.00132.0064.0060.0053.00532.003511左正卿2010021709南昌二中106.50127.00115.5073.0060.0050.00532.003511史博文2010021324南昌二中100.50113.00124.0071.0069.0054.00531.503524邹煜洋2010020920南昌二中105.5090.00118.5069.0093.0055.00531.003541杨泽霖2010021233南昌二中114.50104.00135.0047.0070.0060.00530.503556张之珮2010020214南昌二中106.50109.00129.0056.0078.0052.00530.503556宗紫琪2010020633南昌二中99.50107.00115.5066.0068.0074.00530.003578吴凌云2010020915南昌二中115.5094.00131.5064.0074.0051.00530.003578皇甫逸宁2010020429南昌二中105.00126.00121.0057.0067.0054.00530.003578李钟煜2010020913南昌二中111.00120.00127.0065.0044.0063.00530.003578罗晶晶2010021721南昌二中110.00113.00118.5047.0080.0061.00529.503603范升2010021515南昌二中100.50110.00125.0064.0057.0073.00529.503603胡澜夕2010021734南昌二中105.50113.00127.5051.0071.0061.00529.003628付琪玥2010021105南昌二中112.00119.00127.0068.0056.0047.00529.003628赵晶2010020128南昌二中106.50115.00121.5063.0071.0052.00529.003628熊思琛2010022006南昌二中113.00120.00131.5032.0067.0065.00528.503644熊世豪2010020324南昌二中113.0096.00118.5073.0072.0056.00528.503644李嘉铭2010021213南昌二中103.50107.00127.5067.0059.0064.00528.003669吴以2010020632南昌二中97.00127.00121.0051.0079.0053.00528.003669熊奕臣2010020725南昌二中108.00102.00117.5064.0069.0067.00527.503697杜明2010022002南昌二中104.50103.00122.0080.0072.0046.00527.503697李雅雯2010021102南昌二中105.50110.00133.0058.0066.0055.00527.503697李子萱2010020219南昌二中112.50119.00122.5059.0055.0059.00527.003717滕千惠2010022225南昌二中108.50115.00130.5055.0062.0056.00527.003717徐楷2010021025南昌二中101.50119.00112.5067.0074.0053.00527.003717刘星雨2010021203南昌二中99.50115.00138.0054.0069.0051.00526.503747史佳玮2010020418南昌二中104.50122.00124.5062.0061.0052.00526.003769衷青2010021835南昌二中107.00103.00115.0060.0074.0067.00526.003769陈文萱2010021713南昌二中106.5095.00119.5064.0078.0063.00526.003769蔡盛伟2010021122南昌二中98.00110.00131.5069.0062.0055.00525.503788罗楠2010022230南昌二中106.50125.00111.0053.0066.0064.00525.503788熊乐天2010022202南昌二中121.00104.00125.5054.0055.0066.00525.503788舒雨清2010021410南昌二中111.00104.00124.0051.0059.0076.00525.003807饶羽琴2010020525南昌二中113.50109.00128.5062.0053.0059.00525.003807余振豪2010021903南昌二中94.50125.00112.0073.0065.0055.00524.503829王世威2010020837南昌二中97.00113.00120.5055.0081.0058.00524.503829游智慧2010020405南昌二中101.50116.00118.5054.0073.0061.00524.003855罗正坤2010021537南昌二中109.5095.00114.0087.0062.0056.00523.503875欧阳欣雨2010020824南昌二中92.50116.00127.0052.0079.0057.00523.503875柳纪琛2010020512南昌二中116.00100.00133.5063.0058.0053.00523.503875刘世曜2010021915南昌二中105.00129.00113.5054.0070.0052.00523.503875徐慧玲2010021610南昌二中101.50108.00116.5070.0068.0059.00523.003899彭乐阳2010022214南昌二中94.50106.00115.5063.0082.0062.00523.003899熊惠2010021625南昌二中116.50115.00120.5049.0065.0057.00523.003899张耀宗2010020431南昌二中108.50112.00118.5063.0068.0052.00522.003932彭艺彦2010022136南昌二中106.50114.00120.5069.0055.0057.00522.003932祝瑶瑶2010020826南昌二中114.5099.00115.0060.0075.0058.00521.503948方莺2010020526南昌二中109.50116.00112.0062.0063.0059.00521.503948李书源2010021218南昌二中101.00124.00119.5068.0061.0048.00521.503948雷欣岚2010020605南昌二中111.00110.00110.0050.0076.0064.00521.003969万志豪2010020527南昌二中107.50106.00111.0072.0076.0048.00520.503990陈明晖2010020430南昌二中104.50117.00111.0051.0075.0062.00520.503990。

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