2017-2018 南昌二中高一上学期第一次月考试卷

合集下载

2018届江西省南昌市第二中学高三上学期第一次月考理科

2018届江西省南昌市第二中学高三上学期第一次月考理科

南昌二中2018—2018学年度上学期第一次考试高三数学(理)试卷一、选择题(本大题共12小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的)1.已知函数lg y x =的定义域为集合A ,集合{}01B x x =≤≤,则A B =( )A .(0,)+∞B .[0,1]C .[0,1) D .(0,1]2.已知α为第二象限角,且sin α=35,则tan(π+α)的值是( )A. 43B. 34 C .-43D .-343.下列说法正确的是( )A .命题“若x 2=1,则x =1”的否命题为:“若x 2=1,则x ≠1” B .已知()y f x = 是R 上的可导函数,则“0()0f x '=”是“0x 是函数()y f x =的极值点”的必要不充分条件C .命题“存在x ∈R,使得x 2+x +1<0”的否定是:“对任意x ∈R,均有x 2+x +1<0”D .命题“角α的终边在第一象限角,则α是锐角”的逆否命题为真命题4.已知角α终边上一点P 的坐标是(2sin 2,-2cos 2),则sin α等于( )A .sin 2B .-sin 2C .cos 2D .-cos 25.设21log 3a =,12b e -=,lnc π=,则( )A .c a b <<B .a c b <<C .a b c <<D .b a c <<6.设点P 是曲线3233+-=x x y 上的任意一点,P 点处切线倾斜角α的取值范围A .),65[)2,0[πππ B . ),32[ππ C .),32[)2,0[πππ D .]65,2(ππ7.将函数()sin 23f x x π⎛⎫=+ ⎪⎝⎭向右平移23π个单位,再将所得的函数图象上的各点纵坐标不变,横坐标变为原来的2倍,得到函数()y g x =的图象,则函数()y g x =与2x π=-,3x π=,x 轴围成的图形面积为( )A .12B .32C.1 D.1 8.已知函数25,(1)(),(1)x ax x f x a x x⎧---≤⎪=⎨ >⎪⎩是R 上的增函数,则a 的取值范围是( )A .3-≤a <0B .3-≤a ≤2-C .a ≤2-D .a <09.已知函数()x f y =是定义在R 上的偶函数,且()()11-=+x f x f ,当[]1,0∈x 时,()12-=x x f ,则函数()()ln 2xg x f x =-的零点个数为( )A .3B .4C .5D .610.若βα,都是锐角,且55cos =α,1010)sin(=-βα,则=βcos ( ) A .22B .102 C.22或102-D .22或10211.已知a ≤1-xx+ln x 对任意1[,2]2x ∈恒成立,则a 的最大值为( )A .0B .1C .2D .312.设函数()f x =(21)x e x ax a --+,其中1a <,若存在唯一的整数t ,使得()0f t <,则a 的取值范围是( ) A . 3,12e ⎡⎫-⎪⎢⎣⎭B . 33,24e ⎡⎫-⎪⎢⎣⎭ C . 33,24e ⎡⎫⎪⎢⎣⎭ D . 3,12e ⎡⎫⎪⎢⎣⎭二、填空题(本大题共4小题,每小题5分,共20分.把答案填在答题卡中对应题号后的横线上.)13.已知tan 2α=,则 2sin 2sin 2-αα= .14.已知函数()x f 的导函数为()x f ',且满足()()2'232xf x x f +=,则()'4f = .15. 在ABC ∆中,如果cos()2sin sin 1B A A B ++=,那么△ABC 的形状是________.16. 已知函数()2sin f x x ω=(其中常数0ω>),若存在12,03x π⎡⎫∈-⎪⎢⎣⎭,20,4x π⎛⎤∈ ⎥⎝⎦, 使得()()12f x f x =,则ω的取值范围为 .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10分)已知函数()sin()(,0,0)2f x A x x R πωϕωϕ=+∈><<的部分图象如图所示.(Ⅰ)求函数()f x 的解析式;(Ⅱ)求函数()f x 的单调递增区间.18.(本小题12分)已知函数223()m m f x x -++= ()m Z ∈是偶函数,且()f x 在(0,)+∞上单调递增.(1)求m 的值,并确定()f x 的解析式;(2)2()log [32()]g x x f x =--,求()g x 的定义域和值域。

江西省南昌市第二中学2017-2018学年高三上学期第一次月考数学(文)试题 Word版含答案

江西省南昌市第二中学2017-2018学年高三上学期第一次月考数学(文)试题 Word版含答案

南昌二中2017-2018学年度上学期第一次考试高三数学(文)试卷一、选择题:本大题共12小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的.1.已知全集U ={1,2,3,4,5},集合A ={2,3,4},B ={1,4},则(∁U A )∪B 为( ) A .{1} B .{1,5}C .{1,4}D .{1,4,5}2. 在单位圆中,面积为1的扇形所对的圆心角的弧度数为( ) A.1 B.2C.3D.43. 下列判断正确的是( )A. 若命题p 为真命题,命题q 为假命题,则命题“p q ∧”为真命题B. 命题“若0xy =,则0x =”的否命题为“若0xy =,则0x ≠”C. “1sin 2α=”是“6πα=”的充分不必要条件 D. 命题“,20xx ∀∈>R ”的否定是“ 00,20x x ∃∈≤R ”4. 已知AB =(-1,-2),BC =(-3,-4),则CA =( ) A. (4,6) B. (-4,-6) C. (2,2)D. (-2,-2)5. 已知曲线的一条切线的斜率为,则切点的横坐标为( )A .3B .2C .1D .6. 若△ABC 的三个内角满足sin A ∶sin B ∶sin C =2∶3∶4,则△ABC ( ) A .一定是锐角三角形 B .一定是直角三角形C .一定是钝角三角形D .可能是锐角三角形,也可能是钝角三角形7.已知lg lg 0a b +=,则函数xa x f =)(与x x gb log )(-=的图象可能是( )A B C D 8.已知函数()sin()f x x ωϕ=+(0,2πωϕ><)的部分图像如图所示,则()y f x = 的图象可由cos y x ω= 的图象 ( )A .向右平移3π个长度单位 B .向左平移3π个长度单位 C .向右平移6π个长度单位 D .向左平移6π个长度单位 9.定义运算⎪⎪⎪⎪⎪⎪a b cd =ad -bc .若cos α=17,⎪⎪⎪⎪⎪⎪sin α sin βcos α cos β=3314,0<β<α<π2,则β等于( )A. π12B. π6C. π4D. π3 10.如图,为测得河对岸塔AB 的高,先在河岸上选一点C ,使 C 在塔底B 的正东方向上,测得点A 的仰角为60°,再由点C 沿北偏东15°方向走10m 到位置D ,测得∠BDC=45°,则塔 AB 的高是( )(单位:m ) A .10B .10C .10D .1011.已知函数()sin cos f x a x b x =-(0ab ≠, x R ∈)在4x π=处取得最大值,则函数()4y f x π=-是( )A .偶函数且它的图象关于点(,0)π对称B .奇函数且它的图象关于点3(,0)2π对称 C .偶函数且它的图象关于点3(,0)2π对称 D .奇函数且它的图象关于点 (,0)π对称12.已知a 为常数,函数()()ln f x x x ax =-有两个极值点1212,()x x x x <,则( )A. 121()0,()2f x f x >>- B. 121()0,()2f x f x <<- C. 121()0,()2f x f x ><- D. 121()0,()2f x f x <>-二、填空题:本大题共4小题,每小题5分,共20分.13.已知函数2log ,0,()2,0xx x f x x >⎧=⎨<⎩,则1()4f f ⎛⎫= ⎪⎝⎭. 14. 已知向量(1,2)a =,5a b ⋅=,25a b -=,则||b = .15.已知函数()3sin f x x x x =--+,不等式()()sin cos20f m f θθ++>对任意02πθ⎛⎫∈ ⎪⎝⎭,都成立,则实数m 的取值范围 .16. 已知函数()cos sin 2f x x x =,下列命题中,其中正确命题的序号为(把你认为正确的序号都填上)_______.①()y f x =的图像关于点(,0)π中心对称; ②()y f x =的图像关于直线③()f x 的最大值为; ④()f x 既是奇函数,又是周期函数三、解答题:本大题共6小题,共60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)已知函数)(x f 12x π⎛⎫-⎪⎝⎭,x∈R. (I)求6f π⎛⎫-⎪⎝⎭的值;(II) 在平面直角坐标系中,以Ox 为始边作角θ,它的终边与单位圆相交于点P18.(本小题满分12分)已知ABC ∆的内角C B A ,,所对的边分别为c b a ,,,且3π=C .设向量),,(b a =)sin ,(sin A B n =, )2,2(--=a b p .(I) 若∥,求B ;(II) 若,⊥ABC ∆的面积为3,求边长c .19. (本小题满分12分)已知函数c bx ax x f ++=3)(在点2=x 处取得极值16-c . (I)求b a ,的值;(II)若)(x f 在[]3,3-上有两个零点,求c 的范围.20. (本小题满分12分)如图,在AOB ∆中,,,4,26AOB BAO AB D ππ∠=∠==为线段BA 的中点.AOC ∆由AOB ∆绕直线AO 旋转而成,记,0,2BOC πθθ⎛⎤∠=∈ ⎥⎝⎦.(I )证明:2COD AOB πθ=⊥当时,平面平面;(II )当三棱锥D BOC -的体积为1时,求三棱锥A BOC -的全面积.21. (本小题满分12分)已知()2()2cos()cos 2sin 1026f x x x x ππωωωω⎛⎫=-++-> ⎪⎝⎭,直线12y =与()f x 的图像交点之间最短距离为π.(I) 求()f x 的解析式及单调递增区间;(II)在△ABC 中,角A 、B 、C 所对的边分别是a 、b 、c 若有()2cos cos a c B b C -=,则求角B 的大小以及()f A 的取值范围.22.(本小题满分12分)已知函数(其中常数),(是圆周率) .(I )当时,求函数的单调递增区间;(II )当时,求函数在上的最小值,并探索:是否存在满足条件的实数,使得对任意的,恒成立。

江西省南昌市第二中学2017-2018学年高一上学期第一次月考物理试题 Word版含答案

江西省南昌市第二中学2017-2018学年高一上学期第一次月考物理试题 Word版含答案

南昌二中2017-2018学年度上学期第一次月考高一物理试卷一、本题共12小题,每小题4分,共48分.其中1-7为单选题,8-12为多选题,全部选对的得4分,选不全的得2分,有选错或不答的得0分. 1.关于速度、速度改变量、加速度,正确的说法是: A.速率是速度的大小,平均速率是平均速度的大小 B.物体运动的速度改变量越大,它的加速度一定越大 C.速度很大的物体,其加速度可以很小,也可以为零D.作变速直线运动的物体,加速度方向与运动方向相同,当物体加速度减小时,它的 速度也减小。

2.一辆汽车以20 m/s 的速度行驶,现因故刹车,并最终停止运动,已知汽车刹车过程的加速度大小是5 m/s 2。

则汽车从刹车起经过5s 所通过的距离是: A.60mB.50mC.40mD.30m3.已知心电图记录仪的出纸速度(纸带移动的速度)是2.5cm/s ,如图所示是仪器记录下来的某人的心电图,图中每个小方格的边长为0.5cm ,由此可知A .此人的心率约为75次/分B .此人的心率约为125次/分C .此人心脏每跳动一次所需时间约为0.75sD .此人心脏每跳动一次所需时间约为0.60s4.一个从静止开始作匀加速直线运动的物体,从开始运动起,连续通过三段位移的时间分别是1s 、2s 、3s ,这三段位移的长度之比和这三段位移上的平均速度之比分别是: A.1:22:32,1:2:3 B. 1:2:3,1:1:1 C. 1:23:33,1:22:32 D.1:3:5,1:2:35.某人欲估算飞机着陆时的速度,他假设飞机停止运动前在平直跑道上做匀减速运动,飞机在跑道上滑行的距离为s ,从着陆到停下来所用的时间为t ,则飞机着陆时的速度为: A .stB .2s t C .2s t D .s t到2st 之间的某个值 6.一质点自x 轴原点O 出发,沿正方向以加速度a 运动,经过t o 时间速度变为v 0,接着以–a 加速度运动,当速度变为–v o2 时,加速度又变为a ,直至速度变为v o 4 时,加速度再变为–a ,直至速度变为–v o8 ……,其v -t 图象如图所示,则下列说法中正确的是: A.质点一直沿x 轴正方向运动B.质点将在x 轴上—直运动,永远不会停止C.质点运动过程中离原点的最大距离为v o t oD.质点最终静止时离开原点的距离一定大于v o t o7.图示为一质点作直线运动的v t -图像,下列说法正确的是: A .整个过程中,E 点对应时刻离出发点最远 B .整个过程中,BC 段对应过程的加速度最大 C .在18s ~22s 时间内,质点的位移为24m D .在14s ~18s 时间内,质点的位移为34m8. 下列说法正确的是:A. 任何细小的物体都可以看作质点B. 两个时刻之间的间隔是一段时间C.第3s 末与第4s 初是同一时刻D. 参考系必须选取静止不动的物体9.现在的物理学中加速度的定义式为a = v t -v 0t ,而历史上有些科学家曾把相等位移内速度变化相等的单向直线运动称为“匀变速直线运动”(现称“另类匀变速直线运动”),“另类加速度”定义为A = v t -v 0s ,其中v 0和v t 分别表示某段位移s 内的初速度和末速度。

江西省南昌二中2018学年高一上学期第一次月考数学试卷

江西省南昌二中2018学年高一上学期第一次月考数学试卷

2018-2018学年江西省南昌二中高一(上)第一次月考数学试卷一、选择题(每小题5分,共60分.)1.已知集合A={x|x2﹣1=0},则下列式子表示不正确的是()A.1∈A B.{﹣1}∈A C.∅⊆A D.{1,﹣1}⊆A2.集合A={y|y=,B={x|x2﹣x﹣2≤0},则A∩B=()A.[2,+∞)B.[0,1]C.[1,2]D.[0,2]3.下列各组函数f(x)与g(x)的图象相同的是()A.f(x)=x,g(x)=()2B.与g(x)=x+2C.f(x)=1,g(x)=x0D.f(x)=|x|,g(x)=4.已知映射f:(x,y)→(x+2y,x﹣2y),在映射f下(3,﹣1)的原象是()A.(3,﹣1)B.(1,1)C.(1,5)D.(5,﹣7)5.若函数y=f(x)的定义域是[0,2],则函数g(x)=的定义域是()A.[0,1]B.[0,1)C.[0,1)∪(1,4] D.(0,1)6.已知f()=,则f(x)的解析式可取为()A.B.﹣C.D.﹣7.设函数f(x)=若f(f(t))≤2,则实数t的取值范围是()A.(﹣∞,]B.[,+∞)C.(﹣∞,﹣2]D.[﹣2,+∞)8.函数f(x)=(x∈R)的最小值为()A.2 B.3 C.2D.2.59.幂函数f(x)=(m2﹣4m+4)x在(0,+∞)为减函数,则m的值为()A.1或3 B.1 C.3 D.210.若函数y=x2﹣3x﹣4的定义域为[0,m],值域为[﹣,﹣4],则m的取值范围是()A.(0,4]B. C. D.11.设函数f(x)=,若互不相等的实数x1,x2,x3满足f(x1)=f(x2)=f(x3),则x1+x2+x3的取值范围是()A.(]B.()C.(]D.()12.设f(x)满足f(﹣x)=﹣f(x),且在[﹣1,1]上是增函数,且f(﹣1)=﹣1,若函数f(x)≤t2﹣2at+1对所有的x∈[﹣1,1],当a∈[﹣1,1]时都成立,则t的取值范围是()A.﹣≤t≤B.﹣2≤t≤2C.t≥或t≤﹣或t=0 D.t≥2或t≤﹣2或t=0二、填空题(每小题5分,共20分.)13.集合M={y|y=x2﹣1,x∈R},集合N={x|y=},则(∁R M)∩N=.14.函数y=的增区间为.15.个人取得的劳务报酬,应当交纳个人所得税.每月劳务报酬收入(税前)不超过800 800某人某月劳务报酬应交税款为800元,那么他这个月劳务报酬收入(税前)为元.16.函数f(x)=.给出函数f(x)下列性质:(1)函数的定义域和值域均为[﹣1,1];(2)函数的图象关于原点成中心对称;(3)函数在定义域上单调递增;(4)A、B为函数f(x)图象上任意不同两点,则<|AB|≤2.请写出所有关于函数f(x)性质正确描述的序号.三、解答题(共70分)17.已知函数f(x)=的定义域为集合A,B={x∈Z|2<x<10},C={x∈R|x<a或x>a+1}(1)求A,(∁R A)∩B;(2)若A∪C=R,求实数a的取值范围.18.函数.(1)若f(x)的定义域为R,求实数a的取值范围;(2)若f(x)的定义域为[﹣2,1],求实数a的值.19.已知集合A={x|2x2﹣5x﹣3≤0},函数f(x)=的定义域为集合B.(I)若A∪B=(﹣1,3],求实数a的值;(Ⅱ)若A∩B=∅,求实数a的取值范围.20.已知函数y=f(x)的定义域为[﹣1,1],且f(﹣x)=﹣f(x),f(0)=1,当a,b∈[﹣1,1]且a+b≠0,时>0恒成立.(1)判断f(x)在[﹣1,1]上的单调性并证明结论;(2)解不等式f(x+)<f()21.设二次函数f(x)=ax2+bx+c(a,b,c∈R)满足下列条件:①当x∈R时,f(x)的最小值为0,且f(x﹣1)=f(﹣x﹣1)恒成立;②当x∈(0,5)时,x≤f(x)≤2|x﹣1|+1恒成立.(I)求f(1)的值;(Ⅱ)求f(x)的解析式;(Ⅲ)求最大的实数m(m>1),使得存在实数t,只要当x∈[1,m]时,就有f(x+t)≤x 成立.22.(1)求证:函数y=x+有如下性质:如果常数a>0,那么该函数在(0,]上是减函数,在[,+∞)上是增函数.(2)若f(x)=,x∈[0,1],利用上述性质,求函数f(x)的值域;(3)对于(2)中的函数f(x)和函数g(x)=﹣x﹣2a,若对任意x1∈[0,1],总存在x2∈[0,1],使得g(x2)=f(x1),求实数a的值.2018-2018学年江西省南昌二中高一(上)第一次月考数学试卷参考答案与试题解析一、选择题(每小题5分,共60分.)1.已知集合A={x|x2﹣1=0},则下列式子表示不正确的是()A.1∈A B.{﹣1}∈A C.∅⊆A D.{1,﹣1}⊆A【考点】集合的包含关系判断及应用.【分析】先求出集合的元素,根据集合元素和集合关系进行判断.【解答】解;∵集合A={x|x2﹣1=0}={x|x2=1}={﹣1,1},∴1∈A,{﹣1}⊊A,∅⊆A,{1,﹣1}⊆A,∴B不正确.故选:B.2.集合A={y|y=,B={x|x2﹣x﹣2≤0},则A∩B=()A.[2,+∞)B.[0,1]C.[1,2]D.[0,2]【考点】交集及其运算.【分析】求出A中y的范围确定出A,求出B中不等式的解集确定出B,找出两集合的交集即可.【解答】解:由A中y=≥0,得到A=[0,+∞),由B中不等式变形得:(x﹣2)(x+1)≤0,解得:﹣1≤x≤2,即B=[﹣1,2],则A∩B=[0,2],故选:D.3.下列各组函数f(x)与g(x)的图象相同的是()A.f(x)=x,g(x)=()2B.与g(x)=x+2C.f(x)=1,g(x)=x0D.f(x)=|x|,g(x)=【考点】判断两个函数是否为同一函数.【分析】根据两个函数的定义域相同,对应关系也相同,即可判断它们是相同函数.【解答】解:对于A,函数f(x)=x(x∈R),与g(x)==x(x≥0)的定义域不同,所以不是相同函数;对于B,函数f(x)==x+2(x≠2),与g(x)=x+2(x∈R)的定义域不同,所以不是相同函数;对于C,函数f(x)=1,与g(x)=x0=1(x≠0)的定义域不同,所以不是相同函数;对于D,函数f(x)=|x|(x∈R),与g(x)==|x|(x∈R)的定义域相同,对应关系也相同,所以是相同函数.故选:D.4.已知映射f:(x,y)→(x+2y,x﹣2y),在映射f下(3,﹣1)的原象是()A.(3,﹣1)B.(1,1)C.(1,5)D.(5,﹣7)【考点】映射.【分析】设在映射f下(3,﹣1)的原象为(x,y),由题设条件建立方程组能够求出象(3,﹣1)的原象.【解答】解:设原象为(x,y),则有,解得,则(3,﹣1)在f 下的原象是(1,1).故选B.5.若函数y=f(x)的定义域是[0,2],则函数g(x)=的定义域是()A.[0,1]B.[0,1)C.[0,1)∪(1,4] D.(0,1)【考点】函数的定义域及其求法.【分析】根据f(2x)中的2x和f(x)中的x的取值范围一样得到:0≤2x≤2,又分式中分母不能是0,即:x﹣1≠0,解出x的取值范围,得到答案.【解答】解:因为f(x)的定义域为[0,2],所以对g(x),0≤2x≤2且x≠1,故x∈[0,1),故选B.6.已知f()=,则f(x)的解析式可取为()A.B.﹣C.D.﹣【考点】函数解析式的求解及常用方法.【分析】利用换元法,设,则x=,代入从而化简可得.【解答】解:已知f()=,设,则x=,那么:f()=转化为g(t)==,∴f(x)的解析式可取为f(x)=,故选C.7.设函数f(x)=若f(f(t))≤2,则实数t的取值范围是()A.(﹣∞,]B.[,+∞)C.(﹣∞,﹣2]D.[﹣2,+∞)【考点】分段函数的应用.【分析】运用换元法,令a=f(t),则f(a)≤2,即有或,分别解出它们,再求并集可得a≥﹣2.即有f(t)≥﹣2,则或,分别解出它们,再求并集即可得到.【解答】解:令a=f(t),则f(a)≤2,即有或,即有﹣2≤a≤0或a>0,即为a≥﹣2.即有f(t)≥﹣2,则或,即有t≤0或0<t,即有t≤.则实数t的取值范围是(﹣∞,].故选A.8.函数f(x)=(x∈R)的最小值为()A.2 B.3 C.2D.2.5【考点】函数的最值及其几何意义.【分析】令t=(t≥2),则y=t+在[2,+∞)上单调递增,即可求出结论.【解答】解:令t=(t≥2),则y=t+在[2,+∞)上单调递增,∴t=2,即x=0,函数f(x)=(x∈R)的最小值为2.5,故选D.9.幂函数f(x)=(m2﹣4m+4)x在(0,+∞)为减函数,则m的值为()A.1或3 B.1 C.3 D.2【考点】幂函数的性质.【分析】根据幂函数的定义和单调性求m即可.【解答】解:∵为幂函数∴m2﹣4m+4=1,解得m=3或m=1.由当x∈(0,+∞)时为减函数,则m2﹣6m+8<0,解得2<m<4.∴m=3,故选:C.10.若函数y=x2﹣3x﹣4的定义域为[0,m],值域为[﹣,﹣4],则m的取值范围是()A.(0,4]B. C. D.【考点】二次函数的性质.【分析】根据函数的函数值f()=﹣,f(0)=﹣4,结合函数的图象即可求解【解答】解:∵f(x)=x2﹣3x﹣4=(x﹣)2﹣,∴f()=﹣,又f(0)=﹣4,故由二次函数图象可知:m的值最小为;最大为3.m的取值范围是:[,3],故选:C11.设函数f(x)=,若互不相等的实数x1,x2,x3满足f(x1)=f(x2)=f(x3),则x1+x2+x3的取值范围是()A.(]B.()C.(]D.()【考点】分段函数的解析式求法及其图象的作法.【分析】先作出函数f(x)=的图象,如图,不妨设x1<x2<x3,则x2,x3关于直线x=3对称,得到x2+x3=6,且﹣<x1<0;最后结合求得x1+x2+x3的取值范围即可.【解答】解:函数f(x)=的图象,如图,不妨设x1<x2<x3,则x2,x3关于直线x=3对称,故x2+x3=6,且x1满足﹣<x1<0;则x1+x2+x3的取值范围是:﹣+6<x1+x2+x3<0+6;即x1+x2+x3∈(,6).故选D12.设f(x)满足f(﹣x)=﹣f(x),且在[﹣1,1]上是增函数,且f(﹣1)=﹣1,若函数f(x)≤t2﹣2at+1对所有的x∈[﹣1,1],当a∈[﹣1,1]时都成立,则t的取值范围是()A.﹣≤t≤B.﹣2≤t≤2C.t≥或t≤﹣或t=0 D.t≥2或t≤﹣2或t=0【考点】函数恒成立问题.【分析】有f(﹣1)=﹣1得f(1)=1,f(x)≤t2﹣2at+1对所有的x∈[﹣1,1]都成立,只需要比较f(x)的最大值与t2﹣2at+1即可.【解答】解:若函数f(x)≤t2﹣2at+1对所有的x∈[﹣1,1]都成立,由已知易得f(x)的最大值是1,∴1≤t2﹣2at+1⇔2at﹣t2≤0,设g(a)=2at﹣t2(﹣1≤a≤1),欲使2at﹣t2≤0恒成立,则⇔t≥2或t=0或t≤﹣2.故选:D.二、填空题(每小题5分,共20分.)13.集合M={y|y=x2﹣1,x∈R},集合N={x|y=},则(∁R M)∩N=[﹣,﹣1).【考点】交、并、补集的混合运算.【分析】求出M的补集,从而求出其和N的交集即可.【解答】解:M={y|y=x2﹣1,x∈R}={y|y≥﹣1},故∁R M={y|y<﹣1},集合N={x|y=}={x|﹣≤x≤},则(∁R M)∩N=[﹣,﹣1),故答案为:[﹣,﹣1).14.函数y=的增区间为[﹣5,﹣3] .【考点】复合函数的单调性.【分析】根据复合函数单调性之间的关系即可得到结论.【解答】解:由﹣x2﹣6x﹣5≥0得x2+6x+5≤0,解得﹣5≤x≤﹣1,故函数的定义域为[﹣5,﹣1],设t=﹣x2﹣6x﹣5,则y=为增函数,要求函数的增区间,根据复合函数单调性之间的关系即求t=﹣x2﹣6x﹣5,∵函数t=﹣x2﹣6x﹣5的对称轴为x=﹣3,∴函数t=﹣x2﹣6x﹣5的递增区间为[﹣5,﹣3],故答案为:[﹣5,﹣3]15.个人取得的劳务报酬,应当交纳个人所得税.每月劳务报酬收入(税前)不超过800 800某人某月劳务报酬应交税款为800元,那么他这个月劳务报酬收入(税前)为5000元.【考点】函数模型的选择与应用.【分析】通过设他这个月劳务报酬收入(税前)为x元,通过×20%=640确定x>4000,进而计算可得结论.【解答】解:设他这个月劳务报酬收入(税前)为x元,∵×20%=640,∴x>4000,∴(x﹣4000)×80%×20%=800,解得x=5000,故答案为:5000.16.函数f(x)=.给出函数f(x)下列性质:(1)函数的定义域和值域均为[﹣1,1];(2)函数的图象关于原点成中心对称;(3)函数在定义域上单调递增;(4)A、B为函数f(x)图象上任意不同两点,则<|AB|≤2.请写出所有关于函数f(x)性质正确描述的序号(2).【考点】函数单调性的判断与证明;函数的定义域及其求法;函数的值域.【分析】化简函数f(x),画出函数f(x)的图象,结合图象,对选项中的命题进行分析判断即可.【解答】解:∵函数f(x)=,∴,解得﹣1≤x≤1且x≠0,∴函数f(x)的定义域为[﹣1,0)∪(0,1],(1)错误;∵f(x)==作出函数f(x)图象,如图所示;由图象知函数f(x)的图象关于原点成中心对称,(2)正确;由图象知函数f(x)在[﹣1,0)上为单调增函数,在(0,1]上也是单调增函数,但在定义域[﹣1,0)∪(0,1]上不是增函数,如﹣1<1,但f(﹣1)=f(1)=0,故(3)错误;由图象知图象为两个四分之一个圆弧构成,且半径为1,最大为AB连线且过原点,最大值为2,最小为AB是0,但取不到,即0<|AB|≤2,故(4)错误.综上,正确的命题是(2).故答案为:(2).三、解答题(共70分)17.已知函数f(x)=的定义域为集合A,B={x∈Z|2<x<10},C={x∈R|x<a或x>a+1}(1)求A,(∁R A)∩B;(2)若A∪C=R,求实数a的取值范围.【考点】集合关系中的参数取值问题;交、并、补集的混合运算;函数的定义域及其求法.【分析】(1)先求出集合A,化简集合B,根据根据集合的运算求,(C R A)∩B;(2)若A∪C=R,则可以比较两个集合的端点,得出参数所满足的不等式解出参数的取值范围.【解答】解:(1)由题意,解得7>x≥3,故A={x∈R|3≤x<7},B={x∈Z|2<x<10}═{x∈Z|3,4,5,6,7,8,9},∴(C R A)∩B{7,8,9}(2)∵A∪C=R,C={x∈R|x<a或x>a+1}∴解得3≤a<6实数a的取值范围是3≤a<618.函数.(1)若f(x)的定义域为R,求实数a的取值范围;(2)若f(x)的定义域为[﹣2,1],求实数a的值.【考点】一元二次不等式的应用;函数的定义域及其求法.【分析】(1)要使f(x)的定义域为R,只需(1﹣a2)x2+3(1﹣a)x+6≥0在R上恒成立,然后讨论二次项系数可求出所求;(2)根据f(x)的定义域为[﹣2,1]可知(1﹣a2)x2+3(1﹣a)x+6≥0的解集为[﹣2,1],则(1﹣a2)x2+3(1﹣a)x+6=0的两个根为﹣2,1,然后利用根与系数的关系解之即可.【解答】解:(1)∵f(x)的定义域为R,∴(1﹣a2)x2+3(1﹣a)x+6≥0在R上恒成立当a=1时,6≥0恒成立当a=﹣1时,6x+6≥0在R上不恒成立,故舍去当a≠±1时,解得:﹣≤a<1综上所述:﹣≤a≤1(2)∵f(x)的定义域为[﹣2,1],∴(1﹣a2)x2+3(1﹣a)x+6≥0的解集为[﹣2,1],即(1﹣a2)x2+3(1﹣a)x+6=0的两个根为﹣2,1∴解得a=2故a的值为2.19.已知集合A={x|2x2﹣5x﹣3≤0},函数f(x)=的定义域为集合B.(I)若A∪B=(﹣1,3],求实数a的值;(Ⅱ)若A∩B=∅,求实数a的取值范围.【考点】子集与交集、并集运算的转换.【分析】(I)先化简A,B,利用A∪B=(﹣1,3],分类讨论,即可求实数a的值;(Ⅱ)若A∩B=∅,分类讨论,即可求实数a的取值范围.【解答】解:A={x|2x2﹣5x﹣3≤0}=[﹣,3],B={x|[x﹣(2a+1)][x﹣(a﹣1)]<0}且B≠∅(I)由题意有:①若2a+1=﹣1⇒a=﹣1,则B=(﹣2,﹣1),不符合题意;②若a﹣1=﹣1⇒a=0,则B=(﹣1,1),符合题意;∴a=0(Ⅱ)B≠∅⇒2a+1≠a﹣1⇒a≠﹣2①若2a+1<a﹣1⇒a<﹣2时,或2a+1≥3或a≥1∴a<﹣2②若a﹣1<2a+1⇒a>﹣2时,或a﹣1≥3或a≥4∴或a≥4综上,实数a的取值范围是或a≥4且a≠﹣2.20.已知函数y=f(x)的定义域为[﹣1,1],且f(﹣x)=﹣f(x),f(0)=1,当a,b∈[﹣1,1]且a+b≠0,时>0恒成立.(1)判断f(x)在[﹣1,1]上的单调性并证明结论;(2)解不等式f(x+)<f()【考点】函数单调性的判断与证明.【分析】(1)根据条件即可得出a﹣b≠0时,恒成立,进而得出恒成立,根据增函数定义即可得出f(x)在[﹣1,1]上单调递增;(2)根据(1)得出的f(x)的单调性,便可由得出,解该不等式即可得出原不等式的解集.【解答】解:(1)∵当a,b∈[﹣1,1],且a+b≠0时,恒成立;∴,且f(﹣b)=﹣f(b);∴;∴a<b时,f(a)<f(b);∴f(x)在[﹣1,1]上是单调增函数;(2)∵f(x)在[﹣1,1]上是单调增函数,且;∴;解得;故所求不等式的解集为.21.设二次函数f(x)=ax2+bx+c(a,b,c∈R)满足下列条件:①当x∈R时,f(x)的最小值为0,且f(x﹣1)=f(﹣x﹣1)恒成立;②当x∈(0,5)时,x≤f(x)≤2|x﹣1|+1恒成立.(I)求f(1)的值;(Ⅱ)求f(x)的解析式;(Ⅲ)求最大的实数m(m>1),使得存在实数t,只要当x∈[1,m]时,就有f(x+t)≤x 成立.【考点】函数恒成立问题;函数解析式的求解及常用方法;二次函数的性质.【分析】(1)由当x∈(0,5)时,都有x≤f(x)≤2|x﹣1|+1恒成立可得f(1)=1;(2)由f(﹣1+x)=f(﹣1﹣x)可得二次函数f(x)=ax2+bx+c(a,b,c∈R)的对称轴为x=﹣1,于是b=2a,再由f(x)min=f(﹣1)=0,可得c=a,从而可求得函数f(x)的解析式;(3)可由f(1+t)≤1,求得:﹣4≤t≤0,再利用平移的知识求得最大的实数m.【解答】解:(1)∵x∈(0,5)时,都有x≤f(x)≤2|x﹣1|+1恒成立,∴1≤f(1)≤2|1﹣1|+1=1,∴f(1)=1;(2)∵f(﹣1+x)=f(﹣1﹣x),∴f(x)=ax2+bx+c(a,b,c∈R)的对称轴为x=﹣1,∴﹣=﹣1,b=2a.∵当x∈R时,函数的最小值为0,∴a>0,f(x)=ax2+bx+c(a,b,c∈R)的对称轴为x=﹣1,∴f(x)min=f(﹣1)=0,∴a=c.∴f(x)=ax2+2ax+a.又f(1)=1,∴a=c=,b=.∴f(x)=x2+x+=(x+1)2.(3)∵当x∈[1,m]时,就有f(x+t)≤x成立,∴f(1+t)≤1,即(1+t+1)2≤1,解得:﹣4≤t≤0.而y=f(x+t)=f[x﹣(﹣t)]是函数y=f(x)向右平移(﹣t)个单位得到的,显然,f(x)向右平移的越多,直线y=x与二次曲线y=f(x+t)的右交点的横坐标越大,∴当t=﹣4,﹣t=4时直线y=x与二次曲线y=f(x+t)的右交点的横坐标最大.∴(m+1﹣4)2≤m,∴1≤m≤9,∴m max=9.22.(1)求证:函数y=x+有如下性质:如果常数a>0,那么该函数在(0,]上是减函数,在[,+∞)上是增函数.(2)若f(x)=,x∈[0,1],利用上述性质,求函数f(x)的值域;(3)对于(2)中的函数f(x)和函数g(x)=﹣x﹣2a,若对任意x1∈[0,1],总存在x2∈[0,1],使得g(x2)=f(x1),求实数a的值.【考点】函数恒成立问题;对勾函数.【分析】(1)利用函数的单调性的定义,直接证明即可.(2)转化函数的表达式为(1)的函数的形式,然后求解函数的值域即可.(3)利用函数的值域以及子集关系,列出不等式组求解即可.【解答】解:(1)证明:设,任取x1,x2∈(0,]且x1<x2,,显然,x1﹣x2<0,x1x2>0,x1x2﹣a<0,∴h(x1)﹣h(x2)>0,即该函数在∈(0,]上是减函数;同理,对任意x1,x2∈[,+∞)且x1<x2,h(x1)﹣h(x2)<0,即该函数在[,+∞)上是增函数;(2)解:,设u=2x+1,x∈[0,1],1≤u≤3,则,u∈[1,3].由已知性质得,当1≤u≤2,即时,f(x)单调递减,所以减区间为;同理可得增区间为;由f(0)=﹣3,,,得f(x)的值域为[﹣4,﹣3].(3)g(x)=﹣x﹣2a为减函数,故g(x)∈[﹣1﹣2a,﹣2a],x∈[0,1].由题意,f(x)的值域是g(x)的值域的子集,∴,∴.2018年1月8日。

2017-2018学年江西省南昌市第二中学高一上学期第一次月考化学试题

2017-2018学年江西省南昌市第二中学高一上学期第一次月考化学试题

南昌二中2017—2018学年度上学期第一次月考高一化学试卷(最新)可能用到的相对原子质量:H-1 C-12 N-14 O-16 Al-27 Ba-137 S-32 Mg-24 Cl-35.5 Na-23 Zn-65一、选择题(本题共16小题,每小题3分,共48分。

只有一个选项符合题目要求)1. 在盛放浓硫酸的试剂瓶的标签上应印有下列警示标记的是 ( ) A. B. C. D.2. 下列实验基本操作(或实验注意事项)中,主要是处于实验安全考虑的是( )A.实验剩余的药品不能放回原试剂瓶B.可燃性气体点燃前先验纯C.分液漏斗在使用前先检漏D.滴管不能交叉使用3.下列事故处理方法正确的是( )A .浓 NaOH 溶液溅到皮肤上,立即用水冲洗,然后涂上稀醋酸溶液B .为了使过滤速率加快,可用玻棒在过滤器中轻轻搅拌,加速液体流动C .蒸发食盐溶液时,发生液滴飞溅现象,应立即加水冷却D .制备蒸馏水的实验中,在蒸馏烧瓶中盛约1/3体积的自来水,并放入几粒沸石,收集蒸馏水时,应弃去开始馏出的部分4.下列说法正确的是(N A 代表阿伏加德罗常数的值)( )A.标况下,22.4L 任何气体所含的原子均为1molB.常温常压下,2g 氢气含氢原子数为2N AC.0.3 mol·L -1的Na2SO 4溶液中含有Na +和SO 2-4的总物质的量为0.9 molD. 1mol Mg 转化为MgCl 2必定有N A 个Cl 2分子参加了反应5.下列除杂质的方法可行的是( )A .除去CO 2中混有的少量CO ,通入适量氧气后点燃B .用溶解、过滤的方法分离KNO 3和NaCl 固体的混合物C .除去乙醇中的少量水,可以加入生石灰再蒸馏D .用酒精萃取溴水中的溴单质的操作可选用分液漏斗,而后静置分液6.某硫原子的质量是ag,12C 原子的质量是bg ,若N A 只表示阿伏加德罗常数的数值,则下列说法中正确的是( )①该硫原子的相对原子质量为12a b②mg 该硫原子的物质的量为Am aN mol ③该硫原子的摩尔质量是aN A g ④ag 该硫原子所含的电子数为16N AA .①③B .②④C .①②D .②③7. 下列条件下,两瓶气体所含原子数一定相等的是( )A. 同质量、不同密度的N 2和CO 2B. 同温度、同体积的H 2和N 2C. 同体积、同密度的C 2H 4和C 3H 6D. 同压强、同体积的N 2O 和CO 2 8. 下列溶液与100 mL 0.5 mol·L -1 NaCl 溶液中所含Cl -的物质的量浓度相同的是( )A .100 mL 0.5 mol•L -1 MgCl 2溶液B .200 mL 0.25 mol•L -1AlCl 3溶液 C .50 mL 1 mol•L -1 NaCl 溶液 D .25 mL 0.5 mol•L -1 HCl 溶液 9.标准状况下,a L 气体X 2和b L 气体Y 2恰好完全反应生成c L 气体Z ,若2a =6b =3c ,则Z 的化学式为( )A .XY 2B .X 2YC .X 3YD .XY 310. 三聚氰胺又名蛋白精[分子式:C 3N 3(NH 2)3,相对分子质量:126]是一种低毒性化工产品,婴幼儿大量摄入可引起泌尿系统疾患。

2017-2018学年江西省南昌市第二中学高一上学期第一次月考语文

2017-2018学年江西省南昌市第二中学高一上学期第一次月考语文

2017-2018学年江西省南昌市第二中学高一上学期第一次月考语文一、基础知识(27分,每小题3分)1.下列加点字的读音全部正确的一项是:()A.寥廓(liáo)隽永(jùn)颓圮(qǐ)浪遏飞舟(è)B.百舸(kě)火钵(bō)青荇(xìng)忸怩不安(ní)C.团箕(jī)悄然(qiāo)长篙(gāo)叱咤风云(zhà)D.彷徨(páng)漫溯(sù)淬火(cuì)自怨自艾(yì)2.下列词语完全正确的一组是()A.拜谒堤堰石碾弊帚自珍B. 偏袒凄惋濡缕刎颈之交C 荡漾凌侮沧茫瞋目而视D. 斑斓彘肩刀俎切齿拊心3.下列各句中加点成语的使用,全都正确的一组是()①过去五年间,中国快递市场的规模经历了一场疯狂增长,而国外快递巨头纷纷涌入中国市场后,国内快递业的转型已成燃眉之急。

②海参养殖、加工企业鱼目混珠,导致市场上的海产品品质各异,让人优劣难辨。

③春意甚浓了,但在北方还是五风十雨,春意料峭,一阵暖人心意的春风刚刚吹过,又来了一阵冷雨。

④艺术品是宇宙的精华,也是世界的缩影,是自然界的产物,也是自然界具体而微的表现。

⑤肖邦的前奏曲影响深远,后人为其中的每一首都起了个名字,当然这些名字中有不少穿凿附会的内容,但有些也确实与音乐的意境相吻合。

⑥台湾经济存在许多病灶,新“政府”提出的各项经济与产业政策,说好听是缓不济急,讲难听是望梅止渴、难见成效。

A.①②⑤ B.①④⑥ C.②③⑤ D.③④⑥4.下列各句中,没有语病的一项是( )A.毛泽东的故乡韶山一直以优越的历史地位、优美的自然环境,吸引着成千上万的中外游客,广泛受到不同年龄、不同肤色游客的关注。

B.自马尔克斯的《百年孤独》问世40多年来,曾经影响了中国几代人,莫言坦承,他的不少作品就受到了马尔克斯的影响。

C.实践证明,父母与子女能否消除代沟,关键要靠孩子引导自己的父母来认识和接受那些自己希望父母接受的新观点。

江西省南昌市第二中学2017-2018学年高一上学期第一次月考英语试题Word版含答案

江西省南昌市第二中学2017-2018学年高一上学期第一次月考英语试题Word版含答案

南昌二中 2017-2018 学年度上学期第一次月考高一英语试卷第Ⅰ卷(选择题,共110 分)第一部分听力(共两节,满分30分)听下边 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,并标在试卷的相应地点。

听完每段对话后,你都有10 秒钟的时间往返答相关小题和阅读下一小题。

每段对话仅读一遍。

1.What does the woman need to buy?A. BreadB. VegetablesC. Juice2.What does the man do?A. A teacherB. A doctorC. A weather reporter3.Which place does the man want to go to?A. A restaurantB. A bankC. A library4.What will the woman do tomorrow?A. Have a picnicB. Go to the hospitalC. Visit her uncle’ s house5.Why didn ’ t the man go to the concert last night?A.He was sickB.He was not interested in it.C.He didn ’ t think it worth the money.听下边 5 段对话或独白。

每段对话或独白后有一个小题,从题中所给的A, B,C 三个选项中选出最正确选项,并标在试卷的相应地点。

听每段对话或独白前,你将有时间阅读各个小题,每题 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 段资料,回答第6、 7题6.What will the speakers have at 11:30 tomorrow?A. Volleyball practiceB. Swimming lessonsC. Sailing lessons7.When will the speakers have tennis lessons tomorrow?A. In the morningB. At noonC. In the afternoon听第 7 段资料,回答第8、 9题8.Where does the woman come from?A. MadridB. GetafeC. New York9. What does the man think of living in his hometown?A. ExpensiveB. BoringC. Lucky听第 8 段资料,回答第10、 11、 12 题10. Where are the speakers?A. In a hotelB. In a hospitalC. In the woman’ s house11.How long can the bathroom be used?A. 7 hoursB. 8 hoursC.13 hours12.Where should the man put his wallet?A. In a boxB. Under his bedC. On a table听第 9 段资料,回答第13、14、 15、 16 题13.What are the speakers discussing?A.The way of studyingB.The habit of studying aloneC.The advantage of studying with friends14.What is the g irl ’ s opinion on studying with friends when the work is hard?A. It’ s noisyB. It’ s boringC. It’ s helpful15. What does the boy do with his friends to remember vocabulary?A. Make gamesB. Watch televisionC. Listen to music16.What do we know about the boy?A.He listens to the radio when studying.B.He doesn ’ t have a computer.C.He seldom uses the Internet.听第 10 段资料,回答第17、 18、 19、 20 题17.What is the speaker mainly talking about?A.The film industry in IndiaB.The brief introduction to Bollywood.C.The comparison between Bollywood and Hollywood.18.What does the speaker say about Bollywood?A. It is the biggest film industry in the world.B. It produces 29000 films a year.C. It pays the actors much.19.What do most Indians like doing during their free time?A. Doing sportsB. Singing and dancingC. Watching films20.How much do most Bollywood films cost?A. $ 2,000,000B.$ 5,000,000C. $500,000,000第二部分:英语知识运用(共两节,满分40 分)第一节:单项选择(共10 小题;每题 1 分,满分10 分)21.Now most of the shops in big city ____ open until late into the night.A. leaveB. stayC. is keptD. put22.This publishing firm is planning ______ school text book.A. a series new ofB. new a series ofC. a series newD. a new series of23.To be thought of as honest , you should always ______ .A. clear awayB. keep your words C make sense D make a go of it24.If you want to do international trade successfully, ______ of English is _______.A. good command; a mustB. a good command; a needC. a good command; a mustD. good command; must25.She pretended to be calm but _______she was more than nervous at the time she was beingquestioned.A. in actualB. actuallyC. as matter of factD. in a fact26.Some people are good at _____ voices on the phone while others ______.A. knowing; aren’ tB. recognizing; don’ tC. knowing; don’ tD. recognizing; aren’ t27.We would rather die than ______ difficulty.A. give up toB. give in toC. give way toD. give back to28.The determined look in their eyes told us that nothing could make them ______.A. change their mindB. to change their mindsC. change their mindsD. to change their mind29.The hunter insisted that he _____ a tiger and that a searching team ______ to hunt for it.A. has seen; be set upB. had seen; be set upC. saw; was set upD. should see; was set up30.Green tree frog, _____live in treesA. eventuallyB. unfortunatelyC. even thoughD. as the name suggests第二节完形填空(共20 小题,每题 1.5分,满分 30 分)People always say that the earlier one learns a language, the31it is to do so, in theory it is that,32, in my opinion, that refers to spoken language. Capability( 能力 ) to practice some essential( 基本的 )33of a language and read between the lines can only be trained through proper reading ways and hard work34 .So spending money to help35learn English may36up with disappointment.It is likely that the more you37, the more you are let down.The daughter of one of my friends38English in primary school,39her foreign teacher’blindness40psychology. She did not want to go on41English until middle school,42 a college student studying English slowly43her interest in the language.It is better to have the child learn Chinese than to have some difficulty44learning English for several years. Having been engaged in English education,45find that despite( 尽管 )their excellent46, many students have47command of English words and phrases. So I suggest that children48classical Chinese prose(散文 ), rather than49 them to learn English hurriedly. Otherwise,they may let go the best time to50the language ability of their mother tongue.31.A. easy B. difficult C. easier D. more difficult32.A. but B. however C. though D. yet33.A. opinions B. regards C. requests D. expressions34.A. step by step B. right away C. at once D. quickly35.A. people B. girls C. children D. boys36.A. begin B. start C. finish D. end37.A. pay B. get C. buy D. take38.A. loved B. liked C. disliked D. learned39.A. because of B. because C. instead of D. instead40.A. of B. at C. in D. to41.A. learning B. to learn C. with learning D. for learning42.A. while B. where C. when D. as43.A. introduced B. practiced C. explained D. developed44.A. in B. to C. at D. of45.A. He B. I C. She D. They46.A. pronunciation B. phrase C. language D. writing47.A. few B. less C. little D. fewer48.A. write B. do C. remember D. memorize49.A. have B. let C. cause D. make50.A. study B. improve C. learn D. master第三部分阅读理解(共 20 小题,每题 2 分,满分 40 分)第一节 (共 15 小题;每题 2 分,满分30 分)阅读以下短文,从每题所给的四个选项中,选出最正确选项,并在答题卡大将该项涂黑。

2018-2019学年江西省南昌市第二中学高一上学期第一次月考化学试题 含解析

2018-2019学年江西省南昌市第二中学高一上学期第一次月考化学试题 含解析

2018-2019学年江西省南昌市第二中学高一上学期第一次月考化学试题化学注意事项: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、单选题1.下列叙述正确的是A.摩尔是物质的量的单位B.氢的摩尔质量为2g/molC.NaOH的摩尔质量是40gD.2molH2O的摩尔质量是1molH2O摩尔质量的2倍2.下列叙述中,正确的是A.1molH2O中含有1molH2和1mol OB.等质量的O2和O3中所含的氧原子数相同C.等质量的CO与CO2中所含碳原子数之比为7︰11D.98g H2SO4溶解于500mL水中,所得溶液中硫酸的物质的量浓度为2 mol•L-13.下列装置或操作能达到实验目的的是A.检查装置气密性B.从碘的四氯化碳溶液中分离出碘C.分离甲苯和乙醇(互溶)D.氨气的吸收4.关于粗盐提纯的下列说法中正确的是A.溶解粗盐时,应尽量让溶液稀些,防止食盐不完全溶解B.将制得的食盐晶体转移到新制过滤器中用大量水进行洗涤C.当蒸发到剩余少量液体时,停止加热,利用余热将液体蒸干D.为除去粗盐中的Ca2+、Mg2+、SO42-,可依次向溶液中加入足量的碳酸钠溶液、氢氧化钠溶液、氯化钡溶液、稀盐酸5.下列分离混合物的实验计划中不正确的是A.分离乙酸(沸点77.1℃)与某种液态有机物(沸点120℃)的混合物——蒸馏B.从含有少量NaCl的KNO3溶液中提取KNO3——热水溶解、降温结晶、过滤C.用CCl4萃取水中的碘,待液体分层后——下层液体从下口放出,上层液体从上口倒出D.将溴水中的溴转移到有机溶剂中——加入酒精萃取6.下列叙述错误的是①用试管夹夹持试管时,试管夹从试管底部往上套,夹在离试管管口的1/3处②给盛有液体的体积为1/2容积的试管加热③把鼻孔靠近容器口去闻气体的气味④将试管平放,用纸槽往试管里送入固体粉末后,然后竖立试管⑤取用放在细口瓶中液体时,取下瓶塞倒放在桌面上,倾倒液体时,瓶上标签对着地面⑥将滴管垂直伸进试管内滴加液体⑦用坩埚钳夹取加热后的蒸发皿⑧稀释浓硫酸时,把水迅速倒入盛有浓硫酸的量筒中⑨检验装置的气密性时,把导管的一端浸入水中,用手捂住容器的外壁或用酒精灯微热A.②③⑤⑥⑧B.①④⑦⑨C.①④⑤⑦D.②④⑦⑨7.用N A表示阿伏加德罗常数的值,下列说法错误的是A.常温常压下,48gO2含有的氧原子数为3N AB.1.7g NH3含有的质子数为N AC.标准状况下,11.2L氦气和氢气的混合气含有的分子数为0.5N AD.1L 0.1 mol/LCH3CH2OH水溶液中含H原子数目为0.6N A8.设N A为阿伏加德罗常数的值,下列说法正确的是A.标准状况下,22.4L四氯化碳分子数为N AB.常温下,46g NO2、N2O4组成的混合气体中所含有的原子数为3N A姓名准考证号考场号座位号C.常温下,1mol/L的AlCl3溶液中含有的Cl-离子数为3N AD.18gNH4+离子中所含的电子总数为12N A9.下列溶液中Cl-浓度与50mL1mol/L AlCl3溶液中Cl-浓度相等的是A.150mL1mol/L的NaCl溶液B.75mL2mol/L的NH4Cl溶液C.150mL2mol/L的KCl溶液D.75mL1mol/L的FeCl3溶液10.在11g某化合物X2S中.含S2﹣0.1 mol.则X的相对原子质量为A.23 B.24 C.39 D.4011.等物质的量浓度的KCl、MgCl2、AlCl3三种溶液。

江西省南昌市高二英语上学期第一次月考试题(new)

江西省南昌市高二英语上学期第一次月考试题(new)

2017—2018学年度上学期第一次月考高二英语试卷第一部分听力(共两节,满分 30 分)第一节(共5小题;每小题1分,满分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,并标在试卷的相应位置.听完每段对话后,你都有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 does the man probably do?A。

A shop assistant. B。

A policeman。

C。

A postman。

2. How old is the man’s daughter?A。

Six months old。

B. One year old。

C。

Two years old。

3. When did the woman plan to go to Spain?A。

In spring. B. In summer. C. In autumn。

4. Where will the speakers go first?A。

A restaurant. B. A cinema。

C. A hospital.5。

What does the man think of the lecture?A。

It was interesting.B. It was far beyond his understanding。

C. It was long but easy to understand。

第二节 (共15小题;每小题1分,满分15分)听下面5段对话或独白。

每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。

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 does the man usually do at home?A。

江西省南昌市第二中学-2017-2018学年高三上学期第一次月考化学试题 Word版含答案

江西省南昌市第二中学-2017-2018学年高三上学期第一次月考化学试题 Word版含答案

南昌二中2017-2018学年度上学期第一次考试高三化学试卷需要用到的相对原子质量:H—1 C—12 N—14 O—16 Ne—20 Na—23 Mg—24Al—27 S—32 Cl—35.5 K—39 Fe—56一、选择题(每小题只有一个正确选项,每小题3分,共48分)1.下列我国古代的技术应用中,其不涉及化学反应的是()①非金属氧化物一定是酸性氧化物。

②依据丁达尔现象可将分散系分为溶液、胶体与浊液③利用金属钠可区分乙醇和乙醚④碱性氧化物一定是金属氧化物。

⑤某钾盐溶于盐酸,产生能使澄清石灰水变浑浊的无色无味气体,说明该钾盐是K2CO3⑥欲配制1L1.00mol/L的KCl溶液,可将74.5gKCl溶于1L水中⑦质子数、中子数和电子数都相同的粒子一定是同一种粒子。

A.全部B.①②③④⑦C.①②④D.③④AA.2.0gH218O与D2O的混合物中所含中子数为N AB.1 mol氢氧化铁胶粒所含氢氧化铁的分子数为N AC.标准状况下,5.6LCO2与足量Na2O2反应转移的电子数为0.5 N AD.足量的Fe和1mol Cl2完全反应,转移的电子数为3N A5622慢慢通入CO2至过量,下列有关说法正确的是( )A. 整个过程中共发生四个离子反应。

B. 根据现象可判断酸性强弱为:H2CO3>HAlO2>HCO3-C. 通入3molCO2和通入5molCO2都只产生3mol沉淀D. 整个过程中生成沉淀的物质的量与通入CO 2的体积的图象如右图所示:7.某溶液中可能含有Na +、NH +4、Ba 2+、SO 2-4、I -、S 2-。

分别取样:①用pH 计测试,溶液显弱酸性;②加氯水和淀粉无明显现象。

为确定该溶液的组成,还需检验的离子是 ( )A .Na +B .SO 2-4C .Ba 2+D .NH +48.已知非金属单质在碱性条件下易发生岐化反应,而其生成物在酸性条件下能够发生归中反应。

2018-2019学年江西省南昌市第二中学高一上学期第一次月考数学试题(解析版)

2018-2019学年江西省南昌市第二中学高一上学期第一次月考数学试题(解析版)

2018-2019学年江西省南昌市第二中学高一上学期第一次月考数学试题一、单选题 1.设集合,,则( )A .B .C .D .【答案】D 【解析】 【分析】先解不等式得到集合,然后再求出即可.【详解】由题意得,∴.故选D . 【点睛】本题考查集合的交集运算,考查运算能力,解题的关键是是通过解不等式得到集合,属于基础题.2.已知集合2{|320,},{|05,}A x x x x R B x x x N =-+=∈=<<∈,则满足条件A CB ⊆⊆的集合C 的个数为( )A . 1B . 2C . 3D . 4 【答案】D【解析】试题分析:解得1x =或2x =.所以{}1,2A =.又{}1,2,3,4B =,所以满足A C B ⊆⊆的集合C 可能为{}{}{}{}1,2,1,2,3,1,2,4,1,2,3,4共4个.故D 正确.【考点】集合的运算.3.函数的定义域为,则函数的定义域是()A.B.C.D.【答案】A【解析】【分析】根据函数定义域的概念求解,令,求得的范围后即为函数的定义域.【详解】由题意,令,解得,∴函数的定义域是.故选A.【点睛】解答类似问题时注意两点:(1)函数的定义域时指自变量x的取值范围.(2)若y=f(t)的定义域为(a,b),则解不等式得a<q(x)<b即可求出y=f(q(x))的定义域;若y=f(g(x))的定义域为(a,b),则求出g(x)的值域即为f(t)的定义域.4.已知函数,则()A.0 B.C.1 D.0或1【答案】C【解析】【分析】先求出,再求出即为所求.【详解】由题意得,∴.故选C.【点睛】已知分段函数的解析式求函数值时,首先要判断出自变量所在的范围,然后选择相应的解析式代入求解即可得到所求的函数值.5.点在映射下的对应元素为,则在作用下点的原象是()A.B.C.D.【答案】D【解析】【分析】设原象为,则在映射下的对应元素为,结合题意得到关于的方程组,解方程组可得所求.【详解】设原象为,则该点在映射下的对应元素为,由题意得,解得,∴在作用下点的原象是,故选D.【点睛】解题的关键是弄清映射中的对应关系,然后由此得到方程组,求解方程组后即可得到结果,本题考查对映射概念的理解和运用.6.函数的值域是()A.[0,+∞) B.(-∞,0] C.D.[1,+∞)【答案】C【解析】【分析】用换元法转化为求二次函数的值域求解或根据函数的单调性求解.【详解】方法一:设,则,∴,∴函数在上单调递增,∴,∴函数的值域是.故选C.方法二:由得,∴函数的定义域为,又由题意得函数为增函数,∴,∴函数的值域是.故选C.【点睛】对于一些无理函数,可通过换元转化为有理函数(如二次函数),再利用有理函数求值域的方法解决问题,“换元法”的实质是等价转化的思想方法,解题中要注意新元的范围.7.已知A,B是非空集合,定义,()A.B.(-∞,3] C.( -∞,0)∪(0,3) D.( -∞,3)【答案】A【解析】【分析】根据条件分别求出集合,然后按照定义求出即可.【详解】由题意得,,∴,∴.故选A.【点睛】本题属于集合中的新定义问题,旨在考查接受和处理新信息的能力,解题时要充分理解题目的含义,进行全面分析、灵活处理.8.已知函数则()A.B.C.D.【答案】C【解析】【分析】先求出函数的解析式,然后再求出函数值.【详解】由题意得,∴,∴.故选C.【点睛】解答本题的关键是求出函数的解析式,已知的解析式,求的解析式时,一般用换元法求解,即令,然后用表示出,得到的解析式,再把换为即可,解题中要注意新元的范围.9.已知函数y=ax2+bx+c,如果a>b>c且a+b+c=0,则它的图象可能是( )A.B.C.D.【答案】D【解析】【分析】根据和可得到的符号,然后再根据四个选项中的抛物线的开口方向和图象与y轴的交点进行判断即可得到结论.【详解】∵且,∴,∴抛物线的开口向上,与y轴的交点在负半轴上,∴选项D符合题意.故选D.【点睛】本题考查函数图象的识别,考查分析问题和理解问题的能力,解题的关键是由题意得到的符号,然后再根据抛物线的特征进行判断.10.设M={a,b,c},N={﹣2,0,2},从M到N的映射满足f(a)>f(b)≥f(c),这样的映射f的个数为()A.1 B.2 C.4 D.5【答案】C【解析】【分析】由题意及映射概念逐一写出满足条件的映射后可得答案.【详解】∵,∴a对应2时,b对应0,c对应0或−2,有2个映射;a对应2时,b对应−2,c对应−2,有1个映射;a对应0时,b对应−2,c对应−2,有1个映射.综上,满足条件的映射个数为4个.故选C . 【点睛】本题考查映射的概念,考查理解和运用的能力,解题的关键是根据定义确定出各种对应的情况,通过列举得到结果. 11.已知函数()f x =[)12,2,x x ∈+∞,都有不等式()()21210f x f x x x ->-成立,则实数a 的取值范围是 ( )A . ()0,+∞B . 1,2⎡⎫+∞⎪⎢⎣⎭ C . 10,2⎛⎤⎥⎝⎦ D . 1,22⎡⎤⎢⎥⎣⎦【答案】D【解析】因为函数()fx =对任意两个不相等的实数[)12,2,x x ∈+∞,都有不等式()()21210f x f x x x ->-成立,所以函数()f x =[)2,+∞上第增, 0a =时不合题意,只需2{222560 222a a a a>⨯-⨯-+≥--≤ ,解得122a ≤≤ ,即实数a 的取值范围是1,22⎡⎤⎢⎥⎣⎦,故选D.12.对于实数x ,符号[x]表示不超过x 的最大整数,例如[π]=3,[﹣1.08]=﹣2,定义函数f (x )=x ﹣[x],则下列命题中正确的是①函数f (x )的最大值为1; ②函数f (x )的最小值为0;③方程有无数个根; ④函数f (x )是增函数.A . ②③B . ①②③C . ②D . ③④ 【答案】A 【解析】 【分析】本题考查取整函数问题,在解答时要先充分理解[x]的含义,根据解析式画出函数的图象,结合图象进行分析可得结果. 【详解】画出函数f(x)=x−[x]的图象,如下图所示.由图象得,函数f(x)的最大值小于1,故①不正确;函数f(x)的最小值为0,故②正确;函数每隔一个单位重复一次,所以函数有无数个零点,故③正确;函数f(x)有增有减,故④不正确.故答案为:②③.【点睛】本题难度较大,解题的关键是正确理解所给函数的意义,然后借助函数的图象利用数形结合的方法进行求解.二、填空题13.已知,则函数的单调递增区间是_______.【答案】【解析】【分析】画出函数的图象,根据图象可得结果.【详解】由题意得,画出函数的图象如下图所示.由图象可得,函数的单调递增区间为.(填也可).【点睛】求函数的单调区间时,首先应注意函数的定义域,函数的单调区间都是其定义域的子集;其次掌握一次函数、二次函数等基本初等函数的单调区间.常用方法:根据定义,利用图象和单调函数的性质.14.已知函数的定义域是,则实数的取值范围是_______.【答案】【解析】【分析】由题意得在上恒成立,然后分和两种情况进行分析可得结果.【详解】∵函数的定义域是,∴在上恒成立,即在上恒成立.①当时,,在上恒成立.②当时,由题意得,∴,解得,综上.∴实数的取值范围是.【点睛】解答本题的关键是根据函数的定义域为R得到不等式恒成立,然后再结合分类讨论进行分析求解,考查转化和分析解决问题的能力.15.已知函数,记,则_____________【答案】42【解析】【分析】根据函数的特点先得到,然后将两式相加可得到的值.【详解】由题意得,∴.故答案为42.16.已知函数的定义域为,则可求的函数的定义域为,求实数m的取值范围__________.【答案】【解析】函数的定义域为,,令,则,由题意知,当时,,作出函数的图象,如图所示,由图可得,当或时,,当时,,时,实数的取值范围是,故答案为.三、解答题17.设A={x|2x2+ax+2=0},B={x|x2+3x+2a=0},A∩B={2}.(1)求a的值及集合A、B;(2)设集合U=A∪B,求(CuA)∪(CuB)的所有子集.【答案】(1)a=﹣5,A={2,},B={2,﹣5};(2)见解析【解析】【分析】(1)由题意得2∈A,2∈B,代入方程后可得,然后解方程可得集合A、B;(2)结合(1)中的结论得到(C u A)∪(C u B),然后写出它的所有子集即可.【详解】(1)根据题意得2∈A,2∈B,将x=2代入A中的方程得:8+2a+2=0,解得a=﹣5,∴A={x|2x2﹣5x+2=0}={2,},B={x|x2+3x﹣10=0}={2,﹣5}.(2)由题意得全集U=A∪B={2,,﹣5},A∩B={2},∴(C u A)∪(C u B)=∁U(A∩B)={,﹣5},∴(C u A)∪(C u B)的所有子集为,{﹣5},{},{﹣5,}.【点睛】本题考查集合的基本运算,解题的关键是正确地得到相关集合,再根据要求求解,属于基础题.18.已知二次函数= ,满足条件和=.(1)求函数的解析式.(2)若函数,当时,求函数的最小值.【答案】(1);(2)【解析】【分析】(1)由题意得到,再根据=两边比较系数可得,于是得到解析式.(2)求出函数的解析式,再根据抛物线的开口方向和对称轴与区间的关系,结合图象求解可得最小值.【详解】(1)由题意得,∴= ,∵,∴==,∴,解得,∴.(2)由(1)得,函数图象的①当,即时,函数在上单调递增,∴.②当,即时,函数在上单调递减,在上单调递增,∴.综上可得【点睛】(1)二次函数在闭区间上的最值主要有三种类型:轴定区间定、轴动区间定、轴定区间动,不论哪种类型,解决的关键是考查对称轴与区间的关系,当含有参数时,要依据对称轴与区间的关系进行分类讨论;(2)二次函数的单调性问题则主要依据二次函数图象的对称轴进行分析讨论求解. 19.已知函数()[]9,1,6,.f x x a a x a R x=--+∈∈ (1)若1a =,试判断并用定义证明()f x 的单调性; (2)若8a =,求()f x 的值域. 【答案】(1)单调递增;(2) []6,10【解析】试题分析:(1)当a=1时,由x ∈[1,6],化简f (x ),用单调性定义讨论f (x )的增减性;(2)当()981?6a f x x x ⎛⎫==-+ ⎪⎝⎭时,,利用对勾函数的图象与性质可得()f x 的值域. 试题解析:(1)当1a =时, ()[]9111,6f x x x x =--+∈ 9911x x x x=--+=-递增 证:任取[]12,1,6x x ∈且12x x < 则()()()()122121212112999x x f x f x x x x x x x x x --=--+=--=()2112910x x x x ⎡⎤-+>⎢⎥⎣⎦()()()21f x f x f x ∴>∴在[]1,6上单调递增.(2)当8a =时, ()999888816f x x x x x x x ⎛⎫=--+=--+=-+ ⎪⎝⎭令9t x x=+[]1,6x ∈ []6,10t ∴∈ ()[]166,10f x y t ∴==-∈所以()f x 的值域为[]6,10.点睛:证明函数单调性的一般步骤:(1)取值:在定义域上任取12,x x ,并且12x x >(或12x x <);(2)作差: ()()12f x f x -,并将此式变形(要注意变形到能判断整个式子符号为止);(3)定号:判断()()12f x f x -的正负(要注意说理的充分性),必要时要讨论;(4)下结论:根据定义得出其单调性.20.已知函数.(1)用分段函数的形式表示函数f (x );(2)在平面直角坐标系中画出函数f (x )的图象;在同一平面直角坐标系中,再画出函数的图象(不用列表),观察图象直接写出当x >0时,不等式f (x )>g(x )的解集.【答案】(1);(2)【解析】 【分析】(1)去掉绝对值符号可得所求的解析式;(2)在同一坐标系内画出函数和的图象,结合图象可得所求. 【详解】(1)当x≥0时,f (x )=1; 当x <0时,f (x )=x+1;所以.(2)同一坐标系内画出函数和函数图象,如下图所示.由上图可知当x >1时,f (x )>g (x ),∴不等式f (x )>(x >0)的解集为{x|x >1}. 【点睛】画函数图象的注意点:(1)熟练掌握几种基本函数的图像,如二次函数、反比例函数、指数函数、对数函数、幂函数等;(2)掌握平移变换、伸缩变换、对称变换、翻折变换等常用的方法技巧,可帮助我们简化作图过程.21.设定义在R 上的函数()f x 对于任意实数x y ,,都有()()()2f x y f x f y +=+-成立,且(1)1f =,当0x >时,()2f x <. (1)判断()f x 的单调性,并加以证明;(2)试问:当12x -≤≤时,()f x 是否有最值?如果有,求出最值;如果没有,说明理由;(3)解关于x 的不等式22()()(2)(2)f bx f b x f x f b -<-,其中22b >.【答案】(1)()f x 在R 上是减函数,证明见解析;(2)()f x 的最大值是3,最小值是0;(3)当b >2{|x x b<或}x b >,当b <解集为2{|}x b x b<<.【解析】试题分析:(1)任意实数12x x ,,且12x x <,不妨设21x x m =+,利用差比较法,计算21()()0f x f x -<,所以函数为减函数;(2)()f x 在[1,2]-上单调递减,所以()f x 有最大值(1)f -,有最小值(2)f .利用赋值法求出()(1)3,20f f -==;(3)化简不等式得22(2)(2)f bx b f b x x +<+,由于()f x 为减函数,所以2222bx b b x x +>+,()(2)0x b bx -->.由于22b >,b >b <b >2b b >,不等式的解集为2{|x x b <或}x b >;当b <2b b<,不等式的解集为2{|}x b x b<<.试题解析:(1)()f x 在R 上是减函数,证明如下:对任意实数12x x ,,且12x x <,不妨设21x x m =+,其中m >,则21111()()()()f x f x f x m f x -=+-=+-, ∴21()()f x f x <.故()f x 在R 上单调递减.………………4分(2)∵()f x 在[1,2]-上单调递减,∴1x =-时,()f x 有最大值(1)f -,2x =时,()f x 有最小值(2)f .在()()()f x y f x f y +=+-中,令1y =,得(1)()(1)2f x f x f f x +=+-=-, 故(2)(1)10f f =-=,(1)(0)1(1)2f f f =-=--,所以(1)3f -=. 故当12x -≤≤时,()f x 的最大值是3,最小值是0.………………6分 (3)由原不等式,得22()(2)()(2)f bx f b f b x f x +<+,由已知有22(2)2(2)2f bx b f b x x ++<++,即22(2)(2)f bx b f b x x +<+.∵()f x 在R 上单调递减,∴2222bx b b x x +>+,∴()(2)0x b b x-->.………………9分∵22b >,∴b >b <当b >2b b >,不等式的解集为2{|x x b <或}x b >;当b <2b b <,不等式的解集为2{|}x b x b<<.………………12分【考点】函数的单调性.【方法点晴】本题主要考查抽象函数单调性的证明.证明出单调性后利用单调性求解最值和不等式.对于函数单调区间的求解,一般要根据函数的表达形式来选择合适的方法,对于基本初等函数单调区间的求解,可以在熟记基本初等函数的单调性的基础上进行求解;对于在基本初等函数的基础上进行变化的函数,则可以采用利用函数图象求出相应的单调区间来求得;复合函数的单调区间的求得宜采用复合函数法(同增异减)的方法来求得.抽象函数单调性利用定义法来求解.22.已知函数f(x)是二次函数,不等式f(x)≥0的解集为{x|﹣2≤x≤3},且f(x)在区间[﹣1,1]上的最小值是4.(1)求f(x)的解析式;(2)设g(x)=x+5﹣f(x),若对任意的,均成立,求实数m的取值范围.【答案】(1);(2)【解析】【分析】(1)由不等式的解集为可得二次函数的零点为,进而得到二次函数的零点式解析式,然后再根据最小值为4得到相应的参数,于是可得解析式.(2)由(1)得函数的解析式,然后将不等式恒成立的问题通过分离参数求解,转化为二次函数的最值问题和解二次不等式的问题求解.【详解】(1)由f(x)≥0解集为{x|﹣2≤x≤3},∴二次函数的零点为,于是可设f(x)=a(x+2)(x﹣3)=a(x2﹣x﹣6),且a<0,∴函数图象的对称轴,且在上单调递增,在上单调递减,又,∴,解得,∴.(2)由(1)得g(x)=x+5+x2﹣x﹣6=x2﹣1.在上恒成立,即在上恒成立,整理得在上恒成立,所以在上恒成立.令,记,则y=﹣3t2﹣2t+1,二次函数图象的对称轴为,所以当时,y有最小值,且.所以,整理得(3m2+1)(4m2﹣3)≥0,解得或,故实数m的取值范围为.【点睛】(1)解决恒成立问题的常用方法是通过分离参数,转化为求函数最值的问题处理.(2)二次函数、二次方程与二次不等式统称“三个二次”,它们常结合在一起,而二次函数又是“三个二次”的核心,通过二次函数的图象贯穿为一体.因此,有关二次函数的问题,数形结合,密切联系图象是探求解题思路的有效方法.。

江西省南昌市第二中学2017-2018学年高三上学期第一次月考理数试题 Word版含解析

江西省南昌市第二中学2017-2018学年高三上学期第一次月考理数试题 Word版含解析

2017-2018学年一、选择题(本大题共12小题,每小题5分,共60分.在每小题给出的四个选项中,只有一项是符合题目要求的)1.已知函数lg y x =的定义域为集合A ,集合{}01B x x =≤≤,则A B =( )A .(0,)+∞B .[0,1]C .[0,1)D .(0,1]【答案】D 【解析】试题分析:由题意可知,{}|0A x x =>,则A B =(0,1],故选D.考点:集合的交集.2.已知α为第二象限角,且sin α=35,则tan(π+α)的值是( ) A.43B.34 C . 43-D .34-【答案】D考点:同角的基本关系. 3.下列说法正确的是( )A .“若x 2=1,则x =1”的否为:“若x 2=1,则x ≠1”B .已知()y f x = 是R 上的可导函数,则“0()0f x '=”是“0x 是函数()y f x =的极值点”的必要不充分条件C .“存在x ∈R,使得x 2+x +1<0”的否定是:“对任意x ∈R,均有x 2+x +1<0” D .“角α的终边在第一象限角,则α是锐角”的逆否为真【答案】B考点:1.的真假;2. 常用逻辑关系.4.已知角α终边上一点P 的坐标是(2sin 2,-2cos 2),则sin α等于( ) A .sin 2 B .-sin 2 C .cos 2 D .-cos 2【答案】D 【解析】试题分析:因为 2r =;由任意三角函数的定义:sin cos 2yrα==-,故答案是D. 考点:任意角的三角函数.5.设21log 3a =,12b e -=,lnc π=,则( )A .c a b <<B .a c b <<C .a b c <<D .b a c << 【答案】C 【解析】试题分析:因为1221log 01ln 3a b e c π-=<<=<<=,所以a b c <<.考点:1.对数;2.大小比较.6.设点P 是曲线3233+-=x x y 上的任意一点,P 点处切线倾斜角α的取值范围 A .),65[)2,0[πππ B . ),32[ππ C .),32[)2,0[πππ D .]65,2(ππ 【答案】C 【解析】试题分析:因23y x '=≥,故切线斜率k ≥α的取值范围是20,,23πππ⎡⎫⎡⎫⋃⎪⎪⎢⎢⎣⎭⎣⎭。

江西省南昌市第二中学20182019学年高一数学上学期第一次月考试题

江西省南昌市第二中学20182019学年高一数学上学期第一次月考试题

南昌二中2018—2019学年度上学期第一次月考高一数学试卷一、选择题(每小题5分,共60分.) 1.设集合则( )A .B .C .D .2.已知集合{}{}N x x x B R x x x x A ∈=∈=+-=,<<,,50|023|2,则满足条件B C A ⊆⊆的集合C 的个数为A.1B.2C.3D.43. 函数)(x f 的定义域为]1,0[,则函数)2(-x f 的定义域是( ) A. ]3,2[B. ]1,0[C. ]1,2[--D. ]1,1[-4.已知函数⎪⎩⎪⎨⎧∉∈=Qx Q x x f 01)(,则=)]([πf f ( )A. 0B. πC. 1D. 0或15.点),(y x 在映射f 下的对应元素为),(y x y x -+,则在f 作用下点)0,2(的原象是( ) A. )2,0(- B. )2,2( C. )1,1(- D. )1,1(6.函数的值域是( )A .[0,+∞)B .(-∞,0]C .D .[1,+∞)7. 已知A,B 是非空集合,定义{}2|,,|3A B x x A B x A B A x y x x ⎧⎫⎪⨯=∈∉==⎨⎪-⎩且若, {}|||,=B x x x A B =>-⨯则( )A.(,0)(0,3]-∞ B.(-∞,3]C.( -∞,0)∪(0,3)D.( -∞,3)8.已知函数2211()3,f x x xx+=++则(3)f =( ) .A .8 .B 9 .C 10.D 119.已知函数y =ax 2+bx +c ,如果a >b >c 且a +b +c =0,则它的图象可能是( )10.设M={a ,b ,c},N={﹣2,0,2},从M 到N 的映射满足f (a )>f (b )≥f(c ),这样的映射f 的个数为( ) A .1B .2C .4D .511.已知函数()6522+--=a x ax x f 对任意两个不相等的实数).2[,21∞+∈x x ,都有不等式()()01212>--x x x f x f 成立,则实数a 的取值范围是( )()]2,21[.]21,0(.C ,21[.,0.D B A )+∞+∞ 12.对于实数x ,符号[x]表示不超过x 的最大整数,例如[π]=3,[﹣1.08]=﹣2,定义函数f (x )=x ﹣[x],则下列命题中正确的是①函数f (x )的最大值为1; ②函数f (x )的最小值为0; ③方程()()21-=x f x G 有无数个根; ④函数f (x )是增函数. A. ②③ B.①②③C.②D.③④二、填空题(每小题5分,共20分.)13.已知()|3|-=x x f ,则函数的单调递增区间是_______. 14.已知函数()352-+-=ax ax x f 的定义域是R ,则实数a 的取值范围是_______. 15.已知函数3()1x f x x +=+,记(1)(2)(4)(8)(1024)f f f f f m +++++=1111()()()()2481024f f f f n ++++=,则m n += . 16.已知函数()142-+-x x f 的定义域为],0[m ,则可求的函数()12-x f 的定义域为]2,0[,求实数m 的取值范围__________.三、解答题(共70分)17.(本大题共10分)设A={x |2x 2+ax +2=0},B={x |x 2+3x +2a =0},A∩B={2}. (1)求a 的值及集合A 、B ;(2)设集合U=A∪B ,求(C u A )∪(C u B )的所有子集.18.(本大题共12分)已知二次函数()f x = 2ax bx c ++,满足条件()00f =和()()2f x f x +-=4x . (1)求函数()f x 的解析式.(2)若函数()()22+-=kx x f x g ,当[)1,x ∞∈+时,求函数()g x 的最小值.19.(本大题共12分)已知函数9()||,[1,6],.f x x a a x a R x=--+∈∈ (1)若1a =,试判断并用定义证明()f x 的单调性; (2)若8a =,求()f x 的值域.20.(本大题共12分)已知函数()4|x|x1f -+=x.(1)用分段函数的形式表示函数f(x);(2)在平面直角坐标系中画出函数f(x)的图象;在同一平面直角坐标系中,再画出函数g(x)= (x>0)的图象(不用列表),观察图象直接写出当x>0时,不等式f(x)>g(x)的解集.21.(本大题共12分)设定义在R 上的函数()f x 对于任意实数x y ,,都有()()()2f x y f x f y +=+-成立,且(1)1f =,当0x >时,()2f x <.(1)判断()f x 的单调性,并加以证明;(2)试问:当12x -≤≤时,()f x 是否有最值?如果有,求出最值;如果没有,说明理由;(3)解关于x 的不等式22()()(2)(2)f bx f b x f x f b -<-,其中22b >.22.(本大题共12分)已知函数f (x )是二次函数,不等式f (x )≥0的解集为{x|﹣2≤x≤3},且f (x )在区间[﹣1,1]上的最小值是4.(1)求f (x )的解析式;(2)设g (x )=x +5﹣f (x ),若对任意的]43,(--∞∈x ,()()()]m g x g m [41-x g m x g 2+≤-⎪⎭⎫⎝⎛均成立,求实数m 的取值范围.南昌二中2018—2019学年度上学期第一次月考高一数学试卷参考答案DDACD CACDC DA13.),3(+∞ 或者),3[+∞均可 14.]0,12(- 15. 42 16.[2,4] 17.解:(1)根据题意得:2∈A,2∈B,将x=2代入A 中的方程得:8+2a+2=0,即a=﹣5,则A={x|2x 2﹣5x+2=0}={2,0.5},B={x|x 2+3x ﹣10=0}={2,﹣5};........5分 (2)∵全集U=A∪B={2,0.5,﹣5},A∩B={2}, ∴(C u A )∪(C u B )=∁U (A∩B)={0.5,﹣5};∴(C u A )∪(C u B )的所有子集为∅,{0.5},{﹣5},{0.5,﹣5}.......10分 18.解析:(1)由题意得()()220,22c a x b x ax bx =+++--=442ax a b ++=4x , 即1,2a b ==-,∴()22f x x x =-. ...............6分(2)()1:]2,1[,2)1(22+=∈++-=k x x x k x x g 对称轴为①当()()k g x g k k 211011min -==≤≤+时,时,即②当()()1210112min +--=+=>+<k k k g x g k k 时,时,即综上,()⎩⎨⎧>+--≤-=0120212min k k k k k x g .............12分19.解:(1)当1a =时,9()|1|1[1,6]f x x x x =--+∈9911x x x x=--+=-递增 证:任取12,[1,6]x x ∈且12x x < 则1221212121129()99()()()x x f x f x x x x x x x x x --=--+=--=21129()[1]0x x x x -+>21()()()f x f x f x ∴>∴在[1,6]上单调递增.......6分(2)当8a =时,999()|8|88816()f x x x x x x x=--+=--+=-+令9tx x=+ [1,6]x ∈[6,10]t ∴∈()16[6,10]f x y t ∴==-∈所以()f x 的值域为[6,10]. .........12分 20.解:(Ⅰ)因为当x≥0时,f (x )=1; 当x <0时,f (x )=x+1;所以; .....4分(2)函数图象如图 ....8分由上图可知当x >1时,f (x )>g (x ),∴不等式f (x )>的解集为{x|x >1} ......12分21.解:(1)()f x 在R 上是减函数,证明如下:对任意实数12x x ,,且12x x <,不妨设21x x m =+,其中0m >,则211111()()()()()()2()()20f x f x f x m f x f x f m f x f m -=+-=+--=-<,∴21()()f x f x <.故()f x 在R 上单调递减.………………4分(2)∵()f x 在[1,2]-上单调递减,∴1x =-时,()f x 有最大值(1)f -,2x =时,()f x有最小值(2)f .在()()()2f x y f x f y +=+-中,令1y =,得(1)()(1)2()1f x f x f f x +=+-=-,故(2)(1)10f f =-=,(1)(0)1(1)2f f f =-=--,所以(1)3f -=. 故当12x -≤≤时,()f x 的最大值是3,最小值是0.………………8分 (3)由原不等式,得22()(2)()(2)f bx f b f b x f x +<+,由已知有22(2)2(2)2f bx b f b x x ++<++,即22(2)(2)f bx b f b x x +<+. ∵()f x 在R 上单调递减,∴2222bx b b x x +>+,∴()(2)0x b bx -->.……10分 ∵22b >,∴2b >或2b <-.当2b >时,2b b >,不等式的解集为2{|x x b <或}x b >; 当2b <-时,2b b <,不等式的解集为2{|}x b x b<<.22.解:(Ⅰ)由f (x )≥0解集为{x|﹣2≤x≤3},可设f (x )=a (x+2)(x ﹣3)=a (x 2﹣x ﹣6),且a <0 对称轴,开口向下,f (x )min =f (﹣1)=﹣4a=4,解得a=﹣1,f (x )=﹣x 2+x+6;…(4分)(Ⅱ)g (x )=x+5+x 2﹣x ﹣6=x 2﹣1,恒成立即对恒成立化简,即对恒成立…(8分) 令,记,则y=﹣3t 2﹣2t+1, 二次函数开口向下,对称轴为,当时y max =﹣,故…(10分)所以(3m 2+1)(4m 2﹣3)≥0,解得或…(12分)。

2017-2018学年江西省南昌二中高一(上)第一次月考英语试卷

2017-2018学年江西省南昌二中高一(上)第一次月考英语试卷

2017-2018学年江西省南昌二中高一(上)第一次月考英语试卷第一部分:听力(共两节,满分30分)1.(1.5分)How much will the man pay if he wants two shirts?A.35 dollars.B.60 dollars.C.70 dollars.2.(1.5分)Where are the speakers?A.In a restaurant.B.At a bus﹣stop.C.In a library.3.(1.5分)What does the man wish to do?A.Avoid the rush hour.B.Go to a park.C.Park his car.4.(1.5分)When does the second bus leave on Saturdays?A.At 7:30.B.At 8:30.C.At 9:30.5.(1.5分)What is the man's present job?A.A driver.B.A waiter.C.A businessman.6.(3分)听第6段材料,回答第6、7题.6.When will they play tennis?A.Tomorrow morning.B.Tomorrow evening.C.The day after tomorrow.7.What should the woman remember to bring?A.Some food.B.Some water.C.Extra tennis ball.7.(3分)听第7段材料,回答第8、9题.8.What kind of party will the man attend?A.Disco.B.Jazz.C.Birthday.9.How will the man go to the party?A.By bus.B.By bike.C.By taxi.8.(4.5分)听第8段材料,回答第10至12题.10.When does the conversation probably take place?A.In the morning.B.In the afternoon.C.In the evening.11.Who bought the tickets?A.The woman.B.Mary.C.The man.12.What can we know about the man from the conversation?A.He won't go to bed until he finished writing his report.B.He won't work on his report after the film.C.He will finish his report tomorrow.9.(6分)听第9段材料,回答第13至16题.13.Where are the speakers?A.In a park.B.At school.C.In the countryside.14.What do we know about the woman?A.She is homesick.B.She is angry.C.She is sick.15.What will they do this weekend?A.Drive to the countryside.B.Attend a party.C.Go back home.16.What will the woman bring to the barbecue?A.Some gifts.B.Some drinks.C.Moon cakes.10.(6分)听第10段材料,回答第17至20题.17.What has the speaker been doing recently?A.Helping to decorate.B.Fixing up his house.C.Working in the garden.18.Why will the speaker put in new light fixtures?A.The old lights don't light up the room very well.B.The old lights are all out of fashion.C.There are not enough lights in the house.19.What will the new addition include?A.A family room and a dining area.B.A living room and a toilet.C.A deck and a living room.20.What can we learn from the talk?A.A back door will be added to the house.B.The speaker intends to replace the roof.C.The speaker plans to replace a part of the carpet.第二部分:阅读理解(共两节)第一节(满分30分)阅读下列短文,从每题所给的四个选项(A、B、C和D)中,选出最佳选项,并在答题卡上将该项涂黑.11.(8分)English is a very interesting language.It has borrowed words from many other languages.Immigrants(移民)coming into the United States have contributed many words to the language,which have kept their original pronunciation."Coolie"and"kowtow"were taken from the Chinese language,"kamikaze from the Japanese,"shampoo"from India,"blitz"from German,"amigo"and"Los Angeles"from Spanish and so on.Many students have studied English for years,some as many as eight.However,some students still have difficulty in speaking fluent English.Some know many words but are unable to discern them when native speakers use them.In our Oral English classes we will focus on speaking and listening to native English speakers.For this reason,because we are trying to train your ears to hear English aNd your mouths to speak intelligible (易理解的)English,we will have a rule that ONLY ENGLISH will be spoken in our English classes.Anyone speaking Chinese in class will be required to pay a fine in order to encourage the speaking and understanding of English.If teachers enter a classroom and discover that anyone is speaking Chinese,they will require everyone in the room to pay the fine.It is everyone's job to follow the English ﹣Only rule.It is for your benefit.It is because we want to accustom (使习惯于)your ears to hearing English.Other subjects may be learned only from books but the only way to learn a foreign language is to SPEAK IT!Students are often nervous about speaking in class at first but we hope to make the classEs fun,so you will forget your nervousness and learn to speak out.Enjoy your classes.21.The first paragraph is mainly about.A.the difficulty of learning English.B.how interesting and various English isC.different words in different languagesD.the immigrants'contributions to America22.The underlined word"discern"in Para 2can be replaced by.A.understandB.noticeC.hearD.speak23.In the author's opinion,the only way to learn a language well is to.A.listen to it on the radioB.learn it from booksC.play games with itD.speak it often24.The article is probably aimed at.A.Japanese studentsB.German studentsC.Chinese studentsD.Indian students.12.(6分)Things to Do in Atlanta This WeekendStart the year off by experiencing something new this weekend!Here are our favorite events picked for this weekend.The Glenn Miller Orchestra at Spivey Hall,SaturdayThe music of Glenn Miller has an unusual sound.Combine that with the perfect sound affects at Spivey Hall,and you get a concert that is sure to have your toes tapping.And really,who doesn't love Challanooga Choo Choo?Garden Lights at the Atlanta Botanical Garden,SaturdayThis is the last night for this annual light show at the Atlanta Botanical Garden.It is particularly unusual that the magical sound of frogs in the conservatory(温室)is not from a recording.That magic is coming from the real frogs that live there.If it's no too crowded,stop along the way,close your eyes and just listen.Children's Workshop:Egyptian Hieroglyphs(象形文字),SaturdayDoes your child dream of becoming a historian when he grows up?If you answer"yes",you will want to take him to the Michael C.Carlos Museum.Participants will learn how to read and write ancient Egyptian hieroglyphs from Egyptologist Annie Shanley.Children will have an opportunity to discover themeaning of hieroglyphs on tomb reliefs and statues.Colin Mochrie at Dad's Garage Theatre,Saturday and SundaySet the tone for a great 2017with laughter.Colin Mochrie comes to Dad's Garage Theatre and brings his lightning﹣quick creativity and humor.If you want to look at all the events happening this weekend,check out our full events calendar.25.What makes the light show at the Atlanta Botanical Garden unusual?A.The perfect light effects.B.The sound from real frogs.C.The specially designed conservatory.D.The magical sound from a recording.26.What can tourists do at the Michael C.Carlos Museum?A.Interview Annie Shanley.B.Meet some sinners face to face.C.Learn to read ancient Egyptian writing.D.Carve hieroglyphs on tomb reliefs and statues.27.Which of the following can best describe the event to be held at Dad's Garage Theatre?A.Challenging.B.Educational.C.Historic.D.Amusing.13.(8分)Most people agree that honesty is a good thing.But does Mother Nature agree?Animals can't talk,but can they lie in other ways?Can they lie with their bodies and behavior?Animal experts may not call it lying,but they do agree that many animals,from birds to chimpanzees,behave dishonestly to fool other animals.Why?Dishonesty often helps them survive.Many kinds of birds are very successful at fooling other animals.For example,a bird called the plover sometimes pretends to be hurt in order to protect its young.When a predator(猎食动物)gets close to its nest,the plover leads the predator away from the nest.How?It pretends to have a broken wing.The predator followsthe"hurt"adult,leaving the baby birds safe in the nest.Another kind of bird,the scrub jay,buries its food so it always has something to eat.Scrub jays are also thieves.They watch where others bury their food and steal it.But clever scrub jays seem to know when a thief is watching them.So they go back later,unbury the food,and bury it again somewhere else.Birds called cuckoos have found a way to have babies without doing much work.How?They don't make nests.Instead,they get into other birds'nests secretly.Then they lay their eggs and fly away.When the baby birds come out,their adoptive parents feed them.Chimpanzees,or chimps,can also be sneaky.After a fight,the losing chimp will give its hand to the other.When the winning chimp puts out its hand,too,the chimps are friendly again.But an animal expert once saw a losing chimp take the winner's hand and start fighting again.Chimps are sneaky in other ways,too.When chimps find food that they love,such as bananas,it is natural for them to cry out.Then other chImps come running.But some clever chimps learn to cry very softly when they find food.That way,other chimps don't hear them,and they don't need to share their food.As children,many of us learn the saying"You can't fool Mother Nature."But maybe you can't trust her,either.28.A plover protects its young from a predator by.A.getting closer to its youngB.driving away the adult predatorC.leaving its young in another nestD.pretending to be injured29.By"Chimpanzees,or chimps,can also be sneaky"(paragraph5),the author means.A.chimps are ready to attack othersB.chimps are sometimes dishonestC.chimps are jealous of the winnersD.chimps can be selfish too30.Which of the following is true according to the passage?A.Some chimps lower their cry to keep food away from others.B.The losing chimp won the fight by taking the winner's hand.C.Cuckoos fool their adoptive parents by making no nests.D.Some clever scrub jays often steal their food back.31.Which of the following might be the best title of the passage?A.Do animals lie?B.Does Mother Nature fool animals?C.How do animals learn to lie?D.How does honesty help animals survive?14.(8分)A survey said the average Asian dad spent one minute a day with his children.I was shocked.I mean,a whole minute?Every day?Get real.Once a week maybe.The fact is,many Asian males are terrible at kid﹣related things.In fact,I am one of them.Child﹣rearing (养育)doesn't come naturally to guys.My mother knew the names of our teachers,best friends and crushes.My dad was only vaguely aware there were short people sharing the apartment.My mother bought healthy fresh food at the market every day.My dad would only go shopping when there was nothing in the fridge except a jar of butter.Then he'd buy beer.My mother always knew the right questions to ask our teachers.My dad would ask my English teacher if she could get us a discount on school fees.My mother served kid food to kids.My dad added chili sauce to everything,including our baby food.The truth is,mother have superpowers.My son fell off a wall once and hurt himself all over.I demanded someone bring me a computer so I could google what to do.My wife ignored me and did some sort of chanting(咏诵)phrase such as"Mummy kiss it better,"and cured 17separate injuries in less than 15seconds.Yes,mothers are incredible people,but they are not always correct.Yet honesty forces me to record the fact that mothers only know best 99.99percent of thetime.Here are some famous slip﹣ups.The mother of Bill Gates:"If you're going to drop out of college and hang out with your stupid friends,don't come running to me when you find yourself penniless."The mother of Albert Einstein:"When you grow up,you'll find that sitting around thinking about the nature of time and space won't pay the grocery bills."The mother of George W.Bush:"You'll never be like your dad,who became President of the United States and started his own war."32.The tone for the writer to write the passage is.A.serious.B.humorous.C.disapproving.D.critical.33.In paragraph 2the writer makes a comparison between mothers and fathers to prove that.A.females love kids more than malesB.child﹣rearing is difficult both for females and malesC.my dad is not interested in child﹣rearingD.males are not good at child﹣rearing34.What does the underlined word"slip﹣ups"in paragraph 4probably mean?.A.Stories.B.Shortcomings.C.Mistakes.D.Disadvantages.35.The last paragraph is mainly developed by.A.providing different examplesB.following the order of spaceC.making comparisonsD.analyzing causes.第二节(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.选项中有两项为多余选项.15.(10分)Improve your memoryDo you find yourself forgetting where you left your keys or blanking out information on important tests?(36)Before your next big exam,be sure to check out some of these tried and tested ways of improving memory.Focus your attention on the materials you are studying.Attention is one of the major parts of memory.In order for information to move from short﹣term memory into long﹣term memory,you need to actively attend to this information.(37)Structure and organize the information you are studying.Researchers have found that information is best organized in related groups.(38)Or you can make an outline of your notes and textbook readings to help group related concepts (概念).(39)When you are studying unfamiliar material,take the time to think about how this information relates to things that you already know.By establishing relationships between new ideas and previously existing memories,you can remember the recently lEarned information.Pay extra attention to difficult information.Remembering information at the beginning or end of an article is easy.However,recalling information from its middle can be difficult.(40)Another way is to try restructuring what you have learned so it will be easier to remember.When you come across an especially difficult concept,devote some extra time to memorizing the information.A.Use this approach in your own studies.B.Try grouping similar concepts and terms together.C.Relate new information to things you already know.D.Repeat the information you are studying to improve memory.E.Try to study in a quiet place and concentrate on your learning.F.Fortunately,there are things that you can do to help improve your memory.G.You can overcome this problem by spending extra time memorizing this information.第三部分:英语知识运用(共两节)第一节(满分30分)阅读下面短文,从短文后各题所给的四个选项(A、B、C 和D)中选出可以填入空白处的最佳选项,并在答题卡上将该项涂黑.16.(30分)Nancy had just got a job as a secretary in a company.Monday was the first day she went to work,so she was very (41)and arrived early.She(42)the door open and found nobody there."I am the (43)to arrive."She thought and came to her desk.She was surprised to find a bunch of (44)on it.They were fresh.She(45)them and they were sweet.She looked around for a(46)to put them in."Somebody has sent me flowers the very first day!"she thought (47)."But who could it be?"she began to (48).The day passed quickly and Nancy did everything with (49)interest.For the following days of the (50),the first thing Nancy did was to change water for the followers and then (51)her work.Then came another Monday.(52)she came near her desk she was overjoyed to see a(n)(53)bunch of flowers there.She quickly put them in the vase,(54)the old ones.The same thing happened again the next Monday.Nancy began to think of ways to find out the (55).On Tuesday afternoon,she was sent to hand in a plan to the(56).She waited for his directives(命令)at his secretary's(57).She happened to see on the desk a half﹣opened notebook,which(58):"In order to keep the secretaries(59),the company has decided that every (60) a bunch of fresh flowers should be put on each secretary's desk."Later,she was told that their general manager was a business management psychologist(心理学家).41.A.depressed B.encouraged C.excited D.surprised 42.A.turned B.pushed C.knocked D.forced 43.A.last B.second C.third D.first 44.A.keys B.grapes C.flowers D.bananas 45.A.smelled B.ate C.took D.held 46.A.vase B.room C.glass D.bottle 47.A.angrily B.quietly C.strangely D.happily 48.A.seek B.wonder C.Work D.ask 49.A.low B.little C.great D.general 50.A.month B.period C.year D.week 51.A.set about B.set up C.set out D.set off 52.A.Unless B.When C.Since D.Before 53.A.old B.red C.blue D.new 54.A.covering B.demanding C.replacing D.forbidding 55.A.sender B.receiver C.secretary D.waiter 56.A.assistant B.colleague C.employee D.manager 57.A.notebook B.desk C.office D.house 58.A.said B.written C.printed D.signed 59.A.at home B.on time C.in high spirits D.in lowspirits60.A.Sunday morning B.Monday morning C.Mondayafternoon D.Tuesday afternoon第二节(共5小题;每小题1分,满分5分)从A、B、C 和D四个选项,选出可以填入空白处的最佳选项,并在答题卡上将该项涂黑.17.(1分)You'd better _______ your score and see if you have passed the exam.()A.add up B.add to C.add up to D.add18.(1分)It was ___ that their parents had visited the city.()A.the first time B.for the first timeC.first time D.on the first time19.(1分)If better use is _____ of your spare time,you'll make progress.()A.spent B.made C.taken D.thought20.(1分)You can hardly imagine the _____ they went _____.()A.suffering;through B.concern;throughC.recovery;over D.worry;over21.(1分)Can you explain how it _______ that you missed the morning classes?()A.came across B.came to C.came up D.came about第三节(共10小题;每小题1分,满分10分)根据首字母或中文提示,在空白处填入单词的正确形式.22.(1分)The only clue to the(身份)of the killer was a half﹣smoked cigarette.23.(1分)The extremely weak patient had great trouble r from his serious illness.24.(1分)Hindi and English are the o languages of India.25.(1分)Taking plenty of exercises every day,he is always(充满活力的).26.(1分)It is generally required that a English﹣Chinese translator have a good c of both Chinese and English.27.(1分)The(航行)from America to France used to take two months.28.(1分)The old lady was g to the young man who had helped her.29.(1分)Though I haven't met him for many years,I could(辨认出)him immediately when I saw him in the crowd.30.(1分)The girl used to be shy,but is g getting active in group work and is more willing to express herself.31.(1分)Of these two basketball teams,the former comes from the US;the lcomes from England.第四部分:写作(共两节)第一节短文改错(满分10分)32.(10分)假定英语课上老师要求同桌之间交换修改作文,请你修改你同桌写的以下作文.文中共有10处语言错误,每句中最多有两赴,每处错误仅涉及一个单词的增加、删除或修改.增加:在缺词处加一个漏字符号(八),并在其下面写出该加的词.删除:把多余的词用斜线(\)划掉.修改:在错的词下划一横线,并在该词下面写出修改后的词.注意:1.每处错误及其修改均限一词.2.只允许修改10处,多者(从第11处起)不计分.My father is the man I respect most.Strict as he may be,however he never fails to show her care and consideration.Once I broke a neighbor window.Seeing nobody around,I ran away immediately.When Dad came home,he noticed my uneasiness and asked me what has happened.I could only tell him a truth.Instead of scold,he praised my honesty and then encouraged me to apologize our neighbor.I learned from this incident that not only does Dad take care of my health and he also teaches me what to be a good citizen.How luckily I am to have such a good father!第二节书面表达(满分25分)33.(25分)假定你是李华.你的笔友晓东是某学校一名高一学生.最近他来信说不知道如何学好高中英语,感到苦恼.请你给他写一封回信,提出你的建议.建议内容如下:1 尽力用英语交流,不要怕犯错误.2 坚持每天早晚朗读英语3 多读英文报纸,看英文电影.注意:1 词数100左右; 2 可以适当增加细节,使行文连贯;3 开头和结尾已给出,但不计入总词数.Dear Xiao Dong,I hope you will find these ideas useful.Yours,Li Hua.2017-2018学年江西省南昌二中高一(上)第一次月考英语试卷参考答案与试题解析第一部分:听力(共两节,满分30分)1.(1.5分)How much will the man pay if he wants two shirts?A.35 dollars.B.60 dollars.C.70 dollars.【解答】B2.(1.5分)Where are the speakers?A.In a restaurant.B.At a bus﹣stop.C.In a library.【解答】A3.(1.5分)What does the man wish to do?A.Avoid the rush hour.B.Go to a park.C.Park his car.【解答】C4.(1.5分)When does the second bus leave on Saturdays?A.At 7:30.B.At 8:30.C.At 9:30.【解答】C5.(1.5分)What is the man's present job?A.A driver.B.A waiter.C.A businessman.【解答】A6.(3分)听第6段材料,回答第6、7题.6.When will they play tennis?A.Tomorrow morning.B.Tomorrow evening.C.The day after tomorrow.7.What should the woman remember to bring?A.Some food.B.Some water.C.Extra tennis ball.【解答】AC7.(3分)听第7段材料,回答第8、9题.8.What kind of party will the man attend?A.Disco.B.Jazz.C.Birthday.9.How will the man go to the party?A.By bus.B.By bike.C.By taxi.【解答】BC8.(4.5分)听第8段材料,回答第10至12题.10.When does the conversation probably take place?A.In the morning.B.In the afternoon.C.In the evening.11.Who bought the tickets?A.The woman.B.Mary.C.The man.12.What can we know about the man from the conversation?A.He won't go to bed until he finished writing his report.B.He won't work on his report after the film.C.He will finish his report tomorrow.【解答】CBA9.(6分)听第9段材料,回答第13至16题.13.Where are the speakers?A.In a park.B.At school.C.In the countryside.14.What do we know about the woman?A.She is homesick.B.She is angry.C.She is sick.15.What will they do this weekend?A.Drive to the countryside.B.Attend a party.C.Go back home.16.What will the woman bring to the barbecue?A.Some gifts.B.Some drinks.C.Moon cakes.【解答】BAAC10.(6分)听第10段材料,回答第17至20题.17.What has the speaker been doing recently?A.Helping to decorate.B.Fixing up his house.C.Working in the garden.18.Why will the speaker put in new light fixtures?A.The old lights don't light up the room very well.B.The old lights are all out of fashion.C.There are not enough lights in the house.19.What will the new addition include?A.A family room and a dining area.B.A living room and a toilet.C.A deck and a living room.20.What can we learn from the talk?A.A back door will be added to the house.B.The speaker intends to replace the roof.C.The speaker plans to replace a part of the carpet.【解答】BAAB第二部分:阅读理解(共两节)第一节(满分30分)阅读下列短文,从每题所给的四个选项(A、B、C和D)中,选出最佳选项,并在答题卡上将该项涂黑.11.(8分)English is a very interesting language.It has borrowed words from many other languages.Immigrants(移民)coming into the United States have contributed many words to the language,which have kept their original pronunciation."Coolie"and"kowtow"were taken from the Chinese language,"kamikaze from the Japanese,"shampoo"from India,"blitz"from German,"amigo"and"Los Angeles"from Spanish and so on.Many students have studied English for years,some as many as eight.However,some students still have difficulty in speaking fluent English.Some know many words but are unable to discern them when native speakers use them.In our Oral English classes we will focus on speaking and listening to native English speakers.For this reason,because we are trying to train your ears to hear English aNd your mouths to speak intelligible (易理解的)English,we will have a rule that ONLY ENGLISH will be spoken in our English classes.Anyone speaking Chinese in class will be required to pay a fine in order to encourage the speaking and understanding of English.If teachers enter a classroom and discover that anyone is speaking Chinese,they will require everyone in the room to pay the fine.It is everyone's job to follow the English ﹣Only rule.It is for your benefit.It is because we want to accustom (使习惯于)your ears to hearing English.Other subjects may be learned only from books but the only way to learn a foreign language is to SPEAK IT!Students are often nervous about speaking in class at first but we hope to make the classEs fun,so you will forget your nervousness and learn to speak out.Enjoy your classes.21.The first paragraph is mainly about B.A.the difficulty of learning English.B.how interesting and various English isC.different words in different languagesD.the immigrants'contributions to America22.The underlined word"discern"in Para 2can be replaced by A.A.understandB.noticeC.hearD.speak23.In the author's opinion,the only way to learn a language well is to D.A.listen to it on the radioB.learn it from booksC.play games with itD.speak it often24.The article is probably aimed at C.A.Japanese studentsB.German studentsC.Chinese studentsD.Indian students.【解答】21﹣24 BADC21.C 细节理解题.从第一段的前两句话"English is a very interesting language.It has borrowed words from many other languages."英语是一门非常有趣的语言,它借用了许多其他语言的词汇.可找到答案,因此选B.22.B猜测词义题.根据句子some students still have difficulty in speaking fluent English.有些学生讲流利的英语仍然有困难.Some know many words but are unable to discern them when native speakers use them.句意中一些人知道很多词而不明白它们的意思,"discern"这个词的意思为"仔细识别,明白".可得知答案选A.23.D 细节理解题.从最后一句话中"but the only way to learn a foreign language is to SPEAK IT!"但是学习一门外语的唯一方法是说它.可得知答案选D.24.C 推理判断题.通过第二段中"Anyone speaking Chinese in class will be required to pay a fine in order to encourage the speaking and understanding of English."可知任何人在课上说汉语就会被要求罚款,推断出学英语的是中国学生,他们稍不注意就会把母语说出来.因此答案选C.12.(6分)Things to Do in Atlanta This WeekendStart the year off by experiencing something new this weekend!Here are our favorite events picked for this weekend.The Glenn Miller Orchestra at Spivey Hall,SaturdayThe music of Glenn Miller has an unusual sound.Combine that with the perfect sound affects at Spivey Hall,and you get a concert that is sure to have your toes tapping.And really,who doesn't love Challanooga Choo Choo?Garden Lights at the Atlanta Botanical Garden,SaturdayThis is the last night for this annual light show at the Atlanta Botanical Garden.It isparticularly unusual that the magical sound of frogs in the conservatory(温室)is not from a recording.That magic is coming from the real frogs that live there.If it's no too crowded,stop along the way,close your eyes and just listen.Children's Workshop:Egyptian Hieroglyphs(象形文字),SaturdayDoes your child dream of becoming a historian when he grows up?If you answer"yes",you will want to take him to the Michael C.Carlos Museum.Participants will learn how to read and write ancient Egyptian hieroglyphs from Egyptologist Annie Shanley.Children will have an opportunity to discover the meaning of hieroglyphs on tomb reliefs and statues.Colin Mochrie at Dad's Garage Theatre,Saturday and SundaySet the tone for a great 2017with laughter.Colin Mochrie comes to Dad's Garage Theatre and brings his lightning﹣quick creativity and humor.If you want to look at all the events happening this weekend,check out our full events calendar.25.What makes the light show at the Atlanta Botanical Garden unusual?B A.The perfect light effects.B.The sound from real frogs.C.The specially designed conservatory.D.The magical sound from a recording.26.What can tourists do at the Michael C.Carlos Museum?CA.Interview Annie Shanley.B.Meet some sinners face to face.C.Learn to read ancient Egyptian writing.D.Carve hieroglyphs on tomb reliefs and statues.27.Which of the following can best describe the event to be held at Dad's Garage Theatre?D A.Challenging.B.Educational.C.Historic.D.Amusing.【解答】25﹣27 BCD25.B.细节理解题.根据"It is particularly unusual that the magical sound of frogs in the conservatory(温室)is not from a recording.That magic is coming from the real frogs that live there"可知真实青蛙的声音让Atlanta Botanical Garden的灯光秀不同寻常.故选B.26.C.推理判断题.根据"Participants will learn how to read and write ancient Egyptian hieroglyphs from Egyptologist Annie Shanley."可知在Michael C.Carlos Museum,游客能学会阅读古埃及文字.故选C.27.D.推理判断题.根据"Colin Mochrie comes to Dad's Garage Theatre and brings his lightning﹣quick creativity and humor."可知Dad's Garage Theatre里发生的事情非常有趣.故选D.13.(8分)Most people agree that honesty is a good thing.But does Mother Nature agree?Animals can't talk,but can they lie in other ways?Can they lie with their bodies and behavior?Animal experts may not call it lying,but they do agree that many animals,from birds to chimpanzees,behave dishonestly to fool other animals.Why?Dishonesty often helps them survive.Many kinds of birds are very successful at fooling other animals.For example,a bird called the plover sometimes pretends to be hurt in order to protect its young.When a predator(猎食动物)gets close to its nest,the plover leads the predator away from the nest.How?It pretends to have a broken wing.The predator follows the"hurt"adult,leaving the baby birds safe in the nest.Another kind of bird,the scrub jay,buries its food so it always has something to eat.Scrub jays are also thieves.They watch where others bury their food and steal it.But clever scrub jays seem to know when a thief is watching them.So they go back later,unbury the food,and bury it again somewhere else.Birds called cuckoos have found a way to have babies without doing much work.How?They don't make nests.Instead,they get into other birds'nests secretly.Then they lay their eggs and fly away.When the baby birds come out,their adoptive parents feed them.。

高一化学第一次月考试卷含解析

高一化学第一次月考试卷含解析

南昌二中—上学期第一次考试高一化学试卷可能用到的相对原子质量:H:1 C:12 O:16 Cl:35.5 Na:23 N:14一:选择题(每小题只有一个正确答案,每小题3分,共48分)1.下列实验中,①pH试纸的使用,②过滤,③蒸发,④配制一定物质的量浓度溶液,均用到的仪器是A.蒸发皿B.玻璃棒C.试管D.分液漏斗2.下列仪器常用于物质分离的是①②③④⑤⑥A.①③⑤B.①②⑥C.②④⑤D.②③⑤3.下列从混合物中分离出其中的某一成分,所采取的分离方法正确的是A.利用氯化钾与碳酸钙的溶解性差异,可用溶解、过滤的方法除去碳酸钙B.由于碘在酒精中的溶解度大,所以可用酒精把碘水中的碘萃取出来C.水的沸点是100 ℃,酒精的沸点是78.5 ℃,所以可用加热蒸馏法使含水酒精变为无水酒精D.氯化钠的溶解度随温度的下降而减小,所以可用冷却法从热的含有少量氯化钾的氯化钠浓溶液中得到纯净的氯化钠晶体4.设N A表示阿伏加得德罗常数的数值,下列叙述中正确的是A.0.2 mol/L的Ba(NO3)2溶液1 L,含NO3-数为0.2N AB.常温常压下,22 .4 L氧气所含的原子数为2N AC.常温常压下,48 g O2、O3组成的混合气体所含的氧原子数为3N AD.1 L 0.1 mol/LNaCl溶液中所含的Na+的电子数为1.1N A5.100 mL 0.03 mol·L-1 Na2SO4溶液和50 mL 0.02 mol·L-1 Al2(SO4)3溶液混合,下列说法中正确的是(不考虑混合后溶液体积的变化)A.混合液中c(SO2-4)为0.04 mol·L-1C.混合后的溶液中含有0.003 mol Na+D.混合后溶液中c(Na+)为0.02 mol·L-16.粗食盐中除含有钙离子、镁离子、硫酸根离子等可溶性杂质外,还含有泥沙等不溶性杂质。

为了除去杂质在实验室中可将粗盐溶于水然后进行下列操作:(1)过滤;(2)加入过量的氢氧化钠溶液;(3)加入适量的盐酸;(4)加过量Na2CO3溶液;(5)加过量的BaCl2溶液,以上操作的正确顺序是A.(5)(4)(2)(1)(3)B.(1)(2)(3)(4)(5)C.(5)(4)(3)(2)(1)D.(1)(5)(4)(2)(3)7.在两个容积相同的容器中,一个盛有C2H4气体,另一个盛有N2和CO的混合气体。

江西省南昌市高一数学上学期第一次月考试题-人教版高一全册数学试题

江西省南昌市高一数学上学期第一次月考试题-人教版高一全册数学试题

2017—2018学年度上学期第一次月考高一数学试卷一、选择题(每小题5分,共60分。

) 1.下列给出的命题正确的是( )A.高中数学课本中的难题可以构成集合B.有理数集Q 是最大的数集C.空集是任何非空集合的真子集D.自然数集N 中最小的数是1 2.已知集合},02|{R x x xx M ∈≥-=,},1|{2R x x y y N ∈+==,则=)(N M C R ( )A.]2,0[B. ]2,0(C.)2,(-∞D. ]2,(-∞ 3.下面各组函数中表示同一函数的是( ) A .35x y -= 与 x x y 5-= B .122++=x x y 与 12y 2++=t tC .2)3(x y = 与 x y 3=D .22-•+=x x y 与 ()()22-+=x x y4.函数()0212)(++++=x x x x f 的定义域为( ) A.(-1,+∞) B.(-2,-1) ∪(-1,+∞) C.[-1,+∞) D.[-2,-1)∪(-1,+∞) 5.在映射中N M f →:,(){}Ry x y x y x M ∈>=,,,其中,(){}R y x y x N ∈=,,; )对应到中的元素(y x M ,)中的元素(y x xy N +,,则N 中元素(4,5)的原像为( )A.(4,1)B.(20, 1)C.(7,1)D.(1,4)或(4,1) 6.幂函数()132296m )(+-+-=m m x m x f ()∞+,在0上单调递增,则m 的值为( )A. 2B. 3C. 4D. 2或4 7.函数()[]⎩⎨⎧<+≥-=10,6,10,2)(x x F F x x x F ,则()5F 的值为( )A.10B. 11C. 12D. 138.如果2()(1)1f x mx m x =+-+在区间]1,(-∞上为减函数,则m 的取值X 围( )A .⎪⎭⎫⎢⎣⎡31,0B .⎥⎦⎤ ⎝⎛31,0C .⎪⎭⎫ ⎝⎛31,0 D.10,3⎡⎤⎢⎥⎣⎦9.已知)(x f 的图像关于y 轴对称,且在区间(]0-,∞单调递减,则满足)21()13(f x f <+的实数x 的取值X 围是( )A. [—,21—61)B.(—,21—61) C. [—,31—61) D. (—,31—61)10.若函数)(x f 满足对任意的)](,[m n m n x <∈,都有km x f kn≤≤)( 成立,则称函数)(x f 在区间)](,[m n m n <上是“被K 约束的”。

  1. 1、下载文档前请自行甄别文档内容的完整性,平台不提供额外的编辑、内容补充、找答案等附加服务。
  2. 2、"仅部分预览"的文档,不可在线预览部分如存在完整性等问题,可反馈申请退款(可完整预览的文档不适用该条件!)。
  3. 3、如文档侵犯您的权益,请联系客服反馈,我们会尽快为您处理(人工客服工作时间:9:00-18:30)。

南昌二中2017—2018学年度上学期第一次月考高一数学试卷命题人:孙涛 审题人:曹开文一、选择题(每小题5分,共60分。

) 1.下列给出的命题正确的是( )A.高中数学课本中的难题可以构成集合B.有理数集Q 是最大的数集C.空集是任何非空集合的真子集D.自然数集N 中最小的数是1 2.已知集合},02|{R x x xx M ∈≥-=,},12|{R x y y N x ∈+==,则=)(N M C R ( ) A.]2,0[ B. ]2,0( C.)2,(-∞ D. ]2,(-∞ 3.下面各组函数中表示同一函数的是( )A .35x y -= 与 x x y 5-=B .122++=x x y 与 12y 2++=t t C .2)3(x y = 与 x y 3= D .22-•+=x x y 与 ()()22-+=x x y4.函数()0212)(++++=x x x x f 的定义域为( ) A.(-1,+∞) B.(-2,-1) ∪(-1,+∞) C.[-1,+∞) D.[-2,-1)∪(-1,+∞) 5.在映射中N M f →:,(){}Ry x y x y x M ∈>=,,,其中,(){}R y x y x N ∈=,,; )对应到中的元素(y x M ,)中的元素(y x xy N +,,则N 中元素(4,5)的原像为( )A.(4,1)B.(20,1)C.(7,1)D.(1,4)或(4,1) 6.幂函数()132296m )(+-+-=m m xm x f ()∞+,在0上单调递增,则m 的值为( )A. 2B. 3C. 4D. 2或4 7.函数()[]⎩⎨⎧<+≥-=10,6,10,2)(x x F F x x x F ,则()5F 的值为( )A.10B. 11C. 12D. 138.如果2()(1)1f x mx m x =+-+在区间]1,(-∞上为减函数,则m 的取值范围( )A .⎪⎭⎫⎢⎣⎡31,0B .⎥⎦⎤ ⎝⎛31,0 C .⎪⎭⎫ ⎝⎛31,0 D.10,3⎡⎤⎢⎥⎣⎦9.已知)(x f 的图像关于y 轴对称,且在区间(]0-,∞单调递减,则满足)21()13(f x f <+的实数x 的取值范围是( ) A. [-,21-61) B.(-,21-61)C. [-,31-61)D. (-,31-61) 10.已知函数()()()25,1,1x ax x f x a x x⎧---≤⎪=⎨>⎪⎩是R 上的增函数,则a 的取值范围是( )A .30a -≤<B .2a ≤-C .32a -≤≤-D .0a <11.已知函数()()()21,143,1x x f x x x x ⎧-≤⎪=⎨-+>⎪⎩.若()()0≥m f f ,则实数m 的取值范围是( ) A .[]2,2- B.[)2,24,⎡-++∞⎣ C.2,2⎡-⎣ D .[][)2,24,-+∞12.若函数)(x f 满足对任意的)](,[m n m n x <∈,都有km x f kn≤≤)( 成立,则称函数)(x f 在区间)](,[m n m n <上是“被K 约束的”。

若函数22)(a ax x x f +-=在区间)0](,1[>a a a上是“被2约束的”,则实数a 的取值范围是( ) A .⎥⎦⎤ ⎝⎛3,321, B .]2,1( C . ⎥⎦⎤⎝⎛2323, D .]2,2( 二、填空题(每小题5分,共20分。

)13.已知全集U R =,集合{|2}A y y ==,2{|7120}B x x x =-+≤,则()U A C B = .14.函数()3212--=x x x f 的单调增区间为 .15. 函数212+-+=x x x y 的值域为 .16.给出下列命题:(1)若函数)(x f 的定义域为]2,0[,则函数)2(x f 的定义域为]4,0[;(2)已知集合{}{}1,0,1,,-==Q b a P ,则映射Q P f →:中满足()0=b f 的映射共有3个; (3)函数1()f x x=的单调递减区间是(,0)(0,)-∞+∞;(4)若(12)(12)f x f x +=-,则()f x 的图象关于直线1x =对称;(5)已知1x ,2x 是)(x f 定义域内的两个值,且12x x <,若12()()f x f x >,则)(x f 是减函数; 其中正确命题的序号是 . 三、解答题(共70分)17.已知集合{},62≤≤=x x A {}.213m x m x B <<+-= (1)若2=m ,求A ∪B, ()B A C R ⋂; (2)若A ∩B=A ,求m 的取值范围.18.已知函数()58-41-22+=x x x f .(1) 求函数()x f 的解析式;(2) 若关于x 的不等式()02432>++-t t x f 在[﹣1,2]上有解,求实数t 的取值范围;19.已知定义在()1,1-的函数()21ax bf x x+=+满足:()00=f ,且1225f ⎛⎫= ⎪⎝⎭. (1)求函数()f x 的解析式; (2)证明:()f x 在()1,1-上是增函数.20. 已知函数f (x )=1)1()1(22+++-x a x a ; (1)若f (x )的定义域为 (-∞,+∞), 求实数a 的范围; (2)若f (x )的值域为 [0, +∞), 求实数a 的范围21.已知二次函数c bx ax x f ++=2)(的图象经过点)0,2(-,且不等式221)(22+≤≤x x f x 对一切实数x 都成立. (1)求函数)(x f 的解析式;(2)若对一切]1,1[-∈x ,不等式)2()(x f t x f <+恒成立,求实数t 的取值范围.22.已知定义在区间()∞+,0上的函数()4()5f x t x x=+-,其中常数0t >. (1)若函数)(x f 分别在区间(0,2),(2,)+∞上单调,试求t 的取值范围; (2)当1t =时,方程()m x f =有四个不相等的实根1234,,,x x x x . ①证明:123416x x x x =;②是否存在实数,a b ,使得函数)(x f 在区间[]b a ,单调,且)(x f 的取值范围为[]mb ma ,,若存在,求出m 的取值范围;若不存在,请说明理由.答案1.C 难题不具有确定性,不能构造集合,A 错误; 实数集R 就比有理数集Q 大,B 错误; 空集是任何非空集合的真子集,故C 错误; 自然数集N 中最小的数是0,D 错误; 故选C .2.D (]()+∞∞-=⎭⎬⎫⎩⎨⎧⎩⎨⎧≠-≥-=⎭⎬⎫⎩⎨⎧≥-=,20,020)2(|02| x x x x x x x M ,{}()+∞=∈+==,1,12|R x y x N x ,()+∞=∴,2N M ,(]2,)(∞-=N M C R .3.B 对于A :两函数的值域不同; 对于B:两函数的三要素完全相同,故为同一函数; 对于C:两函数与的定义域不同; 对于D :两函数的定义域不同;故选项为B. 4.B 要使函数有意义,需使{10202≠+≥+≠+x x x 且,解得2->x 且1-≠x 故选B. 5. A 由{45==+xy y x 可得:{14==x y 或{41==x y ;又y x >,则{41==x y ,所以原像为(4,1),选A.6. C 由题意得:01319622>+-=+-m m m m ,且.,42C m m 故选(舍去)或者==⇒ 7.B ()()[]()()[]()1113159115=====F F F F F F F ,故选B.8.D 当0m =时,()1f x x =-,满足在区间(],1-∞上为减函数,当0m ≠时,由于()()211f x mx m x =+-+的图象对称轴为12mx m-=,且函数在区间(],1-∞上为减函数,0112m mm>⎧⎪∴-⎨≥⎪⎩,求得103m <≤,∈∴m 10,3⎡⎤⎢⎥⎣⎦,故选D. 9.B )(x f 的图像关于y 轴对称,且在区间(]0-,∞单调递减,则)(x f 在),0[+∞单调递增函数;再由)21()13(f x f <+,可得2113<+x ,解出即得6121-<<-x ;故选B . 10.C 由题意可得:⎪⎩⎪⎨⎧≤--<≥-aa a a60,12且,得⎩⎨⎧-≥-≤32a a ,即32a -≤≤-.故选C.11.B 因为令()f m n =,则()()0ff m ≥就是()0f n ≥.画出函数()f x 的图象可知,11n -≤≤,或3n ≥,即()11f m -≤≤或()3f m ≥.由11x -=-得,2x =或2x =-.由2431,2x x x -+==±.由2433x x -+=得,0x =或4x =.再根据图象得到[)2,24,m ⎡-+∞⎣∈+,故选D.12.B 据题意得:22122x ax a a a ≤-+≤对任意的x ∈)0](,1[>a a a 都成立.由1a a>得1a >. 22111()12112f a a a a=-+>-=>恒成立. 由222()2f a a aa a a a =-+=≤得2a ≤.因为1a >,所以222211()111f a a a a a=-+<-+=.22()f x x ax a =-+的对称轴为2a x =.由231()242a a f a =≥得a ≥.1<,所以a 的取值范围为]2,1(.选B.13.[2,3) {|2}{|04}A y y y y ==≤≤,2{|7120}{|25}B x x x x x =-+≤=≤≤,所以(){|23}U A C B x x =≤<.14.)1,(--∞ 由0322>--x x ,得1-<x 或3>x .321)(2--=x x x f 可看作t y 1=,322--=x x t 复合而成的,而t y 1=单调递减,要求321)(2--=x x x f 的增区间,只需求322--=x x t 的减区间即可,322--=x x t 的增区间为)1,(--∞,所以321)(2--=x x x f 的增区间为)1,(--∞.15.⎥⎦⎤⎢⎣⎡1,71-由.1,71-⎥⎦⎤⎢⎣⎡∆法可得:值域为者换元构造双勾型函数或 16.(2)(4) (1)因为)(x f 的定义域为]2,0[,由022x ≤≤得01x ≤≤,所以)2(x f 定义域为[]0,1,故(1)错; (2)()0f b =时,()f a 可取的值为1,0,1-,所以满足()0=b f 的映射共有3个,故(2)正确; (3)由反比例函数的图象和性质知,1()f x x=的单调递减区间有两个,(),0-∞和()0,+∞,故(3)错; (4) 因为()()1212f x f x +=-,令()()2,11t x f t f t =+=-,所以函数()f x 的图象自身关于直线1x =对称,故(4)正确;()21,5x x 必须是任意取值,故(5)错误.17.(1)∵m=2∴A={x ∣2≤x ≤6}, {},45<<-=x x B {},62><=x x x A C R 或∴{},65≤<-=⋃x x B A {}.25)(<<-=⋂x x B A C R(2) ∵A ∩B=A ∴B A ⊆ ∴⎩⎨⎧<<+-m m 26213 ∴).,3(,3331+∞∈>⇒⎪⎩⎪⎨⎧>->m m m m 则18.(1)换元法:令t=2x-121+⇒t ()22521821422+-=++•-⎪⎭⎫⎝⎛+=⇒t t t t t f 所以 f (x )=x 2﹣2x+2 .(2)f (x )=x 2﹣2x+2=(x ﹣1)2+1, 对称轴为x=1∈[﹣1,2], 又f (﹣1)=5, f (2)=2, 所以f max (x )=f (﹣1)=5. 关于x 的不等式()02432>++-t t x f 在[﹣1,2]有解,则()max 22-4-3x f t t <07-4-32<⇒t t ()()0173<+-⇒t t 371-<<⇒t 所以实数t 的取值范围为.371,-⎪⎭⎫ ⎝⎛. 19.(1)由()00=f 得:⇒=0b ()21axf x x =+,又1225f ⎛⎫= ⎪⎝⎭,所以1a =;则()21x f x x =+.(2)(),且任取2121,1,1,x x x x <-∈ 则()()()()()()222122122211222*********x x x x x x x x x x x x x f x f +++-+=+-+=-()()()()()()()()()();11111x -22212121222112212121x x x x x x x x x x x x x x f x f ++--=++-+=-⇒又1211x x -<<<,∴221212120,10,10,10x x x x x x -<->+>+>,从而()()120f x f x -<, 即()()12f x f x < 故()f x 在()1,1-上是增函数.20. (1)依题意可得:(a 2-1)x 2+(a +1)x +1≥0对一切x ∈R 恒成立; 当时,012=-a 即;11-==a a 或01:1=a 不符合,舍去;,21012-≥⇒≥+x x :120-=a 1显然符合;,0≥当a 2-1≠0时,即;且11-≠≠a a ()()⎪⎩⎪⎨⎧≤--+=∆>-014101222a a a ⎪⎩⎪⎨⎧≥-≤>-<⇒35111a a a a 或或 ∴a <-1或35≥a . 故).,35[]1,(+∞⋃--∞∈a(2)依题意可得:只要t =(a 2-1)x 2+(a +1)x +1能取到所有的正数; 当时,012=-a 即;11-==a a 或01:1=a 显然符合;,12+=x t :120-=a t=1显然不符合,舍去;, 当a 2-1≠0时,即⇒-≠≠11a a 且⎩⎨⎧≥∆>-0012a ⇒1<a ≤35; 则1≤a ≤35⇒.35,1⎥⎦⎤⎢⎣⎡∈a 21.(1)由题设知,024=+-c b a ① 令22122+=x x ,解得2=x , 由题意可得2221)2(222+⨯≤≤⨯f , 即4)2(4≤≤f ,所以4)2(=f , 即424=++c b a ② 由①、②可得1,42=-=b a c . 又x x f 2)(≥恒成立, 即0)2(2≥+-+c x b ax 恒成立, 所以0>a ,且04)2(2≤--=∆ac b , 即0)42(4)21(2≤---a a ,所以41=a ,从而142=-=a c . 因此函数)(x f 的解析式为 141)(2++=x x x f . (2)由)2()(x f t x f <+得122411)()(4122++⎪⎭⎫ ⎝⎛<++++xx t x t x ,整理得 0)382)(2(<+++t x t x . 当3822+-<-t t 即2>t 时,3822+-<<-t x t , 此不等式对一切]1,1[-∈x 都成立的充要条件是⎪⎩⎪⎨⎧>+--<-138212t t ,此不等式组无解.当3822+-=-t t 即2=t 时,0)2(2<+t x ,矛盾.当3822+->-t t 即2<t 时,t x t 2382-<<+-,此不等式对一切]1,1[-∈x 都成立的充要条件是⎪⎩⎪⎨⎧>--<+-121382t t ,解得2125-<<-t . 综合可知,t 的范围是⎪⎭⎫ ⎝⎛--21,25. 22. (1)设4()()h x t x x =+ ∵0t > ∴()h x 分别在(0,2),(2,)+∞上单调 且()4h x t ≥;要使()f x 分别在区间(0,2),(2,)+∞上单调则只需54504t t -≥⇒≥⇒).,45[+∞∈t(2)①当1t =时,44()5()5x m x m x x +-=⇒+-=或4()5x m x+-=- 即2(5)40x m x -++=或2(5)40x m x +-+=; ∵1234,,,x x x x 为方程()f x m=的四个不相等的实根∴由根与系数的关系得12344416x x x x =⨯= 则得证.②如图,可知01m <<,()f x 在(0,1)、(1,2)、(2,4)、(4,)+∞均为单调函数1[](],0,1a b ⊆: ()f x 在[],a b 上单调递减, 则()()f a mb f b ma=⎧⎨=⎩ 两式相除整理得()(5)0a b a b -+-=∵(],0,1a b ∈ ∴上式不成立 即,a b 无解,m 无取值;2[](],1,2a b ⊆: ()f x 在[],a b 上单调递增, 则()()f a maf b mb=⎧⎨=⎩ 即2451m a a =-+-在(]1,2a ∈有两个不等实根 ; 而令11,12t a ⎡⎫=∈⎪⎢⎣⎭则2245591()4()816t t a a ϕ-+-==--+ 作()t ϕ在1,12⎡⎫⎪⎢⎣⎭的图像可知,19216m ≤< ; 03[](],2,4a b ⊆: ()f x 在[],a b 上单调递减, 则()()f a mb f b ma=⎧⎨=⎩ 两式相除整理得()(5)0a b a b -+-= ∴5a b += ∴5b a a =-> ∴522a <<由45a mb a--+=得24544115255(5)()24a a m a a a a --==+=+---- 则m 关于a 的函数是单调的,而455a a m a--=-应有两个不同的解 ∴此种情况无解,舍去; 04[][),4,a b ⊆+∞: 同01可以解得m 无取值;综上,m 的取值范围为19,216⎡⎫⎪⎢⎣⎭.。

相关文档
最新文档