江西省南昌市2018届高三第一次模拟测试
(完整word)江西省南昌市2018届高三第一次模拟测试(理综)

江西省南昌市2018届高三第一次模拟测试理科综合本试卷分第I 卷(选择题)和第n 卷(非选择题)两部分;共 以下数据可供解题时参考:本试卷参考相对原子质量: H-1 B -11 C 12N-14、选择题:(本大题包括13小题,每小题6分,共78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.下列哪项实验的材料可以和观察DNA 和RNA 在细胞中的分布”的选材相同A .观察细胞中的叶绿体B .观察细胞中的线粒体C .观察洋葱外表皮的质壁分离D .制备细胞膜2•在细胞中,下列哪个生化反应过程需要消耗ATPA .溶酶体中大分子水解成小分子B .呼吸作用中葡萄糖分解成丙酮酸C .光反应中水分解成 02和[H]D .暗反应中C 3化合物的还原 3 •下列可以引起神经元静息电位绝对值降低的是哪一项A .增加细胞外 K +浓度B •增加细胞内K +浓度C .增加细胞内Na +浓度D •降低细胞外 Na +浓度4.某种南瓜矮生突变体可分为两类:激素合成缺陷型突变体和激素不敏感型突变体。
为研究某种矮 生南瓜的矮生突变体属于哪种类型,研究者应用赤霉素和生长素溶液进行了相关实验,结果如 图所示。
下列相关分析正确的是A •由图可看出,赤霉素能促迸正常植株茎的伸长,生长素对正常植株的作用具有两重性B .由图可以判断,该矮生南瓜突变体是生长素和赤霉素不敏感型突变体C .若两种南瓜内生长素和赤霉素的含量都很接近陷型D .正常南瓜茎的伸长对赤霉素的作用更敏感 5.如图表示一片草原上的兔子和狼在一段时间内相对数量变化的趋势,下列相关分析正确的是 A .甲代表狼,乙代表兔子 B .狼的K 值接近B 点对座的数值14页。
时量150分钟,满分 300分。
O 16P-31CI-35.5Ca-40 Cu -64第I 卷(选择题共21题,每小题6分,共126 分)ci- ftw 5 D2 2 11茎伸煮±彳已CJaaa is Ts赤■棄氷就)生 K 4^<(7U M)135,则可以判断该矮生南瓜突变体是激素合成缺C .兔子的K值接近C点对应的数值D .第2年,狼的数量因为缺乏食物而下降6•下图为某家族的遗传系谱图,已知川-4号个体不携带任何致病基因,下列相关分析正确的是图例口世常男械o正常女性□甲it传病男柱惠溝©甲it传M如皐耆KJ乙il裁病期41慧着A .甲病为X染色体上隐性基因控制B .川-2的C . W -4同时携带两种致病基因的概率为0 D .川-3和川7•化学与生产、生活密切相关。
2018届江西省南昌市高三第一次模拟考试英语试题

第一部分听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项。
听完每段对话后,你将有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. What was the weather like yesterday?A. Cloudy.B. Foggy.C. Stormy.2. What is the woman doing now?A. Making a reservation.B. Changing clothes.C. Reading a book.3. Where are the speakers?A. At a tailor s.B. In a theatre.C. At home.4. What are the speakers going to do?A. Make orders.B. Pack boxes. C, Stop working.5. When will the next train for Nanchang leave?A. At 9:30.B. At 11:30.C. At 11:40.第二节(共15小题;每小题1.5分;满分22.5分)听下面5段对话或独白。
每段对话或独白后有几个小题,从感中所给的A、B、C三个选项中选出最佳选项。
听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
听第6段材料,回答第6、7题。
6. What information does the woman provide?A. The address of the hotel.B. Her telephone number.C. The downtown map.7. Which direction will the woman take first?A. Turn left.B. Turn right.C. Go straight ahead.听第7段材料,回答第8、9题。
江西省南昌市2018届高三第一次模拟考试英语试题+Word版含答案

江西省南昌市2018届高三第一次模拟考试英语试题第一部分听力(共两节,满分30分)第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项。
听完每段对话后,你将有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. What was the weather like yesterday?A. Cloudy.B. Foggy.C. Stormy.2. What is the woman doing now?A. Making a reservation.B. Changing clothes.C. Reading a book.3. Where are the speakers?A. At a tailor s.B. In a theatre.C. At home.4. What are the speakers going to do?A. Make orders.B. Pack boxes. C, Stop working.5. When will the next train for Nanchang leave?A. At 9:30.B. At 11:30.C. At 11:40.第二节(共15小题;每小题1.5分;满分22.5分)听下面5段对话或独白。
每段对话或独白后有几个小题,从感中所给的A、B、C三个选项中选出最佳选项。
听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
听第6段材料,回答第6、7题。
6. What information does the woman provide?A. The address of the hotel.B. Her telephone number.C. The downtown map.7. Which direction will the woman take first?A. Turn left.B. Turn right.C. Go straight ahead.听第7段材料,回答第8、9题。
江西省南昌市2018届高三第一次模拟考试理数试题(考试版)

第1页 共4页 ◎ 第2页 共4页绝密★启用前江西省南昌市2018届高三第一次模拟考试数学(理)试题一、单选题 1.已知,,则A . B.C .D .2.已知复数为纯虚数,则 A .B .C . 或D . 3.设命题,则是 A .B .C .D .4.已知平面向量,则 A .B .C .D . 5.已知等比数列的各项均为正数,前项和为,若,则A .B .C .D .6.已知动点满足线性条件,定点,则直线斜率的最大值为A .B .C .D . 7.已知椭圆的左右焦点分别为,过且垂直于长轴的直线交椭圆于两点,则△内切圆的半径为A .B .C .D .8.已知函数,若将函数的图象向右平移个单位后关于轴对称,则下列结论中不正确...的是 A. B. 是图象的一个对称中心 C.D.是图象的一条对称轴9.若向区域内投点,则该点落在由直线与曲线围成区域内的概率为A .B .C .D .10.如图,网格纸上小正方形的边长为,粗线条画出的是一个三棱锥的三视图,则该三棱锥中最长棱的长度为A .B .C .D .11.已知双曲线的左、右焦点分别为,点在双曲线的右支上,且,则双曲线离心率的取值范围是 A .B .C .D .12.若关于的方程存在三个不等实根,则实数的取值范围是 A . B .C .D .二、填空题13.的展开式中含项的系数为___________.14.更相减损术是出自《九章算术》的一种算法.如图所示的程序框图是根据更相减损术写出的,若输入,则输出的值为_____.15.底面是正多边形,顶点在底面的射影是底面中心的棱锥叫正棱锥,已知同底的两个正三棱锥内接于同一个球.已知两个正三棱锥的底面边长为a ,球的半径为R , 设两个正三棱锥的侧面与底面所成的角分别为α,β,则t a n ()αβ+的值是 . 16.在数列中,,且对任意,成等差数列,其公差为,则 ________.第3页 共4页 ◎ 第4页 共4页三、解答题 17.在△中,内角的对边分别为,其面积.(1)求的值; (2) 设内角的平分线交于,,,求 .18.某种植园在芒果临近成熟时,随机从一些芒果树上摘下100个芒果,其质量分别在,,,,,(单位:克)中,经统计得频率分布直方图如图所示.(1)现按分层抽样从质量为,的芒果中随机抽取个,再从这个中随机抽取个,记随机变量表示质量在内的芒果个数,求的分布列及数学期望.(2)以各组数据的中间数代表这组数据的平均值,将频率视为概率,某经销商来收购芒果,该种植园中还未摘下的芒果大约还有个,经销商提出如下两种收购方案: A :所以芒果以元/千克收购;B :对质量低于克的芒果以元/个收购,高于或等于克的以元/个收购. 通过计算确定种植园选择哪种方案获利更多? 19.如图,在直四棱柱中,底面为等腰梯形,.(1)证明:;(2)设是线段上的动点,是否存在这样的点,使得二面角的余弦值为,如果存在,求出的长;如果不存在,请说明理由.20.已知直线过抛物线:的焦点,且垂直于抛物线的对称轴,与抛物线两交点间的距离为. (1)求抛物线的方程; (2)若点,过点的直线与抛物线相交于,两点,设直线与的斜率分别为和.求证:为定值,并求出此定值. 21.已知函数.(1)求证:函数有唯一零点;(2)若对任意,恒成立,求实数的取值范围.22.选修4—4:坐标系与参数方程选讲.已知曲线的参数方程为(为参数),以直角坐标系的原点为极点,轴的正半轴为极轴建立极坐标系,曲线的极坐标方程为.(1)求的普通方程和的直角坐标方程; (2)若过点的直线与交于,两点,与交于两点,求的取值范围.23.选修4—5:不等式选讲. 已知函数. (1)求的解集; (2) 若的最小值为,正数满足,求证:.。
江西省南昌市2018届高三数学第一次模拟考试试题理

江西省南昌市2018届高三数学第一次模拟考试试题 理一、选择题:本大题共12个小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的.1.已知集合{A x N y =∈,{}21,B x x n n Z ==+∈,则A B =( )A.(],4-∞B.{}1,3C.{}1,3,5D.[]1,32.欧拉公式cos sin ix e x i x =+(i 为虚数单位)是由瑞士著名数学家欧拉发现的,它将指数函数的定义域扩大到复数,建立了三角函数和指数函数的关系,它在复变函数论里非常重要,被誉为“数学中的天桥”。
根据欧拉公式可知,3x i e 表示的复数位于复平面中的( ) A.第一象限B.第二象限C.第三象限D.第四象限3.已知角α的终边经过点()sin 47,cos47P °°,则()sin 13α-=°( ) A.12C.12-D. 4.已知奇函数()'f x 是函数()()f x x R ∈是导函数,若0x >时()'0f x >,则( ) A.()()()320log 2log 3f f f >>- B.()()()32log 20log 3f f f >>- C.()()()23log 3log 20f f f ->>D.()()()23log 30log 2f f f ->>5.设不等式组3010350x y x y x y +-≥⎧⎪-+≥⎨⎪--≤⎩表示的平面区域为M ,若直线y kx =经过区域M 内的点,则实数k 的取值范围为( )A.1,22⎛⎤ ⎥⎝⎦B.14,23⎡⎤⎢⎥⎣⎦C.1,22⎡⎤⎢⎥⎣⎦D.4,23⎡⎤⎢⎥⎣⎦6.平面内直角三角形两直角边长分别为,a b,直角顶点到斜边的距离为,空间中三棱锥的三条侧棱两两垂直,三个侧面的面积分别为123,,S S S ,类比推理可( )7.已知圆台和正三棱锥的组合体的正视图和俯视图如图所示,图中网格是单位正方形,那么组合体的侧视图的面积为( )A.6+B.152C.6D.88.执行如图程序框图,则输出的n 等于( )A.1B.2C.3D.49.函数()()()2sin xx e e x f x x e ππ-+=-≤≤的图象大致为( )ABCD10.已知具有线性相关的五个样本点()10,0A ,()22,2A ,()33,2A ,()44,2A ,()56,4A ,用最小二乘法得到回归直线方程1:l y bx a =+,过点1A ,2A 的直线方程2:l y mx n =+,那么下列4个命题中,①,m b a n >>;②直线1l 过点3A ;③()()552211i i i i i i y bx a y mx n ==--≥--∑∑④5511i i i i i i y bx a y mx n ==--≥--∑∑.(参考公式()()()1122211nni iii i i nniii i x ynxy xx y yb xnxxx====---==--∑∑∑∑,a y bx =-)正确命题的个数有( ) A.1个B.2个C.3个D.4个11.设函数()1,121,1x ax a f x x a x a -⎧⎛⎫<+⎪ ⎪=⎨⎝⎭⎪-+-≥+⎩,若()f x 的最大值不超过1,则实数a 的取值范围为( ) A.3,2⎡⎫-+∞⎪⎢⎣⎭B.3,2⎛⎫-+∞ ⎪⎝⎭C.5,04⎡⎫-⎪⎢⎣⎭D.35,24⎡⎫--⎪⎢⎣⎭12.已知椭圆22:12412x y E +=,O 为坐标原点,,A B 是椭圆上两点,,OA OB 的斜率存在并分别记为OA k 、OB k ,且12OA OB k k ⋅=-,则11OA OB +的最小值为( )B.13二、填空题(每题5分,满分20分,将答案填在答题纸上)13.()3121x x ⎛⎫+- ⎪⎝⎭展开式中的常数项为________________.14.平面向量()1,a m =,()4,b m =,若有()()20a ba b -+=,则实数m =________________.15.在圆224x y +=上任取一点,则该点到直线0x y +-的距离[]0,1d ∈的概率为________________.16.已知台风中心位于城市A 东偏北α(α为锐角)度的200公里处,若()24cos 25αβ-=,则v =__________.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.已知等比数列{}n a 的前n 项和为n S ,满足4421S a =-,3321S a =-. (1)求{}n a 的通项公式;(2)记()21log n n n b a a +=⋅,数列{}n b 的前n 项和为n T ,求证:121112nT T T +++<…. 18.某校为了推动数学教学方法的改革,学校将高一年级部分生源情况基本相同的学生分成甲、乙两个班,每班各40人,甲班按原有模式教学,乙班实施教学方法改革.经过一年的教学实验,将甲、乙两个班学生一年来的数学成绩取平均数,两个班学生的平均成绩均在[]50,100,按照区间[)50,60,[)60,70,[)70,80,[)80,90,[]90,100进行分组,绘制成如下频率分布直方图,规定不低于80分(百分制)为优秀.(1) 完成表格,并判断是否有90%以上的把握认为“数学成绩优秀与教学改革有关”;(2)从乙班[)70,80,[)80,90,[]90,100分数段中,按分层抽样随机抽取7名学生座谈,从中选三位同学发言,记来自[)80,90发言的人数为随机变量X ,求X 的分布列和期望.19.如图,四棱锥P ABCD -中,PA ⊥底面ABCD ,ABCD 为直角梯形,AD BC ∥,AD AB ⊥,132AB BC AP AD ====,AC BD O =,过O 点作平面α平行于平面PAB ,平面α与棱BC ,AD ,PD ,PC 分别相交于点E ,F ,G ,H .(1)求GH 的长度;(2)求二面角B FH E --的余弦值.20.已知抛物线()2:20C y px p =>的焦点为F ,准线为l ,过焦点F 的直线交C 于()11,A x y ,()22,B x y 两点,124y y =-.(1)求抛物线方程;(2)点B 在准线l 上的投影为E ,D 是C 上一点,且AD EF ⊥,求ABD △面积的最小值及此时直线AD 的方程.21.已知函数()()ln f x ax bx =+在点()()1,1f 处的切线是0y =. (1)求函数()f x 的极值;(2)当()()210x mx e f x x m e e-≥+<恒成立时,求实数m 的取值范围(e 为自然对数的底数).22.在平面直角坐标系xOy 中,曲线C 的参数方程为2cos 2sin 2x y θθ=⎧⎨=+⎩(θ为参数),以坐标原点为极点,x 轴非负半轴为极轴建立极坐标系. (1)求C 的极坐标方程;(2)若直线12,l l 的极坐标方程分别为()6R πθρ=∈,()2=3R πθρ∈,设直线12,l l 与曲线C 的交点为O ,M ,N ,求OMN △的面积. 23.已知()223f x x a =+.(1)当0a =时,求不等式()23f x x +-≥的解集;(2)对于任意实数x ,不等式()212x f x a +-<成立,求实数a 的取值范围.参考答案一.选择题:本大题共12个小题,每小题5分,共60分,在每小题给出的四个选项中,只有一项是符合题目要求的.13.4 14. 2± 15.1316.100 三.解答题:本大题共6小题,共70分,解答应写出文字说明,证明过程或推演步骤. 17.【解析】(Ⅰ)设{}n a 的公比为q ,由434S S a -=得,43422a a a -=, 所以432a a =, 所以2q =. 又因为3321S a =-, 所以11112481a a a a ++=-, 所以11a =. 所以12n n a -=. (Ⅱ)由(Ⅰ)知1212log ()log (22)21n n n n n b a a n -+=⋅=⨯=-, 所以21(21)2n n T n n +-==, 所以22212111111111+++1121223(1)n T T T n n n+++=<++++创-11111111222231n n n=+-+-++-=-<-. 18.(Ⅰ)依题意得2240(12202820) 3.333 2.70640403248K ⨯⨯-⨯=≈>⨯⨯⨯ 有90%以上的把握认为“数学成绩优秀与教学改革有关”(Ⅱ)从乙班[70,80),[80,90),[90,100]分数段中抽人数分别为2,3,2 …依题意随机变量X 的所有可能取值为0123,,,xOB E2134343377418(0),(1),3535C C C P X P X C C ======1234333377121(2),(3)3535C C C P X P X C C ======所以18121459()123353535357E X =???= 19. 【解析】(Ⅰ)【法一】(Ⅰ)因为//a 平面PAB ,平面a平面ABCD EF =,O EF Î,平面PAB 平面ABCD AB =,所以//EF AB ,同理//,//EH BP FG AP ,因为BC ∥,6,3AD AD BC ==,所以BOC D ∽DOA D ,且12BC CO AD AO ==, 所以12EO OF =,11,23CE CB BE AF ====, 同理13CH EH CO PC PB CA ===, 连接HO ,则有HO ∥PA , 所以HO EO ⊥,1HO =,所以13EH PB ==,同理,223FG PA ==, 过点H 作HN ∥EF 交FG 于N ,则GH 【法二】因为//a 平面PAB ,平面a 平面ABCD EF =,O EF Î,平面PAB平面ABCD AB =,根据面面平行的性质定理,所以//EF AB ,同理//,//EH BP FG AP , 因为//,2BC AD AD BC =,所以BOC DOA ∽D D ,且12BC CO AD OA ==, 又因为COE D ∽AOF D ,AF BE =,所以2BE EC =, 同理2AF FD =,2PG GD =,123,233EF AB EH PB FG AP ====== 如图:作//,,//,HN BC HN PB N GM AD GMPA M ==,所以//,HN GM HN GM =,故四边形GMNH 为矩形,即GH MN =, 在PMN D 中,所以MN =GH =(Ⅱ)建立如图所示空间直角坐标系(3,0,0),(0,2,0),(3,2,0),(2,2,1)B F E H ,(1,2,1),(2,0,1)BH FH =-=, 设平面BFH 的法向量为(,,)n x y z =,2020n BHx y z n FHx z ìï?-++=ïíï?+=ïî,令2z =-,得3(1,,2)2n =-,因为平面//EFGH 平面PAB ,所以平面EFGH 的法向量(0,1,0)m =3cos ,29||||m nm n m n ×===,二面角B FH E -- 20. 【解析】(Ⅰ)依题意(,0)2pF , 当直线AB 的斜率不存在时,2||4,2AB p p =-=-= 当直线AB 的斜率存在时,设:()2pAB y k x =-由22()2y pxpy k x ⎧=⎪⎨=-⎪⎩,化简得2220p y y p k --= 由124y y =-得24p =,2p =,所以抛物线方程24y x =.(Ⅱ)设00(,)D x y ,2(,)4t B t ,则(1,)E t -,又由124y y =-,可得244(,)A t t -因为2EF t k =-,AD EF ⊥,所以2AD k t =,故直线2424:()AD y x t t t+=- 由2248240y xx ty t ⎧=⎪⎨---=⎪⎩, 化简得2216280y ty t ---=,所以10102162,8y y t y y t+==--.所以10|||AD y y =-==设点B 到直线AD 的距离为d,则22222816|4||8|t t t d ---++==所以1||162ABD S AD d ∆=⋅=≥,当且仅当416t =,即2t =± 2:30t AD x y =--=时,, 2:30t AD x y =-+-=时,.21. 【解析】(Ⅰ)因为()ln()f x ax bx =+,所以1()a f x b b ax x¢=+=+, 因为点(1,(1))f 处的切线是0y =,所以(1)10f b ¢=+=,且(1)ln 0f a b =+= 所以,1a e b ==-,即()ln 1f x x x =-+((0,)x ??)所以11()1xf x x x-¢=-=,所以在(0,1)上递增,在(1,)+?上递减 所以()f x 的极大值为(1)ln 10f e =-=,无极小值.(Ⅱ)当21()x mx ef x x e e-?(0)m <在(0,)x ??恒成立时, 由(Ⅰ)()ln 1f x x x =-+,即ln 112x mx x e x e+?+(0)m <在(0,)x ??恒成立, 【法一】设ln 11(),()2e e x mx x g x h x x +==+-,则(1)()e x m x g x -'=,2ln ()xh x x '=-,又因为0m <,所以当01x <<时,()0,()0g x h x ''<>;当1x >时,()0,()0g x h x ''><. 所以()g x 在(0,1)上单调递减,在(1,)+∞上单调递增,min ()(1)e mg x g ==; ()h x 在(0,1)上单调递增,在(1,)+∞上单调递减,max1()(1)1eh x h ==-.所以(),()g x h x 均在1x =处取得最值,所以要使()()g x h x ≥恒成立, 只需min max ()()g x h x ≥,即11e em ≥-,解得1e m ≥-,又0m <, 所以实数m 的取值范围是[10)e ,-. 【法二】设ln 11()2x x mx g x x e e +=--+((0,)x ??),则2ln (1)()xx m x g x x e --¢=+ 当01x << 时,ln 0x ->,10x -<,则2ln 0x x ->,(1)0xm x e->,即()0g x ¢>当1x > 时,ln 0x -<,10x ->,则2ln 0x x -<,(1)0xm x e-<,即()0g x ¢< 所以()g x 在(0,1)x Î上单调递增,在(1,)x ??上单调递减. 所以max 1()(1)120m g x g e e ==-+-?,即11m e e?,又0m < 所以实数m 的取值范围是[10)e ,-. 22. 【解析】(Ⅰ)由参数方程2cos 2sin x y θθ=⎧⎨=+⎩2,得普通方程22(2)4x y -+=,所以极坐标方程2222cos sin 4sin 0r q r q r q +-=,即4sin r q =.(Ⅱ)直线()1π:R 6l q r =?与曲线C 的交点为,O M ,得||4sin 26M OM pr ===,又直线()22π:R 3l q r =?与曲线C 的交点为,O N ,得2||4sin 3N ON pr ===且2MON π∠=,所以11||||222OMN S OM ON D ==创23. 【解析】(Ⅰ)当0a =时,()|2||2||2|3f x x x x +-=+-?,0223x x x ì<ïïíï-+-?ïî 得13x ?;02223x x x ì#ïïíï+-?ïî 得12x #;2223x x x ì>ïïíï+-?ïî 得2x >, 所以()|2|2f x x +-?的解集为1(,][1,)3-?+?. (Ⅱ)对于任意实数x ,不等式|21|()2x f x a +-<成立,即2|21||23|2x x a a +-+<恒成立,又因为222|21||23||2123||31|x x a xx a a +-+?--=-,要使原不等式恒成立,则只需2|31|2a a -<,当0a <时,无解;当03a#时,2132a a -<,解得133a <?;当3a >时,2312a a -<,解得13a <<. 所以实数a 的取值范围是1(,1)3.。
2018届江西省南昌市高三第一次模拟考试卷 数学(理)

2018 届江西省南昌市高三第一次模拟考试卷6.平面内直角三角形两直角边长分别为 a , b ,则斜边长为 a2 b2 ,直角顶点到斜边的距离为ab a b22,空间数学(理)注意 事项: 1 .答题前,先将自己的姓名、准考证号填写在试题卷和答题卡上,并将准考证号条形码粘贴在答2 3 中三棱锥的三条侧棱两两垂直,三个侧面的面积分别为 S1 , S2 , S 3 ,类比推理可得底面积为 S12 S2 ,则 S3三棱锥顶点到底面的距离为( A. 3S1S2 S3 2 3 S12 S2 S3)S1S2 S3 2 3 S12 S2 S3B.C.座位号题卡上的指定位置。
2 .选择题的作答:每小题选出答案后,用 2B 铅笔把答题卡上对应题目的答案标号涂黑,写在试 题卷、草稿纸和答题卡上的非答题区域均无效。
3 .非选择题的作答:用签字笔直接答在答题卡上对应的答题区域内。
写在试题卷、草稿纸和答题 卡上的非答题区域均无效。
4 .考试结束后,请将本试题卷和答题卡一并上交。
一、选择题:本大题共 12 个小题,每小题 5 分,共 60 分.在每小题给出的四个选项中,只有一项是符合题目 要求的. 1.已知集合 A x N y 4 x , B x x 2n 1, n Z ,则 A A. , 4 B. 1,3 C. 1,3,52S1S2 S3 2 3 S12 S2 S3D.3S1S2 S3 2 3 S12 S2 S37.已知圆台和正三棱锥的组合体的正视图和俯视图如图所示,图中网格是单位正方形,那么组合体的侧视图的 面积为( )考场号B ()D. 1,3 A. 6 3 3 42.欧拉公式 eix cos x i sin x ( i 为虚数单位)是由瑞士著名数学家欧拉发现的,它将指数函数的定义域扩大到复 数,建立了三角函数和指数函数的关系,它在复变函数论里非常重要,被誉为“数学中的天桥”。
江西省南昌市2018届高三第一次模拟考试数学(文)试题含答案

(1) 求 x 的值和乙班同学成绩的众数; (2) 完成表格,若有 90% 以上的把握认为“数学成绩优秀与教学改革有关”的话,那么学校将扩大教学改革面,请问学校是否要扩大改 革面?说明理由.
19. 如图,四棱锥 P ABCD 中, PA 底面 ABCD , ABCD 为直角梯形, AC 与 BD 相交于点 O , AD ∥ BC , AD AB ,
) D.125
C.100
二、填空题(每题 5 分,满分 20 分,将答案填在答题纸上) 13.设函数 f x 在 0, 内可导,其导函数为 f ' x ,且 f ln x x ln x ,则 f ' 1 ____________. 14.已知平面向量 a 1, m , b 4, m ,若 2a b a b 0 ,则实数 m ____________. 15.在圆 x y 4 上任取一点,则该点到直线 x y 2 2 0 的距离 d 0,1 的概率为____________. 16.已知函数 f x x sin x ,若 0, , , ,且 f f 2 ,则 cos ________. 4 4 2 2
江西省南昌市 2018 届高三第一次模拟考试数学(文)试题含答案
第一次模拟测试卷 文科数学 一、选择题:本大题共 12 个小题,每小题 5 分,共 60 分.在每小题给出的四个选项中,只有一项是符合题目要求的. 1.已知集合 A x N y A. , 4
4 x , B x x 2n 1, n Z ,则 A B (
,则双曲线 C 的离心率为( 2
B. 2 3
江西省南昌市2018届高三第一次模拟测试(文综参考答案)

江西省南昌市2018届高三第一次模拟测试文科综合参考答案二、非选择题(一)必考题36.(1)水热(气候)适宜(2分),土地丰富(廉价)(2分),劳动力廉价(2分),政策支持(靠近珠江三角洲)(2分)。
(2)原料(甘蔗)(2分)。
甘蔗制糖业原料运输成本高(2分),向广西转移(靠近原料)可减少花费(提高利润)(2分)。
(3)珠江三角洲(工业化)城市化发展(2分),城市人口增加(2分),农副产品需求量增大(2分),农副产品运成本高(宜就近供应,土地资源有限,引起甘蔗种植产业向外转移)(2分)。
37.(1)水量大,对气温的调节作用较强(2分);高山阻挡冷空气,冬季温暖(地处谷地,与周边热量交换少;靠近海洋,气温受海洋调节)(2分);夏季冰川融水注入,水温偏低(2分)(2)湖泊比热容大,气温变化比周边陆地慢(2分);地处谷地,气温变化比周边山地慢(2分)。
(3)夏季午后周边区域气温高(2分),R湖(因湖水稳定)气温偏低(2分),形成热力环流(2分),湖面区域受下沉气流影响(2分)。
(4)(受沉积物阻挡,)落差小,流速慢(2分);水量大,河流吐纳对湖水运动影响小(缓冲作用强)(2分);山地环抱,风浪小(2分)。
38.①政府定价范围缩减,有利于激发市场活力,增强社会投资积极性,促进经济增长,增加就业;(4分)②价格主要由市场竞争形成,能促使企业提高产品质量,更好地满足人民群众的需求;(3分)③能在竞争中降低生产成本和产品价格,提高群众消费水平;(3分)④能源、资源价格的改革,有利于节约资源、保护环境,增强人民群众幸福感、获得感。
(4分)39.原因:①我国网络安全形势严峻,加强网络安全立法是保护公民、法人和其他组织的合法权益的要求。
(2分)②加强网络安全立法是贯彻依法治国的要求,能够为公民、社会组织参与网络活动提供法律准则和依据,能为政府进行有效网络治理提供法律依据;(3分)③加强网络安全立法,建设网络强国,是维护我国网络空间主权和国家安全、促进世界和平与发展的要求。
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江西省南昌市2018届高三第一次模拟测试英语注意事项:1.答题前,考生务必将自己的姓名和考试号写在答题卡相应的位置。
2.全部答案在答题卡上完成,用2B铅笔涂满涂黑,答在试卷上无效。
3.考试结束后,将答题卡上交。
第一部分听力(共两节,满分30分)第一节(共5小题;每小题分,满分分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项。
听完每段对话后,你将有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. What was the weather like yesterdayA. Cloudy.B. Foggy.C. Stormy.2. What is the woman doing nowA. Making a reservation.B. Changing clothes.C. Reading a book.3. Where are the speakersA. At a tailor s.B. In a theatre.C. At home.4. What are the speakers going to doA. Make orders.B. Pack boxes. C, Stop working.5. When will the next train for Nanchang leaveA. At 9:30.B. At 11:30.C. At 11:40.第二节(共15小题;每小题分;满分分)听下面5段对话或独白。
每段对话或独白后有几个小题,从感中所给的A、B、C三个选项中选出最佳选项。
听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
听第6段材料,回答第6、7题。
6. What information does the woman provideA. The address of the hotel.B. Her telephone number.C. The downtown map.7. Which direction will the woman take firstA. Turn left.B. Turn right.C. Go straight ahead.听第7段材料,回答第8、9题。
8. Who is SusanA. A mother.B. A nurse,C. A patient.9. How is Doctor Lee to SusanA. Encouraging.B. Thankful.C. Strict.听第8段材料,回答第10至12题。
10. What are the speakers mainly talking aboutA. Loneliness in the university.B. Pressures of studies.C. Pressures of living expenses.11. What costs the woman mostA. Transportation.B. Clothing.C. Meals.12. What does die woman do in her spare timeA. Go traveling.B. See movies.C. Hang out for a drink.听第9段材料,回答第13至16题。
13. What are the speakers attendingA. A wedding.B. A dinner party.C. An anniversary celebration.14. What does the woman want most right nowA. To dance.B. To eat something.C. To change her clothes.15. What did the man eat when he first arrivedA. Cheese.B. Fish and rice.C. Cakes.16. What probably cost a lot of moneyA. The flowers.B. Matt’s tie.C. Jennie’s clothes.听第10段材料,回答第17至20题。
17. When will the Winter Carnival endA. In January.B. In February.C. In March.18. What will be held in the center of the siteA. A dress competitionsB. A flower show.C. An ice sculpture exhibition.19. How many areas will be used for amusement ridesA. 2.B. 4.C. 6.20. Where will the food shops beA. Next to the amusement rides.B. At the comer of the site.C. Beside the entrance.第二部分阅读理解(共两节,满分40分)第一节(共15小题;每小题2分,满分30分)阅读下列短文,从每题所给的A、B、C、和D四个选项中,选出最佳选项。
AEach applicant to Harvard College is considered with great care. We consider each applicant to Harvard College as a whole person, and put enormous care into evaluating every application. We hope you will explore the information in this section to understand what we look for in our admissions process.How to ApplySubmit your application through the Common Application, the Coalition Application, or the Universal College Application. Each is treated equally by the Admissions Committee. Complete and submit your materials as soon as possible to ensure full and timely consideration of your application. View our Application Tips for step-by-step information.When to ApplyWhat We Look ForWe seek promising students who will contribute to the Harvard community during their college years, and to society throughout their lives.While academic accomplishment is the basic requirement, the Admission Committee considers many other factors—strong personal qualities, special talents or excellences of all kinds, perspectives formed by unusual personal circumstances, and the ability to take advantage of available resources and opportunities.We outline everything you need to apply to Harvard.Click harvard. edu/admissions/apply to get detailed requirements.21. Where can you find step-by-step information of how to applyA. Common Application.B. Coalition Application.C. Universal College Application.D. Application Tips.22. Which is the final day for Financial Aid ApplicationA. November 1.B. January 1.C. March 1.D. May 1.23. What is the basic requirement to apply for Harvard UniversityA. Academic accomplishment.B. Strong personal qualities.C. Special talents.D. Unique perspectives.BWhen I was five or six years old, I remember watching TV and seeing other children suffer in other parts of the world. I would say to myself, “When I grow up, when I can get rich, I will save kids all over the world.”At 17, I started my career here in America, and by the age of 18, I started my first charity organization. I went on to team up with other organizations in the following years, and met, helped, and even lost some of the most beautiful souls, tern six-year-old Jasmina Anema who passed away in 2010 from leukemia (白血病), whose story inspired thousands to volunteer as donors, to 2012 when grandmother lost her battle with cancer, which is the very reason and the driving force behind the Clara Lionel Foundation( CLF). We’re all human. And we all just want a chance: a chance at life, a chance in education, a chance at a future, really. And at CLF, our mission is to impact as many lives as possible, but it starts with just one.People make it seem too hard to do charity work. The truth is, you don’t have to be rich to help others. You don’t need to be famous. You don’t even have to be college -educated. But it starts with your neighbor, the person right next to you, the person sitting next to you in class, the kid down the block in your neighborhood. You just do whatever you can to help in any way that you can. And today, I want to challenge each of you to make a commitment to help one person,one organization,one situation that touches your heart.My grandmother always used to say, “If you’ve got a dollar, there’s plenty to share.”24. What did the author want to do at a young ageA. Watch TV.B. Grow up quickly.C. Become wealthy.D. Help other children.25. What directly caused the author to create and develop the CLFA. A six-year-old kid’s request.B. Her grandmother’s death of cancer.C. Many volunteers’ inspiration.D. Other organizations’ encouragement.26. What does the underlined word “one” in Paragraph 2 refer toA. A chance.B. A task.C. A life.D. An organization.27. What does the author suggest people do in the last paragraphA. Do little things to help those around them.B. Work hard to get a college education.C. Challenge their friends to offer help.D. Do charity work whatever you are.CAlongside air and water, food is a necessity for human beings to survive and thrive. But it’s a lot more than that. As Mariette Dichristina of Scientific American wrote: “The most intimate (亲密的) relationship we will ever have is not with any fellow human being. Instead, it is between our bo dies and our food.”Nowadays, for most people in the worlds wealthiest countries, food is a hobby, an enthusiasm, and even something fashionable.Turn on the TV in the US, UK or France, and you’ll find at least one channel feeding this popular obsession.A nd most of us know at least one person who thinks of themselves as a “foodie”. It’s almost impossible nowadays to check our social media apps without at least two or three photos of delicious meals appearing on our screen.But behind the fancy recipes and social media bragging (夸耀), many of us forget how much we take food for granted. This is why World Food Day is held each year.Take Kenya for example. This east African nation has been suffering terrible droughts. The result is that people are beginning to starve. Children in particular are suffering, with some of them even dying.This may seem shocking to know, especially as many cultures outside of Africa think of food in a completely different way. But even in the UK, families on low incomes are forced to use food banks—European organizations that hand out donated food to those who can’t afford to pay for it themselves.So what can we do on World Food Day One good way to spend it would be to feel humble and appreciate what we have. After all, food is essential for survival, but not everyone is as lucky as we are when it comes to dinner time.28. According to Mariette Dichristina, what has the closest relationship with usA. Air.B. Water.C. Food.D. Human beings.29. What does the underlined word “foodie” in Paragraph 4 probably meanA. Delicious food.B. A person fond of food.C. A social media app.D. A photo of delicious meals.30. Who can get help from food banksA. Poor people in the UK.B. Poor people in Africa.C. Starving children in Kenya.D. People in the drought-stricken.31. What’s the best title for the textA. Treat Food as a HobbyB. Time to Appreciate FoodC. Food Shortage in Some CountriesD. How to Spend World Food DayDTraditionally, robots have been hard, made of metal and other rigid material: But a team of scientists at Harvard University in the US has managed to build an entirely soft robot-one that draws inspiration from an octopus (章鱼).Described in science journal Nature, the “Octobot” cou ld pave the way for more effective autonomous robots that could be used in search,rescue and exploration. “The Octobot is minimal system which may serve as a foundation for a new generation of completely soft, autonomous robots” the study’s authors wrote.Robots built for precise, repetitive movements in a controlled environment don’t do so well on rough terrains (地形) or in unpredictable conditions. And they aren’t especially safe around humans, because they’re made out of hard and heavy parts that could be potentially dangerous to their users.So researchers have been working on building soft robots for decades. They’re taken inspiration from nature, looking to animals from jellyfish to cockroaches, which are often made up of more flexible matter.But creating a completely soft robot remains a challenge. Even if engineers build a silicone (硅酮) body, it’s still a grand challenge to construct flexible versions of essential parts, such as a source of power.“Although soft robotics is still in its early stage, i t holds great promise for several applications, such as search-rescue operations and exploration,” Barbara Mazzolai and Virgilio Mattoli of the Italian Institute of Technology’ Center for Micro-BioRobotics, wrote in a comment. “Soft robots might also open up new approaches to improving wellness and quality of life.”32. What’s the special feature of “Octobot”A. It’s soft.B. It’s made of metal.C. It’s very small.D. It looks like an octopus.33. What’s the disadvantage of traditional robotsA. They’re hard to control.B. They’re too heavy to move.C. They can’t predict conditions.D. They can’t behave well all the time.34. One of the biggest challenges is to build Octobot’s ________.A. silicone bodyB. complex componentsC. precise movementsD. flexible power source35. What’s the possible application of “Octobot”A. Medical research.B. Life rescue.C. Machine operation.D. House cleaning.第二节(共5小题;每小题2分,满分10分)根据短文内容,从短文后的选项中选出能填入空白处的最佳选项。